% The following TeXfile comprises part of

%     MEASURE THEORY, by D.H.Fremlin
%          (University of Essex, Colchester, England;
%                   e-mail  fremdh@essex.ac.uk)

% Please contact me with any difficulties not covered by the remarks
% below.

% The file is set up for processing by plain TeX in the presence of
% files  amstex.tex ,  amssym.tex  and  amsppt.sty  suitably supported by the
% appropriate fonts.   If these are unavailable, a readable version
% can be obtained by removing the comment symbol  %  from the line

%        %   \amstexsupportedfalse

% below.   Some systems do not allow access to files  amstex.tex  and
% amsppt.sty , but instead employ a command "amstex <filename.tex>";
% for such systems it is sometimes right to delete or disable the
% commands

%        \input amstex
%        \documentstyle{amsppt}

% (I have little experience with this and should welcome comments.)

% Various formats of output are obtainable.   The principal variations
% are

%    1. "Standard" output uses the command
%               \smallprint
% and gives output adapted to A4 paper (21 cm x 29.7 cm).

%    2.  Using the command
%               \USsmallprint
% gives output adapted to 8.5" x 11" .

%    3.  Using the command
%               \noproofs
% produces an abridged version, in which proofs are omitted.

%    4.   Using the command
%               \resultsonly
% produces a super-abridged version, omitting comments and exercises
% as well as proofs.

% These commands are grouped together at the end of the introductory
% definitions;  they may be activated by deleting the initial comment
% symbols  % .

% Any diagrams included will be called for by inputting postscript files.
% (You can look for these by hunting for   \picture  in the code below.)
% To get these you will need the relevant  .ps  files with the expected
% names  ( \picture{mt133g}{...}  demands the file  mt133g.ps , for instance).


%(If you can read TeX fluently, you may wish to fast-forward to a line
%beginning   %file starts here   or   \frfilename
%if one of these is present.)

%TeXfiles with this heading are designed for processing with AmSTeX,
%accessed through TeXfiles  amstex.tex (with amssym.tex  and
% amsppt.sty  and possibly  amsppt.sti)  during compilation.   
%If these are unavailable, a readable
%version can (I hope) be produced by deleting the  %  before
% \amstexsupportedfalse  below.

%Some special difficulties arise with the importation of  .ps  files
%in diagrams.   Look for  \picture  below for notes on these.

%The following \count registers are used for special purposes:
%   \count1=\stylenumber  to specify style
%   \count2=\footnotenumber  to count footnotes

%AmSTeX declaration
\newif\ifamstexsupported
     \amstexsupportedtrue
%     \amstexsupportedfalse               %use when AMSTeX unavailable
\newif\ifDHFamstexloaded        %for detection of a non-standard AMSTeX
     \DHFamstexloadedfalse    
\newif\ifDHFamspptloaded        
     \DHFamspptloadedfalse    
\newif\iffourteensavailable
     \fourteensavailabletrue
    \fourteensavailablefalse
\newif\ifathome
     \athometrue     %to cope with a quirk     of my son's Ghostscript
     \athomefalse
\newif\iffirstpage
     \firstpagetrue

%Style assignments
\countdef\stylenumber=1
\countdef\footnotenumber=2
\def\fullsize{\magnification=\magstep1
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=1}
\def\smallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=2}
\def\printersize{\magnification=\magstep1
     \baselineskip24pt
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{8.5truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=3}
\def\refsize{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=4}
\def\rrversion{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=5}
\def\bigsize{\magnification=\magstep2
     \overfullrule0pt
     \def\pagewidth{325pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=6}
\def\bbigsize{\magnification=\magstep3
     \overfullrule0pt
     \def\pagewidth{271pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=7}
\def\USsmallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{482pt}
     \def\pageheight{8.7truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=8}
\def\halfmagsize{\magnification=\magstephalf
     \def\pagewidth{427pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=9}
           % register  stylenumber  is used to control the output format

%addresses, letterheads
\def\deptaddress{\hskip 3truein Department of Mathematical Sciences\par
   \noindent e-mail: {\tt fremdh}\hskip 2.9truein\today\par
   \medskip}
\long\def\univaddress#1{\hskip 3truein Department of Mathematical Sciences\par
   \hskip 3truein University of Essex\par
   \hskip 3truein Colchester CO4 3SQ\par
   \hskip 3truein England\par
   %\hskip 3truein van Vleck Hall\par
   %\hskip 3truein 480 Lincoln Drive\par
   %\hskip 3truein Madison\par
   %\hskip 3truein WI 53706-1388, U.S.A.\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
       %{\tt fremlin\@math.wisc.edu}
   \bigskip}
\long\def\homeaddress#1{\hskip 3truein 25 Ireton Road\par
   \hskip 3truein Colchester CO3 3AT\par
   \hskip 3truein England\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
   \bigskip}
\long\def\letterhead#1#2#3{\univaddress{#1}
   {\parindent=0pt
    #2
   }
   \medskip

   \noindent Dear #3,
     }
\def\shortaddress{\noindent Departmentof Mathematical Sciences, University of Essex,
   Colchester CO4 3SQ, England\par
   \noindent e-mail: david\@fremlin.org}
\def\valediction#1#2{\bigskip
   \hskip 3truein #1,
   \vskip 1truein
   \hskip 3.3truein #2
   \bigskip}

%picture importation
   %It is clear that there is no general standard.   The syntax I
   %use when calling for a  .ps  file is
   %    \def\Caption{<...>}
   %    \picture{<filename>}{<dim>}
   %which is supposed to import  <filename>.ps  at height  <dim> ,
   %centered, and adding a one-line centered caption  <...> .
   %The basic macro listed in  \def\picture  is that which
   %used to work at Essex.   I now generally use the macro in
   % \def\picturewithpsfig , which requires a previous input
   % psfig.tex  or  psfig.sty ;
   %if you interpolate  \atUEssex  early in your file it will
   %give you the latter.
   %   I find that  psfig  is erratic in its centering, especially
   %under magnification, so I have written a routine
   % \centerpicturewithpsfig  (which may be called from
   % \centerpicture ) with the syntax
   %    \centerpicture{<filename>}{<dim>}{<dim>}
   %the first dimension being width and the second height.

   % A second package is called for by  \usegraphicx .   This requires
   % miniltx.tex  and  graphicx.sty ;  miniltx.tex  is currently (April 2016)
   % available from
   % http://mirror.ox.ac.uk/sites/ctan.org/macros/plain/graphics/miniltx.tex
   % The advantage of this is that it seems to allow manipulation of the
   % dvi>pdf stage using  dvipdfmx .

\def\picture#1#2{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at height #2
         with dvitops}
  \vbox{\vskip #2
    \special{dvitops: import #1.ps \pagewidth #2}
    \Centerline{\Caption}
    }
   \medskip
   \def\Caption{}}
%an alternative version, without caption
\def\centerpicture#1#2#3{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with dvitops}
  \vbox{\vskip #3
    \special{dvitops: import #1.ps \pagewidth #3}
  }}
%to put pictures on same line
\def\sideshiftedpicture#1#2#3#4{\bigskip
   \bigskip
   \immediate\write0{importing diagram #1.ps at height #4
        and width #3 with dvitops}
   \vbox{\special{dvitops:  import #1.ps #3 #4}}
   \hskip #2 plus 0pt minus 0pt
   }
\def\startsideshiftedpicture{}
\newcount\scaleup
\newcount\scaledown
\scaleup=1
\scaledown=1
\def\scalesideshiftedpicture#1#2#3#4{
  \dimen4=#4
  \dimen3=#3
  \multiply\dimen4 by\scaleup
  \divide\dimen4 by \scaledown
  \multiply\dimen3 by\scaleup
  \divide\dimen3 by \scaledown
  \sideshiftedpicture{#1}{#2}{\dimen3}{\dimen4}}

\def\picturewithpsfig#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with psfig}
   \vbox{\hsize\pagewidth
   \centerline{\psfig{figure=#1.ps,height=#2}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithpsfig#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with psfig}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\psfig{file=#1.ps,width=\dimen3,height=\dimen4}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithpsfig#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with psfig}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\psfig{figure=#1.ps,height=\dimen4,width=\dimen3}}
   \psfig{figure=#1.ps,height=\dimen4,width=\dimen3}
   \hskip #2 plus 0pt minus 0pt}

\def\picturewithgraphicx#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with graphicx}
   \vbox{\hsize\pagewidth
   \centerline{\includegraphics[height=#2]{#1.ps}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithgraphicx#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with graphicx}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\includegraphics[width=\dimen3,height=\dimen4]{#1.ps}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithgraphicx#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with graphicx}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\includegraphics[height=\dimen4,width=\dimen3]{#1.ps}}
   \includegraphics[height=\dimen4,width=\dimen3]{#1.ps}
   \hskip #2 plus 0pt minus 0pt}

%alternative picture importation code;  I've been told about this
  %but have not used it successfully myself
%\def\Caption{}
%\input epsf
%\epsfverbosetrue
%\def\picture#1#2{\bigskip
%    \bigskip
%    \immediate\write0{importing diagram #1.ps at height #2 with epsf}
%    \epsfysize=#2
%    \centerline{\epsffile{#1.ps}}
%    \medskip
%    \Centerline{\Caption}
%    \medskip
%    }

%in addition it may be necessary to amend the imported  .ps  file
%by replacing the   %%BoundingBox   header by  %%BoundingBox:

\def\atUEssex{
   \input psfig.sty
   \def\picture##1##2{\picturewithpsfig{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithpsfig{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithpsfig{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
   }
\def\usegraphicx{
   \input miniltx
   \def\Gin@driver{dvips.def}
   \input graphicx.sty
   \def\picture##1##2{\picturewithgraphicx{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithgraphicx{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithgraphicx{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
}


%special fonts
\def\biggerscriptfonts{\scriptfont0=\eightrm
     \scriptfont1=\eighti
     \scriptfont2=\eightsy
     \scriptfont3=\eightex
     \scriptfont\bffam=\eightbf
     \scriptfont\msbfam=\eightmsb
     \scriptfont\eufmfam=\eighteufm
     }
\def\Loadcyrillic{\ifamstexsupported
      \input cyracc.def
      \font\tencyr=wncyr10
      \def\cyr{\tencyr\cyracc} \def\itcyr{\tenitcyr\cyracc}
    \else\def\cyr{\smc}\def\cdprime{u\i}\def\Cdprime{UI}\fi}
%http://wkweb3.cableinet.co.uk/linux/LDP/HOWTO/Cyrillic-HOWTO-7.html
%cyrillic.tex

\def\Loadfourteens{\iffourteensavailable
       \font\fourteenrm=cmr14
       \font\fourteenbf=cmbx14
       \font\fourteenit=cmsl14
   \else\font\fourteenrm=cmr10 scaled \magstep2
      \font\fourteenbf=cmbx10 scaled \magstep2
      \font\fourteenit=cmsl10 scaled \magstep2
   \fi}
\def\smallerfonts{\eightpoint
   \textfont\msbfam=\eightmsb
   \textfont\eufmfam=\eighteufm
   \textfont\bffam=\eightbf
   \textfont\eurmfam=\eighteurm
   \textfont\eusmfam=\eighteusm
   }
\font\twelveex=cmex10 scaled \magstep1


%Copyright declarations
\newcount\oldcopyrightdate
\newcount\newcopyrightdate
\oldcopyrightdate=0
\def\copyrightdate#1{\newcopyrightdate=#1
       \ifnum\newcopyrightdate=\oldcopyrightdate\else
       \oldcopyrightdate=#1
       \oldfootnote{}{\eightrm \copyright\ #1 D.~H.~Fremlin}
       \immediate\write0{Copyright #1}
       \fi}

%hyphenation
\hyphenation{equi-ver-id-ical}

%miscellaneous definitions
\def\Aut{\mathop{\text{Aut}}}
\def\backstep#1{\hskip -0.#1em plus 0 pt minus 0 pt}
\def\Bang{\vthsp\hbox{\bf X\hskip0.07em\llap X\hskip0.07em\llap X\rm}}
\def\BanG{\Bang{\hskip 1.111pt plus 3.333pt minus 0pt}}
%\def\BbbN{\mathchoice{\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb N$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb N$\hskip0.04em}}}
%\def\BbbQ{\mathchoice{\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Q$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Q$\hskip0.04em}}}
%\def\BbbR{\mathchoice{\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb R$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb R$\hskip0.04em}}}
%  %do not use with $\Bbb R^X$
%\def\BbbZ{\mathchoice{\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Z$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Z$\hskip0.04em}}}
\def\BbbN{\openBbb{N}}
\def\BbbQ{\openBbb{Q}}
\def\BbbR{\openBbb{R}}
  %do not use with $\Bbb R^X$  do not use in $_{\Bbb R^r}$
\def\BbbZ{\openBbb{Z}}
\def\Bbbone{\hbox{1\hskip0.13em\llap{1}}}
\def\biggerint{\textfont3=\twelveex\intop\nolimits}
  % you can get this size into a displayed formula with
  %  $$\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}$$
  % but it may then seem cramped
  % or try \hbox{$\biggerint_{\Cal S}\,$}
\def\biggeriint{\textfont3=\twelveex\iint}
  % remarkable fact:  this seems to infect all \int within a given
  % $..$ ; but  $\biggerint$$\int$ seems to leave the $\int$ normal
  % really  \textfont3=\twelveex  is more reliable, since  \biggerint
  % as written gets turned off rather easily (e.g. by  \comment,
  % or by an extra  {...})
\def\bover#1#2{{{#1}\over{#2}}}
\def\Bover#1#2{\backstep4\hbox{
     \biggerscriptfonts${\shortstrut#1}\over{\shortstrut#2}$}}
  % when there are superscripts in the denominator, as in \Bover1{2^m}
  % or  \Bover1{\sqrt{2\pi}^r} , these can sometimes come out with better
  % horizontal alignment if you use  \Bover1{{\sshortstrut 2}^m}  or
  % \Bover1{{\sqrt{2\pi}}^r}
\def\bsover#1#2{\textstyle\bover{#1}{#2}}
\def\bsp{/\penalty-100}
  % \penalty seems to behave oddly in a \tt string in \fullsize mode
\def\Caption{}
\def\Center#1{

   \Centerline{#1}

   \noindent}
\def\Centerline#1{\vskip 5pt
     {\def\int{\text{}$$\textfont3=\twelveex\intop\nolimits}
      \def\dashint{\text{}$$\textfont3=\twelveex
         \mskip2mu\vrule height2.65pt width4.5pt depth-2.35pt\mskip-11mu
	 \intop\nolimits\mskip-2mu}
      \def\idashint{\text{}$$\int\mskip-7mu\dashint}	 
      \def\dashiint{\text{}$$\dashint\mskip-6mu\int}	 
      \def\dashidashint{\text{}$$\dashint\mskip-6mu\dashint}	 
      \def\underlineint{
           \text{}$$\textfont3=\twelveex\underline{\intop\nolimits}}
      \def\overlineint{
           \text{}$$\textfont3=\twelveex\overline{\intop\nolimits}}
      \def\iint{\text{}$$\textfont3=\twelveex
           \intop\nolimits\hskip-0.5em\intop\nolimits}
      \def\sum{\text{}$$\sumop}
      \def\prod{\text{}$$\prodop}
      \def\bigcap{\text{}$$\bigcapop}
      \def\bigcup{\text{}$$\bigcupop}
   \centerline{#1}}\vskip 5pt}
   \mathchardef\bigcapop="1354
   \mathchardef\bigcupop="1353
   \mathchardef\prodop="1351
   \mathchardef\sumop="1350
   %in  \Centerline{$\overline{}$, $\cmmnt{}$} must use \sumop etc
\def\cf{\mathop{\text{cf}}}
\def\circumflex#1{{\accent"5E #1}}
\def\closeplus{\hbox{+}}
\long\def\cmmnt#1{\ifresultsonly\else{#1}\fi}
\def\coint#1{\left[#1\right[}
\def\curl{\mathop{\text{curl}}}
\def\dashint{\mathchoice{-\mskip-20mu\intop\nolimits}
   {{\vrule height2.75pt width4.5pt depth-2.45pt\mskip-10mu\intop\nolimits}}
      {-\mskip-16.5mu\intop\nolimits} %not yet used
      {-\mskip-16.5mu\intop\nolimits}} %not yet used
\def\idashint{\mathchoice{\int\mskip-6mu\dashint}
      {\int\mskip-5mu\dashint}
      {\int\mskip-5mu\dashint} %not yet used
      {\int\mskip-5mu\dashint}} %not yet used
\def\dashiint{\mathchoice{\dashint\mskip-10mu\int}
      {\dashint\mskip-6mu\int}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashidashint{\mathchoice{\dashint\mskip-10mu\dashint}
      {\dashint\mskip-6mu\dashint}
      {\dashint\mskip-6mu\dashint} %not yet used
      {\dashint\mskip-6mu\dashint}} %not yet used
\def\dashidashiint{\mathchoice{\dashint\mskip-8mu\dashint\mskip-10mu\int}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashiidashint{\mathchoice{\dashint\mskip-10mu\int\mskip-8mu\dashint}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\idashidashint{\mathchoice{\int\mskip-8mu\dashint\mskip-8mu\dashint}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\diam{\mathop{\text{diam}}}
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
           \else\ifnum\stylenumber=3{\frnewpage}
           \else\ifnum\stylenumber=5{\frnewpage}
           \else\bigskip\fi\fi\fi}
\long\def\discrversionA#1#2{\ifnum\stylenumber=1#1\else#2\fi}
\long\def\discrversionB#1#2{\ifnum\stylenumber=2#1\else#2\fi}
\long\def\discrversionC#1#2{\ifnum\stylenumber=3#1\else#2\fi}
\long\def\discrversionD#1#2{\ifnum\stylenumber=4#1\else#2\fi}
\long\def\discrversionE#1#2{\ifnum\stylenumber=5#1\else#2\fi}
\long\def\discrversionF#1#2{\ifnum\stylenumber=6#1\else#2\fi}
\long\def\discrversionG#1#2{\ifnum\stylenumber=7#1\else#2\fi}
\long\def\discrversionH#1#2{\ifnum\stylenumber=8#1\else#2\fi}
           % facility for special interpolations in particular output
           % versions
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\diverg{\mathop{\text{div}}}
\def\dom{\mathop{\text{dom}}}
\def\eae{=_{\text{a.e.}}}

\def\esssup{\mathop{\text{ess sup}}}
\mathchardef\Exists="3239

\def\family#1#2#3{\langle#3\rangle_{#1\in#2}}
\def\familyi#1#2{\family{i}{#1}{#2}}
\def\familyiI#1{\familyi{I}{#1}}

\def\frfilename#1{\ifnum\stylenumber=1{\noindent filename #1}
               \else\ifnum\stylenumber=6{\noindent{\fiverm filename #1}}
               \else\ifnum\stylenumber=7{\noindent{\fiverm filename #1}}
               \else\fi\fi\fi}
\def\Footnote#1#2{\footnote"#1"{#2}}
\mathchardef\Forall="3238
\def\frakc{\frak c\phantom{.}}
\def\frnewpage{\vskip \pageheight plus 0pt minus\pageheight
           \eject
           \gdef\topparagraph{}
           \gdef\bottomparagraph{}
           }
\def\frsmallcirc{{\hskip 1pt plus 0pt minus 0pt}
     {\raise 1pt\hbox{$\smallcirc$}}{\hskip 1pt plus 0pt minus 0pt}}

\def\geae{\ge_{\text{a.e.}}}
\def\grad{\mathop{\text{grad}}}
\def\grheada{{\bf ($\pmb{\alpha}$)}}
\def\grheadb{{\bf ($\pmb{\beta}$)}}
\def\grheadc{{\bf ($\pmb{\gamma}$)}}
\def\grheadd{{\bf ($\pmb{\delta}$)}}
\def\grheade{{\bf ($\pmb{\epsilon}$)}}
\def\grheadz{{\bf ($\pmb{\zeta}$)}}
\def\grv#1{{\accent"12 #1}}

\def\hfillall{\hskip\pagewidth plus0pt minus\pagewidth}

\def\Imag{\mathop{\Cal I\text{m}}}
\long\def\inset#1{{\narrower\narrower{\vskip 1pt plus 1pt minus 0pt}
     #1
     {\vskip 1pt plus 1pt minus 0pt}}}
\def\Int#1#2{\displaystyle{\int_{#1}^{#2}}}
\def\interior{\mathop{\text{int}}}

\def\leae{\le_{\text{a.e.}}}
\long\def\leaveitout#1{}
\def\LLcorner{\mathchoice{\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height5pt width0.3pt depth0pt
       \vrule height0.3pt width4pt depth0pt}\,}
   {\,\hbox{\vrule height4pt width0.25pt depth0pt
       \vrule height0.25pt width3pt depth0pt}\,}
  }
\def\Ln{\mathop{\text{Ln}}}
\def\Loadfont#1{\font#1}
\def\loibr{\hbox{\hskip0.1em]\hskip0.05em}}
\def\Matrix#1{\pmatrix#1\endpmatrix}
\def\Mu{\text{M}}
\def\noamslogo{\def\amslogo{}}
\def\normalsubgroup{\vartriangleleft}
\def\Nu{\text{N}}
\def\ocint#1{\left]#1\right]}
\def\ofamily#1#2#3{\langle#3\rangle_{#1<#2}}
\let\oldcases=\cases
\let\oldfootnote=\footnote
\def\On{\mathop{\text{On}}}
\def\ooint#1{\left]#1\right[}
\def\openBbb#1{\Bbb#1\mskip1mu}
\def\pd#1#2{\bover{\partial#1}{\partial#2}}
\def\Pd#1#2{\Bover{\partial#1}{\partial#2}}
\def\pdd#1#2{{{\partial^2#1}\over{\partial#2^2}}}
\def\Pdd#1#2{\Bover{\partial^2#1}{\partial#2^2}}
\def\pde#1#2#3{{{\partial^2#1}\over{\partial#2\partial#3}}}
\def\Pde#1#2#3{\Bover{\partial^2#1}{\partial#2\partial#3}}
\def\Prf{\vthsp\hbox{\bf P\hskip0.07em\llap P\hskip0.07em\llap P\rm}}
\def\pushbottom#1{\hbox{\vrule height0pt depth#1 width0pt}}
\def\Qed{\vthsp\hbox{\bf Q\hskip0.07em\llap Q\hskip0.07em\llap Q\rm}}
\def\QeD{\Qed{\hskip 1.111pt plus 3.333pt minus 0pt}}
\def\Quer{\vthsp\hbox{\bf ?\hskip0.05em\llap ?\hskip0.05em\llap
   ?\thinspace\rm}}
\def\query{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{query}}{}}
\def\Real{\mathop{\Cal R\text{e}}}
\def\restr{\mathchoice
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\scriptstyle\restriction$}\mskip 1mu}
  {\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
\def\restrp{\mathchoice
  {\hbox{$\restriction$}\mskip 2mu}
  {\hbox{$\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptstyle\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
    %use in \restrp V \restrp W \restrp Y
    % \restrp\Cal P \restrp\Bbb R \restrp\frak
    % \restrp\Omega \restrp\Tau \restrp\Upsilon
    % \restrp[ \restrp\widehat{}
\def\Rho{\text{P}}
\def\roibr{\hbox{\hskip0.05em[\hskip0.1em}}
\def\RoverC{\hbox{\biggerscriptfonts${{\Bbb R}\atop{\Bbb C}}$}}
\def\sequence#1#2{\family{#1}{\Bbb N}{#2}}
\def\sequencen#1{\sequence{n}{#1}}
\def\setover#1#2{\overset{#1}\to{#2}}
\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}
\def\sshortstrut{\hbox{\vrule height5.8pt depth2.5pt width0pt}}
\def\smallcirc{{\scriptstyle\circ}}
\def\ssbullet{{\scriptscriptstyle\bullet}}
\def\symmdiff{\triangle}
\def\thicken#1{#1\hskip0.03em\llap #1\hskip0.03em\llap
   #1\hskip0.03em\llap #1}
\def\Tau{{\text{T}}}
\def\tbf#1{\text{\bf #1}}
\def\terror#1#2#3{#1:  `#2' for `#3'}
\def\today{\number\day.\number\month.\number\year}
\def\UEMDRR{University of Essex Mathematics Department Research Report}
\def\UEMDWP{University of Essex Mathematics Department Working Paper}
\def\versiondate#1{
    \immediate\write0{Version of #1}
    \def\vsdate{#1}
               \ifnum\stylenumber=1\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=2\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=6\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=7\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=8\hfill Version of #1
                              \medskip
               \else\fi\fi\fi\fi\fi}
\def\versionhead{\ifnum\stylenumber=3 \hfill PRINTER'S VERSION

                        \bigskip
                   \else\ifnum\stylenumber=4 \hfill REFEREE'S VERSION

                        \bigskip
                   \else \fi\fi}
\def\versionref{\ifnum\stylenumber=1\workingref
           \else\ifnum\stylenumber=2\workingref
           \else\ifnum\stylenumber=8\workingref
           \else\printerref\fi\fi\fi}
\def\vthsp{\hskip0.001em}

%Missing AmSTeX can be coped with by using the following lines:

\ifamstexsupported\else
   \immediate\write0{AmSTeX-unsupported version of  fremtex.tex }
   \fourteensavailablefalse
   \def\@{@}
   \def\\{\cr}
   \def\AmSTeX{AmS\TeX}
   \def\Bbb#1{\underline{\hbox{\bf #1}}}
   \def\biggerscriptfonts{}
   \def\bumpeq{=_{\text{approx}}}
   \def\Cal#1{{\cal#1}}
   \def\curlyeqprec{\preceq}
   \def\eightit{\it}
   \def\eightrm{\rm}
   \def\eightsmc{\smc}
   \def\eurm#1{\underline{\cal#1}}
   \def\Footnote#1#2{\footnote#1#2}
   \def\frac#1#2{{#1}\over{#2}}
   \def\frak#1{{\hbox{\bf #1\hskip0.1em\llap #1}}}
   \def\frnewpage{\vfill\eject}
   \def\iint{\int\!\!\!\int}
   \def\iiint{\int\!\!\!\int\!\!\!\int}
   \def\iiiint{\int\!\!\!\int\!\!\!\int\!\!\!\int}
   \def\llcorner{\lfloor}
   \def\loadeurm{}
   \def\loadeusm{}
%   \def\loadeusmx#1{\def\eusm#1{\underline{\cal#1}}}
   \def\ltimes{|\hskip-0.5em\times}
   \def\eusm#1{\underline{\cal#1}}
   \def\ltimes{|\times}
   \def\Loadfont#1{}
   \def\Matrix#1{\pmatrix{#1}}
   \def\noamslogo{}
   \def\nrightarrow{\not\rightarrow}
   \def\preccurlyeq{\preceq}
   \def\query{\discrversionA{\immediate\write0{query}
                             \footnote{$^*$}{query}}{}}
   \def\restriction{|}
   \def\rtimes{\times|}
   \def\setover#1#2{#2^{#1}}
   \def\shortstrut{\strut}
   \def\smallerfonts{\eightrm}
   \def\smallfrown{\wedge}
   \def\smc{}
   \def\spcheck{\check{\phantom{x}}}
   \def\sphat{\hat{\phantom{x}}}
   \def\square{\mathchoice\sqr64\sqr64\sqr43\sqr33}
   \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
          \hbox{\vrule width.#2pt height#1pt \kern#1pt
                  \vrule width.#2pt}
          \hrule height.#2pt}}}}
   \def\pmb#1{\underline{#1}}
   \def\text#1{\hbox{#1}}
   \def\therefore{\ssbullet\raise.9ex\hbox{$\ssbullet$}\ssbullet}
   \def\twelvebf{\bf}
   \def\vartriangleleft{\triangleleft}
   \def\Vdash{|\vdash}
   \def\vDash{|=}
   \fi

%End of lines for missing AmSTeX

\ifamstexsupported\input amstex
%              \documentstyle{amsppt}
               \input amsppt.sty
               \PSAMSFonts
	       \tenpoint
               %\font\eightmsb=msym8
               \font\eightmsb=msbm10
               \font\eighteufm=eufm10 at 8 pt
               \font\eighteurm=eurm8
               \font\eighteusm=eusm8
               \font\varseveneufm=eufm10 at 7 pt
               \scriptfont\eufmfam=\varseveneufm
   \def\.#1{{\accent"5F #1}}
   \def\=#1{{\accent"16 #1}}
   \gdef\nopagenumbers{\NoPageNumbers}
   \else\fi

\newcount\footmarkcount
\ifDHFamstexloaded
    \loadmsBm
    \def\eightBbb#1{{\fam\msBfam#1}}
    \loadmsFm
    \def\eightfrak#1{{\fam\msFfam#1}}
  \else
    \def\eightBbb#1{{\scriptstyle\Bbb#1}}
    \def\eightfrak#1{{\scriptstyle\frak#1}}
  \fi
\ifathome\athome\fi

%interword glue (\tenrm standard):
%  (space) 3.333 + 1.667 - 1.111
%  (,)     3.333 + 2.083 - 0.889
%  (.)     4.444 + 5.    - 0.370

%very odd fact:  \discrversionA{2.}{}  doesn't work,
%   \discrversionA{ 2.}{} does

%The false line feed introduced by my Fortran programs, leading to
%a surplus  ae  at the top of the next page, seems to be associated
%with  Fk(\032)3081  or something similar in the PostScript file.


\immediate\write0{Extract from MEASURE THEORY,
   by D.H.Fremlin, University of Essex, Colchester.
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   http://dsl.org/copyleft/dsl.txt.   This is a development version and the
   source files are not permanently archived.
   For further information contact david\@fremlin.org.}

%hyphenations
\hyphenation{Archi-me-dean}

%new style
\def\smallprintersize{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \nologo
     \def\copyrightdate##1{}
     \def\folio{\ifnum\pageno<0{}\else\number\pageno\fi}
     \def\mtfootline{}
     \stylenumber=10
     \pageno=-1000}

%With-proofs, results-only versions
\newif\ifwithproofs
\withproofstrue
\newif\ifresultsonly
\resultsonlyfalse
\def\noproofs{\withproofsfalse}
\def\resultsonly{\resultsonlytrue\withproofsfalse}
\long\def\proof#1{\ifwithproofs{\medskip

     \noindent{\bf proof} #1}\else\fi}
\long\def\prooflet#1{\ifwithproofs{#1}\else\fi}
%\long\def\Prooflet#1{\prooflet{\Prf\ #1 \Qed\ }}
\long\def\endnotes#1{\ifresultsonly\else{#1}\fi}
\long\def\exercises#1{\ifresultsonly\else{#1}\fi}
\long\def\references#1{\ifresultsonly\else{#1}\fi}
\long\def\dvro#1#2{\ifresultsonly{#1}\else{#2}\fi}

%Headline assignments

\def\topparagraph{}
\def\bottomparagraph{}
\def\newparagraph{}
\newif\ifpageturned

\newdimen\rhoffset
\newdimen\lhoffset
\lhoffset=0cm
\rhoffset=0cm

\newcount\volumeno
\volumeno=5
\newcount\luluvolumeno
\luluvolumeno=0

\let\plainpagebody=\pagebody
\def\pagebody{\ifodd\pageno\global\hoffset=\rhoffset
         \else\global\hoffset=\lhoffset\fi
     \plainpagebody
     \turnpage}
\def\turnpage{
     \iffirstpage\else{\baselineskip=24pt\line{\mtfootline}}\fi
%     \immediate\write0{turning page \folio,
%           top paragraph \topparagraph,   bottom \bottomparagraph,
%           new \newparagraph, pagetotal \the\pagetotal}
     \global\pageturnedtrue
     \global\firstpagefalse
     \global\let\topparagraph=\bottomparagraph
     \global\let\bottomparagraph=\newparagraph
     }

\newdimen\pagetotalplus
\newdimen\pagetotalminus

%wheader parameters:  #1 new paragraph name
 % #2#3#4  \vskip parameters if page break not forced (no. of points)
 % #5  spare height demanded if page break not to be forced
\long\def\wheader#1#2#3#4#5{\gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #5
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
%     \immediate\write0{forcing page break}
     \eject
     \else\vskip #2pt plus #3pt minus #4pt\fi
     \else\vskip #2pt plus #3pt minus #4pt\fi
  \ifpageturned\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
     \fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
%   \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  }
\def\header#1{
     \wheader{#1}{10}{4}{4}{24pt}}
\def\leader#1#2{\immediate\write0{#1 #2}
     \wheader{#1}{14}{7}{5}{24pt}
     {\bf #1 #2}}
\def\vleader#1#2#3{\immediate\write0{#2 #3}
     \wheader{#2}{14}{7}{5}{#1}
     {\bf #2 #3}}
\def\Notesheader#1{\immediate\write0{#1 Notes and comments}
     \wheader{#1 Notes}{16}{8}{5}{48pt}
     \noindent{\bf #1 Notes and comments}}
\def\vNotesheader#1#2{\immediate\write0{#2 Notes and comments}
     \wheader{#2 Notes}{16}{8}{5}{#1}
     \noindent{\bf #2 Notes and comments}}
\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\def\spstheader#1#2#3#4#5{\header{*#1#2#3#4#5}{\bf (#5)}}
\def\vspheader#1#2#3#4#5#6{
     \wheader{#2#3#4#5#6}{10}{4}{4}{#1}{\bf (#6)}}
\def\sqheader#1#2#3#4#5{\header{#1#2#3#4#5}{$\pmb{>}${\bf (#5)}}}
\def\errheader#1#2#3{
     \wheader{#3}{6}{4}{2}{24pt}
     {\bf p #1 l #2}}
\long\def\woddheader#1#2#3#4#5{
   \ifodd\pageno\wheader{#1}{#2}{#3}{#4}{#5}
   \else\vskip #2pt plus #3pt minus #4pt\fi}

\def\indexmedskip{
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by 18 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
      \else\medskip\fi\else\medskip\fi}

\def\indexheader#1{\immediate\write0{#1}\vindexheader{#1}{24}}
\def\vindexheader#1#2{
  \gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #2 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
%  \ifdim\pagetotal>\pagegoal
%      \immediate\write0{warning: pagetotal greater than pagegoal}
%      \immediate\write0{...   last paragraph reached is #1}
%      \immediate\write0{...   check for underfull vbox here}
%      \else\fi
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal{\eject}
      \fi\fi
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
  \ifpageturned{\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
      }\else\fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%    \bottomparagraph}
  }

\def\leftheadline{\rlap{\folio}\hfill{\eightit\chaptername}\hfill
       \llap{\eightbf\topparagraph}}
\def\rightheadline{\rlap{\eightbf\bottomparagraph}\hfill
       {\eightit\headlinesectionname}\hfill\llap{\folio}}

%Footline assignments
\def\mtfootline{\ifodd\pageno\rightfootline\else\leftfootline\fi}
\def\rightfootline{\hfill\eightsmc D.H.Fremlin}
\def\leftfootline{\ifwithproofs{\eightsmc Measure Theory\hfill}
    \else{\eightsmc Measure Theory \eightit (abridged version)\hfill}
    \fi}

%Style assignments
\def\newchapter#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Chap.\ #1 {\it intro.}}
   \gdef\newparagraph{Chap.\ #1 {\it intro.}}
   \footmarkcount=0
   \gdef\sectionname{Introduction}
   \gdef\headlinesectionname{Introduction}
   \centerline{\bf Chapter #1}
   \medskip
   \centerline{\bf \chaptername}
   \medskip
   }
\def\newsection#1{\gdef\bottomparagraph{\S#1 {\it intro.}}
   \noindent{\bf #1 \sectionname}
   \smallskip
   \let\headlinesectionname=\sectionname}
\def\newvolume#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Vol.\ #1 {\it intro.}}
   \Loadfont{\twelvebf=cmbx12}
   \def\chaptername{Volume #1}
   \def\sectionname{\volumename}
   \def\headlinesectionname{\volumename}
   \centerline{\twelvebf Volume #1:  \volumename}
   \bigskip
   }

%font variations
\newif\ifvolwp
\volwpfalse
\def\tenrmstretch#1{\fontdimen3\tenrm=#1}
\font\elevenex=cmex10 at 11pt

%contents and introductions
\newif\iflargelogo
\def\vtmpb#1{}
\def\newvolume#1{}
\def\pagereference#1#2{\hfill\hbox{\ifresultsonly#2\qquad \hskip 1truein
                                                 \else#1\qquad \fi}}
\long\def\section#1#2#3#4#5#6{\ifresultsonly
    \qquad #1 #2\pagereference{#4}{#5}
  \else\vskip 2pt plus 1pt minus 0pt
    \vbox{\qquad #1 #2\pagereference{#4}{#5}\par
    \abstr{#6}
    }\vskip -1pt plus 1pt minus 0pt
  \fi}
\long\def\chapintrosection#1#2#3{\ifresultsonly
    \qquad Introduction\pagereference{#2}{#3}
  \else\vskip 1pt plus 1pt minus 0pt
    \qquad Introduction\pagereference{#2}{#3}
    \vskip 1pt plus 1pt minus 0pt
  \fi}
\def\abstr#1{\ifresultsonly{}
   \else{\advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \vskip1pt plus 0pt minus0pt
     \parindent=0pt
     \smallerfonts #1%
     \hbox{\vrule height0pt depth2pt width0pt}
     \vskip1pt plus 0pt minus0pt}\fi}
     
%language variations
\newif\ifUSEnglish
\USEnglishfalse
%\USEnglishtrue

%"Later editions only"
\newcount\latereditionpage
\newcount\latereditionmark
\latereditionpage=-1
\latereditionmark=0
\newcount\newpageno

%removing the \write0 instruction altogether disorganizes it!
\def\Latereditions{\newpageno=\pageno%
  \ifdim\pagetotal>\pagegoal\advance\newpageno by 1\fi%
     \divide\newpageno by 2%
  \immediate\write0{%
%     "Later editions only";
%      pagetotal \the\pagetotal, pagegoal \the\pagegoal,
%      pageno \number\pageno,\par newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage%
  }\ifnum\latereditionpage=\newpageno%
     \ifDHFamspptloaded$^{\number\latereditionmark}$%
     \else\footnote{Later editions only.}\fi%
  \else\footnote{Later editions only.}%
  \global\latereditionpage=\newpageno%
  \global\latereditionmark=\footmarkcount\fi}%
%  \immediate\write0{newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage}

   
%general definitions
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
                       {\lower-.3ex\hbox{$\ssbullet$}}
                       {\ssbullet}
                       {\ssbullet}}  %see also \varaction
\def\add{\mathop{\text{add}}\nolimits}
\long\def\allowmorestretch#1#2{
  \ifdim\pagewidth<#1pt\else
    \fontdimen3\tenrm=2pt\fontdimen4\tenrm=1.67pt\fi
  #2
    \fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt}
\def\am{atom\discretionary{-}{}{}lessly-measurable}
\def\AmuA{\Aut_{\bar\mu}\mskip-2mu\frak A}
\def\ASint{\mathchoice{\Bbb A\mskip-5mu}
                  {\Bbb A\mskip-3mu}
                  {\Bbb A\mskip-2mu}
                  {\Bbb A\mskip-2mu}\Sint}
\def\Asisfour{$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$}
\def\Asissix{$(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Asisfourt{$(\frak A,\bar\mu,\coint{0,\infty},
   \langle\frak A_t\rangle_{t\ge 0})$}
\def\Asissixt{$(\frak A,\bar\mu,\coint{0,\infty},
   \penalty-100\langle\frak A_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Atop#1#2{{\scriptstyle #1}\atop{\scriptstyle #2}}
\def\Aut{\mathop{\text{Aut}}}

\def\Bcap{\Booleanop{\cap}}
\def\Bcup{\Booleanop{\cup}}
\def\Booleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
 %   {{\mskip2mu\scriptstyle#1\mskip2mu}}  %until 15.10.17
    {{\scriptscriptstyle#1}}  %from 15.10.17
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminus{\Booleanop{\setminus}}
\def\Bsubseteq{\hskip.1em\Booleanop{\subseteq}\hskip.1em}
\def\Bsubset{\Booleanop{\subset}}
\def\Bsupseteq{\Booleanop{\supseteq}}
\def\Bsupset{\Booleanop{\supset}}
\def\Bsymmdiff{\Booleanop{\symmdiff}}
\def\Bcapshort{\Booleanopshort{\cap}}
\def\Bcupshort{\Booleanopshort{\cup}}
\def\Booleanopshort#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hskip-1pt\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminusshort{\Booleanopshort{\setminus}}
\def\BSint{\mathchoice{\Bbb B\mskip-5mu}
                  {\Bbb B\mskip-3mu}
                  {\Bbb B\mskip-2mu}
                  {\Bbb B\mskip-2mu}\Sint}
%\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}\hskip.1em}
\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}}
\def\Bsubsetshort{\Booleanopshort{\subset}}
\def\Bsupseteqshort{\Booleanopshort{\supseteq}}
\def\Bsymmdiffshort{\Booleanopshort{\symmdiff}}

\def\closeBooleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\closeBcup{\closeBooleanop{\cup}}
\def\closeBcap{\closeBooleanop{\cap}}
\def\closeBsetminus{\closeBooleanop{\setminus}}
\def\closeBsymmdiff{\closeBooleanop{\symmdiff}}

\def\biggerHeint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 2.6pt{\sevenrm H}\hss}
   \intop\nolimits}
\def\biggerPfint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}
   \intop\nolimits\hskip 1.2pt}
\def\biggerSint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 3.6pt{\sevenrm S}\hss}
   \intop\nolimits}
\def\bu{\mathop{\text{bu}}}
\def\Bvalue#1{\mathchoice
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$\scriptstyle[\![#1]\!]$}}
   {\hbox{$\scriptscriptstyle[\![#1]\!]$}}}
\def\cadlag{c\`adl\`ag}
\def\CalBa{\Cal{B}\eurm{a}}
   %note: neither  \cal  nor  \eusm  has lower case
\def\CalFr{\Cal F_{\text{Fr}}}
\def\callal{c\`all\`al}
\def\CalNwd{\Cal{N}\hskip-1.2pt\eurm{w}\hskip-.8pt\eurm{d}}

\def\CalRbg{\Cal{R}\mskip-1mu\eurm{b}\mskip-1mu\eurm{g}}
\def\CalSmz{\Cal{S}\mskip-2mu\eurm{mz}}
\def\CalUn{\Cal{U}\eurm{n}}
\def\capacity{\mathop{\text{cap}}\nolimits}
\def\Caratheodory{Cara\-th\'eo\-dory}
\def\Cdlg{C_{\text{dlg}}}
\def\cff{\cf\mskip1mu}
\def\checkpmb#1{\check{\pmb #1}}
\def\ci{\mathop{\text{ci}}}
\def\cl{\mathop{\text{cl}}}
\def\Clll{\hbox{$C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\clMsimp{\overline{M}_{\text{simp}}}
\def\closeuparrow{\mathchar"0222}
\def\closedownarrow{\mathchar"0223}
\def\clstar{\text{cl*}}
\def\cov{\mathop{\text{cov}}}
\def\covar#1#2{\mathchoice
  {[#1{\lower4pt\hbox{${\scriptstyle *}\atop{\vrule height 18pt depth -10pt}$}}#2]}
  {[#1{*\atop{\vrule height 8pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}}
\def\covSh{\mathop{\text{cov}}\nolimits_{\text{Sh}}}
\def\cdownarrow{c\mskip2mu^{\downarrow}}
\def\cycle#1{(\overleftarrow{#1})}
\def\ddaggerloc{$(\ddagger)_{\text{loc}}$}
\def\ddownarrow{d\mskip2mu^{\downarrow}}
%\def\DiniD{\setover{
%  \hbox{\hskip.1em\vrule height-1pt width.5em depth1.4pt\hfil}}D}
%\def\Dinid{\hskip.1em
%  \hbox{\hskip.05em\vrule height-1.3pt width.5em depth1.7pt}\hskip-.65em D}
\def\Dinid{\hbox{$D$\hskip-8.3pt\vrule height-1pt width7pt depth1.85pt}
   \hskip1pt}
\def\DiniD{\hbox{$D$\hskip-8.3pt\vrule height9.2pt width7ptdepth-8.35pt}
   \hskip1pt}

\def\discrcenter#1#2{\ifdim\pagewidth>#1{#2}\else{

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt45.tex}
\versiondate{5.6.09}
\copyrightdate{1998}

\def\chaptername{Perfect measures and disintegrations}
\def\sectionname{Introduction}

\newchapter{45}

One of the most remarkable features of countably additive
measures is that they provide us with a framework for probability
theory, as described in Chapter 27.   The extraordinary achievements of
probability theory since Kolmogorov are to a large extent possible
because of the rich variety of probability measures which can be
constructed.   We have already seen image measures (234C\formerly{1{}12E})
and product measures (\S254).   The former are elementary,
but a glance at the index will confirm that they have many
surprises to offer;  the latter are obviously fundamental to any idea of
what probability theory means.   In this chapter I will look at some
further constructions.   The most important are those associated with
`disintegrations' or `regular conditional probabilities' (\S\S452-453)
and methods for confirming the existence of measures on product spaces
with given images on subproducts (\S454, 455A).   We find that these
constructions have to be based on measure spaces of special types;  the
measures involved in the principal results are the Radon measures of
Chapter 41 (of course), the compact and perfect measures of Chapter 34,
and an intermediate class, the `countably compact' measures of
{\smc Marczewski 53} (451B).   So the first section of this chapter is a
systematic discussion of compact, countably compact and perfect
measures.

A `disintegration', when present, is likely to provide us with a
particularly effective instrument for studying a measure, analogous to
Fubini's theorem for product measures (see 452F).   \S\S452-453
therefore concentrate on theorems guaranteeing the existence of
disintegrations compatible with some pre-existing structure, typically
an \imp\ function (452I, 452O, 453K) or a product structure
(452M).   Both depend on the existence of
suitable liftings, and for the topological version in \S453 we need a
`strong' lifting, so much of that section is devoted to the study of
such liftings.

One of the central concerns of probability theory is to understand
`stochastic processes', that is, models of systems evolving
randomly over time.   If we think of our state space as consisting
of functions, so that a whole possible history is described by a
random function of time, it is natural to think of our functions as
members of some set $\prod_{n\in\Bbb N}Z_n$ (if we think of
observations as being taken at discrete time intervals) or
$\prod_{t\in\coint{0,\infty}}Z_t$ (if we regard our system as
evolving continuously), where $Z_t$ represents the set of possible
states
of the system at time $t$.   We are therefore led to consider
measures on such product spaces, and the new idea is that we may
have some definite intuition concerning the joint distribution of
{\it finite} strings $(f(t_0),\ldots,f(t_n))$ of values of our
random function, that is to say, we may think we know something
about the image measures on finite products
$\prod_{i\le n}Z_{t_i}$.   So we come immediately to a fundamental
question:
given a (probability) measure $\mu_J$ on $\prod_{i\in J}Z_i$ for
each finite $J\subseteq T$, when will there be a measure on
$\prod_{i\in T}Z_i$ compatible with every $\mu_J$?   In \S454 I give the
most important generally applicable existence theorems for such
measures, and in 455A-455E %455A 455C 455E
I show how they can be applied to a general
construction for models of Markov processes.   These models enable us to
discuss the Markov property either in terms of disintegrations or in terms
of conditional expectations (455C, 455O), and for L\'evy processes, in
terms of \imp\ functions (455U).

The abstract theory of \S454 yields measures on product spaces which,
from the point of view of a probabilist, are unnaturally large, often
much larger than intuition suggests.   Some of the most powerful results in
the theory of Markov processes, such as the strong Markov property (455O),
depend on moving to much smaller spaces;  most notably the space of
\cadlag\ functions (455G), but the larger space of \callal\ functions is
also of interest.   The most important example, Brownian motion,
will have to wait for Chapter
47, but I give the basic general theory of L\'evy processes in complete
metric groups.

One of the defining characteristics of Brownian motion is the fact that
all its finite-dimensional marginals are Gaussian distributions.
Stochastic processes with this
property form a particularly interesting class, which I examine in
\S456.   From the point of view of this volume, one of their most
striking properties is Talagrand's theorem
that, regarded as measures on powers $\BbbR^I$, they are $\tau$-additive
(456O).

The next two sections look again at some of the ideas of the
previous sections when interpreted as answers to questions of the form
`can all the measures in such-and-such a family be simultaneously
extended to a single measure?'   If we seek only a {\it finitely}
additive common extension, there is a reasonably convincing general
result (457A);  but countably additive measures remain puzzling even in
apparently simple circumstances (457Z).
In \S458 I introduce `relatively independent' families of
$\sigma$-algebras, with the associated concept of `relative product' of
measures, and the corresponding concepts for probability algebras.
Finally, in \S459, I give some basic results on
symmetric measures and exchangeable random variables, with De Finetti's
theorem (459C) and corresponding theorems on representing
permutation-invariant measures on products as mixtures of product
measures (459E, 459H).

\discrpage

\frfilename{mt451.tex}
\versiondate{8.11.07}
\copyrightdate{2002}

\def\undphi{\underline{\phi}\vthsp}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Perfect, compact and countably compact measures}

\newsection{451}

In \S\S342-343 I introduced `compact' and `perfect' measures as part of
a study of the representation of homomorphisms of measure algebras by
functions between measure spaces.   An intermediate class of `countably
compact' measures\cmmnt{ (the `compact' measures of
{\smc Marczewski 53})} has appeared in the exercises.   It is now time to collect these
ideas together in a more systematic way.   In this section I run through
the standard properties of compact, countably compact and perfect
measures (451A-451J), with a couple of simple examples of their
interaction with topologies (451M-451P).   An example of a
perfect measure space which is not countably compact is in 451U.   Some
new ideas, involving non-trivial set theory, show that measurable
functions from compact totally finite measure spaces to metrizable
spaces have `essentially separable ranges' (451R);  consequently, any
measurable function from a Radon measure space to a metrizable space is
almost continuous (451T).

\cmmnt{
\leader{451A}{}\cmmnt{ Let me begin by recapitulating the principal
facts already covered.

\medskip

}{\bf (a)} A family $\Cal K$ of sets is a {\bf compact class} if
$\bigcap\Cal K'\ne\emptyset$ whenever $\Cal K'\subseteq\Cal K$ has the
finite intersection property.
If $\Cal K\subseteq\Cal PX$, then
$\Cal K$ is a compact class iff there is a compact topology on $X$ for
which every member of $\Cal K$ is closed\cmmnt{ (342D)}.   A subfamily
of a compact class is compact\cmmnt{ (342Ab)}.

\spheader 451Ab A measure on a set $X$ is {\bf compact} if it is inner
regular with respect to some compact class of sets;  equivalently, if it
is inner regular with respect to the closed sets for some compact
topology on $X$\cmmnt{ (342F)}.   All Radon measures are compact
measures\cmmnt{ (416Wa)}.   If $(X,\Sigma,\mu)$ is a semi-finite
compact measure space with measure algebra $\frak A$, $(Y,\Tau,\nu)$ is
a complete strictly localizable measure space with measure algebra
$\frak B$, and $\pi:\frak A\to\frak B$ is an order-continuous Boolean
homomorphism, there is a function $g:Y\to X$ such that
$g^{-1}[E]\in\Tau$ and $g^{-1}[E]^{\ssbullet}=\pi(E^{\ssbullet})$ for
every $E\in\Sigma$\cmmnt{ (343B)}.

\spheader 451Ac A family $\Cal K$ of sets is a {\bf countably compact
class} if $\bigcap_{n\in\Bbb N}K_n\ne\emptyset$ whenever
$\sequencen{K_n}$ is a sequence in $\Cal K$ such that
$\bigcap_{i\le n}K_i\ne\emptyset$ for every $n\in\Bbb N$.   Any
subfamily of a
countably compact class is countably compact.   If $\Cal K$ is a
countably compact class, then there is a countably compact class
$\Cal K^*\supseteq\Cal K$ which is closed under finite unions and
countable intersections\cmmnt{ (413T)}.

\spheader 451Ad A measure space $(X,\Sigma,\mu)$ is {\bf perfect} if
whenever $f:X\to\Bbb R$ is measurable, $E\in\Sigma$ and $\mu E>0$, there
is a compact set $K\subseteq f[E]$ such that $\mu f^{-1}[K]>0$.
A countably separated semi-finite measure space is compact iff it is
perfect\cmmnt{ (343K)}.   A measure space $(X,\Sigma,\mu)$ is
isomorphic to the unit interval with Lebesgue measure iff it is an
atomless complete countably separated perfect probability
space\cmmnt{ (344Ka)}.
}%end of comment

\leader{451B}{}\cmmnt{ Now for the new class of measures.

\medskip

\noindent}{\bf Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $(X,\Sigma,\mu)$, or $\mu$, is {\bf
countably compact} if $\mu$ is inner regular with respect to some
countably compact class of sets.

\cmmnt{Evidently compact measures are
also countably compact.   A simple example of a countably compact
measure which is not compact is the countable-cocountable measure on an
uncountable set (342M).   For an example of a perfect
measure which is not countably compact, see 451U.

Note that if $\mu$ is inner regular with respect to a countably compact
class $\Cal K$, then it is also inner regular with respect to
$\Cal K\cap\Sigma$ (411B), and $\Cal K\cap\Sigma$ is still countably
compact.}

\leader{451C}{Proposition}\cmmnt{ ({\smc Ryll-Nardzewski 53})} Any
semi-finite countably compact measure is perfect.

\proof{ The central idea is the same as in 342L, but we need to refine
the second half of the argument.

\medskip

{\bf (a)} Let $(X,\Sigma,\mu)$ be a countably compact measure space,
$f:X\to\Bbb R$ a measurable function, and $E\in\Sigma$ a set of positive
measure.  Let $\Cal K$ be a countably compact class such that $\mu$ is
inner regular with respect to $\Cal K$;  by 451Ac, we may suppose that
$\Cal K$ is closed under finite unions and countable intersections.

Because $\mu$ is semi-finite, there is a measurable set $F\subseteq E$
such that $0<\mu F<\infty$;  replacing $F$ by a set of the form $F\cap
f^{-1}[[-n,n]]$ if necessary, we may suppose that $f[F]$ is bounded;
finally, we may suppose that $F\in\Cal K$.
Let $\langle\epsilon_q\rangle_{q\in\Bbb Q}$ be a family of strictly
positive real numbers such that $\sum_{q\in\Bbb Q}\epsilon_q<\bover12\mu
F$.   For each $q\in\Bbb Q$,
set $E_q=\{x:x\in F,\,f(x)\le q\}$, $E'_q=\{x:x\in F,\,f(x)>q\}$, and
choose $K_q$, $K'_q\in\Cal K\cap\Sigma$ such that $K_q\subseteq E_q$,
$K'_q\subseteq E'_q$
and $\mu(E_q\setminus K_q)\le\epsilon_q$, $\mu(E'_q\setminus
K'_q)\le\epsilon_q$.   Then $K=\bigcap_{q\in\Bbb Q}(K_q\cup K'_q)\in\Cal
K\cap\Sigma$, $K\subseteq F$ and

\Centerline{$\mu(F\setminus K)\le\sum_{q\in\Bbb Q}\mu(E_q\setminus
K_q)+\mu(E'_q\setminus K'_q)<\mu F$, }

\noindent so $\mu K>0$.

\medskip

{\bf (b)} Take any $t\in\overline{f[K]}$.   Enumerate $\Bbb Q$ as
$\sequencen{q_n}$ and define $\sequencen{L_n}$ in $\Cal K$ by the rule

$$\eqalign{L_n&=K_{q_n}\text{ if }t<q_n,\cr
&=K'_{q_n}\text{ if }t>q_n,\cr
&=F\text{ if }t=q_n.\cr}$$

\noindent Now $\bigcap_{i\le n}L_i\ne\emptyset$ for every $n\in\Bbb N$.
\Prf\ Because $t\in\overline{f[K]}$, there must be some $s\in f[K]$ such
that $s<q_i$ whenever $i\le n$ and $t<q_i$, while $s>q_i$ whenever $i\le
n$ and $t>q_i$.   Let $x\in K$ be such that $f(x)=s$.   Then, for any
$i\le n$,

\inset{{\it either} $t<q_i$, $f(x)<q_i$ so $x\notin K'_{q_i}$ and $x\in
K_{q_i}=L_i$}

\inset{{\it or} $t>q_i$, $f(x)>q_i$ so $x\notin K_{q_i}$ and $x\in
K'_{q_i}=L_i$,}

\inset{{\it or} $t=q_i$ and $x\in F=L_i$.}

\noindent So $x\in\bigcap_{i\le n}L_i$.\ \Qed

As $\Cal K$ is a countably compact class, there must be some
$x\in\bigcap_{n\in\Bbb N}L_n$.   But this means that, for any $n\in\Bbb
N$,

\inset{if $t>q_n$ then $x\in K'_{q_n}$ and $f(x)>q_n$,}

\inset{if $t<q_n$ then $x\in K_{q_n}$ and $f(x)<q_n$.}

\noindent So in fact $f(x)=t$.   Accordingly $t\in f[K]$.

\medskip

{\bf (c)} What this shows is that $\overline{f[K]}\subseteq f[K]$ and
$f[K]$ is closed.   Because (by the choice of $F$) it is also bounded,
it is compact (2A2F).   Of course we now have $f[K]\subseteq f[E]$,
while $\mu f^{-1}[f[K]]\ge\mu K>0$.   As $f$ and $E$ are arbitrary,
$\mu$ is perfect.
}%end of proof of 451C

\leader{451D}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$E\in\Sigma$;  let $\mu_E$ be the subspace measure on $E$.

(a) If $\mu$ is compact, so is $\mu_E$.

(b) If $\mu$ is countably compact, so is $\mu_E$.

(c) If $\mu$ is perfect, so is $\mu_E$.

\proof{{\bf (a)-(b)} Let $\Cal K$ be a (countably) compact class such
that $\mu$ is inner regular with respect to $\Cal K$.   Then $\mu_E$ is
inner regular with respect to $\Cal K$ (412Oa), so is (countably)
compact.

\medskip

{\bf (c)} Suppose that $f:E\to\Bbb R$ is $\Sigma_E$-measurable, where
$\Sigma_E=\Sigma\cap\Cal PE$ is the subspace $\sigma$-algebra, and
$F\subseteq E$ is such that $\mu F>0$.   Set

$$\eqalign{g(x)&=\arctan f(x)\text{ if }x\in E,\cr
&=2\text{ if }x\in X\setminus E.\cr}$$

\noindent Then $g$ is $\Sigma$-measurable, so there is a compact set
$K\subseteq g[F]$ such that $\mu g^{-1}[K]>0$.   Set
$L=\{\tan t:t\in K\}$;  then $L\subseteq f[F]$ is compact and
$f^{-1}[L]=g^{-1}[K]$ has non-zero measure.   As $f$ and $F$ are
arbitrary, $\mu_E$ is perfect.
}%end of proof of 451D

\leader{451E}{Proposition} Let $(X,\Sigma,\mu)$ be a perfect measure
space.

(a) If $(Y,\Tau,\nu)$ is another measure space and $f:X\to Y$ is an
\imp\ function, then $\nu$ is perfect.

(b) In particular, $\mu\restrp\Tau$ is perfect for any
$\sigma$-subalgebra $\Tau$ of $\Sigma$.

\proof{{\bf (a)} Suppose that $g:Y\to\Bbb R$ is $\Tau$-measurable and
$F\in\Tau$ is such that $\nu F>0$.   Then $gf:X\to\Bbb R$ is
$\Sigma$-measurable and $\mu f^{-1}[F]>0$.   So there is a compact set
$K\subseteq(gf)[f^{-1}[F]]$ such that $\mu(gf)^{-1}[K]>0$.   But now
$K\subseteq g[F]$ and $\nu g^{-1}[K]>0$.    As $g$ and $F$ are
arbitrary, $\nu$ is perfect.

\medskip

{\bf (b)} Apply (a) to $Y=X$, $\nu=\mu\restrp\Tau$ and $f$ the identity
function.
}%end of proof of 451E

\cmmnt{\medskip

\noindent{\bf Remark} We shall see in 452R that there is a similar
result for countably compact measures;  but for compact measures, there
is not (342Xf, 451Xh).
}%end of comment

\leader{451F}{Lemma}\cmmnt{ ({\smc Sazonov 66})} Let $(X,\Sigma,\mu)$
be a semi-finite measure
space.   Then the following are equiveridical:

(i) $\mu$ is perfect;

(ii) $\mu\restrp\Tau$ is compact for every
countably generated $\sigma$-subalgebra $\Tau$ of $\Sigma$;

(iii) $\mu\restrp\Tau$ is perfect for every countably generated
$\sigma$-subalgebra $\Tau$ of $\Sigma$;

(iv) for every countable set $\Cal E\subseteq\Sigma$ there is a
$\sigma$-algebra $\Tau\supseteq\Cal E$ such that $\mu\restrp\Tau$ is
perfect.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Suppose that $\mu$ is perfect, and
that $\Tau$ is a countably generated $\sigma$-subalgebra of $\Sigma$.
Let $\sequencen{E_n}$ be a sequence in $\Tau$ which $\sigma$-generates
it, and define $f:X\to\Bbb R$ by setting
$f(x)=\sum_{n=0}^{\infty}3^{-n}\chi E_n(x)$ for every $x\in X$.
Then $f$ is measurable.   Set $\Cal
K=\{f^{-1}[L]:L\subseteq f[X]$ is compact$\}$.   Then $\Cal K$ is a
compact class.   \Prf\ If $\Cal K'\subseteq\Cal K$ is non-empty and has
the finite intersection property, then $\Cal L'=\{L:L\subseteq f[X]$ is
compact, $f^{-1}[L]\in\Cal K'\}$ is also a non-empty family with the
finite intersection property.   So there is an $\alpha\in\bigcap\Cal L'$;
since $\alpha\in f[X]$, there is an $x$ such that $f(x)=\alpha$,
and now $x\in\bigcap\Cal K'$.   As $\Cal K'$ is arbitrary, $\Cal K$ is a
compact class.\ \Qed

Observe next that, for any $n\in\Bbb N$,

\Centerline{$E_n=\{x:\,\exists\,I\subseteq n,\,
  \sum_{i\in I}3^{-i}+3^{-n}\le f(x)<\sum_{i\in I}3^{-i}+3^{-n+1}\}$.}

\noindent So $\Tau'=\{f^{-1}[F]:F\subseteq\Bbb R\}$ contains
every $E_n$;  as it is a
$\sigma$-algebra of subsets of $X$, it includes $\Tau$.

Now $\mu\restrp\Tau$ is inner regular with respect to $\Cal K$.   \Prf\
If $E\in\Tau$ and $\mu E>0$, there is a set $F\subseteq\Bbb R$ such that
$E=f^{-1}[F]$.   Because $f$ is $\Sigma$-measurable
and $\mu$ is perfect, there is a compact set $L\subseteq f[E]$ such that
$\mu f^{-1}[L]>0$.   But now $f^{-1}[L]\in\Cal K\cap\Tau$, and
$f^{-1}[L]\subseteq E$ because $L\subseteq F$.
Because $\Cal K$ is closed under finite unions, this is enough to show
that $\mu\restrp\Tau$ is inner regular with respect to $\Cal K$.\ \Qed

Thus $\Cal K$ witnesses that $\mu\restrp\Tau$ is a compact measure.

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Now suppose that $\mu\restrp\Tau$ is
compact for every countably generated $\sigma$-algebra
$\Tau\subseteq\Sigma$, that $f:X\to\Bbb R$ is a measurable function,
and that $\mu E>0$.   Let $F\subseteq E$ be a
measurable set of non-zero finite measure, and $\Tau$ the
$\sigma$-algebra generated by
$\{F\}\cup\{f^{-1}[\,\ooint{-\infty,q}\,]:q\in\Bbb Q\}$, so that $\Tau$
is countably generated and $f$ is $\Tau$-measurable.   Because
$\mu\restrp\Tau$ is compact, so is the subspace measure
$(\mu\restrp\Tau)_F$ (451Da);  but this is now perfect (342L or 451C),
while $F\in\Tau$ and $\mu F>0$, so there is a compact set
$L\subseteq f[F]\subseteq f[E]$ such that $\mu f^{-1}[L]>0$.   As $f$
and $E$ are arbitrary, $\mu$ is perfect.

\medskip

{\bf (c)(i)$\Rightarrow$(iv)} is trivial.

\medskip

{\bf (d)(iv)$\Rightarrow$(iii)} If (iv) is true, and $\Tau$ is a
countably generated $\sigma$-subalgebra of $\Sigma$, let $\Cal E$ be a
countable set generating it.   Then there is a $\sigma$-algebra
$\Tau_1\supseteq\Cal E$ such that $\mu\restrp\Tau_1$ is perfect.
By 451Eb, $\mu\restrp\Tau=(\mu\restrp\Tau_1)\restrp\Tau$ is compact,
therefore perfect.

\medskip

{\bf (e)(iii)$\Rightarrow$(ii)} If (iii) is true, and $\Tau$ is a
countably generated $\sigma$-subalgebra of $\Sigma$, then
$\mu\restrp\Tau$ is perfect;  but as (i)$\Rightarrow$(ii), and $\Tau$ is
a countably generated $\sigma$-subalgebra of itself, $\mu\restrp\Tau$ is
compact.
}%end of proof of 451F

\leader{451G}{Proposition} Let $(X,\Sigma,\mu)$ be a
measure space.   Let $(X,\hat\Sigma,\hat\mu)$ be its completion and
$(X,\tilde\Sigma,\tilde\mu)$ its c.l.d.\ version.   Then

(a)(i) if $\mu$ is compact, so are $\hat\mu$ and $\tilde\mu$;

\quad(ii) if $\mu$ is semi-finite and either $\hat\mu$ or $\tilde\mu$ is
compact, then $\mu$ is compact.

(b)(i) If $\mu$ is countably compact, so are $\hat\mu$ and $\tilde\mu$;

\quad(ii) if $\mu$ is semi-finite and either $\hat\mu$ or $\tilde\mu$ is
countably compact, then $\mu$ is countably compact.

(c)(i) If $\mu$ is perfect, so are $\hat\mu$ and $\tilde\mu$;

\quad(ii) if $\hat\mu$ is perfect, then $\mu$ is perfect;

\quad(iii) if $\mu$ is semi-finite and $\tilde\mu$ is perfect, then
$\mu$ is perfect.

\proof{{\bf (a)-(b)}  The arguments for $\hat\mu$ and $\tilde\mu$ run
very closely together.   Write $\check\mu$ for either of them, and
$\check\Sigma$ for its domain.

\medskip

\quad{\bf (i)} If $\mu$ is inner regular with respect to $\Cal K$, so is
$\check\mu$ (412Ha).   So if $\mu$ is (countably) compact, so is
$\check\mu$.

\medskip

\quad{\bf (ii)} Now suppose that $\mu$ is semi-finite.   The point is
that if $\Cal K$ is closed under countable intersections and $\check\mu$
is inner regular with respect to $\Cal K$, so is $\mu$.   \Prf\ Suppose
that $E\in\Sigma$ and that $\mu E>\gamma$.   Choose sequences
$\sequencen{E_n}$ in $\Sigma$ and $\Cal K_n$ in $\Cal K$ inductively, as
follows.   $E_0$ is to be such that $E_0\subseteq E$ and
$\gamma<\mu E_0<\infty$.   Given that $\gamma<\mu E_n<\infty$, let
$K_n\in\Cal K\cap\check\Sigma$ be such that $K_n\subseteq E_n$ and
$\check\mu K_n>\gamma$;  now take $E_{n+1}\in\Sigma$ such that
$E_{n+1}\subseteq K_n$ and $\mu E_{n+1}=\check\mu K_n$ (212C or 213Fc),
and continue.
At the end of the induction, $\bigcap_{n\in\Bbb N}K_n=\bigcap_{n\in\Bbb
N}E_n$ is a member of
$\Sigma\cap\Cal K$ included in $E$ and of measure at least $\gamma$.
As $E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with respect
to $\Cal K$.\ \Qed

It follows that if $\check\mu$ is compact or countably compact, so is
$\mu$.  \Prf\ Let $\Cal K$ be a (countably) compact class such that
$\check\mu$ is inner regular with respect to $\Cal K$;  by 451Aa or
451Ac, there is a (countably) compact class $\Cal K^*$, including $\Cal
K$, which is closed under countable intersections, so that $\mu$ is
inner regular with respect to $\Cal K^*$, and is itself (countably)
compact.\ \Qed

\medskip

{\bf (c)(i)}\grheada\ Let $f:X\to\Bbb R$ be $\hat\Sigma$-measurable, and
$E\in\hat\Sigma$ such that $\hat\mu E>0$.   Then there are a
$\mu$-conegligible set $F_0\in\Sigma$ such that $f\restr F_0$ is
$\Sigma$-measurable (212Fa), and an $F_1\in\Sigma$ such that
$F_1\subseteq E$ and $\hat\mu(E\setminus F_1)=0$.  Set $F=F_0\cap F_1$.
By 451Dc, the subspace measure $\mu_F$ is perfect, while $f\restr F$ is
$\Sigma_F$-measurable;  so there is a compact set $K\subseteq f[F]$
such that $\mu(F\cap f^{-1}[K])>0$.   But now $K\subseteq f[E]$ and
$\hat\mu f^{-1}[K]>0$.   As $f$ and $E$ are arbitrary, $\hat\mu$ is
perfect.

\medskip

\qquad\grheadb\ Let $f:X\to\Bbb R$ be $\tilde\Sigma$-measurable,
and $E\in\tilde\Sigma$ such
that $\tilde\mu E>0$.   Then there is a set $F\in\Sigma$ such that
$\mu F<\infty$ and $\hat\mu(F\cap E)$ is defined and greater than $0$
(213D).   In this case, $\hat\mu$ and $\tilde\mu$ induce the same
subspace measure
$\hat\mu_F$ on $F$.   Accordingly $f\restr F$ is
$\hat\Sigma$-measurable.   Because $\hat\mu$ is perfect (by ($\alpha$)
just above), so is $\hat\mu_F$ (451Dc), and there is a compact set
$K\subseteq f[F\cap E]$ such that $\hat\mu_F(f\restr F)^{-1}[K]>0$.
But now, of course, $K\subseteq f[E]$ and $\tilde\mu f^{-1}[K]>0$.   As
$f$ and $E$ are arbitrary, $\tilde\mu$ is perfect.

\medskip

\quad{\bf (ii)} Suppose that $\hat\mu$ is perfect.   Since
$\mu=\hat\mu\restr\Sigma$, $\mu$ is perfect, by 451Eb.

\medskip

\quad{\bf (iii)} Similarly, if $\tilde\mu$ is perfect and $\mu$ is
semi-finite, then $\mu=\tilde\mu\restr\Sigma$, by 213Hc, so
$\mu$ is perfect.
}%end of proof of 451G

\leader{451H}{Lemma} Let $\familyiI{X_i}$ be a family of sets with
product $X$.   Suppose that $\Cal K_i\subseteq\Cal PX_i$ for each $i\in
I$, and set $\Cal K=\{\pi_i^{-1}[K]:i\in I,\,K\in\Cal K_i\}$, where
$\pi_i:X\to X_i$ is the coordinate map for each $i\in I$.   Then

(a) if every $\Cal K_i$ is a compact class, so is $\Cal K$;

(b) if every $\Cal K_i$ is a countably compact class, so is $\Cal K$.

\proof{{\bf (a)} For each $i\in I$, let $\frak T_i$ be a compact
topology on $X_i$ such that every
member of $\Cal K_i$ is closed.   Then the product topology $\frak T$ on
$X$ is compact (3A3J), and every member of $\Cal K$ is $\frak T$-closed,
so $\Cal K$ is a compact class.

\medskip

{\bf (b)} If $\sequencen{K_n}$ is a sequence in
$\Cal K$ such that $\bigcap_{k\le n}K_k\ne\emptyset$ for every
$n\in\Bbb N$, then we must be able to express each $K_n$ as
$\pi_{j_n}^{-1}[L_n]$, where $j_n\in I$ and $L_n\in\Cal K_{j_n}$
for every $n$.   Now, for $i\in I$,
$\Cal L_i=\{K_{j_n}:n\in\Bbb N,\,j_n=i\}$ is a countable subset of
$\Cal K_i$, and any finite
subfamily of $\Cal L_i$ has non-empty intersection.   Since
$K_0\ne\emptyset$, $X_i\ne\emptyset$;  so, whether $\Cal L_i$ is empty
or not, $X_i\cap\bigcap\Cal L_i$ is non-empty.   Accordingly

\Centerline{$\bigcap_{k\in\Bbb N}K_k
=\prod_{i\in I}(X_i\cap\bigcap\Cal L_i)$}

\noindent is not empty.   As $\sequencen{K_n}$ is arbitrary, $\Cal K$ is
countably compact.
}%end of proof of 451H

\leader{451I}{Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with c.l.d.\ product $(X\times Y,\Lambda,\lambda)$.

(a) If $\mu$ and $\nu$ are compact, so is $\lambda$.

(b) If $\mu$ and $\nu$ are countably compact, so is $\lambda$.

(c) If $\mu$ and $\nu$ are perfect, so is $\lambda$.

\proof{{\bf (a)-(b)} Let $\Cal K\subseteq\Cal PX$,
$\Cal L\subseteq\Cal PY$ be
(countably) compact classes such that $\mu$ is inner regular with
respect to $\Cal K$ and $\nu$ is inner regular with respect to $\Cal L$.
Set $\Cal M_0=\{K\times Y:K\in\Cal K\}\cup\{X\times L:L\in\Cal L\}$.
Then $\Cal M_0$ is (countably) compact, by 451H.   By 451Aa/451Ac, there
is a (countably) compact class $\Cal M\supseteq\Cal M_0$ which is closed
under finite unions and countable intersections.   By 412R, $\lambda$ is
inner regular with respect to $\Cal M$, so is (countably) compact.

\medskip

{\bf (c)(i)} Let $f:X\times Y\to\Bbb R$ be $\Lambda$-measurable, and
$V\in\Lambda$ a set of positive measure.   Then there are $G\in\Sigma$,
$H\in\Tau$ such that $\mu G$, $\nu H$ are both finite and
$\lambda(V\cap(G\times H))>0$.   Recall that the subspace measure
$\lambda_{G\times H}$ on $G\times H$ is just the product of the subspace
measures $\mu_G$ and $\mu_H$ (251P(ii-$\alpha$)), and is the completion
of its restriction $\theta$ to the $\sigma$-algebra
$\Sigma_G\tensorhat\Tau_H$ generated by $\{E\times F:E\in\Sigma_G$,
$F\in\Tau_H\}$, where $\Sigma_G$ and $\Tau_H$ are the subspace
$\sigma$-algebras on $G$, $H$ respectively, the domains of $\mu_G$ and
$\mu_H$ (251K).
Next, for any $W\in\Sigma_G\tensorhat\Tau_H$, there are
countable families $\Cal E\subseteq\Sigma_G$, $\Cal F\subseteq\Tau_H$
such that $W$ belongs to the $\sigma$-algebra of subsets of $G\times H$
generated by $\{E\times F:E\in\Cal E,\,F\in\Cal F\}$ (331Gd).

\medskip

\quad{\bf (ii)} The point is that $\theta$ is perfect.   \Prf\ Let
$\Lambda'$ be any countably generated $\sigma$-subalgebra of
$\Sigma_G\tensorhat\Tau_H$;  let $\sequencen{W_n}$ be a sequence in
$\Lambda'$ generating it.   Then there are countable families
$\Cal E\subseteq\Sigma_G$, $\Cal F\subseteq\Tau_H$ such that every $W_n$
belongs to the $\sigma$-algebra generated by
$\{E\times F:E\in\Cal E,\,F\in\Cal F\}$.   Let $\Sigma'$, $\Tau'$ be the
$\sigma$-algebras of
subsets of $G$ and $H$ generated by $\Cal E$ and $\Cal F$ respectively;
then every $W_n$ belongs to $\Sigma'\tensorhat\Tau'$, so
$\Lambda'\subseteq\Sigma'\tensorhat\Tau'$.   Let $\lambda'$ be the
product of the measures $\mu\restr\Sigma'=\mu_G\restr\Sigma'$ and
$\nu\restrp\Tau'$.   Then $\lambda'$ is the completion of its
restriction to $\Sigma'\tensorhat\Tau'$.

Now trace through the results above.   $\mu_G$ and $\nu_H$ are perfect
(451Dc), so $\mu_G\restr\Sigma'$ and $\nu_H\restrp\Tau'$ are compact
(451F), so $\lambda'$ is
compact ((a) of this theorem), so $\lambda'$ is perfect
(342L or 451C again).
But $\theta$ must agree with $\lambda'$ on $\Lambda'$, by Fubini's
theorem (252D), or otherwise, so $\theta\restr\Lambda'$ is a restriction
of $\lambda'$, and is perfect (451Eb).

Thus $\theta\restr\Lambda'$ is perfect for every countably generated
$\sigma$-subalgebra $\Lambda'$ of $\dom\theta$.   By 451F, $\theta$ is
perfect.\ \Qed

\medskip

\quad{\bf (iii)} By 451G(c-i), $\lambda_{G\times H}$ is perfect.   Now
$f\restr G\times H$ is measurable, and $\lambda_{G\times
H}(V\cap(G\times H))>0$, so there is a compact set $K\subseteq
f[V\cap(G\times H)]$ such that
$\lambda_{G\times H}((G\times H)\cap f^{-1}[K])>0$;  in which case
$K\subseteq f[V]$ and $\lambda f^{-1}[K]>0$.

As $f$ and $V$ are arbitary, $\lambda$ is perfect.
}%end of proof of 451I

\leader{451J}{Theorem} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family
of probability spaces, with product $(X,\Sigma,\mu)$.

(a) If every $\mu_i$ is compact, so is $\mu$.

(b) \cmmnt{({\smc Marczewski 53})}
If every $\mu_i$ is countably compact, so is $\mu$.

(c) If every $\mu_i$ is perfect, so is $\mu$.

\proof{ The same strategy as in 451I is again effective.

\medskip

{\bf (a)-(b)} For each $i\in I$, let $\Cal K_i\subseteq\Cal PX_i$ be a
(countably) compact class such that $\mu_i$ is inner regular with
respect to $\Cal K_i$.   Set
$\Cal M_0=\{\pi_i^{-1}[K]:i\in I,\,K\in\Cal K_i\}$, so that
$\Cal M_0$ is (countably) compact.   Let $\Cal M\supseteq\Cal M_0$ be a
(countably)
compact class which is closed under finite unions and countable
intersections.   By 412T, $\mu$ is inner regular with respect to
$\Cal M$, so is (countably) compact.

\medskip

{\bf (c)} Let $\Lambda'$ be a countably generated $\sigma$-subalgebra of
$\Tensorhat_{i\in I}\Sigma_i$, the $\sigma$-algebra of subsets of $X$
generated by the sets $\{x:x(i)\in E\}$ for $i\in I$ and $E\in\Sigma_i$.
Then $\lambda\restr\Lambda'$ is perfect.   \Prf\ For every
$W\in\Tensorhat_{i\in I}\Sigma_i$, we must be able to find countable
subsets $\Cal E_i$ of $\Sigma_i$ such that $W$ is in the
$\sigma$-algebra generated by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$;
so there are in fact countable sets $\Cal E_i\subseteq\Sigma_i$ such
that the $\sigma$-algebra generated by $\{\pi_i^{-1}[E]:i\in
I,\,E\in\Sigma_i\}$ includes $\Lambda'$.   Let $\Tau_i$ be the
$\sigma$-subalgebra of $\Sigma_i$ generated by $\Cal E_i$, so that
$\mu_i\restrp\Tau_i$ is compact.   Let $\lambda'$ be the product of
$\familyiI{\mu_i\restrp\Tau_i}$;  then $\lambda'$ is compact, by (a)
above, therefore perfect.   Now $\lambda$ is an extension of $\lambda'$,
by 254G or otherwise, so $\lambda'$ is an extension of
$\lambda\restr\Lambda'$, and $\lambda\restr\Lambda'$ is perfect.\ \QeD\
As $\Lambda'$ is arbitrary, $\lambda\restr\Tensorhat_{i\in I}\Sigma_i$
is perfect, and its completion $\lambda$ (254Ff) also is perfect.
}%end of proof of 451J

\cmmnt{\medskip

\noindent{\bf Remark} This theorem is generalized in 454Ab.
}

\leader{451K}{}\cmmnt{ The following result is interesting because it
can be reached from an unexpectedly weak hypothesis;  it will be useful in
\S455.

\medskip

\noindent}{\bf Proposition} Let $\familyiI{X_i}$ be a family of sets
with product $X$, and
$\Sigma_i$ a $\sigma$-algebra of subsets of $X_i$ for each $i$.   Let
$\lambda$ be a perfect totally finite measure with domain
$\Tensorhat_{i\in I}\Sigma_i$.   Set $\pi_J(x)=x\restr J$ for
$x\in X$ and $J\subseteq I$.

(a) Let $\Cal K$ be the set
$\{V:V\subseteq X$, $\pi_J[V]\in\Tensorhat_{i\in J}\Sigma_i$ for every
$J\subseteq I\}$.   Then $\lambda$ is inner regular with respect to
$\Cal K$.

(b) Let $\hat\lambda$ be the completion of $\lambda$.

\quad(i) For any $J\subseteq I$, the completion of the image measure
$\lambda\pi_J^{-1}$ on $\prod_{i\in J}X_i$ is the image measure
$\hat\lambda\pi_J^{-1}$.

\quad(ii) If $W$ is measured by $\hat\lambda$ and $W$ is determined by
coordinates in $J\subseteq I$, then there is a
$V\in\Tensorhat_{i\in I}\Sigma_i$ such that
$V\subseteq W$, $V$ is determined by coordinates in $J$ and $W\setminus V$
is $\lambda$-negligible.

\proof{{\bf (a)(i)} Take $W\in\Tensorhat_{i\in I}\Sigma_i$.
Then we can find a family $\familyiI{\Tau_i}$ such
that $\Tau_i$ is a countably generated $\sigma$-subalgebra of $\Sigma_i$
for each $i$ and $W\in\Tensorhat_{i\in I}\Tau_i$.
For each $i\in I$ and $E\in\Tau_i$ set
$\lambda_iE=\lambda\{x:x\in X$, $x(i)\in E\}$;  then $\lambda_i$ is perfect
(451Ea).   Because $\Tau_i$ is countably generated, $\lambda_i$ is compact
(451F);  let $\Cal K_i$ be a compact class such that $\lambda_i$ is inner
regular with respect to $\Cal K_i$.   By 342D, we may suppose that
$\Cal K_i$ is the family of closed sets for a compact topology $\frak T_i$
on $X_i$.

\medskip

\quad{\bf (ii)} Let $\Cal V$ be the family of all sets $V\subseteq X$
expressible in the form

\Centerline{$V
=\bigcap_{n\in\Bbb N}\bigcup_{i\in J_n}\{x:x\in X$, $x(i)\in K_{ni}\}$}

\noindent where $\sequencen{J_n}$ is a sequence of finite subsets of $I$
and $K_{ni}\in\Cal K_i\cap\Tau_i$ whenever $n\in\Bbb N$ and $i\in J_n$.
Given $V$ expressed in this form, set
$V_n=\bigcap_{m\le n}\bigcup_{i\in J_m}\{x:x(i)\in K_{mi}\}$ for each $n$.
Then $\pi_J[V]=\bigcap_{n\in\Bbb N}\pi_J[V_n]$ for every
$J\subseteq I$.   \Prf\ The product topology
$\frak T$ on $X$ is compact, and all the $V_n$ are $\frak T$-closed.
If $z\in\bigcap_{n\in\Bbb N}\pi_J[V_n]$, then for each $n\in\Bbb N$ there
is an $x_n\in V_n$ such that $\pi_J(x)=z$.   Let $x$
be a cluster point of $\sequencen{x_n}$.
The topologies are not Hausdorff, so we do not know at once that
$\pi_J(x)=z$;  but if we define $x'$ by saying that

$$\eqalign{x'(i)
&=z(i)\text{ if }i\in J,\cr
&=x(i)\text{ if }i\in I\setminus J,\cr}$$

\noindent then any neighbourhood $U$
of $x'$ must include a neighbourhood of
the form $\{y:y(i)\in U_i$ for $i\in K\}$ where $K\subseteq I$ is finite
and $U_i$ is a neighbourhood of $x'(i)$ for each $i\in K$.   In this case,
$\{y:y\in U_i$ for $i\in K\setminus J\}$ is a neighbourhood of $x$, so

\Centerline{$\{n:x_n\in U\}
\supseteq\{n:x_n(i)\in U_i$ for $i\in K\setminus J\}$}

\noindent is infinite.   Thus $x'$ also is a cluster point of
$\sequencen{x_n}$, while $\pi_J(x')=z$.   Since
$x'\in\overline{\{x_m:m\ge n\}}\subseteq V_n$ for every $n$, $x\in V$,
and $z\in\pi_J[V]$.   Thus
$\bigcap_{n\in\Bbb N}\pi_J[V_n]\subseteq\pi_J[V]$.   Since surely
$\pi_J[V]\subseteq\bigcap_{n\in\Bbb N}\pi_J[V_n]$, we have equality.\ \Qed

It follows that $V\in\Cal K$.   \Prf\ If $J\subseteq I$ and
$n\in\Bbb N$, then $V_n$ belongs to the  algebra of
subsets of $X$ generated by sets of the form $\{x:x(i)\in H\}$ where
$i\in I$ and $H\in\Sigma_i$, which we can identify with the free product
$\bigotimes_{i\in I}\Sigma_i$ (315Ma\formerly{3{}15L}).
This means that $V_n$ can be
expressed as a finite union of cylinder sets of the form
$C=\prod_{i\in I}H_i$ where $H_i\in\Sigma_i$ for every $i$ and
$\{i:H_i\ne X_i\}$ is finite (315Kb\formerly{3{}15J}).
But in this case $\pi_J[C]$ is
either empty or $\prod_{i\in J}H_i$, and in either case belongs to
$\Tensorhat_{i\in J}\Sigma_i$.   So $\pi_J[V_n]$, being a finite union
of such sets, also belongs to $\Tensorhat_{i\in J}\Sigma_i$.   As this
is true for every $n\in\Bbb N$,
$\pi_J[V]=\bigcap_{n\in\Bbb N}\pi_J[V_n]$ belongs to
$\Tensorhat_{i\in J}\Sigma_i$.  As $J$ is arbitrary, $V\in\Cal K$.\ \Qed

\medskip

\quad{\bf (iii)} Observe next that $\Cal V$ is closed under finite unions.
\Prf\ If $V'$, $V''\in\Cal V$, express them as

\Centerline{$V'
=\bigcap_{n\in\Bbb N}\bigcup_{i\in J'_n}\{x:x(i)\in K'_{ni}\}$}

\Centerline{$V''
=\bigcap_{n\in\Bbb N}\bigcup_{i\in J''_n}\{x:x(i)\in K''_{ni}\}$}

\noindent where, for each $n$, $J'_n$, $J''_n\subseteq I$ are finite,
$K'_{ni}\in\Cal K_{ni}\cap\Sigma_i$ for $i\in J'_n$ and
$K''_{ni}\in\Cal K_{ni}\cap\Sigma_i$ for $i\in J''_n$.
For $m$, $n\in\Bbb N$, set $J_{mn}=J'_m\cup J''_n$ and

$$\eqalign{K_{mni}
&=K'_{mi}\cup K''_{ni}\text{ if }i\in J'_m\cap J''_n,\cr
&=K'_{mi}\text{ if }i\in J'_m\setminus J''_n,\cr
&=K''_{ni}\text{ if }i\in J''_n\setminus J''_m.\cr}$$

\noindent Then

\Centerline{$V'\cap V''
=\bigcap_{m,n\in\Bbb N}\bigcup_{i\in J_{mn}}\{x:x(i)\in K_{mni}\}
\in\Cal V$.  \Qed}

\noindent We see also, immediately from its definition, that $\Cal V$ is
closed under countable intersections.

\medskip

\quad{\bf (iv)} Now consider the family $\Cal A$ of sets of the form
$\{x:x(i)\in E\}$ where $i\in I$ and $E\in\Tau_i$.   If
$A\in\Cal A$ is expressed in this form, then

\Centerline{$\sup\{\lambda V:V\in\Cal V$, $V\subseteq A\}
\ge\sup\{\lambda_iK:K\in\Cal K_i\cap\Tau_i$, $K\subseteq E\}
=\lambda_iE=\lambda A$.}

\noindent By 412C, $\lambda\restr\Tensorhat_{i\in I}\Tau_i$ is inner
regular with respect to $\Cal V$.   In particular, returning to our
original set $W$,

\Centerline{$\mu W=\sup\{\lambda V:V\in\Cal V$, $V\subseteq W\}
=\sup\{\lambda K:K\in\Cal K$, $K\subseteq W\}$.}

\noindent As $W$ is arbitrary, $\lambda$ is inner regular with respect to
$\Cal K$.

\medskip

{\bf (b)(i)} Write $\lambda_J=\lambda\pi_J^{-1}$ and $\hat\lambda_J$
for its completion.
Since $\pi_J:X\to\prod_{i\in J}X_J$ is \imp\ for $\lambda$ and $\lambda_J$,
it is \imp\ for $\hat\lambda$ and $\hat\lambda_J$
(234Ba\formerly{2{}35Hc}), that is,
$\hat\lambda\pi_J^{-1}$ extends $\hat\lambda_J$.   Now suppose that
$V$ is measured by $\hat\lambda\pi_J^{-1}$.   Since $\lambda$ is inner
regular with respect to $\Cal K$, so is $\hat\lambda$ (412Ha again), so

$$\eqalign{\hat\lambda\pi_J^{-1}[V]
&=\sup\{\lambda K:K\in\Cal K,\,K\subseteq\pi_J^{-1}[V]\}\cr
&\le\sup\{\lambda\pi_J^{-1}[\pi_J[K]]:K\in\Cal K,\,
  K\subseteq\pi_J^{-1}[V]\}\cr
&\le\sup\{\lambda\pi_J^{-1}[F]:F\in\Tensorhat_{i\in J}\Sigma_i,\,
   F\subseteq V\}.\cr}$$

\noindent As $V$ is arbitrary,
$\hat\lambda\pi_J^{-1}$ is inner regular with respect to
$\Tensorhat_{i\in J}\Sigma_i$.   By 412Mb (or otherwise),
$\hat\lambda\pi_J^{-1}=\hat\lambda_J$.

\medskip

\quad{\bf (ii)} Because $\pi_J^{-1}[\pi_J[W]]=W$, $\pi_J[W]$ is measured by
$\hat\lambda\pi_J^{-1}=\hat\lambda_J$.   So there is a
$V'\subseteq\pi_J[W]$, measured by $\lambda_J$, such that

\Centerline{$0=\hat\lambda_J(\pi_J[W]\setminus V')
=\hat\lambda(W\setminus\pi_J^{-1}[V'])$,}

\noindent and we can take $V=\pi_J^{-1}[V']$.
}%end of proof of 451K

\leader{*451L}{}\cmmnt{ The next result is sometimes useful, as a
fractionally weaker sufficient condition for compactness or countable
compactness of a measure.

\medskip

\noindent}{\bf Proposition}\cmmnt{
({\smc Borodulin-Nadzieja \& Plebanek 05})}
Let $(X,\Sigma,\mu)$ be a strictly localizable
measure space.   Let us say
that a family $\Cal E\subseteq\Sigma$ is {\bf $\mu$-centered} if
$\mu(\bigcap\Cal E_0)>0$ for every non-empty finite
$\Cal E_0\subseteq\Cal E$.

(i) Suppose that $\mu$ is inner regular with respect to some
$\Cal K\subseteq\Sigma$ such that every $\mu$-centered subset of $\Cal K$
has non-empty intersection.   Then $\mu$ is compact.

(ii) Suppose that $\mu$ is inner regular with respect to some
$\Cal K\subseteq\Sigma$ such that every
countable $\mu$-centered subset of $\Cal K$
has non-empty intersection.   Then $\mu$ is countably compact.

\proof{ I take the two arguments together, as follows.   The case
$\mu X=0$ is trivial;  suppose henceforth that $\mu X>0$.
Let $\hat\mu$ be
the completion of $\mu$.   Then $\hat\mu$ is still strictly localizable
(212Gb) so has a lifting $\phi:\Sigma\to\Sigma$ (341K).   Let $\Cal K_1$ be
the set of all those $K\in\Sigma$ for which there is some sequence
$\sequencen{K_n}$ in $\Cal K$ such that

\Centerline{$K=\bigcap_{n\in\Bbb N}K_n
\subseteq\bigcap_{n\in\Bbb N}\phi K_n$.}

\noindent Then $\mu$ is inner regular with respect to $\Cal K_1$.   \Prf\
Suppose that $E\in\Sigma$ and $0\le\gamma<\mu E$.   Because $\mu$ is
semi-finite, there is an $F\in\Sigma$ such that $F\subseteq E$ and
$\gamma<\mu F<\infty$.   Choose $\sequencen{K_n}$ in $\Cal K$ inductively,
as follows.   $K_0$ is to be such that $K_0\subseteq F$ and
$\mu K_0>\gamma$.   Given that $\mu K_n>\gamma$, then
$\hat\mu(K_n\cap\phi K_n)=\hat\mu K_n>\gamma$;  also $\hat\mu$ is inner
regular with respect to $\Cal K$ (412Ha once more),
so there is a $K_{n+1}\in\Cal K$
such that $K_{n+1}\subseteq K_n\cap\phi K_n$ and
$\mu K_{n+1}=\hat\mu K_{n+1}>\gamma$.   Continue.   At the end of the
induction,

\Centerline{$K=\bigcap_{n\in\Bbb N}K_n
\subseteq\bigcap_{n\in\Bbb N}\phi K_n$}

\noindent belongs to $\Cal K_1$, is included in $E$ and has measure at
least $\gamma$.\ \Qed

Now $\Cal K_1$ is (countably) compact.   \Prf\ Let
$\Cal K'\subseteq\Cal K_1$ be a [countable] set with the finite
intersection
property.   For each $K\in\Cal K'$, let $\Cal E_K\subseteq\Cal K$ be a
countable set such that
$K=\bigcap\Cal E_K\subseteq\bigcap\{\phi E:E\in\Cal E_K\}$;  set
$\Cal E=\bigcup_{K\in\Cal K'}\Cal E_K$.   If $\Cal E_0\subseteq\Cal E$ is
finite and not empty, then
$\phi(\bigcap\Cal E_0)=\bigcap_{E\in\Cal E_0}\phi E$ includes the
intersection of a finite subfamily of $\Cal K'$, so is not empty, and
$\mu(\bigcap\Cal E_0)=\hat\mu(\bigcap\Cal E_0)$ is non-zero.   Thus
$\Cal E\subseteq\Cal K$ is a [countable] $\mu$-centered set and must have
non-empty intersection.   But now $\bigcap\Cal K'=\bigcap\Cal E$ is
non-empty.   As $\Cal K'$ is arbitrary, $\Cal K_1$ is (countably) compact.\
\Qed

So $\Cal K_1$ witnesses that $\mu$ is (countably) compact, as claimed.
}%end of proof of 451L

\leader{451M}{}\cmmnt{ The following is one of the basic ways in which
we can find ourselves with a compact measure.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma)$ be a standard Borel space.
Then any semi-finite measure $\mu$ with domain $\Sigma$ is compact,
therefore perfect.

\proof{ If $\frak T$ is a Polish topology on $X$ with respect to which
$\Sigma$ is the Borel $\sigma$-algebra, then $\mu$ is inner regular with
respect to the family $\Cal K$ of $\frak T$-compact sets (433Ca), which
is a compact class.
}%end of proof of 451M

\vleader{48pt}{451N}{Proposition} Let $(X,\Sigma,\mu)$ be a perfect
measure
space and $\frak T$ a T$_0$ topology on $X$ with a countable network
consisting of measurable sets.
\cmmnt{(For instance, $\mu$ might be a topological measure on a
regular space with a countable network (4A2Ng), or a second-countable
space.   In particular, $X$ might be a separable metrizable space.)}
Then $\mu$ is inner regular with respect to the compact sets.

\proof{ This is a refinement of 343K.
Let $\sequencen{E_n}$ be a sequence in $\Sigma$ running over a
network for $\frak T$.   Define $g:X\to\Bbb R$ by setting
$g=\sum_{n=0}^{\infty}3^{-n}\chi H_n$ (cf.\ 343E).   Then $g$ is
measurable, because every $\chi E_n$ is.
Writing $\alpha_I=\sum_{i\in I}3^{-i}$
for $I\subseteq\Bbb N$, and

\Centerline{$H_n=\bigcup_{I\subseteq n}
  \ooint{\alpha_I+\bover123^{-n},\alpha_I+3^{-n+1}}$,}

\noindent we see that $E_n=g^{-1}[H_n]$ for each $n\in\Bbb N$.
This shows that $g$ is injective, because if $x$, $y$ are distinct
points in $X$ there is an open set containing one but not the other, and
now there is an $n\in\Bbb N$ such that $E_n$ contains that one and not
the other, so that just one of $g(x)$, $g(y)$ belongs to $H_n$.
Also $g^{-1}:g[X]\to X$ is continuous, since
$(g^{-1})^{-1}[E_n]=g[E_n]=H_n\cap g[X]$ is relatively open in $g[X]$
for every $n\in\Bbb N$ (4A2B(a-ii)).

Now suppose that $E\in\Sigma$ and $\mu E>0$.   Then there is a compact
set $K\subseteq g[E]$ such that $\mu g^{-1}[K]>0$.   But as $g$ is
injective, $g^{-1}[K]\subseteq E$, and as $g^{-1}$ is continuous,
$g^{-1}[K]$ is compact.   By 412B, this is enough to show that $\mu$ is
inner regular with respect to the compact sets.
}%end of proof of 451N

\leader{451O}{Corollary}\dvAformerly{4{}51N} Let
$(X,\Sigma,\mu)$ be a complete perfect
measure space, $Y$ a Hausdorff space with a countable network
consisting of Borel sets and $f:X\to Y$ a measurable function.   If
the image measure $\mu f^{-1}$ is locally finite, it is a Radon
measure.

\proof{ Because $f$ is measurable, $\mu f^{-1}$ is a topological
measure;  by 451Ea, it is perfect;  by 451N, it is tight;  and it is
complete because $\mu$ is.
Because $Y$ has a countable network, it is Lindel\"of (4A2Nb), and
$\mu f^{-1}$ is $\sigma$-finite (411Ge), therefore locally determined.
So it is a Radon measure.
}%end of proof of 451O

\leader{451P}{Corollary} Let $(X,\Sigma,\mu)$ be a perfect measure
space, $Y$ a separable metrizable space, and $f:X\to Y$ a measurable
function.

(a) If $E\in\Sigma$ and $\gamma<\mu E$, there is a compact set
$K\subseteq f[E]$ such that $\mu(E\cap f^{-1}[K])\ge\gamma$.

(b) If $\nu=\mu f^{-1}$ is the image measure, then
$\mu_*f^{-1}[B]=\nu_*B$ for every $B\subseteq Y$.

(c) If moreover $\mu$ is $\sigma$-finite, then $\mu^*f^{-1}[B]=\nu^*B$
for every $B\subseteq Y$.

\proof{{\bf (a)} Consider the subspace measure $\mu_E$, the measurable
function
$f\restr E$ from $E$ to the separable metrizable space $f[E]$, and the
image measure $\nuprime=\mu_E(f\restr E)^{-1}$ on $f[E]$.   By
451Dc, 451Ea and 451N, this is tight, while $\nuprime f[E]=\mu E$;
so there is a compact set $K\subseteq f[E]$ such that
$\nuprime K\ge\gamma$, and this serves.

\medskip

{\bf (b)}(i) If $F\in\dom\nu$ and $F\subseteq B$, then

\Centerline{$\nu F=\mu f^{-1}[F]\le\mu_*f^{-1}[B]$;}

\noindent as $F$ is arbitrary, $\mu_*f^{-1}[B]\ge\nu_*B$.  (ii) If
$E\in\Sigma$ and $E\subseteq f^{-1}[B]$ and $\gamma<\mu E$, then (a)
tells us that there is a compact set $K\subseteq f[E]$ such that
$\mu(E\cap f^{-1}[K])\ge\gamma$, in which case

\Centerline{$\nu_*B\ge\nu K\ge\gamma$.}

\noindent As $E$ and $\gamma$ are arbitrary, $\nu_*B\ge\mu_*f^{-1}[B]$.

\medskip

{\bf (c)}(i) If $F\in\dom\nu$ and $F\supseteq B$, then

\Centerline{$\nu F=\mu f^{-1}[F]\ge\mu^*f^{-1}[B]$;}

\noindent as $F$ is arbitrary, $\mu^*f^{-1}[B]\le\nu^*B$.  (ii) If
$\mu^*f^{-1}[B]=\infty$, then of course $\mu^*f^{-1}[B]=\nu^*B$.
Otherwise, because $\mu$ is $\sigma$-finite, we can find a disjoint
sequence $\sequencen{E_n}$ of subsets of $X$ of finite measure, covering
$X$, such that $E_0\supseteq f^{-1}[B]$ and $\mu E_0=\mu^*f^{-1}[B]$.
Let $\epsilon>0$.   For each $n\ge 1$, (a) tells us that there is a
compact set $K_n\subseteq f[E_n]$ such that
$\mu f^{-1}[E_n\setminus K_n]\le 2^{-n}\epsilon$.   Set
$H=Y\setminus\bigcup_{n\ge 1}K_n$;  then $\nu H\le\mu E+\epsilon$, and
$B\subseteq H$.   So

\Centerline{$\nu^*B\le\nu H\le\mu E+\epsilon=\mu^*f^{-1}[B]+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu^*B\le\mu^*f^{-1}[B]$.
}%end of proof of 451P

\leader{451Q}{}\cmmnt{ I turn now to a remarkable extension of the
idea above to general metric spaces $Y$.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a semi-finite compact
measure space, and $\familyiI{E_i}$ a disjoint family of subsets of $X$
such that $\bigcup_{i\in J}E_i\in\Sigma$ for every $J\subseteq I$.
Then $\mu(\bigcup_{i\in I}E_i)=\sum_{i\in I}\mu E_i$.

\proof{{\bf (a)} To begin with (down to the end of part (d) of the
proof) assume that $\mu$ is complete and totally finite and that every $E_i$ is
negligible.
Set $X_0=\bigcup_{i\in I}E_i$, and let $\mu_0$ be the
subspace measure on $X_0$.   Define $f:X_0\to I$
by setting $f(x)=i$ if $i\in I$, $x\in E_i$, and let $\nu$ be the image
measure $\mu_0f^{-1}$, so that $\nu J=\mu(\bigcup_{i\in J}E_i)$ for
$J\subseteq I$;
then $(I,\Cal PI,\nu)$ is a totally finite measure space.

\medskip

{\bf (b)} $\nu$ is purely atomic.   \Prf\Quer\ Suppose, if possible,
otherwise;  that there is a $K\subseteq I$ such that $\nu K>0$ and the
subspace measure $\nu\restrp\Cal PK$ is atomless.   In this case there
is an \imp\ function $g:K\to[0,\gamma]$, where
$\gamma=\nu K$ and $[0,\gamma]$ is given
Lebesgue measure (343Cc);  write $\lambda$ for Lebesgue measure on
$[0,\gamma]$.   Set $X_1=f^{-1}[K]=\bigcup_{i\in K}E_i$ and let $\mu_1$
be the subspace measure on $X_1$.   Now $gf:X_1\to[0,\gamma]$ is \imp\
for $\mu_1$ and $\lambda$.
Because $\mu$ is compact, so is $\mu_1$ (451Da), so $\mu_1$ is perfect
(342L or 451C once more).
By 451O, the image measure $\lambda_1=\mu_1(gf)^{-1}$
is a Radon measure.   But $\lambda_1$ must be an extension of Lebesgue
measure $\lambda$, because $gf$ is \imp\ for $\mu_1$ and $\lambda$, and
$\lambda_1$ and $\lambda$ must agree on all compact sets.   By
416E(b-ii), $\lambda_1$ and
$\lambda$ are identical, and, in particular, have the same domains.
Now for any set $A\subseteq [0,\gamma]$,
$(gf)^{-1}[A]=\bigcup_{i\in J}E_i\in\Sigma$, where
$J=g^{-1}[A]\subseteq I$;  so
$A\in\dom\lambda_1=\dom\lambda$.   But we know from 134D or 419I that
not every subset of $[0,\gamma]$ can be Lebesgue measurable.\ \Bang\Qed

\medskip

{\bf (c)} But $\nu$ is also atomless.  \Prf\Quer\ Suppose, if possible,
that $M\subseteq I$ is an atom
for $\nu$.   Set $\gamma=\nu M=\mu(\bigcup_{i\in M}E_i)$,

\Centerline{$\Cal F=\{F:F\subseteq M,\,\nu(M\setminus F)=0\}$.}

\noindent Because $\nu F$ is defined for every $F\subseteq M$, and $M$
is an atom, $\Cal F$ is an ultrafilter on $M$;  and because $\nu$ is
countably additive, the intersection of any sequence in $\Cal F$ belongs
to $\Cal F$, that is, $\Cal F$ is $\omega_1$-complete (definition:
4A1Ib).   Also $\Cal F$ must be non-principal, because we are supposing
that $\nu\{i\}=0$ for every $i\in M$.   By 4A1K, there are a regular
uncountable cardinal $\kappa$ and a
function $h:M\to\kappa$ such that the image filter $\Cal H=h[[\Cal F]]$
is normal.

For each $\xi<\kappa$, $\kappa\setminus\xi\in\Cal H$, so

\Centerline{$G_{\xi}=(hf)^{-1}[\kappa\setminus\xi]
=\bigcup\{E_i:h(i)\ge\xi\}\in\Sigma$,
\quad$\mu G_{\xi}=\nu h^{-1}[\kappa\setminus\xi]=\gamma>0$.}

\noindent At this point I apply the full strength of the hypothesis that
$\mu$ is a compact measure.   Let $\Cal K\subseteq\Sigma$ be a compact
class such that $\mu$ is inner regular with respect to $\Cal K$, and for
each $\xi<\kappa$ choose $K_{\xi}\in\Cal K$ such that
$K_{\xi}\subseteq
G_{\xi}$ and $\mu K_{\xi}\ge\bover12\gamma$.   Let
$S\subseteq[\kappa]^{<\omega}$ be the family of those finite sets
$L\subseteq\kappa$ such that $\bigcap_{\xi\in L}K_{\xi}=\emptyset$.
Because $\Cal H$ is a normal
ultrafilter, there is an $H\in\Cal H$ such that,
for every $n\in\Bbb N$, $[H]^n$ is either a subset of $S$ or disjoint
from $S$ (4A1L).

If we look at $\{G_{\xi}:\xi\in H\}$, we see that it has empty
intersection,
because $h(f(x))\ge\xi$ for every $x\in G_{\xi}$, and $\sup H=\kappa$.
So $\bigcap_{\xi\in H}K_{\xi}=\emptyset$.   Because all the $K_{\xi}$
belong to
the compact class $\Cal K$, there must be a finite set $L_0\subseteq H$
such that $\bigcap_{\xi\in L_0}K_{\xi}=\emptyset$, that is, $L_0\in S$.
But this means that $[H]^n\cap S\ne\emptyset$, where $n=\#(L_0)$, so
that $[H]^n\subseteq S$, by the choice of $H$.
However, $H$ is surely infinite, so we can find distinct
$\xi_0,\ldots,\xi_{2n}$ in $H$.   If we now look at
$K_{\xi_0},\ldots,K_{\xi_{2n}}$, we see that
$\#(\{i:i\le 2n,\,x\in K_{\xi_i}\})<n$ for every $x\in X$, so

\Centerline{$\sum_{i=0}^{2n}\chi K_{\xi_i}\le(n-1)\chi G_0$,
\quad$\sum_{i=0}^{2n}\int\chi K_{\xi_i}\ge\Bover12\gamma(2n+1)$,}

\noindent which is impossible, because $\mu G_0=\gamma$.\ \Bang\Qed

\medskip

{\bf (d)} Thus $\nu$ is simultaneously atomless and purely atomic, which
means that $\nu I=0$, that is, that
$\mu(\bigcup_{i\in I}E_i)=0=\sum_{i\in I}\mu E_i$.

\medskip

{\bf (e)} Now let us return to the general case.   Of course

\Centerline{$\sum_{i\in I}\mu E_i
=\sup_{J\subseteq I\text{ is finite}}\sum_{i\in J}\mu E_i
\le\mu(\bigcup_{i\in I}E_i)$.}

\noindent\Quer\ Suppose, if possible, that
$\sum_{i\in I}\mu E_i<\mu(\bigcup_{i\in I}E_i)$.
Because $\mu$ is semi-finite, there is
a set $F\subseteq\bigcup_{i\in I}E_i$ such that
$\sum_{i\in I}\mu E_i<\mu F<\infty$.   Set $L=\{i:i\in I,\,\mu E_i>0\}$;
then $L$ must be
countable, so $\mu(\bigcup_{i\in J}E_i)=\sum_{i\in J}\mu E_i<\mu F$, and
$\mu G>0$, where $G=F\setminus\bigcup_{i\in L}E_i$.   Set
$E'_i=G\cap E_i$ for every $i\in I$, and let $\hat\mu_G$ be the
completion of the subspace measure $\mu_G$ on $G$.
Then $\hat\mu_G$ is compact (451Da, 451G(a-i)) and totally finite,
$\hat\mu_GE'_i=0$ for
every $i\in I$, $\bigcup_{i\in J}E'_i=G\cap\bigcup_{i\in J}E_i$ is
measured by
$\hat\mu_G$ for every $J\subseteq I$, every $E'_i$ is
$\hat\mu_G$-negligible,
but $\hat\mu_G(\bigcup_{i\in I}E'_i)=\mu G$ is not zero;
which contradicts the result of (a)-(d) above.\ \Bang

So $\sum_{i\in I}\mu E_i=\mu(\bigcup_{i\in I}E_i)$, as required.
}%end of proof of 451Q

\leader{451R}{Lemma} Let $(X,\Sigma,\mu)$ be a totally finite compact
measure space, $Y$ a metrizable space, and $f:X\to Y$ a measurable
function.  Then there is a closed separable subspace $Y_0$ of $Y$ such
that $f^{-1}[Y\setminus Y_0]$ is negligible.

\proof{{\bf (a)} (Cf.\ 438D.) By 4A2L(g-ii), there is a
$\sigma$-disjoint base $\Cal U$ for the topology of $Y$.   Express
$\Cal U$ as
$\bigcup_{n\in\Bbb N}\Cal U_n$ where $\Cal U_n$ is disjoint for each
$n$.   Then $\family{U}{\Cal U_n}{f^{-1}[U]}$ is disjoint, so
$\sum_{U\in\Cal U_n}\mu f^{-1}[U]\le\mu X$ is finite, and
$\Cal V_n=\{V:V\in\Cal U_n,\,\mu f^{-1}[V]>0\}$ is
countable for each $n$.

If $\Cal W\subseteq\Cal U_n\setminus\Cal V_n$, then

\Centerline{$\mu(\bigcup_{U\in\Cal W}f^{-1}[U])=f^{-1}[\bigcup\Cal W]$}

\noindent is measurable.   By 451Q,

\Centerline{$\mu f^{-1}[\bigcup(\Cal U_n\setminus\Cal V_n)]
=\mu(\bigcup_{U\in\Cal U_n\setminus\Cal V_n}f^{-1}[U])
=\sum_{U\in\Cal U_n\setminus\Cal V_n}\mu f^{-1}[U]=0$.}

Set

\Centerline{$\Cal V=\bigcup_{n\in\Bbb N}\Cal V_n$,
\quad$Y_0=Y\setminus\bigcup(\Cal U\setminus\Cal V)$.}

\noindent Then $Y_0$ is closed, and

\Centerline{$f^{-1}[Y\setminus Y_0]\subseteq
\bigcup_{n\in\Bbb N}f^{-1}[\bigcup(\Cal U_n\setminus\Cal V_n)]$}

\noindent is negligible, so $f^{-1}[Y_0]$ is conegligible.   On the
other hand, $Y_0$ is
separable.   \Prf\ Because $\Cal U$ is a base for the topology of $X$,
$\{Y\cap U:U\in\Cal U\}$ is a base for the topology of $Y$ (4A2B(a-vi)).
But this is included in the countable family
$\{Y\cap V:V\in\Cal V\}\cup\{\emptyset\}$, so $Y$ is second-countable,
therefore separable (4A2Oc).\ \Qed

So we have found an appropriate $Y_0$.
}%end of proof of 451R

\leader{451S}{Proposition} Let $(X,\Sigma,\mu)$ be a semi-finite compact
measure space, $Y$ a metrizable space and $f:X\to Y$ a measurable
function.

(a) The image measure $\nu=\mu f^{-1}$ is tight.

(b) If $\nu$ is locally finite and $\mu$ is complete and locally
determined, $\nu$ is a Radon measure.

\proof{{\bf (a)} Take $F\subseteq Y$ such that $\nu F>0$.   Then
$\mu f^{-1}[F]>0$.   Because $\mu$ is semi-finite, there is an
$E\in\Sigma$
such that $E\subseteq f^{-1}[F]$ and $0<\mu E<\infty$.

Consider the subspace measure $\mu_E$ and the restriction $f\restr E$.
$\mu_E$ is a totally finite compact measure and $f\restr E$ is
measurable,
so 451R tells us that there is a closed separable subspace
$Y_0\subseteq Y$ such that $\mu(E\setminus f^{-1}[Y_0])=0$.   Set
$E_1=E\cap f^{-1}[Y_0]$, so that $\mu E_1>0$.   Again, the subspace
measure $\mu_{E_1}$ is a totally finite compact measure, therefore
perfect, while
$f[E_1]\subseteq Y_0$.   So the image measure
$\mu_{E_1}(f\restr E_1)^{-1}$ on $Y_0$ is perfect (451Ea), therefore
tight (451N), and there is
a compact set $K\subseteq Y_0\cap F$ such that $\nu K=\mu f^{-1}[K]>0$.
By 412B, this is enough to show that $\nu$ is tight.

\medskip

{\bf (b)} $\nu$ is complete because $\mu$ is.   Now
suppose that $H\subseteq Y$ is such that $H\cap F$ belongs to the domain
$\Tau$ of $\nu$ whenever $\mu F<\infty$.   In this case $\mu$ is inner
regular with respect to
$\Cal E=\{E:E\in\Sigma,\,E\cap f^{-1}[H]\in\Sigma\}$.   \Prf\ Suppose
that $E\in\Sigma$ and that $\mu E>0$.   Applying (a) to $\mu_E$ and
$f\restr E$,
there is a compact set $K\subseteq f[E]$ such that $\mu f^{-1}[K]>0$.
Now $\nu K<\infty$, because $\nu$ is locally finite, so $K\cap H\in\Tau$
and $f^{-1}[K]\cap f^{-1}[H]\in\Sigma$.   Thus $f^{-1}[K]$ is a
non-negligible
member of $\Cal E$ included in $E$.   Since $\Cal E$ is closed under
finite unions, this is enough to show that $\mu$ is inner regular with
respect to $\Cal E$.\ \Qed

Accordingly $f^{-1}[H]\in\Sigma$, by 412Ja.   As $H$ is
arbitrary, $\nu$ is locally determined, therefore a Radon measure.
}%end of proof of 451S

\leader{451T}{Theorem}\cmmnt{ ({\smc Fremlin 81},
{\smc Koumoullis \& Prikry 83})} Let $(X,\frak T,\Sigma,\mu)$ be a
Radon measure space and $Y$ a metrizable space.
Then a function $f:X\to Y$ is measurable iff it is almost continuous.

\proof{ If $f$ is almost continuous it is surely measurable, by 418E.
Now suppose that $f$ is measurable and that $E\in\Sigma$ and
$\gamma<\mu E$.  Let $E_0\subseteq E$ be such that $E_0\in\Sigma$ and
$\gamma<\mu E_0<\infty$.
Applying 451R to the subspace measure $\mu_{E_0}$ and the restricted
function $f\restr E_0$, we see that there is a closed separable subspace
$Y_0$ of $Y$ such that $\mu(E_0\setminus f^{-1}[Y_0])=0$.   Set
$E_1=E_0\cap f^{-1}[Y_0]$;  then $\mu E_1>\gamma$.   Applying 418J to
$\mu_{E_1}$ and
$f\restr E_1:E_1\to Y_0$, we can find a measurable set $F\subseteq E_1$
such that $f\restr F$ is continuous and $\mu F\ge\gamma$.   As $E$ and
$\gamma$ are arbitrary, $f$ is almost continuous.
}%end of proof of 451T

\leader{451U}{Example}\cmmnt{ ({\smc Vinokurov \& Makhkamov 73},
{\smc Musia{\l} 76})} There is a perfect
completion regular quasi-Radon probability space which is not countably
compact.

\proof{{\bf (a)} Let $\Omega$ be the set of non-zero countable limit
ordinals.   For each $\xi\in\Omega$, let $\sequencen{\theta_{\xi}(n)}$
be a strictly increasing sequence in $\xi$ with supremum $\xi$, and set

\Centerline{$Q_{\xi}=\{x:x\in\{0,1\}^{\omega_1},\,x(\theta_{\xi}(n))=0$
for every $n\in\Bbb N\}$.}

\noindent Write

\Centerline{$X=\{0,1\}^{\omega_1}\setminus\bigcup_{\xi\in\Omega}Q_{\xi}$.}

\noindent Let $\nu_{\omega_1}$ be the usual measure on
$\{0,1\}^{\omega_1}$, and
$\Tau_{\omega_1}$ its domain;  let $\mu$ be the subspace measure on $X$, and
$\Sigma=\dom\mu$.

\medskip

{\bf (b)} It is convenient to note immediately the following fact:  for
every countable set $J\subseteq\omega_1$, the set $\pi_J[X]$ is
conegligible in $\{0,1\}^J$, where $\pi_J(x)=x\restr J$ for
$x\in\{0,1\}^{\omega_1}$.   \Prf\ Set

\Centerline{$A=\{\xi:\xi\in\Omega,\,\theta_{\xi}(n)\in J$ for every
$n\in\Bbb N\}$.}

\noindent Then $A$ is countable, because $\xi\le\sup J$ for every
$\xi\in A$.   So

\Centerline{$D=\bigcup_{\xi\in
A}\{y:y\in\{0,1\}^J,\,y(\theta_{\xi}(n))=0$
for every $n\in\Bbb N\}$}

\noindent is negligible in $\{0,1\}^J$, being a countable union of
negligible sets.   If $y\in\{0,1\}^J\setminus D$, define
$x\in\{0,1\}^{\omega_1}$ by setting $x(\eta)=y(\eta)$ for $\eta\in J$,
$x(\eta)=1$ for $\eta\in\omega_1\setminus J$.   Then $x\notin Q_{\xi}$
for any $\xi\in A$, because $x\restr J=y\restr J$, while $x\notin Q_{\xi}$
for any $\xi\in\Omega\setminus A$ by the definition of $A$.
So $x\in X$.   As $y$ is arbitrary,
$\pi_J[X]\supseteq\{0,1\}^J\setminus D$ is conegligible.\ \Qed

\medskip

{\bf (c)} $\mu$ is a completion regular quasi-Radon measure
because $\nu_{\omega_1}$ is (415E, 415B, 412Pd).   Also $\mu X=1$.
\Prf\  Let $F\in\Tau_{\omega_1}$ be a measurable envelope for $X$.
Then there is a countable
$J\subseteq\omega_1$ such that $\nu_J\pi_J[F]$ is defined and equal to
$\nu_{\omega_1}F$ (254Od), where $\nu_J$ is the usual measure on
$\{0,1\}^J$.   But we know that $\nu_J\pi_J[X]=1$, so

\Centerline{$\mu X=\nu_{\omega_1}^*X=\nu_{\omega_1}F=\nu_J\pi_JF=1$. \Qed}

\medskip

{\bf (d)} $\mu$ is perfect.   \Prf\ Take $E\in\Sigma$ such that
$\mu E>0$, and a measurable function $f:E\to\Bbb R$.   Set
$f_1(x)=\Bover{f(x)}{1+|f(x)|}$ for $x\in E$, $1$ for
$x\in X\setminus E$;  then $f_1:X\to\Bbb R$ is measurable.   Let
$g:\{0,1\}^{\omega_1}\to\Bbb R$ be a measurable function extending
$f_1$.  By 254Pb, there are a countable set $J\subseteq\omega_1$, a
conegligible set
$W\subseteq\{0,1\}^J$, and a measurable $h:W\to\Bbb R$ such that $g$
extends $h\pi_J$.   By (b), $W'=W\cap\pi_J[X]$ is conegligible, while
$W''=\{z:z\in W'$, $h_1(z)<1\}$ is measurable and not
negligible.   Because $W''$ is a non-negligible measurable subset of the
perfect measure space $\{0,1\}^J$, there is a compact set
$K_1\subseteq h[W'']$ such that $\nu_Jh^{-1}[K_1]>0$.   Set
$K=\{\Bover{t}{1-|t|}:t\in K_1\}$;  then $K$ is compact, and we have

\Centerline{$K_1
\subseteq h[W'']
=h[W\cap\pi_J[X]]\cap\ooint{-\infty,1}
\subseteq g[X]\cap\ooint{-\infty,1}
=f_1[X]\cap\ooint{-\infty,1}
=f_1[E]$,}

\Centerline{$K\subseteq f[E]$,}

\noindent while $f_1$, $g$ and $h\pi_J$ all agree on the
$\mu$-conegligible set $X\cap\pi_J^{-1}[W]$, so

$$\eqalignno{\mu f^{-1}[K]
&=\mu f_1^{-1}[K_1]
=\mu(X\cap(h\pi_J)^{-1}[K_1])\cr
&=\nu_{\omega_1}^*(X\cap(h\pi_J)^{-1}[K_1])
=\nu_{\omega_1}(h\pi_J)^{-1}[K_1]\cr
\displaycause{because $\nu_{\omega_1}^*X=1$ and $(h\pi_J)^{-1}[K_1]$ is
measurable}
&=\nu_Jh^{-1}[K_1]
>0.\cr}$$

\noindent As $f$ is arbitrary, $\mu$ is perfect.\ \Qed

\medskip

{\bf (e)} \Quer\ Suppose, if possible, that $\mu$ is countably compact.
Let $\Cal K$ be a countably compact class of sets such that $\mu$ is
inner regular with respect to $\Cal K$;  we may suppose that
$\Cal K\subseteq\Sigma$.

\medskip

\quad{\bf (i)} For $I\subseteq\omega_1$ set

\Centerline{$U(I)=\{x:x\in X,\,x(\eta)=0$ for every $\eta\in I\}$.}

\noindent It will be helpful to know that if $E\in\Sigma$ and $\mu E>0$,
there is a $\gamma<\omega_1$ such that $\mu(E\cap U(I))>0$ for every
finite $I\subseteq\omega_1\setminus\gamma$.   \Prf\ Express $E$ as
$X\cap F$ where $F\in\Tau_{\omega_1}$.
Let $J\subseteq\omega_1$ be a countable
set such that $\nu_{\omega_1}(F'\setminus F)=0$,
where $F'=\pi_J^{-1}[\pi_J[F]]$
(254Od again), and $\gamma<\omega_1$ such that $J\subseteq\gamma$.   If
$I\subseteq\omega_1\setminus\gamma$ is finite, then $I\cap J=\emptyset$,
while $U(I)$ is determined by coordinates in $I$ and $F'$ is determined
by coordinates in $J$;  so

$$\eqalign{\mu(E\cap U(I))
&=\nu_{\omega_1}^*(X\cap F\cap U(I))
=\nu_{\omega_1}(F\cap U(I))\cr
&=\nu_{\omega_1}(F'\cap U(I))
=\nu_{\omega_1} F'\cdot\nu_{\omega_1} U(I)
=\mu E\cdot\nu_{\omega_1} U(I)
>0.\cr}$$

\noindent Thus this $\gamma$ serves.\ \Qed

\medskip

\quad{\bf (ii)} Let $\Cal M$ be the family of countable subsets $M$ of
$\omega_1\cup\Cal K$ such that

\inset{($\alpha$) if $I\subseteq M\cap\omega_1$ is finite there is a
$K\in M\cap\Cal K$ such that $K\subseteq U(I)$ and $\mu K>0$;}

\inset{($\beta$) if $K\in M\cap\Cal K$, $I\subseteq M\cap\omega_1$ is
finite and $\mu(K\cap U(I))>0$, then there is a $K'\in M\cap\Cal K$ such
that $K'\subseteq K\cap U(I)$ and $\mu K'>0$;}

\inset{($\gamma$) if $\gamma\in M\cap\omega_1$ then
$\gamma\subseteq M$;}

\inset{($\delta$) if $K\in M\cap\Cal K$ and $\mu K>0$ then there is a
$\gamma\in M\cap\omega_1$ such that $\mu(K\cap U(I))>0$ whenever
$I\subseteq\omega_1\setminus\gamma$ is finite.}

\noindent Then every countable $M\subseteq\omega_1\cup\Cal K$ is
included in some member $M'$ of $\Cal M$.

\Prf\ Choose $\sequencen{N_n}$ as follows.   $N_0=M$.   Given
that $N_n$ is a countable subset of $\omega_1\cup\Cal K$ then
let $N_{n+1}\subseteq\omega_1\cup\Cal K$ be a countable set such that

\inset{($\alpha$) if $I\subseteq N_n\cap\omega_1$ is finite there is a
$K\in N_{n+1}\cap\Cal K$ such that $K\subseteq U(I)$ and $\mu K>0$;}

\inset{($\beta$) if $K\in N_n\cap\Cal K$, $I\subseteq N_n\cap\omega_1$
is finite and $\mu(K\cap U(I))>0$, then there is a $K'\in
N_{n+1}\cap\Cal K$ such that $K'\subseteq K\cap U(I)$ and $\mu K'>0$;}

\inset{($\gamma$) if $\gamma\in N_n\cap\omega_1$ then
$\gamma\subseteq N_{n+1}$;}

\inset{($\delta$) if $K\in N_n\cap\Cal K$ and $\mu K>0$ then there is a
$\gamma\in N_{n+1}\cap\omega_1$ such that $\mu(K\cap U(I))>0$ whenever
$I\subseteq\omega_1\setminus\gamma$ is finite;}

\inset{($\epsilon$) $N_n\subseteq N_{n+1}$.}

\noindent On completing the induction, set $M'=\bigcup_{n\in\Bbb N}N_n$;
this serves (because every finite subset of $M'$ is a subset of some
$N_n$).\ \Qed

\medskip

\quad{\bf (iii)} Choose a sequence $\sequencen{M_n}$ in $\Cal M$ such
that, for each $n$, $M_n\cup\{\sup(M_n\cap\omega_1)+1\}\subseteq
M_{n+1}$.   Set $\gamma_n=\sup(M_n\cap\omega_1)$ for each $n$.   Note
that $\gamma_n\subseteq M_n$, because if $\eta<\gamma_n$ then there is
some $\xi\in M_n$ such that $\eta<\xi$;  now $\xi\subseteq M_n$
because $M_n\in\Cal M$, so $\eta\in M_n$.   Also
$\gamma_n+1\in M_{n+1}$ for each $n$, so $\sequencen{\gamma_n}$ is
strictly increasing, and $\xi=\sup_{n\in\Bbb N}\gamma_n$ belongs to
$\Omega$.

Set $J=\{\theta_{\xi}(n):n\in\Bbb N\}$.   Then $J\cap\eta$ is finite for
every $\eta<\xi$, and in particular $J\cap\gamma_n$ is finite for every
$n$.   Set $I_0=J\cap\gamma_0$ and
$I_n=J\cap\gamma_n\setminus\gamma_{n-1}$ for $n\ge 1$.   Then
$\bigcap_{n\in\Bbb N}U(I_n)=Q_{\xi}$ is disjoint from $X$.

Choose a sequence $\sequencen{K_n}$ in $\Cal K$ as follows.   Because
$I_0$ is a finite subset of $M_0\cap\omega_1$, there is a $K_0\in
M_0\cap\Cal K$ such that
$K_0\subseteq U(I_0)$ and $\mu K_0>0$.    Given that $K_n\in M_n\cap\Cal
K$ and $\mu K_n>0$, then there is a $\beta\in M_n\cap\omega_1$ such that
$\mu(K_n\cap U(I))>0$ for every finite
$I\subseteq\omega_1\setminus\beta$;  now $\beta\le\gamma_n$ and
$I_{n+1}\cap\gamma_n=\emptyset$, so $\mu(K_n\cap U(I_{n+1}))>0$.   But
$K_n\in M_{n+1}\cap\Cal K$ and $I_{n+1}$ is a finite subset of
$M_{n+1}\cap\omega_1$, so there is a $K_{n+1}\in M_{n+1}\cap\Cal K$ such
that $K_{n+1}\subseteq K_n\cap U(I_{n+1})$ and $\mu K_{n+1}>0$.
Continue.

In this way we find a non-increasing sequence $\sequencen{K_n}$ in $\Cal
K$ such that $K_n\subseteq U(I_n)$ for every $n$ and no $K_n$ is empty.
But in this case $\bigcap_{i\le n}K_i=K_n$ is non-empty for every $n$,
while $\bigcap_{n\in\Bbb N}K_n\subseteq X\cap\bigcap_{n\in\Bbb N}U(I_n)$
is empty.   So $\Cal K$ is not a countably compact class.\ \Bang

\medskip

{\bf (f)} Thus $\mu$ is not countably compact, and has all the
properties claimed.
}%end of proof of 451U (Musial)

\leader{*451V}{Weakly $\alpha$-favourable spaces}\cmmnt{ There is
an interesting variation on the concept of `countably compact' measure
space, as follows.}   For any measure space $(X,\Sigma,\mu)$ we can
imagine an infinite game for two players, whom I will call `Empty' and
`Nonemepty'.   Empty chooses a
non-negligible measurable set $E_0$;  Nonempty chooses a
non-negligible measurable set $F_0\subseteq E_0$;  Empty chooses a
non-negligible measurable set $E_1\subseteq F_0$;  Nonempty chooses a
non-negligible measurable set $F_1\subseteq E_1$, and so on.   At the
end of the game, Empty wins if
$\bigcap_{n\in\Bbb N}E_n=\bigcap_{n\in\Bbb N}F_n$ is empty;  otherwise
Nonempty wins.   (If $\mu X=0$, so that Empty has no legal initial move,
I declare Nonempty the winner by default.)
\cmmnt{If you have seen `Banach-Mazur' games, you will recognise this
as a similar construction, in which open sets are replaced by
non-negligible measurable sets.}

A {\bf strategy} for Nonempty is a\cmmnt{ rule to determine his
moves in terms of the preceding moves for Empty;  that is, a}
function $\sigma:\bigcup_{n\in\Bbb N}(\Sigma\setminus\Cal
N)^{n+1}\to\Sigma\setminus\Cal N$, where $\Cal N$ is the ideal of
negligible sets, such that\cmmnt{ $\sigma(E_0,E_1,\ldots,E_n)\subseteq
E_n$, at least whenever $E_0,\ldots,E_n\in\Sigma\setminus\Cal N$ are
such that $E_{k+1}\subseteq\sigma(E_0,\ldots,E_k)$ for every $k<n$;
since it never matters what Nonempty does if Empty has already
broken the rules, we usually just demand that}
$\sigma(E_0,\ldots,E_n)\subseteq E_n$ for all
$E_0,\ldots,E_n\in\Sigma\setminus\Cal N$.   $\sigma$ is a {\bf winning
strategy} if $\bigcap_{n\in\Bbb N}E_n\ne\emptyset$ whenever
$\sequencen{E_n}$ is a sequence in $\Sigma\setminus\Cal N$ such that
$E_{n+1}\subseteq\sigma(E_0,\ldots,E_n)$ for every $n\in\Bbb N$.
\cmmnt{In terms of the game, we interpret this as saying that Nonempty
will win if he plays $F_n=\sigma(E_0,\ldots,E_n)$ whenever faced
with the position $(E_0,F_0,E_1,F_1,\ldots,F_{n-1},E_n)$.   (Since it is
supposed that Nonempty will use the same strategy throughout the game,
the moves $F_0,\ldots,F_{n-1}$ are determined by $E_0,\ldots,E_{n-1}$
and there is no advantage in taking them separately into account when
choosing $F_n$.)}

Now we say that the measure space $(X,\Sigma,\mu)$ is {\bf weakly
$\alpha$-favourable} if there is\cmmnt{ such} a winning strategy for
Nonempty.

\cmmnt{It turns out that the class of weakly $\alpha$-favourable
spaces behaves in much the same way as the class of countably compact
spaces.   For the moment, however, I leave the details to the exercises
(451Yh-451Yr).   See {\smc Fremlin 00}.}

\exercises{\leader{451X}{Basic exercises (a)}
%\spheader 451Xa
(i) Show that any purely atomic measure space is perfect.  (ii) Show
that any strictly localizable purely atomic measure space is countably
compact.   (iii) Show that the space of 342N is not countably
compact.
%451B

\sqheader 451Xb Show that a compact measure space in which singleton
sets are negligible is atomless.
%451B

\sqheader 451Xc Let $(X,\Sigma,\mu)$ be a measure space, and $\nu$ an
indefinite-integral measure over $\mu$\cmmnt{ (234J\formerly{2{}34B})}.
Show that
$\nu$ is compact, or countably compact, or perfect if $\mu$ is.
%451B

\spheader 451Xd  In 413Xn, show that $\mu$ is a countably compact
measure.   \Hint{show that the algebra $\Sigma$ there is a countably
compact class.}
%451B

\spheader 451Xe Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
measure spaces, with direct sum $(X,\Sigma,\mu)$.   Show that $\mu$ is
compact, or countably compact, or perfect iff every $\mu_i$ is.
%451B

\spheader 451Xf Let $(X,\Sigma,\mu)$ be a measure space and $\Cal K$ a
family of subsets of $X$ such that whenever $E\in\Sigma$ and $\mu E>0$
there is a $K\in\Cal K$ such that $K\subseteq E$ and $\mu_*K>0$.
(i) Show that if $\Cal K$ is a compact class then $\mu$ is a
compact measure.   (ii) Show that if $\Cal K$ is a countably compact class
then $\mu$ is a countably compact measure.
%451B 413T

\spheader 451Xg Let $(X,\Sigma,\mu)$ be a measure space.   For
$A\subseteq X$, write $\mu_A$ for the subspace measure on $A$.  Suppose
that whenever $E\in\Sigma$ and $\mu E>0$ there is a set $A\subseteq X$
such that $\mu_A$ is perfect and $\mu^*(A\cap E)>0$.   Show that $\mu$
is perfect.
%451D

\spheader 451Xh(i) Give an example of a compact probability space
$(X,\Sigma,\mu)$ and a $\sigma$-subalgebra $\Tau$ of $\Sigma$ such that
$\mu\restrp\Tau$ is not compact.   (ii) Give an example of a compact
probability space $(X,\Sigma,\mu)$, a set $Y$ and a function $f:X\to Y$
such that the image measure $\mu f^{-1}$ is not compact.   \Hint{342M,
342Xf, 439Xa.}
%451E

\spheader 451Xi Let $\familyiI{X_i}$ be a family of sets, with product
$X$.   Suppose that $\Cal K_i\subseteq\Cal PX_i$ for each $i$, and set
$\Cal K=\{\prod_{i\in I}K_i:K_i\in\Cal K_i$ for each $i\}$.   (i) Show
that if $\Cal K_i$ is a compact class for each $i$, so is $\Cal K$.
(ii) Show that if $\Cal K_i$ is a countably compact class for each $i$,
so is $\Cal K$.
%451H

\spheader 451Xj Let $A\subseteq[0,1]$ be a set with outer
Lebesgue measure $1$ and inner measure $0$.   Show that there is a
Borel measure $\lambda$ on $A\times[0,1]$ such that
$\lambda$ is not inner regular with respect to sets which have Borel
measurable projections on the factor spaces.
%451K

\spheader 451Xk Let $X$ be a Polish space and $E$ a subset of $X$.
Show that the following are equiveridical:  (i) $E$ is universally
measurable;  (ii) every Borel probability measure on $E$ is perfect;
(iii) every $\sigma$-finite Borel measure on $E$ is compact;  (iv)
$f[E]$ is universally measurable in $\Bbb R$ for every Borel measurable
function $f:X\to\Bbb R$.
%451M

\spheader 451Xl In 451N, show that $\mu$ is a compact measure.
%451N can omit

\spheader 451Xm Find a Radon measure space $(X,\frak T,\Sigma,\mu)$,
a continuous function $f:X\to[0,1]$ and a set
$B\subseteq[0,1]$ such that $\mu^*(f^{-1}[B])<(\mu f^{-1})^*B$.
%451P mt45bits

\spheader 451Xn
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space.   Show that it
is perfect iff whenever $f:X\to\Bbb R$ is measurable there is a
K$_{\sigma}$ set $H\subseteq f[X]$ such that $f^{-1}[H]$ is
conegligible.
%451P

\spheader 451Xo Let $X$ be a metrizable space, and $\mu$ a semi-finite
topological measure on $X$ which (regarded as a measure) is compact.
Show that $\mu$ is $\tau$-additive.
%451Q

\spheader 451Xp Let $(X,\Sigma,\mu)$ be a compact strictly localizable
measure space (e.g., any Radon measure space), $(Y,\Tau,\nu)$ a
$\sigma$-finite measure space, and $f:X\to L^0(\nu)$ a function.   Show
that the following are equiveridical:  (i) $f$ is measurable, when
$L^0(\nu)$ is given its topology of convergence in measure;
(ii) there is a function $h\in\eusm L^0(\lambda)$, where $\lambda$ is
the c.l.d.\ product measure on $X\times Y$, such that
$f(x)=h_x^{\ssbullet}$ for
almost every $x\in X$, where $h_x(y)=h(x,y)$.   \Hint{418R.}
%451R

\sqheader 451Xq Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space.
Show that $\Sigma=\Cal PX$ iff $\mu$ is purely atomic.   \Hint{if
$\Sigma=\Cal PX$, apply 451T with $Y=X$, the discrete topology on $Y$
and the identity function from $X$ to $Y$.}
%451T

\spheader 451Xr Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space
and $U$ a normed space.   Show that if $f$, $g:X\to U$ are measurable
functions, then $f+g$ is measurable.   (Cf.\ 418Xk.)
%451T

%new 2008
\spheader 451Xs Show that in all three of the constructions of 439A, the
measure $\nu$ is
countably compact.   \Hint{for the `third construction', consider
$\{f^{-1}[F]:F\subseteq\{0,1\}^{\frakc}$ is a zero set$\}$.}
%451Xa 451Xd 451B out of order

\leader{451Y}{Further exercises (a)}
%\spheader 451Ya
Show that for any probability space $(X,\Sigma,\mu)$, there is a compact
probability space $(Y,\Tau,\nu)$ with a subspace isomorphic to
$(X,\Sigma,\mu)$.

\spheader 451Yb
Let $\familyiI{X_i}$ be a family of sets, and $\Sigma_i$ a
$\sigma$-algebra of subsets of $X_i$ for each $i$.   Suppose that for
each finite $J\subseteq I$ we are given a finitely additive functional
$\nu_J$ on $X_J=\prod_{i\in J}X_i$, with domain the algebra
$\Tau_J=\bigotimes_{i\in J}\Sigma_i$
generated by sets of the form $\{x:x\in X_J,\,x(i)\in E\}$ for $i\in J$,
$E\in\Sigma_i$, and that ($\alpha$)
$\nu_K\{x:x\in X_K,\,x\restr J\in W\}=\nu_JW$ whenever
$J\subseteq K\in[I]^{<\omega}$ and $W\in\Tau_J$
($\beta$) $\mu_i=\nu_{\{i\}}$ is a countably compact probability measure
for every $i\in I$.   Show that there is a countably compact measure $\mu$
on $X=X_I$ such that $\mu\{x:x\in X,\,x\restr J\in W\}=\nu_JW$ whenever
$J\in[I]^{<\omega}$ and $W\in\Tau_J$.  \Hint{454D.}  (Compare 418M.)
%451J

\spheader 451Yc Describe $\mu$ in the case of 451Yb in which $I=[0,1]$,
$X_i=[0,1]\setminus\{i\}$, $\Sigma_i$ is the algebra of Lebesgue
measurable subsets of $X_i$, and $\nu_JE=\mu_L\{t:z_{Jt}\in E\}$ for
every $E\in\bigotimes_{i\in J}\Sigma_i$, where $z_{Jt}(i)=t$ for
$i\in J$, $t\in [0,1]$.   Contrast this with the difficulty encountered
in 418Xx.
%451J, 451Yb

\spheader 451Yd Let $(X,\Sigma,\mu)$ be a semi-finite compact measure
space, and $\familyiI{E_i}$ a point-finite family of measurable subsets
of $X$ such that $\bigcup_{i\in J}E_i\in\Sigma$ for every
$J\subseteq I$.   Show that $\mu(\bigcup_{i\in I}E_i)
=\sup_{J\subseteq I\text{ is finite}}\mu(\bigcup_{i\in J}E_i)$.
\Hint{438Ya.}
%451Q

\spheader 451Ye Let $X$ be a
hereditarily metacompact space, and $\mu$ a semi-finite
topological measure on $X$ which (regarded as a measure) is compact.
Show that $\mu$ is $\tau$-additive.
%451Yd 451Xo 451Q

\spheader 451Yf
Let $(X,\Sigma,\mu)$ be a compact measure space, $V$ a Banach space and
$f:X\to V$ a measurable function such that $\|f\|:X\to\coint{0,\infty}$
is integrable.   Show that $f$ is Bochner integrable (253Yf).
%451T

\spheader 451Yg Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure
space.   Suppose that $Y$ is a separable
metrizable space and $Z$ is a metrizable space, and that
$f:X\times Y\to Z$ is a function such that $x\mapsto f(x,y)$ is measurable
for every $y\in Y$ and $y\mapsto f(x,y)$ is continuous for every $x\in X$.
Show that $\mu$ is inner regular
with respect to $\{F:F\subseteq X$, $f\restr F\times Y$ is continuous$\}$.
\Hint{418Yk.}
%451T

\spheader 451Yh Show that any purely atomic measure space is weakly
$\alpha$-favourable, so that the space of 342N is weakly
$\alpha$-favourable but not countably compact.
%451V/451Xa

\spheader 451Yi Show that the direct sum of a family of
weakly $\alpha$-favourable measure spaces is weakly $\alpha$-favourable.
%451V/451Xe

\spheader 451Yj Show that an indefinite-integral measure over a
weakly $\alpha$-favourable measure is weakly $\alpha$-favourable.
%451V/451Xc

\spheader 451Yk(i) Show that a countably compact measure space is
weakly $\alpha$-favourable.  (ii) Show that a semi-finite weakly
$\alpha$-favourable measure space is perfect.
%451V/451C

\spheader 451Yl Show that any measurable subspace of a weakly
$\alpha$-favourable measure space is weakly $\alpha$-favourable.
%451V/451D

\spheader 451Ym Let $(X,\Sigma,\mu)$ be a weakly $\alpha$-favourable
measure space, $(Y,\Tau,\nu)$ a semi-finite measure space, and $f:X\to
Y$ a $(\Sigma,\Tau)$-measurable function such that $f^{-1}[F]$ is
negligible whenever $F\subseteq Y$ is negligible.   Show that
$(Y,\Tau,\nu)$ is weakly $\alpha$-favourable.
%451V/451E

\spheader 451Yn(i) Show that a measure space is weakly
$\alpha$-favourable iff its completion is
weakly $\alpha$-favourable.   (ii) Show that a semi-finite measure space
is weakly $\alpha$-favourable iff its c.l.d.\ version is weakly
$\alpha$-favourable.
%451V/451G

\spheader 451Yo Show that the c.l.d.\ product of two weakly
$\alpha$-favourable measure spaces is weakly $\alpha$-favourable.
%451V/451I

\spheader 451Yp Show that the product of
any family of weakly $\alpha$-favourable probability measures is weakly
$\alpha$-favourable.
%451V/451J

\spheader 451Yq Show that the space of 451U is not weakly
$\alpha$-favourable.
%451V/451U

\spheader 451Yr Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\undphi$ a lower density for $\mu$ such that
$\undphi X=X$;  let $\frak T$ be the corresponding density topology
(414P).   Show that $(X,\Sigma,\mu)$ is weakly $\alpha$-favourable iff
$(X,\frak T)$ is weakly $\alpha$-favourable (definition: 4A2A).
%451V

\spheader 451Ys Let $X$ be a set, and $\familyiI{\mu_i}$ a family of weakly
$\alpha$-favourable measures on $X$ with sum $\mu$
(234G\formerly{1{}12Ya}).   Show that if $\mu$ is
semi-finite, it is weakly $\alpha$-favourable.
%451Ym 451V

\spheader 451Yt\dvAnew{2009}
Let $X$ and $Y$ be locally compact Hausdorff groups
and $\phi:X\to Y$ a group homomorphism which is Haar
measurable in the sense of 411L, that is, $\phi^{-1}[H]$ is Haar measurable
for every open $H\subseteq Y$.   Show that $\phi$ is continuous.
%Y_0 an open sigma-compact subgp of  Y ;  451Q  to show that
%\phi^{-1}[Y_0]  not negligible;  hence  \phi^{-1}[V]  not negligible for
%any nhd  V  of identity in  Y  ;  use  443D .
%out of order query

\spheader 451Yu\dvAnew{2015} Let $\mu$ be a quasi-Radon measure on
the Sorgenfrey line (415Xc).   Show that
$\mu$ is weakly $\alpha$-favourable.
%451V

}%end of exercises

\endnotes{
\Notesheader{451} For a useful survey of results on countably compact
and perfect measures, with historical notes, see {\smc Ramachandran 02}.

The concepts of `compact', `countably compact' and
`perfect' measure space can all be regarded as attempts to understand
and classify the special properties of Lebesgue measure on $[0,1]$,
regarded as a measure space.   Because a countably separated perfect
probability space is very nearly isomorphic to Lebesgue measure (451Ad),
we can think of a perfect measure space as one in which the
countably-generated $\sigma$-subalgebras look like Lebesgue measure
(451F).   The arguments of 451Ic and 451Jc already hint at the kind of
results we can hope for.   When we form a product measure, each
measurable set in the product will depend, in effect, on sequences of
measurable sets in the factors, and therefore can be studied in terms of
countably generated subalgebras;  so that many results about products of
perfect measures will be derivable, if we wish to take that route, from
results about products of copies of Lebesgue measure.   Of course my
normal approach in this treatise is to go straight for the general
result;  but like anyone else I often start from a picture based on the
familiar special case.   In the next section we shall have some theorems
for which countable compactness, rather than perfectness, seems to be
the relevant property.

The first half of the section (down to 451P) is essentially a matter of
tidying up the theory of compact and perfect measures, and showing that
the same ideas will cover the new class of countably compact measures.
(You may like to go back to 342G, in which I worked through the basic
properties of compact measures, and contrast the arguments used there
with the slightly more sophisticated ones above.)   In 451Q-451T I enter
new territory, showing that for compact measures (and therefore for
Radon measures) the theory of measurable functions into metric spaces is
particularly simple, without making any assumptions about measure-free
cardinals.
}%end of notes

\discrpage

\frfilename{mt452.tex}
\versiondate{6.11.08}
\copyrightdate{2002}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Integration and disintegration of measures}

\def\undtheta{\underline{\theta}}
\def\undphi{\underline{\phi}}
\def\undpsi{\underline{\psi}}

\newsection{452}

A standard method of defining measures is through a formula

\Centerline{$\mu E=\int\mu_yE\,\nu(dy)$}

\noindent where $(Y,\Tau,\nu)$ is a measure space and
$\family{y}{Y}{\mu_y}$ is a family of measures on another set $X$.   In
practice these constructions commonly involve technical problems
concerning the domain of $\mu$\cmmnt{ (as in 452Xi)}, which is why I
have hardly used them so
far in this treatise.   There are not-quite-trivial examples in 417Yb,
434R and 436F, and the indefinite-integral measures of \S234 can also be
expressed in this way\cmmnt{ (452Xf)};  for a case in which this
approach is worked out fully, see 453N.
%Losert
But when a formula of this kind is valid, as in Fubini's
theorem, it is likely to be so useful that it dominates
further investigation of the topic.   In this section I give one of the
two most important theorems guaranteeing the existence of appropriate
families $\family{y}{Y}{\mu_y}$ when $\mu$ and $\nu$ are given (452I);
the other will follow in the next section\cmmnt{ (453K)}.   They
both suppose that we are provided with a suitable function $f:X\to Y$,
and rely heavily on the Lifting Theorem\cmmnt{ (\S341)} and on
considerations of inner regularity from Chapter 41.

The formal definition of a `disintegration' (which is nearly the same
thing as a `regular conditional probability') is in 452E.   The main
theorem depends, for its full generality, on the concept of `countably
compact measure'\cmmnt{ (451B)}.   It can be strengthened when $\mu$
is actually a Radon measure (452O).

The greater part of the section is concerned with general
disintegrations, in which the measures $\mu_y$ are supposed to be
measures on $X$ and are not necessarily
related to any particular structure on $X$.   However a natural,
and obviously important, class of applications has $X=Y\times Z$ and
each $\mu_y$ based on
the section $\{y\}\times Z$, so that it can be regarded as a measure on
$Z$.   Mostly there is very little more to be said in this case (see
452B-452D);  %452B 452C 452D
but in 452M we find that there is an interesting variation in the way that
countable compactness can be used.

\leader{452A}{Lemma} Let $(Y,\Tau,\nu)$ be a measure space, $X$ a
set, and $\family{y}{Y}{\mu_y}$ a family of measures on $X$.
Let $\Cal A$ be the family of subsets $A$ of $X$ such that
$\theta E=\int\mu_yE\,\nu(dy)$ is defined in $\Bbb R$.   Suppose that
$X\in\Cal A$.

(a) $\Cal A$ is a Dynkin class.

(b) If $\Sigma$ is any $\sigma$-subalgebra of $\Cal A$ then
$\mu=\theta\restr\Sigma$ is a measure on $X$.

(c) Suppose now that every $\mu_y$ is complete.
If, in (b), $\hat\mu$ is the completion of $\mu$ and $\hat\Sigma$ its
domain, then $\hat\Sigma\subseteq\Cal A$ and
$\hat\mu=\theta\restr\hat\Sigma$.

\proof{ For (a) and (b), we have only to look at the definitions of
`Dynkin class' and `measure' and apply the elementary properties of the
integral.    For (c),
if $E\in\hat\Sigma$, then there are $E'$, $E''\in\Sigma$ such that
$E'\subseteq E\subseteq E''$ and $\theta E'=\theta E''$.   So
$\mu_yE'=\mu_yE''$ for
$\nu$-almost every $y$;  since all the $\mu_y$ are supposed to be
complete, $\mu_yE$ is defined and equal to $\mu_yE'$ for almost every
$y$, and
$\theta E$ is defined and equal to $\theta E'=\mu E'=\hat\mu E$.
}%end of proof of 452A

\leader{452B}{Theorem} (a) Let $X$ be a set, $(Y,\Tau,\nu)$ a
measure space, and $\family{y}{Y}{\mu_y}$ a family of
measures on $X$ such that $\int\mu_yX\,\nu(dy)$ is defined and finite.
Let $\Cal E$ be
a family of subsets of $X$, closed under finite intersections, such that
$\int\mu_yE\,\nu(dy)$ is defined in $\Bbb R$ for every $E\in\Cal E$.

\quad(i) If $\Sigma$ is the $\sigma$-algebra of subsets of $X$ generated
by $\Cal E$, we have a totally finite measure $\mu$ on $X$, with domain
$\Sigma$, given by the
formula $\mu E=\int\mu_yE\,\nu(dy)$ for every $E\in\Sigma$.

\quad(ii) If $\hat\mu$ is the completion of $\mu$ and $\hat\Sigma$ its
domain, then $\int\hat\mu_yE\,\nu(dy)$ is defined and equal to
$\hat\mu E$ for every $E\in\hat\Sigma$,
where $\hat\mu_y$ is the completion of $\mu_y$ for each $y\in Y$.

(b) Let $Z$ be a set, $(Y,\Tau,\nu)$ a measure space, and
$\family{y}{Y}{\mu_y}$ a family of measures on $Z$ such that
$\int\mu_yZ\,\nu(dy)$ is defined and finite.   Let
$\Cal H$ be a family of subsets of $Z$,
closed under finite intersections, such that
$\int\mu_yH\,\nu(dy)$ is
defined in $\Bbb R$ for every $H\in\Cal H$.

\quad(i) If $\Upsilon$ is the $\sigma$-algebra of subsets of $Z$
generated by $\Cal H$, we have a
totally finite measure $\mu$ on $Y\times Z$, with domain
$\Tau\tensorhat\Upsilon$, defined by setting
$\mu E=\int\mu_yE[\{y\}]\nu(dy)$ for every
$E\in\Tau\tensorhat\Upsilon$.

\quad(ii) If $\hat\mu$ is the completion of $\mu$ and $\hat\Sigma$ its
domain, then $\int\hat\mu_yE[\{y\}]\nu(dy)$ is defined and equal to
$\hat\mu E$ for every $E\in\hat\Sigma$,
where $\hat\mu_y$ is the completion of $\mu_y$ for each $y\in Y$.

\proof{{\bf (a)} Define $\Cal A\subseteq\Cal PX$ as in 452A.
Then $\Cal E\subseteq\Cal A$, so by the Monotone Class Theorem (136B)
$\Sigma\subseteq\Cal A$ and we have
(i).   Applying 452Ac to $\family{y}{Y}{\hat\mu_y}$ we have (ii).

\medskip

{\bf (b)} Set $X=Y\times Z$.  For $y\in Y$, let $\mu'_y$ be the measure
on $X$ defined by setting $\mu'_yE=\mu_yE[\{y\}]$ whenever this is
defined;  that is, $\mu'_y$ is the
image of $\mu_y$ under the function $z\mapsto(y,z):Z\to X$.   Set
$\Cal E=\{F\times H:F\in\Tau$, $H\in\Cal H\}$.
Then $\Cal E$ is a family of subsets of $X$ closed under
finite intersections, and

\Centerline{$\int\mu'_y(F\times H)\nu(dy)=\int\chi F(y)\mu_yH\,\nu(dy)$}

\noindent is defined whenever $F\in\Tau$ and $H\in\Cal H$.   By (a), we
have a measure $\mu$ on $X$, with domain the $\sigma$-algebra $\Sigma$
generated by $\Cal E$, defined by writing

\Centerline{$\mu E=\int\mu'_yE\,\nu(dy)=\int\mu_yE[\{y\}]\nu(dy)$}

\noindent for every $E\in\Sigma$.   Of course $\Sigma$ includes
$\Tau\tensorhat\Upsilon$ (the set $\{H:Y\times H\in\Sigma\}$ is a
$\sigma$-algebra of subsets of $Z$ including
$\Cal H$, so includes $\Upsilon$) and is therefore equal to
$\Tau\tensorhat\Upsilon$.

\wheader{452B}{0}{0}{0}{30pt}
This proves (i).   If now $E\in\hat\Sigma$, (a-ii) tells us that

\Centerline{$\hat\mu
E=\int\hat\mu'_yE\,\nu(dy)=\int\hat\mu_yE[\{y\}]\nu(dy)$.}
}%end of proof of 452B

\vleader{72pt}{452C}{Theorem} (a) Let $Y$ be a topological space, $\nu$ a
$\tau$-additive topological measure on $Y$, $(X,\frak T)$ a
topological space, and $\family{y}{Y}{\mu_y}$
a family of $\tau$-additive topological measures on $X$ such that
$\int\mu_yX\,\nu(dy)$ is defined and finite.
Suppose that there is a base $\Cal U$ for $\frak T$, closed under finite
unions, such
that $y\mapsto\mu_yU$ is lower semi-continuous for every $U\in\Cal U$.

\quad(i) We can define a $\tau$-additive Borel measure
$\mu$ on $X$ by writing
$\mu E\penalty-100=\penalty-100\int\mu_yE\,\nu(dy)$ for every Borel set
$E\subseteq X$.

\quad(ii) If $\hat\mu$ is the completion of $\mu$ and $\hat\Sigma$ its
domain, then $\int\hat\mu_yE\,\nu(dy)$ is defined and equal to
$\hat\mu E$ for every $E\in\hat\Sigma$,
where $\hat\mu_y$ is the completion of $\mu_y$ for each $y\in Y$.

(b) Let $Y$ be a topological space, $\nu$ a $\tau$-additive topological
measure on $Y$, $(Z,\frak U)$ a topological space, and
$\family{y}{Y}{\mu_y}$
a family of $\tau$-additive topological measures on $Z$ such that
$\int\mu_yZ\,\nu(dy)$ is defined and finite.
Suppose that there is a base $\Cal V$ for $\frak U$, closed under finite
unions, such
that $y\mapsto\mu_yV$ is lower semi-continuous for every $V\in\Cal V$.

\quad(i) We can define a $\tau$-additive Borel measure $\mu$ on
$Y\times Z$ by writing $\mu E=\int\mu_yE[\{y\}]\nu(dy)$ for
every Borel set $E\subseteq Y\times Z$.

\quad(ii) If $\hat\mu$ is the completion of $\mu$ and $\hat\Sigma$ its
domain, then $\int\hat\mu_yE[\{y\}]\nu(dy)$ is defined and equal to
$\hat\mu E$ for every $E\in\hat\Sigma$,
where $\hat\mu_y$ is the completion of $\mu_y$ for each $y\in Y$.

\proof{{\bf (a)} For $A\subseteq X$, set $f_A(y)=\mu_yA$ when this is
defined.   We may suppose that $\emptyset\in\Cal U$.
If $\Cal W\subseteq\Cal U$ is a non-empty upwards-directed set with
union $G$, $\family{W}{\Cal W}{f_W}$ is an upwards-directed family of
lower semi-continuous functions
with supremum $f_G$, because every $\mu_y$ is $\tau$-additive.   So
$f_G$ is lower semi-continuous, and also
$\int f_Gd\nu=\sup_{W\in\Cal W}\int f_Wd\nu$, by 414Ba.
Taking $\Cal E$ to be the family of open subsets of $X$ in 452Ba, we see
that we have a $\tau$-additive Borel measure $\mu$ on $X$ such that
$\mu E=\int\mu_yE\,\nu(dy)$ for every Borel set $E\subseteq X$.
Moreover, if $\Cal G$ is a non-empty upwards-directed family of open
subsets of $X$ with union $G^*$, then
$\Cal W=\{W:W\in\Cal U$, $W\subseteq G$ for some $G\in\Cal G\}$ is an
upwards-directed family with union $G^*$, so

\Centerline{$\mu G^*=\int f_{G^*}d\nu
=\sup_{W\in\Cal W}\int f_Wd\nu\le\sup_{G\in\Cal G}\mu G\le\mu G^*$.}

\noindent As $\Cal G$ is arbitrary, $\mu$ is $\tau$-additive.   This
proves (i);  (ii) follows immediately, as in 452Ba.

\medskip

{\bf (b)} Let $\Cal U$ be the family of sets expressible as
$\bigcup_{i\le n}H_i\times V_i$ where $H_i\subseteq Y$ is open and
$V_i\in\Cal V$ for every $i\le n$.   Because $\Cal V$ is a base for
$\frak U$, $\Cal U$ is a base for the topology of $X=Y\times Z$.   For
$y\in Y$ let $\mu'_y$ be the
measure on $X$ defined by saying that $\mu'_yE=\mu_yE[\{y\}]$ whenever
this is defined.   Then $\mu'_y$ is a $\tau$-additive topological
probability measure on $X$,
by 418Ha or otherwise.   If $U\in\Cal U$, $y\mapsto\mu'_yU$ is lower
semi-continuous.   \Prf\  Express $U$ as $\bigcup_{i\le n}H_i\times V_i$
where $H_i\subseteq Y$
is open and $V_i\in\Cal V$ for each $i$.    Suppose that $y\in Y$ and
$\gamma<\mu_yU$.   Set $I=\{i:i\le n$, $y\in H_i\}$,
$H=Y\cap\bigcap_{i\in I}H_i$
and $V=\bigcup_{i\in I}V_i$.   Then $U[\{y\}]=V\subseteq U[\{y'\}]$ for
every $y'\in H$.   Also $H'=\{y':\mu_{y'}V>\gamma\}$ is a neighbourhood
of $y$.   So $H\cap H'$ is
a neighbourhood of $y$, and $\mu'_{y'}U>\gamma$ for every
$y'\in H\cap H'$.   As $y$ and $\gamma$ are arbitrary, we have the result.\
\Qed

Now applying (a) to $\family{y}{Y}{\mu'_y}$ we see that (b) is true.
}%end of proof of 452C

\leader{452D}{Theorem} (a) Let $(Y,\frak S,\Tau,\nu)$ be a Radon
measure space, $(X,\frak T)$ a topological space, and
$\family{y}{Y}{\mu_y}$
a uniformly tight\cmmnt{ (definition:  437O)} family of Radon
measures on $X$ such that $\int\mu_yX\,\nu(dy)$ is defined and finite.
Suppose that there is a base $\Cal U$ for $\frak T$, closed under finite
unions, such
that $y\mapsto\mu_yU$ is lower semi-continuous for every $U\in\Cal U$.
Then we have a totally finite Radon measure $\tilde\mu$ on $X$ defined by
saying that that $\tilde\mu E=\int\mu_yE\,\nu(dy)$ whenever $\tilde\mu$
measures $E$.

(b) Let $(Y,\frak S,\Tau,\nu)$ be a Radon measure space, $(Z,\frak U)$ a
topological space, and $\family{y}{Y}{\mu_y}$
a uniformly tight family of Radon measures on $Z$ such that
$\int\mu_yZ\,\nu(dy)$ is defined and finite.
Suppose that there is a base $\Cal V$ for $\frak U$, closed under finite
unions, such
that $y\mapsto\mu_yV$ is lower semi-continuous for every $V\in\Cal V$.
Then we have a totally finite Radon measure $\tilde\mu$ on $Y\times Z$
such that $\tilde\mu E=\int\mu_yE[\{y\}]\nu(dy)$ whenever $\tilde\mu$
measures $E$.

\proof{ I take the two parts together.   In (b), write $X$ for
$Y\times Z$.   By 452C we have a
$\tau$-additive Borel measure $\mu$ satisfying the
appropriate formula.
Now for any $\epsilon>0$ there is a compact set $K\subseteq X$ such that
$\mu K\ge\mu X-2\epsilon$.   \Prf\ In (a), take $\eta>0$ such that
$\int\min(\eta,\mu_yX)\nu(dy)\le 2\epsilon$, and $K$ such that
$\mu_y(X\setminus K)\le\eta$ for every $y\in Y$.
In (b), take $\eta>0$
such that $\int\min(\eta,\mu_yZ)\nu(dy)\le\epsilon$.   Now let
$K_1\subseteq Y$ and $K_2\subseteq Z$ be compact sets such that

\Centerline{$\int_{K_1}\mu_yZ\,\nu(dy)\ge\int_Y\mu_yZ\,\nu(dy)-\epsilon$,
\quad$\mu_y(Z\setminus K_2)\le\eta$ for every $y\in Y$.}

\noindent Then $K=K_1\times K_2$ is compact and

$$\eqalign{\mu((Y\times Z)\setminus K)
&\le\int_{Y\setminus K_1}\mu_yZ\,\nu(dy)
  +\int_{K_1}\mu_y(Z\setminus K_2)\nu(dy)\cr
&\le\epsilon+\int_{K_1}\min(\eta,\mu_yZ)\nu(dy)
\le 2\epsilon.  \text{ \Qed}\cr}$$

Since $\mu$ is totally finite it is surely locally finite and
effectively locally finite, so the conditions of 416F(iv) are satisfied
and the c.l.d.\ version
$\tilde\mu$ of $\mu$ is a Radon measure on $X$.   But of course
$\tilde\mu$ is just the completion of $\mu$, so 452C(a-ii) or 452C(b-ii)
tells us that the declared formula also applies to $\tilde\mu$.
}%end of proof of 452D

\leader{452E}{}\cmmnt{ All the constructions above can be thought of
as special cases of the following.

\medskip

\noindent}{\bf Definition}  Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces.   A {\bf disintegration} of $\mu$ over $\nu$ is a family
$\family{y}{Y}{\mu_y}$ of measures on $X$ such that
$\int\mu_yE\,\nu(dy)$
is defined in $[0,\infty]$ and equal to $\mu E$ for every $E\in\Sigma$.
If $f:X\to Y$ is an \imp\ function, a disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$ is {\bf consistent} with $f$
if, for each $F\in\Tau$, $\mu_yf^{-1}[F]=1$ for $\nu$-almost every
$y\in F$.   $\family{y}{Y}{\mu_y}$ is {\bf strongly consistent} with $f$
if, for almost every $y\in Y$, $\mu_y$ is a probability measure for which
$f^{-1}[\{y\}]$ is conegligible.

\cmmnt{A trivial example of a disintegration is when $\nu$ is a
probability measure and $\mu_y=\mu$ for every $y$.   Of course this is
of little interest.   The archetypal disintegration is 452Bb when all
the $\mu_y$ are the same, in which case Fubini's theorem tells us that
we are looking at a product measure on $X=Y\times Z$.
If $\mu$ is a probability measure then this disintegration
is strongly consistent.

The phrase {\bf regular conditional probability} is used for special
types of disintegration;  typically, when $\mu$ and $\nu$ and every
$\mu_y$ are probabilities, and sometimes supposing that every $\mu_y$
has the same domain as $\mu$.   I have seen the word {\bf decomposition}
used for what I call a disintegration.
}%end of comment

\leader{452F}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces and $\family{y}{Y}{\mu_y}$ a disintegration of $\mu$ over
$\nu$.   Then $\iint f(x)\mu_y(dx)\nu(dy)$ is defined and equal to
$\int fd\mu$ for every $[-\infty,\infty]$-valued function $f$ such that
$\int fd\mu$ is defined in $[-\infty,\infty]$.

\proof{{\bf (a)} Suppose first that $f$ is non-negative.   Let
$H\in\Sigma$ be a conegligible set such that $f\restr H$ is
$\Sigma$-measurable.   For $n\in\Bbb N$ set

\Centerline{$E_{nk}=\{x:x\in H,\,2^{-n}k\le f(x)\}$ for $k\ge 1$,
\quad $f_n=2^{-n}\sum_{k=1}^{4^n}\chi E_{nk}$.}

\noindent Then $\sequencen{f_n}$ is a non-decreasing sequence of
functions with $\lim_{n\to\infty}f_n(x)=f(x)$ for every $x\in H$.   Now
$\int\mu_y(X\setminus H)\nu(dy)=0$, so $X\setminus H$ is
$\mu_y$-negligible for almost every $y$.   Set

\Centerline{$V=\{y:\mu_y(X\setminus H)=0$, $E_{nk}\in\dom\mu_y$ for
every $n\in\Bbb N$, $k\ge 1\}$;}

\noindent then $V$ is $\nu$-conegligible.   For $y\in V$,

\Centerline{$\int fd\mu_y
=\lim_{n\to\infty}\int f_nd\mu_y
=\lim_{n\to\infty}2^{-n}$$\sum_{k=1}^{4^n}\mu_yE_{nk}$,}

\noindent while each function $y\mapsto\mu_yE_{nk}$ is $\nu$-virtually
measurable, so $y\mapsto\int fd\mu_y$ is $\nu$-virtually measurable
and

$$\eqalign{\iint fd\mu_y\nu(dy)
&=\lim_{n\to\infty}\iint f_nd\mu_y\nu(dy)
=\lim_{n\to\infty}2^{-n}\sum_{k=1}^{4^n}\int\mu_yE_{nk}\,\nu(dy)\cr
&=\lim_{n\to\infty}2^{-n}\sum_{k=1}^{4^n}\mu E_{nk}
=\lim_{n\to\infty}\int f_nd\mu
=\int fd\mu.\cr}$$

\medskip

{\bf (b)} For general $f$ we now have

$$\eqalign{\iint f(x)\mu_y(dx)\nu(dy)
&=\iint f^+(x)\mu_y(dx)\nu(dy)-\iint f^-(x)\mu_y(dx)\nu(dy)\cr
&=\int f^+d\mu-\int f^-d\mu
=\int fd\mu,\cr}$$

\noindent where $f^+$, $f^-$ are the positive and negative parts of $f$.
}%end of proof of 452F

\medskip

\noindent{\bf Remark} When $X=Y\times Z$ and our disintegration is a
family $\family{y}{Y}{\mu'_y}$ of measures on $X$ defined from a family
$\family{y}{Y}{\mu_y}$ of probability measures
on $Z$\cmmnt{, as in 452Bb,} we can more naturally write
$\int f(y,z)\mu_y(dz)$ in place of $\int f(x)\mu'_y(dx)$, and we get

\Centerline{$\iint f(y,z)\mu_y(dz)\nu(dy)=\int fd\mu$ whenever the
latter is defined in $[-\infty,\infty]$\dvro{.}{}}

\cmmnt{\noindent as in 252B.}

\leader{452G}{}\cmmnt{ The most useful theorems about disintegrations
of course involve some restrictions on their form, most commonly
involving consistency with some kind of projection.   I clear the path
with statements of some elementary facts.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, $f:X\to Y$ an \imp\ function, and $\family{y}{Y}{\mu_y}$
a disintegration of $\mu$ over $\nu$.

(a) If $\family{y}{Y}{\mu_y}$ is consistent with $f$, and $F\in\Tau$,
then $\mu_yf^{-1}[F]=\chi F(y)$ for $\nu$-almost every $y\in Y$;  in
particular, almost every $\mu_y$ is a probability measure.

(b) If $\family{y}{Y}{\mu_y}$ is strongly consistent with $f$ it is
consistent with $f$.

(c) If $\nu$ is countably separated\cmmnt{ (definition: 343D)} and
$\family{y}{Y}{\mu_y}$ is consistent with $f$, then it is strongly
consistent with $f$.

\proof{{\bf (a)} We have $\mu_yf^{-1}[F]=1$ for almost every $y\in F$.
Since also

\Centerline{$\mu_y(X\setminus f^{-1}[F])=\mu_yf^{-1}[Y\setminus F]=1$,
\quad$\mu_yX=\mu_yf^{-1}[Y]=1$}

\noindent for almost every $y\in Y\setminus F$, $\mu_yf^{-1}[F]=0$ for
almost every $y\in X\setminus F$.

\medskip

{\bf (b)} If $F\in\Tau$, then $f^{-1}[F]\supseteq f^{-1}[\{y\}]$ is $\mu_y$-conegligible for almost every $y\in F$;  since we are also
told that $\mu_yX=1$ for almost every $y$, $\mu_yf^{-1}[F]=1$ for almost
every $y\in F$.

\medskip

{\bf (c)} There is a countable $\Cal F\subseteq\Tau$ separating the
points of $Y$;  we may suppose that $Y\in\Cal F$ and that
$Y\setminus F\in\Cal F$ for every $F\in\Cal F$.   Now

\Centerline{$H_F=F\setminus\{y:\mu_yf^{-1}[F]$ is defined and equal to
$1\}$}

\noindent is negligible for every $F\in\Cal F$, so that

\Centerline{$Z
=Y\setminus\bigcup_{F\in\Cal F}H_F$}

\noindent is conegligible.   For $y\in Z$, set
$\Cal F_y=\{F:y\in F\in\Cal F\}$;  then

\Centerline{$\{y\}=\bigcap\Cal F_y$,
\quad$f^{-1}[\{y\}]=\bigcap\{f^{-1}[F]:F\in\Cal F_y\}$,}

\noindent while $\mu_yf^{-1}[F]=1$ for every $F\in\Cal F_y$.   Because
$\Cal F_y$ is countable, $\mu_yf^{-1}[\{y\}]=1$.   This is true for
almost every $y$, so $\family{y}{Y}{\mu_y}$ is strongly consistent with
$f$.
}%end of proof of 452G

\leader{452H}{Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces, and
$T:L^{\infty}(\mu)\to L^{\infty}(\nu)$ a positive linear operator such
that $T(\chi X^{\ssbullet})=\chi Y^{\ssbullet}$ and $\int Tu=\int u$
whenever $u\in L^{\infty}(\mu)^+$.   Let $\Cal K$ be a countably compact
class of subsets of $X$, closed under finite unions and countable
intersections, such that $\mu$ is inner regular with respect to
$\Cal K$.   Then there is a disintegration $\family{y}{Y}{\mu_y}$ of
$\mu$ over $\nu$ such that

(i) $\mu_y$ is a complete probability measure on $X$,
inner regular with respect to $\Cal K$ and measuring every member of
$\Cal K$, for every $y\in Y$;

(ii) setting $h_g(y)=\int g\,d\mu_y$ whenever
$g\in\eusm L^{\infty}(\mu)$ and $y\in Y$ are such that the integral is
defined, $h_g\in\eusm L^{\infty}(\nu)$ and
$T(g^{\ssbullet})=h_g^{\ssbullet}$ for every
$g\in\eusm L^{\infty}(\mu)$.

\proof{{\bf (a)} Completing $\nu$ does not change $\eusm L^{\infty}(\nu)$
or $L^{\infty}(\nu)$, nor does it change the families which are
disintegrations over
$\nu$;  so we may assume throughout that $\nu$ is complete.   It
therefore has a lifting $\theta:\frak B\to\Tau$, where $\frak B$ is the
measure algebra of $\nu$, which gives rise to a Riesz homomorphism $S$
from $L^{\infty}(\nu)\cong L^{\infty}(\frak B)$ to the space
$L^{\infty}(\Tau)$ of bounded $\Tau$-measurable real-valued functions on
$Y$ such that $(Sv)^{\ssbullet}=v$ for every $v\in L^{\infty}(\nu)$
(363I, 363F, 363H).

\medskip

{\bf (b)} For $y\in Y$ and $E\in\Sigma$, set
$\psi_yE=(ST(\chi E^{\ssbullet}))(y)$.   Because
$0\le T(\chi E^{\ssbullet})\le\chi Y^{\ssbullet}$ in
$L^{\infty}(\nu)$, $0\le\psi_yE\le 1$.   The maps

\Centerline{$E\mapsto\chi E\mapsto\chi E^{\ssbullet}
\mapsto T(\chi E^{\ssbullet})\mapsto ST(\chi E^{\ssbullet})$}

\noindent are all additive, so $\psi_y:\Sigma\to[0,1]$ is additive for
each $y\in Y$.   For fixed $E\in\Sigma$,

\Centerline{$\mu E=\int\chi E\,d\mu=\int(\chi E^{\ssbullet})
=\int T(\chi E^{\ssbullet})
=\int ST(\chi E^{\ssbullet})=\int\psi_yE\,\nu(dy)$.}

\medskip

{\bf (c)} Recall that $\mu$ is supposed to be inner
regular with respect to the countably compact class $\Cal K$.   By
413Ua, there is for every $y\in Y$ a complete measure $\mu'_y$ on
$X$ such that $\mu'_yX\le\psi_yX\le 1$, $\Cal K\subseteq\dom\mu'_y$, and
$\mu'_yK\ge\psi_yK$ for every $K\in\Cal K\cap\Sigma$.

\medskip

{\bf (d)} Now, for any fixed $E\in\Sigma$, $\mu'_yE$ is defined and
equal to $\psi_yE$ for almost every $y\in Y$.   \Prf\ Let
$\sequencen{K_n}$, $\sequencen{K'_n}$ be sequences in $\Cal K\cap\Sigma$
such that $K_n\subseteq E$ and $K'_n\subseteq X\setminus E$ for every
$n$, while $\mu E=\sup_{n\in\Bbb N}\mu K_n$ and
$\mu(X\setminus E)=\sup_{n\in\Bbb N}\mu K'_n$.   Set
$L=\bigcup_{n\in\Bbb N}K_n$,
$L'=\bigcap_{n\in\Bbb N}(X\setminus K'_n)$.   Then both $L$ and $L'$
belong to the domain of every $\mu'_y$, and

$$\eqalign{\sup_{n\in\Bbb N}\psi_yK_n
&\le\sup_{n\in\Bbb N}\mu'_yK_n
\le\mu'_yL
\le\mu'_yL'\cr
&\le\inf_{n\in\Bbb N}\mu'_y(X\setminus K'_n)
=\mu'_yX-\sup_{n\in\Bbb N}\mu'_yK'_n
\le 1-\sup_{n\in\Bbb N}\psi_yK'_n\cr}$$

\noindent for every $y$.   On the other hand,

$$\eqalign{\int(1-\sup_{n\in\Bbb N}\psi_yK'_n)\nu(dy)
&\le\nu Y-\sup_{n\in\Bbb N}\int\psi_y(K'_n)\nu(dy)
=\mu X-\sup_{n\in\Bbb N}\mu K'_n
=\mu E\cr
&=\sup_{n\in\Bbb N}\mu K_n
=\sup_{n\in\Bbb N}\int\psi_yK_n\nu(dy)
\le\int\sup_{n\in\Bbb N}\psi_yK_n\nu(dy).\cr}$$

\noindent So

\Centerline{$\sup_{n\in\Bbb N}\psi_yK_n=\mu'_yL=\mu'_yL'
=1-\sup_{n\in\Bbb N}\psi_yK'_n$}

\noindent for almost every $y$.   Because $L\subseteq E\subseteq L'$ and
$\mu'_y$ is complete, $E\in\dom\mu'_y$ and

\Centerline{$\mu'_yE=1-\sup_{n\in\Bbb N}\psi_yK'_n
\ge 1-\psi_y(X\setminus E)\ge\psi_yE$}

\noindent for almost every $y\in Y$.   Similarly,
$\mu'_y(X\setminus E)\ge\psi_y(X\setminus E)$ for almost every $y$.
But as

\Centerline{$\mu'_yE+\mu'_y(X\setminus E)=\mu'_yX\le\psi_yX\le 1$}

\noindent whenever the left-hand side is defined, we must have
$\mu'_yE=\psi_yE$ for almost every $y$, as claimed.\ \Qed

It follows at once that

\Centerline{$\int\mu'_yE\,\nu(dy)=\int\psi_yE\,\nu(dy)=\mu E$}

\noindent for every $E\in\Sigma$, and $\family{y}{Y}{\mu'_y}$ is a
disintegration of $\mu$ over $\nu$.

\medskip

{\bf (e)} At this point observe that

\Centerline{$\int\mu'_yX\,\nu(dy)
=\mu X=\int\chi X^{\ssbullet}=\int T(\chi X^{\ssbullet})
=\nu Y$,}

\noindent so $F_0=\{y:\mu'_yX<1\}$ is negligible.   Taking any $y_0\in
Y\setminus F_0$ and setting

$$\eqalign{\mu_y&=\mu'_{y_0}\text{ for }y\in F_0\cr
&=\mu'_y\text{ for }y\in Y\setminus F_0,\cr}$$

\noindent we find ourselves with a disintegration $\family{y}{Y}{\mu_y}$
of $\mu$ over $\nu$ with the same properties as $\family{y}{Y}{\mu'_y}$,
but now consisting entirely of probability measures.

\medskip

{\bf (f)} For $g\in\eusm L^{\infty}(\mu)$, set $h_g(y)=\int g\,d\mu_y$
whenever $y\in Y$ is such that the integral is defined.   Consider the
set $V$ of those $g\in\eusm L^{\infty}(\mu)$ such that
$h_g\in\eusm L^{\infty}(\nu)$ and $Tg^{\ssbullet}=h_g^{\ssbullet}$ in
$L^{\infty}(\nu)$.   If $E\in\Sigma$, then $h_{\chi E}(y)=\psi_yE$ for
almost every $y$, so

\Centerline{$h_{\chi E}^{\ssbullet}
=(ST(\chi E^{\ssbullet}))^{\ssbullet}
=T(\chi E^{\ssbullet})$;}

\noindent accordingly $\chi E\in V$.   It is easy to check that $V$ is
closed under addition and scalar multiplication, so it contains all
simple functions.   Next, if $\sequencen{g_n}$ is a non-decreasing
sequence of simple functions with limit $g\in\eusm L^{\infty}(\nu)$,
then $h_g=\sup_{n\in\Bbb N}h_{g_n}$ wherever the right-hand side is
defined.   Also $T$ is order-continuous, because it preserves integrals,
so

\Centerline{$Tg^{\ssbullet}=\sup_{n\in\Bbb N}Tg_n^{\ssbullet}
=\sup_{n\in\Bbb N}h_{g_n}^{\ssbullet}=h_g^{\ssbullet}$}

\noindent and $g\in V$.   Finally, if $g\in\eusm L^{\infty}(\mu)$ is
zero almost everywhere, there is a negligible $E\in\Sigma$ such that
$g(x)=0$ for every $x\in X\setminus E$;  $\mu_yE=0$ for almost every
$y$, so $h_g(y)=\int g\,d\mu_y=0$ for almost every $y$ and again
$g\in V$.   Putting these together, we see that
$V=\eusm L^{\infty}(\nu)$, as required by (ii) as stated above.
}%end of proof of 452H

\leader{452I}{Theorem}\cmmnt{ ({\smc Pachl 78})} Let $(X,\Sigma,\mu)$
be a non-empty countably compact
measure space, $(Y,\Tau,\nu)$ a $\sigma$-finite measure space, and
$f:X\to Y$ an \imp\ function.   Then there is a disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$, consistent with $f$, such
that $\mu_y$ is a complete probability measure on $X$
for every $y\in Y$.   Moreover,

(i) if $\Cal K$ is a countably compact class of subsets of $X$ such that
$\mu$ is inner regular with respect to $\Cal K$, then we can arrange
that $\Cal K\subseteq\dom\mu_y$ for every $y\in Y$;

(ii) if, in (i), $\Cal K$ is closed under finite unions and countable
intersections, then we can arrange that $\Cal K\subseteq\dom\mu_y$ and
$\mu_y$ is inner regular with respect to $\Cal K$ for every $y\in Y$.

\proof{{\bf (a)} Consider first the case in which $\nu$ and $\mu$ are
probability measures and we are provided with a class $\Cal K$ as in
(ii).   In this case, for each $u\in L^{\infty}(\mu)$,
$F\mapsto\int_{f^{-1}[F]}u$ is countably additive.   So we have an
operator $T:L^{\infty}(\mu)\to L^{\infty}(\nu)$ defined by saying that
$\int_FTu=\int_{f^{-1}[F]}u$ whenever $u\in L^{\infty}(\mu)$ and
$F\in\Tau$.   Of course $T$ is linear and positive and $\int Tu=\int u$
whenever $u\in L^{\infty}(\mu)$.

By 452H, there is a disintegration $\family{y}{Y}{\mu_y}$ of $\mu$ over
$\nu$ such that

\inset{($\alpha$) for every $y\in Y$, $\mu_yX=1$,
$\Cal K\subseteq\dom\mu_y$ and $\mu_y$ is inner regular with respect to
$\Cal K$;

($\beta$) $T(g^{\ssbullet})=h_g^{\ssbullet}$ whenever
$g\in\eusm L^{\infty}(\mu)$ and $h_g(y)=\int g\,d\mu_y$ when the
integral is defined.}

\noindent If now $F\in\Tau$, set $g=\chi f^{-1}[F]$ in ($\beta$);  then
$Tg^{\ssbullet}$ is defined by saying that

\Centerline{$\int_HTg^{\ssbullet}=\int_{f^{-1}[H]}g
=\mu f^{-1}[F\cap H]=\nu(F\cap H)$}

\noindent for every $H\in\Tau$, so that
$Tg^{\ssbullet}=\chi F^{\ssbullet}$ and we must have $\mu_yf^{-1}[F]=1$
for almost every $y\in F$.   Thus $\family{y}{Y}{\mu_y}$ is a consistent
distribution.

\medskip

{\bf (b)} The theorem is formulated in a way to make it quotable in parts
without committing oneself to a particular class $\Cal K$.   But if we
are given a class satisfying (i), we can extend it to one satisfying
(ii), by 413T;  and if we are told only that $\mu$ is countably
compact, we know from the definition that we shall be able to choose a
countably compact class satisfying (i).

\medskip

{\bf (c)} This proves the theorem on the assumption that $\mu$ and $\nu$
are probability measures.   If $\mu X=\nu Y=0$ then the result is
trivial, as we can take every $\mu_y$ to be the zero measure.
Otherwise, because
$\nu$ is $\sigma$-finite, there is a partition $\sequencen{Y_n}$ of $Y$
into measurable sets of finite measure.   Let $\sequencen{\gamma_n}$ be
a sequence of strictly positive real numbers such that
$\sum_{n=0}^{\infty}\gamma_n\nu Y_n=1$, and write

\Centerline{$\nuprime F=\sum_{n=0}^{\infty}\gamma_n\nu(F\cap Y_n)$ for
$F\in\Tau$,}

\Centerline{$\mu'E=\sum_{n=0}^{\infty}\gamma_n\mu(E\cap X_n)$ for
$E\in\Sigma$.}

\noindent It is easy to check ($\alpha$) that $\nuprime$ and $\mu'$ are
probability measures ($\beta$) that $f$ is \imp\ for $\mu'$ and
$\nuprime$ ($\gamma$) that if $\mu$ is inner regular with respect to
$\Cal K$ so is $\mu'$.   Note that
$\nuprime$ and $\nu$ have the same negligible sets.   By
(a)-(b), $\mu'$ has a disintegration $\family{y}{Y}{\mu_y}$ over
$\nuprime$ which is consistent with $f$, and (if appropriate) has the properties demanded in (i) or (ii).   Now, if $E\in\Sigma$,

$$\eqalignno{\mu E
&=\sum_{n=0}^{\infty}\gamma_n^{-1}\mu'(E\cap X_n)
=\sum_{n=0}^{\infty}\gamma_n^{-1}\int\mu_y(E\cap X_n)\nuprime(dy)\cr
&=\sum_{n=0}^{\infty}\gamma_n^{-1}\int_{Y_n}\mu_yE\,\nuprime(dy)\cr
\noalign{\noindent (because $\mu_yX=1$, $\mu_yX_n=(\chi Y_n)(y)$ for
$\nuprime$-almost every $y$, every $n$)}
&=\sum_{n=0}^{\infty}\int_{Y_n}\mu_yE\,\nu(dy)
=\sum_{n=0}^{\infty}\int\mu_y(E\cap X_n)\nu(dy)
=\int\mu_yE\,\nu(dy).\cr}$$

\noindent So $\family{y}{Y}{\mu_y}$ is a disintegration of $\mu$ over
$\nu$.   If $F\in\Tau$, then $\mu_yf^{-1}[F]=1$ for $\nuprime$-almost
every
$y$, that is, for $\nu$-almost every $y$, so $\family{y}{Y}{\mu_y}$ is
still consistent with $f$ with respect to the measure $\nu$.
}%end of proof of 452I

\cmmnt{
\leader{452J}{Remarks (a)} In the theorem above, I have
carefully avoided making any promises about the domains of the $\mu_y$
beyond that in (i).   If $\Sigma_0$ is the $\sigma$-algebra generated by
$\Cal K\cap\Sigma$, then whenever $E\in\Sigma$ there are $E'$,
$E''\in\Sigma_0$ such that $E'\subseteq E\subseteq E''$ and
$\mu(E''\setminus E')=0$.   (For $\mu$, like $\nu$, must be
$\sigma$-finite, so we can choose $E'$ to be a countable union of
members of $\Cal K\cap\Sigma$, and $E''$ to be the complement of such a
union.)   Thus we shall have a $\sigma$-algebra on which every $\mu_y$
is defined and which will be adequate to describe nearly everything
about $\mu$.   The example of Lebesgue measure on the square
(452E) shows that we cannot ordinarily expect the $\mu_y$ to be defined
on the whole of $\Sigma$ itself.   In many important cases, of course,
we can say more (452Xl).

\spheader 452Jb Necessarily (as remarked in the course of the
proof) $\mu_yX=1$ for almost every $y$.   In some
applications it seems right to change $\mu_y$ for a negligible set of
$y$'s so that every $\mu_y$ is a probability measure.   Of course this
cannot be done if $X=\emptyset\ne Y$, but this case is trivial (we
should have to have $\nu Y=0$).   In other cases, we can make sure that
any new $\mu_y$ is equal to some old one, so that a property required by
(i) or (ii) remains true of the new disintegration.   If we want to have
`$\mu_yf^{-1}[\{y\}]=\mu_yX=1$ for every $y\in Y$', strengthening `strongly consistent',
we shall of course have to begin by checking that $f$ is surjective.

\spheader 452Jc The question of whether `$\sigma$-finite' can
be weakened to `strictly localizable' in the hypotheses of 452I is
related to the Banach-Ulam problem (452Yb).   See also 452O.
}%end of comment

\leader{452K}{Example}\cmmnt{ The hypothesis `countably compact' in
452I is in fact essential (452Ye).   To see at least that it cannot be
omitted, we have the following elementary example.}
Set $Y=[0,1]$, and let $\nu$ be Lebesgue measure on $Y$, with domain
$\Tau$.   Let $X\subseteq[0,1]$ have outer measure $1$ and inner measure
$0$\cmmnt{ (134D, 419I)};  let $\mu$ be the subspace measure on $X$.
Set $f(x)=x$ for $x\in X$.   Then there is no disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$ which is consistent with $f$.

\prooflet{\Prf\Quer\ Suppose, if possible, that
$\family{y}{[0,1]}{\mu_y}$
is such a disintegration.   Then, in particular, the sets

\Centerline{$H_q=[0,q]\setminus\{y:X\cap
[0,q]\in\dom\mu_y,\,\mu_y(X\cap[0,q])=1\}$,}

\Centerline{$H'_q=[q,1]\setminus\{y:X\cap[q,1]\in\dom\mu_y,\,
\mu_y(X\cap[q,1])=1\}$}

\noindent are negligible for every $q\in[0,1]$.   Set
$G=[0,1]\setminus\bigcup_{q\in\Bbb Q\cap[0,1]}(H_q\cup H'_q)$, so that
$G$ is $\nu$-conegligible.   Then there must be some
$y\in G\setminus X$.   Now $\mu_y(X\cap[0,q'])=\mu_y(X\cap[q,1])=1$
whenever $q$,
$q'\in\Bbb Q$ and $0\le q<y<q'\le 1$, so that $\mu_y(X\cap\{y\})=1$.
But $X\cap\{y\}=\emptyset$.\ \Bang\Qed
}%end of prooflet

\leader{452L}{}\cmmnt{ The same ideas as in 452I can be used to prove
a result on the disintegration of measures on product spaces.   It will
help to have a definition.

\medskip

\noindent}{\bf Definition} Let $\familyiI{X_i}$ be a family of sets, and
$\lambda$ a measure on $X=\prod_{i\in I}X_i$.   For each $i\in I$ set
$\pi_i(x)=x(i)$ for $x\in X$.   Then the image measure
$\lambda\pi_i^{-1}$ is the {\bf marginal measure} of $\lambda$ on $X_i$.

\leader{452M}{}\cmmnt{ I return to the context of 452B-452D.

\medskip

\noindent}{\bf Theorem} Let $Y$ and $Z$ be sets and
$\Tau\subseteq\Cal PY$, $\Upsilon\subseteq\Cal PZ\,\,\sigma$-algebras.
Let $\mu$ be a
non-zero totally finite measure with domain $\Tau\tensorhat\Upsilon$,
and $\nu$ the marginal measure of $\mu$ on $Y$.   Suppose that the
marginal measure $\lambda$ of $\mu$ on $Z$ is inner regular with respect
to a
countably compact class $\Cal K\subseteq\Cal PZ$ which is closed under
finite unions and countable intersections.   Then there is a family
$\family{y}{Y}{\mu_y}$ of complete probability measures on $Z$, all
measuring every member of $\Cal K$ and inner regular with respect to
$\Cal K$, such that

\Centerline{$\mu E=\int\mu_yE[\{y\}]\nu(dy)$}

\noindent for every $E\in\Tau\tensorhat\Upsilon$, and

\Centerline{$\int fd\mu=\iint f(y,z)\mu_y(dz)\nu(dy)$}

\noindent whenever $f$ is a $[-\infty,\infty]$-valued function such that
$\int fd\mu$ is defined in $[-\infty,\infty]$.

\proof{{\bf (a)} To begin with, assume that $\mu$ is a probability
measure and that $\nu$ is complete.   Let $\frak B$ be the measure
algebra of $\nu$ and
$\theta:\frak B\to\Tau$ a lifting.   For $H\in\Upsilon$ and $F\in\Tau$
set $\nu_HF=\mu(F\times H)$;  then $\nu_H:\Tau\to[0,1]$ is countably
additive and
$\nu_HF\le\nu F$ for every $F\in\Tau$, so there is a $v_H\in L^1(\nu)$
such that $\int_Fv_H=\nu_HF$ for every $F\in\Tau$ and
$0\le v_H\le\chi 1$.   We can therefore think of $v_H$ as a member of
$L^{\infty}(\nu)\cong L^{\infty}(\frak B)$.   Let
$T:L^{\infty}(\frak B)\to L^{\infty}(\Tau)$ be the Riesz homomorphism
associated with $\theta$, and set $\psi_yH=(Tv_H)(y)$ for every
$y\in Y$.

Each $\psi_y:\Upsilon\to\coint{0,\infty}$ is finitely additive.   So we
have a complete measure $\mu_y$ on $Z$ such that $\mu_yZ\le\psi_yZ=1$,
$\Cal K\subseteq\dom\mu_y$, $\mu_y$ is inner regular with respect to
$\Cal K$ and $\mu_yK\ge\psi_yK$ for every $K\in\Cal K$ (413Ua, as before).

For $H\in\Upsilon$, $F\in\Tau$ we have

\Centerline{$\int_F\psi_yH\,\nu(dy)=\int_FTv_H
=\int_Fv_H=\nu_HF=\mu(F\times H)$.}

\noindent So

\Centerline{$
\underline{\intop}\mu_yK\cdot\chi F(y)\nu(dy)
\ge\int_F\psi_yK\,\nu(dy)=\mu(F\times K)$}

\noindent for every $K\in\Cal K$.   Now note that, for any
$H\in\Upsilon$
and $F\in\Tau$,

$$\eqalign{\mu(F\times H)
  -\sup_{K\in\Cal K,K\subseteq H}\mu(F\times K)
&=\inf_{K\in\Cal K,K\subseteq H}
  \mu(F\times(H\setminus K))\cr
&\le\inf_{K\in\Cal K,K\subseteq H}\lambda(H\setminus K)
=0\cr}$$

\noindent
because $\lambda$ is inner regular with respect to $\Cal K$ (and, like
$\mu$, is a probability measure).   So

$$\eqalign{\underline{\int}(\mu_y)_*H\cdot\chi F(y)\nu(dy)
&\ge\sup_{K\in\Cal K,K\subseteq H}
  \underline{\int}\mu_yK\cdot\chi F(y)\nu(dy)\cr
&\ge\sup_{K\in\Cal K,K\subseteq H}\mu(F\times K)
=\mu(F\times H).\cr}$$

\noindent In particular,

\Centerline{$\underline{\intop}(\mu_y)_*H\,\nu(dy)
\ge\mu(Y\times H)=\lambda H$,}

\noindent and similarly
$\underline{\int}(\mu_y)_*(Z\setminus H)\nu(dy)
\ge\lambda(Z\setminus H)$.

Taking $\nu$-integrable functions $g_1$, $g_2$ such that
$g_1(y)\le(\mu_y)_*H$ and $g_2(y)\le(\mu_y)_*(Z\setminus H)$ for almost
every $y$, $\int g_1d\nu=\underline{\int}(\mu_y)_*H\,\nu(dy)$
and $\int g_2d\nu=\underline{\int}(\mu_y)_*(Z\setminus H)\nu(dy)$
(133Ja), we must have

\Centerline{$g_1(y)+g_2(y)\le(\mu_y)_*H+(\mu_y)_*(Z\setminus H)
\le\mu_yZ\le 1$}

\noindent for almost every $y$, while $\int g_1+g_2\,d\nu\ge 1$;  so that,
for almost all $y$,

\Centerline{$g_1(y)+g_2(y)=(\mu_y)_*H+(\mu_y)_*(Z\setminus H)
=\mu_yZ=1$,}

\noindent and (because $\mu_y$ is complete) $\mu_yH$ is defined
and equal to $g_1(y)$ (413Ec, 413Ef).   It now follows that

\Centerline{$\int_F\mu_yH\,\nu(dy)
=\int_Fg_1(y)\nu(dy)
=\underline{\intop}(\mu_y)_*H\cdot\chi F(y)\nu(dy)
\ge\mu(F\times H)$}

\noindent for every $F\in\Tau$.   But since also

\Centerline{$\int_F\mu_y(Z\setminus H)\nu(dy)
\ge\mu(F\times(Z\setminus H))$,}

\Centerline{$\int_F\mu_yH+\mu_y(Z\setminus H)\,\nu(dy)
\le\nu F=\mu(F\times H)+\mu(F\times(Z\setminus H))$,}

\noindent we must actually have $\int_F\mu_yH\,\nu(dy)=\mu(F\times H)$.

All this is true whenever $F\in\Tau$ and $H\in\Upsilon$.   But now,
setting

\Centerline{$\Cal E
=\{E:E\in\Tau\tensorhat\Upsilon,\,
  \mu E=\int\mu_yE[\{y\}]\nu(dy)\}$,}

\noindent we see that $\Cal E$ is a Dynkin class and includes
$\Cal I=\{F\times H:F\in\Tau$, $H\in\Upsilon\}$, which is closed under
finite intersections;  so that the Monotone Class Theorem tells us that
$\Cal E$ includes the $\sigma$-algebra generated by $\Cal I$, and is the
whole of $\Tau\tensorhat\Upsilon$.

\medskip

{\bf (b)} The rest is just tidying up.
(i) The construction in (a) allows $\mu_yZ$ to be less than
$1$ for a
$\nu$-negligible set of $y$;  but of course all we have to do, if that
happens, is to amend $\mu_y$ arbitrarily on that set to any of the
`ordinary' values of $\mu_y$.
(ii) If the original measure $\nu$ is not complete, let $\hat\mu$ and
$\hat\nu$ be the completions of $\mu$ and $\nu$, and $\hat{\Tau}$ the
domain of $\hat\nu$.   The projection onto $Y$ is \imp\ for $\mu$ and
$\nu$, so is \imp\ for $\hat\mu$ and $\hat\nu$ (234Ba\formerly{2{}35Hc}),
and $\hat\mu$ measures every member of $\hat\Tau\tensorhat\Upsilon$;
set $\mu'=\hat\mu\restr\hat\Tau\tensorhat\Upsilon$.   Next, the marginal
measure of $\mu'$ on $Z$ is still $\lambda$ (since both must have domain
$\Upsilon$).   So we can apply (a) to $\mu'$ to get the result.   (iii)
If the original measure $\mu$ is not a probability measure, apply the
arguments so far to suitable scalar multiples of $\mu$ and $\nu$.

\medskip

{\bf (c)} Thus we have the formula

\Centerline{$\mu E=\int\mu_yE[\{y\}]\nu(dy)$}

\noindent for every $E\in\Tau\tensorhat\Upsilon$.   The second formula
announced follows as in the remark following 452F.
}%end of proof of 452M

\leader{452N}{Corollary} Let $Y$ and $Z$ be sets and
$\Tau\subseteq\Cal PY$, $\Upsilon\subseteq\Cal PZ\,\,\sigma$-algebras.
Let $\mu$ be a
probability measure with domain $\Tau\tensorhat\Upsilon$, and $\nu$ the
marginal measure of $\mu$ on $Y$.   Suppose that

\inset{{\it either} $\Upsilon$ is the Baire $\sigma$-algebra with
respect to a compact Hausdorff topology on $Z$

{\it or} $\Upsilon$ is the Borel $\sigma$-algebra with respect to an
analytic Hausdorff topology on $Z$

{\it or} $(Z,\Upsilon)$ is a standard Borel space.}

\noindent Then there is a family $\family{y}{Y}{\mu_y}$ of probability
measures on $Z$, all with domain $\Upsilon$, such that

\Centerline{$\mu E=\int\mu_yE[\{y\}]\nu(dy)$}

\noindent for every $E\in\Tau\tensorhat\Upsilon$, and

\Centerline{$\int fd\mu=\iint f(y,z)\mu_y(dz)\nu(dy)$}

\noindent whenever $f$ is a $[-\infty,\infty]$-valued function such that
$\int fd\mu$ is defined in $[-\infty,\infty]$.

\proof{ In each case, the marginal measure of $\mu$ on $Z$ is tight
(that is, inner regular with respect to the closed compact sets)
for a Hausdorff topology on
$Z$.   (Use 412D when $\Upsilon$ is the Baire $\sigma$-algebra on a
compact Hausdorff space and 433Ca when it is the Borel $\sigma$-algebra
on an analytic Hausdorff space;  when $(Z,\Upsilon)$ is a standard Borel
space, take any appropriate Polish topology on $Z$ and use 423Ba.)   So
452M tells us that we can achieve the formulae sought with
Radon probability measures $\mu_y$.   Since (in all three cases)
$\dom\mu_y$ will include $\Upsilon$ for every $y$, we can get the result
as stated by replacing each $\mu_y$ by $\mu_y\restr\Upsilon$.
}%end of proof of 452N

\leader{452O}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a Radon
measure space, $(Y,\Tau,\nu)$ a strictly localizable measure space, and
$f:X\to Y$ an \imp\ function.   Then there is a disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$, consistent with $f$, such
that every $\mu_y$ is a Radon measure on $X$.

\proof{{\bf (a)} Let $\familyiI{Y_i}$ be a decomposition of $Y$.   For
each $i\in I$, let $\nu_i$ be the subspace measure on $Y_i$ and
$\lambda_i$ the subspace measure on $X_i=f^{-1}[Y_i]$.   Then
$f_i=f\restr X_i$ is \imp\ for $\lambda_i$ and $\nu_i$.   Let $\Cal K_i$
be the family of compact subsets of $X_i$;  of course $\Cal K_i$ is a
(countably) compact class and $\lambda_i$ is inner regular with respect
to $\Cal K_i$ (412Oa).   By 452I, we can choose, for each
$i\in I$, a disintegration $\family{y}{Y_i}{\tilde\mu_y}$ of $\lambda_i$
over $\nu_i$, consistent with $f\restr X_i$, such that
$\tilde\mu_y$ measures every compact subset of  $X_i$ and is
inner regular with respect to $\Cal K_i$ for every $y\in Y_i$.
Adjusting any which are not probability
measures, and completing them if necessary,
we can suppose that every $\tilde\mu_y$ is a
complete probability measure.   By
412Ja, $\tilde\mu_y$ measures every relatively closed subset of $X_i$
for every $y\in Y_i$.

For $i\in I$ and $y\in Y_i$, set

\Centerline{$\mu_yE=\tilde\mu_y(E\cap X_i)$}

\noindent whenever $E\subseteq X$ and $E\cap X_i$ is measured by
$\tilde\mu_y$.   Then $\mu_y$ is a complete totally finite measure on
$X$;  it is inner regular with respect to $\Cal K_i$ and measures every
closed subset of $X$.   It follows at once that it is tight and measures
every Borel set, that is, is a Radon measure on $X$.

\medskip

{\bf (b)} Now $\mu E=\int\mu_yE\,\nu(dy)$ for every $E\in\Sigma$.  \Prf\
$\bigcup_{i\in J}E\cap X_i=E\cap f^{-1}[\bigcup_{i\in J}Y_i]$  belongs
to $\Sigma$ for every
$J\subseteq I$.   By 451Q, $\mu E=\sum_{i\in I}\mu(E\cap X_i)$.  For
$i\in I$, we have
$\int_{Y_i}\tilde\mu_y(E\cap X_i)\nu_i(dy)=\mu(E\cap X_i)$.   So

$$\eqalign{\mu E
&=\sum_{i\in I}\mu(E\cap X_i)
=\sum_{i\in I}\int_{Y_i}\tilde\mu_y(E\cap X_i)\nu_i(dy)\cr
&=\sum_{i\in I}\int_{Y_i}\mu_yE\,\nu(dy)
=\int\mu_yE\,\nu(dy)\cr}$$

\noindent by 214N.\ \Qed

Thus $\family{y}{Y}{\mu_y}$ is a disintegration of $\mu$ over $\nu$.

\medskip

{\bf (c)} Finally, if $F\in\Tau$ and $i\in I$, then

$$\eqalign{Y_i\cap F\setminus&\{y:\mu_yf^{-1}[F]
  \text{ is defined and equal to }1\}\cr
&=(F\cap Y_i)\setminus\{y:y\in Y_i,\,
  \tilde\mu_yf^{-1}[F\cap Y_i]=1\}\cr}$$

\noindent is negligible for every $i$, so $\mu_yf^{-1}[F]=1$ for almost
every $y$.   Thus $\family{y}{Y}{\mu_y}$ is consistent with $f$.
}%end of proof of 452O

\leader{452P}{Corollary}\cmmnt{ (cf.\ {\smc Blackwell 56})} Let
$(X,\frak T,\Sigma,\mu)$ be a Radon measure space,
$(Y,\frak S,\Tau,\nu)$ an analytic Radon measure space and $f:X\to Y$ an
\imp\ function.   Then there is a disintegration $\family{y}{Y}{\mu_y}$
of $\mu$ over $\nu$, strongly consistent with $f$, such that every
$\mu_y$ is a Radon measure on $X$.

\proof{ By 433B, $\nu$ is countably separated;  now put 452O and 452Gc
together.
}%end of proof of 452P

\leader{452Q}{Disintegrations and conditional
\dvrocolon{expectations}}\cmmnt{ Fubini's theorem provides a
relatively
concrete description of the conditional expectation of a function on a
product of probability spaces with respect to the $\sigma$-algebra
defined by one of the factors, by means of the formula
$g(x,y)=\int f(x,z)dz$ (253H).   This generalizes straightforwardly to
measures with disintegrations, as follows.

\wheader{452Q}{6}{2}{2}{72pt}\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces and $f:X\to Y$ an \imp\ function.   Suppose that
$\family{y}{Y}{\mu_y}$ is a disintegration of $\mu$ over $\nu$ which is
consistent with $f$, and that $g$ is a $\mu$-integrable real-valued
function.

(a) Setting $h_0(y)=\int g\,d\mu_y$ whenever $y\in Y$ and the integral
is defined in
$\Bbb R$, $h_0$ is a Radon-Nikod\'ym derivative of the functional
$F\mapsto\int_{f^{-1}[F]}g\,d\mu:\Tau\to\Bbb R$.

(b) Now suppose that $\nu$ is complete.   Setting
$h_1(x)=\int g\,d\mu_{f(x)}$ whenever $x\in X$ and the integral is
defined in $\Bbb R$, then
$h_1$ is a conditional expectation of $g$ on the $\sigma$-algebra
$\Sigma_0=\{f^{-1}[F]:F\in\Tau\}$.

\proof{{\bf (a)} If $F\in\Tau$, then $f^{-1}[F]$ is $\mu_y$-conegligible
for almost every $y\in F$, and $\mu_y$-negligible for almost every
$y\in Y\setminus F$, so
$\int g\times\chi f^{-1}[F]d\mu_y=h_0(y)\times\chi F(y)$ for almost
every $y$, and

\Centerline{$\int_Fh_0d\nu
=\iint g\times\chi f^{-1}[F]d\mu_y\nu(dy)
=\int_{f^{-1}[F]}g\,d\mu$}

\noindent (452F).   As $F$ is arbitrary, we have the result.

\medskip

{\bf (b)} Of course $\Sigma_0$ is a $\sigma$-algebra (111Xd), and it is
included in $\Sigma$ because $f$ is \imp.   By 452F,
$Y_0=\{y:g$ is $\mu_y$-integrable$\}$ is conegligible, so
$\dom h_1=f^{-1}[Y_0]$ is conegligible.   If $\alpha\in\Bbb R$, then

\Centerline{$F=\{y:y\in Y_0$, $\int g\,d\mu_y\ge\alpha\}$}

\noindent belongs to $\Tau$
because $y\mapsto\int g\,d\mu_y$ is $\nu$-virtually measurable
and $\nu$ is complete.   So

\Centerline{$\{x:x\in\dom h_1$, $h_1(x)\ge\alpha\}=f^{-1}[F]$}

\noindent belongs to $\Sigma_0$, and $h_1$ is $\Sigma_0$-measurable.
If $F\in\Tau$, then

$$\eqalignno{\int_{f^{-1}[F]}h_1\,d\mu
&=\int_{f^{-1}[F]}\int g\,d\mu_{f(x)}\mu(dx)
=\int_F\int g\,d\mu_y\nu(dy)\cr
\displaycause{235G\formerly{2{}35I}}
&=\int_Fh_0d\nu
=\int_{f^{-1}[F]}g\,d\mu\cr}$$

\noindent as in (a).
As $F$ is arbitrary, $h_1$ is a conditional expectation of $g$
on $\Sigma_0$, as claimed.
}%end of proof of 452Q

\leader{*452R}{}\cmmnt{ I take the opportunity to interpolate an
interesting result about countably compact measures.   It demonstrates
the power of 452I to work in unexpected ways.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Pachl 79})} Let $(X,\Sigma,\mu)$
be a countably compact
measure space, $(Y,\Tau,\nu)$ a strictly localizable measure space, and
$f:X\to Y$ an \imp\ function.   Then $\nu$ is countably compact.

\proof{{\bf (a)} For most of the proof (down to the end of (b) below) I
suppose that $\mu$ and $\nu$ are totally finite.

Let $Z$ be the Stone space of the Boolean algebra $\Tau$.   (I am {\it
not} using the measure algebra here!)   For $F\in\Tau$, let $F^*$ be the
corresponding open-and-closed subset of $Z$.   For each $y\in Y$, the
map $F\mapsto\chi F(y)$ is a Boolean homomorphism from $\Tau$ to
$\{0,1\}$, so belongs to $Z$;  define $g:Y\to Z$ by saying that
$g(y)(F)=\chi F(y)$ for $y\in Y$, $F\in\Tau$, that is,
$g^{-1}[F^*]=F$ for every $F\in\Tau$.   Let $\Cal Z$ be the family of
zero sets in $Z$, and $\Lambda$ the Baire $\sigma$-algebra of $Z$.

The set

\Centerline{$\{W:W\subseteq Z,\,g^{-1}[W]\in\Tau\}$}

\noindent is a $\sigma$-algebra of subsets of $Z$ containing all the
open-and-closed sets, so contains every zero set (4A3Od) and
includes $\Lambda$.   Set $\lambda W=\nu g^{-1}[W]$ for $W\in\Lambda$.
Then $\lambda$ is a Baire measure on $Z$, so is inner regular with
respect to $\Cal Z$ (412D).

Set $h=gf:X\to Z$.   Then $h$ is a composition of \imp\ functions, so is
\imp.   By 452I, there is a disintegration $\family{z}{Z}{\mu_z}$ of
$\mu$ over $\lambda$ which is consistent with $h$.

\medskip

{\bf (b)} Let $\Cal K\subseteq\Cal PY$ be the family of sets

\Centerline{$\{g^{-1}[V]:V\in\Cal Z,\,\mu_zh^{-1}[V]=\mu_z X=1$ for
every $z\in V\}$.}

\medskip

\quad{\bf (i)} $\Cal K$ is a countably compact class of sets.   \Prf\
Let $\sequencen{K_n}$ be a sequence in $\Cal K$ such that $\bigcap_{i\le
n}K_i\ne\emptyset$ for each $n\in\Bbb N$.   For each $n\in\Bbb N$, let
$V_n\in\Cal Z$ be such that $K_n=g^{-1}[V_n]$ and
$\mu_zh^{-1}V_n=\mu_zX=1$ for every $z\in V_n$.   Then

\Centerline{$g^{-1}[\bigcap_{i\le n}V_i]=\bigcap_{i\le n}K_i
\ne\emptyset$}

\noindent for every $n\in\Bbb N$, so $\{V_n:n\in\Bbb N\}$ has the finite
intersection property and (because $Z$ is compact) there is a
$z\in\bigcap_{n\in\Bbb N}V_n$.   Now

\Centerline{$\mu_zh^{-1}[V_n]=\mu_zX=1$}

\noindent for every $n\in\Bbb N$, so

\Centerline{$\emptyset\ne\bigcap_{n\in\Bbb N}h^{-1}[V_n]
=f^{-1}[\bigcap_{n\in\Bbb N}K_n]$.}

\noindent Thus $\bigcap_{n\in\Bbb N}K_n$ is non-empty.   As
$\sequencen{K_n}$ is arbitrary, $\Cal K$ is a countably compact
class.\ \Qed

\medskip

\quad{\bf (ii)} $\nu$ is inner regular with respect to $\Cal K$.   \Prf\
Suppose that $F\in\Tau$ and $\gamma<\nu F$.   Choose a sequence
$\sequencen{V_n}$ in $\Cal Z$ as follows.   Start with $V_0=F^*$, so
that

\Centerline{$\lambda V_0=\nu g^{-1}[V_0]=\nu F>\gamma$.}

\noindent  Given that $V_n\in\Cal Z$ and $\lambda V_n>\gamma$, then we
know that $\mu_zh^{-1}[V_n]=\mu_zX=1$ for $\lambda$-almost every
$z\in V_n$;
because $\lambda$ is inner regular with respect to $\Cal Z$, there is a
$V_{n+1}\in\Cal Z$ such that $V_{n+1}\subseteq V_n$, $\lambda
V_{n+1}>\gamma$ and $\mu_zh^{-1}[V_n]=\mu_zX=1$ for every $z\in
V_{n+1}$.   Continue.

At the end of the induction, set $V=\bigcap_{n\in\Bbb N}V_n$.   Then
$V\in\Cal Z$.   If $z\in V$, then

\Centerline{$\mu_zh^{-1}[V]
=\lim_{n\to\infty}\mu_zh^{-1}[V_n]=1=\mu_zX$,}

\noindent so $g^{-1}[V]\in\Cal K$.   Because $V\subseteq V_0=F^*$,
$g^{-1}[V]\subseteq F$, and

\Centerline{$\nu g^{-1}[V]
=\lambda V
=\lim_{n\to\infty}\lambda V_n
\ge\gamma$.}

\noindent As $F$ and $\gamma$ are arbitrary, $\nu$ is inner regular with
respect to $\Cal K$.\ \Qed

Thus $\Cal K$ witnesses that $\nu$ is countably compact.

\medskip

{\bf (c)} For the general case, let $\familyiI{Y_i}$ be a decomposition
of $Y$.   For each $i\in I$, set $X_i=f^{-1}[Y_i]$;  let $\mu_i$ be the
subspace measure on $X_i$ and $\nu_i$ the subspace measure on $Y_i$.
Then $\mu_i$ is countably compact (451Db) and $f\restr X_i:X_i\to Y_i$
is \imp\ for $\mu_i$ and $\nu_i$, so $\nu_i$ is countably compact, by
(a)-(b) above.   Let $\Cal K_i\subseteq\Cal PY_i$ be a countably compact
class such that $\nu_i$ is inner regular with respect to $\Cal K_i$.
Then $\Cal K=\bigcup_{i\in I}\Cal K_i$ is a countably compact class
(because any sequence in $\Cal K$ with the finite intersection property
must lie within a single $\Cal K_i$).   By 413T, there is a
countably compact class $\Cal K^*\supseteq\Cal K$ which is closed under
finite unions;  by 412Aa, $\nu$ is inner regular with respect to
$\Cal K^*$, so is countably compact.   This completes the proof.
}%end of proof of 452R

\leader{*452S}{Corollary}\cmmnt{ ({\smc Pachl 78})} If
$(X,\Sigma,\mu)$ is a countably compact totally finite measure space,
and $\Tau$ is any $\sigma$-subalgebra of $\Sigma$, then $\mu\restrp\Tau$
is countably compact.

\leader{452T}{}\cmmnt{ In 452E, I remarked in passing that Fubini's theorem
on a product space $X=Y\times Z$ can be thought of as giving us a
disintegration of the product measure on $X$ over the factor measure on
$Y$.   There are other contexts in
which we find that a canonical disintegration is provided for a structure
$(X,\mu,Y,\nu)$ without calling
on the Lifting Theorem.   Here I will describe
an important case arising naturally in the theory of group actions.

\medskip

\noindent}{\bf Theorem} Let $X$ be a locally compact Hausdorff space, $G$ a
compact Hausdorff topological group and $\action$ a continuous action of
$G$ on $X$.   Suppose that $\mu$ is a $G$-invariant
Radon probability measure on $X$.
For $x\in X$, write $f(x)$ for the corresponding
orbit $\{a\action x:a\in G\}$ of the
action.   Let $Y=f[X]$ be the set of orbits, with the topology
$\{W:W\subseteq Y$, $f^{-1}[W]$ is open in $X\}$.
Write $\nu$ for the image measure $\mu f^{-1}$ on $Y$.

(a) $Y$ is locally compact and
Hausdorff, and $\nu$ is a Radon probability measure.

(b) For each $\pmb{y}\in Y$, there is a unique $G$-invariant Radon
probability $\mu_{\pmb{y}}$ on $X$ such that $\mu_{\pmb{y}}(\pmb{y})=1$.

(c) $\family{\pmb{y}}{Y}{\mu_{\pmb{y}}}$ is a
disintegration of $\mu$ over $\nu$, strongly consistent with $f$.

\proof{{\bf (a)} By 4A5Ja, $Y$ is locally compact and Hausdorff, and $f$ is
an open map.   By 418I, $\nu$ is a Radon measure.

\medskip

{\bf (b)} Let $\lambda$ be the unique Haar probability measure on $G$
(442Id).   By 443Ub-443Ud, applied to the action
$\action\restr G\times\pmb{y}$ of $G$ on $\pmb{y}$, we have a unique
$G$-invariant Radon probability measure $\mu'_{\pmb{y}}$ on $\pmb{y}$
defined by saying that $\mu'_{\pmb{y}}E=\lambda\{g:g\action x\in E\}$ for
every $x\in\pmb{y}$ and Borel set $E\subseteq\pmb{y}$.
Now $\mu_{\pmb{y}}$ must be the unique extension of
$\mu'_{\pmb{y}}$ to $X$.   Of course we still have
$\mu_{\pmb{y}}E=\lambda\{g:g\action x\in E\}$ for
every $x\in\pmb{y}$ and Borel set $E\subseteq X$.

\medskip

{\bf (c)(i)} Let $V\subseteq X$ be an open set, and set
$h_V(\pmb{y})=\mu_{\pmb{y}}V$ for $\pmb{y}\in Y$.   Then $h_V$ is
lower semi-continuous.   \Prf\ Suppose that $\pmb{y}\in Y$ and
$\alpha\in\Bbb R$ are such that $h_V(\pmb{y})>\alpha$.   Then there is a
compact set $K\subseteq V$ such that $\mu_{\pmb{y}}K>\alpha$.
Fix $x\in\pmb{y}$, and set $L=\{g:g\action x\in K\}$, so that $L$ is a
compact subset of $G$ and $\lambda L>\alpha$.   The set
$\{(g,x'):g\in L$, $x'\in X$, $g\action x'\notin V\}$ is closed in
$L\times X$, so its projection
$\{x':\,\Exists g\in L$, $g\action x'\notin V\}$ is closed
(4A2Gm) and
$U=\{x':g\action x'\in V$ for every $g\in L\}$ is open in $X$.   Now
$f[U]$ is open in $Y$, because $f$ is an open map.   Of course
$x\in U$ and $\pmb{y}\in f[U]$.   But if
$\pmb{y}'\in f[U]$, there is an $x'\in U$ such that
$f(x')=\pmb{y}'$, and now

\Centerline{$h_V(\pmb{y}')=\mu_{\pmb{y}'}V
=\lambda\{g:g\action x'\in V\}\ge\lambda L>\alpha$.}

\noindent As $\pmb{y}$ and $\alpha$ are arbitrary, $h_V$ is lower
semi-continuous.\ \Qed

\medskip

\quad{\bf (ii)} In particular, $h_V$ is Borel measurable;  because $f$ is
\imp\ for $\mu$ and $\nu$,

$$\eqalignno{\int h_V\,d\nu
&=\int h_V(f(x))\mu(dx)\cr
\displaycause{235G again}
&=\int\lambda\{g:g\action x\in V\}\mu(dx)
=\int\mu\{x:g\action x\in V\}\lambda(dg)\cr
\displaycause{by 417G,
because $\mu$ and $\lambda$ are totally finite Radon measures
and $\{(g,x):g\action x\in V\}$ is an open set in $G\times X$}
&=\int\mu(g^{-1}\action V)\lambda(dg)
=\int\mu V\lambda(dg)\cr
\displaycause{because $\mu$ is $G$-invariant}
&=\mu V.\cr}$$

By the Monotone Class Theorem, as usual, it follows that
$\int\mu_{\pmb{y}}E\,\nu(d\pmb{y})=\mu E$ for every Borel set
$E\subseteq X$
(apply 136C to $\mu$ and $E\mapsto\int\mu_{\pmb{y}}E\,\nu(d\pmb{y})$),
and therefore (because every $\mu_{\pmb{y}}$ is complete and $\mu$ is the
completion of a Borel measure) for every $E\in\dom\mu$.   %452Xg
So $\family{\pmb{y}}{Y}{\mu_{\pmb{y}}}$ is a
disintegration of $\mu$ over $\nu$.   Since

\Centerline{$\mu_{\pmb{y}}f^{-1}[\{\pmb{y}\}]
=\mu_{\pmb{y}}(\pmb{y})=1$}

\noindent for every $\pmb{y}\in Y$, the disintegration
is strongly consistent with $f$.
}%end of proof of 452T

\exercises{
\leader{452X}{Basic exercises (a)}
%\spheader 452Xa
Let $Y$ be a first-countable topological space, $\nu$ a topological
probability measure on $Y$, $Z$ a topological space, and
$\family{y}{Y}{\mu_y}$ a family of topological probability measures on
$Z$ such that $y\mapsto\mu_yV$ is lower semi-continuous for every open
set $V\subseteq Z$.
Show that there is a Borel probability measure $\mu$ on $Y\times Z$ such
that $\mu E=\int\mu_yE[\{y\}]\nu(dy)$ for every Borel set
$E\subseteq Y\times Z$.   \Hint{434R.}
%452C

\spheader 452Xb Let $(Y,\Tau,\nu)$ be a probability space, $Z$ a
topological space and $P$ the set of topological probability measures on
$Z$ with its narrow
topology (437Jd).   Let $y\mapsto\mu_y:Y\to P$ be a function which is
measurable in the sense of 411L.   Show that, writing $\Cal B(Z)$ for
the Borel $\sigma$-algebra of $Z$,
we have a probability measure $\mu$ defined on $\Tau\tensorhat\Cal B(Z)$
such that $\mu E=\int\mu_yE[\{y\}]\nu(dy)$ for every
$E\in\Tau\tensorhat\Cal B(Z)$.
%452Xa 452C

\spheader 452Xc Let $(Y,\Tau,\nu)$ be a probability space, $Z$ a
topological space and $P_{\CalBa}$ the set of Baire probability measures
on $Z$ with its vague
topology (437Jc).   Let $y\mapsto\mu_y:Y\to P_{\CalBa}$ be a measurable
function.
Show that, writing $\CalBa(Z)$ for the Baire $\sigma$-algebra of $Z$,
we have a probability measure $\mu$ defined on $\Tau\tensorhat\CalBa(Z)$
such that $\mu E=\int\mu_yE[\{y\}]\nu(dy)$ for every
$E\in\Tau\tensorhat\CalBa(Z)$.
%452Xa 452C 452Xb

\spheader 452Xd
Let $(Y,\frak S,\Tau,\nu)$ be a Radon probability space, $(X,\frak T)$ a
topological space, and $\family{y}{Y}{\mu_y}$
a family of Radon probability measures on $X$.
Suppose that (i) there is a base $\Cal U$ for $\frak T$, closed under
finite unions, such
that $y\mapsto\mu_yU$ is lower semi-continuous for every $U\in\Cal U$
(ii) $\nu$ is inner regular with respect to the family
$\{K:K\subseteq Y$, $\{\mu_y:y\in K\}$ is uniformly tight$\}$.
Show that we have a Radon probability measure $\tilde\mu$ on $X$
such that $\tilde\mu E=\int\mu_yE\nu(dy)$ whenever $\tilde\mu$
measures $E$.
%452D

\spheader 452Xe Let $(Y,\frak S,\Tau,\nu)$ be a Radon probability space,
$(Z,\frak U)$ a Prokhorov Hausdorff space (437U), and $P$ the space of
Radon probability measures
on $Z$ with its narrow topology.   Suppose that $y\mapsto\mu_y:Y\to P$
is almost continuous.
Show that we have a Radon probability measure $\tilde\mu$ on $Y\times Z$
such that $\tilde\mu E=\int\mu_yE[\{y\}]\nu(dy)$ whenever $\tilde\mu$
measures $E$.
%452D

\spheader 452Xf
Let $(X,\Tau,\nu)$ be a measure space, and $\mu$ an indefinite-integral
measure over $\nu$ (234J\formerly{2{}34B}).
Show that there is a disintegration
$\family{x}{X}{\mu_x}$ of $\mu$ over $\nu$ such that $\mu_x\{x\}=\mu_xX$
for every $x\in X$.
%452E

\sqheader 452Xg Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces and $\family{y}{Y}{\mu_y}$ a disintegration of $\mu$ over $\nu$.
Show that $\family{y}{Y}{\hat\mu_y}$ is a disintegration of $\hat\mu$
over $\nu$, where $\hat\mu_y$ and $\hat\mu$ are the completions of
$\mu_y$ and $\mu$ respectively.
%452E proofs of 452T, 455E

\sqheader 452Xh Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\nuprime$ an indefinite-integral measure over $\nu$,
defined
from a $\nu$-virtually measurable function $g:Y\to\coint{0,\infty}$.
Suppose that
$\family{y}{Y}{\mu_y}$ is a disintegration of $\mu$ over $\nuprime$.
Show
that $\family{y}{Y}{g(y)\mu_y}$ is a disintegration of $\mu$ over $\nu$.
%452E

\spheader 452Xi Let $(Y,\Tau,\nu)$ be a probability space, $X$ a set
and $\family{y}{Y}{\mu_y}$ a family of probability measures on $X$.
Set $\theta A=\overline{\int}\mu^*_y(A)\,\nu(dy)$ for every
$A\subseteq X$.   (i) Show that $\theta$ is an outer measure on $X$.
(ii) Let $\mu$ be the measure on $X$ defined from $\theta$ by
\Caratheodory's construction.   Show that $\family{y}{Y}{\mu_y}$ is a
disintegration of $\mu$ over $\nu$.   (iii) Suppose that $X=[0,1]^2$,
$\nu$ is Lebesgue measure on $[0,1]=Y$ and $\mu_yE=\nu\{x:(x,y)\in E\}$
whenever this is defined.   Show that, for any $E$ measured by $\mu$,
$\mu_yE\in\{0,1\}$ for $\nu$-almost every $y$.
%452E

\spheader 452Xj Explore connexions between 452F and the formula
$\int fd\mu=\iint fd\nu_z\lambda(dz)$ of 443Qe.
%452F

\spheader 452Xk Let $(X,\Sigma,\mu)$ be a countably compact
$\sigma$-finite measure space, $(Y,\Tau,\nu)$ a $\sigma$-finite measure
space, and $f:X\to Y$ a $(\Sigma,\Tau)$-measurable function such that
$f^{-1}[F]$ is $\mu$-negligible whenever $F\subseteq Y$ is
$\nu$-negligible.   Show that there is a disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$ such that, for each
$F\in\Tau$, $\mu_y(X\setminus f^{-1}[F])=0$ for almost every $y\in F$.
\Hint{Reduce to the case in which $\mu$ is totally finite, and
disintegrate $\mu$ over $\nuprime=(\mu f^{-1})\restrp\Tau$.}
%452I, 452Xh

\sqheader 452Xl Let $(X,\Sigma,\mu)$ be a non-empty countably compact
measure space such that $\Sigma$ is countably generated (as
$\sigma$-algebra), $(Y,\Tau,\nu)$ a $\sigma$-finite measure space,
and $f:X\to Y$ an \imp\ function.   (i) Show that there is a
disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$, consistent with $f$, such
that every
$\mu_y$ is a probability measure with domain $\Sigma$.   (ii) Show
that if $\family{y}{Y}{\mu'_y}$ is any other disintegration of $\mu$
over $\nu$ which is consistent with $f$, then $\mu_y=\mu'_y\restr\Sigma$
for almost every $y$.
%452J

\spheader 452Xm Let $(X,\Sigma)$ be a non-empty standard Borel space,
$\mu$ a measure with domain $\Sigma$, $(Y,\Tau,\nu)$ a $\sigma$-finite
measure space,
and $f:X\to Y$ an \imp\ function.   (i) Show that there is a
disintegration
$\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$, consistent with $f$, such
that every
$\mu_y$ is a probability measure with domain $\Sigma$.   (ii) Show
that if $\family{y}{Y}{\mu'_y}$ is any other disintegration of $\mu$
over $\nu$ which is consistent with $f$, then $\mu_y=\mu'_y\restr\Sigma$
for almost every $y$.
%452Xl 452J omittable

\spheader 452Xn Let $(X,\Sigma,\mu)$ be a totally finite countably
compact measure space and $\Tau\subseteq\Sigma$ a countably-generated
$\sigma$-algebra;  set $\nu=\mu\restrp\Tau$.   Show that there is a
disintegration $\family{x}{X}{\mu_x}$ of $\mu$ over $\nu$ such that
$\mu_xH_x=\mu_xX=1$ for every $x\in X$, where
$H_x=\bigcap\{F:x\in F\in\Tau\}$ for every $x$.   \Hint{apply 452I
with $Y=\{H_x:x\in X\}$.}
%452J

\spheader 452Xo Show that 452I can be deduced from 452M.   \Hint{start
with the case $\nu Y=1$;  set $\lambda W=\mu\{x:(x,f(x))\in W\}$ for
$W\in\Sigma\tensorhat\Tau$.}
%452M

\spheader 452Xp Show that, in 452M, we shall have
$\hat\mu E=\int\mu_y E[\{y\}]\nu(dy)$ whenever the completion
$\hat\mu$ of $\mu$ measures $E$.
%452M

\sqheader 452Xq Let $\Tau$ be the Borel $\sigma$-algebra of $[0,1]$,
$\nu$ the restriction of Lebesgue measure to $\Tau$, $Z\subseteq[0,1]$ a
set with inner measure
$0$ and outer measure $1$, and $\Upsilon$ the Borel $\sigma$-algebra of
$Z$.   Show that there is a probability measure $\mu$ on $[0,1]\times Z$
defined by setting
$\mu E=\nu^*\{y:(y,y)\in E\}$ for $E\in\Tau\tensorhat\Upsilon$.   Show
that there is no disintegration of $\mu$ over $\nu$ which is consistent
with the projection
$(y,z)\mapsto y$.
%452M

\sqheader 452Xr Let $(X,\Sigma,\mu)$ be a complete totally finite
countably compact measure space and $\Tau$ a
$\sigma$-subalgebra of $\Sigma$ containing all negligible sets.   Show
that there is a
family $\family{x}{X}{\mu_x}$ of probability measures on $X$ such that
(i) $x\mapsto\mu_xE$ is $\Tau$-measurable and
$\int\mu_xE\,\mu(dx)=\mu E$ for every $E\in\Sigma$ (ii) if $F\in\Tau$,
then $\mu_xF=1$ for almost
every $x\in F$.   Show that if $g$ is any $\mu$-integrable real-valued
function, then $g$ is $\mu_x$-integrable for almost every $x$, and
$x\mapsto\int g\,d\mu_x$ is a conditional expectation of $g$ on $\Tau$.
%452Q

\spheader 452Xs\dvAnew{2010}
Let $(X_0,\Sigma_0,\mu_0)$ and $(X_1,\Sigma_1,\mu_1)$ be
$\sigma$-finite measure spaces.
For each $i$, let $(Y_i,\Tau_i,\nu_i)$ be a measure space and
$\family{y}{Y_i}{\mu^{(i)}_{y}}$ a disintegration of $\mu_i$ over $\nu_i$.
Show that
$\family{(y_0,y_1)}{Y_0\times Y_1}{\mu^{(0)}_{y_0}\times\mu^{(1)}_{y_1}}$
is a disintegration of $\mu_0\times\mu_1$ over $\nu_0\times\nu_1$, where
each product here is a c.l.d.\ product measure.
%452E out of order query

\spheader 452Xt\dvAnew{2011}
In 452M, suppose that $Z$ is a metrizable space and $\Cal K$ is the family
of compact subsets of $Z$, and let $(Y,\hat\Tau,\hat\nu)$ be the completion
of $(Y,\Tau,\nu)$.   Show that $y\mapsto\mu_y$ is a $\hat\Tau$-measurable
function from $Y$ to the set of Radon probability measures on $Z$ with
its narrow topology.   \Hint{437Rh.}
%452M out of order query

\spheader 452Xu\dvAnew{2012} $SU(r)$, for $r\ge 2$,
is the set of $r\times r$ matrices $T$ with complex coefficients such that
$\det T=1$ and $TT^*=I$, where $T^*$ is the complex conjugate of the
transpose of $T$.   (i) Show that under the natural action
$(T,u)\mapsto Tu:SU(r)\times\Bbb C^r\to\Bbb C^r$ the orbits are the
spheres $\{u:u\dotproduct\bar u=\gamma\}$, for $\gamma>0$, together with
$\{0\}$.   (ii) Show that if a Borel set $C\subseteq\Bbb C^r$ is such that
$\gamma C\subseteq C$ for every $\gamma>0$, and $\mu_0$, $\mu_1$ are two
$SU(r)$-invariant Radon probability measures on $\Bbb C^r$ such that
$\mu_0\{0\}=\mu_1\{0\}$, than $\mu_0C=\mu_1C$.\footnote{I am grateful to
G.Vitillaro for bringing this to my attention.}
%452T

\spheader 452Xv\dvAnew{2016}
Let $\familyiI{X_i}$ be a family of compact Hausdorff spaces with product $X$,
and $\mu$ a completion regular topological measure on $X$.   Show that all the
marginal measures of $\mu$ are completion regular.   \Hint{434U.}

\leader{452Y}{Further exercises (a)}
%\spheader 452Ya
Let $Z$ be a set, $(Y,\Tau,\nu)$ a measure space, and
$\family{y}{Y}{\mu_y}$ a family of measures on $Z$.   Let $\Upsilon$
be a $\sigma$-algebra of subsets of $Z$ such that, for every
$H\in\Upsilon$, $y\mapsto\mu_yH:Y\to[0,\infty]$ is defined $\nu$-a.e.\
and is $\nu$-virtually measurable.
For $F\in\Tau$, set $\Cal H_F=\{H:H\in\Upsilon$, $\mu_yH$ is defined for
every $y\in F$ and $\sup_{y\in F}\mu_yH<\infty\}$.   Show that there is
a measure $\mu$
on $Y\times Z$, with domain $\Tau\tensorhat\Upsilon$, defined by setting

$$\eqalign{\mu E
&=\sup\{\sum_{i=0}^n\int_{F_i}\mu_y(E[\{y\}]\cap H_i)\nu(dy):
   F_0,\ldots,F_n\in\Tau\text{ are disjoint},\cr
&\hskip13em \nu F_i<\infty\text{ and }H_i\in\Cal H_{F_i}\text{ for every
}i\le n\}\cr}$$

\noindent for $E\in\Tau\tensorhat\Upsilon$.

\spheader 452Yb
Let $(X,\Sigma,\mu)$ be a
semi-finite countably compact measure space, $(Y,\Tau,\nu)$ a strictly
localizable measure space, and $f:X\to Y$ an \imp\ function.   Suppose
that the magnitude of $\nu$ (definition:  332Ga) is finite or a
measure-free cardinal (definition:  438A).   Show that there
is a disintegration $\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$ which is
consistent with $f$.
%452O

\spheader 452Yc Give an example to show that the phrase `strictly
localizable' in the statements of 452O and 452Yb cannot be dispensed
with.
%452O

\spheader 452Yd Give an example to show that, in 452M, we cannot
always arrange that $\Upsilon\subseteq\dom\mu_y$ for $\nu$-almost every
$y\in Y$.
%452M  %mt45bits

\spheader 452Ye Let $(X,\Sigma,\mu)$ be a probability space such that
whenever $(Y,\Tau,\nu)$ is a probability space and $f:X\to Y$ is an
\imp\ function, there is a disintegration $\family{y}{Y}{\mu_y}$ of
$\mu$ over $\nu$ which is consistent with $f$.
Show that $\mu$ is countably compact.   \Hint{452R, or {\smc Pachl 78}.}
%452R

\spheader 452Yf Let $X$ be a K-analytic Hausdorff space and $\mu$ a totally
finite measure on $X$
which is inner regular with respect to the closed sets.
Show that $\mu$ is countably compact.   \Hint{432D.}
%452S

%new 2008
\spheader 452Yg Let $X$ be a set, and $\familyiI{\mu_i}$ a family of
countably compact measures on $X$ with sum $\mu$
(234G\formerly{1{}12Ya}).   Show that if $\mu$ is
semi-finite, it is countably compact.
%452S

\spheader 452Yh\dvAnew{2012}
Let $X$ be a locally compact Hausdorff space, $G$ a
compact Hausdorff group, and $\action$ a continuous action of $G$ on
$X$.   Let $H$ be another group and $\varaction$ a continuous
action of $H$ on $X$ which commutes with $\action$ in the sense that
$g\action(h\varaction x)=h\varaction(g\action x)$ for all
$g\in G$, $h\in H$ and $x\in X$.   (i) Show that
$((g,h),x)\to g\action(h\varaction x):(G\times H)\times X\to X$ is a
continuous
action of the product group $G\times H$ on $X$.   (ii) Suppose that the
action in (i) is transitive.   Show that if $\mu$, $\mu'$ are $G$-invariant
Radon probability measures on $X$ and $E\subseteq X$ is a Borel set such
that $h\varaction E=E$ for every $h\in H$, then $\mu E=\mu'E$.
%452T 452Xu

}%end of exercises

\endnotes{\Notesheader{452}
452B and 452C correspond respectively to the ordinary and
$\tau$-additive product measures of \S\S251 and 417.   I have not
attempted to find a suitable general formulation
for the constructions when the measures involved are not totally finite.
In 452Ya I set out a possible version which at least agrees
with the c.l.d.\ product measure
when all the $\mu_y$ are the same.   Any product measure which has an
associated Fubini theorem can be expected to be generalizable in the
same way;  for instance, 434R becomes 452Xa.

The hypotheses in 452B are closely matched with the conclusion, and
clearly cannot be relaxed substantially if the theorem is to remain
true.   452C and 452D are a rather different matter.
While the condition `$y\mapsto\mu_yV$ is lower semi-continuous' is a
natural one, and plainly necessary for the argument given, the
integrated measure
$\mu$ can be $\tau$-additive or Radon for other reasons.   In
particular, the most interesting specific example in this book of a
Radon measure constructed through these
formulae (453N below) does not satisfy the lower semi-continuity
condition for the section measures.

Early theorems on disintegrations
concentrated on cases in which all the measure spaces involved were
`standard' in that the measures were defined on standard Borel algebras,
or were the completions of such measures.   Theorem 452I here
is the end (so far) of a long search for ways to escape from topological
considerations.   As usual, of course, the most important applications
(in probability theory) are still rooted in the standard case.   Being
countably separated, such spaces automatically yield disintegrations
which are concentrated on fibers, in the sense that
$\mu_yf^{-1}[\{y\}]=\mu_yX=1$ for almost every $y$ (452P).   The general
question of when we can expect to find disintegrations of this type is
an important one to which I will return in the next section.

452I and 452O, as stated, assume that the functions $f:X\to Y$
controlling the disintegrations are \imp.   In fact it is easy to weaken
this assumption (452Xk).
Note the constructions for conditional expectations in 452Q and 452Xr.

Obviously 452I and 452M are nearly the same theorem;  but I write out
formally independent proofs because the constructions needed to move
between them are not quite trivial.   In fact I think it is easier to
deduce 452I from 452M than the other way about (452Xo).   The point of
452N is that the spaces $(X,\Sigma)$ there have the `countably compact
measure property', that is, any totally finite measure with domain
$\Sigma$ is countably compact.   I will return to this in the exercises
to \S454 (454Xf {\it et seq.}).

The method of 452R, due to J.Pachl, may have inspired the proof of
(vi)$\Rightarrow$(i) in 343B.   In the general introduction to this work
I wrote `I have very little confidence in anything I have ever read
concerning the history of ideas'.   We have here a case indicating the
difficulties a historian faces.   I
proved 343B in the winter of 1996-97, while a guest of the University of
Wisconsin at Madison.   Around that time I was renewing my acquaintance
with {\smc Pachl 78}.   I know I ran my eye over the proof of 452R,
without, I may say, understanding it, as became plain when I came to
write the first draft of the present section in the summer of 1997;
whether I had
understood it twenty years earlier I do not know.   It is entirely
possible that a subterranean percolation of Pachl's idea was what
dislodged an obstacle to my attempts to prove 343B, but I was not at the
time conscious of any connexion.
}%end of notes

\discrpage

\frfilename{mt453.tex} 
\versiondate{22.3.10/23.3.10} 
\copyrightdate{1998} 
      
\def\chaptername{Perfect measures, disintegrations and processes} 
\def\sectionname{Strong liftings} 
      
\def\undtheta{\underline{\theta}} 
\def\undphi{\underline{\phi}} 
\def\undpsi{\underline{\psi}} 
\def\lti{left-\vthsp translation-\vthsp invariant} 
      
\newsection{453} 
      
The next step involves the concept of `strong' lifting on a 
topological measure space (453A);  I devote a few pages to describing 
the principal cases in which strong liftings are known to exist 
(453B-453J).   When we have {\it Radon} measures $\mu$ and $\nu$, with 
an {\it almost continuous} \imp\ function between them, and a {\it 
strong} lifting for $\nu$, we 
can hope for a disintegration $\family{y}{Y}{\mu_y}$ such that (almost) 
every $\mu_y$ lives on the appropriate fiber.   This is 
the content of 453K.   I end the section with a note on the relation 
between strong liftings and Stone spaces (453M) and with V.Losert's 
example of a space with no strong lifting (453N). 
      
\cmmnt{Much of the work here is based on ideas in 
{\smc Ionescu Tulcea \& Ionescu Tulcea 69}.} 
      
\leader{453A}{}\cmmnt{ The proof of the first disintegration theorem I 
presented, 452H, depended on two 
essential steps:  the use of a lifting for $(Y,\Tau,\nu)$ to define the 
finitely additive functionals $\psi_y$, and the use of a countably 
compact class to convert these into countably additive functionals.   In 
452O I observed that if our countably compact class is the family of 
compact sets in a Hausdorff space, we can get Radon measures in our 
disintegration.   Similarly, if 
we have a lifting of a special type, we can hope for special properties 
of the disintegration.   A particularly important kind of lifting, in 
this context, is the following. 
      
\medskip 
      
\noindent}{\bf Definition} Let $(X,\frak T,\Sigma,\mu)$ be a topological 
measure space.   A lifting $\phi:\Sigma\to\Sigma$ is {\bf strong} or 
{\bf of local type} if $\phi G\supseteq G$ for every open set 
$G\subseteq X$, that is, if $\phi F\subseteq F$ for every closed set 
$F\subseteq X$.   I will say that $\phi$ is {\bf almost strong} if 
$\bigcup_{G\in\frak T}\,G\setminus\phi G$ is negligible. 
      
Similarly, if $\frak A$ is the measure algebra of $\mu$, a lifting 
$\theta:\frak A\to\Sigma$ is {\bf strong} if  
$\theta G^{\ssbullet}\supseteq G$ for every open set $G\subseteq X$, and {\bf almost strong} if  
$\bigcup_{G\in\frak T}G\setminus\theta G^{\ssbullet}$ is negligible. 
      
\cmmnt{ Obviously a strong lifting is 
almost strong.} 
 
%what about the definition:  $\phi$ is strong if  
%$\phi E\subseteq\overline{E}$ for every $E\in\Sigma$?   
%works for non-topological measures? 
      
\leader{453B}{}\cmmnt{ We already have the machinery to describe a 
particularly striking class of strong liftings. 
      
\medskip 
      
\noindent}{\bf Theorem} Let $X$ be a topological group with a Haar 
measure $\mu$, and $\Sigma$ its algebra of Haar measurable sets. 
      
(a) If $\phi:\Sigma\to\Sigma$ is 
a \lti\ lifting\cmmnt{, in the sense of 447A}, then $\phi$ is strong. 
      
(b) $\mu$ has a strong lifting. 
      
\proof{{\bf (a)} Apply 447B with $Y=\{e\}$ and $\undphi=\phi$. 
      
\medskip 
      
{\bf (b)} For there is a \lti\ lifting (447J). 
}%end of proof of 453B 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} In particular, translation-invariant liftings on 
$\BbbR^r$ or $\{0,1\}^I$ (\S345) are strong. 
}%end of comment 
      
\leader{453C}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a topological 
measure space and $\phi:\Sigma\to\Sigma$ a lifting.   Write 
$\eusm L^{\infty}$ for the space of bounded $\Sigma$-measurable 
real-valued 
functions on $X$, so that $\eusm L^{\infty}$ can be identified with 
$L^{\infty}(\Sigma)$\cmmnt{ (363H)} and the Boolean homomorphism 
$\phi:\Sigma\to\Sigma$ gives rise to a Riesz homomorphism 
$T:\eusm L^{\infty}\to\eusm L^{\infty}$\cmmnt{ (363F)}. 
      
(a) If $\phi$ is a strong lifting, then $Tf=f$ for every bounded 
continuous function $f:X\to\Bbb R$. 
      
(b) If $(X,\frak T)$ is completely regular and 
$Tf=f$ for every $f\in C_b(X)$, then 
$\phi$ is strong. 
      
\proof{{\bf (a)} Suppose first that 
$f\ge 0$.   For $\alpha\in\Bbb R$, set $G_{\alpha}=\{x:x\in 
X,\,f(x)>\alpha\}$;  then $G_{\alpha}$ is open, so $\phi 
G_{\alpha}\supseteq G_{\alpha}$.   We have $f\ge\alpha\chi G_{\alpha}$, 
so 
      
\Centerline{$Tf\ge\alpha T(\chi G_{\alpha}) 
=\alpha\chi(\phi G_{\alpha})\ge\alpha\chi G_{\alpha}$,} 
      
\noindent that is, $(Tf)(x)\ge\alpha$ whenever $f(x)>\alpha$.   As 
$\alpha$ is arbitrary, $Tf\ge f$.   At the same time, setting 
$\gamma=\|f\|_{\infty}$, we have 
      
\Centerline{$T(\gamma\chi X-f)\ge\gamma\chi X-f$, 
\quad $T(\gamma\chi X)=\gamma\chi(\phi X)=\gamma\chi X$,} 
      
\noindent so $Tf\le f$ and $Tf=f$. 
      
For general $f\in C_b(X)$, 
      
\Centerline{$Tf=T(f^+-f^-)=Tf^+-Tf^-=f^+-f^-=f$,} 
      
\noindent where $f^+$ and $f^-$ are the positive and negative parts of 
$f$. 
      
\medskip 
      
{\bf (b)} Let $G\subseteq X$ be open and $x$ any 
point of $G$.   Then there is an $f\in C_b(X)$ such 
that $f\le\chi G$ and $f(x)=1$.   In this case 
      
\Centerline{$f=Tf\le T(\chi G)=\chi(\phi G)$,} 
      
\noindent so $x\in\phi G$.   As $x$ is arbitrary, $G\subseteq\phi 
G$;  as $G$ is arbitrary, $\phi$ is strong. 
}%end of proof of 453C 
      
\leader{453D}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a topological 
measure space. 
      
(a) If $\mu$ has a strong lifting it is strictly 
positive\cmmnt{ (definition: 411Nf)}. 
      
(b) If $\mu$ is strictly positive and complete, and has an almost strong 
lifting, it has a strong lifting. 
      
(c) If $\mu$ has an almost strong lifting it is $\tau$-additive, 
so has a support. 
      
(d) If $\mu$ is complete and $\mu X>0$ and the subspace measure $\mu_E$ 
has an almost strong lifting for some conegligible set $E\subseteq X$, 
then $\mu$ has an almost strong lifting. 
      
\proof{{\bf (a)} If $\phi:\Sigma\to\Sigma$ is a strong lifting, then 
$G\subseteq\phi G=\emptyset$ whenever $G$ is a negligible open set, so 
$\mu$ is strictly positive. 
      
\medskip 
      
{\bf (b)} If $\mu$ is strictly positive and complete and 
$\phi:\Sigma\to\Sigma$ is an almost strong lifting, set 
$A=\bigcup_{G\in\frak T}G\setminus\phi G$.   For each $x\in A$, let 
$\Cal I_x$ be the ideal of subsets of $X$ generated by 
      
\Centerline{$\{F:F\subseteq X$ is closed, $x\notin F\}\cup\{B:B\subseteq 
X$ is negligible$\}$.} 
      
\noindent Then $X\notin\Cal I_x$, because $\mu$ is strictly positive, so 
a closed set not containing $x$ cannot be conegligible.   There is 
therefore a Boolean homomorphism $\psi_x:\Cal PX\to\{0,1\}$ such that 
$\psi_xF=0$ for every $F\in\Cal I_x$ (311D).   Set 
      
\Centerline{$\tilde\phi E 
=(\phi E\setminus A)\cup\{x:x\in A,\,\psi_xE=1\}$} 
      
\noindent for $E\in\Sigma$.   It is easy to check 
that $\tilde\phi:\Sigma\to\Cal PX$ is a Boolean homomorphism.   (Compare 
the proof of 341J.)   If $E\in\Sigma$, then 
      
\Centerline{$E\symmdiff\tilde\phi E\subseteq(E\symmdiff\phi E)\cup A$} 
      
\noindent is negligible, so (because $\mu$ is complete) $\tilde\phi 
E\in\Sigma$.   If $E$ is negligible, then 
$E\in\Cal I_x$ and $\psi_xE=0$ for every $x\in A$, so $\tilde\phi E=\phi 
E=\emptyset$.   Thus $\tilde\phi$ is a lifting.   Now suppose that $x\in 
G\in\frak T$.   If $x\in A$, then $X\setminus G\in\Cal I_x$, 
so $\psi_x(X\setminus G)=0$, $\psi_xG=1$ and $x\in\tilde\phi G$.   If 
$x\notin A$, then $x\in\phi G$ and again $x\in\tilde\phi G$.   As $x$ 
and $G$ are arbitrary, $\tilde\phi$ is a strong lifting. 
      
\medskip 
      
{\bf (c)} Suppose that $\phi:\Sigma\to\Sigma$ is an almost strong 
lifting.   Let $\Cal G$ be a non-empty upwards-directed family of open 
sets with union $H$.   If $\sup_{G\in\Cal G}\mu G=\infty$, this is 
surely equal to $\mu H$.   Otherwise, there is a non-decreasing sequence 
$\sequencen{G_n}$ in $\Cal G$ such that $G\setminus G^*$ is negligible 
for every $G\in\Cal G$, where $G^*=\bigcup_{n\in\Bbb N}G_n$ (215Ab).   
Then $\phi G\subseteq\phi G^*$ 
for every $G\in\Cal G$.   This means that 
      
\Centerline{$H\setminus\phi G^* 
\subseteq\bigcup_{G\in\Cal G}G\setminus\phi G$} 
      
\noindent is negligible, because $\phi$ is almost strong, and 
      
\Centerline{$\mu H\le\mu(\phi G^*)=\mu G^* 
=\lim_{n\to\infty}\mu G_n=\sup_{G\in\Cal G}\mu G$.} 
      
\noindent As $\Cal G$ is arbitrary, $\mu$ is $\tau$-additive. 
By 411Nd, it has a support. 
      
\medskip 
      
{\bf (d)} Now suppose that $\mu$ is complete, that $\mu X>0$ and that 
there is a conegligible $E\subseteq X$ such that $\mu_E$ has an almost 
strong lifting $\phi$.   Let $\psi:\Cal PX\to\{\emptyset,X\}$ be any 
Boolean homomorphism such that $\psi A=\emptyset$ whenever $A$ is 
negligible.   (This is where I use the hypothesis that $X$ is not 
negligible.)   Define $\tilde\phi:\Sigma\to\Cal PX$ by setting 
      
\Centerline{$\tilde\phi F=\phi(E\cap F)\cup(\psi F\setminus E)$.} 
      
\noindent Then $\tilde\phi$ is a Boolean homomorphism because $\phi$ and 
$\psi$ are; 
      
\Centerline{$F\symmdiff\tilde\phi F 
  \subseteq((E\cap F)\symmdiff\phi(E\cap F))\cup(X\setminus E)$} 
      
\noindent is negligible, so $\tilde\phi F\in\Sigma$, for every 
$F\in\Sigma$, because $\mu$ is complete;  and if $F$ is negligible, then 
$\phi(E\cap F)=\psi F=\emptyset$ so $\tilde\phi F=\emptyset$.   Thus 
$\tilde\phi$ is a lifting.   Finally, 
      
\Centerline{$\bigcup_{G\in\frak T}G\setminus\tilde\phi G 
\subseteq(X\setminus E)\cup\bigcup_{G\in\frak T}((G\cap 
E)\setminus\phi(G\cap E))$} 
      
\noindent is negligible because $\phi$ is almost strong and $E$ is 
conegligible. 
}%end of proof of 453D 
      
\leader{453E}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a complete 
strictly localizable topological measure space with an almost strong 
lifting, and $A\subseteq X$ a non-negligible set.   Then the subspace 
measure $\mu_A$ has an almost strong lifting. 
      
\proof{ Let $\phi:\Sigma\to\Sigma$ be an almost strong lifting. 
Because $\mu$ is strictly localizable, $A$ has a measurable envelope $W$ 
say (put 213J and 213L together).   Write $\Sigma_A$ for the 
subspace $\sigma$-algebra on $A$.   Let 
$\psi:\Sigma_A\to\{\emptyset,A\}$ be any Boolean homomorphism such that 
$\psi H=\emptyset$ for every negligible set $H\subseteq A$. 
      
If $E$, $F\in\Sigma$ and $E\cap A=F\cap A$, then $\phi E\cap\phi W=\phi 
F\cap\phi W$.   \Prf\ 
      
\Centerline{$\mu((E\symmdiff F)\cap W)=\mu^*((E\symmdiff F)\cap A)=0$,} 
      
\noindent so 
      
\Centerline{$(\phi E\cap\phi W)\symmdiff(\phi F\cap\phi W) 
=\phi((E\symmdiff F)\cap W)=\emptyset$.   \Qed} 
      
We can therefore define a function $\tilde\phi:\Sigma_A\to\Cal PA$ by 
setting 
      
\Centerline{$\tilde\phi H=(\phi E\cap\phi W\cap A)\cup(\psi 
H\setminus\phi W)$} 
      
\noindent whenever $E\in\Sigma$ and $H=E\cap A$.   It is easy to check 
that $\tilde\phi$ is a Boolean homomorphism.   If $E\in\Sigma$ then 
      
\Centerline{$(E\cap A)\symmdiff\tilde\phi(E\cap A) 
\subseteq(E\symmdiff\phi E)\cup(A\setminus\phi W) 
\subseteq(E\symmdiff\phi E)\cup(W\setminus\phi W)$} 
      
\noindent is negligible, so $\tilde\phi(E\cap A)\in\Sigma_A$ (because 
$\mu$ and $\mu_A$ are complete).   If $H\in\Sigma_A$ is negligible, then 
      
\Centerline{$\tilde\phi H\subseteq\phi H\cup\psi H=\emptyset$,} 
      
\noindent so $\tilde\phi$ is a lifting for $\mu_A$. 
      
Now set $B=(A\setminus\phi W)\cup\bigcup_{G\in\frak T}G\setminus\phi G$. 
Because $\phi$ is almost strong, $B$ is negligible.   If $H\subseteq 
A$ is relatively open, then $H\setminus\tilde\phi H\subseteq B$.   \Prf\ 
Take $x\in H\setminus\tilde\phi H$.   Express $H$ as $G\cap A$ where 
$G\subseteq X$ is open.   If $x\in\phi W$, then $x\notin\phi G$ so $x\in 
B$;  if $x\notin\phi W$, then of course $x\in B$.\ \QeD\  Thus 
      
\Centerline{$\bigcup\{H\setminus\tilde\phi H:H\subseteq A$ is relatively 
open$\}\subseteq B$} 
      
\noindent is negligible and $\tilde\phi$ is almost strong. 
}%end of proof of 453E 
      
\leader{453F}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a complete 
strictly localizable topological measure space. 
      
(a) If $\frak T$ has a countable network, 
any lifting for $\mu$ is almost strong. 
      
(b) Suppose that $\mu X>0$ and $\mu$ is inner regular with respect to 
      
\Centerline{$\Cal K=\{K:K\in\Sigma,\,\mu_K$ has an almost strong 
lifting$\}$,} 
      
\noindent where $\mu_K$ is the subspace measure on $K$.   Then 
$\mu$ has an almost strong lifting. 
      
\proof{{\bf (a)} Let $\Cal E$ be a countable network for 
$\frak T$, and $\phi:\Sigma\to\Sigma$ a lifting.   For each  
$E\in\Cal E$, let $\hat E$ be a measurable envelope of $E$ (213J/213L
again).   Then 
      
$$\eqalignno{\bigcup_{G\in\frak T}G\setminus\phi G 
&=\bigcup_{G\in\frak T,E\in\Cal E,E\subseteq G}E\setminus\phi G
\subseteq\bigcup_{G\in\frak T,E\in\Cal E,E\subseteq G} 
  \hat E\setminus\phi\hat E\cr 
\displaycause{because if $E\subseteq G\in\frak T$, then $G\in\Sigma$, so $\mu(\hat E\setminus G)=0$ and $\phi\hat E\subseteq\phi G$} 
&\subseteq\bigcup_{E\in\Cal E}\hat E\setminus\phi\hat E\cr}$$ 
      
\noindent is negligible, so $\phi$ is almost strong. 
      
\medskip 
      
{\bf (b)} Let $\Cal L\subseteq\Cal K$ be a disjoint family such that 
$\mu^*A=\sum_{L\in\Cal L}\mu^*(A\cap L)$ for every $A\subseteq X$ 
(412Ib).   For each $L\in\Cal L$, let $\Sigma_L$ be the corresponding 
subspace $\sigma$-algebra and $\phi_L:\Sigma_L\to\Sigma_L$ an almost 
strong lifting.   Set $E=\bigcup\Cal L$;  then 
      
\Centerline{$\mu^*(X\setminus E)=\sum_{L\in\Cal L}\mu(L\setminus E)=0$,} 
      
\noindent so $E$ is conegligible.   For $F\in\Sigma_E$ set $\phi 
F=\bigcup_{L\in\Cal L}\phi_L(F\cap L)$;  then $\phi$ is a Boolean 
homomorphism from $\Sigma_E$ to $\Cal PE$.   If $F\in\Sigma_E$, then 
      
\Centerline{$\mu^*(F\symmdiff\phi F) 
=\sum_{L\in\Cal L}\mu^*(L\cap(F\symmdiff\phi F)) 
=\sum_{L\in\Cal L}\mu^*((F\cap L)\symmdiff\phi_L(F\cap L)) 
=0$,} 
      
\noindent while if $\mu F=0$ then $\phi_L(F\cap L)=\emptyset$ for every 
$L$, so $\phi F=\emptyset$.   Thus $\phi$ is a lifting.   Now set 
      
\Centerline{$A=\bigcup\{H\setminus\phi H:H\subseteq E$ is relatively 
open$\}$.} 
      
\noindent If $L\in\Cal L$, then 
      
\Centerline{$A\cap L 
=\bigcup\{(H\cap L)\setminus\phi_L(H\cap L):H\subseteq E$ is relatively 
open$\}$} 
      
\noindent is negligible, because $\phi_L$ is almost strong;  thus $\phi$ 
is an almost strong lifting for $\mu_E$.   By 453Dd, $\mu$ also has an 
almost strong lifting. 
}%end of proof of 453F 
      
\leader{453G}{Corollary} (a) A non-zero quasi-Radon measure on a 
separable metrizable space has an almost strong lifting. 
      
(b) A non-zero Radon measure $\mu$ on an analytic Hausdorff space $X$ 
has an almost strong lifting. 
      
\proof{{\bf (a)} A quasi-Radon measure is complete and strictly 
localizable (415A), so, if non-zero, has a lifting (341K).   A separable 
metrizable space has a countable network (4A2P(a-iii)),  
so this lifting must be almost strong. 
      
\medskip 
      
{\bf (b)} If $K\subseteq X$ is compact and non-negligible, it is 
metrizable (423Dc), so that the subspace measure $\mu_K$ 
has an almost strong lifting, by (a);  as $\mu$ is tight (that is, inner 
regular with respect to the closed 
compact sets), it has an almost strong lifting, by 453Fb. 
}%end of proof of 453G 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} In particular, Lebesgue measure on $\BbbR^r$ has 
an almost strong lifting and therefore, by 453Db, a strong lifting, as 
already noted in 453B. 
}%end of comment 
      
\leader{453H}{Lemma} Let $(X,\Sigma,\mu)$ be a complete locally 
determined measure space and $\frak T$ a topology on $X$ generated by a 
family $\Cal U\subseteq\Sigma$.   Suppose that $\phi:\Sigma\to\Sigma$ is 
a lifting such that $\phi U\supseteq U$ for every $U\in\Cal U$.   Then 
$\mu$ is a $\tau$-additive topological measure, and $\phi$ is a strong 
lifting. 
      
\proof{ Of course $\phi$ is a lower density, and $\phi X=X$, so by 414P 
we have a density topology 
      
\Centerline{$\frak T_d=\{E:E\in\Sigma,\,E\subseteq\phi E\}$} 
      
\noindent with respect to which $\mu$ is a $\tau$-additive topological 
measure.   But our hypothesis is that $\Cal U\subseteq\frak T_d$, so 
$\frak T\subseteq\frak T_d$ and $\mu$ is a $\tau$-additive topological 
measure with respect to $\frak T$.   Also, of course, 
$\phi G\supseteq G$ for every $G\in\frak T$, so $\phi$ is a strong 
lifting. 
}%end of proof of 453H 
      
\leader{453I}{Proposition} Let 
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a family of topological 
probability spaces such that every $\frak T_i$ has a countable network  
and every $\mu_i$ is strictly positive.   Let $\lambda$ be the 
(ordinary) complete product measure on $X=\prod_{i\in I}X_i$.   Then 
$\lambda$ is a $\tau$-additive topological measure and has a strong 
lifting. 
      
\proof{{\bf (a)} The strategy of the proof is as follows.   We may 
suppose that $I=\kappa$ is a cardinal.   Write $\Lambda$ for the domain 
of $\lambda$, and for each $\xi\le\kappa$ let $\Lambda_{\xi}$ be the 
$\sigma$-algebra of members of $\Lambda$ determined by coordinates less than $\xi$;  write $\pi_{\xi}:X\to X_{\xi}$ for the canonical map.   I seek to define a lifting $\phi:\Lambda\to\Lambda$ such that 
$\phi W\supseteq W$ for every open set $W\in\Lambda$.   This will be the 
last in a family $\langle\phi_{\xi}\rangle_{\xi\le\kappa}$ of partial 
liftings, constructed inductively as in the proof of 341H, with 
$\dom\phi_{\xi}=\Lambda_{\xi}$ for each $\xi$.   The inductive 
hypothesis will be that $\phi_{\xi}$ extends $\phi_{\eta}$ whenever 
$\eta\le\xi$, and 
$\phi_{\xi}\pi_{\eta}^{-1}[G]\supseteq\pi_{\eta}^{-1}[G]$ for every 
$\eta<\xi$ and every open $G\subseteq X_{\eta}$. 
      
The induction starts with $\Lambda_0=\{\emptyset,X\}$, 
$\phi_0\emptyset=\emptyset$, $\phi_0X=X$.   For $\xi\le\kappa$, set 
$\frak B_{\xi}=\{W^{\ssbullet}:W\in\Lambda_{\xi}\}$. 
      
\medskip 
      
{\bf (b)} {\it Inductive step to a successor ordinal $\xi+1$} Suppose 
that $\phi_{\xi}$ has been defined, where $\xi<\kappa$. 
      
\medskip 
      
\quad{\bf (i)} By 341Nb, there is a lifting 
$\phi'_{\xi}:\Lambda\to\Lambda$ extending $\phi_{\xi}$.   Let 
$\Cal E_{\xi}$ be a countable network for 
$\frak T_{\xi}$.   
For each $E\in\Cal E_{\xi}$ let $\hat E$ be a measurable envelope of $E$.   Set 
      
\Centerline{$Q 
=\bigcup\{\pi_{\xi}^{-1}[\hat E] 
  \setminus\phi'_{\xi}(\pi_{\xi}^{-1}[\hat E]): 
E\in\Cal E_{\xi}\}$;} 
      
\noindent then $Q$ is negligible. 
      
\medskip 
      
\quad{\bf (ii)} For $x\in Q$, let $\Cal I_x\subseteq\Lambda$ be the 
ideal generated by 
      
\Centerline{$\{W:W\in\Lambda_{\xi},\,x\notin\phi_{\xi}W\} 
\cup\{\pi_{\xi}^{-1}[F]:F\subseteq X_{\xi}$ is closed, 
$\pi_{\xi}(x)\notin F\} 
\cup\{W:\lambda W=0\}$.} 
      
\noindent Then $X\notin\Cal I_x$.   \Prf\Quer\ Otherwise, there are a 
$W\in\Lambda_{\xi}$, a closed $F\subseteq X_{\xi}$ and a negligible $W'\in\Lambda$ such that $W\cup W'\cup\pi_{\xi}^{-1}[F]=X$ while 
$x\notin\phi_{\xi}W\cup\pi_{\xi}^{-1}[F]$.   But in this case 
      
$$\eqalign{0 
&=\lambda W' 
\ge\lambda((X\setminus W)\cap(X\setminus\pi^{-1}_{\xi}[F]))\cr 
&=\lambda(X\setminus W)\cdot\lambda(X\setminus\pi_{\xi}^{-1}[F]) 
=\lambda(X\setminus W)\cdot\mu_{\xi}(X_{\xi}\setminus F) 
>0\cr}$$ 
      
\noindent because $\mu_{\xi}$ is strictly positive and 
$\phi_{\xi}W\ne X$.\ \Bang\Qed 
      
There is therefore a Boolean homomorphism $\psi_x:\Lambda\to\{0,1\}$ 
which is zero on $\Cal I_x$. 
      
\medskip 
      
\quad{\bf (iii)} Set 
      
\Centerline{$\phi_{\xi+1}W=(\phi'_{\xi}W\setminus Q)\cup\{x:x\in 
Q,\,\psi_xW=1\}$} 
      
\noindent for every $W\in\Lambda_{\xi+1}$.   Then $\phi_{\xi+1}$ is a 
Boolean homomorphism from $\Lambda_{\xi+1}$ to $\Cal PX$.   Because 
$\phi_{\xi+1}W\symmdiff\phi'_{\xi}W\subseteq Q$ is negligible, 
$\phi_{\xi+1}W\in\Lambda$ and $W\symmdiff\phi_{\xi+1}W$ is negligible for every $W\in\Lambda_{\xi+1}$.   If $\lambda 
W=0$ then $\phi'_{\xi}W=\emptyset$ and $\psi_xW=0$ for every $x\in Q$, 
so $\phi_{\xi+1}W=\emptyset$;  thus 
$\phi_{\xi+1}:\Lambda_{\xi+1}\to\Lambda$ is a partial lifting. 
If $W\in\Lambda_{\xi}$, then, for $x\in Q$, 
      
$$\eqalign{x\in\phi_{\xi}W 
&\Longrightarrow x\notin\phi_{\xi}(X\setminus W) 
\Longrightarrow X\setminus W\in\Cal I_x\cr 
&\Longrightarrow \psi_x(X\setminus W)=0 
\Longrightarrow \psi_xW=1\iff x\in\phi_{\xi+1}W\cr 
&\Longrightarrow W\notin\Cal I_x 
\Longrightarrow x\in\phi_{\xi}W,\cr}$$ 
      
\noindent so $\phi_{\xi+1}W=\phi_{\xi}W$.   Thus $\phi_{\xi+1}$ extends 
$\phi_{\xi}$. 
      
\medskip 
      
\quad{\bf (iv)} Suppose that $\eta\le\xi$ and $G\subseteq X_{\eta}$ is 
open. 
If $\eta<\xi$ then 
      
\Centerline{$\phi_{\xi+1}(\pi_{\eta}^{-1}[G]) 
=\phi_{\xi}(\pi_{\eta}^{-1}[G]) 
\supseteq \pi_{\eta}^{-1}[G]$} 
      
\noindent by the inductive hypothesis.   If $\eta=\xi$, take any 
$x\in\pi_{\xi}^{-1}[G]$.   If $x\in Q$, then 
$X\setminus\pi^{-1}_{\xi}[G]\in\Cal I_x$, so 
$\psi_x(\pi_{\xi}^{-1}[G])=1$ and $x\in\phi_{\xi+1}(\pi_{\xi}^{-1}[G])$. 
If $x\notin Q$, there is an $E\in\Cal E_{\xi}$ such that  
$x(\xi)\in E\subseteq G$.   In this case, $x(\xi)\in\hat E$, so 
      
\Centerline{$x\in\pi_{\xi}^{-1}[\hat E]\setminus Q 
\subseteq\phi'_{\xi}(\pi_{\xi}^{-1}[\hat E])\setminus Q 
\subseteq\phi_{\xi+1}(\pi_{\xi}^{-1}[\hat E]) 
\subseteq\phi_{\xi+1}(\pi_{\xi}^{-1}[G])$} 
      
\noindent because $\hat E\setminus G$ and  
$\pi_{\xi}^{-1}[\hat E]\setminus\pi_{\xi}^{-1}[G]$ are negligible.   As $x$ is arbitary,  
$\pi^{-1}_{\eta}[G]\subseteq\phi_{\xi+1}(\pi^{-1}_{\eta}[G])$ in this case also. 
      
Thus the induction continues. 
      
\medskip 
      
{\bf (c)} {\it Inductive step to a non-zero limit ordinal $\xi$ of 
countable cofinality} Suppose that $0<\xi\le\kappa$, that 
$\cf\xi=\omega$ and that $\phi_{\eta}$ has been defined for every 
$\eta<\xi$.   Let $\sequencen{\zeta_n}$ be a non-decreasing sequence in 
$\xi$ with limit $\xi$.   Then $\frak B_{\xi}$ is the closed subalgebra 
of $\frak A$ generated by $\bigcup_{n\in\Bbb N}\frak B_{\zeta_n}$ (using 
254N and 254Fe, or otherwise).   By 341G, there is a partial lower 
density $\undphi:\Lambda_{\xi}\to\Lambda$ extending every 
$\phi_{\zeta_n}$, and 
therefore extending $\phi_{\eta}$ for every $\eta<\xi$.   By 341Jb 
(applied to $\lambda\restrp\widehat{\Lambda}_{\xi}$, where 
$\widehat{\Lambda}_{\xi}$ is the $\sigma$-subalgebra of $\Lambda$ 
generated by $\Lambda_{\xi}\cup\{W:\lambda W=0\}$), there is a partial 
lifting $\phi_{\xi}:\Lambda_{\xi}\to\Lambda$ such that 
$\undphi W\subseteq\phi_{\xi}W$ for every $W\in\Lambda_{\xi}$. 
      
If $\eta<\xi$ and $W\in\Lambda_{\eta}$, then 
      
\Centerline{$\phi_{\eta}W=\undphi W\subseteq\phi_{\xi}W$, 
\quad$X\setminus\phi_{\eta}W=\phi_{\eta}(X\setminus W) 
\subseteq\phi_{\xi}(X\setminus W)=X\setminus\phi_{\xi}W$,} 
      
\noindent so $\phi_{\xi}$ extends $\phi_{\eta}$.   If $\eta<\xi$ and $G\subseteq X_{\eta}$ is open, 
      
\Centerline{$\phi_{\xi}(\pi_{\eta}^{-1}[G]) 
=\phi_{\eta+1}(\pi_{\eta}^{-1}[G]) 
\supseteq\pi_{\eta}^{-1}[G]$.} 
      
\noindent So again the induction continues. 
      
\medskip 
      
{\bf (d)} {\it Inductive step to a limit ordinal $\xi$ of uncountable 
cofinality} In this case, 
$\frak B_{\xi}=\bigcup_{\eta<\xi}\frak B_{\eta}$, as in the proof of 
341H;  so there will be a unique 
partial lifting $\phi_{\xi}:\Lambda_{\xi}\to\Lambda$ extending 
$\phi_{\eta}$ for every $\eta<\xi$ (set $\phi_{\xi}W=\phi_{\eta}W'$ 
whenever $W\in\Lambda_{\xi}$, $\eta<\xi$, $W'\in\Lambda_{\eta}$ and 
$W\symmdiff W'$ is negligible).   As in (c), we again have 
      
\Centerline{$\phi_{\xi}(\pi_{\eta}^{-1}[G]) 
=\phi_{\eta+1}(\pi_{\eta}^{-1}[G]) 
\supseteq \pi_{\eta}^{-1}[G]$} 
      
\noindent whenever $\eta<\xi$ and $G\subseteq X_{\eta}$ is open. 
      
\medskip 
      
{\bf (e)} At the end of the induction, we have a lifting 
$\phi=\phi_{\kappa}$ of $\Lambda$ such that $\phi U\supseteq U$ for 
every $U\in\Cal U$, where 
$\Cal U=\{\pi_{\xi}^{-1}[G]:\xi<\kappa,\,G\in\frak T_{\xi}\}$. 
By 453H, 
$\lambda$ is a $\tau$-additive topological measure and $\phi$ is a 
strong lifting. 
}%end of proof of 453I 
      
\leader{453J}{Corollary} Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ 
be a family of quasi-Radon probability spaces such that every 
$\frak T_i$ has a countable network consisting of measurable sets and 
every $\mu_i$ is strictly positive.   Then the ordinary product measure 
$\lambda$ on $X=\prod_{i\in I}X_i$ is quasi-Radon and has a strong 
lifting. 
If every $X_i$ is compact and Hausdorff, then $\lambda$ is a Radon 
measure. 
      
\proof{ We have just seen that $\lambda$ is a $\tau$-additive 
topological measure with a strong lifting; 
but also it is inner regular with respect to the closed sets, by 412Ua, 
so it is a quasi-Radon measure. 
If all the $X_i$ are compact and Hausdorff, so is $X$, so $\lambda$ is a 
Radon measure (416G). 
}%end of proof of 453J 
%for compact Hausdorff X_i, would it be enough here if every 
%(X_i,\Sigma_i) were countably separated?  yes, see 453Xf 
      
\leader{453K}{}\cmmnt{ We come now to the construction of 
disintegrations from strong liftings. 
      
\medskip 
      
\noindent}{\bf Theorem} Let $(X,\frak T,\Sigma,\mu)$ and 
$(Y,\frak S,\Tau,\nu)$ be Radon measure spaces and $f:X\to Y$ an almost 
continuous 
\imp\ function.   Suppose that $\nu$ has an almost strong lifting. 
Then there is a disintegration $\family{y}{Y}{\mu_y}$ of $\mu$ over 
$\nu$ such that every $\mu_y$ is a Radon measure and 
$\mu_yX=\mu_yf^{-1}[\{y\}]=1$ for almost every $y\in Y$. 
      
\proof{{\bf (a)(i)}  
Suppose first that $X$ is compact, $\mu$ is a probability 
measure and that $f$ is continuous. 
      
Turn back to the proofs of 452H-452I.   In part (a) of the proof 
of 452H, suppose that the lifting $\theta:\frak B\to\Tau$ 
corresponds to an almost strong 
lifting $\phi:\Tau\to\Tau$ (see 341Ba).   Set 
$B=\bigcup_{H\in\frak S}H\setminus\phi H$, so that $B$ is negligible. 
In part (c) of the proof of 452H, take $\Cal K$ to be 
the family of compact subsets of $X$.   Then all the $\mu_y$, as 
constructed in 452H, will be Radon probability measures. 
For every $y$, $f^{-1}[\{y\}]$ is a closed set, so is 
necessarily measured by $\mu_y$.   But also it is $\mu_y$-conegligible 
for every $y\in Y\setminus B$. 
\Prf\ Let $K\subseteq X\setminus f^{-1}[\{y\}]$ be a compact set.   Then 
$f[K]$ is a compact set not containing $y$.   Because $Y$ is 
Hausdorff, there is an open set $H$ 
containing $y$ such that $\overline{H}\cap f[K]=\emptyset$ (4A2F(h-i)). 
Now 
      
\Centerline{$y\in H\setminus B\subseteq\phi H 
\subseteq\phi\overline{H}$.} 
      
\noindent Let $E$ be the compact set $f^{-1}[\overline{H}]$.   Taking  
$T:L^{\infty}(\mu)\to L^{\infty}(\nu)$ as in part (a) of the proof of 452I, 
$T(\chi E^{\ssbullet})=\chi\overline{H}^{\ssbullet}$, so 
      
\Centerline{$\psi_yE=(ST(\chi E^{\ssbullet}))(y) 
=(S(\chi\overline{H}^{\ssbullet}))(y)=(\chi(\phi\overline{H}))(y)=1$.} 
      
\noindent Because $E\in\Cal K$, $\mu_yE\ge\psi_yE$;  since we always 
have $\mu_yX=1$, $E$ is $\mu_y$-conegligible.   But  
$K\cap E=\emptyset$, so $\mu_yK=0$.   As $K$ is arbitrary,  
$\mu_y(X\setminus f^{-1}[\{y\}])=0$. \Qed 
      
Thus $\mu_yf^{-1}[\{y\}]=1$ for almost every $y\in Y$, while
$\mu_yX\le 1$ for every $y$.
 
\medskip 
 
\quad{\bf (ii)} The result for totally finite $\mu$ and $\nu$ and 
continuous $f$ follows at once. 
      
\medskip 
      
{\bf (b)} Now suppose that $\mu$ and $\nu$ are totally finite, and that $f$ 
is almost continuous. 
      
\medskip 
      
\quad{\bf (i)} Let $\Cal K$ be the family of subsets $K\subseteq X$ such 
that 
      
\inset{$K$ is compact and $f\restr K$ is continuous,} 
      
\inset{whenever $F\in\Tau$ and $\nu(F\cap f[K])>0$ then 
$\mu(K\cap f^{-1}[F])>0$,} 
      
\inset{either $K=\emptyset$ or $\mu K>0$.} 
      
\noindent Take any $E\in\Sigma$ such that $\mu E>0$.   Then there is a 
$K\in\Cal K$ such that $K\subseteq E$ and $\mu K>0$.   \Prf\ Let 
$K_0\subseteq E$ 
be a compact set such that $f\restr K_0$ is continuous and $\mu K_0>0$. 
Let $\delta>0$ be such that $\mu K_0-\delta\nu Y>0$.   For compact sets 
$K\subseteq K_0$ set $q(K)=\mu K-\delta\nu f[K]$.   Choose 
$\sequencen{\alpha_n}$, $\langle K_n\rangle_{n\ge 1}$ as follows. 
Given that $K_n$ is a compact subset of $K_0$, where $n\in\Bbb N$, set 
      
\Centerline{$\alpha_n=\sup\{q(K):K\subseteq K_n$ is compact$\}$,} 
      
\noindent and choose a compact subset $K_{n+1}$ of $K_n$ such that 
$q(K_{n+1})\ge\max(q(K_n),\alpha_n-2^{-n})$.   Continue. 
      
Set $K=\bigcap_{n\in\Bbb N}K_n$.   We have 
      
$$\eqalign{q(K) 
&=\mu K-\delta\nu f[K]\cr 
&\ge\lim_{n\to\infty}\mu K_n-\delta\inf_{n\in\Bbb N}\nu f[K_n] 
=\lim_{n\to\infty}q(K_n) 
=\sup_{n\in\Bbb N}q(K_n)\cr}$$ 
      
\noindent because $\sequencen{q(K_n)}$ is non-decreasing.   Of course 
$K\subseteq E$, 
      
\Centerline{$\mu K\ge q(K)\ge q(K_0)>0$,} 
      
\noindent and $f\restr K$ is continuous because $K\subseteq K_0$. 
      
\Quer\ If there is an $F\in\Tau$ such that $\nu(F\cap f[K])>0$ but 
$\mu(K\cap f^{-1}[F])=0$, take a compact set $K'\subseteq K\setminus 
f^{-1}[F]$ such that $\mu K'>\mu K-\delta\nu(F\cap f[K])$.   Then 
$f[K']\subseteq f[K]\setminus F$, so 
      
\Centerline{$q(K') 
=\mu K'-\delta\nu f[K'] 
\ge\mu K'-\delta(\nu f[K]-\nu(F\cap f[K])) 
>\mu K-\delta\nu f[K] 
=q(K)$.} 
      
\noindent Let $n\in\Bbb N$ be such that $q(K')>q(K)+2^{-n}$;  then $K'$ 
is a compact subset of $K_n$, so 
      
\Centerline{$\alpha_n\ge q(K')>q(K)+2^{-n} 
\ge q(K_{n+1})+2^{-n}\ge\alpha_n$,} 
      
\noindent which is impossible.\ \BanG\  
Thus $K$ belongs to $\Cal K$ and will serve.\ \Qed 
      
\medskip 
      
\quad{\bf (ii)} By 342B, there is a countable disjoint set 
$\Cal K_0\subseteq\Cal K$ 
such that $\mu(X\setminus\bigcup\Cal K_0)=0$.   Enumerate $\Cal K_0$ as 
$\langle K_n\rangle_{n<\#(\Cal K_0)}$;  for convenience of notation, if 
$\Cal K_0$ is finite, set $K_n=\emptyset$ for $n\ge\#(\Cal K_0)$, so 
that every $K_n$ belongs to $\Cal K$ and $\mu 
E=\sum_{n=0}^{\infty}\mu(E\cap K_n)$ for every $E\in\Sigma$. 
      
\medskip 
      
\quad{\bf (iii)} For each $n\in\Bbb N$, define $\lambda_n:\Tau\to\Bbb R$ 
by setting $\lambda_nF=\mu(K_n\cap f^{-1}[F])$ for every $F\in\Tau$. 
Then $\lambda_n$ is a measure dominated by $\nu$, so 
there is a $\Tau$-measurable $g_n:Y\to[0,1]$ such that 
$\lambda_nF=\int_Fg_n$ for every 
$F\in\Tau$, by the Radon-Nikod\'ym theorem.   Because 
$\lambda_n(Y\setminus f[K_n])=0$, we may suppose that $g_n(y)=0$ for 
$y\notin f[K_n]$.   We have 
      
\Centerline{$\int_F\sum_{n=0}^{\infty}g_n 
=\sum_{n=0}^{\infty}\int_Fg_n 
=\sum_{n=0}^{\infty}\mu(K_n\cap f^{-1}[F]) 
=\mu f^{-1}[F] 
=\nu F$} 
      
\noindent for every $F\in\Tau$, so $\sum_{n=0}^{\infty}g_n(y)=1$ for 
$\nu$-almost every $y$.   Reducing the $g_n$ further on a set of measure 
zero, if need be, we may suppose that $\sum_{n=0}^{\infty}g_n(y)\le 1$ 
for 
every $y$. 
      
\medskip 
      
\quad{\bf (iv)} For each $n\in\Bbb N$, let $\tilde\lambda_n$ be the 
subspace measure on $f[K_n]$ induced by $\lambda_n$, and $\tilde\mu_n$ 
the subspace measure on $K_n$ induced by $\mu$.   Then $f\restr K_n$ is 
\imp\ for $\tilde\mu_n$ and $\tilde\lambda_n$.   Also, $\tilde\lambda_n$ 
has an almost strong lifting.   \Prf\ If $K_n=\emptyset$, this is 
trivial.   Otherwise, $\nu f[K_n]\ge\mu K_n>0$, so the subspace measure 
$\tilde\nu_n$ induced by $\nu$ on $f[K_n]$ has an almost strong lifting, 
by 453E.   But $\tilde\nu_n$ and $\tilde\lambda_n$ have the same domain 
$\Tau\cap\Cal P(f[K_n])$ and the same null ideal, because 
$K_n\in\Cal K$;  so an almost strong lifting for $\tilde\nu_n$ is an 
almost strong lifting for $\tilde\lambda_n$.\ \Qed 
      
By (a) above, we can find a disintegration 
$\family{y}{f[K_n]}{\mu_{ny}}$ of $\tilde\mu_n$ over $\tilde\lambda_n$ 
such that every $\mu_{ny}$ is a Radon measure on $K_n$, 
$\mu_{ny}K_n\le 1$ for every $y$ and 
      
\Centerline{$\mu_{ny}\{x:x\in K_n,\,f(x)=y\}=1$} 
      
\noindent for $\tilde\lambda_n$-almost every $y\in f[K_n]$, that is, for 
$\nu$-almost every $y\in f[K_n]$.   For $y\in Y\setminus f[K_n]$, let 
$\mu_{ny}$ be the zero measure on $K_n$. 
      
\wheader{453K}{6}{2}{2}{48pt} 
      
\quad{\bf (v)} Now, for $y\in Y$, set 
      
\Centerline{$\mu_yE=\sum_{n=0}^{\infty}g_n(y)\mu_{ny}(E\cap K_n)$} 
      
\noindent for all those $E\subseteq X$ such that the sum is defined. 
Then $\mu_y$ is a Radon measure and $\mu_yX\le 1$.   \Prf\ Because every 
$\mu_{ny}$ is a complete measure, so is $\mu_y$.   We have 
      
\Centerline{$\mu_yX=\sum_{n=0}^{\infty}g_n(y)\mu_{ny}K_n 
\le\sum_{n=0}^{\infty}g_n(y)\le 1$} 
      
\noindent by the choice of the $g_n$.   If $G\subseteq X$ is open then 
$\mu_{ny}$ measures $G\cap K_n$ for every $n$, so $\mu_y$ measures $G$; 
accordingly $\mu_y$ measures every compact set.   If $\mu_yE>0$, there 
is some $n\in\Bbb N$ such that $g_n(y)>0$ and $\mu_{ny}(E\cap K_n)>0$; 
now there is a compact set $K\subseteq E\cap K_n$ such that 
$\mu_{ny}K>0$, in which case $\mu_yK>0$.   By 412B, $\mu_y$ is tight, 
and is a Radon measure.\ \Qed 
      
\medskip 
      
\quad{\bf (vi)} $\family{y}{Y}{\mu_y}$ is a disintegration of $\mu$ over 
$\nu$.   \Prf\ If $E\in\Sigma$ then 
      
$$\eqalignno{\mu E 
&=\sum_{n=0}^{\infty}\mu(E\cap K_n)\cr 
\displaycause{by the choice of the $K_n$ in (ii) above} 
&=\sum_{n=0}^{\infty}\tilde\mu_n(E\cap K_n) 
=\sum_{n=0}^{\infty} 
  \int_{f[K_n]}\mu_{ny}(E\cap K_n)\tilde\lambda_n(dy)\cr 
\noalign{\noindent (because $\family{y}{f[K_n]}{\mu_{ny}}$ is a 
disintegration of $\tilde\mu_n$ over $\tilde\lambda_n$)} 
&=\sum_{n=0}^{\infty} 
  \int_{f[K_n]}\mu_{ny}(E\cap K_n)\lambda_n(dy) 
=\sum_{n=0}^{\infty}\int\mu_{ny}(E\cap K_n)\lambda_n(dy)\cr 
\noalign{\noindent (because $\lambda_n(Y\setminus f[K_n])=0$)} 
&=\sum_{n=0}^{\infty}\int g_n(y)\mu_{ny}(E\cap K_n)\nu(dy)\cr 
\noalign{\noindent (235A)} 
&=\int\sum_{n=0}^{\infty}g_n(y)\mu_{ny}(E\cap K_n)\nu(dy) 
=\int\mu_yE\,\nu(dy).\text{ \Qed}\cr}$$ 
      
\woddheader{453K}{4}{2}{2}{40pt}
      
\quad{\bf (vii)} It follows that
$\mu_yf^{-1}[\{y\}]=1$ for almost every $y$.   \Prf\ 
      
$$\eqalign{\{y:\mu_yf^{-1}[\{y\}]\ne 1\} 
&\subseteq\{y:\mu_yX\ne 1\} 
  \cup\{y:\mu^*_y(X\setminus f^{-1}[\{y\}])>0\}\cr 
&\subseteq\{y:\mu_yX\ne 1\} 
  \cup\bigcup_{n\in\Bbb N} 
  \{y:y\in f[K_n],\,\mu_{ny}(K_n\setminus f^{-1}[\{y\}])>0\}\cr}$$ 
      
\noindent is negligible.\ \Qed 
      
\medskip 
      
{\bf (c)} Now let us turn to the general case.   This proceeds just as 
in 452O.   Let $\familyiI{Y_i}$ be a decomposition of $Y$.   For each 
$i\in I$, take $X_i$, $\lambda_i$ and $\nu_i$ as in the proof of 452O. 
Note that $\lambda_i$ and $\nu_i$ are Radon measures, so that we can 
apply (b) above to find a disintegration $\family{y}{Y_i}{\tilde\mu_y}$ 
of $\lambda_i$ over $\nu_i$ such that every $\tilde\mu_y$ is a Radon 
measure and $\tilde\mu_yX_i=\tilde\mu_yf^{-1}[\{y\}]=1$ for 
$\nu_i$-almost every $y\in Y_i$.   Just as in 452O, we can set 
      
\Centerline{$\mu_yE=\tilde\mu_y(E\cap X_i)$} 
      
\noindent whenever $y\in Y_i$ and $\mu_y$ measures $E\cap X_i$, to 
obtain a disintegration $\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$ in 
which every $\mu_y$ is a Radon measure and $\mu_yX=1$ for almost every 
$y$;  and this time 
      
\Centerline{$\{y:y\in Y,\,\mu_yf^{-1}[\{y\}]\ne 1\} 
=\bigcup_{i\in I}\{y:y\in Y_i,\,\tilde\mu_yf^{-1}[\{y\}]\ne 1\}$} 
      
\noindent is negligible.   So we have a disintegration of the required 
type. 
}%end of proof of 453K 
      
\cmmnt{ 
\leader{453L}{Remark} If $f$ is surjective, we can arrange that 
every $\mu_y$ is a Radon probability measure for which 
$X_y=f^{-1}[\{y\}]$ is $\mu_y$-conegligible, just by changing some of the 
$\mu_y$ to Dirac measures.   If $f$ is not surjective, then we can still 
(if $X$ itself is not empty) arrange that every $\mu_y$ is a Radon 
probability measure;  but it might be more appropriate to make some of 
the $\mu_y$ the zero measure, so that $X_y$ is always 
$\mu_y$-conegligible. 
      
I have continued to express this theorem in terms of measures $\mu_y$ on 
the whole space $X$.   Of course, if we take it that $X_y$ 
is to be $\mu_y$-conegligible for every $y$, it will sometimes be easier to 
think of $\mu_y$ as a measure on $X_y$;  this is very much what we do in 
the case of Fubini's theorem, where all the $X_y$ are, in effect, the 
same. 
}%end of comment 
      
\leader{453M}{Strong liftings and Stone spaces} Let  
$(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure space, and  
$(Z,\frak S,\Tau,\nu)$ the Stone space of the measure algebra  
$(\frak A,\bar\mu)$ 
of $\mu$.   For $E\in\Sigma$ let $E^*\subseteq Z$ be the open-and-closed 
set corresponding to the equivalence class $E^{\ssbullet}\in\frak A$. 
Let $R$ be the relation 
      
\Centerline{$\bigcap_{F\subseteq X\text{ is closed}}
\{(z,x):z\in Z\setminus F^*
  \text{ or }x\in F\}\subseteq Z\times X$\dvro{.}{}} 
      
\noindent\cmmnt{(415Q).  }For every lifting $\phi:\Sigma\to\Sigma$ we 
have a unique function $g_{\phi}:X\to Z$ such that 
$\phi E=g_{\phi}^{-1}[E^*]$ for every $E\in\Sigma$\cmmnt{ (see 341P)}. 
\cmmnt{Now we have the following easy facts.} 
      
\spheader 453Ma $\phi$ is strong iff $(g_{\phi}(x),x)\in R$ for every 
$x\in X$.   \prooflet{\Prf\ 
      
$$\eqalign{(g_{\phi}(x),x)\in R&\text{ for every }x\in X\cr 
&\iff x\in F\text{ whenever }F\text{ is closed and }g_{\phi}(x) 
  \in F^*\cr 
&\iff g_{\phi}^{-1}[F^*]\subseteq F 
  \text{ for every closed set }F\subseteq X\cr 
&\iff \phi F\subseteq F\text{ for every closed set }F\subseteq X\cr 
&\iff \phi\text{ is strong.\ \Qed}\cr}$$ 
}%end of prooflet 
      
\spheader 453Mb If $\frak T$ is Hausdorff, so that $R$ is the graph of a 
function $f$\cmmnt{ (415Ra)}, then $\phi$ is strong iff 
$fg_{\phi}(x)=x$ for every $x\in X$.   \prooflet{(For 
$(g_{\phi}(x),x)\in R$ iff $fg_{\phi}(x)=x$.)} 
      
\leader{453N}{Losert's example}\cmmnt{ ({\smc Losert 79})} There is a 
compact Hausdorff space with a strictly positive 
completion regular Radon probability measure which has no strong 
lifting. 
      
\proof{{\bf (a)} Let $\nu$ be the usual measure on $\{0,1\}^{\Bbb N}=Y$. 
Let $M\subseteq Y$ be a closed nowhere dense set such that $\nu M>0$ 
(cf.\ 419B), and $\nu_1$ a Radon probability measure on $Y$ such that 
$\nu_1M=1$ (e.g., a Dirac measure concentrated at some point of $M$). 
      
Let $I$ be any set with cardinal at least $\omega_2$ such that 
$I\cap(I\times I)=\emptyset$.   Let $\lambda$ be the product measure on 
$Y^I$, giving each factor the measure $\nu$;  of course $\lambda$ can be 
identified with the usual measure on 
$\{0,1\}^{\Bbb N\times I}$ (254N).   Note that $\lambda$ and $\nu$ are 
both strictly positive.   For $i\in I$ write 
$M_i=\{z:z\in Y^I,\,z(i)\in M\}$;  then $M_i$ is closed in $Y^I$. 
      
Set $A=\{(i,j):i,\,j\in I,\,i\ne j\}$.   For $z\in Y^I$ and $(i,j)\in A$ 
let $\nu^{(z)}_{ij}$ be the Radon probability measure on $Y$ given by 
setting 
      
$$\eqalign{\nu^{(z)}_{ij} 
&=\nu_1\text{ if }z\in M_i\cap M_j,\cr 
&=\nu\text{ otherwise}.\cr}$$ 
      
\noindent Now, for $z\in Y^I$, let $\lambda_z$ be the Radon product 
measure of $\family{(i,j)}{A}{\nu^{(z)}_{ij}}$ on $Y^A$. 
      
\medskip 
      
{\bf (b)} Let $\Cal U$ be the family of sets $U\subseteq Y^A$ of the 
form $\{u:u(i,j)\in U_{ij}$ for $(i,j)\in B\}$, where $B\subseteq A$ is 
finite and $U_{ij}\subseteq Y$ is open-and-closed for every 
$(i,j)\in B$.   Then the function $z\mapsto\lambda_zU:Y^I\to[0,1]$ is 
Borel measurable for every $U\in\Cal U$.   \Prf\ Express $U$ in the 
given form.   For $C\subseteq B$ set 
      
\Centerline{$E_C=\{z:z\in Y^I$, 
$C=\{(i,j):(i,j)\in B,\,z\in M_i\cap M_j\}\}$,} 
      
\noindent so that $\langle E_C\rangle_{C\subseteq B}$ is a partition of 
$Y^I$ into Borel sets.   For any $C\subseteq B$, 
      
$$\lambda_zU 
=\prod_{(i,j)\in B}\nu^{(z)}_{ij}(U_{ij}) 
=\prod_{(i,j)\in C}\nu_1U_{ij} 
  \cdot\prod_{(i,j)\in B\setminus C}\nu U_{ij}$$ 
      
\noindent is constant for $z\in E_C$.\ \Qed 
      
\medskip 
      
{\bf (c)} There is a Radon measure $\mu$ on $X=Y^I\times Y^A$ 
specified by the formula 
      
\Centerline{$\mu E=\int\lambda_zE[\{z\}]\lambda(dz)$} 
      
\noindent for every Baire set $E\subseteq X$.   \Prf\ Let $\Cal E$ be 
the class of those sets $E\subseteq X$ such that 
$\int\lambda_zE[\{z\}]\lambda(dz)$ is defined.   Then $\Cal E$ is closed 
under monotone limits of sequences, and $E\setminus E'\in\Cal E$ 
whenever $E$, $E'\in\Cal E$ and $E'\subseteq E$;  also $\Cal E$ contains 
all the basic open-and-closed sets in $X$ of the form 
$V\times U$, where $V\subseteq Y^I$ is open-and-closed and $U\in\Cal U$. 
By the Monotone Class Theorem (136B), $\Cal E$ includes the 
$\sigma$-algebra generated by such sets, which is the Baire 
$\sigma$-algebra $\CalBa$ of $X$ (4A3Of).   Of course 
$E\mapsto\int\lambda_zE[\{z\}]\lambda(dz)$ is countably additive on 
$\CalBa$, so is a Baire measure on $X$, and has a unique extension to 
a Radon measure, by 432F.\ \Qed 
      
$\mu$ is strictly positive.  \Prf\ Let $W\subseteq X$ be any non-empty 
open set.   Then it includes an open set of the form $V\times U$ where 
$V=\{z:z\in Y^I,\,z(i)\in V_i$ for every $i\in J\}$, 
$U=\{u:u\in Y^A,\,u(j,k)\in U_{jk}$ for every $(j,k)\in B\}$, 
$J\subseteq I$ and 
$B\subseteq A$ are finite sets, and $V_i$, $U_{jk}\subseteq Y$ are 
non-empty open sets for every $i\in J$ and $(j,k)\in B$.   Now $\nu$ is 
strictly positive, so $\lambda V'>0$, where 
      
\Centerline{$V'=\{z:z\in V,\,z\notin M_j$ whenever $(j,k)\in B\}$.} 
      
\noindent (This is where we need to know that the $M_j$ are nowhere 
dense.)   But if $z\in V'$ then $\nu_{jk}^{(z)}=\nu$ for every 
$(j,k)\in B$, so 
      
\Centerline{$\lambda_zU=\prod_{(j,k)\in B}\nu U_{jk}>0$.} 
      
\noindent Accordingly 
      
\Centerline{$\mu W\ge\int_{V'}\lambda_zU\lambda(dz)>0$.} 
      
\noindent As $W$ is arbitrary, $\mu$ is strictly positive.\ \Qed 
      
Write $\Sigma$ for the domain of $\mu$. 
      
\medskip 
      
{\bf (d)} Fix on a self-supporting compact set $K\subseteq X$.   I seek 
to show that, regarded as a subset of $Y^{I\cup A}$, $K$ is determined 
by coordinates in some countable set. 
      
\medskip 
      
\quad{\bf (i)} There is a zero set $L\supseteq K$ such that 
$\mu L=\mu K$.   \Prf\ Let $\sequencen{K_n}$ be a sequence of compact 
subsets of 
$X\setminus K$ such that $\lim_{n\to\infty}\mu K_n=\mu(X\setminus K)$. 
For each $n\in\Bbb N$ there is a continuous function $f_n:X\to[0,1]$ 
which is zero on $K$ and $1$ on $K_n$;  now $L=\{x:f_n(x)=0$ for every 
$n\in\Bbb N\}$ is a zero set including $K$ and of the same measure as 
$K$.\ \Qed 
      
\medskip 
      
\quad{\bf (ii)} By 4A3Nc, $L$ is determined by coordinates in a 
countable subset of $I\cup A$, that is, there are countable sets 
$J_0\subseteq I$, 
$B_0\subseteq A$ such that whenever $(z,u)\in L$, $(z',u')\in X$, 
$z\restr J_0=z'\restr J_0$ and $u\restr B_0=u'\restr B_0$ we shall have 
$(z',u')\in L$.   Set 
      
\Centerline{$J=J_0\cup\{i:(i,j)\in B_0\}\cup\{j:(i,j)\in B_0\}$, 
\quad$B=A\cap(J\times J)$;} 
      
\noindent  then $J\supseteq J_0$ and $B\supseteq B_0$ are still 
countable, and $L$ is determined by coordinates in $J\cup B$. 
      
\medskip 
      
\quad{\bf (iii)} Take any $(z_0,u_0)\in X\setminus K$.   Because $K$ is 
closed, we can find finite sets $J_1\subseteq I$ and $B_1\subseteq A$, 
open-and-closed sets $V_i\subseteq Y$ for $i\in J_1$, and 
open-and-closed sets $U_{ij}\subseteq Y$ for $(i,j)\in B_1$, such that 
      
\Centerline{$W=\{(z,u):z(i)\in V_i$ for every $i\in J_1$, 
$u(i,j)\in U_{ij}$ for every $(i,j)\in B_1\}$} 
      
\noindent contains $(z_0,u_0)$ and is disjoint from $K$.   Set 
      
$$\eqalign{W_1=\{(z,u):(z,u)\in X,\,&z(i)\in V_i\text{ for every }i\in 
J_1\cap J,\cr 
&u(i,j)\in U_{ij}\text{ for every }(i,j)\in B_1\cap B\},\cr}$$ 
      
\Centerline{$Q=\{z:z\in Y^I,\,\lambda_z((L\cap W_1)[\{z\}])>0\}$,} 
      
\noindent so that $W_1$ is an open-and-closed set in $X$ and $Q$ is a 
Borel set in $Y^I$ ((b) above).   Now $Q$ is determined by coordinates 
in $J$. 
\Prf\ Suppose that $z\in Q$, $z'\in Y^I$ and $z\restr J=z'\restr J$. 
Because both $L$ and $W_1$ are determined by coordinates in $J\cup B$, 
$(L\cap W_1)[\{z\}]=(L\cap W_1)[\{z'\}]=H$ say, and $H$ is determined by 
coordinates in $B$.   At the same time, for any $(i,j)\in B$, 
$M_i\cap M_j$ is determined by coordinates in $J$, so contains $z$ iff 
it contains $z'$, and $\nu^{(z)}_{ij}=\nu^{(z')}_{ij}$.   This means 
that, writing $\lambda'_z$ and $\lambda'_{z'}$ for the products of 
$\family{(i,j)}{B}{\nu^{(z)}_{ij}}$ and 
$\family{(i,j)}{B}{\nu^{(z')}_{ij}}$ on $Y^B$, 
$\lambda'_z=\lambda'_{z'}$.   So 
      
\Centerline{$\lambda_{z'}((L\cap W_1)[\{z'\}]) 
=\lambda_{z'}H=\lambda'_{z'}H'=\lambda'_zH'=\lambda_zH 
=\lambda_z((L\cap W_1)[\{z\}])>0$,} 
      
\noindent where $H'=\{u\restr B:u\in H\}$ (254Ob), and $z'\in Q$.\ \Qed 
      
\medskip 
      
\quad{\bf (iv)} Set 
      
\Centerline{$J_2 
=(\{i:(i,j)\in B_1\setminus B\}\cup\{j:(i,j)\in B_1\setminus B\}) 
  \setminus J$.} 
      
\noindent Then $J_2$ is a finite subset of $I\setminus J$, and 
$B_1\subseteq(J\cup J_2)\times(J\cup J_2)$.   Set 
      
\Centerline{$G=\{z:z\in Y^I,\,z(i)\notin M$ for every $i\in J_2\}$,} 
      
\noindent so that $G$ is a dense open subset of $Y^I$. 
%3{}A3Ea, 3{}A3Ie 
Set 
      
\Centerline{$G_1=\{z:z\in Y^I,\,z(i)\in V_i$ for every $i\in 
J_1\setminus J\}$.} 
      
\noindent Then $G_1$ is a non-empty open set, so $G\cap G_1\ne\emptyset$ 
and $\lambda(G\cap G_1)>0$. 
      
\medskip 
      
\quad{\bf (v)} Set 
      
\Centerline{$U=\{u:u\in Y^A,\,u(i,j)\in U_{ij}$ for every $(i,j)\in 
B_1\setminus B\}$.} 
      
\noindent If $z\in G$, then $z\notin M_i\cap M_j$ whenever $(i,j)\in 
B_1\setminus B$, so $\nu_{ij}^{(z)}=\nu$ for every $(i,j)\in 
B_1\setminus B$, and 
      
\Centerline{$\lambda_zU=\prod_{(i,j)\in B_1\setminus B}\nu U_{ij}>0$.} 
      
\medskip 
      
\quad{\bf (vi)} \Quer\ Suppose, if possible, that $\lambda Q>0$. 
Because $Q$ is determined by coordinates in $J$ and $G\cap G_1$ is 
determined by coordinates in $J_2\cup(J_1\setminus J)$, 
      
\Centerline{$\lambda(Q\cap G\cap G_1) 
=\lambda Q\cdot\lambda(G\cap G_1)>0$.} 
      
\noindent If $z\in Q\cap G\cap G_1$, 
      
$$\eqalignno{\lambda_z((L\cap W)[\{z\}]) 
&=\lambda_z(U\cap(L\cap W_1)[\{z\}])\cr 
\noalign{\noindent (because $W=W_1\cap(Y^I\times U)\cap(G_1\times Y^A)$, 
and $z\in G_1$)} 
&=\lambda_zU\cdot\lambda_z((L\cap W_1)[\{z\}])\cr 
\noalign{\noindent (because $(L\cap W_1)[\{z\}]$ is determined by 
coordinates in $B$, while $U$ is determined by coordinates in 
$B_1\setminus B$, and $\lambda_z$ is a product measure)} 
&>0\cr}$$ 
      
\noindent because $z\in G\cap Q$.   But this means that 
      
\Centerline{$0<\int\lambda_z((L\cap W)[\{z\}])\lambda(dz) 
=\mu(L\cap W)=\mu(K\cap W)=\mu\emptyset$,} 
      
\noindent which is absurd.\ \Bang 
      
Thus $\lambda Q$ must be zero. 
      
\medskip 
      
\quad{\bf (vii)} Consequently 
      
\Centerline{$\mu(K\cap W_1) 
=\mu(L\cap W_1)=\int\lambda_z((L\cap W_1)[\{z\}])\lambda(dz) 
=0$;} 
      
\noindent because $K$ is self-supporting, $K\cap W_1=\emptyset$.   And 
$W_1$ contains $(z_0,u_0)$ and is determined by coordinates in $J\cup 
B$. 
      
\medskip 
      
\quad{\bf (viii)} What this means is that there can be no $(z,u)\in K$ 
such that $z\restr J=z_0\restr J$ and $u\restr B=u_0\restr B$.   At this 
point, recall that $(z_0,u_0)$ was an arbitrary point of $X\setminus K$. 
So what must be happening is that $K$ is determined by coordinates in 
the countable set $J\cup B$.   By 4A3Nc again, in the other 
direction, $K$ is a zero set. 
      
\medskip 
      
{\bf (e)} Part (d) shows that every self-supporting compact subset of 
$X$ is a zero set.   Since $\mu$ is certainly inner regular with respect 
to the self-supporting compact sets, it is inner regular with respect to 
the zero sets, that is, is completion regular. 
      
It follows that whenever $E\in\Sigma$ there is an $E'\subseteq E$, 
determined by coordinates in a countable subset of $I\cup A$, such that 
$E\setminus E'$ is negligible.   (Take $E'$ to be a countable union of 
self-supporting compact sets.) 
      
\medskip 
      
{\bf (f)} \Quer\ Now suppose, if possible, that we could find a strong 
lifting $\phi$ for $\mu$.   For each $i\in I$, take a set 
$E_i\subseteq\phi(M_i\times Y^A)$ such that 
$\mu E_i=\mu\phi(M_i\times Y^A)$ and $E_i$ is determined by coordinates 
in $J_i\cup B_i$, where $J_i\subseteq I$ and $B_i\subseteq A$ are 
countable.   Set 
      
\Centerline{$J^*_i=\{j:(j,k)\in B_i\}\cup\{k:(j,k)\in B_i\}$,} 
      
\noindent so that $J^*_i$ also is countable.   Because 
$\#(I)\ge\omega_2$, there are distinct $i$, $j\in I$ such that 
$i\notin J^*_j$ and $j\notin J^*_i$ (4A1Ea).   So 
$(i,j)\notin B_i\cup B_j$. 
      
Set 
      
\Centerline{$F=\{u:u\in Y^A,\,u(i,j)\in M\}$.} 
      
\noindent Then $\mu((M_i\cap M_j)\times(Y^A\setminus F))=0$.   \Prf\ If 
$z\in M_i\cap M_j$, then 
      
\Centerline{$\lambda_z(Y^A\setminus F)=\nu^{(z)}_{ij}(Y\setminus M)=0$.} 
      
\noindent But $(M_i\cap M_j)\times(Y^A\setminus F)$ is a Baire set, so 
      
$$\eqalign{\mu((M_i\cap M_j)\times(Y^A\setminus F)) 
&=\int\lambda_z((M_i\cap M_j)\times(Y^A\setminus F))[\{z\}]\lambda(dz)\cr 
&=\int_{M_i\cap M_j}\lambda_z(Y^A\setminus F)\lambda(dz) 
=0.\text{ \Qed}\cr}$$ 
      
\noindent Accordingly 
      
$$\eqalignno{E_i\cap E_j 
&\subseteq\phi(M_i\times Y^A)\cap\phi(M_j\times Y^A) 
=\phi((M_i\cap M_j)\times Y^A) 
\subseteq\phi(Y^I\times F)\cr 
\noalign{\noindent (because 
$((M_i\cap M_j)\times Y^A)\setminus(Y^I\times F)$ is negligible)} 
&\subseteq Y^I\times F\cr}$$ 
      
\noindent because $Y^I\times F$ is closed and $\phi$ is supposed to be 
strong.   However, $E_i\cap E_j$ is determined by coordinates in 
$J_i\cup J_j\cup B_i\cup B_j$, while $Y^I\times F$ is determined by 
coordinates in $\{(i,j)\}$, which does not meet $B_i\cup B_j$. 
So either $E_i\cap E_j$ is empty or $F=Y^A$.   But $F\ne Y^A$ because 
$M\ne Y$, while 
      
$$\eqalign{\mu(E_i\cap E_j) 
&=\mu(\phi(M_i\times Y^A)\cap\phi(M_j\times Y^A)) 
=\mu((M_i\cap M_j)\times Y^A)\cr 
&=\lambda(M_i\cap M_j) 
=(\nu M)^2 
>0,\cr}$$ 
      
\noindent so $E_i\cap E_j\ne\emptyset$.\ \Bang 
      
\medskip 
      
{\bf (g)} Thus $\mu$ has no strong lifting, as claimed. 
}%end of proof of 453N (Losert) 
      
\exercises{ 
\leader{453X}{Basic exercises $\pmb{>}$(a)} 
%\spheader 453Xa 
Let $(\frak A,\bar\mu)$ be a measure algebra and 
$(Z,\frak T,\Sigma,\mu)$ its Stone space.   Show that the canonical 
lifting for $\mu$ (341O) is strong. 
%453A 
      
\spheader 453Xb Show that there is a strong lifting for Lebesgue measure on 
the Sorgenfrey line (415Xc).
\Hint{set $\undphi E=\{x:\lim_{\delta\downarrow 0} 
\Bover1{\delta}\mu(E\cap[x,x+\delta])=1\}$, and use 341Jb.} 
%453A 
      
\sqheader 453Xc Let $\mu$ be the usual measure on the split interval 
(343J, 419L).   Show that $\mu$ has a strong lifting. 
%453Xb, 453A 
      
\spheader 453Xd Let $(X,\frak T,\Sigma,\mu)$ be a complete locally 
determined topological measure space such that $\mu$ is inner regular 
with respect to the closed sets, and $\phi:\Sigma\to\Sigma$ a strong 
lifting.   Show that $\mu$ is a quasi-Radon measure with respect to the 
lifting topology $\frak T_l$ (414Q).   Show that if $\frak T$ is regular 
then $\frak T_l\supseteq\frak T$. 
%453A 
      
\spheader 453Xe Let $(X,\frak T,\Sigma,\mu)$ be a topological measure 
space which has an almost strong lifting.   Show that any non-zero
indefinite-integral measure over $\mu$ (234J\formerly{2{}34B})  
has an almost strong lifting. 
%453E 
      
\spheader 453Xf Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space 
such that $(X,\Sigma,\mu)$ is countably separated and $\mu X>0$;  for 
example, 
$(X,\frak T)$ could be an analytic space (433B).   Show that $\mu$ is 
inner regular with respect to the compact metrizable subsets of $X$, so 
has an almost strong lifting.   \Hint{there is an injective measurable 
$f:X\to\Bbb R$, which must be almost continuous.} 
%453F 
      
\spheader 453Xg Let $(X,\frak T,\Sigma,\mu)$ be a complete locally 
determined topological measure space such that $\mu$ is effectively 
locally finite and inner regular with respect to the closed sets, and 
$\undphi:\Sigma\to\Sigma$ a lower density such that 
$\undphi G\supseteq G$ for every open $G\subseteq X$. 
Show that $\mu$ is a quasi-Radon measure with respect to both $\frak T$ 
and the density topology associated with $\undphi$. 
%453H 
      
\spheader 453Xh Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon 
measure space and $\undphi:\Sigma\to\Sigma$ a lower density such that 
$\undphi G\supseteq G$ for every open $G\subseteq X$.   Let 
$\family{x}{X}{G_x}$ be a family of open sets in $X$ such that 
$x\notin\undphi(X\setminus G_x)$ for every $x\in X$.   (i) Show that 
$A\setminus\bigcup_{x\in A}(G_x\cap U_x)$ is negligible whenever 
$A\subseteq X$ and $U_x$ is a neighbourhood of $x$ for every $x\in A$. 
(ii) Let $\frak S$ be the topology on $X$ generated by 
$\frak T\cup\{\{x\}\cup G_x:x\in X\}$.   Show that $\mu$ is quasi-Radon 
with respect to $\frak S$. 
%453Xg 
      
\spheader 453Xi Let $X$ and $Y$ be Hausdorff spaces, and $\mu$ a Radon 
probability measure on $X\times Y$;  set $\pi(x,y)=y$ for $x\in X$, 
$y\in Y$, and let $\nu$ be the image measure $\mu\pi^{-1}$.   Suppose 
that $\nu$ has an almost strong lifting.   Show 
that there is a family $\family{y}{Y}{\mu_y}$ of Radon probability 
measures on $X$ such that $\mu E=\int\mu_y(E^{-1}[\{y\}])\nu(dy)$ for 
every $E\in\dom\mu$. 
%453K 
      
\spheader 453Xj Use 453Xe to simplify part (b) of the proof of 453K. 
%453K 
      
\spheader 453Xk In 453N, show that $\int\lambda_zE[\{z\}]\lambda(dz)$ is 
defined and equal to $\mu E$ whenever $\mu$ measures $E$. 
%453N 
      
      
\leader{453Y}{Further exercises (a)} 
%\spheader 453Ya 
Let $(Y,\frak S,\Tau,\nu)$ be a Radon measure space such 
that $\nu Y>0$ and whenever $(X,\frak T,\Sigma,\mu)$ is a Radon measure 
space and $f:X\to Y$ is an almost continuous \imp\ function, then there 
is a disintegration $\family{y}{Y}{\mu_y}$ of $\mu$ over $\nu$ such that 
$\mu_yf^{-1}[\{y\}]=1$ for almost every $y$.   Show that $\nu$ has an 
almost strong lifting.   \Hint{Start with the case in which $Y$ is 
compact.   Take $f:X\to Y$ to be the function described in 415R, 416V 
and 453Mb.   Set $\undphi E=\{y:\mu_yE^*=\mu_yX=1\}$.} 
%453M 
      
\leader{453Z}{Problems (a)} 
If $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$ are compact Radon 
measure spaces with strong liftings, does their product necessarily have 
a strong lifting?   What if they are both Stone spaces? 
      
\spheader 453Zb If $(X,\frak T,\Sigma,\mu)$ is a Radon probability space 
with countable Maharam type, must it have an almost strong lifting? 
}%end of exercises 
      
\endnotes{\Notesheader{453} 
As I noted in \S452, early theorems on disintegrations concentrated on 
cases in which all the measure spaces involved were `standard' in 
that the measures were defined on standard Borel spaces (\S424), or were 
the completions of such 
measures.   Under these conditions the distinction between 452I and 453K 
becomes blurred;  measures (when completed) have to be Radon measures 
(433Cb), liftings have to be almost strong (453F) and disintegrations 
have to be concentrated on fibers (452Gc). 
Theorem 453K provides disintegrations concentrated on fibers without any 
limitation on the size of the spaces involved, though making strong 
topological assumptions. 
      
The strength of 453K derives from the remarkable variety of the (Radon) 
measure spaces which have strong liftings, as in 453F, 453G, 453I and 
453J.   For some ten years there were hopes that every strictly positive
Radon measure had a strong lifting, which were finally dashed by 
{\smc Losert 79};  I give a version of the example in 453N.   This 
is a special construction, and it remains unclear whether some much more 
direct approach might yield another example (453Za).   I should perhaps 
remark straight away that if the continuum hypothesis is true, then any 
strictly positive Radon measure with Maharam type at most $\omega_1$ 
has a strong lifting (see 535I in Volume 5).   In 
particular, subject to the continuum hypothesis, $Z\times Z$ has a 
strong lifting, where $Z$ is the Stone space of the Lebesgue measure 
algebra, and we have a positive answer to 453Zb. 
}%end of notes 
      
\discrversionA{ 
\bigskip 
      
{\bf question} If $\mu$ is a Radon probability measure, does it have a 
lifting $\phi$ such that $\mu$ is inner regular with respect to 
$\{\phi E:E\in\Sigma\}$? 
      
If there is such a lifting, then there are lots of closed $F$ such that  
$\phi F\subseteq F$.   But it is not clear that enough of them are 
self-supporting.   What about Losert's example? 
}{} 
      
\discrpage 
      
\frfilename{mt454.tex}
\versiondate{19.5.16/5.8.22}
\copyrightdate{2001}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Measures on product spaces}

\newsection{454}

A central concern of probability theory is the study of `processes',
that is, families $\family{t}{T}{X_t}$ of random variables thought of as
representing the evolution of a system in time.   Kolmogorov's successful
representation of such processes as
measurable functions on an abstract probability space was one of the
foundations on which the modern concept of `random variable' was built.
In this
section I give a version of Kolmogorov's theorem on the extension of
consistent families of measures on subproducts to a measure on the whole
product (454D).   It turns out that some restriction on the marginal
measures is necessary, and `perfectness' seems to be an
appropriate hypothesis, necessarily satisfied if the factor spaces are
standard Borel spaces or the marginal measures are Radon measures.   If
we have marginal measures with stronger properties then we shall be
able to infer corresponding properties of the measure on the product
space (454A, generalizing 451J).

The apparatus here makes it easy to describe joint distributions of
arbitrary families of real-valued random variables (454J-454P),
extending the ideas of \S271.
For the sake of the theorem that almost all Brownian paths are
continuous (477B)
I briefly investigate measures on $C(T)$, where $T$ is a Polish space
(454Q-454S).  %454Q 454R 454S

\leader{454A}{Theorem} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
non-empty family of totally finite measure spaces.   Set
$X=\prod_{i\in I}X_i$ and let $\mu$ be a measure on $X$ which is inner
regular with
respect to the $\sigma$-algebra $\Tensorhat_{i\in I}\Sigma_i$ generated
by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$, where $\pi_i:X\to X_i$ is
the coordinate map for each $i\in I$.   Suppose that every $\pi_i$ is
\imp.

(a) If $\Cal K\subseteq\Cal PX$ is a family of sets which is closed
under finite unions and countable intersections, and $\mu_i$ is inner
regular with respect to
$\Cal K_i=\{K:K\subseteq X_i,\,\pi_i^{-1}[K]\in\Cal K\}$ for every
$i\in I$, then $\mu$ is inner regular with respect to $\Cal K$.

(b)(i) If every $\mu_i$ is a compact measure, so is $\mu$;

\quad(ii) if every $\mu_i$ is a countably compact measure, so is $\mu$;

\quad(iii) if every $\mu_i$ is a perfect measure, so is $\mu$.

\proof{ If $X$ is empty this is all trivial, so we may suppose that
$X\ne\emptyset$.

\medskip

{\bf (a)} Set $\Cal A=\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$.   If
$A\in\Cal A$, $V\in\Sigma$ and $\mu(A\cap V)>0$, there is a
$K\in\Cal K\cap\Cal A$ such that $K\subseteq A$ and $\mu(K\cap V)>0$.
\Prf\ Express $A$ as $\pi_i^{-1}[E]$, where $E\in\Sigma_i$;  take
$L\in\Cal K_i$ such that $L\subseteq E$ and
$\mu_iL>\mu_iE-\mu(A\cap V)$, and set $K=\pi_i^{-1}[L]$.\ \Qed

By 412C, $\mu\restr\Tensorhat_{i\in I}\Sigma_i$ is inner regular with
respect to $\Cal K$;  by 412Ab, so is $\mu$.

\medskip

{\bf (b)(i)-(ii)} Suppose that every $\mu_i$ is (countably) compact.
Then for each $i\in I$ we can find a (countably) compact class
$\Cal K_i\subseteq\Cal PX_i$ such that $\mu_i$ is inner regular with
respect
to $\Cal K_i$.   Set $\Cal L=\{\pi_i^{-1}[K]:i\in I,\,K\in\Cal K_i\}$.
Then $\Cal L$ is (countably) compact (451H).   So there is a
(countably) compact $\Cal K\supseteq\Cal L$ which is closed under finite
unions and countable intersections (342Da, 413T).   Now $\mu$ is inner
regular with respect to $\Cal K$, by (a), and therefore (countably)
compact.

\medskip

\quad{\bf (iii)} Let $\Tau_0$ be a countably generated
$\sigma$-subalgebra of $\Tensorhat_{i\in I}\Sigma_i$.   Then there must
be some countable subfamily $\Cal E$ of
$\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$ such that $\Tau_0$ is included
in the $\sigma$-algebra generated by $\Cal E$ (use 331Gd).   Set
$\Cal E_i=\{E:E\in\Sigma_i,\,\pi_i^{-1}[E]\in\Cal E\}$ for each $i$, so
that $\Cal E_i$ is countable, and let $\Sigma'_i$ be the
$\sigma$-algebra
generated by $\Cal E_i$.   Then $\mu_i\restr\Sigma'_i$ is compact
(451F(ii)).   Applying (i), we see that
$\mu\restr\Tensorhat_{i\in I}\Sigma'_i$ is compact, therefore perfect;
while $\Tau_0\subseteq\Tensorhat_{i\in I}\Sigma'_i$.   As $\Tau_0$ is
arbitrary, $\mu\restr\Tensorhat_{i\in I}\Sigma_i$ is perfect (451F(i)).
But as the completion of $\mu$ is exactly the completion of
$\mu\restr\Tensorhat_{i\in I}\Sigma_i$, $\mu$ also is perfect, by 451Gc.
}%end of proof of 454A

\leader{454B}{Corollary} Let $\familyiI{X_i}$ be a family of Polish
spaces with product $X$.   Then any totally finite Baire measure on $X$
is a compact measure.

\proof{ If $\mu$ is a Baire measure on $X$, then its domain
$\CalBa(X)$ is $\Tensorhat_{i\in I}\Cal B(X_i)$, where $\Cal B(X_i)$
is the Borel $\sigma$-algebra of $X_i$ for each $i\in I$ (4A3Na).
So each
image measure $\mu_i$ on $X_i$ is a Borel measure, therefore tight (that
is, inner regular with respect to the closed compact sets, 433Ca),
and by 454A(b-i) $\mu$ is compact.
}%end of proof of 454B

\leader{454C}{Theorem}\cmmnt{ ({\smc Marczewski \& Ryll-Nardzewski 53})}
Let $(X,\Sigma,\mu)$ be a perfect totally finite measure
space and $(Y,\Tau,\nu)$ any totally finite measure space.   Let
$\Sigma\otimes\Tau$ be the algebra of subsets of $X\times Y$ generated
by $\{E\times F:E\in\Sigma$, $F\in\Tau\}$.   If
$\lambda_0:\Sigma\otimes\Tau\to\coint{0,\infty}$ is a non-negative
finitely additive functional such that $\lambda_0(E\times Y)=\mu E$ and
$\lambda_0(X\times F)=\nu F$ whenever $E\in\Sigma$ and $F\in\Tau$, then
$\lambda_0$ has a unique extension to a measure defined on\cmmnt{ the
$\sigma$-algebra} $\Sigma\tensorhat\Tau$\cmmnt{ generated by
$\Sigma\otimes\Tau$}.

\proof{{\bf (a)} By 413Lb, it will be enough to show that
$\lim_{n\to\infty}\lambda_0W_n=0$ for every non-increasing sequence
$\sequencen{W_n}$ in $\Sigma\otimes\Tau$ with empty intersection.   Take
such a sequence.   Each $W_n$ must belong to the algebra generated by
some finite subset of $\{E\times F:E\in\Sigma,\,F\in\Tau\}$, so there
must be a countable set $\Cal E\subseteq\Sigma$ such that every $W_n$
belongs to the algebra generated by $\{E\times F:E\in\Cal
E,\,F\in\Tau\}$;  let $\Sigma_0$ be the $\sigma$-subalgebra of $\Sigma$
generated by $\Cal E$, so that every $W_n$ belongs to
$\Sigma_0\tensorhat\Tau$.

\medskip

{\bf (b)} By 451F, $\mu\restr\Sigma_0$ is a compact measure;  let
$\Cal K\subseteq\Cal PX$ be a compact class such that
$\mu\restr\Sigma_0$ is
inner regular with respect to $\Cal K$.   We may suppose that $\Cal K$
is the family of closed sets for a compact topology on $X$ (342Da).
Let $\Cal W$ be the family of those elements $W$ of
$\Sigma_0\otimes\Tau$
such that every horizontal section $W^{-1}[\{y\}]$ belongs to $\Cal K$.
Then $\Cal W$ is closed under finite unions and intersections.

\medskip

{\bf (c)} If $W\in\Sigma_0\otimes\Tau$ and $\epsilon>0$, then there is a
$W'\in\Cal W$ such that $W'\subseteq W$ and
$\lambda_0(W\setminus W')\le\epsilon$.   \Prf\ Express $W$ as
$\bigcup_{i\le n}E_i\times F_i$, where $E_i\in\Sigma_0$ and $F_i\in\Tau$
for each $i\le n$.   (Cf.\ 315Kb.)
For each $i\le n$, take
$K_i\in\Cal K\cap\Sigma_0$ such that
$\mu(E_i\setminus K_i)\le\bover1{n+1}\epsilon$, and set
$W'=\bigcup_{i\le n}K_i\times F_i$.   Then $W'\in\Cal W$,
$W'\subseteq W$ and

$$\eqalign{\lambda_0(W\setminus W')
&\le\sum_{i=0}^n\lambda_0((E_i\times F_i)\setminus(K_i\times F_i))
\le\sum_{i=0}^n\lambda_0((E_i\setminus K_i)\times Y)\cr
&=\sum_{i=0}^n\mu_0(E_i\setminus K_i)
\le\epsilon.\text{ \Qed}\cr}$$

\medskip

{\bf (d)} Take any $\epsilon>0$.   Then for each $n\in\Bbb N$ we can
find $W'_n\in\Cal W$ such that $W'_n\subseteq W_n$ and
$\lambda_0(W_n\setminus W'_n)\le 2^{-n}\epsilon$.   Set
$V_n=\bigcap_{i\le n}W'_i$, so that $V_n\in\Cal W$ and


\Centerline{$\lambda_0(W_n\setminus V_n)
\le\sum_{i=0}^n\lambda_0(W_i\setminus W'_i)
\le 2\epsilon$}

\noindent for each $n$, and $\sequencen{V_n}$ is non-increasing, with
empty intersection.

Because $V_n\in\Sigma_0\otimes\Tau$, its projection $H_n=V_n[X]$ belongs
to $\Tau$, for each $n$.   Of course $\sequencen{H_n}$ is
non-increasing;  also $\bigcap_{n\in\Bbb N}H_n=\emptyset$.   \Prf\ If
$y\in Y$, then $\sequencen{V_n^{-1}[\{y\}]}$ is a non-increasing
sequence in $\Cal K$ with empty intersection, because $\bigcap_{n\in\Bbb
N}V_n\subseteq\bigcap_{n\in\Bbb N}W_n$ is empty.  But $\Cal K$ is a
compact class, so there must be some $n$ such that $V_n^{-1}[\{y\}]$ is
empty, that is, $y\notin H_n$.\ \QeD\  Accordingly $\lim_{n\to\infty}\nu
H_n=0$.   But as $V_n\subseteq X\times H_n$,
$\lim_{n\to\infty}\lambda_0V_n=0$.

This means that $\lim_{n\to\infty}\lambda_0W_n\le 2\epsilon$.   But as
$\epsilon$ is arbitrary, $\lim_{n\to\infty}\lambda_0W_n=0$, as required.
}%end of proof of 454C

\leader{454D}{Theorem}\cmmnt{ ({\smc Kolmogorov 1933}, \S III.4)} Let
$\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family
of totally finite perfect measure spaces.   Set $X=\prod_{i\in I}X_i$,
and write $\bigotimes_{i\in I}\Sigma_i$ for the algebra of subsets of
$X$ generated by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$, where
$\pi_i:X\to X_i$ is the coordinate map for each $i\in I$.   Suppose that
$\lambda_0:\bigotimes_{i\in I}\Sigma_i\to\coint{0,\infty}$ is a
non-negative finitely additive functional such that
$\lambda_0\pi_i^{-1}[E]=\mu_iE$ whenever $i\in I$ and $E\in\Sigma_i$.
Then $\lambda_0$ has a unique extension to a measure $\lambda$ with
domain $\Tensorhat_{i\in I}\Sigma_i$, and $\lambda$ is perfect.

\proof{{\bf (a)} The argument follows the same pattern as that of 454C.
This time, take a non-increasing sequence $\sequencen{W_n}$ in
$\bigotimes_{i\in I}\Sigma_i$ with empty intersection.   Each $W_n$
belongs to the algebra generated by some finite subset of
$\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$, so we can find countable sets
$\Cal E_i\subseteq\Sigma_i$ such that every $W_n$ belongs to the
subalgebra generated by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Cal E_i\}$.   Let
$\Tau_i$ be the $\sigma$-subalgebra of $\Sigma_i$ generated by
$\Cal E_i$, so that every $W_n$ belongs to $\bigotimes_{i\in I}\Tau_i$.

\medskip

{\bf (b)} For each $i\in I$, $\mu_i\restrp\Tau_i$ is compact (451F);
let $\frak T_i$ be a compact topology on $X_i$ such that
$\mu_i\restrp\Tau_i$
is inner regular with respect to the closed sets (342F).   Let $\frak T$
be the product topology on $X$, so that $\frak T$ is compact (3A3J).
Let $\Cal W$ be the family of closed sets in $X$ belonging to
$\bigotimes_{i\in I}\Tau_i$.

\medskip

{\bf (c)} If $W\in\bigotimes_{i\in I}\Tau_i$ and $\epsilon>0$, there is
a $W'\in\Cal W$ such that $W'\subseteq W$ and
$\lambda_0(W\setminus W')\le\epsilon$.   \Prf\ We can express $W$ as
$\bigcup_{k\le n}\bigcap_{i\in J_k}\pi_i^{-1}[E_{ki}]$ where each $J_k$
is a finite
subset of $I$ and $E_{ki}\in\Sigma_i$ for $k\le n$, $i\in J_k$ (again as in
315Kb).   Let $\langle\epsilon_{ki}\rangle_{k\le n,i\in J_k}$ be a
family of strictly positive numbers with sum at most $\epsilon$.   For
each $k\le n$, $i\in J_k$ take a closed set $K_{ki}\in\Tau_i$ such that
$K_{ki}\subseteq E_{ki}$ and
$\mu_i(E_{ki}\setminus K_{ki})\le\epsilon_{ki}$, and set
$W'=\bigcup_{k\le n}\bigcap_{i\in J_k}\pi_i^{-1}[K_{ki}]$.\ \Qed

\medskip

{\bf (d)} Take any $\epsilon>0$.   Then for each $n\in\Bbb N$ we can
find $W'_n\in\Cal W$ such that $W'_n\subseteq W_n$ and
$\lambda_0(W_n\setminus W'_n)\le 2^{-n}\epsilon$.   Set
$V_n=\bigcap_{i\le n}W'_i$.   Then $\sequencen{V_n}$ is a non-increasing
sequence of closed sets in the compact space $X$, and has empty
intersection, so there is some $n$ such that $V_n$ is empty, and

\Centerline{$\lambda_0W_n\le\sum_{i=0}^n\lambda_0(W_i\setminus W'_i)
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lim_{n\to\infty}\lambda_0W_n=0$.

\medskip

{\bf (e)} As $\sequencen{W_n}$ is arbitrary, $\lambda_0$ has a unique
countably additive extension to $\Tensorhat_{i\in I}\Sigma_i$, by 413Lb,
as before.   Of course the extension is perfect, by 454A(b-iii).
}%end of proof of 454D

\leader{454E}{Corollary} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
family of perfect measure spaces.
Let $\Cal C$ be the family of subsets of $X=\prod_{i\in I}X_i$
expressible in the form $X\cap\bigcap_{i\in J}\pi_i^{-1}[E_i]$ where
$J\subseteq I$ is finite and $E_i\in\Sigma_i$ for every $i\in I$,
writing $\pi_i(x)=x(i)$ for $x\in X$, $i\in I$.   Suppose that
$\lambda_0:\Cal C\to\Bbb R$ is a functional such that (i)
$\lambda_0\pi_i^{-1}[E]=\mu_iE$ whenever $i\in I$ and $E\in\Sigma_i$
(ii) $\lambda_0C
=\lambda_0(C\cap\pi_i^{-1}[E])+\lambda_0(C\setminus\pi_i^{-1}[E])$
whenever $C\in\Cal C$, $i\in I$ and $E\in\Sigma_i$.   Then $\lambda_0$
has a unique extension to a measure on $\Tensorhat_{i\in I}\Sigma_i$,
which is necessarily perfect.

\proof{ By 326E, $\lambda_0$ has an extension to an additive functional
on $\bigotimes_{i\in I}\Sigma_i$, so we can apply 454D.
}%end of proof of 454E

\leader{454F}{Corollary} Let $\familyiI{(X_i,\Sigma_i)}$ be a family of
standard Borel spaces.   Set $X=\prod_{i\in I}X_i$, and let
$\bigotimes_{i\in I}\Sigma_i$ be the algebra of subsets of $X$ generated
by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$, where $\pi_i:X\to X_i$ is
the coordinate map for each $i$.   Let
$\lambda_0:\bigotimes_{i\in I}\Sigma_i\to\coint{0,\infty}$ be a
non-negative finitely
additive functional such that all the marginal functionals
$E\mapsto\lambda_0\pi_i^{-1}[E]:\Sigma_i\to\coint{0,\infty}$ are
countably additive.   Then $\lambda_0$ has a unique extension to a
measure defined on $\Tensorhat_{i\in I}\Sigma_i$, which is a compact
measure.

\proof{ This follows immediately from 454D and 454A if we note that all
the measures $\lambda_0\pi_i^{-1}$ are necessarily compact, therefore
perfect (451M).
}%end of proof of 454F

\vleader{48pt}{454G}{Corollary} Let $\familyiI{X_i}$ be a family of
sets, and
$\Sigma_i$ a $\sigma$-algebra of subsets of $X_i$ for each $i\in I$.
Suppose that for each finite set $J\subseteq I$ we are given a totally
finite measure $\mu_J$ on $Z_J=\prod_{i\in J}X_i$ with domain
$\Tensorhat_{i\in J}\Sigma_i$ such that (i) whenever $J$, $K$ are finite
subsets of $I$ and $J\subseteq K$, then the canonical projection from
$Z_K$ to $Z_J$ is \imp\ (ii) every marginal measure $\mu_{\{i\}}$ on
$Z_{\{i\}}\cong X_i$ is perfect.   Then there is a unique measure $\mu$
defined on $\Tensorhat_{i\in I}\Sigma_i$ such that the canonical
projection $\tilde\pi_J:\prod_{i\in I}X_i\to Z_J$ is \imp\ for every
finite $J\subseteq I$.

\proof{ All we need to observe is that

\Centerline{$\bigotimes_{i\in I}\Sigma_i
=\{\tilde\pi_J^{-1}[V]:J\in[I]^{<\omega},\,
  V\in\bigotimes_{i\in J}\Sigma_i\}$.}

\noindent Because all the canonical projections from $X_K$ onto $X_J$
are \imp, we have $\mu_JV=\mu_KV'$ whenever $J$, $K$ are finite subsets
of $I$, $V\in\bigotimes_{i\in J}\Sigma_i$,
$V'\in\bigotimes_{i\in K}\Sigma_i$ and
$\tilde\pi_J^{-1}[V]=\tilde\pi_K^{-1}[V']$.   So we have
a functional $\lambda_0:\bigotimes_{i\in I}\Sigma_i\to\coint{0,\infty}$
such that $\lambda_0\tilde\pi_J^{-1}[V]=\mu_JV$ whenever $J\subseteq I$
is finite and $V\in\bigotimes_{i\in J}\Sigma_i$.   It is easy to check
that $\lambda_0$ is finitely additive and satisfies the conditions of
454D.   So $\lambda_0$ can be extended to a measure $\mu$ defined on
$\Tensorhat_{i\in I}\Sigma_i$.

If $J\subseteq I$ is finite, then $\mu_J$ and $\mu\tilde\pi_J^{-1}$
agree on $\bigotimes_{i\in J}\Sigma_J$ and therefore (by the Monotone
Class Theorem, 136C) on $\Tensorhat_{i\in J}\Sigma_i$;  that is,
$\tilde\pi_J$
is \imp.   To see that $\mu$ itself is unique, observe that the
conditions define its values on $\bigotimes_{i\in I}\Sigma_i$ and
therefore on
$\Tensorhat_{i\in I}\Sigma_i$, by the Monotone Class Theorem once more.
}%end of proof of 454G

\leader{454H}{Corollary} Let $\sequencen{(X_n,\Sigma_n)}$ be a sequence
of standard Borel spaces.   For each $n\in\Bbb N$ set
$Z_n=\prod_{i<n}X_i$ and $\Tau_n=\Tensorhat_{i<n}\Sigma_i$.
\cmmnt{(For $n=0$, we have $Z_0=\{\emptyset\}$,
$\Tau_0=\{\emptyset,Z_0\}$.)}   For $n\in\Bbb N$, $W\in\Tau_{n+1}$ and
$z\in Z_n$ write $W[\{z\}]=\{\xi:\xi\in X_n,\,(z,\xi)\in W\}$;  set
$X=\prod_{n\in\Bbb N}X_n$ and write $\tilde\pi_n$ for the canonical
projection of $X$ onto $Z_n$.   Suppose that for each $n\in\Bbb N$ and
$z\in Z_n$ we are given a probability measure $\nu_z$ on $X_n$ with
domain $\Sigma_n$ such that $z\mapsto\nu_z(E)$ is $\Tau_n$-measurable
for every $E\in\Sigma_n$.

(a) We have a sequence $\sequencen{\mu_n}$ of probability measures
such that, for each $n\in\Bbb N$, $\mu_n$ has domain $\Tau_n$ and

\Centerline{$\mu_{n+1}(W)=\biggerint\nu_zW[\{z\}]\mu_n(dz)$}

\noindent for every $W\in\Tau_{n+1}$, and

$$\eqalign{\int fd\mu_{n+1}
&=\iint\ldots\iint f(\xi_0,\ldots,\xi_n)
  \nu_{(\xi_0,\ldots,\xi_{n-1})}(d\xi_n)\cr
&\qquad\qquad\qquad\qquad\nu_{(\xi_0,\ldots,\xi_{n-2})}(d\xi_{n-1})
  \ldots\nu_{\xi_0}(d\xi_1)\nu_{\emptyset}(d\xi_0)\cr}$$

\noindent for every $n\in\Bbb N$ and $\mu_{n+1}$-integrable
real-valued function $f$.

(b) There is a unique probability measure
$\mu$ on $X=\prod_{n\in\Bbb N}X_n$, with domain
$\Tensorhat_{n\in\Bbb N}\Sigma_n$, such that
$\mu_n$ is the image measure $\mu\tilde\pi_n^{-1}$ on $Z_n$ for
every $n\in\Bbb N$

\proof{{\bf (a)} Of course this is
an induction on $n$.

\medskip

\quad{\bf (i)} $\mu_0$ must be the unique probability
measure on
the singleton set $Z_0$.   Given that $\dom\mu_n=\Tau_n$, the
class $\Cal W$ of sets $W\subseteq Z_{n+1}$ for which
$\int\nu_zW[\{z\}]\mu_n(dz)$ is defined
will contain all sets of the form $\prod_{i\le n}E_i$ where
$E_i\in\Sigma_i$ for every $i\le n$, just because the function
$z\mapsto\nu_zE_n$ is $\Tau_n$-measurable.   Since $\Cal W$ is
closed under increasing sequential unions and differences of comparable
sets, the Monotone Class Theorem (136B) tells us that it includes the
$\sigma$-algebra generated by the cylinder sets, which is $\Tau_{n+1}$.

\medskip

\quad{\bf (ii)} As for the integrals, we start with the elementary case
$Z_1=X_0$, $\mu_1=\nu_{\emptyset}$ and

\Centerline{$\int fd\mu_1=\int f(\xi_0)\nu_{\emptyset}(d\xi_0)$}

\noindent for $f:X_0\to\Bbb R$.   For the inductive step to $n\ge 1$,
$\family{z}{Z_n}{\nuprime_z}$ is a disintegration of
$\mu_{n+1}$ over $\mu_n$, where $\nuprime_z$ is the measure
with
domain $\Tau_{n+1}$ defined by writing $\nuprime_z(W)=\nu_zW[\{z\}]$ for
$W\in\Tau_{n+1}$, $z\in Z_n$.   By 452F,

$$\eqalign{\int_{Z_{n+1}}fd\mu_{n+1}
&=\int_{Z_n}\int_{Z_{n+1}}f(w)\nuprime_z(dw)\mu_n(dz)\cr
&=\int_{Z_n}\int_{X_n}f(z,\xi_n)
   \nu_z(d\xi_n)\mu_n(dz)\cr}$$

\noindent for every $\mu_{n+1}$-integrable function $f$.   (The
second
equality can be regarded as an application of the change-of-variable
formula 235Gb applied to the $(\nu_z,\nuprime_z)$-\imp\ function
$\xi\mapsto (z,\xi):X_n\to Z_{n+1}$.)

\medskip

{\bf (b)(i)} The canonical maps from $Z_{n+1}$ to $Z_n$ are all \imp, just
because every $\nu_z$ is a probability measure.   We therefore have a
well-defined functional
$\lambda_0:\bigotimes_{n\in\Bbb N}\Sigma_n\to[0,1]$ defined by setting
$\lambda_0\tilde\pi_n^{-1}[W]=\mu_nW$ whenever $n\in\Bbb N$ and
$W\in\bigotimes_{i<n}\Sigma_i$, and this $\lambda_0$ is finitely
additive;  moreover, each marginal measure $\lambda_0\pi_n^{-1}$, where
$\pi_n:X\to X_n$ is the coordinate map, is countably additive, because
it is expressible as an image measure of $\mu_{n+1}$ on $X_n$.

\medskip

\quad{\bf (ii)} Everything so far has been valid for any sequence
$\sequencen{(X_n,\Sigma_n)}$ of sets with attached $\sigma$-algebras.
But at this point we note that every marginal measure
$\lambda_0\pi_n^{-1}$ must be perfect, because $(X_n,\Sigma_n)$ is a
standard Borel space.   So Theorem 454D gives the result.
}%end of proof of 454H

\cmmnt{
\leader{454I}{Remarks} In 454F and 454H the hypotheses call
for `standard Borel spaces' $(X_i,\Sigma_i)$.   As the proofs make
clear, what is needed in each case is that `every totally finite measure
with domain $\Sigma_i$ must be perfect'.   We have already seen other
ways in which this can be true:  for instance, if $X$ is any Radon
Hausdorff space (434C), and $\Sigma$ its Borel $\sigma$-algebra.
Further examples are in 454Xd, 454Xh-454Xi %454Xh 454Xe 454Xi
and 454Yb-454Yc.
Indeed, even weaker hypotheses can be fully adequate.   In 454H, for
instance, it will be quite enough if all the marginal measures on the
factors $X_n$ are perfect;  in view of 454A and 451E, this will be so
iff all the measures $\tilde\mu_n$ on the partial products $Z_n$ are
perfect.   It may be difficult to be sure of this unless either
we have some argument from the nature of the factor spaces
$(X_n,\Sigma_n)$, as
suggested above, or a clear understanding of the marginal measures.   In
applications such as 455A below, however, there may be other approaches
available.
}%end of comment

\vleader{72pt}{454J}{Distributions of random \dvrocolon{processes}}\cmmnt{
For the next few paragraphs I shift to probabilists' notation.

\medskip

\noindent}{\bf Proposition} Let $(\Omega,\Sigma,\mu)$ be a probability
space and $\familyiI{X_i}$ a family of real-valued random variables on
$\Omega$\cmmnt{ (see \S271)}.

(i) There is a unique complete probability measure $\nu$ on $\BbbR^I$,
measuring every Baire set and inner regular with respect to the zero
sets, such that

\Centerline{$\nu\{x:x\in\BbbR^I$,
  $x(i_r)\le\alpha_r$ for every $r\le n\}
=\Pr(X_{i_r}\le\alpha_r$ for every $r\le n)$}

\noindent whenever $i_0,\ldots,i_n\in I$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$.

(ii) If $i_0,\ldots,i_n\in I$ and $\tilde\pi(x)=(x(i_0),\ldots,x(i_n))$
for $x\in\BbbR^I$, then the image measure $\nu\tilde\pi^{-1}$ on
$\BbbR^{n+1}$ is the joint distribution of
$X_{i_0},\ldots,X_{i_n}$\cmmnt{ as defined in 271C}.

(iii) $\nu$ is a compact measure.   If $I$ is countable then $\nu$ is a
Radon measure.

(iv) If every $X_i$ is defined everywhere on $\Omega$, then the function
$\omega\mapsto\familyiI{X_i(\omega)}:\Omega\to\BbbR^I$ is \imp\ for
$\hat\mu$ and $\nu$, where $\hat\mu$ is the completion of $\mu$.

\proof{{\bf (a)} Completing $\mu$, and adjusting the $X_i$ on negligible
sets, does not change any of the joint distributions of families
$X_{i_0},\ldots,X_{i_n}$ (271Ad), so we may suppose henceforth that
$\mu$ is complete and that every $X_i$ is defined on the whole of
$\Omega$.   Set $\phi(\omega)=\familyiI{X_i(\omega)}$ for
$\omega\in\Omega$.   Then
$\{F:F\subseteq\BbbR^I$, $\phi^{-1}[F]\in\Sigma\}$ is a $\sigma$-algebra
of subsets of $\BbbR^I$ containing $\{x:x(i)\le\alpha\}$ whenever
$i\in I$ and $\alpha\in\Bbb R$, so includes the Baire $\sigma$-algebra
$\CalBa(\BbbR^I)$ of $\BbbR^I$ (4A3Na again).
If we define $\nu_0F=\mu\phi^{-1}[F]$ for $F\in\CalBa(\BbbR^I)$, $\nu_0$
is a Baire measure on $\BbbR^I$ for which $\phi$ is \imp.   We are
supposing that $\mu$ is complete, so $\phi$ is still \imp\ for $\mu$ and
the completion $\nu$ of $\nu_0$ (234Ba).
Since $\nu_0$ is inner
regular with respect to the zero sets (412D), so is $\nu$ (412Ha), and
of course $\nu$ measures every Baire set.   By 454B, $\nu_0$ is compact,
so $\nu$ also is (451Ga).

\medskip

{\bf (b)} If $i_0,\ldots,i_n\in I$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$, then

$$\eqalign{\Pr(X_{i_r}\le\alpha_r\text{ for every }r\le n)
&=\mu\{\omega:X_{i_r}(\omega)\le\alpha_r\text{ for }r\le n\}\cr
&=\mu\{\omega:\phi(\omega)(i_r)\le\alpha_r\text{ for }r\le n\}\cr
&=\nu\{x:x(i_r)\le\alpha_r\text{ for }r\le n\}.\cr}$$

\medskip

{\bf (c)} If $i_0,\ldots,i_n\in I$ and we set
$\tilde\pi(x)=(x(i_0),\ldots,x(i_n))$ for $x\in\BbbR^I$, then
$\nu\tilde\pi^{-1}$ is a Radon measure (451O).   Since it agrees with
the distribution of $X_{i_0},\ldots,X_{i_n}$ on all sets of the form
$\{z:z(r)\le\alpha_r$ for $r\le n\}$, it must be exactly the
distribution of $X_{i_0},\ldots,X_{i_n}$ (271Ba).

\medskip

{\bf (d)} If $I$ is countable, then $\BbbR^I$ is Polish, so $\nu$ is a
Radon measure (433Cb).

\medskip

{\bf (e)} The only point I have not covered is the uniqueness of $\nu$.
But suppose that $\nuprime$ is another measure on $\BbbR^I$ with the
properties described in (i).   If $i_0,\ldots,i_n\in I$ and
$\tilde\pi(x)=(x(i_0),\ldots,x(i_n))$ for $x\in\BbbR^I$, then the image
measures $\nu\tilde\pi^{-1}$ and $\nuprime\tilde\pi^{-1}$ on
$\BbbR^{n+1}$ are both the distribution of $X_{i_0},\ldots,X_{i_n}$, by
the argument of (c) above.   This means that $\nu$ and $\nuprime$ agree
on the algebra of subsets of $\BbbR^I$ generated by sets of the form
$\{x:x(i)\in E\}$ where $i\in I$ and $E\subseteq\Bbb R$ is Borel.   By
4A3Na and 454D, they agree on all zero sets, and must be equal (412M).
}%end of proof of 454J

\leader{454K}{Definition} In the context of 454J, I will call $\nu$
the ({\bf joint}) {\bf distribution} of the process $\familyiI{X_i}$.

\cmmnt{Note that if $I=n\in\Bbb N\setminus\{0\}$, then $\nu$ is a
Radon measure on $\BbbR^n$, so is the distribution of
$\ofamily{i}{n}{X_i}$ in the sense of 271C.}

\leader{454L}{\dvrocolon{Independence}}\cmmnt{ With this
extension of
the notion of `distribution' we have a straightforward reformulation of the
characterization of independence in 272G.

\medskip

\noindent}{\bf Theorem} Let $(\Omega,\Sigma,\mu)$ be a probability
space and $\familyiI{X_i}$ a family of real-valued random variables on
$\Omega$, with distribution $\nu$ on $\BbbR^I$.   Then $\familyiI{X_i}$ is
independent iff $\nu$ is the c.l.d.\
product of the marginal measures on $\Bbb R$.

\proof{{\bf (a)} For $i\in I$, write $\nu_i$ for the marginal measure
$\mu\pi_i^{-1}$ on $\Bbb R$, taking $\pi_i(x)=x(i)$ as usual.
If $J\subseteq I$ is finite, and $\tilde\pi_J(x)=x\restr J$, then
$\nu\tilde\pi_J^{-1}$ is the distribution (in the sense of Chapter 27) of
$\family{i}{J}{X_i}$, by 454J(iii).   In particular, $\nu_i$ is the
distribution of $X_i$ for each $i$.

\medskip

{\bf (b)} If $\nu$ is the product measure $\prod_{i\in I}\nu_i$, and
$J\subseteq I$ is finite, then
$\nu\tilde\pi_J^{-1}$ is the product measure $\prod_{i\in J}\nu_i$
(254Oa), so $\family{i}{J}{X_i}$ is independent (272G).   As $J$ is
arbitrary, $\familyiI{X_i}$ is independent (272Bb).

\medskip

{\bf (c)} Conversely, if $\familyiI{X_i}$ is independent, then
$\nu$ agrees with $\lambda=\prod_{i\in I}\nu_i$ on all sets of the form
$\{x:x(i)\le\alpha_i$ for $i\in J\}$ where $J\subseteq I$ is finite and
$\family{i}{J}{\alpha_i}\in\BbbR^J$.   By the uniqueness assertion in
454J(i), $\nu=\lambda$.
}%end of proof of 454L

\leader{454M}{}\cmmnt{ The fundamental existence theorem
454G takes a more direct form in this context.

\medskip

\noindent}{\bf Proposition} Let $I$ be a set, and suppose that for each
finite $J\subseteq I$ we are given a Radon probability measure $\nu_J$ on
$\BbbR^J$ such that whenever $K$ is a finite subset of $I$ and
$J\subseteq K$, then the canonical projection from $\BbbR^K$ to $\BbbR^J$
is \imp.   Then there is a unique complete probability measure $\nu$ on
$\BbbR^I$, measuring every Baire set and inner regular with respect to the
zero sets, such that the canonical projection from $\BbbR^I$ to $\BbbR^J$
is \imp\ for every finite $J\subseteq I$.

\proof{ For finite $J\subseteq I$, let $\mu_J$ be the restriction of
$\nu_J$ to the Borel $\sigma$-algebra $\Cal B(\BbbR^J)$.   Then the
canonical projection from $\BbbR^K$ to $\BbbR^J$
is \imp\ for $\mu_K$ and $\mu_J$ whenever $J\subseteq K$ are finite subsets
of $I$.   Moreover, $\mu_{\{i\}}$ is a Borel measure on $\Bbb R$, therefore
perfect, for every $i\in I$.   By 454G, we have a
unique Baire probability measure
$\mu$ on $\BbbR^I$ such that the projections $\BbbR^I\to\BbbR^J$ are
$(\mu,\mu_J)$-\imp\ for all finite $J\subseteq I$.   Let $\nu$ be the
completion of $\mu$;  then the projections are $(\nu,\nu_J)$-\imp\ because
$\nu_J$ is always the completion of $\mu_J$.   Finally, $\nu$ is unique
because $\nu\restr\CalBa(\BbbR^I)$ must have the defining property for
$\mu$.
}%end of proof of 454M

\leader{454N}{}\cmmnt{ We know that Radon measures are often
determined
by the integrals they give to continuous functions (415I).   If we look at
distributions we get a stronger result for probability measures.

\medskip

\noindent}{\bf Proposition} Let $\Omega$ be a Hausdorff space, $\mu$ and
$\nu$ two Radon probability measures on $\Omega$, and $\familyiI{X_i}$ a
family of continuous functions separating the points of $\Omega$.   If
$\mu$ and $\nu$ give $\familyiI{X_i}$ the same distribution, they are
equal.

\proof{{\bf (a)} If $K$ and $L$ are disjoint compact subsets of $\Omega$,
there is an open set $G$ such that $K\subseteq G\subseteq X\setminus L$
and $\mu G=\nu G$.   \Prf\
$W_i=\{(\omega,\omega'):X_i(\omega)\ne X_i(\omega')\}$ is an open subset of
$\Omega\times\Omega$, and $\bigcup_{i\in I}W_i$ includes the compact set
$K\times L$.   So there is a finite set $J\subseteq I$ such that
$K\times L\subseteq\bigcup_{i\in J}W_i$.   Define $f:\Omega\to\BbbR^J$ by
setting $f(\omega)(i)=X_i(\omega)$ for $\omega\in\Omega$ and $i\in J$;
then $f$ is continuous, and $f[K]\cap f[L]=\emptyset$.   Also the image
measures $\mu f^{-1}$ and $\nu f^{-1}$ must be the same, because they are
both the common distribution of $\family{i}{J}{X_i}$.   Set
$G=\Omega\setminus f^{-1}[L]$;  this works.\ \Qed

\medskip

{\bf (b)} Now if $E\subseteq\Omega$ is a Borel set, and $\epsilon>0$,
there are compact sets $K\subseteq E$, $L\subseteq\Omega\setminus E$ such
that $\mu K\ge\mu E-\epsilon$ and
$\nu L\ge\nu(\Omega\setminus E)-\epsilon$.   Let $G$ be an open set such
that $\mu G=\nu G$ and $K\subseteq G\subseteq\Omega\setminus L$.   Then

$$\eqalign{\mu E
&\le\epsilon+\mu K\le\epsilon+\mu G
=\epsilon+\nu G
\le\epsilon+\nu(\Omega\setminus L)\cr
&=\epsilon+1-\nu L
\le 2\epsilon+1-\nu(\Omega\setminus E)
=2\epsilon+\nu E.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\mu E\le\nu E$;  similarly,
$\nu E\le\mu E$.   As $E$ is arbitrary, $\mu$ and $\nu$ agree on the Borel
sets and must coincide.
}%end of proof of 454N

\leader{454O}{What distributions\dvrocolon{ determine}}
\dvArevised{2018}\cmmnt{ Suppose
that we have two families $\familyiI{X_i}$, $\familyiI{X'_i}$ on
possibly different
probability spaces, and we are told that they have the same distribution.
Then $f(\familyiI{X_i})$ and $f(\familyiI{X'_i})$
will have the same distribution
for any Baire measurable function $f:\BbbR^I\to\Bbb R$.   More generally, we
have the following.

\medskip

\noindent}{\bf Proposition} Let
$(\Omega,\Sigma,\mu)$, $(\Omega',\Sigma',\mu')$ probability
spaces, $\familyiI{X_i}$ a family of random variables on $\Omega$ and
$\familyiI{X'_i}$ a family of random variables on $\Omega'$, both with
the same distribution $\nu$ on $R^I$.   Suppose that
$\family{j}{J}{I_j}$ is a family of countable subsets of $I$, and that for
each $j\in I$ we have a Borel measurable function $f_j:\BbbR^{I_j}$ to $\Bbb R$.
For $j\in J$ define $Y_j$, $Y'_j$ by saying that

\Centerline{$Y_j(\omega)=f_j(\family{i}{I_j}{X_i(\omega)})$
for $\omega\in\Omega\cap\bigcap_{i\in I_j}\dom X_i$,}

\Centerline{$Y'_j(\omega')=f_j(\family{i}{I_j}{X'_i(\omega')})$
for $\omega'\in\Omega'\cap\bigcap_{i\in I_j}\dom X'_i$.}

\noindent Then $\family{j}{J}{Y_j}$ and $\family{j}{J}{Y'_j}$ have the same
distribution.

\proof{ For each $j\in J$ the domain $\Omega\cap\bigcap_{i\in I_j}\dom X_i$ is a
conegligible subset of $\Omega$ and $Y_j$ is $\Sigma$-measurable (418Bd), so
$Y_j$ is a real-valued random variable on $\Omega$;  similarly, every
$Y'_j$ is a real-valued random variable on $\Omega'$, and we can speak of the
distributions of $\family{j}{J}{Y_j}$ and $\family{j}{J}{Y'_j}$.   Let
$\nu$ be the common distribution of $\family{i}{I}{X_i}$ and
$\family{i}{I}{X'_i}$.   For each $i\in I$ let $\hat X_i$ be any extension of
$X_i$ to a function from $\Omega$ to $\Bbb R$, and write $\hat\mu$ for the
completion of $\mu$.  If $j_0,\ldots,j_n\in J$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$,

$$\eqalignno{\Pr(Y_{j_r}\le\alpha_r\text{ for every }r\le n)&
=\mu\{\omega:
  \omega\in\Omega\cap\bigcap_{r\le n}\bigcap_{i\in I_{j_r}}\dom X_i,\cr&
\mskip100mu
  f_{j_r}(\family{i}{I_j}{X_i(\omega)})\le\alpha_r\text{ for every }r\le n\}\cr&
=\hat\mu\{\omega:\omega\in\Omega,
  f_{j_r}(\family{i}{I_j}{\hat X_i(\omega)})\le\alpha_r
      \text{ for every }r\le n\}\cr&
=\nu\{x:x\in\BbbR^I,\,
   f_{j_r}(x\restr I_{j_r})\le\alpha_r\text{ for every }r\le n\}
\Displaycause{454J(iv)}
=\Pr(Y'_{j_r}\le\alpha_r\text{ for every }r\le n).\cr}$$

\noindent By 454J(i), the distributions of
$\family{j}{J}{Y_j}$ and $\family{j}{J}{Y'_j}$ coincide.
}%end of proof of 454O

\leader{454P}{Theorem} Let $I$ be a set.

(a) Let $\nu$ and $\nuprime$ be Baire probability measures on $\BbbR^I$
such that $\int e^{if(x)}\nu(dx)=\int e^{if(x)}\nuprime(dx)$ for every
continuous linear functional $f:\BbbR^I\to\Bbb R$.   Then
$\nu=\nuprime$.

(b) Let $\family{j}{I}{X_j}$ and $\family{j}{I}{Y_j}$ be two families of
random variables such that

\Centerline{$\Expn(\exp(i\sum_{r=0}^n\alpha_rX_{j_r}))
=\Expn(\exp(i\sum_{r=0}^n\alpha_rY_{j_r}))$}

\noindent whenever $j_0,\ldots,j_n\in I$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$.   Then $\family{j}{I}{X_j}$ and
$\family{j}{I}{Y_j}$ have the same distribution.

\proof{{\bf (a)} For each finite set $J\subseteq I$, write
$\tilde\pi_J(x)=x\restr J$ for $x\in X$.   Then we have Radon
probability measures $\mu_J$ and $\mu'_J$ on $\BbbR^J$ defined by saying that
$\mu_JF=\nu\tilde\pi_J^{-1}[F]$,
$\mu'_JF=\nuprime\tilde\pi_J^{-1}[F]$ for Borel sets
$F\subseteq\BbbR^J$.
If $\family{j}{J}{\alpha_j}\in\BbbR^J$, then

$$\eqalign{\int\exp(i\sum_{j\in J}\alpha_jz(j))\mu_J(dz)
&=\int\exp(i\sum_{j\in J}\alpha_jx(j))\nu(dx)\cr
&=\int\exp(i\sum_{j\in J}\alpha_jx(j))\nuprime(dx)
=\int\exp(i\sum_{j\in J}\alpha_jz(j))\mu'_J(dz),\cr}$$

\noindent so $\mu_J$ and $\mu'_J$ have the same characteristic function,
therefore are equal (285M).   This is true for every $J$, so $\nu$ and
$\nuprime$ are equal, by 454D.

\medskip

{\bf (b)} Taking $\nu$ and $\nuprime$ to be the two distributions, (a)
(with
454O) tells us that their restrictions to the Baire $\sigma$-algebra of
$\BbbR^I$ are the same, so they must be identical.
}%end of proof of 454P

\leader{454Q}
{Continuous \dvrocolon{processes}}\cmmnt{ The original,
and still by far the most important, context for 454D is when every
$(X_i,\Sigma_i)$ is $\Bbb R$ with its Borel $\sigma$-algebra, so that
$X=\prod_{i\in I}X_i$ can be identified with $\BbbR^I$.   In the
discussion so far, the set $I$ has been an abstract set, except in the
very special case of 454H.   But some of the most important applications
(to which I shall come in \S455) involve index sets carrying
a topological structure;  for instance, $I$ could be the unit interval
$[0,1]$ or the half-line $\coint{0,\infty}$.   In such a case, we have a
wide variety of subspaces of $\BbbR^I$ (for instance, the space of
continuous functions) marked out as special, and it is important to know
when, and in what sense, our measures on the product space $\BbbR^I$
can be regarded as, or replaced by, measures on the subspace of
interest.   In the next few paragraphs I look briefly at spaces of
continuous functions on Polish spaces.

\medskip

\noindent}{\bf Lemma} Let $T$ be a separable metrizable space and
$(X,\Sigma,\mu)$ a semi-finite measure space.   Let $\frak T$ be a
topology on $X$ such that $\mu$ is inner regular with respect to the
closed sets.

(a) Let $\phi:X\times T\to\Bbb R$ be a function such that (i) for each
$x\in X$, $t\mapsto\phi(x,t)$ is continuous (ii) for each $t\in T$,
$x\mapsto\phi(x,t)$ is $\Sigma$-measurable.   Then $\mu$ is inner
regular with respect to $\Cal K=\{K:K\subseteq X,\,\phi\restr K\times T$
is continuous$\}$.

(b) Let $\theta:X\to C(T)$ be a function such that
$x\mapsto\theta(x)(t)$ is $\Sigma$-measurable for every $t\in T$.   Give
$C(T)$ the topology $\frak T_c$ of uniform convergence on compact
subsets of $T$.   Then $\theta$ is almost continuous.

\proof{ The result is trivial if $T$ is empty, so we may suppose that
$T\ne\emptyset$.

\medskip

{\bf (a)} Take $E\in\Sigma$ and $\gamma<\mu E$;  take
$F\in\Sigma$ such that $F\subseteq E$ and $\gamma<\mu F<\infty$.   Let
$\Cal U$ be a countable base for the topology of $T$ consisting of
non-empty sets, $D$ a countable dense subset of $T$ and $\Cal V$ a
countable base for the topology of $\Bbb R$.   For $U\in\Cal U$,
$V\in\Cal V$ set

\Centerline{$E_{UV}=\{x:\phi(x,t)\in V$ for every $t\in U\cap D\}$;}

\noindent then $E_{UV}\in\Sigma$.   Let
$\langle\epsilon_{UV}\rangle_{U\in\Cal U,V\in\Cal V}$ be a family of
strictly positive numbers with sum at most $\mu F-\gamma$.
For each $U\in\Cal U$, $V\in\Cal V$ take a closed set
$F_{UV}\subseteq F\setminus E_{UV}$ such that
$\mu F_{UV}\ge\mu(F\setminus E_{UV})-\epsilon_{UV}$.   Consider

\Centerline{$K
=\bigcap_{U\in\Cal U,V\in\Cal V}F_{UV}\cup(F\cap E_{UV})$.}

\noindent Then $K\subseteq E$ and $\mu K\ge\gamma$.

If $x\in K$, $t\in T$ and $\phi(x,t)\in V_0\in\Cal V$, let $V\in\Cal V$
be such that $\phi(x,t)\in V$ and $\overline{V}\subseteq V_0$.   Then
$\{t':\phi(x,t')\in V\}$ is an open set containing $t$, so there is some
$U\in\Cal U$ such that $t\in U$ and $\phi(x,t')\in V$ for every
$t'\in U$.   This
means that $x\in E_{UV}$, so that $(K\setminus F_{UV})\times U$ contains
$(x,t)$, and is a relatively open set in $K\times T$.   If
$(x',t')\in(K\setminus F_{UV})\times U$, then $x'\in E_{UV}$, so
$\phi(x',t'')\in V$ whenever $t''\in U\cap D$;  as $D$ is dense,
$\phi(x',t'')\in\overline{V}$ whenever $t''\in U$;  in particular,
$\phi(x',t')\in\overline{V}\subseteq V_0$.   This shows that
$(K\times T)\cap\phi^{-1}[V_0]$ is relatively open in $K\times T$;  as
$V_0$ is arbitrary, $\phi\restr K\times T$ is continuous.

So $K\in\Cal K$.   As $E$ and $\gamma$ are arbitrary, $\mu$ is inner
regular with respect to $\Cal K$.

\medskip

{\bf (b)} Set $\phi(x,t)=\theta(x)(t)$ for $x\in X$, $t\in T$.   Because
$\theta(x)\in C(T)$ for every $x$, $\phi$ is continuous in the second
variable;  and the hypothesis on $\theta$ is just that $\phi$ is
measurable in the first variable.   So $\mu$ is inner regular with
respect to $\Cal K$ as described in (a).   But $\theta\restr K$ is
continuous for every $K\in\Cal K$, by 4A2G(g-ii).   So $\theta$ is
almost continuous.
}%end of proof of 454Q

\leader{454R}{Proposition} Let $T$ be an analytic
metrizable
space\cmmnt{ (e.g., a Polish space, or any Souslin-F subset of a
Polish space)}, and $\mu$ a
probability measure on $C(T)$ with domain the $\sigma$-algebra $\Sigma$
generated by the evaluation functionals $f\mapsto f(t):C(T)\to\Bbb R$
for $t\in T$.   Give $C(T)$ the topology $\frak T_c$ of uniform
convergence on compact subsets of $T$.   Then the completion of $\mu$ is
a $\frak T_c$-Radon measure.

\proof{ If $T$ is empty this is trivial, so let us suppose henceforth
that $T\ne\emptyset$.

\medskip

{\bf (a)} Let $D$ be a countable dense subset of $T$.   Let
$\pi:C(T)\to\BbbR^D$ be the restriction map.
Set $X=\pi[C(T)]\subseteq\BbbR^D$;  then $X$, with the topology it
inherits from $\BbbR^D$, is a separable metrizable space.   Note that,
because $D$ is dense, $\pi$ is injective.

We need to know that $\pi$ is an isomorphism between $(C(T),\Sigma)$ and
$(X,\Cal B)$, where $\Cal B$ is the Borel $\sigma$-algebra of $X$.
\Prf\ Since
the Borel $\sigma$-algebra of $\BbbR^D$ is just the $\sigma$-algebra
generated
by the functionals $g\mapsto g(t):\BbbR^D\to\Bbb R$ as $t$ runs over
$D$ (4A3Dc/4A3E), $\Cal B$ is the $\sigma$-algebra of subsets of $X$
generated by the functionals $g\mapsto g(t):X\to\Bbb R$ for $t\in D$.
So $\pi$ is surely $(\Sigma,\Cal B)$-measurable.   On the other hand, if
$t\in X$, there is a sequence $\sequencen{t_n}$ in $D$ converging to
$t$, so that $\pi^{-1}(g)(t)=\lim_{n\to\infty}g(t_n)$ for every
$g\in X$, and $g\mapsto\pi^{-1}(g)(t):X\to\Bbb R$ is $\Cal B$-measurable.
Accordingly $\pi^{-1}$ is $(\Cal B,\Sigma)$-measurable.\ \Qed

\medskip

{\bf (b)} Let $\sequencen{U_n}$ be a sequence running over a base for
the topology of $T$, with no $U_n$ empty.   For each $n\in\Bbb N$,
$g\in\BbbR^D$ set

\Centerline{$\omega_n(g)=\sup_{t,u\in U_n\cap D}\min(1,g(t)-g(u))$,}

\noindent so that $\omega_n:\BbbR^D\to[0,1]$ is $\Tau$-measurable,
where $\Tau$ is the Borel (or Baire) algebra of $\BbbR^D$.   For
$g\in\BbbR^D$, $g\in X$ iff $g$ has an extension to a continuous
function on $T$, that is,

\Centerline{for every $t\in T$, $k\in\Bbb N$ there is an $n\in\Bbb N$
such that $t\in U_n$ and $\omega_n(g)\le 2^{-k}$.}

\noindent Turning this round, $\BbbR^D\setminus X$ is the projection
onto the first coordinate of the set

\Centerline{$Q=\bigcup_{k\in\Bbb N}\bigcap_{n\in\Bbb N}
  \{(g,t)$: either $t\notin
U_n$ or $\omega_n(g)>2^{-k}\}\subseteq\BbbR^D\times T$.}

\noindent But (because every $U_n$ is an open set and every $\omega_n$
is Borel measurable) $Q$ is a Borel set in the
analytic space $\BbbR^D\times T$.   So $Q$ and
$\BbbR^D\setminus X$ are analytic (423Eb, 423Bb).   Since $\BbbR^D$, being
Polish (4A2Qc), is a Radon space (434Kb), $X$ is a Radon space (434Fd).

\medskip

{\bf (c)} The image measure $\nu=\mu\pi^{-1}$ on $X$ is a Borel
probability measure.   Because $X$ is a Radon space, $\nu$ is tight, and
its completion $\hat\nu$
is a Radon measure.

By 454Qb, $\pi^{-1}:X\to C(T)$ is almost continuous if we give $C(T)$
the topology $\frak T_c$.   So the image measure
$\lambda=\hat\nu(\pi^{-1})^{-1}$ is a Radon measure for $\frak T_c$
(418I).   But of course $\lambda$ is the completion of $\mu$, just
because $\pi$ is a bijection and $\hat\nu$ is the completion of $\nu$.
}%end of proof of 454R

\leader{454S}{Corollary} Let $T$ be an analytic metrizable space.

(a) $C(T)$, with either the topology $\frak T_p$ of uniform convergence
on finite subsets of $T$ or the topology $\frak T_c$ of uniform
convergence on compact subsets of $T$, is a measure-compact Radon space.

(b) Let $\mu$ be a Baire probability measure on $\BbbR^T$ such that
$\mu^*C(T)=1$.   Then the subspace measure $\hat\mu_C$ on $C(T)$ induced by
the completion of $\mu$ is a Radon measure on $C(T)$ if $C(T)$ is given
either $\frak T_p$ or $\frak T_c$.
$\mu$ itself is $\tau$-additive and has a unique
extension $\tilde\mu$ which is a Radon measure on $\BbbR^T$;
$\hat\mu_C$ is the subspace measure on $C(T)$ induced by
$\tilde\mu$.

\proof{{\bf (a)} Let $\mu$ be a probability measure on $C(T)$ which is
either a Baire measure or a Borel measure with respect to either
$\frak T_p$ or $\frak T_c$.    Let $\tilde\mu$ be the completion of
$\mu\restr\Sigma$, where $\Sigma$ is the $\sigma$-algebra generated by
the functionals $f\mapsto f(t)$;  because these are
$\frak T_p$-continuous, $\Sigma$ is certainly included in the Baire
$\sigma$-algebra for $\frak T_p$, so that $\Sigma\subseteq\dom\mu$.
454R tells us that $\tilde\mu$ is
a Radon measure for $\frak T_c$.   Because $\frak T_p$ is a coarser
Hausdorff topology, $\tilde\mu$ is also a Radon measure for $\frak T_p$.
Also $\tilde\mu$ must extend $\mu$, because its domain includes that of
$\mu$ and the completion of $\mu$ must extend $\tilde\mu$ (in fact, of
course, this means that $\tilde\mu$ is actually the completion of
$\mu$).   Now $\tilde\mu$ is $\tau$-additive (for either topology), so
$\mu$ also is;  as $\mu$ is arbitrary, $C(T)$ is measure-compact (for
either topology).   On the other hand, if $\mu$ is a Borel measure (for
either topology), it must be tight for that topology;  so that $C(T)$ is
a Radon space.

\medskip

{\bf (b)} Write $\mu_C$ for the subspace measure on $C(T)$.   Recall
that the domain $\Sigma$ of $\mu$ is just the $\sigma$-algebra generated
by the functionals $f\mapsto f(t):\BbbR^T\to\Bbb R$, as $t$ runs over
$T$ (4A3Na once more), so that the domain $\Sigma_C$ of $\mu_C$ is the
$\sigma$-algebra of subsets of $C(T)$ generated by the functionals
$f\mapsto f(t):C(T)\to\Bbb R$.   By 454R again, the completion of
$\mu_C$ is a Radon measure on $C(T)$ if we give $C(T)$ the topology
$\frak T_c$ of uniform convergence on compact subsets of $T$, and therefore
also for the coarser Hausdorff topology $\frak T_p$.   Because
the $\mu_C$-negligible sets for $\mu_C$ are just the intersections of
$C(T)$ with $\mu$-negligible sets (214Cb), the completion of $\mu_C$ is the
subspace measure $\hat\mu_C$ induced by the completion of $\mu$
(214Ib).

The embedding $C(T)\embedsinto\BbbR^T$ is of course continuous for
$\frak T_c$ and the product topology on $\BbbR^T$, so we have a Radon
image measure $\tilde\mu$ on $\BbbR^T$ defined by saying that
$\tilde\mu E=\hat\mu_C(E\cap C(T))$ whenever $E\cap C(T)$ is measured by
$\hat\mu_C$.   If $E\in\Sigma$, then

\Centerline{$\tilde\mu E=\hat\mu_C(E\cap C(T))
=\mu_C(E\cap C(T))=\mu^*(E\cap C(T))=\mu E$}

\noindent because $\mu^*C(T)=1$, so $\tilde\mu$ extends $\mu$.   Of
course $\tilde\mu C(T)=1$ and the subspace measure on $C(T)$ induced by
$\tilde\mu$ is just $\hat\mu_C$.

Finally, because $\mu$ has an extension to a Radon measure, it must
itself be $\tau$-additive.   Because $\Sigma$ includes a base for
the topology of $\BbbR^T$, $\mu$ can have only one extension to a Radon
measure on $\BbbR^T$ (415H(iv)).
}%end of proof of 454S

\leader{*454T}{Convergence of distributions}\dvAnew{2018}
{\bf (a)} Let $I$ be a set.
Write $M$ for the set of distributions on $\Bbb R^I$, that is, the set of
completions of probability measures with domain $\CalBa(\BbbR^I)$.
For any $\nu\in M$, the integral $\int f\,d\nu$ is defined for
every bounded continuous function $f:\BbbR^I\to\Bbb R$\cmmnt{, just because such
functions are Baire measurable}.
I will say that the {\bf vague topology} on $M$ is the topology generated by
the functionals $\nu\mapsto\int fd\nu$ as $f$ runs over the space
$C_b(\BbbR^I)$ of bounded continuous real-valued functions on $\BbbR^I$.
\cmmnt{(Compare 437Jc.)}

\spheader 454Tb The vague topology on $M$ is
Hausdorff.   \prooflet{\Prf\ If $\nu$, $\nu'\in M$ are different, then
$\nu\restr\CalBa(X)\ne\nu'\restr\CalBa(\BbbR^I)$.
$\CalBa(\BbbR^I)$ is the $\sigma$-algebra generated by the family $\Cal Z$ of
zero subsets of $\BbbR^I$ (4A3Kb);  by the Monotone Class Theorem (136C),
there is an $F\in\Cal Z$ such that $\nu F\ne\nu'F$.   Suppose that
$\nu F<\nu'F$.   Let $f:\BbbR^I\to\Bbb R$ be a continuous function such that
$F=\{x:x\in\BbbR^I$, $f(x)=0\}$.   Then there is a $\delta>0$ such that
$\nu\{x:x\in\BbbR^I$, $|f(x)|<\delta\}<\nu'F$.  Set
$g(x)=\min(\delta,|f(x)|)$ for $x\in\BbbR^I$, so that $g$ is a bounded
continuous function.   Now

\Centerline{$\int g\,d\nu'\le\delta\nu'\{x:|f(x)|>0\}
<\delta\nu\{x:|f(x)|\ge\delta\}\le\int g\,d\nu$}

\noindent and $\nu$, $\nu'$ are separated by the vague topology.\ \Qed}

\leader{*454U}{Theorem}\dvAnew{2018}
Let $(\Omega,\Sigma,\mu)$ be a probability space, and
$I$ a set.   Let $M$ be the set of distributions on $\BbbR^I$;
for a family $\pmb X=\familyiI{X_i}$ of real-valued random variables on
$\Omega$, let $\nu_{\pmb X}$ be its distribution\cmmnt{ (454K)}.
Then the function
$\pmb X\mapsto\nu_{\pmb X}:\eusm L^0(\mu)^I\to M$ is continuous for the
product topology on
$\eusm L^0(\mu)^I$ corresponding to the topology of convergence in measure on
$\eusm L^0(\mu)$\cmmnt{ (245A)} and the vague topology on
$M$\cmmnt{ (454Ta)}.
%274Yf

\proof{ By the definition of `vague topology' we need to prove that
$\pmb X\mapsto\int fd\nu_{\pmb X}:\eusm L^0(\mu)^I\to\Bbb R$ is
continuous for every bounded continuous function $f:\BbbR^I\to\Bbb R$.

\medskip

{\bf (a)} Consider first the case in which
$I$ is countable and we are given $\pmb Y=\familyiI{Y_i}$ and a
sequence $\sequencen{\pmb X_n}=\sequencen{\familyiI{X_{ni}}}$ in
$\eusm L^0(\mu)^I$ such that $\sequencen{X_{ni}}$
converges a.e.\ to $Y_i$ for every $i\in I$.
Then $\nu_{\pmb Y}=\lim_{n\to\infty}\nu_{\pmb X_n}$.
\Prf\ As $I$ is countable, the set

\Centerline{$F=\{\omega:
\omega\in\bigcap_{n\in\Bbb N,i\in I}\dom X_{ni}\cap\dom Y_i$,
$Y_i(\omega)=\lim_{n\to\infty}X_{ni}(\omega)\}$}

\noindent is $\mu$-conegligible.

Now if $J\subseteq I$ is finite, $E_i\subseteq\Bbb R$ is a Borel set for each
$i\in J$ and $W=\{z:z\in\BbbR^I,\,z(i)\in E_i\text{ for every }i\in J\}$,

\Centerline{$\nu_{\pmb Y}W
=\hat\mu(\Omega\cap\bigcap_{i\in J}\{\omega:Y_i(\omega)\in E_i\})
=\hat\mu Y^{-1}[W]$}

\noindent where $Y(\omega)=\familyiI{Y_i(\omega)}$ for $\omega\in F$ and
$\hat\mu$ is the completion of $\mu$.
By the Monotone Class Theorem, $\nu_{\pmb Y}W=\hat\mu Y^{-1}[W]$ for every
$W\in\Tensorhat_{i\in I}\Cal B(\Bbb R)=\CalBa(\BbbR^I)$.
So if $f:\BbbR^I\to\Bbb R$ is a bounded continuous
function, $\int fd\nu_{\pmb Y}=\int fYd\mu$.

Similarly, if we set $X_n(\omega)=\familyiI{X_{ni}(\omega)}$ for
$\omega\in\Omega$,
$\int fd\nu_{\pmb X_n}=\int fX_nd\mu$ for every $n$.   But now observe that
$Y(\omega)=\lim_{n\to\infty}X_n(\omega)$ in $\BbbR^I$ for every $\omega\in F$,
so
$fY(\omega)=\lim_{n\to\infty}fX_n(\omega)$ for every $\omega\in F$, and

\Centerline{$\int fd\nu_{\pmb Y}
=\int fYd\mu
=\lim_{n\to\infty}fX_nd\mu
=\lim_{n\to\infty}\int fd\nu_{\pmb X_n}$}

\noindent by Lebesgue's Dominated Convergence Theorem.   As $f$ is arbitrary,
$\nu_{\pmb Y}=\lim_{n\to\infty}\nu_{\pmb X_n}$.\ \Qed

\medskip

{\bf (b)} Next suppose that
$I$ is countable and we are given $\pmb Y=\familyiI{Y_i}$ and a
sequence $\sequencen{\pmb X_n}=\sequencen{\familyiI{X_{ni}}}$ in
$\eusm L^0(\mu)^I$ such that $\sequencen{X_{ni}}$
converges in measure to $Y_i$ for every $i\in I$.
Then $\nu_{\pmb Y}=\lim_{n\to\infty}\nu_{\pmb X_n}$.   \Prf\Quer\ Otherwise,
$I$ is surely not empty and there are a continuous bounded
$f:\BbbR^I\to\Bbb R$ and an $\epsilon>0$ such that
$J=\{n:n\in\Bbb N$, $|\int fd\nu_{\pmb X_n}-\int fd\nu_{\pmb Y}|\ge\epsilon\}$
is infinite.   Let $\sequence{k}{i_k}$ be a sequence running over
$I$, and $\sequencen{m(n)}$ a strictly increasing sequence in $J$ such
that

\Centerline{$\hat\mu\{\omega:|X_{m(n),i_k}(\omega)-Y_{i_k}(\omega)|\ge 2^{-n}\}
\le 2^{-n}$}

\noindent whenever $k\le n\in\Bbb N$.
Now $Y_i\eae\lim_{n\to\infty}X_{m(n),i}$
for every $i\in I$, so $\nu_{\pmb Y}=\lim_{n\to\infty}\nu_{\pmb X_{m(n)}}$,
by (a);  but
$|\int fd\nu_{\pmb X_{m(n)}}-\int fd\nu_{\pmb Y}|\ge\epsilon$ for every $n$.\
\Bang\Qed

\medskip

{\bf (c)} This shows that if $I$ is countable, $\pmb X\mapsto\nu_{\pmb X}$ is
sequentially continuous.   But as the topology of convergence in measure
on $\eusm L^0(\mu)$ is
pseudometrizable (see 245Eb), the product topology on
$\eusm L^0(\mu)^I$ is also pseudometrizable (4A2Lh\query), and sequentially
continuous functions on $\eusm L^0(\mu)^I$ are continuous (4A2Ld).   So in this
case $\pmb X\mapsto\nu_{\pmb X}$ is continuous.

\medskip

{\bf (d)} This deals with the case of countable $I$.   For the general case,
given $I$ and a continuous bounded $f:\BbbR^I\to\Bbb R$, there are a countable
set $J\subseteq I$ and a continuous $g:\BbbR^J\to\Bbb R$ such that
$f=g\pi_J$ where $\pi_J(x)=x\restr J$ for every $x\in\BbbR^I$
(put 4A2E(a-iii) and 4A2F(b-ii)
together).   For any $\pmb X=\familyiI{X_i}\in\eusm L^0(\Sigma)^I$ with
distribution $\nu_{\pmb X}$, write
$\pmb X'$ for $\family{i}{J}{X_i}$ and $\nu_{\pmb X'}$ for its distribution.
Then $\pi_J:\BbbR^I\to\BbbR^J$ is \imp\ for $\nu_{\pmb X}$ and $\nu_{\pmb X'}$.
\Prf\ If $K\subseteq J$ is finite and $E_i\in\Cal B(\Bbb R)$ for $i\in K$,

$$\eqalign{\nu_{\pmb X}\pi_J^{-1}\{y:y\in\Bbb R^J,\,y(i)\in E_i
   \text{ for }i\in K\}&
=\nu_{\pmb X}\{x:x\in\Bbb R^I,\,x(i)\in E_i\text{ for }i\in K\}\cr&
=\mu\{\omega:\omega\in\Omega,\,X_i(\omega)\in E_i\text{ for }i\in K\}\cr&
=\nu_{\pmb X'}\{y:y\in\Bbb R^J,\,y(i)\in E_i\text{ for }i\in K\}.\cr}$$

\noindent By the Monotone Class Theorem (136C),
$\nu_{\pmb X}\pi_J^{-1}[H]=\nu_{\pmb X'}H$ for every $H\in\CalBa(\BbbR^J)$;
as $\nu_{\pmb X}$ is complete and $\nu_{\pmb X'}$ is the completion of its
restriction to $\CalBa(\BbbR^J)$, $\pi_J$ is \imp.\ \Qed

It follows that

\Centerline{$\int fd\nu_{\pmb X}=\int g\pi_Jd\nu_{\pmb X'}
=\int g\,d\nu_{\pmb X'}$}

\noindent (235G), and this is true for every $\pmb X\in\eusm L^0(\mu)^I$.

Now observe that
$\pmb X\mapsto\pmb X'=\pmb X\restr J:L^0(\mu)^I\to L^0(\mu)^J$ is continuous,
while $\pmb X'\to\int g\,d\nu_{\pmb X'}$ is continuous by (a)-(c) above;
so $\pmb X\mapsto\int fd\nu_{\pmb X}$ is continuous.   As $f$ is arbitrary.
$\pmb X\mapsto\nu_{\pmb X}$ is continuous, and the result is true in this case
also.
}%end of proof of 454U

\leader{*454V}{}\dvAnew{2018}\cmmnt{ In this
volume I am deliberately leaving some of the
central concerns of Volume 3 to one side.   But the concept of `joint
distribution' has a natural, and in some contexts important, alternative
expression in the language of \S364, as follows.

\medskip

}{\bf (a)(i)} If $\frak A$ is a Dedekind $\sigma$-complete
Boolean algebra, $I$ is a set, and $u\in L^0(\frak A)^I$,
we have a sequentially order-continuous Boolean homomorphism
$E\mapsto\Bvalue{u\in E}:\CalBa(\Bbb R^I)\to\frak A$ defined by saying that

\Centerline{$\Bvalue{u\in\{x:x\in\BbbR^I$, $x(i)\le\alpha\}}
=\Bvalue{u(i)\le\alpha}$}

\noindent whenever $i\in I$ and
$\alpha\in\Bbb R$.  \prooflet{\Prf\ It is enough to consider the case in which
$\frak A=\Sigma/\Cal I$, where $\Sigma$ is a $\sigma$-algebra of subsets of a
set $X$ and $\Cal I$ is a $\sigma$-ideal of $\Sigma$.   In this case
each $u(i)$ can be identified with the equivalence class of a
$\Sigma$-measurable function $f_i:X\to\Bbb R$ (364C);  setting
$f(x)=\familyiI{f_i(x)}$ for $x\in X$, $f$ is
$(\Sigma,\CalBa(\BbbR^I))$-measurable (4A3Ne) and we have a corresponding
function $E\mapsto f^{-1}[E]^{\ssbullet}:\CalBa(\BbbR^I)\to\frak A$ which has
the required properties.   Since $\CalBa(\BbbR^I)$ is the $\sigma$-algebra
generated by $\Cal E=\{\{x:x(i)\le\alpha\}:i\in I$, $\alpha\in\Bbb R\}$,
there is only one sequentially order-continuous homomorphism with the right
values on $\Cal E$.\ \Qed}

\medskip

\quad{\bf (ii)} If $h:\BbbR^I\to\Bbb R$ is a Baire measurable
function, there is a function $\bar h:L^0(\frak A)^I\to L^0(\frak A)$
defined by saying that $\Bvalue{\bar h(u)\in E}=\Bvalue{u\in h^{-1}[E]}$
for every Borel set $E\subseteq\Bbb R$.   \prooflet{\Prf\
$E\mapsto\Bvalue{u\in h^{-1}[E]}$ is a sequentially order-continuous Boolean
homomorphism so we can use 364F.\ \Qed}

\spheader 454Vb Suppose that
$(\frak A,\bar\mu)$ is a probability algebra, $I$ is a set and
$u\in L^0(\frak A)^I$.   Then there is a unique complete
probability measure $\nu$ on $\BbbR^I$,
measuring every Baire set and inner regular with respect to the zero
sets, such that

\Centerline{$\nu\{x:x\in\BbbR^I$,
  $x(i)\in E_i$ for every $i\in J\}
=\bar\mu(\inf_{i\in J}\Bvalue{u(i)\in E_i})$}

\noindent whenever $J\subseteq I$ is finite and
$E_i\subseteq\Bbb R$ is a Borel set for every $i\in J$.
\prooflet{\Prf\ Express
$(\frak A,\bar\mu)$ as the measure algebra of a probability space
$(\Omega,\Sigma,\mu)$ (321J), and for each $i\in I$ choose a measurable
function $X_i:\Omega\to\Bbb R$ such that $u(i)$ can be identified with
$X_i^{\ssbullet}\in L^0(\mu)$ (364Ic);  now the distribution $\nu$ of
$\familyiI{X_i}$, as defined in 454J-454K, has the required properties.
The argument of part (e) of the proof of 454J still applies, so $\nu$ is
unique.\ \Qed}

\spheader 454Vc In this context, I will call $\nu$ the {\bf (joint)
distribution} of $u$.
\cmmnt{(Compare 364Gb\formerly{3{}64Xd}.)}

\spheader 454Vd\cmmnt{ Translating 454O into this language,
we get the following.} Let $(\frak A,\bar\mu)$ and $(\frak A',\bar\mu')$ be
probability algebras, and $u\in L^0(\frak A)^I$,
$u'\in L^0(\frak A')^I$ families with the same distribution.
Suppose that $\family{j}{J}{h_j}$ is a family of Baire measurable functions
from $\BbbR^I$ to $\Bbb R$.   Then
$\family{j}{J}{\bar h_j(u)}$ and
$\family{j}{J}{\bar h_j(u')}$ have the same distribution.   \prooflet{\Prf\ If
$J\subseteq I$ is non-empty and finite and $E_j$ is a Borel subset of $\Bbb R$
for $j\in J$,

$$\eqalignno{\bar\mu(\inf_{j\in J}\Bvalue{h_j(u)\in E_j})&
=\bar\mu(\Bvalue{u\in\bigcap_{j\in J}h_j^{-1}[E_j]})
=\nu(\bigcap_{j\in J}h_j^{-1}[E_j])
\Displaycause{where $\nu$ is the common distribution of $u$ and $u'$}
=\bar\mu'(\inf_{j\in J}\Bvalue{h_j(u')\in E_j}).  \text{ \Qed}\cr}$$
}

\spheader 454Ve Similarly, if $(\frak A,\bar\mu)$ is a probability algebra,
$I$ a set, and we write $\nu_u$ for the distribution of $u\in L^0(\frak A)^I$,
$u\mapsto\nu_u$ is continuous for the product topology on
$L^0(\frak A)^I$ corresponding to the topology of convergence in measure on
$L^0(\frak A)$\cmmnt{ (367L)} and the vague topology on the space $M$ of
distributions on $\BbbR^I$.   \prooflet{\Prf\ It is enough to consider the case
in which $(\frak A,\bar\mu)$ is the measure algebra of a probability space
$(\Omega,\Sigma,\mu)$, so that $L^0(\frak A)$ can be identified with $L^0(\mu)$.
Let $\phi:L^0(\frak A)\to\eusm L^0(\mu)$ be any function such that
$\phi(w)^{\ssbullet}=w$ for every $w\in L^0(\frak A)$.   Then $\phi$ is
continuous for the topologies of convergence in measure on
$L^0(\frak A)\cong L^0(\mu)$ and $\eusm L^0(\mu)$ (see 245B) and the
corresponding map
$u\mapsto\phi u=\familyiI{\phi(u(i))}:L^0(\frak A)^I\to\eusm L^0(\mu)^I$ is
continuous.   As $\Bvalue{w\in E}=\{\omega:\phi(w)(\omega)\in E\}^{\ssbullet}$ 
for any
$w\in L^0(\frak A)$ and $E\in\Cal B(\Bbb R)$, the distributions $\nu_u$,
$\nu_{\phi u}$ of $u$ and
$\phi u$ are the same for every $u\in L^0(\frak A)^I$.   Since
$\pmb X\mapsto\nu_{\pmb X}:\eusm L^0(\mu)^I\to M$ is continuous (454U), so is
$u\mapsto\nu_u:L^0(\frak A)^I\to M$.\ \Qed}

\exercises{\leader{454X}{Basic exercises $\pmb{>}$(a)}
%\spheader 454Xa
Let $\mu$ be Lebesgue measure on $[0,1]$, and $\Sigma$ its domain.   Let
$X_0$, $X_1\subseteq[0,1]$ be disjoint sets of full outer measure.
For each $i$, let $\Sigma_i$ be the relative $\sigma$-algebra on
$X_i$.   Show that we have
a finitely additive functional $\lambda$ defined on
$\Sigma_0\otimes\Sigma_1$ by the formula

\Centerline{$\lambda((E\cap X_0)\times(F\cap X_1))=\mu(E\cap F)$ for all
$E$, $F\in\Sigma$,}

\noindent and that $\lambda$ has no extension to a measure on
$X_0\times X_1$.
%454F

\spheader 454Xb Adapt the example of 419K to provide a counter-example
for 454G if we omit the hypothesis that the marginal measures
$\mu_{\{i\}}$ must be perfect.
%454G

\spheader 454Xc Adapt the example of 419K/454Xb to provide a
counter-example for 454H if we omit the hypothesis that the
$(X_n,\Sigma_n)$ must be standard Borel spaces.   \Hint{if
$z\in\prod_{i\le n}X_i$, try $\nu_z(E)=1$ if $z(n)\in E$, $0$
otherwise.}
%454H

\sqheader 454Xd Let $X$ be a set and $\Sigma$ a $\sigma$-algebra of
subsets of $X$.   Let us say that $(X,\Sigma)$ has the {\bf perfect
measure property} if every totally finite measure with domain $\Sigma$
is perfect.   Show that (i) if $(X,\Sigma)$ has the perfect measure
property, so does $(E,\Sigma_E)$ for any $E\in\Sigma$, where $\Sigma_E$
is the subspace $\sigma$-algebra on $E$ (ii) if
$\familyiI{(X_i,\Sigma_i)}$ is a family of spaces with the perfect
measure property, then $(\prod_{i\in I}X_i,\Tensorhat_{i\in I}\Sigma_i)$
has the perfect measure property.
%454I

\spheader 454Xe Let $(X,\Sigma)$ be a space with the perfect measure
property, and $\Tau$ the smallest $\sigma$-algebra including $\Sigma$
and closed under Souslin's operation.   Show that $(X,\Tau)$ has the
perfect measure property.
%454I

\spheader 454Xf Let $X$ be a set and $\Sigma$ a $\sigma$-algebra of
subsets of $X$.   Let us say that $(X,\Sigma)$ has the {\bf countably
compact measure property} if every totally finite measure with domain
$\Sigma$ is countably compact.   Show that (i) if $(X,\Sigma)$ has the
countably compact measure property it has the perfect measure property
(ii) if $(X,\Sigma)$ has the countably compact measure property so does
$(E,\Sigma_E)$ for every $E\in\Sigma$, where $\Sigma_E$ is the subspace
$\sigma$-algebra on $E$
(iii) if $\familyiI{(X_i,\Sigma_i)}$ is a family of spaces with the
countably compact measure property, then
$(\prod_{i\in I}X_i,\Tensorhat_{i\in I}\Sigma_i)$
has the countably compact measure property.
%454I

\spheader 454Xg Suppose that $(X,\Sigma)$ has the countably compact
measure property.   (i) Let $\mu$ be a totally finite measure with
domain $\Sigma$, $(Y,\Tau,\nu)$ a measure space, and $f:X\to Y$ an \imp\
function.   Show that $\mu$ has a disintegration $\family{y}{Y}{\mu_y}$
over $\nu$ which is consistent with $f$.
(ii) Let $Y$ be any set, $\Tau$ a $\sigma$-algebra of subsets of $Y$,
and $\lambda$ a probability measure with domain $\Sigma\tensorhat\Tau$.
Show that there is a family $\family{y}{Y}{\mu_y}$ of probability
measures on $X$ such that
$\lambda W=\int\mu_yW^{-1}[\{y\}]\nu(dy)$ for every
$W\in\Sigma\tensorhat\Tau$, where $\nu$ is the marginal measure of
$\lambda$ on $Y$.   \Hint{452M.}
%454I

\spheader 454Xh(i) Let $X$ be any set, and $\Sigma$ the
countable-cocountable algebra on $X$.   Show that $(X,\Sigma)$ has the
countably compact measure property.  (ii) Show that any standard Borel
space has the countably compact measure property.
%454I

\spheader 454Xi Let $X$ be a Radon Hausdorff space, and
$\Sigma_{\text{um}}$ the
algebra of universally measurable sets in $X$ (434D).   Show that
$(X,\Sigma_{\text{um}})$ has the countably compact measure property.
%454I

\sqheader 454Xj Let $\familyiI{X_i}$ be an independent family of normal random
variables.   Show that its distribution is
a quasi-Radon measure on $\BbbR^I$.   \Hint{415E.}
%454L

\spheader 454Xk Give an example of a metrizable space $\Omega$ with a
continuous injective function $X:\Omega\to[0,1]$ and two different
quasi-Radon probability measures $\mu$, $\nu$ on $\Omega$ giving the same
distribution to the random variable $X$.
%454N

\spheader 454Xl Let $I$ be a set and $\nu$, $\nuprime$ two quasi-Radon
measures on $\BbbR^I$ such that
$\int e^{if(x)}\nu(dx)=\int e^{if(x)}\nuprime(dx)$ for every continuous
linear functional $f:\BbbR^I\to\Bbb R$.   Show that $\nu=\nuprime$.
%454P

\sqheader 454Xm Let $\Sigma$ be the $\sigma$-algebra of subsets of
$C(\coint{0,\infty})$ generated by the functionals $f\mapsto f(t)$ for
$t\ge 0$.   Give $C(\coint{0,\infty})$ the topology $\frak T_c$ of
uniform convergence on compact sets.   (i) Show that $\frak T_c$ is
Polish, and that $\Sigma\cap\frak T_c$ is a base for $\frak T_c$ which
generates $\Sigma$ as $\sigma$-algebra.   (ii) Use this to give a quick
proof of 454R in this case.
%454R

\spheader 454Xn Let $T$ be a Polish space, and $\frak T_c$ the topology
on $C(T)$ of uniform convergence on compact sets.   Show that if
$\frak T$ is any Hausdorff topology on $C(T)$, coarser than $\frak T_c$,
such that all the functionals $f\mapsto f(t)$, for $t\in T$, are Baire
measurable for $\frak T$, then $(C(T),\frak T)$ is a measure-compact
Radon space.
%454S

\leader{454Y}{Further exercises (a)}
%\spheader 454Ya
In 454Ab, show that $\mu$ is weakly $\alpha$-favourable (definition:
451V) if every $\mu_i$ is.
%454A

\spheader 454Yb
Let $\Sigma$ be the algebra of Lebesgue measurable
subsets of $\Bbb R$.   Show that $(\BbbR,\Sigma)$ has the perfect
measure property (454Xd) iff $\frak c$ is measure-free.
%454I

\spheader 454Yc Let $\Cal B$ be the Borel $\sigma$-algebra of $\omega_1$
with its
order topology.   Show that $(\omega_1,\Cal B)$ has the perfect measure
property.   \Hint{439Xn.}
%454I

\spheader 454Yd Let $(X,\Sigma,\mu)$ be a semi-finite measure space with
a topology such that $\mu$ is inner regular with respect to the closed
sets, $T$ a second-countable space and $Y$ a separable metrizable space.
Suppose that $\phi:X\times T\to Y$ is continuous in the second variable
and measurable in the first, as in 454Q.   Show that $\mu$ is inner
regular with respect to $\Cal K=\{K:K\subseteq X,\,\phi\restr K\times T$
is continuous$\}$.
%454Q
}%end of exercises

\endnotes{
\Notesheader{454} 454A generalizes Theorem 451J, which gave the same
result (with essentially the same proof) for product measures.   One of
the themes of this section is the idea that we can deduce properties of
measures on product spaces from properties of their marginal measures,
that is, the image measures on the factors.   The essence of
`compactness', `countable compactness' and `perfectness' is that we can
find enough points in the measure space to do what we want.   (See, for
instance, the characterization of local compactness in 343B, or Pachl's
characterization of countable compactness in 452Ye.)   Since the
canonical feature of a product space is that we put in every point the
Axiom of Choice provides us with, it's perhaps not surprising that such
properties can be inherited by measures on product spaces.

Theorems 454C and 454D can be regarded as further variations on the same
theme.   A finitely additive non-negative functional on an algebra of
sets will have an extension to a measure if, and only if, it is
sequentially smooth in the sense that the measures of a decreasing
sequence of sets with empty intersection converge to zero (413L).   If
we have a decreasing sequence of sets, with measures bounded away from
zero, but with empty intersection, one interpretation of the phenomenon
is that some points which ought to have been present got left out of the
sets.   What 454D tells us is that perfectness (and countable
additivity) of the marginal measures is enough to ensure that there are
enough points in the product to stop this happening.   In effect, 454C
tells us that it will be enough if every marginal but one is perfect.

These results are of course associated with the projective limit
constructions in 418M-418Q.  %418M 418O 418P 418Q
In the theorems there we had Radon
measures, so that they were actually compact rather than perfect;  in
return for the stronger hypothesis on the measures, we could handle
projective limits corresponding to rather small subsets of the product
spaces (see the formulae in 418O-418Q).  %418O 418P 418Q
Just as in \S418, the patterns
change when we have countable rather than uncountable families to deal
with (418P-418Q, 454H).

In 454J-454P, I insist rather arbitrarily that `the' joint
distribution of a family $\familyiI{X_i}$ of real-valued random
variables is the completion of a Baire measure on $\BbbR^I$.   Of course
all the ideas can also be expressed in terms of the Baire measure
itself, but I have sought a formulation which is consistent with the
rules set out in \S271.   When $I$ is countable, we get a Radon measure
(454J(iii)), as in the finite-dimensional case.   There are other cases
in which the distribution is a quasi-Radon measure (454Xj).   As
always, we can ask whether the distribution is $\tau$-additive;  in this
case it will have a canonical extension to a quasi-Radon measure (415N).
Important examples of this phenomenon are described in 454Xj, 455H and 456O.
Because $\BbbR^I$ has a linear topological space structure, we have a
notion of `characteristic function' for any probability measure on
$\BbbR^I$ measuring the zero sets, and the characteristic function of a
Baire measure determines that measure (454P, 454Xl).

In 454R, $C(T)$, with $\frak T_c$, has a
countable network (4A2Oe), so the subspace measure $\mu_C$
induced by $\mu$ on $C(T)$ must be a $\tau$-additive topological measure
with respect to $\frak T_c$ (414O) and has a unique extension to a
quasi-Radon
measure on $C(T)$ (415M).   The hard bit is the next step, showing that
$C(T)$, under $\frak T_c$, is a Radon space;  this is the real point of
454Q-454R.   For the most important case, in which $T=\coint{0,\infty}$,
we have a useful simplification, because $\frak T_c$ is actually Polish
(454Xm).   Even in this case, however, we need to observe that the
measure we are seeking is a little more complicated than a simple
completion of a measure on $\BbbR^T$.
We must complete the {\it subspace} measure on $C(T)$, and $C(T)$ is far
from being a measurable set.   The measure $\tilde\mu$ of 454S will not
as a rule be completion regular, for instance.   Spaces of continuous
functions are so important that it is worth noticing that the results
here will be valid for various topologies on $C(T)$ (454Xn).

I suppose that pretty well every result on distributions in Chapter 27
corresponds to some significant development expressible in the language of this
section.   454T-454U take up the idea of Exercise 274Yf.   Looking at the facts
here from the point of view of Volume 3 we get the alternative versions in 454V.
}%end of notes

\discrpage

\frfilename{mt455.tex}
\versiondate{18.1.09/12.10.18}
\copyrightdate{2007}

\def\mucdlg{\mu_{\text{cdlg}}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Markov and L\'evy processes}

\newsection{455}

For a `Markov' process, in which the evolution of the system after a
time $t$ depends only on the state at time $t$, the general theory of
\S454 leads
to a straightforward existence theorem (at least for random variables
taking values in standard Borel spaces) dependent only on a natural
consistency condition on the transitional probabilities (455A, 455E).
The formulation leads naturally to descriptions of the `Markov property'
(for stopping times taking only countably many values) in
terms of disintegrations and conditional expectations (455C, 455Ec).
With appropriate continuity conditions, we find that the process can be
represented either by a Radon measure (455H) or
by a measure on the set of \cadlag\ paths (455Gc) for which we
have a formulation of the strong Markov property (for general
stopping times) in terms of disintegrations
(455O).   These conditions are satisfied by L\'evy processes
(455P-455R).  %455P 455Q 455R
For these, we have an alternative expression of the strong Markov property
in terms of \imp\ functions (455U).
By far the most important example of a continuous-time Markov process is
Brownian motion, but I defer discussion of this to \S477.

\vleader{72pt}{455A}{Theorem} Let $T$ be a totally ordered set with least
element $t^*$, and for each $t\in T$ let $\Omega_t$ be a
non-empty set and $\Tau_t$ a
$\sigma$-algebra of subsets of $\Omega_t$ containing all singleton subsets
of $\Omega_t$.   Set $\Omega=\prod_{t\in T}\Omega_t$ and for $t\in T$,
$\omega\in\Omega$ set $X_t(\omega)=\omega(t)$.
Fix $x^*\in\Omega_{t^*}$.   Suppose that we are given, for each
pair $s<t$ in $T$, a family $\family{x}{\Omega_s}{\nu^{(s,t)}_x}$ of
perfect probability measures on $\Omega_t$, all with domain $\Tau_t$,
and suppose that

\inset{($\dagger$) whenever $s<t<u$ in $T$ and $x\in\Omega_s$,
then $\family{y}{\Omega_t}{\nu^{(t,u)}_y}$ is a disintegration of
$\nu^{(s,u)}_x$ over $\nu^{(s,t)}_x$.}

\noindent For $J\subseteq T$ write $\pi_J$ for the canonical map from
$\Omega$ onto $Z_J=\prod_{t\in J}\Omega_t$.
Then there is a unique probability measure $\mu$ on $\Omega$,
with domain $\Tensorhat_{t\in T}\Tau_t$, such that, writing
$\lambda_J$ for the image measure $\mu\pi_J^{-1}$,

$$\eqalign{\int fd\lambda_J
&=\int f(\omega(t^*),\omega(t_1),\ldots,\omega(t_n))\mu(d\omega)\cr
&=\int\ldots\iint f(x^*,x_1,\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\qquad\qquad\qquad\qquad
  \nu_{x_{n-2}}^{(t_{n-2},t_{n-1})}(dx_{n-1})
  \ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr}$$

\noindent whenever $t^*<t_1<\ldots<t_n$, $J=\{t^*,t_1,\ldots,t_n\}$ and
$f$ is $\lambda_J$-integrable.   $\mu$ is perfect, and the marginal measure
$\mu_t=\mu X_t^{-1}$ is equal to $\nu_{x^*}^{(t^*,t)}$, if $t>t^*$,
while $\mu_{t^*}\{x^*\}=1$.

\proof{{\bf (a)} For $I\subseteq T$, write
$\Tau_I=\Tensorhat_{t\in I}\Tau_t$.   If $I=\{t_0,t_1,\ldots,t_n\}$ is
a finite subset of $T$ with $t^*=t_0<t_1<\ldots<t_n$, then we have a
probability measure $\lambda_I$ on $Z_I$ with domain $\Tau_I$ such that

$$\eqalign{\int fd\lambda_I
&=\int\ldots\iint f(x^*,x_1,\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\qquad\qquad\qquad\qquad
  \nu_{x_{n-2}}^{(t_{n-2},t_{n-1})}(dx_{n-1})
  \ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr}$$

\noindent for every $\lambda_I$-integrable function $f$.   \Prf\ Use
454Ha on the finite sequence $(\Omega_{t_0},\ldots,\Omega_{t_n})$.
The measures
$\nu_z$ required by 454H must be constructed by the rule

\Centerline{$\nu_z=\nu_{z(t_m)}^{(t_m,t_{m+1})}$}

\noindent for $m<n$, $z\in\prod_{i\le m}\Omega_{t_i}$, while of course
$\nu_{\emptyset}\{x^*\}=1$.   Having a finite sequence rather than an
infinite one clearly makes things easier.)\ \Qed

When $I=\{t^*\}$, so that $Z_I$ can be identified with $\Omega_{t^*}$,
I mean to interpret these formulae in such a way that
$\lambda_I\{x^*\}=1$.   When $J=\{t^*,t\}$, with $t^*<t$,
and $E\in\Tau_t$, then we can apply the formula above to the function
$z\mapsto\chi E(z(t))$ to get
$\lambda_J\{z:z(t)\in E\}=\nu_{x^*}^{(t^*,t)}(E)$.

\medskip

{\bf (b)} Of course the point of this is that these measures $\lambda_I$
form a consistent family;  if $t^*\in I\subseteq J\in[T]^{<\omega}$, then
the canonical projection $\pi_{IJ}:Z_J\to Z_I$ is \imp.   \Prf\ It is
enough to consider the case in which $J$ has just one more point than
$I$, since then we can induce on $\#(J\setminus I)$.   In this case,
express $J$ as $\{t_0,\ldots,t_n\}$ where $t^*=t_0<\ldots<t_n$, and
suppose that $I=J\setminus\{t_m\}$.   If $W\in\Tau_I$, then

$$\eqalignno{\lambda_J\pi_{IJ}^{-1}[W]
&=\int\ldots\iint\ldots\int\chi W(x^*,x_1,\ldots,x_{m-1},x_{m+1},\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\qquad\qquad\qquad\qquad
   \ldots\nu_{x_m}^{(t_m,t_{m+1})}(dx_{m+1})
   \nu_{x_{m-1}}^{(t_{m-1},t_m)}(dx_m)
   \ldots\nu^{(t^*,t_1)}_{x^*}(dx_1)\cr
&=\int\ldots\iint g_{(x_1,\ldots,x_{m-1})}(x_{m+1})
    \nu_{x_m}^{(t_m,t_{m+1})}(dx_{m+1})&(^*)\cr
&\qquad\qquad\qquad\qquad\qquad\qquad
  \nu_{x_{m-1}}^{(t_{m-1},t_m)}(dx_m)\ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr}$$

\noindent where

$$\eqalign{g_{(x_1,\ldots,x_{m-1})}(x_{m+1})
&=\int\ldots\int\chi W(x^*,x_1\ldots,x_{m-1},x_{m+1},\ldots,x_n)\cr
&\qquad\qquad\qquad\qquad
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
   \ldots\nu_{x_{m+1}}^{(t_{m+1},t_{m+2})}(dx_{m+2}).\cr}$$

\noindent Here, of course, we use the hypothesis ($\dagger$);  since
$\family{y}{\Omega_{t_m}}{\nu_y^{(t_m,t_{m+1})}}$ is a disintegration of
$\nu_{x_{m-1}}^{(t_{m-1},t_{m+1})}$ over $\nu_{x_{m-1}}^{(t_{m-1},t_m)}$,
and $g_{(x_1,\ldots,x_{m-1})}$ is bounded and
$\nu^{(t_{m-1},t_{m+1})}_{x_{m-1}}$-integrable (by 454H),

$$\eqalign{\int g_{(x_1,\ldots,x_{m-1})}&(x_{m+1})
  \nu_{x_{m-1}}^{(t_{m-1},t_{m+1})}(dx_{m+1})\cr
&=\iint g_{(x_1,\ldots,x_{m-1})}(x_{m+1})
  \nu_{x_m}^{(t_m,t_{m+1})}(dx_{m+1})
  \nu_{x_{m-1}}^{(t_{m-1},t_m)}(dx_m)\cr}$$

\noindent (452F).   Substituting this into (*) above,

$$\eqalignno{\lambda_J\pi_{IJ}^{-1}[W]
&=\int\ldots\iint g_{(x_1,\ldots,x_{m-1})}(x_{m+1})\cr
&\qquad\qquad\qquad\qquad
  \nu_{x_m}^{(t_m,t_{m+1})}(dx_{m+1})
  \nu_{x_{m-1}}^{(t_{m-1},t_m)}(dx_m)\ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr
&=\int\ldots\int g_{(x_1,\ldots,x_{m-1})}(x_{m+1})
  \nu_{x_{m-1}}^{(t_{m-1},t_{m+1})}(dx_{m+1})\ldots
  \nu_{x^*}^{(t^*,t_1)}(dx_1)\cr
&=\int\ldots\int\ldots\int
      \chi W(x^*,x_1,\ldots,x_{m-1},x_{m+1},\ldots,x_n)\cr
&\qquad\qquad\qquad\qquad
  \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\ldots
  \nu_{x_{m-1}}^{(t_{m-1},t_{m+1})}(dx_{m+1})\ldots
  \nu_{x^*}^{(t^*,t_1)}(dx_1)\cr
&=\lambda_IW,\cr}$$

\noindent applying the formula in (a) again.\ \Qed

(Some of the formulae here are inappropriate if $m=n>1$.   In this case,
of course,

$$\eqalignno{\lambda_J\pi_{IJ}^{-1}[W]
&=\int\ldots\int\chi W(x^*,x_1,\ldots,x_{n-1})
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr
&=\int\ldots\int\chi W(x^*,x_1,\ldots,x_{n-1})
  \nu_{x_{n-2}}^{(t_{n-2},t_{n-1})}(dx_{n-1})\ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)
 =\lambda_IW.\cr}$$

\noindent If $m=1<n$, there is a collapse of a different kind;  we must
look at

$$\eqalign{\lambda_J\pi_{IJ}^{-1}[W]
&=\int\ldots\iint\chi W(x^*,x_2,\ldots,x_n)
  \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
     \ldots\nu_{x_1}^{(t_1,t_2)}(dx_2)\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr
&=\int\ldots\iint\chi W(x^*,x_2,\ldots,x_n)
  \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
     \ldots\nu_{x^*}^{(t^*,t_2)}(dx_2)
=\lambda_IW.\cr}$$

\noindent If $m=n=1$ then

\Centerline{$\lambda_J\pi_{IJ}^{-1}[W]
=\int\chi W(x^*)\nu_{x^*}^{(t^*,t_1)}(dx_1)
=\chi W(x^*)
=\lambda_IW$.)}

\medskip

{\bf (c)} Part (b) tells us that we have a consistent family of measures
on the finite products $Z_J$, and therefore have a functional $\lambda$
on $\bigotimes_{t\in T}\Tau_t$ defined by setting
$\lambda\pi_J^{-1}[W]=\lambda_JW$ for every finite $J\subseteq T$
containing $t^*$ and $W\in\bigotimes_{t\in J}\Tau_t$.
$\lambda$ is finitely
additive, and its images $\mu_t=\lambda X_t^{-1}$ are all countably
additive and perfect because $\mu_t=\nu_{x^*}^{(t^*,t)}$ for $t>t^*$,
while $\mu_{t^*}$ is concentrated at $\{x^*\}$.

By 454D, we have a perfect
measure $\mu$ extending $\lambda$.   We have to
check that each $\lambda_J$ is the image
measure $\mu\pi_J^{-1}$;  but this is true because they agree on
$\bigotimes_{t\in J}\Tau_t$ (using the Monotone Class Theorem in the
form 136C, as always).   So the integral formula sought for
$\lambda_J$ is just that obtained in part (a).
By the last remark in (a), we have the declared formulae for the marginal
measures $\mu_t$.
}%end of proof of 455A

\leader{455B}{Lemma} Suppose that $T$, $t^*$,
$\family{t}{T}{(\Omega_t,\Tau_t)}$, $\Omega$, $x^*$ and
$\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_s}$ are as in
455A.

(a) Suppose that $\mu$ is constructed from $x^*$
and $\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_s}$ as in 455A.
If $F\in\Tensorhat_{t\in T}\Tau_t$ is determined by
coordinates in $[t^*,t_0]$ and
$H^*=\{\omega:\omega(t_i)\in E_i$ for $1\le i\le n\}$ where
$t_0<t_1\ldots<t_n$ and $E_i\in\Tau_{t_i}$ for $1\le i\le n$, then

$$\eqalignno{\mu(H^*\cap F)
&=\int_F\int\ldots\int\chi H(y_1,\ldots,y_n)
   \nu_{y_{n-1}}^{(t_{n-1},t_n)}(dy_n)
   \ldots\nu_{\omega(t_0)}^{(t_0,t_1)}(dy_1)\mu(d\omega)&(*)}$$

\noindent where $H=\prod_{1\le i\le n}E_i$.

(b) Suppose that $\omega\in\Omega$ and $a\in T\cup\{\infty\}$, where
$\infty$ is taken to be greater than every element of $T$.   For
$s<t$ in $T$ and $x\in\Omega_s$ set

$$\eqalign{\nu^{(s,t)}_{\omega ax}
&=\nu_x^{(s,t)}\text{ if }a<s,\cr
&=\nu_{\omega(a)}^{(a,t)}\text{ if }s\le a<t,\cr
&=\delta^{(t)}_{\omega(t)}\text{ if }t\le a,\cr}$$

\noindent here writing $\delta^{(t)}_x$ for the
probability measure with domain $\Tau_t$ such that
$\delta^{(t)}_x(\{x\})=1$.

\quad(i) $\nu^{(s,t)}_{\omega ax}$ is always a
perfect probability measure with domain $\Tau_t$, and
$\family{y}{\Omega_t}{\nu^{(t,u)}_{\omega ay}}$ is a disintegration of
$\nu^{(s,u)}_{\omega ax}$ over $\nu^{(s,t)}_{\omega ax}$ whenever $s<t<u$
in $T$ and $x\in\Omega_s$.

\quad(ii) Taking $\mu_{\omega a}$ to be the measure on $\Omega$ defined from
$\omega(t^*)$ and $\langle\nu^{(s,t)}_{\omega ax}\rangle_{s<t,x\in\Omega_t}$
by the method of 455A, then
$\{\omega':\omega'\in\Omega$, $\omega'\restr D=\omega\restr D\}$ is
$\mu_{\omega a}$-conegligible for every countable
$D\subseteq T\cap[t^*,a]$.

\quad(iii) If $\omega$, $\omega'\in\Omega$ and
$\omega\restr[t^*,a]=\omega'\restr[t^*,a]$ then
$\mu_{\omega a}=\mu_{\omega'a}$.

\proof{{\bf (a)(i)} Suppose first that $F$ is of the form
$\{\omega:\omega(s_i)\in F_i$ for $i\le m\}$ where
$t^*=s_0<\ldots<s_m=t_0$.   For $x\in\Omega_{t_0}$ set

\Centerline{$f(x)=\int\ldots\int\chi H(y_1,\ldots,y_n)
   \nu_{y_{n-1}}^{(t_{n-1},t_n)}(dy_n)
   \ldots\nu_x^{(t_0,t_1)}(dy_1)$.}

\noindent Writing $G=\prod_{i\le m}F_i$, we have

$$\eqalignno{\mu(H^*\cap F)
&=\int\ldots\iint\ldots\int\chi G(x^*,x_1,\ldots,x_m)
   \chi H(y_1,\ldots,y_n)\nu_{y_{n-1}}^{(t_{n-1},t_n)}(dy_n)\cr
&\mskip100mu
   \ldots\nu_{x_m}^{(s_m,t_1)}(dy_1)
   \nu_{x_{m-1}}^{(s_{m-1},s_m)}(dx_m)
   \ldots\nu_{x^*}^{(t^*,s_1)}(dx_1)\cr
&=\int\ldots\int\chi G(x^*,x_1,\ldots,x_m)f(x_m)
   \nu_{x_{m-1}}^{(s_{m-1},s_m)}(dx_m)
   \ldots\nu_{x^*}^{(t^*,s_1)}(dx_1)\cr
&=\int g\,d\lambda_J\cr
\displaycause{where $J=\{t^*,s_1,\ldots,s_m\}$,
$g(z)=\chi G(z(t^*),\ldots,z(s_m))f(z(s_m))$ for
$z\in\prod_{s\in J}\Omega_s$, and $\lambda_J$ is defined as in 455A}
&=\int g\pi_Jd\mu
=\int_Ff(\omega(t_0))\mu(d\omega)\cr
\displaycause{because $g\pi_J(\omega)=f(\omega(s_m))=f(\omega(t_0))$
if $\omega\in F$, $0$ otherwise}
&=\int_F\int\ldots\int\chi H(y_1,\ldots,y_n)
   \nu_{y_{n-1}}^{(t_{n-1},t_n)}(dy_n)
   \ldots\nu_{\omega(t_0)}^{(t_0,t_1)}(dy_1)\mu(d\omega).\cr}$$

\medskip

\quad{\bf (ii)} Let $\Cal I$ be the family of sets $F$ of the type dealt with in
(a).   Since the intersection of two members of $\Cal I$ belongs to
$\Cal I$, the Monotone Class Theorem tells us
that (*) is true for all sets in the $\sigma$-algebra $\Tau$
generated by $\Cal I$.   But any member of
$\Tensorhat_{t\in T}\Tau_t$ determined by coordinates in $[t^*,t_0]$
belongs to $\Tau$.
\Prf\ Fix $v\in\prod_{s\in T\setminus[t^*,t_0]}\Omega_s$.   For
$\omega\in\Omega$ define $f(\omega)\in\Omega$ by setting

$$\eqalign{f(\omega)(s)
&=\omega(s)\text{ if }s\le t_0,\cr
&=v(s)\text{ if }s>t_0.\cr}$$

\noindent Then $\Tau'=\{F:F\subseteq\Omega$, $f^{-1}[F]\in\Tau\}$ is a
$\sigma$-algebra of
subsets of $\Omega$ containing $\{\omega:\omega(t)\in E\}$ whenever
$t\in T$ and $E\in\Tau_t$, so includes $\Tensorhat_{t\in T}\Tau_t$.
If $F\in\Tensorhat_{t\in T}\Tau_t$ and $F$ is determined by coordinates
in $[t^*,t_0]$, then $F=f^{-1}[F]\in\Tau$.\ \Qed

So (*) is true of every $F\in\Tensorhat_{t\in T}\Tau_t$, as claimed.

\woddheader{455B}{4}{2}{2}{108pt}

{\bf (b)(i)} Of course every $\nu^{(s,t)}_{\omega ax}$ is a perfect
probability measure with domain $\Tau_t$.
If $s<t<u$ and $E\in\Tau_u$, then

$$\eqalign{
\int_{\Omega_t}\nu^{(t,u)}_{\omega ay}(E)\nu^{(s,t)}_{\omega ax}(dy)
&=\int_{\Omega_t}\nu^{(t,u)}_y(E)\nu^{(s,t)}_x(dy)
   =\nu_x^{(s,u)}(E)
   =\nu_{\omega ax}^{(s,u)}(E)\cr
&\mskip250mu   \text{ if }a<s,\cr
&=\int_{\Omega_t}\nu^{(t,u)}_y(E)
   \nu^{(a,t)}_{\omega(a)}(dy)
   =\nu_{\omega(a)}^{(a,u)}(E)
   =\nu_{\omega ax}^{(s,u)}(E)\cr
&\mskip250mu
   \text{ if }s\le a<t,\cr
&=\int_{\Omega_t}\nu^{(t,u)}_{\omega(t)}(E)\delta^{(t)}_{\omega(t)}(dy)
   =\nu^{(t,u)}_{\omega(t)}(E)
   =\nu_{\omega ax}^{(s,u)}(E)\cr
&\mskip250mu
   \text{ if }a=t,\cr
&=\int_{\Omega_t}\nu^{(a,u)}_{\omega(a)}(E)
   \delta^{(t)}_{\omega(t)}(dy)
   =\nu^{(a,u)}_{\omega(a)}(E)
   =\nu_{\omega ax}^{(s,u)}(E)\cr
&\mskip250mu
   \text{ if }t<a<u,\cr
&=\int_{\Omega_t}\delta^{(u)}_{\omega(u)}(E)
   \delta^{(t)}_{\omega(t)}(dy)
   =\delta^{(u)}_{\omega(u)}(E)
   =\nu_{\omega ax}^{(s,u)}(E)\cr
&\mskip250mu
   \text{ if }u\le a.\cr}$$

\medskip

\quad{\bf (ii)}
Consider first the case $D=\{t\}$, where $t^*<t\le a$.   Then

\Centerline{$\mu_{\omega a}\{\omega':\omega'(t)=\omega(t)\}
=\nu^{(t^*,t)}_{\omega,a,\omega(t^*)}\{\omega(t)\}
=\delta^{(t)}_{\omega(t)}\{\omega(t)\}
=1$.}

\noindent As for $D=\{t^*\}$, $\mu_{\omega a}$ starts at $\omega(t^*)$, so
(as noted in the last clause of the statement of 455A)
$\mu_{\omega a}\{\omega':\omega'(t^*)=\omega(t^*)\}=1$.

For general $D$, we have an intersection of countably many sets
of these types, which will be $\mu_{\omega a}$-conegligible.

\medskip

\quad{\bf (iii)} Looking at the definition, we see that
$\nu^{(s,t)}_{\omega' ax}=\nu^{(s,t)}_{\omega ax}$
for all $s$, $t$ and $x$, and of course $\omega'(t^*)=\omega(t^*)$, so
$\mu_{\omega'a}=\mu_{\omega a}$.
}%end of proof of 455B

\leader{455C}{Theorem} Suppose that $T$, $t^*$,
$\family{t}{T}{(\Omega_t,\Tau_t)}$, $\Omega$, $x^*$,
$\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_s}$ and $\mu$ are as in
455A.   Adjoin a point $\infty$ to $T$ above any point of $T$,
and let $\tau:\Omega\to T\cup\{\infty\}$ be a
function taking countably many
values and such that $\{\omega:\tau(\omega)\le s\}$
belongs to $\Tensorhat_{t\in T}\Tau_t$ and is determined by
coordinates in $[t^*,s]$ for every $s\in T$.

(a) For $\omega\in\Omega$ define $\nu^{(s,t)}_{\omega,\tau(\omega),x}$, for
$s<t$ and $x\in\Omega_s$, as in 455Bb, and let $\mu_{\omega,\tau(\omega)}$
be the corresponding measure on $\Omega$.   Then
$\family{\omega}{\Omega}{\mu_{\omega,\tau(\omega)}}$ is a disintegration of
$\mu$ over itself.

(b) Let $\Sigma_{\tau}$ be the set of those
$E\in\Tensorhat_{t\in T}\Tau_t$ such that
$E\cap\{\omega:\tau(\omega)\le t\}$ is determined by
coordinates in $[t^*,t]$ for every $t\in T$.
Then $\Sigma_{\tau}$ is a $\sigma$-subalgebra of
$\Tensorhat_{t\in T}\Tau_t$.   If
$f$ is any $\mu$-integrable real-valued function, and we set
$g_f(\omega)=\int fd\mu_{\omega,\tau(\omega)}$ when this is defined
in $\Bbb R$, then $g_f$ is a
conditional expectation of $f$ on $\Sigma_{\tau}$.

\proof{{\bf (a)(i)}
Set $F_t=\{\omega:\omega\in\Omega$, $\tau(\omega)=t\}$ for
$t\in T\cup\{\infty\}$;  note that $F_t\in\Tensorhat_{t\in T}\Tau_t$ for
every $t\in T\cup\{\infty\}$, and that $F_t$ is determined by coordinates
in $[t^*,t]$ for $t\in T$.

\medskip

\quad{\bf (ii)} Consider first the case in which $\tau$ takes only
finitely many values.   Suppose that
$J\subseteq T$ is a finite set including
$\{t^*\}\cup(T\cap\tau[\Omega])$.
Enumerate $J$ as $\langle t_i\rangle_{i\le n}$.
Suppose that $E_i\in\Tau_{t_i}$ for $i\le n$ and set
$H^*=\{\omega:\omega(t_i)\in E_i$ for every $i\le n\}$.
We need to calculate
$\int_{\Omega}\mu_{\omega,\tau(\omega)}(H^*)\mu(d\omega)$.

Set $H=\prod_{i\le n}E_i$,

\Centerline{$H_j=\prod_{j<i\le n}E_j$,
\quad$H^*_j=\{\omega:\omega\in\Omega$,
$\omega(t_i)\in E_i$ for $j<i\le n\}$,}

\Centerline{$G^*_j=\{\omega:\omega(t_i)\in E_i$ for $i\le j\}$}

\noindent for $j\le n$.
If $i<n$, $j\le n$, $\omega\in F_{t_j}$ and $x\in\Omega_{t_i}$, then

$$\eqalign{\nu_{\omega,\tau(\omega),x}^{(t_i,t_{i+1})}
&=\nu_x^{(t_i,t_{i+1})}\text{ if }i>j,\cr
&=\nu^{(t_j,t_{j+1})}_{\omega(t_j)}\text{ if }i=j,\cr
&=\delta^{(t_{i+1})}_{\omega(t_{j+1})}\text{ if }i<j.\cr}$$

So if $j\le n$ and $\omega\in F_{t_j}$,

$$\eqalign{\mu_{\omega,\tau(\omega)}(H^*)
&=\int\ldots\int\chi H(\omega(t^*),x_1,\ldots,x_n)
   \nu_{\omega,\tau(\omega),x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
   \ldots\nu_{\omega,\tau(\omega),\omega(t^*)}^{(t^*,t_1)}(dx_1)\cr
&=\iint\ldots\int\chi H(x_0,\ldots,x_n)
   \nu_{\omega,\tau(\omega),x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
   \ldots\nu_{\omega,\tau(\omega),x_0}^{(t^*,t_1)}(dx_1)
   \delta^{(t^*)}_{\omega(t^*)}(dx_0)\cr
&=\iint\ldots\iiint\ldots\int\chi H(x_0,\ldots,x_n)
   \nu_{\omega,\tau(\omega),x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\mskip100mu
   \ldots\nu_{\omega,\tau(\omega),x_{j+1}}^{(t_{j+1},t_{j+2})}(dx_{j+2})
   \nu_{\omega,\tau(\omega),x_j}^{(t_j,t_{j+1})}(dx_{j+1})
   \nu_{\omega,\tau(\omega),x_{j-1}}^{(t_{j-1},t_j)}(dx_j)\cr
&\mskip100mu  \ldots\delta^{(t_1)}_{\omega(t_1)}(dx_1)
    \delta^{(t^*)}_{\omega(t^*)}(dx_0)\cr
&=\int\ldots\iiint\ldots\int\chi H(x_0,\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\mskip100mu   \ldots\nu_{x_j}^{(t_j,t_{j+1})}(dx_{j+1})
   \nu_{\omega(t_j)}^{(t_j,t_{j+1})}(dx_{j+1})
   \delta^{(t_j)}_{\omega(t_j)}(dx_j)
   \ldots
   \delta^{(t^*)}_{\omega(t^*)}(dx_0)\cr
&=\int\ldots\int\chi H(\omega(t^*),\ldots,\omega(t_j),x_{j+1},\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\mskip100mu
   \ldots\nu_{\omega(t_j)}^{(t_j,t_{j+1})}(dx_{j+1})\cr
&=\int\ldots\int\chi H_j(x_{j+1},\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
   \ldots\nu_{\omega(t_j)}^{(t_j,t_{j+1})}(dx_{j+1})\cr
&\mskip200mu\text{ if }\omega\in G^*_j,\cr
&=0\text{ otherwise}.\cr}$$

\noindent As noted in 455B(b-ii),
$\mu_{\omega,\tau(\omega)}(H^*)=\chi H^*(\omega)$ if $\tau(\omega)=\infty$.

Now

$$\eqalignno{\int_{\Omega}\mu_{\omega,\tau(\omega)}(H^*)\mu(d\omega)
&=\sum_{j=0}^n\int_{F_{t_j}}\mu_{\omega,\tau(\omega)}(H^*)\mu(d\omega)
  +\int_{F_{\infty}}\mu_{\omega,\tau(\omega)}(H^*)\mu(d\omega)\cr
&=\sum_{j=0}^n\int_{F_{t_j}\cap G^*_j}\int\ldots\int
\chi H_j(x_{j+1},\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\mskip100mu
   \ldots\nu_{\omega(t_j)}^{(t_j,t_{j+1})}(dx_{j+1})\mu(d\omega)
   +\mu(F_{\infty}\cap H^*)\cr
&=\sum_{j=0}^n\mu(F_{t_j}\cap G^*_j\cap H^*_j)
   +\mu(H^*\cap F_{\infty})\cr
\displaycause{by 455Ba}
&=\sum_{j=0}^n\mu(F_{t_j}\cap H^*)
   +\mu(F_{\infty}\cap H^*)
=\mu H^*.\cr}$$

Thus we have the formula we need when $E$ is of the special form
$\{\omega:\omega(t)\in E_t$ for every $t\in J\}$, $J\subseteq T$
being a finite set and $E_t$ being a member of $\Tau_t$ for every
$t\in J$.   By the Monotone Class Theorem (136B), we shall have
$\int\mu_{\omega,\tau(\omega)}(E)\mu(d\omega)=\mu E$ for every
$E\in\Tensorhat_{t\in T}\Tau_t$, so that
$\family{\omega}{\Omega}{\mu_{\omega,\tau(\omega)}}$ is a disintegration of $\mu$ over
itself.

\medskip

\quad{\bf (iii)} If $\tau$ takes infinitely many values, enumerate them as
$\sequencen{t_n}$, and for $n\in\Bbb N$ define
$\tau_n:T\to T\cup\{\infty\}$ by setting

$$\eqalign{\tau_n(\omega)
&=t_i\text{ if }i\le n\text{ and }\tau(\omega)=t_i,\cr
&=\infty\text{ if }\tau(\omega)\notin\{t_i:i\le n\}.\cr}$$

\noindent Then $\tau_n$ takes only finitely many values, and
$\{\omega:\tau_n(\omega)\le t\}\in\Tensorhat_{t\in T}\Tau_t$
is determined by coordinates in $[0,t]$ for every $t\in T$.
So we shall have

\Centerline{$\int\mu_{\omega,\tau_n(\omega)}(E)\mu(d\omega)=\mu E$}

\noindent for every $E\in\Tensorhat_{t\in T}\Tau_t$.   Now observe that
$\mu_{\omega,\tau_n(\omega)}=\mu_{\omega,\tau(\omega)}$ whenever
$\tau(\omega)=\tau_n(\omega)$.
So, for each $\omega$,
$\mu_{\omega,\tau_n(\omega)}=\mu_{\omega,\tau(\omega)}$ for all but
finitely many $n$.   This means that, for every
$E\in\Tensorhat_{t\in T}\Tau_t$,

\Centerline{$\mu_{\omega,\tau(\omega)}(E)
=\lim_{n\to\infty}\mu_{\omega,\tau_n(\omega)}(E)$}

\noindent for every $\omega\in\Omega$, and

\Centerline{$\int\mu_{\omega,\tau(\omega)}(E)\mu(d\omega)
=\lim_{n\to\infty}\int\mu_{\omega,\tau_n(\omega)}(E)\mu(d\omega)=\mu E$,}

\noindent as required.

\medskip

{\bf (b)(i)} Since $\{\omega:\tau(\omega)\le t\}$ is determined by
coordinates in $[t^*,t]$ for every $t\in T$, $\Omega\in\Sigma_{\tau}$, and it
is now elementary to confirm that $\Sigma_{\tau}$ is a $\sigma$-algebra.

\medskip

\quad{\bf (ii)} I had better note that $g_f$ is defined almost everywhere;
this is because, by (a) above and 452F,

\Centerline{$\int g_fd\mu=\iint fd\mu_{\omega,\tau(\omega)}\mu(d\omega)
=\int fd\mu$.}

\medskip

\quad{\bf (iii)} If $\omega$, $\omega'\in\Omega$ and
$\omega'\restr[t^*,\tau(\omega)]=\omega\restr[t^*,\tau(\omega)]$, then
$g_f(\omega)=g_f(\omega')$ if either is defined.   \Prf\ Since
$F_{\tau(\omega)}$ is determined by coordinates in $[t^*,\tau(\omega)]$,
$\tau(\omega')=\tau(\omega)$.   By 455B(b-ii),
$\mu_{\omega',\tau(\omega')}=\mu_{\omega,\tau(\omega)}$, so
$g_f(\omega)=g_f(\omega')$ if either is defined.\ \Qed

\medskip

\quad{\bf (iv)} If $F\in\Sigma_{\tau}$ and $\omega\in\Omega$, then
$\mu_{\omega,\tau(\omega)}F=1$ if $\omega\in F$, $0$ otherwise.   \Prf\
Setting $b=\tau(\omega)$, $F\cap F_b$ and $F_b\setminus F$
are determined by coordinates
in a countable subset of $T\cap[t^*,b]$, so by 455B(b-ii) we have
$\mu_{\omega b}F=1$ if $\omega\in F_b\cap F$ and
$\mu_{\omega b}(F_b\setminus F)=1$ if $\omega\in F_b\setminus F$.\ \Qed

It follows that if $f$ is $\mu$-integrable and $F\in\Sigma_{\tau}$, then
$g_{f\times\chi F}=g_f\times\chi F$.   \Prf\ If $\omega\in F$, then
$\mu_{\omega,\tau(\omega)}F=1$ and

\Centerline{$g_{f\times\chi F}(\omega)
=\int_Ffd\mu_{\omega,\tau(\omega)}=\int fd\mu_{\omega,\tau(\omega)}$;}

\noindent if $\omega\notin F$ then $\mu_{\omega,\tau(\omega)}F=0$ and

\Centerline{$g_{f\times\chi F}(\omega)
=\int_Ffd\mu_{\omega,\tau(\omega)}=0$.   \Qed}

\medskip

\quad{\bf (v)} Now let $f$ be any $\mu$-integrable real-valued function.
Then there
is a $\Sigma_{\tau}$-measurable function
$g'_f:\Omega\to\ocint{-\infty,\infty}$
such that $g'_f\eae g_f$, $g'_f(\omega)\le g_f(\omega)$ for every
$\omega\in\dom g_f$, and $g'_f(\omega)=-\infty$ for every
$\omega\in\Omega\setminus\dom g_f$.   \Prf\ For $q\in\Bbb Q$, set
$W_q=\{\omega:\omega\in\dom g_f$, $g_f(\omega)\ge q\}$.
For $q\in\Bbb Q$ and $b\in\tau[\Omega]$, consider $W_{bq}=W_q\cap F_b$.
$W_{bq}$ is measured by the completion $\hat\mu$ of $\mu$,
and is determined by coordinates in $T\cap[t^*,b]$, by (iii).
By 451K(b-ii) there is a
$W'_{bq}\in\Tensorhat_{t\in T}\Tau_t$ such that
$W'_{bq}\subseteq W_{bq}$, $W_{bq}\setminus W'_{bq}$ is negligible and
$W'_{bq}$ is determined by coordinates in $T\cap [t^*,b]$.

Having defined the family
$\langle W'_{bq}\rangle_{b\in\tau[\Omega],q\in\Bbb Q}$, set
$W'_q=\bigcup_{b\in\tau[\Omega]}W'_{bq}$ for $q\in\Bbb Q$.   Then
$W'_q\in\Tensorhat_{t\in T}\Tau_t$ and
$W'_q\cap F_b=W'_{bq}$ is determined by
coordinates in $T\cap[t^*,b]$ for every $b\in\tau[\Omega]$, so
$W'_q\in\Sigma_{\tau}$.   Also
$W'_q\subseteq W_q$ and $W_q\setminus W'_q$ is negligible.

Set

\Centerline{$g'_f(\omega)
=\sup\{q:q\in\Bbb Q$, $\omega\in W'_q\}$}

\noindent for $\omega\in\Omega$, counting $\sup\emptyset$ as
$-\infty$.   Then $g'_f$ is $\Sigma_{\tau}$-measurable,
$g'_f(\omega)=-\infty$ for $\omega\notin\dom g_f$,
$g'_f(\omega)\le g_f(\omega)$ for $\omega\in\dom g_f$, and
$g'_f=g_f$ on $\dom g_f\setminus\bigcup_{q\in\Bbb Q}W_q\setminus W'_q$, so
$g'_f\eae g_f$.\ \Qed

\medskip

\quad{\bf (vi)} Continuing from (v), we find that $g'_f$ is a conditional
expectation of $f$ on $\Sigma_{\tau}$.\footnote{The definition of
`conditional expectation' in 233D was directed towards real-valued
functions, and $g'_f$ is permitted to take the values $\pm\infty$.
So what I really mean here
is that the restriction of $g'_f$ to the set on which
it is finite is a conditional expectation of $f$.}
\Prf\ I have already shown that
$g'_f$ is $\Sigma_{\tau}$-measurable.   If $F\in\Sigma_{\tau}$ then

$$\eqalignno{\int_Fg'_fd\mu
&=\int g'_f\times\chi F\,d\mu
=\int g_f\times\chi F\,d\mu\cr
\displaycause{because $g_f\eae g'_f$}
&=\int g_{f\times\chi F}d\mu\cr
\displaycause{by (iv)}
&=\iint f\times\chi F\,d\mu_{\omega,\tau(\omega)}\mu(d\omega)
=\int f\times\chi F\,d\mu\cr
\displaycause{452F once again}
&=\int_Ffd\mu.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (vii)} Similarly, or applying the arguments of (v)-(vi) to $-f$,
we see that for any $\mu$-integrable function $f$ there is a conditional
expectation $g''_f$ of $f$ on $\Sigma_{\tau}$ such that
$g''_f(\omega)\ge g_f(\omega)$ when $\omega\in\dom g_f$ and
$g''_f(\omega)=\infty$ when $g_f(\omega)$ is undefined.   Now
$g'_f\eae g''_f$ and both are $\Sigma_{\tau}$-measurable.   It follows that
$g_f$ is defined, and equal to both $g'_f$ and $g''_f$,
$(\mu\restr\Sigma_{\tau})$-a.e.;  so that $g_f$ itself is also a
conditional expectation of $f$ on $\Sigma_{\tau}$.
}%end of proof of 455C

\cmmnt{
\leader{455D}{Remarks (a)}
The idea of the construction in 455A
is that $\family{t}{T}{X_t}$ is a family of random
variables, and that we start from the assurance that `history is
irrelevant';  if, at time $b$, we wish to make guesses about the
behaviour of $X_t$, the state of the system at a future time $t$, then
we expect that it will be useful to look at the current state $X_b$, but
once we know the value
of $X_b$ then any further information about $X_s$ for $s<b$ will
tell us nothing more about $X_t$.   We are given the {\bf transitional
probabilities}
$\nu_x^{(s,t)}$, which can be thought of as the conditional
distributions
of $X_t$ given that $X_s=x$.   The condition ($\dagger$) of 455A
is plainly
necessary if the system is going to make sense at all;  the content of
the theorem is that it is also sufficient, at least when all the
conditional expectations are perfect measures, to ensure
that the system as a whole can indeed be represented as a family of
random variables, in Kolmogorov's sense, on a suitable probability
space.

\spheader 455Db The statement
`$\family{\omega}{\Omega}{\mu_{\omega,\tau(\omega)}}$
is a disintegration of $\mu$ over itself' in 455Ca
is not obviously a target worth
working very hard for.   But the point of this particular family is that
not only does $\mu_{\omega,\tau(\omega)}$ follow $\omega$ up to
and including time $\tau(\omega)$ (455B(b-ii)), but also
$\mu_{\omega,\tau(\omega)}=\mu_{\omega',\tau(\omega')}$ whenever
$\omega'\restr[t^*,\tau(\omega)]=\omega\restr[t^*,\tau(\omega)]$, as noted
in (b-iii) of the proof of 455B.

If we take $\tau$ in 455C to be constant, with value $b\in T$, then we get
a precise description of what it means for `history to be irrelevant'.
In this case, we can take the measures $\mu_{\omega b}$, and
project them
onto $\prod_{t\ge b}\Omega_t$;  let
$\lambda^{(\omega)}_{\coint{b,\infty}}$
be the image measure.   Then it is easy to check that
$\lambda^{(\omega)}_{\coint{b,\infty}}$ is the measure defined from the
point $\omega(b)$ and the family
$\langle\nu^{(s,t)}_x\rangle_{b\le s<t,x\in\Omega_s}$ by the method of
455A;  so that
$\lambda^{(\omega)}_{\coint{b,\infty}}
=\lambda^{(\omega')}_{\coint{b,\infty}}$ whenever $\omega(b)=\omega'(b)$.

\spheader 455Dc I have called 455C a `theorem', and there are certainly
enough ideas in it to warrant the title.   But the restriction to stopping
times taking only countably many values means that we are a large step away
from a result which is really useful in continuous time.
The calculations with sets
$\{\omega:\tau(\omega)=b\}$ in the proofs of 455C and 455E
are a clear sign that we are not yet
ready for continuous stopping times, in which
$\{\omega:\tau(\omega)=b\}$ will usually be negligible for every $b$,
except perhaps $b=\infty$.   Of course we can
use 455C with $T=\Bbb N$;  but it must be obvious that there are better
and cleaner expressions of the result in this case.   In the work below,
455C is going to function as a lemma, the first stage in much stronger
results (starting with 455O)
which depend on special properties of the measures $\nu^{(s,t)}_x$.

\spheader 455Dd In the context of 455A, it seemed to involve
fewer explanations to take a fixed $\sigma$-algebra $\Tau_t$ for each $t$
and to define $\mu$ on
$\Tensorhat_{t\in T}\Tau_t$.   As you know, I ordinarily have a strong
prejudice in favour of completing measures.   In the situations most
important to us,
this is perfectly straightforward, if a touch laborious;  I present a
version in the next theorem.
}%end of comment

\vleader{72pt}{455E}{Theorem} Let $T$ be a totally ordered set with least
element $t^*$.   Let $\family{t}{T}{\Omega_t}$ be a family of Hausdorff
spaces;  suppose that we are given an $x^*\in\Omega_{t^*}$
and, for each
pair $s<t$ in $T$, a family $\family{x}{\Omega_s}{\nu^{(s,t)}_x}$ of
Radon probability measures on $\Omega_t$ such that

\inset{$\family{y}{\Omega_s}{\nu_y^{(t,u)}}$ is a disintegration of
$\nu_x^{(s,u)}$ over $\nu_x^{(s,t)}$
whenever $s<t<u$ in $T$ and $x\in\Omega_s$.}

\noindent Write $\Omega=\prod_{t\in T}\Omega_t$;
for $t\in T$ let
$\Cal B(\Omega_t)$ be the Borel $\sigma$-algebra of $\Omega_t$, and
$X_t:\Omega\to\Omega_t$ the canonical map;
for $J\subseteq T$ write $\pi_J$ for the canonical map from
$\Omega$ onto $\prod_{t\in J}\Omega_t$.   For $t\in T$ and $x\in\Omega_t$
let $\delta^{(t)}_x$ be the Dirac measure on $\Omega_t$ concentrated at
$x$.

(a) There is a unique complete probability measure
$\hat\mu$ on $\Omega$, inner regular with respect to
$\Tensorhat_{t\in T}\Cal B(\Omega_t)$,
such that, writing $\hat\lambda_J$ for the image measure
$\hat\mu\pi_J^{-1}$,

$$\eqalign{\int fd\hat\lambda_J
&=\int f(\omega(t^*),\omega(t_1),\ldots,\omega(t_n))\hat\mu(d\omega)\cr
&=\int\ldots\iint f(x^*,x_1,\ldots,x_n)
   \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\qquad\qquad\qquad\qquad
  \nu_{x_{n-2}}^{(t_{n-2},t_{n-1})}(dx_{n-1})
  \ldots\nu_{x^*}^{(t^*,t_1)}(dx_1)\cr}$$

\noindent
whenever $t^*<t_1<\ldots< t_n$ in $T$, $J=\{t^*,t_1,\ldots,t_n\}$ and
$f$ is $\hat\lambda_J$-integrable.
In particular, the image measure $\hat\mu X_t^{-1}$ is equal
to $\nu_{x^*}^{(t^*,t)}$ if $t>t^*$, and to $\delta^{(t^*)}_{x^*}$ if
$t=t^*$.

(b)(i) For $\omega\in\Omega$ and $a\in T\cup\{\infty\}$ define
$\langle\nu^{(s,t)}_{\omega ax}\rangle_{s<t,x\in\Omega_s}$
by setting

$$\eqalign{\nu^{(s,t)}_{\omega ax}
&=\nu_x^{(s,t)}\text{ if }a<s,\cr
&=\nu_{\omega(a)}^{(a,t)}\text{ if }s\le a<t,\cr
&=\delta^{(t)}_{\omega(t)}\text{ if }t\le a.\cr}$$

\noindent The family
$\langle\nu^{(s,t)}_{\omega ax}\rangle_{s<t,x\in\Omega_s}$,
together with the point $\omega(t^*)\in\Omega_{t^*}$,
satisfy the conditions of (a), so can
be used to define a complete measure $\hat\mu_{\omega a}$ on $\Omega$.

\quad(ii) If $\omega\in\Omega$ and $D\subseteq T\cap[t^*,a]$
is countable, then
$\hat\mu_{\omega a}\{\omega':\omega'\restr D=\omega\restr D\}=1$.

\quad(iii) If $\omega$, $\omega'\in\Omega$ and
$\omega'\restr[t^*,a]=\omega\restr[t^*,a]$, then
$\hat\mu_{\omega' a}=\hat\mu_{\omega a}$.

(c) Let $\Sigma$ be the domain of $\hat\mu$.
Suppose that $\tau:\Omega\to T\cup\{\infty\}$ is a
function taking countably many
values and such that $\{\omega:\tau(\omega)\le t\}$ belongs to
$\Sigma$ and is determined by
coordinates in $[t^*,t]$ for every $t\in T$.

\quad(i) $\family{\omega}{\Omega}{\hat\mu_{\omega,\tau(\omega)}}$
is a disintegration of $\hat\mu$ over itself.

\quad(ii) Let $\Sigma_{\tau}$ be the set

$$\eqalign{\{E:E\in\Sigma&
,\,E\cap\{\omega:\tau(\omega)\le t\}\text{ is determined by
coordinates in }[t^*,t]\cr&
\mskip220mu\text{ for every }t\in T\}.\cr}$$

\noindent Then $\Sigma_{\tau}$ is a $\sigma$-subalgebra of
$\Sigma$.   If
$f$ is any $\hat\mu$-integrable real-valued function, and we set
$g_f(\omega)=\int fd\hat\mu_{\omega,\tau(\omega)}$ when this is defined
in $\Bbb R$, then $g_f$ is a
conditional expectation of $f$ on $\Sigma_{\tau}$.

\proof{ My aim is to apply 455A-455C %455A 455B 455C
to the Borel measures
$\grave\nu_x^{(s,t)}=\nu_x^{(s,t)}\restr\Cal B(\Omega_t)$, and
take $\hat\mu$ to be the completion of the Baire measure
$\mu$ produced by the
method of 455A.   The essential discipline is to check carefully that
almost every measure $\zeta$ is
the completion of an appropriate measure $\grave\zeta$.

\medskip

{\bf (a)} At the start, every Radon probability measure is the completion
of the corresponding Borel measure, so that the
$\nu^{(s,t)}_x$ are indeed the completions of the
$\grave\nu^{(s,t)}_x$ defined from them.    Since completing a measure does not
affect the associated integration (212Fb), the condition

\inset{whenever $s<t<u$ in $T$, $x\in\Omega_s$ and
$E\subseteq\Omega_u$ is a Borel
set, then $\grave\nu_s^{(s,u)}(E)=\int\grave\nu_y^{(t,u)}(E)\grave\nu_x^{(s,t)}(dy)$}

\noindent follows at once from

\inset{whenever $s<t<u$ in $T$ and $x\in\Omega_s$,
then $\family{y}{\Omega_s}{\nu_y^{(t,u)}}$ is a disintegration of
$\nu_x^{(s,u)}$ over $\nu_x^{(s,t)}$.}

\noindent Also the $\grave\nu_x^{(s,t)}$, being tight Borel measures,
are all perfect (342L/451C).
So we can indeed form a measure $\mu$ on
$\Omega$ with domain $\Tensorhat_{t\in T}\Cal B(\Omega_t)$
by the process in 455A, and complete it.

The next step has a little more content in it:  I need to show that for any
$J\subseteq T$, the image measure $\hat\mu\pi_J^{-1}$ on
$\prod_{t\in J}\Omega_t$ is the completion of the image measure
$\mu\pi_J^{-1}$.   But here we just have to recall that $\mu$ is perfect
(454D), so that we can use 451Kb.   For finite $J\subseteq T$
we can therefore write
$\hat\lambda_J$ indifferently for the completion of
$\lambda_J=\mu\pi_J^{-1}$ and for $\hat\mu\pi_J^{-1}$,
and the formula for $\int fd\hat\lambda_J$ can be read off
from 455A, since it deals only with integrals, which are unaffected by
completions.

\medskip

{\bf (b)} This follows 455Bb.   This time we must start by noting that
every $\nu^{(s,t)}_{\omega ax}$ is a Radon probability measure.

\medskip

\quad{\bf (i)} The formulae of part (i) of the proof of 455Bb can still be
applied to show that

\Centerline{$\int_{\Omega_t}\nu^{(t,u)}_{\omega ay}(E)
   \nu^{(s,t)}_{\omega ax}(dy)
=\nu_{\omega ax}^{(s,u)}(E)$}

\noindent whenever $s<t<u$, $x\in\Omega_s$ and $E\in\Cal B(\Omega_u)$.
Since any set measured by $\nu_{\omega ax}^{(s,u)}$ can be
approximated internally and externally by Borel sets, we see that
$\family{y}{\Omega_t}{\nu^{(t,u)}_{\omega ay}}$ is a
disintegration of $\nu^{(s,u)}_{\omega ax}$ over
$\nu^{(s,t)}_{\omega ax}$.   (Cf.\ 452Xg.)

\medskip

\quad{\bf (ii)} Similarly, the argument of part (ii) of the proof of 455Bb
can still be used to show that whenever $\omega\in\Omega$ and
$D\subseteq T\cap[t^*,a]$ is countable, then
$\omega'\restr D=\omega\restr D$ for
$\hat\mu_{\omega a}$-almost every $\omega'\in\Omega$.

\medskip

\quad{\bf (iii)} Once again, we can use the argument
from 455B;  if $\omega'\restr[t^*,a]=\omega\restr[t^*,a]$, then
$\nu^{(s,t)}_{\omega' ax}=\nu^{(s,t)}_{\omega ax}$ for all
$x$, $s$ and $t$, and $\hat\mu_{\omega'a}=\hat\mu_{\omega a}$.

\medskip

{\bf (c)(i)}\grheada\
The key step here is to observe that there is a function
$\grave\tau:\Omega\to T\cup\{\infty\}$ which satisfies the properties
required in 455C and is equal $\hat\mu$-almost everywhere to $\tau$.
\Prf\
For each $a\in T\cap\tau[\Omega]$, $F_a=\tau^{-1}[\{a\}]$ belongs to
$\Sigma$ and is determined by coordinates in $[t^*,a]$.
By 451K(b-ii) again, there is an
$F'_a\in\Tensorhat_{t\in T}\Cal B(\Omega_t)$ such that
$F'_a\subseteq F_a$, $F'_a$ is determined by coordinates in $[t^*,a]$ and
$\hat\mu(F_a\setminus F'_a)=0$.   Define $\grave\tau$ by setting

$$\eqalign{\grave\tau(\omega)
&=a\text{ if }a\in T\cap\tau[\Omega]\text{ and }\omega\in F'_a,\cr
&=\infty
  \text{ if }\omega\in\Omega\setminus\bigcup_{a\in T\cap\tau[\Omega]}F'_a.
\cr}$$

\noindent It is easy to check that this $\grave\tau$ will serve.\ \Qed

\medskip

\qquad\grheadb\ For $\omega\in\Omega$ and $a\in T$, define
$\langle\grave\nu^{(s,t)}_{\omega ax}
\rangle_{s<t,x\in\Omega_t}$ and
$\family{\omega}{\Omega}{\mu_{\omega a}}$
from $\langle\grave\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_t}$ and
$\grave\tau$ as in 455Bb.   If $\tau(\omega)=\grave\tau(\omega)$ then
$\grave\nu^{(s,t)}_{\omega,\grave\tau(\omega),x}
=\nu^{(s,t)}_{\omega,\tau(\omega),x}\restr\Cal B(\Omega_t)$
for all $s$, $t$ and $x$, so that
$\hat\mu_{\omega,\tau(\omega)}$ is the completion of
$\mu_{\omega,\grave\tau(\omega)}$.
This is true for almost all $\omega$.   Now we know from 455Ca that
$\family{\omega}{\Omega}{\mu_{\omega,\grave\tau(\omega)}}$
is a disintegration of $\mu$ over itself,
and therefore also over $\hat\mu$.   It follows that
$\family{\omega}{\Omega}{\hat\mu_{\omega,\grave\tau(\omega)}}$ is a
disintegration of $\hat\mu$ over $\hat\mu$, by 452B(ii).   But
$\hat\mu_{\omega,\grave\tau(\omega)}=\hat\mu_{\omega,\tau(\omega)}$ for
$\hat\mu$-almost every
$\omega$, so $\family{\omega}{\Omega}{\hat\mu_{\omega,\tau(\omega)}}$
also is a disintegration of $\hat\mu$ over itself.

\medskip

\quad{\bf (ii)}\grheada\ Just as in part (b-i) of the proof of 455C,
$\Sigma_{\tau}$ is a $\sigma$-algebra because it contains $\Omega$.

\medskip

\qquad\grheadb\ Recall the $F_a$, $F'_a$ in (i-$\alpha$) above.
Set $F_{\infty}=\tau^{-1}[\{\infty\}]$, and take
$F'_{\infty}\in\Tensorhat_{t\in T}\Cal B(\Omega_t)$ such that
$F'_{\infty}\subseteq F_{\infty}$ and $F_{\infty}\setminus F'_{\infty}$ is
negligible.   Then $F^*=\bigcup_{a\in\tau[\Omega]}F'_a$ is conegligible in
$\Omega$.   Write $\grave\Sigma_{\grave\tau}$ for the
set of those $F\in\Tensorhat_{t\in T}\Cal B(\Omega_t)$ such that
$F\cap F'_a$ is determined by coordinates in $[t^*,a]$ for every
$a\in T\cap\grave\tau[\Omega]$.   Then
$F^*\in\grave\Sigma_{\grave\tau}\cap\Sigma_{\tau}$ because
$F^*\cap F'_a=F^*\cap F_a=F'_a$ for every $a\in\tau[\Omega]$.
In fact we have more.   First, $\tau\restr F^*=\grave\tau\restr F^*$.
Next, if $F\subseteq F^*$ and $F\in\grave\Sigma_{\grave\tau}$, then
$F\in\Sigma_{\tau}$.   \Prf\ For any $a\in T\cap\tau[\Omega]$,
$F\cap F_a=F\cap F'_a$ belongs to
$\Tensorhat_{t\in T}\Cal B(\Omega_t)\subseteq\Sigma$
and is determined by coordinates in $[t^*,a]$.\ \QeD\  And thirdly, if
$F\in\Sigma_{\tau}$, there is a
$G\in\grave\Sigma_{\grave\tau}$ such that $G\subseteq F$ and
$F\setminus G$ is negligible.   \Prf\ As in (iv-$\alpha$), we can find for
each $a\in\tau[\Omega]$ a set $G_a\in\Tensorhat_{t\in T}\Cal B(\Omega_t)$,
determined by coordinates in $T\cap[t^*,a]$, such that
$G_a\subseteq F\cap F_a$ and
$(F\cap F_a)\setminus G_a$ is negligible.   Set
$G=\bigcup_{a\in\tau[\Omega]}G_a$.\ \Qed

\medskip

\qquad\grheadc\ Now take a $\hat\mu$-integrable function $f$.   Then it is
$\mu$-integrable.   By 455Cb,
$\grave g_f$ is a conditional expectation of $f$ on
$\grave\Sigma_{\grave\tau}$, where

\Centerline{$\grave g_f(\omega)
=\int fd\mu_{\omega,\grave\tau(\omega)}
=\int fd\hat\mu_{\omega,\grave\tau(\omega)}$}

\noindent whenever the integral is defined in $\Bbb R$.
We know that there is a $\grave\Sigma_{\grave\tau}$-measurable
function $g':\Omega\to\Bbb R$ equal to
$\grave g_f$ except perhaps on a negligible set $H$ belonging to
$\grave\Sigma_{\grave\tau}$.    Replacing $g'$ by $g'\times\chi F^*$
and $H$ by $H\cup(\Omega\setminus F^*)$ if
necessary, we can suppose that $g'$ is zero outside $F^*$
and that $\Omega\setminus H\subseteq F^*$.   In this case,
$g'$ is $\Sigma_{\tau}$-measurable.   \Prf\ For any $\alpha\in\Bbb R$,

$$\eqalign{\{\omega:g'(\omega)\ge\alpha\}
&=\{\omega:\omega\in F^*,\,g'(\omega)\ge\alpha\}\cup(\Omega\setminus F^*)
   \text{ if }\alpha\le 0,\cr
&=\{\omega:\omega\in F^*,\,g'(\omega)\ge\alpha\}
   \text{ if }\alpha>0,\cr}$$

\noindent and in either case belongs to $\Sigma_{\tau}$, by
($\beta$).\ \QeD\  At the same time, we note that $H\in\Sigma_{\tau}$.

If $\omega\in\Omega\setminus H$, then $\omega\in F^*$,
$\tau(\omega)=\grave\tau(\omega)$,
$\hat\mu_{\omega,\tau(\omega)}=\hat\mu_{\omega,\grave\tau(\omega)}$ and

\Centerline{$g_f(\omega)=\int fd\hat\mu_{\omega,\tau(\omega)}
=\int fd\hat\mu_{\omega,\grave\tau(\omega)}
=\int fd\mu_{\omega,\grave\tau(\omega)}
=\grave g_f(\omega)=g'(\omega)$.}

\noindent So $g_f$ is defined and equal to $g'$ and $\grave g_f$
except perhaps on the
negligible set $H$ belonging to $\Sigma_{\tau}$;  consequently $g_f$ is
defined $(\hat\mu\restr\Sigma_{\tau})$-a.e.\ and is
$(\hat\mu\restr\Sigma_{\tau})$-virtually measurable.

If $F\in\Sigma_{\tau}$,
there is a $G\in\grave\Sigma_{\grave\tau}$ such
that $G\subseteq F$ and $F\setminus G$ is negligible,
by the last remark in ($\beta$).   So

$$\eqalignno{\int_Ffd\hat\mu
&=\int_Gfd\mu
=\int_G\grave g_fd\mu\cr
\displaycause{because $\grave g_f$ is a conditional expectation of $f$ on
$\grave\Sigma_{\grave\tau}$}
&=\int_F\grave g_fd\hat\mu
=\int_Fg_fd\hat\mu
=\int_Fg_fd(\hat\mu\restr\Sigma_{\tau}).\cr}$$

\noindent As $F$ is arbitrary, $g_f$ is a conditional expectation of
$f$ on $\Sigma_{\tau}$, and the proof is complete.
}%end of proof of 455E

\leader{455F}{}\cmmnt{ Of course the leading example for the work above
is the case in which $T=\coint{0,\infty}$ and $\Omega_t=\Bbb R$ for every
$t\ge 0$.   Moving towards this, a natural intermediate stage is when
$T=\coint{0,\infty}$ and all the $\Omega_t$ are the same, so that we can
regard an element of $\prod_{t\in T}\Omega_t$ as the path of a moving
point.   In this case we can begin to think about paths which are more or
less continuous.   The next theorem gives a widely applicable condition for
existence of many paths which are one-sidedly continuous.
It depends on a fairly strong continuity property for the transitional
probabilities.

\medskip

\noindent}{\bf Definitions (a)} Let $U$ be a Hausdorff space and
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ a family of Radon
probability measures on $U$.   I will say that
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is
{\bf narrowly continuous} if
it is continuous, as a function from $\{(s,t):0\le s<t\}\times U$
to the set of
Radon probability measures on $U$, when the latter is given its narrow
topology\cmmnt{ (437Jd)}.

\cmmnt{\medskip

\noindent{\bf Remark} I speak of the `narrow' topology here partly
because, in the present treatise, this has become the standard topology on
spaces of Radon measures, and partly because the phrase `vaguely
continuous' seems inappropriate.   But, as will appear, all the results
below will rely on the fact that the vague topology (437Jc)
is coarser than the narrow topology.   In the present context, in which we
have Radon measures on a completely regular Hausdorff space, the two
topologies actually coincide (437L).   So
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is narrowly continuous iff
$(s,t,x)\mapsto\int fd\nu^{(s,t)}_x$ is continuous for every bounded
continuous $f:\Omega\to\Bbb R$.}

\medskip

{\bf (b)} Let $(U,\rho)$ be a metric
space, and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ a family of
Radon probability measures on $U$.   I will say that
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is
{\bf uniformly time-continuous on the right} if for every $\epsilon>0$
there is a
$\delta>0$ such that $\nu^{(s,t)}_xB(x,\epsilon)\ge 1-\epsilon$ whenever
$x\in U$ and $0\le s<t\le s+\delta$.

%slightly stronger than "Feller process", which means I think "uniformly
%time-cts on the right on cpct sets",
%but used only for locally compact spaces

\leader{455G}{Theorem} Let $(U,\rho)$ be a complete metric
space and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ a family of
Radon probability measures on $U$, uniformly time-continuous on the right,
such that $\family{y}{U}{\nu_y^{(t,u)}}$ is a disintegration of
$\nu_x^{(s,u)}$ over $\nu_x^{(s,t)}$ whenever $0\le s<t<u$ and $x\in U$.
Take a point $\tilde\omega$ in $\Omega=U^{\coint{0,\infty}}$, and
$a\in[0,\infty]$.   Let $\hat\mu_{\tilde\omega a}$ be the completed
probability measure on $\Omega$ defined from
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$, $\tilde\omega$ and $a$
as in 455Eb.

(a) For $\hat\mu_{\tilde\omega a}$-almost every $\omega\in\Omega$,
$\lim_{q\in\Bbb Q,q\downarrow t}\omega(q)$ and
$\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)$ are defined in $U$ for every
$t>a$.

(b)(i) If $a\le t<\infty$, then
$\omega(t)=\lim_{q\in\Bbb Q,q\downarrow t}\omega(q)$
for $\hat\mu_{\tilde\omega a}$-almost every $\omega\in\Omega$.

\quad(ii) If $a<t<\infty$, then
$\omega(t)=\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)$
for $\hat\mu_{\tilde\omega a}$-almost every $\omega\in\Omega$.

(c)(i) Let $\Clll$ be the set of \callal\ functions from
$\coint{0,\infty}$ to $U$\cmmnt{ (438S)}.
If $\tilde\omega\in\Clll$,
$\Clll$ has full outer measure for $\hat\mu_{\tilde\omega a}$.

\quad(ii) Let $\Cdlg$ be the set of \cadlag\ functions from
$\coint{0,\infty}$ to $U$.   If $\tilde\omega\in\Cdlg$,
$\Cdlg$ has full outer measure for $\hat\mu_{\tilde\omega a}$.

\cmmnt{\medskip

\noindent{\bf Remark} In this result and the ones
to follow, I have not spelt
out separately what it means if $a=0$;  but of course this is the case in
which we are starting the process at time $t^*=0$ and value
$x^*=\tilde\omega(0)$,
just as in the original construction 455A.
}

\proof{{\bf (a)} Of course we can assume in this part of the proof that $a$
is finite.

\medskip

\quad{\bf (i)} Suppose that $\eta\in\ooint{0,1}$ and $\epsilon$, $\delta>0$ are
such that $\nu^{(s,t)}_xB(x,\epsilon)\ge 1-\eta$ whenever
$x\in U$ and $0\le s<t\le s+\delta$.   Then

\Centerline{$\hat\mu_{\tilde\omega a}\{\omega:\omega\in\Omega,\,
\diam\omega[D]\le 4\epsilon\}\ge\Bover{1-2\eta}{1-\eta}$}

\noindent whenever $D\subseteq\coint{a,\infty}$ is a
countable set of diameter at most $\delta$.

\medskip

\Prf\ \grheada\ For finite $D$, I seek to induce on $\#(D)$.
If $\#(D)\le 1$ then of course
$\diam\omega[D]\le 4\epsilon$ for every $\omega$ and we can stop.   So
suppose that $D=\{t_0,\ldots,t_n\}$ where $n\ge 1$ and
$a\le t_0<\ldots<t_n$.    To begin with, I go through the formulae when
$t_0>0$.

For $k\le n$ set

\Centerline{$E_k=\{\omega:\rho(\omega(t_k),\omega(t_0))>2\epsilon$,
$\rho(\omega(t_i),\omega(t_0))\le 2\epsilon$ for $i<k\}$,}

\Centerline{$F_k=\{\omega:\omega\in E_k$,
$\rho(\omega(t_n),\omega(t_k))\le\epsilon\}$,}

\Centerline{$G_k=\{(x_0,\ldots,x_k):\rho(x_k,x_0)>2\epsilon$,
$\rho(x_i,x_0)\le 2\epsilon$ for $i<k\}\subseteq U^{k+1}$.}

\noindent If $1\le k<n$ then

$$\eqalignno{\hat\mu_{\tilde\omega a}F_k
&=\lambda_{\{0,t_0,\ldots,t_k,t_n\}}\{(x,x_0,\ldots,x_k,x_n):
\rho(x_i,x_0)\le 2\epsilon\text{ for }i<k,\cr
&\mskip300mu
  \rho(x_0,x_k)>2\epsilon,\,\rho(x_k,x_n)\le\epsilon\}\cr
\displaycause{defining $\lambda_J$ as the image measure of
$\hat\mu_{\tilde\omega a}$ on $U^J$, as in
455E}
&=\int\ldots\int\chi G_k(x_0,\ldots,x_k)\chi B(x_k,\epsilon)(x_n)
\nu^{(t_k,t_n)}_{\tilde\omega ax_k}(dx_n)
\ldots\nu^{(0,t_0)}_{\tilde\omega(0)}(dx_0)\cr
\displaycause{were $\nu^{(s,t)}_{\tilde\omega ax}$ is defined as in 455Eb}
&=\int\ldots\int\chi G_k(x_0,\ldots,x_k)\chi B(x_k,\epsilon)(x_n)
\nu^{(t_k,t_n)}_{x_k}(dx_n)
\ldots\nu^{(0,t_0)}_{\tilde\omega(0)}(dx_0)\cr
\displaycause{because $a\le t_0<\ldots<t_n$}
&=\int\ldots\int\chi G_k(x_0,\ldots,x_k)
\nu^{(t_k,t_n)}_{x_k}(B(x_k,\epsilon))\nu^{(t_{k-1},t_k)}_{x_{k-1}}(dx_k)
\ldots\nu^{(0,t_0)}_{\tilde\omega(0)}(dx_0)\cr
&\ge\int\ldots\int(1-\eta)\chi G_k(x_0,\ldots,x_k)
\nu^{(t_{k-1},t_k)}_{x_{k-1}}(dx_k)
\ldots\nu^{(0,t_0)}_{\tilde\omega(0)}(dx_0)\cr
\displaycause{because $t_k<t_n\le t_k+\delta$,
so $\nu^{(t_k,t_n)}_xB(x,\epsilon)\ge 1-\eta$ for every $x$}
&=(1-\eta)\lambda_{\{0,t_0,\ldots,t_k\}}
\{(x,x_0,\ldots,x_k,x_n):
\rho(x_i,x_0)\le 2\epsilon\text{ for }i<k,\cr
&\mskip420mu \rho(x_0,x_k)>2\epsilon\}\cr
&=(1-\eta)\hat\mu_{\tilde\omega a} E_k.\cr}$$

\noindent If $k=n$, then of course $F_k=E_k$, so again
$\hat\mu_{\tilde\omega a} F_k\ge(1-\eta)\hat\mu_{\tilde\omega a} E_k$.
Accordingly

$$\eqalign{(1-\eta)\sum_{k=1}^n\hat\mu_{\tilde\omega a} E_k
&\le\sum_{k=1}^n\hat\mu_{\tilde\omega a} F_k
\le\hat\mu_{\tilde\omega a}\{\omega:\rho(\omega(t_n),\omega(t_0))>\epsilon\}\cr
&=\lambda_{\{0,t_0,t_n\}}\{(x,x_0,x_n):\rho(x_0,x_n)>\epsilon\}\cr
&=\int\nu^{(t_0,t_n)}_{x_0}(U\setminus B(x_0,\epsilon))
   \nu^{(0,t_0)}_{\tilde\omega(0)}(dx_0)
\le\eta\cr}$$

\noindent because $t_n-t_0\le\delta$ so
$\nu^{(t_0,t_n)}_x(U\setminus B(x,\epsilon))\le\eta$ for every $x$.

But now we have

$$\eqalign{\hat\mu_{\tilde\omega a}\{\omega:\omega\in\Omega,\,\diam\omega[D]\le 4\epsilon\}
&\ge\hat\mu_{\tilde\omega a}\{\omega:\rho(\omega(t_k),\omega(t_0))\le 2\epsilon
 \text{ for }1\le k\le n\}\cr
&=\hat\mu_{\tilde\omega a}(\Omega\setminus\bigcup_{1\le k\le n}E_k)
\ge 1-\Bover{\eta}{1-\eta}
=\Bover{1-2\eta}{1-\eta}\cr}$$

\noindent as required.

\medskip

\qquad\grheadb\ If $t_0=a=0$ the formulae simplify slightly, but the ideas
are the same.   We have $\omega(0)=\tilde\omega(0)$ for
$\hat\mu_{\tilde\omega 0}$-almost every $\omega$, so

$$\eqalign{\hat\mu_{\tilde\omega 0}F_k
&=\lambda_{0,t_1,\ldots,t_k,t_n}
   \{(\tilde\omega(0),x_1,\ldots,x_k,x_n):
      \rho(x_i,\tilde\omega(0))\le 2\epsilon\text{ for }i<k,\cr
&\mskip300mu
      \rho(\tilde\omega(0),x_k)>2\epsilon,\,\rho(x_k,x_n)\le\epsilon\}\cr
&=\int\ldots\int\chi G_k(\tilde\omega(0),x_1,\ldots,x_k)
   \chi B(x_k,\epsilon)(x_n)\nu^{(t_k,t_n)}_{\tilde\omega ax_k}(dx_n)
   \ldots\nu^{(0,t_1)}_{\tilde\omega(0)}(dx_1)\cr
&\ge(1-\eta)\int\ldots\int\chi G_k(\tilde\omega(0),x_1,\ldots,x_k)
   \nu^{(t_{k-1},t_k)}_{\tilde\omega ax_{k-1}}(dx_k)
   \ldots\nu^{(0,t_1)}_{\tilde\omega(0)}(dx_1)\cr
&=(1-\eta)\hat\mu_{\tilde\omega 0}E_k\cr}$$

\noindent for $1\le k<n$,

$$\eqalign{(1-\eta)\sum_{k=1}^n\hat\mu_{\tilde\omega 0}E_k
&\le\lambda_{\{0,t_n\}}\{(\tilde\omega(0),x_n):
   \rho(\tilde\omega(0),x_n)>\epsilon\}\cr
&=\int\nu^{(t_0,t_n)}_{\tilde\omega(0)}
   (U\setminus B(\tilde\omega(0),\epsilon))
\le\eta,\cr}$$

\noindent and the final calculation is unchanged.

\medskip

\qquad\grheadc\ For countably infinite $D$, let $\sequencen{I_n}$ be a
non-decreasing sequence of finite sets with union $D$;  then
$\sequencen{\{\omega:\omega\in\Omega,\,\diam\omega[I_n]\le 4\epsilon\}}$
is a non-increasing sequence with intersection
$\{\omega:\omega\in\Omega,\,\diam\omega[D]\le 4\epsilon\}$, so the measure
of the limit is the limit of the measures, and is at most
$\Bover{1-2\eta}{1-\eta}$.\ \Qed

\wheader{455G}{4}{2}{2}{36pt}

\quad{\bf (ii)}
For $m\in\Bbb N$, $\epsilon>0$ and $A\subseteq\coint{0,\infty}$
let $G(A,\epsilon,m)$ be

$$\eqalign{\{\omega:\omega\in\Omega,
  \text{ there are }&s_0<s'_0\le s_1<s'_1\le\ldots\le s_m\le s'_m
\text{ in }A\cr
&\text{ such that }\rho(\omega(s'_i),\omega(s_i))>4\epsilon
\text{ for every }i\le m\}.\cr}$$

\noindent Let $\delta>0$ be such that
$\nu_x^{(s,t)}B(x,\epsilon)\ge\bover45$
whenever $x\in U$ and $s<t\le s+\delta$.
Then $\hat\mu_{\tilde\omega a} G(D,\epsilon,m)\le 2^{-m}$
whenever $m\in\Bbb N$ and
$D\subseteq\coint{a,\infty}$ is a countable
set of diameter at most $\delta$.

\medskip

\Prf\ \grheada\ As in (i), first consider finite $D$.
For these, we can induce on $m$.   If $m=0$ then
$G(D,\epsilon,0)=\{\omega:\diam\omega[D]>4\epsilon\}$ so
(i), with $\eta=\bover15$, tells us that
$\hat\mu_{\tilde\omega a}G(D,\epsilon,0)\le\Bover{2\eta}{1-\eta}
=\Bover12$.   For the
inductive step to $m+1$, define $\tau:\Omega\to[0,\infty]$ by setting

$$\eqalign{\tau(\omega)
&=\min\{t:t\in D,\,\omega\in G(D\cap[a,t],\epsilon,m)\}
\text{ if }\omega\in G(D,\epsilon,m),\cr
&=\infty\text{ otherwise}.\cr}$$

\noindent Then $\tau$ takes only finitely many values, all strictly
greater than $a$, and $\{\omega:\tau(\omega)=t\}$ belongs to
$\Tensorhat_{\coint{0,\infty}}\Cal B(U)
=\Tensorhat_{t\in\coint{0,\infty}}\Cal B(U)$
and is determined by coordinates in $[0,t]$ for every $t\ge 0$.
We can therefore apply 455E(b)-(c).

For each $\omega\in\Omega$, define
$\langle\nu^{(s,t)}_{\omega,\tau(\omega),x}\rangle_{s<t,x\in U}$
from $\langle\nu^{(s,t)}_x\rangle_{s<t,x\in U}$ as in 455Eb;  let
$\langle\tilde\nu^{(s,t)}_{\omega,\tau(\omega),x}\rangle_{s<t,x\in U}$
be the family defined in the same way from
$\langle\nu^{(s,t)}_{\tilde\omega ax}\rangle_{s<t,x\in U}$.
Let $\hat\mu_{\omega,\tau(\omega)}$ be defined from
$\omega(0)$ and
$\langle\nu^{(s,t)}_{\omega,\tau(\omega),x}\rangle_{s<t,x\in U}$,
and $\hat\mu'_{\omega,\tau(\omega)}$ from
$\omega(0)$ and
$\langle\tilde\nu^{(s,t)}_{\omega,\tau(\omega),x}\rangle_{s<t,x\in U}$,
again as in 455Eb.   Then 455E(c-i) tells us that
$\family{\omega}{\Omega}{\hat\mu'_{\omega,\tau(\omega)}}$ is a
disintegration of $\hat\mu_{\tilde\omega a}$ over itself.   But now
observe that, for any $\omega\in\Omega$ and $x\in U$,

$$\eqalignno{\nu^{(s,t)}_{\omega,\tau(\omega),x}
&=\nu^{(s,t)}_x=\nu^{(s,t)}_{\tilde\omega ax}
   =\tilde\nu^{(s,t)}_{\omega,\tau(\omega),x}
   \text{ if }\tau(\omega)<s<t,\cr
\displaycause{because $a<\tau(\omega)$}
&=\nu^{(\tau(\omega),t)}_{\omega(\tau(\omega))}
  =\nu^{(\tau(\omega),t)}_{\tilde\omega,a,\omega(\tau(\omega))}
  =\tilde\nu^{(s,t)}_{\omega,\tau(\omega),x}
  \text{ if }s\le\tau(\omega)<t,\cr
&=\delta^{(t)}_{\omega(t)}
   =\tilde\nu^{(s,t)}_{\omega,\tau(\omega),x}
  \text{ if }s<t\le\tau(\omega),\cr}$$

\noindent so
$\hat\mu_{\omega,\tau(\omega)}=\hat\mu'_{\omega,\tau(\omega)}$.
Accordingly
$\family{\omega}{\Omega}{\hat\mu_{\omega,\tau(\omega)}}$ is a
disintegration of $\hat\mu_{\tilde\omega a}$ over itself.

Now, for $\omega\in\Omega$, consider

\Centerline{$H_{\omega}=\{\omega':\omega'\in G(D,\epsilon,m+1)$,
  $\omega'\restr D\cap[0,\tau(\omega)]
  =\omega\restr D\cap[0,\tau(\omega)]\}$.}

\noindent If $\omega\notin G(D,\epsilon,m)$ then $\tau(\omega)=\infty$
and $H_{\omega}=\emptyset$, because
$G(D,\epsilon,m+1)\subseteq G(D,\epsilon,m)$ are determined by coordinates
in $D$.   If
$\omega\in G(D,\epsilon,m)$ and $\tau(\omega)=b$, then

\Centerline{$H_{\omega}
=\{\omega':\omega'\restr D\cap[0,b]=\omega\restr D\cap[0,b]
  \text{ and }\diam(\omega'[D\cap\coint{b,\infty}\,])>4\epsilon\}$,}

\noindent so that $\hat\mu_{\omega,\tau(\omega)}H_{\omega}\le\bover12$ by
(i), again with $\eta=\bover15$.

So

$$\eqalignno{\hat\mu_{\tilde\omega a} G(D,\epsilon,m+1)
&=\int\hat\mu_{\omega,\tau(\omega)}G(D,\epsilon,m+1)
   \hat\mu_{\tilde\omega a}(d\omega)
=\int\hat\mu_{\omega,\tau(\omega)}H_{\omega}
   \hat\mu_{\tilde\omega a}(d\omega)\cr
\displaycause{using 455E(b-ii)}
&=\int_{G(D,\epsilon,m)}\hat\mu_{\omega,\tau(\omega)}H_{\omega}
   \hat\mu_{\tilde\omega a}(d\omega)
\le\Bover12\hat\mu_{\tilde\omega a}G(D,\epsilon,m)
\le 2^{-m-1}\cr}$$

\noindent by the inductive hypothesis.   Thus the induction proceeds.

\medskip

\qquad\grheadb\ Now, for countably infinite $D$, again express $D$ as the
union of a non-decreasing sequence $\sequencen{I_n}$ of finite sets, and
observe that $\sequencen{G(I_n,\epsilon,m)}$ is a non-decreasing sequence
with union $G(D,\epsilon,m)$;  so

\Centerline{$\hat\mu_{\tilde\omega a} G(D,\epsilon,m)
=\lim_{n\to\infty}\hat\mu_{\tilde\omega a} G(I_n,\epsilon,m)\le 2^{-m}$}

\noindent for every $m\in\Bbb N$.\ \Qed

\medskip

\quad{\bf (iii)} For $n\in\Bbb N$, let $\delta_n>0$ be such that
$\nu_x^{(s,t)}B(x,2^{-n})\ge\bover45$ whenever $x\in U$ and
$s<t\le s+\delta_n$.   Consider the set

\Centerline{$E=\bigcup_{n,k\in\Bbb N}\bigcap_{m\in\Bbb N}
G(\Bbb Q\cap[a+k\delta_n,a+(k+1)\delta_n],2^{-n+2},m)$.}

\noindent Then $\hat\mu_{\tilde\omega a}E=0$.   Suppose that
$\omega\in\Omega\setminus E$ and $t>a$.
\Quer\ If $\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)$ is undefined,
then (because $U$ is
complete under $\rho$) there must be an $n\in\Bbb N$ and a strictly
increasing sequence $\sequence{i}{q_i}$ in $\Bbb Q$, with supremum
$t$, such that $\rho(\omega(q_{i+1}),\omega(q_i))\ge 2^{-n+2}$ for every
$i\in\Bbb N$.   Let $k\in\Bbb N$ be such
that $t\in\ocint{a+k\delta_n,a+(k+1)\delta_n}$;  let $l\in\Bbb N$ be such
that $q_l\ge a+k\delta_n$.   Then, for every $m\in\Bbb N$,
$(q_l,q_{l+1},q_{l+1},q_{l+2},\ldots,q_{m-1},q_m)$ witnesses that
$\omega\in G(\Bbb Q\cap[a+k\delta_n,a+(k+1)\delta_n],2^{-n+2},m)$;
which is
impossible.\ \BanG\   So $\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)$ is
defined;  similarly, $\lim_{q\in\Bbb Q,q\downarrow t}\omega(q)$ is defined.

As $E$ is $\hat\mu_{\tilde\omega a}$-negligible, this proves (a).

\medskip

{\bf (b)(i)} This is actually easier.
Consider part (a-i) of the proof above.   Given $n\in\Bbb N$, we
see that there is a $\delta_n>0$ such that
$\hat\mu_{\tilde\omega a}\{\omega:\diam\omega[D]\le 2^{-n}\}
\ge 1-2^{-n}$ whenever $D\subseteq\coint{a,\infty}$ is a countable set of
diameter at most $\delta_n$.   Set

\Centerline{$D_n=\{t\}\cup(\Bbb Q\cap[t,t+\delta_n])$,
\quad$E_n=\{\omega:\diam\omega[D_n]\le 2^{-n}\}$}

\noindent for each $n\in\Bbb N$, and
$E=\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}E_m$.   Then
$\hat\mu_{\tilde\omega a}E=1$, and for $\omega\in E$ we have an
$n\in\Bbb N$ such that
$\rho(\omega(t),\omega(q))\le 2^{-m}$ whenever $m\ge n$ and
$q\in\Bbb Q\cap[t,t+\delta_m]$, so that
$\omega(t)=\lim_{q\in\Bbb Q,q\downarrow t}\omega(q)$.

\medskip

\quad{\bf (ii)} If $t>a$, the same argument applies on the other side of
$t$, taking $D_n=\{t\}\cup(\Bbb Q\cap[\max(a,t-\delta_n),t])$, to see that
$\omega(t)=\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)$ for
$\hat\mu_{\tilde\omega a}$-almost every $\omega$.

\medskip

{\bf (c)}\grheada\ Suppose that $E\subseteq\Omega$ and
$\hat\mu_{\tilde\omega a}E>0$.   Then there is an $\omega^*\in E$ such that

\Centerline{$\omega^*(t)=\tilde\omega(t)$ for every $t\le a$,}

\Centerline{$\omega^*(t)=\lim_{s\downarrow t}\omega^*(s)$ for every
$t\ge a$,}

\Centerline{$\lim_{s\uparrow t}\omega^*(t)$ is defined for every $t>a$.}

\noindent\Prf\ Let $E'\in\Tensorhat_{\coint{0,\infty}}\Cal B(U)$ be such
that $E'\subseteq E$ and $\hat\mu_{\tilde\omega a}E'>0$.   Let
$D\subseteq\coint{0,\infty}$ be a countable set such that $E'$ is
determined by coordinates in $D$;  we can suppose that $a\in D$ if $a$ is
finite.   Let $F$ be the set of those $\omega\in\Omega$ such that

\Centerline{$\lim_{q\in\Bbb Q,q\downarrow t}\omega(q)$ and
$\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)$ are defined in $U$ for every
$t>a$,}

\Centerline{$\omega(t)=\lim_{q\in\Bbb Q,q\downarrow t}\omega(t)$
for every $t\in D\cap\coint{a,\infty}$,}

\Centerline{$\omega(t)=\tilde\omega(t)$ for every
$t\in D\cap[0,a]$.}

\noindent Then (a) and (b), with 455E(b-ii), tell us that $F$ is
$\hat\mu_{\tilde\omega a}$-conegligible.   So there is an
$\omega\in E\cap F$.    Define $\omega^*\in\Omega$ by setting

$$\eqalign{\omega^*(t)&=\tilde\omega(t)\text{ if }t\le a,\cr
&=\lim_{q\in\Bbb Q,q\downarrow t}\omega(t)\text{ if }t\ge a;\cr}$$

\noindent note that the definitions of $\omega^*(a)$ are consistent if $a$
is finite, and that $\omega^*\restr D=\omega\restr D$, so that
$\omega^*\in E'\subseteq E$.

If $t\le a$, then of course $\omega^*(t)=\tilde\omega(t)$.   If $t\ge a$
and $\epsilon>0$, there is a $\delta>0$ such that
$\rho(\omega(q),\omega^*(t))\le\epsilon$ whenever
$q\in\Bbb Q\cap\ocint{t,t+\delta}$;  in which case
$\rho(\omega^*(s),\omega^*(t))\le\epsilon$ whenever
$s\in\coint{t,t+\delta}$;  as $\epsilon$ is arbitrary,
$\omega^*(t)=\lim_{s\downarrow t}\omega^*(s)$.
If $t>a$ and $\epsilon>0$, there is a $\delta>0$ such that
$\rho(\omega(q),\omega(q'))\le\epsilon$ whenever
$q\in\Bbb Q\cap\coint{t-\delta,t}$;  in which case
$\rho(\omega^*(s),\omega^*(s'))\le\epsilon$ whenever
$s\in\coint{t-\delta,t}$;  as $\epsilon$ is arbitrary and $U$ is complete,
$\lim_{s\uparrow t}\omega^*(s)$ is defined in $U$.   So we have an
appropriate $\omega^*$.\ \Qed

\medskip

\quad\grheadb\ Suppose, in ($\alpha$), that $\tilde\omega\in\Clll$.
Then $\omega^*\in\Clll$.   \Prf\

\Centerline{$\lim_{s\uparrow t}\omega^*(s)
=\lim_{s\uparrow t}\tilde\omega(s)$ is defined whenever $0<t\le a$,}

\Centerline{$\lim_{s\downarrow t}\omega^*(s)
=\lim_{s\downarrow t}\tilde\omega(s)$ is defined whenever $0\le t<a$,}

\Centerline{if $a>0$,
$\lim_{s\downarrow 0}\omega^*(s)
=\lim_{s\downarrow 0}\tilde\omega(s)=\tilde\omega(0)=\omega^*(0)$,}

\inset{\noindent
if $0<t<a$, then $\omega^*(t)=\tilde\omega(t)$ is equal to at least one of
$\lim_{s\uparrow t}\omega^*(s)=\lim_{s\uparrow t}\tilde\omega(s)$,
$\lim_{s\downarrow t}\omega^*(s)=\lim_{s\downarrow t}\tilde\omega(s)$.}

\noindent Since we already know that

\Centerline{$\omega^*(t)=\lim_{s\downarrow t}\omega^*(s)$ for every
$t\ge a$,}

\Centerline{$\lim_{s\uparrow t}\omega^*(t)$ is defined for every $t>a$,}

\noindent $\omega^*$ is \callal.\ \Qed

As $E$ is arbitrary,
it follows that if $\tilde\omega\in\Clll$ then $\Clll$ meets every
non-negligible
$\hat\mu_{\tilde\omega a}$-measurable set, so that
$\hat\mu_{\tilde\omega a}^*\Clll=1$, as required by (i).

\medskip

\quad\grheadc\ Similarly, if $\tilde\omega\in\Cdlg$, then any $\omega^*$
with the properties described in ($\alpha$) also belongs to $\Cdlg$.
\Prf\ This time, we have

\Centerline{if $0\le t<a$, then $\omega^*(t)=\tilde\omega(t)
=\lim_{s\downarrow t}\omega^*(s)=\lim_{s\downarrow t}\tilde\omega(s)$,}

\noindent which with the other properties listed is enough to ensure that
$\omega^*\in\Cdlg$.\ \QeD\  Since $E$ is arbitrary,
$\hat\mu_{\tilde\omega a}^*\Cdlg=1$.

This completes the proof of part (c).
}%end of proof of 455G

\leader{455H}{Corollary} Let $(U,\rho)$ be a complete metric
space and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ a family of
Radon probability measures on $U$, uniformly time-continuous on the right,
such that $\family{y}{U}{\nu_y^{(t,u)}}$ is a disintegration of
$\nu_x^{(s,u)}$ over $\nu_x^{(s,t)}$
whenever $0\le s<t<u$ and $x\in U$.   Let $\Clll(U)$ be the set of
\callal\ functions from
$\coint{0,\infty}$ to $U$.   Suppose that $\tilde\omega\in\Clll(U)$, and
$a\in[0,\infty]$;  let $\hat\mu_{\tilde\omega a}$ be the completed
probability measure on
$\Omega=U^{\coint{0,\infty}}$ defined from
$\tilde\omega$, $a$ and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ as in
455Eb.   Then $\hat\mu_{\tilde\omega a}$ has a unique
extension to a Radon measure $\tilde\mu_{\tilde\omega a}$ on $\Omega$, and
$\tilde\mu_{\tilde\omega a}\Clll(U)=1$.

\proof{{\bf (a)} In the language of 455E(b-i),
$\nu^{(0,t)}_{\tilde\omega ax}$ is a Radon measure
whenever $t>0$ and $x\in U$, so
the image measure defined from $\hat\mu_{\tilde\omega a}$ and the map
$\omega\mapsto\omega(t)$ is always a Radon measure on $U$, and there
there is a $\sigma$-compact set $H_t\subseteq U$
such that $\omega(t)\in H_t$ for
$\hat\mu_{\tilde\omega a}$-almost every $\omega$.
Set $U_0=\overline{\bigcup_{q\in\Bbb Q\cap\coint{0,\infty}}H_q}$;
then $U_0$ is separable and $\hat\mu_{\tilde\omega a}E=1$, where
$E=\{\omega:\omega(q)\in U_0$ for every $q\in\Bbb Q\cap\coint{0,\infty}\}$.
By 455G(c-i), $E\cap\Clll(U)$ has full outer measure;
and if $\omega\in E\cap\Clll(U)$, then
$\omega(t)\in U_0$ for every $t\ge 0$.

\medskip

{\bf (b)}
Thus $E\cap\Clll(U)$ is included in $\Clll(U_0)$, the set of \callal\
functions from $\coint{0,\infty}$ to the Polish space $U_0$.
So $\hat\mu_{\tilde\omega a}^*\Clll(U_0)=1$.
Let $\hat\mu_C$ be the subspace probability measure on $\Clll(U_0)$.

Since $\hat\mu_{\tilde\omega a}$ is inner regular with respect to
$\Tensorhat_{\coint{0,\infty}}\Cal B(U)$,
$\hat\mu_C$ is inner regular with respect to the $\sigma$-algebra
$\Sigma=\{E\cap\Clll(U_0):E\in\Tensorhat_{\coint{0,\infty}}\Cal B(U)\}$
(412Ob).   But $\Sigma$ is just the $\sigma$-algebra generated by the maps
$\omega\mapsto\omega(t):\Clll(U_0)\to U_0$ for $t\ge 0$, which is the Baire
$\sigma$-algebra of $\Clll(U_0)$ (4A3Na, 4A3Nd).  Accordingly
$\hat\mu_C\restr\Sigma$ is a Baire measure and is inner regular with
respect to the closed sets (412D);  it follows that its completion
$\hat\mu_C$ is inner regular with respect to the closed sets
(412Ha).

At this point, recall that $\Clll(U_0)$ is K-analytic (438Sc).
So $\hat\mu_C$ has an extension to a Radon measure $\tilde\mu_C$
on $\Clll(U_0)$ (432D).   Now $\tilde\mu_C$ has an extension to a Radon
probability measure $\tilde\mu_{\tilde\omega a}$ on $\Omega$ such that
$\tilde\mu_{\tilde\omega a}\Clll(U)=\tilde\mu_{\tilde\omega a}\Clll(U_0)=1$.
And if $\hat\mu_{\tilde\omega a}$ measures $E$, then

\Centerline{$\tilde\mu_{\tilde\omega a} E
=\tilde\mu_C(E\cap\Clll(U_0))
=\hat\mu_C(E\cap\Clll(U_0))
=\hat\mu_{\tilde\omega a}^*(E\cap\Clll(U_0))
=\hat\mu_{\tilde\omega a}E$,}

\noindent so $\tilde\mu_{\tilde\omega a}$ extends
$\hat\mu_{\tilde\omega a}$.

\medskip

{\bf (c)} As for uniqueness, observe that
$\dom\hat\mu_{\tilde\omega a}$ includes a base for the topology
of $\Omega$, so by 415H there can be at most one Radon measure extending
$\hat\mu_{\tilde\omega a}$.
}%end of proof of 455H

\leader{455I}{}\cmmnt{ In fact we can go farther;  the Radon measure
$\tilde\mu$ is much more closely related to the completed Baire measure
it extends than one might expect.

\medskip

\noindent}{\bf Lemma} Let $(U,\rho)$ be a complete
separable metric space
and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ a family of
Radon probability measures on $U$, uniformly time-continuous on the right,
such that $\family{y}{U}{\nu_y^{(t,u)}}$ is a disintegration of
$\nu_x^{(s,u)}$ over $\nu_x^{(s,t)}$ whenever $0\le s<t<u$ and $x\in U$.
Suppose that $\tilde\omega\in\Omega$, and
$a\in[0,\infty]$;   let $\hat\mu_{\tilde\omega a}$ be the completed
probability measure on $\Omega=U^{\coint{0,\infty}}$ defined from
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$, $\tilde\omega$ and $a$
as in 455Eb.

(a) Suppose that $0\le q_0<q_1$ and $\epsilon>0$.
For $\omega\in\Omega$, I will say that
$\ooint{q_0,q_1}$ is an {\bf$\epsilon$-shift interval}
of $\omega$ with {\bf$(q_0,q_1,\epsilon)$-shift point} $t$ if
$\rho(\omega(q_0),\omega(q_1))>2\epsilon$ and

$$\eqalign{t
&=\sup\{q:q\in\Bbb Q\cap\ooint{q_0,q_1},\,
\rho(\omega(q),\omega(q_0))\le\epsilon\}\cr
&=\inf\{q:q\in\Bbb Q\cap\ooint{q_0,q_1},\,
\rho(\omega(q),\omega(q_1))\le\epsilon\}.\cr}$$

\noindent Let $E$ be the set of such $\omega$.

\quad(i) $E\in\CalBa(\Omega)=\Tensorhat_{\coint{0,\infty}}\Cal B(U)$.

\quad(ii) The function
$f:E\to\ooint{q_0,q_1}$ which takes each $\omega\in E$ to its
$(q_0,q_1,\epsilon)$-shift point is $\CalBa(\Omega)$-measurable.

\quad(iii) If $q_0\ge a$,
the set $\{\omega:\omega\in E$, $f(\omega)=t\}$ is
$\hat\mu_{\tilde\omega a}$-negligible for every $t\in\ooint{q_0,q_1}$.

\quad(iv) If $q_0$, $q_1\in\Bbb Q$, $\omega\in E$, $\omega'\in\Omega$ and
$\omega'\restr\Bbb Q=\omega\restr\Bbb Q$, then $\omega'\in E$ and
$f(\omega')=f(\omega)$.

(b) Suppose that $\langle q_i\rangle_{i\le n}$,
$\langle q'_i\rangle_{i\le n}$, $\langle\le_i\rangle_{i\le n}$,
$\epsilon>0$, $E\in\CalBa(\Omega)$
and $\langle f_i\rangle_{i\le n}$ are such that, for every $i\le n$,

\Centerline{$q_i$, $q'_i\in\Bbb Q$,
\quad$q_i<q'_i$,
\quad$\le_i$ is either $\le$ or $\ge$,}

\Centerline{$\ooint{q_i,q'_i}$ is an $\epsilon$-shift interval of $\omega$
with
$(q_i,q'_i,\epsilon)$-shift point $f_i(\omega)$, for every $\omega\in E$,}

\noindent and also

\Centerline{$a\le q_0$,
\quad$q'_i\le q_{i+1}$ for every $i<n$,}

\Centerline{whenever $\omega$, $\omega'\in E$ there is an $i\le n$ such
that $f_i(\omega')\le_if_i(\omega)$.}

\noindent Then $E$ is $\hat\mu_{\tilde\omega a}$-negligible.

(c) Suppose that $\langle q_i\rangle_{i\le n}$,
$\langle q'_i\rangle_{i\le n}$, $\langle\le_i\rangle_{i\le n}$,
$\epsilon>0$, $E\in\CalBa(\Omega)$
and $\langle f_i\rangle_{i\le n}$ are such that, for every $i\le n$,

\Centerline{$q_i$, $q'_i\in\Bbb Q$,
\quad$q_i<q'_i$,
\quad$\le_i$ is either $\le$ or $\ge$,}

\Centerline{$\ooint{q_i,q'_i}$
is an $\epsilon$-shift interval of $\omega$ with
$(q_i,q'_i,\epsilon)$-shift point $f_i(\omega)$, for every $\omega\in E$,}

\noindent and also

\Centerline{$a\le q_0$,
\quad$q'_i\le q_{i+1}$ for every $i<n$.}

\noindent Then for $\hat\mu_{\tilde\omega a}$-almost every $\omega\in E$
there is an $\omega'\in E$ such that $f_i(\omega')<_if_i(\omega)$ for every
$i\le n$.

\proof{{\bf (a)(i)} Note that by 4A3Na we can identify
$\Tensorhat_{\coint{0,\infty}}\Cal B(U)$ with the Baire $\sigma$-algebra
$\CalBa(\Omega)$ of $\Omega$.   If $s$, $t\ge 0$, then
$\omega\mapsto(\omega(s),\omega(t)):\Omega\to U^2$ is
$\CalBa(\Omega)$-measurable, by 418Bb;  so
$\omega\mapsto\rho(\omega(s),\omega(t))$ is $\CalBa(\Omega)$-measurable.
For $\omega\in\Omega$, $\omega\in E$ iff
($\alpha$) $\rho(\omega(q_0),\omega(q_1))>2\epsilon$
($\beta$) whenever $q$, $q'\in\Bbb Q\cap\ooint{q_0,q_1}$,
$\rho(\omega(q),\omega(q_0))\le\epsilon$ and
$\rho(\omega(q'),\omega(q_1))\le\epsilon$ then $q\le q'$ ($\gamma$) for
every $n\in\Bbb N$ there are $q$, $q'\in\Bbb Q\cap\ooint{q_0,q_1}$
such that $\rho(\omega(q),\omega(q_0))\le\epsilon$,
$\rho(\omega(q'),\omega(q'_0))\le\epsilon$ and $q'\le q+2^{-n}$.
So $E\in\CalBa(\Omega)$.

\medskip

\quad{\bf (ii)} Now, for any $t$,

\Centerline{$\{\omega:\omega\in E$, $f(\omega)>t\}
=\bigcup_{q\in\Bbb Q\cap\ooint{t,q_1}}\{\omega:\omega\in E$,
   $\rho(\omega(q),\omega(q_0))\le\epsilon\}$}

\noindent belongs to $\CalBa(\Omega)$, so $f$ is
$\CalBa(\Omega)$-measurable.

\medskip

\quad{\bf (iii)} Consider the set $E'$ of those $\omega\in\Omega$ such that

\Centerline{$\lim_{q\in\Bbb Q,q\uparrow t}\omega(q)
  =\omega(t)=\lim_{q\in\Bbb Q,q\downarrow t}\omega(q)$.}

\noindent If $\omega\in E\cap E'$,
at least one of $\rho(\omega(t),\omega(q_0))$,
$\rho(\omega(t),\omega(q_1))$ must be greater than $\epsilon$;  in the
first case, $t$ cannot be
$\sup\{q:q\in\Bbb Q\cap\ooint{q_0,q_1}$,
$\rho(\omega(q),\omega(q_0))\le\epsilon\}$;
in the second case, $t$ cannot be
$\inf\{q:q\in\Bbb Q\cap\ooint{q_0,q_1}$,
$\rho(\omega(q),\omega(q_1))\le\epsilon\}$;  so in either case $f(\omega)$
cannot be equal to $t$.   Now $\hat\mu_{\tilde\omega a}E'=1$, by 455Gb, so
$\{\omega:\omega\in E$, $f(\omega)=t\}\subseteq\Omega\setminus E'$ is
$\hat\mu_{\tilde\omega a}$-negligible.

\medskip

\quad{\bf (iv)} Immediate from the definitions.

\medskip

{\bf (b)} Induce on $n$.   Of course we need consider only the case
$E\ne\emptyset$.

\medskip

\quad{\bf (i)} If $n=0$, $f_0$ must be constant on $E$, so
$E$ must be negligible, by (a-iii).

\medskip

\quad{\bf (ii)} For the inductive step to $n\ge 1$, set
$E_t=\{\omega:\omega\in E$, $f_0(\omega)=t\}$ for
$t\in\ooint{q_0,q'_0}$;  by (a-ii),
$E_t\in\CalBa(\Omega)$.

\medskip

\qquad\grheada\ There is a countable
set $J\subseteq\ooint{q_0,q'_0}$ such that whenever
$t\in[q_0,q'_0]\setminus J$ and $\omega$, $\omega'\in E_t$ then
there is an $i$ such that $1\le i\le n$ and $f_i(\omega)\le_if_i(\omega')$.
\Prf\ Let $\Cal W$ be a countable base for the topology of
$\prod_{1\le i\le n}\ooint{q_i,q'_i}$.   For $\omega\in E$ set
$g(\omega)=\langle f_i(\omega)\rangle_{1\le i\le n}$;  note that
$g:E\to\prod_{1\le i\le n}\ooint{q_i,q'_i}$ is
$\CalBa(\Omega)$-measurable.   For $W\in\Cal W$, set

\Centerline{$A_W=\{t:t\in\ooint{q_0,q'_0}$
and there is an $\omega\in E_t$ such that $g(\omega)\in W\}$.}

\noindent Set

\Centerline{$J
=\{t:t\in\ooint{q_0,q'_0}$, $t$ is either $\inf A_W$ or $\sup A_W$
for some $W\in\Cal W\}$.}

\noindent Then $J$ is a countable subset of $\ooint{q_0,q'_0}$.
\Quer\ Suppose that $t\in\ooint{q_0,q'_0}\setminus J$ and
$\omega$, $\omega'\in E_t$ are such that $f_i(\omega')<_if_i(\omega)$ for
$1\le i\le n$.   Let $W\in\Cal W$ be such that
$g(\omega')\in W$ and $z(i)<_if_i(\omega)$
whenever $1\le i\le n$ and $z\in W$.   Then $\omega'$ witnesses that
$t\in A_W$;  since $t$ is neither the greatest nor the least element of
$A_W$, there is a $t'\in A_W$ such that $t'<_0t$;
take $\omega''\in E_{t'}$ such that $g(\omega'')\in W$.   Then

\Centerline{$f_0(\omega'')=t'<_0t=f_0(\omega)$,}

\Centerline{$f_i(\omega'')=g(\omega'')(i)<_if_i(\omega)$
for $1\le i\le n$,}

\noindent which is impossible.\ \BanG\  Thus
$J$ has the required property.\ \Qed

\medskip

\qquad\grheadb\  Now consider the family
$\family{\omega}{\Omega}{\hat\mu_{\omega q_1}}$.   Because $q_1>a$,
this is a disintegration of $\hat\mu_{\tilde\omega a}$ over itself.
\Prf\ As in part (a-ii-$\alpha$) of the proof of 455G,
we can think of each $\hat\mu_{\omega q_1}$ as defined either from
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ or from
$\langle\nu^{(s,t)}_{\tilde\omega ax}\rangle_{a\le s<t,x\in U}$;
and in the latter form we can apply 455E(c-i).\ \Qed

Consider $\mu_{\omega q_1}(E)$ for $\omega\in\Omega$.   This time, note
that
$\{\omega':\omega'\restr[0,q_1]\cap\Bbb Q=\omega\restr[0,q_1]\cap\Bbb Q\}$
is $\mu_{\omega q_1}$-conegligible.
In particular, $\mu_{\omega q_1}(E)=0$ unless $\ooint{q_0,q'_0}$ is an
$\epsilon$-shift interval of $\omega$.   Next,

\Centerline{$\{\omega:\ooint{q_0,q'_0}$
is an $\epsilon$-shift interval of $\omega$ with
$(q_0,q'_0,\epsilon)$-shift point in $J\}$}

\noindent is $\hat\mu_{\tilde\omega a}$-negligible, by (a-iii) again.
Finally, suppose that
$\omega\in\Omega$ is such that $\ooint{q_0,q'_0}$ is an $\epsilon$-shift
interval of $\omega$ with $(q_0,q'_0,\epsilon)$-shift
point $t\in\ooint{q_0,q'_0}\setminus J$.   Then

$$\eqalign{\mu_{\omega q_1}E
&=\mu_{\omega q_1}\{\omega':\omega'\in E,\,
   \omega'\restr\Bbb Q\cap[0,q_1]=\omega\restr\Bbb Q\cap[0,q_1]\}\cr
&=\mu_{\omega q_1}\{\omega':\omega'\in E_t\}.
\cr}$$

\noindent But the choice of $J$ in ($\alpha$) ensured that
$E_t$ would be a set of the same type as $E$, one level down, determined by
intervals starting from $q_1$,
so that $\hat\mu_{\omega q_1}E_t=0$, by the inductive hypothesis applied
to $\omega$ and $q_1$ in place of $\tilde\omega$ and $a$.

\medskip

\qquad\grheadc\ So we see that $\hat\mu_{\omega q_1}E=0$ for
$\hat\mu_{\tilde\omega a}$-almost every
$\omega$, and $\hat\mu_{\tilde\omega a}E=0$.   Thus the induction proceeds.

\medskip

{\bf (c)} Let $F$ be the set of those $\omega\in E$ for which there
is no $\omega'\in E$ such that $f_i(\omega')<_if_i(\omega)$
for every $i\le n$.   Then $F\in\CalBa(\Omega)$.
\Prf\ For each $\omega\in E$ set
$f(\omega)=\langle f_i(\omega)\rangle_{i\le n}$ and

\Centerline{$W_{\omega}
=\{z:z\in\prod_{i\le n}\ooint{q_i,q'_i}$, $f_i(\omega)<_iz(i)$
for every $i\le n\}$,}

\noindent so that $W_{\omega}$ is open in
$\prod_{i\le n}\ooint{q_i,q'_i}$.   Set
$W=\bigcup_{\omega\in E}W_{\omega}$.   Then $W$ is open and
$F=\{\omega:\omega\in E$, $f(\omega)\notin W\}$ belongs to
$\CalBa(\Omega)$.\ \Qed

If $\omega$, $\omega'\in F$ then there is surely some $i\le n$ such that
$f_i(\omega')\le_if_i(\omega)$.   By (b), $\hat\mu_{\tilde\omega a}F=0$.
}%end of proof of 455I

\leader{455J}{Theorem} Let $(U,\rho)$ be a complete separable metric space
and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ a family of
Radon probability measures on $U$, uniformly time-continuous on the right,
such that $\family{y}{U}{\nu_y^{(t,u)}}$ is a disintegration of
$\nu_x^{(s,u)}$ over $\nu_x^{(s,t)}$ whenever $0\le s<t<u$ and $x\in U$.
Write $\Clll$ for the set of \callal\ functions from
$\coint{0,\infty}$ to $U$.
Suppose that $\tilde\omega\in\Clll$, and
$a\in[0,\infty]$;   let $\hat\mu_{\tilde\omega a}$ be the completed
probability measure on $\Omega=U^{\coint{0,\infty}}$ defined from
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$, $\tilde\omega$ and $a$
as in 455Eb, and $\tilde\mu_{\tilde\omega a}$ its extension to a Radon
measure on $\Omega$, as in 455H.   Then $\tilde\mu_{\tilde\omega a}$ is
inner regular with
respect to sets of the form $F\cap\Clll$ where $F\subseteq\Omega$ is a zero
set.

\proof{{\bf (a)} As in 455I, $\Tensorhat_{\coint{0,\infty}}\Cal B(U)$
is the Baire $\sigma$-algebra of $\Omega$.
Let $D$ be
$((\{a\}\cup\Bbb Q)\cap\coint{0,\infty})
\cup\{t:t\ge 0$, $\tilde\omega$ is not continuous at $t\}$;
then $D$ is countable (438S(a-i)).   Let $E^*$ be
$\{\omega:\omega\in\Omega$,
$\omega\restr D\cap[0,a]=\tilde\omega\restr D\cap[0,a]\}$;  then
$E^*$ is $\hat\mu_{\tilde\omega a}$-conegligible, by 455E(b-ii) once again.

\medskip

{\bf (b)} Let $\Cal G$ be a countable base for the topology of
$U$.   Let $\Cal W$ be the family of open subsets of $\Omega$ of the form
$\{\omega:\omega(q)\in G_q$ for every $q\in J\}$ where $J\subseteq D$
is finite and $G_q\in\Cal G$ for every $q\in J$.
Let $\Theta$ be the set of all strings

\Centerline{$\theta=(q_0,q'_0,\ldots,q_n,q'_n,
\le_0,\ldots,\le_n,k,W)$}

\noindent such that

\Centerline{$q_0,\ldots,q'_n\in\Bbb Q$,
\quad$a\le q_0<q'_0\le q_1<q'_1\le\ldots\le q_n<q'_n$,}

\Centerline{for each $i\le n$, $\le_i$ is either $\le$ or $\ge$,}

\Centerline{$k\in\Bbb N$,
\quad$W\in\Cal W$;}

\noindent then $\Theta$ is countable.

\medskip

{\bf (c)} Let $K\subseteq E^*\cap\Clll$ be compact.
Set $L=\pi_D^{-1}[\pi_D[K]]$,
where $\pi_D(\omega)=\omega\restr D$ for $\omega\in\Omega$;
then $L$ is a Baire subset of $\Omega$, because $\pi_D[K]$ is a
compact subset of the metrizable space $U^D$.

\medskip

\quad{\bf (i)} For

\Centerline{$\theta=(q_0,q'_0,\ldots,q_n,q'_n,
\le_0,\ldots,\le_n,k,W)\in\Theta$}

\noindent let $E_{\theta}$ be the set of those
$\omega\in L\cap W$ such that, for each $i\le n$,
$\ooint{q_i,q'_i}$ is a $2^{-k}$-shift interval of $\omega$
(definition:  455Ia).   For $\omega\in E_{\theta}$ and
$i\le n$ let $f_i(\theta,\omega)$ be the $(q_i,q'_i,2^{-k})$-shift
point of $\omega$.   By
455Ia, $E_{\theta}$ is a Baire subset of $\Omega$ and
$\omega\mapsto f_i(\theta,\omega)$ is Baire measurable.   Let
$F_{\theta}$ be the set of those $\omega\in E_{\theta}$ such that
there is no $\omega'\in E_{\theta}$ with
$f_i(\theta,\omega')<_if_i(\theta,\omega)$ for every $i\le n$;  by
455Ic, $F_{\theta}$ is $\hat\mu_{\tilde\omega a}$-negligible.
So $F^*=\bigcup_{\theta\in\Theta}F_{\theta}$ is
$\hat\mu_{\tilde\omega a}$-negligible.

\medskip

\quad{\bf (ii)} Suppose that $\omega\in K\setminus F^*$.
Let $A$ be the set of points in $\ooint{a,\infty}$
at which $\omega$ is discontinuous.   If $J\subseteq A$ is finite and
$\epsilon_t\in\{-1,1\}$ for each $t\in J$,
there is an $\omega'\in K$ such that
$\omega'\restr D=\omega\restr D$ and
$\omega'$ is continuous on the right
at every point $t$ of $J$ such that $\epsilon_t=1$, while
$\omega'$ is continuous on the left
at every point $t$ of $J$ such that $\epsilon_t=-1$.
\Prf\ This is trivial if
$J$ is empty.   Otherwise, enumerate $J$ in ascending order as
$t_0<t_1<\ldots<t_n$.   Set $x_i=\lim_{t\uparrow t_i}\omega(t)$,
$y_i=\lim_{t\downarrow t_i}\omega(t)$;  because $\omega\in\Clll$ these are
defined, and because $\omega$ is not continuous at $t$ they are different.

Let $k\in\Bbb N$ be such that $\rho(x_i,y_i)>2^{-k+1}$ for each
$i\le n$.   For $i\le n$, let
let $q_i$, $q'_i\in\Bbb Q$ be such that
$q_i<t_i<q'_i$, $\rho(\omega(t),x_i)\le 2^{-k-1}$ for
$t\in\coint{q_i,t_i}$,
and $\rho(\omega(t),y_i)\le 2^{-k-1}$ for $t\in\ocint{t_i,q'_i}$.
Of course we
can suppose that $a\le q_0$ and that $q'_i\le q_{i+1}$ for $i<n$.   Observe
that this will ensure that every $\ooint{q_i,q'_i}$ is a
$2^{-k}$-shift interval of $\omega$ with
$(q_i,q'_i,2^{-k})$-shift point $t_i$.

Let $\le_i$ be $\le$ if $\epsilon_{t_i}=1$, $\ge$ if $\epsilon_{t_i}=-1$.
For each $W\in\Cal W$ containing $\omega$,
let $\theta_W\in\Theta$ be
$(q_0,\ldots,q'_n,\le_0,\ldots,\le_n,k,W)$.   Then
$\omega\in E_{\theta_W}$, and $f_i(\theta_W,\omega)=t_i$ for each
$i\le n$.   Because $\omega\notin F_{\theta_W}$, there is
an $\omega_W\in E_{\theta_W}$ such that
$f_i(\theta_W,\omega_W)<_if_i(\theta_W,\omega)=t_i$ for every $i\le n$.
Let $\omega'_W\in K$ be such that $\omega'_W\restr D=\omega_W\restr D$;
then $\omega'_W\in E_{\theta_W}$ and

\Centerline{$f_i(\theta_W,\omega'_W)=f_i(\theta_W,\omega_W)
<_it_i$}

\noindent for every $i\le n$ (455I(a-iv)).

If $i\le n$ and $\epsilon_{t_i}=1$,

\Centerline{$\rho(\omega'_W(q),\omega'_W(q'_i))\le 2^{-k}$}

\noindent for every rational $q\in\ocint{f_i(\theta_W,\omega'_W),q'_i}$;
because $\omega'_W\in\Clll$ and
$t_i\in\ocint{f_i(\theta_W,\omega'_W),q'_i}$
$\rho(\omega'_W(t_i),\omega'_W(q'_i))\le 2^{-k}$.   Similarly, if
$\epsilon_{t_i}=-1$, $\rho(\omega'_W(t_i),\omega'_W(q_i))\le 2^{-k}$.

Let $\Cal F$ be an ultrafilter on $\Cal W$ containing all sets of the form
$\{W:\omega\in W\subseteq W_0\}$ where $\omega\in W_0\in\Cal W$,
and set $\omega'=\lim_{W\to\Cal F}\omega'_W\in K$.
Then $\omega'\restr D=\omega\restr D$, because $\omega_W$ and $\omega'_W$
belong to $W$ whenever $\omega\in W\in\Cal W$.
If $i\le n$ and $\epsilon_{t_i}=1$, then

\Centerline{$\rho(\omega'(t_i),\omega'(q'_i))
=\lim_{W\to\Cal F}\rho(\omega'_W(t_i),\omega'_W(q'_i))\le 2^{-k}$,}

\Centerline{$\rho(\omega'(q_i),\omega'(q'_i))
=\rho(\omega(q_i),\omega(q'_i))>2^{-k+1}$,}

\noindent so $\rho(\omega'(q_i),\omega'(t_i))>2^{-k}$.
On the other hand,

\Centerline{$\rho(\omega'(q_i),\omega'(q))=\rho(\omega(q_i),\omega(q))
\le 2^{-k}$}

\noindent for every rational $q\in\coint{q_i,t_i}$.   So $\omega'$ cannot
be continuous on the left at $t_i$;  because $\omega'\in\Clll$, it
must be continuous on the right at $t_i$.   Similarly, if $i\le n$ and
$\epsilon_{t_i}=-1$, $\omega'$ cannot be continuous on the right at $t_i$
and must be continuous on the left at $t_i$.
But this is what we need to know.\ \Qed

\medskip

\quad{\bf (iii)} Suppose that $\omega\in K\setminus F^*$,
$\omega'\in\Clll$ and $\omega\restr D=\omega'\restr D$.
Then $\omega'\in K$.   \Prf\ Let $A$ be the set of points in
$\ooint{a,\infty}$ where $\omega$ is
not continuous, and for $t\in A$ let $\epsilon_t$ be $1$ if $\omega'$ is
continuous on the right at $t$, $-1$ if $\omega'$ is continuous on the left
at $t$.   For each finite $J\subseteq A$, (ii) tells us that there is an
$\omega_J\in K$ such that
$\omega_J\restr D=\omega\restr D=\omega'\restr D$ and,
for $t\in J$, $\omega_J$ is continuous on the right at $t$ if
$\epsilon_t=1$, and continuous on the left at $t$ if $\epsilon_t=-1$.
As both $\omega_J$ and $\omega'$ are \callal,
this means that $\omega_J(t)=\omega'(t)$ for $t\in J$.   Taking a
cluster point $\omega^*\in K$ of $\omega_J$ as $J$ increases through the
finite subsets of $A$, we see that
$\omega^*\restr(A\cup D)=\omega'\restr(A\cup D)$.

Now recall that $\omega\in E^*$, so that

\Centerline{$\omega'\restr D\cap[0,a]
=\omega\restr D\cap[0,a]=\tilde\omega\restr D\cap[0,a]$.}

\noindent Since both $\omega'$ and $\tilde\omega$ are \callal,
$\tilde\omega$ is discontinuous at any point
of $\coint{0,a}$ at which $\omega'$ is discontinuous.   Since I arranged
that $a$ (if finite) would be in $D$, $D\cup A$ contains every point at
which $\omega'$ is discontinuous.   But this means that $\omega^*=\omega'$
(438S(a-ii)).   So $\omega'\in K$.\ \Qed

\medskip

\quad{\bf (iv)} Suppose that $\tilde\mu_{\tilde\omega a}K>\gamma\ge 0$.
Then there is a
zero set $F\subseteq\Omega$ such that $F\cap\Clll\subseteq K$ and
$\tilde\mu_{\tilde\omega a}(F\cap\Clll)\ge\gamma$.   \Prf\ Because
$\tilde\mu_{\tilde\omega a}F^*=\hat\mu_{\tilde\omega a}F^*=0$,
there is a compact
$K'\subseteq K\setminus F^*$ such that
$\tilde\mu_{\tilde\omega a}K'\ge\gamma$.
Set $F=\pi_D^{-1}[\pi_D[K']]$;  $F$ is a zero set in $\Omega$ because
$\pi_D[K']$ is a zero set in $U^D$.   By (iii),
$F\cap\Clll\subseteq K$;  and

\Centerline{$\tilde\mu_{\tilde\omega a}(F\cap\Clll)
\ge\tilde\mu_{\tilde\omega a} K'\ge\gamma$.\ \Qed}

\medskip

{\bf (c)} Since $E^*$ and $\Clll$ are
$\tilde\mu_{\tilde\omega a}$-conegligible, the Radon measure
$\tilde\mu_{\tilde\omega a}$ is certainly inner regular with respect to the
compact subsets of $E^*\cap\Clll$;   by (b-iv), $\tilde\mu_{\tilde\omega a}$
is inner regular with respect to the
intersections of $\Clll$ with zero sets.
}%end of proof of 455J

\leader{455K}{Corollary} Suppose, in 455J, that $\tilde\omega\in\Cdlg$,
the space of \cadlag\ functions from $\coint{0,\infty}$ to $U$.
Then the subspace measure $\ddot\mu_{\tilde\omega a}$
on $\Cdlg$ induced by $\hat\mu_{\tilde\omega a}$ is a completion regular
quasi-Radon measure.

\proof{ The point is that the outer measures
$\tilde\mu_{\tilde\omega a}^*$ and $\hat\mu^*_{\tilde\omega a}$
agree on subsets of $\Cdlg$.   \Prf\ Since $\tilde\mu_{\tilde\omega a}$
extends $\hat\mu_{\tilde\omega a}$,
$\tilde\mu_{\tilde\omega a}^*A\le\hat\mu_{\tilde\omega a}^*A$
for every $A\subseteq\Omega$.   On the other hand, if $A\subseteq\Cdlg$ and
$\tilde\mu_{\tilde\omega a}^*A<\gamma$, there is an $E\supseteq A$ such that
$\tilde\mu_{\tilde\omega a} E<\gamma$.   By 455J, there is a zero set
$F\subseteq\Omega$ such that $E\cap F\cap\Clll=\emptyset$ and
$\tilde\mu_{\tilde\omega a}(F\cap\Clll)\ge 1-\gamma$.   Now

$$\eqalignno{\hat\mu_{\tilde\omega a}^*A
&\le\hat\mu_{\tilde\omega a}^*(\Cdlg\setminus F)
=\hat\mu_{\tilde\omega a}(\Omega\setminus F)\cr
\displaycause{because $\hat\mu_{\tilde\omega a}^*\Cdlg=1$, by 455G(c-ii)}
&=\tilde\mu_{\tilde\omega a}(\Omega\setminus F)
=\tilde\mu_{\tilde\omega a}(\Clll\setminus F)
\le\gamma.\cr}$$

\noindent As $\gamma$ is arbitrary,
$\hat\mu_{\tilde\omega a}^*A\le\tilde\mu_{\tilde\omega a}^*A$.\ \Qed

Write $\ddot{\tilde\mu}_{\tilde\omega a}$
for the subspace measure on $\Cdlg$ induced by
$\tilde\mu_{\tilde\omega a}$.   By 214Cd, the outer measures
$\ddot{\tilde\mu}\vthsp_{\tilde\omega a}^*
=\tilde\mu_{\tilde\omega a}^*\restr\Cal P\Cdlg$ and
$\ddot\mu_{\tilde\omega a}^*$ are the same.   Because
$\ddot{\tilde\mu}_{\tilde\omega a}$ and
$\ddot\mu_{\tilde\omega a}$ are both complete
probability measures, they must be identical (213C).
Because $\tilde\mu_{\tilde\omega a}$ is a Radon measure,
$\ddot{\tilde\mu}_{\tilde\omega a}=\ddot\mu_{\tilde\omega a}$
is quasi-Radon (415B).   Because
$\hat\mu_{\tilde\omega a}$ is the completion of a Baire measure, therefore
inner regular with respect to the zero sets in $\Omega$ (412D, 412Ha),
$\ddot\mu_{\tilde\omega a}$ is inner regular
with respect to the zero sets in $\Cdlg$, by 412Pd,
and is completion regular.
}%end of proof of 455K

\leader{455L}{Stopping times}\cmmnt{ We need the continuous-time
version of the concept of `stopping time' introduced in \S275.}
Let $\Omega$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $\Omega$ and
$\langle\Sigma_t\rangle_{t\ge 0}$ a non-decreasing family of
$\sigma$-subalgebras of $\Sigma$.   (Such a family is called a
{\bf filtration}.)  For $t\ge 0$, set
$\Sigma^+_t=\bigcap_{s>t}\Sigma_s$, so that
$\langle\Sigma^+_t\rangle_{t\ge 0}$ also is a non-decreasing family of
$\sigma$-algebras.   Of course $\Sigma^+_t=\bigcap_{s>t}\Sigma^+_s$ for
every $t\ge 0$.

\spheader 455La A function
$\tau:\Omega\to[0,\infty]$ is a {\bf stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$} if
$\{\omega:\omega\in\Omega$, $\tau(\omega)\le t\}$ belongs to $\Sigma_t$
for every $t\ge 0$.

\cmmnt{Note that in this case }$\tau$ will be $\Sigma$-measurable.

\spheader 455Lb A function $\tau:\Omega\to[0,\infty]$ is a stopping time
adapted to $\langle\Sigma^+_t\rangle_{t\ge 0}$ iff
$\{\omega:\tau(\omega)<t\}\in\Sigma_t$ for every $t\ge 0$.
\prooflet{\Prf\ (i) If $\tau$ is adapted to
$\langle\Sigma^+_t\rangle_{t\ge 0}$ and $t\ge 0$, then
$\{\omega:\tau(\omega)\le q\}\in\Sigma^+_q\subseteq\Sigma_t$ whenever
$0\le q<t$, so

\Centerline{$\{\omega:\tau(\omega)<t\}
=\bigcup_{q\in\Bbb Q\cap\coint{0,t}}\{\omega:\tau(\omega)\le q\}
\in\Sigma_t$.}

\noindent Thus $\tau$ satisfies the condition.   (ii) If $\tau$ satisfies
the condition and $t\ge 0$, set $t_n=t+2^{-n}$ for each $n$.
Then

\Centerline{$\{\omega:\tau(\omega)<t_n\}
\in\Sigma_{t_n}\subseteq\Sigma_{t_m}$}

\noindent whenever $m\le n$, so

\Centerline{$\{\omega:\tau(\omega)\le t\}
=\bigcap_{n\ge m}\{\omega:\tau(\omega)<t_n\}\in\Sigma_{t_m}$}

\noindent for every $m$, and

\Centerline{$\{\omega:\tau(\omega)\le t\}
\in\bigcap_{m\in\Bbb N}\Sigma_{t_m}
=\Sigma^+_t$.}

\noindent As $t$ is arbitrary, $\tau$ is adapted to
$\langle\Sigma^+_t\rangle_{t\ge 0}$.\ \Qed}

\spheader 455Lc{\bf (i)}
Constant functions on $\Omega$ are\cmmnt{ of course} stopping
times.

\medskip

\quad{\bf (ii)} If $\tau$ and $\tau'$ are stopping times adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$, so is
$\tau+\tau'$.   \prooflet{\Prf\

\Centerline{$\{\omega:\tau(\omega)+\tau'(\omega)\le t\}
=\bigcap_{q\in\Bbb Q\cap[0,t]}
  \{\omega:\tau(\omega)\le q\}\cup\{\omega:\tau'(\omega)\le t-q\}
\in\Sigma_t$}

\noindent for every $t\ge 0$.\ \Qed}

\medskip

\quad{\bf (iii)}\cmmnt{ (Compare 455Cb and 455E(c-ii)).} If $\tau$ is a
stopping time adapted to $\langle\Sigma_t\rangle_{t\ge 0}$, then

\Centerline{$\Sigma_{\tau}
=\{E:E\in\Sigma$, $E\cap\{\omega:\tau(\omega)\le t\}\in\Sigma_t$ for every
$t\ge 0\}$}

\noindent is a $\sigma$-subalgebra of $\Sigma$.   \prooflet{(The check is
elementary.)}

\medskip

\quad{\bf (iv)} If $\familyiI{\tau_i}$ is a countable
family of stopping times adapted to $\langle\Sigma_t\rangle_{t\ge 0}$,
then $\tau=\sup_{i\in I}\tau_i$ is adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$.   \prooflet{\Prf\ For any $t\ge 0$,

\Centerline{$\{\omega:\tau(\omega)\le t\}
=\bigcap_{i\in I}\{\omega:\tau_i(\omega)\le t\}
\in\Sigma_t$.  \Qed}}

\medskip

\quad{\bf (v)} If $\familyiI{\tau_i}$ is a countable family of stopping
times adapted to $\langle\Sigma^+_t\rangle_{t\ge 0}$,
then $\tau=\inf_{i\in I}\tau_i$ is adapted to
$\langle\Sigma^+_t\rangle_{t\ge 0}$\prooflet{, because

\Centerline{$\{\omega:\tau(\omega)<t\}
=\bigcup_{i\in I}\{\omega:\tau_i(\omega)<t\}\in\Sigma_t$}

\noindent for every $\tau\ge 0$}.

\spheader 455Ld Now suppose that $Y$ is a topological space and
we have a family
$\langle X_t\rangle_{t\ge 0}$ of functions from $\Omega$ to $Y$, and that
$\tau:\Omega\to[0,\infty]$ is any $\Sigma$-measurable function.
Set $X_{\tau}(\omega)=X_{\tau(\omega)}(\omega)$ when $\tau(\omega)<\infty$.
If $(t,\omega)\mapsto X_t(\omega):\coint{0,\infty}\times\Omega\to Y$
is $\Cal B(\coint{0,\infty})\tensorhat\Sigma$-measurable, where
$\Cal B(\coint{0,\infty})$ is the Borel $\sigma$-algebra of
$\coint{0,\infty}$, then $X_{\tau}:\{\omega:\tau(\omega)<\infty\}\to Y$
is $\Sigma$-measurable.   \prooflet{\Prf\ Setting
$\Omega_0=\{\omega:\tau(\omega)<\infty\}$, the map
$\omega\mapsto(\tau(\omega),\omega):
\Omega_0\to\coint{0,\infty}\times\Omega$
is $(\Sigma,\Cal B(\coint{0,\infty})\tensorhat\Sigma)$-measurable (4A3Bc),
so $X_{\tau}$ is the composition of a
$(\Sigma,\Cal B(\coint{0,\infty})\tensorhat\Sigma)$-measurable function
with a $\Cal B(\coint{0,\infty})\tensorhat\Sigma$-measurable function and
is $\Sigma$-measurable, by 4A3Bb.\ \Qed}

%\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\header{455Le}{\bf *(e)} Again take a topological space $Y$, a family
$\langle X_t\rangle_{t\ge 0}$ of functions from $\Omega$ to $Y$, and a
stopping time $\tau:\Omega\to[0,\infty]$ adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$.   This time, suppose that
$\langle X_t\rangle_{t\ge 0}$ is {\bf progressively measurable}, that is,
that $(s,\omega)\mapsto X_s(\omega):[0,t]\times\Omega\to Y$ is
$\Cal B([0,t])\tensorhat\Sigma_t$-measurable for every $t\ge 0$, and
moreover that $\Sigma_t$ is closed under Souslin's operation\cmmnt{ (421B)}
for every $t$.   Then $X_{\tau}$, as defined in (d), will be
$\Sigma_{\tau}$-measurable.
\prooflet{\Prf\ Suppose that $H\subseteq Y$ is open, and set
$E=\{\omega:\omega\in\dom X_{\tau}$, $X_{\tau}(\omega)\in H\}$.
Of course $\langle X_t\rangle_{t\ge 0}$ satisfies the condition of (d), so
$E\in\Sigma$.   Take any $t\ge 0$.   Then

$$\eqalign{\{(s,\omega):0\le s\le t,\,\tau(\omega)=s\}
&=\bigcap_{n\in\Bbb N}
  \bigcup_{i\in\Bbb N}\{s:2^{-n}(i-1)<s\le\min(t,2^{-n}i)\}\cr
&\mskip80mu  \times\{\omega:2^{-n}(i-1)<\tau(\omega)
   \le\min(t,2^{-n}i)\}\cr
&\in\Cal B([0,t])\tensorhat\Sigma_t,\cr}$$

\Centerline{$\{(s,\omega):s\le t$, $X_s(\omega)\in H\}
\in\Cal B([0,t])\tensorhat\Sigma_t$,}

\noindent so

\Centerline{$W=\{(s,\omega):s\le t$, $\tau(\omega)=s$,
$X_s(\omega)\in H\}$}

\noindent also belongs to $\Cal B([0,t])\tensorhat\Sigma_t$.
Consequently the projection of $W$ onto $\Omega$ belongs to
$\Cal S(\Sigma_t)=\Sigma_t$ (423O).   But this is just

\Centerline{$\{\omega:\tau(\omega)\le t$, $X_{\tau(\omega)}\in H\}
=E\cap\{\omega:\tau(\omega)\le t\}$.}

\noindent As $t$ is arbitrary, $E\in\Sigma_{\tau}$;  as $H$ is arbitrary,
$X_{\tau}$ is $\Sigma_{\tau}$-measurable.\ \Qed}

\header{455Lf}{\bf *(f)}\cmmnt{ There are some technical
points
concerning stopping times which are perhaps worth noting here.

\medskip

\quad}{\bf (i)} Suppose
that $\mu$ is a probability measure with domain $\Sigma$ and null ideal
$\Cal N(\mu)$.   Then we can
form the completion $\hat\mu$ with domain $\hat\Sigma$.   If we now
set $\hat\Sigma_t=\{E\symmdiff A:E\in\Sigma_t$, $A\in\Cal N(\mu)\}$,
$\langle\hat\Sigma_t\rangle_{t\ge 0}$ and
$\langle\hat\Sigma_t^+\rangle_{t\ge 0}$ are filtrations, where
$\hat\Sigma_t^+=\bigcap_{s>t}\hat\Sigma_s$ for $t\ge 0$.

\medskip

\quad{\bf (ii)} We find that
$\hat\Sigma_t^+=\{E\symmdiff A:E\in\Sigma_t^+$, $A\in\Cal N(\mu)\}$ for
every $t\ge 0$.   \prooflet{\Prf\ Of course

\Centerline{$\{E\symmdiff A:E\in\Sigma_t^+$, $A\in\Cal N(\mu)\}
\subseteq\bigcap_{s>t}
   \{E\symmdiff A:E\in\Sigma_s$, $A\in\Cal N(\mu)\}
=\hat\Sigma_t^+$.}

\noindent If $F\in\hat\Sigma_t^+$, then for every $q\in\Bbb Q$ such that
$q>t$ there is an $E_q\in\Sigma_q$ such that $F\symmdiff E_q$ is
negligible.   Set

\Centerline{$E
=\bigcup_{q\in\Bbb Q,q>t}\bigcap_{q'\in\Bbb Q,t<q'\le q}E_{q'}$,
\quad$A=F\symmdiff E$;}

\noindent then $E\in\Sigma_t^+$, $A\in\Cal N(\mu)$ and $F=E\symmdiff A$.\
\Qed}

\medskip

\quad{\bf (iii)} Of course every stopping time adapted to
$\langle\Sigma_t^+\rangle_{t\ge 0}$ is adapted to
$\langle\hat\Sigma_t^+\rangle_{t\ge 0}$.   Conversely, if
$\tau:\Omega\to[0,\infty]$ is a stopping time adapted to
$\langle\hat\Sigma_t^+\rangle_{t\ge 0}$, there is a stopping time
$\tau'$, adapted to $\langle\Sigma_t^+\rangle_{t\ge 0}$, such that
$\tau\eae\tau'$.   \prooflet{\Prf\
For each $q\in\Bbb Q\cap\coint{0,\infty}$, set
$F_q=\{\omega:\tau(\omega)<q\}$;  by (b), $F_q\in\hat\Sigma_q$ and
there is an $E_q\in\Sigma_q$ such that $F_q\symmdiff E_q$ is negligible.
For $\omega\in\Omega$,
set $\tau'(\omega)=\inf\{q:q\in\Bbb Q\cap\coint{0,\infty}$,
$\omega\in E_q\}$, counting $\inf\emptyset$ as $\infty$.   Then
$\{\omega:\tau'(\omega)<t\}
=\bigcup_{q\in\Bbb Q\cap\coint{0,t}}E_q$ belongs to $\Sigma_t$ for
every $t$, so $\tau'$ is adapted to $\langle\Sigma_t^+\rangle_{t\ge 0}$.
And $\{\omega:\tau'(\omega)\ne\tau(\omega)\}
\subseteq\bigcup_{q\in\Bbb Q\cap\coint{0,\infty}}E_q\symmdiff F_q$ is
negligible.\ \Qed}

\medskip

\quad{\bf (iv)} Continuing from (iii) just above, we find that, defining
$\hat\Sigma_{\tau}^+$ from $\langle\hat\Sigma_t^+\rangle_{t\ge 0}$ and
$\tau$ and $\Sigma_{\tau'}^+$ from
$\langle\Sigma_t^+\rangle_{t\ge 0}$ and
$\tau'$ by the formula in (c-iii), then
$\hat\Sigma_{\tau}^+=\{F\symmdiff A:F\in\Sigma_{\tau'}^+$,
$A\in\Cal N(\mu)\}$.   \prooflet{\Prf\ Let $A_0$ be the negligible set
$\{\omega:\tau(\omega)\ne\tau'(\omega)$.   ($\alpha$) If
$E\in\Sigma_{\tau'}^+$, then for every $t\ge 0$ we have

\Centerline{$E\cap\{\omega:\tau'(\omega)\le t\}
\in\Sigma^+_t$,}

\Centerline{$(E\cap\{\omega:\tau(\omega)\le t\})\symmdiff
   (E\cap\{\omega:\tau'(\omega)\le t\})\subseteq A_0\in\Cal N(\mu)$,}

\noindent so (using (ii))
$E\cap\{\omega:\tau(\omega)\le t\}\in\hat\Sigma^+_t$;  as $t$ is arbitrary,
$E\in\hat\Sigma_{\tau}^+$.   ($\beta$) If $F\in\hat\Sigma_{\tau}^+$, then
for every $q\in\Bbb Q\cap\coint{0,\infty}$ the sets
$F\cap\{\omega:\tau(\omega)\le q\}$ and
$F\cap\{\omega:\tau'(\omega)\le q\}$ belong to $\hat\Sigma^+_q$, so there
is an $E_q\in\Sigma^+_q$ such that
$E_q\symmdiff(F\cap\{\omega:\tau'(\omega)\le q\})$ is negligible.
Set $E'_q=\bigcup_{r\in\Bbb Q\cap[0,q]}E_r$ for
$q\in\Bbb Q\cap\coint{0,\infty}$;  then $E'_q\in\Sigma^+_q$ and
$E'_q\symmdiff(F\cap\{\omega:\tau'(\omega)\le q\})$ is negligible for
each $q$, while $E'_q\subseteq E'_r$ if $q\le r$ in
$\Bbb Q\cap\coint{0,\infty}$.   It follows that

\Centerline{$\bigcap_{q\in\Bbb Q\cap\coint{t,\infty}}E'_q
=\bigcap_{q\in\Bbb Q\cap[t,s]}E'_q\in\Sigma^+_s$}

\noindent whenever $t<s$ in $\coint{0,\infty}$, so that
$\bigcap_{q\in\Bbb Q\cap\coint{t,\infty}}E'_q\in\Sigma^+_t$ for every $t$.
Set

\Centerline{$E=\bigcap_{q\in\Bbb Q\cap\coint{0,\infty}}
   E'_q\cup\{\omega:\tau'(\omega)>q\}$.}

\noindent Then

$$\eqalign{&E\cap\{\omega:\tau'(\omega)\le t\}\cr
&\mskip50mu
=\{\omega:\tau'(\omega)\le t\}\cap\bigcap_{q\in\Bbb Q\cap\coint{0,\infty}}
   (E'_q\cup\{\omega:\tau'(\omega)>q\})\cr
&\mskip50mu
=\{\omega:\tau'(\omega)\le t\}\cap\bigcap_{q\in\Bbb Q\cap\coint{0,t}}
   (E'_q\cup\{\omega:\tau'(\omega)>q\})
   \cap\bigcap_{q\in\Bbb Q\cap\coint{t,\infty}}E'_q\cr
&\mskip50mu\in\Sigma^+_t\cr}$$

\noindent for any $t\ge 0$, so $E\in\Sigma^+_{\tau'}$.
If we look at $(E\symmdiff F)\cap\{\omega:\tau'(\omega)<\infty\}$,
we see that this is included in the negligible set

\Centerline{$\bigcup_{q\in\Bbb Q\cap\coint{0,\infty}}
   E'_q\symmdiff(F\cap\{\omega:\tau'(\omega)\le q\})$}

\noindent because $\{\omega:\tau'(\omega)<\infty\}\cap F$ is just

\Centerline{$\{\omega:\tau'(\omega)<\infty\}
  \cap\bigcap_{q\in\Bbb Q\cap\coint{0,\infty}}
      (F\cap\{\omega:\tau'(\omega)\le q\})\cup\{\omega:\tau'(\omega)>q\}$.}

\noindent As for the set $H=\{\omega:\tau'(\omega)=\infty\}$, this belongs
to $\Sigma$, and every subset of $H$ belonging to $\Sigma$ also belongs to
$\Sigma^+_{\tau'}$.   Let $H'\in\Sigma$ be such that $H'\subseteq H$ and
$H'\symmdiff(F\cap H)$ is negligible;  then $E'=H'\cup(E\setminus H)$
belongs to $\Sigma^+_{\tau'}$ and differs from $F$ by a negligible set.\
\Qed}

\leader{455M}{Hitting \dvrocolon{times}}\cmmnt{ I mention a class of
stopping times which is particularly important in applications, and also
very helpful in giving an idea of the concept.

\medskip

\noindent}{\bf Proposition}  Let $U$ be a Polish
space and $\Cdlg$ the set of \cadlag\ functions from $\coint{0,\infty}$ to
$U$.   Let $A\subseteq U$ be an analytic set, and define
$\tau:\Cdlg\to[0,\infty]$ by setting

\Centerline{$\tau(\omega)=\inf\{t:\omega(t)\in A\}$}

\noindent for $\omega\in\Cdlg$, counting $\inf\emptyset$ as $\infty$.

(a) Let $\Sigma$ be a $\sigma$-algebra of subsets of $\Cdlg$ closed
under Souslin's operation and including the algebra generated by the
functionals $\omega\mapsto\omega(t)$ for $t\ge 0$.   Then $\tau$ is
$\Sigma$-measurable.

(b) For $t\ge 0$ let $\Sigma_t$ be

\Centerline{$\{F:F\in\Sigma,\,\omega'\in F$ whenever $\omega$,
$\omega'\in\Cdlg$, $\omega\in F$ and
$\omega\restr[0,t]=\omega'\restr[0,t]\}$,}

\noindent and $\Sigma_t^+=\bigcap_{s>t}\Sigma_t$.   Then $\tau$ is a
stopping time adapted to $\langle\Sigma^+_t\rangle_{t\ge 0}$.

(c) If $A$ is closed, then $\tau$ is adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$.
%this bit seems to need hitting time to be  \inf_{t\ge 0}  rather than
% \inf_{t>0}

\proof{{\bf (a)(i)} It will help to recall from 4A3Q that there is a Polish
topology $\frak S$ on $\Cdlg$ such that the corresponding Borel
$\sigma$-algebra $\Cal B(\Cdlg)$
is just the $\sigma$-algebra generated by the coordinate functions
$\omega\mapsto\omega(t)$, so is included in $\Sigma$.   
In this case, every $\frak S$-analytic set,
being $\frak S$-Souslin-F (423Eb), belongs to $\Sigma$.

\medskip

\quad{\bf (ii)} The set

\Centerline{$W
=\{(\omega,t,x):\omega\in\Cdlg$, $t\ge 0$, $x\in U$, $\omega(t)=x\}$}

\noindent is a Borel subset of $\Cdlg\times\coint{0,\infty}\times U$.
\Prf\ If $\rho$ is a metric on $U$ inducing its topology and $D$ is a
countable dense subset of $U$,

$$\eqalign{W
&=\bigcap_{n\in\Bbb N}\bigcup_{q\in\Bbb Q,q\ge 0}\bigcup_{y\in D}
    \{(\omega,t,x):t\in[q-2^{-n},q],\cr
&\mskip200mu  \rho(\omega(q),y)\le 2^{-n},\,
    \rho(x,y)\le 2^{-n}\}.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)} Since $\Cdlg$, $\coint{0,\infty}$ and $U$ are all
Polish, $W$ is an analytic set.   Now, for any $t\ge 0$,

\Centerline{$W'
=\{(\omega,s,x):s\in\coint{0,t}$, $x\in A$, $(\omega,s,x)\in W\}$}

\noindent is analytic and its projection

\Centerline{$\{\omega:\tau(\omega)<t\}
=\{\omega:$ there are $s$, $x$ such that $(\omega,s,x)\in W'\}$}

\noindent is analytic and belongs to $\Sigma$.   As $t$ is arbitrary,
$\tau$ is $\Sigma$-measurable.

\medskip

{\bf (b)} Now, given $t\ge 0$, $F=\{\omega:\tau(\omega)<t\}$ belongs to
$\Sigma$, and if $\omega\in F$, $\omega'\in\Cdlg$ are such that
$\omega'\restr[0,t]=\omega\restr[0,t]$, there is an $s<t$ such that
$\omega'(s)=\omega(s)\in A$, so $\tau(\omega')<t$ and $\omega'\in F$.
Thus $F\in\Sigma_t$.   As $t$ is arbitrary, $\tau$ is a stopping time
adapted to $\langle\Sigma_t^+\rangle_{t\ge 0}$, by 455Lb.

\medskip

{\bf (c)} As $A$ is closed and every member of $\Cdlg$ is continuous on the
right, $\omega(\tau(\omega))\in A$ whenever $\tau(\omega)<\infty$.   So
if $\omega$, $\omega'\in\Cdlg$, $\tau(\omega)\le t$ and
$\omega'\restr[0,t]=\omega\restr[0,t]$, then
$\omega'(\tau(\omega))\in A$ and
$\tau(\omega')\le t$.   Thus $\{\omega:\tau(\omega)\le t\}\in\Sigma_t$ for
every $t$, and $\tau$ is adapted to $\langle\Sigma_t\rangle_{t\ge 0}$.
}%end of proof of 455M

\leader{455N}{}\cmmnt{ We need an elementary fact about narrow (more
properly, vague) convergence.

\medskip

\noindent}{\bf Lemma} Let $(U,\rho)$ be a metric space, $n\in\Bbb N$
and $f:U^{n+1}\to\Bbb R$ a bounded uniformly continuous function.   Let
$\langle\nu^{(k)}_x\rangle_{k<n,x\in U}$ be a family of topological
probability measures on $U$ such that $x\mapsto\nu^{(k)}_x$ is continuous
for the narrow topology for every $k<n$.   Then

\Centerline{$y
\mapsto\iint\ldots\int f(y,x_1,\ldots,x_n)\nu^{(n-1)}_{x_{n-1}}(dx_n)
    \ldots\nu_{x_1}^{(1)}(dx_2)\nu_y^{(0)}(dx_1)$}

\noindent is defined everywhere on $U$ and continuous.

\proof{ Induce on $n$.   If $n=0$ the formula is just
$y\mapsto f(y)$, so the result is trivial.
For the inductive step to $n\ge 1$, set

\Centerline{$g(y,x_1)
=\int\ldots\int f(y,x_1,\ldots,x_n)
  \nu^{(n-1)}_{x_{n-1}}(dx_n)\ldots\nu_{x_1}^{(1)}(dx_2)$}

\noindent for $y$, $x_1\in U$;  by the inductive hypothesis this is
well-defined and $x_1\mapsto g(y,x_1)$ is continuous.   Note that
$g$ is bounded because $f$ is.   It follows that
$h(y)=\int g(y,x_1)\nu_y^{(0)}(dx_1)$ is defined for every
$y$.   I need to show that $h$ is continuous.   Take any
$y\in U$ and $\epsilon>0$.   Then there is a $\delta_0>0$ such that
$|f(y',x_1,\ldots,x_n)-f(y,x_1,\ldots,x_n)|\le\epsilon$ whenever
$\rho(y',y)\le\delta_0$ and $x_1,\ldots,x_n\in U$;  so that
$|g(y',x_1)-g(y,x_1)|\le\epsilon$ whenever $\rho(y',y)\le\delta_0$ and
$x_1\in U$.   Next, because $x\mapsto\nu_x^{(0)}$ is narrowly continuous,
$x\mapsto\int g(y,x_1)\nu_x^{(0)}(dx_1)$ is continuous (437Jf/437Kb),
and there is a $\delta\in\ocint{0,\delta_0}$ such that
$|\int g(y,x_1)\nu_{y'}^{(0)}(dx_1)
-\int g(y,x_1)\nu_y^{(0)}(dx_1)|\le\epsilon$ whenever
$\rho(y',y)\le\delta$.   So if $\rho(y',y)\le\delta$,

$$\eqalign{|h(y')-h(y)|
&\le|\int g(y',x_1)\nu_{y'}^{(0)}(dx_1)-\int g(y,x_1)\nu_{y'}^{(0)}(dx_1)|
  \cr
&\mskip100mu
 +|\int g(y,x_1)\nu_{y'}^{(0)}(dx_1)-\int g(y,x_1)\nu_{y}^{(0)}(dx_1)|\cr
&\le\int|g(y',x_1)-g(y,x_1)|\nu_{y'}^{(0)}(dx_1)
  +\epsilon
\le 2\epsilon.\cr}$$

\noindent As $y$ and $\epsilon$ are arbitrary, $h$ is continuous and the
induction proceeds.
}%end of proof of 455N

\leader{455O}{}\cmmnt{ If both the continuity conditions in 455F are
satisfied, we have a version of 455C/455Eb which is much more to the point.

\medskip

\noindent}{\bf Theorem} Suppose that $(U,\rho)$ is a complete metric space,
$x^*$ is a point of $U$,
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is a family of Radon
probability measures on $U$ which is both narrowly continuous and uniformly
time-continuous on the right, and
that $\family{y}{U}{\nu^{(t,u)}_y}$ is a disintegration of $\nu^{(s,u)}_x$
over $\nu^{(s,t)}_x$ whenever $x\in U$ and $s<t<u$.
Let $\hat\mu$ be the
corresponding completed measure on $\Omega=U^{\coint{0,\infty}}$,
as in 455E.
Let $\Cdlg$ be the set of \cadlag\ functions from $\coint{0,\infty}$ to
$U$, $\ddot\mu$ the subspace measure on
$\Cdlg$, and $\ddot\Sigma$ its domain.   For $t\ge 0$, let $\ddot\Sigma_t$
be

\Centerline{$\{F:F\in\ddot\Sigma$,
$\omega'\in F$ whenever $\omega\in F$, $\omega'\in\Cdlg$ and
$\omega'\restr[0,t]=\omega\restr[0,t]\}$,}

\noindent and $\ddot\Sigma^+_t=\bigcap_{s>t}\ddot\Sigma_t$.

For $\omega\in\Omega$ and $a\ge 0$ let $\hat\mu_{\omega a}$ be the
completed measure on $\Omega$ built from $\omega(0)$ and
$\langle\nu^{(s,t)}_{\omega ax}\rangle_{0\le s<t,x\in U}$
as in 455Eb;  let $\ddot\mu_{\omega a}$ be the subspace measure on $\Cdlg$.
Let $\tau:\Cdlg\to[0,\infty]$ be a stopping time adapted to
$\langle\ddot\Sigma^+_t\rangle_{t\ge 0}$.

(a) $\family{\omega}{\Cdlg}{\ddot\mu_{\omega,\tau(\omega)}}$
is a disintegration of $\ddot\mu$ over itself.

(b) Set

\Centerline{$\ddot\Sigma^+_{\tau}
=\{F:F\in\ddot\Sigma,\,F\cap\{\omega:\tau(\omega)\le t\}
   \in\ddot\Sigma^+_t\text{ for every }t\ge 0\}$.}

\noindent Then $\ddot\Sigma^+_{\tau}$ is a $\sigma$-algebra of subsets of
$\Cdlg$.   For a $\ddot\mu$-integrable function
$f$ on $\Cdlg$, write
$\ddot g_f(\omega)=\int_{\Cdlg}fd\ddot\mu_{\omega,\tau(\omega)}$
when this is defined in $\Bbb R$.
Then $\ddot g_f$ is a conditional expectation of $f$ on
$\ddot\Sigma^+_{\tau}$.

(c) If $\tau$ is adapted to $\langle\ddot\Sigma_t\rangle_{t\ge 0}$, set

\Centerline{$\ddot\Sigma_{\tau}
=\{F:F\in\ddot\Sigma,\,F\cap\{\omega:\tau(\omega)\le t\}
   \in\ddot\Sigma_t\text{ for every }t\ge 0\}$.}

\noindent Then $\ddot\Sigma_{\tau}$ is a $\sigma$-algebra of subsets of
$\Cdlg$, and $\ddot g_f$ is a conditional expectation of $f$ on
$\ddot\Sigma_{\tau}$, for every $f\in\eusm L^1(\ddot\mu)$.

\proof{{\bf (a)(i)} I had better begin by checking that the ground is
clear.   By 455G, $\hat\mu^*\Cdlg=\hat\mu^*_{\omega a}\Cdlg=1$
for every $\omega\in\Cdlg$ and $a\ge 0$, so that $\ddot\mu$
and $\ddot\mu_{\omega a}$ (for $\omega\in\Cdlg$) are all
probability measures.
\leaveitout{Because $\hat\mu$ is the completion of its restriction to
$\Tensorhat_{\coint{0,\infty}}\Cal B(U)$, $\ddot\mu$ is the completion of
its restriction to the subspace $\sigma$-algebra $\Tau$ induced by
$\Tensorhat_{\coint{0,\infty}}\Cal B(U)$ on $\Cdlg$
(214Ib\formerly{2{}14Xb});
now this is the $\sigma$-algebra generated by the functionals
$\omega\mapsto\omega(t)$ for $t\ge 0$, so that $(\Cdlg,\Tau)$ is a standard
Borel space (4A3N, 4A3Q).}

Of course $\langle\ddot\Sigma_t\rangle_{t\ge 0}$ is a non-decreasing family of
$\sigma$-subalgebras of $\ddot\Sigma$, so that
$\langle\ddot\Sigma^+_t\rangle_{t\ge 0}$ is another such family, and
we are in the territory explored in 455L.

\medskip

\quad{\bf (ii)} Write $\Sigma$ for the domain of $\hat\mu$, and for
$t\ge 0$ set

\Centerline{$\Sigma_t
=\{E:E\in\Sigma$, $E$ is determined by coordinates in
$[0,t]\}$.}

\noindent Then $\ddot\Sigma_t=\{E\cap\Cdlg:E\in\Sigma_t\}$.
\Prf\ If $E\in\Sigma_t$, then $E\cap\Cdlg\in\ddot\Sigma$ and clearly
$E\cap\Cdlg\in\ddot\Sigma_t$.  If $F\in\ddot\Sigma_t$,
let $E\in\Sigma$ be such that $E\cap\Cdlg=F$.
Applying 455Ec to the stopping time with constant value $t$, we have

\Centerline{$\hat\mu E
=\int_{\Omega}\hat\mu_{\omega t}(E)\hat\mu(d\omega)$.}

\noindent Set

\Centerline{$E^*=\{\omega:\omega\in\Omega$, $\hat\mu_{\omega t}(E)$ is
defined$\}$,}

\Centerline{$E_0=\{\omega:\omega\in E^*$, $\hat\mu_{\omega t}(E)=0\}$,
\quad$E_1=\{\omega:\omega\in E^*$, $\hat\mu_{\omega t}(E)=1\}$.}

\noindent Then $E^*$, $E_0$ and $E_1$ are measured by $\hat\mu$ and are
determined by coordinates in $[0,t]$ (by 455E(b-iii)),
and $\hat\mu E^*=1$.

If $\omega\in E^*\cap\Cdlg$, then $\hat\mu_{\omega t}^*\Cdlg=1$, so

\Centerline{$\ddot\mu_{\omega t}(F)
=\hat\mu^*_{\omega t}(E\cap\Cdlg)
=\hat\mu_{\omega t}(E)$.}

\noindent If $\omega\in\Cdlg$, let $D$ be a countable dense subset of
$[0,t]$ containing $t$;  then

$$\eqalign{1
&=\hat\mu_{\omega t}\{\omega':\omega'\in\Omega,\,
   \omega'\restr D=\omega\restr D\}
=\ddot\mu_{\omega t}\{\omega':\omega'\in\Cdlg,\,
   \omega'\restr D=\omega\restr D\}\cr
&=\ddot\mu_{\omega t}\{\omega':\omega'\in\Cdlg,\,
   \omega'\restr[0,t]=\omega\restr[0,t]\}.\cr}$$

\noindent So if $\omega\in E^*\cap\Cdlg$,

\Centerline{$\hat\mu_{\omega t}(E)
=\ddot\mu_{\omega t}(F)
=\ddot\mu_{\omega t}\{\omega':\omega'\in F,\,
  \omega'\restr[0,t]=\omega\restr[0,t]\}
=\chi F(\omega)\in\{0,1\}$}

\noindent because $F$ is determined (relative to $\Cdlg$)
by coordinates in $[0,t]$.   This means
that $E_1\cap\Cdlg\subseteq F$ and $E_0\cap F=\emptyset$, while
$E_0\cup E_1$ is $\hat\mu$-conegligible.   So if we take

\Centerline{$E'
=E_1\cup\{\omega:\omega\Omega\setminus E_1$
and there is an $\omega'\in F$ such that
$\omega'\restr[0,t]=\omega\restr[0,t]\}$,}

\noindent $E'\cap\Cdlg=F$, $E'$ is determined by coordinates in $[0,t]$,
$E_1\subseteq E'\subseteq\Omega\setminus E_0$, $\hat\mu$ measures $E'$ and
$E'\in\Sigma_t$.   Thus $\ddot\Sigma_t=\{E\cap\Cdlg:E\in\Sigma_t\}$,
as claimed.\ \Qed

\medskip

\quad{\bf (iii)} Take $n\in\Bbb N$, and set $D_n=\{2^{-n}i:i\in\Bbb N\}$.
Suppose that $\tau:\Cdlg\to D_n\cup\{\infty\}$ is a
stopping time adapted to $\langle\ddot\Sigma_t\rangle_{t\ge 0}$.   Then
$\family{\omega}{\Cdlg}{\hat\mu_{\omega,\tau(\omega)}}$
is a disintegration of $\hat\mu$ over $\ddot\mu$.   \Prf\ For each
$i\in\Bbb N$, $F_i=\tau^{-1}[\{2^{-n}i\}]$ belongs to
$\ddot\Sigma_{2^{-n}i}$, so there is an $E_i\in\Sigma$,
determined by coordinates in
$[0,2^{-n}i]$, such that $F_i=E_i\cap\Cdlg$.   For $\omega\in\Omega$, set

\Centerline{$\grave\tau(\omega)
=\inf\{2^{-n}i:i\in\Bbb N,\,\omega\in E_i\}$,}

\noindent counting $\inf\emptyset$ as $\infty$.   Then
$\grave\tau\restr\Cdlg=\tau$.  Also
$\grave\tau[\Omega]\subseteq D_n\cup\{\infty\}$ is
countable, and $\grave\tau^{-1}[\{b\}]\in\Sigma$ is determined by
coordinates in $[0,b]$ for every $b\in D_n$.   By 455Ec,
$\family{\omega}{\Omega}{\hat\mu_{\omega,\grave\tau(\omega)}}$ is a
disintegration of $\hat\mu$ over itself.

Now take any $E\in\Sigma$.   Then

$$\eqalignno{\hat\mu E
&=\int_{\Omega}\hat\mu_{\omega,\grave\tau(\omega)}(E)\hat\mu(d\omega)
=\int_{\Cdlg}\hat\mu_{\omega,\grave\tau(\omega)}(E)\ddot\mu(d\omega)\cr
\displaycause{214F}
&=\int_{\Cdlg}\hat\mu_{\omega,\tau(\omega)}(E)\ddot\mu(d\omega).
\text{ \Qed}\cr}$$

It follows that
$\family{\omega}{\Cdlg}{\ddot\mu_{\omega,\tau(\omega)}}$
is a disintegration of $\ddot\mu$ over itself.   \Prf\ If
$F\in\ddot\Sigma$, there is an $E\in\Sigma$ such that $F=E\cap\Cdlg$.
Now

$$\eqalign{\ddot\mu F
&=\hat\mu E
=\int_{\Cdlg}\hat\mu_{\omega,\tau(\omega)}(E)\ddot\mu(d\omega)\cr
&=\int_{\Cdlg}\ddot\mu_{\omega,\tau(\omega)}(E\cap\Cdlg)\ddot\mu(d\omega)
=\int_{\Cdlg}\ddot\mu_{\omega,\tau(\omega)}(F)\ddot\mu(d\omega).
\text{ \Qed}\cr}$$

\medskip

\quad{\bf (iv)} Now let $\tau:\Cdlg\to[0,\infty]$ be any stopping time adapted to
$\langle\ddot\Sigma^+_t\rangle_{t\ge 0}$.   For each $n\in\Bbb N$, define
$\tau_n:\Cdlg\to D_n\cup\{\infty\}$ by setting

$$\eqalign{\tau_n(\omega)
&=2^{-n}(i+1)\text{ if }i\in\Bbb N\text{ and }
   2^{-n}i\le\tau(\omega)<2^{-n}(i+1),\cr
&=\infty\text{ if }\tau(\omega)=\infty.\cr}$$

\noindent By 455Lb, $\{\omega:\tau_n(\omega)=t\}\in\ddot\Sigma_t$ for every
$t\in D_n$.   So (iii) tells us that
$\langle\ddot\mu_{\omega,\tau_n(\omega)}\rangle_{\omega\in\Cdlg}$ is a
disintegration of $\ddot\mu$ over itself.

\medskip

\quad{\bf (v)} Suppose that $k\in\Bbb N$, $0=t_0<t_1<\ldots<t_k$,
$h:U^{k+1}\to\Bbb R$ is bounded and uniformly continuous,
and $\omega\in\Cdlg$.   Then

\Centerline{$\int_{\Omega}h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau(\omega)}(d\omega')
=\lim_{n\to\infty}\int_{\Omega}h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau_n(\omega)}(d\omega')$.}

\noindent\Prf\ Recall from 455E that

$$\eqalign{\int_{\Omega}h(\omega'(t_0),\ldots\,&,\omega'(t_k))
   \hat\mu_{\omega,\tau(\omega)}(d\omega')\cr
&=\int_U\ldots\int_Uh(\omega(0),x_1,\ldots,x_k)
   \nu_{\omega,\tau(\omega),x_{k-1}}^{(t_{k-1},t_k)}(dx_k)
   \ldots\nu^{(0,t_1)}_{\omega(0)}(dx_1),\cr}$$

\noindent and similarly for each $\tau_n$.
If $\tau(\omega)\ge t_k$, then

\Centerline{$\nu^{(t_{i-1},t_i)}_{\omega,\tau_n(\omega),x}
=\delta_{\omega(t_i)}=\nu^{(t_{i-1},t_i)}_{\omega,\tau(\omega),x}$}

\noindent for $1\le i\le k$, $n\in\Bbb N$ and $x\in U$, so the result is
trivial.   If $j\le k$ is such that $t_{j-1}\le\tau(\omega)<t_j$, then

$$\eqalign{\nu^{(t_{i-1},t_i)}_{\omega,\tau(\omega),x}
&=\delta_{\omega(t_i)}\text{ if }i<j,\cr
&=\nu^{(\tau(\omega),t_j)}_{\omega(\tau(\omega))}
   \text{ if }i=j,\cr
&=\nu_x^{(t_{i-1},t_i)}\text{ if }j<i<k.\cr}$$

\noindent So

$$\eqalign{\int_{\Omega}h(\omega'(t_0),\ldots\,&,\omega'(t_k))
   \hat\mu_{\omega,\tau(\omega)}(d\omega')\cr
&=\int_U\int_U\ldots\int_Uh(\omega(0),\ldots,\omega(j-1),x_j,\ldots,x_k)\cr
&\mskip150mu
   \nu_{x_{k-1}}^{(t_{k-1},t_k)}(dx_k)
   \ldots
   \nu_{x_j}^{(t_j,t_{j+1})}(dx_{j+1})
   \nu_{\omega(\tau(\omega))}^{(\tau(\omega),t_j)}(dx_j).\cr}$$

\noindent Moreover, there is some $n_0$ such that $\tau_n(\omega)<t_j$ for
every $n\ge n_0$, so that we can use this formula for all such $n$.
Setting

\Centerline{$g(x)
=\int_U\ldots\int_Uh(\omega(0),\ldots,\omega(j-1),x,x_{j+1},\ldots,x_k)
   \nu_{x_{k-1}}^{(t_{k-1},t_k)}(dx_k)
   \ldots
   \nu_x^{(t_j,t_{j+1})}(dx_{j+1})$}

\noindent for $x\in U$, we see from 455N that $g$ is continuous, while of
course it is also bounded, because $h$ is bounded.   At this point, recall
that $\omega$ is supposed to be continuous on the right, while
the system of transitional probabilities is jointly continuous, so that

\Centerline{$\nu_{\omega(\tau(\omega))}^{(\tau(\omega),t_j)}
=\lim_{n\to\infty}\nu_{\omega(\tau_n(\omega))}^{(\tau_n(\omega),t_j)}$}

\noindent for the narrow topology, and

$$\eqalign{\lim_{n\to\infty}
   \int_{\Omega}h(\omega'(t_0),\ldots\,&,\omega'(t_k))
   \hat\mu_{\omega,\tau_n(\omega)}(d\omega')\cr
&=\lim_{n\to\infty}
  \int_U\int_U\ldots\int_Uh(\omega(0),\ldots,\omega(j-1),x_j,\ldots,x_k)\cr
&\mskip100mu
   \nu_{x_{k-1}}^{(t_{k-1},t_k)}(dx_k)
   \ldots
   \nu_{x_j}^{(t_j,t_{j+1})}(dx_{j+1})
   \nu_{\omega(\tau_n(\omega))}^{(\tau_n(\omega),t_j)}(dx_j)\cr
&=\lim_{n\to\infty}\int_Ug(x_j)
   \nu_{\omega(\tau_n(\omega))}^{(\tau_n(\omega),t_j)}(dx_j)\cr
&=\int_Ug(x_j)
   \nu_{\omega(\tau(\omega))}^{(\tau(\omega),t_j)}(dx_j)\cr
&=\int_{\Omega}h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau(\omega)}(d\omega'),\cr}$$

\noindent as claimed.\ \Qed

\medskip

\quad{\bf (vi)} Again suppose that $0=t_0<t_1<\ldots<t_k$.   If
$h:U^{k+1}\to\Bbb R$ is bounded and uniformly continuous, then

\Centerline{$\int_{\Cdlg}\int_{\Omega}h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau(\omega)}(d\omega')\ddot\mu(d\omega)
=\int_{\Omega}h(\omega(t_0),\ldots,\omega(t_k))\hat\mu(d\omega)$.}

\noindent\Prf\ The point here is that

$$\eqalign{\int_{\Cdlg}\int_{\Omega}h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau_n(\omega)}(d\omega')\ddot\mu(d\omega)
&=\int_{\Omega}h(\omega(t_0),\ldots,\omega(t_k))
   \hat\mu(d\omega)\cr}$$

\noindent is defined for every $n\in\Bbb N$, by (iii) and 452F, as usual.
Now the integrands

\Centerline{$\omega
\mapsto\int_{\Omega}h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau_n(\omega)}(d\omega')$}

\noindent converge at every point of $\Cdlg$, by (v), and are uniformly
bounded, because $h$ is, so that

$$\eqalign{\int_{\Cdlg}\int_{\Omega}h(\omega'(t_0),\ldots\,&,\omega'(t_k))
   \hat\mu_{\omega,\tau(\omega)}(d\omega')\ddot\mu(d\omega)\cr
&=\lim_{n\to\infty}\int_{\Cdlg}\int_{\Omega}
    h(\omega'(t_0),\ldots,\omega'(t_k))
   \hat\mu_{\omega,\tau_n(\omega)}(d\omega')\ddot\mu(d\omega)\cr
&=\int_{\Omega}h(\omega(t_0),\ldots,\omega(t_k))(d\omega).
\text{  \Qed}\cr}$$

If $G\subseteq U^{k+1}$ is open, there is a non-decreasing sequence
$\sequence{m}{h_m}$ of uniformly continuous functions from $U^{k+1}$ to
$[0,1]$ such that $\chi G=\sup_{m\in\Bbb N}h_m$, in which case

$$\eqalign{\int_{\Cdlg}
  \hat\mu_{\omega,\tau(\omega)}
  \{\omega':(\omega'(t_0)&,\ldots,\omega'(t_k))\in G\}
 \ddot\mu(d\omega)\cr
&=\lim_{m\to\infty}
  \int_{\Cdlg}\int_{\Omega}h_m(\omega'(t_0),\ldots,\omega'(t_k))
  \hat\mu_{\omega,\tau(\omega)}(d\omega')\ddot\mu(d\omega)\cr
&=\lim_{m\to\infty}
  \int_{\Omega}h_m(\omega(t_0),\ldots,\omega(t_k))\hat\mu(d\omega)\cr
&=\hat\mu\{\omega:(\omega(t_0),\ldots,\omega(t_k))\in G\}.\cr}$$

\noindent By the Monotone Class Theorem, we get

$$\eqalign{\int_{\Cdlg}
  \hat\mu_{\omega,\tau(\omega)}
  \{\omega':(\omega'(t_0)&,\ldots,\omega'(t_k))\in E\}
 \ddot\mu(d\omega)\cr
&=\hat\mu\{\omega:(\omega(t_0),\ldots,\omega(t_k))\in E\}\cr}$$

\noindent for every Borel set $E\subseteq U^{k+1}$.   Now recall that
$t_0,\ldots,t_k$ were any strictly increasing sequence starting at $0$, so
we can use the Monotone Class Theorem yet again to see that

\Centerline{$\int_{\Cdlg}
  \hat\mu_{\omega,\tau(\omega)}(E)\ddot\mu(d\omega)
=\hat\mu(E)$}

\noindent for every $E\in\Tensorhat_{\coint{0,\infty}}\Cal B(U)$
and therefore for every $E\in\Sigma$.

\medskip

\quad{\bf (vii)} Finally, if $F\in\ddot\Sigma$,
there is an $E\in\Sigma$ such that $F=E\cap\Cdlg$, so that

\Centerline{$\ddot\mu(F)=\hat\mu(E)
=\int_{\Cdlg}\hat\mu_{\omega,\tau(\omega)}(E)\ddot\mu(d\omega)
=\int_{\Cdlg}\ddot\mu_{\omega,\tau(\omega)}(F)\ddot\mu(d\omega)$;}

\noindent which is what we set out to prove.

\medskip

{\bf (b)(i)} By 455L(c-iii), $\ddot\Sigma^+_{\tau}$ is a $\sigma$-algebra.
If $f$ is a $\ddot\mu$-integrable real-valued function, then
$\int_{\Cdlg}\ddot g_f\ddot\mu=\int_{\Cdlg}f\ddot\mu$, by (a) and 452F.
For $\alpha\in\Bbb R$ set

\Centerline{$E(f,\alpha)
=\{\omega:\omega\in\Cdlg$, $\ddot g_f(\omega)$ is defined in $\Bbb R$ and
$\ddot g_f(\omega)\le\alpha\}$,}

\noindent so that $E(f,\alpha)\in\ddot\Sigma$.   For $t\ge 0$, set

\Centerline{$H_t=\{\omega:\omega\in\Cdlg$, $\tau(\omega)\le t\}$,
\quad$H'_t=\{\omega:\omega\in\Cdlg$, $\tau(\omega)<t\}$,}

\noindent so that $H_t\in\ddot\Sigma_t^+$ and $H'_t\in\ddot\Sigma_t$
(455Lb).

\medskip

\quad{\bf (ii)} If $\omega$, $\omega'\in\Cdlg$ and $s>\tau(\omega)$
are such that
$\omega'\restr[0,s]=\omega\restr[0,s]$, then
$\tau(\omega')=\tau(\omega)$.   \Prf\ $H_{\tau(\omega)}$,
$H'_{\tau(\omega)}$ and their difference belong to $\ddot\Sigma_s$,
so are determined (relative to $\Cdlg$) by coordinates in $[0,s]$;
since $H_{\tau(\omega)}\setminus H'_{\tau(\omega)}$ contains $\omega$, it
also contains $\omega'$, and $\tau(\omega')=\tau(\omega)$.\ \Qed

\medskip

\quad{\bf (iii)} If $f$ is $\ddot\mu$-integrable, $\alpha\in\Bbb R$ and
$s>0$, then $E(f,\alpha)\cap H'_s\in\ddot\Sigma_s$.
\Prf\ Certainly $E(f,\alpha)\cap H'_s\in\ddot\Sigma$.   If $\omega$,
$\omega'\in\Cdlg$ and $\omega\restr[0,s]=\omega'\restr[0,s]$,
then

$$\eqalignno{\omega\in E(f,\alpha)\cap H'_s
&\Longrightarrow\tau(\omega)<s\text{ and }
  \int_{\Cdlg}fd\ddot\mu_{\omega,\tau(\omega)}\le\alpha\cr
&\Longrightarrow\tau(\omega')=\tau(\omega)<s\text{ and }
  \int_{\Cdlg}fd\ddot\mu_{\omega,\tau(\omega')}\le\alpha\cr
\displaycause{by (ii)}
&\Longrightarrow\omega'\in E(f,\alpha)\cap H'_s.\cr}$$

\noindent So $E(f,\alpha)\cap H'_s$ is determined (relative to $\Cdlg$) by
coordinates in $[0,s]$ and belongs to $\ddot\Sigma_s$.\ \Qed

Consequently $E(f,\alpha)\cap H_t\in\ddot\Sigma^+_t$ for every $t\ge 0$.
\Prf\ $H_t=\bigcap_{n\in\Bbb N}H'_{t_n}$ where $t_n=t+2^{-n}$ for each $n$,
so

\Centerline{$E(f,\alpha)\cap H_t
=\bigcap_{n\ge m}E(f,\alpha)\cap H'_{t_n}$}

\noindent belongs to $\ddot\Sigma_{t_m}$ for every $m\in\Bbb N$, and
$E(f,\alpha)\cap H_t\in\ddot\Sigma^+_t$.\ \Qed

Thus $E(f,\alpha)\in\ddot\Sigma^+_{\tau}$ for every $\alpha$.   As $\alpha$ is
arbitrary, $\dom\ddot g_f\in\ddot\Sigma^+_{\tau}$ and $\ddot g_f$ is
$\ddot\Sigma^+_{\tau}$-measurable.

\medskip

\quad{\bf (iv)} Define $\sequencen{\tau_n}$ as in (a-iv) above, so that
each $\tau_n$ is a stopping time adapted to
$\langle\ddot\Sigma_t\rangle_{t\ge 0}$ and $\sequencen{\tau_n(\omega)}$ is a
non-increasing sequence with limit $\tau(\omega)$ for every $\omega$.
For a $\ddot\mu$-integrable real-valued function $f$ on $\Cdlg$,
$\omega\in\Cdlg$ and $n\in\Bbb N$, set

\Centerline{$\ddot g^{(n)}_f(\omega)
=\int_{\Cdlg}fd\ddot\mu_{\omega,\tau_n(\omega)}$}

\noindent whenever the right-hand side is defined in $\Bbb R$.
By (a), $\int_{\Cdlg}\ddot g^{(n)}_fd\ddot\mu=\int_{\Cdlg}fd\ddot\mu$.
We have seen also, in
(a-iii), that each $\tau_n$ has an extension $\grave\tau_n$ which is a
stopping time on $\Omega$ of the type considered in 455Ec.   So if we take
a $\hat\mu$-integrable function $\tilde f$ extending $f$, and set

\Centerline{$g^{(n)}_{\tilde f}(\omega)
=\int_{\Omega}\tilde fd\hat\mu_{\omega,\grave\tau_n(\omega)}$}

\noindent whenever $\omega\in\Omega$ is such that the integral is defined
in $\Bbb R$, $g^{(n)}_{\tilde f}$ will be a conditional expectation
of $\tilde f$ on $\Sigma_{\grave\tau_n}$, the algebra of sets
$E\in\Sigma$ such that
$E\cap\{\omega:\grave\tau_n(\omega)\le t\}$ is determined by coordinates in
$[0,t]$ for every $t\ge 0$.

If $\omega\in\Cdlg$, then $\Cdlg$ has full outer measure for
$\hat\mu_{\omega,\tau_n(\omega)}=\hat\mu_{\omega,\grave\tau_n(\omega)}$, so

\Centerline{$g^{(n)}_{\tilde f}(\omega)
=\int_{\Omega}\tilde fd\hat\mu_{\omega,\grave\tau_n(\omega)}
=\int_{\Cdlg} fd\ddot\mu_{\omega,\tau_n(\omega)}
=\ddot g^{(n)}_f(\omega)$}

\noindent whenever either is defined.

\medskip

\quad{\bf (v)} Set

\Centerline{$\ddot\Sigma_{\tau_n}
=\{F:F\in\ddot\Sigma$, $F\cap\{\omega:\tau_n(\omega)\le t\}\in\ddot\Sigma_t$
for every $t\ge 0\}$.}

\noindent Then every $F\in\ddot\Sigma_{\tau_n}$ is of the form
$\tilde F\cap\Cdlg$ where $\tilde F\in\Sigma_{\grave\tau_n}$.
\Prf\ Recall that $\tau_n$ and $\grave\tau_n$ take values in
$D_n\cup\{\infty\}$, where $D_n=\{2^{-n}i:i\in\Bbb N\}$.   For each
$i\in\Bbb N$, set $F_i=\{\omega:\omega\in F$, $\tau_n(\omega)=2^{-n}i\}$;
then $F_i\in\ddot\Sigma_{2^{-n}i}$, so there is an
$E_i\in\Sigma_{2^{-n}i}$ such that $F_i=E_i\cap\Cdlg$ (a-ii).
Let $E_{\infty}\in\Sigma$ be such that
$E_{\infty}\cap\Cdlg=\{\omega:\tau_n(\omega)=\infty\}$, and try

\Centerline{$\tilde F
=\bigcup_{i\in\Bbb N}(E_i\cap\grave\tau_n^{-1}[\{2^{-n}i\}])
   \cup(E_{\infty}\cap\grave\tau_n^{-1}[\{\infty\}])$.}

\noindent Then $\tilde F\cap\Cdlg=F$ (because $\grave\tau_n$ extends
$\tau_n$) and $\tilde F\in\Sigma_{\grave\tau_n}$ (because

\Centerline{$\tilde F\cap\grave\tau_n^{-1}[\{2^{-n}i\}]
=E_i\cap\grave\tau_n^{-1}[\{2^{-n}i\}]\in\Sigma_{2^{-n}i}$}

\noindent for every $i$).\ \Qed

\medskip

\quad{\bf (vi)} If $f$ is $\ddot\mu$-integrable, then $\ddot g^{(n)}_f$ is a
conditional expectation of $f$ on $\ddot\Sigma_{\tau_n}$ for every $n$.
\Prf\ Take $F\in\ddot\Sigma_{\tau_n}$.   Then there are an
$\tilde F\in\Sigma_{\grave\tau_n}$ such that
$F=\tilde F\cap\Cdlg$, and a $\hat\mu$-integrable $\tilde f$ such that
$f=\tilde f\restr\Cdlg$.   So

$$\eqalignno{\int_Ffd\ddot\mu
&=\int_{\tilde F}\tilde fd\hat\mu
=\int_{\tilde F}g^{(n)}_{\tilde f}d\hat\mu\cr
\displaycause{455E(c-ii)}
&=\int_{\tilde F}\int_{\Omega}
  \tilde fd\hat\mu_{\omega,\grave\tau_n(\omega)}
  \hat\mu(d\omega)
=\int_F\int_{\Omega}
  \tilde fd\hat\mu_{\omega,\tau_n(\omega)}\ddot\mu(d\omega)\cr
\displaycause{because $\hat\mu^*\Cdlg=1$, $F=\tilde F\cap\Cdlg$ and
$\tau_n=\grave\tau_n\restr\Cdlg$}
&=\int_F\int_{\Cdlg}fd\ddot\mu_{\omega,\tau_n(\omega)}\ddot\mu(d\omega)
=\int_F\ddot g^{(n)}_fd\ddot\mu.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (vii)} Let $\Phi$ be the set of those $\ddot\mu$-integrable
real-valued functions $f$ such that
\penalty-100$\lim_{n\to\infty}
\int_{\Cdlg}|\ddot g_f-\ddot g^{(n)}_f|d\ddot\mu=0$.
For $J\subseteq\coint{0,\infty}$ let $\pi_J:\Omega\to U^J$ be the
restriction map.
By (a-v), $f=h\pi_J\restr\Cdlg$ belongs to $\Phi$ whenever
$J\subseteq\coint{0,\infty}$ is finite and
$h:U^J\to\Bbb R$ is bounded and uniformly continuous, since in this case

$$\eqalign{\ddot g_f(\omega)
&=\int_{\Cdlg}h\pi_Jd\ddot\mu_{\omega,\tau(\omega)}
=\int_{\Omega}h\pi_Jd\hat\mu_{\omega,\tau(\omega)}
=\lim_{n\to\infty}\int_{\Omega}h\pi_Jd\hat\mu_{\omega,\tau_n(\omega)}\cr
&=\lim_{n\to\infty}\int_{\Cdlg}h\pi_Jd\ddot\mu_{\omega,\tau_n(\omega)}
=\lim_{n\to\infty}\ddot g^{(n)}_f(\omega)\cr}$$

\noindent for every
$\omega\in\Cdlg$.   Next, $\ddot g_{\alpha f}\eae\alpha\ddot g_f$,
$\ddot g_{f+f'}\eae\ddot g_f+\ddot g_{f'}$ and

\Centerline{$\int_{\Cdlg}|\ddot g_f-\ddot g_{f'}|d\ddot\mu
=\int_{\Cdlg}|\ddot g_{f-f'}|d\ddot\mu
\le\int_{\Cdlg}\ddot g_{|f-f'|}d\ddot\mu
=\int_{\Cdlg}|f-f'|d\ddot\mu$}

\noindent for all $f$, $f'\in\eusm L^1(\ddot\mu)$ and $\alpha\in\Bbb R$;
and we have similar expressions for every $\ddot g^{(n)}_f$.   So $f+f'\in\Phi$
and $\alpha f\in\Phi$ whenever $f$, $f'\in\Phi$, and moreover
$f\in\Phi$ whenever $f\in\eusm L^1(\ddot\mu)$ and there is a sequence
$\sequence{f}{f_k}$ in $\Phi$ such that
$\lim_{k\to\infty}\int_{\Cdlg}|f-f_k|d\ddot\mu=0$.

If $J\subseteq\coint{0,\infty}$ is finite and $G\subseteq U^J$,
then $(\chi G)\pi_J\restr\Cdlg\in\Phi$.   \Prf\ There is a
non-decreasing sequence $\sequence{k}{h_k}$ of bounded uniformly continuous
functions on $U^J$ with limit $\chi G$;  now $h_k\pi_J\restr\Cdlg\in\Phi$ and
$(\chi G)\pi_J\restr\Cdlg=\lim_{k\to\infty}h_k\pi_J\restr\Cdlg$.\ \QeD\  By the Monotone Class
Theorem, $(\chi E)\pi_J\restr\Cdlg\in\Phi$
whenever $J\subseteq\coint{0,\infty}$ is
finite and $E\in\Cal B(U^J)$.   By the Monotone Class Theorem again,
$\chi(E\cap\Cdlg)\in\Phi$ whenever
$E\in\Tensorhat_{\coint{0,\infty}}\Cal B(U)$.   Since we surely have
$f'\in\Phi$ whenever $f\in\Phi$ and $f'=f\,\,\ddot\mu$-a.e.,
$\chi(E\cap\Cdlg)\in\Phi$ whenever $E\in\Sigma$, that is,
$\chi E\in\Phi$ for every $E\in\ddot\Sigma$.   It follows at once that
$\Phi=\eusm L^1(\ddot\mu)$.

\medskip

\quad{\bf (viii)} We are nearly home.   Suppose that
$f\in\eusm L^1(\ddot\mu)$ and $F\in\ddot\Sigma^+_{\tau}$.   If $n\in\Bbb N$,
then $F\in\ddot\Sigma_{\tau_n}$.   \Prf\ For any $t>0$,

\Centerline{$F\cap\{\omega:\tau(\omega)<t\}
=\bigcup_{q\in\Bbb Q,q<t}F\cap\{\omega:\tau(\omega)\le q\}
\in\ddot\Sigma_t$.}

\noindent So, for any $i\in\Bbb N$,

\Centerline{$F\cap\{\omega:\tau_n(\omega)\le 2^{-n}i\}
=F\cap\{\omega:\tau(\omega)<2^{-n}i\}
\in\ddot\Sigma_{2^{-n}i}$.  \text{\Qed}}

\noindent So $\int_F\ddot g^{(n)}_fd\ddot\mu=\int_Ffd\ddot\mu$.
But $f\in\Phi$, so

\Centerline{$\int_F\ddot g_fd\ddot\mu=\lim_{n\to\infty}\int_F\ddot g^{(n)}_fd\ddot\mu
=\int_Ffd\ddot\mu$.}

\noindent Since we already know, from (iii) above,
that $\dom\ddot g_f\in\ddot\Sigma^+_{\tau}$ and
$\ddot g_f$ is $\ddot\Sigma^+_{\tau}$-measurable,
$\ddot g_f$ is a conditional expectation
of $f$ on $\ddot\Sigma^+_{\tau}$, as claimed.

\medskip

{\bf (c)(i)} By 455L(c-iii) again,
$\ddot\Sigma_{\tau}$ is a $\sigma$-algebra.

\medskip

\quad{\bf (ii)} If $\omega$, $\omega'\in\Cdlg$ and
$\omega'\restr[0,\tau(\omega)]=\omega\restr[0,\tau(\omega)]$ then
$\ddot\mu_{\omega',\tau(\omega')}=\ddot\mu_{\omega,\tau(\omega)}$.   \Prf\
Set $t=\tau(\omega)$.   This time, $H_t$ and $H'_t$, defined as in
(b-i), belong to $\ddot\Sigma_t$, so their difference belongs to
$\ddot\Sigma_t$ and
is determined (relative to $\Cdlg$) by coordinates in $[0,t]$;  so
$\omega'\in H_t\setminus H'_t$ and $\tau(\omega')=t$.  Now, reading off the
definition in 455Eb, $\nu^{(s,u)}_{\omega'tx}=\nu^{(s,u)}_{\omega tx}$ for
all $s$, $u$ and $x$, so $\ddot\mu_{\omega't}=\ddot\mu_{\omega t}$.\ \Qed

\medskip

\quad{\bf (iii)} It follows that if $f\in\eusm L^1(\ddot\mu)$
and $\alpha\in\Bbb R$ then
$F=\{\omega:\omega\in\Cdlg$, $\ddot g_f(\omega)$ is defined and at most
$\alpha\}$ belongs to $\ddot\Sigma_{\tau}$.
\Prf\ We know from (b-iii) that $F\in\ddot\Sigma$.   If $t\ge 0$,
$\omega\in F$, $\omega'\in\Cdlg$,
$\tau(\omega)\le t$ and $\omega'\restr[0,t]=\omega\restr[0,t]$ then
$\ddot\mu_{\omega',\tau(\omega')}=\ddot\mu_{\omega,\tau(\omega)}$, so
$\ddot g_f(\omega')=\ddot g_f(\omega)$ and $\omega'\in F$.   Thus
$F\cap\{\omega:\tau(\omega)\le t\}$ is determined (relative to $\Cdlg$)
by coordinates in $[0,t]$
and belongs to $\ddot\Sigma_t$.\ \Qed

\medskip

\quad{\bf (iv)} Thus $\dom\ddot g_f\in\ddot\Sigma_{\tau}$ and $\ddot g_f$ is
$\ddot\Sigma_{\tau}$-measurable.   As we already know that it is a
conditional
expectation of $f$ on $\ddot\Sigma_{\tau}^+\supseteq\ddot\Sigma_{\tau}$, it is a
conditional expectation of $f$ on $\ddot\Sigma_{\tau}$.
}%end of proof of 455O

\leader{455P}{}\cmmnt{ The eventual objective of this section is to
provide a foundation for study of the original,
and still by far the most important, example of a
continuous-time Markov process, Brownian motion.   In the language
developed above, we shall
have $U=\Bbb R$ (or, when we come to the applications
in \S\S477-479, $U=\BbbR^r$), and all the transitional probabilities
$\nu_x^{(s,t)}$ will be Gaussian.   But the techniques so far developed
can tell us a great deal about much more general processes with
some of the same features.

\medskip

\noindent}{\bf Theorem} Let $U$ be a metrizable
topological group which is complete
under a \rti\ metric $\rho$ inducing its topology.
Let $\langle\lambda_t\rangle_{t>0}$ be a family of Radon probability measures
on $U$ such that the convolution $\lambda_s*\lambda_t$\cmmnt{ (444A)} 
is equal to $\lambda_{s+t}$ for all $s$, $t>0$.
Suppose that $\lim_{t\downarrow 0}\lambda_tG=1$ for every open
neighbourhood $G$ of the identity in $U$.
For $x\in U$ and $0\le s<t$, let $\nu^{(s,t)}_x$ be the Radon probability
measure on $U$ defined by saying that
$\nu^{(s,t)}_x(E)=\lambda_{t-s}(Ex^{-1})$ whenever $\lambda_{t-s}$ measures
$Ex^{-1}$.

(a) $\family{y}{U}{\nu^{(t,u)}_y}$ is a disintegration of $\nu^{(s,u)}_x$
over $\nu^{(s,t)}_x$ whenever $x\in U$ and $0\le s<t<u$.

(b) $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is narrowly continuous
and uniformly time-continuous on the right.

(c)(i) We can define a complete measure
$\hat\mu$ on $U^{\coint{0,\infty}}$ by the method of 455E applied to
$x^*=e$ and $\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$.

\quad(ii) If $\Cdlg$ is the space of \cadlag\ functions from
$\coint{0,\infty}$ to $U$, then $\hat\mu^*\Cdlg=1$, and the subspace
measure $\ddot\mu$ on $\Cdlg$ will have the properties described in 455O, with
$\omega(0)=e$ for $\ddot\mu$-almost every $\omega\in\Cdlg$.

\quad(iii) $\hat\mu$ has a unique extension to a Radon measure $\tilde\mu$
on $U^{\coint{0,\infty}}$.

\proof{{\bf (a)} Note first that $y\mapsto yx$ is \imp\ for $\lambda_{t-s}$
and $\nu^{(s,t)}_x$, so that
$\int f(y)\nu_x^{(s,t)}(dy)=\int f(yx)\lambda_{t-s}(dy)$
for any real-valued function on $U$ for which either is defined
(235Gb).   If $E\subseteq U$ is measured by $\nu^{(s,u)}_x$, then

$$\eqalignno{\nu^{(s,u)}_x(E)
&=\lambda_{u-s}(Ex^{-1})
=(\lambda_{u-t}*\lambda_{t-s})(Ex^{-1})
=\int\lambda_{u-t}(Ex^{-1}y^{-1})\lambda_{t-s}(dy)\cr
\displaycause{444A}
&=\int\nu_{yx}^{(t,u)}(E)\lambda_{t-s}(dy)
=\int\nu_y^{(t,u)}(E)\nu_x^{(s,t)}(dy);\cr}$$

\noindent as $E$ is arbitrary,
$\family{y}{U}{\nu^{(t,u)}_y}$ is a disintegration of $\nu^{(s,u)}_x$
over $\nu^{(s,t)}_x$.

\medskip

{\bf (b)(i)}\grheada\
Suppose that $x\in U$ and $0\le s<t$;  set $u=t-s$.
Let $f:U\to\Bbb R$ be
a bounded continuous function and set $M=\|f\|_{\infty}$.   Take
$\epsilon\in\ooint{0,\bover12}$.   Let $K\subseteq U$ be a compact set such that
$\lambda_uK\ge 1-\epsilon$.   Then there is a
symmetric open neighbourhood $V$
of the identity $e$ of $U$ such that
$|f(wyx^{-1})-f(wx^{-1})|\le 2\epsilon$
whenever $w\in K$ and $y\in V^2$.   \Prf\ For each $w\in K$ there is a
neighbourhood $W_w$ of $e$ such that $|f(wyx^{-1})-f(wx^{-1})|\le\epsilon$
whenever $y\in W_w^2$.   Because $K$ is compact, there are
$w_0,\ldots,w_n\in K$ such that $K\subseteq\bigcup_{i\le n}w_iW_{w_i}$;
set $W=\bigcap_{i\le n}W_{w_i}$.   If $w\in K$ and $y\in W$, there is an
$i\le n$ such that $w\in w_iW_{w_i}$, in which case

$$\eqalign{|f(wyx^{-1})-f(wx^{-1})|
&\le|f(w_i(w_i^{-1}wy)x^{-1})-f(w_ix^{-1})|\cr
&\mskip200mu
     +|f(w_ix^{-1})-f(w_i(w_i^{-1}w)x^{-1})|\cr
&\le 2\epsilon\cr}$$

\noindent because both $w_i^{-1}wy$ and $w_i^{-1}w$ belong to $W_{w_i}^2$.
So if we take a symmetric open neighbourhood $V$ of $e$ such that
$V^2\subseteq W$, this will serve.\ \Qed

\medskip

\qquad\grheadb\
Let $\delta>0$ be such that $\lambda_vV\ge 1-\epsilon$ whenever
$0<v\le 2\delta$.   It will be worth noting that
$\lambda_v(KV)\ge 1-2\epsilon$ whenever $0<v<u$ and $u-v\le 2\delta$.
\Prf\ In this case, $\lambda_u=\lambda_v*\lambda_{u-v}$.   Now
$U\setminus K\supseteq(U\setminus KV)V^{-1}$.   So

\Centerline{$\epsilon
\ge\lambda_u(U\setminus K)
\ge\lambda_v(U\setminus KV)\lambda_{u-v}(V^{-1})
\ge(1-\epsilon)\lambda_v(U\setminus KV)$}

\noindent and

\Centerline{$\lambda_v(KV)\ge 1-\Bover{\epsilon}{1-\epsilon}
\ge 1-2\epsilon$.  \Qed}

\medskip

\qquad\grheadc\
Suppose that $0\le s'<t'$ and $y\in U$ are such that
$y^{-1}x\in W$, $|s'-s|\le\delta$ and $|t'-t|\le\delta$.   Then
$|\int fd\nu_y^{(s',t')}-\int fd\nu_x^{(s,t)}|\le(6M+4)\epsilon$.
\Prf\ Set $u'=t'-s'$, so that $|u-u'|\le 2\delta$.   We have

\Centerline{$|\int fd\nu_y^{(s',t')}-\int fd\nu_x^{(s,t)}|
=|\int f(wy^{-1})\lambda_{u'}(dw)-\int f(wx^{-1})\lambda_u(dw)|$.}

\medskip

{\bf case 1} Suppose that $u'<u$.   Then
$\lambda_u=\lambda_{u'}*\lambda_{u-u'}$, so

$$\eqalignno{|\int fd\nu_y^{(s',t')}-\int fd\nu_x^{(s,t)}|
&=|\int f(wy^{-1})\lambda_{u'}(dw)
   -\int f(wx^{-1})(\lambda_{u'}*\lambda_{u-u'})(dw)|\cr
&=|\int f(wy^{-1})\lambda_{u'}(dw)
   -\iint f(wzx^{-1})\lambda_{u-u'}(dz)\lambda_{u'}(dw)|\cr
\displaycause{444C}
&\le\int|f(wy^{-1})-\int f(wzx^{-1})\lambda_{u-u'}(dz)|\lambda_{u'}(dw)\cr
&\le 4M\epsilon
   +\sup_{w\in KV}|f(wy^{-1})-\int f(wzx^{-1})\lambda_{u-u'}(dz)|\cr
\displaycause{because $\lambda_{u'}(U\setminus KV)\le 2\epsilon$, by
($\beta$), and
$|f(wy^{-1})-\int f(wzx^{-1})\lambda_{u-u'}(dz)|\le 2M$ for every $w$}
&\le 4M\epsilon
   +\sup_{w\in KV}\int|f(wy^{-1})-\int f(wzx^{-1})|\lambda_{u-u'}(dz)\cr
&\le 6M\epsilon
   +\sup_{w\in KV,z\in V}|f(wy^{-1})-\int f(wzx^{-1})|\cr
\displaycause{because $\lambda_{u-u'}V\ge 1-\epsilon$}
&\le 6M\epsilon
   +\sup_{w\in K,v\in V,z\in V}|f(wvy^{-1}xx^{-1})-f(wvzx^{-1})|\cr
&\le 6M\epsilon
   +\sup_{w\in K,v\in V,z\in V}\bigl(|f(wvy^{-1}xx^{-1})-f(wx^{-1})|\cr
&\mskip220mu
         +|f(wvzx^{-1})-f(wx^{-1})|\bigr)\cr
&\le 6M\epsilon+4\epsilon\cr}$$

\noindent by the choice of $V$, because $y^{-1}x\in V$.

\medskip

{\bf case 2} Suppose that $u'=u$.   Then

$$\eqalign{|\int fd\nu_y^{(s',t')}-\int fd\nu_x^{(s,t)}|
&\le\int|f(wy^{-1})-f(wx^{-1})|\lambda_u(dw)\cr
&\le 2M\epsilon+\sup_{w\in K}|f(wy^{-1}xx^{-1})-f(wx^{-1})|\cr
&\le 2M\epsilon+\epsilon.\cr}$$

\medskip

{\bf case 3} Suppose that $u'>u$.   Then
$\lambda_{u'}=\lambda_u*\lambda_{u'-u}$, so

$$\eqalignno{|\int fd\nu_y^{(s',t')}-\int fd\nu_x^{(s,t)}|
&=|\iint f(wzy^{-1})\lambda_{u'-u}(dz)\lambda_u(dw)
   -\int f(wx^{-1})\lambda_u(dw)|\cr
&\le\int|\int f(wzy^{-1})\lambda_{u'-u}(dz)-f(wx^{-1})|\lambda_u(dw)\cr
&\le 2M\epsilon+\sup_{w\in K}
   |\int f(wzy^{-1})\lambda_{u'-u}(dz)-f(wx^{-1})|\cr
&\le 2M\epsilon+2M\epsilon
   +\sup_{w\in K,z\in V}|f(wzy^{-1})-f(wx^{-1})|\cr
&\le 4M\epsilon+2\epsilon.\cr}$$

\noindent So we have the result in all cases.\ \Qed

\medskip

\qquad\grheadd\ As $s$, $t$, $x$, $\epsilon$ and $f$ are arbitrary,
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is narrowly (= vaguely)
continuous.

\medskip

\quad{\bf (ii)} Given $\epsilon>0$, there is a $\delta>0$ such that
$\lambda_t\{x:\rho(x,e)<\epsilon\}\ge 1-\epsilon$ whenever $0<t\le\delta$.
Now suppose that $x\in U$ and $0\le s<t\le s+\delta$.   Then

$$\eqalignno{\nu^{(s,t)}_xB(x,\epsilon)
&=\lambda_{t-s}(B(x,\epsilon)x^{-1})
=\lambda_{t-s}B(e,\epsilon)\cr
\displaycause{because $\rho$ is \rti}
&\ge 1-\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ is uniformly
time-continuous on the right.

\medskip

{\bf (c)} This is now just a matter of putting 455O and 455H
together, and recalling from 455E that $\omega(0)=e$ for $\hat\mu$-almost every
$\omega\in\Omega$.
}%end of proof of 455P

\leader{455Q}{L\'evy processes}
\cmmnt{If we approach as
probabilists, without prejudices in favour of any particular realization,
the processes in 455P manifest themselves as follows.}
Let $U$ be a separable metrizable
topological group with identity $e$, and consider the following list of
properties of a family
$\langle X_t\rangle_{t\ge 0}$ of $U$-valued random variables:

\inset{$X_0=e$ almost everywhere,}

\inset{$\Pr(X_tX_s^{-1}\in F)=\Pr(X_{t-s}\in F)$ whenever $0\le s<t$ and
$F\subseteq U$ is Borel}

\noindent (the process is {\bf stationary}),

\inset{whenever $0\le t_0<t_1<\ldots<t_n$, then
$X_{t_1}X_{t_0}^{-1},X_{t_2}X_{t_1}^{-1},\ldots,X_{t_n}X_{t_{n-1}}^{-1}$
are independent}

\noindent\cmmnt{ in the sense of 418U }(the process has
{\bf independent increments}),

\inset{$X_t\to e$ in measure as $t\downarrow 0$}

\noindent (that is, $\lim_{t\downarrow 0}\Pr(X_t\in G)=1$ for every
neighbourhood $G$ of the identity).
\cmmnt{I say here that $U$ should be
separable and metrizable
in order to ensure that all the functions $X_tX_s^{-1}$
should be measurable (of course it will be enough if $U$ is metrizable and
of measure-free weight, as in 438E).}   Such a family I will call a
{\bf L\'evy process}.

\leader{455R}{Theorem} Let $U$ be a Polish group with identity $e$
which is complete under a \rti\ metric inducing its topology.
A family $\langle X_t\rangle_{t\ge 0}$
of $U$-valued random variables is a L\'evy process
iff there is a family $\langle\lambda_t\rangle_{t>0}$ of Radon probability
measures on
$U$, satisfying the conditions of 455P, such that if we start from $x^*=e$
and build the measure $\hat\mu$ on $U^{\coint{0,\infty}}$ as in 455Pc, then

\Centerline{$\Pr(X_{t_i}\in F_i$ for every $i\le n)
=\hat\mu\{\omega:\omega(t_i)\in F_i$ for every $i\le n\}$}

\noindent whenever $t_0,\ldots,t_n\in\coint{0,\infty}$ and $F_i\subseteq U$
is a Borel set for every $i\le n$.

\proof{{\bf (a)} Suppose we have a family
$\langle\lambda_t\rangle_{t>0}$ of Radon probability measures on $U$
such that $\lambda_s*\lambda_t=\lambda_{s+t}$ for all $s$, $t>0$ and
$\lim_{t\downarrow 0}\lambda_tG=1$ for every open
neighbourhood $G$ of $e$ in $U$.   Define
$\langle\nu_x^{(s,t)}\rangle_{0\le s<t,x\in U}$ as in 455P,
and let $\hat\mu$ be the corresponding completed measure on
$\Omega=U^{\coint{0,\infty}}$ as in 455Pc.   
Set $X_t(\omega)=\omega(t)$ for $t\ge 0$ and $\omega\in\Omega$.
Then $X_0=e$ a.e.\ (455Ea) and

\Centerline{$\Pr(X_t\in F)=\hat\mu X_t^{-1}[F]=\nu^{(0,t)}_0F=\lambda_tF$}

\noindent for $t>0$ and $F\in\Cal B(U)$ (455Ea again).   In particular,

\Centerline{$\lim_{t\downarrow 0}\Pr(X_t\in G)
=\lim_{t\downarrow 0}\lambda_tG=1$}

\noindent for every neighbourhood $G$ of the identity.
If $0<s<t$ and $F\in\Cal B(U)$, set
$H=\{(e,x,y):yx^{-1}\in F\}\subseteq U^3$.   Then

$$\eqalignno{\Pr(X_tX_s^{-1}\in F)
&=\hat\mu\{\omega:(\omega(0),\omega(s),\omega(t))\in H\}\cr
&=\iint\chi H(e,x,y)\nu_x^{(s,t)}(dy)\nu_e^{(0,s)}(dx)\cr
\displaycause{455E}
&=\iint\chi H(e,x,yx)\lambda_{t-s}(dy)\lambda_s(dx)\cr
&=\iint\chi F(y)\lambda_{t-s}(dy)\lambda_s(dx)
=\lambda_{t-s}(F)
=\Pr(X_{t-s}\in F).\cr}$$

\noindent If $0=s<t$ then $X_tX_s^{-1}\eae X_t=X_{t-s}$, of course.
If $0=t_0<t_1<\ldots<t_n$ and
$F_0,\ldots,F_{n-1}\in\Cal B(U)$, set

\Centerline{$E_k
=\{\omega:\omega\in\Omega$, $\omega(t_{i+1})\omega(t_i)^{-1}\in F_i$
  for every $i<k\}$,}

\Centerline{$H_k=\{(x_0,\ldots,x_k):x_{i+1}x_i^{-1}\in F_i$ for every
$i<k\}\subseteq U^{k+1}$}

\noindent for $k\le n$.   Then

\Centerline{$\hat\mu E_1
=\hat\mu\{\omega:\omega(t_1)\omega(0)^{-1}\in F_0\}
=\hat\mu\{\omega:\omega(t_1)\in F_0\}
=\nu_e^{(0,t_1)}F_0
=\lambda_{t_1}F_0$,}

\noindent and for $k\ge 2$

$$\eqalignno{\Pr(X_{t_{i+1}}X_{t_i}^{-1}&\in F_i\text{ for every }i<k)\cr
&=\hat\mu E_k
=\int\ldots\int\chi H_k(e,x_1,\ldots,x_k)
    \nu_{x_{k-1}}^{(t_{k-1},t_k)}(dx_k)
    \ldots\nu_e^{(0,t_1)}(dx_1)\cr
\displaycause{455E}
&=\int\ldots\iint\chi H_k(e,x_1,\ldots,x_{k-1},x_kx_{k-1})
    \lambda_{t_k-t_{k-1}}(dx_k)\cr
&\mskip100mu
    \nu_{x_{k-2}}^{(t_{k-2},t_{k-1})}(dx_{k-1})
    \ldots\nu_e^{(0,t_1)}(dx_1)\cr
&=\int\ldots\iint\chi H_{k-1}(e,x_1,\ldots,x_{k-1})
    \chi F_{k-1}(x_k)
    \lambda_{t_k-t_{k-1}}(dx_k)\cr
&\mskip100mu
    \nu_{x_{k-2}}^{(t_{k-2},t_{k-1})}(dx_{k-1})
    \ldots\nu_e^{(0,t_1)}(dx_1)\cr
&=\lambda_{t_k-t_{k-1}}(F_{k-1})
    \int\ldots\int\chi H_{k-1}(e,x_1,\ldots,x_{k-1})\cr
&\mskip100mu
    \nu_{x_{k-2}}^{(t_{k-2},t_{k-1})}(dx_{k-1})
    \ldots\nu_e^{(0,t_1)}(dx_1)\cr
&=\lambda_{t_k-t_{k-1}}(F_{k-1})\cdot\hat\mu E_{k-1}.\cr}$$

\noindent So

$$\eqalign{\Pr(X_{t_{i+1}}X_{t_i}^{-1}&\in F_i\text{ for every }i<n)
=\hat\mu E_n
=\prod_{i=0}^{n-1}\lambda_{t_i-t_{i-1}}F_i\cr
&=\prod_{i=0}^{n-1}\Pr(X_{t_i-t_{i-1}}\in F_i)
=\prod_{i=0}^{n-1}\Pr(X_{t_i}X_{t_{i-1}}^{-1}\in F_i).\cr}$$

\noindent As $F_0,\ldots,F_{n-1}$ are arbitrary,
$X_{t_1}X_{t_0}^{-1},X_{t_2}X_{t_1}^{-1},\ldots,X_{t_n}X_{t_{n-1}}^{-1}$
are independent.   Thus all the conditions of 455Q are satisfied.

\medskip

{\bf (b)(i)} In the other direction, given a family
$\langle X_t\rangle_{t\ge 0}$ with the properties listed in 455Q, then for
each $t>0$ there is a Radon measure
$\lambda_t$ on $U$ such that $\lambda_tF=\Pr(X_t\in F)$
for every $F\in\Cal B(U)$, for each $t>0$.   \Prf\ $U$ is Polish, therefore
analytic, and we can apply 433Cb to the Borel measure
$F\mapsto\Pr(X_t\in F)$.\ \QeD\   If $s$, $t>0$, then
the distribution of $X_{s+t}X_s^{-1}$ is the same as the distribution of
$X_t$, so is $\lambda_t$.

If $s$, $t>0$ then $\lambda_{s+t}=\lambda_s*\lambda_t$.   \Prf\ If
$F_1$, $F_2\in\Cal B(U)$ then

$$\eqalignno{\Pr((X_t,X_{s+t}X_t^{-1})\in F_1\times F_2)
&=\Pr(X_t\in F_1,X_{s+t}X_t^{-1}\in F_2)\cr
&=\Pr(X_t\in F_1)\Pr(X_{s+t}X_t^{-1}\in F_2)\cr
\displaycause{because $X_t$ and $X_{s+t}X_t^{-1}$ are independent}
&=\lambda_tF_1\cdot\lambda_sF_2
=(\lambda_t\times\lambda_s)(F_1\times F_2).\cr}$$

\noindent By the Monotone Class Theorem, or otherwise,

\Centerline{$\Pr((X_t,X_{s+t}X_t^{-1})\in H)
=(\lambda_t\times\lambda_s)H$}

\noindent for every $H\in\Cal B(U)\tensorhat\Cal B(U)=\Cal B(U^2)$.   So if
$F\in\Cal B(U)$ we shall have

$$\eqalign{(\lambda_s*\lambda_t)(F)
&=(\lambda_s\times\lambda_t)\{(x,y):xy\in F\}
=(\lambda_t\times\lambda_s)\{(y,x):xy\in F\}\cr
&=\Pr(X_{s+t}X_t^{-1}X_t\in F)
=\lambda_{s+t}F,\cr}$$

\noindent and $\lambda_s*\lambda_t=\lambda_{s+t}$.  (Cf.\
272T\formerly{2{}72S}.)\ \Qed

Next, for any neighbourhood $G$ of $e$,

\Centerline{$\lim_{t\downarrow 0}\lambda_tG
=\lim_{t\downarrow 0}\Pr(X_t\in G)
=1$.}

\noindent So $\langle\lambda_t\rangle_{t>0}$ satisfies the conditions of
455P.   Let $\hat\mu$ be the corresponding completed measure on
$U^{\coint{0,\infty}}$ as in 455Pc.

\medskip

\quad{\bf (ii)} $\Pr(X_{t_i}\in F_i$ for every $i\le k)
=\hat\mu\{\omega:\omega(t_i)\in F_i$ for every $i\le k\}$
whenever $t_0,\ldots,t_n\in\coint{0,\infty}$ and $F_i\in\Cal B(U)$
for every $i\le k$.   \Prf\ It is enough to consider the case
$0=t_0<t_1<\ldots<t_n$.   In this case, whenever
$E_0,\ldots,E_n\in\Cal B(U)$,

$$\eqalignno{
\Pr((X_{t_0},&X_{t_1}X_{t_0}^{-1},\ldots,X_{t_n}X_{t_{n-1}}^{-1})\in
  E_0\times\ldots\times E_n)\cr
&=\Pr(X_{t_0}\in E_0)\Pr(X_{t_1}X_{t_0}^{-1}\in E_1)\ldots
  \Pr(X_{t_n}X_{t_{n-1}}^{-1})\in E_n)\cr
&=\delta_e(E_0)\lambda_{t_1-t_0}(E_1)\ldots\lambda_{t_n-t_{n-1}}(E_n)\cr
\displaycause{where $\delta_e$ is the Dirac measure concentrated at $e$}
&=\hat\mu\{\omega:\omega(t_0)\in E_0\}
  \hat\mu\{\omega:\omega(t_1)\omega(t_0)^{-1}\in E_1\}\ldots\cr
&\mskip200mu
  \hat\mu\{\omega:\omega(t_n)\omega(t_{n-1})^{-1}\in E_n\}\cr
&=\hat\mu\{\omega:\omega(t_0)\in E_0,\,
  \omega(t_1)\omega(t_0)^{-1}\in E_1,\,
  \ldots,\omega(t_n)\omega(t_{n-1})^{-1}\in E_n\}\cr
\displaycause{by (a) above}
&=\hat\mu\{\omega:(\omega(t_0),\omega(t_1)\omega(t_0)^{-1},\ldots,
  \omega(t_n)\omega(t_{n-1})^{-1})\in E_0\times\ldots\times E_n\}.\cr}$$

\noindent So in fact

$$\eqalign{\Pr((X_{t_0},X_{t_1}X_{t_0}^{-1},
  &\ldots,X_{t_n}X_{t_{n-1}}^{-1})\in H)\cr
&=\hat\mu\{\omega:(\omega(t_0),\omega(t_1)\omega(t_0)^{-1},\ldots,
  \omega(t_n)\omega(t_{n-1})^{-1})\in H\}\cr}$$

\noindent for every Borel set $H\subseteq U^{n+1}$.

Set

\Centerline{$\phi(x_0,\ldots,x_n)
=(x_0,x_1x_0,x_2x_1x_0,\ldots,x_nx_{n-1}\ldots x_1x_0)$}

\noindent for $x_0,\ldots,x_n\in U$, so that $\phi:U^{n+1}\to U^{n+1}$ is
continuous.   If $H\in\Cal B(U^{n+1})$, then

$$\eqalign{\Pr(X_{t_0},\ldots,X_{t_n}\in H)
&=\Pr(\phi(X_{t_0},X_{t_1}X_{t_0}^{-1},\ldots,X_{t_n}X_{t_{n-1}}^{-1})
  \in H)\cr
&=\Pr((X_{t_0},X_{t_1}X_{t_0}^{-1},\ldots,X_{t_n}X_{t_{n-1}}^{-1})
  \in\phi^{-1}[H])\cr
&=\hat\mu\{\omega:((\omega(t_0),\omega(t_1)\omega(t_0)^{-1},\ldots,
  \omega(t_n)\omega(t_{n-1})^{-1})\in\phi^{-1}[H]\}\cr
&=\hat\mu\{\omega:(\omega(t_0),\ldots,\omega(t_n))\in H\}.\cr}$$

\noindent Taking $H=F_0\times\ldots\times F_n$ we have the result.\ \Qed
}%end of proof of 455R

\leader{455S}{Lemma} Let $U$ be a metrizable
topological group which is complete
under a \rti\ metric inducing its topology.
Let $\langle\lambda_t\rangle_{t>0}$
be a family of Radon probability measures
on $U$ such that $\lambda_s*\lambda_t=\lambda_{s+t}$ for all $s$, $t>0$
and $\lim_{t\downarrow 0}\lambda_tG=1$ for every open
neighbourhood $G$ of the identity $e$ in $U$.   For $x\in U$ and
$0\le s<t$, let $\nu_x^{(s,t)}$ be the Radon probability measure on $U$
defined by saying that $\nu_x^{(s,t)}(E)=\lambda_{t-s}(Ex^{-1})$ whenever
$\lambda_{t-s}$ measures $Ex^{-1}$.

(a) If $0\le t_0<t_1<\ldots<t_n$, $z\in U$ and $f:\BbbR^J\to\Bbb R$ is a
bounded Borel measurable function, where $J=\{t_0,\ldots,t_n\}$, then

$$\eqalign{
&\iint\ldots\int f(z,x_1,\ldots,x_n)\nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
\ldots\nu_{x_1}^{(t_1,t_2)}(dx_2)\nu_z^{(t_0,t_1)}(dx_1)\cr
&\mskip50mu
=\iint\ldots\int f(z,x_1z,\ldots,x_nz)\nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)\cr
&\mskip200mu
\ldots\nu_{x_1}^{(t_1,t_2)}(dx_2)\nu_e^{(t_0,t_1)}(dx_1).\cr}$$

(b) Take $\omega\in U^{\coint{0,\infty}}$ and $a\ge 0$.
Let $\hat\mu$ and $\hat\mu_{\omega a}$
be the measures on $U^{\coint{0,\infty}}$ defined
from $\langle\nu^{(s,t)}_x\rangle_{s<t,x\in U}$ by the method of 455E,
starting from $x^*=e$.   Define
$\phi_{\omega a}:U^{\coint{0,\infty}}\to U^{\coint{0,\infty}}$ by setting

$$\eqalign{\phi_{\omega a}(\omega')(t)
&=\omega(t)\text{ if }t<a,\cr
&=\omega'(t-a)\omega(a)\text{ if }t\ge a.\cr}$$

\noindent Then $\hat\mu_{\omega a}$ is the image measure
$\hat\mu\phi_{\omega a}^{-1}$.

(c) In (b), suppose that $\omega$ belongs to the set $\Cdlg$ of \cadlag\
functions from $\coint{0,\infty}$ to $U$.
Then $\phi_{\omega a}(\omega')\in\Cdlg$ for every
$\omega'\in\Cdlg$, and $\phi_{\omega a}:\Cdlg\to\Cdlg$ is \imp\ for
the subspace measures $\ddot\mu$ and $\ddot\mu_{\omega a}$ on $\Cdlg$.

\proof{{\bf (a)(i)} If $x\in U$ and $0\le s<t$, then
$\nu_x^{(s,t)}(E)=\nu_e^{(s,t)}(Ex^{-1})$ for any $E\subseteq U$ such that
either is defined;  so
$\int f(y)\nu_x^{(s,t)}(dy)=\int f(yx)\nu_e^{(s,t)}(dy)$ for any function
$f:U\to\Bbb R$ for which either is defined.   More generally,

\Centerline{$\int f(y)\nu_{xz}^{(s,t)}(dy)=\int f(yxz)\nu_e^{(s,t)}(dy)
=\int f(yz)\nu_x^{(s,t)}(dy)$}

\noindent whenever $f$ is such that any of the three integrals is defined.

\medskip

\quad{\bf (ii)} Now induce on $n$.   For the case $n=0$, the natural
interpretation of both sides of the formula presented
is $f(z)$.   For the inductive step to $n+1$,
we have

$$\eqalignno{
&\int\ldots\int f(z,x_1,x_2,\ldots,x_{n+1})
   \nu_{x_n}^{(t_n,t_{n+1})}(dx_{n+1})\ldots\nu_z^{(t_0,t_1)}(dx_1)\cr
&\mskip50mu
=\int\ldots\int f(z,x_1z,\ldots,x_nz,x_{n+1})
   \nu_{x_nz}^{(t_n,t_{n+1})}(dx_{n+1})
%\cr&\mskip200mu
\ldots\nu_e^{(t_0,t_1)}(dx_1)\cr
\displaycause{by the inductive hypothesis applied to
$(x_0,x_1,\ldots,x_n)
\mapsto\int f(x_0,\ldots,x_n,x_{n+1})\nu_{x_n}^{(t_n,t_{n+1})}(dx_{n+1})$}
&\mskip50mu
=\int\ldots\int f(z,x_1z,\ldots,x_nz,x_{n+1}z)
   \nu_{x_n}^{(t_n,t_{n+1})}(dx_{n+1})
%\cr&\mskip200mu
\ldots\nu_e^{(t_0,t_1)}(dx_1)\cr}$$

\noindent by (i) applied to the functions
$y\mapsto f(z,x_1z,\ldots,x_nz,y)$ for each $x_1,\ldots,x_n$.

\medskip

{\bf (b)(i)} Suppose that $J\subseteq\coint{0,\infty}$ is a finite set
containing both $0$ and $a$, enumerated in increasing order as
$(t_0,\ldots,t_n)$ with $a=t_j$.   Set $z=\omega(a)$.
Let $f:\BbbR^J\to\Bbb R$ be a function.
Then $f\pi_J\phi_{\omega a}=g\pi_K$ where $K=\{0,t_{j+1}-a,\ldots,t_n-a\}$
and $g(x_j,\ldots,x_n)
=f(\omega(0),\omega(t_1),\ldots,\omega(t_{j-1}),x_jz,\ldots,x_nz)$ for
$x_j,\ldots,x_n\in U$.
\Prf\ For $\omega'\in U^{\coint{0,\infty}}$,

$$\eqalign{f\pi_J\phi_{\omega a}(\omega')
&=(f(\phi_{\omega a}(\omega')(t_0)),\ldots,f(\phi_{\omega a}(\omega')(t_n))
   \cr
&=(f(\omega(0)),\ldots,f(\omega(t_{j-1})),f(\omega'(0)z),
   \ldots f(\omega'(t_n-a)z))\cr
&=g(\omega'(0),\ldots,\omega'(t_n-a))
=g\pi_K(\omega').  \text{\Qed}\cr}$$

\medskip

{\bf (ii)} Again suppose that $J\subseteq\coint{0,\infty}$ is a finite set
containing both $0$ and $a$, enumerated in increasing order as
$(t_0,\ldots,t_n)$ with $a=t_j$, and set $z=\omega(a)$.
This time, let $f:\BbbR^J\to\Bbb R$ be a bounded Borel measurable function.
Then

$$\eqalignno{\int f\pi_Jd\hat\mu_{\omega a}
&=\int\ldots\int
    f(e,x_1,\ldots,x_n)\nu_{\omega ax_{n-1}}^{(t_{n-1},t_n)}(dx_n)
  \ldots\nu_{\omega a0}^{(0,t_1)}(dx_1)\cr
&=\int\ldots\iint\ldots\int
    f(e,x_1,\ldots,x_{j-1},x_j,x_{j+1},\ldots,x_n)\cr
&\mskip100mu
    \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
      \ldots\nu_z^{(a,t_{j+1})}(dx_{j+1})
      \delta_z(dx_j)\ldots\delta_{\omega(t_1)}(dx_1)\cr
\displaycause{reading from the formulae in 455E;  here each $\delta_x$ is a
Dirac measure on $U$}
&=\int\ldots\int
    f(e,\omega(t_1),\ldots,z,x_{j+1},\ldots,x_n)\cr
&\mskip200mu
    \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
  \ldots\nu_z^{(a,t_{j+1})}(dx_{j+1})\cr
&=\int\ldots\int
    f(e,\omega(t_1),\ldots,z,x_{j+1}z,\ldots,x_nz)\cr
&\mskip200mu
    \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
  \ldots\nu_e^{(a,t_{j+1})}(dx_{j+1})\cr
\displaycause{applying (a) to the function $(y_0,\ldots,y_{n-j})
\mapsto f(e,\ldots,\omega(t_{j-1}),y_0,\ldots,y_{n-j})$}
&=\int\ldots\int
    g(e,x_{j+1},\ldots,x_n)
    \nu_{x_{n-1}}^{(t_{n-1},t_n)}(dx_n)
  \ldots\nu_e^{(a,t_{j+1})}(dx_{j+1})\cr
\displaycause{where $g(x_j,\ldots,x_n)
=f(\omega(0),\omega(t_1),\ldots,\omega(t_{j-1}),x_jz,\ldots,x_nz)$ for
$x_j,\ldots,x_n\in U$}
&=\int\ldots\int g(e,x_{j+1},\ldots,x_n)
    \nu_{x_{n-1}}^{(t_{n-1}-a,t_n-a)}(dx_n)
  \ldots\nu_e^{(0,t_{j+1}-a)}(dx_{j+1})\cr
\displaycause{because $\nu_x^{(s-a,t-a)}E=\lambda_{t-s}(Ex^{-1})
=\nu_x^{(s,t)}E$ whenever $E\subseteq\BbbR^K$ is Borel, $x\in U$ and
$a\le s<t$}
&=\int g\pi_Kd\hat\mu\cr
\displaycause{where $K=\{0,t_{j+1}-a,\ldots,t_n-a\}$}
&=\int f\pi_J\phi_{\omega a}d\hat\mu
=\int f\pi_Jd(\hat\mu\phi_{\omega a}^{-1}).\cr}$$

\noindent As $f$ and $J$ are arbitrary, $\hat\mu_{\omega a}$ and
$\hat\mu\phi_{\omega a}^{-1}$ agree on the algebra
$\bigotimes_{\coint{0,\infty}}\Cal B(U)$ generated by sets of the form
$\{\omega:\omega(t)\in E\}$ for $t\ge 0$ and Borel sets $E\subseteq U$.
By the Monotone Class Theorem, the measures agree on the $\sigma$-algebra
$\Tensorhat_{\coint{0,\infty}}\Cal B(U)$
generated by $\bigotimes_{\coint{0,\infty}}\Cal B(U)$;  because they are
both defined as complete measures inner regular with respect to this
$\sigma$-algebra, they are identical.

\medskip

{\bf (c)} The defining formula for $\phi_{\omega a}$ makes it plain that
$\phi_{\omega a}(\omega')$ is \cadlag\ whenever $\omega$, $\omega'$ are
\cadlag.
If $W$ is measured by $\ddot\mu_{\omega a}$, there is a
$W'\in\dom\hat\mu_{\omega a}$ such that $W=W'\cap\Cdlg$.   In this case,
$\phi_{\omega a}^{-1}[W]\cap\Cdlg=\phi_{\omega a}^{-1}[W']\cap\Cdlg$
while $\phi_{\omega a}^{-1}[W']\in\dom\hat\mu$;  so
$\ddot\mu\phi_{\omega a}^{-1}[W]$ is defined and equal to

\Centerline{$\hat\mu\phi_{\omega a}^{-1}[W']
=\hat\mu_{\omega a}W'=\ddot\mu_{\omega a}W$.}
}%end of proof of 455S

\leader{455T}{Corollary} Let $U$ be a metrizable
topological group which is complete
under a \rti\ metric inducing its topology.
Let $\langle\lambda_t\rangle_{t>0}$
be a family of Radon probability measures
on $U$ such that $\lambda_s*\lambda_t=\lambda_{s+t}$ for all $s$, $t>0$
and $\lim_{t\downarrow 0}\lambda_tG=1$ for every open
neighbourhood $G$ of the identity $e$ in $U$;
let $\hat\mu$ be the measure on $U^{\coint{0,\infty}}$ defined
from $\langle\lambda_t\rangle_{t>0}$ by the method of 455Pc.   
Let $\Cdlg$ be the set of \cadlag\ functions from
$\coint{0,\infty}$ to $U$, $\ddot\mu$ the subspace measure on $\Cdlg$
and $\ddot\Sigma$ its domain.   For $t\ge 0$, let $\ddot\Sigma_t$ be

\Centerline{$\{F:F\in\ddot\Sigma$, $\omega'\in F$ whenever $\omega\in F$,
   $\omega'\in\Cdlg$ and $\omega'\restr[0,t]=\omega\restr[0,t]\}$}

\noindent and
$\hat{\ddot\Sigma}_t=\{F\symmdiff A:F\in\ddot\Sigma_t$, $\ddot\mu A=0\}$.
Then $\hat{\ddot\Sigma}_t=\bigcap_{s>t}\hat{\ddot\Sigma}_s$ includes
$\ddot\Sigma^+_t=\bigcap_{s>t}\ddot\Sigma_s$.

\proof{{\bf (a)} I show first that
$\ddot\Sigma^+_t\subseteq\hat{\ddot\Sigma}_t$.
\Prf\ Take $E\in\ddot\Sigma^+_t$.
Let $\tau:\Cdlg\to[0,\infty]$
be the constant stopping time with value $t$,
and $f$ the characteristic function $\chi E$.   Set
$g(\omega)=\int_{\Cdlg}fd\ddot\mu_{\omega t}$ when this is defined in
$\Bbb R$, where $\ddot\mu_{\omega t}$ is defined as in 455O;  then
$g$ is a conditional expectation of $f$ on $\ddot\Sigma_{\tau}$
(455Ob).   Since

$$\eqalign{\ddot\Sigma_{\tau}^+
&=\{H:H\in\ddot\Sigma,\,H\cap\{\omega:\tau(\omega)\le s\}\in\ddot\Sigma_s^+
\text{ for every }s\ge 0\}\cr
&=\{H:H\in\ddot\Sigma,\,H\in\ddot\Sigma_s^+
\text{ for every }s\ge t\}
=\ddot\Sigma_t^+\cr}$$

\noindent contains $E$, $g\eae\chi E$.
Setting $F=\{\omega:\omega\in\dom g$, $g(\omega)=1\}$, $F\in\ddot\Sigma$
(remember that $\ddot\mu$ is complete), and $E\symmdiff F$ is negligible.

Now 455Sc, with 235Gb, tells us that

\Centerline{$g(\omega)=\int_{\Cdlg}fd\ddot\mu_{\omega t}
=\int_{\Cdlg}f\phi_{\omega t}d\ddot\mu$}

\noindent whenever either integral is defined in $\Bbb R$, where

$$\eqalign{\phi_{\omega t}(\omega')(s)
&=\omega(s)\text{ if }s<t,\cr
&=\omega'(s-t)\omega(t)\text{ if }s\ge t.\cr}$$

\noindent If $\omega_0$, $\omega_1\in\Cdlg$ and
$\omega_0\restr[0,t]=\omega_1\restr[0,t]$, then
$\phi_{\omega_0t}=\phi_{\omega_1t}$ so $g(\omega_0)=g(\omega_1)$ if either
is defined.   It follows that $\omega_0\in F$ iff $\omega_1\in F$.
As $\omega_0$ and $\omega_1$ are arbitrary, $F\in\ddot\Sigma_t$ and
$E\in\hat{\ddot\Sigma}_t$.\ \Qed

\medskip

{\bf (b)} Of course $\hat{\ddot\Sigma}_t\subseteq\hat{\ddot\Sigma}_s$
whenever $s>t$.   Putting (a) and 455L(f-ii) together,

\Centerline{$\bigcap_{s>t}\hat{\ddot\Sigma}_s
=\{E\symmdiff A:E\in\ddot\Sigma_t^+$, $\ddot\mu A=0\}
\subseteq\{E\symmdiff A:E\in\hat{\ddot\Sigma}_t$, $\ddot\mu A=0\}
=\hat{\ddot\Sigma}_t$}

\noindent and we have equality.
}%end of proof of 455T

\leader{455U}{Theorem}\dvArevised{2018} Let $U$ be a metrizable
topological group which is complete
under a \rti\ metric inducing its topology.
Let $\langle\lambda_t\rangle_{t>0}$
be a family of Radon probability measures
on $U$ such that $\lambda_s*\lambda_t=\lambda_{s+t}$ for all $s$, $t>0$
and $\lim_{t\downarrow 0}\lambda_tG=1$ for every open
neighbourhood $G$ of the identity $e$ in $U$;
let $\hat\mu$ be the measure on $U^{\coint{0,\infty}}$ defined
from $\langle\lambda_t\rangle_{t>0}$ by the method of 455Pc.  
Let $\Cdlg$ be the set of \cadlag\ functions from
$\coint{0,\infty}$ to $U$, $\ddot\mu$ the subspace measure on $\Cdlg$
and $\ddot\Sigma$ its domain.   For $t\ge 0$, let $\ddot\Sigma_t$ be

\Centerline{$\{F:F\in\ddot\Sigma$, $\omega'\in F$ whenever $\omega\in F$,
   $\omega'\in\Cdlg$ and $\omega'\restr[0,t]=\omega\restr[0,t]\}$,}

\noindent and 
$\hat{\ddot\Sigma}_t=\{F\symmdiff A:F\in\ddot\Sigma_t$, $\ddot\mu A=0\}$;
let $\tau:\Cdlg\to[0,\infty]$ be a stopping time adapted to
$\langle\hat{\ddot\Sigma}_t\rangle_{t\ge 0}$.   Define
$\phi_{\tau}:\Cdlg\times\Cdlg\to\Cdlg$ by setting

$$\eqalign{\phi_{\tau}(\omega,\omega')(t)
&=\omega'(t-\tau(\omega))\omega(\tau(\omega))
   \text{ if }t\ge\tau(\omega),\cr
&=\omega(t)\text{ otherwise}.\cr}$$

\noindent Then $\phi_{\tau}$ is \imp\ for the product measure
$\ddot\mu\times\ddot\mu$ on $\Cdlg\times\Cdlg$ and $\ddot\mu$ on $\Cdlg$.

\proof{{\bf (a)} To begin with (down to the end of (c) below), suppose that
$\tau$ is adapted to $\langle\ddot\Sigma^+_t\rangle_{t\ge 0}$, where
$\ddot\Sigma^+_t=\bigcap_{s>t}\ddot\Sigma_s$ for $t\ge 0$.
In this case we know from 455P
that the conditions of 455O are satisfied.   I aim to apply 455Oa, using
455S to give a description of the measures
$\ddot\mu_{\omega,\tau(\omega)}$.
Now if $f$ is $\ddot\mu$-integrable, we have, in the notation of 455O and
455S,

$$\eqalignno{\int_{\Cdlg}fd\ddot\mu
&=\int_{\Cdlg}\int_{\Cdlg}fd\ddot\mu_{\omega,\tau(\omega)}\ddot\mu(d\omega)\cr
\displaycause{455Oa}
&=\int_{\Cdlg}\int_{\Cdlg}f\phi_{\omega,\tau(\omega)}(\omega')
  \ddot\mu(d\omega')\ddot\mu(d\omega)\cr
\displaycause{455Sc}
&=\int_{\Cdlg}\int_{\Cdlg}f\phi_{\tau}(\omega,\omega')
  \ddot\mu(d\omega')\ddot\mu(d\omega)\cr.}$$

\medskip

{\bf (b)} To convert the repeated integral into the product measure,
we have still to check for measurability.    The point is that, writing
$\Lambda$ for the domain of the product measure $\ddot\mu\times\ddot\mu$,
$\phi_{\tau}$ is
$(\Lambda,\Tensorhat_{\coint{0,\infty}}\Cal B(U))$-measurable.

\medskip

\Prf{\bf (i)} Consider first the case in which $U$ is separable.
Take $t\ge 0$.   Then
$E_t=\{\omega:\omega\in\Cdlg$, $\tau(\omega)\le t\}$ belongs to
$\ddot\Sigma$.   The function

\Centerline{$\omega\mapsto t-\tau(\omega):E_t\to\coint{0,\infty}$}

\noindent is $(\ddot\Sigma,\Cal B(\coint{0,\infty}))$-measurable;
the function

\Centerline{$(\omega',s)\mapsto\omega'(s):\Cdlg\times\coint{0,\infty}
\to U$}

\noindent is
$(\ddot\Sigma\tensorhat\Cal B(\coint{0,\infty}),\Cal B(U))$-measurable, by
4A3Qc, because $U$ is Polish;   so the function

\Centerline{$(\omega,\omega')\mapsto\omega'(t-\tau(\omega)):
   E_t\times\Cdlg\to U$}

\noindent is $(\ddot\Sigma\tensorhat\ddot\Sigma,\Cal B(U))$-measurable.   Next,
similarly,

\Centerline{$\omega\mapsto\omega(\tau(\omega)):E_t\to U$}

\noindent is $(\ddot\Sigma,\Cal B(U))$-measurable, while

\Centerline{$(y,z)\mapsto yz:U\times U\to U$}

\noindent is $(\Cal B(U)\tensorhat\Cal B(U),\Cal B(U))$-measurable,
because $U$ is a second-countable topological group.   But this means
that

\Centerline{$(\omega,\omega')
  \mapsto\omega'(t-\tau(\omega))\omega(\tau(\omega)):
  E_t\times\Cdlg\to U$}

\noindent is $(\ddot\Sigma\tensorhat\ddot\Sigma,\Cal B(U))$-measurable.
On the other hand, of course,

\Centerline{$(\omega,\omega')\mapsto\omega(t):
   (\Cdlg\setminus E_t)\times\Cdlg\to U$}

\noindent is $(\ddot\Sigma\tensorhat\ddot\Sigma,\Cal B(U))$-measurable.
Putting these together,

\Centerline{$(\omega,\omega')\mapsto\phi_{\tau}(\omega,\omega')(t):
   \Cdlg\times\Cdlg\to U$}

\noindent is $(\ddot\Sigma\tensorhat\ddot\Sigma,\Cal B(U))$-measurable.
This is true for every $t\ge 0$, so $\phi_{\tau}\restr\Cdlg\times\Cdlg$ is
$(\ddot\Sigma\tensorhat\ddot\Sigma,
  \Tensorhat_{\coint{0,\infty}}\Cal B(U))$-measurable.

\medskip

\quad{\bf (ii)} For the general case, we can use the trick in 455H.
There is a separable subgroup $U'$ of $U$ such that
$\nu_e^{(0,q)}U'=1$ for every rational $q\ge 0$.
We can suppose that $U'$ is
a closed subgroup of $U$.  Because $U'$ is closed,

$$\eqalign{\Cdlg'
&=\{\omega:\omega\in\Cdlg,\,\omega(t)\in U'\text{ for every }t\ge 0\}\cr
&=\{\omega:\omega\in\Cdlg,\,\omega(q)\in U'
   \text{ for every }q\in\Bbb Q\cap\coint{0,\infty}\}\cr}$$

\noindent is $\ddot\mu$-conegligible in $\Cdlg$, and
because $U'$ is a subgroup,
$\phi_{\tau}(\omega,\omega')\in\Cdlg'$ for all $\omega$,
$\omega'\in\Cdlg'$.   Now the argument of (i) shows that
$\phi_{\tau}\restr\Cdlg'\times\Cdlg'$ is
$(\ddot\Sigma\tensorhat\ddot\Sigma,
   \Tensorhat_{\coint{0,\infty}}\Cal B(U'))$-measurable, therefore
$(\ddot\Sigma\tensorhat\ddot\Sigma,
   \Tensorhat_{\coint{0,\infty}}\Cal B(U))$-measurable.
Since $\Cdlg'\times\Cdlg'$ is
$(\ddot\mu\times\ddot\mu)$-conegligible,
$\phi_{\tau}$ is
$(\Lambda,\Tensorhat_{\coint{0,\infty}}\Cal B(U))$-measurable.\ \Qed

\medskip

{\bf (c)} It follows that if $E\in\Tensorhat_{\coint{0,\infty}}\Cal B(U)$
then, setting $f=\chi(E\cap\Cdlg)$ in (a),

\Centerline{$(\ddot\mu\times\ddot\mu)\phi_{\tau}^{-1}[E\cap\Cdlg]
=\int f(\phi_{\tau}(\omega,\omega'))\ddot\mu(d\omega')\ddot\mu(d\omega)
=\int fd\ddot\mu
=\ddot\mu(E\cap\Cdlg)$}

\noindent by Fubini's theorem.
But $\ddot\mu$ is the subspace measure generated by the
completion of a
measure with domain $\Tensorhat_{\coint{0,\infty}}\Cal B(U)$,
so is inner regular with respect to sets of the form $E\cap\Cdlg$ with
$E\in\Tensorhat_{\coint{0,\infty}}\Cal B(U)$;  by 412K,
$\phi_{\tau}$ is \imp.

\medskip

{\bf (d)} Now suppose only that $\tau$ is adapted to
$\langle\hat{\ddot\Sigma}_t\rangle_{t\ge 0}$.   By 455L(f-iii),
there is a stopping time
$\tau':\Cdlg\to[0,\infty]$, adapted to $\langle\ddot\Sigma_t^+\rangle_{t\ge 0}$,
such that $\tau'\eae\tau$.   
Now we see from (a)-(c) that $\phi_{\tau'}$ is \imp, while

\Centerline{$\{(\omega,\omega'):
  \phi_{\tau'}(\omega,\omega')\ne\phi_{\tau}(\omega,\omega')\}
\subseteq\{(\omega,\omega'):\tau'(\omega)\ne\tau(\omega)\}$}

\noindent is $(\ddot\mu\times\ddot\mu)$-negligible, so 
$\phi_{\tau}$ also is \imp.
}%end of proof of 455U

\exercises{\leader{455X}{Basic exercises (a)}
%\spheader 455Xa
Let $\langle A_n\rangle_{n\ge 1}$ be a non-increasing sequence of subsets
of $[0,1]$, all with Lebesgue outer measure $1$, and with empty
intersection.   Set $T=\{0\}\cup\{\bover1n:n\ge 1\}$, $\Omega_0=\{0\}$,
$\Omega_{1/n}=A_n$ for $n\ge 1$;  for $t\in T$ let $\Tau_t$ be the Borel
$\sigma$-algebra of $\Omega_t$.   For $s<t$ in $T$ and $x\in\Omega_s$
define a Borel measure $\nu^{(s,t)}_x$ on $\Omega_t$ by saying that

\inset{if $n\ge 1$,
then $\nu^{(0,1/n)}_0(E\cap A_n)$ is the Lebesgue measure
of $E$ for every Borel set $E\subseteq[0,1]$,

if $0<s<t$, then $\nu^{(s,t)}_x\{x\}=1$.}

\noindent Show that $\family{y}{\Omega_t}{\nu^{(t,u)}_y}$ is a
disintegration of $\nu^{(s,u)}_x$ over $\nu^{(s,t)}_x$ whenever
$s<t<u$ in $T$ and $x\in\Omega_s$.   Taking $x^*=0$,
show that there is no measure $\mu$ on
$\prod_{t\in T}\Omega_t$ with the properties listed in 455A.
%455A

\spheader 455Xb Let $T$, $t^*$,
$\family{t}{T}{(\Omega_t,\Tau_t)}$, $x^*$,
$\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_s}$ and $\mu$ be as in 455A.
Suppose that we are given a family
$\family{t}{T}{(\Omega'_t,\Tau'_t,\pi_t)}$ such
that ($\alpha$) $\Omega'_t$ is a set, $\Tau'_t$ is a $\sigma$-algebra of
subsets of $\Omega'_t$ and $\pi_t:\Omega_t\to\Omega'_t$ is a surjective
$(\Tau_t,\Tau'_t)$-measurable function for every $t\in T$
($\beta$) whenever $s<t$ in $T$ and $x$, $x'\in\Omega_s$ are such that
$\pi_s(x)=\pi_s(x')$, then $\nu^{(s,t)}_x$ and $\nu^{(s,t)}_{x'}$ agree on
$\{\pi_t^{-1}[F]:F\in\Tau'_t\}$.   (i) Show that
if we set $\acute\nu^{(s,t)}_w(F)=\nu^{(s,t)}_x\pi_t^{-1}[F]$
whenever $s<t$ in $T$, $x\in\Omega_s$, $w=\pi_s(x)$ and $F\in\Tau'_t$,
then every $\acute\nu^{(s,t)}_w$ is a
perfect probability measure, and
$\family{z}{\Omega'_t}{\acute\nu^{(t,u)}_z}$ is a disintegration of
$\acute\nu^{(s,u)}_w$ over $\acute\nu^{(s,t)}_w$ whenever $s<t<u$ in $T$
and $w\in\Omega'_s$.   (ii) Let $\mu'$ be the measure on
$\Omega'=\prod_{t\in T}\Omega'_t$ defined by the method of 455A from
$\pi_{t^*}(x^*)$ and
$\langle\acute\nu^{(s,t)}_w\rangle_{s<t,w\in\Omega'_s}$.   Show that
$\pi:\Omega\to\Omega$ is \imp\ for $\mu$ and $\mu'$, where
$\pi(\omega)(t)=\pi_t(\omega(t))$ for $\omega\in\Omega$ and $t\in T$.
%455A

\spheader 455Xc
In 455E, set $T=\{-1\}\cup\coint{0,\infty}$, let each
$\Omega_t$ be $\Bbb R$, and for $x\in\Bbb R$,
$0\le s<t$ let $\nu_x^{(s,t)}$ be the Dirac measure on $\Bbb R$
concentrated at $\psi(x,t-s)$ on $\Bbb R$, where

$$\eqalign{\psi(x,t)
&=\Bover{x}{1-xt}\text{ if }xt\ne 1\text{ and }x\ne 0,\cr
&=0\text{ if }xt=1,\cr
&=-\Bover1t\text{ if }x=0.\cr}$$

\noindent Let $\nu$ be any atomless Radon probability measure on
$\Bbb R$, and complete the definition by setting
$\nu_x^{(-1,t)}(E)=\nu\{y:\psi(y,t)\in E\}$ whenever $t\ge 0$ and this is
defined;  set $x^*=0$.
Show that the conditions of 455E are satisfied, that the
measure $\hat\mu$ constructed in 455E is a distribution on $\BbbR^T$, and
that $\hat\mu$ is not $\tau$-additive.   \Hint{setting $\phi(y)(-1)=0$,
$\phi(y)(t)=\psi(y,t)$ for $t\ge 0$ and $y\in\Bbb R$,
show that $\phi:\Bbb R\to\BbbR^T$ is \imp\ for
$\nu_0$ and $\hat\mu$, and that every point of
$\BbbR^T$ has a neighbourhood of zero measure.}
%455E

\spheader 455Xd Let $T$, $t^*$, $\family{t}{T}{\Omega_t}$, $x^*$,
$\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_s}$ and
$\hat\mu$ be as in 455E.   Suppose that we are given a family
$\family{t}{T}{(\Omega'_t,\pi_t)}$ such
that ($\alpha$) $\Omega'_t$ is a Hausdorff space and
$\pi_t:\Omega_t\to\Omega'_t$ is a continuous surjective
function for every $t\in T$
($\beta$) whenever $s<t$ in $T$ and $x$, $x'\in\Omega_s$ are such that
$\pi_s(x)=\pi_s(x')$, then the image measures $\nu^{(s,t)}_x\pi_t^{-1}$
and $\nu^{(s,t)}_{x'}\pi_t^{-1}$ on $\Omega'_t$ are the same.
(i) Show that if we set $\acute\nu^{(s,t)}_w=\nu^{(s,t)}_x\pi_t^{-1}$
whenever $s<t$ in $T$, $x\in\Omega_s$ and $w=\pi_s(x)$,
then every $\acute\nu^{(s,t)}_w$ is a Radon probability measure, and
$\family{z}{\Omega'_t}{\acute\nu^{(t,u)}_z}$ is a disintegration of
$\acute\nu^{(s,u)}_w$ over $\acute\nu^{(s,t)}_w$ whenever $s<t<u$ in $T$
and $w\in\Omega'_s$.   (ii) Let $\hat\mu'$ be the measure on
$\Omega'=\prod_{t\in T}\Omega'_t$ defined by the method of 455E from
$\pi_{t^*}(x^*)$ and
$\langle\acute\nu^{(s,t)}_w\rangle_{s<t,w\in\Omega'_s}$.   Show that
$\pi:\Omega\to\Omega$ is \imp\ for $\hat\mu$ and $\hat\mu'$, where
$\pi(\omega)(t)=\pi_t(\omega(t))$ for $\omega\in\Omega$ and $t\in T$.
%455E

\spheader 455Xe Let $U$ be a locally compact metrizable group and $\nu$
any
Radon probability measure on $U$.   For $t>0$ let $\lambda_t$ be the Radon
probability measure

\Centerline{$e^{-t}(\delta_e+t\nu+\Bover{t^2}{2!}\nu*\nu
+\Bover{t^3}{3!}\nu*\nu*\nu+\ldots)$,}

\noindent where $\delta_e$ is the Dirac measure on $U$ concentrated at the
identity $e$ of $U$, and the sum is defined as in 234G\formerly{1{}12Ya}.
Show that $\langle\lambda_t\rangle_{t>0}$ satisfies the
conditions of 455P (with respect to an appropriate metric on $U$).
\Hint{4A5Mb, 4A5Q(iv).}
%455P

\spheader 455Xf Let $U$, $\langle\lambda_t\rangle_{t>0}$ and $\hat\mu$ be
as in 455Pc.   Let $V$ be a Hausdorff space, $z^*$ a point of $V$ and
$\action$ a continuous action of $U$ on $V$;  set
$\pi(x)=x\action z^*$ for $x\in U$.   Define
$\langle\nu^{(s,t)}_x\rangle_{0\le s<t,x\in U}$ from
$\langle\lambda_t\rangle_{t>0}$ as in 455P.   (i) Show that if $x$,
$x'\in U$ and $\pi(x)=\pi(x')$, then the image measures
$\nu^{(s,t)}_x\pi^{-1}$, $\nu^{(s,t)}_{x'}\pi^{-1}$ on $V$ are equal
whenever $s<t$.   (ii) Let $\hat\mu'$ be the measure on
$V^{\coint{0,\infty}}$ defined as in 455Xd.   Show that if we define
$\tilde\pi:U^{\coint{0,\infty}}\to V^{\coint{0,\infty}}$ by setting
$\tilde\pi(\omega)(t)=\omega(t)\action z^*$ for every
$\omega\in U^{\coint{0,\infty}}$ and $t\ge 0$, $\tilde\pi$ is \imp\ for
$\hat\mu$ and $\hat\mu'$ and also for the Radon measures extending them;
moreover, that the restriction of $\tilde\pi$ to $\Cdlg(U)$, the space
of \cadlag\ functions from $\coint{0,\infty}$ to $U$, is \imp\ for the
subspace measures on $\Cdlg(U)$ and $\Cdlg(V)$.
%455P 455Xd

\sqheader 455Xg For $t>0$, let $\lambda_t$ be the normal distribution on
$\Bbb R$ with expectation $0$ and variance $t$.
Show that $\langle\lambda_t\rangle_{t>0}$ satisfies the conditions of
455Q.
%455Q

\sqheader 455Xh For $t>0$, let $\lambda_t$ be the Poisson
distribution with expectation $t$, that is,
$\lambda_t(E)=e^{-t}\sum_{m\in E\cap\Bbb N}t^m/m!$ for $E\subseteq\Bbb R$
(cf.\ 285Q, 285Xr).
(i) Show that $\langle\lambda_t\rangle_{t>0}$ satisfies the conditions of
455Q.
(ii) Show that if $\tilde\mu$ is the Radon measure defined from
$\langle\lambda_t\rangle_{t>0}$ as in 455Pc,
then $\omega$ is non-decreasing
and $\omega[\,\coint{0,\infty}\,]=\Bbb N$ for $\tilde\mu$-almost every
$\omega\in\BbbR^{\coint{0,\infty}}$.
%455Q

\sqheader 455Xi For $t>0$, let $\lambda_t$ be the Cauchy
distribution with centre $0$ and scale parameter $t$, that is, the
distribution with probability density function
$x\mapsto\Bover{t}{\pi(x^2+t^2)}$ (285Xp).
(i) Show that $\langle\lambda_t\rangle_{t>0}$ satisfies the conditions of
455Q.   (ii) Show that if $\hat\mu$ is the corresponding distribution on
$\BbbR^{\coint{0,\infty}}$, then $C(\coint{0,\infty})$ is
$\hat\mu$-negligible.
\Hint{estimate $\Pr(|X_{(i+1)/n}-X_{i/n}|\le\epsilon$ for every $i<n)$.}
(iii) Suppose that $\alpha>0$.   Define
$T_{\alpha}:\BbbR^{\coint{0,\infty}}\to\BbbR^{\coint{0,\infty}}$ by setting
$(T_{\alpha}\omega)(t)=\Bover1{\alpha}\omega(\alpha t)$ for $t\ge 0$ and
$\omega\in\BbbR^{\coint{0,\infty}}$.   Show that $T_{\alpha}$ is \imp\ for
$\hat\mu$.   
%(iv) For $\omega\in\BbbR^{\coint{0,\infty}}$ set
%$(R\omega)(t)=t^2\omega(\bover1t)$ if $t>0$, $\omega(0)$ if $t=0$.
%Show that $R:\BbbR^{\coint{0,\infty}}\to\BbbR^{\coint{0,\infty}}$ is \imp\
%for $\hat\mu$.   (Compare 477E below.)
%455Q

\sqheader 455Xj(i) The {\bf standard gamma distribution} with
parameter $t$ is the probability distribution $\lambda_t$ on
$\Bbb R$ with probability density function
$x\mapsto\Bover1{\Gamma(t)}x^{t-1}e^{-x}$ for $x>0$.   Show
that its expectation is $t$.  \Hint{225Xh(iv).}   Show that its
variance is $t$.
(ii) Show that $\langle\lambda_t\rangle_{t>0}$ satisfies the conditions of
455Q.    \Hint{272U\formerly{2{}72T}, 252Yf.}   (iii) Show that
$\lim_{t\downarrow 0}t\Gamma(t)=1$, so that
$\lim_{t\downarrow 0}\bover1t\lambda_t\coint{1,\infty}
=\int_1^{\infty}\bover1xte^{-x}dx>0$.
(iv) Show that if $\tilde\mu$ is the Radon measure on
$\BbbR^{\coint{0,\infty}}$ defined from $\langle\lambda_t\rangle_{t>0}$ as
in 455Pc, then $\{\omega:\omega$ is strictly increasing and not
continuous$\}$ is $\tilde\mu$-conegligible.
%455Q

\spheader 455Xk Let $U$ be an abelian Hausdorff topological group.   Let
$\langle\lambda'_t\rangle_{t>0}$, $\langle\lambda''_t\rangle_{t>0}$ be two
families of Radon probability measures on $U$ and set
$\lambda_t=\lambda'_t*\lambda''_t$ for $t>0$.   (i) Show that if
$\lambda'_{s+t}=\lambda'_s*\lambda'_t$ and
$\lambda''_{s+t}=\lambda''_s*\lambda''_t$ for all $s$, $t>0$, then
$\lambda_{s+t}=\lambda_s*\lambda_t$ for all $s$, $t>0$.   (ii) Show that if
$\lim_{t\downarrow 0}\lambda'_tG=\lim_{t\downarrow 0}\lambda''_tG=1$ for
every open set containing the identity $e$ of $U$, then
$\lim_{t\downarrow 0}\lambda_tG=1$ for every open set $G$ containing $e$.
(iii) Now suppose that $U$ is metrizable and complete under a \rti\ metric
inducing its topology.   Let $\hat\mu'$, $\hat\mu''$ and $\hat\mu$ be the
measures on $U^{\coint{0,\infty}}$ defined from
$\langle\lambda'_t\rangle_{t>0}$, $\langle\lambda''_t\rangle_{t>0}$ and
$\langle\lambda_t\rangle_{t>0}$ as in 455Pc.   Set
$\theta(\omega,\omega')(t)=\omega(t)\omega'(t)$ for $\omega$,
$\omega'\in U^{\coint{0,\infty}}$ and $t\ge 0$.   Show that $\theta:
U^{\coint{0,\infty}}\times U^{\coint{0,\infty}}\to U^{\coint{0,\infty}}$
is \imp\ for $\hat\mu'\times\hat\mu''$ and $\hat\mu$.
(iv) Repeat (iii) for the subspace measures on the space of
\cadlag\ functions from $\coint{0,\infty}$ to $U$.
%455U

\leader{455Y}{Further exercises (a)}
%\spheader 455Ya
Let $\family{n}{\Bbb Z}{X_n}$ be a double-ended sequence
of real-valued random variables such that (i) for each $n\in\Bbb Z$,
$Y_n=X_{n+1}-X_n$ is independent of $\{X_i:i\le n\}$ (ii)
$\family{n}{\Bbb Z}{Y_n}$ is identically distributed.   Show that the
$Y_n$ are essentially constant.   \Hint{285Yc.}
%455B

\spheader 455Yb For $0\le s<t$ and $x\in\Bbb R$ define a Radon
probability measure
$\nu^{(s,t)}_x$ on $\Bbb R$ by saying that

$$\eqalign{\nu^{(s,t)}_x
&=\Bover{1-t}{1-s}\delta_0+\Bover{t-s}{1-s}\lambda_{[s,t]}
\text{ if }x=0\text{ and }t\le 1,\cr
&=\Bover{t-1}{1-s}\delta_0+\Bover{2-t-s}{1-s}\lambda_{[s,2-t]}
\text{ if }x=0\text{ and }1\le t<2-s,\cr
%&=\delta_0\text{ if }x=0\text{ and }2-s\le t\cr
&=\delta_0\text{ if }0<x<1\text{ and }2-x\le t,\cr
&=\delta_x\text{ otherwise},\cr}$$

\noindent writing $\delta_x$ for the Dirac measure on $\Bbb R$ concentrated
at $x$, and $\lambda_{[s,t]}$ for the uniform distribution based on the
interval $[s,t]$.   (i) Show that
$\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Bbb R}$ satisfies the conditions of
455E.   (ii) Starting from $x^*=0$,
let $\hat\mu$ be the corresponding measure on
$\BbbR^{\coint{0,\infty}}$.   Show that $\hat\mu^*\Cdlg=1$, where $\Cdlg$
is the space of \cadlag\ functions from $\coint{0,\infty}$ to $\Bbb R$.
(iii) Show that $\hat\mu$ has a unique extension to a Radon measure
$\tilde\mu$ on $\BbbR^{\coint{0,\infty}}$.   (iv) Show that
$\tilde\mu\Cdlg=0$.   (v) Show that the subspace measure on $\Cdlg$ induced
by $\hat\mu$ is not $\tau$-additive.
%455K
%the intention is to put the measure on $\omega_{\alpha}$, for
%$0\le\alpha\le 1$, where $\omega_{\alpha}(t)=\alpha$ for
%$\alpha\le t<2-\alpha$, $0$ elsewhere.

\spheader 455Yc A probability distribution $\lambda$ on $\Bbb R$ is {\bf
infinitely divisible} if for every $n\ge 1$ it is expressible as a
convolution $\nu*\ldots*\nu$ of $n$ copies of a probability distribution.
Let $\phi$ be the characteristic function (\S285) of an infinitely
divisible distribution $\lambda$.   (i) Show that for each $n\ge 1$
there is a
characteristic function $\phi_n$ such that $\phi_n^n=\phi$.   (ii) Show
that if $\delta>0$ is such that $\phi(y)\ne 0$ for $|y|\le\delta$, then
$\lim_{n\to\infty}\phi_n(y)=1$ for $|y|\le\delta$.   (iii) Show that
$\lim_{n\to\infty}\phi_n(y)=1$ for every $y\in\Bbb R$.   \Hint{for any
characteristic function $\psi$, $4\Real\psi(y)\le 3+\Real\psi(2y)$ for
every $y$.}   (iv) Show that $\phi$ is never zero, and that there is a
unique family $\langle\lambda_t\rangle_{t>0}$ of distributions satisfying
the conditions in 455P and such that $\lambda_1=\lambda$.   (v) Show that
if $\lambda$ has finite expectation, then $\Expn(\lambda_t)$ is defined and
equal to $t\Expn(\lambda)$ for every $t>0$.
%272Xi 285Ga
(vi) Show that if
$\lambda$ has finite variance, then $\Var(\lambda_t)$ is defined and equal
to $t\Var(\lambda)$ for every $t>0$.
%455P

\spheader 455Yd Let $U$ be  a Hausdorff topological group and
$\langle\lambda_t\rangle_{t>0}$ a family of Radon probability measures on
$U$ such that $\lambda_{s+t}=\lambda_s*\lambda_t$ whenever $s$, $t>0$.
(i) Show that we can define a family
$\langle\nu_x^{(s,t)}\rangle_{0\le s<t,x\in U}$ as in 455P, and that
$\family{y}{U}{\nu_y^{(t,u)}}$ is a disintegration of $\nu_x^{(s,u)}$ over
$\nu_x^{(s,t)}$ whenever $x\in U$ and $s<t<u$, so that, starting from
$x^*=e$ the identity of $U$, we can apply 455E to obtain a measure
$\hat\mu$ on $U^{\coint{0,\infty}}$.   (ii) Now suppose that $U=\BbbR^r$
where $r\ge 1$.
For $t>0$, $E\subseteq U$ set $\Reverse{\lambda}_tE=\lambda_t(-E)$
whenever  $\lambda_t$ measures $-E$;  now set
$\lambda^{\#}_t=\lambda_t*\Reverse{\lambda}_t$.   Show that
$\lambda^{\#}_{s+t}=\lambda^{\#}_s*\lambda^{\#}_t$ for all $s$, $t>0$, and
that $\lim_{t\downarrow 0}\lambda^{\#}_tG=1$ for every
open neighbourhood $G$ of $0$.   Show that $\hat\mu$ has an extension to a
Radon measure on $(\BbbR^r)^{\coint{0,\infty}}$.
%455P

\spheader 455Ye Let $Y$ be a metrizable
space and $\Cdlg$ the set of \cadlag\ functions from $\coint{0,\infty}$ to
$Y$.   For $\omega\in\Cdlg$ and $t\ge 0$ set $X_t(\omega)=\omega(t)$.
Let $\Sigma$ be a $\sigma$-algebra of subsets of $\Cdlg$ such that
$X_t:\Cdlg\to Y$ is measurable for every $t\ge 0$.
For $t\ge 0$ let $\Sigma_t$ be

\Centerline{$\{F:F\in\Sigma,\,\omega'\in F$ whenever $\omega$,
$\omega'\in\Cdlg$, $\omega\in F$ and
$\omega\restr[0,t]=\omega'\restr[0,t]\}$.}

\noindent Show that
$\langle X_t\rangle_{t\ge 0}$ is progressively measurable with respect
to $\langle\Sigma_t\rangle_{t\ge 0}$.
%455M out of order query
}%end of exercises

\endnotes{
\Notesheader{455}
This section has grown into the longest in this treatise.
There are some big theorems here.
I am trying to do two rather different things:  sketch the
fundamental properties of Markov processes, and work through
the details of particular realizations of them.   I remarked in the
introduction to Chapter 27 that probability theory is not really about
measure spaces and measurable functions.   It is much more about
distributions, and by `distribution' here I do not really mean a Radon
probability measure on $\BbbR^r$, let alone a completed Baire measure on
$\BbbR^I$, as in 454K.   I mean rather the family of probabilities of
the type $\Pr(X_i\le\alpha_i\Forall i\le n)$;  everything else is formal
structure, offering proofs and (I hope) some kinds of deeper understanding,
but essentially secondary.   The appalling formulae above
($\nu^{(t_j,t_{j+1})}_{\omega,\tau(\omega),x_j}(dx_{j+1})$,
$\ddot\Sigma\tensorhat\ddot\Sigma$ and so on)
arise from my attempts to distinguish
clearly among the host of probability spaces which present themselves to us
as relevant.

However one of the messages of this section is that for many stochastic
processes it is possible to identify semi-canonical realizations.   We
already have a crude one in 454J;  starting from any family
$\familyiI{X_i}$ of real-valued
random variables on any probability space, we can move to a measure
on $\BbbR^I$ which is in some sense unique and carries the probabilistic
content of the original family.   I noted in \S454 that when
this measure is $\tau$-additive we have a canonical extension to a
quasi-Radon measure, just as good regarded as a realization of the abstract
process, and possibly with useful further properties.   In 455H we find
that many of the most important processes can be represented by Radon
measures;  I do not think these Radon measures have been much studied,
except, of course, in the case of Brownian motion.
But 455O and 455U show that for some purposes we are better off
with quasi-Radon measures on the set of \cadlag\ functions.
The most important stopping times are the hitting times of 455M, which are
adapted to families of the form $\langle\Sigma_t^+\rangle_{t\ge 0}$;  and
for such a stopping time to be approximated by discrete stopping times, as
in parts (a-vi) and (b-vii) of the proof of 455O, we need to know that
our paths are continuous on the right.

It is of course true that when the complete metric space $U$, in 455O or later,
is separable, then we have a standard Borel structure on the space $\Cdlg$
of \cadlag\ functions (4A3Qb), so that the measures $\ddot\mu$ are Radon
measures for appropriate Polish topologies on $\Cdlg$.
\leaveitout{{\smc Billingsley 99};
 http://eom.springer.de/S/s110200.htm;
 http://eom.springer.de/s/s130370.htm.
 No apparent hope of good topological behaviour of the functions
 $\phi_{\tau}$ -- examples 455Xi, 455Xj where we have atomless marginals
 $\nu^{(0,t)}_{x^*}$ but discontinuous processes;  in this case we can have
 continuous stopping times but $\omega\mapsto\phi_{\tau}(\omega,\omega')$
 seriously discontinuous.}

Returning to the detailed exposition, 455A is an attempt at a
continuous-time version of 454H.   I
use the letters $t$, $T$ to suggest the probabilistic intuitions behind
these results;  we think of the spaces $\Omega_t$ in 455A as being the sets
of possible states of a system at `time' $t$, so that the measures
$\nu_x^{(s,t)}$ are descriptions of how we believe the system is likely
to evolve between times $s$ and $t$, having observed that
it is in state $x$ at
time $s$.   In the case of `discrete time', when we observe the system
only at clearly separated moments, it is easy to handle non-Markov
processes, in which evolution between times $n-1$ and $n$ can depend on
the whole history up to time $n-1$;  thus in 454H the measures
$\nu_z=\nu_z^{(n-1,n)}$ are defined for every $z\in\prod_{i<n}X_i$, but
we make no attempt to describe measures $\nu_z^{(n-1,m)}$ for any $m>n$.
In `continuous time' we need to say something about arbitrary time
steps, and it is hard to formulate a consistency condition to fill the
place of ($\dagger$) in 455A without limiting the kind of process being
examined.   At the cost of an appalling increase in complexity, of
course, the formulae of 455A can
sometimes be adapted to general processes, if we
replace the `current' state space $\Omega_t$ by the `historical' state
space $\prod_{t^*\le s\le t}\Omega_s$.
(For we can hope that
$(\prod_{t^*\le s\le t}\Omega_s,\Tensorhat_{t^*\le s\le t}\Tau_s)$
will have the
`perfect measure property' of 454Xd.)   We should finish up with a
measure on $\prod_{t\in T}(\prod_{t^*\le s\le t}\Omega_s)$.
But the important
applications, even when not Markov, are open to more economical and more
enlightening approaches.
We really do need a least element $t^*$ of $T$;  see 455Ya.

I have not yet come to the reason why this section is such hard work.
This is in its attempt to analyze the `Markov property' of the
distributions being examined here.   The point about the families
$\langle\nu^{(s,t)}_x\rangle_{s<t,x\in\Omega_s}$ of transitional
probabilities is that they not only give us stochastic processes, as in
455A, but also recipes for conditional expectations, derived from the
truncated families $\langle\nu^{(s,t)}_x\rangle_{a\le s<t,x\in\Omega_s}$.
These lead to measures $\mu'_{ax}$ on $\prod_{t\ge a}\Omega_t$
which can be thought of as distributions of future paths given
that we have reached the point $x$ at time $a$.   It is no surprise that
these should
provide straightforward descriptions of conditional expectations on
algebras of the form

\Centerline{$\{F:F$ is determined by coordinates in $[t^*,t]\}$.}

\noindent Without much more trouble,
we can extend this to suitable algebras
defined from simple `stopping times', as in 455C.
The arguments there have
some technical features which you may find annoying (and I invite you to
find your own way past the complications),
but are essentially elementary, as they have to be in such a
general context.   It is interesting that we can move to stopping times
taking countably many values without further difficulty.

However, we are still only seven pages into the section, and not everything
to come is as straightforward as the completion processes described in
455E.   An essential aspect of continuous-time Markov processes is the
possibility of stopping times which take a continuum of values, as is
typically the case in the examples provided by 455M.   These are
much harder to deal with, and we have to restrict sharply the
class of processes we examine.
The particular restriction I have chosen is
described by the definitions in 455F.   I should of course say that these,
particularly 455Fb (`uniformly time-continuous on the right') are more
limiting than is strictly necessary;  in `Feller processes'
({\smc Rogers \& Williams 94}, III.6) we have a slightly different approach
to the same intuitive target.   The aim is to find sufficient conditions
for the `strong Markov property', in which we can find disintegrations and
conditional expectations associated with general stopping times, as in
455O.   To do this, we have to abandon the set
$\Omega=U^{\coint{0,\infty}}$ and move to the correct
set of full outer measure, the set $\Cdlg$ of `\cadlag' functions, which
dominates the central part of this section.   The first thing the
definition 455Fb must do is to ensure that $\Cdlg$ has full outer
measure not only for the distribution on $\Omega$ but also for the
conditional distributions we shall be using (455G).
If $U$ is a Polish space, $\Cdlg$ has a standard Borel structure (4A3Qb),
which is comforting.

I hope that you are becoming resigned to the view that the
notational complexities of this section are not
solely due to an inconsiderate disregard for the
reader's eyesight.   The original probability measures $\nu^{(s,t)}_x$
of 455A really do form a three-parameter family, the conversion of these
into finite-dimensional distributions $\lambda_J$ really is a multiple
repeated integral, the derived probabilities $\nu^{(s,t)}_{\omega ax}$
in 455B are
a five-parameter system.   Without wishing to insist on my use of
grave accents in the proof of 455E, it is surely safer to have a way of
distinguishing between completed and uncompleted measures, and while the
result may be `obvious', I think there are some twists on the way which not
everyone would foresee.   Again, if you wish to dispense with
the double-dotted symbols from 455O on, you will have to find some other
way of reminding yourself that we are looking at a new representation of
the process on a new probability space.

This treatise as a whole is theory-heavy and example-light.   I assure you
that all the theory here is in fact example-driven.   You should start with
the four examples of L\'evy processes in
455Xg-455Xj.   %455Xg 455Xh 455Xi 455Xj
Of these, 455Xg is {\bf Brownian motion}, the starting point
of the whole theory;  I will return to this in \S477.   A problem with the
formalization in 455A is that we have to start with an exact description of
the transitional probabilities $\nu^{(s,t)}_x$.   It does not help at all
in establishing the existence of such families matching
some probabilistic intuition.   Only in rather special cases do we have
elegant formulae for these systems.   In 455Xb, 455Xd and 455Xf I try to
show how the general theory gives us methods of using one
system to build others.

I suppose that 455O is the summit;  from here on the going is easier.
In 455P I introduce `L\'evy processes', a particularly interesting class
intermediate in generality between the continuous processes of 455O and
Brownian motion.   These have of course mostly been considered in the case
$U=\Bbb R$, but the extension to
Banach spaces $U$ is an obvious one, and we
can even manage non-abelian groups if we are careful.   (For an elementary
example of a process which can really exploit a non-abelian group,
see 455Xe.)   The `Poisson process' in 455Xh is by some way the most
important example after Brownian motion itself.
L\'evy processes on $\Bbb R$ are well understood;  the family
$\langle\lambda_t\rangle_{t>0}$ is determined by $\lambda_1$, any
`infinitely divisible' distribution can be taken for $\lambda_1$
(455Yc), and
a complete description of infinitely divisible distributions is provided by
the L\'evy-Khintchine representation theorem ({\smc Fristedt \& Gray 97},
16.3).
%Khintchine the spelling used in his Roman-alphabet papers? yes it seems
As a final result in the general theory, I give an alternative version of
the strong Markov property in 455U.   For L\'evy processes, we can
re-start, following any of the usual stopping times,
with an exact copy of the process,
and this corresponds to a true \imp\ function from $\Cdlg^2$ to $\Cdlg$.

A comment on 455T.   The idea behind the $\sigma$-algebras $\Sigma_t$,
$\ddot\Sigma_t$  of 455M, 455O and later is that they consist of events
`observable at time $t$', that is, determined by the path taken up to and
including time $t$.   We quickly find ourselves forced to consider
augmented algebras $\Sigma_t^+=\bigcap_{s>t}\Sigma_s$, where somehow we are
allowed infinitesimal intuitions into the immediate future.   (A typical
situation is that of 455Mb when the set $A$ is open, so that if
$\omega(t)\in\overline{A}$ we can expect that there will be paths which
continue immediately into $A$, and others which do not, and it may not be
obvious which, if either, should be regarded as typical.)   The question
is, whether $\Sigma_t$ is really different from $\Sigma_t^+$.
The claim of 455T is that
$\ddot\Sigma_t^+$ is included in a kind of completion $\hat{\ddot\Sigma_t}$
of $\ddot\Sigma_t$.
Of course the completion is in terms of the measure $\ddot\mu$ on the
whole space $\Cdlg$ of \cadlag\ paths;  we need advance knowledge of which
subsets of $\Cdlg$ are negligible.   But if we are interested in the
measure algebra $\frak A$ of $\ddot\mu$ and its closed subalgebras
$\frak A_t=\{E^{\ssbullet}:E\in\ddot\Sigma_t\}$, 455T tells us that (in the
context of L\'evy processes) we can expect to have
$\frak A_t=\bigcap_{s>t}\frak A_s$.   Turning to the definition of
$\ddot\mu$ in 455O as a subspace measure, we see that $\frak A$ can be
regarded as the measure algebra of the measure $\hat\mu$ on
$U^{\coint{0,\infty}}$ defined by the formulae of 455E;  and even that
$\frak A_t$ can be identified with

\Centerline{$\{E^{\ssbullet}:E\in\dom\hat\mu$, $E$ is determined by
coordinates in $[0,t]\}$}

\noindent (see part (a-ii) of the proof of 455O).   But I think that this
last step will not usually be helpful, because (as noted above) $\ddot\mu$
will commonly be a Radon measure for an appropriate topology, while
$\hat\mu$ is likely at best to be the completion of a Baire measure.

I have cast the second half of the section in terms of measures on
$\Cdlg$, because it is reasonably well adapted to L\'evy processes in
general.   When we come to look at particular processes, we often find
that there is a smaller class of functions
(e.g., continuous functions in the case of
Brownian motion, or non-decreasing $\Bbb N$-valued functions in the case of
the Poisson process) which is fully adequate and easier to focus on.
For the detailed study of such processes, as in \S477 below, I think it
will usually be helpful to make the shift.   But there may be rival
conegligible subsets of $\Cdlg$ with different virtues, as in 477Ef.
}%end of notes

\discrpage

\frfilename{mt456.tex}
\versiondate{19.5.10}
\copyrightdate{2010}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Gaussian distributions}

\newsection{456}

Uncountable powers of $\Bbb R$ are not as a rule
measure-compact\cmmnt{ (439P, 455Xc;  see also 533J in Volume 5)}.
Accordingly distributions, in
the sense of 454K, need not be $\tau$-additive.   But some, at least,
of the distributions most important to us
are indeed $\tau$-additive, and therefore have interesting canonical
extensions.
This section is devoted to a remarkable result, taken from
{\smc Talagrand 81}, concerning a class of distributions which are of
great importance in probability theory.   It demands a combination of
techniques from classical probability theory and from the topological
measure theory of this volume.   I begin with the definition and
fundamental properties of what I call `centered Gaussian distributions'
(456A-456I).  %456A 456B 456C 456D 456E 456F 456G 456H 456I
These are fairly straightforward
adaptations of the classical finite-dimensional theory, and
will be useful in \S477 when we come to study Brownian motion.
Another relatively easy idea is that of `universal' Gaussian
distribution (456J-456L).  %456J 456K 456L
In 456M we come to a much deeper result, a step towards classifying the
ways in which a Gaussian family of $n$-dimensional random variables can
accumulate at $0$.   The ideas are combined in
456N-456O to complete the proof of Talagrand's theorem that Gaussian
distributions on powers of $\Bbb R$ are $\tau$-additive.

\leader{456A}{Definitions (a)} Write $\mu_G$ for the Radon probability
measure on $\Bbb R$ which is the distribution of a standard normal
random variable\cmmnt{, that is, the probability distribution with
density function $x\mapsto\Bover1{\sqrt{2\pi}}e^{-x^2/2}$ (274A)}.   For any set
$I$, write $\mu_G^{(I)}$ for the measure on $\BbbR^I$ which is the
product of copies of $\mu_G$;  this is
always quasi-Radon\cmmnt{ (415E/453J)};  if $I$ is countable, it is
Radon\cmmnt{ (417Q)};  if
$I=n\in\Bbb N\setminus\{0\}$, it is the probability distribution with
density function
$x\mapsto(2\pi)^{-n/2}e^{-x\dotproduct x/2}$\cmmnt{ (272I)};
if $I=\emptyset$, it is the unique probability measure on the singleton
set $\BbbR^{\emptyset}$.

\spheader 456Ab I will use the phrase {\bf centered Gaussian
distribution} to mean a measure $\mu$ on a power $\BbbR^I$ of $\Bbb R$
such that $\mu$ is the completion of a Baire measure\cmmnt{ (that is,
is a distribution in the sense of 454K)} and every
continuous linear functional $f:\BbbR^I\to\Bbb R$ is either zero almost
everywhere or is a normal random variable with zero
expectation.   \cmmnt{(Note that I call the
distribution concentrated at the point $0$ in $\BbbR^I$ a
`Gaussian distribution'.)}

\spheader 456Ac If $I$ is a set and $\mu$ is a centered Gaussian
distribution on $\BbbR^I$, its {\bf covariance matrix} is the
family $\langle\sigma_{ij}\rangle_{i,j\in I}$ where
$\sigma_{ij}=\int x(i)x(j)\mu(dx)$ for $i$, $j\in I$.   \cmmnt{(The
integral is always defined and finite because each function
$x\mapsto x(i)$ is either essentially constant or normally distributed,
and in either case is square-integrable.)}

\leader{456B}{}\cmmnt{ I start with some fundamental facts about
Gaussian distributions.

\medskip

\noindent}{\bf Proposition} (a) Suppose that $I$ and $J$ are sets, $\mu$
is a centered Gaussian distribution on $\BbbR^I$, and
$T:\BbbR^I\to\BbbR^J$ is a continuous linear operator.   Then there is a
unique centered Gaussian distribution on $\BbbR^J$ for which $T$ is
\imp;  if $J$ is countable, this is the image measure $\mu T^{-1}$.

(b) Let $I$ be a set, and $\mu$, $\nu$ two centered Gaussian
distributions on $\BbbR^I$.   If they have the same covariance matrices
they are equal.

(c) For any set $I$, $\mu_G^{(I)}$ is the centered Gaussian distribution
on $\BbbR^I$ with the identity matrix for its covariance matrix.

(d) Suppose that $I$ is a countable set.   Then a measure $\mu$ on
$\BbbR^I$ is a centered Gaussian distribution iff it is of the form
$\mu_G^{(\Bbb N)}T^{-1}$ where $T:\BbbR^{\Bbb N}\to\Bbb R^I$ is a
continuous linear operator.

(e) Suppose $\family{j}{J}{I_j}$ is a disjoint family of sets with union
$I$, and that for each $j\in J$ we have a centered Gaussian distribution
$\nu_j$ on $\BbbR^{I_j}$.   Then the product $\nu$ of the measures
$\nu_j$, regarded as a measure on $\BbbR^I$, is a centered Gaussian
distribution.

(f) Let $I$ be any set, $\mu$ a centered Gaussian distribution on
$\BbbR^I$ and $E\subseteq\BbbR^I$ a set such that $\mu$ measures $E$.
Writing $-E=\{-x:x\in E\}$, $\mu(-E)=\mu E$.

\proof{{\bf (a)(i)} For Baire sets $F\subseteq\BbbR^J$, set
$\nu F=\mu T^{-1}[F]$;  this is always defined because $T$ is continuous
(4A3Kc).   This makes $\nu$ a Baire probability measure on $\BbbR^J$ for
which $T$ is \imp.   Because $\mu$ is complete, $T$ is still \imp\ for
the completion $\hat\nu$ of $\nu$ (234Ba\formerly{2{}35Hc}).
If $g:\BbbR^J\to\Bbb R$ is
a continuous linear functional, so is $gT:\BbbR^I\to\Bbb R$;  now
$\nu\{y:g(y)\le\alpha\}=\mu\{x:gT(x)\le\alpha\}$ for every $\alpha$, so
$g$ and $gT$ have the same distribution, and are both either zero a.e.\
or normal random variables.   As $g$ is arbitrary, $\hat\nu$ is a
centered Gaussian distribution as defined in 456Ab.   Of course it is
the only such distribution on $\BbbR^J$ for which $T$ is \imp.

\medskip

\quad{\bf (ii)} Now suppose that $J$ is countable.   Then $\BbbR^J$
is Polish (4A2Qc), so $\nu$ is a Borel measure and $\hat\nu$ is a Radon
measure (433Cb).
$\BbbR^J$ has a countable network consisting of Borel sets, $\mu$ is
perfect (454A(b-iii)) and totally finite, and $T$ is measurable (418Bd),
so $\mu T^{-1}$ is a Radon measure (451O).   Thus $\hat\nu$ and
$\mu T^{-1}$ are Radon measures agreeing on the Borel sets and must be
equal.

\medskip

{\bf (b)} The point is that $\mu f^{-1}=\nu f^{-1}$ for every continuous
linear functional $f:\BbbR^I\to\Bbb R$.   \Prf\ By (a),
$\mu f^{-1}$ and $\nu f^{-1}$ are Radon measures on $\Bbb R$, and by the
definition of `Gaussian distribution' each is either a normal
distribution with expectation zero, or is concentrated at $0$.   By
4A4Be, we
can express $f$ in the form $f(x)=\sum_{i\in I}\beta_ix(i)$ for every
$x\in\BbbR^I$, where $\{i:\beta_i\ne 0\}$ is finite.   In this case

$$\eqalignno{\int t^2(\mu f^{-1})(dt)
&=\int f(x)^2\mu(dx)\cr
\displaycause{235G\formerly{2{}35I}}
&=\sum_{i,j\in I}\beta_i\beta_j\int x(i)x(j)\mu(dx)\cr
&=\sum_{i,j\in I}\beta_i\beta_j\int x(i)x(j)\nu(dx)
=\int t^2(\nu f^{-1})(dt)\cr}$$

\noindent because $\mu$ and $\nu$ have the same covariance matrices.
But this means that $\mu f^{-1}$ and $\nu f^{-1}$ have the same
variance;  if this is zero, they both give measure $1$ to $\{0\}$;
otherwise, they are normal distributions with the same expectation and
the same variance, so again are equal.\ \Qed

By 454P, $\mu=\nu$.

\medskip

{\bf (c)} Being a completion regular quasi-Radon probability measure
(415E), $\mu^{(I)}_G$ is the completion of a Baire probability measure
on $\BbbR^I$.   If $f:\BbbR^I\to\Bbb R$ is a continuous linear
functional, then it is expressible in the form
$f(z)=\sum_{i\in I}\beta_iz(i)$, where $J=\{i:\beta_i\ne 0\}$ is finite.
I need to show that $f$ is either zero a.e.\ or a normal random variable
with expectation $0$.   If $J=\emptyset$ then $f=0$ everywhere and we
can stop.
Otherwise, $f=\sum_{i\in J}\beta_i\pi_i$, where $\pi_i(x)=x(i)$ for
$i\in I$ and $x\in\BbbR^I$.   Now, with respect to the measure
$\mu_G^{(I)}$, $\family{i}{J}{\pi_i}$ is an independent family of normal
random variables with zero expectation (272G).   So
$\family{i}{J}{\beta_i\pi_i}$ is independent (272E), and $\beta_i\pi_i$
is normal for $i\in J$ (274Ae).   By 274B,
$f=\sum_{i\in J}\beta_i\pi_i$ is normal, and of course it has zero
expectation.
As $f$ is arbitrary, $\mu_G^{(I)}$ is a centered Gaussian distribution.

We have

\Centerline{$\int x(i)x(i)\mu_G^{(I)}(dx)=\int t^2\mu_G(dt)=1$}

\noindent for $i\in I$, and

\Centerline{$\int x(i)x(j)\mu_G^{(I)}(dx)
=\int t\mu_G(dt)\cdot\int t\mu_G(dt)=0$}

\noindent if $i$, $j\in I$ are distinct.   So the covariance matrix of
$\mu_G^{(J)}$ is the identity matrix.   By (b), it is the only centered
Gaussian distribution with this covariance matrix.

\medskip

{\bf (d)(i)} It follows from (c) and (a) that if
$\mu=\mu_G^{(\Bbb N)}T^{-1}$, where $T:\BbbR^{\Bbb N}\to\BbbR^I$ is a
continuous linear operator, then $\mu$ is a centered Gaussian
distribution.

\medskip

\quad{\bf (ii)} Now suppose that $\mu$ is a centered Gaussian
distribution on $\BbbR^I$.   Set $\pi_i(x)=x(i)$ for $i\in I$ and
$x\in\BbbR^I$;  for
$i\in I$, set $u_i=\pi_i^{\ssbullet}$ in $L^2=L^2(\mu)$.   By 4A4Jh,
there is a countable orthonormal family $\family{j}{J}{v_j}$ in $L^2$
such that every $v_j$ is a linear combination of the $u_i$, and every
$u_i$ is a linear combination of the $v_j$.   We may suppose that
$J\subseteq\Bbb N$.   For $i\in I$, express $u_i$ as
$\sum_{j\in J}\alpha_{ij}v_j$, where $\{j:\alpha_{ij}\ne 0\}$ is finite.
Define $T:\BbbR^{\Bbb N}\to\BbbR^I$ by setting
$(Tz)(i)=\sum_{j\in J}\alpha_{ij}z(j)$ for every $z\in\BbbR^{\Bbb N}$ and
$i\in I$.   Then $T$ is a continuous linear functional.   Set
$\nu=\mu_G^{(\Bbb N)}T^{-1}$, so that $\nu$ is a centered Gaussian
distribution on $\BbbR^I$, by (a).   Because $\BbbR^I$ is Polish, both
$\mu$ and $\nu$ must be Radon measures.

\medskip

\quad{\bf (iii)} Now $\mu$ and $\nu$ have the same covariance matrices.
\Prf\ If $i$, $i'\in I$ then

$$\eqalign{\int x(i)x(i')\mu(dx)
&=\innerprod{u_i}{u_{i'}}
=\sum_{j,j'\in J}\alpha_{ij}\alpha_{i'j'}\innerprod{v_j}{v_{j'}}\cr
&=\sum_{j\in J}\alpha_{ij}\alpha_{i'j}
=\sum_{j,j'\in J}\alpha_{ij}\alpha_{i'j'}
  \int z(j)z(j')\mu_G^{(\Bbb N)}(dz)\cr
&=\int (Tz)(i)(Tz)(i')\mu_G^{(\Bbb N)}(dz)
=\int x(i)x(i')\nu(dx). \text{ \Qed}\cr}$$

\noindent By (b), $\mu=\nu$ is of the required form.

\medskip

{\bf (e)} We must first confirm that $\nu$ is the completion of a Baire
measure.   \Prf\ If we write $\CalBa(\BbbR^{I_j})$ for the Baire
$\sigma$-algebra of $\BbbR^{I_j}$, then each $\nu_j$ is the completion
of its restriction $\nu_j\restr\CalBa(\BbbR^{I_j})$, so $\nu$ is also
the product of the measures $\nu_j\restr\CalBa(\BbbR^{I_j})$ (254I), and
is therefore the completion of its restriction to
$\Tensorhat_{j\in J}\CalBa(\BbbR^{I_j})$ (254Ff).   But as
$\CalBa(\BbbR^{I_j})=\Tensorhat_{i\in I_j}\CalBa(\Bbb R)$ for every $j$
(4A3Na), $\Tensorhat_{j\in J}\CalBa(\BbbR^{I_j})$ can be identified with
$\Tensorhat_{i\in I}\CalBa(\Bbb R)=\CalBa(\BbbR^I)$, so that $\nu$ is
indeed the completion of $\nu\restr\CalBa(\BbbR^I)$.\ \Qed

Now suppose that $f:\BbbR^I\to\Bbb R$ is a continuous linear functional.
Then we can express $f$ in the form $f(x)=\sum_{i\in K}\alpha_ix(i)$ for
every $x\in\BbbR^I$, where $K\subseteq I$ is finite.   Set
$L=\{j:K\cap I_j\ne\emptyset\}$ and $K_j=K\cap I_j$ for $j\in L$, so
that $L$ and every $K_j$ are finite;  for $j\in L$ and $x\in\BbbR^I$ set
$f_j(x)=\sum_{i\in K_j}\alpha_ix(i)$.   Now $f=\sum_{j\in L}f_j$.

If we set $g_j(y)=\sum_{i\in K_j}\alpha_iy(i)$ for $y\in\BbbR^{I_j}$,
then $g_j$ is either zero a.e.\ or a normal random variable with respect
to the probability measure $\nu_j$.   Since

\Centerline{$\nu\{x:f_j(x)\le\alpha\}=\nu\{x:g_j(x\restr I_j)\le\alpha\}
=\nu_j\{y:g_j(y)\le\alpha\}$}

\noindent for every $\alpha\in\Bbb R$, $f_j$ (regarded as a random
variable on $(\BbbR^I,\nu)$) has the same distribution as $g_j$
(regarded as a random variable on $(\BbbR^{I_j},\nu_j)$).   This is true
for every $j\in L$.   Moreover, the different $f_j$, as $j$ runs over
$L$, are independent.   So $f=\sum_{j\in L}f_j$ is the sum of
independent random variables which are all either normal or essentially
constant.   By 274B again, $f$ also is either normal or essentially
constant.   And of course its expectation is zero.   As $f$ is
arbitrary, this shows that $\nu$ is a centered Gaussian distribution.

\medskip

{\bf (f)} Set $Tx=-x$ for $x\in\BbbR^I$, so that $T$ is a continuous
linear operator and we have a unique centered Gaussian distribution
$\nu$ on $\BbbR^I$ such that $T$ is \imp\ for $\mu$ and $\nu$, by (a).
For any $i$, $j\in I$,

\Centerline{$\int x(i)x(j)\nu(dx)
=\int(Tx)(i)(Tx)(j)\mu(dx)
=\int x(i)x(j)\mu(dx)$,}

\noindent so $\mu$ and $\nu$ have the same covariance matrices and are
equal, by (b).   Accordingly

\Centerline{$\mu(-E)=\mu T^{-1}[E]=\nu E=\mu E$}

\noindent whenever $\mu$ measures $E$.
}%end of proof of 456B

\leader{456C}{}\cmmnt{ Since a Gaussian distribution is determined by
its covariance matrix (456Bb), we naturally seek descriptions of which
matrices can arise.

\medskip

\noindent}{\bf Theorem}\dvAformerly{4{}56Xa.}
Let $I$ be a set and $\langle\sigma_{ij}\rangle_{i,j\in I}$ a family of
real numbers.   Then the following are equiveridical:

(i) $\langle\sigma_{ij}\rangle_{i,j\in I}$ is the covariance matrix of a
centered Gaussian distribution on $\BbbR^I$;

(ii) there are a (real) Hilbert
space $U$ and a family $\familyiI{u_i}$ in $U$ such that
$\innerprod{u_i}{u_j}=\sigma_{ij}$ for all $i$, $j\in I$;

(iii) for every finite $J\subseteq I$,
$\langle\sigma_{ij}\rangle_{i,j\in J}$ is the covariance matrix of a
centered Gaussian distribution on $\BbbR^J$;

(iv) $\langle\sigma_{ij}\rangle_{i,j\in I}$ is
symmetric and positive semi-definite
in the sense that $\sigma_{ij}=\sigma_{ji}$ for all $i$, $j\in I$ and
$\sum_{i,j\in J}\alpha_i\alpha_j\sigma_{ij}\ge 0$ whenever
$J\subseteq I$ is finite and $\family{i}{J}{\alpha_i}\in\Bbb R^J$.

\proof{{\bf (i)$\Rightarrow$(ii)} If $\mu$ is a centered Gaussian
distribution on $\BbbR^I$ with covariance matrix
$\langle\sigma_{ij}\rangle_{i,j\in I}$, then $L^2(\mu)$ is a Hilbert space.
Setting $X_i(x)=x(i)$ for $x\in\BbbR^I$,
$u_i=X_i^{\ssbullet}$ belongs to the Hilbert space $L^2(\mu)$ for
every $i\in I$, and

\Centerline{$\innerprod{u_i}{u_j}=\int X_i\times X_jd\mu
=\int x(i)x(j)\mu(dx)=\sigma_{ij}$}

\noindent for all $i$, $j\in I$.

\medskip

{\bf (ii)$\Rightarrow$(iv)} In this context,

\Centerline{$\sigma_{ij}=\innerprod{u_i}{u_j}=\innerprod{u_j}{u_i}
=\sigma_{ji}$,}

\Centerline{$\sum_{i,j\in J}\alpha_i\alpha_j\sigma_{ij}
=\sum_{i,j\in J}\alpha_i\alpha_j\innerprod{u_i}{u_j}
=\|\sum_{i\in J}\alpha_iu_i\|^2
\ge 0$.}

\medskip

{\bf (iv)$\Rightarrow$(iii)} Here we have to know something about symmetric
matrices.   Given a family $\langle\sigma_{ij}\rangle_{i,j\in I}$
satisfying the conditions of (iv), and a
finite set $J\subseteq I$, we have a linear operator $T:\BbbR^J\to\BbbR^J$
defined by saying that $(Tz)(i)=\sum_{j\in J}\sigma_{ij}z(j)$ for
$z\in\BbbR^J$ and $i\in J$.   Give $\BbbR^J=\ell^2(J)$ its usual inner
product, so that $w\dotproduct z=\sum_{j\in J}w(j)z(j)$ for $w$,
$z\in\BbbR^J$;  then $\BbbR^J$ is a Hilbert space and

\Centerline{$Tw\dotproduct z
=\sum_{i\in J}\sum_{j\in J}\sigma_{ij}w(j)z(i)
=\sum_{j\in J}\sum_{i\in J}\sigma_{ji}z(i)w(j)
=w\dotproduct Tz$}

\noindent for all $w$, $z\in\BbbR^J$, so that $T$ is self-adjoint.
Moreover, if $z\in\BbbR^J$,

\Centerline{$Tz\dotproduct z
=\sum_{i,j\in J}\sigma_{ij}z(i)z(j)\ge 0$}

\noindent by the other condition on
$\langle\sigma_{ij}\rangle_{i,j\in I}$.

By 4A4M\footnote{Or, rather, its finite-dimensional special case, which
is easier;  you may know it under the slogan `symmetric matrices are
diagonalisable'.},
$\BbbR^J$ has an orthonormal basis consisting of eigenvectors for $T$;
if $\#(J)=n$, we have a basis $\ofamily{k}{n}{u_k}$ and a family
$\ofamily{k}{n}{\gamma_k}$ of real numbers such that $Tu_k=\gamma_ku_k$ for
each $k<n$.   We need to know that $\sum_{k<n}u_k(i)u_k(j)=1$ if $i=j$, $0$
otherwise.   \Prf\ Let $\family{i}{J}{v_i}$
be the standard basis of $\BbbR^J$,
so that $v_i(j)=1$ if $i=j$, $0$ if $i\ne j$.   Then

\Centerline{$v_i=\sum_{k<n}(v_i\dotproduct u_k)u_k
=\sum_{k<n}u_k(i)u_k$}

\noindent for $i\in J$, so

$$\eqalign{\sum_{k<n}u_k(i)u_k(j)
=\sum_{k,l<n}u_k(i)u_l(j)u_k\dotproduct u_l
=v_i\dotproduct v_j
&=1\text{ if }i=j,\cr
&=0\text{ otherwise.  \Qed}\cr}$$

Now $\gamma_k=Tu_k\dotproduct u_k\ge 0$, so $\sqrt{\gamma_k}$
is defined for each $k$, and we have a linear operator
$S:\BbbR^n\to\BbbR^J$ defined by setting $Se_k=\sqrt{\gamma_k}u_k$ for each
$k$, where $\ofamily{k}{n}{e_k}$ is the standard basis of $\BbbR^n$,
defined by saying that $e_k(l)=1$ if $k=l$, $0$ otherwise.

Taking $\mu_G^{(n)}$ to be the standard Gaussian distribution on $\BbbR^n$,
$\mu=\mu_G^{(n)}S^{-1}$ is a centered Gaussian distribution on $\BbbR^J$,
by 456Ba.   For $i$, $j\in J$,

$$\eqalign{\int w(i)w(j)\mu(dw)
&=\int (Sz)(i)(Sz)(j)\mu_G^{(n)}(dz)\cr
&=\int\sum_{k<n}\sqrt{\gamma_k}z(k)u_k(i)
  \cdot\sum_{l<n}\sqrt{\gamma_l}z(l)u_l(j)
   \mu_G^{(n)}(dz)\cr
&=\sum_{k<n}\sum_{l<n}\sqrt{\gamma_k\gamma_l}u_k(i)u_l(j)
  \int z(k)z(l)\mu_G^{(n)}(dz)\cr
&=\sum_{k<n}\gamma_k u_k(i)u_k(j)
=\sum_{k<n}Tu_k(i)u_k(j)\cr
&=\sum_{k<n,l\in J}\sigma_{il}u_k(l)u_k(j)
=\sum_{l\in J}\sigma_{il}\sum_{k<n}u_k(l)u_k(j)
=\sigma_{ij}.\cr}$$

\noindent So $\mu$ is the distribution we are looking for.

\medskip

{\bf (iii)$\Rightarrow$(i)} I seek to apply 454M.
For each finite $J\subseteq I$, let $\mu_J$ be
a centered Gaussian distribution on $\BbbR^J$ with covariance matrix
$\langle\sigma_{ij}\rangle_{i,j\in J}$;  by 456Bb, it is unique.   If
$K\subseteq I$ is finite and $J\subseteq K$, set $T_{KJ}z=z\restr J$ for
$z\in\BbbR^K$;  then $\mu_KT_{KJ}^{-1}$ is a centered Gaussian distribution on
$\BbbR^J$, by 456Ba, and its covariance matrix is that of $\mu_J$, so
$\mu_J=\mu_KT_{KJ}^{-1}$.
By 454M, we have a distribution $\mu$ on $\BbbR^I$, the completion of a
Baire probability measure, such that $\mu_J=\mu T_J^{-1}$ for every finite
$J\subseteq I$, setting $T_Jx=x\restr J$ for $x\in\BbbR^I$.

Applying this with $J=\{i,j\}$, we see that
$\int x(i)x(j)\mu(dx)=\sigma_{ij}$ for all $i$, $j\in I$.   To see that
$\mu$ is a centered Gaussian distribution in the sense of 456Ab, take
a continuous linear functional $f:\BbbR^I\to\Bbb R$.   Then there is a
finite family $\family{i}{J}{\beta_i}$ in $\Bbb R$
such that $f(x)=\sum_{i\in J}\beta_ix(i)$ for each $x\in\Bbb R^I$.
Setting $g(z)=\sum_{i\in J}\beta_iz(i)$ for $z\in\Bbb R^J$, we have
$f=gT_J$, so that the image distribution $\mu f^{-1}$ on $\Bbb R$ is
just $\mu_Jg^{-1}$, and (because $\mu_J$ is a centered Gaussian
distribution) is either normal or concentrated at $0$.   As $f$ is
arbitrary, $\mu$ itself is a centered Gaussian distribution.
}%end of proof of 456C

\leader{456D}{Gaussian \dvrocolon{processes}}\cmmnt{ I take a page
to spell out the connexion between centered Gaussian distributions,
and the processes considered in 454J-454K.

\medskip

\noindent}{\bf Definition} A family $\familyiI{X_i}$ of real-valued
random variables on a probability space is a
{\bf centered Gaussian process} if its distribution\cmmnt{ (454J)}
is a centered Gaussian distribution.

\leader{456E}{Independence and \dvrocolon{correlation}}\cmmnt{ We have an
important characterization of independence of families forming a Gaussian
process.   The essential idea is in (a) below.   I give the more elaborate
version (b) for the sake of an application in \S477.

\medskip

\noindent}{\bf Proposition}
(a)\dvAformerly{4{}56Xh} Let $\familyiI{X_i}$ be a centered
Gaussian process.   Then $\familyiI{X_i}$ is independent iff
$\Expn(X_i\times X_j)=0$ for all distinct $i$, $j\in I$.

(b)\dvAnew{2007}
Let $\familyiI{X_i}$ be a centered Gaussian process on a complete
probability space $(\Omega,\Sigma,\mu)$, and
$\Cal J$ a disjoint family of subsets of $I$;  for $J\in\Cal J$ let
$\Sigma_J$ be the $\sigma$-algebra of subsets of $\Omega$ generated by
$\{X_i^{-1}[F]:i\in J$, $F\subseteq\Bbb R$ is Borel$\}$.
Suppose that $\Expn(X_i\times X_j)=0$
whenever $J$, $J'$ are distinct members of $\Cal J$, $i\in J$ and
$j\in J'$.   Then $\family{J}{\Cal J}{\Sigma_J}$ is independent.

\proof{{\bf (a)(i)} If $\familyiI{X_i}$ is independent, and
$i$, $j\in I$ are
distinct, then $\Expn(X_i\times X_j)=\Expn(X_i)\Expn(X_j)=0$, by
272R\formerly{2{}72Q}.

\medskip

\quad{\bf (ii)}
If $\Expn(X_i\times X_j)=0$ for all distinct $i$, $j\in I$, let
$\mu$ be the distribution of $\familyiI{X_i}$ and
$\langle\sigma_{ij}\rangle_{i,j\in I}$ its covariance matrix.   Then
$\sigma_{ij}=0$ whenever $i\ne j$.   So if we take $\nu_i$ to be the
normal distribution on $\Bbb R$ with expectation $0$ and variance
$\sigma_{ii}$ (or the distribution concentrated at $0$ if $\sigma_{ii}=0$),
the product $\nu=\prod_{i\in I}\nu_i$ will be a centered
Gaussian distribution on $\BbbR^I$ (456Be) also with covariance matrix
$\langle\sigma_{ij}\rangle_{i,j\in I}$, and is equal to $\mu$, by 456Bb.
Thus $\mu$ is a product measure and $\familyiI{X_i}$ is independent
(454L).

\medskip

{\bf (b)} Set $K=\bigcup\Cal J$.
For each $J\in\Cal J$, let $\nu_J$ be the distribution of
$\family{i}{J}{X_i}$, and let $\nu=\prod_{J\in\Cal J}\nu_J$ be the product
measure on $\prod_{J\in\Cal J}\BbbR^J$, which we can identify with
$\BbbR^K$.   Then $\nu$ is a centered Gaussian
distribution, and its covariance matrix
$\langle\sigma_{ij}\rangle_{i,j\in K}$ is such that

$$\eqalign{\sigma_{ij}
&=\Expn(X_i\times X_j)\text{ if }i,\,j\text{ belong to the same member of }
   \Cal J,\cr
&=\Expn(X_i)\Expn(X_j)=0\text{ otherwise};\cr}$$

\noindent that is, it is the covariance matrix of the process
$\family{i}{K}{X_i}$.   Let $f:\Omega\to\BbbR^K$ be a function such that
$f(\omega)(i)=X_i(\omega)$ whenever $i\in K$ and $\omega\in\dom X_i$;
because $\mu$ is complete, $f$ is $(\Sigma,\CalBa(\BbbR^K))$-measurable.
For $J\in\Cal J$ and $\omega\in\Omega$
set $f_J(\omega)=f(\omega)\restr J$.

Now suppose that $\Cal J_0\subseteq\Cal J$ is non-empty
and finite and that $E_J\in\Sigma_J$ for each $J\in\Cal J_0$.   Then for
each $J\in\Cal J_0$ there is a Baire set $F_J\subseteq\BbbR^J$ such that
$E_J\symmdiff f_J^{-1}[F_J]$ is $\mu$-negligible, and
$\mu E_J=\nu_JF_J$.   Next, the distribution of $\family{i}{K}{X_i}$ is
a centered Gaussian distribution on $\BbbR^K$, and has covariance matrix
$\langle\Expn(X_i\times X_j)\rangle_{i,j\in K}
=\langle\sigma_{ij}\rangle_{i,j\in K}$, so it must be $\nu$.   But this
means that, setting
$F=\{x:x\in\BbbR^K$, $x\restr J\in F_J$ for every $J\in\Cal J_0\}
\in\CalBa(\BbbR^K)$,

$$\eqalign{\mu(\bigcap_{J\in\Cal J}E_J)
&=\mu(\bigcap_{J\in\Cal J}f_J^{-1}[F_J])
=\mu f^{-1}[F]\cr
&=\nu F
=\prod_{J\in\Cal J}\nu_JF_J
=\prod_{J\in\Cal J}\mu E_J.\cr}$$

\noindent As $\family{J}{\Cal J_0}{E_J}$ is arbitrary,
$\family{J}{\Cal J}{\Sigma_J}$ is independent.
}%end of proof of 456E

\leader{456F}{Proposition} Let $\familyiI{X_i}$ be
a family of random variables on a probability space
$(\Omega,\Sigma,\mu)$.   Then the following are equiveridical:

(i) the distribution of $\familyiI{X_i}$\cmmnt{, in the sense of
454K,} is a centered Gaussian distribution;

(ii) whenever $i_0,\ldots,i_n\in I$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$ then $\sum_{r=0}^n\alpha_rX_{i_r}$
is either zero a.e.\ or a normal random variable with zero expectation;

(iii) whenever $i_0,\ldots,i_n\in I$ then the joint distribution of
$X_{i_0},\ldots,X_{i_n}$\cmmnt{, in the sense of 271C,} is a centered
Gaussian distribution;

(iv) whenever $J\subseteq I$ is finite then there is an independent
family $\family{k}{K}{Y_k}$ of standard normal random variables on
$\Omega$ such that each $X_i$, for $i\in J$, is almost everywhere equal
to a linear combination of the $Y_k$.

\proof{{\bf (ii)$\Leftrightarrow$(i)} is immediate from the definition in
456Ab and Proposition 454O.

\medskip

{\bf (i)$\Leftrightarrow$(iii)} is also direct from 456Ab and the
identification of the two concepts of `distribution' (454K).

\medskip

{\bf (iv)$\Rightarrow$(ii)} is direct from 274A-274B.

\medskip

{\bf (i)$\Rightarrow$(iv)} For $i\in J$ set $u_i=X_i^{\ssbullet}$ in
$L^2(\mu)$.   By 4A4Jh again there is an orthonormal family
$\family{k}{K}{v_k}$ in $L^2(\mu)$ such that each $v_k$ is a linear
combination of the $u_i$ and each $u_i$ is a linear combination of the
$v_k$.   Take $Y_k$ such that $Y_k^{\ssbullet}=v_k$ for each $k$;  then
each $X_i$ is equal almost everywhere to a linear combination of the
$Y_k$, while each $Y_k$ is equal almost everywhere to a linear
combination of the $X_i$.   As $\#(K)$ must be the dimension of the
linear span of $\{u_i:i\in J\}$, $K$ is finite.   Any linear combination
of the $Y_k$ is equal almost everywhere to a linear combination of the
$X_k$, so is either zero a.e.\ or a normal random variable with zero
expectation.   Because (ii)$\Rightarrow$(iii), $\family{k}{K}{Y_k}$ has
a centered Gaussian distribution $\nu$ say.   Each $Y_k$ has variance
$\innerprod{v_k}{v_k}=1$, so is a standard normal random variable.

The covariance matrix of $\nu$ is given by

$$\eqalignno{\int y(j)y(k)\nu(dy)
=\Expn(Y_j\times Y_k)
=\innerprod{v_j}{v_k}
&=1\text{ if }j=k,\cr
&=0\text{ otherwise}.\cr}$$

\noindent By 456E, $\family{k}{K}{Y_k}$ is independent, so we have found a
suitable family.
}%end of proof of 456F

\leader{456G}{}\cmmnt{ Now I start work on material for the main theorem
of this section.

\medskip

\noindent}{Lemma} Let $I$ be a finite set and $\mu$ a centered
Gaussian distribution on $\BbbR^I$.   Suppose that $\gamma\ge 0$ and
$\alpha=\mu\{x:\sup_{i\in I}|x(i)|\ge\gamma\}$.   Then
$\mu\{x:\sup_{i\in I}|x(i)|\ge\bover12\gamma\}\ge 2\alpha(1-\alpha)^3$.

\proof{{\bf (a)}
The case $\gamma=0$, $\alpha=1$ is trivial;  suppose that
$\gamma>0$.   We may suppose that $I$ has a total order $\le$.   Give
$\BbbR^{I\times 4}\cong(\BbbR^I)^4$ the product $\lambda$ of four
copies of $\mu$;  then $\lambda$ is a centered Gaussian distribution
(456Be).   Define $T:\BbbR^{I\times 4}\to\BbbR^I$ by setting
$(Ty)(i)=\bover12\sum_{r=0}^3y(i,r)$ for $i\in I$,
$y\in\BbbR^{I\times 4}$;  then $\lambda T^{-1}$ is a centered Gaussian
distribution on $\BbbR^I$ (456Ba).   Now $\lambda T^{-1}$ has the same
covariance matrix as $\mu$.   \Prf\ If $i$, $j\in I$ then

$$\eqalignno{\int x(i)x(j)(\lambda T^{-1})(dx)
&=\int (Ty)(i)(Ty)(j)\lambda(dy)
=\Bover14\sum_{r=0}^3\sum_{s=0}^3\int y(i,r)y(j,s)\lambda(dy)\cr
&=\Bover14\sum_{r=0}^3\int y(i,r)y(j,r)\lambda(dy)
=\Bover14\sum_{r=0}^3\int x(i)x(j)\mu(dx)\cr
\displaycause{because the map
$y\mapsto\familyiI{y(i,r)}:\BbbR^{I\times 4}\to\BbbR^I$ is \imp\ for
each $r$}
&=\int x(i)x(j)\mu(dx).  \text{ \Qed}\cr}$$

\noindent So $\lambda T^{-1}=\mu$, by 456Bb.

\medskip

{\bf (b)} Define

\Centerline{$E_{ir}=\{y:y\in\BbbR^{I\times 4}$,
  $|y(i,r)|\ge\gamma\}$}

\noindent for $r<4$ and $i\in I$, and

\Centerline{$E_r=\bigcup_{i\in I}E_{ir}$}

\noindent for  $r<4$, so that

\Centerline{$\lambda E_r=\lambda\{y:\sup_{i\in I}|y(i,r)|\ge\gamma\}
=\mu\{x:\sup_{i\in I}|x(i)|\ge\gamma\}=\alpha$.}

\noindent Set $E'_r=E_r\setminus\bigcup_{s\ne r}E_s$, so that

$$\eqalignno{\lambda E'_r
&=\lambda\{y:\sup_{i\in I}|y(i,r)|\ge\gamma,\,
  \sup_{i\in I}|y(i,s)|<\gamma\text{ for }s\ne r)\cr
&=\lambda\{y:\sup_{i\in I}|y(i,r)|\ge\gamma\}
  \cdot\prod_{s\ne r}\lambda\{y:\sup_{i\in I}|y(i,s)|<\gamma\}\cr
\displaycause{because these are independent events}
&=\alpha(1-\alpha)^3.\cr}$$

\medskip

{\bf (c)} Next, for $i\in I$ and $r<4$, set
$E'_{ir}=E_{ir}\setminus(\bigcup_{j<i}E_{jr}\cup\bigcup_{s\ne r}E_s)$,
so that $E'_r=\bigcup_{i\in I}E'_{ir}$.   Observe that
$\langle E'_{ir}\rangle_{i\in I,r<4}$ is disjoint.   Set

\Centerline{$F_{ir}=\{y:y\in E'_{ir}$,
$y(i,r)\sum_{s\ne r}y(i,s)\ge 0\}$.}

\noindent Then $\nu F_{ir}\ge\bover12\nu E'_{ir}$.   \Prf\ We can think
of $\BbbR^{I\times 4}$ as a product $\BbbR^J\times\BbbR^K$, where
$J=I\times\{r\}$ and $K=I\times(4\setminus\{r\})$.   In this case,
$\lambda$ becomes identified with a product $\lambda_J\times\lambda_K$,
where $\lambda_J$ and $\lambda_K$ are centered Gaussian distributions on
$\BbbR^J$ and $\BbbR^K$ respectively, and $E'_{ir}$ is of the form
$V\times W$, where

\Centerline{$V=\{v:v\in\BbbR^J$, $|v(i,r)|\ge\gamma$,
$|v(j,r)|<\gamma$ for $j<i\}$,}

\Centerline{$W=\{w:w\in\BbbR^K$, $|w(j,s)|<\gamma$ for every $j\in I$,
$s\ne r\}$.}

\noindent In the same representation, $F_{ir}$ becomes
$(V^+\times W^+)\cup(V^-\times W^-)$, where

\Centerline{$V^+=\{v:v\in V$, $v(i,r)\ge\gamma\}$,
\quad$V^-=\{v:v\in V$, $v(i,r)\le-\gamma\}$,}

\Centerline{$W^+=\{w:w\in W$, $\sum_{r\ne s}w(i,s)\ge 0\}$,
\quad$W^-=\{w:w\in W$, $\sum_{r\ne s}w(i,s)\le 0\}$.}

\noindent By 456Bf, $\lambda_JV^+=\lambda_JV^-$ and
$\lambda_KW^+=\lambda_KW^-$;  since $V^-=V\setminus V^+$, while
$W^+\cup W^-=W$, we have

\Centerline{$\lambda_JV^+=\lambda_JV^-=\Bover12\lambda_JV$,
\quad$\lambda_KW^+=\lambda_KW^-\ge\Bover12\lambda_KW$.}

\noindent But this means that

$$\eqalignno{\lambda F_{ir}
&=\lambda(V^+\times W^+)+\lambda(V^-\times W^-)\cr
&=\lambda_JV^+\cdot\lambda_KW^++\lambda_JV^-\cdot\lambda_KW^-
\ge\Bover12\lambda_JV\cdot\lambda_KW
=\Bover12\lambda E'_{ir},\cr}$$

\noindent as claimed.\ \Qed

\medskip

{\bf (d)} At this point, observe that if $y\in F_{ir}$ then
$|\sum_{s=0}^3y(i,s)|\ge|y(i,r)|\ge\gamma$.   So

$$\eqalign{\mu\{x:\sup_{i\in I}|x(i)|\ge\Bover12\gamma\})
&=\lambda T^{-1}[\{x:\sup_{i\in I}|x(i)|\ge\Bover12\gamma\})]\cr
&=\lambda\{y:\sup_{i\in I}
   |\Bover12\sum_{r=0}^3y(i,r)|\ge\Bover12\gamma\})\cr
&=\lambda\{y:\sup_{i\in I}|\sum_{r=0}^3y(i,r)|\ge\gamma)\cr
&\ge\lambda(\bigcup_{i\in I,r<4}F_{ir})
=\sum_{i\in I,r<4}\lambda F_{ir}\cr
&\ge\Bover12\sum_{r<4}\sum_{i\in I}\lambda E'_{ir}
\ge\Bover12\sum_{r<4}\lambda E'_r
=2\alpha(1-\alpha)^3,\cr}$$

\noindent which is what we set out to prove.
}%end of proof of 456G

\leader{456H}{The support of a Gaussian distribution:  Proposition} Let
$I$ be a set and $\mu$ a centered Gaussian distribution on $\BbbR^I$.
Write $Z$ for the set of those $x\in\BbbR^I$ such that $f(x)=0$ whenever
$f:\BbbR^I\to\Bbb R$ is a continuous linear functional and $f=0$ a.e.
Then $Z$ is a self-supporting closed linear subspace of $\BbbR^I$ with
full outer measure.   If $I$ is countable $Z$ is the support of $\mu$.

\proof{{\bf (a)} Being the intersection of a family of closed linear
subspaces, of course $Z$ is a closed linear subspace.

\medskip

{\bf (b)} $Z$ has full outer measure.   \Prf\ Let $F\subseteq\BbbR^I$
be a non-negligible zero set.   Let $J\subseteq I$ be a countable set
such that $F$ is determined by coordinates in $J$.   For $i\in I$ and
$x\in\BbbR^I$ set $\pi_i(x)=x(i)$;  then each $\pi_i$ is either normally
distributed or zero almost everywhere, so is square-integrable;  set
$u_i=\pi_i^{\ssbullet}$ in $L^2=L^2(\mu)$.
Let $\family{k}{K}{v_k}$ be a countable orthonormal family in $L^2$ such
that every $v_k$ is a linear combination of the $u_i$, for $i\in J$, and
every $u_i$, for $i\in J$, is a linear combination of the $v_k$ (4A4Jh
once more).
Extend $\family{k}{K}{v_k}$ to a Hamel basis $\family{l}{L}{v_l}$ of
$L^2$.   For every $i\in I$, we can express $u_i$ as
$\sum_{l\in L}\alpha_{il}v_l$, where $\{l:\alpha_{il}\ne 0\}$ is finite;
and the construction ensures that $\alpha_{il}=0$ if $i\in J$ and
$l\in L\setminus K$.

Consider the linear operator $T_0:\BbbR^K\to\BbbR^J$ defined by setting
$(T_0z)(i)=\sum_{k\in K}\alpha_{ik}z(k)$ for $z\in\BbbR^K$ and $i\in J$.
If we give $\BbbR^K$ the product measure $\mu_G^{(K)}$, then the image
measure $\mu_G^{(K)}T_0^{-1}$ is a Gaussian distribution (456Ba), with
covariance matrix

$$\eqalign{\sigma_{ii'}
&=\int x(i)x(i')(\mu_G^{(K)}T_0^{-1})dx
=\int(T_0z)(i)(T_0z)(i')\mu_G^{(K)}(dz)\cr
&=\sum_{k,k'\in K}\alpha_{ik}\alpha_{i'k'}z(k)z(k')\mu_G^{(K)}(dz)
=\sum_{k\in K}\alpha_{ik}\alpha_{i'k}\cr
&=\sum_{k,k'\in K}\alpha_{ik}\alpha_{i'k}\innerprod{v_k}{v_{k'}}
=\innerprod{u_i}{u_{i'}}
=\int x(i)x(i')\mu(dx).\cr}$$

\noindent But this means that $\mu_G^{(K)}T_0^{-1}$ has the same
covariance matrix as $\mu\tilde\pi_J^{-1}$, where
$\tilde\pi_Jx=x\restr J$ for $x\in\BbbR^I$.   Since this also is a
centered Gaussian distribution, the two measures must be equal (456Bb).
We know that $\tilde\pi_J[F]$ has non-zero measure, so there is a
$z_0\in\BbbR^K$ such that $T_0z_0\in\tilde\pi_J[F]$.   Extend $z_0$
arbitrarily to $z_1\in\BbbR^L$.

Set $x_1(i)=\sum_{l\in L}\alpha_{il}z_1(l)$ for $i\in I$.   Then
$\tilde\pi_Jx_1=T_0z_0\in\tilde\pi_J[F]$, so $x_1\in F$, because $F$ is
determined by coordinates in $J$.   If a continuous linear functional
$f:\BbbR^I\to\Bbb R$ is zero a.e., it can be expressed in the form
$f(x)=\sum_{i\in I}\beta_ix(i)$ where $\{i:\beta_i\ne 0\}$ is finite
(4A4Be again).   In this case,

\Centerline{$0=f^{\ssbullet}=\sum_{i\in I}\beta_iu_i
=\sum_{l\in L}\sum_{i\in I}\beta_i\alpha_{il}v_l$}

\noindent in $L^2$.   Since $\family{l}{L}{v_l}$ is linearly
independent, $\sum_{i\in I}\beta_i\alpha_{il}=0$ for every $l\in L$.
But this means that

\Centerline{$f(x_1)=\sum_{i\in I,l\in L}\beta_i\alpha_{il}z_1(l)=0$.}

\noindent As $f$ is arbitrary, $x_1\in Z$ and $Z\cap F\ne\emptyset$.
As $F$ is arbitrary, and $\mu$ is inner regular with respect to the zero
sets, $Z$ has full outer measure.\ \Qed

\medskip

{\bf (c)} $Z$ is self-supporting.   \Prf\ If $W\subseteq\BbbR^I$ is an
open set meeting $Z$, there is an open set $V$, depending on coordinates
in a finite set $J\subseteq I$, such that $V\subseteq W$ and
$V\cap Z\ne\emptyset$.   Write $\tilde\pi_J(x)=x\restr J$ for
$x\in\BbbR^I$, and $\nu_J$ for the image measure $\mu\tilde\pi_J^{-1}$
on $\BbbR^J$;  by 456Ba, this is a centered Gaussian distribution.   By
456Bd, there is a continuous linear operator $T:\BbbR^{\Bbb
N}\to\BbbR^J$ such that $\nu_J=\mu_G^{(\Bbb N)}T^{-1}$.   Since the
support of $\mu_G$ is
$\Bbb R$, the support of $\mu_G^{(\Bbb N)}$ is $\BbbR^{\Bbb N}$
(417E(b-iii), or otherwise), and the support of $\nu_J$ is
$Z_1=\overline{T[\BbbR^{\Bbb N}]}$ (411Ne).

Write $Q$ for the set of linear functionals $g:\BbbR^J\to\Bbb R$
(necessarily continuous, because $J$ is finite) which
are zero on $Z_1$.   If $g\in Q$, then $gT=0$, so $g=0\,\,\nu_J$-a.e.\
and $g\tilde\pi_J=0\,\,\mu$-a.e.   This means that $g\tilde\pi_J(x)=0$
for every $x\in Z$, that is, $g(y)=0$ for every $y\in\tilde\pi_J[Z]$.
Because $Z_1$ is a linear subspace of $\BbbR^J$, this is enough to show
that $\tilde\pi_J[Z]\subseteq Z_1$.

Now recall that $V\cap Z\ne\emptyset$ so
$Z_1\cap\tilde\pi_J[V]\ne\emptyset$, while
$V=\tilde\pi_J^{-1}[\tilde\pi_J[V]]$.   Since $\tilde\pi_J$ is an open
map (4A2B(f-i)), $\tilde\pi_J[V]$ is open and

\Centerline{$\mu^*(W\cap Z)=\mu W\ge\mu V=\nu_J\tilde\pi_J[V]>0$,}

\noindent because $Z_1$ is the support of $\nu_J$.
\Qed

\medskip

{\bf (d)} If $I$ is countable, $\mu$ is a topological measure so
measures $Z$, and $Z$ is the support of $\mu$.
}%end of proof of 456H

\leader{456I}{Remarks (a)} In the context of 456H, I will call $Z$ the
{\bf support} of the centered Gaussian distribution $\mu$, even though
$\mu$ need not be a topological measure\cmmnt{,
so the definition 411Nb is not immediately applicable}.
\cmmnt{In 456P we shall see that $Z$ really is the support of a canonical extension of $\mu$.}

\spheader 456Ib\cmmnt{ It is worth making one elementary point at once.}   If
$I$ and $J$ are sets, $\mu$ and $\nu$ are centered Gaussian
distributions on $\BbbR^I$ and $\BbbR^J$ respectively with supports $Z$
and $Z'$, and $T:\BbbR^I\to\BbbR^J$ is an \imp\ continuous linear
operator, then $Tz\in Z'$ for every $z\in Z$.   \prooflet{\Prf\ If
$g:\BbbR^J\to\Bbb R$ is a continuous linear functional which is zero
$\nu$-a.e., then $gT:\BbbR^I\to\Bbb R$ is a continuous linear functional
which is zero $\mu$-a.e., so $g(Tz)=(gT)(z)=0$.\ \Qed}

\leader{456J}{Universal Gaussian distributions:  Definition} A centered
Gaussian distribution on $\BbbR^I$ is {\bf universal} if its covariance
matrix $\langle\sigma_{ij}\rangle_{i,j\in I}$ is the inner product for a
Hilbert space structure on $I$.   \cmmnt{(See 456Xe.)}

\leader{456K}{Proposition} Let $I$ be any set, and $\mu$ a centered
Gaussian distribution on $I$.   Then there are a set $J$, a universal
centered Gaussian distribution $\nu$ on $\BbbR^J$, and a continuous
\imp\ linear operator $T:\BbbR^J\to\BbbR^I$.

\proof{{\bf (a)} Set $J=L^2(\mu)$.   Then for any finite
$K\subseteq J$ there is a centered Gaussian distribution $\mu_K$ on
$\BbbR^K$ such that $\int x(u)x(v)\mu(dx)=\innerprod{u}{v}$ for all $u$,
$v\in K$.   \Prf\ If $K=\emptyset$ or $K=\{0\}$ this is trivial, as we
take $\mu$ to be the trivial distribution concentrated at $0$.
Otherwise, let $\ofamily{i}{n}{w_i}$ be an orthonormal basis for the
linear subspace of $J$ generated by $K$.   For each $u\in K$, express
it as $\sum_{i=0}^{n-1}\alpha_{ui}w_i$.   Define $T:\BbbR^n\to\BbbR^K$
by setting $(Tz)(u)=\sum_{i=0}^{n-1}\alpha_{ui}z(i)$ for $z\in\BbbR^n$
and $u\in K$.   Set $\mu_K=\mu_G^{(n)}T^{-1}$.   Then $\mu_K$ is a
centered Gaussian distribution, by 456Ba, and its covariance matrix is
given by

$$\eqalign{\int x(u)x(v)\mu_K(dx)
&=\int(Ty)(u)(Ty)(v)\mu_G^{(n)}(dy)
=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\alpha_{ui}\alpha_{vj}
  \int y(i)y(j)\mu_G^{(n)}(dy)\cr
&=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\alpha_{ui}\alpha_{vj}
  \innerprod{w_i}{w_j}
=\innerprod{u}{v}\cr}$$

\noindent for all $u$, $v\in K$.   So $\mu_K$ is the distribution we
seek.\ \Qed

\medskip

{\bf (b)} If $K\subseteq J$ is finite and $L\subseteq K$, then
$\mu_L=\mu_K\pi_{KL}^{-1}$, where $\pi_{KL}(x)=x\restr L$ for
$x\in\BbbR^K$.   \Prf\ Since $\mu_K\pi_{KL}^{-1}$ is a centered Gaussian
distribution, all we have to do is to check its covariance matrix.   But
if $u$, $v\in L$ then

$$\eqalign{\int y(u)y(v)(\mu_K\pi_{KL}^{-1})(dy)
&=\int(\pi_{KL}x)(u)(\pi_{KL}x)(v)\mu_K(dx)\cr
&=\int x(u)x(v)\mu_K(dx)
=\innerprod{u}{v}
=\int y(u)y(v)\mu_L(dy).\cr}$$

\noindent By 456Bb, $\mu_L=\mu_K\pi_{KL}^{-1}$.\ \Qed

\medskip

{\bf (c)} By 454G, there is a Baire measure $\nuprime$ on $\BbbR^J$
such that $\nuprime\pi_{JK}^{-1}[E]=\mu_KE$ for every finite
$K\subseteq J$ and every Borel set $E\subseteq\BbbR^K$.
Take $\nu$ to be the
completion of $\nuprime$.   Then $\pi_{JK}$ is \imp\ for $\nu$ and
$\mu_K$, for every finite $K\subseteq J$.   If
$f:\BbbR^{J}\to\Bbb R$ is a continuous linear functional, there are a
finite $K\subseteq J$ and a linear functional $g:\BbbR^K\to\Bbb R$
such that $f=\pi_{JK}g$, so that

\Centerline{$\nu\{x:f(x)\le\alpha\}=\mu_K\{x:g(x)\le\alpha\}$}

\noindent for every $\alpha$, and $f$ and $g$ have the same
distribution;  as $g$ is either normal with zero expectation or zero
a.e., so
is $f$.   As $f$ is arbitrary, $\nu$ is a centered Gaussian
distribution.

\medskip

{\bf (d)} $\nu$ is universal.   \Prf\ If $u$, $v\in J$ set
$K=\{u,v\}$.   Then

$$\eqalign{\int x(u)x(v)\nu(dx)
&=\int(\pi_{JK}x)(u)(\pi_{JK}x)(v)\nu(dx)\cr
&=\int y(u)y(v)\mu_K(dy)
=\innerprod{u}{v}.\cr}$$

\noindent Thus the covariance matrix of $\nu$ is just the inner product
of the standard Hilbert space structure of $J$.\ \Qed

\medskip

{\bf (e)} For $i\in I$, let $u_i\in J$ be the equivalence class of the
square-integrable function $x\mapsto x(i):\BbbR^I\to\Bbb R$.   Define
$T:\BbbR^J\to\BbbR^I$ by setting $(Ty)(i)=y(u_i)$ for every
$i\in I$ and $y\in\BbbR^J$.   Then there is a centered Gaussian
distribution $\mu'$ on $\BbbR^I$ such that $T$ is \imp\ for $\nu$ and
$\mu'$.   Now the covariance matrix of $\mu'$ is defined by

$$\eqalign{\int x(i)x(j)\mu'(dx)
&=\int(Ty)(i)(Ty)(j)\nu(dy)
=\int y(u_i)y(u_j)\nu(dy)\cr
&=\innerprod{u_i}{u_j}
=\int x(i)x(j)\mu(dx)\cr}$$

\noindent for all $i$, $j\in I$.   So $\mu$ and $\mu'$ are equal and $T$
is \imp\ for $\nu$ and $\mu$.
}%end of proof of 456K

\leader{456L}{Lemma} Let $\mu$ be a universal centered Gaussian
distribution on $\BbbR^I$;  give $I$ a corresponding Hilbert space
structure such that $\int x(i)x(j)\mu(dx)=\innerprod{i}{j}$ for all $i$,
$j\in I$.   Let $F\in\dom\mu$ be a set determined by coordinates
in $J$, where $J\subseteq I$ is a closed linear subspace for the Hilbert
space structure of $I$.   Let $W$ be the union of all the open subsets
of $\BbbR^I$ which meet $F$ in a negligible set, and $W'$ the union of
the open subsets of $\BbbR^I$ which meet $F$ in a negligible set and are
determined by coordinates in $J$.   If $F\subseteq W$ then
$F\subseteq W'$.

\proof{{\bf (a)} Let $Z$ be the support of $\mu$ in the sense of 456H.
We need to know that $Z$ is just the set of all linear functionals from
$I$ to $\Bbb R$.   \Prf\ If $K\subseteq I$ is finite and
$\family{i}{K}{\alpha_i}\in\BbbR^K$ and
$f(x)=\sum_{i\in I}\alpha_ix(i)$ for $x\in\BbbR^I$, then

\Centerline{$\|f\|_2^2
=\sum_{i,j\in K}\alpha_i\alpha_j\innerprod{i}{j}
=\|\sum_{i\in K}\alpha_ii\|^2$.}

\noindent So, for $x\in\BbbR^I$,

$$\eqalign{x\in Z
&\iff f(x)=0\text{ whenever }f\in(\BbbR^I)^*\text{ and }\|f\|_2=0\cr
&\iff \sum_{i\in K}\alpha_ix(i)=0\text{ whenever }K\subseteq I
  \text{ is finite and }\sum_{i\in K}\alpha_ii=0\text{ in }I\cr
&\iff x:I\to\Bbb R\text{ is linear.  \Qed}\cr}$$

\medskip

{\bf (b)} Let $J^{\perp}$ be the orthogonal complement of $J$ in $I$, so
that $I=J\oplus J^{\perp}$ (4A4Jf).   Give $\BbbR^J$ and
$\BbbR^{J^{\perp}}$ the centered Gaussian
distributions $\mu_J$, $\mu_{J^{\perp}}$ induced by $\mu$ and the
projections
$x\mapsto x\restr J$, $x\mapsto x\restr J^{\perp}$.   Then the product
measure $\lambda$ on $\BbbR^J\times\BbbR^{J^{\perp}}$ is also a centered
Gaussian distribution (456Be).   Define
$T:\BbbR^J\times\BbbR^{J^{\perp}}\to\BbbR^I$ by setting
$T(u,v)(j+k)=u(j)+v(k)$ whenever $j\in J$, $k\in J^{\perp}$,
$u\in\BbbR^J$ and $v\in\BbbR^{J^{\perp}}$.   Then $T$ is \imp\ for
$\lambda$ and $\mu$.   \Prf\ $T$ is a continuous linear operator so we
have a centered Gaussian distribution $\mu'$ on $\BbbR^I$ such that $T$
is \imp\ for $\lambda$ and $\mu'$.   If $j$, $j'\in J$ and $k$,
$k'\in J^{\perp}$,

$$\eqalign{\int x(j+k)x(j'+k')\mu'(dx)
&=\int T(u,v)(j+k)T(u,v)(j'+k')\lambda(d(u,v))\cr
&=\int (u(j)+v(k))(u(j')+v(k'))\lambda(d(u,v))\cr
&=\int u(j)u(j')\mu_J(du)+\int v(k)v(k')\mu_{J^{\perp}}(dv)\cr
&=\int x(j)x(j')\mu(dx)+\int x(k)x(k')\mu(dx)\cr
&=\innerprod{j}{j'}+\innerprod{k}{k'}
=\innerprod{j+k}{j'+k'}\cr
&=\int x(j+k)x(j'+k')\mu(dx).\cr}$$

\noindent Thus $\mu$ and $\mu'$ have the same covariance matrix and are
equal, and $T$ is \imp\ for $\lambda$ and $\mu$.\ \Qed

\medskip

{\bf (c)} Take any $z\in W\cap Z$.   Then there is an open set $V$,
determined by coordinates in a finite set $K_0\subseteq I$, such that
$z\in V$ and $\mu(V\cap F)=0$.   Let $\epsilon>0$ be such that $y\in V$
whenever $x\in\BbbR^I$ and $|x(i)-z(i)|<2\epsilon$ for every $i\in K_0$.
Express each $k\in K_0$ as $k'+k''$
where $k'\in J$ and $k''\in J^{\perp}$.   Set

\Centerline{$V'
=\{x:x\in\BbbR^I,\,|x(k')-z(k')|<\epsilon$ for every $k\in K_0\}$.}

\noindent Then $V'$ is an open set, determined by coordinates in $J$,
and contains $z$.   Also $\mu(V'\cap F)=0$.   \Prf\ Set $V''
=\{x:x\in\BbbR^I$, $|x(k'')-z(k'')|<\epsilon$ for every $k\in K_0\}$.
Then $V\supseteq V'\cap V''\cap Z$.   Since $Z$ has full outer measure
(456H),

\Centerline{$\mu(V'\cap V''\cap F)
=\mu^*(V'\cap V''\cap F\cap Z)\le\mu(V\cap F)=0$.}

Now

$$\eqalignno{0
&=\mu(V'\cap V''\cap F)
=\lambda T^{-1}[V'\cap F\cap V'']\cr
&=\lambda\{(x\restr J,x\restr J^{\perp}):x\in V'\cap F\cap V''\}\cr
\displaycause{because $V'\cap F\cap V''$ is determined by coordinates in
$J\cup J^{\perp}$}
&=\mu_J\{x\restr J:x\in V'\cap F\}
  \cdot\mu_{J^{\perp}}\{x\restr J^{\perp}:x\in V''\}\cr}$$

\noindent because $V'\cap F$ is determined by coordinates in $J$, while
$V''$ is determined by coordinates in $J^{\perp}$.   However,
$z\in V''$, and $z\restr J^{\perp}$ belongs to the support $Z'$ of
$\mu_{J^{\perp}}$, by 456Ib;   since $Z'$ is self-supporting, and
$\{x\restr J^{\perp}:x\in V''\}$ is open,
$\mu_{J^{\perp}}\{x\restr J^{\perp}:x\in V''\}>0$.   We conclude that

\Centerline{$0=\mu_J\{x\restr J:x\in V'\cap F\}=\mu(V'\cap F)$.  \Qed}

\medskip

{\bf (d)} This shows that $z\in W'$.   As $z$ is arbitrary,
$W\cap Z\subseteq W'$.

\Quer\ Suppose, if possible, that there is a point
$z_0\in F\setminus W'$.   If $i$, $j\in J$ and $\alpha\in\Bbb R$, then

\Centerline{$\{x:x(i+j)\ne x(i)+x(j)\}$,
\quad$\{x:x(\alpha i)\ne\alpha x(i)\}$}

\noindent are negligible open sets determined by coordinates in $J$, so
are included in $W'$ and do not contain $z_0$.   Thus
$z_0\restr J:J\to\Bbb R$ is linear.   Let $z:I\to\Bbb R$ be a linear
functional extending $z_0\restr J$.   Then $z\in Z$ and
$z\restr J=z_0\restr J$;  as both $F$ and $W'$ are determined by
coordinates in $J$, $z\in F\setminus W'$.   But this means that
$z\in Z\cap W\setminus W'$, which is impossible.\ \Bang

So $F\subseteq W'$, as claimed.
}%end of proof of 456L

\leader{456M}{Cluster sets:  Lemma} Let $I$ be a countable set, $n\ge 1$
an integer and
$\mu$ a centered Gaussian distribution on $\BbbR^{I\times n}$.
For $\epsilon>0$ set

\Centerline{$I_{\epsilon}
=\{i:i\in I$, $\int|x(i,r)|^2\mu(dx)\le\epsilon^2$ for every $r<n\}$;}

\noindent suppose that no $I_{\epsilon}$ is empty.

(a) There is a closed set $F\subseteq\BbbR^n$ such that

\Centerline{$F=\bigcap_{\epsilon>0}
\overline{\{\ofamily{r}{n}{x(i,r)}:i\in I_{\epsilon}\}}$}

\noindent for almost every $x\in\BbbR^{I\times n}$.

(b) If $z\in F$ and $-1\le\alpha\le 1$, then $\alpha z\in F$.

(c) If $F$ is bounded, then there is some $\epsilon>0$ such that
$\sup_{i\in I_{\epsilon},r<n}|x(i,r)|<\infty$ for almost every
$x\in\BbbR^{I\times n}$.

\proof{{\bf (a)(i)} For $x\in\BbbR^{I\times n}$ and $i\in I$ set
$S_i(x)=\ofamily{r}{n}{x(i,r)}\in\BbbR^n$.   For $x\in\BbbR^{I\times n}$
set

\Centerline{$F_x
=\bigcap_{\epsilon>0}\overline{\{S_ix:i\in I_{\epsilon}\}}$,}

\noindent so that $F_x$ is a closed subset of $\BbbR^n$.   For
$A\subseteq\BbbR^n$ set
$E_A=\{x:x\in\BbbR^{I\times n}$, $A\cap F_x\ne\emptyset\}$.

\medskip

\quad{\bf (ii)} By 456Bd, there is a continuous linear operator
$T:\BbbR^{\Bbb N}\to\BbbR^{I\times n}$ such that
$\mu=\mu_G^{(\Bbb N)}T^{-1}$.   Set $T_i=S_iT$ for $i\in I$;  then
$T_i:\BbbR^{\Bbb N}\to\BbbR^n$ is a continuous linear operator.   For
$y\in\BbbR^{\Bbb N}$, set

\Centerline{$\tilde F_y=F_{T(y)}
=\bigcap_{\epsilon>0}\overline{\{T_i(y):i\in I_{\epsilon}\}}$.}

\noindent For $A\subseteq\BbbR^n$ set

\Centerline{$\tilde E_A=T^{-1}[E_A]
=\{y:y\in\BbbR^{\Bbb N}$, $A\cap\tilde F_y\ne\emptyset\}$.}

\medskip

\quad{\bf (iii)} If $K\subseteq\BbbR^n$ is compact, then
$\tilde E_K$ is a Borel subset of $\BbbR^{\Bbb N}$.   \Prf\
Let $\Cal V$ be a countable base for the topology of $\BbbR^n$, and for
$k\ge 1$ let $\Cal V_k$ be the set of members of $\Cal V$ with diameter
at most $1/k$ which meet $K$.   Then $T_i^{-1}[V]$ is a Borel set for
every $V\in\Cal V$ and $i\in I$, so

\Centerline{$E'
=\bigcap_{k\ge 1}\bigcup_{V\in\Cal V_k}\bigcup_{i\in I_{1/k}}
  T_i^{-1}[V]$}

\noindent is a Borel set.

If $y\in\tilde E_K$, take $k\ge 1$.   There is a
$z\in\tilde F_y\cap K$.   Let $V\in\Cal V$ be such that $z\in V$ and
$\diam V\le 1/k$;  in this case $V\in\Cal V_k$.   Because
$z\in\overline{\{T_i(y):i\in I_{1/k}\}}$, there is an
$i\in I_{1/k}$ such that $T_i(y)\in V$.   As $k$ is arbitrary, this
shows that $y\in E'$;  thus $\tilde E_K\subseteq E'$.

If $y\notin\tilde E_K$, then $K$ is a compact set disjoint from the
closed set $\tilde F_y$.   There is therefore some $\epsilon>0$ such
that $K\cap\overline{\{T_i(y):i\in I_{\epsilon}\}}=\emptyset$ (since
these form a downwards-directed family of compact sets with empty
intersection).   Next, there is a $\delta>0$ such that
$B(z,\delta)\cap\overline{\{T_i(y):i\in I_{\epsilon}\}}=\emptyset$ for
every $z\in K$ (2A2Ed).   Let $k\ge 1$ be such that
$1/k\le\min(\epsilon,\delta)$.   If $V\in\Cal V_k$ and
$i\in I_{1/k}$, there is some $z\in K\cap V$ so $V\subseteq B(z,\delta)$
and $T_i(y)\notin V$.  This shows that $y\notin E'$.   As $y$ is
arbitrary, $E'\subseteq\tilde E_K$.

So $\tilde E_K=E'$ is a Borel set.\ \Qed

It follows at once that $\tilde E_H$ is a Borel set for every
K$_{\sigma}$ set $H$, in particular, for any open or closed set $H$.

\medskip

\quad{\bf (iv)} We need a simple estimate on the coefficients of the
linear operators $T_i$.   Let $\alpha_{irj}$ be such that
$T_i(y)=\ofamily{r}{n}{\sum_{j=0}^{\infty}\alpha_{irj}y(j)}$ for
$i\in I$, $r<n$ and $y\in\BbbR^{\Bbb N}$.   (Of course
$\{j:\alpha_{irj}\ne 0\}$ is finite for each $i$ and $r$.)   Then

\Centerline{$\biggerint x(i,r)^2\mu(dx)
=\int T_i(y)(r)^2\mu_G^{(\Bbb N)}(dy)
=\sum_{j=0}^{\infty}\alpha_{irj}^2$,}

\noindent so $|\alpha_{irj}|\le\epsilon$ whenever $i\in I_{\epsilon}$,
$r<n$ and $j\in\Bbb N$.

\medskip

\quad{\bf (v)} Now suppose that $H\subseteq\BbbR^n$ is open, that
$K\subseteq H$ is compact, and that $\mu_G^{(\Bbb N)}\tilde E_K>0$.
Then $\mu_G^{(\Bbb N)}\tilde E_H=1$.   \Prf\ Let $\epsilon>0$.   Let
$\sequence{j}{\epsilon_j}$ be a sequence of strictly positive real
numbers such that $\sum_{j=0}^{\infty}\epsilon_j
\le\bover12\min(\epsilon,\mu_G^{(\Bbb N)}\tilde E_K)$, and for each
$j\in\Bbb N$ let $\gamma_j\ge 0$ be such that
$\mu_G[-\gamma_j,\gamma_j]\ge 1-\epsilon_j$.   Then

\Centerline{$\tilde E=\{y:y\in\tilde E_K$, $|y(j)|\le\gamma_j$ for every
$j\in\Bbb N\}$}

\noindent has measure at least
$\mu_G^{(\Bbb N)}\tilde E_K-\sum_{j=0}^{\infty}\epsilon_j>0$.   By
254Sb, there are an $m\in\Bbb N$ and a set $\tilde E'$, of measure at
least $1-\bover12\epsilon$, such that for every $y'\in\tilde E'$ there
is a $y\in\tilde E$ such that $y(j)=y'(j)$ whenever
$j>m$.   Set $\tilde E''
=\{y:y\in\tilde E'$, $|y(j)|\le\gamma_j$ for every $j\in\Bbb N\}$, so
that $\mu_G^{(\Bbb N)}\tilde E''\ge 1-\epsilon$.

Let $\delta>0$ be such that $z'\in H$ whenever $z\in K$ and
$\|z-z'\|\le 2\delta$.   Let $\eta>0$ be such that
$2\eta\sqrt{n}\sum_{j=0}^m\gamma_j\le\delta$.   If $y'\in\tilde E''$ and
$i\in I_{\eta}$, there is a $y\in\tilde E$ such that
$y(j)=y'(j)$ for $j>m$.   Also
$|y(j)-y'(j)|\le 2\gamma_j$ for $j\le m$, so

\Centerline{$|T_i(y)(r)-T_i(y')(r)|
\le\sum_{j=0}^{\infty}|\alpha_{irj}||y(j)-y'(j)|
\le\sum_{j=0}^m2\eta\gamma_j
\le\Bover{\delta}{\sqrt n}$}

\noindent for every $r<n$, and $\|T_i(y)-T_i(y')\|\le\delta$.

Now $\tilde F_y\cap K\ne\emptyset$;  take $z\in\tilde F_y\cap K$.   For
every $\zeta>0$, there is an $i\in I_{\min(\eta,\zeta)}$ such that
$\|z-T_i(y)\|\le\delta$, so that $\|z-T_i(y')\|\le 2\delta$.
This means that $B(z,2\delta)\cap\{T_i(y'):i\in I_{\zeta}\}$ is not empty.
As $B(z,2\delta)$ is compact, it must meet $\tilde F_{y'}$.   But this
means that $H\cap\tilde F_{y'}\ne\emptyset$, by the choice of $\delta$.
As $y'$ is arbitrary, $\tilde E''\subseteq\tilde E_H$, while
$\mu_G^{(\Bbb N)}\tilde E''\ge 1-\epsilon$.

This works for every $\epsilon>0$.   So $\tilde E_H$ is conegligible, as
claimed.\ \Qed

\medskip

\quad{\bf (vi)} If $H\subseteq\BbbR^n$ is open and
$\mu_G^{(\Bbb N)}\tilde E_H>0$, then (because $H$ is $\sigma$-compact)
there is a compact set $K\subseteq H$ such that
$\mu_G^{(\Bbb N)}\tilde E_K>0$, and (e) tells us that
$\mu_G^{(\Bbb N)}\tilde E_H=1$.

Set

\Centerline{$\Cal V_0
=\{V:V\in\Cal V$, $\mu_G^{(\Bbb N)}\tilde E_V=0\}
=\{V:V\in\Cal V$, $\mu_G^{(\Bbb N)}\tilde E_V<1\}$.}

\noindent Then we see that

\Centerline{$\Cal V_0=\{V:V\in\Cal V$, $\tilde F_y\cap V=\emptyset\}$}

\noindent for almost every $y\in\BbbR^{\Bbb N}$, that is,

\Centerline{$\Cal V_0=\{V:V\in\Cal V$, $F_x\cap V=\emptyset\}$}

\noindent for almost every $x\in\BbbR^{I\times n}$.   But as every $F_x$
is closed, we have
$F_x=\BbbR^n\setminus\bigcup\Cal V_0$ for almost every $x$.   So we can
set $F=\BbbR^n\setminus\bigcup\Cal V_0$.

\medskip

{\bf (b)(i)} Give $\BbbR^{I\times 2n}\cong(\BbbR^{I\times n})^2$ the
measure $\lambda$ corresponding to the product measure $\mu\times\mu$;
by 456Be, this is a centered Gaussian distribution.   For
$(x_1,x_2)\in(\BbbR^{I\times n})^2$, set

\Centerline{$F'_{x_1x_2}=\bigcap_{\epsilon>0}
  \overline{\{(S_ix_1,S_ix_2):i\in I_{\epsilon}\}}$.}

\noindent By (a), we have a closed set $F'\subseteq\BbbR^{2n}$ such that
$F'=F'_{x_1x_2}$ for almost all $x_1$, $x_2$.   (Of course
$I_{\epsilon}=\{i:\int|x_j(i,r)|^2\lambda(dx)\le\epsilon^2$ for every
$j\in\{1,2\}$, $r<n\}$ whenever $\epsilon>0$.)

\medskip

\quad{\bf (ii)} Now $(z,0)\in F'$.   \Prf\ Take
$x_1\in\BbbR^{I\times n}$ such that $F=F_{x_1}$ and
$E=\{x_2:F'=F'_{x_1x_2}\}$ is conegligible.   (Almost every point of
$\BbbR^{I\times n}$ has these properties.)   For $k\in\Bbb N$,
$z\in\overline{\{S_ix_1:i\in I_{2^{-k}}\}}$;  let
$i_k\in I_{2^{-k}}$ be such that
$\sum_{r<n}|z(r)-x_1(i_k,r)|\le 2^{-k}$.   Next,

\Centerline{$\sum_{k=0}^{\infty}\sum_{r=0}^{n-1}
  $$\biggerint|x(i_k,r)|^2\mu(dx)
\le$$\sum_{k=0}^{\infty}2^{-2k}n<\infty$,}

\noindent so $\sum_{k=0}^{\infty}\sum_{r=0}^{n-1}|x(i_k,r)|^2$ is finite
for almost every $x\in\BbbR^I$, and there must be an
$x_2\in E$ such that
$\sum_{k=0}^{\infty}\sum_{r=0}^{n-1}|x_2(i_k,r)|^2$ is finite.   But in
this case $\lim_{k\to\infty}x_2(i_k,r)=0$ for every $r$, while
$\lim_{k\to\infty}x_1(i_k,r)=z(r)$.   Accordingly
$(z,0)\in F'_{x_1x_2}=F'$.\ \Qed

\medskip

\quad{\bf (iii)} Set $\beta=\sqrt{1-\alpha^2}$ and define
$\tilde T:(\BbbR^{I\times n})^2\to\BbbR^{I\times n}$ by setting
$\tilde T(x_1,x_2)=\alpha x_1+\beta x_2$ for $x_1$,
$x_2\in\BbbR^{I\times n}$.    Then $\tilde T$ is a continuous linear
operator, so the image measure $\lambda\tilde T^{-1}$ is a centered
Gaussian distribution on $\BbbR^{I\times n}$ (456Ba).   Moreover, it has
the same covariance matrix as $\mu$.   \Prf\ If $i$, $j\in I$ then

$$\eqalign{\int x(i)x(j)(\lambda\tilde T^{-1})(dx)
&=\int\tilde T(x_1,x_2)(i)\tilde T(x_1,x_2)(j)\lambda(d(x_1,x_2))\cr
&=\int(\alpha x_1(i)+\beta x_2(i))
  (\alpha x_1(j)+\beta x_2(j))\lambda(d(x_1,x_2))\cr
&=\alpha^2\int x_1(i)x_1(j)\lambda(d(x_1,x_2))
  +\beta^2\int x_2(i)x_2(j)\lambda(d(x_1,x_2))\cr
&=(\alpha^2+\beta^2)\int x(i)x(j)\mu(dx)
=\int x(i)x(j)\mu(dx).  \text{ \Qed}\cr}$$

\noindent So $\lambda\tilde T^{-1}=\mu$ (456Bb).

\medskip

\quad{\bf (iv)} If $x_1$, $x_2\in\BbbR^I$ are such that
$(z,0)\in F'_{x_1x_2}$, then $\alpha z\in F_{\tilde T(x_1,x_2)}$.
\Prf\ For every $\epsilon>0$ there is an $i\in I_{\epsilon}$ such that
$|z(r)-x_1(i,r)|\le\epsilon$ and $|x_2(i,r)|\le\epsilon$ for every
$r<n$.   But now $|\alpha z(r)-\tilde T(x_1,x_2)(r)|\le 2\epsilon$ for
every $r<n$.\ \QeD\   So

\Centerline{$\tilde T^{-1}[\{x:\alpha z\in F_x\}]
=\{(x_1,x_2):\alpha z\in F_{\tilde T(x_1,x_2)}\}
\supseteq\{(x_1,x_2):(z,0)\in F'_{x_1x_2}\}$}

\noindent is $\lambda$-conegligible, and $\alpha z\in F_x$ for
$\mu$-almost every $x$, that is, $\alpha z\in F$, as claimed.

\medskip

{\bf (c)} Suppose now that $F$ is bounded.

\medskip

\quad{\bf (i)} For $L\subseteq I$, $\alpha\ge 0$ set

\Centerline{$Q(L,\alpha)
=\bigcup_{i\in L,r<n}\{x:|x(i,r)|\ge\alpha\}$.}

\noindent By 456G, applied to the image of $\mu$ under the map
$x\mapsto x\restr L\times n:\BbbR^{I\times n}\to\BbbR^{L\times n}$,

\Centerline{$\mu Q(L,\bover12\alpha)
\ge 2\mu Q(L,\alpha)(1-\mu Q(L,\alpha))^3$}

\noindent for every finite $L\subseteq I$ and every $\alpha\ge 0$.

Let $\beta>0$ be such that $\delta=2\beta(1-(n+1)\beta)^3-\beta>0$, and let $\alpha_0>0$ be such that $\alpha_0^2\beta\ge 1$ and $\|z\|<\bover12\alpha_0$ for every $z\in F$, so that
$\mu\{x:|x(i,r)|\ge\alpha_0\}\le\beta$ whenever $i\in I_1$ and $r<n$,
and $\mu Q(\{i\},\alpha_0)\le n\beta$ for every $i\in I_1$.   Set
$K=\{z:z\in\BbbR^n$,
$\bover12\alpha_0\le\max_{r<n}|z(r)|\le\alpha_0\}$, so that $K$ is a
compact set disjoint from $F$.   For almost every $x$,

\Centerline{$\emptyset=K\cap F=K\cap F_x
=\bigcap_{k\ge 1}K\cap\overline{\{S_ix:i\in I_{1/k}\}}$,}

\noindent so there is a $k\ge 1$ such that
$S_ix\notin K$ for every $i\in I_{1/k}$.   Since the sets

\Centerline{$\{x:S_i(x)\in K$ for some $i\in I_{1/k}\}$}

\noindent form a non-increasing sequence of measurable sets with
negligible intersection, there is a $k\ge 1$ such that

\Centerline{$\mu\{x:S_i(x)\in K$ for some $i\in I_{1/k}\}
<\delta$.}

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that

\Centerline{$\mu Q(I_{1/k},\alpha_0)>\beta$.}

\noindent Let $L\subseteq I_{1/k}$ be a finite set of minimal size such
that $\gamma=\mu Q(L,\alpha_0)\ge\beta$.   Since
$\mu Q(\{i\},\alpha_0)\le n\beta$ for any $i\in L$,
and $L$ is minimal, we must have

\Centerline{$\beta\le\gamma\le(n+1)\beta$.}

\noindent Now this means that

$$\eqalignno{\mu Q(L,\Bover12\alpha_0)
&\ge 2\gamma(1-\gamma)^3\cr
\displaycause{see (i) above}
&\ge 2\gamma(1-(n+1)\beta)^3
=\gamma(1+\Bover{\delta}{\beta})
\ge\gamma+\delta,\cr}$$

\noindent so that
$\mu(Q(L,\bover12\alpha_0)\setminus Q(L,\alpha_0))\ge\delta$.
But if $x\in Q(L,\bover12\alpha_0)\setminus Q(L,\alpha_0)$ there is some
$i\in L$ such that $\max_{r<n}|x(i,r)|\ge\bover12\alpha_0$ while
$\max_{r<n}|x(i,r)|<\alpha_0$, in which case $S_i(x)\in K$.   So we get

$$\eqalign{\delta
&\le\mu(Q(L,\Bover12\alpha_0)\setminus Q(L,\alpha_0))
\le\mu\{x:S_i(x)\in K\text{ for some }i\in L\}\cr
&\le\mu\{x:S_i(x)\in K\text{ for some }i\in I_{1/k}\}
<\delta\cr}$$

\noindent which is absurd.\ \Bang

\medskip

\quad{\bf (iii)} Thus $\mu Q(I_{1/k},\alpha_0)\le\beta$.   For
$\alpha\ge 0$, set $f(\alpha)=\mu Q(I_{1/k},\alpha)$;  then $f$ is
non-increasing.   Also $f(\bover12\alpha)\ge 2f(\alpha)(1-f(\alpha))^3$
for every $\alpha$.   \Prf\Quer\ Otherwise, because

\Centerline{$f(\alpha)
=\sup\{\mu Q(L,\alpha):L\subseteq I_{1/k}$ is finite$\}$,}

\noindent there is a finite $L\subseteq I_{1/k}$ such that
$f(\bover12\alpha)<2\gamma(1-\gamma)^3$, where
$\gamma=\mu Q(L,\alpha)$.   But in this case
$\mu Q(L,\bover12\alpha)\le f(\bover12\alpha)<2\gamma(1-\gamma)^3$,
which is impossible, as remarked in (i).\ \Bang\Qed

Set $\zeta=\lim_{\alpha\to\infty}f(\alpha)$.   Then

\Centerline{$\zeta=\lim_{\alpha\to\infty}f(\bover12\alpha)
\ge 2\zeta(1-\zeta)^3$.}

\noindent But we also know, from (ii), that
$\zeta\le f(\alpha_0)\le\beta$.   So
$(1-\zeta)^3\ge(1-\beta)^3>\bover12$ and $\zeta$ must be $0$.

What this means is that if we set $\epsilon=\bover1k$ then

\Centerline{$\lim_{\alpha\to\infty}
  \mu\{x:\sup_{i\in I_{\epsilon},r<n}|x(i,r)|>\alpha\}
=0$,}

\noindent that is, $\sup_{i\in I_{\epsilon},r<n}|x(i,r)|$ is finite for
almost every $x\in\BbbR^{I\times n}$, as claimed.
}%end of proof of 456M

\leader{456N}{Lemma} Let $J$ be a set and $\mu$ a centered Gaussian
distribution on $\BbbR^J$.   Let $M$ be the linear subspace of
$L^2(\mu)$ generated by $\{\pi_j^{\ssbullet}:j\in J\}$, where
$\pi_j(x)=x(j)$ for $x\in\BbbR^J$ and $j\in J$.   If $M$ is
separable\cmmnt{ (for the norm topology)} then $\mu$ is
$\tau$-additive.

\proof{ Suppose, if possible, otherwise.

\medskip

{\bf (a)} There is an upwards-directed family $\Cal G$ of open Baire
sets in $\BbbR^J$ such that $W_0=\bigcup\Cal G$ is a
Baire set and
$\mu W_0>\sup_{G\in\Cal G}\mu G$.   Let $\Cal G_0\subseteq\Cal G$ be a
countable upwards-directed set such that
$\sup_{G\in\Cal G_0}\mu G=\sup_{G\in\Cal G}\mu G$, and set
$W_1=W_0\setminus\bigcup\Cal G_0$;   then $\mu W_1>0$ and
$\mu(W_1\cap G)=0$ for every $G\in\Cal G$.   Let $W$ be a non-negligible
zero set included in $W_1$.

For each $n\in\Bbb N$, let $\Cal V_n$ be a countable base for the
topology of $\BbbR^n$ consisting of open balls.   Let $\Cal G_n^*$ be
the family of open sets of $\BbbR^J$ of the form $T^{-1}[V]$, where
$T:\BbbR^J\to\BbbR^n$ is a continuous linear operator,
$V\subseteq\BbbR^n$ is open and
$\mu(W\cap T^{-1}[V])=0$.   Of course $\Cal G_0^*=\emptyset$.

\medskip

{\bf (b)} For $n\ge 1$ and $V\in\Cal V_n$, let $\Cal T_{nV}$ be the
family of continuous linear operators $T:\BbbR^J\to\BbbR^n$ such that
$W\cap T^{-1}[V]$ is negligible, but not included in
$\bigcup\Cal G_{n-1}^*$.   Index $\Cal T_{nV}$ as
$\family{i}{I(n,V)}{T_i}$;  it will be convenient to do this in such a
way that all the $I(n,V)$ are disjoint.   Define $f_{ir}$, for
$i\in I(n,V)$ and $r<n$, by saying that
$T_i(x)=\ofamily{r}{n}{f_{ir}(x)}$ for $x\in\BbbR^J$.
Define $\phi_n:\bigcup_{V\in\Cal V_n}I(n,V)\to M^n$ by setting
$\phi_n(i)=\ofamily{r}{n}{f_{ir}^{\ssbullet}}$ for each
$i\in\bigcup_{V\in\Cal V_n}I(n,V)$.   Because $M$ is separable (in its
norm topology), $M^n$ is separable in its product topology (4A2P(a-v)).
Fix a countable set $I'(n,V)\subseteq I(n,V)$ such that
$\{\phi_n(i):i\in I'(n,V)\}$ is dense in $\{\phi_n(i):i\in I(n,V)\}$.
Set $\rho_n(\ofamily{r}{n}{u_r},\ofamily{r}{n}{v_r})
=\max_{r<n}\|u_r-v_r\|_2$ for  $\ofamily{r}{n}{u_r}$,
$\ofamily{r}{n}{v_r}\in M^n$, so that $\rho_n$ is a metric defining the
product topology of $M^n$.

\medskip

{\bf (c)} If $j\in I(n,V)$, then there is a $\delta>0$ such that

\Centerline{$\{T_i(x):i\in I'(n,V)$,
$\rho_n(\phi_n(i),\phi_n(j))\le\delta\}$}

\noindent is bounded for almost every $x\in\BbbR^J$.   \Prf\
Define $S:\BbbR^J\to\BbbR^{I'(n,V)\times n}$ by setting
$(Sx)(i,r)=T_i(x,r)-T_j(x,r)$ for $x\in\BbbR^J$, $i\in I'(n,V)$ and $r<n$.
By 456Ba, the image measure $\lambda=\nu S^{-1}$ is a centered Gaussian
distribution on $\BbbR^{I'(n,V)\times n}$.   For $\delta>0$, set

$$\eqalignno{I'_{\delta}
&=\{i:i\in I'(n,V),\,\int|y(i,r)|^2\lambda(dy)\le\delta^2
  \text{ for every }r<n\}\cr
&=\{i:i\in I'(n,V),\,\int|Sx(i,r)|^2\nu(dx)\le\delta^2
  \text{ for every }r<n\}\cr
&=\{i:i\in I'(n,V),\,\int|T_i(x)(r)-T_j(x)(r)|^2\nu(dx)\le\delta^2
  \text{ for every }r<n\}\cr
&=\{i:i\in I'(n,V),\,\rho_n(\phi_n(i),\phi_n(j))\le\delta\}.\cr}$$

\noindent By 456Ma, there is a closed set $F\subseteq\BbbR^n$ such that
$F=\bigcap_{\delta>0}\overline{\{y(i,r):i\in I'_{\delta}\}}$ for
$\lambda$-almost every $y\in\BbbR^{I'(n,V)\times n}$, so that
$F=\bigcap_{\delta>0}\overline{\{Sx(i,r):i\in I'_{\delta}\}}$ for
$\nu$-almost every $x\in\BbbR^J$.

By 456Mb,
$\alpha z\in F$ whenever $z\in F$ and $|\alpha|\le 1$.   \Quer\ If $F$
is not bounded, then it must include a line $L$ through $0$.   (The sets
$\{\Bover1nz:z\in F$, $\|z\|=n\}$, for $n\ge 1$, form a non-increasing
sequence of non-empty compact sets, so there is a point $z_0$ belonging
to them all;  take $L$ to be the set of multiples of $z_0$.)
Let $D\subseteq L$ be a countable dense set.   For $z\in D$ and
$k\in\Bbb N$ we know that

\inset{for every $i\in I(n,V)$, $T_i(x)\notin V$ for almost every
$x\in W$,}

\inset{for almost every $x\in\BbbR^J$ there is an $i\in I'(n,V)$ such
that $\|T_i(x)-T_j(x)-z\|\le 2^{-k}$}

\noindent and therefore

\inset{for almost every $x\in W$, $T_i(x)\notin V$ for every
$i\in I'(n,V)$, but there is an $i\in I'(n,V)$ such that
$\|T_i(x)-T_j(x)-z\|\le 2^{-k}$}

\noindent so that

\inset{for almost every $x\in W$, the distance from $T_j(x)+z$ to the
closed set $\BbbR^n\setminus V$ is at most $2^{-k}$.}

\noindent This is true for every $k\in\Bbb N$, so we get

\inset{$T_j(x)+z\notin V$ for almost every $x\in W$.}

\noindent And {\it this} is true for every $z\in D$, so we get

\inset{for almost every $x\in W$, $T_j(x)+z\notin V$ for every $z\in D$,
so $T_j(x)\notin V+L$.}

Let $S_0:\BbbR^n\to\BbbR^{n-1}$ be a linear operator with kernel $L$, and
set $V'=S_0[V]$.   Then $V'\subseteq\BbbR^{n-1}$ is open, and
$W\cap(S_0T_j)^{-1}[V']=W\cap T_j^{-1}[V+L]$ is negligible.   But this
means that $T_j^{-1}[V]\subseteq(S_0T_j)^{-1}[V']\in\Cal G_{n-1}^*$ and
$T_j^{-1}[V]$ is included in $\bigcup\Cal G_{n-1}^*$;  which
contradicts the definition of $\Cal T_{nV}$.\ \Bang

So $F$ is bounded.   By 456Mc, there is some $\delta>0$ such that
$\sup_{i\in I'_{\delta},r<n}|y(i,r)|<\infty$ for $\lambda$-almost every
$y\in\BbbR^{I'(n,V)\times n}$, in which case
$\sup_{i\in I'_{\delta},r<n}|Sx(i,r)|<\infty$ for $\nu$-almost every
$x\in\BbbR^J$, that is,
$\{T_i(x)-T_j(x):i\in I'_{\delta}\}$ is bounded for $\nu$-almost every
$x$.   Of course this means that
$\{T_i(x):i\in I(n,V),\,\rho_n(\phi_n(i),\phi_n(j))\le\delta\}$ is
bounded for almost every $x\in\BbbR^J$.\ \Qed

\medskip

{\bf (d)} Accordingly $\phi_n[I(n,V)]$ is covered by the family
$\Cal U_{nV}$ of open sets $U\subseteq M^n$ such that
$\{T_i(x):i\in I'(n,V)$, $\phi_n(i)\in U\}$ is bounded for almost every
$x$.   Because $M^n$ is separable and metrizable, it is
hereditarily Lindel\"of (4A2P(a-iii)), so there is a sequence
$\sequence{k}{U_{nVk}}$ in $\Cal U_{nV}$ covering $\phi_n[I(n,V)]$.
For each $k$, set
$I_{nVk}=I'(n,V)\cap\phi_n^{-1}[U_{nVk}]$.   Then
$\{T_i(x):i\in I_{nVk}\}$
is bounded for almost every $x$.  Because $\phi_n[I'(n,V)]$ is dense in
$\phi_n[I(n,V)]$, $\phi_n[I_{nVk}]=\phi_n[I'(n,V)]\cap U_{nVk}$ is dense
in $\phi_n[I(n,V)]\cap U_{nVk}$.
So for every $i\in I(n,V)$ there is a
$k\in\Bbb N$ such that $\phi_n(i)\in\overline{\phi_n[I_{nVk}]}$.

\woddheader{456N}{4}{2}{2}{30pt}

{\bf (e)} Recall that $W\subseteq\BbbR^J$ is a non-negligible zero set
included in

\Centerline{$\bigcup_{n\ge 1}\bigcup\Cal G_n^*
=\bigcup_{n\ge 1}\bigcup_{V\in\Cal V_n}
  \bigcup_{i\in I(n,V)}T_i^{-1}[V]$.}

\noindent Let $J_0\subseteq J$ be a countable set such that $W$ is
determined by coordinates in $J_0$.

\ifdim\pagewidth>467pt\fontdimen3\tenrm=2pt\fi
\ifdim\pagewidth>467pt\fontdimen4\tenrm=1.67pt\fi
Let $\family{j}{J_0}{\epsilon_j}$ and
$\langle\epsilon'_{nVk}\rangle_{n\ge 1,V\in\Cal V_n,k\in\Bbb N}$
be families of strictly positive real numbers such that
$\sum_{n=1}^{\infty}\sum_{V\in\Cal V_n}
\sum_{k=0}^{\infty}\epsilon'_{nVk}$
and $\sum_{j\in J_0}\epsilon_j$
are both at most $\bover13\mu W$.   Let $\family{j}{J_0}{\gamma_j}$
and $\langle\gamma'_{nVk}\rangle_{n\ge 1,V\in\Cal V_n,k\in\Bbb N}$ be
such that
\fontdimen3\tenrm=1.67pt
\fontdimen4\tenrm=1.11pt

\Centerline{$\mu\{x:x\in\BbbR^J$, $|x(j)|\ge\gamma_j\}\le\epsilon_j$ for
every $j\in J_0$,}

\Centerline{$\mu\{x:x\in\BbbR^J$,
$\sup_{i\in I_{nVk}}\|T_i(x)\|\ge\gamma'_{nVk}\}
\le\epsilon'_{nVk}$ for every $n\ge 1$, $V\in\Cal V_n$, $k\in\Bbb N$.}

\noindent Set
$W'=\{x:x\in W$, $|x(j)|\le\gamma_j$ for every $j\in J_0\}$;  then
$\mu W'\ge\bover23\mu W$ and $W'$ is of the form
$C\times\BbbR^{J\setminus J_0}$, where $C\subseteq\BbbR^{J_0}$ is
compact.   Set

\Centerline{$W''=\{x:x\in W'$, $\|T_i(x)\|\le\gamma'_{nVk}$ whenever
$n\ge 1$, $V\in\Cal V_n$, $k\in\Bbb N$ and $i\in I_{nVk}\}$;}

\noindent then $\mu W''\ge\bover13\mu W$.

\medskip

{\bf (f)} Set $I=\bigcup_{n\ge 1,V\in\Cal V_n}I(n,V)\times n$.   If
$K\subseteq I$ is finite, then

$$\eqalign{W'_K
&=\{x:x\in W',\,\|f_{ir}(x)\|\le\gamma'_{nVk}\text{ whenever }n\ge 1,
  \,V\in\Cal V_n,\,k\in\Bbb N,\cr
&\mskip250mu(i,r)\in K
  \text{ and }\phi_n(i)\in\overline{\phi_n[I_{nVk}]}\}\cr}$$

\noindent has measure at least $\bover13\mu W$.   \Prf\ For each
quintuple $(i,r,n,V,k)$ with $n\ge 1$, $V\in\Cal V_n$, $k\in\Bbb N$,
$(i,r)\in K$ and $\phi_n(i)\in\overline{\phi_n[I_{nVk}]}$, there is a
sequence $\sequence{m}{i_m}$ in $I_{nVk}$ such that
$\rho_n(\phi_n(i),\phi_n(i_m))\le 2^{-m}$ for every $m$;  so that
$\|f_{ir}-f_{i_mr}\|_2\le 2^{-m}$ for every $m$.   But this means that
$f_{ir}(x)=\lim_{m\to\infty}f_{i_mr}(x)$ for almost every $x\in\BbbR^J$.
Accordingly $|f_{ir}(x)|\le\gamma'_{nVk}$ for almost every $x\in W''$.
Since there are only countably many such quintuples $(i,r,n,V,k)$, we
see that $W''\setminus W'_K$ is negligible, so
$\mu W'_K\ge\mu W''\ge\bover13\mu W$.\ \Qed

\medskip

{\bf (g)} For $x\in\BbbR^J$, define $Tx\in\BbbR^I$ by setting
$(Tx)(i,r)=f_{ir}(x)$ for $i\in I(n,V)$ and $r<n$.
Then $T:\BbbR^J\to\BbbR^I$ is a continuous linear operator.
By 4A4H, $T[W']$ is closed.

For finite $K\subseteq I$, let $\Cal H_K$ be the family of open subsets
$H$ of $\BbbR^K$ such that $\mu\{x:x\in W$, $Tx\restr K\in H\}=0$.
Then $\Cal H_K$ is closed under countable unions so has a largest member
$H_K$.   Now there is a $K\in[I]^{<\omega}$ such that
$Tx\restr K\in H_K$ for every $x\in W'_K$.   \Prf\Quer\ Otherwise,
choose for each $K\in[I]^{<\omega}$ an $x_K\in W'_K$ such that
$Tx_K\restr K\notin H_K$.   Let $\Cal F$ be an ultrafilter on
$[I]^{<\omega}$ containing $\{K:L\subseteq K\in[I]^{<\omega}\}$ for
every finite $L\subseteq I$.   If $(i,r)\in I$, there are
$n\ge 1$, $V\in\Cal V_n$ and $k\in\Bbb N$ such that $r<n$ and
$\phi_n(i)\in\overline{\phi_n[I_{nVk}]}$, in which case
$|f_{ir}(x_K)|\le\gamma'_{nVk}$ whenever $K\in[I]^{<\omega}$ contains
$(i,r)$.
This means that $\lim_{K\to\Cal F}f_{ir}(x_K)$ must be defined in
$[-\gamma'_{nVk},\gamma'_{nVk}]$;  consequently
$y^*=\lim_{K\to\Cal F}Tx_K$ is defined in $\BbbR^I$.   Since $x_K\in W'$
for every $K$, $y^*\in\overline{T[W']}=T[W']$.

Let $x^*\in W'$ be such that $Tx^*=y^*$.   Since $x^*\in W$, there are
$n\ge 1$, $V\in\Cal V_n$ and $i\in I(n,V)$ such that $T_i(x^*)\in V$.
Set $L=\{(i,r):r<n\}$,
$H=\{z:z\in\BbbR^L$, $\ofamily{r}{n}{z(i,r)}\in V\}$;  then
$\{x:Tx\restr L\in H\}=T_i^{-1}[V]$.   Since
$y^*\restr L=Tx^*\restr L$ belongs to $H$, and $H$ is open, there must
be a $K\supseteq L$ such that $Tx_K\restr L\in H$.   But in this case
$H'=\{z:z\in\BbbR^K$, $z\restr L\in H\}$ is an open subset of $\BbbR^K$
and

\Centerline{$\{x:Tx\restr K\in H'\}=\{x:Tx\restr L\in H\}
=\{x:T_i(x)\in V\}$}

\noindent meets $W$ in a negligible set, and $H'\subseteq H_K$.   But
this means that $Tx_K\restr K\in H_K$, contrary to the choice of $x_K$.\
\Bang\Qed

\medskip

{\bf (h)} Putting (f) and (g) together, we find ourselves trying to
believe simultaneously that $\mu W'_K>0$ and that $Tx\restr K\in H_K$
for every $x\in W'_K$ and that $W'_K\subseteq W$ and that
$\{x:x\in W$, $Tx\restr K\in H_K\}$ is negligible.   Faced with this we
have to abandon the original supposition that $\mu$ is not
$\tau$-additive.
}%end of proof of 456N

\leader{456O}{}\cmmnt{ We now have all the ideas needed for the main
theorem of this section.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Talagrand 81})} Every centered
Gaussian distribution is $\tau$-additive.

\proof{\Quer\ Suppose, if possible, that $\mu$ is a centered Gaussian
distribution on a set $\BbbR^I$ which is not $\tau$-additive.

\medskip

{\bf (a)} By 456K, there are a set $J$ and a universal centered Gaussian
distribution $\nu$ on $\BbbR^J$ and a continuous linear operator
$T:\BbbR^J\to\BbbR^I$ which is \imp\ for $\nu$ and $\mu$.   By
418Ha, $\nu$ is not $\tau$-additive.

\medskip

{\bf (b)} As in part (a) of the proof of 456N, there are a
non-negligible zero set $W\subseteq\BbbR^J$ and a family $\Cal G$ of
open sets, covering $W$, such that $\nu(W\cap G)=0$ for every
$G\in\Cal G$.   Give $J$ a Hilbert space structure such that
$\int x(i)x(j)\nu(dx)=\innerprod{i}{j}$ for all $i$, $j\in J$.   Let
$K_0\subseteq J$ be a countable set such that $W$ is determined by
coordinates in $K_0$, and let $K$ be the closed linear subspace of $J$
generated by $K_0$.
Let $\Cal G'$ be the family of open sets determined by coordinates in
$K$ which meet $W$ in negligible sets.   Then
$W\subseteq\bigcup\Cal G'$, by 456L.

Let $\lambda$ be the centered Gaussian distribution on $\BbbR^K$ for
which the map $\tilde\pi_K=x\mapsto x\restr K:\BbbR^J\to\BbbR^K$ is
\imp.   Then $\tilde\pi_K[W]$ is a zero set in $\BbbR^K$,
$\lambda\tilde\pi_K[W]=\nu W>0$, $\{\tilde\pi_K[G]:G\in\Cal G'\}$ is a
family of open sets in $\BbbR^K$ covering $\tilde\pi_K[W]$, and
$\lambda(\tilde\pi_K[W]\cap\tilde\pi_K[G])=\nu(W\cap G)=0$ for every
$G\in\Cal G'$;  so $\lambda$ is not $\tau$-additive.   However, $K$,
regarded as a normed space, is separable (see 4A4Bg);  and if we set
$\pi_j(y)=y(j)$ for $y\in\BbbR^K$ and $j\in K$, then
$\|\pi_i^{\ssbullet}-\pi_j^{\ssbullet}\|_2=\|i-j\|$ for all $i$,
$j\in K$.   So $\{\pi_j^{\ssbullet}:j\in K\}$ is separable in
$L^2(\lambda)$.   And this is impossible, by 456N.\ \Bang

Thus every centered Gaussian distribution must be $\tau$-additive.
}%end of proof of 456O

\leader{456P}{Corollary} If $\mu$ is a centered Gaussian distribution on
$\BbbR^I$, there is a unique quasi-Radon measure $\tilde\mu$ on
$\BbbR^I$ extending $\mu$.   The support of $\mu$ as defined in 456H is
the support of $\tilde\mu$ as defined in 411N.

\proof{ By 415L, $\mu$ has a unique extension to a quasi-Radon measure
$\tilde\mu$.   Now the support $Z$ of $\mu$ is a closed set, so
$\tilde\mu Z=\mu^*Z$ (415L(i)).   Also $Z$ is self-supporting for $\mu$.
If $G\subseteq\BbbR^I$ is an open set meeting $Z$, then there is a
cozero set $H\subseteq G$ which also meets $Z$, and $\mu^*(Z\cap H)>0$.
It follows that $\mu^*(Z\setminus H)<1$;  as $\tilde\mu$ extends $\mu$,
$\tilde\mu(Z\setminus H)<1$ and $\tilde\mu(Z\cap G)>0$.   This shows
that $Z$ is self-supporting for $\tilde\mu$, so must be the support of
$\tilde\mu$ in the standard sense.
}%end of 456P

\leader{456Q}{Proposition}\dvAnew{2009}
Let $I$ be a set and $R$ the set of functions
$\sigma:I\times I\to\Bbb R$ which are symmetric and positive semi-definite
in the sense of 456C;  give $R$ the subspace topology induced by the usual
topology of $\BbbR^{I\times I}$.   Let $P_{\text{qR}}(\BbbR^I)$
be the space of
quasi-Radon probability measures on $\BbbR^I$ with its narrow
topology\cmmnt{ (437Jd)}.   For $\sigma\in R$, let $\mu_{\sigma}$ be the
centered Gaussian distribution on $\BbbR^I$ with covariance matrix
$\sigma$\cmmnt{ (456C)}, and $\tilde\mu_{\sigma}$ the quasi-Radon measure
extending $\mu_{\sigma}$\cmmnt{ (456P)}.   Then $R$ is a closed subset of
$\BbbR^{I\times I}$ and the function
$\sigma\mapsto\tilde\mu_{\sigma}:R\to P_{\text{qR}}(\BbbR^I)$ is
continuous.

\proof{{\bf (a)} From 456C(iv) we see at once that $R$ is closed.   So the
rest of this proof will be devoted to showing that
$\sigma\mapsto\tilde\mu_{\sigma}$ is continuous.

\medskip

{\bf (b)} I had better begin with the one-dimensional case.   If
$I=\{j\}$ is a singleton, and we identify $\BbbR^I$ with $\Bbb R$, then
$\tilde\mu_{\sigma}$ is the ordinary normal distribution with mean $0$ and
variance $\sigma(j,j)$, counting the Dirac measure centered at $0$ as a
normal distribution with zero variance.   If $H\subseteq\Bbb R$ is open
and $\gamma\in\Bbb R$, set

\Centerline{$G=\{\alpha:\alpha>0$,
$\Bover1{\sqrt{2\pi\alpha}}\int_He^{-t^2/\alpha}dt>\gamma\}$;}

\noindent then $G$ is open.   If $0\notin H$, then

\Centerline{$\{\sigma:\tilde\mu_{\sigma}H>\gamma\}
=\{\sigma:\sigma(i,i)\in G\}$}

\noindent is open.   If $0\in H$ and $\gamma\ge 1$, then
$\{\sigma:\tilde\mu_{\sigma}H>\gamma\}$ is empty;  if
$0\in H$ and $\gamma<1$, then

\Centerline{$\{\sigma:\tilde\mu_{\sigma}H>\gamma\}
=\{\sigma:\sigma(i,i)\in G\}\cup\{0\}$}

\noindent is open because there is an $\eta>0$ such that
$[-\eta,\eta]\subseteq H$ and $\alpha\in G$ whenever $\alpha>0$ and

\Centerline{$\Bover1{\sqrt{2\pi}}
  \int_{-\eta/\sqrt{\alpha}}^{\eta/\sqrt{\alpha}}
  e^{-t^2/2}dt>\gamma$.}

\noindent As $H$ is arbitrary, $\sigma\mapsto\tilde\mu_{\sigma}$ is
continuous.

\medskip

{\bf (c)} Now suppose that $I$ is finite.   Let $\sequencen{\sigma_n}$ be a
sequence in $R$ with limit $\sigma\in R$.   Let $\varphi_n$, $\varphi$ be the
characteristic functions of $\tilde\mu_{\sigma_n}$,
$\tilde\mu_{\sigma}$ respectively
(\S285).   If $y\in\BbbR^I$, set $f(x)=\varinnerprod{x}{y}$ for
$x\in\BbbR^I$;  then

\Centerline{$\varphi(y)
=\int e^{i f(x)}\tilde\mu_{\sigma}(dx)
=\int e^{it}(\tilde\mu_{\sigma}f^{-1})(dt)$,}

\noindent writing $\tilde\mu_{\sigma}f^{-1}$ for the image Radon measure on
$\Bbb R$.   Now $\tilde\mu_{\sigma}f^{-1}$ is the
one-dimensional Gaussian distribution
with variance $\sum_{j,k\in I}\sigma(j,k)y(j)y(k)$ (see part (b) of the
proof of 456B).   But since

\Centerline{$\sum_{j,k\in I}\sigma(j,k)y(j)y(k)
=\lim_{n\to\infty}\sum_{j,k\in I}\sigma_n(j,k)y(j)y(k$,}

\noindent (a) tells us that $\tilde\mu_{\sigma}f^{-1}
=\lim_{n\to\infty}\tilde\mu_{\sigma_n}f^{-1}$ for the narrow topology on
$P_{\text{qR}}(\Bbb R)$, therefore also for the vague topology (437L),
and $\varphi(y)=\lim_{n\to\infty}\varphi_n(y)$.   By 285L,
$\tilde\mu_{\sigma}=\lim_{n\to\infty}\tilde\mu_{\sigma_n}$ for the vague
topology, therefore also for the narrow topology.

Thus $\sigma\mapsto\tilde\mu_{\sigma}$ is sequentially continuous.   As
$I$ is countable, $R$ is metrizable, and
$\sigma\mapsto\tilde\mu_{\sigma}$ is continuous.

\medskip

{\bf (d)} For the general case, suppose that $H\subseteq\BbbR^I$ is an open
set and that $\gamma\in\Bbb R$.   Set
$G_{H\gamma}=\{\sigma:\sigma\in R$, $\tilde\mu_{\sigma}H>\gamma\}$.

\medskip

\quad{\bf (i)} If $H$ is determined by coordinates in a finite set
$J\subseteq I$ then $G_{H\gamma}$ is open in $R$.   \Prf\
Let $R_J$ be the set of symmetric positive semi-definite functions on
$\BbbR^{J\times J}$;  write $h(\sigma)=\sigma\restr J\times J$ for
$\sigma\in R$, and $\tilde\pi_J(x)=x\restr J$ for $x\in\BbbR^I$.
Of course $h(\sigma)\in R_J$ for $\sigma\in R$, and
$h:R\to R_J$ is continuous.   For $\sigma\in R$,
we know that there is a centered Gaussian distribution $\nu$ on $\BbbR^J$
such that $\tilde\pi_J$ is \imp\ for $\mu_{\sigma}$ and $\nu$, by
456Ba;  the covariance matrix of $\nu$
is of course $h(\sigma)$, so we can call it $\mu_{h(\sigma)}$.
Next, there is a quasi-Radon measure $\tilde\nu$
on $\BbbR^J$ such that $\tilde\pi_J$ is \imp\ for $\tilde\mu_{\sigma}$ and
$\tilde\nu$ (418Hb);  as $\tilde\nu$ must extend the Baire measure
$\nu$, it is the unique quasi-Radon measure extending
$\nu$, and we can call it $\tilde\mu_{h(\sigma)}$.

Because $H$ is determined by coordinates in $J$, $H=\tilde\pi_J^{-1}[H']$
where $H'=\tilde\pi_J[H]$ is open in $\BbbR^J$ (4A2B(f-i) again).   So
$G'=\{\tau:\tau\in R_J$, $\tilde\mu_{\tau}H'>\gamma\}$ is open in $R_J$, by
(b), and

\Centerline{$G_{H\gamma}
=\{\sigma:(\tilde\mu_{\sigma}\tilde\pi_J^{-1})(H')>\gamma\}
=\{\sigma:\tilde\mu_{h(\sigma)}H'>\gamma\}
=h^{-1}[G']$}

\noindent is open in $R$.\ \Qed

\medskip

\quad{\bf (ii)} In fact $G_{H\gamma}$ is open in $R$ for any open set
$H\subseteq\BbbR^I$ and $\gamma\in\Bbb R$.   \Prf\ Take any
$\sigma\in G_{H\gamma}$.   Because $\tilde\mu_{\sigma}$ is
$\tau$-additive, and the family

\Centerline{$\Cal V
=\{V:V\subseteq\BbbR^I$ is open and determined by coordinates in a finite
set$\}$}

\noindent is a base for the topology of $\BbbR^I$ closed under finite
unions, there is a $V\in\Cal V$ such that $V\subseteq H$ and
$\tilde\mu_{\sigma}V>\gamma$.   Now
$\sigma\in G_{V\gamma}\subseteq G_{H\gamma}$;  by (i),
$G_{V\gamma}$ is open, so $\sigma\in\interior G_{H\gamma}$;  as $\sigma$ is
arbitrary, $G_{H\gamma}$ is open.\ \QeD\  But this is just what we need to
know to see that $\sigma\mapsto\tilde\mu_{\sigma}$ is continuous for the
narrow topology on $P_{\text{qR}}(\BbbR^I)$, and the proof is complete.
}%end of proof of 456Q

\exercises{\leader{456X}{Basic exercises (a)}%
%\spheader 456Xa
\dvAnew{2009} Let $I$ be any set.
(i) Show that if $y\in\ell^1(I)$ then
$\int\sum_{i\in I}|y(i)x(i)|\mu^{(I)}_G(dx)=\Bover2{\sqrt{2\pi}}\|y\|_1$.
\Hint{start by evaluating $\Expn(|Z|)$ where $Z$ is a standard normal
random variable.}
(ii) Show that if $y\in\ell^2(I)$ then
$\int\sum_{i\in I}|y(i)x(i)|^2\mu^{(I)}_G(dx)=\|y\|_2^2$.
%456B

\spheader 456Xb
Let $n\ge 1$ be an integer.   (i) Show that
$\mu_G^{(n)}T^{-1}=\mu_G^{(n)}$ for any orthogonal linear operator
$T:\BbbR^n\to\BbbR^n$.   (ii) Set $p(x)=\Bover1{\|x\|}x$ for
$x\in\BbbR^n\setminus\{0\}$;  take $p(0)$ to be any point of $S^{n-1}$.
Show that $\mu_G^{(n)}p^{-1}$ is a multiple of $(n-1)$-dimensional
Hausdorff measure on $S^{n-1}$.   \Hint{443U.}
%456B

\spheader 456Xc\dvAnew{2009}
Let $G$ be a group, and $h:G\to\Bbb R$ a real positive definite
function (definition: 445L).   (i) Show that we have a centered Gaussian
distribution $\mu$ on $\BbbR^G$ with covariance matrix
$\langle h(a^{-1}b)\rangle_{a,b\in G}$.
(ii) Show that $\mu$ is invariant under
the left shift action $\action_l$ of $G$ on $\BbbR^G$ (4A5Cc).
%456C

\spheader 456Xd Let $I$ be a countable set, $\mu$ a centered Gaussian
distribution on $\BbbR^I$, and $\gamma\ge 0$.   Set
$\alpha=\mu\{x:\sup_{i\in I}|x(i)|\ge\gamma\}$.   Show that
$\mu\{x:\sup_{i\in I}|x(i)|\ge\bover12\gamma\}\ge 2\alpha(1-\alpha)^3$.
%456G

\spheader 456Xe Let $I$ be a set and
$\langle\sigma_{ij}\rangle_{i,j\in I}$ a family of real numbers.   Show
that there is at most one inner product space structure on $I$ for which
$\sigma_{ij}=\innerprod{i}{j}$ for all $i$, $j\in I$.
%456J

\spheader 456Xf Let $\sequencen{X_n}$ be an independent sequence of
standard normal random variables, and $\sequencen{\alpha_n}$ a
square-summable real sequence.   (i) Show that for any
permutation $\pi:\Bbb N\to\Bbb N$, $X=\sum_{n=0}^{\infty}\alpha_nX_n$ and
$\sum_{n=0}^{\infty}\alpha_{\pi(n)}X_{\pi(n)}$ are finite and equal a.e.
\Hint{273B.}  (ii) Show that $X$ is normal, with mean $0$ and variance
$\sum_{n=0}^{\infty}\alpha_n^2$.
%456K

\sqheader 456Xg For any set $I$, I will say that a
{\bf centered Gaussian quasi-Radon measure} on $\BbbR^I$ is a quasi-Radon
measure $\mu$ on $\BbbR^I$ such that
every continuous linear functional $f:\BbbR^I\to\Bbb R$ is either zero a.e.\ or is normally distributed with zero expectation.   Show that

\quad(i) there is a one-to-one correspondence between centered Gaussian quasi-Radon measures $\mu$ on $\BbbR^I$ and centered Gaussian distributions $\nu$ on $\BbbR^I$
obtained by matching $\mu$ with $\nu$ iff they agree on the zero sets of $\BbbR^I$;

\quad(ii) if $\mu$, $\nu$ are centered Gaussian quasi-Radon measures on $\BbbR^I$ and $\int x(i)x(j)\mu(dx)=\int x(i)x(j)\nu(dx)$ for all $i$, $j\in I$, then $\mu=\nu$;

\quad(iii) the support of a centered Gaussian quasi-Radon measure on $\BbbR^I$ is a linear subspace of $\BbbR^I$;

\quad(iv) if $\family{j}{J}{I_j}$ is a disjoint family of sets with union $I$, and $\mu_j$ is a centered Gaussian quasi-Radon measure on $\BbbR^{I_j}$ for each $j\in J$,
then the quasi-Radon product of $\family{j}{J}{\mu_j}$, regarded as a measure on $\BbbR^I$, is a centered Gaussian quasi-Radon measure.
%456P

\spheader 456Xh\dvAnew{2011}
Let $I$ be a set, and let $H$ be a Hilbert space with
orthonormal basis $\familyiI{e_i}$.   For $i\in I$, $x\in\Bbb R^I$ set
$f_i(x)=x(i)$.    Show that there is a bounded
linear operator $T:H\to L^1(\mu_G^{(I)})$ such that $Te_i=f_i^{\ssbullet}$
for every $i\in I$, and that $\|Tu\|_1=\Bover{2}{\sqrt{2\pi}}\|u\|_2$
for every $u\in H$.
%456Xf 456K out of order query

\leader{456Y}{Further exercises (a)}
%\spheader 456Ya
Let $(\Omega,\Sigma,\mu)$ be a probability space with measure algebra
$(\frak A,\bar\mu)$, and $\familyiI{u_i}$ a family in
$L^2(\mu)\cong L^2(\frak A,\bar\mu)$ which is a centered
Gaussian process in the
sense that whenever $X_i\in\eusm L^2(\mu)$ is such that
$X_i^{\ssbullet}=u_i$ for every $i$, then $\familyiI{X_i}$ is a centered
Gaussian process.   Suppose that $\gamma\ge 0$ and that
$\alpha=\bar\mu(\sup_{i\in I}\Bvalue{|u_i|\ge\gamma})$.   Show that
$\bar\mu(\sup_{i\in I}\Bvalue{|u_i|\ge\bover12\gamma})
\ge 2\alpha(1-\alpha)^3$.
%456Xd 456G

\spheader 456Yb Let $U$ be a Hilbert space with an orthonormal basis
$\family{j}{J}{u_j}$, and $\mu$ the universal centered Gaussian
distribution on $\BbbR^U$ with covariance matrix defined by the inner
product of $U$.   Show that there is a function $T:\BbbR^J\to\BbbR^U$,
\imp\ for $\mu_G^{(J)}$ and $\mu$, such that whenever $\sequencen{j_n}$
is a sequence of distinct elements of $J$ and $\sequencen{\alpha_n}$ is
a square-summable sequence in $\Bbb R$, then
$(Tx)(\sum_{n=0}^{\infty}\alpha_nu_{j_n})
=\sum_{n=0}^{\infty}\alpha_nx(j_n)$ for almost every $x\in\BbbR^J$.
%456Xf 456Ea 456K

\spheader 456Yc Let $U$ be an infinite-dimensional Hilbert space and
$\mu$ the universal centered Gaussian distribution on $\BbbR^U$ with
covariance matrix defined by the inner product of $U$.   Show that
$\mu C=0$ for every compact set $C\subseteq\BbbR^U$.
%456K mt45bits

\spheader 456Yd Let $I$ be a set and
$\mu$ be a centered Gaussian distribution on $\BbbR^I$.   Show
that the following are equiveridical:  (i) $\mu$ has countable Maharam
type;  (ii) $L^2(\mu)$ is separable;  (iii) $I$ is separable under the
pseudometric $(i,j)\mapsto\sqrt{\int(x(i)-x(j))^2\mu(dx)}$.
%456N
}%end of exercises

\endnotes{
\Notesheader{456} This section has aimed for a direct route to Talagrand's
theorem 456O, leaving most of the real reasons for studying Gaussian
processes (see {\smc Fernique 97}) to one side.   It should
nevertheless be clear from
such fragments as 252Xi, 456Bb, 456G and the exercises here that
they are one of the many concepts of probability theory which are both
significant and delightful.   Very much the most important Gaussian
processes are those associated with Brownian motion, which will be treated
in \S477 {\it et seq.}

You will of course have observed that the methods used here are entirely
different from those in \S455, even though one of the concerns of
that section was a check for $\tau$-additive distributions and
corresponding quasi-Radon versions, as in 455K.   However the results of
\S455 were based on the fact that in the most important cases the
distributions there have extensions to Radon measures (455H).   Gaussian
distributions need not be like this at all, even when they have countable
Maharam type;  see 456Yc.
}%end of notes

\discrversionA{
\bigskip

{\bf question}  if $\mu$ is a centered Gaussian quasi-Radon measure on
$\BbbR^I$ and $T:\BbbR^I\to\BbbR^J$ is a continuous linear operator,
must $\mu T^{-1}$ be a quasi-Radon
measure?   (certainly it is $\tau$-additive, is it inner regular with
respect to the closed sets?)
}{}

\discrpage

\frfilename{mt457.tex}
\versiondate{18.1.13}
\copyrightdate{2010}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Simultaneous extension of measures}

\newsection{457}

The questions addressed in \S\S451, 454 and 455 can all be regarded as
special cases of a general class of problems:  given a set $X$ and a
family $\familyiI{\nu_i}$ of (probability) measures on $X$, when can we
expect to find a measure on $X$ extending every $\nu_i$?   An
alternative formulation, superficially more general, is to ask:  given a
set $X$, a family $\familyiI{(Y_i,\Tau_i,\nu_i)}$ of probability spaces,
and functions $\phi_i:X\to Y_i$ for each $i$, when can we find a measure
on $X$ for which every $\phi_i$ is \imp?   Even the simplest non-trivial
case, when $X=\prod_{i\in I}Y_i$ and every $\phi_i$ is the coordinate
map, demands a significant construction (the product measures of Chapter
25).   In this section I bring together a handful of important further
cases which are accessible by the methods of this chapter.   I begin
with a discussion of extensions of finitely additive measures
(457A-457D), which are much easier, before considering the problems
associated with countably additive measures (457E-457G), with examples
(457H-457J).   In 457K-457M %457L
I look at a pair of optimisation problems.

\leader{457A}{}\cmmnt{ It is helpful to start with a widely applicable
result on common extensions of finitely additive measures.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Boolean algebra and
$\familyiI{\frak B_i}$ a non-empty family of subalgebras of $\frak A$.
For each $i\in I$, we may identify $L^{\infty}(\frak B_i)$ with the
closed linear subspace of $L^{\infty}(\frak A)$ generated by
$\{\chi b:b\in\frak B_i\}$\cmmnt{ (363Ga)}.
Suppose that for each $i\in I$ we are given a finitely additive
functional $\nu_i:\frak B_i\to[0,1]$ such that $\nu_i1=1$;  write
$\dashint\ldots d\nu_i$ for the
corresponding positive linear functional on
$L^{\infty}(\frak B_i)$\cmmnt{ (363L)}.   Then the following
are equiveridical:

\quad(i) there is an additive functional $\mu:\frak A\to[0,1]$ extending
every $\nu_i$;

\quad(ii) whenever $i_0,\ldots,i_n\in I$, $a_k\in\frak B_{i_k}$ for
$k\le n$, and $\sum_{k=0}^n\chi a_k\ge m\chi 1$ in $S(\frak A)$, where
$m\in\Bbb N$, then $\sum_{k=0}^n\nu_{i_k}a_k\ge m$;

\quad(iii) whenever $i_0,\ldots,i_n\in I$, $a_k\in\frak B_{i_k}$ for
$k\le n$, and $\sum_{k=0}^n\chi a_k\le m\chi 1$, where $m\in\Bbb N$,
then $\sum_{k=0}^n\nu_{i_k}a_k\le m$;

\ifdim\pagewidth>467pt\fontdimen3\tenrm=2pt\fi
\ifdim\pagewidth>467pt\fontdimen4\tenrm=1.67pt\fi
\quad(iv)\dvArevised{2013} whenever $i_0,\ldots,i_n\in I$ are distinct,
$u_k\in L^{\infty}(\frak B_{i_k})$ for every $k\le n$, and
$\sum_{k=0}^nu_k\ge\chi 1$, then
$\sum_{i=0}^n\dashint u_kd\nu_{i_k}\ge 1$;
\fontdimen3\tenrm=1.67pt
\fontdimen4\tenrm=1.11pt

\quad(v)\dvArevised{2013} whenever $i_0,\ldots,i_n\in I$ are distinct,
$u_k\in L^{\infty}(\frak B_{i_k})$ for every $k\le n$, and
$\sum_{k=0}^nu_k\le\chi 1$, then
$\sum_{i=0}^n\dashint u_kd\nu_{i_k}\le 1$.

%maybe we could use versions with non-negative u_k?
%or with i_k declared distinct?

\proof{{\bf (a)} It is elementary to check that if (i) is true then
(ii)-(v) are all true, simply because we have a positive linear
functional $\dashint\,d\mu$ extending all the functionals
$\dashint\,d\nu_i$.

\wheader{457A}{6}{2}{2}{24pt}

{\bf (b)(ii)$\Rightarrow$(iii)} Given that $a_k\in\frak B_{i_k}$ and
$\sum_{k=0}^n\chi a_k\le m\chi 1$, then

\Centerline{$\sum_{k=0}^n\chi(1\Bsetminus a_k)
=(n+1)\chi 1-\sum_{k=0}^n\chi a_k
\ge(n+1-m)\chi 1$,}

\noindent so

\Centerline{$\sum_{k=0}^n\nu_{i_k}a_k
=n+1-\sum_{k=0}^n\nu_{i_k}(1\Bsetminus a_k)
\le n+1-(n+1-m)=m$,}

\noindent as required by (iii).

\medskip

{\bf (c)(iii)$\Rightarrow$(i)} Assume (iii).   Set
$\psi a=\sup\{\nu_ia:i\in I,\,a\in\frak B_i\}$ for $a\in\frak A$
(interpreting $\sup\emptyset$ as $0$, as usual in such contexts).   Then
$\psi$ satisfies the condition (ii) of 391F.
\Prf\Quer\ Otherwise,
there is a finite indexed family $\family{k}{K}{a_k}$ in
$\frak A$ such that $\inf_{k\in J}a_k=0$ whenever $J\subseteq K$ and
$\#(J)\ge\sum_{k\in K}\psi a_i$.   The general hypothesis of the lemma
implies that $\frak A\ne\{0\}$, so $\inf\emptyset=1\ne 0$ and $K$ is
non-empty.   Taking $K$ to be of minimal size, we get
an example in which $\psi a_k>0$ for every $k\in K$.
Set
$m=\|\sum_{k\in K}\chi a_k\|_{\infty}$;  then $m\in\Bbb N$ and
$m<\sum_{k\in K}\psi a_k$, so we can find for each $k\in K$ an $i_k\in I$
such that $a_k\in\frak B_{i_k}$ and $m<\sum_{k\in K}\nu_{i_k}a_k$.   But
this contradicts our hypothesis (iii).\ \Bang\Qed

By 391F, there is
a non-negative finitely additive functional $\mu$ such that $\mu 1=1$
and $\mu a\ge\psi a$ for every $a\in\frak A$, that is, $\mu b\ge\nu_ib$
whenever $i\in I$ and $b\in\frak B_i$.   But observe now
that, because $\mu 1=\nu_i1$ and
$\mu(1\Bsetminus b)\ge\nu_i(1\Bsetminus b)$, we actually have
$\mu b=\nu_ib$ for every $b\in\frak B_i$, so that $\mu$ extends $\nu_i$,
for every $i\in I$.

\medskip

{\bf (d)(iv)$\Rightarrow$(ii)} Suppose that (iv) is true, and that
$i_0,\ldots,i_n\in I$, $a_k\in\frak B_{i_k}$ for
$k\le n$, and $\sum_{k=0}^n\chi a_k\ge m\chi 1$ in $S(\frak A)$, where
$m\in\Bbb N$.   If $m=0$ then of course $\sum_{k=0}^n\nu_{i_k}a_k\ge m$.
Otherwise, set $J=\{i_k:k\le n\}$ and enumerate $J$ as
$\langle j_l\rangle_{l\le r}$.   For $l\le r$ set
$u_l=\bover1m\sum_{k\le n,i_k=j_l}\chi a_k$.   Then
$u_l\in S(\frak B_{j_l})$ for each $l$, and

\Centerline{$\sum_{l=0}^ru_l
=\Bover1m\sum_{l=0}^r\sum_{k\le n,i_k=j_l}\chi a_k
=\Bover1m\sum_{k=0}^n\chi a_k\ge\chi 1$.}

\noindent As $j_0,\ldots,j_l$ are distinct,

\Centerline{$\sum_{l=0}^r\dashint u_ld\nu_{j_l}
=\Bover1m\sum_{k=0}^n\nu_{i_k}a_k\ge 1$.}

\noindent So (ii) is true.

\medskip

{\bf (e)(v)$\Rightarrow$(iii)} Use the same argument as in (d) above.
}%end of proof of 457A

\leader{457B}{Corollary} Let $X$ be a set and $\familyiI{Y_i}$ a family
of sets.   Suppose that for each $i\in I$ we have an algebra $\Cal E_i$
of subsets of $Y_i$, an additive functional $\nu_i:\Cal E_i\to[0,1]$
such that $\nu_iY_i=1$, and a function $f_i:X\to Y_i$.   Then the
following are equiveridical:

\quad(i) there is an additive functional $\mu:\Cal PX\to[0,1]$ such that
$\mu f_i^{-1}[E]=\nu_iE$ whenever $i\in I$ and $E\in\Cal E_i$;

\quad(ii) whenever $i_0,\ldots,i_n\in I$ and
$E_k\in\Cal E_{i_k}$ for $k\le n$, then there is an $x\in X$ such that
$\sum_{k=0}^n\nu_{i_k}E_k\le\#(\{k:k\le n,\,f_{i_k}(x)\in E_k\})$.

\proof{{\bf (i)$\Rightarrow$(ii)} is elementary;  if
$m=\lceil\sum_{k=0}^n\nu_{i_k}E_k\rceil-1$, then
$\sum_{k=0}^n\mu f_{i_k}^{-1}[E_k]>m\mu X$, so
$\sum_{k=0}^n\chi f_{i_k}^{-1}E_k\not\le m\chi X$, that is, there is an
$x\in X$ such that

\Centerline{$\#(\{k:f_{i_k}(x)\in E_k\})
=\sum_{k=0}^n(\chi f_{i_k}^{-1}[E_k])(x)
\ge m+1\ge\sum_{k=0}^n\nu_{i_k}E_k$.}

\medskip

{\bf (ii)$\Rightarrow$(i)} Now suppose that (ii) is true.   For $i\in I$
set $\frak B_i=\{f_i^{-1}[E]:E\in\Cal E_i\}$.   Note that if
$E\in\Cal E_i$ and $\nu_iE>0$, then (applying (ii) with $n=0$, $i_0=i$
and $E_0=E$) $f_i^{-1}[E]$ cannot be empty;  accordingly we have an
additive functional $\nuprime_i:\frak B_i\to[0,1]$ defined by setting
$\nuprime_if^{-1}[E]=\nu_iE$ for every $E\in\Cal E_i$, and
$\nuprime_iX=1$.
If $i_0,\ldots,i_n\in I$,
$H_0\in\frak B_{i_0},\ldots,H_n\in\frak B_{i_n}$ and $m\in\Bbb N$ are
such that $\sum_{k=0}^n\chi H_k\le m\chi X$, express each $H_k$ as
$f_{i_k}^{-1}[E_k]$, where $E_k\in\Cal E_{i_k}$;  then there is an
$x\in X$ such that

\Centerline{$\sum_{k=0}^n\nuprime_{i_k}H_k=\sum_{k=0}^n\nu_{i_k}E_k
\le\#(\{k:f_k(x)\in E_k\})=\sum_{k=0}^m\chi H_k(x)\le m$.}

\noindent But this means that the condition of 457A(iii) is satisfied,
with $\frak A=\Cal PX$, so 457A(i) and (i) here are also true.
}%end of proof of 457B

\leader{457C}{Corollary} (a) Let $\frak A$ be a Boolean algebra and
$\frak B_1$,
$\frak B_2$ two subalgebras of $\frak A$ with finitely additive
functionals $\nu_i:\frak B_i\to[0,1]$ such that $\nu_11=\nu_21=1$.
Then the following are equiveridical:

\quad(i) there is an additive functional
$\mu:\frak A\to[0,1]$ extending both the $\nu_i$;

\quad(ii) whenever $b_1\in\frak B_1$, $b_2\in\frak B_2$ and
$b_1\Bcup b_2=1$, then $\nu_1b_1+\nu_2b_2\ge 1$;

\quad(iii) whenever $b_1\in\frak B_1$, $b_2\in\frak B_2$ and
$b_1\Bcap b_2=0$, then $\nu_1b_1+\nu_2b_2\le 1$.

(b) Let $X$, $Y_1$, $Y_2$ be sets, and for $i\in\{1,2\}$ let $\Cal E_i$
be an algebra of subsets of $Y_i$, $\nu_i:\Cal E_i\to[0,1]$ an additive
functional such that $\nu_iY_i=1$, and $f_i:X\to Y_i$ a function.   Then
the following are equiveridical:

\quad(i) there is an additive functional
$\mu:\Cal PX\to[0,1]$ such that $\mu f_i^{-1}[E]=\nu_iE$ whenever
$i\in\{1,2\}$ and $E\in\Cal E_i$;

\quad(ii) $f_1^{-1}[E_1]\cap f_2^{-1}[E_2]\ne\emptyset$ whenever
$E_1\in\Cal E_1$, $E_2\in\Cal E_2$ and $\nu_1E_1+\nu_2E_2>1$;

\quad(iii) $\nu_1E_1\le\nu_2E_2$ whenever $E_1\in\Cal E_1$,
$E_2\in\Cal E_2$ and $f_1^{-1}[E_1]\subseteq f_2^{-1}[E_2]$.

\proof{{\bf (a)(i)$\Rightarrow$(iii)} is elementary (and is a special
case of 457A(i)$\Rightarrow$457A(iii)).

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} If (iii) is true, and
$b_1\in\frak B_1$, $b_2\in\frak B_2$ are such that $b_1\Bcup b_2=1$,
then $(1\Bsetminus b_1)\Bcap(1\Bsetminus b_2)=0$, so

\Centerline{$\nu_1b_1+\nu_2b_2
=2-\nu_1(1\Bsetminus b_1)-\nu_2(1\Bsetminus b_2)\ge 1$.}

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} The point is that (ii) here implies (ii)
of 457A.   \Prf\
Suppose that $i_0,\ldots,i_n\in\{1,2\}$, $a_k\in\frak B_{i_k}$ for
$k\le n$ and $\sum_{k=0}^n\chi a_k\ge m\chi 1$ in $S(\frak A)$, where
$m\in\Bbb N$.   Set $K_j=\{k:k\le n,\,i_k=j\}$ for each $j$,
$u=\sum_{k\in K_1}\chi a_k\in S(\frak B_1)$,
$v=\sum_{k\in K_2}\chi a_k\in S(\frak B_2)$.   Then we can express $u$
as $\sum_{j=0}^{m_1}\chi c_j$ where $c_j\in\frak B_1$ for each
$j\le m_1$ and $c_0\Bsupseteq c_1\Bsupseteq\ldots\Bsupseteq c_{m_1}$
(see the proof of 361Ec).   Taking $c_j=0$ for $m_1<j\le m$ if
necessary, we may suppose that $m_1\ge m$.   Similarly,
$v=\sum_{j=0}^{m_2}\chi d_j$ where $m_2\ge m$, $d_j\in\frak B_2$ for
each $j\le m_2$ and $d_0\Bsupseteq\ldots\Bsupseteq d_{m_2}$.

For $j<m$, set $b_j=1\Bsetminus(c_j\Bcup d_{m-j-1})$.
Then, because $b_j\Bcap c_j=0$,

\Centerline{$u\times\chi b_j
=\sum_{r=0}^{m_1}\chi(c_r\Bcap b_j)
=\sum_{r=0}^{j-1}\chi(c_r\Bcap b_j)
\le j\chi b_j$,}

\noindent and similarly $v\times\chi b_j\le(m-j-1)\chi b_j$, so

\Centerline{$m\chi b_j\le(u+v)\times\chi b_j
=u\times\chi b_j+v\times\chi b_j\le(m-1)\chi b_j$,}

\noindent and $b_j$ must be $0$.

Thus $c_j\Bcup d_{m-j-1}=1$ for every $j<m$.   But this means that
$\nu_1c_j+\nu_2d_{m-j-1}\ge 1$ for every $j<m$, so that

$$\eqalign{\sum_{k=0}^n\nu_{i_k}a_k
&=\sum_{k\in K_1}\nu_1a_k+\sum_{k\in K_2}\nu_2a_k
=\dashint u\,d\nu_1+\dashint v\,d\nu_2\cr
&=\sum_{j=0}^{m_1}\nu_1c_j+\sum_{j=0}^{m_2}\nu_2d_j
\ge\sum_{j=0}^{m-1}\nu_1c_j+\nu_2d_{m-1-j}
\ge m,\cr}$$

\noindent as required.\ \Qed

Because 457A(ii) implies 457A(i), we have the result.

\medskip

{\bf (b)} We can convert (i) and (ii) here into (a-i) and (a-iii) just
above by the same translation as in 457B.   So (i) and (ii) are
equiveridical.   As for (iii), this corresponds exactly to replacing
$E_2$ by $Y_2\setminus E_2$ in (ii).
}%end of proof of 457C

\leader{*457D}{}\cmmnt{ The proof of 457A is based, at some remove,
on the Hahn-Banach theorem, as applied in the proof of 391E-391F.   An
alternative proof uses the max-flow min-cut theorem of graph theory.
To show the power of this method I apply it to an elaboration of 457C,
as follows.

\woddheader{457D}{4}{2}{2}{72pt}

\noindent}{\bf Proposition}\cmmnt{ ({\smc Strassen 65})} Let $\frak A$
be a Boolean algebra and $\frak B_1$, $\frak B_2$ two subalgebras of
$\frak A$.   Suppose that $\nu_i:\frak B_i\to[0,1]$ are finitely
additive functionals such that $\nu_11=\nu_21=1$, and
$\theta:\frak A\to\coint{0,\infty}$ another additive functional.
Then the following are equiveridical:

(i) there is an additive functional $\mu:\frak A\to\coint{0,\infty}$
extending both the $\nu_i$, and such that $\mu a\le\theta a$ for every
$a\in\frak A$;

(ii) $\nu_1b_1+\nu_2b_2\le 1+\theta(b_1\Bcap b_2)$ whenever
$b_1\in\frak B_1$ and $b_2\in\frak B_2$.

\proof{{\bf (a)} As usual in this context, (i)$\Rightarrow$(ii) is
elementary;  if $\mu\le\theta$ extends both $\nu_j$, and $b_j\in\frak
B_j$ for both $j$, then

\Centerline{$\nu_1b_1+\nu_2b_2
=\mu b_1+\mu b_2
=\mu(b_1\Bcup b_2)+\mu(b_1\Bcap b_2)
\le 1+\theta(b_1\Bcap b_2)$.}

\medskip

{\bf (b)} For the reverse implication, suppose to begin with (down to
the end of (d) below) that $\frak A$ is finite.   Let $I$, $J$ and $K$
be the sets of atoms of $\frak B_1$, $\frak B_2$ and $\frak A$
respectively.   Consider the transportation network $(V,E,\gamma)$ where

\Centerline{$V=\{(0,0)\}\cup\{(b,1):b\in I\}\cup\{(d,2):d\in
K\}\cup\{(c,3):c\in J\}\cup\{(1,4)\},$}

\Centerline{$E=\{e^0_b:b\in I\}\cup\{e^1_d:d\in K\}\cup\{e^2_d:d\in
K\}\cup\{e^3_c:c\in J\}$,}

\noindent where

\inset{for $b\in I$, $e^0_b$ runs from $(0,0)$ to $(b,1)$,

for $d\in K$, $e^1_d$ runs from $(b,1)$ to $(d,2)$, where $b$ is the
member of $I$ including $d$,

for $d\in K$, $e^2_d$ runs from $(d,2)$ to $(c,3)$, where $c$ is the
member of $J$ including $d$,

for $c\in J$, $e^3_c$ runs from $(c,3)$ to $(1,4)$.}

\noindent Define the capacity $\gamma(e)$ of each link by setting

\Centerline{$\gamma(e^0_b)=\nu_1b$ for $b\in I$,}

\Centerline{$\gamma(e^1_d)=\gamma(e^2_d)=\theta d$ for $d\in K$,}

\Centerline{$\gamma(e^3_c)=\nu_2c$ for $c\in J$.}

By the max-flow min-cut theorem (4A4N), there are a flow $\phi$ and a
cut $X$ of the same value;  that is, we have a function
$\phi:E\to\coint{0,\infty}$ and a set $X\subseteq E$ such that

\Centerline{$\sum_{e\text{ starts from }v}\phi(e)
=\sum_{e\text{ ends at }v}\phi(e)$}

\noindent for every $v\in V\setminus\{(0,0),(1,4)\}$,

\Centerline{$\phi(e)\le\gamma(e)$}

\noindent for every $e\in E$,

\Centerline{$\sum_{e\text{ starts from }(0,0)}\phi e
=\sum_{e\text{ ends at }(1,4)}\phi e
=\sum_{e\in X}\gamma(e)$,}

\noindent and there is no path from $(0,0)$ to $(1,4)$ using only links
in $E\setminus X$.

Now, for any $d\in K$, there is exactly one link $e^1_d$ ending at $d$
and exactly one link $e^2_d$ starting from $d$.   So
$\phi(e^1_d)=\phi(e^2_d)$, and we may define an additive functional
$\mu$ on $\frak A$ by setting

\Centerline{$\mu a=\sum_{d\in K,d\Bsubseteq a}\phi(e^1_d)
=\sum_{d\in K,d\Bsubseteq a}\phi(e^2_d)$}

\noindent for every $a\in\frak A$.

\medskip

{\bf (c)(i)} $\mu b\le\nu_1b$ for every $b\in\frak B_1$.   \Prf\ Because
$I$ is the set of atoms of the finite Boolean algebra $\frak B_1$, it is
enough to show that $\mu b\le\nu_1b$ for every $b\in I$.   Now, for such
$b$,

$$\eqalign{\mu b
&=\sum_{d\in K,d\Bsubseteq b}\phi(e^1_d)
=\sum_{e\text{ starts from }(b,1)}\phi(e)\cr
&=\sum_{e\text{ ends at }(b,1)}\phi(e)
=\phi(e^0_b)
\le\gamma(e^0_b)
=\nu_1b,\cr}$$

\noindent because the only link ending at $(b,1)$ is $e^0_b$.\ \Qed

\medskip

\quad{\bf (ii)} Similarly, because the only link starting at $(c,3)$ has
capacity $\nu_2c$, $\mu c\le\nu_2c$ for every $c\in J$.   But this means
that $\mu c\le\nu_2c$ for every $c\in\frak B_2$.

\medskip

\quad{\bf (iii)} In third place, because

\Centerline{$\mu d=\phi(e^1_d)\le\gamma(e^1_d)
=\theta d$}

\noindent for every $d\in K$, $\mu a\le\theta a$ for every
$a\in\frak A$.

\medskip

{\bf (d)} (The key.)  $\mu 1\ge 1$.  \Prf\ We have

$$\eqalign{\mu 1
&=\sum_{d\in K}\mu d
=\sum_{d\in K}\phi(e^1_d)\cr
&=\sum_{b\in I}\sum_{d\in K,d\Bsubseteq b}\phi(e^1_d)
=\sum_{b\in I}\sum_{e\text{ starts from }(b,1)}\phi(e)\cr
&=\sum_{b\in I}\sum_{e\text{ ends at }(b,1)}\phi(e)
=\sum_{e\text{ starts from }(0,0)}\phi(e)
=\sum_{e\in X}\gamma(e).\cr}$$

\noindent Set

$$\eqalign{b^*&=\sup\{b:b\in I,\,e^0_b\in X\}\in\frak B_1,\cr
a_1^*&=\sup\{d:d\in K,\,e^1_d\in X\},\cr
a_2^*&=\sup\{d:d\in K,\,e^2_d\in X\},\cr
c^*&=\sup\{c:c\in J,\,e^3_c\in X\}\in\frak B_2.\cr}$$

\noindent For any $d\in K$, we have a four-link path
$e^0_b,e^1_d,e^2_d,e^3_c$ from $(0,0)$ to $(1,4)$, where $b\in I$,
$c\in J$ are the atoms of $\frak B_1$, $\frak B_2$ including $d$.   At
least one of the links in this path must belong to $X$, so that $d$ is
included in $b^*\Bcup a_1^*\Bcup a_2^*\Bcup c^*$.   Thus, writing
$a=(1\Bsetminus b^*)\Bcap(1\Bsetminus c^*)$,
$a\Bsubseteq a_1^*\Bcup a_2^*$ and
$\theta a\le\theta a_1^*+\theta a_2^*$.
But this means that

$$\eqalignno{\mu 1
&=\sum_{e\in X}\gamma(e)\cr
&=\sum_{b\in I,e^0_b\in X}\gamma(e^0_b)
  +\sum_{d\in K,e^1_d\in X}\gamma(e^1_d)
  +\sum_{d\in K,e^2_d\in X}\gamma(e^2_d)
  +\sum_{c\in J,e^3_c\in X}\gamma(e^3_c)\cr
&=\sum_{b\in I,e^0_b\in X}\nu_1b
  +\sum_{d\in K,e^1_d\in X}\theta d
  +\sum_{d\in K,e^2_d\in X}\theta d
  +\sum_{c\in J,e^3_c\in X}\nu_2c\cr
&=\nu_1b^*+\theta a_1^*+\theta a_2^*+\nu_2c^*\cr
\displaycause{remember that $\theta$ is additive}
&\ge\nu_1b^*+\theta((1\Bsetminus b^*)\Bcap(1\Bsetminus c^*))+\nu_2c^*
\ge\nu_1b^*+\nu_1(1\Bsetminus b^*)+\nu_2(1\Bsetminus c^*)-1+\nu_2c^*\cr
\displaycause{applying the hypothesis (ii)}
&=1,\cr}$$

\noindent as claimed.\ \Qed

Since we already know that $\nu_11=1$ and that $\mu b\le\nu_1b$ for
every $b\in\frak B_1$, we must have $\mu 1=1$ and $\mu b=\nu_1b$ for
every $b\in\frak B$, so that $\mu$ extends $\nu_1$.   Similarly, $\mu$
extends $\nu_2$.

\medskip

{\bf (e)} Thus the proposition is proved in the case in which $\frak A$
is finite.   In the general case, for each finite subset $K$ of $\frak
A$ write $\frak A_K$ for the subalgebra of $\frak A$ generated by $K$.
Then (b)-(d) tell us that there is a non-negative additive functional
$\mu_K$ on $\frak A_K$, dominated by $\theta$ on $\frak A_K$, agreeing
with $\nu_1$ on $\frak A_K\cap\frak B_1$ and agreeing with $\nu_2$ on
$\frak A_K\cap\frak B_2$.   Let $\mu$ be any cluster point of the
$\mu_K$ in $[0,1]^{\frak A}$ as $K$ increases through the finite subsets
of $\frak A$;  then $\mu$ will be a non-negative additive functional on
$\frak A$, dominated by $\theta$, and extending $\nu_1$ and $\nu_2$.

This proves the result.
}%end of proof of 457D

\leader{457E}{Proposition} Let $X$ be a non-empty set and
$\familyiI{\nu_i}$ a family of probability measures on $X$ satisfying
the conditions of Lemma 457A, taking $\frak A=\Cal PX$ and
$\frak B_i=\dom\nu_i$ for each $i$.   Suppose that there is a countably
compact class $\Cal K\subseteq\Cal PX$ such that every $\nu_i$ is inner
regular with respect to $\Cal K$.   Then there is a probability measure
$\mu$ on $X$ extending every $\nu_i$.

\proof{ If $I=\emptyset$ this is trivial.   Otherwise, by 457A, there is
a finitely additive functional $\nu$ on $\Cal PX$ extending every
$\nu_i$.   Now 413Ua tells us that there is a complete measure $\mu$ on
$X$ such that $\mu X\le\nu X$ and $\mu K\ge\nu K$ for every
$K\in\Cal K$.   In this case, for any $i\in I$ and
$E\in\Tau_i=\dom\nu_i$, we must have

$$\eqalign{\mu_*E
&\ge\sup_{K\in\Cal K,K\subseteq E}\mu K
\ge\sup_{K\in\Cal K\cap\dom\nu_i,K\subseteq E}\mu K\cr
&\ge\sup_{K\in\Cal K\cap\dom\nu_i,K\subseteq E}\nu K
=\sup_{K\in\Cal K\cap\dom\nu_i,K\subseteq E}\nu_iK
=\nu_iE.\cr}$$

\noindent In particular, $\mu X\ge\nu_iX=1$.   Also $\mu X\le\nu X=1$, so

\Centerline{$\mu^*E=1-\mu_*(X\setminus E)
\le 1-\nu_i(X\setminus E)=\nu_iE$}

\noindent for any $E\in\Tau_i$;  as $\mu$ is complete, $\mu E$ is
defined and equal to $\nu_iE$ for every $E\in\Tau_i$, and $\mu$ extends
$\nu_i$, as required.
}%end of proof of 457E

\leader{457F}{Proposition} (a) Let $(X,\Sigma,\mu)$ be a perfect
probability space and $(Y,\Tau,\nu)$ any probability space.   Write
$\Sigma\otimes\Tau$ for the algebra of subsets of $X\times Y$ generated
by $\{E\times F:E\in\Sigma,\,F\in\Tau\}$.   Suppose that
$Z\subseteq X\times Y$ is such that

\inset{(i) $Z$ is expressible as the intersection of a sequence in
$\Sigma\otimes\Tau$,

(ii) $Z\cap(E\times F)\ne\emptyset$ whenever $E\in\Sigma$, $F\in\Tau$
are such that $\mu E+\nu F>1$.}

\noindent Then there is a probability measure $\lambda$ on $Z$ such that
the maps $(x,y)\mapsto x:Z\to X$ and $(x,y)\mapsto y:Z\to Y$ are both
\imp.

(b) Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of perfect
probability spaces.   Write $\bigotimes_{i\in I}\Sigma_i$ for the
algebra of subsets of $X=\prod_{i\in I}X_i$ generated by
$\{\{x:x\in X,\,x(i)\in E\}: i\in I,\,E\in\Sigma_i\}$.
Suppose that $Z\subseteq X$ is such that

\inset{(i) $Z$ is expressible as the intersection of a sequence in
$\bigotimes_{i\in I}\Sigma_i$,

(ii) whenever $i_0,\ldots,i_n\in I$ and $E_k\in\Sigma_{i_k}$ for
$k\le n$, there is a $z\in Z$ such that
$\#(\{k:k\le n, z(i_k)\in E_k\})\ge\sum_{k=0}^n\mu_{i_k}E_k$.}

\noindent Then there is a perfect probability measure $\lambda$ on $Z$
such that $z\mapsto z(i):Z\to X_i$ is \imp\ for every $i\in I$.

\proof{{\bf (a)} Apply 457Cb to the coordinate maps $f_1:Z\to X$ and
$f_2:Z\to Y$.   The condition (ii) here shows that 457C(b-ii) is
satisfied, so
there is an additive functional $\theta:\Cal PZ\to[0,1]$ such that
$\theta f_1^{-1}[E]=\mu E$ for every $E\in\Sigma$ and
$\theta f_2^{-1}[F]=\nu F$ for every $F\in\Tau$.

Define $\theta':\Sigma\otimes\Tau\to[0,1]$ by setting
$\theta'W=\theta(Z\cap W)$ for every $W\in\Sigma\otimes\Tau$.
Then $\theta'(E\times Y)=\mu E$ for every $E\in\Sigma$ and
$\theta'(X\times F)=\nu F$ for every $F\in\Tau$.   Because $\mu$ is
perfect, $\theta'$ has an extension to a measure $\tilde\lambda$ defined
on $\Sigma\tensorhat\Tau$ (454C).   Now $Z$ is supposed to be
expressible as $\bigcap_{n\in\Bbb N}W_n$ where $W_n\in\Sigma\otimes\Tau$
for every $n$;  since

\Centerline{$\tilde\lambda W_n=\theta'W_n
=\theta(Z\cap W_n)=\theta Z=1$}

\noindent for every $n$, $\tilde\lambda Z=1$.   So if we take $\lambda$
to be the subspace measure on $Z$ induced by $\tilde\lambda$, $\lambda$
will be a probability measure on $Z$.   If $E\in\Sigma$, then

$$\eqalign{\lambda(Z\cap(E\times Y))
&=\tilde\lambda(Z\cap(E\times Y))
=\tilde\lambda(E\times Y)\cr
&=\theta'(E\times Y)
=\theta(Z\cap(E\times Y))
=\mu E.\cr}$$

\noindent So $f_1:Z\to X$ is \imp\ for $\lambda$ and $\mu$.
Similarly, $f_2:Z\to Y$ is \imp\ for $\lambda$ and $\nu$.

\medskip

{\bf (b)} We use the same ideas, but appealing to 457B and 454D instead
of 457Cb and 454C.   Taking $f_i:X\to X_i$ to be the coordinate map for
each $i\in I$, (ii) here, with 457B, tells us that there is an additive
functional $\theta:\Cal PZ\to[0,1]$ such that
$\theta f_i^{-1}[E]=\mu_iE$ whenever $i\in I$ and $E\in\Sigma_i$.

Define $\theta':\bigotimes_{i\in I}\Sigma_i\to[0,1]$ by setting
$\theta'W=\theta(Z\cap W)$ for every $W\in\bigotimes_{i\in I}\Sigma_i$.
Then

\Centerline{$\theta'\{x:x\in X,\,x(i)\in E\}
=\theta\{z:z\in Z,\,z(i)\in E\}=\mu_iE$}

\noindent whenever $i\in I$ and $E\in\Sigma_i$.   Because every $\mu_i$
is perfect, $\theta'$ has an extension to a perfect measure
$\tilde\lambda$ defined on $\Tensorhat_{i\in I}\Sigma_i$ (454D).   Now
$Z$ is supposed to be expressible as $\bigcap_{n\in\Bbb N}W_n$ where
$W_n\in\bigotimes_{i\in I}\Sigma_i$ for every $n$;  since

\Centerline{$\tilde\lambda W_n=\theta'W_n
=\theta(Z\cap W_n)=\theta Z=1$}

\noindent for every $n$, $\tilde\lambda Z=1$.   So if we take $\lambda$
to be the subspace measure on $Z$ induced by $\tilde\lambda$, $\lambda$
will be a probability measure on $Z$;  by 451Dc, $\lambda$ is perfect.
If $i\in I$ and $E\in\Sigma_i$, then

$$\eqalign{\lambda\{z:z\in Z,\,z(i)\in E\}
&=\tilde\lambda\{x:x\in X,\,x(i)\in E\}
=\theta'\{x:x\in X,\,x(i)\in E\}\cr
&=\theta\{z:z\in Z,\,z(i)\in E\}
=\mu_iE.\cr}$$

\noindent So $z\mapsto z(i):Z\to X_i$ is \imp\ for $\lambda$ and $\mu_i$
for every $i\in I$, as required.
}%end of proof of 457F

\leader{457G}{Theorem} Let $X$ be a set and $\familyiI{\mu_i}$ a
family of probability measures on $X$ which is upwards-directed in the
sense that for any $i$, $j\in I$ there is a $k\in I$ such that $\mu_k$
extends both $\mu_i$ and $\mu_j$.   Suppose that for any countable
$J\subseteq I$ there is a measure on $X$ extending $\mu_i$ for every
$i\in J$.   Then there is a measure on $X$ extending $\mu_i$ for every
$i\in I$.

\proof{ Set $\Sigma_i=\dom\mu_i$ for each $i\in I$.   Because
$\familyiI{\mu_i}$ is upwards-directed,
$\Tau=\bigcup_{i\in I}\Sigma_i$ is an algebra of subsets of $X$, and we
have a finitely additive functional $\nu:\Tau\to[0,1]$ defined by saying
that $\nu E=\mu_iE$ whenever $i\in I$ and $E\in\Sigma_i$.   Now if
$\sequencen{E_n}$ is any non-increasing sequence in $\Tau$ with empty
intersection, there is a countable set $J\subseteq I$ such that
$E_n\in\bigcup_{i\in J}\Sigma_i$ for every $n\in\Bbb N$.   We are
told that there is a measure $\lambda$ on $X$ extending $\mu_i$ for
every $i\in J$;  now $\nu E_n=\lambda E_n$ for every $n\in\Bbb N$, so
$\lim_{n\to\infty}\nu E_n=0$.   By 413Lb, $\nu$ has an extension to a
measure on $X$, which of course extends every $\mu_i$.
}%end of proof of 457G

\leader{457H}{Example} Set $X=\{(x,y):0\le x<y\le 1\}\subseteq[0,1]^2$.
Write $\pi_1$, $\pi_2:X\to\Bbb R$ for the coordinate maps, and $\mu_L$
for Lebesgue measure on $[0,1]$, with $\Sigma_L$ its domain.

\spheader 457Ha There is a finitely additive functional
$\nu:\Cal PX\to[0,1]$ such that $\nu\pi_i^{-1}[E]=\mu_LE$ whenever
$i\in\{1,2\}$ and $E\in\Sigma_L$.   \prooflet{\Prf\ If $E_1$,
$E_2\in\Sigma_L$ and $\mu_LE_1+\mu_LE_2>1$, then neither is empty and
$\inf E_1<\sup E_2$, so there are $x\in E_1$, $y\in E_2$ such that
$x<y$, and
$(x,y)\in\pi_1^{-1}[E_1]\cap\pi_2^{-1}[E_2]$.   So the result follows by
457Cb.\ \Qed}

\spheader 457Hb However, there is no measure $\mu$ on $X$ for which both
$\pi_1$ and $\pi_2$ are \imp.
\prooflet{\Prf\Quer\ If there were,

\Centerline{$\int\pi_1(x,y)\mu(d(x,y))
=\int x\mu_L(dx)
=\int y\mu_L(dy)
=\int\pi_2(x,y)\mu(d(x,y))$}

\noindent by 235G;  but $\pi_1(x,y)<\pi_2(x,y)$ for every $(x,y)\in X$,
so this is impossible.\ \Bang\Qed}

\cmmnt{\spheader 457Hc If we write
$\Tau_i=\{\pi_i^{-1}[E]:E\subseteq[0,1]$ is Borel$\}$ for each $i$, then
we have a measure $\nu_i$ with domain $\Tau_i$ defined by setting
$\nu_i\pi_i^{-1}[E]=\mu_LE$ for each Borel set $E\subseteq[0,1]$.   Now
$\nu_1$ and $\nu_2$ have no common extension to a Borel measure on $X$,
even though $X$ is a Polish space and each
$\nu_i$ is a compact measure, being inner regular with respect to the
compact class $\Cal K_i=\{\pi_i^{-1}[K]:K\subseteq\ooint{0,1}$ is
compact$\}$.   (The trouble is that $\Cal K_1\cup\Cal K_2$ is {\it not}
compact, so we cannot apply 457E.)}

\leader{457I}{Example} Let $\mu_L$ be Lebesgue measure on $[0,1]$ and
$\Sigma_L$ its domain.   Set

\Centerline{$X=\{(\xi_1,\xi_2,\xi_3):0\le\xi_i\le 1$ for each $i$,
$\sum_{i=1}^3\xi_i\le\Bover32$, $\sum_{i=1}^3\xi_i^2\le 1\}$.}

\noindent For $1\le i\le 3$ set $\pi_i(x)=\xi_i$ for
$x=(\xi_1,\xi_2,\xi_3)\in X$.

\spheader 457Ia If $E_i\in\Sigma_L$ for $i\le 3$, then there is an
$x\in X$ such that $\#(\{i:\pi_i(x)\in E_i\})\ge\sum_{i=1}^3\mu_LE_i$.
\prooflet{\Prf\ Set $\alpha_i=\inf(E_i\cup\{1\})$ for each $i$, and set

\Centerline{$m=\lceil\sum_{i=1}^3\mu_LE_i\rceil
\le\lceil\sum_{i=1}^31-\alpha_i\rceil
=3-\lfloor\sum_{i=1}^3\alpha_i\rfloor$,}

\noindent so that $\sum_{i=1}^3\alpha_i<4-m$.   Take
$\xi_i\in E_i\cup\{1\}$ such that $\sum_{i=1}^3\xi_i<4-m$.
It will be enough to consider the case in which $\xi_1\le\xi_2\le\xi_3$.

\medskip

\quad{\bf (i)} If $m=1$, then $\sum_{i=1}^3\xi_i<3$ so $\xi_1<1$ and
$\xi_1\in E_1$.   Set $x=(\xi_1,0,0)$;  then $x\in X$ and

\Centerline{$\#(\{i:\pi_i(x)\in E_i\})\ge 1\ge\sum_{i=1}^3\mu_LE_i$.}

\medskip

\quad{\bf (ii)} If $m=2$, then $\sum_{i=1}^3\xi_i<2$ so $\xi_2<1$ and
$\xi_1\in E_1$, $\xi_2\in E_2$.   Set $x=(\xi_1,\xi_2,0)$.   We have
$\xi_1+\xi_2\le\bover43\le\bover32$.   Also

\Centerline{$\xi_2\le\Bover12(\xi_2+\xi_3)\le 1-\Bover12\xi_1$,}

\noindent so

\Centerline{$\xi_1^2+\xi_2^2\le\xi_1^2+(1-\Bover12\xi_1)^2
=1-\xi_1+\Bover54\xi_1^2\le 1$}

\noindent because $\xi_1\le\bover23\le\bover45$.   So $x\in X$ and

\Centerline{$\#(\{i:\pi_i(x)\in E_i\})\ge 2\ge\sum_{i=1}^3\mu_LE_i$.}

\medskip

\quad{\bf (iii)} If $m=3$ then $\sum_{i=1}^3\xi_i<1$ so $\xi_i\in E_i$
for every $i$;  set $x=(\xi_1,\xi_2,\xi_3)$.   Since
$\sum_{i=1}^3\xi_i^2\le\sum_{i=1}^3\xi_i\le 1$, $x\in X$ and

\Centerline{$\#(\{i:\pi_i(x)\in E_i\})=3\ge\sum_{i=1}^3\mu_LE_i$.}

Putting these together, we have the result.\ \Qed}

\spheader 457Ib There is no finitely additive functional $\nu$ on $X$
such that $\nu\pi_i^{-1}[E]=\mu_LE$ for each $i$ and every
$E\in\Sigma_L$.
\prooflet{\Prf\Quer\ Suppose there were.   Set
$\Tau_i=\{\pi_i^{-1}[E]:E\in\Sigma_L\}$ and $\nu_i=\nu\restrp\Tau_i$ for
each $i$.   Then $\nu_i$ is a probability measure on $X$;  moreover,
because $X$ is compact, $\pi_i^{-1}[K]$ is compact for every compact
$K\subseteq[0,1]$, so $\nu_i$ is inner regular with respect to the
compact subsets of $X$.   By 457E, the $\nu_i$ have a common extension
to a countably additive measure $\mu$.   Now

\Centerline{$\int_X\xi_1+\xi_2+\xi_3\,\mu(dx)=3\int_0^1t\,dt
=\Bover32$,}

\noindent so we must have $\xi_1+\xi_2+\xi_3=\Bover32$ for $\mu$-almost
every $x$;  similarly,

\Centerline{$\int_X\xi_1^2+\xi_2^2+\xi_3^2\,\mu(dx)
=3\int_0^1t^2\,dt=1$,}

\noindent so we must have $\xi_1^2+\xi_2^2+\xi_3^2=1$ for $\mu$-almost
every $x$.   Since

\Centerline{$(\Bover32-\xi_3)^2
=(\xi_1+\xi_2)^2\le 2(\xi_1^2+\xi_2^2)\le 2(1-\xi_3^2)$}

\noindent for almost every $x$, $\xi_3-\xi_3^2\ge\bover1{12}$ for almost
every $x$, which is impossible, since
$\mu\{x:\xi_3\le\Bover12-\Bover1{\sqrt 6}\}>0$.\
\Bang\Qed}%end of prooflet

\leader{457J}{Example} There are a set $X$ and a family
$\familyiI{\mu_i}$ of probability measures on $X$ such that (i) for
every countable set $J\subseteq I$ there is a measure on $X$ extending
$\mu_i$ for every $i\in J$ (ii) there is no measure on $X$ extending
$\mu_i$ for every $i\in I$.

\proof{ By 439Fc, there is an uncountable universally negligible subset
of $[0,1]$.   Because $[0,1]$ and $\Cal P\Bbb N$ are uncountable Polish
spaces, they have isomorphic Borel structures (424Cb), so there is an
uncountable universally negligible set
$X_0\subseteq\Cal P\Bbb N$.   The map $a\mapsto\Bbb N\setminus a$ is an
autohomeomorphism of $\Cal P\Bbb N$, so
$X_1=\{\Bbb N\setminus a:a\in X_0\}$ is universally negligible, and
$X=X_0\cup X_1$ is universally negligible (439Cb).

For $n\in\Bbb N$, set $E_n=\{a:n\in a\in X\}$ and
$\Sigma_n=\{\emptyset,E_n,X\setminus E_n,X\}$;  note that, because $X$
is closed under complementation, neither $E_n$ nor $X\setminus E_n$ is
empty, and we have a probability measure $\mu_n$ with domain $\Sigma_n$
defined by setting $\mu_nE_n=\mu_n(X\setminus E_n)=\bover12$.   Next,
for $a\in X$, set $\Sigma'_a=\{\emptyset,\{a\},X\setminus\{a\},X)$, and
let $\mu'_a$ be the probability measure with domain $\Sigma'_a$ defined
by setting $\mu'_a\{a\}=0$.

If $J\subseteq X$ is countable, then there is a probability measure on
$X$ extending $\mu_n$ for every $n\in\Bbb N$ and $\mu'_a$ for every
$a\in J$.   \Prf\ Because $X_0$ is uncountable, there is a $b\in X_0$
such that neither $b$ nor $b'=\Cal P\Bbb N\setminus b$ belongs to $J$.
Let $\mu$ be the probability measure with domain $\Cal PX$ defined by
setting $\mu\{b\}=\mu\{b'\}=\bover12$;  this extends all the $\mu_n$ and
all the $\mu'_a$ for $a\in J$.\ \Qed

\Quer\ Suppose, if possible, that $\mu$ is a measure on $X$ extending
every $\mu_n$ and every $\mu'_a$.   In this case, because $\mu$ extends
every $\mu_n$, its domain includes the Borel $\sigma$-algebra $\Cal B$
of $X$, and $\mu\restr\Cal B$ is a Borel probability measure on $X$.
Since $X$ is universally negligible, there is a point $a\in X$ such that
$\mu\{a\}>0$;  in which case $\mu$ cannot extend $\mu'_a$.\ \Bang

Thus the $\mu_n$, $\mu'_a$ constitute a family of the kind required.
}%end of proof of 457J

\leader{457K}{}\cmmnt{ In addition to existence, we can ask for
solutions to simultaneous-extension problems
which are optimal in some sense;
some transportation problems can be interpreted as questions of this kind.
In this direction I give just one result, which is also connected to
the ideas of \S437.\footnote{I am indebted to J.Pachl for leading me to
this material.}

\medskip

\noindent}{\bf Definition}\dvAnew{2009}\cmmnt{ ({\smc Bogachev 07},
\S8.10(viii))}
Let $(X,\rho)$ be a metric space.   For
quasi-Radon probability measures $\mu$, $\nu$ on $X$, set

\Centerline{$\rho_{\text{W}}(\mu,\nu)
=\sup\{|\int u\,d\mu-\int u\,d\nu|:
u:X\to\Bbb R$ is bounded and $1$-Lipschitz$\}$.}

\cmmnt{\noindent (Compare the metric $\rhoKR$ of 437Qb.  $\rho_{\text{W}}$
is sometimes called the `Wasserstein metric'.)}
%see "Wasserstein metric" or "Monge-Wasserstein metric"
% or"Vasershtein metric" ({\smc Dudley 76},
%Wikipedia
%R.L. Dobrushin,
%"Prescribing a system of random variables by conditional distributions"
%Theor. Probab. Appl. , 15  (1970)  pp. 458-486)
%Vasershtein 69
%Hutchinson 81, \S4.3 calls  \rho_{\text{W}}  the  "L-metric"
%thanks to J.Pachl for the lead

\leader{457L}{Theorem}\dvAnew{2009} Let $(X,\rho)$ be a metric space and
$P_{\text{qR}}$ the set of
quasi-Radon probability measures on $X$;  define $\rho_{\text{W}}$ as in
457K.

(a) For all $\mu$, $\nu$ and $\lambda$ in $P_{\text{qR}}$,

\Centerline{$\rho_{\text{W}}(\mu,\nu)=\rho_{\text{W}}(\nu,\mu)$,
\quad$\rho_{\text{W}}(\mu,\lambda)
  \le\rho_{\text{W}}(\mu,\nu)+\rho_{\text{W}}(\nu,\lambda)$,}

\Centerline{$\rho_{\text{W}}(\mu,\nu)=0$ iff $\mu=\nu$.}

(b)\cmmnt{ (cf.\ {\smc Vasershtein 69})} If $\mu$,
$\nu\in P_{\text{qR}}$, then $\rho_{\text{W}}(\mu,\nu)
=\inf_{\lambda\in Q(\mu,\nu)}\int\rho(x,y)\lambda(d(x,y))$, where
$Q(\mu,\nu)$ is the set of quasi-Radon probability measures on $X\times X$
with marginal measures $\mu$ and $\nu$.
%"Kantorovich-Rubinstein theorem" ({\smc Dudley 76}, 11.8.2)

(c) In (b), if $\mu$ and $\nu$ are Radon measures, $Q(\mu,\nu)$ is included
in $P_{\text{R}}(X\times X)$, the space of Radon probability measures on
$X\times X$, and is compact for the narrow topology on
$P_{\text{R}}(X\times X)$;  and there
is a $\lambda\in Q(\mu,\nu)$ such that
$\rho_{\text{W}}(\mu,\nu)=\int\rho(x,y)\lambda(d(x,y))$.

(d) If $\rho$ is bounded, then  $\rho_{\text{W}}$ is a metric on
$P_{\text{qR}}$ inducing
the narrow topology\cmmnt{ (definition:  437Jd)}.

\proof{{\bf (a)} The first two clauses are immediate from the definition.
For the third, observe that if $\mu\ne\nu$ then
$\rho_{\text{W}}(\mu,\nu)\ge\rhoKR(\mu,\nu)>0$ by 437R.

\medskip

{\bf (b)} Write $\zeta\in[0,\infty]$ for $\rho_{\text{W}}(\mu,\nu)$,
$\eusm L^{\infty}_{\dom\mu}$ for the space of bounded
$\dom\mu$-measurable functions from $X$ to $\Bbb R$ and
$\eusm L^{\infty}_{\dom\nu}$ for the space of bounded
$\dom\nu$-measurable functions from $X$ to $\Bbb R$.

\medskip

\quad{\bf (i)} We have

$$\eqalign{\zeta
&=\sup\{\int u\,d\mu+\int v\,d\nu:
   u\in\eusm L^{\infty}_{\dom\mu},\,v\in\eusm L^{\infty}_{\dom\nu},\cr
&\mskip150mu u(x)+v(y)\le\rho(x,y)\text{ for all }x,\,y\in X\}.\cr}$$

\noindent\Prf\grheada\ Suppose that $u\in\eusm L^{\infty}_{\dom\mu}$,
$v\in\eusm L^{\infty}_{\dom\nu}$ and
$u(x)+v(y)\le\rho(x,y)$ for all $x$, $y\in X$.   Set

\Centerline{$w(x)=\inf_{y\in X}\rho(x,y)-v(y)$}

\noindent for $x\in X$.   Then $u(x)\le w(x)$ and $w(x)+v(x)\le 0$ for
every $x$, so $u\le w\le -v$ and $w$ is bounded;  also $w$ is
$1$-Lipschitz, because if $x$, $x'\in X$ then

\Centerline{$w(x)-\rho(x,x')=\inf_{y\in X}\rho(x,y)-v(y)-\rho(x,x')
\le\inf_{y\in X}\rho(x',y)-v(y)=w(x')$.}

\noindent Accordingly

\Centerline{$\int u\,d\mu+\int v\,d\nu
\le\int w\,d\mu-\int w\,d\nu\le\zeta$.}

\noindent\grheadb\ In the other direction, given $\gamma<\zeta$,
there is a
bounded $1$-Lipschitz function $u:X\to\Bbb R$ such that
$|\int u\,d\mu-\int u\,d\nu|\ge\gamma$.   Replacing $u$ by $-u$ if
necessary, we can arrange that $\int u\,d\mu-\int u\,d\nu\ge\gamma$.
Now set $v=-u$;  then
$u(x)+v(y)\le\rho(x,y)$ for all $x$, $y$, and
$\int u\,d\mu+\int v\,d\nu\ge\gamma$.\ \Qed

It follows that if $u\in\eusm L^{\infty}_{\dom\mu}$,
$v\in\eusm L^{\infty}_{\dom\nu}$ and
$u(x)+v(y)\le\beta\rho(x,y)$ for
all $x$, $y\in X$, where $\beta>0$, then

\Centerline{$\int u\,d\mu+\int v\,d\nu
=\beta(\int\Bover1{\beta}u\,d\mu+\int\Bover1{\beta}v\,d\nu)\le\beta\zeta$.}

\medskip

\quad{\bf (ii)} $\int\rho\,d\lambda\ge\zeta$ for every
$\lambda\in Q(\mu,\nu)$.   \Prf\ If
$u\in\eusm L^{\infty}_{\dom\mu}$,
$v\in\eusm L^{\infty}_{\dom\nu}$
and $u(x)+v(y)\le\rho(x,y)$ for all $x$, $y\in X$, then

$$\eqalignno{\int u\,d\mu+\int v\,d\nu
&=\int u(x)\lambda(d(x,y))+\int v(y)\lambda(d(x,y))\cr
\displaycause{235G}
&\le\int\rho\,d\lambda}$$

\noindent so (i) gives us the result.\ \Qed

If $\zeta=\infty$, we can stop;  so henceforth suppose that $\zeta$ is
finite.

\medskip

\quad{\bf (iii)} Define $p:\ell^{\infty}(X\times X)\to\coint{0,\infty}$ by
setting

\Centerline{$p(w)=\inf\{\alpha+\beta\zeta:\alpha$, $\beta>0$,
$w(x,y)\le\alpha+\beta\rho(x,y)$ for all $x$, $y\in X\}$.}

\noindent Then $p(w+w')\le p(w)+p(w')$ and $p(\alpha w)=\alpha p(w)$
whenever $w$, $w'\in\ell^{\infty}(X\times X)$ and
$\alpha\in\coint{0,\infty}$.   For $u$, $v\in\BbbR^X$ define
$u\otimes v\in\BbbR^{X\times X}$ by setting $(u\otimes v)(x,y)=u(x)v(y)$
for all $x$, $y\in X$ (cf.\ 253B);  set

\Centerline{$V=\{(u\otimes\chi X)+(\chi X\otimes v):
u\in\eusm L^{\infty}_{\dom\mu}$,
$v\in\eusm L^{\infty}_{\dom\nu}\}$.}

\noindent Let $\mu\times\nu$ be the quasi-Radon product measure on
$X\times X$ (417R).
Then we have a linear functional $h_0:V\to\Bbb R$ defined by saying
that $h_0(w)=\int w\,d(\mu\times\nu)$ for $w\in V$.   The point is that
$h_0(w)\le p(w)$ for every $w\in V$.   \Prf\ We have
$u\in\eusm L^{\infty}_{\dom\mu}$,
$v\in\eusm L^{\infty}_{\dom\nu}$ such
that $w(x,y)=u(x)+v(y)$ for all $x$, $y\in X$.   If $\alpha$, $\beta>0$
are such that $w(x,y)\le\alpha+\beta\rho(x,y)$ for all $x$, $y\in X$, set
$u_0(x)=u(x)-\alpha$ for every $x$;  then $u_0(x)+v(y)\le\beta\rho(x,y)$
for all $x$ and $y$, so

$$\eqalign{h_0(w)
&=\int u\otimes\chi X\,d(\mu\times\nu)
  +\int\chi X\otimes v\,d(\mu\times\nu)
=\int u\,d\mu+\int v\,d\nu\cr
&=\alpha+\int u_0\,d\mu+\int v\,d\nu
\le\alpha+\beta\zeta\cr}$$

\noindent by the last remark in (i).
As $\alpha$ and $\beta$ are arbitrary, $h_0(w)\le p(w)$.\ \Qed

\medskip

\quad{\bf (iv)} By the Hahn-Banach theorem (3A5Aa), there is a linear
functional $h:\ell^{\infty}(X\times X)\to\Bbb R$, extending $h_0$, such
that $h(w)\le p(w)$ for every $w\in\ell^{\infty}(X\times X)$.
In this case, $h$ must be a positive linear functional, because if $w\ge 0$
then $p(-w)=0$, so $h(-w)\le 0$.   Since also

\Centerline{$h(\chi(X\times X))
=h_0(\chi(X\times X))=(\mu\times\nu)(X\times X)=1$,}

\noindent $\|h\|=1$ in $\ell^{\infty}(X\times X)^*$.   If $u$,
$v\in C_b(X)$ then

\Centerline{$h(u\otimes\chi X)=h_0(u\otimes\chi X)=\int u\,d\mu$,
\quad$h(\chi X\otimes v)=h_0(\chi X\otimes v)=\int v\,d\nu$.}

\noindent
%If $\gamma\ge 0$, then
%$h(\rho\wedge\gamma\chi(X\times X))\le p(\rho)\le 1$.
Let $\theta:\Cal P(X\times X)\to[0,1]$ be the additive functional
defined by setting $\theta W=h(\chi W)$ for $W\subseteq X\times X$.
Observe that $\theta(E\times X)=\mu E$ for every $E\in\dom\mu$
and $\theta(X\times E)=\nu E$ for every $E\in\dom\nu$.

\medskip

\quad{\bf (v)} Because both $\mu$ and $\nu$ are inner regular with
respect to the totally bounded sets (434L), there is a separable subset $Y$
of $X$ such that $\mu Y=\nu Y=1$, and we can take $Y$ to be a Borel set.
Now let $\epsilon>0$.   Then we have a
countable partition $\familyiI{E_i}$ of $Y$ into non-empty Borel sets of
diameter at most $\epsilon$.   For $i$, $j\in I$, set

$$\eqalign{\alpha_{ij}
&=\Bover{\theta(E_i\times E_j)}{\mu E_i\,\nu E_j}
\text{ if }\mu E_i\cdot\nu E_j>0,\cr
&=0\text{ otherwise}.\cr}$$

\noindent Since $\theta(E_i\times E_j)\le\min(\mu E_i,\nu E_j)$,
$\theta(E_i\times E_j)=\alpha_{ij}\mu E_i\nu E_j$.   If $i\in I$ is
such that $\mu E_i>0$, then $\sum_{j\in I}\alpha_{ij}\nu E_j=1$.   \Prf\
For any $\eta>0$ there is a finite $K_0\subseteq I$ such that
$\nu(X\setminus\bigcup_{j\in K_0}E_j)\le\eta$.   Now

$$\eqalign{|1-\sum_{j\in K}\alpha_{ij}\nu E_j|\mu E_i
&=|\mu E_i-\sum_{j\in K}\theta(E_i\times E_j)|
=|\theta(E_i\times X)-\theta(E_i\times\bigcup_{j\in K}E_j)|\cr
&=\theta(E_i\times(X\setminus\bigcup_{j\in K}E_j))
\le\theta(X\times(X\setminus\bigcup_{j\in K}E_j))\cr
&=\nu(X\setminus\bigcup_{j\in K}E_j)
\le\eta\cr}$$

\noindent whenever $K$ is a finite subset of $I$ including $K_0$;
as $\eta$ is arbitrary,
$\mu E_i\cdot\sum_{j\in I}\alpha_{ij}\nu E_j=\mu E_i$
and $\sum_{j\in I}\alpha_{ij}\nu E_j=1$.\ \QeD\  Similarly,
$\sum_{i\in I}\alpha_{ij}\mu E_i=1$ whenever $\nu E_j>0$.

\medskip

\quad{\bf (vi)} Define a Borel measurable function
$w_0:X\times X\to\coint{0,\infty}$ by setting

$$\eqalign{w_0(x,y)
&=\alpha_{ij}\text{ if }i,\,j\in I,\,x\in E_i\text{ and }y\in E_j,\cr
&=0\text{ if }(x,y)\in(X\times X)\setminus(Y\times Y).\cr}$$

\noindent Let $\lambda$ be the indefinite-integral measure
over $\mu\times\nu$ defined by $w_0$;  then $\lambda$ is a quasi-Radon
probability
measure with marginals $\mu$, $\nu$.   \Prf\ If $E\in\dom\mu$, then

$$\eqalignno{\lambda(E\times X)
&=\int_{E\times X}w_0d(\mu\times\nu)
=\sum_{i,j\in I}\int_{(E\cap E_i)\times E_j}w_0d(\mu\times\nu)\cr
&=\sum_{i,j\in I}\alpha_{ij}\mu(E\cap E_i)\cdot\nu E_j
=\sum_{i\in I}\mu(E\cap E_i)\cr
\displaycause{because $\sum_{j\in I}\alpha_{ij}\nu E_j=1$ whenever
$\mu E_i>0$}
&=\mu E.\cr}$$

\noindent In particular, $\lambda(X\times X)=1$, so $\lambda$ is a
probability measure, and is quasi-Radon by 415Ob;  and the coordinate
projection $(x,y)\mapsto x$ is \imp\ for $\lambda$ and $\mu$.   To see that
$\mu$ is exactly the image measure, observe that if $E\subseteq X$ is such
that $\lambda(E\times X)$ is defined, then $(E\cap E_i)\times E_j$ must
be measured by $\mu\times\nu$ whenever $\alpha_{ij}>0$.   For any $i\in I$
such that $\mu E_i>0$, there is surely some $j$ such that $\alpha_{ij}>0$,
in which case $E\cap E_i\in\dom\mu$;  since $\bigcup_{i\in I}E_i$ is
$\mu$-conegligible (and $\mu$ is complete and $I$ is countable),
$E\in\dom\mu$.   Thus $\mu$ is the marginal of $\lambda$ on the first
coordinate.   Similarly, $\nu$ is the marginal of $\lambda$ on the second
coordinate.\ \Qed

For $i$, $j\in I$ we have

\Centerline{$\lambda(E_i\times E_j)
=\alpha_{ij}\mu E_i\cdot\nu E_j=\theta(E_i\times E_j)$.}

\medskip

\quad{\bf (vii)} $\int\rho\,d\lambda\le\zeta+2\epsilon$.   \Prf\
For $i$, $j\in I$, set

\Centerline{$\beta_{ij}=\inf_{x\in E_i,y\in E_j}\rho(x,y)$;}

\noindent set

$$\eqalign{w(x,y)
&=\beta_{ij}\text{ if }i,\,j\in I,\,x\in E_i\text{ and }y\in E_j,\cr
&=0\text{ if }(x,y)\in(X\times X)\setminus(Y\times Y).\cr}$$

\noindent Then

\Centerline{$w\le\rho\times\chi(Y\times Y)
\le w+2\epsilon\chi(X\times X)$,}

\noindent so

$$\eqalign{\int\rho\,d\lambda
&=\int_{Y\times Y}\rho\,d\lambda
\le 2\epsilon+\int w\,d\lambda\cr
&=2\epsilon+\sum_{i,j\in I}\beta_{ij}\lambda(E_i\times E_j)
=2\epsilon+\sum_{i,j\in I}\beta_{ij}\theta(E_i\times E_j).\cr}$$

\noindent Now, for any finite $K\subseteq I$,

\Centerline{$\sum_{i,j\in K}\beta_{ij}\theta(E_i\times E_j)
=h(w\times\chi(\bigcup_{i,j\in K}E_i\times E_j))
\le h(\rho)
\le p(\rho)\le\zeta$}

\noindent by the definition of $p$.
So $\int\rho\,d\lambda\le 2\epsilon+\zeta$, as claimed.\ \Qed

\medskip

\quad{\bf (viii)} As $\epsilon$ is arbitrary,

\Centerline{$\inf_{\lambda\in Q(\mu,\nu)}\int\rho\,d\lambda\le\zeta$.}

\noindent With (ii), this completes the proof of (b).

\medskip

{\bf (c)} For every $\epsilon>0$, there is a compact set $K\subseteq X$
such that $\mu(X\setminus K)+\nu(X\setminus K)\le\epsilon$.   In this case
$\lambda((X\times X)\setminus(K\times K))\le\epsilon$ for every
$\lambda\in Q(\mu,\nu)$.   In the first place, this shows that if
$\lambda\in Q(\mu,\nu)$, then $\lambda$ is a Radon measure, by 416C(iv).
Thus $Q(\mu,\nu)\subseteq P_{\text{R}}(X\times X)$.   Next, we
see also that $Q(\mu,\nu)$ is uniformly tight (437O), therefore relatively
compact in the space $M^+_{\text{R}}(X\times X)$ of totally finite Radon
measures on $X\times X$ (437P).

Writing $\pi_1$, $\pi_2$ for the coordinate projections from $X\times X$ to
$X$, we see that

\Centerline{$Q(\mu,\nu)
=\{\lambda:\lambda\in M_{\text{R}}^+(X\times X)$, $\lambda\pi_1^{-1}=\mu$
and $\lambda\pi_2^{-1}=\nu\}$.}

\noindent Since the functions $\lambda\mapsto\lambda\pi_1^{-1}$ and
$\lambda\mapsto\lambda\pi_2^{-1}$ from $M^+_{\text{R}}(X\times X)$ to
$M^+_{\text{R}}(X)$ are continuous (437N), and $M^+_{\text{R}}(X)$ is
Hausdorff in its narrow topology (437R(a-ii)), $Q(\mu,\nu)$ is closed in
$M^+_{\text{R}}(X\times X)$, therefore compact.

Finally, the function $\lambda\mapsto\int\rho\,d\lambda$
from $M^+_{\text{R}}(X\times X)$ to $[0,\infty]$ is lower
semi-continuous (437Jg), and must
attain its infimum on the compact set $Q(\mu,\nu)$ (4A2B(d-viii)).
But (b) tells us that this infimum is just $\rho_{\text{W}}(\mu,\nu)$.

\medskip

{\bf (d)(i)} Suppose first that $\rho(x,y)\le 2$ for all $x$, $y\in X$.
Then $\rho_{\text{W}}
=\rhoKR\restr P_{\text{qR}}\times P_{\text{qR}}$.
\Prf\ As already noted in (a),
$\rho_{\text{W}}(\mu,\nu)\ge\rhoKR(\mu,\nu)$ for all $\mu$, $\nu\in P_{\text{qR}}$.
In the other direction, if $\mu$, $\nu\in P_{\text{qR}}$ and $u:X\to\Bbb R$ is
$1$-Lipschitz, then $|u(x)-u(y)|\le 2$ for all $x$, $y\in X$, so there is
an $\alpha\in\Bbb R$ such that $|u(x)-\alpha|\le 1$ for all $x\in X$.
Set $v(x)=u(x)-\alpha$ for every $x$;  then $v:X\to[-1,1]$ is
$1$-Lipschitz, so

$$\eqalignno{|\int u\,d\mu-\int u\,d\nu|
&=|\int v\,d\mu-\int v\,d\nu|\cr
\displaycause{because $\mu X=\nu X$}
&\le\rhoKR(\mu,\nu).\cr}$$

\noindent As $u$ is arbitrary,
$\rho_{\text{W}}(\mu,\nu)\le\rhoKR(\mu,\nu)$ and
the two metrics are equal.\ \Qed

\medskip

\quad{\bf (ii)} In general, take $\gamma>0$ such that
$\rho(x,y)\le 2\gamma$ for all
$x$, $y\in X$.   Set $\sigma=\Bover1{\gamma}\rho$, so that $\sigma$
is a metric on $X$ equivalent to $\rho$.   Now $\sigma_{\text{KR}}$
defines the narrow topology on $P_{\text{qR}}$, by 437R(g-i), so
$\rho_{\text{W}}=\gamma\sigma_{\text{W}}
=\gamma\sigma_{\text{KR}}\restr P_{\text{qR}}\times P_{\text{qR}}$
also does.
}%end of proof of 457L

\leader{457M}{}\cmmnt{ If we relax our demands, and look for measures
dominated by each measure in a family rather than extending them,
similar methods give further results.

\medskip

\noindent}{\bf Theorem}\dvAnew{2010}\cmmnt{ (see {\smc Kellerer 84})}
Let $X$ be a Hausdorff space and $\familyiI{\nu_i}$ a non-empty
finite family of locally finite
measures on $X$ all inner regular with respect to the closed sets.

(a) For $A\subseteq X\times\coint{0,\infty}$, set

$$\eqalign{c(A)
&=\inf\{\sum_{i\in I}\int h_id\nu_i:
h_i:X\to[0,\infty]\text{ is }\dom\nu_i\text{-measurable
for each }i\in I,\cr
&\mskip250mu \alpha\le\sum_{i\in I}h_i(x)\text{ whenever }
  (x,\alpha)\in A\}.\cr}$$

\quad(i) $c$ is a Choquet capacity\cmmnt{ (definition:  432J)}.

\quad(ii)
For every $A\subseteq X\times\coint{0,\infty}$, the infimum in the
definition of $c(A)$ is attained.

(b) Let $f:X\to\coint{0,\infty}$ be a function such that
$\{x:f(x)\ge\alpha\}$ is K-analytic for every $\alpha>0$.   Then

$$\eqalignno{&\inf\{\sum_{i\in I}\int h_id\nu_i:
h_i:X\to[0,\infty]\text{ is }\dom\nu_i\text{-measurable
for each }i\in I,\,f\le\sum_{i\in I}h_i\}\cr
&\mskip75mu=\sup\{\int f\,d\mu:\mu\text{ is a Radon measure on }X
  \text{ and }\mu\le\nu_i\text{ for every }i\in I\}\dvro{.}{,}\cr}$$

\cmmnt{\noindent where `$\mu\le\nu_i$' here is to be interpreted in
the sense of 234P.}

\proof{{\bf (a)(i)}\grheada\ For $f:X\to[0,\infty]$ set

\Centerline{$\Omega_f=\{(x,\alpha):x\in X$, $\alpha\le f(x)\}$,
\quad$\Omega'_f=\{(x,\alpha):x\in X$, $\alpha<f(x)\}$}

\noindent as in 252N.

It will be convenient to amalgamate the $\nu_i$ into a
single measure, as follows.   Let $(Y,\Tau,\nu)$ be the direct sum of the
family
$\familyiI{(X_i,\nu_i)}$ in the sense of 214L, so that
$Y=X\times I$ and $\nu E=\sum_{i\in I}\nu_i\{x:(x,i)\in E\}$ for those
$E\subseteq Y$ for which the sum is defined.   Give $Y$ its disjoint-union
topology, that is, the product topology if $I$ is given the discrete
topology;  then it is easy to check that
$\nu$ is locally finite (see 411Xh) and inner regular with respect to
the closed sets (see 412Xp).   For $h\in[0,\infty]^Y$ and $x\in X$ set
$(Th)(x)=\sum_{i\in I}h(x,i)$;  observe that $T(h+h')=Th+Th'$ and
$T(\alpha h)=\alpha Th$ for all $h$, $h':Y\to[0,\infty]$ and $\alpha\ge 0$.
Now, for any $A\subseteq X\times\coint{0,\infty}$, we have

$$\eqalignno{c(A)
&=\inf\{\int h\,d\nu:
  h:Y\to[0,\infty]\text{ is }\Tau\text{-measurable},\cr
&\mskip200mu \alpha\le Th(x)\text{ whenever }(x,\alpha)\in A\}\cr
\displaycause{because $\int h\,d\nu=\sum_{i\in I}\int h(x,i)\nu_i(dx)$
for non-negative $h$, by 214M}
&=\inf\{\int h\,d\nu:
  h:Y\to[0,\infty]\text{ is }\Tau\text{-measurable},\,
  A\subseteq\Omega_{Th}\}.\cr}$$

\medskip

\qquad\grheadb\ Of course $c:\Cal P(X\times\coint{0,\infty})\to[0,\infty]$
is non-decreasing.   To see that it is sequentially order-continuous on the
left, I show in fact that if $\sequencen{A_n}$ is a
non-decreasing sequence of subsets of $X\times[0,\infty]$ with union
$A$, and $\gamma=\sup_{n\in\Bbb N}c(A_n)$ is finite, then there is
a $\Tau$-measurable $h:Y\to[0,\infty]$ such that $\alpha\le Th(x)$ whenever
$(x,\alpha)\in A$ and $\int h\,d\nu=\gamma$.   \Prf\
Surely $c(A)\ge\gamma$.   For each $n\in\Bbb N$
we have a $\Tau$-measurable $h_n:Y\to[0,\infty]$ such that
$\int h_nd\nu\le\gamma+2^{-n}$ and $A_n\subseteq\Omega_{Th_n}$.   By
Koml\'os's theorem (276H), there is a strictly increasing sequence
$\sequence{k}{n(k)}$ in $\Bbb N$ such that
$\lim_{m\to\infty}\Bover1{m+1}\sum_{k=0}^mh_{n(k)}$ is defined $\nu$-a.e.;
set $h=\limsup_{m\to\infty}\Bover1{m+1}\sum_{k=0}^mh_{n(k)}$.   Then
$h:Y\to[0,\infty]$ is $\Tau$-measurable, and
$h\eae\liminf_{m\to\infty}\Bover1{m+1}\sum_{k=0}^mh_{n(k)}$.
By Fatou's Lemma,

\Centerline{$\int h\,d\nu
\le\liminf_{m\to\infty}\Bover1{m+1}\sum_{k=0}^m\int h_{n(k)}d\nu
\le\gamma$,}

\noindent while if $j\in\Bbb N$ and $(x,\alpha)\in A_j$.
$\alpha\le Th_{n(k)}(x)$ for every $k\ge j$, so

$$\eqalignno{\alpha
&\le\liminf_{m\to\infty}\Bover1{m+1}\sum_{k=0}^mTh_{n(k)}(x)
\le\limsup_{m\to\infty}\sum_{i\in I}\Bover1{m+1}\sum_{k=0}^mh_{n(k)}(x,i)\cr
&\le\sum_{i\in I}\limsup_{m\to\infty}
  \Bover1{m+1}\sum_{k=0}^mh_{n(k)}(x,i)\cr
\displaycause{because $I$ is finite}
&=\sum_{i\in I}h(x,i)
=Th(x).\cr}$$

\noindent Thus $A\subseteq\Omega_{Th}$, so

\Centerline{$c(A)\le\int h\,d\nu\le\gamma\le c(A)$}

\noindent and we have equality.\ \Qed

\medskip

\qquad\grheadc\ Now suppose that $K\subseteq X\times\coint{0,\infty}$ is
compact, and $\epsilon>0$.   Set $L=\pi_1[K]$, where
$\pi_1:X\times\coint{0,\infty}\to X$ is the canonical map;  then
$L\subseteq X$ and $L\times I\subseteq Y$ are compact.   Because $\nu$ is
locally finite, there is an open set $H\subseteq Y$ such that
$L\times I\subseteq H\in\Tau$ and $\nu H$ is finite (see 411Ga).
Let $\nu_H$ be the subspace measure induced by $\nu$ on $H$, and $\Tau_H$
its domain;  then $\nu_H$
is totally finite and inner regular with respect to the closed sets
(412Pc), therefore outer regular with respect to the open sets (411D).
Let $h:Y\to[0,\infty]$ be a $\Tau$-measurable function such that
$A\subseteq\Omega_{Th}$ and $\int h\,d\nu\le c(K)+\epsilon$.   Set
$h_1(y)=h(y)+\Bover{\epsilon}{\nu H}$ for $y\in H$;  then
$\int_Hh_1d\nu_H\le c(K)+2\epsilon$.   By 412Wa, there is a lower
semi-continuous $\Tau_H$-measurable $g_1:H\to[0,\infty]$ such that
$h_1\le g_1$ and $\int_Hg_1d\nu_H\le c(K)+3\epsilon$.   Extend $g_1$ to a
function $g:Y\to[0,\infty]$ by setting $g(y)=0$ for $y\in Y\setminus H$;
then $g$ is $\Tau$-measurable and lower semi-continuous and
$\int g\,d\nu\le c(K)+3\epsilon$.   Moreover, if $(x,\alpha)\in K$, then

$$\eqalignno{Tg(x)
&>T(h\times\chi H)(x)
=Th(x)\cr
\displaycause{because $\{x\}\times I\subseteq H$}
&\ge\alpha,\cr}$$

\noindent so $K\subseteq\Omega'_{Tg}$.

The point is that $\Omega'_{Tg}$ is open in $X\times\coint{0,\infty}$.
\Prf\ If $x\in X$ and
$0\le\alpha<Tg(x)=\sum_{i\in I}g(x,i)$, let $\familyiI{\alpha_i}$ be such
that $0\le\alpha_i<g(x,i)$ for each $i\in I$ and
$\sum_{i\in I}\alpha_i=\alpha'>\alpha$.
Set $G=\bigcap_{i\in I}\{z:z\in X$, $g(z,i)>\alpha_i\}$;  then $G$ is an
open subset of $X$, and
$(x,\alpha)\in G\times\coint{0,\alpha'}\subseteq\Omega'_{Tg}$.   Thus
$(x,\alpha)\in\interior\Omega'_g$;  as $(x,\alpha)$ is arbitrary,
$\Omega'_{Tg}$ is open.\ \Qed

Since $c(\Omega'_{Tg})$ is surely less than or equal to $\int g\,d\nu$,
and $\epsilon$ is arbitrary, we have

\Centerline{$c(K)=\inf\{c(U):U\subseteq X\times\coint{0,\infty}$ is open
and $K\subseteq U\}$.}

\noindent Thus all the conditions of 432Ja are satisfied, and $c$ is a
Choquet capacity.

\medskip

\quad{\bf (ii)} We need consider only the case $c(A)<\infty$,
which is dealt with in (i-$\beta$) above, if we
take $A_n=A$ for every $n$.

\medskip

{\bf (b)(i)} For $g:X\to[-\infty,\infty]$, set

$$\eqalignno{p(g)
&=\inf\{\sum_{i\in I}\int h_id\nu_i:
h_i\in[0,\infty]^X\text{ is }\dom\nu_i\text{-measurable
for each }i\in I,\cr
&\mskip400mu |g|\le\sum_{i\in I}h_i\}\cr
&=\inf\{\int h\,d\nu:h:Y\to[0,\infty]\text{ is }\Tau\text{-measurable},\,
  \Omega_{|g|}\subseteq\Omega_{Th}\}
%&=\inf\{\int h\,d\nu:h:Y\to[0,\infty]\text{ is }\Tau\text{-measurable},\,
%  \Omega'_{|g|}\subseteq\Omega_{Th}\}\cr
=c(\Omega_{|g|})%
%=c(\Omega'_{|g|})
.\cr}$$

\noindent Then $p(\alpha g)=|\alpha|p(g)$
whenever $g\in[-\infty,\infty]^X$ and $\alpha\in\Bbb R$,
$p(g_1)\le p(g_2)$ whenever $|g_1|\le|g_2|$,
and $p(g_1+g_2)\le p(g_1)+p(g_2)$ for
all $g_1$, $g_2:X\to[-\infty,\infty]$;  so if we set
$V=\{g:g\in\BbbR^X$, $p(g)<\infty\}$, $V$ is a
solid linear subspace of $\BbbR^X$
and $p\restr V$ is a seminorm.

Suppose that $\mu$ is a Radon measure on $X$ and $\mu\le\nu_i$
for every $i\in I$.   Then $\int fd\mu\le p(f)$.
\Prf\ Because $\mu$ measures every K-analytic set (432A),
$\int fd\mu$ is defined.   If $p(f)=\infty$ then of course
$\int fd\mu\le p(f)$.
Otherwise, for any $\gamma>p(f)$, we have $\dom\nu_i$-measurable functions
$h_i:X\to[0,\infty]$ such that $f\le\sum_{i\in I}h_i$ and
$\sum_{i\in I}\int h_id\nu_i\le\gamma$.   But now $\int h_id\mu$ is defined
and less than or equal to $\int h_id\nu_i$ for each $i$
(234Qc), so
$\int fd\mu\le\sum_{i\in I}\int h_id\mu\le\gamma$.   As $\gamma$ is
arbitrary, $\int fd\mu\le p(f)$.\ \Qed

\medskip

\quad{\bf (ii)}\grheada\
In the other direction, suppose that $\gamma<p(f)$, and
set $A=\{(x,\alpha):0<\alpha<f(x)\}$;   then

\Centerline{$A=\bigcup_{q\in\Bbb Q}\{(x,\alpha):f(x)\ge q>\alpha>0\}$}

\noindent is K-analytic (422Ge,
%countable prod of K-analytic is K-analytic
422Hc,
%K-analytics closed under Souslin opn
423Bb,
%cts image of analytic is analytic
423C).
%analytic iff K-analytic + ctble network
On the other hand,
for any $h:Y\to[0,\infty]$, $A\subseteq\Omega_{Th}$ iff
$\Omega_f\subseteq\Omega_{Th}$.   So

\Centerline{$c(A)=c(\Omega_f)=p(f)>\gamma$.}

\noindent By Choquet's theorem 432K, there is a compact set $K\subseteq A$
such that $c(K)>\gamma$.   Set

\Centerline{$f_1(x)=\sup(\{0\}\cup K[\{x\}])$}

\noindent for $x\in X$.   As in
(a-i-$\gamma$) above, we have for any $i\in I$ an open set $G$ including
$L_0=\pi_1[K]$ such that $\nu_iG$ is
defined and finite, so $\chi L_0$ and $f_1$ belong to $V$.   By the
Hahn-Banach theorem (4A4Da), there is a linear functional
$\theta:V\to\Bbb R$ such that $|\theta(g)|\le p(g)$ for every $g\in V$ and
$\theta(f_1)=p(f_1)$.   Since $|\theta(g)|\le p(g_0)$ whenever
$|g|\le g_0$, $\theta$ is
order-bounded, and if $\theta^+$ is its positive part (355Eb),
we shall still have $\theta^+(g)\le p(g)$ for every $g\in V$ and
$\theta^+(f_1)=p(f_1)$.

\medskip

\qquad\grheadb\ Set $\mu_0C=\theta^+(\chi(C\cap L_0))$ for $C\subseteq X$.   Then
$\mu_0:\Cal PX\to\coint{0,\infty}$ is additive.   By 416K, there is a Radon
measure $\mu$ on $X$ such that $\mu L\ge\mu_0L$ for every compact
$L\subseteq X$ and $\mu G\le\mu_0G$ for every open $G\subseteq X$.
Now $\dom\nu_i\subseteq\dom\mu$ for every $i\in I$.
\Prf\ Suppose that $E\in\dom\nu_i$.    Let $L\subseteq X$ be
compact.   Then there is an open set $G_0\supseteq L$ such that $\nu_iG_0$
is defined and finite.   Take any $\delta>0$.
Because the subspace measure induced by $\nu_i$ on $G_0$ is
totally finite and inner regular with respect to the closed sets,
there are a closed set $F$ and an open set $G$, both
measured by $\nu_i$, such that $F\subseteq E\cap G_0\subseteq G$ and
$\nu_i(G\setminus F)\le\delta$.  In this case

\Centerline{$\mu(G\setminus F)\le\mu_0(G\setminus F)
=\theta^+(\chi(L_0\cap G\setminus F))
\le p(\chi(G\setminus F))
\le\int\chi(G\setminus F)d\nu_i\le\delta$.}

\noindent So

\Centerline{$\mu^*(E\cap G_0)
\le\mu G
\le\mu F+\delta
\le\mu_*(E\cap G_0)+\delta$;}

\noindent as $\delta$ is arbitrary, $\mu^*(E\cap G_0)=\mu_*(E\cap G_0)$
and $\mu$ measures $E\cap G_0$ (Ef), and therefore also measures
$E\cap L=E\cap G_0\cap L$.   As
$L$ is arbitrary, $\mu$ measures $E$ (412Ja).\ \Qed

In fact, $\mu\le\nu_i$.   \Prf\ If $\nu_i$ measures $E$ and $L\subseteq X$
is compact, the arguments just above show that for any $\delta>0$
there is an open set
$G\supseteq E\cap L$ such that $\nu_iG\le\nu_iE+\delta$, so
that

\Centerline{$\mu(E\cap L)\le\mu G\le\mu_0G=\theta^+(\chi(G\cap L_0))
\le p(\chi(G\cap L_0))\le\nu_iG\le\nu_iE+\delta$.}

\noindent As $L$ and $\delta$ are arbitrary, $\mu E\le\nu_iE$.\ \Qed

\medskip

\qquad\grheadc\ To estimate $\int fd\mu$, recall that
$\theta^+(f_1)>\gamma$, while $\theta^+(\chi L_0)$ is finite.
There is therefore an $\eta>0$ such that
$\eta\theta^+(\chi L_0)\le\theta^+(f_1)-\gamma$ and
$\theta^+(f_2)\ge\gamma$, where
$f_2=(f_1-\eta\chi L_0)^+=(f_1-\eta\chi X)^+$.
For $k\in\Bbb N$
set $F_k=\pi_1[K\cap\coint{(k+1)\eta,\infty}\,]$, so that each
$F_k$ is a compact subset of $L_0$ and
$f_2\le\sum_{k=0}^m\eta\chi F_k\le f$,
where $m\in\Bbb N$ is such that
$K\subseteq X\times[0,m\eta]$.   Now

$$\eqalign{\gamma
&\le\theta^+(f_2)
\le\theta^+(\sum_{k=0}^m\eta\chi F_k)
=\eta\sum_{k=0}^m\theta^+(\chi F_k)\cr
&=\eta\sum_{k=0}^m\mu_0F_k
\le\eta\sum_{k=0}^m\mu F_k
\le\int fd\mu.\cr}$$

\qquad\grheadd\ As $\gamma$ is arbitrary,

\Centerline{$\sup\{\int fd\mu:
\mu\text{ is a Radon measure on }X,\,\mu\le\nu_i
\text{ for every }i\in I\}
\ge p(f)$}

\noindent and we must have equality.   This completes the proof.
}%end of proof of 457M

\cmmnt{
\leader{457N}{Remarks} It may not be quite obvious how close the domination
requirement `$\mu\le\nu_i$ for every $i\in I$' is to the marginal
requirement `$\nu_i=\mu\pi_i^{-1}$ for every $i\in I$', so I spell out the
correspondence.   Let
$\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of probability
spaces, $X=\prod_{i\in I}X_i$, and $\pi_i:X\to X_i$ the canonical map
for each $i$.

\spheader 457Na For each $i\in I$ we have a (unique)
pull-back probability measure $\nu_i$ on $X$ with domain
$\{\pi_i^{-1}[E]:E\in\Sigma_i\}$ such that the image measure
$\nu_i\pi_i^{-1}$ is $\mu_i$ (see
234F).   Now it is elementary to check that, for a measure
$\mu$ on $X$, $\mu\le\nu_i$ iff $\mu\pi_i^{-1}\le\mu_i$;  and if $\mu$ is
required to be a probability measure, then $\mu\le\nu_i$ iff $\mu$ extends
$\nu_i$ iff $\mu\pi_i^{-1}$ extends $\mu_i$.

\spheader 457Nb We find also that if $\mu\le\nu_i$ for
every $i$, then there is a probability measure $\mu'$ on $X$ such that
$\mu\le\mu'$ and $\mu'$ extends $\nu_i$ for every $i$.
\Prf\ Set $\gamma=\mu X$.   If
$\gamma=1$, set $\mu'=\mu$.   Otherwise, for each
$i\in I$, set $\lambda_iE=\Bover1{1-\gamma}(\mu_iE-\mu\pi_i^{-1}[E])$ for
$E\in\Sigma_i$.   Then $\lambda_i$ is a probability measure on $X_i$;  let
$\lambda=\prod_{i\in I}\lambda_i$ be the product measure, and set
$\mu'=\mu+(1-\gamma)\lambda$.   Then

\Centerline{$\mu'\pi_i^{-1}
=\mu\pi_i^{-1}+(1-\gamma)\lambda\pi_i^{-1}
=\mu\pi_i^{-1}+(1-\gamma)\lambda_i=\mu_i$}

\noindent and $\mu'$ extends $\nu_i$ for each $i$.\ \Qed

\spheader 457Nc In the simplest intended applications, therefore, in which
we have two Radon probability spaces $(X_1,\Sigma_1,\mu_1)$ and
$(X_2,\Sigma_2,\mu_2)$ and a profit function
$f:X\to\coint{0,\infty}$, and we are looking for a Radon
probability measure $\mu$ on $X=X_1\times X_2$, with marginals $\mu_1$ and
$\mu_2$, maximising $\int fd\mu$, then we can seek to apply 457Mb with the
pull-back measures $\nu_1$ and $\nu_2$ of (a) here to see that the optimum
is

\Centerline{$\inf\{\int h_1d\mu_1+\int h_2d\mu_2:
f(x_1,x_2)\le h_1(x_1)+h_2(x_2)\Forall x_1\in X_1,\,x_2\in X_2\}$.}

\noindent If the process of part (b-ii) of the proof of 457M leads to a
more or less optimal
measure $\mu$ which is not itself a probability measure, we can increase
it to $\mu'$ with $\mu'\pi_i^{-1}$ extending $\mu_i$ for each $i$;  and in
this case we shall have $\mu'\pi_i^{-1}=\mu_i$ for each $i$, by 418I and
416E, as usual.   Of course we shall need to confirm that $\int fd\mu'$ is
defined, but in the context of 457Mb, this will automatically be so.

\spheader 457Nd There is an obvious parallel between the formulae of 457M
and that in part (b-i) of the proof of 457L.   Allowing for the change of
direction, where an infimum in 457L corresponds to a supremum in 457M,
the pattern of the duality is the same in both cases, and there is some
overlap (457Xq).   But the arguments of the two
theorems -- in particular, the proofs that we can get countably
additive measures from the finitely additive measures provided by the
Hahn-Banach theorem -- are rather different.
}%end of comment

\exercises{\leader{457X}{Basic exercises (a)}
%\spheader 457Xa
Let $X$ be a non-empty set and $\familyiI{\nu_i}$ a family of
probability measures on $X$ satisfying the conditions of Lemma 457A,
taking $\frak A=\Cal PX$ and $\frak B_i=\dom\nu_i$ for each $i$.
Suppose that there is a totally finite measure $\theta$ on $X$ such that
$\theta E$ is defined and greater than or equal to $\nu_iE$ whenever
$i\in I$ and $\nu_i$ measures $E$.   Show that there is a measure on $X$
extending every $\nu_i$.   \Hint{391E.}
%457A

\spheader 457Xb Find a set $X$ and non-negative additive functionals
$\mu_1$, $\mu_2$ defined on subalgebras of $\Cal PX$ which agree on
$\dom\mu_1\cap\dom\mu_2$ but have
no common extension to a non-negative additive functional.   \Hint{take
$\#(X)=3$.}
%457C

\spheader 457Xc Let $\frak A$ be a Boolean algebra and
$\familyiI{\nu_i}$ a family of non-negative finitely additive
functionals, each $\nu_i$ being defined on a subalgebra $\frak B_i$ of
$\frak A$.   Show that if any finite number of the $\nu_i$ have a common
extension to an additive functional on a subalgebra of $\frak A$, then
the whole family has a common extension to an additive functional on the
whole algebra $\frak A$.
%457Xb 457C

\spheader 457Xd Set $X=\{0,1,2\}$ and in the algebra $\Cal PX$ let
$\frak B_i$ be the subalgebra $\{\emptyset,\{i\},X\setminus\{i\},X\}$ for
each $i$.   Let $\nu_i:\frak B_i\to[0,1]$ be the additive functional such
that $\nu_i\{i\}=\bover12$, $\nu_iX=1$.   Show that any pair of $\nu_0$,
$\nu_1$, $\nu_2$ have a common extension to an additive functional on
$\Cal PX$, but that the three together have no such extension.
%457C

\spheader 457Xe Let $\frak A$ be a Boolean algebra, $\frak B$ a
subalgebra of $\frak A$, and $\nu:\frak B\to\coint{0,\infty}$,
$\theta:\frak A\to\coint{0,\infty}$ additive functionals such that
$\nu b\le\theta b$ for every $b\in\frak B$.   Show directly, without
using either 457D or 391F, that there is an additive functional
$\mu:\frak A\to\coint{0,\infty}$, extending $\nu$, such that
$\mu a\le\theta a$ for every $a\in\frak A$.   \Hint{first consider the
case in which $\frak A$ is the algebra generated by $\frak B\cup\{c\}$.}
%457D

\sqheader 457Xf Let $(Y_1,\frak S_1,\Tau_1,\nu_1)$ and
$(Y_2,\frak S_2,\Tau_2,\nu_2)$ be Radon probability spaces and
$X\subseteq Y_1\times Y_2$ a closed set.   Show that the following are
equiveridical:  (i) there is a measure on $X$ such that the coordinate
map from $X$ to $Y_i$ is \imp\ for both $i$;  (ii) there is a Radon
measure on $X$ such that the coordinate map from $X$ to $Y_i$ is \imp\
for both $i$;  (iii) for every compact $K\subseteq Y_1$,
$\nu_1K\le\nu^*_2(X[K])$.   \Hint{for (iii)$\Rightarrow$(ii), use 457C
to show that there is a finitely additive functional $\nu$ on $\Cal PX$
of the required type;  now observe that $\nu$ must give large mass to
compact subsets of $X$, and apply 413U.}
%457E

\sqheader 457Xg Suppose that $\frak A$ is a Boolean algebra, $\frak B$
is a subalgebra of $\frak A$ and $I\subseteq\frak A$ a finite set;  let
$\frak C$ be the subalgebra of $\frak A$ generated by $I\cup\frak B$ and
$\nu:\frak C\to\coint{0,\infty}$ a finitely additive functional.   (i)
Show that if
$\nu\restr\frak B$ is completely additive then $\nu$ is completely
additive.
(ii) Show that if $\frak A$ is Dedekind $\sigma$-complete,
$\frak B$ is a $\sigma$-subalgebra and
$\nu\restr\frak B$ is countably additive then $\nu$ is countably additive.
%457E

\spheader 457Xh Let $(X,\Sigma,\mu)$ be a probability space, $\Cal A$ a
finite family of subsets of $X$ and $\Tau$ the subalgebra of $\Cal PX$
generated by $\Sigma\cup\Cal A$.   Show that if $\nu:\Tau\to[0,1]$ is a
finitely additive functional extending $\mu$, then $\nu$ is countably
additive.
%457Xg 457E

\spheader 457Xi Let $(X,\Sigma,\mu)$ be a probability space,
$\familyiI{A_i}$ a partition of $X$ and $\familyiI{\alpha_i}$ a
family in $[0,1]$ summing to $1$.   Show that the following are
equiveridical:  (i) there is a measure $\nu$ on $X$, extending $\mu$, such
that $\nu A_i=\alpha_i$ for every $i\in I$;
(ii) there is a finitely additive functional $\nu:\Cal PX\to[0,1]$,
extending $\mu$, such that $\nu A_i=\alpha_i$ for every $i\in I$;
(iii) $\mu_*(\bigcup_{i\in J}A_i)\le\sum_{i\in J}\alpha_i$ for every
$J\subseteq I$;  (iv) $\mu^*(\bigcup_{i\in J}A_i)\ge\sum_{i\in J}\alpha_i$
for every finite $J\subseteq I$.   \Hint{for (ii)$\Rightarrow$(i) use
457Xh.}
%457Xhm 457E

\spheader 457Xj Let $X\subseteq[0,1]^2$ be a Lebesgue measurable set
such that $X\cap(E\times F)$ is not negligible for any non-negligible
sets $E$, $F\subseteq[0,1]$.   (For the construction of such sets, see
the notes to \S325.)   Show that there is a Radon measure on $X$ such
that both the coordinate projections from $X$ to $[0,1]$ are \imp, where
$[0,1]$ is given Lebesgue measure.   \Hint{show that there is a
measure-preserving bijection $\phi$ between conegligible subsets of
$[0,1]$ which is covered by $X$;   $\phi$ can be taken to be of the form
$\phi(x)=x-\alpha_n$ for $x\in E_n$.}
%457H

\spheader 457Xk Set
$X=\{(t,2t):0\le t\le\bover12\}\cup\{(t,2t-1):\bover12\le t\le 1\}$.
Show that there is a Radon measure on $X$ for which both the coordinate
maps onto $[0,1]$ are \imp, but that $X$ does not include the graph of
any measure-preserving bijection between conegligible subsets of
$[0,1]$.
%457H

\spheader 457Xl Let $X$ be the eighth-sphere
$\{x:x\in[0,1]^3,\,\|x\|=1\}$.   Show that there is a measure on $X$
such that all three coordinate maps from $X$ onto $[0,1]$ are \imp.
\Hint{265Xe.}
%457I

\spheader 457Xm Set $X=\{x:x\in[0,1]^3,\,\xi_1+\xi_2+\xi_3=\bover32\}$.
Show that there is a measure on $X$ such that all the coordinate maps
from $X$ onto $[0,1]$ are \imp.   \Hint{note that $X$ is a regular
hexagon;  try one-dimensional Hausdorff measure on its boundary.}
%457I

\spheader 457Xn Explain how to adapt the example in 457J to provide a
family $\familyiI{\mu_i}$ of probability measures on a set $X$ such that
(i) $\familyiI{\mu_i}$ is upwards-directed, in the sense of 457G
(iii) there is no measure on
$X$ extending $\mu_i$ for every $i\in I$.
%457J

\spheader 457Xo\dvAnew{2009} Let $X$ be a topological space and
$P_{\text{qR}}$ the set of quasi-Radon probability measures on $X$.
For $\mu$,
$\nu\in P_{\text{qR}}$, write $Q(\mu,\nu)$ for the set of quasi-Radon
probability measures on $X\times X$ which have marginal measures
$\mu$ on the first copy of $X$, $\nu$ on the second.   (i) For
a bounded continuous pseudometric $\rho$ on $X$, set
$\rho_{\text{W}}(\mu,\nu)
=\inf\{\int\rho(x,y)\lambda(d(x,y)):\lambda\in Q(\mu,\nu)\}$.   Show that
$\rho_{\text{W}}$ is a pseudometric on $P_{\text{qR}}$.
(ii) Show that if $X$ is completely regular and $\Rho$ is a family of
bounded pseudometrics defining
the topology of $X$, then $\{\rho_{\text{W}}:\rho\in\Rho\}$ defines the
narrow topology of $P_{\text{qR}}$.
%457L

\spheader 457Xp\dvAnew{2010} Suppose that $X$, $\familyiI{\nu_i}$ and
$c:\Cal P(X\times\coint{0,\infty})\to[0,\infty]$ are as in 457M.
(i) Show that $c$ is a
submeasure.   (ii) Show that if every $\nu_i$ is outer
regular with respect to the open sets, then $c$ is an outer regular Choquet
capacity.
%457M

\spheader 457Xq\dvAnew{2010} Show that if the metric $\rho$ is bounded,
then 457Lc can be deduced from 457Mb and part (b-i) of the proof of 457L.
%457N

\spheader 457Xr\dvAnew{2010} Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\le n}$ be a finite
family of Radon probability spaces,
$X=\prod_{i\in I}X_i$, and $f:X\to\Bbb R$ a bounded Baire measurable
function.   Show that

$$\eqalign{&\inf\{\int fd\mu:\mu\text{ is a Radon measure on }X
\text{ with marginal measure }\mu_i\text{ on each }X_i\}\cr
&\mskip50mu
=\sup\{\sum_{i=0}^n\int h_id\mu_i:h_i\in\ell^{\infty}(X_i)\text{ is
}\Sigma_i\text{-measurable for each }i,\cr
&\mskip200mu
\sum_{i=0}^nh_i(\xi_i)\le f(x)\text{ whenever }
x=(\xi_0,\ldots,\xi_n)\in X\}.\cr}$$

\noindent\Hint{reduce to the case in which every $X_i$ is K$_{\sigma}$.}
%457N

\leader{457Y}{Further exercises (a)}
%\spheader 457Ya
Show that for any $n\ge 2$ there are a finite set $X$
and a family $\langle\mu_i\rangle_{i\le n}$ of measures on $X$ such that
$\{\mu_i:i\le n,\,i\ne j\}$ have a common extension to a measure on $X$
for every $j\le n$, but the whole family $\{\mu_i:i\le n\}$ has no such
extension.
%457Xb 457C

\spheader 457Yb Show that the example in 457H has the property:  if
$f_i$ is a $\nu_i$-integrable real-valued function for each $i$, and
$\int f_1d\nu_1+\int f_2d\nu_2<1$, then there is an
$(x,y)\in\dom f_1\cap\dom f_2$ such that $f_1(x,y)+f_2(x,y)<1$.
%457H

\spheader 457Yc Suppose we replace the set $X$ in 457H with
$X'=X\cup\{(x,x):x\in[0,\bover12]\}$, and write $\nuprime_i$ for the
measures on $X'$ defined by the coordinate projections.   Show that
(i) if $f_i$ is a $\nuprime_i$-integrable real-valued function on $X'$
for
each $i$, and $\int f_1d\nuprime_1+\int f_2d\nuprime_2\le 1$, then there
is an
$(x,y)\in\dom f_1\cap\dom f_2$ such that $f_1(x,y)+f_2(x,y)\le 1$
(ii) there is no measure on $X'$ extending both $\nuprime_i$.
%457H

\spheader 457Yd In 457Xm, show that there are many Radon measures on
$X$ such that all the coordinate maps from $X$ onto $[0,1]$ are \imp.
%457Xm, 457I

\spheader 457Ye\dvAnew{2010}
Give an example of a compact Hausdorff space $X$, a
sequence $\sequencen{\nu_n}$ of tight probability measures on $X$,
and a K$_{\sigma}$ set $E\subseteq X$ such that

\Centerline{$\inf\{\sum_{n=0}^{\infty}\int h_nd\nu_n:
\chi E\le\sum_{n=0}^{\infty}h_n\}=1$,}

\Centerline{$\sup\{\mu E:\mu\text{ is a Radon measure on }X
  \text{ and }\mu\le\nu_n\text{ for every }n\in\Bbb N\}\le\Bover12$.}
%mt45bits 457M
}%end of exercises

\leaveitout{if $(Z,\nu)$ is the Stone space of the measure algebra
of Lebesgue measure, is the product measure on $Z\times Z$ inner regular
with respect to $\Cal PZ\tensorhat\Cal PZ$?}

\leader{457Z}{Problems} Give $[0,1]$ Lebesgue measure.

\spheader 457Za Characterize the sets $X\subseteq[0,1]^2$
for which there is a measure on $X$ such that both the projections from
$X$ to $[0,1]$ are \imp.

\spheader 457Zb Set $X=\{x:x\in[0,1]^3,\,\|x\|=1\}$.   Is there more
than one Radon measure on $X$ for which all three coordinate maps from $X$
onto $[0,1]$ are \imp?   \cmmnt{(See 457Xl, 457Yd.)}

\endnotes{
\Notesheader{457}
In the context of this section, as elsewhere (compare
391E-391G %391E 391F 391G
and 391J),
finitely additive extensions, as in 457A-457D, generally present
easier problems than countably additive extensions.   So techniques for
turning additive functionals into measures (391D, 413L, 413U, 416K,
454C, 454D, 457E, 457G, 457Lb, 457Mb, 457Xi) are very valuable.
Note that 457D offers possibilities in this
direction:  if $\theta$ there is countably additive, $\mu$ also will be
(457Xa).

457H and 457J demonstrate obstacles which can arise when seeking countably
additive extensions even when finitely additive extensions give no
difficulty.   For finitely additive extensions a problem can arise at
any finite number of measures (see 457Ya), but there is no further
obstruction with infinite families (457Xc).   For countably additive
measures we have a positive result (457G) only under very restricted
circumstances;  relaxing any of the hypotheses can lead to failure
(457J, 457Xn).   Even in the apparently concrete case in which we have
an open or closed set $X\subseteq[0,1]^2$ and we are seeking a measure
on $X$ with
prescribed image measures on each coordinate, there can be surprises
(457H, 457Xj, 457Xk), and I know of no useful description of the sets
for which such a measure can be found (457Za).

The two-dimensional case has a special feature:  when verifying the
conditions (ii) or (iii) in 457A, or the condition (ii) of 457B, it is
enough to consider only one set associated with each coordinate (457C).
Put another way, in conditions (iv) and (v) of 457A it is enough to
examine indicator functions.   This is not the case as soon as we
have three coordinates (457I).   Compare 457A(ii)-(iii) with
the definition of `intersection number'
of an indexed family in a Boolean algebra (391H), where we had to allow
repetitions for essentially the same reason.

In 457K-457L, we can of course work with
$\tau$-additive Borel measures in place of quasi-Radon measures, as in
437M.   The essential content of 457L
is already displayed in the case of separable $X$, in which case all Borel
measures are $\tau$-additive, and we can fractionally simplify our
hypotheses;  indeed this is true whenever $X$ has measure-free weight
(438J).

The functional $\rho_{\text{W}}$ of 457K-457L
is a kind of $[0,\infty]$-valued metric;  see
4A2T for another occasion on which it would have saved explanation if the
definition of `metric' allowed infinite distances.   In 457Lb we think of
the metric $\rho$ as representing a cost to be minimised, and in 457Mb we
think of $f$ as a profit to be maximised;  since both arguments rely on the
functions being non-negative, they cannot be simply
inverted unless $\rho$ or $f$ is bounded above (as in 457Xq), and there
is a further complication from the asymmetric nature of the condition
`$\{x:f(x)\ge\alpha\}$ is K-analytic' in 457M.   However, for the primary
applications, as in 457Xr, this is not a problem.   Observe that the same
pattern has already appeared in 457A(iv)-(v).
}%end of notes

\discrpage

\frfilename{mt458.tex}
\versiondate{20.11.17}
\copyrightdate{2007}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Relative independence and relative products}

\newsection{458}

Stochastic independence is one of the central concepts of probability
theory, and pervades measure theory.   I come now to a generalization
of great importance.   If $X_1$, $X_2$ and $Y$ are random
variables, we may find that $X_1$ and $X_2$ are `relatively independent
over $Y$', or `independent when conditioned on $Y$', in the sense that
if we know the value of $Y$, then we learn nothing further about one of
the $X_i$ if we are told the value of the other.   For any stochastic
process, where information comes to us piecemeal, this idea is likely to
be fundamental.   In this section I set out a general framework for
discussion of relative independence (458A), introducing relative
distributions (458I) and relative independence in measure algebras
(458L-458M).   In the last third of the section I look at
`relative product measures'
(458N, 458Q), giving the basic existence theorems (458O, 458S, 458T).

\leader{458A}{Relative independence} Let
$(X,\Sigma,\mu)$ be a probability space and $\Tau$ a $\sigma$-subalgebra
of $\Sigma$.

\spheader 458Aa I say that a family $\familyiI{E_i}$ in $\Sigma$ is
{\bf relatively (stochastically)
independent} over $\Tau$ if whenever $J\subseteq I$ is finite and not
empty, and $g_i$ is a conditional expectation of
$\chi E_i$ on $\Tau$ for each $i\in J$,
then $\mu(F\cap\bigcap_{i\in J}E_i)=\int_F\prod_{i\in J}g_id\mu$ for every
$F\in\Tau$;  that is, $\prod_{i\in J}g_i$ is a conditional expectation of
$\chi(\bigcap_{i\in J}E_i)$ on $\Tau$.
\cmmnt{(Note that this does not depend on which
conditional expectations $g_i$ we take, since any two conditional
expectations of $\chi E_i$ must be equal almost everywhere.)}
A family $\familyiI{\Sigma_i}$ of
subalgebras of $\Sigma$ is {\bf relatively independent} over
$\Tau$ if $\familyiI{E_i}$ is relatively independent over $\Tau$
whenever $E_i\in\Sigma_i$ for every $i\in I$.

\spheader 458Ab I say that a family $\familyiI{f_i}$ in
$\eusm L^0(\mu)$\cmmnt{ (the space of almost-everywhere-defined
virtually measurable real-valued functions, or `random variables')} is
{\bf relatively independent} over $\Tau$ if
$\familyiI{\Sigma_i}$ is relatively independent over $\Tau$ with
respect to the completion of $\mu$, where $\Sigma_i$ is the
$\sigma$-algebra defined by $f_i$\cmmnt{ in the sense of 272C, that
is, the $\sigma$-algebra generated by
$\{f_i^{-1}[F]:F\subseteq\Bbb R$ is a Borel set$\}$}.

\spheader 458Ac I remark at once that a family of subalgebras or
random variables is relatively independent iff every finite subfamily
is\cmmnt{ (cf.\ 272Bb)}.

\spheader 458Ad\cmmnt{ It will be convenient to have a shorthand referring
to  lattices of $\sigma$-algebras of sets.}   If $\Sigma$, $\Tau$ are
algebras of subsets of a set $X$, I will write $\Sigma\vee\Tau$
for the $\sigma$-algebra
of subsets of $X$ generated by $\Sigma\cup\Tau$;  \cmmnt{similarly,}
if $\familyiI{\Sigma_i}$ is a family of algebras of
subsets of $X$, then $\bigvee_{i\in I}\Sigma_i$ will be the
$\sigma$-algebra generated by $\bigcup_{i\in I}\Sigma_i$.
\cmmnt{Note that the functions
$\vee$, $\bigvee$ here are always supposed to yield
$\sigma$-algebras, even if we start with algebras which are not closed
under countable unions, so that
$\Sigma\vee\Sigma$ could in principle be strictly larger than $\Sigma$.
As will become evident in 458D and 458G, the difference between a
$\sigma$-algebra and a simple algebra of sets is relatively unimportant just
here.}

\leader{458B}{}\cmmnt{ There are some surprising results at the very
beginning of the theory of relative independence;  see 458Xa, for instance.
On the positive side, we have the following facts.

\medskip

\noindent}{\bf Lemma} Let
$(X,\Sigma,\mu)$ be a probability space, $\Tau$ a $\sigma$-subalgebra
of $\Sigma$, and $\familyiI{\Sigma_i}$ a family of subalgebras of
$\Sigma$ such that $\Tau\subseteq\bigcup_{i\in I}\Sigma_i$.
Suppose that whenever $J\subseteq I$ is finite and not
empty, $E_i\in\Sigma_i$ and $g_i$ is a conditional expectation of
$\chi E_i$ on $\Tau$ for each $i\in J$,
then $\mu(\bigcap_{i\in J}E_i)=\int\prod_{i\in J}g_id\mu$.   Then
$\familyiI{\Sigma_i}$ is relatively independent over $\Tau$.

\proof{ Take $F\in\Tau$, a finite non-empty $J\subseteq I$ and
$E_i\in\Sigma_i$ for $i\in J$.   Let $j\in I$ be such that $F\in\Sigma_j$.
Set $K=J\cup\{j\}$;  if $j\notin J$, set $E_j=X$.   Now set
$E_j'=E_j\cap F$ and $E_i'=E_i$ for $i\in K\setminus\{j\}$.

For $i\in K$, let $g_i$ be a conditional expectation of $\chi E_i$ on
$\Tau$.   Set $g'_j=g_j\times\chi F$ and $g'_i=g_i$ for
$i\in K\setminus\{j\}$;  then $g'_i$ is a conditional expectation of
$\chi E'_i$ for each $i\in K$.   So we have

\Centerline{$\mu(F\cap\bigcap_{i\in J}E_i)
=\mu(\bigcap_{i\in K}E'_i)
=\int\prod_{i\in K}g'_id\mu
=\int_F\prod_{i\in J}g_id\mu$.}

\noindent As $F$ and $\family{i}{J}{E_i}$ are arbitrary,
$\familyiI{\Sigma_i}$ is relatively independent over $\Tau$.
}%end of proof of 458B

\leader{458C}{Proposition}
Let $(X,\Sigma,\mu)$ be a probability space, $\Bbb T$ a non-empty
upwards-directed family of subalgebras of $\Sigma$,
and $\familyiI{\Sigma_i}$ a family of
$\sigma$-subalgebras of $\Sigma$ which is relatively independent over
$\Tau$ for every $\Tau\in\Bbb T$.   Then $\familyiI{\Sigma_i}$ is
relatively independent over $\bigvee\Bbb T$.

%no such general result for downwards-directed families, see
%458Ya

\proof{{\bf (a)} Suppose first that $\Bbb T$ is countable;
because it is upwards-directed, there is a
non-decreasing sequence $\sequencen{\Tau_n}$ in $\Bbb T$ such
that $\bigcup\Bbb T=\bigcup_{n\in\Bbb N}\Tau_n$ and
$\bigvee\Bbb T=\bigvee_{n\in\Bbb N}\Tau_n$.
Take a non-empty finite set $J\subseteq I$
and $E_i\in\Sigma_i$ for $i\in J$;  set $E=\bigcap_{i\in J}E_i$.
For $i\in J$, let $g_{ni}$ be a conditional
expectation of $\chi E_i$ on $\Tau_n$ for each $n$;  then
$g_i=\lim_{n\to\infty}g_{ni}$ is a conditional expectation of $\chi E_i$
on $\bigvee\Bbb T$ (275I).   Similarly, if $h_n$ is a
conditional expectation of $\chi E$ on $\Tau_n$ for each $n$,
$h=\lim_{n\to\infty}h_n$ is a
conditional expectation of $\chi E$ on $\bigvee\Bbb T$.
Since $\family{i}{J}{E_i}$ is relatively independent over $\Tau_n$,
$h_n\eae\prod_{i\in J}g_{ni}$ for each $n$;  accordingly
$h\eae\prod_{i\in J}h_i$, and $\prod_{i\in J}h_i$ is a conditional
expectation of $\chi E$ on $\bigvee\Bbb T$.
As $\family{i}{J}{E_i}$ is arbitrary,
$\familyiI{\Sigma_i}$ is relatively independent over $\bigvee\Bbb T$.

\medskip

{\bf (b)} For the general case, take a non-empty finite
$J\subseteq I$ and $E_i\in\Sigma_i$ for $i\in J$;  set
$E=\bigcap_{i\in I}E_i$.   For each $i\in J$, let
$g_i:X\to[0,1]$ be a $\bigvee\Bbb T$-measurable conditional expectation of
$\chi E_i$ on $\bigvee\Bbb T$, and $g:X\to[0,1]$ a
$\bigvee\Bbb T$-measurable
conditional expectation of $\chi E$ on $\bigvee\Bbb T$.
Then for every $i\in J$ and $q\in\Bbb Q$ there is
a countable set $\Bbb T_{iq}\subseteq\Bbb T$ such that
$\{x:g_i(x)\ge q\}\in\bigvee\Bbb T_{iq}$;
similarly, there is for each $q\in\Bbb Q$ a
countable set $\Bbb T'_q\subseteq\Bbb T$ such that
$\{x:g(x)\ge q\}\in\bigvee\Bbb T'_q$.
Let $\tilde{\Bbb T}$ be a countable upwards-directed subset of $\Bbb T$
including
$\bigcup_{i\in J,q\in\Bbb Q}\Bbb T_{iq}\cup\bigcup_{q\in\Bbb Q}\Bbb T'_q$.
Then every $g_i$ is $\bigvee{\tilde\Bbb T}$-measurable,
so is a conditional expectation of $\chi E_i$ on $\bigvee{\tilde\Bbb T}$;
similarly, $g$ is a conditional expectation of $\chi E$ on
$\bigvee{\tilde\Bbb T}$.   By (i), $g\eae\prod_{i\in J}g_i$, so
$\prod_{i\in J}g_i$ is a conditional expectation of $\chi E$ on
$\bigvee\Bbb T$.   As $\family{i}{J}{E_i}$ is arbitrary,
$\familyiI{\Sigma_i}$ is relatively independent over
$\bigvee\Bbb T$, as claimed.
}%end of proof of 458C

\leader{458D}{Proposition} Let $(X,\Sigma,\mu)$ be a probability space,
$\Tau$ a $\sigma$-subalgebra of $\Sigma$ and $\familyiI{\Sigma_i}$ a
family of subalgebras of $\Sigma$ which is relatively independent over
$\Tau$.

(a) If $J\subseteq I$ and $\Sigma'_i$ is a subalgebra of $\Sigma_i$ for
$i\in J$, then $\family{i}{J}{\Sigma'_j}$ is relatively independent over
$\Tau$.

(b) Set $\Sigma^*_i=\Sigma_i\vee\Tau$ for $i\in I$.
Then $\familyiI{\Sigma^*_i}$ is relatively independent over $\Tau$.

(c) If $\Cal E\subseteq\bigcup_{i\in I}\Sigma_i$, then
$\familyiI{\Sigma_i}$ is relatively independent over
the $\sigma$-algebra generated by $\Tau\cup\Cal E$.

\proof{{\bf (a)} Immediate from the definition in 458Aa.

\medskip

{\bf (b)(i)} Suppose that $F_0\in\Tau$ and that $\Sigma'_i$ is the algebra
generated by $\Sigma_i\cup\{F_0\}$ for each $i\in I$.   Then
$\familyiI{\Sigma'_i}$ is relatively independent over $\Tau$.
\Prf\ Suppose that $J\subseteq I$ is finite and not empty, and that
$E_i\in\Sigma'_i$ for each $i\in J$.   For $i\in I$, we can express $E_i$
as $(G_i\cap F_0)\cup(H_i\setminus F_0)$, where $G_i$, $H_i\in\Sigma_i$.   Let
$g_i$, $h_i$ be conditional expectations of $\chi G_i$, $\chi H_i$ on
$\Tau$;  then $f_i=g_i\times\chi F_0+h_i\times\chi(X\setminus F_0)$ is a
conditional expectation of $\chi E$ on $\Tau$.   Now, for any
$F\in\Tau$, we have

$$\eqalignno{\int_F\prod_{i\in J}f_i
&=\int_F\prod_{i\in J}g_i\times\chi F_0
  +\prod_{i\in J}h_i\times\chi(X\setminus F_0)\cr
&=\int_{F\cap F_0}\prod_{i\in J}g_i
  +\int_{F\setminus F_0}\prod_{i\in J}h_i
=\mu(F\cap\bigcap_{i\in J}G_i\cap F_0)
  +\mu(F\cap\bigcap_{i\in J}H_i\setminus F_0)\cr
\displaycause{because the families
$\family{i}{J}{G_i}$ and $\family{i}{J}{H_i}$ are
both relatively independent over $\Tau$}
&=\mu(F\cap\bigcap_{i\in J}E_i).\cr}$$

\noindent As $\family{i}{J}{E_i}$ is arbitrary, $\familyiI{\Sigma'_i}$ is
relatively independent over $\Tau$.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $\Cal E\subseteq\Tau$ is finite, and that
$\Sigma'_i$ is the algebra generated by $\Sigma_i\cup\Cal E$ for each $i$.
Then $\familyiI{\Sigma'_i}$ is relatively independent over $\Tau$.   \Prf\
Induce on $\#(\Cal E)$, using (i) for the inductive step.\ \Qed

\medskip

\quad{\bf (iii)} Suppose that $\Sigma'_i$ is the algebra generated by
$\Sigma_i\cup\Tau$ for
each $i\in I$.   Then $\familyiI{\Sigma'_i}$ is
relatively independent over $\Tau$.   \Prf\ If $J\subseteq I$ is finite and
not empty, and
$E_i\in\Sigma'_i$ for each $i\in J$, then there is a finite set
$\Cal E\subseteq\Tau$ such that $E_i$ belongs to the algebra $\Sigma''_i$
generated by $E_i\cup\Cal E$ for every $i\in J$.   By (ii),
$\familyiI{\Sigma''_i}$ is relatively independent over $\Tau$, so
$\family{i}{J}{E_i}$ is relatively independent over $\Tau$;  as
$\family{i}{J}{E_i}$ is arbitrary, $\familyiI{\Sigma'_i}$ is
relatively independent over $\Tau$.\ \Qed

\medskip

\quad{\bf (iv)}
Finally, suppose that $J\subseteq I$ is finite and not
empty, that $E_i\in\Sigma^*_i$ for each $i\in J$, that $F\in\Tau$
and that $\epsilon>0$.
For $i\in J$, let $\Sigma'_i$ be the algebra generated by
$\Sigma_i\cup\Tau$;  then there is an $E'_i\in\Sigma'_i$ such that
$\mu(E'_i\symmdiff E_i)\le\epsilon$ (136H).
Let $g_i$, $g'_i$ be conditional expectations of $\chi E_i$, $\chi E'_i$ on
$\Tau$;  we can arrange that they are all defined on the whole of $X$ and
take values in $[0,1]$.   Then

$$\eqalignno{|\mu(F\cap\bigcap_{i\in J}E_i)-\int_F\prod_{i\in J}g_i|
&\le\sum_{i\in J}\mu(E_i\symmdiff E'_i)
  +|\mu(F\cap\bigcap_{i\in J}E'_i)-\int_F\prod_{i\in J}g'_i|\cr
&\mskip250mu  +\int_F|\prod_{i\in J}g'_i-\prod_{i\in J}g_i|\cr
&\le\epsilon\#(J)+0
  +\int_F\sum_{i\in J}|g'_i-g_i|\cr
\displaycause{(iii) above and 285O}
&\le\epsilon\#(J)+\sum_{i\in J}\int|g'_i-g_i|\cr
&\le\epsilon\#(J)+\sum_{i\in J}\int|\chi E'_i-\chi E_i|\cr
\displaycause{233J or 242Je}
&=\epsilon\#(J)+\sum_{i\in J}\mu(E'_i\symmdiff E_i)
\le 2\epsilon\#(J).\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\mu(F\cap\bigcap_{i\in J}E_i)=\int_F\prod_{i\in J}g_i$.}

\noindent As $\family{i}{J}{E_i}$ and $F$ are arbitrary,
$\familyiI{\Sigma^*_i}$ is relatively independent.

\medskip

{\bf (c)} For any $\Cal E\subseteq\bigcup_{i\in I}\Sigma_i$,
write $\Tau_{\Cal E}$ for the
$\sigma$-algebra generated by $\Tau\cup\Cal E$.

\medskip

\quad{\bf (i)}
Suppose that $i$, $j\in I$ are distinct, $E\in\Sigma_i$, $g$ is a
conditional expectation of $\chi E$ on $\Tau$, and $H\in\Sigma_j$.   Then
$g$ is a conditional expectation of $\chi E$ on $\Tau_{\{H\}}$.
\Prf\ Let $h$ be a conditional expectation of $\chi H$ on
$\Tau$.   If $F\in\Tau$, then

$$\eqalignno{\mu(F\cap H\cap E_i)
&=\int_Fg\times h\cr
\displaycause{because $\Sigma_j$ and $\Sigma_i$ are relatively independent
over $\Tau$}
&=\int_Fg\times\chi H\cr
\displaycause{because $g\times h$ is a conditional expectation of
$g\times\chi H$ on $\Tau$, see 233Eg}
&=\int_{F\cap H}g.\cr}$$

\noindent Similarly, $\mu(F\cap E_i\setminus H)=\int_{F\setminus H}g$.
Now any $G\in\Tau_{\{H\}}$ is expressible as
$(F_1\cap H)\cup(F_2\setminus H)$ where $F_1$, $F_2\in\Tau$, so that

\Centerline{$\mu(G\cap E)=\mu(F_1\cap E\cap H)+\mu(F_2\cap E\setminus H)
=\int_{F_1\cap H}g+\int_{F_2\setminus H}g=\int_Gg$,}

\noindent as required.\ \Qed

\medskip

\quad{\bf (ii)} If $j\in I$ and $H\in\Sigma_j$, $\familyiI{\Sigma_i}$ is
relatively independent over $\Tau_{\{H\}}$.

\medskip

\Prf\ \grheada\ Let $J\subseteq I$ be a non-empty finite set
containing $j$, and
$\family{i}{J}{E_i}$ a family such that $E_i\in\Sigma_i$ for $i\in J$.
Set $K=J\setminus\{j\}$.
For $i\in K$, let $g_i:X\to[0,1]$ be a $\Tau$-measurable conditional
expectation of $\chi E_i$ on $\Tau$.
Then $g_i$ is a conditional expectation of $\chi E_i$ on $\Tau_{\{H\}}$, by
(i).   Let $g_j$ be a conditional expectation of $\chi E_j$ on
$\Tau_{\{H\}}$, and $g'_j$ a conditional
expectation of $\chi(E_j\cap H)$ on $\Tau$.   Then, for any $F\in\Tau$,

$$\eqalignno{\mu(F\cap H\cap\bigcap_{i\in J}E_i)
&=\mu(F\cap(E_j\cap H)\cap\bigcap_{i\in K}E_i)
=\int_Fg'_j\times\prod_{i\in K}g_i\cr
\displaycause{because $\family{i}{J}{\Sigma_i}$ is relatively independent
over $\Tau$}
&=\int_F\chi(E_j\cap H)\times\prod_{i\in K}g_i\cr
\displaycause{233Eg again, because $\prod_{i\in K}g_i$ is bounded and
$\Tau$-measurable}
&=\int_{F\cap H}\chi E_j\times\prod_{i\in K}g_i
=\int_{F\cap H}g_j\times\prod_{i\in K}g_i\cr}$$

\noindent (because $\prod_{i\in K}g_i$ is bounded and
$\Tau_{\{H\}}$-measurable).   Similarly,

\Centerline{$\mu(F\cap\bigcap_{i\in J}E_i\setminus H)
=\int_{F\setminus H}g_j\times\prod_{i\in K}g_i
=\int_{F\setminus H}\prod_{i\in J}g_i$}

\noindent for every $F\in\Tau$;   putting these together, as in (i),

\Centerline{$\mu(G\cap\bigcap_{i\in J{}}E_i)
=\int_G\prod_{i\in J}g_i$}

\noindent for every $G\in\Tau_{\{H\}}$, and $\prod_{i\in J}g_i$ is a
conditional expectation of $\chi(\bigcap_{i\in J}E_i)$ on $\Tau_{\{H\}}$.

\medskip

\qquad\grheadb\
This is not exactly the formula demanded by the definition
in 458Aa, because I supposed that $j\in J$;  but if we have a non-empty
finite $J\subseteq I\setminus\{j\}$ and
$\family{i}{J}{E_j}\in\prod_{i\in J}\Sigma_j$, set $J'=J\cup\{j\}$ and
$E_j=X$ to see that there is a family $\family{i}{J'}{g_i}$ such that $g_i$
is a conditional expectation of $\chi E_i$ on $\Tau_{\{H\}}$ for every $i$,
and

\Centerline{$\mu(G\cap\bigcap_{i\in J}E_i)
=\mu(G\cap\bigcap_{i\in J'}E_i)
=\int_G\prod_{i\in J'}g_i
=\int_G\prod_{i\in J}g_i$}

\noindent for every $G\in\Tau_{\{H\}}$.
So $\familyiI{\Sigma_i}$ really is relatively
independent over $\Tau_{\{H\}}$.\ \Qed

\medskip

\quad{\bf (iii)} Inducing on $\#(\Cal E)$, we see that
$\familyiI{\Sigma_i}$
is relatively independent over $\Tau_{\Cal E}$ whenever
$\Cal E\subseteq\bigcup_{i\in I}\Sigma_i$ is finite.
By 458C,
$\familyiI{\Sigma_i}$
is relatively independent over $\Tau_{\Cal E}$ for every
$\Cal E\subseteq\bigcup_{i\in I}\Sigma_i$.
}%end of proof of 458D

\cmmnt{\medskip

\noindent{\bf Remark} Putting (a) and (b) above together, we see that
if $\familyiI{\Sigma_i}$ is a family of subalgebras of $\Sigma$, and
$\hat\Sigma_i$ is the $\sigma$-algebra generated by $\Sigma_i$ for each $i$,
then $\familyiI{\Sigma_i}$ is relatively independent over $\Tau$ iff
$\familyiI{\hat\Sigma_i}$ is relatively independent over $\Tau$.
}

\leader{458E}{Example}\cmmnt{ The simplest examples of relatively
independent $\sigma$-algebras arise as follows.}  Let $(X,\Sigma,\mu)$
be a probability space, $\familyiI{\Tau_i}$ an independent family of
$\sigma$-subalgebras of $\Sigma$\cmmnt{, as in 272Ab}, and $\Tau$ a
$\sigma$-subalgebra of $\Sigma$ which is independent of
$\bigvee_{i\in I}\Tau_i$.   For each
$i\in I$, let $\Sigma_i$ be $\Tau\vee\Tau_i$.
Then $\familyiI{\Sigma_i}$ is relatively independent over $\Tau$.

\proof{ In view of 458Db, it is enough to show that $\familyiI{\Tau_i}$ is
relatively independent over $\Tau$.   But if we have a non-empty finite
$J\subseteq I$ and $E_i\in\Tau_i$ for $i\in I$, then
$\mu(E_i\cap F)=\mu E_i\cdot\mu F$ for $F\in\Tau$, so
$f_i=\mu E_i\cdot\chi X$ is a
conditional expectation of $\chi E_i$ on $\Tau$, for each $i$.
Similarly, setting $E=\bigcap_{i\in J}E_i$, $\mu E\cdot\chi X$
is a conditional
expectation of $\chi E$ on $\Tau$.   Since $\mu E=\prod_{i\in J}\mu E_j$,
$\prod_{i\in J}f_i$ is a conditional expectation of $\chi E$ on $\Tau$,
which is what we need to know.
}%end of proof of 458E

\leader{458F}{}\cmmnt{ The following facts are elementary but occasionally
useful.

\wheader{458F}{4}{2}{2}{60pt}

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a probability space and
$\Tau$ a $\sigma$-subalgebra of $\Sigma$.

(a) Let $\familyiI{f_i}$ be a family of
non-negative $\mu$-integrable functions on $X$ which is relatively
independent over $\Tau$.   For each $i\in I$ let
$g_i$ be a conditional expectation of $f_i$ on $\Tau$.   Then for any
$F\in\Tau$ and $i_0,\ldots,i_n\in I$,

\Centerline{$\int_F\prod_{j=0}^ng_{i_j}\le\int_F\prod_{j=0}^nf_{i_j}$}

\noindent with equality if all the $i_j$ are distinct.

(b) Suppose that $\Sigma_1$, $\Sigma_2$ are $\sigma$-subalgebras of
$\Sigma$ which are relatively independent over $\Tau$, and that
$f\in\eusm L^1(\mu\restr\Sigma_1)$.   If $g$ is a conditional expectation
of $f$ on $\Tau$, then it is a conditional expectation of $f$ on
$\Tau\vee\Sigma_2$.

\proof{{\bf (a)}
Let $\Sigma_i$ be the $\sigma$-algebra generated by $f_i$ for
each $i$, so that $\familyiI{\Sigma_i}$ is relatively independent (with respect
to the completion of $\mu$) over $\Tau$.

\medskip

\quad{\bf (i)} To begin with, suppose that $i_0,\ldots,i_n$ are all
different.

\medskip

\qquad\grheada\
If $f_i=\chi E_i$ for each $i\in I$, where $E_i\in\Sigma_i$,
the result is just the definition of `relative independence' in 458Aa.

\medskip

\qquad\grheadb\ Because both sides of the desired equality are
multilinear expressions of the inputs, and conditional expectation is an
essentially linear operation, the same is true if all the $f_i$ are simple
functions.

\medskip

\qquad\grheadc\ For general non-negative
integrable random variables $f_i$, let
$\sequence{k}{f_{ik}}$ be a non-decreasing sequence
of non-negative $\Sigma_i$-simple
functions converging almost everywhere to $f_i$ for each $i$, and $g_{ik}$
a conditional expectation of $f_{ik}$ for all $i$ and $k$.   Then
$\sequence{k}{g_{ik}}$ is non-decreasing almost everywhere and converges
a.e.\ to the given conditional expectation $g_i$ of $f_i$.   So

\Centerline{$\int_F\prod_{j=0}^ng_{i_j}
=\lim_{k\to\infty}\int_F\prod_{j=0}^ng_{i_jk}
=\lim_{k\to\infty}\int_F\prod_{j=0}^nf_{i_jk}
=\int_F\prod_{j=0}^nf_{i_j}$,}

\noindent as required.

\medskip

\quad{\bf (ii)}\grheada\
Now suppose that the $i_0,\ldots,i_n$ are not all distinct,
but that all the $f_{i_j}$ are bounded.   Let $l_0,\ldots,l_m$
enumerate
$\{i_0,\ldots,i_n\}$ and for $j\le m$ set $k_j=\#(\{r:i_r=l_j\})$.
For each $j\le m$, let $h_j$ be a conditional expectation of
$f_{l_j}^{k_j}=|f_{l_j}|^{k_j}$.   Because $t\mapsto|t|^{k_j}$ is convex,
$g_{l_j}^{k_j}\leae h_j$ (233J).   So

$$\eqalignno{\int_F\prod_{j=0}^ng_{i_j}
&=\int_F\prod_{j=0}^mg_{l_j}^{k_j}
\le\int_F\prod_{j=0}^mh_j
=\int_F\prod_{j=0}^mf_{l_j}^{k_j}\cr
\displaycause{by part (i), because each $f_{l_j}^{k_j}$ is
$\Sigma_{l_j}$-measurable}
&=\int_F\prod_{j=0}^nf_{i_j},\cr}$$

\noindent as required.

\medskip

\qquad\grheadb\ Finally, for the general case, take simple functions
$f_{ik}$ and conditional expectations $g_{ik}$ as in
(a-iii) above.   Then

\Centerline{$\int_F\prod_{j=0}^ng_{i_j}
=\lim_{k\to\infty}\int_F\prod_{j=0}^ng_{i_jk}
\le\lim_{k\to\infty}\prod_{j=0}^n\int_Ff_{i_jk}
=\int_F\prod_{j=0}^nf_{i_j}$,}

\noindent and the proof is complete.

\medskip

{\bf (b)} Adjusting $f$ on a negligible set if necessary, we may
suppose that $f$ is $\Sigma_1$-measurable.
Take any $F\in\Sigma_2\vee\Tau$, and let $h\ge 0$ be
a conditional expectation of $\chi F$ on $\Tau$.
By 458Db and 458Da, $\Sigma_1$ and
$\Sigma_2\vee\Tau$ are relatively independent over $\Tau$, so $f$ and
$\chi F$ are relatively independent over $\Tau$.   Accordingly

$$\eqalignno{\int_Ff
&=\int f\times\chi F
=\int g\times h\cr
\displaycause{applying (a) to the positive and negative parts of $f$}
&=\int g\times\chi F\cr
\displaycause{233K}
&=\int_Fg.\cr}$$

\noindent As $F$ is arbitrary and

\Centerline{$g\in\eusm L^1(\mu\restr\Tau)
\subseteq\eusm L^1(\mu\restr\Sigma_2\vee\Tau)$,}

\noindent $g$ is a conditional expectation of $f$ on
$\Sigma_2\vee\Tau$.
}%end of proof of 458F

\cmmnt{\medskip

\noindent{\bf Remark} In (a),
I have avoided speaking of conditional expectations
of products $\prod_{j=0}^nf_{i_j}$ because these need not be integrable
functions.   But when $\prod_{j=0}^nf_{i_j}$ is integrable and has a
conditional expectation $g$, then we must have
$\prod_{j=0}^ng_{i_j}\leae g$, with equality almost everywhere
when the $i_j$ are distinct.
}

%can probably do without the next bit
\leader{*458G}{}\cmmnt{ It is sometimes useful to know that
`relative independence' can be defined without using the apparatus of
conditional expectations;  indeed, we have a formulation which can be used
with finitely additive functionals rather than measures.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a probability space,
$\Tau$ a $\sigma$-subalgebra of $\Sigma$, and $\familyiI{\Sigma_i}$ a
family of subalgebras of $\Sigma$.
Let $\Bbb T$ be the family of finite
subalgebras of $\Tau$.   For $\Lambda\in\Bbb T$ write $\Cal A_{\Lambda}$
for the set of non-negligible atoms in $\Lambda$.   For non-empty finite
$J\subseteq I$, $\family{i}{J}{E_i}\in\prod_{i\in J}\Sigma_i$ and
$F\in\Tau$, set

\Centerline{$\phi_{\Lambda}(F,\family{i}{J}{E_i})
=\sum_{H\in\Cal A_{\Lambda}}\mu(H\cap F)
  \cdot\prod_{i\in J}\Bover{\mu(E_i\cap H)}{\mu H}$.}

\noindent Then $\familyiI{\Sigma_i}$ is relatively independent over
$\Tau$ iff $\lim_{\Lambda\in\Bbb T,\Lambda\uparrow}
\phi_{\Lambda}(F,\family{i}{J}{E_i})=\mu(F\cap\bigcap_{i\in J}E_i)$
whenever
$J\subseteq I$ is finite and not empty, $E_i\in\Sigma_i$ for every
$i\in J$ and $F\in\Tau$.

\proof{{\bf (a)} The point is just that if $J\subseteq I$ is finite and
not empty, $E_i\in\Sigma_i$ for $i\in J$, $g_i$ is a conditional
expectation of $\chi E_i$ on $\Tau$ for each $i$, and $F\in\Tau$, then
$\int_F\prod_{i\in J}g_id\mu
=\lim_{\Lambda\uparrow}\phi_{\Lambda}(F,\family{i}{J}{E_i})$.   \Prf\
Adjusting each $g_i$ on a negligible set if necessary, we may suppose
that it is $\Tau$-measurable, defined
everywhere on $X$ and takes values between $0$ and $1$.

Fix $n\in\Bbb N$ for the moment.   Let $\Lambda_n$ be the finite
subalgebra
of $\Tau$ generated by sets of the form $\{x:g_i(x)\le 2^{-n}k\}$ for
$i\in J$ and $k\le 2^n$, and $\Lambda$ any finite
subalgebra of $\Sigma_0$ including $\Lambda_n$.   If $H\in\Cal A_{\Lambda}$
then there are integers $k_{iH}$, for $i\in J$,
such that
$2^{-n}k_{iH}\le g_i(x)<2^{-n}(k_{iH}+1)$ for every $i\in J$ and $x\in H$.
So

\Centerline{$2^{-n}k_{iH}\le\Bover{\mu_i(E\cap H)}{\mu H}<2^{-n}(k_{iH}+1)$}

\noindent for each $i$.   Accordingly

$$\eqalign{\sum_{H\in\Cal A_{\Lambda}}\mu(H\cap F)
   \cdot\prod_{i\in J}2^{-n}k_{iH}
&\le\phi_{\Lambda}(F,\family{i}{J}{E_i})\cr
&\le\sum_{H\in\Cal A_{\Lambda}}\mu(H\cap F)\cdot\prod_{i\in J}
   \min(1,2^{-n}(k_{iH}+1)),\cr}$$

\noindent that is,

\Centerline{$\int_F\prod_{i\in J}g'_{in}d\mu
\le\phi_{\Lambda}(F,\family{i}{J}{E_i})
\le\int_F\prod_{i\in J}g''_{in}d\mu$,}

\noindent where $g'_{in}(x)=2^{-n}k$, $g''_{in}(x)=\min(1,2^{-n}(k+1))$
when $2^{-n}k\le g_i(x)<2^{-n}(k+1)$.   But this means that

$$\eqalignno{|\phi_{\Lambda}(F,\family{i}{J}{E_i})
  -\int_F\prod_{i\in J}g_id\mu|
&\le\int\prod_{i\in J}g''_{in}-\prod_{i\in J}g'_id\mu
\le\int\sum_{i\in J}g''_{in}-g'_id\mu\cr
\displaycause{because all the $g'_{in}$, $g''_{in}$ take values
in $[0,1]$}
&\le 2^{-n}\#(J).\cr}$$

\noindent Since this is true for every $\Lambda\supseteq\Lambda_n$ and
every $n\in\Bbb N$,
$\lim_{\Lambda\uparrow}\phi_{\Lambda}(F,\family{i}{J}{E_i})
=\int_F\prod_{i\in J}g_id\mu$.\ \Qed

\medskip

{\bf (b)} Accordingly the condition given exactly matches the
definition in 458A.
}%end of proof of 458G

\leader{458H}{}\cmmnt{ All the fundamental theorems concerning
stochastic independence have relativized forms.   A simple one is the
following.

\medskip

\noindent}{\bf Proposition}\cmmnt{ (Compare 272K.)}  Let
$(X,\Sigma,\mu)$ be a probability space and $\Tau$ a $\sigma$-subalgebra of
$\Sigma$.
Let $\familyiI{\Sigma_i}$ be a family of $\sigma$-subalgebras of
$\Sigma$ which is relatively independent over $\Tau$.   Let
$\family{j}{J}{I_j}$ be a partition of $I$, and for each $j\in J$ let
$\tilde\Sigma_j$ be $\bigvee_{i\in I_j}\Sigma_i$.

(a) If $\familyiI{\Sigma_i}$ is relatively independent over $\Tau$,
then $\family{j}{J}{\tilde\Sigma_j}$ is
relatively independent over $\Tau$.

(b) Suppose that
$\family{j}{J}{\tilde\Sigma_j}$ is relatively independent over $\Tau$ and
that $\family{i}{I_j}{\Sigma_i}$ is relatively independent over $\Tau$ for
every $j\in J$.   Then $\familyiI{\Sigma_i}$ is relatively independent
over $\Tau$.

\proof{ For each $E\in\Sigma$ let $f_E$ be a conditional expectation of
$\chi E$ on $\Tau$.

\medskip

{\bf (a)} Take any finite $K\subseteq J$, and let $\pmb{W}$
be the set of families $\family{j}{K}{W_j}$ such that
$W_j\in\tilde\Sigma_j$ for each $j\in K$ and
$\mu(F\cap\bigcap_{j\in K}W_j)=\int_F\prod_{j\in K}f_{W_j}d\mu$ for
every $F\in\Tau$.   For each $j\in K$, let $\Cal C_j$ be the family of
measurable cylinders
expressible as $W=X\cap\bigcap_{i\in L}E_i$ where $L\subseteq I_j$
is finite and $E_i\in\Sigma_i$ for $i\in L$.   Note that in this case

\Centerline{$\mu(F\cap W)
=\mu(F\cap\bigcap_{i\in L}E_i)=\int_F\prod_{i\in L}f_{E_i}d\mu$}

\noindent for every $F\in\Tau$, so $f_W\eae\prod_{i\in L}f_{E_i}$,
taking the product to be $\chi X$ if $L$ is empty.

If $W_j\in\Cal C_j$ for each $j\in K$, then
$\family{j}{K}{W_j}\in\pmb{W}$.   \Prf\ Express $W_j$ as
$X\cap\bigcap_{i\in L_j}E_i$ where $L_j\subseteq I_j$ is finite and
$E_i\in\Sigma_i$ whenever $j\in K$ and $i\in L_j$.   Then,
setting $L=\bigcup_{j\in K}L_j$,

$$\eqalignno{\mu(F\cap\bigcap_{j\in K}W_j)
&=\mu(F\cap\bigcap_{i\in L}E_i)
=\int_F\prod_{i\in L}f_{E_i}d\mu\cr
\displaycause{because $\familyiI{\Sigma_i}$ is relatively independent}
&=\int_F\prod_{j\in K}\prod_{i\in L_j}f_{E_i}d\mu
=\int_F\prod_{j\in K}f_{W_i}d\mu\cr}$$

\noindent for every $F\in\Tau$.\ \Qed

Observe next that if we fix $k\in K$, and a family
$\family{j}{K\setminus\{k\}}{W_j}$, then the set of those
$W_k\in\tilde\Sigma_k$ such that $\family{j}{K}{W_j}\in\pmb{W}$ is a
Dynkin class, so if it includes $\Cal C_k$ it must include the
$\sigma$-algebra generated by $\Cal C_k$, viz., $\tilde\Sigma_k$.   Now
an easy induction on $n$ shows that if
$\family{j}{K}{W_j}\in\prod_{j\in K}\tilde\Sigma_j$ and
$\#(\{j:W_j\notin\Cal C_j\})=n$, then $\family{j}{K}{W_j}\in\pmb{W}$.
Taking $n=\#(K)$ we see that
$\prod_{j\in K}\tilde\Sigma_j\subseteq\pmb{W}$.

As this is true for every finite $K\subseteq J$,
$\family{j}{J}{\tilde\Sigma_j}$ is relatively independent over $\Tau$,
as claimed.

\medskip

{\bf (b)} This time, let $K\subseteq I$ be a
non-empty finite set, and $E_i\in\Sigma_i$ for $i\in K$.   Set
$L=\{j:j\in J$, $K\cap I_j\ne\emptyset\}$, and for $j\in L$ set
$G_j=\bigcap_{i\in K\cap I_j}E_i$;  set
$E=\bigcap_{i\in J}E_i=\bigcap_{j\in L}G_j$.
Because $\family{i}{I_j}{\Sigma_i}$ is
relatively independent over $\Tau$,
$f_{G_j}\eae\prod_{i\in K\cap I_j}f_{E_i}$.   Because
$\family{j}{J}{\tilde\Sigma_j}$ is relatively independent over
$\Tau$,

\Centerline{$f_E\eae\prod_{j\in L}f_{G_j}\eae\prod_{i\in K}f_{E_i}$.}

\noindent As $\family{i}{K}{E_i}$ is arbitrary, we have the result.
}%end of proof of 458H

\leader{458I}{}\cmmnt{ For the next result, we need
a concept of `relative probability distribution', as follows.

\medskip

\noindent}{\bf Definition} Let $(X,\Sigma,\mu)$ be a probability space,
$\Tau$ a $\sigma$-subalgebra of $\Sigma$, and
$f\in\eusm L^0(\mu)$.   Then a {\bf relative distribution} of $f$ over
$\Tau$ will be a family $\family{x}{X}{\nu_x}$ of Radon probability
measures on $\Bbb R$ such that
$x\mapsto\nu_xH$ is a conditional expectation of $\chi f^{-1}[H]$ on
$\Tau$ for every Borel set $H\subseteq\Bbb R$.

\leader{458J}{Theorem} Let $(X,\Sigma,\mu)$ be a probability space,
$\Tau$ a $\sigma$-subalgebra of $\Sigma$, and $f\in\eusm L^0(\mu)$.
Then there is a relative distribution of $f$ over $\Tau$, which is
essentially unique in the sense that if
$\family{x}{X}{\nu_x}$ and $\family{x}{X}{\nuprime_x}$ are two such
relative distributions, 
then $\nu_x=\nuprime_x$ for $\mu\restrp\Tau$-almost every $x$.

\proof{{\bf (a)} Write $\mu_0$ for the restriction of $\mu$ to $\Tau$,
$\hat\mu$ for the completion of $\mu$, $\hat\Sigma$ for the domain of
$\hat\mu$, and $\Cal B$ for the Borel $\sigma$-algebra of $\Bbb R$.
Then the function $x\mapsto (x,f(x)):\dom f\to X\times\Bbb R$ is
$(\hat\Sigma,\Tau\tensorhat\Cal B)$-measurable, just because
$F\cap f^{-1}[H]\in\hat\Sigma$ for every $F\in\Tau$ and $H\in\Cal B$.
So we have a probability measure $\nu$ on $X\times\Bbb R$ defined by
setting
$\nu W=\hat\mu\{x:(x,f(x))\in W\}$ for every $W\in\Tau\tensorhat\Cal B$.
The marginal measure on $\Bbb R$ is tight
just because it is a Borel probability measure (433Ca).
By 452M, we have a family $\family{x}{X}{\nu_x}$ of Radon
probability measures on $\Bbb R$ such that
$\nu W=\int\nu_xW[\{x\}]\mu_0(dx)$ for every
$W\in\Tau\tensorhat\Cal B$.   In particular, if $H\in\Cal B$,

\Centerline{$\int_F\chi f^{-1}[H]d\mu=\hat\mu(F\cap f^{-1}[H])
=\nu(F\times H)=\int_F\nu_xH\,\mu(dx)$}

\noindent for every $F\in\Tau$, and $x\mapsto\nu_xH$ is a conditional
expectation of $\chi f^{-1}[H]$ on $\Tau$, that is, $\family{x}{X}{\nu_x}$
is a relative distribution of $f$ over $\Tau$.

\medskip

{\bf (b)} Now suppose that $\family{x}{X}{\nuprime_x}$ is another
relative
distribution of $f$ over $\Tau$.   Then for each $H\in\Cal B$ we have
$\int_F\nu_xH\mu(dx)=\int_F\nuprime_xH\mu(dx)$ for every $F\in\Tau$, so
that
$\nu_xH=\nuprime_xH$ for $\mu_0$-almost every $x$.   But this means that
for $\mu_0$-almost every $x$, we have $\nu_xH=\nuprime_xH$ for every
interval $H$ with rational endpoints;  and for such $x$ we must have
$\nu_x=\nuprime_x$ (415H(v)).
}%end of proof of 458J

\leader{458K}{}\cmmnt{ Now we can state and prove a result
corresponding to 272G.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a probability space,
$\Tau$ a $\sigma$-subalgebra of $\Sigma$, and
$\familyiI{f_i}$ a family in $\eusm L^0(\mu)$.   For each $i\in I$, let
$\family{x}{X}{\nu_{ix}}$ be a relative distribution of $f_i$ over
$\Tau$, and $\tilde f_i:X\to\Bbb R$ an arbitrary extension of $f_i$ to
the whole of $X$.   Then the following are equiveridical:

(i) $\familyiI{f_i}$ is relatively independent over $\Tau$;

(ii) for any Baire set $W\subseteq\BbbR^I$ and any $F\in\Tau$,

\Centerline{$\hat\mu(F\cap\pmb{f}^{-1}[W])
=\int_F\lambda_xW\mu(dx)$,}

\noindent where $\hat\mu$ is the completion of $\mu$,
$\pmb{f}(x)=\familyiI{\tilde f_i(x)}$ for $x\in X$, and $\lambda_x$ is
the product of $\familyiI{\nu_{ix}}$ for each $x$;

(iii) for any non-negative Baire measurable function $h:\BbbR^I\to\Bbb R$
and any $F\in\Tau$,

\Centerline{$\int_Fh\pmb{f}d\mu
=\int_F\int h\,d\lambda_x\mu(dx)$.}


\proof{{\bf (a)} Note first that if $i\in I$ and $H\subseteq\Bbb R$ is a
Borel set, then $x\mapsto\nu_{ix}H$ is a conditional expectation of
$\chi f_i^{-1}[H]$ on $\Tau$, so that
$\int_F\nu_{ix}H\mu(dx)=\hat\mu(F\cap f_i^{-1}[H])$ for
every $F\in\Tau$.

Suppose that $\familyiI{f_i}$ is relatively independent, and $F\in\Tau$.
Let $\Cal C$ be the family of Baire measurable cylinders
of $\BbbR^I$ expressible in the
form $C=\{z:z\in\BbbR^I,\,z(i)\in H_i$ for every $i\in J\}$ where
$J\subseteq I$ is finite and $H_i\subseteq\BbbR$ is a Borel set for each
$i\in J$.   For such a set $C$,

$$\eqalignno{\hat\mu(F\cap\pmb{f}^{-1}[C])&
=\hat\mu(F\cap\bigcap_{i\in J}\tilde f_i^{-1}[H_i])
=\hat\mu(F\cap\bigcap_{i\in J}f_i^{-1}[H_i])
=\int_F\prod_{i\in J}\nu_{ix}H_i\mu(dx)
\Displaycause{interpreting an empty product as $\chi X$}
=\int_F\lambda_xC\mu(dx).\cr}$$

\noindent So

\Centerline{$\Cal W=\{W:W\subseteq\BbbR^I$,
  $\hat\mu(F\cap\pmb{f}^{-1}[W])=\biggerint\lambda_xW\mu(dx)\}$}

\noindent includes $\Cal C$;  since it
is a Dynkin class, it contains every Baire subset of $\BbbR^I$
(by the Monotone Class Theorem, 136B), and (ii) is true.

\medskip

{\bf (b)} Now suppose that (ii) is true.   Let $\Sigma_i$ be the
$\sigma$-algebra defined by $f_i$ for each $i$.   If $J\subseteq I$ is
finite and $E_i\in\Sigma_i$
for each $i\in J$, then there are Borel sets $H_i\subseteq\Bbb R$ such
that $E_i\symmdiff f_i^{-1}[H_i]$ is negligible for each $i$, so that
$x\mapsto\nu_{ix}H_i$ is a conditional expectation of $\chi E_i$ on
$\Tau$.   Now by the same equations as before, in the opposite
direction,

$$\eqalignno{\int_F\prod_{i\in J}\nu_{ix}H_i\mu(dx)
&=\int_F\lambda_xC\mu(dx)\cr
\displaycause{where $C=\{z:z(i)\in H_i$ for $i\in J\}$}
&=\hat\mu(F\cap\pmb{f}^{-1}[C])
=\hat\mu(F\cap\bigcap_{i\in J}f_i^{-1}[H_i])
=\hat\mu(F\cap\bigcap_{i\in J}E_i)\cr}$$

\noindent for every $F\in\Tau$.   As $\family{i}{J}{E_i}$ is arbitrary,
$\familyiI{\Sigma_i}$ and $\familyiI{f_i}$ are
relatively independent.

\medskip

{\bf (c)} Thus (i)$\Leftrightarrow$(ii).   For (ii)$\Rightarrow$(iii), observe that
(ii) covers the case in which $h$ is an indicator function $\chi W$;
for the general case, express $h$ as the supremum of a non-decreasing
sequence of linear combinations of indicator functions, as usual.
And (iii)$\Rightarrow$(ii) is trivial.
}%end of proof of 458K

\cmmnt{\medskip

\noindent{\bf Remarks} Of course the ungainly shift to $\tilde f_i$ is
unnecessary if $I$ is countable;  but for uncountable $I$ the
intersection $\bigcap_{i\in I}\dom f_i$, which is the only suitable
domain for $\pmb{f}$, may not be conegligible.

I said that $\lambda_x$ should be `the product of
$\familyiI{\nu_{ix}}$'.   Since the $\nu_{ix}$ are Radon probability
measures, we have two possible interpretations of this:  either the
`ordinary' product measure of \S254 or the `quasi-Radon' product measure
of \S417.   But as we are interested only in the values of $\lambda_xW$
for Baire sets $W$, it makes no difference which we use.
}%end of comment

\leader{458L}{Measure algebras}\cmmnt{ We can look at
the same ideas
in the context of measure algebras.    Let $(\frak A,\bar\mu)$ be a
probability algebra and $\frak C$ a closed subalgebra of $\frak A$.

\medskip

} {\bf (a)} If $a\in\frak A$, then we can say that
$u\in L^{\infty}(\frak C)$ is the conditional expectation of $\chi a$ on
$\frak C$ if $\int_cu=\bar\mu(c\Bcap a)$ for every
$c\in\frak C$\cmmnt{ (365Q\formerly{3{}65R})}.
Now we can say that a family
$\familyiI{b_i}$ in $\frak A$ is {\bf relatively (stochastically)
independent over $\frak C$} if
$\bar\mu(c\Bcap\inf_{i\in J}b_i)=\int_c\prod_{i\in J}u_i$
whenever $J\subseteq I$ is a non-empty finite set
and $u_i$ is the conditional expectation of $\chi b_i$ on $\frak C$ for
every $i\in J$;  while a family
$\familyiI{\frak B_i}$ of subalgebras of $\frak A$ is {\bf relatively
(stochastically) independent over
$\frak C$} if $\familyiI{b_i}$ is relatively independent over $\frak C$
whenever $b_i\in\frak B_i$ for every $i\in I$.

Corresponding to 458Ab, we can say that a family $\familyiI{w_i}$ in
$L^0(\frak A)$ is {\bf relatively (stochastically) independent over
$\frak C$} if $\familyiI{\frak B_i}$ is relatively stochastically
independent, where $\frak B_i$ is the closed subalgebra of $\frak A$
generated by $\{\Bvalue{w_i>\alpha}:\alpha\in\Bbb R\}$ for each $i$.

Returning to the original form of these ideas, we say that a family
$\familyiI{b_i}$ in $\frak A$ is {\bf (stochastically) independent} if
it is relatively independent over $\{0,1\}$, that is, if
$\bar\mu(\inf_{i\in J}b_i)=\prod_{i\in J}\bar\mu b_i$ whenever
$J\subseteq I$ is finite.   Similarly, a family $\familyiI{\frak B_i}$ of
subalgebras of $\frak A$ is (stochastically) independent, in the sense of
325L, iff it is relatively independent over $\{0,1\}$ in the sense here.

\spheader 458Lb
Let $(X,\Sigma,\mu)$ be a probability space and
$(\frak A,\bar\mu)$ its measure algebra.   Let $\familyiI{E_i}$,
$\familyiI{\Sigma_i}$ and $\familyiI{f_i}$ be, respectively, a family in
$\Sigma$, a family of subalgebras of $\Sigma$, and a family of
$\mu$-virtually measurable real-valued functions defined almost everywhere
on $X$;  let $\Tau$ be a $\sigma$-subalgebra of $\Sigma$.   For $i\in I$,
set $a_i=E_i^{\ssbullet}\in\frak A$,
$\frak B_i=\{E^{\ssbullet}:E\in\Sigma_i\}$, and
$w_i=f_i^{\ssbullet}\in L^0(\frak A)$, identified with
$L^0(\mu)$\cmmnt{ (364Ic)}.
Set $\frak C=\{F^{\ssbullet}:F\in\Tau\}$.   Then

\inset{$\familyiI{a_i}$ is relatively independent over $\frak C$ iff
$\familyiI{E_i}$ is relatively independent over $\Tau$,

$\familyiI{\frak B_i}$ is relatively independent over $\frak C$ iff
$\familyiI{\Sigma_i}$ is relatively independent over $\Tau$,

$\familyiI{w_i}$ is relatively independent over $\frak C$ iff
$\familyiI{f_i}$ is relatively independent over $\Tau$.}

\prooflet{\noindent\Prf\ The point is that if $f\in\eusm L^1(\mu)$ (in
particular, if $f=\chi E$ for some $E\in\Sigma$), and
$g\in\eusm L^1(\mu\restrp\Tau)\subseteq\eusm L^1(\mu)$ is a conditional
expectation of $f$ on $\Tau$, then $g^{\ssbullet}$ is a conditional
expectation of $f^{\ssbullet}$ on $\frak C$;  see 242J and 365Q.\ \Qed}

\spheader 458Lc
Corresponding to 458B, we see that if
$\familyiI{\frak A_i}$ is a family of subalgebras of $\frak A$ such that
$\frak C\subseteq\bigcup_{i\in I}\frak A_i$, and
$\int\prod_{i\in J}u_id\bar\mu=\bar\mu(\inf_{i\in J}a_i)$ whenever
$J\subseteq I$ is finite and not empty and $a_i\in\frak A_i$ and
$u_i\in L^{\infty}(\frak C)$ is a conditional expectation of $\chi a_i$ on
$\frak C$ for each $i\in J$, then $\familyiI{\frak A_i}$ is relatively
independent over $\frak C$.

\spheader 458Ld
Corresponding to 458Db, we see that if $\familyiI{\frak B_i}$ is a
family of subalgebras of $\frak A$  which is relatively independent over
$\frak C$, and $\frak B_i^*$ is the closed subalgebra of $\frak A$
generated by $\frak B_i\cup\frak C$ for each $i$, then
$\familyiI{\frak B_i^*}$ is relatively independent over $\frak C$.
\prooflet{The most natural proof, from where we are now
standing, is to express $(\frak A,\bar\mu)$ as the measure algebra of a
probability space $(X,\Sigma,\mu)$, set
$\Tau=\{F:F^{\ssbullet}\in\frak C\}$ and
$\Sigma_i=\{E:E^{\ssbullet}\in\frak B_i\}$ for each $i\in I$, and use
458D.}   

Corresponding to 458Dc, we see that if $\familyiI{\frak B_i}$ is a family
of subalgebras of $\frak A$ which is relatively independent over $\frak C$,
$D_i\subseteq\frak B_i$ for every $i\in I$, and $\frak D$ is the closed
subalgebra of $\frak A$ generated by $\frak C\cup\bigcup_{i\in I}D_i$, then
$\familyiI{\frak B_i}$ is relatively independent over $\frak D$.

\spheader 458Le Following 458H, we have the result that if
$\familyiI{\frak B_i}$ is relatively independent over $\frak C$, and
$\family{j}{J}{I_j}$ is a partition of $I$, and $\tilde{\frak B}_j$ is
the closed subalgebra of $\frak A$ generated by
$\bigcup_{i\in I_j}\frak B_i$ for every $j\in J$, then
$\family{j}{J}{\tilde{\frak B}_j}$ is relatively independent over
$\frak C$.

\spheader 458Lf Note that\cmmnt{ if $a\in\frak A$ and $u$ is the
conditional expectation of $\chi a$ on $\frak C$, then
$\Bvalue{u>0}=\upr(a,\frak C)$, by 365Qc.   So} if
$\familyiI{\frak B_i}$ is a family of subalgebras of $\frak A$ which is
relatively independent over $\frak C$, and $J\subseteq I$ is finite, and
$b_i\in\frak B_i$ for each $i\in J$, then $\inf_{i\in J}b_i=0$ iff
$\inf_{i\in J}\upr(b_i,\frak C)=0$.   \prooflet{(If $u_i$ is a
conditional expectation of $\chi b_i$ on $\frak C$ for each $i$, then

\Centerline{$\inf_{i\in J}\upr(b_i,\frak C)
=\inf_{i\in J}\Bvalue{u_i>0}
=\Bvalue{\prod_{i\in J}u_i>0}$}

\noindent is zero iff
$\bar\mu(\inf_{i\in J}b_i)=\int\prod_{i\in J}u_i=0$.)}

\spheader 458Lg\cmmnt{ We have a straightforward
version of 458E, as follows.}   If $\familyiI{\frak C_i}$
is a stochastically independent family of closed subalgebras of
$\frak A$, $\frak C$ is independent of
the algebra generated by $\bigcup_{i\in I}\frak C_i$, and $\frak B_i$ is
the closed subalgebra of $\frak A$ generated by $\frak C\cup\frak C_i$
for each $i$, then $\familyiI{\frak B_i}$ is relatively independent over
$\frak C$.   \prooflet{(Either repeat the proof of 458E, looking at
$\frak B_{i0}=\frak C_i$ and $\frak B_{i1}=\frak C$ for each $i$, or
move to a measure space representing $\frak A$ and quote 458E.)}

\spheader 458Lh\cmmnt{ Similarly, we can translate 458F into this
language.}  Let
$P:L^1(\frak A,\bar\mu)\to L^1(\frak C,\bar\mu\restrp\frak C)
\subseteq L^1(\frak A,\bar\mu)$
be the conditional expectation operator associated with
$\frak C$\cmmnt{ (365Q again)}.   Suppose that
$\familyiI{\frak B_i}$ is a family of closed subalgebras of $\frak A$ which
is relatively independent over $\frak C$.   Then

\Centerline{$\int_c\prod_{j=0}^nPu_j\le\int_c\prod_{j=0}^nu_j$}

\noindent whenever $c\in\frak C$, $i_0,\ldots,i_n\in I$ and
$u_j\in L^1(\frak B_{i_j},\bar\mu\restrp\frak B_{i_j})^+$ for each
$j\le n$, with equality if $i_0,\ldots,i_n$ are distinct.

\leader{458M}{Proposition}
Let $(\frak A,\bar\mu)$ be a probability algebra and
$\frak B$, $\frak C$ closed subalgebras of $\frak A$.   Write
$P_{\frak B}$, $P_{\frak C}$ and $P_{\frak B\cap\frak C}$
for the conditional expectation operators associated with
$\frak B$, $\frak C$ and $\frak B\cap\frak C$.   Then the following are
equiveridical:

(i) $\frak B$ and $\frak C$ are relatively independent over
$\frak B\cap\frak C$;

(ii) $P_{\frak B\cap\frak C}(v\times w)
=P_{\frak B\cap\frak C}v\times P_{\frak B\cap\frak C}w$ whenever
$v\in L^{\infty}(\frak B)$ and $w\in L^{\infty}(\frak C)$;

(iii) $P_{\frak B}P_{\frak C}=P_{\frak B\cap\frak C}$;

(iv) $P_{\frak B}P_{\frak C}=P_{\frak C}P_{\frak B}$;

(v) $P_{\frak B}u\in L^0(\frak C)$
for every $u\in L^1(\frak C,\bar\mu\restrp\frak C)$.

\proof{ Write $P$ for $P_{\frak B\cap\frak C}$.

\medskip

{\bf (i)$\Rightarrow$(ii)} If (i) is true,
$v\in L^{\infty}(\frak B)$ and $w\in L^{\infty}(\frak C)$, then
$Pv\times Pw$ certainly belongs to
$L^{\infty}(\frak B\cap\frak C)$, and if $d\in\frak B\cap\frak C$,
$\int_dPv\times Pw=\int_dv\times w$ by 458Lh.   So
$Pv\times Pw=P(v\times w)$.

\medskip

{\bf (ii)$\Rightarrow$(i)} If (ii) is true, $b\in\frak B$, $c\in\frak C$
and $d\in\frak B\cap\frak C$, then

\Centerline{$\bar\mu(d\Bcap b\Bcap c)=\int_d\chi b\times\chi c
=\int_dP(\chi b\times\chi c)=\int_dP(\chi b)\times P(\chi c)$}

\noindent as required by the definition in 458La.

\medskip

{\bf (ii)$\Rightarrow$(iii)} Suppose that (ii) is true.
First note that if $w\in L^{\infty}(\frak C)$ then
$Pw=P_{\frak B}w$.   \Prf\ Of course
$Pw\in L^{\infty}(\frak B\cap\frak C)\subseteq L^{\infty}(\frak B)$.
If $b\in\frak B$, then

$$\eqalignno{\int_bw
&=\int\chi b\times w
=\int P(\chi b\times w)
=\int P\chi b\times Pw
=\int\chi b\times Pw\cr
\displaycause{365Qa}
&=\int_bPw,\cr}$$

\noindent so that $Pw$ possesses the defining properties of
$P_{\frak B}w$.\ \Qed

But this means that if $u\in L^{\infty}(\frak A)$,
$P_{\frak B}P_{\frak C}u=PP_{\frak C}u$, which in turn is equal to
$Pu$ just because $\frak B\cap\frak C\subseteq\frak C$ (see 233Eh).
As $u$ is arbitrary, $P_{\frak B}P_{\frak C}$ agrees with $P$ on
$L^{\infty}(\frak A)$;  but $L^{\infty}(\frak A)$ is $\|\,\|_1$-dense in
$L^1(\frak A,\bar\mu)$, and $P_{\frak B}P_{\frak C}$ and $P$ are both
$\|\,\|_1$-continuous, so they agree everywhere on
$L^1(\frak A,\bar\mu)$ and are equal, as required by (iii).

\medskip

{\bf (iii)$\Rightarrow$(ii)} Suppose that (iii) is true, and that
$v\in L^{\infty}(\frak B)$, $w\in L^{\infty}(\frak C)$ and
$d\in\frak B\cap\frak C$.   Then

$$\eqalignno{\int_dPv\times Pw
&=\int\chi d\times Pv\times Pw
=\int\chi d\times v\times Pw\cr
\displaycause{because $\chi d\times Pw\in L^{\infty}(\frak B\cap\frak C)$}
&=\int\chi d\times v\times P_{\frak B}w\cr
\displaycause{because $P_{\frak C}w=w$}
&=\int\chi d\times v\times w\cr
\displaycause{because $\chi d\times v\in L^{\infty}(\frak B)$}
&=\int_dv\times w.\cr}$$

\noindent As $d$ is arbitrary and
$Pv\times Pw\in L^{\infty}(\frak B\cap\frak C)$,
$Pv\times Pw=P(v\times w)$.

\medskip

{\bf (i)$\Rightarrow$(iv)} follows immediately from (i)$\Rightarrow$(iii)
and the symmetry of the relation `$\frak B$ and $\frak C$ are
relatively independent over $\frak B\cap\frak C$'.

\medskip

{\bf (iv)$\Rightarrow$(v)} If (iv) is true and
$u\in L^1(\frak C,\bar\mu\restrp\frak C)$, then

\Centerline{$P_{\frak B}u=P_{\frak B}P_{\frak C}u
=P_{\frak C}P_{\frak B}u\in L^0(\frak C)$,}

\noindent so (v) is true.

\medskip

{\bf (v)$\Rightarrow$(iii)} If (v) is true, and
$u\in L^1_{\bar\mu}$, then
$P_{\frak C}u\in L^1(\frak C,\bar\mu\restrp\frak C)$, so
$P_{\frak B}P_{\frak C}u$ belongs to
$L^0(\frak C)\cap L^0(\frak B)=L^0(\frak B\cap\frak C)$, and of course

\Centerline{$\int_dP_{\frak B}P_{\frak C}u
=\int_dP_{\frak C}u=\int_du$}

\noindent for every
$d\in\frak B\cap\frak C$.   So $P_{\frak B}P_{\frak C}u=Pu$.
As $u$ is arbitrary, (iii) is true.
}%end of proof of 458M

\leader{458N}{Relative free products of probability algebras:
Definition} Let
$\familyiI{(\frak A_i,\bar\mu_i)}$ be a family of probability algebras
and $(\frak C,\bar\nu)$ a probability algebra, and suppose that we are
given a measure-preserving Boolean homomomorphism
$\pi_i:\frak C\to\frak A_i$
for each $i\in I$.   A {\bf relative free product} of
$\familyiI{(\frak A_i,\bar\mu_i,\pi_i)}$ over
$(\frak C,\bar\nu)$ is a probability algebra
$(\frak A,\bar\mu)$, together with a measure-preserving Boolean
homomorphism $\phi_i:\frak A_i\to\frak A$ for each $i\in I$, such that

\inset{$\frak A$ is the closed subalgebra of itself generated by
$\bigcup_{i\in I}\phi_i[\frak A_i]$,

$\phi_i\pi_i=\phi_j\pi_j:\frak C\to\frak A$ for all $i$,
$j\in I$,

writing $\frak D$ for the common value of the $\phi_i[\pi_i[\frak C]]$,
$\familyiI{\phi_i[\frak A_i]}$ is relatively independent over
$\frak D$.}

\cmmnt{\medskip

\noindent{\bf Remark} The homomorphisms $\pi_i$ and $\phi_i$ are essential
for the formal content of this definition, and will necessarily appear in
the basic result 458O.   But conceptually they are a nuisance;  we should
much prefer to think of every $\frak A_i$ as a subalgebra of $\frak A$, and
of $\frak C$ as actually equal to $\bigcap_{i\in I}\frak A_i=\frak D$.
It may help if I spell out
the key condition `$\familyiI{\phi_i[\frak A_i]}$ is relatively independent
over $\frak D$' in terms of $\frak C$ and the $\pi_i$.

The common value $\pi$ of the
$\phi_i\pi_i$ is a measure-preserving isomorphism between $\frak C$ and
$\frak D$, so gives rise to an $f$-algebra isomorphism
$S:L^0(\frak C)\to L^0(\frak D)$ such that
$S(\chi c)=\chi(\pi c)$ for every $c\in\frak C$ (364P);
note that $S[L^{\infty}(\frak C)]=L^{\infty}(\frak D)$ and
$\int Su\,d\bar\mu=\int u\,d\bar\nu$ for every $u\in L^1(\frak C)$
(365N\formerly{3{}65O}).
If $u\in L^{\infty}(\frak C)$ and $d\in\frak D$, then

$$\eqalign{\int_dSu\,d\bar\mu
&=\int Su\times\chi d\,d\bar\mu
=\int Su\times S(\chi(\pi^{-1}d))d\bar\mu\cr
&=\int S(u\times\chi(\pi^{-1}d))d\bar\mu
=\int u\times\chi(\pi^{-1}d)\,d\bar\nu
=\int_{\pi^{-1}d}u\,d\bar\nu.\cr}$$

Next, for $i\in I$ and
$a\in\frak A_i$, we have a completely additive functional
$c\mapsto\bar\mu_i(a\Bcap\pi_ic):\frak C\to[0,1]$;  let
$u_{ia}\in L^{\infty}(\frak C)$ be a corresponding Radon-Nikod\'ym
derivative, so that $\int_cu_{ia}d\bar\nu=\bar\mu_i(a\Bcap\pi_ic)$ for
every $c\in\frak C$ (365E).   (Thus $u_{ia}\in L^{\infty}(\frak C)$
corresponds to the conditional expectation of $\chi a$ on the algebra
$\pi_i[\frak C]\subseteq\frak A_i$.)    The image $Su_{ia}$ in
$L^{\infty}(\frak D)$ is defined by the property

\Centerline{$\int_dSu_{ia}d\bar\mu
=\int_{\pi^{-1}d}u_{ia}d\bar\nu
=\bar\mu_i(a\Bcap\pi_i(\phi_i\pi_i)^{-1}d)
=\bar\mu_i(a\Bcap\phi_i^{-1}d)
=\bar\mu(\phi_ia\Bcap d)$}

\noindent for every $d\in\frak D$;  that is, $Su_{ia}$ is the conditional
expectation of $\chi(\phi_ia)$ on $\frak D$.

Note also that $\frak D\subseteq\phi_i[\frak A_i]$ for every $i\in I$.
So we can use the criterion of 458B/458Lc to see that

$$\eqalign{\familyiI{\phi_i[\frak A_i]}&\text{ is relatively independent
  over }\frak D\cr
&\text{iff }\bar\mu(\inf_{i\in J}\phi_ia_i)
  =\int\prod_{i\in J}Su_{i,a_i}d\bar\mu\cr
&\mskip50mu\text{ whenever }J\subseteq I
  \text{ is finite and not empty and }a_i\in\frak A_i\text{ for }i\in J\cr
&\text{iff }\bar\mu(\inf_{i\in J}\phi_ia_i)
  =\int\prod_{i\in J}u_{i,a_i}d\bar\nu\cr
&\mskip50mu\text{ whenever }J\subseteq I
  \text{ is finite and not empty and } a_i\in\frak A_i
  \text{ for }i\in J\cr}$$

\noindent because $S$ is multiplicative, so we always have

\Centerline{$\int\prod_{i\in J}Su_{i,a_i}d\bar\mu
=\int S(\prod_{i\in J}u_{i,a_i})d\bar\mu
=\int\prod_{i\in J}u_{i,a_i}d\bar\nu$.}
}%end of comment

\leader{458O}{Theorem} Let
$\familyiI{(\frak A_i,\bar\mu_i)}$ be a family of probability algebras,
$(\frak C,\bar\nu)$ a probability algebra and
$\pi_i:\frak C\to\frak A_i$ a
measure-preserving Boolean homomomorphism for each $i\in I$.  Then
$\familyiI{(\frak A_i,\bar\mu_i,\pi_i)}$ has an essentially unique
relative free product over $(\frak C,\bar\nu)$.

\proof{{\bf (a)(i)} Let $\frak B$ be the free product of
$\familyiI{\frak A_i}$ (315I);  write
$\varepsilon_i:\frak A_i\to\frak B$ for the canonical embedding of
$\frak A_i$ in $\frak B$.   For each $i\in I$, $a\in\frak A_i$ let
$u_{ia}\in L^{\infty}(\frak C)$ be such that
$\int_cu_{ia}d\bar\nu=\bar\mu_i(a\Bcap\pi_ic)$ for
every $c\in\frak C$ (365E again).

Because the map
$a\mapsto u_{ia}:\frak A_i\to L^{\infty}(\frak C)$ is additive for each
$i$, 326E tells us that there is a unique additive
functional $\lambda:\frak B\to[0,1]$ such that

\Centerline{$\lambda(\inf_{i\in J}\varepsilon_ia_i)
=\biggerint\prod_{i\in J}u_{i,a_i}d\bar\nu$}

\noindent whenever $J\subseteq I$ is a non-empty finite set and
$a_i\in\frak A_i$ for every $i\in J$.   Of course $u_{i1}=\chi 1$ in
$L^{\infty}(\frak C)$ for every $i\in I$, interpreting the `$1$' in
$u_{i1}$ in the Boolean algebra $\frak A_i$, and the `$1$' in $\chi 1$
in the Boolean algebra $\frak C$;  so (this time interpreting `$1$' in
$\frak B$) $\lambda 1=1$ (the final `$1$' being a real number).

We see also that $u_{i,\pi_ic}=\chi c$ whenever $i\in I$ and $c\in\frak C$.
\Prf

\Centerline{$\int_du_{i,\pi_ic}d\bar\nu=\bar\mu_i(\pi_ic\Bcap\pi_id)
=\bar\nu(c\Bcap d)=\int_d\chi c\,d\bar\nu$}

\noindent for every $d\in\frak D$.\ \Qed
\medskip

\quad{\bf (ii)} By 392I, there are a probability algebra
$(\frak A,\bar\mu)$ and a
Boolean homomorphism $\phi:\frak B\to\frak A$ such that
$\lambda=\bar\mu\phi$.   We can of course suppose that $\frak A$ is the
order-closed subalgebra of itself generated by $\phi[\frak B]$ (which is
in fact automatically the case if we use the construction in the proof
of 392I).

For each $i\in I$, set $\phi_i=\phi\varepsilon_i:\frak A_i\to\frak A$.
This is a Boolean homomorphism because $\phi$ and $\varepsilon_i$ are.
If $a\in\frak A_i$, then

\Centerline{$\bar\mu\phi_ia
=\bar\mu(\phi\varepsilon_ia)=\lambda\varepsilon_ia
=\int u_{ia}d\bar\nu
=\bar\mu_i(a\Bcap\pi_i1)=\bar\mu_ia$,}

\noindent so $\phi_i$ is measure-preserving.

If $i$, $j\in I$ and $c\in\frak C$ then

$$\eqalign{\bar\mu(\phi_i\pi_ic\Bsymmdiff\phi_j\pi_jc)
&=\lambda(\varepsilon_i\pi_ic\Bsymmdiff\varepsilon_j\pi_jc)\cr
&=\lambda(\varepsilon_i\pi_ic)+\lambda(\varepsilon_j\pi_jc)
  -2\lambda(\varepsilon_i\pi_ic\Bcap\varepsilon_j\pi_jc)\cr
&=\int u_{i,\pi_ic}d\bar\nu+\int u_{j,\pi_jc}d\bar\nu
  -2\int u_{i,\pi_ic}\times u_{j,\pi_jc}d\bar\nu\cr
&=\int\chi c\,d\bar\nu+\int\chi c\,d\bar\nu
  -2\int\chi c\times\chi c\,d\bar\nu
=0.\cr}$$

\noindent So $\phi_i\pi_i=\phi_j\pi_j$.  Let $\frak D$ be the common value
of $\phi_i[\pi_i[\frak C]]$.   (In the trivial case $I=\emptyset$, take
$\frak D=\frak A=\{0,1\}$.)

\medskip

\quad{\bf (iii)} Suppose that
$J\subseteq I$ is finite and not empty and
that $a_i\in\frak A_i$ for each $i\in J$.   Then

$$\bar\mu(\inf_{i\in J}\phi_ia_i)
=\bar\mu(\inf_{i\in J}\phi\varepsilon_ia_i)
=\bar\mu\phi(\inf_{i\in J}\varepsilon_ia_i)
=\lambda(\inf_{i\in J}\varepsilon_ia_i)
=\int\prod_{i\in J}u_{i,a_i}d\bar\nu.$$

\noindent But this is precisely the condition described in 458N, so
$\familyiI{\phi_i[\frak A_i]}$ is relatively independent over $\frak D$,
and $(\frak A,\bar\mu,\familyiI{\phi_i})$ is a relative free product of
$\familyiI{(\frak A_i,\bar\mu_i,\pi_i)}$ over $(\frak C,\bar\nu)$.

\medskip

{\bf (b)} Now suppose that $(\frak A',\bar\mu',\familyiI{\phi'_i})$ is
another relative free product of $\familyiI{(\frak A_i,\bar\mu_i,\pi_i)}$
over $(\frak C,\bar\nu)$.   Then we have a Boolean
homomorphism $\psi:\frak B\to\frak A'$ such that
$\phi'_i=\psi\varepsilon_i$ for every $i\in I$ (315J).   In this case,
$\bar\mu'\psi=\lambda$.   \Prf\ Let $\pi'$ be the common value of
$\phi'_i\pi_i$ for $i\in I$ and set $\frak D'=\pi'[\frak C]$
If $J\subseteq I$ is finite and not empty, and $a_i\in\frak A_i$ for
$i\in J$, then

\Centerline{$\bar\mu'(\psi(\inf_{i\in J}\varepsilon_ia_i))
=\bar\mu'(\inf_{i\in J}\phi'_ia_i)
=\int\prod_{i\in J}u_{i,a_i}d\bar\nu
=\lambda(\inf_{i\in J}\varepsilon_ia_i)$.}

\noindent Because $\lambda$ is the only additive functional on $\frak B$
taking the right values on elements of this form,
$\bar\mu'\psi=\lambda$.\ \Qed

In particular, $\psi b=0$ whenever $b\in\frak B$ and $\lambda b=0$.  It
follows that $\psi b=\psi b'$ whenever $b$, $b'\in\frak B$ and
$\phi b=\phi b'$, since in this case
$\lambda(b\Bsymmdiff b')=\bar\mu(\phi b\Bsymmdiff\phi b')$ is zero.
So we have a function $\theta:\phi[\frak B]\to\frak A'$ defined by
setting $\theta(\phi b)=\psi b$ for every $b\in\frak B$, and of course
$\theta$ is a Boolean homomorphism;  moreover,

\Centerline{$\bar\mu'\theta(\phi b)=\bar\mu'\psi b=\lambda b
=\bar\mu\phi b$}

\noindent for every $b$, so $\theta$ is measure-preserving and an
isometry for the
measure metrics of $\frak A$ and $\frak A'$.   If $i\in I$ and
$a\in\frak A$, then

\Centerline{$\theta\phi_ia=\theta\phi\varepsilon_ia=\psi\varepsilon_ia
=\phi_i'a$,}

\noindent so $\theta\phi_i=\phi'_i$ for every $i$.   Because $\frak A$
and $\frak A'$ are the closed subalgebras generated by
$\bigcup_{i\in I}\phi_i[\frak A_i]$ and
$\bigcup_{i\in I}\phi'_i[\frak A_i]$ respectively, $\phi[\frak B]$ and
$\psi[\frak B]$ are dense (323J).   The isometry $\theta$ therefore
extends uniquely to a measure algebra isomorphism
$\hat\theta:\frak A\to\frak A'$ which must be the unique isomorphism
such that $\hat\theta\phi_i=\phi'_i$ for every $i$.   Thus
$(\frak A,\bar\mu,\familyiI{\phi_i})$ and
$(\frak A',\bar\mu',\familyiI{\phi'_i})$ are isomorphic, and the
relative free product is essentially unique.
}%end of proof of 458O

\leader{458P}{}\cmmnt{ Developing the argument of the last part of
the proof of 458O, we have the following.

\medskip

\noindent}{\bf Theorem} Let
$\familyiI{(\frak A_i,\bar\mu_i)}$, $\familyiI{(\frak A'_i,\bar\mu'_i)}$
be two families of probability algebras, and
$\psi_i:\frak A_i\to\frak A'_i$ a measure-preserving Boolean
homomorphism for each $i$.   Let $(\frak C,\bar\nu)$,
$(\frak C',\bar\nu')$ be probability algebras and
$\pi_i:\frak C\to\frak A_i$, $\pi'_i:\frak C'\to\frak A'_i$
measure-preserving Boolean homomomorphisms for each $i\in I$;  suppose
that we have a measure-preserving isomorphism $\psi:\frak C\to\frak C'$
such that $\pi'_i\psi=\psi_i\pi_i:\frak C\to\frak A'_i$ for each $i$.
Let $(\frak A,\bar\mu,\familyiI{\phi_i})$ and
$(\frak A',\bar\mu',\familyiI{\phi'_i})$ be relative free products of
$\familyiI{(\frak A_i,\bar\mu_i,\pi_i)}$,
$\familyiI{(\frak A'_i,\bar\mu'_i,\pi'_i)}$ over $(\frak C,\bar\nu)$,
$(\frak C',\bar\nu')$ respectively.   Then there is a unique
measure-preserving Boolean homomorphism $\hat\psi:\frak A\to\frak A'$
such that $\hat\psi\phi_i=\phi'_i\pi_i:\frak A_i\to\frak A'$ for every
$i\in I$.

\proof{ By the uniqueness assertion of 458O, we may suppose that
$(\frak A,\bar\mu,\familyiI{\phi_i})$ has been constructed by the method
of part (a) of the proof of 458O.

\medskip

{\bf (a)} For $i\in A$, $a\in\frak A$ and $a'\in\frak A'$ let
$u_{ia}\in L^{\infty}(\frak C)$, $u'_{ia'}\in L^{\infty}(\frak C')$ be
defined as in (a-i) of the proof of 458O, so that

\Centerline{$\int_cu_{ia}d\bar\nu=\bar\mu_i(a\Bcap\pi_ic)$,
\quad$\int_{c'}u'_{ia'}d\bar\nu'=\bar\mu'_i(a'\Bcap\pi'_ic')$}

\noindent whenever $c\in\frak C$ and $c'\in\frak C'$.   Let
$T:L^0(\frak C)\to L^0(\frak C')$ be the $f$-algebra isomorphism
such that $T(\chi c)=\chi(\psi c)$ for every
$c\in\frak C$.   Now $u'_{i,\psi_ia}=Tu_{ia}$ whenever
$i\in I$ and $a\in\frak A_i$.   \Prf\ If $c\in\frak C$, then

$$\eqalignno{\int_{\psi c}Tu_{ia}d\bar\nu'
&=\int_cu_{ia}d\bar\nu
=\bar\mu_i(a\Bcap\pi_ic)\cr
&=\bar\mu_i'\psi_i(a\Bcap\pi_ic)
=\bar\mu_i'(\psi_ia\Bcap\pi'_i\psi c)
=\int_{\psi c}u'_{i,\psi_ia}d\bar\nu'.\cr}$$

\noindent Because $\psi$ is surjective, it follows that
$Tu_{ia}=u'_{i,\psi_ia}$.\ \Qed

\medskip

{\bf (b)} Let $\frak B$ be the free product of $\familyiI{\frak A_i}$
and $\varepsilon_i:\frak A_i\to\frak B$ the canonical embedding for each
$i$;  let $\lambda$ be the functional on $\frak B$ defined by the
process of (a-i) in the proof of 458O.   By 315J, there is a Boolean
homomorphism $\theta:\frak B\to\frak A'$ such that
$\theta\varepsilon_i=\phi'_i\psi_i:\frak A_i\to\frak A'$ for every $i$.
Now $\bar\mu'\theta=\lambda$.   \Prf\ If $J\subseteq I$ is finite and
$a_i\in\frak A_i$ for every $i\in J$, then

$$\eqalignno{\bar\mu'\theta(\inf_{i\in J}\varepsilon_ia_i)
&=\bar\mu'(\inf_{i\in J}\phi'_i\psi_ia_i)
=\int\prod_{i\in J}u'_{i,\psi_ia_i}d\bar\nu'\cr
&=\int\prod_{i\in J}Tu_{i,a_i}d\bar\nu'
=\int T(\prod_{i\in J}u_{i,a_i})d\bar\nu'\cr
&=\int\prod_{i\in J}u_{i,a_i}d\bar\nu
=\lambda(\inf_{i\in J}\varepsilon_ia_i).\cr}$$

\noindent As $\lambda$, $\theta$ and $\bar\nu$ are all additive,
$\lambda=\bar\mu'\theta$ (using 315Kb).\ \Qed

\medskip

{\bf (c)} Let $\phi:\frak B\to\frak A$ be the map described in (a-ii) of
the proof of 458O.   Then $\bar\mu(\phi b)=\lambda b=\bar\mu'(\theta b)$
for every $b\in\frak B$;  in particular,

\Centerline{$\phi b=0\mskip8mu\Longrightarrow\mskip8mu\bar\mu(\phi b)=0
\mskip8mu\Longrightarrow\mskip8mu\bar\mu'(\theta b)=0
\mskip8mu\Longrightarrow\mskip8mu\theta b=0$.}

\noindent There is therefore a Boolean homomorphism
$\tilde\theta:\phi[\frak B]\to\frak A'$ such that
$\tilde\theta\phi=\theta$, and $\tilde\theta$ is measure-preserving on
$\phi[\frak B]$.   Since $\phi[\frak B]$ is topologically dense in
$\frak A$ (use 323J), $\tilde\theta$ has an extension to a
measure-preserving Boolean homomorphism $\hat\psi:\frak A\to\frak A'$
(324O).   Now, for $i\in I$ and $a\in\frak A_i$,

\Centerline{$\hat\psi\phi_ia
=\hat\psi\phi\varepsilon_ia
=\tilde\theta\phi\varepsilon_ia
=\theta\varepsilon_ia
=\phi'_i\psi_ia$,}

\noindent as required.

\medskip

{\bf (d)} To see that $\hat\psi$ is unique, we need observe only that
the given formula defines it on the subalgebra $\phi[\frak B]$ and that
this is topologically dense in $\frak A$, while $\hat\psi$, being
measure-preserving, must be continuous.
}%end of proof of 458P

\leader{458Q}{Relative product measures:  Definitions (a)}
Let $\familyiI{X_i}$ be a family of sets, $Y$ a set, and
$\pi_i:X_i\to Y$ a function for each $i\in I$.   The {\bf fiber product}
of $\familyiI{(X_i,\pi_i)}$ is the set
$\Delta=\{x:x\in\prod_{i\in I}X_i$, $\pi_ix(i)=\pi_jx(j)$ for all $i$,
$j\in I\}$.

\spheader 458Qb  Let
$\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of probability spaces
and $(Y,\Tau,\nu)$ a probability space, and suppose that we are given an
\imp\ function $\pi_i:X_i\to Y$ for each $i\in I$;  let $\Delta$ be the
fiber product of $\familyiI{(X_i,\pi_i)}$.   A {\bf relative
product measure} on $\Delta$ is a probability measure $\mu$ on $\Delta$
such that

\inset{($\dagger$) whenever $J\subseteq I$ is finite and not empty and
$E_i\in\Sigma_i$ for $i\in J$, and $g_i$ is a Radon-Nikod\'ym derivative
with respect to $\nu$ of the
functional $F\mapsto\mu_i(E\cap\pi_i^{-1}[F]):\Tau\to[0,1]$ for each
$i\in J$, then $\mu\{x:x\in\Delta$, $x(i)\in E_i$ for every $i\in J\}$
is defined and equal to $\int\prod_{i\in J}g_id\nu$;

($\ddagger$) for every $W\in\Sigma$ there is a $W'$ in the
$\sigma$-algebra generated by
$\{\{x:x\in\Delta$, $x(i)\in E\}:i\in I$, $E\in\Sigma_i\}$ such that
$\mu(W\symmdiff W')=0$.}

\medskip

\noindent{\bf Remark} If $\mu$ is a
relative product measure of $\familyiI{(\mu_i,\pi_i)}$ over
$\nu$, then all the functions $x\mapsto x(i):\Delta\to X_i$ are \imp.
\prooflet{\Prf\ The condition ($\dagger$) tells us that if
$E\in\Sigma_i$ and $g$ is any Radon-Nikod\'ym derivative of
$F\mapsto\mu_i(E\cap\pi_i^{-1}[F])$, then

\Centerline{$\mu\{x:x(i)\in E\}=\int g\,d\nu=\mu_iE$.\ \Qed}

\noindent}It follows that if $I$ is not empty then we have an \imp\
function $\pi:\Delta\to Y$ defined by setting $\pi x=\pi_ix(i)$ whenever
$x\in\Delta$ and $i\in I$.

\cmmnt{Note that when verifying ($\dagger$) we need check the equality
$\mu\{x:x\in\Delta$, $x(i)\in E_i$ for every $i\in J\}
=\int\prod_{i\in J}g_id\nu$ for
only one representative family $\family{i}{J}{g_i}$ of Radon-Nikod\'ym
derivatives for any given $\family{i}{J}{E_i}$.}

\vleader{36pt}{458R}{Proposition} Suppose that
$\familyiI{(X_i,\Sigma_i,\mu_i)}$ is a family of probability spaces,
$(Y,\Tau,\nu)$ a probability space, $\pi_i:X_i\to Y$ an \imp\
function for each $i\in I$, $\Delta$ the fiber product of
$\familyiI{(X_i,\pi_i)}$ and $\mu$ a relative product measure of
$\familyiI{(\mu_i,\pi_i)}$.   Let $(\frak A_i,\bar\mu_i)$,
$(\frak C,\bar\nu)$ and $(\frak A,\bar\mu)$
be the measure algebras of $\mu_i$, $\nu$ and
$\mu$ respectively, and for $i\in I$ define
$\bar\pi_i:\frak C\to\frak A_i$ and $\bar\phi_i:\frak A_i\to\frak A$ by
setting $\bar\pi_iF^{\ssbullet}=\pi_i^{-1}[F]^{\ssbullet}$,
$\bar\phi_iE^{\ssbullet}=\{x:x\in\Delta$, $x(i)\in E\}^{\ssbullet}$
whenever $F\in\Tau$ and
$E\in\Sigma_i$.   Then $(\frak A,\bar\mu,\familyiI{\bar\phi_i})$ is a
relative free product of $\familyiI{(\frak A_i,\bar\mu_i,\bar\pi_i)}$
over $(\frak C,\bar\nu)$.

\proof{ The case $I=\emptyset$ is trivial (if you care to follow through
the definitions to the letter, $\Delta=\prod_{i\in I}X_i=\{\emptyset\}$
and $\frak A$ is the two-point algebra).   So I will take it that $I$ is
not empty.   For $i\in I$ define $\phi_i:\Delta\to X_i$ by setting
$\phi_i(x)=x(i)$ for $x\in\Delta$.

\medskip

{\bf (a)} Of course we have to check that all the $\bar\pi_i$ and
$\bar\phi_i$ are measure-preserving Boolean homomorphisms between the
appropriate algebras, but in view of the remark following the definition
458Q, this is elementary.   The condition that $\frak A$ should be the
closed subalgebra generated by
$\bigcup_{i\in I}\bar\phi_i[\frak A_i]$ is just a translation of the
condition ($\ddagger$).

\medskip

{\bf (b)} As $I$ is not empty, we have a well-defined \imp\ map
$\pi:\Delta\to Y$ given by the formula $\pi(x)=\pi_ix(i)$ whenever
$x\in\Delta$ and $i\in I$.   Let
$\bar\pi:\frak C\to\frak A$ be the corresponding measure-preserving
homomorphism, so that $\bar\pi=\bar\phi_i\bar\pi_i$ for every $i$.   Set
$\frak D=\bar\pi[\frak C]\subseteq\frak A$, and let
$T:L^{\infty}(\frak C)\to L^{\infty}(\frak D)$ be the $f$-algebra
isomorphism corresponding to $\bar\pi$ (363F).   For $i\in I$ and
$E\in\Sigma_i$, let $g_{iE}$ be a Radon-Nikod\'ym derivative with respect
to $\nu$ of the functional $F\mapsto\mu_i(E\cap\pi_i^{-1}[F])$,
and set $u_{iE}=Tg_{iE}^{\ssbullet}\in L^{\infty}(\frak D)$.
Then $u_{iE}$ is the conditional expectation of
$\chi\{x:x(i)\in E\}^{\ssbullet}$ on $\frak D$.   \Prf\ If
$d\in\frak D$, it is of the form
$\bar\pi F^{\ssbullet}=\bar\phi_i\bar\pi_iF^{\ssbullet}$
where $F\in\Tau$, so that $\chi d=T(\chi F)^{\ssbullet}$ and

$$\eqalign{\int_du_{iE}d\bar\mu
&=\int u_{iE}\times\chi d\,d\bar\mu
=\int T(g_{iE}^{\ssbullet}\times\chi F^{\ssbullet})d\bar\mu\cr
&=\int g_{iE}^{\ssbullet}\times\chi F^{\ssbullet}d\bar\nu
=\int_Fg_{iE}d\nu\cr
&=\mu_i(E\cap\pi_i^{-1}[F])
=\mu(\phi_i^{-1}[E]\cap\phi_i^{-1}[\pi_i^{-1}[F]])
=\bar\mu(d\Bcap\phi_i^{-1}[E]^{\ssbullet}).\cr}$$

\noindent As $d$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (c)} It follows that if $J\subseteq I$ is finite and not empty, and
$a_i\in\bar\phi_i[\frak A_i]$ and $v_i$ is the conditional expectation
of $\chi a_i$ on $\frak D$ for each $i\in J$, then
$\bar\mu(\inf_{i\in J}a_i)=\int\prod_{i\in J}v_id\bar\mu$.   \Prf\
Express $a_i$ as $\phi_i^{-1}[E_i]^{\ssbullet}$, where $E_i\in\Sigma_i$,
so that $v_i=u_{i,E_i}$ for each $i$.   Then

$$\eqalign{\int\prod_{i\in J}v_id\bar\mu
&=\int\prod_{i\in J}Tg_{i,E_i}^{\ssbullet}d\bar\mu
=\int T(\prod_{i\in J}g_{i,E_i}^{\ssbullet})d\bar\mu
=\int\prod_{i\in J}g_{i,E_i}^{\ssbullet}d\bar\nu\cr
&=\int\prod_{i\in J}g_{i,E_i}d\nu
=\mu(\bigcap_{i\in J}\phi_i^{-1}[E_i])
=\bar\mu(\inf_{i\in J}a_i).  \text{ \Qed}\cr}$$

\noindent But this is exactly what we need to know to see that
$\familyiI{\bar\phi_i[\frak A_i]}$ is relatively independent over
$\frak D$, completing the proof that
$(\frak A,\bar\mu,\familyiI{\bar\phi_i})$ is a relative free product of
$\familyiI{(\frak A_i,\bar\mu_i,\bar\pi_i)}$ over $(\frak C,\bar\nu)$.
}%end of proof of 458R

\leader{458S}{}\cmmnt{ There is no general result on relative product
measures to match 458O (see 458Xj-458Xm). %458Xj 458Xk 458Xl 458Xm
The general question
of when we can expect relative product measures to exist seems
interesting (458Yf, 458Yg).   Here I give a couple of sample results
dealing with important special cases.

\medskip

\noindent}{\bf Proposition} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
family of probability spaces, $(Y,\Tau,\nu)$ a probability space, and
$\pi_i:X_i\to Y$ an \imp\ function for each $i$.   Suppose that for each
$i$ we have a disintegration $\family{y}{Y}{\mu_{iy}}$ of $\mu_i$ such
that $\mu_{iy}^*\pi_i^{-1}[\{y\}]=\mu_{iy}X_i=1$ for every $y\in Y$.
Let $\Delta\subseteq\prod_{i\in I}X_i$ be the fiber product of
$\familyiI{(X_i,\pi_i)}$, and $\Upsilon$ the subspace $\sigma$-algebra
on $\Delta$ induced by
$\Tensorhat_{i\in I}\Sigma_i$.   For $y\in Y$, let $\lambda_y$ be the
product of
$\familyiI{\mu_{iy}}$, $(\lambda_y)_{\Delta}$ the subspace measure on
$\Delta$ and $\lambda'_y$ its restriction to $\Upsilon$.   Then
$\mu W=\int\lambda'_yW\nu(dy)$ is defined for every
$W\in\Upsilon$, and $\mu$ is a
relative product measure of $\familyiI{(\mu_i,\pi_i)}$ over $\nu$.

\proof{ If $y\in Y$, then

\Centerline{$\lambda'_y\Delta=\lambda^*_y\Delta
=\lambda_y^*(\prod_{i\in I}\pi_i^{-1}[\{y\}])=1$}

\noindent (254Lb).   For $i\in I$ and $E\in\Sigma_i$ set
$g_{iE}(y)=\mu_{iy}E$ when this is defined;  then
$g_{iE}$ is a Radon-Nikod\'ym derivative of
$F\mapsto\mu_i(E\cap\pi_i^{-1}[F]):\Tau\to[0,1]$ (452Qa).   Write $X$
for $\prod_{i\in I}X_i$;  for
$i\in I$ and $x\in X$ set $\phi_i(x)=x(i)$.   If
$J\subseteq I$ is finite and $E_i\in\Sigma_i$ for each $i\in J$, then

$$\eqalignno{\int\lambda'_y(\Delta
  \cap\bigcap_{i\in J}\phi_i^{-1}[E_i])\nu(dy)
&=\int(\lambda_y)_{\Delta}(\Delta
  \cap\bigcap_{i\in J}\phi_i^{-1}[E_i])\nu(dy)\cr
&=\int\lambda_y(X\cap\bigcap_{i\in J}\phi_i^{-1}[E_i])\nu(dy)\cr
\displaycause{because $\lambda_y^*\Delta=1$ and $\lambda_y$ measures
every $\phi_i^{-1}[E_i]$ for almost every $y$}
&=\int\prod_{i\in J}\mu_{iy}E_i\nu(dy)
=\int\prod_{i\in J}g_{i,E_i}d\nu.\cr}$$

\noindent In particular,
$\int\lambda'_y(\Delta\cap\bigcap_{i\in J}\phi_i^{-1}[E_i])\nu(dy)$ is
defined.

The set
$\{W:W\subseteq X$, $\int\lambda'_y(W\cap\Delta)\nu(dy)$ is defined$\}$
is a Dynkin class of subsets of $X$ containing
$\bigcap_{i\in J}\phi_i^{-1}[E_i]$ whenever $J\subseteq I$ is
finite and
not empty and $E_i\in\Sigma_i$ for each $i\in J$;  by the Monotone Class
Theorem, it includes $\Tensorhat_{i\in I}\Sigma_i$.   So
$\mu W=\int\lambda'_yW\nu(dy)$ is defined for every $W\in\Upsilon$.
Moreover, the formula displayed above tells us that
$\mu(\Delta\cap\bigcap_{i\in I}\phi_i^{-1}[E_i])
=\int\prod_{i\in J}g_{i,E_i}d\nu$
whenever $J\subseteq I$ is finite and $E_i\in\Sigma_i$ for each
$i\in I$.   Thus ($\dagger$) of 458Q is satisfied.   And ($\ddagger$) is
true by the choice of $\Upsilon$.
}%end of proof of 458S

\leader{458T}{}\cmmnt{ The latitude I have permitted in the definition
of `relative product' makes it possible to look for relative product
measures with further properties, as in the following.

\medskip

\noindent}{\bf Proposition} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a family of compact Radon
probability spaces, $(Y,\frak S,\Tau,\nu)$ a Radon probability space,
and $\pi_i:X_i\to Y$ a continuous \imp\ function for each $i$.   Then
$\familyiI{(\mu_i,\pi_i)}$ has a relative product measure $\mu$
over $\nu$ which is a Radon measure for the topology on the fiber
product of $\familyiI{(X_i,\pi_i)}$ induced by the product topology on
$\prod_{i\in I}X_i$.

\proof{{\bf (a)} For $i\in I$ and $E\in\Sigma_i$ let $g_{iE}$ be a
Radon-Nikod\'ym derivative of
$F\mapsto\mu_i(E\cap\pi_i^{-1}[F]):\Tau\to[0,1]$.   Let $\Cal C$ be the
family of measurable cylinders in $X=\prod_{i\in I}X_i$;  write
$\phi_i(x)=x(i)$ for $x\in X$ and $i\in I$.   We have a
functional $\lambda_0:\Cal C\to[0,1]$ defined by setting

\Centerline{$\lambda_0(\bigcap_{i\in J}\phi_i^{-1}[E_i])
=\int\prod_{i\in J}g_{i,E_i}d\nu$}

\noindent whenever $J\subseteq I$ is finite and not empty and
$E_i\in\Sigma_i$ for every $i\in I$.   It is easy to check that
$\lambda_0$ is additive in the sense required by 454E
so (because every $\mu_i$
is perfect, by 416Wa) it has an extension to a measure
$\lambda$ on $X$ with domain $\Tensorhat_{i\in I}\Sigma_i$.   By 454Aa,
with $\Cal K$ the family of compact subsets of $X$, $\lambda$ is inner
regular with respect to the compact sets.   By 416N, there is a Radon
measure $\tilde\lambda$ on $X$ extending $\lambda$.

\medskip

{\bf (b)} Let $\Delta$ be the fiber product of $\familyiI{(X_i,\pi_i)}$.
Now the point is that
$\Delta$ is $\tilde\lambda$-conegligible.   \Prf\ Because every $\pi_i$
is continuous, $\Delta$ is closed.   \Quer\ If it is not conegligible,
then, because $\tilde\lambda$ is $\tau$-additive, there must be a basic
open set of non-zero measure disjoint from $\Delta$;  express such a set
as $W=\prod_{i\in J}\phi_i^{-1}[G_i]$ where $J\subseteq I$ is finite and
$G_i\subseteq X_i$ is
open for each $i\in J$.   Because $\tilde\lambda$ is inner regular with
respect to the compact sets, there is a compact set $K\subseteq W$ such
that $\tilde\lambda K>0$;  setting $K_i=\phi_i[K]$,
$K_i\subseteq G_i$ is compact for each $i$ and
$W'=\bigcap_{i\in J}\phi_i^{-1}[K_i]$ is non-negligible.   Now we have

\Centerline{$0<\tilde\lambda(\prod_{i\in J}\phi_i^{-1}[K_i])
=\lambda_0(\prod_{i\in J}\phi_i^{-1}[K_i])
=\int\prod_{i\in J}g_{i,K_i}d\nu$,}

\noindent so $F=\{y:y\in Y$, $g_{i,K_i}(y)>0$ for every $i\in J\}$ is
non-negligible.   On the other hand, for each $i\in J$ we have

\Centerline{$\int_{Y\setminus\pi_i[K_i]}g_{i,K_i}d\nu
=\mu_i(K_i\Bcap\pi_i^{-1}[Y\setminus K_i])=0$}

\noindent so that $F\setminus\pi_i[K_i]$ is negligible.   Accordingly
$\bigcap_{i\in J}\pi_i[K_i]$ is non-negligible, and must meet the
support $Y_0$ of $Y$;  let $y$ be any point of the intersection.   For
$i\in J$, choose $x(i)\in K_i$ such that $\pi_ix(i)=y$.   For
$i\in I\setminus J$, $\pi_i[X_i]$ is a compact subset of $Y$, and
$\nu\pi_i[X_i]=\mu_i\pi_i^{-1}[\pi_i[X_i]]=1$, so
$Y_0\subseteq\pi_i[X_i]$ and we can therefore choose $x(i)\in X_i$ with
$\pi_ix(i)=y$.   This defines $x\in\Delta$.   But as $x(i)\in K_i$ for
$i\in J$, we also have

\Centerline{$x\in\bigcap_{i\in J}\phi_i^{-1}[K_i]
\subseteq\bigcap_{i\in J}\phi_i^{-1}[G_i]\subseteq X\setminus\Delta$,}

\noindent which is impossible.\ \Bang

Thus $\Delta$ is $\tilde\lambda$-conegligible, as claimed.\ \Qed

\medskip

{\bf (c)} Let $\mu$ be the subspace measure on $\Delta$ induced by
$\tilde\lambda$, and $\Sigma$ its domain, so that $\mu$ is a Radon
probability measure on $\Delta$ with its subspace topology
(416Rb).   Concerning ($\dagger$) of 458Q, if $J\subseteq I$ is finite
and not empty and $E_i\in\Sigma_i$ for $i\in J$, then

\Centerline{$\mu(\Delta\cap\bigcap_{i\in J}\phi_i^{-1}[E_i])
=\tilde\lambda(\bigcap_{i\in J}\phi_i^{-1}[E_i])
=\lambda_0(\bigcap_{i\in J}\phi_i^{-1}[E_i])
=\int\prod_{i\in J}g_{i,E_i}d\nu$,}

\noindent as required.   Finally, for ($\ddagger$), the $\sigma$-algebra
$\Upsilon$ of subsets of $\Delta$ generated by
$\{\Delta\cap\phi_i^{-1}[E]:i\in I$, $E\in\Sigma_i\}$ is just the subspace
$\sigma$-algebra induced by $\Tensorhat_{i\in I}\Sigma_i$.   Let
$\frak A$ be the measure algebra of $\tilde\lambda$ and
$\frak B\subseteq\frak A$ the set
$\{W^{\ssbullet}:W\in\Tensorhat_{i\in I}\Sigma_i\}$.   Then $\frak B$ is
a closed subalgebra of $\frak A$.   If $W\subseteq X$ is open, then for
every $\epsilon>0$ there is a $W_0\in\Tensorhat_{i\in I}\Sigma_i$ such
that $W_0\subseteq W$ and $\tilde\lambda(W\setminus W_0)\le\epsilon$, so
$W^{\ssbullet}\in\frak B$;  accordingly $\{W:W^{\ssbullet}\in\frak B\}$
contains every open set and every Borel set and must be the whole of
$\dom\tilde\lambda$.   Returning to the measure $\mu$, we see that if
$W\in\Sigma$ there must be a $W_0\in\Tensorhat_{i\in I}\Sigma_i$ such
that $\tilde\lambda(W\symmdiff W_0)=0$;  now $W_0\cap\Delta\in\Upsilon$
and
$\mu(W\symmdiff(W_0\cap\Delta))=0$.    So ($\ddagger$) also is true, and
we have a relative product measure of the declared type.
}%end of proof of 458T

\leader{458U}{}\cmmnt{ We can of course make a general search through
theorems about product measures, looking for ways of re-presenting them
as theorems about relative product measures.   There is an associative
law, for instance (458Xr).   To give an idea of what is to be expected,
I offer a result corresponding to 253D.

\medskip

\noindent}{\bf Proposition} Let $(X_1,\Sigma_1,\mu_1)$,
$(X_2,\Sigma_2,\mu_2)$ and $(Y,\Tau,\nu)$ be probability spaces, and
$\pi_1:X_1\to Y$, $\pi_2:X_2\to Y$ \imp\ functions.   Let $\Delta$ be
the fiber product of $(X_1,\pi_1)$ and $(X_2,\pi_2)$, and suppose that
$\mu$ is a relative product measure of $(\mu_1,\pi_1)$ and
$(\mu_2,\pi_2)$ over $\nu$;  set $\pi x=\pi_1x(1)=\pi_2x(2)$ for
$x\in\Delta$.   Take $f_1\in\eusm L^1(\mu_1)$ and
$f_2\in\eusm L^2(\mu_2)$, and set
$(f_1\otimes f_2)(x)=f_1(x(1))f_2(x(2))$ when
$x\in\Delta\cap(\dom f_1\times\dom f_2)$.   For $i=1$, $2$ let
$g_i\in\eusm L^1(\nu)$ be a Radon-Nikod\'ym derivative of
$H\mapsto\int_{\pi_i^{-1}[H]}f_id\mu_i:\Tau\to\Bbb R$.
Then $\int_Fg_1\times g_2d\nu=\int_{\pi^{-1}[F]}f_1\otimes f_2d\mu$ for
every $F\in\Tau$.

\proof{{\bf (a)} Suppose first that $f_1=\chi E_1$ and $f_2=\chi E_2$ where
$E_1\in\Sigma_1$ and $E_2\in\Sigma_2$.   For each $i$, set
$f'_i=\chi(E_i\cap\pi_i^{-1}[F])$ and
$g'_i=g_i\times\chi F$;  then

\Centerline{$\int_Hg'_id\nu=\int_{H\cap F}g_id\nu
=\int_{\pi_i^{-1}[H\cap F]}f_id\mu_i=\int_{\pi_i^{-1}[H]}f'_id\mu_i$}

\noindent for every $H\in\Tau$.   Now

$$\eqalignno{\int_Fg_1\times g_2\,d\nu&
=\int g'_1\times g'_2\,d\nu
=\mu\{x:x\in\Delta,\,x(i)\in E'_i\text{ for both }i\}
\Displaycause{$(\dagger)$ of the definition 458Qb}
=\mu\{x:x\in\pi^{-1}[F],\,x(i)\in E_i\text{ for both }i\}
=\int_{\pi^{-1}[F]}f_1\times f_2d\mu.\cr}$$

\noindent So we have the result in this case.   (Cf.\ 458B.)

\medskip
the
functional $F\mapsto\mu_i(E\cap\pi_i^{-1}[F]):\Tau\to[0,1]$
{\bf (b)} Generally, the
formula for $g_i$ corresponds to a linear operator from $L^1(\mu_i)$
to $L^1(\nu)$, so the result is true for simple functions $f_1$ and
$f_2$.   If $f_1$ and $f_2$ are almost everywhere limits of
non-decreasing sequences $\sequencen{f_{1n}}$, $\sequencen{f_{2n}}$ of
non-negative simple functions, then the corresponding sequences
$\sequencen{g_{1n}}$, $\sequencen{g_{2n}}$ will also be non-decreasing
and non-negative and convergent to $g_1$, $g_2\,\,\nu$-a.e.;  moreover,
because $x\mapsto x(1)$ and $x\mapsto x(2)$ are \imp,
$f_1\otimes f_2=\lim_{n\to\infty}f_{1n}\otimes f_{2n}\,\,\mu$-a.e.   So
in this case we shall have

$$\eqalign{\int_{\pi^{-1}[F]}f_1\otimes f_2d\mu
&=\lim_{n\to\infty}\int_{\pi^{-1}[F]}f_{1n}\otimes f_{2n}d\mu\cr
&=\lim_{n\to\infty}\int_Fg_{1n}\times g_{2n}d\nu
=\int_Fg_1\times g_2d\nu\cr}$$

\noindent for every $F\in\Tau$.   Finally, considering positive and
negative parts, we can extend the result to general integrable $f_1$ and
$f_2$.
}%end of proof of 458U

\exercises{\leader{458X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 458Xa
Find an example of a probability space $(X,\Sigma,\mu)$
with $\sigma$-subalgebras $\Sigma_1$, $\Sigma_2$ and $\Tau$ of $\Sigma$
such that $\Sigma_1$ and $\Sigma_2$ are independent but are not
relatively independent over $\Tau$.
%458A

\spheader 458Xb Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$,
$\Sigma_1$ and $\Sigma_2\,\,\sigma$-subalgebras of $\Sigma$.   Show that if
$\Sigma_1\subseteq\Tau$ then $\Sigma_1$ and $\Sigma_2$ are relatively
independent over $\Tau$.
%458A

\sqheader 458Xc Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$ a
subalgebra of $\Sigma$.   Let $\familyiI{\Cal E_i}$ be a family of
subsets of $\Sigma$ such that (i) each $\Cal E_i$ is closed under finite
intersections (ii) $\familyiI{E_i}$ is relatively independent over $\Tau$
whenever $E_i\in\Cal E_i$ for every $i$.   For each $i\in I$,
let $\Sigma_i$ be the
$\sigma$-subalgebra of $\Sigma$ generated by $\Cal E_i$.   Show that
$\familyiI{\Sigma_i}$ is relatively independent over $\Tau$.
%458D

\sqheader 458Xd Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$ a
$\sigma$-subalgebra of $\Sigma$.   Let $f_1$, $f_2$ be $\mu$-integrable
real-valued functions which are relatively independent over $\Tau$, and
suppose that $f_1\times f_2$ is integrable.
Let $g_1$, $g_2$ be conditional expectations of $f_1$, $f_2$ on $\Tau$.
Show that $g_1\times g_2$ is a conditional expectation of
$f_1\times f_2$ on $\Tau$.
%458F  

\spheader 458Xe In 458I, show that (writing $\hat\mu$ for the completion
of $\mu$) $\hat\mu(F\cap f^{-1}[H])=\int_F\nu_xH\mu(dx)$ for every
$F\in\Tau$ and every universally measurable $H\subseteq\Bbb R$.
%458I

\spheader 458Xf Let $(X,\Sigma,\mu)$ be a probability space and $\Sigma_1$,
$\Sigma_2$ and $\Tau\,\,\sigma$-subalgebras of $\Sigma$.
Show that the following are
equiveridical:  (i) $\Sigma_1$ and $\Sigma_2$ are relatively independent
over $\Tau$;  (ii) whenever $f\in\eusm L^1(\mu\restr\Sigma_1)$ and $g$ is a
conditional expectation of $f$ on $\Tau$, then $g$ is a conditional
expectation of $f$ on $\Sigma_2\vee\Tau$.
%458H

\spheader 458Xg Prove 458Ld directly from 313G, without appealing to 458D.
%458L

\spheader 458Xh Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a family of
probability algebras.   Show that their probability algebra free product
(325K) can be identified with their relative free product over
$(\frak C,\bar\nu)$ if $\frak C$ is the two-element
Boolean algebra, $\bar\nu$ its unique probability measure, and
$\pi_i:\frak C\to\frak A_i$ the trivial Boolean homomorphism for every
$i$.
%458O

\sqheader 458Xi Let $Y$ be a set, $\familyiI{Z_i}$ a family of sets, and
$\pi_i:Y\times Z_i\to Y$ the canonical map for each $i$.   Show that the
fiber product of $\familyiI{(Y\times Z_i,\pi_i)}$ can be identified with
$Y\times\prod_{i\in I}Z_i$.
%458Q

\spheader 458Xj Let $\nu$ be Lebesgue
measure on $[0,1]$, and $X_1$,
$X_2\subseteq[0,1]$ disjoint sets with outer measure $1$.   For each
$i\in\{1,2\}$ let $\mu_i$ be the
subspace measure on $X_i$ and
$\pi_i:X_i\to[0,1]$ the identity map.   Show that $(\mu_1,\pi_1)$ and
$(\mu_2,\pi_2)$ have no relative product measure over $\nu$.
%458Q

\spheader 458Xk Let $\nu$ be the usual measure on the split interval
$I^{\|}$ (343J), and $\mu$ Lebesgue measure on
$[0,1]$.   Set $\pi_1(t)=t^+$, $\pi_2(t)=t^-$ for
$t\in[0,1]$.   Show that $(\mu,\pi_1)$ and
$(\mu,\pi_2)$ have no relative product measure over $\nu$.
%458Q

\spheader 458Xl Let $\nu$ be Lebesgue measure on $[0,1]$.   For each
$t\in[0,1]$, set $X_t=[0,1]\setminus\{t\}$;  let
$\mu_t$ be the subspace measure on $X_t$ and
$\pi_t:X_t\to[0,1]$ the identity map.   Show that
$\family{t}{[0,1]}{(\mu_t,\pi_t)}$ has no relative product measure
over $\nu$.
%458Q

\spheader 458Xm(i) Show that there is a set $X\subseteq[0,1]^2$ with outer
planar Lebesgue measure $1$ and just one point in each vertical section.
\Hint{419H-419I.}  (ii) Set $X_1=X_2=X$ and
$\mu_1=\mu_2$ the subspace measure on $X$;
let $(Y,\Tau,\nu)$ be $[0,1]$ with Lebesgue measure, and $\pi_1=\pi_2$ the
first-coordinate projection from $X$ to $Y$.   Show that
$(\mu_1,\pi_1)$ and $(\mu_2,\pi_2)$ have no
relative product measure over $\nu$.
%458Q

\spheader 458Xn Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a
$\sigma$-subalgebra of $\Sigma$, and $\familyiI{\Sigma_i}$ a family of
$\sigma$-subalgebras of $\Sigma$, all including $\Tau$.   Set
$\pi_i(x)=x$ for every
$x\in X$.   Show that $\familyiI{\Sigma_i}$ is relatively independent
over $\Tau$ iff  $\mu\restr\Sigma^*$ is a
relative product measure of $\familyiI{(\mu\restr\Sigma_i,\pi_i)}$
over $\mu\restrp\Tau$, where $\Sigma^*=\bigvee_{i\in I}\Sigma_i$.
%458Q

\spheader 458Xo(i) Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces and $(X,\Sigma,\mu)$ their ordinary probability space
product.   Show that $\mu$ is a relative
product measure of $\familyiI{(\mu_i,\pi_i)}$ over
$\nu$ where $Y$ is a singleton set, $\nu$ its
unique probability measure, and $\pi_i:X_i\to Y$ the unique function for
each $i$.   (ii) Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of quasi-Radon probability spaces and $(X,\frak T,\Sigma,\mu)$
their quasi-Radon probability space product (417R).   Show that
$\mu$ is a relative product measure of
$\familyiI{\mu_i}$ in the same sense as in (i).
%458Q

\spheader 458Xp Suppose that $\familyiI{(X_i,\Sigma_i,\mu_i)}$ is a
family of probability spaces and $(Y,\Tau,\nu)$ is a probability space,
and that for each $i\in I$ we are given an \imp\ function
$\pi_i:X_i\to Y$.   Write $\hat\mu_i$ and
$\hat\nu$ for the completions of $\mu_i$, $\nu$ respectively.   Show
that $\familyiI{(\mu_i,\pi_i)}$ has a relative product measure over
$\nu$ iff $\familyiI{(\hat\mu_i,\pi_i)}$ has a relative product measure
over $\hat\nu$.
%458Q

\spheader 458Xq Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
family of probability spaces, $(Y,\Tau,\nu)$ a probability space, and
$\pi_i:X_i\to Y$ an \imp\ function for each $i\in I$.   Show that if
$\familyiI{(\mu_i,\pi_i)}$ has a relative product measure over $\nu$, so
does $\family{i}{J}{(\mu_i,\pi_i)}$ for any $J\subseteq I$.
%458Q

\spheader 458Xr Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
family of probability spaces, $(Y,\Tau,\nu)$ a probability space, and
$\pi_i:X_i\to Y$ an \imp\ function for each $i\in I$.   Let
$\family{k}{K}{J_k}$ be a partition of $I$ into non-empty sets.   For
each $k\in K$, let $\Delta_k$ be the fiber product of
$\family{i}{J_k}{(X_i,\pi_i)}$;  suppose that $\tilde\mu_k$ is a
relative product measure of $\family{i}{J_k}{(\mu_i,\pi_i)}$.
Define $\tilde\pi_k:\Delta_k\to Y$ by
setting $\tilde\pi_k(x)=\pi_ix(i)$ whenever $x\in\Delta_k$ and
$i\in J_k$, so that $\tilde\pi_k$ is \imp.  Suppose that $\mu$ is a
relative product measure of $\family{k}{K}{(\tilde\mu_k,\tilde\pi_k)}$
over $\nu$.   Show that $\mu$ can be regarded as a relative product
measure of $\familyiI{(\mu_i,\pi_i)}$ over $\nu$.
%458Q

\spheader 458Xs Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ and
$\familyiI{(X'_i,\Sigma'_i,\mu'_i)}$ be two families of
probability spaces, $(Y,\Tau,\nu)$ a probability
spaces, and $f_i:X_i\to X'_i$, $\pi_i:X'_i\to Y'$ \imp\ functions for
each $i$.   Show that if
there is a relative product measure of $\familyiI{(\mu_i,\pi_if_i)}$
over $\nu$, then there is a relative product measure of
$\familyiI{(\mu'_i,\pi_i)}$ over $\nu$.
%458R

\spheader 458Xt
Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a countable
family of probability spaces, $(Y,\Tau,\nu)$ a probability space, and
$\pi_i:X_i\to Y$ an \imp\ function for each $i$.   Suppose that for each
$i$ we have a disintegration of $\mu_i$ over $\nu$ which is strongly
consistent with $\pi_i$.   Show that
$\familyiI{(\mu_i,\pi_i)}$ has a relative product measure over
$\nu$.
%458S

\spheader 458Xu Let $(X,\Sigma,\mu)$, $(X',\Sigma',\mu')$ and
$(Y,\Tau,\nu)$ be probability spaces.   Suppose that $\pi:X\to Y$
and $\pi':X'\to Y$ are \imp\ functions, and that $\mu'$ has a
disintegration $\family{y}{Y}{\mu'_y}$ over $(Y,\Tau,\nu)$ which is
strongly consistent with $\pi'$.   Show that $(\mu,\pi)$ and
$(\mu',\pi')$
have a relative product measure over $\nu$.   \Hint{set
$\lambda W=\int\mu'_{\pi(x)}W[\{x\}]\mu(dx)$ for every
$W\in\Sigma\tensorhat\Sigma'$.}
%458S

\sqheader 458Xv Let $Y$ be a Hausdorff space, $\familyiI{Z_i}$ a family
of Hausdorff spaces, $\mu_i$ a Radon probability measure on
$Z_i\times Y$ and $\pi_i:Y\times Z_i\to Y$ the canonical map for each
$i$.   Suppose that all the image measures $\mu_i\pi_i^{-1}$ on $Y$ are
the same, and that all but countably many of the $Z_i$ are compact.
Show that there is a Radon probability measure $\mu$ on
$Y\times\prod_{i\in I}Z_i$ such that $\mu_i=\mu\phi_i^{-1}$ for each
$i$, where $\phi_i(y,z)=(y,z(i))$ for $y\in Y$, $z\in\prod_{j\in I}Z_j$.
%458T 416O

\spheader 458Xw Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
countable family of Radon
probability spaces, $(Y,\frak S,\Tau,\nu)$ a Radon probability space,
and $\pi_i:X_i\to Y$ an almost continuous \imp\ function for each $i$.
Show that $\familyiI{(\mu_i,\pi_i)}$ has a relative product measure
over $\nu$ which is a Radon measure for the topology on the fiber
product of $\familyiI{(X_i,\pi_i)}$ induced by the product topology on
$\prod_{i\in I}X_i$.   Discuss the relation of this result to 418Q.
%458T

\leader{458Y}{Further exercises (a)}%
%\spheader 458Ya
(i) Let $(X,\Sigma,\mu)$ be a probability space,
$\sequencen{\Tau_n}$ a non-increasing sequence of $\sigma$-subalgebras of
$\Sigma$ with intersection $\Tau$, and $\familyiI{\Sigma_i}$ a family of
subalgebras of $\Sigma$.   Suppose that $\familyiI{\Sigma_i}$ is relatively
independent over $\Tau_n$ for every $n$.   Show that it is relatively
independent over $\Tau$.   \Hint{275K.}
(ii) Give an example of a probability space
$(X,\Sigma,\mu)$, a downwards-directed
family $\Bbb T$ of $\sigma$-subalgebras of $\Sigma$,
and a family $\familyiI{E_i}$ in $\Sigma$ which is
relatively independent over $\Tau$ for every $\Tau\in\Bbb T$, but not over
$\bigcap\Bbb T$.
%458C mt45bits

\spheader 458Yb Let $(X,\Sigma,\mu)$ be a probability space
and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Let
$\familyiI{\Cal E_i}$ be a family of subsets of $\Sigma$ such that (i)
$E\cap F\in\Cal E_i$ whenever $i\in I$ and $E$, $F\in\Cal E_i$ (ii)
$\familyiI{E_i}$ is relatively independent over $\Tau$ whenever
$E_i\in\Cal E_i$ for every $i\in I$.   For each $i\in I$,
let $\Sigma_i$ be the $\sigma$-algebra generated by $\Cal E_i$.   Show that
$\familyiI{\Sigma_i}$ is relatively independent over $\Tau$.
%458D 

\spheader 458Yc Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a
$\sigma$-subalgebra of $\Sigma$, and $\sequencen{\Sigma_n}$ a sequence
of $\sigma$-subalgebras of $\Sigma$ which is relatively independent over
$\Tau$.   Show that for every
$E\in\bigcap_{n\in\Bbb N}\bigvee_{m\ge n}\Sigma_n$
there is an $F\in\Tau$ such that $E\symmdiff F$ is
negligible.   (Compare 272O.)
%relative zero-one law
%458H

\spheader 458Yd Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a
$\sigma$-subalgebra of $\Sigma$, and $f$, $g\in\eusm L^0(\mu)$
relatively independent over $\Tau$;  suppose that $\family{x}{X}{\nu_x}$
and $\family{x}{X}{\nuprime_x}$ are relative distributions of $f$ and
$g$ over $\Tau$.   Show that $\family{x}{X}{\nu_x*\nuprime_x}$ is a
relative distribution of $f+g$ over $\Tau$.   (Compare 272T.)
%relative convolutions
%458H

\spheader 458Ye Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a
$\sigma$-subalgebra of $\Sigma$, and $\sequencen{f_n}$ a sequence in
$\eusm L^2(\mu)$ such that $\sequencen{f_n}$ is relatively independent
over $\Tau$ and $\int_Ff_nd\mu=0$ for every $n\in\Bbb N$ and every
$F\in\Tau$.   (i) Suppose that $\sequencen{\beta_n}$ is a non-decreasing
sequence in $\ooint{0,\infty}$, diverging to $\infty$, such that
$\sum_{n=0}^{\infty}\Bover1{\beta_n^2}\|f_n\|_2^2<\infty$.   Show that
$\lim_{n\to\infty}\Bover1{\beta_n}\sum_{i=0}^nf_i=0$ a.e.   (ii)
Suppose that $\sup_{n\in\Bbb N}\|f_n\|_2<\infty$.   Show that
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf_i=0$ a.e.   (Compare 273D.)
%relative strong law
%458H

\spheader 458Yf Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces, $(Y,\Tau,\nu)$ a probability space, and $\pi_i:X_i\to Y$ an
\imp\ function for each $i\in I$.   For $i\in I$ and $E\in\Sigma_i$ let
$g_{iE}$ be a Radon-Nikod\'ym derivative of the functional
$F\mapsto\mu_i(E\cap\pi_i^{-1}[F])$.   Let $\Cal C$ be the family of
measurable cylinders in $X=\prod_{i\in I}X_i$.   If
$C=\{x:x\in X$, $x(i)\in E_i$ for every $i\in J\}$ where
$J\subseteq I$
is finite and not empty and $E_i\in\Sigma_i$ for $i\in J$, set
$\lambda_0C=\int\prod_{i\in J}g_{i,E_i}d\nu$.   Let $\Delta\subseteq X$
be the fiber product of $\familyiI{(X_i,\pi_i)}$.   Show that the
following are equiveridical:  (i) $\familyiI{(\mu_i,\pi_i)}$ has a
relative product measure over $\nu$ (ii) whenever
$\sequencen{C_n}$ is a sequence in $\Cal C$ covering $\Delta$,
$\sum_{n=0}^{\infty}\lambda_0C_n\ge 1$.
%458Q

\spheader 458Yg Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces, $(Y,\Tau,\nu)$ a probability space, and
$\pi_i:X_i\to Y$ a
surjective \imp\ function for each $i\in I$.   Suppose that
$\family{i}{J}{(\mu_i,\pi_i)}$ has a relative product measure over $\nu$
for every countable $J\subseteq I$.   Show that
$\family{i}{I}{(\mu_i,\pi_i)}$ has a relative product measure over $\nu$.
%458Yf 458Q

\spheader 458Yh Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a countable
family of perfect
probability spaces, $(Y,\Tau,\nu)$ a countably separated
probability space, and
$\pi_i:X_i\to Y$ an \imp\ function for each $i\in I$.   Show that
$\familyiI{(\mu_i,\pi_i)}$ has a relative product measure over $\nu$.
%458Yf 458S

\spheader 458Yi Let $(\frak A,\bar\mu)$ be a probability algebra and
$\frak C$ a closed subalgebra of $\frak A$.   Let
$\frak C_0\subseteq\frak C$ be the core subalgebra of countable Maharam
type described in the
canonical form of such structures given in 333N.   Show that there is a
closed subalgebra $\frak B$ of $\frak A$, including $\frak C_0$, such that
$\frak B$ and $\frak C$ are relatively independent over $\frak C_0$, and
$\frak A$ is the closed subalgebra of itself generated by
$\frak B\cup\frak C$.
%+
}%end of exercises

\endnotes{
\Notesheader{458}
The elementary theory of relative independence has two aspects.   First,
there is the matter of systematically formulating and verifying appropriate
variations on standard results on stochastic independence;  458F, 458H,
458J, 458K, 458Xd, 458Yc-458Ye %458Yc 458Yd 458Ye
come under this heading.   More interestingly, we study the new phenomena
associated with changes in the core $\sigma$-algebras, as in 458C, 458D
and 458Xa.

At a couple of points in Volume 3 (Dye's theorem, in \S388, and
Kawada's theorem, in \S395) I took the trouble to generalize standard
theorems to `non-ergodic' forms.   In both 388L and 395P the results are
complicated
by potentially non-trivial closed subalgebras of the probability algebra
we are studying.   I remarked on both occasions that the generalization
is only a matter of technique, but I do not suppose that it was obvious
just why this must be so.   It is however a fundamental theorem of the
topic of `random reals' in the theory of forcing that {\it any} theorem
about probability algebras must have a relativized form as a theorem
about probability algebras with arbitrary closed subalgebras.   The
concept of `relative Maharam type' from \S333, for instance, is what
matches `Maharam type' for simple algebras;  the concept of
`exchangeable' sequence (definition:  459C)
is what matches `independent identically
distributed' sequence.   (In probability theory, the keyword
is `mixture'.)   In this section I present another example in the idea
of `relatively independent' closed subalgebras (458L-458M).   I should
emphasize that the forcing method, when we eventually come to it
in \S556 in Volume 5,
will not as a rule apply directly to measure spaces;  it
deals with measure algebras.   But of course the ideas generated by this
theory can often be profitably applied to constructions in measure
spaces, and this is what I am seeking to do with relatively independent
$\sigma$-algebras and relative product measures.

Just as independent $\sigma$-algebras are associated with product spaces
(272J), relatively independent algebras are associated with relative
products (458Xn).   The archetype of a relative product measure is
458S;  it is a kind of disintegrated product.   It is frequently
profitable to express the `relative' concepts of measure theory in terms
of disintegrations.

I introduce `relative free products' of probability algebras before
proceeding
to measure spaces because the uniqueness property proved in 458O shows
that we have an unambiguous definition.   For measure spaces it seems
for the moment better to leave ourselves a bit of freedom, not (for
instance) favouring one product construction over another (458Xo).
The requirement that a relative product measure be carried by the fiber
product is seriously limiting
(458Xj-458Xl, %458Xj 458Xk 458Xl
458Yf), and forces us to seek strongly consistent disintegrations
(458S), at least for uncountable products (see 458Xt).   However, as we
might hope, the special case of compact spaces with Radon measures and
continuous functions is amenable to a different approach (458T);  and we
have a one-sided method for the product of two spaces (458Xu) which is
reminiscent of 454C and 457F.

There are corresponding complications when we
come to look at maps between different relative products.   For
measure algebras, we have a natural theorem (458P), based on the
same algebraic considerations as the corresponding theorems in \S\S315
and 325;  the only possibly surprising feature is the need to assume
that $\psi:\frak C\to\frak C'$ is actually an isomorphism.   For measure
spaces there is a similar result (458Xs).

}%end of notes

\discrpage

%}}}}}

\frfilename{mt459.tex}
\versiondate{7.12.10}
\copyrightdate{2003}

\def\chaptername{Perfect measures, disintegrations and processes}
\def\sectionname{Symmetric measures and exchangeable random variables}

\newsection{459}

Among the relatively independent families of random variables discussed
in 458K,
it is natural to give extra attention to those which are `relatively
identically distributed'.   It turns out that these have a particularly
appealing characterization as the `exchangeable' families (459C).   In
the same way, among the measures on a product space $X^I$ there is a
special place for those which are invariant under permutations of
coordinates (459E, 459H).   A more abstract kind of permutation-invariance
is examined in 495L-495M.

\leader{459A}{}\cmmnt{ The following elementary fact seems to have
gone unmentioned so far.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces and $\phi:X\to Y$ an \imp\ function;  set
$\Sigma_0=\{\phi^{-1}[F]:F\in\Tau\}$.   Let $\Tau_1$ be a
$\sigma$-subalgebra of $\Tau$ and
$\Sigma_1=\{f^{-1}[F]:F\in\Tau_1\}$.   If $g\in\eusm L^1(\nu)$ and $h$
is a conditional expectation of $g$ on $\Tau_1$, then $h\phi$ is a
conditional expectation of $g\phi$ on $\Sigma_1$.

\proof{ $h$ is $\nu\restr\Tau_1$-integrable and $\phi$ is \imp\ for
$\mu\restr\Sigma_1$ and $\nu\restr\Tau_1$, so $h\phi$ is
$\mu\restr\Sigma_1$-integrable (235G).
If $E\in\Sigma_1$ then there is an
$F\in\Tau_1$ such that $E=\phi^{-1}[F]$, and now

\Centerline{$\int_Eg\phi\,d\mu=\int_{f^{-1}[F]}g\phi\,d\mu=\int_Fg\,d\nu
=\int_Fh\,d\nu=\int_Eh\phi\,d\mu$.}

\noindent As $E$ is arbitrary, $h\phi$ is a conditional expectation of
$g\phi$ on $\Sigma_1$.
}%end of proof of 459A

\vleader{72pt}{459B}{Theorem} Let $(X,\Sigma,\mu)$ be
a probability space, $Z$ a set, $\Upsilon$ a $\sigma$-algebra of subsets
of $Z$ and $\familyiI{f_i}$ an infinite family of
$(\Sigma,\Upsilon)$-measurable functions from $X$ to $Z$.   For each
$i\in I$, set $\Sigma_i=\{f_i^{-1}[H]:H\in\Upsilon\}$.
Then the following are equiveridical:

\inset{(i) whenever $i_0,\ldots,i_r\in I$ are distinct,
$j_0,\ldots,j_r\in I$ are distinct, and $H_k\in\Upsilon$ for each
$k\le r$, then
$\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k])$;

(ii) there is a $\sigma$-subalgebra $\Tau$ of $\Sigma$ such that

\quad($\alpha$) $\familyiI{\Sigma_i}$ is relatively independent over
$\Tau$,

\quad($\beta$) whenever $i$, $j\in I$, $H\in\Upsilon$ and $F\in\Tau$,
then $\mu(F\cap f_i^{-1}[H])=\mu(F\cap f_j^{-1}[H])$.}

\noindent Moreover, if $I$ is totally ordered by $\le$, we can add

\inset{(iii) whenever $i_0<\ldots<i_r\in I$, $j_0<\ldots<j_r\in I$ and
$H_k\in\Upsilon$ for each $k\le r$, then
$\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k])$.}

\proof{{\bf (a)} Since there is always some total
order on $I$, we may assume that we have one from the start.
Of course (i)$\Rightarrow$(iii).   Also (ii)$\Rightarrow$(i).
\Prf\ Suppose that (ii) is true.   Then (ii-$\beta$) tells us that for
each $H\in\Upsilon$ there is a $\Tau$-measurable function
$g_H:X\to[0,1]$ which is a conditional expectation of
$\chi(f_i^{-1}[H])$ on $\Tau$ for every $i\in I$.   Now suppose that
$i_0,\ldots,i_r\in I$ are distinct, $j_0,\ldots,j_r\in I$ are distinct,
and $H_k\in\Upsilon$ for each $k\le r$.   Then

\Centerline{$\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
=\int(\prod_{k=0}^rg_{H_k})d\mu
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k])$}

\noindent because $\familyiI{\Sigma_i}$ is relatively independent over
$\Tau$.   So (i) is true.\ \Qed

So henceforth I will
suppose that (iii) is true and seek to prove (ii).

\medskip

{\bf (b)} Suppose first that $I=\Bbb N$ with its usual ordering.

\medskip

\quad\grheada\
For each $n$, $r\in\Bbb N$, let $\Sigma_{nr}$ be the $\sigma$-subalgebra
of $\Sigma$ generated by $\bigcup_{n\le i\le n+r}\Sigma_i$;  let
$\Tau_n$ be the
$\sigma$-algebra generated by $\bigcup_{r\in\Bbb N}\Sigma_{nr}$, and
$\Tau=\bigcap_{n\in\Bbb N}\Tau_n$.   For $n\in\Bbb N$ and
$H\in\Upsilon$, let $g_{nH}:X\to\Bbb R$ be a $\Tau$-measurable
function which is a conditional expectation of
$\chi f_n^{-1}[H]$ on $\Tau$.

\medskip

\quad\grheadb\ (The key.)  For any $n\in\Bbb N$ and Borel set
$H\subseteq\Bbb R$, $g_{nH}$ is a conditional expectation of
$\chi f_n^{-1}[H]$ on $\Tau_{n+1}$.
\Prf\ For $m$, $r\in\Bbb N$, let $h_{mr}:X\to[0,1]$ be a
$\Sigma_{mr}$-measurable function which is a conditional
expectation of $\chi f_n^{-1}[H]$ on $\Sigma_{mr}$;  for $m\in\Bbb N$,
set $h_m=\lim_{r\to\infty}h_{mr}$ where this is defined.
By L\'evy's martingale theorem (275I) $h_m$ is defined almost everywhere
and is a conditional expectation of $\chi f_n^{-1}[H]$ on $\Tau_m$.

For $m$, $r\in\Bbb N$, define $F_{mr}:X\to Z^{r+2}$ by setting
$F_{mr}(x)=(f_n(x),f_m(x),f_{m+1}(x),\ldots,\penalty-100f_{m+r}(x))$ for
$x\in X$.   At this point, examine the hypothesis (iii).   This implies
that if $m>n$ and $r\in\Bbb N$ then

$$\eqalign{\mu(F_{mr}^{-1}[H'\times H_0\times\ldots\times H_r])
&=\mu(f_n^{-1}[H']\cap\bigcap_{k\le r}f_{m+k}^{-1}[H_k])\cr
&=\mu(f_n^{-1}[H']\cap\bigcap_{k\le r}f_{m+1+k}^{-1}[H_k])\cr
&=\mu(F_{m+1,r}^{-1}[H'\times H_0\times\ldots\times H_r])\cr}$$

\noindent for all $H'$, $H_0,\ldots,H_r\in\Upsilon$.   By the Monotone
Class Theorem (136C), the image measures $\mu F_{mr}^{-1}$ and
$\mu F_{m+1,r}^{-1}$ agree on the $\sigma$-algebra
$\Tensorhat_{r+2}\Upsilon$ of subsets of $Z^{r+2}$ generated by
measurable cylinders;  set

\Centerline{$\lambda=\mu F_{mr}^{-1}\restr\Tensorhat_{r+2}\Upsilon
=\mu F_{m+1,r}^{-1}\restr\Tensorhat_{r+2}\Upsilon$.}

\noindent Let $\Lambda$ be the $\sigma$-subalgebra of
$\Tensorhat_{r+2}\Upsilon$ generated by sets of the form
$Z\times H_0\times\ldots\times H_r$ where $H_0,\ldots,H_r\in\Upsilon$,
and let $h$ be a conditional
expectation of $\chi(H\times Z^{r+1})$ on $\Lambda$ with respect to
$\lambda$.   Then 459A tells us that $hF_{mr}$ is a conditional
expectation of
$\chi(f_n^{-1}[H])$ on $\Sigma_{mr}$, and is therefore equal almost
everywhere to $h_{mr}$.
Similarly, $hF_{m+1,r}\eae h_{m+1,r}$, and this is true for every
$r\in\Bbb N$.   But as $F_{mr}$ and $F_{m+1,r}$
are both \imp\ for $\mu$ and $\lambda$, this means that $h_{mr}$, $h$
and $h_{m+1,r}$ all have the same distribution.
In particular, $\int h_{mr}^2d\mu=\int h_{m+1,r}^2d\mu$.   Now
$\sequence{r}{h_{mr}}$ and $\sequence{r}{h_{m+1,r}}$ converge almost
everywhere to $h_m$ and $h_{m+1}$ respectively, so

\Centerline{$\int h_m^2d\mu=\lim_{r\to\infty}\int h_{mr}^2d\mu
=\lim_{r\to\infty}\int h_{m+1,r}^2d\mu=\int h_{m+1}^2d\mu$.}

\noindent On the other hand, $\Tau_{m+1}\subseteq\Tau_m$, so
$h_{m+1}$ is a conditional expectation of
$h_m$ on $\Tau_{m+1}$ (233Eh).   This means that

\Centerline{$\int h_m\times h_{m+1}d\mu
=\int h_{m+1}\times h_{m+1}d\mu$}

\noindent (233Eg).   A direct calculation tells us that
$\int(h_m-h_{m+1})^2d\mu=0$, so that $h_m\eae h_{m+1}$.   Inducing on
$r$, we see that $h_m\eae h_r$  whenever $n<m\le r$.

Now the reverse martingale theorem (275K) tells us that
$\lim_{m\to\infty}h_m$ is
defined almost everywhere and is a conditional expectation of
$\chi f_n^{-1}[H]$ on $\Tau$, that is, is equal almost everywhere to
$g_{nH}$.   Since the $h_m$, for $m>n$, are equal almost everywhere,
they are all equal to $g_{nH}$ a.e.   In particular, $g_{nH}$ is equal
a.e.\ to $h_{n+1}$, and is a
conditional expectation of $\chi f_n^{-1}[H]$ on $\Tau_{n+1}$.\ \Qed

\medskip

\quad\grheadc\ If $n\in\Bbb N$ and $H_0,\ldots,H_r\in\Upsilon$, then
$\prod_{i=0}^rg_{n+i,H_i}$ is a conditional expectation of
$\chi(\bigcap_{i\le r}f_{n+i}^{-1}[H_i])$ on $\Tau$.   \Prf\ Induce on
$r$.   For $r=0$ this is just the definition of $g_{nH_0}$.   For
the inductive step to $r\ge 1$, observe that
$g_{nH_0}\times\prod_{i=1}^r\chi f_{n+i}^{-1}[H_i]$ is a
conditional expectation of $\prod_{i=0}^r\chi f_{n+i}^{-1}[H_i]$
on $\Tau_{n+1}$, by 233Eg or 233K, because $g_{nH_0}$ is a
conditional expectation of $\chi f_n^{-1}[H_0]$ on $\Tau_{n+1}$ and
$\prod_{i=1}^r\chi f_{n+i}^{-1}[H_i]$ is
$\Tau_{n+1}$-measurable.   But as (by the inductive hypothesis)
$\prod_{i=1}^rg_{n+i,H_i}$ is a conditional expectation of
$\prod_{i=1}^r\chi f_{n+i}^{-1}[H_i]$ on $\Tau$, while $g_{nH_0}$ is
$\Tau$-measurable,
$\prod_{i=0}^rg_{n+i,H_i}$ is a conditional expectation of
$\prod_{i=0}^r\chi f_{n+i}^{-1}[H_i]$ on $\Tau$, by 233Eg/233K
again.\ \Qed

\medskip

\quad\grheadd\ In particular, $\prod_{i=0}^rg_{iH_i}$ is a conditional
expectation of $\chi(\bigcap_{i\le r}f_i^{-1}[H_i])$ on $\Tau$ for every
$r\in\Bbb N$ and $H_0,\ldots,H_r\in\Upsilon$.   This
shows that $\sequencen{\Sigma_n}$ is relatively independent over $\Tau$.

\medskip

\quad\grheade\ Now consider part ($\beta$) of the condition (ii).   For
this, observe
that if $m>0$, $H\in\Upsilon$ and $H_i\in\Upsilon$ for $i\le r$, then

\Centerline{$\mu(f_0^{-1}[H]\cap\bigcap_{i\le r}f_{m+i+1}^{-1}[H_i])
=\mu(f_m^{-1}[H]\cap\bigcap_{i\le r}f_{m+i+1}^{-1}[H_i])$.}

\noindent By the Monotone Class Theorem,

\Centerline{$\mu(F\cap f_0^{-1}[H])=\mu(F\cap f_m^{-1}[H])$}

\noindent for any $F\in\Tau_{m+1}$ and in particular for any
$F\in\Tau$.
Thus (ii) is true.

\medskip

{\bf (c)} Now suppose that there is a strictly increasing sequence
$\sequence{k}{j_k}$ in $I$.   For each $n$, let $\Tau_n$ be the
$\sigma$-algebra generated by $\bigcup_{k\ge n}\Sigma_{j_k}$, and set
$\Tau=\bigcap_{n\in\Bbb N}\Tau_n$.   Then (b), applied to
$\sequence{k}{f_{j_k}}$, tells us that $\sequence{k}{\Sigma_{j_k}}$ is
relatively independent over $\Tau$ and that for each $H\in\Upsilon$
there is a function $g_H$ which is a conditional expectation of
$\chi(f_{j_k}^{-1}[H])$ on $\Tau$ for every $k\in\Bbb N$.

\medskip

\quad\grheada\ If $i_0,\ldots,i_r\in I$ are distinct and
$H_0,\ldots,H_r\in\Upsilon$, then
$\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k])$.   \Prf\ Let $\rho$ be the
permutation of $\{0,\ldots,r\}$ such that
$i_{\rho(0)}<i_{\rho(1)}<\ldots<i_{\rho(r)}$.   Then

$$\eqalign{\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
&=\mu(\bigcap_{k\le r}f_{i_{\rho(k)}}^{-1}[H_{\rho(k)}])
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_{\rho(k)}])\cr
&=\int(\prod_{k=0}^rg_{H_{\rho(k)}})d\mu
=\int(\prod_{k=0}^rg_{H_k})d\mu
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k]).  \text{ \Qed}\cr}$$

\medskip

\quad\grheadb\ Now suppose that $i_0,\ldots,i_r\in I$ are distinct.
Then there is some $m\in\Bbb N$ such that
$j_k\ne i_l$ for any $l\le r$ and $k\ge m$.   In this case, consider the
sequence $f_{i_0},\ldots,f_{i_r},f_{j_m},f_{j_{m+1}},\ldots$.   By
($\alpha$) here, this sequence satisfies the condition (iii).   We can
therefore apply the construction of (b).   But observe that the tail
$\sigma$-algebra obtained from $f_{i_0},\ldots,f_{i_r},f_{j_m},\ldots$
is precisely $\Tau$, as defined from $\sequence{k}{f_{j_k}}$ just above.
So $\langle\Sigma_{i_k}\rangle_{k\le r}$ is relatively independent over
$\Tau$.   As $i_0,\ldots,i_r$ are arbitrary, $\familyiI{\Sigma_i}$ is
relatively independent over $\Tau$.   At the same time we see that if
$H\in\Upsilon$ then all the $\chi(f_{i_k}^{-1}[H])$ have the same
conditional expectations over $\Tau$ as $\chi(f_{j_m}^{-1}[H])$.   So
(ii-$\beta$) is satisfied.

\medskip

{\bf (d)} Finally, if there is no strictly increasing sequence in $I$,
then $(I,\ge)$ is well-ordered;  since $I$ is infinite, the
well-ordering starts with an initial segment of order type $\omega$, that is,
a sequence $\sequence{k}{j_k}$ such that $j_0>j_1>\ldots$.   But note
now that (iii) tells us that
$\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k])$ whenever
$i_0>\ldots>i_r$ and $j_0>\ldots>j_r$ and $H_k\in\Upsilon$ for every
$k$.   So we can apply (c) to
$(I,\ge)$ to get the result in this case also.
}%end of proof of 459B

\vleader{72pt}{459C}{Exchangeable random \dvrocolon{variables}}\cmmnt{ I
spell out the leading special case of this theorem.

\medskip

\noindent}{\bf De Finetti's theorem} Let $(X,\Sigma,\mu)$ be
a probability space, and $\familyiI{f_i}$ an infinite family in
$\eusm L^0(\mu)$.   Then the following are equiveridical:

\inset{(i) the joint distribution of $(f_{i_0},f_{i_1},\ldots,f_{i_r})$
is the same as the joint distribution of
$(f_{j_0},f_{j_1},\ldots,f_{j_r})$
whenever $i_0,\ldots,i_r\in I$ are distinct and $j_0,\ldots,j_r\in I$
are distinct;
%overfull hbox reported;  no problem

(ii) there is a $\sigma$-subalgebra $\Tau$ of $\Sigma$ such that
$\familyiI{f_i}$ is relatively independent over $\Tau$ and all the
$f_i$ have the same relative distribution over $\Tau$.}

\noindent Moreover, if $I$ is totally ordered by $\le$, we can add

\inset{(iii) the joint distribution of
$(f_{i_0},f_{i_1},\ldots,f_{i_r})$ is the
same as the joint distribution of $(f_{j_0},f_{j_1},\ldots,f_{j_r})$
whenever $i_0<\ldots<i_r$ and $j_0<\ldots<j_r$ in $I$.}
%overfull hbox reported;  no problem

\cmmnt{\medskip

\noindent{\bf Remark} Families of random variables
satisfying the condition in (i) are called {\bf exchangeable}.   The
equivalence of (i) and (ii) can be expressed by
saying that `an exchangeable family of random variables is a mixture
of independent identically distributed families'.
}%end of comment

\proof{ Changing each $f_i$ on a negligible set will
not change either their joint distributions (271De) or their relative
distributions over $\Tau$ or their relative independence;   so we may
suppose that every $f_i$ is a $\Sigma$-measurable
function from $X$ to $\Bbb R$.   Now look at 459B, taking
$(Z,\Upsilon)$ to be $\Bbb R$ with its Borel $\sigma$-algebra.   The
condition 459B(i) reads

\inset{\noindent whenever $i_0,\ldots,i_r\in I$ are distinct,
$j_0,\ldots,j_r\in I$ are distinct, and $H_k\in\Upsilon$ for each
$k\le r$, then
$\mu(\bigcap_{k\le r}f_{i_k}^{-1}[H_k])
=\mu(\bigcap_{k\le r}f_{j_k}^{-1}[H_k])$,}

\noindent matching (i) here, by 271B;  similarly, (iii) of 459B
matches (iii) here.   Equally, condition (ii) here is just a
re-phrasing of 459B(ii) in the language of 458A and 458I-458J.
So 459B gives the result.
}%end of proof of 459C

\leader{459D}{}\cmmnt{ Specializing 459B in another direction, we
have the case in which $X$ is actually the product $Z^I$.   In this
case, the condition 459B(i) corresponds to a strong kind of symmetry in
the measure $\mu$.   It now makes sense to look for subsets of $X=Z^I$
which are essentially invariant under permutations, and we have the
following result.

\medskip

\noindent}{\bf Proposition} Let $Z$ be a set, $\Upsilon$ a
$\sigma$-algebra of subsets of $Z$, $I$ an infinite set and $\mu$ a
measure on $Z^I$ with domain the $\sigma$-algebra $\Tensorhat_I\Upsilon$
generated by
$\{\pi_i^{-1}[H]:i\in I$, $H\in\Upsilon\}$, taking $\pi_i(x)=x(i)$ for
$x\in Z^I$ and $i\in I$.   For each permutation $\rho$ of $I$, define
$\hat\rho:Z^I\to Z^I$ by setting $\hat\rho(x)=x\rho$ for $x\in Z^I$.
Suppose that $\mu=\mu\hat\rho^{-1}$ for every $\rho$.   Let $\Cal E$ be
the family of those sets $E\in\Tensorhat_I\Upsilon$ such that
$\mu(E\symmdiff\hat\rho^{-1}[E])=0$ for every permutation $\rho$ of $I$,
and $\Cal V$ the family of those sets $V\in\Tensorhat_I\Upsilon$ such
that $V$ is determined by coordinates in $I\setminus\{i\}$ for every
$i\in I$.

(a) $\Cal E$ is a $\sigma$-subalgebra of $\Tensorhat_I\Upsilon$.

(b) $\Cal V$ is a $\sigma$-subalgebra of $\Cal E$.

(c) If $E\in\Cal E$ and $J\subseteq I$ is infinite, then there is a
$V\in\Cal V$, determined by coordinates in $J$, such that
$\mu(E\symmdiff V)=0$.

(d) Setting $\Sigma_i=\{\pi_i^{-1}[H]:H\in\Upsilon\}$ for each $i\in I$,

\inset{($\alpha$) $\familyiI{\Sigma_i}$ is relatively independent over
$\Cal E$,

($\beta$) for every $H\in\Upsilon$ there is an $\Cal E$-measurable
function $g_H:Z^I\to[0,1]$ which is a conditional expectation of
$\chi(\pi_i^{-1}[H])$ on $\Cal E$ for every $i\in I$.}

\proof{{\bf (a)} is elementary.

\medskip

{\bf (b)} Let $V\in\Cal V$.   Suppose that $\rho:I\to I$ is a
permutation, $J\subseteq I$ is finite and
$H_j\in\Upsilon$ for
every $j\in J$.   Then there is a permutation $\sigma:I\to I$ such that
$\sigma(j)=\rho(j)$ for every $j\in J$ and $J'=\{i:\sigma(i)\ne i\}$
is finite.   By 254Ta, $V$ is determined by coordinates in
$I\setminus J'$, so $\hat\sigma^{-1}[V]=V$.   Now

$$\eqalign{\mu(\hat\rho^{-1}[V]\cap\bigcap_{j\in J}\pi_j^{-1}[H_j])
&=\mu(V\cap\bigcap_{j\in J}\pi_{\rho(j)}^{-1}[H_j])
=\mu(V\cap\bigcap_{j\in J}\pi_{\sigma(j)}^{-1}[H_j])\cr
&=\mu(\hat\sigma^{-1}[V]\cap\bigcap_{j\in J}\pi_j^{-1}[H_j])
=\mu(V\cap\bigcap_{j\in J}\pi_j^{-1}[H_j]).\cr}$$

\noindent By the Monotone Class Theorem, as usual,
$\mu(E\cap\hat\rho^{-1}[V])=\mu(E\cap V)$ for every
$E\in\Tensorhat_I\Upsilon$.   In particular, taking $E=V$ and $E=Z^I\setminus V$, we see that
$V\symmdiff\hat\rho^{-1}[V]$ is negligible.   As $\rho$ is arbitrary,
$V\in\Cal E$.

This shows that $\Cal V\subseteq\Cal E$.   Of course $\Cal V$ is a
$\sigma$-algebra, since it is just the intersection of the
$\sigma$-algebras $\{V:V\in\Tensorhat_I\Sigma$, $V$ is determined by
coordinates in $I\setminus\{i\}\}$.

\medskip

{\bf (c)} For each $n\in\Bbb N$, there is a set
$E_n\in\Tensorhat_I\Sigma$, determined by a finite set $J_n$ of
coordinates, such that $\mu(E\symmdiff E_n)\le 2^{-n}$.   Choose
permutations $\rho_n$ of $I$ such that $\sequencen{\rho_n[J_n]}$ is a
disjoint sequence of subsets of $J$.   Set $F_n=\hat\rho_n^{-1}[E_n]$;
then $F_n$ is determined by coordinates in $\rho_n[J_n]$ for each
$n\in\Bbb N$, so $V=\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}F_m$ belongs to
$\Cal V$ and is determined by coordinates in $J$.   Also

\Centerline{$\mu(E\symmdiff F_n)=\mu(\hat\rho[E]\symmdiff E_n)
=\mu(E\symmdiff E_n)\le 2^{-n}$}

\noindent for each $n$, so $\mu(E\symmdiff V)=0$, as required.

\medskip

{\bf (d)} Let $\sequencen{j_n}$ be any sequence of distinct points of
$I$.   Set $J=\{j_n:n\in\Bbb N\}$.   For $n\in\Bbb N$ let $\Tau_n$ be
the $\sigma$-algebra generated by $\bigcup_{k\ge n}\Sigma_{j_k}$, and
set $\Tau=\bigcap_{n\in\Bbb N}\Tau_n$, so that $\Tau=\{V:V\in\Cal V$,
$V$ is determined by coordinates in $J\}$.   \Prf\ Of course
$\Tau\subseteq\Cal V$ and every member of $\Tau$ is determined by
coordinates in $J$, because every member of $\Tau_0$ is.   On the other hand, if $V\in\Cal V$ is determined by coordinates in $J$, then
fix some $w\in Z^{I\setminus J}$.   In this case, identifying $Z^I$ with
$Z^J\times Z^{I\setminus J}$, the set $V_1=\{z:z\in Z^J$, $(z,w)\in V\}$
must belong to $\Tensorhat_J\Upsilon$, so $V=V_1\times Z^{I\setminus J}$
belongs to $\Tau_0$.   Applying the same idea to
$J\setminus\{j_k:k<n\}$, we see that $V\in\Tau_n$ for every $n$, so that
$V\in\Tau$.\ \Qed

Part (c) of the proof of 459B tells us that $\familyiI{\Sigma_i}$ is
relatively independent over $\Tau$ and that for every $H\in\Upsilon$
there is a $\Tau$-measurable $g_H$ which is a conditional expectation of
$\chi(\pi_i^{-1}[H])$ on $\Tau$ for every $i\in I$.   Now (c) here tells
us that $g_H$ is a conditional expectation of $\chi(\pi_i^{-1}[H])$ on
$\Cal E$;  and examining the definition in 458Aa, we see that
$\familyiI{\Sigma_i}$ is relatively independent over $\Cal E$, as
claimed.
}%end of proof of 459D

\leader{459E}{}\cmmnt{ If $\mu$ is countably
compact, we have a strong disintegration theorem, as follows.

\medskip

\noindent}{\bf Theorem} Let $Z$ be a set, $\Upsilon$ a
$\sigma$-algebra of subsets of $Z$, $I$ an infinite set, and $\mu$ a
countably compact probability measure on $Z^I$ with domain the
$\sigma$-algebra $\Tensorhat_I\Upsilon$\cmmnt{ generated by
$\{\pi_i^{-1}[H]:i\in I$, $H\in\Upsilon\}$, taking $\pi_i(x)=x(i)$ for
$x\in Z^I$ and $i\in I$}.   Then the following are equiveridical:

\inset{(i) for every permutation $\rho$ of $I$,
$x\mapsto x\rho:Z^I\to Z^I$ is \imp\ for $\mu$;

(ii) for every transposition $\rho$ of two elements of $I$,
$x\mapsto x\rho:Z^I\to Z^I$ is \imp\ for $\mu$;

(iii) for each $n\in\Bbb N$ and any two injective functions $p$,
$q:n\to I$ the maps $x\mapsto xp:Z^I\to Z^n$, $x\mapsto xq:Z^I\to Z^n$
induce the same measure on $Z^n$;

(iv) there are a probability space $(Y,\Tau,\nu)$ and a family
$\family{y}{Y}{\lambda_y}$ of probability measures on $Z$ such that
$\family{y}{Y}{\lambda_y^I}$ is a disintegration of $\mu$ over $\nu$,
writing $\lambda_y^I$ for the product of copies of $\lambda_y$.}

\noindent Moreover, if $I$ is totally ordered, we can add

\inset{(v) for each $n\in\Bbb N$ and any two strictly increasing
functions $p$, $q:n\to I$ the maps $x\mapsto xp:Z^I\to Z^n$,
$x\mapsto xq:Z^I\to Z^n$ induce the same measure on $Z^n$.}

\noindent If the conditions (i)-(v) are satisfied, then there is a
countably compact measure $\lambda$, with domain $\Upsilon$, which is
the common marginal measure of $\mu$ on every coordinate;  and
if $\Cal K$ is a countably compact class of subsets of $Z$, closed under
finite unions and countable intersections, such that $\lambda$ is inner
regular with respect to $\Cal K$, then

\inset{(iv)$'$ there are a probability space $(Y,\Tau,\nu)$ and a family
$\family{y}{Y}{\lambda_y}$ of complete probability measures on $Z$, all
with domains including $\Cal K$ and inner regular with respect to
$\Cal K$, such that $\family{y}{Y}{\lambda_y^I}$ is a disintegration of
$\mu$ over $\nu$.}

\proof{{\bf (a)} Since any set $I$ can be totally ordered, we may
suppose from the outset that we have been given a total ordering $\le$
of $I$.   I start with the easy bits.

\medskip

\quad{\bf (iv)$'\mskip-5mu\Rightarrow$(iv)} is trivial, at least if
there is a common countably compact marginal measure on $Z$.

\medskip

\quad{\bf (iv)$\Rightarrow$(i)} If (iv) is true and $\rho:I\to I$ is a
permutation, take any $E\in\Tensorhat_I\Sigma$ and set
$E'=\{x:x\in Z^I$, $x\rho\in E\}$.   For any $y\in Y$, $x\mapsto x\rho$
is an isomorphism of the measure space $(Z^I,\lambda_y^I)$, so

\Centerline{$\mu E'=\int\lambda_y^IE'\,\nu(dy)
=\int\lambda_y^IE\,\nu(dy)=\mu E$.}

\noindent As $E$ is arbitrary, (i) is true.

\medskip

\quad{\bf (i)$\Rightarrow$(ii)} is trivial.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} There is a permutation $\rho$ of $I$
such that $q=\rho p$ and $\rho$ moves only finitely many points of $I$,
that is, $\rho$ is a product of transpositions.   By (ii),
$x\mapsto x\rho$ and $x\mapsto x\rho^{-1}$ are \imp\ for $\mu$, that is, are isomorphisms of $(Z^I,\mu)$.   But this means that
$x\mapsto xp$ and $x\mapsto x\rho p=xq$ must induce the same measure on
$Z^n$.

\medskip

\quad{\bf (iii)$\Rightarrow$(v)} is trivial.

\medskip

{\bf (b)} So for the rest of the proof I assume that (v) is true.
Taking $n=1$ in the statement of (v), we see that there is a common
image measure $\lambda=\mu\pi_i^{-1}$ for every $i\in I$.   By 452R,
$\lambda$ is countably compact.   Let $\Cal K\subseteq\Cal PZ$ be a
countably compact class, closed under finite unions and countable
intersections, such that $\lambda$ is inner regular with respect to
$\Cal K$.

In 459B, set $X=Z^I$ and $\Sigma=\Tensorhat_I\Upsilon$ and
$f_i=\pi_i:X\to Z$ for $i\in I$.   Then (v) here corresponds to (iii) of
459B, so (translating (ii) of 459B) we have a $\sigma$-subalgebra
$\Tau$ of $\Tensorhat_I\Upsilon$ and a family
$\family{H}{\Upsilon}{g_H}$ of $\Tau$-measurable functions from $Z^I$ to
$[0,1]$ such that

\Centerline{$\mu(\bigcap_{i\in J}\pi_i^{-1}[H_i])
=\int(\prod_{i\in J}g_{H_i})d\mu$}

\noindent whenever $J\subseteq I$ is finite and not empty and
$H_i\in\Upsilon$ for $i\in J$.
In particular, $g_H$ is a conditional expectation of
$\chi(\pi_i^{-1}[H])$ on $\Tau$ whenever $H\in\Upsilon$ and $i\in I$.

Fix $i^*\in I$ for the moment.   Set $\nu=\mu\restr\Tau$.
The \imp\ function $\pi_{i^*}$ from $(X,\mu)$ to $(Z,\lambda)$ gives us
an integral-preserving Riesz homomorphism
$T_0:L^{\infty}(\lambda)\to L^{\infty}(\mu)$ defined by setting
$T_0h^{\ssbullet}=(h\pi_{i^*})^{\ssbullet}$ for every
$h\in\eusm L^{\infty}(\lambda)$.   Let $P:L^1(\mu)\to L^1(\nu)$ be the
conditional expectation operator;  then
$T=PT_0:L^{\infty}(\lambda)\to L^{\infty}(\nu)$ is an
integral-preserving positive linear operator, and
$T(\chi Z^{\ssbullet})=\chi X^{\ssbullet}$.

By 452H, we have a family $\family{x}{X}{\lambda_x}$ of complete
probability measures on $Z$, all with domains including $\Cal K$ and
inner regular with respect to $\Cal K$, such that
$\int_Fh\pi_{i^*}d\mu=\int_F\int_Zh\,d\lambda_x\nu(dx)$ for every
$h\in\eusm L^{\infty}(\lambda)$ and $F\in\Tau$.   In particular, setting
$g'_H(x)=\lambda_xH$ whenever $H\in\Upsilon$ and $x\in X$ are such that
$H\in\dom\lambda_x$, then $g'_H$ will be a conditional expectation of
$\chi\pi_{i^*}^{-1}[H]$ on $\Tau$, and will be equal $\nu$-almost
everywhere to $g_H$.

This means that if $J\subseteq I$ is finite and not empty and
$H_i\in\Upsilon$ for $i\in J$,

$$\eqalign{\int_X\lambda_x^I(\bigcap_{i\in J}\pi_i^{-1}[H_i])\nu(dx)
&=\int_X\prod_{i\in J}\lambda_xH_i\,\nu(dx)
=\int_X\prod_{i\in J}g'_{H_i}d\nu\cr
&=\int_X\prod_{i\in J}g_{H_i}d\nu
=\int_X\prod_{i\in J}g_{H_i}d\mu
=\mu(\bigcap_{i\in J}\pi_i^{-1}[H_i]).\cr}$$

\noindent Thus the family $\Cal W$ of sets $E\subseteq X$ such that
$\int\lambda_x^IE\,\nu(dx)$ and $\mu E$ are defined and equal contains
all measurable cylinders.   As $\Cal W$ is a Dynkin class it includes
$\Tensorhat_I\Upsilon$.   But this says exactly that
$\family{x}{X}{\lambda_x^I}$ is a disintegration of $\mu$ over $\nu$, as
required by (iv)$'$.

Thus (v)$\Rightarrow$(iv)$'$ and the proof is complete.
}%end of proof of 459E

\leader{459F}{Lemma}\dvAnew{2010}
Let $X$ be a Hausdorff space and $P_{\text{R}}(X)$
the space of Radon probability measures on $X$ with its narrow
topology\cmmnt{ (definition:  437Jd)}.
If $\sequencen{K_n}$ is a disjoint sequence of compact subsets of $X$,
then $A=\{\mu:\mu\in P_R(X)$, $\mu(\bigcup_{n\in\Bbb N}K_n)=1\}$ is a
K-analytic subset of $P_R(X)$.

\proof{ (Recall that $P_{\text{R}}(X)$ is Hausdorff, by 437R(a-ii).)
For each $n\in\Bbb N$, let $C_n$ be the set of Radon measures on
$K_n$ with magnitude at most $1$;  by 437R(f-ii), $C_n$ is compact in its
narrow topology.   Let $C$ be the compact space $\prod_{n\in\Bbb N}C_n$;
for the rest of this
proof, I will use the formula $\pmb{\mu}=\sequencen{\mu_n}$ to describe
the coordinates of members of $C$.   Define
$\psi:C\to[0,1]^{\Bbb N}$ by setting $\psi(\pmb{\mu})(n)=\mu_nK_n$ for
$n\in\Bbb N$ and $\pmb{\mu}\in C$.   Then $\psi$ is continuous.   Since
$B=\{\sequencen{\alpha_n}:\sum_{n=0}^{\infty}\alpha_n=1\}$ is a
Borel subset of $[0,1]^{\Bbb N}$, therefore a Baire set (4A3Kb),
$D=\psi^{-1}[B]$ is a Baire subset of $C$ (4A3Kc), therefore
Souslin-F (421L) and K-analytic (422Hb).

For $\pmb{\mu}\in D$, define a function $\phi(\pmb{\mu})$ by saying that

\Centerline{$\phi(\pmb{\mu})(E)
=\sum_{n=0}^{\infty}\mu_n(E\cap K_n)\text{ if }E\subseteq X\text{ and }
  \mu_n\text{ measures }E\cap K_n\text{ for every }n$}

\noindent and is undefined otherwise.   It is easy to check that
$\phi(\pmb{\mu})\in P_{\text{R}}(X)$.   Also $\phi:D\to P_{\text{R}}(X)$
is continuous.   \Prf\ If $G\subseteq X$ is open, then
$\nu\mapsto\nu(G\cap K_n):C_n\to[0,1]$ and therefore
$\pmb{\mu}\mapsto\mu_n(G\cap K_n):D\to[0,1]$ are lower semi-continuous for
each $n$ (4A2B(d-ii)), so
$\pmb{\mu}\mapsto\phi(\pmb{\mu})(G)$ is lower semi-continuous (4A2B(d-iii),
4A2B(d-v)), and $\{\pmb{\mu}:\phi(\pmb{\mu})(G)>\alpha\}$ is open for every
$\alpha$;  by 4A2B(a-iii), $\phi$ is continuous.\ \Qed

Consequently $A=\phi[D]$ is K-analytic (422Gd).
}%end of proof of 459F

\leader{459G}{Lemma}\dvAnew{2010}
Let $X$ be a topological space,
$(Y,\frak S,\Tau,\nu)$ a totally finite quasi-Radon measure space,
$y\mapsto\mu_y$ a continuous function from $Y$ to the space
$M^+_{\text{qR}}(X)$ of totally finite quasi-Radon measures on $X$ with its
narrow topology, and $\Cal U$ a base for the topology of $X$,
containing $X$ and closed under finite intersections.   If
$\mu\in M^+_{\text{qR}}(X)$ is such that $\mu U=\int\mu_yU\,\nu(dy)$ for
every $U\in\Cal U$, then $\family{y}{Y}{\mu_y}$ is a disintegration of
$\mu$ over $\nu$.

\proof{{\bf (a)} Let $\Cal E$ be the family of subsets $E$ of $X$ such that
$\mu E$ and $\int\mu_yE\,\nu(dy)$ are defined and equal.   Because
$X\in\Cal E$, $\Cal E$ is
a Dynkin class;  as $\Cal U$ is included in $\Cal E$ and is closed under
finite intersections, the $\sigma$-algebra of sets generated by $\Cal U$ is
included in $\Cal E$, and in particular any finite union of members of
$\Cal U$ belongs to $\Cal E$.

\medskip

{\bf (b)} In fact every open subset of $X$ belongs to $\Cal E$.   \Prf\ If
$G\subseteq X$ is open, set
$\Cal H=\{H:H\subseteq G$ is a finite union of members of $\Cal U\}$.
Then $\Cal H$ is upwards-directed and has union $G$.   Set
$f_H(y)=\mu_yH$ for $y\in Y$ and $H\in\Cal H$.   Since
$\lambda\mapsto\lambda H:M^+_{\text{qR}}(X)\to\Bbb R$ is lower
semi-continuous (by the definition of the narrow topology) and
$y\mapsto\mu_y$ is continuous, $f_H:Y\to\Bbb R$ is lower semi-continuous
(4A2B(d-ii) again).
Moreover, $\{f_H:H\in\Cal H\}$ is an upwards-directed
family of functions with supremum $f_G$, where $f_G(y)=\mu_yG$ for each
$y$, because every $\mu_y$ is $\tau$-additive.   Now

$$\eqalignno{\mu G
&=\sup_{H\in\Cal H}\mu H=\sup_{H\in\Cal H}\int f_Hd\nu=\int f_Gd\nu\cr
\displaycause{414Ba}
&=\int\mu_yG\,\nu(dy)\cr}$$

\noindent and $G\in\Cal E$.\ \Qed

\medskip

{\bf (c)} It follows that every Borel subset of $X$ belongs to $\Cal E$,
that is, that $\family{y}{Y}{\mu_y}$ is a disintegration of the restriction
$\mu_{\Cal B}$ to the Borel $\sigma$-algebra of $X$.   Since every $\mu_y$
is complete, $\family{y}{Y}{\mu_y}$ is also a disintegration over
$\nu$ of the completion of $\mu_{\Cal B}$ (452B(a-ii)), which is
$\mu$.
}%end of proof of 459G

\leader{459H}{Theorem} Let $Z$ be a Hausdorff space, $I$ an infinite
set, and $\tilde\mu$ a quasi-Radon probability measure on $Z^I$ such
that the marginal measures on each copy of $Z$ are Radon measures.
%is this the right way of putting it?  we just need a filling sequence
%of compact sets.
Write $P_{\text{R}}(Z)$ for the set of Radon probability measures on $Z$
with its narrow topology.   Then the following are equiveridical:

\inset{(i) for every permutation $\rho$ of $I$,
$w\mapsto w\rho:Z^I\to Z^I$ is \imp\ for $\tilde\mu$;

(ii) for every transposition $\rho$ of two elements of $I$,
$w\mapsto w\rho:Z^I\to Z^I$ is \imp\ for $\tilde\mu$;

(iii) for each $n\in\Bbb N$ and any two injective functions $p$,
$q:n\to I$ the maps $w\mapsto wp:Z^I\to Z^n$ and
$w\mapsto wq:Z^I\to Z^n$ induce the same measure on $Z^n$;

(iv) there are a probability space $(Y,\Tau,\nu)$ and a family
$\family{y}{Y}{\mu_y}$ of $\tau$-additive Borel probability measures on
$Z$ such that $\family{y}{Y}{\tilde\mu_y^I}$ is a disintegration of
$\tilde\mu$ over $\nu$, writing $\tilde\mu_y^I$ for the
$\tau$-additive product of copies of $\mu_y$;

(v) there is a Radon probability measure $\tilde\nu$ on
$P_{\text{R}}(Z)$ such that
$\family{\theta}{P_{\text{R}}(Z)}{\tilde\theta^I}$ is disintegration of
$\tilde\mu$ over $\tilde\nu$, writing $\tilde\theta^I$ for the
quasi-Radon product of copies of $\theta$.}

\noindent Moreover, if $I$ is totally ordered, we can add

\inset{(vi) for each $n\in\Bbb N$ and any two strictly increasing
functions $p$, $q:n\to I$ the maps $w\mapsto wp:Z^I\to Z^n$ and
$w\mapsto wq:Z^I\to Z^n$ induce the same measure on $Z^n$.}

\proof{{\bf (a)} As in 459E, we need consider only the case in which
$I$ is totally ordered, and the implications

\Centerline{(v)$\Rightarrow$(iv)$\Rightarrow$(i)$
\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(vi)}

\noindent are elementary.   So henceforth I will suppose that (vi) is
true and seek to prove (v).

\medskip

{\bf (b)} We are going to need a second topology on the set $Z$, so I will
call the original topology $\frak T$, and for the rest of this proof
I will declare the
topology on which each topological concept or construction is based.
Write $\mu$ for
$\tilde\mu\restr\Tensorhat_I\Cal B(Z,\frak T)$, where $\Cal B(Z,\frak T)$
is the Borel $\sigma$-algebra of $Z$ for the topology $\frak T$.
Then (vi) is also true of $\mu$.
(Strictly speaking, we ought to check that the different images of
$\mu$ all have the same domain.   But this is true, because the
image of $\mu$ corresponding to a strictly increasing function
$p:r\to I$ has domain $\Tensorhat_r\Cal B(Z,\frak T)$.)
The (unique) marginal
measure $\lambda$ of $\mu$ is the restriction to $\Cal B(Z,\frak T)$ of
the $\frak T$-Radon measure $\tilde\lambda$ which is the marginal of
$\tilde\mu$, so is a $\frak T$-tight $\frak T$-Borel
measure, therefore countably compact.   By 454A(b-ii), $\mu$ is countably
compact.   So 459E, with $\Cal K$ the family of $\frak T$-compact subsets
of $Z$,
tells us that there are a probability space $(Y_0,\Tau_0,\nu_0)$ and a
family $\family{y}{Y_0}{\mu_y}$ in $P_{\text{R}}(Z,\frak T)$ such that
$\family{y}{Y_0}{\mu_y^I}$ is a disintegration of $\mu$ over
$\nu_0$, writing $\mu_y^I$ for the ordinary product of copies of
$\mu_y$.   We can of course suppose that $\nu_0$ is complete.
Note also that $\family{y}{Y_0}{\mu_y}$ is a
disintegration of $\lambda$;  this is clearly achieved by the proof of
459E, and it is necessarily true if $\family{y}{Y_0}{\mu_y^I}$ is to be
a disintegration of $\mu$.   Because every $\mu_y$ is complete,
$\family{y}{Y_0}{\mu_y}$ is also a
disintegration of the completion $\tilde\lambda$ of $\lambda$.

\medskip

{\bf (c)}
Let $\sequencen{K_n}$ be a disjoint sequence of $\frak T$-compact
subsets of $Z$ such that $\sum_{n=0}^{\infty}\tilde\lambda K_n=1$
(412Aa).   Let $\frak S$ be

\Centerline{$\{H:H\subseteq Z$, $Z\setminus(H\cap K_n)\in\frak T$ for every
$n\in\Bbb N\}$.}

\noindent Then $\frak S$ is a locally compact topology on $Z$ finer than
$\frak T$.   (If you like, $\frak S$ is the disjoint union topology
corresponding to the partition
$\{K_n:n\in\Bbb N\}\cup\{\{z\}:z\in Z\setminus\bigcup_{n\in\Bbb N}K_n\}$.)
Note that the subspace topologies on any $K_n$ induced by $\frak S$ and
$\frak T$ are the same, so that a $\frak T$-compact subset of $K_n$ is
$\frak S$-compact.   Because $\frak S$ is finer than $\frak T$,
$P_{\text{R}}(Z,\frak S)\subseteq P_{\text{R}}(Z,\frak T)$ (use 418I).
If $\theta\in P_{\text{R}}(Z,\frak T)$ and
$\theta(\bigcup_{n\in\Bbb N}K_n)=1$,
then, from the standpoint of the topology $\frak S$, $\theta$ is a
complete topological probability
measure inner regular with respect to the compact sets, so belongs to
$P_{\text{R}}(Z,\frak S)$.   In particular,
$\tilde\lambda\in P_{\text{R}}(Z,\frak S)$.

We shall need to know that the family $\Cal V$
of $\frak T$-Borel $\frak S$-cozero subsets of $Z$
is a base for $\frak S$.   \Prf\ If $z\in H\in\frak S$, then if
$z\notin\bigcup_{n\in\Bbb N}K_n$ the singleton $\{z\}$ belongs to
$\Cal V$.   If $n\in\Bbb N$ and $z\in K_n$, then $H\cap K_n\in\frak S$;
as $\frak S$ is locally compact, there is an $\frak S$-cozero set $G$ such
that $z\in G\subseteq H\cap K_n$, and now $G$ is $\frak T$-relatively open
in the $\frak T$-compact set $K_n$, so $G$ is $\frak T$-Borel.\ \QeD\

\medskip

{\bf (d)} We know that

\Centerline{$\int\mu_y(\bigcup_{n\in\Bbb N}K_n)\nu_0(dy)
=\tilde\lambda(\bigcup_{n\in\Bbb N}K_n)=1$;}

\noindent since $\mu_yZ=1$ for every $y$, the set
$Y=\{y:y\in Y_0,\,\mu_y(\bigcup_{n\in\Bbb N}K_n)=1\}$ must be
$\nu_0$-conegligible.   Let $\nu$ be the subspace measure induced by
$\nu_0$ on $Y$.   Then $\family{y}{Y}{\mu_y}$ is a disintegration of
$\tilde\lambda$ over $\nu$, and $\mu_y\in P_{\text{R}}(Z,\frak S)$ for
every $y\in Y$, by (c).

\medskip

{\bf (e)} By 459F, the set

\Centerline{$A=\{\theta:\theta\in P_{\text{R}}(Z,\frak S)$,
$\theta(\bigcup_{n\in\Bbb N}K_n=1\}$}

\noindent is K-analytic in its narrow
topology, while $\mu_y\in A$ for every $y\in Y$.
If $G\in\Cal V$ and $\alpha>0$,
$\{y:y\in Y_0$, $\mu_yG>\alpha\}\in\Tau_0$, so
$\{y:y\in Y$, $\mu_yG>\alpha\}$ is measured by $\nu$.   By 432I,
applied to the map $y\mapsto\mu_y:Y\to A$, there is a
Radon probability measure $\tilde\nu_A$ on $A$ such that

\Centerline{$\int h\,d\tilde\nu_A=\int h(\mu_y)\nu(dy)$}

\noindent for every bounded continuous $h:A\to\Bbb R$.

\medskip

{\bf (f)}
Now suppose that $f:Z\to\Bbb R$ is bounded and $\frak S$-continuous.
Then $\theta\mapsto\int fd\theta:P_{\text{R}}(Z,\frak S)\to\Bbb R$ is
continuous (437K), so that

\Centerline{$\iint fd\theta\,\tilde\nu_A(d\theta)
=\iint fd\mu_y\nu(dy)$.}

\noindent If $G\in\Cal V$, there is a
non-decreasing sequence $\sequencen{f_n}$ of non-negative
$\frak S$-continuous functions with supremum $\chi G$, so

$$\eqalign{\int\theta G\,\tilde\nu_A(d\theta)
&=\sup_{n\in\Bbb N}\iint f_nd\theta\,\tilde\nu_A(d\theta)
=\sup_{n\in\Bbb N}\iint f_nd\mu_y\nu(dy)\cr
&=\int\mu_yG\,\nu(dy)
=\lambda G
=\tilde\lambda G.\cr}$$

\noindent So we can apply 459G to the identity map from
$A$ to itself and the family
$\family{\theta}{A}{\theta}$ to see that
$\family{\theta}{A}{\theta}$ is a disintegration of
$\tilde\lambda$ over $\tilde\nu_A$.

It follows that if $E\subseteq Z$ is $\tilde\lambda$-negligible, then
$\theta E=0$ for $\tilde\nu_A$-almost every $\theta$.
Moreover, since $\family{y}{Y}{\mu_y}$ is a disintegration of
$\tilde\lambda$ over $\nu$, $\mu_yE=0$ for $\nu$-almost every $y$.

\medskip

{\bf (g)} If $J\subseteq I$ is finite, $G_j\in\frak T$ for
$j\in J$,  and $W=\{w:w\in Z^I$, $w(j)\in G_j$ for $j\in J\}$, then

\Centerline{$\tilde\mu W=\int\theta^IW\tilde\nu_A(d\theta)$.}

\Prf\ Because $\Cal V$ is a base for $\frak S$ closed under countable
unions, and $\tilde\lambda$ is $\frak S$-Radon, there is for each $j\in J$
a $G'_j\in\Cal V$, included in $G_j$, such that
$\tilde\lambda G'_J=\tilde\lambda G_j$.
Set $W'=\{w:w\in Z^I$, $w(j)\in G'_j$ for $j\in J\}$.
We have

\Centerline{$W\setminus W'
\subseteq\bigcup_{j\in J}\{w:w(j)\in G_j\setminus G'_j\}$,}

\noindent while

\Centerline{$\tilde\mu\{w:w(j)\in G_j\setminus G'_j\}
=\tilde\lambda(G_j\setminus G'_j)=0$}

\noindent for each $j$, so $\tilde\mu W'$ is defined and equal to
$\tilde\mu W=\mu W$.   Note that the same calculation shows that
$\theta^IW=\theta^IW'$ whenever $\theta\in A$ is such
that $\theta G'_j=\theta G_j$ for every $j$, that is, for
$\tilde\nu_A$-almost every $\theta$.   Now, for each $j\in J$,
we have a non-decreasing sequence $\sequencen{f_{jn}}$ of non-negative
$\frak S$-continuous real-valued functions with supremum $\chi G'_j$.
Set $g_n(w)=\prod_{j\in J}f_{jn}(w(j))$ for $w\in Z^I$ and $n\in\Bbb N$.
(I suppose you should take $g_n(w)=1$ if $J$ is empty.)   Then each
$g_n$ is $\frak S^I$-continuous, so if we set
$h_n(\theta)=\int g_nd\theta^I$ for $\theta\in A$,
$h_n$ is continuous (put 437Mb and 437Kb together).
Also $\sequencen{g_n}$ is a
non-decreasing sequence with supremum $\chi W'$, so

\Centerline{$\theta^IW'=\sup_{n\in\Bbb N}\int g_nd\theta^I
=\sup_{n\in\Bbb N}h_n(\theta)$}

\noindent for $\theta\in A$.   Accordingly

$$\eqalign{\tilde\mu W
&=\tilde\mu W'
=\int\mu_y^IW'\nu(dy)
=\sup_{n\in\Bbb N}\int h_n(\mu_y)\nu(dy)\cr
&=\sup_{n\in\Bbb N}\int h_nd\tilde\nu_A
=\sup_{n\in\Bbb N}\iint g_nd\theta^I\,\tilde\nu_A(d\theta)\cr
&=\int\theta^IW'\,\tilde\nu_A(d\theta)
=\int\theta^IW\,\tilde\nu_A(d\theta),\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (h)} We are nearly ready to dispense with the topology $\frak S$.
Since the embeddings
$A\embedsinto P_{\text{R}}(Z,\frak S)\embedsinto P_{\text{R}}(Z,\frak T)$
are continuous
(437Jh), we have an image Radon probability measure $\tilde\nu$ on
$P_{\text{R}}(Z,\frak T)$, and

\Centerline{$\int_{P_{\text{R}}(Z,\frak T)}h\,d\tilde\nu
=\int_Ah\,d\tilde\nu_A$}

\noindent for every $h:P_{\text{R}}(Z,\frak T)\to\Bbb R$ such that
$\int_Ah\,d\tilde\nu_A$ is defined.

In particular, if we take $\Cal W$ to be the family of
$\frak T^I$-open cylinder sets expressible as
$\{w:w\in Z^I$, $w(j)\in G_j$ for $j\in J\}$ where $J\subseteq I$ is finite
and $G_j\in\frak T$ for each $j$, (g) tells us that

\Centerline{$\tilde\mu W
=\int\theta^IW\,\tilde\nu_A(d\theta)
=\int\theta^IW\,\tilde\nu(d\theta)
=\int\tilde\theta^IW\,\tilde\nu(d\theta)$}

\noindent for every $W\in\Cal W$,
where I now write $\frak T^I$ for the product topology on $Z^I$
corresponding to the topology $\frak T$ on $Z$, and
$\tilde\theta^I$ for the
$\frak T^I$-quasi-Radon product measure on $Z^I$ corresponding to the
$\frak T$-Radon measure $\theta$ (417R).   Now turn again to 437Mb and
459G;  $\theta\mapsto\tilde\theta^I$ is a continuous function from
$P_{\text{R}}(Z,\frak T)$ to the space $P_{\text{qR}}(Z^I,\frak T^I)$ of
$\frak T^I$-quasi-Radon probability measures on $Z^I$, and $\Cal W$
is a base for the topology $\frak T^I$, so
$\family{\theta}{P_{\text{R}}(Z,\frak T)}{\tilde\theta^I}$ is a
disintegration of $\tilde\mu$ over $\tilde\nu$, which is what I set out to
prove.
}%end of proof of 459H

\leader{459I}{}\cmmnt{ I come now to a lemma based on ideas in
{\smc Tao 07}.   It is in a form more elaborate than is required for the
elementary application here (459J), but which will be needed in \S497.

\medskip

\noindent}{\bf Lemma}\dvAnew{2010}
Let $(X,\Sigma,\mu)$ be a probability space and
$I$ a set.   For a family $\Bbb T$ of subalgebras of $\Cal PX$,
write $\bigvee\Bbb T$ for the $\sigma$-algebra generated by
$\bigcup\Bbb T$\cmmnt{, as in 458Ad}.
Let $G$ be the group of permutations $\phi$ of $I$ such that
$\{i:\phi(i)\ne i\}$ is finite.   Suppose that $\action$ is an action of
$G$ on $X$ such that $x\mapsto\phi\action x$ is \imp\ for each $\phi\in G$;
set $\phi\action A=\{\phi\action x:x\in A\}$ for $\phi\in G$ and
$A\subseteq X$\cmmnt{, as in 441Aa and 4A5Bc}.
Let $\langle\Sigma_J\rangle_{J\subseteq I}$ be a family of
$\sigma$-subalgebras of $\Sigma$ such that

\inset{(i) for every $J\subseteq I$,
$\Sigma_J$ is the $\sigma$-algebra generated by
$\bigcup_{K\subseteq J\text{ is finite}}\Sigma_K$;

(ii) if $J\subseteq I$, $E\in\Sigma_J$ and $\phi\in G$, then
$\phi\action E\in\Sigma_{\phi[J]}$;

(iii) if $J\subseteq I$, $E\in\Sigma_J$ and $\phi\in G$ is such that
$\phi(i)=i$ for every $i\in J$, then $\phi\action E=E$.}

\noindent Suppose that $\Cal J^*$ is a filter on $I$ not containing any
infinite set, and that $K\subseteq I$, $\Cal K\subseteq\Cal PI$ and
$\Cal J\subseteq\Cal J^*$ are such that for every $K'\in\Cal K$ there is a
$J\in\Cal J$ such that $K\cap K'\subseteq J$.
Then $\Sigma_K$ and $\bigvee_{K'\in\Cal K}\Sigma_{K'}$
are relatively independent over $\bigvee_{J\in\Cal J}\Sigma_J$.

\proof{{\bf (a)(i)} Let us note straight away that condition (i)
above implies
that $\Sigma_K\subseteq\Sigma_J$ whenever $K\subseteq J\subseteq I$.

\medskip

\quad{\bf (ii)}
For any $\sigma$-subalgebra $\Tau$ of $\Sigma$, I will
(slightly abusing notation, as in 242Jh)
write $L^2(\mu\restr\Tau)$ for the $\|\,\|_2$-closed linear
subspace of $L^2(\mu)$ consisting of
equivalence classes of $\mu$-square-integrable $\Tau$-measurable
real-valued functions defined on $X$, and
$P_{\Tau}:L^2(\mu)\to L^2(\mu\restr\Tau)$ for the corresponding
conditional-expectation operator (244M).   Note that $P_{\Tau}$ is an
orthogonal projection (244Nb).

\medskip

\quad{\bf (iii)} We have an action of $G$ on $L^2(\mu)$,
defined by saying that

\Centerline{$(\phi\action f)(x)=f(\phi^{-1}\action x)$ for $\phi\in G$,
$x\in X$ and $f\in\Bbb R^X$}

\noindent(4A5C(c-i)),

\Centerline{$\phi\action f^{\ssbullet}=(\phi\action f)^{\ssbullet}$ for
$\phi\in G$ and $f\in\eusm L^2(\mu)\cap\Bbb R^X$}

\noindent(441Kc).

\medskip

\quad{\bf (iv)} If $\Bbb T$ is the family of $\sigma$-algebras of subsets
of $X$, we have an action of $G$ on $\Bbb T$ defined by setting

\Centerline{$\phi\action\Tau=\{\phi\action E:E\in\Tau\}$}

\noindent for $\Tau\in\Bbb T$ and $\phi\in G$.   If
$\family{\gamma}{\Gamma}{\Tau_{\gamma}}$ is a family in $\Bbb T$, then
$\phi\action\bigvee_{\gamma\in\Gamma}\Tau_{\gamma}
=\bigvee_{\gamma\in\Gamma}\phi\action\Tau_{\gamma}$ for every
$\phi\in G$, just because $E\mapsto\phi\action E$ is an automorphism of the
Boolean algebra $\Cal PX$.

\medskip

\quad{\bf (v)} If $\phi\in G$ and $L\subseteq I$, then
$\phi\action\Sigma_L=\Sigma_{\phi[L]}$.   \Prf\ Condition (ii) of this
lemma says just that
$\phi\action\Sigma_L=\{\phi\action E:E\in\Sigma_L\}$ is included in
$\Sigma_{\phi[L]}$;  and now of course

\Centerline{$\Sigma_{\phi[L]}=\phi\action\phi^{-1}\action\Sigma_L
\subseteq\phi\action\Sigma_{\phi^{-1}[\phi[L]]}
=\phi\action\Sigma_L$.\  \Qed}

\medskip

\quad{\bf (vi)} If $\phi\in G$ and $\Tau$ is a $\sigma$-subalgebra of
$\Sigma$, then $\phi\action(P_{\Tau}u)=P_{\phi\action\Tau}(\phi\action u)$
for every $u\in L^2(\mu)$.    \Prf\ I should of course note that
$\phi\action\Sigma=\Sigma$ because $x\mapsto\phi\action x$ is an
automorphism of $(X,\Sigma,\mu)$, so $\phi\action\Tau\subseteq\Sigma$ and
we can speak of $P_{\phi\action\Tau}$.   Let $f:X\to\Bbb R$ be a
$\Sigma$-measurable function such that $f^{\ssbullet}=u$, and
$g:X\to\Bbb R$ a $\Tau$-measurable function which is a conditional
expectation of $f$ on $\Tau$.   In this case, for any $\alpha\in\Bbb R$,

$$\eqalign{\{x:(\phi\action g)(x)>\alpha\}
&=\{x:g(\phi^{-1}\action x)>\alpha\}
=\{\phi\action x:g(x)>\alpha\}\cr
&=\phi\action\{x:g(x)>\alpha\}
\in\phi\action\Tau,\cr}$$

\noindent so $\phi\action g$ is $(\phi\action\Tau)$-measurable.   Next, for
any $F\in\phi\action\Tau$,

$$\eqalignno{\int_F\phi\action g\,d\mu
&=\int_Fg(\phi^{-1}\action x)\mu(dx)
=\int_{\phi^{-1}\action F}g(x)\mu(dx)\cr
\displaycause{applying 235G to the \imp\ function
$x\mapsto\phi\action x:X\to X$ and the integrable function
$x\mapsto g(\phi^{-1}\action x)$}
&=\int_{\phi^{-1}\action F}f\,d\mu\cr
\displaycause{because $\phi^{-1}\action F\in\Tau$}
&=\int_F\phi\action fd\mu.}$$

\noindent As $F$ is arbitrary, $\phi\action g$ is a conditional expectation
of $\phi\action f$ on $\phi\action\Tau$, and

\Centerline{$\phi\action(P_{\Tau}u)=\phi\action g^{\ssbullet}
=(\phi\action g)^{\ssbullet}
=P_{\phi\action\Tau}(\phi\action f)^{\ssbullet}
=P_{\phi\action\Tau}(\phi\action u)$. \Qed}

\medskip

{\bf (b)(i)}
Let $\family{\gamma}{\Gamma}{J_{\gamma}}$ be a non-empty finite
family of subsets of $I$ with infinite intersection, and set
$\Lambda=\bigvee_{\gamma\in\Gamma}\Sigma_{J_{\gamma}}$.   Suppose that $K$,
$\family{\gamma}{\Gamma}{K_{\gamma}}$ are such that

\Centerline{$K\in[I]^{<\omega}$,
\quad$K_{\gamma}\in[I]^{<\omega}$ and
$K\cap K_{\gamma}\subseteq J_{\gamma}$ for every $\gamma\in\Gamma$.}

\noindent Take $E\in\Sigma_K$ and
$F_{\gamma}\in\Sigma_{K_{\gamma}}$ for every $\gamma\in\Gamma$, and set
$F=\bigcap_{\gamma\in\Gamma}F_{\gamma}$.   Let $g$, $h:X\to[0,1]$ be
$\Lambda$-measurable functions which are conditional expectations of
$\chi E$, $\chi F$ respectively on $\Lambda$.   Let $\epsilon>0$.

\medskip

\quad{\bf (ii)} For $L\subseteq I$ set
$\Lambda_L=\bigvee_{\gamma\in\Gamma}\Sigma_{J_{\gamma}\cap L}
\subseteq\Lambda$.   For any
$u\in L^2(\mu)$ there is a finite $L\subseteq I$ such that
$\|P_{\Tau}u-P_{\Lambda}u\|_2\le\epsilon$ whenever $\Tau$ is a
$\sigma$-subalgebra of $\Lambda$ including $\Lambda_L$.
\Prf\ By condition (i) of this lemma,
$\Lambda$ is the $\sigma$-algebra generated by

\Centerline{$\bigcup_{L\subseteq I\text{ is finite}}
\bigcup_{\gamma\in\Gamma}\Sigma_{J_{\gamma}\cap L}$,}

\noindent so $\{\Lambda_L:L\in[I]^{<\omega}\}$ is an upwards-directed
family of $\sigma$-algebras whose union $\sigma$-generates $\Lambda$, and
$\bigcup_{L\subseteq I\text{ is finite}}L^2(\mu\restr\Lambda_L)$
is norm-dense in
$L^2(\mu\restr\Lambda)$.   There are therefore a finite
$L\subseteq I$ and a $v\in L^2(\mu\restr\Lambda_L)$ such that
$\|v-P_{\Lambda}u\|_2\le\epsilon$.  If now
$\Lambda_L\subseteq\Tau\subseteq\Lambda$,
$v\in L^2(\mu\restr\Tau)$, while
$P_{\Tau}$ is the orthogonal projection onto $L^2(\mu\restr\Tau)$, so

\Centerline{$\|P_{\Tau}u-P_{\Lambda}u\|_2
=\|P_{\Tau}P_{\Lambda}u-P_{\Lambda}u\|_2\le\|v-P_{\Lambda}u\|_2
\le\epsilon$.  \Qed}

\medskip

\quad{\bf (iii}
Set $u=\chi E^{\ssbullet}$ and $v=\chi F^{\ssbullet}$,
so that $g^{\ssbullet}=P_{\Lambda}u$ and $h^{\ssbullet}=P_{\Lambda}v$.
By (b), there is an $L_0\in[I]^{<\omega}$ such that

\Centerline{$\|P_{\Tau}u-P_{\Lambda}u\|_2\le\epsilon$,
\quad$\|P_{\Tau}v-P_{\Lambda}v\|_2\le\epsilon$,
\quad$\|P_{\Tau}(u\times v)-P_{\Lambda}(u\times v)\|_2\le\epsilon$}

\noindent whenever $\Tau$ is a $\sigma$-subalgebra of $\Lambda$ including
$\Lambda_{L_0}$.   We can suppose that
$L_0\supseteq K\cup\bigcup_{\gamma\in\Gamma}K_{\gamma}$.   Write $\Tau_0$
for $\Lambda_{L_0}$.   We have

$$\eqalignno{\|P_{\Lambda}u\times P_{\Lambda}v
  &-P_{\Tau_0}u\times P_{\Tau_0}v\|_2
  \cr
&\le\|P_{\Lambda}u\times(P_{\Lambda}v-P_{\Tau_0}v)\|_2
   +\|(P_{\Lambda}u-P_{\Tau_0}u)\times P_{\Tau_0}v\|_2\cr
&\le\|P_{\Lambda}v-P_{\Tau_0}v\|_2
   +\|P_{\Lambda}u-P_{\Tau_0}u\|_2\cr
\displaycause{because $\|P_{\Lambda}u\|_{\infty}$ and
$\|P_{\Tau_0}v\|_{\infty}$ are both at most $1$}
&\le 2\epsilon.\cr}$$

\medskip

\quad{\bf (iv)} Let
$L_1\subseteq\bigcap_{\gamma\in\Gamma}J_{\gamma}\setminus L_0$
be a set of size $\#(L_0\setminus K)$;  let $\phi\in G$ be such that
$\phi[L_0\setminus K]=L_1$, $\phi^2$ is the identity and $\phi(i)=i$ for
$i\in I\setminus(L_1\cup(L_0\setminus K))$.   In this case,
$\phi(i)=i$ for $i\in K$, so
$\phi[L]\subseteq(L\cap K)\cup\bigcap_{\gamma\in\Gamma}J_{\gamma}$
for every $L\subseteq L_0$.   Setting
$M_{\gamma}=(L_0\cap J_{\gamma})\cup\phi[L_0\cap J_{\gamma}]$, we have

\Centerline{$L_0\cap J_{\gamma}\subseteq M_{\gamma}=\phi[M_{\gamma}]
\subseteq J_{\gamma}$,
\quad$\phi[K_{\gamma}]\subseteq J_{\gamma}$}

\noindent for each $\gamma\in\Gamma$.   (This is where we need to know that
$K\cap K_{\gamma}\subseteq J_{\gamma}$.)

Now

\Centerline{$\phi\action u
=\phi\action(\chi E^{\ssbullet})=\chi(\phi\action E)^{\ssbullet}
=\chi E^{\ssbullet}=u$}

\noindent by condition (iii) of this lemma;  also

\Centerline{$\|\phi\action(P_{\Tau_0}u)-P_{\Lambda}u\|_2
\le 3\epsilon$.}

\noindent\Prf\ By (a-iv) and (a-v),

$$\eqalign{\phi\action\Tau_0
&=\phi\action\bigvee_{\gamma\in\Gamma}\Sigma_{L_0\cap J_{\gamma}}
=\bigvee_{\gamma\in\Gamma}\phi\action\Sigma_{L_0\cap J_{\gamma}}\cr
&=\bigvee_{\gamma\in\Gamma}\Sigma_{\phi[L_0\cap J_{\gamma}]}
\subseteq\bigvee_{\gamma\in\Gamma}\Sigma_{M_{\gamma}}
\subseteq\bigvee_{\gamma\in\Gamma}\Sigma_{J_{\gamma}}
=\Lambda.\cr}$$

\noindent Set $\Tau=\Tau_0\vee\phi\action\Tau_0$;  then
$\Tau_0\subseteq\Tau=\phi[\Tau]\subseteq\Lambda$.   But now

\Centerline{$\phi\action(P_{\Tau}u)=P_{\phi\action\Tau}(\phi\action u)
=P_{\Tau}u$}

\noindent(see (a-vi)), so

$$\eqalign{\|\phi\action(P_{\Tau_0}u)-P_{\Lambda}u\|_2
&\le\|\phi\action(P_{\Tau_0}u)-\phi\action(P_{\Tau}u)\|_2
    +\|P_{\Tau}u-P_{\Lambda}u\|_2\cr
&=\|P_{\Tau}u-P_{\Tau_0}u\|_2+\|P_{\Tau}u-P_{\Lambda}u\|_2\cr
&\le\|P_{\Lambda}u-P_{\Tau_0}u\|_2+2\|P_{\Tau}u-P_{\Lambda}u\|_2
\le 3\epsilon.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (v)} Set

\Centerline{$\Tau^*
=\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}\cup M_{\gamma}}$.}

\noindent Because $L_0\cap J_{\gamma}\subseteq M_{\gamma}$ for every
$\gamma$, $\Tau^*$ and

\Centerline{$\phi\action\Tau^*
=\bigvee_{\gamma\in\Gamma}\Sigma_{\phi[K_{\gamma}]\cup M_{\gamma}}$}

\noindent include $\Lambda_{L_0}=\Tau_0$,
while $\phi\action\Tau^*\subseteq\Lambda$ because
$\phi[K_{\gamma}]\cup M_{\gamma}\subseteq J_{\gamma}$ for every $\gamma$.
Also $F\in\Tau^*$, because
$F_{\gamma}\in\Sigma_{K_{\gamma}}\subseteq\Tau^*$ for every $\gamma$.
Now

\noindent and

$$\eqalignno{\|P_{\Tau_0}(u\times v)
  -P_{\Tau_0}u\times P_{\Tau_0}v\|_2
&=\|P_{\Tau_0}P_{\Tau^*}(u\times v)
  -P_{\Tau_0}u\times P_{\Tau_0}v\|_2\cr
\displaycause{because $T_0\subseteq T^*$}
&=\|P_{\Tau_0}(v\times P_{\Tau^*}u)
  -P_{\Tau_0}(v\times P_{\Tau_0}u)\|_2\cr
\displaycause{because $v\in L^2(\mu\restr\Tau^*)$ and
$P_{\Tau_0}u\in L^2(\mu\restr\Tau_0)$, see 242L}
&\le\|v\times P_{\Tau^*}u-v\times P_{\Tau_0}u\|_2
\le\|P_{\Tau^*}u-P_{\Tau_0}u\|_2\cr
\displaycause{because $\|v\|_{\infty}\le 1$}
&=\|\phi\action(P_{\Tau^*}u)-\phi\action(P_{\Tau_0}u)\|_2\cr
&=\|P_{\phi\action\Tau^*}(\phi\action u)
  -\phi\action(P_{\Tau_0}u)\|_2\cr
&\le\|P_{\phi\action\Tau^*}u-P_{\Lambda}u\|_2
  +\|P_{\Lambda}u-\phi\action(P_{\Tau_0}u)\|_2\cr
&\le\epsilon+3\epsilon
=4\epsilon.\cr}$$

\medskip

\quad{\bf (vi)} Putting these together,

$$\eqalign{\|P_{\Lambda}(u\times v)-P_{\Lambda}u\times P_{\Lambda}v\|_2
&\le\|P_{\Lambda}(u\times v)-P_{\Tau_0}(u\times v)\|_2\cr
&\mskip100mu
  +\|P_{\Tau_0}(u\times v)-P_{\Tau_0}u\times P_{\Tau_0}v\|_2\cr
&\mskip100mu
  +\|P_{\Lambda}u\times P_{\Lambda}v-P_{\Tau_0}u\times P_{\Tau_0}v\|_2\cr
&\le\epsilon+4\epsilon+2\epsilon
=7\epsilon.\cr}$$

\medskip

\quad{\bf (vii)} As $\epsilon$ is arbitrary,
$P_{\Lambda}(u\times v)=P_{\Lambda}u\times P_{\Lambda}v$, that is,
$g\times h$ is a conditional expectation of $\chi(E\cap F)$ on $\Lambda$,
and $E$ and $F$ are relatively independent over $\Lambda$.

\medskip

{\bf (c)} It follows that $\Sigma_K$ and
$\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$ are relatively independent
over $\Lambda$.  \Prf\ Suppose that $E\in\Sigma_K$, and consider the set

\Centerline{$\Cal E=\{F:F\in\Sigma$,
$P_{\Lambda}\chi(E\cap F)^{\ssbullet}
=P_{\Lambda}(\chi E^{\ssbullet})\times P_{\Lambda}(\chi F^{\ssbullet})\}$.}

\noindent Then $\Cal E$ is a Dynkin class, and by (b) above it contains

\Centerline{$\Cal E_0
=\{\bigcap_{\gamma\in\Gamma}F_{\gamma}:F_{\gamma}\in\Sigma_{K_{\gamma}}$
for every $\gamma\in\Gamma\}$,}

\noindent which is closed under $\cap$.
Accordingly $\Cal E$ includes the $\sigma$-algebra generated by
$\Cal E_0$, which is $\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$.
Thus

\Centerline{$P_{\Lambda}\chi(E\cap F)^{\ssbullet}
=P_{\Lambda}(\chi E^{\ssbullet})\times P_{\Lambda}(\chi F^{\ssbullet})$}

\noindent for every $E\in\Sigma_K$ and
$F\in\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$, and
$\Sigma_K$ and
$\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$ are relatively independent
over $\Lambda$.\ \Qed

\medskip

{\bf (d)} Now suppose that $\family{\gamma}{\Gamma}{J_{\gamma}}$ is a
non-empty
finite family of subsets of $I$ with infinite intersection.   As before,
write $\Lambda$ for $\bigvee_{\gamma\in\Gamma}\Sigma_{J_{\gamma}}$.
Suppose that $K\subseteq I$ and that
$\family{\gamma}{\Gamma}{K_{\gamma}}$ is a family of subsets of $I$ such
that $K\cap K_{\gamma}\subseteq J_{\gamma}$ for every $\gamma\in\Gamma$.
Then $\Sigma_K$ and $\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$ are
relatively independent over $\Lambda$.
\Prf\ Set $\Tau=\bigcup\{\Sigma_L:L\in[K]^{<\omega}\}$ and
for $\gamma\in\Gamma$ set
$\Tau_{\gamma}=\bigcup\{\Sigma_L:L\in[K_{\gamma}]^{<\omega}\}$.
Then (b)-(c) tell us that $\Tau$ and the algebra
$\Tau'\,\,\sigma$-generated by
$\bigcup_{\gamma\in\Gamma}\Tau_{\gamma}$ are relatively independent over
$\Lambda$.
Since $\Sigma_K$ is the $\sigma$-algebra
generated by $\Tau$, while $\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$
is the $\sigma$-algebra generated by $\Tau'$,
458Da-458Db tell us that $\Sigma_K$ and
$\bigvee_{\gamma\in\Gamma}\Sigma_{K_{\gamma}}$ are relatively independent
over $\Lambda$.\ \Qed

\medskip

{\bf (e)} At last we are ready to approach the sets $K$, $\Cal K$ and
$\Cal J$ of the statement of this lemma.   The case $\Cal J=\emptyset$
is trivial (as then $\Cal K$ must also be empty), so suppose that $\Cal J$
is non-empty.

\medskip

\quad{\bf (i)} To begin with, suppose that $\Cal J$ and $\Cal K$ are
finite.   In this case, we can find finite families
$\family{\gamma}{\Gamma}{J_{\gamma}}$ and
$\family{\gamma}{\Gamma}{K_{\gamma}}$ running over $\Cal J$,
$\Cal K\cup\{\emptyset\}$
respectively such that $K\cap K_{\gamma}\subseteq J_{\gamma}$ for every
$\gamma$.   So (d) tells us that $\Sigma_K$ and
$\bigvee_{K'\in\Cal K}\Sigma_{K'}\vee\Sigma_{\emptyset}$ are relatively
independent over $\bigvee_{J\in\Cal J}\Sigma_J$.

\medskip

\quad{\bf (ii)} If $\Cal K$ is finite but $\Cal J$ is infinite, then
let $\Cal J_0\subseteq\Cal J$ be a finite set such that for every
$K'\in\Cal K$ there is a $J\in\Cal J_0$ including $K\cap K'$.
Then for any finite $\Cal J'\subseteq\Cal J$ including $\Cal J_0$,
$\Sigma_K$ and $\bigvee_{K'\in\Cal K}\Sigma_{K'}$
are relatively independent over $\bigvee_{J\in\Cal J'}\Sigma_J$.   Since

\Centerline{$\{\bigvee_{J\in\Cal J'}\Sigma_J:
\Cal J_0\subseteq\Cal J'\in[\Cal J]^{<\omega}\}$}

\noindent is an upwards-directed family of $\sigma$-algebras whose union
$\sigma$-generates $\bigvee_{J\in\Cal J}\Sigma_J$, 458C tells us that
$\Sigma_K$ and $\bigvee_{K'\in\Cal K}\Sigma_{K'}$ are relatively independent
over $\bigvee_{J\in\Cal J}\Sigma_J$.

\medskip

\quad{\bf (iii)} Finally, for the general case, (ii) tells us that
$\Sigma_K$ and $\bigvee_{K'\in\Cal K'}\Sigma_{K'}$
are relatively independent
over $\bigvee_{J\in\Cal J}\Sigma_J$ for every finite
$\Cal K'\subseteq\Cal K$, so $\Sigma_K$ and
$\bigvee_{K'\in\Cal K}\Sigma_{K'}$ are relatively independent
over $\bigvee_{J\in\Cal J}\Sigma_J$, by 458D again.
}%end of proof of 459I

\leader{459J}{Corollary}\dvAnew{2010}
Let $(X,\Sigma,\mu)$ be a probability space and
$I$ a set.   Let $G$ be the group of permutations $\phi$ of $I$ such that
$\{i:\phi(i)\ne i\}$ is finite.   Suppose that $\action$ is an action of
$G$ on $X$ such that $x\mapsto\phi\action x$ is \imp\ for each $\phi\in G$.
Let $\langle\Sigma_J\rangle_{J\subseteq I}$ be a family of
$\sigma$-subalgebras of $\Sigma$ such that

\inset{(i) for every $J\subseteq I$,
$\Sigma_J$ is the $\sigma$-algebra generated by
$\bigcup_{K\subseteq J\text{ is finite}}\Sigma_K$;

(ii) if $J\subseteq I$, $E\in\Sigma_J$ and $\phi\in G$, then
$\phi\action E\in\Sigma_{\phi[J]}$;

(iii) if $J\subseteq I$, $E\in\Sigma_J$ and $\phi\in G$ is such that
$\phi(i)=i$ for every $i\in J$, then $\phi\action E=E$.}

\noindent Then if $J\subseteq I$ is infinite and
$\family{\gamma}{\Gamma}{K_{\gamma}}$ is a family of subsets of $I$ such
that $K_{\gamma}\cap K_{\delta}\subseteq J$ for all distinct $\gamma$,
$\delta\in\Gamma$, $\family{\gamma}{\Gamma}{\Sigma_{K_{\gamma}}}$ is
relatively independent over $\Sigma_J$.

\proof{ By 459I, $\Sigma_{K_{\gamma}}$ and
$\bigvee_{\delta\in\Delta}\Sigma_{K_{\delta}}$ are
relatively independent over $\Sigma_J$ whenever $\Delta\subseteq\Gamma$ and
$\gamma\in\Gamma\setminus\Delta$.   Now
458Hb tells us that we can induce on $\#(\Delta)$ to see that
$\family{\gamma}{\Delta}{\Sigma_{K_{\gamma}}}$ is relatively independent
over $\Sigma_J$ for every finite $\Delta\subseteq\Gamma$, and it follows at
once that $\family{\gamma}{\Gamma}{\Sigma_{K_{\gamma}}}$
is relatively independent over $\Sigma_J$, as remarked in 458Ac.
}%end of proof of 459J
\proof{ Note first that if $G$ is the group of permutations $\phi$ of $I$
such that $\{i:\phi(i)\ne i\}$ is finite, then
any $\phi\in G$ is expressible as the
product of finitely many transpositions, so $w\mapsto w\phi$ is an
automorphism of $(X^I,\mu)$.   Let $\action$ be the action of $G$ on $X^I$
defined by saying that $\phi\action w=w\phi^{-1}$ for $x\in X^I$ and
$\phi\in G$.   Then $w\mapsto\phi\action w$ is \imp\ for every $\phi$.

If $L\subseteq I$ then $\Sigma_L$ is the $\sigma$-algebra of subsets of
$X^I$ generated by sets of the form $\{x:x(i)\in E\}$ where $i\in L$ and
$E\in\Sigma$.   So $\Sigma_L$ is the $\sigma$-algebra generated by
$\bigcup\{\Sigma_K:K\in[L]^{<\omega}\}$.

If $i\in I$, $E\in\Sigma$ and $\phi\in G$, then

\Centerline{$\phi\action\{x:x(i)\in E\}=\{\phi\action x:x(i)\in E\}
=\{x:(\phi^{-1}\action x)(i)\in E\}
=\{x:x(\phi(i))\in E\}$.}

\noindent So if $L\subseteq I$ and $\phi\in G$,
$\{W:\phi\action W\in\Sigma_{\phi[L]}\}$ is a $\sigma$-algebra of subsets
of $X^I$ containing $\{x:x(i)\in E\}$ whenever $i\in L$ and $E\in\Sigma$,
therefore including $\Sigma_L$;  that is,
$\phi\action W\in\Sigma_{\phi[L]}$ whenever $W\in\Sigma_L$.

If $L\subseteq I$ and $\phi\in G$ is such that $\phi(i)=i$
for every $i\in L$, then $\{W:\phi\action W=W\}$ is a $\sigma$-algebra of
subsets of $X^I$ containing $\{x:x(i)\in E\}$ whenever $i\in L$ and
$E\in\Sigma$, so $\phi\action W=W$ for every $W\in\Sigma_L$.

Thus the conditions of 459I are satisfied, and the result follows at once.
}%end of proof of 459J

\leader{459K}{}\cmmnt{ Following the results of \S452 (especially
452Ye), we do not generally expect to find disintegrations of measures
which are not countably compact.   It may however illuminate the
constructions here if I give a specific example related to the contexts
of 459E and 459H.

\medskip

\noindent}{\bf Example}\cmmnt{ ({\smc Dubins \& Freedman 79})}
There are a separable metrizable space $Z$ and a quasi-Radon measure on
$Z^{\Bbb N}$, invariant under permutations of coordinates, which cannot
be disintegrated into powers of measures on $Z$.

\proof{{\bf (a)} Let $\lambda$ be Lebesgue measure on $[0,1]$.
$Q=[0,1]\times[0,1]^{\Bbb N}$, with its usual topology, is a compact
metrizable space, so has just $\frak c$ Borel sets (4A3F).   Let
$\ofamily{\xi}{\frakc}{W_{\xi}}$ enumerate the Borel subsets of
$Q$ with non-zero measure for the product
measure $\lambda\times\lambda^{\Bbb N}$.
(Remember that $\lambda\times\lambda^{\Bbb N}$ is a Radon measure,
by 416U.)   For each $\xi$, we have
$0<(\lambda\times\lambda^{\Bbb N})(W_{\xi})
=\int\lambda^{\Bbb N}(W_{\xi}[\{t\}])\lambda(dt)$, so
$A_{\xi}=\{t:W_{\xi}[\{t\}]\ne\emptyset\}$ has cardinal $\frak c$
(419H);  we can therefore choose $\ofamily{\xi}{\frakc}{t_{\xi}}$ in
$[0,1]$ such that $t_{\xi}\in A_{\xi}\setminus\{t_{\eta}:\eta<\xi\}$
for every $\xi<\frak c$.   Now choose $t_{\xi n}$, for $\xi<\frak c$ and
$n\in\Bbb N$, such that $(t_{\xi},\sequencen{t_{\xi n}})\in W_{\xi}$.
Set
$Z=\{(t_{\xi},t_{\xi n}):\xi<\frak c$, $n\in\Bbb N\}\subseteq[0,1]^2$.

\medskip

{\bf (b)} Set $X=([0,1]^2)^{\Bbb N}$ and define $\phi:Q\to X$ by setting
$\phi(t,\sequencen{t_n})=\sequencen{(t,t_n)}$ for $t$, $t_n\in[0,1]$.
Then $\phi$ is a homeomorphism between $Q$ and $\phi[Q]$, so there is a
unique Radon measure $\mu^{\#}$ on $X$ such that $\phi$ is \imp\ for
$\lambda\times\lambda^{\Bbb N}$ and $\mu^{\#}$.   Now $\mu^{\#}$ is
invariant under permutations of coordinates, because if
$\rho:\Bbb N\to\Bbb N$ is a permutation and $\hat\rho(x)=x\rho$ for
$x\in X$, then $\hat\rho\phi=\phi\bar\rho$, where
$\bar\rho(t,\sequencen{t_n})=(t,\sequencen{t_{\rho(n)}})$;  and as
$\bar\rho:Q\to Q$ is \imp, so is $\hat\rho:X\to X$.

Also $Z^{\Bbb N}$ has full outer measure for $\mu^{\#}$.   \Prf\ If
$\mu^{\#}W>0$, then $(\lambda\times\lambda^{\Bbb N})\phi^{-1}[W]>0$, so
there is some $\xi<\frak c$ such that $W_{\xi}\subseteq\phi^{-1}[W]$.
Now $\sequencen{(t_{\xi},t_{\xi n})}\in Z^{\Bbb N}\cap W$.\ \QeD\
Accordingly the subspace measure $\tilde\mu$ on $Z$ is a probability
measure.   Because $\mu^{\#}$ is invariant under permutations of
coordinates, so is $\tilde\mu$;  because $\mu^{\#}$ is a Radon measure,
$\tilde\mu$ is a quasi-Radon measure (416Ra).

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that there are a probability
space $(Y,\Tau,\nu)$ and a family $\family{y}{Y}{\mu_y}$ of probability
measures on $Z$ such that $\tilde\mu E=\int\mu_y^{\Bbb N}E\,\nu(dy)$ for
every Borel set $E\subseteq Z^{\Bbb N}$.   (The argument to follow will not depend on which product measure is used in forming the
$\mu_y^{\Bbb N}$.)   Looking at sets of the form
$(Z\cap H)\times Z\times Z\times\ldots$, where
$H\subseteq[0,1]^2$ is a Borel set, we see that $\mu_y(Z\cap H)$ must be
defined for almost every $y$;  as $Z$ is second-countable, $\mu_y$ must
be a topological measure for almost every $y$.
Looking at sets of the form
$(Z\cap(G_0\times[0,1]))\times(Z\cap(G_1\times[0,1]))\times
Z\times\ldots$, where $G_0$ and $G_1$ are disjoint Borel subsets of
$[0,1]$, we see that
$\mu_y(Z\cap(G_0\times[0,1]))\cdot\mu_y(Z\cap(G_1\times[0,1]))=0$ for
almost every $y$;  as $[0,1]$ is second-countable and Hausdorff, there
must be, for almost every $y\in Y$, an $s_y\in[0,1]$ such that
$\mu_y(Z\cap(\{s_y\}\times[0,1]))=1$.

Next, if $G\subseteq[0,1]$ is a Borel set, then
$\mu_y(Z\cap([0,1]\times G))=\lambda G$ for almost every $y$.   \Prf\

\Centerline{$h(y)
=\mu_y^{\Bbb N}((Z\cap([0,1]\times G))\times Z\times\ldots)
=\mu_y(Z\cap([0,1]\times G))$}

\noindent is defined for almost every $y$, and $h$ is $\nu$-integrable,
with

\Centerline{$\int h\,d\nu
=\tilde\mu((Z\cap([0,1]\times G))\times Z\times\ldots)
=\mu^{\#}(([0,1]\times G)\times[0,1]^2\times\ldots)=\lambda G$.}

\noindent At the same time,

$$\eqalign{\int h(y)(1-h(y))\nu(dy)
&=\tilde\mu((Z\cap([0,1]\times G))
\times(Z\cap([0,1]\times([0,1]\setminus G)))\times Z\times\ldots)\cr
&=\mu^{\#}(([0,1]\times G)
\times([0,1]\times([0,1]\setminus G))\times[0,1]^2\times\ldots)\cr
&=\lambda G(1-\lambda G).\cr}$$

\noindent Rearranging, we see that $\int h^2d\nu=(\int h)^2$.   But this
means that $\int(h(y)-\int h)^2\nu(dy)=0$ and $h(y)=\lambda G$ for
almost every $y$.\ \Qed

It follows that, for at least some $y$,
$\mu_y(Z\cap(\{s_y\}\times G))=\lambda G$ for every interval
$G\subseteq[0,1]$ with rational endpoints.   But this is impossible,
because all the vertical sections of $Z$ are countable.\ \Bang

Thus there is no such disintegration, as claimed.
}%end of proof of 459K

\exercises{\leader{459X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 459Xa
Let $(X,\Sigma,\mu)$ be a probability space and
$\sequencen{f_n}$ an exchangeable sequence of real-valued random
variables on $X$ all with finite expectation.   Use 459C and 273I to
show that $\sequencen{\Bover1{n+1}\sum_{i=0}^nf_i}$ converges a.e.
(Compare 276Xg\formerly{2{}76Xe}.)
%459C

\spheader 459Xb Let $(X,\Sigma,\mu)$ be a probability space and
$\sequencen{f_n}$ an exchangeable sequence of real-valued random
variables on $X$ all with finite variance, such that
$\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf_i=0$ a.e.   Show that
$\sequencen{\Pr(\sum_{i=0}^nf_i\ge\alpha\sqrt{n+1})}$ is
convergent for every $\alpha\in\Bbb R$.   \Hint{274I.}
%459C

\spheader 459Xc\dvAformerly{4{}59F}
Let $X$ be a completely regular topological space,
$(Y,\frak S,\Tau,\nu)$ a totally finite quasi-Radon measure space, and
$y\mapsto\mu_y$ a continuous function from $Y$ to the space
$M^+_{\text{qR}}(X)$ of totally finite quasi-Radon measures on $X$ with its
narrow topology.   Show that if
$\mu\in M^+_{\text{qR}}(X)$ is such that
$\int fd\mu=\iint fd\mu_y\,\nu(dy)$ for
every $f\in C_b(X)$, then $\family{y}{Y}{\mu_y}$ is a disintegration of
$\mu$ over $\nu$.
%459G

\sqheader 459Xd
({\smc Diaconis \& Freedman 80}) Let $Z$ be a non-empty
compact Hausdorff space and $I$ an infinite set including $\Bbb N$.
Let $\tilde\mu$ be a Radon probability measure on $Z^I$ invariant
under permutations of $I$.   For $k\le n$ let $D_{nk}\subseteq n^k$ be the set of injective functions from $k$ to $n$ and $\Omega_{nk}$ the set
$Z^I\times n^k\times D_{nk}$, endowed with the product $\lambda_{nk}$ of
$\tilde\mu$ and the uniform probability measures on the finite sets
$n^k$ and $D_{nk}$.   Define $\phi_{nk}:\Omega_{nk}\to Z^k$ and
$\psi_{nk}:\Omega_{nk}\to Z^k$ by setting

$$\eqalign{\phi_{nk}(w,p,q)&=wp,\cr
\psi_{nk}(w,p,q)&=wp\text{ if }p\in D_{nk},\cr
&=wq\text{ otherwise.}\cr}$$

\noindent (i) Show that there is a disintegration
$\family{w}{Z^I}{\mu_{nw}^k}$ of the image measure
$\lambda_{nk}\phi_{nk}^{-1}$
over $\tilde\mu$ where each $\mu_{nw}$ is a suitable point-supported
measure on $Z^k$.
(ii) Show that the image measure $\lambda_{nk}\psi_{nk}^{-1}$
is the image measure $\tilde\mu_k=\tilde\mu\tilde\pi_k^{-1}$, where
$\tilde\pi_k(w)=w\restr k$ for $w\in Z^I$.    (iii) Show that if $n>0$ then
$|\tilde\mu_kW-\int\mu_{nw}^kW\,\tilde\mu(dw)|
\le\Bover{k(k-1)}{2n}$ for every Baire set $W\subseteq Z^k$.
(iv) Show that there is a Radon probability measure $\tilde\nu_n$ on
$P_{\text{R}}(Z)$ for which $w\mapsto\mu_{nw}$ is \imp.   (v) Show that
if $\tilde\nu$ is any cluster point of
$\sequencen{\tilde\nu_n}$ in $P_{\text{R}}(Z)$ then
$\family{\theta}{P_{\text{R}}(Z)}{\tilde\theta^I}$ is a
disintegration of $\tilde\mu$ over $\tilde\nu$,
writing $\tilde\theta^I$
for the Radon product of copies of any $\theta\in P_{\text{R}}(Z)$.
%459H

\sqheader 459Xe ({\smc Hewitt \& Savage 55}) Let $X$ be a
non-empty compact Hausdorff space and $I$
an infinite set.   Let $Q$ be the set of Radon probability measures on
$X^I$ which are invariant under permutations of $I$.   Show that (i) $Q$
is a closed convex subset of the set $P_{\text{R}}(X^I)$ of all Radon
probability measures on $X^I$ with its narrow topology;  (ii) $Q$ is
isomorphic, as topological convex structure, to
$P_{\text{R}}(P_{\text{R}}(X))$;  (iii) the
extreme points of $Q$ are just the powers of Radon probability measures
on $X$.
%459H

\spheader 459Xf Let $X$, $I$ be sets, $\Sigma$ a
$\sigma$-algebra of subsets of $X$ and $\mu$ a probability measure with
domain $\Tensorhat_I\Sigma$ which is transposition-invariant
in the sense that for every
transposition $\tau:I\to I$ the function $x\mapsto x\tau:X^I\to X^I$ is
\imp.   For $J\subseteq I$, let $\Sigma_J$ be the $\sigma$-algebra

\Centerline{$W:W\in\Tensorhat_I\Sigma$,
$W$ is determined by coordinates in $J\}$.}

\noindent Show that if $J\subseteq I$ is infinite and
$\family{\gamma}{\Gamma}{K_{\gamma}}$ is a family of subsets of $I$ such
that $K_{\gamma}\cap K_{\delta}\subseteq J$ for all distinct $\gamma$,
$\delta\in\Gamma$, $\family{\gamma}{\Gamma}{\Sigma_{K_{\gamma}}}$ is
relatively independent over $\Sigma_J$ (i) using 459D (ii) using 459J.
%459D 459J

\leaveitout{\spheader 459X?
Let $X$ be a non-empty Hausdorff space.   Let $Q$ be the
set of Radon probability measures on
$X^{\Bbb N}$ which are invariant under permutations of $\Bbb N$.   Show
that (i) $Q$ is a closed convex subset of the set
$P_{\text{R}}(X^{\Bbb N})$ of all Radon
probability measures on $X^{\Bbb N}$ with its narrow topology;  (ii) $Q$
is isomorphic, as topological convex structure, to
$P_{\text{R}}(P_{\text{R}}(X))$;  (iii) the
extreme points of $Q$ are just the powers of Radon probability measures
on $X$.
%459H 459Xe

\query is (ii) true? (see 459Z) and if so, should it be in 459Y?  see
459Ya.
}%end of leaveitout

\leader{459Y}{Further exercises (a)}%
%\spheader 459Ya
\dvAnew{2010} Let $X$ be a
topological space and $I$ an infinite set.   Write $P_{\tau}(X)$,
$P_{\tau}(X^I)$ and
$P_{\tau}(P_{\tau}(X))$ for the spaces of $\tau$-additive Borel probability
measures in $X$, $X^I$ and $P_{\tau}(X)$ respectively,
with their narrow topologies.
(i) For $\theta\in P_{\tau}(X)$ write
$\tilde\theta^I$ for the $\tau$-additive Borel measure on $X^I$
corresponding to $\theta$, that is, the restriction to the Borel
$\sigma$-algebra of $X^I$ of the $\tau$-additive product measure described
in 417F.   Show that
$\theta\mapsto\tilde\theta^I:P_{\tau}(X)\mapsto P_{\tau}(X^I)$ is
continuous.   (ii) Show that
if $\nu\in P_{\tau}(P_{\tau}(X))$ there is a unique
$\mu_{\nu}\in P_{\tau}(X^I)$ such that
$\family{\theta}{P_{\tau}{(X)}}{\tilde\theta^I}$ is a disintegration of
$\mu_{\nu}$ over $\nu$, where $\tilde\theta^I$ is the $\tau$-additive
Borel product
measure on $X^I$ corresponding to $\theta\in P_{\tau}(X)$.
(iii) Show that $\nu\mapsto\mu_{\nu}$ is a homeomorphism between
$P_{\tau}(P_{\tau}(X))$ and its image in $P_{\tau}(X^I)$.
%mt45bits 459H

\spheader 459Yb
Discuss the problems which arise in
459B, 459C, 459E and 459H if the index set $I$ is finite.
%459H
}%end of exercises

\leaveitout{
\leader{459Z}{Problem}\dvAnew{2010} Let $X$ be a topological space,
$P_{\text{qR}}(X)$ the space of quasi-Radon
probability measures on $X$ with its narrow
topology, and $\nu$ a quasi-Radon probability measure on
$P_{\text{qR}}(X)$.   Is there necessarily a quasi-Radon probability
measure $\mu$ on $X$ such that $\family{\theta}{P_{\text{qR}}(X)}{\theta}$
is a disintegration of $\mu$ over $\nu$?
%459Ya (Yes if $X$ is regular.)
}

\endnotes{
\Notesheader{459} As I have presented this material, the centre of the
argument of 459A-459H %459A 459B 459C 459D 459E 459F 459G 459H
lies in the martingales in part (b-$\beta$) of the proof of 459B.
We are
trying to resolve the functions $f_i$ into `common' and `independent'
parts.   The `common' part is given by the conditional expectations of
the $f_i$ over an appropriate $\sigma$-algebra $\Tau$, and we approach
these by looking at the conditional expectations of each $f_i$ on
$\sigma$-algebras $\Tau_n$ generated by `distant' $f_j$.   All the most
important ideas are already exhibited when the index set $I$ is equal to
$\Bbb N$.   Note in particular that in the basic hypothesis that all
finite strings $(f_{i_0},\ldots,f_{i_r})$ have the same joint
distribution, it is enough to look at increasing strings.   But there is
a striking phenomenon which appears in sharper relief with uncountable
sets $I$:  any sequence $\sequence{k}{j_k}$ of distinct elements of $I$
can be used to generate an adequate $\sigma$-algebra, because while the
tail $\sigma$-algebra of sets depends on the choice of the $j_k$, they
all lead to the same closed subalgebra of the measure algebra (459D).

Perhaps I should emphasize at this point that $I$ really does have to be
infinite, though for large finite $I$ there are approximations to the
results here.

The proof of 459B is one of the standard proofs of De Finetti's theorem,
with trifling modifications.   In the case of real-valued random
variables we have a notion of relative distribution (458I) which gives a
quick way of saying that all the $f_i$ have the same conditional
expectations over $\Tau$, as in 459C(ii).   For variables taking values
in other spaces the situation may be different (459K), unless (as in
\S452) we have a countably compact measure (459E).

Specializing to the case $X=Z^I$ in 459B, we find ourselves examining
symmetric measures on infinite product spaces, which are of great
interest in themselves.   Note that while in the hypothesis of 459E I
have asked for the measure $\mu$ on the product space $Z^I$ to be
countably compact, what is actually necessary is that the marginal
measure on $Z$ should be countably compact.   By 454Ab, this comes to the
same thing.

As in 452O, we can look for a disintegration consisting of Radon
measures, provided of course that the marginal measure is a Radon
measure.   What we have to work harder for is a direct expression in
terms of an integral $\int\tilde\theta^I\tilde\nu(d\theta)$ where
$\tilde\nu$ is itself a Radon probability measure on the space of Radon
probability measures $\theta$ (459H).   But most of the extra work
consists of finding the correct reduction to the case of
locally compact spaces.
For compact spaces we can approach by a completely different route
(459Xd).   I will not go farther with this idea here, but I note that
the method can be used in a wide variety of problems involving symmetric
structures.

Lemma 459I is entirely different.   I include it here because it gives
another approach to relative independence and looks at
permutation-invariant measures, though in a more abstract setting which
does not bind us to the product spaces which are their most natural
expressions.   Its power lies precisely in the fact that in its hypotheses
we do {\it not} suppose that
$\Sigma_{J\cup K}=\Sigma_J\vee\Sigma_K$ for $J$, $K\subseteq I$, so the
$\sigma$-algebras $\bigvee_{K'\in\Cal K}\Sigma_{K'}$ and
$\bigvee_{J\in\Cal J}\Sigma_J$ have to be handled with special care.
}%end of notes


\discrpage


\frfilename{mt45conc.tex}
\versiondate{27.2.04}
\copyrightdate{2004}
     
\def\chaptername{Concordance}
\def\sectionname{Concordance}
     
\centerline{\bf Concordance for Volume 4}
     
\bigskip
     
I list here the section and paragraph numbers which have (to my knowledge) appeared in print in references to this volume, and which have since been changed.
     
\medskip
     
\leaveitout{Fremlin 00:   Recall that a measure space $(X,\Sigma,\mu)$ is {\bf (countably)
compact} if there is a (countably) compact class $\Cal K\subseteq\Sigma$
such that $\mu$ is inner regular with respect to $\Cal K$, in the sense
that $\mu E=\sup\{\mu K:K\in\Cal K,\,K\subseteq E\}$ for every
$E\in\Sigma$ (\MaEC\ and \RoIB, or \FrII, \S3{}42 and
\S4{}51).
}%end of leaveitout
     
\leaveitout{\S45{}1 451Ad 451B {\smc Fremlin 00} countably 
compact measure, perfect measure
}
     
\leaveitout{45{}1E 451E {\smc Fremlin 00} If $(X,\Sigma,\mu)$ is perfect, 
$(Y,\Tau,\nu)$ any measure space and $f:X\to Y$ is \imp, then $\nu$ is 
perfect.
}
     
\leaveitout{4{}51F {\smc Fremlin 00} It is worth noting that a semi-finite measure space $(X,\Sigma,\mu)$ is
perfect iff $(X,\Sigma_0,\mu\restr\Sigma_0)$ is compact for every
countably-generated $\sigma$-algebra $\Sigma_0\subseteq\Sigma$ iff
$(X,\Sigma_0,\mu\restr\Sigma_0)$ is countably compact for every
countably-generated $\sigma$-algebra $\Sigma_0\subseteq\Sigma$.
451F}
     
\leaveitout{45{}1I 451I {\smc Fremlin 00} the product of finitely many compact, or countably compact, or perfect measure spaces of any magnitude is again compact or countably compact or perfect
}
     
\leaveitout{45{}1J 451J {\smc Fremlin 00} The product of compact probability spaces is compact, the product of countably compact probability spaces
is countably compact, and the product of perfect
probability spaces is perfect
}

\leaveitout{4{}52E {\smc Bianchini \& Caravenna 09} definition of
`consistent disintegration'}
     
\leader{45{}2I}{} In {\smc Fremlin 00} I quote Pachl's result that if $(X,\Sigma,\mu)$ is countably compact,
$(Y,\Tau,\nu)$ is strictly localizable and $f:X\to Y$ is \imp, 
then $\nu$ is countably compact;  this is now in 452R.

\leaveitout{45{}3Dc 453Dc {\smc Burke Macheras Strauss p22}}
     
\leaveitout{45{}3F 453F {\smc Burke Macheras Strauss p22}}
     
\leaveitout{45{}3M 453M 4{}43N 453N {\smc Burke Macheras Strauss p22}}
     
\leaveitout{45{}3Xd 453Xd {\smc Burke Macheras Strauss p22}}

\leaveitout{45{}3Z 453Za {\smc Burke Macheras Strauss p22}}
     
\leaveitout{45{}4A 454A {\smc Fremlin 00} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a non-empty family of measure spaces.   Let
$\mu$ be a probability measure on $X$, inner regular with respect to the
$\sigma$-algebra $\Tensorhat_{i\in I}\Sigma_i$ generated by
$\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$, where $\pi_i:X\to X_i$ is the
canonical map for each $i\in I$.   For each $i\in I$ set
$\Cal C_i=\{\pi_i^{-1}[E]:E\in\Sigma_i\}$, and write
$\Cal C=\bigcup_{i\in I}\Cal C_i$.   Let $\Cal K$ be the set of
non-negligible subsets of $X$ expressible in the form $K
=\bigcap_{n\in\Bbb N}\bigcup_{r\le k_n}\bigcap_{i\in J_{nr}}C_{nri}$,
where $k_n\in\Bbb N$ for $n\in\Bbb N$, $J_{nr}$ is a finite
subset of $I$ for $n\in\Bbb N$, $r\le k_n$, and $C_{nri}\in\Cal C_i$ for
every $n\in\Bbb N$, $r\le k_n$, $i\in J_{nr}$.   Then $\mu$ is inner regular with respect to $\Cal K$.
}
     
\leaveitout{45{}4A 454A {\smc Fremlin 00} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a non-empty family of totally finite measure spaces.   Set $X=\prod_{i\in I}X_i$ and let $\mu$ be a measure on $X$ which is inner regular with
respect to the $\sigma$-algebra $\Tensorhat_{i\in I}\Sigma_i$ generated
by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$, where $\pi_i:X\to X_i$ is
the coordinate map for each $i\in I$.   Suppose that every $\pi_i$ is
\imp.  Then
(a) If $\Cal K\subseteq\Cal PX$ is a family of sets which is closed
under finite unions and countable intersections, and $\mu_i$ is inner
regular with respect to
$\Cal K_i=\{K:K\subseteq X_i,\,\pi_i^{-1}[K]\in\Cal K\}$ for every 
$i\in I$, then $\mu$ is inner regular with respect to $\Cal K$.
(b)(i) if every $\mu_i$ is a compact measure, so is $\mu$;
\quad(ii) if every $\mu_i$ is a countably compact measure, so is $\mu$;
\quad(iii) if every $\mu_i$ is a perfect measure, so is $\mu$.
}
     
\leader{4{}55D}{} The material on Brownian motion in \S4{}55, 
mentioned in {\smc K\"onig 04} and {\smc K\"onig 06},
has been moved to \S477.
% Brownian motion as a Radon measure.  
%\S454 also mentioned

\leader{4{}58Yd}{} This exercise (on the strong law of large numbers for
relatively independent sequences), referred to in the 2008 and 2015 printings of
Volume 5, %556M
is now 458Ye.

\bigskip
     
\noindent{\bf References}

\leaveitout{
Bianchini S. \& Caravenna L.\ [09] `On the extremality, uniqueness and
optimality of transference plans', Bull.\ Inst.\ Mathematics Academica
Sinica (N.S.) 4 (2009) 353-454.

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p22] `Density topologies and the marginal strong 
lifting problem for hyperstonian spaces', preprint, 2022.
%BMS.2022.pdf
}
     
Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces', 
Fundamenta Math.\ 165 (2000) 67-94.
     
K\"onig H.\ [04] `Projective limits via inner premeasures and the true 
Wiener measure', Mediterranean J.\ Math.\ 1 (2004) 1-40.

K\"onig H.\ [06] `Stochastic processes in terms of inner premeasures', 
Note di Matematica 25 (2006) 1-30.


\frnewpage

\references{
\frfilename{mt4r.tex}
\versiondate{9.2.13}
\copyrightdate{2002}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\Loadcyrillic

\Centerline{\bf References for Volume 4}


\medskip%A

Aldaz J.M.\ [95] `On compactness and Loeb measures', Proc.\ Amer.\
Math.\ Soc.\ 123 (1995) 173-175.
\cmmnt{[439Xa.]}

Aldaz J.M.\ [97] `Borel extensions of Baire measures', Fund.\ Math.\ 154
(1997) 275-293.
\cmmnt{[434Yn.]}

Aldaz J.M.\ \& Render H.\ [00] `Borel measure extensions of measures
defined on sub-$\sigma$-algebras', Advances in Math.\ 150 (2000)
233-263.
\cmmnt{[432D.]}

Alexandroff P.S.\ \& Urysohn P.S.\ [1929] {\it  M\'emoire sur les espaces
topologiques compacts},
Verhandelingen Akad.\ Wetensch.\ Amsterdam 14 (1929) 1-96.
\cmmnt{[419L.]}

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P.,
1987.
\cmmnt{[4A4N.]}

Andretta A.\ \& Camerlo R.\ [13] `The descriptive set theory of the
Lebesgue density theorem', Advances in Math.\ 234 (2013).
%{\tt http://arxiv.org/abs/1105.3355}.
\cmmnt{[475Yg.]}

Arkhangel'skii A.V.\ [92] {\it Topological Function Spaces.}  Kluwer,
1992.
\cmmnt{[\S462 {\it intro.}, \S467 {\it notes\/}.]}

\leaveitout{
Armitage D.H.\ \& Gardiner S.J.\ [01] {\it Classical Potential Theory.}
Springer, 2001.
478Yd}

Aronov B., Basu S., Pach J.\ \& Sharir M.\ [03] (eds.)
{\it Discrete and Computational Geometry;  the Goodman-Pollack
Festschrift,}  Springer, 2003.

Asanov M.O.\ \& Velichko N.V.\ [81] `Compact sets in $C_p(X)$', Comm.\
Math.\ Helv.\ 22 (1981) 255-266.
\cmmnt{[462Ya.]}

Austin T.\ [10a] `On the norm convergence of nonconventional ergodic
averages', Ergodic Theory and Dynamical Systems 30 (2010) 321-338.
%({\tt http://{\penalty-100}arxiv.org/abs/{\penalty-100}0805.0320}).
\cmmnt{[497M.]}

Austin T.\ [10b] `Deducing the multidimensional Szemer\'edi theorem from
the infinitary hypergraph removal lemma',
J.\ d'Analyse Math.\ 111 (2010) 131-150.
%({\tt http://arxiv.org{\bsp}abs/0808.2267}).
\cmmnt{[497M.]}

\leaveitout{
Avil\'es A., Plebanek G.\ \& Rodr\'{\i}guez J.\ [p11] `Measurability in
$C(2^{\kappa})$ and Kunen cardinals', 2011.
%material on  C(\{0,1\}^{\kappa}) ;  Baire and Borel \sigma-algebras
%in weak and pointwise topologies
}

\leaveitout{Axler S. \& Ramey W.\ [95]
`Harmonic polynomials and Dirichlet-type problems',
Proc.\ Amer.\  Math.\ Soc.\ 123 (1995) 3765-3773.
http://www.axler.net/HarmonicPolynomials.html
478Yd}

\medskip%B

Balogh Z.\ [96] `A small Dowker space in ZFC', Proc.\ Amer.\ Math.\
Soc.\ 124 (1996) 2555-2560.
\cmmnt{[439O.]}

Banach S.\ [1923] `Sur le probl\`eme de la mesure',
Fundamenta Math.\ 4 (1923) 7-33.
\cmmnt{[449O.]}

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996 (London Math.\ Soc.\
Lecture Note Series 232).  \cmmnt{[424H, 448P, \S448 {\it notes\/}.]}

Bellow A.\ {\it see} Ionescu Tulcea A.

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).
%4r

Bergelson V.\ [85] `Sets of recurrence of $\Bbb Z^m$-actions and properties of
sets of differences in $\Bbb Z^m$',
J.\ London Math.\ Soc.\ (2) 31 (1985) 295-304.
%http://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Bergelson%20-%20Sets%20of%20recurrence%20of%20Zm-actions%20and%20properties%20of%20sets%20of%20differences%20in%20Zm%20(1985).pdf
\cmmnt{[491Yt.]}

Bergman G.M.\ [06] `Generating infinite symmetric groups', Bull.\ London
Math.\ Soc.\ 38 (2006) 429-440.
%http://math.berkeley.edu/~gbergman/papers/
%({\tt http:/{\bsp}arxiv.org{\bsp}abs{\bsp}math{\bsp}0401304v2}).
\cmmnt{[\S494 {\it notes}.]}

Billingsley P.\ [99] {\it Convergence of Probability Measures.}  Wiley,
1999.
\cmmnt{[\S4A3 {\it notes}.]}
%second edition ISBN 0-471-19745-9

Blackwell D.\ [56] `On a class of probability spaces',
pp.\ 1-6 in {\smc Neyman 56}.
\cmmnt{[419K, 452P.]}

Bochner S.\ [1933] `Monotone Funktionen, Stieltjessche Integrale, und
harmonische Analyse', Math.\ Ann.\ 108 (1933) 378-410.
\cmmnt{[445N.]}

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.
\cmmnt{[437Q, 437Xs, 437Yo, 437Yu, 457K.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[4A1G, 4A4N.]}

Bongiorno B., Piazza L.di \ \& Preiss D.\ [00]
`A constructive minimal integral which includes Lebesgue integrable
functions and derivatives', J.\ London Math.\ Soc.\ (2) 62 (2000) 117-126.
\cmmnt{[483Yi.]}

Borodulin-Nadzieja P.\ \& Plebanek G.\ [05] `On compactness of measures on
Polish spaces', Illinois J.\ Math.\ 49 (2005) 531-545.
\cmmnt{[451L.]}

Bourbaki N.\ [65] {\it Int\'egration,} chaps.~1-4.   Hermann, 1965
(Actualit\'es Scientifiques et Industrielles 1175).
\cmmnt{[\S416 {\it notes\/}, \S436 {\it notes\/}.]}

Bourbaki N.\ [66] {\it Espaces Vectoriels Topologiques}, chaps.~1-2,
$2^{\text{e}}$ \'ed.
Hermann, 1966 (Actualit\'es Scientifiques et Industrielles 1189).
\cmmnt{[4A2J.]}
%Store A QA 37.B6 no 7792 0070 5

Bourbaki N.\ [69] {\it Int\'egration,} chap.~9.   Hermann, 1969
(Actualit\'es Scientifiques et Industrielles 1343).
\cmmnt{[\S436 {\it notes.}]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S461 {\it notes\/}, \S4A4.]}

\indexheader{Bourgain}
Bourgain J., Fremlin D.H.\ \& Talagrand M.\ [78] `Pointwise compact
sets of Baire-measurable functions', Amer.\ J.\ Math.\ 100 (1978)
845-886.
\cmmnt{[462Ye.]}

\indexheader{Brodski\v{\i}}
Brodski\v{\i} M.L.\ [1949] `On some properties of sets of positive measure',
Uspehi Matem.\ Nauk (N.S.) 4 (1949) 136-138.
\cmmnt{[498B.]}

Brook R.B.\ [70] `A construction of the greatest ambit',
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\cmmnt{[449D.]}

Burke D.K.\ \& Pol R.\ [05] `Note on separate continuity and the Namioka
property', Top.\ Appl.\ 152 (2005) 258-268.
\cmmnt{[\S463 {\it notes.}]}

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p21] `The strong marginal lifting
problem for hyperstonian spaces', preprint (2021).
%bms5s.pdf

\medskip%C

Carrington D.C.\ [72] `The generalised Riemann-complete integral', PhD
thesis, Cambridge, 1972.
\cmmnt{[481L.]}

\v{C}ech E.\ [66] {\it Topological Spaces.}  Wiley, 1966.
\cmmnt{[\S4A2, 4A3S.]}

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[494F.]}

\indexheader{Choquet}
Choquet G.\ [55] `Theory of capacities', Ann.\ Inst.\ Fourier (Grenoble)
5 (1955) 131-295.
\cmmnt{[432K.]}

\indexheader{Chung}
Chung K.L.\ [95] {\it Green, Brown and Probability.}   World Scientific,
1995.
% 978-9810225339  HA33.4.C5  479Xt

\indexheader{Ciesielski}
Ciesielski K.\ \& Pawlikowski J.\ [03]
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\cmmnt{[498C.]}

Cs\'asz\'ar \'A.\ [78] {\it General Topology.}  Adam Hilger, 1978.
\cmmnt{[\S4A2, \S4A5.]}

\medskip%D

Davies R.O.\ [70] `Increasing sequences of sets and Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 20 (1970) 222-236.
\cmmnt{[471G.]}

Davies R.O.\ [71] `Measures not approximable or not specifiable by means
of balls', Mathematika 18 (1971) 157-160.
\cmmnt{[\S466 {\it notes\/}.]}

Davis W.J., Ghoussoub N.\ \& Lindenstrauss J.\ [81] `A lattice-renorming
theorem and applications to vector-valued processes', Trans.\ Amer.\
Math.\ Soc.\ 263 (1981) 531-540.
\cmmnt{[\S467 {\it notes\/}.]}

\indexheader{de Caux}
de Caux P.\ [76] `A collectionwise normal, weakly $\theta$-refinable
Dowker space which is neither irreducible nor realcompact', Topology
Proc.\ 1 (1976) 67-77.
\cmmnt{[439O.]}

Dellacherie C.\ [80] `Un cours sur les ensembles analytiques', pp.\
183-316 in {\smc Rogers 80}.
\cmmnt{[\S432 {\it notes\/}.]}

de Maria J.L.\ \& Rodriguez-Salinas B.\ [91] `The space
$(\ell_{\infty}/c_0)$,weak) is not a Radon space', Proc.\ Amer.\ Math.\
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\cmmnt{[466I.]}

Deville R., Godefroy G.\ \& Zizler V.\ [93] {\it Smoothness and
Renormings in Banach Spaces.}  Pitman, 1993.
\cmmnt{[\S467 {\it intro.}, 467Ye, \S467 {\it notes\/}.]}

Diaconis, P.\ \& Freedman, D.\ [80] `Finite exchangeable sequences.'
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\cmmnt{[459Xd.] }

Droste M., Holland C.\ \& Ulbrich G.\ [08]
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uncountable cofinalities',
Bull.\ London Math. Soc.\ 40 (2008) 463-472.
%{\tt http://web.mac.com/chollan1{\bsp}iWeb/Site{\bsp}publications.html}.
\cmmnt{[494Q.]}

Dubins L.E.\ \& Freedman D.A.\ [79]
`Exchangeable processes need not be mixtures of independent,
identically distributed random variables',
Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 48 (1979) 115-132.
\cmmnt{[459K.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S4A4.]}

Du Plessis, N.\ [70] {\it Introduction to Potential Theory.}  Oliver \&
Boyd, 1970.
%used for \S478

\medskip%E

Eggleston H.G.\ [54] `Two measure properties of Cartesian product sets',
Quarterly J.\ Math.\ (2) 5 (1954) 108-115.
\cmmnt{[498B.]}

Emerson W.\ \& Greenleaf F.\ [67] `Covering properties and F{\o}lner
conditions for locally compact groups',
Math.\ Zeitschrift 102 (1967) 370-384.
\cmmnt{[449J.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[4A1A.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[Vol.\ 4 {\it intro.}, 415Yb, \S434 {\it notes\/},
\S435 {\it notes\/}, 462L, \S4A2, 4A4B.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[Chap.\ 47 {\it intro.}, \S473 {\it intro.}, \S473 {\it notes\/},
\S474 {\it notes\/}, \S475 {\it notes\/}.]}

\medskip%F

Falconer K.J.\ [90] {\it Fractal Geometry.}  Wiley, 1990.
\cmmnt{[\S471 {\it intro.}]}

Federer H.\ [69] {\it Geometric Measure Theory.}
Springer, 1969 (reprinted 1996).
\cmmnt{[\S441 {\it notes\/}, 442Ya,
\S443 {\it notes\/}, Chap.\ 47 {\it intro.}, 471Yk,
\S471 {\it notes\/}, \S472 {\it notes\/}, \S475 {\it notes\/}.]}

Fernique X.\ [97] {\it Fonctions al\'eatoires gaussiennes,
vecteurs al\'eatoires gaussiens}, CRM Publications, 1997.
\cmmnt{[\S456 {\it notes\/}.]}

Folland G.B.\ [95] {\it A Course in Abstract Harmonic Analysis.}   CRC
Press, 1995
\cmmnt{[\S4A5, \S4A6.]}

\indexheader{F{\o}lner}
F{\o}lner E.\ [55] `On groups with full Banach mean value',
Math.\ Scand.\ 3 (1955) 243-254.
\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
Ford L.R.\ \& Fulkerson D.R.\ [56] `Maximal flow through a network',
Canadian J.\ Math.\ 8 (1956) 399-404.
\cmmnt{[4A4N.]}

Frankl P.\ \& R\"odl V.\ [02] `Extremal problems on set systems',
Random Structures Algorithms 20 (2002) 131-164.
\cmmnt{[497L.]}

\indexheader{Fremlin}
Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
Manuscripta Math.\ 15 (1975) 219-242.
\cmmnt{[463K.]}

Fremlin D.H.\ [75b] `Topological measure spaces:  two counter-examples',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 78 (1975) 95-106.
\cmmnt{[419C, 419D.]}

Fremlin D.H.\ [76] `Products of Radon measures:  a counterexample',
Canadian Math.\ Bull.\ 19 (1976) 285-289.
\cmmnt{[419E.]}

Fremlin D.H.\ [81] `Measurable functions and almost continuous
functions', Manuscripta Math.\ 33 (1981) 387-405.
\cmmnt{[451T.]}

Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S434 {\it notes\/}, 439Yd, 4A2E.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S438 {\it notes\/}.]}

Fremlin D.H.\ [95] `The generalized McShane integral',
Illinois J.\ Math.\ 39 (1995) 39-67.
\cmmnt{[481N.]}

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
Fundamenta Math.\ 165 (2000) 67-94.
\cmmnt{[451V.]}

Fremlin D.H.\ [n05] `Strictly localizable Borel measures', note of 7.5.05
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[434Yr.]}

Fremlin D.H.\ [n15] `The Sorgenfrey line is not a Prokhorov space',
note of 5.8.15
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[439Yk.]}

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fund.\ Math.\ 147 (1995) 27-37.
\cmmnt{[434Q.]}

Fremlin D.H.\ \& Talagrand M.\ [78] `On the representation of increasing
linear functionals on Riesz spaces by measures',
Mathematika 25 (1978) 213-215.
\cmmnt{[439I.]}

Fremlin D.H.\ \& Talagrand M.\ [79] `A decomposition theorem for
additive set functions and applications to Pettis integral and ergodic
means', Math.\ Zeitschrift 168 (1979) 117-142.
\cmmnt{[\S464.]}  %{\it 464C}, 464I, 464M, 464N, \S464 {\it notes\/}

Fristedt B.\ \& Gray L.\ [97] {\it A Modern Approach to Probability
Theory.}  Birkh\"auser, 1997.
\cmmnt{[\S455 {\it notes\/}, \S495 {\it notes\/}.]}
%HA33.4.F7

Frol{\'\i}k Z.\ [61] `On analytic spaces', Bull.\ Acad.\ Polon.\ Sci.\ 
%8 (1961) 747-750.
9 (1961) 721-726.   %Paulo Monteiro, email 8.9.22
\cmmnt{[422F.]}

Frol\'\i k Z.\ [88] (ed) {\it General Topology and its Relations to
Modern Analysis and Algebra VI} (Proc.\ Sixth Prague Topological
Symposium), Heldermann, 1988.

Furstenberg H.\ [77]
`Ergodic behavior of diagonal measures and a theorem of Szemer\'edi
on arithmetic progressions', J. d'Analyse Math.\ 31 (1977), 204-256.
\cmmnt{[\S497 {\it intro.}, \S497 {\it notes\/}.]}

Furstenberg H.\ [81]
{\it Recurrence in Ergodic Theory and Combinatorial Number Theory.}
Princeton U.P., 1981.
\cmmnt{[\S497 {\it intro.}, 497N.]}
% QA614

Furstenberg H.\ \& Katznelson Y.\ [85]
`An ergodic Szemer\'edi theorem for IP-systems and combinatorial theory',
J.\ d'Analyse Math.\ 45 (1985) 117-168.
\cmmnt{[\S497 {\it intro.}]}

\medskip%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S4A2.]}

Gardner R.J.\ [75] `The regularity of Borel measures and Borel
measure-compactness', Proc.\ London Math.\ Soc.\ (3) 30 (1975) 95-113.
\cmmnt{[438M.]}

Gardner R.J.\ \& Pfeffer W.F.\ [84] `Borel measures', pp. 961-1043 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[434G.]}

Giordano T.\ \& Pestov V.\ [02] `Some extremely amenable groups',
C.R.A.S.\ Paris (S\'er.\ I) 334 (2002) 273-278.
\cmmnt{[494I.]}

Glasner S.\ [98] `On minimal actions of Polish groups', Topology and its Applications 85 (1998) 119-125.
\cmmnt{[\S493 {\it notes\/}.]}

Gordon R.A.\ [94] {\it The Integrals of Lebesgue, Denjoy, Perron and
Henstock.}   Amer.\ Math.\ Soc., 1994 (Graduate Studies in Mathematics
4).
\cmmnt{[481Q, \S483 {\it notes\/}.]}

Greenleaf F.P.\ [69] {\it Invariant Means on Topological Groups and Their
Applications.}   van Nostrand, 1969.
\cmmnt{[\S449 {\it notes.}]}

Grothendieck A. [92] {\it Topological Vector Spaces.}  Gordon \& Breach,
1992.
\cmmnt{[\S462 {\it notes\/}.]}

\indexheader{Grzegorek}
Grzegorek E.\ [81] `On some results of Darst and Sierpi\'nski concerning
universal null and universally negligible sets', Bull.\ Acad.\ Polon.\
Sci.\ (Math.) 29 (1981) 1-5.
\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
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\cmmnt{[494E.]}

Halmos P.R.\ [50] {\it Measure Theory.}   Springer, 1974.
\cmmnt{[\S441 {\it notes\/}, \S442 {\it notes\/}, 443M, \S443 {\it notes\/}.]}

Halmos P.R.\ [56] {\it Lectures on ergodic theory.}  Chelsea, 1956.
\cmmnt{[494A.]}

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Banach spaces', Serdica Math.\ J.\ 27 (2001) 1-66.
\cmmnt{[466D.]}

Hart J.E.\ \& Kunen K.\ [99] `Orthogonal continuous functions',
Real Analysis Exchange 25 (1999) 653-659.
\cmmnt{[416Yh.]}

Hartman S.\ \& Mycielski J.\ [58] `On the imbedding of topological groups into connected topological groups', Colloq.\ Math.\ 5 (1958) 167-169.
\cmmnt{[493Ya.]}

Haydon R.\ [74] `On compactness in spaces of measures and
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\cmmnt{[438J.]}

Henry J.P.\ [69] `Prolongement des mesures de Radon', Ann.\ Inst.\
Fourier 19 (1969) 237-247.
\cmmnt{[416N.]}

Henstock R.\ [63] {\it Theory of Integration.}  Butterworths, 1963.
\cmmnt{[Chap.\ 48 {\it intro.}, 481J.]}

Henstock R.\ [91] {\it The General Theory of Integration.}   Oxford
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\cmmnt{[\S481 {\it notes\/}.]}

Herer W.\ \& Christensen J.P.R.\ [75] `On the existence of pathological submeasures and the construction of exotic topological groups', Math.\ Ann.\ 213 (1975) 203-210.
\cmmnt{[\S493 {\it notes\/}.]}

Herglotz G.\ [1911] `\"Uber Potenzreihen mit positivem reellen Teil im
Einheitskreis', Leipziger Berichte 63 (1911) 501-511.
\cmmnt{[445N.]}

Hewitt E.\ \& Ross K.A.\ [63] {\it Abstract Harmonic Analysis I.}
Springer, 1963 and 1979.
\cmmnt{[\S442 {\it notes\/}, \S444 {\it intro.}, \S444 {\it notes\/},
\S4A5, \S4A6.]}

Hewitt E.\ \& Savage L.J.\ [55]
`Symmetric measures on Cartesian products',
Trans.\ Amer.\ Math.\ Soc.\ 80 (1955) 470-501.
\cmmnt{[459Xe.]}

%Holick\'y P.\ \& Spurn\'y J.\ [03] `Perfect images of absolute Souslin
%and absolute Borel Tychonoff spaces', Topology and its Appl.\ 131 (2003)
%281-294.
%\cmmnt{[434Yr.]}

\indexheader{Howroyd}
Howroyd J.D.\ [95] `On dimension and on the existence of sets of finite
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\cmmnt{[471R, 471S.]}

\medskip%I

\indexheader{Ionescu Tulcea}
Ionescu Tulcea A.\ [73] `On pointwise convergence, compactness and
equicontinuity I', Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 26
(1973) 197-205.
\cmmnt{[463C.]}

Ionescu Tulcea A.\ [74] `On pointwise convergence, compactness, and
equicontinuity II', Advances in Math.\ 12 (1974) 171-177.
\cmmnt{[463G.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [67] `On the existence of a
lifting commuting with the left translations of an arbitrary locally
compact group', pp.\ 63-97 in {\smc LeCam \& Neyman 67}.
\cmmnt{[\S447 {\it intro.}, 447I, \S447 {\it notes\/}.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.
\cmmnt{[\S453 {\it intro.}]}

\indexheader{Ismail}
Ismail M.\ \& Nyikos P.J.\ [80] `On spaces in which countably compact
sets are closed, and hereditary properties', Topology and its Appl.\ 11
(1980) 281-292.
\cmmnt{[\S434 {\it notes\/}.]}

\medskip%J

%Jackson S.\ \& Mauldin R.D.\ [99] `On the $\sigma$-class generated by
%open balls', Math.\ Proc.\ Cambridge Phil.\ Soc.\ 127 (1999) 99-108.

Jameson G.J.O.\ [74] {\it Topology and Normed Spaces.}  Chapman \& Hall,
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\cmmnt{[\S4A4.]}

\indexheader{Jayne}
Jayne J.E.\ [76] `Structure of analytic Hausdorff spaces', Mathematika
23 (1976) 208-211.
\cmmnt{[422Yc.]}

Jayne J.E.\ \& Rogers C.A.\ [95] `Radon measures on Banach spaces with
their weak topologies', Serdica Math.\ J.\ 21 (1995) 283-334.
\cmmnt{[466H, \S466 {\it notes\/}.]}

\indexheader{Jech}
Jech T.\ [78] {\it Set Theory.}  Academic, 1978.
\cmmnt{[423S, \S4A1.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003.
%ISBN 3540440852
\cmmnt{[\S4A1.]}

Johnson R.A.\ [82] `Products of two Borel measures', Trans.\ Amer.\
Math.\ Soc.\ 269 (1982) 611-625.
\cmmnt{[434R.]}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
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Juh\'asz I., Kunen K.\ \& Rudin M.E.\ [76] `Two more hereditarily
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\cmmnt{[\S437 {\it notes\/}, 439K, \S439 {\it notes\/}.]}

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}  Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[4A1A.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}  Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[\S4A1.]}

\medskip%K

Kampen, E.R.van [1935] `Locally bicompact abelian groups and their character groups',
Ann.\ of Math.\ (2) 36 (1935) 448-463.
\cmmnt{[445U.]}

\indexheader{Kaplansky}
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Univ.\ of Chicago Press, 1971.
\cmmnt{[\S446 {\it notes.}]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
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\cmmnt{[Chap.\ 42 {\it intro.}, 423T, \S423 {\it notes\/},
\S448 {\it notes\/}, \S4A2, 4A3S, 4A3T.]}

%Kechris A.S.\ [p09] {\it Global Aspects of Ergodic Group Actions.}
%Amer.\ Math.\ Soc.\

Kellerer H.G.\ [84] `Duality theorems for marginal problems', Zeitschrift
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\cmmnt{[457M.]}

Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[438R.]}

Kelley J.L.\ \& Srinivasan T.P.\ [71] `Pre-measures on lattices of sets',
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\cmmnt{[413Xr.]}

Kittrell J.\ \& Tsankov T.\ [09] `Topological properties of full groups',
Ergodic Theory and Dynamical Systems 2009 (doi:10.1017/S0143385709000078).
\cmmnt{[494O, \S494 {\it notes}.]}

Kolmogorov A.N.\ [1933] {\it Grundbegriffe der
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\cmmnt{[454D.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[\S462 {\it notes\/}, \S4A4, 4A6C.]}

Koumoullis G.\ \& Prikry K.\ [83] `The Ramsey property and measurable
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\cmmnt{[451T.]}

Kuipers L.\ \& Niederreiter H.\ [74] {\it Uniform Distribution of
Sequences.}   Wiley, 1974.
\cmmnt{[491Z.]}

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[\S4A1, 4A2E.]}

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.
%4r

Kuratowski K.\  [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[423T, 462Ye, \S4A2, 4A3S.]}

Kurzweil J.\ [57] `Generalized ordinary differential equations and
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\cmmnt{[Chap.\ 48 {\it intro.}]}

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%{\tt http://www.math.uiuc.edu/\penalty-100$^{\sim}$ssolecki/papers.html}.
\cmmnt{[\S448 {\it notes\/}.]}

\medskip%L

Laczkovich M.\ [02] `Paradoxes in measure theory', pp.\ 83-123 in
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\cmmnt{[\S449 {\it notes\/}.]}

LeCam L.M.\ \& Neyman J.\ [67] (eds.) {\it Proc.\ Fifth Berkeley
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Levy A.\ [79] {\it Basic Set Theory.}  Springer, 1979.
\cmmnt{[\S4A1.]}

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\cmmnt{[\S495 {\it notes\/}.]}

\indexheader{Losert}
Losert V.\ [79] `A measure space without the strong lifting property',
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\cmmnt{[453N, \S453 {\it notes\/}.]}

\indexheader{Luke\v{s}}
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\cmmnt{[414Ye.]}

\medskip%M

\indexheader{Mackey}
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\cmmnt{[448S.]}

Marczewski E.\ [53] `On compact measures', Fundamenta Math.\ 40 (1953)
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\cmmnt{[413T, Chap.\ 45 {\it intro.}, \S451 {\it intro.}, 451J.]}

Marczewski E.\ \& Ryll-Nardzewski C.\ [53] `Remarks on the compactness
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\cmmnt{[454C.]}

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\cmmnt{[435C.]}

Mattila P.\ [95] {\it Geometry of Sets and Measures in Euclidean
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\cmmnt{[Chap.\ 47 {\it intro.}, \S471 {\it intro.}, \S472 {\it notes\/},
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%QA311.M2 9520 6456 1

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\cmmnt{[418T.]}

Maurey B.\ [79] `Construction des suites sym\'etriques', C.R.A.S.\ (Paris) 288 (1979) 679-681.
\cmmnt{[492H.]}
  % one m in sym\'etrique

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\cmmnt{[481M.]}

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\cmmnt{[\S495 {\it notes\/}.]}

Milman V.D.\ \& Schechtman G.\ [86]
{\it Asymptotic Theory of Finite Dimensional Normed Spaces.}
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\cmmnt{[492G, \S492 {\it notes\/}.]}

\leaveitout{Mohammed D.\ \& Saadi K.\ [p09#]
`Probl\`emes de mesurabilit\'es sur
l'espace des fonctions continues sur un compact', 2009.
\cmmnt{[466Y?.]}
%refereeing for Canad Math Bull;  rejected
}

Molt\'o A., Orihuela J., Troyanski S.\ \& Valdivia M.\ [09] {\it A
Non-linear Transfer Technique for Renorming.}   Springer, 2009 (Lecture
Notes in Mathematics 1951).
\cmmnt{[\S467 {\it notes.}]}

Montgomery D.\ \& Zippin L.\ [55] {\it Topological Transformation
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\cmmnt{[\S446 {\it notes.}]}

\indexheader{Moran}
Moran W.\ [68] `The additivity of measures on completely regular
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\cmmnt{[439P.]}

Moran W.\ [69] `Measures and mappings on topological spaces', Proc.\
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\cmmnt{[435Xk.]}

Moran W.\ [70] `Measures on metacompact spaces', Proc.\ London Math.\
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\cmmnt{[438J.]}

\indexheader{Moschovakis}
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\cmmnt{[Chap.\ 42 {\it intro.}, \S423 {\it notes\/}.]}

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\cmmnt{[481P, \S482 {\it notes\/}.]}

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%ISBN 5-900975-01-0

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\cmmnt{[451U.]}

\medskip%N

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\cmmnt{[497J.]}

%Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.


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\medskip%O
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\cmmnt{[434Yq.]}

\medskip%P

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\cmmnt{[452I, 452S, 452Ye, \S452 {\it notes\/}.]}

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\cmmnt{[452R.]}

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\cmmnt{[449K, \S449 {\it notes\/}.]}
% QA 403.P2 store A

Perron O.\ [1914] `\"Uber den Integralbegriff', Sitzungsberichte der
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\cmmnt{[\S493 {\it notes\/}.]}

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\cmmnt{[484Xb.]}

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\cmmnt{[484B.]}

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\cmmnt{[\S461 {\it notes\/}.]}

\indexheader{Plewik}
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\cmmnt{[443Yl.]}

\indexheader{Pontryagin}
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\cmmnt{[445U.]}

\indexheader{Preiss}
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Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 27 (1973) 109-116.
\cmmnt{[439S.]}
%ILL 31.7.15

Preiss D.\ \& Ti\v{s}er J.\ [91]
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\cmmnt{[\S466 {\it notes\/}.]}

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\cmmnt{[462B, 462C.]}

\medskip%Q\medskip%R

Radon J.\ [1913] `Theorie und Anwendungen der absolut additivien
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\cmmnt{[\S416 {\it notes\/}.]}

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\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
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\cmmnt{[419F.]}

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\cmmnt{[424C.]}

\indexheader{Rao M.}
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%used for \S477

\indexheader{Ressel}
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\cmmnt{[417C, 437M.]}

Riesz F.\ \& Sz.-Nagy B.\ [55] {\it Functional Analysis.}  Frederick Ungar,
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\cmmnt{[\S494 {\it notes\/}.]}

Rogers C.A.\ [70] {\it Hausdorff Measures.}   Cambridge, 1970.
\cmmnt{[\S471 {\it notes\/}.]}

Rogers C.A.\ [80] (ed.) {\it Analytic Sets.}   Academic, 1980.
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Rogers L.C.J.\ \& Williams D.\ [94] {\it Diffusions, Markov Processes
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\cmmnt{[\S455 {\it notes\/}.]}
%HA33.4.W5

Rokhlin V.A.\ [1948] `A ``general'' measure-preserving transformation is not
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\cmmnt{[494E.]}

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\cmmnt{[494Ye.]}

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\cmmnt{[439O.]}

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\cmmnt{[439O.]}

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\cmmnt{[\S4A4, \S4A6.]}

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\cmmnt{[451C.]}

\medskip%S

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\cmmnt{[451F.]}

Schachermayer W.\ [77] `Eberlein-compacts et espaces de Radon',
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\cmmnt{[467P.]}

Schaefer H.H.\ [71] {\it Topological Vector Spaces.}  Springer, 1971.
\cmmnt{[\S4A4.]}

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\indexheader{Sierpi\'nski}
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\indexheader{Solecki}
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\cmmnt{[444Ye.]}

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\cmmnt{[497M.]}

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\cmmnt{[434Ya, \S439 {\it notes\/}.]}

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\cmmnt{[441C.]}

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\cmmnt{[449Yi.]}

\indexheader{Szemer\'edi}
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\cmmnt{[497L.]}
\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

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\cmmnt{[467M.]}

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\cmmnt{[466I.]}

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\cmmnt{[466Za.]}

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\cmmnt{[465M.]}

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\cmmnt{[463Ya, 463Yd, 463Za, \S465 {\it intro.}, 465B, 465N,
465R-465T, %465R 465S 465T
466I, \S466 {\it notes\/}.]}

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\cmmnt{[\S465 {\it intro.}, 465L, 465M, {\it 465Yd},
\S465 {\it notes\/}.]}

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\cmmnt{[465U.]}

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\cmmnt{[\S465 {\it notes\/}.]}

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\cmmnt{[\S465 {\it notes\/}.]}
%465Xm

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\cmmnt{[459I, \S497.]}

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\cmmnt{[\S4A4.]}

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\cmmnt{[477L.]}

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\cmmnt{[413J, 416K.]}

Tops{\o}e F.\ [70b] {\it Topology and Measure.}  Springer, 1970 (Lecture
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\cmmnt{[437J.]}
%used in chap 41

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%({\tt http://arxiv.org/abs/0909.3869}).
\cmmnt{[425D, 425Ya.]}

\medskip%U

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\cmmnt{[4A5T.]}

\medskip%V

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\cmmnt{[457L.]}

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{\cyr Vinokurov V.G.} \& {\cyr Ma{kh}kamov B.M.\ [73] `O prostranstva{kh}
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Nau{ch}ye zapiski Ta{sh}ken{t}skogo instituta narodnogo
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\cmmnt{[451U.]}
% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

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\cmmnt{[\S434 {\it notes\/}.]}

Wagon S.\ [85] {\it The Banach-Tarski Paradox.}  Cambridge U.P., 1985.
\cmmnt{[449Yj.]}

Weil A.\ [1940] {\it L'int\'egration dans les groupes topologiques et ses
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\cmmnt{[445N.]}

\indexheader{Wheeler}
Wheeler R.F.\ [83] `A survey of Baire measures and strict topologies',
Expositiones Math.\ 1 (1983) 97-190.
  % printer's error "2" for "1" at head of paper refers to part, not
%volume
\cmmnt{[\S435 {\it notes\/}, \S437 {\it notes\/}.]}

%\medskip%X\medskip%Y\medskip%Z

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\frnewpage
}%end of references


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