% 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{mt49.tex}
\versiondate{31.8.09}
\copyrightdate{2009}

\def\chaptername{Further topics}
\newchapter{49}

I conclude the volume with notes on six almost unconnected special
topics.   In \S491 I look at equidistributed sequences
and the ideal $\Cal Z$ of sets with asymptotic density zero.
I give the principal theorems on the existence of equidistributed
sequences in abstract topological measure spaces,
and examine the way in which an equidistributed sequence can induce an
embedding of a measure algebra in the quotient algebra
$\Cal P\Bbb N/\Cal Z$.
The next three sections are linked.
In \S492 I present some forms of `concentration of measure'
which echo ideas from \S476 in combinatorial, rather than geometric,
contexts, with theorems of Talagrand and
Maurey on product measures and the Haar measure of a permutation group.
In \S493 I show how the ideas
of \S\S449, 476 and 492 can be put together in the theory of
`extremely amenable' topological groups.
Some of the important examples of extremely amenable groups are full groups
of measure-preserving automorphisms of measure algebras, previously
treated in \S383;   these are the
subject of \S494\dvAnew{2009}, where I look also at some striking algebraic
properties of these groups.
In \S495, I move on to Poisson point processes, with notes on
disintegrations and some special cases in which they can be represented
by Radon measures.
In \S496\dvAnew{2008}, I revisit the Maharam submeasures of Chapter 39,
showing that various ideas from the present
volume can be applied in this more general context.
In \S497\dvAnew{2009}, I give a version of Tao's proof of
Szemer\'edi's theorem on arithmetic progressions, based on a deep analysis
of relative independence, as introduced in \S458.
Finally, in \S498 I give a pair of simple, but
perhaps surprising, results on subsets of
sets of positive measure in product spaces.

\discrpage

%496 much less weighty than 497
\frfilename{mt491.tex}
\versiondate{1.5.24}
\copyrightdate{2002}

\def\chaptername{Further topics}
\def\sectionname{Equidistributed sequences}

\newsection{491}

In many of the most important topological probability spaces, starting
with Lebesgue measure\cmmnt{ (491Eb, 491Xg)}, there are sequences which are
equidistributed in the sense that, in
the limit, they spend the right proportion of their time in each part
of the space\cmmnt{ (491Yf)}.   I give the basic results on existence
of equidistributed sequences in
491E-491H,  %491E 491F 491G 491H
491Q and 491R.   Investigating such sequences, we are led to some
interesting properties of the asymptotic
density ideal $\Cal Z$ and the quotient algebra
$\frak Z=\Cal P\Bbb N/\Cal Z$ (491A, 491I-491K,  %491I 491J 491K
491P).   For `effectively regular' measures (491L-491M), equidistributed
sequences lead to embeddings of measure algebras in $\frak Z$ (491N).

\leader{491A}{The asymptotic density ideal (a)} If $I$ is a subset of
$\Bbb N$, its {\bf upper asymptotic density} is
$d^*(I)=\limsup_{n\to\infty}\bover1n(I\cap n)$, and its
{\bf asymptotic density} is
$d(I)=\lim_{n\to\infty}\bover1n\#(I\cap n)$ if this is defined.
\cmmnt{It is easy to check that $d^*$ is a submeasure on
$\Cal P\Bbb N$\cmmnt{ (definition: 392A)}, so that}

\Centerline{$\Cal Z=\{I:I\subseteq\Bbb N$,
$d^*(I)=0\}=\{I:I\subseteq\Bbb N$, $d(I)=0\}$}

\noindent is an ideal, the {\bf asymptotic density ideal}.

\spheader 491Ab Note that

\Centerline{$\Cal Z=\{I:I\Bsubseteq\Bbb N$,
$\lim_{n\to\infty}2^{-n}\#(I\cap 2^{n+1}\setminus 2^n)=0\}$.}

\prooflet{\noindent\Prf\ If $I\subseteq\Bbb N$ and $d^*(I)=0$, then

\Centerline{$2^{-n}\#(I\cap 2^{n+1}\setminus 2^n)
\le 2\cdot 2^{-n-1}\#(I\cap 2^{n+1})\to 0$}

\noindent as $n\to\infty$.   In the other direction, if
$\lim_{n\to\infty}2^{-n}\#(I\cap 2^{n+1}\setminus 2^n)=0$, then for
any $\epsilon>0$ there is an $m\in\Bbb N$ such that
$\#(I\cap 2^{k+1}\setminus 2^k)\le 2^k\epsilon$ for every $k\ge m$.
In this case, for $n\ge 2^m$, take $k_n$ such that
$2^{k_n}\le n<2^{k_n+1}$, and see that

\Centerline{$\Bover1n\#(I\cap n)
\le 2^{-k_n}(\#(I\cap 2^m)+\sum_{k=m}^{k_n}2^k\epsilon)
\le 2^{-k_n}\#(I\cap 2^m)+2\epsilon\to 2\epsilon$}

\noindent as $n\to\infty$, and $d^*(I)\le 2\epsilon$;  as $\epsilon$
is arbitrary, $I\in\Cal Z$.\ \Qed}

\spheader 491Ac Writing $\Cal D$ for the domain of $d$,

$$\eqalign{\Cal D
&=\{I:I\subseteq\Bbb N,\,\limsup_{n\to\infty}\Bover1n\#(I\cap n)
=\liminf_{n\to\infty}\Bover1n\#(I\cap n)\}\cr
&=\{I:I\subseteq\Bbb N,\,d^*(I)=1-d^*(\Bbb N\setminus I)\},\cr}$$

\Centerline{$\Bbb N\in\Cal D$,
\quad if $I$, $J\in\Cal D$ and $I\subseteq J$ then
$J\setminus I\in\Cal D$,}

\Centerline{if $I$, $J\in\Cal D$ and $I\cap J=\emptyset$ then
$I\cup J\in\Cal D$ and $d(I\cup J)=d(I)+d(J)$.}

\noindent It follows that if $\Cal I\subseteq\Cal D$ and
$I\cap J\in\Cal I$ for all $I$, $J\in\Cal I$, then the subalgebra of
$\Cal P\Bbb N$ generated by $\Cal I$ is included in
$\Cal D$\cmmnt{ (313Ga)}.   \cmmnt{But note that $\Cal D$ itself
is {\it not} a subalgebra of $\Cal P\Bbb N$ (491Xa).}

\spheader 491Ad \cmmnt{ The following elementary fact will be
useful.}
If $\sequencen{l_n}$ is a strictly increasing sequence in $\Bbb N$
such
that $\lim_{n\to\infty}l_{n+1}/l_n=1$,
and $I\subseteq\Bbb R$, then

\Centerline{$d^*(I)\le\limsup_{n\to\infty}
\Bover1{l_{n+1}-l_n}\#(I\cap l_{n+1}\setminus l_n)$.}

\prooflet{\noindent\Prf\   Set
$\gamma=\limsup_{n\to\infty}
\Bover1{l_{n+1}-l_n}\#(I\cap l_{n+1}\setminus l_n)$, and take
$\epsilon>0$.   Let $n_0$ be such that
$\#(I\cap l_{n+1}\setminus l_n)\le(\gamma+\epsilon)(l_{n+1}-l_n)$ and
$l_{n+1}-l_n\le\epsilon l_n$ for every $n\ge n_0$, and write $M$ for
$\#(I\cap l_{n_0})$.
If $m>l_{n_0}$, take $k$ such that $l_k\le m<l_{k+1}$;  then

$$\eqalign{\#(I\cap m)
&\le M+\sum_{n=n_0}^{k-1}\#(I\cap l_{n+1}\setminus l_n)+(m-l_k)\cr
&\le M+\sum_{n=n_0}^{k-1}(\gamma+\epsilon)(l_{n+1}-l_n)+l_{k+1}-l_k
\le M+m(\gamma+\epsilon)+\epsilon m,\cr}$$

\noindent so

\Centerline{$\Bover1m\#(I\cap m)\le\Bover{M}{m}+\gamma+2\epsilon$.}

\noindent  Accordingly $d^*(I)\le\gamma+2\epsilon$;  as $\epsilon$ is
arbitrary, $d^*(I)\le\gamma$.\ \Qed}%end of prooflet

\header{491Ae}{\bf *(e)}\cmmnt{ The following remark will not be used
directly in this section, but is one of the fundamental properties of
the ideal $\Cal Z$.}   If $\sequencen{I_n}$ is any sequence in $\Cal Z$,
there is an $I\in\Cal Z$ such that $I_n\setminus I$ is finite for
every
$n$.   \prooflet{\Prf\ Set $J_n=\bigcup_{j\le n}I_j$ for each $n$, so
that $\sequencen{J_n}$ is a non-decreasing sequence in $\Cal Z$.   Let
$\sequencen{l_n}$ be a strictly increasing
sequence in $\Bbb N$ such that, for each $n$,
$\#(J_n\cap k)\le 2^{-n}k$ for every $k\ge l_n$.
Set $I=\bigcup_{n\in\Bbb N}J_n\setminus l_n$.   Then
$I_n\setminus I\subseteq l_n$ is finite for each $n$.   Also, if
$n\in\Bbb N$ and $l_n\le k<l_{n+1}$,

\Centerline{$\#(I\cap k)\le\#(J_n\cap k)\le 2^{-n}k$,}

\noindent so $I\in\Cal Z$.\ \Qed}

\leader{491B}{Equidistributed sequences}
Let $X$ be a topological space and $\mu$ a probability measure on $X$.
I say that a sequence $\sequence{i}{x_i}$ in $X$
is {\bf (asymptotically) equidistributed}
if $d^*(\{i:x_i\in F\})\le\mu F$ for every measurable closed set
$F\subseteq X$;   equivalently, if
$\liminf_{n\to\infty}\bover1n\#(\{i:i<n$, $x_i\in G\})\ge\mu G$ for
every measurable open set $G\subseteq X$.

\cmmnt{\medskip

\noindent{\bf Remark} Equidistributed sequences are often called
{\bf uniformly distributed}.   Traditionally, such sequences have been
defined in terms of their action on continuous functions, as in 491Cf.
I have adopted the
definition here in order to deal both with Radon measures on spaces
which are not completely regular (so that we cannot identify the
measure with an integral) and with
Baire measures (so that there may be closed sets which are not
measurable).   Note that we cannot demand that the sets
$\{i:x_i\in F\}$ should have well-defined densities (491Yu).
}%end of comment

\leader{491C}{}\cmmnt{ I work through a list of basic facts.   The
technical
details (if we do not specialize immediately to metrizable or compact
spaces) are not quite transparent, so I set them out carefully.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological space, $\mu$ a
probability measure on $X$ and $\sequence{i}{x_i}$ a sequence in $X$.

(a) $\sequence{i}{x_i}$ is equidistributed iff
$\liminf_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)\ge\int fd\mu$
for every measurable bounded lower semi-continuous function
$f:X\to\Bbb
R$.

(b) If $\mu$ measures every zero set and $\sequence{i}{x_i}$ is
equidistributed, then
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$ for every
$f\in C_b(X)$.

(c) Suppose that $\mu$ measures every zero set in $X$.
If $\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$ for
every $f\in C_b(X)$, then $d^*(\{i:x_i\in F\})\le\mu F$ for every zero
set $F\subseteq X$.

(d) Suppose that $X$ is normal and that $\mu$ measures every zero set
and is inner regular with respect to the closed sets.
If $\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$ for
every $f\in C_b(X)$, then
$\sequence{i}{x_i}$ is equidistributed.

(e) Suppose that $\mu$ is $\tau$-additive and there is a base $\Cal G$
for the topology of $X$, consisting of measurable sets and closed
under finite unions, such that
$\liminf_{n\to\infty}\bover1{n+1}\#(\{i:i\le n$, $x_i\in G\})\ge\mu G$
for every $G\in\Cal G$.   Then $\sequence{i}{x_i}$ is equidistributed.

(f) Suppose that $X$ is completely regular and that $\mu$ measures
every zero set and is $\tau$-additive.
Then $\sequence{i}{x_i}$ is equidistributed iff the limit
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)$ is defined and equal to
$\int fd\mu$ for every $f\in C_b(X)$.

(g) Suppose that $X$ is metrizable and that $\mu$ is a topological
measure.   Then $\sequence{i}{x_i}$ is equidistributed iff the limit
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)$ is defined and equal to
$\int fd\mu$ for every $f\in C_b(X)$.

(h) Suppose that $X$ is compact, Hausdorff and zero-dimensional, and
that $\mu$ is a Radon measure on $X$.   Then $\sequence{i}{x_i}$ is
equidistributed iff
$d(\{i:x_i\in G\})=\mu G$ for every open-and-closed subset $G$ of $X$.

\proof{{\bf (a)(i)} Suppose that $\sequence{i}{x_i}$ is equidistributed.
Let $f:X\to[0,1]$ be a measurable lower semi-continuous function and
$k\ge 1$.   For each $j\le k$ set
$G_j=\{x:f(x)>\bover{j}{k}\}$.   Then

\Centerline{$\liminf_{n\to\infty}\Bover1{n+1}\sum_{j=1}^n\chi G_j(x_i)
=\liminf_{n\to\infty}\Bover1{n+1}\#(\{i:i\le n,\,x_i\in G_j\})
\ge\mu G_j$}

\noindent because $\sequence{i}{x_i}$ is equidistributed and $G_j$ is a
measurable open set.
Also $f-\bover1k\chi X\le\bover1k\sum_{j=1}^k\chi G_j\le f$, so

$$\eqalign{\int fd\mu-\Bover1k
&\le\Bover1k\sum_{j=1}^k\mu G_j
\le\Bover1k\sum_{j=1}^k
  \liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n\chi G_j(x_i)\cr
&\le\Bover1k
  \liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n\sum_{j=1}^k\chi G_j(x_i)
\le\liminf_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(x_i).\cr}$$

\noindent As $k$ is arbitrary,
$\liminf_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)\ge\int fd\mu$.

The argument above depended on $f$ taking values in $[0,1]$.   But
multiplying by an appropriate positive scalar we see that
$\liminf_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)\ge\int fd\mu$
for every bounded measurable lower semi-continuous
$f:X\to\coint{0,\infty}$, and adding a multiple of $\chi X$ we see
that
the same formula is valid for all bounded
measurable lower semi-continuous $f:X\to\Bbb R$.

\medskip

\quad{\bf (ii)}  Conversely, if
$\liminf_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)\ge\int fd\mu$
for every bounded measurable lower semi-continuous $f:X\to\Bbb R$, and
$G\subseteq X$ is a
measurable open set, then $\chi G$ is lower semi-continuous, so
$\liminf_{n\to\infty}\bover1{n+1}\#(\{i:i\le n,\,x_i\in G\})\ge\mu G$.
As $G$ is arbitrary, $\sequence{i}{x_i}$ is equidistributed.

\medskip

{\bf (b)} Apply (a) to the lower semi-continuous functions $f$ and $-f$.
(Recall that if $\mu$ measures every zero set, then every
bounded continuous real-valued function is integrable, by 4A3L.)

\medskip

{\bf (c)} Let $F\subseteq X$ be a zero set, and
$\epsilon>0$.   Then there is a continuous $f:X\to\Bbb R$ such that
$F=f^{-1}[\{0\}]$.   Let $\delta>0$ be such that
$\mu\{x:0<|f(x)|\le\delta\}\le\epsilon$, and set
$g=(\chi X-\Bover1\delta|f|)^+$.   Then $g:X\to[0,1]$ is continuous
and
$\chi F\le g$, so

$$\eqalign{d^*(\{i:x_i\in F\})
&\le\limsup_{n\to\infty}\Bover1{n+1}\sum_{i=0}^ng(x_i)\cr
&=\int g\,d\mu\le\mu\{x:|f(x)|\le\delta\}
\le\mu F+\epsilon.\cr}$$

\noindent As $\epsilon$ and $F$ are arbitrary, we have the result.

\medskip

{\bf (d)} Let $F\subseteq X$ be a measurable closed set and
$\epsilon>0$.   Because $\mu$ is inner regular with respect to the
closed sets, there is a measurable closed set
$F'\subseteq X\setminus F$ such that
$\mu F'\ge\mu(X\setminus F)-\epsilon$.   Because $X$ is normal, there
is
a continuous function $f:X\to[0,1]$ such that
$\chi F\le f\le\chi(X\setminus F')$.   Now

\Centerline{$d^*(\{i:x_i\in
F\})\le\limsup_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(x_i)=$$\biggerint
fd\mu\le\mu(X\setminus F')\le\mu F+\epsilon$.}

\noindent As $F$ and $\epsilon$ are arbitrary, $\sequence{i}{x_i}$ is
equidistributed.

\medskip

{\bf (e)} Let $G\subseteq X$ be a measurable open set, and $\epsilon>0$.
Then $\Cal H=\{H:H\in\Cal G$, $H\subseteq G\}$
is upwards-directed and has union
$G$;  since $\mu$ is $\tau$-additive, there is an $H\in\Cal H$ such that
$\mu H\ge\mu G-\epsilon$.   Now

$$\eqalign{\mu G
\le\epsilon+\mu H
&\le\epsilon+\liminf_{n\to\infty}\Bover1n\#(\{i:i<n,\,x_i\in H\})\cr
&\le\epsilon+\liminf_{n\to\infty}\Bover1n\#(\{i:i<n,\,x_i\in G\});\cr}$$

\noindent as $\epsilon$ and $G$ are arbitrary, $\sequence{i}{x_i}$ is
equidistributed.

\medskip

{\bf (f)}(i) If $\sequence{i}{x_i}$ is equidistributed then (b) tells us
that $\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$ for
every $f\in C_b(X)$.
(ii) Suppose that
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$ for every
$f\in C_b(X)$.
If $G\subseteq X$ is a cozero set, we can apply (c) to its complement
to see that
$\liminf_{n\to\infty}\bover1{n+1}\#(\{i:i\le n$, $x_i\in G\})\ge\mu G$.
So applying (e) with $\Cal G$ the family of cozero sets we see that
$\sequence{i}{x_i}$ is equidistributed.

\medskip

{\bf (g)} Because every closed set is a zero set, this follows at once
from (b) and (c).

\medskip

{\bf (h)} If $\sequence{i}{x_i}$ is equidistributed and $G\subseteq X$
is
open-and-closed, then $d^*(\{i:x_i\in G\})\le\mu G$ because $G$ is
closed and $d^*(\{i:x_i\notin G\})\le 1-\mu G$
because $G$ is open;  so $d(\{i:x_i\in G\})=\mu G$.   If the condition
is satisfied, then (e) tells us that $\sequence{i}{x_i}$ is
equidistributed.
}%end of proof of 491C

\leader{491D}{}\cmmnt{ The next lemma provides a useful general
criterion for the existence of equidistributed sequences.

\medskip

\noindent}{\bf Lemma} Let $X$ be a topological space and $\mu$ a
probability measure on $X$.   Suppose that there is a sequence
$\sequencen{\nu_n}$ of point-supported
probability measures on $X$ such that
$\limsup_{n\to\infty}\nu_nF\le\mu F$ for every measurable closed set
$F\subseteq X$.   Then $\mu$ has an equidistributed sequence.

\proof{ For each $n\in\Bbb N$, let $q_n:X\to[0,1]$ be such that
$\nu_nE=\sum_{x\in E}q_n(x)$ for every $E\subseteq X$.   Let
$q'_n:X\to[0,1]$ be such that $\sum_{x\in X}q'_n(x)=1$,
$K_n=\{x:q'_n(x)>0\}$ is finite, $q'_n(x)$ is rational for every $x$,
and $\sum_{x\in X}|q_n(x)-q'_n(x)|\le 2^{-n}$;  then
$\limsup_{n\to\infty}\nu'_nF\le\mu F$ for every measurable
closed $F$, where $\nu'_n$ is defined from $q'_n$.   For each $n$, let
$s_n\ge 1$ be such that $r_n(x)=q'_n(x)s_n$ is an integer for every
$x\in K_n$.
Let $\ofamily{i}{s_n}{x_{ni}}$ be a family in $K_n$ such that
$\#(\{i:i<s_n$, $x_{ni}=x\})=r_n(x)$ for each $x\in K_n$;  then
$\nu'_nE=\Bover1{s_n}\#(\{i:i<s_n$, $x_{ni}\in E\})$ for
every $E\subseteq X$.

Let $\sequence{k}{m_k}$ be such that
$s_{k+1}\le 2^{-k}\sum_{j=0}^km_js_j$ for each $k$.   Set $l_0=0$.
Given $l_n$, take
the largest $k$ such that $\sum_{j=0}^{k-1}m_js_j\le l_n$;  set
$l_{n+1}=l_n+s_k$ and $x_i=x_{k,i-l_n}$ for $l_n\le i<l_{n+1}$;
continue.
By the choice of the $m_k$, $l_{n+1}/l_n\to 1$ as $n\to\infty$.   For
any $E\subseteq X$,
$\#(\{i:l_n\le i<l_{n+1},\,x_i\in E\})=\#(\{j:j<s_k,\,x_{kj}\in E\})$
whenever $\sum_{j=0}^{k-1}m_js_j\le l_n<\sum_{j=0}^km_js_j$.   So for
any measurable closed set $F\subseteq X$,

$$\eqalignno{d^*(\{i:x_i\in F\})
&\le\limsup_{k\to\infty}\Bover1{s_k}\#(\{j:j<s_k,\,x_{kj}\in F\})\cr
\displaycause{491Ad}
&=\limsup_{k\to\infty}\nu'_kF
\le\mu F.\cr}$$

\noindent As $F$ is arbitrary, $\sequence{i}{x_i}$ is an equidistributed
sequence for $\mu$.
}%end of proof of 491D

\leader{491E}{Proposition} (a)(i) Suppose that $X$ and $Y$ are
topological spaces, $\mu$ is a probability measure on $X$ and
$f:X\to Y$ is a continuous function.   If $\sequence{i}{x_i}$
is a sequence in $X$ which is equidistributed with respect to $\mu$,
then $\sequence{i}{f(x_i)}$ is equidistributed with respect to the image
measure $\mu f^{-1}$.

\quad(ii) Suppose that $(X,\mu)$ and $(Y,\nu)$ are topological
probability spaces and $f:X\to Y$ is a continuous \imp\ function.   If
$\sequence{i}{x_i}$
is a sequence in $X$ which is equidistributed with respect to $\mu$,
then $\sequence{i}{f(x_i)}$ is equidistributed with respect to $\nu$.

(b) Let $X$ be a topological space and $\mu$ a probability measure on
$X$, and suppose that $X$ has a countable network consisting of sets
measured by $\mu$.   Let $\lambda$ be the
ordinary product measure on $X^{\Bbb N}$.   Then $\lambda$-almost
every sequence in $X^{\Bbb N}$ is $\mu$-equidistributed.

\proof{{\bf (a)(i)} Let $F\subseteq Y$ be a closed set which is measured
by $\mu f^{-1}$.   Then $f^{-1}[F]$ is a closed set in $X$ measured by
$\mu$.   So

\Centerline{$d^*(\{i:f(x_i)\in F\})=d^*(\{i:x_i\in f^{-1}[F]\})
\le\mu f^{-1}[F]$.}

\medskip

\quad{\bf (ii)} Replace `$\mu f^{-1}$' above by `$\nu$'.
\medskip

{\bf (b)} Let $\Cal A$ be a countable network for the given topology
$\frak S$ of $X$ consisting of measurable sets, and let $\Cal E$ be
the countable subalgebra of $\Cal PX$ generated
by $\Cal A$.   Let $\frak T\supseteq\frak S$ be the second-countable
topology generated by $\Cal E$;  then $\mu$ is a $\tau$-additive
topological measure with respect to $\frak T$ (4A2Nb, 414O),
and $\Cal E$ is a base for $\frak T$ closed under finite unions.
If $E\in\Cal E$, then $d(\{i:x_i\in E\})=\mu E$ for $\lambda$-almost
every sequence $\sequence{i}{x_i}$ in $X$, by the strong law of large
numbers (273J).   So

\Centerline{$d(\{i:x_i\in E\})=\mu E$ for every $E\in\Cal E$}

\noindent for $\lambda$-almost every $\sequence{i}{x_i}$.   Now 491Ce
tells us that any such sequence is equidistributed with respect to
$\frak T$ and therefore with respect to
$\frak S$.
}%end of proof of 491E

\leader{491F}{Theorem} Let
$\family{\alpha}{A}{(X_{\alpha},\frak T_{\alpha},\Sigma_{\alpha},\mu_{\alpha})}$
be a family of
$\tau$-additive topological probability spaces,
each of which has an equidistributed sequence.
If $\#(A)\le\frak c$, then the $\tau$-additive product measure $\lambda$
on $X=\prod_{\alpha\in A}X_{\alpha}$\cmmnt{ (definition:  417F)} has an
equidistributed sequence.

\proof{{\bf (a)} For the time being (down to the end of (f)), let us
suppose that $A=\Cal P\Bbb N$ and that every $\mu_{\alpha}$ is inner regular
with respect to the Borel
sets.   (This will simplify the formulae and make it possible to use the
theorems of \S417, in particular 417G and 417J.)

For each $\alpha\subseteq\Bbb N$, let $\sequence{i}{t_{\alpha i}}$ be an
equidistributed
sequence in $X_{\alpha}$;  for $n\in\Bbb N$, let $\nu_{\alpha n}$ be the
point-supported measure on $X_{\alpha}$ defined by setting
$\nu_{\alpha n}E=\bover1{n+1}\#(\{i:i\le n$, $t_{\alpha i}\in E\})$ for
$E\subseteq X_{\alpha}$.   For each finite set $I\subseteq\Cal P\Bbb N$,
set $Y_I=\prod_{\alpha\in I}X_{\alpha}$ and
$\pi_I(x)=x\restr I\in Y_I$ for $x\in X$.
Let $\lambda_I$ be the $\tau$-additive product of
$\family{\alpha}{I}{\mu_I}$ and, for each $n$, let
$\check\nu_{In}$ be the product of the measures
$\family{\alpha}{I}{\nu_{\alpha n}}$.
(Because $I$ is finite, this is a point-supported
probability measure, as in 251Xu.
I do not say `$\tau$-additive product' here
because I do not wish to assume that all singleton sets are Borel, so
the $\nu_{\alpha n}$ may not be inner regular with respect to the Borel sets.)
%}%end of allowmorestretch

\medskip

{\bf (b)} Suppose that $I\subseteq\Cal P\Bbb N$ is finite and that
$W\subseteq Y_I$ is an open set.
Then $\lambda_IW\le\liminf_{n\to\infty}\check\nu_{In}W$.
\Prf\ Induce on $\#(I)$.   If $I=\emptyset$, $Y_I$
is a singleton and the result is trivial.   For the inductive step, if
$I\ne\emptyset$, take any $\alpha\in I$ and set
$I'=I\setminus\{\alpha\}$.   Then we can identify
$Y_I$ with $Y_{I'}\times X_{\alpha}$, $\lambda_I$ with the
$\tau$-additive product of $\lambda_{I'}$ and $\mu_{\alpha}$ (417J), and
each $\check\nu_{In}$ with the product of $\check\nu_{I'n}$
and $\nu_{\alpha n}$.

Let $\Cal V$ be the family of those subsets $V$ of $Y_I$ which
are expressible as a finite union of sets of the form $U\times H$
where $U\subseteq Y_{I'}$ and $H\subseteq X_{\alpha}$
are open.   Then $\Cal V$ is a base for the topology of $Y_I$
closed under finite unions.   Let $\epsilon>0$.   Because
$\lambda_I$ is $\tau$-additive, there is a $V\in\Cal V$ such
that $\lambda_IV\ge\lambda W-\epsilon$.
The function $t\mapsto\lambda_{I'}V[\{t\}]:X_{\alpha}\to[0,1]$ is lower
semi-continuous (417Ba), so 491Ca tells us that

$$\eqalign{\lambda_IV
&=\int\lambda_{I'}V[\{t\}]\mu_{\alpha}(dt)\cr
&\le\liminf_{n\to\infty}\Bover1{n+1}
  \sum_{i=0}^n\lambda_{I'}V[\{t_{\alpha i}\}]
=\liminf_{n\to\infty}\int\lambda_{I'}V[\{t\}]\nu_{\alpha n}(dt).\cr}$$

\noindent At the same time, there are only finitely many sets of the
form $V[\{t\}]$, and for each of these we have
$\lambda_{I'}V[\{t\}]
\le\liminf_{n\to\infty}\check\nu_{I'n}V[\{t\}]$,
by the inductive hypothesis.   So there is an $m\in\Bbb N$ such that
$\lambda_{I'}V[\{t\}]\le\check\nu_{I'n}V[\{t\}]+\epsilon$
for every $n\ge m$ and every $t\in X_{\alpha}$.   We must therefore have

$$\eqalign{\lambda_IW
&\le\lambda_IV+\epsilon
\le\liminf_{n\to\infty}
   \int\lambda_{\Cal I'}V[\{t\}]\nu_{\alpha n}(dt)+\epsilon\cr
&\le\liminf_{n\to\infty}
   \int\check\nu_{\Cal I'n}V[\{t\}]\nu_{\alpha n}(dt)+2\epsilon\cr
&=\liminf_{n\to\infty}\check\nu_{In}V+2\epsilon
\le\liminf_{n\to\infty}\check\nu_{In}W+2\epsilon.\cr}$$

\noindent As $\epsilon$ and $W$ are arbitrary, the induction
proceeds.\ \Qed

\medskip

{\bf (c)} For $K\subseteq n\in\Bbb N$, set
$A_{nK}=\{\alpha:\alpha\subseteq\Bbb N$, $\alpha\cap n=K\}$
and $Z_{nK}=\prod_{\alpha\in A_{nK}}X_{\alpha}$.
Then for each $n\in\Bbb N$ we can identify $X$ with
the finite product $\prod_{K\subseteq n}Z_{nK}$.
For $K\subseteq n\in\Bbb N$ and $i\in\Bbb N$, define $z_{nKi}\in Z_{nK}$
by setting $z_{nKi}(\alpha)=t_{\alpha i}$ for $\alpha\in A_{mK}$;  let
$\tilde\nu_{nK}$ be the point-supported measure
on $Z_{nK}$ defined by setting
$\tilde\nu_{nK}W=\bover1{n+1}\#(\{i:i\le n$, $z_{nKi}\in W\})$ for
each $W\subseteq Z_{nK}$.
For $n\in\Bbb N$ let $\tilde\nu_n$ be the measure on $X$ which is the
product of the measures $\tilde\nu_{nK}$ for $K\subseteq n$;  this
too is point-supported.

\medskip

{\bf (d)} If $I\subseteq\Cal P\Bbb N$ is finite, there is an $m\in\Bbb N$
such that $\pi_I:X\to Y_I$ is $(\tilde\nu_n,\check\nu_{In})$-\imp\ for every
$n\ge m$.
\Prf\ Let $m$ be such that $\alpha\cap m\ne\alpha'\cap m$ for all distinct
$\alpha$,
$\alpha'\in I$.   If $n\ge m$, then $\tilde\nu_n$ is the product of the
$\tilde\nu_{nK}$
for $K\subseteq n$.   Now $\pi_I$, interpreted as a function
from $\prod_{K\subseteq n}Z_{nK}$ onto $Y_I$, is of the form
$\pi_I(\langle z_K\rangle_{K\subseteq n})
=\family{\alpha}{I}{z_{\alpha\cap n}(\alpha)}$.   If $\alpha\in I$ and
$E\subseteq X_{\alpha}$, then

\Centerline{$\{z:z\in\prod_{K\subseteq n}Z_{nK}$, $\pi_I(z)(\alpha)\in E\}
=\{z:z\in\prod_{K\subseteq n}Z_{nK}$, $z_{\alpha\cap n}(\alpha)\in E\}$,}

\noindent so

$$\eqalign{\tilde\nu_n\{z:\pi_I(z)(\alpha)\in E\}&
=\tilde\nu_{n,\alpha\cap n,n}\{y:y\in Z_{n,\alpha\cap n},\,y(\alpha)\in E\}\cr&
=\Bover1{n+1}\#(\{i:i\le n,\,z_{n,\alpha\cap n,i}\in E\})\cr&
=\Bover1{n+1}\#(\{i:i\le n,\,t_{\alpha i}\in E\})
=\nu_{\alpha n}E.\cr}$$

\noindent If $W\subseteq Y_I$ is of the form
$\{y:y(\alpha)\in E_{\alpha}$ for every $\alpha\in I\}$, where
$E_{\alpha}\subseteq X_{\alpha}$ for each $\alpha\in I$, then

$$\eqalignno{\tilde\nu_n\pi_I^{-1}[W]&
=\tilde\nu_n(\bigcap_{\alpha\in I}\{z:z\in\prod_{K\subseteq n}Z_{nK},\,
   z(\alpha\cap n)(\alpha)\in E_{\alpha}\})\cr&
=\prod_{\alpha\in I}\tilde\nu_n\{z:z(\alpha\cap n)(\alpha)\in E_{\alpha}\}
\Displaycause{because $\{z:z(\alpha\cap n)(\alpha)\in E_{\alpha}\}$ is
determined by coordinates in $\{\alpha\cap n\}$ for each $\alpha\in I$, and
$\alpha\mapsto\alpha\cap n:I\to\Cal Pn$ is injective}
=\prod_{\alpha\in I}\nu_{\alpha n}E_{\alpha}
=\check\nu_{In}W.\cr}$$

\noindent In particular, $\tilde\nu_n\pi_I^{-1}[\{y\}]=\check\nu_{In}\{y\}$ for
every $y\in Y_I$.   Consequently $\tilde\nu_n\pi_I^{-1}[D]=\check\nu_{In}D$
for any countable set $D\subseteq Y_I$.
But if $W$ is any subset of $Y_I$, there are countable
subsets $D$, $D'$ of $W$ and $Y_I$ respectively such that

\Centerline{$\check\nu_{In}D=\check\nu_nW$,
\quad$\check\nu_{In}D'=\check\nu_{In}(Y_I\setminus W)$,
\quad$\check\nu_{In}D+\check\nu_{In}D'=1$}

\noindent because $\check\nu_n$ is point-supported, and now

\Centerline{$\pi_I^{-1}[D]\subseteq\pi_I^{-1}[W]$,
\quad$\pi_I^{-1}[D']\subseteq X\setminus\pi_I^{-1}[W]$,
\quad$\tilde\nu_n\pi_I^{-1}[D]+\tilde\nu_n\pi_I^{-1}[D']=1$,}

\noindent so $\tilde\nu_n\pi_I^{-1}[W]$ is defined and equal to
$\tilde\nu_n\pi_I^{-1}[D]=\check\nu_{In}W$
because $\tilde\nu_n$ is complete.   So $\pi_I$ is
$(\tilde\nu_n\pi_I^{-1},\check\nu_{In})$-\imp.\ \Qed

\medskip

{\bf (e)} Let $\Cal W$ be the family of those open sets
$W\subseteq X$ expressible in the form $\pi_I^{-1}[W']$ for
some finite $I\subseteq\Cal P\Bbb N$ and
some open $W'\subseteq Y_I$.   If $W\in\Cal W$, then
$\lambda W\le\liminf_{n\in\Bbb N}\tilde\nu_nW$.   \Prf\ Take
$I\in[\Cal P\Bbb N]^{<\omega}$ and an open $W'\subseteq Y_I$
such that $W=\pi_I^{-1}[W']$.   Then

$$\eqalignno{\lambda W
&=\lambda_IW'\cr
\displaycause{417K}
&\le\liminf_{n\to\infty}\check\nu_{In}W'\cr
\displaycause{by (b) above}
&=\liminf_{n\to\infty}\tilde\nu_n\pi_I^{-1}[W']\cr
\displaycause{by (d)}
&=\liminf_{n\to\infty}\tilde\nu_nW. \text{ \Qed}\cr}$$

\medskip

{\bf (f)} If now $F\subseteq X$ is any closed set and $\epsilon>0$, then
(because $\Cal W$ is a base for the topology of $X$ closed under finite
unions) there is a $W\in\Cal W$
such that $W\subseteq X\setminus F$ and
$\lambda W\ge 1-\lambda F-\epsilon$.   In this case

\Centerline{$\limsup_{n\to\infty}\tilde\nu_nF
\le 1-\liminf_{n\to\infty}\tilde\nu_nW
\le 1-\lambda W
\le\lambda F+\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\limsup_{n\to\infty}\tilde\nu_nF\le\lambda F$;  as $F$ is arbitrary,
491D tells us that there is an equidistributed sequence in
$X$.

\medskip

{\bf (g)} All this was done while assuming that $A=\Cal P\Bbb N$ and every
$\mu_{\alpha}$ is inner
regular with respect to the Borel sets.   For the superficially more
general case enunciated, given only
that $\#(A)\le\frak c$ and each $\mu_{\alpha}$ is a
$\tau$-additive topological measure with an
equidistributed sequence, we can of course take it that $A$ is a subset of
$\Cal P\Bbb N$.   Now let $\mu'_{\alpha}$ be the restriction of $\mu_{\alpha}$
to the Borel $\sigma$-algebra of $X_{\alpha}$ for each $\alpha\in A$, and for
$\alpha\in\Cal P\Bbb N\setminus A$ take $X_{\alpha}$ to be a singleton set,
$\frak T_{\alpha}$ its only topology and $\mu'_{\alpha}$ the only probability
measure on $X_{\alpha}$.   Every $\mu'_{\alpha}$ is now
$\tau$-additive, and for $\alpha\in A$ any
equidistributed sequence for $\mu_{\alpha}$ is of
course equidistributed for $\mu'_{\alpha}$, while for
$\alpha\in\Cal P\Bbb N\setminus A$ the only sequence in $X_{\alpha}$ is
equidistributed for $\mu'_{\alpha}$.   If we take $\lambda'$ to be the
$\tau$-additive product of
$\langle\mu'_{\alpha}\rangle_{\alpha\subseteq\Bbb N}$
on $X'=\prod_{\alpha\subseteq\Bbb N}X'_{\alpha}$, then (a)-(f) show that
$\lambda'$ has an equidistributed sequence $\sequence{i}{x_i}$ say.

Let $\pi_A:X'\to X$ be the restriction map $x\mapsto x\restr A$.   This is
continuous, so $\sequence{i}{\pi_A(x_i)}$ is equidistributed with respect to
$\lambda'\pi_A^{-1}$, by 491Ea.   And $\lambda'\pi_A^{-1}$ agrees with $\lambda$
on the open subsets of $X$.   \Prf\ If $I\subseteq A$ is finite and
$H_{\alpha}\subseteq X_{\alpha}$ is open for $\alpha\in I$, then

$$\eqalign{(\lambda'\pi_A^{-1})
  \{x:x\in X,\,&x(\alpha)\in H_{\alpha}\text{ for }\alpha\in I\}\cr&
=\lambda'\{x:x\in X,\,x(\alpha)\in H_{\alpha}\text{ for }\alpha\in I\}\cr&
=\prod_{\alpha\in I}\mu_{\alpha}H_{\alpha}
=\lambda\{x:x\in X,\,x(\alpha)\in H_{\alpha}\text{ for }\alpha\in I\}.\cr}$$

\noindent So $\lambda'\pi_A^{-1}$ and $\lambda$ agree on the family $\Cal V_0$
of open cylinder subsets of $X$.   But $\Cal V_0$ is
closed under finite intersections, so the probability measures
$\lambda'\pi_A^{-1}$ and $\lambda$ agree on the $\sigma$-algebra of subsets of
$X$ generated by $\Cal V_0$, by the Monotone Class Theorem (136C).
In particular, they agree on the family $\Cal V_1$ of sets expressible as finite
unions of members of $\Cal V_0$, which is a base for the topology of $X$ closed
under finite unions.   If $W\subseteq X$ is open, then
$\{V:V\in\Cal V_1$, $V\subseteq W\}$ is upwards-directed and has union $W$, so

$$\eqalignno{(\lambda'\pi_A^{-1})W&
=\sup_{V\in\Cal V_1,V\subseteq W}(\lambda'\pi_A^{-1})(V)
\Displaycause{because $\lambda'\pi_A^{-1}$ is $\tau$-additive, by 411Gj}
=\sup_{V\in\Cal V_1,V\subseteq W}\lambda V
=\lambda W.  \text{ \Qed}\cr}$$

At the same time, $\sequence{i}{\pi_A(x_i)}$ is equidistributed for
$\lambda'\pi_A^{-1}$, by 491E(a-i).   Directly from the definition in 491B,
we see that $\sequence{i}{\pi_A(x_i)}$ is also equidistributed for
$\lambda$, and $\lambda$ has an equidistributed sequence in this case also.
}%end of proof of 491F

\leader{491G}{Corollary} The usual measure of $\{0,1\}^{\frakc}$ has an
equidistributed sequence.
%491P

\proof{ The usual measure of $\{0,1\}$ of course has an equidistributed
sequence (just set $x_i=0$ for even $i$, $x_i=1$ for odd $i$), so
491F gives the result at once.
}%end of proof of 491G

\leader{491H}{Theorem}\cmmnt{ ({\smc Veech 71})} Any separable compact
Hausdorff topological group has an equidistributed sequence for its Haar
probability measure.

\proof{ Let $X$ be a separable compact Hausdorff topological group.
Recall that $X$ has exactly one Haar probability measure $\mu$,
which is both a left Haar measure and a right Haar measure (442Ic).

\medskip

{\bf (a)} We need some elementary facts about convolutions.

\medskip

\quad{\bf (i)} If $\nu_1$ and $\nu_2$ are point-supported probability
measures on $X$, then $\nu_1*\nu_2$ is point-supported.   \Prf\ If
$\nu_1E=\sum_{x\in E}q_1(x)$
and $\nu_2E=\sum_{x\in E}q_2(x)$ for every $E\subseteq X$, then

$$\eqalignno{(\nu_1*\nu_2)(E)&=(\nu_1\times\nu_2)\{(x,y):xy\in E\}
\Displaycause{444A}
=\sum_{xy\in E}q_1(x)q_2(y)
=\sum_{z\in E}q(z)\cr}$$

\noindent where $q(z)=\sum_{x\in X}q_1(x)q_2(x^{-1}z)$ for $z\in X$.\
\Qed

\medskip

\quad{\bf (ii)} Let $\nu$, $\lambda$ be Radon probability measures on
$X$.   Suppose that $f\in C(X)$, $\alpha\in\Bbb R$ and $\epsilon>0$
are such that
$|\int f(yxz)\nu(dx)-\alpha|\le\epsilon$ for every $y$, $z\in X$.
Then $|\int f(yxz)(\lambda*\nu)(dx)-\alpha|\le\epsilon$ and
$|\int f(yxz)(\lambda*\nu)(dx)-\alpha|\le\epsilon$ for every $y$, $z\in X$.
\Prf\

$$\eqalignno{|\int f(yxz)(\lambda*\nu)&(dx)-\alpha|\cr
&=|\iint f(ywxz)\nu(dx)\lambda(dw)-\alpha|\cr
\displaycause{444C}
&\le\int|\int f(ywxz)\nu(dx)-\alpha|\lambda(dw)
\le\int\epsilon\lambda(dw)
=\epsilon,\cr
|\int f(yxz)(\nu*\lambda)&(dx)-\alpha|\cr
&=|\iint f(yxwz)\nu(dx)\lambda(dw)-\alpha|\cr
&\le\int|\int f(yxwz)\nu(dx)-\alpha|\lambda(dw)
\le\int\epsilon\lambda(dw)
=\epsilon.  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} Let $A\subseteq X$ be a countable dense set.   Let $\Nu$ be
the set of point-supported probability measures $\nu$ on $X$ which are
defined by functions $q$ such that
$\{x:q(x)>0\}$ is a finite subset of $A$ and $q(x)$ is rational for
every $x$.   Then $\Nu$ is countable.   Now, for every $f\in C(X)$ and
$\epsilon>0$, there is a
$\nu\in\Nu$ such that $|\int f(yxz)\nu(dx)-\int fd\mu|\le\epsilon$ for
all $y$, $z\in X$.   \Prf\ Because $X$ is compact, $f$ is uniformly
continuous for the right uniformity of
$X$ (4A2Jf), so there is a neighbourhood $U$ of the identity $e$ such
that $|f(x')-f(x)|\le\bover12\epsilon$ whenever $x'x^{-1}\in U$.
Next, again because $X$ is compact, there
is a neighbourhood $V$ of $e$ such that $yxy^{-1}\in U$ whenever
$x\in V$ and $y\in X$ (4A5Ej).   Because $A$ is dense,
$V^{-1}x\cap A\ne\emptyset$ for every $x\in X$, that is,
$VA=X$;  once more because $X$ is compact, there are
$x_0,\ldots,x_n\in A$ such that $X=\bigcup_{i\le n}Vx_i$.   Set
$E_i=Vx_i\setminus\bigcup_{j<i}Vx_j$ for each $i\le n$.
Let $\alpha_0,\ldots,\alpha_n\in[0,1]\cap\Bbb Q$ be such that
$\sum_{i=0}^n\alpha_i=1$ and
$\|f\|_{\infty}\sum_{i=0}^n|\alpha_i-\mu E_i|\le\bover12\epsilon$,
and define $\nu\in\Nu$ by
setting $\nu E=\sum\{\alpha_i:i\le n$, $x_i\in E\}$ for every
$E\subseteq X$.

Let $y$, $z\in X$.   If $i\le n$ and $x\in E_i$, then $xx_i^{-1}\in V$
so $(yxz)(yx_iz)^{-1}=yxx_i^{-1}y^{-1}\in U$ and
$|f(yxz)-f(yx_iz)|\le\bover12\epsilon$.   Accordingly

$$\eqalignno{|\int f(yxz)\nu(dx)&-\int f(x)\mu(dx)|\cr&
=|\sum_{i=0}^n\alpha_i f(yx_iz)-\int f(yx)\mu(dx)|
\Displaycause{441Ac}
=|\sum_{i=0}^n\alpha_i f(yx_iz)-\Delta(z)\int f(yxz)\mu(dx)|
\Displaycause{where $\Delta$ is the left modular function of $X$, by 442Kc}
=|\sum_{i=0}^n\alpha_i f(yx_iz)-\Delta(z)\int f(yxz)\mu(dx)|
\Displaycause{because $X$ is unimodular, by 442Ic}
\le\sum_{i=0}^n|\alpha_i f(yx_iz)-\int_{E_i}f(yxz)\mu(dx)|\cr&
\le\sum_{i=0}^n|\alpha_i-\mu E_i||f(yx_iz)|
  +\sum_{i=0}^n|f(yx_iz)\mu E_i-\int_{E_i}f(yxz)\mu(dx)|\cr&
\le\|f\|_{\infty}\sum_{i=0}^n|\alpha_i-\mu E_i|
  +\sum_{i=0}^n\int_{E_i}|f(yxz)-f(yx_iz)|\mu(dx)\cr&
\le\Bover{\epsilon}2+\sum_{i=0}^n\Bover{\epsilon}2\mu E_i
=\epsilon.  \text{ \Qed}\cr}$$

\medskip

{\bf (c)} Let $\sequencen{\nu_n}$ be a sequence running over $\Nu$,
and set $\lambda_n=\nu_0*\nu_1*\ldots*\nu_n$ for each $n$.   (Recall from
444B that convolution is associative.)   Then each
$\lambda_n$ is a point-supported probability
measure on $X$, by (a-i).   Also
$\lim_{n\to\infty}\int fd\lambda_n=\int fd\mu$ for every $f\in C(X)$.
\Prf\ If $f\in C(X)$ and $\epsilon>0$, then (b) tells us that
there is an $m\in\Bbb N$ such that
$|\int f(yxz)\nu_m(dx)-\int fd\mu|\le\epsilon$ for all $y$, $z\in X$.
For any $n\ge m$, $\lambda_n$ is of the form $\lambda'*\nu_m*\lambda''$.
By (a-ii), used in both parts successively,
$|\int fd\lambda_n-\int fd\mu|\le\epsilon$.
As $\epsilon$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (d)} If $F\subseteq X$ is closed, then

$$\eqalign{\mu F
&=\inf\{\int fd\mu:\chi F\le f\in C(X)\}\cr
&=\inf_{\chi F\le f}\limsup_{n\to\infty}\int fd\lambda_n
\ge\limsup_{n\to\infty}\lambda_nF.\cr}$$

\noindent By 491D, $\mu$ has an equidistributed sequence.
}%end of proof of 491H

\leader{491I}{The quotient $\Cal P\Bbb N/\Cal Z$}\cmmnt{ I now return
to the asymptotic density ideal
$\Cal Z$, moving towards a striking relationship between the
corresponding quotient algebra and equidistributed
sequences.}    Since
$\Cal Z\normalsubgroup\Cal P\Bbb N$, we can form the quotient
{\bf asymptotic density algebra}
$\frak Z=\Cal P\Bbb N/\Cal Z$.   The functional $d^*$ descends
naturally to $\frak Z$ if we set

\Centerline{$\bar d^*(I^{\ssbullet})=d^*(I)$ for every $I\subseteq\Bbb N$.}

\spheader 491Ia $\bar d^*$ is a strictly positive submeasure on
$\frak Z$.   \prooflet{\Prf\ $\bar d^*$ is a submeasure on $\frak Z$
because $d^*$ is a
submeasure on $\Cal P\Bbb N$.   $\bar d^*$ is strictly positive
because $\Cal Z\supseteq\{I:d^*(I)=0\}$.\ \Qed}

\spheader 491Ib Let $\bar\rho$ be the metric on $\frak Z$ defined by
saying that $\bar\rho(a,b)=\bar d^*(a\Bsymmdiff b)$ for
all $a$, $b\in\frak Z$.   Under $\bar\rho$, the Boolean operations
$\Bcup$, $\Bcap$, $\Bsymmdiff$ and $\Bsetminus$ and the function
$\bar d^*:\frak Z\to[0,1]$
are uniformly continuous\cmmnt{ (392Hb)}, and $\frak Z$ is complete.
\prooflet{\Prf\ Let $\sequencen{c_n}$ be a sequence in $\frak Z$ such
that $\bar\rho(c_{n+1},c_n)\le 2^{-n}$ for every $n\in\Bbb N$;
then $\bar\rho(c_r,c_i)\le 2^{-i+1}$ for $i\le r$.
For each $n\in\Bbb N$ choose $C_n\subseteq\Bbb N$ such that
$C_n^{\ssbullet}=c_n$;  then $d^*(C_r\symmdiff C_i)\le 2^{-i+1}$ for
$i\le r$.   Choose a strictly increasing sequence
$\sequencen{k_n}$ in $\Bbb N$ such that $k_{n+1}\ge 2k_n$ for every
$n$ and, for each $n\in\Bbb N$,

\Centerline{$\Bover1m\#((C_n\symmdiff C_i)\cap m)\le 2^{-i+2}$
whenever $i\le n$, $m\ge k_n$.}

Set $C=\bigcup_{n\in\Bbb N}C_n\cap k_{n+1}\setminus k_n$, and
$c=C^{\ssbullet}\in\frak Z$.   If $n\in\Bbb N$ and $m\ge k_{n+1}$,
then take $r>n$ such that $k_r\le m<k_{r+1}$;  in this case
$k_i\le 2^{i-r}m$ for $i\le r$, so

$$\eqalignno{\#((C\symmdiff C_n)\cap m)
&\le k_n+\sum_{i=n}^{r-1}\#((C\symmdiff C_n)\cap k_{i+1}\setminus k_i)
    +\#((C\symmdiff C_n)\cap m\setminus k_r)\cr
&=k_n+\sum_{i=n}^{r-1}\#((C_i\symmdiff C_n)\cap k_{i+1}\setminus k_i)
    +\#((C_r\symmdiff C_n)\cap m\setminus k_r)\cr
&\le k_n+\sum_{i=n+1}^{r-1}\#((C_i\symmdiff C_n)\cap k_{i+1})
    +\#((C_r\symmdiff C_n)\cap m)\cr
&\le k_n+\sum_{i=n+1}^{r-1}2^{-n+2}k_{i+1}
    +2^{-n+2}m\cr
&\le k_n+\sum_{i=n+1}^{r-1}2^{-n+2}2^{i+1-r}m
    +2^{-n+2}m\cr
&\le k_n+2^{-n+3}m+2^{-n+2}m.\cr}$$

\noindent But this means that

$$\eqalign{\bar\rho(c,c_n)
=d^*(C\symmdiff C_n)
\le\lim_{m\to\infty}\Bover{k_n}m+2^{-n+3}+2^{-n+2}
\le 2^{-n+4}\cr}$$

\noindent for every $n$, and $\sequencen{c_n}$ converges to $c$ in
$\frak Z$.\ \Qed}%end of prooflet

\cmmnt{For the rest of this section, I will take it that $\frak Z$ is endowed
with the metric $\bar\rho$.}

%\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\header{491Ic}{\bf *(c)} If $\sequencen{a_n}$ is a non-increasing
sequence in $\frak Z$, there is an $a\in\frak Z$ such that
$a\Bsubseteq a_n$ for every $n$ and
$\bar d^*(a)=\inf_{n\in\Bbb N}\bar d^*(a_n)$.   \prooflet{\Prf\ For
each $n\in\Bbb N$, choose $I_n\subseteq\Bbb N$ such that
$I_n^{\ssbullet}=a_n$;  replacing $I_n$ by
$\bigcap_{j\le n}I_j$ if necessary, we can arrange that
$I_{n+1}\subseteq I_n$ for every $n$.   Set
$\gamma=\inf_{n\in\Bbb N}\bar d^*(a_n)=\inf_{n\in\Bbb N}d^*(I_n)$.
Let $\sequencen{k_n}$ be a strictly increasing sequence in $\Bbb N$
such that
$\#(I_n\cap k_n)\ge(\gamma-2^{-n})k_n$ for every $n$.   Set
$I=\bigcup_{n\in\Bbb N}I_n\cap k_n$ and $a=I^{\ssbullet}\in\frak Z$.   Then
$\#(I\cap k_n)\ge(\gamma-2^{-n})k_n$ for every $n$, so
$\bar d^*(a)=d^*(I)\ge\gamma$.
Also $I\setminus I_n\subseteq k_n$ is finite, so $a\Bsubseteq a_n$,
for every $n$.   Of course it follows
at once that $\bar d^*(a)=\gamma$ exactly, as required.\ \Qed}

\header{491Id}{\bf *(d)} $\bar d^*$ is a Maharam submeasure on
$\frak Z$.   \prooflet{(Immediate from (c).)}

\leader{491J}{Lemma} Let $\sequencen{a_n}$ be a non-decreasing
sequence in $\frak Z=\Cal P\Bbb N/\Cal Z$ such that
$\lim_{n\to\infty}\bar d^*(a_n)+\bar d^*(1\Bsetminus a_n)=1$.
Then $\sequencen{a_n}$ is topologically convergent to a member $a$ of $\frak Z$;
$a=\sup_{n\in\Bbb N}a_n$ in $\frak Z$ and
$d^*(a)+d^*(1\Bsetminus a)=1$.

\proof{{\bf (a)} The point is that if $m\le n$ then
$\bar d^*(a_n\Bsetminus a_m)\le\bar d^*(a_n)+\bar d^*(1\Bsetminus a_m)-1$.
\Prf\ Let $I$, $J\Bsubseteq\Bbb N$ be such that $I^{\ssbullet}=a_m$
and $J^{\ssbullet}=a_n$.   For any $k\ge 1$,

\Centerline{$\Bover1k\#(k\cap J)+\Bover1k\#(k\setminus I)
=\Bover1k\#(k\cap J\setminus I)+\Bover1k\#(k\setminus(I\setminus J))$,}

\noindent so

$$\eqalign{d^*(J)+d^*(\Bbb N\setminus I)
&=\limsup_{k\to\infty}\Bover1k\#(k\cap J)
   +\limsup_{k\to\infty}\Bover1k\#(k\setminus I)\cr
&\ge\limsup_{k\to\infty}
   \bigl(\Bover1k\#(k\cap J)+\Bover1k\#(k\setminus I)\bigr)\cr
&=\limsup_{k\to\infty}\bigl(\Bover1k\#(k\cap J\setminus I)
   +\Bover1k\#(k\setminus(I\setminus J))\bigr)\cr
&\ge\limsup_{k\to\infty}\Bover1k\#(k\cap J\setminus I)
   +\liminf_{k\to\infty}\Bover1k\#(k\setminus(I\setminus J))
=d^*(J\setminus I)+1\cr}$$

\noindent because $a_m\Bsubseteq a_n$, so $I\setminus J\in\Cal Z$.
But this means that

\Centerline{$\bar d^*(a_n\Bsetminus a_m)
=d^*(J\setminus I)
\le d^*(J)+d^*(\Bbb N\setminus I)-1
=\bar d^*(a_n)+\bar d^*(1\Bsetminus a_m)-1$.  \Qed}

\medskip

{\bf (b)} Accordingly

$$\eqalignno{\limsup_{m\to\infty}\sup_{n\ge m}\bar\rho(a_m,a_n)
&=\limsup_{m\to\infty}\sup_{n\ge m}\bar d^*(a_n\Bsetminus a_m)\cr
&\le\limsup_{m\to\infty}
   (\sup_{n\ge m}\bar d^*(a_n)+\bar d^*(1\Bsetminus a_m)-1)\cr
&=\limsup_{m\to\infty}\sup_{n\ge m}\bar d^*(a_n)-\bar d^*(a_m)\cr
\displaycause{because
$\lim_{m\to\infty}\bar d^*(a_m)+\bar d^*(1\Bsetminus a_m)=1$}
&=0,\cr}$$

\noindent and $\sequencen{a_n}$ is a Cauchy sequence in $\frak Z$.

\medskip

{\bf (c)} Because $\frak Z$ is complete, $a=\lim_{n\to\infty}a_n$ is
defined in $\frak Z$ (491Ib).   For each $m\in\Bbb N$,
$a_m\Bsetminus a=\lim_{n\to\infty}a_m\Bsetminus a_n=0$
(because $\Bsetminus$ is continuous), so $a_m\subseteq a$;  thus $a$
is an upper bound of $\{a_n:n\in\Bbb N\}$.   If $b$ is any upper bound of
$\{a_n:n\in\Bbb N\}$, then
$a\Bsetminus b=\lim_{n\to\infty}a_n\Bsetminus b=0$;  so
$a=\sup_{n\in\Bbb N}a_n$.   Finally,

\Centerline{$\bar d^*(a)+\bar d^*(1\Bsetminus a)=\lim_{n\to\infty}\bar
d^*(a_n)+\bar d^*(1\Bsetminus a_n)=1$.}
}%end of proof of 491J

\leader{491K}{Corollary} Set
$D=\{a:a\in\frak Z$, $\bar d^*(a)+\bar d^*(1\Bsetminus a)=1\}$, and write
$\bar d$ for $\bar d^*\restr D$.

(a) If $I\subseteq\Bbb N$ then its asymptotic density $d(I)$ is defined iff
$I^{\ssbullet}\in D$, and in this case $d(I)=\bar d(I^{\ssbullet})$.

(b) If $a\in D$ then its complement $1\Bsetminus a$ in $\frak Z$ belongs to
$D$;   if $a$, $b\in D$ and $a\Bcap b=0$, then $a\Bcup b\in D$ and
$\bar d(a\Bcup b)=\bar d(a)+\bar d(b)$;
if $a$, $b\in D$ and $a\Bsubseteq b$ then $b\Bsetminus a\in D$ and
$\bar d(b\Bsetminus a)=\bar d(b)-\bar d(a)$.

(c) $D$ is a topologically closed subset of $\frak Z$.

(d) If $A\subseteq D$ is upwards-directed, then $\sup A$ is defined in
$\frak Z$ and belongs to $D$;
moreover there is a sequence in $A$ with the same supremum
as $A$, and $\sup A$ belongs to the topological closure of $A$.

(e) Let $\frak B\subseteq D$ be a subalgebra of $\frak Z$.   Then the
following are equiveridical:

\quad (i) $\frak B$ is topologically closed in $\frak Z$;

\quad (ii) $\frak B$ is order-closed in $\frak Z$;

\quad (iii) setting $\bar\nu=\bar d^*\restr\frak B=\bar d\restr\frak B$,
$(\frak B,\bar\nu)$ is a probability algebra.

\noindent In this case, $\frak B$ is regularly embedded in $\frak Z$.

(f) If $I\subseteq D$ is closed under either $\Bcap$ or $\Bcup$, then
the topologically closed subalgebra of $\frak Z$ generated by $I$,
which is also the
order-closed subalgebra of $\frak Z$ generated by $I$, is included in
$D$.

\proof{{\bf (a)}

$$\eqalign{I^{\ssbullet}\in D&
\iff\bar d^*(I^{\ssbullet})+\bar d^*(1\Bsetminus I^{\ssbullet})=1\cr&
\iff d^*(I)+d^*(\Bbb N\setminus I)=1
\iff d(I)\text{ is defined}\cr}$$

\noindent by 491Ac, and in this case

\Centerline{$d(I)=d^*(I)=\bar d^*(I^{\ssbullet})=\bar d(I^{\ssbullet})$.}

\medskip

{\bf (b)} These all follow directly from the corresponding
results concerning $\Cal P\Bbb N$ and $d$ (491Ac).

\medskip

{\bf (c)} All we have to know is that $a\mapsto\bar d^*(a)$,
$a\mapsto 1\Bsetminus a$ are continuous (392Hb);  so that
$\{a:\bar d^*(a)+\bar d^*(1\Bsetminus a)=1\}$ is closed.

\medskip

{\bf (d)} Because $A$ is upwards-directed, and $\bar d^*$ is a
non-decreasing functional on $\frak Z$, there is a non-decreasing
sequence $\sequencen{a_n}$ in $A$ such that
$\lim_{n\to\infty}\bar d^*(a_n)=\sup_{a\in A}\bar d^*(a)=\gamma$ say.
By 491J, $b=\lim_{n\to\infty}a_n=\sup_{n\in\Bbb N}a_n$ is defined in
$\frak Z$ and belongs to $D$.
If $a\in A$ and $\epsilon>0$, there is an $n\in\Bbb N$ such that
$\bar d^*(a_n)\ge\gamma-\epsilon$.   Let $a'\in A$ be a common upper
bound of $a$ and $a_n$.   Then

\Centerline{$\bar d^*(a\Bsetminus b)\le\bar d^*(a'\Bsetminus a_n)
=\bar d^*(a')-\bar d^*(a_n)\le\gamma-\bar d^*(a_n)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $a\Bsubseteq b$;  as $a$ is
arbitrary, $b$ is an upper bound of $A$;  as $b=\sup_{n\in\Bbb N}a_n$,
$b$ must be the supremum of $A$.

\medskip

{\bf (e)(i)$\Rightarrow$(ii)} Suppose that $\frak B$ is topologically
closed.   If $A\subseteq\frak B$ is a non-empty upwards-directed
subset with supremum $b\in\frak Z$, then
(d) tells us that $b\in\overline{A}\subseteq\frak B$.   It follows
that $\frak B$ is order-closed in $\frak Z$ (313E(a-i)).

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} Suppose that $\frak B$ is
order-closed in $\frak Z$.   If $A\subseteq\frak B$ is non-empty, then
$A'=\{a_0\Bcup\ldots\Bcup a_n:a_0,\ldots,a_n\in A\}$
is non-empty and upwards-directed, so has a supremum in $\frak Z$,
which must belong to $\frak B$, and must be the least upper bound of $A$
in $\frak B$.   Thus $\frak B$ is Dedekind ($\sigma$-)complete.   Now
let $\sequencen{a_n}$ be a disjoint sequence in $\frak B$ and set
$b_n=\sup_{i\le n}a_i$
for each $n$.   Then $\sequencen{b_n}$ is a non-decreasing sequence in
$D$ so has a limit and supremum $b\in D$, and $b\in\frak B$.   Also
$\bar d^*(b_n)=\sum_{i=0}^n\bar d^*(a_i)$ for each $n$ (induce on
$n$), so

\Centerline{$\bar\nu b=\bar d^*(b)=\lim_{n\to\infty}\bar d^*(b_n)
=\sum_{i=0}^{\infty}\bar d^*(a_i)=\sum_{i=0}^{\infty}\bar\nu a_i$.}

\noindent Since certainly $\bar\nu 0=0$, $\bar\nu 1=1$ and $\bar\nu b>0$
whenever $b\in\frak B\setminus\{0\}$, $(\frak B,\bar\nu)$ is a
probability algebra.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} Suppose that $(\frak B,\bar\nu)$ is a
probability algebra.   Then it is complete under its measure metric
(323Gc), which agrees on $\frak B$ with
the metric $\bar\rho$ of $\frak Z$;  so $\frak B$ must be
topologically closed in $\frak Z$.

We see also that $\frak B$ is regularly embedded in $\frak Z$.  \Prf\
(Compare 323H.)  If $A\subseteq\frak B$ is
non-empty and downwards-directed and has infimum $0$ in $\frak B$, and
$b\in\frak Z$ is any lower bound of $A$ in $\frak Z$, then

\Centerline{$\bar d^*(b)\le\inf_{a\in A}\bar d^*(a)
=\inf_{a\in A}\bar\nu a=0$}

\noindent (321F), so $b=0$.   Thus $\inf A=0$ in $\frak Z$.   As $A$ is
arbitrary, this is enough to show that the identity map from $\frak B$
to $\frak Z$ is
order-continuous (313Lb), that is, that $\frak B$ is regularly
embedded in $\frak Z$.\ \Qed

\medskip

{\bf (f)} Let $\frak B$ be the
order-closed subalgebra of $\frak Z$ generated by $I$.
If $I$ is closed under $\Bcap$, then (b), (d) and 313Gc tell us that
$\frak B\subseteq D$.   If $I$ is closed under $\Bcup$,
then $I'=\{1\Bsetminus a:a\in I\}$
is a subset of $D$ closed under $\Bcap$, while $\frak B$ is the
order-closed subalgebra generated by $I'$, so again $\frak B\subseteq D$.
By (e), $\frak B$ is in either case topologically closed.   So we see that
the topologically closed subalgebra generated by $I$ is included in $D$;
by (e) again, it is equal to $\frak B$.
}%end of proof of 491K

\leader{491L}{Effectively regular measures}\cmmnt{ The examples 491Xf
and 491Yc show that the definition in 491B is drawn a little too wide
for comfort, and allows some uninteresting
pathologies.   These do not arise in the measure spaces we care most
about, and the following definitions provide a fire-break.}
Let $(X,\Sigma,\mu)$ be a measure space, and $\frak T$ a topology on
$X$.

\spheader 491La I will say that
a measurable subset $K$ of $X$ of finite measure is {\bf regularly
enveloped} if for every $\epsilon>0$ there are an open measurable set
$G$ and a closed measurable set $F$ such that $K\subseteq G\subseteq F$
and $\mu(F\setminus K)\le\epsilon$.

\spheader 491Lb Note that the family\cmmnt{ $\Cal K$} of
regularly enveloped
measurable sets of finite measure is closed under finite unions and
countable intersections.   \prooflet{\Prf\
(i) If $K_1$, $K_2\in\Cal K$ and $*$ is either $\cup$ or $\cap$, let
$\epsilon>0$.   Take measurable open sets $G_1$, $G_2$ and measurable
closed sets $F_1$, $F_2$ such that $K_i\subseteq G_i\subseteq F_i$ and
$\mu(F_i\setminus K_i)\le\bover12\epsilon$ for both $i$.   Then
$G_1*G_2$ is a measurable open set, $F_1*F_2$ is a measurable closed
set, $K_1*K_2\subseteq G_1*G_2\subseteq F_1*F_2$ and
$\mu((F_1*F_2)\setminus(K_1*K_2))\le\epsilon$.   As $\epsilon$ is
arbitrary, $K_1*K_2\in\Cal K$.   (ii) If $\sequencen{K_n}$ is a
non-increasing sequence in $\Cal K$ with intersection $K$ and
$\epsilon>0$, let $n\in\Bbb N$ be such that $\mu K_n<\mu K+\epsilon$.
Then we can find a measurable open set $G$ and a measurable
closed set $F$ such that $K_n\subseteq G\subseteq F$ and
$\mu F\le\mu K+\epsilon$.   As $\epsilon$ is arbitrary, $K\in\Cal K$.
Together with (i), this is enough to show that
$\Cal K$ is closed under countable intersections.\ \Qed}

\spheader 491Lc \cmmnt{Now I say that} $\mu$ is {\bf effectively
regular} if it
is inner regular with respect to the regularly enveloped sets of
finite measure.

\leader{491M}{Examples (a)} Any totally finite Radon measure is
effectively regular.   \prooflet{\Prf\ Let $(X,\frak T,\Sigma,\mu)$
be a totally finite Radon measure space.
If $K\subseteq X$ is compact and $\epsilon>0$, let
$L\subseteq X\setminus K$ be a compact set such that
$\mu L\ge\mu X-\mu K+\epsilon$.   Let $G$, $H$ be disjoint open sets
including $K$, $L$ respectively (4A2F(h-i)).   Then
$K\subseteq G\subseteq X\setminus H$, $G$ is open, $X\setminus H$
is closed, both $G$ and $X\setminus H$ are measurable, and
$\mu((X\setminus H)\setminus K)\le\epsilon$.   This shows that every
compact set is regularly enveloped,
and $\mu$ is effectively regular.\ \Qed}

\spheader 491Mb Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space such that $\frak T$ is a regular topology.   Then $\mu$ is
effectively regular.   \prooflet{\Prf\ Let
$E\in\Sigma$ and take $\gamma<\mu E$.   Choose sequences
$\sequencen{E_n}$ and $\sequencen{G_n}$ inductively, as follows.
$E_0\subseteq E$ is to be any measurable set
such that $\gamma<\mu E_0<\infty$.   Given that $\mu E_n>\gamma$, let
$G$ be an open set of finite measure such that $\mu(E_n\cap G)>\gamma$
(414Ea), and $F\subseteq G\setminus E_n$
a closed set such that $\mu F\ge\mu(G\setminus E_n)-2^{-n}$.   Let
$\Cal H$ be the family of open sets $H$ such that
$\overline H\subseteq G\setminus F$.   Then $\Cal H$ is
upwards-directed and covers $E_n$ (because $\frak T$ is regular), so
there is a $G_n\in\Cal H$ such that $\mu(E_n\cap G_n)>\gamma$ (414Ea
again).   Now $\mu(\overline{G}_n\setminus E_n)\le 2^{-n}$.   Set
$E_{n+1}=E_n\cap G_n$, and continue.

At the end of the induction, set $K=\bigcap_{n\in\Bbb N}E_n$.   For
each $n$, $K\subseteq G_n\subseteq\overline{G}_n$ and

\Centerline{$\lim_{n\to\infty}\mu(\overline{G}_n\setminus K)
\le\lim_{n\to\infty}2^{-n}+\mu(E_n\setminus K)=0$,}

\noindent so $K$ is regularly enveloped.   At the same time,
$K\subseteq E$ and $\mu K\ge\gamma$.   As $E$ and $\gamma$ are
arbitrary, $\mu$ is effectively regular.\ \Qed}

\spheader 491Mc Any totally finite Baire measure is effectively regular.
\prooflet{\Prf\ Let $\mu$ be a totally finite Baire measure on a
topological space $X$.   If $F\subseteq X$
is a zero set, let $f:X\to\Bbb R$ be a continuous function such that
$F=f^{-1}[\{0\}]$.   For each $n\in\Bbb N$, set
$G_n=\{x:|f(x)|<2^{-n}\}$, $F_n=\{x:|f(x)|\le 2^{-n}\}$;  then
$G_n$ is a measurable open set, $F_n$ is a measurable closed set,
$F\subseteq G_n\subseteq F_n$ for every $n$ and
$\lim_{n\to\infty}\mu F_n=\mu F$ (because $\mu$ is totally finite).
This shows that every zero set is regularly enveloped;  as $\mu$ is
inner regular with respect to the zero sets (412D), $\mu$ is
effectively regular.\ \Qed}

\spheader 491Md A totally finite completion regular topological measure
is effectively regular.   \prooflet{(As in (c), all zero sets are
regularly enveloped.)}

\leader{491N}{Theorem} Let $X$ be a topological space and $\mu$ an
effectively regular probability measure on $X$, with measure algebra
$(\frak A,\bar\mu)$.
Suppose that $\sequence{i}{x_i}$ is an equidistributed sequence in $X$.
Then we have a unique order-continuous Boolean
homomorphism $\pi:\frak A\to\frak Z=\Cal P\Bbb N/\Cal Z$ such that
$\pi G^{\ssbullet}\Bsubseteq\{i:x_i\in G\}^{\ssbullet}$ for every
measurable open set $G\subseteq X$,
and $\bar d^*(\pi a)=\bar\mu a$ for every $a\in\frak A$.

\proof{{\bf (a)} Define $\theta:\Cal PX\to\frak Z$ by setting
$\theta A=\{i:x_i\in A\}^{\ssbullet}$ for $A\subseteq X$;  then
$\theta$ is a Boolean homomorphism.
If $F\subseteq X$ is closed and measurable, then
$\bar d^*(\theta F)\le\mu F$, because $\sequence{i}{x_i}$ is
equidistributed.   Write $\Cal K$ for the family of
regularly enveloped measurable sets.

If $K\in\Cal K$, then $\pi_0K=\inf\{\theta G:K\subseteq G\in\Sigma\cap\frak T\}$
is defined in $\frak Z$, $\bar d^*(\pi_0K)=\mu K$ and $\pi_0K\in D$ as
defined in 491K.   \Prf\ For each $n\in\Bbb N$, let $G_n$,
$F_n\in\Sigma$ be such that $K\subseteq G_n\subseteq F_n$,
$G_n$ is open, $F_n$ is closed and $\mu(F_n\setminus K)\le 2^{-n}$.
Set $H_n=X\setminus\bigcap_{i\le n}G_i$.   Then

$$\eqalign{\bar d^*(\theta H_n)+\bar d^*(1\Bsetminus\theta H_n)
&\le\bar d^*(\theta H_n)+\bar d^*(\theta(\bigcap_{i\le n}F_i))
\le\mu H_n+\mu(\bigcap_{i\le n}F_i)\cr
&\le\mu(X\setminus K)+\mu F_n
\le 1+2^{-n}.\cr}$$

\noindent Also $\sequencen{\theta H_n}$ is a non-decreasing sequence in
$\frak Z$.   By 491J,
$a=\lim_{n\to\infty}\theta H_n=\sup_{n\in\Bbb N}\theta H_n$ is defined
in $\frak Z$ and belongs to $D$.   Set

\Centerline{$b=1\Bsetminus a
=\lim_{n\to\infty}1\Bsetminus\theta H_n
=\lim_{n\to\infty}\theta(\bigcap_{i\le n}G_i)$,}

\noindent so that $b$ also belongs to $D$.   If
$K\subseteq G\in\Sigma\cap\frak T$, then

\Centerline{$b\Bsetminus\theta G
=\lim_{n\to\infty}\theta(\bigcap_{i\le n}G_i)\Bsetminus\theta G
=\lim_{n\to\infty}\theta(\bigcap_{i\le n}G_i\setminus G)$}

\noindent and

$$\eqalign{\bar d^*(b\Bsetminus\theta G)
&=\lim_{n\to\infty}\bar d^*(\theta(\bigcap_{i\le n}G_i\setminus G))\cr
&\le\lim_{n\to\infty}\bar d^*(\theta(\bigcap_{i\le n}F_i\setminus G))
\le\lim_{n\to\infty}\mu(\bigcap_{i\le n}F_i\setminus G)
=0.\cr}$$

\noindent This shows that $b\subseteq\theta G$ whenever
$K\subseteq G\in\Sigma\cap\frak T$.   On the other hand, any lower
bound of $\{\theta G:K\subseteq G\in\Sigma\cap\frak T\}$ is also a
lower bound of $\{\theta(\bigcap_{i\le n}G_i):n\in\Bbb N\}$, so is
included in $b$.   Thus
$b=\inf\{\theta G:K\subseteq G\in\Sigma\cap\frak T\}$ and $\pi_0(K)=b$ is
defined.

To compute $\bar d^*(b)$, observe first that
$b\Bsubseteq 1\Bsetminus\theta H_n\Bsubseteq\theta F_n$ for every $n$, so

\Centerline{$\bar d^*(b)\le\inf_{n\in\Bbb N}\bar d^*(\theta F_n)
\le\inf_{n\in\Bbb N}\mu F_n=\mu K$.}

\noindent On the other hand,

\Centerline{$\bar d^*(\theta(\bigcap_{i\le n}G_i))
\ge 1-\bar d^*(\theta H_n)\ge 1-\mu H_n\ge\mu K$}

\noindent for every $n$, so

\Centerline{$\bar d^*(b)
=\lim_{n\to\infty}\bar d^*(\theta(\bigcap_{i\le n}G_i))\ge\mu K$.}

\noindent Accordingly $\bar d^*(b)=\mu K$, and $\pi_0K$ has the required
properties.\ \Qed

\medskip

{\bf (b)} If $K$, $L\in\Cal K$, then $\pi_0(K\cap L)=\pi_0K\Bcap\pi_0L$.
\Prf\ We know that $K\cap L\in\Cal K$ (491Lb).   And

$$\eqalignno{\pi_0K\Bcap\pi_0L
&=\inf\{\theta G:K\subseteq G\in\frak T\}
   \Bcap\inf\{\theta H:L\subseteq H\in\frak T\}\cr
&=\inf\{\theta G\Bcap\theta H:
  K\subseteq G\in\frak T,\,L\subseteq H\in\frak T\}\cr
&=\inf\{\theta(G\cap H):
  K\subseteq G\in\frak T,\,L\subseteq H\in\frak T\}
\Bsupseteq\pi_0(K\cap L).\cr}$$

\noindent Now suppose that $U\supseteq K\cap L$ is a measurable open
set
and $\epsilon>0$.   Let $G$, $G'$ be measurable open sets and $F$,
$F'$
measurable closed sets such that
$K\subseteq G\subseteq F$, $L\subseteq G'\subseteq F'$,
$\mu(F\setminus
K)\le\epsilon$ and $\mu(F'\setminus L)\le\epsilon$.   Then

$$\eqalign{\bar d^*(\pi_0K\Bcap\pi_0L\Bsetminus\theta U)
&\le\bar d^*(\theta G\Bcap\theta G'\Bsetminus\theta U)
=\bar d^*(\theta(G\cap G'\setminus U))\cr
&\le\bar d^*(\theta(F\cap F'\setminus U))
\le\mu(F\cap F'\setminus U)
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\pi_0K\Bcap\pi_0L\Bsubseteq\theta U$;  as $U$ is arbitrary,
$\pi_0K\Bcap\pi_0L\Bsubseteq\pi_0(K\cap L)$.\ \Qed

This means that $\{\pi_0K:K\subseteq X$ is a regularly embedded
measurable set$\}$ is a subset of $D$ closed under $\Bcap$.
By 491Kf, the topologically closed subalgebra $\frak B$ of $\frak Z$
generated by this family is included in $D$;
by 491Ke, $\frak B$ is order-closed and regularly embedded in $\frak Z$,
and $(\frak B,\bar d^*\restrp\frak B)$ is a probability algebra.

\medskip

{\bf (c)} Now observe that if we set
$Q=\{K^{\ssbullet}:K\in\Cal K\}\subseteq\frak A$, we have a function
$\pi:Q\to\frak B$ defined by setting $\pi K^{\ssbullet}=\pi_0K$
whenever $K\in\Cal K$.   \Prf\ Suppose that $K$, $L\in\Cal K$
and $\mu(K\symmdiff L)=0$.   Then

$$\eqalignno{\bar d^*(\pi_0K\Bsymmdiff\pi_0L)
&=\bar d^*(\pi_0K)+\bar d^*(\pi_0L)-2\bar d^*(\pi_0K\Bcap\pi_0L)\cr
\displaycause{because $\pi_0K$ and $\pi_0L$ belong to $\frak B\subseteq D$}
&=\bar d^*(\pi_0K)+\bar d^*(\pi_0L)-2\bar d^*(\pi_0(K\cap L))\cr
&=\mu K+\mu L-2\mu(K\cap L)
=0.\cr}$$

\noindent So $\pi_0K=\pi_0L$ and either can be used to define
$\pi K^{\ssbullet}$.\ \QeD\  Next, the same formulae show that
$\pi:Q\to\frak B$ is an isometry
when $Q$ is given the measure metric of $\frak A$, since if $K$,
$L$ belong to $\Cal K$,

\Centerline{$\bar\rho(\pi K^{\ssbullet},\pi L^{\ssbullet})
=\bar d^*(\pi_0K\Bsymmdiff\pi_0L)=\mu K+\mu L-2\mu(K\cap L)
=\mu(K\symmdiff L)=\bar\mu(K^{\ssbullet}\Bsymmdiff L^{\ssbullet})$.}

\noindent As $Q$ is dense in $\frak A$ (412N), there is a unique
extension of $\pi$ to an isometry from $\frak A$ to $\frak B$.

\medskip

{\bf (d)} Because

\Centerline{$\pi(K^{\ssbullet}\Bcap L^{\ssbullet})
=\pi((K\cap L)^{\ssbullet})=\pi_0(K\cap L)=\pi_0K\Bcap\pi_0L
=\pi K^{\ssbullet}\Bcap\pi L^{\ssbullet}$}

\noindent for all $K$, $L\in\Cal K$, $\pi(a\Bcap a')=\pi a\Bcap\pi a'$
for all $a$, $a'\in\frak A$.
It follows that $\pi$ is a Boolean homomorphism.   \Prf\ The point is
that $\bar d^*(\pi a)=\bar\mu a$ for every $a\in Q$, and therefore for
every $a\in\frak A$.   Now if $a\in\frak A$, $\pi(1\Bsetminus a)$ must
be disjoint from $\pi a$ (since certainly $\pi 0=0$), and has the same
measure as $1\Bsetminus\pi a$ (remember that we know that
$(\frak B,\bar d^*\restrp\frak B)$ is a measure algebra), so must be
equal to $1\Bsetminus\pi a$.
By 312H(ii), $\pi$ is a Boolean homomorphism.\ \Qed

By 324G, $\pi$ is order-continuous when regarded as a function from
$\frak A$ to $\frak B$.   Because $\frak B$ is regularly embedded in
$\frak Z$, $\pi$ is order-continuous when regarded as a function from
$\frak A$ to $\frak Z$.

\medskip

{\bf (e)} Let $G\in\Sigma\cap\frak T$.   For any $\epsilon>0$, there
is
a $K\in\Cal K$ such that $K\subseteq G$ and $\mu(G\setminus
K)\le\epsilon$.
In this case, $\pi K^{\ssbullet}=\pi_0K\subseteq\theta G$.   So

\Centerline{$\bar d^*(\pi G^{\ssbullet}\Bsetminus\theta G)
\le\bar d^*(\pi G^{\ssbullet}\Bsetminus\pi K^{\ssbullet})
=\bar\mu(G^{\ssbullet}\Bsetminus K^{\ssbullet})=\mu(G\setminus K)
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\pi G^{\ssbullet}\subseteq\theta G$.

\medskip

{\bf (f)} This shows that we have a homomorphism $\pi$ with the
required
properties.   To see that $\pi$ is unique, suppose that
$\pi':\frak A\to\frak Z$ is also a
homomorphism of the same kind.   In this case

\Centerline{$\bar d^*(1\Bsetminus\pi'a)=\bar d^*(\pi'(1\Bsetminus
a))=\bar\mu(1\Bsetminus a)=1-\bar\mu a=1-\bar d^*(\pi'a)$,}

\noindent so $\pi'a\in D$, for every $a\in\frak A$.   If $K\in\Cal K$,
then $\pi'K^{\ssbullet}\subseteq\theta G$ whenever $K\subseteq
G\in\Sigma\cap\frak T$, so
$\pi'K^{\ssbullet}\Bsubseteq\pi_0K=\pi K^{\ssbullet}$.   As both $\pi
K^{\ssbullet}$ and $\pi'K^{\ssbullet}$ belong to $D$,

\Centerline{$\bar d^*(\pi K^{\ssbullet}\Bsetminus\pi'K^{\ssbullet})
=\bar d^*(\pi K^{\ssbullet})-\bar d^*(\pi'K^{\ssbullet})
=\mu K-\mu K=0$,}

\noindent and $\pi K^{\ssbullet}=\pi'K^{\ssbullet}$.   As
$\{K^{\ssbullet}:K\in\Cal K\}$ is topologically dense in $\frak A$,
and both $\pi$ and $\pi'$ are continuous,
they must be equal.
}%end of proof of 491N

\leader{491O}{Proposition} Let $X$ be a topological space and $\mu$ an
effectively regular probability measure on $X$ which measures every zero
set, and suppose that
$\sequence{i}{x_i}$ is an equidistributed sequence in $X$.   Let $\frak A$
be the measure algebra of $\mu$ and
$\pi:\frak A\to\frak Z=\Cal P\Bbb N/\Cal Z$ the regular embedding
described in 491N;  let
$T_{\pi}:L^{\infty}(\frak A)\to L^{\infty}(\frak Z)$ be the corresponding
order-continuous Banach algebra embedding\cmmnt{ (363F)}.   Let
$S:\ell^{\infty}(X)\to\ell^{\infty}$ be the Riesz homomorphism defined
by setting $(Sf)(i)=f(x_i)$ for
$f\in\ell^{\infty}(X)$ and $i\in\Bbb N$, and
$R:\ell^{\infty}\to L^{\infty}(\frak Z)$ the Riesz homomorphism
corresponding to the Boolean homomorphism
$I\mapsto I^{\ssbullet}:\Cal P\Bbb N\to\frak Z$.   For
$f\in\eusm L^{\infty}(\mu)$ let $f^{\ssbullet}$ be the corresponding
member of $L^{\infty}(\mu)\cong L^{\infty}(\frak A)$
\cmmnt{ (363I)}.   Then $T_{\pi}(f^{\ssbullet})=RSf$ for every
$f\in C_b(X)$.

\proof{ To begin with, suppose that $f:X\to[0,1]$ is continuous and
$k\ge 1$.   For each $i\le k$ set $G_i=\{x:f(x)>\bover{i}{k}\}$,
$F_i=\{x:f(x)\ge\bover{i}{k}\}$.
Then $\bover1k\sum_{i=1}^k\chi F_i\le f\le\bover1k\sum_{i=0}^k\chi G_i$.
So

\Centerline{$\Bover1k\sum_{i=1}^k\chi(\pi F_i^{\ssbullet})
\le T_{\pi}f^{\ssbullet}\le\Bover1k\sum_{i=0}^k\chi(\pi G_i^{\ssbullet})$,}

\Centerline{$\Bover1k\sum_{i=1}^k\chi(\theta F_i)
\le RSf\le\Bover1k\sum_{i=0}^k\chi(\theta G_i)$}

\noindent where $\theta:\Cal PX\to\frak Z$ is the Boolean homomorphism
described in the proof of 491N, because
$RS:\ell^{\infty}(X)\to L^{\infty}(\frak Z)$
is the Riesz homomorphism corresponding
to $\theta$ (see 363Fa, 363Fg).   Now 491N tells us that
$\pi G^{\ssbullet}\Bsubseteq\theta G$ for every cozero set
$G\subseteq X$, so

$$\eqalign{T_{\pi}f^{\ssbullet}
&\le\Bover1k\sum_{i=0}^k\chi(\pi G_i^{\ssbullet})
\le\Bover1k\sum_{i=0}^k\chi(\theta G_i)\cr
&\le\Bover1k\sum_{i=0}^k\chi(\theta F_i)
=\Bover1ke+\sum_{i=1}^k\chi(\theta F_i)
\le\Bover1ke+RSf\cr}$$

\noindent where $e$ is the standard order unit of the $M$-space
$L^{\infty}(\frak Z)$.   But looking at complements we see that we
must have $\pi F^{\ssbullet}\Bsupseteq\theta F$
for every zero set $F\subseteq X$, so

$$\eqalign{RSf
&\le\Bover1k\sum_{i=0}^k\chi(\theta G_i)
\le\Bover1k\sum_{i=0}^k\chi(\theta F_i)\cr
&\le\Bover1k\sum_{i=0}^k\chi(\pi F_i^{\ssbullet})
=\Bover1ke+\sum_{i=1}^k\chi(\pi F_i^{\ssbullet})
\le\Bover1ke+T_{\pi}f^{\ssbullet}.\cr}$$

\noindent This means that $|T_{\pi}f^{\ssbullet}-RSf|\le\bover1ke$ for
every $k\ge 1$, so that $T_{\pi}f^{\ssbullet}=RSf$.   This is true
whenever $f\in C_b(X)$ takes values in
$[0,1]$;  as all the operators here are linear, it is true for every
$f\in C_b(X)$.
}%end of proof of 491O

\leader{491P}{Proposition} Any probability algebra $(\frak A,\bar\mu)$
with cardinal at most $\frak c$ can be regularly embedded as a
subalgebra of $\frak Z=\Cal P\Bbb N/\Cal Z$
in such a way that $\bar\mu$ is identified with the restriction of the
submeasure $\bar d^*$ to the image of $\frak A$.

\proof{ The usual measure of $\{0,1\}^{\frak c}$ is a totally finite
Radon measure (416Ub), so is effectively regular (491Ma).   It has an
equidistributed sequence
(491G), so its measure algebra $(\frak B_{\frak c},\bar\nu_{\frak c})$
can be regularly embedded in $\frak Z$ in a way which matches
$\bar\nu_{\frak c}$ with $\bar d^*$ (491N).
Now if $(\frak A,\bar\mu)$ is any probability algebra with cardinal at
most $\frak c$, it can be regularly embedded (by a measure-preserving
homomorphism) in
$(\frak B_{\frak c},\bar\nu_{\frak c})$ (332N), and therefore in
$(\frak Z,\bar d^*)$.
}%end of proof of 491P

\leader{491Q}{Corollary} Every Radon probability measure on
$\{0,1\}^{\frak c}$ has an equidistributed sequence.

\proof{ Let $\mu$ be a Radon probability measure on $\{0,1\}^{\frak c}$,
and $(\frak A,\bar\mu)$ its measure algebra.
For $\xi<\frak c$ set $E_{\xi}=\{x:x\in\{0,1\}^{\frak c}$, $x(\xi)=1\}$
and $e_{\xi}=E_{\xi}^{\ssbullet}\in\frak A$.

\medskip

{\bf (a)} $\#(\frak A)\le\frak c$.
\Prf\ The $\sigma$-algebra
generated by $\{E_{\xi}:\xi<\frak c\}$ is the Baire $\sigma$-algebra
$\CalBa(\{0,1\}^{\frak c})$ (4A3Na) and $E^{\ssbullet}$ belongs to the
$\sigma$-subalgebra
$\frak B$ of $\frak A$ generated by $\{e_{\xi}:\xi<\frak c\}$ whenever
$E\in\CalBa(\{0,1\}^{\frak c})$.   Now if $G\subseteq\{0,1\}^{\frak c}$ is open, then
$\Cal H=\{H:H\subseteq G$ is determined by a finite set of coordinates$\}$ is an
upwards-directed family of open sets with union $G$, while $H$ is a Baire set
(see 4A3Nb) and
$H^{\ssbullet}\in\frak B$ for every $H\in\Cal H$.   If $\epsilon>0$, there is
an $H\in\Cal G$ such that $\mu G\le\mu H+\epsilon$, because $\mu$ is
$\tau$-additive;  so we have a sequence $\sequencen{H_n}$ in $\Cal H$ such that
$\mu(G\symmdiff\bigcup_{n\in\Bbb N}H_n)=0$ and
$G^{\ssbullet}=\sup_{n\in\Bbb N}H_n^{\ssbullet}\in\frak B$.   It follows that
$F^{\ssbullet}\in\frak B$ for every closed $F\subseteq\{0,1\}^{\frak c}$.
Next, if $E\subseteq\{0,1\}^{\frak c}$ is measured by $\mu$, there is a sequence
$\sequencen{F_n}$ of closed subsets of $E$ such that
$\mu(E\setminus\bigcup_{n\in\Bbb N}F_n)=0$, because $\mu$ is inner regular with
respect to the closed sets, so that
$E^{\ssbullet}=\sup_{n\in\Bbb N}F_n^{\ssbullet}\in\frak B$;
thus $\frak A=\frak B$. But $\frak B$ has cardinal at most $\frak c$, by 4A1O.\
\Qed

\medskip

{\bf (b)} By 491P, there is a measure-preserving embedding
$\pi:\frak A\to\frak Z$, and $\pi[\frak A]\subseteq D$ as defined in
491K.   For $\xi<\frak c$ let $I_{\xi}\subseteq\Bbb N$ be such that
$I_{\xi}^{\ssbullet}=\pi e_{\xi}$ in $\frak Z$.   Define $x_i(\xi)$, for
$i\in\Bbb N$ and $\xi<\frak c$, by setting $x_i(\xi)=1$ if $i\in I_{\xi}$,
$0$ otherwise.
Now suppose that $H\subseteq\{0,1\}^{\frak c}$ is a basic open set of
the form $\{x:x(\xi)=1$ for $\xi\in K$, $0$ for $\xi\in L\}$, where
$K$, $L\subseteq\frak c$
are finite.   Set $b=\pi H^{\ssbullet}$ in $\frak Z$,

\Centerline{$I=\{i:x_i\in H\}
=\Bbb N\cap\bigcap_{\xi\in K}I_{\xi}\setminus\bigcup_{\xi\in L}I_{\xi}$.}

\noindent Then

$$\eqalign{b
&=\pi H^{\ssbullet}
=\pi(\inf_{\xi\in K}a_{\xi}\Bsetminus\sup_{\xi\in L}a_{\xi})\cr
&=\inf_{\xi\in K}\pi a_{\xi}\Bsetminus\sup_{\xi\in L}\pi a_{\xi}
=\inf_{\xi\in K}I_{\xi}^{\ssbullet}
  \Bsetminus\sup_{\xi\in L}I_{\xi}^{\ssbullet}
=I^{\ssbullet}.\cr}$$

\noindent Since $b\in D$, $d(I)$ is defined and is equal to
$\bar d^*(b)=\bar\mu H^{\ssbullet}=\mu H$.

If we now take $E$ to be an open-and-closed subset of
$\{0,1\}^{\frak c}$, it can be expressed as a disjoint union of
finitely
many basic open sets of the type just considered;  because $d$ is
additive on disjoint sets, $d(\{i:x_i\in E\})$ is defined and equal to
$\mu E$.   But this is enough to ensure that $\sequence{i}{x_i}$ is
equidistributed, by 491Ch.
}%end of proof of 491Q

\leader{491R}{}\cmmnt{ In this section I have been looking at
probability measures with equidistributed sequences.   A standard
line of investigation is to ask which of our ordinary constructions,
applied to such measures, lead to others of the same kind, as in
491Ea and 491F.   We find that the language developed here enables us
to express another result of this type.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological space, $\mu$ an
effectively regular topological probability measure on $X$ which has
an equidistributed sequence, and $\nu$ a probability measure on $X$
which is an indefinite-integral measure over $\mu$.   Then $\nu$ has
an equidistributed sequence.

\proof{ Let $\Cal K$ be the family of regularly enveloped measurable
sets.

\medskip

{\bf (a)} Consider first the case in which $\nu$ has Radon-Nikod\'ym
derivative of the form $\Bover1{\mu K}\chi K$ for some $K\in\Cal K$
of non-zero measure.    For each $m\in\Bbb N$, we have an open set
$G_m\supseteq K$ such that $\mu(\overline{G}_m\setminus K)\le 2^{-m}$;
of course we can arrange that $G_{m+1}\subseteq G_m$ for
each $m$.   Set $F_m=\overline{G}_m$ for each $m\in\Bbb N$.
Let $\sequence{i}{x_i}$ be an equidistributed sequence for
$\mu$.   Then there is an $I\subseteq\Bbb N$ such that $d(I)=\mu K$
and $\{i:i\in I$, $x_i\notin G_m\}$ is finite for every $m$.   \Prf\ For
each $m\in\Bbb N$, set $I_m=\{i:x_i\in G_m\}$.   We know that
$\liminf_{n\to\infty}\Bover1n\#(I_m\cap n)\ge\mu G_m\ge\mu K$ for
each $m$, so we can find a strictly increasing sequence
$\sequence{m}{k_m}$ such that
$\Bover1n\#(I_m\cap n)\ge\mu K-2^{-m}$ whenever $m\in\Bbb N$ and
$n>k_m$.   Set

\Centerline{$I=\bigcup_{m\in\Bbb N}I_m\cap k_{m+1}
\subseteq\bigcap_{m\in\Bbb N}(I_m\cup k_{m+1})$.}

\noindent   If $k_m<n\le k_{m+1}$,

\Centerline{$\Bover1n\#(I\cap n)\ge\Bover1n\#(I_m\cap n)
\ge\mu K-2^{-m}$;}

\noindent so $\liminf_{n\to\infty}\Bover1n\#(I\cap n)\ge\mu K$.   On
the other hand, for any $m\in\Bbb N$,

\Centerline{$\{i:i\in I$, $x_i\notin F_m\}\subseteq I\setminus I_m
\subseteq k_{m+1}$}

\noindent is finite, so

$$\eqalign{\limsup_{n\to\infty}\Bover1n\#(I\cap n)
&\le\limsup_{n\to\infty}\Bover1n\#(\{i:i<n,\,x_i\in F_m\})\cr
&\le\mu F_m\le\mu K+2^{-m}.\cr}$$

\noindent Accordingly $\limsup_{n\to\infty}\Bover1n\#(I\cap n)\le\mu K$ and
$d(I)$ is defined and equal to $\mu K$.\ \Qed

Let $\sequencen{j_n}$ be the increasing enumeration of $I$, and set
$y_n=x_{j_n}$ for each $n$.   Then $\sequencen{y_n}$ is
equidistributed for $\nu$.   \Prf\ Note first that

\Centerline{$\lim_{n\to\infty}\Bover{n}{j_n}
=\lim_{n\to\infty}\Bover1{j_n}\#(I\cap j_n)=\mu K$.}

\noindent Let $F\subseteq X$ be closed.
Then $\nu F=\Bover{\mu(F\cap K)}{\mu K}$.   On the other hand, for any
$m\in\Bbb N$,

$$\eqalignno{d^*(\{n:y_n\in F\})
&=\limsup_{n\to\infty}\Bover1n\#(\{i:i<n,\,x_{j_i}\in F\})\cr
&=\limsup_{n\to\infty}\Bover1n\#(\{i:i<j_n,\,i\in I,\,x_i\in F\})\cr
&=\limsup_{n\to\infty}
  \Bover1n\#(\{i:i<j_n,\,i\in I,\,x_i\in F\cap G_m\})\cr
\displaycause{because $\{i:i\in I$, $x_i\notin G_m\}$ is finite}
&\le\limsup_{n\to\infty}
  \Bover{j_n}{n}\Bover1{j_n}\#(\{i:i<j_n,\,x_i\in F\cap F_m\})\cr
&=\Bover1{\mu K}\limsup_{n\to\infty}
  \Bover1{j_n}\#(\{i:i<j_n,\,x_i\in F\cap F_m\})\cr
&\le\Bover1{\mu K}\limsup_{n\to\infty}
  \Bover1{n}\#(\{i:i<n,\,x_i\in F\cap F_m\})\cr
&\le\Bover1{\mu K}\mu(F\cap F_m)
\le\Bover1{\mu K}(\mu(F\cap K)+2^{-m})
=\nu F+\Bover1{2^m\mu K};\cr}$$

\noindent as $m$ is arbitrary, $d^*(\{n:y_n\in F\})\le\nu F$;  as $F$
is arbitrary, $\sequencen{y_n}$ is equidistributed for $\nu$.\ \Qed

\medskip

{\bf (b)} Now turn to the general case.   Let $f$ be a
Radon-Nikod\'ym derivative of $\nu$;  we may suppose that $f$ is
measurable and non-negative and defined everywhere in $X$.
Then there is a sequence $\sequence{m}{K_m}$ in $\Cal K$
such that $f\eae\sum_{m=0}^{\infty}\bover1{m+1}\chi K_m$.
\Prf\ Choose $f_m$, $K_m$ inductively, as follows.   $f_0=f$.
Given that $f_m\ge 0$ is measurable, set
$E_m=\{x:f_m(x)\ge\bover1{m+1}\}$ and let $K_m\in\Cal K$ be such that
$K_m\subseteq E_m$ and $\mu(E_m\setminus K_m)\le 2^{-m}$;  set
$f_{m+1}=f_m-\bover1{m+1}\chi K_m$.   Then $\sequence{m}{f_m}$ is
non-increasing;  set $g=\lim_{m\to\infty}f_m$.   \Quer\ If $g$ is not
zero almost everywhere, let $r\in\Bbb N$ be such that $\mu
E>2^{-r+1}$ where $E=\{x:g(x)\ge\bover1{r+1}\}$.   Then $E\subseteq
E_m$ for every $m\ge r$, so $\mu(E\setminus K_m)\le 2^{-m}$ for every
$m\ge r$ and $F=E\cap\bigcap_{m\ge r}K_m$ is not empty.   Take $x\in
F$;  then
$f_{m+1}(x)\le f_m(x)-\bover1{m+1}$ for every $m\ge r$, which is
impossible.\ \BanG\  So $g=0$ a.e.\ and
$f\eae\sum_{m=0}^{\infty}\bover1{m+1}\chi K_m$.\ \Qed

By (a), we have for each $m$ a sequence $\sequencen{y_{mn}}$ in $X$
such that

\Centerline{$\mu(F\cap K_m)
\ge\mu K_m\cdot\limsup_{n\to\infty}\Bover1n\#(\{i:i<n,\,y_{mi}\in F\})$}

\noindent for
every closed $F\subseteq X$.   For $n\in\Bbb N$, let $\nu_n$ be the
point-supported measure on $X$ defined by setting

\Centerline{$\nu_nA=\sum_{m=0}^{\infty}
  \Bover{\mu K_m}{(n+1)(m+1)}\#(\{i:i\le n,\,y_{mi}\in A\})$}

\noindent for $A\subseteq X$;  because
$\sum_{m=0}^{\infty}\Bover{\mu K_m}{m+1}=\int fd\mu=1$, $\nu_n$ is a
probability measure.   If $F\subseteq X$ is closed,

$$\eqalignno{\limsup_{n\to\infty}\nu_nF
&\le\sum_{m=0}^{\infty}\Bover{\mu K_m}{m+1}
  \limsup_{n\to\infty}\Bover1{n+1}\#(\{i:i\le n,\,y_{mi}\in F\})\cr
\displaycause{because $\sum_{m=0}^{\infty}\Bover{\mu
K_m}{m+1}<\infty$}
&\le\sum_{m=0}^{\infty}\Bover1{m+1}\mu(F\cap K_m)
=\int_Ffd\mu
=\nu F.\cr}$$

\noindent So 491D tells us that there is an equidistributed sequence
for $\nu$, as required.
}%end of proof of 491R

\leader{491S}{The asymptotic density
filter}\cmmnt{ Corresponding to the
asymptotic density ideal, of course we have a filter.
It is not surprising that convergence along this filter, in the sense of
2A3Sb, should be interesting and sometimes important.

\medskip

}{\bf (a)} Set

\Centerline{$\Cal F_d=\{\Bbb N\setminus I:I\in\Cal Z\}
=\{I:I\subseteq\Bbb N$, $\lim_{n\to\infty}\Bover1n\#(I\cap n)=1\}$.}

\noindent\cmmnt{ Then }$\Cal F_d$ is a filter on $\Bbb N$,
the {\bf (asymptotic) density filter}.

\spheader 491Sb For a
bounded sequence $\sequencen{\alpha_n}$ in $\Bbb C$,
$\lim_{n\to\Cal F_d}\alpha_n=0$ iff
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n|\alpha_k|=0$.
\prooflet{\Prf\
Set $M=\sup_{k\in\Bbb N}|\alpha_k|$, and for $\epsilon>0$ set
$I_{\epsilon}=\{n:|\alpha_n|\le\epsilon\}$.   Then, for any $n\ge 1$,

\Centerline{$\Bover{\epsilon}{n+1}\#((n+1)\setminus I_{\epsilon})
\le\Bover1{n+1}\sum_{k=0}^n|\alpha_k|
\le\epsilon+\Bover{M}{n+1}\#((n+1)\setminus I_{\epsilon})$.}

\noindent So if
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n|\alpha_k|=0$, then
$\lim_{n\to\infty}\Bover1{n+1}\#((n+1)\setminus I_{\epsilon})=0$, that is,
$\Bbb N\setminus I_{\epsilon}\in\Cal Z$ and $I_{\epsilon}\in\Cal F_d$;  as
$\epsilon$ is arbitrary, $\lim_{n\to\Cal F_d}\alpha_n=0$.   While if
$\lim_{n\to\Cal F_d}\alpha_n=0$ then
$\Bbb N\setminus I_{\epsilon}\in\Cal Z$ and
$\limsup_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n|\alpha_k|\le\epsilon$;
again, $\epsilon$ is arbitrary, so
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n|\alpha_k|=0$.\ \Qed}

\spheader 491Sc For any $m\in\Bbb N$ and $A\subseteq\Bbb N$,
$A+m\in\Cal F_d$ iff $A\in\Cal F_d$.   \prooflet{\Prf\ For any $n\ge m$,
$\#(n\cap(A+m))=\#((n-m)\cap A)$, so

\Centerline{$d(A+m)
=\lim_{n\to\infty}\Bover1n\#(n\cap(A+m))
=\lim_{n\to\infty}\Bover1n\#(n\cap A)
=d(A)$}

\noindent if either $d(A+m)$ or $d(A)$ is defined, in particular, if
either $A+m$ or $A$ belongs to $\Cal F_d$.\ \Qed}
Hence, or otherwise,
for any (real or complex) sequence $\sequencen{\alpha_n}$,
$\lim_{n\to\Cal F_d}\alpha_n=\lim_{n\to\Cal F_d}\alpha_{m+n}$ if either is
defined.




\exercises{\leader{491X}{Basic exercises (a)}%\spheader 491Xa
(i) Show that if $I$, $J\in\Cal D=\dom d$ as defined in 491A, then
$I\cup J\in\Cal D$ iff $I\cap J\in\Cal D$ iff $I\setminus J\in\Cal D$
iff $I\symmdiff J\in\Cal D$.
(ii) Show that if $\Cal E\subseteq\Cal D$ is an algebra of sets, then
$d\restr\Cal E$ is additive.
(iii) Find $I$, $J\in\Cal D$ such that $I\cap J\notin\Cal D$.
%491A

\sqheader 491Xb Suppose that $I\subseteq\Bbb N$ and that
$f:\Bbb N\to\Bbb N$ is strictly increasing.   Show that
$d^*(f[I])\le d^*(I)d^*(f[\Bbb N])$, with equality if either $I$ or
$f[\Bbb N]$ has asymptotic density.
%491A

\spheader 491Xc Suppose that $I\subseteq\Bbb N$.
(i)\dvAnew{2024} Show that there is a
$J\subseteq I$ such that $d^*(J)=d^*(I\setminus J)=d^*(I)$.   (ii) Show that
if $0\le\alpha\le d^*(I)$ there is a $J\subseteq I$ such that
$d^*(J)=\alpha$ and $d^*(I\setminus J)=d^*(I)-\alpha$.   (iii)\dvAnew{2024}
Show that if $d(I)$ is defined and $0\le\alpha\le d(I)$ there is a 
$J\subseteq I$ such that $d(J)$ is defined and equal to $\alpha$.
%491A 491Xb

\spheader 491Xd(i)($\alpha$) Show that if $I$, 
$K\subseteq\Bbb N$ are such that $d(I)$
and $d(K)$ are defined, there is a $J\subseteq\Bbb N$ such that $d(J)$ is
defined and $d(J)=d^*(J\cap I)=d^*(J\cap K)=\min(d(I),d(K))$. 
% use (c-iii) to reduce to case $d(I)=d(K)$;  take 
%$J=(I\cap\bigcup_{j\in\Bbb N}(k_{2j+1}\setminus k_{2j}))
%    \cup(K\cap\bigcup_{j\in\Bbb N}(k_{2j+2}\setminus k_{2j+1}))
($\beta$) Let $I\subseteq\Bbb N$ be such that
$d(J)=d^*(J\cap I)+d^*(J\setminus I)$ for every $J\subseteq\Bbb N$
such that $d(J)$ is defined.
Show that either $I\in\Cal Z$ or $\Bbb N\setminus I\in\Cal Z$.  (ii)
Show that for every $\epsilon>0$ there is an $I\subseteq\Bbb N$ such
that $d^*(I)=\epsilon$ but $d(J)=1$ whenever $J\supseteq I$ and $d(J)$
is defined.
%491A 491Xb
%(ii)  take $I=\bigcup_{n\in\Bbb N}\coint{(1-\epsilon)k_n,k_n}$

\spheader 491Xe Let $(X,\Sigma,\mu)$ be a probability space and
$\sequencen{E_n}$ a sequence in $\Sigma$.   For $x\in X$, set
$I_x=\{n:n\in\Bbb N$, $x\in E_n\}$.   Show that
$\int d^*(I_x)\mu(dx)\ge\liminf_{n\to\infty}\mu E_n$.
%491A

\spheader 491Xf Let $(X,\frak T,\Sigma,\mu)$ be a compact Radon
probability space.   Take any point $\infty$ not belonging to $X$, and
give $X\cup\{\infty\}$ the topology generated by
$\{G\cup\{\infty\}:G\in\frak T\}$.   Show that $X\cup\{\infty\}$ is
compact and that the image measure $\mu_{\infty}$
of $\mu$ under the identity map from $X$ to $X\cup\{\infty\}$ is a
quasi-Radon measure, inner regular with respect to the compact sets.
Show that if we set $x_i=\infty$ for every $i$, then $\sequence{i}{x_i}$
is equidistributed for $\mu_{\infty}$.
%491B

\sqheader 491Xg(i) Show that a sequence $\sequence{i}{t_i}$ in $[0,1]$ is
equidistributed (with respect to Lebesgue measure) iff
$\lim_{n\to\infty}\bover1{n+1}\penalty-100\#(\{i:i\le n$,
$t_i\le\beta\})=\beta$ for every $\beta\in[0,1]$.
(ii) Show that if $\alpha\in\Bbb R$ is irrational then the sequence
$\sequence{i}{\fraction{i\alpha}}$ of fractional parts of
multiples of $\alpha$ is equidistributed.  \Hint{281N.}
(iii) Show that a function $f:[0,1]\to\Bbb R$ is Riemann
integrable (134K) iff $\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(t_i)$ is
defined in $\Bbb R$ for every equidistributed sequence $\sequence{i}{t_i}$
in $[0,1]$.   (iv)\dvAnew{2024} Show that a
sequence $\sequence{i}{t_i}$ in $[0,1]$ is equidistributed iff
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(t_i)$ is defined and equal to
$\Rint_0^1 f$ for every Riemann integrable function $f:[0,1]\to\Bbb R$.
%491C

\sqheader 491X?\dvAnew{2024}  Let $X$ be a topological space, 
$\mu$ a probability
measure on $X$ measuring every zero set, and $\sequence{i}{x_i}$ an
equidistributed sequence in $X$.   Show that
$\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(x_i)$ is defined and equal to 
$\int f\,d\mu$ for every bounded $f:X\to\Bbb R$ which is continuous almost
everywhere.   (Cf.\ 134L.)
%491Xg 491Cb

\spheader 491Xh Show that the usual measure on the split interval
(419L) has an equidistributed sequence.

\spheader 491Xj Let $X$ be a metrizable space, and $\mu$ a quasi-Radon
probability measure on $X$.   (i) Show that there is an
equidistributed
sequence for $\mu$.   (ii) Show that if the
support of $\mu$ is not compact, and $\sequence{i}{x_i}$ is an
equidistributed sequence for $\mu$, then there is a continuous
integrable function $f:X\to\Bbb R$ such that
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\infty$.
%491E

\spheader 491Xk Let $\phi:\frak c\to\Cal P\Bbb N$ be an injective
function.   For each $n\in\Bbb N$ let $\lambda_n$ be the uniform
probability measure on $\Cal P(\Cal Pn)$, giving measure $2^{-2^n}$
to each singleton.
Define $\psi_n:\Cal P(\Cal Pn)\to\{0,1\}^{\frak c}$ by setting
$\psi_n(\Cal I)(\xi)=1$ if
$\phi(\xi)\cap n\in\Cal I$, $0$ otherwise, and let $\nu_n$ be the
image measure $\lambda_n\psi_n^{-1}$.   Show that $\nu_nE$ is the usual
measure of $E$ whenever
$E\subseteq\{0,1\}^{\frak c}$ is determined by coordinates in a finite
set on which the map $\xi\mapsto\phi(\xi)\cap n$ is injective.   Use
this with 491D to prove 491G.
%491G

\sqheader 491Xl(i) Let $Z$ be the Stone space of the measure algebra of
Lebesgue measure on $[0,1]$, with its usual measure.
Show that there is no equidistributed sequence in $Z$.   \Hint{meager
sets in $Z$ have negligible closures.}
(ii) Show that Dieudonn\'e's measure on $\omega_1$ (411Q) has no
equidistributed sequence.   (iii) Show that if $\#(I)>\frak c$ then the
usual measure on $\{0,1\}^I$ has no equidistributed sequence.   \Hint{if
$\sequence{i}{x_i}$ is any
sequence in $\{0,1\}^I$, there is an infinite $J\subseteq I$ such that
$\sequence{i}{x_i(\eta)}=\sequence{i}{x_i(\xi)}$ for all $\eta$, $\xi\in J$.}
(iv) Show that if $X$ is a topological group with a Haar probability
measure $\mu$, and $X$ is not separable, then $\mu$ has no
equidistributed sequence.   \Hint{use
443D to show that every separable subset is negligible.}
%491G

\spheader 491Xm Let $X$ be a compact Hausdorff abelian topological
group and $\mu$ its Haar probability measure.   Show that a sequence
$\sequence{i}{x_i}$ in $X$ is equidistributed for $\mu$ iff
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^n\chi(x_i)=0$ for every
non-trivial character $\chi:X\to S^1$.   \Hint{281G.}
%491H

\spheader 491Xn(i) Let $\sequencen{a_n}$ be a non-decreasing sequence
in $\frak Z=\Cal P\Bbb N/\Cal Z$.   Show that there is an $a\in\frak Z$
such that $a_n\Bsubseteq a$ for every $n\in\Bbb N$ and
$\bar d^*(a)=\sup_{n\in\Bbb N}\bar d^*(a_n)$.
(ii) Show that $\frak Z$ is not Dedekind $\sigma$-complete.
\Hint{393Bc}.
%491I  491Xc(i) n99527

\spheader 491Xo Let $\frak Z$, $\bar d^*$ and $D$ be as in 491K.   Show
that if $a\in D\setminus\{0\}$ and $\frak Z_a$ is the principal ideal of
$\frak Z$ generated by $a$, then $(\frak Z_a,\bar d^*\restrp\frak Z_a)$
is isomorphic, up to a scalar multiple of the submeasure, to
$(\frak Z,\bar d^*)$.
%491K 491Xb

\spheader 491Xp Let $(X,\Sigma,\mu)$ be a semi-finite measure space
and
$\frak T$ a topology on $X$.   Show that $\mu$ is effectively regular
iff whenever $E\in\Sigma$,
$\mu E<\infty$ and $\epsilon>0$ there are a measurable open set $G$
and
a measurable closed set $F\supseteq G$ such that $\mu(F\setminus
E)+\mu(E\setminus G)\le\epsilon$.
%491L
52
\spheader 491Xq Let $X$ be a normal topological space and $\mu$ a
topological measure on $X$ which is inner regular with respect to the
closed sets and effectively locally finite.
Show that $\mu$ is effectively regular.
%491M

\spheader 491Xr Let $X$ be a topological space and $\mu$ an
effectively
regular measure on $X$.   (i) Show that the completion and c.l.d.\
version of $\mu$ are also effectively regular.   (ii) Show that if
$Y\subseteq X$ then the subspace measure on $Y$ is again effectively regular.
(iii) Show that any totally
finite indefinite-integral measure over $\mu$ is effectively regular.
%491M

\spheader 491Xs(i) Let $X_1$, $X_2$ be topological spaces with
effectively regular measures $\mu_1$, $\mu_2$.   Show that the c.l.d.\
product measure on $X_1\times X_2$ is
effectively regular with respect to the product topology.   \Hint{412R.}
(ii) Let $\familyiI{X_i}$ be a family of topological spaces and $\mu_i$
an effectively regular probability
measure on $X_i$ for each $i$.   Show that the product probability
measure on $\prod_{i\in I}X_i$ is effectively regular.
%491M

\spheader 491Xt Give $[0,1]$ the topology $\frak T$ generated by the
usual topology and $\{[0,1]\setminus A:A\subseteq\Bbb Q\}$.   Let
$\mu_L$ be Lebesgue measure on $[0,1]$,
and $\Sigma$ its domain.   For $E\in\Sigma$ set
$\mu E=\mu_LE+\#(E\cap\Bbb Q)$ if $E\cap\Bbb Q$ is finite, $\infty$
otherwise.   Show that $\mu$ is a $\sigma$-finite
quasi-Radon measure with respect to the topology $\frak T$, but is not
effectively regular.
%491M

\spheader 491Xu Let $\frak A$ be a countable Boolean algebra and $\nu$ a
finitely additive functional on $\frak A$ such that $\nu 1=1$.   Show
that there is a Boolean
homomorphism $\pi:\frak A\to\Cal P\Bbb N$ such that $d(\pi a)$ is
defined and equal to $\nu a$ for every $a\in\frak A$ (i) using 491Xc
(ii) using 392H, 491P and 341Xc.
%491P

\spheader 491Xv Let $X$ be a dyadic space.   (i) Show that there is a
Radon probability measure on $X$ with support $X$.
(ii) Show that the following are equiveridical:  ($\alpha$)
$w(X)\le\frak c$;  ($\beta$) every Radon probability measure on $X$ has an
equidistributed sequence;  ($\gamma$) $X$ is separable.
\Hint{4A2Dd, 418L.}
%491Q

\spheader 491Xw Give an example of a Radon probability space
$(X,\mu)$ with a closed conegligible set $F\subseteq X$ such that
$\mu$ has an equidistributed sequence but the subspace measure
$\mu_F$ does not.   \Hint{the Stone space of the measure algebra of
Lebesgue measure embeds into $\{0,1\}^{\frak c}$.}   
%491R  query:  example for (ii) if \mu not eff reg

\spheader 491Xx Let $X$ be a topological space.   A sequence
$\sequencen{x_n}$ in $X$
is called {\bf statistically convergent} to $x\in X$ if
$d(\{i:x_i\in G\})=1$ for every open set $G$ containing $x$.
(i) Show that if
$X$ is first-countable then $\sequencen{x_n}$ is statistically convergent
to $x$ iff there is a set $I\subseteq\Bbb N$ such that $d(I)=1$ and
$\family{n}{I}{x_n}$ converges to $x$ in the ordinary sense that
$\{n:n\in I$, $x_n\notin G\}$ is finite for every open set $G$ containing
$x$.  (ii) Show that a bounded sequence $\sequencen{\alpha_n}$ in $\Bbb R$
is statistically convergent to $\alpha$ iff
$\lim_{n\to\infty}\Bover1n\sum_{i=0}^{n-1}|\alpha_i-\alpha|=0$.
%491+

\leader{491Y}{Further exercises (a)}
%\spheader 491Ya
Show that every subset $A$ of $\Bbb N$ is expressible in
the form $I_A\symmdiff J_A$ where $d(I_A)=d(J_A)=\bover12$ (i) by a
direct construction, with
$A\mapsto I_A$ a continuous function (ii) using 443D.
%491A

\spheader 491Yt\dvAnew{2021} (cf.\ {\smc Bergelson 85})
Let $\frak A$ be a Boolean algebra, $A\subseteq\frak A\setminus\{0\}$ a
non-empty set and $\alpha\in[0,1]$.
Show that the following are equiveridical:  (i) there is a
finitely additive functional $\nu:\frak A\to[0,1]$ such that $\nu a\ge\alpha$
for every $a\in A$ (ii) for every sequence $\sequencen{a_n}$ in $A$ there is a
set $I\subseteq\Bbb N$ such that $d^*(I)\ge\alpha$ and
$\inf_{i\in I\cap n}a_i\ne 0$ for every $n\in\Bbb N$.
%mt49bits  491Xe 491A  out of order query

\spheader 491Yb Let $\frak A$ be a Boolean algebra, and
$\nu:\frak A\to[0,\infty]$ a submeasure.   Show that $\nu$ is uniformly
exhaustive iff whenever $\sequencen{a_n}$ is a sequence in $\frak A$ such
that $\inf_{n\in\Bbb N}\nu a_n>0$, there is a set $I\subseteq\Bbb N$ such
that $d^*(I)>0$ and $\inf_{i\in I\cap n}a_i\ne 0$ for every $n\in\Bbb N$.
%491Xe 491A 491Yt 392E

\sqheader 491Yu\dvAformerly{4{}91Xi} Show that if $X$ is a Hausdorff space and
$f:\Bbb N\to X$ is injective, then there is an open set
$G\subseteq X$ such that $f^{-1}[G]$ does not have asymptotic density.
%491C mt49bits

\spheader 491Yc
Find a topological space $X$ with a $\tau$-additive probability
measure
$\mu$ on $X$, a sequence $\sequence{i}{x_i}$ in $X$ and a base $\Cal G$
for the topology of
$X$, consisting of measurable sets and closed under finite
intersections, such that
$\mu G\le\liminf_{n\to\infty}\bover1{n+1}\#(\{i:i\le n$, $x_i\in G\})$
for every $G\in\Cal G$ but $\sequence{i}{x_i}$ is not equidistributed.
\Hint{take $\#(X)=4$.}
%491C mt49bits

\spheader 491Yd Let $X$ be a compact Hausdorff space on which every
Radon probability measure has an equidistributed sequence.   Show that
the cylindrical $\sigma$-algebra of
$C(X)$ is the $\sigma$-algebra generated by sets of the form
$\{f:f\in C(X)$, $f(x)>\alpha\}$ where $x\in X$ and $\alpha\in\Bbb R$.
\Hint{436J, 491Cb.}
%491C

\spheader 491Ye Give $\omega_1+1$ and $[0,1]$ their usual compact
Hausdorff topologies.   Let $\sequence{i}{t_i}$ be a sequence in
$[0,1]$
which is equidistributed for
Lebesgue measure $\mu_L$, and set $Q=\{t_i:i\in\Bbb N\}$,
$X=(\omega_1\times([0,1]\Bsetminus Q))\cup(\{\omega_1\}\times Q)$,
with the subspace topology inherited from $(\omega_1+1)\times[0,1]$.
(i) Set $F=\{\omega_1\}\times Q$.   Show that
$F$ is a closed Baire set in the completely regular Hausdorff space
$X$.
(ii) Show that if $f\in C_b(X)$ then there are a $g_f\in C([0,1])$
and a
$\zeta<\omega_1$ such that
such that $f(\xi,t)=g_f(t)$ whenever $(\xi,t)\in X$ and
$\zeta\le\xi\le\omega_1$.   (iii) Show that there is a Baire measure
$\mu$ on $X$ such that $\int fd\mu=\int g_fd\mu_L$ for every
$f\in C_b(X)$.   (iv) Show that $\mu F=0$.   (v) Show that
$\int fd\mu=\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(\omega_1,t_i)$
for every $f\in C_b(X)$, but that $\sequence{i}{(\omega_1,t_i)}$ is not
equidistributed with respect to $\mu$.
%491C

\spheader 491Yf Let $(X,\frak T,\Sigma,\mu)$ be a topological measure
space.   Let $\Cal E$ be the Jordan algebra of $X$ (411Yc).
(i) Suppose that $\mu$ is a complete probability measure on $X$ and
$\sequence{i}{x_i}$ an equidistributed sequence in $X$.
Show that the asymptotic density $d(\{i:x_i\in E\})$ is defined and
equal to $\mu E$ for every $E\in\Cal E$.  (ii) Suppose that $\mu$ is a
probability measure on $X$
and that $\sequence{i}{x_i}$ is a sequence in $X$ such that
$d(\{i:x_i\in E\})$ is defined and equal to $\mu E$ for every
$E\in\Cal E$.   Show that
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$ for every
$f\in C_b(X)$.
%491C

\spheader 491Yg Show that a sequence $\sequence{i}{x_i}$ in $[0,1]$
is equidistributed for Lebesgue measure iff there is some $r_0\in\Bbb N$
such that $\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nx_i^r=\bover1{r+1}$
for every $r\ge r_0$.
%491C

\spheader 491Yh Let $Z$, $\mu$, $X=Z\times\{0,1\}$ and $\nu$ be as
described in 439K, so that $\mu$ is a Radon probability
measure on the compact metrizable space $Z$, $X$ has a compact
Hausdorff topology finer than the product topology,
and $\nu$ is a measure on $Z$ extending $\mu$.   (i) Show that if
$f\in C(X)$, then
$\{t:t\in Z$, $f(t,0)\ne f(t,1)\}$ is countable.   \Hint{4A2F(h-vii).}
(ii) Show that $\int f(t,0)\mu(dt)=\int f(t,1)\nu(dt)$ for every
$f\in C(X)$.
(iii) Let $\lambda$ be the measure $\nu g^{-1}$ on $Z$, where
$g(t,0)=g(t,1)=(t,1)$ for $t\in Z$.   Show that there is a
sequence $\sequence{i}{x_i}$ in $Z\times\{0\}$ such that
$\int fd\lambda=\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)$ for
every $f\in C(X)$, but that $\sequence{i}{x_i}$ is not
$\lambda$-equidistributed.
%491C

\spheader 491Yi(i)\dvAformerly{4{}91Yg} Show that a Radon
probability measure on an extremally disconnected
compact Hausdorff space has an equidistributed sequence iff it is
point-supported.   \Hint{see the hint for 326Yo.}
(ii) Show that there is a separable compact Hausdorff space with a Radon
probability measure which has no equidistributed sequence.
%491G
%(ii) Stone space of regular open algebra of \Bbb R

\spheader 491Yj Show that there is a countable dense set
$D\subseteq[0,1]^{\frakc}$ such that no sequence in $D$ is
equidistributed for the usual measure on $[0,1]^{\frakc}$.
%491G 49bits

\spheader 491Yk Let $\frak Z=\Cal P\Bbb N/\Cal Z$ and
$\bar d^*:\frak Z\to[0,1]$ be as in 491I.   Show that $\bar d^*$ is
{\bf order-continuous on the left} in the sense that whenever
$A\subseteq\frak Z$ is non-empty and upwards-directed and has a supremum
$c\in\frak Z$, then $\bar d^*(c)=\sup_{a\in A}\bar d^*(a)$.
%491K n99527

\spheader 491Yl(i) Show that $\frak Z$ is \wsid.
(ii) Show that $\frak Z\cong\frak Z^{\Bbb N}$.
(iii) Show that $\frak Z$ has the $\sigma$-interpolation property, but is
not Dedekind $\sigma$-complete.
(iv) Show that $\frak Z$ has many involutions in the sense of 382O.
%491K 491Yk n99527

\spheader 491Ym Let $(X,\rho)$ be a separable metric space and $\mu$ a
Borel probability measure on $X$.   (i) Show that there is an
equidistributed sequence in $X$.
(ii) Show that if $\sequence{i}{x_i}$ is an equidistributed sequence in
$X$, and $\sequence{i}{y_i}$ is a sequence in $X$ such that
$\lim_{i\to\infty}\rho(x_i,y_i)=0$, then
$\sequence{i}{y_i}$ is equidistributed.
(iii) Show that if $f:X\to\Bbb R$ is a bounded function, then
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)-f(y_i)=0$ for all
equidistributed
sequences $\sequence{i}{x_i}$, $\sequence{i}{y_i}$ in $X$ iff
$\{x:f$ is continuous at $x\}$ is conegligible, and in this case
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nf(x_i)=\int fd\mu$
for every equidistributed sequence $\sequence{i}{x_i}$ in $X$.
\Hint{$\{x:f$ is continuous at $x\}=\bigcap_{m\in\Bbb N}G_m$, where
$G_m=\bigcup\{G:G\subseteq X$ is open,
$\sup_{x,y\in G}|f(x)-f(y)|\le 2^{-m}\}$.}   Compare 491Xg.
%491M

\spheader 491Yn Let $(X,\frak T)$ be a topological space, $\mu$ a
probability measure on $X$, and $\phi:X\to X$ an \imp\ function.
(i) Suppose that $\frak T$ has a countable network
consisting of measurable sets, and that $\phi$ is ergodic.   Show that
$\sequencen{\phi^n(x)}$ is equidistributed for almost every $x\in X$.
\Hint{372Qb.}   (ii)
Suppose that $\mu$ is {\it either} inner regular with respect to the
closed sets {\it or}
effectively regular, and that $\{x:\sequencen{\phi^n(x)}$ is
equidistributed$\}$ is not negligible.   Show that $\phi$ is ergodic.
%491M 491Xe

\spheader 491Yo Let $\ofamily{\xi}{\frak c}{X_{\xi}}$ be a family of
topological spaces with countable networks consisting of Borel sets, and
$\mu$ a $\tau$-additive topological probability measure on
$X=\prod_{\xi<\frak c}X_{\xi}$.   Show that $\mu$ has an equidistributed
sequence.
%491Q
%basic facts needed:  (i) if $\frak C\subseteq\frak Z$ a countable
%subalgebra, it has a lifting to a subalgebra of $\Cal P\Bbb N$ (ii)
%if $\Cal E$ a countable
% algebra of subsets of $X$, $\pi:\Cal E\to\Cal P\Bbb N$ an injective
%Boolean homomorphism, we can find $\sequence{i}{x_i}$ in $X$ such that
%$\{i:x_i\in\pi E\}\symmdiff E$
% is finite for every $E\in\Cal E$

\spheader 491Yp(i)
Show that there is a family $\ofamily{\xi}{\frakc}{a_{\xi}}$ in
$\frak Z$ such that
$\inf_{\xi\in I}a_{\xi}=0$ and $\sup_{\xi\in I}a_{\xi}=1$ for every
infinite $I\subseteq\frak c$.   (ii) Show that if
$B\subseteq\frak Z\setminus\{0\}$ has cardinal less than $\frak c$
then
there is an $a\in\frak Z$ such that $b\Bcap a$ and $b\Bsetminus a$ are
non-zero for every $b\in B$.
%491P

\spheader 491Yq Let $(X,\frak T,\Sigma,\mu)$ be a $\tau$-additive
topological probability space.   A sequence $\sequence{i}{x_i}$ in $X$ is
{\bf completely equidistributed} if, for every $r\ge 1$, the sequence
$\sequencen{\ofamily{i}{r}{x_{n+i}}}$ is equidistributed for some
(therefore any) $\tau$-additive extension of the c.l.d.\ product measure
$\mu^r$ on $X^r$.   (i) Show that if there is an equidistributed sequence
in $X$, then there is a completely equidistributed sequence in $X$.   (ii)
Show that if $\frak T$ is second-countable, then $\mu^{\Bbb N}$-almost
every sequence in $X$ is completely equidistributed.   (iii) Show that
if $X$ has two disjoint open sets of non-zero measure, then no sequence
which is well-distributed in the sense of 281Ym can
be completely equidistributed.

\spheader 491Yr Suppose,
in 491O, that $\mu$ is a topological measure.   Show
that $T_{\pi}f^{\ssbullet}\le RSf$ for every bounded lower semi-continuous
$f:X\to\Bbb R$.
%491O

\spheader 491Ys (M.Elekes)
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure
space and $\sequencen{E_n}$ a sequence in $\Sigma$ such that
$\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m$ is conegligible.
Show that there is an $I\in\Cal Z$ such that
$\bigcap_{n\in\Bbb N}\bigcup_{m\in I\setminus n}E_m$ is conegligible.
%491A out of order query
}%end of exercises

\leader{491Z}{Problem} It is known that for almost every $x>1$ the
sequence $\sequence{i}{\fraction{x^i}}$ of fractional parts of powers of $x$ is
equidistributed for Lebesgue measure on
$[0,1]$\cmmnt{ ({\smc Kuipers \& Niederreiter 74}, p.\ 35)}.   But is
$\sequencen{\fraction{(\bover32)^n}}$ equidistributed?

% question:  must a Radon probability measure on a hereditarily
% separable space have an equidistributed sequence?

% possibly useful fact:  if $\mu$ is inner regular for $\{E:\mu_E$ has
% equidistr seq for normalized subspace measure$\}$, then $\mu$ has
% equidistributed sequence

\endnotes{
\Notesheader{491} The notations $d^*$, $d$ (491A) are standard, and
usefully suggestive.   But coming from measure theory we have to
remember that $d^*$, although a submeasure,
is not an outer measure, the domain of $d$ is not an algebra of sets
(491Xa), and $d$ and $d^*$ are related by only one of the formulae we
expect to connect a
measure with an outer measure (491Ac, 491Xd).   In 491C, the limit
$\lim_{n\to\infty}\bover1n\sum_{i=0}^nf(x_i)$, when it is defined,
is the {\bf Ces\`aro mean} of the sequence $\sequence{i}{f(x_i)}$.
The delicacy of the arguments here arises from the fact that the family
of (bounded) sequences with Ces\`aro means, although a norm-closed linear
subspace of $\ell^{\infty}$, is neither a sublattice nor a subalgebra.
When we turn to the quotient algebra $\frak Z=\Cal P\Bbb N/\Cal Z$, we
find ourselves with a natural submeasure to which we can apply ideas
from \S392 to good effect (491I;  see also 491Yk and 491Yl).
What is striking is that
equidistributed sequences induce regular embeddings of measure algebras
in $\frak Z$ which can be thought of as measure-preserving (491N).

Most authors have been content to define an `equidistributed sequence'
to be one such that the integrals of bounded continuous functions are
correctly specified (491Cf, 491Cg);
that is, that the point-supported measures
$\bover1{n+1}\sum_{i=0}^n\delta_{x_i}$ converge to $\mu$ in the vague
topology on an appropriate class of measures (437J).
I am going outside this territory in order to cover some ideas I find
interesting.   491Yh shows that it makes a difference;  there are Borel
measures on compact Hausdorff spaces
which have sequences which give the correct Ces\`aro means for continuous
functions, but lie within negligible closed sets;  and the same can
happen with Baire measures (491Ye).
It seems to be difficult, in general, to determine whether
a topological probability space -- even a compact Radon probability
space -- has an equidistributed sequence.   In the proofs of
491D-491G %491D 491E 491F 491H 491G
I have tried to collect the principal techniques for showing that spaces
do have equidistributed sequences.   In the other direction, it is
obviously impossible for a space to
have an equidistributed sequence if every separable subspace is
negligible (491Xl).   For an example of a separable compact Hausdorff
space with a Radon measure which does not have
an equidistributed sequence, we seem to have to go deeper (491Yi).

491Z is a famous problem.   It is not clear that it is a problem in
measure theory, and there is no reason to suppose that any of the ideas
of this treatise beyond 491Xg(i) are
relevant.   I mention it because I think everyone should know that it
is there.
}%end of notes

\discrpage

\frfilename{mt492.tex}
\versiondate{30.5.16}
\copyrightdate{2001}

\def\chaptername{Further topics}
\def\sectionname{Combinatorial concentration of measure}

\newsection{492}

`Concentration of measure' takes its most dramatic forms in the
geometrically defined notions of concentration explored in \S476.   
But the phenomenon is observable in many other contexts, if
we can devise the right abstract geometries to capture it.   In this
section I present one of Talagrand's theorems on the concentration of
measure in product spaces, using the Hamming metric (492D), and Maurey's
theorem on concentration of measure in permutation groups (492H).

\leader{492A}{Lemma} Let $(X,\Sigma,\mu)$ be a totally finite measure
space, $\alpha<\beta$ in $\Bbb R$, $\phi:[\alpha,\beta]\to\Bbb R$ a
convex function, and $f:X\to[\alpha,\beta]$ a $\Sigma$-measurable
function.   Then

\Centerline{$\biggerint\phi(f(x))\mu(dx)
\le\Bover{\phi(\beta)-\phi(\alpha)}{\beta-\alpha}\int fd\mu
  +\Bover{\beta\phi(\alpha)-\alpha\phi(\beta)}{\beta-\alpha}\mu X$.}

\proof{ %Solecki
If $t\in[\alpha,\beta]$ then
$t=\Bover{t-\alpha}{\beta-\alpha}\beta
+\Bover{\beta-t}{\beta-\alpha}\alpha$, so

\Centerline{$\phi(t)
\le\Bover{t-\alpha}{\beta-\alpha}\phi(\beta)
+\Bover{\beta-t}{\beta-\alpha}\phi(\alpha)
=\Bover{\phi(\beta)-\phi(\alpha)}{\beta-\alpha}t
+\Bover{\beta\phi(\alpha)-\alpha\phi(\beta)}{\beta-\alpha}$.}

\noindent Accordingly

\Centerline{$\phi(f(x))
\le\Bover{\phi(\beta)-\phi(\alpha)}{\beta-\alpha}f(x)
+\Bover{\beta\phi(\alpha)-\alpha\phi(\beta)}{\beta-\alpha}$}

\noindent for every $x\in X$;  integrating with respect to $x$, we have the result.
}%end of proof of 492A

\leader{492B}{Corollary} Let $(X,\Sigma,\mu)$ be a probability space
and $f:X\to[\alpha,1]$ a measurable function, where $0<\alpha\le 1$.
Then $\biggerint\Bover1fd\mu\cdot\int fd\mu\le\Bover{(1+\alpha)^2}{4\alpha}$.

\proof{ Set $\gamma=\int fd\mu$, so that $\alpha\le\gamma\le 1$.   By
492A, with $\phi(t)=\Bover1t$,

\Centerline{$\biggerint\Bover1fd\mu
\le\Bover{\gamma}{1-\alpha}(1-\Bover1{\alpha})
   +\Bover1{1-\alpha}(\Bover1{\alpha}-\alpha)
=\Bover{1+\alpha-\gamma}{\alpha}$.}

\noindent Now $\Bover{1+\alpha-\gamma}{\alpha}\cdot\gamma$ takes its
maximum value $\Bover{(1+\alpha)^2}{4\alpha}$ when
$\gamma=\Bover{1+\alpha}2$, so this is also a bound for $\int\bover1f\int f$.
}%end of proof of 492B

\leader{492C}{Lemma} $\Bover12(1+\cosh t)\le e^{t^2/4}$ for every $t\in\Bbb R$.

\proof{ For $k\ge 1$, $4^kk!\le 2(2k)!$ (induce on $k$), so

\Centerline{$1+\cosh t
=2+\sum_{k=1}^{\infty}\Bover{t^{2k}}{(2k)!}
\le 2+\sum_{k=1}^{\infty}\Bover{2t^{2k}}{4^kk!}
=2e^{t^2/4}$.}
}%end of proof of 492C

\leader{492D}{Theorem}\cmmnt{ ({\smc Talagrand 95})} Let
$\ofamily{i}{n}{(X_i,\Sigma_i,\mu_i)}$ be a non-empty finite family of
probability
spaces with product $(X,\Lambda,\lambda)$.   Let $\rho$ be the {\bf
normalized Hamming metric} on
$X$ defined by setting $\rho(x,y)=\Bover1n\#(\{i:i<n,\,x(i)\ne y(i)\})$
for $x$, $y\in X$.   If $W\in\Lambda$ and $\lambda W>0$, then

\Centerline{$\overlineint e^{t\rho(x,W)}\lambda(dx)
\le\Bover1{\lambda W}e^{t^2/4n}$}

\noindent for every $t\ge 0$.

\proof{ The formulae below will go much more smoothly if we work with
the simple Hamming metric $\sigma(x,y)=\#(\{i:x(i)\ne y(i)\})$ instead of
$\rho$.   In this case, we can make sense of the case $n=0$, and this
will be useful.   In terms of $\sigma$, our target is to prove that
if $W\in\Lambda$ and $\lambda W>0$, then

\Centerline{$\overlineint e^{t\sigma(x,W)}\lambda(dx)
\le\Bover1{\lambda W}e^{nt^2/4}$}

\noindent for every $t\ge 0$.


\medskip

{\bf (a)} To begin with, suppose that every
$X_i=Z=\{0,1\}^{\Bbb N}$, every $\mu_i$ is a Borel measure, and $W$ is
compact.   Note that in this case $\lambda$ is a Radon measure (because
the $X_i$ are compact and metrizable), and

\Centerline{$\{x:\sigma(x,W)\le m\}
=\bigcup_{I\subseteq n,\#(I)\le m}\{x:\,\Exists y\in W,\,
   x\restr n\setminus I=y\restr n\setminus I\}$}

\noindent is compact for every $m$, so the function
$x\mapsto\sigma(x,W)$ is measurable.

Induce on $n$.   If $n=0$ we must have $W=X=\{\emptyset\}$ and
$\sigma(x,W)=0$ for every $x$, so the result is trivial.   For the
inductive step to $n+1$, we have
$W\subseteq X\times X_n$, where $X=\prod_{i<n}X_i$, and we are looking
at $\iint e^{t\sigma((x,\xi),W)}\lambda(dx)\mu_n(d\xi)$.   Now, setting
$V_{\xi}=\{x:(x,\xi)\in W\}$ for $\xi\in X_n$,

\Centerline{$V=\bigcup_{\xi\in X_n}V_{\xi}
=\{x:\,\Exists \xi\in X_n,\,(x,\xi)\in W\}$,}

\noindent we have

\Centerline{$\sigma((x,\xi),W)\le\min(\sigma(x,V_{\xi}),1+\sigma(x,V))$}

\noindent for all $x$ and $\xi$, counting $\sigma(x,\emptyset)$ as $\infty$
if $V_{\xi}$ is empty.   So, for any $\xi\in X_n$,

$$\eqalign{\int e^{t\sigma((x,\xi),W)}\lambda(dx)
&\le\min(\int e^{t\sigma(x,V_{\xi})}\lambda(dx),
  e^t\int e^{t\sigma(x,V)}\lambda(dx))\cr
&\le e^{nt^2/4}
  \min(\Bover1{\lambda V_{\xi}},\Bover{e^t}{\lambda V})\cr}$$

\noindent by the inductive hypothesis, counting
$\min(\Bover10,\Bover{e^t}{\lambda V})$ as $\Bover{e^t}{\lambda V}$.

It follows that if we set
$f(\xi)=\max(e^{-t},\Bover{\lambda V_{\xi}}{\lambda V})$ for
$\xi\in X_n$,

$$\eqalignno{(\lambda\times\mu_n)(W)
  &\cdot\int e^{t\sigma((x,\xi),W)}\lambda(dx)\mu_n(d\xi)\cr
&\le\int\lambda V_{\xi}\mu_n(d\xi)\cdot e^{nt^2/4}
\int\min(\Bover1{\lambda V_{\xi}},\Bover{e^t}{\lambda V})\mu_n(d\xi)\cr
&=e^{nt^2/4}\int\Bover{\lambda V_{\xi}}{\lambda V}\mu_n(d\xi)
  \cdot\int\min(e^t,\Bover{\lambda V}{\lambda V_{\xi}})\mu_n(d\xi)\cr
&\le e^{nt^2/4}\int f(\xi)\mu_n(d\xi)
  \cdot\int\Bover1{f(\xi)}\mu_n(d\xi)\cr
&\le e^{nt^2/4}\cdot\Bover{(1+e^{-t})^2}{4e^{-t}}\cr
\displaycause{492B}
&=\Bover12 e^{nt^2/4}(1+\cosh t)
\le e^{(n+1)t^2/4}\cr}$$

\noindent by 492C, and

\Centerline{$\biggerint e^{t\sigma((x,\xi),W)}\lambda(dx)\mu_n(d\xi)
\le\Bover1{(\lambda\times\mu_n)(W)}e^{(n+1)t^2/4}$,}

\noindent so the induction continues.

\medskip

{\bf (b)} Now turn to the general case.   If $W\in\Lambda$, there is a
$W_1\subseteq W$ such that $W_1\in\Tensorhat_{i<n}\Sigma_i$ and
$\lambda W_1=\lambda W$ (251Wf).   There must be countably-generated
$\sigma$-subalgebras $\Sigma'_i$ of $\Sigma_i$ such that
$W_1\in\Tensorhat_{i<n}\Sigma'_i$.   For each $i<n$, let
$\sequence{k}{E_{ik}}$ be a sequence in $\Sigma_i$ generating
$\Sigma'_i$, and let $h_i:X_i\to Z$ be the corresponding
Marczewski functional, so that $h_i(\xi)=\sequence{k}{\chi E_{ik}(\xi)}$
for $\xi\in X_i$.   Let $\mu'_i$ be the Borel measure on
$Z$ defined by setting $\mu'_iF=\mu_ih_i^{-1}[F]$ for
every Borel set $F\subseteq Z$, and let $\nu$ be the
product of the measures $\mu'_i$ on $Y=Z^n$.   If we
set $h(x)=\ofamily{i}{n}{h_i(x(i))}$ for $x\in X$, then $h:X\to Y$ is
\imp\ for $\lambda$ and $\nu$ (254H).   Moreover, by the choice of the
$E_{ik}$, $W_1=h^{-1}[V]$ for some Borel set $V\subseteq Y$.

Because $Y$ is a compact metrizable space, $\nu$ is the completion of a
Borel measure and is a Radon measure (433Cb).   For each $I\subseteq n$,
write $\nu_I$ for the product measure on $Z^I$, and set

\Centerline{$V_I=\{u:u\in Z^{n\setminus I},\,
  \nu_I\{v:v\in Z^I,\,(u,v)\in V\}>0\}$,}

\Centerline{$V_I'=\{y:y\in Y,\,y\restr n\setminus I\in V_I\}$.}

\noindent Then $\nu(V\setminus V'_I)=0$ for every $I\subseteq n$, so if
we set $V'=\bigcap_{I\subseteq n}V'_I$ then $\nu V'=\nu V$.   (Of course
$V'\subseteq V'_{\emptyset}=V$.)

Take any $\gamma\in\ooint{0,\lambda W}=\ooint{0,\nu V'}$.
Let $K\subseteq V'$ be a compact set such that $\nu K\ge\gamma$.   Set
$g(y)=e^{t\sigma(y,K)}$ for $y\in Y$, where I write $\sigma$ for the
Hamming metric on $Y$ (regarded as a product of $n$ factor spaces).
Then $g:Y\to\Bbb R$
is Borel measurable and $gh:X\to\Bbb R$ is $\Lambda$-measurable.   Also,
for any $x\in X$, $\sigma(x,W)\le\sigma(h(x),K)$.   \Prf\ Take $y\in K$
such
that $\sigma(h(x),y)=\sigma(h(x),K)$, and set

\Centerline{$I=\{i:h(x)(i)\ne y(i)\}$,
\quad$u=h(x)\restr n\setminus I=y\restr n\setminus I$.}

\noindent Because $y\in V'$, $u\in V_I$ and $\nu_IH>0$, where
$H=\{v:v\in Z^I,\,(u,v)\in V\}$.   But if we write $\lambda_I$ for
the product measure on $\prod_{i\in I}X_i$, and
$h_I(z)=\familyiI{h_i(z(i))}$ for $z\in\prod_{i\in I}X_i$, then $h_I$ is
\imp\ for $\lambda_I$ and $\nu_I$;  in particular, $h_I^{-1}[H]$ is
non-empty.   This means that we can find an $x'\in X$ such that
$x'\restr n\setminus I=x\restr n\setminus I$ and
$x'\restr I\in h_I^{-1}[H]$.   In this case, $h(x')\in V$, so
$x'\in W_1\subseteq W$, and

\Centerline{$\sigma(x,W)\le\sigma(x,x')\le\#(I)=\sigma(h(x),K)$.  \Qed}

Accordingly

\Centerline{$e^{t\sigma(x,W)}\le e^{t\sigma(h(x),K)}=g(h(x))$}

\noindent for every $x\in X$, and

$$\eqalignno{\overline{\int}e^{t\sigma(x,W)}\lambda(dx)
&\le\int gh\,d\lambda
=\int g\,d\nu\cr
\displaycause{because $g$ is $\nu$-integrable and $h$ is \imp, see 235G}
&\le\Bover1{\nu K}e^{nt^2/4}\cr
\displaycause{by (a)}
&\le\Bover1{\gamma}e^{nt^2/4}.\cr}$$

\noindent As $\gamma$ is arbitrary,

\Centerline{$\overlineint e^{t\sigma(x,W)}\lambda(dx)
\le\Bover1{\lambda W}e^{nt^2/4}$,}

\noindent as claimed.
}%end of proof of 492D

\leader{492E}{Corollary} %\cmmnt{ ({\smc Talagrand 95})}
Let $\ofamily{i}{n}{(X_i,\Sigma_i,\mu_i)}$ be a non-empty finite family of
probability spaces with product $(X,\Lambda,\lambda)$.

(a) Let $\rho$ be the normalized Hamming metric on $X$.   If $W\in\Lambda$ and $\lambda W>0$, then

\Centerline{$\lambda^*\{x:\rho(x,W)\ge\gamma\}
  \le\Bover1{\lambda W}e^{-n\gamma^2}$}

\noindent for every $\gamma\ge 0$.

(b) If $W$, $W'\in\Lambda$ and $\gamma>0$ are such that
$e^{-n\gamma^2}<\lambda W\cdot\lambda W'$ then there are $x\in W$,
$x'\in W'$ such that $\#(\{i:i<n,\,x(i)\ne x'(i)\})<n\gamma$.

\proof{{\bf (a)} Set $t=2n\gamma$.   By 492D, there is a measurable
function $f:X\to\Bbb R$ such that $f(x)\ge e^{t\rho(x,W)}$ for every
$x\in X$ and $\int fd\lambda\le\Bover1{\lambda W}e^{t^2/4n}$.   So

$$\eqalign{\lambda^*\{x:\rho(x,W)\ge\gamma\}
&\le\lambda\{x:f(x)\ge e^{t\gamma}\}
\le e^{-t\gamma}\int fd\lambda\cr
&\le\Bover1{\lambda W}e^{-t\gamma+t^2/4n}
=\Bover1{\lambda W}e^{-n\gamma^2}.\cr}$$

\medskip

{\bf (b)} By (a), $\lambda^*\{x:\rho(x,W')\ge\gamma\}<\lambda W$, so
there must be an $x\in W$ such that $\rho(x,W')<\gamma$.
}%end of proof of 492E

\leader{492F}{}\cmmnt{ The next theorem concerns concentration of
measure in permutation groups.   I approach this through a general
result about slowly-varying martingales (492G).

\medskip

\noindent}{\bf Lemma} $e^t\le t+e^{t^2}$ for every $t\in\Bbb R$.

\proof{ If $t\ge 1$ then $t\le t^2$ so

\Centerline{$e^t\le e^{t^2}\le t+e^{t^2}$.}

\noindent If $0\le t\le 1$ then

$$\eqalign{e^t
&=1+t+\sum_{n=2}^{\infty}\Bover{t^n}{n!}
\le 1+t
  +\sum_{k=1}^{\infty}(\Bover{1}{(2k)!}+\Bover{1}{(2k+1)!})t^{2k}\cr
&\le 1+t
  +\sum_{k=1}^{\infty}\Bover{t^{2k}}{k!}
=t+e^{t^2}.\cr}$$

\noindent If $t\le 0$ then

$$\eqalign{e^t
&=1+t+\sum_{n=2}^{\infty}\Bover{t^n}{n!}
\le 1+t+\sum_{k=1}^{\infty}\Bover{t^{2k}}{(2k)!}\cr
&\le 1+t+\sum_{k=1}^{\infty}\Bover{t^{2k}}{k!}
=t+e^{t^2}.\cr}$$
}%end of proof of 492F

\leader{492G}{Lemma}\cmmnt{ ({\smc Milman \& Schechtman 86})} Let
$(X,\Sigma,\mu)$ be a probability space, and
$\sequencen{f_n}$ a martingale on $X$.
Suppose that $f_n\in\eusm L^{\infty}(\mu)$ for every $n$, and that
$\alpha_n\ge\esssup|f_n-f_{n-1}|$ for $n\ge 1$.   Then for any
$n\ge 1$ and $\gamma\ge 0$,

\Centerline{$\Pr(f_n-f_0\ge\gamma)
\le\exp(-\gamma^2/4\sum_{i=1}^n\alpha_i^2)$,}

\noindent at least if $\sum_{i=1}^n\alpha_i^2>0$.

\proof{ Let $\sequencen{\Sigma_n}$ be a non-decreasing
sequence of $\sigma$-subalgebras of $\Sigma$ to which $\sequencen{f_n}$
is adapted.   

\medskip

{\bf (a)} I show first that

\Centerline{$\Expn(\exp(\lambda(f_n-f_0)))
\le\exp(\lambda^2\sum_{i=1}^n\alpha_i^2)$}

\noindent for any $n\ge 0$ and any $\lambda>0$.   \Prf\ Induce on $n$.
For $n=0$, interpreting $\sum_{i=1}^0$ as $0$, this is trivial.   For
the inductive step to $n+1$, set $g=f_n-f_{n-1}$ and let $g_1$, $g_2$ be
conditional expectations of $\exp(\lambda g)$ and $\exp(\lambda^2g^2)$
on $\Sigma_{n-1}$.   Because $|g|\le\alpha_n$ a.e.,
$\exp(\lambda^2g^2)\le\exp(\lambda^2\alpha_n^2)$ a.e.\ and
$g_2\le\exp(\lambda^2\alpha_n^2)$ a.e.   Because
$\exp(\lambda g)\le\lambda g+\exp(\lambda^2g^2)$ wherever $g$ is defined
(492F), and $0$ is a conditional expectation of $g$ on $\Sigma_{n-1}$,
$g_1\le g_2\le\exp(\lambda^2\alpha_n^2)$ a.e.

Now observe that $f_{n-1}-f_0$ is $\Sigma_{n-1}$-measurable, so that
$\exp(\lambda(f_{n-1}-f_0))\times g_1$ is a conditional expectation of
$\exp(\lambda(f_{n-1}-f_0))\times\exp(\lambda g)\eae\exp(\lambda(f_n-f_0))$
on $\Sigma_{n-1}$ (233Eg).   Accordingly

$$\eqalignno{\Expn(\exp(\lambda(f_n-f_0)))
&=\Expn(\exp(\lambda(f_{n-1}-f_0))\times g_1)\cr
&\le\esssup|g_1|\cdot\Expn(\exp(\lambda(f_{n-1}-f_0)))\cr
&\le\exp(\lambda^2\alpha_n^2)
  \exp(\lambda^2\sum_{i=1}^{n-1}\alpha_i^2)\cr
\displaycause{by the inductive hypothesis}
&=\exp(\lambda^2\sum_{i=1}^n\alpha_i^2)\cr}$$

\noindent and the induction continues.\ \Qed

\medskip

{\bf (b)} Now take $n\ge 1$ such that $\sum_{i=1}^n\alpha_i^2>0$
and $\gamma\ge 0$.
Set $\lambda=\gamma/2\sum_{i=1}^n\alpha_i^2$.   Then

$$\eqalignno{\Pr(f_n-f_0\ge\gamma)
&=\Pr(\exp(\lambda(f_n-f_0))\ge e^{\lambda\gamma})\cr
&\le e^{-\lambda\gamma}\Expn(\exp(\lambda(f_n-f_0)))
\le e^{-\lambda\gamma}\exp(\lambda^2\sum_{i=1}^n\alpha_i^2)\cr
\displaycause{by (a) above}
&=e^{-\lambda\gamma/2}
=\exp(-\gamma^2/4\sum_{i=1}^n\alpha_i^2)\cr}$$

\noindent as claimed.
}%end of proof of 492G

\leader{492H}{Theorem}\cmmnt{ ({\smc Maurey 79})} Let $X$ be a
non-empty finite set and $G$ the
group of all permutations of $X$ with its discrete topology.   For
$\pi$, $\phi\in G$ set

\Centerline{$\rho(\pi,\phi)
=\Bover{\#(\{x:x\in X,\,\pi(x)\ne\phi(x)\})}{\#(X)}$.}

\noindent Then $\rho$ is a metric on $G$.   Give $G$ its Haar
probability measure, and let $f:G\to\Bbb R$ be a $1$-Lipschitz function.
Then

\Centerline{$\Pr(f-\Expn(f)\ge\gamma)\le\exp(-\Bover{\gamma^2\#(X)}{16})$}

\noindent for any $\gamma\ge 0$.

\proof{{\bf (a)} We may suppose that $X=n=\{0,\ldots,n-1\}$ where $n=\#(X)$.
For $m\le n$, $p:m\to n$ set $A_p=\{\pi:\pi\in G,\,\pi\restr m=p\}$, and
let $\Sigma_m$ be the subalgebra of $\Cal PG$ generated by
$\{A_p:p\in n^m\}$, and $f_m$ 
the (unique) conditional expectation of $f$ on $\Sigma_m$.   
Then

\Centerline{$f_m(\pi)=\Bover1{\#(A_p)}\sum_{\phi\in A_p}f(\phi)$}

\noindent whenever $\pi\in G$ and $p=\pi\restr m$, while

\Centerline{$\{\emptyset,G\}
=\Sigma_0\subseteq\Sigma_1\subseteq\ldots\subseteq\Sigma_{n-1}
=\Sigma_n=\Cal PG$.}

\medskip

{\bf (b)} 
$|f_m(\pi)-f_{m-1}(\pi)|\le\Bover2n$ whenever $1\le m\le n$ and $\pi\in G$.
\Prf\ Set $p=\pi\restr m-1$ and
$k=\pi(m-1)$.   Set $J=p[m-1]=\{\pi(i):i<m-1\}$, and for
$j\in n\setminus J$ let $p_j=p^{\smallfrown}\fraction{j}$ be that function from $m$
to $n$ which extends $p$ and takes the value $j$ at $m-1$;  let
$\alpha_j$ be the common value of $f_m(\phi)$ for $\phi\in A_{p_j}$, so
that $f_m(\pi)=\alpha_k$.   Now, for each $j\in n\setminus(J\cup\{k\})$,
the function $\phi\mapsto\cycle{j\,k}\phi$ is a bijection from
$A_{p_k}$ to $A_{p_j}$, where $\cycle{j\,k}\in G$ is the transposition
which exchanges $j$ and $k$.   But this means that

$$\eqalign{|\alpha_j-\alpha_k|
&=\bigl|\Bover1{(n-m)!}\sum_{\phi\in A_{p_j}}f(\phi)
  -\Bover1{(n-m)!}\sum_{\phi\in A_{p_k}}f(\phi)\bigr|\cr
&=\Bover1{(n-m)!}\bigl|\sum_{\phi\in A_{p_j}}f(\phi)
  -f(\cycle{j\,k}\phi)\bigr|\cr
&\le\sup_{\phi\in A_{p_j}}\bigl|f(\phi)-f(\cycle{j\,k}\phi)\bigr|
\le\Bover2n\cr}$$

\noindent because $f$ is $1$-Lipschitz and
$\rho(\phi,\cycle{j\,k}\phi)=\bover2n$ for every $\phi$.
And this is true for every $j\in n\setminus(J\cup\{k\})$.

Accordingly

$$\eqalign{|f_m(\pi)-f_{m-1}(\pi)|
&=\bigl|\alpha_k-\Bover1{(n-m+1)!}\sum_{\phi\in A_p}f(\phi)\bigr|
=\bigl|\alpha_k-\Bover1{n-m+1}\sum_{j\in n\setminus J}\alpha_j\bigr|\cr
&\le\Bover1{n-m+1}\sum_{j\in n\setminus J}|\alpha_k-\alpha_j|
\le\Bover1{n-m+1}\sum_{j\in n\setminus J}\Bover2n
=\Bover2n,\cr}$$

\noindent as claimed.\ \Qed

\medskip

{\bf (c)} Now observe that $f=f_{n-1}$ and that $f_0$ is the constant
function with value $\Expn(f)$, so that

$$\eqalignno{\Pr(f-\Expn(f)\ge\gamma)
&=\Pr(f_{n-1}-f_0\ge\gamma)
\le\exp(-\gamma^2/4\sum_{i=1}^{n-1}(\Bover2n)^2)\cr
\displaycause{492G}
&=\exp(-\Bover{n\gamma^2}{16}),\cr}$$

\noindent which is what we were seeking to prove.
}%end of proof of 492H

\vleader{48pt}{492I}{Corollary} 
Let $X$ be a non-empty finite set, with $\#(X)=n$, and $G$ the
group of all permutations of $X$.   Let $\mu$ be the
Haar probability measure of $G$ when given its discrete topology.
Suppose that $A\subseteq G$ and $\mu A\ge\bover12$.   Then

\Centerline{$\mu\{\pi:\pi\in G,\,\Exists\phi\in A,\,
  \#(\{x:x\in X,\,\pi(x)\ne\phi(x)\})\le k\}
\ge 1-\exp(-\Bover{k^2}{64n})$}

\noindent for every $k\le n$.

\proof{ If $\exp(-\Bover{k^2}{64n})\ge\bover12$, this is trivial, since the
left-hand-side of the inequality is surely at least $\bover12$.
Otherwise, set
$g(\pi)=\Bover1n\min_{\phi\in A}\#(\{x:x\in X,\,\pi(x)\ne\phi(x)\})$ for
$\pi\in G$, so that $g$ is $1$-Lipschitz for the metric $\rho$ of 492H.
Applying 492H to $f=-g$, we see that

\Centerline{$\Pr(\Expn(g)-g\ge\Bover{k}{2n})
\le\exp(-\Bover{k^2}{64n})<\Bover12$,}

\noindent and there must be some $\pi\in A$ such that
$\Expn(g)-g(\pi)<\Bover{k}{2n}$, so that $\Expn(g)<\Bover{k}{2n}$.
This means that

$$\eqalign{\mu\{\pi:\pi\in G,\,\Exists\phi\in A,\,
&\#(\{x:x\in X,\,\pi(x)\ne\phi(x)\})\le k\}\cr
&=1-\mu\{\pi:\pi\in G,\,g(\pi)>\Bover{k}{n}\}\cr
&\ge 1-\Pr(g-\Expn(g)\ge\Bover{k}{2n})
\ge 1-\exp(-\Bover{k^2}{64n}),\cr}$$

\noindent applying 492H to $g$ itself.
}%end of proof of 492I

\exercises{\leader{492X}{Basic exercises (a)}
%\spheader 492Xa
Let $(X,\Sigma,\mu)$ be a probability space, and
$\sequencen{f_n}$ a martingale on $X$.
Suppose that $f_n\in\eusm L^{\infty}(\mu)$ for every $n$, and that
$\sigma=\sqrt{\sum_{n=1}^{\infty}\alpha_n^2}$ is finite and not zero,
where $\alpha_n=\esssup|f_n-f_{n-1}|$ for $n\ge 1$.   Show that
$f=\lim_{n\to\infty}f_n$ is defined a.e., and that
$\Pr(f-f_0\ge\gamma)\le\exp(-\gamma^2/4\sigma^2)$ for every
$\gamma\ge 0$.
\Hint{show first that $\|f_n\|_1\le\|f_n\|_2\le\sigma+\|f_0\|_2$ for
every $n$, so that we can apply 275G.}
%492G

\spheader 492Xb Let $(X,\rho)$ be a metric space and $\mu$ a topological
probability measure on $X$.   Suppose that $\gamma$, $\epsilon>0$ are
such that
$\Pr(f-\Expn(f)\ge\gamma)\le\epsilon$ whenever $f:X\to[0,1]$ is
$1$-Lipschitz.   Show that if $\mu F\ge\bover12$ then
$\mu\{x:\rho(x,F)\ge 2\gamma\}\le\epsilon$.
%492I

\spheader 492Xc Let $(X,\rho)$ be a metric space and $\mu$ a topological
probability measure on $X$.   Suppose that $\gamma$, $\epsilon>0$ are
such that $\mu\{x:\rho(x,F)>\gamma\}\le\epsilon$ whenever
$\mu F\ge\bover12$.   Show that if $f:X\to[0,1]$ is a $1$-Lipschitz
function then $\Pr(f-\Expn(f)>2\gamma+\epsilon)\le\epsilon$.
%492I

\spheader 492Xd Use 492G to show that if
$\ofamily{i}{n}{(X_i,\Sigma_i,\mu_i)}$ is a non-empty finite family of
probability spaces with product $(X,\Lambda,\lambda)$, and $X$ is given
its normalized Hamming metric, and $f\in\eusm L^{\infty}(\lambda)$ is
$1$-Lipschitz, then $\Pr(f-\Expn(f)\ge\gamma)\le e^{-n\gamma^2/4}$ for
every $\gamma\ge 0$.   \Hint{if $\Sigma_k\subseteq\Lambda$ is the
$\sigma$-algebra of subsets of $X$ determined by coordinates in $k$, and
$f_k$ is a conditional expectation of $f$ on $\Sigma_k$, then
$\esssup|f_{k+1}-f_k|\le\bover1n$.}
%492G

%\leader{492Y}{Further exercises (a)}

}%end of exercises

\endnotes{
\Notesheader{492}
In metric
spaces, we can say that a probability measure is `concentrated' if every
Lipschitz function $f$ is almost constant in the sense that, for some
$\alpha$, the sets $\{x:|f(x)-\alpha|\ge\gamma\}$ have small measure.
What is astonishing is that this does not mean that the measure itself
is concentrated on a small set.   In 492H, the measure is the Haar
probability measure, spread as evenly as it well could be.   Of course,
when I say that
$\{x:|f(x)-\alpha|\ge\gamma\}$ has `small' measure, I have to let some
other parameter -- in 492H, the size of $X$ -- vary, while $\gamma$
itself is fixed.   Also the shapes of the formulae depend on which
normalizations we choose (observe the effect of moving from $\rho$ to
$\sigma$ in the proof of 492D).   But the value of 492H is that it gives
a strong bound which is independent of the particular function $f$,
provided that it is $1$-Lipschitz.   This kind of concentration of
measure can be described either in terms of the variation of Lipschitz
functions from their means or in terms of the measures of neighbourhoods
of sets of measure $\bover12$ (492Xb-492Xc).   The latter, in a more
abstract context, is what is described by the concentration functions of
measures on uniform spaces;  there is an example of this in 493C.

The martingale method can be used to prove a version of 492E (492Xd).
The method of 492D gives a better exponent ($e^{-n\gamma^2}$ in place of
$e^{-n\gamma^2/4}$) and also information of a slightly different kind,
in that
it can be applied directly to sets $W$ of small measure, at least
provided that
$\gamma>\Bover1{\sqrt n}$ in 492E.   We also need a little more measure
theory here, since sets which are measured by product measures can be
geometrically highly irregular, and our Lipschitz functions
$x\mapsto\rho(x,W)$ need not be measurable.

In the proof of 492G we have an interesting application of the idea of
`martingale'.   The inequality here is quite different from the standard
martingale inequalities like 275D or 275F or 275Yd-275Ye.   
It gives a very
strong inequality concerning the difference $f_n-f_0$, at the cost of
correspondingly strong hypotheses on the differences $f_i-f_{i-1}$;  but
since we need control of $\sum_i\esssup|f_i-f_{i-1}|^2$, not of
$\sum_i\esssup|f_i-f_{i-1}|$, there is scope for applications like
492H.   What the inequality tells us is that most of the time the
differences $f_i-f_{i-1}$ cancel out, just as in the Central Limit
Theorem, and that once again we have a vaguely Gaussian sum $f_n-f_0$.

Concentration of measure, in many forms, has been studied intensively in
the context of the geometry of normed spaces, 
as in {\smc Milman \& Schechtman 86}, from which 492F-492I %492F 492G 492H 492I
are taken.
}%end of notes

\discrpage

\frfilename{mt493.tex}
\versiondate{4.1.13}
\copyrightdate{2001}

\def\Isom{\mathop{\text{Isom}}}

\def\chaptername{Further topics}
\def\sectionname{Extremely amenable groups}

\newsection{493}

A natural variation on the idea of `amenable group'\cmmnt{ (\S449)} is
the concept of
`extremely amenable' group (493A).   Expectedly, most of the ideas of
449C-449E %449C 449D 449E
can be applied to extremely amenable groups
(493B);  unexpectedly, we find not only that there are interesting
extremely amenable groups, but that we need some of the central ideas of
measure theory to study them.   I give a criterion for extreme
amenability of a group in terms of the existence of suitably
concentrated measures (493C) before turning to three examples:  measure
algebras under symmetric difference (493D), $L^0$ spaces (493E)
and isometry groups
of spheres in infinite-dimensional Hilbert spaces (493G).

\leader{493A}{Definition} Let $G$ be a topological group.   Then $G$ is
{\bf extremely amenable} or has the {\bf fixed point on compacta
property} if every continuous action of $G$ on a compact
Hausdorff space has a fixed point.

\leader{493B}{Proposition} (a) Let $G$ and $H$ be topological groups
such that there is a continuous surjective homomorphism from $G$ onto
$H$.   If $G$ is extremely amenable, so is $H$.

(b) Let $G$ be a topological group and suppose that there is a dense
subset $A$ of $G$ such that every finite subset of $A$ is included in an
extremely amenable subgroup of $G$.    Then $G$ is extremely
amenable.

(c) Let $G$ be a topological group with an extremely amenable normal
subgroup $H$ such that $G/H$ is extremely amenable.   Then $G$ is
extremely amenable.

(d) The product of any family of extremely amenable topological groups
is extremely amenable.

(e)\dvAformerly{4{}93C} Let $G$ be a topological group.   Then $G$ is
extremely amenable iff there is a point in the greatest ambit $Z$ of
$G$\cmmnt{ (definition:  449D)} which is fixed by the action of $G$ on
$Z$.

(f)\dvAformerly{4{}93Xa} Let $G$ be an extremely amenable topological group.
Then every dense subgroup of $G$ is extremely amenable.

\proof{ We can use the same arguments as in 449C-449F, %449C 449D 449E 449F
with some simplifications.

\medskip

{\bf (a)} As in 449Ca, let $\phi:G\to H$ be a continuous surjective
homomorphism, $X$ a non-empty compact Hausdorff space and
$\action:H\times X\to X$ a continuous action.   Let $\action_1$ be the
continuous action of $G$ on $X$ defined by the formula
$a\action_1x=\phi(a)\action x$.   Then any fixed point for $\action_1$
is a fixed point for $\action$.

\medskip

{\bf (b)} Let $X$ be a non-empty compact Hausdorff space and
$\action$ a continuous action of $G$ on $X$.   For $I\in[A]^{<\omega}$
let $H_I$ be an extremely amenable subgroup of $G$ including $I$.   The
restriction of the action to $H_I\times X$ is
a continuous action of $H_I$ on $X$, so

\Centerline{$\{x:a\action x=x$ for every $a\in I\}
\supseteq\{x:a\action x=x$ for every $a\in H_I\}$}

\noindent is closed and non-empty.   Because $X$ is compact, there is an
$x\in X$ such that $a\action x=x$ for every $a\in A$.
Now $\{a:a\action x=x\}$ includes the dense set $A$, so is the whole
of $G$, and $x$ is fixed under the action of $G$.   As $X$ and $\action$
are arbitrary, $G$ is extremely amenable.

\medskip

{\bf (c)} Let $X$ be a compact Hausdorff space and
$\action$ a continuous action of $G$ on $X$.   Set
$Q=\{x:x\in X,\,a\action x=x$ for every $a\in H\}$;  then $Q$ is a
closed subset of $X$ and, because $H$ is extremely amenable, is
non-empty.   Next, $b\action x\in Q$ for every $x\in Q$ and $b\in G$.
\Prf\ If $a\in H$, then $b^{-1}ab\in H$ and

\Centerline{$a\action(b\action x)=(ab)\action x
=(bb^{-1}ab)\action x=b\action((b^{-1}ab)\action x)=b\action x$.}

\noindent As $a$ is
arbitrary, $b\action x\in Q$.\ \QeD\    Accordingly we have a continuous
action of $G$ on the compact Hausdorff space $Q$.

If $b\in G$, $a\in H$ and $x\in Q$, then $(ba)\action x=b\action x$.
So we have an action of $G/H$ on $Q$ defined by saying that
$b^{\ssbullet}\action x=b\action x$ for every $b\in G$ and $x\in Q$, and
this is continuous for the quotient topology on $G/H$, as in the proof
of 449Cc.   Because $G/H$ is extremely
amenable, there is a point $x$ of $Q$ which is fixed
under the action of $G/H$.   So $b\action x=b^{\ssbullet}\action x=x$
for every
$b\in G$, and $x$ is fixed under the action of $G$.
As $X$ and $\action$ are arbitrary, $G$ is extremely amenable.

\medskip

{\bf (d)} By (c), the product of two extremely amenable topological
groups is extremely amenable, since each can be regarded as a normal
subgroup of the product.   It follows that the product of finitely
many extremely amenable topological groups is extremely amenable.   Now
let $\familyiI{G_i}$
be any family of extremely amenable topological groups with product $G$.
For finite $J\subseteq I$ let $H_J$ be the set of those $a\in G$ such
that $a(i)$ is the identity in $G_i$ for every $i\in I\setminus J$.
Then $H_J$ is isomorphic (as topological group) to $\prod_{i\in J}G_i$,
so is extremely amenable.   Since $\{H_J:J\in[I]^{<\omega}\}$ is an
upwards-directed
family of subgroups of $G$ with dense union, (b) tells us that $G$ is
extremely amenable.

\medskip

{\bf (e)} Repeat the arguments of 449E(i)$\Leftrightarrow$(ii),
noting that if
$z_0\in Z$ is a fixed point under the action of $G$ on $Z$, then its
images under the canonical maps $\phi$ of 449Dd will be fixed for other
actions.

\medskip

{\bf (f)} Again, the idea is to repeat the argument of 449F(a-ii).
As there, let $H$ be a dense subgroup of $G$,
$U$ the space of bounded real-valued functions on $G$ which are uniformly
continuous for the right uniformity, and $V$ the space of bounded
real-valued functions on $H$ which are uniformly continuous for the right
uniformity.   As in 449F(a-ii), we have an extension operator $T:V\to U$
defined
by saying that $Tg$ is the unique continuous extension of $g$ for every
$g\in V$;  and $b\action_lTg=T(b\action_lg)$ for every $b\in H$ and
$g\in V$.   Now $T$ is a Riesz homomorphism.   So if $z\in Z$ is fixed by
the action of $G$, that is, $z(a\action_lf)=z(f)$ for every $a\in G$ and
$f\in U$, then $zT:V\to\Bbb R$ is a Riesz homomorphism, with
$z(T\chi H)=1$, and $z(T(b\action_lg))=z(b\action_lTg)=z(Tg)$ whenever
$g\in V$ and $b\in H$.   Thus $zT$ is a fixed point of the greatest ambit
of $H$, and $H$ is extremely amenable.
}%end of proof of 493B

\leader{493C}{Theorem}\discrversionA{\footnote{Formerly 4{}93D;
revised 2009.}}{}
Let $G$ be a topological group and $\Cal B$ its
Borel $\sigma$-algebra.   Suppose that
for every $\epsilon>0$, open neighbourhood $V$ of the identity of $G$,
finite set $I\subseteq G$ and finite family $\Cal E$ of zero sets in $G$
there is a finitely additive functional $\nu:\Cal B\to[0,1]$ such that
$\nu G=1$ and

(i) $\nu(VF)\ge 1-\epsilon$ whenever $F\in\Cal E$ and
$\nu F\ge\bover12$,

%(ii) for every $a\in I$ there is a $b\in Va$ such that
(ii) for every $a\in I$ there is a $b\in aV$ such that
$|\nu(bF)-\nu F|\le\epsilon$ for every $F\in\Cal E$.

\noindent Then $G$ is extremely amenable.

\proof{{\bf (a)} Write $P$ for the set of finitely additive functionals
$\nu:\Cal B\to[0,1]$ such that $\nu G=1$.   If $V$ is an open
neighbourhood of the identity $e$ of $G$, $\epsilon>0$,
$I\in[D]^{<\omega}$ and $\Cal E$ is a finite family of zero sets in $G$,
let $A(V,\epsilon,I,\Cal E)$ be the set of those $\nu\in P$ satisfying
(i) and (ii) above.   Our hypothesis is that none of these sets
$A(V,\epsilon,I,\Cal E)$ are empty;  since
$A(V,\epsilon,I,\Cal E)\subseteq A(V',\epsilon',I',\Cal E')$ whenever
$V\subseteq V'$, $\epsilon\le\epsilon'$, $I\supseteq I'$ and
$\Cal E\supseteq\Cal E'$, there is an ultrafilter $\Cal F$ on $P$
containing all these sets.

Let $U$ be the space of bounded real-valued functionals on $G$ which are
uniformly continuous for the right uniformity on $G$.   If we identify
$L^{\infty}(\Cal B)$ with the space of bounded Borel measurable
real-valued functions on $G$ (363H), then $U$ is a Riesz subspace of
$L^{\infty}(\Cal B)$.   For each $\nu\in P$ we have a positive linear
functional $\dashint\,d\nu:L^{\infty}(\Cal B)\to\Bbb R$ (363L).   For
$f\in U$ set $z(f)=\lim_{\nu\to\Cal F}\dashint fd\nu$.

\medskip

{\bf (b)} $z:U\to\Bbb R$ is a Riesz homomorphism, and $z(\chi G)=1$.
\Prf\ Of course $z$ is a positive linear functional taking the value
$1$ at $\chi G$, just because all the integrals $\dashint\,d\nu$ are.   Now
suppose that $f_0$, $f_1\in U$ and $f_0\wedge f_1=0$.   Take any
$\epsilon>0$.   Then there is an open neighbourhood $V$ of $e$ such that
$|f_i(x)-f_i(y)|\le\epsilon$ whenever $xy^{-1}\in V$ and $i\in\{0,1\}$.
Set $F_i=\{x:f_i(x)=0\}$, $E_i=VF_i$ for each
$i$.   Then $F_0\cup F_1=X$, so $\nu F_0+\nu F_1\ge 1$ for every
$\nu\in P$, and there is a $j\in\{0,1\}$ such that
$A_0=\{\nu:\nu F_j\ge\bover12\}\in\Cal F$.   Next,

\Centerline{$A_1=\{\nu$:
 if $\nu F_j\ge\Bover12$ then $\nu(VF_j)\ge 1-\epsilon\}$}

\noindent belongs to $\Cal F$.   Accordingly
$\lim_{\nu\to\Cal F}\nu E_j\ge 1-\epsilon$.   As $f_j(x)\le\epsilon$ for
every $x\in E_j$,

\Centerline{$z(f_j)=\lim_{\nu\to\Cal F}\dashint f_jd\nu
\le\epsilon(1+\|f_j\|_{\infty})$.}

\noindent This shows that
$\min(z(f_0),z(f_1))\le\epsilon(1+\|f_1\|_{\infty}+\|f_2\|_{\infty})$.
As $\epsilon$ is arbitrary, $\min(z(f_0),z(f_1))=0$;  as $f_0$ and $f_1$
are arbitrary, $z$ is a Riesz homomorphism (352G(iv)).\ \Qed

Thus $z$ belongs to the greatest ambit $Z$ of $G$.

\medskip

{\bf (c)} $\lim_{\nu\to\Cal F}\dashint (a^{-1}\action_lf)d\nu
=\lim_{\nu\to\Cal F}\dashint fd\nu$ for every
non-negative $f\in U$ and $a\in G$.   \Prf\ Take any $\epsilon>0$.   Let
$V$ be an open neighbourhood of $e$ such that $|f(x)-f(y)|\le\epsilon$
whenever $x\in Vy$;  then

\Centerline{$\|a^{-1}\action_lf-b^{-1}\action_lf\|_{\infty}
=\sup_{x\in G}|f(ax)-f(bx)|\le\epsilon$}

\noindent whenever $b\in Va$.   For $n\in\Bbb N$ set
$F_n=\{x:x\in G,\,f(x)\ge n\epsilon\}$.   Set
$m=\lfloor\bover1\epsilon\|f\|_{\infty}\rfloor$, so that $F_n=\emptyset$
for every $n>m$.   Set $\delta=\bover1{m+1}$,

$$\eqalign{A
&=\{\nu:\text{ there is a }b\in Va\text{ such that }
|\nu(b^{-1}F_n)-\nu F_n|\le\delta\text{ for every }n\le m\}\cr
&=\{\nu:\text{ there is a }c\in a^{-1}V^{-1}\text{ such that }
|\nu(cF_n)-\nu F_n|\le\delta\text{ for every }n\le m\}
\in\Cal F.\cr}$$

\noindent Take any $\nu\in A$ and $b\in Va$
such that $|\nu(b^{-1}F_n)-\nu F_n|\le\delta$ for every $n\le m$.
Then, setting $g=\sum_{n=1}^m\epsilon\chi F_n$, we have
$g\in L^{\infty}(\Cal B)$ and $g\le f\le g+\epsilon\chi G$.   Since
$b^{-1}\action_lg$ (in the language of 4A5Cc)
% (a\action_lf)(y)=f(a^{-1}y)
is just $\sum_{n=1}^m\epsilon\chi(b^{-1}F_n)$, we have

$$\eqalignno{|\dashint a^{-1}\action_lf\,d\nu-\dashint f\,d\nu|
&\le\epsilon+|\dashint b^{-1}\action_lf\,d\nu-\dashint f\,d\nu|
\le 3\epsilon+|\dashint b^{-1}\action_lg\,d\nu-\dashint g\,d\nu|\cr
\displaycause{because $\|b^{-1}\action_lg-b^{-1}\action_lf\|_{\infty}
=\|g-f\|_{\infty}\le\epsilon$}
&\le 3\epsilon+\epsilon\sum_{n=1}^m|\nu F_n-\nu(b^{-1}F_n)|
\le 3\epsilon+m\epsilon\delta
\le 4\epsilon.\cr}$$

\noindent As $A\in\Cal F$,

\Centerline{$|\lim_{\nu\to\Cal F}
  \dashint a^{-1}\action_lfd\nu-\dashint fd\nu|
\le 5\epsilon$;}

\noindent as $\epsilon$ is arbitrary,
$\lim_{\nu\to\Cal F}\dashint a^{-1}\action_lf\,d\nu
=\lim_{\nu\to\Cal F}\dashint fd\nu$.\ \Qed

\medskip

{\bf (d)} Thus, for any $a\in G$,

\Centerline{$(a\action z)(f)
=z(a^{-1}\action_lf)
=\lim_{\nu\to\Cal F}\dashint a^{-1}\action_lf\,d\nu
=\lim_{\nu\to\Cal F}\dashint f\,d\nu
=z(f)$}

\noindent for every non-negative $f\in U$ and therefore for every $f\in U$,
and $a\action z=z$.   So
$z\in Z$ is fixed under the action of $G$ on $Z$;  by
493Ba, this is enough to ensure that $G$ is extremely amenable.
}%end of proof of 493C

\leader{493D}{}\dvAformerly{4{}93E}\cmmnt{ I
turn now to examples of extremely amenable
groups.   The first three are groups which we have already studied for
other reasons.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be an atomless measure
algebra.   Then $\frak A$, with the group operation $\Bsymmdiff$ and the
measure-algebra topology\cmmnt{ (definition:  323A)}, is an extremely
amenable group.

\proof{{\bf (a)} To begin with let us suppose that $(\frak A,\bar\mu)$
is a probability algebra;  write $\sigma$ for the measure metric of
$\frak A$, so that $\sigma(a,a')=\bar\mu(a\Bsymmdiff a')$ for $a$,
$a'\in\frak A$.   I seek to apply 493C.

\medskip

\quad{\bf (i)} Let $V$ be an open neighbourhood of $0$ in $\frak A$,
$\epsilon\in\ooint{0,3}$,
$I\in[\frak A]^{<\omega}$ and $\Cal E$ a finite family of
zero sets in $\frak A$.   Let $\gamma>0$ be such that
$V\supseteq\{a:\bar\mu a\le 2\gamma\}$.   Let $\frak B_0$ be the finite
subalgebra of $\frak A$ generated by
$I$ and $B_0$ the set of atoms in $\frak B_0$.   Set

\Centerline{$t=\Bover1{\gamma}\ln\Bover3{\epsilon}$,
\quad$n=\lceil\max(t^2,\sup_{b\in B_0}\Bover1{\bar\mu b})\rceil$.}

\noindent Because $\frak A$ is atomless, we can split any member of
$\frak A\setminus\{0\}$ into two parts of equal measure (331C);  if,
starting from the disjoint set $B_0$, we successively split the largest
elements until we have a disjoint set $B$ with just $n$ elements, then
we shall have $\bar\mu b\le\Bover2n$ for every $b\in B$.   We have a
natural identification between $\{0,1\}^B$ and the subalgebra $\frak B$
of $\frak A$ generated by $B$, matching $x\in\{0,1\}^B$ with
$f(x)=\sup\{b:b\in B$, $x(b)=1\}$.   Writing $\rho$ for the normalized
Hamming metric on $\{0,1\}^B$ (492D), we have
$\sigma(f(x),f(y))\le 2\rho(x,y)$ for all $x$,
$y\in\{0,1\}^B$.   \Prf\ Set $J=\{b:b\in B,\,x(b)\ne y(b)\}$, so that

\Centerline{$\sigma(f(x),f(y))
=\bar\mu(f(x)\Bsymmdiff f(y))=\bar\mu(\sup J)
=\sum_{b\in J}\bar\mu b
\le\Bover2n\#(J)=2\rho(x,y)$.   \Qed}

\medskip

\quad{\bf (ii)} Let $\nu_B$ be the usual measure on $\{0,1\}^B$ and
set $\lambda E=\nu_B f^{-1}[E]$ for every Borel set $E\subseteq\frak A$.
Then $\lambda$ is a probability measure.   Note that
$f:\{0,1\}^B\to\frak B$
is a group isomorphism if we give $\{0,1\}^B$ the addition $+_2$
corresponding to its identification with $\Bbb Z_2^B$, and $\frak B$ the
operation $\Bsymmdiff$.   Because $\nu_B$ is translation-invariant for
$+_2$, its copy, the subspace measure $\lambda_{\frak B}$ on the
$\lambda$-conegligible finite set $\frak B$, is translation-invariant for
$\Bsymmdiff$.   But this means that
$\lambda\{b\Bsymmdiff d:d\in F\}=\lambda F$
whenever $b\in\frak B$ and $F\subseteq\frak B$, and therefore that
$\lambda\{b\Bsymmdiff d:d\in F\}=\lambda F$ whenever $b\in I$ and
$F\in\Cal E$.
This shows that $\lambda$ satisfies condition (ii) of 493C.

\medskip

\quad{\bf (iii)} Now suppose that $F\in\Cal E$ and that
$\lambda F\ge\bover12$.   Set $W=f^{-1}[F]$, so that $\nu_B W\ge\bover12$.
By 492D,

\Centerline{$\int e^{t\rho(x,W)}\nu_B(dx)\le 2e^{t^2/4n}
\le 2e^{1/4}\le 3$,}

\noindent so

\Centerline{$\nu_B\{x:\rho(x,W)\ge\gamma\}
=\nu_B\{x:t\rho(x,W)\ge\ln\bover3{\epsilon}\}
=\nu_B\{x:e^{t\rho(x,W)}\ge\bover3{\epsilon}\}
\le\epsilon$.}

\noindent Accordingly

$$\eqalignno{\lambda\{a\Bsymmdiff d:a\in V,\,d\in F\}
&\ge\lambda\{a:\sigma(a,F)\le 2\gamma\}
=\nu_B\{x:\sigma(f(x),F)\le 2\gamma\}\cr
&\ge\nu_B\{x:\sigma(f(x),f[W])\le 2\gamma\}
\ge\nu_B\{x:\rho(x,W)\le\gamma\}\cr
\displaycause{because $f$ is $2$-Lipschitz}
&\ge 1-\epsilon.\cr}$$

\noindent So $\lambda$ also satisfies (i) of 493C.

\medskip

\quad{\bf (iv)} Since $V$, $\epsilon$, $I$ and $\Cal E$ are arbitrary,
493C tells us that $\frak A$ is an extremely amenable group, at least
when $(\frak A,\bar\mu)$ is an atomless probability algebra.

\medskip

{\bf (b)} For the general case, observe first that if
$(\frak A,\bar\mu)$ is atomless and totally finite then
$(\frak A,\Bsymmdiffshort)$ is an extremely amenable group;
this is trivial
if $\frak A=\{0\}$, and otherwise there is a probability measure on
$\frak A$ which induces the same topology, so we can apply (a).   For a
general atomless measure algebra $(\frak A,\bar\mu)$, set
$\frak A^f=\{c:c\in\frak A,\,\bar\mu c<\infty\}$ and for $c\in\frak A^f$
let $\frak A_c$ be the principal ideal generated by $c$.   Then
$\frak A_c$ is a subgroup of $\frak A$ and the measure-algebra topology
of $\frak A_c$, regarded as a measure algebra in itself, is the subspace
topology induced by the measure-algebra topology of $\frak A$.   So
$\{\frak A_c:c\in\frak A^f\}$ is an upwards-directed family of extremely
amenable subgroups of $\frak A$ with union which is dense in $\frak A$,
so $\frak A$ itself is extremely amenable, by 493Bb.   This completes
the proof.
}%end of proof of 493D

\leader{493E}{Theorem}\dvAformerly{4{}93F}\cmmnt{ ({\smc Pestov 02})} Let
$(X,\Sigma,\mu)$ be an atomless measure
space.   Then $L^0(\mu)$, with the group operation $+$ and the topology
of convergence in measure, is an extremely amenable group.

\proof{ It will simplify some of the formulae if we move at once to the
space $L^0(\frak A)$, where $(\frak A,\bar\mu)$ is the measure algebra
of $(X,\Sigma,\mu)$;  for the identification of $L^0(\frak A)$ with
$L^0(\mu)$ see 364Ic;  for a note on convergence in measure in
$L^0(\frak A)$, see 367L;
of course $\frak A$ is atomless if $(X,\Sigma,\mu)$ is (322Bg).

\medskip

{\bf (a)} I seek to prove that $S(\frak A)$, with the
group operation of addition and the topology of convergence in measure,
is extremely amenable.
As in 493D, I start with the case in which $(\frak A,\bar\mu)$
is a probability algebra, and use 493C.

\medskip

\quad{\bf (i)} Take an open neighbourhood $V$ of $0$ in $S(\frak A)$, an
$\epsilon\in\ooint{0,3}$,
a finite set $I\subseteq S(\frak A)$ and a finite family
$\Cal E$ of zero sets in $S(\frak A)$.   Let $\gamma>0$ be such that
$u\in V$ whenever $u\in S(\frak A)$ and
$\bar\mu\Bvalue{u\ne 0}\le 2\gamma$.   Let $\frak B_0$ be a finite
subalgebra of $\frak A$ such that $I$ is included in the linear subspace
of $S(\frak A)$ generated by $\{\chi b:b\in\frak B_0\}$, and $B_0$ the
set of atoms of $\frak B_0$.   As in the proof of 493D, set

\Centerline{$t=\Bover1{\gamma}\ln\Bover3{\epsilon}$,
\quad$n=\lceil\max(t^2,\sup_{b\in B_0}\Bover1{\bar\mu b})\rceil$,}

\noindent and let $B\subseteq\frak A\setminus\{0\}$ be a partition of
unity with $n$ elements, refining $B_0$, such that
$\bar\mu b\le\bover2n$ for every $b\in B$.   We have a natural
identification between $\BbbR^B$ and the linear subspace of $S(\frak A)$
generated by $\{\chi b:b\in B\}$, matching $x\in\BbbR^B$ with
$f(x)=\sum_{b\in B}x(b)\chi b$, which is continuous if $\BbbR^B$ is
given its product topology.   Writing $\rho$ for the normalized Hamming
metric on $\BbbR^B$, we have

\Centerline{$\bar\mu\Bvalue{f(x)\ne f(y)}=\sum_{x(b)\ne y(b)}\bar\mu b
\le\Bover2n\#(\{b:x(b)\ne y(b)\})=2\rho(x,y)$}

\noindent for all $x$, $y\in\BbbR^B$.

\medskip

\quad{\bf (ii)} Set $\beta=\sup_{v\in I}\|v\|_{\infty}$ (if
$I=\emptyset$, take $\beta=0$).   Let $M>0$ be so large that
$(M+\beta)^n\le(1+\bover12\epsilon)M^n$.   On $\Bbb R$, write $\mu_L$
for Lebesgue measure and $\mu_L'$ for the
indefinite-integral measure over $\mu_L$ defined by the function
$\Bover1{2M}\chi[-M,M]$, so that $\mu_L'E=\Bover1{2M}\mu_L(E\cap[-M,M])$
whenever $E\subseteq\Bbb R$ and $E\cap[-M,M]$ is Lebesgue measurable.
Let $\lambda$, $\lambda'$ be the product measures on $\BbbR^B$ defined
from $\mu_L$ and $\mu'_L$.   Let $\nu$ be the Borel probability measure
on $S(\frak A)$ defined by setting $\nu F=\lambda'f^{-1}[F]$ for every
Borel set $F\subseteq S(\frak A)$.

Now $|\nu(v+F)-\nu F|\le\epsilon$ for every $v\in I$ and Borel set
$F\subseteq S(\frak A)$.   \Prf\ Because $B$ refines $B_0$, $v$ is
expressible as $f(y)$ for some $y\in\BbbR^B$;  because
$\|v\|_{\infty}\le\beta$, $|y(b)|\le\beta$ for every $b\in B$.   Because
$f:\BbbR^B\to S(\frak A)$ is linear, $f^{-1}[v+F]=y+f^{-1}[F]$.   Now

$$\eqalignno{|\nu(v+F)-\nu F|
&=|\lambda'f^{-1}[v+F]-\lambda'f^{-1}[F]|\cr
&=\Bover1{(2M)^n}
  \bigl|\lambda(f^{-1}[v+F]\cap[-M,M]^n)
    -\lambda(f^{-1}[F]\cap[-M,M]^n)\bigr|\cr
\displaycause{use 253I, or otherwise}
&=\Bover1{(2M)^n}
  \bigl|\lambda((y+f^{-1}[F])\cap[-M,M]^n)
      -\lambda(f^{-1}[F]\cap[-M,M]^n)\bigr|\cr
&=\Bover1{(2M)^n}
  \bigl|\lambda(f^{-1}[F]\cap([-M,M]^n-y))
      -\lambda(f^{-1}[F]\cap[-M,M]^n)\bigr|\cr
&\le\Bover1{(2M)^n}
  \lambda(([-M,M]^n-y)\symmdiff [-M,M]^n)\cr
&=\Bover2{(2M)^n}
  \lambda(([-M,M]^n-y)\setminus [-M,M]^n)\cr
&\le\Bover2{(2M)^n}
  \lambda(([-M-\beta,M+\beta]^n\setminus [-M,M]^n)\cr
&=\Bover2{M^n}((M+\beta)^n-M^n)
\le\epsilon.  \text{ \Qed}\cr}$$

\noindent So $\nu$ satisfies (ii) of 493C.

\medskip

\quad{\bf (iii)} Now suppose that $F\in\Cal E$ and $\nu F\ge\bover12$.
Set $W=f^{-1}[F]$, so that $\lambda'W\ge\bover12$.   Just as in the
proof of 493D,
$\int e^{t\rho(x,W)}\lambda'(dx)\le 2e^{t^2/4n}\le 3$, so

\Centerline{$\lambda'\{x:\rho(x,W)\ge\gamma\}
=\lambda'\{x:e^{t\rho(x,W)}\ge\bover3{\epsilon}\}\le\epsilon$,}

\noindent and

$$\eqalignno{\nu\{v+u:v\in V,\,u\in F\}
&\ge\nu\{w:\,\Exists u\in F,\,\bar\mu\Bvalue{u\ne w}\le 2\gamma\}\cr
&\ge\lambda'\{x:\rho(x,W)\le\gamma\}
\ge 1-\epsilon.\cr}$$

\noindent So $\nu$ also satisfies (i) of 493C.

\medskip

\quad{\bf (iv)} Since $V$, $\epsilon$, $I$ and $\Cal E$ are arbitrary,
493C tells us that $S(\frak A)$ is an extremely amenable group, at least
when $(\frak A,\bar\mu)$ is an atomless probability algebra.

\medskip

{\bf (b)} The rest of the argument is straightforward, as in 493D.
First, $S(\frak A)$ is extremely amenable whenever $(\frak A,\bar\mu)$
is an atomless totally finite measure algebra.   For a general atomless
measure algebra
$(\frak A,\bar\mu)$, set $\frak A^f=\{c:\bar\mu c<\infty\}$.   For each
$c\in\frak A^f$, let $\frak A_c$ be the corresponding principal ideal of
$\frak A$.   Then we can identify $S(\frak A_c)$, as topological group,
with the linear subspace of $L^0(\frak A)$ generated by
$\{\chi a:a\in\frak A_c\}$, and it is extremely amenable.   Since
$\{S(\frak A_c):c\in\frak A^f\}$ is an upwards-directed family of
extremely amenable subgroups of $L^0(\frak A)$ with dense union in
$L^0(\frak A)$, $L^0(\frak A)$ itself is extremely amenable, by 493Bb,
as before.
}%end of proof of 493E

\leader{493F}{}\dvAformerly{4{}93I}\cmmnt{ Returning to the ideas of
\S476, we find another remarkable example of an extremely amenable
topological group.   I recall the notation of 476I.    Let $X$ be a
(real) inner product space.   $S_X$ will be the unit sphere
$\{x:x\in X,\,\|x\|=1\}$.   Let $H_X$ be the isometry group of $S_X$
with its topology of pointwise convergence.
When $X$ is finite-dimensional, it is isomorphic, as inner product
space, to $\BbbR^r$, where $r=\dim X$.   In this case $S_X$ is
compact, so (if $r\ge 1$) has a unique $H_X$-invariant Radon probability
measure $\nu_X$, which is strictly positive, and is a multiple of
$(r-1)$-dimensional Hausdorff measure;  also $H_X$ is compact\cmmnt{
(441Gb)}, so has a unique Haar probability measure $\lambda_X$.

\medskip

\noindent}{\bf Lemma} For any $m\in\Bbb N$ and any $\epsilon>0$, there
is an $r(m,\epsilon)\ge 1$ such that whenever $X$ is a
finite-dimensional inner product space over $\Bbb R$ of dimension at least
$r(m,\epsilon)$,
$x_0,\ldots,x_{m-1}\in S_X$, $Q_1$, $Q_2\subseteq H_X$
are closed sets and $\min(\lambda_XQ_1,\lambda_XQ_2)\ge\epsilon$, then
there are $f_1\in Q_1$, $f_2\in Q_2$ such that
$\|f_1(x_i)-f_2(x_i)\|\le\epsilon$ for every $i<m$.

\proof{ Induce on $m$.
For $m=0$, the result is trivial.   For the
inductive step to $m+1$, take $r(m+1,\epsilon)>r(m,\bover13\epsilon)$
such that whenever $r(m+1,\epsilon)\le\dim X<\omega$ and $A_1$,
$A_2\subseteq S_X$ and $\min(\nu^*_XA_1,\nu^*_XA_2)\ge\bover12\epsilon$
then there are $x\in A_1$, $y\in A_2$ such that
$\|x-y\|\le\bover13\epsilon$;  this is possible by 476L.

Now take any inner product space $X$ over $\Bbb R$ of finite dimension
$r\ge r(m+1,\epsilon)$, closed sets $Q_1$, $Q_2\subseteq H_X$ such that
$\min(\lambda_XQ_1,\lambda_XQ_2)\ge\epsilon$, and
$x_0,\ldots,x_m\in S_X$.
Let $Y$ be the $(r-1)$-dimensional subspace
$\{x:x\in X,\,\innerprod{x}{x_m}=0\}$, so that
$\dim Y\ge r(m,\bover13\epsilon)$, and for $i<m$ let $y_i\in Y$ be a
unit vector such that $x_i$ is a linear combination of $y_i$ and $x_m$.
Set $H'_Y=\{f:f\in H_X,\,f(x_m)=x_m\}$;  then $f\mapsto f\restr S_Y$ is
a topological group isomorphism from $H'_Y$ to $H_Y$.   \Prf\ (i) If
$f\in H'_Y$ and $x\in S_X$, then

\Centerline{$x\in S_Y\iff\innerprod{x}{x_m}=0
\iff\innerprod{f(x)}{f(x_m)}=0
\iff f(x)\in S_Y$,}

\noindent so $f\restr S_Y$ is a permutation of $S_Y$ and
belongs to $H_Y$.   (ii) If $g\in H_Y$, we can define $f\in H'_Y$ by
setting $f(\alpha x+\beta x_m)=\alpha g(x)+\beta x_m$ whenever
$x\in S_Y$ and $\alpha^2+\beta^2=1$, and $f\restr S_Y=g$.   (iii) Note
that $H'_Y$ is a closed subgroup of $H_X$, so in itself is a compact
Hausdorff topological group.   Since the map
$f\mapsto f\restr S_Y:H'_Y\to H_Y$ is a bijective continuous group
homomorphism between compact Hausdorff topological groups, it is a
topological group isomorphism.\ \Qed

Let $\lambda'_Y$ be the Haar probability measure of $H'_Y$.   Then
$\lambda_XQ_1=\int\lambda'_Y(H'_Y\cap f^{-1}Q_1)\lambda_X(df)$ (443Ue),
so $\lambda_XQ'_1\ge\bover12\epsilon$, where
$Q'_1=\{f:\lambda'_Y(H'_Y\cap f^{-1}Q_1)\ge\bover12\epsilon\}$.
Similarly, setting
$Q'_2=\{f:\lambda'_Y(H'_Y\cap f^{-1}Q_2)\ge\bover12\epsilon\}$,
$\lambda_XQ'_2\ge\bover12\epsilon$.   Next, setting $\theta(f)=f(x_m)$
for $f\in H_X$, $\lambda_X\theta^{-1}$ is an $H_X$-invariant Radon
probability measure on $S_X$ (443Ub), so must be equal to $\nu_X$.
Accordingly

\Centerline{$\nu_X(\theta[Q'_j])=\lambda_X(\theta^{-1}[\theta[Q'_j]])
\ge\lambda_XQ'_j\ge\Bover12\epsilon$}

\noindent for both $j$.

We chose $r(m+1,\epsilon)$ so large that we can be sure that there are
$z_1\in\theta[Q_1']$, $z_2\in\theta[Q_2']$ such that
$\|z_1-z_2\|\le\bover13\epsilon$.   Let $h_1\in Q'_1$, $h_2\in Q'_2$ be
such that $h_1(x_m)=\theta(h_1)=z_1$ and $h_2(x_m)=z_2$.
Let $h\in H_X$ be such that $h(z_1)=z_2$ and
$\|h(x)-x\|\le\bover13\epsilon$ for every $x\in S_X$ (4A4Jg).   Set
$\tilde h_2=hh_1$, so that $\tilde h_2(x_m)=z_2$ and
$\|h_1(x)-\tilde h_2(x)\|\le\bover13\epsilon$ for every $x\in S_X$.
Note that $\tilde h_2^{-1}h_2\in H'_Y$, so that $\tilde h_2$ and $h_2$
belong to the same left coset of $H'_Y$, and

\Centerline{$\lambda'_Y(H'_Y\cap\tilde h_2^{-1}Q_2)
=\lambda'_Y(H'_Y\cap h_2^{-1}Q_2)\ge\Bover12\epsilon$}

\noindent by 443Qa.
%shows that we have a measure indep of choice of representative

At this point, recall that $\dim Y\ge r(m,\bover13\epsilon)$, and that
$\lambda'_Y$ is a copy of $\lambda_Y$, the Haar probability measure on
$Y$.   So we have
$g_1\in H'_Y\cap h_1^{-1}Q_1$, $g_2\in H'_Y\cap\tilde h_2^{-1}Q_2$ such
that $\|g_1(y_i)-g_2(y_i)\|\le\bover13\epsilon$ for every $i<m$.   We
have $f_1=h_1g_1\in Q_1$ and $f_2=\tilde h_2g_2\in Q_2$.   For any
$i<m$,

$$\eqalign{\|f_1(y_i)-f_2(y_i)\|
&\le\|h_1g_1(y_i)-h_1g_2(y_i)\|
  +\|h_1g_2(y_i)-\tilde h_2g_2(y_i)\|\cr
&\le\|g_1(y_i)-g_2(y_i)\|+\Bover13\epsilon
\le\Bover23\epsilon.\cr}$$

\noindent Also $g_1(x_m)=g_2(x_m)=x_m$, so

\Centerline{$\|f_1(x_m)-f_2(x_m)\|
=\|h_1(x_m)-\tilde h_2(x_m)\|\le\Bover13\epsilon$.}

\noindent If $i<m$, then
$x_i=\innerprod{x_i}{x_m}x_m+\innerprod{x_i}{y_i}y_i$, so
$f_j(x_i)=\innerprod{x_i}{x_m}f_j(x_m)+\innerprod{x_i}{y_i}f_j(y_i)$ for
both $j$ (476J) and

$$\eqalign{\|f_1(x_i)-f_2(x_i)\|
&\le\Bover13\epsilon|\innerprod{x_i}{x_m}|
  +\Bover23\epsilon|\innerprod{x_i}{y_i}| \cr
&\le\sqrt{(\Bover13\epsilon)^2+(\Bover23\epsilon)^2}
  \sqrt{\innerprod{x_i}{x_m}^2+\innerprod{x_i}{y_i}^2}
\le\epsilon.\cr}$$

\noindent So $f_1$ and $f_2$ witness that the induction proceeds.
}%end of proof of 493F

\leader{493G}{Theorem}\dvAformerly{4{}93J} Let
$X$ be an infinite-dimensional inner product
space over $\Bbb R$.
Then the isometry group $H_X$ of its unit sphere $S_X$, with
its topology of pointwise convergence, is extremely amenable.

\proof{{\bf (a)}
Let $\Cal Y$ be the family of finite-dimensional subspaces of
$X$.   For $Y\in\Cal Y$, write $Y^{\perp}$ for the orthogonal complement
of $Y$, so that $X=Y\oplus Y^{\perp}$ (4A4Jf).   For $q\in H_Y$
define $\theta_Y(q):S_X\to S_X$ by saying that
$\theta_Y(q)(\alpha y+\beta z)=\alpha q(y)+\beta z$ whenever $y\in S_Y$,
$z\in S_{Y^{\perp}}$ and $\alpha^2+\beta^2=1$.   Then
$\theta_Y:H_Y\to H_X$ is a injective group homomorphism.   Also it is
continuous, because $q\mapsto\alpha q(y)+\beta z$ is continuous for all
relevant $\alpha$, $\beta$, $y$ and $z$.

If $Y$, $W\in\Cal Y$ and $Y\subseteq W$ then
$\theta_Y[H_Y]\subseteq\theta_W[H_W]$.   \Prf\ For any $q\in H_Y$ we can
define $q'\in H_W$ by saying that
$q'(\alpha y+\beta x)=\alpha q(y)+\beta x$ whenever $y\in S_Y$,
$x\in S_{W\cap Y^{\perp}}$ and $\alpha^2+\beta^2=1$.   Now
$\theta_Y(q)=\theta_W(q')\in\theta_W[H_W]$.\ \Qed

Set $G^*=\bigcup_{Y\in\Cal Y}\theta_Y[H_Y]$, so that $G^*$ is a subgroup
of $H_X$.

\medskip

{\bf (b)} Let $V$ be an open neighbourhood of the identity in $G^*$
(with the subspace topology inherited from the topology of pointwise
convergence on $H_X$), $\epsilon>0$ and
$I\subseteq G^*$ a finite set.   Then there is a Borel probability
measure $\lambda$ on $G^*$ such that

\quad(i) $\lambda(fQ)=\lambda Q$ for every $f\in I$ and every closed set
$Q\subseteq G^*$,

\quad(ii) $\lambda(VQ)\ge 1-\epsilon$ whenever $Q\subseteq G^*$ is
closed and $\lambda Q\ge\bover12$.

\noindent\Prf\ We may suppose that $\epsilon\le\bover12$.   Let
$J\in[S_X]^{<\omega}$ and $\delta>0$ be such that
$f\in V$ whenever $f\in G^*$ and
$\|f(x)-x\|\le\delta$ for every $x\in J$.   We may suppose that $J$ is
non-empty;  set $m=\#(J)$.
Let $Y\in\Cal Y$ be such that $J\subseteq Y$ and
$I\subseteq\theta_Y[H_Y]$
and $\dim Y=r\ge r(m,\epsilon)$, as chosen in 493F.   (This is
where we need to know that $X$ is infinite-dimensional.)    Set
$\lambda F=\lambda_Y\theta_Y^{-1}[F]$ for every Borel set
$F\subseteq G^*$, where $\lambda_Y$ is the Haar probability measure of
$H_Y$, as before.

If $f\in I$ and $F\subseteq G^*$ is closed, then

\Centerline{$\lambda(fF)=\lambda_Y\theta_Y^{-1}[fF]
=\lambda_Y(\theta_Y^{-1}(f)\theta_Y^{-1}[F])
=\lambda_Y\theta_Y^{-1}[F]=\lambda F$.}

\noindent So $\lambda$ satisfies condition (i).

\Quer\ Suppose, if possible, that $Q_1\subseteq G^*$ is a closed set
such that $\lambda Q_1\ge\bover12$ and $\lambda(VQ_1)<1-\epsilon$;  set
$Q_2=G^*\setminus VQ_1$.   Then
$\theta_Y^{-1}[Q_1]$ and $\theta_Y^{-1}[Q_2]$ are subsets of $H_Y$ both
of measure at least $\epsilon$.   Set
$R_j=\{q:q\in H_Y,\,q^{-1}\in\theta_Y^{-1}[Q_j]\}$ for each $j$;
because
$H_Y$ is compact, therefore unimodular,

\Centerline{$\lambda_YR_j
=\lambda_Y\theta_Y^{-1}[Q_j]=\lambda Q_j\ge\epsilon$}

\noindent for both $j$.   Because $\dim Y\ge r(m,\epsilon)$,
there are $q_1\in R_1$, $q_2\in R_2$ such that
$\|q_1(x)-q_2(x)\|\le\epsilon$ for $x\in J$.   Set
$f=\theta_Y(q_2^{-1}q_1)$.   If $x\in J$, then

\Centerline{$\|f(x)-x\|=\|q_2^{-1}q_1(x)-x\|
=\|q_1(x)-q_2(x)\|\le\epsilon$.}

\noindent As this is true whenever $x\in J$ and $f\in V$.   On the other
hand, $\theta_Y(q_1^{-1})\in Q_1$ and $\theta_Y(q_2^{-1})\in Q_2$ and
$f\theta_Y(q_1^{-1})=\theta_Y(q_2^{-1})$, so
$\theta_Y(q_2^{-1})\in VQ_1\cap Q_2$, which is impossible.\ \Bang

Thus $\lambda$ satisfies (ii).\ \Qed

\medskip

{\bf (c)} By 493C, $G^*$ is extremely amenable.   But $G^*$ is dense in
$H_X$.   \Prf\ If $f\in H_X$ and $I\subseteq S_X$ is finite and not
empty, let $Y_1$ be the linear subspace of $X$ generated by $I$, and let
$(y_1,\ldots,y_m)$ be an orthonormal basis of $Y_1$.   Set $z_j=f(y_j)$
for each $j$, so that $(z_1,\ldots,z_m)$ is orthonormal (476J);  let $Y$
be the linear subspace of $X$ generated by
$y_1,\ldots,y_m,z_1,\ldots,z_m$.   Set $r=\dim Y$ and extend the
orthonormal sets $(y_1,\ldots,y_m)$ and $(z_1,\ldots,z_m)$ to
orthonormal bases $(y_1,\ldots,y_r)$ and $(z_1,\ldots,z_r)$ of $Y$.
Then we have an isometric linear operator $T:Y\to Y$ defined by saying
that $Ty_i=z_i$ for each $i$;  set $q=T\restr S_Y\in H_Y$.   By 476J,
$q(x)=f(x)$ for every $x\in I$, so $\theta_Y(q)$ agrees with $f$ on $I$,
while $\theta_Y(q)\in G^*$.   As $f$ and $I$ are arbitrary, $G^*$ is
dense in $G$.\ \Qed

So 493Bb tells us that $H_X$ is extremely amenable, and the proof is
complete.
}%end of proof of 493G

\leader{493H}{}\dvAformerly{4{}93K}\cmmnt{ The
following result shows why extremely amenable groups did not appear in
Chapter 44.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Veech 77})} If $G$ is a locally
compact Hausdorff topological group with more than one element, it is
not extremely amenable.
%Veech, th. 2.2.1

\proof{ If $G$ is compact, this is trivial, since the left action of $G$
on itself has no fixed point;  so let us assume henceforth that $G$ is
not compact.

\medskip

{\bf (a)} Let $Z$ be the greatest ambit of $G$, $a\mapsto\hat a:G\to Z$
the canonical map, and $U$ the space of
bounded right-uniformly continuous real-valued functions on $G$.   (I
aim to show that the action
of $G$ on $Z$ has no fixed point.)   Take any $z^*\in Z$.   Let $V_0$ be
a compact neighbourhood of the identity $e$ in $G$, and let
$B_0\subseteq G$ be a maximal set such that $V_0b\cap V_0c=\emptyset$
for all distinct $b$, $c\in B_0$.   Then for any $a\in G$ there is a
$b\in B_0$ such that $V_0a\cap V_0b\ne\emptyset$, that is,
$a\in V_0^{-1}V_0B_0$.   So if we set
$Y_0=\overline{\{\hat b:b\in B_0\}}\subseteq Z$,
$\{a\action y:a\in V_0^{-1}V_0,\,y\in Y_0\}$ is a compact subset of $Z$
including $\{\hat a:a\in G\}$, and is therefore the whole of $Z$
(449Dc).   Let $a_0\in V_0^{-1}V_0$, $y_0\in Y_0$ be such that
$a_0\action y_0=z^*$, and set $B_1=a_0B_0$, $V_1=a_0V_0a_0^{-1}$;  then
$z^*\in\overline{\{\hat b:b\in B_1\}}$ and $V_1b\cap V_1c=\emptyset$ for
all distinct $b$, $c\in B_1$.

\medskip

{\bf (b)} Because $V_1$ is compact and $G$ is not compact, there is an
$a_1\in G\setminus V_1$.   Let $V_2\subseteq V_1$ be a neighbourhood of
$e$ such that $a_1^{-1}V_2V_2^{-1}a_1\subseteq V_1$.   Then we can
express $B_1$ as $D_0\cup D_1\cup D_2$ where
$a_1D_i\cap V_2D_i=\emptyset$ for all $i$.   \Prf\ Consider
$\{(b,c):b,\,c\in B_1,\,a_1b\in V_2c\}$.   Because
$V_2c\cap V_2c'\subseteq V_1c\cap V_1c'=\emptyset$ for all distinct $c$,
$c'\in B_1$, this is the graph of a function $h:D\to B_1$ for some
$D\subseteq B_1$.   \Quer\ If $h$ is not injective, we have distinct
$b$, $c\in B_1$ and $d\in B_1$ such that $a_1b$ and $a_1c$ both belong
to $V_2d$.   But in this case $b$ and $c$ both belong to $a_1^{-1}V_2d$
and $bc^{-1}\in a_1^{-1}V_2dd^{-1}V_2^{-1}a_1\subseteq V_1$ and
$b\in V_1c$, which is impossible.\ \BanG\  At the same time, if
$b\in B_1$, then
$a_1b\notin V_2b$ because $a_1\notin V_2$, so $h(b)\ne b$ for every
$b\in D$.

Let $D_0\subseteq D$ be a maximal set such that
$h[D_0]\cap D_0=\emptyset$, and set $D_1=h[D_0]$,
$D_2=B_1\setminus(D_0\cup D_1)$.   Then $h[D_0]\cap D_0=\emptyset$ by
the choice of $D_0$;  $h[D\cap D_1]\cap D_1=\emptyset$ because $h$ is
injective and $D_1\subseteq h[D\setminus D_1]$;  and
$h[D\cap D_2]\subseteq D_0$ because if $b\in D\cap D_2$ there must have
been some reason why we did not put $b$ into $D_0$, and it wasn't
because
$b\in h[D_0]$ or because $h(b)=b$.   So $h[D_i]\cap D_i=\emptyset$ for
all $i$, which is what was required.\ \Qed

\medskip

{\bf (c)} Since $z^*\in\overline{\{\hat b:b\in B_1\}}$, there must be
some $j\le 2$ such that $z^*\in\overline{\{\hat b:b\in D_j\}}$.   Now
recall that the right uniformity on $G$, like any uniformity, can be
defined by some family of pseudometrics (4A2Ja).  There is therefore a
pseudometric $\rho$ on $G$ such that
$W_{\epsilon}=\{(a,b):a,\,b\in G,\,\rho(a,b)\le\epsilon\}$ is a member
of the right uniformity on $G$ for every $\epsilon>0$ and
$W_1\subseteq\{(a,b):ab^{-1}\subseteq V_2\}$.   If now we set

\Centerline{$f(a)=\min(1,\rho(a,D_j))
=\min(1,\inf\{\rho(a,b):b\in D_j\})$}

\noindent for $a\in G$, $f:G\to\Bbb R$ is bounded and uniformly
continuous for the right uniformity, so belongs to $U$.   On the other
hand, if $b$, $c\in D_j$, then $a_1b\notin V_2c$, that is,
$a_1bc^{-1}\notin V_2$ and $\rho(a_1b,c)>1$;  as $c$ is arbitrary,
$f(a_1b)=1$.

\medskip

{\bf (d)} Now $\hat b(f)=f(b)=0$ for every $b\in D_j$, so $z^*(f)=0$.
On the other hand, because $z\mapsto a_1\action z$ is continuous,

\Centerline{$a_1\action z^*\in\overline{\{a_1\action\hat b:b\in D_j\}}
=\overline{\{\widehat{a_1b}:b\in D_j\}}$,}

\noindent so

\Centerline{$(a_1\action z^*)(f)\ge\inf_{b\in D_j}\widehat{a_1b}(f)
=\inf_{b\in D_j}f(a_1b)=1$,}

\noindent and $a_1\action z^*\ne z^*$.   As $z^*$ is arbitrary, this
shows that the action of $G$ on $Z$ has no fixed point, and $G$ is not
extremely amenable.
}%end of proof of 493H

\exercises{\leader{493X}{Basic exercises (a)}
%\spheader 493Xa
Let $G$ be a Hausdorff topological group, and $\widehat{G}$
its completion with respect to its bilateral uniformity.   Show that $G$
is extremely amenable iff $\widehat{G}$ is.
%493Bb 493Bf 449Xi

\sqheader 493Xb Let $X$ be a set with more than one member and $\rho$
the zero-one metric on $X$.
Let $G$ be the isometry group of $X$ with the topology of
pointwise convergence.   Show that $G$ is not extremely
amenable.   \Hint{give $X$ a total
ordering $\le$, and let $x$, $y$ be any two points of $X$.   For
$a\in G$ set $f(a)=1$ if $a^{-1}(x)<a^{-1}(y)$, $-1$ otherwise.   Show
that, in the language of 449D, $f\in U$.   Show that if $\cycle{x\,y}$
is the transposition exchanging $x$ and $y$ then
$\cycle{x\,y}\action_l f=-f$, while $|z(f)|=1$ for every $z$ in the
greatest ambit of $G$.} (Compare 449Xh.)
%493Be 449Xh

\spheader 493Xc Show that under the conditions of 493C there is a
finitely additive functional $\nu:\Cal B\to[0,1]$ such that
$\nu(aF)=\nu F$ for every $a\in G$ and every zero set $F\subseteq G$,
while $\nu(VF)=1$ whenever $V$ is a neighbourhood of the identity, $F$
is a zero set and $\nu F>\bover12$.
%493C

\spheader 493Xd\dvAnew{2013} Prove 493G for infinite-dimensional
inner product spaces over $\Bbb C$.
%493G

\spheader 493Xe Let $X$ be any (real or complex)
inner product space.   Show that the
isometry group of $X$, with its topology of pointwise convergence, is
amenable.   \Hint{449Cd.}
%493G

\spheader 493Xf Let $X$ be a separable Hilbert space.   (i) Show that the
isometry group $G$ of its unit sphere, with its topology of pointwise
convergence, is a Polish group.   (ii) Show that if $X$ is
infinite-dimensional, then every countable discrete group can be embedded
as a closed subgroup of $G$, so that $G$ is an extremely amenable Polish
group with a closed subgroup which is not amenable.   (Cf.\ 449K.)
%493G

\spheader 493Xg\dvAnew{2010} If $X$ is a (real or complex)
Hilbert space, a bounded linear
operator $T:X\to X$ is {\bf unitary} if it is an invertible isometry.
Show that the set of unitary operators on $X$, with its strong operator
topology (3A5I),
is an extremely amenable topological group.
%493G out of order

\spheader 493Xh Let $G$ be a topological group carrying Haar measures.
Show that it is extremely amenable iff its topology is the indiscrete
topology.   \Hint{443L.}
%493H

\leader{493Y}{Further exercises (a)}
%\spheader 493Ya
For a Boolean algebra $\frak A$ and a group $G$ with
identity $e$, write $S(\frak A;G)$ for the set of partitions of unity
$\family{g}{G}{a_g}$ in $\frak A$ such that $\{g:a_g\ne 0\}$ is finite.
For $\family{g}{G}{a_g}$, $\family{g}{G}{b_g}\in S(A)$, write
$\family{g}{G}{a_g}\cdot\family{g}{G}{b_g}=\family{g}{G}{c_g}$ where
$c_g=\sup\{a_h\Bcap b_{h^{-1}g}:h\in G\}$ for $g\in G$.   (i) Show that
under this operation $S(\frak A;G)$ is a group.   (ii) Show that if we
write $h\chi a$ for the member $\family{g}{G}{a_g}$ of $S(\frak A;G)$
such that $a_h=a$ and $a_g=0$ for other $g\in G$, then
$g\chi a\cdot h\chi b=(gh)\chi(a\Bcap b)$, and $S(\frak A;G)$ is
generated by $\{g\chi a:g\in G,\,a\in\frak A\}$.   (iii) Show that if
$\frak A=\Sigma/\Cal I$ where $\Sigma$ is an algebra of subsets of a set
$X$ and $\Cal I$ is an ideal of $\Sigma$, then $S(\frak A;G)$ can be
identified with a space of equivalence classes in a suitable subgroup of
$G^X$.   (iv) Devise a universal mapping theorem for the construction
$S(\frak A;G)$ which matches 361F in the case $(G,\cdot)=(\Bbb R,+)$.
(v) Now suppose that
$(\frak A,\bar\mu)$ is a measure algebra and that $G$ is a topological
group.   Show that we have a topology on $S(\frak A;G)$,
making it a topological group,
for which basic neighbourhoods of the identity
$e\chi 1$ are of the form $V(c,\epsilon,U)=\{\family{g}{G}{a_g}:
\bar\mu(c\Bcap\sup_{g\in G\setminus U}a_g)\le\epsilon\}$ with
$\bar\mu c<\infty$, $\epsilon>0$ and $U$ a neighbourhood of the identity
in $G$.   (vi) Show that if $G$ is an amenable locally compact Hausdorff
group and $(\frak A,\bar\mu)$ is an atomless measure algebra, then
$S(\frak A;G)$ is extremely amenable.   \Hint{{\smc Pestov 02}.}
*(vi) Explore possible constructions of spaces $L^0(\frak A;G)$.
(See {\smc Hartman \& Mycielski 58}.)
%493E

}%end of exercises

\endnotes{
\Notesheader{493} In writing this section I have relied heavily on
{\smc Pestov 99} and {\smc Pestov 02}, where you may find many further
examples of extremely amenable groups.   It is a striking fact that
while the theories of locally compact groups and extremely amenable
groups are necessarily almost entirely separate (493H), both are
dependent on measure theory.   Curiously, what seems to have been the
first non-trivial extremely amenable group to be described was found in
the course of investigating the Control Measure Problem
({\smc Herer \& Christensen 75}).

The theory of locally compact groups has for seventy years now been a
focal point for measure theory.   Extremely amenable groups have not yet
had such an influence.   But they encourage us to look again at
concentration-of-measure theorems, which are of the highest importance
for quite separate reasons.   In all the principal examples of
this section, and again in the further example to come in \S494,
we need concentration of measure in product spaces
(493D-493E and 494J), permutation groups (494I) or on spheres in
Euclidean space
(493G).   493D and 493E are special cases of a general result in
{\smc Pestov 02} (493Ya(vi)) which itself extends an idea from
{\smc Glasner 98}.   I note that 493D needs only concentration of measure
in $\{0,1\}^I$, while 493E demands something rather closer to the full
strength of Talagrand's theorem 492D.

I have expressed 493G as a theorem about the isometry groups of spheres
in infinite-dimensional inner product spaces;  of course these are
isomorphic to the orthogonal groups of the whole spaces with their
strong operator topologies (476Xd).   Adapting the basic
concentration-of-measure theorem 476K
to the required lemma 493F involves an instructive application of ideas
from \S443.
}%end of notes

\discrpage

\frfilename{mt494.tex}
\versiondate{17.5.13}

\copyrightdate{2009}

\def\glitem#1{{\it #1}\enskip}
\def\mag{\mathop{\text{mag}}\nolimits}

\def\chaptername{Further topics}
\def\sectionname{Groups of measure-preserving automorphisms}

\newsection{494\dvAnew{2009}}

I return to the study of automorphism groups of measure algebras,
as in Chapter 38 of Volume 3, but this time with the
intention of exploring possible topological group structures.
Two topologies in particular have attracted interest,
the `weak' and `uniform' topologies
(494A).   After a brief account of their basic properties
(494B-494C) %494B 494C
I begin work on the four main theorems.   The first is the Halmos-Rokhlin
theorem that if $(\frak A,\bar\mu)$ is the Lebesgue
probability algebra the set of weakly mixing measure-preserving
automorphisms of $\frak A$ which are not mixing is comeager for the weak
topology on $\AmuA$ (494E).   This depends on a striking characterization
of weakly mixing automorphisms of a probability algebra
in terms of eigenvectors of the corresponding operators on
the complex Hilbert space $L^2_{\Bbb C}(\frak A,\bar\mu)$
(494D\cmmnt{, 494Xj(i)}).
It turns out that there is an elegant example of a weakly
mixing automorphism which is not mixing which can be described in terms of
a Gaussian distribution of the kind introduced in \S456, so I give it here
(494F).

We need a couple of preliminary results on fixed-point
subalgebras (494G-494H) before approaching the other three theorems.
If $(\frak A,\bar\mu)$ is an atomless
probability algebra, then $\AmuA$ is extremely amenable under its weak
topology (494L);  if $\AmuA$ is given its uniform topology, then every
group homomorphism from $\AmuA$ to a Polish group is continuous (494O);
finally, there is no strictly increasing sequence of subgroups with union
$\AmuA$ (494Q).   All these results have wide-ranging extensions to
full subgroups of $\AmuA$ subject to certain restrictions on the
fixed-point subalgebras.

\cmmnt{The work of this section will rely heavily on concepts and results
from
Volume 3 which have hardly been mentioned so far in the present volume.
I hope that the cross-references, and the brief remarks in 494Ac-494Ad,
will be adequate.}

\leader{494A}{Definitions}\cmmnt{ ({\smc Halmos 56})}
Let $(\frak A,\bar\mu)$ be a measure algebra,
and $\AmuA$ the group of measure-preserving automorphisms of
$\frak A$\cmmnt{ (see \S383)}.
Write $\frak A^f$ for $\{c:c\in\frak A$, $\bar\mu c<\infty\}$.

\spheader 494Aa I will say that the
{\bf weak topology} on $\AmuA$ is that generated by the pseudometrics
$(\pi,\phi)\mapsto\bar\mu(\pi c\Bsymmdiff\phi c)$
as $c$ runs over $\frak A^f$.
%Tornquist 11:  "weak topology"

\spheader 494Ab I will say that the {\bf uniform topology} on $\AmuA$
is that generated by the pseudometrics

\Centerline{$(\pi,\phi)
\mapsto\sup_{a\in\frak A}\bar\mu(c\Bcap(\pi a\Bsymmdiff\phi a))$}

\noindent as $c$ runs over $\frak A^f$.

\cmmnt{\spheader 494Ac I recall some notation from
Volume 3.   For any Boolean algebra $\frak A$ and $a\in\frak A$,
$\frak A_a$ will be the principal ideal
of $\frak A$ generated by $a$ (312D).
I will generally use the symbol $\iota$
for the identity in the automorphism group $\Aut\frak A$ of $\frak A$.
If $\pi\in\Aut\frak A$ and $a\in\frak A$, $a$ supports $\pi$ if $\pi d=d$
whenever $d\Bcap a=0$;  the support

\Centerline{$\supp\pi=\sup\{a\Bsymmdiff\pi a:a\in\frak A\}$}

\noindent of $\pi$ is the smallest member of $\frak A$ supporting $\pi$,
if this is defined (381Bb, 381Ei).   A subgroup $G$
of $\Aut\frak A$ is `full' if $\phi\in G$ whenever $\phi\in\Aut\frak A$
and there are $\familyiI{a_i}$, $\familyiI{\pi_i}$ such that
$\familyiI{a_i}$ is a partition of unity in $\frak A$ and $\pi_i\in G$ and
$\phi d=\pi_id$ whenever $i\in I$ and $d\Bsubseteq a_i$ (381Be).

If $a$, $b\in\frak A\setminus\{0\}$ are disjoint
and $\pi\in\Aut\frak A$ is such that
$\pi a=b$, then $\cycle{a\,_{\pi}\,b}\in\Aut\frak A$ will be the exchanging
involution defined by saying that

$$\eqalign{\cycle{a\,_{\pi}\,b}(d)
&=\pi d\text{ if }d\Bsubseteq a,\cr
&=\pi^{-1}d\text{ if }d\Bsubseteq b,\cr
&=d\text{ if }d\Bsubseteq 1\Bsetminus(a\Bcup b)\cr}$$

\noindent (381R).

\spheader 494Ad In addition, I will repeatedly use the following ideas.
Suppose that $(\frak A,\bar\mu)$ is a probability algebra (322Aa),
$\frak C$ is a closed subalgebra of $\frak A$ (323I), and
$L^{\infty}(\frak C)$ the $M$-space defined in \S363.   Then for each
$a\in\frak A$ we have a conditional expectation
$u_a\in L^{\infty}(\frak C)$ of $\chi a$ on $\frak C$, so that
$\int_cu_a=\bar\mu(a\Bcap c)$ for every $c\in\frak C$
(365Q\formerly{3{}65R}).

If $\frak A$ is relatively atomless over $\frak C$ (331A), $a\in\frak A$,
and $v\in L^{\infty}(\frak C)$ is such that $0\le v\le u_a$, there is a
$b\in\frak A$ such that $b\Bsubseteq a$
and $v=u_b$ (apply Maharam's lemma 331B
to the functional $c\mapsto\int_cv:\frak C\to[0,1]$).
Elaborating on this, we see that if $\sequencen{v_n}$ is a sequence in
$L^{\infty}(\frak C)^+$ and $\sum_{i=0}^nv_i\le u_a$ for every $n$,
there are disjoint $b_0,\ldots\Bsubseteq a$ such that $v_i=u_{b_i}$ for
every $i$ (choose the $b_i$ inductively).}

\leader{494B}{Proposition}
Let $(\frak A,\bar\mu)$ be a measure algebra, and give $\AmuA$
its weak topology.

(a) $\AmuA$ is a topological group.

(b) $(\pi,a)\mapsto\pi a:\AmuA\times\frak A\to\frak A$ is continuous for
the weak topology on $\AmuA$ and the measure-algebra topology on
$\frak A$.

(c) If $(\frak A,\bar\mu)$ is semi-finite\cmmnt{ (definition:  322Ad)},
$\AmuA$ is Hausdorff.

(d) If $(\frak A,\bar\mu)$ is localizable\cmmnt{ (definition:  322Ae)},
$\AmuA$ is complete under its bilateral uniformity.

(e)\dvAformerly{4{}93Xe}
If $(\frak A,\bar\mu)$ is $\sigma$-finite\cmmnt{ (definition:  322Ac)}
and $\frak A$ has countable
Maharam type\cmmnt{ (definition:  331Fa)}, then $\AmuA$ is a Polish group.

\proof{{\bf (a)}  (Compare 441G.)
Set $\rho_c(\pi,\phi)=\bar\mu(\pi c\Bsymmdiff\phi c)$ for
$\pi$, $\phi\in\AmuA$ and $c\in\frak A^f$;  it is elementary to check that
$\rho_c$ is always a pseudometric, so 494Aa is a proper definition of a
topology.   If $\pi$, $\phi\in\AmuA$ and $c\in\frak A^f$, then for any
$\pi'$, $\phi'\in\AmuA$ we have

$$\eqalign{\rho_c(\pi'\phi',\pi\phi)
&=\bar\mu(\pi'\phi'c\Bsymmdiff\pi\phi c)
\le\bar\mu(\pi'\phi'c\Bsymmdiff\pi'\phi c)
  +\bar\mu(\pi'\phi c\Bsymmdiff\pi\phi c)\cr
&=\bar\mu(\phi'c\Bsymmdiff\phi c)
  +\rho_{\phi c}(\pi',\pi)
=\rho_c(\phi',\phi)+\rho_{\phi c}(\pi',\pi);\cr}$$

\noindent as $c$ is arbitrary, $(\pi',\phi')\mapsto\pi'\phi'$ is continuous
at $(\pi,\phi)$;  thus multiplication is continuous.
If $\phi\in\AmuA$ and $c\in\frak A^f$, then for any $\pi\in\AmuA$

$$\eqalign{\rho_c(\pi^{-1},\phi^{-1})
&=\bar\mu(\pi^{-1}c\Bsymmdiff\phi^{-1}c)
=\bar\mu(c\Bsymmdiff\pi\phi^{-1}c)\cr
&=\bar\mu(\phi\phi^{-1}c\Bsymmdiff\pi\phi^{-1}c)
=\rho_{\phi^{-1}c}(\pi,\phi);\cr}$$

\noindent as $c$ is arbitrary, $\pi\mapsto\pi^{-1}$ is continuous at
$\phi$;  thus inversion is continuous and $\AmuA$ is a topological
group.

\medskip

{\bf (b)} Suppose that $\phi\in\AmuA$, $b\in\frak A$ and that $V$ is a
neighbourhood of $\phi b$ in $\frak A$.   Then there are  $c\in\frak A^f$ and
$\epsilon>0$ such that $V$ includes
$\{d:\bar\mu(c\Bcap(d\Bsymmdiff\phi b))\le 4\epsilon\}$.   In this case,
because inversion in $\AmuA$ is continuous,

\Centerline{$U=\{\pi:\pi\in\AmuA$,
$\bar\mu(\pi^{-1}c\Bsymmdiff\phi^{-1}c)\le\epsilon$,
$\bar\mu(\pi(\phi^{-1}c\Bcap b)
   \Bsymmdiff\phi(\phi^{-1}c\Bcap b))\le\epsilon\}$,}

\Centerline{$V'=\{a:a\in\frak A$,
$\bar\mu(\phi^{-1}c\Bcap(a\Bsymmdiff b))\le\epsilon\}$}

\noindent are neighbourhoods of $\phi$, $b$ respectively.   If
$\pi\in U$ and $a\in V'$, then

$$\eqalign{\bar\mu(c\Bcap(\pi a\Bsymmdiff\phi b))
&\le\bar\mu(c\Bcap(\pi a\Bsymmdiff\pi b))
    +\bar\mu(c\Bcap(\pi b\Bsymmdiff\phi b))\cr
&=\bar\mu(\pi^{-1}c\Bcap(a\Bsymmdiff b))
    +\bar\mu((c\Bcap\pi b)\Bsymmdiff(c\Bcap\phi b))\cr
&\le\bar\mu(\pi^{-1}c\Bsymmdiff\phi^{-1}c)
    +\bar\mu(\phi^{-1}c\Bcap(a\Bsymmdiff b))
    +\bar\mu(\pi(\pi^{-1}c\Bcap b)\Bsymmdiff\phi(\phi^{-1}c\Bcap b))\cr
&\le\epsilon+\epsilon
    +\bar\mu(\pi(\pi^{-1}c\Bcap b)\Bsymmdiff\pi(\phi^{-1}c\Bcap b))\cr
&\mskip100mu
    +\bar\mu(\pi(\phi^{-1}c\Bcap b)\Bsymmdiff\phi(\phi^{-1}c\Bcap b))\cr
&\le 2\epsilon
    +\bar\mu((\pi^{-1}c\Bcap b)\Bsymmdiff(\phi^{-1}c\Bcap b))
    +\epsilon\cr
&\le 3\epsilon
    +\bar\mu(\pi^{-1}c\Bsymmdiff\phi^{-1}c)
\le 4\epsilon,\cr}$$

\noindent and $\pi a\in V$.   As $V$, $\phi$ and $b$ are arbitrary,
$(\pi,a)\mapsto\pi a$ is continuous.

\medskip

{\bf (c)} Because $(\frak A,\bar\mu)$ is semi-finite, the measure-algebra
topology on $\frak A$ is Hausdorff (323Ga), so the product topology on
$\frak A^{\frak A}$ is Hausdorff.   Now
$\pi\mapsto\family{a}{\frak A}{\pi a}:\AmuA\to\frak A^{\frak A}$ is
injective, and by (b) it is continuous, so the topology of $\AmuA$ must be
Hausdorff.

\medskip

{\bf (d)(i)} For $c\in\frak A^f$ and $\pi\in\AmuA$, set
$\theta_c(\pi)=\pi c$;  then $\theta_c:\AmuA\to\frak A^f$
is uniformly continuous for the bilateral uniformity of $\AmuA$ and the
measure metric $\rho$ of $\frak A^f$ (323Ad).   \Prf\ We have
$\rho(d,d')=\bar\mu(d\Bsymmdiff d')$ for $d$, $d'\in\frak A^f$.
Let $\epsilon>0$;  then $U=\{\pi:\rho_c(\pi,\iota)\le\epsilon\}$ is a
neighbourhood of $\iota$ in $\AmuA$,
and $W=\{(\pi,\phi):\phi^{-1}\pi\in U\}$ belongs
to the bilateral uniformity.   If $(\pi,\phi)\in W$, then

\Centerline{$\rho(\theta_c(\pi),\theta_c(\phi))
=\bar\mu(\pi c\Bsymmdiff\phi c)
=\bar\mu(\phi^{-1}\pi c\Bsymmdiff c)
=\rho_c(\phi^{-1}\pi,\iota)\le\epsilon$;}

\noindent as $\epsilon$ is arbitrary, $\theta_c$ is uniformly continuous.\
\Qed

\medskip

\quad{\bf (ii)} Let $\Cal F$ be a filter on $\AmuA$ which is Cauchy for the
bilateral uniformity on $\AmuA$.   If $c\in\frak A^f$, the image filter
$\theta_c[[\Cal F]]$ is Cauchy for the measure metric on
$\frak A^f$ (4A2Ji).   Because
$\frak A^f$ is complete under its measure metric
(323Mc), $\theta_c[[\Cal F]]$
converges to $\psi_0c$ say for the measure metric.

If $c$, $d\in\frak A^f$ and $*$ is either of the
Boolean operations $\Bcap$, $\Bsymmdiff$, then

$$\eqalignno{\psi_0(c*d)
&=\lim_{\pi\to\Cal F}\pi(c*d)
=\lim_{\pi\to\Cal F}\pi c*\pi d
=\lim_{\pi\to\Cal F}\pi c*\lim_{\pi\to\Cal F}\pi d\cr
\displaycause{because $*$ is continuous for the measure metric,
see 323Ma}
&=\psi_0c*\psi_0d.\cr}$$

\noindent So $\psi_0:\frak A^f\to\frak A^f$ is a
ring homomorphism.   Next, if $c\in\frak A^f$, then

\Centerline{$\bar\mu\psi_0c
=\bar\mu(\lim_{\pi\to\Cal F}\pi c)
=\lim_{\pi\to\Cal F}\bar\mu\pi c
=\bar\mu c$}

\noindent because $\bar\mu:\frak A^f\to\coint{0,\infty}$ is continuous
(323Mb).

Now recall that $\pi\mapsto\pi^{-1}:\AmuA\to\AmuA$ is uniformly continuous
for the bilateral uniformity (4A5Hc).   So if we set
$\theta'_c(\pi)=\pi^{-1}c$ for $\pi\in\AmuA$, we can apply the argument
just above to $\theta'$ to find a ring homomorphism
$\psi_0':\frak A^f\to\frak A^f$ such that
$\psi_0'c=\lim_{\pi\to\Cal F}\pi^{-1}c$
for every $c\in\frak A^f$.   To relate $\psi_0$
and $\psi_0'$, we can argue as follows.   Given $c\in\frak A^f$,

$$\eqalign{\bar\mu(c\Bsymmdiff\psi_0\psi'_0c)
&=\bar\mu(c\Bsymmdiff\lim_{\phi\to\Cal F}\phi\psi'_0c)
=\lim_{\phi\to\Cal F}\bar\mu(c\Bsymmdiff\phi\psi'_0c)
=\lim_{\phi\to\Cal F}\bar\mu(\phi^{-1}c\Bsymmdiff\psi'_0c)\cr
&=\bar\mu(\lim_{\phi\to\Cal F}(\phi^{-1}c\Bsymmdiff\psi'_0c))
=\bar\mu((\lim_{\phi\to\Cal F}\phi^{-1}c)\Bsymmdiff\psi'_0c)
=\bar\mu(\psi'_0c\Bsymmdiff\psi'_0c)
=0;\cr}$$

\noindent as $c$ is arbitrary, $\psi_0\psi'_0$ is the identity on
$\frak A^f$.   Similarly, $\psi_0'\psi_0$ is the
identity on $\frak A^f$.   Thus $\psi_0$, $\psi'_0$ are the two halves of a
measure-preserving ring isomorphism of $\frak A^f$.

If we give $\frak A$ its measure-algebra uniformity (323Ab), then
$\psi_0$ is uniformly continuous for the induced uniformity on $\frak A^f$.
\Prf\ If $c$, $d_1$, $d_2\in\frak A^f$, then

\Centerline{$\bar\mu(c\Bcap(\psi_0d_1\Bsymmdiff\psi_0d_2))
=\bar\mu(\psi_0^{-1}c\Bcap(d_1\Bsymmdiff d_2))$.\ \Qed}

\noindent Since $\frak A^f$ is dense in $\frak A$ for the measure-algebra
topology on $\frak A$ (323Bb), and $\frak A$ is complete
for the measure-algebra uniformity (323Gc), there is a unique
extension of $\psi_0$ to a uniformly continuous function
$\psi:\frak A\to\frak A$ (3A4G).   Since the Boolean operations
$\Bsymmdiff$, $\Bcap$ on $\frak A$ are continuous for the measure-algebra
topology (323Ba), $\psi$ is a ring homomorphism.   Similarly, we have a
unique continuous $\psi':\frak A\to\frak A$ extending $\psi'_0$;  since
$\psi\psi'$ and $\psi'\psi$ are continuous functions agreeing with the
identity operator $\iota$ on $\frak A^f$, they are both $\iota$, and
$\psi\in\Aut\frak A$.   To see that $\psi$ is measure-preserving, note
just that if $a\in\frak A$ then

$$\eqalignno{\bar\mu\psi a
&=\sup\{\bar\mu c:c\in\frak A^f,\,c\Bsubseteq\psi a\}
=\sup\{\bar\mu\psi_0c:c\in\frak A^f,\,\psi_0c\Bsubseteq\psi a\}\cr
\displaycause{because $\psi_0$ is a permutation of $\frak A^f$}
&=\sup\{\bar\mu c:c\in\frak A^f,\,\psi c\Bsubseteq\psi a\}
=\sup\{\bar\mu c:c\in\frak A^f,\,c\Bsubseteq a\}
=\bar\mu a.\cr}$$

\noindent Thus $\psi\in\AmuA$.

Finally, $\Cal F\to\psi$.   \Prf\ If $c\in\frak A^f$ and $\epsilon>0$,
there is an $F\in\Cal F$ such that
$\bar\mu(\pi c\Bsymmdiff\phi c)\le\epsilon$ whenever $\pi$, $\phi\in F$.
We have

\Centerline{$\bar\mu(\pi c\Bsymmdiff\psi c)
=\bar\mu(\pi c\Bsymmdiff\lim_{\phi\to\Cal F}\phi c)
=\lim_{\phi\to\Cal F}\bar\mu(\pi c\Bsymmdiff\phi c)
\le\epsilon$}

\noindent for every $\pi\in F$.   As $c$ and $\epsilon$ are arbitrary,
$\Cal F\to\psi$ for the weak topology on $\AmuA$.\ \QeD\   As $\Cal F$ is
arbitrary, the bilateral uniformity is complete.

\medskip

{\bf (e)(i)} The point is that $\frak A^f$ is separable for the measure
metric.   \Prf\ Because $\frak A$ has countable Maharam type, there is a
countable subalgebra $\frak B$ of $\frak A$ which $\tau$-generates
$\frak A$;  by 323J, $\frak B$ is dense in $\frak A$ for the measure
algebra topology.   Next, there is a non-decreasing sequence
$\sequencen{c_n}$ in $\frak A$ with supremum $1$.   Set
$D=\{b\Bcap c_n:b\in\frak B$, $n\in\Bbb N\}$.   Then $D$ is a countable
subset of $\frak A^f$.   If $c\in\frak A^f$ and $\epsilon>0$, there are an
$n\in\Bbb N$ such that $\bar\mu(c\Bsetminus c_n)\le\epsilon$, and a
$b\in\frak B$ such that $\bar\mu(c_n\Bcap(c\Bsymmdiff b))\le\epsilon$.
Now $d=b\Bcap c_n$ belongs to $D$, and

$$\eqalign{\bar\mu(c\Bsymmdiff d)
&\le\bar\mu(c\Bsymmdiff(c\Bcap c_n))
   +\bar\mu((c\Bcap c_n)\Bsymmdiff(b\Bcap c_n))\cr
&=\bar\mu(c\Bsetminus c_n)
   +\bar\mu(c_n\Bcap(c\Bsymmdiff b))
\le 2\epsilon.\cr}$$

\noindent As $c$ and $\epsilon$ are arbitrary, $D$ is dense in $\frak A^f$
and $\frak A^f$ is separable.\ \Qed

\medskip

%regular Hausdorff second-countable > separable & metrizable 4{}A2Pb
%countable subbase > second-countable 4{}A2Oa

\quad{\bf (ii)} Let $D$ be a countable dense subset of $\frak A^f$, and
$\Cal U$ the family of sets of the form

\Centerline{$\{\pi:\pi\in\AmuA$, $\bar\mu(d\Bsymmdiff\pi d')<2^{-n}\}$}

\noindent where $d$, $d'\in D$ and $n\in\Bbb N$.   All these sets are open
for the weak topology.
\Prf\ If $U=\{\pi:\bar\mu(d\Bsymmdiff\pi d')<2^{-n}\}$ and $\phi\in U$,
set $\eta=\bover13(2^{-n}-\bar\mu(d\Bsymmdiff\phi d'))$.   Then
$V=\{\pi:\bar\mu(d\Bcap(\pi d'\Bsymmdiff\phi d'))\le\eta\}$ is a
neighbourhood of $\phi$.   If $\pi\in V$, then

\Centerline{$\bar\mu(d\Bsymmdiff\pi d')
\le\bar\mu(d\Bsymmdiff\phi d')+\bar\mu(\phi d'\Bsymmdiff\pi d')
<2^{-n}$}

\noindent and $\pi\in U$.   Thus $\phi\in\interior U$;  as $\phi$ is
arbitrary, $U$ is open.\ \Qed

\medskip

\quad{\bf (iii)} In fact $\Cal U$ is a subbase for the weak topology on
$\AmuA$.   \Prf\ If $W\subseteq\AmuA$ is open and $\phi\in W$, there are
$c_0,\ldots,c_n\in\frak A^f$ and $k\in\Bbb N$
such that $W$ includes
$\{\pi:\bar\mu(\pi c_i\Bsymmdiff\phi c_i)\le 2^{-k}$ for every
$i\le n\}$.   Let $d_0,\ldots,d_n,d_0',\ldots,d'_n\in D$ be such that
$\bar\mu(d_i\Bsymmdiff c_i)<2^{-k-2}$,
$\bar\mu(d'_i\Bsymmdiff\phi c_i)<2^{-k-2}$ for each $i\le n$.   Set
$U_i=\{\pi:\bar\mu(d'_i\Bsymmdiff\pi d_i)<2^{-k-1}\}$;  then
$U_i\in\Cal U$ and $\phi\in U_i$ for each $i\le n$, because

\Centerline{$\bar\mu(d'_i\Bsymmdiff\phi d_i)
\le\bar\mu(d'_i\Bsymmdiff\phi c_i)+\bar\mu(\phi c_i\Bsymmdiff\phi d_i)
=\bar\mu(d'_i\Bsymmdiff\phi c_i)+\bar\mu(c_i\Bsymmdiff d_i)
<2^{-k-1}$.}

\noindent If $\pi\in U_i$, then

\Centerline{$\bar\mu(\pi c_i\Bsymmdiff\phi c_i)
\le\bar\mu(\pi c_i\Bsymmdiff\pi d_i)+\bar\mu(\pi d_i\Bsymmdiff d'_i)
   +\bar\mu(d'_i\Bsymmdiff\phi c_i)
\le 2^{-k}$,}

\noindent so $\bigcap_{i\le n}U_i\subseteq W$.   As $W$ and $\phi$ are
arbitrary, $\Cal U$ is a subbase for the topology.\ \Qed

\medskip

\quad{\bf (iv)} Since $\Cal U$ is countable, the weak topology is
second-countable (4A2Oa).   Since the weak topology is a group
topology, it is regular (4A5Ha, or otherwise);  by (c) above it is
Hausdorff;  so by 4A2Pb it is separable and metrizable.   Accordingly the
bilateral uniformity is metrizable (4A5Q(v));  by (d) above, $\AmuA$ is
complete under the bilateral uniformity, so its topology is Polish.
}%end of proof of 494B

\leader{494C}{Proposition}
Let $(\frak A,\bar\mu)$ be a measure algebra, and give
$\AmuA$ its uniform topology.

(a) $\AmuA$ is a topological group.

(b) For $c\in\frak A^f$ and $\epsilon>0$, set

\Centerline{$U(c,\epsilon)=\{\pi:\pi\in\AmuA$, $\pi$ is supported by an
$a\in\frak A$ such that $\bar\mu(c\Bcap a)\le\epsilon\}$.}

\noindent Then $\{U(c,\epsilon):c\in\frak A^f$, $\epsilon>0\}$ is a base of
neighbourhoods of $\iota$.

(c) The set of
periodic measure-preserving automorphisms of $\frak A$ with supports of
finite measure is dense in $\AmuA$.

(d) The weak topology on $\AmuA$ is coarser than the uniform
topology.

(e) If $(\frak A,\bar\mu)$ is semi-finite, $\AmuA$ is Hausdorff.

(f) If $(\frak A,\bar\mu)$ is localizable,
$\AmuA$ is complete under its bilateral uniformity.

(g) If $(\frak A,\bar\mu)$ is semi-finite and
$G$ is a full subgroup of $\AmuA$,
then $G$ is closed.

(h) If $(\frak A,\bar\mu)$ is $\sigma$-finite, then $\AmuA$ is
metrizable.

(i) Suppose that $(\frak A,\bar\mu)$ is $\sigma$-finite and $\frak A$ has
countable Maharam type.   If $D\subseteq\AmuA$ is countable, then the full
subgroup $G$ of $\AmuA$ generated by $D$, with its induced topology,
is a Polish group.

\proof{{\bf (a)} For $\pi$, $\phi\in\AmuA$ and $c\in\frak A^f$, set
$\rho'_c(\pi,\phi)
=\sup_{a\in\frak A}\bar\mu(c\Bcap(\pi a\Bsymmdiff\phi a))$;  as in part (a)
of the proof of 494B, it is
elementary that every $\rho'_c$ is a pseudometric, so the uniform topology
$\frak T_u$ is properly defined.
If $\pi$, $\phi$, $\pi'$, $\phi'\in\AmuA$,
$c\in\frak A^f$ and $a\in\frak A$, then

$$\eqalign{\bar\mu(c\Bcap(\pi'\phi'a\Bsymmdiff\pi\phi a))
&\le\bar\mu(c\Bcap(\pi'\phi'a\Bsymmdiff\pi\phi'a))
  +\bar\mu(c\Bcap(\pi\phi'a\Bsymmdiff\pi\phi a))\cr
&\le\rho'_c(\pi',\pi)
  +\bar\mu(\pi^{-1}c\Bcap(\phi'a\Bsymmdiff\phi a))\cr
&\le\rho'_c(\pi',\pi)
  +\rho'_{\pi^{-1}c}(\phi',\phi);\cr}$$

\noindent as $a$ is arbitrary,
$\rho'_c(\pi'\phi',\pi\phi)
\le\rho'_c(\pi',\pi)+\rho'_{\pi^{-1}c}(\phi',\phi)$;  as $c$ is arbitrary,
$(\pi',\phi')\mapsto\pi'\phi'$ is continuous at $(\pi,\phi)$;  thus
multiplication is continuous.   If $\pi$, $\phi\in\AmuA$, $c\in\frak A^f$
and $a\in\frak A$, then

\Centerline{$\bar\mu(c\Bcap(\pi^{-1}a\Bsymmdiff\phi^{-1}a))
=\bar\mu(\phi c\Bcap(\phi\pi^{-1}a\Bsymmdiff\pi\pi^{-1}a)
\le\rho'_{\phi c}(\phi,\pi)$;}

\noindent thus $\rho'_c(\pi^{-1},\phi^{-1})\le\rho'_{\phi c}(\pi,\phi)$,
$\pi\mapsto\pi^{-1}$ is continuous at $\phi$, and inversion is continuous.
So once more we have a topological group.

\medskip

{\bf (b)(i)} If $c\in\frak A^f$, $\pi\in\AmuA$ and $\epsilon>0$ are such
that $\rho'_c(\pi,\iota)\le\bover13\epsilon$, then $\pi\in U(c,\epsilon)$.
\Prf\ Consider
$A=\{a:a\in\frak A_c$, $a\Bcap\pi a=0\}$.   Then

\Centerline{$\bar\mu a\le\bar\mu(c\Bcap(\pi a\Bsymmdiff a))
\le\rho'_c(\pi,\iota)\le\Bover13\epsilon$}

\noindent for
every $a\in A$.   If $B\subseteq A$ is upwards-directed, then
$b^*=\sup B$ is defined in $\frak A$, and
$\bar\mu b^*=\sup_{b\in B}\bar\mu b$ (321C).   Now
$\pi b^*=\sup_{b\in B}\pi b$, so
$b^*\Bcap\pi b^*=\sup_{b\in B}b\Bcap\pi b=0$, and $b^*\in A$.
By Zorn's Lemma, $A$ has a maximal element $a^*$.
Suppose that $d\in\frak A_c$ is
disjoint from $\pi^{-1}a^*\Bcup a^*\Bcup\pi a^*$.   Then
$a^*\Bcup(d\Bsetminus\pi d)\in A$;  by the maximality of $a^*$,
$d\Bsubseteq\pi d$ and $d=\pi d$ (because $\bar\mu d=\bar\mu\pi d<\infty$).
Thus
$(1\Bsetminus c)\Bcup(\pi^{-1}a^*\Bcup a^*\Bcup\pi a^*)$ supports $\pi$
and witnesses that $\pi\in U(c,\epsilon)$.\ \Qed

So every $U(c,\epsilon)$ is a $\frak T_u$-neighbourhood of $\iota$.

\medskip

\quad{\bf (ii)} Conversely, if $c\in\frak A^f$, $\epsilon>0$ and
$\pi\in U(c,\epsilon)$, then $\rho'_c(\pi,\iota)\le\epsilon$.   \Prf\
Let $d\in\frak A$ be such that $\pi$ is supported by $d$ and
$\bar\mu(c\Bcap d)\le\epsilon$.   Then, for any $a\in\frak A$,
$a\Bsymmdiff\pi a\Bsubseteq d$, so
$\bar\mu(c\Bcap(a\Bsymmdiff\pi a))\le\epsilon$;  which is what we need to
know.\ \Qed

So $\{U(c,\epsilon):c\in\frak A^f$, $\epsilon>0\}$ is a base of
neighbourhoods of $\iota$ for $\frak T_u$.

\medskip

{\bf (c)} Take a non-empty open subset $U$ of $\AmuA$ and
$\phi\in U$.

\medskip

\quad{\bf (i)} By (b), there are a $c\in\frak A^f$ and an $\epsilon>0$
such that $U(c,3\epsilon)U(c,3\epsilon)\subseteq U^{-1}\phi$.   Now there
is a $\psi\in\AmuA$ such that $\psi^{-1}\phi\in U(c,3\epsilon)$
and $\psi$ is supported by $e=c\Bcup\phi c$.
\Prf\ By 332L, applied to $\frak A_e$ and $\phi\restrp\frak A_c$,
there is a measure-preserving automorphism $\psi_0:\frak A_e\to\frak A_e$
agreeing with $\phi$ on $\frak A_c$;  now set

\Centerline{$\psi a=\psi_0(a\Bcap e)\Bcup(a\Bsetminus e)$}

\noindent for every $a\in\frak A$ to get $\psi\in\AmuA$ agreeing with
$\phi$ on $\frak A_c$ and supported by $e$.   As
$\psi^{-1}\phi a=a$ for $a\Bsubseteq c$, $\psi^{-1}\phi$ is supported by
$1\Bsetminus c$ and belongs to $U(c,3\epsilon)$.\ \Qed

\medskip

\quad{\bf (ii)} By 381H, applied to $\psi\restrp\frak A_e$,
there is a partition
$\langle c_m\rangle_{1\le m\le\omega}$ of unity in $\frak A_e$ such that
$\psi c_m\Bsubseteq c_m$ for every $m$, $\psi\restrp\frak A_{c_m}$ is
periodic with period $m$ for every $m\in\Bbb N\setminus\{0\}$, and
$\psi\restrp\frak A_{c_{\omega}}$ is aperiodic.   Of course $\psi c_m=c_m$
for every $m$, just because $\bar\mu\psi c_m=\bar\mu c_m$.
Let $n\ge 1$ be such that $\bar\mu c_{\omega}\le n!\epsilon$ and
$\bar\mu(\sup_{n<m<\omega}c_m)\le\epsilon$.
By the Halmos-Rokhlin-Kakutani lemma (386C), applied to
$\psi\restrp\frak A_{c_{\omega}}$, there is a $b\Bsubseteq c_{\omega}$
such that $b$, $\psi b,\ldots,\psi^{n!-1}b$ are disjoint and
$\bar\mu(c_{\omega}\Bsetminus\sup_{i<n!}\psi^ib)\le\epsilon$.
Note that $\bar\mu b$ is also at most $\epsilon$.

Set

\Centerline{$d=\sup_{n<m<\omega}c_m
  \Bcup(c_{\omega}\Bsetminus\sup_{i<n!}\psi^ib)$,
\quad$d'=d\Bcup\psi^{n!-1}b$,}

\noindent and let $\pi:\frak A\to\frak A$ be the measure-preserving Boolean
automorphism such that

$$\eqalign{\pi a
&=\psi a\text{ if }1\le m\le n\text{ and }a\Bsubseteq c_m,\cr
&=\psi a\text{ if }0\le i\le n!-2\text{ and }a\Bsubseteq\psi^ib,\cr
&=\psi^{-n!+1}a\text{ if }a\Bsubseteq\psi^{n!-1}b,\cr
&=a\text{ if }a\Bsubseteq d\Bcup(1\Bsetminus e).\cr}$$

\noindent Then

$$\eqalign{\pi^{n!}a
&=\psi^{n!}a=a\text{ if }1\le m\le n\text{ and }a\Bsubseteq c_m,\cr
&=\psi^{n!}\psi^{-n!}a=a
   \text{ if }0\le i<n!\text{ and }a\Bsubseteq\psi^ib,\cr
&=a\text{ if }a\Bsubseteq d\cup(1\Bsetminus e),\cr}$$

\noindent so $\pi^{n!}=\iota$ and $\pi$ is periodic.
Since $\pi$ is supported by $e$, and $\frak A_e$ is Dedekind complete,
$\pi$ has a support of finite measure.   On the other hand
$\pi a=\psi a$ whenever $a\Bcap d'=0$, so
$\pi^{-1}\psi$ is supported by $d'$ and
belongs to $U(c,\bar\mu d')\subseteq U(c,3\epsilon)$.

Now

\Centerline{$\pi^{-1}\phi
=\pi^{-1}\psi\psi^{-1}\phi
\in U(c,3\epsilon)U(c,3\epsilon)
\subseteq U^{-1}\phi$,
\quad$\pi\in U$;}

\noindent as $U$ is arbitrary,
the set of periodic automorphisms with supports of finite measure
is dense in $\AmuA$.

\medskip

{\bf (d)}
Let $V$ be a neighbourhood of the identity $\iota$ for the weak topology
$\frak T_w$ on $\AmuA$.   Then there are $c_0,\ldots,c_k\in\frak A^f$
and $\epsilon_0,\ldots,\epsilon_k>0$ such that

\Centerline{$V\supseteq\{\pi:
  \bar\mu(c_i\Bsymmdiff\pi c_i)\le\epsilon_i$ for every $i\le k\}$.}

\noindent Set $c=\sup_{i\le k}c_i$,
$\epsilon=\bover12\min_{i\le k}\epsilon_i$.   If $\pi\in U(c,\epsilon)$ as
defined in (b), there is an $a\in\frak A$, supporting $\pi$, such that
$\bar\mu(c\Bcap a)\le\epsilon$.   In this case, for each $i\le k$,
$c_i\Bsetminus\pi c_i\Bsubseteq c\Bcap a$, so

\Centerline{$\bar\mu(c_i\Bsymmdiff\pi c_i)
=2\bar\mu(c_i\Bsetminus\pi c_i)
\le 2\bar\mu(c\Bcap a)
\le\epsilon_i$.}

\noindent Thus $V\supseteq U(c,\epsilon)$ and $V$ is a neighbourhood of
$\iota$ for $\frak T_u$.   As $\AmuA$ is a topological group
under either topology, it follows that
$\frak T_u$ is finer than $\frak T_w$ (4A5Fb).

\medskip

{\bf (e)} Because $(\frak A,\bar\mu)$ is semi-finite,
the weak topology is Hausdorff (494Bc), so the uniform topology,
being finer, must also be Hausdorff.

\medskip

{\bf (f)} Let $\Cal F$ be a Cauchy filter for the
$\frak T_u$-bilateral uniformity on $\AmuA$.   Because the identity map
from $(\AmuA,\frak T_u)$ to $(\AmuA,\frak T_w)$ is continuous ((d) above),
it is uniformly continuous for the corresponding bilateral uniformities
(4A5Hd), and $\Cal F$ is Cauchy for the $\frak T_w$-bilateral
uniformity (4A2Ji).   It follows that $\Cal F$ has a $\frak T_w$-limit
$\psi$ say (494Bd), in which case $\psi a$ is the limit
$\lim_{\pi\to\Cal F}\pi a$, for the measure-algebra topology of $\frak A$,
for every $a\in\frak A$ (494Bb).
But $\psi$ is also the $\frak T_u$-limit of $\Cal F$.
\Prf\ Suppose that $c\in\frak A^f$ and $\epsilon>0$.
Set $V(c,\epsilon)=\{\pi:\rho'_c(\pi,\iota)\le\epsilon\}$, where
$\rho'_c$ is defined as (a) above.   Then
$V(c,\epsilon)$ is a $\frak T_u$-neighbourhood of
$\iota$, so $\{(\pi,\phi):\phi\pi^{-1}\in V(c,\epsilon)\}$ belongs to the
$\frak T_u$-bilateral uniformity, and there is an $F\in\Cal F$ such that
$\phi\pi^{-1}\in V(c,\epsilon)$ whenever $\pi$, $\phi\in F$.

Now if $\phi\in F$ and $a\in\frak A$,

$$\eqalignno{\bar\mu(c\Bcap(\phi a\Bsymmdiff\psi a))
&=\bar\mu(c\Bcap(\phi a\Bsymmdiff\lim_{\pi\to\Cal F}\pi a))
=\lim_{\pi\to\Cal F}\bar\mu(c\Bcap(\phi a\Bsymmdiff\pi a))\cr
\displaycause{because $b\mapsto\bar\mu(c\Bcap(\phi a\Bsymmdiff b))$ is
continuous}
&=\lim_{\pi\to\Cal F}\bar\mu(c\Bcap(\phi\pi^{-1}\pi a\Bsymmdiff\pi a))
\le\sup_{\pi\in F,b\in\frak A}\bar\mu(c\Bcap(\phi\pi^{-1}b\Bsymmdiff b))
\le\epsilon.\cr}$$

\noindent Thus $\rho'_c(\phi,\psi)\le\epsilon$ for every $\phi\in F$.
As $c$ and $\epsilon$ are arbitrary, $\Cal F$ is $\frak T_u$-convergent to
$\psi$.\ \Qed

As $\Cal F$ is arbitrary, $\AmuA$ is complete for the $\frak T_u$-bilateral
uniformity.

\medskip

{\bf (g)(i)} Suppose that $\phi$ belongs to the closure of $G$ in $\AmuA$.
Let $B$ be the set of those $b\in\frak A$ for which there is a $\pi\in G$
such that $\pi$ and $\phi$ agree on the principal ideal $\frak A_b$.
Then $B$ is order-dense
in $\frak A$.   \Prf\ Suppose that $a\in\frak A\setminus\{0\}$.
Because $(\frak A,\bar\mu)$ is semi-finite, there is a non-zero
$c\in\frak A^f$ such that $c\Bsubseteq a$.   Take
$\epsilon\in\ooint{0,\bar\mu c}$.   Then there is a $\pi\in G$ such that
$\pi^{-1}\phi\in U(c,\epsilon)$.   Let $d\in\frak A$ be such that $d$
supports $\pi^{-1}\phi$ and $\bar\mu(c\Bcap d)\le\epsilon$.   Set
$b=c\Bsetminus d$.   If $b'\Bsubseteq b$, then
$\pi^{-1}\phi b'=b'$, that is, $\phi b'=\pi b'$;  so $\pi$ and $\phi$ agree
on $\frak A_b$ and $b\in B$, while $0\ne b\Bsubseteq a$.\ \Qed

\medskip

\quad{\bf (ii)} There is therefore a partition $\familyiI{b_i}$ of unity
consisting of members of $B$.   For each $i\in I$ take $\pi_i\in G$ such
that $\pi_i$ and $\phi$ agree on $\frak A_{b_i}$;  because $G$ is full,
$\familyiI{(b_i,\pi_i)}$ witnesses that $\phi\in G$.   As $\phi$ is
arbitrary, $G$ is closed.

\medskip

{\bf (h)} Let $\sequencen{c_n}$ be a non-decreasing sequence in $\frak A^f$
with supremum $1$.   Then $\{U(c_n,2^{-n}):n\in\Bbb N\}$ is a base of
neighbourhoods of $\iota$.   \Prf\ If $c\in\frak A^f$
and $\epsilon>0$, there is an $n\in\Bbb N$ such that
$\bar\mu(c\Bsetminus c_n)+2^{-n}\le\epsilon$.   If $\pi\in U(c_n,2^{-n})$,
there is an $a\in\frak A$, supporting $\pi$, such that
$\bar\mu(c_n\Bcap a)\le 2^{-n}$;  in which case
$\bar\mu(c\Bcap a)\le\epsilon$ and $\pi\in U(c,\epsilon)$.   Thus we have
found an $n$ such that $U(c_n,2^{-n})\subseteq U(c,\epsilon)$.\ \Qed

By 4A5Q, $\AmuA$ is metrizable.

\medskip

{\bf (i)}\grheada\ By (h), $\AmuA$ and therefore $G$ are metrizable;  the
bilateral uniformity of $\AmuA$ is therefore metrizable (4A5Q(v)).
By (f), $\AmuA$ is complete under its bilateral uniformity;  by
(g), $G$ is closed, so is complete under the induced uniformity.   So
there is a metric on $G$, inducing its topology, under which $G$ is
complete, and all I have to show is that $G$ is separable.

\medskip

\quad\grheadb\ Since the subgroup of $\AmuA$ generated by $D$ is again
countable, we may suppose that $D$ is itself a subgroup of $\AmuA$.
Let $\sequencen{c_n}$ be a non-decreasing sequence in $\frak A^f$
with supremum $1$, and $\frak B$ a countable subalgebra of $\frak A$,
which $\tau$-generates $\frak A$;  by 323J again,
$\frak B$ is dense in $\frak A$ for the measure-algebra topology.   For
$m$, $n\in\Bbb N$, $\pi_0,\ldots,\pi_m\in D$ and
$b_0,\ldots,b_m\in\frak B$, write
$E(m,n,\pi_0,\ldots,\pi_m,b_0,\ldots,b_m)$ for

\Centerline{$\{\pi:\pi\in G$,
$\sum_{i=0}^m\bar\mu(c_n\Bcap b_i\Bcap\supp(\pi^{-1}\pi_i))\le 2^{-n}\}$.}

\noindent (The supports are defined because $\frak A$ is Dedekind complete;
see 381F.)   Let $D'\subseteq G$ be a countable set such that
$D'\cap E(m,n,\pi_0,\ldots,\pi_m,b_0,\ldots,b_m)$ is non-empty whenever
$m$, $n\in\Bbb N$, $\pi_0,\ldots,\pi_m\in D$ and
$b_0,\ldots,b_m\in\frak B$ are
such that $E(m,n,\pi_0,\ldots,\pi_m,b_0,\ldots,b_m)$ is non-empty.

Suppose that $\pi\in G$, $c\in\frak A^f$ and $\epsilon>0$.
Let $n\in\Bbb N$ be such that
$\bar\mu(c\Bsetminus c_n)+2^{-n+2}<\epsilon$.   We have a family
$\family{j}{J}{(a_j,\pi_j)}$ such that
$\family{j}{J}{a_j}$ is a partition
of unity in $\frak A$ consisting of elements of finite measure, and, for
each $j\in J$, $\pi_j\in D$ and $\pi$ agrees with $\pi_j$ on
$\frak A_{a_j}$ (381Ia),
that is, $a_j\Bcap\supp(\pi^{-1}\pi_j)=0$.
Let $j_0,\ldots,j_m\in J$ be such that
$\bar\mu(c_n\Bsetminus\sup_{i\le m}a_{j_i})\le 2^{-n}$;  for each
$i\le m$, let $b_i\in\frak B$ be such that
$\bar\mu(c_n\Bcap(b_i\Bsymmdiff a_{j_i}))\le\Bover{2^{-n}}{m+1}$.  In
this case, $\pi\in E(m,n,\pi_{j_0},\ldots,\pi_{j_m},b_0,\ldots,b_m)$, so
there is a
$\tilde\pi\in D'\cap E(m,n,\pi_{j_0},\ldots,\pi_{j_m},b_0,\ldots,b_m)$.
Consider $d=\supp(\pi^{-1}\tilde\pi)$.   If we set
$d_i=\supp(\pi^{-1}\pi_{j_i})\Bcup\supp(\tilde\pi^{-1}\pi_{j_i})$, then
$\pi$ and $\tilde\pi$ both agree with $\pi_{j_i}$ on $1\Bsetminus d_i$,
so $d\Bsubseteq d_i$.   Now

$$\eqalign{\bar\mu(c_n\Bcap d)
&\le\bar\mu(c_n\Bsetminus\sup_{i\le m}b_i)
   +\sum_{i=0}^m\bar\mu(c_n\Bcap b_i\Bcap d)\cr
&\le\bar\mu(c_n\Bsetminus\sup_{i\le m}a_{j_i})
   +\sum_{i=0}^m\bar\mu(a_{j_i}\Bsetminus b_i)
   +\sum_{i=0}^m\bar\mu(c_n\Bcap b_i\Bcap d_i)\cr
&\le 2^{-n}+2^{-n}
   +\sum_{i=0}^m\bar\mu(c_n\Bcap b_i\Bcap\supp(\pi^{-1}\pi_{j_i}))
   +\sum_{i=0}^m\bar\mu(c_n\Bcap b_i\Bcap\supp(\tilde\pi^{-1}\pi_{j_i}))\cr
&\le 4\cdot 2^{-n}=2^{-n+2},\cr}$$

\noindent and $\bar\mu(c\Bcap d)<\epsilon$.
But this means that $\pi^{-1}\tilde\pi\in U(c,\epsilon)$;  as $c$,
$\epsilon$ and $\pi$ are arbitrary, $D'$ is dense in $G$ and $G$ is
separable.
}%end of proof of 494C

\leader{494D}{Lemma}\dvAnew{2011}
Let $(\frak A,\bar\mu)$ be a probability algebra and
$\phi\in\Aut_{\bar\mu}\frak A$.   Let
$T=T_{\phi}:L^2_{\Bbb C}\to L^2_{\Bbb C}$ be
the corresponding operator on the complex Hilbert space
$L^2_{\Bbb C}=L^2_{\Bbb C}(\frak A,\bar\mu)$\cmmnt{ (366M)}.
Then the following are equiveridical:

\inset{($\alpha$) $\phi$ is weakly
mixing\cmmnt{ (definition:  372Ob)};

($\beta$) $\inf_{k\in\Bbb N}|\innerprod{T^kw}{w}|<1$ whenever
$w\in L^2_{\Bbb C}$, $\|w\|_2=1$ and $\int w=0$;

($\gamma$) $\inf_{k\in\Bbb N}|\innerprod{T^kw}{w}|=0$ whenever
$w\in L^2_{\Bbb C}$, $\|w\|_2=1$ and $\int w=0$.}

\proof{{\bf (a)}
Regarding $\Bbb Z$, with addition and its discrete
topology, as a topological group, its dual group is the circle group
$S^1=\{z:z\in\Bbb C$, $|z|=1\}$ with multiplication and its usual topology
(445Bb-445Bc);  the duality being given by the functional
$(k,z)\mapsto z^k:\Bbb Z\times S^1\to S^1$.
For $u\in L^2_{\Bbb C}$, define $h_u:\Bbb Z\to\Bbb C$ by
setting $h_u(k)=\innerprod{T^ku}{u}$ for $k\in\Bbb Z$.   Then $h_u$ is
positive definite in the sense of 445L.   \Prf\ If
$\zeta_0,\ldots,\zeta_n\in\Bbb C$ and $m_0,\ldots,m_n\in\Bbb Z$, then

$$\eqalignno{\sum_{j,k=0}^n\zeta_j\bar\zeta_kh_u(m_j-m_k)
&=\sum_{j,k=0}^n\zeta_j\bar\zeta_k\innerprod{T^{m_j-m_k}u}{u}
=\sum_{j,k=0}^n\zeta_j\bar\zeta_k\innerprod{T^{m_j}u}{T^{m_k}u}\cr
\displaycause{366Me}
&=\innerprod{\sum_{j=0}^n\zeta_jT^{m_j}u}
  {\sum_{k=0}^n\zeta_kT^{m_k}u}
\ge 0.  \text{ \Qed}\cr}$$

\noindent By Bochner's theorem (445N),
there is a Radon probability measure $\nu_u$ on $S^1$
such that

\Centerline{$\int z^k\nu_u(dz)=h_u(k)=\innerprod{T^ku}{u}$}

\noindent for every $k\in\Bbb Z$.   Note that

\Centerline{$\nu_u(S^1)=\int z^0d\nu_u=\innerprod{u}{u}=\|u\|_2^2$.}

\medskip

{\bf (b)(i)} Let $P\subseteq C(S^1;\Bbb C)$ be the set of
functions which are expressible in the form

\Centerline{$p(z)=\sum_{k\in\Bbb Z}\zeta_kz^k$ for every $z\in S^1$}

\noindent where $\zeta_k\in\Bbb C$ for every $k\in\Bbb Z$ and
$\{k:\zeta_k\ne 0\}$ is finite.    Then $P$ is a linear subspace of
the complex Banach space $C(S^1;\Bbb C)$, closed under multiplication.
Also, if $p\in P$, then $\bar p\in P$, where
$\bar p(z)=\overline{p(z)}$ for every $z\in S^1$.
\Prf\ If $p(z)=\sum_{k\in\Bbb Z}\zeta_kz^k$, then

\Centerline{$\bar p(z)=\sum_{k\in\Bbb Z}\bar\zeta_kz^{-k}
=\sum_{k\in\Bbb Z}\bar\zeta_{-k}z^k$}

\noindent for every $z\in S^1$.\ \QeD\
Of course $P$ contains the constant function $z\mapsto z^0$ and the
identity function $z\mapsto z$, so by the Stone-Weierstrass theorem (281G)
$P$ is $\|\,\|_{\infty}$-dense in $C(S^1;\Bbb C)$.

\medskip

\quad{\bf (ii)} For any $p\in P$ the coefficients of the
corresponding expression
$p(z)=\sum_{k\in\Bbb Z}\zeta_kz^k$ are uniquely defined, since

\Centerline{$\zeta_k=\Bover1{2\pi}\int_{-\pi}^{\pi}e^{-ikt}p(e^{it})dt$}

\noindent for every $k$.   So we can define $u_p$, for every
$u\in L^2_{\Bbb C}$, by saying that $u_p=\sum_{k\in\Bbb Z}\zeta_kT^ku$.
Now we have

\Centerline{$\innerprod{u_p}{u}
=\sum_{k\in\Bbb Z}\zeta_k\innerprod{T^ku}{u}
=\sum_{k\in\Bbb Z}\zeta_k\int z^k\nu_u(dz)
=\int p\,d\nu_u$.}

\noindent We also see that

$$\eqalign{\int u_p
&=\innerprod{u_p}{\chi 1}
=\sum_{k\in\Bbb Z}\zeta_k\innerprod{T^ku}{\chi 1}\cr
&=\sum_{k\in\Bbb Z}\zeta_k\innerprod{u}{T^{-k}\chi 1}
=\sum_{k\in\Bbb Z}\zeta_k\innerprod{u}{\chi 1}
=p(1)\int u.\cr}$$

It is elementary to check that
if $p\in P$ and $q(z)=zp(z)$ for every $z\in S^1$, then $u_q=Tu_p$.
Note also that $p\mapsto u_p:P\to L^2_{\Bbb C}$ is linear.

\medskip

\quad{\bf (iii)} For any $p\in P$ and $u\in L^2_{\Bbb C}$,
$\|u_p\|_2\le\|u\|_2\|p\|_{\infty}$.   \Prf\
If $p(z)=\sum_{k\in\Bbb Z}\zeta_kz^k$ for $z\in S^1$, set

\Centerline{$q(z)=p(z)\overline{p(z)}
=\sum_{j,k\in\Bbb Z}\zeta_j\bar\zeta_kz^{j-k}$}

\noindent for $z\in S^1$.   Then

$$\eqalign{\|u_p\|_2^2
&=\innerprod{\sum_{j\in\Bbb Z}\zeta_jT^ju}{\sum_{k\in\Bbb Z}\zeta_kT^ku}
=\sum_{j,k\in\Bbb Z}\zeta_j\bar\zeta_k\innerprod{T^ju}{T^ku}
=\sum_{j,k\in\Bbb Z}\zeta_j\bar\zeta_k\innerprod{T^{j-k}u}{u}\cr
&=\innerprod{u_q}{u}
=\int q\,d\nu_u
\le\|q\|_{\infty}\nu_u(S^1)
=\|p\|_{\infty}^2\|u\|_2^2.  \text{ \Qed}}$$

\medskip

{\bf (c) Case 1} Suppose that $\nu_u\{z\}=0$ whenever $u\in L^2_{\Bbb C}$,
$z\in S^1$ and $\int u=0$.

\medskip

\quad{\bf (i)} If $u\in L^2_{\Bbb C}$ and $\int u=0$, then
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n|\innerprod{T^ku}{u}|^2=0$.
\Prf\ For any $n\in\Bbb N$,

$$\eqalign{\Bover1{n+1}\sum_{k=0}^n|\innerprod{T^ku}{u}|^2
&=\Bover1{n+1}\sum_{k=0}^n\innerprod{T^ku}{u}\innerprod{u}{T^ku}
=\Bover1{n+1}\sum_{k=0}^n\innerprod{T^ku}{u}\innerprod{T^{-k}u}{u}\cr
&=\Bover1{n+1}\sum_{k=0}^n\int z^k\nu_u(dz)\int w^{-k}\nu_u(dw)\cr
&=\Bover1{n+1}\sum_{k=0}^n\int z^kw^{-k}\nu_u^2(d(z,w))\cr}$$

\noindent where $\nu_u^2$ is the product measure on $(S^1)^2$.   But observe
that

\Centerline{$|\Bover1{n+1}\sum_{k=0}^nz^kw^{-k}|\le 1$}

\noindent for all $z$, $w\in S^1$, while for $w\ne z$ we have

\Centerline{$\Bover1{n+1}\sum_{k=0}^nz^kw^{-k}
=\Bover{1-(w^{-1}z)^{n+1}}{(n+1)(1-w^{-1}z)}\to 0$.}

\noindent Since

\Centerline{$\nu_u^2\{(w,z):w=z\}=\int\nu_u\{z\}\nu_u(dz)=0$,}

\noindent Lebesgue's Dominated Convergence Theorem tells us that

\Centerline{
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^{n}|\innerprod{T^ku}{u}|^2
=\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n\int z^kw^{-k}\nu_u^2(d(z,w))
=0$.\ \Qed}

\medskip

\quad{\bf (ii)} Write $\Cal F_d$ for the asymptotic density filter on
$\Bbb N$ (491S).   If $u\in L^2_{\Bbb C}$ and $\int u=0$, then
$\lim_{k\to\Cal F_d}|\innerprod{T^ku}{u}|^2=0$, by (i) above and 491Sb.
It follows at once that $\lim_{k\to\Cal F_d}\innerprod{T^ku}{u}=0$.

In fact $\lim_{k\to\Cal F_d}\innerprod{T^ku}{v}=0$ whenever
$\int u=\int v=0$.   \Prf\ We have

$$\eqalignno{\lim_{k\to\Cal F_d}\innerprod{T^ku}{v}+\innerprod{T^kv}{u}
&=\lim_{k\to\Cal F_d}\innerprod{T^k(u+v)}{u+v}
   -\innerprod{T^ku}{v}-\innerprod{T^kv}{v}
=0,&(*)\cr}$$

\noindent and similarly

\Centerline{$\lim_{k\to\Cal F_d}i\innerprod{T^ku}{v}-i\innerprod{T^kv}{u}
=\lim_{k\to\Cal F_d}\innerprod{T^k(iu)}{v}+\innerprod{T^kv}{iu}
=0$,}

\noindent so
$\lim_{k\to\Cal F_d}\innerprod{T^ku}{v}-\innerprod{T^kv}{u}=0$;
adding this to (*), $\lim_{k\to\Cal F}\innerprod{T^ku}{v}=0$.\ \Qed

\medskip

\quad{\bf (iii)} Now take any $a$, $b\in\frak A$ and set
$u=\chi a-(\bar\mu a)\chi 1$, $v=\chi b-(\bar\mu b)\chi 1$.
In this case, $\int u=\int v=0$ and

$$\eqalign{\innerprod{T^ku}{v}
&=\innerprod{\chi(\phi^ka)-(\bar\mu a)\chi 1}{\chi b-(\bar\mu b)\chi 1}\cr
&=\bar\mu(b\Bcap\phi^ka)-\bar\mu a\cdot\bar\mu b
   -\bar\mu(\phi^ka)\cdot\bar\mu b+\bar\mu a\cdot\bar\mu b
=\bar\mu(b\Bcap\phi^ka)-\bar\mu a\cdot\bar\mu b\cr}$$

\noindent for every $k$, so
$\lim_{k\to\Cal F_d}\bar\mu(b\Bcap\phi^ka)-\bar\mu a\cdot\bar\mu b=0$ and

\Centerline{$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n
  |\bar\mu(b\Bcap\phi^ka)-\bar\mu a\cdot\bar\mu b|=0$,}

\noindent by 491Sb in the other direction.
As $a$ and $b$ are arbitrary, $\phi$ is weakly mixing.

\medskip

{\bf (d) Case 2} Suppose there are $u\in L^2_{\Bbb C}$ and
$t\in\ocint{-\pi,\pi}$ such that $\int u=0$ and $\nu_u\{e^{it}\}>0$.

\medskip

\quad{\bf (i)} For $n\in\Bbb N$, set
set $f_n(z)=\max(0,1-2^n|z-e^{it}|)$ for $z\in S^1$.   Then

$$\eqalign{|zf_n(z)-e^{it}f_n(z)|
&\le 2^{-n}\text{ if }|z-e^{it}|\le 2^{-n},\cr
&=0\text{ for other }z\in S^1.\cr}$$

\noindent Because $P$ is
$\|\,\|_{\infty}$-dense in $C(S^1;\Bbb C)$, there is a $p_n\in P$
such that $\|p_n-f_n\|_{\infty}\le 2^{-n}$, in which case

\Centerline{$|zp_n(z)-e^{it}p_n(z)|\le 3\cdot 2^{-n}$}

\noindent for every $z\in S^1$.   Set $q_n(z)=zp_n(z)$ for $z\in S^1$;
then

$$\eqalignno{\|Tu_{p_n}-e^{-it}u_{p_n}\|_2
&=\|u_{q_n}-e^{-it}u_{p_n}\|_2
\le\|u\|_2\|q_n-e^{-it}p_n\|_{\infty}\cr
\displaycause{by (b-iii)}
&\le 3\cdot 2^{-n}\|u\|_2,\cr}$$

\noindent while

\Centerline{$\|u_{p_n}\|_2\le\|u\|_2\|p_n\|_{\infty}\le 2\|u\|_2$.}

\medskip

\quad{\bf (ii)}
Let $\Cal F$ be any non-principal ultrafilter on $\Bbb N$.
Then $v=\lim_{n\to\Cal F}u_{p_n}$ is defined for the weak topology of the
complex Hilbert space $L^2_{\Bbb C}$ (4A4Ka).   Also

\Centerline{$\innerprod{v}{u}=\lim_{n\to\Cal F}\innerprod{u_{p_n}}{u}
=\lim_{n\to\infty}\int p_nd\nu_u
=\lim_{n\to\infty}\int f_nd\nu_u
=\nu_u\{e^{it}\}>0$}

\noindent so $v\ne 0$.   But we also have

\Centerline{$\int v=\innerprod{v}{\chi 1}
=\lim_{n\to\Cal F}\innerprod{u_{p_n}}{\chi 1}
=\lim_{n\to\Cal F}p_n(1)\int u=0$,}

\noindent and, taking limits in the weak topology on $L^2_{\Bbb C}$,

$$\eqalignno{Tv
&=\lim_{n\to\Cal F}Tu_{p_n}\cr
\displaycause{because $T$ is continuous for the weak topology,
see 4A4Bd}
&=\lim_{n\to\Cal F}e^{it}u_{p_n}
=e^{it}v.\cr}$$

\noindent Set $w=\Bover1{\|v\|_2}v$;  then $\|w\|_2=1$, $\int w=0$,

\Centerline{$\inf_{k\in\Bbb N}|\innerprod{T^kw}{w}|
=\inf_{k\in\Bbb N}|e^{ikt}\innerprod{w}{w}|
=1$}

\noindent and ($\beta$) is false.

\medskip

{\bf (e)} Putting (c) and (d) together, we see that either ($\alpha$) is
true or ($\beta$) is false, that is, that ($\beta$) implies ($\alpha$).

\medskip

{\bf (f)} On the other hand, ($\alpha$) implies ($\gamma$).
\Prf\ Suppose that $\phi$ is weakly mixing.   Then

\Centerline{$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n
  |\bar\mu(b\Bcap\phi^ka)-\bar\mu a\cdot\bar\mu b|=0$}

\noindent for all $a$, $b\in\frak A$;  by 491Sb again,

\Centerline{$\lim_{k\to\Cal F_d}
\bar\mu(b\Bcap\phi^ka)-\bar\mu a\cdot\bar\mu b=0$,}

\noindent that is,

\Centerline{$\lim_{k\to\Cal F_d}\innerprod{T^k\chi a}{\chi b}
=\innerprod{\chi a}{\chi 1}\cdot\innerprod{\chi 1}{\chi b}$,}

\noindent whenever
$a$, $b\in\frak A$.   Because $\innerprod{\,}{\,\,}$ is sesquilinear,

\Centerline{$\lim_{k\to\Cal F_d}\innerprod{T^ku}{v}
=\innerprod{u}{\chi 1}\cdot\innerprod{\chi 1}{v}$}

\noindent whenever $u$, $v$ belong to
$S_{\Bbb C}=S_{\Bbb C}(\frak A)$, the complex
linear span of $\{\chi a:a\in\frak A\}$.
Because $S_{\Bbb C}$ is norm-dense in $L^2_{\Bbb C}$ (366Mb),
and $\{T^ku:k\in\Bbb N\}$ is norm-bounded, we shall have

\Centerline{$\lim_{k\to\Cal F_d}\innerprod{u}{T^{-k}v}
=\lim_{k\to\Cal F_d}\innerprod{T^ku}{v}
=\innerprod{u}{\chi 1}\cdot\innerprod{\chi 1}{v}$}

\noindent whenever $u\in S_{\Bbb C}$ and $v\in L^2_{\Bbb C}$;
now $\{T^{-k}v:k\in\Bbb N\}$ is norm-bounded, so

\Centerline{$\lim_{k\to\Cal F_d}\innerprod{T^ku}{v}
=\lim_{k\to\Cal F_d}\innerprod{u}{T^{-k}v}
=\innerprod{u}{\chi 1}\cdot\innerprod{\chi 1}{v}$}

\noindent for all $u$, $v\in L^2_{\Bbb C}$.   In particular, if $\|w\|_2=1$
and $\int w=0$,

\Centerline{$\inf_{k\in\Bbb N}|\innerprod{T^kw}{w}|
\le\lim_{k\to\Cal F_d}|\innerprod{T^kw}{w}|
=|\innerprod{w}{\chi 1}|^2=0$,}

\noindent as required.\ \Qed

\medskip

{\bf (g)} Since ($\gamma$) obviously implies ($\beta$), the three
conditions are indeed equiveridical.
}%end of proof of 494D

\leader{494E}{Theorem}\dvAnew{2011}\cmmnt{ ({\smc Halmos 1944},
{\smc Rokhlin 1948})} Let $(\frak A,\bar\mu)$ be a probability algebra,
and give $\AmuA$ its weak topology.

(a) If $\frak A\ne\{0,1\}$,
the set of mixing measure-preserving Boolean automorphisms is meager in
$\AmuA$.

(b) If $\frak A$ is atomless and homogeneous, the set of
two-sided Bernouilli shifts on $(\frak A,\bar\mu)$\cmmnt{ (definition:
385Qb)} is dense in $\AmuA$.

(c) If $\frak A$ has countable Maharam type, the set of weakly mixing
measure-preserving Boolean automorphisms is a G$_{\delta}$ subset of
$\AmuA$.

(d) If $\frak A$ is atomless and has countable Maharam type, the set
of weakly mixing measure-preserving Boolean automorphisms which are not
mixing is comeager in $\AmuA$, and is not empty.

\proof{{\bf (a)} Take $a\in\frak A\setminus\{0,1\}$.   Let $\delta>0$ be
such that $\bar\mu a>\delta+(\bar\mu a)^2$, and consider

\Centerline{$F_n=\{\pi:\pi\in\AmuA$,
$\bar\mu(a\Bcap\pi^ka)\le\delta+(\bar\mu a)^2$ for every $k\ge n\}$.}

\noindent Because $\pi\mapsto\bar\mu(a\Bcap\pi^ka)$ is continuous for every
$k$ (494Bb), every $F_n$ is closed.   Because $F_n$ cannot contain any
periodic automorphism, $(\AmuA)\setminus F_n$ is dense for the uniform
topology on $\AmuA$ (494Cc) and therefore for the weak topology (494Cd).
Accordingly $\bigcup_{n\in\Bbb N}F_n$ is meager;  and every mixing
measure-preserving automorphism belongs to $\bigcup_{n\in\Bbb N}F_n$.

\medskip

{\bf (b)} Suppose that $\phi\in\AmuA$, $A\subseteq\frak A$ is finite
and $\epsilon>0$.

\medskip

\quad{\bf (i)} By 494Cc, there is a periodic $\psi\in\AmuA$ such that
$\bar\mu(\phi a\Bsymmdiff\psi a)\le\epsilon$ for every $a\in\frak A$.
Let $\frak B$ be the subalgebra of $\frak A$ generated by
$\{\psi^ka:k\in\Bbb Z$, $a\in A\}$;  then $\frak B$ is finite (because
$\{\psi^k:k\in\Bbb Z\}$ is finite).   Let $B$ be the set of
atoms of $\frak B$.   Since $\psi[\frak B]=\frak B$, $\psi[B]=B$ and
$\psi\restr B$ is a permutation of $B$.   Let $B_0\subseteq B$ be such that
$B_0$ meets each orbit of $\psi\restr B$ in just one point;  enumerate
$B_0$ as $\ofamily{j}{n}{b_j}$.

Let $r\ge 1$ be such that $\#(B)+1\le\epsilon r$.   For each $j<n$,
let $m_j$ be the size of the orbit of $\psi\restr B$
containing $b_j$, and $p_j=\lceil r\bar\mu b_j\rceil-1$;  set
$M=\sum_{j=0}^{n-1}m_jp_j$.   Because $\frak A$
is atomless, we can find a disjoint family
$\ofamily{l}{p_j}{c_{jl}}$ in $\frak A_{b_j}$
such that $\bar\mu c_{jl}=\bover1r$ for every $l<p_j$.
Because
$\langle\psi^kb_j\rangle_{j<n,k<m_j}$ is disjoint, so is
$\langle\psi^kc_{jl}\rangle_{j<n,l<p_j,k<m_j}$.   Set

\Centerline{$C=\{\psi^kc_{jl}:j<n$, $l<p_j$, $k<m_j\}$,
\quad$c=\sup C$;}

\noindent then

$$\eqalign{\bar\mu c
&=\Bover{M}r
=\Bover1r\sum_{j=0}^{n-1}p_jm_j
\ge\Bover1r\sum_{j=0}^{n-1}m_j(r\bar\mu b_j-1)\cr
&=1-\Bover1r\sum_{j=0}^{n-1}m_j
=1-\Bover{\#(B)}r
\ge 1-\epsilon.\cr}$$

\noindent We shall need to know later that

\Centerline{$\Bover{M}r
=\Bover1r\sum_{j=0}^{n-1}p_jm_j
<\Bover1r\sum_{j-0}^{n-1}rm_j\bar\mu b_j=1$.}

\medskip

\quad{\bf (ii)}
Let $f:C\to C$ be the cyclic permutation defined by setting

$$\eqalign{f(\psi^kc_{jl})
&=\psi^{k+1}c_{jl}\text{ if }j<n,\,l<p_j,\,k\le m_j-2,\cr
&=c_{j,l+1}\text{ if }j<n,\,l\le p_j-2,\,k=m_j-1,\cr
&=c_{j+1,0}\text{ if }j\le n-2,\,l=p_j-1,\,k=m_j-1,\cr
&=c_{00}\text{ if }j=n-1,\,l=p_j-1,\,k=m_j-1.\cr}$$

\noindent Set

\Centerline{$C'=\{c:c\in C$, $f(c)$ and $\psi(c)$ are included in different
members of $B\}$.}

\noindent Then $\#(C')\le n$.   \Prf\ If $c\in C'$, express it as
$\psi^kc_{jl}$ where $j<n$, $l<p_j$ and $k<m_j$.   We surely have
$f(c)\ne\psi c$, so $k$ must be $m_j-1$.   In this case,

\Centerline{$\psi c=\psi^{m_j}c_{jl}\Bsubseteq\psi^{m_j}b_j=b_j$,}

\noindent so $f(c)\notBsubseteq b_j$ and $l$ must be $p_j-1$.
Thus $c=\psi^{m_j-1}c_{j,p_j-1}$ for some $j<n$, and there are only $n$
objects of this form.\ \Qed

\medskip

\quad{\bf (iii)} We know that there is a two-sided Bernouilli shift $\pi_0$
on $(\frak A,\bar\mu)$ (385Sb).   Now $\pi_0$ is mixing
(385Se), therefore ergodic (372Qb) and aperiodic (386D).
We know that $\bover{M}r<1$, so by 386C again
there is a $d_0\in\frak A$ such that
$d_0,\pi_0d_0,\ldots,\pi_0^{M-1}d_0$ are disjoint and
$\bar\mu d_0=\bover1r$.
Because $\bar\mu f^i(c_{00})=\bar\mu\pi_0^id_0=\bover1r$ for every $i<M$
and $\frak A$ is homogeneous,
there is a $\theta\in\AmuA$ such that $\theta(\pi_0^id_0)=f^i(c_{00})$ for
every $i<M$.   Set $\pi=\theta\pi_0\theta^{-1}$;  then $\pi$
is a two-sided Bernouilli shift (385Sg).   Now

\Centerline{$\pi f^i(c_{00})
=\theta\pi_0\theta^{-1}f^i(c_{00})
=\theta\pi_0\pi_0^id_0
=f^{i+1}(c_{00})$}

\noindent whenever $i\le M-2$.   So

$$\eqalign{C''
&=\{c:c\in C,\,\pi c\text{ and }\psi(c)\text{ are included in
different members of }B\}\cr
&\subseteq C'\cup\{f^{M-1}(c_{00})\}\cr}$$

\noindent has at most $n+1$ members.

Because $B$ is disjoint,
$e=\sup_{b\in B}\pi b\Bsymmdiff\psi b$ is disjoint from

\Centerline{$\sup_{b\in B}\pi b\Bcap\psi b\Bsupseteq\sup(C\setminus C'')$}

\noindent and has measure at most

\Centerline{$\bar\mu(\sup C'')+\bar\mu(1\Bsetminus c)
\le\Bover{n+1}r+\epsilon
\le 2\epsilon$.}

\noindent If $a\in A$, then $a$ is the supremum of the members of $B$ it
includes, so $\pi a\Bsymmdiff\psi a\Bsubseteq e$ and

\Centerline{$\bar\mu(\pi a\Bsymmdiff\phi a)
\le\bar\mu(\pi a\Bsymmdiff\psi a)+\bar\mu(\psi a\Bsymmdiff\phi a)
\le 3\epsilon$.}

\medskip

\quad{\bf (iv)} Thus, given $\phi\in\AmuA$, $A\in[\frak A]^{<\omega}$
and $\epsilon>0$, we can find a two-sided Bernouilli shift $\pi$ such that
$\bar\mu(\pi a\Bsymmdiff\phi a)\le 3\epsilon$ for every $a\in A$;  as
$\phi$, $A$ and $\epsilon$ are arbitrary, the two-sided Bernouilli shifts
are dense in $\AmuA$.

\medskip

{\bf (c)(i)} The point is that
$L^2_{\Bbb C}=L^2_{\Bbb C}(\frak A,\bar\mu)$ is separable in its norm
topology.   \Prf\ By 331O, there is a countable set
$A\subseteq\frak A$ which is dense for the measure-algebra topology of
$\frak A$.   Let $C$ be a countable dense subset of $\Bbb C$, and

\Centerline{$D
=\{\sum_{j=0}^n\zeta_j\chi a_j:\zeta_0,\ldots,\zeta_n\in C$,
$a_0,\ldots,a_n\in A\}$,}

\noindent so that $D$ is a countable subset of $L^2_{\Bbb C}$.
Because the function

\Centerline{$(\zeta_0,\ldots,\zeta_n,a_0,\ldots,a_n)
\mapsto\sum_{j=0}^n\zeta_j\chi a_j:\Bbb C^{n+1}\times\frak A^{n+1}
\to L^2_{\Bbb C}$}

\noindent is continuous for each $n$, $\overline{D}$ contains
$\sum_{j=0}^n\zeta_j\chi a_j$ whenever
$\zeta_0,\ldots,\zeta_n\in\Bbb C$ and $a_0,\ldots,a_n\in\frak A$,
that is, $S_{\Bbb C}=S_{\Bbb C}(\frak A)\subseteq\overline{D}$.   But
$S_{\Bbb C}$ is norm-dense in $L^2_{\Bbb C}$, so
$D$ also is dense and $L^2_{\Bbb C}$ is separable.\ \Qed

\medskip

\quad{\bf (ii)} For $\pi\in\AmuA$, let
$T_{\pi}:L^2_{\Bbb C}\to L^2_{\Bbb C}$ be the corresponding linear
operator, as in 494D.   We need to know that the function
$\pi\mapsto T_{\pi}v:\AmuA\to L^2_{\Bbb C}$ is continuous for every
$v\in L^2$.   \Prf\ It is elementary to check that
$a\mapsto\chi a:\frak A\to L^2_{\Bbb C}$ is continuous for the
measure-algebra topology on $\frak A$, so
$(\pi,a)\mapsto T_{\pi}\chi a=\chi\pi a$ is continuous (494Ba-494Bb), and
$\pi\mapsto T_{\pi}\chi a$ is continuous, for every $a\in\frak A$.
Because addition and scalar multiplication are continuous on
$L^2_{\Bbb C}$, $\pi\mapsto T_{\pi}v$ is continuous for every
$v\in S_{\Bbb C}$.   Now if $v$ is any member of $L^2_{\Bbb C}$,
$\phi\in\AmuA$ and $\epsilon>0$, there is a $v'\in S_{\Bbb C}$ such that
$\|v-v'\|_2\le\epsilon$, in which case

\Centerline{$\{\pi:\|T_{\pi}v-T_{\phi}v\|_2\le 3\epsilon\}
\supseteq\{\pi:\|T_{\pi}v'-T_{\phi}v'\|\le\epsilon\}$}

\noindent is a neighbourhood of $\phi$.   Thus $\pi\mapsto T_{\pi}v$ is
continuous for arbitrary $v\in L^2_{\Bbb C}$.\ \Qed

\medskip

\quad{\bf (iii)} It follows from (i) that the set
$V=\{v:v\in L^2_{\Bbb C}$, $\|v\|_2=1$, $\int v=0\}$ is separable
(4A2P(a-iv)).   Let $D'$ be a countable dense subset of $V$.
For $v\in D'$, set

\Centerline{$F_v
=\{\pi:|\innerprod{T_{\pi}^kv}{v}|\ge\Bover12$ for every $k\in\Bbb N\}$.}

\noindent Since the maps

\Centerline{$\pi\mapsto\pi^k\mapsto T_{\pi^k}v=T_{\pi}^kv
\mapsto\innerprod{T_{\pi}^kv}{v}$}

\noindent are all continuous (494Ba and (ii) just above), $F_v$ is closed.
Consider $E=\AmuA\setminus\bigcup_{v\in D'}F_v$.
If $\pi\in\AmuA$ is weakly mixing, then ($\alpha$)$\Rightarrow$($\gamma$)
of
494D tells us that $\pi\in E$.   On the other hand, if $\pi\in\AmuA$ is not
weakly mixing, ($\beta$)$\Rightarrow$($\alpha$) of 494D tells us that
there is
a $w\in V$ such that $\inf_{k\in\Bbb N}|\innerprod{T_{\pi}^kw}{w}|\ge 1$.
Let $v\in D'$ be such that $\|v-w\|_2\le\Bover14$.   Then, for any
$k\in\Bbb N$,

$$\eqalign{|\innerprod{T_{\pi}^kv}{v}|
&\ge|\innerprod{T_{\pi}^kw}{v}|-\|T_{\pi}^kw-T_{\pi}^kv\|_2\|v\|_2
\ge|\innerprod{T_{\pi}^kw}{v}|-\Bover14\cr
&\ge|\innerprod{T_{\pi}^kw}{w}|-\|T_{\pi}^kw\|_2\|v-w\|_2-\Bover14
\ge\Bover12.\cr}$$

\noindent So $\pi\in F_v\subseteq(\AmuA)\setminus E$.
Thus the set of weakly mixing automorphisms is
precisely $E$, and is a G$_{\delta}$ set.

\medskip

{\bf (d)} We know that every
two-sided Bernouilli shift is weakly mixing (385Se, 372Qb), so the set
$E$ of
weakly mixing automorphisms is dense, by (b) here, and G$_{\delta}$, by
(c), therefore comeager.   By (a), the set $E'$ of weakly mixing
automorphisms which are not mixing is also comeager.
By 494Be, $\AmuA$ is a Polish space, so $E'$ is non-empty.
}%end of proof of 494E

\leader{494F}{}\dvAnew{2011}\cmmnt{ 494Ed tells us that `many'
automorphisms of the
Lebesgue probability algebra are weakly mixing but not mixing.   It is
another matter to give an explicit description of one.   Bare-handed
constructions (e.g., {\smc Chacon 69}) demand ingenuity and determination.
I prefer to show you an example taken from {\smc Tao l08}, Lecture 12,
Exercises 5 and 8,
although it will take some pages in the style of this book,
as it gives practice in using ideas already presented.

\medskip

\noindent}{\bf Example} (a) There is a Radon
probability measure $\nu$ on $\Bbb R$, zero on singletons, such that

\Centerline{$\int\cos(2\pi\cdot 3^jt)\,\nu(dt)=\int\cos 2\pi t\,\nu(dt)>0$}

\noindent for every $j\in\Bbb N$.

(b) Set $\sigma_{jk}=\int\cos 2\pi(k-j)t\,\nu(dt)$ for $j$, $k\in\Bbb Z$.
Then there is a centered Gaussian distribution $\mu$ on $X=\BbbR^{\Bbb Z}$
with covariance matrix $\langle\sigma_{jk}\rangle_{j,k\in\Bbb Z}$.

(c) Let $S:X\to X$ be the shift operator defined by saying that
$(Sx)(j)=x(j+1)$ for $x\in X$ and $j\in\Bbb Z$.   Then $S$ is an
automorphism of $(X,\mu)$.

(d) Let $(\frak A,\bar\mu)$ be the measure
algebra of $\mu$ and $\phi\in\AmuA$ the automorphism represented by
$S$.   Then $\phi$ is not mixing.

(e) $\phi$ is weakly mixing.

\proof{{\bf (a)(i)} Let $\tilde\nu$ be the usual measure on
$\Cal P\Bbb N$ (254Jb, 464A).   Define $h:\Cal P\Bbb N\to\Bbb R$ by setting

\Centerline{$h(I)=\Bover23\sum_{j\in I}3^{-j}$}

\noindent for $I\subseteq\Bbb N$.   Then $h$ is continuous,
so the image measure $\nu=\tilde\nu h^{-1}$ is a Radon probability
measure on $\Bbb R$ (418I).
Also $h$ is injective, so $\nu$, like $\tilde\nu$,
is zero on singletons.

\medskip

\quad{\bf (ii)} The function $t\mapsto\fraction{3t}=3t-\lfloor 3t\rfloor$
is \imp\ for $\nu$.   \Prf\ Set $\psi_0(I)=\{j:j+1\in I\}$
for $I\subseteq\Bbb N$,
$\psi_1(t)=\fraction{3t}$ for $t\in\Bbb R$.   Then
$\psi_0:\Cal P\Bbb N\to\Cal P\Bbb N$ is \imp\ for $\tilde\nu$,
because

\Centerline{$\tilde\nu\{I:\psi_0(I)\cap J=K\}
=\tilde\nu\{I:I\cap(J+1)=K+1\}=2^{-\#(J)}$}

\noindent whenever $K\subseteq J\in[\Bbb N]^{<\omega}$.   Next,
for any  $I\in\Cal P\Bbb N\setminus\{\Bbb N,\Bbb N\setminus\{0\}\}$,

\Centerline{$\psi_1(h(I))
=\fraction{2\sumop_{j\in I}3^{-j}}
=2\sum_{j\in I\setminus\{0\}}3^{-j}
=2\sum_{j+1\in I}3^{-j-1}
=h(\psi_0(I))$.}

\noindent So $\psi_1h\eae h\psi_0$, and

\Centerline{$\nu\psi_1^{-1}=\tilde\nu h^{-1}\psi_1^{-1}
=\tilde\nu\psi_0^{-1}h^{-1}=\tilde\nu h^{-1}=\nu$.  \Qed}

Similarly, if we set

$$\eqalign{\theta(t)
&=\Bover13-t\text{ if }0\le t\le\Bover13,\cr
&=\Bover53-t\text{ if }\Bover23\le t\le 1,\cr
&=t\text{ otherwise},\cr}$$

\noindent then $\theta h(I)=h(I\symmdiff(\Bbb N\setminus\{0\}))$ for every
$I\subseteq\Bbb N$, and $\nu\theta^{-1}=\nu$.

\medskip

\quad{\bf (iii)} Consequently, for any $m\in\Bbb N$,

\Centerline{$\int\cos(2\pi\cdot 3mt)\,\nu(dt)
=\int\cos(2\pi m\fraction{3t})\,\nu(dt)
=\int\cos 2\pi mt\,\nu(dt)$}

\noindent (235G).   Inducing on $j$, we see that

\Centerline{$\int\cos(2\pi\cdot 3^jt)\,\nu(dt)=\int\cos 2\pi t\,\nu(dt)$}

\noindent for every $j\in\Bbb N$.

\medskip

\quad{\bf (iv)} As for $\int\cos 2\pi t\,\nu(dt)$, this is
equal to $\int\cos 2\pi\theta(t)\,\nu(dt)$.   Now

$$\eqalign{\cos 2\pi t+\cos 2\pi\theta(t)
&=\cos 2\pi t+\cos 2\pi(\Bover13-t)\cr
&\mskip100mu
=2\cos\Bover{\pi}3\cos 2\pi(t-\Bover16)>0
   \text{ if }0\le t\le\Bover13,\cr
&=\cos 2\pi t+\cos 2\pi(\Bover53-t)\cr
&\mskip100mu
=2\cos\Bover{5\pi}3\cos 2\pi(t-\Bover56)>0
   \text{ if }\Bover23\le t\le 1;\cr}$$

\noindent but $h[\Cal P\Bbb N]\subseteq[0,\bover13]\cup[\bover23,1]$,
so $[0,\bover13]\cup[\bover23,1]$ is $\nu$-conegligible, and
$\cos 2\pi t+\cos 2\pi\theta(t)>0$ for $\nu$-almost every $t$.
Accordingly

\Centerline{$\int\cos 2\pi t\,\nu(dt)
=\Bover12\int\cos 2\pi t+\cos 2\pi\theta(t)\nu(dt)
>0$.}

\medskip

{\bf (b)} $\sigma_{jk}=\sigma_{kj}$ for all
$j$, $k\in\Bbb Z$.   If $J\subseteq\Bbb Z$ is finite
and $\family{j}{J}{\gamma_j}\in\BbbR^J$, then

$$\eqalignno{\sum_{j,k\in J}\gamma_j\gamma_k\sigma_{jk}
&=\sum_{j,k\in J}\gamma_j\gamma_k\int\cos 2\pi(k-j)t\,\nu(dt)\cr
&=\sum_{j,k\in J}\gamma_j\gamma_k\int\cos 2\pi kt\cos 2\pi jt
   +\sin 2\pi kt\sin 2\pi jt\,\nu(dt)\cr
&=\int\sum_{j,k\in J}\gamma_j\gamma_k\cos 2\pi kt\cos2\pi jt\,\nu(dt)\cr
&\mskip100mu
   +\int\sum_{j,k\in J}\gamma_j\gamma_k\sin 2\pi kt\sin 2\pi jt\,\nu(dt)\cr
&=\int\sum_{j\in J}\gamma_j\cos 2\pi jt\,
     \sum_{k\in J}\gamma_k\cos 2\pi kt\,\nu(dt)\cr
&\mskip100mu
   +\int\sum_{j\in J}\gamma_j\sin 2\pi jt\,
      \sum_{k\in J}\gamma_k\sin 2\pi kt\,\nu(dt)\cr
&\ge 0.\cr}$$

\noindent By 456C(iv), we have a Gaussian distribution of the
right kind.

\medskip

{\bf (c)} Of course $S$ is linear, and $\Bbb Z$ is countable,
so the image measure $\mu S^{-1}$ is a centered Gaussian distribution
(456Ba).   Since

$$\eqalign{\int x(j)x(k)(\mu S^{-1})(dx)
&=\int (Sx)(j)(Sx)(k)\mu(dx)\cr
&=\int x(j+1)x(k+1)\mu(dx)
=\sigma_{j+1,k+1}
=\sigma_{jk}\cr}$$

\noindent for all $j$, $k\in\Bbb Z$, $\mu S^{-1}$ and $\mu$ have the same
covariance matrix, and are equal (456Bb).
Thus the bijection $S$ is an automorphism of $(X,\mu)$.

\medskip

{\bf (d)} Write $L^2$ for $L^2(\frak A,\bar\mu)$, and
$T_{\phi}:L^2\to L^2$ for the linear operator associated
with the automorphism $\phi$.   For $k\in\Bbb Z$, set $f_k(x)=x(k)$ for
$x\in X$ and $u_k=f_k^{\ssbullet}\in L^2$.
Then $f_kS=f_{k+1}$ so $T_{\phi}u_k=u_{k+1}$, by 364Qd.
Consider

$$\eqalign{\innerprod{T_{\phi}^{3^j}u_0}{u_0}
&=\int u_{3^j}\times u_0
=\int x(3^j)x(0)\mu(dx)\cr
&=\sigma_{3^j,0}
=\int\cos(2\pi\cdot 3^jt)\nu(dt)
=\int\cos 2\pi t\,\nu(dt)
\ne 0,}$$

\noindent for every $j$, while

\Centerline{$\int u_0=\int x(0)\mu(dx)=0$.}

\noindent By 372Q(a-iv), $\pi$ is not mixing.

\medskip

{\bf (e)(i)} $\lim_{n\to\infty}\Bover1{n+1}
\sum_{k=0}^n|\int e^{2\pi ikt}\nu(dt)|^2=0$.   \Prf\ For any $n\in\Bbb N$,

$$\eqalign{\Bover1{n+1}\sum_{k=0}^n|\int e^{2\pi ikt}\nu(dt)|^2
&=\Bover1{n+1}\sum_{k=0}^n
   \int e^{2\pi iks}\nu(ds)\int e^{-2\pi ikt}\nu(dt)\cr
&=\int\Bover1{n+1}\sum_{k=0}^n e^{2\pi ik(s-t)}\nu^2(d(s,t))\cr}$$

\noindent where $\nu^2$ is the product measure on $\BbbR^2$.   Now, for any
$s$, $t\in\Bbb R$,
$|\Bover1{n+1}\sum_{k=0}^n e^{2\pi ik(s-t)}|\le 1$ for every $n$, while
if $s-t$ is not an integer,

\Centerline{$\Bover1{n+1}\sum_{k=0}^n e^{2\pi ik(s-t)}
=\Bover{1-\exp(2\pi i(n+1)(s-t))}{(n+1)(1-\exp(2\pi i(s-t)))}
\to 0$}

\noindent as $n\to\infty$.   As $\nu$ is zero on singletons,

\Centerline{$\nu^2\{(s,t):s-t\in\Bbb Z\}
=\int\nu\{s:s\in t+\Bbb Z\}\nu(ds)=0$.}

\noindent So

\Centerline{$\lim_{n\to\infty}
  \Bover1{n+1}\sum_{k=0}^n|\int e^{2\pi ikt}\nu(dt)|^2
=\lim_{n\to\infty}
  \int\Bover1{n+1}\sum_{k=0}^n e^{2\pi ik(s-t)}\nu^2(d(s,t))
=0$}

\noindent by Lebesgue's dominated convergence theorem.\ \Qed

Consequently, as in (c-ii) of the proof of 494D,

$$\eqalign{0
&=\lim_{k\to\Cal F_d}|\int e^{2\pi ikt}\nu(dt)|^2
=\lim_{k\to\Cal F_d}\int e^{2\pi ikt}\nu(dt)\cr
&=\lim_{k\to\Cal F_d}\Real\int e^{2\pi ikt}\nu(dt)
=\lim_{k\to\Cal F_d}\int \cos(2\pi kt)\nu(dt).\cr}$$

\noindent By 491Sc,

\Centerline{$\lim_{k\to\Cal F_d}\sigma_{jk}
=\lim_{k\to\Cal F_d}\sigma_{j,j+k}
=\lim_{k\to\Cal F_d}\int\cos(2\pi kt)\nu(dt)
=0$.}

\medskip

\quad{\bf (ii)} Suppose that $f$, $g:X\to\Bbb R$ are functions such
that, for some finite $J\subseteq\Bbb Z$, there are continuous bounded
functions $f_0$, $g_0:\BbbR^J\to\Bbb R$ such that
$f(x)=f_0(x\restr J)$ and $g(x)=g_0(x\restr J)$ for every
$x\in\BbbR^X$.
Then $\lim_{n\to\Cal F_d}\int fS^n\times g\,d\mu=\int fd\mu\int g\,d\mu$.

\Prf\ For any $n\in\Bbb N$, define $ R_n:X\to\BbbR^{J\times\{0,1\}}$ by
setting

\Centerline{$ R_n(x)(j,0)=x(j)$,
\quad$ R_n(x)(j,1)=x(j+n)$}

\noindent for $x\in X$ and $j\in J$;  then $ R_n$ is linear, so
$\mu R_n^{-1}$ is a centered Gaussian distribution on
$\BbbR^{J\times\{0,1\}}$.
The covariance matrix $\sigma^{(n)}$ of $\mu R_n^{-1}$ is given by

$$\eqalign{\sigma^{(n)}_{(j,\epsilon),(k,\epsilon')}
&=\int z(j,\epsilon)z(k,\epsilon')\mu R_n^{-1}(dz)
=\int ( R_nx)(j,\epsilon)( R_nx)(k,\epsilon')\mu(dx)\cr
&=\int x(j)x(k)\mu(dx)=\sigma_{jk}\text{ if }\epsilon=\epsilon'=0,\cr
&=\int x(j)x(k+n)\mu(dx)=\sigma_{j,k+n}
   \text{ if }\epsilon=0,\,\epsilon'=1,\cr
&=\int x(j+n)x(k)\mu(dx)=\sigma_{j+n,k}=\sigma_{k,j+n}
   \text{ if }\epsilon=1,\,\epsilon'=0,\cr
&=\int x(j+n)x(j+n)\mu(dx)=\sigma_{j+n,k+n}
=\sigma_{jk}\text{ if }\epsilon=\epsilon'=1\cr}$$

\noindent for all $j$, $k\in J$.   So

$$\eqalign{\lim_{n\to\Cal F_d}\sigma^{(n)}_{(j,\epsilon),(k,\epsilon')}
&=\sigma_{jk}\text{ if }\epsilon=\epsilon',\cr
&=0\text{ if }\epsilon\ne\epsilon'.\cr}$$

\noindent Let $\tilde\mu$ be the centered Gaussian distribution
$\tilde\mu$ on $\BbbR^{J\times\{0,1\}}$ with covariance matrix $\tau$ where

$$\eqalign{\tau_{(j,\epsilon),(k,\epsilon')}
&=\sigma_{jk}\text{ if }\epsilon=\epsilon',\cr
&=0\text{ if }\epsilon\ne\epsilon'\cr}$$

\noindent for any $j$, $k\in J$.   By 456Q, there is such a distribution
and $\tilde\mu=\lim_{n\to\Cal F_d}\mu R_n^{-1}$ for the narrow topology.

Next observe that, for $x\in X$ and $z\in\BbbR^{J\times\{0,1\}}$,

$$\eqalign{ R_n(x)=z
&\Longrightarrow x(j+n)=z(j,1)\text{ for every }j\in J\cr
&\Longleftrightarrow (S^nx)(j)=z(j,1)\text{ for every }j\in J\cr
&\Longrightarrow f(S^nx)=f'_0(z),\cr
 R_n(x)=z
&\Longrightarrow x(j)=z(j,0)\text{ for every }r<m\cr
&\Longrightarrow g(x)=g'_0(z),\cr}$$

\noindent where we set

\Centerline{$f'_0(z)=f_0(\family{j}{J}{z(j,1)})$,
\quad$g'_0(z)=g_0(\family{j}{J}{z(j,0)}$}

\noindent for $z\in\BbbR^{J\times\{0,1\}}$.
So $fS^n=f'_0 R_n$, $g=g'_0 R_n$,

\Centerline{$\int fS^n\times g\,d\mu
=\int(f'_0 R_n)\times(g'_0 R_n)d\mu
=\int f'_0\times g'_0d(\mu R_n^{-1})$}

\noindent for every $n$, and

\Centerline{$\lim_{n\to\Cal F_d}\int fS^n\times g\,d\mu
=\int f'_0\times g'_0d\tilde\mu$}

\noindent because $f'_0\times g'_0$ is a bounded continuous function
(437Mb).

Since $\tau_{(j,0),(k,1)}=0$ whenever $j$, $k\in J$,
the $\sigma$-algebras $\Sigma_0$, $\Sigma_1$ generated by
coordinates in $J\times\{0\}$, $J\times\{1\}$ respectively are
$\tilde\mu$-independent (456Eb).
Since $f'_0$ is $\Sigma_0$-measurable and
$g'_0$ is $\Sigma_1$-measurable,

$$\eqalignno{\lim_{n\to\Cal F_d}\int fS^n\times g\,d\mu
&=\int f'_0\times g'_0\,d\tilde\mu
=\int f'_0d\tilde\mu\int g'_0d\tilde\mu\cr
\displaycause{272D, 272R}
&=\lim_{n\to\Cal F_d}\int f'_0d(\mu R_n^{-1})
   \cdot\lim_{n\to\Cal F_d}\int g'_0d(\mu R_n^{-1})\cr
&=\lim_{n\to\Cal F_d}\int f'_0 R_nd\mu
   \int g'_0 R_nd\mu\cr
&=\lim_{n\to\Cal F_d}\int fS^nd\mu\int g\,d\mu\cr
&=\lim_{n\to\Cal F_d}\int fd\mu\int g\,d\mu
=\int fd\mu\cdot\int g\,d\mu,\cr}$$

\noindent as required.\ \Qed

\medskip

\quad{\bf (iii)} If $F$, $F'\subseteq X$ are compact, then
$\lim_{n\to\Cal F_d}\mu(S^{-n}[F]\cap F')=\mu F\cdot\mu F'$.   \Prf\
Let $\epsilon>0$.
For $k\in\Bbb N$, set $J_k=\{j:j\in\Bbb Z$, $|j|\le k\}$ and
$F_k=\{x\restr J_k:x\in F\}$.
Set $f^{(0)}_k(z)=\max(0,1-2^k\rho_k(z,F_k))$ for $z\in\Bbb R^{J_k}$, where
$\rho_k$ is Euclidean distance in $\BbbR^{J_k}$, and
$f_k(x)=f^{(0)}_k(x\restr J_k)$ for $x\in X$.   Then
$\sequence{k}{f_k(x)}\to\chi F(x)$ for every $x\in X$.   So there is a
$k\in\Bbb N$ such that $\int|f_k-\chi F|d\mu\le\epsilon$.   Set $f=f_k$;
then $f$ is a continuous function from $X$ to $[0,1]$,
$\int|f-\chi F|\le\epsilon$, and $f$ factors
through the continuous function $f^{(0)}_k:\BbbR^{J_k}\to[0,1]$.

Similarly, there is a continuous function $g:X\to[0,1]$ such that
$\int|g-\chi F'|d\mu\le\epsilon$ and $g$ factors through a continuous
function on $\BbbR^{J_l}$ for some $l$.   Setting $J=J_k\cup J_l$,
we see that $f$ and $g$ satisfy the conditions of (ii) and

\Centerline{$\lim_{n\to\Cal F_d}\int fS^n\times g\,d\mu
=\int fd\mu\int g\,d\mu$.}

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

$$\eqalign{|\mu(S^{-n}[F]\cap F')-\int fS^n\times g|
&\le\int|fS^n-\chi S^{-n}[F]|+|g-\chi F'|d\mu\cr
&=\int|f-\chi F|+|g-\chi F'|d\mu
\le 2\epsilon,\cr}$$

\Centerline{$|\mu F\cdot\mu F'-\int fd\mu\int g\,d\mu|
\le\int|f-\chi F|+|g-\chi F'|d\mu
\le 2\epsilon$,}

\noindent so

$$\eqalign{&\limsup_{n\to\Cal F_d}
   |\mu(S^{-n}[F]\cap F')-\mu F\cdot\mu F'|\cr
&\mskip100mu
\le 4\epsilon+
\lim_{n\to\Cal F_d}|\int fS^n\times g\,d\mu-\int fd\mu\int g\,d\mu|
=4\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\lim_{n\to\Cal F_d}\mu(S^{-n}[F]\cap F')=\mu F\cdot\mu F'$.\ \Qed

\medskip

\quad{\bf (iv)} Now suppose that $a$, $b\in\frak A$ and $\epsilon>0$.
Because $\mu$ is a Radon measure (454J(iii)), there are compact sets
$F_0$, $F_1\subseteq X$ such that
$\bar\mu(a\Bsymmdiff F_0^{\ssbullet})+\bar\mu(b\Bsymmdiff F_1^{\ssbullet})
\le\epsilon$.   Now, for any $n\in\Bbb N$,

$$\eqalign{|\bar\mu(\phi^na\Bcap b)-\mu(S^{-n}[F_0]\cap F_1)|
&=|\bar\mu(\phi^n a\Bcap b)
   -\bar\mu(\phi^nF_0^{\ssbullet}\Bcap F_1^{\ssbullet})|\cr
&\le\bar\mu(\phi^n a\Bsymmdiff\phi^nF_0^{\ssbullet})
  +\bar\mu(b\Bsymmdiff F_1^{\ssbullet})\cr
&=\bar\mu(a\Bsymmdiff F_0^{\ssbullet})
  +\bar\mu(b\Bsymmdiff F_1^{\ssbullet})
\le\epsilon,\cr}$$

\Centerline{$|\bar\mu a\cdot\bar\mu b-\mu F_0\cdot\mu F_1|
\le|\bar\mu a-\mu F_0|+|\bar\mu b-\mu F_1|
\le\epsilon$.}

\noindent So

$$\eqalign{&\limsup_{n\to\Cal F_d}|\bar\mu(\phi^na\Bcap b)
   -\bar\mu a\cdot\bar\mu b|\cr
&\mskip100mu
\le 2\epsilon
  +\lim_{n\to\Cal F_d}|\mu(S^{-n}[F_0]\cap F_1)-\mu F_0\cdot\mu F_1|
=2\epsilon\cr}$$

\noindent by (iii).
As $\epsilon$, $a$ and $b$ are arbitrary, $\phi$ is weakly
mixing (using 491Sb once more).
}%end of proof of 494R

\cmmnt{\medskip

\noindent{\bf Remark} Of course the measure $\nu$ of part (a) is
Cantor measure (256Hc, 256Xk).}

\leader{494G}{Proposition} Let $(\frak A,\bar\mu)$ be a
measure algebra and $G$ a full subgroup of $\AmuA$, with fixed-point
subalgebra $\frak C$\cmmnt{ (definition:  395Ga)}.

(a) If $a\in\frak A^f$ and $\pi\in G$, there is a
$\phi\in G$, supported by
$a\Bcup\pi a$, such that $\phi d=\pi d$ for every $d\Bsubseteq a$.

(b) If $(\frak A,\bar\mu)$ is localizable
and $a$, $b\in\frak A^f$, then the following are equiveridical:

\quad(i) there is a $\pi\in G$ such that $\pi a\Bsubseteq b$;

\quad(ii) $\bar\mu(a\Bcap c)\le\bar\mu(b\Bcap c)$ for every $c\in\frak C$.

(c) If $(\frak A,\bar\mu)$ is localizable
and $a$, $b\in\frak A^f$, then the following are equiveridical:

\quad(i) there is a $\pi\in G$ such that $\pi a=b$;

\quad(ii) $\bar\mu(a\Bcap c)=\bar\mu(b\Bcap c)$ for every $c\in\frak C$.

(d) If $(\frak A,\bar\mu)$ is totally finite\cmmnt{ (definition:  322Ab)}
and $\familyiI{a_i}$,
$\familyiI{b_i}$ are disjoint families in $\frak A$ such that
$\bar\mu(a_i\Bcap c)=\bar\mu(b_i\Bcap c)$ for every $i\in I$ and
$c\in\frak C$, there is a $\pi\in G$ such that $\pi a_i=b_i$ for every
$i\in I$.

(e) If $(\frak A,\bar\mu)$ is localizable and
$H=\{\pi:\pi\in\AmuA$, $\pi c=c$ for every $c\in\frak C\}$,
then $H$ is the closure of $G$ for the weak topology of $\AmuA$.

\proof{{\bf (a)} Let $\familyiI{(a_i,n_i,b_i)}$ be a maximal family such
that

----- $\familyiI{a_i}$ is a disjoint family in
$\frak A_{\pi a\Bsetminus a}\setminus\{0\}$,

----- $\familyiI{b_i}$ is a disjoint family in
$\frak A_{a\Bsetminus\pi a}$;

----- for every $i\in I$, $n_i\in\Bbb Z$ and $\pi^{n_i}a_i=b_i$.

\noindent Because $\bar\mu a<\infty$, $I$ is countable.   Set

\Centerline{$a'=(\pi a\Bsetminus a)\Bsetminus\sup_{i\in I}a_i$,
\quad$b'=(a\Bsetminus\pi a)\Bsetminus\sup_{i\in I}b_i$;}

\noindent then

\Centerline{$\bar\mu a'
=\bar\mu\pi a-\bar\mu(a\Bcap\pi a)-\sum_{i\in I}\bar\mu a_i
=\bar\mu a-\bar\mu(a\Bcap\pi a)-\sum_{i\in I}\bar\mu b_i
=\bar\mu b'$.}

\Quer\ If $a'\ne 0$, set $c=\sup_{n\in\Bbb Z}\pi^na'$.   Then $\pi c=c$, so

\Centerline{$\bar\mu(c\Bcap b_i)
=\bar\mu(c\Bcap\pi^{n_i}a_i)
=\bar\mu(\pi^{n_i}(c\Bcap a_i))
=\bar\mu(c\Bcap a_i)$}

\noindent for every $i\in I$, and

$$\eqalign{\bar\mu(c\Bcap b')
&=\bar\mu(c\Bcap a\Bsetminus\pi a)-\sum_{i\in I}\bar\mu(c\Bcap b_i)
=\bar\mu(c\Bcap a)-\bar\mu(c\Bcap a\Bcap\pi a)
  -\sum_{i\in I}\bar\mu(c\Bcap b_i)\cr
&=\bar\mu(c\Bcap\pi a)-\bar\mu(c\Bcap a\Bcap\pi a)
  -\sum_{i\in I}\bar\mu(c\Bcap a_i)
=\bar\mu(c\Bcap\pi a\Bsetminus a)-\sum_{i\in I}\bar\mu(c\Bcap a_i)\cr
&=\bar\mu(c\Bcap a')
=\bar\mu a'
>0,\cr}$$

\noindent and $c\Bcap b'\ne 0$.   There is therefore an
$n\in\Bbb Z$ such that $\pi^na'\Bcap b'\ne 0$.   But now, setting
$d=a'\Bcap\pi^{-n}b'$, $d\ne 0$ and we ought to have added
$(d,n,\pi^nd)$ to $\familyiI{(a_i,n_i,b_i)}$.\ \Bang

Thus $\sup_{i\in I}a_i=\pi a\Bsetminus a$ and
$\sup_{i\in I}b_i=a\Bsetminus\pi a$.
Now we can define $\phi\in\Aut\frak A$ by the formula

$$\eqalign{\phi d
&=\pi d\text{ if }d\Bsubseteq a,\cr
&=\pi^{n_i}d\text{ if }i\in I\text{ and }d\Bsubseteq a_i,\cr
&=d\text{ if }d\Bcap(a\Bcup\pi a)=0\cr}$$

\noindent (381C, because $I$ is countable and $\frak A$ is Dedekind
$\sigma$-complete).
Because $G$ is full, $\phi\in G$;  $\phi$ is supported by
$a\Bcup\pi a$, and $\phi$ agrees with $\pi$ on $\frak A_a$, as required.

\medskip

{\bf (b)(i)$\Rightarrow$(ii)} If $\pi a\Bsubseteq b$ and $c\in\frak C$,
then

\Centerline{$\bar\mu(a\Bcap c)
=\bar\mu\pi(a\Bcap c)
=\bar\mu(\pi a\Bcap\pi c)
\le\bar\mu(b\Bcap c)$.}

\medskip

\quad{\bf(ii)$\Rightarrow$(i)} Now suppose that
$\bar\mu(a\Bcap c)\le\bar\mu(b\Bcap c)$ for every $c\in\frak C$.

\medskip

\qquad\grheada\ Consider first the case in which $a\Bcap b=0$.
Let $\familyiI{(a_i,\pi_i,b_i)}$ be a maximal family
such that

----- $\familyiI{a_i}$ is a disjoint family in $\frak A_a\setminus\{0\}$,

----- $\familyiI{b_i}$ is a disjoint family in $\frak A_b$,

----- for every $i\in I$, $\pi_i\in G$ and $\pi_ia_i=b_i$.

\noindent Set $a'=\sup_{i\in I}a_i$,
$b'=\sup_{i\in I}b_i$ and

\Centerline{$c=\upr(b\Bsetminus b',\frak C)
=\sup_{\pi\in G}\pi(b\Bsetminus b')\in\frak C$}

\noindent (395G, because $\frak A$ is Dedekind complete).

$a\Bcap c=a'\Bcap c$.   \Prf\Quer\ Otherwise,
$a\Bsetminus a'$ meets $c$, so there is a $\pi\in G$ such that
$(a\Bsetminus a')\Bcap\pi(b\Bsetminus b')\ne 0$, in which case we ought to
have added

\Centerline{$((a\Bsetminus a')\Bcap\pi(b\Bsetminus b'),\pi^{-1},
\pi^{-1}(a\Bsetminus a')\Bcap b\Bsetminus b')$}

\noindent to our family
$\familyiI{(a_i,\pi_i,b_i)}$.\ \Bang\Qed

Now note that $1\Bsetminus c\in\frak C$, so

$$\eqalign{\bar\mu(a\Bsetminus c)
&\le\bar\mu(b\Bsetminus c)
=\bar\mu(b'\Bsetminus c)
=\sum_{i\in I}\bar\mu(b_i\Bsetminus c)\cr
&=\sum_{i\in I}\bar\mu\pi_i(a_i\Bsetminus c)
=\sum_{i\in I}\bar\mu(a_i\Bsetminus c)
=\bar\mu(a'\Bsetminus c),\cr}$$

\noindent so $a\Bsetminus c=a'\Bsetminus c$ and $a=a'$.

Accordingly we can define $\pi\in\Aut\frak A$ by setting

$$\eqalign{\pi d
&=\pi_id\text{ if }i\in I\text{ and }d\Bsubseteq a_i,\cr
&=\pi_i^{-1}d\text{ if }i\in I\text{ and }d\Bsubseteq b_i,\cr
&=d\text{ if }d\Bsubseteq 1\Bsetminus(a\Bcup b')\cr}$$

\noindent(381C again).   Because $G$ is full, $\pi\in G$, and

\Centerline{$\pi a=\pi(\sup_{i\in I}a_i)
=\sup_{i\in I}\pi a_i=\sup_{i\in I}b_i=b'\Bsubseteq b$.}

\medskip

\qquad\grheadb\ For the general case, we have

\Centerline{$\bar\mu(c\Bcap a\Bsetminus b)
=\bar\mu(c\Bcap a)-\bar\mu(c\Bcap a\Bcap b)
\le\bar\mu(c\Bcap b)-\bar\mu(c\Bcap a\Bcap b)
=\bar\mu(c\Bcap b\Bsetminus a)$}

\noindent for every $c\in\frak C$,
so ($\alpha$) tells us that there is a $\pi_0\in G$ such that
$\pi_0(a\Bsetminus b)\Bsubseteq b\setminus a$.   Now if we set
$\pi=\cycle{a\Bsetminus b\,_{\pi_0}\,\pi_0(a\Bsetminus b)}$,
$\pi\in G$ (because $G$ is full, see 381Sd), and $\pi a\Bsubseteq b$,
as required.

\medskip

{\bf (c)} If $\pi\in G$ and $\pi a=b$, then
$\pi a\Bsubseteq b$ and $\pi^{-1}b\Bsubseteq a$, so (b) tells us that
$\bar\mu(a\Bcap c)=\bar\mu(b\Bcap c)$ for every $c\in\frak C$.
If $\bar\mu(a\Bcap c)=\bar\mu(b\Bcap c)$ for every $c\in\frak C$,
then (b) tells us that there is a $\pi\in G$ such that
$\pi a\Bsubseteq b$;  but as $\bar\mu\pi a=\bar\mu a=\bar\mu b$,
we have $\pi a=b$.

\medskip

{\bf (d)} Let $j$ be any object not belonging to $I$ and set
$a_j=1\Bsetminus\sup_{i\in I}a_i$, $b_j=1\Bsetminus\sup_{i\in I}b_i$.
Then

\Centerline{$\bar\mu(a_j\Bcap c)
=\bar\mu c-\sum_{i\in I}\bar\mu(a_i\Bcap c)
=\bar\mu c-\sum_{i\in I}\bar\mu(b_i\Bcap c)
=\bar\mu(b_j\Bcap c)$}

\noindent for every $c\in\frak C$.   Set $J=I\cup\{j\}$.
By (c), there is for each
$i\in J$ a $\pi_i\in G$ such that $\pi_ia_i=b_i$.
Now $\family{i}{J}{a_i}$ and
$\family{i}{J}{b_i}$ are partitions of unity in $\frak A$, so there is a
$\pi\in\Aut\frak A$ such that $\pi d=\pi_id$ whenever $i\in J$ and
$d\Bsubseteq a_j$;  because $G$ is full, $\pi\in G$, and has the property
we seek.

\medskip

{\bf (e)(i)} If $a\in\frak A$, then
$U=\{\pi:\pi\in\AmuA$, $\pi a\notBsubseteq a\}$ is
open for the weak topology.   \Prf\ The functions

\Centerline{$\pi\mapsto\pi a:\AmuA\to\frak A$,
\quad$b\mapsto b\Bsetminus a:\frak A\to\frak A$}

\noindent are continuous (494Bb and 323Ba), and
$c\mapsto\bar\mu c:\frak A\to[0,\infty]$ is lower semi-continuous
(323Cb, because $\frak A$ is semi-finite), so
$\pi\mapsto\bar\mu(\pi a\Bsetminus a)$ is lower semi-continuous
(4A2B(d-ii)) and $U=\{\pi:\bar\mu(\pi a\Bsetminus a)>0\}$ is open.\ \Qed

Consequently, $\{\pi:\pi c\Bsubseteq c$ for every $c\in\frak C\}$ is
closed.   But if $\pi c\Bsubseteq c$ for every $c\in\frak C$, then
$\pi c=c$ for every $c\in\frak C$.  So $H$ is closed.   Of course
$H$ includes $G$, so $\overline{G}\subseteq H$.

\medskip

\quad{\bf (ii)}
Suppose that $\pi\in H$ and that $U$ is an open neighbourhood of $\pi$.
Then there are $a_0,\ldots,a_n\in\frak A^f$ and
$\delta>0$ such that $U$ includes
$\{\phi:\phi\in\AmuA$, $\bar\mu(\pi a_i\Bsymmdiff\phi a_i)\le\delta$
for every $i\le n\}$.
Set $e=\sup_{i\le n}a_i$;
let $\frak B$ be the finite subalgebra of $\frak A_e$
generated by
$\{e\Bcap a_i:i\le n\}$, and $B$ the set of its
atoms\cmmnt{ (definition:  316K)}.
If $b\in B$, then
$\bar\mu(\pi b\Bcap c)=\bar\mu(b\Bcap c)$
for every $c\in\frak C$, so there is a $\phi_b\in G$ such that
$\phi_bb=\pi b$, by (c) above.
Equally, there is a $\phi\in G$ such that $\phi e=\pi e$.   Now we can
define $\psi\in\AmuA$ by saying that

$$\eqalign{\psi d
&=\phi_bd\text{ if }b\in B\text{ and }d\Bsubseteq b,\cr
&=\phi d\text{ if }d\Bsubseteq 1\Bsetminus e;\cr}$$

\noindent as usual, $\psi\in G$, while $\psi b=\pi b$ for every $b\in B$.
But this means that $\psi a_i=\pi a_i$ for every $i\le n$, so
$\psi\in G\cap U$.   As $U$ is arbitrary, $\pi\in\overline{G}$;
as $\pi$ is arbitrary, $G$ is dense in $H$ and $H=\overline{G}$.
}%end of proof of 494G

\leader{494H}{Proposition} Let $\frak A$ be a Boolean algebra, $G$ a
full subgroup of $\Aut\frak A$, and $a\in\frak A$.   Set
$G_a=\{\pi:\pi\in G$, $\pi$ is supported by $a\}$,
$H_a=\{\pi\restrp\frak A_a:\pi\in G_a\}$.

(a) $G_a$ is a full subgroup of $\Aut\frak A$ and
$H_a$ is a full subgroup of $\Aut\frak A_a$, for every $a\in\frak A$.

(b) Suppose that $\frak A$ is Dedekind complete, and that the fixed-point
subalgebra of $G$ is $\frak C$.
Then the fixed-point subalgebra of $H_a$ is $\{a\Bcap c:c\in\frak C\}$.

\proof{{\bf (a)(i)} By 381Eb and 381Eh, $G_a$
is a subgroup of $G$,
and $\pi\mapsto\pi\restrp\frak A_a$ is a group homomorphism from $G_a$ to
$\Aut\frak A_a$, so its image $H_a$ is a subgroup of $\Aut\frak A_a$.

\medskip

\quad{\bf (ii)} Suppose that $\phi\in\Aut\frak A$ and that
$\familyiI{(a_i,\pi_i)}$ is a family in
$\frak A\times G_a$ such that $\familyiI{a_i}$ is a partition of unity in
$\frak A$ and $\pi_id=\phi d$ whenever
$i\in I$ and $d\Bsubseteq a_i$.   Then $\phi\in G$, because $G$ is full;
and

\Centerline{$\phi d
=\sup_{i\in I}\pi_i(d\Bcap a_i)=\sup_{i\in I}d\Bcap a_i=d$}

\noindent whenever $d\Bcap a=0$, so $\phi$ is supported by $a$ and belongs
to $G_a$.

\medskip

\quad{\bf (iii)} Suppose that $\phi\in\Aut\frak A_a$ and that
$\familyiI{(a_i,\pi_i)}$ is a family in
$\frak A_a\times H_a$ such that $\familyiI{a_i}$ is a partition of unity in
$\frak A_a$ and $\pi_id=\phi d$ whenever
$i\in I$ and $d\Bsubseteq a_i$.   For each $i\in I$, there is a
$\pi'_i\in G_a$ such that $\pi_i=\pi'_a\restrp\frak A_a$.
Take $j\notin I$ and set
$J=I\cup\{j\}$, $a_j=1\Bsetminus a$, $\pi'_j=\iota$;  define
$\psi\in\Aut\frak A$ by setting $\psi d=\phi d$ for $d\Bsubseteq a$,
$d$ for $d\Bsubseteq 1\Bsetminus a$.   Then
$\family{j}{J}{a_j}$ is a partition of unity in $\frak A$ and
$\psi d=\pi'_jd$ whenever $j\in J$ and $d\Bsubseteq a_j$, so
$\psi\in G$.   Also $\psi$ is supported by $a$,
so $\phi=\psi\restrp\frak A_a$ belongs to $H_a$.
As $\phi$ and $\familyiI{(a_i,\pi_i)}$ are arbitrary, $H_a$ is full.

\medskip

{\bf (b)(i)} If $c\in\frak C$, then
$\pi(a\Bcap c)=\pi a\Bcap\pi c=a\Bcap c$ whenever $\pi\in G$ and $\pi a=a$,
so $a\Bcap c$ belongs to the fixed-point subalgebra of $H_a$.

\medskip

\quad{\bf (ii)} In the other direction, take any $b$ in the fixed-point
subalgebra of $H_a$.   Set $c=\upr(b,\frak C)=\sup_{\pi\in G}\pi b$
(395G once more).
Of course $b\Bsubseteq a\Bcap c$.   \Quer\ If $b\ne a\Bcap c$, set
$e=a\Bcap c\Bsetminus b$.   Then there is a $\pi\in G$ such that
$e_1=e\Bcap\pi b\ne 0$;  set $e_2=\pi^{-1}e_1\Bsubseteq b$ and
$\phi=\cycle{e_2\,_{\pi}\,e_1}$.
Then $\phi\in G$ (381Sd again) and $\phi$ is supported by
$e_1\Bcup e_2\Bsubseteq a$, so
$\phi\restrp\frak A_a\in H_a$;  but $\phi b\ne b$, so this is impossible.\
\BanG\  Thus $b$ is expressed as the intersection of $a$ with a member of
$\frak C$, as required.
}%end of proof of 494H

\leader{494I}{}\dvAformerly{4{}93G}\cmmnt{ I take the proof of the next
theorem in a series of lemmas, the first being the leading special case.

\medskip

\noindent}{\bf Lemma}\cmmnt{ ({\smc Giordano \& Pestov 92})}
Let $(\frak A,\bar\mu)$ be an atomless homogeneous probability
algebra\cmmnt{ (definitions:  316Kb, 316N)}.
Then $\AmuA$, with its weak topology, is extremely amenable.

\proof{ I seek to apply 493C.

\medskip

{\bf (a)} Take $\epsilon>0$, a neighbourhood $V$ of the identity in
$\AmuA$,
a finite set $I\subseteq\AmuA$ and a finite family $\Cal A$ of zero sets in
$\AmuA$.
Let $\delta>0$ and $K\in[\frak A]^{<\omega}$ be such that $\pi\in V$
whenever $\pi\in\AmuA$ and $\bar\mu(a\Bsymmdiff\pi a)\le\delta$ for every
$a\in K$.   Let $C$ be the set of atoms of the finite subalgebra $\frak C$
of $\frak A$ generated by $K$, and $D$ the set of atoms of the subalgebra
$\frak D$ generated by $K\cup\bigcup_{\pi\in I}\pi[K]$;
set $k=\#(C)$ and
$k'=\#(D)$.   Let $m\in\Bbb N$ be so large that $2kk'\le m\delta$ and
$(m\delta-1)^2\ge 64m\ln\Bover1{\epsilon}$;
set $r=\lfloor m\delta\rfloor$, so
that $\exp(-\Bover{r^2}{64m})\le\epsilon$.

\medskip

{\bf (b)} For each $d\in D$ let $E_d$ be a maximal disjoint family in
$\frak A_d$ such that $\bar\mu e=\bover1m$ for every $e\in E_d$;  let $E$
be a partition of unity in $\frak A$, including
$\bigcup_{d\in D}E_d$, such that $\bar\mu e=\bover1m$ for every $e\in E$.
Let $H$ be the group of permutations of $E$.   Then we have a group
homomorphism $\theta:H\to\AmuA$ such that $\theta(\psi)\restr E=\psi$ for
every $\psi\in H$.   \Prf\ Fix $e_0\in E$.   Then for each
$e\in E$ there is a measure-preserving isomorphism
$\phi_e:\frak A_{e_0}\to\frak A_e$, because $\frak A$ is homogeneous
(331I).   For $\psi\in H_E$, $E$ and
$\family{e}{E}{\psi e}$ are partitions of unity in $\frak A$, so we can
define $\theta(\psi)\in\AmuA$ by the formula

\Centerline{$\theta(\psi)(a)=\phi_{\psi e}\phi_e^{-1}a$
whenever $a\Bsubseteq e\in E$.}

\noindent It is easy to see that $\theta(\psi)(e)=\psi e$ for every
$e\in E$.   If $\psi$, $\psi'\in H_E$, then

$$\eqalign{\theta(\psi\psi')(a)
&=\phi_{\psi\psi'e}\phi_e^{-1}a\cr
&=\phi_{\psi\psi'e}\phi_{\psi'e}^{-1}\phi_{\psi'e}\phi_e^{-1}a
=\theta(\psi)\theta(\psi')(a)\cr}$$

\noindent whenever $a\Bsubseteq e\in E$;  so $\theta$ is a group
homomorphism.\ \Qed

Write $G$ for $\theta[H]$, so that $G$ is a subgroup of $\AmuA$.

\medskip

{\bf (c)} $I\subseteq GV^{-1}$.   \Prf\ Take $\pi\in I$.   For $c\in C$,
set

\Centerline{$E'_c=\bigcup\{E_d:d\in D$, $d\Bsubseteq c\}$,
\quad$E''_c=\bigcup\{E_d:d\in D$, $d\Bsubseteq\pi c\}$.}

\noindent Since $\sup E_d\subseteq d$ and
$\bar\mu(d\Bsetminus\sup E_d)\le\bover1m$ for every $d\in D$,
$\sup E'_c\Bsubseteq c$ and $\bar\mu(c\Bsetminus\sup E'_c)\le\Bover{k'}m$;
so $m\bar\mu c-k'\le\#(E'_c)\le m\bar\mu c$.
Similarly, $\sup E''_c\Bsubseteq\pi c$ and

\Centerline{$m\bar\mu c-k'=m\bar\mu\pi c-k'\le\#(E''_c)\le m\bar\mu\pi c
=m\bar\mu c$.}

\noindent
Let $\tilde E'_c\subseteq E'_c$, $\tilde E''_c\subseteq E''_c$ be sets of
size $\min(\#(E'_c),\#(E''_c))\ge m\bar\mu c-k'$.
Setting $c'=\sup\tilde E'_c$ and $c''=\sup\tilde E''_c$ we have

\Centerline{$c'\Bsubseteq c$,
\quad$\bar\mu(c\Bsetminus c')
=\Bover1m(m\bar\mu c-\#(\tilde E'_c))\le\Bover{k'}m$,}

\noindent and similarly $c''\Bsubseteq\pi c$ and
$\bar\mu(\pi c\Bsetminus c'')\le\Bover{k'}m$.

Because $\family{c}{C}{\tilde E'_c}$ and $\family{c}{C}{\tilde E''_c}$ are
both disjoint, there is a $\psi\in H$ such that
$\psi[\tilde E'_c]=\tilde E''_c$ for every $c\in C$.   Set
$\phi=\theta(\psi)$;  then $\phi\in G$ and $\phi c'=c''$ for every
$c\in C$.   Now this means that

$$\eqalign{\bar\mu(c\Bsymmdiff\pi^{-1}\phi c)
&=\bar\mu(\pi c\Bsymmdiff\phi c)
\le\bar\mu(\pi c\Bsymmdiff c'')+\bar\mu(c''\Bsymmdiff\phi c)\cr
&=\bar\mu(\pi c\Bsymmdiff c'')+\bar\mu(c'\Bsymmdiff c)
\le\Bover{2k'}m\cr}$$

\noindent for every $c\in C$.   Consequently

\Centerline{$\bar\mu(a\Bsymmdiff\pi^{-1}\phi a)\le\Bover{2kk'}m\le\delta$}

\noindent for every $a\in K$, and $\pi^{-1}\phi\in V$.   Accordingly
$\pi\in\phi V^{-1}\subseteq GV^{-1}$;  as $\pi$ is arbitrary,
$I\subseteq GV^{-1}$.\ \Qed

\medskip

{\bf (d)} I am ready to introduce the functional $\nu$ demanded by the
hypotheses of 493C.   Let $\lambda$ be the Haar probability measure on the
finite group $H$, and $\nu$ the image measure $\lambda\theta^{-1}$,
regarded as a measure on $\AmuA$.   If $\pi\in I$, then (c) tells us that
there is a $\psi\in H$ such that
$\phi=\theta(\psi)$ belongs to $\pi V$.   In this case, for any
$A\in\Cal A$,

\Centerline{$\nu(\phi A)=\lambda\theta^{-1}[\phi A]
=\lambda(\psi\theta^{-1}[A])=\lambda\theta^{-1}[A]=\nu A$.}

\noindent So $\nu$ satisfies condition (ii) of 493C.

\medskip

{\bf (e)} As for condition (i) of 493C, consider
$W=\{\psi:\psi\in H$, $\#(\{e:e\in E$, $\psi e\ne e\})\le r\}$.
Then $\theta[W]\subseteq V$.   \Prf\ If $\psi\in W$, then
$\theta(\psi)(d)=d$ whenever $d\Bsubseteq e\in E$ and $\psi e=e$.   So
$\theta(\psi)$ is supported by $b=\sup\{e:e\in E$, $\psi e\ne e\}$.   Now
$\bar\mu b\le\bover{r}m\le\delta$.   So
$\bar\mu(a\Bsymmdiff\phi a)\le\bar\mu b\le\delta$ for every $a\in\frak A$,
and $\phi\in V$.\ \Qed

Now suppose that $F\subseteq\AmuA$ and $\nu F\ge\bover12$.   Then

$$\eqalignno{\nu(VF)
&=\lambda\theta^{-1}[VF]
\ge\lambda(W\theta^{-1}[F])\cr
\displaycause{because $\theta[W]\subseteq V$}
&\ge 1-\exp(-\Bover{r^2}{64m})\cr
\displaycause{by 492I, because $\lambda\theta^{-1}[F]=\nu F\ge\bover12$}
&\ge 1-\epsilon.\cr}$$

\noindent So $\nu$ satisfies the first condition in 493C.

\medskip

{\bf (f)} As $\epsilon$, $V$, $I$ and $\Cal A$ are arbitrary, $\AmuA$ is
extremely amenable.
}%end of proof of 494I

\vleader{48pt}{494J}{Lemma} Let $(\frak C,\bar\lambda)$ be a totally finite
measure algebra, $(\frak B,\bar\nu)$ a probability
algebra, and $(\frak A,\bar\mu)$ the localizable measure algebra
free product $(\frak C,\bar\lambda)\tensorhat(\frak B,\bar\nu)$\cmmnt{
(325E)}.
Give $\AmuA$ its weak topology, and let $G$ be the subgroup
$\{\pi:\pi\in\AmuA$, $\pi(c\otimes 1)=c\otimes 1$ for every
$c\in\frak C\}$.   Suppose that $\frak B$ is either finite, with all its
atoms of the same measure, or
homogeneous.   Then $G$ is amenable, and if either $\frak B$
is homogeneous or $\frak C$ is atomless, $G$ is extremely amenable.

\proof{{\bf (a)} Let $\Cal E$ be the family of finite partitions of unity
in $\frak C$ not containing $\{0\}$.   Then for any $E\in\Cal E$
we have a function
$\theta_E:(\Aut_{\bar\nu}\frak B)^E\to G$ defined by saying that

\Centerline{$\theta_E(\pmb{\phi})(c\otimes b)
=\sup_{e\in E}(c\Bcap e)\otimes\phi_eb$}

\noindent whenever
$\pmb{\phi}=\family{e}{E}{\phi_e}\in(\Aut_{\bar\nu}\frak B)^E$,
$c\in\frak C$ and $b\in\frak B$.
\Prf\ For each $e\in E$, the defining universal mapping theorem
325Da tells us that there is a unique measure-preserving Boolean
homomorphism $\psi_e:\frak A\to\frak A$ such that
$\psi_e(c\otimes 1)=c\otimes 1$ and $\psi_e(1\otimes b)=1\otimes\phi_eb$
for all $b\in\frak B$ and $c\in\frak C$.   To see that $\psi_e$ is
surjective, note that $\psi_e[\frak A]$ must be a closed subalgebra
including $\frak C\otimes\frak B$ (324Kb), which is dense (325D(c-i)).   So
$\psi_e\in\AmuA$.   Now
$\family{e}{E}{e\otimes 1}$ is a partition of unity in $\frak A$, and
$\psi_e(e\otimes 1)=e\otimes 1$ for every $e$, so we have a $\pi\in\AmuA$
defined by saying that $\pi a=\sup_{e\in E}\psi_e(a\Bcap e)$ for every
$a\in\frak A$.   Because $G$ is full, $\pi\in G$.  So we can set
$\theta_E(\pmb{\phi})=\pi$.   Of course $\pi$ is the only automorphism
satisfying the given formula for $\theta_E(\pmb{\phi})$.\ \Qed

\medskip

{\bf (b)(i)} It is easy to check that if $E\in\Cal E$ then
$\theta_E$ is a group homomorphism
from $(\Aut_{\bar\nu}\frak B)^E$ to $G$;
write $G_E$ for its set of values.   Because $0\notin E$, $\theta_E$ is
injective, and $G_E$ is a subgroup of $G$ isomorphic to the group
$(\Aut_{\bar\nu}\frak B)^E$.   Give $\Aut_{\bar\nu}\frak B$ its weak
topology, $(\Aut_{\bar\nu}\frak B)^E$ the product topology and $G$ the
topology induced by the weak topology of $\AmuA$.

\medskip

\qquad{\bf (ii)}  $\theta_E$ is continuous.
\Prf\ If $U$ is a neighbourhood of the identity in $G_E$, there are
$a_0,\ldots,a_n\in\frak A$ and $\epsilon>0$ such that $U$ includes
$\{\pi:\pi\in G_E$, $\bar\mu(a_i\Bsymmdiff\pi a_i)\le 3\epsilon$ for
every $i\le n\}$.   For each $i\le n$, there is an
$a'_i\in\frak C\otimes\frak B$ such that
$\bar\mu(a_i\Bsymmdiff a'_i)\le\epsilon$.  Let $\frak B_0$ be a
finite subalgebra of $\frak B$ such that $a'_i\in\frak C\otimes\frak B_0$
for every $i\le n$.   Let $\delta>0$ be such that
$\delta\bar\lambda 1\le\epsilon$.
Then there is a neighbourhood $V$ of the identity in
$\Aut_{\bar\nu}\frak B$ such that
$\bar\nu(b\Bsymmdiff\phi b)\le\delta$ whenever $\phi\in V$ and
$b\in\frak B_0$.   If now $\pmb{\phi}=\family{e}{E}{\phi_e}$ belongs to
$V^E$, then for each $i\le n$ we can express $a'_i$ as
$\sup_{j\le m_i}c_{ij}\otimes b_{ij}$ where
$\langle c_{ij}\rangle_{j\le m_i}$ is a partition of unity in $\frak C$ and
$b_{ij}\in\frak B_0$ for every $j\le m_i$ (315Oa).   So

$$\eqalignno{\bar\mu(a'_i\Bsymmdiff\theta_E(\pmb{\phi})a'_i)
&\le\sum_{\Atop{j\le m_i}{e\in E}}\bar\mu(((c_{ij}\Bcap e)\otimes b_{ij})
    \Bsymmdiff\theta_E(\pmb{\phi})((c_{ij}\Bcap e)\otimes b_{ij}))\cr
\displaycause{because
$\langle(c_{ij}\cap e)\otimes b_{ij}\rangle_{j\le m_i,e\in E}$ is a
disjoint family with supremum $a'_i$}
&=\sum_{\Atop{j\le m_i}{e\in E}}\bar\mu(((c_{ij}\Bcap e)\otimes b_{ij})
    \Bsymmdiff((c_{ij}\Bcap e)\otimes\phi_e b_{ij}))\cr
&=\sum_{\Atop{j\le m_i}{e\in E}}\bar\mu((c_{ij}\Bcap e)
   \otimes(b_{ij}\Bsymmdiff\phi_e b_{ij}))\cr
&=\sum_{\Atop{j\le m_i}{e\in E}}\bar\lambda(c_{ij}\Bcap e)
  \cdot\bar\nu(b_{ij}\Bsymmdiff\phi_e b_{ij})
\le\delta\sum_{\Atop{j\le m_i}{e\in E}}\bar\lambda(c_{ij}\Bcap e)
\le\epsilon,\cr}$$

\noindent and

\Centerline{$\bar\mu(a_i\Bsymmdiff\theta_E(\pmb{\phi})a_i)
\le\bar\mu(a_i\Bsymmdiff a'_i)
  +\bar\mu(a'_i\Bsymmdiff\theta_E(\pmb{\phi})a'_i)
  +\bar\mu(\theta_E(\pmb{\phi})a'_i\Bsymmdiff\theta_E(\pmb{\phi})a_i)
\le 3\epsilon$.}

\noindent This is true for every $i\le n$, so $\theta_E(\pmb{\phi})\in U$
whenever $\pmb{\phi}\in V^E$.   As $U$ is arbitrary, $\theta_E$
is continuous.\ \Qed

\medskip

\qquad{\bf (iii)} $\theta_E^{-1}$ is continuous.
\Prf\ Let $V$ be a neighbourhood of the identity in
$\Aut_{\bar\nu}\frak B$.   Then there are $\epsilon>0$ and
$b_0,\ldots,b_n\in\frak B$ such that $\phi\in V$ whenever
$\phi\in\Aut_{\bar\nu}\frak B$ and
$\bar\nu(b_i\Bsymmdiff\phi b_i)\le\epsilon$ for every $i\le n$.   Let
$\delta>0$ be such that $\delta\le\epsilon\bar\lambda e$ for every
$e\in E$, and let $U$ be

\Centerline{$\{\pi:\pi\in G_E$,
$\bar\mu((e\otimes b_i)\Bsymmdiff\pi(e\otimes b_i))\le\delta$ whenever
$e\in E$ and $i\le n\}$.}

\noindent Then $U$ is a neighbourhood of the identity in $G_E$.   If
$\pmb{\phi}=\family{e}{E}{\phi_e}\in(\Aut_{\bar\nu}\frak B)^E$ is such that
$\theta_E(\pmb{\phi})\in U$, then for every $e\in E$ and $i\le n$ we have

\Centerline{$\bar\nu(b_i\Bsymmdiff\phi_eb_i)
=\Bover1{\bar\lambda e}
  \bar\mu((e\otimes b_i)\Bsymmdiff\theta_E(\pmb{\phi})(e\otimes b_i))
\le\Bover{\delta}{\bar\lambda e}
\le\epsilon$,}

\noindent so $\pmb{\phi}\in V^E$.   As $V$ is arbitrary, $\theta_E^{-1}$ is
continuous.\ \Qed

\medskip

\qquad{\bf (iv)} Putting these together, $\theta_E$ is a topological group
isomorphism.

\medskip

{\bf (c)} The next step is to show that $\bigcup_{E\in\Cal E}G_E$ is dense
in $G$.

\medskip

\quad{\bf (i)} Note first that there is an upwards-directed
family $\Bbb D$ of finite
subalgebras $\frak D$ of $\frak B$ such that if $\frak D\in\Bbb D$ then
every atom of $\frak D$ has the same measure, and $\bigcup\Bbb D$ is dense
in $\frak B$ (for the measure-algebra topology of $\frak B$).
\Prf\ If $\frak B$ is finite, with all its atoms of the same measure,
this is trivial;  take $\Bbb D=\{\frak B\}$.
Otherwise, because $\frak B$ is homogeneous, $(\frak B,\bar\nu)$ must be
isomorphic to the measure algebra $(\frak B_{\kappa},\bar\nu_{\kappa})$
of the usual measure on $\{0,1\}^{\kappa}$ for some infinite
cardinal $\kappa$, and we can take $\Bbb D$ to be the family of subalgebras
determined by finite subsets of $\kappa$.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $\pi\in G$,
$a_0,\ldots,a_n\in\frak A$ and $\epsilon>0$.
Let $\frak A_0$ be the
subalgebra of $\frak A$ generated by $a_0,\ldots,a_n$ and $A$ the set of
its atoms;  let $\eta>0$ be such that $12\eta\#(A)\le\epsilon$.
Consider subalgebras of
$\frak A$ of the form $\frak C_0\otimes\frak D$ where $\frak C_0$ is a
finite subalgebra of $\frak C$ and $\frak D\in\Bbb D$.   This is an
upwards-directed family of subalgebras, and the closure of its union
includes $c\otimes b$ whenever $c\in\frak C$ and $b\in\frak B$,
so is the whole of $\frak A$.   There must therefore be a finite subalgebra
$\frak C_0$ of $\frak C$, a $\frak D\in\Bbb D$, and
$a'$, $a''\in\frak C_0\otimes\frak D$, for each $a\in A$, such that
$\bar\mu(a\Bsymmdiff a')\le\eta$ and
$\bar\mu(\pi a\Bsymmdiff a'')\le\eta$ for every $a\in A$.
Note that this implies that

\Centerline{$|\bar\mu a'-\bar\mu a''|
\le|\bar\mu a'-\bar\mu a|+|\bar\mu a-\bar\mu a''|
\le 2\eta$}

\noindent for every $a\in A$.

\medskip

\quad{\bf (iii)}
Let $E\in\Cal E$ be the set of atoms of $\frak C_0$,
$D$ the set of atoms of
$\frak D$, and $\gamma$ the common measure of the members of $D$.   For
$e\in E$ and $a\in A$, set

\Centerline{$D_{ea}'=\{d:d\in D$, $\bar\mu((e\otimes d)\Bcap a)
>\Bover12\bar\mu(e\otimes d)\}$,
\quad$b_{ea}'=\sup D_{ea}'$,}

\Centerline{$D_{ea}''=\{d:d\in D$, $\bar\mu((e\otimes d)\Bcap\pi a)
>\Bover12\bar\mu(e\otimes d)\}$,
\quad$b_{ea}''=\sup D_{ea}''$.}

\noindent Note that as $A$ is disjoint,
$\family{a}{A}{D_{ea}'}$ is disjoint, for each $e$;  and similarly
$\family{a}{A}{D_{ea}''}$ is disjoint for each $e$, because
$\family{a}{A}{\pi a}$ is disjoint.   Next, for $a\in A$ and $e\in E$,
set $D_{ea}=\{d:d\in D$, $e\otimes d\Bsubseteq a'\}$.   Then, for each
$a\in A$,

$$\eqalign{\bar\mu(a'\Bsymmdiff\sup_{e\in E}e\otimes b_{ea}')
&=\sum_{\Atop{e\in E}{d\in D'_{ea}\setminus D_{ea}}}\bar\mu(e\otimes d)
   +\sum_{\Atop{e\in E}{d\in D_{ea}\setminus D'_{ea}}}
      \bar\mu(e\otimes d)\cr
&\le\sum_{\Atop{e\in E}{d\in D'_{ea}\setminus D_{ea}}}
       2\bar\mu((e\otimes d)\Bcap a)
   +\sum_{\Atop{e\in E}{d\in D_{ea}\setminus D'_{ea}}}
      2\bar\mu((e\otimes d)\Bsetminus a)\cr
&\le\sum_{\Atop{e\in E}{d\in D\setminus D_{ea}}}
      2\bar\mu((e\otimes d)\Bcap a)
   +\sum_{\Atop{e\in E}{d\in D_{ea}}}2\bar\mu((e\otimes d)\Bsetminus a)\cr
&=2\bar\mu(a\Bsetminus a')+2\bar\mu(a'\Bsetminus a)
=2\bar\mu(a\Bsymmdiff a')
\le 2\eta,\cr}$$

\noindent and
$\bar\mu(a\Bsymmdiff\sup_{e\in E}e\otimes b_{ea}')\le 3\eta$.
Similarly, passing through $a''$ in place of $a'$, we see that
$\bar\mu(\pi a\Bsymmdiff\sup_{e\in E}e\otimes b_{ea}'')\le 3\eta$.

Consequently, for any $a\in A$,

$$\eqalignno{
\sum_{e\in E}\bar\lambda e\cdot\gamma|\#(D_{ea}')-\#(D_{ea}'')|
&=\sum_{e\in E}|\bar\mu(e\otimes b_{ea}')-\bar\mu(e\otimes b_{ea}'')|\cr
&\le\sum_{e\in E}
  \bar\mu((e\otimes b_{ea}')\Bsymmdiff((e\otimes 1)\Bcap a))\cr
&\mskip100mu
  +|\bar\mu((e\otimes 1)\Bcap a)-\bar\mu((e\otimes 1)\Bcap\pi a)|\cr
&\mskip100mu
  +\bar\mu(((e\otimes 1)\Bcap\pi a)\Bsymmdiff(e\otimes b_{ea}''))\cr
&=\sum_{e\in E}
 \bar\mu((e\otimes b_{ea}')\Bsymmdiff((e\otimes 1)\Bcap a))\cr
&\mskip100mu
  +\bar\mu(((e\otimes 1)\Bcap\pi a)\Bsymmdiff(e\otimes b_{ea}''))\cr
\displaycause{because $\pi\in G$, so
$(e\otimes 1)\Bcap\pi a=\pi((e\otimes 1)\Bcap a)$ for every $e$}
&=\bar\mu(a\Bsymmdiff\sup_{e\in E}e\otimes b_{ea}')
     +\bar\mu(\pi a\Bsymmdiff\sup_{e\in E}e\otimes b_{ea}'')
\le 6\eta.\cr}$$

\medskip

\quad{\bf (iv)} Fix $e\in E$ for the moment.    For each
$a\in A$, take $\tilde D_{ea}'\subseteq D_{ea}'$,
$\tilde D_{ea}''\subseteq D_{ea}''$ such that
$\#(\tilde D_{ea}')=\#(\tilde D_{ea}'')=\min(\#(D_{ea}'),\#(D_{ea}''))$.
As $\family{a}{A}{\tilde D_{ea}'}$ and $\family{a}{A}{\tilde D_{ea}''}$ are
always disjoint families, there is a permutation $\psi_e:D\to D$ such that
$\psi_e[\tilde D_{ea}']=\tilde D_{ea}''$ for every $a\in A$.   Because
$(\frak B,\bar\nu)$ is homogeneous, there is a
$\phi_e\in\Aut_{\bar\nu}\frak B$ such that $\phi_ed=\psi_ed$ for every
$d\in D$.

\medskip

\quad{\bf (v)}
This gives us a family $\pmb{\phi}=\family{e}{E}{\phi_e}$.   Consider
$\theta_E(\pmb{\phi})$.   For each $a\in A$,

$$\eqalign{\bar\mu(\pi a\Bsymmdiff\theta_E(\pmb{\phi})(a))
&\le\bar\mu(\pi a\Bsymmdiff\sup_{e\in E}e\otimes b_{ea}'')
   +\bar\mu((\sup_{e\in E}e\otimes b_{ea}'')
      \Bsymmdiff\theta_E(\pmb{\phi})(\sup_{e\in E}e\otimes b_{ea}'))\cr
&\mskip220mu
   +\bar\mu(\theta_E(\pmb{\phi})(\sup_{e\in E}e\otimes b_{ea}')
              \Bsymmdiff\theta_E(\pmb{\phi})(a))\cr
&\le 3\eta
   +\sum_{e\in E}\bar\mu((e\otimes b_{ea}'')
                   \Bsymmdiff(e\otimes\phi_eb_{ea}'))
   +\bar\mu((\sup_{e\in E}e\otimes b_{ea}')\Bsymmdiff a)\cr
&\le 3\eta
   +\sum_{e\in E}\bar\lambda e
      \cdot\gamma\#(D_{ea}''\symmdiff\psi_e[D_{ea}'])
   +3\eta\cr
&\le 6\eta+\sum_{e\in E}\bar\lambda e
  \cdot\gamma(\#(D_{ea}''\setminus\tilde D_{ea}'')
         +\#(D_{ea}'\setminus\tilde D_{ea}'))\cr
&=6\eta+\sum_{e\in E}\bar\lambda e
  \cdot\gamma|\#(D_{ea}'')-\#(D_{ea}')|
\le 12\eta.\cr}$$

\medskip

\quad{\bf (vi)} Now, for each $i\le n$, set
$A_i=\{a:a\in A$, $a\Bsubseteq a_i\}$;  then

\Centerline{$\bar\mu(\pi a_i\Bsymmdiff\theta_E(\pmb{\phi})(a_i))
\le\sum_{a\in A_i}\bar\mu(\pi a\Bsymmdiff\theta_E(\pmb{\phi})(a))
\le 12\eta\#(A)\le\epsilon$,}

\noindent while $\theta_E{\pmb{\phi}}\in G_E$.
As $\phi$, $a_0,\ldots,a_n$ and $\epsilon$ are arbitrary,
$\bigcup_{E\in\Cal E}G_E$ is dense in $G$.

\medskip

\quad{\bf (vii)} Note also that if $E$, $E'\in\Cal E$, there is an
$F\in\Cal E$ such that $G_F\supseteq G_E\cup G_{E'}$.   \Prf\ Set
$F=\{e\Bcap e':e\in E$, $e'\in E'\}\setminus\{0\}$.   If
$\pmb{\phi}=\family{e}{E}{\phi_e}\in(\Aut_{\bar\nu}\frak B)^E$, define
$\family{f}{F}{\psi_f}\in(\Aut_{\bar\nu}\frak B)^F$ by saying that
$\psi_f=\phi_e$ whenever $f\in F$, $e\in E$ and $f\Bsubseteq e$.
Then it is easy to check that
$\theta_F(\family{f}{F}{\psi_f})=\theta_E(\pmb{\phi})$.   So
$G_F\supseteq G_E$;  similarly, $G_F\supseteq G_{E'}$.\ \Qed

So $\{G_E:E\in\Cal E\}$ is an upwards-directed family of subgroups of $G$
with dense union in $G$.

\medskip

{\bf (d)} At this point, we start looking at the rest of the hypotheses.

\medskip

\quad{\bf (i)} Suppose that $\frak B$ is atomless.   Then 494I tells us
that $\Aut_{\bar\nu}\frak B$ is extremely amenable.   So all the products
$(\Aut_{\bar\nu}\frak B)^E$ are extremely amenable (493Bd), all the $G_E$
are extremely amenable, and $G$ is extremely amenable by (c) and 493Bb.

\medskip

\quad{\bf (ii)} Suppose that $\frak B$ is finite.   Then
$\Aut_{\bar\nu}\frak B$ is finite, therefore amenable (449Cg);
all the products $(\Aut_{\bar\nu}\frak B)^E$ are amenable (449Ce),
and $G$ is amenable (449Cb).

\medskip

{\bf (e)} I have still to finish the case in which $\frak C$ is atomless
and $\frak B$ is finite.   If $\frak B=\{0\}$ then of course $G=\{\iota\}$
is extremely amenable, so we may take it that $\bar\lambda 1>0$.

\medskip

\quad{\bf (i)} Take
$\epsilon>0$, a neighbourhood $V$ of the identity in $G$, a finite set
$I\subseteq G$ and a finite family $\Cal A$ of zero sets in $G$.
Let $V_1$ be a neighbourhood of the identity in $G$ such that
$V_1^2\subseteq V^{-1}$.   By (c), there is an $E'\in\Cal E$ such that
$I\subseteq G_{E'}V_1$.   Set $k=\#(E')$.
$V_1$ is a neighbourhood of the identity for
the uniform topology on $G$ (494Cd), so there is a $\delta>0$ such that
$\pi\in V_1$ whenever $\pi\in G$ and the support of
$\pi$ has measure at most $\delta$ (494Cb).   Let $m$ be so large that
$m\delta\ge k\bar\lambda 1$ and
$(\Bover{m\delta}{\bar\lambda 1}-1)^2\ge m\ln(\Bover{2}{\epsilon})$;  set
$r=\lfloor\Bover{m\delta}{\bar\lambda 1}\rfloor$, so that
$2\exp(-\Bover{r^2}m)\le\epsilon$.

\medskip

\quad{\bf (ii)} For each $e\in E'$ let $D_e$ be a maximal disjoint set of
elements of measure $\bover1m\bar\lambda 1$ in $\frak C_e$;
let $E\supseteq\bigcup_{e\in E'}D_e$ be a maximal disjoint
set of elements of measure $\bover1m\bar\lambda 1$ in $\frak C$.   Note that
$c=1\Bsetminus\sup_{e\in E'}\sup D_e$ has
measure at most $\bover{k}m\bar\lambda 1\le\delta$.
Consequently $G_{E'}\subseteq G_EV_1$.   \Prf\
If $\pi'\in G_{E'}$, express $\theta_{E'}^{-1}(\pi')$ as
$\family{e}{E'}{\phi'_e}$.   Let
$\family{e}{E}{\phi_e}\in(\Aut_{\bar\nu}\frak B)^E$
be such that
$\phi_e=\phi'_{e'}$ whenever $e'\in E'$ and $e\in D_{e'}$, and set
$\pi=\theta_E(\family{e}{E}{\phi_e})$.   Then
$\pi a=\pi'a$ for every $a\Bsubseteq(1\Bsetminus c)\otimes 1$,
so $\pi^{-1}\pi'$ is
supported by $c\otimes 1$ and belongs to $V_1$.
Thus $\pi'\in\pi V_1$;  as $\pi'$
is arbitrary, $G_{E'}\subseteq G_EV_1$.\ \QeD\  It follows that
$I\subseteq G_EV_1^2\subseteq G_EV^{-1}$.

\medskip

\quad{\bf (iii)} Set $H=\Aut_{\bar\nu}\frak B$, and let $\lambda_0$ be the
Haar probability measure on $H$, that is, the uniform probability measure.
Let $\lambda$ be the product measure on $H^E$, so that $\lambda$ is the
Haar probability measure on $H^E$.
Let $\nu$ be the image measure $\lambda\theta_E^{-1}$ on
$G$.   If $\pi\in I$, then $G_E$ meets $\pi V$, so there is a
$\pmb{\phi}\in H^E$ such that $\theta_E(\pmb{\phi})\in\pi V$;  now

\Centerline{$\nu(\theta_E(\pmb{\phi})F)
=\lambda\theta_E^{-1}[\theta_E(\pmb{\phi})F]
=\lambda(\pmb{\phi}\theta_E^{-1}[F])
=\lambda\theta_E^{-1}[F]
=\nu F$}

\noindent for every $F\subseteq G$, and in particular for every
$F\in\Cal A$.   Thus $\nu$ satisfies condition (ii) of 493C.

\medskip

\quad{\bf (iv)} Set

\Centerline{$U
=\{\family{e}{E}{\phi_e}:\phi_e\in\Aut_{\bar\nu}\frak B$ for every $e\in
E$, $\#(\{e:\phi_e$ is not the identity$\})\le r\}$.}

\noindent Then $\theta_E[U]\subseteq V$.   \Prf\ If
$\pmb{\phi}=\family{e}{E}{\phi_e}$ belongs to $U$, then
$b=\sup\{e:e\in E$, $\phi_e$ is not the identity$\}$ has measure at most
$\bover{r}{m}\bar\lambda 1\le\delta$, while $b$ supports
$\theta_E(\pmb{\phi})$.   So
$\bar\lambda(a\Bsymmdiff\theta_E(\pmb{\phi})(a))\le\delta$ for every
$a\in\frak A$, and $\theta_E(\pmb{\phi})\in V$.\ \Qed

Let $\rho$ be the normalized Hamming metric on $H^E$ (492D).
If $\pmb{\phi}=\family{e}{E}{\phi_e}$ and
$\pmb{\psi}=\family{e}{E}{\psi_e}$ belong to $H^E$ and
$\rho(\pmb{\phi},\pmb{\psi})\le\bover{r}{m}$, then
$\{e:\phi_e\psi^{-1}_e$ is not the identity$\}$ has at most $r$ members,
and $\pmb{\phi}\pmb{\psi}^{-1}\in U$, that is, $\pmb{\phi}\in U\pmb{\psi}$.
So if $W\subseteq H^E$ is such that $\lambda W\ge\bover12$,

$$\eqalignno{\lambda(UW)
&\ge\lambda\{\pmb{\phi}:\rho(\pmb{\phi},W)\le\Bover{r}m\}
\ge 1-2\exp(-m(\Bover{r}m)^2)\cr
\displaycause{492Ea}
&\ge 1-\epsilon\cr}$$

\noindent by the choice of $m$ and $r$.   Transferring this to $G$,
remembering that $\theta_E:H^E\to G$ is an injective homomorphism, we get

\Centerline{$\nu(VF)
=\lambda\theta_E^{-1}[VF]
\ge\lambda(U\theta_E^{-1}[F])
\ge 1-\epsilon$}

\noindent whenever $F\subseteq G$ and $\nu F\ge\bover12$.   So $\nu$
satisfies the first condition of 493C.

\medskip

\quad{\bf (v)} As $\epsilon$, $V$, $I$ and $\Cal A$ were arbitrary, 493C
tells us that $G$ is extremely amenable.   This completes the proof.
}%end of proof of 494J

\leader{494K}{Lemma} Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra, and give $\AmuA$ its weak topology.   Let $G$  be a subgroup
of $\AmuA$ with fixed-point subalgebra $\frak C$, and suppose that
$G=\{\pi:\pi\in\AmuA$, $\pi c=c$ for every $c\in\frak C\}$.
Then $G$ is amenable, and if every atom of $\frak A$ belongs to $\frak C$,
then $G$ is extremely amenable.

\proof{{\bf (a)} We need the structure theorems of \S333;  the final one
333R is the best adapted to our purposes here.   I repeat some of the
special notation used in that theorem.
For $n\in\Bbb N$, set $\frak B_n=\Cal P(n+1)$ and let $\bar\nu_n$ be the
uniform probability measure on $n+1$, so that $\frak B_n$ has $n+1$ atoms
of the same measure;  for an infinite cardinal $\kappa$, let
$(\frak B_{\kappa},\bar\nu_{\kappa})$ be the measure algebra of the usual
measure on $\{0,1\}^{\kappa}$.   Then
333R tells us that there are a
partition of unity $\langle c_i\rangle_{i\in I}$ in $\frak C$, where $I$
is a countable set of cardinals, and a measure-preserving isomorphism
$\theta:\frak A\to\frak A'=\prod_{i\in I}\frak C_{c_i}\tensorhat\frak B_i$
such that
$\theta c=\familyiI{(c\Bcap c_i)\otimes 1}$ for every $c\in\frak C$.
In particular, for any $i\in I$,

$$\eqalign{(\theta c_i)(j)
&=c_i\otimes 1\text{ if }j=i,\cr
&=0\text{ otherwise},\cr}$$

\noindent that is, $\theta[\frak A_{c_i}]$ is just
the principal ideal of $\frak A'$ corresponding to the factor
$\frak C_{c_i}\tensorhat\frak B_i$.   Thus we have an
isomorphism
$\theta_i:\frak A_{c_i}\to\frak C_{c_i}\tensorhat\frak B_i$
such that $\theta_ic=c\otimes 1$ for every $c\in\frak C_{c_i}$.

\medskip

{\bf (b)} For each $i\in I$, set
$H_i=\{\pi\restrp\frak A_{c_i}:\pi\in G\}$.   Because $\pi c_i=c_i$ for
every $\pi\in G$, $H_i$ is a subgroup of $\Aut\frak A_{c_i}$, and
$\pi\mapsto\pi\restrp\frak A_{c_i}$ is a group homomorphism from $G$ to
$H_i$.   Set $\Theta(\pi)=\familyiI{\pi\restrp\frak A_{c_i}}$ for
$\pi\in G$.   Then $\Theta:G\to\prod_{i\in I}H_i$ is a group homomorphism.
Because $\familyiI{c_i}$ is a partition of unity in $\frak A$,
$\Theta$ is injective.   In the other
direction, suppose that $\pmb{\phi}=\familyiI{\phi_i}$ is such that every
$\phi_i$ is a measure-preserving automorphism of $\frak A_{c_i}$ and
$\phi_ic=c$ for every $c\in\frak C_{c_i}$.   Then we have a $\pi\in\AmuA$
such that $\pi a=\phi_ia$ whenever $i\in I$ and $a\Bsubseteq c_i$;  it is
easy to check that $\pi\in G$ and now $\Theta(\pi)=\pmb{\phi}$.
Thus

\Centerline{$H_i=\{\phi:\phi\in\Aut\frak A_{c_i}$ is measure-preserving,
$\phi c=c$ for every $c\in\frak C_{c_i}\}$}

\noindent for each $i$, and $\Theta$ is
a group isomorphism between $G$ and $\prod_{i\in I}H_i$.

\medskip

{\bf (c)} As in part (b) of the proof of 494J, the next step is to confirm
that $\Theta$ is a homeomorphism for
the weak topologies.   The argument is very similar.

\medskip

\quad{\bf (i)} If $U$ is a neighbourhood of the identity in $G$,
then there are a finite set $K\subseteq A$ and an $\epsilon>0$ such that
$U$ includes $\{\pi:\pi\in G$, $\bar\mu(a\Bsymmdiff\pi a)\le2\epsilon$
for every $a\in K\}$.   Let $J\subseteq I$ be a finite set such that
$\sum_{i\in I\setminus J}\bar\mu c_i\le\epsilon$, and set

$$\eqalign{V
&=\{\familyiI{\phi_i}:\phi_i\in H_i\text{ for every }i\in I,\,\cr
&\mskip50mu
\bar\mu((a\Bcap c_i)\Bsymmdiff\phi_i(a\Bcap c_i))
\le\Bover{\epsilon}{1+\#(J)}\text{ for every }i\in J\text{ and }a\in K\}.
\cr}$$

\noindent Then $V$ is a neighbourhood of the identity in
$\prod_{i\in I}H_i$.
If $\pmb{\phi}=\familyiI{\phi_i}$ belongs to $V$, and
$\pi=\Theta^{-1}(\pmb{\phi})$, then, for
$a\in K$,

$$\eqalign{\bar\mu(a\Bsymmdiff\pi a)
&=\sum_{i\in I}\bar\mu((a\Bcap c_i)\Bsymmdiff\pi(a\Bcap c_i))
=\sum_{i\in I}\bar\mu((a\Bcap c_i)\Bsymmdiff\phi_i(a\Bcap c_i))\cr
&\le\sum_{i\in J}\bar\mu((a\Bcap c_i)\Bsymmdiff\phi_i(a\Bcap c_i))
   +\sum_{i\in I\setminus J}\bar\mu c_i
\le\Bover{\epsilon\#(J)}{\#(J)+1}+\epsilon
\le 2\epsilon,\cr}$$

\noindent and $\pi\in U$.   As $U$ is arbitrary, $\Theta^{-1}$ is
continuous.

\medskip

\quad{\bf (ii)} If $V$ is a neighbourhood of the identity in
$\prod_{i\in I}H_i$, then there are a finite $J\subseteq I$, finite sets
$K_j\subseteq\frak A_{c_j}$ for $j\in J$, and an $\epsilon>0$ such that
$\pmb{\phi}=\familyiI{\phi_i}$ belongs to $V$ if
$\pmb{\phi}\in\prod_{i\in I}H_i$ and
$\bar\mu(a\Bsymmdiff\phi_ja)\le\epsilon$
whenever $j\in J$ and $a\in K_j$.   In this case.

\Centerline{$U=\{\pi:\pi\in G$, $\bar\mu(a\Bsymmdiff\pi a)\le\epsilon$
whenever $a\in\bigcup_{j\in J}K_j\}$}

\noindent is a neighbourhood of the identity in $G$, and $\Theta(\pi)\in V$
whenever $\pi\in U$.   As $V$ is arbitrary, $\Theta$ is continuous.

\medskip

{\bf (d)} Now observe that under the isomorphism $\theta_i$ the
group $H_i$ corresponds to the group of measure-preserving automorphisms of
$\frak C_{c_i}\tensorhat\frak B_i$ fixing $c\otimes 1$ for every
$c\in\frak C_{c_i}$.   By 494J, $H_i$ is amenable.
By (b)-(c) and 449Ce, $G$ is amenable.

\medskip

{\bf (e)} Finally, suppose that every atom of $\frak A$ belongs to
$\frak C$, and look more closely at the algebras
$\frak C_{c_i}\tensorhat\frak B_i$ and the groups $H_i$.   If
$i\in I$ is an infinite cardinal, then $\frak B_i$ is homogeneous and 494J
tells us that $H_i$ is
extremely amenable.   If $0\in I$, then $\frak B_0=\{0,1\}$ and
$\frak C_{c_0}\tensorhat\frak B_0$ is isomorphic to $\frak C_{c_0}$;  in
this case, $H_0$ consists of the identity alone, and is surely extremely
amenable.   If $i\in I$ is finite and not $0$, then $\frak B_i$ is finite;
and also $\frak C_{c_i}$ is atomless.   \Prf\Quer\ If $c\in\frak C_{c_i}$
is an atom, take an atom $b$ of $\frak B_i$;  then $c\otimes b$ is an atom
of $\frak C_{c_i}\tensorhat\frak B_i$, and $\theta^{-1}(c\otimes b)$ is an
atom of $\frak A$ not belonging to $\frak C$.\ \Bang\QeD\  So in this case
again, 494J tells us that $H_i$ is extremely amenable.   Thus $G$ is
isomorphic to a product of extremely amenable groups and is extremely
amenable (493Bd).
}%end of proof of 494K

\leader{494L}{Theorem}\discrversionA{\footnote{Expands and replaces the
former 4{}93H.}}{} Let $(\frak A,\bar\mu)$ be a measure algebra and $G$
a full subgroup of $\AmuA$, with the topology induced by the weak
topology of $\AmuA$.   Then $G$ is amenable.
If every atom of $\frak A$ with finite measure
belongs to the fixed-point subalgebra of $G$, then $G$ is
extremely amenable.

%given that G has many involutions, do we also have to suppose that it's
%full?

\proof{{\bf (a)} To begin with, suppose that $(\frak A,\bar\mu)$ is totally
finite.   Let $\frak C$ be the fixed-point subalgebra of $G$, and
$G'\supseteq G$ the subgroup $\{\pi:\pi\in\AmuA$, $\pi c=c$ for every
$c\in\frak C\}$.   Then $G$ is dense in $G'$, by 494Ge.
$\frak C$ is of course the fixed-point subalgebra of $G'$, so $G'$ is
amenable (494K) and $G$ is amenable (449F(a-ii)).
If every atom of $\frak A$ belongs to $\frak C$, then $G'$ and $G$ are
extremely amenable, by 494K and 493Bf.

\medskip

{\bf (b)} Now for the general case.

\medskip

\quad{\bf (i)} For each $a\in\frak A^f$, set

\Centerline{$G_a=\{\pi:\pi\in G$, $\pi$ is supported by $a\}$,
\quad$H_a=\{\pi\restrp\frak A_a:\pi\in G_a\}$.}

\noindent Then $H_a$ is a full subgroup of
$\Aut_{\bar\mu\restrp\frak A_a}\frak A_a$ (494Ha), and is isomorphic
to $G_a$;  moreover, the isomorphism is a homeomorphism for the weak
topologies.   \Prf\ Set $\theta(\pi)=\pi\restrp\frak A_a$ for
$\pi\in G_a$.   ($\alpha$) If $V$ is a neighbourhood of the identity in
$H_a$, let $\delta>0$ and $K\in[\frak A_a]^{<\omega}$ be such that

\Centerline{$V
\supseteq\{\phi:\phi\in H_a$, $\bar\mu(b\Bsymmdiff\phi b)\le\delta$
for every $b\in K\}$;}

\noindent then

\Centerline{$U
=\{\pi:\pi\in G_a$, $\bar\mu(b\Bsymmdiff\pi b)\le\delta$
for every $b\in K\}$}

\noindent is a neighbourhood of the identity in $G_a$, and
$\theta(\pi)\in V$
for every $\pi\in U$.   So $\theta$ is continuous.   ($\beta$) If
$U$ is a neighbourhood of the identity in $G_a$, let
$\delta>0$ and $K\in[\frak A]^{<\omega}$ be such that

\Centerline{$U
\supseteq\{\pi:\pi\in G_a$, $\bar\mu(b\Bsymmdiff\pi b)\le\delta$
for every $b\in K\}$;}

\noindent then

\Centerline{$V
=\{\phi:\phi\in G_a$,
  $\bar\mu((b\Bcap a)\Bsymmdiff\phi(b\Bcap a))\le\delta$
  for every $b\in K\}$}

\noindent is a neighbourhood of the identity in $G_a$, and
$\theta^{-1}(\phi)\in U$ for every $\phi\in V$, because

\Centerline{$\theta^{-1}(\phi)(b)=\phi(b\Bcap a)\Bcup(b\Bsetminus a)$,
\quad$b\Bsymmdiff\theta^{-1}(\phi)(b)=(b\cap a)\Bsymmdiff\phi(b\Bcap a)$}

\noindent for every $\phi\in G_a$ and $b\in\frak A$.    Thus $\theta^{-1}$
is continuous.\ \Qed

By (a), $H_a$, and therefore $G_a$, is amenable.

\medskip

\quad{\bf (ii)} $H=\bigcup_{a\in\frak A^f}G_a$ is dense in $G$.   \Prf\
If $\pi\in G$, $a_0,\ldots,a_n\in\frak A^f$ and $\epsilon>0$, set
$a=\sup_{i\le n}a_i$ and $b=a\Bcup\pi a$.   Then there is a
$\phi\in G$ such that $\phi$ agrees with $\pi$ on $\frak A_a$ and
$\phi$ is supported by $b$ (494Ga).   In this case, $\phi\in G_b$ and
$\bar\mu(\pi a_i\Bsymmdiff\phi a_i)=0\le\epsilon$ for every $i\le n$.\ \Qed

Since $\family{a}{\frak A^f}{G_a}$ is upwards-directed, and every $G_a$ is
amenable, $G$ is amenable (449Cb).

\medskip

\quad{\bf (iii)} If every atom of $\frak A$ of finite measure is fixed
under the action of $G$, then every atom of $\frak A_a$ is fixed under the
action of $H_a$, for every $a\in\frak A^f$.   So every $H_a$ and
every $G_a$ is extremely amenable, and $G$ is extremely amenable, by
493Bb.
}%end of proof of 494L

\leader{494M}{Lemma} Let $\frak A$ be a Boolean algebra, $G$ a full
subgroup of $\Aut\frak A$, and $V\subseteq G$ a symmetric set.
Let $\sim_G$ be the orbit equivalence relation on $\frak A$
induced by the action of $G$, so that $a\sim_Gb$ iff there is a $\phi\in G$
such that $\phi a=b$.   Suppose that $a\in\frak A$ and
$\pi$, $\pi'\in G$ are such that

\inset{$\pi=\cycle{b\,_{\pi}\,c}$ and $\pi'=\cycle{b'\,_{\pi'}\,c'}$ are
exchanging involutions supported by $a$,

$b\sim_Gb'$ and $a\Bsetminus(b\Bcup c)\sim_Ga\Bsetminus(b'\Bcup c')$,

$\pi\in V$,

whenever $\phi\in G$ is supported by $a$ there is a $\psi\in V$ agreeing
with $\phi$ on $\frak A_a$.}

\noindent Then $\pi'\in V^3$.

\proof{{\bf (a)} There is a $\phi\in G$, supported by $a$, such that
$\phi\pi'=\pi\phi$.   \Prf\ We know that there are $\phi_0$, $\phi_1\in G$
such that $\phi_0(a\Bsetminus(b'\Bcup c'))=a\Bsetminus(b\Bcup c)$ and
$\phi_1b'=b$.   Set $\phi_2=\pi\phi_1\pi'$;  then $\phi_2\in G$,
$\pi\phi_2=\phi_1\pi'$, $\phi_2\pi'=\pi\phi_1$ and $\phi_2c'=c$.   Because
$(a\Bsetminus(b'\Bcup c'),b',c',1\Bsetminus a)$ and
$(a\Bsetminus(b\Bcup c),b,c,1\Bsetminus a)$ are partitions of unity in
$\frak A$, there is a $\phi\in\Aut\frak A$ such that

$$\eqalign{\phi d
&=\phi_0d\text{ if }d\Bsubseteq a\Bsetminus(b'\Bcup c'),\cr
&=\phi_1d\text{ if }d\Bsubseteq b',\cr
&=\phi_2d\text{ if }d\Bsubseteq c',\cr
&=d\text{ if }d\Bsubseteq 1\Bsetminus a\cr}$$

\noindent (381C once more);  because $G$ is full, $\phi\in G$.   Of course
$\phi$ is supported by $a$.   Now

$$\eqalignno{\phi\pi'd
&=\phi d=\pi\phi d
\text{ if }d\Bsubseteq a\Bsetminus(b'\Bcup c')\cr
\noalign{\noindent(because $\phi d=\phi_0d$ is disjoint from $b\Bcup c$),}
&=\phi_2\pi'd=\pi\phi_1d=\pi\phi d
\text{ if }d\Bsubseteq b',\cr
&=\phi_1\pi'd=\pi\phi_2d=\pi\phi d
\text{ if }d\Bsubseteq c',\cr
&=\phi d=d=\pi\phi d
\text{ if }d\Bsubseteq 1\Bsetminus a.\cr}$$

\noindent So $\phi\pi'=\pi\phi$.\ \Qed

\medskip

{\bf (b)} By our hypothesis, there is a $\psi\in V$ agreeing with $\phi$ on
$\frak A_a$.   In this case,

$$\eqalign{\psi\pi'd
&=\phi\pi'd=\pi\phi d=\pi\psi d\text{ if }d\Bsubseteq a,\cr
&=\psi d=\pi\psi d\text{ if }d\Bsubseteq 1\Bsetminus a.\cr}$$

\noindent So $\psi\pi'=\pi\psi$ and

\Centerline{$\pi'=\psi^{-1}\pi\psi\in V^3$}

\noindent because $V$ is symmetric and $\pi$, $\psi\in V$.
}%end of proof of 494M

\leader{494N}{Lemma}
Let $(\frak A,\bar\mu)$ be a probability algebra and
$G\subseteq\AmuA$ a full subgroup with fixed-point subalgebra $\frak C$.
Suppose that $\frak A$ is relatively atomless over $\frak C$.
For $a\in\frak A$, let $u_a\in L^{\infty}(\frak C)$ be the conditional
expectation of $\chi a$ on $\frak C$, and
let $G_a$ be $\{\pi:\pi\in G$ is supported by $a\}$.   Suppose that
$a\Bsubseteq e$ in $\frak A$ and $V\subseteq G$ are such that

\inset{$V$ is symmetric\cmmnt{, that is, $V=V^{-1}$},

for every $\phi\in G_e$ there is a $\psi\in V$ such that
$\phi$ and $\psi$ agree on $\frak A_e$,

there is an involution in $V$ with support $a$,

$u_a\le\bover23 u_e$.}

\noindent Then $G_a
\subseteq V^{18}\cmmnt{\mskip5mu=\{\pi_1\ldots\pi_{18}:
   \pi_1,\ldots,\pi_{18}\in V\}}$.

\proof{{\bf (a)(i)} Note that 494Gb, in the language of conditional
expectations, tells us that if $b$, $c\in\frak A$ then
$b\sim_Gc$ in the notation of 494M iff $u_b=u_c$.
Similarly, 494Gd tells us that if
$\familyiI{b_i}$ and $\familyiI{c_i}$ are disjoint families in $\frak A$
and $u_{b_i}=u_{c_i}$ for every $i\in I$, there is a $\pi\in G$ such that
$\pi b_i=c_i$ for every $i$.

\medskip

\quad{\bf (ii)} It follows that every non-zero $b\in\frak A$ is the
support of an involution in $G$.   \Prf\ Because $\frak A$ is relatively
atomless over $\frak C$, there is a $c\Bsubseteq b$ such
that $u_c=\bover12u_b$ (494Ad);  now there is a $\phi\in G$ such that
$\phi c=b\Bsetminus c$, and
$\pi=\cycle{c\,_{\phi}\,b\Bsetminus c}$ is an
involution, belonging to $G$ (381Sd once more), with support $b$.\ \Qed

\medskip

{\bf (b)} If $\pi'\in G$ is an involution with
support $a'\Bsubseteq e$ and $u_{a'}=u_a$, then
$\pi'\in V^3$.   \Prf\ Let $\pi_0\in V$ be an involution with support
$a$.   Because $\frak A$ is Dedekind complete, $\pi_0$ is an exchanging
involution (382Fa);  express it as $\cycle{b_0\,_{\pi_0}\,c_0}$ and
$\pi'$ as $\cycle{b'\,_{\pi'}\,c'}$.   Because $\pi b_0=c_0$,
$u_{b_0}=u_{c_0}$, while $u_{b_0}+u_{c_0}=u_{b_0\Bcup c_0}=u_a$;
so $u_{b_0}=\bover12u_a$.
Similarly, $u_{b'}=\bover12u_{a'}=u_{b_0}$.   On the other hand,

\Centerline{$u_{e\Bsetminus a}=u_e-u_a=u_e-u_{a'}=u_{e\Bsetminus a'}$}

\noindent and $e\Bsetminus a\sim_Ge\Bsetminus a'$.   So the conditions of
494M are satisfied and $\pi'\in V^3$.\ \Qed

\medskip

{\bf (c)} Now suppose that $\pi$ is any involution in $G$
with support included in $a$.   Then $\pi\in V^6$.
\Prf\ Let $b$, $c$ be such that
$\pi=\cycle{b\,_{\pi}\,c}$.   Once again,

\Centerline{$u_b=u_c\le\Bover12u_a$,
\quad$u_{e\Bsetminus a}=u_e-u_a\ge\Bover12u_a$,}

\noindent so we can find $d\Bsubseteq e\Bsetminus a$ such that
$u_d=u_c$, while there is also a $b_1\Bsubseteq b$ such that
$u_{b_1}=\bover12u_b$.   Set

\Centerline{$c_1=\pi b_1$,
\quad$b_2=b\Bsetminus b_1$,
\quad$c_2=\pi b_2$,
\quad$\pi_1=\cycle{b_1\,_{\pi}\,c_1}$,
\quad$\pi_2=\cycle{b_2\,_{\pi}\,c_2}$;}

\noindent then $\pi_1$ and $\pi_2$ are involutions, with supports
$b_1\Bcup c_1$ and $b_2\Bcup c_2$ respectively, belonging to $G$
and such that $\pi=\pi_1\pi_2$.
Next, (a-iii) tells us that there are involutions $\pi_3$, $\pi_4\in G$
with supports $d$ and $a\Bsetminus(b\Bcup c)$ (if $a=b\Bcup c$, set
$\pi_4=\iota$).
Since $\pi_1$, $\pi_2$, $\pi_3$ and $\pi_4$ have disjoint supports,
they commute (381Ef).   Consequently
$\pi_1\pi_3\pi_4$, $\pi_2\pi_3\pi_4$ are involutions, belonging to $G$,
with supports
$a_1=b_1\Bcup c_1\Bcup(a\Bsetminus(b\Bcup c))\Bcup d$,
$a_2=b_2\Bcup c_2\Bcup(a\Bsetminus(b\Bcup c))\Bcup d$ respectively.
But now observe that

\Centerline{$u_{a_1}
=u_{b_1}+u_{c_1}+u_a-u_b-u_c+u_d
=u_b+u_a-u_b=u_a$,}

\noindent and similarly $u_{a_2}=u_a$.   By (b), both
$\pi_1\pi_3\pi_4$ and $\pi_2\pi_3\pi_4$ belong to $V^3$.
But this means that

\Centerline{$\pi=\pi_1\pi_2
=\pi_1\pi_2\pi_3^2\pi_4^2=\pi_1\pi_3\pi_4\pi_2\pi_3\pi_4$}

\noindent belongs to $V^6$, as claimed.\ \Qed

\medskip

{\bf (d)} By 382N, every member of $G_a$ is
expressible as the product of at most three involutions belonging to
$G_a$, so belongs to $V^{18}$.
}%end of proof of 494N

\leader{494O}{Theorem}\cmmnt{ ({\smc Kittrell \& Tsankov 09})}
Suppose that $(\frak A,\bar\mu)$ is an atomless probability algebra and
$G\subseteq\AmuA$ is a full ergodic subgroup\cmmnt{ (definition:
395Ge)}, with the topology induced by the uniform topology of $\AmuA$.

(a) If $V\subseteq G$ is symmetric
and $G$ can be covered by countably many left translates of $V$ in $G$,
then $V^{38}=\{\pi_1\pi_2\ldots\pi_{38}:\pi_1,\ldots,\pi_{38}\in V\}$
is a neighbourhood of the identity in $G$.

(b) If $H$ is a topological group such that for every neighbourhood
$W$ of the identity in $H$ there is a countable set $D\subseteq H$ such
that $H=DW$,
and $\theta:G\to H$ is a group homomorphism, then $\theta$ is continuous.

\proof{{\bf (a)(i)} Let $\sequencen{\psi_n}$ be a sequence in $G$ such that
$G=\bigcup_{n\in\Bbb N}\psi_nV$.   It may help if I note straight away that
$\iota\in V^2$.   \Prf\ There is an $n\in\Bbb N$ such that
$\iota\in\psi_nV$, that is, $\psi_n^{-1}\in V$;  as $V$ is symmetric,
$\psi_n\in V$ and $\iota=\psi_n\psi_n^{-1}$ belongs to $V^2$.\ \Qed

\medskip

\quad{\bf (ii)} As before, set
$G_a=\{\pi:\pi\in G$, $\pi$ is supported by  $a\}$ for
$a\in\frak A$.   Now there is a non-zero $e\in\frak A$ such that for every
$\pi\in G_e$ there is a $\phi\in V^2$ agreeing with $\pi$ on
$\frak A_e$.   \Prf\ Because $\frak A$ is atomless, there is a disjoint
sequence $\sequencen{b_n}$ in $\frak A\setminus\{0\}$.   \Quer\ Suppose, if
possible, that for every $n\in\Bbb N$ there is a
$\pi_n\in G_{b_n}$ such that there is no
$\phi\in V^2$ agreeing with $\pi_n$ on $\frak A_{b_n}$.
If $n\in\Bbb N$, then
$V^2=(\psi_nV)^{-1}\psi_nV$ and $\pi_n=\iota^{-1}\pi_n$,
so there must be a $\pi'_n\in G_{b_n}$, either $\iota$ or $\pi_n$,
not agreeing with
$\phi$ on $\frak A_{b_n}$ for any $\phi\in\psi_nV$.   Define
$\psi\in\AmuA$ by the formula

$$\eqalign{\psi d
&=\pi'_nd\text{ if }n\in\Bbb N\text{ and }d\Bsubseteq b_n,\cr
&=d\text{ if }d\Bcap\sup_{n\in\Bbb N}b_n=0.\cr}$$

\noindent Because $G$ is full, $\psi\in G$ and
there is an $m\in\Bbb N$ such that
$\psi\in\psi_mV$.   But now $\pi'_m$ agrees with $\psi$ on
$\frak A_{b_m}$, contrary to the choice of $\pi'_m$.\ \BanG\  So one of the
$b_n$ will serve for $e$.\ \Qed

\medskip

\quad{\bf (iii)} There is an involution $\pi\in V^2$, supported by $e$,
such that $\bar\mu(\supp\pi)\le\bover23\bar\mu e$.   \Prf\ Take disjoint
$b$, $b'\Bsubseteq e$ such that $\bar\mu b=\bar\mu b'=\bover12\bar\mu e$.
Because $G$ is full and ergodic, there is a $\phi\in G$ such that
$\phi b=b'$.   (By 395Gf, the fixed-point subalgebra of $G$
is $\{0,1\}$, so we can apply 494Gc.)
For every $d\in\frak A_b$, set $\phi_d=\cycle{d\,_{\phi}\,\phi d}$.
Because $G$ is full, $\phi_d\in G$.   Observe that

\Centerline{$\phi_c\phi_d
=\phi_{c\Bsetminus d}\phi_{c\Bcap d}\phi_{c\Bcap d}\phi_{d\Bsetminus c}
=\phi_{c\Bsetminus d}\phi_{d\Bsetminus c}
=\phi_{c\Bsymmdiff d}$}

\noindent for all $c$, $d\Bsubseteq b$.   Set
$A_n=\{d:d\in\frak A_b$, $\phi_d\in\psi_nV\}$ for each $n\in\Bbb N$.
Since $\frak A_b$ is complete under its measure metric, there is an
$n\in\Bbb N$ such that $A_n$ is non-meager;  because $\frak A$ is atomless,
$\frak A_b$ has no isolated points;  so there are $d_0$, $d_1\in A_n$
such that $0<\bar\mu(d_0\Bsymmdiff d_1)\le\bover13\bar\mu e$.   Set
$d=d_0\Bsymmdiff d_1$.   Then

\Centerline{$\phi_d
=\phi_{d_0}\phi_{d_1}=\phi_{d_0}^{-1}\phi_{d_1}
\in V^{-1}\psi_n^{-1}\psi_nV=V^2$,}

\noindent and we can take $\phi_d$ for $\pi$.\ \Qed

\medskip

\quad{\bf (iv)} Taking $a=\supp\pi$ in (iv), $a$ and $e$ satisfy the
conditions of 494N with respect to $V^2$ and $\frak C=\{0,1\}$, so
$G_a\Bsubseteq(V^2)^{18}=V^{36}$.

\medskip

\quad{\bf (v)}  Finally, there is a $\delta>0$ such that, in the language
of 494C,
$G\cap U(1,\delta)\Bsubseteq V^{38}$.   \Prf\Quer\ Otherwise, we can find
for each $n\in\Bbb N$ a
$\pi_n\in G\cap U(1,2^{-n-1}\bar\mu a)\setminus V^{38}$.   Set
$\pi'_n=\psi_n\pi_n\psi_n^{-1}$, $b_n=\supp\pi'_n$;  then
$\bar\mu b_n=\bar\mu(\supp\pi_n)\le 2^{-n-1}\bar\mu a$ for each $n$
(381Gd).   So
$b=\sup_{n\in\Bbb N}b_n$ has measure at most $\bar\mu a$, and there is a
$\phi\in G$ such that $\phi b\Bsubseteq a$.
In this case,
there is an $n\in\Bbb N$ such that $\phi^{-1}\in\psi_nV$, that is,
$\phi\psi_n\in V^{-1}=V$.   Now
$\pi=\phi\psi_n\pi_n\psi_n^{-1}\phi^{-1}$ has support
$\phi b_n\Bsubseteq a$, so belongs to $V^{36}$.   But this means that
$\pi_n=\psi_n^{-1}\phi^{-1}\pi\phi\psi_n$ belongs to $V^{38}$, contrary to
the choice of $\pi_n$.\ \Bang\Qed

So $V^{38}$ is a neighbourhood of $\iota$ in $G$, as claimed.

\medskip

{\bf (b)} Let $W$ be a neighbourhood of the identity in $H$.   Then there
is a symmetric neighbourhood $W_1$ of the identity in $H$ such that
$W_1^{38}\subseteq W$.   Set $V=\theta^{-1}[W_1]$.   Let $W_2$ be a
neighbourhood of the identity in $H$ such that
$W_2^{-1}W_2\subseteq W_1$, and
$\sequencen{y_n}$ a sequence in $H$ such that
$H=\bigcup_{n\in\Bbb N}y_nW_2$.
For each $n\in\Bbb N$, choose $\psi_n\in G$ such that
$\theta(\psi_n)\in y_nW_2$ whenever $\theta[G]$ meets $y_nW_2$.
If $\pi\in G$, there is an $n\in\Bbb N$ such that
$\theta(\pi)\in y_nW_2$;   in this case, $\theta(\psi_n)\in y_nW_2$, so

\Centerline{$\theta(\psi_n^{-1}\pi)\in W_2^{-1}y_n^{-1}y_nW_2
\subseteq W_1$}

\noindent and
$\psi_n^{-1}\pi\in V$.   Thus $\pi\in\psi_nV$;  as $\pi$ is arbitrary,
$G=\bigcup_{n\in\Bbb N}\psi_nV$.   By (a), $V^{38}$ is a neighbourhood
of $\iota$;  but
$V^{38}\subseteq\theta^{-1}[W_1^{38}]\subseteq\theta^{-1}[W]$, so
$\theta^{-1}[W]$ is a neighbourhood of $\iota$.   As $W$ is
arbitrary, $\theta$ is continuous (4A5Fa).
}%end of proof of 494O

\leader{494P}{Remark} Note that if a topological group $H$ is either
Lindel\"of or ccc, it satisfies the condition of (b) above.
\prooflet{\Prf\
Let $W$ be an open neighbourhood of the identity in $H$.
($\alpha$) If $H$ is Lindel\"of,
the result follows immediately from the fact that $\{yW:y\in H\}$
is an open cover of $H$.   ($\beta$) If $H$ is ccc, let $W_1$ be an open
neighbourhood of the identity such that $W_1W_1^{-1}\subseteq W$, and
$D\subseteq H$ a maximal set such that $\family{y}{D}{yW_1}$ is disjoint.
Then $D$ is countable.
If $x\in H$, there is a $y\in D$ such that $xW_1\cap yW_1\ne\emptyset$,
that is, $x\in yW_1W_1^{-1}\subseteq yW$;  thus $H=DW$.\ \Qed}
\cmmnt{  See also 494Yh.}

%for Lindel\"of but not ccc, or the other way round, see mt49bits

\leader{494Q}{}\cmmnt{ Some of the same ideas lead to an interesting
group-theoretic property of the automorphism groups here.

\medskip

\noindent}{\bf Theorem}\cmmnt{ (see {\smc Droste Holland \& Ulbrich 08})}
Let $(\frak A,\bar\mu)$ be a probability algebra and $G$ a full
subgroup of $\AmuA$ such that
$\frak A$ is relatively atomless over the fixed-point subalgebra $\frak C$
of $G$.   Let $\sequencen{V_n}$ be a non-decreasing
sequence of subsets of $G$ such that $V_n^2\subseteq V_{n+1}$ for
every $n$ and $G=\bigcup_{n\in\Bbb N}V_n$.   Then there is an $n\in\Bbb N$
such that $G=V_n$.

\proof{{\bf (a)} For the time being (down to the end of (e) below), suppose
that every $V_n$ is symmetric.
As in 494N, for each $a\in\frak A$ write $u_a$ for the
conditional expectation of $\chi a$ on $\frak C$, and set
$G_a=\{\pi:\pi\in G$, $\pi$ is supported by $a\}$.

\medskip

{\bf (b)} There are an $\alpha_1>0$, an $a_1\in\frak A$ and an
$n_0\in\Bbb N$ such that $u_{a_1}=\alpha_1\chi 1$ and
for every $\pi\in G_{a_1}$ there is a $\phi\in V_{n_0}$ agreeing with
$\pi$ on $\frak A_{a_1}$.   \Prf\Quer\ Otherwise, because $\frak A$ is
relatively atomless, we can choose inductively a disjoint sequence
$\sequencen{b_n}$ in $\frak A$ such that $u_{b_n}=2^{-n-1}\chi 1$ for each
$n$ (use 494Ad).   For each $n\in\Bbb N$ there must be a $\pi_n\in G_{b_n}$
such that there is no $\phi\in V_n$ agreeing with $\pi_n$ on
$\frak A_{b_n}$.   Because $G$ is full,
there is a $\phi\in G$ agreeing with $\pi_n$ on $\frak A_{b_n}$ for every
$n$.   But now $\phi\notin\bigcup_{n\in\Bbb N}V_n=G$.\ \Bang\Qed

\medskip

{\bf (c)} There is an
$a_0\Bsubseteq a_1$ such that $u_{a_0}=\bover23\alpha_1\chi 1$
and there is an involution $\pi\in G$ with support $a_0$.
\Prf\ Take disjoint $a$, $a'\Bsubseteq a_1$ such that
$u_a=u_{a'}=\bover13\alpha_1\chi 1$ (494Ad again).
Set $a_0=a\Bcup a'$, so that
$u_{a_0}=\bover23\alpha_1\chi 1$.   There is a $\phi\in G$ such
that $\phi a=a'$, and $\pi=\cycle{a\,_{\phi}\,a'}$ is an involution in $G$
with support $a_0$.\ \QeD\
Let $n_1\ge n_0$ be such that $\pi\in V_{n_1}$.

\medskip

{\bf (d)} By 494N, $G_{a_0}\subseteq V_{n_1}^{18}\subseteq V_{n_1+5}$.
Taking $k\ge\Bover{3}{\alpha_1}$, 494Ad once more gives us a disjoint
family $\ofamily{i}{k}{d_i}$ in $\frak A$ such that
$u_{d_i}=\bover1k\chi 1$ for every $i<k$;  since
$\sum_{i=0}^{k-1}\bar\mu d_i=1$, $\ofamily{i}{k}{d_i}$ is a partition of
unity, while $u_{d_i}\le\bover13\alpha_1\chi 1$ for every $i$.
For $i$, $j<k$, let
$\phi_{ij}\in G$ be such that $\phi_{ij}(d_i\Bcup d_j)\Bsubseteq a_0$
(494Gc).
Let $n_2\ge n_1+5$ be such that $\phi_{ij}\in V_{n_2}$ for all $i$, $j<k$.
Then any involution in $G$ belongs to $V_{n_2}^{3k^2}$.   \Prf\ Let
$\pi\in G$ be an involution;  by 382Fa again, we can
express it as $\cycle{e\,_{\pi}\,e'}$.
For $i$, $j<k$, set $e_{ij}=e\Bcap d_i\Bcap\pi d_j$,
$e'_{ij}=\pi e_{ij}=e'\Bcap\pi d_i\Bcap d_j$;  set
$\pi_{ij}=\cycle{e_{ij}\,_{\pi}\,e'_{ij}}$.
In this case, because all the $e_{ij}$ and $e'_{ij}$ are disjoint,
$\langle\pi_{ij}\rangle_{i,j<k}$ is a commuting family, and we can talk of
$\prod_{i,j<k}\pi_{ij}$, which of course is equal to $\pi$.   Now, for each
$i$, $j<k$,

\Centerline{$\phi_{ij}\pi_{ij}\phi_{ij}^{-1}
=\cycle{\phi_{ij}e_{ij}\,_{\phi_{ij}\pi\phi_{ij}^{-1}}\,\phi_{ij}e'_{ij}}$}

\noindent (381Sb) belongs to
$G_{a_0}\subseteq V_{n_1+5}\subseteq V_{n_2}$.   So

\Centerline{$\pi_{ij}
=\phi_{ij}^{-1}\phi_{ij}\pi_{ij}\phi_{ij}^{-1}\phi_{ij}$}

\noindent belongs to $V_{n_2}^3$ and $\pi=\prod_{i,j<k}\pi_{ij}$ belongs to
$V_{n_2}^{3k^2}$. \Qed

\medskip

{\bf (e)} Since, by 382N again, every member of $G$ is expressible as
a product of at most three involutions belonging to $G$,
$G\subseteq V_{n_2}^{9k^2}\subseteq V_n$, where
$n=n_2+\lceil\log_2(9k^2)\rceil$.

\medskip

{\bf (f)} This completes the proof on the assumption that every $V_n$ is
symmetric.   For the general case, set $W_n=V_n\cap V_n^{-1}$ for every
$n$.   Then $\sequencen{W_n}$ is a non-decreasing sequence of symmetric
sets with union $G$, and

\Centerline{$W_n^2\subseteq V_n^2\cap V_n^{-2}=V_n^2\cap(V_n^2)^{-1}
\subseteq V_{n+1}\cap V_{n+1}^{-1}=W_{n+1}$}

\noindent for every $n$, so there is an $n\in\Bbb N$ such that
$G=W_n=V_n$.
}%end of proof of 494Q

\leader{494R}{}\cmmnt{ There are many alternative versions of 494Q;
see, for instance, 494Xm.   Rather than attempt a portmanteau
result to cover them all, I give one which can be applied to the measure
algebra of Lebesgue measure on $\Bbb R$ and
indicates some of the new techniques required.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be an atomless
localizable measure algebra, and $G$ a full ergodic
subgroup of $\AmuA$.   Let $\sequencen{V_n}$ be a
non-decreasing sequence of subsets of $G$, covering $G$, with
$V_n^2\subseteq V_{n+1}$ for every $n$.
Then there is an $n\in\Bbb N$ such that $G=V_n$.

\proof{{\bf (a)} My aim is to mimic the proof of 494Q.   We have a
simplification because $G$ is ergodic, but 494M will be applied in a
different way.   As before, it will be enough to consider the case in which
every $V_n$ is symmetric;  as before, I will write
$G_a$ for $\{\pi:\pi\in G$, $\pi$ is supported by $a\}$.

Because $G$ is ergodic, $\frak A$ must be quasi-homogeneous (374G);  as it
is also atomless, there is an infinite cardinal $\kappa$ such
that $\frak A_a$ is homogeneous, with Maharam type $\kappa$, for every
$a\in\frak A^f\setminus\{0\}$ (374H).
If $(\frak A,\bar\mu)$ is totally finite,
then the result is immediate from 494Q, normalizing
the measure if necessary.   So I will assume that $(\frak A,\bar\mu)$ is
not totally finite.   In this case, the orbits of $G$ can be described
in terms of `magnitude' (332Ga).   If $a\in\frak A^f$, $\mag a=\bar\mu a$;
otherwise, $\mag a$ is the cellularity of $\frak A_a$, and there will be
a disjoint family in $\frak A_a$ of this cardinality (332F).   Set
$\lambda=\mag 1\ge\omega$;  then whenever $a\in\frak A$ and
$\mag a=\lambda$, there is a partition of unity $\sequencen{a_n}$ in
$\frak A_a$ such that $\mag a_n=\lambda$ for every $n$.

\medskip

{\bf (b)} The key fact, corresponding to 494Gd, is as follows:  if
$\familyiI{a_i}$ and $\familyiI{b_i}$ are partitions of unity in
$\frak A$ such that $\mag a_i=\mag b_i$ for every $i\in I$, then there is a
$\phi\in G$ such that $\phi a_i=b_i$ for every $i\in I$.

\medskip

\Prf{\bf (i)} Consider first the case in which all the $a_i$, $b_i$ have
finite measure.   In this case, let $\family{j}{J}{(c_j,\pi_j,d_j)}$ be a
maximal family such that

\inset{----- $\family{j}{J}{c_j}$
is a disjoint family in $\frak A^f\setminus\{0\}$,

----- $\family{j}{J}{d_j}$ is a disjoint family in $\frak A$,

----- for every $j\in J$, $\pi_j\in G$, $\pi_jc_j=d_j$ and there is an
$i\in I$ such that $c_j\Bsubseteq a_i$ and $d_j\Bsubseteq b_i$.}

\noindent Set $a=1\Bsetminus\sup_{j\in J}c_j$ and
$b=1\Bsetminus\sup_{j\in J}d_j$.   \Quer\ If $a\ne 0$, there is an
$i\in I$ such that $a\Bcap a_i\ne 0$.   In this case,

$$\eqalign{\sum_{j\in J}\bar\mu(b_i\Bcap d_j)
&=\sum_{j\in J,d_j\Bsubseteq b_i}\bar\mu d_j
=\sum_{j\in J,d_j\Bsubseteq b_i}\bar\mu\pi_j^{-1}d_j\cr
&=\sum_{j\in J,c_j\Bsubseteq a_i}\bar\mu c_j
<\bar\mu a_i
=\bar\mu b_i,\cr}$$

\noindent and $b\Bcap b_i\ne 0$.   Because $G$ is ergodic, there is a
$\pi\in G$ such that $\pi(a\Bcap a_i)\Bcap(b\Bcap b_i)\ne 0$.
Setting $d=a\Bcap a_i\Bcap\pi^{-1}(b\Bcap b_i)$, we ought to have added
$(d,\pi,\pi d)$ to $\family{j}{J}{(c_j,\pi_j,d_j)}$.\ \Bang

Thus $a=0$;  similarly, $b=0$ and $\family{j}{J}{c_j}$,
$\family{j}{J}{d_j}$ are partitions of unity in $\frak A$.   Because
$\frak A$ is Dedekind complete, there is a $\phi\in\Aut\frak A$ such that
$\phi d=\pi_jd$ whenever $j\in J$ and $d\Bsubseteq c_j$, and now
$\phi\in G$ and $\phi a_i=b_i$ for every $i\in I$.

\medskip

\quad{\bf (ii)} For the general case, refine the partitions
$\familyiI{a_i}$ and $\familyiI{b_i}$ as follows.   For each $i\in I$, if
$\bar\mu a_i=\bar\mu b_i$ is finite, take $\lambda_i=1$, $c_{i0}=a_i$
and
$d_{i0}=b_i$;  otherwise, take $\lambda_i=\mag a_i=\mag b_i$, and let
$\ofamily{\xi}{\lambda_i}{c_{i\xi}}$, $\ofamily{\xi}{\lambda_i}{d_{i\xi}}$
be partitions of unity in $\frak A_{a_i}$, $\frak A_{b_i}$ respectively
with $\bar\mu c_{i\xi}=\bar\mu d_{i\xi}=1$ for every $\xi<\lambda_i$
(332I).   Now (i) tells us that there is a $\phi\in G$ such that
$\phi c_{i\xi}=d_{i\xi}$ whenever $i\in I$ and $\xi<\lambda_i$, in which
case $\phi a_i$ will be equal to $b_i$ for every $i\in I$.\ \Qed

\medskip

{\bf (c)} There are an $a_1\in\frak A$ and an $n_0\in\Bbb N$
such that $\mag a_1=\mag(1\Bsetminus a_1)=\lambda$ and whenever
$\pi\in G_{a_1}$ there is a
$\phi\in V_{n_0}$ agreeing with $\pi$ on $\frak A_{a_1}$.
\Prf\Quer\ Otherwise, let
$\sequencen{b_n}$ be a partition of unity in $\frak A$ such that
$\mag b_n=\lambda$ for every $n$.   For each $n\in\Bbb N$ there must be a
$\pi_n\in G_{b_n}$
such that there is no $\phi\in V_n$ agreeing with $\pi_n$ on
$\frak A_{b_n}$.   Now there is a
$\phi\in G$ agreeing with $\pi_n$ on $\frak A_{b_n}$ for every
$n$.   But in this case $\phi\notin\bigcup_{n\in\Bbb N}V_n=G$.\ \Bang\Qed

\medskip

{\bf (d)} There is an
$a_0\Bsubseteq a_1$ such that $\mag a_0=\mag(a_1\Bsetminus a_0)=\lambda$
and there is an involution $\pi_0\in G$ with support $a_0$.
\Prf\ Take disjoint $a$, $a'$, $a''\Bsubseteq a_1$ all of magnitude
$\lambda$;  by (b), there is a $\phi\in G$ such
that $\phi a=a'$, and $\pi_0=\cycle{a\,_{\phi}\,a'}$ is an involution
with support $a_0=a\Bcup a'$ of magnitude $\lambda$, while
$a_1\Bsetminus a_0\Bsupseteq a''$ also has magnitude $\lambda$.\ \QeD\
Let $n_1\ge n_0$ be such that $\pi_0\in V_{n_1}$.

\medskip

{\bf (e)} If $\pi\in G$ is an involution with
support $b_1\Bsubseteq a_1$ and
$\mag b_1=\mag(a_1\Bsetminus b_1)=\lambda$, then
$\pi\in V_{n_1}^3$.   \Prf\ Express $\pi_0$ and $\pi$ as
$\cycle{a\,_{\pi_0}\,a'}$ and $\cycle{b\,_{\pi}\,b'}$ respectively, and set
$\tilde a=a_1\Bsetminus(a\Bcup a')$, $\tilde b=a_1\Bsetminus b_1$;
then $a$, $a'$, $b$, $b'$, $\tilde a$ and $\tilde b$ must all have
magnitude $\lambda$.   By (b), there is a $\phi_0\in G$ such that

\Centerline{$\phi_0b=a$,
\quad$\phi_0b'=a'$,
\quad$\phi_0\tilde b=\tilde a$,
\quad$\phi_0(1\Bsetminus a_1)=1\Bsetminus a_1$.}

\noindent In particular, $a\sim_Gb$ and $\tilde a\sim_G\tilde b$, in the
language of 494M, and (c) tells us that the final hypothesis of 494M is
satisfied;  so $\pi\in V_{n_1}^3$.\ \Qed

\medskip

{\bf (f)} Now suppose that $\pi$ is any involution in $G_{a_0}$.
Then $\pi\in V_{n_1}^6$.   \Prf\ Let $b$, $b'$ be such that
$\pi=\cycle{b\,_{\pi}\,b'}$.   Next take
disjoint $c$, $c'\Bsubseteq a_1\Bsetminus a_0$ such that
$\mag c=\mag c'=\mag(a_1\Bsetminus(a_0\Bcup c\Bcup c'))=\lambda$.
Then there there is an involution $\pi'\in G$ exchanging $c$ and $c'$,
and $\pi'$, $\pi\pi'$ are both involutions in $G_{a_1}$ satisfying the
conditions of (e).   So both belong to $V_{n_1}^3$ and
$\pi=\pi\pi'\pi'\in V_{n_1}^6$.\ \Qed

\medskip

{\bf (g)} Set $d_0=a_0$, $d_1=a_1\Bsetminus a_0$ and $d_2=1\Bsetminus a_1$.
Then $d_0$, $d_1$ and $d_2$ all have magnitude $\lambda$, so for all $i$,
$j<3$ there is a $\phi_{ij}\in G$ such that
$\phi_{ij}(d_i\Bcup d_j)=d_0$.
Let $n_2\ge n_1+3$ be such that $\phi_{ij}\in V_{n_2}$ for all $i$, $j<3$.
Then any involution in $G$ belongs to $V_{n_2}^{27}$.   \Prf\ Let
$\pi\in G$ be an involution;  express it as $\cycle{e\,_{\pi}\,e'}$.
For $i$, $j<3$, set $e_{ij}=e\Bcap d_i\Bcap\pi d_j$,
$e'_{ij}=\pi e_{ij}=e'\Bcap\pi d_i\Bcap d_j$;  set
$\pi_{ij}=\cycle{e_{ij}\,_{\pi}\,e'_{ij}}$.
In this case, because all the $e_{ij}$ and $e'_{ij}$ are disjoint,
$\langle\pi_{ij}\rangle_{i,j<3}$ is a commuting family, and we can talk of
$\prod_{i,j<3}\pi_{ij}$, which of course is equal to $\pi$.   Now, for each
pair $i$, $j<3$,

\Centerline{$\phi_{ij}\pi_{ij}\phi_{ij}^{-1}
=\cycle{\phi_{ij}e_{ij}\,_{\phi_{ij}\pi\phi_{ij}^{-1}}\,\phi_{ij}e'_{ij}}$}

\noindent is an involution in $G_{a_0}$, so belongs to
$V_{n_1}^6\subseteq V_{n_1+3}\subseteq V_{n_2}$.   So

\Centerline{$\pi_{ij}
=\phi_{ij}^{-1}\phi_{ij}\pi_{ij}\phi_{ij}^{-1}\phi_{ij}$}

\noindent belongs to $V_{n_2}^3$ and $\pi=\prod_{i,j<3}\pi_{ij}$ belongs to
$V_{n_2}^{27}$. \Qed

\medskip

{\bf (h)} Since every member of $G$ is expressible as
a product of at most three involutions in $G$ (382N once more),
$G=V_{n_2}^{81}=V_{n_2+7}$.
}%end of proof of 494R

\exercises{\leader{494X}{Basic exercises (a)}
%\spheader 494Xa
Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.
Show that the natural action of $\AmuA$ on $\frak A^f$ identifies $\AmuA$
with a subgroup of the isometry group $G$ of $\frak A^f$ when $\frak A^f$
is given its measure metric, and that the weak topology on $\AmuA$
corresponds to the topology of pointwise convergence
on $G$  as described in 441G.
%494A

\spheader 494Xb Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.
Show
that the following are equiveridical:  (i) $\frak A$ is purely atomic and
has at most finitely many atoms of any fixed measure;
(ii) $\AmuA$ is locally compact in its weak topology;
(iii) $\AmuA$ is compact in its uniform topology;  (iv) $\AmuA$ has a Haar
measure for its weak topology.
%494A

\spheader 494Xc
Let $(\frak A,\bar\mu)$ be a measure algebra.   Show that the weak topology
on $\AmuA$ is that induced by the product topology on $\frak A^{\frak A}$
if $\frak A$ is given its measure-algebra topology.
%494B
% \bar\mu(\pi c\Bsymmdiff\phi c)
%  \le 2\bar\mu(\phi c\Bsymmdiff(\pi c\Bsymmdiff\phi c))

\spheader 494Xd Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra,
$G$ a subgroup of $\AmuA$ and $\overline{G}$ its closure for the weak
topology on $\AmuA$.   Show that $G$ is
ergodic iff $\overline{G}$ is ergodic.
%494B

\spheader 494Xe
Let $I$ be a set, $\nu_I$ the usual measure on
$\{0,1\}^I$, and $(\frak B_I,\bar\nu_I)$ its measure algebra.
Let $\Psi$ be the group of
measure space automorphisms $g$ of $\{0,1\}^I$ for which there is a
countable set $J\subseteq I$ such that for every $x\in\{0,1\}^I$ there is a
finite set $K\subseteq J$ such that $g(x)(i)=x(i)$ for every
$i\in I\setminus K$.   For $g\in\Psi$, let
$\pi_g\in\Aut_{\bar\nu_I}\frak B_I$ be the corresponding automorphism
defined by saying that
$\pi_g(E^{\ssbullet})=g^{-1}[E]^{\ssbullet}$ whenever $\nu_I$ measures $E$.
(i) Show that
$G=\{\pi_g:g\in\Psi\}$ is a full subgroup of $\Aut_{\bar\nu_I}\frak B_I$.
(ii) Show that $G$ is ergodic and
dense in $\Aut_{\bar\nu_I}\frak B_I$ for the weak topology on
$\Aut_{\bar\nu_I}\frak B_I$.
%3{}83Xl 494B

\spheader 494Xf\dvAformerly{4{}93Yd}
Let $(\frak A,\bar\mu)$ be the measure algebra of
Lebesgue measure on $[0,1]$, and give $\AmuA$ its weak topology.   (i)
Show that the entropy function $h$ of 385M is Borel measurable.
\Hint{385Xj.}   (ii) Show that the set of ergodic measure-preserving
automorphisms is a dense G$_{\delta}$ set.
\Hint{let $D\subseteq\frak A$ be a countable dense set.   Show that
$\pi\in\AmuA$ is ergodic iff
$\inf_{n\in\Bbb N}\|\Bover1{n+1}
\sum_{i=0}^n\chi(\pi^id)-\bar\mu d\cdot\chi 1\|_1=0$ for every
$d\in D$.}
%494B

\spheader 494Xg Let $(\frak A,\bar\mu)$ be an atomless semi-finite
measure algebra.
(i) Show that $\AmuA$ is metrizable under its weak topology iff
$(\frak A,\bar\mu)$ is $\sigma$-finite and has countable Maharam type.
%iff $\{\iota\}$ is G$_{\delta}$
(ii) Show that $\AmuA$ is metrizable under its uniform topology iff
$(\frak A,\bar\mu)$ is $\sigma$-finite.
%iff $\{\iota\}$ is G$_{\delta}$
%494Bc 494Cg

\spheader 494Xh Let $(\frak A,\bar\mu)$ be a localizable measure algebra.
Show that $\supp:\AmuA\to\frak A$ is continuous for the uniform topology on
$\AmuA$ and the measure-algebra topology on $\frak A$.
%494C

\spheader 494Xi Let $(\frak A,\bar\mu)$ be an atomless homogeneous
probability algebra.   Show that there is a weakly mixing
measure-preserving automorphism of $\frak A$ which is not mixing.
\Hint{372Yj}.
%494E

\spheader 494Xj Let $(\frak A,\bar\mu)$ be a probability algebra,
$\pi\in\AmuA$ and $T$ the corresponding operator on
$L^2_{\Bbb C}=L^2_{\Bbb C}(\frak A,\bar\mu)$.   (i)
Show that $\pi$ is not ergodic iff there is a non-zero
$v\in L^2_{\Bbb C}$ such that $\int v=0$ and $Tv=v$.   (ii) Show that
$\pi$ is not weakly mixing iff there is a non-zero $v\in L^2_{\Bbb C}$
such that $\int v=0$ and $Tv$ is a multiple of $v$.   (iii)
Let $(\frak A\tensorhat\frak A,\bar\lambda)$ be the probability algebra
free product of $(\frak A,\bar\mu)$ with itself (definition:  325K), and
$\tilde\pi\in\Aut_{\bar\lambda}(\frak A\tensorhat\frak A)$ the automorphism
such that $\tilde\pi(a\otimes b)=\pi a\otimes\pi b$ for all $a$,
$b\in\frak A$.   Show that $\pi$ is weakly mixing iff $\tilde\pi$ is
ergodic iff $\tilde\pi$ is weakly mixing.
\Hint{consider $T_{\tilde\pi}(v\otimes\bar v)$.}
%494D %494Xi

\spheader 494Xk Let $(\frak A,\bar\mu)$ be a measure algebra.   For
$\pi\in\AmuA$ let $T_{\pi}$ be the corresponding operator on
$L^2_{\Bbb C}=L^2_{\Bbb C}(\frak A,\bar\mu)$.   Show that
$(\pi,v)\mapsto T_{\pi}v:\AmuA\times L^2_{\Bbb C}\to L^2_{\Bbb C}$ is
continuous if $\AmuA$ is given its weak topology and $L^2_{\Bbb C}$ its
norm topology.
%494E

\spheader 494Xl\dvAformerly{4{}93Yb}
Let $(\frak A,\bar\mu)$ be an atomless probability
algebra;  give $\frak A$ its measure metric.   Show that the
isometry group of $\frak A$, with its topology of pointwise convergence,
is extremely amenable.   \Hint{every
isometry of $\frak A$ is of the form $a\mapsto c\Bsymmdiff\pi a$, where
$c\in\frak A$ and $\pi\in\AmuA$;  now use 493Bc.}
%494L

\spheader 494Xm Let $\frak A$ be a homogeneous Dedekind
complete Boolean algebra,
and $\sequencen{V_n}$ a non-decreasing sequence of subsets of
$\Aut\frak A$, covering $\Aut\frak A$, with $V_n^2\subseteq V_{n+1}$ for
every $n$.   Show that there is an $n\in\Bbb N$ such that
$\Aut\frak A=V_n$.
%494R

\leader{494Y}{Further exercises (a)}
%\spheader 494Ya
Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.
For $\pi\in\AmuA$, let
$T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ be the Riesz space automorphism
such that $T_{\pi}(\chi a)=\chi(\pi a)$ for every $a\in\frak A$
(364P).   Take any $p\in\coint{1,\infty}$ and write $L^p_{\bar\mu}$ for
$L^p(\frak A,\bar\mu)$ as defined in 366A.   Set
$G_p=\{T_{\pi}\restr L^p_{\bar\mu}:\pi\in\AmuA\}$.
(i) Show that
$\pi\mapsto T_{\pi}\restr L^p_{\bar\mu}$ is a topological group isomorphism
between $\AmuA$, with its weak topology, and $G_p$, with the strong
operator topology from
$\eurm B(L^p_{\bar\mu};L^p_{\bar\mu})$ (3A5I).
(ii) Show that $G_p$ is closed in $\eurm B(L^p_{\bar\mu};L^p_{\bar\mu})$.
(iiI) Show that if $(\frak A,\bar\mu)$ is totally finite, then
$\pi\mapsto T_{\pi}\restr L^1_{\bar\mu}$ is a topological group isomorphism
between $\AmuA$, with its uniform topology, and $G_1$ with
the topology of uniform convergence on weakly compact subsets of
$L^1_{\bar\mu}$.
%494A 494Xk useful but function spaces off-topic a bit

\spheader 494Yb Let $\frak A$ be any Boolean algebra.   For
$I\subseteq\frak A$, set
$U_I=\{\pi:\pi\in\Aut\frak A$, $\pi a=a$ for every $a\in I\}$.
(i) Show that
$\{U_I:I\in[\frak A]^{<\omega}\}$ is a base of neighbourhoods of the
identity for a Hausdorff topology on $\Aut\frak A$ under which
$\Aut\frak A$ is a topological group.   (ii) Show that if $\frak A$ is
countable then $\Aut\frak A$, with this topology, is a Polish group.
%494A

\spheader 494Yc Let $(\frak A,\bar\mu)$ be a
semi-finite measure algebra.   Show
that $\AmuA$, with its weak topology, is weakly $\alpha$-favourable.
%494B

\spheader 494Yd Let $\bar\mu$ be counting measure on $\Bbb N$.   (i) Show
that if we identify $\Aut_{\bar\mu}\Cal P\Bbb N$ with the set of
permutations on $\Bbb N$, the weak topology of $\Aut_{\bar\mu}\Cal P\Bbb N$
is the
topology induced by the usual topology of $\NN$.   (ii) Show that there is
a comeager conjugacy class in $\Aut_{\bar\mu}\Cal P\Bbb N$.
%494B

\spheader 494Ye ({\smc Rosendal 09})
Let $(\frak A,\bar\mu)$ be the measure algebra of Lebesgue
measure on $[0,1]$, and give $\AmuA$ its weak topology.
Let $\Cal V$ be a countable base of
open neighbourhoods of $\iota$ in $\AmuA$.
(i) Show that if $I\subseteq\Bbb N$ is infinite and $V\in\Cal V$, then
$\{\pi:\pi\in\AmuA$, $\pi^n=\iota$ for some $n\in I\}$ is dense in $\AmuA$,
and that $B(I,V)=\{\pi:\pi^n\in V$ for some $n\in I\}$ is dense and open.
(ii) Show that if $I\subseteq\Bbb N$ is infinite then
$C(I)=\bigcap_{V\in\Cal V}B(I,V)$ is comeager, and is a union of
conjugacy classes.   (iii) Show that
$\bigcap\{C(I):I\in[\Bbb N]^{\omega}\}=\{\iota\}$.   (iv) Show that every
conjugacy class in $\AmuA$ is meager.
%494Yd 494B

\spheader 494Yf Let $(\frak A,\bar\mu)$ be a measure algebra.   Suppose
that $\pi\in\AmuA$ is aperiodic.   Show that the set of conjugates of $\pi$
in $\AmuA$ is dense for the weak topology on $\AmuA$.
%494E mt49bits

\spheader 494Yg Let $(\frak A,\bar\mu)$ be a probability algebra.   Show
that the set of weakly mixing automorphisms, with the subspace topology
inherited from the weak topology of $\AmuA$, is weakly $\alpha$-favourable.
%494E mt49bits

\spheader 494Yh Let $G$ be a Hausdorff
topological group.   Show that the following
are equiveridical:  (i) for every neighbourhood $V$ of the identity in $G$
there is a countable set $D\subseteq G$ such that $G=DV$;  (ii)
there is a family $\familyiI{H_i}$ of Polish groups such that
$G$ is isomorphic, as topological group, to a subgroup of
$\prod_{i\in I}H_i$.
%494O mt49bits

\spheader 494Yi Let $(\frak A,\bar\mu)$ be an atomless
$\sigma$-finite measure algebra, and $G$ a full ergodic subgroup of
$\AmuA$.   Let $V\subseteq\AmuA$ be a symmetric set such that
countably many left translates of $V$ cover $\AmuA$.   Show that $V^{228}$
is a neighbourhood of $\iota$ for the uniform topology on $\AmuA$.
%494O mt49bits

\spheader 494Yj Let $(\frak A,\bar\mu)$ be a purely atomic probability
algebra with two atoms of measure $2^{-n-2}$ for each $n\in\Bbb N$;  give
$\AmuA$ its uniform topology.   (i) Show that $\AmuA\cong\Bbb Z_2^{\Bbb N}$
is compact, therefore not extremely amenable, and can be regarded as a
linear space over the field $\Bbb Z_2$.
(ii) Show that there is a strictly increasing sequence of
subgroups of $\AmuA$ with union $\AmuA$.
(iii) Show that there is a subgroup $V$ of $\AmuA$, not open,
such that $\AmuA$ is covered
by countably many translates of $V$.   (iv) Show that there is a
discontinuous homomorphism from $\AmuA$ to a Polish group.
%494Q $\Bbb Z_2^{\Bbb N}$ has a Hamel basis

\spheader 494Yk Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   Show that it is not the union of a strictly increasing sequence
of subalgebras.
%494Q


}%end of exercises

\leader{494Z}{Problems} For $k\in\Bbb N$, say that a topological
group $G$ is {\bf $k$-Steinhaus} if whenever $V\subseteq G$ is a symmetric
set, containing the identity, such that countably many left translates of
$V$ cover $G$, then $V^k$
is a neighbourhood of the identity.   For your favourite groups, determine
the smallest $k$, if any, for which they are $k$-Steinhaus.
\cmmnt{(See {\smc Rosendal \& Solecki 07}.)}

\endnotes{
\Notesheader{494}
In 494B-494C %494B 494C
I run through properties of the weak and uniform topologies
of $\AmuA$ in parallel.   The effect is to emphasize their similarities,
but they are of course very different -- for instance, consider 494Xg, or
the contrast between 494Cg and 494Ge.   Both have expressions in terms of
standard topologies on spaces of linear operators (494Ya), and the weak
topology corresponds to the pointwise topology of an isometry group
(494Xa).   There are other more or less
natural topologies which can be considered (e.g., that of 494Yb), but at
present the two examined in this section seem to be the most important.
I spell out 494Be and 494Ci to show that the groups
here provide interesting examples of Polish groups with striking
properties.

The formulation of 494D is specifically designed for the application in
the proof of 494E(b-ii);
the version in 494Xj(ii) is much closer to the real strength of the idea,
and takes us directly to one of the important reasons for being interested
in weakly mixing automorphisms in 494Xj(iii).   The proof of 494D
through Bochner's theorem saves space here, but fails to signal
the concept of `spectral resolution' of a unitary operator on a Hilbert
space ({\smc Riesz \& Sz.-Nagy 55}, \S109),
which is an important tool in understanding
operators $T_{\pi}$ and hence automorphisms $\pi$.

While 494H and 494G are of some interest in themselves, their function here
is to prepare the way to 494L, 494O and 494Q.   The first
belongs to the series in \S493;  like the results in that section, it
depends on concentration-of-measure theorems, quoted in part (e) of the
proof of 494I and again in part (e) of the proof of 494J.   In addition,
for the generalization from ergodic full groups to arbitrary full groups,
we need the structure theory for closed subalgebras developed in
\S333.

494O and 494Q-494R break new ground.   The former, following
{\smc Kittrell \& Tsankov 09}, examines a curious
phenomenon identified by {\smc Rosendal \& Solecki 07} in the course of
a search
for automatic-continuity results.   We cannot dispense entirely with the
hypotheses that $\frak A$ should be atomless and $G$ ergodic (494Yj),
though perhaps they can be relaxed.
%what if \frak A is prob alg relatively atomless over \frak C?
Many examples are now known of
$k$-Steinhaus groups (494O, 494Yi), but as far as I am aware there are no
non-trivial cases in which the critical value of $k$ has been determined
(494Z).   The automatic-continuity corollary in 494Ob is really a result
about homomorphisms into Polish groups (see 494Yh), but applies in many
other cases (494P).

The phenomenon of 494Q, which we might call a (negative) `algebraic
cofinality' result, has attracted attention with regard to many
algebraic structures, starting with {\smc Bergman 06}.   Apart from the
variations of 494Q in 494R and 494Xm, there is a simple example
in 494Yk.   494Yj again indicates one of the limits of the result.
}%end of notes

\leaveitout{
{\smc Kechris p09}, I.4.1:  if $(\frak A,\bar\mu)$ is the Lebesgue
probability algebra and $G$, $H\subseteq\AmuA$ are ergodic full subgroups
which are isomorphic as groups, then $G$ and $H$ are conjugate in $\AmuA$
(?).
}

\discrpage

\leaveitout{
\leader{494yL}{Theorem}\cmmnt{ ({\smc Kittrell \& Tsankov 09})}
Let $(\frak A,\bar\mu)$ be a probability algebra and
$G\subseteq\AmuA$ a full subgroup with fixed-point subalgebra $\frak C$.
Suppose that $\frak A$ is relatively atomless over $\frak C$.
Give $G$ the topology induced by the uniform topology of $\AmuA$.

(a) If $V\subseteq G$ is symmetric
and $G$ can be covered by countably many left translates of $V$ in $G$,
then $V^{38}=\{\pi_1\pi_2\ldots\pi_{38}:\pi_1,\ldots,\pi_{38}\in V\}$
is a neighbourhood of the identity in $G$.

(b) If $H$ is a topological group such that for every neighbourhood
$W$ of the identity in $H$ there is a countable set $D\subseteq H$ such
that $H=DW$,
and $\theta:G\to H$ is a group homomorphism, then $\theta$ is continuous.

\proof{{\bf (a)(i)} Let $\sequencen{\psi_n}$ be a sequence in $G$ such that
$G=\bigcup_{n\in\Bbb N}\psi_nV$.   It may help if I note straight away that
$\iota\in V^2$.   \Prf\ There is an $n\in\Bbb N$ such that
$\iota\in\psi_nV$, that is, $\psi_n^{-1}\in V$;  as $V$ is symmetric,
$\psi_n\in V$ and $\iota=\psi_n\psi_n^{-1}$ belongs to $V^2$.\ \Qed

\medskip

\quad{\bf (ii)} For $a\in\frak A$, set
$G_a=\{\pi:\pi\in G$ is supported by $a\}$, and let $u_a$ be the
conditional expectation of $\chi a$ on $\frak C$.
There is a non-zero $e\in\frak A$ such that $u_e$ is a multiple of $\chi 1$
and for every
$\pi\in G_e$ there is a $\phi\in V^2$ agreeing with $\pi$ on
$\frak A_e$.   \Prf\ Because $\frak A$ is relatively atomless over
$\frak C$, we can choose inductively a disjoint
sequence $\sequencen{b_n}$ in $\frak A\setminus\{0\}$ such that
$u_{b_n}=2^{-n-1}\chi 1$ for every $n$ (use 494Ad, as usual);  then
$\sum_{n=0}^{\infty}\bar\mu b_n=1$, so $\sup_{n\in\Bbb N}b_n=1$ in
$\frak A$.
\Quer\ Suppose, if possible, that for every $n\in\Bbb N$ there is a
$\pi_n\in G_{b_n}$ such that there is no
$\phi\in V^2$ agreeing with $\pi_n$ on $\frak A_{b_n}$.
If $n\in\Bbb N$, then
$V^2=(\psi_nV)^{-1}\psi_nV$ and $\pi_n=\iota^{-1}\pi_n$,
so there must be a $\pi'_n\in G_{b_n}$, either $\iota$ or $\pi_n$,
not agreeing with
$\phi$ on $\frak A_{b_n}$ for any $\phi\in\psi_nV$.   Define
$\psi\in\AmuA$ by the formula

\Centerline{$\psi d
=\pi'_nd\text{ if }n\in\Bbb N\text{ and }d\Bsubseteq b_n$.}

\noindent Because $G$ is full, $\psi\in G$ and
there is an $m\in\Bbb N$ such that
$\psi\in\psi_mV$.   But now $\pi'_m$ agrees with $\psi$ on
$\frak A_{b_m}$, contrary to the choice of $\pi'_m$.\ \BanG\  So one of the
$b_n$ will serve for $e$.\ \Qed

\medskip

\quad{\bf (iii)} There is an involution $\pi\in V^2$, supported by $e$,
such that $u_{\supp\pi}\le\bover23u_e$.   \Prf\ Take disjoint
$b$, $b'\Bsubseteq e$ such that $u_b=u_{b'}=\bover13u_e$.
Because $G$ is full, there is a $\phi\in G$ such that
$\phi b=b'$ (494Gb).   For $d\Bsubseteq b$, set
$\phi_d=\cycle{d\,_{\phi}\,\phi d}$, so that $\phi_d$ is either $\iota$ or
an involution belonging to $G$.   Since $\frak A_b$ is surely uncountable
(for instance, there is for every $\alpha\in[0,1]$ a $d\in\frak A_b$ such
that $u_d=\alpha u_b$), there must be an $n\in\Bbb N$ and distinct $d$,
$d'\in\frak A_b$ such that $\phi_d$ and $\phi_{d'}$ both belong to
$\psi_nV$.   In this case

\Centerline{$\pi=\phi_{d\Bsymmdiff d'}=\phi_d^{-1}\phi_{d'}$}

\noindent is an involution belonging to $(\psi_nV)^{-1}\psi_nV=V^2$,
while $\supp\pi\Bsubseteq u_b\Bcup u_{b'}$ so
$u_{\supp\pi}\le\bover23u_e$.\ \Qed

\medskip

\quad{\bf (iv)} Taking $a=\supp\pi$ in (iii), $a$ and $e$ satisfy the
conditions of 494N with respect to $V^2$, so
$G_c\Bsubseteq(V^2)^{18}=V^{36}$.

\medskip

\quad{\bf (v)}  Finally, there is a $\delta>0$ such that, in the language
of 494Cb,
$G\cap U(1,\delta)\Bsubseteq V^{38}$.   \Prf\Quer\ Otherwise, we can find
for each $n\in\Bbb N$ a
$\pi_n\in G\cap U(1,2^{-n-1}\bar\mu a)\setminus V^{38}$.   Set
$\pi'_n=\psi_n\pi_n\psi_n^{-1}$, $b_n=\supp\pi'_n$;  then
$\bar\mu b_n=\bar\mu(\supp\pi_n)$ for each $n$, by 381Gd again.   So
$b'=\sup_{n\in\Bbb N}b_n$ has measure at most $\bar\mu c$, and there is a
$\phi\in G$ such that $\phi b'\Bsubseteq c$.   In this case,
there is an $n\in\Bbb N$ such that $\phi^{-1}\in\psi_nV$, that is,
$\phi\psi_n\in V^{-1}=V$.   Now
$\pi=\phi\psi_n\pi_n\psi_n^{-1}\phi^{-1}$ has support
$\phi b_n\Bsubseteq c$, so belongs to $V^{36}$.   But this means that
$\pi_n=\psi_n^{-1}\phi^{-1}\pi\phi\psi_n$ belongs to $V^{38}$, contrary to
the choice of $\pi_n$.\ \Bang\Qed

So $V^{38}$ is a neighbourhood of $\iota$ in $G$, as claimed.

\medskip

{\bf (b)} Let $W$ be a neighbourhood of $e$ in $H$.   Then there
is a symmetric neighbourhood $W_1$ of the identity in $H$ such that
$W_1^{38}\subseteq W$.   Set $V=\theta^{-1}[W_1]$.   Let $W_2$ be a
neighbourhood of the identity in $H$ such that
$W_2^{-1}W_2\subseteq W_1$, and
$\sequencen{y_n}$ a sequence in $H$ such that
$H=\bigcup_{n\in\Bbb N}y_nW_2$.
For each $n\in\Bbb N$, choose $\psi_n\in G$ such that
$\theta(\psi_n)\in y_nW_2$ whenever $\theta[G]$ meets $y_nW_2$.
If $\pi\in G$, there is an $n\in\Bbb N$ such that
$\theta(\pi)\in y_nW_2$;   in this case, $\theta(\psi_n)\in y_nW_2$, so

\Centerline{$\theta(\psi_n^{-1}\pi)\in W_2^{-1}y_n^{-1}y_nW_2
\subseteq W_1$}

\noindent and
$\psi_n^{-1}\pi\in V$.   Thus $\pi\in\psi_nV$;  as $\pi$ is arbitrary,
$G=\bigcup_{n\in\Bbb N}\psi_nV$.   By (b), $V^{38}$ is a neighbourhood
of $\iota$;  but
$V^{38}\subseteq\theta^{-1}[W_1^{38}]\subseteq\theta^{-1}[W]$, so
$\theta^{-1}[W]$ is a neighbourhood of $\iota$.   As $W$ is
arbitrary, $\theta$ is continuous (4A5Fa again).
}%end of proof of 494O
}

\frfilename{mt495.tex}
\versiondate{20.12.08/1.1.19}
\copyrightdate{2003}

\def\chaptername{Further topics}
\def\sectionname{Poisson point processes}

\newsection{495}

A classical challenge in probability theory is to formulate a consistent
notion of `random set'.   Simple geometric considerations lead us to a
variety of measures which are both interesting and important.
All these are manifestly special constructions.   Even in the most
concrete structures, we have to make choices which come to seem
arbitrary as soon as we are conscious of the many alternatives.   There
is however one construction which has a claim to pre-eminence because it
is both robust under the transformations of abstract measure theory and
has striking properties when applied to familiar measures (to the point,
indeed, that it is relevant to questions in physics and chemistry).
This gives the `Poisson point processes' of 495D-495E.   In this section
I give a brief introduction to the measure-theoretic aspects of this
construction.

\leader{495A}{Poisson distributions}\cmmnt{ We need a little of the
elementary theory of Poisson distributions.

\medskip

}{\bf (a)} The {\bf Poisson distribution} with parameter $\gamma>0$
is the point-supported Radon probability measure $\nu_{\gamma}$ on
$\Bbb R$ such that $\nu_{\gamma}\{n\}=\Bover{\gamma^n}{n!}e^{-\gamma}$
for every $n\in\Bbb N$.   \cmmnt{(See 285Q and 285Xr.)}   Its
expectation is
$\cmmnt{\sum_{n=1}^{\infty}\Bover{\gamma^n}{(n-1)!}e^{-\gamma}
=\mskip5mu}\gamma$.
\cmmnt{Since
$\nu_{\gamma}\Bbb N=1$, }$\nu_{\gamma}$ can be identified with the
corresponding subspace measure on $\Bbb N$.
It will be convenient to allow $\gamma=0$, so that the Dirac
measure on $\Bbb R$ or $\Bbb N$ concentrated at $0$ becomes a `Poisson
distribution with expectation $0$'.

\spheader 495Ab The convolution of two Poisson distributions is a
Poisson distribution.   \prooflet{\Prf\ If $\alpha$, $\beta>0$ then

$$\eqalignno{(\nu_{\alpha}*\nu_{\beta})(\{n\})
&=\int\nu_{\beta}(\{n\}-t)\nu_{\alpha}(dt)\cr
\displaycause{444A}
&=\sum_{i=0}^n\Bover{\beta^{n-i}}{(n-i)!}e^{-\beta}
  \cdot\Bover{\alpha^i}{i!}e^{-\alpha}\cr
&=\Bover1{n!}
   \sum_{i=0}^n\Bover{n!}{i!(n-i)!}\alpha^i\beta^{n-i}e^{-\alpha-\beta}
=\Bover{(\alpha+\beta)^n}{n!}e^{-\alpha-\beta}\cr}$$

\noindent for every $n\in\Bbb N$, so
$\nu_{\alpha}*\nu_{\beta}=\nu_{\alpha+\beta}$.\ \QeD}
So if $f$ and $g$ are independent random variables with Poisson
distributions then $f+g$ has a Poisson distribution\cmmnt{
(272T\formerly{2{}72S})}.

\spheader 495Ac If $\familyiI{f_i}$ is a countable independent family of
random variables with Poisson distributions, and
$\alpha=\sum_{i\in I}\Expn(f_i)$ is finite, then $f=\sum_{i\in I}f_i$ is
defined a.e.\ and has a Poisson distribution with expectation $\alpha$.
\prooflet{\Prf\ For finite $I$ we can induce on $\#(I)$, using (b) (and
272L) for the inductive step.   For the infinite case we can suppose
that $I=\Bbb N$.   In this case $f_i\ge 0$ a.e.\ for each $i$ so
$f=\sum_{i=0}^{\infty}f_i$ is defined a.e.\ and has expectation
$\alpha$, by B.Levi's theorem.   Setting $g_n=\sum_{i=0}^nf_i$ for each
$n$, so that $g_n$ has a Poisson distribution with expectation
$\beta_n=\sum_{i=0}^n\alpha_i$, we have

$$\Pr(f\le\gamma)
=\lim_{n\to\infty}\Pr(g_n\le\gamma)
=\lim_{n\to\infty}\sum_{i=0}^{\lfloor\gamma\rfloor}
  \Bover{\beta_n^i}{i!}e^{-\beta_n}
=\sum_{i=0}^{\lfloor\gamma\rfloor}\Bover{\alpha^i}{i!}e^{-\alpha}$$

\noindent for every $\gamma\ge 0$, so $f$ has a Poisson distribution
with expectation $\alpha$.\ \Qed}

\spheader 495Ad
\cmmnt{I find myself repeatedly calling on the simple fact that}
$1-e^{-\gamma}(1+\gamma)=\nu_{\gamma}(\Bbb N\setminus\{0,1\})$ is at most
$\bover12\gamma^2$ for every $\gamma\ge 0$\prooflet{;  this is because
$\bover{d}{dt}(\bover12t^2+e^{-t}(1+t))=t(1-e^{-t})\ge 0$ for $t\ge 0$}.

\leader{495B}{Theorem} Let $(X,\Sigma,\mu)$ be a measure
space.   Set $\Sigma^f=\{E:E\in\Sigma$, $\mu E<\infty\}$.   Then for any
$\gamma>0$ there
are a probability space $(\Omega,\Lambda,\lambda)$ and a family
$\family{E}{\Sigma^f}{g_E}$ of random variables on $\Omega$ such that

\inset{(i) for every $E\in\Sigma^f$, $g_E$ has a Poisson distribution
with expectation $\gamma\mu E$;

(ii) whenever $\familyiI{E_i}$ is a disjoint family in $\Sigma^f$, then
$\familyiI{g_{E_i}}$ is stochastically independent;

(iii) whenever $\sequence{i}{E_i}$ is a disjoint sequence in $\Sigma^f$
with union $E\in\Sigma^f$, then $g_E\eae\sum_{i=0}^{\infty}g_{E_i}$.}

\proof{{\bf (a)} Let
$\Cal H\subseteq\{H:H\in\Sigma$, $0<\mu H<\infty\}$ be a maximal family
such that $H\cap H'$ is negligible for all distinct $H$, $H'\in\Cal H$.
For $H\in\Cal H$, let $\mu'_H$ be the normalized subspace measure
defined by setting $\mu'_HE=\mu E/\mu H$ for $E\in\Sigma\cap\Cal PH$,
and $\lambda_H$ the corresponding product probability measure on
$H^{\Bbb N}$.   Next, for
$H\in\Cal H$, let $\nu_H$ be the Poisson distribution with expectation
$\gamma\mu H$,
regarded as a probability measure on $\Bbb N$.   Let $\lambda$ be the
product measure on
$\Omega=\prod_{H\in\Cal H}(\Bbb N\times H^{\Bbb N})$, giving each
$\Bbb N\times H^{\Bbb N}$ the product measure $\nu_H\times\lambda_H$.
For $\omega\in\Omega$, write $m_H(\omega)$, $x_{Hj}(\omega)$ for its
coordinates, so that
$\omega=\family{H}{\Cal H}{(m_H(\omega),\sequence{j}{x_{Hj}(\omega)})}$.

\medskip

{\bf (b)} For $H\in\Cal H$ and $E\in\Sigma$, set
$g_{HE}(\omega)=\#(\{j:j<m_H(\omega)$, $x_{Hj}(\omega)\in E\})$
when this is finite.
Then $g_{HE}$ is measurable and has a Poisson distribution with
expectation
$\gamma\mu(H\cap E)$;  moreover, if $E_0,\ldots,E_r\in\Sigma$ are
disjoint,
then $g_{HE_0},\ldots,g_{HE_r}$ are independent.   \Prf\ It is enough to
examine the case in which the $E_i$ cover $X$.   Then for any
$n_0,\ldots,n_r\in\Bbb N$ with sum $n$,

$$\eqalignno{\lambda\{\omega:&\mskip5mu g_{HE_i}(\omega)=n_i
   \text{ for every }i\le r\}\cr
&=\lambda\{\omega:\#(\{j:j<m_H(\omega),\,x_{Hj}\in E_i\})=n_i
   \text{ for every }i\le r\}\cr
&=\lambda\{\omega:m_H(\omega)=n,\,\#(\{j:j<n,\,x_{Hj}\in E_i\})=n_i
   \text{ for every }i\le r\}\cr
&=\sum_{\Atop{J_0,\ldots,J_r\text{ partition }n}
   {\#(J_i)=n_i\text{ for each }i\le r}}
   \lambda\{\omega:m_H(\omega)=n,\,
   x_{Hj}\in E_i\text{ whenever }i\le r,\,j\in J_i\}\cr
&=\sum_{\Atop{J_0,\ldots,J_r\text{ partition }n}
   {\#(J_i)=n_i\text{ for each }i\le r}}
   \Bover{(\gamma\mu H)^n}{n!}e^{-\gamma\mu H}
     \prod_{i=0}^r\bigl(\Bover{\mu(H\cap E_i)}{\mu H}\bigr)^{n_i}\cr
&=\Bover{n!}{n_0!\ldots n_r!}\Bover1{n!}e^{-\gamma\mu H}
     \prod_{i=0}^r(\gamma\mu(H\cap E_i))^{n_i}\cr
&=\prod_{i=0}^r\Bover{(\gamma\mu(H\cap E_i))^{n_i}}{n_i!}
     e^{-\gamma\mu(H\cap E_i)},\cr}$$

\noindent which is just what we wanted to know.\ \Qed

Obviously $g_{HE}=\sum_{i=0}^{\infty}g_{HE_i}$ whenever
$\sequence{i}{E_i}$ is a disjoint sequence in $\Sigma$ with union
$E$, and $g_{HE}=0$ a.e.\ if $\mu(H\cap E)=0$.

\medskip

{\bf (c)} Suppose that $H_0,\ldots,H_m\in\Cal H$ are distinct and
$E_0,\ldots,E_r\in\Sigma$ are disjoint.   Then the random variables
$g_{H_jE_i}$ are independent.   \Prf\ For each $j\le m$, $g_{H_jE_i}$ is
$\Lambda_{H_j}$-measurable, where $\Lambda_{H_j}$ is the
$\sigma$-algebra of subsets of $\Omega$ which are measured by $\lambda$
and determined by the single coordinate $H_j$ in the product
$\prod_{H\in\Cal H}(\Bbb N\times H^{\Bbb N})$.   Now the
$\sigma$-algebras $\Lambda_{H_j}$ are independent (272Ma).   So if we
have any family $\langle n_{ij}\rangle_{i\le r,j\le m}$ in $\Bbb N$,

$$\eqalign{\lambda\{\omega:g_{H_jE_i}(\omega)=n_{ij}
&  \text{ for every }i\le r,j\le m\}\cr
&=\prod_{j=0}^m\lambda\{\omega:g_{H_jE_i}(\omega)=n_{ij}
  \text{ for every }i\le r\}\cr
&=\prod_{j=0}^m\prod_{i=0}^r
  \lambda\{\omega:g_{H_jE_i}(\omega)=n_{ij}\}\cr}$$

\noindent by (b);  and this is what we need to know.\ \Qed

\medskip

{\bf (d)} For $E\in\Sigma^f$, set
$\Cal H_E=\{H:H\in\Cal H$, $\mu(E\cap H)>0\}$;  then $\Cal H_E$ is
countable, because $\Cal H$ is almost disjoint, and
$\mu E=\sum_{H\in\Cal H_E}\mu(H\cap E)$, because $\Cal H$ is maximal.
Set $g_E(\omega)=\sum_{H\in\Cal H_E}g_{HE}(\omega)$ when this is finite.
Then $g_E$ is defined a.e.\ and has a Poisson distribution with
expectation
$\gamma\mu E$ (495Ac).   Also $\familyiI{g_{E_i}}$ are independent
whenever $\familyiI{E_i}$ is a disjoint family in $\Sigma^f$.   \Prf\ It
is enough to deal with the case of finite $I$ (272Bb).    Set
$\Cal H^*=\bigcup_{i\in I}\Cal H_{E_i}$, so that $\Cal H^*$ is
countable, and for $i\in I$ set $g'_i=\sum_{H\in\Cal H^*}g_{HE_i}$.
Then each $g'_i$ is equal almost everywhere to the corresponding
$g_{E_i}$, and
$\familyiI{g'_i}$ is independent, by 272K.   (The point is that each
$g'_i$ is $\Lambda^*_i$-measurable, where $\Lambda^*_i$ is the
$\sigma$-algebra generated by $\{g_{HE_i}:H\in\Cal H\}$, and 272K, with (c)
above,
assures us that the $\Lambda^*_i$ are independent.)   It follows at once
that $\familyiI{g_{E_i}}$ is independent (272H).\ \QeD\  This proves (i)
and (ii).

\medskip

{\bf (e)} Similarly, if $\sequence{i}{E_i}$ is a disjoint sequence in
$\Sigma^f$ with union $E\in\Sigma^f$, set
$\Cal H^*=\Cal H_E\cup\bigcup_{i\in\Bbb N}\Cal H_{E_i}$.   For each
$i\in\Bbb N$, set $g'_i=\sum_{H\in\Cal H^*}g_{HE_i}$;  then
$g'_i\eae g_{E_i}$.   Now

$$\sum_{i=0}^{\infty}g_{E_i}
=_{\text{a.e.}}\sum_{i=0}^{\infty}g'_i
=\sum_{H\in\Cal H^*}\sum_{i=0}^{\infty}g_{HE_i}
=\sum_{H\in\Cal H^*}g_{HE}
=_{\text{a.e.}}g_E,$$

\noindent as required by (iii).
}%end of proof of 495B

\vleader{48pt}{495C}{Lemma} Let $X$ be a set and $\Cal E$ a subring of
the Boolean algebra $\Cal PX$.   Let $\Cal H$ be the family of sets
of the form

\Centerline{$\{S:S\subseteq X$, $\#(S\cap E_i)=n_i$ for every
$i\in I\}$}

\noindent where $\familyiI{E_i}$ is a finite disjoint family in $\Cal E$
and $n_i\in\Bbb N$ for every $i\in I$.   Then the Dynkin class
$\Tau\subseteq\Cal P(\Cal PX)$ generated by $\Cal H$\cmmnt{ (136A)}
is the $\sigma$-algebra of subsets of $\Cal PX$ generated by $\Cal H$.

\proof{ Let $Q$ be the set of functions $q$ from finite subsets
of $\Cal E$ to $\Bbb N$, and for $q\in Q$ set

\Centerline{$H_q=\{S:S\subseteq X$, $\#(S\cap E)=q(E)$ for every
$E\in\dom q\}$.}

\noindent Our family $\Cal H$ is just
$\{H_q:q\in Q$, $\dom q$ is disjoint$\}$.

If $q\in Q$ and $\dom q$ is a subring of $\Cal E$, then $H_q\in\Tau$.
\Prf\ Being a finite Boolean ring, $\dom q$ is a Boolean algebra;  let
$\Cal A$ be the set of its atoms.   Then $H_q$ is either empty or equal
to $H_{q\restr\Cal A}$;  in either case it belongs to $\Tau$.\ \Qed

If $q$ is any member of $Q$, then $H_q\in\Tau$.   \Prf\ Let $\Cal E'$ be
the subring of $\Cal PX$ generated by $\dom q$.   Then
$H_q=\bigcup_{q\subseteq q'\in Q,\dom q'=\Cal E'}H_{q'}$ is the union of
a countable disjoint family in $\Tau$, so belongs to $\Tau$.\ \Qed

Now observe that $\Cal H_1=\{H_q:q\in Q\}\cup\{\emptyset\}$ is a subset
of $\Tau$ closed under finite intersections, so by the Monotone Class
Theorem (136B) $\Tau$ includes the $\sigma$-algebra generated by
$\Cal H_1$, and must be precisely the $\sigma$-algebra generated by
$\Cal H$.
}%end of proof of 495C

\leader{495D}{Theorem} Let $(X,\Sigma,\mu)$ be an atomless measure
space.   Set $\Sigma^f=\{E:E\in\Sigma$, $\mu E<\infty\}$;  for
$E\in\Sigma^f$, set $f_E(S)=\#(S\cap E)$ when $S\subseteq X$ meets $E$
in a finite set.   Let $\Tau$ be the $\sigma$-algebra of subsets of
$\Cal PX$ generated by sets of the form $\{S:f_E(S)=n\}$ where
$E\in\Sigma^f$ and $n\in\Bbb N$.   Then for any $\gamma>0$ there
is a unique probability measure $\nu$ with domain $\Tau$ such that

(i) for every $E\in\Sigma^f$, $f_E$ is
measurable and has a Poisson distribution with expectation
$\gamma\mu E$;

(ii) whenever $\familyiI{E_i}$ is a disjoint family in $\Sigma^f$, then
$\familyiI{f_{E_i}}$ is stochastically independent.

\proof{{\bf (a)} Let $\Cal H$, $\family{H}{\Cal H}{\nu_H}$,
$\family{H}{\Cal H}{\mu_H}$,
$\family{H}{\Cal H}{\mu'_H}$,
$\family{H}{\Cal H}{\lambda_H}$, $\Omega$, $\lambda$,
$\family{E}{\Sigma^f}{\Cal H_E}$ and $\family{E}{\Sigma^f}{g_E}$ be as
in the proof of 495B.   Note that all the $\mu'_H$ are atomless
(234Nf\formerly{2{}34F}).
Define $\phi:\Omega\to\Cal PX$ by setting

\Centerline{$\phi(\omega)
=\{x_{Hj}(\omega):H\in\Cal H$, $j<m_H(\omega)\}$}

\noindent for $\omega\in\Omega$.

\medskip

{\bf (b)} For $E\in\Sigma^f$, let $A_E$ be the set of those
$\omega\in\Omega$ such that

\inset{{\it either} there are $H\in\Cal H\setminus\Cal H_E$,
$j\in\Bbb N$ such that $x_{Hj}(\omega)\in E$

{\it or} there are distinct $H$, $H'\in\Cal H_E$ and $j\in\Bbb N$ such
that $x_{Hj}(\omega)\in H'$

{\it or} there is an $H\in\Cal H$ such that the
$x_{Hj}(\omega)$, for $j\in\Bbb N$, are not all distinct.}

\noindent Then for any sequence $\sequence{i}{E_i}$ in $\Sigma^f$,
$\lambda_*(\bigcup_{i\in\Bbb N}A_{E_i})=0$.    \Prf\ Set
$\Cal H^*=\bigcup_{i\in\Bbb N}\Cal H_{E_i}$, so that $\Cal H^*$ is a
countable subset of $\Cal H$.   For $H\in\Cal H$, set

\Centerline{$F_H=H\setminus(\bigcup\{E_i:i\in\Bbb N$, $H\cap E_i$ is
negligible$\}\cup\bigcup\{H':H'\in\Cal H^*$, $H'\ne H\})$,}

\Centerline{$W_H=\{\pmb{x}:\pmb{x}\in H^{\Bbb N}$ is injective$\}$,}

\noindent so that $F_H$ is $\mu'_H$-conegligible and $W_H$ is
$\lambda_H$-conegligible (because $\mu'_H$ is atomless, see 254V).
Now

\Centerline{$\Omega\setminus\bigcup_{k\in\Bbb N}A_{E_k}
\supseteq\prod_{H\in\Cal H}
  (\Bbb N\times(W_H\cap F_H^{\Bbb N}))$}

\noindent has full outer measure in $\Omega$, by 254Lb, and its
complement has zero inner measure (413Ec).\ \Qed

It follows that there is a probability measure $\tilde\lambda$ on
$\Omega$, extending $\lambda$, such that $\tilde\lambda A_E=0$ for every
$E\in\Sigma^f$ (417A).
Let $\nu_0$ be the image measure $\tilde\lambda\phi^{-1}$.

\medskip

{\bf (c)} If $E\in\Sigma^f$ and $\omega\in\Omega\setminus A_E$,
then $f_E(\phi(\omega))=g_E(\omega)$ if either is defined.   \Prf\
If $H\in\Cal H$, then all the $x_{Hj}(\omega)$ are distinct;  if
$H\in\Cal H\setminus\Cal H_E$, no $x_{Hj}(\omega)$ can belong to $E$;
if $H$, $H'\in\Cal H_E$ are distinct, then no $x_{Hj}(\omega)$ can
belong to $H'$.   So all the $x_{Hj}(\omega)$, $x_{H'k}(\omega)$ for
$H$, $H'\in\Cal H_E$ and $j$, $k\in\Bbb N$ must be distinct, and

$$\eqalign{f_E(\phi(\omega))
&=\#(\{x_{Hj}(\omega):H\in\Cal H,\,j<m_H(\omega),\,
   x_{Hj}(\omega)\in E\})\cr
&=\#(\{(H,j):H\in\Cal H_E,\,j<m_H(\omega),\,
   x_{Hj}(\omega)\in E\})\cr
&=\sum_{H\in\Cal H_E}g_{HE}(\omega)
=g_E(\omega)}$$

\noindent if any of these is finite.\ \QeD\   It follows at once that if
$E_0,\ldots,E_r\in\Sigma^f$ are disjoint, then
$\{\omega:f_{E_i}(\phi(\omega))=g_{E_i}(\omega)$ for every $i\le r\}$ is
$\tilde\lambda$-conegligible, so that if $n_0,\ldots,n_r\in\Bbb N$ then

$$\eqalign{\nu_0\{S:f_{E_i}(S)=n_i\text{ for every }i\le r\}
&=\tilde\lambda\{\omega:f_{E_i}(\phi(\omega))=n_i
  \text{ for every }i\le r\}\cr
&=\tilde\lambda\{\omega:g_{E_i}(\omega)=n_i\text{ for every }i\le r\}\cr
&=\lambda\{\omega:g_{E_i}(\omega)=n_i\text{ for every }i\le r\}\cr
&=\prod_{i=0}^r\Bover{(\gamma\mu E_i)^{n_i}}{n_i!}
  e^{-\gamma\mu E_i}.\cr}$$

\noindent Thus every $f_{E_i}$ is finite $\nu_0$-a.e., belongs to
$\eusm L^0(\nu_0)$ and has a Poisson
distribution with the appropriate expectation, and they are independent.

\medskip

{\bf (d)} As $\Tau$ is defined to be the $\sigma$-algebra generated by
the family $\{f_E:E\in\Sigma^f\}$, it is included in the domain of
$\nu_0$.   Set $\nu=\nu_0\restr\Tau$;  then $\nu$ has the properties (i)
and (ii).   To see that it is unique, observe that if $\nuprime$ also
has these properties,
then $\{A:\nu A=\nuprime A\}$ is a Dynkin class containing every set of
the form

\Centerline{$\{S:f_{E_i}(S)=n_i$ for $i\le r\}$}

\noindent where $E_0,\ldots,E_r\in\Sigma^f$ are disjoint and
$n_0,\ldots,n_r\in\Bbb N$.   By 495C it contains the $\sigma$-algebra
generated by this family, which is $\Tau$.   So $\nu$ and $\nuprime$
agree on $\Tau$, and are equal.
}%end of proof of 495D

\leader{495E}{Definition} In the context of 495D, I will call the
completion of $\nu$ the
{\bf Poisson point process} on $X$ with {\bf intensity} or {\bf density}
$\gamma$.

\cmmnt{Note that the Poisson point process on $(X,\mu)$ with intensity
$\gamma>0$ is identical with the Poisson point process on
$(X,\gamma\mu)$ with intensity $1$.   There would therefore be no real
loss of generality in the main theorems of this section if I spoke only
of point processes with intensity $1$.   I retain the extra parameter
because applications frequently demand it, and the formulae will be more
useful with the
$\gamma$s in their proper places;  moreover, there are important ideas
associated with variations in $\gamma$, as in 495Xe.
}%end of comment

\leader{495F}{Proposition}\dvAformerly{4{}95Xc(i)}
Let $(X,\Sigma,\mu)$ be an atomless measure space,
$\familyiI{X_i}$ a countable partition of $X$ into measurable sets and
$\gamma>0$.   Let $\nu$ be the Poisson point process of $(X,\Sigma,\mu)$
with intensity $\gamma$;  for
$i\in I$ let $\nu_i$ be the Poisson point process of
$(X_i,\Sigma_i,\mu_i)$ with intensity $\gamma$, where
$\mu_i$ is the subspace measure on $X_i$ and $\Sigma_i$ its domain.
For
$S\subseteq X$ set $\phi(S)=\familyiI{S\cap X_i}\in\prod_{i\in I}\Cal PX_i$.
Then $\phi$ is an isomorphism between $\nu$ and the product measure
$\lambda=\prod_{i\in I}\nu_i$ on $Z=\prod_{i\in I}\Cal PX_i$.
%out of order query

\proof{{\bf (a)} Because $\familyiI{X_i}$ is a partition of $X$,
$\phi$ is a bijection.

\medskip

{\bf (b)} For $E\in\Sigma^f$ and $S\subseteq X$, set
$f_E(S)=\#(S\cap E)$ if this is finite, $\infty$ otherwise.
As in 495D, take $\Tau$ to be the $\sigma$-algebra of subsets of $X$
generated by
$\{f_E:E\in\Sigma^f\}$;  similarly, take $\Tau_i$ to be the $\sigma$-algebra of
subsets of $X_i$ generated by $\{f_F\restr\Cal PX_i:F\in\Sigma_i^f\}$
for each $i\in I$.

If $E\in\Sigma^f$, $f_E(S)=\sum_{i\in I}f_{E\cap X_i}(S\cap X_i)$
for every $S\subseteq X$.   Consequently $f_E$ is measurable with respect to the
$\sigma$-algebra $\Tau'$ generated by $\bigcup_{i\in I}\{f_F:F\in\Sigma_i^f\}$;
as $\Tau'\subseteq\Tau$ and $E$ is arbitrary, $\Tau'=\Tau$.   Next, the
$\sigma$-algebra $\Tensorhat_{i\in I}\Tau_i\subseteq\Cal PZ$
is generated by the family of subsets of $Z$ of the form

\Centerline{$W_{jFm}=\{\familyiI{S_i}:S_i\subseteq X_i\text{ for every }i\in I,\,
f_F(S_j)=m\}$}

\noindent with $j\in I$, $F\in\Sigma_j^f$ and $m\in\Bbb N$.   Since

\Centerline{$\phi^{-1}[W_{jFm}]=\{S:S\subseteq X$, $f_F(S)=m\}\in\Tau$.}

\noindent $\phi$ is $(\Tau,\Tensorhat_{i\in I}\Tau_i)$-measurable.   On the
other hand, the $\sigma$-algebra of subsets of $\Cal PX$ generated by
$\{\phi^{-1}[W_{jFm}]:j\in I$, $F\in\Sigma_j$, $m\in\Bbb N\}$ is $\Tau'=\Tau$.
So $\phi$ is actually an isomorphism between $(\Cal PX,\Tau)$ and
$(Z,\Tensorhat_{i\in I}\Tau_i)$.

\medskip

{\bf (c)} I repeat the idea of 495C in a more complex form.   This time,
let $Q$ be the set of functions from finite subsets of
$\{E:E\in\Sigma^f$, $\{i:E\cap X_i\ne\emptyset\}$ is finite$\}$ to $\Bbb N$,
and for $q\in Q$ set

\Centerline{$H_q=\{S:S\subseteq X$, $\#(S\cap E)=q(E)$ for every
$E\in\dom q\}$;}

\noindent let $\Cal H$ be

\Centerline{$\{H_q:q\in Q$, $\dom q\subseteq\bigcup_{i\in I}\Sigma_i^f$
is disjoint$\}$,}

\noindent and
$\Tau''\subseteq\Cal P(\Cal PX)$ the Dynkin class generated by $\Cal H$.

If $q\in Q$ then $H_q\in\Tau''$.   \Prf\
$J=\bigcup_{E\in\dom q}\{i:E\cap X_i\ne\emptyset\}$ is finite, and
$F=\bigcup\dom q$ belongs to $\Sigma^f$ and is included in
$\bigcup_{i\in J}X_i$.   Let $\Cal A$ be the set of atoms of the subring
$\Cal E$ generated by $\dom q\cup\{F\cap X_i:i\in J\}$.   If $q'\in Q$ and
$\dom q'=\Cal E$ then $H_{q'}$ is either empty or equal to
$H_{q'\restr\Cal A}\in\Cal H$.   Now

\Centerline{$H_q=\bigcup_{q\subseteq q'\in Q,\dom q'=\Cal E}H_{q'}$}

\noindent is the union of a countable disjoint family in $\Tau''$, so belongs to
$\Tau''$.\ \Qed

Accordingly $\Cal H_1=\{H_q:q\in Q\}\cup\{\emptyset\}$ is a subset of $\Tau''$
closed under finite intersections so $\Tau''$ includes the $\sigma$-algebra
generated by $\Cal H_1$ and is the $\sigma$-algebra generated by $\Cal H$.

\medskip

{\bf (d)} In (c), if $q\in Q$ and $\dom q\subseteq\bigcup_{i\in I}\Sigma_i^f$
is disjoint, then $\lambda\phi[H_q]=\nu H_q$.   \Prf\ If $\emptyset\in\dom q$
and $q(\emptyset)\ne 0$ then $H_q$ is empty and we can stop.   Otherwise,
set $\Cal A=\dom q\setminus\{\emptyset\}$;  then
$H_q=\bigcap_{E\in\Cal A}\{S:S\subseteq X$, $f_E(S)=q(E)\}$, so

\Centerline{$\nu H_q
=\prod_{E\in\Cal A}\Bover{(\gamma\mu E)^{q(E)}}{q(E)!}e^{-\gamma\mu E}$}

\noindent because $\Cal A$ is disjoint.   On the other hand, setting
$\Cal A_i=\{E:E\in\Cal A$, $E\subseteq X_i\}$ and
$H_{qi}=X_i\cap\bigcap_{E\in\Cal A_i}\{S:S\subseteq X_i$, $f_E(S)=q(E)\}$
for $i\in I$, $\phi[H_q]=\prod_{i\in I}H_{qi}$ and

$$\eqalign{\lambda\phi[H_q]&
=\prod_{i\in I}\nu_iH_{qi}
=\prod_{i\in I}\prod_{E\in\Cal A_i}
   \Bover{(\gamma\mu E)^{q(E)}}{q(E)!}e^{-\gamma\mu E}\cr&
=\prod_{E\in\Cal A}\Bover{(\gamma\mu E)^{q(E)}}{q(E)!}e^{-\gamma\mu E}
=\nu H_q.  \text{ \Qed}\cr}$$

\medskip

{\bf (e)} Of course $\{H:H\in\Tau$, $\lambda\phi[H]=\nu H\}$ is a Dynkin class;
as it includes $\Cal H$, it includes $\Tau''$.
But $\Cal H$ contains
$\{S:S\subseteq X$, $\#(S\cap E)=m\}$ whenever $E\in\Sigma^f$ is included in
some $X_i$ and $m\in\Bbb N$.   Because $I$ is countable and $\Tau''$ is a
$\sigma$-algebra, $\{S:S\subseteq X$, $\#(S\cap E)=m\}$ belongs to $\Tau''$
whenever $E\in\Sigma^f$ and $m\in\Bbb N$, just because
$f_E(S)=\sum_{i\in I}f_{E\cap X_i}(S)$ is a countable sum of $\Tau''$-measurable
functions.   But now we see that $\Tau''=\Tau$ and
$\phi$ is an isomorphism between $(\Cal PX,\Tau,\nu\restr\Tau)$ and
$(Z,\lambda,\lambda\restr\Tensorhat_{i\in I}\Tau_i)$.

\medskip

{\bf (f)} Finally, $\nu$ was defined as the completion of $\nu\restr\Tau$,
while $\lambda$ is the completion of $\lambda\restr\Tensorhat_{i\in I}\Tau_i$
because $\nu_i$ is always the completion of $\nu\restr\Tau_i$.   So
$\phi$ is an isomorphism between $(\Cal PX,\nu)$ and $(Z,\lambda)$, as required.
}%end of proof of 495F

\leader{495G}{Proposition}\dvAformerly{4{}95F} 
Let $(X,\Sigma,\mu)$ be a perfect atomless
measure space, and $\gamma>0$.   Then the Poisson point process on $X$
with intensity $\gamma$ is a perfect probability measure.

\proof{ I refer to the construction in 495B-495D.   In (b) of the proof
of 495D, use
the construction set out in the proof of 417A, so that the domain
$\tilde\Lambda$ of $\tilde\lambda$ is precisely the family of sets of
the form
$W\symmdiff A$ where $W$ belongs to the domain $\Lambda$ of the product
measure $\lambda$ and $A$ belongs to the $\sigma$-ideal $\Cal A^*$
generated by $\{A_E:E\in\Sigma^f\}$.   Then $\tilde\lambda$ is perfect.
\Prf\ Let $h:\Omega\to\Bbb R$ be a $\tilde\Lambda$-measurable function
and $W\in\tilde\Lambda$ a set of non-zero measure.   Then there are a
$W'\in\Lambda$ and an $A\in\Cal A^*$ such that
$W\symmdiff W'\subseteq A$ and $h\restr\Omega\setminus A$ is
$\Lambda$-measurable;  let $\sequencen{E_n}$ be a sequence in $\Sigma^f$
such that $A\subseteq\bigcup_{n\in\Bbb N}A_{E_n}$, and
$h_1:\Omega\to\Bbb R$ a $\Lambda$-measurable function agreeing with $h$
on $\Omega\setminus A$.   Set
$\Cal H^*=\bigcup_{n\in\Bbb N}\Cal H_{E_n}$, so that $\Cal H^*$ is
countable.   As in the proof of 495D, set

\Centerline{$F_H=H\setminus(\bigcup\{E_n:n\in\Bbb N$, $H\cap E_n$ is
negligible$\}\cup\bigcup\{H':H'\in\Cal H^*$, $H'\ne H\})$,}

\Centerline{$W_H=\{\pmb{x}:\pmb{x}\in H^{\Bbb N}$ is injective$\}$,}

\noindent for $H\in\Cal H$, so that
$W'_H=W_H\cap F_H^{\Bbb N}$ is $\lambda_H$-conegligible.
Set $\Omega'=\prod_{H\in\Cal H}(\Bbb N\times W'_H)$.   This is disjoint
from every $A_{E_n}$ (as in 495D) and therefore from $A$.   The subspace
measure $\lambda_{\Omega'}$ on $\Omega'$ induced by $\lambda$ is just
the product of the
measures on $\Bbb N\times W'_H$ (254La).   All of these are perfect
(451Jc, 451Dc), so $\lambda_{\Omega'}$ also is perfect (451Jc again).
Now

\Centerline{$\lambda_{\Omega'}(W\cap\Omega')
=\lambda_{\Omega'}(W'\cap\Omega')=\lambda W'=\tilde\lambda W>0$.}

\noindent It follows that there is a compact set
$K\subseteq h_1[W\cap\Omega']$ such that
$\lambda_{\Omega'}(h_1^{-1}[K]\cap\Omega')>0$.   As $h$ and $h_1$ agree
on $\Omega'$, $K\subseteq h[W]$, while

$$\eqalignno{\tilde\lambda h^{-1}[K]
&=\tilde\lambda h_1^{-1}[K]\cr
\displaycause{because $\tilde\lambda A=0$}
&=\lambda h_1^{-1}[K]
=\lambda_{\Omega'}(h^{-1}[K]\cap\Omega')
>0.\cr}$$

\noindent As $W$ and $h$ are arbitrary, $\tilde\lambda$ is perfect.\
\Qed

It follows at once that the image measure $\tilde\lambda\phi^{-1}$
and its restriction to $\Tau$ are perfect (451Ea);  finally, the
completion is perfect, by 451Gc.
}%end of proof of 495G

\leader{495H}{Lemma}\dvAnew{2018} Let $(X,\Sigma,\mu)$ be an atomless
$\sigma$-finite measure space, and $\gamma>0$;  let $\nu$ be the
Poisson point process on $X$ with intensity $\gamma$.
Suppose that $f:X\to\Bbb R$ is a
$\Sigma$-measurable function such that $\mu f^{-1}[\{\alpha\}]=0$ for every
$\alpha\in\Bbb R$.   Then
$\nu\{S:S\subseteq X$, $f\restr S$ is injective$\}=1$.

\proof{ If $E\subseteq X$ has finite measure then
$\{S:S\subseteq X$, $f\restr S\cap E$ is injective$\}$ is
$\nu$-conegligible.   \Prf\ Set $\beta=\mu E$.   For $\alpha\in[-\infty,\infty]$
set
$h(\alpha)=\mu\{x:x\in E$, $f(x)\le\alpha\}$;  then
$h:[-\infty,\infty]\to[0,\beta]$ is
non-decreasing, continuous and surjective.   Take any $m\ge 1$;  then there are
$\alpha_0<\ldots<\alpha_m$ in $[-\infty,\infty]$ such that $\alpha_0=-\infty$,
$\alpha_m=\infty$ and
$h(\alpha_{i+1})-h(\alpha_i)=\Bover{\beta}{m}$ for each $i<m$.   Set
$F_{ni}=\{x:\alpha_i\le f(x)<\alpha_{i+1}\}$ for $i<m$.   Then

$$\eqalign{\nu_*\{S:S\subseteq X,\,f\restr S\cap E\text{ is injective}\}
&\ge\nu\{S:\#(S\cap F_{ni})\le 1\text{ for every }i<m\}\cr&
=\prod_{i<m}\nu\{S:\#(S\cap F_{ni})\le 1\}\cr&
=\prod_{i<m}e^{-\gamma\beta/m}(1+\Bover{\gamma\beta}m)\cr&
=e^{-\gamma\beta}(1+\Bover{\gamma\beta}m)^m
\to 1\cr}$$

\noindent as $m\to\infty$.   So
$\{S:f\restr S\cap E$ is injective$\}$ has inner measure $1$ and is
conegligible.\ \Qed

Now we are supposing that there is a non-decreasing sequence
$\sequencen{E_n}$ of measurable sets of finite measure covering $X$, so that

\Centerline{$\{S:f\restr S$ is injective$\}
=\bigcap_{n\in\Bbb N}\{S:f\restr S\cap E_n$ is injective$\}$}

\noindent has measure
$\lim_{n\to\infty}\nu\{S:f\restr S\cap E_n$ is injective$\}=1$.
}%end of proof of 495H

\leader{495I}{Proposition}\dvAformerly{4{}95Xa}
Let $(X,\Sigma,\mu)$ be an atomless countably separated measure
space\cmmnt{ (definition:  343D)} and $\gamma>0$.   Let $\nu'$ be a complete
probability measure on $\Cal PX$ such that
$\nu'\{S:S\subseteq X$, $S\cap E=\emptyset\}$ is defined and equal to
$e^{-\gamma\mu E}$ whenever $E\in\Sigma$ has finite measure.   Then
$\nu'$ extends the Poisson point process $\nu$ on $X$ with intensity $\gamma$.

\proof{{\bf (a)} Write $\Tau$ for $\dom\nu$, and
for $E\in\Sigma^f$, write $A(E)$ for
$\{S:S\subseteq X$, $S\cap E\ne\emptyset\}\in\Tau$.   If
$E_0,\ldots,E_n\in\Sigma^f$ are disjoint, then $A(E_0),\ldots,A(E_n)$ are
$\nu'$-independent.   \Prf\ For $I\subseteq\{0,\ldots,n\}$,

$$\eqalign{\nu'(\Cal PX\cap\bigcap_{i\in I}A(E_i))&
=1-\nu'\{S:S\cap\bigcup_{i\in I}E_i=\emptyset\}
=\exp(-\gamma\mu(\bigcup_{i\in I}E_i))\cr&
=\prod_{i\in I}\exp(-\gamma\mu E_i)
=\prod_{i\in I}\nu'A(E_i).  \text{ \Qed}\cr}$$

\medskip

{\bf (b)}
If $E_0,\ldots,E_k\in\Sigma^f$ are disjoint and $n_0,\ldots,n_k\in\Bbb N$, then
$\nu'\{S:S\subseteq X$, $\#(S\cap E_j)=n_j$ for every $j\le k\}$
is defined and equal to
$\nu\{S:S\subseteq X$, $\#(S\cap E_j)=n_j$ for every $j\le k\}$.
\Prf\ Let $\sequence{m}{F_m}$ be a sequence
in $\Sigma$ separating the points of $X$.   For each $m\in\Bbb N$ let
$\Cal E_m$ be the subalgebra of $\Cal PX$ generated by
$\{E_j:j\le k\}\cup\{F_i:i\le m\}$,
$\Cal A_m$ the set of atoms of $\Cal E_m$ included in $E=\bigcup_{j\le k}E_j$,

\Centerline{$\frak I_m=\{\Cal I:\Cal I\subseteq\Cal A_m$,
$\#(\{A:A\in\Cal I$, $A\subseteq E_j\})=n_j$ for every $j\le k\}$}

\noindent and

\Centerline{$C_m
=\{S:S\subseteq X$, $\{A:A\in\Cal A_m$, $S\cap A\ne\emptyset\}\in\frak I_m\}$.}

\noindent Then

$$\eqalignno{\nu' C_m&
=\sum_{\Cal I\in\frak I_m}
          \nu'\{S:S\cap A\ne\emptyset\text{ for }A\in \Cal I,
              S\cap E\setminus\bigcup \Cal I=\emptyset\}\cr&
=\sum_{\Cal I\in\frak I_m}
          e^{-\gamma\mu(E\setminus\bigcup\Cal I)}
             \prod_{A\in\Cal I}(1-e^{-\gamma\mu(E\cap A)})
\Displaycause{by (a)}
=\sum_{\Cal I\in\frak I_m}
          \nu\{S:S\cap A\ne\emptyset\text{ for }A\in\Cal I,
              S\cap E\setminus\bigcup\Cal I=\emptyset\}\cr&
=\nu C_m.\cr}$$

\noindent Now, for $S\subseteq X$,

$$\eqalign{\lim_{m\to\infty}\chi C_m(S)
&=1\text{ if }\#(S\cap E_j)=n_j\text{ for every }j\le k,\cr
&=0\text{ otherwise}\cr}$$

\noindent because $\{F_m:m\in\Bbb N\}$ separates the points of $X$.
By Lebesgue's Dominated Convergence Theorem (123C),
$\nu'\{S:\#(S\cap E_j)=n_j$ for every $j\le k\}$ is defined and equal to

\Centerline{$\lim_{m\to\infty}\nu'C_m
=\lim_{m\to\infty}\nu C_m=\nu\{S:\#(S\cap E_j)=n_j$ for every $j\le k\}$.  \Qed}

\medskip

{\bf (c)} Now

\Centerline{$\{H:H\subseteq\Cal PX$, $\nu'H$ and $\nu H$ are defined and
equal$\}$}

\noindent is a Dynkin class including the family $\Cal H$ of 495C, so includes
the $\sigma$-algebra $\Tau$ of subsets of
$\Cal PX$ generated by $\{\{S:\#(S\cap E)=n\}:E\in\Sigma^f$, $n\in\Bbb N\}$.
Because $\nu$ is defined to be the completion of
$\nu\restr\Tau$ and $\nu'$ is complete,
$\nu'H$ is defined and equal to $\nu H$ whenever $H$ is measured by $\nu$.
}%end of proof of 495I

\leader{495J}{Proposition}\dvAformerly{4{}95G}
Let $(X_1,\Sigma_1,\mu_1)$ and
$(X_2,\Sigma_2,\mu_2)$ be
atomless measure spaces, and $f:X_1\to X_2$ an \imp\ function.   Let
$\gamma>0$, and let $\nu_1$, $\nu_2$ be the Poisson point processes on
$X_1$, $X_2$ respectively with intensity $\gamma$.   Then
$S\mapsto f[S]:\Cal PX_1\to\Cal PX_2$ is \imp\ for $\nu_1$ and $\nu_2$;
in particular, $\Cal PA$ has full outer measure for $\nu_2$ whenever
$A\subseteq X_2$ has full outer measure for $\mu_2$.

\proof{ Set $\psi(S)=f[S]$ for $S\subseteq X_1$.

\medskip

{\bf (a)} If $F\in\Sigma_2^f$, then
$\{S:S\subseteq X_1$, $f\restr f^{-1}[F]\cap S$ is not injective$\}$ is
$\nu_1$-negligible.   \Prf\ Let $n\in\Bbb N$.   Set
$\alpha=\bover1{n+1}\mu_2F$.   Because $\mu_2$ is atomless, we can find
a partition of $F$ into sets $F_0,\ldots,F_n$ of measure $\alpha$.
Now

\Centerline{$\{S:f\restr f^{-1}[F]\cap S$
  is not injective$\}
\subseteq\bigcup_{i\le n}\{S:\#(S\cap f^{-1}[F_i])>1\}$}

\noindent has $\nu_1$-outer measure at most

\Centerline{$(n+1)(1-e^{-\gamma\alpha}(1+\gamma\alpha))
\le\Bover12(n+1)\alpha^2\gamma^2=\Bover1{2(n+1)}(\gamma\mu_2F)^2$.}

\noindent As $n$ is arbitrary,
$\{S:f\restr f^{-1}[F]\cap S$ is not injective$\}$ is negligible.\ \Qed

\medskip

{\bf (b)} It follows that, for any $F\in\Sigma_2^f$ and $n\in\Bbb N$,

\Centerline{$\{S:\#(f[S]\cap F)=n\}
\symmdiff\{S:\#(S\cap f^{-1}[F])=n\}$}

\noindent is $\nu_1$-negligible and $\{S:\#(f[S]\cap F)=n\}$ is measured
by $\nu_1$.   So if $\Tau_2$ is the
$\sigma$-algebra of subsets of
$\Cal PX_2$ generated by sets of the form $\{T:\#(F\cap T)=n\}$ for
$F\in\Sigma_2^f$ and $n\in\Bbb N$, then $\nu_1$ measures
$\psi^{-1}[H]$ for every $H\in\Tau_2$.   Next,
if $\familyiI{F_i}$ is a finite disjoint family in $\Sigma_2^f$ and
$n_i\in\Bbb N$ for $i\in I$,

$$\eqalignno{\nu_1\{S:\#(f[S]\cap F_i)=n_i&\text{ for every }i\in I\}\cr
&=\nu_1\{S:\#(S\cap f^{-1}[F_i])=n_i\text{ for every }i\in I\}\cr
&=\prod_{i\in I}\Bover{(\gamma\mu_1f^{-1}[F_i])^n}{n!}
  e^{-\gamma\mu_1f^{-1}[F_i]}\cr
\displaycause{because $\familyiI{f^{-1}[F_i]}$ is a disjoint family in
$\Sigma_1^f$}
&=\prod_{i\in I}\Bover{(\gamma\mu_2F_i)^n}{n!}e^{-\gamma\mu_2F_i}.\cr}$$

\noindent So the image measure $\nu_1\psi^{-1}$ satisfies (i) and (ii)
of 495D, and must agree with $\nu_2$ on $\Tau_2$;  that is, $\psi$ is
\imp\ for $\nu_1$ and $\nu_2\restr\Tau_2$.   As $\nu_1$ is complete,
$\psi$ is \imp\ for $\nu_1$ and $\nu_2$ (234Ba\formerly{2{}35Hc}).

\medskip

{\bf (c)} If $A\subseteq X_2$ has full outer measure, then we can take
$\mu_1$ to be the subspace measure on $X_1=A$ and $f(x)=x$ for $x\in A$.
In this case, $\Cal PA=\psi[\Cal PA]$ must have full outer measure for
$\nu_2$.
}%end of proof of 495J

\leader{495K}{Lemma}\dvAformerly{4{}95H}
Let $(\tilde X,\tilde\Sigma,\tilde\mu)$ be an
atomless $\sigma$-finite measure space, and $\gamma>0$.   Write $\mu_L$
for Lebesgue measure on $[0,1]$, $\mu'$ for the product measure on
$X'=\tilde X\times[0,1]$, and $\lambda'$ for the product measure on
$\Omega'=[0,1]^{\tilde X}$.   Let $\tilde\nu$, $\nuprime$ be the Poisson
point processes on $\tilde X$, $X'$ respectively with intensity $\gamma$.
For $T\subseteq\tilde X$ define $\psi_T:\Omega'\to\Cal PX'$ by setting
$\psi_T(z)=\{(t,z(t)):t\in T\}$ for $z\in\Omega'$;  let $\nuprime_T$ be
the
image measure $\lambda'\psi_T^{-1}$ on $\Cal PX'$.   Then
$\langle\nuprime_T\rangle_{T\subseteq\tilde X}$ is a disintegration of
$\nuprime$ over $\tilde\nu$\cmmnt{ (definition:  452E)}.

\proof{{\bf (a)} Let $E\subseteq X'$ be a measurable set with finite
measure, and write $H_E=\{S:S\cap E\ne\emptyset\}$.   Then
$\nuprime H_E=1-e^{-\gamma\mu'E}\le\gamma\mu'E$;  but
also $\overline{\int}\nuprime_T(H_E)\tilde\nu(dT)\le 2\gamma\mu'E$.
\Prf\ We know that $\int\mu_LE[\{t\}]\tilde\mu(dt)=\mu'E$ (252D).   Let
$Y\subseteq\tilde X$ be a conegligible set such that $E[\{t\}]$ is
measurable for every $t\in Y$ and $t\mapsto\mu_LE[\{t\}]:Y\to[0,1]$ is
measurable.   Set $F_i=\{t:t\in Y$, $2^{-i-1}<\mu_LE[\{t\}]\le 2^{-i}\}$
for each $i\in\Bbb N$;  let $\sequence{i}{F'_i}$ be a sequence of sets
of finite measure with union $\tilde X\setminus\bigcup_{i\in\Bbb N}F_i$.
Let $W$ be the set of those $T\subseteq\tilde X$ such that
$T\cap(\tilde X\setminus Y)$ is empty and $T\cap F_i$, $T\cap F'_i$ are
finite for every $i\in\Bbb N$;  then $W$ is $\tilde\nu$-conegligible.

For any $T\in W$,

$$\eqalign{\psi_T^{-1}[H_E]
&=\{z:\psi_T(z)\cap E\ne\emptyset\}\cr
&=\bigcup_{i\in\Bbb N}\bigcup_{t\in T\cap F_i}\{z:z(t)\in E[\{t\}]\}
   \cup\bigcup_{i\in\Bbb N}\bigcup_{t\in T\cap F'_i}
   \{z:z(t)\in E[\{t\}]\}\cr}$$

\noindent is measured by $\lambda'$ and has measure at most
$\sum_{i=0}^{\infty}2^{-i}\#(T\cap F_i)$, because $\mu_LE[\{t\}]$ has
measure at most $2^{-i}$ if $t\in T\cap F_i$, and is zero if
$t\in T\cap F'_i$.   So

$$\eqalignno{\overline{\int}\nuprime_T(H_E)\tilde\nu(dT)
&=\overline{\int}\lambda'\psi_T^{-1}[H_E]\tilde\nu(dT)
\le\int\sum_{i=0}^{\infty}2^{-i}\#(T\cap F_i)\tilde\nu(dT)\cr
&=\sum_{i=0}^{\infty}2^{-i}\int\#(T\cap F_i)\tilde\nu(dT)
=\sum_{i=0}^{\infty}2^{-i}\gamma\tilde\mu F_i\cr
\displaycause{because $T\mapsto\#(T\cap F_i)$ has expectation
$\gamma\tilde\mu F_i$}
&\le 2\gamma\int\mu_LE[\{t\}]\tilde\nu(dt)
=2\gamma\mu'E.  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} Suppose that $\ofamily{j}{s}{F_j}$,
$\langle C_{ij}\rangle_{i<r,j<s}$ and
$\langle n_{ij}\rangle_{i<r,j<s}$ are such that

\inset{$r$, $s\in\Bbb N$,

$n_{ij}\in\Bbb N$ for $i<r$, $j<s$,

$\ofamily{j}{s}{F_j}$ is a disjoint family of subsets of $\tilde X$ with
finite measure,

for each $j<s$, $\ofamily{i}{r}{C_{ij}}$ is a disjoint family of
measurable subsets of $[0,1]$.}

\noindent Set $E_{ij}=F_j\times C_{ij}$ for $i<r$ and $j<s$, and
$H=\{S:S\subseteq X'$,
$\#(S\cap E_{ij})=n_{ij}$ for every $i<r$, $j<s\}$.
Then $\int\nuprime_T(H)\tilde\nu(dT)=\nuprime H$.

\medskip

\Prf\ {\bf (i)} To begin with, suppose that
$\bigcup_{i<r}C_{ij}=[0,1]$ for every $j$.   Set
$n_j=\sum_{i=0}^{r-1}n_{ij}$ for each $j$, and let $W$
be the set of those $T\subseteq\tilde X$ such that $\#(T\cap F_j)=n_j$
for every $j$.   Then
$\tilde\nu W=\prod_{j=0}^{s-1}
\Bover{(\gamma\tilde\mu F_j)^{n_j}}{n_j!}e^{-\gamma\tilde\mu F_j}$.
If $T\subseteq\tilde X$ and $z\in\psi_T^{-1}[H]$, then for each $j<s$ we
must have

$$\eqalign{\#(T\cap F_j)
&=\#(\{t:t\in T\cap F_j,\,z(t)\in\bigcup_{i<r}C_{ij}\})\cr
&=\sum_{i=0}^{r-1}\#(\{t:t\in T\cap F_j,\,z(t)\in C_{ij}\})\cr
&=\sum_{i=0}^{r-1}\#(\psi_T(z)\cap E_{ij})
=\sum_{i=0}^{r-1}n_{ij}
=n_j.\cr}$$

\noindent Turning this round, we see that if $T\notin W$ then
$\psi_T^{-1}[H]=\emptyset$ and $\nuprime_TH=0$.

If $T\in W$, let $Q$ be the set of all
$q=\langle q(i,j)\rangle_{i<r,j<s}$ such that $\ofamily{i}{r}{q(i,j)}$
is a disjoint family of subsets of $T\cap F_j$ for each $j$ and
$\#(q(i,j))=n_{ij}$ for all $i$ and $j$.   Then
$\#(Q)=\prod_{j=0}^{s-1}\Bover{n_j!}{\prod_{i=0}^{r-1}n_{ij}!}$.
Accordingly

$$\eqalignno{\nuprime_TH
&=\lambda'\{z:\psi_T(z)\in H\}\cr
&=\lambda'\{z:\#(\{t:t\in T\cap F_j,\,z(t)\in C_{ij}\})=n_{ij}
   \text{ for all }i,\,j\}\cr
&=\sum_{q\in Q}\lambda\{z:z(t)\in C_{ij}
   \text{ whenever }i<r,\,j<s\text{ and }t\in q(i,j)\}\cr
&=\sum_{q\in Q}\prod_{\Atop{i<r,j<s}{t\in q(i,j)}}\mu_LC_{ij}
=\sum_{q\in Q}\prod_{i<r,j<s}(\mu_LC_{ij})^{n_{ij}}
=\prod_{j=0}^{s-1}\bigl(n_j!\prod_{i=0}^{r-1}
   \bover{(\mu_LC_{ij})^{n_{ij}}}{n_{ij}!}\bigr).\cr}$$

\noindent It follows that

$$\eqalignno{\int\nuprime_T(H)\tilde\nu(dT)
&=\tilde\nu W\cdot\prod_{j=0}^{s-1}\bigl(n_j!
  \prod_{i=0}^{r-1}\bover{(\mu_LC_{ij})^{n_{ij}}}{n_{ij}!}\bigr)\cr
&=\prod_{j=0}^{s-1}
  \bover{(\gamma\tilde\mu F_j)^{n_j}}{n_j!}e^{-\gamma\tilde\mu F_j}
  \cdot\prod_{j=0}^{s-1}\bigl(n_j!
  \prod_{i=0}^{r-1}\bover{(\mu_LC_{ij})^{n_{ij}}}{n_{ij}!}\bigr)\cr
&=\prod_{j=0}^{s-1}\bigl(
  (\gamma\tilde\mu F_j)^{n_j}e^{-\gamma\tilde\mu F_j}
  \prod_{i=0}^{r-1}\bover{(\mu_LC_{ij})^{n_{ij}}}{n_{ij}!}\bigr)\cr
&=\prod_{j=0}^{s-1}\prod_{i=0}^{r-1}e^{-\gamma\mu'E_{ij}}
  \bover{(\gamma\mu'E_{ij})^{n_{ij}}}{n_{ij}!}
=\nuprime H,\cr}$$

\noindent as required.

\medskip

\quad{\bf (ii)} For the general case, set
$C_{rj}=[0,1]\setminus\bigcup_{i<r}C_{ij}$,
$E_{rj}=F_j\times C_{rj}$ for each $j<s$.
For $\sigma\in\Bbb N^{(r+1)\times s}$, set

\Centerline{$H_{\sigma}=\{S:S\subseteq X'$, $\#(S\cap E_{ij})=\sigma(i,j)$
for every $i\le r$ and $j<s\}$.}

\noindent By (i), we have
$\nuprime H_{\sigma}=\int\nuprime_T(H_{\sigma})\tilde\nu(dT)$ for every
$\sigma\in\Bbb N^{(r+1)\times s}$.

Set

\Centerline{$J=\{\sigma:\sigma\in\Bbb N^{(r+1)\times s}$,
$\sigma(i,j)=n_{ij}$ for $i<r$, $j<s\}$,
\quad$K=\Bbb N^{(r+1)\times s}\setminus J$,}

\Centerline{$H'_1=\bigcup_{\sigma\in J}H_{\sigma}$,
\quad$H'_2=\bigcup_{\sigma\in K}H_{\sigma}$.}

\noindent Then $H'_1\subseteq H$, $H'_2\cap H=\emptyset$ and

\Centerline{$H'_1\cup H'_2
=\{S:S\cap E_{ij}$ is finite for all $i\le r$, $j<s\}$}

\noindent is $\nuprime$-conegligible.   Accordingly we have

$$\eqalignno{\underline{\int}(\nuprime_T)_*(H)\tilde\nu(dT)
&\ge\underline{\int}(\nuprime_T)_*(H'_1)\tilde\nu(dT)
\ge\sum_{\sigma\in J}\int\nuprime_T(H_{\sigma})\tilde\nu(dT)\cr
&=\sum_{\sigma\in J}\nuprime H_{\sigma}
=1-\sum_{\sigma\in K}\nuprime H_{\sigma}\cr
\displaycause{because $\bigcup_{\sigma\in J\cup K}H_{\sigma}$ is
$\nuprime$-conegligible}
&=1-\sum_{\sigma\in K}\int\nuprime_T(H_{\sigma})\tilde\nu(dT)
=\int 1-\sum_{\sigma\in K}\nuprime_T(H_{\sigma})\,\tilde\nu(dT)\cr
&=\int\nuprime_T(\Cal PX'\setminus H'_2)\tilde\nu(dT)
\ge\overline{\int}(\nuprime_T)^*(H)\tilde\nu(dT).\cr}$$

\noindent But this means, first, that $(\nuprime_T)_*(H)=(\nuprime_T)^*(H)$
for $\tilde\nu$-almost every $T$;  since $\nuprime_T$, being an image of
the complete measure $\lambda'$, is always complete, $\nuprime_T(H)$ is
defined for $\tilde\nu$-almost every $T$.   Finally,

\Centerline{$\nuprime H=\nuprime H'_1=\sum_{\sigma\in J}\nuprime H_{\sigma}
=\int\nuprime_T(H)\tilde\nu(dT)$,}

\noindent as required.\ \Qed

\medskip

{\bf (c)} Suppose that $\ofamily{i}{r}{E_i}$ is a disjoint family in
$\tilde\Sigma\otimes\Sigma_L$ such that all the projections of the $E_i$
onto $\tilde X$ have finite measure, and $n_i\in\Bbb N$ for each $i<r$.
Set $H=\{S:S\subseteq X'$,
$\#(S\cap E_i)=n_i$ for every $i<r\}$.
Then $\int\nuprime_T(H)\tilde\nu(dT)=\nuprime H$.

\Prf\ Let $\Cal E$ be a finite subalgebra of $\tilde\Sigma$ such that
every $E_i$ belongs to
$\Cal E\otimes\Sigma_L$, and let $\ofamily{j}{s}{F_j}$ enumerate the
atoms of $\Cal E$ of finite measure;  extend this to an enumeration
$\ofamily{j}{s'}{F_j}$ of all the atoms of $\Cal E$.   Then we can
express each $E_i$ as $\bigcup_{j<s'}F_j\times C_{ij}$ where each
$C_{ij}\in\Sigma_L$;  but as the projection of $E_i$ has finite measure,
$C_{ij}$ must be empty for every $j\ge s$, so
$E_i=\bigcup_{j<s}F_j\times C_{ij}$.   Let $Q$ be the set of all
$q\in\BbbN^{r\times s}$ such that $\sum_{j=0}^{s-1}q(i,j)=n_i$ for every
$i<r$.   For $q\in Q$ set

\Centerline{$H_q=\{S:S\subseteq X'$,
$\#(S\cap(F_j\times C_{ij}))=q(i,j)$ for every $i<r$, $j<s\}$.}

\noindent Then $\family{q}{Q}{H_q}$ is disjoint and has union $H$, so

\Centerline{$\biggerint\nuprime_T(H)\tilde\nu(dT)
=$$\sum_{q\in Q}$$\biggerint\nuprime_T(H_q)\tilde\nu(dT)
=$$\sum_{q\in Q}\nuprime H_q
=\nuprime H$,}

\noindent using (b) for the middle equality.\ \Qed

\medskip

{\bf (d)} Now let $\ofamily{i}{r}{E_i}$ be a finite disjoint family of
subsets of $X'$ of finite measure, and $\ofamily{i}{r}{n_i}$ a family in
$\Bbb N$.   Set $H=\{S:S\subseteq X'$, $\#(S\cap E_i)=n_i$ for every
$i<r\}$.   Then $\int\nuprime_T(H)\tilde\nu(dT)$ is defined and equal to
$\nuprime H$.

\Prf\ Let $\epsilon>0$.   For each $i<r$ we can find an
$E'_i\in\tilde\Sigma\otimes\Sigma_L$ such that
$\mu'(E_i\symmdiff E'_i)\le\epsilon$ (251Ie).   Discarding a negligible
set from $E'_i$ if
necessary, we may suppose that the projection of $E'_i$ on $\tilde X$
has finite measure.   Set
$\hat E_i=E'_i\setminus\bigcup_{k<i}E'_k$ for each $i$, so that
$\ofamily{i}{r}{\hat E_i}$ is a disjoint family in
$\tilde\Sigma\otimes\Sigma_L$, and the projections of the $\hat E_i$ are
still of finite measure.   Set
$\hat H=\{S:S\subseteq X'$, $\#(S\cap\hat E_i)=n_i$ for every $i<r\}$.
Then (c) tells us that
$\int\nuprime_T(\hat H)\tilde\nu(dT)=\nuprime\hat H$.

Set $E=\bigcup_{i<r}(E_i\symmdiff E'_i)$.   Then $\mu'E\le r\epsilon$,
while $E$ includes $E_i\symmdiff\hat E_i$ for every $i$, so

\Centerline{$\hat H\setminus H_E\subseteq H\subseteq\hat H\cup H_E$,}

\noindent where $H_E=\{S:S\cap E\ne\emptyset\}$ as in (a).   Accordingly

$$\eqalignno{\nuprime H-3r\gamma\epsilon
&\le\nuprime\hat H-2r\gamma\epsilon\cr
\displaycause{by (a)}
&=\int\nuprime_T(\hat H)\tilde\nu(dT)-2r\gamma\epsilon\cr
\displaycause{by (c)}
&\le\int\nuprime_T(\hat H)-\nuprime_T(H_E)\tilde\nu(dT)\cr
\displaycause{by the other part of (a)}
&\le\underline{\int}(\nuprime_T)_*(H)\tilde\nu(dT)
\le\overline{\int}(\nuprime_T)^*(H)\tilde\nu(dT)\cr
&\le\int\nuprime_T(\hat H)+\nuprime_T(H_E)\tilde\nu(dT)
\le\nuprime\hat H+2r\gamma\epsilon
\le\nuprime H+3r\gamma\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$
\nuprime H=\underline{\intop}(\nuprime_T)_*(H)\tilde\nu(dT)
=\overline{\intop}(\nuprime_T)^*(H)\tilde\nu(dT)$.}

\noindent As in (c-ii) above, it follows that
$\int\nuprime_T(H)\tilde\nu(dT)$ is defined and equal to
$\nuprime H$.\ \Qed

\medskip

{\bf (e)} So if we write $\Cal H$ for the family of subsets $H$ of
$\Cal PX'$ such that $\int\nuprime_T(H)\tilde\nu(dT)$ is defined and
equal
to $\nuprime H$, and $\Cal H_0$ for the family of sets of the form
$H=\{S:S\subseteq X'$, $\#(S\cap E_i)=n_i$ for every $i<r\}$
where $\ofamily{i}{r}{E_i}$ is a disjoint family of sets of finite
measure and $n_i\in\Bbb N$ for $i<r$, we have $\Cal H\supseteq\Cal H_0$.
But $\Cal H$ is a Dynkin class, so includes the $\sigma$-algebra $\Tau'$
generated by $\Cal H_0$, by 495C.   Since every $\nuprime$-negligible
set is
included in a $\nuprime$-negligible member of $\Tau'$, $\Cal H$ contains
every $\nuprime$-negligible set, and therefore every set measured by
$\nuprime$;  which is what we need to know.
}%end of proof of 495K

\leader{495L}{Theorem}\dvAformerly{4{}95I} 
Let $(X,\Sigma,\mu)$ and
$(\tilde X,\tilde\Sigma,\tilde\mu)$ be atomless $\sigma$-finite measure
spaces and $\gamma>0$.   Let $\nu$, $\tilde\nu$ be the Poisson point
processes on $X$, $\tilde X$ respectively with intensity $\gamma$.
Suppose that $f:X\to\tilde X$ is \imp\ and that
$\family{t}{\tilde X}{\mu_t}$ is a disintegration of $\mu$ over
$\tilde\mu$ consistent with $f$\cmmnt{ (definition:  452E)} such that
every $\mu_t$ is a probability measure.   Write $\lambda$ for the
product measure $\prod_{t\in\tilde X}\mu_t$ on
$\Omega=X^{\tilde X}$, and for $T\subseteq\tilde X$ define
$\phi_T:\Omega\to\Cal PX$ by setting $\phi_T(z)=z[T]$ for $z\in\Omega$;
let $\nu_T$ be the image measure $\lambda\phi_T^{-1}$ on $\Cal PX$.
Then $\langle\nu_T\rangle_{T\subseteq\tilde X}$ is a disintegration of
$\nu$ over $\tilde\nu$.   Moreover

(i) setting $\tilde f(S)=f[S]$ for $S\subseteq X$,
$\langle\nu_T\rangle_{T\subseteq\tilde X}$ is consistent with
$\tilde f:\Cal PX\to\Cal P\tilde X$;

(ii) if $\family{t}{\tilde X}{\mu_t}$ is strongly consistent with $f$, then
$\langle\nu_T\rangle_{T\subseteq\tilde X}$ is strongly consistent with
$\tilde f$.

\proof{{\bf (a)} For $T\subseteq\tilde X$ let $V_T$ be the set of those
$z\in\Omega$ such that $fz\restr T$ is injective.   We need to know that

\Centerline{$W=\{T:T\subseteq\tilde X$, $T$ is countable, $V_T$ is
$\lambda$-conegligible$\}$}

\noindent is $\tilde\nu$-conegligible.   \Prf\ Write
$\tilde\Sigma^f=\{F:F\in\tilde\Sigma$, $\tilde\mu F<\infty\}$.   Because
$\tilde\mu$ is atomless and $\sigma$-finite, there is a countable
subalgebra $\Cal E$ of $\tilde\Sigma$ such that for every
$\epsilon>0$ there is a cover of $\tilde X$ by members of $\Cal E$ of
measure at most $\epsilon$.   Set

\Centerline{$Y=\{t:t\in\tilde X$, $\mu_tf^{-1}[F]=(\chi F)(t)$ for
every $F\in\Cal E\}$,}

\noindent so that $Y$ is $\tilde\mu$-conegligible and $\Cal PY$ is
$\tilde\nu$-conegligible.   For $F\in\Cal E$, let $W_F$ be the set of
those $T\subseteq Y$ such that for every
$t\in T\cap F$ there is an $F'\in\Cal E$ such that $T\cap F'=\{t\}$.
Now, given $F\in\Cal E\cap\tilde\Sigma^f$ and $\epsilon>0$, there is a
partition $\familyiI{F_i}$ of $F$ into members of $\Cal E$ of measure
at most $\epsilon$.   Then

$$\eqalign{\tilde\nu^*(\Cal PY\setminus W_F)&
\le\tilde\nu\{T:\#(T\cap F_i)>1\text{ for some }i\in I\}\cr&
\le\sum_{i\in I}1-e^{-\gamma\tilde\mu F_i}(1+\gamma\tilde\mu F_i)
\le\sum_{i\in I}\Bover12(\gamma\tilde\mu F_i)^2\cr&
\le\Bover12\epsilon\gamma^2\sum_{i\in I}\tilde\mu F_i
=\Bover12\epsilon\gamma^2\tilde\mu F.\cr}$$

\noindent As $\epsilon$ is arbitary, $W_F$ is $\tilde\nu$-conegligible;
accordingly $W'=\bigcap\{W_F:F\in\Cal E\cap\tilde\Sigma^f\}$ is
$\tilde\nu$-conegligible.

Now suppose that $T\in W'$.   Because $\tilde X$ is covered by
$\Cal E\cap\tilde\Sigma^f$, we see that for every $t\in T$ there is an
$F\in\Cal E\cap\tilde\Sigma^f$ containing $t$, and now there is an
$F'\in\Cal E$ such that $T\cap F'=\{t\}$.   In particular, $T$ is
countable, so

\Centerline{$U_T
=\{z:z(t)\in f^{-1}[F]$ whenever $F\in\Cal E$ and $t\in T\cap F\}$}

\noindent is $\lambda$-conegligible.   Take $z\in U_T$.   If $t$, $t'$
are distinct points of $T$, there is an $F\in\Cal E$ containing $t$ but
not $t'$, and now $F$ contains $f(z(t))$ but not $f(z(t'))$.   So
$fz\restr T$ is injective.   Thus $U_T\subseteq V_T$ and $V_T$ is
$\lambda$-conegligible.   This is true for every $T\in W'$, so
$W\supseteq W'$ is $\tilde\nu$-conegligible.\ \Qed

\medskip

{\bf (b)} Suppose that $\ofamily{i}{r}{E_i}$ is a disjoint family of
subsets of $X$ with finite measure, and $n_i\in\Bbb N$ for $i<r$.   Set
$H=\{S:S\subseteq X$, $\#(S\cap E_i)=n_i$ for every $i<r\}$.   Then
$\int\nu_T(H)\tilde\nu(dT)$ is defined and equal to $\nu H$.   \Prf\
As in 495K, set $X'=\tilde X\times[0,1]$ with the product measure
$\mu'$, and write $\lambda'$ for the product measure on
$\Omega'=[0,1]^{\tilde X}$.   Let $\nuprime$ be the Poisson point
process on
$X'$ with intensity $\gamma$.   For $T\subseteq\tilde X$ define
$\psi_T:\Omega'\to\Cal PX'$ by setting $\psi_T(z)=\{(t,z(t)):t\in T\}$
for $z\in\Omega'$ and let $\nuprime_T$ be the image measure
$\lambda'\psi_T^{-1}$ on $\Cal PX'$.   By 495K,
$\langle\nuprime_T\rangle_{T\subseteq\tilde X}$ is a disintegration of
$\nuprime$ over $\tilde\nu$.

For each $i<r$, $\int\mu_t(E_i)\tilde\mu(dt)=\mu E_i$;  set
$Y_1=\{t:\mu_tE_i$ is defined for every $i<r\}$, so that
$Y_1\subseteq\tilde X$ is $\tilde\nu$-conegligible.   Set
$g_i(t)=\sum_{j<i}\mu_tE_j$ for $t\in Y_1$ and $i\le r$, and

\Centerline{$E'_i
=\{(t,\alpha):t\in Y_1$, $g_i(t)\le\alpha<g_{i+1}(t)\}$}

\noindent for $i<r$.   Then
$\mu'E'_i=\int g_{i+1}-g_i\,d\tilde\nu=\mu E_i$ for each $i$, by 252N.

Set

\Centerline{$H'
=\{S:S\subseteq X'$, $\#(S'\cap E_i')=n_i$ for every $i<r\}$,}

\Centerline{$W_1
=\{T:T\in W$, $T\subseteq Y_1$, $\nuprime_TH'$ is defined$\}$,}

\noindent so $H'$ is measured by $\nu'$ and
$W_1$ is $\tilde\nu$-conegligible.   Let $T$ be any
member of $W_1$.   Let $Q$ be the set of partitions
$q=\langle q(i)\rangle_{i\le r}$ of $T$ such that $\#(q(i))=n_i$ for
every $i<r$;  because $T$ is countable, so is $Q$.   Set
$E_r=X\setminus\bigcup_{i<r}E_i$ and
$E'_r=X'\setminus\bigcup_{i<r}E'_i$.   Then

\Centerline{$\mu_tE_r=1-\sum_{i=0}^{r-1}\mu_tE_i
=1-g_r(t)=\mu_LE'_r[\{t\}]$}

\noindent for every $t\in Y_1$.   Now

$$\eqalignno{\nuprime_TH'
&=\lambda'\{z:z\in\Omega',\,\psi_T(z)\in H'\}\cr
&=\lambda'\{z:z\in\Omega',\,\#(\{t:t\in T,\,z(t)\in E'_i[\{t\}]\})=n_i
       \text{ for every }i<r\}\cr
&=\sum_{q\in Q}\lambda'\{z:z\in\Omega',\,z(t)\in E'_i[\{t\}]
       \text{ whenever }i\le r\text{ and }t\in q(i)\}\cr
&=\sum_{q\in Q}\prod_{i=0}^r\prod_{t\in q(i)}\mu_LE'_i[\{t\}]
=\sum_{q\in Q}\prod_{i=0}^r\prod_{t\in q(i)}\mu_tE_i\cr
&=\sum_{q\in Q}\lambda\{z:z\in\Omega,\,z(t)\in E_i
       \text{ whenever }i\le r\text{ and }t\in q(i)\}\cr
&=\lambda\{z:z\in\Omega,\,\#(\{t:t\in T,\,z(t)\in E_i\})=n_i
       \text{ for every }i<r\}\cr
&=\lambda\{z:z\in V_T,\,\#(\{t:t\in T,\,z(t)\in E_i\})=n_i
       \text{ for every }i<r\}\cr
&=\lambda\{z:z\in V_T,\,\#(z[T]\cap E_i)=n_i
       \text{ for every }i<r\}\cr
&=\lambda\{z:z\in V_T,\,\phi_T(z)\in H\}
=\nu_TH.\cr}$$

\noindent Since this is true for $\tilde\nu$-almost every $T$,

$$\eqalignno{\int\nu_T(H)\tilde\nu(dT)
&=\int\nuprime_T(H')\tilde\nu(dT)
=\nuprime H'\cr
&=\prod_{i<r}\Bover{(\gamma\mu'E'_i)^{n_i}}{n_i!}e^{-\gamma\mu'E'_i}
=\prod_{i<r}\Bover{(\gamma\mu E_i)^{n_i}}{n_i!}e^{-\gamma\mu E_i}
=\nu H.  \text{ \Qed}\cr}$$

\noindent Now, just as in part (e) of the proof of 495K, 495C tells us
that $\int\nu_T(H)\tilde\nu(dT)=\nu H$ whenever $\nu$ measures $H$, so
that $\langle\nu_T\rangle_{T\subseteq\tilde X}$ is a disintegration of
$\nu$ over $\tilde\nu$.

\medskip

{\bf (c)} Let $\ofamily{i}{r}{F_i}$ be a disjoint family in
$\tilde\Sigma^f$, and take $n_i\in\Bbb N$ for $i<r$.   Set

\Centerline{$F_r=\tilde X\setminus\bigcup_{i<r}F_i$,}

\Centerline{$Y_2=\{t:t\in\tilde X$,
$\mu_tf^{-1}[F_i]=(\chi F_i)(t)$ for every $i\le r\}$,}

\Centerline{$W_2=\{T:T\in W$, $T\subseteq Y_2\}$,}

\noindent so that $Y_2$ is $\tilde\mu$-conegligible and $W_2$ is
$\tilde\nu$-conegligible.   Set

\Centerline{$\tilde H
=\{T:T\in W_2$, $\#(T\cap F_i)=n_i$ for every $i<r\}$,}

\Centerline{$H=\tilde f^{-1}[\tilde H]
=\{S:S\subseteq X$, $f[S]\in\tilde H\}$.}

\noindent Then $\nu_TH=\chi\tilde H(T)$ for every
$T\in\tilde H$.   \Prf\ For $i\le r$ and $t\in T\cap F_i$, we
have

\Centerline{$\lambda\{z:z(t)\in f^{-1}[F_i]\}=\mu_tf^{-1}[F_i]=1$}

\noindent because $T\subseteq Y_2$.   So

\Centerline{$V=\{z:z\in V_T$,
$f(z(t))\in F_i$ whenever $i\le r$ and $t\in T\cap F_i\}$}

\noindent is $\lambda$-conegligible.   But if $z\in V$ then $fz\restr T$
is injective, so

\Centerline{$\#(f[z[T]]\cap F_i)=\#(T\cap(fz)^{-1}[F_i])
=\#(T\cap F_i)$}

\noindent for every $i<r$, and $z[T]\in H$ iff $T\in\tilde H$.
Thus

$$\eqalign{\nu_TH
=\lambda\{z:z[T]\in H\}
&=\lambda V=1\text{ if }T\in\tilde H,\cr
&=\lambda(\Omega\setminus V)=0\text{ otherwise.  \Qed}\cr}$$

Setting

\Centerline{$\Cal H=\{\tilde H:\tilde H\subseteq\tilde X$,
$\nu_T\tilde f^{-1}[\tilde H]=\chi\tilde H(T)$ for $\tilde\nu$-almost
every $T\in\tilde H\}$,}

\noindent it is easy to check that $\Cal H$ is a Dynkin class containing
all sets of the form
$\{T:T\subseteq\tilde X$, $\#(T\cap F_i)=n_i$ for every $i<r\}$ where
$\ofamily{i}{r}{F_i}$ is a disjoint family in $\tilde\Sigma^f$, and
therefore including the $\sigma$-algebra generated by such sets, by
495C.   But as $\Cal H$ also contains any subset of a negligible set
belonging to $\Cal H$ (remember that all the $\nu_T$ are complete
probability measures, like $\lambda$), it includes the domain of
$\tilde\nu$, and $\langle\nu_T\rangle_{T\subseteq\tilde X}$ is
consistent with $\tilde f$.

\medskip

{\bf (d)} Now suppose that $\family{t}{\tilde X}{\mu_t}$ is strongly
consistent with $f$.   Set $Y_3=\{t:\mu_tf^{-1}[\{t\}]=1\}$ and
$W_3=\{T:T\in W$, $T\subseteq Y_3\}$.   Then
$\nu_T\tilde f^{-1}[\{T\}]=1$ for every $T\in W_3$.   \Prf\ Set
$V'_T=\{z:z\in\Omega$, $f(z(t))=t$ for every $t\in T\}$.   For
each $t\in T$,

\Centerline{$\lambda\{z:f(z(t))=t\}=\mu_tf^{-1}[\{t\}]=1$,}

\noindent because $T\subseteq W_3$.   As $T$ is countable, $V'_T$ is
$\lambda$-conegligible.   But now

\Centerline{$\nu_T\tilde f^{-1}[\{T\}]=\lambda\{z:f[z[T]]=T\}
\ge\lambda V'_T=1$.  \Qed}

As $W_3$ is $\tilde\nu$-conegligible,
$\langle\nu_T\rangle_{T\subseteq\tilde X}$ is strongly consistent with
$\tilde f$.
}%end of proof of 495L

\vleader{72pt}{495M}{Proposition}\dvAformerly{4{}95J}
Let $(\frak A,\bar\mu)$ be a measure algebra,
and $\gamma>0$.   Then there are a probability algebra
$(\frak B,\bar\lambda)$ and a function $\theta:\frak A\to\frak B$ such
that

(i) $\theta(\sup A)=\sup\theta[A]$ for every non-empty
$A\subseteq\frak A$ such that $\sup A$ is defined in $\frak A$;

(ii) $\bar\lambda\theta(a)=1-e^{-\gamma\bar\mu a}$ for every
$a\in\frak A$, interpreting $e^{-\infty}$ as $0$;

(iii) whenever $\familyiI{a_i}$ is a disjoint family in
$\frak A$ and $\frak C_i$ is the closed subalgebra of $\frak B$
generated by $\{\theta(a):a\Bsubseteq a_i\}$ for
each $i$, then $\familyiI{\frak C_i}$ is stochastically independent.

\proof{{\bf (a)} We may suppose that $(\frak A,\bar\mu)$ is the measure
algebra of a measure space $(X,\Sigma,\mu)$ (321J).   Set
$\Sigma^f=\{E:E\in\Sigma$, $\mu E<\infty\}$ and
$\frak A^f=\{a:a\in\frak A$, $\bar\mu a<\infty\}$.   Let
$(\Omega,\Lambda,\lambda)$ and $\family{E}{\Sigma^f}{g_E}$ be as in
495B, and take $(\frak B,\bar\lambda)$ to be the measure algebra of
$(\Omega,\Lambda,\lambda)$.   Note that if $E$, $F\in\Sigma^f$ and
$\mu(E\symmdiff F)=0$, then $g_{E\setminus F}$ and $g_{F\setminus E}$
have Poisson distributions with expectation $0$, so are zero almost
everywhere, while $g_E\eae g_{E\cap F}+g_{E\setminus F}$ and
$g_F\eae g_{E\cap F}+g_{F\setminus E}$;  so that $g_E\eae g_F$.   This
means that we can define
$\theta:\frak A^f\to\frak B$ by setting
$\theta(E^{\ssbullet})=\{\omega:g_E(\omega)\ne 0\}^{\ssbullet}$ whenever
$E\in\Sigma^f$, and we shall have
$\bar\lambda(\theta a)=1-e^{-\gamma\bar\mu a}$ because $g_E$ has a
Poisson distribution with expectation $\bar\mu a$ whenever
$E\in\Sigma^f$ and $E^{\ssbullet}=a$.   For $a\in\frak A\setminus\frak
A^f$ set $\theta(a)=1_{\frak B}$.

\medskip

{\bf (b)} If $a$, $b\in\frak A^f$ are disjoint, they can be represented
as $E^{\ssbullet}$, $F^{\ssbullet}$ where $E$, $F\in\Sigma^f$ are
disjoint.   In this case, $g_{E\cup F}\eae g_E+g_F$, so
$\theta(a\Bcup b)=\theta(a)\Bcup\theta(b)$.   Of course the same is true
if $a$, $b\in\frak A$ are disjoint and either has infinite measure.   It
follows at once that for any $a$, $b\in\frak A$,

\Centerline{$\theta(a\Bcup b)
=\theta(a\Bsetminus b)\Bcup\theta(a\Bcap b)\Bcup\theta(b\Bsetminus a)
=\theta(a)\Bcup\theta(b)$.}

\noindent Consequently $\theta(\sup A)=\sup\theta[A]$ for any finite set
$A\Bsubseteq\frak A$.   If $A\Bsubseteq\frak A$ is an infinite set with
supremum $a^*$, then $A'=\{\sup B:B\in[A]^{<\omega}\}$ is an
upwards-directed set with supremum $a^*$, so there is a non-decreasing
sequence $\sequencen{a_n}$ in $A'$ such that
$\lim_{n\to\infty}\bar\mu a_n=\bar\mu a^*$ (321D).   In this case,
$b^*=\sup_{n\in\Bbb N}\theta(a_n)$ is defined in $\frak B$ and

\Centerline{$\bar\lambda b^*=\lim_{n\to\infty}\bar\lambda\theta(a_n)
=\lim_{n\to\infty}1-e^{-\gamma\bar\mu a_n}
=1-e^{-\gamma\bar\mu a^*}
=\bar\lambda\theta(a^*)$.}

\noindent So $b^*=\theta(a^*)$;  since $\theta(a^*)$ is certainly an
upper bound of $\theta[A']$, it must actually be the supremum of
$\theta[A']$ and therefore (because $\theta$ preserves finite suprema)
of $\theta[A]$.

\medskip

{\bf (c)} Thus $\theta$ satisfies (i) and (ii).   As for (iii), note
first that if
$\familyiI{a_i}$ is a finite disjoint family in $\frak A$, then
$\bar\lambda(\inf_{i\in I}\theta(a_i))
=\prod_{i\in I}\bar\lambda\theta(a_i)$.  \Prf\ Set
$J=\{i:i\in I$, $\bar\mu a_i<\infty\}$.   For $i\in J$, represent $a_i$
as $E_i^{\ssbullet}$ where $\family{i}{J}{E_i}$ is a disjoint family in
$\Sigma^f$.   Then $\family{i}{J}{g_{E_i}}$ is independent, so

$$\eqalign{\bar\lambda(\inf_{i\in I}\theta(a_i))
&=\bar\lambda(\inf_{i\in J}\theta(a_i))
=\lambda(\Omega\cap\bigcap_{i\in J}\{\omega:g_{E_i}(\omega)=0\})\cr
&=\prod_{i\in J}\lambda\{\omega:g_{E_i}(\omega)=0\}
=\prod_{i\in J}\bar\lambda\theta(a_i)
=\prod_{i\in I}\bar\lambda\theta(a_i).  \text{ \Qed}\cr}$$

Now suppose that $\familyiI{a_i}$ is a finite disjoint family in
$\frak A$ and that $\frak D_i$ is the subalgebra of $\frak B$ generated
by $D_i=\{\theta(a):a\Bsubseteq a_i\}$ for each $i$.   We know that each
$D_i$ is closed under $\Bcup$ (by (i)) and that
$\bar\lambda(\inf_{i\in J}d_i)=\prod_{i\in J}\bar\lambda d_i$ whenever
$J\subseteq I$ and $d_i\in D_i$ for each $i\in J$, that is, that
$\familyiI{d_i}$ is stochastically independent whenever $d_i\in D_i$ for
each $i$.   Setting $D'_i=\{1\Bsetminus d:d\in D_i\}\cup\{0\}$, we see
that $D'_i$ is closed under $\Bcap$ and that $\familyiI{d_i}$ is
stochastically independent whenever $d_i\in D'_i$ for each $i$ (as in
272F).   An induction on $\#(J)$, using 313Ga for the inductive step,
shows that if $J\subseteq I$, $d_i\in\frak D_i$ for $i\in J$, and
$d_i\in D'_i$ for $i\in I\setminus J$, then
$\bar\lambda(\inf_{i\in I}d_i)=\prod_{i\in I}\bar\lambda d_i$.   At the
end of the induction, we see that
$\bar\lambda(\inf_{i\in I}d_i)=\prod_{i\in I}\bar\lambda d_i$ whenever
$d_i\in\frak D_i$ for each $i$, and therefore whenever $d_i$ belongs to
the topological closure of $\frak D_i$ for each $i$, where $\frak B$ is
given its measure-algebra topology (\S323).

Finally, suppose that $\familyiI{a_i}$ is any disjoint family in
$\frak A$, and $\frak C_i$ is the closed subalgebra of $\frak B$
generated by $D_i=\{\theta(a):a\Bsubseteq a_i\}$ for each $i$.   Take a
finite set $J\subseteq I$ and $c_i\in\frak C_i$ for each $i\in J$.   By
323J, $\frak C_i$ is the topological closure of the subalgebra
$\frak D_i$ of $\frak B$ generated by $\{\theta(a):a\Bsubseteq a_i\}$;
so $\bar\lambda(\inf_{i\in J}c_i)=\prod_{i\in J}\bar\lambda c_i$.   As
$\family{i}{J}{c_i}$ is arbitrary, $\familyiI{\frak C_i}$ is
independent.
}%end of proof of 495M

\leader{495N}{Proposition}\dvAformerly{4{}95K} 
Let $U$ be any $L$-space.
Then there are a probability space $(\Omega,\Lambda,\lambda)$ and a
positive linear operator $T:U\to L^1(\lambda)$ such that
$\|Tu\|_1=\|u\|_1$ whenever $u\in L^1(\mu)^+$ and $\familyiI{Tu_i}$ is
stochastically independent in $L^0(\lambda)$ whenever $\familyiI{u_i}$
is a disjoint family in $L^1(\mu)$.

\medskip

\noindent{\bf Remarks}\cmmnt{ Recall that a family $\familyiI{u_i}$ in
a Riesz space is `disjoint' if $|u_i|\wedge|u_j|=0$ for all distinct
$i$, $j\in I$ (352C).}   \cmmnt{A family} $\familyiI{v_i}$ in
$L^0(\lambda)$ is `independent' if $\familyiI{g_i}$ is an independent
family of random variables whenever $g_i\in\eusm L^0(\lambda)$
represents $v_i$ for each $i$\cmmnt{;  compare 367W}.

\proof{{\bf (a)} By Kakutani's theorem, there is a measure algebra
$(\frak A,\bar\mu)$ such that $U$ is isomorphic, as Banach lattice, to
$L^1(\frak A,\bar\mu)$;  now $(\frak A,\bar\mu)$ can be represented as
the measure algebra of a measure space $(X,\Sigma,\mu)$, and we
can identify $U$ and $L^1(\frak A,\bar\mu)$ with $L^1(\mu)$ (365B).
Set $\Sigma^f=\{E:E\in\Sigma$, $\mu E<\infty\}$ and
$\frak A^f=\{a:a\in\frak A$, $\bar\mu a<\infty\}$ as usual.   Take
$(\Omega,\Lambda,\lambda)$ and $\family{E}{\Sigma^f}{g_E}$ from
495B, with $\gamma=1$.   As in the proof of 495M, we have $g_E\eae g_F$
whenever $E$, $F\in\Sigma^f$ and $\mu(E\symmdiff F)=0$;  consequently we
can define $\psi:\frak A^f\to L^1(\lambda)$ by setting
$\psi a=g_E^{\ssbullet}$ whenever $E\in\Sigma^f$ and $E^{\ssbullet}=a$.
Again as in 495M, $g_{E\cup F}\eae g_E+g_F$ whenever $E$, $F\in\Sigma^f$
are disjoint, so $\psi$ is additive.   Also

\Centerline{$\|\psi a\|_1=\biggerint g_Ed\lambda=\mu E=\bar\mu a$}

\noindent whenever $E\in\Sigma^f$ represents $a\in\frak A^f$.   By
365I, there is a
unique bounded linear operator $T:L^1(\frak A,\bar\mu)\to L^1(\lambda)$
such that $T(\chi a)=\psi a$ for every $a\in\frak A^f$.   By
365Ja\formerly{3{}65Ka}, $T$
is a positive operator.   The set
$\{u:u\in L^1(\frak A,\bar\mu)^+, \|Tu\|_1=\|u\|_1\}$ is closed under
addition, norm-closed and contains $\alpha\chi a$ for every
$a\in\frak A^f$ and $\alpha\ge 0$,
so is the whole of $L^1(\frak A,\bar\mu)^+$, by 365F.

Note that if $\familyiI{a_i}$ is a disjoint family in $\frak A^f$, then
$\familyiI{\psi a_i}$ is stochastically independent, by
495B(ii).

\medskip

{\bf (b)} Now let $\familyiI{u_i}$ be a disjoint family in
$L^1(\frak A,\bar\mu)$.   Then $\familyiI{Tu_i}$ is independent.
\Prf\Quer\ Otherwise, there are a finite set $J\subseteq I$ and a family
$\family{i}{J}{V_i}$ such that $V_i$ is a neighbourhood of $Tu_i$ in the
topology of convergence in measure on $L^0(\mu)$ for each $i\in J$, and
$\family{i}{J}{v_i}$ is not independent whenever $v_i\in V_i$ for each
$i$ (367W).   Because the embedding
$L^1(\lambda)\embedsinto L^0(\lambda)$ is continuous for the norm topology
on $L^1(\lambda)$ and the topology of convergence in measure (245G),
there is a $\delta>0$ such that $Tu'_i\in V_i$ whenever $i\in J$,
$u'_i\in L^1(\frak A,\bar\mu)$ and $\|u'_i-u_i\|_1\le\delta$.   Now we
can find such $u'_i\in S(\frak A^f)$ with $|u'_i|\le|u_i|$ (365F).

Express each $u'_i$ as $\sum_{k=0}^{n_i}\alpha_{ik}\chi a_{ik}$ where
$\langle a_{ik}\rangle_{k\le n_i}$ is a disjoint family in
$\frak A^f$ and no $\alpha_{ik}$ is zero (361Eb).   In this case, all
the $a_{ik}$, for $i\in J$ and $k\le n_i$, are disjoint, so all the
$\psi(a_{ik})$ are independent.   But this means that
$\family{i}{J}{Tu'_i}
=\family{i}{J}{\sum_{k=0}^{n_i}\alpha_{ik}\psi(a_{ik})}$ is independent
(272K);  which is impossible, because $Tu'_i\in V_i$ for every
$i\in J$.\ \Bang\Qed

So $T$, regarded as a function from $U$ to $L^1(\lambda)$, has the
required properties.
}%end of proof of 495N

\leader{495O}{}\cmmnt{ The following is a more concrete expression of
the same ideas.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}95L}
Let $(X,\Sigma,\mu)$ be an atomless measure
space, and $\nu$ the Poisson point process on $X$ with intensity
$\gamma>0$.

(a) If $f\in\eusm L^1(\mu)$, $Q_f(S)=\sum_{x\in S\cap\dom f}f(x)$ is
defined and finite for $\nu$-almost every $S\subseteq X$, and
$\int Q_fd\nu=\gamma\int fd\mu$.

(b)\dvAnew{2018} If $f\in\eusm L^1(\mu)\cap\eusm L^2(\mu)$,
$\int Q_f^2d\nu$ is defined and
equal to $\gamma\int f^2d\mu+(\gamma\int fd\mu)^2$.

(c) We have a positive linear operator $T:L^1(\mu)\to L^1(\nu)$
defined by setting $T(f^{\ssbullet})=Q_f^{\ssbullet}$ for every
$f\in\eusm L^1(\mu)$.

(d) $\|Tu\|_1=\gamma\|u\|_1$ whenever $u\in L^1(\mu)^+$ and
$\familyiI{Tu_i}$ is stochastically independent in $L^0(\lambda)$
whenever $\familyiI{u_i}$ is a disjoint family in $L^1(\mu)$.

\proof{{\bf (a)} In the language of 495D, $Q_{\chi E}=f_E$ for every
$E\in\Sigma^f$.   So $Q_{\chi E}\in\eusm L^1(\nu)$ and
$\int Q_{\chi E}d\nu=\gamma\mu E$ for every $E\in\Sigma^f$.   If
$f=\sum_{i=0}^r\alpha_i\chi E_i$ is a simple function on $X$, then
$Q_f\eae\sum_{i=0}^r\alpha_iQ_{\chi E_i}\in\eusm L^1(\nu)$ and
$\int Q_fd\nu=\gamma\int f\,d\mu$.   If $f\in\eusm L^1(\mu)$ is zero a.e.,
then $\{S:S\subseteq f^{-1}[\{0\}]\}$ is $\nu$-conegligible, so
$Q_f=0$ a.e.   It follows that if $f\in\eusm L^1(\mu)$ is non-negative,
and $\sequencen{f_n}$ is a non-decreasing sequence of simple functions
converging to $f$ almost everywhere, then
$Q_f\eae\lim_{n\to\infty}Q_{f_n}$, while $Q_{f_n}\leae Q_{f_{n+1}}$ for
every $n$;  so $Q_f$ is $\nu$-integrable and

\Centerline{$\int Q_fd\nu=\lim_{n\to\infty}\int Q_{f_n}d\nu
=\lim_{n\to\infty}\gamma\int f_nd\mu=\gamma\int fd\mu$.}

\medskip

{\bf (b)} We can use the same ideas, with some further twists.

\medskip

\quad{\bf (i)} If $E\in\Sigma^f$, then

$$\eqalign{\int f_E^2d\nu-\gamma\mu E&
=\int f_E^2d\nu-\gamma\int f_Ed\nu\cr&
=\sum_{n=0}^{\infty}(n^2-n)\nu\{S:\#(S\cap E)=n\}\cr&
=\sum_{n=2}^{\infty}n(n-1)\Bover{(\gamma\mu E)^n}{n!}e^{-\gamma\mu E}\cr&
=(\gamma\mu E)^2
   \sum_{n=0}^{\infty}\Bover{(\gamma\mu E)^n}{n!}e^{-\gamma\mu E}
=(\gamma\mu E)^2\cr}$$

\noindent so $\int f_E^2d\nu=\gamma\mu E+(\gamma\mu E)^2$.

\medskip

\quad{\bf (ii)} If $f=\sum_{i=0}^n\alpha_i\chi E_i$ where
$E_0,\ldots,E_n\in\Sigma^f$ are
disjoint and $\alpha_0,\ldots,\alpha_n\in\Bbb R$,
then $f_{E_0},\ldots,f_{E_n}$ are $\nu$-independent (495D), so

$$\eqalignno{\int Q_f^2d\nu&
=\int(\sum_{i=0}^n\alpha_if_{E_i})^2d\nu\cr&
=\sum_{i,j\le n}\alpha_i\alpha_j\int f_{E_i}\times f_{E_j}d\nu\cr&
=\sum_{i=0}^n\alpha_i^2\int f_{E_i}^2d\nu
   +\sum_{i,j\le n,i\ne j}
      \alpha_i\alpha_j\int f_{E_i}d\nu\cdot\int f_{E_j}d\nu
\Displaycause{272R}
=\sum_{i=0}^n\alpha_i^2(\gamma\mu E_i+(\gamma\mu E_i)^2)
   +(\int Q_fd\nu)^2-\sum_{i=0}^n\alpha_i^2(\int f_{E_i}d\nu)^2\cr&
=\gamma\sum_{i=0}^n\alpha_i^2\mu E_i
   +\sum_{i=0}^n\alpha_i^2(\gamma\mu E_i)^2
   +(\gamma\int fd\mu)^2-\sum_{i=0}^n\alpha_i^2(\gamma\mu E_i)^2\cr&
=\gamma\int f^2d\mu+(\gamma\int fd\mu)^2.\cr}$$

\medskip

\quad{\bf (iii)} If $f\in\eusm L^1(\mu)\cap\eusm L^2(\mu)$ is non-negative,
there is a non-decreasing sequence $\sequencen{f_n}$ of
non-negative simple functions converging $\mu$-almost everywhere to $f$.
Now $\sequencen{Q_{f_n}}$ is non-decreasing and converges
$\nu$-almost everywhere to $Q_f$, and

$$\eqalign{\int Q_f^2d\nu
&=\lim_{n\to\infty}\int Q_{f_n}^2d\nu
=\lim_{n\to\infty}\gamma\int f_n^2d\mu+(\gamma\int f_nd\mu)^2\cr&
=\gamma\int f^2d\mu+(\gamma\int fd\mu)^2.\cr}$$

\medskip

\quad{\bf (iv)} Generally, if $f\in\eusm L^1(\mu)\cap\eusm L^2(\mu)$, it is
equal almost everywhere to a difference $f_1-f_2$ of non-negative functions
in $\eusm L^1(\mu)\cap\eusm L^2(\mu)$ such that there is an $E\in\Sigma$
for which $f_2(x)=0$ whenever $x\in E$ and $f_1(x)=0$ whenever
$x\in X\setminus E$.   In this case, identifying $\Cal PX$ with
$\Cal PE\times\Cal P(X\setminus E)$, $Q_{f_1}$ depends on the first coordinate
and $Q_{f_2}$ on the second, so they are $\nu$-independent
(495F).   Consequently

$$\eqalign{\int Q_f^2d\nu&
=\int Q_{f_1}^2d\nu-2\int Q_{f_1}\times Q_{f_2}d\nu+\int Q_{f_2}^2d\nu\cr&
=\gamma\int f_1^2d\mu+(\gamma\int f_1d\mu)^2
   -\int Q_{f_1}d\nu\cdot\int Q_{f_2}d\mu
   +\gamma\int f_2^2d\mu+(\gamma\int f_2d\mu)^2\cr&
=\gamma\int f_1^2d\mu+(\gamma\int f_1d\mu)^2
   -\gamma^2\int f_1d\mu\cdot\int f_2d\mu
   +\gamma\int f_2^2d\mu+(\gamma\int f_2d\mu)^2\cr&
=\gamma\int f^2d\mu+(\gamma\int fd\mu)^2,\cr}$$

\noindent as claimed.

\medskip

{\bf (c)} Since $Q_f\eae Q_{f'}$ whenever $f\eae f'$ in
$\eusm L^1(\mu)$, we can define $T:L^1(\mu)\to L^1(\nu)$ by setting
$T(f^{\ssbullet})=(Q_f)^{\ssbullet}$ for every $f\in\eusm L^1(\mu)$;
because $Q_{\alpha f}\eae\alpha Q_f$ and $Q_{f+f'}\eae Q_f+Q_{f'}$
whenever $f$, $f'\in\eusm L^1(\mu)$ and $\alpha\in\Bbb R$, $T$ is
linear;  because $Q_f\ge 0$ a.e.\ whenever $f\ge 0$ a.e., $T$ is positive.

\medskip

{\bf (d)} Because $\int Q_fd\nu=\gamma\int f\,d\mu$ for every
$f\in\eusm L^1(\mu)$, and $T$ is positive, $\|Tu\|_1=\gamma\|u\|_1$ for
every $u\in L^1(\mu)^+$.   Finally, if $\familyiI{u_i}$ is a finite
disjoint family in $L^1(\mu)$, we can find a family $\familyiI{f_i}$ of
measurable functions from $X$ to $\Bbb R$ such that
$f_i^{\ssbullet}=u_i$ for each $i$ and the sets $E_i=\{x:f_i(x)\ne 0\}$
are disjoint.   For each $i\in I$, let $\Tau_i$ be the $\sigma$-algebra
of subsets of $\Cal PX$ generated by sets of the form
$\{S:\#(S\cap E)=n\}$ where $E\subseteq E_i$ has finite measure and
$n\in\Bbb N$.   Then $\familyiI{\Tau_i}$ is independent (as in part (c)
of the proof of 495M), and each $Q_{f_i}$ is $\Tau_i$-measurable, so
$\familyiI{Q_{f_i}}$ is independent and $\familyiI{Tu_i}$ is
independent.
}%end of proof of 495O

\leader{495P}{}\cmmnt{ We can identify the characteristic functions of
the random variables $Q_f$ as defined above.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}95M} 
Let $(X,\Sigma,\mu)$ be an atomless measure
space, and $\nu$ the Poisson point process on $X$ with intensity
$\gamma>0$.   For $f\in\eusm L^1(\mu)$ set
$Q_f(S)=\sum_{x\in S\cap\dom f}f(x)$ when $S\subseteq X$ and the sum is
defined in $\Bbb R$.   Then

\Centerline{$\biggerint_{\Cal PX}e^{iyQ_f}d\nu
=\exp\bigl(\gamma\int_X(e^{iyf}-1)d\mu\bigr)$}

\noindent for any $y\in\Bbb R$.

\proof{ Note that $Q_f$ is defined $\nu$-almost everywhere, by 495Oa.

\medskip

{\bf (a)} Consider first the case in which $f$ is a simple function,
expressed as $\sum_{j=0}^n\alpha_j\chi F_j$ where
$\langle F_j\rangle_{j\le n}$ is a disjoint family of sets of finite
measure and $\alpha_j\in\Bbb R$ for each $j$.   Then
$Q_f(S)=\sum_{j=0}^n\alpha_j\#(S\cap F_j)$ for
$\nu$-almost every $S$, so

$$\eqalignno{\biggerint e^{iyQ_f}d\nu
&=\int\prod_{j=0}^ne^{iy\alpha_j\#(S\cap F_j)}\nu(dS)
=\prod_{j=0}^n\int e^{iy\alpha_j\#(S\cap F_j)}\nu(dS)\cr
\displaycause{because the functions $S\mapsto\#(S\cap F_j)$ are
independent}
&=\prod_{j=0}^n\sum_{k=0}^{\infty}\bover{(\gamma\mu F_j)^k}{k!}
   e^{-\gamma\mu F_j}e^{iy\alpha_jk}
=\prod_{j=0}^ne^{-\gamma\mu F_j}\sum_{k=0}^{\infty}
  \bover{(e^{iy\alpha_j}\gamma\mu F_j)^k}{k!}\cr
&=\prod_{j=0}^n\exp\bigl((e^{iy\alpha_j}-1)\gamma\mu F_j)\bigr)
=\exp\bigl(\gamma\sum_{j=0}^n(e^{iy\alpha_j}-1)\mu F_j\bigr)\cr
&=\exp\bigl(\gamma\int(e^{iyf}-1)d\mu\bigr).\cr}$$

\medskip

{\bf (b)} Now suppose that $f$ is any integrable function.   Then there
is a sequence $\sequencen{f_n}$ of simple functions such that
$|f_n|\leae|f|$ for every $n$ and $\lim_{n\to\infty}f_n\eae f$.   Write
$q_n$, $q$ for $Q_{f_n}$, $Q_f$.   Set

\Centerline{$D=\{x:x\in\dom f$, $|f_n(x)|\le|f(x)|$ for every $n$ and
$\lim_{n\to\infty}f_n(x)=f(x)\}$,}

\noindent so that $D$ is $\mu$-conegligible.   If $S\subseteq D$ and
$Q_{|f|}(S)$ is defined, then $q(S)=\lim_{n\to\infty}q_n(S)$, and this
is true for $\nu$-almost every $S$;  so
$\int e^{iyq}d\nu=\lim_{n\to\infty}e^{iyq_n}d\nu$, by Lebesgue's
Dominated Convergence Theorem.   On the other hand,

\Centerline{$|e^{i\alpha}-1|
=|\biggerint_0^{\alpha}\Bover1ie^{it}dt|\le\alpha$,
\quad$|e^{-i\alpha}-1|
=|\biggerint_0^{\alpha}\Bover1ie^{-it}dt|\le\alpha$}

\noindent for every $\alpha\ge 0$.   So if we set
$g(x)=e^{iyf(x)}-1$, $g_n(x)=e^{iyf_n(x)}-1$ when these are defined, we
have $|g_n|\leae|yf_n|\leae|yf|$ for every $n$.   Accordingly

\Centerline{$\biggerint(e^{iyf}-1)d\mu
=\int\lim_{n\to\infty}(e^{iyf_n}-1)d\mu
=\lim_{n\to\infty}\int(e^{iyf_n}-1)d\mu$}

\noindent by Lebesgue's theorem again.   It follows that

$$\eqalignno{\int e^{iyQ_f}d\nu
&=\lim_{n\to\infty}\int e^{iyq_n}d\nu
=\lim_{n\to\infty}
  \exp\bigl(\gamma\int(e^{iyf_n}-1)d\mu\bigr)\cr
\displaycause{by (a)}
&=\exp\bigl(\gamma\lim_{n\to\infty}
  \int(e^{iyf_n}-1)d\mu\bigr)
=\exp\bigl(\gamma\int(e^{iyf}-1)d\mu\bigr),\cr}$$

\noindent as claimed.
}%end of proof of 495P

\cmmnt{\medskip

\allowmorestretch{468}{
\noindent{\bf Remark} Recall that a Poisson random variable with
expectation $\gamma$ has characteristic function
$y\mapsto\exp(\gamma(e^{iy}-1))$ (part (a) of the proof of 285Q),
corresponding to the case $f=\chi F$ where $\mu F=1$.   The random
variables $Q_f$ have {\bf compound Poisson} distributions.
}%end of allowmorestretch
}%end of comment

\leader{495Q}{}\cmmnt{ If our underlying measure is a Radon measure,
we can look for Radon measures on $\Cal PX$ to represent the Poisson
point processes on $X$.   There seem to be difficulties in general, but
I can offer the following.   See also 495Yd.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}95N}
Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure
space such that $\mu$ is outer regular with respect to the open sets,
and $\gamma>0$.
Give the space $\Cal C$ of closed subsets of $X$ its
Fell topology\cmmnt{ (4A2T)}.

(a) There is a unique quasi-Radon probability measure
$\tilde\nu$ on $\Cal C$ such that

\Centerline{$\tilde\nu\{C:\#(C\cap E)=\emptyset\}=e^{-\gamma\mu E}$}

\noindent whenever $E\subseteq X$ is a measurable set of
finite measure.

(b) If $E_0,\ldots,E_r$ are disjoint sets of finite measure, none
including any singleton set of non-zero measure, and $n_i\in\Bbb N$ for
$i\le r$, then

\Centerline{$\tilde\nu\{C:\#(C\cap E_i)=n_i$ for every $i\le r\}
=\prod_{i=0}^r\Bover{(\gamma\mu E_i)^{n_i}}{n_i!}e^{-\gamma\mu E_i}$.}

(c) Suppose that $\mu$ is atomless and $\nu$ is the Poisson point process on $X$
with intensity $\gamma$.

\quad(i) $\Cal C$ has full outer measure for $\nu$, and
$\tilde\nu$ extends the subspace measure $\nu_{\Cal C}$.

\quad(ii) If moreover $\mu$ is $\sigma$-finite, then $\Cal C$ is
$\nu$-conegligible.

(d) If $X$ is locally compact then $\tilde\nu$ is a Radon measure.

(e) If $X$ is second-countable and $\mu$ is atomless
then $\tilde\nu=\nu_{\Cal C}$.
%new 2008

\proof{{\bf (a)(i)} Set $\Sigma^f=\{E:\mu E<\infty\}$.
There is a disjoint family $\Cal H$ of non-empty self-supporting
measurable subsets of $X$ of finite measure such that
$\mu E=\sum_{H\in\Cal H}\mu(E\cap H)$ for every $E\in\Sigma$ (412I);
so if $G\subseteq X$ is an open set of finite measure,
$\{H:H\in\Cal H$, $G\cap H\ne\emptyset\}$ is countable.   If $E$ is any set
of finite measure, it is included in an open set of finite measure, because
$\mu$ is outer regular with respect to the open sets;  so once again
$\{H:H\in\Cal H$, $E\cap H\ne\emptyset\}$ is countable.

Build $\Omega=\prod_{H\in\Cal H}\Bbb N\times H^{\Bbb N}$,
$\langle g_{HE}\rangle_{H\in\Cal H,E\in\Sigma}$ and the product
measure $\lambda$ on $\Omega$ as in the proof of
495B;  as in the proof of 495D, set

\Centerline{$\phi(\omega)=\{x_{Hj}(\omega):H\in\Cal H$, $j<m_H(\omega)\}$}

\noindent for $\omega\in\Omega$.

\medskip

\quad{\bf (ii)} If $E\in\Sigma^f$,
$\lambda\{\omega:E\cap\phi(\omega)=\emptyset\}=e^{-\gamma\mu E}$.
\Prf\ $\Cal H'=\{H:H\in\Cal H$, $E\cap H\ne\emptyset\}$ is countable.
Now

\Centerline{$\{\omega:E\cap\phi(\omega)=\emptyset\}
=\bigcap_{H\in\Cal H'}\{\omega:x_{Hj}\notin E$ for every $j<m_H(\omega)\}$}

\noindent has measure

\Centerline{$\prod_{H\in\Cal H'}\lambda\{\omega:g_{HE}(\omega)=0\}
=\prod_{H\in\Cal H'}e^{-\gamma\mu(H\cap E)}
=e^{-\gamma\mu E}$}

\noindent because $\mu E=\sum_{H\in\Cal H'}\mu(H\cap E)$.\ \Qed

Let $\Tau_0$ be the $\sigma$-algebra of subsets of $\Cal PX$
generated by sets of the form $\{S:S\cap E=\emptyset\}$ where
$E\in\Sigma^f$.   By the Monotone Class Theorem (136B),
$\lambda$ measures $\phi^{-1}[W]$ for every
$W\in\Tau_0$;  set $\nu_0W=\lambda\phi^{-1}[W]$ for $W\in\Tau_0$, so that
if $E\in\Sigma$ then $\nu_0\{S:S\cap E=\emptyset\}=e^{-\gamma\mu E}$.

\medskip

\quad{\bf (iii)} Give $\Cal PX$ the topology $\frak S$ generated by sets of
the form

\Centerline{$\{S:S\cap G\ne\emptyset\}$,
\quad$\{S:S\cap K=\emptyset\}$}

\noindent for open sets $G\subseteq X$ and compact sets $K\subseteq X$.
(Thus the Fell topology on $\Cal C$ is the subspace topology induced by
$\frak S$.)   Then $\Cal PX$ is compact.   \Prf\ Follow the proof of
4A2T(b-iii) word for word, but replacing every $\Cal C$ with $\Cal PX$.\
\Qed

\medskip

\quad{\bf (iv)} $\nu_0$ is inner regular with respect to the
$\frak S$-closed sets.   \Prf\ Write $\Cal L$ for the family of
$\frak S$-closed sets belonging to $\Tau_0$.   Of course $\Cal L$ is closed
under finite unions and countable intersections.

($\alpha$) Suppose that $E\in\Sigma^f$ and
$W=\{S:S\cap E\ne\emptyset\}$.   Let $\epsilon>0$.   Then there is a
compact set $K\subseteq E$ such that $\mu(E\setminus K)\le\epsilon$.
Set $V=\{S:S\cap K\ne\emptyset\}$;  then $V\in\Cal L$, $V\subseteq W$ and

\Centerline{$\nu_0(W\setminus V)
\le\nu_0\{S:S\cap E\setminus K\ne\emptyset\}
\le 1-e^{-\gamma\epsilon}$.}

\noindent As $\epsilon$ is arbitrary,
$\nu_0W=\sup\{\nu_0V:V\in\Cal L$, $V\subseteq W\}$.

($\beta$) Suppose that $E\in\Sigma^f$ and
$W=\{S:S\cap E=\emptyset\}$.   Let $\epsilon>0$.   Then there is an
open set $G\supseteq E$ such that $\mu(G\setminus E)\le\epsilon$.
Set $V=\{S:S\cap G=\emptyset\}$;  then $V\in\Cal L$, $V\subseteq W$ and

\Centerline{$\nu_0(W\setminus V)
\le\nu_0\{S:S\cap G\setminus E\ne\emptyset\}
\le 1-e^{-\gamma\epsilon}$.}

\noindent As $\epsilon$ is arbitrary,
$\nu_0W=\sup\{\nu_0V:V\in\Cal L$, $V\subseteq W\}$.

($\gamma$) By 412C, $\nu_0$ is inner regular with respect to the
$\frak S$-closed sets.\ \Qed

\medskip

\quad{\bf (v)} Since $\frak S$ is a compact topology, the family of
$\frak S$-closed sets is a compact class, so 413P tells us that $\nu_0$ has
an extension to a complete topological measure $\tilde\nu_0$ on
$\Cal PX$, inner regular with respect to the closed sets.
Of course $\tilde\nu_0$, being a probability measure, is
effectively locally finite and locally determined, so it is a quasi-Radon
measure with respect to the topology $\frak S$.   Consequently the subspace
measure $\tilde\nu$ on $\Cal C$ is a quasi-Radon measure for the
Fell topology on $\Cal C$ (415B).

\medskip

\quad{\bf (vi)} $\Cal C$ has full outer measure for $\tilde\nu_0$.
\Prf\Quer\ Otherwise, there is a non-empty closed set
$V\subseteq\Cal PX\setminus\Cal C$.
Consider the family $\Cal U$ of subsets of $\Cal PX$ of the form

\Centerline{$\{S:S\cap K=\emptyset$, $S\cap G_i\ne\emptyset$ for
$i<r\}$}

\noindent where $K\subseteq X$ is compact and $G_i\subseteq X$ is an open
set of finite measure for every $i<r$.
Because $\mu$ is locally finite, this is a base for $\frak S$.   So
$\Cal U'=\{U:U\in\Cal U$, $U\cap W=\emptyset\}$ is a cover of
$\Cal PX\setminus W\supseteq\Cal C$.   Of course $U\cap\Cal C$ is open in
the Fell topology for every $U\in\Cal U$;  because $\Cal C$ is compact,
there are $U_0,\ldots,U_m\in\Cal U'$ covering $\Cal C$.

Express each $U_j$ as
$\{S:S\cap K_j=\emptyset$, $S\cap G_{ji}\ne\emptyset$ for
$i<r_j\}$, where the $K_j$ are all compact and the $G_{ji}$ are all open.
Because $\bigcup_{j\le m}U_j$ is disjoint from $V$, there is an
$S\subseteq\Cal PX$ which does not belong to any $U_j$.   Let $\Cal E$ be
the finite algebra of subsets of $X$ generated by
$\{K_j:j\le m\}\cup\{G_{ji}:j\le m$, $i<r_j\}$;  then there is a finite set
$C\subseteq S$ such that $C\cap E\ne\emptyset$ whenever $E\in\Cal E$ and
$S\cap E\ne\emptyset$.   In this case,
$C\in\Cal C\setminus\bigcup_{j\le m}U_j$;  which is supposed to be
impossible.\ \Bang\Qed

\medskip

\quad{\bf (vii)} Consequently $\tilde\nu$ is a probability measure.
If $E\in\Sigma^f$, then

$$\eqalign{\tilde\nu\{C:C\in\Cal C,\,C\cap E=\emptyset\}
&=\tilde\nu(\Cal C\cap\{S:S\subseteq X,\,S\cap E=\emptyset\})\cr
&=\tilde\nu_0\{S:S\cap E=\emptyset\}
=\nu_0\{S:S\cap E=\emptyset\}
=e^{-\gamma\mu E}.\cr}$$

\medskip

\quad{\bf (viii)} To see that $\tilde\nu$ is uniquely defined, let
$\tilde\nuprime$ be another quasi-Radon probability measure on $\Cal C$
with the same property.

\medskip

\qquad\grheada\ Suppose that $E_0,\ldots,E_r\subseteq X$ are disjoint
measurable sets of finite measure, and

\Centerline{$W
=\{C:C\in\Cal C$, $C\cap E_0=\emptyset$, $C\cap E_i\ne\emptyset$ for
$1\le i\le r\}$.}

\noindent Then $\tilde\nu W=\tilde\nu'W$.   \Prf\ Induce on $r$.   If
$r=0$ the result is immediate.   For the inductive step to $r\ge 1$,
consider
$\{C:C\cap E_0=\emptyset$, $C\cap E_i\ne\emptyset$ for $1\le i<r\}$ and
$\{C:C\cap(E_0\cup E_r)=\emptyset$,
$C\cap E_i\ne\emptyset$ for $1\le i<r\}$.   By the inductive hypothesis,
$\tilde\nu$ and $\tilde\nu'$ agree on these two sets, and therefore on
their difference
$\{C:C\in\Cal C$, $C\cap E_0=\emptyset$, $C\cap E_i\ne\emptyset$ for
$1\le i\le r\}$.\ \Qed

\medskip

\qquad\grheadb\
Suppose that we have a compact set $K\subseteq X$
and open sets $G_i\subseteq X$ of finite measure, for $i<r$, and set

\Centerline{$V=\{C:C\in\Cal C$, $C\cap K=\emptyset$,
$C\cap G_i\ne\emptyset$ for every $i<r\}$.}

\noindent Then $\tilde\nu V=\tilde\nuprime V$.   \Prf\ Let $\Cal E$
be the finite subalgebra of $\Cal PX$ generated
by $\{G_i:i<r\}\cup\{K\}$, and $\Cal A$ the set of atoms of $\Cal E$
included in $K\cup\bigcup_{i<r}G_i$.
For $\Cal I\subseteq\Cal A$ set

\Centerline{$V_{\Cal I}=\{C:C\in\Cal C$, $C\cap E\ne\emptyset$ for
$E\in\Cal I$, $C\cap E=\emptyset$ for $E\in\Cal A\setminus\Cal I\}$.}

\noindent Then $V=\bigcup_{\Cal I\in\frak I}V_{\Cal I}$, where

$$\eqalign{\frak I
&=\{\Cal I:\Cal I\subseteq\Cal A,\,A\cap K=\emptyset
  \text{ for every }A\in\Cal I,\cr
&\mskip100mu\text{ for every }i<r\text{ there is an }A\in\Cal I
  \text{ such that }A\subseteq G_i\}.\cr}$$

\noindent Now ($\alpha$) shows that
$\tilde\nu V_{\Cal I}=\tilde\nuprime V_{\Cal I}$ for every
$\Cal I\subseteq\Cal A$, so that $\tilde\nu V=\tilde\nuprime V$.
Since sets $V$ of the type described form a base for the Fell
topology closed under finite intersections, $\tilde\nu=\tilde\nuprime$
(415H(v)).\ \Qed

This completes the proof of (a).

\medskip

{\bf (b)(i)} In the construction of 495B and (a-i) above, all the
normalized subspace measures $\mu'_H$ are Radon measures (416Rb), while of
course all the Poisson distributions $\nu_H$ are Radon measures, so the
product measure $\lambda$ on
$\Omega=\prod_{H\in\Cal H}\Bbb N\times H^{\Bbb N}$ has an extension to a
Radon measure $\tilde\lambda$ (417Q).
Let $\Cal W$ be the family of those sets $W\subseteq\Cal PX$
such that $\tilde\nu_0W$ and $\tilde\lambda\phi^{-1}[E]$ are defined and
equal.   Then $\Cal W$ is a Dynkin class.   So if
$\Cal W_0\subseteq\Cal W$ is closed under finite intersections,
the $\sigma$-algebra of subsets of $\Cal PX$ generated by $\Cal W_0$ is
included in $\Cal W$.   By (a-ii), $\Tau_0\subseteq\Cal W$.

\medskip

\quad{\bf (ii)} Let $\frak S_0$ be the topology on $\Cal PX$ generated by
sets of the form $\{S:S\cap G\ne\emptyset\}$ where $G\subseteq X$ is open.
(So $\frak S_0$ is coarser than the topology $\frak S$ of (a-iii) above.)
Then $\phi:\Omega\to\Cal PX$ is continuous for the product topology
$\frak U$ on $\Omega$ and $\frak S_0$ on $\Cal PX$.   \Prf\ If
$G\subseteq X$ is open, then

\Centerline{$\phi^{-1}[\{S:S\cap G\ne\emptyset\}]
=\Omega\cap\bigcap_{i<r}\bigcup_{H\in\Cal H,j\in\Bbb N}
  \{\omega:j<m_H(\omega),\,x_{Hj}(\omega)\in G_i\}$}

\noindent is open;  by 4A2B(a-ii), this is enough.\ \Qed

\medskip

\quad{\bf (iii)} $\frak S_0\subseteq\Cal W$.   \Prf\ Because $\mu$ is
locally finite, the family $\Cal U$ of sets of the form

\Centerline{$\{S:S\cap G_i\ne\emptyset$ for $i<r\}$,}

\noindent where $G_i\subseteq X$ is an open set of finite measure for each
$i<r$, is a base for $\frak S_0$;  and
$\Cal U\subseteq\Tau_0\subseteq\Cal W$.   So if $W\in\frak S_0$,
$\Cal V=\{V:V\in\frak S_0\cap\Tau_0$, $V\subseteq W\}$ is an
upwards-directed family of sets with union $W$.   Since $\tilde\nu_0$ and
$\tilde\lambda$ are both $\tau$-additive, and $\phi^{-1}[V]$ is open for
every $V\in\Cal V$,

\Centerline{$\tilde\lambda\phi^{-1}[W]
=\sup_{V\in\Cal V}\tilde\lambda\phi^{-1}[V]
=\sup_{V\in\Cal V}\tilde\nu_0V
=\tilde\nu_0W$,}

\noindent and $W\in\Cal W$.\ \Qed

\medskip

\quad{\bf (iv)} If $G\subseteq X$ is open and $n\in\Bbb N$,
$W=\{S:\#(S\cap G)\ge n\}$ belongs to $\frak S_0$.   \Prf\

$$\eqalign{W
&=\bigcup\{\{S:S\cap G_i\ne\emptyset\text{ for every }i<n\}:
  \ofamily{i}{n}{G_i}\text{ is a disjoint family}\cr
&\mskip100mu
  \text{of open subsets of }G\text{ of finite measure}\}.
  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (v)} If $E_0,\ldots,E_r\subseteq X$ are sets of finite measure,
and $n_0,\ldots,n_r\in\Bbb N$, then

\Centerline{$V=\{S:\#(S\cap E_i)\ge n_i$ for $i\le r\}$}

\noindent belongs to $\Cal W$.   \Prf\ Let $\epsilon>0$.   Then there is a
$\delta>0$ such that $1-e^{-\gamma\delta}\le\epsilon$.   Let
$G_0,\ldots,G_r$ be open sets such that $E_i\subseteq G_i$ for $i\le r$ and
$\sum_{i=0}^r\mu(G_i\setminus E_i)\le\delta$.   Set

\Centerline{$W=\{S:\#(S\cap G_i)\ge n_i$ for $i\le r\}$,
\quad$W_0=\{S:S\cap\bigcup_{i\le r}G_i\setminus E_i\ne\emptyset\}$;}

\noindent then $W\in\frak S_0$ and $W_0\in\Tau_0$, so both belong to
$\Cal W$, while

\Centerline{$\tilde\nu_0W_0=\tilde\lambda\phi^{-1}[W_0]
=1-\exp(-\gamma\mu(\bigcup_{i\le r}G_i\setminus E_i))\le\epsilon$.}

Now

\Centerline{$W\setminus W_0\subseteq V\subseteq W$,
\quad$\phi^{-1}[W]\setminus\phi^{-1}[W_0]\subseteq\phi^{-1}[V]
\subseteq\phi^{-1}[W]$.}

\noindent So

\Centerline{$\tilde\nu_0^*V-(\tilde\nu_0)_*V\le\epsilon$,
\quad$\tilde\lambda^*(\phi^{-1}[V])-\tilde\lambda_*(\phi^{-1}[V])
\le\epsilon$,
\quad$|\tilde\nu_0^*V-\tilde\lambda^*(\phi^{-1}[V])|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary (and $\tilde\nu_0$, $\tilde\lambda$
are complete), $V$ is measured by $\tilde\nu_0$, $\phi^{-1}[V]$ is measured
by $\tilde\lambda$, and

\Centerline{$|\tilde\nu_0V-\tilde\lambda\phi^{-1}[V]|
=|\tilde\nu_0^*V-\tilde\lambda^*(\phi^{-1}[V])|
=0$.  \Qed}

\medskip

\quad{\bf (vi)} If $E_0,\ldots,E_r\subseteq X$ are sets of finite measure,
$n_0,\ldots,n_r\in\Bbb N$ and $j\le r$, then

\Centerline{$\{S:\#(S\cap E_i)=n_i$ for $i<j$,
$\#(S\cap E_i)\ge n_i$ for $j\le i\le r\}$}

\noindent belongs to $\Cal W$.   \Prf\ Induce on $j$.   For $j=0$ we just
have the case of (v).   For the inductive step to $j+1$, we have

$$\eqalign{\{S:\#&(S\cap E_i)=n_i\text{ for }i\le j,\,
\#(S\cap E_i)\ge n_i\text{ for }j<i\le r\}\cr
&=\{S:\#(S\cap E_i)=n_i\text{ for }i<j,\,
\#(S\cap E_i)\ge n_i\text{ for }j\le i\le r\}\cr
&\mskip50mu
  \setminus\{S:\#(S\cap E_i)=n_i\text{ for }i<j,\,
    \#(S\cap E_j)\ge n_j+1,\cr
&\mskip308mu
    \#(S\cap E_i)\ge n_i\text{ for }j<i\le r\}\cr
&\in\Cal W\cr}$$

\noindent because $\Cal W$ is a Dynkin class.\ \Qed

\medskip

\quad{\bf (vii)}
If $E\in\Sigma$ has finite measure and does not include any
non-negligible singleton, then $\#(E\cap\phi(\omega))=g_E(\omega)$, as
defined in 495B, for
$\lambda$-almost every $\omega\in\Omega$.
\Prf\ Let $A_E$ be the set of those $\omega\in\Omega$ such that

\inset{{\it either} there are an $H\in\Cal H$ and $j\in\Bbb N$ such that
$\mu(H\cap E)=0$ and $x_{Hj}(\omega)\in E$

{\it or} there are an $H\in\Cal H$ and distinct $i$, $j\in\Bbb N$
such that $x_{Hi}(\omega)=x_{Hj}(\omega)\in E$.}

\noindent As observed in (a-i) above,
$\{H:H\in\Cal H$, $H\cap E\ne\emptyset\}$ is countable;  while
for any $H\in\Cal H$ and distinct $i$, $j\in\Bbb N$ the set
$\{\omega:x_{Hi}(\omega)=x_{Hj}(\omega)\in E\}$ is negligible because the
subspace measure on $E$ is atomless (414G/416Xa), so the
diagonal $\{(x,x):x\in E\}$ is negligible in $X^2$.   Consequently
$\lambda A_E=0$.   But $\#(E\cap\phi(\omega))=g_E(\omega)$ for every
$\omega\in\Omega\setminus A_E$.\ \Qed

\medskip

\quad{\bf (viii)} Now suppose that $E_0,\ldots,E_r\subseteq X$ are disjoint
sets of finite measure, none including any non-negligible singleton, and
$n_0,\ldots,n_r\in\Bbb N$.   Then

\Centerline{$V=\{S:S\subseteq X$,
$\#(S\cap E_i)=n_i$ for every $i\le r\}$}

\noindent belongs to $\Cal W$, by (vi).   Next,

\Centerline{$\phi^{-1}[V]=\{\omega:\#(E_i\cap\phi(\omega))=n_i$
for every $i\le r\}$,}

\noindent so

$$\eqalignno{\tilde\nu_0V
&=\tilde\lambda\phi^{-1}[V]
=\tilde\lambda\{\omega:g_{E_i}(\omega)=n_i\text{ for every }i\le r\}\cr
\displaycause{by (vii)}
&=\prod_{i=0}^r\tilde\lambda\{\omega:g_{E_i}(\omega)=n_i\}
=\prod_{i=0}^r\Bover{(\gamma\mu E_i)^{n_i}}{n_i!}e^{-\gamma\mu E_i}.\cr}$$

Finally, because $\tilde\nu_0^*\Cal C=1$ and $\tilde\nu$ is the subspace
measure on $\Cal C$,

$$\eqalign{\tilde\nu\{C:C\in\Cal C,\,\#(C\cap E_i)=n_i
\text{ for every }i\le r\}
&=\tilde\nu(V\cap\Cal C)
=\tilde\nu_0V\cr
&=\prod_{i=0}^r\Bover{(\gamma\mu E_i)^{n_i}}{n_i!}e^{-\gamma\mu E_i}.\cr}$$

\noindent This completes the proof of (b).

\medskip

{\bf (c)(i)} Taking $\Tau\supseteq\Tau_0$ to be the $\sigma$-algebra of
subsets of $\Cal PX$ generated by sets of the form
$\{S:\#(S\cap E)=n\}$ where $E\in\Sigma^f$ and $n\in\Bbb N$,
(b-viii) tells us that $\tilde\nu_0\restr\Tau$ satisfies the conditions
of 495D, so its completion $\nu$
is the Poisson point process as defined in 495E.   Because $\tilde\nu_0$
is complete, it extends $\nu$.  (The identity map from $\Cal PX$ to itself
is \imp\ for $\tilde\nu_0$ and $\tilde\nu_0\restr\Tau$, therefore also for
their completions $\tilde\nu_0$ and $\nu$.)   Since $\Cal C$ has full outer
measure for $\tilde\nu_0$, by (a-v), it has full outer measure for $\nu$,
and

\Centerline{$\nu_{\Cal C}(V\cap\Cal C)=\nu V=\tilde\nu_0V
=\tilde\nu(V\cap\Cal C)$}

\noindent whenever $\nu$ measures $V$, so $\tilde\nu$ extends
$\nu_{\Cal C}$.

\medskip

\quad{\bf (ii)} If $\mu$ is $\sigma$-finite, then there is a sequence
$\sequencen{H_n}$ of open sets of finite measure covering $X$.   For each
$n\in\Bbb N$, $\{S:S\subseteq X$, $S\cap H_n$ is finite$\}$ is
$\nu$-conegligible.   So $W=\{S:S\cap H_n$ is finite for every $n\}$ is
$\nu$-conegligible.   But $W\subseteq\Cal C$, so $\Cal C$ is
$\nu$-conegligible.

\medskip

{\bf (d)} If $X$ is locally compact then $\Cal C$ is Hausdorff
(4A2T(e-ii));  so $\tilde\nu$, being a quasi-Radon probability
measure on a compact Hausdorff space, is a Radon measure (416G).

\medskip

{\bf (e)} Now suppose that $X$ is second-countable.

\medskip

\quad{\bf (i)} $\Cal C$ has a
countable network consisting of sets in $\Tau_{\Cal C}$, the subspace
$\sigma$-algebra induced by the $\sigma$-algebra $\Tau$ of (c-i).
\Prf\ Let $\Cal U$ be a countable base for $\frak T$,
closed under finite unions, consisting of sets of finite measure.
For $U_0\in\Cal U$ and finite $\Cal U_0\subseteq\Cal U$, set

\Centerline{$V(U_0,\Cal U_0)
=\{C:C\in\Cal C$, $C\cap U_0=\emptyset$,
$C\cap U\ne\emptyset$ for every $U\in\Cal U\}\in\Tau_{\Cal C}$.}

\noindent If $W\subseteq\Cal C$ is open for the Fell topology and
$C_0\in W$, there are a compact set $K\subseteq X$ and a finite family
$\Cal G\subseteq\frak T$ such that

\Centerline{$C_0
\in\{C:C\in\Cal C$, $C\cap K=\emptyset$, $C\cap G\ne\emptyset$ for every
$G\in\Cal G\}$.}

\noindent For $G\in\Cal G$ let $y_G$ be a point of $C_0\cap G$.
Now there are a $U_0\in\Cal U$ such that
$K\subseteq U_0\subseteq X\setminus C$ and a family
$\family{G}{\Cal G}{U_G}$ in $\Cal U$ such that $x_G\in U_G\subseteq G$ for
every $G\in\Cal G$.   In this case,

\Centerline{$C_0\in V(U_0,\{U_G:G\in\Cal G\})\subseteq W$.}

\noindent As $C_0$ and $W$ are arbitrary, the countable set
$\{V(U_0,\Cal U_0):U_0\in\Cal U$, $\Cal U_0\in[\Cal U]^{<\omega}\}$ is a
network for the topology of $\Cal C$.\ \Qed

\medskip

\quad{\bf (ii)} Since $\nu_{\Cal C}$ measures every set in this countable
network, it is a topological measure.
Since it is also complete, and $\tilde\nu$, being a quasi-Radon probability
measure, is the completion of its restriction to the Borel $\sigma$-algebra
of $\Cal C$, $\nu_{\Cal C}$ extends $\tilde\nu$, and the two must be equal.
}%end of proof of 495Q

\leader{495R}{Proposition}\dvAformerly{4{}95O}
Let $(X,\frak T)$ be a $\sigma$-compact
locally compact Hausdorff space and $M^{\infty+}_{\text{R}}(X)$ the set
of Radon measures on $X$.   Give $M^{\infty+}_{\text{R}}(X)$ the topology
generated by sets of the form $\{\mu:\mu G>\alpha\}$ and
$\{\mu:\mu K<\alpha\}$ where $G\subseteq X$ is open, $K\subseteq X$ is
compact and $\alpha\in\Bbb R$.   Let $\Cal C$ be the space of closed
subsets of $X$ with its Fell topology, and
$P_{\text{R}}(\Cal C)$ the set of Radon probability measures on
$\Cal C$ with its narrow topology\cmmnt{ (definition:  437Jd)}.   For
$\mu\in M^{\infty+}_{\text{R}}(X)$ and $\gamma>0$ let
$\tilde\nu_{\mu,\gamma}$ be the Radon measure on
$\Cal C$ defined from $\mu$ and $\gamma$ as in 495Q.   Then the function
$(\mu,\gamma)\mapsto\tilde\nu_{\mu,\gamma}:
M^{\infty+}_{\text{R}}(X)\times\ooint{0,\infty}\to P_{\text{R}}(\Cal C)$ is
continuous.

\proof{{\bf (a)} Note that because $X$ is $\sigma$-compact, every Radon
measure on $X$ is $\sigma$-finite, therefore outer regular with respect to
the open sets (412Wb), and we can apply 495Q to build the measures
$\tilde\nu_{\mu,\gamma}$.
Just as in 495E for ordinary Poisson point processes, the
uniqueness assertion in 495Qa assures us that
$\tilde\nu_{\mu,\gamma}=\tilde\nu_{\gamma\mu,1}$ for all $\gamma$ and
$\mu$.   Of course the sets

\Centerline{$\{(\mu,\gamma):\gamma\mu G>\alpha\}$,
\quad$\{(\mu,\gamma):\gamma\mu K<\alpha\}$}

\noindent where $G\subseteq X$ is open,
$K\subseteq X$ is compact and $\alpha\in\Bbb R$, are all open in
$M^{\infty+}_{\text{R}}(X)\times\ooint{0,\infty}$;  so the map
$(\mu,\gamma)\mapsto\gamma\mu$ is continuous.
It will therefore be enough to show that the map
$\mu\mapsto\tilde\nu_{\mu,1}:M^{\infty+}_{\text{R}}(X)\to P_{\text{R}}(\Cal C)$ is
continuous.   Write $\tilde\nu_{\mu}$ for $\tilde\nu_{\mu,1}$.

\medskip

{\bf (b)} Fix an open set $W_0\subseteq\Cal C$, $\alpha_0>0$ and
$\mu_0\in M^{\infty+}_{\text{R}}(X)$ such that $\tilde\nu_{\mu_0}W_0>\alpha_0$.
Let $\Cal E$ be the family of relatively compact Borel subsets $E$ of
$X$ such that $\mu_0(\partial E)=0$.   Then $\Cal E$ is a subring of
$\Cal PX$ (4A2Bi).   Also
$\mu\mapsto\mu E:M^{\infty+}_{\text{R}}(X)\to\coint{0,\infty}$ is
continuous at $\mu_0$ for every $E\in\Cal E$.   \Prf\ If $E\in\Cal E$
and $\epsilon>0$, then

$$\eqalign{\{\mu:\mu_0E-\epsilon&<\mu E<\mu_0E+\epsilon\}\cr
&\supseteq\{\mu:\mu(\interior E)>\mu_0(\interior E)-\epsilon,\,
  \mu\overline{E}<\mu_0\overline{E}+\epsilon\}\cr}$$

\noindent is a neighbourhood of $\mu_0$.\ \Qed

\medskip

{\bf (c)} Next, $\Cal U=\Cal E\cap\frak T$ is a base for $\frak T$
(411Gi).   It follows that the family $\Cal V$ of sets of the form

\Centerline{$\{C:C\in\Cal C$, $C\cap U_i\ne\emptyset$ for $i<r$,
$C\cap\overline{U}=\emptyset\}$,}

\noindent where $U$, $U_0,\ldots\in\Cal U$, is a base for the
Fell topology on $\Cal C$.   \Prf\
If $W\subseteq\Cal C$ is open for the Fell topology and
$C_0\in W$, there are $r\in\Bbb N$, open sets $G_i\subseteq X$ for
$i<r$ and a compact set $K\subseteq X$ such that

\Centerline{$C_0\in\{C:C\cap G_i\ne\emptyset$ for each $i<r$,
$C\cap K=\emptyset\}\subseteq W$.}

\noindent For each $i<r$ choose $x_i\in C_0\cap G_i$ and
$U_i\in\Cal U$
such that $x_i\in U_i\subseteq G_i$.   Because $X$ is locally compact
and Hausdorff, it is regular, so every point of $K$ belongs to a member
of $\Cal U$ with closure disjoint from $C_0$;  because $\Cal U$ is
closed under finite unions, there is a $U\in\Cal U$ such that
$K\subseteq U$ and $C_0\cap\overline{U}=\emptyset$.   Now

\Centerline{$\{C:C\in\Cal C$, $C\cap U_i\ne\emptyset$ for $i<r$,
$C\cap\overline{U}=\emptyset\}$}

\noindent belongs to $\Cal V$, contains $C_0$ and is included in
$W$.   As $C_0$ and $W$ are arbitrary, $\Cal V$ is a base for the
Fell topology on $\Cal C$.\ \Qed

\medskip

{\bf (d)} If $V\in\Cal V$, then
$\mu\mapsto\tilde\nu_{\mu}V:M^{\infty+}_{\text{R}}(X)\to[0,1]$ is
continuous at $\mu_0$.   \Prf\ Express $V$ as
$\{C:C\in\Cal C$, $C\cap U_i\ne\emptyset$ for $i<r$,
$C\cap\overline{U}=\emptyset\}$, where $U_i$, $U\in\Cal U$.
Let $\Cal A$ be the set of atoms of the finite subring of $\Cal E$
generated by $\{U_i:i<r\}\cup\{\overline{U}\}$.   For
$\Cal I\subseteq\Cal A$ set

\Centerline{$V_{\Cal I}=\{C:C\in\Cal C$,
$\Cal I=\{A:A\in\Cal A$, $C\cap A\ne\emptyset\}\}$.}

\noindent Let $\frak I$ be the set of those $\Cal I\subseteq\Cal A$ such
that $A\cap\overline{U}=\emptyset$ for every $A\in\Cal I$ and for every
$i<r$ there is an $A\in\Cal I$ such that $A\subseteq U_i$.   Then
$\family{\Cal I}{\frak I}{V_{\Cal I}}$ is a partition of $V$.
Moreover, for any $\mu\in M^{\infty+}_{\text{R}}(X)$ and
$\Cal I\subseteq\Cal A$,

\Centerline{$\tilde\nu_{\mu}V_{\Cal I}
=\prod_{A\in\Cal A\setminus\Cal I}e^{-\mu A}
  \cdot\prod_{A\in\Cal I}(1-e^{-\mu A})$.}

\noindent Since each $\mu\mapsto\mu A$ is continuous at $\mu_0$, by (a),
so are the functionals $\mu\mapsto\tilde\nu_{\mu}V_{\Cal I}$, for
$\Cal I\subseteq\Cal A$, and $\mu\mapsto\tilde\nu_{\mu}V
=\sum_{\Cal I\in\frak I}\tilde\nu_{\mu}V_{\Cal I}$.\ \Qed

\medskip

{\bf (e)} Let $\Cal V^*$ be the family of Borel subsets $V$ of $\Cal C$
such that
$\mu\mapsto\tilde\nu_{\mu}V:M^{\infty+}_{\text{R}}(X)\to\coint{0,\infty}$
is continuous at $\mu_0$.   Then $\Cal V\subseteq\Cal V^*$ (by (c)),
$\Cal C\in\Cal V^*$ and $V\setminus V'\in\Cal V^*$ whenever $V$,
$V'\in\Cal V^*$ and $V'\subseteq V$.   Because $\Cal V$ is closed under
finite intersections, it follows that $\Cal V^*$ includes the algebra of
subsets of $\Cal C$ generated by $\Cal V$ (313Ga);  in particular, any
finite union of members of $\Cal V$ belongs to $\Cal V^*$.

\medskip

{\bf (f)} Let us return to the open set $W_0\subseteq\Cal C$ and the
$\alpha_0\in\Bbb R$ of part (a).   Because $\tilde\nu_{\mu_0}$ is
$\tau$-additive and $\Cal V$ is a base for the topology of $\Cal C$ ((b)
above), there is a finite family $\Cal V_0\subseteq\Cal V$ such that
$V_0=\bigcup\Cal V_0$ is included in $W_0$ and
$\tilde\nu_{\mu_0}V_0>\alpha_0$.   But this means that

\Centerline{$\{\mu:\mu\in M^{\infty+}_{\text{R}}(X)$,
$\tilde\nu_{\mu}W_0>\alpha_0\}
\supseteq\{\mu:\mu\in M^{\infty+}_{\text{R}}(X)$,
$\tilde\nu_{\mu}V_0>\alpha_0\}$}

\noindent is a neighbourhood of $\mu_0$.   As $\mu_0$ is arbitrary,
$\{\mu:\tilde\nu_{\mu}W_0>\alpha_0\}$ is open;  as $W_0$ and $\alpha_0$
are arbitrary, $\mu\mapsto\tilde\nu_{\mu}$ is continuous.
}%end of proof of 495R

\leader{495S}{}\cmmnt{ There are many constructions which, in
particular cases, can be used as an alternative to the method of
495B-495D in setting up Poisson point processes.   I give one which
applies to the half-line $\coint{0,\infty}$ with Lebesgue measure.

\medskip

\noindent}{\bf Theorem}\dvAformerly{4{}95P} 
Let $\gamma>0$, and let $\nu$ be the Poisson
point process on $\coint{0,\infty}$, with Lebesgue measure, with intensity
$\gamma$.   Let $\lambda_0$ be the exponential distribution with
expectation $1/\gamma$, regarded as a Radon probability measure on
$\ooint{0,\infty}$, and $\lambda$  the corresponding product measure on
$\ooint{0,\infty}^{\Bbb N}$.   Define
$\phi:\ooint{0,\infty}^{\Bbb N}\to\Cal P(\coint{0,\infty})$ by setting
$\phi(x)=\{\sum_{i=0}^nx(i):n\in\Bbb N\}$ for
$x\in\ooint{0,\infty}^{\Bbb N}$.   Then $\phi$ is a measure space
isomorphism between $\ooint{0,\infty}^{\Bbb N}$ and a $\nu$-conegligible
subset of $\Cal P(\coint{0,\infty})$.

\cmmnt{\medskip

\noindent{\bf Remark} As I seem not to have mentioned exponential
distributions earlier in this treatise, I remark now that the
{\bf exponential distribution} with parameter $\gamma$ has distribution
function

\Centerline{$F(t)=0$ if $t<0$, $1-e^{-\gamma t}$ if $t\ge 0$,}

\noindent and probability density function

\Centerline{$f(t)=0$ if $t\le 0$, $\gamma e^{-\gamma t}$ if $t>0$;}

\noindent its expectation is

\Centerline{$\biggerint_0^{\infty}\gamma te^{-\gamma t}dt
=-\int_0^{\infty}\Bover{d}{dt}
  \bigl(\Bover{\gamma t+1}{\gamma}e^{-\gamma t}\bigr)dt
=\Bover1{\gamma}$.}

\noindent Because (when regarded as a Radon probability measure on
$\Bbb R$, following my ordinary rule set out in \S271) it gives
measure zero to $\ocint{-\infty,0}$, it can be identified with the
subspace measure on $\ooint{0,\infty}$, as here.
}%end of comment

\proof{{\bf (a)} For each $n\in\Bbb N$, $\#(S\cap[0,n])$ is finite for
$\nu$-almost every $S$;  so the set

\Centerline{$Q_0
=\{S:S\subseteq\coint{0,\infty}$, $\#(S\cap[0,n])$ is finite for every
$n\}$}

\noindent is $\nu$-conegligible.   Next, the sets
$\{S:S\cap\coint{n,n+1}\ne\emptyset\}$ are $\nu$-independent and have
measure $1-e^{-\gamma}>0$, so

\Centerline{$\{S:S\cap\coint{n,n+1}\ne\emptyset$ for infinitely many
$n\}$}

\noindent is $\nu$-conegligible (273K).   Finally, $\nu\{S:0\in S\}=0$,
so $Q=\{S:S\in Q_0$, $0\notin S$, $S$ is infinite$\}$ is
$\nu$-conegligible.   For
$S\in Q$, let $\sequencen{g_n(S)}$ be the increasing enumeration of $S$.
Let $\Tau$ be the $\sigma$-algebra of subsets of
$\Cal P(\coint{0,\infty})$ generated by sets of the form
$\{S:\#(S\cap E)=n\}$ where $E\subseteq\coint{0,\infty}$ has finite
measure and $n\in\Bbb N$.   Then, for $n\in\Bbb N$ and $\alpha\ge 0$,
$\{S:g_n(S)\le\alpha\}=\{S:\#(S\cap[0,\alpha])\ge n+1\}$ belongs to
$\Tau$, so $g_n$ is $\Tau$-measurable.   Set $h_0(S)=g_0(S)$,
$h_n(S)=g_n(S)-g_{n-1}(S)$ for $n\ge 1$,
and $h(S)=\sequencen{h_n(S)}$;  then $h:Q\to\ooint{0,\infty}^{\Bbb N}$
is a bijection, and its inverse is $\phi$.

\medskip

{\bf (b)} For each $k\in\Bbb N$,
$I_S=\{i:i\in\Bbb N$, $S\cap\coint{2^{-k}i,2^{-k}(i+1)}\ne\emptyset\}$
is infinite for every $S\in Q$.   So we can define
$g_{kn}:Q\to\ooint{0,\infty}$, for each $n$, by taking
$g_{kn}(S)=2^{-k}(j+1)$ if $j\in I_S$ and $\#(I_S\cap j)=n$.   Because
all the sets $\{S:j\in I_S\}$ belong to $\Tau$, each $g_{kn}$ is
$\Tau$-measurable, and $\sequence{k}{g_{kn}}$ is a non-increasing
sequence with limit $g_n$.   Set $h_{k0}(S)=g_{k0}(S)$ and
$h_{kn}(S)=g_{kn}(S)-g_{k,n-1}(S)$ for $n\ge 1$.   Then
$h_n=\lim_{k\to\infty}h_{kn}$.

\medskip

{\bf (c)} For any $n\in\Bbb N$, $j_0,\ldots,j_n\in\Bbb N$, $k\in\Bbb N$,
set $j'_r=\sum_{i=0}^rj_i$ for $r\le n$.   Then

$$\eqalignno{\nu\{S:\mskip5mu&S\in Q,\,h_{ki}(S)
=2^{-k}(j_i+1)\text{ for every }i\le n\}\cr
&=\nu\{S:S\in Q,\,g_{kr}(S)=2^{-k}(r+1+j'_r)
   \text{ for every }r\le n\}\cr
&=\nu\{S:S\cap\coint{2^{-k}(r+j'_r),2^{-k}(r+1+j'_r)}\ne\emptyset
     \text{ for every }r\le n,\cr
&\mskip70mu
  S\cap\coint{0,2^{-k}j_0}=\emptyset,\cr
&\mskip70mu
  S\cap\coint{2^{-k}(r+1+j'_r),2^{-k}(r+1+j'_{r+1})}=\emptyset
     \text{ for every }r<n\}\cr
&=(1-\exp(-2^{-k}\gamma))^{n+1}\exp(-2^{-k}\gamma j_0)
  \prod_{r<n}\exp(-2^{-k}\gamma(j'_{r+1}-j'_r))\cr
&=\prod_{i=0}^n(1-\exp(-2^{-k}\gamma))\exp(-2^{-k}\gamma j_i).\cr}$$

\noindent This means that the $h_{ki}$, for $i\in\Bbb N$ are
independent, with

\Centerline{$\Pr(h_{ki}=2^{-k}(j+1))
=(1-e^{-2^{-k}\gamma})e^{-2^{-k}\gamma j}$}

\noindent for each $j$.
Since $h_{ki}\to h_i\,\,\nu$-a.e.\ for each $i$, $\sequence{i}{h_i}$ is
also independent (367W).   Now, for any $\alpha>0$,

$$\eqalignno{\Pr(h_{ki}\le\alpha)
&=\sum_{2^{-k}(j+1)\le\alpha}
  (1-\exp(-2^{-k}\gamma))\exp(-2^{-k}\gamma j)\cr
&=1-\exp(-2^{-k}\gamma\lfloor 2^k\alpha\rfloor)
\to 1-e^{-\gamma\alpha}\cr}$$

\noindent as $k\to\infty$.   So

$$\eqalignno{\Pr(h_i\le\alpha)
&=\inf_{\beta>\alpha}\liminf_{k\to\infty}\Pr(h_{ki}\le\beta)\cr
\displaycause{271L}
&=\inf_{\beta>\alpha}1-e^{-\gamma\beta}
=1-e^{-\gamma\alpha}\cr}$$

\noindent for every $\alpha\ge 0$ and every $i\in\Bbb N$.

\medskip

{\bf (d)} Accordingly $\sequence{i}{h_i}$ is an independent sequence of
random variables, each exponentially distributed with expectation
$1/\gamma$.   It follows that
$h:Q\to\ooint{0,\infty}^{\Bbb N}$ is \imp\ for the subspace measure
$\nu_Q$ and $\lambda$ (254G).

Observe next that if $E\subseteq\coint{0,\infty}$ is Lebesgue measurable
and $n\in\Bbb N$, then

\Centerline{$\{x:x\in\ooint{0,\infty}^{\Bbb N}$, $\#(\phi(x)\cap E)=n\}
=\bigcup_{I\in[\Bbb N]^n}\{x:\sum_{i=0}^jx(i)\in E\iff j\in I\}
\in\Lambda$,}

\noindent writing $\Lambda$ for the domain of $\lambda$.   So $\phi$
is $(\Lambda,\Tau)$-measurable.   Now, for any $W\in\Tau$,

\Centerline{$\lambda\phi^{-1}[W]=\nu(h^{-1}[\phi^{-1}[W]])
=\nu(W\cap Q)=\nu W$.}

\noindent So $\phi$ is \imp\ for $\lambda$ and for $\nu\restr\Tau$.
Since $\lambda$ is complete and $\nu$ is defined as the completion of
its restriction to $\Tau$, $\phi$ is \imp\ for $\lambda$ and $\nu$.
Thus $\phi$ and $h$ are the two halves of an isomorphism between
$(\ooint{0,\infty}^{\Bbb N},\lambda)$ and the subspace $(Q,\nu_Q)$, as
claimed.
}%end of proof of 495S

%\leader{495T}{}
%\leader{495U}{}
%\leader{495V}{}
%\leader{495W}{}

\exercises{\leader{495X}{Basic exercises}
\spheader 495Xb
Let $(X,\Sigma,\mu)$ be an atomless measure space, and
$\nu$ a Poisson point process on $X$.   (i)
Show that $[X]^{\le\omega}$ has full outer measure for $\nu$.   (ii)
Show that if $\mu$ is semi-finite then $[X]^{\le\omega}$ is
conegligible iff $\mu$ is $\sigma$-finite.   (iii) Show that if $\mu$ is
semi-finite, then $[X]^{<\omega}$ is non-negligible iff $[X]^{<\omega}$
is conegligible iff $\mu$ is totally finite.
%495E

\spheader 495Xc Let $(X,\Sigma,\mu)$ be a strictly localizable measure space and
$\familyiI{X_i}$ a decomposition of $X$.
Let $\nu$ be the Poisson point process on $X$ with intensity
$1$, and for each $i\in I$ let $\nu_i$ be the Poisson point
process on $X_i$ with intensity $1$ corresponding to the subspace
measure $\mu_{X_i}$ on $X_i$. Let $\lambda$ be the product of the family
$\familyiI{\nu_i}$.   Show that the map
$S\mapsto\familyiI{S\cap X_i}:\Cal PX\to\prod_{i\in I}\Cal PX_i$
is \imp\ for $\nu$ and $\lambda$.
%495F

\sqheader 495Xd Let $(X,\Sigma,\mu)$ be an atomless
measure space and $\frak T$ a topology on $X$ such that $X$ is covered
by a sequence of open sets of finite outer measure.   Let $\nu$
be a Poisson point process on $X$.   Show that
$\nu$-almost every set $S\subseteq X$ is locally finite in the sense
that $X$ is covered by the open sets meeting $S$ in finite sets;  in
particular, if $X$ is T$_1$, then
$\nu$-almost every subset of $X$ is closed.
%495E

\sqheader 495Xe(i) Let $(X,\Sigma,\mu)$ be an atomless measure space, and
for $\gamma>0$ let $\nu_{\gamma}$ be the Poisson point process on $X$
with intensity $\gamma$.   Show that for any $\gamma$, $\delta>0$ the map
$(S,T)\mapsto S\cup T:\Cal PX\times\Cal PX\to\Cal PX$ is \imp\ for the
product measure $\nu_{\gamma}\times\nu_{\delta}$ and
$\nu_{\gamma+\delta}$.   (ii) Let $X$ be a set, $\Sigma$ a $\sigma$-algebra
of subsets of $X$, and $\familyiI{\mu_i}$ a
countable family of measures with domain
$\Sigma$ such that $\mu=\sum_{i\in I}\mu_i$ is atomless.   Let $\nu$,
$\nu_i$ be the Poisson point processes with intensity $1$ corresponding to
the measures $\mu$, $\mu_i$.   Show that the map
$\familyiI{S_i}\mapsto\bigcup_{i\in I}S_i:(\Cal PX)^I\to\Cal PX$ is \imp\
for the product measure $\prod_{i\in I}\nu_i$ and $\nu$.
(iii) Compare with 495Xc(i).
%495E

\spheader 495Xf Let $(X,\Sigma,\mu)$ be an atomless semi-finite measure
space, and $\nu$ the Poisson point process on $X$ with intensity $1$.
Show that $\nu$ is perfect iff $\mu$ is.
%495G

\spheader 495Xg Let $(\frak A,\bar\mu)$ be a measure algebra.   Show
that there is a probability measure $\lambda$ on $\BbbR^{\frak A^f}$
such that (i) for every $a\in\frak A^f$ the corresponding marginal
measure on $\Bbb R$ is the Poisson distribution with expectation
$\bar\mu a$ (ii) whenever $a_0,\ldots,a_n\in\frak A^f$ are disjoint, the
functions $z\mapsto z(a_i):\BbbR^{\frak A^f}\to\Bbb R$ are
stochastically independent with respect to $\lambda$.   \Hint{prove the
result for finite $\frak A$ and use 454D.}   Use this to prove 495M.
%495M

\spheader 495Xh Let $U$ be a Hilbert space.   Show
that there are a probability algebra $(\frak B,\bar\lambda)$ and a
linear operator $T:U\to L^2(\frak B)$ such that (i)
for every $u\in U$, $Tu$ has a normal distribution
with expectation $0$ and variance $\|u\|_2^2$ (ii) if $\familyiI{u_i}$
is an orthogonal family in $U$ then $\familyiI{Tu_i}$
is $\bar\lambda$-independent.   \Hint{see the proof of 456K.}
%495N

\spheader 495Xi Let $\nu$ be the Poisson point process with intensity $1$
on $\coint{0,\infty}$ with Lebesgue measure.   Set
$Q_0=\{S:S\subseteq\coint{0,\infty}$, $S\cap[0,n]$ is finite for every
$n\}$ and for $S\in Q_0$ set $\psi(S)(t)=\#(S\cap[0,t])$ for
$t\in\coint{0,\infty}$.   Show that $\psi$ is \imp\ for the subspace
measure $\nu_{Q_0}$ and the distribution on $\BbbR^{\coint{0,\infty}}$
corresponding to the Poisson process of 455Xh.
%495N

\spheader 495Xj Let $(Y,\Tau,\nu)$ be a probability space, and
$\lambda_0$ the exponential distribution with expectation $1$, regarded
as a Radon measure on $\ooint{0,\infty}$.   Let
$\lambda$ be the product measure $\lambda_0^{\Bbb N}\times\nu^{\Bbb N}$
on $\ooint{0,\infty}^{\Bbb N}\times Y^{\Bbb N}$.   Set
$\phi(x,y)=\{(\sum_{i=0}^nx(i),y(n)):n\in\Bbb N\}$ for
$x\in\ooint{0,\infty}^{\Bbb N}$ and $y\in Y^{\Bbb N}$.   Show that
$\phi:\ooint{0,\infty}^{\Bbb N}\times Y^{\Bbb N}
\to\Cal P(\ooint{0,\infty}\times Y)$ is a measure space isomorphism
between $(\ooint{0,\infty}^{\Bbb N}\times Y^{\Bbb N},\lambda)$ and a
conegligible set for the Poisson point process on
$\ooint{0,\infty}\times Y$ with intensity $1$ for the c.l.d.\ product
measure $\mu_L\times\nu$, where $\mu_L$ is Lebesgue measure.
%495N

\spheader 495Xk Let $\Cal C$ be the family of closed subsets of
$\coint{0,\infty}$.   Let $\rho$ be the usual metric on
$\coint{0,\infty}$ and $\tilde\rho$ the corresponding Hausdorff metric
on $\Cal C\setminus\{\emptyset\}$ (4A2T).
Let $\nu$ be the Poisson point process
on $\coint{0,\infty}$ with intensity $1$ over Lebesgue measure.   Show that
every member of $\Cal C\setminus\{\emptyset\}$ has a $\nu$-negligible
$\tilde\rho$-neighbourhood.
%495Q

\spheader 495Xl Show that the topology on $M_{\text{R}}^+(X)$
described in 495R is just the topology induced by the natural embedding
of $M_{\text{R}}(X)$ into $C_k(X)^{\sim}$ (436J) and the weak topology
$\frak T_s(C_k(X)^{\sim},C_k(X))$, where $C_k(X)$ is the Riesz space
of continuous real-valued functions on $X$ with compact support.

\spheader 495Xm Let $\Cal C$ be the set of closed subsets of
$\coint{0,\infty}$ with its Fell topology.   For
$\delta\in\ocint{0,1}$ let $\lambda_{\delta}$ be the measure on
$\{0,1\}^{\Bbb N}$ which is the product of copies of the measure on
$\{0,1\}$ in which $\{1\}$ is given measure $\delta$.   Define
$\phi_{\delta}:\{0,1\}^{\Bbb N}\to\Cal C$ by setting
$\phi_{\delta}(x)=\{n\delta:n\in\Bbb N$, $x(n)=1\}$, and let
$\tilde\nu_{\delta}$ be the Radon measure
$\lambda_{\delta}\phi_{\delta}^{-1}$ on $\Cal C$.   Show that the Radon
measure on $\Cal C$ representing the Poisson
point process on $\coint{0,\infty}$ with intensity $1$ over Lebesgue
measure is the limit
$\lim_{\delta\downarrow 0}\tilde\nu_{\delta}$ for the narrow
topology on the space of Radon probability measures on $\Cal C$.
%495R

\spheader 495Xn Show that the standard gamma distribution with
expectation $1$ is the exponential distribution with expectation $1$.
%495S

\spheader 495Xo Let $r\ge 1$ be an integer;  let $\mu$ be Lebesgue
measure on $\BbbR^r$ and $\beta_r$ the volume of the unit ball in
$\BbbR^r$.   Set $\psi(t)=(t/\beta_r)^{1/r}$ for $t\ge 0$, so that the
volume of a ball of radius $\psi(t)$ is $t$.   Let $S_{r-1}$ be the unit
sphere in $\BbbR^r$ and $\theta$ the invariant Radon probability measure
on $S_{r-1}$, so that $\theta$ is a multiple of $(r-1)$-dimensional
Hausdorff measure (see 476I).   Let $\lambda_0$ be the exponential
distribution with expectation $1$, regarded as a Radon probability
measure on $\ooint{0,\infty}$, and $\lambda$ the product measure
$\lambda_0^{\Bbb N}\times\theta^{\Bbb N}$ on
$\ooint{0,\infty}^{\Bbb N}\times S_{r-1}^{\Bbb N}$.   Set

\Centerline{$\phi(x,z)=\{\psi(\sum_{i=0}^nx(i))z(n):n\in\Bbb N\}$}

\noindent for $x\in\ooint{0,\infty}^{\Bbb N}$ and
$z\in S_{r-1}^{\Bbb N}$.   Show that
$\phi:\ooint{0,\infty}^{\Bbb N}\times S_{r-1}^{\Bbb N}
\to\Cal P(\BbbR^r)$ is a measure space isomorphism between
$\ooint{0,\infty}^{\Bbb N}\times S_{r-1}^{\Bbb N}$ and a conegligible
set for the Poisson point process on $\BbbR^r$ with intensity $1$.
%495Xj 495S


\leader{495Y}{Further exercises (a)}
%\spheader 495Ya
Let $U$ be an $L$-space.   Show that there are a
probability algebra $(\frak B,\bar\lambda)$ and a linear operator
$T:U\to L^0(\frak B)$ such that (i) for every
$u\in U$, $Tu$ has a Cauchy distribution with centre
$0$ and scale parameter $\|u\|$ (ii) if $\familyiI{u_i}$ is a disjoint
family in $U$ then $\familyiI{Tu_i}$ is $\bar\lambda$-independent.
%495N

\spheader 495Yb Let $U$ be an $L$-space.   Show
that there are a probability algebra $(\frak B,\bar\lambda)$ and a
linear operator $T:U\to L^1(\frak B,\bar\lambda)$ such that (i)
for every $u\in U^+$, $Tu$ has a standard gamma distribution (definition:
455Xj) with expectation $\|u\|$ (ii) if
$\familyiI{u_i}$ is a disjoint family in $U$ then
$\familyiI{Tu_i}$ is $\bar\lambda$-independent.
%495N

\spheader 495Yc Let $(\frak A,\bar\mu)$ be a measure algebra.   For
$\alpha$, $y\in\Bbb R$ set $h_y(\alpha)=e^{iy\alpha}$, and let
$\bar h_y:L^0(\frak A)\to L^0_{\Bbb C}(\frak A)$ (definition:
366M\formerly{3{}64Yn}) be
the corresponding operator (to be defined, following the ideas of
364H\formerly{3{}64I}
or otherwise).    Show that there are a probability algebra
$(\frak B,\bar\lambda)$ and a positive linear operator
$T:L^1(\frak A,\bar\mu)\to L^1(\frak B,\bar\lambda)$ such that
(i) $\|Tu\|_1=\|u\|_1$ whenever $u\in L^1(\frak A,\bar\mu)^+$ (ii)
$\familyiI{Tu_i}$ is $\bar\lambda$-independent in $L^0(\frak B)$
whenever $\familyiI{u_i}$ is a disjoint family in
$L^1(\frak A,\bar\mu)$ (iii)
$\int\bar h_y(Tu)d\bar\lambda=\exp(\int(\bar h_y(u)-\chi 1)d\bar\mu)$
for every $u\in L^1(\frak A,\bar\mu)$ and $y\in\Bbb R$.
%495P

\spheader 495Yd
\discrversionA{\footnote{Revised 2008.}}{}
Let $(X,\rho)$ be a totally bounded metric space,
$\mu$ a Radon measure on $X$ and $\gamma>0$.   Let
$\Cal C$ be the set of closed subsets of $X$, and $\tilde\nu$ the
quasi-Radon measure of 495Q;  let $\tilde\rho$ be the
Hausdorff metric on $\Cal C\setminus\{\emptyset\}$.   Show that the
subspace measure on $\Cal C\setminus\{\emptyset\}$ induced by $\tilde\nu$
is a Radon measure for the topology induced by $\tilde\rho$.
%495Q
}%end of exercises

\endnotes{
\Notesheader{495} The underlying fact on which this section relies is
that the Poisson distributions form a one-parameter semigroup of
infinitely divisible distributions, with
$\nu_{\alpha}*\nu_{\beta}=\nu_{\alpha+\beta}$ for all $\alpha$,
$\beta>0$.   Other well-known families with this property are normal
distributions, Cauchy distributions and gamma distributions;  for each
of these we have results corresponding to 495B and 495N (495Xh, 495Ya,
495Yb).   The same distributions appeared, for the same reason, in the
L\'evy processes of \S455. % 455Xh 455Xi 455Xj
Observe that the version for the normal distribution is related to the
Gaussian processes of \S456.   The `compound Poisson' distributions of
495P provide further examples, which approach the general form of
infinitely divisible distributions ({\smc Lo\`eve 77}, \S23, or
{\smc Fristedt \& Gray 97}, \S16.3).

The special feature of the Poisson point process, in this context, is
the fact that (for atomless measure spaces $(X,\mu)$) it can be
represented by a measure on $\Cal PX$ rather than on some abstract
auxiliary space (495D);  so that we have a notion of `random subset',
and can discuss the expected topological properties of subsets of $X$
(495Xb, 495Xd).   In Euclidean spaces the geometric properties of these
random subsets are also of great interest;  see
{\smc Meester \& Roy 96}.   Here I look at the relations between this construction and others
which have been prominent in this book, such as \imp\ functions (495J)
and disintegrations (495K-495L).   In the latter we find ourselves in an
interesting difficulty.   If, as in 495K, we have a measure space
$X=\tilde X\times[0,1]$, where $\tilde X$ is an atomless measure space,
then it is natural to suppose that our Poisson process on $X$ can be
represented by picking a random subset $T$ of $\tilde X$ and then, for
each $t\in T$, a random $(t,\alpha)\in X$.   The obvious model for this
idea is the map $(T,z)\mapsto\{(t,z(t)):t\in T\}:
\Cal P\tilde X\times[0,1]^{\tilde X}\to\Cal PX$.   The problem with this
model is that the map is simply not measurable for the standard
$\sigma$-algebras on $\Cal P\tilde X$,
$\Cal P\tilde X\times[0,1]^{\tilde X}$ and $\Cal PX$.   When we have a
canonical ordering in order type $\omega$ of almost every subset of
$\tilde X$ (`almost every' with respect to the Poisson point process on
$\tilde X$, of course), as in 495Xo, there can be a way around this,
cutting $[0,1]^{\tilde X}$ down to a countable product and re-inventing
the representation of pairs $(T,z)$ as subsets of $X$.   But in the
general case it seems that we have to set up a disintegration of the
Poisson point process on $X$ over the Poisson point process on
$\tilde X$ which does not correspond to any measure on a product
$\Cal P\tilde X\times\Omega$.

Following my usual custom, I have expressed the theorems of this section
in terms of arbitrary (atomless) measure spaces.   The results are not
quite without interest when applied to totally finite measures, but
their natural domain is the class of non-totally-finite $\sigma$-finite
measures, as in 495Q-495S.   There is an unavoidable obstacle if we wish
to extend the ideas to measure spaces which are not atomless.   The
functions $S\mapsto\#(S\cap E)$ may no longer have Poisson
distributions, since if $E$ is a singleton of positive measure then we
shall have a non-trivial two-valued random variable.   In 495Q-495R I
take one of the possible resolutions of this, with measures $\tilde\nu$
on spaces of subsets for which at least the sets
$\{S:S\cap E=\emptyset\}$, for disjoint $E$, are independent.   An
alternative which is sometimes appropriate is to work with functions
$h:X\to\Bbb N$ and $\sum_{x\in E}h(x)$ in place of subsets $S$ of $X$
and $\#(S\cap E)$;  see {\smc Fristedt \& Gray 97}, \S29.

In 495M-495O we have a little cluster of results which are relevant to
rather different questions, to which I will return in Chapter 52 of
Volume 5.   % 526D 528H
The objective here is to connect the structure of a measure
algebra or Banach lattice of arbitrarily large cellularity with
something which can be realized in a probability space.   In each case,
disjointness is transformed into stochastic independence.
Once again, the special feature of the Poisson point process is that we
have a concrete representation of a linear operator which can also be
described in a more abstract way (495O).

The construction of 495B-495D seems to be the most straightforward way
to generate Poisson point processes.   It fails however to give a direct
interpretation of one of the most important approaches to these
processes, as limits of purely atomic processes in which sets are chosen
by including or excluding individual points independently
(495Xm).   In order to make sense of the limit here it seems that
we need to put some further structure onto the underlying measure space,
and `$\sigma$-finite locally compact Radon measure space' is sufficient
to give a positive result (495R).

}%end of notes

\discrpage

\frfilename{mt496.tex}
\versiondate{27.5.09}

\def\chaptername{Further topics}
\def\sectionname{Maharam submeasures}

\newsection{496\dvAnew{2008}}

The old problem of characterizing measurable algebras led, among other
things, to the concepts of `Maharam submeasure' and `Maharam algebra'
(\S393).   It is known that these can be very different from measures
(\S394), but the differences are not well understood.   In this section I
will continue the work of \S393 by showing that some, at least, of the
ways in which topologies and measures interact apply equally to Maharam
submeasures.   The most important of these interactions are associated with
the concept of `Radon measure', so the first step is to find a
corresponding notion of `Radon submeasure' (496C, 496Y).   In
496D-496K %496D 496E 496F 496G 496H 496I 496J 496K
I run
through a handful of theorems which parallel results in \S\S416 and
431-433.  %431 432 433
Products of submeasures remain problematic, but something can be
done (496L-496M).

\leader{496A}{Definitions}\cmmnt{ As
we have hardly had `submeasures' before
in this volume, I repeat the essential definitions from Chapter 39.}
If $\frak A$ is a Boolean algebra, a {\bf submeasure} on $\frak A$ is a
functional $\mu:\frak A\to[0,\infty]$
such that $\mu 0 = 0$ and $\mu a\le\mu(a\Bcup b)\le\mu a+\mu b$ for all
$a$, $b\in\frak A$\cmmnt{ (392A)}.
$\mu$ is {\bf strictly positive} if $\mu a>0$ for every
$a\in\frak A\setminus\{0\}$\cmmnt{ (392Ba)}, {\bf exhaustive} if
$\lim_{n\to\infty}\mu a_n=0$ for every disjoint sequence $\sequencen{a_n}$
in $\frak A$\cmmnt{ (392Bb)},
{\bf totally finite} if $\mu 1<\infty$\cmmnt{ (392Bd)},
a {\bf Maharam submeasure} if it is totally finite and
$\lim_{n\to\infty}\mu a_n=0$ for every
non-increasing sequence $\sequencen{a_n}$ in $\frak A$ with zero
infimum\cmmnt{ (393A)}.
A Maharam submeasure is sequentially
order-continuous\cmmnt{ (393Ba)}.
If $\mu$ and $\nu$ are two submeasures on a Boolean algebra $\frak A$, then
$\mu$ is {\bf absolutely continuous} with respect to $\nu$ if for every
$\epsilon>0$ there is a $\delta>0$ such that $\mu a\le\epsilon$ whenever
$\nu a\le\delta$\cmmnt{ (392Bg)}.
A {\bf Maharam algebra} is a Dedekind $\sigma$-complete Boolean algebra
which carries a strictly positive Maharam
submeasure\cmmnt{ (393E)}.

\leader{496B}{Basic facts }\cmmnt{I list some elementary ideas for future
reference.

\medskip

}{\bf (a)} Let $\mu$ be a submeasure on a Boolean algebra $\frak A$.

\medskip

\quad{\bf (i)} Set $I=\{a:a\in\frak A$, $\mu a=0\}$.
\cmmnt{Clearly }$I$ is an
ideal of $\frak A$;  write $\frak C$ for the quotient Boolean algebra
$\frak A/I$.   Then we have a strictly positive
submeasure $\bar\mu$ on $\frak C$ defined by setting
$\bar\mu a^{\ssbullet}=\mu a$ for every $a\in\frak A$.   \prooflet{\Prf\
If $a^{\ssbullet}=b^{\ssbullet}$ then

\Centerline{$\mu(a\Bsetminus b)=\mu(b\Bsetminus a)=\mu(a\Bsymmdiff b)=0$,
\quad$\mu a=\mu(a\Bcap b)=\mu b$;}

\noindent so $\bar\mu$ is well-defined.   The formulae defining
`submeasure' transfer directly from $\mu$ to $\bar\mu$.   If
$\bar\mu a^{\ssbullet}=0$ then $\mu a=0$, $a\in I$ and $a^{\ssbullet}=0$,
so $\bar\mu$ is strictly positive.\ \Qed}

\medskip

\quad{\bf (ii)} If $\mu$ is exhaustive, so is $\bar\mu$.
\prooflet{\Prf\ If $\sequencen{a_n}$ is a
sequence in $\frak A$ such that $\sequencen{a_n^{\ssbullet}}$ is disjoint
in $\frak A/I$, set $b_n=a_n\Bsetminus\sup_{i<n}a_i$ for each $n$;  then
$\sequencen{b_n}$ is disjoint so

\Centerline{$\lim_{n\to\infty}\bar\mu a_n^{\ssbullet}
=\lim_{n\to\infty}\bar\mu b_n^{\ssbullet}
=\lim_{n\to\infty}\mu b_n=0$;}

\noindent thus $\bar\mu$ is exhaustive.\ \Qed}

\medskip

\quad{\bf (iii)} If $\frak A$ is Dedekind $\sigma$-complete and $\mu$ is a
Maharam submeasure, then $\frak C$
is a Maharam algebra.  \prooflet{\Prf\ As $\mu$ is sequentially
order-continuous, $I$ is a $\sigma$-ideal and $\frak C$ is Dedekind
$\sigma$-complete (314C).   Now suppose that $\sequencen{a_n}$ is a
sequence in $\frak A$ such that $\sequencen{a_n^{\ssbullet}}$ is
non-increasing and has zero infimum in $\frak C$.   Set
$b_n=\inf_{i\le n}a_i$ for each $n$, and $a=\inf_{n\in\Bbb N}a_n$;  then
$a^{\ssbullet}=0$ so $\mu a=0$ and (again because $\mu$ is sequentially
order-continuous)

\Centerline{$\lim_{n\to\infty}\bar\mu a_n^{\ssbullet}
=\lim_{n\to\infty}\bar\mu b_n^{\ssbullet}
=\lim_{n\to\infty}\mu b_n
=\mu a=0$.}

\noindent Since we already know that $\bar\mu$ is a strictly positive
submeasure, it is a strictly positive Maharam submeasure and $\frak C$ is a
Maharam algebra.\ \Qed}

In this context I will say that $\frak C$ is
{\bf the Maharam algebra of} $\mu$.

\spheader 496Bb If $\mu$ is a strictly positive totally finite submeasure
on a Boolean algebra $\frak A$, there is an associated metric
$(a,b)\mapsto\mu(a\Bsymmdiff b)$\cmmnt{
(392H)};
the corresponding metric completion $\widehat{\frak A}$ admits a continuous
extension of $\mu$ to a strictly positive submeasure $\hat\mu$ on
$\widehat{\frak A}$.   If $\mu$ is exhaustive, then $\hat\mu$ is a Maharam
submeasure and $\widehat{\frak A}$ is a Maharam
algebra\cmmnt{ (393H)}.
A Maharam algebra is ccc, therefore Dedekind
complete, and \wsid\cmmnt{ (393Eb)}.

\spheader 496Bc If $\mu$ is a submeasure defined on an algebra
$\Sigma$ of subsets of a set $X$, I will say that the {\bf null ideal}
$\Cal N(\mu)$ of $\mu$ is
the ideal of subsets of $X$ generated by $\{E:E\in\Sigma$, $\mu E=0\}$.
If $\Cal N(\mu)\subseteq\Sigma$ I will say that $\mu$ is {\bf complete}.
Generally,
the {\bf completion} of $\mu$ is the functional $\hat\mu$ defined by saying
that $\hat\mu(E\symmdiff A)=\mu E$ whenever $E\in\Sigma$ and
$A\in\Cal N(\mu)$;\cmmnt{  it is elementary to check that} $\hat\mu$ is a
complete submeasure.

\spheader 496Bd If $\frak A$ is a Maharam algebra, and $\mu$, $\nu$ are two
strictly positive Maharam submeasures on $\frak A$, then each is absolutely
continuous with respect to the other\cmmnt{ (393F)}.
Consequently the metrics associated with them\cmmnt{ are uniformly
equivalent, and} induce the same topology, the {\bf Maharam-algebra
topology} of $\frak A$\cmmnt{ (393G)}.

\leader{496C}{Radon submeasures} Let $X$ be a Hausdorff space.
A {\bf totally finite Radon submeasure} on $X$
is a complete totally finite submeasure
$\mu$ defined on a $\sigma$-algebra $\Sigma$ of subsets of $X$
such that (i) $\Sigma$ contains every open set (ii)
$\inf\{\mu(E\setminus K):K\subseteq E$ is compact$\}=0$
for every $E\in\Sigma$.

In this context I
will say that a set $E\in\Sigma$ is {\bf self-supporting} if
$\mu(E\cap G)>0$ whenever $G\subseteq X$ is open and $G\cap E\ne\emptyset$.

\leader{496D}{Proposition}
Let $\mu$ be a totally finite Radon submeasure on a Hausdorff space $X$ with domain
$\Sigma$.

(a) $\mu$ is a Maharam submeasure.

(b) $\inf\{\mu(G\setminus E):G\supseteq E$ is open$\}=0$ for every
$E\in\Sigma$.

(c) If $E\in\Sigma$ there is a relatively closed $F\subseteq E$ such that
$F$ is self-supporting and $\mu(E\setminus F)=0$.

(d) If $E\in\Sigma$ and $\epsilon>0$ there is a compact self-supporting
$K\subseteq E$ such that $\mu(E\setminus K)\le\epsilon$.

\proof{{\bf (a)}
Let $\sequencen{E_n}$ be a non-increasing
sequence in $\Sigma$ with empty intersection.   \Quer\ If
$\inf_{n\in\Bbb N}\mu E_n=\epsilon>0$, then for each $n\in\Bbb N$ choose a
compact set $K_n\subseteq E_n$ such that
$\mu(E_n\setminus K_n)\le 2^{-n-2}\epsilon$.   For each $n\in\Bbb N$,

\Centerline{$\mu(E_n\setminus\bigcap_{i\le n}K_i)
\le\sum_{i=0}^n\mu(E_i\setminus K_i)<\epsilon\le\mu E_n$,}

\noindent so $\bigcap_{i\le n}K_i\ne\emptyset$.   There is therefore a
point in $\bigcap_{n\in\Bbb N}K_n\subseteq\bigcap_{n\in\Bbb N}E_n$.\ \BanG\
As $\sequencen{E_n}$ is arbitrary, $\mu$ is a Maharam submeasure.

\medskip

{\bf (b)} We have only to observe that

$$\eqalign{&\inf\{\mu(G\setminus E):G\supseteq E\text{ and }G
  \text{ is open}\}\cr
&\mskip150mu
\le\inf\{\mu((X\setminus E)\setminus K):K\subseteq X\setminus E
  \text{ is compact}\}
=0.\cr}$$

\medskip

{\bf (c)} Let $\Cal G$ be the family of open subsets $G$ of $X$ such that
$\mu(E\cap G)=0$, and $H=\bigcup\Cal G$.   Then $\Cal G$ is
upwards-directed.   If $\epsilon>0$ there is a compact set
$K\subseteq E\cap H$ such that $\mu(E\cap H\setminus K)\le\epsilon$;  now
there is a $G\in\Cal G$ such that $K\subseteq G$ and
$\mu(E\cap H)\le\epsilon+\mu K=\epsilon$.   As $\epsilon$ is arbitrary,
$H\in\Cal G$;  set $F=E\setminus H$.

\medskip

{\bf (d)} There is a compact $K_0\subseteq E$ such that
$\mu(E\setminus K_0)\le\epsilon$;  by (c), there is a closed
self-supporting $K\subseteq K_0$ such that $\mu(K_0\setminus K)=0$.
}%end of proof of 496D

\vleader{84pt}{496E}{Theorem} Let $X$ be a Hausdorff space and $\Cal K$ the
family of
compact subsets of $X$.   Let $\phi:\Cal K\to\coint{0,\infty}$ be a
bounded functional such that

\inset{($\alpha$) $\phi\emptyset=0$ and
$\phi K\le\phi(K\cup L)\le\phi K+\phi L$ for all $K$,
$L\in\Cal K$;

($\beta$) whenever $K\in\Cal K$ and $\epsilon>0$ there is an $L\in\Cal K$
such that $L\subseteq X\setminus K$ and $\phi K'\le\epsilon$ whenever
$K'\in\Cal K$ is disjoint from $K\cup L$;

($\gamma$) whenever $K$, $L\in\Cal K$ and $K\subseteq L$ then
$\phi L\le\phi K+\sup\{\phi K':K'\in\Cal K$, $K'\subseteq L\setminus K\}$.
}

\noindent Then there is a unique totally finite Radon submeasure on $X$ extending $\phi$.

\proof{{\bf (a)} For $A\subseteq X$ write
$\phi_*A=\sup\{\phi K:K\subseteq A$ is compact$\}$.   Then $\phi_*$ extends
$\phi$, by ($\alpha$).  Also
$\phi_*(\bigcup_{n\in\Bbb N}G_n)\le\sum_{n=0}^{\infty}\phi_*G_n$ for every
sequence $\sequencen{G_n}$ of open subsets of $X$.
\Prf\ If $K\subseteq\bigcup_{n\in\Bbb N}G_n$ is compact, it is
expressible as $\bigcup_{i\le n}K_i$ where $n\in\Bbb N$
and $K_i\subseteq G_i$ is compact for every $i\le n$ (4A2Fj).   Now

\Centerline{$\phi K\le\sum_{i=0}^n\phi K_i
\le\sum_{i=0}^{\infty}\phi_*G_i$.}

\noindent As $K$ is arbitrary,
$\phi_*(\bigcup_{n\in\Bbb N}G_n)\le\sum_{n=0}^{\infty}\phi_*G_n$.\ \QeD\
In particular, because $\phi\emptyset=0$,
$\phi_*(G\cup H)\le\phi_*G+\phi_*H$ for all open $G$, $H\subseteq X$.

\medskip

{\bf (b)} Let $\Sigma$ be the family of subsets $E$
of $X$ such that for every $\epsilon>0$ there is a $K\subseteq X$ such that
$K\cap E$ and $K\setminus E$ are both compact and
$\phi_*(X\setminus K)\le\epsilon$.   Then $\Sigma$ is an algebra of subsets
of $X$ including $\Cal K$.   \Prf\ (i) Of course $X\setminus E\in\Sigma$
whenever $E\in\Sigma$.   (ii) If $E$, $F\in\Sigma$ and $\epsilon>0$, let
$K$, $L\subseteq X$ be such that $K\cap E$, $K\setminus E$, $L\cap F$ and
$L\setminus F$ are all compact and $\phi_*(X\setminus K)$,
$\phi_*(X\setminus L)$ are both at most $\bover12\epsilon$.   Then
$(K\cap L)\cap(E\cup F)$ and $(K\cap L)\setminus(E\cup F)$ are both
compact, and

\Centerline{$\phi_*(X\setminus(K\cap L))
\le\phi_*(X\setminus K)+\phi_*(X\setminus L)\le\epsilon$.}

\noindent As $\epsilon$ is
arbitrary, $E\cup F\in\Sigma$.   (iii) By hypothesis ($\beta$),
$\Cal K\subseteq\Sigma$.\ \Qed

\medskip

{\bf (c)} $\Sigma$ is a $\sigma$-algebra of subsets of $X$.   \Prf\ Let
$\sequencen{E_n}$ be a sequence in $\Sigma$ with intersection $E$, and
$\epsilon>0$.   For each $n\in\Bbb N$ let $K_n\subseteq X$ be such that
$K_n\cap E_n$ and $K_n\setminus E_n$ are compact and
$\phi_*(X\setminus K_n)\le 2^{-n}\epsilon$;  set
$K=\bigcap_{n\in\Bbb N}K_n$.   Set $L=\bigcap_{n\in\Bbb N}K_n\cap E_n$, so
that $L\subseteq E$ is compact, and let $L'\subseteq X\setminus L$ be a
compact set such that $\phi_*(X\setminus(L\cup L'))\le\epsilon$;  set
$K'=K\cap(L\cup L')$.   Then $\phi_*(X\setminus K')\le 3\epsilon$.
As $L'\cap L=\emptyset$ there is an $n\in\Bbb N$ such that
$L'\cap\bigcap_{i\le n}K_i\cap E_i$ is empty.   Now

\Centerline{$K\cap L'
\subseteq\bigcup_{i\le n}(X\setminus(K_i\cap E_i))\cap\bigcap_{i\le n}K_i
\subseteq\bigcup_{i\le n}X\setminus E_i
\subseteq X\setminus E$,}

\noindent so $K'\cap E=K\cap L$ and $K'\setminus E=K\cap L'$ are compact.
As $\epsilon$ is arbitrary, $E\in\Sigma$.\ \Qed

\medskip

{\bf (d)} Set $\mu=\phi_*\restr\Sigma$.   Then $\mu$ is subadditive.
\Prf\ Suppose that $E$, $F\in\Sigma$ and $K\subseteq E\cup F$ is compact.
Let $\epsilon>0$.  Then there are $L_1$, $L_2\in\Cal K$ such that
$L_1\cap E$, $L_1\setminus E$, $L_2\cap F$ and $L_2\setminus F$ are all
compact, while $\phi_*(X\setminus L_1)$ and $\phi_*(X\setminus L_2)$ are
both at most $\epsilon$.   Set $K_1=L_1\cap E$ and
$K_2=L_2\cap F$, so that

$$\eqalignno{\phi K
&\le\phi(K\cup K_1\cup K_2)
\le\phi(K_1\cup K_2)+\phi_*(K\setminus(K_1\cup K_2))\cr
\displaycause{by hypothesis ($\gamma$)}
&\le\phi K_1+\phi K_2+\phi_*(X\setminus(L_1\cap L_2))
\le\phi_*E+\phi_*F+2\epsilon.\cr}$$

\noindent As $\epsilon$ and $K$ are
arbitrary, $\phi_*(E\cup F)\le\phi_*E+\phi_*F$.\ \Qed

\medskip

{\bf (e)} Every open set belongs to $\Sigma$.   \Prf\ Let $G\subseteq X$ be
open, and $\epsilon>0$.   Applying ($\beta$) with $K=\emptyset$ we have an
$L\in\Cal K$ such that $\phi_*(X\setminus L)\le\epsilon$.   Next, there is
an $L'\in\Cal K$, disjoint from $L\setminus G$, such that
$\phi_*(X\setminus((L\setminus G)\cup L'))\le\epsilon$.   Set
$L''=L\cap((L\setminus G)\cup L')$.   Then $L''\cap G=L\cap L'$ and
$L''\setminus G=L\setminus G$ are compact and
$\phi_*(X\setminus L'')\le 2\epsilon$.\ \Qed

\medskip

{\bf (f)} If $E\subseteq F\in\Sigma$ and $\mu F=0$ then $E\in\Sigma$.
\Prf\ Let $\epsilon>0$.   Let $K\subseteq X$ be such that $K\cap F$ and
$K\setminus F$ are both compact and $\phi_*(X\setminus K)\le\epsilon$.
Then $(K\setminus F)\cap E$ and $(K\setminus F)\setminus E$
are both compact, and

\Centerline{$\phi_*(X\setminus(K\setminus F))
=\mu(X\setminus(K\setminus F))
\le\mu(X\setminus K)+\mu F=\phi_*(X\setminus K)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $E\in\Sigma$.\ \Qed

\medskip

{\bf (g)} If
$E\in\Sigma$ and $\epsilon>0$, there is a compact $K\subseteq E$ such that
$\mu(E\setminus K)\le\epsilon$.   \Prf\ Let $K_0\subseteq X$ be such that
$K_0\cap E$
and $K_0\setminus E$ are both compact and
$\phi_*(X\setminus K_0)\le\epsilon$.   Set $K=E\cap K_0$.
If $L\in\Cal K$ and $L\subseteq E\setminus K$ then
$\phi L\le\phi_*(X\setminus K_0)\le\epsilon$;  so
$\mu(E\setminus K)\le\epsilon$.\ \Qed

\medskip

{\bf (h)} So $\mu$ is a totally finite Radon submeasure.
To see that it is unique, let
$\mu'$ be another totally finite Radon submeasure with the same
properties, and $\Sigma'$
its domain.   By condition (ii) of 496C, $\mu'=\phi_*\restr\Sigma'$.
If $E\in\Sigma$ there are sequences $\sequencen{K_n}$,
$\sequencen{L_n}$ of compact sets such that $K_n\subseteq E$,
$L_n\subseteq X\setminus E$ and
$\mu(E\setminus K_n)+\mu((X\setminus E)\setminus L_n)\le 2^{-n}$
for every $n$.   Set $F=\bigcup_{n\in\Bbb N}K_n$ and
$F'=\bigcup_{n\in\Bbb N}L_n$;  then $F\cup F'$ belongs to
$\Sigma\cap\Sigma'$ and

$$\eqalign{\mu'(X\setminus(F\cup F'))
&=\phi_*(X\setminus(F\cup F'))
=\mu(X\setminus(F\cup F'))\cr
&\le\inf_{n\in\Bbb N}\mu(X\setminus(K_n\cup L_n))
=0.\cr}$$

\noindent Consequently $E\setminus F\in\Sigma'$ and $E\in\Sigma'$.

The same works with $\mu$ and $\mu'$ interchanged, so $\Sigma=\Sigma'$ and
$\mu'=\phi_*\restr\Sigma=\mu$.
}%end of proof of 496E

\leader{496F}{Theorem} Let $X$ be a zero-dimensional compact Hausdorff
space
and $\Cal E$ the algebra of open-and-closed subsets of $X$.   Let
$\nu:\Cal E\to\coint{0,\infty}$ be an exhaustive submeasure.   Then there
is a unique totally finite Radon submeasure on $X$ extending $\nu$.

\proof{{\bf (a)} Let $\Cal K$ be the family of compact subsets of
$X$ and for $K\in\Cal K$ set $\phi K=\inf\{\nu E:K\subseteq E\in\Cal E\}$.
Then $\phi$ satisfies the conditions of 496E.

\medskip

\Prf\grheada\
Of course $\phi\emptyset=0$ and $\phi K\le\phi L$ whenever $K\subseteq L$
in $\Cal K$.   If $K\subseteq E\in\Cal E$ and $L\subseteq F\in\Cal E$,
then $K\cup L\subseteq E\cup F\in\Cal E$ and
$\nu(E\cup F)\le\nu E+\nu F$, so $\phi$ is subadditive.

\medskip

\quad\grheadb\ The point is that for every $K\in\Cal K$ and $\epsilon>0$
there is an $E\in\Cal E$ such that $K\subseteq E$ and $\nu F\le\epsilon$
whenever $F\in\Cal E$ and $F\subseteq E\setminus K$;  since otherwise we
could find a disjoint sequence $\sequencen{F_n}$ in $\Cal E$ with
$\nu F_n\ge\epsilon$ for every $n$.   But now $L=X\setminus E$ is compact
and disjoint from $K$, and every compact subset of
$X\setminus(K\cup L)=E\setminus K$ is included in a member of $\Cal E$
included in $E\setminus K$;  so
$\sup\{\phi K':K'\subseteq X\setminus(K\cup L)$ is compact$\}\le\epsilon$.

\medskip

\quad\grheadc\ If $K$ and $L$ are compact and $K\subseteq L$ and
$\epsilon>0$, take
$E\in\Cal E$ such that $K\subseteq E$ and $\nu E\le\phi K+\epsilon$.
Set $K'=L\setminus E$.   If $F\in\Cal E$ and $F\supseteq K'$, then
$E\cup F\supseteq L$, so

\Centerline{$\phi L\le\nu(E\cup F)\le\nu E+\nu F\le\phi K+\epsilon+\nu F$.}

\noindent As $F$ is arbitrary, $\phi L\le\phi K+\phi K'+\epsilon$.\ \Qed

\medskip

There is therefore a totally finite Radon submeasure $\mu$ extending $\phi$ and
$\nu$.

\medskip

{\bf (b)} If $\mu'$ is another totally finite Radon submeasure extending $\nu$, then
$\mu'\restr\Cal K=\phi$.   \Prf\ Of course $\mu'K\le\phi K$ for every
$K\in\Cal K$.   \Quer\ If $K\in\Cal K$ and $\epsilon>0$ and
$\mu'K+\epsilon<\phi K$, let $E\in\Cal E$ be such that $K\subseteq E$ and
$\phi L\le\epsilon$ whenever $L\subseteq E\setminus K$ is compact, as in
(a-$\beta$) above.   Then

$$\eqalign{\mu'(E\setminus K)
&=\sup\{\mu'L:L\subseteq E\setminus K\text{ is compact}\}\cr
&\le\sup\{\phi L:L\subseteq E\setminus K\text{ is compact}\}
\le\epsilon\cr}$$

\noindent and

\Centerline{$\nu E=\mu'E\le\epsilon+\mu'K<\mu K\le\mu E=\nu E$.  \Bang\Qed}

\noindent By the guarantee of uniqueness in 496E, $\mu'=\mu$.
}%end of proof of 496F

\vleader{48pt}{496G}{Theorem} Let $\frak A$ be a Maharam algebra, and
$\mu$ a strictly positive Maharam submeasure on $\frak A$.
Let $Z$ be the Stone space
of $\frak A$, and write $\widehat{a}$ for the open-and-closed subset of $Z$
corresponding to each $a\in\frak A$.
Then there is a unique totally finite Radon
submeasure $\nu$ on $Z$ such that
$\nu\widehat{a}=\mu a$ for every $a\in\frak A$.
The domain of $\nu$ is the Baire-property algebra $\widehat{\Cal B}$
of $Z$, and the null ideal of $\nu$ is the nowhere dense ideal of $Z$.

\proof{ Let $\Cal E$ be the algebra of open-and-closed subsets of $Z$, and
$\Cal M$ the ideal of meager subsets of $Z$.
Because $\frak A$ is Dedekind complete (393Eb/496Bb),
$\Cal E$ is the regular open algebra of $Z$ (314S).   By 496R(b-ii),
$\widehat{\Cal B}=\{E\symmdiff F:E\in\Cal E$, $F\in\Cal M\}$.

For $a\in\frak A$, let $\widehat{a}$ be the corresponding member of
$\Cal E$.   By 314M, we have an isomorphism
$\theta:\frak A\to\widehat{\Cal B}/\Cal M$ defined by setting
$\theta(a)=\widehat{a}^{\ssbullet}$ for every $a\in\frak A$.
For $E\in\widehat{\Cal B}$, set $\nu E=\mu(\theta^{-1}E^{\ssbullet})$.
Because
$E\mapsto E^{\ssbullet}$ is a sequentially order-continuous Boolean
homomorphism (313P(b-ii)), $\nu$ is a Maharam submeasure on
$\widehat{\Cal B}$.   Because $\mu$ is strictly positive, the null ideal of
$\nu$ is $\Cal M$.

Because $\frak A$ is \wsid\ (393Eb/496Bb), $\Cal M$ is the ideal of nowhere
dense subsets of $Z$ (316I).   If $E\in\widehat{\Cal B}$,
consider $B=\{b:b\in\frak A$, $\widehat{b}\subseteq E\}$;  set $a=\sup B$
in $\frak A$.
Now $E\setminus\widehat{a}$ is nowhere dense.   \Prf\Quer\ Otherwise, there
is a non-zero $c\in\frak A$ such that
$F=\widehat{c}\setminus(E\setminus\widehat{a})$ is nowhere dense.   In this
case, the non-empty open set $\widehat{c}\setminus\overline{F}$
is included in $E\setminus\widehat{a}$ and there is a non-zero
$b\in\frak A$ such that $\widehat{b}\subseteq E\setminus\widehat{a}$.
But in this case $b\in B$ and $\widehat{b}\subseteq\widehat{a}$, which is
absurd.\ \Bang\Qed

Set $D=\{a\Bsetminus b:b\in B\}$.   Then $D$ is downwards-directed and
has infimum $0$.   Because $\mu$ is sequentially order-continuous and
$\frak A$ is ccc, $\mu$ is order-continuous (316Fc), and
$\inf_{d\in D}\mu d=0$.   Accordingly

$$\eqalign{\inf\{\nu(E\setminus K):K\subseteq E\text{ is compact}\}
&\le\inf_{b\in B}\nu(E\setminus\widehat{b})
=\inf_{b\in B}\nu(\widehat{a}\setminus\widehat{b})\cr
&=\inf_{b\in B}\mu(a\Bsetminus b)
=0.\cr}$$

\noindent Thus condition (ii) of 496C is satisfied and $\nu$ is a totally finite Radon
measure.

By 496F, $\nu$ is unique.
}%end of proof of 496G

\leader{496H}{Theorem} Let $X$ be a Hausdorff space, $\Sigma_0$ an algebra
of subsets of $X$, and $\mu_0:\Sigma_0\to\coint{0,\infty}$ an exhaustive
submeasure such that $\inf\{\mu_0(E\setminus K):K\in\Sigma_0$ is compact,
$K\subseteq E\}=0$ for every $E\in\Sigma_0$.
Then $\mu_0$ has an extension to a totally finite Radon
submeasure $\mu_1$ on $X$.

\proof{{\bf (a)}
Let $P$ be the set of all submeasures $\mu$, defined on algebras of
subsets of $X$, which extend $\mu_0$, and have the properties

\inset{($\alpha$) $\inf\{\mu(E\setminus K):K\in\dom\mu$ is compact,
$K\subseteq E\}=0$ for every $E\in\dom\mu$,

(*) for every $E\in\dom\mu$ and $\epsilon>0$ there is an $F\in\Sigma_0$
such that $\mu(E\symmdiff F)\le\epsilon$.}

\noindent Order $P$ by extension of functions, so that $P$ is a partially
ordered set.

\medskip

{\bf (b)} If $\mu\in P$, then $\mu$ is exhaustive.   \Prf\Quer\ Otherwise,
let $\sequencen{E_n}$ be a disjoint sequence in $\dom\mu$ such that
$\epsilon=\inf_{n\in\Bbb N}\mu E_n$ is greater than $0$.   For each
$n\in\Bbb N$, let $F_n\in\Sigma_0$ be such that
$\mu(E_n\symmdiff F_n)\le 2^{-n-2}\epsilon$;  set
$G_n=F_n\setminus\bigcup_{i<n}F_i$ for each $n$.   Then

\Centerline{$E_n\subseteq G_n\cup\bigcup_{i\le n}(E_i\symmdiff F_i)$,
\quad$\epsilon\le\mu G_n+\sum_{i=0}^n2^{-i-2}\epsilon
\le\mu_0G_n+\Bover12\epsilon$}

\noindent and $\mu_0G_n\ge\bover12\epsilon$ for every $n$.   But
$\sequencen{G_n}$ is disjoint and $\mu_0$ is supposed to be exhaustive.\
\Bang\Qed

\medskip

{\bf (c)} Suppose that $\mu\in P$ has domain $\Sigma$, and that
$V\subseteq X$ is such that

\doubleinset{$\ddagger(V,\mu)$:  for every $\epsilon>0$ there is a
$K\in\Sigma$ such that $K\cap V$ is compact and
$\mu(X\setminus K)\le\epsilon$.}

\medskip

\quad{\bf (i)} Set $\Cal H=\{H:V\subseteq H\in\Sigma\}$.   Then
$\Cal H$ is downwards-directed.   If $\epsilon>0$ there is an $H\in\Cal H$
such that $\mu(H\setminus H')\le\epsilon$ for every $H'\in\Cal H$.
\Prf\Quer\ Otherwise, there would be a non-increasing sequence
$\sequencen{H_n}$ in $\Cal H$ such that
$\mu(H_n\setminus H_{n+1})\ge\epsilon$ for every $n$;  but $\mu$ is
exhaustive, by (b).\ \Bang\Qed

\medskip

\quad{\bf (ii)} Let $\Cal F$ be the filter on $\Cal H$
generated by sets of the form $\{H':H'\in\Cal H$, $H'\subseteq H\}$ for
$H\in\Cal H$.   Then
$\lim_{H\to\Cal F}\mu((E\cap H)\cup(F\setminus H))$ is defined for all
$E$, $F\in\Sigma$.   \Prf\ Given $\epsilon>0$, then (i) tells us that there
is an $H_0\in\Cal H$ such that
$\mu(H\symmdiff H')\le\mu(H_0\setminus(H\cap H'))\le\epsilon$ whenever $H$,
$H'\in\Cal H$ are included in $H_0$.   Now, for such $H$ and $H'$,

\Centerline{
$((E\cap H)\cup(F\setminus H))\symmdiff((E\cap H')\cup(F\setminus H'))
\subseteq H\symmdiff H'$,}

\noindent so

\Centerline{$|\mu((E\cap H)\cup(F\setminus H))
-\mu((E\cap H')\cup(F\setminus H'))|\le\epsilon$.  \Qed}

\medskip

\quad{\bf (iii)} If $E$, $F$, $E'$, $F'\in\Sigma$ and
$(E\cap V)\cup(F\setminus V)=(E'\cap V)\cup(F'\setminus V)$, then

\Centerline{$\lim_{H\to\Cal F}\mu((E\cap H)\cup(F\setminus H))
=\lim_{H\to\Cal F}\mu((E'\cap H)\cup(F'\setminus H))$.}

\noindent\Prf\ Given $\epsilon>0$, there is an $H_0\in\Cal H$ such that
$\mu G\le\epsilon$ whenever $G\in\Sigma$ and $G\subseteq H_0\setminus V$,
by (i).   Now if $H\in\Cal H$ and $H\subseteq H_0$,

\Centerline{$G=((E\cap H)\cup(F\setminus H))
\symmdiff((E'\cap H)\cup(F'\setminus H))
\subseteq H\setminus V$,}

\noindent so

\Centerline{$|\mu((E\cap H)\cup(F\setminus H))
-\mu((E'\cap H)\cup(F'\setminus H))|\le\mu G\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, the limits are equal.\ \Qed

\medskip

\quad{\bf (iv)} Consequently, taking $\Sigma'$ to be the algebra
$\{(E\cap V)\cup(F\setminus V):E$, $F\in\Sigma\}$ of subsets of $X$
generated by $\Sigma\cup\{V\}$, we have a
functional $\mu':\Sigma'\to\coint{0,\infty}$ defined by saying that

\Centerline{$\mu'((E\cap V)\cup(F\setminus V))
=\lim_{H\to\Cal F}\mu((E\cap H)\cup(F\setminus H))$}

\noindent whenever $E$, $F\in\Sigma$.

\medskip

\quad{\bf (v)} $\mu'$ is a submeasure extending $\mu$.
\Prf\ If $E\in\Sigma$, then

\Centerline{$\mu'E=\mu'((E\cap V)\cup(E\setminus V))
=\lim_{H\to\Cal F}\mu((E\cap H)\cup(E\setminus H))=\mu E$,}

\noindent so $\mu'$ extends $\mu$.   If $E_1$, $E_2$, $F_1$,
$F_2\in\Sigma$, set $E=E_1\cup E_2$, $F=F_1\cup F_2$;  then

\Centerline{$((E_1\cap A)\cup(F_1\setminus A))
\cup((E_2\cap A)\cup(F_2\setminus A))
=((E\cap A)\cup(F\setminus A))$}

\noindent for every set $A$, so

$$\eqalign{\mu'\bigl(((E_1\cap V)\cup(&F_1\setminus V))
\cup((E_2\cap V)\cup(F_2\setminus V))\bigr)\cr
&=\mu'(((E\cap V)\cup(F\setminus V))\cr
&=\lim_{H\to\Cal F}\mu((E\cap H)\cup(F\setminus H))\cr
&=\lim_{H\to\Cal F}\mu\bigl(((E_1\cap H)\cup(F_1\setminus H))
   \cup((E_2\cap H)\cup(F_2\setminus H))\bigr)\cr
&\le\lim_{H\to\Cal F}\mu((E_1\cap H)\cup(F_1\setminus H))
   +\mu((E_2\cap H)\cup(F_2\setminus H))\cr
&=\mu'((E_1\cap V)\cup(F_1\setminus V))
   +\mu'((E_2\cap V)\cup(F_2\setminus V)).\cr}$$

\noindent Thus $\mu'$ is subadditive;  monotonicity is easier.\ \Qed

\medskip

\quad{\bf (vi)} $\mu'$ has the property ($\alpha$).   \Prf\ Suppose that
$E$, $F\in\Sigma$ and that $\epsilon>0$.   Let $H_0\in\Cal H$ be such that
$\mu(H_0\setminus H)\le\epsilon$ whenever $H\in\Cal H$ and
$H\subseteq H_0$.   Let $K_0\in\Sigma$ be such that
$\mu(X\setminus K_0)\le\epsilon$ and $K_0\cap V$ is compact.
Let $K_1\subseteq E$ and $K_2\subseteq F\setminus H_0$
be compact sets, belonging to $\Sigma$, such that
$\mu(E\setminus K_1)\le\epsilon$ and
$\mu((F\setminus H_0)\setminus K_2)\le\epsilon$.   Set
$K=(K_1\cap K_0\cap V)\cup K_2$, so that $K$ is a compact set belonging
to $\Sigma'$
and $K\subseteq(E\cap V)\cup(F\setminus V)$.   Now if $H\in\Cal H$ and
$H\subseteq H_0$,

$$\eqalign{\mu\bigl(((E\setminus(&K_1\cap K_0))\cap H)
   \cup((F\setminus K_2)\setminus H)\bigr)\cr
&\le\mu(E\setminus K_1)+\mu(X\setminus K_0)
   +\mu((F\setminus H_0)\setminus K_2)+\mu(H_0\setminus H)
\le 4\epsilon.\cr}$$

\noindent Taking the limit along $\Cal F$,

\Centerline{$\mu'\bigl(((E\cap V)\cup(F\setminus V))\setminus K\bigr)
=\mu'\bigl(((E\setminus(K_1\cap K_0))\cap V)
   \cup((F\setminus K_2)\setminus V)\bigr)
\le 4\epsilon$.}

\noindent As $E$, $F$ and $\epsilon$ are arbitrary, we have the result.\
\Qed

\medskip

\quad{\bf (vii)} $\mu'$ has the property (*).   \Prf\
Suppose that
$E$, $F\in\Sigma$ and that $\epsilon>0$.   Let $H_0\in\Cal H$ be such that
$\mu(H_0\setminus H)\le\epsilon$ whenever $H\in\Cal H$ and
$H\subseteq H_0$.   Set $G=(E\cap H_0)\cup(F\setminus H_0)\in\Sigma$.
Then

\Centerline{$((E\cap V)\cup(F\setminus V))\symmdiff G
\subseteq H_0\setminus V$,}

\noindent so

\Centerline{$\mu'\bigl(((E\cap V)\cup(F\setminus V))\symmdiff G\bigr)
\le\mu'(H_0\setminus V)
=\lim_{H\to\Cal F}\mu(H_0\setminus H)\le\epsilon$.  \Qed}

\medskip

{\bf (d)(i)} If $\mu\in P$ and $V\in\Cal N(\mu)$, then $\ddagger(V,\mu)$
is true.   \Prf\ Let $\epsilon>0$.
There is an $E\in\dom\mu$, including $V$, such that
$\mu E=0$;   now there is a compact $K\in\dom\mu$, included in
$X\setminus E$, such that

\Centerline{$\epsilon\ge\mu((X\setminus E)\setminus K)=\mu(X\setminus K)$,}

\noindent while $K\cap V=\emptyset$ is compact.\ \Qed

\medskip

\quad{\bf (ii)} If $\mu\in P$ and $V\subseteq X$ is closed, then
$\ddagger(V,\mu)$ is true.   \Prf\ For every $\epsilon>0$, there is a
compact $K\in\dom\mu$ such that $\mu(X\setminus K)\le\epsilon$,
and now $K\cap V$ is compact.\ \Qed

\medskip

\quad{\bf (iii)} Now suppose that $\mu\in P$ is such that every compact
subset of $X$ belongs to the domain $\Sigma$ of $\mu$, and that
$\sequencen{E_n}$ is a sequence in $\Sigma$ with intersection $V$.   Then
$\ddagger(V,\mu)$ is true.   \Prf\ Let $\epsilon>0$.   For each
$n\in\Bbb N$, there are compact sets
$K_n\subseteq E_n$, $K'_n\subseteq X\setminus E_n$ such that

\Centerline{$\mu(E_n\setminus K_n)+\mu((X\setminus E_n)\setminus K'_n)
\le 2^{-n-1}\epsilon$.}

\noindent Set $K=\bigcap_{n\in\Bbb N}K_n\cup K'_n$;  then $K$ is compact,
so belongs to $\Sigma$.   If $L\subseteq X\setminus K$ is compact, then
there is an $n\in\Bbb N$ such that $L\cap\bigcap_{i\le n}K_i\cup K'_i$
is empty, so that

\Centerline{$\mu L\le\sum_{i=0}^n\mu(X\setminus(K_i\cup K'_i))
\le\sum_{i=0}^n2^{-n-1}\epsilon\le\epsilon$.}

\noindent As $L$ is arbitrary, $\mu(X\setminus K)\le\epsilon$.   Finally,

\Centerline{$K\cap V=\bigcap_{n\in\Bbb N}(K_n\cup K'_n)\cap E_n
=\bigcap_{n\in\Bbb N}K_n$}

\noindent is compact.\ \Qed

\medskip

{\bf (e)} If $Q\subseteq P$ is a non-empty totally ordered subset of $P$,
$\bigcup Q\in P$.   So $P$ has a maximal element $\mu_1$,
which is a submeasure,
satisfying ($\alpha$), and extending $\mu_0$.
Setting $\Sigma_1=\dom\mu_1$, (c) tells us that $V\in\Sigma_1$ whenever
$V\subseteq X$ and $\ddagger(V,\mu_1)$ is true.   By (d-i),
$\Cal N(\mu_1)\subseteq\Sigma_1$ and $\mu_1$ is complete.   By (d-ii),
every closed set, and therefore every open set, belongs to $\Sigma_1$.
So (d-iii) tells us that
$\bigcap_{n\in\Bbb N}E_n\in\Sigma_1$ for every sequence $\sequencen{E_n}$
in $\Sigma_1$, and $\Sigma_1$ is a $\sigma$-algebra.   Putting these
together, all the conditions of 496C are satisfied, and $\mu_1$ is a totally finite Radon
submeasure.
}%end of proof of 496H

\leader{496I}{Theorem} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$, and $\mu$ a complete Maharam submeasure on
$\Sigma$.

(a) $\Sigma$ is closed under Souslin's operation.

(b) If $A$ is the kernel of a Souslin scheme
$\family{\sigma}{S}{E_{\sigma}}$ in $\Sigma$, and $\epsilon>0$, there is a
$\psi\in\NN$ such that

\Centerline{$\mu(A\setminus\bigcup_{\phi\in\NN,\phi\le\psi}
   \bigcap_{n\ge 1}E_{\phi\restr n})\le\epsilon$.}

\proof{{\bf (a)} Let $\Cal N(\mu)$ be the null ideal of $\mu$.
Because $\mu$ is exhaustive, every disjoint sequence in
$\Sigma\setminus\Cal N(\mu)$ is countable, so 431G tells us that
$\Sigma$ is closed under Souslin's operation.

\medskip

{\bf (b)} The argument of 431D applies, with trifling modifications in its
expression.   For $\sigma\in S=\bigcup_{k\in\Bbb N}\BbbN^k$, set
$A_{\sigma}
=\bigcup_{\sigma\subseteq\phi\in\NN}\bigcap_{n\ge 1}E_{\phi\restr n}$;
then $A_{\sigma}\in\Sigma$, by (a).   Given $\epsilon>0$, let
$\family{\sigma}{S}{\epsilon_{\sigma}}$ be a family of strictly positive
real numbers such that $\sum_{\sigma\in S}\epsilon_{\sigma}\le\epsilon$.
For each $\sigma\in S$, let $m_{\sigma}$ be such that
$\mu(A_{\sigma}\setminus\bigcup_{i\le m_{\sigma}}
  A_{\sigma^{\smallfrown}\fraction{i}})\le\epsilon_{\sigma}$.
Set

\Centerline{$\psi(k)
=\max\{m_{\sigma}:\sigma\in\BbbN^k$, $\sigma(i)\le\psi(i)$
for every $i<k\}$}

\noindent for $k\in\Bbb N$;  then

\Centerline{$A\setminus\bigcup_{\phi\in\NN,\phi\le\psi}
    \bigcap_{n\ge 1}E_{\phi\restr n}
\subseteq\bigcup_{\sigma\in S}\bigl(A_{\sigma}
  \setminus\bigcup_{i\le m_{\sigma}}A_{\sigma^{\smallfrown}\fraction{i}}
  \bigr)$}

\noindent has submeasure at most $\epsilon$.
}%end of proof of 496I

\leader{496J}{Theorem} Let $X$ be a K-analytic
Hausdorff space and $\mu$ a
Maharam submeasure defined on the Borel $\sigma$-algebra of $X$.   Then

\Centerline{$\inf\{\mu(X\setminus K):K\subseteq X$ is compact$\}=0$.}

\proof{ Again, we have only to re-use the ideas of 432B.
Let $\hat\mu$ be the completion of $\mu$ (496A) and $\Sigma$ the
domain of $\hat\mu$.   Let $R\subseteq\NN\times X$ be an usco-compact
relation such that $R[\NN]=X$.   For
$\sigma\in S^*=\bigcup_{n\ge 1}\BbbN^n$
set $I_{\sigma}=\{\phi:\sigma\subseteq\phi\in\NN\}$,
$F_{\sigma}=\overline{R[I_{\sigma}]}$;  then $X$ is the kernel of the
Souslin scheme $\family{\sigma}{S^*}{F_{\sigma}}$.   Now, given $\epsilon>0$,
496Ib tells us that
there is a $\psi\in\NN$ such that $\mu(X\setminus F)\le\epsilon$, where
$F=\bigcup_{\phi\in\NN,\phi\le\psi}\bigcap_{n\ge 1}F_{\phi\restr n}$;
but $F=R[K]$ where $K$ is the compact set
$\{\phi:\phi\in\NN$, $\phi\le\psi\}$, so $F$ is compact.
}%end of proof of 496J

\leader{496K}{Proposition} Let $\mu$ be a Maharam submeasure
on the Borel $\sigma$-algebra of an analytic Hausdorff space $X$.
Then the completion of $\mu$ is a totally finite Radon submeasure on $X$.

\proof{ If $E\subseteq X$ is Borel, then it is K-analytic (423Eb);
applying 496J to $\mu\restr\Cal PE$,
we see that $\inf\{\mu(E\setminus K):K\subseteq E$ is compact$\}=0$.
Consequently, writing $\Sigma$ for the domain of the completion $\hat\mu$
of $\mu$,
$\inf\{\hat\mu(E\setminus K):K\subseteq E$ is compact$\}=0$ for every
$E\in\Sigma$.   Condition (i) of the definition 496C is surely
satisfied by $\hat\mu$, so $\hat\mu$ is a totally finite Radon submeasure.
}%end of proof of 496K

\leader{496L}{Free products of Maharam algebras} If $\frak A$, $\frak B$
are Boolean algebras with submeasures $\mu$, $\nu$ respectively, we have a
submeasure $\mu\ltimes\nu$ on the free product
$\frak A\otimes\frak B$\cmmnt{ (392K)}.
It is easy to see\cmmnt{, in 392K,} that
if $\mu$ and $\nu$ are strictly positive so is $\mu\ltimes\nu$;  moreover,
if $\mu$ and $\nu$ are exhaustive so is $\mu\ltimes\nu$\cmmnt{ (392Ke)}.

Now suppose that $\familyiI{\frak A_i}$ is a family of
Maharam algebras, where $I$ is a finite totally ordered set.   Then
we can take a strictly positive Maharam submeasure $\mu_i$ on
each $\frak A_i$, form an exhaustive submeasure $\lambda$ on
$\frak C_I=\bigotimes_{i\in I}\frak A_i$, and use $\lambda$ to
construct a metric completion $\widehat{\frak C}_I$ which is a
Maharam algebra\cmmnt{, as in 393H/496Bb}.
\cmmnt{(If $I=\{i_0,\ldots,i_n\}$ where $i_0<\ldots<i_n$, then
$\lambda=(.(\mu_{i_0}\ltimes\mu_{i_1})\ltimes\ldots)\ltimes\mu_{i_n}$
(392Kf).   By 392Kc, the product is associative, so the
arrangement of the brackets is immaterial.)}
If we change each $\mu_i$ to $\mu'_i$, where $\mu'_i$ is
another strictly positive Maharam submeasure on
$\frak A_i$, then every $\mu'_i$ is absolutely continuous with
respect to $\mu_i$\cmmnt{ (393F/496Bd)},
so the corresponding $\lambda'$ will be
absolutely continuous with respect to $\lambda$, and vice
versa\cmmnt{ (392Kd)};
in which case the metrics on $\frak C_I$ are uniformly equivalent
and we get the same metric completion
$\widehat{\frak C}_I$ up to Boolean algebra isomorphism.
We can therefore think of $\widehat{\frak C}_I$ as `the' {\bf Maharam
algebra free product} of the family $\familyiI{\frak A_i}$ of Boolean
algebras;  \cmmnt{as in 392Kf,} we shall have an isomorphism between
$\widehat{\frak C}_{J\cup K}$ and the Maharam algebra
free product of $\widehat{\frak C}_J$ and $\widehat{\frak C}_K$
whenever $J$, $K\subseteq I$ and $j<k$ for every $j\in J$ and $k\in K$.

\dvro{If}{From 392Kg we see that if} $(\frak A,\mu)$ and $(\frak B,\nu)$
are probability algebras, then
their Maharam algebra free product, regarded as a Boolean algebra, is
isomorphic to their probability algebra free product\cmmnt{ as defined in
\S325}.

\leader{496M}{}{\bf Representing products of Maharam algebras:
Theorem} Let $X$
and $Y$ be sets, with $\sigma$-algebras $\Sigma$ and $\Tau$ and Maharam
submeasures $\mu$ and $\nu$ defined on $\Sigma$, $\Tau$ respectively.
Let $\frak A$, $\frak B$ be their Maharam algebras and write $\bar\mu$,
$\bar\nu$ for the strictly positive Maharam submeasures on $\frak A$ and
$\frak B$ induced by $\mu$ and $\nu$\cmmnt{ as in 496Ba above}.   Let
$\Sigma\tensorhat\Tau$ be the $\sigma$-algebra of subsets of $X\times Y$
generated by $\{E\times F:E\in\Sigma$, $F\in\Tau\}$.

(a)\cmmnt{ (Compare 418T.)}
Give $\frak B$ its Maharam-algebra topology\cmmnt{ (393G/496Bd)}.
If $W\in\Sigma\tensorhat\Tau$ then $W[\{x\}]\in\Tau$ for every $x\in X$
and the function $x\mapsto W[\{x\}]^{\ssbullet}:X\to\frak B$ is
$\Sigma$-measurable and has separable range.   \cmmnt{Consequently}
$x\mapsto\nu W[\{x\}]:X\to\coint{0,\infty}$ is $\Sigma$-measurable.

(b) For $W\in\Sigma\tensorhat\Tau$ set

\Centerline{$\lambda W
=\inf\{\epsilon:\epsilon>0$,
  $\mu\{x:\nu W[\{x\}]>\epsilon\}\le\epsilon\}$.}

\noindent Then $\lambda$ is a Maharam submeasure on $\Sigma\tensorhat\Tau$,
and

\Centerline{$\lambda^{-1}[\{0\}]
=\{W:W\in\Sigma\tensorhat\Tau$,
   $\{x:W[\{x\}]\notin\Cal N(\nu)\}\in\Cal N(\mu)\}$.}

(c) Let $\frak C$ be the Maharam algebra of $\lambda$.   Then
$\frak A\otimes\frak B$ can be embedded in $\frak C$ by mapping
$E^{\ssbullet}\otimes F^{\ssbullet}$ to $(E\times F)^{\ssbullet}$ for all
$E\in\Sigma$ and $F\in\Tau$.

(d) This embedding identifies $(\frak C,\bar\lambda)$ with the metric
completion of $(\frak A\otimes\frak B,\bar\mu\ltimes\bar\nu)$.

\proof{{\bf (a)} Write $\Cal W$ for the set of those
$W\subseteq X\times Y$ such that $W[\{x\}]\in\Tau$ for every $x\in X$ and
$x\mapsto W[\{x\}]^{\ssbullet}:X\to\frak B$ is
$\Sigma$-measurable and has separable range.
Then $\Sigma\otimes\Tau$
(identified with the algebra of subsets of $X\times Y$ generated by
$\{E\times F:E\in\Sigma$, $F\in\Tau\}$) is included in $\Cal W$.

If $\sequencen{W_n}$ is a non-decreasing sequence in $\Cal W$ with union
$W$, then $W\in\Cal W$.   \Prf\ Of course
$W[\{x\}]=\bigcup_{n\in\Bbb N}W_n[\{x\}]$ belongs to $\Tau$ for every
$x\in X$.  Set $f_n(x)=W_n[\{x\}]^{\ssbullet}$ for $n\in\Bbb N$ and
$x\in X$.   For each $x\in X$, $W[\{x\}]\setminus W_n[\{x\}]$ is a
non-increasing sequence with empty intersection, so
$\lim_{n\to\infty}\nu(W[\{x\}]\setminus W_n[\{x\}])=0$ and
$\sequencen{f_n(x)}$
converges to $f(x)=W[\{x\}]^{\ssbullet}$ in $\frak B$.
By 418Ba, $f$ is measurable.  Also
$D=\overline{\{f_n(x):x\in X,\,n\in\Bbb N\}}$
is a separable subspace of $\frak B$
including $f[X]$.   So $W\in\Cal W$.\ \Qed

Similarly, $\bigcap_{n\in\Bbb N}W_n\in\Cal W$ for any non-increasing
sequence $\sequencen{W_n}$ in $\Cal W$.   $\Cal W$ therefore includes the
$\sigma$-algebra generated
by $\Sigma\otimes\Tau$ (136G), which is $\Sigma\tensorhat\Tau$.

Now $x\mapsto\nu W[\{x\}]=\bar\nu W[\{x\}]^{\ssbullet}$ is measurable
because $\bar\nu:\frak B\to\Bbb R$ is continuous.

\medskip

{\bf (b)} Of course $\lambda\emptyset=0$ and
$\lambda W\le\lambda W'$ if $W$, $W'\in\Sigma\tensorhat\Tau$
and $W\subseteq W'$.   If $W_1$, $W_2\in\Sigma\tensorhat\Tau$ have union
$W$, $\lambda W_1=\alpha_1$ and
$\lambda W_2=\alpha_2$, then

\Centerline{$\{x:\nu W[\{x\}]>\alpha_1+\alpha_2\}
\subseteq\{x:\nu W_1[\{x\}]>\alpha_1\}\cup\{x:\nu W_2[\{x\}]>\alpha_2\}$,}

\noindent so, setting $\alpha=\alpha_1+\alpha_2$,

\Centerline{$\mu\{x:\nu W[\{x\}]>\alpha\}
\le\mu\{x:\nu W_1[\{x\}]>\alpha_1\}+\mu\{x:\nu W_2[\{x\}]>\alpha_2\}
\le\alpha_1+\alpha_2=\alpha$,}

\noindent and $\lambda W\le\alpha$.   Thus $\lambda$ is monotonic and subadditive.

If now $\sequencen{W_n}$ is a non-increasing sequence in $\Sigma\tensorhat\Tau$ with
empty intersection, and $\epsilon>0$, set $E_n=\{x:\nu W_n[\{x\}]\ge\epsilon\}$ for
each $n$.   Then $\sequencen{E_n}$ is non-increasing;  moreover, for any $x\in X$,
$\sequencen{W_n[\{x\}]}$ is a non-increasing sequence in $\Tau$ with empty
intersection, so $\lim_{n\to\infty}\nu W_n[\{x\}]=0$ and
$x\notin\bigcap_{n\in\Bbb N}E_n$.   There is therefore an $n$ such that
$\mu E_n\le\epsilon$ and $\lambda W_n\le\epsilon$.   As $\sequencen{W_n}$ and
$\epsilon$ are arbitrary, $\lambda$ is a Maharam submeasure.

Finally, for $W\in\Sigma\tensorhat\Tau$,

$$\eqalign{\lambda W=0
&\iff\mu\{x:\nu W[\{x\}]\ge 2^{-n}\}\le 2^{-n}\text{ for every }n\in\Bbb N\cr
&\iff\mu\{x:\nu W[\{x\}]\ge 2^{-m}\}\le 2^{-n}\text{ for every }m,\,n\in\Bbb N\cr
&\iff\mu\{x:\nu W[\{x\}]>0\}\le 2^{-n}\text{ for every }n\in\Bbb N\cr
&\iff\mu\{x:\nu W[\{x\}]>0\}=0
\iff\{x:W[\{x\}]\notin\Cal N(\nu)\}\in\Cal N(\mu).\cr}$$

\medskip

{\bf (c)} If $E\in\Sigma$, then $\lambda(E\times F)=\min(\mu E,\nu F)$ for
all $E\in\Sigma$ and $F\in\Tau$.   So $\lambda(E\times F)=0$ iff
$E^{\ssbullet}\otimes F^{\ssbullet}=0$ in $\frak A\otimes\frak B$.
Consequently we have injective Boolean homomorphisms
from $\frak A$ to $\frak C$ and from $\frak B$ to $\frak C$ defined by the
formulae

\Centerline{$E^{\ssbullet}\mapsto(E\times Y)^{\ssbullet}$ for $E\in\Sigma$,
\quad$F^{\ssbullet}\mapsto(X\times F)^{\ssbullet}$ for $F\in\Tau$;}

\noindent by 315J and 315Kb\formerly{3{}15I-3{}15J},
we have an injective Boolean homomorphism from
$\frak A\otimes\frak B$ to $\frak C$ which maps
$E^{\ssbullet}\otimes F^{\ssbullet}$ to $(E\times F)^{\ssbullet}$ whenever
$E\in\Sigma$ and $F\in\Tau$.

\medskip

{\bf (d)} $\bar\lambda(\phi e)=(\mu\ltimes\nu)(e)$ for
every $e\in\frak A\otimes\frak B$.   \Prf\ Express $e$ as
$\sup_{i\in I}a_i\otimes b_i$ where $\familyiI{a_i}$ is a finite partition of unity in
$\frak A$ and $b_i\in\frak B$ for each $i$.   For each $i$, we can express
$a_i$, $b_i$ as $E_i^{\ssbullet}$, $F_i^{\ssbullet}$ where $E_i\in\Sigma$
and $F_i\in\Tau$;  moreover, we can do this in such a way that
$\familyiI{E_i}$ is a
partition of $X$.   In this case, $\phi e=W^{\ssbullet}$ where
$W=\bigcup_{i\in I}E_i\times F_i$, so that, for $\epsilon>0$,

\Centerline{$\mu\{x:\nu W[\{x\}]>\epsilon\}
=\mu(\bigcup\{E_i:i\in I,\,\nu F_i>\epsilon\})
=\bar\mu(\sup\{a_i:i\in I,\,\bar\nu b_i>\epsilon\})$.}

\noindent Accordingly

$$\eqalign{(\mu\ltimes\nu)(e)
&=\inf\{\epsilon:\bar\mu(\sup\{a_i:i\in I,\,\bar\nu b_i>\epsilon\})\le\epsilon\}\cr
&=\inf\{\epsilon:\mu\{x:\nu W[\{x\}]>\epsilon\}\le\epsilon\}
=\lambda W
=\bar\lambda W^{\ssbullet}
=\bar\lambda(\phi e). \text{ \Qed}\cr}$$

Next, $\phi[\frak A\otimes\frak B]$ is dense in $\frak C$ for the metric induced by
$\bar\lambda$.   \Prf\ Let $\frak D$ be the metric closure of
$\phi[\frak A\otimes\frak B]$ and set
$\Cal V=\{V:V\in\Sigma\otimes\Tau$, $V^{\ssbullet}\in\frak D\}$.   Then
$\Cal V$ includes $\Sigma\otimes\Tau$ and is closed under unions and
intersections of monotonic sequences, so is the whole of
$\Sigma\tensorhat\Tau$, and $\frak D=\frak C$, as required.\ \QeD\   But
this means that we can identify $\frak C$ with the metric
completions of $\phi[\frak A\otimes\frak B]$ and
$\frak A\otimes\frak B$.
}%end of proof of 496M

\exercises{\leader{496X}{Basic exercises (a)}
%\spheader 496Xa
Let $X$ and $Y$ be Hausdorff spaces, $\mu$ a totally finite Radon
submeasure on $X$, and $f:X\to Y$ a function which is almost continuous in
the sense that for every $\epsilon>0$ there is a compact $K\subseteq X$
such that $f\restr K$ is continuous and
$\mu(X\setminus K)\le\epsilon$.   Show that the image submeasure
$\mu f^{-1}$, defined on $\{F:F\subseteq Y$, $f^{-1}[F]\in\dom\mu\}$,
is a totally finite Radon submeasure on $Y$.
%496C

\spheader 496Xb Let $X$ be a Hausdorff space and $\mu$ a totally finite
Radon submeasure on $X$.   For $A\subseteq X$, set
$\mu^*A=\inf\{\mu E:A\subseteq E\in\dom\mu\}$.   Show that $\mu^*$ is an
outer regular Choquet capacity on $X$.
%496C

\spheader 496Xc Let $X$ and $Y$ be compact Hausdorff spaces,
$f:X\to Y$ a continuous surjection, and $\nu$ a totally finite Radon
submeasure on $Y$.
Show that there is a totally finite Radon submeasure $\mu$ on $X$ such
that $\nu$ is the image submeasure $\mu f^{-1}$.
%496H

\spheader 496Xd Let $X$ be a
regular K-analytic Hausdorff space, and $\mu$ a
Maharam submeasure on the Borel $\sigma$-algebra of $X$
which is $\tau$-additive in the sense that whenever $\Cal G$ is a non-empty
upwards-directed family of open sets in $X$ with union $H$, then
$\inf_{G\in\Cal G}\mu(H\setminus G)=0$.   Show that the completion of $\mu$
is a totally finite Radon submeasure on $X$.  \Hint{let $\Sigma_0$ be the algebra of
subsets of $X$ generated by the compact sets;  show that there is a totally finite Radon
submeasure extending $\mu\restr\Sigma_0$.}
%496H, 496J

\leader{496Y}{Further exercises}
In the following exercises, I will say that a {\bf Radon submeasure} is a
complete submeasure $\mu$ on a
Hausdorff space $X$ such that (i) the domain $\Sigma$ of $\mu$ is a
$\sigma$-algebra of subsets of $X$ containing every open set (ii) every
point of $X$ belongs to an open set $G$ such that $\mu G<\infty$
(iii)($\alpha$) $\mu E=\sup\{\mu K:K\subseteq E$ is compact$\}$ for every
$E\in\Sigma$ ($\beta$)
$\inf\{\mu(E\setminus K):K\subseteq E$ is compact$\}=0$ whenever
$E\in\Sigma$ and $\mu E<\infty$ (iv) if $E\subseteq X$ is such that
$E\cap K\in\Sigma$ for every compact $K\subseteq X$, then $E\in\Sigma$.

\spheader 496Ya Let $\mu$ be a Radon submeasure with domain $\Sigma$ and
null ideal $\Cal N(\mu)$.   Show that $\Sigma/\Cal N(\mu)$ is Dedekind
complete.

\spheader 496Yb
Let $X$ be a Hausdorff space, $Y$ a metrizable space,
$\mu$ a Radon submeasure on $X$ with domain $\Sigma$, and $f:X\to Y$ a
$\Sigma$-measurable
function.   Let $\Cal H$ be the family of those $H\in\Sigma$ such that
$f\restr H$ is continuous.   Show that ($\alpha$)
$\mu E=\sup\{\mu H:H\in\Cal H$, $H\subseteq E\}$ for every $E\in\Sigma$
($\beta$) $\inf\{\mu(E\setminus H):H\in\Cal H$, $H\subseteq E\}=0$ whenever
$E\in\Sigma$ and $\mu E<\infty$.
%496C

\spheader 496Yc
Let $X$ and $Y$ be Hausdorff spaces, $\mu$ a Radon
submeasure on $X$ with domain $\Sigma$, and $f:X\to Y$ a function.
Let $\Cal F$ be the family of those $F\in\Sigma$ such that
$f\restr F$ is continuous, and suppose that ($\alpha$)
$\mu E=\sup\{\mu F:F\in\Cal F$, $F\subseteq E\}$ for every $E\in\Sigma$
($\beta$) $\inf\{\mu(E\setminus F):F\in\Cal F$, $F\subseteq E\}=0$ whenever
$E\in\Sigma$ and $\mu E<\infty$.
(i) Show that the image submeasure
$\nu=\mu f^{-1}$, defined on $\{F:F\subseteq Y$, $f^{-1}[F]\in\Sigma\}$,
is a submeasure on $Y$ defined on a $\sigma$-algebra of sets
containing every open subset of $Y$.
(ii) Show that if $\nu$ is locally finite in the sense that
$Y=\bigcup\{H:H\subseteq Y$ is open, $\nu H<\infty\}$, then $\nu$ is a
Radon submeasure.
%496Xa 496C

\spheader 496Yd Let $X$ be a Hausdorff space and $\mu$ a Radon submeasure
on $X$ which is either submodular or supermodular.   Show that there is a
Radon measure on $X$ with the same domain and null ideal as $\mu$.
\Hint{413Yh.}
%496C

\spheader 496Ye Let $X$ be a topological space, $\Cal G$ the family of
cozero subsets of $X$, $\CalBa$ the Baire $\sigma$-algebra of $X$ and
$\psi:\Cal G\to\coint{0,\infty}$ a functional.   Show that $\psi$ can be
extended to a Maharam submeasure with domain $\CalBa$ iff

\inset{($\alpha$)
$\psi G\le\psi H$ whenever $G$, $H\in\Cal G$ and $G\subseteq H$,

($\beta$)
$\psi(\bigcup_{n\in\Bbb N}G_n)\le\sum_{n=0}^{\infty}\psi G_n$ for every
sequence $\sequencen{G_n}$ in $\Cal G$,

($\gamma$) $\lim_{n\to\infty}\psi G_n=0$ for every non-increasing
sequence $\sequencen{G_n}$ in $\Cal G$ with empty intersection,}

\noindent and that in this case the extension is unique.   \Hint{consider
the family of sets $E\subseteq X$ such that for every $\epsilon>0$ there
are a cozero set $G\supseteq E$ and a zero set $F\subseteq E$ such that
$\psi(G\setminus F)\le\epsilon$.}
%496E mt49bits

\spheader 496Yf Let $X$ be a Hausdorff space and $\Cal K$ the
family of
compact subsets of $X$.   Let $\phi:\Cal K\to\coint{0,\infty}$ be a
functional such that

\inset{($\alpha$) $\phi\emptyset=0$ and
$\phi K\le\phi(K\cup L)\le\phi K+\phi L$ for all $K$,
$L\in\Cal K$;

($\beta$) for every disjoint sequence $\sequencen{K_n}$ in $\Cal K$, either
$\lim_{n\to\infty}\phi K_n=0$ or
$\lim_{n\to\infty}\phi(\bigcup_{i\le n}K_i)=\infty$;

($\gamma$) whenever $K$, $L\in\Cal K$ and $K\subseteq L$ then
$\phi L\le\phi K+\sup\{\phi K':K'\in\Cal K$, $K'\subseteq L\setminus K\}$;

($\delta$) for every $x\in X$ there is an open set $G$ containing $x$ such
that $\sup\{\phi K:K\in\Cal K$, $K\subseteq G\}$ is finite.
}

\noindent Show that there is a unique
Radon submeasure on $X$ extending $\phi$.
%496E
}%end of exercises

\endnotes{
\Notesheader{496} `Submeasures' turn up in all sorts of places, if you are
looking out for them;  so, as always, I have tried to draw my definitions
as wide as practicable.   When we come to `Maharam' and `Radon'
submeasures, however, we certainly want to begin with results corresponding
to the familiar properties of totally finite measures, and the new language
is complex enough without troubling with infinite submeasures.   For the
main part of this section, therefore,
I look only at totally finite submeasures.

I have tried here to give a sample of the ideas from
the present volume which can be applied to submeasures as well as to
measures.   I think they go farther than most of us would take for granted.
One key point concerns the definition of inner regularity:  to the familiar
`$\mu E=\sup\{\mu K:K\in\Cal K$, $K\subseteq E\}$' we need to add
`if $\mu E$ is finite, then
$\inf\{\mu(E\setminus K):K\in\Cal K$, $K\subseteq E\}=0$' (496C, 496Y;  see
also condition ($\beta$) of 496Yf).   Using this refinement, we
can repeat a good proportion of the arguments of measure theory which are
based on topology and orderings rather than on arithmetic identities.
}%end of endnotes

\discrpage

\frfilename{mt497.tex}
\versiondate{7.12.10}
\copyrightdate{2009}

\def\TaoIG{{\smc Tao 07}}

\def\chaptername{Further topics}
\def\sectionname{Tao's proof of Szemer\'edi's theorem}

\newsection{497\dvAnew{2009}}

Szemer\'edi's celebrated theorem on arithmetic
progressions (497L) is not obviously part of measure theory.
Remarkably, however, it has stimulated significant developments in
the subject.   The first was Furstenberg's multiple recurrence
theorem\cmmnt{ ({\smc Furstenberg 77}, {\smc Furstenberg 81},
{\smc Furstenberg \& Katznelson 85})}.   In this section I will give an
account of an approach due to T.Tao\cmmnt{ (\TaoIG)} which
introduces another
phenomenon of great interest from a measure-theoretic point of view.

\leader{497A}{Definitions (a)} Let $(X,\Sigma,\mu)$ be a probability space,
$\Tau$ a subalgebra of $\Sigma$\cmmnt{ ({\it not} necessarily a
$\sigma$-subalgebra)} and
$\familyiI{\Sigma_i}$ a family of $\sigma$-subalgebras of $\Sigma$.
\cmmnt{I will say that }$\familyiI{\Sigma_i}$ has
{\bf $\Tau$-removable intersections} if whenever $J\subseteq I$ is finite
and not empty, $E_i\in\Sigma_i$ for $i\in J$,
$\mu(\bigcap_{i\in J}E_i)=0$ and $\epsilon>0$, there is a family
$\family{i}{J}{F_i}$ such that $F_i\in\Tau\cap\Sigma_i$ and
$\mu(E_i\setminus F_i)\le\epsilon$ for each $i\in J$, and
$\bigcap_{i\in J}F_i=\emptyset$.
\cmmnt{(This is a stronger version of what \TaoIG\ calls the
`uniform intersection property'.)}

\spheader 497Ab
If $X$ is a set and $\Sigma$, $\Sigma'$ are two $\sigma$-algebras
of subsets of $X$, $\Sigma\vee\Sigma'$ will be the $\sigma$-algebra
generated by $\Sigma\cup\Sigma'$.   If $\familyiI{\Sigma_i}$ is a family of
$\sigma$-algebras of subsets of $X$, I will write
$\bigvee_{i\in I}\Sigma_i$
for the $\sigma$-algebra generated by $\bigcup_{i\in I}\Sigma_i$.

\spheader 497Ac
If $(X,\Sigma,\mu)$ is a probability space and
$\Cal A\subseteq\Cal E\subseteq\Sigma$,\cmmnt{ I will say that} $\Cal A$
is {\bf metrically dense} in $\Cal E$ if for every $E\in\Cal E$ and
$\epsilon>0$
there is an $F\in\Cal A$ such that $\mu(E\symmdiff F)\le\epsilon$\cmmnt{;
that is, if $\{F^{\ssbullet}:F\in A\}$ is dense in
$\{E^{\ssbullet}:E\in\Cal E\}$ for the measure-algebra topology on the
measure algebra of $\mu$ (323A)}.   Note
that a subalgebra of $\Sigma$ is metrically dense
in the $\sigma$-algebra it generates\cmmnt{ (compare 323J)}.

\leader{497B}{Lemma}
Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$ a
subalgebra of $\Sigma$.   Let $\familyiI{\Sigma_i}$ be a family of
$\sigma$-subalgebras of $\Sigma$.

(a) $\familyiI{\Sigma_i}$
has $\Tau$-removable intersections iff $\family{i}{J}{\Sigma_i}$ has
$\Tau$-removable intersections for every finite $J\subseteq I$.

(b) Suppose that $\familyiI{\Sigma_i}$ has $\Tau$-removable intersections
and that $\Tau\cap\Sigma_i$ is metrically dense in $\Sigma_i$ for every
$i$.   Let $J$ be any set and $f:J\to I$ a function.   Then
$\family{j}{J}{\Sigma_{f(j)}}$ has $\Tau$-removable intersections.

(c) Suppose that, for each $i\in I$, we are given a $\sigma$-subalgebra
$\Sigma'_i$ of $\Sigma_i$ such that for every $E\in\Sigma_i$ there is an
$E'\in\Sigma'_i$ such that $E\symmdiff E'$ is negligible.   If
$\familyiI{\Sigma'_i}$ has $\Tau$-removable intersections,
so has $\familyiI{\Sigma_i}$.

\proof{{\bf (a)} is trivial.

\medskip

{\bf (b)} Suppose that $K\subseteq J$ is finite and not empty, that
$\family{j}{K}{E_j}\in\prod_{j\in K}\Sigma_{f(j)}$ is such that
$\mu(\bigcap_{j\in K}E_j)=0$, and $\epsilon>0$.   Set $n=\#(K)$ and
$\eta=\bover{\epsilon}{n+2}>0$.
Set $E'_i=\bigcap_{j\in K,f(j)=i}E_j\in\Sigma_i$ for $i\in f[K]$;  then
$\bigcap_{i\in f[K]}E'_i=\bigcap_{j\in K}E_j$ is negligible, so we have
$F'_i\in\Tau\cap\Sigma_i$, for $i\in f[K]$, such that
$\bigcap_{i\in f[K]}F'_i=\emptyset$ and $\mu(E'_i\setminus F'_i)\le\eta$
for every $i\in f[K]$.   As $\Tau\cap\Sigma_i$ is metrically dense in
$\Sigma_i$ for each $i$, we can find $G_j\in\Tau\cap\Sigma_{f(j)}$
such that $\mu(E_j\symmdiff G_j)\le\eta$ for each $j\in K$.   Set
$G'_i=\bigcap_{j\in K,f(j)=i}G_j$ for $i\in f[K]$.   Then

\Centerline{$\mu(G'_i\setminus F'_i)
\le\mu(G'_i\setminus E'_i)+\mu(E'_i\setminus F'_i)
\le\sum_{j\in K,f(j)=i}\mu(G_j\setminus E_j)+\eta
\le(n+1)\eta$.}

\noindent Note that $G'_i\in\Tau\cap\Sigma_i$ for each $i$.   Now set
$F_j=G_j\setminus(G'_{f(j)}\setminus F'_{f(j)})$ for $j\in K$.
Then $F_j\in\Tau\cap\Sigma_{f(j)}$ and

\Centerline{$\mu(E_j\setminus F_j)
\le\mu(E_j\setminus G_j)+\mu(G'_{f(j)}\setminus F'_{f(j)})
\le(n+2)\eta=\epsilon$.}

\noindent Also

$$\eqalign{\bigcap_{j\in K}F_j
&=\bigcap_{i\in f[K]}\bigcap_{\Atop{j\in K}{f(j)=i}}
    G_j\setminus(G'_i\setminus F'_i)\cr
&=\bigcap_{i\in f[K]}G'_i\setminus(G'_i\setminus F'_i)
\subseteq\bigcap_{i\in f[K]}F'_i
=\emptyset.\cr}$$

\noindent As $\family{j}{K}{E_j}$ and $\epsilon$ are arbitrary,
$\family{j}{J}{\Sigma_{f(j)}}$ has $\Tau$-removable intersections.

\medskip

{\bf (c)} If $J\subseteq I$ is finite and not empty,
$\family{j}{J}{E_j}\in\prod_{j\in J}\Sigma_j$,
$\bigcap_{j\in J}E_j$ is negligible and $\epsilon>0$, then for each
$j\in J$ let $E'_j\in\Sigma'_j$ be such that $E'_j\symmdiff E_j$ is
negligible.   In this case, $\bigcap_{j\in J}E'_j$ is negligible, so
there are $F_j\in\Tau\cap\Sigma'_j$, for $j\in J$, such that
$\mu(E'_j\setminus F_j)\le\epsilon$ for every $j\in J$ and
$\bigcap_{j\in J}F_j$ is empty.   Now $\mu(E_j\setminus F_j)\le\epsilon$
for every $j$.   As $\family{j}{J}{E_j}$ and $\epsilon$ are arbitrary,
$\familyiI{\Sigma_i}$ has $\Tau$-removable intersections.
}%end of proof of 497B

\leader{497C}{Lemma} Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$ a
subalgebra of $\Sigma$.   Let $I$ be a set, $A$ an upwards-directed set,
and $\langle\Sigma_{\alpha i}\rangle_{\alpha\in A,i\in I}$ a family
of $\sigma$-subalgebras of $\Sigma$ such that, setting
$\Sigma_i=\bigvee_{\alpha\in A}\Sigma_{\alpha i}$ for each $i$,

\inset{(i) $\Sigma_{\alpha i}\subseteq\Sigma_{\beta i}$ whenever $i\in I$
and $\alpha\le\beta$ in $A$,

(ii) $\familyiI{\Sigma_{\alpha i}}$ has $\Tau$-removable intersections for
every $\alpha\in A$,

(iii) $\Sigma_i$ and $\bigvee_{j\in I}\Sigma_{\alpha j}$ are relatively
independent over $\Sigma_{\alpha i}$ for every $i\in I$ and $\alpha\in A$.}

\noindent Then $\familyiI{\Sigma_i}$ has $\Tau$-removable intersections.
%Tau 6.13

\proof{ Take a non-empty finite set $J\subseteq I$, a family
$\family{i}{J}{E_i}$ such that $E_i\in\Sigma_i$ for every $i\in J$ and
$\bigcap_{i\in J}E_i$ is negligible, and $\epsilon>0$.   Set
$\delta=\Bover1{\#(J)+1}$,
$\eta=\delta\sqrt{\bover{\epsilon}2}>0$.
For each $i\in J$ there are an $\alpha\in A$ and
an $E'_i\in\Sigma_{\alpha i}$ such that $\mu(E_i\symmdiff E'_i)\le\eta^2$
(because $A$ is upwards-directed, so
$\bigcup_{\alpha\in A}\Sigma_{\alpha i}$ is a subalgebra of $\Sigma$ and
is metrically dense in $\Sigma_i$);
we can suppose that it is the same $\alpha$ for each $i$.   Let
$g_i:X\to[0,1]$ be a $\Sigma_{\alpha i}$-measurable function which is a
conditional expectation of $\chi E_i$ on $\Sigma_{\alpha i}$;   then

\Centerline{$\|\chi E_i-g_i\|_2\le\|\chi E_i-\chi E'_i\|_2\le\eta$}

\noindent(cf.\  244Nb).   Set
$E''_i=\{x:g_i(x)\ge 1-\delta\}\in\Sigma_{\alpha i}$;
then

\Centerline{$\mu(E_i\setminus E''_i)
=\mu\{x:\chi E_i(x)-g_i(x)>\delta\}
\le\Bover{\eta^2}{\delta^2}=\Bover{\epsilon}{2}$.}

Set $E=\bigcap_{i\in J}E''_i$.   Then $\mu E=0$.   \Prf\
Since $\mu(\bigcap_{i\in J}E_i)=0$,
$\mu E\le\sum_{i\in J}\mu(E\setminus E_i)$.
For $i\in J$, set
$H_i=X\cap\bigcap_{j\in J\setminus\{i\}}E''_j$ and let $h_i$ be a
conditional
expectation of $\chi H_i$ on $\Sigma_{\alpha i}$.   Then
$X\setminus E_i\in\Sigma_i$ and
$H_i\in\bigvee_{j\in I}\Sigma_{\alpha j}$ are relatively independent over
$\Sigma_{\alpha i}$, while $\chi X-g_i$ is a
conditional expectation
of $\chi(X\setminus E_i)$ on $\Sigma_{\alpha i}$, so

$$\eqalignno{\mu(E\setminus E_i)
&=\mu((E''_i\setminus E_i)\cap H_i)
=\int_{E''_i}\chi(X\setminus E_i)
  \times\chi H_id\mu
=\int_{E''_i}(\chi X-g_i)\times h_id\mu\cr
\displaycause{by the definition of `relative independence', 458Aa}
&=\int(\chi X-g_i)\times\chi E''_i\times h_id\mu
\le\int\delta\chi E''_i\times h_id\mu\cr
\displaycause{by the definition of $E''_i$}
&=\delta\int_{E''_i}h_id\mu
=\delta\mu(E''_i\cap H_i)
=\delta\mu E.\cr}$$

Summing, we have

\Centerline{$\mu E\le\delta\#(J)\mu E$;}

\noindent but $\delta\#(J)<1$, so $\mu E=0$.\ \Qed

Because $\familyiI{\Sigma_{\alpha i}}$ has $\Tau$-removable intersections, there are $F_i\in\Tau\cap\Sigma_{\alpha i}\subseteq\Tau\cap\Sigma_i$, for $i\in J$,
such that $\bigcap_{i\in I}F_i=\emptyset$ and
$\mu(E''_i\setminus F_i)\le\bover{\epsilon}2$ for each $i$;  in which case
$\mu(E_i\setminus F_i)\le\epsilon$ for each $i$.
As $\family{i}{J}{E_i}$ and $\epsilon$ are arbitrary,
$\familyiI{\Sigma_i}$ has $\Tau$-removable intersections.
}%end of proof of 497C

\leader{497D}{Lemma} Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a
subalgebra of $\Sigma$, and $\familyiI{\Sigma_i}$ a finite family of
$\sigma$-subalgebras of $\Sigma$ which has $\Tau$-removable intersections;
suppose that $\Tau\cap\Sigma_i$ is metrically dense in $\Sigma_i$ for each
$i$.   Set $\Sigma^*=\bigvee_{i\in I}\Sigma_i$.
Suppose that we have a finite set $\Gamma$, a function $g:\Gamma\to I$ and a family
$\family{\gamma}{\Gamma}{\Lambda_{\gamma}}$ of $\sigma$-subalgebras of $\Sigma$ such that

\inset{$\family{\gamma}{\Gamma}{\Lambda_{\gamma}}$
is relatively independent over $\Sigma^*$,

for each $\gamma\in\Gamma$, $\Lambda_{\gamma}$ and $\Sigma^*$ are relatively independent
over $\Sigma_{g(\gamma)}$,

for each $\gamma\in\Gamma$, $\Tau\cap\Lambda_{\gamma}$ is metrically dense in $\Lambda_{\gamma}$.}

\noindent Let $A$ be a finite
set and $f:A\to I$, $\phi:A\to\Cal P\Gamma$ functions
such that $\Sigma_{g(\gamma)}\subseteq\Sigma_{f(\alpha)}$ whenever $\alpha\in A$ and
$\gamma\in\phi(\alpha)$.   Suppose that

\inset{for each $\alpha\in A$, $\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$ and
$\Sigma^*\vee\bigvee_{\gamma\in\Gamma\setminus\phi(\alpha)}\Lambda_{\gamma}$
are relatively independent over $\Sigma_{f(\alpha)}$.}

\noindent Set $\tilde\Sigma_{\alpha}
=\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$
for $\alpha\in A$.   Then $\family{\alpha}{A}{\tilde\Sigma_{\alpha}}$ has
$\Tau$-removable intersections.
%compare Tao 6.12

\proof{ Of course we can suppose that $A$ is non-empty, and
that $\Gamma=\bigcup_{\alpha\in A}\phi(\alpha)$.

\medskip

{\bf (a)} To begin with, suppose that every $\Lambda_{\gamma}$ is actually a
finite subalgebra of $\Tau$.

\medskip

\quad{\bf (i)} Take a non-empty set $B\subseteq A$, a family
$\family{\alpha}{B}{E_{\alpha}}\in\prod_{\alpha\in B}\tilde\Sigma_{\alpha}$
such that $\bigcap_{\alpha\in B}E_{\alpha}$ is negligible, and
$\epsilon>0$.   Set $\Delta=\bigcup_{\alpha\in B}\phi(\alpha)$.
Let $\Cal A$ be the set of atoms of $\bigvee_{\gamma\in\Delta}\Lambda_{\gamma}$
and set $\eta=\Bover{\epsilon}{\#(\Cal A)}>0$.

\medskip

\quad{\bf (ii)} For each $H\in\Cal A$ and $\alpha\in B$,
let $C(H,\alpha)$ be the atom of $\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$
including $H$.   Then there is a family
$\family{\alpha}{B}{F_{H\alpha}}$, with empty intersection, such that
$F_{H\alpha}\in\Tau\cap\Sigma_{\alpha}$ and
$\mu(E_{\alpha}\cap C(H,\alpha)\setminus F_{H\alpha})\le\eta$ for each
$\alpha\in B$.   \Prf\ For
each $\gamma\in\Delta$, let $H_{\gamma}$ be the atom of $\Lambda_{\gamma}$ including $H$,
$h_{\gamma}:X\to[0,1]$ a $\Sigma_{g(\gamma)}$-measurable function which is a
conditional
expectation of $\chi H_{\gamma}$ on $\Sigma_{g(\gamma)}$, and $G_{\gamma}=\{x:h_{\gamma}(x)>0\}$.
Note that

\Centerline{$C(H,\alpha)=X\cap\bigcap_{\gamma\in\phi(\alpha)}H_{\gamma}$ for every
$\alpha\in B$,}

\Centerline{$H=X\cap\bigcap_{\gamma\in\Delta}H_{\gamma}=\bigcap_{\alpha\in B}C(H,\alpha)$.}

Because $\Lambda_{\gamma}$ and $\Sigma^*$ are relatively independent over
$\Sigma_{g(\gamma)}$, and $\Sigma_{g(\gamma)}\subseteq\Sigma^*$,
$h_{\gamma}$ is a conditional expectation of $\chi H_{\gamma}$ on $\Sigma^*$ for each
$\gamma$ (458Fb).
Because $\family{\gamma}{\Delta}{\Lambda_{\gamma}}$ is relatively independent over
$\Sigma^*$, $h=\prod_{\gamma\in\Delta}h_{\gamma}$ is a conditional expectation of
$\chi H=\chi(X\cap\bigcap_{\gamma\in\Delta}H_{\gamma})$ on $\Sigma^*$.
(For the trivial case in which $\Delta=\emptyset$, take $h=\chi X$.)
For each $\alpha\in B$ we have
$E_{\alpha}\in\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$,
so there is an an $E'_{\alpha}\in\Sigma_{f(\alpha)}$ such
that $E_{\alpha}\cap C(H,\alpha)=E'_{\alpha}\cap C(H,\alpha)$.
Now $\bigcap_{\alpha\in B}E'_{\alpha}\in\Sigma^*$, so

\Centerline{$\int_{\bigcapop_{\alpha\in B}E'_{\alpha}}h
=\mu(\bigcap_{\alpha\in B}E'_{\alpha}\cap\bigcap_{\gamma\in\Delta}H_{\gamma})
\le\mu(\bigcap_{\alpha\in B}(E_{\alpha}\cap C(H,\alpha)))
=0$;}

\noindent accordingly
$\bigcap_{\alpha\in B}E'_{\alpha}\cap\bigcap_{\gamma\in\Delta}G_{\gamma}$ is negligible.
Set $E''_{\alpha}=E'_{\alpha}\cap\bigcap_{\gamma\in\phi(\alpha)}G_{\gamma}$ for each
$\alpha\in B$;  then $E''_{\alpha}\in\Sigma_{f(\alpha)}$, because
$G_{\gamma}\in\Sigma_{g(\gamma)}\subseteq\Sigma_{f(\alpha)}$ whenever
$\gamma\in\phi(\alpha)$.
Also $\bigcap_{\alpha\in B}E''_{\alpha}$ is negligible.

Because $\familyiI{\Sigma_i}$ has $\Tau$-removable intersections and
$\Tau\cap\Sigma_i$ is metrically dense in $\Sigma_i$ for each $i$,
$\family{\alpha}{B}{\Sigma_{f(\alpha)}}$ has $\Tau$-removable intersections
(497Bb).
So we have $F_{H\alpha}\in\Tau\cap\Sigma_{f(\alpha)}$, for
$\alpha\in B$, such that $\bigcap_{\alpha\in A}F_{H\alpha}=\emptyset$ and
$\mu(E''_{\alpha}\setminus F_{H\alpha})\le\eta$ for every $\alpha$.

If $\alpha\in B$ and $\gamma\in\phi(\alpha)$,

\Centerline{$0=\int_{E'_{\alpha}\setminus G_{\gamma}}h_{\gamma}
=\mu(H_{\gamma}\cap E'_{\alpha}\setminus G_{\gamma})$}

\noindent because $h_{\gamma}$ is a conditional expectation of $\chi H_{\gamma}$ on
$\Sigma^*$.   So if $\alpha\in B$,

$$\eqalignno{\mu(E_{\alpha}\cap C(H,\alpha)\setminus F_{H\alpha})
&=\mu(E'_{\alpha}\cap C(H,\alpha)\setminus F_{H\alpha})\cr
&\le\mu(E''_{\alpha}\setminus F_{H\alpha})
  +\sum_{\gamma\in\phi(\alpha)}\mu(E'_{\alpha}\cap H_{\gamma}\setminus G_{\gamma})\cr
\displaycause{because $C(H,\alpha)=X\cap\bigcap_{\gamma\in\phi(\alpha)}H_{\gamma}$}
&\le\eta,\cr}$$

\noindent as required.\ \Qed

\medskip

\quad{\bf (iii)}
For $\alpha\in B$ let $\Cal A_{\alpha}$ be the set of atoms of
$\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$ and set

\Centerline{$F_{\alpha}
=\bigcup_{G\in\Cal A_{\alpha}}\bigcap_{H\in\Cal A,H\subseteq G}
   F_{H\alpha}$.}

\noindent Then $F_{\alpha}\in\Tau\cap\tilde\Sigma_{\alpha}$ and

$$\eqalign{\mu(E_{\alpha}\setminus F_{\alpha})
&=\sum_{G\in\Cal A_{\alpha}}\mu(E_{\alpha}\cap G\setminus F_{\alpha})\cr
&\le\sum_{G\in\Cal A_{\alpha}}\sum_{\Atop{H\in\Cal A}{H\subseteq G}}
   \mu(E_{\alpha}\cap G\setminus F_{H\alpha})\cr
&=\sum_{H\in\Cal A}
   \mu(E_{\alpha}\cap C(H,\alpha)\setminus F_{H\alpha})
\le\eta\#(\Cal A)=\epsilon.\cr}$$

\noindent If $H\in\Cal A$ then $H\subseteq C(H,\alpha)\in\Cal A_{\alpha}$
and

\Centerline{$H\cap F_{\alpha}
\subseteq F_{\alpha}\cap C(H,\alpha)
\subseteq F_{H\alpha}$,}

\noindent for each $\alpha$.   So $H\cap\bigcap_{\alpha\in A}F_{\alpha}$ is
empty.   But $X=\bigcup\Cal A$ so
$\bigcap_{\alpha\in A}F_{\alpha}=\emptyset$.   As
$\family{\alpha}{B}{E_{\alpha}}$ and $\epsilon$ are arbitrary,
$\family{\alpha}{A}{\tilde\Sigma_{\alpha}}$ has
$\Tau$-removable intersections.

\medskip

{\bf (b)} Next, suppose that each $\Lambda_{\gamma}$ is the
$\sigma$-algebra generated by $\Tau\cap\Lambda_{\gamma}$.

\medskip

\quad{\bf (i)} For $L\in[\Tau]^{<\omega}$, $\gamma\in\Gamma$ and $\alpha\in A$
write $\Lambda_{L\gamma}$ for the algebra $\sigma$-generated by
$\Lambda_{\gamma}\cap L$ and $\tilde\Sigma_{L\alpha}
=\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{L\gamma}$.
Then

\Centerline{$\tilde\Sigma_{\Delta\alpha}\subseteq\tilde\Sigma_{L\alpha}$ whenever
$\alpha\in A$ and $\Delta\subseteq L\in[\Tau]^{<\omega}$,}

\Centerline{$\bigvee_{L\in[\Tau]^{<\omega}}\tilde\Sigma_{L\alpha}
=\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}
   \bigvee_{L\in[\Tau]^{<\omega}}\Lambda_{L\gamma}
=\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}
=\tilde\Sigma_{\alpha}$}

\noindent because each $\Lambda_{\gamma}$ is the $\sigma$-algebra generated by
$\Tau\cap\Lambda_{\gamma}=\bigcup_{L\in[\Tau]^{<\omega}}\Lambda_{L\gamma}$.
By (a), $\family{\alpha}{A}{\tilde\Sigma_{L\alpha}}$ has $\Tau$-removable
intersections for every $L\in[\Tau]^{<\omega}$.

\medskip

\quad{\bf (ii)} Suppose that $\alpha\in A$ and $L\in[\Tau]^{<\omega}$.
Then $\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$ and
$\Sigma^*\vee\bigvee_{\gamma\in\Gamma\setminus\phi(\alpha)}\Lambda_{\gamma}$ are
relatively independent over $\Sigma_{f(\alpha)}$, by hypothesis.   So
$\tilde\Sigma_{\alpha}
=\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{\gamma}$ and
$\Sigma^*\vee\bigvee_{\gamma\in\Gamma\setminus\phi(\alpha)}\Lambda_{\gamma}$ are
relatively independent over $\Sigma_{f(\alpha)}$ (458Db).
Because
$\Sigma_{f(\alpha)}\subseteq\tilde\Sigma_{L\alpha}
\subseteq\tilde\Sigma_{\alpha}$,
$\tilde\Sigma_{\alpha}$ and
$\Sigma^*\vee\bigvee_{\gamma\in\Gamma\setminus\phi(\alpha)}\Lambda_{\gamma}$ are
relatively independent over $\tilde\Sigma_{L\alpha}$ (458Dc).   Because
$\bigvee_{\gamma\in\phi(\alpha)}\Lambda_{L\gamma}\subseteq\tilde\Sigma_{L\alpha}$,
$\tilde\Sigma_{\alpha}$ and
$\Sigma^*\vee\bigvee_{\gamma\in\Gamma}\Lambda_{L\gamma}$ 
are relatively independent over $\tilde\Sigma_{L\alpha}$ (458Db again).
So $\tilde\Sigma_{\alpha}$ and

\Centerline{$\bigvee_{\beta\in A}\tilde\Sigma_{L\beta}
=\bigvee_{\beta\in A}
  (\Sigma_{f(\beta)}\vee\bigvee_{\gamma\in\phi(\beta)}\Lambda_{L\gamma})
\subseteq\Sigma^*\vee\bigvee_{\gamma\in\Gamma}\Lambda_{L\gamma}$}

\noindent are relatively independent over $\tilde\Sigma_{L\alpha}$.



\medskip

\quad{\bf (iii)} With (i), this shows that the family
$\langle\tilde\Sigma_{L\alpha}\rangle_{L\in[\Tau]^{<\omega},\alpha\in A}$
satisfies the conditions of 497C, and
$\family{\alpha}{A}{\tilde\Sigma_{\alpha}}$ has $\Tau$-removable
intersections.

\medskip

{\bf (c)} Finally, for the general case, let $\Lambda'_{\gamma}$ be the
$\sigma$-algebra generated by $\Lambda_{\gamma}\cap\Tau$ for $\gamma\in\Gamma$, and
$\Sigma'_{\alpha}
=\Sigma_{f(\alpha)}\vee\bigvee_{\gamma\in\phi(\alpha)}\Lambda'_{\gamma}$ for
$\alpha\in A$.   If $\gamma\in\Gamma$ and $F\in\Lambda_{\gamma}$, there is an
$F'\in\Lambda'_{\gamma}$ such that $F\symmdiff F'$ is negligible;  so if
$\alpha\in A$ and $E\in\tilde\Sigma_{\alpha}$, there is an
$E'\in\Sigma'_{\alpha}$ such that $E\symmdiff E'$ is negligible.   By
(b), $\family{\alpha}{A}{\Sigma'_{\alpha}}$ has $\Tau$-removable
intersections;  by 497Bc,
$\family{\alpha}{A}{\tilde\Sigma_{\alpha}}$ has $\Tau$-removable
intersections.
}%end of proof of 497D

\leader{497E}{Theorem}\cmmnt{ (\TaoIG)} % Theorem 4.2
Let $(X,\Sigma,\mu)$ be a
probability space, and $\Tau$ a subalgebra of $\Sigma$.   Let $\Gamma$ be a
partially ordered set such that $\gamma\wedge\delta=\inf\{\gamma,\delta\}$
is defined in $\Gamma$ for all
$\gamma$, $\delta\in\Gamma$, and $\family{\gamma}{\Gamma}{\Sigma_{\gamma}}$
a family of $\sigma$-subalgebras of $\Sigma$ such that

\inset{(i) $\Tau\cap\Sigma_{\gamma}$ is metrically dense in
$\Sigma_{\gamma}$ for every $\gamma\in\Gamma$,

(ii) if $\gamma$, $\delta\in\Gamma$ and $\gamma\le\delta$ then
$\Sigma_{\gamma}\subseteq\Sigma_{\delta}$,

(iii) if $\gamma\in\Gamma$ and $\Delta$, $\Delta'$ are
finite subsets of
$\Gamma$ such that $\delta\wedge\gamma\in\Delta'$ for every
$\delta\in\Delta$, then $\Sigma_{\gamma}$ and
$\bigvee_{\delta\in\Delta}\Sigma_{\delta}$ are relatively independent over
$\bigvee_{\delta\in\Delta'}\Sigma_{\delta}$.}

\noindent Then $\family{\gamma}{\Gamma}{\Sigma_{\gamma}}$
has $\Tau$-removable intersections.

\proof{{\bf (a)} To begin with (down to the end of (d) below)
suppose that $\Gamma$ is finite.
In this case, we have a rank function
$r:\Gamma\to\Bbb N$ such that $r(\gamma)
=\min\{n:n\in\Bbb N$, $r(\delta)<n$ for every $\delta<\gamma\}$
for each $\gamma\in\Gamma$.   For $a\subseteq\Gamma$ set
$\tilde\Sigma_a=\bigvee_{\gamma\in a}\Sigma_{\gamma}$;  note that
$\Tau\cap\tilde\Sigma_a$ is always metrically dense in $\tilde\Sigma_a$.

Let $A$ be the family of those sets
$a\subseteq\Gamma$ such that $\gamma\in a$ whenever $\gamma\le\delta\in a$.
For $n\in\Bbb N$ set $\Gamma_n=\{\gamma:r(\gamma)=n\}$ and
$A_n=\{a:a\in A$, $r(\gamma)<n$ for every $\gamma\in a\}$

\medskip

{\bf (b)} Suppose that $a$, $b$, $c$ are subsets of
$\Gamma$ and that $\gamma\wedge\delta\in c$ whenever
$\gamma\in a$ and
$\delta\in b\cup(a\setminus\{\gamma\})$.   Then

\inset{(i) $\family{\gamma}{a}{\Sigma_{\gamma}}$ is relatively
independent over $\tilde\Sigma_c$,

(ii) $\tilde\Sigma_a$ and $\tilde\Sigma_b$ are relatively independent
over $\tilde\Sigma_c$.}

\noindent\Prf\ Induce on $\#(a)$.   If $a=\emptyset$ then
$\tilde\Sigma_a=\{\emptyset,X\}$ and the result is trivial.   For the
inductive step, take $\gamma_0\in a$ and set $a'=a\setminus\{\gamma_0\}$.
Then the inductive hypothesis tells us that
$\family{\gamma}{a'}{\Sigma_{\gamma}}$ is relatively
independent over $\tilde\Sigma_c$ and that
$\tilde\Sigma_{a'}$ and $\tilde\Sigma_b$ are relatively independent
over $\tilde\Sigma_c$.   We also see that $\gamma_0\wedge\delta\in c$
whenever $\delta\in a'$, so that
$\Sigma_{\gamma_0}$ and $\tilde\Sigma_{a'}$
are relatively independent over $\tilde\Sigma_c$, by condition (iii) of
this theorem.  But this means that
$\family{\gamma}{a}{\Sigma_{\gamma}}$ is relatively independent over
$\tilde\Sigma_c$ (458Hb).   Similarly, because in fact
$\gamma_0\wedge\delta\in c$ for every $\delta\in a'\cup b$,
$\Sigma_{\gamma_0}$ and $\tilde\Sigma_{a'}\vee\tilde\Sigma_b$ are
relatively independent over $\tilde\Sigma_c$;  so the triple
$\Sigma_{\gamma_0}$, $\tilde\Sigma_{a'}$ and $\tilde\Sigma_b$ are
relatively independent over $\tilde\Sigma_c$ (458Hb again),
and $\tilde\Sigma_a=\Sigma_{\gamma_0}\vee\tilde\Sigma_{a'}$ and
$\tilde\Sigma_b$ are relatively independent over $\tilde\Sigma_c$
(458Ha).   Thus the induction continues.\ \Qed

\medskip

{\bf (c)} For each $n\in\Bbb N$, $\family{a}{A_n}{\tilde\Sigma_a}$ has
$\Tau$-removable intersections.   \Prf\ Induce on $n$.   If $n=0$ then
$A_n=\{\emptyset\}$ and the result is trivial.   For the inductive step to
$n+1\ge 1$, apply 497D, as follows.   The inductive hypothesis tells
us that $\family{a}{A_n}{\tilde\Sigma_{a}}$ has $\Tau$-removable
intersections, and we know that
$\Tau\cap\tilde\Sigma_{a}$ is always metrically dense in
$\tilde\Sigma_a$.   Set

\Centerline{$\Sigma^*=\bigvee_{a\in A_n}\tilde\Sigma_a=\tilde\Sigma_d$ }

\noindent where $d=\bigcup_{m<n}\Gamma_m$
is the largest member of $A_n$.   Define $g:\Gamma_n\to A_n$ by setting
$g(\gamma)=\{\delta:\delta<\gamma\}$.   Then $\gamma\wedge\delta\in d$ for
all distinct $\gamma$, $\delta\in\Gamma_n$, so
$\family{\gamma}{\Gamma_n}{\Sigma_{\gamma}}$ is relatively independent over
$\Sigma^*$, by (b-i) just above.   If $\gamma\in\Gamma_n$ and
$\delta\in d$, $\gamma\wedge\delta\in g(\gamma)$, so
$\Sigma_{\gamma}=\tilde\Sigma_{\{\gamma\}}$ and
$\Sigma^*=\tilde\Sigma_d$ are relatively independent over
$\tilde\Sigma_{g(\gamma)}$, by (b-ii).
Of course $\Tau\cap\Sigma_{\gamma}$ is
metrically dense in $\Sigma_{\gamma}$ for every $\gamma\in\Gamma_n$.

For $a\in A_{n+1}$, set $\phi(a)=a\cap\Gamma_n$ and

\Centerline{$f(a)=a\setminus\phi(a)=a\cap\bigcup_{m<n}\Gamma_m\in A_n$.}

\noindent If $\gamma\in\phi(a)$
then $g(\gamma)\subseteq a$, by the definition of $A$, so
$g(\gamma)\subseteq f(a)$ and
$\tilde\Sigma_{g(\gamma)}\subseteq\tilde\Sigma_{f(a)}$.   Finally, by
(b-ii), $\bigvee_{\gamma\in\phi(a)}\Sigma_{\gamma}$ and
$\Sigma^*\vee\bigvee_{\gamma\in\Gamma_n\setminus\phi(a)}\Sigma_{\gamma}$
are relatively independent over $\tilde\Sigma_{f(a)}$, because if
$\gamma\in\phi(a)$ and $\delta\in d\cup(\Gamma_n\setminus\phi(a))$ then
$\gamma\wedge\delta\in g(\gamma)\subseteq f(a)$.

So all the hypotheses of 497D are satisfied, and

\Centerline{$\family{a}{A_{n+1}}
{\tilde\Sigma_{f(a)}\vee\bigvee_{\gamma\in\phi(a)}\Sigma_{\gamma}}
=\family{a}{A_{n+1}}{\tilde\Sigma_{a}}$}

\noindent has $\Tau$-removable intersections.   Thus the induction
proceeds.\ \Qed

\medskip

{\bf (d)} Because $\Gamma$ is finite, there is some $n$ such that $A=A_n$.
Now, for each $\gamma\in\Gamma$, set
$e_{\gamma}=\{\delta:\delta\le\gamma\}$;  then $e_{\gamma}\in A$ and
$\Sigma_{\gamma}=\tilde\Sigma_{e_{\gamma}}$.
By 497Bb, or otherwise,
$\family{\gamma}{\Gamma}{\Sigma_{\gamma}}$ has $\Tau$-removable
intersections, as required.

\medskip

{\bf (e)} Thus the theorem is true when $\Gamma$ is finite.   For the
general case, take any finite $\Gamma_0\subseteq\Gamma$ and set
$\Gamma'=\{\inf a:a\subseteq\Gamma_0$ is non-empty$\}$.   Then
$\Gamma'$ is finite and closed under $\wedge$,
and $\family{\gamma}{\Gamma'}{\Sigma_{\gamma}}$ satisfies the conditions of
the theorem.   So $\family{\gamma}{\Gamma'}{\Sigma_{\gamma}}$ and
$\family{\gamma}{\Gamma_0}{\Sigma_{\gamma}}$ have $\Tau$-removable
intersections.   As $\Gamma_0$ is arbitrary,
$\family{\gamma}{\Gamma}{\Sigma_{\gamma}}$ has $\Tau$-removable
intersections (497Ba), and the proof is complete.
}%end of proof of 497E

\leader{497F}{Invariant measures on $\Cal P([I]^{<\omega})$ (a)}
Let $I$ be a set.   Then $\Cal P([I]^{<\omega})$
is a compact Hausdorff space, if we give it its usual topology, generated
by sets of the form $\{R:a\in R\subseteq[I]^{<\omega}$, $b\notin R\}$ for
finite sets $a$, $b\subseteq I$.
\cmmnt{(You should perhaps fix on the case $I=\Bbb N$
for the first reading of this paragraph,
so that $[I]^{<\omega}$ will be a relatively familiar
countable set, and you can remember that $\Cal P([I]^{<\omega})$
is homeomorphic to the Cantor set.)}
Let $G_I$ be the set of permutations of $I$,
and for $\phi\in G_I$, $R\subseteq[I]^{<\omega}$ set

\Centerline{$\phi\action R=\{\phi[a]:a\in R\}
=\{a:a\in[I]^{<\omega}$, $\phi^{-1}[a]\in R\}$,}

\noindent so that $\action$ is an action of $G_I$ on
$\Cal P([I]^{<\omega})$, and $R\mapsto\phi\action R$ is a homeomorphism for
every $\phi\in G_I$.
Let $P_I$ be the set of Radon probability measures on
$\Cal P([I]^{<\omega})$.   Then we have an action of $G_I$ on $P_I$ defined
by saying that

\Centerline{$\phi\action E=\{\phi\action R:R\in E\}$}

\noindent for $\phi\in G_I$ and $E\subseteq\Cal P([I]^{<\omega})$, and

\Centerline{$(\phi\action\mu)(E)=\mu(\phi^{-1}\action E)$}

\noindent for $\phi\in G_I$, $\mu\in P_I$ and Borel sets
$E\subseteq\Cal P([I]^{<\omega})$.
Because $R\mapsto\phi\action R$ is a homeomorphism, the map
$\mu\mapsto\phi\action\mu$ is a homeomorphism when $P_I$ is given its
narrow topology\cmmnt{, corresponding to the weak* topology on
$C(\Cal P([I]^{<\omega}))^*$ (437J, 437Kc)}.

\spheader 497Fb If $\mu\in P_I$,\cmmnt{ I will say that} $\mu$ is
{\bf permutation-invariant} if $\mu=\phi\action\mu$ for every
$\phi\in G_I$.

\spheader 497Fc
For $R\subseteq[I]^{<\omega}$ and $J\subseteq I$\cmmnt{ I} write
$R\lceil J$ for the trace $R\cap\Cal PJ\subseteq[J]^{<\omega}$
of $R$ on $J$.   Let $\Cal V$ be the
family of sets of the form
$\cmmnt{V_{JS}=}\{R:R\subseteq[I]^{<\omega}$, $R\lceil J=S\}$
where $J\subseteq I$ is
finite and $S\subseteq\Cal PJ$.   If $\mu$, $\nu\in P_I$ agree on
$\Cal V$, they are equal.   \prooflet{\Prf\
If $E\subseteq\Cal P([I])^{<\omega}$ is
open-and-closed, it is determined by coordinates in some finite subset
$\Cal K$ of $[I]^{<\omega}$, in the sense that if $R\in E$,
$R'\subseteq[I]^{<\omega}$ and $R\cap\Cal K=R'\cap\Cal K$, then
$R'\in E$.   Let $J\subseteq I$ be a finite set such that
$\Cal K\subseteq[J]^{<\omega}$, and set
$\Cal S=\{R\lceil J:R\in E\}$.   Now $\family{S}{\Cal S}{V_{JS}}$ is a
disjoint family in $\Cal V$ with union $E$, so

\Centerline{$\mu E=\sum_{S\in\Cal S}\mu V_{JS}=\nu E$.}

\noindent As $E$ is arbitrary, $\mu=\nu$ (416Qa).\ \Qed}

%Observe also that if we take $\Cal W$ to be the
%family of sets of the form
%$\{R:R\subseteq[I]^{<\omega}$, $R\lceil J\in\Cal S\}$
%where $J\subseteq I$ is countable and $\Cal S\subseteq\Cal PJ$, then
%$\Cal W$ is a $\sigma$-algebra of subsets of $\Cal P([I]^{<\omega})$
%including $\Cal V$, so contains every Baire set.

\spheader 497Fd
If $I$, $J$ are sets and $f:I\to J$ is a function,\cmmnt{ I} define
$\tilde f:\Cal P([J]^{<\omega})\to\Cal P([I]^{<\omega})$ by setting
$\tilde f(R)=\{a:a\in[I]^{<\omega}$, $f[a]\in R\}$ for
$R\subseteq[J]^{<\omega}$.   \cmmnt{Note that} $\tilde f$ is
continuous\prooflet{, since
$\{R:a\in\tilde f(R)\}=\{R:f[a]\in R\}$ is
a basic open-and-closed set in $\Cal P([J]^{<\omega})$ for every
$a\in[I]^{<\omega}$}.
%if $f:I\to J$ and $g:J\to K$ then $(gf)^{\sim}(R)=\{a:g[f[a]]\in R\}
%=\{a:f[a]\in\tilde g(R)\}=\tilde f(\tilde g(R))$, so
%(f)^{\sim}=\tilde f\tilde g
If $I\subseteq J$ and $f$ is the identity function, then
$\tilde f(R)=R\lceil I$ for every $R\subseteq[J]^{<\omega}$.
Observe that when $\phi\in G_I$ and $R\subseteq[I]^{<\omega}$
then $\tilde\phi(R)=\phi^{-1}\action R$.

\leader{497G}{Theorem}\cmmnt{ (\TaoIG)}
Let $I$ be an infinite set and $\Cal J$ a filter on $I$ not containing
any finite set.   Let $\Tau$ be the algebra of open-and-closed subsets of
$\Cal P([I]^{<\omega})$, and $\mu\in P_I$ a
permutation-invariant measure.   For $J\subseteq I$, write $\Sigma_J$ for
the $\sigma$-algebra of subsets of $\Cal P([I]^{<\omega})$ generated by
sets of the form $E_a=\{R:a\in R\subseteq[I]^{<\omega}\}$ where
$a\in[J]^{<\omega}$.   Then
$\family{J}{\Cal J}{\Sigma_J}$ has $\Tau$-removable intersections with
respect to $\mu$.

\proof{ I seek to apply 497E with $\Gamma=\Cal J$, ordered by $\subseteq$.
If $J\in\Cal J$ and $a\in[J]^{<\omega}$
then $\{R:a\in R\subseteq[I]^{<\omega}\}$ belongs to $\Tau\cap\Sigma_J$;
accordingly $\Sigma_J$ is the
$\sigma$-algebra generated by $\Tau\cap\Sigma_J$ and $\Tau\cap\Sigma_J$
is metrically dense in $\Sigma_J$.
Condition (ii) of 497E is obviously satisfied.
As for condition (iii), we can use 459I, as follows.   Taking
$X=\Cal P([I]^{<\omega})$, we have the action $\action$ of $G_I$ on $X$
described in 497Fa, and $R\mapsto\phi\action R$ is \imp\ for each $\phi$
because $\mu$ is permutation-invariant.   Now we see easily that

\inset{----- for every $J\subseteq I$,
$\bigcup_{K\in[J]^{<\omega}}\Sigma_K$ contains $E_a$ for every
$a\in[J]^{<\omega}$, so $\sigma$-generates $\Sigma_J$;

----- if $a\in[I]^{<\omega}$ and $\phi\in G_I$,

\Centerline{$E_{\phi[a]}
=\{R:\phi[a]\in R\}=\{\phi\action R:\phi[a]\in\phi\action R\}
=\{\phi\action R:a\in R\}=\phi\action E_a$;}

----- if
$J\subseteq I$, then $\{E:\phi\action E\in\Sigma_{\phi[J]}\}$ is a
$\sigma$-algebra of sets containing
$\phi^{-1}\action E_{\phi[a]}=E_a$ whenever
$a\in[J]^{<\omega}$, so it includes $\Sigma_J$, and
$\phi\action E\in\Sigma_{\phi[J]}$ for every $E\in\Sigma_J$;

----- if $J\subseteq I$ and $\phi\in G_I$ is such that
$\phi(i)=i$ for every $i\in J$, then $\{E:\phi\action E=E\}$ is a
$\sigma$-algebra of sets containing $E_a$ for every $a\in[J]^{<\omega}$, so
it includes $\Sigma_J$, and $\phi\action E=E$ for every $E\in\Sigma_J$.}

\noindent Thus the conditions of 459I are satisfied.
So if $J\in\Cal J$ and $\Cal K$, $\Cal K'$ are (finite) subsets of $\Cal J$
such that $J\cap K\in\Cal K'$ for every $K\in\Cal K$, 459I tells us that
$\Sigma_J$ and $\bigvee_{K\in\Cal K}\Sigma_K$ are relatively independent
over $\bigvee_{K\in\Cal K'}\Sigma_K$, as required by (iii) of 497E.

So 497E gives the result we seek.
}%end of proof of 497G

\leader{497H}{}\cmmnt{ I come now to the next essential ingredient of the
proof.

\woddheader{497H}{4}{2}{2}{72pt}

\noindent}{\bf Construction} Suppose we are given a sequence
$\sequencen{(m_n,T_n)}$ and a non-principal
ultrafilter $\Cal F$ on $\Bbb N$ such that

\inset{($\alpha$) $\sequencen{m_n}$ is a sequence in
$\Bbb N\setminus\{0\}$ and $\lim_{n\to\Cal F}m_n=\infty$,

($\beta$) $T_n\subseteq\Cal Pm_n$ for each $n$.}

\noindent Then for any set $I$ there is a permutation-invariant
$\mu\in P_I$ such that

\Centerline{$\mu\{R:R\lceil K=S\}
=\lim_{n\to\Cal F}\Bover1{m_n^{\#(K)}}
\#(\{z:z\in m_n^K$, $\tilde z(T_n)=S\})$}

\noindent whenever $K\subseteq I$ is finite and $S\subseteq\Cal PK$.

\proof{{\bf (a)}
For each $n\in\Bbb N$ let $\nu_n$ be the usual measure
on $m_n^I$, the product of $I$ copies of the uniform probability measure on
the finite set $m_n$.   The function
$w\mapsto\tilde w(T_n):m_n^I\to\Cal P([I]^{<\omega})$ is continuous, since
for any $a\in[I]^{<\omega}$ the set
$\{w:a\in\tilde w(T_n)\}=\{w:w[a]\in T_n\}$ is determined by coordinates in
the finite set $a$.   So we have a corresponding Radon
probability measure $\mu_n$ on
$\Cal P([I]^{<\omega})$ defined by saying that
$\mu_nE=\nu_n\{w:\tilde w(T_n)\in E\}$ for every set
$E\subseteq\Cal P([I]^{<\omega})$ such that $\nu_n$ measures
$\{w:\tilde w(T_n)\in E\}$ (418I).
If $K\subseteq I$ is finite and $w\in m_n^I$, then

\Centerline{$\tilde w(T_n)\lceil K
=\{a:a\subseteq K,\,w[a]\in T_n\}
=\{a:a\subseteq K,\,(w\restr K)[a]\in T_n\}
=(w\restr K)^{\sptilde}(T_n)$.}

\noindent So if $S\subseteq\Cal PK$, then

$$\eqalign{\mu_n\{R:R\lceil K=S\}
&=\nu_n\{w:w\in m_n^I,\,\tilde w(T_n)\lceil K=S\}\cr
&=\nu_n\{w:w\in m_n^I,\,(w\restr K)^{\sptilde}(T_n)=S\}\cr
&=\Bover1{m_n^{\#(K)}}\#(\{z:z\in m_n^K,\,\tilde z(T_n)=S\}).\cr}$$

\noindent Let $\mu$ be the limit $\lim_{n\to\Cal F}\mu_n$ in the narrow
topology on $P_I$;  then

$$\eqalignno{\mu\{R:R\lceil K=S\}
&=\lim_{n\to\Cal F}\mu_n\{R:R\lceil K=S\}\cr
\displaycause{because $\{R:R\lceil K=S\}$ is open-and-closed;  see 437Jf}
&=\lim_{n\to\Cal F}\Bover1{m_n^{\#(K)}}
  \#(\{z:z\in m_n^K,\,\tilde z(T_n)=S\})\cr}$$

\noindent whenever $K\subseteq I$ is finite and $S\subseteq\Cal PK$.

\medskip

{\bf (b)} Now let $\phi:I\to I$ be any permutation and
$\tilde\phi:\Cal P([I]^{<\omega})\to\Cal P([I]^{<\omega})$ the
corresponding permutation.   Then for any finite $K\subseteq I$,

$$\eqalign{\tilde\phi(R)\lceil K
&=\{a:a\subseteq K,\,a\in\tilde\phi(R)\}\cr
&=\{a:a\subseteq K,\,\phi[a]\in R\}
=\{a:a\in[I]^{<\omega},\,\phi[a]\in R\lceil\phi[K]\}.\cr}$$

\noindent Fix $n\in\Bbb N$ for the moment.   If $K\subseteq I$ is finite,
and $S\subseteq\Cal PK$, then

$$\eqalignno{\mu_n\tilde\phi^{-1}\{R:R\lceil K=S\}
&=\mu_n\{R:\tilde\phi(R)\lceil K=S\}\cr
&=\mu_n\{R:S=\{a:a\in[I]^{<\omega},\,\phi[a]\in R\lceil\phi[K]\}\}\cr
&=\mu_n\{R:R\lceil\phi[K]=\{\phi[a]:a\in S\}\}\cr
&=\nu_n\{w:\tilde w(T_n)\lceil\phi[K]=\{\phi[a]:a\in S\}\}\cr
&=\nu_n\{w:\{a:a\subseteq\phi[K],\,w[a]\in T_n\}
  =\{\phi[a]:a\in S\}\}\cr
&=\nu_n\{w:\{\phi[a]:a\subseteq K,\,w[\phi[a]]\in T_n\}
  =\{\phi[a]:a\in S\}\}\cr
&=\nu_n\{w:\{a:a\subseteq K,\,(w\phi)[a]\in T_n\}=S\}\cr
&=\nu_n\{w:\{a:a\subseteq K,\,w[a]\in T_n\}=S\}\cr
\displaycause{because $w\mapsto w\phi:m_n^I\to m_n^I$ is an automorphism
for the measure $\nu_n$}
&=\nu_n\{w:\tilde w(T_n)\lceil K=S\}
=\mu_n\{R:R\lceil K=S\}.\cr}$$

\noindent So $\mu_n$ and $\mu_n\tilde\phi^{-1}$ agree on the family
$\Cal V$ of basic open-and-closed sets described in 497F.
As this is true
for every $n$, we also have $\mu V=\mu\tilde\phi^{-1}[V]$ for every
$V\in\Cal V$, and $\mu=\mu\tilde\phi^{-1}$.   As $\phi$ is
arbitrary, $\mu$ is permutation-invariant.
}%end of proof of 497H

\leader{497I}{Definition} If $I$, $J$ are sets, $R\subseteq\Cal PI$ and
$S\subseteq\Cal PJ$,\cmmnt{ I will say for the purposes of the next two
results that} an
{\bf embedding} of $(I,R)$ in $(J,S)$ is an injective function $f:I\to J$
such that $f[a]\in S$ for every $a\in R$\cmmnt{, that is (when
$S\subseteq[J]^{<\omega}$), $R\subseteq\tilde f(S)$}.

\leader{497J}{Theorem}\cmmnt{ ({\smc Nagle R\"odl \& Schacht 06})}
% Theorem 3)
Let $L$ be a finite set with $r$ members, and
$T\subseteq\Cal PL$.   Then for every $\epsilon>0$
there is a $\delta>0$ such that whenever $I$ is a non-empty finite set,
$R\subseteq\Cal PI$ and the number of embeddings of $(L,T)$ in
$(I,R)$ is at most $\delta\#(I)^r$,
there is an $S\subseteq\Cal PI$
such that $\#(S\cap[I]^k)\le\epsilon\#(I)^k$ for every $k$
and there is no embedding of $(L,T)$ in $(I,R\setminus S)$.

\proof{ (\TaoIG) \Quer\ Suppose, if possible,
otherwise.

\medskip

{\bf (a)} We have a sequence $\sequencen{(m_n,T_n)}$ such that

\inset{\noindent $m_n\in\Bbb N\setminus\{0\}$, $T_n\subseteq\Cal Pm_n$;
the number of embeddings of $(L,T)$ in $(m_n,T_n)$
is at most $2^{-n}m_n^r$;
if $S\subseteq\Cal Pm_n$ and
$\#(S\cap[m_n]^k)\le\epsilon m_n^k$ for every $k$
then there is an embedding of $(L,T)$ in $(m_n,T_n\setminus S)$}

\noindent for every $n\in\Bbb N$.   Of course $(L,T)$ always has at
least one embedding in $(m_n,T_n)$ so $\lim_{n\to\infty}m_n=\infty$.
Let $I$ be an infinite set including $L$ and $\Cal F$ a non-principal
ultrafilter on $\Bbb N$.   Let $\mu\in P_I$ be the
permutation-invariant measure defined from
$\sequencen{(m_n,T_n)}$ and $\Cal F$ by the process of 497H.

\medskip

{\bf (b)} For $c\subseteq L$ set $J_c=c\cup(I\setminus L)$, so that
$\Sigma_{J_c}$, in the notation of 497G, is the $\sigma$-algebra of subsets of
$\Cal P([I]^{<\omega})$ generated by sets of the form
$E_a=\{R:a\in R\subseteq[I]^{<\omega}\}$ where
$a\in[c\cup(I\setminus L)]^{<\omega}$.   Note that every member of
$\Sigma_{J_c}$ is determined by coordinates in $\Cal PJ_c$,
in the sense that if $R\in E\in\Sigma_{J_c}$,
$R'\subseteq\Cal P([I]^{<\omega})$ and
$R\cap\Cal PJ_c)=R'\cap\Cal PJ_c$, then $R'\in E$.

By 497G, applied to the filter $\Cal J$ on $I$
generated by $\{I\setminus L\}$,
$\langle\Sigma_{J_c}\rangle_{c\subseteq L}$
has $\Tau$-removable intersections with respect to $\mu$, where $\Tau$ is
the algebra of open-and-closed subsets of $\Cal P([I]^{<\omega})$.
{\it A fortiori}, $\family{c}{T}{\Sigma_{J_c}}$
has $\Tau$-removable intersections with respect to $\mu$.

\medskip

{\bf (c)} $E_c\in\Sigma_{J_c}$ for every $c\in T$, and

$$\eqalignno{\mu(\bigcap_{c\in T}E_c)
&=\mu\{R:T\subseteq R\}
=\sum_{T\subseteq T'\subseteq\Cal PL}\mu\{R:R\lceil L=T'\}\cr
&=\sum_{T\subseteq T'\subseteq\Cal PL}
   \lim_{n\to\Cal F}\Bover1{m_n^r}
   \#(\{z:z\in m_n^L,\,\tilde z(T_n)=T'\})\cr
&=\lim_{n\to\Cal F}\Bover1{m_n^r}
   \sum_{T\subseteq T'\subseteq\Cal PL}
   \#(\{z:z\in m_n^L,\,\tilde z(T_n)=T'\})\cr
&=\lim_{n\to\Cal F}\Bover1{m_n^r}
   \#(\{z:z\in m_n^L,\,T\subseteq\tilde z(T_n)\})\cr
&=\lim_{n\to\Cal F}\Bover1{m_n^r}
   \#(\{z:z\in m_n^L\text{ is injective},\,T\subseteq\tilde z(T_n)\})\cr
\displaycause{because $\lim_{n\to\infty}\Bover{k_n}{m_n^r}=0$, where
$k_n=m_n^r-\Bover{m_n!}{(m_n-r)!}$ is the number
of non-injective functions from $L$ to $m_n$}
&=\lim_{n\to\Cal F}\Bover1{m_n^r}
   \#(\{z:z\text{ is an embedding of }(L,T)\text{ in }(m_n,T_n)\})
=0.}$$

\medskip

{\bf (d)} Take $\eta>0$ such that $2\eta\#(T)<\epsilon$, and
$\family{c}{T}{F_c}$ such that
$\bigcap_{c\in T}F_c=\emptyset$ and
$F_c\in\Tau\cap\Sigma_{J_c}$ and
$\mu(E_c\setminus F_c)\le\eta$ for every $c\in T$.
Every $F_c$ is open-and-closed, so
there is an $M\in[I]^{<\omega}$ such that $L\subseteq M$ and
every $F_c$ is determined by coordinates in $\Cal PM$.   In this case,
each $F_c$ is determined by coordinates in
$\Cal PM\cap\Cal PJ_c=\Cal P(c\cup(M\setminus L))$.   Setting

\Centerline{$F'_c=\{R\lceil M:R\in F_c\}$,
\quad$E'_c=\{R\lceil M:R\in E_c\}=\{R:c\in R\subseteq\Cal PM\}$,}

\noindent we have

\Centerline{$F_c=\{R:R\subseteq[I]^{<\omega}$, $R\lceil M\in F'_c\}$,
\quad$E_c=\{R:R\subseteq[I]^{<\omega}$, $R\lceil M\in E'_c\}$,}

\noindent while both $E'_c$ and $F'_c$, and therefore
$E'_c\setminus F'_c$, regarded as subsets of $\Cal P(\Cal PM)$,
are determined by coordinates in
$\Cal P(c\cup(M\setminus L))$.   Because
$\bigcap_{c\in T}F_c$ is empty, so is $\bigcap_{c\in T}F'_c$.

\wheader{497J}{4}{2}{2}{48pt}

{\bf (e)} Let $n\ge r$ be such that

$$\eqalign{\Bover1{m_n^{\#(M)}}\#(\{z:z\in m_n^M,\,
   \tilde z(T_n)\in E'_c\setminus F'_c\})
&\le\eta+\mu\{R:R\lceil M\in E'_c\setminus F'_c\}\cr
&=\eta+\mu\{R:R\in E_c\setminus F_c\}
\le 2\eta\cr}$$

\noindent for every $c\in T$.   For $c\in T$ set

\Centerline{$Q_c=\{z:z\in m_n^M$, $\tilde z(T_n)\in E'_c\setminus F'_c\}$,}

\noindent so that $\#(Q_c)\le 2\eta m_n^{\#(M)}$.   Since

$$\eqalign{\sum_{\Atop{c\in T}{w\in m_n^{M\setminus L}}}
   \#(\{z:w\subseteq z\in Q_c\})
&=\sum_{c\in T}\#(Q_c)
\le 2\eta\#(T)m_n^{\#(M)}\cr
&\le\epsilon m_n^{\#(M)}
=\epsilon\#(m_n^{M\setminus L})m_n^r,\cr}$$

\noindent there must be a $w\in m_n^{M\setminus L}$ such that

\Centerline{$\sum_{c\in T,w\in m_n^{M\setminus L}}
   \#(\{z:w\subseteq z\in Q_c\})
\le \epsilon m_n^r$;}

\noindent set

\Centerline{$Q'_c=\{z:w\subseteq z\in Q_c$, $z\restr c$ is injective$\}$}

\noindent for $c\in T$, so that $\sum_{c\in T}\#(Q'_c)\le\epsilon m_n^r$.

If $c\in T$ and $\#(c)=k$, then

\Centerline{$\#(\{z[c]:z\in Q'_c\})=\Bover1{m_n^{r-k}}\#(Q'_c)$.}

\noindent\Prf\ If $z$, $z'\in m_n^M$ and
$z\restr(c\cup(M\setminus L))=z'\restr(c\cup(M\setminus L))$, then
$\tilde z(T_n)\lceil(c\cup(M\setminus L))
=\tilde z'(T_n)\lceil(c\cup(M\setminus L))$, so
$\tilde z(T_n)\in E'_c\setminus F'_c$ iff
$\tilde z'(T_n)\in E'_c\setminus F'_c$, that is, $z\in Q_c$ iff
$z'\in Q_c$.   So if $a=z[c]$ for some $z\in Q'_c$, then

\Centerline{$\{z':z'\in Q'_c$, $z'[c]=a\}
=\{z':z'\in m_n^M$,
$z'\restr(c\cup(M\setminus L))=z\restr(c\cup(M\setminus L))\}$}

\noindent has just $\#(m_n^{L\setminus c})=m_n^{r-k}$ members.\ \Qed

\medskip

{\bf (f)} Consider

\Centerline{$S=\{z[c]:c\in T$, $z\in Q'_c\}$.}

\noindent Then

$$\eqalignno{\#(S\cap[m_n]^k)
&=\#([m_n]^k\cap\{z[c]:c\in T,\,z\in Q'_c\})\cr
&=\#(\{z[c]:c\in T\cap[L]^k,\,z\in Q'_c\})\cr
\displaycause{because every member of $Q'_c$ is injective on $c$}
&\le\sum_{\Atop{c\in T}{\#(c)=k}}\Bover1{m_n^{r-k}}\#(Q'_c)\cr
\displaycause{by the last remark in (e)}
&\le\Bover1{m_n^{r-k}}\epsilon m_n^r
=\epsilon m_n^k\cr}$$

\noindent for every $k$.   So by the choice of $(m_n,T_n)$
there is an embedding $v$ of $(L,T)$ in
$(m_n,T_n\setminus S)$;
take $z=v\cup w$, so that $w\subseteq z\in m_n^M$ and
$z\restr L=v$ is injective and $z[c]\notin S$ for every $c\in T$.
However, there is some $c\in T$ such that
$\tilde z(T_n)\notin F'_c$.   As $c\in\tilde z(T_n)$,
$\tilde z(T_n)\in E'_c$.   But now
$z\in Q'_c$ and $z[c]\in S$.\ \Bang

This contradiction proves the theorem.
}%end of proof of 497J

\leader{497K}{Corollary: the Hypergraph Removal Lemma}
For every $\epsilon>0$ and $r\ge 1$
there is a $\delta>0$ such that whenever $I$ is a finite set,
$R\subseteq[I]^r$ and
$\#(\{J:J\in[I]^{r+1}$, $[J]^r\subseteq R\})\le\delta\#(I)^{r+1}$,
there is an $S\subseteq[I]^r$
such that $\#(S)\le\epsilon\#(I)^r$ and there is no
$J\in[I]^{r+1}$ such that $[J]^r\subseteq R\setminus S$.

\proof{ In 497J, take $L$ to be a set with $r+1$ members, and set
$T=[L]^r$ in 497J.   Then there is a $\delta_0>0$ such that whenever
$I$ is a finite set, $R\subseteq[I]^r$ and the number of embeddings of
$(L,[L]^r)$ in $(I,R)$ is at most $\delta_0\#(I)^{r+1}$, there is an
$S\subseteq[I]^r$ such that $\#(S)\le\epsilon\#(I)^r$ and there
is no embedding of $(L,[L^r])$ in $(I,R\setminus S)$.   Try
$\delta=\Bover1{(r+1)!}\delta_0$.   If $I$ is finite, $R\subseteq[I]^r$
and $\Cal J=\{J:J\in[I]^{r+1}$, $[J]^r\subseteq R\}$ has at most
$\delta\#(I)^{r+1}$ members,
then an embedding of $(L,[L]^r)$ in $(I,R)$ is an injective function
$f:L\to I$ such that $f[J]\in R$ for every $J\in[L]^r$, that is,
$f[L]\in\Cal J$.   So the number of such embeddings is
$(r+1)!\#(\Cal J)\le\delta_0\#(I)^{r+1}$.   There is therefore an
$S\subseteq[I]^r$ such that $\#(S)\le\epsilon\#(I)^r$ and there is
no embedding of $(L,[L]^r)$ in $(I,R\setminus S)$, that is, there is no
$J\in[I]^{r+1}$ such that $[J]^r\subseteq R\setminus S$.
}%end of proof of 497K

\leader{497L}{Corollary:  Szemer\'edi's
Theorem}\cmmnt{ ({\smc Szemer\'edi 75})}
For every $\epsilon>0$ and $r\ge 2$ there is an $n_0\in\Bbb N$
such that whenever $n\ge n_0$, $A\subseteq n$ and $\#(A)\ge\epsilon n$
there is an arithmetic progression of length $r+1$ in $A$.

\proof{ ({\smc Frankl \& R\"odl 02})
Set $\eta=\Bover1{r!}(\Bover{\epsilon}{2r!})^r$.
Take $\delta>0$ such that
whenever $I$ is a finite set, $R\subseteq[I]^r$ and
$\#(\{J:J\in[I]^{r+1}$, $[J]^r\subseteq R\})\le\delta\#(I)^{r+1}$,
there is an $S\subseteq[I]^r$
such that $\#(S)\le\Bover{\eta}{2(r+1)^r}\#(I)^r$
and there is
no $J\in[I]^{r+1}$ such that $[J]^r\subseteq R\setminus S$.
Let $n_0$ be such that $\epsilon n\ge 2r\cdot r!$ and
$n(r+1)^{r+1}\delta\ge 1$
whenever $n\ge n_0$.   Take $n\ge n_0$ and
$A\subseteq n$ such that $\#(A)\ge\epsilon n$.

Let $C\subseteq n^r$ be the set

\Centerline{$\{(i_0,i_1,\ldots,i_{r-1}):\sum_{j=0}^{r-1}(j+1)i_j
\in A\}$.}

\noindent Then $\#(C)\ge\eta n^r$.\footnote{For the rest of this proof, and also in
497M and 497N below, I will use the formula $n^r$
both for the set of functions from $r=\{0,\ldots,r-1\}$ to
$n=\{0,\ldots,n-1\}$ and for its
cardinal interpreted as a real number;  I trust that this will not lead to
any confusion.}   \Prf\ For $m<r!$ set
$A_m=\{i:i\in A$, $i\equiv m\mod r!\}$.   Then there is an $m$ such that
$\#(A_m)\ge\Bover{\epsilon n}{r!}$.   Now we have an injection
$\phi:[A_m]^r\to C$ given by saying that if $l_0<\ldots<l_{r-1}$ in $A_m$
then

$$\eqalign{\phi(\{l_0,\ldots,l_{r-1}\})(j)
&=l_0\text{ if }j=0\cr
&=\Bover1{j+1}(l_j-l_{j-1})\text{ if }0<j<r.\cr}$$

\noindent So

\Centerline{$\#(C)\ge\#([A_m]^r)
\ge\Bover1{r!}(\Bover{\epsilon n}{r!}-r)^r
\ge\Bover1{r!}(\Bover{\epsilon n}{2r!})^r
=\eta n^r$.  \Qed}

Let $I$ be $n\times(r+1)$ and for $c=(i_0,\ldots,i_{r-1})\in C$ set

\Centerline{$J_c
=\{(i_j,j):j<r\}\cup\{(\sum_{j=0}^{r-1}i_j,r)\}\in[I]^{r+1}$.}

\noindent Observe that if $c$, $c'\in C$ are distinct, then
$[J_c]^r\cap[J_{c'}]^r=\emptyset$, since given any face of
the $r$-simplex $J_c$ we can read off all but at most
one of the coordinates of $c$ and calculate the last.
Set $R=\bigcup_{c\in C}[J_c]^r\subseteq[I]^r$.

\Quer\ Suppose, if possible, that the only $r$-simplices $J\in[I]^{r+1}$
such that $[J]^r\subseteq R$ are of the form $J_c$ for some $c\in C$.
Then there are at most

\Centerline{$\#(C)\le n^r\le n^r\cdot n(r+1)^{r+1}\delta
=\delta\#(I)^{r+1}$}

\noindent such simplices; by the choice of $\delta$, there is an
$S\subseteq[I]^r$ such that $R\setminus S$ covers no $r$-simplices and

\Centerline{$\#(S)\le\Bover{\eta}{2(r+1)^r}\#(I)^r
=\Bover{\eta}2n^r<\#(C)$.}

\noindent
But every $J_c$ must have a face in $S$, and no two $J_c$ share a face, so
this is impossible.\ \Bang

So we have an $r$-simplex $J\in[I]^{r+1}$, which is not of the form $J_c$
where $c\in C$, such that $[J]^r\subseteq R$.   Now since the only faces
put into $R$ come from the $J_c$, and therefore meet each of the $r+1$
levels $n\times\{k\}$ in at most one point, $J$ must be of the form
$\{(i_j,j):j<r\}\cup\{(l,r)\}$.   Since $\{(i_j,j):j<r\}$ is a face of some
$J_c$, $c=(i_0,\ldots,i_{r-1})\in C$.   Set $l'=i_0+\ldots+i_{r-1}$;  then
$l'\ne l$ because $J\ne J_c$.   For each $k<r$,
$J\setminus\{(i_k,k)\}$ is a face of $J$ and therefore of $J_{c'}$ for some
$c'\in C$;  now $J_{c'}$ must be

\Centerline{$(J\setminus\{(i_k,k)\})\cup\{(l-\sum_{j<r,j\ne k}i_j,k)\}
=(J\setminus\{(i_k,k)\})\cup\{(i_k+l-l',k)\}$}

\noindent and

\Centerline{$\sum_{j=0}^r(j+1)i_j+(k+1)(l-l')$}

\noindent belongs to $A$.   Since this is true for every $k<r$,
and we also have $\sum_{j=0}^r(j+1)i_j\in A$ because $c\in C$,
we have an arithmetic progression in $A$ of length $r+1$, as required.
}%end of proof of 497L

\leader{497M}{}\cmmnt{ For a full-strength version of the multiple
recurrence theorem it seems that the ideas described above are
inadequate;  for an adaptation which goes farther,
see {\smc Austin 10a} and
{\smc Austin 10b}.   However the methods here can reach the following.

\medskip

\noindent}{\bf Lemma} (cf.\ {\smc Solymosi 03})
Suppose that $r\ge 1$ and $n\in\Bbb N$.   For $0\le j$,
$k<r$ set $e_j(k)=1$ if $k=j$, $0$ otherwise.   For
$z\in n^r$ and $C\subseteq n^r$ write

\Centerline{$\Delta(z,C)
=\{k:k\in\Bbb Z$, $z+ke_i\in C$ for every $i<r\}$,
\quad$q(z,C)=\#(\Delta(z,C))$.}

\noindent Then for every $\epsilon>0$ there is a $\delta>0$ such that
$\#(\{z:z\in n^r$, $q(z,C)\ge\delta n\})\ge\delta n^r$
whenever $n\in\Bbb N$, $C\subseteq n^r$ and $\#(C)\ge\epsilon n^r$.

\proof{ We can use some of the same ideas as in 497L.
Let $\epsilon'>0$ be such that $2^rr^r\epsilon'<\epsilon$.
Let $\delta>0$ be such that whenever $I$ is
finite, $R\subseteq[I]^r$ and
$\#(\{J:J\in[I]^{r+1}$, $[J]^r\subseteq R\})\le\delta\#(I)^{r+1}$
there is an $S\subseteq R$ such that
$\#(S)\le\epsilon'\#(I)^r$ and there is no $J\in[I]^{r+1}$ such
that $[J]^r\subseteq R\setminus S$ (497K).

Take $n\in\Bbb N$ and $C\subseteq n^r$ such that $\#(C)\ge\epsilon n^r$.
Set $I=(n\times r)\cup(nr\times\{r\})$, so that $\#(I)=2nr$.
For $c\in C$ set

\Centerline{$J_c
=\{(c(i),i):i<r\}\cup\{(\sum_{i=0}^{r-1}c(i),r)\}\in[I]^{r+1}$;}

\noindent set $R=\bigcup_{c\in C}[J_c]^r$.   Observe that if $c$, $c'\in C$
are distinct then $[J_c]^r$ and $[J_{c'}]^r$ are disjoint.
If $S\subseteq R$ and $\#(S)\le\epsilon'\#(I)^r$, then $\#(S)<\epsilon n^r$
and there must be a $c\in C$ such that $[J_c]^r\cap S=\emptyset$ and
$[J_c]^r\subseteq R\setminus S$.  Consequently

\Centerline{$\Cal K=\{K:K\in[I]^{r+1}$, $[K]^r\subseteq R\}$}

\noindent must have more than
$\delta\#(I)^{r+1}\ge 2\delta n^{r+1}$ members, by the choice of $\delta$.

Next, $\#(\Cal K)=\sum_{z\in n^r}q(z,C)$.   \Prf\
Set $B=\{(z,k):z\in n^r$, $k\in\Bbb Z$, $z+ke_i\in C$ for every $i<r\}$;
then $\#(B)=\sum_{z\in n^r}q(z,C)$.
For any $K\in\Cal K$, there must be a $c_K\in C$ such that
$(c_K(i),i)\in K$ for every $i<r$ while
$(k_K+\sum_{i=0}^{r-1}c_K(i),r)\in K$ for
some $k_K$;  in this case, $c_K+k_Ke_i\in C$ for every $i<r$ and
$(c_K,k_K)\in B$.   Conversely, starting from $(z,k)\in B$,
$\{(z(i),i):i<r\}\cup\{(k+\sum_{i=0}^{r-1}z(i),r)\}$ belongs to $K$.
So $K\mapsto(c_K,k_K)$ is a bijection from $\Cal K$ to $B$ and
$\#(\Cal K)=\#(B)$.\ \Qed

Thus $\sum_{z\in n^r}q(z,C)\ge 2\delta n^{r+1}$.
Of course

\Centerline{$q(z,C)\le\#(\{k:z+ke_0\in n^r\}\le n$}

\noindent for every $z\in n^r$.   So setting
$D=\{z:z\in n^r$, $q(z,C)\ge\delta n\}$, we have

\Centerline{$2\delta n^{r+1}\le n\#(D)+\delta n\cdot n^r
\le n\#(D)+\delta n^{r+1}$}

\noindent and $\#(D)\ge\delta n^r$, as claimed.
}%end of proof of 497M

\leader{497N}{Theorem}\cmmnt{ ({\smc Furstenburg 81})} %7.13
Let $(\frak A,\bar\mu)$ be a probability algebra and
$\ofamily{i}{r}{\pi_i}$
a non-empty finite commuting family of measure-preserving
Boolean homomorphisms from $\frak A$ to itself.
If $a\in\frak A\setminus\{0\}$, there is an $\eta>0$ such that

\Centerline{$\sum_{k=0}^{n-1}\bar\mu(\inf_{i<r}\pi_i^ka)\ge\eta n$}

\noindent for every $n\in\Bbb N$.

\proof{{\bf (a)} To being with, suppose that every $\pi_i$ is an
automorphism.
Doubling $r$ if necessary, we can suppose that for every $i<r$
there is a $j<r$ such that $\pi_j=\pi_i^{-1}$.   Let $(Z,\Sigma,\mu)$ be
the Stone space of $(\frak A,\bar\mu)$ (321K), and set $E=\widehat{a}$, the
open-and-closed subset of $Z$ corresponding to $a\in\frak A$.
For each $i<r$ let
$T_i:Z\to Z$ be the homeomorphism corresponding to
$\pi_i:\frak A\to\frak A$, so that $T_i^{-1}[\widehat{b}]=\widehat{\pi_ib}$
for every $b\in\frak A$ (312Q\formerly{3{}12P});  note that
$T_iT_j$ corresponds to $\pi_j\pi_i$ (312R\formerly{3{}12Q}) and
$T_iT_j=T_jT_i$ (because the representations in 312Q are unique), for all
$i$, $j<r$.

In 497M, set $\epsilon=\bover12\mu E=\bover12\bar\mu a$ and take a
corresponding $\delta>0$;  set $\eta=\bover12\delta^2\epsilon$.
Now, given $n\ge 1$, then for $z\in n^r$ set
$\tilde T_z=\prod_{i<r}T_i^{z(i)}$.   (We can speak of the product
without inhibitions because the $T_i$ commute.)   Consider the set
$W=\{(x,z):z\in n^r$, $\tilde T_z(x)\in E\}$.
Then $W^{-1}[\{z\}]$ has measure $\mu E$ for
every $z$, so if we set $F=\{x:x\in E$, $\#(W[\{x\}])\ge\epsilon n^r\}$
we have

\Centerline{$n^r\mu E\le n^r\mu F+\epsilon n^r$,
\quad$\mu F\ge\epsilon$.}

In the notation of 497M, set

\Centerline{$V=\{(x,z):(x,z)\in W$, $q(z,W[\{x\}])\ge\delta n\}$;}

\noindent then for any $x\in F$ we have $\#(V[\{x\}])\ge\delta n^r$, by
the choice of $\delta$.   There must therefore be a $z\in n^r$ such that
$\mu V^{-1}[\{z\}]\ge\delta\mu F\ge\delta\epsilon$.
Take any $x\in V^{-1}[\{z\}]$ and $k\in\Delta(z,W[\{x\}])$.
Setting $e_i(i)=1$ and $e_i(j)=0$ for $i<r$ and $j\in r\setminus\{i\}$,
$z+ke_i\in W[\{x\}]$, that is, $T_i^k\tilde T_z(x)\in E$, for every
$i<r$.   Also $|k|<n$.   Set $G=\tilde T_z[V^{-1}[\{z\}]]$, so that
$\mu G\ge\delta\epsilon$ and for every $y\in G$ we have
$\#(\{k:|k|<n$, $T_i^k(y)\in E$ for every $i<r\})\ge\delta n$.   But
this means that

$$\eqalignno{\sum_{k=0}^{n-1}\bar\mu(\inf_{i<r}\pi_i^ka)
&=\sum_{k=0}^{n-1}\mu\{x:T_i^k(x)\in E\text{ for every }i<r\}\cr
&\ge\Bover12\sum_{|k|<n}\mu\{x:T_i^k(x)\in E\text{ for every }i<r\}\cr
\displaycause{because if $T_i^k(x)\in E$ for every $i<r$ then
$T_i^{|k|}(x)\in E$ for every $i<r$}
&=\Bover12\int\#(\{k:|k|<n,\,T_i^k(x)\in E
  \text{ for every }i<r\})\mu(dx)\cr
&\ge\Bover12\delta n\mu G
\ge\Bover12\delta^2\epsilon n
=\eta n,\cr}$$

\noindent as required.

\medskip

{\bf (b)} For the general case, 328J tells us that there are
a probability algebra $(\frak C,\bar\lambda)$, a measure-preserving Boolean
homomorphism $\pi:\frak A\to\frak C$ and a commuting family
$\ofamily{i}{r}{\tilde\pi_i}$ of measure-preserving automorphisms of
$\frak C$ such that $\tilde\pi_i\pi=\pi\pi_i$ for every $i<r$.
Now $\pi a\in\frak C\setminus\{0\}$, so there is an $\eta>0$ such that

\Centerline{$\eta n
\le\sum_{k=0}^{n-1}\bar\lambda(\inf_{i<r}\tilde\pi_i^k\pi a)
=\sum_{k=0}^{n-1}\bar\lambda(\inf_{i<r}\pi\pi_i^ka)
=\sum_{k=0}^{n-1}\bar\mu(\inf_{i<r}\pi_i^ka)$}

\noindent for every $n\in\Bbb N$.
}%end of proof of 497N

\exercises{\leader{497X}{Basic exercises (a)}
%\spheader 497Xa
Let $(X,\Sigma,\mu)$ be a probability space,
$\familyiI{\Sigma_i}$ an independent family of $\sigma$-subalgebras of
$\Sigma$, and $\Tau$ a subalgebra of $\Sigma$ such that $\Tau\cap\Sigma_i$
is metrically dense in $\Sigma_i$ for every $i\in I$.   For $J\subseteq I$
set $\tilde\Sigma_J=\bigvee_{i\in J}\Sigma_i$.   Show that
$\langle\tilde\Sigma_J\rangle_{J\subseteq I}$ has $\Tau$-removable
intersections.
%497E

\spheader 497Xb Let $I$ be a set, $G_I$ the
group of permutations of $I$ with its topology of pointwise convergence
(441G, 449Xh), and $\action$ the action of $G_I$ on
$\Cal P([I]^{<\omega})$ described in 497F.   Show that $\action$ is
continuous.
%497F

\spheader 497Xc In 497F, show that
$\{\mu:\mu\in P_I$ is permutation-invariant$\}$ is a
closed subset of $P_I$.
%497F

\leader{497Y}{Further exercises (a)}
%\spheader 497Ya
(i) Show that if $A\subseteq\Bbb N$ has non-zero upper
asymptotic density then there is a translation-invariant additive
functional $\nu:\Cal P\Bbb Z\to[0,1]$ such that $\nu A>0$.
(ii) Consider the statement

\inset{($\dagger$) If $\epsilon>0$ and $A\subseteq\Bbb N$ are such that
$\#(A\cap n)\ge\epsilon n$ for every $n$ then $A$ includes arithmetic
progressions of all finite lengths.}

\noindent Use Theorem 497N to prove ($\dagger$).
(iii) Find a direct proof that ($\dagger$) implies Szemer\'edi's
theorem.
}%end of exercises

\endnotes{\Notesheader{497} I am grateful to T.D.Austin for introducing me
to a preprint of \TaoIG, on which this section is based.

Regarded as a proof of Szemer\'edi's theorem, the argument
above has the virtues of reasonable brevity and
(I hope) of completeness and correctness.   It depends, of course, on
non-trivial ideas from measure theory, which for anyone except a measure
theorist will compromise the claim of `brevity';  and even measure
theorists may find that the proofs here demand close attention.
There are further,
more significant, defects.   The outstanding problem associated with
Szemer\'edi's theorem is the estimation of $n_0$ as a function of $r$ and
$\epsilon$;  and while in a theoretical sense it must be possible to trace
through the arguments above to establish rigorous bounds, the methods are
not well adapted to such an exercise, and one would not expect the bounds
obtained to be good.
There is also the point that I have made uninhibited use of the axiom of
choice.   The ultrafilter in 497J can easily be replaced by an appropriate
sequence, but all standard treatments of measure theory assume at least
the countable axiom of choice, and Szemer\'edi's theorem is clearly
true in significantly weaker theories than ordinary ZF.

The first `measure-theoretic' proof of Szemer\'edi's theorem was due to
{\smc Furstenburg 77}, and relied on a deep analysis of the structure of
measure-preserving transformations.   While the methods described here do
not seem to give us any information on this structure, it is apparently a
folklore result that the hypergraph removal lemma provides a quick
proof of the basic theorem used in Furstenburg's approach (497N, 497Ya).

The value of the work here, therefore, lies less in its applications to the
hypergraph removal lemma and Szemer\'edi's theorem, than in the idea of
`removable intersections', where Theorems 497E and 497G give us two
remarkable results, and useful exercises in the theory of relative
independence from \S458.
We also have an instructive example of a more general
phenomenon.   Given a sequence of finite objects with quantititative
aspects, it is often profitable to seek a measure $\mu$
reflecting the asymptotic behaviour of this sequence;  this is the idea of
the construction in 497H.   The `quantitative aspects' here, as
developed in 497J, are the proportion of functions from $L$ to $m_n$ which
are embeddings of $(L,T)$ in $(m_n,T_n)$, and the proportion of simplices
in $[m_n]^k$ which must be removed from $T_n$
in order to destroy all these embeddings.
The measure $\mu$ is set up to describe the limits of
these proportions as measures of appropriate sets.

Returning to the definition 497Aa, most of its clauses
can be expressed in terms of the
measure algebra of the measure $\mu$;  but the final
`$\bigcap_{i\in J}F_i=\emptyset$' has to be taken literally, and makes
sense only in terms of the measure space itself.   In the key application
(part (d) of the proof of 497J), the original sets $E_c$, with negligible
intersection, already belong to the algebra $\Tau$, but the adjustment
to sets $F_c$ with empty intersection is still non-trivial, because of the
requirement that each $F_c$ must belong to the prescribed $\sigma$-algebra
$\Sigma_{J_c}$.

I said in 497F that you could note that $\Cal P([\Bbb N]^{<\omega})$ is
homeomorphic to the Cantor set, so that $P_{\Bbb N}$ is isomorphic
to the space of Radon probability measures on $\{0,1\}^{\Bbb N}$.
However the point of the construction there
is that we are looking at a particular action of the symmetric group
$G_{\Bbb N}$ on $\Cal P([\Bbb N]^{<\omega})$;  and this has very little
to do with the natural actions of $G_{\Bbb N}$ on $\Cal P\Bbb N$ or
$\{0,1\}^{\Bbb N}$, as studied in 459E and 459H, for instance.
In particular, permutation-invariant measures, in the
sense of 497Fb, will not normally be invariant under the much larger group
derived from all permutations of $[\Bbb N]^{<\omega}$ rather than
just those corresponding to permutations of $\Bbb N$.

I express 497N in terms of measure-preserving
automorphisms of probability algebras in order to connect it with the
treatment of ergodic theory in Chapter 38, but you will observe that the
proof presented
immediately shifts to a more traditional formulation in terms of
probability spaces.   This is only one of many
multiple recurrence theorems, some of them much
stronger (and, it seems, deeper) than 497N or, indeed, 497J.
}%end of notes

\discrpage
\frfilename{mt498.tex}
\versiondate{25.3.22}
\copyrightdate{2002}

\def\chaptername{Further topics}
\def\sectionname{Cubes in product spaces}

\newsection{498}

I offer a brief note on a special property of (Radon) product measures.

\leader{498A}{Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra with its measure-algebra topology\cmmnt{ (323A)}.
Suppose that $A\subseteq\frak A$ is an uncountable analytic set.   Then
there is a compact set $L\subseteq A$, homeomorphic to
$\{0,1\}^{\Bbb N}$, such that $\inf L\ne 0$ in $\frak A$.

\proof{ $A\setminus\{0\}$ is still an uncountable analytic subset of
$\frak A$.   By 423K, it has a subset homeomorphic to
$\{0,1\}^{\Bbb N}\cong\Cal P\Bbb N$;  let
$f:\Cal P\Bbb N\to A\setminus\{0\}$ be an injective continuous
function.   Because $(\frak A,\bar\mu)$ is semi-finite,
there is an $a\Bsubseteq f(\emptyset)$ such that $0<\bar\mu a<\infty$;
set $\delta=\bover12\bar\mu a$.   Note that
$(I,J)\mapsto\bar\mu(a\Bcap f(I)\Bsetminus f(J)):
(\Cal P\Bbb N)^2\to\Bbb R$ is continuous.   Choose a sequence
$\sequencen{k_n}$ in $\Bbb N$ inductively, as follows.
Given $\ofamily{i}{n}{k_i}$, set $K_n=\{k_i:i<n\}$.   For each
$J\subseteq K_n$ we have
$\lim_{r\to\infty}\bar\mu(f(J)\Bsetminus f(J\cup\{r\}))=0$,
so there is a $k_n$, greater than $k_i$ for every $i<n$, such that
$\bar\mu(f(J)\Bsetminus f(J\cup\{k_n\})\le 2^{-2n-1}\delta$
for every $J\subseteq K_n$;  continue.

Now

\Centerline{$\bar\mu(a\Bcap\inf_{J\subseteq K_n}f(J))
\ge\delta(1+2^{-n})$}

\noindent for every $n\in\Bbb N$.   \Prf\ Induce on $n$.   If $n=0$,
then $a\Bcap\inf_{J\subseteq K_n}f(J)=a\Bcap f(\emptyset)$ has measure
$2\delta=\delta(1+2^{-0})$.   For the inductive step to
$n+1\ge 1$, observe that

$$\eqalignno{\bar\mu(a\Bcap\inf_{J\subseteq K_{n+1}}f(J))
&=\bar\mu(a\Bcap\inf_{J\subseteq K_n}f(J)\Bcap f(J\cup\{k_n\}))\cr
&\ge\bar\mu(a\Bcap\inf_{J\subseteq K_n}f(J))
 -\sum_{J\subseteq K_n}\bar\mu(f(J)\Bsetminus f(J\cup\{k_n\}))\cr
&\ge\delta(1+2^{-n})-\sum_{J\subseteq K_n}2^{-2n-1}\delta\cr
\displaycause{by the inductive hypothesis and the choice of $k_n$}
&=\delta(1+2^{-n-1}).\cr}$$

\noindent So the induction proceeds.\ \Qed

Set $K=\{k_i:i\in\Bbb N\}$, $c=\inf\{f(J):J\subseteq K$ is finite$\}$.
Then

\Centerline{$\bar\nu(a\Bcap c)
=\inf_{n\in\Bbb N}\bar\mu(a\Bcap\inf_{J\subseteq K_n}f(J))\ge\delta$,}

\noindent and $c\ne 0$.   But now observe that $L=f[\Cal PK]$ is a subset of $A$
homeomorphic to $\Cal PK$ and therefore to $\{0,1\}^{\Bbb N}$.
Also $\{b:b\Bsupseteq c\}$ is closed (323D(d-i)), so
$C=\{J:f(J)\Bsupseteq c\}$ is closed in $\Cal PK$;  as it
includes the dense set $[K]^{<\omega}$, $C=\Cal PK$ and
$\inf L\Bsupseteq c$ is non-zero.
}%end of proof of 498A

\leader{498B}{Proposition}\cmmnt{ (see {\smc Brodski\v{\i} 1949},
{\smc Eggleston 54})}
Let $(X,\frak T,\Sigma,\mu)$ be an atomless Radon measure space,
$(Y,\frak S,\Tau,\nu)$ an effectively locally finite $\tau$-additive
topological measure space
and $\tilde\lambda$ the $\tau$-additive product measure on
$X\times Y$\cmmnt{ (417C, 417F)}.   Then if $W\subseteq X\times Y$ is
closed and $\tilde\lambda W>0$ there are a
non-scattered compact set $K\subseteq X$ and a closed set $F\subseteq Y$ of
positive measure such that $K\times F\subseteq W$.

\proof{{\bf (a)} To begin with (down to the end of (c)), let us suppose
that both $\mu$ and $\nu$ are totally finite.   Let $(\frak B,\bar\nu)$
be the measure algebra of $\nu$.   Writing $\lambda$ for the c.l.d.\ product
measure on $X\times Y$, there is a $W'\supseteq W$ such that
$\lambda W'$ is defined and equal to $\tilde\lambda W$
(apply 417C(b-v) to the complement of $W$).
By 418Tb, there is a $\mu$-conegligible set $X_0$ such
that $W'[\{x\}]\in\Tau$ for every $x\in X_0$,
$B=\{W'[\{x\}]^{\ssbullet}:x\in X_0\}$ is separable for the
measure-algebra topology of $\frak B$, and
$x\mapsto W'[\{x\}]^{\ssbullet}:X_0\to\frak B$ is measurable.   Now
Fubini's theorem, applied in the form of 252D to $\lambda$ and in the
form of 417Ga to $\tilde\lambda$, tells us that

\Centerline{$\int\nu W'[\{x\}]\mu(dx)=\lambda W'=\tilde\lambda W
=\int\nu W[\{x\}]\mu(dx)$.}

\noindent So
$X_1=\{x:x\in X_0$, $W'[\{x\}]^{\ssbullet}=W[\{x\}]^{\ssbullet}\}$ is
$\mu$-conegligible.
Since the topology of $\frak B$ is metrizable (323Ad or 323Gb),
$B$ is separable
and metrizable, and $x\mapsto W[\{x\}]^{\ssbullet}:X_1\to B$ is almost
continuous (418J, applied to the subspace measure on $X_1$).   Let
$K^*\subseteq X_1$ be a compact set of non-zero measure such that
$x\mapsto W[\{x\}]^{\ssbullet}:K^*\to\frak B$ is continuous.

\medskip

{\bf (b)} There is a non-zero $c\in\frak B$ such that
$K_c=\{x:x\in K^*$, $c\Bsubseteq W[\{x\}]^{\ssbullet}\}$ is compact and
not scattered.   \Prf\ Because $x\mapsto W[\{x\}]^{\ssbullet}$ is
continuous on $K^*$, $B^*=\{W[\{x\}]^{\ssbullet}:x\in K^*\}$ is compact
and every $K_c$ is compact.   (i) If $B^*$ is countable, then
$K^*=\bigcup_{b\in B^*}K_b$, so there is some
$c\in B^*$ such that $\mu K_c>0$.   Let $E$ be a non-negligible
self-supporting subset of $K_c$;  then (because $\mu$ is atomless,
therefore zero on singletons) $E$ has no isolated points.   So $K_c$ is
not scattered.   (ii) If $B^*$ is uncountable, then by
498A there is a set $D\subseteq B^*$, homeomorphic to
$\{0,1\}^{\Bbb N}$, with a non-zero lower bound $c$ in $\frak B$.   Now
$\{W[\{x\}]^{\ssbullet}:x\in K_c\}$ includes $D$, so $\{0,1\}^{\Bbb N}$
and therefore $[0,1]$ are continuous images of closed subsets of $K_c$
and $K_c$ is not scattered (4A2G(j-iv)).\ \Qed

\medskip

{\bf (c)} Set $K=K_c$.   Then 414Ac tells us that

\Centerline{$(\bigcap_{x\in K}W[\{x\}])^{\ssbullet}
=\inf_{x\in K}W[\{x\}]^{\ssbullet}\Bsupseteq c$}

\noindent is non-zero, so $F=\bigcap_{x\in K}W[\{x\}]$ is
non-negligible;  while $K\times F\subseteq W$.   Since every
section $W[\{x\}]$ is closed, so is $F$.   So we have found
appropriate sets $K$ and $F$, at least when $\mu$ and $\nu$ are totally
finite.

\medskip

{\bf (d)} For the general case, we need observe only that by 417C(b-iii)
there are $X'\in\Sigma$ and $Y'\in\Tau$, both of finite measure, such
that $\tilde\lambda(W\cap(X'\times Y'))>0$.   Now the subspace measure
$\mu_{X'}$ on $X'$ is atomless and Radon (214Ka, 416Rb), the subspace
measure $\nu_{Y'}$ on $Y'$ is $\tau$-additive (414K), and the
$\tau$-additive product of $\mu_{X'}$ and $\nu_{Y'}$ is the subspace
measure on $X'\times Y'$ induced by $\tilde\lambda$ (417I), while
$W'=W\cap(X'\times Y')$ is relatively closed in $X'\times Y'$.   So we can
apply (a)-(c) to $\mu_{X'}$ and $\nu_{Y'}$ to see that there are a
non-scattered compact set $K\subseteq X'$ and a non-negligible
relatively closed set $F'\subseteq Y'$ such that $K\times F'\subseteq W'$.
Now the closure $F=\overline{F'}$ of $F'$ in $Y$ is closed and
$K\times F=\overline{K\times F'}\subseteq W$.
}%end of proof of 498B

\leader{498C}{Proposition}\cmmnt{ (see
{\smc Ciesielski \& Pawlikowski 03})} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a countable family of
atomless Radon probability spaces, and $\tilde\lambda$ the product Radon
probability measure on $X=\prod_{i\in I}X_i$\cmmnt{ (417Q, 417R)}.
If $W\subseteq X$ and $\tilde\lambda W>0$, there is a family
$\familyiI{K_i}$ such that $K_i\subseteq X_i$ is a non-scattered compact
set for each $i\in I$ and $\prod_{i\in I}K_i\subseteq W$.

\proof{{\bf (a)} To begin with, let us suppose that $I=\Bbb N$.   As
$\tilde\lambda$ is inner regular with respect to the closed sets, it is enough
to deal with the case in which $W$ is closed.   For
each $n\in\Bbb N$, set $Y_n=\prod_{i\ge n}X_i$ and let $\tilde\lambda_n$
be the product Radon probability measure on $Y_n$, so that
$\tilde\lambda_0=\tilde\lambda$ and $\tilde\lambda_n$ can be identified
with the Radon product of $\mu_n$ and $\tilde\lambda_{n+1}$ (417J).   Using
498B repeatedly, we can find non-scattered compact sets
$K_n\subseteq X_n$ and closed non-negligible sets $W_n\subseteq Y_n$
such that $W_0=W$ and $K_n\times W_{n+1}\subseteq W_n$ for
every $n$.   In this case,
$\prod_{i<n}K_i\times W_n\subseteq W_0$ for every $n$.   If
$x\in\prod_{i\in\Bbb N}K_i$, then there is for each $n\in\Bbb N$ an
$x_n\in(\prod_{i<n}K_i)\times W_n$ such that $x_n\restr n=x\restr n$, just
because $W_n$ is not empty.   But now every $x_n$ belongs to $W$ and so does
$x=\lim_{n\to\infty}x_n$.
$x$ is arbitrary, $\prod_{i\in\Bbb N}K_i\subseteq W$.

\medskip

{\bf (b)} For the general case, we may suppose that $I\subseteq\Bbb N$.
For $i\in\Bbb N\setminus I$, take $(X_i,\frak T_i,\Sigma_i,\mu_i)$ to be
$[0,1]$ with Lebesgue measure.   Set
$\tilde W=\{x:x\in\prod_{i\in\Bbb N}X_i$, $x\restr I\in W\}$.   By (a),
there are non-scattered compact sets $K_i\subseteq X_i$ such that
$\prod_{i\in\Bbb N}K_i\subseteq\tilde W$, in which case
$\prod_{i\in I}K_i\subseteq W$, as required.
}%end of proof of 498C

\exercises{\leader{498X}{Basic exercises (a)}
%\spheader 498Xa
Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space,
$(Y,\frak S,\Tau,\nu)$ an effectively locally finite $\tau$-additive
topological measure space, and $\tilde\lambda$ the $\tau$-additive
product measure on
$X\times Y$.   Show that if $W\subseteq X\times Y$ is closed and
$\tilde\lambda W>0$ there are a
compact set $K\subseteq X$ and a closed set $F\subseteq Y$ of positive measure
such that $K\times F\subseteq W$ and $K$ is either non-scattered or
non-negligible.

\spheader 498Xb
Let $(X,\frak T,\Sigma,\mu)$ be an atomless Radon measure space,
$(Y,\Tau,\nu)$ any measure space, and $\lambda$ the c.l.d.\ product
measure on $X\times Y$.   Show that if $W\subseteq X\times Y$ and
$\lambda W>0$ there are a non-scattered
compact set $K\subseteq X$ and a set $F\subseteq Y$ of positive measure
such that $K\times F\subseteq W$.   \Hint{reduce to the case in which
$\nu$ is totally finite and $\Tau$ is countably generated, so that the
completion of $\nu$ is a quasi-Radon measure for an appropriate
second-countable topology.}

\spheader 498Xc Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be any
family of atomless Radon probability spaces, and $\lambda$ the ordinary
product measure on $X=\prod_{i\in I}X_i$.   Show that if $W\subseteq X$
and $\lambda W>0$ then there are non-scattered compact sets
$K_i\subseteq X_i$ for $i\in I$ such that
$\prod_{i\in I}K_i\subseteq W$.
%498C

\spheader 498Xd Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a countable family of
atomless Radon probability spaces, and $W\subseteq\prod_{i\in I}X_i$ a
set with positive measure for the Radon product of $\familyiI{\mu_i}$.
Show that there are atomless Radon probability measures $\nu_i$ on $X_i$
such that $W$ is conegligible for the Radon product of
$\familyiI{\nu_i}$.   \Hint{439Xh(vii).}

\leader{498Y}{Further exercises (a)}
%\spheader 498Ya
Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a family of atomless
perfect probability spaces, and $\lambda$ the ordinary product measure
on $X=\prod_{i\in I}X_i$.   Show that if $W\subseteq X$ and
$\lambda W>0$ then there are sets
$K_i\subseteq X_i$ for $i\in I$, all with cardinal $\frak c$, such that
$\prod_{i\in I}K_i\subseteq W$.
%498C
}%end of exercises

\endnotes{
\Notesheader{498} I have previously noted (325Yd) that a set $W$ of
positive measure in a product space need not include the product of two
sets of positive measure;  this fact is also the basis of 419E.   Here,
however, we see that if one of the factors is a Radon measure space then
$W$ does include the product of a
non-trivial compact set and a set of positive measure.   There are many
possible variations on the result, corresponding to different product
measures (498B, 498Xb) and different notions of `non-trivial' (498Xa,
498Ya).
The most important of the latter seems to be the idea of a
`non-scattered' compact set $K$;  this is a quick way of saying that
$[0,1]$ is a continuous image of $K$, which is a little stronger than
saying that $\#(K)\ge\frak c$,
and arises naturally from the proof of 498B.
}%end of notes

\discrpage

\leaveitout{
\frfilename{mt49conc.tex}
\versiondate{10.8.08}
\copyrightdate{2008}
     
\def\chaptername{Further topics}
\def\sectionname{Concordance}
     
\centerline{\bf Concordance for Chapter 49}
     
\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.

\leaveitout{Farah 06:  $\frak B_{\frakc}$ is regularly embedded in
$\frak Z=\Cal P\Bbb N/\Cal Z$ (4{}91P) %491P
}

\bigskip
     
\noindent{\bf Reference}
     
Farah I.\ [06] `Analytic Hausdorff gaps II: the density zero ideal', 
Israel J.\ Math.\ 154 (2006) 235-246.


\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',
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\cmmnt{[449O.]}

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
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Bellow A.\ {\it see} Ionescu Tulcea A.

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Springer, 1976 (Lecture Notes in Mathematics 541).
%4r

Bergelson V.\ [85] `Sets of recurrence of $\Bbb Z^m$-actions and properties of
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%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
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%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,
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\cmmnt{[\S4A3 {\it notes}.]}
%second edition ISBN 0-471-19745-9

Blackwell D.\ [56] `On a class of probability spaces',
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\cmmnt{[419K, 452P.]}

Bochner S.\ [1933] `Monotone Funktionen, Stieltjessche Integrale, und
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\cmmnt{[445N.]}

\indexheader{Bogachev}
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\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
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\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,
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\cmmnt{[4A2J.]}
%Store A QA 37.B6 no 7792 0070 5

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\cmmnt{[\S436 {\it notes.}]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
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\indexheader{Bourgain}
Bourgain J., Fremlin D.H.\ \& Talagrand M.\ [78] `Pointwise compact
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\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.]}

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\cmmnt{[449D.]}

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\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]
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\cmmnt{[494F.]}

\indexheader{Choquet}
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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

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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}
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\cmmnt{[439O.]}

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\cmmnt{[466I.]}

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\cmmnt{[459Xd.] }

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\cmmnt{[494Q.]}

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\cmmnt{[\S4A4.]}

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%used for \S478

\medskip%E

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Quarterly J.\ Math.\ (2) 5 (1954) 108-115.
\cmmnt{[498B.]}

Emerson W.\ \& Greenleaf F.\ [67] `Covering properties and F{\o}lner
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\cmmnt{[449J.]}

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\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.}
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\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}
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\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
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\cmmnt{[4A4N.]}

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\cmmnt{[497L.]}

\indexheader{Fremlin}
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\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
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\cmmnt{[463K.]}

Fremlin D.H.\ [75b] `Topological measure spaces:  two counter-examples',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 78 (1975) 95-106.
\cmmnt{[419C, 419D.]}

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Canadian Math.\ Bull.\ 19 (1976) 285-289.
\cmmnt{[419E.]}

Fremlin D.H.\ [81] `Measurable functions and almost continuous
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\cmmnt{[451T.]}

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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
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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.\
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\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
mixing', Annals of Math.\ 45 (1944) 786-792.
\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
U.P., 1991.
\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
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\cmmnt{[463C.]}

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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
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\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
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(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.\
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\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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\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',
Math.\ Annalen 190 (1971) 233-241.
\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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\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}}
Luke\v{s} J., Mal\'y J.\ \& Zaj\'\i\v{c}ek L.\ [86]
{\it Fine topology methods
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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

Mauldin R.D. \& Stone A.H.\ [81] `Realization of maps', Michigan Math.\
J.\ 28 (1981) 369-374.
\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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Monthly 80 (1973) 349-359.
\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
Groups.}   Interscience, 1955.
\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.\
London Math.\ Soc.\ (3) 19 (1969) 493-508.
\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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%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{[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{[484B.]}

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\indexheader{Plewik}
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\indexheader{Pontryagin}
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\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

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\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
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\cmmnt{[419F.]}

Rao B.V. \& Srivastava S.M.\ [94] `Borel isomorphism theorem',
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\cmmnt{[424C.]}

\indexheader{Rao M.}
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{\tt http://www1.essex.ac.uk/maths/people{\bsp}fremlin{\bsp}rao.htm}.
%used for \S477

\indexheader{Ressel}
Ressel P.\ [77] `Some continuity and measurability results on spaces of
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\cmmnt{[417C, 437M.]}

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\cmmnt{[\S494 {\it notes\/}.]}

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\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
and Martingales,} vol.\ I.   Wiley, 1994.
\cmmnt{[\S455 {\it notes\/}.]}
%HA33.4.W5

Rokhlin V.A.\ [1948] `A ``general'' measure-preserving transformation is not
mixing', Doklady Akademii Nauk SSSR 60 (1948) 349-351.
\cmmnt{[494E.]}

Rosendal C.\ [09]
`The generic isometry and measure preserving homeomorphism are conjugate
to their powers',  Fundamenta Math.\ 205 (2009) 1-27.
%({\tt http://www.math.uic.edu{\bsp}$\sptilde$rosendal/}).
\cmmnt{[494Ye.]}

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fixed points on metric compacta', Israel J.\ Math.\ 162 (2007) 349-371.
\cmmnt{[494Z, \S494 {\it notes}.]}

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normal', Fund.\ Math.\ 73 (1971) 179-186.
\cmmnt{[439O.]}

Rudin M.E.\ [84] `Dowker spaces', pp.\ 761-780 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[439O.]}

Rudin W.\ [67] {\it Fourier Analysis on Groups.} Wiley, 1967 and 1990.
\cmmnt{[\S445 {\it intro\/}.]}

Rudin W.\ [91] {\it Functional Analysis.}  McGraw-Hill, 1991.
\cmmnt{[\S4A4, \S4A6.]}

Ryll-Nardzewski C.\ [53] `On quasi-compact measures', Fundamenta Math.\
40 (1953) 125-130.
\cmmnt{[451C.]}

\medskip%S

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\indexheader{Sierpi\'nski}
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\indexheader{Szemer\'edi}
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\cmmnt{[497L.]}
\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

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\cmmnt{[463Ya, 463Yd, 463Za, \S465 {\it intro.}, 465B, 465N,
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466I, \S466 {\it notes\/}.]}

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%465Xm

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\cmmnt{[\S4A4.]}

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%used in chap 41

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\cmmnt{[425D, 425Ya.]}

\medskip%U

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\medskip%V

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% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

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\cmmnt{[\S434 {\it notes\/}.]}

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\indexheader{Wheeler}
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  % 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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