% 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{mt4.tex}
\versiondate{14.12.06}
\copyrightdate{2001}

\def\volumename{Topological Measure Spaces}

\newvolume{4}

I return in this volume to the study of measure {\it spaces} rather than
measure {\it algebras}.   For fifty years now measure theory has been
intimately connected with general topology.   Not only do a very large
proportion of the measure spaces arising in applications carry
topologies related in interesting ways to their measures, but many
questions in abstract measure theory can be effectively studied by
introducing suitable topologies.   Consequently any course in measure
theory at this level must be frankly dependent on a substantial
knowledge of topology.   With this proviso, I hope that the present
volume will be accessible to graduate students, and will lead them to
the most important ideas of modern abstract measure theory.

The first and third chapters of the volume seek to provide a thorough
introduction into the ways in which topologies and measures can
interact.   They are divided by a short chapter on descriptive set
theory, on the borderline between set theory, logic, real analysis and
general topology, which I single out for detailed exposition because I
believe that it forms an indispensable part of the background of any
measure theorist.   Chapter 41 is dominated by the concepts of inner
regularity and $\tau$-additivity, coming together in Radon measures
(\S416).   Chapter 43 concentrates rather on questions concerning
properties of a
topological space which force particular relationships with measures on
that space.   But plenty of side-issues are treated in both, such as
Lusin measurability (\S418), the definition of measures from linear
functionals (\S436) and
measure-free cardinals (\S438).   Chapters 45 and 46 continue some of
the same themes, with particular investigations into `disintegrations'
or regular conditional probabilities (\S\S452-453),
stochastic processes (\S\S454-456), %\S454 \S455 \S456
Talagrand's theory of
stable sets (\S465) and the theory of measures on normed spaces (\S\S466-467).

In contrast with the relatively amorphous structure of Chapters 41, 43,
45 and 46, four chapters of this volume have definite topics.
I have already said that Chapter 42 is an introduction to descriptive
set theory;  like Chapters 31 and 35 in the preceding volume, it is a
kind of
appendix brought into the main stream of the argument.   Chapter 44
deals with topological groups.   Most of it is of course devoted to Haar
measure, giving the Pontryagin-van Kampen duality theorem (\S445) and
the Ionescu Tulcea theorem on the existence of translation-invariant
liftings (\S447).   But there are also sections on Polish groups
(\S448) and amenable groups (\S449), and some of the general theory of
measures on measurable groups
(\S444).   Chapter 47 is a second excursion, after Chapter 26, into
geometric measure theory.   It starts with Hausdorff measures (\S471),
gives a proof of the Di Giorgio-Federer
Divergence Theorem (\S475), and then examines a number of examples of
`concentration of measure' (\S476).   In the second half of the chapter,
\S\S477-479, I decribe Brownian motion and use it as a basis of
the theory of Newtonian capacity.
In Chapter 48, I set out the elementary theory of gauge integrals, with
sections on the Henstock and Pfeffer integrals (\S\S483-484).
Finally, in Chapter 49, I give notes on seven special topics:
equidistributed sequences (\S491),
combinatorial forms of concentration of measure (\S492),
extremely amenable groups and groups of measure-preserving automorphisms
(\S\S493-494),
Poisson point processes (\S495),
submeasures (\S496), Szemer\'edi's theorem (\S497) and
subproducts in product spaces (\S498).

I had better mention prerequisites, as usual.   To embark on this
material you will certainly need a solid foundation in measure theory.
Since I do of course use my own exposition as my principal source of
references to the elementary ideas, I advise readers to ensure that they
have easy access to all three previous volumes before starting serious
work on this one.   But you may not need to read very much of them.   It
might be prudent to glance through the detailed contents of Volume 1 and
the first five chapters of Volume 2 to
check that most of the material there is more or less familiar.   I think
\S417 might be difficult to read without at least the results-only version
of Chapter 25 to hand.   But
Volume 3, and the last three chapters of Volume 2, can probably be left
on one side for the moment.   Of course you will need the
Lifting Theorem (Chapter 34) for \S\S447, 452 and 453, and Chapter
26 is essential background for Chapter 47, while Chapter 28 (on Fourier
analysis) may help to make sense of Chapter 44, and parts of Chapter 27
(on probability theory) are necessary for \S\S455-456 and 458-459.
You will certainly need some Fourier analysis for \S479.
And measure algebras
are mentioned in every chapter except (I think) Chapter 48;  but I hope
that the cross-references are precise enough to lead you to what
you need to know at any particular point.   Even Maharam's theorem is
hardly used in this volume.

What you will need, apart from any knowledge of measure theory, is a
sound background in general topology.   This volume calls on a great
many miscellaneous facts from general topology, and the list in \S4A2 is
not a good place to start if continuity and compactness and the
separation axioms are unfamiliar.   My primary reference for topology is
{\smc Engelking 89}.   I do not insist that you should have read this
book (though of course I hope you will do so sometime);  but I do think
you should make sure that you can use it.

In the general introduction to this treatise, I wrote `I make no attempt
to describe the history of the subject', and I have generally been
casual -- some would say negligent -- in my attributions of results to
their discoverers.   Through much of the first three volumes I did at
least have the excuse that the history exists in print in far
more detail than I am qualified to describe.   In the present volume I
find my position more uncomfortable, in that I have been watching
the evolution of the subject relatively closely over the last forty
years, and ought to be able to say something about it.   Nevertheless I
remain reluctant to make definite
statements crediting one person rather than another with originating an
idea.   My more intimate knowledge of the topic makes me even more
conscious than elsewhere of the danger of error and of the breadth of
reading that would be necessary to produce a balanced account.   In some
cases I do attach a result to a specific published paper, but these
attributions should never be regarded as an assertion that any
particular author has priority;  at most, they declare that a historian
should examine the source cited before coming to any decision.   I
assure my friends and colleagues that my omissions are not intended to
slight either them or those we all honour.   What I have tried to do is
to include in
the bibliography to this volume all the published work which (as far as
I am consciously aware) has influenced me while writing it, so that
those who wish to go into the matter will have somewhere to start their
investigations.

\bigskip

\noindent{\bf Note on second printing}

\medskip

I fear that there were even more errors, not all of them trivial,
in the first printing of this
volume than there were in previous volumes.   I have tried to correct
those which I have noticed;  many surely remain.   Apart from these, there
are many minor expansions and elaborations, and a couple of new results,
but few new ideas and no dramatic rearrangements.
Details may be found in
{\tt http://www1.essex.ac.uk/maths/people/fremlin\penalty-100/mterr4.03.pdf}.

Both printers and readers found that the 945-page format of the first
printing was hard to handle.   I have therefore divided the volume into two
parts for the second printing.   I hope you will find that the additional
convenience is worth the the increase in cost.

\bigskip

\noindent{\bf Note on second (`Lulu') edition}

\medskip

I was right about many errors remaining (particularly in \S458, on relative
independence), and I hope I have cleared some of
them out of the way.   There are substantial additions in the new edition,
the most important being a vastly expanded \S455 on L\'evy processes, an
account of Brownian motion and Newtonian potential in \S\S477-479, and
Tao's proof of Szemer\'edi's theorem in \S497.   I have included
theorems of A.T\"ornquist and G.W.Mackey on the realization of group
actions on measure algebras, some material on a version of the
Kantorovich-Rubinstein distance between two measures, and a section on
Maharam submeasures (\S496).

\frnewpage

\frfilename{mt41.tex} 
\versiondate{17.4.10} 
\copyrightdate{1997} 
      
\def\chaptername{Topologies and Measures I} 
\newchapter{41} 
\def\chaptername{Topologies and measures I} 
      
I begin this volume with an introduction to some of the most important 
ways in which topologies and measures can interact, and with a 
description of 
the forms which such constructions as subspaces and product spaces take 
in such contexts.   By far the most important concept is that of Radon 
measure (411Hb, \S416).   In Radon measure spaces we find both the 
richest combinations of ideas and the most important applications. 
But, as usual, we are led both by analysis of these ideas and by other 
interesting examples to consider wider classes of topological measure 
space, and the 
greater part of the chapter, by volume, is taken up by a description of 
the many properties of Radon measures individually and in partial 
combinations. 
      
I begin the chapter with a short section of definitions (\S411), 
including a handful of more or less elementary examples.   The two 
central properties 
of a Radon measure are `inner regularity' (411B) and `$\tau$-additivity' 
(411C).   The former is an idea of great versatility which I look at in 
an abstract setting in \S412.   I take a section (\S413) to describe 
some methods of constructing measure spaces, extending the rather 
limited range of constructions offered in earlier volumes.   There are 
two sections on $\tau$-additive measures, \S\S414 and 417;  the former 
covers the elementary ideas, and the latter looks at product measures, 
where it turns out that we need a new technique to supplement the purely 
measure-theoretic constructions of Chapter 25.   On the way to Radon 
measures in \S416, I 
pause over `quasi-Radon' measures (411Ha, \S415), where inner regularity 
and $\tau$-additivity first come effectively together. 
      
The possible interactions of a topology and a measure on the same space 
are so varied that even a brief account makes a long chapter;  and this 
is with hardly any mention of results associated with particular types 
of topological space, most of which must wait for later chapters.   But 
I include one section on the two most important classes of functions 
acting between topological measure spaces (\S418), and another 
describing some examples to demonstrate special phenomena (\S419). 

\discrpage 
\frfilename{mt411.tex}
\versiondate{31.12.08}
\copyrightdate{2002}

\def\chaptername{Topologies and measures I}
\def\sectionname{Definitions}

\newsection{411}

In something of the spirit of \S211, but this time without apologising,
I start this volume with a list of definitions.   The rest of Chapter 41
will be devoted to discussing these definitions and relationships
between them, and integrating the new ideas into the concepts and
constructions
of earlier volumes;  I hope that by presenting the terminology now
I can give you a sense of the directions the following sections will
take.   I ought to remark immediately that there are many cases in which
the exact phrasing of a definition is important in ways which may
not be immediately apparent.

\leader{411A}{}\cmmnt{ I begin with a phrase which will be a useful
shorthand for the context in which most, but not all, of the theory
here will be developed.

\medskip

\noindent}{\bf Definition} A {\bf topological measure space} is a
quadruple $(X,\frak T,\Sigma,\mu)$ where $(X,\Sigma,\mu)$ is a measure
space and $\frak T$ is a topology on $X$ such that
$\frak T\subseteq\Sigma$\cmmnt{, that is, every open set (and
therefore every Borel set) is measurable}.

\leader{411B}{}\cmmnt{ Now I come to what are in my view the two most
important concepts to master;  jointly they will dominate the chapter.

\medskip

\noindent}{\bf Definition} Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K$ a family of sets.   I say that $\mu$ is {\bf inner
regular with respect to $\Cal K$} if

\Centerline{$\mu E=\sup\{\mu K:K\in\Sigma\cap\Cal K,\,K\subseteq E\}$}

\noindent for every $E\in\Sigma$.   \cmmnt{(Cf.\ 256Ac, 342Aa.)}

\cmmnt{\medskip

\noindent{\bf Remark} Note that in this definition I do not assume that
$\Cal K\subseteq\Sigma$, nor even that $\Cal K\subseteq\Cal PX$.   But
of course $\mu$ will be inner regular with
respect to $\Cal K$ iff it is inner regular with respect to
$\Cal K\cap\Sigma$.

It is convenient in this context to interpret $\sup\emptyset$ as $0$, so
that we have to check the definition only when $\mu E>0$, and need not
insist that $\emptyset\in\Cal K$.
}%end of comment

\leader{411C}{Definition} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$.   I say that $\mu$ is
{\bf $\tau$-additive} (the phrase {\bf $\tau$-regular} has also been used)
if whenever $\Cal G$ is a non-empty upwards-directed family of open sets
such that $\Cal G\subseteq\Sigma$ and $\bigcup\Cal G\in\Sigma$ then
$\mu(\bigcup\Cal G)=\sup_{G\in\Cal G}\mu G$.

\cmmnt{\medskip

\noindent{\bf Remark} Note that in this definition I do not assume that
every open set is measurable.   Consequently we cannot take it for
granted that an extension of a $\tau$-additive measure will be
$\tau$-additive;  on
the other hand, the restriction of a $\tau$-additive measure to any
$\sigma$-subalgebra will be $\tau$-additive.

}%end of comment

\leader{411D}{}\cmmnt{ Complementary to 411B we have the following.

\medskip

\noindent}{\bf Definition} Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal H$ a family of subsets of $X$.   Then $\mu$ is
{\bf outer regular with respect to $\Cal H$} if

\Centerline{$\mu E=\inf\{\mu H:H\in\Sigma\cap\Cal H,\,H\supseteq E\}$}

\noindent for every $E\in\Sigma$.

Note that a totally finite measure on a topological space is
inner regular with respect to the family of
closed sets iff it is outer regular with
respect to the family of open sets.

\leader{411E}{}\cmmnt{ I delay discussion of most of the relationships
between the concepts here to later in the chapter.   But it will be
useful to have a basic fact set out immediately.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$.   If $\mu$ is inner regular with respect to
the compact sets, it is $\tau$-additive.

\proof{ Let $\Cal G$ be a non-empty upwards-directed family of
measurable open sets such that $H=\bigcup\Cal G\in\Sigma$.   If
$\gamma<\mu H$, there is a compact set $K\subseteq H$ such that
$\mu K\ge\gamma$;  now there must be a $G\in\Cal G$ which includes $K$,
so that $\mu G\ge\gamma$.  As $\gamma$ is arbitrary,
$\sup_{G\in\Cal G}\mu G=\mu H$.
}%end of proof of 411E

\leader{411F}{}\cmmnt{ In order to deal efficiently with measures
which are not totally finite, I think we need the following ideas.

\medskip

\noindent}{\bf Definitions} Let $(X,\Sigma,\mu)$ be a measure space
and $\frak T$ a topology on $X$.

\spheader 411Fa I say that $\mu$ is {\bf locally finite} if every point
of $X$ has a neighbourhood of finite measure\cmmnt{, that is, if the
open sets of finite outer measure cover $X$}.

\spheader 411Fb I say that $\mu$ is {\bf effectively locally finite} if
for every non-negligible measurable set $E\subseteq X$ there is
a measurable open set $G\subseteq X$ such that $\mu G<\infty$ and
$E\cap G$ is not negligible.

\cmmnt{Note that an effectively locally finite measure
must measure many open sets, while a locally finite measure need not.
}%end of comment

\spheader 411Fc \cmmnt{This seems a convenient moment at which to
introduce the following term.}   A real-valued function $f$ defined on a
subset of $X$ is {\bf locally integrable} if for every $x\in X$ there
is an open set $G$ containing $x$ such that $\int_Gf$ is
defined\cmmnt{ (in the sense of 214D)} and finite.

\leader{411G}{Elementary facts (a)}
%\spheader 411Ga
If $\mu$ is a locally finite measure on a topological space $X$, then
$\mu^*K<\infty$ for every compact set $K\subseteq X$.   \prooflet{\Prf\
The family $\Cal G$ of open sets of finite outer measure is
upwards-directed and covers $X$,
so there must be some $G\in\Cal G$ including
$K$, in which case $\mu^*K\le\mu^*G$ is finite.\ \Qed}

\spheader 411Gb A measure $\mu$ on $\BbbR^r$ is locally finite
iff every bounded set has finite outer measure\cmmnt{ (cf.\ 256Ab)}.
\prooflet{\Prf\ (i) If every bounded set has finite outer measure then,
in particular, every open ball has finite outer measure, so that $\mu$
is locally finite.   (ii) If $\mu$ is locally finite and
$A\subseteq\Bbb R^r$ is bounded, then its closure $\overline{A}$ is
compact (2A2F), so
that $\mu^*A\le\mu^*\overline{A}$ is finite, by (a) above.\ \Qed}

\cmmnt{\spheader 411Gc I should perhaps remark immediately that a
locally finite topological measure need not be effectively locally
finite (419A), and an effectively locally finite measure need not be
locally finite (411P).
}%end of comment

\spheader 411Gd An effectively locally finite measure must be
semi-finite.

\spheader 411Ge A locally finite measure on a Lindel\"of space
$X$ is $\sigma$-finite.
\prooflet{\Prf\ Let $\Cal G$ be the family of open sets of finite
outer measure.   Because $\mu$ is locally finite, $\Cal G$ is a cover
of $X$.   Because $X$
is Lindel\"of, there is a sequence $\sequencen{G_n}$ in $\Cal G$
covering $X$.   For each $n\in\Bbb N$, there is a measurable set
$E_n\supseteq G_n$
of finite measure, and now $\sequencen{E_n}$ is a sequence of sets of
finite measure covering $X$.\ \Qed}

\spheader 411Gf Let $(X,\frak T,\Sigma,\mu)$ be a topological measure
space such that $\mu$ is locally
finite and inner regular with respect to the compact sets.   Then
$\mu$ is effectively locally finite.   \prooflet{\Prf\ Suppose that
$\mu E>0$.   Then there is a measurable compact set $K\subseteq E$ such
that $\mu K>0$.   As in the argument for (a) above, there is an open set
$G$ of finite measure including $K$, so that $\mu(E\cap G)>0$.\ \Qed}

\spheader 411Gg \cmmnt{Corresponding to (a) above, we have the
following fact.}   If $\mu$ is a measure on a topological space and
$f\in\eusm L^0(\mu)$ is locally
integrable, then $\int_Kfd\mu$ is finite for every compact
$K\subseteq X$\prooflet{, because $K$ can be covered by a finite family
of open sets $G$ such that $\int_G|f|d\mu<\infty$}.

\spheader 411Gh If $\mu$ is a locally finite measure on
a topological space $X$, and $f\in\eusm L^p(\mu)$ for some
$p\in[1,\infty]$, then $f$ is locally integrable\cmmnt{;  this is
because $\int_G|f|\le\int_E|f|\le\|f\|_p\|\chi E\|_q$ is finite whenever
$G\subseteq E$ and $\mu E<\infty$, where $\bover1p+\bover1q=1$, by
H\"older's inequality (244Eb)}.

\spheader 411Gi If $(X,\frak T)$ is a completely regular space and $\mu$
is a locally finite topological measure on $X$, then the collection of open
sets with negligible boundaries is a base for $\frak T$.
\prooflet{\Prf\ If $x\in G\in\frak T$, let $H\subseteq G$ be an open set
of finite measure containing $x$, and $f:X\to[0,1]$ a continuous
function such that $f(x)=1$ and $f(y)=0$ for $y\in X\setminus H$.   Then
$\{f^{-1}[\{\alpha\}]:0<\alpha<1\}$ is an uncountable disjoint family of
measurable subsets of $H$, so there must be some $\alpha\in\ooint{0,1}$
such that $f^{-1}[\{\alpha\}]$ is negligible.   Set
$U=\{y:f(y)>\alpha\}$;  then $U$ is an open neighbourhood of $x$
included in $G$ and its boundary
$\partial U\subseteq f^{-1}[\{\alpha\}]$ is negligible.\ \Qed}

\spheader 411Gj\dvAnew{2021} Let $X$ and $Y$ be topological spaces, 
$f:X\to Y$ a continuous function, $\mu$ a measure on $X$ and $\mu f^{-1}$ the
image measure on $Y$.   Then if $\mu$ is a topological measure, 
so is $\mu f^{-1}$, and if $\mu$ is $\tau$-additive, so is $\mu f^{-1}$.
\prooflet{(Immediate from the definitions.)}

\leader{411H}{}\cmmnt{ Two particularly important combinations of the
properties above are the following.

\medskip

\noindent}{\bf Definitions (a)} A {\bf quasi-Radon measure space} is a
topological measure space $(X,\frak T,\Sigma,\mu)$ such that (i)
$(X,\Sigma,\mu)$ is complete and locally determined (ii) $\mu$ is
$\tau$-additive, inner regular with respect to the closed sets and
effectively locally finite.

\spheader 411Hb  A {\bf Radon measure space} is a
topological measure space $(X,\frak T,\Sigma,\mu)$ such that (i)
$(X,\Sigma,\mu)$ is
complete and locally determined (ii) $\frak T$ is Hausdorff (iii) $\mu$
is locally finite and inner regular with respect to the compact sets.

\leader{411I}{Remarks}\cmmnt{ {\bf (a)} You may like to seek your own
proof that a Radon measure space is always quasi-Radon, before looking
it up in \S416 below.

\medskip

{\bf (b)}} Note that a measure on Euclidean space $\BbbR^r$ is a
Radon measure on the definition above iff it is a Radon measure as
described in 256Ad.   \prooflet{\Prf\ In 256Ad, I said that a measure
$\mu$ on $\BbbR^r$ is `Radon' if it is a locally finite complete
topological
measure, inner regular with respect to the compact sets.   (The
definition of `locally
finite' in 256A was not the same as the one above, but I have already
covered this point in 411Gb.)   So the only thing to add is that $\mu$
is necessarily locally determined, because it is $\sigma$-finite
(256Ba).\ \Qed
}%end of prooflet

\leader{411J}{}\cmmnt{ The following special types of inner regularity
are of sufficient importance to have earned separate names.

\medskip

\noindent}{\bf Definitions (a)} If $(X,\frak T)$ is a topological space,
I will say that a measure $\mu$ on $X$ is {\bf tight} if it is inner
regular with respect to the closed compact sets.

\spheader 411Jb If $(X,\frak T,\Sigma,\mu)$ is a topological
measure space, I will say that $\mu$ is {\bf completion regular} if it
is inner regular with respect to the zero sets\cmmnt{ (definition:
3A3Qa)}.

\leader{411K}{Borel and Baire measures}
If $(X,\frak T)$ is a topological space, I will call a measure with
domain\cmmnt{ (exactly)} the Borel $\sigma$-algebra of
$X$\cmmnt{ (4A3A)} a {\bf Borel measure}
on $X$, and a measure with domain\cmmnt{ (exactly)} the Baire
$\sigma$-algebra of $X$\cmmnt{ (4A3K)} a {\bf Baire measure} on $X$.

\cmmnt{Of course a Borel measure is a topological measure in the
sense of 411A.   On a metrizable space,
the Borel and Baire measures coincide (4A3Kb).
The most important measures in this chapter will be c.l.d.\ versions
of Borel measures.}

\leader{411L}{}\cmmnt{ When we come to look at functions defined on
a topological measure space, we shall have to relate ideas of continuity
and measurability.   Two basic concepts are the following.

\medskip

\noindent}{\bf Definition} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra
of subsets of $X$ and $(Y,\frak S)$ a topological space.   I will say
that a function $f:X\to Y$ is {\bf measurable} if $f^{-1}[G]\in\Sigma$
for every open set $G\subseteq Y$.

\cmmnt{\medskip

\noindent{\bf Remarks (a)} Note that a function $f:X\to\Bbb R$ is
measurable on this definition (when $\Bbb R$ is given its usual
topology) iff it is measurable according to the familiar definition in
121C, which
asks only that sets of the form $\{x:f(x)<\alpha\}$ should be measurable
(121Ef).

\medskip

{\bf (b)} For any topological space $(Y,\frak S)$, a function $f:X\to Y$
is measurable iff $f$ is $(\Sigma,\Cal B(Y))$-measurable,
where $\Cal B(Y)$ is the Borel $\sigma$-algebra of $Y$ (4A3Cb).
}%end of comment

\leader{411M}{Definition} Let $(X,\Sigma,\mu)$ be a measure space,
$\frak T$ a topology on $X$, and $(Y,\frak S)$ another topological
space.   I will say that a function $f:X\to Y$ is
{\bf almost continuous}\cmmnt{ or {\bf Lusin measurable}} if $\mu$
is inner regular with respect to the
family of subsets $A$ of $X$ such that $f\restr A$ is continuous.

\leader{411N}{}\cmmnt{ Finally, I introduce some terminology to
describe ways in which (sometimes) measures can be located in one part
of a topological space rather than another.

\medskip

\noindent}{\bf Definitions} Let $(X,\Sigma,\mu)$ be a measure space
and $\frak T$ a topology on $X$.

\spheader 411Na I will call a set $A\subseteq X$ {\bf self-supporting}
if $\mu^*(A\cap G)>0$ for every open set $G$ such that $A\cap G$ is
non-empty.   \cmmnt{(Such sets are sometimes called {\bf of positive
measure everywhere}.)}

\spheader 411Nb A {\bf support} of $\mu$ is a closed self-supporting
set $F$ such that $X\setminus F$ is negligible.

\spheader 411Nc\cmmnt{ Note that} $\mu$ can have at most one support.
\prooflet{\Prf\ If $F_1$, $F_2$ are supports then
$\mu^*(F_1\setminus F_2)\le\mu^*(X\setminus F_2)=0$ so $F_1\setminus F_2$
must be empty.
Similarly, $F_2\setminus F_1=\emptyset$, so $F_1=F_2$.\ \Qed}

\spheader 411Nd If $\mu$ is a $\tau$-additive topological measure it
has a support.   \prooflet{\Prf\ Let $\Cal G$ be the family
of negligible open sets, and $F$ the closed set
$X\setminus\bigcup\Cal G$.   Then $\Cal G$ is an upwards-directed family
in $\frak T\cap\Sigma$ and $\bigcup\Cal G\in\frak T\cap\Sigma$, so

\Centerline{$\mu(X\setminus F)=\mu(\bigcup\Cal G)
=\sup_{G\in\Cal G}\mu G=0$.}

\noindent If $G$ is open and $G\cap F\ne\emptyset$ then $G\notin\Cal G$
so $\mu^*(G\cap F)=\mu(G\cap F)=\mu G>0$;  thus $F$ is self-supporting
and is the support of $\mu$.\ \Qed}

\spheader 411Ne Let $X$ and $Y$ be topological spaces with topological
measures $\mu$, $\nu$ respectively and a continuous \imp\ function
$f:X\to Y$.   Suppose that $\mu$ has a support $E$.   Then
$\overline{f[E]}$ is the support of $\nu$.   \prooflet{\Prf\ We have
only to observe that for an open set $H\subseteq Y$

$$\eqalign{\nu H>0
&\iff\mu f^{-1}[H]>0
\iff f^{-1}[H]\cap E\ne\emptyset\cr
&\iff H\cap f[E]\ne\emptyset
\iff H\cap\overline{f[E]}\ne\emptyset. \text{ \Qed}\cr}$$}

\spheader 411Nf $\mu$ is {\bf strictly positive}\cmmnt{ (with
respect
to $\frak T$)} if $\mu^*G>0$ for every non-empty open set
$G\subseteq X$\cmmnt{, that is, $X$ itself is the support of $\mu$}.

\header{*411Ng}{\bf *(g)} If $(X,\frak T)$ is a topological space, and
$\mu$ is a strictly positive $\sigma$-finite measure on $X$ such that
the domain\cmmnt{ $\Sigma$} of $\mu$ includes a
$\pi$-base\cmmnt{ $\Cal U$} for $\frak T$, then $X$ is ccc.
\prooflet{\Prf\ Let $\sequencen{E_n}$
be a sequence of sets of finite measure covering $X$.   Let $\Cal G$ be
a disjoint family of non-empty open sets.   For each $G\in G$, take
$U_G\in\Cal U\setminus\{\emptyset\}$ such that $U_G\subseteq G$;  then
$\mu U_G>0$, so there is an $n(G)$ such that $\mu(E_{n(G)}\cap U_G)>0$.
Now $\sum_{G\in G,n(G)=k}\mu(E_k\cap U_G)\le\mu E_k$ is finite for every
$k$, so $\{G:n(G)=k\}$ must be countable and $\Cal G$ is countable.\
\Qed}

\leader{411O}{Example} Lebesgue measure on $\BbbR^r$ is a Radon
measure\cmmnt{ (256Ha)};  \cmmnt{in particular,} it is locally
finite and tight.   It is\cmmnt{ therefore} $\tau$-additive and
effectively locally finite\cmmnt{ (411E, 411Gf)}.   It is completion
regular\cmmnt{ (because every compact set is a zero set, see
4A2Lc)}, outer regular with respect to the open
sets\cmmnt{ (134Fa)} and strictly positive.

\leader{411P}{Example:  Stone spaces (a)} Let $(Z,\frak T,\Sigma,\mu)$
be the Stone space of a semi-finite measure algebra
$(\frak A,\bar\mu)$, so that $(Z,\frak T)$ is a
zero-dimensional compact Hausdorff space, $(Z,\Sigma,\mu)$ is complete
and semi-finite, the open-and-closed sets are measurable, the negligible
sets are the nowhere dense sets, and every measurable set differs by a
nowhere dense set from an open-and-closed set\cmmnt{ (311I, 321K,
322Bd, 322Ra\formerly{3{}22Qa})}.
%311I:  X compact 0-dim
%321K:  measurable iff (open&closed\symmdiff meager),
%       negligible iff meager, \mu complete
%322Bd:  semi-finite
%322Ra:  meager iff nwd

\spheader 411Pb $\mu$ is inner regular with respect to the
open-and-closed sets\cmmnt{ (322Ra)};  \cmmnt{in particular,} it
is completion regular and tight.   Consequently it is
$\tau$-additive\cmmnt{ (411E)}.

\spheader 411Pc $\mu$ is strictly positive\cmmnt{, because the
open-and-closed sets form a base for $\frak T$ (311I) and a non-empty
open-and-closed set has non-zero measure}.   $\mu$ is effectively
locally finite.   \prooflet{\Prf\ Suppose that $E\in\Sigma$ is not
negligible.
There is a measurable set $F\subseteq E$ such that $0<\mu F<\infty$;
now there is a non-empty open-and-closed
set $G$ included in $F$, in which case $\mu G<\infty$ and
$\mu(E\cap G)>0$.\ \Qed}

\wheader{411Pd}{0}{0}{0}{60pt}

\spheader 411Pd The following are equiveridical, that is, if one is true
so are the others:

\quad (i) $(\frak A,\bar\mu)$ is localizable;

\quad(ii) $\mu$ is strictly localizable;

\quad(iii) $\mu$ is locally determined;

\quad(iv) $\mu$ is a quasi-Radon measure.

\prooflet{\noindent\Prf\ The equivalence of (i)-(iii) is Theorem
322O\formerly{3{}22N}.
(iv)$\Rightarrow$(iii) is trivial.   If one, therefore all, of (i)-(iii)
are true, then $\mu$ is a topological measure, because if
$G\subseteq Z$ is open, then $\overline{G}$ is open-and-closed, by 314S,
therefore measurable, and $\overline{G}\setminus G$ is nowhere dense,
therefore also measurable.   We know already that $\mu$ is complete,
effectively locally finite and $\tau$-additive, so that if it is also
locally determined it is a quasi-Radon measure.\ \Qed}

\spheader 411Pe The following are equiveridical:

\quad(i) $\mu$ is a Radon measure;

\quad(ii) $\mu$ is totally finite;

\quad (iii) $\mu$ is locally finite;

\quad (iv) $\mu$ is outer regular with respect to the open sets.

\prooflet{\noindent\Prf\ (ii)$\Rightarrow$(iv) If $\mu$ is totally
finite and $E\in\Sigma$, then for any $\epsilon>0$ there is a closed
set $F\subseteq Z\setminus E$ such that
$\mu F\ge\mu(Z\setminus E)-\epsilon$, and now
$G=Z\setminus F$ is an open set including $E$ with
$\mu G\le\mu E+\epsilon$.   (iv)$\Rightarrow$(iii) Suppose that $\mu$
is outer regular with respect to the
open sets, and $z\in Z$.   Because $Z$ is Hausdorff, $\{z\}$ is
closed.   If it is open it is
measurable, and because $\mu$ is semi-finite it must have finite
measure.   Otherwise it is nowhere dense, therefore negligible, and
must be included
in open sets of arbitrarily small measure.   Thus in both cases $z$
belongs to an open set of finite measure;  as $z$ is arbitrary, $\mu$
is locally finite.  (iii)$\Rightarrow$(ii) Because $Z$ is compact, this
is a consequence of 411Ga.   (i)$\Rightarrow$(iii)
is part of the definition of `Radon measure'.   Finally,
(ii)$\Rightarrow$(i), again directly from the definition and the
facts set out in (a)-(b) above.\ \Qed}

\spheader 411Pf Let $W\subseteq Z$ be the union of all the open
subsets of $Z$ with finite measure.   \cmmnt{Because $\mu$ is
effectively locally finite,} $W$ has full outer measure, so
$(\frak A,\bar\mu)$ can be identified with the measure algebra of the
subspace measure $\mu_W$\cmmnt{ (322Jb)}.   \cmmnt{By the
definition of $W$,} $\mu_W$ is locally finite.   If
$(\frak A,\bar\mu)$ is localizable, then $\mu_W$ is a Radon measure.
\prooflet{\Prf\ Every open subset of $W$ belongs to $\Sigma$, by (d),
and therefore to the domain of $\mu_W$, and $\mu_W$ is a topological
measure.   By 214Ka, $\mu_W$ is complete and locally determined.
Because $\mu$ is inner regular with respect to the compact sets, so
is $\mu_W$.\ \Qed}


\leader{411Q}{Example:  Dieudonn\'e's measure} Recall that a set
$E\subseteq\omega_1$ is a Borel set iff either $E$ or its complement
includes a cofinal closed set\cmmnt{ (4A3J)}.   So we may define
a Borel measure $\mu$ on $\omega_1$ by saying that $\mu E=1$ if $E$
includes a cofinal closed set and $\mu E=0$ if $E$ is disjoint from a
cofinal closed set.   \cmmnt{If $E$ is disjoint from some
cofinal closed set, so is any subset of $E$, so} $\mu$ is complete.
\cmmnt{Since} $\mu$\cmmnt{ takes only the values $0$ and $1$, it}
is a purely atomic probability measure.

$\mu$ is a topological measure;  \cmmnt{being totally finite,} it
is\cmmnt{ surely} locally finite and effectively locally finite.
It is inner regular with respect to
the closed sets\cmmnt{ (because if $\mu E>0$, there is a cofinal
closed set $F\subseteq E$, and now $F$ is a closed set with
$\mu F=\mu E$)}, therefore outer regular with respect to the open sets.
It is not $\tau$-additive\cmmnt{ (because $\xi=\coint{0,\xi}$ is an open
set of zero measure for every $\xi<\omega_1$, and the union of these
sets is a measurable open set of measure $1$)}.

$\mu$ is not completion regular\cmmnt{, because the set of countable
limit ordinals is a closed set (4A1Bb) which does not include
any uncountable zero set (see 411R below)}.

The only self-supporting subset of $\omega_1$ is the empty
set\cmmnt{ (because there is a cover of $\omega_1$ by negligible
open sets)}.   \cmmnt{In particular,} $\mu$ does not have a support.

\cmmnt{\medskip

\noindent{\bf Remark} There is a measure of this type on any ordinal
of uncountable cofinality;  see 411Xj.
}%end of comment

\leader{411R}{Example:  The Baire $\sigma$-algebra of $\omega_1$}
The Baire $\sigma$-algebra\cmmnt{ $\CalBa(\omega_1)$} of $\omega_1$ is the
countable-cocountable algebra\cmmnt{ (4A3P)}.   The
countable-cocountable measure $\mu$ on $\omega_1$
is\cmmnt{ therefore} a Baire measure\cmmnt{ on the definition of
411K}.   \cmmnt{Since all sets of
the form $\ooint{\xi,\omega_1}$ are zero sets,} $\mu$ is inner regular
with respect to the zero sets and outer regular with respect to the
cozero sets.   \cmmnt{Since sets of
the form $\coint{0,\xi}$ ($=\xi$) form a cover of $\omega_1$ by
measurable open sets of zero measure,} $\mu$ is not $\tau$-additive.

\exercises{
\leader{411X}{Basic exercises $\pmb{>}$(a)}
%\spheader 411Xa
Let $(X,\Sigma,\mu)$ be a totally finite measure space and $\frak T$ a
topology on $X$.   Show that $\mu$ is inner regular with respect to the
closed sets iff it is outer regular with respect to the open sets, and
is inner regular with respect to the zero sets iff it is outer regular
with respect to the cozero sets.
%411D used in 457M

\spheader 411Xb Let $\mu$ be a Radon measure on $\BbbR^r$, where
$r\ge 1$, and $f\in\eusm L^0(\mu)$.   Show that $f$ is locally
integrable in the sense of 411Fc iff it is locally integrable in the
sense of 256E, that is, $\int_Efd\nu<\infty$ for every bounded set
$E\subseteq\BbbR^r$.
%411F

\spheader 411Xc Let $\mu$ be a measure on a topological space, $\hat\mu$
its completion and $\tilde\mu$ its c.l.d.\ version.   Show that $\mu$
is locally finite iff $\hat\mu$ is locally finite, and in this case
$\tilde\mu$ is locally finite.
%411G

\sqheader 411Xd Let $\mu$ be an effectively locally finite measure on
a topological space $X$.   (i) Show that the completion and c.l.d.\
version of $\mu$ are effectively locally finite.   (ii) Show that if
$\mu$ is complete and locally determined, then the union of the
measurable open sets of finite measure is conegligible.   (iii) Show
that if $X$ is hereditarily Lindel\"of then $\mu$ must be
$\sigma$-finite.
%411G

\spheader 411Xe Let $X$ be a topological space and $\mu$ a measure on
$X$.   Let $U\subseteq L^0(\mu)$ be the set of equivalence classes of
locally integrable functions in $\eusm L^0(\mu)$.   Show that $U$ is a
solid linear subspace of $L^0(\mu)$.   Show that if $\mu$ is locally
finite then $U$ includes $L^p(\mu)$ for every $p\in[1,\infty]$.
%411G

\spheader 411Xf Let $X$ be a topological space.   (i) Let $\mu$, $\nu$
be two totally finite Borel measures which agree on the closed sets.
Show that they are equal.   \Hint{136C.}   (ii) Let $\mu$, $\nu$ be
two totally finite Baire measures which agree on the zero sets.   Show
that they are equal.
%411K

\spheader 411Xg Let $(X,\frak T)$ be a topological space, $\mu$ a
measure on $X$, and $Y$ a subset of $X$;  let $\frak T_Y$, $\mu_Y$ be
the subspace
topology and measure.    Show that if $\mu$ is a topological measure,
or locally finite, or a Borel measure, so is $\mu_Y$.

\spheader 411Xh Let $\langle (X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be
a family of measure spaces, with direct sum $(X,\Sigma,\mu)$;  suppose
that we are given a topology $\frak T_i$ on each $X_i$, and let
$\frak T$ be the disjoint union topology on $X$ (definition:  4A2A).
Show that $\mu$ is a topological
measure, or locally finite, or effectively locally finite, or a Borel
measure, or a Baire measure, or strictly positive, iff every $\mu_i$ is.

\spheader 411Xi Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two measure
spaces, with c.l.d.\ product measure $\lambda$ on $X\times Y$.   Suppose
we are given topologies $\frak T$, $\frak S$ on $X$, $Y$ respectively,
and give $X\times Y$ the product topology.   Show that $\lambda$ is
locally finite, or effectively locally finite, if $\mu$ and $\nu$ are.

\spheader 411Xj Let $\alpha$ be any ordinal of uncountable cofinality
with its order topology (definitions:  3A1Fb, 4A2A).   Show that there is a complete topological
probability measure $\mu$ on $\alpha$ defined by saying that $\mu E=1$
if $E$ includes a cofinal closed set in $\alpha$, $0$ if $E$ is
disjoint from some
cofinal closed set.   Show that $\mu$ is inner regular with respect to
the closed sets but is not completion regular.
%411Q

\spheader 411Xk\dvAnew{2013} Let $\familyiI{(X_i,\frak T_i)}$ be a family
of topological spaces, and $\mu_i$ a strictly positive probability measure
on $X_i$ for each $i$.   Show that the product measure on 
$\prod_{i\in I}X_i$ is strictly positive.
%411N out of order query

\leader{411Y}{Further exercises (a)}
%\spheader 411Ya
Let $r$, $s\ge 1$ be integers.
Show that a function $f:\BbbR^r\to\BbbR^s$ is measurable
iff it is almost continuous (where $\BbbR^r$ is endowed with Lebesgue
measure and its usual topology, of course).   \Hint{256F.}
%411M

\spheader 411Yb Let $(X,\rho)$ be a metric space, $r\ge 0$, and write
$\mu_{Hr}$ for $r$-dimensional Hausdorff measure on $X$ (264K,
471A).   (i) Show that $\mu_{Hr}$ is a topological measure, outer
regular with respect to the
Borel sets.  (ii) Show that the c.l.d.\ version $\tilde\mu_{Hr}$ of
$\mu_{Hr}$ is inner regular with respect to the closed totally bounded
sets.   (iii) Show that $\tilde\mu_{Hr}$ is completion regular.
(iv) Show that if $X$ is complete then $\tilde\mu_{Hr}$ is tight.
%411J

\spheader 411Yc Let $(X,\frak T,\Sigma,\mu)$ be a topological measure
space.   Set $\Cal E=\{E:E\subseteq X,\,\mu(\partial E)=0\}$, where
$\partial E$ is the boundary of $A$.   (i) Show that $\Cal E$ is a
subalgebra of $\Cal PX$, and that every member of $\Cal E$ is measured
by the completion of $\mu$.   ($\Cal E$ is sometimes called the {\bf
Jordan algebra} of $(X,\frak T,\Sigma,\mu)$.   Do not confuse with the
`Jordan algebras' of abstract algebra.)   (ii) Suppose that $\mu$
is totally finite and inner regular with respect to the
closed sets, and that $\frak T$ is normal.   Show that
$\{E^{\ssbullet}:E\in\Cal E\cap\Sigma\}$ is dense in the measure algebra of
$\mu$ endowed with its usual topology.
(iii) Suppose that $\mu$ is a quasi-Radon measure
and $\frak T$ is completely regular.
Show that $\{E^{\ssbullet}:E\in\Cal E\}$ is dense in the measure
algebra of $\mu$.   \Hint{414Aa.}
%411G 411H

\spheader 411Yd\dvAnew{2011}
Let $(X,\frak T,\Sigma,\mu)$ be a second-countable
atomless topological probability space with a strictly positive measure, 
$\Cal E$ the
Jordan algebra of $\mu$ as defined in 411Yc, $(\frak A,\bar\mu)$ the
measure algebra of $\mu$ and $\frak E$ the image
$\{E^{\ssbullet}:E\in\Cal E\}\subseteq\frak A$.   Let $\frak B$ be a
Boolean algebra and $\nu:\frak B\to[0,1]$ a finitely additive functional.
Show that $(\frak B,\nu)\cong(\frak E,\bar\mu\restrp\frak E)$ iff
($\alpha$) 
$\nu$ is strictly positive and properly atomless in the sense of
326F\formerly{3{}26Ya}, and $\nu 1=1$ 
($\beta$) there is a countable subalgebra $\frak B_0$ of $\frak B$
such that $\nu b=\sup\{\nu c:c\in\frak B_0$, $c\Bsubseteq b\}$ for every
$b\in\frak B$ 
($\gamma$) whenever $A$, $B\subseteq\frak B$ are upwards-directed
sets such that $a\Bcap b=0$ for every $a\in A$ and $b\in B$ and
$\sup\{\nu(a\Bcup b):a\in A$, $b\in B\}=1$, then $\sup A$ is defined in
$\frak B$.
%411Yc 411H

}%end of exercises

\cmmnt{\Notesheader{411} Of course the list above can give only a
rough idea of the ways in which topologies and measures can interact.
In particular I have rather arbitrarily given a sort of priority to
three particular relationships between the domain $\Sigma$ of a measure
and the topology:  `topological measure space' (in which $\Sigma$
includes the
Borel $\sigma$-algebra), `Borel measure' (in which $\Sigma$ is precisely
the Borel $\sigma$-algebra) and `Baire measure' (in which $\Sigma$ is
the Baire $\sigma$-algebra).

Abstract topological measure theory is a relatively new subject, and
there are many technical questions on which different authors take
different views.   For instance, the phrase `Radon measure' is commonly
used to mean what I would call a `tight locally finite Borel measure'
(cf.\ 416F);  and some writers
enlarge the definition of `topological measure' to include Baire
measures as defined above.

I give very few examples at this stage, two drawn from the constructions
of Volumes 1-3 (Lebesgue measure and Stone spaces, 411O-411P) and one
new one (`Dieudonn\'e's measure', 411Q), with a glance at the
countable-cocountable measure of $\omega_1$ (411R).   The most glaring
omission is that of the product measures on $\{0,1\}^I$ and $[0,1]^I$.
I pass these by at the moment because a proper study of them requires
rather more preparation than can be slipped into a parenthesis.   (I
return to them in 416U.)   I have also omitted
any discussion of `measurable' and `almost continuous' functions, except
for a reference to a theorem in Volume 2 (411Ya), which will have
to be repeated and amplified later on (418K).
There is an obvious complementarity between the notions of `inner' and
`outer' regularity (411B, 411D), but it works well only for totally
finite spaces (411Xa);  in other cases it may not be obvious what will
happen (411O, 411Pe, 412W).

}%end of notes

\discrpage
\frfilename{mt412.tex}
\versiondate{17.6.16}
\copyrightdate{2000}

\def\chaptername{Topologies and measures I}
\def\sectionname{Inner regularity}

\newsection{412}

As will become apparent as the chapter progresses, the concepts
introduced in \S411 are synergic;  their most interesting manifestations
are in combinations of various kinds.   Any linear account of their
properties will be more than usually like a space-filling curve.   But I
have to start somewhere, and enough results can be expressed in terms of
inner regularity, more or less by itself, to be a useful beginning.

After a handful of elementary basic facts (412A) and a list of standard
applications (412B), I give some useful sufficient conditions for inner
regularity of topological and Baire measures (412D, 412E, 412G), based
on an important
general construction (412C).   The rest of the section amounts to a
review of ideas from Volume 2 and Chapter 32 in the light of the new
concept here.   I touch on completions (412H), c.l.d.\ versions and
complete locally determined spaces (412H, 412J, 412M), strictly
localizable spaces (412I), \imp\ functions (412K, 412L), measure
algebras (412N), subspaces (412O, 412P), indefinite-integral measures
(412Q) and product measures (412R-412V),
  %412R 412S 412T 412U 412V
with a brief mention of outer regularity (412W);  most of the hard work
has already been done in Chapters 21 and 25.

\leader{412A}{}\cmmnt{ I begin by repeating a lemma from Chapter
34, with some further straightforward facts.

\medskip

\noindent}{\bf Lemma} (a) Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K$ a family of sets such that

\inset{whenever $E\in\Sigma$ and $\mu E>0$ there is a
$K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and $\mu K>0$.}

\noindent Then whenever $E\in\Sigma$ there is a countable disjoint
family $\familyiI{K_i}$ in $\Cal K\cap\Sigma$ such that $K_i\subseteq E$
for every $i$ and $\sum_{i\in I}\mu K_i=\mu E$.   If moreover

\inset{($\dagger$) $K\cup K'\in\Cal K$ whenever $K$, $K'$ are disjoint
members of $\Cal K$,}

\noindent then $\mu$ is inner regular with respect to $\Cal K$.   If
$\bigcup_{i\in I}K_i\in\Cal K$ for every countable disjoint family
$\familyiI{K_i}$ in $\Cal K$, then for every $E\in\Sigma$ there is a
$K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and $\mu K=\mu E$.

(b) Let $(X,\Sigma,\mu)$ be a measure space, $\Tau$ a
$\sigma$-subalgebra of $\Sigma$, and $\Cal K$ a family of sets.   If
$\mu$ is inner regular with respect to $\Tau$ and $\mu\restrp\Tau$ is
inner regular with respect to $\Cal K$, then $\mu$ is inner regular with
respect to $\Cal K$.

(c) Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\sequencen{\Cal K_n}$ a sequence of families of sets such that
$\mu$ is inner regular with respect to $\Cal K_n$ and

\inset{($\ddagger$) if $\sequence{i}{K_i}$ is a
non-increasing sequence in $\Cal K_n$, then
$\bigcap_{i\in\Bbb N}K_i\in\Cal K_n$}

\noindent for every $n\in\Bbb N$.   Then $\mu$ is inner regular with
respect to $\bigcap_{n\in\Bbb N}\Cal K_n$.

\proof{{\bf (a)} This is 342B-342C.

\medskip

{\bf (b)} If $E\in\Sigma$ and $\gamma<\mu E$, there are an $F\in\Tau$
such that $F\subseteq E$ and $\mu F>\gamma$, and a $K\in\Cal K\cap\Tau$
such that $K\subseteq F$ and $\mu K\ge\gamma$.

\medskip

{\bf (c)} Suppose that $E\in\Sigma$ and that $0\le\gamma<\mu E$.
Because $\mu$ is semi-finite, there is an $F\in\Sigma$ such that
$F\subseteq E$ and $\gamma<\mu F<\infty$ (213A).   Choose
$\sequence{i}{K_i}$ inductively, as
follows.   Start with $K_0=F$.   Given that $K_i\in\Sigma$ and
$\gamma<\mu
K_i$, then let $n_i\in\Bbb N$ be such that $2^{-n_i}(i+1)$ is an odd
integer, and choose $K_{i+1}\in\Cal K_{n_i}\cap\Sigma$ such that
$K_{i+1}\subseteq K_i$ and $\mu K_{i+1}>\gamma$;
this will be possible because $\mu$ is
inner regular with respect to $\Cal K_{n_i}$.   Consider
$K=\bigcap_{i\in\Bbb N}K_i$.   Then $K\subseteq E$ and
$\mu K=\lim_{i\to\infty}\mu K_i\ge\gamma$.   But also

\Centerline{$K=\bigcap_{j\in\Bbb N}K_{2^n(2j+1)}\in\Cal K_n$}

\noindent because $\sequence{j}{K_{2^n(2j+1)}}$ is a non-increasing
sequence in $\Cal K_n$, for each $n$.   So
$K\in\bigcap_{n\in\Bbb N}\Cal K_n$.   As
$E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with respect to
$\bigcap_{n\in\Bbb N}\Cal K_n$.
}%end of proof of 412A

\vleader{60pt}{412B}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$.   Suppose that $\Cal K$ is

\inset{{\it either} the family of Borel subsets of $X$}

\inset{{\it or} the family of closed subsets of $X$}

\inset{{\it or} the family of compact subsets of $X$}

\inset{{\it or} the family of zero sets in $X$,}

\noindent and suppose that whenever $E\in\Sigma$ and $\mu E>0$ there is
a $K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and $\mu K>0$.   Then
$\mu$ is inner regular with respect to $\Cal K$.

\proof{ In every case, $\Cal K$ satisfies the condition ($\dagger$) of
412Aa.
}%end of proof of 412B

\leader{412C}{}\cmmnt{ The next lemma provides a particularly useful
method of proving that measures are inner regular with respect to
`well-behaved' families of sets.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space, and suppose that $\Cal A\subseteq\Sigma$ is such that

\inset{$\emptyset\in\Cal A\subseteq\Sigma$,

$X\setminus A\in\Cal A$ for every $A\in\Cal A$.}

\noindent Let $\Tau$ be the $\sigma$-subalgebra of $\Sigma$ generated by
$\Cal A$.
Let $\Cal K$ be a family of subsets of $X$ such that

\inset{($\dagger$) $K\cup K'\in\Cal K$ whenever $K$, $K'\in\Cal K$,

($\ddagger$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ for every
sequence $\sequencen{K_n}$ in $\Cal K$,

whenever $A\in\Cal A$, $F\in\Sigma$ and $\mu(A\cap F)>0$, there
is a $K\in\Cal K\cap\Tau$ such that $K\subseteq A$ and
$\mu(K\cap F)>0$.}

\noindent Then $\mu\restrp\Tau$ is inner regular with respect to $\Cal K$.

\proof{{\bf (a)} Write $\frak A$ for the measure algebra of
$(X,\Sigma,\mu)$, and
$\Cal L=\Cal K\cap\Tau$, so that $\Cal L$ also is closed under finite
unions and countable intersections.   Set

\Centerline{$\Cal H
=\{E:E\in\Sigma,\,\sup_{L\in\Cal L,L\subseteq E}L^{\ssbullet}
    =E^{\ssbullet}\}$ in $\frak A$,}

\Centerline{$\Tau'=\{E:E\in\Cal H,\,X\setminus E\in\Cal H\}$,}

\noindent so that the last condition tells us that
$\Cal A\subseteq\Cal H$ and therefore that $\Cal A\subseteq\Tau'$.

\medskip

{\bf (b)} The intersection of any sequence in $\Cal H$ belongs to
$\Cal H$.   \Prf\ Let $\sequencen{H_n}$ be a sequence in $\Cal H$ with
intersection $H$.   Write $A_n$ for
$\{L^{\ssbullet}:L\in\Cal L,\,L\subseteq H_n\}
\subseteq\frak A$ for each $n\in\Bbb N$.   Since $\mu$ is semi-finite,
$\frak A$ is \wsid\ (322F).   As $A_n$ is upwards-directed and
$\sup A_n=H_n^{\ssbullet}$ for each $n\in\Bbb N$,

$$\eqalignno{H^{\ssbullet}
&=\inf_{n\in\Bbb N}H_n^{\ssbullet}\cr
\displaycause{because $F\mapsto F^{\ssbullet}:\Sigma\to\frak A$ is
sequentially order-continuous, by 321H}
&=\inf_{n\in\Bbb N}\sup A_n
=\sup\{\inf_{n\in\Bbb N}a_n:a_n\in A_n\text{ for every }n\in\Bbb N\}\cr
\displaycause{316H(iv)}
&=\sup\{(\bigcap_{n\in\Bbb N}L_n)^{\ssbullet}:L_n\in\Cal L,
  \,L_n\subseteq H_n\text{ for every }n\in\Bbb N\}\cr
&\Bsubseteq\sup\{L^{\ssbullet}:L\in\Cal L,\,L\subseteq H\}\cr
\displaycause{by ($\ddagger$)}
&\Bsubseteq H^{\ssbullet},\cr}$$

\noindent and $H\in\Cal H$.\ \Qed

\medskip

{\bf (c)} The union of any sequence in $\Cal H$ belongs to $\Cal H$.
\Prf\ If $\sequencen{H_n}$ is a sequence in $\Cal H$
with union $H$ then

\Centerline{$\sup_{L\in\Cal L,L\subseteq H}L^{\ssbullet}
\Bsupseteq\sup_{n\in\Bbb N}\sup_{L\in\Cal L,L\subseteq E_n}
  L^{\ssbullet}
=\sup_{n\in\Bbb N}H_n^{\ssbullet}
=H^{\ssbullet}$,}

\noindent so $H\in\Cal H$.\ \Qed

\medskip

{\bf (d)} $\Tau'$ is a $\sigma$-subalgebra of $\Sigma$.
\Prf\ (i) $\emptyset$ and $X$ belong to $\Cal A\subseteq\Cal H$, so
$\emptyset\in\Tau'$.
(ii) Obviously $X\setminus E\in\Tau'$ whenever $E\in\Tau'$.
(iii) If $\sequencen{E_n}$ is a sequence in $\Tau'$ with union $E$ then
$E\in\Cal H$, by (c);  but also
$X\setminus E=\bigcap_{n\in\Bbb N}(X\setminus E_n)$ belongs to $\Cal H$,
by (b).   So $E\in\Tau'$.\ \Qed

\medskip

{\bf (e)} Accordingly $\Tau\subseteq\Tau'$, and
$E^{\ssbullet}=\sup_{L\in\Cal L,L\subseteq E}L^{\ssbullet}$ for every
$E\in\Tau$.   It follows at once that if $E\in\Tau$ and $\mu E>0$, there
must be an $L\in\Cal L$ such that $L\subseteq E$ and $\mu L>0$;  since
($\dagger$) is true, and $\Cal L\subseteq\Tau$, we can apply 412Aa to
see that $\mu\restrp\Tau$ is inner regular with respect to $\Cal L$,
therefore with respect to $\Cal K$.
}%end of proof of 412C

\leader{412D}{}\cmmnt{ As corollaries of the last lemma I give
two-and-a-half basic theorems.

\medskip

\noindent}{\bf Theorem} Let $(X,\frak T)$ be a topological space and
$\mu$ a semi-finite Baire measure on $X$.   Then $\mu$ is inner regular
with respect to the zero sets.

\proof{ Write $\Sigma$ for the Baire $\sigma$-algebra of $X$, the domain
of $\mu$, $\Cal K$ for the family of zero sets, and $\Cal A$ for
$\Cal K\cup\{X\setminus K:K\in\Cal K\}$.   Since the union of two zero
sets is a zero set (4A2C(b-ii)), the intersection of a sequence of zero
sets is a zero set (4A2C(b-iii)), and the complement of a zero set is
the union of a sequence of zero sets (4A2C(b-vi)), the conditions of
412C are satisfied;  and as the $\sigma$-algebra generated by $\Cal A$
is just $\Sigma$, $\mu$ is inner regular with respect to $\Cal K$.
}%end of proof of 412D

\leader{412E}{Theorem} Let $(X,\frak T)$ be a perfectly normal
topological space\cmmnt{ (e.g., any metrizable space)}.   Then any
semi-finite Borel measure on $X$ is inner regular with respect to the
closed sets.

\proof{ Because the Baire and Borel $\sigma$-algebras are the same
(4A3Kb), this is a special case of 412D.
}%end of proof of 412E

\leader{412F}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$ such that $\mu$ is effectively locally
finite with respect to $\frak T$.   Then

\Centerline{$\mu E=\sup\{\mu(E\cap G):G$ is a measurable open set of
finite measure$\}$}

\noindent for every $E\in\Sigma$.

\proof{ Apply 412Aa with $\Cal K$ the family of subsets of measurable
open sets of finite measure.
}%end of proof of 412F

\vleader{72pt}{412G}{Theorem} Let $(X,\Sigma,\mu)$ be a measure space
with a
topology $\frak T$ such that $\mu$ is effectively locally finite with
respect to $\frak T$ and $\Sigma$ is the $\sigma$-algebra generated by
$\frak T\cap\Sigma$.   If

\Centerline{$\mu G=\sup\{\mu F:F\in\Sigma$ is closed, $F\subseteq G\}$}

\noindent for every measurable open set $G$ of finite measure, then
$\mu$ is inner regular with respect to the closed sets.

\proof{ In
412C, take $\Cal K$ to be the family of measurable closed subsets of
$X$, and $\Cal A$ to be the family of measurable sets which are
{\it either} open {\it or} closed.   If $G\in\Sigma\cap\frak T$,
$F\in\Sigma$ and $\mu(G\cap F)>0$, then there is an open set $H$ of
finite measure such that
$\mu(H\cap G\cap F)>0$, because $\mu$ is effectively locally finite;
now there is a $K\in\Cal K$ such that $K\subseteq H\cap G$ and
$\mu K>\mu(H\cap G)-\mu(H\cap G\cap F)$, so that $\mu(K\cap F)>0$.
This is the only non-trivial item in the list of hypotheses in 412C, so
we can conclude that $\mu\restrp\Tau$ is inner regular with respect to
$\Cal K$, where $\Tau$ is the $\sigma$-algebra generated by $\Cal A$;
but of course this is just $\Sigma$.
}%end of proof of 412G

\cmmnt{\medskip

\noindent{\bf Remark} There is a similar result in 416F(iii) below.
}%end of comment

\leader{412H}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K$ a family of sets.

(a) If $\mu$ is inner regular with respect to
$\Cal K$, so are its completion $\hat\mu$\cmmnt{ (212C)} and
c.l.d.\ version $\tilde\mu$\cmmnt{ (213E)}.

(b) Now suppose that $\mu$ is semi-finite and that

\inset{($\ddagger$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ whenever
$\sequencen{K_n}$ is a non-increasing sequence in $\Cal K$.}

\noindent If either $\hat\mu$ or $\tilde\mu$ is inner regular with respect to
$\Cal K$ then $\mu$ is inner regular with respect to $\Cal K$.

\proof{{\bf (a)} If $F$ belongs to the domain of $\hat\mu$, then there
is an $E\in\Sigma$ such that $E\subseteq F$ and
$\hat\mu(F\setminus E)=0$.
So if $0\le\gamma<\hat\mu F=\mu E$, there is a $K\in\Cal K\cap\Sigma$
such that $K\subseteq E\subseteq F$ and $\hat\mu K=\mu K\ge\gamma$.

If $H$ belongs to the domain of $\tilde\mu$ and
$0\le\gamma<\tilde\mu H$, there is an $E\in\Sigma$ such that
$\mu E<\infty$ and $\hat\mu(E\cap H)>\gamma$ (213D).   Now there is a
$K\in\Cal K\cap\Sigma$ such that
$K\subseteq E\cap H$ and $\mu K\ge\gamma$.   As $\mu K<\infty$,
$\tilde\mu K=\mu K\ge\gamma$.

\medskip

{\bf (b)} Write $\check\mu$ for whichever of $\hat\mu$, $\tilde\mu$ is
supposed to be inner regular with respect to $\Cal K$.   Then
$\check\mu$ is inner regular with respect to $\Sigma$ (212Ca, 213Fc), so
is inner regular with respect to $\Cal K\cap\Sigma$ (412Ac).   Also
$\check\mu$ extends $\mu$ (212D, 213Hc).
Take $E\in\Sigma$ and $\gamma<\mu E=\check\mu E$.   Then there is a
$K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and
$\gamma<\check\mu K=\mu K$.   As $E$ and $\gamma$ are arbitrary, $\mu$
is inner regular with respect to $\Cal K$.
}%end of proof of 412H

\leader{412I}{Lemma} Let $(X,\Sigma,\mu)$ be a strictly localizable
measure space and $\Cal K$ a family of sets such that whenever
$E\in\Sigma$ and $\mu E>0$ there is a $K\in\Cal K\cap\Sigma$ such that
$K\subseteq E$ and $\mu K>0$.

(a) There is a decomposition $\familyiI{X_i}$ of $X$ such that at most
one $X_i$ does not belong to $\Cal K$, and that exceptional one, if any,
is negligible.

(b) There is a disjoint family $\Cal L\subseteq\Cal K\cap\Sigma$ such
that $\mu^*A=\sum_{L\in\Cal L}\mu^*(A\cap L)$ for every $A\subseteq X$.

(c) If $\mu$ is $\sigma$-finite then the family $\familyiI{X_i}$ of (a)
and the set $\Cal L$ of (b) can be taken to be countable.

\proof{{\bf (a)} Let $\family{j}{J}{E_j}$ be any decomposition of $X$.
For each $j\in J$, let $\Cal K_j$ be a maximal disjoint subset of

\Centerline{$\{K:K\in\Cal K\cap\Sigma,\,K\subseteq E_j,\,\mu K>0\}$.}

\noindent Because $\mu E_j<\infty$, $\Cal K_j$ must be countable.   Set
$E_j'=E_j\setminus\bigcup\Cal K_j$.   By the maximality of $\Cal K_j$,
$E'_j$ cannot include any non-negligible set in $\Cal K\cap\Sigma$;  but
this means that $\mu E'_j=0$.   Set $X'=\bigcup_{j\in J}E'_j$.   Then

\Centerline{$\mu X'=\sum_{j\in J}\mu(X'\cap E_j)
=\sum_{j\in J}\mu E'_j=0$.}

Note that if $j$, $j'\in J$ are distinct, $K\in\Cal K_j$ and
$K'\in\Cal K_{j'}$, then $K\cap K'=\emptyset$;  thus
$\Cal L=\bigcup_{j\in J}\Cal K_j$ is disjoint.
Let $\familyiI{X_i}$ be any indexing of $\{X'\}\cup\Cal L$.   This is a
partition (that is, disjoint cover) of $X$ into sets of finite measure.
If $E\subseteq X$ and
$E\cap X_i\in\Sigma$ for every $i\in I$, then for every $j\in J$

\Centerline{$E\cap E_j
=(E\cap X'\cap E_j)\cup\bigcup_{K\in\Cal K_j}E\cap K$}

\noindent belongs to $\Sigma$, so that $E\in\Sigma$ and

\Centerline{$\mu E=\sum_{j\in J}\mu(E\cap E_j)
=\sum_{j\in J}\sum_{K\in\Cal K_j}\mu(E\cap K)
=\sum_{i\in I}\mu(E\cap X_i)$.}

\noindent Thus $\familyiI{X_i}$ is a decomposition of $X$, and it is of
the right type because every $X_i$ but one belongs to
$\Cal L\subseteq\Cal K$.

\medskip
\wheader{412I}{0}{0}{0}{36pt}

{\bf (b)} If now $A\subseteq X$ is any set,

\Centerline{$\mu^*A=\mu_AA=\sum_{i\in I}\mu_A(A\cap X_i)
=\sum_{i\in I}\mu^*(A\cap X_i)$}

\noindent by 214Ia, writing $\mu_A$ for the subspace measure on $A$.
So we have

\Centerline{$\mu^*A=\mu^*(A\cap X')+\sum_{L\in\Cal L}\mu^*(A\cap L)
=\sum_{L\in\Cal L}\mu^*(A\cap L)$,}

\noindent while $\Cal L\subseteq\Cal K$ is disjoint.

\medskip

{\bf (c)} If $\mu$ is $\sigma$-finite we can take $J$ to be countable,
so that $I$ and $\Cal L$ will also be countable.
}%end of proof of 412I

\leader{412J}{Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, and $\Cal K$ a family of sets such that $\mu$
is inner regular with respect to $\Cal K$.

(a) If $E\subseteq X$ is such that $E\cap K\in\Sigma$ for every
$K\in\Cal K\cap\Sigma$, then $E\in\Sigma$.

(b) If $E\subseteq X$ is such that $E\cap K$ is negligible for every
$K\in\Cal K\cap\Sigma$, then $E$ is negligible.

(c) For any $A\subseteq X$,
$\mu^*A=\sup_{K\in\Cal K\cap\Sigma}\mu^*(A\cap K)$.

(d) Let $f$ be a non-negative $[0,\infty]$-valued function defined on a
subset of $X$.   If $\int_Kf$ is defined in $[0,\infty]$ for every
$K\in\Cal K$, then $\int f$ is defined and equal to
$\sup_{K\in\Cal K}\int_Kf$.

(e) If $f$ is a $\mu$-integrable function and $\epsilon>0$, there is a
$K\in\Cal K$ such that $\int_{X\setminus K}|f|\le\epsilon$.

\cmmnt{\medskip

\noindent{\bf Remark} In (c), we must interpret $\sup\emptyset$ as $0$
if $\Cal K\cap\Sigma=\emptyset$.
}

\proof{{\bf (a)} If $F\in\Sigma$ and $\mu F<\infty$, then $E\cap
F\in\Sigma$.   \Prf\ If $\mu F=0$, this is trivial, because $\mu$ is
complete and $E\cap F$ is negligible.   Otherwise, there is a sequence
$\sequencen{K_n}$ in $\Cal K\cap\Sigma$ such that $K_n\subseteq F$ for
each $n$ and $\sup_{n\in\Bbb N}\mu K_n=\mu F$.   Now
$E\cap F\setminus\bigcup_{n\in\Bbb N}K_n$ is negligible, therefore
measurable,
while $E\cap K_n$ is measurable for every $n\in\Bbb N$, by hypothesis;
so $E\cap F$ is measurable.\ \QeD\   As $\mu$ is locally determined,
$E\in\Sigma$, as claimed.

\medskip

{\bf (b)} By (a), $E\in\Sigma$;  and because $\mu$ is inner regular with
respect to $\Cal K$, $\mu E$ must be $0$.

\medskip

{\bf (c)} Let $\mu_A$ be the subspace measure on $A$.   Because $\mu$ is
complete and locally determined, $\mu_A$ is semi-finite (214Id).   So if
$0\le\gamma<\mu^*A=\mu_AA$, there is an $H\subseteq A$ such that
$\mu_AH$ is defined, finite and greater than $\gamma$.   Let
$E\in\Sigma$ be a
measurable envelope of $H$ (132Ee), so that $\mu E=\mu^*H>\gamma$.
Then there is a $K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and
$\mu K\ge\gamma$.   In this case

\Centerline{$\mu^*(A\cap K)\ge\mu^*(H\cap K)=\mu(E\cap K)
=\mu K\ge\gamma$.}

\noindent As $\gamma$ is arbitrary,

\Centerline{$\mu^*A\le\sup_{K\in\Cal K\cap\Sigma}\mu^*(A\cap K)$;}

\noindent but the reverse inequality is trivial, so we have the result.

\medskip

{\bf (d)} Applying (b) with $E=X\setminus\dom f$, we see that $f$ is
defined almost everywhere in $X$.   Applying (a) with
$E=\{x:x\in\dom f,\,f(x)\ge\alpha\}$ for each $\alpha\in\Bbb R$, we see
that $f$ is measurable.   So $\int f$ is defined in $[0,\infty]$, and of
course $\int f\ge\sup_{K\in\Cal K}\int_Kf$.   If $\gamma<\int f$, there
is a non-negative simple function $g$ such that $g\leae f$ and
$\int g>\gamma$;   taking $E=\{x:g(x)>0\}$, there is a $K\in\Cal K$ such
that $K\subseteq E$ and
$\mu(E\setminus K)\|g\|_{\infty}\le\int g-\gamma$, so that
$\int_Kf\ge\int_Kg\ge\gamma$.   As $\gamma$ is arbitrary,
$\int f=\sup_{K\in\Cal K}\int_Kf$.

\medskip

{\bf (e)} By (d), there is a $K\in\Cal K$ such that
$\int_K|f|\ge\int|f|-\epsilon$.
}%end of proof of 412J

\cmmnt{\medskip

\noindent{\bf Remark} See also 413F below.
}%end of comment

\leader{412K}{Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, $(Y,\Tau,\nu)$ a measure space and
$f:X\to Y$ a function.   Suppose that $\Cal K\subseteq\Tau$ is such
that

\inset{(i) $\nu$ is inner regular with respect to $\Cal K$;

(ii) $f^{-1}[K]\in\Sigma$ and $\mu f^{-1}[K]=\nu K$ for every
$K\in\Cal K$;

(iii) whenever $E\in\Sigma$ and $\mu E>0$ there is a $K\in\Cal K$
such that $\nu K<\infty$ and $\mu(E\cap f^{-1}[K])>0$.}

\noindent Then $f$ is \imp\ for $\mu$ and $\nu$.

\proof{{\bf (a)} If $F\in\Tau$, $E\in\Sigma$ and $\mu E<\infty$, then
$E\cap f^{-1}[F]\in\Sigma$.   \Prf\ Let $H_1$, $H_2\in\Sigma$ be
measurable envelopes for $E\cap f^{-1}[F]$ and $E\setminus f^{-1}[F]$
respectively.   \Quer\ If $\mu(H_1\cap H_2)>0$, there is a
$K\in\Cal K$ such that $\nu K$ is finite and $\mu(H_1\cap H_2\cap
f^{-1}[K])>0$.   Because $\nu$ is inner regular with respect to
$\Cal K$, there are $K_1$, $K_2\in\Cal K$ such that $K_1\subseteq
K\cap F$, $K_2\subseteq K\setminus F$ and

$$\eqalign{\nu K_1+\nu K_2
&>\nu(K\cap F)+\nu(K\setminus F)-\mu(H_1\cap H_2\cap f^{-1}[K]) \cr
&=\nu K-\mu(H_1\cap H_2\cap f^{-1}[K]).\cr}$$

\noindent Now

\Centerline{$\mu(H_1\cap f^{-1}[K_2])
=\mu^*(E\cap f^{-1}[F]\cap f^{-1}[K_2])=0$,}

\Centerline{$\mu(H_2\cap f^{-1}[K_1])
=\mu^*(E\cap f^{-1}[K_1]\setminus f^{-1}[F])=0$,}

\noindent so
$\mu(H_1\cap H_2\cap f^{-1}[K_1\cup K_2])=0$ and

$$\eqalign{\mu(H_1\cap H_2\cap f^{-1}[K])
&\le\mu(f^{-1}[K]\setminus f^{-1}[K_1\cup K_2])\cr
&=\mu f^{-1}[K]-\mu f^{-1}[K_1]-\mu f^{-1}[K_2]\cr
&=\nu K-\nu K_1-\nu K_2
<\mu(H_1\cap H_2\cap f^{-1}[K]),\cr}$$

\noindent which is absurd.\ \Bang

Now $(E\cap H_1)\setminus(E\cap f^{-1}[F])\subseteq H_1\cap H_2$ is
negligible, therefore measurable (because $\mu$ is complete), and
$E\cap f^{-1}[F]\in\Sigma$, as claimed.\ \Qed

\medskip

{\bf (b)} It follows (because $\mu$ is locally determined) that
$f^{-1}[F]\in\Sigma$ for every $F\in\Tau$.

\medskip

{\bf (c)} If $F\in\Tau$ and $\nu F=0$ then $\mu f^{-1}[F]=0$.
\Prf\Quer\ Otherwise, there is a $K\in\Cal K$ such that $\nu
K<\infty$ and

\Centerline{$0<\mu(f^{-1}[F]\cap f^{-1}[K])=\mu f^{-1}[F\cap K]$.}

\noindent  Let $K'\in\Cal K$ be such that $K'\subseteq K\setminus F$
and $\nu K'>\nu K-\mu f^{-1}[F\cap K]$.   Then
$f^{-1}[K']\cap f^{-1}[F\cap K]=\emptyset$, so

\Centerline{$\nu K
=\mu f^{-1}[K]
\ge\mu f^{-1}[K']+\mu f^{-1}[F\cap K]
>\nu K'+\nu K-\nu K'=\nu K$,}

\noindent which is absurd.\ \Bang\Qed

\medskip

{\bf (d)} Finally, $\mu f^{-1}[F]=\nu F$ for every $F\in\Tau$.
\Prf\ Let $\familyiI{K_i}$ be a countable disjoint family in $\Cal K$
such that $K_i\subseteq F$ for every $i$ and
$\sum_{i\in I}\nu K_i=\nu F$ (412Aa).   Set
$F'=F\setminus\bigcup_{i\in I}K_i$.
Then

\Centerline{$\mu f^{-1}[F]
=\mu f^{-1}[F']
 +\sum_{i\in I}\mu f^{-1}[K_i]
=\mu f^{-1}[F']
 +\sum_{i\in I}\nu K_i
=\mu f^{-1}[F']+\nu F$.}

\noindent If $\nu F=\infty$ then surely $\mu f^{-1}[F]=\infty=\nu F$.
Otherwise, $\nu F'=0$ so $\mu f^{-1}[F']=0$ (by (c)) and again
$\mu f^{-1}[F]=\nu F$.\ \Qed

Thus $f$ is \imp.
}%end of proof of 412K

\leader{412L}{Corollary} Let $(X,\Sigma,\mu)$ be a complete probability
space, $(Y,\Tau,\nu)$ a probability space and $f:X\to Y$ a function.
Suppose that whenever $F\in\Tau$ and $\nu F>0$ there is a $K\in\Tau$
such that $K\subseteq F$, $\nu K>0$, $f^{-1}[K]\in\Sigma$ and
$\mu f^{-1}[K]\ge\nu K$.   Then $f$ is \imp.

\proof{ Set
$\Cal K^*=\{K:K\in\Tau,\,f^{-1}[K]\in\Sigma,\,\mu f^{-1}[K]\ge\nu K\}$.
Then $\Cal K^*$ is closed under countable
disjoint unions and includes $\Cal K$, so for every $F\in\Tau$ there is
a $K\in\Cal K^*$ such that $K\subseteq F$ and $\nu K=\nu F$, by 412Aa.
But this means that $\mu f^{-1}[K]=\nu K$ for every $K\in\Cal K^*$.
\Prf\ There is a $K'\in\Cal K^*$ such that $K'\subseteq Y\setminus K$
and $\nu K'=1-\nu K$;  but in this case

\Centerline{$\mu f^{-1}[K']+\mu f^{-1}[K]\le 1=\nu K'+\nu K$,}

\noindent so $\mu f^{-1}[K]$ must be equal to $\nu K$.\ \QeD\  Moreover,
there is a $K^*\in\Cal K^*$ such that $\nu K^*=\nu Y=1$, so
$\mu f^{-1}[K^*]=\mu X=1$ and
$\mu(E\cap f^{-1}[K^*])>0$ whenever $\mu E>0$.   Applying 412K to
$\Cal K^*$ we have the result.
}%end of proof of 412L

\leader{412M}{Proposition} Let $X$ be a set and $\Cal K$ a family of
subsets of $X$.   Suppose
that $\mu$ and $\nu$ are two complete locally determined measures on $X$,
with domains including $\Cal K$, and both inner
regular with respect to $\Cal K$.

(a) If $\mu K\le\nu K$ for every $K\in\Cal K$, then $\mu\le\nu$ in the sense of
234P\cmmnt{, that is, $\dom\nu\subseteq\dom\mu$ and $\mu E\le\nu E$ for every
$E\in\dom\nu$}.

(b) If $\mu K=\nu K$ for every $K\in\Cal K$, then $\mu=\nu$.

\proof{{\bf (a)(i)} Write $\Sigma$ for the domain of $\mu$ and $\Tau$ for the
domain of $\nu$.   If $E\in\Sigma\cap\Tau$ then of course

\Centerline{$\mu E=\sup_{K\in\Cal K,K\subseteq E}\mu K
\le\sup_{K\in\Cal K,K\subseteq E}\nu K=\nu E$.}

\noindent So we have only to show that $\Tau\subseteq\Sigma$.

\medskip

\quad{\bf (ii)} Suppose that $F\in\Tau$, $E\in\Sigma$ and $\mu E<\infty$.
Let $E'$, $E''\subseteq E$ be measurable envelopes of $E\cap F$, $E\setminus F$
with respect to $\mu$.   Take any $K\in\Cal K$ such that
$K\subseteq E'\cap E''$,   Then $K\cap F$ and $K\setminus F$ belong to
$\Tau$.   Let $\sequencen{L_n}$, $\sequencen{L'_n}$ be sequences in $\Cal K$
such that $L_n\subseteq K\cap F$ and $L'_n\subseteq K\setminus F$ for every
$n$, $\lim_{n\to\infty}\nu L_n=\nu(K\cap F)$ and
$\lim_{n\to\infty}\nu L'_n=\nu(K\setminus F)$.  Because $\nu K<\infty$,
$\nu(K\setminus\bigcup_{n\in\Bbb N}(L_n\cup L'_n))=0$ and therefore
$\mu(K\setminus\bigcup_{n\in\Bbb N}(L_n\cup L'_n))=0$, since
$K\setminus\bigcup_{n\in\Bbb N}(L_n\cup L'_n)\in\Sigma\cap\Tau$.
On the other hand, for
each $n$, $L_n\in\Sigma$ and $L_n\subseteq E''\cap F$ so
$\mu L_n=\mu^*(L_n\setminus F)=0$;
similarly, $\mu L'_n=0$.   We conclude that $\mu K=0$.   As $K$ is arbitrary,
$\mu(E'\cap E'')=0$;  as $\mu$ is complete,
$F\cap E''$ and $F\cap E=(E\setminus E'')\cup(F\cap E'')$ belong to $\Sigma$;
as $E$ is arbitrary and $\mu$ is locally determined, $F\in\Sigma$;
as $F$ is arbitrary, $\Tau\subseteq\Sigma$ and $\mu\le\nu$.

\medskip

{\bf (b)} By (a), we have $\mu\le\nu$ and $\nu\le\mu$, so $\mu=\nu$ (234Qa).
}%end of proof of 412M

\vleader{48pt}{412N}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K$ a family of subsets of $X$ such that $\mu$ is inner regular
with respect to $\Cal K$.   Then

\Centerline{$E^{\ssbullet}
=\sup\{K^{\ssbullet}:K\in\Cal K\cap\Sigma,\,K\subseteq E\}$}

\noindent in the measure algebra $\frak A$ of $\mu$, for every
$E\in\Sigma$.
\cmmnt{In particular, }$\{K^{\ssbullet}:K\in\Cal K\cap\Sigma\}$ is order-dense
in $\frak A$;  and if $\Cal K$ is closed under finite unions, then
$\{K^{\ssbullet}:K\in\Cal K\cap\Sigma\}$ is topologically dense in
$\frak A$ for the measure-algebra topology.

\proof{ \Quer\ If $E^{\ssbullet}
\ne\sup\{K^{\ssbullet}:K\in\Cal K\cap\Sigma,\,K\subseteq E\}$, there is
a non-zero $a\in\frak A$ such
that $a\Bsubseteq E^{\ssbullet}\Bsetminus K^{\ssbullet}$ whenever
$K\in\Cal K\cap\Sigma$ and $K\subseteq E$.   Express $a$ as
$F^{\ssbullet}$ where
$F\subseteq E$.   Then $\mu F>0$, so there is a $K\in\Cal K\cap\Sigma$
such that $K\subseteq F$ and $\mu K>0$.   But in this case
$0\ne K^{\ssbullet}\Bsubseteq a$, while $K\subseteq E$.\ \BanG\

It follows at once that $D=\{K^{\ssbullet}:K\in\Cal K\cap\Sigma\}$ is
order-dense.   If $\Cal K$ is closed under finite unions, and
$a\in\frak A$, then $D_a=\{d:d\in D$, $d\Bsubseteq a\}$ is
upwards-directed and has supremum $a$, so
$a\in\overline{D}_a\subseteq\overline{D}$ (323D(a-ii)).

}%end of proof of 412N

\leader{412O}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K$ a family of subsets of $X$ such that $\mu$ is inner regular
with respect to $\Cal K$.

(a) If $E\in\Sigma$, then the subspace measure
$\mu_E$\cmmnt{ (131B)} is inner regular with respect to $\Cal K$.

(b) Let $Y\subseteq X$ be any set such that the
subspace measure $\mu_Y$\cmmnt{ (214B)} is semi-finite.   Then
$\mu_Y$ is inner regular with respect to
$\Cal K_Y=\{K\cap Y:K\in\Cal K\}$.

\proof{{\bf (a)} This is elementary.

\medskip

{\bf (b)} Suppose that $F$ belongs to the domain $\Sigma_Y$ of
$\mu_Y$ and
$0\le\gamma<\mu_YF$.   Because $\mu_Y$ is semi-finite there is an
$F'\in\Sigma_Y$ such that $F'\subseteq F$ and $\gamma<\mu_YF'<\infty$.
Let $G\in\Sigma$ be such that $F'=G\cap Y$, and let $E\subseteq G$
be a measurable envelope for $F'$ with respect to $\mu$, so that

\Centerline{$\mu E=\mu^*F'=\mu_YF'>\gamma$.}

\noindent There is a $K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and
$\mu K\ge\gamma$, in which case $K\cap Y\in\Cal K_Y\cap\Sigma_Y$ and

\Centerline{$\mu_Y(K\cap Y)=\mu^*(K\cap Y)=\mu^*(K\cap F')
=\mu(K\cap E)=\mu K\ge\gamma$.}

\noindent As $F$ and $\gamma$ are arbitrary, $\mu_Y$ is inner regular
with respect to $\Cal K_Y$.
}%end of proof of 412O

\cmmnt{\medskip

\noindent{\bf Remark} Recall from 214Ic that if $(X,\Sigma,\mu)$ has
locally determined negligible sets (in particular, is either strictly
localizable or complete and locally
determined), then all its subspaces are semi-finite.
}%end of comment

\leader{412P}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
$\frak T$ a topology on $X$ and $Y$ a subset of $X$;  write $\frak T_Y$
for the subspace topology of $Y$ and $\mu_Y$ for the subspace measure on
$Y$.   Suppose that {\it either} $Y\in\Sigma$ {\it or}
$\mu_Y$ is semi-finite.

(a) If $\mu$ is a topological measure, so is $\mu_Y$.

(b) If $\mu$ is inner regular with respect to the Borel sets, so is
$\mu_Y$.

(c) If $\mu$ is inner regular with respect to the closed sets, so is
$\mu_Y$.

(d) If $\mu$ is inner regular with respect to the zero sets, so is
$\mu_Y$.

(e) If $\mu$ is effectively locally finite, so is $\mu_Y$.

\proof{(a) is an immediate consequence of the definitions of `subspace
measure', `subspace topology' and `topological measure'.   The other
parts follow directly from 412O if we recall that

(i) a subset of $Y$ is Borel for $\frak T_Y$ whenever it is expressible
as $Y\cap E$ for some Borel set $E\subseteq X$ (4A3Ca);

(ii) a subset of $Y$ is closed in $Y$ whenever it is expressible as
$Y\cap F$ for some closed set $F\subseteq X$;

(iii) a subset of $Y$ is a zero set in $Y$ whenever it is expressible as
$Y\cap F$ for some zero set $F\subseteq X$ (4A2C(b-v));

(iv) $\mu$ is effectively locally finite iff it is inner regular with
respect to subsets of open sets of finite measure.
}%end of proof of 412P

\leader{412Q}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$\nu$ an indefinite-integral measure over $\mu$\cmmnt{ (definition:
234J)}.   If $\mu$ is inner regular with respect to a family $\Cal K$ of
sets, so is $\nu$.

\proof{ Because $\mu$ and its completion $\hat\mu$ give the same
integrals, $\nu$ is an indefinite-integral measure over $\hat\mu$
(234Ke);  and
as $\hat\mu$ is still inner regular with respect to $\Cal K$ (412Ha), we
may suppose that $\mu$ itself is complete.   Let $f$ be a
Radon-Nikod\'ym derivative of $\nu$ with respect to $\mu$;  by
234Ka, we may suppose that $f:X\to\coint{0,\infty}$ is $\Sigma$-measurable.

Suppose that $F\in\dom\nu$ and that
$\gamma<\nu F$.   Set $G=\{x:f(x)>0\}$, so that $F\cap G\in\Sigma$
(234La).   For $n\in\Bbb N$, set
$H_n=\{x:x\in F,\,2^{-n}\le f(x)\le 2^n\}$, so that $H_n\in\Sigma$ and

\Centerline{$\nu F
=\int f\times\chi F d\mu
=\lim_{n\to\infty}\int f\times\chi H_nd\mu$.}

\noindent Let $n\in\Bbb N$ be such that
$\int f\times\chi H_nd\mu>\gamma$.

If $\mu H_n=\infty$, there is a $K\in\Cal K$ such that $K\subseteq H_n$
and $\mu K\ge 2^n\gamma$, so that
$\nu K\ge\gamma$.   If $\mu H_n$ is finite, there is a
$K\in\Cal K$ such that $K\subseteq H_n$ and
$2^n(\mu H_n-\mu K)\le\int f\times\chi H_nd\mu-\gamma$, so that
$\int f\times\chi(H_n\setminus K)d\mu+\gamma\le\int f\times\chi H_nd\mu$ and
$\nu K=\int f\times\chi K\,d\mu\ge\gamma$.
Thus in either case we have a $K\in\Cal K$ such that $K\subseteq F$ and
$\nu K\ge\gamma$;  as $F$ and $\gamma$ are arbitrary, $\nu$ is inner
regular with respect to $\Cal K$.
}%end of proof of 412Q

\leader{412R}{Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, with c.l.d.\ product
$(X\times Y,\Lambda,\lambda)$\cmmnt{ (251F)}.   Suppose that
$\Cal K\subseteq\Cal PX$, $\Cal L\subseteq\Cal PY$ and
$\Cal M\subseteq\Cal P(X\times Y)$ are such that

\quad(i) $\mu$ is inner regular with respect to $\Cal K$;

\quad(ii) $\nu$ is inner regular with respect to $\Cal L$;

\quad(iii) $K\times L\in\Cal M$ for all $K\in\Cal K$, $L\in\Cal L$;

\quad(iv) $M\cup M'\in\Cal M$ whenever $M$, $M'\in\Cal M$;

\quad(v) $\bigcap_{n\in\Bbb N}M_n\in\Cal M$ for every sequence
$\sequencen{M_n}$ in $\Cal M$.

\noindent Then $\lambda$ is inner regular with respect to $\Cal M$.

\proof{ Write
$\Cal A=\{E\times Y:E\in\Sigma\}\cup\{X\times F:F\in\Tau\}$.   Then the
$\sigma$-algebra of subsets of $X\times Y$ generated by $\Cal A$ is
$\Sigma\tensorhat\Tau$.   If
$V\in\Cal A$, $W\in\Lambda$ and $\lambda(W\cap V)>0$, there is an
$M\in\Cal M\cap(\Sigma\tensorhat\Tau)$ such that $M\subseteq W$ and
$\lambda(M\cap V)>0$.   \Prf\ Suppose that $V=E\times Y$ where
$E\in\Sigma$.   There must be $E_0\in\Sigma$ and $F_0\in\Tau$, both of
finite measure, such that $\lambda(W\cap V\cap(E_0\times F_0))>0$
(251F).   Now there are $K\in\Cal K\cap\Sigma$ and $L\in\Cal L\cap\Tau$
such that $K\subseteq E\cap E_0$, $L\subseteq F\cap F_0$ and

\Centerline{$\mu((E\cap E_0)\setminus K)\cdot\nu F_0
+\mu E_0\cdot\nu((F\cap F_0)\setminus L)
<\lambda(W\cap V\cap(E_0\times F_0))$;}

\noindent but this means that
$M=K\times L$ is included in $V$ and $\mu(W\cap M)>0$, while
$M\in\Cal M\cap(\Sigma\tensorhat\Tau)$.   Reversing the
roles of the coordinates, the same argument deals with the case in which
$V=X\times F$ for some $F\in\Tau$.\ \Qed

By 412C, $\lambda\restr\Sigma\tensorhat\Tau$ is inner regular
with respect to $\Cal M$.   But $\lambda$ is inner regular with respect
to $\Sigma\tensorhat\Tau$ (251Ib) so is also inner regular
with respect to $\Cal M$ (412Ab).
}%end of proof of 412R

\leader{412S}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with c.l.d.\ product
$(X\times Y,\Lambda,\lambda)$.   Let $\frak T$, $\frak S$ be topologies
on $X$ and $Y$ respectively, and give $X\times Y$ the product topology.

(a) If $\mu$ and $\nu$ are inner regular with respect to the closed
sets, so is $\lambda$.

(b) If $\mu$ and $\nu$ are tight\cmmnt{ (that is, inner regular with
respect to the closed compact sets)}, so is $\lambda$.

(c) If $\mu$ and $\nu$ are inner regular with respect to the zero sets,
so is $\lambda$.

(d) If $\mu$ and $\nu$ are inner regular with respect to the Borel sets,
so is $\lambda$.

(e) If $\mu$ and $\nu$ are effectively locally finite, so is $\lambda$.

\proof{ We have only to read the conditions (i)-(v) of 412R carefully
and check that they apply in each case.   (In part (e), recall that
`effectively locally finite' is the same thing as `inner regular with
respect to the subsets of open sets of finite measure'.)
}%end of proof of 412S

\leader{412T}{Lemma} Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of probability spaces, with product probability space
$(X,\Lambda,\lambda)$\cmmnt{ (\S254)}.   Suppose that
$\Cal K_i\subseteq\Cal PX_i$, $\Cal M\subseteq\Cal PX$ are such that

\quad (i) $\mu_i$ is inner regular with respect to $\Cal K_i$ for each
$i\in I$;

\quad (ii) $\pi_i^{-1}[K]\in\Cal M$ for every $i\in I$ and $K\in\Cal K_i$,
writing $\pi_i(x)=x(i)$ for $x\in X$;

\quad (iii) $M\cup M'\in\Cal M$ whenever $M$, $M'\in\Cal M$;

\quad (iv) $\bigcap_{n\in\Bbb N}M_n\in\Cal M$ for every sequence
$\sequencen{M_n}$ in $\Cal M$.

\noindent Then $\lambda$ is inner regular with respect to $\Cal M$.

\proof{ (Compare 412R.)
Write $\Cal A=\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$.   If
$V\in\Cal A$, $W\in\Lambda$ and $\lambda(W\cap V)>0$, express $V$ as
$\pi_i^{-1}[E]$, where $i\in I$ and $E\in\Sigma_i$, and take
$K\in\Cal K_i$ such that $K\subseteq E$ and
$\mu_i(E\setminus K)<\lambda(W\cap V)$;  then $M=\pi_i^{-1}[K]$ belongs
to $\Cal M\cap\Cal A$, is included in $V$, and meets $W$ in a
non-negligible set.   So the conditions of 412C are met by $\Cal A$ and
$\Cal M$.

It follows that $\lambda_0=\lambda\restr\Tensorhat_{i\in I}\Sigma_i$ is
inner regular with respect to $\Cal M$.   But $\lambda$ is the
completion of $\lambda_0$ (254Fd, 254Ff), so is also inner regular with
respect to $\Cal M$ (412Ha).
}%end of proof of 412T

\leader{412U}{Proposition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces, with product probability space
$(X,\Lambda,\lambda)$.   Suppose that we are given a topology
$\frak T_i$ on each $X_i$, and let $\frak T$ be the product topology on
$X$.

(a) If every $\mu_i$ is inner regular with respect to the closed sets,
so is $\lambda$.

(b) If every $\mu_i$ is inner regular with respect to the zero sets, so
is $\lambda$.

(c) If every $\mu_i$ is inner regular with respect to the Borel sets, so
is $\lambda$.

\proof{ This follows from 412T just as 412S follows from 412R.
}%end of proof of 412U

\leader{412V}{Corollary} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces, with product probability space
$(X,\Lambda,\lambda)$.   Suppose that we are given a Hausdorff topology
$\frak T_i$ on each $X_i$, and let $\frak T$ be the product topology on
$X$.   Suppose that every $\mu_i$ is tight, and that $X_i$ is compact
for all but countably many $i\in I$.   Then $\lambda$ is tight.

\proof{ By 412Ua, $\lambda$ is inner regular with respect to the closed
sets.   If $W\in\Lambda$ and $\gamma<\lambda W$, let $V\subseteq W$ be a
measurable closed set such that $\lambda V>\gamma$.   Let $J$
be the set of those $i\in I$ such that $X_i$ is not compact;  we are
supposing that $J$ is countable.   Let $\family{i}{J}{\epsilon_i}$ be a
family of strictly positive real numbers such that
$\sum_{i\in J}\epsilon_j\le\lambda V-\gamma$ (4A1P).   For each
$i\in J$, let $K_i\subseteq X_i$ be a compact measurable set such that
$\mu_i(X_i\setminus K_i)\le\epsilon_i$;  and for
$i\in I\setminus J$, set $K_i=X_i$.   Then
$K=\prod_{i\in I}K_i$ is a compact measurable subset of $X$, and

\Centerline{$\lambda(X\setminus K)
\le\sum_{i\in J}\mu_i(X_i\setminus K_i)\le\lambda V-\gamma$,}

\noindent so $\lambda(K\cap V)\ge\gamma$;  while $K\cap V$ is a compact
measurable subset of $W$.   As $W$ and $\gamma$ are arbitrary, $\lambda$
is tight.
}%end of proof of 412V

\leader{*412W}{Outer \dvrocolon{regularity}}\cmmnt{ I have already
mentioned the complementary notion of `outer regularity' (411D).   In
this book it will not be given much prominence.   It is however a useful
tool when dealing with Lebesgue measure (see, for instance, the proof of
225K), for reasons which the next proposition will make clear.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$.

(a) Suppose that $\mu$ is outer regular with respect to the open sets.
Then for any integrable function $f:X\to[0,\infty]$ and $\epsilon>0$,
there is a lower semi-continuous measurable function
$g:X\to[0,\infty]$ such that $f\le g$ and $\int g\le\epsilon+\int f$.

(b) Now suppose that there is a sequence of measurable open sets of
finite measure covering $X$.   Then the following are equiveridical:

\quad(i) $\mu$ is inner regular with respect to the closed sets;

\quad(ii) $\mu$ is outer regular with respect to the open sets;

\quad(iii) for any measurable set $E\subseteq X$ and $\epsilon>0$, there
are a measurable closed set $F\subseteq E$ and a measurable open set
$H\supseteq E$ such that $\mu(H\setminus F)\le\epsilon$;

\quad(iv) for every measurable function $f:X\to\coint{0,\infty}$ and
$\epsilon>0$, there is a lower semi-continuous measurable function
$g:X\to[0,\infty]$ such that $f\le g$ and $\int g-f\le\epsilon$;

\quad(v) for every measurable function $f:X\to\Bbb R$ and $\epsilon>0$,
there is a lower semi-continuous measurable function
$g:X\to\ocint{-\infty,\infty}$ such that $f\le g$ and
$\mu\{x:g(x)\ge f(x)+\epsilon\}\le\epsilon$.

\proof{{\bf (a)} Let $\eta\in\ocint{0,1}$ be such that
$\eta(9+\int fd\mu)\le\epsilon$.
For $n\in\Bbb Z$, set $E_n=\{x:(1+\eta)^n\le f(x)<(1+\eta)^{n+1}\}$, and
let $E'_n\in\Sigma$ be a measurable envelope of $E_n$;  let
$G_n\supseteq E'_n$ be a measurable open set such that
$\mu G_n\le 3^{-|n|}\eta+\mu E'_n$.
Next, let $E$ be a measurable envelope of the negligible set
$\{x:f(x)=\infty\}$, and for each $n$ let $H_n\supseteq E$ be a measurable open
set such that $\mu H_n\le 2^{-n}$.   Set

\Centerline{$g=\sum_{n=-\infty}^{\infty}(1+\eta)^{n+1}\chi G_n
+\sum_{n=0}^{\infty}\chi H_n$.}

\noindent Then $g$ is lower
semi-continuous (4A2B(d-iii), 4A2B(d-v)), $f\le g$ and

$$\eqalign{\int g\,d\mu
&=\sum_{n=-\infty}^{\infty}(1+\eta)^{n+1}\mu G_n+\sum_{n=0}^{\infty}\mu H_n\cr&
\le(1+\eta)\sum_{n=-\infty}^{\infty}(1+\eta)^n\mu E'_n
   +\sum_{n=-\infty}^{\infty}(1+\eta)^{n+1}3^{-|n|}\eta
   +\sum_{n=0}^{\infty}2^{-n}\eta\cr
&\le(1+\eta)\int fd\mu+9\eta
\le\int fd\mu+\epsilon,\cr}$$

\noindent as required.

\medskip

{\bf (b)} Let $\sequencen{G_n}$ be a sequence of open sets of finite
measure covering $X$;  replacing it by $\sequencen{\bigcup_{i<n}G_i}$ if
necessary, we may suppose that $\sequencen{G_n}$ is non-decreasing and
that $G_0=\emptyset$.

\medskip

\quad{\bf (i)$\Rightarrow$(iii)} Suppose that $\mu$ is inner regular
with respect to the closed sets, and that $E\in\Sigma$, $\epsilon>0$.
For each $n\in\Bbb N$ let $F_n\subseteq G_n\setminus E$ be a measurable
closed set such that $\mu F_n\ge\mu(G_n\setminus E)-2^{-n-2}\epsilon$.
Then $H=\bigcup_{n\in\Bbb N}(G_n\setminus F_n)$ is a measurable open set
including $E$ and $\mu(H\setminus E)\le\bover12\epsilon$.   Applying the
same argument to $X\setminus E$, we get a closed set $F\subseteq E$ such
that $\mu(E\setminus F)\le\bover12\epsilon$, so that
$\mu(H\setminus F)\le\epsilon$.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} The same idea works.   Suppose that
$\mu$ is outer regular with respect to the open sets, and that
$E\in\Sigma$, $\epsilon>0$.    For each $n\in\Bbb N$, let
$H_n\supseteq G_n\cap E$ be an open set such that
$\mu(H_n\setminus E)\le 2^{-n-2}\epsilon$;  then
$H=\bigcup_{n\in\Bbb N}H_n$ is a measurable open set including $E$, and
$\mu(H\setminus E)\le\bover12\epsilon$.   Now repeat the argument on
$X\setminus E$ to find a measurable closed set $F\subseteq E$ such that
$\mu(E\setminus F)\le\bover12\epsilon$.

\medskip

\quad{\bf (iii)$\Rightarrow$(iv)} Assume (iii), and let
$f:X\to\coint{0,\infty}$ be a measurable function, $\epsilon>0$.   Set
$\eta_n=2^{-n}\epsilon/(16+4\mu G_n)$ for each $n\in\Bbb N$.   For
$k\in\Bbb N$ set
$E_k=\bigcup_{n\in\Bbb N}\{x:x\in G_n,\,k\eta_n\le f(x)<(k+1)\eta_n\}$,
and choose an open set $H_k\supseteq E_k$ such that
$\mu(H_k\setminus E_k)\le 2^{-k}$.    Set

\Centerline{$g=\sup_{k,n\in\Bbb N}(k+1)\eta_n\chi(G_n\cap H_k)$.}

\noindent Then $g:X\to[0,\infty]$ is lower semi-continuous (4A2B(d-v)
again).   Since

\Centerline{$\sup_{k,n\in\Bbb N}k\eta_n\chi(G_n\cap E_k)
\le f\le\sup_{k,n\in\Bbb N}(k+1)\eta_n\chi(G_n\cap E_k)$,}

\noindent $f\le g$ and

\Centerline{$g-f
\le\sup_{k,n\in\Bbb N}(k+1)\eta_n\chi(G_n\cap H_k\setminus E_k)
  +\sup_{k,n\in\Bbb N}\eta_n\chi(G_n\cap E_k)$}

\noindent has integral at most

\Centerline{$\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}(k+1)\eta_n2^{-k}
  +\sum_{n=0}^{\infty}\eta_n\mu G_n
\le\epsilon$.}

\medskip

\quad{\bf (i)$\Rightarrow$(v)} Assume (i), and suppose that
$f:X\to\Bbb R$ is
measurable and $\epsilon>0$.   For each $n\in\Bbb N$, let
$\alpha_n\ge 0$ be such that $\mu E_n<2^{-n-1}\epsilon$, where
$E_n=\{x:x\in G_{n+1}\setminus G_n,\,f(x)\le -\alpha_n\}$.   Let
$F_n\subseteq(G_{n+1}\setminus G_n)\setminus E_n$ be a measurable closed
set such that
$\mu((G_{n+1}\setminus G_n)\setminus F_n)\le 2^{-n-2}\epsilon$.
Because $\sequencen{F_n}$ is disjoint,
$h=\sum_{n=0}^{\infty}\alpha_n\chi F_n$ is defined as a function from
$X$ to $\coint{0,\infty}$.   $\{F_n:n\in\Bbb N\}$ is locally finite, so
$\{x:h(x)\ge\alpha\}=\bigcup_{n\in\Bbb N,\alpha_n\ge\alpha}F_n$ is
closed for every $\alpha>0$ (4A2B(h-i)), and $h$ is upper
semi-continuous.   Now $f_1=f+h$ is a real-valued measurable function.   Since
(i)$\Rightarrow$(iii)$\Rightarrow$(iv), there is a measurable lower
semi-continuous function $g_1:X\to[0,\infty]$ such that
$f_1^+\le g_1$ and $\int g_1-f_1^+\le\bover12\epsilon^2$, where
$f_1^+=\max(0,f_1)$.   But if we now set
$g=g_1-h$, $g$ is lower semi-continuous, $f\le g$ and

$$\eqalign{\{x:f(x)+\epsilon\le g(x)\}
&\subseteq\{x:f_1^+(x)+\epsilon\le g_1(x)\}
  \cup\{x:f_1(x)<0\}\cr
&\subseteq\{x:f_1^+(x)+\epsilon\le g_1(x)\}
  \cup\bigcup_{n\in\Bbb N}(G_{n+1}\setminus G_n)\setminus F_n\cr}$$

\noindent has measure at most $\epsilon$, as required.

\medskip

\quad{\bf (iv)$\Rightarrow$(ii)} and {\bf (v)$\Rightarrow$(ii)} Suppose
that either (iv) or (v) is true, and that $E\in\Sigma$, $\epsilon>0$.
Then there is a measurable lower semi-continuous function
$g:X\to\ocint{0,\infty}$ such that $\chi E\le g$ and
$\mu\{x:\chi E(x)+\bover12\le g(x)\}\le\epsilon$, since this is
certainly true if $\int g-\chi E\le\bover12\epsilon$.   Set
$G=\{x:g(x)>\bover12\}$;  then $E\subseteq G$ and
$\mu(G\setminus E)\le\epsilon$.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} is trivial.   Assembling these
fragments, the proof is complete.
}%end of proof of 412W


\exercises{
\leader{412X}{Basic exercises (a)}
%\spheader 412Xa
Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\sequencen{f_n}$ a sequence in $\eusm L^0(\mu)$ which converges almost
everywhere to $f\in\eusm L^0(\mu)$.   Show that $\mu$ is inner regular
with respect to $\{E:\sequencen{f_n\restr E}$ is uniformly convergent$\}$.
(Cf.\ 215Yb.)
%412-

\spheader 412Xb
Let $(X,\Sigma,\mu)$ be a measure space and $\frak T$ a
topology on $X$ such that $\mu$ is inner regular with respect to the
closed sets {\it and} with respect to the compact sets.   Show that $\mu$
is tight.
%412A

\spheader 412Xc Explain how 213A is a special case of 412Aa.
%412A

\spheader 412Xd Let $X$ be a set and $\Cal K$ a family of sets.
Suppose that $\mu$ and $\nu$ are two semi-finite measures on $X$
with the same domain and the same null ideal.   Show that if one is
inner regular with respect to $\Cal K$, so is the other.    \Hint{show that if
$\nu F<\infty$ then $\nu F=\sup\{\nu E:E\subseteq F$, $\mu E<\infty\}$.}
%412A

\sqheader 412Xe Let $(X,\Sigma,\mu)$ be a measure space, and $\Sigma_0$
a $\sigma$-subalgebra of $\Sigma$ such that $\mu$ is inner regular with
respect to $\Sigma_0$.   Show that if $1\le p<\infty$ then every member
of $L^p(\mu)$ is of the form $f^{\ssbullet}$ for some
$\Sigma_0$-measurable $f:X\to\Bbb R$.
%412Xc, 412A

\spheader 412Xf Let $\familyiI{\mu_i}$ be a family of measures on a set
$X$, with sum $\mu$ (234G).   Suppose that $\Cal K\subseteq\dom\mu$ is a
family of sets such that $K\cup K'\in\Cal K$ whenever $K$, $K'\in\Cal K$
are disjoint and every $\mu_i$ is inner regular with respect to
$\Cal K$.   Show that $\mu$ is inner regular with respect to
$\Cal K$.
%412A

\spheader 412Xg Let $X$ be a topological space, and $\mu$ a tight topological
measure on $X$.   Suppose that $\Cal F$ is a
non-empty downwards-directed family of
closed compact subsets of $X$ with intersection $F_0$, and that
$\gamma=\inf_{F\in\Cal F}\mu F$ is finite.   Show that $\mu F_0=\gamma$.
%412A

\sqheader 412Xh Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\Cal A\subseteq\Sigma$ an algebra of sets such that the
$\sigma$-algebra generated by $\Cal A$ is $\Sigma$.   Write $\Cal K$ for
$\{\bigcap_{n\in\Bbb N}E_n:E_n\in\Cal A$ for every $n\in\Bbb N\}$.
Show that $\mu$ is inner regular with respect to $\Cal K$.
%412C

\spheader 412Xi Let $(X,\frak T,\Sigma,\mu)$ be an effectively locally
finite Hausdorff topological measure space such that $\mu$ is inner
regular with respect to the Borel sets.   Suppose that
$\mu G=\sup\{\mu K:K\subseteq G$ is compact$\}$ for every open set
$G\subseteq X$.   Show that $\mu$ is tight.
%412G

\spheader 412Xj Let $(X,\frak T)$ be a topological space such that every
open set is an F$_{\sigma}$ set.   Show that any effectively locally
finite Borel measure on $X$ is inner regular with respect to the closed
sets.
%412G

\spheader 412Xk Let $(X,\frak T)$ be a normal topological space and
$\mu$ a topological measure on $X$ which is inner regular with respect
to the closed sets.   Show that
$\mu G=\max\{\mu H:H\subseteq G$ is a cozero set$\}$ for every open set
$G\subseteq X$.   Show that if $\mu$ is totally finite, then
$\mu F=\min\{\mu H:H\supseteq F$ is a zero set$\}$ for every closed set
$F\subseteq X$.
%412G

\spheader 412Xl Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, and suppose that $\mu$ is inner regular with respect to a
family $\Cal K$ of sets.   Let $\Sigma_0$ be the $\sigma$-algebra of
subsets of
$X$ generated by $\Cal K\cap\Sigma$.   (i) Show that $\mu$ is the
c.l.d.\ version of $\mu\restr\Sigma_0$.   \Hint{412J-412M.}   (ii) Show
that if $\mu$ is
$\sigma$-finite, it is the completion of $\mu\restr\Sigma_0$.
%412M

\sqheader 412Xm(i) Let $(X,\Sigma,\mu)$ be a $\sigma$-finite
measure space and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Show that
if $\mu$ is inner regular
with respect to $\Tau$ then the completion of $\mu\restr\Tau$ extends
$\mu$, so that $\mu$ and $\mu\restr\Tau$ have the same negligible sets.
(ii) Show that if $\mu$ is a $\sigma$-finite topological measure which
is inner regular
with respect to the Borel sets, then every $\mu$-negligible set is
included in a $\mu$-negligible Borel set.
%412Xl 412M

\spheader 412Xn Devise a direct proof of 412Mb by (i)
showing that $\mu^*(A\cap K)=\nu^*(A\cap K)$ whenever $A\subseteq X$ and
$K\in\Cal K$ (ii) showing that $\mu^*=\nu^*$ (iii) quoting 213C.
%412M

\spheader 412Xo Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, $Y$ a set and $f:X\to Y$ a function.   Show that the
following are equiveridical:  (i) $\mu$ is inner regular with respect to
$\{f^{-1}[B]:B\subseteq Y\}$ (ii) $f^{-1}[f[E]]\setminus E$ is
negligible for every $E\in\Sigma$.
%412L

\spheader 412Xp Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
measure spaces, with direct sum $(X,\Sigma,\mu)$.   Suppose that for
each $i\in I$ we are given a topology $\frak T_i$ on $X_i$, and let
$\frak T$ be
the corresponding disjoint union topology on $X$.   Show that (i) $\mu$
is inner regular with respect to the closed sets iff every $\mu_i$ is
(ii) $\mu$ is inner regular with respect to the compact sets iff every
$\mu_i$ is (iii) $\mu$ is inner regular with respect to the zero sets
iff every $\mu_i$ is (iv) $\mu$ is inner regular with respect to the
Borel sets iff every $\mu_i$ is.
%412P

\spheader 412Xq Use 412M and 412Q to shorten the proof of 253I.
%412Q

\spheader 412Xr Let $\familyiI{X_i}$ be a family of sets, and
suppose that we are given, for each $i\in I$, a $\sigma$-algebra
$\Sigma_i$ of subsets of $X_i$ and a topology $\frak T_i$ on $X_i$.
Let $\frak T$ be the product topology on $X=\prod_{i\in I}X_i$, and
$\Sigma=\Tensorhat_{i\in I}\Sigma_i$.   Let $\mu$ be a totally finite
measure with domain
$\Sigma$, and set $\mu_i=\mu\pi_i^{-1}$ for each $i\in I$, where
$\pi_i(x)=x(i)$ for $i\in I$, $x\in X$.   (i) Show that $\mu$ is
inner regular with respect to the family $\Cal K$ of sets expressible as
$X\setminus\bigcup_{n\in\Bbb N}\prod_{i\in I}E_{ni}$ where
$E_{ni}\in\Sigma_i$ for every $n$, $i$ and $\{i:E_{ni}\ne X_i\}$ is
finite for each $n$.
(ii) Show that if every $\mu_i$ is inner regular with respect to the
closed sets, so is $\mu$.
(iii) Show that if every $\mu_i$ is inner regular with respect to the
zero sets, so is $\mu$.
(iv) Show that if every $\mu_i$ is inner regular with respect to the
Borel sets, so is $\mu$.
(v) Show that if every $\mu_i$ is tight, and all but countably many of
the $X_i$ are compact, then $\mu$ is tight.
%412V 412C

\spheader 412Xs Let $(X,\Sigma,\mu)$ be a measure space and $\frak T$ a
Lindel\"of topology on $X$ such that $\mu$ is locally finite.   (i) Show
that $\mu$ is
$\sigma$-finite.   (ii) Show that $\mu$ is inner regular with respect to
the closed sets iff it is outer regular with respect to the open sets.
%412W

\spheader 412Xt Let $X$ be a topological space and $\mu$ a measure on
$X$ which is outer regular with respect to the open sets.   Show that
for any $Y\subseteq X$ the subspace measure on $Y$ is outer regular with
respect to the open sets.
%412W

\spheader 412Xu Let $X$ be a topological space and $\mu$ a measure on
$X$ which is inner regular with respect to the closed sets and
outer regular with respect to the open sets.   Show that if
$f:X\to[-\infty,\infty]$ is integrable and $\epsilon>0$ then there is an
integrable lower
semi-continuous $g:X\to\ocint{-\infty,\infty}$ such that $f\le g$ and
$\int g\le\epsilon+\int f$.
%412W

\sqheader 412Xv Let $X$ be a topological space and $\mu$ a measure on $X$ which is
effectively locally finite and inner regular with respect to the closed sets.
(i) Show that if $\mu E<\infty$ and $\epsilon>0$ there is a measurable open set $G$
such that $\mu(E\symmdiff G)\le\epsilon$.   (ii) Show that if $f$ is a
non-negative integrable function and $\epsilon>0$ there is a measurable lower
semi-continuous function $g:X\to\coint{0,\infty}$ such that $\int|f-g|\le\epsilon$.
(iii) Show that if $f$ is an integrable
real-valued function there are measurable lower semi-continuous functions
$g_1$, $g_2:X\to[0,\infty]$ such that $f\eae g_1-g_2$ and
$\int g_1+g_2\le\int|f|+\epsilon$.   (iv) Now suppose that $\mu$ is $\sigma$-finite.
Show that for every measurable $f:X\to\Bbb R$ there are measurable lower
semi-continuous functions $g_1$, $g_2:X\to[0,\infty]$ such that $f\eae g_1-g_2$.
%412W

\leader{412Y}{Further exercises (a)}
%\spheader 412Ya
In 216E, give $\{0,1\}^I$ its usual compact Hausdorff
topology.   Show that the measure $\mu$ described there is inner regular
with respect to the zero sets.
%412-

\spheader 412Yb
Let $\Cal K$ be the family of subsets of $\Bbb R$ which are homeomorphic
to the Cantor set.   Show that Lebesgue measure is inner regular with
respect to $\Cal K$.
%\Hint{show that if
%$F\subseteq\Bbb R\setminus\Bbb Q$ is an uncountable compact set, then
%$\{x:[x-\delta,x+\delta]\cap F$ is uncountable for every $\delta>0\}$
%belongs to $\Cal K$.}
%-%

\spheader 412Yc(i) Show that if $X$ is a perfectly normal space
then any semi-finite topological measure on $X$ which is inner
regular with respect to the Borel sets is inner regular
with respect to the closed sets.   (ii) Show that any subspace of a
perfectly normal space is perfectly normal.  (iii) Show
that $\omega_1$, with its order topology, is completely regular, normal
and Hausdorff, but
not perfectly normal.   (iv) Show that $[0,1]^I$ is perfectly normal iff
$I$ is countable.
%412E

\spheader 412Yd\dvAnew{2016} Give an example of a measure space $(X,\Sigma,\mu)$
and a family $\Cal K$ of sets such that

\inset{($\ddagger$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ whenever
$\sequencen{K_n}$ is a non-increasing sequence in $\Cal K$}

\noindent and the completion of $\mu$ is inner regular with respect to $\Cal K$,
but $\mu$ is not.
%412H mt41bits

\spheader 412Ye Let $(X,\Sigma,\mu)$ be a measure space, and suppose
that $\mu$ is inner regular with respect to
$\Cal K\subseteq\Cal PX$.
Write $\Sigma^f$ for $\{E:E\in\Sigma$, $\mu E<\infty\}$.    Show that
$\{E^{\ssbullet}:E\in\Cal K\cap\Sigma^f\}$ is dense in
$\{E^{\ssbullet}:E\in\Sigma^f\}$ for the strong measure-algebra
topology.
%412N

\spheader 412Yf Let $(X,\Sigma,\mu)$ be $[0,1]$ with Lebesgue measure,
and $Y=[0,1]$ with counting measure $\nu$;  give $X$ its usual topology
and $Y$ its discrete topology, and let $\lambda$ be the c.l.d.\ product
measure on $X\times Y$.   (i) Show that $\mu$, $\nu$ and $\lambda$ are
all tight (for the appropriate
topologies) and therefore completion regular.
(ii) Let $\lambda_0$ be the primitive product measure on
$X\times Y$ (definition:  251C).   Show that $\lambda_0$ is not tight.
\Hint{252Yk.}
{\it Remark\/}:  it is undecidable in ZFC whether $\lambda_0$ is inner
regular with respect to the closed sets.
%412S %mt41bits

\spheader 412Yg Give an example of a Hausdorff topological measure space
$(X,\frak T,\Sigma,\mu)$ such that $\mu$ is complete, strictly
localizable and outer regular with respect to the open sets, but not
inner regular with respect to the closed sets.
%412W mt41bits

}%end of exercises

\endnotes{\Notesheader{412} In this volume we are returning to
considerations which have been left on one side for almost the whole of
Volume 3 -- the exceptions being in Chapter 34, where I looked at
realization of homomorphisms of measure algebras by functions between
measure spaces, and was necessarily dragged into an investigation of
measure spaces which had enough points to be adequate codomains (343B).
The idea of `inner regularity' is to distinguish families $\Cal K$ of
sets which will be large enough to describe the measure entirely, but
whose members will be of familiar types.   For an example of this
principle see 412Yb.   Of course we cannot always find a single type of
set adequate to fill a suitable family $\Cal K$, though this happens
oftener than one might expect, but it is surely easier to think about an
arbitrary zero set (for instance) than an arbitrary measurable set, and
whenever a measure is inner regular with respect to a recognisable class
it is worth knowing about it.

I have tried to use the symbols $\dagger$ and $\ddagger$ (412A, 412C)
consistently enough for them to act as a guide to some of the ideas
which will be used repeatedly in this chapter.   Note the emphasis on
disjoint unions and countable intersections;  I mentioned
similar conditions in 136Xi-136Xj.
You will recognise 412Aa as an exhaustion principle;  observe that it is
enough to use disjoint unions, as in 313K.   In the examples of this
section this disjointness is not important.   Of course inner regularity
has implications for the measure algebra (412N), but it is important to
recognise that `$\mu$ is inner regular with respect to $\Cal K$' is
saying much more than `$\{K^{\ssbullet}:K\in\Cal K\}$ is order-dense in
the measure algebra';  the latter formulation tells us only that
whenever
$\mu E>0$ there is a $K\in\Cal K$ such that $K\setminus E$ is negligible
and $\mu K>0$, while the former tells us that we can take $K$ to be
actually a subset of $E$.

412D, 412E and 412G are all of great importance.   412D looks striking,
but of course the reason it works is just that the Baire
$\sigma$-algebra
is very small.   In 412E the Baire and Borel
$\sigma$-algebras coincide, so it is nothing but a special case of 412D;
but as metric spaces are particularly important it is worth having it
spelt out explicitly.   In 412D and 412E the hypothesis `semi-finite' is
sufficient, while in 412G we need `effectively locally finite';  this is
because in both 412D and 412E the open sets we are looking at are
countable unions of measurable closed sets.   There are interesting
non-metrizable spaces in
which the same thing happens (412Yc).   As you know, I am strongly
biased in favour of complete and locally determined measures, and
the Baire and Borel measures dealt with in these three results are
rarely complete;  but they can still be applied to completions and
c.l.d.\ versions of these measures, using 412Ab or 412H.

412O-412V are essentially routine.   For subspace measures, the only
problem we need to come to terms with is the fact that subspaces of
semi-finite measure spaces need not be semi-finite (216Xa).   For
product measures the point is that the c.l.d.\ product of two measure
spaces, and the product of any family of probability spaces, as I
defined them in Chapter 25, are inner regular with respect to the
$\sigma$-algebra
of sets generated by the cylinder sets.   This is not in general true of
the `primitive' product measure (412Yf), which is one of my reasons for
being prejudiced against it.   I should perhaps warn you of a trap in
the language I use here.   I say that if the factor measures are inner
regular with respect to the closed sets, so is the c.l.d.\ product
measure.   But I
do not say that all closed sets in the product are measured by the
product measure, even if closed sets in the factors are measured by
the factor measures.   So the path is open for a different product
measure to exist, still inner regular with respect to the closed sets;
and indeed I
shall be going down that path in \S417.   The uniqueness result in
412M specifically refers to complete locally determined measures defined
on all sets of the family $\Cal K$.

There is one special difficulty in 412V:  in order to ensure that there
are enough compact measurable sets in $X=\prod_{i\in I}X_i$, we need to
know that all but countably many of the $X_i$ are actually compact.
When we come to look more closely at products of Radon probability
spaces we shall need to consider this point again (417Q, 417Xq).

In fact some of the ideas of 412U-412V are not restricted to the product
measures considered there.   Other measures on the product space will
have inner regularity properties if their images on the factors, their
`marginals' in the language of probability theory, are inner regular;
see 412Xr.   I will return to this in \S454.

This section is almost exclusively concerned with {\it inner}
regularity.   The complementary notion of {\it outer} regularity is not
much use except in $\sigma$-finite spaces (415Xi), and not always then
(416Ya).   In totally finite spaces, of course, and some others, any
version of inner regularity corresponds to a version of outer
regularity, as in 412Wb(i)-(ii);  and when we have something as strong
as 412Wb(iii) available it is worth knowing about it.
}%end of notes

\discrpage

\frfilename{mt413.tex}
\versiondate{1.1.17}
\copyrightdate{2000}

\def\chaptername{Topologies and measures I}
\def\sectionname{Inner measure constructions}

\newsection{413}

I now turn in a different direction, giving some basic results on
the construction of inner regular measures.   The first step is to
describe `inner measures' (413A) and a construction corresponding to the
\Caratheodory\ construction of measures from outer measures (413C).
Just as every measure gives rise to an outer measure, it gives rise to
an inner measure (413D).   Inner measures form an effective tool for
studying complete locally determined measures (413F).

The most substantial results of the section concern the construction of
measures as extensions of functionals defined on various classes
$\Cal K$ of sets.   Typically, $\Cal K$ is closed under finite unions
and countable
intersections, though we can sometimes relax the hypotheses a bit.
The methods here make it possible to distinguish
arguments which produce finitely additive functionals (413I, 413O, 413R,
413S) from
the succeeding steps to countably additive measures (413J, 413P, 413U).
413I-413N investigate conditions on a functional
$\phi:\Cal K\to\coint{0,\infty}$ sufficient to produce a measure
extending $\phi$, necessarily
unique, which is inner regular with respect to $\Cal K$ or
$\Cal K_{\delta}$, the
set of intersections of sequences in $\Cal K$.   413O-413P look instead
at functionals defined on sublattices of the class $\Cal K$ of interest,
and at sufficient conditions to ensure the existence of a measure, not
normally unique, defined on the whole of $\Cal K$, inner regular with
respect to
$\Cal K$ and extending the given functional.   Finally, 413R-413U are
concerned with majorizations rather than extensions;  we seek a measure
$\mu$ such that $\mu K\ge\phi K$ for $K\in\Cal K$, while $\mu X$ is
as small as possible.

\leader{413A}{}\cmmnt{ I begin with some material from the exercises
of earlier volumes.

\medskip

\noindent}{\bf Definition} Let $X$ be a set.   An {\bf inner measure} on
$X$ is a functional $\phi:\Cal PX\to[0,\infty]$ such that

\inset{$\phi\emptyset=0$;}

\inset{($\alpha$) $\phi(A\cup B)\ge\phi A+\phi B$ for all disjoint $A$,
$B\subseteq X$;}

\inset{($\beta$) if $\sequencen{A_n}$ is a non-increasing sequence of
subsets of $X$ and $\phi A_0<\infty$ then
$\phi(\bigcap_{n\in\Bbb N}A_n)=\inf_{n\in\Bbb N}\phi A_n$;}

\inset{($*$) $\phi A=\sup\{\phi B:B\subseteq A,\,\phi B<\infty\}$ for
every $A\subseteq X$.}

\leader{413B}{}\cmmnt{ The following fact will be recognised as an
element of \Caratheodory's method.   There will be an application later
in which it
will be useful to know that it is not confined to proving countable
additivity.

\medskip

\noindent}{\bf Lemma} Let $X$ be a set and $\phi:\Cal PX\to[0,\infty]$ any
functional such that $\phi\emptyset=0$.   Then

\Centerline{$\Sigma
=\{E:E\subseteq X,\,\phi A=\phi(A\cap E)+\phi(A\setminus E)$
for every $A\subseteq X\}$}

\noindent is an algebra of subsets of $X$, and
$\phi(E\cup F)=\phi E+\phi F$ for all disjoint $E$, $F\in\Sigma$.

\proof{ The symmetry of the definition of $\Sigma$ ensures that
$X\setminus E\in\Sigma$ whenever $E\in\Sigma$.   If $E$, $F\in\Sigma$ and
$A\subseteq X$, then

$$\eqalignno{\phi(A\cap(E\cup &F))+\phi(A\setminus(E\cup F))\cr
&=\phi(A\cap(E\cup F)\cap E)+\phi(A\cap(E\cup F)\setminus E)
   +\phi(A\setminus(E\cup F))\cr
&=\phi(A\cap E)+\phi((A\setminus E)\cap F)
   +\phi((A\setminus E)\setminus F)\cr
&=\phi(A\cap E)+\phi(A\setminus E)
=\phi A.\cr}$$

\noindent As $A$ is arbitrary, $E\cup F\in\Sigma$.   Finally, if
$A\subseteq X$,

\Centerline{$\phi(A\cap\emptyset)+\phi(A\setminus\emptyset)
=\phi\emptyset+\phi A=\phi A$}

\noindent because $\phi\emptyset=0$;  so $\emptyset\in\Sigma$.

Thus $\Sigma$ is an algebra of sets.   If $E$, $F\in\Sigma$ and
$E\cap F=\emptyset$, then

\Centerline{$\phi(E\cup F)
=\phi((E\cup F)\cap E)+\phi((E\cup F)\setminus E)=\phi E+\phi F$.}
}%end of proof of 413B

\leader{413C}{Measures from inner \dvrocolon{measures}}\cmmnt{ I come
now to a construction corresponding to \Caratheodory's method of
defining measures from outer measures.

\medskip

\noindent}{\bf Theorem} Let $X$ be a set and
$\phi:X\to[0,\infty]$ an inner measure.   Set

\Centerline{$\Sigma=\{E:E\subseteq X,\,\phi(A\cap E)+\phi(A\setminus
E)=\phi A$ for every $A\subseteq X\}$.}

\noindent Then $(X,\Sigma,\phi\restr\Sigma)$ is a complete measure
space.

\proof{ (Compare 113C.)

\medskip

{\bf (a)} The first step is to note that if $A\subseteq B\subseteq X$
then

\Centerline{$\phi B\ge\phi A+\phi(B\setminus A)\ge\phi A$.}

\noindent Next, a subset $E$ of $X$
belongs to $\Sigma$ iff $\phi A\le\phi(A\cap E)+\phi(A\setminus E)$
whenever $A\subseteq X$ and $\mu A<\infty$.
\Prf\ Of course any element of $\Sigma$ satisfies the condition.   If
$E$ satisfies the condition and $A\subseteq X$, then

$$\eqalign{\phi A
&=\sup\{\phi B:B\subseteq A,\,\phi B<\infty\}\cr
&\le\sup\{\phi(B\cap E)+\phi(B\setminus E):B\subseteq A\}\cr
&=\phi(A\cap E)+\phi(A\setminus E)
\le\phi A,\cr}$$

\noindent so $E\in\Sigma$.\ \Qed

\medskip

{\bf (b)} By 413B, $\Sigma$ is an algebra of subsets of $X$.   Now
suppose
that $\langle E_n\rangle_{n\in\Bbb N}$ is a non-decreasing
sequence in $\Sigma$, with union $E$.   If $A\subseteq X$ and
$\phi A<\infty$, then

\Centerline{$\phi(A\setminus E)=\inf_{n\in\Bbb N}\phi(A\setminus E_n)
=\lim_{n\to\infty}\phi(A\setminus E_n)$}

\noindent because $\sequencen{A\setminus E_n}$ is non-increasing and
$\phi(A\setminus E_0)$ is finite;  so

\Centerline{$\phi(A\cap E)+\phi(A\setminus E)
\ge\lim_{n\to\infty}\phi(A\cap E_n)+\phi(A\setminus E_n)
=\phi A$.}

\noindent By (a), $E\in\Sigma$.   So $\Sigma$ is a $\sigma$-algebra.

\medskip

{\bf (c)} If $E$, $F\in\Sigma$ and $E\cap F=\emptyset$ then
$\phi(E\cup F)=\phi E+\phi F$, by 413B.   If $\sequencen{E_n}$
is a disjoint sequence in $\Sigma$ with union $E$, then

\Centerline{$\mu E\ge\mu(\bigcup_{i\le n}E_i)=\sum_{i=0}^n\mu E_i$}

\noindent for every $n$, so $\mu E\ge\sum_{i=0}^{\infty}\mu E_i$.
\Quer\ If $\mu E>\sum_{i=0}^{\infty}\mu E_i$, there is an $A\subseteq E$
such that $\sum_{i=0}^{\infty}\mu E_i<\phi A<\infty$.   But now, setting
$F_n=\bigcup_{i\le n}E_i$ for each $n$, we have
$\lim_{n\to\infty}\phi(A\setminus F_n)=0$, so that

\Centerline{$\phi A
=\lim_{n\to\infty}\phi(A\cap F_n)+\phi(A\setminus F_n)
=\sum_{i=0}^{\infty}\phi(A\cap E_i)<\phi A$,}

\noindent which is absurd.\ \BanG\    Thus
$\mu E=\sum_{i=0}^{\infty}\mu E_i$.   As $\sequencen{E_n}$ is arbitrary,
$\mu$ is a measure.

\medskip

{\bf (d)} Finally, suppose that $B\subseteq E\in\Sigma$ and $\mu E=0$.
Then for any $A\subseteq X$ we must have

\Centerline{$\phi(A\cap B)+\phi(A\setminus B)\ge\phi(A\setminus E)
=\phi(A\cap E)+\phi(A\setminus E)=\phi A$,}

\noindent so $B\in\Sigma$.   Thus $\mu$ is complete.
}%end of proof of 413C

\cmmnt{\medskip

\noindent{\bf Remark} For a simple example see 213Yd.
}%end of comment

\leader{413D}{The inner measure defined by a measure} Let
$(X,\Sigma,\mu)$ be any measure space.
\cmmnt{ Just as} $\mu$\cmmnt{ has an associated outer
measure $\mu^*$ defined by the formula

\Centerline{$\mu^*A=\inf\{\mu E:A\subseteq E\in\Sigma\}$}

\noindent (132A-132B), it} gives rise to an inner measure $\mu_*$
defined by the formula

\Centerline{$\mu_*A=\sup\{\mu E:E\in\Sigma^f,\,E\subseteq A\}$,}

\noindent where I write $\Sigma^f$ for $\{E:E\in\Sigma,\,\mu E<\infty\}$.
\prooflet{\Prf\ $\mu_*\emptyset=\mu\emptyset=0$.   ($\alpha$) If
$A\cap B=\emptyset$, and $E\subseteq A$, $F\subseteq B$ belong to
$\Sigma^f$, then
$E\cup F\subseteq A\cup B$ also has finite measure, so

\Centerline{$\mu_*(A\cup B)\ge\mu(E\cup F)=\mu E+\mu F$;}

\noindent taking the supremum over $E$ and $F$,
$\mu_*(A\cup B)\ge\mu_*A+\mu_*B$.   ($\beta$) If $\sequencen{A_n}$ is a
non-increasing sequence of sets with intersection $A$ and
$\mu_*A_0<\infty$, then for each $n\in\Bbb N$ we can find an
$E_n\subseteq A_n$ such that $\mu E_n\ge\mu_*A_n-2^{-n}$.   In this case,

\Centerline{$\mu(\bigcup_{n\in\Bbb N}E_n)\le\mu_*A_0<\infty$.}

\noindent Set

\Centerline{$E=\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m\subseteq A$.}

\noindent Then $E\in\Sigma^f$, so

\Centerline{$\mu_*A\ge\mu E\ge\limsup_{n\to\infty}\mu E_n
=\lim_{n\to\infty}\mu_*A_n\ge\mu_*A$.}

\noindent ($*$) If $A\subseteq X$ and $\mu_*A=\infty$ then

\Centerline{$\sup\{\mu_*B:B\subseteq A,\,\mu_*B<\infty\}
\ge\sup\{\mu E:E\in\Sigma^f,\,E\subseteq A\}=\infty$.  \Qed}
}%end of prooflet

\cmmnt{\medskip

\noindent{\bf Warning} Many authors use the formula

\Centerline{$\mu_*A=\sup\{\mu E:A\supseteq E\in\Sigma\}$.}

\noindent In `ordinary' cases, when $(X,\Sigma,\mu)$ is semi-finite,
this agrees with my usage (413Ed);  but for non-semi-finite spaces there
is a difference.   See 413Yd.
}%end of comment

\vleader{72pt}{413E}{}\cmmnt{ I note the following simple facts
concerning inner measures defined from measures.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space.
Write $\Sigma^f$ for $\{E:E\in\Sigma,\,\mu E<\infty\}$.

(a) For every $A\subseteq X$ there is an $E\in\Sigma$ such that
$E\subseteq A$ and $\mu E=\mu_*A$.

(b) $\mu_*A\le\mu^*A$ for every $A\subseteq X$.

(c) If $E\in\Sigma$ and $A\subseteq X$, then
$\mu_*(E\cap A)+\mu^*(E\setminus A)\le\mu E$, with equality if either
(i) $\mu E<\infty$ or (ii) $\mu$ is semi-finite.

(d) \cmmnt{In particular,} $\mu_*E\le\mu E$ for every $E\in\Sigma$, with
equality if
either $\mu E<\infty$ or $\mu$ is semi-finite.

(e) If $\mu$ is inner regular with respect to $\Cal K$, then
(counting the supremum of the empty set as $0$)
$\mu_*A=\sup\{\mu K:K\in\Cal K\cap\Sigma^f,\,K\subseteq A\}$ for every
$A\subseteq X$.

(f) If $A\subseteq X$ is such that $\mu_*A=\mu^*A<\infty$, then $A$ is
measured by the completion of $\mu$.

(g) If $\hat\mu$, $\tilde\mu$ are the completion and c.l.d.\ version of
$\mu$, then $\hat\mu_*=\tilde\mu_*=\mu_*$.

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

\Centerline{$\mu^*(f^{-1}[B])\le\nu^*B$,
\quad$\mu_*(f^{-1}[B])\ge\nu_*B$}

\noindent for every $B\subseteq Y$, and

\Centerline{$\nu^*(f[A])\ge\mu^*A$}

\noindent for every $A\subseteq X$.

(i) Suppose that $\mu$ is semi-finite.   If $A\subseteq E\in\Sigma$,
then $E$ is a measurable envelope of $A$ iff $\mu_*(E\setminus A)=0$.

\proof{{\bf (a)} There is a sequence $\sequencen{E_n}$ in $\Sigma^f$
such that $E_n\subseteq A$ for each $n$ and
$\lim_{n\to\infty}\mu E_n=\mu_*A$;
now set $E=\bigcup_{n\in\Bbb N}E_n$.

\medskip

{\bf (b)} If $E$, $F\in\Sigma$ and
$E\subseteq A\subseteq F$ we must have $\mu E\le\mu F$.

\medskip

{\bf (c)} If $F\subseteq E\cap A$ and $F\in\Sigma^f$, then

\Centerline{$\mu F+\mu^*(E\setminus A)\le\mu F+\mu(E\setminus F)=\mu
E$;}

\noindent taking the supremum over $F$, $\mu_*(E\cap A)+\mu^*(E\setminus
A)\le\mu E$.   If $\mu E<\infty$, then

$$\eqalign{\mu_*(E\cap A)
&=\sup\{\mu F:F\in\Sigma,\,F\subseteq E\cap A\}\cr
&=\mu E-\inf\{\mu(E\setminus F):F\in\Sigma,\,F\subseteq E\cap A\}\cr
&=\mu E-\inf\{\mu F:F\in\Sigma,\,E\setminus A\subseteq F\subseteq E\}
=\mu E-\mu^*(E\setminus A).\cr}$$

\noindent If $\mu$ is semi-finite, then

$$\eqalign{\mu_*(E\cap A)+\mu^*(E\setminus A)
&\ge\sup\{\mu_*(F\cap A)+\mu^*(F\setminus A):F\in\Sigma^f,\,F\subseteq
E\}\cr
&=\sup\{\mu F:F\in\Sigma^f,\,F\subseteq E\}
=\mu E.\cr}$$

\medskip

{\bf (d)} Take $A=E$ in (c).

\medskip

{\bf (e)}

$$\eqalign{\mu_*A
&=\sup\{\mu E:E\in\Sigma^f,\,E\subseteq A\}\cr
&=\sup\{\mu K:K\in\Cal K\cap\Sigma,\,\exists\,E\in\Sigma^f,\,
  K\subseteq E\subseteq A\}\cr
&=\sup\{\mu K:K\in\Cal K\cap\Sigma^f,\,K\subseteq A\}.\cr}$$

\medskip

{\bf (f)} By (a) above and 132Aa, there are $E$, $F\in\Sigma$ such that
$E\subseteq A\subseteq F$ and

\Centerline{$\mu E=\mu_*A=\mu^*A=\mu F<\infty$;  }

\noindent now $\mu(F\setminus E)=0$, so $F\setminus A$ and $A$ are
measured by the completion of $\mu$.

\medskip

{\bf (g)} Write $\check\mu$ for either $\hat\mu$ or $\tilde\mu$, and
$\check\Sigma$ for its domain, and let $A\subseteq X$.   (i) If
$\gamma<\mu_*A$, there is an $E\in\Sigma$ such that $E\subseteq A$ and
$\gamma\le\mu E<\infty$;  now $\check\mu E=\mu E$ (212D, 213Fa), so
$\check\mu_*A\ge\gamma$.   As $\gamma$ is arbitrary,
$\mu_*A\le\check\mu_*A$.   (ii) If $\gamma<\check\mu_*A$, there is an
$E\in\check\Sigma$ such that $E\subseteq A$ and
$\gamma\le\check\mu E<\infty$.   Now there is an $F\in\Sigma$ such that
$F\subseteq E$ and $\mu F=\check\mu E$ (212Cb, 213Fc), so that
$\mu_*A\ge\gamma$.   As $\gamma$ is arbitrary, $\mu_*A\ge\check\mu_*A$.

\medskip

{\bf (h)} This is elementary;  all we have to note is that if $F$,
$F'\in\Tau$ and $F\subseteq B\subseteq F'$, then
$f^{-1}[F]\subseteq f^{-1}[B]\subseteq f^{-1}[F']$, so that

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

\noindent As $F$ and $F'$ are arbitrary,

\Centerline{$\nu_*B\le\le\mu_*f^{-1}[B]$,
\quad$\mu^*f^{-1}[B]\le\nu^*B$.}

\noindent Now, for $A\subseteq X$,

\Centerline{$\mu^*A\le\mu^*(f^{-1}[f[A]])\le\nu^*(f[A])$.}

\medskip

{\bf (i)}(i) If $E$ is a measurable envelope of $A$ and $F\in\Sigma$ is
included in $E\setminus A$, then

\Centerline{$\mu F=\mu(F\cap E)=\mu^*(F\cap A)=0$;}

\noindent as $F$ is arbitrary, $\mu_*(E\setminus A)=0$.   (ii) If $E$ is
not a measurable envelope of $A$, there is an $F\in\Sigma$ such that
$\mu^*(F\cap A)<\mu(F\cap E)$.   Let $G\in\Sigma$ be such that
$F\cap A\subseteq G$ and $\mu G=\mu^*(F\cap A)$.   Then
$\mu(F\cap E\setminus G)>0$;  because $\mu$ is semi-finite,
$\mu_*(E\setminus A)\ge\mu_*(F\cap E\setminus G)>0$.
}%end of proof of 413E

\leader{413F}{}\cmmnt{ The language of 413D makes it easy to express
some useful facts about complete locally determined measure spaces,
complementing 412J.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space and $\Cal K$ a family of subsets of $X$ such
that $\mu$ is inner regular with respect to $\Cal K$.
Then for $E\subseteq X$ the following are equiveridical:

\quad(i) $E\in\Sigma$;

\quad(ii) $E\cap K\in\Sigma$ whenever $K\in\Sigma\cap\Cal K$;

\quad(iii) $\mu^*(K\cap E)+\mu^*(K\setminus E)=\mu^*K$ for every
$K\in\Cal K$;

\quad(iv) $\mu_*(K\cap E)+\mu_*(K\setminus E)=\mu_*K$ for every
$K\in\Cal K$;

\quad(v) $\mu^*(E\cap K)=\mu_*(E\cap K)$ for every
$K\in\Cal K\cap\Sigma$;

\quad(vi) $\min(\mu^*(K\cap E),\mu^*(K\setminus E))<\mu K$ whenever
$K\in\Cal K\cap\Sigma$ and $0<\mu K<\infty$;

\quad(vii) $\max(\mu_*(K\cap E),\mu_*(K\setminus E))>0$ whenever
$K\in\Cal K\cap\Sigma$ and $\mu K>0$.

\proof{{\bf (a)} Assume (i).   Then of course
$E\cap K\in\Sigma$ for every $K\in\Sigma\cap\Cal K$, and (ii) is true.
For any $K\in\Cal K$ there is an $F\in\Sigma$ such that $F\supseteq K$
and $\mu F=\mu^*K$ (132Aa again);  now

\Centerline{$\mu^*K\le\mu^*(K\cap E)+\mu^*(K\setminus E)
\le\mu(F\cap E)+\mu(F\setminus E)=\mu F=\mu^*K$,}

\noindent so (iii) is true.   Next, for any $K\in\Cal K$,

$$\eqalignno{\mu_*(K\cap E)+\mu_*(K\setminus E)
&\le\mu_*K
=\sup\{\mu F:F\in\Sigma^f,\,F\subseteq K\}\cr
\noalign{\noindent (writing $\Sigma^f$ for
$\{F:F\in\Sigma,\,\mu F<\infty\}$)}
&=\sup\{\mu(F\cap E)+\mu(F\setminus E):F\in\Sigma^f,\,F\subseteq K\}\cr
&\le\mu_*(K\cap E)+\mu_*(K\setminus E).\cr}$$

\noindent So (iv) is true.   If $K\in\Cal K\cap\Sigma$, then

\Centerline{$\mu_*(E\cap K)
=\sup\{\mu F:F\in\Sigma^f,\,F\subseteq E\cap K\}
=\mu(E\cap K)=\mu^*(E\cap K)$}

\noindent because $\mu$ is semi-finite.   So (v) is true.   Since
(iii)$\Rightarrow$(vi) and (iv)$\Rightarrow$(vii), we see that all the
conditions are satisfied.

\medskip

{\bf (b)} Now suppose that $E\notin\Sigma$;  I have to show that
(ii)-(vii) are all false.   Because $\mu$ is locally determined, there
is an $F\in\Sigma^f$ such that $E\cap F\notin\Sigma$.
Take measurable envelopes $H$, $H'$ of $F\cap E$ and $F\setminus E$
respectively (132Ee).   Then
$F\setminus H'\subseteq F\cap E\subseteq F\cap H$, so

\Centerline{$G=(F\cap H)\setminus(F\setminus H')=F\cap H\cap H'$}

\noindent cannot be negligible.
Take $K\in\Cal K\cap\Sigma$ such that $K\subseteq G$ and $\mu K>0$.   As
$G\subseteq F$, $\mu K<\infty$.   Now

\Centerline{$\mu^*(K\cap E)=\mu^*(K\cap F\cap E)=\mu(K\cap H)=\mu K$,}

\Centerline{$\mu^*(K\setminus E)=\mu^*(K\cap F\setminus E)
=\mu(K\cap H')=\mu K$.}

\noindent But this means that

\Centerline{$\mu_*(K\cap E)=\mu K-\mu^*(K\setminus E)=0$,
\quad$\mu_*(K\setminus E)=\mu K-\mu^*(K\cap E)=0$}

\noindent by 413Ec.   Now we see that this $K$ witnesses that (ii)-(vii)
are all false.
}%end of proof of 413F

\leader{413G}{}\cmmnt{ The ideas of 413F can be used to give criteria
for measurability of real-valued functions.   I spell out one which is
particularly useful.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space and suppose that $\mu$ is inner regular with
respect to $\Cal K\subseteq\Sigma$.   Suppose that $f:X\to\Bbb R$ is a
function, and for $\alpha\in\Bbb R$ set
$E_{\alpha}=\{x:f(x)\le\alpha\}$, $F_{\alpha}=\{x:f(x)\ge\beta\}$.
Then $f$ is $\Sigma$-measurable iff

\Centerline{$\min(\mu^*(E_{\alpha}\cap K),\mu^*(F_{\beta}\cap K))
<\mu K$}

\noindent whenever $K\in\Cal K$, $0<\mu K<\infty$ and $\alpha<\beta$.

\proof{{\bf (a)} If $f$ is measurable, then

\Centerline{$\mu^*(E_{\alpha}\cap K)+\mu^*(F_{\beta}\cap K)
=\mu(E_{\alpha}\cap K)+\mu(F_{\beta}\cap K)
\le\mu K$}

\noindent whenever $K\in\Sigma$ and $\alpha<\beta$, so if
$0<\mu K<\infty$ then we must have
$\min(\mu^*(E_{\alpha}\cap K),\mu^*(F_{\beta}\cap K))<\mu K$.

\medskip

{\bf (b)} If $f$ is not measurable, then there is some $\alpha\in\Bbb R$
such that $E_{\alpha}$ is not measurable.   413F(vi) tells us that
there is a $K\in\Cal K$ such that $0<\mu K<\infty$ and
$\mu^*(E_{\alpha}\cap K)=\mu^*(K\setminus E_{\alpha})=\mu K$.   Note
that $K$ is a measurable envelope of $K\cap E_{\alpha}$ (132Eb).   Now
$\sequencen{K\cap F_{\alpha+2^{-n}}}$ is a non-decreasing sequence with
union $K\setminus E_{\alpha}$, so there is some $\beta>\alpha$ such that
$K\cap F_{\beta}$ is not negligible.   Let $H\subseteq K$ be a
measurable envelope of $K\cap F_{\beta}$, and $K'\in\Cal K$ such that
$K'\subseteq H$ and $\mu K'>0$;  then

\Centerline{$\mu^*(K'\cap E_{\alpha})
=\mu^*(K'\cap K\cap E_{\alpha})=\mu(K'\cap K)=\mu K'$,}

\Centerline{$\mu^*(K'\cap F_{\beta})
=\mu^*(K'\cap H\cap F_{\beta})=\mu(K'\cap H)=\mu K'$,}

\noindent so $K'$, $\alpha$ and $\beta$ witness that the condition is
not satisfied.
}%end of proof of 413G
%compare 213Ye

\leader{413H}{}\dvAformerly{4{}13T}\cmmnt{ The following fact is interesting
and not quite obvious.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete totally
finite measure space, $(Y,\Tau,\nu)$ a measure space, and $\frak S$ a
Hausdorff topology on $Y$ such that $\nu$ is inner
regular with respect to the closed sets.   Let $\sequencen{f_n}$ be a
sequence of
\imp\ functions from $X$ to $Y$.   If $f(x)=\lim_{n\to\infty}f_n(x)$ is
defined in $Y$ for every $x\in X$, then $f$ is \imp.

\proof{ Let $\mu_*$ be the inner measure associated with $\mu$
(413D).   If $F\in\Tau$ is closed then
$\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}f_m^{-1}[F]\subseteq f^{-1}[F]$, so

\Centerline{$\mu_*f^{-1}[F]
\ge\mu(\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}f_m^{-1}[F])
\ge\liminf_{n\to\infty}\mu f_n^{-1}[F]=\nu F$.}

\noindent So if $H$ is any member of $\Tau$,

$$\eqalign{\mu_*f^{-1}[H]
&\ge\sup\{\mu_*f^{-1}[F]:
  F\in\Tau,\,F\subseteq H\text{ and }F\text{ is closed}\}\cr
&\ge\sup\{\nu F:
  F\in\Tau,\,F\subseteq H\text{ and }F\text{ is closed}\}
=\nu H.\cr}$$

\noindent Taking complements,

$$\eqalignno{\mu^*f^{-1}[H]
&=\mu X-\mu_*f^{-1}[Y\setminus H]\cr
\displaycause{413Ec}
&\le\nu Y-\nu(Y\setminus H)
=\nu H\cr}$$

\noindent (of course $\nu Y=\mu f_0^{-1}[Y]=\mu X$).   So
$\mu^*f^{-1}[H]=\mu_*f^{-1}[H]=\nu H$.   Because $\mu$ is complete,
$\mu f^{-1}[H]$ is defined and equal to $\nu H$ (413Ef).  As $H$ is
arbitrary, $f$ is \imp.
}%end of proof of 413H

%doesn't work for $\sigma$-finite case, try $X=Y=\Bbb R\cup\{\infty}$ and
% $f_n(x)=x+n$

\leader{413I}{}\cmmnt{ Inner measure constructions based on 413C are
important
because they offer an efficient way of setting up measures which are
inner regular with respect to given families of sets.   Two of the
fundamental
results are 413J and 413K.   I proceed by means of a lemma on finitely
additive functionals.

\medskip

\noindent}{\bf Lemma} Let $X$ be a set and $\Cal K$ a family of subsets
of $X$ such that

\inset{$\emptyset\in\Cal K$,}

\inset{($\dagger$) $K\cup
K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are disjoint,}

\inset{($\ddagger$) $K\cap K'\in\Cal K$ for all $K$, $K'\in\Cal K$.}

\noindent Let $\phi_0:\Cal K\to\coint{0,\infty}$ be a
functional such that

\inset{($\alpha$) $\phi_0 K
=\phi_0 L+\sup\{\phi_0 K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$
whenever $K$, $L\in\Cal K$ and $L\subseteq K$.}

\noindent Set

\Centerline{$\phi A=\sup\{\phi_0K:K\in\Cal K,\,K\subseteq A\}$ for
$A\subseteq X$,}

\Centerline{$\Sigma=\{E:E\subseteq X,\,\phi A=\phi(A\cap
E)+\phi(A\setminus
E)$ for every $A\subseteq X\}$.}

\noindent Then $\Sigma$ is an algebra of subsets of $X$, including
$\Cal K$,
and $\phi\restr\Sigma:\Sigma\to[0,\infty]$ is an additive functional
extending $\phi_0$.

\proof{{\bf (a)} To see that $\Sigma$ is an algebra of subsets and
$\phi\restr\Sigma$ is additive, all we need to know is that
$\phi\emptyset=0$ (413B);  and this is because, applying hypothesis
($\alpha$) with $K=L=\emptyset$,
$\phi_0\emptyset=\phi_0\emptyset+\phi_0\emptyset$, so
$\phi_0\emptyset=0$.
($\alpha$) also assures us that $\phi_0L\le\phi_0K$ whenever $K$,
$L\in\Cal
K$ and $L\subseteq K$, so $\phi K=\phi_0K$ for every $K\in\Cal K$.

\medskip

{\bf (b)} To check that $\Cal K\subseteq\Sigma$, we have a little more
work
to do.   First, observe that ($\dagger$) and ($\alpha$) together tell us
that $\phi_0(K\cup K')=\phi_0K+\phi_0K'$ for all disjoint $K$,
$K'\in\Cal
K$.   So if $A$, $B\subseteq X$ and $A\cap B=\emptyset$ then

$$\phi A+\phi B
=\sup_{\Atop{K\in\Cal K}{K\subseteq A}}
  \phi_0K+\sup_{\Atop{L\in\Cal K}{L\subseteq B}}\phi_0L
=\sup_{\Atop{K,L\in\Cal K}{\Atop{K\subseteq A}{L\subseteq B}}}
   \phi_0(K\cup L)
\le\phi(A\cup B).$$

\medskip

{\bf (c)} $\Cal K\subseteq\Sigma$.   \Prf\ Take $K\in\Cal K$ and
$A\subseteq X$.   If $L\in\Cal K$ and $L\subseteq A$, then

\Centerline{$\phi_0L
=\phi_0(K\cap L)
  +\sup\{\phi_0L':L'\in\Cal K,\,L'\subseteq L\setminus K\}
\le\phi(A\cap K)+\phi(A\setminus K)$.}

\noindent (Note the use of the hypothesis ($\ddagger$).)   As $L$ is
arbitrary,
$\phi A\le\phi(A\cap K)+\phi(A\setminus K)$.   We already know that
$\phi(A\cap K)+\phi(A\setminus K)\le\phi A$;
as $A$ is arbitrary, $K\in\Sigma$.\ \Qed

This completes the proof.
}%end of proof of 413I

\vleader{60pt}{413J}{Theorem}\cmmnt{ ({\smc Tops{\o}e 70a})}
%Theorem 1 p 198
Let $X$ be a set and $\Cal K$ a family of subsets of $X$ such that

\inset{$\emptyset\in\Cal K$,}

\inset{($\dagger$) $K\cup
K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are disjoint,}

\inset{($\ddagger$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ whenever
$\sequencen{K_n}$ is a sequence in $\Cal K$.}

\noindent Let $\phi_0:\Cal K\to\coint{0,\infty}$ be a
functional such that

\inset{($\alpha$) $\phi_0 K
=\phi_0 L+\sup\{\phi_0 K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$
whenever $K$, $L\in\Cal K$ and $L\subseteq K$,}

\inset{($\beta$) $\inf_{n\in\Bbb N}\phi_0K_n=0$ whenever
$\sequencen{K_n}$ is
a non-increasing sequence in $\Cal K$ with empty intersection.}

\noindent Then there is a unique complete locally determined measure
$\mu$ on $X$ extending $\phi_0$ and inner regular with respect to
$\Cal K$.

\proof{{\bf (a)} Set

\Centerline{$\phi A=\sup\{\phi_0K:K\in\Cal K,\,K\subseteq A\}$ for
$A\subseteq X$,}

\Centerline{$\Sigma=\{E:E\subseteq X,\,\phi A=\phi(A\cap
E)+\phi(A\setminus
E)$ for every $A\subseteq X\}$.}

\noindent Then 413I tells us that $\Sigma$ is an algebra of subsets of
$X$,
including $\Cal K$, and $\mu=\phi\restr\Sigma$ is an additive functional
extending $\phi_0$.

\medskip

{\bf (b)} Now $\mu(\bigcap_{n\in\Bbb N}K_n)=\inf_{n\in\Bbb
N}\mu K_n$ whenever
$\sequencen{K_n}$ is a non-increasing sequence in $\Cal K$.   \Prf\ Set
$L=\bigcap_{n\in\Bbb N}K_n$.   Of course $\mu L\le\inf_{n\in\Bbb
N}\mu K_n$.   For the reverse inequality, take $\epsilon>0$.   Then
($\alpha$) tells us that there is a $K'\in\Cal K$ such that $K'\subseteq
K_0\setminus L$ and
$\mu K_0\le\mu L+\mu K'+\epsilon$.   Since $\sequencen{K_n\cap K'}$
is a non-increasing sequence in $\Cal K$ with empty intersection,
($\beta$)
tells us that there is an
$n\in\Bbb N$ such that $\mu(K_n\cap K')\le\epsilon$.   Now

$$\eqalign{\mu K_0-\mu L
&=\mu(K_0\setminus L)
=\mu(K_0\setminus(K'\cup L))+\mu K'\cr
&\le\epsilon+\mu(K_n\cap K')+\mu(K'\setminus K_n)
\le 2\epsilon+\mu(K_0\setminus K_n)
=2\epsilon+\mu K_0-\mu K_n.\cr}$$

\noindent (These calculations depend, of course, on the additivity of
$\mu$
and the finiteness of $\mu K_0$.)   So
$\mu L\ge\mu K_n-2\epsilon$.   As $\epsilon$ is arbitrary, $\mu
L=\inf_{n\in\Bbb N}\mu K_n$.\ \Qed

\medskip

{\bf (c)} If $\sequencen{A_n}$ is a non-increasing sequence of
subsets of $X$, with intersection $A$, and $\phi A_0<\infty$, then $\phi
A=\inf_{n\in\Bbb N}\phi A_n$.   \Prf\ Of course $\phi A\le\phi A_n$ for
every $n$.   Given $\epsilon>0$, then for each $n\in\Bbb N$ choose
$K_n\in\Cal K$ such that $K_n\subseteq A_n$ and
$\phi_0K_n\ge\phi A_n-2^{-n}\epsilon$ (this is where I use the
hypothesis
that $\phi A_0$ is finite);  set $L_n=\bigcap_{i\le n}K_i$ for each $n$,
and $L=\bigcap_{n\in\Bbb N}L_n$.   Then we have

$$\eqalign{\phi A_{n+1}-\mu L_{n+1}
&=\phi A_{n+1}-\mu(K_{n+1}\cap L_n)\cr
&=\phi A_{n+1}-\mu K_{n+1}-\mu L_n+\mu(K_{n+1}\cup L_n)\cr
&\le 2^{-n-1}\epsilon-\mu L_n+\phi A_n\cr}$$

\noindent because $K_{n+1}\subseteq A_{n+1}\subseteq A_n$ and
$L_n\subseteq
K_n\subseteq A_n$.   Inducing on $n$, we see that
$\mu L_n\ge\phi A_n-2\epsilon+2^{-n}\epsilon$ for every $n$.   So

\Centerline{$\phi A\ge\mu L=\inf_{n\in\Bbb N}\mu L_n\ge\inf_{n\in\Bbb
N}\phi A_n-2\epsilon$,}

\noindent using (b) above for the middle equality.   As $\epsilon$ is
arbitrary, $\phi A=\inf_{n\in\Bbb N}\phi A_n$.\ \Qed

\medskip

{\bf (d)} It follows that $\phi$ is an inner measure.
\Prf\ The arguments of parts (a) and (b) of the proof of 413I tell us
that
$\phi\emptyset=0$ and $\phi(A\cup B)\le\phi A+\phi B$ whenever $A$,
$B\subseteq X$ are disjoint.   We have just seen that
$\phi(\bigcap_{n\in\Bbb N}A_n)=\inf_{n\in\Bbb N}\phi A_n$ whenever
$\sequencen{A_n}$ is a non-increasing sequence of sets and $\phi
A_0<\infty$.
Finally, $\phi K=\phi_0K$ is finite for every $K\in\Cal
K$, so $\phi A=\sup\{\phi B:B\subseteq A,\,\phi B<\infty\}$ for every
$A\subseteq X$.   Putting these together, $\phi$ is an inner measure.\
\Qed

\medskip

{\bf (e)} So 413C tells us that $\mu$ is a complete measure, and of
course it is inner regular with respect to $\Cal K$, by the definition
of $\phi$.   It is semi-finite because $\mu K=\phi_0K$ is finite for
every $K\in\Cal K$.   Now suppose that
$E\subseteq X$ and that $E\cap F\in \Sigma$ whenever $\mu F<\infty$.
Take any $A\subseteq X$.   If $L\in\Cal K$ and $L\subseteq A$, we have
$L\in\Sigma$ and $\mu L<\infty$, so

\Centerline{$\phi_0L=\mu L=\mu(L\cap E)+\mu(L\setminus E)
=\phi(L\cap E)+\phi(L\setminus E)\le\phi(A\cap E)+\phi(A\setminus E)$;}

\noindent taking the supremum over $L$,
$\phi A\le\phi(A\cap E)+\phi(A\setminus E)$.   As $A$ is arbitrary,
$E\in\Sigma$;  as $E$ is arbitrary, $\mu$ is locally determined.

\medskip

{\bf (f)} Finally, $\mu$ is unique by 412Mb.
}%end of proof of 413J

\vleader{60pt}{413K}{Theorem}\dvAformerly{4{}13J} Let $X$ be a set and 
$\Cal K$ a family of subsets of $X$ such that

\inset{$\emptyset\in\Cal K$,}

\inset{($\dagger$) $K\cup
K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are disjoint,}

\inset{($\ddagger$) $K\cap K'\in\Cal K$ whenever $K$, $K'\in\Cal K$.}

\noindent Let $\phi_0:\Cal K\to\coint{0,\infty}$ be a
functional such that

\inset{($\alpha$) $\phi_0 K
  =\phi_0 L+\sup\{\phi_0 K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$
whenever $K$, $L\in\Cal K$ and $L\subseteq K$,}

\inset{($\beta$) $\inf_{n\in\Bbb N}\phi_0K_n=0$ whenever
$\sequencen{K_n}$ is
a non-increasing sequence in $\Cal K$ with empty intersection.}

\noindent Then there is a unique complete locally determined measure
$\mu$ on $X$ extending $\phi_0$ and inner regular with respect to $\Cal
K_{\delta}$, the family of sets expressible as intersections of
sequences in $\Cal K$.

\proof{{\bf (a)} Set

\Centerline{$\psi A=\sup\{\phi_0K:K\in\Cal K,\,K\subseteq A\}$ for
$A\subseteq X$,}

\Centerline{$\Tau
=\{E:E\subseteq X,\,\psi A=\psi (A\cap E)+\psi(A\setminus E)$
for every $A\subseteq X\}$.}

\noindent Then 413I tells us that $\Tau$ is an algebra of subsets of $X$,
including $\Cal K$, and $\nu =\psi \restrp\Tau$ is an additive functional
extending $\phi_0$.

\medskip

{\bf (b)} Write $\Tau^f$ for $\{E:E\in\Tau,\,\nu E<\infty\}$.
If $\sequencen{E_n}$ is a non-increasing sequence in $\Tau^f$ with empty
intersection, $\lim_{n\to\infty}\nu E_n=0$.   \Prf\
Given $\epsilon>0$, we can choose a sequence $\sequencen{K_n}$ in
$\Cal K$ such that $K_n\subseteq E_n$ and

\Centerline{$\nu K_n=\phi_0K_n\ge\nu E_n-2^{-n}\epsilon$}

\noindent for each $n$.   Set $L_n=\bigcap_{i\le n}K_i$ for each $n$;
then

\Centerline{$\lim_{n\to\infty}\nu L_n=\lim_{n\to\infty}\phi_0L_n=0$}

\noindent by hypothesis ($\beta$).   But also, for each $n$,

\Centerline{$\nu E_n
\le\nu L_n+\sum_{i=0}^n\nu(E_i\setminus K_i)
\le\nu L_n+2\epsilon$,}

\noindent because $\nu$ is additive and non-negative and $E_n\subseteq
L_n\cup\bigcup_{i\le n}(E_i\setminus K_i)$.
So $\limsup_{n\to\infty}\nu E_n\le 2\epsilon$;  as $\epsilon$ is
arbitrary, $\lim_{n\to\infty}\nu E_n=0$.\ \Qed

\medskip

{\bf (c)} Write $\Tau^f_{\delta}$ for the family of sets expressible as
intersections of sequences in $\Tau^f$, and for $H\in\Tau^f_{\delta}$
set $\phi_1H=\inf\{\nu E:H\subseteq E\in\Tau\}$.
Note that because $E\cap F\in\Tau^f$ whenever $E$, $F\in\Tau^f$, every
member of $\Tau^f_{\delta}$ can be expressed as the intersection of a
non-increasing sequence in $\Tau^f$.

\medskip

\quad{\bf (i)} If $\sequencen{E_n}$ is a non-increasing sequence in
$\Tau^f$ with intersection $H\in\Tau^f_{\delta}$, $\phi_1
H=\lim_{n\to\infty}\nu E_n$.   \Prf\ Of course

\Centerline{$\phi_1 H\le\inf_{n\in\Bbb N}\nu E_n
=\lim_{n\to\infty}\nu E_n$.}

\noindent On the other hand, if $H\subseteq E\in\Tau$, then
$\sequencen{E_n\setminus E}$ is a non-increasing sequence in $\Tau^f$
with empty intersection, and

\Centerline{$\nu E\ge\lim_{n\to\infty}\nu(E_n\cap E)
=\lim_{n\to\infty}\nu E_n-\lim_{n\to\infty}\nu(E_n\setminus E)
=\lim_{n\to\infty}\nu E_n$}

\noindent by (b) above.   As $E$ is arbitrary,
$\phi_1(\bigcap_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\nu E_n$.\ \Qed

\medskip

\quad{\bf (ii)} Because $\Cal K\subseteq\Tau^f$,
$\Cal K_{\delta}\subseteq\Tau^f_{\delta}$.   Now for any
$H\in\Tau^f_{\delta}$,
$\phi_1H=\sup\{\phi_1L:L\in\Cal K_{\delta},\,L\subseteq H\}$.   \Prf\
Express $H$ as $\bigcap_{n\in\Bbb N}E_n$ where $\sequencen{E_n}$ is a
non-increasing sequence in $\Tau^f$.   Given $\epsilon>0$, we can choose
a sequence $\sequencen{K_n}$ in $\Cal K$ such that $K_n\subseteq E_n$ and
$\nu K_n\ge\nu E_n-2^{-n}\epsilon$ for each $n$.   Setting
$L_n=\bigcap_{i\le n}K_i$ for each $n$ and $L=\bigcap_{n\in\Bbb N}L_n$,
we have $L\in\Cal K_{\delta}$, $L\subseteq H$ and

\Centerline{$\phi_1H=\lim_{n\to\infty}\nu E_n
\le\lim_{n\to\infty}(\nu L_n+\sum_{i=0}^n\nu(E_i\setminus K_i))
\le\phi_1L+2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, this gives the result.\ \Qed

\medskip

{\bf (d)} We find that $\Tau^f_{\delta}$ and $\phi_1$ satisfy the
conditions of
413J.   \Prf\ Of course $\emptyset\in\Tau^f_{\delta}$.   If $G$,
$H\in\Tau^f_{\delta}$ and $G\cap H=\emptyset$, express them as
$\bigcap_{n\in\Bbb N}E_n$, $\bigcap_{n\in\Bbb N}F_n$ where
$\sequencen{E_n}$, $\sequencen{F_n}$ are non-increasing sequences in
$\Tau^f$.   Then

\Centerline{$G\cup H=\bigcap_{n\in\Bbb N}E_n\cup F_n$}

\noindent belongs to $\Tau^f_{\delta}$, and

$$\eqalignno{\phi_1(G\cup H)
&=\lim_{n\to\infty}\nu(E_n\cup F_n)
=\lim_{n\to\infty}\nu E_n+\nu F_n-\nu(E_n\cap F_n)\cr
&=\lim_{n\to\infty}\nu E_n+\nu F_n\cr
\noalign{\noindent (by (b))}
&=\phi_1G+\phi_1H.\cr}$$

The definition of $\Tau^f_{\delta}$ as the set of intersections of
sequences
in $\Tau^f$ ensures that the intersection of any sequence in
$\Tau^f_{\delta}$ will belong to $\Tau^f_{\delta}$.

Now suppose that $G$, $H\in\Tau^f_{\delta}$ and that $G\subseteq H$.
Express them as intersections $\bigcap_{n\in\Bbb N}E_n$,
$\bigcap_{n\in\Bbb N}F_n$ of non-increasing sequences in $\Tau^f$, so that
$\phi_1G=\lim_{n\to\infty}\nu E_n$ and $\phi_1H=\lim_{n\to\infty}\nu F_n$.
For each $n$, set $H_n=\bigcap_{m\in\Bbb N}F_m\setminus E_n$, so that
$H_n\in\Tau^f_{\delta}$, $H_n\subseteq H\setminus G$, and

$$\eqalign{\phi_1H_n
&=\lim_{m\to\infty}\nu(F_m\setminus E_n)
=\lim_{m\to\infty}\nu F_m-\nu(F_m\cap E_n)\cr
&\ge\lim_{m\to\infty}\nu F_m-\nu E_n
=\phi_1H-\nu E_n.\cr}$$

\noindent Accordingly

\Centerline{$\sup\{\phi_1G':G'\in\Tau^f_{\delta},\,G'\subseteq
H\setminus
G\}\ge\sup_{n\in\Bbb N}\phi_1H-\nu E_n=\phi_1H-\phi_1G$.}

\noindent On the other hand, if $G'\in\Tau^f_{\delta}$ and $G'\subseteq
H\setminus G$, then

\Centerline{$\phi_1G+\phi_1G'=\phi_1(G\cup G')\le\phi_1H$}

\noindent because of course $\phi_1$ is non-decreasing, as well as being
additive on disjoint sets.   So

\Centerline{$\sup\{\phi_1G':G'\in\Tau^f_{\delta},\,G'\subseteq
H\setminus G\}=\phi_1H-\phi_1G$}

\noindent as required by condition ($\alpha$) of 413J.
Finally, suppose that $\sequencen{H_n}$ is a non-increasing sequence in
$\Tau^f_{\delta}$ with empty intersection.   For each $n\in\Bbb N$, let
$\sequence{i}{E_{ni}}$ be a non-increasing sequence in $\Tau^f$ with
intersection $H_n$, and set $F_m=\bigcap_{i,j\le m}E_{ji}$ for each $m$.
Then $\sequence{m}{F_m}$ is a non-increasing sequence in $\Tau^f$ with
empty intersection, while $H_m\subseteq F_m$ for each $m$, so

\Centerline{$\lim_{m\to\infty}\phi_1H_m
\le\lim_{m\to\infty}\nu F_m=0$.}

\noindent Thus condition 413J($\beta$) is satisfied, and we have the
full
list.\ \Qed

\medskip

{\bf (e)} By 413J, we have a complete locally determined measure $\mu$,
extending $\phi_1$, and inner regular with respect to $\Tau^f_{\delta}$.
Since $\phi_1K=\nu K=\phi_0K$ for $K\in\Cal K$, $\mu$ extends $\phi_0$.
If $G$ belongs to the domain of $\mu$, and $\gamma<\mu G$, there is an
$H\in\Tau^f_{\delta}$ such that $H\subseteq G$ and
$\gamma<\mu H=\phi_1H$;
by (c-ii), there is an $L\in\Cal K_{\delta}$ such that $L\subseteq H$
and $\gamma\le\phi_1L=\mu L$.   Thus $\mu$ is inner regular with respect
to $\Cal K_{\delta}$.
To see that $\mu$ is unique, observe that if $\mu'$ is any other measure
with these properties, and $L\in\Cal K_{\delta}$, then $L$ is
expressible as
$\bigcap_{n\in\Bbb N}K_n$ where $\sequencen{K_n}$ is a sequence in
$\Cal K$.   Now

\Centerline{$\mu L
=\lim_{n\to\infty}\mu(\bigcap_{i\le n}K_i)
=\lim_{n\to\infty}\phi_0(\bigcap_{i\le n}K_i)
=\mu'L$.}

\noindent So $\mu$ and $\mu'$ must agree on $\Cal K_{\delta}$, and by
412Mb again they are identical.
}%end of proof of 413K

\leader{413L}{Corollary} (a) Let $X$ be a set, $\Sigma$ a subring of
$\Cal PX$, and $\nu:\Sigma\to\coint{0,\infty}$
a non-negative finitely additive functional such
that $\lim_{n\to\infty}\nu E_n=0$ whenever $\sequencen{E_n}$ is a
non-increasing sequence in $\Sigma$ with empty intersection.   Then
$\nu$ has a unique extension to a complete locally determined measure on
$X$ which is inner regular with respect to the family $\Sigma_{\delta}$
of intersections of sequences in $\Sigma$.

(b) Let $X$ be a set, $\Sigma$ a subalgebra of $\Cal PX$, and
$\nu:\Sigma\to\coint{0,\infty}$ a non-negative finitely additive functional such
that $\lim_{n\to\infty}\nu E_n=0$ whenever $\sequencen{E_n}$ is a
non-increasing sequence in $\Sigma$ with empty intersection.   Then
$\nu$ has a unique extension to a measure defined on the
$\sigma$-algebra of subsets of $X$ generated by $\Sigma$.

\proof{{\bf (a)} Take $\Sigma$, $\nu$ in place of $\Cal K$, $\phi_0$ in
413K.

\medskip

{\bf (b)} Let $\nu_1$ be the complete extension as in (a), and let
$\nu_1'$ be the restriction of $\nu_1$ to the $\sigma$-algebra $\Sigma'$
generated by $\Sigma$;  this is the extension required here.   To see
that $\nu_1'$ is unique, use the Monotone Class Theorem (136C).
}%end of proof of 413L

\cmmnt{\medskip

\noindent{\bf Remark} These are versions of the {\bf Hahn extension
theorem}.   You will sometimes see (b) above stated as `an
additive functional on an algebra of sets extends to a measure iff it is
countably additive'.   But this formulation depends on a different
interpretation
of the phrase `countably additive' from the one used in this book;  see
the note after the definition in 326I.}

\leader{413M}{}\cmmnt{ It will be useful to have a definition extending
an idea in \S342.

\medskip

\noindent}{\bf Definition} A {\bf countably compact class}\cmmnt{ (or
{\bf semicompact paving})} is a family
$\Cal K$ of sets such that $\bigcap_{n\in\Bbb N}K_n\ne\emptyset$
whenever $\sequencen{K_n}$ is a sequence in $\Cal K$ such that
$\bigcap_{i\le n}K_i\ne\emptyset$ for every $n\in\Bbb N$.

\leader{413N}{Corollary} Let $X$ be a set and $\Cal K$ a countably
compact class of subsets of $X$ such that

\inset{$\emptyset\in\Cal K$,}

\inset{($\dagger$) $K\cup
K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are disjoint,}

\inset{($\ddagger$) $K\cap K'\in\Cal K$ whenever $K$, $K'\in\Cal K$.}

\noindent Let $\phi_0:\Cal K\to\coint{0,\infty}$ be a
functional such that

\inset{($\alpha$) $\phi_0 K
  =\phi_0 L+\sup\{\phi_0 K':K'\in\Cal K,\,K'\subseteq
K\setminus L\}$ whenever $K$, $L\in\Cal K$ and $L\subseteq K$.}

\noindent Then there is a unique complete locally determined measure
$\mu$ on $X$ extending $\phi_0$ and inner regular with respect to
$\Cal K_{\delta}$, the family of sets expressible as intersections of
sequences in $\Cal K$.

\proof{ The point is that the hypothesis ($\beta$) of 413K is
necessarily satisfied:  if $\sequencen{K_n}$ is
a non-increasing sequence in $\Cal K$ with empty intersection, then,
because $\Cal K$ is countably compact, there must be some $n$ such that
$K_n=\emptyset$.   Since hypothesis ($\alpha$) here is already enough to
ensure that $\phi_0\emptyset=0$ and $\phi_0 K\ge 0$ for every $K\in\Cal K$,
we must have $\inf_{n\in\Bbb N}\phi_0K_n=0$.   So we apply 413K to get
the result.
}%end of proof of 413N

\leader{413O}{}\cmmnt{ I now turn to constructions of a different
kind, being extension theorems in which the extension is not uniquely
defined.   Again I start with a theorem on finitely additive
functionals.

\wheader{413O}{4}{2}{2}{60pt}

\noindent}{\bf Theorem} Let $X$ be a set, $\Tau_0$ a subring of
$\Cal PX$,
and $\nu_0:\Tau_0\to\coint{0,\infty}$ a finitely additive functional.
Suppose that $\Cal K\subseteq\Cal PX$ is a family of sets such that

\inset{($\dagger$) $K\cup K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are disjoint,}

\inset{($\ddagger$) $K\cap K'\in\Cal K$ for all $K$, $K'\in\Cal K$,}

\inset{every member of $\Cal K$ is included in some
member of $\Tau_0$,}

\noindent and $\nu_0$ is inner regular with respect to $\Cal K$ in the
sense that

\inset{($\alpha$) $\nu_0E=\sup\{\nu_0K:K\in\Cal K\cap\Tau_0,\,K\subseteq
E\}$ for every $E\in\Tau_0$.}

\noindent Then $\nu_0$ has an extension to a
non-negative finitely additive functional $\nu_1$, defined on a subring
$\Tau_1$ of $\Cal PX$ including $\Tau_0\cup\Cal K$, inner regular with
respect to $\Cal K$, and such that whenever $E\in\Tau_1$ and $\epsilon>0$
there is an $E_0\in\Tau_0$ such that $\nu_1(E\symmdiff E_0)\le\epsilon$.

\proof{{\bf (a)} Let $P$ be the set of all non-negative additive
real-valued
functionals $\nu$, defined on subrings of $\Cal PX$, inner regular with
respect to $\Cal K$, and such that

\inset{($*$) whenever $E\in\dom\nu$ and $\epsilon>0$ there is an
$E_0\in\Tau_0$
such that $\nu(E\symmdiff E_0)\le\epsilon$.}

\noindent Order $P$ by extension of functions, so that $P$ is a
partially
ordered set.

\medskip

{\bf (b)} It will be convenient to borrow some notation from the theory
of countably additive functionals.   If $\Tau$ is a subring of $\Cal PX$
and $\nu:\Tau\to\coint{0,\infty}$ is a non-negative additive functional,
set

\Centerline{$\nu^*A=\inf\{\nu E:A\subseteq E\in\Tau\}$,
\quad$\nu_*A=\sup\{\nu E:A\supseteq E\in\Tau\}$}

\noindent for every $A\subseteq X$ (interpreting $\inf\emptyset$ as
$\infty$ if necessary).   Now if $A\subseteq X$ and $E$, $F\in\Tau$ are
disjoint,

\Centerline{$\nu^*(A\cap(E\cup F))=\nu^*(A\cap E)+\nu^*(A\cap F)$,}

\Centerline{$\nu_*(A\cap(E\cup F))=\nu_*(A\cap E)+\nu_*(A\cap F)$.}

$$\eqalignno{\text{\Prf\ }\nu^*(A\cap(E\cup F))
&=\inf\{\nu G:G\in\Tau,\,A\cap(E\cup F)\subseteq G\}\cr
&=\inf\{\nu G:G\in\Tau,\,A\cap(E\cup F)\subseteq G\subseteq E\cup F\}\cr
&=\inf\{\nu(G\cap E)+\nu(G\cap F):G\in\Tau,\,A\cap(E\cup F)\subseteq
G\subseteq E\cup F\}\cr
&=\inf\{\nu G_1+\nu G_2:G_1,\,G_2\in\Tau,\,A\cap E\subseteq G_1\subseteq
E,\,A\cap F\subseteq G_2\subseteq F\}\cr
&=\inf\{\nu G_1:G_1\in\Tau,\,A\cap E\subseteq G_1\subseteq E\}\cr
 &\qquad\qquad\qquad
  +\inf\{\nu G_2:G_2\in\Tau,\,A\cap F\subseteq G_2\subseteq F\}\cr
&=\nu^*(E\cap A)+\nu^*(F\cap A),\cr
\noalign{\vskip 12pt}
\nu_*(A\cap(E\cup F))
&=\sup\{\nu G:G\in\Tau,\,A\cap(E\cup F)\supseteq G\}\cr
&=\sup\{\nu(G\cap E)+\nu(G\cap F):G\in\Tau,\,A\cap(E\cup F)\supseteq
G\}\cr
&=\sup\{\nu G_1+\nu G_2:G_1,\,G_2\in\Tau,\,A\cap E\supseteq G_1,\,A\cap
F\supseteq G_2\}\cr
&=\sup\{\nu G_1:G_1\in\Tau,\,A\cap E\supseteq G_1\}\cr
  &\qquad\qquad\qquad+\sup\{\nu G_2:G_2\in\Tau,\,A\cap F\supseteq
G_2\}\cr
&=\nu_*(E\cap A)+\nu_*(F\cap A).   \text{ \Qed}\cr}$$

\medskip

{\bf (c)} The key to the proof is the following fact:  if $\nu\in P$ and
$M\in\Cal K$, there is a $\nuprime\in P$ such that $\nuprime$ extends $\nu$ and
$M\in\dom\nuprime$.   \Prf\ Set
$\Tau=\dom\nu$, $\Tau'=\{(E\cap M)\cup(F\setminus M):E,\,F\in\Tau\}$.
For $H\in\Tau'$, set

\Centerline{$\nuprime H=\nu^*(H\cap M)+\nu_*(H\setminus M)$.}

\noindent Now we have to check the following.

\medskip

\quad{\bf (i)} $\Tau'$ is a subring of $\Cal PX$, because if $E$, $F$,
$E'$, $F'\in\Tau$ then

\Centerline{$((E\cap M)\cup(F\setminus M))
  \star((E'\cap M)\cup(F'\setminus M))
=((E\star E')\cap M)\cup((F\star F')\setminus M)$}

\noindent for both the Boolean operations $\star=\symmdiff$ and $\star=\cap$.
$\Tau'\supseteq\Tau$ because $E=(E\cap M)\cup(E\setminus M)$ for every
$E\in\Tau$.   (Cf.\ 312N.)   $M\in\Tau'$ because there is some
$E\in\Tau_0$ such that
$M\subseteq E$, so that $M=(E\cap M)\cup(\emptyset\setminus M)\in\Tau'$.

\medskip

\quad{\bf (ii)} $\nuprime$ is finite-valued because if $H=(E\cap
M)\cup(F\setminus M)$, where $E$, $F\in\Tau$, then
$\nuprime H\le\nu E+\nu F$.
If $H$, $H'\in\Tau$ are disjoint, they can be expressed as $(E\cap
M)\cup(F\setminus M)$, $(E'\cap M)\cup(F'\setminus M)$ where $E$, $F$,
$E'$,
$F'$ belong to $\Tau$;  replacing $E'$, $F'$ by $E'\setminus E$
and $F'\setminus F$ if necessary, we may suppose that $E\cap E'=F\cap
F'=\emptyset$.   Now

$$\eqalignno{\nuprime(H\cup H')
&=\nu^*((E\cup E')\cap M)+\nu_*((F\cup F')\cap(X\setminus M))\cr
&=\nu^*(E\cap M)+\nu^*(E'\cap M)
   +\nu_*(F\cap(X\setminus M))+\nu_*(F'\cap(X\setminus M))\cr
\noalign{\noindent (by (b) above)}
&=\nuprime H+\nuprime H'.\cr}$$

\noindent Thus $\nuprime$ is additive.

\medskip

\quad{\bf (iii)} If $E\in\Tau$, then

$$\eqalign{\nu_*(E\setminus M)
&=\sup\{\nu F:F\in\Tau,\,F\subseteq E\setminus M\}\cr
&=\sup\{\nu E-\nu(E\setminus F):F\in\Tau,\,F\subseteq E\setminus M\}\cr
&=\sup\{\nu E-\nu F:F\in\Tau,\,E\cap M\subseteq F\subseteq E\}\cr
&=\nu E-\inf\{\nu F:F\in\Tau,\,E\cap M\subseteq F\subseteq E\}
=\nu E-\nu^*(E\cap M).\cr}$$

\noindent So

\Centerline{$\nuprime E=\nu^*(E\cap M)+\nu_*(E\setminus M)=\nu E$.}

\noindent Thus $\nuprime$ extends $\nu$.

\medskip

\quad{\bf (iv)} If $H\in\Tau'$ and $\epsilon>0$, express $H$ as $(E\cap
M)\cup(F\setminus M)$, where $E$, $F\in\Tau$.   Then we can find
($\alpha$)
a $K\in\Cal K\cap\Tau$ such that $K\subseteq E$ and $\nu(E\setminus
K)\le\epsilon$ ($\beta$) an $F'\in\Tau$ such that $F'\subseteq
F\setminus
M$
and $\nu F'\ge\nu_*(F\setminus M)-\epsilon$ ($\gamma$) a $K'\in\Cal
K\cap\Tau$ such that $K'\subseteq F'$ and $\nu K'\ge\nu F'-\epsilon$.
Set
$L=(K\cap M)\cup K'\in\Tau'$;  by the hypotheses ($\dagger$) and
($\ddagger$), $L\in\Cal K$.   Now $L\subseteq H$ and

$$\eqalign{\nuprime L
&=\nuprime(K\cap M)+\nuprime K'
=\nuprime(E\cap M)-\nuprime((E\setminus K)\cap M)+\nu K'\cr
&=\nu^*(H\cap M)-\nu^*((E\setminus K)\cap M)+\nu K'
\ge\nu^*(H\cap M)-\nu(E\setminus K)+\nu F'-\epsilon\cr
&\ge\nu^*(H\cap M)+\nu_*(F\setminus M)-3\epsilon
=\nuprime H-3\epsilon.\cr}$$

\noindent As $H$ and $\epsilon$ are arbitrary, $\nu$ is inner regular
with
respect to $\Cal K$.

\medskip
\quad{\bf (v)} Finally, given $H\in\Tau'$ and $\epsilon>0$, take $E$,
$F\in\Tau$ such that $H\cap M\subseteq E$, $F\subseteq H\setminus M$,
$\nu
E\le\nu^*(H\cap M)+\epsilon$ and $\nu F\ge\nu_*(H\setminus M)-\epsilon$.
In this case,

\Centerline{$\nuprime(E\setminus(H\cap M))=\nuprime E-\nuprime(H\cap M)
=\nu E-\nu^*(H\cap M)\le\epsilon$,}

\Centerline{$\nuprime((H\setminus M)\setminus F)=\nuprime(H\setminus M)-\nuprime F
=\nu_*(H\setminus M)-\nu F\le\epsilon$.}

\noindent But as

\Centerline{$H\symmdiff(E\cup F)\subseteq(E\setminus(H\cap
M))\cup((H\setminus M)\setminus F)$,}

\noindent $\nuprime(H\symmdiff(E\cup F))\le 2\epsilon$.   Now $\nu$
satisfies
the condition ($*$), so there is an $E_0\in\Tau_0$ such that $\nu((E\cup
F)\symmdiff E_0)\le\epsilon$, and $\nuprime(H\symmdiff E_0)\le 3\epsilon$.
As
$H$ and $\epsilon$ are arbitrary, $\nuprime$ satisfies ($*$).

This completes the proof that $\nuprime$ is a member of $P$ extending
$\nu$.\ \Qed

\medskip

{\bf (d)} It is easy to check that if $Q\subseteq P$ is a
non-empty totally ordered subset, the smallest common extension $\nuprime$
of the functions in $Q$ belongs to $P$.   (To see that $\nuprime$ is inner
regular with respect to $\Cal K$, observe that if $E\in\dom\nuprime$ and
$\gamma<\nuprime E$,
there is some $\nu\in Q$ such that $E\in\dom\nu$;  now there is a
$K\in\Cal K\cap\dom\nu$ such that $K\subseteq E$ and $\nu K\ge\gamma$,
so that $K\in\Cal K\cap\dom\nuprime$ and $\nuprime K\ge\gamma$.)
And of course $P$ is
not empty, because $\nu_0\in P$.   So by Zorn's Lemma $P$ has a maximal
element $\nu_1$ say;  write $\Tau_1$ for the domain of $\nu_1$.   If
$M\in\Cal K$ there is an element of $P$, with a domain containing $M$,
extending $\nu_1$; as $\nu_1$ is maximal, this must be $\nu_1$ itself,
so $M\in\Tau_1$.   Thus
$\Cal K\subseteq\Tau_1$, and $\nu_1$ has all the required properties.
}%end of proof of 413O

\vleader{72pt}{413P}{Corollary} Let $(X,\Tau,\nu)$ be a measure space and
$\Cal K$ a countably compact class of subsets of $X$ such that

\inset{($\dagger$) $K\cup K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are
disjoint,}

\inset{($\ddagger$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ for every
sequence $\sequencen{K_n}$ in $\Cal K$,}

\inset{$\nu^*K<\infty$ for every $K\in\Cal K$,}

\inset{$\nu$ is inner regular with respect to $\Cal K$.}

\noindent Then $\nu$ has an extension to a complete locally determined
measure $\mu$, defined on every member of $\Cal K$, inner regular with
respect to $\Cal K$, and such that whenever $E\in\dom\mu$ and $\mu E<\infty$
there is an $F\in\Tau$ such that $\mu(E\symmdiff F)=0$.

\proof{{\bf (a)} Set $\Tau^f=\{E:E\in\Tau,\,\nu E<\infty\}$.
Then $\Tau^f$ and $\nu\restr\Tau^f$
satisfy the conditions of 413O;   take $\nu_1$ extending $\nu$ to
$\Tau_1\supseteq\Tau^f\cup\Cal K$ as in 413O.
If $K$, $L\in\Cal K$ and $L\subseteq K$, then

\Centerline{$\nu_1L+\sup\{\nu_1K':K'\in\Cal K,\,K'\subseteq K\setminus
L\}
=\nu_1L+\nu_1(K\setminus L)=\nu_1K$.}

\noindent So $\nu_1\restr\Cal K$ satisfies the conditions of 413N and
there is a complete locally determined measure $\mu$, extending
$\nu_1\restr\Cal
K$, and inner regular with respect to $\Cal K$.

\medskip

{\bf (b)} Write $\Sigma$ for the domain of $\mu$.   Then
$\Tau_1\subseteq\Sigma$.   \Prf\ If $E\in\Tau_1$ and $K\in\Cal K$,

$$\eqalign{\mu_*(K\cap E)&+\mu_*(K\setminus E)\cr
&\ge\sup\{\mu K':K'\in\Cal K,\,K'\subseteq K\cap E\}
  +\sup\{\mu K':K'\in\Cal K,\,K'\subseteq K\setminus E\}\cr
&=\sup\{\nu_1 K':K'\in\Cal K,\,K'\subseteq K\cap E\}
  +\sup\{\nu_1 K':K'\in\Cal K,\,K'\subseteq K\setminus E\}\cr
&=\nu_1(K\cap E)+\nu_1(K\setminus E)
=\nu_1 K
=\mu K.\cr}$$

\noindent By 413F(iv), $E\in\Sigma$.\ \QeD\  It follows at once that
$\mu$ extends $\nu_1$, since if $E\in\Tau_1$

\Centerline{$\nu_1E
=\sup\{\nu_1K:K\in\Cal K,\,K\subseteq E\}
=\sup\{\mu K:K\in\Cal K,\,K\subseteq E\}
=\mu E$.}

\medskip

{\bf (c)} In particular, $\mu$ agrees with $\nu$ on $\Tau^f$.   Now in
fact $\mu$ extends $\nu$.   \Prf\  Take $E\in\Tau$.   If $K\in\Cal K$,
there is an $F\in\Tau$ such that $K\subseteq F$ and
$\nu F<\infty$.   Since $E\cap F\in\Tau^f\subseteq\Sigma$,
$E\cap K=E\cap F\cap K\in\Sigma$.   As $K$ is arbitrary, $E\in\Sigma$, by
413F(ii).   Next, because every member of $\Cal K$ is included in a
member of $\Tau^f$,

$$\eqalign{\nu E
&=\sup\{\nu K:K\in\Cal K\cap\Tau,\,K\subseteq E\}
=\sup\{\nu(E\cap F):F\in\Tau^f\}\cr
&=\sup\{\mu(E\cap F):F\in\Tau^f\}
=\sup\{\mu K:K\in\Cal K,\,K\subseteq E\}
=\mu E.  \text{ \Qed}\cr}$$

\medskip

{\bf (d)} Finally, suppose that $E\in\Sigma$ and $\mu E<\infty$.   For each
$n\in\Bbb N$ we
can find $K_n\in\Cal K$ and $F_n\in\Tau$ such that $K_n\subseteq E$,
$\mu(E\setminus K_n)\le 2^{-n}$ and $\nu_1(K_n\symmdiff F_n)\le 2^{-n}$.
In this case $\sum_{n=0}^{\infty}\mu(E\symmdiff F_n)<\infty$, so
$\mu(E\symmdiff F)=0$, where
$F=\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}F_m\in\Tau$.

Thus $\mu$ has all the required properties.
}%end of proof of 413P

\leader{413Q}{Definitions} Let $P$ be a lattice and
$f:P\to\coint{-\infty,\infty}$ a function.

\spheader 413Qa\dvAformerly{4{}13P} $f$ is {\bf supermodular} if
$f(p\vee q)+f(p\wedge q)\ge f(p)+f(q)$ for all $p$, $q\in P$.

\spheader 413Qb\dvAformerly{4{}32Jc} $f$ is {\bf submodular} if
$f(p\vee q)+f(p\wedge q)\le f(p)+f(q)$ for all $p$, $q\in P$.
\cmmnt{The phrase `strongly subadditive' is used by many
authors in similar contexts.}
%"strongly subadditive" in Choquet 69, Zapletal p06, Kechris 95.

\spheader 413Qc\dvAformerly{4{}13Xq} $f$ is {\bf modular} if
$f(p\vee q)+f(p\wedge q)=f(p)+f(q)$ for all $p$, $q\in P$.

\leader{413R}{}\cmmnt{ I now describe an alternative route to some of
the applications of 413O.   As before, I do as much as possible in the
context of finitely additive functionals.

\medskip

\noindent}{\bf Lemma} Let $X$ be a set and $\Cal K$ a sublattice of
$\Cal PX$ containing $\emptyset$.
Let $\phi:\Cal K\to\Bbb R$ be a bounded supermodular functional
such that $\phi\emptyset=0$.   Then there is a finitely additive functional
$\nu:\Cal PX\to\coint{0,\infty}$ such that

\Centerline{$\nu X=\sup_{K\in\Cal K}\phi K$,
\quad$\nu K\ge\phi K$ for every $K\in\Cal K$.}

\proof{{\bf (a)} Let us consider first the case in which $\Cal K$ is finite and
$\phi$ is non-decreasing.   I
induce on $n=\#(\Cal K)$.   If $n=1$ then $\Cal K=\{\emptyset\}$ and $\nu$ must be the
zero functional.   For the inductive step to $n>1$,
let $K_0$ be a minimal member of $\Cal K\setminus\{\emptyset\}$.
If $K\in\Cal K$ then $K\cap K_0$ is a member of $\Cal K$ included in $K_0$,
so is either empty or
$K_0$, that is, either $K\cap K_0=\emptyset$ or $K\supseteq K_0$.   Set
$Y=X\setminus K_0$ and $\Cal L=\{K\setminus K_0:K\in\Cal K\}$.   Then
$\Cal L$ is a sublattice of $\Cal PY$ containing $\emptyset$, and
$K\mapsto K\setminus K_0:\Cal K\to\Cal L$ is surjective but not injective, so
$\#(\Cal L)<n$.

For $L\in\Cal L$, observe that $L\cup K_0\in\Cal K$.
\Prf\ There is a $K\in\Cal K$ such that $L=K\setminus K_0$.
Now If $K$ is disjoint from
$K_0$, then $L\cup K_0=K\cup K_0$ belongs to $\Cal K$;
if $K$ includes $K_0$ then
$L\cup K_0=K$ belongs to $\Cal K$.\ \Qed

We can therefore define $\phi':\Cal L\to\coint{0,\infty}$ by setting

\Centerline{$\phi'L=\phi(L\cup K_0)-\phi K_0$}

\noindent for every $L\in\Cal L$.   Of course $\phi'\emptyset=0$ and
$\phi'L\le\phi'L'$ whenevery $L\subseteq L'$.   If $L$, $L'\in\Cal L$ then

$$\eqalign{\phi'(L\cup L')+\phi'(L\cap L')
&=\phi(L\cup L'\cup K_0)+\phi((L\cap L')\cup K_0)-2\phi K_0\cr
&=\phi((L\cup K_0)\cup(L'\cup K_0))+\phi((L\cup K_0)\cap(L'\cup K_0))
    -2\phi K_0\cr
&\ge\phi(L\cup K_0)+\phi(L'\cup K_0)-2\phi K_0
=\phi'L+\phi'L'.\cr}$$

\noindent So $\phi'$ is supermodular,
and by the inductive hypothesis there is a finitely additive functional
$\nuprime:\Cal PY\to\coint{0,\infty}$ such that

\Centerline{$\nuprime Y=\sup_{L\in\Cal L}\phi'L$,
\quad$\nuprime L\ge\phi'L$ for every $L\in\Cal L$.}

Fix any $x_0\in K_0$ and define $\nu:\Cal PX\to\coint{0,\infty}$ by setting

$$\eqalign{\nu A
&=\phi K_0+\nuprime(A\cap Y)\text{ if }x_0\in A\subseteq X,\cr
&=\nuprime(A\cap Y)\text{ for other }A\subseteq X.\cr}$$

\noindent Then $\nu$ is additive.   If $K\in\Cal K$ is disjoint from $K_0$ then

\Centerline{$\nu K=\nuprime K\ge\phi'K=\phi(K\cup K_0)-\phi K_0
\ge\phi K-\phi(K\cap K_0)=\phi K$.}

\noindent If $K\in\Cal K$ includes $K_0$ then

\Centerline{$\nu K=\phi K_0+\nuprime(K\setminus K_0)
\ge\phi K_0+\phi'(K\setminus K_0)
=\phi K_0+\phi K-\phi K_0=\phi K$.}

\noindent Finally,

$$\eqalign{\nu X
&=\phi K_0+\nuprime Y
=\phi K_0+\sup_{L\in\Cal L}\phi'L\cr
&=\phi K_0+\sup_{K\in\Cal K}(\phi(K\cup K_0)-\phi K_0)
=\sup_{K\in\Cal K}\phi(K\cup K_0)
=\sup_{K\in\Cal K}\phi K.\cr}$$

\noindent So $\nu$ has the required properties and the induction continues.

\medskip

{\bf (b)} Now suppose only that $\phi$ is non-decreasing.
Set $\gamma=\sup_{K\in\Cal K}\phi K$.
We need to know that every finite subset of $\Cal K$ is included in a
finite sublattice of $\Cal K$;  this is because it is included in a
finite subalgebra $\Cal E$ of $\Cal PX$ and $\Cal K\cap\Cal E$ is a
sublattice.   Let $\Nu$ be the set of all finitely additive functionals
$\nu:\Cal PX\to[0,\gamma]$.   Then $\Nu$ is a closed subset of
$[0,\gamma]^{\Cal PX}$, so is compact.   For each $K\in\Cal K$ set
$\Nu_K=\{\nu:\nu\in\Nu$, $\nu K\ge\phi K\}$.   Then $\Nu_K$ is a
closed subset of $\Nu$.   If $\Cal K_0\subseteq\Cal K$ is finite, there
is a finite sublattice $\Cal L$ of $\Cal K$ including
$\Cal K_0\cup\{\emptyset\}$, and now (a) tells us that there is a
$\nu\in\bigcap_{K\in\Cal L}\Nu_K$.   Thus $\{\Nu_K:K\in\Cal K\}$ has the
finite intersection property and there is a
$\nu\in\bigcap_{K\in\Cal K}\Nu_K$.   In this case,
$\nu:\Cal PX\to[0,\gamma]$ is a finitely additive functional dominating
$\phi$;  it follows that $\nu X=\gamma$ and the proof is complete.

\medskip

{\bf (c)} Finally, for the general case, set
$\phi'K=\sup_{L\in\Cal K,L\subseteq K}\phi L$ for $K\in\Cal K$.
Then $\phi'$ is non-decreasing, $\phi'\emptyset=0$ and
$\sup_{K\in\Cal K}\phi'K=\sup_{K\in\Cal K}\phi K$ is finite.   If
$K$, $K'\in\Cal K$ then

$$\eqalign{\phi'K+\phi'K'&
=\sup_{\Atop{L\in\Cal K}{L\subseteq K}}\phi L
     +\sup_{\Atop{L'\in\Cal K}{L'\subseteq K'}}\phi L'
=\sup_{\Atop{L,L'\in\Cal K}{\Atop{L\subseteq K}{L'\subseteq K'}}}
   \phi L+\phi L'\cr&
\le\sup_{\Atop{L,L'\in\Cal K}{\Atop{L\subseteq K}{L'\subseteq K'}}}
   \phi(L\cup L')+\phi(L\cap L')
\le\phi'(K\cup K')+\phi'(K\cap K'),\cr}$$

\noindent so $\phi'$ is supermodular.   By (b), there is an additive
$\nu:\Cal PX\to\coint{0,\infty}$ such that $\nu K\ge\phi'K\ge\phi K$ for
every $K\in\Cal K$ and
$\nu X=\sup_{K\in\Cal K}\phi'K=\sup_{K\in\Cal K}\phi K$.
}%end of proof of 413R

\vleader{72pt}{413S}{Theorem} Let $X$ be a set and $\Cal K$ a sublattice of
$\Cal PX$ containing $\emptyset$.
Let $\Sigma$ be the algebra of subsets of $X$ generated by $\Cal K$, and
$\nu_0:\Sigma\to\coint{0,\infty}$ a finitely additive functional.
Then there is a finitely additive functional
$\nu:\Sigma\to\coint{0,\infty}$ such that

(i) $\nu X=\sup_{K\in\Cal K}\nu_0K$,

(ii) $\nu K\ge\nu_0K$ for every $K\in\Cal K$,

(iii) $\nu$ is inner regular with respect to $\Cal K$
in the sense that $\nu E=\sup\{\nu K:K\in\Cal K,\,K\subseteq E\}$ for
every $E\in\Sigma$.

\proof{{\bf (a)} Set $\gamma=\sup_{K\in\Cal K}\nu_0K$.   Let $P$ be the
set of all supermodular functionals $\phi:\Cal K\to[0,\gamma]$.
Give $P$ the natural partial
order inherited from $\BbbR^{\Cal K}$.   Note that $\nu_0\restr\Cal K$ is
actually modular, so
belongs to $P$.   If $Q\subseteq P$ is non-empty and upwards-directed,
then $\sup Q$, taken in $\BbbR^{\Cal K}$, belongs to $P$;  so there is
a maximal $\phi\in P$ such that $\nu_0\restr\Cal K\le\phi$.   By
413R, there is a non-negative additive functional $\nu$ on $\Cal PX$ such that
$\nu K\ge\phi K$ for every $K\in\Cal K$ and $\nu X=\gamma$.   Since
$\nu\restr\Cal K$ also belongs to $P$, we
must have $\nu K=\phi K$ for every $K\in\Cal K$.

\medskip

{\bf (b)} Now for any $K_0\in\Cal K$,

\Centerline{$\nu K_0
  +\sup\{\nu L:L\in\Cal K,\,L\subseteq X\setminus K_0\}
=\gamma$.}

\medskip

\Prf\ {\bf (i)} Set
$\Cal L=\{L:L\in\Cal K,\,L\subseteq X\setminus K_0\}$.
For $A\subseteq X$, set $\theta_0A=\sup_{L\in\Cal L}\nu(A\cap L)$.
Because
$\Cal L$ is upwards-directed, $\theta_0:\Cal PX\to\Bbb R$ is additive,
and
of course $0\le\theta_0\le\nu$.   Set $\theta_1=\nu-\theta_0$, so that
$\theta_1$ is another additive functional, and write

\Centerline{$\phi'K=\theta_0K+\sup\{\theta_1M:M\in\Cal K,\,
M\cap K_0\subseteq K\}$}

\noindent for $K\in\Cal K$.

\medskip

\quad{\bf (ii)} If $K$, $K'\in\Cal K$ and $\epsilon>0$, there are $M$,
$M'\in\Cal K$ such that $M\cap K_0\subseteq K$, $M'\cap K_0\subseteq K'$
and

\Centerline{$\theta_0K+\theta_1M\ge\phi'K-\epsilon$,
\quad$\theta_0K'+\theta_1M'\ge\phi'K'-\epsilon$.}

\noindent Now

\Centerline{$M\cup M'\in\Cal K$,\quad $M\cap M'\in\Cal K$,}

\Centerline{$(M\cup M')\cap K_0\subseteq K\cup K'$,
\quad$(M\cap M')\cap K_0\subseteq K\cap K'$,}

\noindent so

$$\eqalign{\phi'(K\cup K')+\phi'(K\cap K')
&\ge\theta_0(K\cup K')+\theta_1(M\cup M')
  +\theta_0(K\cap K')+\theta_1(M\cap M')\cr
&=\theta_0K+\theta_1M+\theta_0K'+\theta_1M'
\ge\phi'K+\phi'K'-2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, 
$\phi'(K\cup K')+\phi'(K\cap K')\ge\phi K+\phi K'$.

\medskip

\quad{\bf (iii)} Suppose that $K$, $M\in\Cal K$ are such that $M\cap
K_0\subseteq K$.   If $L\in\Cal L$, then

$$\eqalign{\nu(K\cap L)+\theta_1M
&=\nu(K\cap L)+\nu M-\theta_0M\cr
&=\nu(M\cap K\cap L)+\nu(M\cup(K\cap L))-\theta_0M
\le\gamma\cr}$$

\noindent because $K\cap L\in\Cal L$;   taking the supremum over $L$ and
$M$, $\phi'K\le\gamma$.   As $K$ is arbitrary, $\phi'\in P$.

\medskip

\quad{\bf (iv)} If $K\in\Cal K$, then of course $K\cap K_0\subseteq K$,
so

\Centerline{$\phi'K\ge\theta_0K+\theta_1K=\nu K=\phi K$.}

\noindent Thus $\phi'\ge\phi$.   Because $\phi$ is maximal,
$\phi'=\phi$.   But this means that

\Centerline{$\phi K_0=\phi'K_0
=\theta_0K_0+\sup\{\theta_1M:M\in\Cal K,\,M\cap K_0\subseteq K_0\}
=\sup_{M\in\Cal K}\theta_1M$.}

Now given $\epsilon>0$ there is an $M\in\Cal K$ such that

\Centerline{$\gamma-\epsilon\le\nu_0M\le\phi M=\nu M$,}

\noindent so that

\Centerline{$\nu K_0=\phi K_0\ge\theta_1M=\nu M-\theta_0M
\ge\gamma-\epsilon-\theta_0M
\ge\gamma-\epsilon-\sup_{L\in\Cal L}\nu L$,}

\noindent and $\nu K_0+\sup_{L\in\Cal L}\nu L\ge\gamma-\epsilon$.   As
$\epsilon$ is arbitrary, $\nu K_0+\sup_{L\in\Cal L}\nu L\ge\gamma$.
But of course $\nu K_0+\nu L\le\nu X=\gamma$ for every $L\in\Cal L$, so
$\nu K_0+\sup_{L\in\Cal L}\nu L=\gamma$, as claimed.\ \Qed

\medskip

{\bf (c)} It follows that if $K$, $L\in\Cal K$ and $L\subseteq K$,

\Centerline{$\nu K=\nu L+\sup\{\nu K':K'\in\Cal K,\,K'\subseteq
K\setminus L\}$.}

\noindent\Prf\ Because $\nu$ is additive and non-negative, we surely
have

\Centerline{$\nu K\ge\nu L+\sup\{\nu K':K'\in\Cal K,\,K'\subseteq
K\setminus L\}$.}

\noindent On the other hand, given $\epsilon>0$, there is an $M\in\Cal
K$ such that $M\subseteq X\setminus L$ and
$\nu L+\nu M\ge\gamma-\epsilon$, so that $M\cap K\in\Cal K$,
$M\cap K\subseteq K\setminus L$
and

\Centerline{$\nu L+\nu(M\cap K)=\nu L+\nu K+\nu M-\nu(M\cup K)
\ge\nu K+\gamma-\epsilon-\gamma=\nu K-\epsilon$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\nu K
\le\nu L+\sup\{\nu K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$}

\noindent and we have equality.\ \Qed

\medskip

{\bf (d)} By 413I, we have an additive functional
$\nuprime:\Sigma\to\coint{0,\infty}$ such that
$\nuprime E=\sup\{\nu K:K\in\Cal K,\,K\subseteq E\}$ for every $E\in\Sigma$.
Using 313Ga, it is easy to show that
$\nuprime$ and $\nu$ must agree on $\Sigma$, but even without doing so we
can see that
$\nuprime$ has the properties (i)-(iii) required in the theorem.
}%end of proof of 413S

\leader{413T}{}\cmmnt{ The following lemma on countably compact
classes, corresponding to 342Db, will be useful.

\medskip

\noindent}{\bf Lemma}\cmmnt{ ({\smc Marczewski 53})} Let $X$ be a set
and $\Cal K$ a countably compact class of subsets of $X$.
Then there is a countably compact class
$\Cal K^*\supseteq\Cal K\cup\{\emptyset,X\}$ such
that $K\cup L\in\Cal K^*$ and $\bigcap_{n\in\Bbb N}K_n\in\Cal K^*$
whenever $K$, $L\in\Cal K^*$ and $\sequencen{K_n}$ is a sequence in $\Cal K^*$.

\proof{{\bf (a)} Of course $\Cal K\cup\{\emptyset,X\}$ is still a countably
compact class of sets, so we can suppose from the beginning that $\emptyset$ and
$X$ belong to $\Cal K$.  Write $\Cal K_s$ for
$\{K_0\cup\ldots\cup K_n:K_0,\ldots,K_n\in\Cal K\}$.
Then $\Cal K_s$ is countably compact.
\Prf\ Let $\sequencen{L_n}$ be a sequence in $\Cal K_s$ such that
$\bigcap_{i\le n}L_i\ne\emptyset$ for each $n\in\Bbb N$.   Then there is
an ultrafilter $\Cal F$ on $X$ containing every $L_n$.
For each $n$, $L_n$ is a finite union of members of $\Cal K$, so there
must be a $K_n\in\Cal K$ such that $K_n\subseteq L_n$ and $K_n\in\Cal
F$.   Now $\bigcap_{i\le n}K_i\ne\emptyset$ for every $n$, so
$\bigcap_{n\in\Bbb N}K_n\ne\emptyset$ and $\bigcap_{n\in\Bbb
N}L_n\ne\emptyset$.   As $\sequencen{L_n}$ is arbitrary, $\Cal K_s$ is
countably compact.\ \Qed

Note that $L\cup L'\in\Cal K_s$ for all $L$, $L'\in\Cal K_s$.

\medskip

{\bf (b)} Write $\Cal K^*$ for

\Centerline{$\{\bigcap\Cal L_0:\Cal L_0\subseteq\Cal K_s$ is non-empty
and countable$\}$.}

\noindent Then $\Cal K^*$ is countably compact.   \Prf\ If
$\sequencen{M_n}$ is any sequence in $\Cal K^*$ such that
$\bigcap_{i\le n}M_i\ne\emptyset$ for every $n\in\Bbb N$, then for each
$n\in\Bbb N$
let $\Cal L_n\subseteq\Cal K_s$ be a countable non-empty set such that
$M_n=\bigcap\Cal L_n$.   Let $\sequencen{L_n}$ be a sequence running
over $\bigcup_{n\in\Bbb N}\Cal L_n$;  then
$\bigcap_{i\le n}L_i\ne\emptyset$ for every $n$, so
$\bigcap_{n\in\Bbb N}L_n=\bigcap_{n\in\Bbb N}M_n$ is non-empty.   As
$\sequencen{M_n}$ is arbitrary, $\Cal K^*$ is countably compact.\ \Qed

\medskip

{\bf (c)} Of course $\Cal K\subseteq\Cal K_s\subseteq\Cal K^*$.   It is
immediate from the definition of $\Cal K^*$ that it is closed under
countable intersections.   Finally, if $M_1$, $M_2\in\Cal K^*$, let
$\Cal L_1$, $\Cal L_2\subseteq\Cal K_s$ be countable sets such that
$M_1=\bigcap\Cal L_1$ and $M_2=\bigcap\Cal L_2$;  then
$\Cal L=\{L_1\cup L_2:L_1\in\Cal L_1,\,L_2\in\Cal L_2\}$ is a countable
subset of $\Cal K_s$, so $M_1\cup M_2=\bigcap\Cal L$ belongs to
$\Cal K^*$.
}%end of proof of 413T

\leader{413U}{Corollary} Let $X$ be a set and $\Cal K$ a countably
compact class of subsets of $X$.
Let $\Tau$ be a subalgebra of $\Cal PX$ and $\nu:\Tau\to\Bbb R$
a non-negative finitely additive functional.

(a) There is a complete measure $\mu$ on $X$ such that $\mu X\le\nu X$,
$\Cal K\subseteq\dom\mu$ and
$\mu K\ge\nu K$ for every $K\in\Cal K\cap\Tau$.

(b) If $\emptyset\in\Cal K$ and

\inset{($\dagger$) $K\cup K'\in\Cal K$ whenever $K$, $K'\in\Cal K$,}

\inset{($\ddagger$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ for every
sequence $\sequencen{K_n}$ in $\Cal K$,}

\noindent we may arrange that $\mu$ is inner regular with respect to $\Cal K$.

\proof{ By 413T, there is always a countably compact class
$\Cal K^*\supseteq\Cal K\cup\{\emptyset\}$
satisfying ($\dagger$) and ($\ddagger$);  for
case (b), take $\Cal K^*=\Cal K$.
By 391G, there is an extension of $\nu$ to an additive
functional $\nuprime:\Cal PX\to\coint{0,\infty}$.
Let $\Tau_1$ be the subalgebra of $\Cal PX$ generated by $\Cal K^*$.
By 413S, there is a non-negative additive functional
$\nu_1:\Tau_1\to\Bbb
R$ such that $\nu_1X\le\nuprime X=\nu X$, $\nu_1K\ge\nuprime K=\nu K$ for every
$K\in\Cal K^*\cap\Tau$ and $\nu_1E=\sup\{\nu_1K:K\in\Cal
K^*,\,K\subseteq
E\}$ for every $E\in\Tau_1$.   In particular, if $K$, $L\in\Cal K^*$,

\Centerline{$\nu_1L
   +\sup\{\nu_1K':K'\in\Cal K^*,\,K'\subseteq K\setminus L\}
=\nu_1L+\nu_1(K\setminus L)=\nu_1K$.}

\noindent So $\Cal K^*$ and $\nu_1\restr\Cal K^*$ satisfy the hypotheses
of 413N.   Accordingly we have a complete measure $\mu$ extending
$\nu_1\restr\Cal K^*$ and inner regular with respect to
$\Cal K^*=\Cal K^*_{\delta}$;  in
which case

\Centerline{$\mu K=\nu_1K\ge\nu K$}

\noindent for every $K\in\Cal K\cap\Tau$, and

\Centerline{$\mu X=\sup_{K\in\Cal K^*}\mu K
=\sup_{K\in\Cal K^*}\nu_1 K\le\nu_1X\le\nu X$,}

\noindent as required.
}%end of proof of 413U

\exercises{
\leader{413X}{Basic exercises (a)}
%\spheader 413Xa
Define $\phi:\Cal P\Bbb N\to\coint{0,\infty}$ by setting
$\phi A=0$ if $A$ is finite, $\infty$ otherwise.   Check that $\phi$
satisfies conditions ($\alpha$) and ($\beta$) of 413A, but that if we
attempt to reproduce the construction of 413C then we obtain
$\Sigma=\Cal P\Bbb N$ and $\mu=\phi$, so that $\mu$ is not countably
additive.
%413C

\spheader 413Xb Let $\phi_1$, $\phi_2$ be two inner measures on a set
$X$, inducing measures $\mu_1$ and $\mu_2$ by the method of 413C.   (i)
Show that $\phi=\phi_1+\phi_2$ is an inner measure.   (ii) Show that the
measure $\mu$
induced by $\phi$ extends the measure $\mu_1+\mu_2$ defined on
$\dom\mu_1\cap\dom\mu_2$.
%413C

\sqheader 413Xc Let $X$ be a set, $\phi$ an inner measure on $X$, and
$\mu$ the measure constructed from it by the method of 413C.   (i) Let
$Y$ be a subset of $X$.   Show that $\phi\restrp\Cal PY$ is an inner
measure on $Y$,
and that the measure on $Y$ defined from it extends the subspace measure
$\mu_Y$ induced on $Y$ by $\mu$.   (ii) Now suppose that $\phi X$ is
finite.   Let $Y$ be a set and $f:X\to Y$
a function.   Show that $B\mapsto\phi f^{-1}[B]$ is an inner measure on
$Y$, and that it defines a measure on $Y$ which extends the image
measure $\mu f^{-1}$.
%413C

\spheader 413Xd Let $(X,\Sigma,\mu)$ be a measure space.   Set
$\theta A=\bover12(\mu^*A+\mu_*A)$ for every $A\subseteq X$.   Show that
$\theta$ is
an outer measure on $X$, and that if $\mu$ is semi-finite then
the measure defined from $\theta$ by
\Caratheodory's method extends $\mu$.   (See also 438Ym below and 543Xd in
the next volume.)
%413D

\sqheader 413Xe Show that there is a partition $\sequencen{A_n}$ of
$[0,1]$ such that $\mu_*(\bigcup_{i\le n}A_i)=0$ for every $n$, where
$\mu_*$ is
Lebesgue inner measure.   \Hint{set $A_n=(A+q_n)\cap[0,1]$ where
$\sequencen{q_n}$ is an enumeration of $\Bbb Q$ and $A$ is a suitable
set;  cf.\ 134B.}
%413D

\spheader 413Xf Let $(X,\Sigma,\mu)$ be a measure space.   (i) Show that
$\mu_*\restr\Sigma$ is the semi-finite version $\mu_{\text{sf}}$ of $\mu$ as
constructed in 213Xc.   (ii) Show that if $A$ is any subset of $X$, and
$\Sigma_A$ the
subspace $\sigma$-algebra, then $\mu_*\restr\Sigma_A$ is a semi-finite
measure on $A$.
%413E

\sqheader 413Xg Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two measure
spaces, and $\lambda$ the c.l.d.\ product measure on $X\times Y$.   Show
that $\lambda_*(A\times B)=\mu_*A\cdot\nu_*B$ for all $A\subseteq X$ and
$B\subseteq Y$.   \Hint{use Fubini's theorem to show that
$\lambda_*(A\times B)\le\mu_*A\cdot\nu_*B$.}
%413E

\spheader 413Xh(i) Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure
space and $f:X\to\Bbb R$ a function such that $\overline{\int}fd\mu$ is
finite.   Show that for every $\epsilon>0$ there is a measure $\nu$ on
$X$ extending $\mu$ such that
$\underline{\int}fd\nu\ge\overline{\int}fd\mu-\epsilon$.
\Hint{215B(viii),
%strictly positive g_0 with \int g_0\le\epsilon
133Ja,
% f\le g\le f+g_0 on set A of full outer measure
% \int g=\overline{\int}f
417Xa.}
% \nu(X\setminus A)=0
(ii) Let $(X,\Sigma,\mu)$ be a totally finite
measure space and $f:X\to\Bbb R$ a bounded function.   Show that there is
a finitely additive functional $\nu:\Cal PX\to\coint{0,\infty}$,
extending $\mu$, such that $\dashint fd\nu$, defined as in 363Lf, is
equal to $\overline{\int}fd\mu$.
%413E

\spheader 413Xi Let $X$ be a set and $\mu$, $\nu$ two complete locally determined measures on $X$ with domains $\Sigma$, $\Tau$ respectively, both inner
regular with respect to $\Cal K\subseteq\Sigma\cap\Tau$.   Suppose that,
for $K\in\Cal K$, $\mu K=0$ iff $\nu K=0$.   Show that $\Sigma=\Tau$ and
that $\mu$ and $\nu$ have the same null ideals.
%413F

\sqheader 413Xj Let $(X,\Sigma,\mu)$ be a measure space.   (i) Show that
the measure constructed by the method of 413C from the inner measure
$\mu_*$ is the c.l.d.\ version of $\mu$.   (ii) Set
$\Cal K=\{E:E\in\Sigma,\,\mu E<\infty\}$, $\phi_0=\mu\restr\Cal K$.   Show
that $\Cal K$ and $\phi_0$ satisfy the conditions of 413J, and that the
measure constructed by the method there is again the c.l.d.\ version of
$\mu$.
%413J

\sqheader 413Xk Let $\Cal K$ be the family of subsets of $\Bbb R$
expressible as disjoint finite unions of bounded closed intervals.
(i) Show from first principles that there is a unique functional
$\phi_0:\Cal K\to\coint{0,\infty}$ such that
$\phi_0[\alpha,\beta]=\beta-\alpha$
whenever $\alpha\le\beta$ and $\phi_0$ satisfies the conditions of 413K.
(ii) Show that the measure on $\Bbb R$ constructed from $\phi_0$ by the
method of 413K is Lebesgue measure.
%413K

\spheader 413Xl Let $X$ be a set, $\Sigma$ a subring of $\Cal PX$, and
$\nu:\Sigma\to\coint{0,\infty}$ a non-negative additive functional such
that $\lim_{n\to\infty}\nu E_n=0$ whenever $\sequencen{E_n}$ is a
non-increasing sequence in $\Sigma$ with empty intersection, as in 413L.
Define $\theta:\Cal PX\to[0,\infty]$ by setting

\Centerline{$\theta A=\inf\{\sum_{n=0}^{\infty}\nu E_n:\sequencen{E_n}$
is a sequence in $\Sigma$ covering $A\}$}

\noindent for $A\subseteq X$, interpreting $\inf\emptyset$ as $\infty$
if necessary.   Show that $\theta$ is an outer measure.   Let
$\mu_{\theta}$ be
the measure defined from $\theta$ by \Caratheodory's method.   Show
that the measure defined from $\nu$ by the process of 413L is the
c.l.d.\ version
of $\mu_{\theta}$.   \Hint{the c.l.d.\ version of $\mu_{\theta}$ is
inner regular with respect to $\Sigma_{\delta}$.}
%413L

\sqheader 413Xm Let $X$ be a set, $\Sigma$ a subring of $\Cal PX$, and
$\nu:\Sigma\to\coint{0,\infty}$ a non-negative additive functional.
Show that the following are equiveridical:  (i) $\nu$ has an extension to a measure on $X$;  (ii) $\lim_{n\to\infty}\nu E_n=0$ whenever
$\sequencen{E_n}$ is a
non-increasing sequence in $\Sigma$ with empty intersection;  (iii)
$\nu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\nu E_n$ whenever
$\sequencen{E_n}$ is a disjoint sequence in $\Sigma$ such that
$\bigcup_{n\in\Bbb N}E_n\in\Sigma$.
%413L

\sqheader 413Xn Let $\sequencen{(X_n,\Sigma_n,\mu_n)}$ be a sequence of
probability spaces, and $\Cal F$ a non-principal ultrafilter on
$\Bbb N$.   For $x$, $y\in\prod_{n\in\Bbb N}X_n$, write $x\sim y$ if
$\{n:x(n)=y(n)\}\in\Cal F$.   (i) Show that $\sim$ is an equivalence
relation;  write $X$ for the set of equivalence classes, and
$x^{\ssbullet}\in X$ for the equivalence class of
$x\in\prod_{n\in\Bbb N}X_n$.  (Compare 351M.)   (ii) Let $\Sigma$ be the
set of subsets of $X$ expressible in the form
$Q(\sequencen{E_n})=\{x^{\ssbullet}:x\in\prod_{n\in\Bbb N}E_n\}$, where
$E_n\in\Sigma_n$ for each $n\in\Bbb N$.   Show that $\Sigma$ is an
algebra of subsets of $X$, and that there is a well-defined additive
functional $\nu:\Sigma\to[0,1]$ defined by setting
$\nu(Q(\sequencen{E_n}))=\lim_{n\to\Cal F}\mu_nE_n$.   (iii) Show that
for any non-increasing sequence $\sequence{i}{H_i}$ in $\Sigma$ there is
an $H\in\Sigma$ such that $H\subseteq\bigcap_{i\in\Bbb N}H_i$ and $\nu
H=\lim_{n\to\infty}\nu H_n$.   \Hint{express each $H_i$ as
$Q(\sequencen{E_{in}})$.   Do this in such a way that
$E_{i+1,n}\subseteq
E_{in}$ for all $i$, $n$.   Take a decreasing sequence
$\sequence{i}{J_i}$
in $\Cal F$, with empty intersection, such that
$\nu H_i\le\mu E_{in}+2^{-i}$ for $n\in J_i$.   Set $E_n=E_{in}$ for
$n\in J_i\setminus J_{i+1}$.}   (iv) Show that there is a unique
extension of
$\nu$ to a complete probability measure $\mu$ on $X$ which is inner regular with
respect to $\Sigma$.   (This is a kind of {\bf Loeb measure}.   Compare 328B.)
%413L

\spheader 413Xo Let $\frak A$ be a Boolean algebra and
$K\subseteq\frak A$ a sublattice containing $0$.   Suppose that
$\phi:K\to\coint{0,\infty}$
is a bounded supermodular functional such that $\phi 0=0$.
Show that there is a non-negative additive
functional $\nu:\frak A\to\Bbb R$ such that $\nu a\ge\phi a$ for
every $a\in K$ and $\nu 1=\sup_{a\in K}\phi a$.
%413R

\spheader 413Xp Let $X$ be a set and $\Cal K$ a sublattice of $\Cal PX$
containing $\emptyset$.   Let $\phi:\Cal K\to\Bbb R$ be an
order-preserving modular function .   Show that there is a non-negative additive functional
$\nu:\Cal PX\to[0,\infty]$ extending $\phi$.   \Hint{start with the case
$X\in\Cal K$.}
%413R

\spheader 413Xq Let $X$ be a set and $\Cal K$ a sublattice of
$\Cal PX$.   Let $\phi:\Cal K\to[0,1]$ be a submodular functional
such that

\Centerline{$\phi K\le\phi K'$ whenever $K$, $K'\in\Cal K$ and
$K\subseteq K'$,
\quad$\inf_{K\in\Cal K}\phi K=0$.}

\noindent Show that there is a finitely
additive functional $\nu:\Cal PX\to[0,1]$ such that

\Centerline{$\nu X=\sup_{K\in\Cal K}\phi K$,
\quad$\nu K\le\phi K$ for every $K\in\Cal K$.}
%413R

\spheader 413Xr Let $X$ be a set and $\Cal K$ a sublattice of $\Cal PX$
containing $\emptyset$.   Let $\phi:\Cal K\to\coint{0,\infty}$ be such
that $\phi K\le\sum_{n=0}^{\infty}(\phi K_n-\phi L_n)$ whenever
$K\in\Cal K$ and $\sequencen{K_n}$, $\sequencen{L_n}$ are sequences in
$\Cal K$ such that $L_n\subseteq K_n$ for every $n$ and
$\sequencen{K_n\setminus L_n}$ is a disjoint cover of
$K$.   Show that there is a measure on $X$ extending $\phi$.
\Hint{Show that $\phi$ is modular.
Show that if $\Tau$ is the ring of subsets of $X$ generated by $\Cal K$,
every member of $\Tau$ is a finite union of differences of members of
$\Cal K$.   Now apply 413La.   See {\smc Kelley \& Srinivasan 71}.}
%413Xp 413L 413R

\leader{413Y}{Further exercises (a)}
%\spheader 413Ya
\dvAnew{2016} Let $\frak A$ be a Boolean algebra, $(S,+)$ a commutative
semigroup with identity $e$ and $\phi:\frak A\to S$ a
function such that $\phi 0=e$.   Show that

\Centerline{$\frak B
=\{b:b\in\frak A,\,\phi a=\phi(a\Bcap b)+\phi(a\Bsetminus b)$
for every $a\in\frak A\}$}

\noindent is a subalgebra of of $\frak A$, and that
$\phi(a\Bcup b)=\phi a+\phi b$ for all disjoint $a$, $b\in\frak B$.
%413B

\spheader 413Yb
Give an example of two inner measures $\phi_1$, $\phi_2$ on a set $X$
such that the measure defined by $\phi_1+\phi_2$ strictly extends the
sum of the measures defined by $\phi_1$ and $\phi_2$.
%413C, 413Xb mt41bits

\spheader 413Yc Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any
family of probability spaces, and $\lambda$ the product measure on
$X=\prod_{i\in I}X_i$.   Show that
$\lambda_*(\prod_{i\in I}A_i)\le\prod_{i\in I}(\mu_i)_*A_i$ whenever
$A_i\subseteq X_i$ for every $i$, with equality if $I$ is countable.
%413E

\spheader 413Yd
Find a measure space $(X,\Sigma,\mu)$, with $\mu X>0$,
and a sequence
$\sequencen{X_n}$ of subsets of $X$, covering $X$, such that
whenever $E\in\Sigma$, $n\in\Bbb N$ and $\mu E>0$, there is an $F\in\Sigma$
such that $F\subseteq E\setminus X_n$ and $\mu F=\mu E$.
(ii) For $A\subseteq X$ set
$\phi A=\sup\{\mu E:E\in\Sigma$, $E\subseteq A\}$.   Set
$\Tau=\{G:G\subseteq X$, $\phi A=\phi(A\cap G)+\phi(A\setminus G)$ for
every $A\subseteq X\}$.   Show that $\phi\restr\Tau$ is not a measure.
%413Xj 413J mt41bits

\spheader 413Ye Let $(X,\Sigma,\mu)$ be a totally finite measure space,
and $Z$ the Stone space of the Boolean algebra $\Sigma$.   For
$E\in\Sigma$ write $\widehat{E}$ for the corresponding open-and-closed
subset of $Z$.   Show that there is a unique function $f:X\to Z$ such
that $f^{-1}[\widehat{E}]=E$ for every $E\in\Sigma$.   Show that there
is a measure $\nu$ on $Z$, inner regular with respect to the
open-and-closed sets,
such that $f$ is \imp\ with respect to $\mu$ and $\nu$, and that $f$
represents an isomorphism between the measure algebras of $\mu$ and
$\nu$.   Use this construction to prove (vi)$\Rightarrow$(i) in Theorem
343B without appealing to the Lifting Theorem.
%413K

\spheader 413Yf Let $X$ be a set, $\Tau$ a subalgebra of $\Cal PX$, and
$\nu:\Tau\to\coint{0,\infty}$ a finitely additive functional.   Suppose
that there is a set $\Cal K\subseteq\Tau$, containing $\emptyset$, such
that (i) $\mu F=\sup\{\mu K:K\in\Cal K,\,K\subseteq F\}$ for every
$F\in\Tau$ (ii) $\Cal K$ is {\bf monocompact}, that is,
$\bigcap_{n\in\Bbb N}K_n\ne\emptyset$ for every non-increasing sequence
in $\Cal K$.   Show that $\nu$ extends to a measure on $X$.
%413L

\spheader 413Yg(i) Let $X$ be a topological space.   Show that the
family of closed countably compact subsets of $X$ is a countably compact
class.   (ii) Let $X$ be a Hausdorff space.   Show that the family of
sequentially compact subsets of $X$ is a countably compact class.
%413M

\spheader 413Yh
Let $\frak A$ be a Boolean algebra and $\nu$ a totally finite
submeasure on $\frak A$ which is {\it either} supermodular
{\it or} exhaustive and submodular.
Show that $\nu$ is uniformly exhaustive.
%413R  n04624  1K

\spheader 413Yi\dvAformerly{4{}13Yh} 
Let $X$ be a set, $\Cal K$ a sublattice of
$\Cal PX$ containing $\emptyset$, and $f:\Cal K\to\Bbb R$ a modular
functional such that $f(\emptyset)=0$.   Show that there is an additive
functional $\nu:\Cal PX\to\Bbb R$ extending $f$.
%413Xp 413R mt41bits

\spheader 413Yj\dvAnew{2017}
Let $(X,\Sigma,\mu)$ be a semi-finite measure space, $\lambda$ the 
c.l.d.\ product measure on $X\times\Bbb R$ when $\Bbb R$ is given Lebesgue
measure, and $\lambda_*$ the associated inner measure.
Show that for any $f:X\to\coint{0,\infty}$,

\Centerline{$\underlineint f\,d\lambda
=\lambda_*\{(x,\alpha):x\in X$, $0\le\alpha<f(x)\}
=\lambda_*\{(x,\alpha):x\in X$, $0\le\alpha\le f(x)\}$.}
% 413E out of order query
}%end of exercises

\cmmnt{\Notesheader{413}
I gave rather few methods of constructing measures in the first three
volumes of this treatise;  in the present volume I shall have to make up
for lost time.   In particular I used \Caratheodory's construction
for Lebesgue measure (Chapter 11), product measures (Chapter 25) and
Hausdorff measures (Chapter 26).   The first two, at least, can be
tackled in quite different ways if we choose.   The first alternative
approach I offer is the `inner measure' method of 413C.   Note the exact
definition in 413A;  I do not think it is an obvious one.   In
particular, while ($\alpha$) seems to have something to do with
subadditivity, and ($\beta$) is a kind of sequential order-continuity,
there is no straightforward way in which to associate an outer measure
with an inner measure, unless they both happen to be derived from
measures (132B, 413D), even when they are finite-valued;  and for an
inner measure which
is allowed to take the value $\infty$ we have to add the semi-finiteness
condition ($*$) of 413A (see 413Xa).

Once we have got these points right, however, we have a method which
rivals \Caratheodory's in scope, and in particular is especially well
adapted to the construction of inner regular measures.   As an almost
trivial example, we have a route to the c.l.d.\ version of a measure
$\mu$ (413Xj(i)), which can be derived from the inner measure $\mu_*$
defined from
$\mu$ (413D).   Henceforth $\mu_*$ will be a companion to the familiar
outer measure $\mu^*$, and many calculations will be a little easier
with both available, as in 413E-413F.

The intention behind 413J-413K is to find a minimal set of properties of
a functional $\phi_0$ which will ensure that it has an extension to a
measure. Indeed it is easy to see that, in the context of 413J, given a
family $\Cal K$ with the properties ($\dagger$) and ($\ddagger$) there,
a functional $\phi_0$ on $\Cal K$ can have an extension to an inner
regular measure only if it satisfies the conditions ($\alpha$) and
($\beta$), so in
this sense 413J is the best possible result.   Note that while
\Caratheodory's construction is liable to produce wildly infinite
measures (like Hausdorff measures, or primitive product measures), the
construction here always gives us locally determined measures, provided
only that $\phi_0$ is finite-valued.

We have to work rather hard for the step from 413J to 413K.   Of course
413J is a special case of 413K, and I could have saved a little space by
giving a direct proof of the latter result.   But I do not think that
this would have made it easier;  413K really does require an extra step,
because somehow we have to extend the functional $\phi_0$ from $\Cal K$
to $\Cal K_{\delta}$.   The method I have chosen uses 413B and 413I to
cast as much of the argument as possible into the context of algebras of
sets with additive functionals, where I hope the required manipulations
will seem natural.   (But perhaps I should insist that you must not take
them too much
for granted, as some of the time we have a finitely additive
functional taking infinite values, and must take care not to
subtract illegally, as well as not to take limits in the wrong places.)
Note that the progression $\phi_0\to\phi_1\to\mu$ in the proof of 413K
involves first an approximation from outside (if $K\in\Cal K_{\delta}$,
then $\phi_1K$ will be $\inf\{\phi_0K':K\subseteq K'\in\Cal K\}$) and
then an approximation from inside (if $E\in\Sigma$, then $\mu
E=\sup\{\phi_1K:K\in\Cal K_{\delta},\,K\subseteq E\}$).   The essential
difficulty in the proof is just that we have to take successive
non-exchangeable limits.
I have slipped 413L in as a corollary of 413K;  but it can be regarded
as one of the fundamental results of measure theory.   A non-negative
finitely additive functional $\nu$ on an algebra $\Sigma$ of sets can be
extended to
a countably additive measure iff it is `relatively countably additive'
in the sense
that $\nu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$ whenever
$\sequencen{E_n}$ is a disjoint sequence in $\Sigma$ such that
$\bigcup_{n\in\Bbb N}E_n\in\Sigma$ (413Xm).   Of course the same result
can easily be got from an outer measure construction (413Xl).   Note
that the outer measure construction also has repeated limits, albeit
simpler ones:  in the formula

\Centerline{$\theta A=\inf\{\sum_{n=0}^{\infty}\nu E_n:\sequencen{E_n}$
is a sequence in $\Sigma$ covering $A\}$}

\noindent the sum $\sum_{n=0}^{\infty}\nu E_n=\sup_{n\in\Bbb
N}\sum_{i=0}^n\nu E_i$ can be regarded as a crude approximation from
inside, while the infimum
is an approximation from outside.   To get the result as stated in 413L,
of course, the outer measure construction needs a third limiting
process, to
obtain the c.l.d.\ version automatically provided by the inner measure
method, and the inner regularity with respect to $\Sigma_{\delta}$,
while easily checked, also demands a few words of argument.

Many applications of the method of 413J-413K pass through 413N;  if the
family $\Cal K$ is a countably compact class then the sequential
order-continuity hypothesis ($\beta$) of 413J or 413K becomes a
consequence
of the other hypotheses.   The essence of the method is the inner
regularity hypothesis ($\alpha$).   I have tried to use the labels
$\dagger$, $\ddagger$, $\alpha$ and $\beta$ consistently enough to
suggest the currents which I think are flowing in this material.

In 413O we strike out in a new direction.   The object here is to build
an extension which is not going to be unique, and for which choices will
have to be made.   As with any such argument, the trick is to specify
the allowable intermediate stages, that is, the partially ordered set
$P$ to which we shall apply Zorn's Lemma.   But here the form of the
theorem makes
it easy to guess what $P$ should be:  it is the set of functionals
satisfying the hypotheses of the theorem which have not wandered outside
the boundary set by the conclusion, that is, which satisfy the condition
($*$) of part (a) of the proof of 413O.   The finitistic nature of the
hypotheses makes it easy to check that totally ordered subsets of $P$
have upper bounds
(that is to say, if we did this by transfinite induction there would be
no problem at limit stages), and all we have to prove is that maximal
elements of $P$ are defined on adequately large domains;  which amounts
to showing
that a member of $P$ not defined on every element of $\Cal K$ has a
proper extension, that is, setting up a construction for the step to a
successor ordinal in the parallel transfinite induction (part (c) of the
proof).

Of course the principal applications of 413O in this book will be in the
context of countably additive functionals, as in 413P.

It is clear that 413O and 413S overlap to some extent.   I include both
because they have different virtues.   413O provides actual extensions of
functionals in a way that 413S, as given, does not;
but its chief advantage,
from the point of view of the work to come, is the approximation of
members of $\Tau_1$, in measure, by members of $\Tau_0$.   This will
eventually enable us to retain control of the Maharam types of measures
constructed by
the method of 413P.   In 413U we have a different kind of control;  we
can specify a lower bound for the measure of each member of our basic
class $\Cal K$, provided only that our specifications are consistent
with some {\it finitely} additive functional.

}%end of notes

\discrpage

\frfilename{mt414.tex}
\versiondate{26.1.10}
\copyrightdate{2002}

\def\chaptername{Topologies and measures I}
\def\sectionname{$\tau$-additivity}
\def\IMPLY#1#2{{\bf (#1)$\Rightarrow$(#2)}}
\def\undphi{\underline{\phi}}
\def\undpsi{\underline{\psi}}

\newsection{414}

The second topic I wish to treat is that of `$\tau$-additivity'.   Here
I collect results which do not depend on any strong kind of inner
regularity.   I begin with what I think of as the most characteristic
feature of $\tau$-additivity, its effect on the properties of
semi-continuous functions (414A), with a variety of corollaries, up to
the behaviour of subspace measures
(414K).   A very important property of $\tau$-additive topological
measures is that they are often strictly localizable (414J).

The theory of inner regular
$\tau$-additive measures belongs to the next section, but here I give
two introductory results:  conditions under which a $\tau$-additive
measure will be inner regular with respect to closed sets (414M) and
conditions under which a measure which is inner regular
with respect to closed sets will be $\tau$-additive (414N).   I end the
section with notes on `density' and `lifting' topologies (414P-414R).

\leader{414A}{Theorem} Let $(X,\frak T)$ be a topological space and
$\mu$ an effectively locally finite $\tau$-additive measure on $X$ with
domain $\Sigma$ and measure algebra $\frak A$.

(a) Suppose that $\Cal G$ is a non-empty family in $\Sigma\cap\frak T$
such that $H=\bigcup\Cal G$ also belongs to $\Sigma$.   Then
$\sup_{G\in\Cal G}G^{\ssbullet}=H^{\ssbullet}$ in $\frak A$.

(b) Write $\eusm L$ for the family of
$\Sigma$-measurable lower semi-continuous functions from $X$ to
$\Bbb R$.   Suppose that
$\emptyset\ne A\subseteq\eusm L$ and set $g(x)=\sup_{f\in A}f(x)$ for
every $x\in X$.   If $g$ is $\Sigma$-measurable and finite almost
everywhere, then $\tilde g^{\ssbullet}=\sup_{f\in A}f^{\ssbullet}$ in
$L^0(\mu)$, where $\tilde g(x)=g(x)$ whenever $g(x)$ is finite.

(c) Suppose that $\Cal F$ is a non-empty family of measurable closed
sets such that $\bigcap\Cal F\in\Sigma$.   Then
$\inf_{F\in\Cal F}F^{\ssbullet}=(\bigcap\Cal F)^{\ssbullet}$ in
$\frak A$.

(d) Write $\eusm U$ for the family of
$\Sigma$-measurable upper semi-continuous functions from $X$ to
$\Bbb R$.   Suppose that $A\subseteq\eusm U$ is non-empty and set
$g(x)=\inf_{f\in A}f(x)$ for every $x\in X$.   If $g$ is
$\Sigma$-measurable and finite almost everywhere, then
$\tilde g^{\ssbullet}=\inf_{f\in A}f^{\ssbullet}$ in $L^0(\mu)$, where
$\tilde g(x)=g(x)$ whenever $g(x)$ is finite.

\proof{{\bf (a)} \Quer\ If
$H^{\ssbullet}\ne\sup_{G\in\Cal G}G^{\ssbullet}$, there is a non-zero
$a\in\frak A$ such that $a\Bsubseteq H^{\ssbullet}$ but
$a\Bcap G^{\ssbullet}=0$ for every $G\in\Cal G$.   Express $a$ as
$E^{\ssbullet}$
where $E\in\Sigma$ and $E\subseteq H$.   Because $\mu$ is effectively
locally finite, there is a measurable open set $H_0$ of finite measure
such that $\mu(H_0\cap E)>0$.   Now $\{H_0\cap G:G\in\Cal G\}$ is an
upwards-directed family of measurable open sets with union
$H_0\cap H\supseteq H_0\cap E$;  as $\mu$ is $\tau$-additive, there is a
$G\in\Cal G$
such that $\mu(H_0\cap G)>\mu H_0-\mu(H_0\cap E)$.   But in this case
$\mu(G\cap E)>0$, which is impossible, because
$G^{\ssbullet}\Bcap E^{\ssbullet}=0$.\ \Bang

\medskip

{\bf (b)} For any $\alpha\in\Bbb R$,

\Centerline{$\{x:g(x)>\alpha\}=\bigcup_{f\in A}\{x:f(x)>\alpha\}$,}

\noindent and these are all measurable open sets.   Identifying
$\{x:g(x)>\alpha\}^{\ssbullet}\in\frak A$ with
$\Bvalue{\tilde g^{\ssbullet}>\alpha}$ (364Ib\formerly{3{}64Jb}),
we see from (a) that
$\Bvalue{\tilde g^{\ssbullet}>\alpha}
=\sup_{f\in A}\Bvalue{f^{\ssbullet}>\alpha}$ for every $\alpha$.   But
this means that $\tilde g^{\ssbullet}=\sup_{f\in A}f^{\ssbullet}$, by
364L(a-ii)\formerly{3{}64Mb}.

\medskip

{\bf (c)} Apply (a) to $\Cal G=\{X\setminus F:F\in\Cal F\}$.

\medskip

{\bf (d)} Apply (b) to $\{-f:f\in A\}$.
}%end of proof of 414A

\leader{414B}{Corollary} Let $X$ be a topological space and $\mu$ an
effectively locally finite $\tau$-additive topological measure on $X$.

(a) Suppose that $A$ is a non-empty upwards-directed family of lower
semi-continuous functions from $X$ to $[0,\infty]$.   Set
$g(x)=\sup_{f\in A}f(x)$ in $[0,\infty]$ for every $x\in X$.   Then
$\int g=\sup_{f\in A}\int f$ in $[0,\infty]$.

(b) Suppose that $A$ is a non-empty downwards-directed family of
non-negative continuous real-valued functions on $X$, and that
$g(x)=\inf_{x\in A}f(x)$ for every $x\in X$.   If any member of $A$ is
integrable, then $\int g=\inf_{f\in A}\int f$.

\proof{{\bf (a)} Of course all the $f\in A$, and also $g$, are
measurable functions.   Set $g_n=g\wedge n\chi X$ for every
$n\in\Bbb N$.   Then

\Centerline{$g_n(x)=\sup_{f\in A}(f\wedge n\chi X)(x)$}

\noindent for every $x\in X$, so
$g_n^{\ssbullet}=\sup_{f\in A}(f\wedge n\chi X)^{\ssbullet}$, by 414Ab,
and

\Centerline{$\int g_n=\int g_n^{\ssbullet}
=\sup_{f\in A}\int(f\wedge n\chi X)^{\ssbullet}
=\sup_{f\in A}\int f\wedge n\chi X$}

\noindent by 365Dh.   But now, of course,

\Centerline{$\int g=\sup_{n\in\Bbb N}\int g_n
=\sup_{n\in\Bbb N,f\in A}\int f\wedge n\chi X=\sup_{f\in A}\int f$,}

\noindent as claimed.

\medskip

{\bf (b)} Take an integrable $f_0\in A$, and apply (a) to
$\{(f_0-f)^+:f\in A\}$.
}%end of proof of 414B

\leader{414C}{Corollary} Let $(X,\frak T,\Sigma,\mu)$ be an effectively
locally finite $\tau$-additive
topological measure space and $\Cal F$ a non-empty
downwards-directed family of closed sets.   If $\inf_{F\in\Cal F}\mu F$
is finite, this is the measure of $\bigcap\Cal F$.

\proof{ Setting $F_0=\bigcap\Cal F$, then
$F_0^{\ssbullet}=\inf_{F\in\Cal F}F^{\ssbullet}$, by 414Ac;  now

\Centerline{$\mu F_0=\bar\mu F_0^{\ssbullet}
=\inf_{F\in\Cal F}\bar\mu F^{\ssbullet}=\inf_{F\in\Cal F}\mu F$}

\noindent by 321F.
}%end of 414C

\leader{414D}{Corollary} Let $\mu$ be an effectively locally finite
$\tau$-additive measure on a topological space $X$.   If $\nu$ is a
totally finite measure with the same domain as $\mu$, truly continuous
with respect to $\mu$, then $\nu$ is $\tau$-additive.   In particular,
if $\mu$ is $\sigma$-finite and $\nu$ is absolutely continuous with
respect to $\mu$, then $\nu$ is $\tau$-additive.

\proof{ We have a functional $\bar\nu:\frak A\to\coint{0,\infty}$, where
$\frak A$ is the measure algebra of $\mu$, such that
$\bar\nu E^{\ssbullet}=\nu E$ for every $E$ in the common domain
$\Sigma$ of $\mu$ and $\nu$.   Now $\bar\nu$ is continuous for the
measure-algebra topology of
$\frak A$ (327Cd), therefore completely additive (327Ba), therefore
order-continuous (326Oc\formerly{3{}26Kc}).
So if $\Cal G$ is an upwards-directed family
of open sets belonging to $\Sigma$ with union $G_0\in\Sigma$,

\Centerline{$\sup_{G\in\Cal G}\nu G
=\sup_{G\in\Cal G}\bar\nu G^{\ssbullet}
=\bar\nu G_0^{\ssbullet}
=\nu G_0$}

\noindent because $G_0^{\ssbullet}=\sup_{G\in\Cal G}G^{\ssbullet}$.

The last sentence follows at once, because on a $\sigma$-finite space an
absolutely continuous countably additive functional is truly continuous
(232Bc).
}%end of proof of 414D

\leader{414E}{Corollary} Let $(X,\frak T,\Sigma,\mu)$ be an effectively
locally finite $\tau$-additive topological measure space.   Suppose that
$\Cal G\subseteq\frak T$ is non-empty and upwards-directed, and
$H=\bigcup\Cal G$.   Then

(a) $\mu(E\cap H)=\sup_{G\in\Cal G}\mu(E\cap G)$ for every
$E\in\Sigma$;

(b) if $f$ is a non-negative virtually measurable real-valued function
defined almost everywhere in $X$, then $\int_Hf=\sup_{G\in\Cal G}\int_Gf$
in $[0,\infty]$.

\proof{{\bf (a)} In the measure algebra $(\frak A,\bar\mu)$ of $\mu$,

$$\eqalign{(E\cap H)^{\ssbullet}
&=E^{\ssbullet}\Bcap H^{\ssbullet}
=E^{\ssbullet}\Bcap\sup_{G\in\Cal G}G^{\ssbullet}\cr
&=\sup_{G\in\Cal G}E^{\ssbullet}\Bcap G^{\ssbullet}
=\sup_{G\in\Cal G}(E\cap G)^{\ssbullet},\cr}$$

\noindent using 414Aa and the distributive law 313Ba.   So

\Centerline{$\mu(E\cap H)
=\bar\mu(E\cap H)^{\ssbullet}
=\sup_{G\in\Cal G}\bar\mu(E\cap G)^{\ssbullet}
=\sup_{G\in\Cal G}\mu(E\cap G)$}

\noindent by 321D, because $\Cal G$ and
$\{(E\cap G)^{\ssbullet}:G\in\Cal G\}$ are upwards-directed.

\wheader{414E}{4}{2}{2}{30pt}

{\bf (b)} For each $G\in\Cal G$,

\Centerline{$\int_Gf=\int f\times\chi G
=\int(f\times\chi G)^{\ssbullet}
=\int f^{\ssbullet}\times\chi G^{\ssbullet}$,}

\noindent where $\chi G^{\ssbullet}$ can be interpreted either as
$(\chi G)^{\ssbullet}$ (in $L^0(\mu)$) or as $\chi(G^{\ssbullet})$ (in
$L^0(\frak A)$, where $\frak A$ is the measure algebra of $\mu$);  see
364J\formerly{3{}64K}.
Now $H^{\ssbullet}=\sup_{G\in\Cal G}G^{\ssbullet}$ (414Aa);
since $\chi$ and $\times$ are order-continuous (364Jc,
364N\formerly{3{}64P}),
$f^{\ssbullet}\times\chi H^{\ssbullet}
=\sup_{G\in\Cal G}f^{\ssbullet}\times\chi G^{\ssbullet}$;  so

\Centerline{$\int_Hf
=\int f^{\ssbullet}\times\chi H^{\ssbullet}
=\sup_{G\in\Cal G}\int f^{\ssbullet}\times\chi G^{\ssbullet}
=\sup_{G\in\Cal G}\int_Gf$}

\noindent by 365Dh again.
}%end of proof of 414E

\leader{414F}{Corollary} Let $(X,\frak T,\Sigma,\mu)$ be an effectively
locally finite $\tau$-additive topological measure space.   Then for
every $E\in\Sigma$ there is a unique relatively closed self-supporting
set $F\subseteq E$ such that $\mu(E\setminus F)=0$.

\proof{ Let $\Cal G$ be the set $\{G:G\in\frak T,\,\mu(G\cap E)=0\}$.
Then $\Cal G$ is upwards-directed, so
$\mu(E\cap G^*)=\sup_{G\in\Cal G}\mu(E\cap G)=0$, where
$G^*=\bigcup\Cal G$.   Set $F=E\setminus G^*$.   Then $F\subseteq E$ is
relatively closed, and $\mu(E\setminus F)=0$.   If $H\in\frak T$ and
$H\cap F\ne\emptyset$, then $H\notin\Cal G$ so
$\mu(F\cap H)=\mu(E\cap H)>0$;  thus $F$ is self-supporting.   If
$F'\subseteq E$ is another self-supporting relatively closed set such
that $\mu(E\setminus F')=0$, then
$\mu(F\setminus F')=\mu(F'\setminus F)=0$;  but
as $F\setminus F'$ is relatively open in $F$, and $F'\setminus F$ is
relatively open in $F'$, these must both be empty, and $F=F'$.
}%end of proof of 414F

\leader{414G}{Corollary} If $(X,\frak T,\Sigma,\mu)$ is a Hausdorff
effectively locally finite $\tau$-additive topological measure space and
$E\in\Sigma$ is an atom for $\mu$\cmmnt{ (definition:  211I)}, then
there is an $x\in E$ such that $E\setminus\{x\}$ is negligible.

\proof{ Let $F\subseteq E$ be a self-supporting set such that
$\mu(E\setminus F)=0$.   Since $\mu F=\mu E>0$, $F$ is not empty;  take
$x\in F$.   \Quer\ If $F\ne\{x\}$, let $y\in F\setminus\{x\}$.   Because
$\frak T$ is Hausdorff, there are disjoint open sets $G$, $H$ containing
$x$, $y$ respectively;  and in this case $\mu(E\cap G)=\mu(F\cap G)$ and
$\mu(E\cap H)=\mu(F\cap H)$ are both non-zero, which is impossible,
since $E$ is an atom.\ \Bang

So $F=\{x\}$ and $E\setminus\{x\}$ is negligible.
}%end of proof of 414G

\leader{414H}{Corollary} If $(X,\frak T,\Sigma,\mu)$ is an effectively
locally finite $\tau$-additive topological measure space and $\nu$ is an
indefinite-integral measure over $\mu$\cmmnt{ (definition:
234J\footnote{Formerly 2{}34B.})},
then $\nu$ is a $\tau$-additive topological measure.

\proof{ Because $\nu$ measures every set in $\Sigma$
(234La\footnote{Formerly 2{}34D.}), it is a
topological measure.   To see that it is $\tau$-additive, apply 414Eb to
a Radon-Nikod\'ym derivative of $\nu$.
}%end of proof of 414H

\leader{414I}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a complete
locally determined effectively locally finite $\tau$-additive
topological measure space.   If
$E\subseteq X$ and $\Cal G\subseteq\frak T$ are such that
$E\subseteq\bigcup\Cal G$ and $E\cap G\in\Sigma$ for every $G\in\Cal G$,
then $E\in\Sigma$.

\proof{ Set $\Cal K=\{K:K\in\Sigma$, $E\cap K\in\Sigma\}$.   Then
whenever $F\in\Sigma$ and $\mu F>0$ there is a $K\in\Cal K$ included in
$F$ with $\mu K>0$.   \Prf\ Set $K_1=F\setminus\bigcup\Cal G$.   Then
$K_1$ is a member of $\Cal K$ included in $F$.   If $\mu K_1>0$ then we
can stop.   Otherwise,
$\Cal G^*=\{G_0\cup\ldots\cup G_n:G_0,\ldots,G_n\in\Cal G\}$ is an
upwards-directed family of open sets, and

\Centerline{$\sup_{G\in\Cal G^*}\mu(F\cap G)
=\mu(F\cap\bigcup\Cal G^*)
=\mu F>0$,}

\noindent by 414Ea.   So there is a $G\in\Cal G^*$ such that
$\mu(F\cap G)>0$;  but now $E\cap G\in\Sigma$ so $F\cap G\in\Cal K$.\
\Qed

By 412Aa, $\mu$ is inner regular with respect to $\Cal K$;  by 412Ja,
$E\in\Sigma$.
}%end of proof of 414I

\leader{414J}{Theorem} Let $(X,\frak T,\Sigma,\mu)$ be a complete
locally determined effectively locally finite $\tau$-additive
topological measure space.   Then $\mu$ is strictly localizable.

% question: will "locally finite" do in place of "eff loc fin"?
% at least to make $\mu$ localizable

\proof{ Let $\Cal F$ be a maximal disjoint family of self-supporting
measurable sets of
finite measure.   Then whenever $E\in\Sigma$ and $\mu E>0$, there is an
$F\in\Cal F$ such that $\mu(E\cap F)>0$.   \Prf\Quer\ Otherwise, let $G$
be an open set of finite measure such that $\mu(G\cap E)>0$, and set
$\Cal F_0=\{F:F\in\Cal F,\,F\cap G\ne\emptyset\}$.   Then
$\mu(F\cap G)>0$ for every $F\in\Cal F_0$, while $\mu G<\infty$ and
$\Cal F_0$ is disjoint, so
$\Cal F_0$ is countable and $\bigcup\Cal F_0\in\Sigma$.   Set
$E'=E\setminus\bigcup\Cal F_0$;  then
$E\setminus E'=E\cap\bigcup\Cal F_0$ is negligible, so
$\mu(G\cap E')>0$.   By 414F, there is a self-supporting set
$F'\subseteq G\cap E'$ such that $\mu F'>0$.   But in this case
$F'\cap F=\emptyset$ for
every $F\in\Cal F$, so we ought to have added $F'$ to
$\Cal F$.\ \Bang\Qed

This means that $\Cal F$ satisfies the criterion of 213Oa.   Because
$(X,\Sigma,\mu)$ is complete and locally determined, it is strictly
localizable.
}%end of proof of 414J

\leader{414K}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$, and $Y\subseteq X$ a subset such that the
subspace measure $\mu_Y$ is semi-finite\cmmnt{ (see the remark
following 412O)}.
If $\mu$ is an effectively locally finite $\tau$-additive
topological measure, so is $\mu_Y$.

\proof{ By 412Pe, $\mu_Y$ is an effectively locally finite topological
measure.   Now
suppose that $\Cal H$ is a non-empty upwards-directed family in
$\frak T_Y$ with union $H^*$.   Set

\Centerline{$\Cal G=\{G:G\in\frak T,\,G\cap Y\in\Cal H\}$,
\quad $G^*=\bigcup\Cal G$,}

\noindent so that $\Cal G$ is upwards-directed and $H^*=Y\cap G^*$.
Let $\Cal K$ be the family of sets $K\subseteq X$ such that
$K\cap G^*\setminus G=\emptyset$ for some $G\in\Cal G$.   If
$E\in\Sigma$,

$$\eqalignno{\mu E
&=\mu(E\setminus G^*)+\mu(E\cap G^*)
=\mu(E\setminus G^*)+\sup_{G\in\Cal G}\mu(E\cap G)\cr
\displaycause{414Ea}
&=\sup_{G\in\Cal G}\mu(E\setminus(G^*\setminus G)),\cr}$$

\noindent so $\mu$ is inner regular with respect to $\Cal K$.   By
412Ob, $\mu_Y$ is inner regular with respect to
$\{K\cap Y:K\in\Cal K\}$.   So if $\gamma<\mu_YH^*$, there is a
$K\in\Cal K$ such that $K\cap Y\subseteq H^*$ and
$\mu_Y(K\cap Y)\ge\gamma$.   But now there is a $G\in\Cal G$ such that
$K\cap G^*\setminus G=\emptyset$, so that
$K\cap Y\subseteq G\cap Y\in\Cal H$ and
$\sup_{H\in\Cal H}\mu H\ge\gamma$.   As $\Cal H$ and $\gamma$ are
arbitrary, $\mu$ is $\tau$-additive.
}%end of proof of 414K

\cmmnt{\medskip

\noindent{\bf Remarks} Recall from 214Ic that if $(X,\Sigma,\mu)$ has
locally determined negligible sets (in particular, is either strictly
localizable or complete and locally
determined), then all its subspaces are semi-finite.   In 419C below I
describe a tight locally finite Borel measure with a subset on which the
subspace measure is not
semi-finite, therefore not effectively locally finite or
$\tau$-additive.   In 419A I describe a $\sigma$-finite locally finite
$\tau$-additive topological measure, inner regular with respect to the
closed sets, with a closed subset on which the subspace measure is
totally finite but not $\tau$-additive.
}%end of comment

\leader{414L}{Lemma} Let $(X,\frak T)$ be a topological space, and
$\mu$, $\nu$ two effectively locally finite Borel measures on $X$ which
agree on the open sets.   Then they are equal.

\proof{ Write $\frak T^f$ for the family of open sets of finite measure.
(I do not need to specify which measure I am using here.)   For
$G\in\frak T^f$, set $\mu_GE=\mu(G\cap E)$, $\nu_GE=\nu(G\cap E)$ for
every Borel
set $E$.   Then $\mu_G$ and $\nu_G$ are totally finite Borel measures
which agree on $\frak T$.   By the Monotone Class Theorem (136C),
$\mu_G$ and $\nu_G$ agree on the $\sigma$-algebra generated by
$\frak T$, that is, the Borel $\sigma$-algebra $\Cal B$.   Now, for any
$E\in\Cal B$,

\Centerline{$\mu E=\sup_{G\in\frak T^f}\mu_GE=\sup_{G\in\frak T^f}\nu_GE
=\nu E$,}

\noindent by 412F.   So $\mu=\nu$.
}%end of proof of 414L

\leader{414M}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space with
a regular topology $\frak T$ such that $\mu$ is effectively locally
finite and $\tau$-additive and $\Sigma$ includes a base for $\frak T$.

(a) $\mu G=\sup\{\mu F:F\in\Sigma$ is closed, $F\subseteq G\}$ for every
open set $G\in\Sigma$.

(b) If $\mu$ is inner regular with respect to the $\sigma$-algebra
generated by $\frak T\cap\Sigma$, it is inner regular with respect to
the closed sets.

\proof{{\bf (a)} For $U\in\Sigma\cap\frak T$, the set

\Centerline{$\Cal H_U=\{H:H\in\Sigma\cap\frak T$,
$\overline{H}\subseteq U\}$}

\noindent is an upwards-directed family of open sets, and
$\bigcup\Cal H_U=U$ because $\frak T$ is regular and $\Sigma$ includes a
base for $\frak T$.   Because $\mu$ is $\tau$-additive,
$\mu U=\sup\{\mu H:H\in\Cal H_U\}$.
Now, given $\gamma<\mu G$, we can choose $\sequencen{U_n}$ in
$\Sigma\cap\frak T$ inductively, as follows.   Start by taking
$U_0\subseteq G$ such that $\gamma<\mu U_0<\infty$ (using the hypothesis
that $\mu$ is
effectively locally finite).   Given $U_n\in\Sigma\cap\frak T$ and
$\mu U_n>\gamma$, take $U_{n+1}\in\Sigma\cap\frak T$ such that
$\overline{U}_{n+1}\subseteq U_n$ and $\mu U_{n+1}>\gamma$.   On
completing the induction, set

\Centerline{$F=\bigcap_{n\in\Bbb N}U_n
=\bigcap_{n\in\Bbb N}\overline{U}_n$;}

\noindent then $F$ is a closed set belonging to $\Sigma$, $F\subseteq G$
and $\mu F\ge\gamma$.   As $\gamma$ is arbitrary, we have the result.

\medskip

{\bf (b)} Let $\Sigma_0$ be the $\sigma$-algebra generated by
$\Sigma\cap\frak T$ and set $\mu_0=\mu\restr\Sigma_0$.   Then
$\Sigma_0\cap\frak T=\Sigma\cap\frak T$ is still a base for $\frak T$
and $\mu_0$ is still $\tau$-additive and effectively locally finite, so
by (a) and 412G it is inner regular with respect to the closed sets.
Now we are supposing that $\mu$ is inner regular with respect to
$\Sigma_0$, so $\mu$ is inner regular with respect to the closed sets,
by 412Ab.
}%end of proof of 414M

\leader{414N}{Proposition} Let $(X,\Sigma,\mu)$ be a measure
space and $\frak T$ a topology on $X$.
Suppose that (i) $\mu$ is semi-finite and inner regular with respect
to the closed sets (ii) whenever $\Cal F$ is a
non-empty downwards-directed family of measurable closed sets with empty
intersection and $\inf_{F\in\Cal F}\mu F<\infty$, then $\inf_{F\in\Cal
F}\mu F=0$.   Then $\mu$ is $\tau$-additive.

\proof{ Let $\Cal G$ be a non-empty upwards-directed family of
measurable open sets with measurable union $H$.   Take any
$\gamma<\mu H$.    Because $\mu$ is semi-finite, there is a measurable
set $E\subseteq H$ such that $\gamma<\mu E<\infty$.   Now there is a
measurable closed set $F\subseteq E$ such that $\mu F\ge\gamma$.
Consider $\Cal F=\{F\setminus G:G\in\Cal G\}$.   This is a
downwards-directed family of closed sets of finite measure with empty
intersection.   So $\inf_{G\in\Cal G}\mu(F\setminus G)=0$, that is,

\Centerline{$\gamma\le\mu F=\sup_{G\in\Cal G}\mu(F\cap G)
\le\sup_{G\in\Cal G}\mu G$.}

\noindent As $\gamma$ is arbitrary, $\mu H=\sup_{G\in\Cal G}\mu G$;  as
$\Cal G$ is arbitrary, $\mu$ is $\tau$-additive.
}%end of proof of 414N

\leader{414O}{}\cmmnt{ The following elementary result is worth
noting.

\medskip

\noindent}{\bf Proposition} If $X$ is a hereditarily Lindel\"of
space\cmmnt{ (e.g., if it is separable and metrizable)}
then every measure on $X$ is $\tau$-additive.

\proof{ If $\mu$ is a measure on $X$, with domain $\Sigma$, and
$\Cal G\subseteq\Sigma$ is a non-empty upwards-directed family of
measurable open sets, then there is a sequence $\sequencen{G_n}$ in
$\Cal G$ such that $\bigcup\Cal G=\bigcup_{n\in\Bbb N}G_n$.   Now

\Centerline{$\mu(\bigcup\Cal G)
=\lim_{n\to\infty}\mu(\bigcup_{i\le n}G_i)\le\sup_{G\in\Cal G}\mu G$.}

\noindent As $\Cal G$ is arbitrary, $\mu$ is $\tau$-additive.
}%end of proof of 414O

\leader{414P}{Density \dvrocolon{topologies}}\cmmnt{ Recall that a
lower density for a measure space $(X,\Sigma,\mu)$ is a function
$\undphi:\Sigma\to\Sigma$ such that $\undphi E=\undphi F$ whenever $E$,
$F\in\Sigma$ and $\mu(E\symmdiff F)=0$,  $\mu(E\symmdiff\undphi E)=0$
for every $E\in\Sigma$, $\undphi\emptyset=\emptyset$ and
$\undphi(E\cap F)=\undphi E\Bcap\undphi F$ for all $E$, $F\in\Sigma$
(341C).

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space and $\undphi:\Sigma\to\Sigma$ a lower density
such that $\undphi X=X$.   Set

\Centerline{$\frak T=\{E:E\in\Sigma,\,E\subseteq\undphi E\}$.}

\noindent Then $\frak T$ is a topology on $X$, the {\bf density
topology} associated with $\undphi$, and $(X,\frak T,\Sigma,\mu)$ is an
effectively
locally finite $\tau$-additive topological measure space;  $\mu$ is
strictly positive and inner regular with respect to the open sets.

\proof{{\bf (a)(i)} For any $E\in\Sigma$,
$\undphi(E\cap\undphi E)=\undphi E$ because $E\setminus\undphi E$ is
negligible;  consequently $E\cap\undphi E\in\frak T$.   In particular,
$\emptyset=\emptyset\cap\undphi\emptyset$ and
$X=X\cap\undphi X$ belong to $\frak T$.   If $E$, $F\in\frak T$ then

\Centerline{$\undphi(E\cap F)=\undphi E\cap\undphi F\supseteq E\cap F$,}

\noindent so $E\cap F\in\frak T$.

\medskip

\quad{\bf (ii)} Suppose that $\Cal G\subseteq\frak T$ and
$H=\bigcup\Cal G$.   By 341M, $\mu$ is (strictly) localizable, so
$\Cal G$ has an essential supremum $F\in\Sigma$ such that
$F^{\ssbullet}=\sup_{G\in\Cal G}G^{\ssbullet}$ in the measure algebra
$\frak A$ of $\mu$;  that is,
for $E\in\Sigma$, $\mu(G\setminus E)=0$ for every $G\in\Cal G$ iff
$\mu(F\setminus E)=0$.   Now $F\setminus H$ is negligible, by 213K.  On
the other hand,

\Centerline{$G\subseteq\undphi G=\undphi(G\cap F)\subseteq\undphi F$}

\noindent for every $G\in\Cal G$, so $H\subseteq\undphi F$, and
$H\setminus F\subseteq\undphi F\setminus F$ is negligible.   But as
$\mu$ is complete, this means that $H\in\Sigma$.   Also
$\undphi H=\undphi F\supseteq H$, so $H\in\frak T$.
Thus $\frak T$ is closed under arbitrary unions and is a topology.

\medskip

{\bf (b)} By its definition, $\frak T$ is included in $\Sigma$, so $\mu$
is a topological measure.   If $E\in\Sigma$ then $E\cap\undphi E$
belongs to $\frak T$, is
included in $E$ and has the same measure as $E$;  so $\mu$ is inner
regular with respect to the open sets.
As $\mu$ is semi-finite, it is inner regular with respect to the open sets
of finite measure, and is effectively locally finite.
If $E\in\frak T$ is non-empty,
then $\undphi E\supseteq E$ is non-empty, so $\mu E>0$;  thus $\mu$ is
strictly positive.   Finally, if $\Cal G$ is a
non-empty upwards-directed family
in $\frak T$, then the argument of (a-ii) shows that
$(\bigcup\Cal G)^{\ssbullet}=\sup_{G\in\Cal G}G^{\ssbullet}$ in
$\frak A$, so that
$\mu(\bigcup\Cal G)=\sup_{G\in\Cal G}\mu G$.   Thus $\mu$ is
$\tau$-additive.
}%end of proof of 414P

\leader{414Q}{Lifting topologies} Let $(X,\Sigma,\mu)$ be a measure
space and $\phi:\Sigma\to\Sigma$ a lifting\cmmnt{, that is, a Boolean
homomorphism such that $\phi E=\emptyset$ whenever $\mu E=0$ and
$\mu(E\symmdiff\phi E)=0$ for every $E\in\Sigma$ (341A)}.   The {\bf
lifting topology} associated with $\phi$ is the topology generated by
$\{\phi E:E\in\Sigma\}$.   \cmmnt{Note that $\{\phi E:E\in\Sigma\}$ is
a topology base, so is a base for the lifting topology.}

\leader{414R}{Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space and $\phi:\Sigma\to\Sigma$ a lifting with
lifting topology $\frak S$ and density topology $\frak T$.   Then
$\frak S\subseteq\frak T\subseteq\Sigma$, and $\mu$ is $\tau$-additive,
effectively locally finite and strictly positive with respect to
$\frak S$.   Moreover, $\frak S$ is zero-dimensional.

\proof{ Of course $\phi$ is a lower density, so we can talk of its
density topology, and since $\phi^2E=\phi E$, $\phi E\in\frak T$ for
every $E\in\Sigma$, so $\frak S\subseteq\frak T$.   Because $\mu$ is
$\tau$-additive and strictly positive with respect to $\frak T$, it must
also be $\tau$-additive and strictly positive with respect to $\frak S$.
If $E\in\Sigma$ and
$\mu E>0$ there is an $F\subseteq E$ such that $0<\mu F<\infty$, and now
$\phi F$ is an $\frak S$-open set of finite measure meeting $E$ in a
non-negligible set;
so $\mu$ is effectively locally finite with respect to $\frak S$.   Of
course $\frak S$ is
zero-dimensional because $\phi[\Sigma]$ is a base for $\frak S$
consisting of open-and-closed sets.
}%end of proof of 414R

%\leader{414S}
%\leader{414T}
%\leader{414U}
%\leader{414V}
%\leader{414W}

\exercises{
\leader{414X}{Basic exercises $\pmb{>}$(a)}
%\spheader 414Xa
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces with
topologies $\frak T$ and $\frak S$, and $f:X\to Y$ a continuous \imp\
function.   Show
that if $\mu$ is $\tau$-additive with respect to $\frak T$ then $\nu$ is
$\tau$-additive with respect to $\frak S$.   Show that if $\nu$ is
locally finite, so is $\mu$.
%414/
%useful in 452(?)

\spheader 414Xb Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
measure spaces, with direct sum $(X,\Sigma,\mu)$;  suppose that we are
given a
topology $\frak T_i$ on $X_i$ for each $i$, and let $\frak T$ be the
disjoint union topology on $X$.   Show that $\mu$ is $\tau$-additive iff
every $\mu_i$ is.
%414/

\sqheader 414Xc Let $(X,\frak T)$ be a topological space and $\mu$ a
totally finite measure on $X$ which is inner regular with respect to the
closed sets.   Suppose that $\mu X=\sup_{G\in\Cal G}\mu G$ whenever
$\Cal G$ is an upwards-directed family of measurable open sets covering
$X$.   Show that $\mu$ is $\tau$-additive.
%414/

\spheader 414Xd Let $\mu$ be an effectively locally finite
$\tau$-additive $\sigma$-finite measure on a topological space $X$, and
$\nu:\dom\mu\to\coint{0,\infty}$ a countably additive functional which is
absolutely continuous with respect to $\mu$.   Show from first
principles that $\nu$ is $\tau$-additive.
%414D

\spheader 414Xe Give an example
of an indefinite-integral measure over Lebesgue measure on $\Bbb R$
which is not effectively locally finite.   \Hint{arrange for every
non-trivial interval to have infinite measure.}
%414D

\spheader 414Xf Let $(X,\frak T)$ be a topological space and 
$\mu$ a complete locally determined effectively locally finite
$\tau$-additive topological measure on $X$.
Show that if $f$ is a real-valued function, defined on a subset of $X$,
which is locally integrable in the sense of 411Fc, then $f$ is measurable.
%414E

\spheader 414Xg Let $(X,\frak T)$ be a topological space and $\mu$ an
effectively locally finite $\tau$-additive measure on $X$.   Let $\Cal
G$ be a cover of $X$ consisting of measurable open sets, and $\Cal K$
the ideal of subsets of $X$ generated by $\Cal G$.   Show that $\mu$ is
inner regular with respect to $\Cal K$.
%414I

\spheader 414Xh Let $(X,\frak T,\Sigma,\mu)$ be a complete locally
determined effectively locally finite $\tau$-additive topological
measure space, and $A$ a subset of $X$.   Suppose that for every $x\in A$
there is an open set $G$ containing
$x$ such that $A\cap G$ is negligible.   Show that $A$ is negligible.
%414I

\spheader 414Xi Give an alternative proof of 414K based on the fact
that the canonical map from the measure algebra of $\mu$ to the measure
algebra of $\mu_Y$ is order-continuous (322Yd).
%414K

\sqheader 414Xj(i) If $\mu$ is an effectively locally finite
$\tau$-additive Borel measure on a regular topological space, show that
the c.l.d.\ version of $\mu$ is a quasi-Radon measure.
(ii) If $\mu$ is a locally finite, effectively locally finite
$\tau$-additive Borel measure on a locally compact Hausdorff space, show
that $\mu$ is tight, so that
the c.l.d.\ version of $\mu$ is a Radon measure.
%414M

\sqheader 414Xk Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\undphi$ a lower density for $\mu$ such that
$\undphi X=X$;  let $\frak T$ be the corresponding density topology.
(i) Show that a
dense open subset of $X$ must be conegligible.   (ii) Show that a subset
of $X$ is nowhere dense for $\frak T$ iff it is negligible iff it is
meager for $\frak T$.   (iii) Show that a function $f:X\to\Bbb R$ is
$\Sigma$-measurable iff it is $\frak T$-continuous at almost every point
of $X$.   \Hint{if $f$ is measurable, set $E_q=\{x:f(x)>q\}$,
$F_q=\{x:f(x)<q\}$;  show that $f$ is continuous at every point of
$X\setminus\bigcup_{q\in\Bbb Q}((E_q\setminus\undphi E_q)
\cup(F_q\setminus\undphi F_q))$.}  (iv) Show that $\Sigma$ is both the
Borel $\sigma$-algebra of $(X,\frak T)$ and the Baire-property algebra of
$(X,\frak T)$.   (v) Show that $(X,\frak T)$ is a Baire space.
%414P

\spheader 414Xl Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\undphi:\Sigma\to\Sigma$ a lower density such that
$\undphi X=X$, with density topology $\frak T$.   Show that if
$A\subseteq X$ and $E$ is a measurable envelope of $A$ then the
$\frak T$-closure of $A$ is just
$A\cup(X\setminus\undphi(X\setminus E))$.
%414P

\spheader 414Xm Let $\mu$ be Lebesgue measure on $\BbbR^r$, $\Sigma$
its domain, $\intstar:\Sigma\to\Sigma$ lower Lebesgue density (341E) and
$\frak T$ the corresponding density topology.   (i) Show that $\frak T$ is
finer than the usual Euclidean topology of $\BbbR^r$.
(ii) Show that for any
$A\subseteq\Bbb R$, the closure of $A$ for $\frak T$ is just
$A\cup\{x:\limsup_{\delta\downarrow 0}
 \Bover{\mu^*(A\cap B(x,\delta))}{\mu B(x,\delta)}>0\}$, and the
interior is $A\cap\{x:\lim_{\delta\downarrow 0}
 \Bover{\mu_*(A\cap B(x,\delta))}{\mu B(x,\delta)}=1\}$.
%414P

\spheader 414Xn Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\undphi:\Sigma\to\Sigma$ a lower density such that
$\undphi X=X$;  let $\frak T$ be the associated density topology.   Let
$A$ be a subset of $X$ and $E$ a measurable envelope of $A$;  let
$\Sigma_A$ be
the subspace $\sigma$-algebra and $\mu_A$ the subspace measure on $A$.
(i) Show that we have a lower density
$\hbox{$\undphi$}_A:\Sigma_A\to\Sigma_A$ defined by setting
$\hbox{$\undphi$}_A(F\cap A)=A\cap\undphi(E\cap F)$ for every
$F\in\Sigma$.   (ii) Show that $\hbox{$\undphi$}_AA=A$ iff
$A\subseteq\undphi E$, and that in this case the density topology on $A$
derived from $\hbox{$\undphi$}_A$ is just the subspace topology.
%414P

\spheader 414Xo Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\phi:\Sigma\to\Sigma$ a lifting, with density
topology $\frak T$ and lifting topology $\frak S$.   (i) Show that

\Centerline{$\frak T=\{H\cap G:G\in\frak S,\,H$ is conegligible$\}
=\{H\cap\phi E:E\in\Sigma,\,H$ is conegligible$\}$.}

\noindent (ii) Show that if $A\subseteq X$ and $E$ is a measurable
envelope of $A$ then the $\frak T$-closure of $A$ is $A\cup\phi E$.
%414R

\spheader 414Xp Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\phi:\Sigma\to\Sigma$ a lifting;  let $\frak S$ be
its lifting topology.   Let $A$ be a subset of $X$ such that
$A\subseteq\phi E$ for some (therefore any) measurable
envelope $E$ of $A$.  Let $\Sigma_A$ be the
subspace $\sigma$-algebra and $\mu_A$ the subspace measure on $A$.   (i)
Show that we have a lifting $\phi_A:\Sigma_A\to\Sigma_A$ defined by
setting $\phi_A(F\cap A)=A\cap\phi F$ for every $F\in\Sigma$.   (ii)
Show that the lifting topology on $A$ derived from $\phi_A$ is just the
subspace topology.
%414R

\spheader 414Xq Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
complete locally determined
measure spaces and $f:X\to Y$ an \imp\ function.
Suppose that we have lower densities $\undphi:\Sigma\to\Sigma$ and
$\undpsi:\Tau\to\Tau$ such that $\undphi X=X$, $\undpsi Y=Y$ and
$\undphi f^{-1}[F]=f^{-1}[\undpsi F]$ for every $F\in\Tau$.   (i) Show that
$f$ is continuous for the density topologies of $\undphi$ and $\undpsi$.
(ii) Show that if $\undphi$ and $\undpsi$ are liftings then $f$ is
continuous for the lifting topologies.
%414R

\spheader 414Xr Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\phi:\Sigma\to\Sigma$ a lifting, with associated
lifting topology $\frak S$.   Show that a function $f:X\to\Bbb R$ is
$\Sigma$-measurable iff there is a conegligible set $H$ such that
$f\restr H$ is $\frak S$-continuous.   (Compare 414Xk, 414Xt.)
%414Xk, 414R

\spheader 414Xs Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $\phi:\Sigma\to\Sigma$ a lifting.   Let $(Z,\Tau,\nu)$
be the Stone space of the measure algebra of $\mu$, and $f:X\to Z$ the
\imp\ function associated with $\phi$ (341P).   Show that the lifting
topology on $X$ is just $\{f^{-1}[G]:G\subseteq Z$ is open$\}$.
%414R

\spheader 414Xt Let $(X,\Sigma,\mu)$ be a strictly localizable measure
space and $\phi:\Sigma\to\Sigma$ a lifting.   Write $\eusm L^{\infty}$
for the Banach lattice of bounded $\Sigma$-measurable real-valued
functions on $X$, identified with $\eusm L^{\infty}(\Sigma)$ (363H);
let $T:\eusm L^{\infty}\to\eusm L^{\infty}$ be the Riesz homomorphism
associated with $\phi$ (363F).   (i) Show that $T^2=T$.  (ii) Show that
if $X$ is given the lifting topology $\frak S$ defined by $\phi$, then
$T[\eusm L^{\infty}]$ is precisely the space of bounded continuous
real-valued functions on $X$.   (iii) Show that if
$f\in\eusm L^{\infty}$, $x\in X$ and $\epsilon>0$ there is an
$\frak S$-open set $U$ containing $x$ such that
$|(Tf)(x)-\Bover1{\mu V}\int_Vfd\mu|\le\epsilon$ for every
non-negligible measurable set $V$ included in $U$.
%414R

\spheader 414Xu\dvAformerly{4{}16Xf} 
Let $(X,\frak T,\Sigma,\mu)$ be a complete locally determined
effectively locally finite $\tau$-additive topological measure space.
Show that there is a decomposition $\familyiI{X_i}$ for $\mu$ 
in which every $X_i$ is expressible
as the intersection of a closed set with an open set.   \Hint{enumerate
the open sets of finite measure as
$\langle G_{\xi}\rangle_{\xi<\kappa}$, and set
$\Cal F=\{G_{\xi}\setminus\bigcup_{\eta<\xi}G_{\eta}:\xi<\kappa\}$.} 
%414J 213O

\leader{414Y}{Further exercises (a)}
%\spheader 414Ya
Let $(X,\frak T,\Sigma,\mu)$ be a totally finite topological
measure space.   For $E\in\Sigma$ set

\Centerline{$\mu_{\tau}E
=\inf\{\sup_{G\in\Cal G}\mu(E\cap G):\Cal G\subseteq\frak T$
  is an upwards-directed set with union $X\}$.}

\noindent Suppose {\it either} that $\mu$ is inner regular with respect
to the closed sets {\it or} that $\frak T$ is regular.   Show that
$\mu_{\tau}$ is a $\tau$-additive measure, the
largest $\tau$-additive measure with domain $\Sigma$ which is dominated
by $\mu$.
%/

\spheader 414Yb Let $X$ be a set, $\Sigma$ an algebra of subsets of $X$,
and $\frak T$ a topology on $X$.   Let $M$ be the $L$-space of bounded
finitely additive real-valued functionals on $\Sigma$ (362B).   Let
$N\subseteq M$ be the set of those functionals $\nu$ such that
$\inf_{G\in\Cal G}|\nu|(H\setminus G)=0$ whenever
$\Cal G\subseteq\frak T\cap\Sigma$ is a non-empty upwards-directed
family with union
$H\in\Sigma$.   Show that $N$ is a band in $M$.   (Cf.\ 362Xi.)
%414Ya

\spheader 414Yc Find a probability space $(X,\Sigma,\mu)$ and a topology
$\frak T$ on $X$ such that $\Sigma$ includes a base for $\frak T$ and
$\mu$ is $\tau$-additive, but there is a set $E\in\Sigma$ such that the
subspace measure $\mu_E$ is not $\tau$-additive.
%414K %mt41bits

\spheader 414Yd Let $\intstar$ be lower Lebesgue density on $\BbbR^r$,
and $\frak T$ the associated density topology.   Show that every
$\frak T$-Borel set is an F$_{\sigma}$ set for $\frak T$.
%414P, 414Xm

\spheader 414Ye Let $(X,\rho)$ be a metric space and $\mu$ a strictly
positive locally finite quasi-Radon measure on $X$;  write $\frak T$ for
the topology of $X$ and $\Sigma$ for the domain of $\mu$.   For
$E\in\Sigma$ set $\undphi(E)=\{x:x\in X$, $\lim_{\delta\downarrow 0}
\Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}=1\}$.   Suppose that
$E\setminus\undphi(E)$ is
negligible for every $E\in\Sigma$ (cf.\ 261D, 472D).   (i) Show that
$\undphi$ is a lower density for $\mu$, with $\undphi(X)=X$.   Let
$\frak T_d$ be the associated density topology.   (ii) Suppose that
$H\in\frak T_d$ and that $K\subseteq H$ is $\frak T$-closed and
$\rho$-totally bounded.   Show that there is a $\frak T$-closed,
$\rho$-totally bounded $K'\subseteq H$ such that $K$ is included in the
$\frak T_d$-interior of $K'$.   (iii) Show that $\frak T_d$ is
completely regular.   \Hint{{\smc Luke\v{s} Mal\'y \& Zaj\'\i\v{c}ek 86}.}
%414P mt41bits

\spheader 414Yf Show that the density topology on $\Bbb R$ associated
with lower Lebesgue density is not normal.
%414P 414Ye mt41bits

\spheader 414Yg Let $\mu$ be Lebesgue measure on $\BbbR^r$, $\Sigma$
its domain, $\intstar:\Sigma\to\Sigma$ lower Lebesgue density and
$\frak T$ the corresponding density topology.   (i) Show that if
$f:\Bbb R^r\to\Bbb R^r$ is a permutation such that $f$ and $f^{-1}$ are
both differentiable
everywhere, with continuous derivatives, then $f$ is a homeomorphism for
$\frak T$.   \Hint{263D.}  (ii) Show that if $\phi:\Sigma\to\Sigma$ is a
lifting and $\frak S$ the corresponding lifting topology, then
$x\mapsto -x$ is not a homeomorphism for $\frak S$.   \Hint{345Xc.}
%414R
}%end of exercises

\cmmnt{\Notesheader{414} I have remarked before that it is one of the
abiding frustrations of measure theory, at least for anyone ambitious to
apply the power of modern general topology to measure-theoretic
problems, that the basic convergence theorems are irredeemably
confined to
sequences.   In Volume 3 I showed that if we move to measure algebras
and function spaces, we can hope that the countable chain condition or
the countable sup property will enable us to replace arbitrary directed
sets with monotonic sequences, thereby giving theorems which apply to
apparently more general types of convergence.   In 414A and its
corollaries we come to a
quite different context in which a measure, or integral, behaves like an
order-continuous functional.   Of course the theorems here depend
directly on the hypothesis of $\tau$-additivity, which rather begs the
question;  but we shall see in the rest of the chapter that this
property does indeed often appear.   For the moment, I remark only that
as Lebesgue measure is $\tau$-additive we certainly have a non-trivial
example to work with.

The hypotheses of the results above move a touch awkwardly between those
with the magic phrase `topological measure' and those without.   The
point is that (as in 412G, for instance) it is sometimes useful to be
able to apply these ideas to Baire measures on completely regular
spaces, which are defined on a base for the topology but may not be
defined on every open set.   The device I have used in the definition of
$\tau$-additivity (411C) makes this possible, at the cost of occasional
paradoxical phenomena like 414Yc.

I hope that no confusion will arise between the two topologies
associated with a lifting on a complete locally determined space.   I
have called
them the `density topology' and the `lifting topology' because the
former can be defined directly from a lower density;  but it would be
equally reasonable to call them the `fine' and `coarse' lifting
topologies.   The
density topology has the apparent advantage of giving us a measure which
is inner regular with respect to the Borel sets, but at the cost of
being rather odd regarded as a topological space (414P, 414Xk, 414Ye,
414Yf).   It has the
important advantage that there are densities (like the Lebesgue lower
density) which have some claim to be called canonical, and others with
useful special properties, as in \S346, while liftings are always
arbitrary and invariance properties for them sometimes unachievable.
So, for instance, the Lebesgue density topology on $\BbbR^r$ is
invariant under diffeomorphisms, which no lifting topology can be
(414Yg).  The lifting topology is well-behaved as a topology, but only
in special circumstances (as in 453Xd) is the measure inner regular with
respect to its Borel sets, and even the closure of a set can be
difficult to determine.

As with inner regularity, $\tau$-additivity can be associated with the
band structure of the space of bounded additive functionals on an
algebra (414Yb);  there will therefore be corresponding decompositions
of measures into $\tau$-additive and `purely non-$\tau$-additive' parts
(cf.\ 414Ya).
}%end of notes

\discrpage

\frfilename{mt415.tex}
\versiondate{16.5.17}
\copyrightdate{2000}

\def\chaptername{Topologies and measures I}
\def\sectionname{Quasi-Radon measure spaces}
\def\undphi{\underline{\phi}}

\newsection{415}

We are now I think ready to draw together the properties of inner
regularity and $\tau$-additivity.   Indeed, this section will unite
several of the
themes which have been running through the treatise so far:  (strict)
localizability, subspaces and products as well as the new concepts of
this chapter.   In
these terms, the principal results are that a quasi-Radon space is
strictly localizable (415A), any subspace of a quasi-Radon space is
quasi-Radon (415B), and the product of a family of strictly positive
quasi-Radon probability measures on separable metrizable spaces is
quasi-Radon (415E).
I describe a basic method of constructing quasi-Radon measures
(415K), with details of one of the standard ways of applying it (415L,
415N) and some notes on how to specify a quasi-Radon measure uniquely
(415H-415I).   I spell out useful results on
indefinite-integral measures (415O) and $L^p$ spaces (415P), and end the
section with a discussion of the Stone space $Z$ of a localizable
measure algebra $\frak A$ and an important relation in
$Z\times X$ when $\frak A$ is the measure algebra of a quasi-Radon
measure space $X$ (415Q-415R).

It would be fair to say that the study of quasi-Radon spaces for their
own sake is a minority interest.   If you are not already well
acquainted
with Radon measure spaces, it would make good sense to read this section
in parallel with the next.   In particular, the constructions of 415K
and 415L derive much of their importance from the corresponding
constructions in \S416.

\leader{415A}{Theorem} A quasi-Radon measure space is strictly
localizable.

\proof{ This is a special case of 414J.
}%end of proof of 415A

\leader{415B}{Theorem} Any subspace of a quasi-Radon measure space is
quasi-Radon.

\proof{ Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure space and
$(Y,\frak T_Y,\Sigma_Y,\mu_Y)$ a subspace with the induced topology and
measure.   Because $\mu$ is complete, locally determined and localizable
(by 415A), so is $\mu_Y$ (214Ie).   Because $\mu_Y$ is semi-finite and
$\mu$ is an effectively locally finite $\tau$-additive topological
measure, so is
$\mu_Y$ (414K).   Because $\mu$ is inner regular with respect to the
closed sets and $\mu_Y$ is semi-finite, $\mu_Y$ is inner regular with
respect to the relatively closed subsets of $Y$ (412Pc).   So $\mu_Y$ is
a quasi-Radon measure.
}%end of proof of 415B

\leader{415C}{}\cmmnt{ In regular topological spaces, the condition
`inner regular with respect to the closed sets' in the definition of
`quasi-Radon measure' can be weakened or omitted.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T)$ be a regular topological
space.

(a) If $\mu$ is a complete locally determined effectively locally finite
$\tau$-additive topological measure on $X$, inner regular with respect
to the Borel sets, then it is a quasi-Radon measure.

(b) If $\mu$ is an effectively locally finite $\tau$-additive Borel
measure on $X$, its c.l.d.\ version is a quasi-Radon measure.

\proof{{\bf (a)} By 414Mb, $\mu$ is inner regular with respect to the
closed sets, which is the only feature missing from the given
hypotheses.

\medskip

{\bf (b)} The c.l.d.\ version of $\mu$ satisfies the hypotheses of
(a).
}%end of proof of 415C

\leader{415D}{}\cmmnt{ In separable metrizable spaces, among others, we
can even omit $\tau$-additivity.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T)$ be a regular hereditarily
Lindel\"of topological space\cmmnt{;  e.g., a separable metrizable
space (4A2P(a-iii)), indeed any regular space with a countable network (4A2Nb)}.

(i) If $\mu$ is a complete effectively locally finite measure on $X$,
inner regular with respect to the Borel sets, and its domain includes
a base for $\frak T$, then it is a quasi-Radon measure.

(ii) If $\mu$ is an effectively locally finite Borel measure on $X$,
then its completion is a quasi-Radon measure.

(iii) Any quasi-Radon measure on $X$ is $\sigma$-finite.

(iv) Any quasi-Radon measure on $X$ is completion regular.

\proof{{\bf (a)} The basic fact we need is that if $\Cal G$ is any
family of open sets in $X$, then there is a countable
$\Cal G_0\subseteq\Cal G$
such that $\bigcup\Cal G_0=\bigcup\Cal G$ (4A2H(c-i)).
Consequently any effectively locally finite measure $\mu$ on $X$ is
$\sigma$-finite.   \Prf\ Let $\Cal G$ be the family of measurable open
sets of finite measure.   Let $\Cal G_0\subseteq\Cal G$ be a countable
set with the same union as $\Cal G$.   Then
$E=X\setminus\bigcup\Cal G_0$ is
measurable, and $E\cap G=\emptyset$ for every $G\in\Cal G$, so
$\mu E=0$; accordingly $\Cal G_0\cup\{E\}$ is a countable cover of $X$
by sets of finite measure, and $\mu$ is $\sigma$-finite.\ \Qed

Moreover, any measure on $X$ is $\tau$-additive.   \Prf\ If $\Cal G$ is
a non-empty upwards-directed family of open measurable sets, there is a
sequence $\sequencen{G_n}$ in $\Cal G$ with union $\bigcup\Cal G$.   If
$n\in\Bbb N$ there is a $G\in\Cal G$ such that
$\bigcup_{i\le n}G_i\subseteq G$, so

\Centerline{$\mu(\bigcup\Cal G)=\mu(\bigcup_{n\in\Bbb N}G_n)
=\sup_{n\in\Bbb N}\mu(\bigcup_{i\le n}G_i)\le\sup_{G\in\Cal G}\mu G$.}

\noindent As $\Cal G$ is arbitrary, $\mu$ is $\tau$-additive.\ \Qed

\medskip

{\bf (b)(i)} Now let $\mu$ be a complete effectively locally finite
measure on $X$, inner regular with respect to the Borel sets, and with
domain $\Sigma$ including a base for the topology of $X$.   If
$H\in\frak T$, then
$\Cal G=\{G:G\in\Sigma\cap\frak T,\,G\subseteq H\}$ has union $H$,
because $\Sigma\cap\frak T$ is a base for $\frak T$;  but in this case
there is a
countable $\Cal G_0\subseteq\Cal G$ such that $H=\bigcup\Cal G_0$, so
that $H\in\Sigma$.   Thus $\mu$ is a topological measure.   We know also
from (a) that it is $\tau$-additive and $\sigma$-finite, therefore
locally determined.   By 415Ca, it is a quasi-Radon measure.

\medskip

\quad{\bf (ii)} If $\mu$ is an effectively locally finite Borel measure
on $X$, then its completion $\hat\mu$ satisfies the conditions of (i),
so is a quasi-Radon measure.

\medskip

\quad{\bf (iii)} If $\mu$ is a quasi-Radon measure on $X$, it is surely
effectively locally finite, therefore $\sigma$-finite.

\medskip

\quad{\bf (iv)} Every closed set is a zero set
(4A2H(c-ii)), so any measure which is inner regular with respect to the
closed sets is completion regular.
}%end of proof of 415D

\leader{415E}{}\cmmnt{ I am delaying most of the theory of products of
(quasi-\nobreak)Radon measures to \S417.   However, there is one result
which is
so important that I should like to present it here, even though some of
the ideas will have to be repeated later.

\medskip

\noindent}{\bf Theorem} Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
be a family of separable metrizable quasi-Radon probability spaces such
that every $\mu_i$ is strictly positive, and $\lambda$ the product
measure on $X=\prod_{i\in I}X_i$.   Then

(i) $\lambda$ is a completion regular quasi-Radon measure;

(ii) if $F\subseteq X$ is a closed self-supporting set, there is a
countable set $J\subseteq I$ such that $F$ is determined by coordinates
in $J$, so $F$ is a zero set.

\proof{{\bf (a)} Write $\Lambda$ for the domain of $\lambda$, and
$\Cal U$ for the family of subsets of $X$ of the form
$\prod_{i\in I}G_i$ where $G_i\in\frak T_i$ for every $i\in I$ and
$\{i:G_i\ne X_i\}$
is finite.   Then $\Cal U$ is a base for the topology of $X$, included
in $\Lambda$.
For $J\subseteq I$ let $\lambda_J$ be the product measure on
$X_J=\prod_{i\in J}X_i$ and $\Lambda_J$ its domain.

\medskip

{\bf (b)} Consider first the case in which $I$ is countable.   In this
case $X$ also is separable and metrizable (4A2P(a-v)), while $\Lambda$
includes a base for its topology.   Also $\lambda$ is a complete
probability measure
and inner regular with respect to the closed sets (412Ua), so must be
a quasi-Radon measure, by 415D(i).

\medskip

{\bf (c)} Now consider uncountable $I$.   The key to the proof is the
following fact:  if $\Cal V\subseteq\Cal U$ has union $W$, then
$W\in\Lambda$ and $\lambda W=\sup_{V\in\Cal V^*}\lambda V$, where
$\Cal V^*$ is the set of unions of finite subsets of $\Cal V$.

\medskip

\Prf{\bf (i)} By 215B(iv), there is a countable set
$\Cal V_1\subseteq\Cal V$ such that $\lambda(U\setminus W_1)=0$ for
every $U\in\Cal V$, where $W_1=\bigcup\Cal V_1$.
Every member of $\Cal U$ is determined by coordinates in some finite set
(see 254M for this concept), so there is a countable set $J\subseteq I$
such that every member of $\Cal V_1$ is determined by coordinates in
$J$, and $W_1$ also is determined by coordinates in $J$.
Let $\pi_J:X\to X_J$ be the canonical
map.   Because it is an open map (4A2B(f-i)), $\pi_J[W]$ and
$\pi_J[W_1]$ are open in $X_J$, and belong to $\Lambda_J$, by (b).

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that
$\lambda_J\pi_J[W]>\lambda_J\pi_J[W_1]$.   Since
$\pi_J[W]=\bigcup\{\pi_J[U]:U\in\Cal V\}$, while $\lambda_J$ is
quasi-Radon and all the sets $\pi_J[U]$ are open, there must
be some $U\in\Cal V$
such that $\lambda_J(\pi_J[U]\setminus\pi_J[W_1])>0$ (414Ea).   Now
$\pi_J$ is \imp\ (254Oa), so

\Centerline{$0<
\lambda\pi_J^{-1}[\pi_J[U]\setminus\pi_J[W_1]]
=\lambda(\pi_J^{-1}[\pi_J[U]]\setminus\pi_J^{-1}[\pi_J[W_1]])
=\lambda(\pi_J^{-1}[\pi_J[U]]\setminus W_1)$,}

\noindent because $W_1$ is determined by coordinates in $J$.

At this point note that $U$ is of the form $\prod_{i\in I}G_i$, where
$G_i\in\frak T_i$ for each $I$, so we can express $U$ as $U'\cap U''$,
where $U'=\pi_J^{-1}[\pi_J[U]]$ and
$U''=\pi_{I\setminus J}^{-1}[\pi_{I\setminus J}[U]]$.
$U'$ is determined by coordinates in $J$ and $U''$ is determined by
coordinates in $I\setminus J$.   In this case

\Centerline{$\lambda(U\setminus W_1)
=\lambda(U''\cap U'\setminus W_1)
=\lambda U''\cdot\lambda(U'\setminus W_1)$,}

\noindent because $U''$ is determined by coordinates in $I\setminus J$
and $U'\setminus W_1$ is determined by coordinates in $J$,
and we can identify $\lambda$ with the product
$\lambda_{I\setminus J}\times\lambda_J$
(254N).   But now recall that every $\mu_i$ is strictly positive.
Since $U$ is
surely not empty, no $G_i$ can be empty and no $\mu_iG_i$ can be $0$.
Consequently $\prod_{i\in I}\mu_iG_i>0$ (because only finitely
many terms in the product are less than $1$) and $\lambda U>0$;  more to
the point, $\lambda U''>0$.   Since we chose $U$ so that
$\lambda(U'\setminus W_1)>0$, we have $\lambda(U\setminus W_1)>0$.   But
this contradicts the first sentence of (i) just above.\ \Bang

\medskip

\quad{\bf (iii)} Thus $\lambda_J\pi_J[W]=\lambda_J\pi_J[W_1]$.   But
this means that $\lambda\pi_J^{-1}[\pi_J[W]]=\lambda W_1$.   Since
$\lambda$ is complete and $W_1\subseteq W\subseteq\pi_J^{-1}[\pi_J[W]]$,
$\lambda W$ is defined and equal to $\lambda W_1$.

Taking $\sequencen{V_n}$ to be a sequence running over
$\Cal V_1\cup\{\emptyset\}$, we have

\Centerline{$\lambda W=\lambda W_1
=\lambda(\bigcup_{n\in\Bbb N}V_n)
=\sup_{n\in\Bbb N}\lambda(\bigcup_{i\le n}V_i)
\le\sup_{V\in\Cal V^*}\lambda V
\le\lambda W,$}

\noindent so $\lambda W=\sup_{V\in\Cal V^*}\lambda V$, as
required.\ \Qed

\medskip

{\bf (d)} Thus we see that $\lambda$ is a topological measure.   But it
is also $\tau$-additive.   \Prf\ If $\Cal W$ is an upwards-directed
family of open sets in $X$ with union $W^*$, set

\Centerline{$\Cal V=\{U:U\in\Cal U,\,\exists\,W\in\Cal W,\,U\subseteq
W\}$.}

\noindent Then $W^*=\bigcup\Cal V$, so $\lambda W^*=\sup_{V\in\Cal
V^*}\lambda V$, where $\Cal V^*$ is the set of finite unions of members
of $\Cal V$.   But because $\Cal W$ is upwards-directed, every member of
$\Cal V^*$ is included in some member of $\Cal W$, so

\Centerline{$\lambda W^*=\sup_{V\in\Cal V^*}\lambda V
\le\sup_{W\in\Cal W}\lambda W\le\lambda W^*$.}

\noindent As $\Cal W$ is arbitrary, $\lambda$ is $\tau$-additive.\ \Qed

\medskip

{\bf (e)} As in (b) above, we know that $\lambda$ is a complete
probability
measure and is inner regular with respect to the closed sets, so it is a
quasi-Radon measure.   Because $\lambda$ is inner regular with respect
to the zero sets (412Ub), it is completion regular.

\medskip

{\bf (f)} Now suppose that $F\subseteq X$ is a closed self-supporting
set.   By 254Oc, there is a set $W\subseteq X$, determined by
coordinates in some countable set $J\subseteq I$, such that
$W\symmdiff F$ is negligible.   \Quer\ Suppose, if possible, that
$x\in F$ and $y\in X\setminus F$ are such that $x\restr J=y\restr J$.
Then there is a $U\in\Cal U$ such that $y\in U\subseteq X\setminus F$.
As in (b-ii) above, we can express $U$ as $U'\cap U''$ where $U'$,
$U''\in\Cal U$ are determined by coordinates in $J$ and $I\setminus J$
respectively.   In this case,

$$\eqalign{\lambda(F\cap U)
&=\lambda(W\cap U)
=\lambda(W\cap U')\cdot\lambda U''\cr
&=\lambda(F\cap U')\cdot\lambda U''
>0,\cr}$$

\noindent because $x\in F\cap U'$ and $F$ is self-supporting, while
$U''\ne\emptyset$ and $\lambda$ is strictly positive.   But $F\cap
U=\emptyset$, so this is impossible.\ \Bang

Thus $F$ is determined by coordinates in the countable set $J$.
Consequently it is of the form $\pi_J^{-1}[\pi_J[F]]$.   But
$\pi_J[X\setminus F]$ is open (4A2B(f-i) again),
so its complement $\pi_J[F]$
is closed.   Now $X_J$ is metrizable (4A2P(a-v)), so $\pi_J[F]$ is a
zero set (4A2Lc) and $F$ is a zero set (4A2C(b-iv)).
}%end of proof of 415E

\leader{415F}{Corollary} (a) If $Y$ is either
$\coint{0,1}$ or $\ooint{0,1}$, endowed with Lebesgue measure, and $I$
is any set, then $Y^I$, with the product topology and measure, is a
quasi-Radon measure space.

(b) If $\familyiI{\nu_i}$ is a family of probability distributions on
$\Bbb R$\cmmnt{, in the sense of \S271 (that is, Radon probability
measures)}, and every $\nu_i$ is strictly positive, then the product
measure on $\BbbR^I$ is a quasi-Radon measure.

\cmmnt{\medskip

\noindent{\bf Remark} See also 416U below, and 453I, where there is
an alternative proof of the main step in 415E, applicable to some
further cases.   Yet another approach, most immediately applicable to
$\coint{0,1}^I$, is in 443Xp.   For further facts about these product
measures, see \S417, particularly 417M.
}%end of comment

\leader{415G}{Comparing quasi-Radon measures:  Proposition} Let $X$ be a
topological space, and $\mu$, $\nu$ two quasi-Radon measures on $X$.
Then the following are equiveridical:

(i) $\mu F\le\nu F$ for every closed set $F\subseteq X$;

(ii) $\mu\le\nu$ in the sense of 234P.

\noindent If $\nu$ is locally finite, we can add

(iii) $\mu G\le\nu G$ for every open set $G\subseteq X$;

(iv) there is a base $\Cal U$ for the topology of $X$ such that
$G\cup H\in\Cal U$ for all $G$, $H\in\Cal U$ and $\mu G\le\nu G$ for
$G\in\Cal U$.

\proof{{\bf (a)} Of course (ii)$\Rightarrow$(i), and (i)$\Rightarrow$(ii) by
412Ma.

\medskip

{\bf (b)} Evidently
(ii)$\Rightarrow$(iii), and (iii)$\Rightarrow$(iv) is trivial.   If (iv)
is true and
$G\subseteq X$ is open, then $\Cal V=\{V:V\in\Cal U,\,V\subseteq G\}$ is
upwards-directed and has union $G$, so

\Centerline{$\mu G
=\sup_{V\in\Cal V}\mu V
\le\sup_{V\in\Cal V}\nu V
=\nu G$.}

\noindent Thus (iv)$\Rightarrow$(iii).

Now assume that $\nu$ is locally finite and that (iii) is true.
\Quer\ Suppose, if possible, that $F\subseteq X$ is a closed set such
that $\nu F<\mu F$.   Then $\Cal H$, as defined in part (a) of the
proof, is upwards-directed and has union $X$, so there is an
$H\in\Cal H$ such that $\nu F<\mu(F\cap H)$.   Now there is a closed set
$F'\subseteq H\setminus F$ such that

\Centerline{$\nu F'>\nu(H\setminus F)-\mu(F\cap H)+\nu F
\ge\nu H-\mu(F\cap H)$.}

\noindent Set $G=H\setminus F'$, so that $F\cap H\subseteq G$ and

\Centerline{$\nu G
=\nu H-\nu F'
<\mu(F\cap H)
\le\mu G$,}

\noindent which is impossible.\ \Bang

This shows that (provided that $\nu$ is locally finite)
(iii)$\Rightarrow$(i).
}%end of proof of 415G

\leader{415H}{Uniqueness of quasi-Radon measures:  Proposition} Let
$(X,\frak T)$ be a topological space and $\mu$, $\nu$ two quasi-Radon
measures on $X$.   Then the following are equiveridical:

(i) $\mu=\nu$;

(ii) $\mu F=\nu F$ for every closed set $F\subseteq X$;

(iii) $\mu G=\nu G$ for every open set $G\subseteq X$;

(iv) there is a base $\Cal U$ for the topology of $X$ such that
$G\cup H\in\Cal U$ for every $G$, $H\in\Cal U$ and
$\mu\restr\Cal U=\nu\restr\Cal U$;

(v) there is a base $\Cal U$ for the topology of $X$ such that
$G\cap H\in\Cal U$ for every $G$, $H\in\Cal U$ and
$\mu\restr\Cal U=\nu\restr\Cal U$.

\proof{ Of course (i) implies all the others.   (ii)$\Rightarrow$(i) is
immediate from 415G (see also 412M).   If (iii) is true, then, for any
closed set $F\subseteq X$,

$$\eqalign{\mu F
&=\sup\{\mu(G\cap F):G\in\frak T,\,\mu G<\infty\}\cr
&=\sup\{\mu G-\mu(G\setminus F):G\in\frak T,\,\mu G<\infty\}\cr
&=\sup\{\nu G-\nu(G\setminus F):G\in\frak T,\,\nu G<\infty\}
=\nu F;\cr}$$

\noindent so (iii)$\Rightarrow$(ii).   (iv)$\Rightarrow$(iii) by the
argument of (iv)$\Rightarrow$(iii) in the proof of 415G.

Finally, suppose that (v) is true.   Then
$\mu(G_0\cup\ldots\cup G_n)=\nu(G_0\cup\ldots\cup G_n)$ for all
$G_0,\ldots,G_n\in\Cal U$.
\Prf\ Induce on $n$.   For the inductive step to $n\ge 1$, if any $G_i$
has infinite measure (for either measure) the result is trivial.
Otherwise,

$$\eqalign{\mu(G_0\cup\ldots\cup G_n)
&=\mu(\bigcup_{i<n}G_i)+\mu G_n-\mu(G_n\cap\bigcup_{i<n}G_i)\cr
&=\nu(\bigcup_{i<n}G_i)+\nu G_n-\nu(G_n\cap\bigcup_{i<n}G_i)
=\nu(G_0\cup\ldots\cup G_n).\text{ \Qed}\cr}$$

\noindent So $\mu$ and $\nu$ agree on the base
$\{G_0\cup\ldots\cup G_n:G_0,\ldots,G_n\in\Cal U\}$, and (iv) is true.
}%end of proof of 415H

\leader{415I}{Proposition} Let $X$ be a completely regular topological
space and $\mu$, $\nu$ two quasi-Radon measures on $X$ such that
$\int fd\mu=\int fd\nu$ whenever $f:X\to\Bbb R$ is a
bounded continuous function
integrable with respect to both measures.   Then $\mu=\nu$.

\proof{ \Quer\ Otherwise, there is an open set $G\subseteq X$ such that
$\mu G\ne\nu G$;  suppose $\mu G<\nu G$.   Because $\nu$ is effectively
locally finite, there is an open set $G'\subseteq G$ such that
$\mu G<\nu G'<\infty$.   Now the cozero sets form a base for the
topology of
$X$, so $\Cal H=\{H:H\subseteq G'$ is a cozero set$\}$ has union $G'$;
as $\nu$ is $\tau$-additive, there is an $H\in\Cal H$ such that
$\nu H>\mu G$.   Express $H$ as $\{x:g(x)>0\}$ where
$g:X\to\coint{0,\infty}$
is continuous.   For each $n\in\Bbb N$, set $f_n=ng\wedge\chi X$;  then
$\sequencen{f_n}$ is a non-decreasing sequence with limit $\chi H$, so
there is an $n\in\Bbb N$ such that $\int f_nd\nu>\mu G\ge\int f_nd\mu$.
But $f_n$ is both $\mu$-integrable and $\nu$-integrable because $\mu G$
and $\nu H$ are both finite.\ \Bang
}%end of proof of 415I

\leader{415J}{Proposition} Let $X$ be a regular topological space, $Y$ a
subspace of $X$, and $\nu$ a quasi-Radon measure on $Y$.
Then there is a quasi-Radon measure $\mu$ on $X$ such that
$\mu E=\nu(E\cap Y)$ whenever $\mu$ measures $E$, that is, $Y$ has full
outer measure in $X$ and $\nu$ is the subspace measure on $Y$.

\proof{ Write $\Cal B$ for the Borel $\sigma$-algebra of $X$, and set
$\mu_0E=\nu(E\cap Y)$ for every $E\in\Cal B$.   Then it is easy to see
that $\mu_0$ is a $\tau$-additive Borel measure on $X$.   Moreover,
$\mu_0$ is effectively locally finite.   \Prf\ If $E\in\Cal B$ and
$\mu_0E>0$, there is a relatively open set $H\subseteq Y$ such that
$\nu H<\infty$ and $\nu(H\cap E\cap Y)>0$.   Now $H$ is of the form
$G\cap Y$ where $G\subseteq X$ is open, and we have
$\mu_0G=\nu H<\infty$, $\mu_0(E\cap G)=\nu(H\cap E\cap Y)>0$.\ \Qed

By 415Cb, the c.l.d.\ version $\mu$ of $\mu_0$ is a quasi-Radon measure
on $X$.   Because $\mu_0$ is semi-finite (411Gd),
$\mu$ extends $\mu_0$ (213Hc), and $\mu F=\mu_0F=\nu(F\cap Y)$ for every
closed $F\subseteq X$.   In particular, $\mu F=0$ whenever $F\subseteq X$ is
closed and $F\cap Y=\emptyset$;  as $\mu$ is inner regular with respect to the
closed sets, $Y$ has full outer measure in $X$.   So if $H\subseteq Y$ is
relatively closed,

\Centerline{$\nu H=\nu(\bar H\cap Y)=\mu\bar H=\mu^*(\bar H\cap Y)=\mu_YH$}

\noindent where $\mu_Y$ is the subspace measure on $Y$ induced by $\mu$.
Now we know that $\mu_Y$ is a quasi-Radon measure on $Y$ (415B);  as it agrees
with the quasi-Radon measure $\nu$ on the relatively closed subsets of $Y$,
$\mu_Y=\nu$ (415H(ii)), as required.
}%end of proof of 415J

\leader{415K}{}\cmmnt{ I come now to a couple of basic results on the
construction of quasi-Radon measures.  The first follows 413K.

\medskip

\noindent}{\bf Theorem} Let $X$ be a topological space
and $\Cal K$ a family of closed subsets of $X$ such that

\inset{$\emptyset\in\Cal K$,

($\dagger$) $K\cup K'\in\Cal K$ whenever $K$, $K'\in\Cal K$ are
disjoint,

($\ddagger$) $F\in\Cal K$ whenever $K\in\Cal K$ and
$F\subseteq K$ is closed.}

\noindent Let $\phi_0:\Cal K\to\coint{0,\infty}$ be a
functional such that

\inset{($\alpha$) $\phi_0 K=\phi_0 L+\sup\{\phi_0 K':K'\in\Cal K,\,
K'\subseteq K\setminus L\}$ whenever $K$, $L\in\Cal K$ and
$L\subseteq K$,

($\beta$) $\inf_{K\in\Cal K'}\phi_0K=0$ whenever
$\Cal K'$ is a non-empty downwards-directed subset of $\Cal K$ with
empty intersection,

($\gamma$) whenever $K\in\Cal K$ and $\phi_0K>0$, there is an
open set $G$ such that the supremum
$\sup_{K'\in\Cal K,K'\subseteq G}\phi_0K'$ is finite,
while $\phi_0K'>0$ for some $K'\in\Cal K$ such that
$K'\subseteq K\cap G$.}

\noindent Then there is a unique quasi-Radon measure on $X$ extending
$\phi_0$ and inner regular with respect to $\Cal K$.

\proof{ By 413K, there is a complete locally determined measure $\mu$ on
$X$, inner regular with respect to $\Cal K$, and
extending $\phi_0$;  write $\Sigma$ for the domain of $\mu$.   If
$F\subseteq X$ is closed, then $K\cap F\in\Cal K\subseteq\Sigma$ for
every $K\in\Cal K$, so $F\in\Sigma$, by 413F(ii);   accordingly every
open set is measurable.   Because $\mu$ is inner regular with respect to
$\Cal K$ it is surely inner regular with respect to the closed sets.
If $E\in\Sigma$ and $\mu E>0$, there is a $K\in\Cal K$ such that
$K\subseteq E$ and $\mu K>0$;  now ($\gamma$) tells us that
there is an open set $G$ such that $\mu G<\infty$ and $\mu(G\cap K)>0$,
so that $\mu(G\cap E)>0$.   As $E$ is arbitrary, $\mu$ is effectively
locally finite.   Now suppose that $\Cal G$ is a non-empty
upwards-directed family of open sets with union $H$, and that
$\gamma<\mu H$.   Then there is a $K\in\Cal K$ such that $K\subseteq H$
and $\mu K>\gamma$.   Applying the hypothesis ($\beta$) to
$\Cal K'=\{K\setminus G:G\in\Cal G\}$, we see that
$\inf_{G\in\Cal G}\mu(K\setminus G)=0$, so that

\Centerline{$\sup_{G\in\Cal G}\mu G\ge\sup_{G\in\Cal G}\mu(K\cap G)
=\mu K\ge\gamma$.}

\noindent As $\Cal G$ and $\gamma$ are arbitrary, $\mu$ is
$\tau$-additive.   So $\mu$ is a quasi-Radon measure.
}%end of proof of 415K

\leader{415L}{Proposition} Let $(X,\Sigma_0,\mu_0)$ be a measure space
and $\frak T$ a topology on $X$ such that
$\Sigma_0$ includes a base for $\frak T$ and $\mu_0$ is $\tau$-additive,
effectively locally finite and inner regular with respect to the closed
sets.   Then $\mu_0$ has a unique extension to a quasi-Radon measure $\mu$ on
$X$.   Moreover,

\inset{(i) $\mu F=\mu_0^*F$ whenever $F\subseteq X$ is closed and
$\mu_0^*F<\infty$,}

\inset{(ii) $\mu G=(\mu_0)_*G$ whenever $G\subseteq X$ is open,}

\inset{(iii) the embedding $\Sigma_0\embedsinto\Sigma$ identifies the
measure algebra $(\frak A_0,\bar\mu_0)$ of $\mu_0$ with an order-dense
subalgebra
of the measure algebra $(\frak A,\bar\mu)$ of $\mu$, so that the
subrings
$\frak A_0^f$, $\frak A^f$ of elements of finite measure coincide, and
$L^p(\mu_0)$ may be identified with $L^p(\mu)$ for $1\le p<\infty$,}

\inset{(iv) whenever $E\in\Sigma$ and $\mu E<\infty$, there is an
$E_0\in\Sigma_0$ such that $\mu(E\symmdiff E_0)=0$,}

\inset{(v) for every $\mu$-integrable real-valued function $f$ there is
a
$\mu_0$-integrable function $g$ such that $f=g\,\,\mu$-a.e.}

\noindent If $\mu_0$ is complete and locally determined, then we have

\inset{(i)$'$ $\mu F=\mu_0^*F$ for every closed $F\subseteq X$.}
\noindent If $\mu_0$ is localizable, then we have

\inset{(iii)$'$ $\frak A_0=\frak A$, so that $L^0(\mu)\cong L^0(\mu_0)$ and
$L^{\infty}(\mu)\cong L^{\infty}(\mu_0)$,}

\inset{(iv)$'$ for every $E\in\Sigma$ there is an $E_0\in\Sigma_0$ such
that
$\mu(E\symmdiff E_0)=0$,}

\inset{(v)$'$ for every $\Sigma$-measurable real-valued function $f$
there
is a $\Sigma_0$-measurable real-valued function $g$ such that
$f=g\,\,\mu$-a.e.}

\proof{{\bf (a)} Let $\Cal K$ be the set of closed subsets of $X$ of
finite outer measure for $\mu_0$.   Note that $\mu_0$ is inner regular
with respect to $\Cal K$, because it is inner regular with respect to
the closed sets and also with respect to the sets of finite measure.

It is obvious from its definition that $\Cal K$ satisfies ($\dagger$)
and ($\ddagger$) of 415K.   For $K\in\Cal K$, set $\phi_0K=\mu_0^*K$.
Then $\phi_0$ satisfies ($\alpha$)-($\gamma$) of 415K.

\medskip

\Prf\ \grheada\ If $K$, $L\in\Cal K$ and $L\subseteq K$, take measurable
envelopes $E_0$, $E_1\in\Sigma_0$ of $K$, $L$ respectively.   (i) Let
$\epsilon>0$.   Because $\mu_0$ is inner regular with respect to the
closed sets, there is a closed set $F\in\Sigma_0$ such that
$F\subseteq E_0\setminus E_1$ and
$\mu_0F\ge\mu_0(E_0\setminus E_1)-\epsilon$.   Set $K'=F\cap K$.   Then
$K'\in\Cal K$ and

\Centerline{$\phi_0K'=\mu_0^*(F\cap K)=\mu_0(F\cap E_0)=\mu_0 F
\ge\mu_0E_0-\mu_0E_1-\epsilon=\phi_0K-\phi_0L-\epsilon$.}

\noindent As $\epsilon$ is arbitrary, we have

\Centerline{$\phi_0K
\le\phi_0L+\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$.}

\noindent (ii) On the other hand, \Quer\ suppose, if possible, that
there is a closed set $K'\subseteq K\setminus L$ such that
$\mu_0^*L+\mu_0^*K'>\mu_0^*K$.   Let $E_2$ be a measurable envelope of
$K'$, so that $\mu_0E_1+\mu_0E_2>\mu_0E_0$;  since

\Centerline{$\mu_0(E_1\setminus
E_0)=\mu_0^*(L\setminus E_0)=\mu_0^*\emptyset=0$,
\quad$\mu_0(E_2\setminus E_0)=\mu_0^*(K'\setminus E_0)=0$,}

\noindent $\mu_0(E_1\cap E_2)>0$.   Because $\mu_0$
is effectively locally finite, there is a measurable open set $G_0$, of
finite measure, such that $\mu_0(G_0\cap E_1\cap E_2)>0$.   Set

\Centerline{$\Cal G=\{G\cup G':G$, $G'\in\Sigma_0\cap\frak T$,
$G\subseteq G_0\setminus L$, $G'\subseteq G_0\setminus K'\}$.}

\noindent Then $\Cal G$ is an upwards-directed family of measurable open
sets, and because $\Sigma_0$ includes a base for the topology of $X$,
its
union is $(G_0\setminus L)\cup(G_0\setminus K')=G_0$.   So there is an
$H\in\Cal G$ such that
$\mu_0H>\mu_0G_0-\mu_0(E_1\cap E_2)$, that is, there are open sets
$G$, $G'\in\Sigma_0$ such that $G\subseteq G_0\setminus L$, $G'\subseteq
G_0\setminus K'$ and $\mu_0((G\cup G')\cap E_1\cap E_2))>0$.
But we must have

\Centerline{$\mu_0(G\cap E_1)=\mu_0^*(G\cap L)=0$,
\quad$\mu_0(G'\cap E_2)=\mu_0^*(G'\cap K')=0$,}

\noindent so this is impossible.\ \Bang

Accordingly

\Centerline{$\phi_0K\ge\phi_0L+\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq
K\setminus L\}$,}

\noindent so that $\phi_0$ satisfies condition $(\alpha)$ of 415K.

\medskip

\quad\grheadb\ Let $\Cal K'\subseteq\Cal K$ be a non-empty
downwards-directed family with empty intersection.   Fix $K_0\in\Cal K'$
and $\epsilon>0$.   Let $E_0$ be a measurable envelope of $K_0$ and
$G_0$ a measurable open set of finite measure such that $\mu_0(G_0\cap
E_0)\ge\mu_0E_0-\epsilon$.   Then

\Centerline{$\Cal G=\{G:G\in\Sigma_0\cap\frak T$, $G\subseteq
G_0\setminus
K$ for some $K\in\Cal K'$ such that $K\subseteq K_0\}$}

\noindent is an upwards-directed family of measurable open sets, and its
union is $G_0\setminus\bigcap\Cal K'=G_0$, again because $\Sigma_0$
includes
a base for the topology $\frak T$.   So there is a $G\in\Cal G$ such
that
$\mu_0G\ge\mu_0G_0-\epsilon$.   Let $K\in\Cal K'$ be such that
$K\subseteq K_0$ and $G\cap K=\emptyset$;  then

\Centerline{$\phi_0K=\mu_0^*K\le\mu_0(E_0\setminus
G)\le\mu_0(E_0\setminus
G_0)+\mu_0(G_0\setminus G)\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\inf_{K\in\Cal K'}\phi_0K=0$.

\medskip

\quad\grheadc\ If $K\in\Cal K$ and $\phi_0K>0$, let $E_0$ be a
measurable envelope of $K$.   Then there is a measurable open set $G$ of
finite measure such that $\mu_0(G\cap E_0)>0$.   Of course
$\sup_{K'\in\Cal
K,K'\subseteq G}\phi_0K'\le\mu_0G<\infty$;  but also there is a
measurable closed set $K'\subseteq G\cap E_0$ such that $\mu_0K'>0$, in
which case $\phi_0(K\cap K')=\mu_0(E_0\cap K')>0$.   So $\phi_0$
satisfies condition ($\gamma$).\ \Qed

\medskip

{\bf (b)} By 415K, $\phi_0$ has an extension to a quasi-Radon measure
$\mu$
on $X$ which is inner regular with respect to $\Cal K$.   Write $\Sigma$
for the domain
of $\mu$.   Note that, for $K\in\Cal K$,

\Centerline{$\mu K=\phi_0K=\mu_0^*K$,}
\noindent so we can already be sure that the conclusion (i) of this
theorem is satisfied.  Now $\mu$ extends $\mu_0$.

\medskip
\Prf{\bf (i)} Take any $K\in\Cal K$.
Let $E_0\in\Sigma_0$ be a measurable envelope of $K$ for the measure
$\mu_0$.   If $E\in\Sigma_0$, then surely

$$\eqalign{\mu_*(K\cap E)
&=\sup\{\mu K':K'\in\Cal K,\,K'\subseteq K\cap E\}\cr
&=\sup\{\mu_0^*K':K'\in\Cal K,\,K'\subseteq K\cap E\}
\le\mu_0^*(K\cap E).\cr}$$

\noindent On the other hand, given
$\gamma<\mu_0^*(K\cap E)=\mu_0(E_0\cap E)$, there is a closed set
$F\in\Sigma_0$ such that $F\subseteq E_0\cap E$
and $\mu_0F\ge\gamma$, so that

\Centerline{$\mu_*(K\cap E)
\ge\mu(K\cap F)
=\mu_0^*(K\cap F)
=\mu_0(E_0\cap F)\ge\gamma$.}

\noindent Thus $\mu_*(K\cap E)=\mu_0^*(K\cap E)$ for every $K\in\Cal K$ and
$E\in\Sigma_0$.

\woddheader{415L}{4}{2}{2}{36pt}

\quad{\bf (ii)} If $K\in\Cal K$ and $E\in\Sigma_0$ then

\Centerline{$\mu_*(K\cap E)+\mu_*(K\setminus E)
=\mu_0^*(K\cap E)+\mu_0^*(K\setminus E)=\mu_0^*K=\mu K$.}

\noindent Because $\mu$ is complete and locally determined and inner
regular with respect to $\Cal K$, $E\in\Sigma$ (413F(iv)).   Thus
$\Sigma_0\subseteq\Sigma$.

\medskip

\quad{\bf (iii)} For any $E\in\Sigma_0$, we now have

$$\eqalignno{\mu E
&=\sup\{\mu K:K\in\Cal K,\,K\subseteq E\}
=\sup\{\mu_0^*K:K\in\Cal K,\,K\subseteq E\}\cr
&\le\mu_0E
=\sup\{\mu_0K:K\in\Cal K\cap\Sigma_0,\,K\subseteq E\}
\le\mu E.\cr}$$

\noindent As $E$ is arbitrary, $\mu$ extends $\mu_0$.\ \Qed

\medskip

{\bf (c)} Because $\Sigma_0\cap\frak T$ is a base for $\frak T$, closed
under finite unions, $\mu$ is unique, by 415H(iv).

\medskip

{\bf (d)} Now for the conditions (i)-(v).   I have already noted that
(i) is guaranteed by the construction.   Concerning (ii), if
$G\subseteq X$ is
open, we surely have $(\mu_0)_*G\le\mu_*G=\mu G$ because $\mu$ extends
$\mu_0$.   On the other hand, writing
$\Cal G=\{G':G'\in\Sigma_0\cap\frak T,\,G'\subseteq G\}$, $\Cal G$ is
upwards-directed and has union $G$, so

\Centerline{$\mu G=\sup_{G'\in\Cal G}\mu G'
=\sup_{G'\in\Cal G}\mu_0G'\le(\mu_0)_*G$.}

\noindent So (ii) is true.

Because $\mu$ extends $\mu_0$, the embedding $\Sigma_0\embedsinto\Sigma$
corresponds to a measure-preserving embedding of $\frak A_0$ as a
$\sigma$-subalgebra of $\frak A$.   To see that $\frak A_0$ is
order-dense in $\frak A$, take any
non-zero $a\in\frak A$.   This is expressible as $E^{\ssbullet}$ for some
$E\in\Sigma$ with $\mu E>0$.   Now there is a $K\in\Cal K$ such that
$K\subseteq E$ and $\mu K>0$.   There is an $E_0\in\Sigma_0$ which is a
measurable envelope for $K$ with respect to $\mu_0$, so that

\Centerline{$\mu E_0=\mu_0E_0=\mu_0^*K=\mu K$.}

\noindent But this means that

\Centerline{$0\ne E_0^{\ssbullet}=K^{\ssbullet}
\Bsubseteq E^{\ssbullet}=a$}

\noindent in $\frak A$, while $E_0^{\ssbullet}\in\frak A_0$.   As $a$ is
arbitrary, $\frak A_0$ is order-dense in $\frak A$.

If $a\in\frak A^f$, then $B=\{b:b\in\frak A_0,\,b\Bsubseteq a\}$ is
upwards-directed and $\sup_{b\in B}\bar\mu_0b\le\bar\mu a$ is finite;
accordingly $B$ has a supremum in $\frak A_0$ (321C), which must also be
its supremum in
$\frak A$, which is $a$ (313O, 313K).   So $a\in\frak A_0$.   Thus
$\frak A^f$ can be identified with $\frak A_0^f$.   But this means that,
for any $p\in\coint{1,\infty}$, $L^p(\mu)\cong L^p(\frak A,\bar\mu)$ is
identified
with $L^p(\frak A_0,\bar\mu_0)\cong L^p(\mu)$ (366H(d-iv)).   This proves
(iii).

Of course (iv) and (v) are just translations of this.   If $E\in\Sigma$
and $\mu E<\infty$, then $E^{\ssbullet}\in\frak A^f\subseteq\frak A_0$,
that is, there is an $E_0\in\Sigma_0$ such that $\mu(E\symmdiff E_0)=0$.
If $f$ is $\mu$-integrable, then $f^{\ssbullet}\in L^1(\mu)=L^1(\mu_0)$,
that is, there is a $\mu_0$-integrable function $f_0$ such that
$f=f_0\,\,\mu$-a.e.

\medskip

{\bf (e)} If $\mu_0$ is complete and locally determined and
$F\subseteq X$ is an arbitrary closed set, then

\Centerline{$\mu_0^*F=\sup_{K\in\Cal K}\mu_0^*(F\cap K)
=\sup_{K\in\Cal K}\mu(F\cap K)=\sup_{K\in\Cal K,K\subseteq F}\mu K
=\mu F$}

\noindent by 412Jc, because $\mu$ and $\mu_0$ are both inner regular
with respect to $\Cal K$.

\medskip

{\bf (f)} If $\mu_0$ is localizable, $\frak A_0$ is Dedekind complete;
as it is order-dense in $\frak A$, the two must coincide (314Ib).
Consequently

\Centerline{$L^0(\mu)\cong L^0(\frak A)=L^0(\frak A_0)\cong L^0(\mu_0)$,
\quad$L^{\infty}(\mu)\cong L^{\infty}(\frak A)
=L^{\infty}(\frak A_0)\cong L^{\infty}(\mu_0)$.}

\noindent So (iii)$'$ is true;  now (iv)$'$ and (v)$'$ follow at once.
}%end of proof of 415L

\leader{415M}{Corollary} Let $(X,\frak T)$ be a regular topological
space and $\mu_0$ an effectively locally finite $\tau$-additive measure
on $X$, defined on the $\sigma$-algebra generated by a base
for $\frak T$.   Then $\mu_0$ has a unique extension to a quasi-Radon
measure on $X$.

\proof{ By 414Mb, $\mu_0$ is inner regular with respect to the closed
sets.   So 415L gives the result.
}%end of proof of 415M

\leader{415N}{Corollary} Let $(X,\frak T)$ be a completely regular
topological space, and $\mu_0$ a $\tau$-additive effectively locally
finite Baire measure on $X$.   Then $\mu_0$ has a unique extension to a
quasi-Radon measure on $X$.

\proof{ This is a special case of 415M, because the domain $\CalBa(X)$ of
$\mu_0$ is generated by the family of
cozero sets, which form a base for $\frak T$ (4A2Fc).
}%end of proof of 415N

\leader{415O}{Proposition} (a) Let $(X,\frak T)$ be a topological space,
and $\mu$, $\nu$ two quasi-Radon measures on $X$.   Then $\nu$ is an
indefinite-integral measure over $\mu$\cmmnt{ (definition:
234J)} iff $\nu F=0$ whenever $F\subseteq X$ is closed and $\mu F=0$.

(b) Let $(X,\frak T,\Sigma,\mu)$ be a
quasi-Radon measure space, and $\nu$ an indefinite-integral measure over
$\mu$.   If $\nu$ is effectively locally
finite it is a quasi-Radon measure.

\proof{{\bf (a)} If $\nu$ is an indefinite-integral measure over $\mu$,
then of course it is zero on all $\mu$-negligible closed sets.   So let
us suppose that the condition is satisfied.   Write $\Sigma=\dom\mu$ and
$\Tau=\dom\nu$.

\medskip

\quad{\bf (i)} If $E\subseteq X$ is a $\mu$-negligible Borel set it is
$\nu$-negligible, because every closed subset of $E$ must be
$\mu$-negligible, therefore $\nu$-negligible, and $\nu$ is inner regular
with respect to the closed sets.
In particular, taking $U^*$ to be the union of the family
$\Cal U=\{U:U\in\frak T$, $\mu U<\infty\}$,
$\nu(X\setminus U^*)=\mu(X\setminus U^*)=0$ because $\mu$ is effectively
locally finite.
Also, of course, taking $V^*$ to be the union of the family
$\Cal V=\{V:V\in\frak T$, $\nu V<\infty\}$, $\nu(X\setminus V^*)=0$
because $\nu$ is effectively locally finite.   Setting
$\Cal G=\Cal U\cap\Cal V$ and $G^*=\bigcup\Cal G$, we have
$G^*=U^*\cap V^*$, so $G^*$ is $\nu$-conegligible.

\medskip

\quad{\bf (ii)} In fact, every $\mu$-negligible set $E$ is
$\nu$-negligible.   \Prf\Quer\ Otherwise, $\nu^*(E\cap G^*)>0$.
Because the subspace measure $\nu_E$ is quasi-Radon (415B), there is a
$G\in\Cal G$ such that $\nu^*(E\cap G)>0$.   But there is an
F$_{\sigma}$ set
$H\subseteq G\setminus E$ such that $\mu H=\mu(G\setminus E)$, and now
$E\cap G$ is included in the $\mu$-negligible Borel set $G\setminus H$,
so that $\nu(E\cap G)=\nu(G\setminus H)=0$.\ \Bang\Qed

\medskip

\quad{\bf (iii)} Let $\Cal K$ be the family of closed subsets $F$ of $X$
such that either $F$ is included in some member of
$\Cal G$ or $F\cap G^*=\emptyset$.   If $E\in\dom\mu$ and
$\mu E>0$, then there is an $F\in\Cal K$ such that $F\subseteq E$ and
$\mu F>0$.   \Prf\ If $\mu(E\setminus G^*)>0$ take any closed set
$F\subseteq E\setminus G^*$ with $\mu F>0$.   Otherwise,
$\mu(E\cap G^*)>0$.   Because the subspace measure
$\mu_E$ is quasi-Radon, there is a $G\in\Cal G$ such that $\mu(E\cap
G)>0$;  and now we can find a closed set
$F\subseteq E\cap G$ with $\mu F>0$, and $F\in\Cal K$.\ \Qed

\medskip

\quad{\bf (iv)} By 412Ia, there is a decomposition $\familyiI{X_i}$ for
$\mu$ such that every $X_i$ except perhaps one belongs to $\Cal K$ and
that exceptional one, if any, is $\mu$-negligible.   Now
$\familyiI{X_i}$ is a decomposition for $\nu$.   \Prf\ Every $X_i$ is
measured by $\nu$ because it is either closed or $\mu$-negligible, and
of finite measure for $\nu$ because it is either $\nu$-negligible
or included in a member of $\Cal G$.   If
$E\subseteq X$ and $\nu E>0$, then $\nu(E\cap G^*)>0$, so there must be
some $G\in\Cal G$ such that
$\nu(E\cap G)>0$.   Now $J=\{i:i\in I$, $\mu(X_i\cap G)>0\}$ is
countable, and $\nu(G\setminus\bigcup_{i\in J}X_i)
=\mu(G\setminus\bigcup_{i\in J}X_i)=0$, so there is an $i\in J$ such
that $\nu(X_i\cap E)>0$.   By 213Ob, $\familyiI{X_i}$ is a decomposition
for $\nu$.\ \Qed

\medskip

\quad{\bf (v)} It follows that $\Sigma\subseteq\Tau$.   \Prf\ If
$E\in\Sigma$, then for every $i\in I$ there is an F$_{\sigma}$ set
$H\subseteq E\cap X_i$ such that $E\cap X_i\setminus H$ is
$\mu$-negligible, therefore $\nu$-negligible, and $E\cap X_i\in\Tau$.
As $i$ is arbitrary, $E\in\Tau$.\ \QeD\   In fact, $\nu$ is the
completion of $\nu\restr\Sigma$.   \Prf\ If $F\in\Tau$, then for every
$i\in I$ there is an F$_{\sigma}$ set $H_i\subseteq F\cap X_i$ such that
$F\cap X_i\setminus H_i$ is $\nu$-negligible.   Set
$H=\bigcup_{i\in I}H_i$;  because $H\cap X_i=H_i$ belongs to $\Sigma$
for every $i$, $H\in\Sigma$;  and
$\nu(F\setminus H)=\sum_{i\in I}\nu(F\cap X_i\setminus H)=0$.
Similarly, there is an $H'\in\Sigma$ such that $H'\subseteq X\setminus
F$ and $\nu((X\setminus F)\setminus H')=0$, so that $H\subseteq
F\subseteq X\setminus H'$ and $\nu((X\setminus H')\setminus H)=0$.   So
$F$ is measured by the completion of $\nu\restr\Sigma$.   Since $\nu$
itself is complete, it must be the completion of $\nu\restr\Sigma$.\
\Qed

\medskip

\quad{\bf (vi)} By (iv), $\nu$ is inner regular with respect to
$\{E:E\in\Sigma$, $\mu E<\infty\}$.   By 234O,
$\nu$ is an indefinite-integral measure over $\mu$.

\medskip

{\bf (b)} Let $f\in\eusm L^0(\mu)$ be a non-negative function such that
$\nu F=\int f\times\chi F\,d\mu$ whenever this is defined.   Because
$\mu$ is complete and locally determined, so is $\nu$.
Because $\mu$ is an effectively locally finite $\tau$-additive
topological measure, $\nu$ is a $\tau$-additive topological measure
(414H).   Because $\mu$ is inner regular with respect to the closed
sets, so is $\nu$ (412Q).   Since we are assuming in the hypotheses that
$\nu$ is effectively locally finite, it is a quasi-Radon measure.
}%end of proof of 415O

\leader{415P}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon
measure space.

(a) Suppose that $(X,\frak T)$ is completely regular.   If
$1\le p<\infty$ and $f\in\eusm L^p(\mu)$, then for any $\epsilon>0$
there is a bounded continuous
function $g:X\to\Bbb R$ such that $\mu\{x:g(x)\ne 0\}<\infty$ and
$\|f-g\|_p\le\epsilon$.

(b) Suppose that $(X,\frak T)$ is regular and Lindel\"of.   Let
$f\in\eusm L^0(\mu)$ be locally integrable.   Then for any $\epsilon>0$
there is a continuous function $g:X\to\Bbb R$ such that
$\|f-g\|_1\le\epsilon$.

\proof{{\bf (a)} Write $\eusm C$ for the set of bounded continuous
functions $g:X\to\Bbb R$ such that $\{x:g(x)\ne 0\}$ has finite measure.
Then $\eusm C$ is a linear subspace of $\Bbb R^X$ included in
$\eusm L^p=\eusm L^p(\mu)$.
Let $\eusm U$ be the closure of $\eusm C$ in $\eusm L^p$, that is, the
set of $h\in\eusm L^p$ such that for every $\epsilon>0$ there is a
$g\in\eusm C$ such that $\|h-g\|_p\le\epsilon$.   Then $\eusm U$ is
closed under addition and scalar multiplication.   Also
$\chi E\in\eusm U$ whenever $\mu E<\infty$.   \Prf\ Let $\epsilon>0$.
Set $\delta=\bover14\epsilon^{1/p}$.   Write $\Cal G$ for the family of
open sets of finite measure.   Because $\mu$ is effectively locally
finite, there is a $G\in\Cal G$ such that $\mu(E\setminus G)\le\delta$
(412Aa).   Let $F\subseteq G\setminus E$ be a closed set such that
$\mu F\ge\mu(G\setminus E)-\delta$;  then
$\mu(E\symmdiff(G\setminus F))\le 2\delta$.   Write
$\Cal H$ for the family of cozero sets.   Because $\frak T$ is
completely regular, $\Cal H$ is a base
for $\frak T$;  because $\Cal H$ is closed under finite unions
(4A2C(b-iii)) and $\mu$ is $\tau$-additive, there is an $H\in\Cal H$
such that
$H\subseteq G\setminus F$ and $\mu H\ge\mu(G\setminus F)-\delta$, so
that $\mu(E\symmdiff H)\le 3\delta$.   Express $H$ as $\{x:g(x)>0\}$
where $g:X\to\Bbb R$ is a continuous function.   For each $n\in\Bbb N$,
set $g_n=ng\wedge\chi X\in\eusm C$;  then

\Centerline{$|\chi E-g_n|^p\le\chi(E\symmdiff H)+(\chi H-g_n)^p$}

\noindent for every $n$, so

\Centerline{$\biggerint|\chi E-g_n|^p
\le\mu(E\symmdiff H)+\int(\chi H-g_n)^p
\to\mu(E\symmdiff H)$}

\noindent as $n\to\infty$, because $g_n\to\chi H$.   So there is an
$n\in\Bbb N$ such that
$\int|\chi E-g_n|^p\le 4\delta$, that is, $\|\chi E-g_n\|_p\le\epsilon$.
As $\epsilon$ is arbitrary, $\chi E\in\eusm U$.\ \Qed

Accordingly every simple function belongs to $\eusm U$.   But if
$f\in\eusm L^p$ and $\epsilon>0$, there is a simple function $h$ such
that $\|f-h\|_p\le\bover12\epsilon$ (244Ha);  now there is a
$g\in\eusm C$ such that $\|h-g\|_p\le\bover12\epsilon$ and
$\|f-g\|_p\le\epsilon$,
as claimed.

\medskip

{\bf (b)} This time, write $\Cal G$ for the family of open subsets of
$X$ such that $\int_Gf$ is finite, so that $\Cal G$ is an open cover of
$X$.   As $X$ is paracompact (4A2H(b-i)), there is a locally finite
family $\Cal G_0\subseteq\Cal G$ covering $X$, which must be countable
(4A2H(b-ii)).

Let $\family{G}{\Cal G_0}{\epsilon_G}$ be a family
of strictly positive real numbers such that
$\sum_{G\in\Cal G_0}\epsilon_G\le\epsilon$ (4A1P).   Since $X$ is
completely regular (4A2H(b-i)), we can apply (a) to see that, for each
$G\in\Cal G_0$, there is a continuous function $g_G:X\to\Bbb R$ such
that $\int|g_G-f\times\chi G|\le\epsilon_G$.   Next, because $X$ is
normal (4A2H(b-i)), there is a family $\family{G}{\Cal G_0}{h_G}$ of
continuous
functions from $X$ to $[0,1]$ such that $h_G\le\chi G$ for every
$G\in\Cal G_0$ and $\sum_{G\in\Cal G_0}h_G(x)=1$ for every $x\in X$
(4A2F(d-viii)).

Set $g(x)=\sum_{G\in\Cal G_0}g_G(x)h_G(x)$ for every $x\in X$.   Because
$\Cal G_0$ is locally finite, $g:X\to\Bbb R$ is continuous (4A2Bh).
Now

$$\eqalign{\int|f-g|
&=\int|\sum_{G\in\Cal G_0}(f-g_G)\times h_G|
\le\sum_{G\in\Cal G_0}\int|(f-g_G)\times h_G|\cr
&\le\sum_{G\in\Cal G_0}\int_G|f-g_G|
\le\sum_{G\in\Cal G_0}\epsilon_G
\le\epsilon,\cr}$$

\noindent as required.
}%end of proof of 415P

\allowmorestretch{468}{
\leader{415Q}{}\cmmnt{ Recall (411P) that if $(\frak A,\bar\mu)$ is a
localizable measure algebra, with Stone space $(Z,\frak S,\Tau,\nu)$,
then $\nu$ is a strictly positive completion regular quasi-Radon
measure, inner regular with respect to the open-and-closed sets (which
are all compact).   Moreover, subsets of $Z$ are negligible iff they are nowhere
dense, and every measurable set differs by a nowhere dense set from an
open-and-closed set.    The following construction is primarily important
for Radon measure spaces
(see 416V), but is also of interest for general quasi-Radon spaces.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a
quasi-Radon measure space and $(\frak A,\bar\mu)$ its measure algebra.
Let $(Z,\frak S,\Tau,\nu)$ be the Stone space of $(\frak A,\bar\mu)$.
For $E\in\Sigma$ let $E^*\subseteq Z$ be the open-and-closed set
corresponding to the image $E^{\ssbullet}$ of $E$ in $\frak A$.   Define
$R\subseteq Z\times X$ by saying that $(z,x)\in R$ iff $x\in F$ whenever
$F\subseteq X$ is closed and $z\in F^*$.   Set $Q=R^{-1}[X]$.
}%end of allowmorestretch

(a) $R$ is a closed subset of $Z\times X$.

(b) For any $E\in\Sigma$, $R[E^*]$ is the smallest closed set such that
$\mu(E\setminus R[E^*])=0$.
In particular, if $F\subseteq X$ is closed then $R[F^*]$ is the
self-supporting closed set included in $F$ such that
$\mu(F\setminus R[F^*])=0$;  and $R[Z]$ is the support of $\mu$.

(c) $Q$ has full outer measure for $\nu$.

(d) For any $E\in\Sigma$, $R^{-1}[E]\symmdiff(Q\cap E^*)$ is negligible;
consequently $\nu^*R^{-1}[E]=\mu E$ and
$R^{-1}[E]\cap R^{-1}[X\setminus E]$ is negligible.

(e) For any $A\subseteq X$, $\nu^*R^{-1}[A]=\mu^*A$.

(f) If $(X,\frak T)$ is regular, then $R^{-1}[G]$ is relatively open in
$Q$ for every open set $G\subseteq X$, $R^{-1}[F]$ is relatively closed
in $Q$ for every closed set $F\subseteq X$ and
$R^{-1}[X\setminus E]=Q\setminus R^{-1}[E]$ for every Borel set $E\subseteq X$.

\proof{{\bf (a)}

$$R=\bigcap_{F\subseteq X\text{ is closed}}((Z\setminus F^*)\times
X)\cup(Z\times F)$$

\noindent is an intersection of closed sets, therefore closed.

\medskip

{\bf (b)} Let $\Cal G$ be the family of open sets $G\subseteq X$ such
that $\mu(E\cap G)=0$, and $G_0=\bigcup\Cal G$;  then $G_0\in\Cal G$
(414Ea).   Set $F_0=X\setminus G_0$, so that $F_0$ is the smallest
closed set such that $E\setminus F_0$ is negligible, and
$F_0^*\supseteq E^*$.   If $(z,x)\in R$ and $z\in E^*$ we must have
$x\in F_0$.   Thus $R[E^*]\subseteq F_0$.
On the other hand, if $x\in F_0$, and $G$ is an open set containing $x$,
then $G\notin\Cal G$ so $\mu(G\cap E)>0$ and $(E\cap G)^*\ne\emptyset$.
Accordingly $\{(G\cap E)^*:x\in G\in\frak T\}$ is a downwards-directed
family of non-empty open-and-closed sets in the compact space $Z$ and
has non-empty
intersection, containing a point $z$ say.   If $H\subseteq X$ is closed
and $z\in H^*$, then $X\setminus H$ is open and $z\notin(X\setminus
H)^*$, so $x$ cannot belong to $X\setminus H$, that is, $x\in H$;  as
$H$ is arbitrary, $(z,x)\in R$ and $x\in R[E^*]$;  as $x$ is arbitrary,
$R[E^*]=F_0$, as claimed.

Of course, when $E$ is actually closed, $R[E^*]=F_0\subseteq E$.
Taking $E=X$ we see that $R[Z]=R[X^*]$ is the support of $\mu$.

\medskip

{\bf (c)} If $W\in\Tau$ and $\nu W>0$, there is a non-empty
open-and-closed set $U\subseteq W$, by 322Ra, which must be of the form
$E^*$ for some $E\in\Sigma$.   By (b), $R[E^*]$ cannot be empty;  but
$E^*\subseteq W$, so $R[W]\ne\emptyset$, that is, $W\cap Q\ne\emptyset$.
As $W$ is arbitrary, $Q$ has full outer measure in $Z$.

\medskip

{\bf (d)(i)} Let $\Cal F$ be the set of closed subsets of $X$ included
in $E$.   Then $\sup_{F\in\Cal F}F^{\ssbullet}=E^{\ssbullet}$ in
$\frak A$ (412N), so $E^*\setminus\bigcup_{F\in\Cal F}F^*$ is nowhere
dense and negligible.   Now for each $F\in\Cal F$, $R[F^*]\subseteq F$,
so $Q\cap F^*\subseteq R^{-1}[F]\subseteq R^{-1}[E]$.   Accordingly

\Centerline{$Q\cap E^*\setminus R^{-1}[E]
\subseteq E^*\setminus\bigcup_{F\in\Cal F}F^*$}

\noindent is nowhere dense and negligible.

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if
possible, that $\nu^*(R^{-1}[E]\setminus E^*)>0$.   Then there is an
open-and-closed set $U$ of finite measure such that
$\nu^*(R^{-1}[E]\cap U\setminus E^*)>0$ (use 412Jc).   Express
$U$ as $H^*$, where $\mu H<\infty$, and let $F\subseteq H\setminus E$ be
a closed set such that
$\mu((H\setminus E)\setminus F)<\nu^*(R^{-1}[E]\cap H^*\setminus E^*)$.
Then we must have $\nu^*(R^{-1}[E]\cap F^*)>0$.
But $R[F^*]\subseteq F\subseteq X\setminus E$ so
$F^*\cap R^{-1}[E]=\emptyset$,
which is impossible.\ \BanG\

\medskip

\quad{\bf (iii)} Putting these together,
$(Q\cap E^*)\symmdiff R^{-1}[E]$ is negligible.

\medskip

\quad{\bf (iv)} It follows at
once that (because $Z$ is a measurable envelope for $Q$)

\Centerline{$\nu^*R^{-1}[E]=\nu^*(Q\cap E^*)=\nu E^*=\mu E$.}

\noindent Moreover, applying the result to $X\setminus E$,

\Centerline{$R^{-1}[X\setminus E]\cap R^{-1}[E]
\subseteq(R^{-1}[X\setminus E]\symmdiff(Q\cap(X\setminus E)^*))
\cup(R^{-1}[E]\symmdiff(Q\cap E^*))$}

\noindent is negligible.

\medskip

{\bf (e)(i)} Take $E\in\Sigma$ such that $A\subseteq E$ and
$\mu E=\mu^*A$;  then $R^{-1}[A]\subseteq R^{-1}[E]$, so

\Centerline{$\nu^*R^{-1}[A]\le\nu^*R^{-1}[E]=\mu E=\mu^*A$.}

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, that
$\nu^*R^{-1}[A]<\mu^*A$.   Let $W\in\Tau$ be a measurable envelope of
$R^{-1}[A]$, and take
$F\in\Sigma$ such that $\nu(W\symmdiff F^*)=0$.   Since

\Centerline{$\mu F=\nu F^*=\nu W<\mu^*A$,}

\noindent $\mu^*(A\setminus F)>0$;  let $G$ be a measurable envelope of
$A\setminus F$ disjoint from $F$ (213L).   Then $G^*\cap F^*=\emptyset$ so

\Centerline{$\nu(G^*\setminus W)=\nu G^*=\mu G>0$}

\noindent and there is a non-empty open-and-closed
$V\subseteq G^*\setminus W$;  let $H\in\Sigma$ be such that
$H\subseteq G$ and $V=H^*$.   In this case, $R[V]$ is closed and
$\mu(H\setminus R[V])=0$, by (b), so that $H\cap R[V]$ is measurable,
not negligible, and included in $G$.   But $H\cap R[V]\cap A$ is empty,
because $V\cap R^{-1}[A]$ is empty, so
$\mu^*(H\cap R[V]\cap A)<\mu(H\cap R[V])$, and $G$ cannot be a
measurable envelope of $A\setminus F$.\ \Bang

Thus $\nu^*R^{-1}[A]=\mu^*A$, as claimed.

\medskip

{\bf (f)} Suppose now that $(X,\frak T)$ is regular.

\medskip

\quad{\bf (i)} If $G\subseteq X$ is open, $R^{-1}[G]\cap
R^{-1}[X\setminus G]=\emptyset$.   \Prf\ If $z\in R^{-1}[G]$, then there
is an $x\in G$ such that $(z,x)\in R$.   Let $H$ be an open set
containing $x$
such that $\overline{H}\subseteq G$.   Then $x\notin X\setminus H$ so
$z\notin(X\setminus H)^*$, that is, $z\in H^*$.   But

\Centerline{$R[H^*]\subseteq R[\overline{H}^*]\subseteq
\overline{H}\subseteq G$,}

\noindent so $H^*\cap R^{-1}[X\setminus G]=\emptyset$ and
$z\notin R^{-1}[X\setminus G]$.\ \Qed

\medskip

\quad{\bf (ii)} It is easy to check that

$$\eqalign{\Cal E
&=\{E:E\subseteq X,\,R^{-1}[E]\cap R^{-1}[X\setminus E]=\emptyset\}\cr
&=\{E:E\subseteq X,\,R^{-1}[X\setminus E]=Q\setminus R^{-1}[E]\}\cr}$$

\noindent is a $\sigma$-algebra of subsets of $X$ (indeed, an algebra
closed under arbitrary unions), just because $R\subseteq Z\times X$ and
$R^{-1}[X]=Q$.   Because it contains all open sets, $\Cal E$ must
contain
all Borel sets.

\medskip

\quad{\bf (iii)} Now suppose once again that $G\subseteq X$ is open and
that $z\in R^{-1}[G]$.   As in (i) above, there is an open set
$H\subseteq
G$ such that $z\in H^*\subseteq Z\setminus R^{-1}[X\setminus G]$, so
that $z\in H^*\cap Q\subseteq R^{-1}[G]$.   As $z$ is arbitrary,
$R^{-1}[G]$ is relatively open in $Q$.

\medskip

\quad{\bf (iv)} Finally, if $F\subseteq X$ is closed,
$R^{-1}[F]=Q\setminus
R^{-1}[X\setminus F]$ is relatively closed in $Q$.
}%end of proof of 415Q

\leader{415R}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a Hausdorff
quasi-Radon measure space and $(Z,\frak S,\Tau,\nu)$ the Stone space of
its measure algebra.   Let $R\subseteq Z\times X$ be the relation
described in 415Q.   Then

(a) $R$ is (the graph of) a function $f$;

(b) $f$ is \imp\ for the subspace measure $\nu_Q$ on $Q=\dom f$, and in
fact $\mu$ is the image measure $\nu_Qf^{-1}$;

(c) if $(X,\frak T)$ is regular, then $f$ is continuous.

\proof{{\bf (a)} If $z\in Z$ and $x$, $y\in X$ are distinct, let $G$,
$H$ be disjoint open sets containing $x$, $y$ respectively.   Then

\Centerline{$(X\setminus G)^*\cup(X\setminus H)^*
=((X\setminus G)\cup(X\setminus H))^*=Z$,}

\noindent defining $^*$ as in 415Q, so $z$ must belong to at least one
of $(X\setminus G)^*$, $(X\setminus H)^*$.   In the former case
$(z,x)\notin R$ and in the latter case $(z,y)\notin R$.   This shows
that $R$ is a
function; to remind us of its new status I will henceforth call it $f$.
The domain of $f$ is just $Q=R^{-1}[X]$.

\medskip

{\bf (b)} By 415Qd, $f$ is \imp\ for $\nu_Q$ and $\mu$.   Suppose that
$A\subseteq X$ and $f^{-1}[A]$ is in the domain $\Tau_Q$ of $\nu_Q$,
that is, is of the form $Q\cap U$ for some $U\in\Tau$.   Take any
$E\in\Sigma$
such that $\mu E>0$;  then either $\nu(E^*\cap U)>0$ or
$\nu(E^*\setminus U)>0$.   ($\alpha$) Suppose that $\nu(E^*\cap U)>0$.
Because $\nu$ is inner regular with respect to the open-and-closed sets,
there is an
$H\in\Sigma$ such that $H^*\subseteq E^*\cap U$ and $\mu H=\nu H^*>0$.
Now there is a closed set $F\subseteq E\cap H$ with $\mu F>0$.   In this
case, $f[F^*]\subseteq F\subseteq E$, by 415Qb,
while $F^*\cap Q\subseteq U\cap Q=f^{-1}[A]$,
so $f[F^*]\subseteq E\cap A$.   But this means that

\Centerline{$\mu_*(E\cap A)\ge\mu f[F^*]=\mu F>0$.}

\noindent ($\beta$) If $\nu(E^*\setminus U)>0$, then the same arguments
show that $\mu_*(E\setminus A)>0$.   ($\gamma$) Thus
$\mu_*(E\cap A)+\mu_*(E\setminus A)>0$ whenever $\mu E>0$.   Because
$\mu$ is complete and locally determined, $A\in\Sigma$ (413F(vii)).

Thus we see that $\{A:A\subseteq X,\,f^{-1}[A]\in\Tau_Q\}$ is included
in $\Sigma$, and $\mu$ is the image measure $\nu_Qf^{-1}$.

\medskip

{\bf (c)} If $\frak T$ is regular, then 415Qf tells us that $f$ is
continuous.
}%end of proof of 415R

\exercises{
\leader{415X}{Basic exercises $\pmb{>}$(a)}
%\spheader 415Xa
Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure space and
$E\in\Sigma$ an atom for the measure.   Show that there is a closed set
$F\subseteq E$
such that $\mu F>0$ and $F$ is an atom of $\Sigma$, in the sense that
the only measurable subsets of $F$ are $\emptyset$ and $F$.
\Hint{414G.}   Show that $\mu$ is atomless iff all countable subsets of
$X$ are negligible.
%- %  maybe look for other versions, e.g. for \tau-additive measures

\spheader 415Xb Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any family of
quasi-Radon measure spaces.   Show that the direct
sum measure on $X=\{(x,i):i\in I,\,x\in X_i\}$ is a
quasi-Radon measure when $X$ is given its disjoint union topology.
%- %

\sqheader 415Xc The {\bf right-facing Sorgenfrey
topology} or {\bf lower limit topology} on $\Bbb R$ is the
topology generated by the half-open intervals of
the form $\coint{\alpha,\beta}$;  I will use the phrase {\bf Sorgenfrey line}
to mean $\Bbb R$ with this topology.   Show that Lebesgue measure on the
Sorgenfrey line is completion regular and quasi-Radon.
\Hint{114Yj or 221Yb, or 419L.}
%- %

\spheader 415Xd Let $X$ be a topological space and $\mu$ a complete
measure on $X$, and suppose that there is a conegligible closed
measurable set
$Y\subseteq X$ such that the subspace measure on $Y$ is quasi-Radon.
Show that $\mu$ is quasi-Radon.
%- %

\spheader 415Xe Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space.   Show that $\mu$ is inner regular with respect to the family of
self-supporting closed sets included in open sets of finite measure.
%- %

\spheader 415Xf Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space.   Show that whenever
$E\in\Sigma$ and $\epsilon>0$ there is an open
set $G$ such that $\mu G\le\mu E+\epsilon$ and $E\setminus G$ is
negligible.
%- %

\spheader 415Xg Describe a compact Hausdorff quasi-Radon measure space which
is not $\sigma$-finite.
%- one-pt cpctification of uncountable set

\spheader 415Xh Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space, $(\frak A,\bar\mu)$ its measure algebra, and $\frak A^f$ the ideal
$\{a:a\in\frak A$, $\bar\mu a<\infty\}$.   Show that
$\{G^{\ssbullet}:G\in\frak T$, $\mu G<\infty\}$ is dense in $\frak A^f$ for
the strong measure-algebra topology (323Ad).
%- could be used in 443B

\spheader 415Xi Let $(X,\frak T,\Sigma,\mu)$ be an atomless quasi-Radon
measure space which is outer regular with respect to the open sets.
Show that it is $\sigma$-finite.   \Hint{take a decomposition
$\familyiI{X_i}$ in which every $X_i$ except one is self-supporting, and
a set $A$ meeting every $X_i$ in just one point.}
%415A

\spheader 415Xj Let $(X,\frak T,\Sigma,\mu)$ be a $\sigma$-finite
quasi-Radon measure space with $\mu X>0$.   Show that there is a
quasi-Radon probability measure on $X$ with the same measurable sets and the same negligible sets as $\mu$.
%415B

\spheader 415Xk Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space
in which $\Sigma$ is countably generated as a $\sigma$-algebra.   Show
that, for a suitable topology on $X$, the completion of $\mu$ is a
quasi-Radon measure.   \Hint{take the topology generated by a countable
subalgebra of $\Sigma$, and use the arguments of 415D.}
%415D

\spheader 415Xl Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
be a family of quasi-Radon probability spaces such
that every $\mu_i$ is strictly positive, and $\lambda$ the product
measure on $X=\prod_{i\in I}X_i$.   Show that if every $\frak T_i$ has a
countable network, $\lambda$ is a quasi-Radon measure.
%415E

\spheader 415Xm Let $\familyiI{X_i}$ be a family of separable metrizable
spaces, and $\mu$ a quasi-Radon measure on $X=\prod_{i\in I}X_i$.   Show
that $\mu$ is completion regular iff every self-supporting closed set in
$X$ is determined by coordinates in a countable set.  \Hint{4A2Eb.}
%415E

\spheader 415Xn Find two quasi-Radon measures $\mu$, $\nu$ on the unit
interval such that $\mu G\le\nu G$ for every open set $G$ but there is a
closed set $F$ such that $\nu F<\mu F$.
%415G

\spheader 415Xo Let $X$ be a topological space and $\mu$, $\nu$ two
quasi-Radon measures on $X$.   (i) Suppose that $\mu F=\nu F$ whenever
$F\subseteq X$ is closed and both $\mu F$ and $\nu F$ are finite.   Show
that $\mu=\nu$.
(ii) Suppose that $\mu G=\nu G$ whenever $G\subseteq X$ is open and both
$\mu G$ and $\nu G$ are finite.   Show that $\mu=\nu$.
%415H

\spheader 415Xp
Find a second-countable Hausdorff topological space $X$
with a $\tau$-additive Borel probability measure which is not inner
regular with respect to the closed sets.   \Hint{starting from a Radon
probability measure, declare a set with full outer measure and zero inner
measure to be open and conegligible.}
%415J

\spheader 415Xq\dvAformerly{4{}15Yf}
Find a second-countable Hausdorff space $X$, a subset $Y$ and a quasi-Radon
probability measure on $Y$ which is not the subspace measure induced by
any quasi-Radon measure on $X$.
%415J 415Xp

\spheader 415Xr Let $(X,\frak T,\Sigma,\mu)$ be a $\sigma$-finite
paracompact Hausdorff quasi-Radon measure space, and
$f\in\eusm L^0(\mu)$ a locally integrable function.   Show that for any
$\epsilon>0$ there is a continuous function $g:X\to\Bbb R$ such that
$\int|f-g|\le\epsilon$.
%415P

\sqheader 415Xs Let $(X,\frak T,\Sigma,\mu)$ be a $\sigma$-finite
completely regular quasi-Radon measure space.   (i) Show that for every
$E\in\Sigma$ there is an $F$ in the Baire $\sigma$-algebra $\CalBa(X)$ of $X$ such that $\mu(E\symmdiff F)=0$.   \Hint{start with an open set $E$ of finite measure.}   (ii) Show that for every $\Sigma$-measurable
function $f:X\to\Bbb R$ there is a $\CalBa(X)$-measurable function equal
almost everywhere to $f$.
%+

\spheader 415Xt Let $(X,\Sigma,\mu)$ be a measure space and $f$ a
$\mu$-integrable real-valued function.   Show that there is a unique
quasi-Radon measure $\lambda$ on $\Bbb R$ such that $\lambda\{0\}=0$ and
$\lambda\coint{\alpha,\infty}=\mu^*\{x:x\in\dom f$, $f(x)\ge\alpha\}$,
$\lambda\ocint{-\infty,-\alpha}=\mu^*\{x:x\in\dom f$, $f(x)\le-\alpha\}$
whenever $\alpha>0$;  and that $\int h\,d\lambda=\int hfd\mu$ whenever
$h\in\eusm L^0(\lambda)$ and $h(0)=0$ and either integral is defined in
$[-\infty,\infty]$.   \Hint{set $\lambda E=\mu^*f^{-1}[E\setminus\{0\}]$
for Borel sets $E\subseteq\Bbb R$, and use 414Mb, 414O and 235Gb.}
%+


\leader{415Y}{Further exercises (a)}
%\spheader 415Ya
Give an example of two quasi-Radon measures $\mu$, $\nu$
on $\Bbb R$ such that their sum, as defined in 234G,
is not effectively locally finite, therefore not a quasi-Radon measure.
%- %

\spheader 415Yb
Show that any quasi-Radon measure space is isomorphic, as topological
measure space, to a subspace of a compact quasi-Radon measure space.
\Hint{if $X$ is a T$_1$ quasi-Radon measure space, let $\hat{X}$ be its
Wallman compactification ({\smc Engelking 89}, 3.6.21).}
%- %mt41bits

\spheader 415Yc Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space.   Show that the following are equiveridical:  (i) $\mu$ is outer
regular with respect to the open sets;  (ii) every negligible subset of
$X$ is included in an open set of finite measure;   (iii)
$\{x:\mu\{x\}=0\}$ can be covered by a sequence of open sets of finite
measure.
%- %

\spheader 415Yd Show that $+:\Bbb R\times\Bbb R\to\Bbb R$ is continuous
for the right-facing Sorgenfrey topology.
%415Xc

\spheader 415Ye Let $r\ge 1$.   On $\BbbR^r$ let $\frak S$ be the
topology generated by the half-open intervals $\coint{a,b}$ where $a$,
$b\in\BbbR^r$ (definition:  115Ab).   (i) Show that $\frak S$ is the
product topology if each
factor is given the right-facing Sorgenfrey topology.   (ii)
Show that Lebesgue measure is quasi-Radon for $\frak S$.   \Hint{induce
on $r$.   See also 417Yi.}
%415Xc %mt41bits

\spheader 415Yf Find a base $\Cal U$ for the topology of
$X=\{0,1\}^{\Bbb N}$ and two totally finite (quasi-\nobreak)Radon
measures
$\mu$, $\nu$ on $X$ such that $G\cap H\in\Cal U$ for all $G$,
$H\in\Cal U$, $\mu G\le\nu G$ for every $G\in\Cal U$, but $\nu X<\mu X$.
%415G mt41bits

\spheader 415Yg Let $X$ be a topological space and $\Cal G$ an open
cover of $X$.   Suppose that for each $G\in\Cal G$ we are given a
quasi-Radon measure $\mu_G$ on $G$ such that $\mu_G(U)=\mu_H(U)$
whenever $G$, $H\in\Cal G$ and $U\subseteq G\cap H$ is open.   Show that
there is a unique quasi-Radon measure on $X$ such that each $\mu_G$ is
the subspace measure on $G$.   \Hint{if $\family{G}{\Cal G}{\mu_G}$ is a
maximal family with the given properties, then $\Cal G$ is
upwards-directed.}
%415H

\spheader 415Yh Let $(X,\Sigma,\mu)$ be a measure space and $\frak T$ a
topology on $X$ such that $\mu$ is inner regular with respect to the closed
sets, and suppose that there is a family
$\Cal U\subseteq\Sigma\cap\frak T$ such that

\inset{$\mu U<\infty$ for
every $U\in\Cal U$,

for every $U\in\Cal U$,
$\frak T\cap\Sigma\cap\Cal PU$
is a base for the subspace topology of $U$,

if $\Cal G$ is an
upwards-directed family in $\frak T\cap\Sigma$ and
$\bigcup\Cal G\in\Cal U$, then
$\mu(\bigcup\Cal G)=\sup_{G\in\Cal G}\mu G$,

if $E\in\Sigma$ and
$\mu E>0$ then there is a $U\in\Cal U$ such that $\mu(E\cap U)>0$.}

\noindent Show that $\mu$ has an extension to a quasi-Radon measure on $X$.
%415L

\spheader 415Yi Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space such that $\frak T$ is normal (but not necessarily Hausdorff or
regular).   Show that if $1\le p<\infty$, $f\in\eusm L^p(\mu)$ and
$\epsilon>0$, there is a bounded continuous function $g:X\to\Bbb R$ such
that $\|f-g\|_p\le\epsilon$ and $\{x:g(x)\ne 0\}$ has finite measure.
%415P

\spheader 415Yj Let $(X,\frak T,\Sigma,\mu)$ be a completely regular
quasi-Radon measure space and suppose that we are given a uniformity
defining the topology $\frak T$.
Show that if $1\le p<\infty$, $f\in\eusm L^p(\mu)$ and $\epsilon>0$,
there is a bounded uniformly continuous function $g:X\to\Bbb R$ such
that $\|f-g\|_p\le\epsilon$ and $\{x:g(x)\ne 0\}$ has finite measure.
%415P

\spheader 415Yk Let $(X,\frak T,\Sigma,\mu)$ be a completely regular
quasi-Radon measure space and $\tau$ an extended Fatou norm on
$L^0(\mu)$ (definition:  369F) such that (i) $\tau\restr L^{\tau}$ is an
order-continuous norm (ii) whenever $E\in\Sigma$ and $\mu E>0$ there is
an open set $G$ such that $\mu(E\cap G)>0$ and
$\tau(\chi G^{\ssbullet})<\infty$.   Show that
$L^{\tau}\cap\{f^{\ssbullet}:f:X\to\Bbb R$ is continuous$\}$ is
norm-dense in $L^{\tau}$.
%415P

\spheader 415Yl Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space.   Show that $\mu$ is a compact measure (definition:  342Ac or
451Ab) iff there is a locally compact topology $\frak S$ on $X$ such that
$(X,\frak S,\Sigma,\mu)$ is quasi-Radon.
%+
}%end of exercises

\endnotes{\Notesheader{415}
415B is particularly important because a very high proportion of the
quasi-Radon measure spaces we study are actually subspaces of Radon
measure spaces.   I would in fact go so far as to say that when you have
occasion to
wonder whether all quasi-Radon measure spaces have a property, you
should as a matter of habit look first at subspaces of Radon measure
spaces;  if
the answer is affirmative for them, you will have most of what you want,
even if the generalization to arbitrary quasi-Radon spaces gives
difficulties.
Of course the reverse phenomenon can also occur.   Stone spaces (411P)
can be thought of as quasi-Radon compactifications of Radon measure
spaces (416V).   But this is relatively rare.   Indeed the reason why I
give so few examples of quasi-Radon spaces at this point is just that
the natural ones arise from Radon measure spaces.   Note however that
the quasi-Radon product of an uncountable family of Radon
probability spaces need not be
Radon (see 417Xq), so that 415E here and 417O below are sources of
non-Radon quasi-Radon
measure spaces.   Density and lifting topologies can also provide us
with quasi-Radon measure spaces (453Xd, 453Xg).

415K is the second in a series of inner-regular-extension theorems;
there will be a third in 416J.

I have been saying since Volume 1 that the business of measure theory,
since Lebesgue's time, has been to measure as many sets and integrate as
many functions as possible.   I therefore take seriously any theorem
offering a
canonical extension of a measure.   415L and its corollaries can be
regarded as improvement theorems, showing that a good measure can be
made even better.   We have already had such improvement theorems in
Chapter 21:  the completion and c.l.d.\ version of a measure (212C,
213E).   In all such
theorems we need to know exactly what effect our improvement is having
on the other constructions we are interested in;  primarily, the measure
algebra and the function spaces.   The machinery of Chapter 36 shows
that if we understand the measure algebra(s) involved then the function
spaces will give us no further surprises.   Completion of a measure does
not affect the
measure algebra at all (322Da).   Taking the c.l.d.\ version does not
change $\frak A^f=\{a:\bar\mu a<\infty\}$ or $L^1$ (213Fc, 213G, 322D(b-i),
366H), but can affect the rest of the measure algebra and
therefore $L^0$ and $L^{\infty}$. In this respect, what we might call
the `quasi-Radon version' behaves like
the c.l.d.\ version (as could be expected, since the quasi-Radon version
must itself be complete and locally determined).
The archetypal application of 415L is 415N.   We shall see later how
Baire measures arise naturally when studying Banach spaces of continuous
functions (436E).   415N will be one of the keys to applying the
general theory of
topological measure spaces in such contexts.   A virtue of Baire
measures is that inner regularity with respect to closed sets comes
almost free
(412D);  but there can be unsurmountable difficulties if we wish to
extend them to Borel measures (439M), and it is important to know
that $\tau$-additivity, even in the relatively weak form allowed by the
definition I use here (411C), is often enough to give a canonical
extension to
a well-behaved measure defined on every Borel set.   In 415C we have
inner regularity for a different reason, and the measure is already
known to be
defined on every Borel set, so in fact the quasi-Radon version of the
measure is just the c.l.d.\ version.

One interpretation of the Lifting Theorem is that for a complete
strictly localizable measure space $(X,\Sigma,\mu)$ there is a function
$g:X\to Z$, where $Z$ is the Stone space of the measure algebra of
$\mu$, such
that $E\symmdiff g^{-1}[E^*]$ is negligible for every $E\in\Sigma$,
where $E^*\subseteq Z$ is the open-and-closed set corresponding to the
image of $E$ in the measure algebra (341Q).   For a Hausdorff
quasi-Radon measure
space we have a function $f:Q\to X$, where $Q$ is a dense subset of $Z$,
such that $(Q\cap E^*)\symmdiff f^{-1}[E]$ is negligible for every
$E\in\Sigma$ (415Qd, 415R);  moreover, there is a canonical construction
for this function.   For completeness' sake, I have given the result for
general, not necessarily Hausdorff, spaces $X$ (415Q);  but evidently it
will be of greatest interest for regular Hausdorff spaces (415Rc).
Perhaps I should remark that in the most important applications, $Q$ is
the whole
of $Z$ (416Vc).   Of course the question arises, whether $fg$ can be
the identity.   ($Z$ typically has larger cardinal than $X$, so asking
for $gf$ to be the identity is a bit optimistic.)   This is in fact an
important question;  I will return to it in 453M-453N.
}%end of notes

\discrpage

\frfilename{mt416.tex}
\versiondate{28.12.17/3.7.20}
\copyrightdate{2002}

\def\chaptername{Topologies and measures I}
\def\sectionname{Radon measure spaces}

\newsection{416}

We come now to the results for which the chapter so far has been
preparing.   The centre of topological measure theory is the theory of
`Radon' measures\cmmnt{ (411Hb)},
measures inner regular with respect to compact sets.   Most of the
section is devoted to pulling the earlier work together, and in
particular to
re-stating theorems on quasi-Radon measures in the new context.   Of
course this has to begin with a check that Radon measures are
quasi-Radon
(416A).   It follows immediately that Radon measures are (strictly)
localizable (416B).   After presenting a miscellany of elementary facts,
I turn to the constructions of \S413, which take on simpler and more
dramatic forms in this context (416J-416P).   I proceed to investigate
subspace measures (416R-416T) and some special product measures
(416U).   I end the section
with further notes on the forms which earlier theorems on Stone spaces
(416V) and compact measure spaces (416W) take when applied to Radon
measure spaces.

\leader{416A}{Proposition} A Radon measure space is quasi-Radon.

\proof{ Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space.   Because
$\frak T$ is Hausdorff, every compact set is closed, so $\mu$ is inner
regular with respect to the closed sets.   By 411E, $\mu$ is
$\tau$-additive;  by 411Gf, it is effectively locally finite.   Thus all
parts of condition (ii) of 411Ha are satisfied, and $\mu$ is a
quasi-Radon measure.
}%end of proof of 416A

\leader{416B}{Corollary} A Radon measure space is strictly localizable.

\proof{ Put 416A and 415A or 414J together.}

\leader{416C}{}\cmmnt{ In order to use the results of \S415
effectively, it will be helpful to spell out elementary conditions
ensuring that a quasi-Radon measure is Radon.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a locally
finite Hausdorff quasi-Radon measure space.   Then the following are
equiveridical:

\inset{(i) $\mu$ is a Radon measure;}

\inset{(ii) whenever $E\in\Sigma$ and $\mu E>0$ there is a compact set
$K$ such that $\mu(E\cap K)>0$;}

\inset{(iii) $\sup\{K^{\ssbullet}:K\subseteq X$ is compact$\}=1$ in the
measure algebra of $\mu$.}

\noindent If $\mu$ is totally finite we can add

\inset{(iv) $\sup\{\mu K:K\subseteq X$ is compact$\}=\mu X$.}

\proof{(i)$\Rightarrow$(ii) and (ii)$\Leftrightarrow$(iii) are trivial.   For
(ii)$\Rightarrow$(i), observe that if $E\in\Sigma$ and $\mu E>0$ there
is a compact set $K\subseteq E$ such that $\mu K>0$.   \Prf\ There is a
compact set $K'$ such that $\mu(E\cap K')>0$, by hypothesis;  now there
is a closed
set $K\subseteq E\cap K'$ such that $\mu K>0$, because $\mu$ is inner
regular with respect to the closed sets, and $K$ is compact.\ \QeD\   By
412Aa, $\mu$ is inner regular with respect to the compact sets.   Being a
complete, locally determined, locally finite topological measure, it is
a Radon measure.

When $\mu X<\infty$, of course, we also have (iii)$\Leftrightarrow$(iv).
}%end of proof of 416C

\leader{416D}{}\cmmnt{ Some further elementary facts are worth writing
out plainly.

\medskip

\noindent}{\bf Lemma} (a) In a Radon measure space, every compact set
has finite measure.

(b) Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space, and
$E\subseteq X$ a set such that $E\cap K\in\Sigma$ for every compact
$K\subseteq X$.   Then $E\in\Sigma$.

(c) A Radon measure is inner regular with respect to the
self-supporting compact sets.

(d) Let $X$ be a Hausdorff space and $\mu$ a tight locally finite complete
locally determined measure on $X$.   If $\mu$ measures
every compact set, $\mu$ is a Radon measure.

(e) Let $X$ be a Hausdorff space and $\familyiI{\mu_i}$ a family of Radon
measures on $X$.   Let $\mu=\sum_{i\in I}\mu_i$ be their
sum\cmmnt{ (definition: 234G)}.
If $\mu$ is locally finite, it is a Radon measure.

\proof{{\bf (a)} 411Ga.

\medskip

{\bf (b)} We have only to remember that $\mu$ is complete, locally
determined and tight, and apply 413F(ii).

\medskip

{\bf (c)} If $(X,\frak T,\Sigma,\mu)$ is a Radon measure space,
$E\in\Sigma$ and $\gamma<\mu E$, there is a compact set $K\subseteq E$
such that $\mu K\ge\gamma$.   By 414F, there is a self-supporting
relatively closed set $L\subseteq K$ such that $\mu L=\mu K$;  but now
of course $L$ is compact, while $L\subseteq E$ and $\mu L\ge\gamma$.

\medskip

{\bf (d)} Let $\Cal K$ be the family of compact subsets of $X$;  write
$\Sigma$ for the domain of $\mu$.   If $F\subseteq X$ is closed, then
$F\cap K\in\Cal K\subseteq\Sigma$ for every $K\in\Cal K$;  accordingly
$F\in\Sigma$, by 412Ja.   But this means that every closed set,
therefore every open set, belongs to $\Sigma$, and $\mu$ is a Radon
measure.

\medskip

{\bf (e)} Because every $\mu_i$ is a topological measure,
so is $\mu$;  because every $\mu_i$ is complete, so is $\mu$
(234Ha).   By hypothesis, $\mu$ is locally finite.
If $\mu E>0$, then there is some $i\in I$ such that $\mu_iE>0$;
now there is a
compact $K\subseteq E$ such that $0<\mu_iK\le\mu K$.   So $\mu$ is inner
regular with respect to the compact sets.

Now suppose that $E\subseteq X$ is such that $\mu$ measures $E\cap F$
whenever $\mu F<\infty$.   Then, in particular, $E\cap K$ is measured by
$\mu$, therefore measured by every $\mu_i$, whenever
$K\subseteq X$ is compact.   By 413F(ii) again, $\mu_i$ measures $E$ for every
$i$, so $\mu$ measures $E$.   As $E$ is
arbitrary, $\mu$ is locally determined and is a Radon measure.
}%end of proof of 416D

\cmmnt{\medskip

\noindent{\bf Remark} In (e) above, note that if $I$ is finite then
$\mu$ is necessarily locally finite.}

\leader{416E}{Specification of Radon \dvrocolon{measures}}\cmmnt{ In
415H I described some conditions which enable us to be sure that
two quasi-Radon measures on a given topological space are the same.   In
the case of Radon measures we have a similar list.   This time I include
a note on the natural ordering of Radon measures.

\medskip

\noindent}{\bf Proposition} Let $X$ be a Hausdorff space and $\mu$,
$\nu$ two Radon measures on $X$.

(a) The following are equiveridical:

\inset{(i) $\nu\le\mu$ in the sense of 234P\cmmnt{, that
is, $\nu E$ is defined and $\nu E\le\mu E$ whenever $\mu$ measures $E$};

(ii) $\mu K\le\nu K$ for every compact set $K\subseteq X$;

(iii) $\mu G\le\nu G$ for every open set $G\subseteq X$;

(iv) $\mu F\le\nu F$ for every closed set $F\subseteq X$.}

\noindent If $X$ is locally compact, we can add

\inset{(v) $\int fd\mu\le\int fd\nu$ for every non-negative continuous
function $f:X\to\Bbb R$ with compact support.}

(b) The following are equiveridical:

\inset{(i) $\mu=\nu$;

(ii) $\mu K=\nu K$ for every compact set $K\subseteq X$;

(iii) $\mu G=\nu G$ for every open set $G\subseteq X$;

(iv) $\mu F=\nu F$ for every closed set $F\subseteq X$.}

\noindent If $X$ is locally compact, we can add

\inset{(v) $\int fd\mu=\int fd\nu$ for every continuous function
$f:X\to\Bbb R$ with compact support.}

\proof{{\bf (a)(i)$\Rightarrow$(iv)$\Rightarrow$(ii)} and
{\bf (i)$\Rightarrow$(iii)} are trivial, if we recall that $\nu\le\mu$ when
$\dom\nu\supseteq\dom\mu$ and $\nu E\le\mu E$ for every $E\in\dom\mu$.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} is a special case of 412Mb.

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} The point is that if $K\subseteq X$ is
compact, then
$\mu K=\inf\{\mu G:G\subseteq X$ is open, $K\subseteq G\}$.
\Prf\ Because $X=\bigcup\{\mu G:G\subseteq X$ is open, $\mu G<\infty\}$,
there is an open set $G_0$ of finite measure including $K$.   Now, for
any $\gamma>\mu K$, there is a compact set $L\subseteq G_0\setminus K$
such that $\mu L\ge\mu G_0-\gamma$, so that $\mu G\le\gamma$, where
$G=G_0\setminus L$ is an open set including $K$.\ \Qed

The same is true for $\nu$.   So, if (iii) is true,

\Centerline{$\mu K=\inf_{G\supseteq K\text{ is open}}\mu G
\le\inf_{G\supseteq K\text{ is open}}\nu G=\nu K$}

\noindent for every compact $K\subseteq X$, and (ii) is true.

\medskip

\quad{\bf (iii)$\Rightarrow$(v)} If (iii) is true and
$f:X\to\coint{0,\infty}$ is a non-negative continuous function, then

$$\eqalignno{\int fd\mu
&=\int_0^{\infty}\mu\{x:f(x)>t\}dt\cr
\displaycause{252O}
&\le\int_0^{\infty}\nu\{x:f(x)>t\}dt
=\int fd\nu.\cr}$$

\medskip

\quad{\bf (v)$\Rightarrow$(iii)} If $X$ is locally compact and (v) is
true, take any open set $G\subseteq X$, and consider
the set $A$ of continuous functions $f:X\to[0,1]$
with compact support such that $f\le\chi G$.
Then $A$ is upwards-directed and $\sup_{f\in A}f(x)=\chi G(x)$
for every $x\in X$, by 4A2G(e-i).   So

\Centerline{$\mu G=\sup_{f\in A}\int fd\mu
\le\sup_{f\in A}\int fd\nu=\nu G$}

\noindent by 414Ba.   As $G$ is arbitrary, (iii) is true.

\medskip

{\bf (b)} now follows at once, or from 415H-415I.
}%end of proof of 416E

\leader{416F}{Proposition} Let $X$ be a Hausdorff space and $\mu$ a
Borel measure on $X$.   Then the following are equiveridical:

\quad(i) $\mu$ has an extension to a Radon measure on $X$;

\quad(ii) $\mu$ is locally finite and tight;

\quad(iii) $\mu$ is locally finite and effectively locally finite, and
$\mu G=\sup\{\mu K:K\subseteq G$ is compact$\}$ for every open set
$G\subseteq X$;

\quad(iv) $\mu$ is locally finite,
effectively locally finite and $\tau$-additive,
and $\mu G=\sup\{\mu(G\cap K):K\subseteq X$ is compact$\}$ for every
open set $G\subseteq X$.

In this case the extension is unique;  it is the c.l.d.\ version
of $\mu$.

\proof{{\bf (a)(i)$\Rightarrow$(iv)} If $\mu=\tilde\mu\restr\Cal B(X)$
where $\tilde\mu$ is a Radon measure and $\Cal B(X)$ is the Borel
$\sigma$-algebra of $X$, then of course $\mu$ is locally finite and
effectively locally finite and $\tau$-additive because $\tilde\mu$ is (see 416A)
and every
open set belongs to $\Cal B(X)$.   Also

\Centerline{$\mu G=\sup\{\mu K:K\subseteq G$ is compact$\}
\le\sup\{\mu(G\cap K):K\subseteq X$ is compact$\}\le\mu G$}

\noindent for every open set $G\subseteq X$, because $\tilde\mu$ is tight
and compact sets belong to $\Cal B(X)$.

\medskip

{\bf (b)(iv)$\Rightarrow$(iii)} Suppose that (iv) is true.
Of course $\mu$ is locally finite and effectively locally finite.
Suppose that $G\subseteq X$ is open and that $\gamma<\mu G$.
Then there is a compact $K\subseteq X$ such that $\mu(G\cap K)>\gamma$.
By 414K, the (totally finite) subspace measure $\mu_K$ is $\tau$-additive.
Now $K$ is a compact Hausdorff space, therefore regular.
By 414Ma there is a closed set $F\subseteq G\cap K$ such that
$\mu_KF\ge\gamma$.   Now $F$ is compact,
$F\subseteq G$ and $\mu F\ge\gamma$.
As $G$ and $\gamma$ are arbitrary, (iii) is true.

\medskip

{\bf (c)(iii)$\Rightarrow$(ii)} I have to show that if $\mu$ satisfies
the conditions of (iii) it is tight.
Let $\Cal K$ be the family of compact subsets of $X$ and
$\Cal A$ the family of subsets of $X$ which are either open or closed.
Then whenever $A\in\Cal A$, $F\in\Sigma$ and $\mu(A\cap F)>0$, there is
a $K\in\Cal K$ such that $K\subseteq A$ and $\mu(K\cap F)>0$.   \Prf\
Because $\mu$ is effectively locally finite, there is an open set $G$ of
finite measure such that $\mu(G\cap A\cap F)>0$.   ($\alpha$) If $A$ is
open, then there will be a compact set $K\subseteq G\cap A$ such that
$\mu K>\mu(G\cap A)-\mu(G\cap A\cap F)$, so that $\mu(K\cap F)>0$.
($\beta$) If $A$ is closed, then let $L\subseteq G$ be a compact set
such that $\mu L>\mu G-\mu(G\cap A\cap F)$;  then $K=L\cap A$ is compact
and $\mu(K\cap F)>0$.\ \Qed

By 412C, $\mu$ is inner regular with respect to $\Cal K$, as required.

\medskip

{\bf (d)(ii)$\Rightarrow$(i)} If $\mu$ is locally finite and tight,
let $\tilde\mu$ be the c.l.d.\ version
of $\mu$.   Then $\tilde\mu$ is complete, locally determined, locally
finite (because $\mu$ is), a topological measure (because $\mu$ is) and
tight (because $\mu$ is, using 412Ha);  so is a Radon
measure.   Every compact set has finite measure for $\mu$, so $\mu$ is
semi-finite and $\tilde\mu$ extends $\mu$ (213Hc).

\medskip

{\bf (e)} By 416Eb there can be at most one Radon measure extending
$\mu$, and we have observed in (d) above that in the present case it is
the c.l.d.\ version of $\mu$.
}%end of proof of 416F

\leader{416G}{}\cmmnt{ One of the themes of \S434 will be the
question:  on which Hausdorff spaces is every locally finite quasi-Radon
measure a Radon measure?   I do not think we are ready for a general
investigation of this, but I can give one easy special
result.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T)$ be a locally compact
Hausdorff space and $\mu$ a locally finite quasi-Radon measure on $X$.
Then $\mu$ is a Radon measure.

\proof{ $\mu$ satisfies condition (ii) of 416C.   \Prf\ Take
$E\in\dom\mu$
such that $\mu E>0$.   Let $\Cal G$ be the family of relatively compact
open subsets of $X$;  then $\Cal G$ is
upwards-directed and has union $X$.   By 414Ea, there is a $G\in\Cal G$
such that $\mu(E\cap G)>0$.   But now $\overline{G}$ is compact and
$\mu(E\cap\overline{G})>0$.\ \QeD\   By 416C, $\mu$ is a Radon measure.
}%end of proof of 416G

\leader{416H}{Corollary} Let $(X,\frak T)$ be a locally compact
Hausdorff space, and $\mu$ a locally finite, effectively locally
finite, $\tau$-additive Borel measure on $X$.   Then $\mu$ is tight and
its c.l.d.\ version
is a Radon measure, the unique Radon measure on $X$ extending $\mu$.

\proof{ By 415Cb, the c.l.d.\ version $\tilde\mu$ of $\mu$ is a
quasi-Radon measure extending $\mu$.   Because $\mu$ is locally
finite, so is $\tilde\mu$;  by 416G, $\tilde\mu$ is a Radon measure.
By 416Eb, the extension is unique.   Now

\Centerline{$\mu E=\tilde\mu E
=\sup_{K\subseteq E\text{ is compact}}\tilde\mu K
=\sup_{K\subseteq E\text{ is compact}}\mu K$}

\noindent for every Borel set $E\subseteq X$, so $\mu$ itself is tight.
}%end of proof of 416H

\leader{416I}{}\cmmnt{ While on the subject of locally compact spaces,
I mention an important generalization of a result from Chapter 24.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a locally
compact Radon measure space.   Write $C_k$ for the space of continuous
real-valued functions on $X$ with compact supports.   If
$1\le p<\infty$, $f\in\eusm L^p(\mu)$ and $\epsilon>0$, there is a
$g\in C_k$ such that $\|f-g\|_p\le\epsilon$.

\proof{ By 415Pa, there is a bounded continuous function
$h_1:X\to\Bbb R$ such that $G=\{x:h_1(x)\ne 0\}$ has finite measure and
$\|f-h_1\|_p\le\bover12\epsilon$.   Let $K\subseteq G$ be a compact set
such that
$\|h_1\|_{\infty}(\mu(G\setminus K))^{1/p}\le\bover12\epsilon$, and let
$h_2\in C_k$ be such that
$\chi K\le h_2\le\chi G$ (4A2G(e-i) again).   Set $g=h_1\times h_2$.   Then
$g\in C_k$ and

\Centerline{$\int|h_1-g|^p\le\int_{G\setminus K}|h_1|^p
\le\mu(G\setminus K)\|h_1\|_{\infty}^p$,}

\noindent so $\|h_1-g\|_p\le\bover12\epsilon$ and
$\|f-g\|_p\le\epsilon$, as required.
}%end of proof of 416I

\leader{416J}{}\cmmnt{ I turn now to constructions of Radon measures
based on ideas in \S413.

\medskip

\noindent}{\bf Theorem} Let $X$ be a Hausdorff space.   Let $\Cal K$ be
the family of compact subsets of $X$ and
$\phi_0:\Cal K\to\coint{0,\infty}$ a functional such that

\inset{($\alpha$) $\phi_0K
=\phi_0L+\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$
whenever $K$, $L\in\Cal K$ and $L\subseteq K$,}

\inset{($\gamma$) for every $x\in X$ there is an open set $G$ containing
$x$ such that $\sup\{\phi_0K:K\in\Cal K,\,K\subseteq G\}$ is finite.}

\noindent Then there is a unique Radon measure on $X$ extending
$\phi_0$.

\proof{ By 413N, there is a unique complete locally determined measure
$\mu$ on $X$, extending $\phi_0$, which is inner regular with respect to
$\Cal K$.   By ($\gamma$), $\mu$ is locally finite;  by 416Dd, it is a
Radon measure.
}%end of proof of 416J

\leader{416K}{Proposition}\cmmnt{ (see {\smc Tops{\o}e 70a})}
%Theorems 2-3 p 201
Let $X$ be a Hausdorff space,
$\Tau$ a subring of
$\Cal PX$ such that $\Cal H=\{G:G\in\Tau$ is open$\}$ covers $X$, and
$\nu:\Tau\to\coint{0,\infty}$ a finitely additive functional.   
Then there is a Radon measure $\mu$ on $X$ such
that $\mu K\ge\nu K$ for every compact $K\in\Tau$ and $\mu G\le\nu G$ for
every open $G\in\Tau$.

\proof{{\bf (a)} For $H\in\Cal H$ set $\Tau_H=\Tau\cap\Cal PH$;  then
$\Tau_H$ is an algebra of subsets of $H$, and $\nu_H=\nu\restr\Tau_H$ is
additive.   By 391G, there is an additive functional
$\tilde\nu_H:\Cal PH\to\coint{0,\infty}$ extending $\nu_H$.   Let $\frak F$
be an ultrafilter on $\Cal H$ containing $\{H:H_0\subseteq H\in\Cal H\}$
for every $H_0\in\Cal H$, and $\tilde\Tau$ the ideal of subsets of $X$
generated by $\Cal H$.   If $A\in\tilde\Tau$ then there is an
$H_0\in\Cal H$ including $A$, and now

\Centerline{$\tilde\nu_H(A\cap H)=\tilde\nu_H(A\cap H_0)
\le\nu_H(H\cap H_0)=\nu(H\cap H_0)\le\nu H_0$}

\noindent for every $H\in\Cal H$, so
$\tilde\nu A=\lim_{H\to\frak F}\nu_H(A\cap H)$ is defined in
$\coint{0,\infty}$.   Note that if $A\in\Tau\cap\tilde\Tau$ then there is
an $H_0\in\Cal H$ including $A$, so that $\nu_HA=\nu A$ whenever
$H\in\Cal H$ and $H\supseteq H_0$, and $\tilde\nu A=\nu A$.
Also $\tilde\nu:\tilde\Tau\to\coint{0,\infty}$ is
additive because all the functionals $A\mapsto\tilde\nu_H(A\cap H)$ are.

\medskip

{\bf (b)} Let $\Cal K$ be the family of compact subsets of $X$.
Because $X=\bigcup\Cal H$, $\Cal K\subseteq\tilde\Tau$.
For $K\in\Cal K$, set

\Centerline{$\phi_0K=\inf\{\tilde\nu G:G\in\tilde\Tau$ is open,
$K\subseteq G\}$.}

\noindent Then $\phi_0$ satisfies the conditions of 416J.   \Prf\
($\alpha$) Take $K$, $L\in\Cal K$ such that $L\subseteq K$, and
$\epsilon>0$.   Then there are open sets $G_0$, $H_0\in\tilde\Tau$ such
that

\Centerline{$K\subseteq G_0$,
\quad$\tilde\nu G_0\le\phi_0K+\epsilon$,
\quad$L\subseteq H_0$,
\quad$\tilde\nu H_0\le\phi_0L+\epsilon$.}

\noindent(i) If $K'\in\Cal K$ is such that
$K'\subseteq K\setminus L$, there are disjoint open sets $H$,
$H'\subseteq X$ such that $L\subseteq H$ and $K'\subseteq H'$ (4A2F(h-i)).   So

\Centerline{$\phi_0L+\phi_0K'
\le\tilde\nu(G_0\cap H)+\tilde\nu(G_0\cap H')\le\tilde\nu G_0
\le\phi_0K+\epsilon$.}

\noindent (ii) In the other direction, consider $K_1=K\setminus H_0$.
Then there is an open set $H_1\in\tilde\Tau$ such that $K_1\subseteq H_1$
and $\tilde\nu H_1\le\phi_0K_1+\epsilon$, so that

\Centerline{$\phi_0K\le\tilde\nu(H_0\cup H_1)\le\tilde\nu H_0+\tilde\nu H_1
\le\phi_0L+\phi_0K_1+2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\phi_0K
=\phi_0L+\sup\{\phi_0K':K'\in\Cal K$, $K'\subseteq K\setminus L\}$}

\noindent as required by 416J($\alpha$).   ($\gamma$) If $x\in X$ there is
an $H_0\in\Cal H$ containing $x$, and now

\Centerline{$\sup\{\phi_0K:K\in\Cal K$, $K\subseteq H_0\}
\le\tilde\nu H_0$}

\noindent is finite.   So the second hypothesis also is satisfied.\ \Qed

\medskip

{\bf (c)} By 416J, we have a Radon measure $\mu$ on $X$ extending $\phi_0$.
If $K\in\Tau$ is compact, then $\mu K\ge\nu K$.   \Prf\ Since
$K\in\Tau\cap\tilde\Tau$, $\tilde\nu K=\nu K$.   Now

\Centerline{$\mu K=
\phi_0K=\inf\{\tilde\nu G:G\in\tilde\Tau$ is open, $K\subseteq G\}
\ge\tilde\nu K=\nu K$.  \Qed}

\noindent If $G\in\Tau$ is open, then
$\mu G=\sup\{\mu K:K\subseteq G$ is compact$\}$.   But if $K\subseteq G$ is
compact then

\Centerline{$\mu K=\phi_0K\le\tilde\nu G=\nu G$,}

\noindent so $\mu G\le\nu G$.

Thus $\mu$ has the required properties.
}%end of proof of 416K

\leader{416L}{Proposition} Let $X$ be a regular Hausdorff space.   Let
$\Cal K$ be the family of compact subsets of $X$, and
$\phi_0:\Cal K\to\coint{0,\infty}$ a functional such that

\inset{($\alpha_1$) $\phi_0K\le\phi_0(K\cup L)\le\phi_0K+\phi_0L$ for
all $K$, $L\in\Cal K$,}

\inset{($\alpha_2$) $\phi_0(K\cup L)=\phi_0K+\phi_0L$ whenever $K$,
$L\in\Cal K$ and $K\cap L=\emptyset$,}

\inset{($\gamma$) for every $x\in X$ there is an open set $G$ containing
$x$ such that $\sup\{\phi_0 K:K\in\Cal K,\, K\subseteq G\}$ is finite.}

\noindent Then there is a unique Radon measure $\mu$ on $X$ such that

\Centerline{$\mu K=\inf_{G\subseteq X\text{ is open},K\subseteq G}
  \sup_{L\subseteq G\text{ is compact}}\phi_0L$}

\noindent for every $K\in\Cal K$.

\proof{{\bf (a)} For open sets $G\subseteq X$ set

\Centerline{$\psi G=\sup_{L\in\Cal K,L\subseteq G}\phi_0L$,}

\noindent and for compact sets $K\subseteq X$ set

\Centerline{$\phi_1K
=\inf\{\psi G:G\subseteq X$ is open, $K\subseteq G\}$.}

\noindent  Evidently $\psi G\le\psi H$ whenever $G\subseteq H$.   We
need to know that $\psi(G\cup H)\le\psi G+\psi H$ for all open sets $G$,
$H\subseteq X$.   \Prf\ If $L\subseteq G\cup H$ is compact, then the
disjoint compact sets
$L\setminus G$, $L\setminus H$ can be separated by disjoint open sets
$H'$, $G'$ (4A2F(h-i) again);  now $L\setminus G'\subseteq H$,
$L\setminus H'\subseteq G$ are compact and cover $L$, so

\Centerline{$\phi_0L
\le\phi_0(L\setminus G')+\phi_0(L\setminus H')
\le\psi H+\psi G$.}

\noindent As $L$ is arbitrary, $\psi(G\cup H)\le\psi G+\psi H$.\ \Qed

Moreover, $\psi(G\cup H)=\psi G+\psi H$ if $G\cap H=\emptyset$.   \Prf\
If $K\subseteq G$, $L\subseteq H$ are compact, then

\Centerline{$\phi_0K+\phi_0L
=\phi_0(K\cup L)
\le\psi(G\cup H)$.}

\noindent As $K$ and $L$ are arbitrary, $\psi G+\psi H\le\psi(G\cup
H)$.\ \Qed

\medskip

{\bf (b)} It follows that $\phi_1K$ is finite for every compact
$K\subseteq X$.   \Prf\ Set $\Cal G=\{G:G\subseteq X$ is open,
$\psi G<\infty\}$.   Then (a) tells us that $\Cal G$ is
upwards-directed.
But also we are supposing that $\Cal G$ covers $X$, by ($\gamma$).
So if $K\subseteq X$ is compact
there is a member of $\Cal G$ including $K$ and $\phi_1K<\infty$.\ \Qed

\medskip

{\bf (c)} Now $\phi_1$ satisfies the conditions of 416J.

\medskip

\Prf\grheada\ Suppose that $K$, $L\in\Cal K$ and $L\subseteq K$.   Set
$\gamma=\sup\{\phi_1M:M\in\Cal K,\,M\subseteq K\setminus L\}$.   Take
any $\epsilon>0$.

Let $G$ be an open set such that $K\subseteq G$ and
$\psi G\le\phi_1K+\epsilon$.   If $M\in\Cal K$ and
$M\subseteq K\setminus L$,
there are disjoint open sets $U$, $V$ such that $L\subseteq U$ and
$M\subseteq V$ (4A2F(h-i) once more);  we can arrange that
$U\cup V\subseteq G$.   In this case,

$$\eqalignno{\phi_1L+\phi_1M
&\le\psi U+\psi V
=\psi(U\cup V)\cr
\noalign{\noindent (by the second part of (a) above)}
&\le\psi G
\le\phi_1K+\epsilon.\cr}$$


\noindent As $M$ is arbitrary, $\gamma\le\phi_1K-\phi_1L+\epsilon$.

On the other hand, there is an open set $H$ such that $L\subseteq H$ and
$\psi H\le\phi_1L+\epsilon$.   Set $F=K\setminus H$, so that $F$ is a
compact subset of $K\setminus L$.   Then there is an open set $V$ such
that $F\subseteq V$ and $\psi V\le\phi_1F+\epsilon$.   In this case
$K\subseteq H\cup V$, so

$$\eqalign{\phi_1K
&\le\psi(H\cup V)
\le\psi H+\psi V\cr
&\le\phi_1L+\phi_1F+2\epsilon
\le\phi_1L+\gamma+2\epsilon,\cr}$$

\noindent so $\gamma\ge\phi_1K-\phi_1L-2\epsilon$.

As $\epsilon$ is arbitrary, $\gamma=\phi_1K-\phi_1L$;  as $K$ and $L$
are arbitrary, $\phi_1$ satisfies condition ($\alpha$) of 416J.

\medskip

\quad\grheadc\ Any $x\in X$ is contained in an open set $G$ such that
$\psi G<\infty$;  but now $\sup\{\phi_1K:K\in\Cal K,\,K\subseteq
G\}\le\psi G$ is finite.   So $\phi_1$ satisfies condition ($\gamma$) of
416J.\ \Qed

\medskip

{\bf (d)} By 416J, there is a unique Radon measure on $X$ extending
$\phi_1$, as claimed.
}%end of proof of 416L

\leader{416M}{Corollary} Let $X$ be a locally compact Hausdorff space.
Let $\Cal K$ be the family of compact subsets of $X$, and
$\phi_0:\Cal K\to\coint{0,\infty}$ a functional such that

\inset{$\phi_0K\le\phi_0(K\cup L)\le\phi_0K+\phi_0L$ for all $K$,
$L\in\Cal K$,}

\inset{$\phi_0(K\cup L)=\phi_0K+\phi_0L$ whenever $K$, $L\in\Cal K$
and $K\cap L=\emptyset$.}

\noindent Then there is a unique Radon measure $\mu$ on $X$ such that

\Centerline{$\mu K
=\inf\{\phi_0K':K'\in\Cal K,\,K\subseteq\interior K'\}$}

\noindent for every $K\in\Cal K$.

\proof{ Observe that $\phi_0$ satisfies the conditions of 416L;
416L($\gamma$) is true because $X$ is locally compact.
Define $\psi$, $\phi_1$ as in the proof of 416L, and set

\Centerline{$\phi_1'K=\inf\{\phi_0K':K'\in\Cal K,\,K\subseteq\interior
K'\}$}

\noindent for every $K\in\Cal K$.   Then $\phi_1'=\phi_1$.   \Prf\ Suppose that
$K\in\Cal K$ and $\epsilon>0$.   (i) There is an open set $G\subseteq X$
such that $K\subseteq G$ and $\psi G\le\phi_1K+\epsilon$.   Now the
relatively compact open subsets with closures included in $G$ form an
upwards-directed cover of $K$, so there is a $K'\in\Cal K$ such that
$K\subseteq\interior K'$ and $K'\subseteq G$.   Accordingly

\Centerline{$\phi_1'K\le\phi_0K'\le\psi G\le\phi_1K+\epsilon$.}

\noindent (ii) There is an $L\in\Cal K$ such that
$K\subseteq\interior L$ and $\phi_0L\le\phi_1'K+\epsilon$, so that

\Centerline{$\phi_1K\le\psi(\interior
L)\le\phi_0L\le\phi'_1K+\epsilon$.}

\noindent (iii) As $K$ and $\epsilon$ are arbitrary,
$\phi'_1=\phi_1$.\ \Qed

Now 416L tells us that there is a unique Radon measure extending
$\phi_1$, and this is the measure we seek.
}%end of proof of 416M

\leader{416N}{}\cmmnt{ The extension theorems in the second half of
\S413 also have important applications to Radon measures.

\medskip

\noindent}{\bf Henry's theorem}\cmmnt{ ({\smc Henry 69})} Let $X$ be a
Hausdorff space and $\mu_0$
a measure on $X$ which is locally finite and tight.   Then $\mu_0$ has an
extension to a Radon measure
$\mu$ on $X$,  and the extension may be made in such a way that whenever
$\mu E<\infty$ there is an $E_0\in\Sigma_0$ such that
$\mu(E\symmdiff E_0)=0$.

\proof{ All the work has been done in \S413;  here we need to check only
that the family $\Cal K$ of compact subsets of $X$ and the measure
$\mu_0$ satisfy the hypotheses of 413P.   But ($\dagger$) and
($\ddagger$) there
are elementary, and $\mu_0^*K<\infty$ for every $K\in\Cal K$ by 411Ga.

Now take the measure $\mu$ from 413P.   It is complete, locally
determined and inner regular with respect to $\Cal K$;  also
$\Cal K\subseteq\dom\mu$.   Because $\mu_0$ is locally finite and $\mu$
extends $\mu_0$, $\mu$ is locally finite.   By 416Dd, $\mu$ is a Radon
measure.   And the construction of 413P ensures that every set of finite
measure for
$\mu$ differs from a member of $\Sigma_0$ by a $\mu$-negligible set.
}%end of proof of 416N

\leader{416O}{Theorem} Let $X$ be a Hausdorff space and $\Tau$ a
subalgebra of $\Cal PX$.   Let $\nu:\Tau\to\coint{0,\infty}$ be a
finitely additive functional such that

\Centerline{$\nu E=\sup\{\nu F:F\in\Tau,\,F\subseteq E,\,F$ is closed$\}$
for every $E\in\Tau$,}

\Centerline{$\nu X
=\sup_{K\subseteq X\text{ is compact}}\inf_{F\in\Tau,F\supseteq K}\nu F$.}

\noindent Then there is a Radon measure $\mu$ on $X$ extending $\nu$.

\proof{{\bf (a)} For $A\subseteq X$, write

\Centerline{$\nu^*A=\inf_{F\in\Tau,F\supseteq A}\nu F$.}

\noindent Let $\sequencen{K_n}$ be a sequence of compact subsets of $X$
such that $\lim_{n\to\infty}\nu^*K_n=\nu X$;  replacing $K_n$ by
$\bigcup_{i<n}K_i$ if necessary, we may suppose that $\sequencen{K_n}$
is non-decreasing and that $K_0=\emptyset$.   For each $n\in\Bbb N$, set

\Centerline{$\nuprime_nE=\nu^*(E\cap K_n)$}

\noindent for every $E\in\Tau$.   Then $\nuprime_n$ is additive.   \Prf\
(I copy from the proof of 413O.)
If $E$, $F\in\Tau$ and $E\cap F=\emptyset$,

$$\eqalignno{\nuprime_n(E\cup F)
&=\inf\{\nu G:G\in\Tau,\,K_n\cap(E\cup F)\subseteq G\}\cr
&=\inf\{\nu G:
  G\in\Tau,\,K_n\cap(E\cup F)\subseteq G\subseteq E\cup F\}\cr
&=\inf\{\nu(G\cap E)+\nu(G\cap F):
  G\in\Tau,\,K_n\cap(E\cup F)\subseteq G\subseteq E\cup F\}\cr
&=\inf\{\nu G_1+\nu G_2:
  G_1,\,G_2\in\Tau,\,K_n\cap E\subseteq G_1\subseteq E,\,
  K_n\cap F\subseteq G_2\subseteq F\}\cr
&=\inf\{\nu G_1:G_1\in\Tau,\,K_n\cap E\subseteq G_1\subseteq E\}\cr
&\qquad\qquad\qquad
  +\inf\{\nu G_2:G_2\in\Tau,\,K_n\cap F\subseteq G_2\subseteq F\}\cr
&=\nuprime_nE+\nuprime_nF.\cr}$$

\noindent As $E$ and $F$ are arbitrary, $\nuprime_n$ is additive.\ \Qed

\medskip

{\bf (b)} For each $n\in\Bbb N$, set
$\nu_nE=\nuprime_{n+1}E-\nuprime_nE$ for
every $E\in\Tau$;  then $\nu_n$ is additive.   Because
$K_{n+1}\supseteq K_n$, $\nu_n$ is non-negative.

If $E\in\Tau$ and $E\cap K_{n+1}=\emptyset$, then
$\nu_nE=\nuprime_{n+1}E=0$.   So if we set
$\Tau_n=\{E\cap K_{n+1}:E\in\Tau\}$, we have an additive functional
$\tilde\nu_n:\Tau_n\to\coint{0,\infty}$ defined by setting
$\tilde\nu_n(E\cap K_{n+1})=\nu_nE$ for every $E\in\Tau$.
Also $\tilde\nu_nH=\sup\{\tilde\nu_nK:K\in\Tau_n$, $K\subseteq H$, $K$
is compact$\}$ for every $H\in\Tau_n$.   \Prf\ Express $H$ as
$E\cap K_{n+1}$ where $E\in\Tau$.   Given $\epsilon>0$, there is a closed set
$F\in\Tau$ such that $F\subseteq E$ and $\nu F\ge\nu E-\epsilon$;  but
now $K=F\cap K_{n+1}\in\Tau_n$ is a compact subset of $H$, and

\Centerline{$\tilde\nu_n(H\setminus K)
=\nu_n(E\setminus F)
\le\nuprime_{n+1}(E\setminus F)
\le\nu(E\setminus F)
\le\epsilon$,}

\noindent so $\tilde\nu_nK\ge\tilde\nu_nH-\epsilon$.\ \Qed

\medskip

{\bf (c)} For each $n\in\Bbb N$, we have a Radon measure $\mu_n$ on
$K_{n+1}$, with domain $\Sigma_n$ say, such that
$\mu_nK_{n+1}\le\tilde\nu_nK_{n+1}$ and $\mu_nK\ge\tilde\nu_nK$ for
every compact set $K\subseteq K_{n+1}$ (416K).   Since $K_{n+1}$ is
itself compact, we must have $\mu_nK_{n+1}=\tilde\nu_nK_{n+1}$.   But
this means that $\mu_n$ extends $\tilde\nu_n$.   \Prf\ If $H\in\Tau_n$ and
$\epsilon>0$ there is a compact set $K\in\Tau_n$ such that $K\subseteq H$
and $\tilde\nu_nK\ge\tilde\nu_nH-\epsilon$, so that
$(\mu_n)_*H\ge\mu_nK\ge\tilde\nu_nH-\epsilon$;  as $\epsilon$ is
arbitrary, $(\mu_n)_*H\ge\tilde\nu_nH$.   So there is an
$F_1\in\Sigma_n$ such that $F_1\subseteq H$ and
$\mu_nF_1\ge\tilde\nu_nH$.   Similarly, there is an $F_2\in\Sigma_n$
such that $F_2\subseteq K_{n+1}\setminus H$ and
$\mu_nF_2\ge\tilde\nu_n(K_{n+1}\setminus H)$.   But in this case
$H\setminus F_1\subseteq K_{n+1}\setminus(F_1\cup F_2)$ is
$\mu_n$-negligible, because

\Centerline{$\mu_nF_1+\mu_nF_2
\ge\tilde\nu_nH+\tilde\nu_n(K_{n+1}\setminus H)
=\tilde\nu_nK_{n+1}
=\mu_nK_{n+1}$.}

\noindent So $H\setminus F_1$ and $H$ belong to $\Sigma_n$ and
$\mu_nH=\mu_nF_1=\tilde\nu_nH$.\ \Qed

\medskip

{\bf (d)} Set

\Centerline{$\Sigma=\{E:E\subseteq X,\,E\cap K_{n+1}\in\Sigma_n$ for
every $n\in\Bbb N\}$,}

\Centerline{$\mu E=\sum_{n=0}^{\infty}\mu_n(E\cap K_{n+1})$ for every
$E\in\Sigma$.}

\noindent Then $\mu$ is a Radon measure on $X$ extending $\nu$.

\medskip

\Prf\ {\bf (i)} It is easy to check that $\Sigma$ is a $\sigma$-algebra
of subsets of $X$ including $\Tau$, just because each $\Sigma_n$ is a
$\sigma$-algebra of subsets of $K_{n+1}$ including $\Tau_n$;  and that
$\mu$ is a complete measure because every $\mu_n$ is.

\medskip

\quad{\bf (ii)} If $E\in\Tau$, then

$$\eqalign{\mu E
&=\sum_{n=0}^{\infty}\mu_n(E\cap K_{n+1})
=\sum_{n=0}^{\infty}\tilde\nu_n(E\cap K_{n+1})
=\sum_{n=0}^{\infty}\nu_nE
=\lim_{n\to\infty}\sum_{i=0}^n\nu_iE\cr
&=\lim_{n\to\infty}\sum_{i=0}^n
  \nu^*(E\cap K_{i+1})-\nu^*(E\cap K_i)
=\lim_{n\to\infty}\nu^*(E\cap K_n)
\le\nu E.\cr}$$

\noindent On the other hand,

\Centerline{$\mu X=\lim_{n\to\infty}\nu^*K_{n+1}=\nu X$,}

\noindent so in fact $\mu E=\nu E$ for every $E\in\Tau$, that is, $\mu$
extends $\nu$.   In particular, $\mu$ is totally finite, therefore
locally determined and locally finite.

\medskip

\quad{\bf (iii)} If $G\subseteq X$ is open, then $G\cap
K_{n+1}\in\Sigma_n$ for every $n$, so $G\in\Sigma$;  thus $\mu$ is a
topological measure.   If $\mu E>0$, there is some $n\in\Bbb N$ such
that $\mu_n(E\cap K_{n+1})>0$;  now there is a compact set
$K\subseteq
E\cap K_{n+1}$ such that $\mu_nK>0$, so that $\mu K>0$.   This shows
that $\mu$ is tight, so is a Radon measure, as required.\ \Qed
}%end of proof of 416O

\cmmnt{\medskip

\noindent{\bf Remark} Observe that in this construction

$$\eqalign{\mu K_{n+1}
&=\sum_{i=0}^{\infty}\mu_i(K_{n+1}\cap K_{i+1})
=\sum_{i=0}^{\infty}\tilde\nu_i(K_{n+1}\cap K_{i+1})
=\sum_{i=0}^{\infty}\nu_i(K_{n+1}\cap K_{i+1})\cr
&=\sum_{i=0}^{\infty}\nu'_{i+1}(K_{n+1}\cap K_{i+1})-\nu'_i(K_{n+1}\cap K_{i+1})\cr
&=\sum_{i=0}^{\infty}\nu^*(K_{n+1}\cap K_{i+1})-\nu^*(K_{n+1}\cap K_i)\cr
&=\sum_{i=0}^n\nu^*(K_{n+1}\cap K_{i+1})-\nu^*(K_{n+1}\cap K_i)
=\nu^*K_{n+1}\cr}$$

\noindent for every $n\in\Bbb N$.
What this means is that if instead of the hypothesis

\Centerline{$\nu X
=\sup_{K\subseteq X\text{ is compact}}
   \inf_{F\in\Tau,F\supseteq K}\nu F$}

\noindent we are presented with a specified non-decreasing sequence
$\sequencen{L_n}$ of compact subsets of $X$ such that
$\nu X=\sup_{n\in\Bbb N}\nu^*L_n$,
then we can take $K_{n+1}=L_n$ in the argument above
and we shall have $\mu L_n=\nu^*L_n$ for every $n$.
}%end of comment

\vleader{72pt}{416P}{Theorem} Let $X$ be a Hausdorff space and $\mu$ a
locally finite measure on $X$ which is inner regular with respect to the
closed sets.   Then the following are equiveridical:

\inset{(i) $\mu$ has an extension to a Radon measure on $X$;

(ii) for every non-negligible measurable set $E\subseteq X$ there is a
compact set $K\subseteq E$ such that $\mu^*K>0$.}

\noindent If $\mu$ is totally finite, we can add

\inset{(iii) $\sup\{\mu^*K:K\subseteq X$ is compact$\}=\mu X$.}

\proof{ Write $\Sigma$ for the domain of $\mu$.

\medskip

{\bf (a)(i)$\Rightarrow$(ii)} If $\lambda$ is a Radon measure
extending $\mu$, and $\mu E>0$,
then $\lambda E>0$, so there is a compact set $K\subseteq E$ such that
$\lambda K>0$;  but now, because $\lambda$ is an extension of $\mu$,

\Centerline{$\mu^*K\ge\lambda^*K=\lambda K>0$.}

\medskip

{\bf (b)(ii)$\Rightarrow$(i) \& (iii)}\grheada\ Let $\Cal E$ be the family
of measurable envelopes of compact sets.   Then $\mu E<\infty$ for every
$E\in\Cal E$.   \Prf\ If $E\in\Cal E$, there is a compact set $K$ such
that $E$ is a measurable envelope of $K$.   Now $\mu E=\mu^*K$ is finite
by 411Ga, as usual.\ \Qed

Next, $\Cal E$ is closed under finite unions, by 132Ed.
The hypothesis (ii) tells us that if $\mu E>0$ then there is some
$F\in\Cal E$ such that $F\subseteq E$ and $\mu F>0$;  for there is a
compact set $K\subseteq E$ such that $\mu^*K>0$, $K$ has a measurable
envelope $F_0$, and $F=E\cap F_0$ is still a measurable envelope of $K$.
So in fact $\mu$ is inner regular with respect to $\Cal E$ (412Aa).
In particular, $\mu$ is semi-finite.

If $\gamma<\mu X$ there is an $F\in\Cal E$ such that $\mu F\ge\gamma$,
and now there is a compact set $K$ such that $F$ is a measurable
envelope of $K$, so that $\mu^*K=\mu F\ge\gamma$.   As $\gamma$ is
arbitrary, (iii) is true.

\medskip

\quad\grheadb\ Because $\mu$ is inner regular with respect to $\Cal E$,
$D=\{E^{\ssbullet}:E\in\Cal E\}$ is order-dense in the measure algebra
$(\frak A,\bar\mu)$ of $\mu$ (412N), so there is a family
$\familyiI{d_i}$ in $D$ which is a partition of unity in $\frak A$
(313K).   For each $i\in I$, take $E_i\in\Cal E$ such that
$E_i^{\ssbullet}=d_i$.   Then

$$\eqalignno{\sum_{i\in I}\mu(E\cap E_i)
&=\sum_{i\in I}\bar\mu(E^{\ssbullet}\Bcap d_i)
=\bar\mu E^{\ssbullet}\cr
\displaycause{321E}
&=\mu E\cr}$$

\noindent for every $E\in\Sigma$.

\medskip

\quad\grheadc\ For each $i\in I$, let $\mu_i$ be the subspace measure on
$E_i$.   Then there is a Radon measure $\lambda_i$ on $E_i$ extending
$\mu_i$.   \Prf\ Because $\mu$ is inner regular with respect to the
closed sets, $\mu_i$ is inner regular with respect to the relatively
closed subsets of $E_i$ (412Oa).   Also there is a compact subset
$K\subseteq E_i$ such that

\Centerline{$\mu_iE_i=\mu E_i=\mu^*K=\mu_i^*K$,}

\noindent so $\mu_i$ satisfies the conditions of 416O and has an
extension to a Radon measure.\ \Qed

\medskip

\quad\grheadd\ Define

\Centerline{$\lambda E=\sum_{i\in I}\lambda_i(E\cap E_i)$}

\noindent whenever $E\subseteq X$ is such that $\lambda_i$ measures
$E\cap E_i$ for every $i\in I$.   Then $\lambda$ is a Radon measure on
$X$ extending $\mu$.   \Prf\ It is easy to check that it is a measure,
just because every $\lambda_i$ is a measure, and it extends $\mu$ by
($\beta$) above.   If $G\subseteq X$ is open, then $G\cap E_i$ is
relatively open for every $i\in I$, so $\lambda$ measures $G$;  thus
$\lambda$ is a topological measure.   If $\lambda E=0$ and
$A\subseteq E$, then
$\lambda_i(A\cap E_i)\le\lambda(E\cap E_i)=0$ for every $i$, so
$\lambda A=0$;  thus $\lambda$ is complete.   For all distinct $i$,
$j\in I$,

\Centerline{$\lambda_i(E_i\cap E_j)=\mu_i(E_i\cap E_j)
=\mu(E_i\cap E_j)=\bar\mu(d_i\Bcap d_j)=0$,}

\noindent so $\lambda E_i=\lambda_iE_i=\mu_iE_i$ is finite.   This means
that if $E\subseteq X$ is such that $\lambda$ measures $E\cap F$
whenever
$\lambda F<\infty$, then $\lambda$ must measure $E\cap E_i$ for every
$i$, and $\lambda$ measures $E$;  thus $\lambda$ is locally determined.
If $\lambda E>0$ there are an $i\in I$ such that $\lambda_i(E\cap
E_i)>0$ and a compact $K\subseteq E\cap E_i$ such that
$0<\lambda_iK=\lambda K$;  consequently $\lambda$ is tight.   Finally, if
$x\in X$, there is an $E\in\Sigma$ such that $x\in\interior E$ and
$\lambda E=\mu E<\infty$, so $\lambda$ is locally finite.   Thus
$\lambda$ is a Radon measure.\ \Qed

So (i) is true.

\medskip

{\bf (c)} Finally, suppose that $\mu$ is totally finite and (iii) is
true.   Then we can appeal directly to 416O to see that (i) is true.
}%end of proof of 416P

\leader{416Q}{Proposition} (a) Let $X$ be a compact Hausdorff space and
$\Cal E$ the algebra of open-and-closed subsets of $X$.   Then any
non-negative finitely additive functional from $\Cal E$ to $\Bbb R$ has
an extension to a Radon measure on $X$.   If $X$ is zero-dimensional
then the extension is unique.

(b) Let $\frak A$ be a Boolean algebra, and $Z$ its Stone space.   Then
there is a one-to-one correspondence between non-negative additive
functionals $\nu$ on $\frak A$ and Radon measures $\mu$ on $Z$ given by
the formula

\Centerline{$\nu a=\mu\widehat{a}$ for every $a\in\frak A$,}

\noindent where for $a\in\frak A$ I write $\widehat{a}$ for the
corresponding open-and-closed subset of $Z$.

\proof{{\bf (a)} Let $\nu:\Cal E\to\coint{0,\infty}$ be a non-negative
additive functional.   Then $\nu$ satisfies the conditions of 416O
(because every member of $\Cal E$ is closed, while $X$ is compact), so
has an extension to a Radon measure $\mu$.   If $X$ is zero-dimensional,
$\Cal E$ is a base for the topology of $X$ closed under finite unions
and intersections, so $\mu$ is unique, by 415H(iv) or 415H(v).

\medskip

{\bf (b)} The map $a\mapsto\widehat{a}$ is a Boolean isomorphism between
$\frak A$ and the algebra $\Cal E$ of open-and-closed subsets of $Z$, so
we have a one-to-one correspondence between non-negative additive
functionals $\nu$ on $\frak A$ and non-negative additive functionals
$\nuprime$ on $\Cal E$, defined by the formula $\nuprime\widehat{a}=\nu a$.
Now $Z$ is compact, Hausdorff and zero-dimensional, so $\nuprime$ has a
unique
extension to a Radon measure on $Z$, by part (a).   And of course every
Radon measure $\mu$ on $Z$ gives us a non-negative additive functional
$\mu\restr\Cal E$ on $\Cal E$, corresponding to a non-negative additive
functional on $\frak A$.
}%end of proof of 416Q

\leader{416R}{Theorem} (a) Any subspace of a Radon measure space is
a quasi-Radon measure space.

(b) A measurable subspace of a Radon measure space is a Radon measure
space.

(c) If $(X,\frak T,\Sigma,\mu)$ is a Hausdorff complete locally
determined
topological measure space, and $Y\subseteq X$ is such that the subspace
measure $\mu_Y$ on $Y$ is a Radon measure, then $Y\in\Sigma$.

\proof{{\bf (a)} Put 416A and 415B together.

\medskip

{\bf (b)} Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space, and
$(E,\frak T_E,\Sigma_E,\mu_E)$ a member of $\Sigma$ with the induced
topology and measure.   Because $\mu$ is complete and locally
determined,
so is $\mu_E$ (214Ka).   Because $\frak T$ is Hausdorff, so is
$\frak T_E$ (3A3Bh).   Because $\mu$ is locally finite, so is
$\mu_E$.   Because $\mu$ is tight (and a
subset of $E$ is compact for $\frak T_E$ whenever it is compact for
$\frak T$), $\mu_E$ is tight (412Oa again).

\medskip

{\bf (c)} \Quer\ If $Y\notin\Sigma$, then there is a set $F\in\Sigma$
such that $\mu_*(Y\cap F)<\mu^*(Y\cap F)$ (413F(v)).   But now
$\mu^*(Y\cap F)=\mu_Y(Y\cap F)$,
so there is a compact set $K\subseteq Y\cap F$ such
that $\mu_YK>\mu_*(Y\cap F)$.   When regarded as a subset of $X$, $K$ is
still compact;  because $\frak T$ is Hausdorff, $K$ is closed, so
belongs to $\Sigma$, and

\Centerline{$\mu_*(Y\cap F)\ge\mu K=\mu_YK>\mu_*(Y\cap F)$,}

\noindent which is absurd.\ \Bang
}%end of proof of 416R

\leader{416S}{}\cmmnt{ Corresponding to 415O, we have the following.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a Radon
measure space.

(a) If $\nu$ is a locally finite indefinite-integral measure over
$\mu$, it is a Radon measure.

(b) If $\nu$ is a Radon measure on $X$ and $\nu K=0$ whenever
$K\subseteq X$ is compact and $\mu K=0$, then $\nu$ is an
indefinite-integral measure over $\mu$.

\proof{{\bf (a)}
Because $\mu$ is complete and locally determined, so is $\nu$
(234Nb).   Because $\mu$ is tight, so is $\nu$ (412Q).
So if $\nu$ is also locally finite, it is a Radon measure.

\medskip

{\bf (b)} Write $\Tau$ for the domain of $\nu$.

\medskip

\quad{\bf (i)} If $E\in\Sigma\cap\Tau$ and $\mu E=0$, then
$\nu K=\mu K=0$ for every compact $K\subseteq E$, so $\nu E=0$.


\medskip

\quad{\bf (ii)}
$\Tau\supseteq\Sigma$.   \Prf\ If $E\in\Sigma$ and $K\subseteq X$ is
compact, there are Borel sets $F$, $F'$ such that
$F\subseteq E\cap K\subseteq F'\subseteq K$ and $\mu(F'\setminus F)=0$.
Consequently $\nu(F'\setminus F)=0$ and
$E\cap K\in\Tau$, because $\nu$ is complete.   By 416Db, $E\in\Tau$;
as $E$ is arbitrary, $\Sigma\subseteq\Tau$.\ \Qed

\medskip

\quad{\bf (iii)} If $E\in\Tau$, there is an $F\in\Sigma$ such that
$E\subseteq F$ and $\nu(F\setminus E)=0$.   \Prf\ By 416Dc and
412Ia, there is a
decomposition $\family{i}{I}{X_i}$ of $X$ for the measure $\mu$ such that
at most one of the $X_i$ is not a compact
$\mu$-self-supporting set, and that exceptional one, if any,
is $\mu$-negligible.   For each $i$, let $F_i$ be such that

\inset{----- if $X_i$ is a compact $\mu$-self-supporting set,
then $F_i$ is a Borel subset of $X_i$,
$F_i\supseteq E\cap X_i$ and $\nu(F_i\setminus E)=0$,

----- if $X_i$ is not a compact $\mu$-self-supporting set, $F_i=X_i$.}

\noindent Then $F_i\in\Sigma$ for every $i$ so $F=\bigcup_{i\in I}F_i$
belongs to $\Sigma$.   We also have $\nu(F_i\setminus E)=0$
for every $i$, because if
$X_i$ is not compact and $\mu$-self-supporting then $\nu X_i=\mu X_i=0$.
Of course $E\subseteq F$.   If $K\subseteq X$ is
compact, there is an open set $G\supseteq K$ such that $\mu G<\infty$;
consequently
$\{i:\mu(X_i\cap G)\ge\epsilon\}$ is finite for every $\epsilon>0$, so
$\{i:X_i\cap K\ne\emptyset\}$ is countable
and $\nu(K\cap F\setminus E)=0$.   By 412Jb, $\nu(F\setminus E)=0$.\
\Qed

Applying the same argument to $X\setminus E$, we can get an $F'\in\Sigma$
such that $F'\subseteq E$ and $\nu(E\setminus F')=0$.   As $E$ is
arbitrary, $\nu$ is the completion of its restriction to $\Sigma$.

\medskip

\quad{\bf (iv)} Now look at the conditions of 234O.   We know that
$\mu$ is localizable and $\nu$ is semi-finite.   We saw in (ii) above that
$\Tau\supseteq\Sigma$ and in (i) that $\nu$ is zero on $\mu$-negligible
sets.   In (iii) we saw that $\nu$ is the completion
of $\nu\restr\Sigma$.   And if $\nu E>0$ there is a compact $K\subseteq E$
such that $\nu K>0$, while $\mu K<\infty$.   So 234O tells us that
$\nu$ is an indefinite-integral measure over $\mu$.
}%end of proof of 416S

\leader{416T}{}\cmmnt{ I said in the notes to \S415 that the most
important quasi-Radon measure spaces are subspaces of Radon measure
spaces.   I do not know of a useful necessary and sufficient condition,
but the following deals with completely regular spaces.

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a locally
finite completely regular Hausdorff quasi-Radon measure space.   Then it
is isomorphic, as topological measure space, to a subspace of a locally
compact Radon measure space.

\proof{{\bf (a)}  Write $\beta X$ for the Stone-\v{C}ech compactification
of $X$ (4A2I);  I will take it that $X$ is actually a subspace of
$\beta X$.   Let $\Cal U$ be the set of those open subsets $U$ of
$\beta X$ such
that $\mu(U\cap X)<\infty$;  then $\Cal U$ is upwards-directed and
covers $X$, because $\mu$ is locally finite.   Set
$W=\bigcup\Cal U\supseteq X$.
Then $W$ is an open subset of $\beta X$, so is locally compact.

\medskip

{\bf (b)} Let $\Cal B(W)$ be the Borel $\sigma$-algebra of $W$.   Then
$V\cap X$ is
a Borel subset of $X$ for every $V\in\Cal B(W)$ (4A3Ca), so we have a
measure $\nu:\Cal B(W)\to[0,\infty]$ defined by setting
$\nu V=\mu(X\cap V)$ for every
$V\in\Cal B(W)$.   Now $\nu$ satisfies the conditions of 415Cb.   \Prf\
($\alpha$) If $\nu V>0$, then, because $\mu$ is effectively locally
finite, there is an open set $G\subseteq X$ such that $\mu(G\cap V)>0$
and $\mu G<\infty$.   There is an open set $U\subseteq\beta X$ such that
$U\cap X=G$, in which case $U\subseteq W$, $\nu U<\infty$ and
$\nu(U\cap V)>0$.   Thus $\nu$ is effectively locally finite.
($\beta$) If $\Cal U$ is an upwards-directed family of open subsets of
$W$, then $\{U\cap X:U\in\Cal U\}$ is an upwards-directed family of open
subsets of $X$, so

$$\eqalign{\nu(\bigcup\Cal U)
&=\mu(X\cap\bigcup\Cal U)
=\mu(\bigcup\{U\cap X:U\in\Cal U\})\cr
&=\sup_{U\in\Cal U}\mu(X\cap U)
=\sup_{U\in\Cal U}\nu U.\cr}$$

\noindent Thus $\nu$ is $\tau$-additive.\ \QeD\
So the c.l.d.\ version $\tilde\nu$ of $\nu$ is a quasi-Radon
measure on $W$ (415Cb).

\medskip

{\bf (c)} The construction of $W$ ensures that $\nu$ and $\tilde\nu$ are
locally finite.   By 416G, $\tilde\nu$ is a Radon measure.   So the
subspace measure $\tilde\nu_X$ is a quasi-Radon measure on $X$ (416Ra).
But $\tilde\nu_XG=\mu G$ for every open set $G\subseteq X$.
\Prf\ Note first that as $\nu$ effectively locally finite, therefore
semi-finite, $\tilde\nu$ extends $\nu$ (213Hc again).   If $K\subseteq W$ is a
compact set not meeting $X$, then

\Centerline{$\tilde\nu K=\nu K=\mu(K\cap X)=0$;}

\noindent accordingly $\tilde\nu_*(W\setminus X)=0$, by 413Ee.
Now there is an open set $U\subseteq W$ such that $G=X\cap U$, and

$$\eqalignno{\tilde\nu_XG
=\tilde\nu^*G
&\le\tilde\nu U
=\nu U
=\mu(U\cap X)
=\mu G\cr
&=\tilde\nu^*(U\cap X)+\tilde\nu_*(U\setminus X)\cr
\noalign{\noindent (by 413E(c-ii), because $\tilde\nu$ is semi-finite)}
&\le\tilde\nu^*(U\cap X)+\tilde\nu_*(W\setminus X)
=\tilde\nu^*G.\text{  \Qed}\cr}$$

\noindent So 415H(iii) tells us that $\mu=\tilde\nu_X$ is the subspace
measure induced by $\nu$.
}%end of proof of 416T

\leader{416U}{Theorem} (a) If $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
is a family of compact metrizable Radon probability spaces such that
every $\mu_i$ is strictly positive, the product measure on
$X=\prod_{i\in I}X_i$ is a completion regular Radon measure.

(b) In particular, the usual measures on $\{0,1\}^I$ and $[0,1]^I$
and $\Cal PI$ are completion regular Radon measures, for any set $I$.

\proof{{\bf (a)} By 415E,
it is a completion regular quasi-Radon probability
measure;  but $X$ is a compact Hausdorff space, so it is a Radon
measure, by 416G or otherwise.

\medskip

{\bf (b)} follows at once.
(I suppose it is obvious that by the `usual measure
on $[0,1]^I$' I mean the product measure when each copy of $[0,1]$ is
given Lebesgue measure.   Recall also that the `usual measure on
$\Cal PI$' is just the copy of the usual measure on $\{0,1\}^I$ induced
by the standard bijection $A\leftrightarrow\chi A$ (254Jb), which is a
homeomorphism (4A2Ud).
}%end of proof of 416U

\leader{416V}{Stone \dvrocolon{spaces}}\cmmnt{ The results of
415Q-415R become simpler and more striking in the present context.

\medskip

\noindent}{\bf Theorem} Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure
space, and $(Z,\frak S,\Tau,\nu)$ the Stone space of its measure algebra
$(\frak A,\bar\mu)$.   For $E\in\Sigma$ let $E^*$ be the open-and-closed
set in $Z$ corresponding to the image $E^{\ssbullet}$ of $E$ in
$\frak A$.   Define $R\subseteq Z\times X$ by saying that $(z,x)\in R$
iff $x\in F$ whenever $F\subseteq X$ is closed and $z\in F^*$.

(a) $R$ is the graph of a function $f:Q\to X$, where $Q=R^{-1}[X]$.   If
we set $W=\bigcup\{K^*:K\subseteq X$ is compact$\}$, then $W\subseteq Q$
is a $\nu$-conegligible open set, and the subspace measure $\nu_W$ on
$W$ is a Radon measure.

(b) Setting $g=f\restrp W$, $g$ is continuous and
$\mu$ is the image measure $\nu_Wg^{-1}$.

(c) If $X$ is compact, $W=Q=Z$ and $\mu=\nu g^{-1}$.

\proof{{\bf (a)} By 415Ra, $R$ is the graph of a function.   If
$K\subseteq X$ is compact and $z\in K^*$, then
$\Cal F=\{F:F\subseteq X$ is closed, $z\in F^*\}$ is a family of
non-empty closed subsets of $X$, closed
under finite intersections, and containing the compact set $K$;  so it
has non-empty intersection, and there is an $x\in K$ such that
$(z,x)\in R$, that is, $z\in Q$ and $f(z)\in K$.   Thus $W\subseteq Q$.
Of course $W$ is an open set, being the union of a family of
open-and-closed sets;  but it is also conegligible, because
$\sup\{K^{\ssbullet}:K\subseteq X$ is compact$\}=1$ in $\frak A$ (412N again),
so $Z\setminus W$ must be nowhere
dense, therefore negligible.   Now the subspace measure $\nu_W$ is
quasi-Radon because $\nu$ is (411P(d-iv), 415B);  but $W$ is a union of
compact open sets of
finite measure, so $\nu_W$ is locally finite and $W$ is locally compact;
by 416G, $\nu_W$ is a Radon measure.

\medskip

{\bf (b)} $g$ is continuous.   \Prf\ Let $G\subseteq X$ be an open set
and $z\in g^{-1}[G]$.   Let $K\subseteq X$ be a compact set such that
$z\in K^*$.   As remarked above, $g(z)=f(z)$ belongs to $K$.   $K$,
being a compact Hausdorff space, is regular (3A3Bb), so there is an open
set $H$ containing $g(z)$ such that $L=\overline{H\cap K}$ is included in $G$.
Note that $L$ is compact, so $L^*\subseteq W$.   Now $g(z)$ does not
belong to the
closed set $X\setminus H$, so $z\notin(X\setminus H)^*$ and $z\in H^*$;
accordingly $z\in(H\cap K)^*\subseteq L^*$.   If $w\in L^*$,
$g(w)\in L\subseteq G$;  so $L^*\subseteq g^{-1}[G]$, and
$z\in\interior g^{-1}[G]$.   As $z$ is arbitrary, $g^{-1}[G]$ is open;
as $G$ is arbitrary, $g$ is continuous.\ \Qed

By 415Rb, we know that $\mu=\nu_Qf^{-1}$, where $\nu_Q$ is the
subspace measure on $Q$.   But as $\nu$ is complete and both $Q$ and $W$
are conegligible,  we have

\Centerline{$\nu_Qf^{-1}[A]=\nu f^{-1}[A]=\nu g^{-1}[A]=\nu_Wg^{-1}[A]$}

\noindent whenever $A\subseteq X$ and any of the four terms is defined,
so that $\mu=\nu_Qf^{-1}=\nu_Wg^{-1}$.

\medskip

{\bf (c)} If $X$ is compact, then $Z=X^*\subseteq W$, so $W=Q=Z$ and
$\nu g^{-1}=\nu_Wg^{-1}=\mu$.
}%end of proof of 416V

\leader{416W}{Compact measure \dvrocolon{spaces}}\cmmnt{ Recall that a
semi-finite measure space $(X,\Sigma,\mu)$ is `compact' (as a measure
space) if there is a family $\Cal K\subseteq\Sigma$ such that $\mu$ is
inner regular with respect to $\Cal K$ and $\bigcap\Cal K'\ne\emptyset$
whenever $\Cal K'\subseteq\Cal K$ has the finite intersection property
(342Ac);  while $(X,\Sigma,\mu)$ is `perfect' if whenever $f:X\to\Bbb R$
is measurable and $\mu E>0$, there is a compact set $K\subseteq f[E]$
such that $\mu f^{-1}[K]>0$ (342K).   In \S342 I introduced these
concepts in order to study the realization of homomorphisms between
measure algebras.   The following result is now very easy.

\medskip

\noindent}{\bf Proposition} (a) Any Radon measure space is a compact
measure space, therefore perfect.

(b) Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure
space, with measure algebra $(\frak A,\bar\mu)$, and $(Y,\Tau,\nu)$ a
complete strictly localizable measure space, with measure algebra
$(\frak B,\bar\nu)$.   If $\pi:\frak A\to\frak B$ is an order-continuous
Boolean homomorphism, there is a function $f:Y\to X$ such that
$f^{-1}[E]\in\Tau$
and $f^{-1}[E]^{\ssbullet}=\pi E^{\ssbullet}$ for every $E\in\Sigma$.
If $\pi$ is measure-preserving, $f$ is \imp.

\proof{{\bf (a)} If ($X,\frak T,\Sigma,\mu)$ is a Radon measure space,
$\mu$ is inner regular with respect to the compact class consisting of
the compact subsets of $X$, so $(X,\Sigma,\mu)$ is a compact measure
space.   By 342L, it is perfect.

\medskip

{\bf (b)} Use (i)$\Rightarrow$(v) of Theorem 343B.   (Of course $f$ is
\imp\ iff $\pi$ is measure-{\vthsp}preserving.)
}%end of proof of 416W

\exercises{
\leader{416X}{Basic exercises $\pmb{>}$(a)}
%\spheader 416Xa
Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space, and $E\in\Sigma$
an atom for the measure.   Show that there is a point $x\in E$ such that
$\mu\{x\}=\mu E$.
%414G

\spheader 416Xb
Let $X$ be a Hausdorff space and $\mu$ a point-supported measure on
$X$, as described in 112Bd.   (i) Show that $\mu$ is tight, so is a
Radon measure iff
it is locally finite.   In particular, show that if $X$ has its discrete
topology then counting measure on $X$ is a Radon measure.   (ii) Show
that every purely atomic Radon measure is of this type.
%/

\sqheader 416Xc Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
Radon measure spaces, with direct sum $(X,\Sigma,\mu)$
(214L).   Give $X$ its disjoint union topology.   Show that $\mu$ is a
Radon measure.
%/

\spheader 416Xd Let $(X,\frak T,\Sigma,\mu)$ be a $\sigma$-finite
Radon measure space with $\mu X>0$.   Show that there is a
Radon probability measure on $X$ with the same measurable sets and the
same negligible sets as $\mu$.
%416A

\spheader 416Xe Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space.
Show that $\mu$ has a decomposition $\langle X_i\rangle_{i\in I}$ in
which every $X_i$ except at most one is a self-supporting compact set,
and the exceptional one, if any, is negligible.
%416B

\spheader 416Xf Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space.
Show that $\alpha\mu$, defined on $\Sigma$, is a
Radon measure for any $\alpha>0$.
%416D

\spheader 416Xg Let $X$ be a Hausdorff space and $\mu$, $\nu$ two Radon
measures on $X$ such that $\nu G=\mu G$ whenever $G\subseteq X$ is open
and of finite measure for both.   Show that $\mu=\nu$.
%416E

\spheader 416Xh Let $(X,\frak T)$ be a completely regular Hausdorff
space and $\mu$ a locally finite topological measure on $X$ which is
inner regular with respect to the closed sets.   Show that

$$\eqalign{\mu K
&=\inf\{\mu G:G\supseteq K\text{ is a cozero set}\}\cr
&=\inf\{\mu F:F\supseteq K\text{ is a zero set}\}
=\inf\{\int fd\mu:\chi K\le f\in C(X)\}\cr}$$

\noindent for every compact set $K\subseteq X$.
%416I

\spheader 416Xi Let $(X,\frak T,\Sigma,\mu)$ be a locally compact Radon
measure space, and $C_k$ the space of continuous real-valued functions
on $X$ with compact supports.   Show that $\{f^{\ssbullet}:f\in C_k\}$
is dense in $L^0(\mu)$ for the topology of convergence in measure.
%416I

\spheader 416Xj Let $(X,\frak T)$ be a Hausdorff space,
$\Sigma\supseteq\frak T$ a $\sigma$-algebra of subsets of $X$, and
$\nu:\Sigma\to\coint{0,\infty}$ a finitely additive functional such that
$\nu E=\sup\{\nu K:K\subseteq E$ is compact$\}$ for every $E\in\Sigma$.
Show that $\nu$ is countably additive and that its completion is a Radon
measure on $X$.
%416K

\spheader 416Xk Let $X$ be a completely regular Hausdorff space and
$\nu$ a locally finite Baire measure on $X$.   (i) Show that
$\nu^*K=\inf\{\nu G:G\subseteq X$ is a cozero set, $K\subseteq G\}$ for
every compact set $K\subseteq X$.   (ii) Show that there
is a Radon measure $\mu$ on $X$ such that $\mu K=\nu^*K$ for every
compact set $K\subseteq X$.
%416L

\spheader 416Xl Let $\sequencen{X_n}$ be a sequence of Hausdorff spaces
with product $X$;  write $\Cal B(X_n)$ for the Borel $\sigma$-algebra of
$X_n$.
Let $\Tau$ be the $\sigma$-algebra $\Tensorhat_{n\in\Bbb N}\Cal B(X_n)$
(definition:  254E).   Let $\nu:\Tau\to\coint{0,\infty}$ be a
finitely additive functional such that
$E\mapsto\nu\pi_n^{-1}[E]:\Cal B(X_n)\to\coint{0,\infty}$ is countably
additive and tight for each $n\in\Bbb N$, writing
$\pi_n(x)=x(n)$ for $x\in X$ and $n\in\Bbb N$.   Show that there is
a unique Radon measure on $X$ extending $\nu$.   \Hint{416O.}
%416O

\spheader 416Xm Set $S_2=\bigcup_{n\in\Bbb N}\{0,1\}^n$, and let
$\phi:S_2\to\coint{0,\infty}$ be a functional such that
$\phi(\sigma)=\phi(\sigma^{\smallfrown}\fraction{0})
 +\phi(\sigma^{\smallfrown}\fraction{1})$
for every $\sigma\in S_2$, writing $\sigma^{\smallfrown}\fraction{0}$ and
$\sigma^{\smallfrown}\fraction{1}$ for the two members of $\{0,1\}^{n+1}$ extending
any $\sigma\in\{0,1\}^n$.   Show that there is a unique Radon measure
$\mu$ on $\{0,1\}^{\Bbb N}$ such that
$\mu\{x:x\restr\{0,\ldots,n-1\}=\sigma\}
=\phi(\sigma)$ whenever $n\in\Bbb N$, $\sigma\in\{0,1\}^{\Bbb N}$.
%416Xl, 416O. 416Q

\spheader 416Xn Let $X$ be a Hausdorff space, $\mu$ a complete locally
finite measure on $X$, and $Y$ a conegligible subset of $X$.   Show that
$\mu$ is a Radon measure iff the subspace measure on $Y$ is a Radon
measure.
%416R

\spheader 416Xo Let $X$ be a Hausdorff space, $Y$ a subset of $X$, and
$\nu$ a Radon measure on $Y$.   Define a measure $\mu$ on $X$ by setting
$\mu E=\nu(E\cap Y)$ whenever $\nu$ measures $E\cap Y$.   Show that if
{\it either} $Y$ is closed {\it or} $\nu$ is totally finite, $\mu$ is a
Radon measure on $X$.   (Cf.\ 418I.)
%416R

\spheader 416Xp Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space
and $\Cal E\subseteq\Sigma$ a non-empty upwards-directed family.   Set
$\nu F=\sup_{E\in\Cal E}\mu(E\cap F)$ whenever $F\subseteq X$ is such
that $\mu$ measures $E\cap F$ for every $E\in\Cal E$.   Show that $\nu$
is a Radon measure on $X$.
%416R

\spheader 416Xq Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space.
Show that a measure $\nu$
on $X$ is an indefinite-integral measure over $\mu$ iff ($\alpha$) $\nu$
is a complete locally determined tight topological measure ($\beta$) $\nu K=0$
whenever $K\subseteq X$ is compact and $\mu K=0$.
%416S

\spheader 416Xr Let $(X,\frak T,\Sigma,\mu)$ be a compact Hausdorff
quasi-Radon measure space.   Let $W\subseteq X$ be the union of the open
subsets of $X$ of finite measure.   Show that the subspace measure on
$W$ is a Radon measure.
%416T

\spheader 416Xs Let $(X,\frak T,\Sigma,\mu)$ be a completely regular
Radon measure space.   Show that it
is isomorphic, as topological measure space,
to a measurable subspace of a locally compact Radon measure space.
%416T

\spheader 416Xt Let $(X,\frak T,\Sigma,\mu)$ be a compact Radon measure
space and $(Z,\frak S,\Tau,\nu)$ the Stone space of its measure algebra.
For $E\in\Sigma$ let $E^*$ be the corresponding open-and-closed subset
of $Z$, as in 416V.   Show that the function described in 416V is the
unique continuous function $h:Z\to X$ such that
$\nu(E^*\symmdiff h^{-1}[E])=0$ for every $E\in\Sigma$.   \Hint{415Qd.}
%416V 415Q

\spheader 416Xu Show that the Sorgenfrey line (415Xc, 439Q),
with Lebesgue measure, is a
quasi-Radon measure space which, regarded as a measure space, is
compact, but, regarded as a topological measure space, is not a Radon
measure space.
%416W

\spheader 416Xv Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra and $(Z,\frak T,\Sigma,\mu)$ its Stone space.
Show that if $\nu$ is a strictly positive Radon measure on $Z$ then $\mu$
is an indefinite-integral measure over $\nu$.
%416W 416Xq

\leader{416Y}{Further exercises (a)}
%\spheader 416Ya 
Let $X\subseteq\beta\Bbb N$ be the union of all those
open sets $G\subseteq\beta\Bbb N$ such that
$\sum_{n\in G\cap\Bbb N}\bover1{n+1}$ is finite.   For $E\subseteq X$
set $\mu E=\sum_{n\in E\cap\Bbb N}\bover1{n+1}$.   Show that $\mu$ is a
$\sigma$-finite Radon
measure on the locally compact Hausdorff space $X$.   Show that $\mu$ is
not outer regular with respect to the open sets.
%412W 416Xb

\spheader 416Yb
Let $X$ be a Hausdorff space and $\nu$ a countably additive real-valued
functional defined on a $\sigma$-algebra $\Sigma$ of subsets of $X$.
Show that the following are equiveridical:  (i)
$|\nu|:\Sigma\to\coint{0,\infty}$,
defined as in 362B, is a Radon measure on $X$;  (ii) $\nu$ is
expressible as $\mu_1-\mu_2$, where $\mu_1$, $\mu_2$ are Radon measures
on $X$ and $\Sigma=\dom\mu_1\cap\dom\mu_2$.
%416D

\spheader 416Yc Let $X$ be a topological space and $\mu_0$ a semi-finite
measure on $X$ which is inner regular with respect to the family
$\Cal K_{\text{ccc}}$ of closed countably compact sets.   Show that $\mu_0$
has an extension to a complete locally determined topological measure
$\mu$ on $X$, still inner regular with respect to $\Cal K_{\text{ccc}}$;
and that the extension may be done in such a way that whenever
$\mu E<\infty$ there
is an $E_0\in\Sigma_0$ such that $\mu(E\symmdiff E_0)=0$.
%\Hint{use
%the argument of 416N, but with
%$\Cal K=\{K:K\in\Cal K_{\text{ccc}},\,\mu_0^*K<\infty\}$.}
%416N

\spheader 416Yd Let $X$ be a topological space and $\mu_0$ a semi-finite
measure on $X$ which is inner regular with respect to the family
$\Cal K_{\text{sc}}$ of sequentially compact sets.   Show that $\mu_0$
has an extension to a complete locally determined topological measure
$\mu$ on $X$, still inner regular with respect to $\Cal K_{\text{sc}}$;
and that the extension may be done in such a way that whenever
$\mu E<\infty$
there is an $E_0\in\Sigma_0$ such that $\mu(E\symmdiff E_0)=0$.
%416Yc, 416N

\spheader 416Ye Set $S=\bigcup_{n\in\Bbb N}\BbbN^n$, and let
$\phi:S\to\coint{0,\infty}$ be a functional such that
$\phi(\sigma)=\sum_{i=0}^{\infty}\phi(\sigma^{\smallfrown}\fraction{i})$
for every
$\sigma\in S$, writing $\sigma^{\smallfrown}\fraction{i}$ for the
members of
$\BbbN^{n+1}$ extending any $\sigma\in\BbbN^n$.   Show that there is a
unique Radon measure $\mu$ on $\BbbN^{\Bbb N}$ such that
$\mu I_{\sigma}=\phi(\sigma)$ for every $\sigma\in S$, where
$I_{\sigma}=\{x:x\restr\{0,\ldots,n-1\}=\sigma\}$ for any $n\in\Bbb N$,
$\sigma\in\BbbN^{\Bbb N}$.
%\Hint{set
%$\theta A=\inf\{\sum_{\sigma\in R}\phi(\sigma):R\subseteq S$,
%$A\subseteq\bigcup_{\sigma\in R}I_{\sigma}\}$ for every
%$A\subseteq\BbbN^{\Bbb N}$, and use \Caratheodory's method.}
%416Xm, 416Xl, 416Q

\spheader 416Yf\dvAnew{2017} In 416Qb, show that $\mu$ is atomless iff
$\nu$ is properly atomless in the sense of 326F.
%416Q 416Xb

\spheader 416Yg Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space.
Show that a measure $\nu$ on $X$ is an indefinite-integral measure over
$\mu$ iff (i) there is a topology $\frak S$ on $X$, including $\frak T$,
such that $\nu$ is a Radon measure with respect to $\frak S$
(ii) $\nu K=0$ whenever $K$ is a $\frak T$-compact set and $\mu K=0$.
%416S, 416Xq

\spheader 416Yh Let $\sequencen{x_n}$ enumerate a dense subset of
$X=\{0,1\}^{\frakc}$ (4A2B(e-ii)).   Let $\nu_{\frakc}$ be the usual
measure on $X$, and set
$\mu E=\bover12\nu_{\frakc} E+\sum_{n=0}^{\infty}2^{-n-2}\chi E(x_n)$ for
$E\in\dom\nu_{\frakc}$.   (i) Show that $\mu$ is a strictly positive Radon
probability measure on $X$ with Maharam type $\frak c$.   (ii) Let
$I\in[\frak c]^{\le\omega}$ be such that $x_m\restr I\ne x_n\restr I$
whenever $m\ne n$.   Set $Z=\{0,1\}^I$ and let $\pi:X\to Z$ be the
canonical map.   Show that if $f\in C(X)$ is such that
$\int f\times g\pi\,d\mu=0$ for every $g\in C(Z)$, then $f=0$.
\Hint{otherwise, take $n\in\Bbb N$ such that
$|f(x_n)|\ge\bover12\|f\|_{\infty}$, and let $g\ge 0$ be such that
$g(\pi x_n)=1$ and
$\int g\,d(\mu\pi^{-1})<3\cdot 2^{-n-3}$;  show that
$\int f\times g\pi\,d\mu>0$.}   (iii) Show that there is no orthonormal
basis for $L^2(\mu)$ in $\{f^{\ssbullet}:f\in C(X)\}$.
(See {\smc Hart \& Kunen 99}.)
%416I 416U

\spheader 416Yi In 254Ye, show that if we start from a continuous
\imp\ $f:[0,1]\to[0,1]^2$, as in 134Yl, we get a continuous \imp\
surjection $g:[0,1]\to[0,1]^{\Bbb N}$.
%416U

\spheader 416Yj Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space and
$\Cal A\subseteq\Sigma$ a countable set.   Let $\frak S$ be the topology
generated by $\frak T\cup\Cal A$.   Show that $\mu$ is $\frak S$-Radon.
%416+
}%end of exercises

\endnotes{\Notesheader{416} The original measures studied by Radon
({\smc Radon 1913}) were, in effect, what I call differences of Radon
measures on $\BbbR^r$, as introduced in \S256.
Successive generalizations moved first to Radon measures on general
compact Hausdorff spaces, then to locally compact Hausdorff spaces, and
finally to
arbitrary Hausdorff spaces, as presented in this section.   I ought
perhaps to remark that, following {\smc Bourbaki 65}, many authors use
the term `Radon measure' to describe a linear
functional on a space of continuous functions;  I will discuss the
relationship between such functionals and the measures of this chapter
in \S436.   For the moment, observe that by 415I a Radon measure on
a completely regular space can be determined from the integrals it
assigns to continuous functions.   It is also common for the phrase
`Radon measure' to be used for what I would call a tight Borel measure;
you have to check each author to see whether local
finiteness is also assumed.   In my usage, a Radon measure is necessarily
the c.l.d.\ version of a Borel measure.   The Borel measures which
correspond to Radon measures are described in 416F.

In \S256, I discussed Radon measures on $\BbbR^r$ as a preparation for
a discussion of convolutions of measures.   It should now be coming
clear why I felt that it was impossible, in that context, to give you a
proper idea of what a Radon measure, in the modern form, `really' is.
In Euclidean space, too many concepts coincide.   As a trivial example,
the simplest definition of `local finiteness' (256Ab) is not the
right formulation in other spaces (411Fa).
Next, because every closed set is a
countable union of compact sets, there is no distinction between `inner
regular with
respect to closed sets' and `inner regular with respect to compact
sets', so one cannot get any intuition for which is important in which
arguments.
(When we come to subspace measures on non-measurable subsets, of course,
this changes;  quasi-Radon measures on subsets of Euclidean space are
important and interesting.)   Third, the fact that the c.l.d.\ product
of two Radon measures on Euclidean space is already a Radon measure
(256K) leaves us with no idea of what to do with a general product of
Radon measures.   (There are real difficulties at this point, which I
will attack
in the next section.   For the moment I offer just 416U.)   And
finally, we simply cannot represent a product of uncountably many Radon
probability measures on Euclidean spaces as a measure on Euclidean
space.

As you would expect, a very large proportion of the results of this
chapter, and many theorems from earlier volumes, were originally proved
for compact Radon measure spaces.   The theory of general totally finite
Radon measures is, in effect, the theory of measurable subspaces of
compact Radon
measure spaces, while the theory of quasi-Radon measures is pretty much
the theory of non-measurable subspaces of Radon measure spaces.   Thus
the theorem that
every quasi-Radon measure space is strictly localizable is almost a
consequence of the facts that every Radon measure space is strictly
localizable and any subspace of a strictly localizable space is strictly
localizable.

The cluster of results between 416J and 416Q form only a sample, I hope
a reasonably representative sample, of the many theorems on construction
of Radon measures from functionals on algebras or lattices of sets.
(See also 416Ye.)   The essential simplification,
compared with the theorems in \S413 and \S415, is that we do not need to
mention any $\sigma$- or
$\tau$-additivity condition of the type 413J($\beta$) or 415K($\beta$),
because we are dealing with a `compact class', the family of compact
subsets of a Hausdorff space.   We can use this even at some distance,
as in 416O (where the hypotheses do not require any non-empty compact
set to belong to the domain of the original functional).   The
particular feature of 416O which makes it difficult to prove from such
results as 413K and 413P above is that we have to retain control of the
outer measures of a sequence $\sequencen{K_n}$ of non-measurable sets.
In general this is hard to do, and is possible here principally because
the sequence is non-decreasing, so that we can make sense of the
functionals $E\mapsto\nu^*(E\cap K_{n+1})-\nu^*(E\cap K_n)$;  compare 214P.

In 416De and 416Ea I suggest elements of an algebraic structure on a space
of Radon measures;  for more about this, see 436Xs and 437Yi below.
}%end of notes

\discrpage


\frfilename{mt417.tex}
\versiondate{9.3.10/14.9.21}
\copyrightdate{2000}

\def\chaptername{Topologies and measures I}
\def\sectionname{$\tau$-additive product measures}

\newsection{417}

The `ordinary' product measures introduced in Chapter 25 have served us
well for a volume and a half.   But we come now to a fundamental
obstacle.   If we start with two Radon measure spaces, their product
measure, as
defined in \S251, need not be a Radon measure (419E).   Furthermore, the
counterexample is one of the basic compact measure spaces of the theory;
and while it is dramatically non-metrizable, there is no other reason to
set it aside.   Consequently, if we wish (as we surely do) to create
Radon measure spaces as
products of Radon measure spaces, we need a new construction.   This is
the object of the present section.   It turns out that the construction
can be adapted to work well beyond the special context of Radon measure
spaces;  the methods here apply to general effectively locally finite
$\tau$-additive topological measures (for the product of finitely many
factors) and to $\tau$-additive topological probability measures (for
the product of infinitely many factors).

The fundamental theorems are 417C and 417E, listing the essential
properties of what I call `$\tau$-additive product measures', which are
extensions of the c.l.d.\ product measures and product probability
measures of Chapter 25.   They depend on a straightforward lemma on the
extension of
a measure to make every element of a given class of sets negligible
(417A).   We still have Fubini's theorem for the new product measures
(417G), and the basic operations from \S254 still apply (417J, 417K,
417M).

It is easy to check that if we start with quasi-Radon measures, then the
$\tau$-additive product measure is again quasi-Radon (417N, 417O).   The
$\tau$-additive product of two Radon measures is Radon (417P), and the
$\tau$-additive product of Radon probability measures with compact
supports is Radon (417Q).

In the last part of the section I look at continuous real-valued
functions and Baire $\sigma$-algebras;  it turns out that for these the
ordinary product measures are adequate (417U, 417V).

\leader{417A}{Lemma} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space, and $\Cal A\subseteq\Cal PX$ a family of sets such that the inner measure
$\mu_*(\bigcup_{n\in\Bbb N}A_n)$ is $0$ for every sequence $\sequencen{A_n}$
in $\Cal A$.   Then there is a measure $\mu'$ on $X$, extending $\mu$,
such that

\inset{(i) $\mu'A$ is defined and zero for every $A\in\Cal A$;

(ii) $\mu'$ is complete if $\mu$ is;
%don't complete automatically on account of ref in 495G

(iii) for
every $F$ in the domain $\Sigma'$ of $\mu'$ there is an $E\in\Sigma$
such that $\mu'(F\symmdiff E)=0$;

(iv) whenever $\Cal K$, $\Cal G$ are families of sets such that

\inset{($\alpha$) $\mu$ is inner regular with respect to $\Cal K$,

($\beta$) $K\cup K'\in\Cal K$ for all $K$, $K'\in\Cal K$,

($\gamma$) $\bigcap_{n\in\Bbb N}K_n\in\Cal K$ for every sequence
$\sequencen{K_n}$ in $\Cal K$,

($\delta$) for every $A\in\Cal A$ there is a $G\in\Cal G$, including $A$,
such that $G\setminus A\in\Sigma$,

($\epsilon$) $K\setminus G\in\Cal K$ whenever $K\in\Cal K$ and
$G\in\Cal G$,}

\noindent then $\mu'$ is inner regular with respect to $\Cal K$.}

\noindent In particular, $\mu$ and $\mu'$ have
isomorphic measure algebras, so that $\mu'$ is localizable if $\mu$ is.

\proof{{\bf (a)} Let $\Cal A^*$ be the collection of subsets of $X$ which
can be
covered by a countable subfamily of $\Cal A$.   Then $\Cal A^*$ is a
$\sigma$-ideal of subsets of $X$ and $\mu_*A=0$ for every
$A\in\Cal A^*$. Set

\Centerline{$\Sigma'=\{E\symmdiff A:E\in\Sigma,\,A\in\Cal A^*\}$.}

\noindent Then $\Sigma'$ is a $\sigma$-algebra of subsets of $X$.
\Prf\ (i) $\emptyset=\emptyset\symmdiff\emptyset\in\Sigma'$.   (ii) If
$E\in\Sigma$, $A\in\Cal A^*$ then
$X\setminus(E\symmdiff A)=(X\setminus E)\symmdiff A\in\Sigma'$.   (iii)
If $\sequencen{E_n}$,
$\sequencen{A_n}$ are sequences in $\Sigma$, $\Cal A^*$ respectively,
then

\Centerline{$E=\bigcup_{n\in\Bbb N}E_n\in\Sigma$,
\quad$A=E\symmdiff\bigcup_{n\in\Bbb N}(E_n\symmdiff A_n)
\subseteq\bigcup_{n\in\Bbb N}A_n\in\Cal A^*$,}

\noindent so $\bigcup_{n\in\Bbb N}(E_n\symmdiff A_n)
=E\symmdiff A\in\Sigma'$.\ \Qed

\medskip

{\bf (b)} If $E$, $E'\in\Sigma$, $A$, $A'\in\Cal A^*$ and
$E\symmdiff A=E'\symmdiff A'$, then
$E\symmdiff E'=A\symmdiff A'\in\Cal A^*$ and
$\mu_*(E\symmdiff E')=0$;  because $\mu$ is semi-finite,
$\mu(E\symmdiff E')=0$ and $\mu E=\mu E'$.   Accordingly we can define
$\mu':\Sigma'\to[0,\infty]$ by setting

\Centerline{$\mu'(E\symmdiff A)=\mu E$ whenever $E\in\Sigma$,
$A\in\Cal A^*$.}

\noindent Evidently $\mu'$ extends $\mu$ and $\mu'A=0$ for every
$A\in\Cal A$.   Also $\mu'$ is a measure.   \Prf\ (i)
$\mu'\emptyset=\mu\emptyset=0$.
(ii) If $\sequencen{F_n}$ is a disjoint sequence in $\Sigma'$, with
union $F$, express each $F_n$ as $E_n\symmdiff A_n$ where
$E_n\in\Sigma$, $A_n\in\Cal A^*$;  set $E=\bigcup_{n\in\Bbb N}E_n$, so
that $F\symmdiff E\in\Cal A^*$ and $\mu'F=\mu E$.    If $m\ne n$, then
$E_m\cap E_n\subseteq A_m\cup A_n$, so $\mu(E_m\cap E_n)=0$;
accordingly

\Centerline{$\mu'F=\mu E=\sum_{n=0}^{\infty}\mu E_n
=\sum_{n=0}^{\infty}\mu'F_n$.  \Qed}

\medskip

{\bf (c)}
A subset of $X$ is $\mu'$-negligible iff it can be included in a set of
the form $E\symmdiff A$ where $\mu E=0$ and $A\in\Cal A^*$, so $\mu'$ is
complete if $\mu$ is.
The embedding $\Sigma\embedsinto\Sigma'$ induces a measure-preserving
homomorphism from the measure algebra of $\mu$ to the measure algebra of
$\mu'$ which is an isomorphism just because every member of $\Sigma'$ is
the symmetric difference of a member of $\Sigma$ and a $\mu'$-negligible
set.

\medskip

{\bf (d)} This deals with (i)-(iii) in the statement of the lemma.   Now
suppose that $\Cal K$ and $\Cal G$ are as in (iv).   Take $F\in\Sigma'$ and
$\gamma<\mu'F$.   Take $E\in\Sigma$ and a sequence $\sequencen{A_n}$ in
$\Cal A$ such that $E\symmdiff F\subseteq\bigcup_{n\in\Bbb N}A_n$.
Then $\mu E=\mu'F>\gamma$ so (again because $\mu$ is semi-finite) there is
a $K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and
$\gamma<\mu K<\infty$;  set
$\epsilon=\bover12(\mu K-\gamma)>0$.   For each $n\in\Bbb N$, choose
$G_n\in\Cal G$ and $K_n\in\Cal K\cap\Sigma$ such that

\Centerline{$A_n\subseteq G_n$,
\quad$E_n=G_n\setminus A_n$ belongs to $\Sigma$,}

\Centerline{$K_n\subseteq K\cap E_n$,
\quad$\mu(K\cap E_n\setminus K_n)\le 2^{-n}\epsilon$.}

\noindent Set $L=\bigcap_{n\in\Bbb N}(K\setminus G_n)\cup K_n$.
Putting the hypotheses (iv-$\epsilon$), (iv-$\beta$) and (iv-$\gamma$)
together, we see that $L\in\Cal K$;  moreover, since $G_n=E_n\cup A_n$
belongs to $\Sigma'$ for every $n$, $L\in\Sigma'$.   Next,
setting $H=\bigcap_{n\in\Bbb N}(K\setminus E_n)\cup K_n$,
$L=H\setminus\bigcup_{n\in\Bbb N}A_n$,
so $\mu'L=\mu H$ and $L\subseteq F$.   But
$K\setminus H\subseteq\bigcup_{n\in\Bbb N}(K\cap E_n\setminus K_n)$ so

\Centerline{$\mu'L=\mu H\ge\mu K-\sum_{n=0}^{\infty}2^{-n}\epsilon
=\gamma$.}

\noindent As $F$ and $\gamma$ are arbitrary, $\mu'$ is inner regular with
respect to $\Cal K$.
}%end of proof of 417A

\leader{417B}{Lemma} Let $X$ and $Y$ be topological spaces, and $\nu$ a
$\tau$-additive topological measure on $Y$.

(a) If $W\subseteq X\times Y$ is open, then
$x\mapsto\nu W[\{x\}]:X\to[0,\infty]$ is lower semi-continuous.

(b) If $\nu$ is effectively locally finite and $\sigma$-finite and
$W\subseteq X\times Y$ is a Borel set, then $x\mapsto\nu W[\{x\}]$ is
Borel measurable.

(c) If $f:X\times Y\to[0,\infty]$ is a lower semi-continuous function,
then $x\mapsto\int f(x,y)\nu(dy):X\to[0,\infty]$ is lower
semi-continuous.

(d) If $\nu$ is totally finite and $f:X\times Y\to\Bbb R$ is a bounded
continuous function, then
$x\mapsto\int f(x,y)\nu(dy)$ is continuous.

(e) If $\nu$ is totally finite and $W\subseteq X\times Y$ is a Baire
set, then $x\mapsto\nu W[\{x\}]$ is Baire measurable.

\proof{{\bf (a)} If $x\in X$ and $\nu W[\{x\}]>\alpha$, then

\Centerline{$\Cal H=\{H:H\subseteq Y$ is open, there is an open set $G$
containing $x$ such that $G\times H\subseteq W\}$}

\noindent is an upwards-directed family of open sets with union
$W[\{x\}]$, so there is an $H\in\Cal H$ such that $\nu H\ge\alpha$.
Now there is an open set $G$ containing $x$ such that
$G\times H\subseteq W$, so that $\nu W[\{x'\}]\ge\alpha$ for every
$x'\in G$.

\medskip

{\bf (b)(i)} Suppose to begin with that $\nu$ is totally finite.   In
this case, the set

$$\eqalign{\{W:W\subseteq X\times Y,\,x\mapsto\nu W[\{x\}]&\text{ is a
Borel measurable function}\cr
&\mskip50mu\text{ defined everywhere on }X\}\cr}$$

\noindent is a Dynkin class containing every open set, so contains every
Borel set, by the Monotone Class Theorem (136B).

\medskip

\quad{\bf (ii)} For the general case, let $\sequencen{Y_n}$ be a
disjoint sequence of sets of finite measure covering $Y$, and for
$n\in\Bbb N$ let $\nu_n$ be the subspace measure on $Y_n$.   Then
$\nu_n$ is effectively locally finite and $\tau$-additive (414K).   If
$W\subseteq X\times Y$ is a Borel set, then $W_n=W\cap(X\times Y_n)$ is
a relatively Borel set for each $n$, so that $x\mapsto\nu_nW_n[\{x\}]$
is Borel measurable, by (i).   Since
$\nu W[\{x\}]=\sum_{n=0}^{\infty}\nu_nW_n[\{x\}]$ for every $x$,
$x\mapsto\nu W[\{x\}]$ is Borel measurable.

\medskip

{\bf (c)} For $i$, $n\in\Bbb N$ set $W_{ni}=\{(x,y):f(x,y)>2^{-n}i\}$,
so that $W_{ni}\subseteq X\times Y$ is open.   Set
$f_n=2^{-n}\sum_{i=1}^{4^n}\chi W_{ni}$;  then $\sequencen{f_n}$ is a
non-decreasing sequence with supremum $f$.   For $n\in\Bbb N$ and
$x\in X$,
$\int f_n(x,y)\nu(dy)=2^{-n}\sum_{i=1}^{4^n}\nu W_{ni}[\{x\}]$, so
$x\mapsto\int f_n(x,y)\nu(dy)$ is lower semi-continuous, by (a) and
4A2B(d-iii).   By 414Ba, $\int f(x,y)\nu(dy)$ is the supremum
$\sup_{n\in\Bbb N}\int f_n(x,y)\nu(dy)$ for every
$x$, so $x\mapsto\int f(x,y)\nu(dy)$ is lower semi-continuous
(4A2B(d-v)).

\medskip

{\bf (d)} Applying (c) to $f+\|f\|_{\infty}\chi(X\times Y)$, we see that
$x\mapsto\int f(x,y)\nu(dy)$ is lower semi-continuous.   Similarly,
$x\mapsto-\int f(x,y)\nu(dy)$ is lower semi-continuous, so
$x\mapsto\int f(x,y)\nu(dy)$ is continuous (4A2B(d-vi)).

\medskip

{\bf (e)} Suppose first that $W$ is a cozero set;  let
$f:X\times Y\to[0,1]$ be a continuous function such that
$W=\{(x,y):f(x,y)>0\}$.   For $n\in\Bbb N$ set
$f_n=nf\wedge\chi(X\times Y)$.   Then $\sequencen{f_n}$ is a
non-decreasing sequence of continuous functions with supremum $\chi W$.
By (d), all the functions $x\mapsto\int f_n(x,y)\nu(dy)$ are continuous,
so their limit $x\mapsto\nu W[\{x\}]$ is Baire measurable.

Now

$$\eqalign{\{W:W\subseteq X\times Y,\,x\mapsto\nu W[\{x\}]
&\text{ is a Baire measurable function}\cr
&\mskip50mu\text{ defined everywhere on }X\}\cr}$$

\noindent is a Dynkin class containing every cozero set, so contains
every Baire set, by the Monotone Class Theorem again.
}%end of proof of 417B

\leader{417C}{Theorem}\dvArevised{2021}\cmmnt{ ({\smc Ressel 77})} Let
$(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be effectively locally finite $\tau$-additive
topological measure spaces.

(a) There is a unique complete locally determined effectively locally finite
$\tau$-additive topological measure $\tilde\lambda$ on $X\times Y$ which is
inner regular with respect to the $\sigma$-algebra
$\tilde\Lambda^0=(\Sigma\tensorhat\Tau)\vee\Cal B(X\times Y)$
generated by $\{E\times F:E\in\Sigma$, $F\in\Tau\}
\cup\{W:W\subseteq X\times Y$ is open$\}$ and is such that
$\tilde\lambda(E\times F)$ is defined and equal to $\mu E\cdot\nu F$ whenever
$E\in\Sigma$ and $F\in\Tau$.

(b)(i) $\tilde\lambda$ extends the c.l.d.\ product measure $\lambda$
on $X\times Y$, and if $\tilde Q$ is measured by $\tilde\lambda$, there is a
$Q$ measured by $\lambda$ such that $\tilde\lambda(\tilde Q\symmdiff Q)=0$.

\quad(ii) The support of $\tilde\lambda$ is the product of the supports of $\mu$
and $\nu$.

\quad(iii) If $\tilde Q$ is measured by $\tilde\lambda$,

\Centerline{$\tilde\lambda\tilde Q
=\sup\{\tilde\lambda(\tilde Q\cap(G\times H)):
G\in\frak T$, $H\in\frak S$, $\mu G<\infty$, $\mu H<\infty\}$.}

\quad(iv) If $\Sigma'\subseteq\Sigma$ and $\Tau'\subseteq\Tau$ are
$\sigma$-algebras such that $\mu$ is inner regular with respect to $\Sigma'$ and
$\nu$ is inner regular with respect to $\Tau'$, then $\tilde\lambda$ is inner
regular with respect to $(\Sigma'\tensorhat\Tau')\vee\Cal B(X\times Y)$.

\quad(v) If $W\subseteq X\times Y$ is open, then

\qquad($\alpha$) there is an open set
$W'\in\Lambda$ such that $W'\subseteq W$ and
$\lambda W'=\tilde\lambda W$, so $\tilde\lambda W=\lambda_*W$,

\qquad($\beta$) if $E\in\Sigma$ and $F\in\Tau$ then

\Centerline{$\tilde\lambda(W\cap(E\times F))
=\int_E\nu(W[\{x\}]\cap F)\mu(dx)=\int_F\mu(W^{-1}[\{y\}]\cap E)\nu(dy)$.}

\quad(vi) If $\mu$ and $\nu$ are inner regular with respect to the Borel
sets, so is $\tilde\lambda$.

\quad(vii) If $\mu$ and $\nu$ are inner regular with respect to the closed sets,
so is $\tilde\lambda$.

\quad(viii) If $\mu$ and $\nu$ are tight\cmmnt{ (that is, inner
regular with respect to the closed compact sets)}, so is
$\tilde\lambda$.

\quad(ix) If $\mu$ and $\nu$ are locally finite, so is $\tilde\lambda$.

\cmmnt{\medskip

\noindent{\bf Notation} For the rest of this section, if $\Sigma$, $\Sigma'$ are
$\sigma$-algebras of subsets of a set $X$, $\Sigma\vee\Sigma'$ will be the
$\sigma$-algebra generated by $\Sigma\cup\Sigma'$;  and if $X$ is a topological
space, $\Cal B(X)$ will be its Borel $\sigma$-algebra.}

\proof{ Write

\Centerline{$\Sigma^f=\{E:E\in\Sigma,\,\mu E<\infty\}$,
\quad$\Tau^f=\{F:F\in\Tau,\,\nu F<\infty\}$,}

\Centerline{$\frak T^f=\frak T\cap\Sigma^f$,
\quad$\frak S^f=\frak S\cap\Tau^f$.}

\medskip

{\bf (a)(i)} Let $\lambda$ be the c.l.d.\ product of $\mu$ and $\nu$, and
$\Lambda$ its domain.   Write $\Cal U$ for
$\{G\times H:G\in\frak T^f,\,H\in\frak S^f\}$.   Because
$\frak T\subseteq\Sigma$ and $\frak S\subseteq\Tau$,
$\Cal U\subseteq\Lambda$.   $\Cal U$ need not be a base for the topology
of $X\times Y$, unless $\mu$ and $\nu$ are locally finite, but if an
open subset of $X\times Y$ is included in a member of $\Cal U$ it is the
union of the members of $\Cal U$ it includes.   Moreover, if
$Q\in\Lambda$, then $\lambda Q=\sup_{U\in\Cal U}\lambda(Q\cap U)$.
\Prf\ By 412R, $\lambda$ is inner regular with respect to
$\bigcup_{U\in\Cal U}\Cal PU$.\ \Qed

Write $\Cal U_s$ for the set of finite
unions of members of $\Cal U$, and $\frak V$ for the set of non-empty
upwards-directed families $\Cal V\subseteq\Cal U_s$ such that
$\sup_{V\in\Cal V}\lambda V<\infty$.
For each $\Cal V\in\frak V$, fix on a countable $\Cal V'\subseteq\Cal V$
such that $\sup_{V\in\Cal V'}\lambda V=\sup_{V\in\Cal V}\lambda V$;
because $\Cal V$ is upwards-directed, we may
suppose that $\Cal V'=\{V_n:n\in\Bbb N\}$ for some non-decreasing
sequence $\sequencen{V_n}$ in $\Cal V$.   Set
$A(\Cal V)=\bigcup\Cal V\setminus\bigcup\Cal V'$.

\medskip

\quad{\bf (ii)}\grheada\
For $V\in\Cal U_s$, set $f_V(x)=\nu V[\{x\}]$ for every
$x\in X$.   This is always defined because $V$ is open;  moreover, $f_V$
is lower semi-continuous, by 417Ba.   Because $V$ is a
finite union of products of sets of finite measure,
$\int f_Vd\mu=\lambda V$.

\medskip

\qquad\grheadb\ The key to the proof is the following fact:  for any
$\Cal V\in\frak V$, almost
every vertical section of $A(\Cal V)$ is negligible.   \Prf\
$\family{V}{\Cal V}{f_V}$ is a
non-empty upwards-directed set of lower semi-continuous functions.
Set

\Centerline{$g(x)=\nu(\bigcup_{V\in\Cal V}V[\{x\}])$,
\quad$h(x)=\nu(\bigcup_{V\in\Cal V'}V[\{x\}])$}

\noindent for every $x\in X$.   Because $\Cal V$ is upwards-directed and
$\nu$ is $\tau$-additive,

\Centerline{$g(x)=\sup_{V\in\Cal V}\nu V[\{x\}]
=\sup_{V\in\Cal V}f_V(x)$}

\noindent in $[0,\infty]$ for each $x$, so, by 414Ba again,

\Centerline{$\int g\,d\mu=\sup_{V\in\Cal V}\int f_Vd\mu
=\sup_{V\in\Cal V}\lambda V=\sup_{V\in\Cal V'}\lambda V
=\int h\,d\mu$.}

\noindent Since $h\le g$ and $\sup_{V\in\Cal V}\lambda V$ is finite,
$g(x)=h(x)<\infty$ for $\mu$-almost every $x$.   But for any such $x$,
we must have

\Centerline{$\nu(\bigcup\Cal V)[\{x\}]
=\nu(\bigcup\Cal V')[\{x\}]<\infty$,}

\noindent so that

\Centerline{$A(\Cal V)[\{x\}]
=(\bigcup\Cal V)[\{x\}]\setminus(\bigcup\Cal V')[\{x\}]$}

\noindent is negligible.\ \Qed

\medskip

\quad{\bf (iii)} \Quer\ Suppose, if possible, that there is a sequence
$\sequencen{\Cal V_n}$ in $\frak V$ such that
$\lambda_*(\bigcup_{n\in\Bbb N}A(\Cal V_n))>0$.    Take $W\in\Lambda$
such that
$W\subseteq\bigcup_{n\in\Bbb N}A(\Cal V_n)$ and $\lambda W>0$.   Because
almost every vertical section of every $A(\Cal V_n)$ is negligible,
almost every vertical section of $W$ is negligible.   But this
contradicts Fubini's theorem (252F).\ \Bang

\medskip

\quad{\bf (iv)} Setting $\Cal A=\{A(\Cal V):\Cal V\in\frak V\}$,
let $\lambda'$ be the corresponding
extension of $\lambda$ as described in 417A, $\tilde\lambda$
the c.l.d.\ version of $\lambda'$ (213E), and $\Lambda'$, $\tilde\Lambda$
the domains of $\lambda'$, $\tilde\lambda$ respectively.

\medskip

\qquad\grheada\ If $W\in\Lambda$, then
$W\in\Lambda'\subseteq\tilde\Lambda$.
Also, because $\lambda$ is semi-finite,

$$\eqalign{\lambda'W
&=\lambda W
=\sup\{\lambda W':W'\subseteq W,\,W\in\Lambda,\,\lambda W'<\infty\}\cr
&\le\sup\{\lambda'W':W'\subseteq W,\,W\in\Lambda',\,\lambda'W'<\infty\}
=\tilde\lambda W
\le\lambda'W.\cr}$$

\noindent Thus $\lambda W=\tilde\lambda W$;  as $W$ is arbitrary,
$\tilde\lambda$ extends $\lambda$.   Because $\mu$ and $\nu$ are
semi-finite (411Gd),

\Centerline{$\tilde\lambda(E\times F)=\lambda(E\times F)=\mu E\cdot\nu F$}

\noindent whenever $E\in\Sigma$ and $\nu\in F$ (251J).

\medskip

\qquad\grheadb\ If $\tilde Q\in\tilde\Lambda$ and
$\gamma<\tilde\lambda\tilde Q$, there is
a $U\in\Cal U$ such that $\tilde\lambda(\tilde Q\cap U)\ge\gamma$.   \Prf\
There is a $Q'\in\Lambda'$ such that $Q'\subseteq\tilde Q$ and
$\gamma<\lambda'Q'<\infty$.   By 417A(iii), there is a $Q\in\Lambda$
such that $\lambda'(Q\symmdiff Q')=0$, so that

\Centerline{$\lambda Q=\lambda'Q=\lambda'Q'>\gamma$.}

\noindent There is a $U\in\Cal U$ such that
$\lambda(Q\cap U)\ge\gamma$, by (i).   Now

\Centerline{$\tilde\lambda(\tilde Q\cap U)
\ge\lambda'(Q'\cap U)=\lambda'(Q\cap U)
=\lambda(Q\cap U)\ge\gamma$.  \Qed}

\noindent Consequently $\tilde\lambda$ is effectively locally finite.

\medskip

\qquad\grheadc\ $\tilde\lambda$ is a topological measure.   \Prf\ Let
$W\subseteq X\times Y$ be an open set.   Suppose that
$\tilde Q\in\tilde\Lambda$
and $\tilde\lambda\tilde Q>0$.   By ($\beta$), there is a $U\in\Cal U$ such that
$\tilde\lambda(\tilde Q\cap U)>0$.
Let $\Cal V$ be $\{V:V\in\Cal U_s,\,V\subseteq W\cap U\}$.   Then
$\Cal V\in\frak V$, so $\tilde\lambda A(\Cal V)=\lambda'A(\Cal V)=0$, by
417A(i); since $\bigcup\Cal V'\in\Lambda$,

\Centerline{$W\cap U=\bigcup\Cal V\in\Lambda'\subseteq\tilde\Lambda$.}

\noindent But this means that $\tilde Q\cap U\cap W$ and
$\tilde Q\cap U\setminus W=(\tilde Q\cap U)\setminus(W\cap U)$ belong to
$\tilde\Lambda$ and

\Centerline{$\tilde\lambda_*(\tilde Q\cap W)
   +\tilde\lambda_*(\tilde Q\setminus W)
\ge\tilde\lambda(\tilde Q\cap U\cap W)
  +\tilde\lambda(\tilde Q\cap U\setminus W)
=\tilde\lambda(\tilde Q\cap U)>0$.}

\noindent Because $\tilde\lambda$ is complete and locally determined,
and $\tilde Q$ is arbitrary, this is enough to ensure that
$W\in\tilde\Lambda$ (413F(vii)).\ \Qed

\medskip

\qquad\grheadd\ $\tilde\lambda$ is $\tau$-additive.   \Prf\Quer\
Suppose, if possible, otherwise;  that there is a non-empty
upwards-directed family $\Cal W$ of open sets in $X\times Y$ such that
$\tilde\lambda W^*>\gamma=\sup_{W\in\Cal W}\tilde\lambda W$, where
$W^*=\bigcup\Cal W$.
In this case, we can find a $Q'\in\Lambda'$ such that
$Q'\subseteq W^*$ and $\lambda'Q'>\gamma$, a $Q\in\Lambda$ such that
$\lambda'(Q'\symmdiff Q)=0$, and a $U\in\Cal U$ such that
$\lambda(Q\cap U)>\gamma$ (using (i) again).   Let
$\Cal V\in\frak V$ be the set of those
$V\in\Cal U_s$ such that $V\subseteq W\cap U$ for some $W\in\Cal W$.
Then $\bigcup\Cal V=W^*\cap U$, so

$$\eqalignno{\gamma
&<\lambda(Q\cap U)
=\lambda'(Q\cap U)
=\lambda'(Q'\cap U)\cr
&\le\tilde\lambda(W^*\cap U)
=\tilde\lambda(\bigcup\Cal V)
=\tilde\lambda(\bigcup\Cal V')\cr
\noalign{\noindent (because $\tilde\lambda A(\Cal V)=0$)}
&=\sup_{V\in\Cal V'}\tilde\lambda V\cr
\noalign{\noindent (because $\Cal V'$ is countable and
upwards-directed)}
&\le\sup_{W\in\Cal W}\tilde\lambda W
\le\gamma,\cr}$$

\noindent which is absurd.\ \Bang\Qed

\medskip

\qquad\grheade\ $\tilde\lambda$ is inner regular with respect to
$\tilde\Lambda^0$.   \Prf\ Applying 412R with $\Cal K=\Sigma$ and $\Cal L=\Tau$,
we see that $\lambda$ is inner regular with respect to
$\Sigma\tensorhat\Tau$ and therefore with respect to $\tilde\Lambda^0$.
If $\Cal V\in\frak V$, then

\Centerline{$A(\Cal V)\subseteq\bigcup\Cal V\in\tilde\Lambda^0$,}

\Centerline{$\bigcup\Cal V\setminus A(\Cal V)=\bigcup\Cal V'
\in\Sigma\tensorhat\Tau\subseteq\tilde\Lambda^0$,}

\noindent so 417A(iv) with $\Cal K=\Sigma\tensorhat\Tau$ and
$\Cal G=\{\bigcup\Cal V:\Cal V\in\frak V\}$
assures us that $\lambda'$ is inner regular with
respect to $\tilde\Lambda^0$, so that $\tilde\lambda$ also is (412Ha).\ \Qed

\medskip

\quad{\bf (v)} Thus $\tilde\lambda$ satisfies the conditions listed.   To see
that it is unique, suppose that $\tilde\lambda'$ is another measure on
$X\times Y$ with the same properties.   Then

\Centerline{$\tilde\lambda(E\times F)=\mu E\cdot\nu F
=\tilde\lambda'(E\times F)$}

\noindent whenever $E\in\Sigma^f$ and $F\in\Tau^f$, so
$\tilde\lambda((E\times F)\cap U)
=\tilde\lambda'((E\times F)\cap U)$ whenever $E\in\Sigma^f$, $F\in\Tau^f$ and
$U\in\Cal U_s$.   Consequently $\tilde\lambda((E\times F)\cap W)
=\tilde\lambda'((E\times F)\cap W)$ whenever $E\in\Sigma^f$, $F\in\Tau^f$ and
$W\subseteq X\times Y$ is open.   \Prf\ Set $X_0=\bigcup\frak T^f$,
$Y_0=\bigcup\frak S^f$;  because $\mu$ is an effectively locally finite
topological measure, $X_0\in\Sigma$ is $\mu$-conegligible;  similarly,
$Y_0\in\Tau$ is $\nu$-conegligible.   Consequently $X_0\times Y_0$ is both
$\tilde\lambda$-conegligible and $\tilde\lambda'$-conegligible.   So, setting
$\Cal V=\{U:U\in\Cal U_s$, $U\subseteq W\}$,

$$\eqalignno{\tilde\lambda((E\times F)\cap W)
&=\tilde\lambda((E\times F)\cap(W\cap(X_0\times Y_0)))\cr&
=\tilde\lambda((E\times F)\cap\bigcup\Cal V)
=\sup_{V\in\Cal V}\tilde\lambda((E\times F)\cap V)
\Displaycause{by 414Ea, because $\tilde\lambda$ is an effectively locally finite
$\tau$-additive topological measure}
=\sup_{V\in\Cal V}\tilde\lambda'((E\times F)\cap V)
=\tilde\lambda'((E\times F)\cap W).  \text{ \Qed}\cr}$$

If $E_0\in\Sigma^f$ and $F_0\in\Tau^f$, then the subspace measures
$\tilde\lambda_{E_0\times F_0}$ and $\tilde\lambda'_{E_0\times F_0}$ agree on

\Centerline{$\Cal I=\{((E\cap E_0)\times(F\cap F_0))\cap W:E\in\Sigma$,
$F\in\Tau$, $W\subseteq X\times Y$ is open$\}$}

\noindent which is closed under $\cap$, so by the Monotone Class Theorem (136C)
they agree on the $\sigma$-algebra of subsets of $E_0\times F_0$ generated by
$\Cal I$, which is $\tilde\Lambda^0\cap\Cal P(E_0\times F_0)$.

Next, if $Q\in\tilde\Lambda^0$,

$$\eqalignno{\tilde\lambda'Q
&=\tilde\lambda'(Q\cap(X_0\times Y_0))
=\sup_{U\in\Cal U}\tilde\lambda'(Q\cap U)
\Displaycause{414Ea again}
=\sup_{E\in\Sigma^f,F\in\Tau^f}\tilde\lambda'(Q\cap(E\times F))
=\sup_{E\in\Sigma^f,F\in\Tau^f}\tilde\lambda(Q\cap(E\times F))
=\tilde\lambda Q.\cr}$$

\noindent So $\tilde\lambda$ and $\tilde\lambda'$ agree on $\tilde\Lambda^0$.
By 412Mb they are equal.

\medskip

{\bf (b)(i)}\grheada\ In the construction of $\tilde\lambda$ described in (a-iv) above,
we have $\lambda=\lambda'\restr\Lambda=\tilde\lambda\restr\Lambda$.   So
$\tilde\lambda$ extends $\lambda$.

\medskip

\qquad\grheadb\ Note that $\lambda$ is (strictly) localizable.
\Prf\ Let $\tilde\mu$,
$\tilde\nu$ be the c.l.d.\ versions of $\mu$ and $\nu$.   These are
$\tau$-additive topological measures (because $\mu$ and $\nu$ are),
complete and locally determined (by construction), and are still
effectively locally finite (cf.\ 412Ha), so are strictly localizable
(414J again).
Now $\lambda$ is the c.l.d.\ product of $\tilde\mu$ and $\tilde\nu$
(251T), therefore strictly localizable (251O).\ \Qed

By 417A, $\lambda'$ is localizable.   By 213Hb, there is a $Q'\in\Lambda'$ such
that $\tilde\lambda(\tilde Q\symmdiff Q')=0$;  by 417A(iii), there is a
$Q\in\Lambda$ such that $\lambda'(Q'\symmdiff Q)=0$;  so that
$0=\tilde\lambda(Q'\symmdiff Q)=\tilde\lambda(\tilde Q\symmdiff Q)$.

\medskip

\quad{\bf (ii)} For an open set $W\subseteq X\times Y$,

$$\eqalignno{W\cap\supp\tilde\lambda=\emptyset&
\iff\tilde\lambda W=0\cr&
\iff\tilde\lambda(G\times H)=0\text{ whenever }G\in\frak T,\,H\in\frak S
   \text{ and }G\times H\subseteq W
\Displaycause{because $\tilde\lambda$ is $\tau$-additive}
\iff\mu G\cdot\nu H=0\text{ whenever }G\in\frak T,\,H\in\frak S
   \text{ and }G\times H\subseteq W\cr&
\iff(G\times H)\cap(\supp\mu\times\supp\nu)=\emptyset\cr&
\mskip100mu
   \text{ whenever }G\in\frak T,\,H\in\frak S
   \text{ and }G\times H\subseteq W\cr&
\iff W\cap(\supp\mu\times\supp\nu)=\emptyset,\cr}$$

\noindent so $\supp\tilde\lambda=\supp\mu\times\supp\nu$.

\medskip

\quad{\bf (iii)} This is dealt with in (a-iv-$\beta$).

\medskip

\quad{\bf (iv)} By 412R, $\lambda$ is inner regular with respect to
$\Sigma'\tensorhat\Tau'$.
Applying 417A(iv) with
$\Cal G=\{\sup\Cal V:\Cal V\in\frak V\}$ as defined in (a-i), we see that
$\lambda'$ is inner regular with respect to
$(\Sigma'\tensorhat\Tau')\vee\Cal B(X\times Y)$;  by 412Ha again, so is its
c.l.d.\ version $\tilde\lambda$.

\medskip

\quad{\bf (v)}\grheada\ Set
$\Cal V=\{V:V\in\Cal U_s,\,V\subseteq W\}$.   Then
$\bigcup\Cal V=W\cap(X_0\times Y_0)$, where $X_0=\bigcup\frak T^f$ and
$Y_0=\bigcup\frak S^f$, as in (a-v) above.   Because $X_0\times Y_0$ is
$\lambda$-conegligible, therefore $\tilde\lambda$-conegligible,

\Centerline{$\tilde\lambda W
=\tilde\lambda(W\cap(X_0\times Y_0))
=\sup_{V\in\Cal V}\tilde\lambda V
=\sup_{V\in\Cal V}\lambda V
\le\tilde\lambda W$.}

\noindent Next, if we take a countable upwards-directed
$\Cal V'\subseteq\Cal V$ such that

\Centerline{$\sup_{V\in\Cal V'}\lambda V=\sup_{V\in\Cal V}\lambda V
=\tilde\lambda W$,}

\noindent and set $W'=\bigcup\Cal V'$, then $W'$ is an open set belonging to
$\Lambda$ and included in $W$, and

\Centerline{$\lambda W'=\tilde\lambda W'=\sup_{V\in\Cal V'}\tilde\lambda V
=\sup_{V\in\Cal V'}\lambda V=\tilde\lambda W$.}

\noindent And because $\tilde\lambda$ extends $\lambda$,

\Centerline{$\tilde\lambda_*W\ge\lambda_*W
\ge\sup_{V\in\Cal V}\lambda V=\tilde\lambda W=\tilde\lambda_*W$}

\noindent so $\tilde\lambda W=\lambda_*W$.

\medskip

\qquad\grheadb\ Because $\mu$ and $\nu$ are semi-finite (411Gd) and $E$ and
$F$ are measurable, the subspace measures $\mu_E$ and $\nu_F$ are semi-finite
(214Ka) therefore effectively locally finite and $\tau$-additive (414K).
For open $V\subseteq X\times Y$,
set $f_V(x)=\nu(V[\{x\}]\cap F)=\nu_F(V[\{x\}]\cap F)$ for $x\in E$;
then $f_V$ is lower semi-continuous (417Ba).   Now
$\int_Ef_Vd\mu=\int_Ef_Vd\mu_E=\lambda(V\cap(E\times F))$ for every
$V\in\Cal V$, just because $V\cap(E\times F)$ is a union of measurable
rectangles and $\mu$ and $\nu$ are semi-finite (251J);  and
setting $g(x)=\nu(W[\{x\}]\cap F)$, we have $g(x)=\sup_{V\in\Cal V}f_V(x)$ for
every $x\in X_0$, by 414Ea.    Once more, 414Ba tells us that

$$\eqalign{\int_E\nu(W[\{x\}]\cap F)\mu(dx)&
=\int_Eg\,d\mu_E
=\sup_{V\in\Cal V}\int_Ef_Vd\mu_E\cr&
=\sup_{V\in\Cal V}\lambda(V\cap(E\times F))
=\tilde\lambda(W\cap(E\times F)).\cr}$$

\noindent Similarly,

\Centerline{$\int_F\mu(W^{-1}[\{y\}]\cap E)\nu(dy)
=\tilde\lambda(W\times(E\times F))$.}

\noindent (The point here is that while some of the arguments of
this proof give different roles to $\mu$ and $\nu$, the asserted properties
of the extension in part (a), and the following deductions, are
symmetric between the two factors.)

\medskip

\quad{\bf (vi)} In this case, $\mu$ is inner regular with respect to
$\Cal B(X)$ and $\nu$ is inner regular with respect to $\Cal B(Y)$, so
(iv) tells us that $\tilde\lambda$ is inner regular with respect to
$(\Cal B(X)\tensorhat\Cal B(Y))\vee\Cal B(X\times Y)=\Cal B(X\times Y)$.

\medskip

\quad{\bf (vii)} If $\mu$ and $\nu$ are
inner regular with respect to the closed
sets, $\lambda$ also is, by 412Sa.   This time, we can apply
417A(iv) with $\Cal K$ the family of closed subsets of $X\times Y$ and
$\Cal G=\{\sup\Cal V:\Cal V\in\frak V\}$ to see that $\lambda'$ and
$\tilde\lambda$ are inner regular with respect to $\Cal K$.

\medskip

\quad{\bf (viii)} Repeat the argument of (vii)
with $\Cal K$ the family of closed compact subsets of $X\times Y$, using 412Sb.

\medskip

\quad{\bf (ix)} $\{G\times H:G\in\frak T^f$, $H\in\frak S^f\}$ is a cover of
$X\times Y$ by open sets of finite measure for $\tilde\lambda$.
}%end of proof of 417C

\leader{417D}{Multiple \dvrocolon{products}}\cmmnt{ Just as with the
c.l.d.\ product measure (see 251W), we can apply the construction of
417C repeatedly to obtain measures on the products of finite families of
$\tau$-additive measure spaces.

\medskip

\noindent}{\bf Proposition}\dvArevised{2021} (a) Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a non-empty finite family of
effectively locally finite $\tau$-additive topological measure spaces.
Then there is a unique complete locally determined effectively locally
finite $\tau$-additive topological measure $\tilde\lambda$ on
$X=\prod_{i\in I}X_i$ which is
inner regular with respect to the $\sigma$-algebra
$(\Tensorhat_{i\in I}\Sigma_i)\vee\Cal B(X)$ generated by
$\{\prod_{i\in I}E_i:E_i\in\Sigma_i$ for every $i\in I\}
\cup\{W:W\subseteq X$ is open$\}$ and is such that
$\tilde\lambda(\prod_{i\in I}E_i)$ is defined and equal to
$\prod_{i\in I}\mu_iE_i$ whenever $E_i\in\Sigma_i$ for every $i\in I$.

(b) If now $\family{k}{K}{I_k}$ is a partition of $I$ into non-empty sets, and
$\tilde\lambda_k$ is the product measure defined by the construction of
(a) on $Z_k=\prod_{i\in I_k}X_i$ for each $k\in K$, then the natural
bijection between $X$ and $\prod_{k\in K}Z_k$ identifies $\tilde\lambda$
with the product of the $\tilde\lambda_k$ defined by the construction of
(a).

\proof{{\bf (a)} Of course the idea is to induce on $\#(I)$, but there are some
wrinkles to take care of.

\medskip

\quad{\bf (i)} Suppose that $I=\{j\}$ is a singleton.   Then $\tilde\lambda$
must be the c.l.d.\ version of $\mu_j$;  this is surely a topological measure,
and it is effectively locally finite by 412Ha, taking $\Cal K$ there to be the
family of subsets of open sets of finite measure for $\mu_j$.

\medskip

\quad{\bf (ii)} If $\#(I)>1$, take $j\in I$ and set $J=I\setminus\{j\}$.   By
the inductive hypothesis, we have a complete locally determined effectively
locally finite $\tau$-additive topological measure $\tilde\lambda_J$ on
$Z=\prod_{i\in J}X_i$ which is inner regular with respect to
$(\Tensorhat_{i\in J}\Sigma_i)\vee\Cal B(Z)$ and is such that
$\tilde\lambda_J(\prod_{i\in I}E_i)=\prod_{i\in J}\mu_iE_i$ whenever
$E_i\in\Sigma_i$ for every $i\in J$.   Write $\tilde\Lambda_J$ for the domain
of $\tilde\lambda_J$.   By 417C, there is a
complete locally determined effectively
locally finite $\tau$-additive topological measure $\tilde\lambda$ on
$Z\times X_j$ which is
inner regular with respect to the $\sigma$-algebra
$\tilde\Lambda^{(0)}=(\tilde\Lambda_J\tensorhat\Sigma_j)\vee\Cal B(X)$
and is such that
$\tilde\lambda(W\times F)=\tilde\lambda_JW\times\mu_jF$ whenever
$W\in\tilde\Lambda_J$ and $F\in\Sigma_j$.   Now 417C(b-iv) tells us that
$\tilde\lambda$ is inner regular with respect to the $\sigma$-algebra

\Centerline{$(((\Tensorhat_{i\in J}\Sigma_i)\vee\Cal B(Z))
   \tensorhat\Sigma_j)\vee\Cal B(X)$}

\noindent generated by

$$\eqalign{\{E\times X_j:&E\in\Tensorhat_{i\in J}\Sigma_i\}
    \cup\{V\times X_j:V\in\Cal B(Z)\}\cup\{Z\times F:F\in\Sigma_j\}
    \cup\Cal B(X)\cr&
\subseteq(\Tensorhat_{i\in I}\Sigma_i)\cup\Cal B(X),\cr}$$

\noindent so $\tilde\lambda$ is inner regular with respect to
$(\Tensorhat_{i\in I}\Sigma_i)\vee\Cal B(X)$.   And of course

\Centerline{$\tilde\lambda(\prod_{i\in I}E_i)
=\tilde\lambda_J(\prod_{i\in J}E_i)\cdot\mu_jE_j
=\prod_{i\in J}\mu_iE_i\cdot\mu_jE_j
=\prod_{i\in I}\mu_iE_i$}

\noindent whenever $E_i\in\Sigma_i$ for $i\in I$.

\medskip

\quad{\bf (iii)} As for the uniqueness of $\tilde\lambda$, we can use the same
argument as in (a-v) of the proof of 417C.   Suppose that $\tilde\lambda'$ is
another measure with the given properties.   Taking

\Centerline{$\Cal U=\{\prod_{i\in I}G_i:G_i\in\frak T_i^f$ for $i\in I\}$,}

\Centerline{$\Cal U_s=\{\bigcup\Cal V:\Cal V\subseteq\Cal U_s$ is finite$\}$,}

\noindent $\bigcup\Cal U=\prod_{i\in I}\bigcup\frak T_i^f$ is conegligible for
both $\tilde\lambda$ and $\tilde\lambda'$, while
$\tilde\lambda$ and $\tilde\lambda'$ agree on

\Centerline{$\{(\prod_{i\in I}E_i)\cap U:U\in\Cal U_s$,
$E_i\in\Sigma_i$ for $i\in I\}$}

\noindent and therefore on

\Centerline{$\{(\prod_{i\in I}E_i)\cap W:W\subseteq\bigcup\Cal U$ is open,
$E_i\in\Sigma_i$ for $i\in I\}$}

\noindent and on

\Centerline{$\{(\prod_{i\in I}E_i)\cap W:W\subseteq X$ is open,
$E_i\in\Sigma_i^f$ for $i\in I\}$.}

\noindent By the Monotone Class Theorem they agree on
$\tilde\Lambda^0\cap\Cal P(\prod_{i\in I}E_i)$ whenever
$E_i\in\Sigma_i^f$ for every $i$, and by 414Ea once more they agree on
$\tilde\Lambda^0$, so must be equal.

\medskip

{\bf (b)(i)} I begin with something to match
417C(b-iv):  if, for each $i\in I$, $\Sigma'_i\subseteq\Sigma_i$ is a
$\sigma$-algebra such that $\mu_i$ is inner regular with respect to $\Sigma'_i$,
then $\tilde\lambda$ will be inner regular with respect to
$(\Tensorhat_{i\in I}\Sigma'_i)\vee\Cal B(X)$.   \Prf\ Induce on $\#(I)$ as in
(i)-(ii) of the proof of (a) above.   If $I=\{j\}$ is a singleton,
$\tilde\lambda$ is the c.l.d.\ version of $\mu_j$ and is inner regular with
respect to $\Sigma_j'$ by 412Ha, as always.   For the inductive step, with
$J=I\setminus\{j\}$, the inductive hypothesis tells us that $\tilde\lambda_J$ is
inner regular with respect to $(\Tensorhat_{i\in J}\Sigma'_i)\vee\Cal B(Z)$;
by 417C(b-iv), $\tilde\lambda$, as constructed in (a),
is inner regular with respect to

\Centerline{$(((\Tensorhat_{i\in J}\Sigma'_i)\vee\Cal B(Z))
   \tensorhat\Sigma'_j)\vee\Cal B(X)
\subseteq(\Tensorhat_{i\in I}\Sigma'_i)\vee\Cal B(X)$.  \Qed}

\medskip

\quad{\bf (ii)} Now, given a partition $\family{k}{K}{I_k}$ into non-empty sets
and associated effectively locally finite $\tau$-additive
product measures $\tilde\lambda_k$ on
$Z_k=\prod_{i\in I_k}X_i$ for $k\in I$, let $\tilde\lambda'$ be the
corresponding effectively locally finite $\tau$-additive
product measure on $Z=\prod_{k\in K}Z_k$.   Then (i)
here tells us that $\tilde\lambda'$ is inner regular with respect to

\Centerline{$\Tensorhat_{k\in K}((\Tensorhat_{i\in I_k}\Sigma_i)\vee\Cal B(Z_i))
                   \vee\Cal B(Z)
=(\Tensorhat_{k\in K}\Tensorhat_{i\in I_k}\Sigma_i)\vee\Cal B(Z)$.}

\noindent Copying this into $X$, we get a c.l.d.\ measure
which is inner regular with respect to
$\tilde\Lambda^0=(\Tensorhat_{i\in I}\Sigma_i)\vee\Cal B(X)$, defined on the
whole of $\tilde\Lambda^0$ and agreeing with $\tilde\lambda$ on products of
measurable sets, so it must be $\tilde\lambda$.
}%end of proof of 417D

\vleader{60pt}{417E}{Theorem} Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of $\tau$-additive topological probability spaces, with
product probability space $(X,\Lambda,\lambda)$.

(a) There is a unique complete $\tau$-additive topological
probability measure $\tilde\lambda$ on $X$
which is inner regular with respect to
$\tilde\Lambda^0=(\Tensorhat_{i\in I}\Sigma_i)\vee\Cal B(X)$ and is such that
$\tilde\lambda\{x:x\in X$, $x(i)\in E_i$ for every $i\in J\}$ is defined and
equal to $\prod_{i\in J}\mu_iE_i$ whenever $J\subseteq I$ is finite and
$E_i\in\Sigma_i$ for every $i\in J$.

(b)(i) If $\tilde Q$ is measured by $\tilde\lambda$, there is a
$Q\in\Lambda$ such that $\tilde\lambda(\tilde Q\symmdiff Q)=0$.

\quad(ii) $\tilde\lambda W=\lambda_*W$ for every open set $W\subseteq X$,
and $\tilde\lambda F=\lambda^*F$ for every closed set $F\subseteq X$.

\quad(iii) The support of $\tilde\lambda$ is the product of the supports of the
$\mu_i$.

\quad(iv) If for each $i\in I$ we are given a $\sigma$-subalgebra
$\Sigma_i'\subseteq\Sigma_i$ such that $\mu_i$ is inner regular with respect to
$\Sigma_i'$, then $\tilde\lambda$ is inner
regular with respect to $(\Tensorhat_{i\in I}\Sigma_i')\vee\Cal B(X)$.

\quad(v) If every $\mu_i$ is inner regular with respect to the Borel sets,
so is $\tilde\lambda$.

\quad(vi) If every $\mu_i$ is inner regular with respect to the closed sets, so
is $\tilde\lambda$.

\proof{ The strategy of the proof is the same as in 417C, subject to
some obviously necessary modifications.   The key step, showing that
every union $\bigcup_{n\in\Bbb N}A(\Cal V_n)$ has zero inner measure, is
harder, but we escape a little work because we no longer have to worry
about sets of infinite measure.

\medskip

{\bf (a)(i)} I begin by setting up some machinery.   Let $\Cal C$ be
the family of subsets of $X$ expressible in the form
$\prod_{i\in I}E_i$, where $E_i\in\Sigma_i$ for every $i$ and
$\{i:E_i\ne X_i\}$ is finite.   Let $\Cal U\subseteq\Cal C$ be the
standard basis for
the topology $\frak T$ of $X$, consisting of sets expressible as
$\prod_{i\in I}G_i$ where $G_i\in\frak T_i$ for every $i\in I$ and
$\{i:G_i\ne X_i\}$ is finite.   Write $\Cal U_s$ for the set of finite
unions of members of $\Cal U$, and $\frak V$ for the set of non-empty
upwards-directed families in $\Cal U_s$.   Note that every member of
$\Cal U_s$ is determined by coordinates in some finite subset of $I$
(definition:  254M).

If $J\subseteq I$, write $\lambda_J$ for the product measure on
$\prod_{i\in J}X_i$; we
shall need $\lambda_{\emptyset}$, which is the unique probability
measure on the single-point set
$\{\emptyset\}=\prod_{i\in\emptyset}X_i$.   For
$J\subseteq I$, $v\in\prod_{i\in J}X_i$ and $W\subseteq X$ set

\Centerline{$f_W(v)=\lambda_{I\setminus J}\{w:(v,w)\in W\}$}

\noindent if this is defined, identifying $\prod_{i\in
J}X_i\times\prod_{i\in I\setminus J}X_i$ with $X$.

\medskip

\quad{\bf (ii)} We need two easy facts.

\medskip

\qquad\grheada\ $f_W(v)=\int f_W(v^{\smallfrown}\fraction{t})\mu_j(dt)$
whenever $W\in\Tensorhat_{i\in I}\Sigma_i$, $J\subseteq I$,
$v\in\prod_{i\in J}X_i$ and $j\in I\setminus J$, writing
$v^{\smallfrown}\fraction{t}$ for the member $v\cup\{(j,t)\}$
of $\prod_{i\in J\cup\{j\}}X_i$
extending $v$ and taking the value $t$ at the coordinate $j$.   \Prf\
Let $\Cal A$ be the family of sets $W$ satisfying the property.   Then
$\Cal A$ is a Dynkin class including $\Cal C$, so includes the
$\sigma$-algebra generated by $\Cal C$,
which is $\Tensorhat_{i\in I}\Sigma_i$.\ \Qed

\medskip

\qquad\grheadb\ If $J\subseteq I$, $v\in\prod_{i\in J}X_i$,
$j\in I\setminus J$ and $V\in\Cal U_s$, and we set
$g(t)=f_V(v^{\smallfrown}\fraction{t})$ for
$t\in X_j$, then $g$ is lower semi-continuous.   \Prf\
We can express $V$ as $\bigcup_{n\le m}\prod_{i\in I}G_{ni}$, where
$G_{ni}\subseteq X_i$ is open for every $n\le m$, $i\in I$.   Now if
$t\in X_j$, we shall have

\Centerline{$\{w:(v^{\smallfrown}\fraction{t},w)\in V\}
\subseteq\{w:(v^{\smallfrown}\fraction{t'},w)\in V\}$}

\noindent whenever

\Centerline{$t'\in H=X_j\cap\bigcap\{G_{nj}:n\le m,\,t\in G_{nj}\}$.}

\noindent So $g(t')\ge g(t)$ for every $t'\in H$, which is an open
neighbourhood of $t$.   As $t$ is arbitrary, $g$ is lower
semi-continuous.\ \Qed

\medskip

\quad{\bf (iii)} For each $\Cal V\in\frak V$, fix, for the remainder of this
proof, a countable $\Cal V'\subseteq\Cal V$ such that
$\sup_{V\in\Cal V'}\lambda V=\sup_{V\in\Cal V}\lambda V$;  because
$\Cal V$ is
upwards-directed, we may suppose that $\Cal V'=\{V_n:n\in\Bbb N\}$ for
some non-decreasing sequence $\sequencen{V_n}$ in $\Cal V$.   Set
$A(\Cal V)=\bigcup\Cal V\setminus\bigcup\Cal V'$.

\Quer\ Suppose, if possible, that there is a sequence
$\sequencen{\Cal V_n}$ in $\frak V$ such that
$\lambda_*(\bigcup_{n\in\Bbb N}A(\Cal V_n))>0$.

\medskip

\qquad\grheada\ We have
$\lambda^*(X\setminus\bigcup_{n\in\Bbb N}A(\Cal V_n))<1$;  let
$\sequencen{C_n}$ be a sequence in $\Cal C$ such that

\Centerline{$X\setminus\bigcup_{n\in\Bbb N}A(\Cal V_n)
\subseteq\bigcup_{n\in\Bbb N}C_n$,
\quad$\sum_{n=0}^{\infty}\lambda C_n=\gamma_0<1$}

\noindent (see 254A-254C).   For each $n\in\Bbb N$, express $\Cal V'_n$
as $\{V_{nr}:r\in\Bbb N\}$ where $\sequence{r}{V_{nr}}$ is
non-decreasing,
and set $W_n=\bigcup\Cal V'_n=\bigcup_{r\in\Bbb N}V_{nr}$.   Let
$J\subseteq I$ be a
countable set such that every $C_n$ and every $V_{nr}$ is determined by
coordinates in $J$.   Express $J$ as $\bigcup_{k\in\Bbb N}J_k$ where
$J_0=\emptyset$ and, for each $k$, $J_{k+1}$ is equal either to $J_k$ or
to $J_k$ with one point added.   (As in the proof of 254Fa, I am using a
formulation which will apply equally to finite and infinite $I$, though
of course the case of finite $I$ is elementary once we have 417C.)

\medskip

\qquad\grheadb\ For each $n\in\Bbb N$, set

\Centerline{$W'_n
=\bigcup_{k\in\Bbb N}\{x:x\in X,\,f_{W_n}(x\restr J_k)=1\}$.}

\noindent Then $\lambda(W'_n\setminus W_n)=0$.   \Prf\ For any
$k\in\Bbb N$, if we think of $\lambda$ as the product of $\lambda_{J_k}$
and $\lambda_{I\setminus J_k}$ and of $f_{W_n}$ as a measurable function
on $\prod_{i\in J_k}X_i$, we see that $\{x:f_{W_n}(x\restr J_k)=1\}$ is
of the form $F_k\times\prod_{i\in I\setminus J_k}X_i$, where
$F_k\subseteq\prod_{i\in J_k}X_i$ is measurable;   and

\Centerline{$\lambda((F_k\times\prod_{i\in I\setminus J_k}X_i)
  \setminus W_n)
=\int_{F_k}(1-f_{W_n}(v))\lambda_{J_k}(dv)=0$.}

\noindent Summing over $k$, we see that $W'_n\setminus W_n$ is
negligible.\ \Qed

Observe that every $W'_n$, like $W_n$, is determined by coordinates in
$J$.   So $\bigcup_{n\in\Bbb N}W'_n\setminus W_n$ is of the form
$E\times\prod_{i\in I\setminus J}X_i$ where $\lambda_JE=0$ (254Ob).
There is therefore a sequence $\sequencen{D_n}$ of measurable cylinders
in $\prod_{i\in J}X_i$ such that $E\subseteq\bigcup_{n\in\Bbb N}D_n$ and
$\sum_{n=0}^{\infty}\lambda_JD_n<1-\gamma_0$.   Set
$C'_n=\{x:x\in X,\,x\restr J\in D_n\}\in\Cal C$ for each $n$.  Then
$\bigcup_{n\in\Bbb N}W'_n\setminus W_n
\subseteq\bigcup_{n\in\Bbb N}C'_n$, so

\Centerline{$(X\setminus\bigcup_{n\in\Bbb N}A(\Cal V_n))
  \cup\bigcup_{n\in\Bbb N}W'_n\setminus W_n
\subseteq\bigcup_{n\in\Bbb N}C_n\cup\bigcup_{n\in\Bbb N}C'_n$,}

\Centerline{$\gamma
=\sum_{n=0}^{\infty}\lambda C_n+\sum_{n=0}^{\infty}\lambda C'_n<1$,}

\noindent while each $C_n$ and each $C'_n$ is determined by coordinates
in a finite subset of $J$.

\medskip

\qquad\grheadc\ For $k\in\Bbb N$, let $P_k$ be the set of those
$v\in\prod_{i\in J_k}X_i$ such that

\Centerline{$\sum_{n=0}^{\infty}f_{C_n}(v)+f_{C'_n}(v)\le\gamma$,
\quad$f_V(v)\le f_{W_n}(v)$ whenever $n\in\Bbb N$ and $V\in\Cal V_n$.}

\noindent Our hypothesis is that

\Centerline{$\sum_{n=0}^{\infty}f_{C_n}(\emptyset)+f_{C'_n}(\emptyset)
=\sum_{n=0}^{\infty}\lambda C_n+\lambda C'_n\le\gamma$,}

\noindent and the $\Cal V'_n$ were chosen such that

\Centerline{$f_V(\emptyset)=\lambda V\le\lambda W_n=f_{W_n}(\emptyset)$}

\noindent for every $n\in\Bbb N$, $V\in\Cal V_n$;  that is,
$\emptyset\in P_0$.

\medskip

\qquad\grheadd\ Now if $k\in\Bbb N$ and $v\in P_k$ there is a
$v'\in P_{k+1}$
extending $v$.   \Prf\ If $J_{k+1}=J_k$ we can take $v'=v$.   Otherwise,
$J_{k+1}=J_k\cup\{j\}$ for some $j\in I\setminus J_k$.   Now

$$\eqalignno{\gamma
&\ge\sum_{n=0}^{\infty}f_{C_n}(v)+f_{C'_n}(v)
=\sum_{n=0}^{\infty}
\int f_{C_n}(v^{\smallfrown}\fraction{t})
  +f_{C'_n}(v^{\smallfrown}\fraction{t})\,\mu_j(dt)\cr
\displaycause{($\alpha$) above}
&=\int\sum_{n=0}^{\infty}f_{C_n}(v^{\smallfrown}\fraction{t})
  +f_{C'_n}(v^{\smallfrown}\fraction{t})\,\mu_j(dt),\cr}$$

\noindent so

\Centerline{$H
=\{t:t\in X_j,\,\sum_{n=0}^{\infty}f_{C_n}(v^{\smallfrown}\fraction{t})
  +f_{C'_n}(v^{\smallfrown}\fraction{t})\mu_j(dt)\le\gamma\}$}

\noindent has positive measure.

Next, for $V\in\Cal U_s$, set $g_V(t)=f_V(v^{\smallfrown}\fraction{t})$ for each
$t\in X_j$.   Then $g_V$ is lower semi-continuous, by
($\beta$) above.   For each $n\in\Bbb N$, $\{g_V:V\in\Cal V_n\}$ is an
upwards-directed family of lower semi-continuous functions, so its
supremum $g^*_n$ is lower semi-continuous, and because $\mu_j$ is
$\tau$-additive,

\Centerline{$\int g^*_nd\mu_j=\sup_{V\in\Cal V_n}\int g_Vd\mu_j
=\sup_{V\in\Cal V_n}f_V(v)\le f_{W_n}(v)
=\int f_{W_n}(v^{\smallfrown}\fraction{t})\mu_j(dt)$}

\noindent (using 414B and ($\alpha$) again).   But also, because
$\sequence{r}{V_{nr}}$ is non-decreasing and has union $W_n$,

\Centerline{$f_{W_n}(v^{\smallfrown}\fraction{t})
=\sup_{r\in\Bbb N}f_{V_{nr}}(v^{\smallfrown}\fraction{t})
\le g^*_n(t)$}

\noindent for every $t\in X_j$.   So we must have

\Centerline{$f_{W_n}(v^{\smallfrown}\fraction{t})=g^*_n(t)$ a.e.($t$).}

\noindent And this is true for every $n\in\Bbb N$.

There is therefore a $t\in H$ such that

\Centerline{$f_{W_n}(v^{\smallfrown}\fraction{t})
=g^*_n(v^{\smallfrown}\fraction{t})$ for
every $n\in\Bbb N$.}

\noindent Fix on such a $t$ and set
$v'=v^{\smallfrown}\fraction{t}\in\prod_{i\in J_{k+1}}X_i$;
then $v'\in P_{k+1}$, as required.\ \Qed

\medskip

\qquad\grheade\ We can therefore choose a sequence $\sequence{k}{v_k}$
such that $v_k\in P_k$ and $v_{k+1}$ extends $v_k$ for each $k$.
Choose $x\in X$ such that $x(i)=v_k(i)$ whenever $k\in\Bbb N$ and
$i\in J_k$, and $x(i)$ belongs to the support of $\mu_i$ whenever
$i\in I\setminus J$.    (Once again, 411Nd tells us that
every $\mu_i$ has a support.)

We need to know that if $k$, $n\in\Bbb N$ and $V\in\Cal V_n$ then
$f_{V\setminus W_n}(v_k)=0$.   \Prf\ For any $r\in\Bbb N$ there is a
$V'\in\Cal V_n$ such that $V\cup V_{nr}\subseteq V'$, so

\Centerline{$f_{V\cup V_{nr}}(v_k)\le f_{V'}(v_k)\le f_{W_n}(v_k)$,}

\noindent and

\Centerline{$f_{V\setminus W_n}(v_k)
\le f_{V\setminus V_{nr}}(v_k)
=f_{V\cup V_{nr}}(v_k)-f_{V_{nr}}(v_k)
\le f_{W_n}(v_k)-f_{V_{nr}}(v_k)
\to 0$}

\noindent as $r\to\infty$.\ \Qed

\medskip

\qquad\grheadz\ If $n\in\Bbb N$, then $x\notin C_n\cup C'_n$.   \Prf\
$C_n$ and $C'_n$ are determined by coordinates in a finite subset of
$J$, so must be
determined by coordinates in $J_k$ for some $k\in\Bbb N$.   Now
$f_{C_n}(v_k)+f_{C'_n}(v_k)\le\gamma<1$, so $\{y:y\restr J_k=v_k\}$
cannot be included in $C_n\cup C'_n$, and must be disjoint from it;
accordingly $x\notin C_n\cup C'_n$.\ \Qed

\medskip

\qquad{\bf ($\pmb{\eta}$)} Because

\Centerline{$(X\setminus\bigcup_{n\in\Bbb N}A(\Cal V_n))
  \cup\bigcup_{n\in\Bbb N}W'_n\setminus W_n
\subseteq\bigcup_{n\in\Bbb N}C_n\cup\bigcup_{n\in\Bbb N}C'_n$,}

\noindent there is some $n\in\Bbb N$ such that

\Centerline{$x\in A(\Cal V_n)\setminus(W'_n\setminus W_n)
\subseteq(\bigcup\Cal V_n)\setminus W'_n$,}

\noindent that is, there is some $V\in\Cal V_n$ such that
$x\in V\setminus W'_n$.   Let $U\in\Cal U$ be such that
$x\in U\subseteq V$.   Express
$U$ as $U'\cap U''$ where $U'\in\Cal U$ is determined by coordinates in
a finite subset of $J$ and $U''\in\Cal U$ is determined by coordinates
in a finite subset of $I\setminus J$.   Let $k\in\Bbb N$ be such that
$U'$ is determined by coordinates in $J_k$.   Then

\Centerline{$f_{U\setminus W_n}(v_k)
\le f_{V\setminus W_n}(v_k)=0$}

\noindent by ($\epsilon$) above.   Now

\Centerline{$\{w:w\in\prod_{i\in I\setminus J_k}X_i,\,
  (v_k,w)\in U\setminus W_n\}
=\{w:(v_k,w)\in U''\setminus W_n\}$}

\noindent (because $(v_k,w)=(x\restr J_k,w)\in U'$ for every $w$), while

\Centerline{$\{w:(v_k,w)\in U''\}$,
\quad$\{w:(v_k,w)\in W_n\}$}

\noindent are stochastically independent because the former is determined by
coordinates in $I\setminus J$, while the latter is determined by coordinates in $J\setminus J_k$.   So we must have

$$\eqalign{0
&=f_{U\setminus W_n}(v_k)
=\lambda_{I\setminus J_k}\{w:(v_k,w)\in U\setminus W_n\}\cr
&=\lambda_{I\setminus J_k}\{w:(v_k,w)\in U''\setminus W_n\}\cr
&=\lambda_{I\setminus J_k}\{w:(v_k,w)\in U''\}
(1-\lambda_{I\setminus J_k}\{w:(v_k,w)\in W_n\}).\cr}$$

At this point, recall that $x(i)$ belongs to the support of $\mu_i$ for
every $i\in I\setminus J$, while $x\in U''$.   So if $U''=\{y:y(i)\in
H_i$
for $i\in K\}$, where $K\subseteq I\setminus J$ is finite and
$H_i\subseteq
X_i$ is open for every $i$, we must have $\mu_iH_i>0$ for every $i$, and

\Centerline{$\lambda_{I\setminus J_k}\{w:(v_k,w)\in U''\}
=\prod_{i\in K}\mu_iH_i>0$.}

\noindent On the other hand, we are also supposing that $x\notin W'_n$,
so

\Centerline{$\lambda_{I\setminus J_k}\{w:(v_k,w)\in W_n\}
=f_{W_n}(v_k)=f_{W_n}(x\restr J_k)<1$.}

\noindent But this means that we have expressed $0$ as the product of
two non-zero numbers, which is absurd.\ \Bang

\medskip

\quad{\bf (iv)} Thus $\lambda_*(\bigcup_{n\in\Bbb N}A(\Cal V_n))=0$ for every
sequence $\sequencen{\Cal V_n}$ in $\frak V$.   Accordingly there is an
extension of $\lambda$ to a measure $\tilde\lambda$ on $X$ as in 417A.
By 417A(ii), $\tilde\lambda$ is complete.

Now $\tilde\lambda$ is a topological measure.   \Prf\ If $W\subseteq X$
is open, then $\Cal V=\{V:V\in\Cal U_s,\,V\subseteq W\}$ belongs to
$\frak V$, and $\bigcup\Cal V=W$.   Since $\bigcup\Cal V'\in\Lambda$
(because $\Cal V'$ is countable),

\Centerline{$W=\bigcup\Cal V'\cup A(\Cal V)$}

\noindent is measured by $\tilde\lambda$.\ \Qed

Also, $\tilde\lambda$ is $\tau$-additive.   \Prf\ Let $\Cal W$ be a
non-empty upwards-directed family of open subsets of $X$ with union
$W^*$.   Set

\Centerline{$\Cal V
=\{V:V\in\Cal U_s,\,\,\exists\,W\in\Cal W,\,V\subseteq W\}$.}

\noindent Then $\Cal V\in\frak V$ and $\bigcup\Cal V=W^*$, so
$\tilde\lambda A(\Cal V)=0$ and

\Centerline{$\tilde\lambda W^*=\tilde\lambda(\bigcup\Cal V')
=\sup_{V\in\Cal V'}\tilde\lambda V
\le\sup_{W\in\Cal W}\tilde\lambda W\le\tilde\lambda W^*$}

\noindent (using the fact that $\Cal V'$ is upwards-directed).   As
$\Cal W$ is arbitrary, $\tilde\lambda$ is $\tau$-additive.\ \Qed

\medskip

\quad{\bf (v)} As for the uniqueness of $\tilde\lambda$, if $\tilde\lambda'$ is
another measure with the same properties, then
$\tilde\lambda'(U\cap C)=\tilde\lambda(U\cap C)$
whenever $U\in\Cal U_s$ and $C\in\Cal C$.   Since $\tilde\lambda$ and
$\tilde\lambda'$ are both $\tau$-additive, they agree on sets of the form
$W\cap C$ where $W\subseteq X$ is open and $C\in\Cal C$;
by the Monotone Class Theorem, they
agree on $\tilde\Lambda^0$ and are therefore both the completion of their common
restriction to $\tilde\Lambda^0$.

\medskip

{\bf (b)(i)} Immediate from 417A(iii).

\medskip

\quad{\bf (ii)} Let $W\subseteq X$ be an open set.   Set
$\Cal V=\{V:V\in\Cal U_s,\,V\subseteq W\}$.   Then

\Centerline{$\tilde\lambda W=\sup_{V\in\Cal V}\tilde\lambda V
=\sup_{V\in\Cal V}\lambda V\le\lambda_*W\le\tilde\lambda W$}

\noindent just because $\tilde\lambda$ is a $\tau$-additive extension of
$\lambda$.   Now if $F\subseteq X$ is closed,

\Centerline{$\tilde\lambda F
=1-\tilde\lambda(X\setminus F)
=1-\lambda_*(X\setminus F)
=\lambda^*F$.}

\medskip

\quad{\bf (iii)} For each $i\in I$ write $Z_i$ for the support of
$\mu_i$, and set $Z=\prod_{i\in I}Z_i$.   This is closed because every
$Z_i$ is.   Its complement is covered by the negligible open sets
$\{x:x\in X,\,x(i)\in X_i\setminus Z_i\}$ as $i$ runs over $I$;  as
$\tilde\lambda$ is $\tau$-additive, the union of the negligible open sets
is negligible, and $Z$ is conegligible.
If $W\subseteq X$ is open and $x\in Z\cap W$, let $U\in\Cal U$ be such
that $x\in U\subseteq W$.
Express $U$ as $\prod_{i\in I}G_i$ where $G_i\in\frak T_i$ for every
$i\in I$ and $J=\{i:G_i\ne X_i\}$ is finite.   Then
$x(i)\in G_i\cap Z_i$, so $\mu_iG_i>0$, for every $i$.   Accordingly

\Centerline{$\tilde\lambda(W\cap Z)
=\tilde\lambda W\ge\lambda U
=\prod_{i\in J}\mu_iG_i>0$.}

\noindent Thus $Z$ is self-supporting and is the support of $\tilde\lambda$.

\medskip

\quad{\bf (iv)} (See (b-iv) of the proof of 417C).
By 412T, $\lambda$ is inner regular with respect to
$\Tensorhat_{i\in I}\Sigma'_i$.   Applying 417A(iv) with
$\Cal G=\{\sup\Cal V:\Cal V\in\frak V\}$, we see that $\tilde\lambda$
is inner regular with respect to
$(\Tensorhat_{i\in I}\Sigma'_i)\vee\Cal B(X)$.

\medskip

\quad{\bf (v), (vi)} As in the proof of 417Cb, apply 417A(iv) with
$\Cal G$ the family of open subsets of $X$ and $\Cal K$ either the Borel
$\sigma$-algebra of $X$ or the family of closed subsets of $X$,
this time using 412U to confirm that
$\lambda$ is inner regular with respect to $\Cal K$.
}%end of proof of 417E

\leader{417F}{Notation} In the context of 417C, 417D or 417E, I will call
$\tilde\lambda$ the {\bf $\tau$-additive product measure} on
$\prod_{i\in I}X_i$.

\cmmnt{\medskip

\noindent{\bf Remarks (a)}
Note that the uniqueness assertions in 417D and 417E mean that
for the products of finitely many probability spaces we do not need to
distinguish
between the two constructions.   The latter also shows that we can
relate 415E to the new method:  if every $\frak T_i$ is separable and
metrizable and every $\mu_i$ is strictly positive, then the `ordinary'
product measure $\lambda$ is a complete topological measure.   Since it
is also inner regular with respect to the Borel sets (412Uc), and
$\tau$-additive (because we now know that it has an extension to a
$\tau$-additive measure) it must be
exactly the $\tau$-additive product measure as described here.

\medskip

{\bf (b)} In 417D it seemed simpler to restrict the concept of `product' to
non-empty families;  in 417E, I omitted any reference to the possiblity that
the set $I$ might be empty.   This is because I regard a product
$X=\prod_{i\in I}X_i$ as a set of functions defined on $I$, and if
$I=\emptyset$ there is just one such function, itself the empty set;  so we have
$X=\{\emptyset\}$, with just one topology on $X$ and just one probability
measure defined on $X$, which will do very nicely for the required
$\tilde\lambda$.
}

\leader{417G}{Fubini's theorem for $\tau$-additive product measures}
\dvArevised{2021} Let
$(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be two complete locally determined
effectively
locally finite $\tau$-additive topological measure spaces.
Let $\tilde\lambda$ be the $\tau$-additive product measure on $X\times Y$,
and $\tilde\Lambda$ its domain.

(a) Let $f$ be a $[-\infty,\infty]$-valued function such
that $\int f\,d\tilde\lambda$ is defined in $[-\infty,\infty]$ and
$(X\times Y)\setminus\{(x,y):(x,y)\in\dom f,\,f(x,y)=0\}$
can be covered by a set of
the form $X\times\bigcup_{n\in\Bbb N}Y_n$ where $\nu Y_n<\infty$ for
every $n\in\Bbb N$.   Then the repeated integral
$\iint f(x,y)\nu(dy)\mu(dx)$ is defined and equal to
$\int fd\tilde\lambda$.

(b) Let $f:X\times Y\to[0,\infty]$ be lower semi-continuous.   Then

\Centerline{$\biggeriint f(x,y)\nu(dy)\mu(dx)
=\iint f(x,y)\mu(dx)\nu(dy)=\int fd\tilde\lambda$}

\noindent in $[0,\infty]$.

(c) Let $f$ be a $\tilde\Lambda$-measurable real-valued function defined
$\tilde\lambda$-a.e.\ on $X\times Y$.   If either
$\iint|f(x,y)|\nu(dy)\mu(dx)$ or $\iint|f(x,y)|\mu(dx)\nu(dy)$ is
defined and finite, then $f$ is $\tilde\lambda$-integrable.

\proof{ As in 417C, set
$\tilde\Lambda^0=(\Sigma\tensorhat\Tau)\vee\Cal B(X\times Y)$.

\medskip

{\bf (a)} I use 252B.

\medskip

\quad{\bf (i)} Write $\Cal W$ for the set of those $W\in\tilde\Lambda$
such that $\int\nu W[\{x\}]\mu(dx)$ is defined in $[0,\infty]$ and equal
to $\tilde\lambda W$.   Then open sets belong to $\Cal W$, by
417C(b-v-$\beta$).
Next, if $W\in\tilde\Lambda^0$ is included in an open set $W_0$ of finite
measure, $W\in\Cal W$.   \Prf\
If $W_0$ is an open set of finite measure,
then $\{W:W\subseteq X\times Y,\,W\cap W_0\in\Cal W\}$ is a Dynkin class, and by
417C(b-v-$\beta$), it contains $W\cap(E\times F)$ whenever $E\in\Sigma$,
$F\in\Tau$ and $W\subseteq X\times Y$ is open.   By the Monotone Class Theorem
it includes $\tilde\Lambda^0$.\ \Qed

Now suppose that $W\subseteq X\times Y$ is $\tilde\lambda$-negligible
and included in $X\times\bigcup_{n\in\Bbb N}Y_n$, where $\nu Y_n<\infty$
for every $n$.   Then $W\in\Cal W$.   \Prf\ Set
$A=\{x:x\in X,\,\nu^*W[\{x\}]>0\}$.   For each $n$, let $H_n\subseteq Y$
be an open set of finite measure such
that $\nu(Y_n\setminus H_n)\le 2^{-n}$;  we may arrange that
$H_{n+1}\supseteq H_n$ for each $n$.  Set $H=\bigcup_{n\in\Bbb N}H_n$,
so that $W[\{x\}]\setminus H\subseteq\bigcup_{n\in\Bbb N}Y_n\setminus H$
is negligible for every $x\in X$.

Fix an open set $G\subseteq X$ of finite measure and $n\in\Bbb N$ for
the moment.   Because
$\tilde\lambda$ is inner regular with respect to $\tilde\Lambda^0$, there
is a $V\in\tilde\Lambda^0$ such that $V\subseteq(G\times H_n)\setminus W$
and $\tilde\lambda V=\tilde\lambda((G\times H_n)\setminus W)$, that is,
$\tilde\lambda V'=0$,
where $V'=(G\times H_n)\setminus V\supseteq (G\times H_n)\cap W$.
We know that $V'\in\Cal W$, so

\Centerline{$\int\nu V'[\{x\}]dx=\tilde\lambda V'=0$,}

\noindent and $\nu V'[\{x\}]=0$ for almost every $x\in X$;  but this
means that $H_n\cap W[\{x\}]$ is negligible for almost every $x\in G$.

At this point, recall that $n$ was arbitrary, so
$H\cap W[\{x\}]$ and $W[\{x\}]$ are negligible for almost every
$x\in G$, that is, $A\cap G$ is negligible.   This is true for every
open set $G\subseteq X$ of finite measure.
Because $\mu$ is inner regular with respect to subsets of open sets of
finite measure, and is complete and locally determined, $A$ is
negligible (412Jb).   But this means that $\int\nu W[\{x\}]\mu(dx)$ is
defined and equal to zero, so that $W\in\Cal W$.\ \Qed

\medskip

\quad{\bf (ii)} Now suppose that $\int fd\tilde\lambda$ is defined in
$[-\infty,\infty]$ and that there is a sequence $\sequencen{Y_n}$ of
sets of finite measure in $Y$ such that $f(x,y)$ is defined and zero
whenever $x\in X$ and $y\in Y\setminus\bigcup_{n\in\Bbb N}Y_n$.   Set
$Z=\bigcup_{n\in\Bbb N}Y_n$.   Write $\lambda$ for the c.l.d.\ product
measure on $X\times Y$ and $\Lambda$ for its domain.  Then there is a
$\Lambda$-measurable function $g:X\times Y\to[-\infty,\infty]$ which is
equal $\tilde\lambda$-almost everywhere to $f$.  \Prf\ For $q\in\Bbb Q$
set $W_q=\{(x,y):(x,y)\in\dom f,\,f(x,y)\ge q\}\in\tilde\Lambda$, and
choose $V_q\in\Lambda$ such that $\tilde\lambda(W_q\symmdiff V_q)=0$
(417C(b-i));  set $g(x,y)=\sup\{q:q\in\Bbb Q,\,(x,y)\in V_q\}$ for
$x\in X$ and $y\in Y$, interpreting $\sup\emptyset$ as $-\infty$.\ \QeD\
Adjusting $g$ if necessary, we may suppose that it is zero on
$X\times(Y\setminus Z)$.   Set

\Centerline{$A=(X\times Y)\setminus\{(x,y):f(x,y)=g(x,y)\}$,}

\noindent so that $A$ is $\tilde\lambda$-negligible and included in
$X\times Z$.   By (i), $\nu A[\{x\}]=0$, that is, $y\mapsto f(x,y)$ and
$y\mapsto g(x,y)$ are equal $\nu$-a.e., for $\mu$-almost every $x$.
Write $\lambda_{X\times Z}$ for the subspace measure induced by
$\lambda$ on $X\times Z$;   note that this is the c.l.d.\ product of
$\mu$ with the subspace measure $\nu_Z$ on $Z$, by 251Q(ii-$\alpha$).

Now we have

$$\eqalignno{\int fd\tilde\lambda
&=\int g\,d\tilde\lambda
=\int g\,d\lambda\cr
\displaycause{by 235Gb, because the identity map from
$(X\times Y,\tilde\lambda)$ to $(X\times Y,\lambda)$ is \imp}
&=\int_{X\times Z}g\,d\lambda
=\int_{X\times Z}g\,d\lambda_{X\times Z}
=\iint_Z g(x,y)\nu_Z(dy)\mu(dx)\cr
\displaycause{by 252B, because $\nu_Z$ is $\sigma$-finite}
&=\iint g(x,y)\nu(dy)\mu(dx)\cr
\displaycause{because $g(x,y)=0$ if $y\in Y\setminus Z$}
&=\iint f(x,y)\nu(dy)\mu(dx).\cr}$$

\woddheader{417G}{4}{2}{2}{24pt}

{\bf (b)} If $f$ is non-negative and lower semi-continuous, set

\Centerline{$W_{ni}=\{(x,y):f(x,y)>2^{-n}i\}$}

\noindent for $n$, $i\in\Bbb N$, and

\Centerline{$f_n=2^{-n}\sum_{i=1}^{4^n}\chi W_{ni}$}

\noindent for $n\in\Bbb N$.   Applying 417C(b-v) we see that

\Centerline{$\int f_nd\tilde\lambda=\biggeriint f_n(x,y)dydx
=\iint f_n(x,y)dxdy$}

\noindent in $[0,\infty]$ for every $n$;  taking the limit as
$n\to\infty$,

\Centerline{$\int fd\tilde\lambda=\biggeriint f(x,y)dydx
=\iint f(x,y)dxdy$,}

\noindent because $\sequencen{f_n}$ is a non-decreasing sequence with
limit $f$.

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that $\gamma=\iint|f(x,y)|dydx$
is finite, but that $f$ is not integrable.   Because $\tilde\lambda$ is
semi-finite, there must be a non-negative $\tilde\lambda$-simple
function $g$ such that $g\leae|f|$ and
$\int g\,d\tilde\lambda>\gamma$ (213B).   For each $n\in\Bbb N$, there
are open sets $G_n\subseteq X$, $H_n\subseteq Y$ of finite measure such
that $\tilde\lambda(\{(x,y):g(x,y)\ge 2^{-n}\}\setminus(G_n\times H_n))
\le 2^{-n}$, by 417C(b-iii);  now $\sequencen{g\times\chi(G_n\times H_n)}\to g$
a.e., so there is some $n$ such that
$\int_{G_n\times H_n}g\,d\tilde\lambda>\gamma$.   In this case, setting
$g'(x,y)=\min(g(x,y),|f(x,y)|)$ for $(x,y)\in(G_n\times H_n)\cap\dom f$,
$0$ otherwise, we have $g=g'$ a.e.\ on $G_n\times H_n$, so that
$\int g'd\tilde\lambda>\gamma$.   But we can apply (a) to $g'$ to see
that

\Centerline{$\gamma<\int g'd\tilde\lambda
=\biggeriint g'(x,y)dydx\le\iint|f(x,y)|dydx\le\gamma$,}

\noindent which is absurd.\ \Bang

So if $\iint|f(x,y)|dydx$ is finite, $f$ must be
$\tilde\lambda$-integrable.   Of course the same arguments, reversing
the roles of $X$ and $Y$, show that $f$ is $\tilde\lambda$-integrable
if $\iint|f(x,y)|dxdy$ is defined and finite.
}%end of proof of 417G

\leader{417H}{Corollary}\dvAnew{2021}
Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be two complete locally determined effectively
locally finite $\tau$-additive topological measure spaces.
Let $\tilde\lambda$ be the $\tau$-additive product measure on $X\times Y$,
and $\tilde\Lambda$ its domain.   If $A\subseteq X$, $B\subseteq Y$ are
non-negligible sets such that $A\times B\in\tilde\Lambda$, then
$A\in\Sigma$ and $B\in\Tau$.

\proof{ (Cf.\ 252Xc.)   If $F\in\Tau$ and $\nu F<\infty$, $B\cap F\in\Tau$.
\Prf\ If $\nu F=0$ then $B\cap F\in\Tau$ because $\nu$ is complete.
Otherwise, set $f=\chi(A\times(B\cap F))$.   Then
$\int f\,d\tilde\lambda=\tilde\lambda((A\times B)\cap(X\times F))$ is defined
and $f$ is zero outside $X\times F$.
By 417Ga, $\iint f(x,y)\nu(dy)\mu(dx)$ is
defined, that is, $\int\chi A(x)\times\nu(B\cap F)\mu(dx)$ is defined and
$x\mapsto\chi A(x)\times\nu(B\cap F)$ is defined $\mu$-a.e.   As $A$ is not
negligible, $\nu(B\cap F)$ is defined.\ \QeD\   As $\nu$ is locally determined,
$B\in\Tau$.

As the specification of $\tilde\lambda$ in 417Ca is symmetric between the two
factors, we must also have $A\in\Sigma$.
}%end of proof of 417H

\leader{417I}{}\cmmnt{ The constructions here have most of the
properties one would hope for.   I give several in the exercises
(417Xd-417Xf, %417Xd, 417Xe, 417Xf,
417Xj).   One fact which is particularly
useful, and also has a trap in it, is the following.

\medskip
\wheader{417I}{0}{0}{0}{48pt}

\noindent}{\bf Proposition}\dvArevised{2021} Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be effectively locally finite $\tau$-additive
topological measure spaces,
and $\tilde\lambda$ the $\tau$-additive product measure
on $X\times Y$.   Suppose that $A\subseteq X$ and $B\subseteq Y$, and
write $\mu_A$, $\nu_B$ for the corresponding subspace measures;  assume
that both $\mu_A$ and $\nu_B$ are semi-finite.   Then these are also
effectively locally finite and $\tau$-additive,
and the subspace measure $\tilde\lambda_{A\times B}$ induced by
$\tilde\lambda$ on $A\times B$ is just the $\tau$-additive product
measure of $\mu_A$ and $\nu_B$.

\proof{{\bf (a)} To check that $\mu_A$ and $\nu_B$ are effectively
locally finite and $\tau$-additive, see 414K.
Next, because $\tilde\lambda$ is complete, locally determined, effectively
locally finite and $\tau$-additive (417Ca), it is strictly localizable
(414J), so $\tilde\lambda_{A\times B}$ is also effectively locally finite and
$\tau$-additive, by 414K again.

\medskip

{\bf (b)} Writing $\tilde\Lambda^0$ for
$(\Sigma\tensorhat\Tau)\vee\Cal B(X\times Y)$, as in 417C, we know that
$\tilde\lambda$ is inner regular with respect to $\tilde\Lambda^0$, so
$\tilde\lambda_{A\times B}$ is inner regular with respect to
$\{Q\cap(A\times B):Q\in\tilde\Lambda^0\}$, by 412Ob.   But this will be the
$\sigma$-algebra $\tilde\Lambda^0_{A\times B}$
of subsets of $A\times B$ generated by

$$\eqalign{\{(E\times F)\cap(A\times B):\mskip5mu&
E\in\Sigma,\,F\in\Tau\}\cup\{W\cap(A\times B):W\in\Cal B(X\times Y)\}\cr&
=\{E\times F:E\in\dom\mu_A,\,F\in\dom\nu_B\}\cup\Cal B(A\times B)\}.\cr}$$

\medskip

{\bf (c)} Now if $C\in\dom\mu_A$ and $D\in\dom\nu_B$, then
$\tilde\lambda^*(C\times D)=\mu_AC\cdot\nu_BD$.   \Prf\ ($\alpha$) There
are $E\in\Sigma$, $F\in\Tau$ such that $C\subseteq E$, $D\subseteq F$,
$\mu E=\mu^*C$ and $\nu F=\nu^*D$;  in which case

$$\eqalignno{\tilde\lambda^*(C\times D)
&\le\tilde\lambda(E\times F)
=\lambda(E\times F)
=\mu E\cdot\nu F\cr
\noalign{\noindent (251J)}
&=\mu^*C\cdot\nu^*D=\mu_AC\cdot\nu_BD.\cr}$$

\noindent ($\beta$) If $\gamma<\mu_AC\cdot\nu_BD$ then, because $\mu_A$
and $\nu_B$ are semi-finite, there are $C'\subseteq C$, $D'\subseteq D$
such that both have finite outer measure and
$\mu^*C'\cdot\nu^*D'\ge\gamma$.   In
this case, take $E'\in\Sigma$, $F'\in\Tau$ such that $C'\subseteq E'$,
$D'\subseteq F'$ and both $E'$ and $F'$ have finite measure.   Now if
$W\in\dom\tilde\lambda$ and $C\times D\subseteq W$, we have
$C'\times D'\subseteq W\cap(E\times F)$, so that
$\nu(W\cap(E\times F))[\{x\}]\ge\nu^*D'$ for every $x\in C'$, and

\Centerline{$\tilde\lambda W
\ge\int_E\nu(W\cap(E\times F))[\{x\}]\mu(dx)
\ge\mu^*C'\cdot\nu^*D'
\ge\gamma$,}

\noindent by 417Ga.   As $W$ is arbitrary,
$\tilde\lambda^*(C\times D)\ge\gamma$;  as $\gamma$ is arbitrary,
$\tilde\lambda^*(C\times D)\ge\mu_AC\cdot\nu_BD$.\ \Qed

\medskip

{\bf (d)} Thus $\tilde\lambda_{A\times B}$ agrees with the product of
$\mu_A$ and $\mu_B$ on measurable rectangles, as well as being inner regular
with respect to $\tilde\Lambda^0_{A\times B}$.   So the uniqueness assertion in
417Ca tells us that $\tilde\lambda_{A\times B}$ is the $\tau$-additive product
measure of $\mu_A$ and $\mu_B$.
}%end of proof of 417I

\leader{417J}{}\cmmnt{ In order to use 417G effectively in the theory
of infinite products, we need a generalized associative law corresponding to
254N and 417Db.

\medskip

\noindent}{\bf Theorem}\dvArevised{2021} Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
$\tau$-additive topological probability spaces
and $\Cal K$ a partition of $I$.   For $J\subseteq I$ let
$\tilde\lambda_J$ be the $\tau$-additive product
measure on $Z_J=\prod_{i\in J}X_i$, and write $Z$ for
$\prod_{K\in\Cal K}Z_K$.   Then we have a natural bijection
$\phi:Z\to Z_I$ defined by setting

\Centerline{$\phi(\family{K}{\Cal K}{z_K})=\bigcup_{K\in\Cal K}z_K$}

\noindent which identifies the $\tau$-additive product $\tilde\lambda$ of
the family $\family{K}{\Cal K}{\tilde\lambda_K}$ with $\tilde\lambda_I$.

In particular, if $K\subseteq I$ is any set, then $\tilde\lambda_I$ can be
identified with the $\tau$-additive product of the $\tau$-additive
product measures on $\prod_{i\in K}X_i$ and
$\prod_{i\in I\setminus K}X_i$.

\proof{{\bf (a)}
I am claiming that $\tilde\lambda_I$ is precisely the image measure
$\tilde\lambda\phi^{-1}$ where $\phi(\family{K}{\Cal K}{z_K})$ is the common
extension of the $z_K$ to a function on $I$.
Now we know that $\tilde\lambda$ is a complete
$\tau$-additive topological measure
and that $\phi$ is continuous (indeed, a homeomorphism), so
$\tilde\lambda\phi^{-1}$ is also a complete $\tau$-additive topological
measure (234Eb, 411Gj).

\medskip

{\bf (b)} Suppose that $\familyiI{E_i}\in\prod_{i\in I}\Sigma_i$ is such that
$\{i:i\in I$, $E_i\ne X_i\}$ is finite.   For $K\in\Cal K$, set
$H_K=\prod_{i\in K}E_i$;  then $H_K$ belongs to the domain $\tilde\Lambda_K$ of
$\tilde\lambda_K$,

\Centerline{$\phi^{-1}[\prod_{i\in I}E_i]
=\{\family{K}{\Cal K}{z_K}:z_K(i)\in E_i\text{ whenever }i\in K\in\Cal K\}
=\prod_{K\in\Cal K}H_K$,}

\noindent while

\Centerline{$\{K:K\in\Cal K$, $H_K\ne Z_K\}
=\{K:K\in\Cal K$, $K\cap\{i:E_i\ne X_i\}\ne\emptyset\}$}

\noindent is finite, so

$$\eqalign{\tilde\lambda\phi^{-1}(\prod_{i\in I}E_i)&
=\tilde\lambda(\prod_{K\in\Cal K}H_K)
=\prod_{K\in\Cal K}\tilde\lambda_KH_K\cr&
=\prod_{K\in\Cal K}\prod_{i\in K}\mu_iE_i
=\prod_{i\in I}\mu_iE_i.\cr}$$

\medskip

{\bf (c)} For each $K\in\Cal K$, $\tilde\lambda_K$ is inner regular with respect
to $\tilde\Lambda^0_K=(\Tensorhat_{i\in K}\Sigma_i)\vee\Cal B(Z_K)$.   By
417E(b-iv), $\tilde\lambda$ is inner regular with respect to

\Centerline{$\tilde\Lambda'
=(\Tensorhat_{K\in\Cal K}((\Tensorhat_{i\in K}\Sigma_i)\vee\Cal B(Z_K)))
\vee\Cal B(Z)
=(\Tensorhat_{K\in\Cal K}(\Tensorhat_{i\in K}\Sigma_i))
\vee\Cal B(Z)$.}

\noindent Now $\phi$ identifies
$(\Tensorhat_{K\in\Cal K}(\Tensorhat_{i\in K}\Sigma_i))$ with
$\Tensorhat_{i\in I}\Sigma_i$ in the sense that for $W\subseteq Z_I$,
$W\in\Tensorhat_{i\in I}\Sigma_i$ iff $\phi^{-1}[W]\in
(\Tensorhat_{K\in\Cal K}(\Tensorhat_{i\in K}\Sigma_i))$, while similarly
(because $\phi$ is a homeomorphism) $W\in\Cal B(Z_I)$ iff
$\phi^{-1}[W]\in\Cal B(Z)$.   It follows that
$W\in\tilde\Lambda^0_I$ iff $\phi^{-1}[W]\in\tilde\Lambda'$.   Now if
$W$ is measured by $\tilde\lambda\phi^{-1}$, there is a
$V\in\tilde\Lambda'$ such that $V\subseteq\phi^{-1}[W]$ and
$\tilde\lambda W=\tilde\lambda\phi^{-1}[W]$;  in which case
$\phi[V]\in\tilde\Lambda_I$, $\phi[V]\subseteq W$ and
$\tilde\lambda\phi^{-1}[\phi[V]]=\tilde\lambda\phi^{-1}[W]$.

This shows that $\tilde\lambda\phi^{-1}$ is inner regular with respect to
$\tilde\Lambda^0_I$.   Together with (a) and (b) and the uniqueness promised in
417E, this shows that
$\tilde\lambda\phi^{-1}=\tilde\lambda_I$, as claimed.
}%end of proof of 417J

\leader{417K}{Corollary}\dvArevised{2021} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
be a family of $\tau$-additive topological probability spaces and
$(X,\tilde\Lambda,\tilde\lambda)$ their $\tau$-additive product.
For $J\subseteq I$ let $\tilde\lambda_J$ be the
$\tau$-additive product measure on $X_J=\prod_{i\in J}X_i$, and
$\tilde\Lambda_J$ its domain;  let $\pi_J:X\to X_J$ be the canonical
map.   Then $\tilde\lambda_J$ is the image measure
$\tilde\lambda\pi_J^{-1}$.
In particular, if $W\in\tilde\Lambda$ is determined by coordinates in
$J\subseteq I$, then $\pi_J[W]\in\tilde\Lambda_J$ and
$\tilde\lambda_J\pi_J[W]=\tilde\lambda W$.

\proof{ By 417J, we can identify $\tilde\lambda$ with the $\tau$-additive
product of $\tilde\lambda_J$ and $\tilde\lambda_{I\setminus J}$.   If
$A\subseteq X_J$, then $\pi_J^{-1}[A]=A\times X_{I\setminus J}$.   If
$A\in\tilde\Lambda_J$, $\pi_J^{-1}[A]\in\tilde\Lambda$ by the
definition in 417C;  if $\pi_J^{-1}[A]\in\tilde\Lambda$ then
$A\in\tilde\Lambda_J$ by 417H.   And in either case

\Centerline{$\tilde\lambda\pi^{-1}_J[A]
=\tilde\lambda_JA\cdot\tilde\lambda_{I\setminus J}X_{I\setminus J}
=\tilde\lambda_JA$.}

\noindent So $\tilde\lambda_J=\tilde\lambda\pi_J^{-1}$.
}%end of proof of 417K

\vleader{48pt}{417L}{Corollary}\dvArevised{2021} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
be a family of $\tau$-additive topological probability spaces, and
$(X,\tilde\Lambda,\tilde\lambda)$ their
$\tau$-additive product.   Let $\family{j}{J}{K_j}$ be a disjoint family
of subsets of $I$, and for $j\in J$ write $\tilde\Lambda_j$ for the
$\sigma$-algebra of members of $\tilde\Lambda$ determined by coordinates in
$K_j$.   Then $\family{j}{J}{\tilde\Lambda_j}$ is a stochastically
independent family of $\sigma$-algebras\cmmnt{ (definition: 272Ab)}.

\proof{ It is enough to consider the case in which $J$ is finite
(272Bb), no $K_j$ is empty (since if $K_j=\emptyset$ then
$\tilde\Lambda_j=\{\emptyset,X\}$) and $\bigcup_{j\in J}K_j=I$ (adding
an extra term if necessary).   In this case, if $W_j\in\tilde\Lambda_j$
for each $j$, then the identification between $X$ and
$\prod_{j\in J}\prod_{i\in K_j}X_i$, as described in 417J, matches
$\bigcap_{j\in J}W_j$ with $\prod_{j\in J}\pi_{K_j}[W_j]$, writing
$\pi_{K_j}(x)$ for $x\restr K_j$.   Now if $\tilde\lambda_j$ is the
$\tau$-additive product measure on $Z_j=\prod_{i\in K_j}X_i$, we have
$\tilde\lambda_j\pi_{K_j}[W_j]=\tilde\lambda W_j$, by 417K.   Since
$\tilde\lambda$ can be identified with the $\tau$-additive product of
$\family{j}{J}{\tilde\lambda_j}$ (417J),

\Centerline{$\tilde\lambda(\bigcap_{j\in J}W_j)
=\prod_{j\in J}\tilde\lambda_j\pi_{K_j}[W_j]
=\prod_{j\in J}\tilde\lambda W_j$.}

\noindent As $\family{j}{J}{W_j}$ is arbitrary, $\family{j}{J}{\tilde\Lambda_j}$
is independent.
}%end of proof of 417L

\vleader{60pt}{417M}{Proposition}\dvArevised{2021} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
be a family of $\tau$-additive topological probability spaces such that
every $\mu_i$ is
strictly positive.   For $J\subseteq I$ let $\pi_J$ be the canonical map
from $X$ onto $X_J=\prod_{i\in J}X_i$;  write $\lambda_J$,
$\tilde\lambda_J$
for the ordinary and $\tau$-additive product measures on $X_J$, and
$\Lambda_J$, $\tilde\Lambda_J$ for their domains.   Set
$\tilde\lambda=\tilde\lambda_I$, $\tilde\Lambda=\tilde\Lambda_I$,
$\lambda=\lambda_I$, $\Lambda=\Lambda_I$.

(a) Let $F\subseteq X$ be a closed self-supporting set, and $J$ the
smallest subset of $I$ such that $F$ is determined by coordinates
in $J$\cmmnt{ (4A2B(g-ii))}.   Then

\quad(i) if $W\in\tilde\Lambda$ is such that $W\symmdiff F$ is
$\tilde\lambda$-negligible and $W$ is determined by coordinates in
$K\subseteq I$, then $K\supseteq J$;

\quad(ii) $J$ is countable;

\quad(iii) there is a $W\in\Lambda$, determined by coordinates in $J$,
such that $W\symmdiff F$ is $\tilde\lambda$-negligible.

(b) $\tilde\lambda$ is inner regular with respect to the family of sets
of the form $\bigcap_{n\in\Bbb N}V_n$ where each $V_n\in\tilde\Lambda$
is determined by coordinates in a finite set.

(c) If $W\in\tilde\Lambda$, there are a countable $J\subseteq I$ and
sets $W'$, $W''\in\tilde\Lambda$, determined by coordinates in $J$, such
that $W'\subseteq W\subseteq W''$ and
$\tilde\lambda(W''\setminus W')=0$.   Consequently
$\tilde\lambda\pi_J^{-1}[\pi_J[W]]=\tilde\lambda W$.

\proof{{\bf (a)(i)} \Quer\ Suppose, if possible, otherwise.   Then $F$
is not determined by coordinates in $K$, so there are $x\in F$,
$y\in X\setminus F$ such that $x\restr K=y\restr K$.
Let $U$ be an open set containing $y$, disjoint from $F$, and of
the form $\prod_{i\in I}G_i$, where $G_i\in\frak T_i$ for every $i$ and
$L=\{i:G_i\ne X_i\}$ is finite.   Set

\Centerline{$U'=\{z:z\in X,\,z(i)\in G_i$ for every $i\in L\cap K\}$,}

\Centerline{$U''=\{z:z(i)\in G_i$ for every $i\in L\setminus K\}$.}

\noindent Then $U'\cap W$ is determined by coordinates in $K$, while
$U''$ is determined by coordinates in $I\setminus K$, so

$$\eqalignno{0
&=\tilde\lambda(F\cap U)
=\tilde\lambda(W\cap U)=\tilde\lambda(W\cap U'\cap U'')
=\tilde\lambda(W\cap U')\cdot\tilde\lambda U''\cr
\displaycause{by 417L}
&=\tilde\lambda(F\cap U')\cdot\tilde\lambda U''
=\tilde\lambda(F\cap U')\cdot\prod_{i\in L\setminus K}\mu_iG_i.\cr}$$

\noindent But $y\in U'$, and $x\restr K=y\restr K$, so $x\in F\cap U'$;
as $F$ is self-supporting, $\tilde\lambda(F\cap U')>0$.   Because every
$\mu_i$ is strictly positive, and no $G_i$ is empty,
$\prod_{i\in L\setminus K}\mu_iG_i>0$;  and this is impossible.\ \Bang

\medskip

\quad{\bf (ii)} By 417E(b-i), there is a $W_0\in\Lambda$ such that
$\tilde\lambda(F\symmdiff W_0)=0$.   By 254Oc there is a
$W_1\in\Lambda$, determined by coordinates in a countable subset $K$ of
$I$, such that $\lambda(W_0\symmdiff W_1)=0$.   Now
$\tilde\lambda(F\symmdiff W_1)=0$, so (i) tells us that $J\subseteq K$
is countable.

\medskip

\quad{\bf (iii)} By 417K, $\pi_J[F]\in\tilde\Lambda_J$.   By 417E(b-i) again,
there is a $V\in\Lambda_J$ such that $V\symmdiff\pi_J[F]$ is
$\tilde\lambda_J$-negligible.   Set $W=\pi_J^{-1}[V]$.   Then
$W\in\Lambda$, $W$ is determined by coordinates in $J$, and
$W\symmdiff F=\pi_J^{-1}[V\symmdiff\pi_J[F]]$ is
$\tilde\lambda$-negligible.

\medskip

{\bf (b)(i)} Write $\Cal V$ for the set of those members of
$\tilde\Lambda$
which are determined by coordinates in a finite set, and $\Cal V_{\delta}$
for the set of intersections of sequences in $\Cal V$.   Because
$\Cal V$ is closed under finite unions, so is $\Cal V_{\delta}$;
$\Cal V_{\delta}$
is surely closed under countable intersections, and $\emptyset$, $X$
belong to $\Cal V_{\delta}$.

\medskip

\quad{\bf (ii)} We need to know that every self-supporting closed set
$F\subseteq X$ belongs to $\Cal V_{\delta}$.   \Prf\ By (a), $F$ is
determined by coordinates in a countable set $J$.   Express $J$ as the
union of a non-decreasing sequence $\sequencen{J_n}$ of finite sets.
Then $F_n=\pi_{J_n}^{-1}[\overline{\pi_{J_n}[F]}]\in\Cal V$ for each
$n$, and $F=\bigcap_{n\in\Bbb N}F_n\in\Cal V_{\delta}$.\ \Qed

\medskip

\quad{\bf (iii)} Set
$\tilde\Lambda^0=(\Tensorhat_{i\in I}\Sigma_i)\vee\Cal B(X)$ as in 417E,
and let $\Cal A$ be the family of subsets of $X$ which are
finite disjoint
unions of sets of the form $B\cap\prod_{i\in I}E_i$ where $B\subseteq X$
is either open or closed, $E_i\in\Sigma_i$ for every $i\in I$, and
$\{i:E_i\ne X_i\}$ is finite.   Then
$X\setminus A\in\Cal A$ for every $A\in\Cal A$, and $\tilde\Lambda^0$ is the
$\sigma$-algebra generated by $\Cal A$.
Now if $A\in\Cal A$, $V\in\tilde\Lambda$ and
$\tilde\lambda(A\cap V)>0$, there is a $K\in\Cal V_{\delta}\cap\Cal A$
such that $K\subseteq A$ and $\tilde\lambda(K\cap V)>0$.   \Prf\ There is a
set $A'$ of the form $B\cap\prod_{i\in I}E_i$ where $B\subseteq X$
is either open or closed, $E_i\in\Sigma_i$ for every $i\in I$, and
$\{i:E_i\ne X_i\}$ is finite, such that $A'\subseteq A$ and
$\tilde\lambda(A'\cap V)>0$.
($\alpha$) If $B$ is open, set

\Centerline{$\Cal U=\{U:U\in\Cal V$ is open, $U\subseteq B\}$.}

\noindent Because $\Cal V$ includes a base for the topology of $X$,
$\bigcup\Cal U=B$;  because $\tilde\lambda$ is $\tau$-additive and
$\Cal V$ is closed under finite unions, there is a $U\in\Cal U$ such
that $U\subseteq B$ and
$\tilde\lambda U>\tilde\lambda B-\tilde\lambda(B\cap V)$, so that
$\tilde\lambda(U\cap V)>0$, while $U\in\Cal V$.
($\beta$) If $B$ is closed, then it includes a
self-supporting closed set $F$ of the same measure (414F), which belongs
to $\Cal V_{\delta}$, by (ii) just above, and now
$F\cap\prod_{i\in I}E_i\in\Cal V_{\delta}$,
$F\cap\prod_{i\in I}E_i\subseteq A$ and
$\tilde\lambda(F\cap\prod_{i\in I}E_i\cap V)>0$.\ \Qed

\medskip

\quad{\bf (iv)} By 412C, $\tilde\lambda\restr\tilde\Lambda^0$
is inner regular
with respect to $\Cal V_{\delta}$.   But $\tilde\lambda$
is just the completion of $\tilde\lambda\restr\tilde\Lambda^0$, so it also is
inner regular with respect to $\Cal V_{\delta}$ (412Ha once more).

\medskip

{\bf (c)} By (b), we have sequences $\sequencen{V_n}$,
$\sequencen{V'_n}$ in $\Cal V_{\delta}$ such that $V_n\subseteq W$,
$V'_n\subseteq X\setminus W$,
$\tilde\lambda V_n\ge\tilde\lambda W-2^{-n}$ and
$\tilde\lambda V'_n\ge\tilde\lambda(X\setminus W)-2^{-n}$ for every
$n\in\Bbb N$.
Each $V_n$, $V'_n$ is determined by a coordinates in a countable set, so
there is a single countable set $J\subseteq I$ such that every $V_n$ and
every $V'_n$ is determined by coordinates in $J$.   Set
$W'=\bigcup_{n\in\Bbb N}V_n$, $W''=X\setminus\bigcup_{n\in\Bbb N}V'_n$;
then $W'$, $W''$ are both determined by coordinates in $J$,
$W'\subseteq W\subseteq W''$ and
$\tilde\lambda(W''\setminus W')=0$, as required.
}%end of proof of 417M

\leader{417N}{Theorem} Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be two quasi-Radon measure spaces.   Then the
$\tau$-additive product measure $\tilde\lambda$ on $X\times Y$ is a
quasi-Radon
measure, the unique quasi-Radon measure on $X\times Y$ such that
$\tilde\lambda(E\times F)=\mu E\cdot\nu F$ for every $E\in\Sigma$ and
$F\in\Tau$.

%can collapse 417N-417O query

\proof{ By 417C(b-vii) it is inner regular with respect to the closed sets;
being also a complete, locally determined, $\tau$-additive and effectively
locally finite topological measure, it is quasi-Radon.
By 417Ca it is the only
quasi-Radon measure on $X\times Y$ with the right values on measurable
rectangles.
}%end of proof of 417N

\leader{417O}{Theorem} Let
$\langle (X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
quasi-Radon
probability spaces.   Then the $\tau$-additive product measure
$\tilde\lambda$ on $X=\prod_{i\in I}X_i$ is a quasi-Radon measure, the
unique quasi-Radon measure on $X$ extending the ordinary product
measure.

\proof{ By 417E(b-vi), $\tilde\lambda$ is inner regular with respect to
the closed sets, so is a quasi-Radon measure, which is unique by 417Ea.
}%end of proof of 417O

\leader{417P}{Theorem} Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be Radon measure spaces.   Then the
$\tau$-additive product
measure $\tilde\lambda$ on $X\times Y$ is a Radon measure, the unique
Radon measure on $X\times Y$ such that
$\tilde\lambda(E\times F)=\mu E\cdot\nu F$ whenever $E\in\Sigma$ and
$F\in\Tau$.

%can collapse 417P and 417Q query

\proof{ By 417C(b-viii) and (b-ix), $\tilde\lambda$ is tight and locally
finite;  being also a quasi-Radon measure (417N), it is a Radon measure;
as in 417N, it is uniquely defined by its values on
measurable rectangles.
}%end of proof of 417P

\leader{417Q}{Theorem} Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
Radon probability
spaces, and $\tilde\lambda$ the $\tau$-additive product measure on
$X=\prod_{i\in I}X_i$.   For each $i\in I$, let
$Z_i\subseteq X_i$ be the support of $\mu_i$.   Suppose that
$J=\{i:i\in I,\,Z_i$ is not compact$\}$ is countable.   Then
$\tilde\lambda$ is a
Radon measure, the unique Radon measure on $X$ extending the ordinary
product measure.

\proof{ Of course $X$, being a product of Hausdorff spaces, is
Hausdorff, and $\tilde\lambda$, being totally finite, is locally finite.
Now, given $\epsilon\in\ocint{0,1}$, let
$\langle\epsilon_j\rangle_{j\in J}$ be a
family of strictly positive numbers such that
$\sum_{j\in J}\epsilon_j\le\epsilon$,
and for $j\in J$ choose a compact set $K_j\subseteq X_j$ such that
$\mu_jK_j\ge 1-\epsilon_j$;  for $i\in I\setminus J$, set $K_i=Z_i$, so
that $K_i$ is compact and $\mu_iK_i=1$.   Consider
$K=\prod_{i\in I}K_i$.
Then, using 417E(b-ii) and 254Lb for the two equalities,

\Centerline{$\tilde\lambda K=\lambda^*K
=\prod_{i\in I}\mu_iK_i\ge\prod_{j\in J}1-\epsilon_j
\ge 1-\epsilon$,}

\noindent where $\lambda$ is the ordinary product measure on $X$.   As
$\epsilon$ is arbitrary, $\tilde\lambda$ satisfies the condition (iii) of
416C, and is a Radon measure.   By 417Ea, it is the unique Radon
measure on $X$ extending $\lambda$.
}%end of proof of 417Q

\leader{417R}{Notation} I will use the phrase {\bf quasi-Radon product
measure} for a $\tau$-additive product measure which is in fact a
quasi-Radon measure;   similarly, a {\bf Radon product measure}
is a $\tau$-additive product measure which is a Radon measure.

\leader{417S}{}\cmmnt{ Later I will give an example in which a
$\tau$-additive product measure is different from the corresponding
c.l.d.\ product measure (419E).   In 415E-415F, 415Ye and 416U I
have described cases in which c.l.d.\ measures are $\tau$-additive
product measures.   It remains very unclear when to expect this to
happen.   I can however give a couple of results which show that
sometimes, at least, we can be sure that the two measures coincide.

\medskip

\noindent}{\bf Proposition}\dvArevised{2021}
(a) Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be effectively locally finite $\tau$-additive
topological measure spaces and $\lambda$ the c.l.d.\ product measure on
$X\times Y$. If every open subset of $X\times Y$ is measured by
$\lambda$, then $\lambda$ is the $\tau$-additive product measure on
$X\times Y$.

(b) Let $\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of $\tau$-additive topological probability spaces and
$\lambda$ the ordinary product measure on $X=\prod_{i\in I}X_i$.   If
every open subset of $X$ is measured by $\lambda$, then $\lambda$
is the $\tau$-additive product measure on $X$.

(c) In (b), let $\lambda_J$ be the ordinary product measure on
$X_J=\prod_{i\in J}X_i$ for each $J\subseteq I$, and $\tilde\lambda_J$
the $\tau$-additive product measure.   If $\lambda_J=\tilde\lambda_J$
for every finite $J\subseteq I$, and every $\mu_i$ is strictly positive,
then $\lambda=\tilde\lambda_I$ is the $\tau$-additive product measure on
$X$.

\proof{{\bf (a), (b)} In both cases, $\lambda$ is a complete locally
determined effectively locally finitemeasure
(assembling facts from 251I, 254F and 412Se). %eff loc fin
We know also from 417C and 417E that $\lambda$ has an extension to a
$\tau$-additive measure $\tilde\lambda$
and is therefore itself $\tau$-additive (411C), while it
is inner regular with respect to $\Sigma\tensorhat\Tau$ or
$\Tensorhat_{i\in I}\Sigma_i$ (251Ib, 254Ff) and therefore with respect to
$\tilde\Lambda_0$ as defined in 417C or 417E.   The additional hypothesis here
is that $\lambda$ is a topological measure.   But that means that it satisfies
all the conditions required of a $\tau$-additive product measure and is equal to
$\tilde\lambda$.

\medskip

{\bf (c)(i)} The first step is to note that $\lambda_J=\tilde\lambda_J$
for every countable $J\subseteq I$.   \Prf\ Express $J$ as
$\bigcup_{n\in\Bbb
N}J_n$ where $\sequencen{J_n}$ is a non-decreasing sequence of finite
sets.   If $F\subseteq X_J$ is closed, then it is
$\bigcap_{n\in\Bbb N}\pi_n^{-1}[\overline{\pi_n[F]}]$, where
$\pi_n:X_J\to X_{J_n}$ is the canonical map for each $n$.   But every
$\overline{\pi_n[F]}$ is a closed subset of $X_{J_n}$, therefore
measured by $\lambda_{J_n}$;  because $\pi_n$ is \imp\ (417K),
$\pi_n^{-1}[\overline{\pi_n[F]}]\in\dom\lambda_J$ for each $n$, and
$F\in\dom\lambda_J$.   Thus every closed set, therefore every open set
is measured by $\lambda_J$, and $\lambda_J$ is a topological measure;
by (b), $\lambda_J=\tilde\lambda_J$.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $W\subseteq X$ is open.   By 417M, there
are $W'$, $W''$ measured by $\tilde\lambda$ such that
$W'\subseteq W\subseteq W''$, both $W''$ and $W'$ are determined by
coordinates in a countable set, and
$\tilde\lambda_I(W''\setminus W')=0$.   Let
$J\subseteq I$ be a countable set such that $W'$ and $W''$ are
determined by coordinates in
$J$.   Then $\lambda_J=\tilde\lambda_J$ measures $\pi_J[W']$, by 417K,
so $\lambda$ measures $W'=\pi_J^{-1}[\pi_J[W]]$, by 254Oa.   Similarly,
$\lambda$ measures $W''$.   Now
$\lambda(W''\setminus W')=\tilde\lambda_I(W''\setminus W')=0$, so
$\lambda$ measures $W$.   As $W$ is arbitrary, $\lambda$ is a
topological measure and must be the $\tau$-additive product measure, by
(b).
}%end of proof of 417S

\leader{417T}{Proposition}\dvArevised{2021} Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be effectively locally finite $\tau$-additive
topological measure spaces
and $\lambda$ the c.l.d.\ product measure on
$X\times Y$.   If $X$ has a conegligible subset with a countable
network\cmmnt{ (e.g., if $X$ is metrizable and $\mu$ is $\sigma$-finite)}, then
$\lambda$ is the $\tau$-additive product measure on $X\times Y$.

\proof{{\bf (a)} Suppose to begin with that $\mu$ and $\nu$ are totally
finite, and that $X$ itself
has a countable network;  let $\sequencen{A_n}$ run
over a network for $X$.   Let $\hat\mu$ be the completion of $\mu$ and
$\hat\Sigma$ its domain.   Let $\tilde\lambda$ be the $\tau$-additive
product measure on $X\times Y$.   (We are going to need Fubini's theorem
both for $\lambda$ and for $\tilde\lambda$.   I will use a sprinkling of
references to \S\S251-252 to indicate which parts of the
argument below depend on the properties of $\lambda$.)

Let $W\subseteq X\times Y$ be an open set.   For each $n\in\Bbb N$, set

\Centerline{$H_n=\bigcup\{H:H\in\frak S,\,A_n\times H\subseteq W\}$,}

\noindent so that $H_n$ is open.   Then
$W=\bigcup_{n\in\Bbb N}A_n\times H_n$.   \Prf\ Of course
$A_n\times H_n\subseteq W$ for every $n\in\Bbb N$.   If $(x,y)\in W$,
there are open sets $G\subseteq X$, $H\subseteq Y$ such that
$(x,y)\in G\times H\subseteq W$;  now there is an $n\in\Bbb N$ such that
$x\in A_n\subseteq G$, so that $H\subseteq H_n$ and
$(x,y)\in A_n\times H_n$.\ \Qed

By 417C(b-v-$\alpha$), there is an open set $W_0$ in the domain $\Lambda$ of
$\lambda$ such that
$W_0\subseteq W$ and $\tilde\lambda(W\setminus W_0)=0$.   By 417Ga,
applied to $\chi(W\setminus W_0)$,
$A=\{x:\nu(W[\{x\}]\setminus W_0[\{x\}])>0\}$ is $\mu$-negligible.   For
each $n\in\Bbb N$, $x\in X$ set $f_n(x)=\nu(H_n\cap W_0[\{x\}])$;  then
252B tells us that $\int f_nd\mu$ is defined and equal to
$\lambda(W_0\cap(X\times H_n))$.   In particular, $f_n$ is
$\hat\Sigma$-measurable.   Set $E_n=\{x:f_n(x)=\nu H_n\}\in\hat\Sigma$.
If $x\in A_n$, then $H_n\subseteq W[\{x\}]$, so
$A_n\setminus E_n\subseteq A$.

Now, by 252B again,

$$\eqalign{\lambda((E_n\times H_n)\setminus W_0)
&=\int_{E_n}\nu(H_n\setminus W_0[\{x\}])\mu(dx)\cr
&=\int_{E_n}\nu H_n-\nu(H_n\cap W_0[\{x\}])\mu(dx)
=0.\cr}$$

\noindent So if we set $W_1=\bigcup_{n\in\Bbb N}E_n\times H_n$,
$W_1\setminus W\subseteq W_1\setminus W_0$ is $\lambda$-negligible.
On the other hand,

\Centerline{$W\setminus W_1
\subseteq\bigcup_{n\in\Bbb N}(A_n\setminus E_n)\times H_n
\subseteq A\times Y$}

\noindent is also $\lambda$-negligible.   Because $\lambda$ is complete,
$W\in\Lambda$.   As $W$ is arbitrary, $\lambda$ is a topological measure
and is equal to $\tilde\lambda$, by 417Sa.

\medskip

{\bf (b)} Now consider the general case.   Let $Z$ be a conegligible
subset of $X$ with a countable network;  since any subset of a space
with a countable network again has a countable network (4A2Na), we may
suppose that $Z\in\Sigma$.   Again let $W$ be an open set in
$X\times Y$.   This time, take arbitrary $E\in\Sigma$, $F\in\Tau$ of
finite measure, and consider the subspace measures $\mu_{E\cap Z}$ and
$\nu_F$.   These are still effectively locally finite and
$\tau$-additive (414K), and are now totally finite.   Also $E\cap Z$
has a countable network.   So (a) tells us that the relatively open set
$W\cap((E\cap Z)\times F)$ is measured by the c.l.d.\ product of
$\mu_{E\cap Z}$ and $\nu_F$, which is the subspace measure on
$(E\cap Z)\times F$ induced by $\lambda$ (251Q).   Since $\lambda$
surely measures $E\times F$, it measures
$W\cap(Z\times Y)\cap(E\times F)$.   As $E$ and $F$ are arbitrary,
$\lambda$ measures $W\cap(Z\times Y)$ (251H).   But
$\lambda((X\setminus Z)\times Y)=\mu(X\times Z)\cdot\nu Y=0$ (251Ia), so
$\lambda$ also measures $W$.   As $W$ is arbitrary, $\lambda$ is the
$\tau$-additive product measure.
}%end of proof of 417T

\leader{417U}{Proposition} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
be a family of $\tau$-additive topological probability spaces.   Let
$\lambda$ be the ordinary product probability measure on
$X=\prod_{i\in I}X_i$ and $\Lambda$ its domain.
Then every continuous function $f:X\to\Bbb R$ is
$\Lambda$-measurable, so $\Lambda$ includes the Baire $\sigma$-algebra
of $X$.

\proof{{\bf (a)} Let
$\tilde\lambda$ be a $\tau$-additive topological measure extending
$\lambda$ (417E), and $\tilde\Lambda$ its domain;  then $f$ is
$\tilde\Lambda$-measurable, just because $\tilde\lambda$ is a
topological measure.   For $\alpha\in\Bbb R$, set

\Centerline{$G_{\alpha}=\{x:x\in X,\,f(x)<\alpha\}$,
\quad$H_{\alpha}=\{x:x\in X,\,f(x)>\alpha\}$,}

\Centerline{$F_{\alpha}=\{x:x\in X,\,f(x)=\alpha\}$.}

\noindent Then $\family{\alpha}{\Bbb R}{F_{\alpha}}$ is disjoint, so
$A=\{\alpha:\alpha\in\Bbb R,\,\tilde\lambda F_{\alpha}>0\}$ is
countable, and $A'=\Bbb R\setminus A$ is dense in $\Bbb R$;  let
$Q\subseteq A'$ be a countable dense set.

For each $q\in Q$, let $V_q\subseteq G_q$, $W_q\subseteq H_q$ be such
that

\Centerline{$\lambda V_q=\lambda_*G_q=\tilde\lambda G_q$,
\quad$\lambda W_q=\lambda_*H_q=\tilde\lambda H_q$}

\noindent (413Ea, 417E(b-ii)).   Then

\Centerline{$\lambda^*(G_q\setminus V_q)
\le\lambda(X\setminus(V_q\cup W_q))
=1-\lambda V_q-\lambda W_q
=\tilde\lambda(X\setminus(G_q\cup H_q))
=0$.}

\noindent Because $\lambda$ is complete, $G_q\setminus V_q$ and $G_q$
belong to $\Lambda$.   But now, if $\alpha\in\Bbb R$,

\Centerline{$\{x:f(x)<\alpha\}=\bigcup_{q\in Q,q<\alpha}G_q\in\Lambda$,}

\noindent so $f$ is $\Lambda$-measurable.

\medskip

\quad{\bf (b)} It follows that every zero set belongs to $\Lambda$, so
that $\Lambda$ must include the Baire $\sigma$-algebra of $X$.
}%end of proof of 417U

\leader{417V}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be effectively locally finite $\tau$-additive
topological
measure spaces, and $(X\times Y,\Lambda,\lambda)$ their c.l.d.\ product.
Then every continuous function $f:X\times Y\to\Bbb R$ is
$\Lambda$-measurable, and the Baire
$\sigma$-algebra of $X\times Y$ is included in $\Lambda$.

\proof{ Let $Z\subseteq X\times Y$ be a zero set.   If $E\in\Sigma$,
$F\in\Tau$ are sets of finite measure, then $Z\cap(E\times F)$ is a zero
set for the relative topology of $E\times F$.   Now the subspace
measures $\mu_E$ and $\nu_F$ are $\tau$-additive topological measures
(414K), so $Z\cap(E\times F)$ is measured by the c.l.d.\ product
$\mu_E\times\nu_F$ of $\mu_E$ and $\nu_F$.   \Prf\ If either $\mu E$ or
$\nu F$ is zero, this is trivial.   Otherwise, they have scalar
multiples $\mu'_E$, $\nuprime_F$ which are probability measures, and of
course are still $\tau$-additive topological measures.   By 417U,
$Z\cap(E\times F)$ is measured by
$\mu'_E\times\nuprime_F$.   Since $\mu_E\times\nu_F$ is just a
scalar multiple of $\mu'_E\times\nuprime_F$, $Z\cap(E\times F)$ is
measured
by $\mu_E\times\nu_F$.\ \QeD\   But $\mu_E\times\nu_F$ is the
subspace measure $\lambda_{E\times F}$ (251Q), so
$Z\cap(E\times F)\in\Lambda$.
As $E$ and $F$ are arbitrary, $Z\in\Lambda$ (251H).

Thus every zero set belongs to $\Lambda$;  accordingly $\Lambda$ must
include the Baire $\sigma$-algebra, and every continuous function must
be $\Lambda$-measurable.
}%end of proof of 417V

%\leader{417W}{}

\exercises{
\leader{417X}{Basic exercises (a)}
%\spheader 417Xa
Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\Cal A$ a family of subsets of $X$.   Show that the following are
equiveridical:  (i) there is a measure $\mu'$ on $X$, extending $\mu$,
such that $\mu'A=0$ for every $A\in\Cal A$;
(ii) $\mu_*(\bigcup_{n\in\Bbb N}A_n)=0$ for every sequence
$\sequencen{A_n}$ in $\Cal A$.
%417A

\sqheader 417Xc Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be topological measure spaces such that $\mu$ and $\nu$ are both
effectively locally finite $\tau$-additive Borel measures.   Show that
there is a unique effectively locally finite $\tau$-additive Borel
measure $\lambda'$ on $X\times Y$ such that
$\lambda'(G\times H)=\mu G\cdot\nu H$ for all open sets $G\subseteq X$,
$H\subseteq Y$.
%417D

\sqheader 417Xd Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of topological probability spaces in which every $\mu_i$ is a
$\tau$-additive Borel measure.   Show that there is a unique
$\tau$-additive Borel measure $\lambda'$ on $X=\prod_{i\in I}X_i$ such
that $\lambda'(\prod_{i\in I}F_i)=\prod_{i\in I}\mu_iF_i$ whenever
$F_i\subseteq X_i$ is closed for every $i\in I$.
%417E

\spheader 417Xe Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be effectively locally finite $\tau$-additive topological measure spaces
and $\tilde\lambda$ the $\tau$-additive product measure on $X\times Y$.
Let $\familyiI{X_i}$, $\family{j}{J}{Y_j}$ be decompositions for $\mu$,
$\nu$ respectively (definition: 211E).   Show that
$\langle X_i\times Y_j\rangle_{i\in I,j\in J}$ is a decomposition for
$\tilde\lambda$.   (Cf.\ 251O.)
%417F

\sqheader 417Xf Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
$\tau$-additive topological probability spaces such
that $\mu_i$ is inner regular with respect to the Borel sets for every
$i$, and $\tilde\lambda$ the
$\tau$-additive product measure on $X=\prod_{i\in I}X_i$.   Take
$A_i\subseteq X_i$ for each $i\in I$.   (i) Show that if $\mu_i^*A_i=1$
for every $i$, then the subspace measure induced by $\tilde\lambda$ on
$A=\prod_{i\in I}A_i$ is
just the $\tau$-additive product $\tilde\lambda^{\#}$ of the subspace
measures on the $A_i$.   \Hint{show that if we set
$\lambda'W=\tilde\lambda^{\#}(W\cap A)$ for Borel sets $W\subseteq X$,
then $\lambda'$ satisfies the conditions of 417Xd.}  (ii) Show that in
any case $\tilde\lambda^*A=\prod_{i\in I}\mu^*_iA_i$.   (Cf.\ 254L.)
%417F query drop "i r wrt Borel" from hyothesis?

\spheader 417Xg Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
and $\familyiI{(Y_i,\frak S_i,\Tau_i,\nu_i)}$ be two families of
$\tau$-additive topological probability spaces in which every $\mu_i$
and every $\nu_i$ is inner regular with respect to the Borel sets.   Let
$\tilde\lambda$, $\tilde\lambda'$ be the $\tau$-additive product
measures on $X=\prod_{i\in I}X_i$ and $Y=\prod_{i\in I}Y_i$
respectively.   Suppose that for each $i\in I$ we are given a continuous
\imp\ function $\phi_i:X_i\to Y_i$.   Show that the function
$\phi:X\to Y$ defined by setting $\phi(x)(i)=\phi_i(x(i))$ for $x\in X$,
$i\in I$ is \imp.
%417F   almost cts won't do  query do we need "i r wrt Borel"?

\spheader 417Xh Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be two complete locally determined effectively
locally
finite $\tau$-additive topological measure spaces such that both $\mu$
and $\nu$ are inner regular with respect to the Borel sets.   Let
$\tilde\lambda$ be
the $\tau$-additive product measure on $X\times Y$, and $\tilde\Lambda$
its domain.   Suppose that $\nu$ is $\sigma$-finite.   Show that for any
$W\in\tilde\Lambda$, $W[\{x\}]\in\Tau$ for almost every $x\in X$,
and $x\mapsto\nu W[\{x\}]$ is measurable.
%417G  query do we need "i r wrt Borel"?

\sqheader 417Xi Let $(X,\frak T,\Sigma,\mu)$ be $[0,1]$ with its usual
topology and Lebesgue measure, and let $(Y,\frak S,\Tau,\nu)$ be $[0,1]$
with its discrete topology and counting measure.   (i) Show that both
are Radon measure spaces.   (ii) Show that the c.l.d.\ product measure
on $X\times Y$ is a Radon measure.   \Hint{252Kc, or use 417T and 417P.}
(iii) Show that 417Ga can fail if we omit the hypothesis on
$\{(x,y):f(x,y)\ne 0\}$.
%417G

\spheader 417Xj Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be two effectively locally finite $\tau$-additive topological measure
spaces.   Let $\lambda$ be the c.l.d.\ product measure and
$\tilde\lambda$ the $\tau$-additive product measure on $X\times Y$.
Show that $\lambda^*(A\times B)=\tilde\lambda^*(A\times B)$ for
all sets $A\subseteq X$, $B\subseteq Y$.   \Hint{start with $A$, $B$ of
finite outer measure, so that 417I applies.}
%417I  omissible

\spheader 417Xk Let $\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in
I}$ be a family of $\tau$-additive topological probability spaces with
strictly positive measures, all inner regular with respect to the Borel
sets, and $(X,\frak T,\tilde\Lambda,\tilde\lambda)$ their
$\tau$-additive product.   For $J\subseteq I$ let $\tilde\lambda_J$ be
the $\tau$-additive product measure on $X_J=\prod_{i\in J}X_i$, and
$\tilde\Lambda_J$ its domain.   (i) Show that if $f$ is a real-valued
$\tilde\Lambda$-measurable function defined $\tilde\lambda$-almost
everywhere on $X$, we can find a countable set $J\subseteq I$ and a
$\tilde\Lambda_J$-measurable function $g$, defined
$\tilde\lambda_J$-almost
everywhere on $X_J$, such that $f$ extends $g\pi_J$.   (ii) In (i), show
that $\int fd\tilde\lambda=\int g\,d\tilde\lambda_J$ if either is
defined in $[-\infty,\infty]$.
%417M query do we need "i r wrt Borel"?

\spheader 417Xl Let
$\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
$\tau$-additive topological probability spaces such
that every $\mu_i$ is inner regular with respect to the Borel sets, and
$(X,\frak T,\tilde\Lambda,\tilde\lambda)$ their $\tau$-additive product.
Show that for any $W\in\tilde\Lambda$ there is a smallest set
$J\subseteq I$ for which there is a $W'\in\tilde\Lambda$, determined by
coordinates in $J$, with $\tilde\lambda(W\symmdiff W')=0$.  \Hint{254R.}
%417M query do we need "i r wrt Borel"?

\spheader 417Xm What needs to be added to 417M and 415Xm
% A q-R measure on a product of separable metr spaces is
% completion regular iff closed self-supp sets are determined by
% coordinates in countable sets
to complete a proof of 415E?
%417M

\spheader 417Xn Let $(X,\frak T,\Sigma,\mu)$ be an atomless
$\tau$-additive topological probability space such that $\mu$ is inner
regular with respect to the Borel sets, and $I$ a set with cardinal at
most that of the support of $\mu$.   Show that the set of injective
functions from $I$ to $X$ has full outer measure for the $\tau$-additive
product measure on $X^I$.
%417M query do we need "i r wrt Borel"?

\sqheader 417Xo Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be Radon measure spaces.   Show that the Radon product measure on
$X\times Y$ is the unique Radon measure $\tilde\lambda$ such that
$\tilde\lambda(K\times L)=\mu K\cdot\nu L$ for all compact sets
$K\subseteq X$, $L\subseteq Y$.
%417P

\sqheader 417Xp Let $I$ be an uncountable set, and $\lambda$,
$\tilde\lambda$ be the ordinary and $\tau$-additive product measures on
$X=\{0,1\}^I$ when each factor is given its usual topology and the Dirac
measure concentrated at $1$.   Show that $\tilde\lambda$ properly extends
$\lambda$, and that the support of
$\tilde\lambda$ is not determined by coordinates in any countable
set.   Find a $\tilde\lambda$-negligible open set $W\subseteq X$ such that its
projection onto $\{0,1\}^J$ is conegligible for every proper subset
$J$ of $I$.
%417Q

\spheader 417Xq Let $\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in
I}$ be a family of Radon probability spaces, and $\tilde\lambda$ the
quasi-Radon product measure on $X=\prod_{i\in I}X_i$.   For each $i\in
I$, let $Z_i\subseteq X_i$ be the support of $\mu_i$.   Show that
$\tilde\lambda$ is a Radon measure iff
$\{i:i\in I,\,Z_i$ is not compact$\}$ is countable.
In particular, show that the ordinary product measure on
$\coint{0,1}^I$, where $I$ is uncountable and
each copy of $\coint{0,1}$ is given Lebesgue measure, is a quasi-Radon
measure, but not a Radon measure.
%417Q

\spheader 417Xr Let $\sequencen{(X_n,\frak T_n,\Sigma_n,\mu_n)}$ be a
sequence of Radon probability spaces.   Show that the Radon product
measure on $X=\prod_{n\in\Bbb N}X_n$ is the unique Radon measure
$\tilde\lambda$ on $X$ such that
$\tilde\lambda(\prod_{n\in\Bbb N}K_n)=\prod_{n=0}^{\infty}\mu_nK_n$
whenever $K_n\subseteq X_n$ is compact for every $n$.
%417Q

\spheader 417Xs Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of $\tau$-additive topological probability spaces and $\lambda$,
$\tilde\lambda$ the ordinary and $\tau$-additive product measures on
$X=\prod_{i\in I}X_i$.   Show that if $A\subseteq X$ has
$\tilde\lambda$-negligible boundary, then $A$ is measured by $\lambda$.
%417U

\spheader 417Xt Let us say that a topological space $X$ is
{\bf chargeable} if there is an additive functional
$\nu:\Cal PX\to\coint{0,\infty}$ such that $\nu G>0$ for every non-empty
open set $G\subseteq X$.   (i) Show that if there is a
$\sigma$-finite measure $\mu$ on $X$ such that $\mu_*G>0$ for every
non-empty open set $G$, then $X$ is chargeable.   \Hint{215B(vii),
391G.}  (ii) Show that any separable space is chargeable.   (iii) Show
that $X$ is chargeable iff its regular open algebra is
chargeable in the sense of 391Bb.
\Hint{see the proof of 314P.}   (iv)
Show that any open subspace of a chargeable space is chargeable.   (v)
Show that if $Y\subseteq X$ is dense, then $X$ is chargeable iff $Y$ is
chargeable.   (vi) Show that if $X$ is expressible as the union of
countably many chargeable subspaces, then it is chargeable.   (vii)
Show that any product of chargeable spaces is chargeable.  (Cf.\
391Xb(iii).)
(viii) Show that if $\familyiI{X_i}$ is a family of chargeable spaces
with product $X$, then all regular open subsets of $X$ and all Baire
subsets of $X$ are determined by coordinates in countable sets.
\Hint{4A2Eb, 4A3Mb.}
(ix) Show that a continuous image of a chargeable space is chargeable.
(x) Show that a compact Hausdorff space is chargeable
iff it carries a strictly positive Radon measure.   \Hint{416K.}
%417U

\spheader 417Xu Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be quasi-Radon measure spaces such that $\mu X\cdot\nu Y>0$.   Show that
the quasi-Radon product measure on $X\times Y$ is completion regular iff
it is equal to the c.l.d.\ product measure and $\mu$ and $\nu$ are both
completion regular.   \Hint{412Sc;  if $\mu E$, $\nu F$ are finite
and $Z\subseteq E\times F$ is a zero set of positive measure, use
Fubini's theorem to show that $Z$ has sections of positive measure.}
%417V

\spheader 417Xv Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of quasi-Radon probability spaces.   Show that the
quasi-Radon product measure on $\prod_{i\in I}X_i$ is completion regular
iff it is equal to the ordinary product measure and every $\mu_i$ is
completion regular.
%417Xu, %417V

\spheader 417Xw Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of $\tau$-additive topological probability spaces such that every
$\mu_i$ is inner regular with respect to the Borel sets,
and $\tilde\lambda$ the $\tau$-additive product measure on
$X=\prod_{i\in I}X_i$;  write $\tilde\Lambda$ for its domain.
(i) Show that if $W\in\tilde\Lambda$, $\tilde\lambda W>0$ and
$\epsilon>0$ then there are a finite $J\subseteq I$ and a $W'\in\tilde\Lambda$
such that $\tilde\lambda W'\ge 1-\epsilon$ and for every $x\in W'$ there is a
$y\in W$ such that $x\restr I\setminus J=y\restr I\setminus J$.   (Cf.\
254Sb.)
(ii) Show that if $A\subseteq X$ is determined by coordinates in
$I\setminus\{i\}$ for every $i\in I$ then $\tilde\lambda^*A\in\{0,1\}$.
(Cf.\ 254Sa.)
%+ query do we need "i r wrt Borel"?

\leader{417Y}{Further exercises (a)}
%\spheader 417Ya
(i) Show that if, in 417A, $\mu$ is strictly localizable, then it has a
strictly localizable extension $\mu'$ with the properties (i)-(iv) there.
(ii) Give an example to show that the construction offered in 417A may not
immediately achieve this result.
%417A %mt41bits

\spheader 417Yb Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be effectively locally finite $\tau$-additive topological
measure spaces such that $\mu$ and
$\nu$ are both inner regular with respect to the Borel sets.

\quad(i) Fix open sets $G\subseteq X$, $H\subseteq Y$ of finite measure.
Let $\Cal W_{GH}$ be the set of those $W\subseteq X\times Y$ such that
$\theta_{GH}(W)=\int_G\hat\nu(W[\{x\}]\cap H)dx$ is defined, where
$\hat\nu$ is the completion of $\nu$.   ($\alpha$) Show
that every open set belongs to $\Cal W_{GH}$.   ($\beta$) Show that
$\theta_{GH}$ is countably additive in the sense that
$\theta_{GH}(\bigcup_{n\in\Bbb N}W_n)
=\sum_{n=0}^{\infty}\theta_{GH}(W_n)$
for every disjoint sequence $\sequencen{W_n}$ in $\Cal W_{GH}$, and
$\tau$-additive in the sense that
$\theta_{GH}(\bigcup\Cal V)=\sup_{V\in\Cal V}\theta_{GH}(V)$ for every
non-empty upwards-directed family $\Cal V$
of open sets in $X\times Y$.   ($\gamma$) Show that every Borel set
belongs to
$\Cal W_{GH}$.   \Hint{Monotone Class Theorem.}   ($\delta$) Show that
$\theta_{GH}\restr\Cal B(X\times Y)$ is a
$\tau$-additive Borel measure;  let $\lambda_{GH}$ be its completion.
($\epsilon$) Show that $\lambda_{GH}=\theta_{GH}\restr\Lambda_{GH}$,
where $\Lambda_{GH}=\dom\lambda_{GH}$.   ($\zeta$) Show that
$\lambda_{GH}(E\times F)$ is defined and equal to $\mu E\cdot\nu F$
whenever $E\in\Sigma$, $F\in\Tau$,
$E\subseteq G$ and $F\subseteq H$.   \Hint{start with open $E$ and $F$,
move to Borel $E$ and $F$ with the Monotone Class Theorem.}   ($\eta$)
Writing
$\lambda$ for the c.l.d.\ product measure on $X\times Y$, show that
$\lambda_{GH}(W)$ is defined and equal to $\lambda(W\cap(G\times H))$
whenever $W\in\dom\lambda$.

\quad(ii) Now take $\tilde\Lambda$ to be
$\bigcap\{\Lambda_{GH}:G\in\frak T,\,
H\in\frak S,\,\mu G<\infty,\,\nu H<\infty\}$ and
$\tilde\lambda W=\sup_{G,H}\lambda_{GH}(W)$ for $W\in\tilde\Lambda$.
Show that $\tilde\lambda$ is an extension of $\lambda$ to a complete
locally determined effectively locally finite $\tau$-additive
topological measure on
$X\times Y$ which is inner regular with respect to the Borel sets, so is
the $\tau$-additive product measure as defined in 417F.
%417G query do we need "i r wrt Borel"?

\spheader 417Yc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be complete
measure spaces with topologies $\frak T$, $\frak S$.   Suppose that
$\mu$ and $\nu$
are effectively locally finite and $\tau$-additive and moreover that
their domains include bases for the two topologies.   Show that the
c.l.d.\ product measure on $X\times Y$ has the same properties.
\Hint{start by
assuming that $\mu X$ and $\nu Y$ are both finite.   If $\Cal V$ is an
upwards-directed family of measurable open sets with measurable open
union $W$, look at $g_V(x)=\nu V[\{x\}]$ for $V\in\Cal V$.}
%417Yb, 417G

\spheader 417Yd Let $\langle(X_i,\frak T_i,\Sigma_i,\mu_i)\rangle_{i\in
I}$ be a family of $\tau$-additive topological probability spaces such
that every $\mu_i$ is inner regular with respect to the Borel sets, and
$(X,\frak T,\tilde\Lambda,\tilde\lambda)$ their $\tau$-additive product.
(i) Show that the following are equiveridical:  ($\alpha$) $\mu_i$ is
strictly positive
for all but countably many $i\in I$;  ($\beta$) whenever
$W\in\tilde\Lambda$
there are a countable $J\subseteq I$ and $W_1$, $W_2\in\tilde\Lambda$,
determined by coordinates in $J$, such that $W_1\subseteq W\subseteq
W_2$ and $\tilde\lambda(W_2\setminus W_1)=0$.   (ii) Show that when
these are false, $\tilde\lambda$ cannot be equal to the ordinary product
measure on $X$.
%417M query do we need "i r wrt Borel"?

\spheader 417Ye Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces with Hausdorff topologies $\frak T$, $\frak S$ such that both $\mu$ and
$\nu$ are inner regular with respect to the families of sequentially
compact sets in
each space.   Show that the c.l.d.\ product measure $\lambda$ on
$X\times Y$ is also inner regular with respect to the sequentially
compact sets, so has an extension to a topological measure which is
inner regular with respect to the sequentially compact sets.
\Hint{412R, 416Yd.}
%417P

\spheader 417Yf Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a
family of probability spaces with topologies $\frak T_i$ such that every
$\mu_i$ is inner regular with respect to the family of closed countably
compact sets in $X_i$ and every $X_i$ is compact.   Show that the
ordinary product measure $\lambda$ on $X=\prod_{i\in I}X_i$ is also
inner regular with respect to the closed countably compact sets, so has
an extension to a topological measure $\tilde\lambda$ which is inner
regular with respect to the closed countably compact sets in $X$.   Show
that this can be done in such a way that for every
$W\in\dom\tilde\lambda$ there is a $V\in\dom\lambda$ such that
$\tilde\lambda(W\symmdiff V)=0$.
\Hint{412T, 416Yc.}
%417Q

\spheader 417Yg Let $\langle(X_n,\Sigma_n,\mu_n)\rangle_{n\in\Bbb N}$ be
a sequence of probability spaces with
Hausdorff topologies $\frak T_n$ such that
every $\mu_n$ is inner regular with respect to the family of
sequentially compact
sets in $X_n$.   Show that the ordinary product measure $\lambda$ on
$X=\prod_{n\in\Bbb N}X_n$ is also inner regular with respect to the
sequentially compact sets, so has an extension to a topological measure
$\tilde\lambda$ which is inner regular with respect to the sequentially
compact sets in $X$.   Show that this can be done in such a way that for
every $W\in\dom\tilde\lambda$ there is a $V\in\dom\lambda$ such that
$\tilde\lambda(W\symmdiff V)=0$.
%417Yf, 417Q

\spheader 417Yh Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of quasi-Radon probability spaces, and $\lambda$,
$\tilde\lambda$ the ordinary and quasi-Radon product measures on
$X=\prod_{i\in I}X_i$.   Suppose that all but {\it one} of the
$\frak T_i$ have countable networks and all but {\it countably many} of
the $\mu_i$ are strictly positive.   Show that $\lambda=\tilde\lambda$.
%417T

\spheader 417Yi Let us say that a quasi-Radon measure space
$(X,\frak T,\Sigma,\mu)$ has the {\bf simple product property} if the
c.l.d.\ product measure on $X\times Y$ is equal to the quasi-Radon
product measure for every quasi-Radon measure space
$(Y,\frak S,\Tau,\nu)$.
(i) Show that if $(X,\frak T)$ has a countable network then
$(X,\frak T,\Sigma,\mu)$ has the simple product property.   (ii) Show
that if a quasi-Radon measure space has the simple product property so
do all its subspaces.  (iii) Show that the quasi-Radon product of two
quasi-Radon measure spaces with the simple product property has the
simple product property.   (iv) Show that the quasi-Radon product of any
family of quasi-Radon probability spaces with the simple product
property has the simple product property.   (v) Show that the
Sorgenfrey line (415Xc, 439Q) with Lebesgue measure
has the simple product property.
%417T
}%end of exercises

\endnotes{\Notesheader{417} The general problem of determining just when
a measure can be extended to a
measure with given properties is one which will recur throughout this
volume.   I have more than once mentioned the Banach-Ulam problem;  if
you like, this is the question of whether there can ever be an extension
of the countable-cocountable measure on an uncountable
set $X$ to a measure defined
on the whole algebra $\Cal PX$.   This particular question appears to be
undecidable from the ordinary axioms of set theory;  but for many sets
(for instance, if $X=\omega_1$) it is known that the answer is `no'.
(See 419G and 438C.)   This being so, we have to take each
manifestation of the general question on its own merits.   In 417C and
417E the challenge is to take a product measure $\lambda$ defined in
terms of the factor measures alone, disregarding their topological
properties, and extend it
to a topological measure, preferably $\tau$-additive.   Of course there
are important cases in which $\lambda$ is itself already a topological
measure;   for instance, we know that the c.l.d.\ product of Lebesgue
measure on $\Bbb R$ with itself is Lebesgue measure on $\BbbR^2$
(251N), and other examples are in 415E, 415Ye, 416U, 417S-417T,
417Yh and 453I.   But in general
not every open set in the product belongs to the domain of $\lambda$,
even when we have the product of two Radon measures on compact Hausdorff
spaces (419E).

Once we have resolved to grasp the nettle, however, there is a natural
strategy for the proof.   It is easy to see that if $\lambda$, in 417C
or 417E, is to have an extension to a $\tau$-additive topological
measure $\tilde\lambda$, then we must have $\tilde\lambda A(\Cal V)=0$
for every $\Cal V$ belonging to the class $\frak V$.   Now 417A descibes
a sufficient (and obviously necessary) condition for there to be an
extension of $\lambda$ with this property.   So all we have to do is
check.   The check is not perfectly straightforward;  in 417E it uses
all the resources of the original proof that there is a product measure
on an arbitrary product of probability spaces (which I suppose is to be
expected), with 414B (of course) to apply
the hypothesis that the factor measures are $\tau$-additive, and a
couple of extra wrinkles (the $W'_n$ and $C'_n$ of part (a-iii-$\beta$) of the
proof of 417E, and the use of supports in part (a-iii-$\epsilon$)).   I take
the opportunity to say that the precise form of the results in 417C and
417E, leading to definitions of `$\tau$-additive product measure' in terms of
the $\sigma$-algebras $\tilde\Lambda_0$, was devised in
response to a remark by M.R.Burke.

It is worth noting that (both for finite and for infinite products) the
measure algebras of $\lambda$ and $\tilde\lambda$ are identical
(417C(b-i), 417E(b-i)), so
there is no new work to do in identifying the measure algebra of
$\tilde\lambda$ and the associated function spaces.

An obstacle we face in 417C-417E is the fact that {\it not} every
$\tau$-additive measure $\mu$ has an extension to a $\tau$-additive
topological measure, even when $\mu$ is totally finite and its domain
includes a base for the topology.   (I give an example in 419J.)
Consequently it is not enough, in 417C or 417E, to show that the
ordinary product measure $\lambda$ is $\tau$-additive.   But perhaps I
should remark that if $\lambda$ is inner regular with respect to the
closed sets, this obstacle evaporates (415L).   Accordingly, for the
principal applications (to quasi-Radon and Radon product measures, and
in particular whenever the topological spaces involved are regular) we
have rather easier proofs available, based on the constructions of
\S415.   For completely regular spaces, there is yet another approach,
because the product measures can be described in terms of the integrals
of continuous functions (415I), which by 417U and 417V can be
calculated from the ordinary product measures.   Of course
the proof that $\lambda$ itself is $\tau$-additive is by no means
trivial, especially in the case of infinite products, corresponding to
417E;  but for finite products there are relatively direct arguments,
applying indeed to slightly more general situations (417Yc).   If we
have measures which are inner regular with respect to countably compact
classes of sets, then there may be other ways of approaching the
extension, using theorems from \S413
(see 417Ye-417Yg), %417Ye 417Yf 417Yg
and for compact Radon measure spaces, $\lambda$ becomes tight (412Sb,
412V), so its $\tau$-additivity is elementary.

The arguments of 417C and 417E depend on specific constructions of the
new product measures $\tilde\lambda$.   It is therefore important to develop
descriptions in respect of which they may be considered canonical, as in
417Ca and 417Ea.   With these in hand, we can reasonably expect
`commutative' and `associative' and
`distributive' laws, as in 417Db, 417J and 417Xe.   Subspaces mostly
behave themselves (417I, 417Xf).   If you prefer to restrict your measures to
Borel algebras, you again get canonical product Borel measures (417Xc-417Xd).

Of course extending the product measure means that we get new integrable
functions on the product, so that Fubini's theorem has to be
renegotiated.   Happily, it remains valid, at least in the contexts in
which it was effective before (417Ga);   we still need, in effect, one
of the measures to be $\sigma$-finite.   The theorem still fails for
arbitrary integrable functions on products of
Radon measure spaces, and the same example works as before (417Xi).
In fact this means that we have an
alternative route to the construction of the $\tau$-additive product of
two measures (417Yb).   But note that on this route `commutativity', the
identification of the product measure on $X\times Y$ with that on
$Y\times X$, becomes something which can no longer be taken for granted,
because if we {\it define} $\tilde\lambda W$ to be $\int\nu W[\{x\}]dx$
we have to worry about when, and why, this will be equal to
$\int\mu W^{-1}[\{y\}]dy$.

A version of Tonelli's theorem follows from Fubini's theorem, as before
(417Gc).   We also have results corresponding to most of the theorems
of \S254.   But note that there are two traps.   In the theorem that a
measurable set can be described in terms of a projection onto a
countable subproduct (254O, 417M) we need to suppose that the factor
measures are strictly positive, and in the theorem that a product of
Radon measures is a Radon measure (417Q) we need to suppose that the
factor measures have compact supports.   The basic examples to note in
this context are 417Xp and 417Xq.

It is not well understood when we can expect c.l.d.\ product measures to
be topological measures, even in the case of compact Radon probability
spaces.   Example 419E remains a rather special case, but of course much
more effort has gone into seeking positive results.   Note that the
ordinary product measures of this section are always
effectively locally finite and $\tau$-additive (417C, 417E), so that
they will be equal to the $\tau$-additive products iff they measure
every open set (417S).   Regarding infinite products, the
$\tau$-additive product
measure can fail to be the ordinary product measure in just two ways:
if one of the {\it finite} product measures is not a topological
measure, or if uncountably many of the factor measures are not strictly
positive (417Sc, 417Xp, 417Yd).   So it is finite products which need
to be studied.

Whenever we have a subset $F$ of an infinite product
$X=\prod_{i\in I}X_i$, it is important to know when $F$ is determined by
coordinates in a proper subset of $I$;  in measure theory, we are
particularly interested in sets determined by coordinates in countable
subsets of $I$ (254Mb).   It may happen that there is a {\it smallest}
set $J$ such that $F$ is determined by coordinates in $J$;  for
instance, when we have a topological product and $F$ is closed (4A2Bg).
When we have a product of probability spaces, we sometimes wish to
identify sets $J$ such that $F$ is `essentially' determined by
coordinates in $J$, in the sense that there is an $F'$, determined
by coordinates in $J$, such that $F\symmdiff F'$ is negligible.   In
this context, again, there is a smallest such set (254Rd), which can be
identified in terms of the probability algebra free product of the
measure algebras (325Mb).   In 417Ma the two ideas come together:  under
the conditions there, we get the same smallest $J$ by either route.

In 417Ma, we have a product of strictly positive $\tau$-additive
topological probability measures.   If we keep the `strictly positive'
but abandon everything else, we still have very striking results just
because the product topology is ccc, so that we can apply 4A2Eb.   An
abstract expression of this idea is in 417Xt.
}%end of notes

\discrpage

\frfilename{mt418.tex}
\versiondate{30.11.20/17.9.21}
\copyrightdate{2000}

\def\undphi{\underline{\phi}}
%\def\qch{check\query}

\def\chaptername{Topologies and measures I}
\def\sectionname{Measurable functions and almost continuous functions}

\newsection{418}

In this section I work through the basic properties of measurable and
almost continuous functions, as defined in 411L and 411M.   I give the
results in
the full generality allowed by the terminology so far introduced, but
most of the ideas are already required even if you are interested only
in Radon measure spaces as the domains of the functions involved.
Concerning the
codomains, however, there is a great difference between metrizable
spaces and others, and among metrizable spaces separability is of
essential importance.

I start with the elementary properties of measurable functions
(418A-418C) and almost continuous functions (418D).
Under mild conditions on
the domain space, almost continuous functions are measurable (418E);
for a separable metrizable codomain, we can expect that measurable
functions should be almost continuous (418J).   Before coming to this, I
spend a couple of paragraphs on image
measures:  a locally finite image
measure under a measurable function is Radon if the measure
on the domain is Radon and the function is almost continuous (418I).

418L-418Q are important results on expressing given Radon measures as
image measures associated with continuous functions, first dealing with
ordinary functions $f:X\to Y$ (418L) and then coming to Prokhorov's
theorem on projective limits of probability spaces (418M).

The machinery of the first part of the section can also be used to
investigate representations of vector-valued functions in terms of
product spaces (418R-418T).

\leader{418A}{Proposition} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra
of subsets of $X$, $Y$ a
topological space and $f:X\to Y$ a measurable function.

(a) $f^{-1}[F]\in\Sigma$ for every Borel set $F\subseteq Y$.

(b) If $A\subseteq X$ is any set, endowed with the subspace
$\sigma$-algebra, then $f\restr A:A\to Y$ is measurable.

(c) Let $(Z,\frak T)$ be another topological space.   Then $gf:X\to Z$
is measurable for every Borel measurable function $g:Y\to Z$;  in
particular, for every continuous function $g:Y\to Z$.

\proof{{\bf (a)} The set $\{F:F\subseteq Y,\,f^{-1}[F]\in\Sigma\}$ is a
$\sigma$-algebra of subsets of $Y$ containing every open set, so
contains every Borel subset of $Y$.

\medskip

{\bf (b)} is obvious from the definition of `subspace $\sigma$-algebra'
(121A).

\medskip

{\bf (c)} If $H\subseteq Z$ is open, then $g^{-1}[H]$ is a Borel subset
of $Y$ so $(gf)^{-1}[H]=f^{-1}[g^{-1}[H]]$ belongs to $\Sigma$.
}%end of proof of 418A

\leader{418B}{Proposition} Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.

(a) If $Y$ is a metrizable space and $\sequencen{f_n}$ is a
sequence of measurable functions from $X$ to $Y$ such that
$f(x)=\lim_{n\to\infty}f_n(x)$ is defined in $Y$ for every $x\in X$,
then $f:X\to Y$ is measurable.

(b) If $Y$ is a topological space, $Z$ is a
separable metrizable space and $f:X\to Y$, $g:X\to Z$ are
functions, then $x\mapsto (f(x),g(x)):X\to Y\times Z$ is measurable iff
$f$ and $g$ are measurable.

(c) If $Y$ is a hereditarily Lindel\"of space, $\Cal U$ a
family of open sets generating its topology, and $f:X\to Y$ a
function such that $f^{-1}[U]\in\Sigma$ for every $U\in\Cal U$, then $f$
is measurable.

(d) If $\familyiI{Y_i}$ is a countable family of separable metrizable
spaces, with product $Y$, then a function $f:X\to Y$ is measurable iff
$\pi_if:X\to Y_i$ is measurable for every $i$, writing $\pi_i(y)=y(i)$
for $y\in Y$ and $i\in\Bbb N$.

\proof{{\bf (a)} Let $\rho$ be a metric defining the topology of $Y$.
Let $G\subseteq Y$ be any open set, and for each $n\in\Bbb N$ set

\Centerline{$F_n=\{y:y\in Y,\,\rho(y,z)\ge 2^{-n}$ for every $z\in
Y\setminus G\}$.}

\noindent Then $F_n$ is closed, so $f_i^{-1}[F_n]\in\Sigma$ for every
$n$, $i\in\Bbb N$.   But this implies that

\Centerline{$f^{-1}[G]
=\bigcup_{n\in\Bbb N}\bigcap_{i\ge n}f_i^{-1}[F_i]\in\Sigma$.}

\noindent As $G$ is arbitrary, $f$ is measurable.

\medskip

{\bf (b)(i)} The functions $(y,z)\mapsto y$, $(y,z)\mapsto z$ are
continuous, so if $x\mapsto(f(x),g(x))$ is measurable, so are $f$ and
$g$, by 418Ac.

\medskip

\quad{\bf (ii)} Now suppose that $f$ and $g$ are measurable, and that
$W\subseteq Y\times Z$ is open.   By 4A2P(a-i), the topology of $Z$ has
a countable base $\Cal H$;  let $\sequencen{H_n}$ be a sequence running
over $\Cal H\cup\{\emptyset\}$.   For each $n$, set

\Centerline{$G_n=\bigcup\{G:G\subseteq Y$ is open, $G\times H_n\subseteq
W\}$;}

\noindent then $G_n$ is open and $G_n\times H_n\subseteq W$.
Accordingly
$W\supseteq\bigcup_{n\in\Bbb N}G_n\times H_n$.   But in fact
$W=\bigcup_{n\in\Bbb N}G_n\times H_n$.   \Prf\ If $(y,z)\in W$, there
are open sets $G\subseteq Y$, $H\subseteq Z$ such that
$(y,z)\in G\times H\subseteq W$.   Now there is an $n\in\Bbb N$ such
that $z\in H_n\subseteq H$, in which case $G\times H_n\subseteq W$ and
$G\subseteq G_n$ and $(y,z)\in G_n\times H_n$.\ \QeD\

Accordingly

\Centerline{$\{x:(f(x),g(x))\in W\}
=\bigcup_{n\in\Bbb N}f^{-1}[G_n]\cap g^{-1}[H_n]\in\Sigma$.}

\noindent As $W$ is arbitrary, $x\mapsto (f(x),g(x))$ is measurable.

\medskip

{\bf (c)} This is just 4A3Db.

\medskip

{\bf (d)} If $f$ is measurable, so is every $\pi_if$, by 418Ac.   If
every $\pi_if$ is measurable, set

\Centerline{$\Cal U=\{\pi_i^{-1}[H]:i\in I,\,H\subseteq Y_i$ is
open$\}$.}

\noindent Then $\Cal U$ generates the topology of $Y$, and if
$U=\pi_i^{-1}[H]$ then $f^{-1}[U]=(\pi_if)^{-1}[H]$, so
$f^{-1}[U]\in\Sigma$ for every $U$.   Also $Y$ is hereditarily
Lindel\"of (4A2P(a-iii)), so $f$ is measurable, by (c).
}%end of proof of 418B

\leader{418C}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$Y$ a Polish space.   Let $\sequencen{f_n}$ be a
sequence of measurable functions from $X$ to $Y$.   Then

\Centerline{$\{x:x\in X,\,\lim_{n\to\infty}f_n(x)$ is defined in $Y\}$}

\noindent belongs to $\Sigma$.

\proof{(Compare 121H.) Let $\rho$ be a complete metric on $Y$ defining
the topology of $Y$.

\medskip

{\bf (a)} For $m$, $n\in\Bbb N$ and $\delta>0$, the set
$\{x:\rho(f_m(x),f_n(x))\le\delta\}$ belongs to $\Sigma$.   \Prf\ The
function $x\mapsto (f_m(x),f_n(x)):X\to Y^2$ is measurable, by 418Bb,
and the function $\rho:Y^2\to\Bbb R$ is continuous, so
$x\mapsto\rho(f_m(x),f_n(x))$ is measurable and
$\{x:\rho(f_m(x),f_n(x))\le\delta\}\in\Sigma$.\ \Qed

\medskip
{\bf (b)} Now $\sequencen{f_n(x)}$ is convergent iff it is Cauchy,
because
$Y$ is complete.   But

$$\eqalign{\{x:x\in X,\,&\sequencen{f_n(x)}\text{ is Cauchy}\}
=\bigcap_{n\in\Bbb N}\bigcup_{m\in\Bbb N}\bigcap_{i\ge m}
\{x:\rho(f_i(x),f_m(x))\le 2^{-n}\}\cr}$$

\noindent belongs to $\Sigma$.
}%end of proof of 418C

\leader{418D}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space and
$\frak T$ a topology on $X$.

(a) Suppose that $Y$ is a topological space.   Then any continuous
function from $X$ to $Y$ is almost continuous.

(b) Suppose that $Y$ and $Z$ are topological spaces, $f:X\to Y$ is
almost continuous and $g:Y\to Z$ is continuous.   Then $gf:X\to Z$ is
almost continuous.

(c) Suppose that $(Y,\frak S,\Tau,\nu)$ is a $\sigma$-finite topological
measure space, $Z$ is a topological space, $g:Y\to Z$ is almost
continuous and $f:X\to Y$ is \imp\ and almost continuous.   Then
$gf:X\to Z$ is almost continuous.

(d) Suppose that $\mu$ is semi-finite, and that $\familyiI{Y_i}$ is a
countable family of topological spaces with product $Y$.   Then a
function $f:X\to Y$
is almost continuous iff $f_i=\pi_if$ is almost continuous for every
$i\in I$, writing $\pi_i(y)=y(i)$ for $i\in I$ and $y\in Y$.

\proof{{\bf (a)} is trivial.

\medskip

{\bf (b)} The set $\{A:A\subseteq X,\,gf\restr A$ is continuous$\}$
includes
$\{A:A\subseteq X,\,f\restr A$ is continuous$\}$;  so if $\mu$ is inner
regular with respect to the latter, it is inner regular with respect to
the former.

\medskip

{\bf (c)} Take $E\in\Sigma$, $\gamma<\mu E$ and $\epsilon>0$.   We
have a cover of $Y$ by a non-decreasing sequence $\sequencen{Y_n}$ of
measurable sets of finite measure;  now $\sequencen{f^{-1}[Y_n]}$ is a
non-decreasing sequence covering $E$, so there is an $n\in\Bbb N$ such
that $\mu(E\cap f^{-1}[Y_n])\ge\gamma$.   Because $f$ is \imp, $E\cap
f^{-1}[Y_n]$ has finite measure.   Now we can find measurable sets
$F\subseteq Y_n$, $E_1\subseteq E\cap f^{-1}[Y_n]$ such that $f\restr
E_1$, $g\restr F$ are continuous and
$\nu F\ge\nu Y_n-\epsilon$, $\mu E_1\ge\mu(E\cap f^{-1}[Y_n])-\epsilon$.
In this case $E_0=E_1\cap f^{-1}[F]$ has measure at
least $\gamma-2\epsilon$ and $gf\restr E_0$ is continuous.   As $E$,
$\gamma$ and $\epsilon$ are arbitrary, $gf$ is almost continuous.

\medskip

{\bf (d)(i)} If $f$ is almost continuous, every $f_i$ must be almost
continuous, by (b).

\medskip

\quad{\bf (ii)} Now suppose that every $f_i$ is almost continuous.
Take $E\in\Sigma$ and $\gamma<\mu E$.   There is an
$E_0\subseteq E$ such that $E_0\in\Sigma$ and $\gamma<\mu E_0<\infty$.
Let $\langle\epsilon_i\rangle_{i\in I}$ be a family of strictly positive
real numbers such that $\sum_{i\in I}\epsilon_i\le\mu E_0-\gamma$.   For
each $i\in I$ choose a measurable set $F_i\subseteq E_0$ such that
$\mu F_i\ge\mu E_0-\epsilon_i$ and $f_i\restr F_i$ is continuous.   Then
$F=E_0\cap\bigcap_{i\in I}F_i$ is a subset of $E$ with measure at least
$\gamma$, and $f\restr F$ is continuous because $f_i\restr F$ is
continuous for every $i$ (3A3Ib).
}%end of proof of 418D

\leader{418E}{Theorem} Let $(X,\frak T,\Sigma,\mu)$ be a complete
locally determined topological measure space, $Y$ a topological space,
and $f:X\to Y$ an almost continuous function.   Then $f$ is measurable.

\proof{ Set $\Cal K=\{K:K\in\Sigma,\,f\restr K$ is continuous$\}$;  then
$\mu$ is inner regular with respect to $\Cal K$.   If $H\subseteq Y$ is
open and $K\in\Cal K$, then $K\cap f^{-1}[H]$ is relatively open in $K$,
that is, there is an open set $G\subseteq X$ such that
$K\cap f^{-1}[H]=K\cap G$.
Because $\mu$ is a topological measure, $G\in\Sigma$ so
$K\cap f^{-1}[H]\in\Sigma$.   As $K$ is arbitrary, and $\mu$ is complete
and locally determined, $f^{-1}[H]\in\Sigma$ (412Ja).   As $H$ is
arbitrary, $f$ is measurable.
}%end of proof of 418E

\leader{418F}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a
semi-finite topological measure space, $Y$ a metrizable space, and
$f:X\to Y$ a function.   Suppose there is a sequence $\sequencen{f_n}$
of almost continuous functions from $X$ to $Y$ such that
$f(x)=\lim_{n\to\infty}f_n(x)$ for almost every $x\in X$.   Then $f$ is
almost continuous.

\proof{ Suppose that $E\in\Sigma$, $\gamma<\mu E$ and $\epsilon>0$.
Then there is a measurable set $F\subseteq E$ such that
$\gamma\le\mu F<\infty$;  discarding a negligible set if necessary, we
may arrange that $f(x)=\lim_{n\to\infty}f_n(x)$ for every $x\in F$.
Let $\rho$ be a metric on $Y$ defining its topology.   For each
$n\in\Bbb N$, let $F_n\subseteq F$ be a measurable set such that
$f_n\restr F_n$ is continuous and
$\mu(F_n\setminus F)\le 2^{-n}\epsilon$;  set
$G=\bigcap_{n\in\Bbb N}F_n$, so that $\mu G\ge\gamma-2\epsilon$ and
$f_n\restr G$ is continuous for every $n\in\Bbb N$.

For $m$, $n\in\Bbb N$, the functions $x\mapsto(f_m(x),f_n(x)):G\to Y^2$
and $x\mapsto\rho(f_m(x),f_n(x)):G\to\Bbb R$ are continuous, therefore
measurable, because $\mu$ is a topological measure.   Also
$\sequencen{f_n(x)}$ is a Cauchy sequence for every $x\in G$.   So if we
set $G_{kn}
=\{x:x\in G,\,\rho(f_i(x),f_j(x))\le 2^{-k}$ for all $i$, $j\ge n\}$,
$\sequencen{G_{kn}}$ is a non-decreasing sequence of measurable sets
with union $G$ for each $k\in\Bbb N$, and we can find a strictly
increasing sequence $\sequence{k}{n_k}$ such that $\mu(G\setminus
G_{kn_k})\le 2^{-k}\epsilon$ for every $k$.   Setting
$H=\bigcap_{k\in\Bbb N}G_{kn_k}$,
$\mu H\ge\mu G-2\epsilon\ge\gamma-4\epsilon$ and
$\rho(f_i(x),f_{n_k}(x))\le 2^{-k}$ whenever $x\in H$ and $i\ge n_k$;
consequently $\rho(f(x),f_{n_k}(x))\le 2^{-k}$ whenever $x\in H$ and
$k\in\Bbb N$.   But this means that $\sequence{k}{f_{n_k}}$ converges to
$f$ uniformly on $H$, while every $f_{n_k}$ is continuous on $H$, so
$f\restr H$ is continuous (3A3Nb).   And of course $H\subseteq E$.

As $E$, $\gamma$ and $\epsilon$ are arbitrary, $f$ is almost continuous.
}%end of proof of 418F
%4@43

\leader{418V}{Proposition}\dvAnew{2021} Let $(X,\Sigma,\mu)$ be a
$\sigma$-finite measure space,
$\frak T$ a topology on $X$ such that $\mu$ is inner regular with respect to the
Borel sets, $(Y,\frak S)$ a topological space and $f:X\to Y$
an almost continuous function.   Then
there is a Borel measurable function $g:X\to Y$ which is equal almost
everywhere to $f$.

%out of order query

\proof{ If $Y$ is empty, so is $X$, and the result is trivial;
so suppose that we have a point $y_0\in Y$.   Let $\sequencen{X_n}$ be a
non-decreasing sequence of subsets of $X$ of finite
measure covering $X$.   For each $n\in\Bbb N$, there is a measurable set
$E_n\subseteq X_n$ such that $\mu E_n\ge\mu X_n-2^{-n}$ and $f\restr E_n$ is
continuous, and a measurable Borel set $E'_n\subseteq E_n$ such that
$\mu E'_n\ge\mu E_n-2^{-n}$ and $E'_n$ is Borel.   Now
$E=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}E'_n$ is a coneglible Borel set.
Set $g(x)=f(x)$ for $x\in E$, $y_0$ for $x\in X\setminus E$;  then $g\eae f$.
If $G\subseteq Y$ is open and $n\in\Bbb N$,
then $E_n\cap f^{-1}[G]$ is relatively open in
$E_n$ so $E'_n\cap f^{-1}[G]$ is a Borel subset of $X$.   Accordingly
$E\cap f^{-1}[G]=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}E'_n\cap f^{-1}[G]$ is a
Borel subset of $X$.   But now $g^{-1}[G]$ is either $E\cap f^{-1}[G]$ or
$(E\cap f^{-1}[G])\cup(X\setminus E)$, and in either case is a Borel set.
As $G$ is arbitrary, $g$ is Borel measurable.
}%end of proof of 418V

\leader{418G}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a
$\sigma$-finite quasi-Radon measure space, $Y$ a metrizable space and
$f:X\to Y$ an almost continuous function.   Then there is a conegligible
set $X_0\subseteq X$ such that $f[X_0]$ is separable.

\proof{{\bf (a)} Let $\Cal K$ be the family of self-supporting
measurable sets $K$ of finite measure such that $f\restr K$ is
continuous.   Then $\mu$ is inner regular with respect to $\Cal K$.
\Prf\ If $E\in\Sigma$ and $\gamma<\mu E$, there is an $F\in\Sigma$ such
that $F\subseteq E$ and $\gamma<\mu F<\infty$;  there is an $H\in\Sigma$
such that $H\subseteq F$, $\gamma\le\mu H$ and $f\restr H$ is
continuous;  and there is a measurable self-supporting $K\subseteq H$
with the same measure as $H$ (414F), in which case $K\in\Cal K$ and
$K\subseteq E$ and $\mu K\ge\gamma$.\ \Qed

\medskip

{\bf (b)} Now $f[K]$ is ccc for every $K\in\Cal K$.   \Prf\ If $\Cal G$
is a disjoint family of non-empty relatively open subsets of $f[K]$,
then $\family{G}{\Cal G}{K\cap f^{-1}[G]}$ is a disjoint family of
non-empty relatively open subsets of $K$, because $f\restr K$ is
continuous, and $\sum_{G\in\Cal G}\mu(K\cap f^{-1}[G])\le\mu K$.
Because $K$ is self-supporting, $\mu(K\cap f^{-1}[G])>0$ for every
$G\in\Cal G$;  because $\mu K$ is finite, $\Cal G$ is countable.   As
$\Cal G$ is arbitrary, $f[K]$ is ccc.\ \Qed

Because $Y$ is metrizable, $f[K]$ must be separable (4A2Pd).

\medskip

{\bf (c)} Because $\mu$ is $\sigma$-finite, there is a countable family
$\Cal L\subseteq\Cal K$ such that $X_0=\bigcup\Cal L$ is conegligible
(412Ic).   Now $f[X_0]=\bigcup_{L\in\Cal L}f[L]$ is a countable union of
separable spaces, so is separable (4A2B(e-i)).
}%end of proof of 418G

\leader{418H}{Proposition} (a) Let $X$ and $Y$ be topological spaces,
$\mu$ an effectively locally finite $\tau$-additive
topological measure on $X$, and $f:X\to Y$ an almost continuous
function.   Then the image measure $\mu f^{-1}$ is $\tau$-additive.

(b) Let $(X,\frak T,\Sigma,\mu)$ be a totally
finite quasi-Radon measure space, $(Y,\frak S)$ a regular topological
space, and $f:X\to Y$ an almost continuous function.   Then there is a
unique quasi-Radon measure $\nu$ on $Y$ such that $f$ is \imp\ for $\mu$
and $\nu$.

\proof{{\bf (a)} Let $\Cal H$ be an upwards-directed family of open
subsets of $Y$, all measured by $\mu f^{-1}$, and suppose that
$H^*=\bigcup\Cal H$ also is measurable.
Take any $\gamma<(\mu f^{-1})(H^*)=\mu f^{-1}[H^*]$.   Then there is a
measurable set $E\subseteq f^{-1}[H^*]$ such that $\mu E\ge\gamma$ and
$f\restr E$ is continuous.
Consider $\{E\cap f^{-1}[H]:H\in\Cal H\}$.   This is an upwards-directed
family of relatively open measurable subsets of $E$ with measurable
union $E$.   By 414K,  %needs `topological measure'
the subspace measure on $E$ is $\tau$-additive, so

\Centerline{$\gamma\le\mu E
\le\sup_{H\in\Cal H}\mu(E\cap f^{-1}[H])
\le\sup_{H\in\Cal H}\mu f^{-1}[H]$.}

\noindent As $\gamma$ is arbitrary,
$\mu f^{-1}[H^*]\le\sup_{H\in\Cal H}\mu f^{-1}[H]$;  as $\Cal H$ is
arbitrary, $\mu f^{-1}$ is $\tau$-additive.

\medskip

{\bf (b)} By 418E, $f$ is measurable.   Let $\nu_0$ be the restriction
of $\mu f^{-1}$ to the Borel $\sigma$-algebra of $Y$;  by (a), $\nu_0$
is $\tau$-additive, and $f$ is \imp\ with respect to $\mu$ and $\nu_0$.
Because $Y$ is regular, the completion $\nu$ of $\nu_0$ is a quasi-Radon
measure (415Cb).   Because $\mu$ is complete, $f$ is still \imp\ with
respect to $\mu$ and $\nu$ (234Ba).

To see that $\nu$ is unique, observe that its values on Borel sets are
determined by the requirement that $f$ be \imp, so that 415H gives the
result.
}%end of proof of 418H

\leader{418I}{}\cmmnt{ The next theorem is one of the central
properties of Radon measures.   I have already presented what amounts to
a special case in 256G.

\medskip

\noindent}{\bf Theorem} Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure
space, $Y$ a Hausdorff space, and $f:X\to Y$ an almost continuous
function.   If the image measure $\nu=\mu f^{-1}$ is locally finite, it
is a Radon measure.

\proof{{\bf (a)} By 418E, $f$ is measurable, that is,
$f^{-1}[H]\in\Sigma$
for every open set $H\subseteq Y$;  but this means that the domain
$\Tau$ of $\nu$ contains every open set, and $\nu$ is a topological
measure.

\medskip

{\bf (b)} $\nu$ is inner regular with respect to the
compact sets.
\Prf\ If $F\in\Tau$ and $\nu F>0$, then $\mu f^{-1}[F]>0$, so there is
an $E\subseteq f^{-1}[F]$ such that $\mu E>0$ and $f\restr E$ is
continuous.
Next, there is a compact set $K\subseteq E$ such that $\mu K>0$.   In
this case, $L=f[K]$ is a compact subset of $F$, and

\Centerline{$\nu L=\mu f^{-1}[L]\ge\mu K>0$.}

\noindent As $Y$ is Hausdorff, this is enough to prove that $\nu$ is tight.\
\QeD\   Note that because $\nu$ is
locally finite, $\nu L<\infty$ for every compact $L\subseteq Y$ (411Ga).

\medskip

{\bf (c)} Because $\mu$ is complete, so is $\nu$ (234Eb).
Next, $\nu$ is locally
determined.   \Prf\ Suppose that $H\subseteq Y$ is such that
$H\cap F\in\Tau$ whenever $\nu F<\infty$.   Then, in particular,
$H\cap f[K]\in\Tau$ whenever $K\subseteq X$ is compact and $f\restr K$
is continuous.   But setting

\Centerline{$\Cal K=\{K:K\subseteq X$ is compact, $f\restr K$ is
continuous$\}$,}

\noindent $\mu$ is inner regular with respect to $\Cal K$ (412Ac).
And if $K\in\Cal K$,

\Centerline{$K\cap f^{-1}[H]=K\cap f^{-1}[H\cap f[K]]\in\Sigma$.}

\noindent Because $\mu$ is complete and locally determined, this is
enough to show that $f^{-1}[H]\in\Sigma$ (412Ja again), that is, $H\in\Tau$.
As $H$ is arbitrary, $\nu$ is locally determined.\ \Qed

\medskip

{\bf (d)} Thus $\nu$ is a complete locally determined locally
finite topological measure which is inner regular with respect to the
compact sets;  that is, it is a Radon measure.
}%end of proof of 418I

\leader{418J}{Theorem} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space and $\frak T$ a topology on $X$ such that $\mu$ is inner regular
with respect to the closed sets.   Suppose that $Y$ is a
second-countable space\cmmnt{ (for instance, $Y$ might be separable and
metrizable),} and $f:X\to Y$ is measurable.
Then $f$ is almost continuous.

\proof{ Let $\Cal H$ be a countable base for the topology of $Y$, and
$\sequencen{H_n}$ a sequence running over $\Cal H\cup\{\emptyset\}$.
Take $E\in\Sigma$ and $\gamma<\mu E$.   Choose $\sequencen{E_n}$
inductively, as
follows.   There is an $E_0\in\Sigma$ such that $E_0\subseteq E$ and
$\gamma<\mu E_0<\infty$.   Given $E_n\in\Sigma$ with
$\gamma<\mu E_n<\infty$, $E_n\setminus f^{-1}[H_n]\in\Sigma$,
so there is a closed set $F_n\in\Sigma$ such that

\Centerline{$F_n\subseteq E_n\setminus f^{-1}[H_n]$,
\quad$\mu((E_n\setminus f^{-1}[H_n])\setminus F_n)<\mu E_n-\gamma$;}

\noindent set $E_{n+1}=(E_n\cap f^{-1}[H_n])\cup F_n$, so that

\Centerline{$E_{n+1}\in\Sigma$,
\quad $E_{n+1}\subseteq E_n$,
\quad$\mu E_{n+1}>\gamma$,
\quad$E_{n+1}\setminus f^{-1}[H_n]=F_n$.}

\noindent Continue.

At the end of the induction, set $F=\bigcap_{n\in\Bbb N}E_n$.   Then
$F\subseteq E$, $\mu F\ge\gamma$, and for every $n\in\Bbb N$

\Centerline{$F\cap f^{-1}[H_n]=F\cap E_{n+1}\cap f^{-1}[H_n]
=F\cap E_{n+1}\setminus F_n=F\setminus F_n$}

\noindent is relatively open in $F$.   It follows that $f\restr F$ is
continuous (4A2B(a-ii)).   As $E$, $\gamma$ are arbitrary, $f$ is almost
continuous.
}%end of proof of 418J

\cmmnt{\medskip

\noindent{\bf Remark} For variations on this idea, see 418Yf, 433E and
434Yb;  also 418Yg.
}

\leader{418K}{Corollary} Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon
measure space and $Y$ a separable metrizable space.   Then a function
$f:X\to Y$ is measurable iff it is almost continuous.

\proof{ Put 418E and 418J together.
}%end of proof of 418K

\cmmnt{\medskip

\noindent{\bf Remark} This generalizes 256F.
}%end of comment

\leader{418L}{}\cmmnt{ In all the results above, the measure starts on
the left of the diagram $f:X\to Y$;  in 418H-418I,
it is transferred to an image measure on
$Y$.   If $X$ has enough compact sets, a measure can move in the reverse
direction, as follows.

\medskip

\noindent}{\bf Theorem} Let $(X,\frak T)$ be a Hausdorff space,
$(Y,\frak S,\Tau,\nu)$ a Radon measure space and $f:X\to Y$ a continuous
function such that whenever $F\in\Tau$ and $\nu F>0$ there is a compact
set $K\subseteq X$
such that $\nu(F\cap f[K])>0$.   Then there is a Radon measure $\mu$ on
$X$ such that $\nu$ is the image measure $\mu f^{-1}$ and the \imp\
function $f$ induces an isomorphism between the measure algebras of
$\nu$ and $\mu$.
%would be enough if for every compact L\subseteq Y with \nu L>0 there
%was a compact  K\subseteq X  with  \nu(L\cap f[K])>0
%and presumably enough if  f\restr K  is continuous for every compact
%K\subseteq X

\proof{{\bf (a)} Note first that $\nu$ is inner regular with respect to
$\Cal L=\{f[K]:K\in\Cal K\}$, where $\Cal K$ is the family of compact
subsets of $X$.   \Prf\ If $\nu F>0$, there is a $K\in\Cal K$ such that
$\nu(F\cap f[K])>0$;  now there is a closed set $F'\subseteq F\cap f[K]$
such that $\nu F'>0$, and $K'=K\cap f^{-1}[F']$ is compact, while
$f[K']\subseteq F$ has non-zero measure.   As $\Cal L$ is closed under
finite unions, this is enough to show that $\nu$ is inner regular with
respect to $\Cal L$ (412Aa).\ \Qed

\medskip

{\bf (b)} Consequently there is a disjoint set $\Cal L_0\subseteq\Cal L$
such that every non-negligible $F\in\Tau$ meets some member of $\Cal L_0$
in a non-negligible set (412Ib).   We can express $\Cal L_0$ as
$\{f[K]:K\in\Cal K_0\}$ where $\Cal K_0\subseteq\Cal K$ is disjoint.
Set $X_0=\bigcup\Cal K_0$.

\medskip

{\bf (c)} Set

\Centerline{$\Sigma_0=\{X_0\cap f^{-1}[F]:F\in\Tau\}$.}

\noindent Then $\Sigma_0$ is a $\sigma$-algebra of subsets of $X_0$.
If $F$, $F'\in\Tau$ and $\nu F\ne\nu F'$, then there must be some
$K\in\Cal
K_0$ such that $f[K]\cap(F\symmdiff F')\ne\emptyset$, so that
$X_0\cap f^{-1}[F]\ne X_0\cap f^{-1}[F']$;  we therefore have a
functional
$\mu_0:\Sigma_0\to[0,\infty]$ defined by setting
$\mu_0(X_0\cap f^{-1}[F])=\nu F$ whenever $F\in\Tau$.   It is easy to
check that $\mu_0$ is a measure on $X_0$.   Now $\mu_0$ is inner regular
with respect to $\Cal K$.   \Prf\ If $E\in\Sigma_0$ and $\mu E>0$, there
is an $F\in\Tau$ such that $E=X_0\cap f^{-1}[F]$ and $\nu F>0$.   There
are a $K\in\Cal K_0$ such that $\nu(F\cap f[K])>0$, and a closed set
$F'\subseteq F\cap f[K]$ such that $\nu F'>0$;  now
$K\cap f^{-1}[F']=X_0\cap f^{-1}[F']$ belongs to $\Sigma_0\cap\Cal K$,
is included in $E$ and has measure greater than $0$.   Because $\Cal K$
is closed under finite unions, this is enough to show that $\mu_0$ is
inner regular with respect to $\Cal K$.\ \Qed

\medskip

{\bf (d)} Set

\Centerline{$\Sigma_1=\{E:E\subseteq X,\,E\cap X_0\in\Sigma_0\}$,
\quad$\mu_1E=\mu_0(E\cap X_0)$ for every $E\in\Sigma_1$.}

\noindent Then $\mu_1$ is a measure on $X$ (being the image measure
$\mu_0\iota^{-1}$, where $\iota:X_0\to X$ is the identity map), and is
inner regular with respect to $\Cal K$.   If $F\in\Tau$, then

\Centerline{$\mu_1 f^{-1}[F]=\mu_0(X_0\cap f^{-1}[F])=\nu F$,}

\noindent so $f$ is \imp\ for $\mu_1$ and $\nu$.   Consequently $\mu_1$
is locally finite.   \Prf\ If $x\in X$, there is an open set $H\subseteq Y$
such that $f(x)\in H$ and $\nu H<\infty$;  now $f^{-1}[H]$ is an open
subset of $X$ of finite measure containing $x$.\ \QeD\  In particular,
$\mu_1^*K<\infty$ for every compact $K\subseteq X$ (411Ga again).

\medskip

{\bf (e)} By 413P, there is an extension of $\mu_1$ to a complete
locally determined measure $\mu$ on $X$ which is inner regular with
respect to $\Cal K$, defined on every member of $\Cal K$, and such that
whenever $E$
belongs to the domain $\Sigma$ of $\mu$ and $\mu E<\infty$, there is an
$E_1\in\Sigma_1$ such that $\mu(E\symmdiff E_1)=0$.   Now $\mu$ is
locally finite because $\mu_1$ is, so $\mu$ is a Radon measure;  and $f$
is \imp\ for $\mu$ and $\nu$ because it is \imp\ for $\mu_1$ and $\nu$.

\medskip

{\bf (f)} The image measure $\mu f^{-1}$ extends $\nu$, so is locally
finite, and is therefore a Radon measure (418I);  since it agrees with
$\nu$ on the compact subsets of $Y$, it must be identical with $\nu$.

\medskip

{\bf (g)} I have still to check that the corresponding
measure-preserving
homomorphism $\pi$ from the measure algebra $\frak B$ of $\nu$ to the
measure algebra $\frak A$ of $\mu$ is actually an isomorphism, that is,
is surjective.   If $a\in\frak A$ and $\bar\mu a<\infty$, we can find
$E\in\Sigma$ such that $E^{\ssbullet}=a$ and $E_1\in\Sigma_1$ such that
$\mu(E\symmdiff E_1)=0$.   Now $E_1\cap X_0=f^{-1}[F]\cap X_0$ for some
$F\in\Tau$;  but in this case

\Centerline{$\mu(E_1\symmdiff f^{-1}[F])
=\mu_1(E_1\symmdiff f^{-1}[F])=0$,
\quad$a=E_1^{\ssbullet}=(f^{-1}[F])^{\ssbullet}=\pi F^{\ssbullet}$.}

Accordingly $\pi[\frak B]$ includes $\{a:\bar\mu a<\infty\}$, and is
order-dense in $\frak A$.   But as $\pi$ is injective and $\frak B$ is
Dedekind complete (being the measure algebra of a Radon measure, which
is strictly localizable), it follows that $\pi[\frak B]=\frak A$
(314Ia).   Thus $\pi$ is an isomorphism, as required.
}%end of proof of 418L

\cmmnt{\medskip

\noindent{\bf Remarks} Of course this result is most commonly applied
when $X$ and $Y$ are both compact and $f$ is a surjection, in which case
the condition

\inset{(*) whenever $F\in\Tau$ and $\nu F>0$ there is a compact set
$K\subseteq X$ such that $\nu(F\cap f[K])>0$}

\noindent is trivially satisfied.

Evidently (*) is necessary if there is to be any Radon measure on $X$
for which $f$ is \imp, so in this sense the result is best possible.
In 433D, however, there is a version of the theorem in which $f$ is not
required to be continuous.
}%end of comment

\leader{418M}{Prokhorov's theorem} Suppose that $(I,\le)$,
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$,
$\langle f_{ij}\rangle_{i\le j\in I}$,
$(X,\frak T)$ and $\familyiI{g_i}$ are such that

\inset{($\alpha$)
$(I,\le)$ is a non-empty upwards-directed partially ordered set,}

\inset{($\beta$) every $(X_i,\frak T_i,\Sigma_i,\mu_i)$ is a Radon probability
space,}

\inset{($\gamma$) $f_{ij}:X_j\to X_i$ is an \imp\ function whenever $i\le j$ in
$I$,}

\inset{($\delta$) $(X,\frak T)$ is a Hausdorff space,}

\inset{($\epsilon$) $g_i:X\to X_i$ is a continuous function for every $i\in I$,}

\inset{($\zeta$) $g_i=f_{ij}g_j$ whenever $i\le j$ in $I$.}

\inset{($\eta$) for every $\epsilon>0$ there is a compact set
$K\subseteq X$ such that $\mu_ig_i[K]\ge 1-\epsilon$ for every $i\in I$.}

\noindent Then there is a Radon probability measure $\mu$ on $X$ such
that every $g_i$ is \imp\ for $\mu$.   If moreover

\inset{($\theta$) the family $\familyiI{g_i}$
separates the points of $X$,}

\noindent then $\mu$ is uniquely defined.

\proof{{\bf (a)} Set

\Centerline{$\Tau=\{g_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}\subseteq\Cal PX$.}

\noindent Then $\Tau$ is a subalgebra of $\Cal PX$.   \Prf\ (i) There is
an $i\in I$, so $\emptyset=g_i^{-1}[\emptyset]$ belongs to $\Tau$.
(ii) If $H\in\Tau$ there are $i\in I$, $E\in\Sigma_i$ such that
$H=g_i^{-1}[E]$;
now $X\setminus H=g_i^{-1}[X_i\setminus E]$ belongs to $\Tau$.   (iii)
If $G$, $H\in\Tau$, there are $i$, $j\in I$ and $E\in\Sigma_i$,
$F\in\Sigma_j$
such that $G=g_i^{-1}[E]$ and $H=g_j^{-1}[F]$.   Now $I$ is
upwards-directed, so there is a $k\in I$ such that $i\le k$ and $j\le
k$.   Because $f_{ik}$ and $f_{jk}$ are \imp, $f_{ik}^{-1}[E]$ and
$f_{jk}^{-1}[F]$ belong to $\Sigma_k$, so that

$$\eqalign{G\cap H
&=g_i^{-1}[E]\cap g_j^{-1}[F]
=(f_{ik}g_k)^{-1}[E]\cap (f_{jk}g_k)^{-1}[F]\cr
&=g_k^{-1}[f_{ik}^{-1}[E]\cap f_{jk}^{-1}[F]]
\in\Tau.  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} There is an additive functional $\nu:\Tau\to[0,1]$ defined by
writing $\nu g_i^{-1}[E]=\mu_iE$ whenever $i\in I$ and $E\in\Sigma_i$.

\medskip

\Prf\ {\bf (i)} Suppose that $i$, $j\in I$ and $E\in\Sigma_i$,
$F\in\Sigma_j$ are such that
$g_i^{-1}[E]=g_j^{-1}[F]$.   Let $k\in I$ be such that $i\le k$ and
$j\le k$.   Then

\Centerline{$g_k^{-1}[f_{ik}^{-1}[E]\symmdiff f_{jk}^{-1}[F]]
=g_i^{-1}[E]\symmdiff g_j^{-1}[F]=\emptyset$,}

\noindent so $g_k[X]\cap(f_{ik}^{-1}[E]
\symmdiff f_{jk}^{-1}[F])=\emptyset$.   But now remember that for every
$\epsilon>0$ there is a set $K\subseteq X$ such that
$\mu_kg_k[K]\ge 1-\epsilon$.   This means that $\mu_kg_k[X]$ must be
$1$, so that $f_{ik}^{-1}[E]\symmdiff f_{jk}^{-1}[F]$ must be
negligible, and

\Centerline{$\mu_iE=\mu_kf_{ik}^{-1}[E]=\mu_kf_{jk}^{-1}[F]=\mu_jF$.}

\noindent Thus the proposed formula for $\nu$ defines a function on
$\Tau$.

\medskip

\quad{\bf (ii)} Now suppose that $G$, $H\in\Tau$ are disjoint.   Again,
take $i$, $j\in I$ and $E\in\Sigma_i$, $F\in\Sigma_j$ such that
$G=g_i^{-1}[E]$ and $H=g_j^{-1}[F]$, and $k\in I$ such that $i\le k$ and
$j\le k$.   Then

$$\eqalign{\nu G+\nu H
&=\mu_iE+\mu_jF
=\mu_kf_{ik}^{-1}[E]+\mu_kf_{jk}^{-1}[F]\cr
&=\mu_k(f_{ik}^{-1}[E]\cup f_{jk}^{-1}[F])
  +\mu_k(f_{ik}^{-1}[E]\cap f_{jk}^{-1}[F])\cr
&=\nu g_k^{-1}[f_{ik}^{-1}[E]\cup f_{jk}^{-1}[F]]
  +\nu g_k^{-1}[f_{ik}^{-1}[E]\cap f_{jk}^{-1}[F]]\cr
&=\nu(G\cup H)+\nu(G\cap H).\cr}$$

\noindent But as $\nu\emptyset$ is certainly $0$, we get
$\nu(G\cup H)=\nu G+\nu H$.   As $G$, $H$ are arbitrary, $\nu$ is
additive.\ \Qed

Note that $\nu X=1$.

\medskip

{\bf (c)} $\nu G=\sup\{\nu H:H\in\Tau$, $H\subseteq G$, $H$ is closed$\}$
for every $G\in\Tau$.   \Prf\ If
$\gamma<\nu G$, there are an $i\in I$ and an $E\in\Sigma_i$ such that
$G=g_i^{-1}[E]$.
In this case $\mu_iE=\nu G>\gamma$;  let $L\subseteq E$ be a compact set
such that
$\mu_iL\ge\gamma$;  then $H=g_i^{-1}[L]$ is a closed subset of $G$ and
$\nu H=\mu_iL\ge\gamma$.\ \Qed

$\nu X=\sup_{K\subseteq X\text{ is compact}}
  \inf_{G\in\Tau,G\supseteq K}\nu G$.   \Prf\
If $\epsilon>0$, there is a compact $K\subseteq X$ such that
$\mu_ig_i[K]\ge 1-\epsilon$
for every $i\in I$, by the final hypothesis of this theorem.   If
$G\in\Tau$ and $G\supseteq K$,
there are an $i\in I$ and an $E\in\Sigma_i$ such that $G=g_i^{-1}[E]$,
in which case
$g_i[K]\subseteq E$, so that

\Centerline{$\nu G=\mu_iE\ge\mu_ig_i[K]\ge 1-\epsilon$.}

\noindent Thus $\inf_{G\in\Tau,G\supseteq K}\nu G\ge 1-\epsilon$;  as
$\epsilon$ is arbitrary, we have the result.\ \Qed

This means that the conditions of 416O are satisfied, and there is a
Radon measure $\mu$ on $X$
extending $\nu$.   Of course this means that every $g_i$ is \imp.

\medskip

{\bf (d)} Now suppose that $\familyiI{g_i}$ separates the points of $X$.
Then whenever $K$, $L\subseteq X$ are disjoint there is an $i\in I$ such
that $g_i[K]\cap g_i[L]=\emptyset$.   \Prf\ Set
$V_i=\{(x,y):x\in K$, $y\in L$, $g_i(x)=g_i(y)\}$ for $i\in I$.
Because $g_i$ is continuous and $\frak T_i$ is Hausdorff, $V_i$ is closed.
If $i\le j$ in $I$, then $g_i=f_{ij}g_i$ so $V_j\subseteq V_i$;
accordingly $\familyiI{V_i}$ is downwards-directed.   Because
$\familyiI{g_i}$ separates the points of $X$, $\bigcap_{i\in I}V_i$ is
empty.   As $K\times L$ is compact, there is an $i\in I$ such that
$V_i=\emptyset$, that is, $g_i[K]$ and $g_i[L]$ are disjoint.\ \Qed

Let $\nu$ be any Radon probability
measure on $X$ such that $g_i$ is \imp\ for $\nu$ and $\mu_i$ for
every $i\in I$.   Let $K\subseteq X$ be compact.   \Quer\ If $\mu K<\nu K$
then there is a compact $L\subseteq X\setminus K$ such that
$\mu L+\nu K>1$.   Let $i\in I$ be such that $g_i[K]\cap g_i[L]=\emptyset$;
then

\Centerline{$1<\mu L+\nu K
\le\mu g_i^{-1}[g_i[L]]+\nu g_i^{-1}[g_i[K]]
=\mu_ig_i[L]+\mu_ig_i[K]
\le 1$,}

\noindent which is impossible.\ \BanG\   So $\nu K\le\mu K$.   Similarly,
$\mu K\le\nu K$.   By 416Eb, $\mu=\nu$.   Thus $\mu$ is uniquely determined.
}%end of proof of 418M

\cmmnt{
\leader{418N}{Remarks (a)} Taking $I$ to be a singleton,
we get a version of 418L in which $Y$ is a probability space, and
omitting the check that the function $g$ induces an isomorphism of the
measure algebras.
Taking $I$ to be the family of finite subsets of a set $T$, and every
$X_i$ to be a product $\prod_{t\in i}Z_t$ of Radon probability
spaces with its product Radon measure, we obtain a method of
constructing
products of arbitrary families of compact probability spaces from finite
products.

\spheader 418Nb In the hypotheses of 418M, I asked only that
the $f_{ij}$ should be measurable, and omitted any check on the
compositions
$f_{ij}f_{jk}$ when $i\le j\le k$.   But it is easy to see that every
$f_{ij}$ must in fact be almost continuous, and that $f_{ij}f_{jk}$ must
be equal almost everywhere to $f_{ik}$ (418Xw), just as in 418P below.

\spheader 418Nc In the theorem as written out above, the space
$X$ and the functions $g_i:X\to X_i$ are part of the data.   Of course
in many applications we start with a structure

\Centerline{$(\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)},
\langle f_{ij}\rangle_{i\le j\in I})$,}

\noindent and the first step is to find suitable $X$ and $g_i$, as in
418O and 418P.

\spheader 418Nd There are important questions concerning possible
relaxations of the hypotheses in 418M, especially in the special case
already mentioned, in which $X_i=\prod_{t\in i}Z_t$,
$f_{ij}(x)=x\restr i$ when $i\subseteq j\in[T]^{<\omega}$,
$X=\prod_{t\in T}Z_t$, and
$g_i(x)=x\restr i$ for $x\in X$ and $i\in I$, but there is no suggestion
that the $\mu_i$ are product measures.
For a case in which we can dispense with any mention of
auxiliary topologies on the $X_i$, see 454G.

\spheader 418Ne A typical class of applications of Prokhorov's
theorem is in the theory of stochastic processes, in which we have large
families $\family{t}{T}{X_t}$ of random variables;  for definiteness,
imagine that $T=\coint{0,\infty}$, so that we are looking at a system
evolving over time.   Not infrequently our intuition leads us to a clear
description of the joint distributions $\nu_J$ of finite subfamilies
$\family{t}{J}{X_t}$ without providing any suggestion of a measure space
on which the whole family $\family{t}{T}{X_t}$ might be defined.   (As I
tried to explain in the introduction to Chapter 27, probability spaces
themselves
are often very shadowy things in true probability theory.)   Each
$\nu_J$ can be thought of as a Radon measure on $\BbbR^J$, and for
$I\subseteq J\in[T]^{<\omega}$ we have a natural map
$f_{IJ}:\BbbR^J\to\BbbR^I$,
setting $f_{IJ}(y)=y\restr I$ for $y\in\BbbR^J$.   If our distributions
$\nu_J$ mean anything at all, every $f_{IJ}$ will surely be \imp;  this
is simply saying that $\nu_I$ is the joint distribution of a subfamily
of $\family{t}{J}{X_t}$.   If we can find a Hausdorff space $\Omega$ and
a continuous function $g:\Omega\to\BbbR^T$ such that, for every finite
$J\subseteq T$ and $\epsilon>0$, there is a compact set
$K\subseteq\Omega$
such that $\nu_Jg_J[K]\ge 1-\epsilon$ (where $g_J(x)=g(x)\restr J$),
then Prokhorov's theorem will give us a measure $\mu$ on $\Omega$ which
will then provide us with a suitable realization of $\family{t}{T}{X_t}$
as a family of random variables on a genuine probability space, writing
$X_t(\omega)=g(\omega)(t)$.   That they become continuous functions on a
Radon measure space is a valuable shield against irrelevant
complications.

Clearly, if this can be done at all it can be done with
$\Omega=\BbbR^T$;  but some of the central results of probability theory
are specifically concerned with the possibility of using other sets
$\Omega$ (e.g., $\Omega$ a set of \callal\ functions, as in 455H, or
continuous functions, as in 477B).

\spheader 418Nf In (e) above, we do always have the option of
regarding each $\nu_J$ as a measure on the compact space
$[-\infty,\infty]^J$.   In this case, by 418O below or otherwise, we can be
sure of finding a measure
on $[-\infty,\infty]^T$ to support functions $X_t$, at the cost of
either allowing the values $\pm\infty$ or (as I should myself ordinarily
do) accepting that each $X_t$ would be undefined on a negligible set.
The advantage of this is just that it gives us confidence in applying
the Kolmogorov-Lebesgue theory to the whole family $\family{t}{T}{X_t}$
at once, rather than to finite or countable subfamilies.   For an
example of what can happen if we try to do similar things with
non-compact measures, see 419K.   For an example of the problems which
can arise with uncountable families, see 418Xx.
}%end of comment

\leader{418O}{}\cmmnt{ I
mention two cases in which we can be sure that the projective limit
$(X,\familyiI{g_i})$ required in Prokhorov's theorem will exist.

\medskip

\noindent}{\bf Proposition} Suppose that $(I,\le)$,
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$
and $\langle f_{ij}\rangle_{i\le j\in I}$ are such that

\inset{$(I,\le)$ is a non-empty upwards-directed partially ordered set,}

\inset{every $(X_i,\frak T_i,\Sigma_i,\mu_i)$ is a compact Radon measure
space,}

\inset{$f_{ij}:X_j\to X_i$ is a continuous \imp\ function whenever
$i\le j$ in $I$,}

\inset{$f_{ij}f_{jk}=f_{ik}$ whenever $i\le j\le k$ in $I$.}

\noindent Then there are a compact Hausdorff space $X$ and a
family $\familyiI{g_i}$ such that
$I$, $\familyiI{X_i}$, $\langle f_{ij}\rangle_{i\le j\in I}$,
$X$ and $\familyiI{g_i}$ satisfy all the hypotheses ($\alpha$)-($\theta$)
of 418M.

\proof{ For each $i$, let $F_i$ be the support of $\mu_i$;  because $X_i$
is compact, so is $F_i$.   If $i\le j$,
then $F_i=\overline{f_{ij}[F_j]}=f_{ij}[F_j]$, by 411Ne.   Set

\Centerline{$X=\{x:x\in\prod_{i\in I}F_i,\,f_{ij}x(j)=x(i)$ whenever
$i\le j\in I\}$,}

\Centerline{$g_i(x)=x(i)$ for $x\in X$, $i\in I$.}

\noindent Of course $g_i=f_{ij}g_j$ whenever $i\le j$.   Also
$g_i[X]\supseteq F_i$
for every $i\in I$.   \Prf\ Take any $y\in F_i$.
For each finite set $J\subseteq I$,

\Centerline{$H_J=\{x:x\in\prod_{J\in I}F_j,\,x(i)=y,\,f_{jk}x(k)=x(j)$
whenever $j\le k\in J\}$}

\noindent is a closed set.   $H_J$ is always non-empty, because if $k$
is an upper bound of $J\cup\{i\}$ there is
a $z\in F_k$ such that $f_{ik}(z)=y$, in which case $x\in H_J$ whenever
$x\in X$ and $x(j)=f_{jk}(z)$ for every $j\in J\cup\{i\}$.   Now
$\{H_J:J\in[I]^{<\omega}\}$ is a downwards-directed family of non-empty
closed sets in the compact space $\prod_{j\in I}F_j$, so has non-empty
intersection, and if $x$ is any point of the intersection then $x\in X$
and $g_i(x)=y$.\ \Qed

Accordingly $\mu_ig_i[X]=\mu_iX_i$ for every $i$;  as $X$ itself is
compact, 418M($\eta$) is satisfied.   Finally, because
$X\subseteq\prod_{i\in I}X_i$, $\familyiI{g_i}$ separates the points of $X$.
}%end of proof of 418O

\vleader{72pt}{418P}{Proposition}
Let $(I,\le)$, $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ and
$\langle f_{ij}\rangle_{i\le j\in I}$ be such that

\inset{$(I,\le)$ is a
countable non-empty upwards-directed partially ordered set,}

\inset{every $(X_i,\frak
T_i,\Sigma_i,\mu_i)$ is a Radon probability space,}

\inset{$f_{ij}:X_j\to X_i$ is an \imp\ almost continuous function
whenever $i\le j$ in $I$,}

\inset{$f_{ij}f_{jk}=f_{ik}\,\,\mu_k$-a.e.\ whenever $i\le j\le k$ in
$I$.}

\noindent Then there are a Radon probability space
$(X,\frak T,\Sigma,\mu)$ and continuous \imp\ functions $g_i:X\to X_i$,
separating the points of $X$, such that $g_i=f_{ij}g_j$
whenever $i\le j$ in $I$.

\proof{{\bf (a)} We can use nearly the same formula as in 418O:

\Centerline{$X=\{x:x\in\prod_{i\in I}X_i,\,f_{ij}x(j)=x(i)$ whenever
$i\le j\in I\}$,}

\Centerline{$g_i(x)=x(i)$ for $x\in X$, $i\in I$.}

\noindent As before, the consistency relation $g_i=f_{ij}g_j$ is a
trivial consequence of the definition of $X$.   For the rest, we have to
check that 418M($\eta$) is satisfied.   Fix
$\epsilon\in\ooint{0,1}$.   Start by
taking a family $\langle\epsilon_{ij}\rangle_{i\le j\in I}$ of strictly
positive numbers such that $\sum_{i\le j\in
I}\epsilon_{ij}\le\bover12\epsilon$.   (This is where we need to know
that $I$ is countable.)    Set $\epsilon_j=\sum_{i\le j}\epsilon_{ij}$
for each $j$, so that $\sum_{j\in I}\epsilon_j\le\bover12\epsilon$.

For $i\le j\le k$ in $I$, set

\Centerline{$E_{ijk}=\{x:x\in X_k,\,f_{ik}(x)=f_{ij}f_{jk}(x)\}$,}

\noindent so that $E_{ijk}$ is $\mu_k$-conegligible;  set
$E_k=\bigcap_{i\le j\le k}E_{ijk}$, so that $E_k$ is
$\mu_k$-conegligible.   For $i\le j\in I$, choose compact sets
$K_{ij}\subseteq E_j$ such that
$\mu_jK_{ij}\ge 1-\epsilon_{ij}$ and $f_{ij}\restr K_{ij}$ is
continuous.   Now we seem to need a three-stage construction, as
follows:

\inset{for $j\in I$, set $K_j=\bigcap_{i\le j}K_{ij}$;}

\inset{for $j\in I$, set
$K^*_j=K_j\cap\bigcap_{i\le j}f_{ij}^{-1}[K_i]$;}

\inset{finally, set $K=X\cap\prod_{j\in I}K_j^*$.}

\noindent Let us trace the properties of these sets stage by stage.

\medskip

{\bf (b)} For each $j\in I$, $K_j\subseteq K_{jj}\subseteq E_j$ is
compact and

\Centerline{$\mu_j(X_j\setminus K_j)
\le\sum_{i\le j}\mu_j(X_j\setminus K_{ij})
\le\sum_{i\le j}\epsilon_{ij}
=\epsilon_j$,}

\noindent so that $\mu_jK_j\ge 1-\epsilon_j$.   Note that $f_{ik}$
agrees with $f_{ij}f_{jk}$ on $K_k$ whenever $i\le j\le k$, and that
$f_{ij}\restr K_j$ is continuous whenever $i\le j$.

\medskip

{\bf (c)} Every $K_j^*$ is compact, and if $i\le j\le k$ then $f_{ik}$
agrees with $f_{ij}f_{jk}$ on $K^*_k$, while $f_{ij}\restr K_j^*$ is
always continuous.   Also

$$\eqalign{\mu_j(X_j\setminus K_j^*)
&\le\mu_j(X_j\setminus K_j)
  +\sum_{i\le j}\mu_j(X_j\setminus f_{ij}^{-1}[K_i])\cr
&\le\epsilon_j+\sum_{i\le j}\epsilon_i
\le\epsilon,\cr}$$

\noindent so $\mu_jK_j^*\ge 1-\epsilon$, for every $j\in I$.

The point of moving from $K_j$ to $K^*_j$ is that
$f_{jk}[K_k^*]\subseteq K_j^*$ whenever $j\le k$ in $I$.   \Prf\
$K_k^*\subseteq f_{jk}^{-1}[K_j]$, so $f_{jk}[K_k^*]\subseteq K_j$.   If
$i\le j$, then

\Centerline{$K_k^*
=K_k^*\cap f_{ik}^{-1}[K_i]
=K_k^*\cap f_{jk}^{-1}[f_{ij}^{-1}[K_i]]$}

\noindent because $f_{ij}f_{jk}$ agrees with $f_{ik}$ on $K_k^*$.   So
$f_{jk}[K_k^*]\subseteq f_{ij}^{-1}[K_i]$.   As $i$ is arbitrary,
$f_{jk}[K_k^*]\subseteq K_j^*$.\ \Qed

Again because $f_{ik}$ agrees with $f_{ij}f_{jk}$ on $K_k^*$, we have
$f_{ik}[K_k^*]=f_{ij}[f_{jk}[K_k^*]]\subseteq f_{ij}[K_j^*]$ whenever
$i\le j\le k$.   And because $f_{ij}\restr K_j^*$ is always continuous,
all the sets $f_{ij}[K_j^*]$ are compact.

\medskip

{\bf (d)(i)}  $K$ is compact. \Prf\

\Centerline{$K=\{x:x\in\prod_{i\in I}K_i^*,\,f_{ij}x(j)=x(i)$ whenever
$i\le j\in I\}$}

\noindent is closed in $\prod_{i\in I}K_i^*$ because $f_{ij}\restr
K_j^*$ is always continuous (and every $X_i$ is Hausdorff).   Since
$\prod_{i\in I}K_i^*$ is compact, so is $K$.\ \Qed

\medskip

\quad{\bf (ii)} $\mu_ig_i[K]\ge 1-\epsilon$ for every $i\in I$.   \Prf\
By (c), $f_{ik}[K_k^*]\subseteq f_{ij}[K_j^*]$ whenever $i\le j\le k$.
So $\{f_{ij}[K_j^*]:j\ge i\}$ is a downwards-directed family of compact
sets;  write $L$ for their intersection.   Since

\Centerline{$\mu_if_{ij}[K_j^*]
=\mu_jf_{ij}^{-1}[f_{ij}[K_j^*]]\ge\mu_jK_j^*\ge 1-\epsilon$}

\noindent for every $j\ge i$, $\mu_iL\ge 1-\epsilon$ (414C).   If
$z\in L$, then for every $k\ge i$ the set

\Centerline{$F_k
=\{x:x\in\prod_{j\in I}K_j^*,\,x(i)=z$, $f_{jk}x(k)=x(j)$
  whenever $j\le k\}$}

\noindent is a closed set in $\prod_{j\in I}K_j^*$, while
$F_k\subseteq F_j$ when $j\le k$.   Also $F_k$ is non-empty, because there
is a $t\in K_k^*$ such that $f_{ik}(t)=z$, and now if we take any
$x\in\prod_{j\in I}K_j^*$ such that $x(j)=f_{jk}(t)$ for every $j\le k$,
we shall have $x\in F_k$.   So $\{F_k:k\ge i\}$ is a
downwards-directed family of non-empty closed sets in a compact space,
and has non-empty intersection.   But if $x\in\bigcap_{k\ge i}F_k$, then
$x\in K$ and $x(i)=z$, so $z\in g_i[K]$.   Thus $g_i[K]\supseteq L$ and
$\mu_ig_i[K]\ge 1-\epsilon$.\ \Qed

\medskip

{\bf (e)} As $\epsilon$ is arbitrary, 418M($\eta$) is
satisfied.   But now 418M gives the result.
}%end of proof of 418P

\leader{418Q}{Corollary} Let
$\sequencen{(X_n,\frak T_n,\Sigma_n,\mu_n)}$
be a sequence of Radon probability spaces, and suppose we are given an
\imp\ almost continuous function $f_n:X_{n+1}\to X_n$ for each $n$.
Set

\Centerline{$X=\{x:x\in\prod_{n\in\Bbb N}X_n,\,f_n(x(n+1))=x(n)$ for
every $n\in\Bbb N\}$.}

\noindent Then there is a unique Radon probability measure $\mu$ on $X$
such that all the coordinate maps $x\mapsto x(n):X\to X_n$ are \imp.

\proof{ For $i\le j\in\Bbb N$, define $f_{ij}:X_j\to X_i$ by
writing

$$\eqalign{f_{ii}(x)&=x\text{ for every }x\in X_i,\cr
f_{i,j+1}&=f_{ij}f_j\text{ for every }j\ge i.\cr}$$

\noindent It is easy to check that $f_{ij}f_{jk}=f_{ik}$ whenever $i\le
j\le k$, and that every $f_{ij}$ is \imp\ and almost continuous (using
418Dc).   So we are exactly in the situation of 418P, and we know that
there is a Radon probability measure on $X$ for which every $g_i$ is
\imp;   moreover, the coordinate functionals $g_i$ separate the points of $X$,
so $\mu$ is unique, by the last remark in 418M.
}%end of proof of 418Q

\leader{418R}{}\cmmnt{ I turn now to a special kind of measurable
function, corresponding to a new view of product spaces.

\medskip

\noindent}{\bf Theorem} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$, and $(Y,\Tau,\nu)$ a $\sigma$-finite measure space.
Give $L^0(\nu)$ the topology of convergence in
measure\cmmnt{ (\S245)}.   Write $\eusm L^0_{\Sigma\tensorhat\Tau}$ for
the space of $\Sigma\tensorhat\Tau$-measurable real-valued functions
on $X\times Y$\cmmnt{, where $\Sigma\tensorhat\Tau$ is  the
$\sigma$-algebra of subsets of $X\times Y$ generated by
$\{E\times F:E\in\Sigma,\,F\in\Tau\}$}. Then for a function
$f:X\to L^0(\nu)$ the following are equiveridical:

(i) $f[X]$ is separable and $f$ is measurable;

(ii) there is an $h\in\eusm L^0_{\Sigma\tensorhat\Tau}$ such that
$f(x)=h_x^{\ssbullet}$ for every $x\in X$, where $h_x(y)=h(x,y)$ for
$x\in X$, $y\in Y$.

\proof{ Let $\sequencen{Y_n}$ be a
non-decreasing sequence of subsets of $Y$ of finite measure covering
$Y$.

\medskip

{\bf (a)(i)$\Rightarrow$(ii)} For each $n\in\Bbb N$, let $\rho_n$ be the
continuous pseudometric on $L^0(\nu)$ defined by saying that
$\rho_n(g_1^{\ssbullet},g_2^{\ssbullet})
=\int_{Y_n}\min(1,|g_1-g_2|)d\nu$
for $g_1$, $g_2\in\eusm L^0_{\Tau}$, writing $\eusm L^0_{\Tau}$ for the
space of $\Tau$-measurable real-valued functions on $Y$.   Then
$\{\rho_n:n\in\Bbb N\}$ defines the topology of $L^0(\nu)$ (see the
proof of 245Eb).
Because $f[X]$ is separable, there is a sequence $\sequence{k}{v_k}$ in
$L^0(\nu)$ such that $f[X]\subseteq\overline{\{v_k:k\in\Bbb N\}}$.   For
each $k$, choose $g_k\in\eusm L^0_{\Tau}$ such that
$g_k^{\ssbullet}=v_k$.   For $n$, $k\in\Bbb N$ set

\Centerline{$E_{nk}=\{x:x\in X,\,\rho_n(f(x),v_k)<2^{-n}\}$,}

\Centerline{$H_{nk}=E_{nk}\setminus\bigcup_{i<k}E_{ni}$.}

\noindent Then every $E_{nk}$ belongs to $\Sigma$ (because $f$ is
measurable) and $\bigcup_{k\in\Bbb N}E_{nk}=X$ (because
$\{v_k:k\in\Bbb N\}$ is dense);
so $\sequence{k}{H_{nk}}$ is a partition of $X$ into measurable sets.
Set $h^{(n)}(x,y)=g_k(y)$ whenever $k\in\Bbb N$, $x\in H_{nk}$ and
$y\in Y$;  then $h^{(n)}\in\eusm L^0_{\Sigma\tensorhat\Tau}$.

Fix $x\in X$ for the moment.   Then for each $n\in\Bbb N$ there is a
unique $k_n$ such that $x\in H_{nk_n}$, and
$\rho_n(f(x),v_{k_n})\le 2^{-n}$.   So if $n\le m$,

$$\eqalign{\int_{Y_n}\min(1,|h^{(m+1)}_x-h^{(m)}_x|)
&=\int_{Y_n}\min(1,|g_{k_{m+1}}-g_{k_m}|)
=\rho_n(g_{k_{m+1}}^{\ssbullet},g_{k_m}^{\ssbullet})\cr
&\le\rho_n(g_{k_{m+1}}^{\ssbullet},f(x))
 +\rho_n(f(x),g_{k_m}^{\ssbullet})\cr
&\le\rho_{m+1}(g_{k_{m+1}}^{\ssbullet},f(x))
 +\rho_m(f(x),g_{k_m}^{\ssbullet})
\le 3\cdot 2^{-m-1}.\cr}$$

\noindent But this means that
$\sum_{m=0}^{\infty}\int_{Y_n}\min(1,|h^{(m+1)}_x-h^{(m)}_x|)$ is
finite, so that $\sequence{m}{h^{(m)}_x}$ must be convergent almost
everywhere in $Y_n$.   As this is true for every $n$,
$\sequence{m}{h^{(m)}_x}$ is convergent a.e.\ on $Y$.   Moreover,

\Centerline{$\lim_{m\to\infty}(h^{(m)}_x)^{\ssbullet}
=\lim_{m\to\infty}g_{k_m}^{\ssbullet}
=f(x)$}

\noindent in $L^0(\nu)$.

Since this is true for every $x$,

\Centerline{$W=\{(x,y):\sequence{m}{h^{(m)}(x,y)}$ converges in
$\Bbb R\}$}

\noindent has conegligible vertical sections, while of course
$W\in\Sigma\tensorhat\Tau$ because every $h^{(m)}$ is
$\Sigma\tensorhat\Tau$-measurable (418C).   If we set
$h(x,y)=\lim_{m\to\infty}h^{(m)}(x,y)$ for $(x,y)\in W$, $0$ for other
$(x,y)\in X\times Y$, then $h\in\eusm L^0_{\Sigma\tensorhat\Tau}$, while
(by 245Ca)

\Centerline{$h_x^{\ssbullet}
=\lim_{m\to\infty}(h^{(m)}_x)^{\ssbullet}=f(x)$}

\noindent in $L^0(\nu)$ for every $x\in X$.   So we have a suitable $h$.


\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Let $\Phi$ be the set of those
$h\in\eusm L^0_{\Sigma\tensorhat\Tau}$ such that (i) is satisfied;  that is,
$x\mapsto h_x^{\ssbullet}$ is measurable, and
$\{h_x^{\ssbullet}:x\in X\}$ is separable.

\medskip

\quad\grheada\ $\Phi$ is closed under addition.   \Prf\ If $h$,
$\tilde h$ belong to $\Phi$, set $A=\{h_x^{\ssbullet}:x\in X\}$,
$\tilde A=\{\tilde h_x^{\ssbullet}:x\in X\}$.   Then both $A$ and
$\tilde A$ are
separable metrizable spaces, so $A\times\tilde A$ is separable and
metrizable and
$x\mapsto(h_x^{\ssbullet},\tilde h_x^{\ssbullet}):X\to A\times\tilde A$
is measurable (418Bb).   But addition on $L^0(\nu)$ is
continuous (245Da), so

\Centerline{$x\mapsto h_x^{\ssbullet}+\tilde h_x^{\ssbullet}
=(h+\tilde h)_x^{\ssbullet}$}

\noindent is measurable (418Ac), and

\Centerline{$\{(h+\tilde h)_x^{\ssbullet}:x\in X\}
\subseteq\{u+\tilde u:u\in A,\,\tilde u\in\tilde A\}$}

\noindent is separable (4A2B(e-iii), 4A2P(a-iv)).
Thus $h+\tilde h\in\Phi$.\ \Qed

\medskip

\quad\grheadb\ $\Phi$ is closed under scalar multiplication, just
because $u\mapsto\alpha u:L^0(\nu)\to L^0(\nu)$ is always continuous.

\medskip

\quad\grheadc\ If $\sequencen{h^{(n)}}$ is a sequence in $\Phi$ and
$h(x,y)=\lim_{n\to\infty}h^{(n)}(x,y)$ for all $x\in X$, $y\in Y$, then
$h\in\Phi$.   \Prf\ Setting $A_n=\{(h^{(n)}_x)^{\ssbullet}:x\in X\}$ for
each $n$, then $A=\{h_x^{\ssbullet}:x\in X\}$ is included in
$\overline{\bigcup_{n\in\Bbb N}A_n}$, which is separable (4A2B(e-i)
again), so $A$ is separable (4A2P(a-iv) again);  moreover,
$h_x^{\ssbullet}=\lim_{n\to\infty}(h^{(n)}_x)^{\ssbullet}$ for every
$x\in X$, so $x\mapsto h_x^{\ssbullet}$ is measurable, by
418Ba.\ \Qed

\medskip

\quad\grheadd\ What this means is that if we set
$\Cal W=\{W:W\in\Sigma\tensorhat\Tau,\,\chi W\in\Phi\}$, then
$W\setminus W'\in\Cal W$ whenever $W$, $W'\in\Cal W$ and
$W'\subseteq W$, and
$\bigcup_{n\in\Bbb N}W_n\in\Cal W$ whenever $\sequencen{W_n}$ is a
non-decreasing sequence in $\Cal W$.   Also, it is easy to see that
$E\times F\in\Cal W$ whenever $E\in\Sigma$ and $F\in\Tau$.   By the
Monotone Class Theorem (136B), $\Cal W$ includes the $\sigma$-algebra
generated by $\{E\times F:E\in\Sigma,\,F\in\Tau\}$, that is, is equal to
$\Sigma\tensorhat\Tau$.   It follows at once, from ($\alpha$) and
($\beta$), that $\sum_{i=0}^n\alpha_i\chi W_i\in\Phi$ whenever
$W_0,\ldots,W_n\in\Sigma\tensorhat\Tau$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$, and hence (using ($\gamma$)) that
$\eusm L^0_{\Sigma\tensorhat\Tau}\subseteq\Phi$, which is what we had to
prove.
}%end of proof of 418R

\leader{418S}{Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces with c.l.d.\ product
$(X\times Y,\Lambda,\lambda)$.   Give $L^0(\nu)$ the topology of
convergence in measure.   \cmmnt{Write $\eusm L^0(\lambda)$ for
the space
of $\Lambda$-measurable real-valued functions defined $\lambda$-a.e.\ on
$X\times Y$, as in 241A.}

(a) If $h\in\eusm L^0(\lambda)$, set $h_x(y)=h(x,y)$ whenever this is
defined.   Then

\Centerline{$\{x:f(x)=h_x^{\ssbullet}$ is defined in $L^0(\nu)\}$}

\noindent is $\mu$-conegligible, and includes a conegligible set $X_0$
such that $f:X_0\to L^0(\nu)$ is measurable and $f[X_0]$ is separable.

(b) If $f:X\to L^0(\nu)$ is measurable and there is a conegligible set
$X_0\subseteq X$ such that $f[X_0]$ is separable, then there is an
$h\in\eusm L^0(\lambda)$ such that $f(x)=h_x^{\ssbullet}$ for almost
every $x\in X$.

\proof{{\bf (a)} The point is that $\lambda$ is just the completion of
its restriction to $\Sigma\tensorhat\Tau$ (251K).   So there is a
conegligible set $W\in\Sigma\tensorhat\Tau$ such that $h\restrp W$ is
$\Sigma\tensorhat\Tau$-measurable (212Fa).   Setting
$\tilde h(x,y)=h(x,y)$ for $(x,y)\in W$, $0$ otherwise, and setting
$\tilde f(x)=\tilde h_x^{\ssbullet}$ for every $x\in X$, we see from
418R that
$\tilde f$ is measurable and that $\tilde f[X]$ is separable.   But 252D
tells us that

\Centerline{$X_0=\{x:((X\times Y)\setminus W)[\{x\}]$ is negligible$\}$}

\noindent is conegligible;  and if $x\in X_0$ then
$h_x=\tilde h_x\,\,\nu$-a.e., so that $f(x)$ is defined and equal to
$\tilde f(x)$.   This proves the result.

\medskip

{\bf (b)}  We can suppose that $X_0\in\Sigma$.   Setting $f_1(x)=f(x)$
for $x\in X_0$, $0$ otherwise, $f_1$ satisfies 418R(i).
So there is some $h\in\eusm L^0(\Sigma\tensorhat\Tau)$ such that
$f_1(x)=h_x^{\ssbullet}$ for every $x$, in which case $f(x)=h_x^{\ssbullet}$
for almost every $x$, and (ii) is true.
}%end of proof of 418S

\leader{418T}{Corollary}\cmmnt{ ({\smc Mauldin \& Stone 81})} Let
$(Y,\Tau,\nu)$ be a
$\sigma$-finite measure space, and $(\frak B,\bar\nu)$ its measure
algebra,  with its measure-algebra topology\cmmnt{ (\S323)}.

(a) Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, and $f:X\to\frak B$ a function.
Then the following are equiveridical:

\quad(i) $f[X]$ is separable and $f$ is measurable;

\quad(ii) there is a $W\in\Sigma\tensorhat\Tau$ such that
$f(x)=W[\{x\}]^{\ssbullet}$ for every $x\in X$.

(b) Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space and
$\Lambda$ the domain of the c.l.d.\ product measure $\lambda$ on
$X\times Y$.

\quad(i) Suppose that $\nu$ is complete.   If $W\in\Lambda$, then

\Centerline{$\{x:f(x)=W[\{x\}]^{\ssbullet}$ is defined in $\frak B\}$}

\noindent is $\mu$-conegligible, and includes a conegligible set $X_0$
such that $f:X_0\to \frak B$ is measurable and $f[X_0]$ is separable.

\quad(ii) If $f:X\to\frak B$ is measurable and there is a conegligible
set $X_0\subseteq X$ such that $f[X_0]$ is separable, then there is a
$W\in\Sigma\tensorhat\Tau$ such that $f(x)=W[\{x\}]^{\ssbullet}$ for
almost every $x\in X$.

\proof{ Everything follows directly from 418R and 418S if we observe
that $\frak B$ is homeomorphically embedded in $L^0(\nu)$ by the
function $F^{\ssbullet}\mapsto(\chi F)^{\ssbullet}$ for $F\in\Tau$
(323Xa, 367Ra).   We do need to check, for (i)$\Rightarrow$(ii) of part
(a), that if $h\in\eusm L^0_{\Sigma\tensorhat\Tau}$ and $h_x^{\ssbullet}$
is always of the form $(\chi F)^{\ssbullet}$, then there is some
$W\in\Sigma\tensorhat\Tau$ such that $h_x^{\ssbullet}=(\chi
W[\{x\}])^{\ssbullet}$ for every $x$;  but of course this is true if we
just
take $W=\{(x,y):h(x,y)=1\}$.   Now (b-ii) follows from (a) just as 418Sb
followed from 418R.
}%end of proof of 418T

\leader{*418U}{Independent families of measurable functions} In \S455 we
shall have occasion to look at independent families of random variables
taking values in spaces other than $\Bbb R$.   We can use the same
principle as in \S272:  a family $\familyiI{X_i}$ of random variables is
independent if $\familyiI{\Sigma_i}$ is independent, where $\Sigma_i$ is
the $\sigma$-subalgebra defined by $X_i$ for each $i$\cmmnt{ (272D)}.
\cmmnt{Of
course this depends on agreement about the definition of $\Sigma_i$.
The natural thing to do, in the context of this section, is to follow 272C,
as follows.   }Let $(X,\Sigma,\mu)$ be a probability space, $Y$ a
topological space, and $f$ a $Y$-valued
function defined on a conegligible subset $\dom f$ of $X$, which is
$\mu$-virtually measurable, that is, such that $f$ is measurable
with respect to the subspace $\sigma$-algebra on $\dom f$ induced by
$\hat\Sigma=\dom\hat\mu$, where $\hat\mu$ is the completion of $\mu$.
\cmmnt{(Note that if $Y$ is not second-countable this may not imply that
$f\restr D$ is $\Sigma$-measurable for a conegligible subset $D$ of $X$.)}
The `$\sigma$-algebra defined by $f$' will be

\Centerline{$\{f^{-1}[F]:F\in\Cal B(Y)\}
\cup\{(\Omega\setminus\dom f)\cup f^{-1}[F]:F\in\Cal B(Y)\}
\subseteq\hat\Sigma$,}

\noindent where $\Cal B(Y)$ is the Borel $\sigma$-algebra of $Y$\cmmnt{;
that is, the $\sigma$-algebra of subsets of $X$ generated by
$\{f^{-1}[G]:G\subseteq Y$ is open$\}$}.

Now, given a family $\familyiI{(f_i,Y_i)}$ where each $Y_i$ is a
topological space and each $f_i$ is a $\hat\Sigma$-measurable $Y_i$-valued
function defined on a conegligible subset of $X$, I will say that
$\familyiI{f_i}$ is {\bf independent} if $\familyiI{\Sigma_i}$ is
independent (with respect to $\hat\mu$),
where $\Sigma_i$ is the $\sigma$-algebra defined by $f_i$ for
each $i$.

\cmmnt{Corresponding to 272D, we can use the Monotone Class Theorem to
show that }$\familyiI{f_i}$ is independent iff

\Centerline{$\hat\mu(\bigcap_{j\le n}f_{i_j}^{-1}[G_j])
=\prod_{j\le n}\hat\mu f_{i_j}^{-1}[G_j]$}

\noindent whenever $i_0,\ldots,i_n\in I$ are
distinct and $G_j\subseteq Y_{i_j}$ is open for every $j\le n$.

%\leader{418W}{}

\exercises{
\leader{418X}{Basic exercises $\pmb{>}$(a)}
%\spheader 418Xa
\dvAnew{2014} Let $(X,\Sigma,\mu)$ be a measure space, and
$(X,\hat\Sigma,\hat\mu)$ its completion.   (i) Show that if $Y$ is a
second-countable topological space, a function $f:X\to Y$ is
$\hat\Sigma$-measurable iff there is a $\Sigma$-measurable $g:X\to Y$ such
that $f\eae g$.   (ii) Show that if
$X$ is endowed with a topology, and $Y$ is a topological space, then
a function from $X$ to $Y$ is $\mu$-almost continuous iff it is
$\hat\mu$-almost continuous.
%418A

\spheader 418Xb
Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a set and $h:X\to Y$ a
function;  give $Y$ the image measure $\mu h^{-1}$.   Show that for any
function $g$ from $Y$ to a topological space $Z$, $g$ is measurable iff
$gh:X\to Z$ is measurable.
%418A

\sqheader 418Xc Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets
of $X$, $\sequencen{Y_n}$
a sequence of topological spaces with product $Y$, and $f:X\to Y$ a
function.   Show that $f$ is measurable iff
$\psi_nf:X\to\prod_{i\le n}Y_i$ is measurable for every $n\in\Bbb N$,
where $\psi_n(y)=(y(0),\ldots,y(n))$
for $y\in Y$ and $n\in\Bbb N$.
%418B

\spheader 418Xd Let $(X,\Sigma,\mu)$ be a semi-finite measure space,
$(Y,\frak S)$ a metrizable space, and
$\sequencen{f_n}$ a sequence of measurable functions from $X$ to $Y$
such that $\sequencen{f_n(x)}$ is convergent for almost every $x\in X$.
Show that $\mu$ is inner regular with respect to
$\{E:\sequencen{f_n\restr E}$ is uniformly convergent$\}$.
(Cf.\ 412Xa.)
%412-

\sqheader 418Xe  Set $Y=[0,1]^{[0,1]}$, with the product topology.   For
$n\in\Bbb N$ and $x\in [0,1]$ define $f_n(x)\in Y$ by saying that
$f_n(x)(t)=\max(0,1-2^n|x-t|)$ for $t\in[0,1]$.   Check that (i) each
$f_n$ is continuous, therefore measurable;  (ii)
$f(x)=\lim_{n\to\infty}f_n(x)$
is defined in $Y$ for every $x\in[0,1]$;  (iii) for each $t\in[0,1]$,
the coordinate functional $x\mapsto f(x)(t)$ is continuous except at
$t$, and in particular is almost continuous and measurable;  (iv)
$f\restr F$ is not
continuous for any infinite closed set $F\subseteq[0,1]$, and in
particular $f$ is not almost continuous;  (v) every subset of $[0,1]$ is
of the form
$f^{-1}[H]$ for some open set $H\subseteq Y$;  (vi) $f$ is not
measurable;
(vii) the image measure $\mu f^{-1}$, where $\mu$ is Lebesgue measure on
$[0,1]$, is neither a topological measure nor tight.
%418D

\spheader 418Xf Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space, $Y$ a topological space, and $f:X\to Y$ a function.   Suppose
that for every $x\in X$ there is an open set $G$ containing $x$ such
that $f\restr G$ is almost continuous with respect to the subspace
measure on $G$.   Show that $f$ is almost continuous.
%418D

\spheader 418Xg For $i=1$, $2$ let $(X_i,\frak T_i,\Sigma_i,\mu_i)$ and
$(Y_i,\frak S_i,\Tau_i,\nu_i)$ be quasi-Radon measure spaces, and
$f_i:X_i\to Y_i$ an almost continuous \imp\ function.   Show that
$(x_1,x_2)\mapsto (f_1(x_1),f_2(x_2))$ is almost continuous and
\imp\ for the product topologies and quasi-Radon product measures.
%418D
%Is there an easy example to show that `almost continuous' is
%essential?

\spheader 418Xh Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ and
$\familyiI{(Y_i,\frak T_i,\Sigma_i,\nu_i)}$ be two families of
topological
spaces with $\tau$-additive Borel probability measures, and let $\mu$,
$\nu$ be the $\tau$-additive product measures on $X=\prod_{i\in I}X_i$,
$Y=\prod_{i\in I}Y_i$ (417E, 417F).
Show that if $f_i:X_i\to Y_i$ is almost continuous and \imp\ for each
$i$, then $x\mapsto\familyiI{f_i(x(i))}:X\to Y$ is \imp, but need not be
almost continuous.
%418D 418Xg 417E

\spheader 418Xi Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be quasi-Radon measure spaces, $(Z,\frak U)$ a topological space and
$f:X\times Y\to Z$ a function which is almost continuous with respect to
the quasi-Radon product measure on $X\times Y$.   Suppose that $\nu$ is
$\sigma$-finite.   Show that $y\mapsto f(x,y)$ is almost continuous for
almost every $x\in X$.
%418D

\spheader 418Xj Let $(X,\frak T,\Sigma,\mu)$ be an effectively locally
finite $\tau$-additive topological measure space, $Y$ a topological
space and $f:X\to Y$ an almost continuous function.   (i) Show that the
image measure $\mu f^{-1}$ is $\tau$-additive.   (ii) Show that if $\mu$
is a totally finite quasi-Radon measure and the topology on $Y$ is
regular, then $\mu f^{-1}$ is quasi-Radon.
%418D

\spheader 418Xk Let $(X,\frak T,\Sigma,\mu)$ be a topological measure
space and $U$ a linear topological space.   Show that if $f:X\to U$ and
$g:X\to U$ are almost continuous, then $f+g:X\to U$ is almost
continuous.
%418D

\spheader 418Xl Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be topological measure spaces, and $(Z,\frak U)$ a topological space;
let $f:X\to Y$ be almost continuous and \imp, and $g:Y\to Z$ almost
continuous.   Show that if {\it either} $\nu$ is strictly localizable
{\it or} $\mu$ is a Radon measure and $\nu$ is locally finite
{\it or} $\mu$ is $\tau$-additive and
effectively locally finite and $\nu$ is effectively locally finite, then
$gf:X\to Z$ is almost continuous.   \Hint{show that if $\mu E>0$ there
is a set $F$ such that $\nu F<\infty$ and $\mu(E\cap f^{-1}[F])>0$.}
%418D

\spheader 418Xm Let $\mu$ be Lebesgue measure on $\BbbR^r$, where
$r\ge 1$, $X$ a Hausdorff space and $f:\BbbR^r\to X$ an almost continuous
function.
Show that for almost every $x\in\BbbR^r$ there is a measurable set
$E\subseteq\BbbR^r$ such that $x$ is a density point of $E$ and
$\lim_{y\in E,y\to x}f(y)=f(x)$.
%418D

\spheader 418Xn Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, $\undphi:\Sigma\to\Sigma$ a lower density such that
$\undphi X=X$, and $\frak T$ the associated density topology on $X$
(414P).   Let
$f:X\to\Bbb R$ be a function.   Show that the following are
equiveridical:  (i) $f$ is
measurable;  (ii) $f$ is almost continuous;  (iii) $f$ is continuous at
almost every point;  (iv) there is a conegligible set $H\subseteq X$
such that $f\restr H$ is continuous.   (Cf.\ 414Xk.)
%418E

\spheader 418Xo Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, $\phi:\Sigma\to\Sigma$ a lifting, and $\frak S$ the
lifting topology on $X$ (414Q).   Let $f:X\to\Bbb R$ be a function.
Show that the
following are equiveridical:  (i) $f$ is measurable;  (ii) $f$ is almost
continuous;  (iii) there is a conegligible set $H\subseteq X$ such that
$f\restr H$ is continuous.   (Cf.\ 414Xr.)
%418E

\spheader 418Xp Let $(X,\frak T,\Sigma,\mu)$ be a quasi-Radon measure
space, $(Y,\frak S)$ a regular topological space and $f:X\to Y$ an
almost continuous function.   Show that there is a quasi-Radon measure
$\nu$ on $Y$ such that $f$ is \imp\ for $\mu$ and $\nu$ iff
$\bigcup\{f^{-1}[H]:H\subseteq Y$ is open, $\mu f^{-1}[H]<\infty\}$ is
conegligible in $X$.
%418H

\spheader 418Xq Let $(X,\frak T,\Sigma,\mu)$ be a Radon measure space,
$(Y,\frak S)$ and $(Z,\frak U)$ Hausdorff spaces, $f:X\to Y$ an almost
continuous function such that $\nu=\mu f^{-1}$ is locally finite, and
$g:Y\to Z$ a function.   Show that $g$ is almost continuous with respect
to $\nu$ iff $gf$ is almost continuous with respect to $\mu$.
%418I

\spheader 418Xr Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be topological probability spaces, and $f:X\to Y$ a measurable function
such that $\mu f^{-1}[H]\ge\nu H$ for every $H\in\frak S$.
Show that (i) $\int gfd\mu=\int g\,d\nu$ for every $g\in C_b(Y)$ (ii)
$\mu f^{-1}[F]=\nu F$ for every Baire set $F\subseteq Y$ (iii) if $\mu$
is a Radon measure and $f$ is almost continuous, then
$\mu f^{-1}[F]=\nu F$ for every Borel set $F\subseteq Y$, so that if in
addition $\nu$ is complete and inner regular with respect to the Borel
sets then it is a Radon measure.
%418I

\spheader 418Xs Let $(X,\frak T,\Sigma,\mu)$ be a $\sigma$-finite
topological measure space in which the topology $\frak T$ is normal and
$\mu$ is outer regular with respect to the open sets.   Show that if
$f:X\to\Bbb R$ is a measurable function and $\epsilon>0$ there is a
continuous $g:X\to\Bbb R$ such that $\mu\{x:g(x)\ne f(x)\}\le\epsilon$.
\Hint{4A2Fd.}
%418J 4A2Fd 412Wb

\spheader 418Xt Let $X$ and $Y$ be Hausdorff spaces, $\nu$ a totally
finite Radon measure on $Y$, and $f:X\to Y$ an injective continuous
function.   Show that the following are equiveridical:  (i) there is a
Radon measure $\mu$ on $X$ such that $f$ is \imp;
(ii) $f[X]$ is conegligible and $f^{-1}:f[X]\to X$ is almost continuous.
%418L

\spheader 418Xu Let $(X,\frak T,\Sigma,\mu)$ and $(Y,\frak S,\Tau,\nu)$
be Radon measure spaces and $f:X\to Y$ an almost continuous \imp\
function.   Show that (i) $\mu_*A\le\nu_*f[A]$ for every $A\subseteq X$
(ii) $\nu$ is precisely the image measure $\mu f^{-1}$.
%418L

\spheader 418Xv Let $X$ be a compact Hausdorff space.   Show that there is
an atomless Radon probability measure on $X$ iff $X$ is not scattered.
\Hint{4A2Gj.}
%418L 4A2G(j-iv)

\spheader 418Xw In 418M, show that all the $f_{ij}$ must be almost
continuous.   Show that if $i\le j\le k$ then $f_{ij}f_{jk}=f_{ik}$
almost everywhere in $X_k$.
%418M

\sqheader 418Xx Let $\Cal I$ be the family of finite subsets of $[0,1]$,
and for each $I\in\Cal I$
let $(X_I,\frak T_I,\Sigma_I,\mu_I)$ be $[0,1]\setminus I$ with its
subspace topology and measure induced by Lebesgue measure.   For
$I\subseteq J\in\Cal I$ and $y\in X_J$ set $f_{IJ}(y)=y$.
Show that these $X_I$,
$f_{IJ}$ satisfy nearly all the hypotheses of 418O, but that there are
no $X$, $g_I$ which satisfy the hypotheses of 418M.
%418N

\spheader 418Xy Let $T$ be any set, and $X$ the set of total
orders on $T$.   (i) Regarding each member of $X$ as a subset of
$T\times T$, show that $X$ is a closed subset of $\Cal P(T\times T)$.
(ii) Show that there is a unique Radon measure $\mu$ on $X$ such that
$\Pr(t_1\le t_2\le\ldots\le t_n)=\Bover1{n!}$ for all distinct
$t_1,\ldots,t_n\in T$.   \Hint{for $I\in[T]^{<\omega}$, let $X_I$ be the
set of total orders on $I$ with the uniform probability measure
giving the same measure to each singleton;  show that the natural map
from $X_I$ to $X_J$ is \imp\ whenever $J\subseteq I$.}
%418N

\spheader 418Xz In 418Sb, suppose that $f_1:X\to L^0(\nu)$ and
$f_2:X\to L^0(\nu)$ correspond to $h_1$, $h_2\in\eusm L^0(\lambda)$.
Show that
$f_1(x)\le f_2(x)\,\,\mu$-a.e.($x$) iff $h_1\le h_2\,\,\lambda$-a.e.
Hence show that (if we assign appropriate algebraic operations to the
space of functions from $X$ to $L^0(\nu)$) we have an $f$-algebra
isomorphism between $L^0(\lambda)$ and the space of equivalence classes
of measurable functions from $X$ to $L^0(\nu)$ with separable ranges.
%418S

\leader{418Y}{Further exercises (a)}
%\spheader 418Ya
Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets of $X$, $Y$ a
topological space and $f:X\to Y$ a function.   Set
$\Tau=\{F:F\subseteq Y,\,f^{-1}[F]\in\Sigma\}$.   Suppose that $Y$ is
hereditarily Lindel\"of and its topology is generated by some subset of
$\Tau$.   Show that $f$ is measurable.
%418B

\spheader 418Yb Let $(X,\Sigma,\mu)$ be a measure space, $Y$ and $Z$
topological spaces
and $f:X\to Y$, $g:X\to Z$ measurable functions.   Show that if $Z$ has
a countable network consisting of Borel sets (e.g., $Z$ is
second-countable, or $Z$ is regular and has a countable network), then
$x\mapsto (f(x),g(x)):X\to Y\times Z$ is measurable.
%418B
%what if $Z$ analytic?

\spheader 418Yc Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets
of $X$, and $\familyiI{Y_i}$ a countable family of topological spaces
with product $Y$.   Suppose that every $Y_i$ has a countable network,
and that $f:X\to Y$ is a function such that $\pi_if$ is measurable for
every $i\in I$, writing $\pi_i(y)=y(i)$.   Show that $f$ is measurable.
%418B %418Ya
%Y is hered Lind

\spheader 418Yd Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space
and $\frak T$ a topology on $X$ such that $\mu$ is effectively locally
finite and $\tau$-additive.   Let $Y$ be a topological space and
$f:X\to Y$ an almost continuous function.   Show that there is a
conegligible subset $X_0$ of $X$ such that $f[X_0]$ is ccc.
%418F

\spheader 418Ye Show that if $\mu$ is Lebesgue measure on $\Bbb R$,
$\frak T$ is the usual topology on $\Bbb R$ and
$\frak S$ is the right-facing Sorgenfrey topology (415Xc), then the
identity map from $(\Bbb R,\frak T,\mu)$ to $(\Bbb R,\frak S)$ is
measurable, but not almost continuous, and the image
measure is not a Radon measure.
%418I

\spheader 418Yf Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\frak T$ a topology on $X$ such that $\mu$ is inner regular with
respect to the closed sets.   Suppose that $Y$ is a topological space
with a countable network consisting of Borel sets, and that $f:X\to Y$
is measurable.   Show that $f$ is almost continuous.
%418J

\spheader 418Yg Find a topological probability space
$(X,\frak T,\Sigma,\mu)$ in which $\mu$ is inner regular with respect to
the closed sets, a topological space $Y$ with a countable network and a
measurable function $f:X\to Y$ which is not almost continuous.
%418J 41bite

\spheader 418Yh Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\frak T$ a topology on $X$ such that $\mu$ is inner regular with
respect to the closed sets.   Let $A\subseteq L^{\infty}(\mu)$ be a
norm-compact set.   Show that there is a set $B$ of bounded real-valued
measurable functions on $X$ such that (i) $A=\{f^{\ssbullet}:f\in B\}$
(ii) $B$ is norm-compact in $\ell^{\infty}(X)$ (iii) $\mu$ is inner
regular with respect to
$\{E:f\restr E$ is continuous for every $f\in B\}$.
%418J

\spheader 418Yi Let $\mu$ be Lebesgue measure on $[0,1]$.   For
$t\in[0,1]$ set $u_t=\chi[0,t]^{\ssbullet}\in L^0(\mu)$.   Show that
$A=\{u_t:t\in[0,1]\}$ is norm-compact in $L^p(\mu)$ for every
$p\in\coint{1,\infty}$ and also compact for the topology of convergence
in measure on $L^0(\mu)$.   Show that if $B$ is a set of measurable
functions such that $A=\{f^{\ssbullet}:f\in B\}$ then $\mu$ is not inner
regular with respect to
$\{E:f\restr E$ is continuous for every $f\in B\}$.
%418Yh 418J

\spheader 418Yj Let $\mu$ be Lebesgue measure on $[0,1]$, and
$A\subseteq[0,1]$ a set with inner measure 0 and outer measure 1;  let
$\frak T$ be the usual topology on $[0,1]$.   Let $\Cal I$ be the family
of sets $I\subseteq A$ such that every point of $A$ has a neighbourhood
containing at most one point of $I$.   Show that
$\frak S=\{G\setminus I:G\in\frak T,\,I\in\Cal I\}$ is a topology on
$[0,1]$ with a countable network.   Show that the identity map from
$[0,1]$ to itself, regarded as a map from $([0,1],\frak T,\mu)$ to
$([0,1],\frak S)$, is measurable but not almost continuous.
%418Yf 418J

\spheader 418Yk Let $(X,\Sigma,\mu)$ be a semi-finite measure
space and $\frak T$ a topology on $X$ such that $\mu$ is inner regular
with respect to the closed sets.   Suppose that $Y$ and $Z$ are
separable metrizable spaces,
$f:X\times Y\to Z$ is a function such that $x\mapsto f(x,y)$ is measurable
for every $y\in Y$, and $y\mapsto f(x,y)$ is continuous for every $x\in X$.
Show that $\mu$ is inner regular
with respect to $\{F:F\subseteq X$, $f\restr F\times Y$ is continuous$\}$.
\leaveitout{Let $\rho$, $\sigma$ be metrics defining the topologies of
$Y$ and $Z$.   For $y\in Y$ and $n\in\Bbb N$ set
$g_{yn}(x)=\sup\{s:\sigma(f(x,y'),f(x,y))\le 2^{-n}$ whenever
$\rho(y',y)\le s\}$, $f_y(x)=f(x,y)$.
Show that if $D\subseteq Y$ is dense
and $F\subseteq X$ is such that $g_{yn}\restr F$ and $f_y\restr F$ are
continuous whenever $y\in D$ and $n\in\Bbb N$, then $f\restr F\times y$ is
continuous.}
%418J see Bouziad 07

\spheader 418Yl Suppose that $(I,\le)$,
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ and
$\langle f_{ij}\rangle_{i\le j\in I}$ are such that
($\alpha$) $(I,\le)$ is a non-empty upwards-directed partially ordered
set ($\beta$) every $(X_i,\frak T_i,\Sigma_i,\mu_i)$ is a completely
regular Hausdorff quasi-Radon probability space
($\gamma$) $f_{ij}:X_j\to X_i$ is a
continuous \imp\ function whenever $i\le j$ in $I$
($\delta$) $f_{ij}f_{jk}=f_{ik}$ whenever $i\le j\le k$ in $I$.
Let $X'_i$ be the
support of $\mu_i$ for each $i$;  show that $f_{ij}[X'_j]$ is a dense
subset of $X'_i$ whenever $i\le j$.   Let $Z_i$ be the Stone-\v{C}ech
compactification of $X'_i$ and let $\tilde f_{ij}:Z_j\to Z_i$ be the
continuous extension of $f_{ij}\restr X'_j$ for $i\le j$;  let
$\tilde\mu_i$ be the Radon probability measure on $Z_i$ corresponding to
$\mu_i\restrp\Cal PX'_i$ (416V).   Show that $Z_i$, $\tilde f_{ij}$
satisfy the conditions of 418O, so that we have a projective limit $Z$,
$\familyiI{g_i}$, $\mu$ as in 418M.
%418M

\spheader 418Ym Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets of
$X$ and $(Y,\Tau,\nu)$ a $\sigma$-finite measure space with countable
Maharam type.   (i) Let $f:X\to L^1(\nu)$ be a function such that
$x\mapsto\int_Ff(x)d\nu$ is $\Sigma$-measurable for every $F\in\Tau$.
Show that $f$ is $\Sigma$-measurable for the norm topology on $L^1(\nu)$.
(ii) Let $g:X\times Y\to\Bbb R$ be a function such that
$\int g(x,y)\nu(dy)$ is defined for every $x\in X$, and
$x\mapsto\int_Fg(x,y)\nu(dy)$ is $\Sigma$-measurable for every $F\in\Tau$.
Show that there is an $h\in\eusm L^0_{\Sigma\tensorhat\Tau}$ such that, for
every $x\in X$, $g(x,y)=h(x,y)$ for $\nu$-almost every $y$.
%418R

\spheader 418Yn\dvAformerly{4{}18Yq} Use 418M and 418O to prove 328H.
%418O

\spheader 418Yo Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$, $(Y,\Tau,\nu)$ a $\sigma$-finite measure space and
$W\in\Sigma\tensorhat\Tau$.   Show that there is a $V\subseteq W$ such that
$V\in\Sigma\tensorhat\Tau$, $W[\{x\}]\setminus V[\{x\}]$ is negligible
for every
$x\in X$, and $\bigcap_{x\in I}V[\{x\}]$ is either empty or non-negligible
for every non-empty finite $I\subseteq X$.   \Hint{341Xb.}
%418T mt41bits\query formerly 418Yq

\spheader 418Yp
Let $X$ be a compact Hausdorff space, $Y$ a Hausdorff space, $\nu$ a
Radon probability measure on $Y$ and $R\subseteq X\times Y$ a closed set
such that $\nu^*R[X]=1$.   Show that there is a Radon
probability measure $\mu$ on $X$ such that
$\mu R^{-1}[F]\ge\nu F$ for every closed set $F\subseteq Y$.
%418L
}%end of exercises

\cmmnt{\Notesheader{418}
The message of this section is that measurable functions are dangerous,
but that almost continuous functions behave themselves.   There are
two fundamental problems with measurable functions:  a function
$x\mapsto (f(x),g(x))$
may not be measurable when the components $f$ and $g$ are measurable
(419Xg), and an image measure under a measurable function can lose
tightness, even when both domain and
codomain are Radon measure spaces and the function is \imp\ (419Xh).
(This is the `image
measure catastrophe' mentioned in 235H and the notes to \S343.)
Consequently, as long as we are dealing with measurable functions, we
often have to impose strong conditions on the range
spaces -- commonly, we have to restrict ourselves to separable
metrizable spaces (418B, 418C), or something similar,
which indeed often means that a measurable function is actually almost
continuous (418J, 433E).   Indeed, for functions taking values in
metrizable spaces, `almost continuous' is very close to `measurable with
essentially separable range' (418G, 418J).   The condition `separable
and metrizable' is a
little stronger than is strictly necessary (418Yb, 418Yc, 418Yf), but
covers the principal applications other than 433E.   If we keep the
`metrizable' we can
very substantially relax the `separable' (438E, 438F), and it is in fact
the case that a measurable function from a Radon measure space to any
metrizable space is almost continuous (451T).   These extensions apply
equally to the results in 418R-418T (438Xi-438Xj, 451Xp).   But both
take us deeper into set theory than seems appropriate at the moment.

For almost continuous functions, the two problems mentioned above do not
arise (418Dd, 418I, 418Xu).   Indeed we rather expect almost continuous
functions to behave as if they were continuous.   But we still have to
be careful.   The limit of a sequence of almost continuous functions
need not be almost continuous (418Xe), unless the codomain is metrizable
(418F);  and if we have a function $f$ from a
topological measure space to an uncountable product of topological
spaces, it can happen that every coordinate of $f$ is an almost
continuous function
while $f$ is not (418Xe again).   But for many purposes, intuitions
gained from the study of measurable functions between Euclidean spaces
can be transferred to general almost continuous functions.

Theorems 418L and 418M are of a quite different kind, but seem to
belong here as well
as anywhere.   Even in the simplest application of 418L (when $Y=[0,1]$
with Lebesgue measure, and $X\subseteq[0,1]^2$ is a closed set meeting
every vertical line) it is not immediately obvious that there will be a
measure with the right projection onto the horizontal axis, though there
are at least two proofs which are easier than the general case treated
in 413O-413P-418L.

As I explain in 418N, the really interesting question concerning 418M is
when, given the projective system
\penalty-100$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$,
$\langle f_{ij}\rangle_{i\le j\in I}$, we can expect to find $X$ and
$\familyiI{g_i}$ satisfying the
rest of the hypotheses, and once past the elementary results 418O-418Q
this can be hard to determine.   I describe a method in 418Yl
which can sometimes be used, but (like the trick in 418Nf) it is too
easy and too abstract to be often illuminating.   See 454G below for
something rather deeper.

The results of 418R-418T stand somewhat aside from anything else
considered in this chapter, but they form part of an important
technique.   A special case has already been mentioned in 253Yg.   I do
not discuss vector-valued measurable functions in this book, except
incidentally, but 418R is one of the fundamental results on their
representation;  it means, for instance, that if $V$ is any of the
Banach function spaces of Chapter 36 we can expect to represent Bochner
integrable $V$-valued functions (253Yf) in terms of functions on product
spaces, because $V$ will be continuously embedded in an $L^0$ space
(367O).   The measure-algebra version in 418T will be useful in
Volume 5 when establishing relationships between properties of measure
spaces and corresponding properties of measure algebras.
}%end of notes

\discrpage

\frfilename{mt419.tex}
\versiondate{2.12.05}
\copyrightdate{2002}

\def\chaptername{Topologies and measures I}
\def\sectionname{Examples}

\newsection{419}

In \S216, I went much of the way to describing examples of spaces with
all the possible combinations of the properties considered in Chapter
21.   When we come to topological measure spaces, the number of
properties involved makes it unreasonable to seek any such comprehensive
list.   I therefore content myself with seven examples
%419A 419C 419D 419E 419J 419K 419L
to indicate some of the boundaries of the theory developed here.

The first example (419A) is supposed to show that the hypothesis
`effectively locally finite' which appears in so many of the theorems of
this chapter cannot as a rule be replaced by `locally finite'.   The
next two (419C-419D) address technical questions concerning the
definition of `Radon measure', and show how small variations in the
definition can lead to very different kinds of measure space.   The
fourth example (419E) shows that the $\tau$-additive product measures of
\S417 are indeed new constructions.   419J is there to
show that extension theorems of the types proved in \S415 and \S417
cannot be taken for granted.   The classic example 419K
exhibits one of the obstacles to generalizations of Prokhorov's theorem
(418M, 418Q).   Finally, I return to the split interval (419L) to
describe its standard topology and its relation to the measure
introduced in 343J.

\vleader{60pt}{419A}{Example} There is a locally compact Hausdorff space
$X$ with a complete, $\sigma$-finite, locally finite, $\tau$-additive
topological measure $\mu$, inner regular with respect to the closed
sets, which has a closed subset $Y$, of measure $1$, such that the
subspace measure $\mu_Y$
on $Y$ is not $\tau$-additive.   In particular, $\mu$ is not effectively
locally finite.

\proof{{\bf (a)} Let $Q$ be a countably infinite set, not containing any
ordinal.   Fix an enumeration $\sequencen{q_n}$ of $Q$, and for
$A\subseteq Q$ set $\nu A=\sum\{\bover{1}{n+1}:q_n\in A\}$.   Let $\Cal I$
be the ideal $\{A:A\subseteq Q,\,\nu A<\infty\}$.
For any sets $I$, $J$, say that $I\subseteq^*J$ if $I\setminus J$ is
finite;  then $\subseteq^*$ is a reflexive transitive relation.

Let $\kappa$ be the smallest cardinal of any family
$\Cal K\subseteq\Cal I$ for which there is no $I\in\Cal I$ such that
$K\subseteq^*I$ for
every $K\in\Cal K$.   Then $\kappa$ is uncountable.   \Prf\ (i) Of
course $\kappa$ is infinite.   (ii) If $\sequencen{I_n}$ is any sequence
in $\Cal I$, then for each $n\in\Bbb N$ we can find a finite
$I'_n\subseteq I_n$
such that $\nu(I_n\setminus I'_n)\le 2^{-n}$;  setting
$I=\bigcup_{n\in\Bbb N}I_n\setminus I'_n$, we have $\nu I\le 2<\infty$,
while $I_n\subseteq^* I$ for every $n$.   Thus $\kappa>\omega$.\ \Qed

\medskip

{\bf (b)} There is a family $\langle I_{\xi}\rangle_{\xi<\kappa}$ in
$\Cal I$ such that (i) $I_{\eta}\subseteq^*I_{\xi}$ whenever
$\eta\le\xi<\kappa$ (ii) there is no $I\in\Cal I$ such that
$I_{\xi}\subseteq^*I$ for every
$\xi<\kappa$.   \Prf\ Take a family $\langle K_{\xi}\rangle_{\xi<\kappa}$
in $\Cal I$ such that there is no $I\in\Cal I$ such that
$K_{\xi}\subseteq^*I$ for every $\xi<\kappa$.   Choose
$\langle I_{\xi}\rangle_{\xi<\kappa}$ in $\Cal I$ inductively in such a
way that

\Centerline{$K_{\xi}\subseteq^*I_{\xi}$,
\quad$I_{\eta}\subseteq^*I_{\xi}$ for every $\eta<\xi$.}

\noindent (This can be done because
$\{K_{\xi}\}\cup\{I_{\eta}:\eta<\xi\}$
will always be a subset of $\Cal I$ with cardinal less than $\kappa$.)
If now $I_{\xi}\subseteq^*I$ for every $\xi<\kappa$, then
$K_{\xi}\subseteq^*I$
for every $\xi<\kappa$, so $I\notin\Cal I$.\ \Qed

Hence, or otherwise, we see that $\kappa$ is regular.   \Prf\ If
$A\subseteq\kappa$ and $\#(A)<\kappa$, then there is an $I\in\Cal I$
such that $I_{\zeta}\subseteq^*I$ for every $\zeta\in A$; now there must
be a $\xi<\kappa$ such that $I_{\xi}\not\subseteq^*I$, in which case
$\zeta<\xi$ for every $\zeta\in A$, and $A$ is not cofinal with
$\kappa$.\ \Qed

\medskip

{\bf (c)} Set $X=Q\cup\kappa$.   (This is where it is helpful to have
arranged at the start that no ordinal belongs to $Q$, so that
$Q\cap\kappa=\emptyset$.)   Let $\frak T$ be the family of sets
$G\subseteq X$ such that

\inset{$G\cap\kappa$ is open for the order topology of $\kappa$,

for every $\xi\in G\cap\kappa\setminus\{0\}$ there is an
$\eta<\xi$ such that $I_{\xi}\setminus I_{\eta}\subseteq^*G$,

if $0\in G$ then $I_0\subseteq^*G$.}

\medskip

\quad{\bf (i)} This is a Hausdorff topology on $X$.   \Prf\ ($\alpha$)
It is easy to check that $X\in\frak T$, $\emptyset\in\frak T$ and
$\bigcup\Cal G\in\frak T$ for every $\Cal G\subseteq\frak T$.
($\beta$) Suppose that $G$, $H\in\frak T$.   Then $(G\cap H)\cap\kappa
=(G\cap\kappa)\cap(H\cap\kappa)$ is open for the order topology of
$\kappa$.   If $\xi\in G\cap H\cap\kappa\setminus\{0\}$ there are
$\eta$, $\zeta<\xi$ such that $I_{\xi}\setminus I_{\eta}\subseteq^*G$
and $I_{\xi}\setminus I_{\zeta}\subseteq^*H$, and now

\Centerline{$\alpha=\max(\eta,\zeta)<\xi$,
\quad$I_{\eta}\cup I_{\zeta}\subseteq^*I_{\alpha}$,}

\noindent so

\Centerline{$I_{\xi}\setminus I_{\alpha}\subseteq^*(I_{\xi}\setminus
I_{\eta})\cap(I_{\xi}\setminus I_{\zeta})\subseteq^*G\cap H$.}

\noindent Finally, if $0\in G\cap H$ then $I_0\subseteq^*G\cap H$.   So
$G\cap H\in\frak T$.   Thus $\frak T$ is a topology on $X$.   ($\gamma$)
For any $\xi<\kappa$, the set $E_{\xi}=(\xi+1)\cup I_{\xi}$ is
open-and-closed for $\frak T$;  for any $q\in Q$, $\{q\}$ is
open-and-closed.   Since these sets separate the points of $X$,
$\frak  T$ is Hausdorff.\ \Qed

\medskip

\quad{\bf (ii)} The sets $E_{\xi}$ of the last paragraph are all compact
for $\frak T$.   \Prf\ Let $\Cal F$ be an ultrafilter on $X$ containing
$E_{\xi}$.   ($\alpha$) If a finite set $K$ belongs to $\Cal F$, then
$\Cal F$ must contain $\{x\}$ for some $x\in K$, and converges to $x$.
So suppose henceforth that $\Cal F$ contains no finite set.   ($\beta$)
If $E_0\in\Cal F$, then for any open set $G$ containing $0$,
$E_0\setminus G$ is finite, so does not belong to $\Cal F$, and
$G\in\Cal F$;  as $G$ is
arbitrary, $\Cal F\to 0$.   ($\gamma$) If $E_0\notin\Cal F$, let
$\eta\le\xi$ be the least ordinal such that $E_{\eta}\in\Cal F$.   If
$G$ is an open set containing $\eta$, there are $\zeta'$, $\zeta''<\eta$
such that $I_{\eta}\setminus I_{\zeta'}\subseteq^*G$ and
$\ocint{\zeta'',\eta}\subseteq G$;  so that
$E_{\eta}\setminus E_{\zeta}\subseteq^*G$,
where $\zeta=\max(\zeta',\zeta'')<\eta$.   Now
$E_{\eta}\in\Cal F$, $E_{\zeta}\notin\Cal F$ and $(E_{\eta}\setminus
E_{\zeta})\setminus G\notin\Cal F$, so that $G\in\Cal F$.   As $G$ is
arbitrary, $\Cal F\to\eta$.   ($\delta$) As $\Cal F$ is arbitrary,
$E_{\xi}$ is compact.\ \Qed

\medskip

\quad{\bf (iii)} It follows that $\frak T$ is locally compact.   \Prf\
For $q\in Q$, $\{q\}$ is a compact open set containing $q$;  for
$\xi<\kappa$, $E_{\xi}$ is a compact open set containing $\xi$.\ \Qed

\medskip

\quad{\bf (iv)} The definition of $\frak T$ makes it clear that
$Q\in\frak T$, that is, that $\kappa$ is a closed subset of $X$.   We
need also to check that the subspace topology $\frak T_{\kappa}$ on
$\kappa$ induced by $\frak T$ is just the order topology of $\kappa$.
\Prf\ ($\alpha$) By the definition of $\frak T$, $G\cap\kappa$ is open
for the order topology of $\kappa$ for every $G\in\frak T$.    ($\beta$)
For any $\xi<\kappa$, $E_{\xi}$ is open-and-closed for $\frak T$ so
$\xi+1=E_{\xi}\cap\kappa$ is
open-and-closed for $\frak T_{\kappa}$.   But this means that all sets
of the forms $\coint{0,\xi}=\bigcup_{\eta<\xi}\eta+1$ and
$\ooint{\xi,\kappa}=\kappa\setminus(\xi+1)$ belong to $\frak
T_{\kappa}$;  as these generate the order topology, every open set for
the order topology belongs to $\frak T_{\kappa}$, and the two topologies
are equal.\ \Qed

\medskip

{\bf (d)} Now let $\Cal F$ be the filter on $X$ generated by the cofinal
closed
sets in $\kappa$.   Because the intersection of any sequence of closed
cofinal sets in $\kappa$ is another (4A1Bd), the intersection of any
sequence in $\Cal F$ belongs to $\Cal F$.   So

\Centerline{$\Sigma=\Cal F\cup\{X\setminus F:F\in\Cal F\}$}

\noindent is a $\sigma$-algebra of subsets of $X$, and we have a measure
$\mu_1:\Sigma\to\{0,1\}$ defined by saying that $\mu_1F=1$,
$\mu_1(X\setminus F)=0$ if $F\in\Cal F$.

\medskip

{\bf (e)} Set $\mu E=\nu(E\cap Q)+\mu_1E$ for $E\in\Sigma$.   Then $\mu$
is a measure.   Let us work through the properties called for.

\medskip

\quad{\bf (i)} If $\mu E=0$ and $A\subseteq E$, then $X\setminus
A\supseteq X\setminus E\in\Cal F$, so $A\in\Sigma$.   Thus $\mu$ is
complete.

\medskip

\quad{\bf (ii)} $\mu\kappa=1$ and $\mu\{q\}$ is finite for every $q\in
Q$, so $\mu$ is $\sigma$-finite.

\medskip

\quad{\bf (iii)} If $G\subseteq X$ is open, then $\kappa\setminus G$ is
closed, in the order topology of $\kappa$;  if it is cofinal with
$\kappa$, it belongs to $\Cal F$;  otherwise, $\kappa\cap G\in\Cal F$.
Thus in either case $G\in\Sigma$, and $\mu$ is a topological measure.

\medskip

\quad{\bf (iv)} The next thing to note is that $\mu G=\nu(G\cap Q)$ for
every open set $G\subseteq X$.   \Prf\ If $G\notin\Cal F$ this is
trivial.   If $G\in\Cal F$, then $\kappa\setminus G$ cannot be cofinal
with $\kappa$, so there is a $\xi<\kappa$ such that
$\kappa\setminus\xi\subseteq G$.
\Quer\ If $G\cap Q\in\Cal I$, then $(G\cap Q)\cup I_{\xi}\in\Cal I$.
There must be a least $\eta<\kappa$ such that
$I_{\eta}\not\subseteq^*(G\cap Q)\cup I_{\xi}$;  of course $\eta>\xi$,
so $\eta\in G$.   There is some $\zeta<\eta$ such that
$I_{\eta}\setminus I_{\zeta}\subseteq^*G$;  but as
$I_{\zeta}\subseteq^*G\cup I_{\xi}$, by the choice of $\eta$, we must
also have $I_{\eta}\subseteq^*G\cup I_{\xi}$,
which is impossible.\ \BanG\   Thus $G\cap Q\notin\Cal I$ and $\mu
G=\nu(G\cap Q)=\infty$.\ \Qed

\medskip

\quad{\bf (v)} It follows that $\mu$ is $\tau$-additive.   \Prf\ Suppose
that $\Cal G\subseteq\frak T$ is a non-empty upwards-directed set with
union $H$.   Then

\Centerline{$\mu H=\nu(H\cap Q)=\sup_{G\in\Cal G}\nu(G\cap Q)
=\sup_{G\in\Cal G}\mu G$}

\noindent because $\nu$ is $\tau$-additive (indeed, is a Radon measure)
with respect to the discrete topology on $Q$.\ \Qed

\medskip

\quad{\bf (vi)} $\mu$ is inner regular with respect to the closed sets.
\Prf\ Take $E\in\Sigma$ and $\gamma<\mu E$.   Then
$\gamma-\mu_1E<\nu(E\cap Q)$, so there is a finite
$I\subseteq E\cap Q$ such that $\nu I>\gamma-\mu_1E$.   If
$\mu_1E=0$, then $I\subseteq E$ is already a closed set with
$\mu I>\gamma$.   Otherwise, $E\cap\kappa\in\Cal F$, so there is a
cofinal closed
set $F\subseteq\kappa$ such that $F\subseteq E$;  now $F$ is closed in
$X$ (because $\kappa$ is closed in $X$ and the subspace topology on
$\kappa$ is the order topology), so $I\cup F$ is closed, and
$\mu(I\cup F)>\gamma$.   As $E$ and $\gamma$ are arbitrary,
$\mu$ is inner regular with respect to the closed sets.\ \Qed

\medskip

\quad{\bf (vii)} $\mu$ is locally finite.   \Prf\ For any $\xi<\kappa$,
$E_{\xi}$ is an open set containing $\xi$, and $\mu E_{\xi}=\nu I_{\xi}$
is finite.   For any $q\in Q$, $\{q\}$ is an open set containing $q$,
and $\mu\{q\}=\nu\{q\}$ is finite. \Qed

\medskip

\quad{\bf (viii)} Now consider $Y=\kappa$.   This is surely a closed
set, and $\mu\kappa=1$.   I noted in (c-iv) above that the subspace
topology $\frak T_{\kappa}$ is just the order topology of $\kappa$.
But this means that $\{\xi:\xi<\kappa\}$ is an upwards-directed family
of negligible relatively open sets with union $\kappa$, so that the
subspace measure $\mu_{\kappa}=\mu_1$ is not $\tau$-additive.

\medskip

\quad{\bf (ix)} It follows from 414K that $\mu$ cannot be effectively
locally finite;  but it is also obvious from the work above that
$\kappa$ is a measurable set, of non-zero measure, such that
$\mu(\kappa\cap G)=0$
whenever $G$ is an open set of finite measure.
}%end of proof of 419A

%for a similar example where $\mu$ is not localizable, see
%mt51x


\leader{419B}{Lemma} For any non-empty set $I$, there is a dense
G$_{\delta}$ set in $[0,1]^I$ which is negligible for the usual measure
on $[0,1]^I$.

\proof{ Fix on some $i_0\in I$, and set $\pi(x)=x(i_0)$ for each
$x\in[0,1]^I$, so that $\pi$ is continuous and \imp\ for the usual
topologies and measures on $[0,1]^I$ and $[0,1]$.   For each $n\in\Bbb N$
let $G_n\supseteq[0,1]\cap\Bbb Q$ be an open subset of $[0,1]$ with
measure at most $2^{-n}$, so that $\pi^{-1}[G_n]$ is an open set of
measure at most $2^{-n}$, and $E=\bigcap_{n\in\Bbb N}\pi^{-1}[G_n]$ is a
G$_{\delta}$ set of measure $0$.   If $H\subseteq[0,1]^I$ is any
non-empty open set, its
image $\pi[H]$ is open in $[0,1]$, so contains some rational number, and
meets $\bigcap_{n\in\Bbb N}G_n$;  but this means that
$H\cap E\ne\emptyset$, so $E$ is dense.
}%end of proof of 419B

\leader{419C}{Example}\cmmnt{ ({\smc Fremlin 75b})} There is a
completion regular Radon measure space $(X,\frak T,\Sigma,\mu)$ such
that

(i) there is an $E\in\Sigma$ such that $\mu(F\symmdiff E)>0$ for
every Borel set $F\subseteq X$, that is, not every element of the
measure algebra of $\mu$ can be represented by a Borel set;

(ii) $\mu$ is not outer regular with respect to the Borel sets;

(iii) writing $\nu$ for the restriction of $\mu$ to the Borel
$\sigma$-algebra of $X$, $\nu$ is a locally finite, effectively locally
finite, tight\cmmnt{ (that is, inner regular with respect to the
compact sets)} $\tau$-additive completion regular
topological measure, and there is a set
$Y\subseteq X$ such that the subspace measure $\nu_Y$ is not semi-finite.

\proof{{\bf (a)} For each $\xi<\omega_1$ set
$X_{\xi}=[0,1]^{\omega_1\setminus\xi}$, and take $\mu_{\xi}$ to be the
usual measure on $X_{\xi}$;  write $\Sigma_{\xi}$ for its domain.
Note that $\mu_{\xi}$ is a completion regular Radon measure for the
usual topology $\frak T_{\xi}$ of $X_{\xi}$ (416U).   Set
$X=\bigcup_{\xi<\omega_1}X_{\xi}$, and let $\mu$ be the direct sum
measure on $X$ (214L), that is, write

\Centerline{$\Sigma=\{E:E\subseteq X,\,E\cap X_{\xi}\in\Sigma_{\xi}$ for
every $\xi<\omega_1\}$,}

\Centerline{$\mu E=\sum_{\xi<\omega_1}\mu_{\xi}(E\cap X_{\xi})$ for
every $E\in\Sigma$.}

\noindent Then $\mu$ is a complete locally determined (in fact, strictly
localizable) measure on $X$.   Write $\Sigma$ for its domain.

\medskip

{\bf (b)} For each $\eta<\omega_1$ let
$\langle\beta_{\xi\eta}\rangle_{\xi\le\eta}$ be a summable family of
strictly positive real numbers with $\beta_{\eta\eta}=1$ (4A1P).
Define $g_{\eta}:X\to\Bbb R$ by setting

$$\eqalign{g_{\eta}(x)&=\Bover1{\beta_{\xi\eta}}x(\eta)
  \text{ if }x\in X_{\xi}\text{ where }\xi\le\eta,\cr
&=0\text{ if }x\in X_{\xi}\text{ where }\xi>\eta.\cr}$$

\noindent Now define $f:X\to\omega_1\times\BbbR^{\omega_1}$ by setting

\Centerline{$f(x)=(\xi,\langle g_{\eta}(x)\rangle_{\eta<\omega_1})$}

\noindent if $x\in X_{\xi}$.   Note that $f$ is injective.   Let
$\frak T$ be the topology on $X$ defined by $f$, that is, the family
$\{f^{-1}[W]:W\subseteq\omega_1\times\BbbR^{\omega_1}$ is open$\}$,
where $\omega_1$ and $\BbbR^{\omega_1}$ are given their usual
topologies (4A2S, 3A3K), and their product is given its product
topology.   Because $f$ is injective, $\frak T$ can be identified with
the subspace topology on $f[X]$;  it is Hausdorff and completely
regular.

\medskip

{\bf (c)} For $\xi$, $\eta<\omega_1$, $g_{\eta}\restr X_{\xi}$ is
continuous for the compact topology $\frak T_{\xi}$.   Consequently
$f\restr X_{\xi}$ is continuous, and the subspace topology on $X_{\xi}$
induced by $\frak T$ must be $\frak T_{\xi}$ exactly.   It follows that
$\mu$ is a Radon measure for $\frak T$.   \Prf\ (i) We know already that
$\mu$ is complete and locally determined.   (ii) If $G\in\frak T$ then
$G\cap X_{\xi}\in\frak T_{\xi}\subseteq\Sigma_{\xi}$ for every
$\xi<\omega_1$, so $G\in\Sigma$;  thus $\mu$ is a topological measure.
(iii) If $E\in\Sigma$ and $\mu E>0$, there is a $\xi<\omega_1$ such that
$\mu_{\xi}(E\cap X_{\xi})>0$.   Because $\mu_{\xi}$ is a Radon measure,
there is a $\frak T_{\xi}$-compact set $F\subseteq E\cap X_{\xi}$ such
that $\mu_{\xi}F>0$.   Now $F$ is $\frak T$-compact and $\mu F>0$.   As
$E$ is arbitrary, $\mu$ is tight (using 412B).   (iv) If $x\in X$, take
that $\xi<\omega_1$ such that $x\in X_{\xi}$, and consider

\Centerline{$G
=f^{-1}[(\xi+1)\times\{w:w\in\Bbb R^{\omega_1},\,w(\xi)<2\}]$.}

\noindent Because $\xi+1$ is open in $\omega_1$,
$G\in\frak T$.   Because $g_{\xi}(x)=x(\xi)\le 1$, $x\in G$.   Now for
$\zeta\le\xi$,

$$\eqalign{\mu_{\zeta}(G\cap X_{\zeta})
&=\mu_{\zeta}\{x:x\in X_{\zeta},\,g_{\xi}(x)<2\}\cr
&=\mu_{\zeta}\{x:x\in X_{\zeta},\,\beta_{\zeta\xi}^{-1}x(\xi)<2\}\cr
&=\mu_{\zeta}\{x:x\in X_{\zeta},\,x(\xi)<2\beta_{\zeta\xi}\}
\le 2\beta_{\zeta\xi},\cr}$$

\noindent so

\Centerline{$\mu G=\sum_{\zeta\le\xi}\mu_{\zeta}(G\cap X_{\zeta})
\le 2\sum_{\zeta\le\xi}\beta_{\zeta\xi}<\infty$.}

\noindent As $x$ is arbitrary, $\mu$ is locally finite, therefore a
Radon measure.\ \Qed

We also find that $\mu$ is completion regular.   \Prf\ If $E\subseteq X$
and $\mu E>0$, then there is a $\xi<\omega_1$ such that
$\mu(E\cap X_{\xi})>0$.   Because $\mu_{\xi}$ is completion regular,
there is a set
$F\subseteq E\cap X_{\xi}$, a zero set for $\frak T_{\xi}$, such that
$\mu F>0$.   Now $X_{\xi}$ is a G$_{\delta}$ set in $X$ (being the
intersection of the open sets $\bigcup_{\eta<\zeta<\xi+1}X_{\zeta}$ for
$\eta<\xi$, unless $\xi=0$, in which case $X_{\xi}$ is actually open), so
$F$ is a G$_{\delta}$ set in $X$ (4A2C(a-iv));  being a compact
G$_{\delta}$ set in
a completely regular space, it is a zero set (4A2F(h-v)).

Thus every set of positive measure includes a zero set of positive
measure.   So $\mu$ is inner regular with respect to the zero sets
(412B).\ \Qed

\medskip

{\bf (d)} The key to the example is the following fact:  if
$G\subseteq X$ is open, then {\it either} there is a cofinal closed set
$V\subseteq\omega_1$ such that $G\cap X_{\xi}=\emptyset$ for every
$\xi\in V$ {\it or} $\{\xi:\mu(G\cap X_{\xi})\ne 1\}$ is countable.
\Prf\ Suppose that $A=\{\xi:G\cap X_{\xi}\ne\emptyset\}$ meets every
cofinal closed set, that is, is stationary (4A1C).   Then
$B=A\cap\Omega$ is stationary, where $\Omega$ is the set of non-zero
countable limit ordinals (4A1Bb, 4A1Cb).   Let
$H\subseteq\omega_1\times\BbbR^{\omega_1}$ be
an open set such that $G=f^{-1}[H]$.

For each $\xi\in B$ choose $x_{\xi}\in G\cap X_{\xi}$.   Then
$f(x_{\xi})\in H$, so there must be a $\zeta_{\xi}<\xi$, a finite set
$I_{\xi}\subseteq\omega_1$, and a $\delta_{\xi}>0$ such that $z\in H$
whenever $z=(\gamma,\ofamily{\eta}{\omega_1}{t_{\eta}})
\in\omega_1\times\BbbR^{\omega_1}$,
$\zeta_{\xi}<\gamma\le\xi$ and
$|t_{\eta}-g_{\eta}(x)|<\delta_{\xi}$ for every $\eta\in I_{\xi}$.
Because $\xi$ is a non-zero limit ordinal,
$\zeta'_{\xi}=\sup(\{\zeta_{\xi}\}\cup(I_{\xi}\cap\xi))<\xi$.

By the Pressing-Down Lemma (4A1Cc), there is a $\zeta<\omega_1$
such that $C=\{\xi:\xi\in B,\,\zeta'_{\xi}=\zeta\}$ is uncountable.
\Quer\ Suppose, if possible, that $\zeta<\eta<\omega_1$ and
$\mu(G\cap X_{\eta})<1$.   Then there is a measurable subset $F$ of
$X_{\eta}\setminus G$, determined by coordinates in a countable set
$J\subseteq\omega_1\setminus\eta$, such that $\mu F=\mu_{\eta}F>0$
(254Ff).   Let $\xi\in C$ be
such that $\eta<\xi$ and $J\subseteq\xi$, and take any $y\in F$.   If we
define $y'\in X_{\eta}$ by setting

$$\eqalign{y'(\gamma)&=y(\gamma)
  \text{ for }\gamma\in\xi\setminus\eta\cr
&=x_{\xi}(\gamma)\text{ for }\gamma\in \omega_1\setminus\xi,\cr}$$

\noindent then $y'\in F$.   But also
$\zeta_{\xi}\le\zeta_{\xi}'=\zeta<\eta<\xi$ and
$\xi\setminus\eta\subseteq\xi\setminus\zeta'_{\xi}$ is disjoint from
$I_{\xi}$, so $g_{\gamma}(y')=g_{\gamma}(x_{\xi})$ for every
$\gamma\in I_{\xi}$, since both are zero if $\gamma<\eta$ and otherwise
$y'(\gamma)=x_{\xi}(\gamma)$.   By the choice of $\zeta_{\xi}$ and
$I_{\xi}$ we must have $f(y')\in H$ and $y'\in F\cap G$;  which is
impossible.\ \Bang

Thus $\mu(G\cap X_{\eta})=1$ for every $\eta>\zeta$, as required by the
second alternative.\ \Qed

\medskip

{\bf (e)} For each $\xi<\omega_1$, let $\Cal I_{\xi}$ be the family of
negligible meager subsets of $X_{\xi}$.   Then $\Cal I_{\xi}$ is a
$\sigma$-ideal;  note that it contains every closed negligible set,
because $\mu_{\xi}$ is strictly positive.   Set

\Centerline{$\Tau_{\xi}=\Cal I_{\xi}\cup\{X_{\xi}\setminus F:F\in\Cal
I_{\xi}\}$,}

\noindent so that $\Tau_{\xi}$ is a $\sigma$-algebra of subsets of
$X_{\xi}$, containing every conegligible open set, and
$\mu_{\xi}F\in\{0,1\}$ for every $F\in\Tau_{\xi}$.   Set

\Centerline{$\Tau
=\{E:E\in\Sigma,\,\{\xi:E\cap X_{\xi}\notin\Tau_{\xi}\}$
is non-stationary$\}$.}

\noindent Then $\Tau$ is a $\sigma$-subalgebra of $\Sigma$ (because the
non-stationary sets form a $\sigma$-ideal of subsets of $\omega_1$,
4A1Cb), and contains every open set, by (d);  so includes the Borel
$\sigma$-algebra $\Cal B$ of $X$.

If we set

\Centerline{$E_{\xi}
=\{x:x\in X_{\xi},\,x(\xi)\le\bover12\}$ for each $\xi<\omega_1$,
\quad$E=\bigcup_{\xi<\omega_1}E_{\xi}$,}

\noindent then $E\in\Sigma$.   But if $F\subseteq X$ is a Borel set,
$F\in\Tau$ so $\mu(E\symmdiff F)=\infty$.   This proves the property (i)
claimed for the example.

\medskip

{\bf (f)} Next, for each $\xi<\omega_1$, take a negligible dense
G$_{\delta}$ set $E'_{\xi}\subseteq X_{\xi}$ (419B).  Set
$Y=\bigcup_{\xi<\omega_1}E'_{\xi}$, so that $\mu Y=0$.   If
$F\supseteq Y$ is a Borel set, then
$F\cap X_{\xi}\supseteq E_{\xi}\notin\Cal I_{\xi}$ for every
$\xi<\omega_1$, while $F\in\Tau$, so
$\{\xi:\mu_{\xi}(F\cap X_{\xi})=0\}$ is non-stationary and
$\mu F=\infty$.   Thus $\mu$ is not
outer regular with respect to the Borel sets.   Taking
$\nu=\mu\restr\Cal B$, the subspace measure $\nu_Y$ is not semi-finite.
\Prf\ We have just seen that $\nu_YY=\nu^*Y$ is infinite.   If
$F\in\Cal B$ and $\nu F<\infty$, then
$A=\{\xi:\mu_{\xi}(F\cap X_{\xi})>0\}$ is countable, so
$F_0=\bigcup_{\xi\in A}E'_{\xi}$ and
$F_1=F\setminus\bigcup_{\xi\in A}X_{\xi}$ are negligible Borel sets;
since $F\cap Y\subseteq F_0\cup F_1$, $\nu_Y(F\cap Y)=0$.   But this
means that $\nu_Y$ takes no values in $\ooint{0,\infty}$ and is not
semi-finite.\ \Qed
}%end of proof of 419C

\cmmnt{\medskip

\noindent{\bf Remark} $X$ here is not locally compact.   But as it is
Hausdorff and completely regular, it can be embedded as a subspace of a
locally compact Radon measure space $(X',\frak T',\Sigma',\mu')$ (416T).
Now $\mu'$ still has the properties (i)-(iii).
}%end of comment

\leader{419D}{Example}\cmmnt{ ({\smc Fremlin 75b})} There is a
complete locally determined $\tau$-additive completion regular
topological measure space $(X,\frak T,\Sigma,\mu)$ in which $\mu$ is
tight and compact sets have finite measure, but $\mu$ is not
localizable.

\proof{{\bf (a)} Let $I$ be a set with cardinal greater than $\frak c$.
Set $X=[0,1]^I$.   For $i\in I$, $t\in[0,1]$ set
$X_{it}=\{x:x\in X,\,x(i)=t\}$.   Give $X_{it}$ its natural topology
$\frak T_{it}$ and
measure $\mu_{it}$, with domain $\Sigma_{it}$, defined from the
expression of $X_{it}$ as $[0,1]^{I\setminus\{i\}}\times\{t\}$, each
factor $[0,1]$ being given its usual topology and Lebesgue measure, and
the singleton factor $\{t\}$ being given its unique (discrete) topology
and (atomic) probability measure.   By 416U, $\mu_{it}$ is a completion
regular Radon measure.   Set

\Centerline{$\frak T=\{G:G\subseteq X,\,G\cap X_{it}\in\frak T_{it}$ for
all $i\in I$, $t\in[0,1]\}$,}

\Centerline{$\Sigma=\{E:E\subseteq X,\,E\cap X_{it}\in\Sigma_{it}$ for
all $i\in I$, $t\in[0,1]\}$,}

\Centerline{$\mu E=\sum_{i\in I,t\in[0,1]}\mu_{it}(E\cap X_{it})$ for
every $E\in\Sigma$.}

\noindent (Compare 216D.)   Then it is easy to check that $\frak T$ is a
topology.   $\frak T$ is Hausdorff because it is
finer\cmmnt{(= larger)} than the usual
topology $\frak S$ on $X$;  because each $\frak T_{it}$ is the subspace
topology induced by $\frak S$, it is also the subspace topology induced
by $\frak T$.   Next, the definition of $\mu$ makes it a locally
determined measure;  it is a tight complete topological measure because
every $\mu_{it}$ is.

\medskip

{\bf (b)} If $K\subseteq X$ is compact, $\mu K<\infty$.   \Prf\Quer\
Otherwise, $M=\{(i,t):i\in I,\,t\in[0,1],\,\mu_{it}(K\cap X_{it})>0\}$
must be infinite.   Take any sequence $\sequencen{(i_n,t_n)}$ of
distinct elements of $M$.   Choose a sequence $\sequencen{x_n}$ in $K$
inductively, as follows.   Given $\langle x_m\rangle_{m<n}$, then set
$C_{ni}=\{x_m(i):m<n\}$ for each $i\in I$, and

\Centerline{$A_n
=\{x:x\in X_{i_nt_n},\,x(i)\notin C_{ni}$
  for $i\in I\setminus\{i_n\}\}$;}

\noindent then $\mu_{i_nt_n}^*A_n=1$ (254Lb).   Since
$\mu_{i_nt_n}(X_{it}\cap X_{ju})=0$ whenever $(i,t)\ne(j,u)$, there must
be some $x_n\in K\cap A_n\setminus\bigcup_{m\ne n}X_{i_m,t_m}$.
Continue.

This construction ensures that if $i\in I$ and $m<n$, either $i\ne
i_n$ so $x_n(i)\notin C_{ni}$ and $x_n(i)\ne x_m(i)$, or $i=i_n\ne i_m$
and $x_m\notin X_{i_nt_n}$ so $x_n(i)=t_n\ne x_m(i)$, or $i=i_m=i_n$ and
$x_n(i)=t_n\ne t_m=x_m(i)$.   But this means that $\{x_n:n\in\Bbb N\}$
is an infinite set meeting each $X_{it}$ in at most one point, and is
closed for $\frak T$;  so $\sequencen{x_n}$ has no cluster point for
$\frak T$, which is impossible.\ \Bang\Qed

\medskip

{\bf (c)} $\mu$ is not localizable.   \Prf\ Fix on any $k\in I$ and
consider $\Cal E=\{X_{kt}:t\in[0,1]\}$.   \Quer\ If $E\in\Sigma$ is an
essential supremum for $\Cal E$, then $E\cap X_{kt}$ must be
$\mu_{kt}$-conegligible for every $t\in [0,1]$.   We can therefore find
a countable set $J_t\subseteq I$ and a $\mu_{kt}$-conegligible set
$F_t\subseteq E\cap X_{kt}$, determined by coordinates in $J_t$.   At
this point recall that $\#(I)>\frak c$, so there is some $j\in
I\setminus(\{k\}\cup\bigcup_{t\in[0,1]}J_t)$.   Since $X_{j0}\cap
X_{kt}$ is negligible for every $t\in[0,1]$, $X_{j0}\cap E$ must be
negligible, and $\int_0^1\nu H_tdt=0$, where
\Centerline{$H_t=\{y:y\in[0,1]^{I\setminus\{j,k\}},\,(y,0,t)\in E\}$}

\noindent and $\nu$ is the usual measure on $[0,1]^{I\setminus\{j,k\}}$,
identifying $X$ with $[0,1]^{I\setminus\{j,k\}}\times[0,1]\times[0,1]$.
But because $F_t$ is determined by coordinates in $I\setminus\{j\}$, we
can identify it with $F'_t\times[0,1]\times\{t\}$ where $F'_t$ is a
$\nu$-conegligible subset of $[0,1]^{I\setminus\{j,k\}}$, and
$F'_t\subseteq H_t$, so $\nu H_t=1$ for every $t$, which is absurd.\
\Bang

Thus $\Cal E$ has no essential supremum in $\Sigma$, and $\mu$ cannot be
localizable.\ \Qed

\medskip

{\bf (d)} I have still to check that $\mu$ is completion regular.
\Prf\ If $E\in\Sigma$ and $\mu E>0$, there are $i\in I$, $t\in[0,1]$
such that $\mu_{it}(E\cap X_{it})>0$, and an $F\subseteq E\cap X_{it}$,
a zero set for the subspace topology of $X_{it}$, such that
$\mu_{it}F>0$.   But now
observe that $X_{it}$ is a zero set in $X$ for the usual topology
$\frak S$, so that $F$ is a zero set for $\frak S$ (4A2G(c-i))
and therefore
for the finer topology $\frak T$.   By 412B, this is enough to show that
$\mu$ is inner regular with respect to the zero sets.\ \Qed
}%end of example 419D

\cmmnt{\medskip

\noindent{\bf Remark} It may be worth noting that the topology $\frak T$
here is not regular.   See {\smc Fremlin 75b}, p.\ 106.
}%end of comment

\leader{419E}{Example}\cmmnt{ ({\smc Fremlin 76})} Let
$(Z,\frak S,\Tau,\nu)$ be the Stone space of the
measure algebra of Lebesgue measure on $[0,1]$, so that $\nu$ is a
strictly positive completion regular Radon probability
measure\cmmnt{ (411P)}.   Then the c.l.d.\ product
measure\cmmnt{ $\lambda$} on $Z\times Z$ is not a topological measure,
so is not equal to the $\tau$-additive product measure $\tilde\lambda$,
and $\tilde\lambda$ is not completion regular.

\proof{ Consider the sets $W$, $\tilde W$ described in 346K.   We have
$W\in\Lambda=\dom\lambda$ and $\tilde W=\bigcup\Cal V$, where

\Centerline{$\Cal V=\{G\times H:G,\,H\subseteq Z$ are
open-and-closed, $(G\times H)\setminus W$ is negligible$\}$.}

\noindent $\tilde W$ is a union of open sets, therefore must be open in
$Z^2$.   And $\lambda_*\tilde W\le\lambda W$.   \Prf\Quer\ Otherwise,
there is a $V\in\Lambda$ such that $V\subseteq\tilde W$ and
$\lambda V>\lambda W$.   Now $\lambda$ is tight, by 412Sb, so there is a
compact set $K\subseteq V$ such that $K\in\Lambda$ and
$\lambda K>\lambda W$.   There must be $U_0,\ldots,U_n\in\Cal V$ such
that $K\subseteq\bigcup_{i\le n}U_i$.   But $\lambda(U_i\setminus W)=0$
for every $i$, so $\lambda(K\setminus W)=0$ and
$\lambda K\le\lambda W$.\ \Bang\Qed

However, the construction of 346K arranged that $\lambda^*\tilde W$
should be $1$ and $\lambda W$ strictly less than $1$.   So
$\lambda_*\tilde W<\lambda^*\tilde W$ and $\tilde W\notin\Lambda$.
Accordingly $\lambda$ is not a topological measure and cannot be equal
to the Radon measure $\tilde\lambda$ of 417P.

We know that $\lambda$ is inner regular with respect to the zero sets
(412Sc) and is defined on every zero set (417V), while $\tilde\lambda$
properly extends $\lambda$.   But this means that $\tilde\lambda$ cannot
be inner regular with respect to the zero sets, by 412Mb, that is, cannot
be completion regular.
}%end of proof of 419E

\leader{419F}{Theorem}\cmmnt{ ({\smc Rao 69})}
%also {\smc Kunen 68}
$\Cal P(\omega_1\times\omega_1)
=\Cal P\omega_1\tensorhat\Cal P\omega_1$\cmmnt{, the $\sigma$-algebra
of subsets of
$\omega_1$ generated by $\{E\times F:E,\,F\subseteq\omega_1\}$}.

\proof{{\bf (a)} Because $\omega_1\le\frak c$, there is an injection
$h:\omega_1\to\{0,1\}^{\Bbb N}$;  set $E_i=\{\xi:h(\xi)(i)=1\}$ for each
$i\in\Bbb N$.

\medskip

{\bf (b)} Suppose that $A\subseteq\omega_1$ has countable vertical
sections.   Then $A\in\Cal P\omega_1\tensorhat\Cal P\omega_1$.   \Prf\
Set $B=A^{-1}[\omega_1]$ and for $\xi\in B$ choose a surjection
$f_{\xi}:\Bbb N\to A[\{\xi\}]$.   Set $g_n(\xi)=f_{\xi}(n)$ for $\xi\in B$
and $n\in\Bbb N$, and $A_n=\{(\xi,f_{\xi}(n)):\xi\in B\}$ for
$n\in\Bbb N$.   Then

$$\eqalign{A_n
&=\{(\xi,\eta):\xi\in B,\,\eta=g_n(\xi)\}\cr
&=\{(\xi,\eta):\xi\in B,\,\eta<\omega_1,\,h(g_n(\xi))=h(\eta)\}\cr
&=\bigcap_{i\in\Bbb N}
\bigl(\{(\xi,\eta):\xi\in g_n^{-1}[E_i],\,\eta\in E_i\}
  \cup\{(\xi,\eta):\xi\in B\setminus g_n^{-1}[E_i],
  \,\eta\in\omega_1\setminus E_i\}\bigr)\cr
&\in\Cal P\omega_1\tensorhat\Cal P\omega_1.\cr}$$

\noindent So

\Centerline{$A=\bigcup_{n\in\Bbb N}A_n\in\Cal P\omega_1\tensorhat\Cal
P\omega_1$.   \Qed}

\medskip

{\bf (c)} Similarly, if a subset of $\omega_1\times\omega_1$ has
countable horizontal sections, it belongs to $\Cal
P\omega_1\tensorhat\Cal
P\omega_1$.   But for any $A\subseteq\omega_1\times\omega_1$, $A=A'\cup
A''$ where

$$\eqalign{A'
&=\{(\xi,\eta):(\xi,\eta)\in A,\,\eta\le\xi\}
  \text{ has countable vertical sections},\cr
A''
&=\{(\xi,\eta):(\xi,\eta)\in A,\,\xi\le\eta\}
  \text{ has countable horizontal sections},\cr}$$

\noindent so both $A'$ and $A''$ belong to $\Cal P\omega_1\tensorhat\Cal
P\omega_1$ and $A$ also does.
}%end of proof of 419F

\leader{419G}{Corollary}\cmmnt{ ({\smc Ulam 1930})} Let $Y$ be a set of
cardinal at most $\omega_1$ and $\mu$ a semi-finite measure with
domain $\Cal PY$.   Then $\mu$ is point-supported;  in particular,
if $\mu$ is $\sigma$-finite
there is a countable conegligible set $A\subseteq Y$.

\proof{ \Quer\ Suppose, if possible, otherwise.

\medskip

{\bf (a)} We can suppose that $Y$ is either countable or actually equal
to $\omega_1$.   Let $\mu_0$ be the point-supported part of $\mu$, that
is, $\mu_0A=\sum_{y\in A}\mu\{y\}$ for every
$A\subseteq Y$;  then $\mu_0$ is a point-supported measure (112Bd), so
is not equal to $\mu$.   Let $A\subseteq Y$ be
such that $\mu_0A\ne\mu A$.   Then $\mu_0A<\mu A$;  because $\mu$ is
semi-finite, there is a set $B\subseteq A$ such that $\mu_0A<\mu B<\infty$.
Set $\nu C=\mu(B\cap C)-\mu_0(B\cap C)$ for $C\subseteq Y$;  then
$\nu$ is a non-zero totally finite measure with domain $\Cal PY$,
and is zero on singletons.

\medskip

{\bf (b)} As $\nu C=0$ for every countable $C\subseteq Y$,
$Y$ is uncountable and $Y=\omega_1$.
Let $\lambda=\nu\times\nu$ be the
product measure on $\omega_1\times\omega_1$.   By 419F, the domain of
$\lambda$ is the whole of $\Cal P(\omega_1\times\omega_1)$;  in
particular, it contains
the set $V=\{(\xi,\eta):\xi\le\eta<\omega_1\}$.   Now by Fubini's
theorem

\Centerline{$\lambda V=\biggerint\nu V[\{\xi\}]\nu(d\xi)
=\int\nu(\omega_1\setminus\xi)\nu(d\xi)
=(\nu\omega_1)^2>0$,}

\noindent and also

\Centerline{$\lambda V=\biggerint\nu V^{-1}[\{\eta\}]\nu(d\eta)
=\int\nu(\eta+1)\nu(d\eta)
=0$.  \Bang}
}%end of proof of 419G

\cmmnt{\medskip

\noindent{\bf Remark} I ought to remark that this result, though not
419F, is valid for many other cardinals besides $\omega_1$;  see, in
particular, 438C below.   There will be more on this topic in Chapter 54 of
Volume 5.
}%end of comment

\leader{419H}{}\cmmnt{ For the next two examples
it will be helpful to know
some basic facts about Lebesgue measure which seemed a little advanced
for Volume 1 and for which I have not found a suitable place since.

\medskip

\noindent}{\bf Lemma} If $(X,\frak T,\Sigma,\mu)$ is an atomless
Radon measure space and $E\in\Sigma$ has non-zero measure, then
$\#(E)\ge\frak c$.

\proof{ The subspace measure on $E$ is a Radon measure (416Rb), therefore
compact and perfect (416Wa), and is not purely atomic;
by 344H, there is in fact a
negligible subset of $E$ with cardinal $\frak c$.
}%end of proof of 419H

\leader{419I}{}\cmmnt{ The next result is a strengthening of 134D.

\medskip

\noindent}{\bf Lemma} Let $\mu$ be Lebesgue measure on $\Bbb R$, and $H$
any measurable subset of $\Bbb R$.   Then there is a disjoint family
$\langle A_{\alpha}\rangle_{\alpha<\frak c}$ of subsets of $H$ such that
$H$ is a measurable envelope of every $A_{\alpha}$;  in particular,
$\mu_*A_{\alpha}=0$ and $\mu^*A_{\alpha}=\mu H$ for every
$\alpha<\frak c$.

\proof{ If $\mu H=0$, we can take every $A_{\alpha}$ to be empty;  so
suppose that $\mu H>0$.   Let $\Cal E$ be the family of closed subsets
of $H$ of non-zero measure.   By 4A3Fa, $\#(\Cal E)\le\frak c$;
enumerate $\Cal E\times\frak c$ as
$\langle(F_{\xi},\alpha_{\xi})\rangle_{\xi<\frak c}$
(3A1Ca).   Choose $\langle x_{\xi}\rangle_{\xi<\frak c}$ inductively, as
follows.   Given $\langle x_{\eta}\rangle_{\eta<\xi}$, where
$\xi<\frak c$, $F_{\xi}$ has cardinal (at least) $\frak c$, by 419H, so
cannot be included in $\{x_{\eta}:\eta<\xi\}$;  take any
$x_{\xi}\in F_{\xi}\setminus\{x_{\eta}:\eta<\xi\}$, and continue.

At the end of the induction, set

\Centerline{$A_{\alpha}=\{x_{\xi}:\xi<\frak c,\,\alpha_{\xi}=\alpha\}$.}

\noindent Then the $A_{\alpha}$ are disjoint just because the $x_{\xi}$
are distinct.

\Quer\ Suppose, if possible, that $H$ is not a measurable envelope of
$A_{\alpha}$ for
some $\alpha$.   Then $\mu_*(H\setminus A_{\alpha})>0$ (413Ei), so there
is a non-negligible measurable set $E\subseteq H\setminus A_{\alpha}$.
Now there is an $F\in\Cal E$ such that
$F\subseteq E$.   Let $\xi<\frak c$ be such that $F=F_{\xi}$ and
$\alpha=\alpha_{\xi}$;  then $x_{\xi}\in A_{\alpha}\cap F$, which is
impossible.\ \Bang

Thus $H$ is always a measurable envelope of $A_{\alpha}$.   It follows
from the definition of `measurable envelope' that
$\mu^*A_{\alpha}=\mu H$.   But also, if $\alpha<\frak c$,
$\mu_*A_{\alpha}\le\mu_*(H\setminus A_{\alpha+1})$, which is $0$, as we
have just seen.   So we have a suitable family.
}%end of proof of 419I

\leader{419J}{Example}\dvAformerly{4{}19H}
There is a complete probability space
$(X,\Sigma,\mu)$ with a Hausdorff topology $\frak T$
on $X$ such that $\mu$ is
$\tau$-additive and inner regular with respect to the Borel sets,
$\frak T$ is generated by $\frak T\cap\Sigma$, but $\mu$ has no
extension to a topological measure.

\proof{{\bf (a)} Set $Y=\omega_1+1=\omega_1\cup\{\omega_1\}$.
Let $\Tau$ be the $\sigma$-algebra of subsets of $Y$ generated by
$\{\{\xi\}:\xi<\omega_1\}$.   Let $\nu$ be the
probability measure with domain $\Tau$ defined by the formula

\Centerline{$\nu F=\Bover12\#(F\cap\{0,\omega_1\})$ for every
$F\in\Tau$.}

Set

\Centerline{$\frak S=\{\emptyset,Y\}\cup\{H:0\in H\subseteq\omega_1\}$.}

\noindent This is a topology on $Y$, and every subset of $Y$ is a Borel
set for $\frak S$;  so $\nu$ is surely inner regular with respect to the
Borel sets.

Note that

\Centerline{$\{\{0,\alpha\}:\alpha<\omega_1\}\cup\{Y\}$}

\noindent is a base for $\frak S$ included in $\Tau$.

\medskip

{\bf (b)} Set $Z=Y^{\Bbb N}\times[0,1]$.   Let $\lambda$ be the
product probability measure on $Z$
when each copy of $Y$ is given the measure
$\nu$ and $[0,1]$ is given Lebesgue measure $\mu_L$;  let
$\frak U$ be the product topology when each copy of $Y$ is given the
topology $\frak S$ and $[0,1]$ its usual topology.
Let $\Lambda$ be the domain of $\lambda$ and $\Lambda_0$ the
$\sigma$-algebra generated by sets of the form
$\{(x,t):x(i)\in F$, $t\in[0,1]$, $t<q\}$ where $i\in\Bbb N$,
$F\in\Tau$ and
$q\in\Bbb Q$;  then $\nu$ is inner regular with respect to $\Lambda_0$
(see 254Ff).
Note that $\frak U\cap\Lambda_0$ is a base for $\frak U$ because
$\frak S\cap\Tau$ is a base for $\frak S$, and $\lambda$ is inner regular
with respect to the $\frak U$-Borel sets (412Uc, or otherwise).

Define $\phi:\{0,1\}^{\Bbb N}\to Y^{\Bbb N}$ by setting

$$\eqalign{\phi(u,t)(n)&=0\text{ if }u(n)=0,\cr
&=\omega_1\text{ if }u(n)=1,\cr}$$

\noindent and set $\psi(u,t)=(\phi(u),t)$ for $u\in\{0,1\}^{\Bbb N}$,
$t\in[0,1]$.
Then $\psi$ is continuous, just because $\frak U$ is a
product topology.
Let $\nu_{\omega}\times\mu_L$ be the product measure on
$\{0,1\}^{\Bbb N}\times[0,1]$;
then $\psi$ is \imp\ for $\nu_{\omega}\times\mu_L$ and $\lambda$, by 254H.

If $V\in\Lambda_0$ there is an $\alpha<\omega_1$ such that

\inset{\noindent
if $x$, $y\in Y^{\Bbb N}$, $t\in[0,1]$ and $y(i)=x(i)$ whenever
$\min(x(i),y(i))<\alpha$, then $(x,t)\in V$ iff $(y,t)\in V$.}

\noindent \Prf\ Let $\Cal W$ be the family of sets
$V\subseteq Z$ with this property.   Then $\Cal W$ is a $\sigma$-algebra
of subsets of $Z$ containing every measurable cylinder,
so includes $\Lambda_0$.\ \QeD\

\medskip

{\bf (c)} $\#(\Lambda_0)\le\frak c$.   \Prf\ Set

\Centerline{$A_{i\xi}
=\{(x,t):x\in Y^{\Bbb N}$, $t\in[0,1]$, $x(i)=\xi\}$,
\quad $A'_q
=\{(x,t):x\in Y^{\Bbb N}$, $t\in[0,1]$, $t\le q\}$;}

\noindent for $i\in\Bbb N$,
$\xi<\omega_1$ and $q\in\Bbb Q$, and

\Centerline{$\Cal A=\{A_{i\xi}:i\in\Bbb N$, $\xi<\omega_1\}
\cup\{A'_q:q\in\Bbb Q\}$.}

\noindent Then $\Lambda_0$ is the
$\sigma$-algebra of subsets of $Z$ generated by
$\Cal A$, and $\#(\Cal A)=\omega_1\le\frak c$,
so $\#(\Lambda_0)\le\frak c$ (4A1O).\ \Qed

\medskip

{\bf (d)} There is a family $\ofamily{\xi}{\frak c}{z_{\xi}}$ in
$Z$ such that

\inset{($\alpha$)
whenever $W\in\Lambda$ and $\lambda W>0$ there is a $\xi<\frak c$
such that $z_{\xi}\in W$ and $\lambda(H\cap W)>0$ whenever $H$ is a
measurable open subset of $Z$ containing $z_{\xi}$,

($\beta$)
setting $z_{\xi}=(x_{\xi},t_{\xi})$ for each $\xi$, there is for every
$\xi<\frak c$ a $j\in\Bbb N$ such that $0<x_{\xi}(j)<\omega_1$,

($\gamma$) $t_{\xi}\ne t_{\eta}$ if $\eta<\xi<\frak c$.}

\noindent\Prf\ By 4A3Fa, the set of closed subsets of
$\{0,1\}^{\Bbb N}\times[0,1]$ has
cardinal at most $\frak c$, so there is a family
$\ofamily{\xi}{\frak c}{(K_{\xi},V_{\xi})}$ running over all pairs $(K,V)$
such that $V\in\Lambda_0$ and
$K\subseteq\psi^{-1}[V]$ is a non-negligible compact set.
Choose $\ofamily{\xi}{\frak c}{(x_{\xi},t_{\xi})}$ inductively, as follows.
Given $\ofamily{\eta}{\xi}{t_{\eta}}$ where $\xi<\frak c$, then

\Centerline{$\{t:t\in[0,1]$,
$\nu_{\omega}K_{\xi}^{-1}[\{t\}]>0\}$}

\noindent is a non-negligible measurable subset of $[0,1]$,
so has cardinal $\frak c$ (419H);  let $t_{\xi}$ be a point of this set
distinct from every $t_{\eta}$ for $\eta<\xi$.   Now Lemma 345E tells us
that there are points $u$, $u'\in K_{\xi}^{-1}[\{t_{\xi}\}]$
which differ at exactly one coordinate $j\in\Bbb N$;  suppose that
$u(j)=1$ and $u'(j)=0$.

Let $\alpha<\omega_1$ be such that
if $x$, $y\in Y^{\Bbb N}$, $t\in[0,1]$ and $y(i)=x(i)$ whenever
$\min(x(i),y(i))<\alpha$, then $(x,t)\in V_{\xi}$ iff $(y,t)\in V_{\xi}$.
Define $x_{\xi}\in Y^{\Bbb N}$ by setting $x_{\xi}(j)=\alpha$ and
$x_{\xi}(i)=\phi(u)(i)$ for $i\ne j$.   Then
$z_{\xi}=(x_{\xi},t_{\xi})$ belongs to $V_{\xi}$.   If $H\subseteq Z$ is
any open set containing $z_{\xi}$, we have a sequence
$\sequence{i}{H_i}$ in $\frak S$ such that
$x_{\xi}\in\prod_{i\in\Bbb N}H_i$ and
$\prod_{i\in\Bbb N}H_i\times\{t_{\xi}\}\subseteq H$;  now $H_j\ne\emptyset$
so $0\in H_j$ and $\phi(u')\in\prod_{i\in\Bbb N}H_i$, so that
$(u',t_{\xi})\in K_{\xi}\cap\psi^{-1}[H]$.
Continue.

The construction ensures that ($\beta$) and ($\gamma$) are satisfied.
Now, if $\lambda W>0$, let $V\in\Lambda_0$ be such that $V\subseteq W$ and
$\lambda V>0$.   In this case,
$(\nu_{\omega}\times\mu_L)(\psi^{-1}[V])>0$;  let
$K\subseteq\psi^{-1}[V]$ be a self-supporting non-negligible compact set.
Let $\xi<\frak c$ be such that $(K,V)=(K_{\xi},V_{\xi})$.   Then
$z_{\xi}\in V_{\xi}=V\subseteq W$.   If $H$ is a measurable open subset of
$Z$ containing $z_{\xi}$, then $K\cap\psi^{-1}[H]$ is not empty;  as $\psi$
is continuous and \imp\ and $K$ is self-supporting,

\Centerline{$0<(\nu_{\omega}\times\mu_L)(K\cap\psi^{-1}[H])
\le(\nu_{\omega}\times\mu_L)\psi^{-1}[V\cap H]
=\lambda(V\cap H)\le\lambda(W\cap H)$.}

\noindent So ($\alpha$) is satisfied.\ \Qed

\medskip

{\bf (e)} Set $X=\{z_{\xi}:\xi<\frak c\}$ and let $\mu$ be the subspace
measure on $X$ induced by $\lambda$;  let $\frak T$ be the subspace
topology on $X$.

\medskip

\quad{\bf (i)} $\lambda$ is complete, so $\mu$ also is.   Next,
$\mu X=\lambda^*X=1$.   \Prf\Quer\ Otherwise, there is a $W\in\Lambda$ such
that $\lambda W>0$ and $X\cap W=\emptyset$.   But we know that there is now
some $\xi<\frak c$ such that $z_{\xi}\in W$.\ \Bang\Qed

\medskip

\quad{\bf (ii)} $\frak T$ is Hausdorff because the projection from $X$ to
$[0,1]$ is injective and continuous.   $\frak T$ is generated by
$\frak T\cap\Sigma$ because $\frak U$ is generated by $\frak U\cap\Lambda$.
$\mu$ is inner regular with respect to the Borel sets because $\lambda$ is
(412Pb).

\medskip

\quad{\bf (iii)} $\mu$ is $\tau$-additive.   \Prf\Quer\ Suppose, if
possible, otherwise.   Then there is an upwards-directed family $\Cal G$
of measurable open subsets of $X$ such that $G^*=\bigcup\Cal G$ is
measurable and $\mu G^*>\sup_{G\in\Cal G}\mu G$.   Let $\Cal H$ be the
family of sets $H\in\Lambda\cap\frak U$ such that $H\cap X$ is
included in some member of $\Cal G$;  because $\frak U$ is generated by
$\frak U\cap\Lambda$, $G^*=W\cap X$, where $W=\bigcup\Cal H$.
At the same time, there is a $V\in\Lambda$ such that $G^*=X\cap V$.

Because
$\Cal G$ is upwards-directed, so is $\Cal H$.   Because $X$ has full outer
measure,

\Centerline{$\sup_{H\in\Cal H}\lambda H=\sup_{H\in\Cal H}\mu(X\cap H)
\le\sup_{G\in\Cal G}\mu G<\mu G^*=\lambda V$.}

\noindent Let $\sequencen{H_n}$ be a non-decreasing sequence in $\Cal H$
such that $\sup_{n\in\Bbb N}\lambda H_n=\sup_{H\in\Cal H}\lambda H$, and
set $W_0=\bigcup_{n\in\Bbb N}H_n$;  then $\lambda W_0<\lambda V$ and
$\lambda(H\setminus W_0)=0$ for every $H\in\Cal H$.
However, $\lambda(V\setminus W_0)>0$, so
there is a $z\in X\cap V\setminus W_0$ such that
$\lambda(H\cap V\setminus W_0)>0$ for every measurable open set $H$
containing $z$.   As $z\in X\cap V=X\cap W$,
there must be an $H\in\Cal H$ containing $z$, so this is
impossible.\ \Bang\Qed

\medskip

\quad{\bf (iv)} \Quer\ Suppose, if possible, that there is a
topological measure $\tilde\mu$ on $X$ agreeing with $\mu$
on every open set in the domain of $\mu$.   For each $i\in\Bbb N$, set
$\pi_i(x)=x(i)$ for $(x,t)\in X$.   Every subset of $Y$ is a Borel set for
$\frak S$;  because $\pi_i$ is continuous for $\frak T$ and $\frak S$,
the image measure
$\tilde\mu\pi_i^{-1}$ has domain $\Cal PY$.   Now $\#(Y)=\omega_1$,
so there must be a countable conegligible set (419G), and there must be
some $\alpha_i<\omega_1$ such that
$\tilde\mu\pi_i^{-1}(\omega_1\setminus\alpha_i)=0$.   On the other hand,

\Centerline{$\tilde\mu\pi_i^{-1}(\alpha_i\setminus\{0\})
=\mu\pi_i^{-1}(\alpha_i\setminus\{0\})
=\lambda\{(x,t):0<x(i)<\alpha_i\}
=\nu(\alpha_i\setminus\{0\})=0$,}

\noindent so $\tilde\mu\pi_i^{-1}(\omega_1\setminus\{0\})=0$.

But (d-$\beta$) ensures that

\Centerline{$X=\bigcup_{i\in\Bbb N}\pi_i^{-1}(\omega_1\setminus\{0\})$,}

\noindent so this is impossible.\ \Bang

Thus we have the required example.
}%end of proof of 419J

\cmmnt{\medskip

\noindent{\bf Remark} I note that the topology of $X$ is not regular.
Of course the phenomenon here cannot arise with regular spaces, by
415M.}

\leader{419K}{Example} ({\smc Blackwell 56}) There are sequences
$\sequencen{X_n}$, $\sequencen{\frak T_n}$ and $\sequencen{\nu_n}$ such
that (i) for each $n$,
$(X_n,\frak T_n)$ is a separable metrizable space and $\nu_n$ is a
quasi-Radon probability measure on $Z_n=\prod_{i\le n}X_i$ (ii) for
$m\le n$ the canonical map $\pi_{mn}:Z_n\to Z_m$ is \imp\ (iii) there is
no probability measure on $Z=\prod_{i\in\Bbb N}X_i$ such that all the
canonical maps from $Z$ to $Z_n$ are \imp.

\proof{ Let $\sequencen{A_n}$ be a disjoint sequence of subsets of
$[0,1]$ such that $\mu_*([0,1]\setminus A_n)=0$, that is, $\mu^*A_n=1$
for every $n$, where $\mu$ is Lebesgue measure (using 419I).   Set
$X_n=\bigcup_{i\ge n}A_i$, so that $\sequencen{X_n}$ is a non-increasing
sequence of sets of outer measure $1$ with empty intersection.   For
each $n\ge 1$, we have a map $f_n:X_n\to Z_n$ defined by setting
$f_n(x)(i)=x$ for every $i\le n$, $x\in X_n$.   Let $\nu_n$ be the image
measure $\mu_{X_n}f_n^{-1}$, where $\mu_{X_n}$ is the subspace measure
on $X_n$ induced by $\mu$.   Note that $f_n$ is a homeomorphism between
$X_n$ and the diagonal $\Delta_n=\{z:z\in Z_n,\,z(i)=z(j)$ for all $i$,
$j\le n\}$, which is a closed subset of $Z_n$;  so that $\nu_n$, like
$\mu_{X_n}$, is a quasi-Radon probability measure.

If $m\le n$, then $\pi_{mn}$ is \imp, where $\pi_{mn}(z)(i)=z(i)$ for
$z\in Z_n$ and $i\le m$.   \Prf\ If $W\subseteq Z_m$ is measured by
$\nu_m$, then $f_m^{-1}[W]$ is measured by $\mu_{X_m}$, so is of the
form $X_m\cap E$ where $E$ is Lebesgue measurable.   But in this case
$f_n^{-1}[\pi_{mn}^{-1}[W]]=X_n\cap E$, so that

\Centerline{$\nu_n(\pi_{mn}^{-1}[W])
=\mu_{X_n}(f_n^{-1}[\pi_{mn}^{-1}[W]])
=\mu^*(X_n\cap E)
=\mu E
=\mu^*(X_m\cap E)
=\nu_mW$.   \Qed}

\Quer\ But suppose, if possible, that there is a probability measure
$\nu$ on $Z=\prod_{i\in\Bbb N}X_i$ such that $\pi_n:Z\to Z_n$ is \imp\
for every $n$, where $\pi_n(z)(i)=z(i)$ for $z\in Z$ and $i\le n$.   Then

\Centerline{$\nu\pi_n^{-1}[\Delta_n]=\nu_n\Delta_n
=\mu_{X_n}f_n^{-1}[\Delta_n]=1$}

\noindent for each $n$, so

\Centerline{$1=\nu(\bigcap_{n\in\Bbb N}\pi_n^{-1}[\Delta_n])
=\nu\{z:z\in Z,\,z(i)=z(j)$ for all $i$, $j\in\Bbb N\}
=\nu\emptyset$,}

\noindent because $\bigcap_{n\in\Bbb N}X_n=\emptyset$;  which is
impossible.\ \Bang
}%end of proof of 419K

\leader{419L}{The split interval again (a)}\cmmnt{ For the sake of an
example in \S343, I have already introduced the `split interval' or
`double arrow space'.   As this construction gives us a topological
measure space
of great interest, I repeat it here.}   Let $I^{\|}$ be the set
$\{a^+:a\in[0,1]\}\cup\{a^-:a\in[0,1]\}$.   Order it by saying that

\Centerline{$a^+\le b^+\iff a^-\le b^+\iff a^-\le b^-\iff a\le b$,
\quad$a^+\le b^-\iff a<b$.}

\noindent Then\cmmnt{ it is easy to check that} $I^{\|}$ is a totally
ordered space, and\cmmnt{ that it} is Dedekind complete.
\prooflet{(If $A\subseteq[0,1]$ is a non-empty set, then
$\sup_{a\in A}a^-=(\sup A)^-$,
while $\sup_{a\in A}a^+$ is either $(\sup A)^+$ or $(\sup A)^-$,
depending on whether
$\sup A$ belongs to $A$ or not.)}   Its greatest element is $1^+$ and
its least
element is $0^-$.  Consequently the order topology on $I^{\|}$ is a
compact Hausdorff topology\cmmnt{ (4A2Ri,
{\smc Alexandroff \& Urysohn 1929})}.   \cmmnt{Note
that} $Q=\{q^+:q\in[0,1]\cap\Bbb Q\}\cup\{q^-:q\in[0,1]\cap\Bbb Q\}$ is
dense\prooflet{, because it meets
every non-trivial interval in $I^{\|}$}.   \cmmnt{By 4A2E(a-ii) and
4A2Rn,} $I^{\|}$ is ccc and hereditarily Lindel\"of.

\spheader 419Lb If we define $h:I^{\|}\to[0,1]$ by writing
$h(a^+)=h(a^-)=a$
for every $a\in[0,1]$, then $h$ is continuous\prooflet{, because
$\{x:h(x)<a\}=\{x:x<a^-\}$, $\{x:h(x)>a\}=\{x:x>a^+\}$ for every
$a\in[0,1]$}.
Now\cmmnt{ we can describe the Borel sets of $I^{\|}$, as follows:}  a
set $E\subseteq I^{\|}$ is
Borel iff there is a Borel set $F\subseteq[0,1]$ such that
$E\symmdiff h^{-1}[F]$ is countable.   \prooflet{\Prf\ Write $\Sigma_0$
for the family
of subsets $E$ of $I^{\|}$ such that $E\symmdiff h^{-1}[F]$ is countable
for some Borel set $F\subseteq [0,1]$.   It is easy to check that
$\Sigma_0$ is a $\sigma$-algebra of subsets of $I^{\|}$.
(If $E\symmdiff h^{-1}[F]$ is countable, so is
$(I^{\|}\setminus E)\symmdiff h^{-1}[[0,1]\setminus F]$;  if
$E_n\symmdiff h^{-1}[F_n]$ is countable for every $n$, so is
$(\bigcup_{n\in\Bbb N}E_n)\symmdiff h^{-1}[\bigcup_{n\in\Bbb N}F_n]$.)
Because the topology of $I^{\|}$ is Hausdorff, every singleton set is
closed, so every countable set is Borel.   Also $h^{-1}[F]$ is Borel for
every Borel set $F\subseteq[0,1]$, because $h$ is continuous (4A3Cd).
So if $E\symmdiff h^{-1}[F]$ is countable for some Borel set
$F\subseteq[0,1]$,
$E=h^{-1}[F]\symmdiff(E\symmdiff h^{-1}[F])$ is a Borel set in $I^{\|}$.
Thus $\Sigma_0$ is included in the Borel $\sigma$-algebra $\Cal B$ of
$I^{\|}$.   On the other
hand, if $J\subseteq I^{\|}$ is an interval, $h[J]$ also is an interval,
therefore a Borel set, and $h^{-1}[h[J]]\setminus J$ can contain at most
two points, so $J\in\Sigma_0$.   If $G\subseteq I^{\|}$ is open, it is
expressible
as $\bigcup_{i\in I}J_i$, where $\langle J_i\rangle_{i\in I}$ is a
disjoint family of non-empty open intervals (4A2Rj).   As $X$ is ccc,
$I$ must
be countable.   Thus $G$ is expressed as a countable union of members of
$\Sigma_0$ and belongs to $\Sigma_0$.   But this means that the Borel
$\sigma$-algebra $\Cal B$ must
be included in $\Sigma_0$, by the definition of `Borel algebra'.   So
$\Cal B=\Sigma_0$, as claimed.\ \Qed}%end of prooflet

\spheader 419Lc In 343J I described the standard measure $\mu$ on
$I^{\|}$;  its domain is the set $\Sigma
=\{h^{-1}[F]\symmdiff M:F\in\Sigma_L,\,M\subseteq I^{\|},\,\mu_Lh[M]=0\}$,
where $\Sigma_L$ is the set of Lebesgue measurable
subsets of $[0,1]$ and $\mu_L$ is Lebesgue measure,
and $\mu E=\mu_Lh[E]$ for $E\in\Sigma$.   $h$ is \imp\ for $\mu$ and
$\mu_L$.

\cmmnt{The new fact I wish to mention is:}  $\mu$ is a completion
regular Radon measure.
\prooflet{\Prf\ I noted in 343Ja that it is a complete probability
measure;  {\it a fortiori}, it is locally determined and locally finite.
If $G\subseteq I^{\|}$ is open, then we can express it as
$h^{-1}[F]\symmdiff C$ for some Borel set $F\subseteq [0,1]$ and countable
$C\subseteq I^{\|}$ ((b)
above), so it belongs to $\Sigma$;  thus $\mu$ is a topological measure.
If $E\in\Sigma$ and $\mu E>\gamma$, then
$F=[0,1]\setminus h[I^{\|}\setminus E]$
is Lebesgue measurable, and $\mu E=\mu_LF$.   So there is a compact set
$L\subseteq F$ such that $\mu_LL\ge\gamma$.   But now
$K=h^{-1}[L]\subseteq E$ is closed, therefore compact, and
$\mu K\ge\gamma$.   Moreover, $L$
is a zero set, being a closed set in a metrizable space (4A2Lc), so $K$
is a zero set (4A2C(b-iv)).   As $E$ and $\gamma$ are arbitrary, $\mu$
is
inner regular with respect to the compact zero sets, and is a completion
regular Radon measure.\ \Qed}%end of prooflet

\exercises{
\leader{419X}{Basic exercises (a)}
%\spheader 419Xa
Show that the topological space $X$ of 419A is zero-\vthsp dimensional.
%419A

\spheader 419Xb Give an example of a compact Radon probability space in
which every dense G$_{\delta}$ set is conegligible.   \Hint{411P.}
%419B

\spheader 419Xc In 419E, show that we can start from any atomless
probability measure in place of Lebesgue measure on $[0,1]$.
%419E

\sqheader 419Xd(i) Show that if $E\subseteq\BbbR^2$ is Lebesgue
measurable, with non-zero measure, then it cannot be covered by fewer
than $\frak c$ lines.   \Hint{if $H=\{t:\mu_1E[\{t\}]>0\}$, where
$\mu_1$ is Lebesgue measure on $\Bbb R$, then $\mu_1H>0$, so
$\#(H)=\frak c$.   So if we have a family $\Cal L$ of lines, with
$\#(\Cal L)<\frak c$, there must be a $t\in H$ such that
$L_t=\{t\}\times\Bbb R$ does not belong to $\Cal L$.   Now
$\#(L_t\cap E)=\frak c$ and each member of $\Cal L$ meets $L_t\cap E$ in
at most one point.}   (ii) Show that there is a subset $A$ of $\BbbR^2$,
of full outer measure, which meets every vertical line and every
horizontal line in exactly one point.   \Hint{enumerate $\Bbb R$ as
$\ofamily{\xi}{\frakc}{t_{\xi}}$ and the closed sets of non-zero measure
as $\ofamily{\xi}{\frakc}{F_{\xi}}$.   Choose
$\ofamily{\xi}{\frakc}{I_{\xi}}$ such that every $I_{\xi}$ is finite,
no two points of $I_{\xi}\cup\bigcup_{\eta<\xi}I_{\eta}$ lie on any
horizontal or vertical line, the lines $\{t_{\xi}\}\times\Bbb R$ and
$\Bbb R\times\{t_{\xi}\}$ both meet
$I_{\xi}\cup\bigcup_{\eta<\xi}I_{\eta}$, and $I_{\xi}$ meets $F_{\xi}$.}
(iii) Show that there is a subset $B$ of $\BbbR^2$, of full outer
measure, such that every straight line meets $B$ in exactly two points.
\Hint{enumerate the straight lines in $\BbbR^2$ as
$\ofamily{\xi}{\frakc}{L_{\xi}}$.   Choose
$\ofamily{\xi}{\frakc}{J_{\xi}}$ such that every $J_{\xi}$ is finite,
no three points of $J_{\xi}\cup\bigcup_{\eta<\xi}J_{\eta}$ lie on any
line, $L_{\xi}\cap(J_{\xi}\cup\bigcup_{\eta<\xi}J_{\eta})$ has just two
points and $J_{\xi}\cap F_{\xi}\ne\emptyset$.}
%419H

\spheader 419Xe Show that there is a subset $A$ of the Cantor set $C$
(134G) such that $A+A$ is not Lebesgue measurable.
\Hint{enumerate the closed
non-negligible subsets of $C+C=[0,2]$ as $\ofamily{\xi}{\frak c}{F_{\xi}}$.
Choose $x_{\xi}\in C$, $y_{\xi}\in C$, $z_{\xi}\in F_{\xi}$ so that
$z_{\xi}\notin A_{\xi}+A_{\xi}$ and $x_{\xi}+y_{\xi}\in F_{\xi}$ and
$A_{\xi+1}+A_{\xi+1}$ does not meet $\{z_{\eta}:\eta\le\xi\}$, where
$A_{\xi}=\{x_{\eta}:\eta<\xi\}\cup\{y_{\eta}:\eta<\xi\}$.}
%419H

\spheader 419Xf Let $\BbbR^{\|}$ be the {\bf split line}, that is, the
set $\{a^+:a\in\Bbb R\}\cup\{a^-:a\in\Bbb R\}$, ordered by the rules in
419L. Show that $\BbbR^{\|}$ is a Dedekind complete totally ordered
set, so that
its order topology $\frak T$ is locally compact.   Write $\mu_L$ for
Lebesgue measure
on $\Bbb R$ and $\Sigma_L$ for its domain.   Set $h(a^+)=h(a^-)=a$ for
$a\in\Bbb R$, $\Sigma
=\{E:E\subseteq X,\,h[E]\in\Sigma_L,\,\mu_L(h[E]\cap h[X\setminus E])=0\}$,
$\mu E=\mu_Lh[E]$ for $E\in\Sigma$.   Show that
$\mu$ is a completion regular Radon measure on $\BbbR^{\|}$ and that
$h$ is continuous and \imp\ for $\mu$ and
$\mu_L$.   Show that the set $\{a^+:a\in\Bbb R\}$, with the induced
topology and measure, is isomorphic, as quasi-Radon measure space, to
the Sorgenfrey line (415Xc) with Lebesgue measure.   Show
that $\BbbR^{\|}$ and the Sorgenfrey line are hereditarily Lindel\"of.
%419L

\spheader 419Xg Let $\mu$ be Lebesgue measure on $[0,1]$ and $\Sigma$
its domain.   Let $I^{\|}$ be the split interval.   (i) Show that the
functions $x\mapsto x^+:[0,1]\to I^{\|}$ and
$x\mapsto x^-:[0,1]\to I^{\|}$ are measurable.
\Hint{419Lb.}   (ii) Show that the function
$x\mapsto(x^+,x^-):[0,1]\to (I^{\|})^2$ is not measurable.  \Hint{the
subspace topology on $\{(x^+,x^-):x\in[0,1]\}$ is discrete.}
%419L

\sqheader 419Xh(i) Again writing $I^{\|}$ for the split interval, show
that the function which exchanges $x^+$ and $x^-$ for every $x\in[0,1]$
is a Borel automorphism and an automorphism for the usual Radon measure
$\nu$ on $I^{\|}$, but is not almost continuous.   (ii) Show that
if we set $f(x)=x^+$ for $x\in[0,1]$, then $f$ is \imp\ for Lebesgue
measure $\mu_L$ on $[0,1]$, but the image measure $\mu_Lf^{-1}$ is
not $\nu$ (nor, indeed, a Radon measure).
%419L

\spheader 419Xi Show that the split
interval $I^{\|}$ is perfectly normal, but that
$I^{\|}\times I^{\|}$ is not perfectly normal.
%419L

\leader{419Y}{Further exercises (a)}
%\spheader 419Ya
In the example of 419E, show that there is a Borel set $V\subseteq Z^2$
such that $\tilde\lambda V=0$ and $\lambda^*V=1$.
%419E

\spheader 419Yb
Show that if $\Cal A\subseteq\Cal P\omega_1$ is any family with
$\#(\Cal A)\le\omega_1$, there is a countably generated $\sigma$-algebra
$\Sigma$ of subsets of $\omega_1$ such that $\Cal A\subseteq\Sigma$.
%419F

\spheader 419Yc Show that the split interval
with its usual topology and measure has the simple product
property (417Yi).
%419L

\spheader 419Yd\dvAnew{2015} Give an example of a Lebesgue
measurable function
$\phi:\BbbR^2\to\Bbb R$ such that $\dom\Pd{\phi}{\xi_1}$ is not
measurable.   (Compare 222Yd, 225Yg, 262Yc.)
%419Xd  419H out of order query

}%end of exercises

\endnotes{\Notesheader{419}
The construction of the locally compact space $X$ in 419A from the
family $\langle I_{\xi}\rangle_{\xi<\kappa}$ is a standard device which
has been used many times.   The relation $\subseteq^*$ also appears in
many contexts.   In effect, part of the argument is taking place in the
quotient algebra $\frak A=\Cal PQ/[Q]^{<\omega}$, since $I\subseteq^*J$
iff $I^{\ssbullet}\Bsubseteq J^{\ssbullet}$ in $\frak A$;  setting
$\Cal I^{\#}=\{I^{\ssbullet}:I\in\Cal I\}$, the cardinal $\kappa$ is
$\min\{\#(A):A\subseteq \Cal I^{\#}$ has no upper bound in
$\Cal I^{\#}\}$, the `additivity' of the partially ordered set
$\Cal I^{\#}$.
Additivities of partially ordered sets will be one of the important
concerns of Volume 5.   I remark that we do not need to know whether
(for instance) $\kappa=\omega_1$ or $\kappa=\frak c$.   This is an early
taste of the kind of manoeuvre which has become a staple of
set-theoretic analysis.   It happens that the cardinal $\kappa$ here is
one of the most important cardinals of set-theoretic measure theory;  it
is `the additivity of Lebesgue measure' (529Xe\formerly{5{}29Xc}),
and under that name will appear repeatedly in Volume 5.

Observe that the measure $\mu$ of 419A only just fails to be a
quasi-Radon measure;  it is locally finite instead of being effectively
locally finite.   And it would be a Radon measure if it were inner
regular with respect to the compact sets, rather than just with respect
to the closed sets.

419C and 419D are relevant to the question:  have I given the `right'
definition of Radon measure space?   419C is perhaps more important.
Here we have a Radon measure space (on my definition) for which the
associated Borel measure is not localizable.   (If $\frak A$ is the
measure algebra of the measure $\mu$, and $\frak B$ the measure algebra of
$\mu\restr\Cal B$ where $\Cal B$ is the Borel $\sigma$-algebra of $X$,
then the
embedding $\Cal B\embedsinto\Sigma$ induces an embedding of $\frak B$ in
$\frak A$ which represents $\frak B$ as an order-dense subalgebra of
$\frak A$, just because $\mu$ is inner regular with respect to $\Cal B$.
Property (i) of 419C shows that $\frak B\ne\frak A$, so $\frak B$ cannot
be Dedekind complete in itself, by 314Ib.)   Since (I believe)
localizable versions of measure spaces should almost always be
preferred, I take this
as strong support for my prejudice in favour of insisting that `Radon'
measure spaces should be locally determined as well as complete.
Property (ii) of 419C is not I think of real significance, but is
further evidence, to be added to 415Xi, that outer regularity is like an
exoskeleton:  it may inhibit growth above a certain size.

In 419D I explore the consequences of omitting the condition `locally
finite' from the definition of Radon measure.   Even if we insist
instead that compact sets should have finite measure, we are in danger
of getting a non-localizable measure.   Of course this particular space
is pathological in terms of most of the criteria of this chapter -- for
instance, every non-empty open set has infinite measure, and the
topology is not regular.

Perhaps the most important example in the section is 419E.   The
analysis of $\tau$-additive product measures in \S417 was long and
difficult, and if these were actually equal to the familiar product
measures in all important cases the structure of the theory would be
very different.   But we find
that for one of the standard compact Radon probability spaces of the
theory, the c.l.d.\ product measure on its square is not a Radon
measure, and something has to be done about it.

\wheader{419 Notes}{0}{0}{0}{72pt}

I present 419J here to indicate one of the obstacles to any
simplification of the arguments in 417C and 417E.   It is not
significant in itself, but it offers a
welcome excuse to describe some fundamental facts about $\omega_1$
(419F-419G).   Similarly, 419K asks for some
elementary facts about Lebesgue measure (419H-419I) which seem to have
got left out.   This example really is important in itself, as it
touches on the general problem of representing stochastic processes, to
which I will return in Chapter 45.
}%end of notes

\discrpage  %mt41conc to come

\ifresultsonly\else\frfilename{mt41conc.tex}
\versiondate{1.1.17}
\copyrightdate{2005}

\def\chaptername{Topologies and measures I}
\def\sectionname{Concordance}
\let\headlinesectionname=\sectionname


\noindent{\bf Concordance to Chapter 41}

\medskip

I list here the section and paragraph numbers which have
(to my knowledge) appeared in print in references to this chapter,
and which have since been changed.

\medskip

\leaveitout{\leader{4{}11C}{$\tau$-additive measures} The definition
of `$\tau$-additive' measure, referred to in
{\smc Burke Macheras \& Strauss p10}, is now in 411C.
}

\leaveitout{\leader{4{}11F}{} (definition of `locally finite') mentioned in
{\smc K\"onig p10}
%411F
}

\leaveitout{\leader{4{}11H}{Radon measures} The definition
of `Radon' measure, referred to in
{\smc Burke Macheras \& Strauss p10}, is now in 411H.
{\smc Burke Macheras \& Strauss p20}
}

\leader{4{}11I}{Completion regular measures} The definition of
`completion regular'
measure, referred to in {\smc Fremlin 00}, has been moved to 411J.

\leaveitout{\leader{4{}11Nc, 4{}11Nd}{Supports} %411N
{\smc Pachl 13}, P.27
}

\leaveitout{\leader{4{}11Y}{Jordan algebra} The `Jordan algebras'
of 4{}11Yc, referred to in
{\smc Burke Macheras \& Strauss p10}, are now in 411Yc.
}

\leaveitout{4{}12Aa {\smc Fremlin 00} 412Aa If $\Cal K$ is closed under finite
unions, $\Cal K\setminus\Cal N_{\mu}$ is coinitial with
$\Sigma\setminus\Cal N_{\mu}$ then $\mu$ is inner regular with respect to
$\Cal K$}

\leaveitout{4{}12Ac {\smc Fremlin 00} 412Ac Let $(X,\Sigma,\mu)$ be a
semi-finite measure space and
$\sequencen{\Cal K_n}$ a sequence of families of sets such that
$\mu$ is inner regular with respect to $\Cal K_n$ for every
$n\in\Bbb N$, if $n\in\Bbb N$ and $\sequence{i}{K_i}$ is a
non-increasing sequence in $\Cal K_n$, then
$\bigcap_{i\in\Bbb N}K_i\in\Cal K_n$.  Then $\mu$ is inner regular with respect to $\bigcap_{n\in\Bbb N}\Cal K_n$.
}

\leaveitout{\leader{4{}12C}{Construction of measures} %412C
{\smc Pachl 13}, P.25
}

\leaveitout{\leader{4{}12D}{Construction of measures} %412D
{\smc Pachl 13}, P.25
}


\leaveitout{4{}12C, 4{}12Ab {\smc Fremlin 00} 412C, 412Ab
$\Cal W$ is an algebra of sets, $\Tau$ is the $\sigma$-algebra it
generates,
$\Cal W_{\delta}$ is the set of countable intersections of members of
$\Cal W$, $\mu$ is inner regular with respect to $\Tau$,
so $\mu$ is inner regular with respect to $\Cal W_{\delta}$.
}

\leaveitout{4{}12D {\smc Kojman \& Michalewski p06} 412D Baire measures
i.r. wrt closed
}

\leader{4{}12L}{Uniqueness of measures} Corollary 4{}12L, referred to in the
2008 and 2015 printings of Volume 5, is now 412Ma.
%538I

\leaveitout{4{}12O {\smc Fremlin 00} 412O subspace measure $\mu$ is inner regular with respect to $\Cal K_X=\{K\cap X:K\in\Cal K\}$
}

\leader{4{}13R}{Countably compact classes} Lemma 4{}13R, referred to in
{\smc Fremlin 00}, is now 413T.

\leaveitout{\leader{4{}13F}{Inner regularity} The list of characterizations
of inner regularity,
referred to in {\smc Bogachev 07}, is now
413F.}

\leader{4{}13I}{Inner measure} The constructions in 4{}13H and 4{}13J,
referred to in {\smc K\"onig p09b} and the 2008 and 2015 editions of
Volume 5, are now 413I and 413K.

\leader{4{}13Yf}{Uniform exhaustivity} Exercise 4{}13Yf, referred to in the 2008
and 2015 editions of Volume 5, is now 413Yh.

\leaveitout{\leader{4{}14Bb}{Downwards limits} %414Bb
{\smc Pachl 13}, P.26
}

\leader{4{}14N}{Density topologies} The note on `density topologies',
referred to % 2{}23Yb
in the 2001 edition of Volume 2, has been moved to 414P.

\leaveitout{4{}14P  414P  {\smc Burke Macheras \& Strauss p22}}

\leaveitout{\leader{4{}15B}{} %415B
Subspaces of quasi-Radon spaces are quasi-Radon.
{\smc Fremlin \& Plebanek 03}.}

\leaveitout{4{}15E 415E {\smc Fremlin 00}
The usual measure on $\{0,1\}^A$ is completion regular.
}

\leaveitout{K\"onig p09b:  415K inner measure construction 4{}15K}

\leaveitout{4{}15L 415L {\smc Konig p09} extension of $\tau$-additive
measure to quasi-Radon measure}

\leaveitout{4{}15Q 415Q 4{}15R 415R Burke Macheras & Strauss p22}

\leader{4{}15Xp}{} This exercise, referred to in the 2008 and 2015
printings of Volume 5,
%535B prf (a)
has been moved to 415Xs.

\leader{4{}15Yd}{Sorgenfrey line} This exercise, referred to in the 2001
edition of Volume 2, %in 2{}61Yj
has been moved to 415Ye.

\leaveitout{\leader{4{}16A}{} %416A
Radon measures are quasi-Radon {\smc Fremlin \& Plebanek 03}.}

\leaveitout{\leader{4{}16J}{} %416J
construction of Radon measures, mentioned in {\smc K\"onig p10},
`contained in the famous theorem of Kisy\'nski 1968', but I don't think
it's quite what's in ZBL 157 p 373}

\leaveitout{\leader{4{}16K}{} (Radon measure above on compact sets, below
on open sets) mentioned in
{\smc K\"onig p10}
%416K
}

\leaveitout{\leader{4{}16L}{} (construction of Radon measure) mentioned in
{\smc K\"onig p10}
%416L
}

\leader{4{}16M}{Henry's theorem}, mentioned in {\smc K\"onig 04},
%following 2.5,
has been re-named 416N.
%416N

\leaveitout{\leader{4{}16N}{} %416N
Henry's theorem mentioned in {\smc K\"onig p10}
}

\leaveitout{4{}16O {\smc K\"onig 04}, following 2.5:
Theorem 4{}16O quoted.
Also mentioned in {\smc K\"onig p10}
%416O
}

\leader{4{}16P}{The algebra of open-and-closed subsets} This paragraph,
referred to in the 2002 edition of Volume 3, %notes to 3{}62
has been moved to 416Q.

\leaveitout{\leader{4{}16P}{} (extension to Radon measure) mentioned in
{\smc K\"onig p10}
%416P
}

\leaveitout{\leader{4{}16Q}{Specification of measures on $\{0,1\}^I}
{\smc Fremlin \& Plebanek 03}, 2A.}
%416Q

\leader{4{}16T}{Kakutani's theorem} The description of the usual measure
on $\{0,1\}^{\kappa}$, referred to in {\smc Fremlin \& Plebanek 03},
has been moved to 416U.

\leaveitout{4{}16U usual measure on $\{0,1\}^I$ {\smc Fremlin 03} %416U
}

\leaveitout{4{}16V 416V 4{}16Xw 416Xw Burke Macheras & Strauss p22}


\leaveitout{4{}17B 417B 4{}17C 417C 4{}17D 417D Burke Macheras & Strauss p22}

\leader{4{}17E}{$\tau$-additive product measures} The reference in
{\smc Fremlin 00} to Kakutani's theorem that the product measure on
$\{0,1\}^I$ is completion regular should be directed to 415E or 416U
rather than 4{}17E.

\leaveitout{4{}17G 417F Burke Macheras & Strauss p22}

\leaveitout{4{}17I 417I Burke Macheras & Strauss p22}

\leaveitout{4{}17T 417T Burke Macheras & Strauss p22}

\leaveitout{4{}18M {\smc K\"onig 04}, 4.15:
Prokhorov's theorem  mentioned.  %418M
}

\leader{4{}19H} The example 4{}19H of a measure, inner regular with respect
to the Borel sets but with no extension to a topological measure, mentioned
in {\smc K\"onig p09}, is now 419J.

\leader{4{}19J}{Partitions into sets of full outer measure}
Lemma 4{}19J, mentioned in the 2004 edition of Volume 1,
has been moved to 419I.

\bigskip

\noindent{\bf References}

\leaveitout{
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p10] `Densities and liftings for
derived algebras', 9.10.10.

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p22] `Density topologies and the marginal strong
lifting problem for hyperstonian spaces', preprint, 2022.
%BMS.2022.pdf
}

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
Fundamenta Math.\ 165 (2000) 67-94.
   %n99821.tex

Fremlin D.H.\ \& Plebanek G.\ [03]
`Large families of mutually singular Radon measures',
Bull.\ Acad.\ Pol.\ Sci.\ (Math.) 51 (2003) 169-174.

\leaveitout{
Kojman M.\ \& Michalewski H.\ [p06] `Borel extensions of Baire measures in
ZFC', 23.8.06.
}

\leaveitout{K\"onig H.\ [04] `Projective limits via inner premeasures and the true
Wiener measure', Mediterranean J.\ Math.\ 1 (2004) 3-42.
}

K\"onig H.\ [p09] `Measure and integration:  the basic extension theorems',
2009.

K\"onig H.\ [p09b] `Measure and integration:  characterization
of the new maximal contents and measures', for Proc.\ Conf.\ Vector
Measures, Integration and Applications, KU Eichst\"att, 24-26 September
2008.

\leaveitout{K\"onig H.\ [p10] `Measure theory:  transplantation theorems
for inner premeasures', 15.6.10.}

%Pachl J.\ [13] {\it Uniform Spaces and Measures.}   Springer, 2013.

\frnewpage
\fi

\references{
\frfilename{mt4r.tex}
\versiondate{9.2.13}
\copyrightdate{2002}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\Loadcyrillic

\Centerline{\bf References for Volume 4}


\medskip%A

Aldaz J.M.\ [95] `On compactness and Loeb measures', Proc.\ Amer.\
Math.\ Soc.\ 123 (1995) 173-175.
\cmmnt{[439Xa.]}

Aldaz J.M.\ [97] `Borel extensions of Baire measures', Fund.\ Math.\ 154
(1997) 275-293.
\cmmnt{[434Yn.]}

Aldaz J.M.\ \& Render H.\ [00] `Borel measure extensions of measures
defined on sub-$\sigma$-algebras', Advances in Math.\ 150 (2000)
233-263.
\cmmnt{[432D.]}

Alexandroff P.S.\ \& Urysohn P.S.\ [1929] {\it  M\'emoire sur les espaces
topologiques compacts},
Verhandelingen Akad.\ Wetensch.\ Amsterdam 14 (1929) 1-96.
\cmmnt{[419L.]}

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P.,
1987.
\cmmnt{[4A4N.]}

Andretta A.\ \& Camerlo R.\ [13] `The descriptive set theory of the
Lebesgue density theorem', Advances in Math.\ 234 (2013).
%{\tt http://arxiv.org/abs/1105.3355}.
\cmmnt{[475Yg.]}

Arkhangel'skii A.V.\ [92] {\it Topological Function Spaces.}  Kluwer,
1992.
\cmmnt{[\S462 {\it intro.}, \S467 {\it notes\/}.]}

\leaveitout{
Armitage D.H.\ \& Gardiner S.J.\ [01] {\it Classical Potential Theory.}
Springer, 2001.
478Yd}

Aronov B., Basu S., Pach J.\ \& Sharir M.\ [03] (eds.)
{\it Discrete and Computational Geometry;  the Goodman-Pollack
Festschrift,}  Springer, 2003.

Asanov M.O.\ \& Velichko N.V.\ [81] `Compact sets in $C_p(X)$', Comm.\
Math.\ Helv.\ 22 (1981) 255-266.
\cmmnt{[462Ya.]}

Austin T.\ [10a] `On the norm convergence of nonconventional ergodic
averages', Ergodic Theory and Dynamical Systems 30 (2010) 321-338.
%({\tt http://{\penalty-100}arxiv.org/abs/{\penalty-100}0805.0320}).
\cmmnt{[497M.]}

Austin T.\ [10b] `Deducing the multidimensional Szemer\'edi theorem from
the infinitary hypergraph removal lemma',
J.\ d'Analyse Math.\ 111 (2010) 131-150.
%({\tt http://arxiv.org{\bsp}abs/0808.2267}).
\cmmnt{[497M.]}

\leaveitout{
Avil\'es A., Plebanek G.\ \& Rodr\'{\i}guez J.\ [p11] `Measurability in
$C(2^{\kappa})$ and Kunen cardinals', 2011.
%material on  C(\{0,1\}^{\kappa}) ;  Baire and Borel \sigma-algebras
%in weak and pointwise topologies
}

\leaveitout{Axler S. \& Ramey W.\ [95]
`Harmonic polynomials and Dirichlet-type problems',
Proc.\ Amer.\  Math.\ Soc.\ 123 (1995) 3765-3773.
http://www.axler.net/HarmonicPolynomials.html
478Yd}

\medskip%B

Balogh Z.\ [96] `A small Dowker space in ZFC', Proc.\ Amer.\ Math.\
Soc.\ 124 (1996) 2555-2560.
\cmmnt{[439O.]}

Banach S.\ [1923] `Sur le probl\`eme de la mesure',
Fundamenta Math.\ 4 (1923) 7-33.
\cmmnt{[449O.]}

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996 (London Math.\ Soc.\
Lecture Note Series 232).  \cmmnt{[424H, 448P, \S448 {\it notes\/}.]}

Bellow A.\ {\it see} Ionescu Tulcea A.

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).
%4r

Bergelson V.\ [85] `Sets of recurrence of $\Bbb Z^m$-actions and properties of
sets of differences in $\Bbb Z^m$',
J.\ London Math.\ Soc.\ (2) 31 (1985) 295-304.
%http://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Bergelson%20-%20Sets%20of%20recurrence%20of%20Zm-actions%20and%20properties%20of%20sets%20of%20differences%20in%20Zm%20(1985).pdf
\cmmnt{[491Yt.]}

Bergman G.M.\ [06] `Generating infinite symmetric groups', Bull.\ London
Math.\ Soc.\ 38 (2006) 429-440.
%http://math.berkeley.edu/~gbergman/papers/
%({\tt http:/{\bsp}arxiv.org{\bsp}abs{\bsp}math{\bsp}0401304v2}).
\cmmnt{[\S494 {\it notes}.]}

Billingsley P.\ [99] {\it Convergence of Probability Measures.}  Wiley,
1999.
\cmmnt{[\S4A3 {\it notes}.]}
%second edition ISBN 0-471-19745-9

Blackwell D.\ [56] `On a class of probability spaces',
pp.\ 1-6 in {\smc Neyman 56}.
\cmmnt{[419K, 452P.]}

Bochner S.\ [1933] `Monotone Funktionen, Stieltjessche Integrale, und
harmonische Analyse', Math.\ Ann.\ 108 (1933) 378-410.
\cmmnt{[445N.]}

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.
\cmmnt{[437Q, 437Xs, 437Yo, 437Yu, 457K.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[4A1G, 4A4N.]}

Bongiorno B., Piazza L.di \ \& Preiss D.\ [00]
`A constructive minimal integral which includes Lebesgue integrable
functions and derivatives', J.\ London Math.\ Soc.\ (2) 62 (2000) 117-126.
\cmmnt{[483Yi.]}

Borodulin-Nadzieja P.\ \& Plebanek G.\ [05] `On compactness of measures on
Polish spaces', Illinois J.\ Math.\ 49 (2005) 531-545.
\cmmnt{[451L.]}

Bourbaki N.\ [65] {\it Int\'egration,} chaps.~1-4.   Hermann, 1965
(Actualit\'es Scientifiques et Industrielles 1175).
\cmmnt{[\S416 {\it notes\/}, \S436 {\it notes\/}.]}

Bourbaki N.\ [66] {\it Espaces Vectoriels Topologiques}, chaps.~1-2,
$2^{\text{e}}$ \'ed.
Hermann, 1966 (Actualit\'es Scientifiques et Industrielles 1189).
\cmmnt{[4A2J.]}
%Store A QA 37.B6 no 7792 0070 5

Bourbaki N.\ [69] {\it Int\'egration,} chap.~9.   Hermann, 1969
(Actualit\'es Scientifiques et Industrielles 1343).
\cmmnt{[\S436 {\it notes.}]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S461 {\it notes\/}, \S4A4.]}

\indexheader{Bourgain}
Bourgain J., Fremlin D.H.\ \& Talagrand M.\ [78] `Pointwise compact
sets of Baire-measurable functions', Amer.\ J.\ Math.\ 100 (1978)
845-886.
\cmmnt{[462Ye.]}

\indexheader{Brodski\v{\i}}
Brodski\v{\i} M.L.\ [1949] `On some properties of sets of positive measure',
Uspehi Matem.\ Nauk (N.S.) 4 (1949) 136-138.
\cmmnt{[498B.]}

Brook R.B.\ [70] `A construction of the greatest ambit',
Math.\ Systems Theory 4 (1970) 243-248.
\cmmnt{[449D.]}

Burke D.K.\ \& Pol R.\ [05] `Note on separate continuity and the Namioka
property', Top.\ Appl.\ 152 (2005) 258-268.
\cmmnt{[\S463 {\it notes.}]}

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p21] `The strong marginal lifting
problem for hyperstonian spaces', preprint (2021).
%bms5s.pdf

\medskip%C

Carrington D.C.\ [72] `The generalised Riemann-complete integral', PhD
thesis, Cambridge, 1972.
\cmmnt{[481L.]}

\v{C}ech E.\ [66] {\it Topological Spaces.}  Wiley, 1966.
\cmmnt{[\S4A2, 4A3S.]}

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[494F.]}

\indexheader{Choquet}
Choquet G.\ [55] `Theory of capacities', Ann.\ Inst.\ Fourier (Grenoble)
5 (1955) 131-295.
\cmmnt{[432K.]}

\indexheader{Chung}
Chung K.L.\ [95] {\it Green, Brown and Probability.}   World Scientific,
1995.
% 978-9810225339  HA33.4.C5  479Xt

\indexheader{Ciesielski}
Ciesielski K.\ \& Pawlikowski J.\ [03]
`Covering Property Axiom CPA$_{\text{cube}}$ and its consequences',
Fundamenta Math. 176 (2003) 63-75.
\cmmnt{[498C.]}

Cs\'asz\'ar \'A.\ [78] {\it General Topology.}  Adam Hilger, 1978.
\cmmnt{[\S4A2, \S4A5.]}

\medskip%D

Davies R.O.\ [70] `Increasing sequences of sets and Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 20 (1970) 222-236.
\cmmnt{[471G.]}

Davies R.O.\ [71] `Measures not approximable or not specifiable by means
of balls', Mathematika 18 (1971) 157-160.
\cmmnt{[\S466 {\it notes\/}.]}

Davis W.J., Ghoussoub N.\ \& Lindenstrauss J.\ [81] `A lattice-renorming
theorem and applications to vector-valued processes', Trans.\ Amer.\
Math.\ Soc.\ 263 (1981) 531-540.
\cmmnt{[\S467 {\it notes\/}.]}

\indexheader{de Caux}
de Caux P.\ [76] `A collectionwise normal, weakly $\theta$-refinable
Dowker space which is neither irreducible nor realcompact', Topology
Proc.\ 1 (1976) 67-77.
\cmmnt{[439O.]}

Dellacherie C.\ [80] `Un cours sur les ensembles analytiques', pp.\
183-316 in {\smc Rogers 80}.
\cmmnt{[\S432 {\it notes\/}.]}

de Maria J.L.\ \& Rodriguez-Salinas B.\ [91] `The space
$(\ell_{\infty}/c_0)$,weak) is not a Radon space', Proc.\ Amer.\ Math.\
Soc.\ 112 (1991) 1095-1100.
\cmmnt{[466I.]}

Deville R., Godefroy G.\ \& Zizler V.\ [93] {\it Smoothness and
Renormings in Banach Spaces.}  Pitman, 1993.
\cmmnt{[\S467 {\it intro.}, 467Ye, \S467 {\it notes\/}.]}

Diaconis, P.\ \& Freedman, D.\ [80] `Finite exchangeable sequences.'
Ann. of Probability 8 (1980) 745-764.
\cmmnt{[459Xd.] }

Droste M., Holland C.\ \& Ulbrich G.\ [08]
`On full groups of measure preserving and ergodic transformations with
uncountable cofinalities',
Bull.\ London Math. Soc.\ 40 (2008) 463-472.
%{\tt http://web.mac.com/chollan1{\bsp}iWeb/Site{\bsp}publications.html}.
\cmmnt{[494Q.]}

Dubins L.E.\ \& Freedman D.A.\ [79]
`Exchangeable processes need not be mixtures of independent,
identically distributed random variables',
Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 48 (1979) 115-132.
\cmmnt{[459K.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S4A4.]}

Du Plessis, N.\ [70] {\it Introduction to Potential Theory.}  Oliver \&
Boyd, 1970.
%used for \S478

\medskip%E

Eggleston H.G.\ [54] `Two measure properties of Cartesian product sets',
Quarterly J.\ Math.\ (2) 5 (1954) 108-115.
\cmmnt{[498B.]}

Emerson W.\ \& Greenleaf F.\ [67] `Covering properties and F{\o}lner
conditions for locally compact groups',
Math.\ Zeitschrift 102 (1967) 370-384.
\cmmnt{[449J.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[4A1A.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[Vol.\ 4 {\it intro.}, 415Yb, \S434 {\it notes\/},
\S435 {\it notes\/}, 462L, \S4A2, 4A4B.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[Chap.\ 47 {\it intro.}, \S473 {\it intro.}, \S473 {\it notes\/},
\S474 {\it notes\/}, \S475 {\it notes\/}.]}

\medskip%F

Falconer K.J.\ [90] {\it Fractal Geometry.}  Wiley, 1990.
\cmmnt{[\S471 {\it intro.}]}

Federer H.\ [69] {\it Geometric Measure Theory.}
Springer, 1969 (reprinted 1996).
\cmmnt{[\S441 {\it notes\/}, 442Ya,
\S443 {\it notes\/}, Chap.\ 47 {\it intro.}, 471Yk,
\S471 {\it notes\/}, \S472 {\it notes\/}, \S475 {\it notes\/}.]}

Fernique X.\ [97] {\it Fonctions al\'eatoires gaussiennes,
vecteurs al\'eatoires gaussiens}, CRM Publications, 1997.
\cmmnt{[\S456 {\it notes\/}.]}

Folland G.B.\ [95] {\it A Course in Abstract Harmonic Analysis.}   CRC
Press, 1995
\cmmnt{[\S4A5, \S4A6.]}

\indexheader{F{\o}lner}
F{\o}lner E.\ [55] `On groups with full Banach mean value',
Math.\ Scand.\ 3 (1955) 243-254.
\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
Ford L.R.\ \& Fulkerson D.R.\ [56] `Maximal flow through a network',
Canadian J.\ Math.\ 8 (1956) 399-404.
\cmmnt{[4A4N.]}

Frankl P.\ \& R\"odl V.\ [02] `Extremal problems on set systems',
Random Structures Algorithms 20 (2002) 131-164.
\cmmnt{[497L.]}

\indexheader{Fremlin}
Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
Manuscripta Math.\ 15 (1975) 219-242.
\cmmnt{[463K.]}

Fremlin D.H.\ [75b] `Topological measure spaces:  two counter-examples',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 78 (1975) 95-106.
\cmmnt{[419C, 419D.]}

Fremlin D.H.\ [76] `Products of Radon measures:  a counterexample',
Canadian Math.\ Bull.\ 19 (1976) 285-289.
\cmmnt{[419E.]}

Fremlin D.H.\ [81] `Measurable functions and almost continuous
functions', Manuscripta Math.\ 33 (1981) 387-405.
\cmmnt{[451T.]}

Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S434 {\it notes\/}, 439Yd, 4A2E.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S438 {\it notes\/}.]}

Fremlin D.H.\ [95] `The generalized McShane integral',
Illinois J.\ Math.\ 39 (1995) 39-67.
\cmmnt{[481N.]}

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
Fundamenta Math.\ 165 (2000) 67-94.
\cmmnt{[451V.]}

Fremlin D.H.\ [n05] `Strictly localizable Borel measures', note of 7.5.05
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[434Yr.]}

Fremlin D.H.\ [n15] `The Sorgenfrey line is not a Prokhorov space',
note of 5.8.15
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[439Yk.]}

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fund.\ Math.\ 147 (1995) 27-37.
\cmmnt{[434Q.]}

Fremlin D.H.\ \& Talagrand M.\ [78] `On the representation of increasing
linear functionals on Riesz spaces by measures',
Mathematika 25 (1978) 213-215.
\cmmnt{[439I.]}

Fremlin D.H.\ \& Talagrand M.\ [79] `A decomposition theorem for
additive set functions and applications to Pettis integral and ergodic
means', Math.\ Zeitschrift 168 (1979) 117-142.
\cmmnt{[\S464.]}  %{\it 464C}, 464I, 464M, 464N, \S464 {\it notes\/}

Fristedt B.\ \& Gray L.\ [97] {\it A Modern Approach to Probability
Theory.}  Birkh\"auser, 1997.
\cmmnt{[\S455 {\it notes\/}, \S495 {\it notes\/}.]}
%HA33.4.F7

Frol{\'\i}k Z.\ [61] `On analytic spaces', Bull.\ Acad.\ Polon.\ Sci.\ 
%8 (1961) 747-750.
9 (1961) 721-726.   %Paulo Monteiro, email 8.9.22
\cmmnt{[422F.]}

Frol\'\i k Z.\ [88] (ed) {\it General Topology and its Relations to
Modern Analysis and Algebra VI} (Proc.\ Sixth Prague Topological
Symposium), Heldermann, 1988.

Furstenberg H.\ [77]
`Ergodic behavior of diagonal measures and a theorem of Szemer\'edi
on arithmetic progressions', J. d'Analyse Math.\ 31 (1977), 204-256.
\cmmnt{[\S497 {\it intro.}, \S497 {\it notes\/}.]}

Furstenberg H.\ [81]
{\it Recurrence in Ergodic Theory and Combinatorial Number Theory.}
Princeton U.P., 1981.
\cmmnt{[\S497 {\it intro.}, 497N.]}
% QA614

Furstenberg H.\ \& Katznelson Y.\ [85]
`An ergodic Szemer\'edi theorem for IP-systems and combinatorial theory',
J.\ d'Analyse Math.\ 45 (1985) 117-168.
\cmmnt{[\S497 {\it intro.}]}

\medskip%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S4A2.]}

Gardner R.J.\ [75] `The regularity of Borel measures and Borel
measure-compactness', Proc.\ London Math.\ Soc.\ (3) 30 (1975) 95-113.
\cmmnt{[438M.]}

Gardner R.J.\ \& Pfeffer W.F.\ [84] `Borel measures', pp. 961-1043 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[434G.]}

Giordano T.\ \& Pestov V.\ [02] `Some extremely amenable groups',
C.R.A.S.\ Paris (S\'er.\ I) 334 (2002) 273-278.
\cmmnt{[494I.]}

Glasner S.\ [98] `On minimal actions of Polish groups', Topology and its Applications 85 (1998) 119-125.
\cmmnt{[\S493 {\it notes\/}.]}

Gordon R.A.\ [94] {\it The Integrals of Lebesgue, Denjoy, Perron and
Henstock.}   Amer.\ Math.\ Soc., 1994 (Graduate Studies in Mathematics
4).
\cmmnt{[481Q, \S483 {\it notes\/}.]}

Greenleaf F.P.\ [69] {\it Invariant Means on Topological Groups and Their
Applications.}   van Nostrand, 1969.
\cmmnt{[\S449 {\it notes.}]}

Grothendieck A. [92] {\it Topological Vector Spaces.}  Gordon \& Breach,
1992.
\cmmnt{[\S462 {\it notes\/}.]}

\indexheader{Grzegorek}
Grzegorek E.\ [81] `On some results of Darst and Sierpi\'nski concerning
universal null and universally negligible sets', Bull.\ Acad.\ Polon.\
Sci.\ (Math.) 29 (1981) 1-5.
\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
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.]}

Hansell R.W.\ [01] `Descriptive sets and the topology of non-separable
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
measure-compact spaces', Proc.\ London Math.\ Soc.\ (3) 29 (1974) 1-6.
\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
positive Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 70 (1995)
581-604.
\cmmnt{[471R, 471S.]}

\medskip%I

\indexheader{Ionescu Tulcea}
Ionescu Tulcea A.\ [73] `On pointwise convergence, compactness and
equicontinuity I', Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 26
(1973) 197-205.
\cmmnt{[463C.]}

Ionescu Tulcea A.\ [74] `On pointwise convergence, compactness, and
equicontinuity II', Advances in Math.\ 12 (1974) 171-177.
\cmmnt{[463G.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [67] `On the existence of a
lifting commuting with the left translations of an arbitrary locally
compact group', pp.\ 63-97 in {\smc LeCam \& Neyman 67}.
\cmmnt{[\S447 {\it intro.}, 447I, \S447 {\it notes\/}.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.
\cmmnt{[\S453 {\it intro.}]}

\indexheader{Ismail}
Ismail M.\ \& Nyikos P.J.\ [80] `On spaces in which countably compact
sets are closed, and hereditary properties', Topology and its Appl.\ 11
(1980) 281-292.
\cmmnt{[\S434 {\it notes\/}.]}

\medskip%J

%Jackson S.\ \& Mauldin R.D.\ [99] `On the $\sigma$-class generated by
%open balls', Math.\ Proc.\ Cambridge Phil.\ Soc.\ 127 (1999) 99-108.

Jameson G.J.O.\ [74] {\it Topology and Normed Spaces.}  Chapman \& Hall,
1974.
\cmmnt{[\S4A4.]}

\indexheader{Jayne}
Jayne J.E.\ [76] `Structure of analytic Hausdorff spaces', Mathematika
23 (1976) 208-211.
\cmmnt{[422Yc.]}

Jayne J.E.\ \& Rogers C.A.\ [95] `Radon measures on Banach spaces with
their weak topologies', Serdica Math.\ J.\ 21 (1995) 283-334.
\cmmnt{[466H, \S466 {\it notes\/}.]}

\indexheader{Jech}
Jech T.\ [78] {\it Set Theory.}  Academic, 1978.
\cmmnt{[423S, \S4A1.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003.
%ISBN 3540440852
\cmmnt{[\S4A1.]}

Johnson R.A.\ [82] `Products of two Borel measures', Trans.\ Amer.\
Math.\ Soc.\ 269 (1982) 611-625.
\cmmnt{[434R.]}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

Juh\'asz I., Kunen K.\ \& Rudin M.E.\ [76] `Two more hereditarily
separable non-Lindel\"of spaces', Canadian J.\ Math.\ 29 (1976)
998-1005.
\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}
Kaplansky I.\ [71] {\it Lie Algebras and Locally Compact Groups.}
Univ.\ of Chicago Press, 1971.
\cmmnt{[\S446 {\it notes.}]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
1995.
\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
f\"ur Wahrscheinlichkeitstheorie verw.\ Gebiete 67 (1984) 399-432.
\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
Wahrscheinlichkeitsrechnung.}  Springer, 1933;  translated as {\it
Foundations of Probability Theory,}  Chelsea, 1950.
\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
selectors', J.\ London Math.\ Soc.\ (2) 28 (1983) 203-210.
\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
continuous dependence on a parameter', Czechoslovak Math.\ J.\ 7 (1957)
418-446.
\cmmnt{[Chap.\ 48 {\it intro.}]}

Kwiatowska A. \& Solecki S.\ [11] `Spatial models of Boolean actions and
groups of isometries', Ergodic Theory Dynam.\ Systems 31 (2011) 405-421.
%{\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
{\smc Pap 02}.
\cmmnt{[\S449 {\it notes\/}.]}

LeCam L.M.\ \& Neyman J.\ [67] (eds.) {\it Proc.\ Fifth Berkeley
Symposium in Mathematical Statistics and Probability}, vol.\ II.
Univ.\ of California Press, 1967.

Levy A.\ [79] {\it Basic Set Theory.}  Springer, 1979.
\cmmnt{[\S4A1.]}

Lo\`eve M.\ [77] {\it Probability Theory I.} Springer, 1977.
\cmmnt{[\S495 {\it notes\/}.]}

\indexheader{Losert}
Losert V.\ [79] `A measure space without the strong lifting property',
Math. Ann. 239 (1979) 119-128.
\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
in real analysis and potential theory,} Springer, 1986 (Lecture Notes in
Mathematics 1189).
\cmmnt{[414Ye.]}

\medskip%M

\indexheader{Mackey}
Mackey G.W.\ [62] `Point realizations of transformation
groups', Illinois J.\ Math.\ 6 (1962) 327-335.
\cmmnt{[448S.]}

Marczewski E.\ [53] `On compact measures', Fundamenta Math.\ 40 (1953)
113-124.
\cmmnt{[413T, Chap.\ 45 {\it intro.}, \S451 {\it intro.}, 451J.]}

Marczewski E.\ \& Ryll-Nardzewski C.\ [53] `Remarks on the compactness
and non-direct products of measures', Fundamenta Math.\ 40 (1953)
165-170.
\cmmnt{[454C.]}

Ma\v{r}\'{\i}k J.\ [57] `The Baire and Borel measure', Czechoslovak
Math.\ J.\ 7 (1957) 248-253.
\cmmnt{[435C.]}

Mattila P.\ [95] {\it Geometry of Sets and Measures in Euclidean
Spaces.}   Cambridge U.P., 1995.
\cmmnt{[Chap.\ 47 {\it intro.}, \S471 {\it intro.}, \S472 {\it notes\/},
\S475 {\it notes\/}.]}
%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

McShane E.J. [73] `A unified theory of integration', Amer.\ Math.\
Monthly 80 (1973) 349-359.
\cmmnt{[481M.]}

Meester R.\ \& Roy R.\ [96] {\it Continuum percolation.}  Cambridge
U.P., 1996.
\cmmnt{[\S495 {\it notes\/}.]}

Milman V.D.\ \& Schechtman G.\ [86]
{\it Asymptotic Theory of Finite Dimensional Normed Spaces.}
Springer, 1986 (Lecture Notes in Mathematics 1200).
\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
spaces', J.\ London Math.\ Soc.\ 43 (1968) 633-639.
\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.\
Soc.\ (3) 20 (1970) 507-524.
\cmmnt{[438J.]}

\indexheader{Moschovakis}
Moschovakis Y.N.\ [80] {\it Descriptive Set Theory.}   Springer, 1980.
\cmmnt{[Chap.\ 42 {\it intro.}, \S423 {\it notes\/}.]}

Muldowney P.\ [87] {\it A General Theory of Integration in Function
Spaces.}   Longmans, 1987 (Pitman Research Notes in Mathematics 153).
\cmmnt{[481P, \S482 {\it notes\/}.]}

Mushtari D.H.\ [96] {\it Probabilities and Topologies on Linear Spaces.}
Kazan Mathematics Foundation, 1996.
%ISBN 5-900975-01-0

Musia\l\ K.\ [76] `Inheritness of compactness and perfectness of
measures by thick subsets',  pp.\ 31-42 in {\smc Bellow \& K\"olzow 76}.
\cmmnt{[451U.]}

\medskip%N

Nadkarni M.G.\ [90] `On the existence of a finite invariant measure',
Proc.\ Indian Acad.\ Sci., Math.\ Sci.\ 100 (1990) 203-220.
\cmmnt{[448P, \S448 {\it notes\/}.]}

Nagle B., R\"odl V.\ \& Schacht M.\ [06]
`The counting lemma for regular k-uniform hypergraphs',
Random Structures and Algorithms 28 (2006) 113-179.
\cmmnt{[497J.]}

%Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.


Neyman J.\ [56] (ed.) {\it Proceedings of the Third Berkeley Symposium
on Mathematical Statistics and Probability, 1954-55,} vol.\ II.   Univ.\
of California, 1956.

\medskip%O
Oxtoby J.C.\ [70] `Homeomorphic measures on metric spaces',
Proc.\ Amer.\ Math.\ Soc.\ 24 (1970) 419-423.
\cmmnt{[434Yq.]}

\medskip%P

Pachl J.K.\ [78] `Disintegration and compact measures', Math.\ Scand.\
43 (1978) 157-168.
\cmmnt{[452I, 452S, 452Ye, \S452 {\it notes\/}.]}

Pachl J.K.\ [79] `Two classes of measures', Colloquium Math.\ 42 (1979)
331-340.
\cmmnt{[452R.]}

Pap E.\ [02] (ed.) {\it Handbook of Measure Theory.}  North-Holland, 2002.

Paterson A.L.T.\ [88] {\it Amenability.}   Amer.\ Math.\ Soc., 1988.
\cmmnt{[449K, \S449 {\it notes\/}.]}
% QA 403.P2 store A

Perron O.\ [1914] `\"Uber den Integralbegriff', Sitzungsberichte der
Heidelberger Akad.\ Wiss.\ A14 (1914) 1-16.
\cmmnt{[Chap.\ 48 {\it intro.}, 483J.]}

Pestov V.\ [99] `Topological groups:  where to from here?',
Topology Proc.\ 24 (1999) 421-502.
\cmmnt{[\S493 {\it notes\/}.]}

Pestov V.\ [02] `Ramsey-Milman phenomenon, Urysohn metric spaces,
and extremely amenable groups', Israel J.\ Math.\ 127 (2002) 317-357.
\cmmnt{[493E, 493Ya, \S493 {\it notes\/}.]}

Pfeffer W.F.\ [91a] `The Gauss-Green theorem', Advances in Math.\ 87
(1991) 93-147.
\cmmnt{[484Xb.]}

Pfeffer W.F.\ [91b] `Lectures on geometric integration and the
divergence theorem', Rend.\ Ist.\ Mat.\ Univ.\ Trieste 23 (1991)
263-314.
\cmmnt{[484B.]}

Pfeffer W.F.\ [01] {\it Derivation and Integration.}  Cambridge U.P.,
2001.
\cmmnt{[\S484 {\it notes\/}.]}

Phelps R.R.\ [66] {\it Lectures on Choquet's Theorem.}  van Nostrand,
1966.
\cmmnt{[\S461 {\it notes\/}.]}

\indexheader{Plewik}
Plewik S.\ \& Voigt B.\ [91] `Partitions of reals:  a measurable approach',
J.\ Combinatorial Theory (A) 58 (1991) 136-140.
\cmmnt{[443Yl.]}

\indexheader{Pontryagin}
Pontryagin L.S.\ [1934] `The theory of topological commutative groups',
Ann.\ Math.\ 35 (1934) 361-388.
\cmmnt{[445U.]}

\indexheader{Preiss}
Preiss D.\ [73] `Metric spaces in which Prohorov's theorem is not valid',
Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 27 (1973) 109-116.
\cmmnt{[439S.]}
%ILL 31.7.15

Preiss D.\ \& Ti\v{s}er J.\ [91]
`Measures in Banach spaces are determined by their values on balls',
Mathematika 38 (1991) 391-397.
\cmmnt{[\S466 {\it notes\/}.]}

Pryce J.D.\ [71] `A device of R.J.Whitley's applied to pointwise
compactness in spaces of continuous functions', Proc.\ London Math.\
Soc.\ (3) 23 (1971) 532-546.
\cmmnt{[462B, 462C.]}

\medskip%Q\medskip%R

Radon J.\ [1913] `Theorie und Anwendungen der absolut additivien
Mengenfunktionen,' Sitzungsbereichen der Kais.\ Akad.\ der Wiss.\ in
Wien 122 (1913) 28-33.
\cmmnt{[\S416 {\it notes\/}.]}

Ramachandran D.\ [02] `Perfect measures and related topics', pp.\
765-786 in {\smc Pap 02}.
\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
Rao B.V.\ [69] `On discrete Borel spaces and projective hierarchies',
Bull.\ Amer.\ Math.\ Soc.\ 75 (1969) 614-617.
\cmmnt{[419F.]}

Rao B.V. \& Srivastava S.M.\ [94] `Borel isomorphism theorem',
privately circulated, 1994.
\cmmnt{[424C.]}

\indexheader{Rao M.}
Rao M.\ [77] {\it Brownian motion and classical potential theory.}
Aarhus Universitet Matematisk Institut, 1977 (Lecture Notes Series 47);
{\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
measures', Math.\ Scand.\ 40 (1977) 69-78.
\cmmnt{[417C, 437M.]}

Riesz F.\ \& Sz.-Nagy B.\ [55] {\it Functional Analysis.}  Frederick Ungar,
1955.
\cmmnt{[\S494 {\it notes\/}.]}

Rogers C.A.\ [70] {\it Hausdorff Measures.}   Cambridge, 1970.
\cmmnt{[\S471 {\it notes\/}.]}

Rogers C.A.\ [80] (ed.) {\it Analytic Sets.}   Academic, 1980.
\cmmnt{[Chap.\ 42 {\it intro.}, \S422 {\it notes\/}, \S423 {\it notes\/}.]}

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.]}

Rosendal C.\ \& Solecki S.\ [07] `Automatic continuity of homomorphisms and
fixed points on metric compacta', Israel J.\ Math.\ 162 (2007) 349-371.
\cmmnt{[494Z, \S494 {\it notes}.]}

Rudin M.E.\ [71] `A normal space $X$ for which $X\times I$ is not
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

Sazonov V.V.\ [66] `On perfect measures', A.M.S.\ Translations (2) 48
(1966) 229-254.
\cmmnt{[451F.]}

Schachermayer W.\ [77] `Eberlein-compacts et espaces de Radon',
C.R.A.S.\ (Paris) (A) 284 (1977) 405-407.
\cmmnt{[467P.]}

Schaefer H.H.\ [71] {\it Topological Vector Spaces.}  Springer, 1971.
\cmmnt{[\S4A4.]}

Schwartz L.\ [73] {\it Radon Measures on Arbitrary Topological Spaces
and Cylindrical Measures.}   Oxford U.P., 1973.
\cmmnt{[\S434 {\it notes\/}.]}

Shelah S.\ \& Fremlin D.H.\ [93] `Pointwise compact and stable sets of
measurable functions', J.\ Symbolic Logic 58 (1993) 435-455.
\cmmnt{[\S465 {\it notes\/}.]}

\indexheader{Sierpi\'nski}
Sierpi\'nski W.\ [1945] `Sur une suite infinie de fonctions de classe 1 dont toute fonction d'accumulation est non mesurable', Fundamenta Math.\ 33 (1945) 104-105.
\cmmnt{[464C.]}

\indexheader{Solecki}
Solecki S.\ [01] `Haar null and non-dominating sets',
Fundamenta Math.\ 170 (2001) 197-217.
\cmmnt{[444Ye.]}

Solymosi J.\ [03]
`Note on a generalization of Roth's theorem', pp.\ 825-827 in
{\smc Aronov Basu Pach \& Sharir 03}.
\cmmnt{[497M.]}

Souslin M.\ [1917] `Sur une d\'efinition des ensembles mesurables B sans
nombres infinis', C.R.Acad.Sci.\ (Paris) 164 (1917) 88-91.
\cmmnt{[421D, \S421 {\it notes\/}.]}

Steen L.A.\ \& Seebach J.A.\ [78] {\it Counterexamples in Topology.}
Springer, 1978.
\cmmnt{[434Ya, \S439 {\it notes\/}.]}

Steinlage R.C.\ [75] `On Haar measure in locally compact T$_2$ spaces',
Amer.\ J.\ Math.\ 97 (1975) 291-307.
\cmmnt{[441C.]}

Strassen V.\ [65] `The existence of probability measures with given
marginals', Ann.\ Math.\ Statist.\ 36 (1965) 423-439.
\cmmnt{[457D.]}

Sullivan J.M.\ [94] `Sphere packings give an explicit bound for the
Besicovitch covering theorem', J.\ Geometric Analysis 4 (1994) 219-231.
\cmmnt{[\S472 {\it notes\/}.]}

\'Swierczkowski S.\ [58] `On a free group of rotations of the Euclidean
space', Indagationes Math.\ 20 (1958) 376-378.
\cmmnt{[449Yi.]}

\indexheader{Szemer\'edi}
Szemer\'edi, E.\ [75]
`On sets of integers containing no k elements in arithmetic progression',
Acta Arithmetica 27 (1975) 199-245.
\cmmnt{[497L.]}
\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

Talagrand M.\ [75] `Sur une conjecture de H.H.Corson', Bull.\ Sci.\
Math.\ 99 (1975) 211-212.
\cmmnt{[467M.]}

Talagrand M.\ [78a] `Sur un th\'eor\`eme de L.Schwartz', C.R.Acad.\
Sci.\ Paris 286 (1978) 265-267.
\cmmnt{[466I.]}

Talagrand M.\ [78b] `Comparaison des boreliens pour les topologies
fortes et faibles', Indiana Univ.\ Math.\ J.\ 21 (1978) 1001-1004.
\cmmnt{[466Za.]}

Talagrand M.\ [80] `Compacts de fonctions mesurables et filtres
non-mesurables', Studia Math.\ 67 (1980) 13-43.
\cmmnt{[464C, 464D.]}

Talagrand M.\ [81] `La $\tau$-r\'egularit\'e de mesures gaussiennes',
Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 57 (1981) 213-221.
\cmmnt{[\S456 {\it intro.}, 456O.]}

Talagrand M.\ [82] `Closed convex hull of set of measurable functions,
Riemann-measurable functions and measurability of translations', Ann.\
Insitut Fourier (Grenoble) 32 (1982) 39-69.
\cmmnt{[465M.]}

Talagrand M.\ [84] {\it Pettis integral and measure theory.}
Mem.\ Amer.\ Math.\ Soc.\  307 (1984).
\cmmnt{[463Ya, 463Yd, 463Za, \S465 {\it intro.}, 465B, 465N,
465R-465T, %465R 465S 465T
466I, \S466 {\it notes\/}.]}

Talagrand M.\ [87] `The Glivenko-Cantelli problem', Ann.\ of Probability
15 (1987) 837-870.
\cmmnt{[\S465 {\it intro.}, 465L, 465M, {\it 465Yd},
\S465 {\it notes\/}.]}

Talagrand M.\ [88] `On liftings and the regularization of stochastic
processes', Prob.\ Theory and Related Fields 78 (1988) 127-134.
\cmmnt{[465U.]}

Talagrand M.\ [89] `A ``regular oscillation'' property of stable sets of
measurable functions', Contemporary Math.\ 94 (1989) 309-313.
\cmmnt{[\S465 {\it notes\/}.]}

Talagrand M.\ [95]
`Concentration of measure and isoperimetric inequalities in product spaces',
Publ.\ Math.\ Inst.\ Hautes \'Etudes Scientifiques 81 (1995) 73-205.
\cmmnt{[492D.]}

Talagrand M.\ [96] `The Glivenko-Cantelli problem, ten years later', J.\
Theoretical Probability 9 (1996) 371-384.
\cmmnt{[\S465 {\it notes\/}.]}
%465Xm

Tamanini I.\ \& Giacomelli C.\ [89] `Approximation of Caccioppoli sets,
with applications to problems in image segmentation',
Ann.\ Univ.\ Ferrara, Sez.\ VII (N.S.) 35 (1989) 187-213.
\cmmnt{[484B.]}

Tao T.\ [07] `A correspondence principle between (hyper)graph theory and
probability theory, and the (hyper)graph removal lemma',
J.\ d'Analyse Math.\ 103 (2007) 1-45'
%({\tt http://front.math.ucdavis.edu/\penalty-100math.CO/{\penalty-100}0602037}).
\cmmnt{[459I, \S497.]}

Tao T.\ [l08] {\it Topics in Ergodic Theory,} lecture notes.
{\tt http://en.wordpress.com/tag/254a-ergodic-theory}, 2008.
\cmmnt{[494F.]}

Taylor A.E.\ [64] {\it Introduction to Functional Analysis.}  Wiley,
1964.
\cmmnt{[\S4A4.]}

Taylor S.J.\ [53] `The Hausdorff $\alpha$-dimensional measure of Brownian
paths in $n$-space',
Proc.\ Cambridge Philosophical Soc.\ 49 (1953) 31-39.
\cmmnt{[477L.]}

Tops{\o}e F.\ [70a] `Compactness in spaces of measures', Studia Math.\ 36
(1970) 195-222.
\cmmnt{[413J, 416K.]}

Tops{\o}e F.\ [70b] {\it Topology and Measure.}  Springer, 1970 (Lecture
Notes in Mathematics 133).
\cmmnt{[437J.]}
%used in chap 41

T\"ornquist A.\ [11] `On the pointwise implementation of near-actions',
Trans.\ Amer.\ Math.\ Soc.\ 363 (2011) 4929-4944.
%({\tt http://arxiv.org/abs/0909.3869}).
\cmmnt{[425D, 425Ya.]}

\medskip%U

Ulam S.\ [1930] `Zur Masstheorie in der allgemeinen
Mengenlehre', Fund.\ Math.\ 16 (1930) 140-150.
\cmmnt{[419G, 438C.]}

Uspenski\v\i\ V.V.\ [88] `Why compact groups are dyadic', pp.\ 601-610
in {\smc Frol\'\i k 88}.
\cmmnt{[4A5T.]}

\medskip%V

Vasershtein L.N.\ [69] `Markov processes over denumerable products of
spaces describing large systems of automata', Problems of Information
Transmission 5 (1969) 47-52.
\cmmnt{[457L.]}

Veech W.A.\ [71] `Some questions of uniform distribution', Annals of Math.\ (2) 94 (1971) 125-138.
\cmmnt{[491H.]}

Veech W.A.\ [77] `Topological dynamics', Bull.\ Amer.\ Math.\ Soc.\ 83 (1977) 775-830.
\cmmnt{[493H.]}

{\cyr Vinokurov V.G.} \& {\cyr Ma{kh}kamov B.M.\ [73] `O prostranstva{kh}
s sover{sh}enno\u{i}, no ne kompaktno\u{i} mero\u{i}',
Nau{ch}ye zapiski Ta{sh}ken{t}skogo instituta narodnogo
{kh}oz{ya}\u{i}stva 71 (1973) 97-103.}
\cmmnt{[451U.]}
% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

Wage M.L.\ [80] `Products of Radon spaces', Russian Math. Surveys 35:3
(1980) 185-187.
\cmmnt{[\S434 {\it notes\/}.]}

Wagon S.\ [85] {\it The Banach-Tarski Paradox.}  Cambridge U.P., 1985.
\cmmnt{[449Yj.]}

Weil A.\ [1940] {\it L'int\'egration dans les groupes topologiques et ses
applications,} Hermann, 1940 (Actualit\'es Scientifiques et Industrielles 869).
\cmmnt{[445N.]}

\indexheader{Wheeler}
Wheeler R.F.\ [83] `A survey of Baire measures and strict topologies',
Expositiones Math.\ 1 (1983) 97-190.
  % printer's error "2" for "1" at head of paper refers to part, not
%volume
\cmmnt{[\S435 {\it notes\/}, \S437 {\it notes\/}.]}

%\medskip%X\medskip%Y\medskip%Z

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\frnewpage
}%end of references


%---------------------------------------------------------------%

\end



