% 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{mt46.tex}
\versiondate{26.8.13}
\copyrightdate{2000}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Introduction}

\newchapter{46}

This chapter collects results inspired by problems in functional
analysis.   \S\S461 and 466 look directly at measures on
linear topological spaces.   The primary applications are of course to
Banach spaces, but as usual we quickly find ourselves considering weak
topologies.   In \S461 I look at `barycenters', or centres of mass, of
probability measures, with the basic theorems on existence and location
of barycenters of given measures and the construction of measures with
given barycenters.   In \S466 I examine topological measures on linear
spaces in terms of the classification developed in Chapter 41.   A
special class of normed spaces, those with `Kadec norms', is
particularly important, and in \S467 I sketch the theory of the most
interesting Kadec norms, the `locally uniformly rotund' norms.

In the middle sections of the chapter, I give an account of the theory
of pointwise compact sets of
measurable functions, as developed by A.Bellow, M.Talagrand and myself.
The first step is to examine pointwise compact sets of continuous
functions (\S462);  these have been extensively studied because they
represent an effective tool for investigating weakly compact sets in
Banach spaces, but here I give only results which are important in measure theory,
with a little background material.   In \S463 I present results on the
relationship between the two most important topologies on spaces of
measurable functions, {\it not} identifying functions which are equal
almost everywhere:  the pointwise topology and the topology of
convergence in measure.   These topologies have very different natures
but nevertheless interact in striking ways.   In particular, we have
important theorems giving conditions under which a pointwise compact set
of measurable functions will be compact for the topology of convergence
in measure (463G, 463L).

The remaining two sections are devoted to some remarkable ideas due to
Talagrand.   The first, `Talagrand's measure' (\S464), is a special
measure on $\Cal PI$ (or $\ell^{\infty}(I)$), extending the usual
measure of $\Cal PI$ in a canonical way.   In \S465 I turn to the theory
of `stable' sets of measurable functions, showing how a concept arising
naturally in the theory of pointwise compact sets led to a
characterization of Glivenko-Cantelli classes in the theory of empirical
measures.

\discrpage

\frfilename{mt461.tex}
\versiondate{9.7.08}
\copyrightdate{2000}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Barycenters and Choquet's theorem}

\newsection{461}

One of the themes of this chapter will be the theory of measures on
linear spaces, and the first fundamental concept is that of `barycenter'
of a measure, its centre of mass (461Aa).   The elementary theory
(461B-461E) uses non-trivial results from the theory of locally convex
spaces\cmmnt{ (\S4A4)}, but is otherwise natural and straightforward.
It is not always easy to be sure whether a measure has a barycenter in a
given space, and I give a representative pair of results in this
direction (461F, 461H).   Deeper questions concern the existence and
nature of measures on a given compact set with a given barycenter.   The
Riesz representation theorem is enough to tell us just which points can
be barycenters of measures on compact sets (461I).   A new idea (461K-461L)
shows that the measures can be moved out towards the boundary of the
compact set.   We need a precise definition of `boundary';  the set of
extreme points seems to be the appropriate concept (461M).
In some important cases, such representing measures on boundaries are
unique (461P).   I append a result
identifying the extreme points of a particular class of compact convex sets
of measures (461Q-461R).


\leader{461A}{Definitions (a)} Let $X$ be a Hausdorff locally convex
linear topological space, and $\mu$ a probability measure on a subset
$A$ of $X$.   Then $x^*\in X$ is a {\bf barycenter}\cmmnt{ or
{\bf resultant}} of $\mu$ if
$\int_Ag\,d\mu$ is defined and equal to $g(x^*)$ for every $g\in X^*$.
\cmmnt{Because $X^*$ separates the points of $X$ (4A4Ec),} $\mu$ can
have at most one barycenter\cmmnt{, so we may speak of `the' barycenter
of $\mu$}.

\spheader 461Ab Let $X$ be any linear space over $\Bbb R$,
and $C\subseteq X$ a
convex set\cmmnt{ (definition:  2A5E)}.   Then a function
$f:C\to\Bbb R$ is {\bf convex} if
$f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$ for all $x$,
$y\in C$ and $t\in[0,1]$.   \cmmnt{ (Compare 233G, 233Xd.)}

\spheader 461Ac\cmmnt{ The following elementary remark is useful.}
Let $X$ be a linear space over $\Bbb R$, $C\subseteq X$ a convex set,
and $f:C\to\Bbb R$ a function.   Then $f$ is convex iff the set
$\{(x,\alpha):x\in C$, $\alpha\ge f(x)\}$ is convex in
$X\times\Bbb R$\cmmnt{ (cf.\ 233Xd)}.

\vleader{72pt}{461B}{Proposition} Let $X$ and $Y$ be Hausdorff locally convex
linear topological spaces, and $T:X\to Y$ a continuous linear operator.
Suppose that $A\subseteq X$, $B\subseteq Y$ are such that
$T[A]\subseteq B$, and let $\mu$ be a probability measure on $A$ which
has a barycenter
$x^*$ in $X$.   Then $Tx^*$ is the barycenter of the image measure
$\mu T^{-1}$ on $B$.

\proof{ All we have to observe is that if $g\in Y^*$ then $gT\in X^*$
(4A4Bd), so that

\Centerline{$g(Tx^*)=\int_Ag(Tx)\mu(dx)=\int_Bg(y)\nu(dy)$}

\noindent by 235G\formerly{2{}35I}.
}%end of proof of 461B

\leader{461C}{Lemma} Let $X$ be a Hausdorff locally convex linear
topological space, $C$ a convex subset of $X$, and $f:C\to\Bbb R$ a lower
semi-continuous convex function.   If $x\in C$ and $\gamma<f(x)$, there
is a $g\in X^*$ such that
$g(y)+\gamma-g(x)\le f(y)$ for every $y\in C$.

\proof{ Let $D$ be the convex set
$\{(x,\alpha):x\in C$, $\alpha\ge f(x)\}$ in $X\times\Bbb R$ (461Ac).
Then the closure $\overline{D}$ of $D$ in $X\times\Bbb R$
is also convex (2A5Eb).   Now
$D$ is closed in $C\times\Bbb R$ (4A2B(d-i)), and $(x,\gamma)\notin D$,
so $(x,\gamma)\notin\overline{D}$.

Consequently there is a continuous linear functional
$h:X\times\Bbb R\to\Bbb R$ such that
$h(x,\gamma)<\inf_{w\in\overline{D}}h(w)$
(4A4Eb).   Now there are $g_0\in X^*$, $\beta\in\Bbb R$ such that
$h(y,\alpha)=g_0(y)+\beta\alpha$ for every $y\in X$ and $\alpha\in\Bbb R$
(4A4Be).   So we have

\Centerline{$g_0(x)+\beta\gamma=h(x,\gamma)
<h(y,f(y))=g_0(y)+\beta f(y)$}

\noindent for every $y\in C$.   In particular,
$g_0(x)+\beta\gamma<g_0(x)+\beta f(x)$, so $\beta>0$.   Setting
$g=-\bover1{\beta}g_0$,

\Centerline{$f(y)\ge\bover1{\beta}g_0(x)+\gamma-\bover1{\beta}g(y)
=g(y)+\gamma-g(x)$}

\noindent for every $y\in C$, as required.
}%end or proof of 461C

\leader{461D}{Theorem} Let $X$ be a Hausdorff locally convex linear
topological space,
$C\subseteq X$ a convex set and $\mu$ a probability measure
on a subset $A$ of $C$.   Suppose that $\mu$ has a barycenter $x^*$ in
$X$ which belongs to $C$.
Then $f(x^*)\le\underline{\int}_Af\,d\mu$ for every
lower semi-continuous convex function $f:C\to\Bbb R$.

\proof{ Take any $\gamma<f(x^*)$.   By 461C there
is a $g\in X^*$ such that
$g(y)+\gamma-g(x^*)\le f(y)$ for every $y\in C$.
Integrating with respect to $\mu$,

\Centerline{$\underline{\intop}_Afd\mu
\ge\gamma-g(x^*)+\int_Ag\,d\mu=\gamma$.}

\noindent As $\gamma$ is arbitrary, $f(x^*)\le\underline{\int}_Af\,d\mu$.
}%end of proof of 461D

\cmmnt{\medskip

\noindent{\bf Remark} Of course $\underline{\int}_Af\,d\mu$ might be infinite.
}%end of comment

\leader{461E}{Theorem} Let $X$ be a Hausdorff locally convex linear
topological space, and $\mu$ a probability measure on $X$ such that (i)
the domain of $\mu$ includes the cylindrical $\sigma$-algebra of $X$
(ii) there is a compact convex set $K\subseteq X$ such that $\mu^*K=1$.
Then $\mu$ has a barycenter in $X$, which belongs to $K$.

\proof{ If $g\in X^*$, then $\gamma_g=\sup_{x\in K}|g(x)|$ is finite,
and $\{x:|g(x)|\le\gamma_g\}$ is a measurable set including $K$, so must
be conegligible, and $\phi(g)=\int g\,d\mu$ is defined and finite.   Now
$\phi:X^*\to\Bbb R$ is a linear functional and
$\phi(g)\le\sup_{x\in K}g(x)$ for every $g\in X^*$;  because $K$ is
compact and convex, there is an $x_0\in K$ such that $\phi(g)=g(x_0)$
for every $g\in X^*$ (4A4Ef), so that $x_0$ is the barycenter of $\mu$ in
$X$.
}%end of proof of 461E

\leader{461F}{Theorem} Let $X$ be a complete locally convex
linear topological space, and $A\subseteq X$ a bounded set.   Let $\mu$
be a $\tau$-additive topological probability measure on $A$.   Then
$\mu$ has a barycenter in $X$.

\proof{{\bf (a)} If $g\in X^*$, then $g\restr A$ is continuous and
bounded (3A5N(b-v)), therefore $\mu$-integrable.   For each
neighbourhood $G$ of $0$ in $X$, set

\Centerline{$F_G
=\{y:y\in X,\,|g(y)-\int_Ag\,d\mu|\le 2\tau_G(g)$ for every
$g\in X^*\}$}

\noindent where $\tau_G(g)=\sup_{x\in G}g(x)$ for $g\in X^*$.   Then
$F_G$ is non-empty.   \Prf\ Set
$H=\interior(G\cap(-G))$, so that $H$ is an open neighbourhood of $0$.
Because $A$ is bounded, there is an $m\ge 1$ such that
$A\subseteq mH$.   The set $\{x+H:x\in A\}$ is an open cover of $A$, and
$\mu$ is a $\tau$-additive topological measure, so there are
$x_0,\ldots,x_n\in A$ such that
$\mu(A\setminus\bigcup_{i\le n}(x_i+H))\le\Bover1m$.   Set

\Centerline{$E_i=A\cap(x_i+H)\setminus\bigcup_{j<i}(x_j+H)$}

\noindent for $i\le n$, and $y=\sum_{i=0}^n(\mu E_i)x_i$;  set
$E=A\setminus\bigcup_{i\le n}(x_i+H)$.   Then, for any
$g\in X^*$,

$$\eqalignno{|g(y)-\int_Ag\,d\mu|
&\le\sum_{i=0}^n|\mu E_ig(x_i)-\int_{E_i}g\,d\mu|+\int_E|g|d\mu\cr
&\le\sum_{i=0}^n\int_{E_i}|g(x)-g(x_i)|\mu(dx)+m\tau_G(g)\mu E\cr
\displaycause{because $E\subseteq mH$, so $g(x)\le m\tau_G(g)$ and
$g(-x)\le m\tau_G(g)$ for every $x\in E$}
&\le\sum_{i=0}^n\tau_G(g)\mu E_i+m\tau_G(g)\mu E\cr
\displaycause{because if $i\le n$ and $x\in E_i$, then $x-x_i$ and
$x_i-x$ belong to $G$, so $|g(x)-g(x_i)|=|g(x-x_i)|\le\tau_G(g)$}
&\le 2\tau_G(g).\cr}$$

\noindent As $g$ is arbitrary, $y\in F_G$ and $F_G\ne\emptyset$.\ \Qed

\medskip

{\bf (c)} Since $F_{G\cap H}\subseteq F_G\cap F_H$ for all
neighbourhoods $G$ and $H$ of $0$, $\{F_G:G$ is a neighbourhood of $0\}$
is a filter base and generates a filter $\Cal F$ on $X$.   Now $\Cal F$
is Cauchy.   \Prf\ If $G$ is any neighbourhood of $0$, let
$G_1\subseteq G$ be a closed convex neighbourhood of $0$, and set
$H=\bover14G_1$.   \Quer\ If $y$, $y'\in F_H$ and $y-y'\notin G$, then
there is a $g\in X^*$ such that $g(y-y')>\tau_{G_1}(g)$ (4A4Eb again).   
But now

$$\eqalign{\tau_H(g)
&=\Bover14\tau_{G_1}(g)
<\Bover14(|g(y)-\int_Ag\,d\mu|+|g(y')-\int_Ag\,d\mu|)\cr
&\le\Bover14(2\tau_H(g)+2\tau_H(g)). \text{ \Bang}\cr}$$

\noindent This means that $F_H-F_H\subseteq G$;  as $G$ is arbitrary,
$\Cal F$ is Cauchy.\ \Qed

\medskip

{\bf (d)} Because $X$ is complete, $\Cal F$ has a limit $x^*$ say.
Take any $g\in X^*$.   \Quer\ If $g(x^*)\ne\int_Ag\,d\mu$, set
$G=\{x:|g(x)|\le\bover13|g(x^*)-\int_Ag\,d\mu|\}$.   Then $G$ is a
neighbourhood of $0$ in $X$, and

$$\eqalign{0
&<|g(x^*)-\int_Ag\,d\mu|
=\lim_{x\to\Cal F}|g(x)-\int_Ag\,d\mu|\cr
&\le\sup_{x\in F_G}|g(x)-\int_Ag\,d\mu|
\le 2\tau_G(g)
\le\Bover23|g(x^*)-\int_Ag\,d\mu|.  \text{ \Bang}\cr}$$

\noindent So $g(x^*)=\int_Ag\,d\mu$;  as $g$ is arbitrary, $x^*$ is the
barycenter of $\mu$.
}%end of proof of 461F

\leader{461G}{Lemma} Let $X$ be a normed space, and $\mu$ a probability
measure on $X$ such that every member of the dual $X^*$ of $X$ is
integrable.   Then $g\mapsto\int g\,d\mu:X^*\to\Bbb R$ is a bounded linear
functional on $X^*$.

\proof{ Replacing $\mu$ by its completion if necessary, we may suppose
that $\mu$ is complete, so that every member of $X^*$ is
$\Sigma$-measurable, where $\Sigma$ is the domain of $\mu$.   (The point
is that $\mu$ and its completion give rise to the same integrals, by
212Fb.)    Set $\phi(g)=\int g\,d\mu$ for $g\in X^*$.
For each $n\in\Bbb N$ let $E_n$ be a measurable envelope of
$B_n=\{x:\|x\|\le n\}$;  replacing $E_n$ by $\bigcap_{i\ge n}E_i$ if
necessary, we may suppose that $\sequencen{E_n}$ is non-decreasing.   If
$n\in\Bbb N$ and $g\in X^*$ then $\{x:x\in E_n,\,|g(x)|>n\|g\|\}$ is a
measurable subset of $E_n$ disjoint from $B_n$, so must be negligible,
and $|\int_{E_n}g|\le n\|g\|$.   We therefore have an element $\phi_n$
of $X^{**}$ defined by setting $\phi_n(g)=\int_{E_n}g$ for every $g\in
X^*$.   But also $\phi(g)=\lim_{n\to\infty}\phi_n(g)$ for every $g$,
because $\sequencen{E_n}$ is a non-decreasing sequence of measurable
sets with union $X$.   By the Uniform Boundedness Theorem (3A5Ha),
$\{\phi_n:n\in\Bbb N\}$ is bounded in $X^{**}$, and $\phi\in X^{**}$.
}%end of proof of 461G

\leader{461H}{Proposition} Let $X$ be a reflexive Banach space, and
$\mu$ a probability measure on $X$ such that every member of $X^*$ is
$\mu$-integrable.   Then $\mu$ has a barycenter in $X$.

\proof{ By 461G, $g\mapsto\int g\,d\mu$ is a bounded linear functional on
$X^*$;  but this means that it is represented by a member of $X$, which
is the barycenter of $\mu$.
}%end of proof of 461H

\leader{461I}{Theorem} Let $X$ be a Hausdorff locally convex linear
topological space, and $K\subseteq X$ a compact set.   Then the closed
convex hull of $K$ in $X$ is just the set of barycenters of Radon
probability measures on $K$.

\proof{{\bf (a)} If $\mu$ is a Radon probability measure on $K$ with
barycenter $x^*$, then

\Centerline{$g(x^*)=\int_Kg(x)\mu(dx)
\le\sup_{x\in K}g(x)\le\sup_{x\in\overline{\Gamma(K)}}g(x)$}

\noindent for every $g\in X^*$;
because $\overline{\Gamma(K)}$ is closed and convex, it must
contain $x^*$ (4A4Eb once more).

\medskip

{\bf (b)} Now suppose that $x^*\in\overline{\Gamma(K)}$.   Let
$W\subseteq C(X)$ be the set of functionals of the form
$g+\alpha\chi X$, where $g\in X^*$ and $\alpha\in\Bbb R$.   Set
$U=\{g\restr K:g\in W\}$, so that $U$ is a linear subspace of $C(K)$
containing $\chi K$.

If $g_1$, $g_2\in W$ and $g_1\restr K=g_2\restr K$, then
$\{x:g_1(x)=g_2(x)\}$ is a closed convex set including $K$, so contains
$x^*$, and $g_1(x^*)=g_2(x^*)$;  accordingly we have a functional
$\phi:U\to\Bbb R$ defined by setting $\phi(g\restr K)=g(x^*)$ for every
$g\in W$.   Of course $\phi$ is linear;  moreover,
$\phi(f)\le\sup_{x\in K}f(x)$ for every $f\in U$, by 4A4Eb
yet again.   Applying this to $\pm f$, we see
that $|\phi(f)|\le\|f\|_{\infty}$ for every $f\in U$.   We therefore
have an extension of $\phi$ to a continuous linear functional $\psi$ on
$C(K)$ such that $\|\psi\|\le 1$, by the Hahn-Banach theorem (3A5Ab).   Now

\Centerline{$\psi(\chi K)=\phi(\chi K)=\chi X(x^*)=1$;}

\noindent so if $0\le f\le\chi K$ then

\Centerline{$|1-\psi(f)|
=|\psi(\chi K-f)|\le \|\chi K-f\|_{\infty}\le 1$,}

\noindent and $\psi(f)\ge 0$.   It follows that $\psi(f)\ge 0$ for every
$f\in C(K)^+$.   But this means that there is a Radon probability
measure $\mu$ on
$K$ such that $\psi(f)=\int f\,d\mu$ for every $f\in C(K)$ (436J/436K).
As $\mu K=\psi(\chi K)=1$, $\mu$ is a probability measure;  and for any
$g\in X^*$

\Centerline{$\int_Kg\,d\mu=\psi(g\restr K)=\phi(g\restr K)=g(x^*)$,}

\noindent so $x^*$ is the barycenter of $\mu$, as required.
}%end of proof of 461I

\leader{461J}{Corollary:  \Krein's theorem} Let $X$ be a complete
Hausdorff locally convex linear topological space, and $K\subseteq X$ a
weakly compact set.   Then the closed convex hull $\overline{\Gamma(K)}$
of $K$ is weakly compact.

\proof{ Give $K$ the weak topology induced by $\frak T_s(X,X^*)$.   Let
$P$ be the set of Radon probability measures on $K$,
so that $P$ is compact in its narrow topology (437R(f-ii)).
By 461F, every $\mu\in P$ has a barycenter $b(\mu)$ in
$K$.   If $g\in X^*$, $g(b(\mu))=\int_Kg\,d\mu$, while $g\restr K$ is
continuous, so $\mu\mapsto\int_Kg\,d\mu$ is continuous, by 437Kc.
Accordingly $b:P\to X$ is continuous for the narrow topology on $P$
and the weak topology on $X$, and $b[P]$ is weakly compact.
But $b[P]$ is the weakly closed convex hull of $K$, by 461I applied to
the weak topology on $X$.   By 4A4Ed, $\Gamma(K)$ has the same closure
for the original topology of $X$ as it has for the weak topology, and
$\overline{\Gamma(K)}=b[P]$ is weakly compact.
}%end of proof of 461J

\leader{461K}{Lemma} Let $X$ be a Hausdorff locally convex linear
topological space, $K$ a compact convex subset of $X$, and $P$ the set of
Radon probability measures on $K$.   Define a relation $\preccurlyeq$ on
$P$ by saying that $\mu\preccurlyeq\nu$ if
$\int fd\mu\le\int fd\nu$ for every continuous convex function
$f:K\to\Bbb R$.

(a) $\preccurlyeq$ is a partial order on $P$.

(b) If $\mu\preccurlyeq\nu$ then
$\int fd\mu\le\int fd\nu$ for every lower semi-continuous convex function
$f:K\to\Bbb R$.

(c) If $\mu\preccurlyeq\nu$ then $\mu$ and $\nu$ have the same barycenter.

(d) If we give $P$ its narrow topology, then $\preccurlyeq$ is closed in
$P\times P$.

(e) For every $\mu\in P$ there is a $\preccurlyeq$-maximal $\nu\in P$
such that $\mu\preccurlyeq\nu$.

\proof{{\bf (a)} Write $\Psi$ for the set of continuous convex functions
from $K$ to $\Bbb R$.   Note that if $f$, $g\in\Psi$ and $\alpha\ge 0$ then
$\alpha f$, $f+g$ and $f\vee g$ all belong to $\Psi$.   Consequently
$\Psi-\Psi$ is a Riesz subspace of $C(K)$.   \Prf\ $\Psi-\Psi$ is a linear
subspace because $\Psi$ is closed under addition and multiplication by
positive scalars.   If $f$, $g\in\Psi$ then

\Centerline{$|f-g|=(f-g)\vee(g-f)=2(f\vee g)-(f+g)\in\Psi$;}

\noindent by 352Ic, $\Psi-\Psi$ is a Riesz subspace.\ \Qed

It follows that $\Psi-\Psi$ is $\|\,\|_{\infty}$-dense in $C(K)$.   \Prf\
Constant functions belong to $\Psi$, and if $x$, $y\in K$ are distinct
there is an $f\in X^*$ such that $f(x)\ne f(y)$, in which case
$f\restr K$ belongs to $\Psi$ and separates $y$ from $x$.   By the
Stone-Weierstrass theorem (281A), $\Psi-\Psi$ is dense.\ \Qed

The definition of $\preccurlyeq$ makes it plain that
it is reflexive and transitive.   But it is also antisymmetric.
\Prf\ If $\mu\preccurlyeq\nu$ and $\nu\preccurlyeq\mu$, then
$\int fd\mu=\int fd\nu$ for every $f\in\Psi$, therefore for every
$f\in\Psi-\Psi$, therefore for every $f\in C(K)$, and $\mu=\nu$
by 416E(b-v).\ \Qed

So $\preccurlyeq$ is a partial order.

\medskip

{\bf (b)(i)}
Now suppose that $f:K\to\Bbb R$ is a lower semi-continuous convex
function, and $x\in K$.
Then $f(x)=\sup\{g(x):g\in\Psi$, $g\le f\}$.   \Prf\ If $\gamma<f(x)$ there
is a $g\in X^*$ such that
$g(y)+\gamma-g(x)\le f(y)$ for every $y\in K$, by 461C.   Now $g\restr K$
belongs to $\Psi$, $g\restr K\le f$ and $(g\restr K)(x)=\gamma$.\ \Qed

\medskip

\quad{\bf (ii)} It follows that if $f:K\to\Bbb R$ is
lower semi-continuous and
convex, $\int fd\mu=\sup\{\int g\,d\mu:g\in\Psi$, $g\le f\}$ for every
$\mu\in P$.   \Prf\ Because $\Psi$ is closed under $\vee$,
$A=\{g:g\in\Psi$, $g\le f\}$ is upwards-directed.   Because $\mu$ is
$\tau$-additive and $f=\sup A$, $\int fd\mu=\sup_{g\in A}\int g\,d\mu$ by
414Ab.\ \Qed

So if $\mu$, $\nu\in P$ and $\mu\preccurlyeq\nu$, then

\Centerline{$\int fd\mu
=\sup_{g\in\Psi,g\le f}\int g\,d\mu
\le\sup_{g\in\Psi,g\le f}\int g\,d\nu
=\int fd\nu$;}

\noindent as $f$ is arbitrary, (b) is true.

\medskip

{\bf (c)} By 461E, applied to the Radon probability measure on $X$
extending $\mu$, $\mu$ has a barycenter $x\in K$.   If $g\in X^*$ then
$g\restr K$ belongs to $\Psi$, so $\int_Kg\,d\mu\le\int_Kg\,d\nu$;  but the
same applies to $-g$, so

\Centerline{$g(x)=\int_Kg\,d\mu=\int_Kg\,d\nu$.}

\noindent As $g$ is arbitrary, $x$ is also the barycenter of $\nu$.

\medskip

{\bf (d)} As noted in 437Kc, the narrow topology on $P$ corresponds to the
weak* topology on $C(K)^*$.   So all the functionals
$\mu\mapsto\int fd\mu$, for $f\in\Psi$, are continuous;  it follows at once
that $\preccurlyeq$ is a closed subset of $P\times P$.

\medskip

{\bf (e)} Any non-empty
upwards-directed $Q\subseteq P$ has an upper bound in $P$.   \Prf\ For
$\nu\in Q$ set $V_{\nu}=\{\lambda:\nu\preccurlyeq\lambda\}$.   Then every
$V_{\nu}$ is closed, and because $Q$ is upwards-directed the family
$\{V_{\nu}:\nu\in Q\}$ has the finite intersection property.   Because $P$
is compact (437R(f-ii) again), $\bigcap_{\nu\in Q}V_{\nu}$ is non-empty;
now any member of the
intersection is an upper bound of $Q$.\ \Qed

By Zorn's lemma, every member of $P$ is dominated by a maximal element of
$P$.
}%end of proof of 461K

\leader{461L}{Lemma} Let $X$ be a Hausdorff locally convex linear
topological space, $K$ a compact convex subset of $X$, and $P$ the set of
Radon probability measures on $K$.   Suppose that
$\mu\in P$ is maximal for the partial order $\preccurlyeq$ of 461K.

(a) $\mu(\bover12(M_1+M_2))=0$ whenever $M_1$, $M_2$ are disjoint closed
convex subsets of $K$.

(b) $\mu F=0$ whenever $F\subseteq K$ is a Baire set (for the
subspace topology of $K$) not containing any extreme point of $K$.

\proof{{\bf (a)} Set $M=\{\bover12(x+y):x\in M_1$, $y\in M_2\}$.
Set $q(x,y)=\bover12(x+y)$ for $x\in M_1$, $y\in M_2$, so that
$q:M_1\times M_2\to M$ is continuous.   Let $\mu_M$ be the
subspace measure on
$M$ induced by $\mu$, so that $\mu_M$ is a Radon measure on
$M$ (416Rb).   Let $\lambda$ be a Radon measure on $M_1\times M_2$ such
that $\mu_M=\lambda q^{-1}$ (418L), and define
$\psi:C(K)\to\Bbb R$ by writing

\Centerline{$\psi(f)
=\int_{K\setminus M}f\,d\mu
  +\int_{M_1\times M_2}\bover12(f(x)+f(y))\lambda(d(x,y))$.}

\noindent Then $\psi$ is linear, $\psi(f)\ge 0$ whenever $f\in C(K)^+$,
and

\Centerline{$\psi(\chi K)
=\mu(K\setminus M)+\lambda(M_1\times M_2)
=\mu(K\setminus M)+\mu_M(M)=1$.}

Let $\nu\in P$ be such that $\int fd\nu=\psi(f)$ for every $f\in C(K)$
(436J/436K again).   If $f:K\to\Bbb R$ is continuous and convex, then

$$\eqalignno{\int fd\nu
=\psi(f)
&=\int_{K\setminus M}f\,d\mu
  +\int_{M_1\times M_2}\Bover12(f(x)+f(y))\lambda(d(x,y))\cr
&\ge\int_{K\setminus M}f\,d\mu
  +\int_{M_1\times M_2}f(\Bover12(x+y))\lambda(d(x,y))\cr
&=\int_{K\setminus M}f\,d\mu
  +\int_{M_1\times M_2}fq\,d\lambda
=\int_{K\setminus M}f\,d\mu
  +\int_Mf\,d\mu_M\cr
\noalign{\noindent (235G)}
&=\int_Kf\times\chi(K\setminus M)d\mu
  +\int_Kf\times\chi M\,d\mu\cr
\noalign{\noindent (131Fa)}
&=\int_Kf\,d\mu.\cr}$$

\noindent So $\mu\preccurlyeq\nu$;  as we are assuming that $\mu$ is
maximal, $\mu=\nu$.

Because $M_1$ and $M_2$ are disjoint compact convex sets in
the Hausdorff locally convex space $X$, there are a $g\in X^*$ and an
$\alpha\in\Bbb R$ such that $g(x)<\alpha<g(y)$ whenever $x\in M_1$ and
$y\in M_2$ (4A4Ee).   Set
$f(x)=|g(x)-\alpha|$ for $x\in K$;  then $f$ is a continuous convex
function.   If $x\in M_1$ and $y\in M_2$, then

$$\eqalign{f(\Bover12(x+y))
&=|g(\Bover12(x+y))-\alpha|
=\Bover12|(g(x)-\alpha)+(g(y)-\alpha)|\cr
&<\Bover12(|g(x)-\alpha|+|g(y)-\alpha|)
=\Bover12(f(x)+f(y)).\cr}$$

\noindent Looking at the formulae above for $\psi(f)$, we see that we
have

\Centerline{$\int f\,d\nu
=\int_{K\setminus M}f\,d\mu
  +\int_{M_1\times M_2}\Bover12(f(x)+f(y))\lambda(d(x,y))$,}

\Centerline{$\int f\,d\mu
=\int_{K\setminus M}f\,d\mu
  +\int_{M_1\times M_2}f(\Bover12(x+y))\lambda(d(x,y))$.}

\noindent Since these are equal, and $f(\bover12(x+y))<\bover12(f(x)+f(y))$
for all $x\in M_1$ and $y\in M_2$, we must have
$\mu M=\lambda(M_1\times M_2)=0$, as required.

\medskip

{\bf (b)(i)} Consider first the case in which $F$ is a zero set for the
subspace topology.   Since $F\subseteq K$ is a closed
G$_{\delta}$ set in $K$, $K\setminus F$ is expressible as a union
$\bigcup_{n\in\Bbb N}F_n$ of compact sets.   For any $n\in\Bbb N$,
$z\in F$ and $y\in F_n$, there is a $g\in X^*$ such that $g(z)\ne g(y)$;
since $F$ and $F_n$ are compact, there is a finite set
$\Phi_n\subseteq X^*$ such that whenever $z\in F$ and $y\in F_n$
there is a $g\in\Phi_n$ such that $g(z)\ne g(y)$.   Set
$\Phi=\bigcup_{n\in\Bbb N}\Phi_n\cup\{0\}$;  then
$\Phi$ is countable;  let $\sequencen{g_n}$ be a sequence running over
$\Phi$, and define $T:X\to\BbbR^{\Bbb N}$ by setting $(Tx)(n)=g_n(x)$
for $n\in\Bbb N$, $x\in X$.   Then $T[F]\cap T[F_n]=\emptyset$ for every
$n$, so $F=K\cap T^{-1}[T[F]]$.

Now $T[F]$ is a compact subset of the metrizable
compact convex set $T[K]$,
and does not contain any extreme point of $T[K]$, by 4A4Gc.

Let $\Cal U$ be the set of convex open subsets of
$\BbbR^{\Bbb N}$.   Because the topology of $\BbbR^{\Bbb N}$ is
locally convex, $\Cal U$ is a base
for the topology of $\BbbR^{\Bbb N}$;  because it is separable and
metrizable, $\Cal U$ includes a countable base $\Cal U_0$ (4A2P(a-iii)),
and $\Cal V=\{T[K]\cap U:U\in\Cal U_0\}$ is a countable base
for the topology of $T[K]$ (4A2B(a-vi)).   Set
$\Cal M=\{K\cap T^{-1}[\overline{V}]:V\in\Cal V\}$, so that
$\Cal M$ is a countable family of closed convex subsets of $K$.

If $z\in F$, then there are distinct $u$, $v\in T[K]$ such that
$Tz=\bover12(u+v)$.   Now there must be $V$, $V'\in\Cal V$, with disjoint
closures, such that $u\in V$ and $v\in V'$, so that
$z\in\bover12(M+M')$, where $M=K\cap T^{-1}[\overline{V}]$ and
$M'=K\cap T^{-1}[\overline{V'}]$ are disjoint members of $\Cal M$.   Thus

\Centerline{$F\subseteq\bigcup\{\bover12(M+M'):
M$, $M'\in\Cal M$, $M\cap M'=\emptyset\}$.}

\noindent But (a) tells us that $\bover12(M+M')$ is
$\mu$-negligible whenever $M$, $M'\in\Cal M$ are disjoint.
As $\Cal M$ is countable, $\mu F=0$, as required.

\medskip

\quad{\bf (ii)} Now consider the Baire measure $\mu\restr\CalBa(K)$, where
$\CalBa(K)$ is the Baire $\sigma$-algebra of $K$.   This is inner regular
with respect to the zero sets (412D).   If $F\in\CalBa(K)$ contains no
extremal point of $K$, then (i) tells us that $\mu Z=0$ for every zero set
$Z\subseteq F$, so $\mu F$ must also be $0$.
}%end of proof of 461L

\leader{461M}{Theorem} Let $X$ be a Hausdorff locally convex linear
topological space, $K$ a compact convex subset of $X$ and $E$ the set of
extreme points of $K$.   Let $x\in X$.   Then there is a probability
measure $\mu$ on $E$ with barycenter $x$.
If $K$ is metrizable we can take $\mu$ to be a Radon measure.

\proof{ Let $P$ be the set of Radon probability measures on $K$
and $\preccurlyeq$ the partial order on $P$ described in 461K.
By 461Ke, there is a maximal element $\nu$ of $P$ such that
$\delta_x\preccurlyeq\nu_0$, where $\delta_x\in P$ is the Dirac measure on
$X$ concentrated at $x$.   By 461Kc, $x$ is the barycenter of $\nu$.

Let $\lambda=\nu\restr\CalBa(K)$ be the Baire measure
associated with $\nu$.   By 461Lb, $\lambda^*E=1$.   So the subspace
measure $\lambda_E$ on $E$ is a probability measure on $E$.   Let $\mu$ be
the completion of $\lambda_E$.

If $g\in X^*$ then
$g\restr K$ is continuous, therefore $\CalBa(K)$-measurable, so

$$\eqalignno{g(x)
&=\int_Kg\,d\nu
=\int_Kg\,d\lambda
=\int_Eg\,d\lambda_E\cr
\displaycause{214F}
&=\int_Eg\,d\mu\cr}$$

\noindent (212Fb).   So $x$ is the barycenter of $\mu$.

Now suppose that $K$ is metrizable.   In this case $E$ is a
G$_{\delta}$ set in $K$.   \Prf\ Let $\rho$ be a metric on $K$ inducing its
topology.   Then

\Centerline{$K\setminus E=\bigcup_{n\ge 1}\{tx+(1-t)y:x$, $y\in K$,
$t\in[2^{-n},1-2^{-n}]$, $\rho(x,y)\ge 2^{-n}\}$}

\noindent is K$_{\sigma}$, so its complement in $K$ is a G$_{\delta}$ set
in $K$.\ \QeD\   So $E$ is analytic (423Eb)
and $\mu$ is a Radon measure (433Cb).
}%end of proof of 461M

\leader{461N}{Lemma} Let $X$ be a Hausdorff locally convex linear
topological space, $K$ a compact convex subset of $X$, and $P$ the set of
Radon probability measures on $K$.   Let $E$ be
the set of extreme points of $K$ and suppose that $\mu\in P$ and
$\mu^*E=1$.   Then $\mu$ is maximal in $P$ for
the partial order $\preccurlyeq$ of 461K.

\proof{{\bf (a)} For $\mu\in P$ write $b(\mu)$ for the barycenter of $\mu$.
Then $b:P\to K$ is continuous for the narrow topology of $P$ and the weak
topology of $X$, as in 461J.    For $f\in C(K)$ define $\bar f:K\to\Bbb R$
by setting $\bar f(x)=\sup\{\int fd\mu:\mu\in P$, $b(\mu)=x\}$ for
$x\in K$.

\medskip

\quad{\bf (i)} Taking $\delta_x$ to be the Dirac measure on $X$
concentrated at $x$, we see that

\Centerline{$f(x)=\int fd\delta_x\le\bar f(x)$}

\noindent for any $x\in K$.

\medskip

\quad{\bf (ii)} For any $x\in K$ there is a $\mu\in P$ such that $b(\mu)=x$
and $\int fd\mu=\bar f(x)$.   \Prf\ The set $\{\mu:\mu\in P$, $b(\mu)=x\}$
is compact, so its continuous image $\{\int fd\mu:b(\mu)=x\}$ is compact
and contains its supremum.\ \Qed

$\bar f$ is upper semi-continuous.   \Prf\ For any
$\alpha\in\Bbb R$, the set

\Centerline{$\{(\mu,x):\mu\in P$, $x\in K$, $b(\mu)=x$,
$\int fd\mu\ge\alpha\}$}

\noindent is
compact, so its projection onto the second coordinate is closed;  but this
projection is just $\{x:\bar f(x)\ge\alpha\}$.\ \Qed

\medskip

\quad{\bf (iii)} $\bar f:K\to\Bbb R$ is concave.   \Prf\ Suppose that
$x$, $y\in K$ and $t\in[0,1]$.   Take $\mu$, $\nu\in P$ such that
$b(\mu)=x$, $\bar f(x)=\int fd\mu$, $b(\nu)=y$ and
$\bar f(y)=\int fd\nu$.   Set
$\lambda=t\mu+(1-t)\nu$.   Then

\Centerline{$\int_Kg\,d\lambda=t\int_Kg\,d\mu+(1-t)\int_Kg\,d\nu
=tg(x)+(1-t)g(y)=g(tx+(1-t)y)$}

\noindent for any $g\in X^*$, so $b(\lambda)=tx+(1-t)y$ and

\Centerline{$\bar f(tx+(1-t)y)
\ge\int fd\lambda
=t\int fd\mu+(1-t)\int fd\nu
=t\bar f(x)+(1-t)\bar f(y)$.}

\noindent As $x$, $y$ and $t$ are arbitrary, $\bar f$ is concave.\ \Qed

\medskip

\quad{\bf (iv)} If $x\in K$ and $\bar f(x)>f(x)$ then $x\notin E$.
\Prf\ There
is a $\mu\in P$ such that $b(\mu)=x$ and $\int fd\mu>f(x)$.   We cannot have
$\mu\{x\}=1$ because $\int fd\mu\ne f(x)$, so there is a point $y$ of the
support of $\mu$ such that $y\ne x$.   Let $g\in X^*$ be such that
$g(y)>g(x)$ and set $G=\{z:z\in K$, $g(z)>g(x)\}$, $t=\mu G$.
Then $t>0$;  also
$\int_Gg\,d\mu>tg(x)=t\int g\,d\mu$, so $t\ne 1$.
Define $\nu_1$, $\nu_2\in P$
by setting $\nu_1H=\bover1t\mu(G\cap H)$ whenever $H\subseteq K$ and
$\mu$ measures $G\cap H$, $\nu_2H=\bover1{1-t}\mu(H\setminus G)$
whenever $H\subseteq K$ and $\mu$ measures $H\setminus G$.   Let $x_1$,
$x_2$ be the barycenters of $\nu_1$, $\nu_2$ respectively.   For any
$h\in X^*$,

$$\eqalign{h(tx_1+(1-t)x_2)
&=th(x_1)+(1-t)h(x_2)
=t\int_Kh\,d\nu_1+(1-t)\int_Kh\,d\nu_2\cr
&=\int_Gh\,d\mu+\int_{K\setminus G}h\,d\mu
=h(x).\cr}$$

\noindent So $x=tx_1+(1-t)x_2$.    But both $x_1$ and $x_2$ belong to $K$
and $g(x_1)>g(x)$, so $x_1\ne x$, while $0<t<1$, so $x$ is not an extreme
point of $K$.\ \Qed


\medskip

{\bf (b)} Now take $\mu\in P$ such that $\mu^*E=1$, and $\nu\in P$ such that
$\mu\preccurlyeq\nu$.   For any convex
$f\in C(K)$, $f-\bar f$ is the sum of lower semi-continuous convex functions
so is lower semi-continuous, and
$\{x:f(x)=\bar f(x)\}=\{x:f(x)-\bar f(x)\ge 0\}$ is a G$_{\delta}$ set.
By (a-iv), it includes $E$, so $\int fd\mu=\int\bar fd\mu$.
In addition, $\int fd\mu\le\int fd\nu$ and
$\int\bar fd\nu\le\int\bar fd\mu$, by 461Kb applied to $-\bar f$.
But since $f\le\bar f$, we have

\Centerline{$\int fd\nu\le\int\bar fd\nu\le\int\bar fd\mu=\int fd\mu$;}

\noindent as $f$ is arbitrary, $\nu\preccurlyeq\mu$;
as $\nu$ is arbitrary, $\mu$ is maximal.
}%end of proof of 461N

\leader{461O}{Lemma} Suppose that $X$ is a Riesz space with a Hausdorff
locally convex linear space topology, and $K\subseteq X$ a
compact convex set such that every non-zero member of the positive cone
$X^+$ is uniquely expressible as $\alpha x$ for some $x\in K$ and
$\alpha\ge 0$.
Let $P$ be the set of Radon probability measures on $K$ and $\preccurlyeq$
the partial order described in 461K.   If $\mu$, $\nu\in P$ have the same
barycenter then they have a common upper bound in $P$.

\proof{{\bf (a)} If $X=\{0\}$ then $K$ is either $\{0\}$ or empty
and the result is immediate, so
henceforth suppose that $X$ is non-trivial.   Because each non-zero member
of $X^+$ is uniquely expressible as a multiple of a member of $K$, no
distinct members of $K\setminus\{0\}$ can be multiples of each other;  as
$K$ is convex, $0\notin K$.
If $z_0,\ldots,z_r\in K$,
$\gamma_0,\ldots,\gamma_r\ge 0$ and $z=\sum_{k=0}^r\gamma_kz_k$ belongs to
$K$, then
$\sum_{k=0}^r\gamma_k=1$.   \Prf\ Setting $\gamma=\sum_{k=0}^r\gamma_k$,
$\gamma\ne 0$ and
$\Bover1{\gamma}z=\sum_{k=0}^r\Bover{\gamma_r}{\gamma}z_k$ belongs to $K$;
accordingly $\Bover1{\gamma}z=z$ and $\gamma=1$.\ \Qed

\medskip

{\bf (b)} Take any $x\in K$, and let $P_x$ be the set of elements of $P$
with barycenter $x$.
Write $Q_x$ for the set of members of $P_x$ with finite support.
Then $Q_x$ is dense in $P_x$.   \Prf\ Suppose that
$\mu\in P_x$, $f_0,\ldots,f_n\in C(K)$ and $\epsilon>0$.   Then there is a
finite cover of $K$ by relatively open convex sets on each of which every
$f_i$ has
oscillation at most $\epsilon$;  so we have a partition $\Cal H$ of $K$ into
finitely many non-empty Borel sets $H$ such that every $f_i$ has
oscillation at most $\epsilon$ on the convex hull $\Gamma(H)$.
If $H\in\Cal H$ and $\mu H>0$, let $x_H$ be the barycenter of the measure
$\mu_H\in P$ where $\mu_HF=\Bover1{\mu H}\mu(F\cap H)$ whenever
$F\subseteq K$
and $\mu$ measures $F\cap H$.   For other $H\in\Cal H$, take any point $x_H$
of $H$.   In all cases, $x_H\in\overline{\Gamma(H)}$ so
$|f_i(y)-f_i(x_H)|\le\epsilon$ whenever $i\le n$ and $y\in H$.

Consider $\nu=\sum_{H\in\Cal H}\mu H\cdot\delta_{x_H}$, where
$\delta_{x_H}\in P$ is the Dirac measure on $K$ concentrated at $x_H$.
If $g\in X^*$ then

\Centerline{$\int_Kg\,d\nu
=\sum_{H\in\Cal H}\mu H\cdot g(x_H)
=\sum_{H\in\Cal H}\int_Hg\,d\mu
=\int_Kg\,d\mu
=g(x)$;}

\noindent as $g$ is arbitrary, $\nu\in P_x$ and $\nu\in Q_x$.   Next, for
$i\le n$,

$$\eqalign{|\int f_id\nu-\int f_id\mu|
&\le\sum_{H\in\Cal H}|f_i(x_H)\mu H-\int_Hf_id\mu|\cr
&\le\sum_{H\in\Cal H}\int_H|f_i(x_H)-f_i(y)|\mu(dy)
\le\sum_{H\in\Cal H}\epsilon\mu H
=\epsilon.\cr}$$

\noindent As $f_0,\ldots,f_n$ and $\epsilon$ are arbitrary, $Q_x$ is dense
in $P_x$.\ \Qed

\medskip

{\bf (c)} Suppose that $x\in K$ and $\mu$, $\nu\in Q_x$.
Then they have a common upper bound in $P$.   \Prf\
Express $\mu$, $\nu$ as $\sum_{i=0}^m\alpha_i\delta_{x_i}$,
$\sum_{j=0}^n\beta_j\delta_{y_j}$ respectively,
where all the $\alpha_i$ and
$\beta_j$ are strictly positive, all the $x_i$ and $y_j$ belong to $K$, and
$\sum_{i=0}^m\alpha_i=\sum_{j=0}^n\beta_j=1$.   If $g\in X^*$ then

\Centerline{$g(x)
=\int_Kg\,d\mu=\sum_{i=0}^m\alpha_ig(x_i)=\sum_{j=0}^n\beta_jg(y_j)$,}

\noindent so $x=\sum_{i=0}^m\alpha_ix_i=\sum_{j=0}^n\beta_jy_j$.   By the
decomposition theorem 352Fd there is a family
$\langle w_{ij}\rangle_{i\le m,j\le n}$ in $X^+$ such that
$\alpha_ix_i=\sum_{j=0}^nw_{ij}$ for every $i\le m$ and
$\beta_jy_j=\sum_{i=0}^nw_{ij}$ for every $j\le n$.   Each $w_{ij}$ is
expressible as $\gamma_{ij}z_{ij}$ where $z_{ij}\in K$ and
$\gamma_{ij}\ge 0$.   Now

\Centerline{$\sum_{j=0}^n\Bover{\gamma_{ij}}{\alpha_i}z_{ij}=x_i\in K$;}

\noindent by (a), $\sum_{j=0}^n\gamma_{ij}=\alpha_i$, for every $i\le m$.
Similarly, $\sum_{i=0}^m\gamma_{ij}=\beta_j$ for $j\le n$.   Of course this
means that $\sum_{i=0}^m\sum_{j=0}^n\gamma_{ij}=1$.

Set $\lambda=\sum_{i=0}^m\sum_{j=0}^n\gamma_{ij}\delta_{z_{ij}}\in P$.
If $f:K\to\Bbb R$ is continuous and convex,

$$\eqalign{\int fd\mu
&=\sum_{i=0}^m\alpha_if(x_i)
=\sum_{i=0}^m\alpha_if(\sum_{j=0}^n\bover{\gamma_{ij}}{\alpha_i}z_{ij})\cr
&\le\sum_{i=0}^m\alpha_i\sum_{j=0}^n\Bover{\gamma_{ij}}{\alpha_i}f(z_{ij})
=\sum_{i=0}^m\sum_{j=0}^n\gamma_{ij}f(z_{ij})
=\int fd\lambda.\cr}$$

\noindent So $\mu\preccurlyeq\lambda$.   Similarly,
$\nu\preccurlyeq\lambda$
and we have the required upper bound for $\{\mu,\nu\}$.\ \Qed

\medskip

{\bf (d)} Now consider

\Centerline{$\{(\mu,\nu,\lambda):\mu$, $\nu$, $\lambda\in P$,
$\mu\preccurlyeq\lambda$, $\nu\preccurlyeq\lambda\}$.}

\noindent This is a closed set in the compact set $P\times P\times P$
(461Kd), so its projection

\Centerline{$R=\{(\mu,\nu):\mu$, $\nu$ have a common upper bound in $P\}$}

\noindent is a closed set in $P\times P$.

If $x\in K$ then (c) tells us that $R$
includes $Q_x\times Q_x$, so (b) tells us that $R$ includes
$P_x\times P_x$.
Thus any two members of $P_x$ have a common upper bound in $P$, as
required.
}%end of proof of 461O

\leader{461P}{Theorem} Suppose that $X$ is a Riesz space with a Hausdorff
locally convex linear space topology, and $K\subseteq X$ a
metrizable compact convex set such that every non-zero member of the
positive cone
$X^+$ is uniquely expressible as $\alpha x$ for some $x\in K$ and
$\alpha\ge 0$.   Let $E$ be the set of extreme points of $K$, and $x$ any
point of $K$.   Then there is a unique Radon
probability measure $\mu$ on $E$ such that $x$ is the barycenter of $\mu$.

\proof{ By 461M, there is a Radon probability measure
$\mu$ on $E$ such that $x$ is the barycenter of $\mu$.
Suppose that $\mu_1$ is another measure with the same properties.   Let
$\nu$, $\nu_1$ be the Radon probability measures on $K$ extending $\mu$,
$\mu_1$ respectively.   Then $\nu$ and $\nu_1$ both have barycenter
$x$ and make $E$ conegligible.
By 461N, they are both maximal in $P$.   By 461O, they must have a
common upper bound in $P$, so they are equal.   But this means that
$\mu=\mu_1$.
}%end of proof of 461P

\leader{461Q}{}\cmmnt{ It is a nearly universal rule that when we
encounter a compact convex set we should try to identify its extreme
points.   I look now at some sets which arose naturally in \S437.

\medskip

\noindent}{\bf Proposition (a)}\dvAnew{2011}
Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\pi:\frak A\to\frak A$ a
sequentially order-continuous Boolean
homomorphism.   Let $M_{\sigma}$ be the $L$-space of countably additive
real-valued functionals on $\frak A$, and $Q$ the set

\Centerline{$\{\nu:\nu\in M_{\sigma}$, $\nu\ge 0$, $\nu 1=1$,
$\nu\pi=\nu\}$.}

\noindent If $\nu\in Q$, then the following are equiveridical:
(i) $\nu$ is an extreme point of $Q$;  (ii) $\nu a\in\{0,1\}$
whenever $\pi a=a$;  (iii)
$\nu a\in\{0,1\}$ whenever $a\in\frak A$ is such that
$\nu(a\Bsymmdiff\pi a)=0$.

(b) Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, and $\phi:X\to X$ a
$(\Sigma,\Sigma)$-measurable function.   Let $M_{\sigma}$ be the
$L$-space of countably additive real-valued functionals on $\Sigma$, and
$Q\subseteq M_{\sigma}$ the set of probability measures with domain
$\Sigma$ for which $\phi$ is \imp.   If $\mu\in Q$, then $\mu$ is an
extreme point of $Q$ iff $\phi$ is ergodic with respect to
$\mu$\cmmnt{ (definition:  372Ob\formerly{3{}72Pb})}.

\proof{{\bf (a)} I ought to remark at once that because $\frak A$ is
Dedekind $\sigma$-complete, every countably additive functional on
$\frak A$ is bounded (326M\formerly{3{}26I}),
so that $M_{\sigma}$ is the $L$-space of
bounded countably additive functionals on $\frak A$, as studied in
362A-362B.

\medskip

\quad{\bf (i)$\Rightarrow$(ii)} Suppose that $\nu$ is an extreme point of
$Q$ and that $a\in\frak A$ is such that $\pi a=a$.   Set $\alpha=\nu a$.
\Quer\ If $0<\alpha<1$, define $\nu_1:\frak A\to\Bbb R$ by setting

\Centerline{$\nu_1b=\Bover1{\alpha}\nu(b\Bcap a)$}

\noindent for $b\in\frak A$.
Then $\nu_1$ is a non-negative countably additive
functional, and $\nu_11=1$.   Moreover, for any $b\in\frak A$,

\Centerline{$\nu_1\pi b=\Bover1{\alpha}\nu(\pi b\Bcap a)
=\Bover1{\alpha}\nu(\pi b\Bcap\pi a)
=\Bover1{\alpha}\nu\pi(b\Bcap a)
=\Bover1{\alpha}\nu(b\Bcap a)
=\nu_1b$,}

\noindent so $\nu_1\pi=\nu_1$ and $\nu_1\in Q$.   Since $\nu_1a=1$,
$\nu_1\ne\nu$.   Similarly,
$\nu_2\in Q$, where $\nu_2b=\Bover1{1-\alpha}\nu(b\Bsetminus a)$ for
$b\in\frak A$.   Now $\nu=\alpha\nu_1+(1-\alpha)\nu_2$ is a proper convex
combination of members of $Q$ and is not extreme.\ \BanG\   So
$\nu a\in\{0,1\}$;  as $a$ is arbitrary, (ii) is true.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} Suppose that (ii) is true, and that
$a\in\frak A$ is such that $\nu(a\Bsymmdiff\pi a)=0$.   Then

\Centerline{$\nu(\pi^na\Bsymmdiff\pi^{n+1}a)
=\nu\pi^n(a\Bsymmdiff\pi a)=\nu(a\Bsymmdiff\pi a)=0$}

\noindent for every $n\in\Bbb N$, so $\nu(a\Bsymmdiff\pi^na)=0$ for every
$n\in\Bbb N$.   Set

\Centerline{$b_n=\sup_{m\ge n}\pi^ma$ for $n\in\Bbb N$,
\quad$b=\inf_{n\in\Bbb N}b_n$.}

\noindent Because $\pi$ is sequentially order-continuous and $\nu$ is
countably additive,

\Centerline{$\nu(a\Bsymmdiff b_n)=0$ for every $n\in\Bbb N$,
\quad$\nu(a\Bsymmdiff b)=0$.}

\noindent Now

\Centerline{$\pi b_n=\sup_{m\ge n}\pi^{m+1}a=b_{n+1}\Bsubseteq b_n$}

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

\Centerline{$\pi b=\inf_{n\in\Bbb N}\pi b_n=\inf_{n\in\Bbb N}b_{n+1}=b$.}

\noindent Consequently
$\nu a=\nu b\in\{0,1\}$.   As $a$ is arbitrary, (iii) is true.

\medskip

\quad{\bf(iii)$\Rightarrow$(i)} Suppose that (iii) is true, and that
$\nu=\bover12(\nu_1+\nu_2)$ where $\nu_1$, $\nu_2\in Q$.   For
$\alpha\ge 0$, set $\theta_{\alpha}=\nu_1-\alpha\nu\in M_{\sigma}$.
Then we have a corresponding
element $a_{\alpha}=\Bvalue{\theta_{\alpha}>0}$ in $\frak A$ such that

\Centerline{$\theta_{\alpha}c>0$ whenever $0\ne c\Bsubseteq a_{\alpha}$,
\quad$\theta_{\alpha}c\le 0$ whenever
$c\Bcap a_{\alpha}=0$}

\noindent(326S\formerly{3{}26O}).   Observe that if $b\in\frak A$, then

\Centerline{$\theta_{\alpha}b
=\theta_{\alpha}a_{\alpha}-\theta_{\alpha}(a_{\alpha}\Bsetminus b)
   +\theta_{\alpha}(b\Bsetminus a_{\alpha})
\le\theta_{\alpha}a_{\alpha}$,}

\Centerline{$\theta_{\alpha}\pi b
=\nu_1\pi b-\alpha\nu\pi b=\nu_1b-\alpha\nu b=\theta_{\alpha}b$.}

Now $\nu(a_{\alpha}\Bsetminus\pi a_{\alpha})=0$.
\Prf\Quer\ Otherwise, setting $c=a_{\alpha}\Bsetminus\pi a_{\alpha}$, we
must have $\theta_{\alpha}c>0$, so

\Centerline{$\theta_{\alpha}(c\Bcup\pi a_{\alpha})
>\theta_{\alpha}\pi a_{\alpha}
=\theta_{\alpha}a_{\alpha}$.  \Bang\Qed}

\noindent Consequently

\Centerline{$\nu(a_{\alpha}\Bsymmdiff\pi a_{\alpha})
=\nu\pi a_{\alpha}-\nu a_{\alpha}+2\nu(a_{\alpha}\Bsetminus\pi a_{\alpha})
=0$}

\noindent and $\nu a_{\alpha}\in\{0,1\}$.

If $\alpha\le\beta$ then $\theta_{\beta}\le\theta_{\alpha}$ and
$a_{\beta}\Bsubseteq a_{\alpha}$.   As $\theta_0=\nu_1$,
$\nu_1(1\Bsetminus a_0)=0$,
$\nu_1a_0=1$ and $\nu a_0\ge\bover12$;  accordingly $\nu a_0=1$.
As $\theta_2\le 0$, $a_2=0$ and $\nu a_2=0$.   So

\Centerline{$\beta=\sup\{\alpha:\nu a_{\alpha}=1\}
=\sup\{\alpha:\nu a_{\alpha}>0\}$}

\noindent is defined in $[0,2]$.

Now $\nu_1=\beta\nu$.   \Prf\ Let $c\in\frak A$.
\Quer\ If $\nu_1c>\beta\nu c$, take $\alpha>\beta$ such
that $\nu_1c>\alpha\nu c$.   Then $\nu a_{\alpha}=0$, but

\Centerline{$0<\theta_{\alpha}c\le\theta_{\alpha}a_{\alpha}
\le\nu_1a_{\alpha}\le 2\nu a_{\alpha}$.  \Bang}

\noindent \Quer\ If
$\nu_1c<\beta\nu c$, take $\alpha\in\coint{0,\beta}$ such that
$\nu_1c<\alpha\nu c$.   Then $\nu a_{\alpha}=1$ so
$\nu(1\Bsetminus a_{\alpha})=0$, but

\Centerline{$0>\theta_{\alpha}c\ge\theta_{\alpha}(c\Bsetminus a_{\alpha})
\ge -\alpha\nu(c\Bsetminus a_{\alpha})
\ge -\alpha\nu(1\Bsetminus a_{\alpha})$.  \Bang}

\noindent Thus $\nu_1c=\beta\nu c$;  as
$c$ is arbitrary, $\nu_1=\beta\nu$.\ \Qed

Accordingly $\nu_1$ is a multiple of $\nu$ and must be equal to $\nu$.
Similarly, $\nu_2=\nu$.   As $\nu_1$ and $\nu_2$ were arbitrary,
$\nu$ is an extreme point of $Q$.

\medskip

{\bf (b)} In (a), set $\frak A=\Sigma$ and $\pi E=\phi^{-1}[E]$ for
$E\in\Sigma$;  then `$\phi$ is ergodic with respect to $\mu$' corresponds
to condition (ii) of (a), so we have the result.
}%end of proof of 461Q

\leader{461R}{Corollary} Let $X$ be a compact Hausdorff space and
$\phi:X\to X$ a continuous function.   Let $Q$ be the non-empty
compact convex set of
Radon probability measures $\mu$ on $X$ such that $\phi$ is \imp\ for
$\mu$, with its narrow topology and the convex
structure defined by 234G and 234Xf.\cmmnt{ (See 437T.)}   
Then the extreme points of $Q$ are those for which $\phi$ is ergodic.

\proof{{\bf (a)} If $\mu_0\in Q$ is not extreme, let
$\Cal B=\Cal B(X)$ be the
Borel $\sigma$-algebra of $X$, so that $\phi$ is
$(\Cal B,\Cal B)$-measurable, and write $Q'$ for the set of
$\phi$-invariant Borel probability measures on $X$.   Then 461Q tells us
that the extreme points of $Q'$ are just the measures for which
$\phi$ is ergodic.   If $\mu\in Q$, then
$\mu\restr\Cal B\in Q'$, and of
course the function $\mu\mapsto\mu\restr\Cal B$ is injective (416Eb)
and preserves convex combinations.   So $\mu_0\restr\Cal B$ is not
extreme in $Q'$.   By 461Q, $\phi$ is not
$\mu_0\restr\Cal B$-ergodic, and therefore not $\mu_0$-ergodic.

\medskip

{\bf (b)} If $\mu_0\in Q$ and $\phi$ is not $\mu_0$-ergodic, let
$E\in\dom\mu_0$ be such that $0<\mu_0E<1$ and $\phi^{-1}[E]=E$.   Set
$\alpha=\mu_0E$ and $\beta=1-\alpha$, and let $\mu_1$, $\mu_2$ be the
indefinite-integral measures over $\mu_0$ defined by
$\Bover1{\alpha}\chi E$ and $\Bover1{\beta}\chi(X\setminus E)$.   Then
$\mu_1$ is a Radon probability measure on $X$ (416Sa), so the image
measure $\mu_1\phi^{-1}$ also is a Radon measure (418I).   The argument
of part (b) of the proof of 461Q tells us that $\mu_1\phi^{-1}$ agrees
with $\mu_1$ on Borel sets, so $\mu_1=\mu_1\phi^{-1}$ (416Eb) and
$\mu_1\in Q$.   Similarly, $\mu_2\in Q$, and
$\mu_0=\alpha\mu_1+\beta\mu_2$, so $\mu_0$ is not extreme in $Q$.
}%end of proof of 461R

\exercises{\leader{461X}{Basic exercises $\pmb{>}$(a)}
%\spheader 461Xa
Let $X$ be a Hausdorff locally convex linear topological space,
$C\subseteq X$ a convex set, and $g:C\to\Bbb R$ a function.   Show that
the following are equiveridical:  (i) $g$ is convex and lower
semi-continuous;  (ii) there are a non-empty set $D\subseteq X^*$ and a
family $\family{f}{D}{\beta_f}$ in $\Bbb R$ such that
$g(x)=\sup_{f\in D}f(x)+\beta_f$ for every $x\in C$.   (Compare 233Hb.)
%461C

\sqheader 461Xb Let $X$ be a Hausdorff locally convex linear topological
space, $K\subseteq X$ a compact convex set, and $x$ an extreme point of
$K$.   Let $\mu$ be a probability measure on $X$ such that $\mu^*K=1$
and $x$ is the barycenter of $\mu$.   (i) Show that $\{y:y\in K$, $f(y)\ne f(x)\}$ is
$\mu$-negligible for every $f\in X^*$.   \Hint{461E.}   (ii) Show that
if $\mu$ is a Radon measure then $\mu\{x\}=1$.

\spheader 461Xc For each $n\in\Bbb N$, define $e_n\in\pmb{c}_0$ by
saying that $e_n(n)=1$, $e_n(i)=0$ if $i\ne n$.   Let $\mu$ be the
point-supported Radon
probability measure on $\pmb{c}_0$ defined by saying that
$\mu E=\sum_{n=0}^{\infty}2^{-n-1}\chi E(2^ne_n)$ for every
$E\subseteq\pmb{c}_0$.   (i) Show that every member of $\pmb{c}_0^*$ is
$\mu$-integrable.   \Hint{$\pmb{c}_0^*$ can be identified with
$\ell^1$.}   (ii) Show that $\mu$ has no barycenter in $\pmb{c}_0$.
%461F

\spheader 461Xd Let $I$ be an uncountable set, and
$X=\{x:x\in\ell^{\infty}(I),\,\{i:x(i)\ne 0\}$ is countable$\}$.
(i) Show that $X$ is a closed linear subspace of $\ell^{\infty}(I)$.
(ii) Show
that there is a probability measure $\mu$ on $X$ such that ($\alpha$)
$\mu\{x:\|x\|\le 1\}$ is defined and equal to $1$ ($\beta$)
$\mu\{x:x(i)=1\}=1$ for every $i\in I$.
(iii) Show that $\int fd\mu$ is defined for every $f\in X^*$.   \Hint{for
any $f\in X^{\sim}$, there is a countable set $J\subseteq X$ such that
$f(x)=0$ whenever $x\restr J=0$, so that $f\eae f(\chi J)$.}   (iv)
Show that $\mu$ has no barycenter in $X$.
%461F

\sqheader 461Xe Let $X$ be a complete Hausdorff locally convex linear
topological
space, and $K\subseteq X$ a compact set.   Show that every extreme point
of $\overline{\Gamma(K)}$ belongs to $K$.   \Hint{show that it cannot be
the barycenter of any measure on $K$ which is not supported by a single
point.}
%461I

\sqheader 461Xf Let $X$ be a Hausdorff locally convex linear topological
space, and $K\subseteq X$ a metrizable compact set.   Show that
$\overline{\Gamma(K)}$ is metrizable.   \Hint{we may suppose that $X$ is
complete, so that $\overline{\Gamma(K)}$ is compact.   Show that
$\overline{\Gamma(K)}$ is a continuous image of the space of Radon
probability measures on $K$, and use 437Rf.}
%461J

\spheader 461Xg Let $X$ be a Hausdorff locally convex linear topological
space and $K\subseteq X$ a compact set.   Show that the Baire
$\sigma$-algebra of $K$ is just the subspace $\sigma$-algebra induced by
the cylindrical $\sigma$-algebra of $X$.
%461L

\sqheader 461Xh Let $X$ be a Hausdorff locally convex linear topological
space, and $K\subseteq X$ a compact convex set;  let $E$ be the set of
extreme points of $K$.   Let $\sequencen{f_n}$ be a sequence in $X^*$
such that $\sup_{x\in E,n\in\Bbb N}|f_n(x)|$ is finite and
$\lim_{n\to\infty}f_n(x)=0$ for every $x\in E$.   Show that
$\lim_{n\to\infty}f_n(x)=0$ for every $x\in K$.
%461M

\sqheader 461Xi Let $X$ be a Hausdorff locally convex linear topological
space, and $K\subseteq X$ a metrizable compact convex set.
Show that the algebra of Borel subsets of $K$ is just the
subspace algebra of the cylindrical $\sigma$-algebra of $X$.
%461M

\spheader 461Xj Let $X$ be a Hausdorff locally convex linear topological
space, and $K\subseteq X$ a compact set.   Let us say that a point $x$
of $K$ is {\bf extreme} if the only Radon probability measure on $K$
with barycenter $x$ is the Dirac measure on $K$ concentrated at $x$.
(Cf.\ 461Xb.)
(i) Show that if $X$ is complete, then $x\in X$ is an extreme point of
$K$ iff it is an extreme point of $\overline{\Gamma(K)}$.   (ii)
Writing $E$ for the set of extreme points of $K$, show that any point of
$K$ is the barycenter of some probability measure on $E$.   (iii) Show
that if $K$ is metrizable then $E$ is a G$_{\delta}$ subset of $K$ and
any point of $K$ is the barycenter of some Radon probability measure on
$E$.
%461M

\spheader 461Xk Let $G$ be an abelian group with identity $e$,
and $K$ the set of positive definite functions
$h:G\to\Bbb C$ such that $h(e)=1$.   (i) Show that $K$ is a compact convex
subset of $\Bbb C^G$.   (ii) Show that the extreme points of $K$ are just
the group homomorphisms from $G$ to $S^1$.
(iii) Show that $K$ generates the
positive cone of a Riesz space.   \Hint{445N.}
%461M

\spheader 461Xl Let $X$ be a compact metrizable space and $G$ a subgroup
of the group of autohomeomorphisms of $X$.   Let $M_{\sigma}$ be the space
of signed Borel measures on $X$ with its vague topology, and
$Q\subseteq M_{\sigma}$ the set of $G$-invariant Borel probability measures
on $X$.   Show that every member of $Q$ is uniquely expressible as the
barycenter of a Radon probability
measure on the set of extreme points of $Q$.
%461P

\spheader 461Xm Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets
of $X$, and $P$ the set of probability measures with domain $\Sigma$,
regarded as a
convex subset of the linear space of countably additive functionals on
$\Sigma$.   Show that $\mu\in P$ is an extreme point in $P$ iff it takes
only the values $0$ and $1$.
%461Q

\spheader 461Xn Let $\frak A$ be a Boolean
algebra and $M$ the $L$-space of bounded finitely additive functionals on
$\frak A$, and $\pi:\frak A\to\frak A$ a Boolean homomorphism.
(i) Show that $U=\{\nu:\nu\in M$, $\nu\pi=\nu\}$ is a closed Riesz subspace
of $M$.   (ii) Set $Q=\{\nu:\nu\in U$, $\nu\ge 0$, $\nu 1=1\}$.
Show that if $\mu$, $\nu$ are distinct extreme points of $Q$ then
$\mu\wedge\nu=0$.   (iii) Set $Q_{\sigma}=\{\nu:\nu\in Q$, $\nu$ is
countably additive$\}$.   Show that any extreme point of $Q_{\sigma}$ is an
extreme point of $Q$.
(iv) Set $Q_{\tau}=\{\nu:\nu\in Q$, $\nu$ is
countably additive$\}$.   Show that any extreme point of $Q_{\tau}$ is an
extreme point of $Q$.
%461Q

\sqheader 461Xo Set $S^1=\{z:z\in\Bbb C$, $|z|=1\}$, and let $w\in S^1$
be such that $w^n\ne 1$ for any integer $1$.   Define $\phi:S^1\to S^1$
by setting $\phi(z)=wz$ for every $z\in S^1$.   Show that the only Radon
probability measure on $S^1$ for which $\phi$ is \imp\ is the Haar
probability measure $\mu$ of $S^1$.   \Hint{use 281N to show that
$\lim_{n\to\infty}\bover1{n+1}\sum_{k=0}^nf(w^kz)=\int fd\mu$ for every
$f\in C(S^1)$;  now put 461R and 372H\formerly{3{}72I} together.}
%461R

\spheader 461Xp Set $\phi(x)=2\min(x,1-x)$ for $x\in[0,1]$ (cf.\
372Xp\formerly{3{}72Xm}).
Show that there are many point-supported Radon measures on $[0,1]$ for
which $\phi$ is \imp.
%461R   are there unexpected atomless measures?

\spheader 461Xq
Let $X$ and $Y$ be Hausdorff locally convex linear
topological spaces, $A\subseteq X$ a convex set and
$\phi:A\to Y$ a continuous
function such that $\phi[A]$ is bounded and
$\phi(tx+(1-t)y)=t\phi(x)+(1-t)\phi(y)$ for all $x$, $y\in A$ and
$t\in[0,1]$.   Let $\mu$ be a topological
probability measure on $A$ with a barycenter $x^*\in A$.
Show that $\phi(x^*)$ is the barycenter of the image measure
$\mu\phi^{-1}$ on $Y$.   \Hint{show first that if
$\familyiI{E_i}$ is a finite partition of $A$ into
non-empty convex sets measured by $\mu$,
$\alpha_i=\mu E_i$ for each $i\in I$ and
$C=\{\sum_{i\in I}\alpha_ix_i:x_i\in E_i$ for every $i\in I\}$, then
$x^*\in\overline{C}$.}
%461B

\leader{461Y}{Further exercises (a)}
%\spheader 461Ya
Let $X$ be a Hausdorff locally convex linear topological space.   (i)
Show that if $M_0,\ldots,M_n$ are non-empty compact convex subsets of
$X$ with empty
intersection, then there is a continuous convex function $g:X\to\Bbb R$
such that $g(\sum_{i=0}^n\alpha_ix_i)<\sum_{i=0}^n\alpha_ig(x_i)$
whenever $x_i\in M_i$ and
$\alpha_i>0$ for every $i\le n$.   (ii) Show that if $K\subseteq X$ is
compact and $x^*\in\overline{\Gamma(K)}$ then there is a Radon
probability measure $\mu$ on $K$,
with barycenter $x^*$, such that $\mu(\alpha_0M_0+\ldots+\alpha_nM_n)=0$
whenever $M_0,\ldots,M_n$ are compact convex subsets of $K$ with empty
intersection, $\alpha_i\ge 0$ for every
$i\le n$, and $\sum_{i=0}^n\alpha_i=1$.
%461L

\spheader 461Yb
In $\BbbR^{\coint{0,2}}$ let $K$ be the set of those functions $u$ such
that ($\alpha$) $0\le u(s)\le u(t)\le 1$ whenever $0\le s\le t\le 1$
($\beta$) $|u(t+1)|\le u(s')-u(s)$ whenever $0\le s<t<s'\le 1$.   (i)
Show that $K$ is a compact convex set.   (ii) Show that the set $E$ of
extreme points of $K$ is just the set of functions of the types
$\tbf{0}$, $\chi[0,1]$, $\chi[s,1]\pm\chi\{1+s\}$ and
$\chi\ocint{s,1}\pm\chi\{1+s\}$ for $0<s<1$.   (iii) Set $w(s)=s$ for
$s\in[0,1]$, $0$ for $s\in\ooint{1,2}$.   Show that if $\mu$ is any
Radon probability measure on $K$ with barycenter $w$ then $\mu E=0$.
%461M
%what about Borel measures?  when there is an \am cardinal I suppose

\spheader 461Yc Write $\nu_{\omega_1}$ for the usual measure on
$Z=\{0,1\}^{\omega_1}$.   Fix any $z_0\in Z$, and let $U$ be the linear
space $\{u:u\in C(Z)$, $u(z_0)=\int u\,d\nu_{\omega_1}\}$.   Let $X$ be the
Riesz space of signed tight Borel measures $\mu$ on $Z$ such that
$\mu\{z_0\}=0$,
with the topology generated by the functionals $\mu\mapsto\int u\,d\mu$ as
$u$ runs over $U$.    Let $K\subseteq X$ be the set of tight Borel
probability measures $\mu$ on $Z$ such that $\mu\{z_0\}=0$.
(i) Show that $K$
is compact and convex and that every member of $X^+\setminus\{0\}$ is
uniquely expressible as a positive multiple of a member of $K$.   (ii) Show
that the set $E$ of extreme points of $K$ can be identified, as topological
space, with $Z\setminus\{z_0\}$, so is a Borel subset of $K$ but not a
Baire subset.   (iii) Show that the restriction of $\nu_{\omega_1}$ to the
Borel $\sigma$-algebra of $Z$ is the barycenter of more than
one Baire measure on $E$.
%461P

\spheader 461Yd Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, and $\Phi$ a set of
$(\Sigma,\Sigma)$-measurable functions from $X$ to itself.   Let
$M_{\sigma}$ be the
$L$-space of countably additive real-valued functionals on $\Sigma$, and
$Q\subseteq M_{\sigma}$ the set of probability measures with domain
$\Sigma$ for which every member of $\Phi$ is \imp.   (i) Show that if
$\mu\in Q$, then $\mu$ is an extreme point of $Q$ iff $\mu E\in\{0,1\}$
whenever $E\in\Sigma$ and $\mu(E\symmdiff\phi^{-1}[E])=0$ for every
$\phi\in\Phi$.
(ii) Show that if $\mu\in Q$ and $\Phi$ is countable and commutative,
then $\mu$ is an extreme point of $Q$ iff $\mu E\in\{0,1\}$ whenever
$E\in\Sigma$ and $E=\phi^{-1}[E]$ for every $\phi\in\Phi$.
%461Q

\spheader 461Ye Let $X$ be a non-empty Hausdorff space, and define
$\phi:X^{\Bbb N}\to X^{\Bbb N}$ by setting $\phi(x)(n)=x(n+1)$ for
$x\in X^{\Bbb N}$ and $n\in\Bbb N$.
Let $Q$ be the set of Radon probability measures
on $X^{\Bbb N}$ for which $\phi$ is \imp.   Show that a Radon
probability measure $\lambda$ on $X^{\Bbb N}$ is an extreme point of
$Q$ iff it is a Radon product measure $\mu^{\Bbb N}$ for some Radon
probability measure $\mu$ on $X$.
%461R mt46bits

\spheader 461Yf Let $G$ be a topological group.   Show that the following
are equiveridical:  (i) $G$ is amenable;  (ii) whenever $X$ is a Hausdorff
locally convex linear topological space, and
$\action$ is a continuous action of $G$ on $X$ such that
$x\mapsto a\action x$ is a linear operator for every $a\in G$, and
$K\subseteq X$ is a
non-empty compact convex set such that $a\action x\in K$ whenever
$a\in G$ and $x\in K$, then there is an $x\in K$ such that
$a\action x=x$ for every $a\in G$;
(iii) whenever $X$ is a Hausdorff
locally convex linear topological space, $K\subseteq X$ is a non-empty
compact convex set, and
$\action$ is a continuous action of $G$ on $K$ such that
$a\action(tx+(1-t)y)=t\,a\action x+(1-t)a\action y$ whenever $a\in G$, $x$,
$y\in K$ and $t\in[0,1]$, then there is an $x\in K$ such that
$a\action x=x$ for every $a\in G$.
Use this to simplify parts of the proof of 449C.   \Hint{493B.}
%461Xq

\spheader 461Yg Let $G$ be an amenable topological group, and $\action$ an
action of $G$ on a reflexive Banach space $U$, continuous for the
given topology on $G$ and the weak topology of $U$, such that
$u\mapsto a\action u$ is a linear operator of
norm at most $1$ for every $a\in G$.   Set
$V=\{v:v\in U$, $a\action v=v$ for every $a\in G\}$.   Show that
$\{u+v-a\action u:u\in U$, $v\in V$, $a\in G\}$ is dense in $U$.
}%end of exercises

\endnotes{
\Notesheader{461} The results above are a little unusual in that we have
studied locally convex spaces for several pages without encountering two
topologies on the same space more than once (461J).
In fact some of the most interesting
properties of measures on locally convex spaces concern their
relationships with strong and weak topologies, but I defer these ideas
to later parts of the chapter.   For the moment, we just have the basic
results affirming (i) that barycenters exist (461E, 461F, 461H) (ii)
that points can be represented as barycenters (461I, 461M).   The
last two can be thought of as refinements of the
\Krein-Milman theorem.   Any compact convex set $K$ (in a locally
compact
Hausdorff space) is the closed convex hull of the set $E$ of its extreme
points.   By 461M, given $x\in K$, we can actually find a measure on $E$
with barycenter $x$;  and if $K$ is metrizable we can do this with a
Radon measure.   Of course the second part of 461M is a straightforward
consequence of the first.   But I do not know of
any proof of 461M which does not pass through 461K-461L.

\Krein's theorem (461J) is a fundamental result in the theory of linear
topological spaces.   The proof here, using the Riesz representation
theorem and vague topologies, is a version of the standard one (e.g.,
{\smc Bourbaki 87}, II.4.1), written out to be a little heavier in the
measure theory and a little lighter in the topological linear space
theory than is usual.   There are of course proofs which do not use
measure theory.

In \S437 I have already looked at an archetypal special case of 461I and
461M.   If $X$ is a compact Hausdorff space and
$P$ the compact convex set of Radon probability measures on $X$ with the
narrow (or vague) topology, then the set of extreme points of $P$ can be
identified with the set $\Delta$ of Dirac measures on $X$ (437S, 437Xt).
If we think of $P$ as a subset of $C(X)^*$ with the weak* topology, so
that the dual of the linear topological space $C(X)^*$ can be identified
with $C(X)$, then any $\mu\in P$ is the barycenter of a Baire
probability measure $\nu$ on
$\Delta$.   In fact (because $\Delta$ here is compact) $\mu$ is the
barycenter of a Radon measure
on $\Delta$, and this is just the image measure $\mu\delta^{-1}$.

I have put the phrase `Choquet's theorem' into the title of this section.
Actually it should perhaps be `first steps in Choquet theory', because
while the theory as a whole was dominated for many years by the work of
G.Choquet the exact attribution of the results
presented here is more complicated.
See {\smc Phelps 66} for a much fuller account.   But certainly both the
existence and uniqueness theorems 461M and 461P draw heavily on Choquet's
ideas.

Theorem 461P, demanding an excursion through 461N-461O, seems fairly hard
work for a relatively specialized result.   But it provides a unified
explanation for a good many apparently disparate phenomena.   Of course the
simplest example is when $X=C(Z)^*$ for some compact metrizable space $Z$
and $K$ is the set of positive linear functionals of norm $1$, so that $E$
can be identified with $Z$ and we find ourselves back with the Riesz
representation theorem.   A less familiar case already examined is in
461Xk.   At the next level we have such examples as 461Xl.

Another class of examples arising in \S437
is explored in 461Q-461R, 461Xm-461Xn and 461Yd-461Ye.
It is when we have an explicit listing of the extreme points,
as in 461Yb and 461Ye,
that we can begin to feel that we understand a compact convex set.
}%end of notes

\leaveitout{\Krein's theorem:\prooflet{(\Bourbaki, II.4.1:  \Schaefer,
II.4.3;  \Kothe, \S20.6.)}
%\NB, 5.7.5.
}%end of leaveitout

\discrpage

\frfilename{mt462.tex}
\versiondate{30.6.07}
\copyrightdate{2007}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Pointwise compact sets of continuous functions}

\newsection{462}

In preparation for the main work of this chapter, beginning in the next
section, I offer a few pages on spaces of continuous functions under
their `pointwise' topologies (462Ab).   There is an extensive general
theory of such spaces, described in {\smc Arkhangel'skii 92};  here I
present only those fragments which seem directly relevant to the theory
of measures on normed spaces and spaces of functions.   In particular,
I star the paragraphs 462C-462D, which are
topology and functional analysis
rather than measure theory.   They are here because although this
material is well known, and may be found in many places, I think that
the ideas, as well as the results, are essential for any understanding
of measures on linear topological spaces.

Measure theory enters the section in the proof of 462E, in the form of
an application of the Riesz representation theorem, though 462E itself
remains visibly part of functional analysis.   In the rest of the
section, however, we come to results which are pure measure theory.
For (countably) compact spaces $X$, the Radon measures on $C(X)$ are the
same for the pointwise and norm topologies (462I).
This fact has extensive implications for the theory of separately
continuous functions (462K) and for the theory of convex hulls in linear
topological spaces (462L).

\leader{462A}{Definitions (a)} A regular Hausdorff space $X$ is {\bf
angelic} if whenever $A$ is a subset of $X$ which is relatively
countably compact in $X$, then (i) its closure $\overline{A}$ is compact
(ii) every point of $\overline{A}$ is the limit of a sequence in $A$.

\cmmnt{(A {\bf Fr\'echet-Urysohn} space is a topological space in
which, for any set $A$, every point of the closure of $A$ is a limit of
a sequence in $A$.   So (ii) here may be written as `every compact
subspace of $X$ is a Fr\'echet-Urysohn space'.)}

\spheader 462Ab If $X$ is any set and $A$ a subset of $\Bbb R^X$, then
the topology of {\bf pointwise convergence} on $A$ is that inherited
from the usual product topology of $\Bbb R^X$\cmmnt{;  that is, the
coarsest topology on $A$ for which the map $f\mapsto f(x):A\to\Bbb R$ is
continuous for every $x\in X$}.   I shall commonly use the symbol
$\frak T_p$ for such a
topology.   In this context, I will say that a sequence or filter is
{\bf pointwise convergent} if it is convergent for the topology of
pointwise convergence.   Note that if $A$ is a linear subspace of
$\Bbb R^X$ then
$\frak T_p$ is a linear space topology on $A$\cmmnt{ (4A4Ba)}.

\vleader{48pt}{*462B}{Proposition}\cmmnt{ ({\smc Pryce 71})} Let
$(X,\frak T)$ be an angelic regular Hausdorff space.

(a) Any subspace of $X$ is angelic.

(b) If $\frak S$ is a regular topology on $X$ finer than $\frak T$, then
$\frak S$ is angelic.

(c) Any countably compact subset of $X$ is compact and sequentially
compact.

\proof{{\bf (a)} Let $Y$ be any subset of $X$.   Then of course the
subspace topology on $Y$ is regular and Hausdorff.   If $A\subseteq Y$
is relatively countably compact in $Y$, then $A$ is
relatively countably compact in $X$, so $\overline{A}$, the closure of
$A$ in $X$, is compact.   Now if $x\in\overline{A}$, there is a sequence
$\sequencen{x_n}$ in $A$ converging to $x$;   but $\sequencen{x_n}$ must
have a cluster point in $Y$, and (because $\frak T$ is Hausdorff) this
cluster point can only be $x$.   Accordingly $\overline{A}\subseteq Y$
and is the closure of $A$ in $Y$.   Thus $A$ is relatively compact in
$Y$.   Moreover, any point of $\overline{A}$ is the limit of a sequence
in $A$.   As $A$ is arbitrary, $Y$ is angelic.

\medskip

{\bf (b)} By hypothesis, $\frak S$ is regular, and it is Hausdorff
because it is finer than $\frak T$.   Now suppose that $A\subseteq X$ is
$\frak S$-relatively countably compact.   Because the identity map from
$(X,\frak S)$ to $(X,\frak T)$ is continuous, $A$ is
$\frak T$-relatively countably compact (4A2G(f-iv)), and the
$\frak T$-closure $\overline{A}$ of $A$ is $\frak T$-compact.

Let $\Cal F$ be any ultrafilter on $X$ containing $A$.
Then $\Cal F$ has a $\frak T$-limit $x\in X$.   \Quer\ If $\Cal F$ is
not $\frak S$-convergent to $x$,
there is an $H\in\frak S$ such that $x\in H$ and
$X\setminus H\in\Cal F$, so that $A\setminus H\in\Cal F$.
Now $x$ belongs to the
$\frak T$-closure of $A\setminus H$, because $A\setminus H\in\Cal F$ and
$\Cal F$ is $\frak T$-convergent to $x$;  because $\frak T$ is angelic,
there is a sequence $\sequencen{x_n}$ in $A\setminus H$ which
$\frak T$-converges to $x$.   But now $\sequencen{x_n}$ has a
$\frak S$-cluster point $x'$.   $x'$ must also be a $\frak T$-cluster
point of $\sequencen{x_n}$, so $x'=x$;  but every $x_n$ belongs to the
$\frak S$-closed set $X\setminus H$, so $x'\notin H$, which is
impossible.\ \Bang

Thus every ultrafilter on $X$ containing $A$ is $\frak S$-convergent.
Because $\frak S$ is regular, the $\frak S$-closure $\tilde A$ of $A$ is
$\frak S$-compact (3A3De).

Again because $\frak S$ is finer than $\frak T$, and $\frak T$ is
Hausdorff, the two topologies must agree on $\tilde A=\overline{A}$.
But now every point of $\overline{A}$ is the $\frak T$-limit of a
sequence in $A$, so every point of $\tilde A$ is the $\frak S$-limit of
a sequence in $A$.   As $A$ is arbitrary, $\frak S$ is angelic.

\medskip

{\bf (c)} If $K\subseteq X$ is countably compact, then of course it is
relatively countably compact in its subspace topology, so (being
angelic) must be compact in its subspace topology.   If
$\sequencen{x_n}$ is a sequence in $K$, let $x$ be any cluster point of
$\sequencen{x_n}$.   If $\{n:x_n=x\}$ is infinite, then this immediately
provides us with a subsequence converging to $x$.   Otherwise, take $n$
such that $x\ne x_i$ for $i\ge n$.   Since $x\in\overline{\{x_i:i\ge
n\}}$, and $\{x_i:i\ge n\}$ is relatively countably compact, there is a
sequence $\sequence{i}{y_i}$ in $\{x_i:i\ge n\}$ converging to $x$.
The topology of $X$ being Hausdorff, $\{y_i:i\in\Bbb N\}$ must be
infinite, and $\sequence{i}{y_i}$, $\sequence{i}{x_i}$ have a common
subsequence which converges to $x$.   As $\sequence{i}{x_i}$ is
arbitrary, $K$ is sequentially compact.
}%end of proof of 462B

\vleader{60pt}{*462C}{Theorem}\cmmnt{ ({\smc Pryce 71})} Let $X$ be a
topological space such that there is a sequence $\sequencen{X_n}$ of
relatively
countably compact subsets of $X$, covering $X$, with the property that a
function $f:X\to\Bbb R$ is continuous whenever $f\restr X_n$ is
continuous for every $n\in\Bbb N$.   Then the space $C(X)$ of continuous
real-valued functions on $X$ is angelic in its topology of pointwise
convergence.

\proof{ Of course $C(X)$ is regular and Hausdorff under $\frak T_p$,
because $\Bbb R^X$ is, so we need attend only to the rest of the
definition in 462Aa.   Let $A\subseteq C(X)$ be relatively countably
compact for $\frak T_p$.

\medskip

{\bf (a)} Since $\{f(x):f\in A\}$, being a continuous image of $A$, must
be relatively countably compact
in $\Bbb R$ (4A2G(f-iv)), therefore relatively compact (4A2Le), for
every $x\in X$, the
closure $\overline{A}$ of $A$ in $\Bbb R^X$ is compact, by Tychonoff's
theorem.

\Quer\ Suppose, if possible, that $\overline{A}\not\subseteq C(X)$;  let
$g\in\overline{A}$ be a discontinuous function.   By the hypothesis of
the theorem, there is an $n\in\Bbb N$ such that $g\restr X_n$ is not
continuous;  take $x^*\in X_n$ such that $g\restr X_n$ is discontinuous
at $x^*$.   Let $\epsilon>0$ be such that for every neighbourhood $U$ of
$x^*$ in $X_n$ there is a point $x\in U$ such that
$|g(x)-g(x^*)|\ge\epsilon$.

Choose sequences $\sequence{i}{f_i}$ in $A$ and $\sequence{i}{x_i}$ in
$X_n$ as follows.   Given $\langle f_i\rangle_{i<m}$ and
$\langle x_i\rangle_{i<m}$, choose $x_m\in X_n$ such that
$|f_i(x_m)-f_i(x^*)|\le 2^{-m}$ for every $i<m$ and
$|g(x^*)-g(x_m)|\ge\epsilon$.   Now choose
$f_m\in A$ such that $|f_m(x^*)-g(x^*)|\le 2^{-m}$ and
$|f_m(x_i)-g(x_i)|\le 2^{-m}$ for every $i\le m$.   Continue.

At the end of the induction, take a cluster point $x$ of
$\sequence{i}{x_i}$ in $X$ and a cluster point $f$ of
$\sequence{i}{f_i}$ in $C(X)$.   Because $|f_i(x_m)-f_i(x^*)|\le 2^{-m}$
whenever $i<m$, $f_i(x)=f_i(x^*)$ for every $i$, and $f(x)=f(x^*)$.
Because $|f_m(x^*)-g(x^*)|\le 2^{-m}$ for every $m$, $f(x^*)=g(x^*)$.
Because $|f_m(x_i)-g(x_i)|\le 2^{-m}$ whenever $i\le m$, $f(x_i)=g(x_i)$
for every $i$, $|g(x^*)-f(x_i)|\ge\epsilon$ for every $i$, and
$|g(x^*)-f(x)|\ge\epsilon$;  but this is impossible, because
$f(x)=f(x^*)=g(x^*)$.\ \Bang

Thus the compact set $\overline{A}\subseteq C(X)$ is the closure of $A$
in $C(X)$, and $A$ is relatively compact in $C(X)$.

\medskip

{\bf (b)} Now take any $g\in\overline{A}$.   There are countable sets
$D\subseteq X$, $B\subseteq A$ such that

\inset{whenever $I\subseteq B\cup\{g\}$ is finite, $n\in\Bbb N$,
$\epsilon>0$ and $x\in X_n$, there is a $y\in D\cap X_n$ such that
$|f(y)-f(x)|\le\epsilon$ for every $f\in I$;}

\inset{whenever $J\subseteq D$ is finite and $\epsilon>0$ there is an
$f\in B$ such that $|f(x)-g(x)|\le\epsilon$ for every $x\in J$.}

\noindent\Prf\ For any finite set $I\subseteq\Bbb R^X$ and $n\in\Bbb N$,
the set $Q_{In}=\{\langle f(x)\rangle_{f\in I}:x\in X_n\}$ is a subset
of the separable metrizable space $\BbbR^I$, so is itself separable, and
there is a countable dense set $D_{In}\subseteq X_n$ such that
$Q'_{In}=\{\langle f(x)\rangle_{f\in I}:x\in D_{In}\}$ is dense in
$Q_{In}$.   Similarly, because $g\in\overline{A}$, we can choose for any
finite set $J\subseteq X$ a sequence $\sequence{i}{f_{Ji}}$ in $A$ such
that $\lim_{i\to\infty}f_{Ji}(x)=g(x)$ for every $x\in J$.

Now construct $\sequence{m}{D_m}$, $\sequence{m}{B_m}$ inductively by
setting

\Centerline{$D_m=\bigcup\{D_{In}:n\in\Bbb N,\,
  I\subseteq\{g\}\cup\bigcup_{i<m}B_i$ is finite$\}$,}

\Centerline{$B_m=\{f_{Jk}:k\in\Bbb N,\,J\subseteq\bigcup_{i<m}D_i$ is
finite$\}$.}

\noindent At the end of the induction, set $D=\bigcup_{m\in\Bbb N}D_m$
and $B=\bigcup_{m\in\Bbb N}B_m$.   Since the construction clearly
ensures that $\sequence{m}{D_m}$ and $\sequence{m}{B_m}$ are
non-decreasing sequences of countable sets, $D$ and $B$ are countable,
and we shall have $D_{In}\subseteq D$ whenever $n\in\Bbb N$ and
$I\subseteq B\cup\{g\}$ is finite, while $f_{Ji}\in B$ whenever
$i\in\Bbb N$ and $J\subseteq D$ is finite.   Thus we have suitable sets
$D$ and $B$.\ \Qed

By the second condition on $D$ and $B$, there must be a sequence
$\sequence{i}{f_i}$ in $B$ such that $g(x)=\lim_{i\to\infty}f_i(x)$ for
every $x\in D$.   In fact $g(y)=\lim_{i\to\infty}f_i(y)$ for every $y\in
X$.   \Prf\Quer\ Otherwise, there is an $\epsilon>0$ such that
$J=\{i:|g(y)-f_i(y)|\ge\epsilon\}$ is infinite.   Let $n$ be such that
$y\in X_n$.   For each $m\in\Bbb N$, $I_m=\{f_i:i\le m\}$ is a finite
subset of $B$, so there is an $x_m\in D_{I_mn}$ such that
$|f(x_m)-f(y)|\le 2^{-m}$ for every $f\in I_m\cup\{g\}$.   Let
$x^*\in X$ be a cluster point of $\sequence{m}{x_m}$, and
$h\in C(X)$ a cluster point of $\langle f_i\rangle_{i\in J}$.   Then

\inset{because $g(x)=\lim_{i\to\infty}f_i(x)$ for every $x\in D$,
$g(x_m)=h(x_m)$ for every $m\in\Bbb N$, and $g(x^*)=h(x^*)$;}

\inset{because $|g(y)-f_i(y)|\ge\epsilon$ for every $i\in J$,
$|g(y)-h(y)|\ge\epsilon$;}

\inset{because $|f_i(x_m)-f_i(y)|\le 2^{-m}$ whenever $i\le m$,
$f_i(x^*)=f_i(y)$
for every $i\in\Bbb N$, and $h(x^*)=h(y)$;}

\inset{because $|g(x_m)-g(y)|\le 2^{-m}$ for every $m$, $g(x^*)=g(y)$;}

\noindent but this means that $g(y)=g(x^*)=h(x^*)=h(y)\ne g(y)$, which is
absurd.\ \Bang\Qed

So $g=\lim_{i\to\infty}f_i$ for $\frak T_p$.   As $g$ is arbitrary, $A$
has both properties required in 462Aa;  as $A$ is arbitrary, $C(X)$ is
angelic.
}%end of proof of 462C

\cmmnt{\medskip

\noindent{\bf Remark} For a slight strengthening of this result, see
462Ya.}

\leader{*462D}{Theorem} Let $U$ be any normed space.   Then it is
angelic in its weak topology.

\proof{ Write $X$ for the unit ball of the dual space $U^*$, with its
weak* topology.   Then $X$ is compact (3A5F).   We have
a natural map $u\mapsto\hat u:U\to\Bbb R^X$ defined by setting
$\hat u(x)=x(u)$ for $x\in X$ and $u\in U$.
By the definition of the weak*
topology, $\hat u\in C(X)$ for every $u\in U$.   The weak topology of
$U$ is normally defined in terms of all functionals $u\mapsto f(u)$, for
$f\in U^*$;  but as every member of $U^*$ is a scalar multiple of some
$x\in X$, we can equally regard the weak topology of $U$ as defined just
by the functionals $u\mapsto x(u)=\hat u(x)$, for $x\in X$.   But this
means that the map $u\mapsto\hat u$ is a homeomorphism between $U$, with
its weak topology, and its image $\widehat U$ in $C(X)$, with the
topology of pointwise convergence.

Since $C(X)$ is $\frak T_p$-angelic (462C), so is $\widehat U$ (462Ba),
and $U$ is angelic in its weak topology.
}%end of proof of 462D

\leader{462E}{Theorem} Let $X$ be a locally compact Hausdorff space, and
$C_0(X)$ the Banach lattice of continuous real-valued functions on $X$
which vanish at infinity\cmmnt{ (436I)}.
Write $\frak T_p$ for the topology of pointwise convergence on $C_0(X)$.

(i) $C_0(X)$ is $\frak T_p$-angelic.

(ii) A sequence $\sequencen{u_n}$ in $C_0(X)$ is weakly convergent to
$u\in C_0(X)$ iff it is $\frak T_p$-convergent to $u$ and norm-bounded.

(iii) A subset $K$ of $C_0(X)$ is weakly compact iff it is norm-bounded
and $\frak T_p$-countably compact.

\proof{{\bf (a)} Let $X^*=X\cup\{x_{\infty}\}$ be the one-point
compactification of $X$ (3A3O).   Then $C(X^*)$ is angelic in its topology
$\frak T^*_p$ of pointwise convergence, by 462C.   Set
$V=\{g:g\in C(X^*)$, $g(x_{\infty})=0\}$.   By 462Ba, $V$ is angelic in the
subspace topology induced by $\frak T_p^*$.   Now observe that we have a
natural bijection $g\mapsto g\restr X:V\to C_0(X)$, and that this is a
homeomorphism for the topologies of pointwise convergence on $V$ and
$C_0(X)$.   So $C_0(X)$ is angelic under $\frak T_p$.

\medskip

{\bf (b)} Since all the maps $u\mapsto u(x)$, where $x\in X$, are
bounded linear functionals on $C_0(X)$, $\frak T_p$ is coarser than
the weak topology $\frak T_s$;  so
a $\frak T_s$-convergent sequence is $\frak T_p$-convergent to the same
limit, and a $\frak T_s$-compact set is $\frak T_p$-compact, therefore
$\frak T_p$-countably compact.

\medskip

{\bf (c)} If $K\subseteq C_0(X)$ is $\frak T_s$-compact, then
$f[K]\subseteq\Bbb R$ must be compact, therefore bounded, for every
$f\in C_0(X)^*$;  by the Uniform Boundedness Theorem (3A5Hb), $K$ is
norm-bounded.   Applying this to
$\{u\}\cup\{u_n:n\in\Bbb N\}$, we see that any $\frak T_s$-convergent
sequence $\sequencen{u_n}$ with limit $u$ is norm-bounded.

\medskip

{\bf (d)} Suppose that $\sequencen{u_n}$ is norm-bounded and
$\frak T_p$-convergent to $u$.   By Lebesgue's Dominated Convergence
Theorem,
$\lim_{n\to\infty}\int u_nd\mu=\int u\,d\mu$ for every totally finite
Radon measure $\mu$ on $X$.   But by the Riesz Representation Theorem
(in the form 436K), this says just that $\lim_{n\to\infty}f(u_n)=f(u)$
for every positive linear functional $f$ on $C_0(X)$.   Since every
member of $C_0(X)^*$ is expressible as the difference of two positive
linear functionals (356Dc), $\lim_{n\to\infty}f(u_n)=f(u)$ for every
$f\in U^*$, that is, $\sequencen{u_n}$ is $\frak T_s$-convergent to $u$.

Putting this together with (b) and (c), we see that (ii) is true.

\medskip

{\bf (e)} Now suppose that $K\subseteq C_0(X)$ is norm-bounded and
$\frak T_p$-countably compact.   Any
sequence $\sequencen{u_n}$ in $K$ has a subsequence which is
$\frak T_p$-convergent to a point of $K$ (462Bc), and this subsequence is
also $\frak T_s$-convergent, by (c).   This means that $K$ is
sequentially compact, therefore
countably compact, in $C_0(X)$ for the topology $\frak T_s$.   Since
$\frak T_s$ is angelic (462D), $K$ is $\frak T_s$-compact, by 462Bc again.
}%end of proof of 462E

\leader{462F}{Lemma} Let $X$ be a topological space, and $Q$ a relatively
countably compact subset of $X$.   Suppose that $K\subseteq C_b(X)$ is
$\|\,\|_{\infty}$-bounded and $\frak T_p$-countably
compact, where $\frak T_p$ is the
topology of pointwise convergence on $C_b(X)$.   Then the map
$u\mapsto u\restr Q:K\to C_b(Q)$ is continuous for $\frak T_p$ on $K$ and
the weak topology of the Banach space $C_b(Q)$.

\proof{ We have a natural map $x\mapsto\hat x:X\to\BbbR^K$ defined by
writing $\hat x(u)=u(x)$ for every $u\in K$ and
$x\in X$.   By the definition of $\frak T_p$, $\hat x\in C(K)$ for every
$x\in X$, if we take $C(K)$ to be the space of $\frak T_p$-continuous
real-valued functions on $K$;  and $x\mapsto\hat x:X\to C(K)$ is
continuous for the given topology on $X$ and the topology of pointwise
convergence on $C(K)$ because $K\subseteq C(X)$.   It follows that
$\{\hat x:x\in Q\}$ is relatively countably compact for the topology of
pointwise convergence on $C(K)$ (4A2G(f-iv)).   But now
$Z=\overline{\{\hat x:x\in Q\}}$ must be actually compact for
the topology of pointwise convergence on $C(K)$, by 462C.

Next, consider the natural map $u\mapsto\hat u:K\to\BbbR^Z$ defined by
setting $\hat u(f)=f(u)$ for $f\in Z$ and $u\in K$.   Just as in the last
paragraph, this is a continuous function from $K$ to $C(Z)$, if we give
$K$, $Z$ and $C(Z)$ their topologies of pointwise convergence.   So
$L=\{\hat u:u\in K\}$ is countably compact for the topology of pointwise
convergence on $C(Z)$ (4A2G(f-vi)).
Moreover, it is norm-bounded, because

$$\eqalign{\sup_{\phi\in L}\|\phi\|_{\infty}
&=\sup_{u\in K,f\in Z}|\hat u(f)|
=\sup_{u\in K,f\in Z}|f(u)|
=\sup_{u\in K,x\in Q}|\hat x(u)|\cr
&=\sup_{u\in K,x\in Q}|u(x)|
\le\sup_{u\in K,x\in X}|u(x)|
=\sup_{u\in K}\|u\|_{\infty}\cr}$$

\noindent is finite.   So 462E(iii) tells us that $L$ is weakly compact in
$C(Z)$.   (Note that $C(Z)=C_0(Z)$ because $Z$ is compact.)
Since the weak topology on $C(Z)$ is finer than the pointwise topology,
while the pointwise topology is Hausdorff,
the two topologies on $L$ coincide;  it follows that
$u\mapsto\hat u:K\to C(Z)$ is continuous for $\frak T_p$ and the weak
topology on $C(Z)$.

Now we have an operator $T:C(Z)\to\Bbb R^Q$ defined by setting

\Centerline{$(T\phi)(x)=\phi(\hat x)$}

\noindent for $\phi\in C(Z)$ and $x\in Q$.   Because
$x\mapsto\hat x:Q\to Z$ is continuous, $T\phi\in C(Q)$ for every
$\phi\in C(Z)$, and of
course $T$, regarded as a linear operator from $C(Z)$ to $C_b(Q)$,
has norm at most $1$.   So $T$ is continuous for the weak topologies of
$C(Z)$ and $C_b(Q)$ (2A5If), and
$u\mapsto T\hat u:K\to C_b(Q)$ is continuous for $\frak T_p$ and the weak
topology of $C_b(Q)$.

But if $u\in K$ and $x\in Q$,

\Centerline{$(T\hat u)(x)=\hat u(\hat x)=\hat x(u)=u(x)$,}

\noindent so $T\hat u=u\restr Q$.   Accordingly
$u\mapsto u\restr Q:K\to C_b(Q)$ is continuous for $\frak T_p$ and the weak
topology on $C_b(Q)$.
}%end of proof of 462F

\leader{462G}{Proposition} Let $X$ be a countably compact topological
space.   Then a subset of $C_b(X)$ is weakly compact iff it is
norm-bounded and compact for the topology\cmmnt{ $\frak T_p$} of
pointwise convergence.

\proof{ A weakly compact subset of $C_b(X)$ is
norm-bounded and $\frak T_p$-compact by the same arguments as in (b)-(c)
of the proof of 462E.   In the other direction, taking $Q=X$ in 462F, we
see that a norm-bounded $\frak T_p$-compact set is weakly compact.
}%end of proof of 462G

\leader{462H}{Lemma} Let $X$ be a topological space, $Q$ a
relatively countably
compact subset of $X$, and $\mu$ a totally finite measure on $C_b(X)$ which
is Radon for the topology $\frak T_p$ of pointwise convergence
on $C_b(X)$.   Let
$T:C_b(X)\to C_b(Q)$ be the restriction map.   Then the image measure
$\nu=\mu T^{-1}$ on $C_b(Q)$ is Radon for the norm topology of $C_b(Q)$.

\proof{{\bf (a)} $T$ is almost continuous for $\frak T_p$ and the weak
topology of $C_b(Q)$.   \Prf\ If $E\in\dom\mu$ and $\mu E>\gamma\ge 0$,
then there is a $\frak T_p$-compact set $K\subseteq E$ such that
$\mu K>\gamma$.   Since all the balls
$\{f:f\in C_b(X)$, $\|f\|_{\infty}\le k\}$ are $\frak T_p$-closed, we may
suppose that $K$ is norm-bounded.   Now $T\restr K$ is continuous for
$\frak T_p$ and the weak topology, by 462F.\ \QeD\   By 418I, $\nu$ is a
Radon measure for the weak topology of $C_b(Q)$.

\medskip

{\bf (b)} I show next that if $F\in\dom\nu$, $\nu F>0$ and
$\epsilon>0$, there is some $g\in C_b(Q)$ such that
$\nu(F\cap B(g,\epsilon))>0$, where
$B(g,\epsilon)=\{h:\|h-g\|_{\infty}\le\epsilon\}$.   \Prf\
Since all the balls $B(g,\epsilon)$ are convex
and norm-closed, they are weakly closed (3A5Ee) and measured by $\nu$.
\Quer\ Suppose, if possible, that $F\cap B(g,\epsilon)$ is
$\nu$-negligible for every $g\in C_b(Q)$.   Set $E=T^{-1}[F]$.
As in (a), there is a
$\|\,\|_{\infty}$-bounded $\frak T_p$-compact set $K\subseteq E$
such that $\mu K>0$.
Choose $\sequencen{K_n}$ and $\sequencen{f_n}$ as follows.   $K_0=K$.
Given that $K_n\subseteq E$ is non-negligible and $\frak T_p$-compact,
and that $\langle f_i\rangle_{i<n}$ is a finite sequence
in $C_b(X)$, then the convex hull

\Centerline{$\Gamma_n=\{\sum_{i=0}^{n-1}\alpha_iTf_i:\alpha_i\ge 0$
for every $i<n$, $\sum_{i=0}^{n-1}\alpha_i=1\}$}

\noindent of the finite set $\{Tf_i:i<n\}$ is
norm-compact in $C_b(Q)$, so there is a finite set $D_n\subseteq\Gamma_n$
such that for every $g\in\Gamma_n$ there is a $g'\in D_n$ such that
$\|g-g'\|_{\infty}\le\bover12\epsilon$.   Now

\Centerline{$H_n=\{f:\|Tf-g\|_{\infty}>\epsilon$ for every $g\in D_n\}$}

\noindent is a $\frak T_p$-open set and

\Centerline{$E\setminus H_n
=\bigcup_{g\in D_n}T^{-1}[F\cap B(g,\epsilon)]$}

\noindent is $\mu$-negligible,
so $\mu(K_n\cap H_n)>0$ and we can find a non-negligible
$\frak T_p$-closed set $K_{n+1}\subseteq K_n\cap H_n$;  choose
$f_n\in K_{n+1}$.   Continue.

At the end of the induction, let $f^*\in K$ be a cluster point of
$\sequencen{f_n}$ for $\frak T_p$.
Since $T\restr K:K\to C_b(Q)$ is continuous for
$\frak T_p$ and the weak topology of $C_b(Q)$,
$Tf^*$ is a cluster point of
$\sequencen{Tf_n}$ for the weak topology on $C_b(Q)$.   The set
$\Gamma=\bigcup_{n\in\Bbb N}\Gamma_n$ is convex, so its
norm-closure $\overline{\Gamma}$ is also convex (2A5Eb), therefore
closed for the weak topology (3A5Ee), and contains $Tf^*$.   So
there is a $g\in\Gamma$ such that
$\|Tf^*-g\|_{\infty}\le\bover12\epsilon$.
Now there is some $n$ such that
$g\in\Gamma_n$.   Let $g'\in D_n$ be such that
$\|g-g'\|_{\infty}\le\bover12\epsilon$, so that
$\|Tf^*-g'\|_{\infty}\le\epsilon$.
But $f_i\in K_{n+1}$ for every $i\ge n$,
so $f^*\in K_{n+1}\subseteq H_n$ and $\|Tf^*-g'\|_{\infty}>\epsilon$,
which is impossible.\ \Bang\Qed

\medskip

{\bf (c)} What this means is that if we take $\Cal K_n$ to be the family
of subsets of $C_b(Q)$ which can be covered by finitely many balls of
radius at most $2^{-n}$, then $\nu$ is inner regular with respect
to $\Cal K_n$ (see 412Aa), and therefore with respect to
$\Cal K=\bigcap_{n\in\Bbb N}\Cal K_n$ (412Ac).   But $\Cal K$ is just
the set of subsets of $C_b(Q)$ which are totally bounded for the
norm-metric $\rho$ on $C_b(Q)$.

At the same time, $\nu$ is inner regular with respect to the $\rho$-closed
sets in $C_b(Q)$.   \Prf\ If $\nu F>\gamma$,
there is a $\|\,\|_{\infty}$-bounded $\frak T_p$-compact set
$K\subseteq T^{-1}[F]$ such that $\mu K\ge\gamma$;
now $T[K]$ is weakly compact,
therefore weakly closed and $\rho$-closed in $C_b(Q)$, while
$T[K]\subseteq F$ is measured by $\nu$ and

\Centerline{$\nu T[K]=\mu T^{-1}[T[K]]\ge\mu K\ge\gamma$.  \Qed}

\medskip

{\bf (d)} By 412Ac again, $\nu$ must be inner regular with respect to the
family of $\rho$-closed $\rho$-totally bounded sets;  because $C_b(Q)$ is
$\rho$-complete, these are the $\rho$-compact sets.   Next, every
$\rho$-compact set is weakly compact, therefore weakly closed, and is
measured by $\nu$, by (a);  and $\nu$, being the image of a complete
totally finite measure, is complete and totally finite.   Consequently
every $\rho$-closed set is measured by $\nu$ (use 412Ja) and $\nu$ is a
$\rho$-Radon measure, as claimed.
}%end of proof of 462H

\leader{462I}{Theorem} Let $X$ be a countably compact topological space.
Then the totally finite Radon measures on $C(X)$ are the same for the
topology of pointwise convergence and the norm topology.

\proof{ Write $\frak T_p$ for the topology of
pointwise convergence on $C(X)$ and $\frak T_{\infty}$ for the norm
topology.   Because $\frak T_p\subseteq\frak T_{\infty}$ and $\frak T_p$ is
Hausdorff, every totally
finite $\frak T_{\infty}$-Radon measure is $\frak T_p$-Radon (418I).
On the other hand, 462H, with $Q=X$, tells us that every $\frak T_p$-Radon
measure is $\frak T_{\infty}$-Radon.
}%end of proof of 462I

\leader{462J}{Corollary} Let $X$ be a countably compact Hausdorff space,
and give $C(X)$ its topology of pointwise convergence.   If $\mu$ is any
Radon measure on $C(X)$, it is inner regular with respect to the family
of compact metrizable subsets of $C(X)$.

\proof{ In the language of 462I, $\mu$ is inner regular with respect to
the family of $\frak T_{\infty}$-compact sets;  but as $\frak
T_p\subseteq\frak T_{\infty}$, the two topologies agree on all such
sets, and they are compact and metrizable for $\frak T_p$.
}%end of proof of 462J

\leader{462K}{Proposition} Let $X$ be a topological space, $Y$ a
Hausdorff space, $f:X\times Y\to\Bbb R$ a bounded separately continuous
function, and $\nu$ a totally finite Radon measure on $Y$.   Set
$\phi(x)=\int f(x,y)\nu(dy)$ for every $x\in X$.   Then $\phi\restr Q$
is continuous for every relatively countably compact set $Q\subseteq X$.

\proof{ For $y\in Y$, set $u_y(x)=f(x,y)$ for every $x\in X$.   Then
every $u_y$ is continuous and bounded, because $f$ is
bounded and continuous in the first
variable, and $y\mapsto u_y:Y\to C_b(X)$ is continuous, if we give $C_b(X)$
the topology $\frak T_p$ of pointwise convergence, because $f$ is
continuous in the second variable.   We therefore have a
$\frak T_p$-Radon image measure $\mu$ on $C_b(X)$, by 418I.

Let $T:C_b(X)\to C_b(Q)$ be the restriction map.
By 462H, the image measure $\lambda=\mu T^{-1}$
is a Radon measure for the norm
topology of $C_b(Q)$.   Now recall that $f$ is bounded.   If
$|f(x,y)|\le M$ for all $x\in X$, $y\in Y$, then $\|u_y\|_{\infty}\le M$
for every
$y\in Y$, and the ball $B(0,M)$ in $C_b(Q)$ is $\lambda$-conegligible.
By 461F, applied to the subspace measure on $B(0,M)$,
$\nu$ has a barycenter $h$ in $C_b(Q)$.   Now we can compute $h$ by
the formulae

$$\eqalignno{h(x)
&=\int_{C_b(Q)}g(x)\lambda(dg)\cr
\noalign{\noindent (because $g\mapsto g(x)$ belongs to $C_b(X)^*$)}
&=\int_{C_b(X)}u(x)\mu(du)
=\int_Yu_y(x)\nu(dy)\cr
\noalign{\noindent (by 235G)}
&=\int_Yf(x,y)\nu(dy)
=\phi(x),\cr}$$

\noindent for every $x\in Q$.   So $\phi=h$ is continuous.
}%end of proof of 462K

\leader{462L}{Corollary} Let $X$ be a topological space such that

\inset{\noindent
whenever $h\in\BbbR^X$ is such that $h\restr Q$ is continuous
for every relatively countably compact $Q\subseteq X$, then $h$ is
continuous.}

\noindent Write $\frak T_p$ for the topology of pointwise convergence on
$C(X)$.   Let $K\subseteq C(X)$ be a $\frak T_p$-compact
set such that $\{h(x):h\in K$, $x\in Q\}$ is bounded for any
relatively countably compact set $Q\subseteq X$.   Then the
$\frak T_p$-closed convex hull of $K$, taken in $C(X)$, is
$\frak T_p$-compact.

\proof{ If $K$ is empty, this is trivial;  suppose that $K\ne\emptyset$.
Since $\sup_{h\in K}|h(x)|$ is finite for every $x\in X$,
the closed convex hull $\overline{\Gamma(K)}$ of $K$, taken in
$\Bbb R^X$, is closed and included in a product of closed bounded
intervals, therefore compact.    If
$h\in\overline{\Gamma(K)}$, then there is a Radon probability measure
$\mu$ on $K$ such that $h$ is the barycenter of $\mu$ (461I), so that
$h(x)=\int f(x)\mu(df)$ for every $x\in X$.

If $Q\subseteq X$ is relatively
countably compact, then $h\restr Q$ is continuous.   \Prf\ Of course we may
suppose that $Q$ is non-empty.   Consider its closure $Z=\overline{Q}$.
We have a continuous linear operator $T:\BbbR^X\to\BbbR^Z$ defined by
setting $Tf=f\restr Z$ for every $f\in\BbbR^X$.   $L=T[K]$ is compact in
$\BbbR^Z$, and $L\subseteq C(Z)$;  moreover,

\Centerline{$\sup_{g\in L}\|g\|_{\infty}
=\sup_{f\in K,x\in Z}|f(x)|
=\sup_{f\in K,x\in Q}|f(x)|$}

\noindent is finite.   Since $T\restr K:K\to L$ is continuous, the image
measure $\nu=\mu(T\restr K)^{-1}$ on $L$ is a Radon measure.
If $x\in Z$, then

\Centerline{$h(x)=\int_Kf(x)\mu(df)=\int_K(Tf)(x)\mu(df)
=\int_Lg(x)\nu(dg)$.}

The map $(x,g)\mapsto g(x):Z\times L\to\Bbb R$ is separately continuous,
because $L\subseteq C(Z)$ is being given its topology of pointwise
convergence, and bounded.    Also every sequence in $Q$ has a cluster point
in $X$ which must also belong to $Z$, and $Q$ is relatively countably
compact in $Z$.   By 462K, $h\restr Q$ is continuous, as required.\ \Qed

Thus the $\frak T_p$-compact set $\overline{\Gamma(K)}$ is included in
$C(X)$, and must be the closed convex hull of $K$ in $C(X)$.
}%end of proof of 462L

\cmmnt{\medskip

\noindent{\bf Remark} The hypothesis

\inset{\noindent
whenever $h\in\Bbb R^X$ is such that $h\restr Q$ is continuous
for every relatively countably compact $Q\subseteq X$, then $h$ is
continuous}

\noindent is clumsy, but seems the best way to cover the large number of
potential applications of the ideas here.   Besides the obvious case of
countably compact spaces $X$, we have all first-countable spaces (for
which, of course, the other hypotheses can be relaxed, as in 462Xc), and
all $k$-spaces.   (A {\bf $k$-space} is a topological space $X$ such
that a set $G\subseteq X$ is open iff $G\cap K$ is relatively open in
$K$ for every compact set $K\subseteq X$;  see {\smc Engelking 89},
3.3.18 {\it et seq.}   In particular, all locally compact spaces are
$k$-spaces.)
}%end of comment

\exercises{\leader{462X}{Basic exercises (a)}
%\spheader 462Xa
(i) Show that the Sorgenfrey line (415Xc)
is angelic.   (ii) Show that any metrizable space is angelic.
(iii) Show that the one-point compactification of an angelic locally
compact Hausdorff space is angelic.   (iv) Find a first-countable regular Hausdorff space which is not angelic.
%462B

\sqheader 462Xb Let $X$ be any countably compact topological space.
Show that a norm-bounded sequence in $C_b(X)$ which is pointwise
convergent is weakly convergent.
%462G

\spheader 462Xc Let $X$ be a first-countable topological space,
$(Y,\Tau,\nu)$ a totally finite measure space, and $f:X\times Y\to\Bbb
R$ a bounded function such that $y\mapsto f(x,y)$ is measurable for
every $x\in X$, and $x\mapsto f(x,y)$ is continuous for almost every
$y\in Y$.   Show that $x\mapsto\int f(x,y)\nu(dy)$ is continuous.
%462K

\spheader 462Xd Give an example of a $\frak T_p$-compact subset $K$ of
$C([0,1])$ such that the convex hull of $K$ is not relatively
compact in $C([0,1])$.
%462L

\leader{462Y}{Further exercises (a)}
%\spheader 462Ya
Let $X$ be a topological space such that there is a
sequence $\sequencen{X_n}$ of
relatively countably compact subsets of $X$, covering
$X$, with the property that a function $f:X\to\Bbb R$ is continuous
whenever $f\restr X_n$ is continuous for every $n\in\Bbb N$.   Let
$\frak T_p$ be the topology of pointwise convergence on $C(X)$.   Show
that, for a set $K\subseteq C(X)$, the following are equiveridical:  (i)
$\phi[K]$ is bounded for every $\frak T_p$-continuous function
$\phi:C(X)\to\Bbb R$;  (ii) whenever $\sequencen{f_n}$ is a sequence in
$K$ and $A\subseteq X$ is countable, there is a cluster point of
$\sequencen{f_n\restr A}$ in $C(A)$ for the topology of pointwise
convergence on $C(A)$;  (iii) $K$ is relatively compact in $C(X)$ for
$\frak T_p$.  (See {\smc Asanov \& Velichko 81}.)
%462C

\spheader 462Yb Let $U$ be a metrizable locally convex linear
topological space.   Show that it is angelic in its weak topology.
\Hint{start with the case in which $U$ is complete, using Grothendieck's
theorem and the full strength of 462C, with $X=U^*$.}
%462D

\spheader 462Yc In 462K, show that the conclusion remains valid for any
totally finite $\tau$-additive topological measure $\nu$ on $Y$ which is
inner regular with respect to the relatively countably compact subsets
of $Y$.
%462K

\spheader 462Yd Show that if $X$ is any compact topological space (more
generally, any topological space such that $X^n$ is Lindel\"of for every
$n\in\Bbb N$), then $C(X)$, with its topology of pointwise convergence,
is countably tight.
%462+

\spheader 462Ye(i)
Let $X$ and $Y$ be Polish spaces, and write $B_1(X;Y)$ for
the set of functions $f:X\to Y$ such that $f^{-1}[H]$ is G$_{\delta}$ in
$X$ for every closed set $H\subseteq Y$ ({\smc Kuratowski 66}, \S31).
Show that $B_1(X;Y)$, with the topology of pointwise convergence inherited
from $Y^X$, is angelic.   \Hint{{\smc Bourgain Fremlin \& Talagrand 78}.}
(ii) Let $X$ be a Polish space.   Show that the space $\tildeClll(X)$ of
438P-438Q is angelic.
}%end of exercises

\leader{462Z}{Problem} Let $K$ be a compact Hausdorff space.   Is
$C(K)$, with the topology of pointwise convergence, necessarily a
pre-Radon space?   \cmmnt{(Compare 454S.)}

%Is C(K) Borel-measure-compact?
%Is C(K) Borel-measure-complete?  or even measure-compact?
%Not if K is one-point compactification
%of a discrete space with cardinal not of measure zero, as then C(K)
%has a subspace homeomorphic to K.   But what if every cardinal has
%measure zero?

\endnotes{
\Notesheader{462} The theory of pointwise convergence in spaces of
continuous functions is intimately connected with the theory of
separately continuous functions of two variables.   For if $X$ and $Y$
are topological spaces, and $f:X\times Y\to\Bbb R$ is any separately
continuous function, then we have natural maps $x\mapsto f_x:X\to C(Y)$
and $y\mapsto f^y:Y\to C(X)$, writing $f_x(y)=f^y(x)=f(x,y)$, which are
continuous if $C(X)$ and $C(Y)$ are given their topologies of pointwise
convergence;  and if $X$ is a topological space and $Y$ is any subset of
$C(X)$ with its topology of pointwise convergence, the map
$(x,y)\mapsto y(x):X\times Y\to\Bbb R$ is separately continuous.   I
include a back-and-forth shuffle between $C(X)$ and separately
continuous functions in 462H-462K-462L as a demonstration of the
principle that all the theorems here can be expressed in both languages.

462Yb is a compendium of \v Smulian's theorem with part of Eberlein's
theorem;
462E and 462L can be thought of as the centre of Krein's theorem.
There are many alternative routes to these results, which may be found
in {\smc K\"othe 69} or {\smc Grothendieck 92}.   In particular, 462E
can be proved without using measure theory;  see, for instance, 
{\smc Fremlin 74}, A2F.

Topological spaces homeomorphic to compact 
uniformly bounded subsets of $C(X)$, where $X$
is some compact space and $C(X)$ is given its topology of pointwise
convergence, are called {\bf Eberlein compacta};  see 467O-467P.

A positive answer to A.Bellow's problem (463Za below) would imply a 
positive answer to 462Z;  so if the continuum hypothesis, for instance, 
is true,
then $C(K)$ is pre-Radon in its topology of pointwise convergence for
any compact space $K$.

}%end of notes

\discrpage

\leaveitout{If $X$ and $Y$ are compact Hausdorff spaces, of which $Y$ is
ccc, and $f:X\times Y\to\Bbb R$ a separately continuous function, then
$f$ is Baire measurable.

\proof{{\bf (a)} To begin with, suppose that $X$ itself is countably
compact.   For $y\in Y$, set $u_y(x)=f(x,y)$ for every $x\in X$.   Then
every $u_y$ is continuous, because $f$ is continuous in the first
variable, and $y\mapsto u_y:Y\to C(X)$ is continuous, if we give $C(X)$
the topology $\frak T_p$ of pointwise convergence, because $f$ is
continuous in the second variable.   We therefore have a
$\frak T_p$-Radon image measure $\mu$ on $C(X)$, by 418I again.

By 462I, $\mu$ is a Radon measure for the norm topology of $C(X)$.   Now
recall that $f$ is bounded.   If
$|f(x,y)|\le M$ for all $x\in X$, $y\in Y$, then $\|u_y\|_{\infty}\le M$
for every
$y\in Y$, and the ball $B(0,M)$ in $C(X)$ is $\mu$-conegligible.   By
461F, applied to the subspace measure on $B(0,M)$,
$\mu$ has a barycenter $h$ in $C(X)$.   Now we can compute $h$ by
the formulae

$$\eqalignno{h(x)
&=\int g(x)\mu(dg)\cr
\noalign{\noindent (because $g\mapsto g(x)$ belongs to $C(X)^*$)}
&=\int u_y(x)\nu(dy)\cr
\noalign{\noindent (by 235G)}
&=\int f(x,y)\nu(dy)
=\phi(x),\cr}$$

\noindent for every $x\in X$.   So $\phi=h$ is continuous.

\medskip

{\bf (b)} For the general result, we have only to apply (a) to $f\restr
C$ for each countably compact $C\subseteq X$.
}%end of proof of 462K

}%end of leaveitout

\frfilename{mt463.tex}
\versiondate{1.2.13}
\copyrightdate{2001}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{$\frak T_p$ and $\frak T_m$}

\newsection{463}
\def\headlinesectionname{$\eightfrak T_p$ and
$\eightfrak T_m$}

We are now ready to start on the central ideas of this chapter with an
investigation of sets of measurable functions which are compact for the
topology of pointwise convergence.   Because `measurability' is, from
the point of view of this topology on $\BbbR^X$, a rather arbitrary
condition, we are looking at compact subsets of a topologically
irregular subspace of $\BbbR^X$;  there are consequently relatively few
of them, and (under a variety of special circumstances, to be examined
later in the chapter and also in Volume 5) they have some striking
special properties.

The presentation here is focused on the relationship between the two
natural topologies on any space of measurable functions, the `pointwise'
topology $\frak T_p$ and the topology $\frak T_m$ of convergence in
measure (463A).   In this section I begin with results which apply to
any $\sigma$-finite measure space (463B-463H) before turning to some
which apply to perfect measure spaces (463I-463L) -- in particular, to
Lebesgue measure.   These lead to some interesting properties of
separately continuous functions (463M-463N).

\leader{463A}{Preliminaries} Let $(X,\Sigma,\mu)$ be a measure space, and
$\eusm L^0\cmmnt{\mskip5mu =\eusm L^0(\Sigma)}$ the space of all
$\Sigma$-measurable functions from $X$ to $\Bbb R$\cmmnt{, so that
$\eusm L^0$ is a linear subspace of $\BbbR^X$}.   On $\eusm L^0$ we
shall be concerned with two\cmmnt{ very different} topologies.   The
first is the topology $\frak T_p$ of pointwise
convergence\cmmnt{ (462Ab)};  the second is the topology $\frak T_m$
of\cmmnt{ (local)} convergence in measure\cmmnt{ (245A)}.   Both
are linear space topologies.   \prooflet{\Prf\ For $\frak T_p$ I have
already noted this in 462Ab.   For $\frak T_m$, repeat the argument of
245Da;  $\frak T_m$ is defined by the functionals
$f\mapsto\int_F\min(1,|f|)d\mu$, where $\mu F<\infty$, and these are
F-seminorms 
(definition:  2A5B).\ \QeD}   
$\frak T_p$ is
Hausdorff\cmmnt{ (3A3Id)} and locally convex\cmmnt{ (4A4Ce);  only
in exceptional circumstances is either true of $\frak T_m$}.
\cmmnt{However, $\frak T_m$ can easily be pseudometrizable (if, for
instance, $\mu$ is $\sigma$-finite, as in 245Eb), while $\frak T_p$ is
not, except in nearly trivial cases.}

Associated with the topology of pointwise convergence on $\BbbR^X$ is
the usual topology of $\Cal PX$\cmmnt{ (4A2A)};  the map
$\chi:\Cal PX\to\BbbR^X$ is a homeomorphism between $\Cal PX$ and its
image $\{0,1\}^X\subseteq\BbbR^X$.

\cmmnt{$\frak T_m$ is intimately associated with the topology of
convergence in measure on $L^0=L^0(\mu)$ (\S245).}   A subset of
$\eusm L^0$ is open for $\frak T_m$ iff it is of the form
$\{f:f^{\ssbullet}\in G\}$ for some open set $G\subseteq L^0$;
\cmmnt{consequently,} a
subset $K$ of $\eusm L^0$ is compact, or separable, for $\frak T_m$ iff
$\{f^{\ssbullet}:f\in K\}$ is compact or separable for the topology of
convergence in measure on $L^0$.

\cmmnt{It turns out that the identity map from $(\eusm L^0,\frak T_p)$
to $(\eusm L^0,\frak T_m)$ is sequentially continuous (463B).   Only in
nearly trivial cases is it actually continuous
(463Xa(i)), and it is similarly rare for the reverse map from $(\eusm
L^0,\frak T_m)$ to $(\eusm L^0,\frak T_p)$ to be continuous (463Xa(ii)).
If, however, we relativise both topologies to a
$\frak T_p$-compact subset of $\eusm L^0$, the situation becomes very
different, and there are many important cases in which the topologies
are comparable.}

\leader{463B}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space, and
$\eusm L^0$ the space of $\Sigma$-measurable real-valued functions on
$X$.   Then every pointwise convergent sequence in $\eusm L^0$ is
convergent in measure to the same limit.

\proof{ 245Ca.
}%end of proof of 463B

\leader{463C}{Proposition}\cmmnt{ ({\smc Ionescu Tulcea 73})} Let
$(X,\Sigma,\mu)$ be a measure space, and $\eusm L^0$ the space of
$\Sigma$-measurable real-valued functions on $X$.   Write $\frak T_p$,
$\frak T_m$ for the topologies of pointwise convergence and convergence
in measure on $\eusm L^0$;  for $A\subseteq\eusm L^0$, write
$\frak T_p^{(A)}$, $\frak T_m^{(A)}$ for the corresponding subspace
topologies.

(a) If $A\subseteq\eusm L^0$ and $\frak T_p^{(A)}$ is metrizable, then the
identity map from $A$ to itself is
$(\frak T_p^{(A)},\frak T_m^{(A)})$-continuous.

(b) Suppose that $\mu$ is semi-finite.   Then, for any
$A\subseteq\eusm L^0$, $\frak T_m^{(A)}$ is Hausdorff iff whenever $f$, $g$
are distinct
members of $A$ the set $\{x:f(x)\ne g(x)\}$ is non-negligible.

(c) Suppose that $K\subseteq\eusm L^0$ is such that $\frak T_p^{(K)}$ is
compact and metrizable.   Then $\frak T_p^{(K)}=\frak T_m^{(K)}$ iff
$\frak T_m^{(K)}$ is Hausdorff.

(d) Suppose that $\mu$ is $\sigma$-finite, and that
$K\subseteq\eusm L^0$ is $\frak T_p$-sequentially compact.   Then
$\frak T_p^{(K)}=\frak T_m^{(K)}$ iff $\frak T_m^{(K)}$ is Hausdorff, and in this
case $\frak T_p^{(K)}$ is compact and metrizable.

(e) Suppose that $K\subseteq\eusm L^0$ is such that $\frak T_p^{(K)}$ is
compact and metrizable.   Then whenever $\epsilon>0$ and $E\in\Sigma$ is
a non-negligible measurable set, there is a non-negligible measurable
set $F\subseteq E$ such that $|f(x)-f(y)|\le\epsilon$ whenever $f\in K$
and $x$, $y\in F$.

\proof{{\bf (a)} All we need is to remember that sequentially continuous
functions from metrizable spaces are continuous (4A2Ld), and apply 463B.

\medskip

{\bf (b)} $\frak T_m^{(A)}$ is Hausdorff iff for any distinct $f$, $g\in A$
there is a measurable set $F$ of finite measure such that
$\int_F\min(1,|f-g|)d\mu>0$, that is,
$\mu\{x:x\in F,\,f(x)\ne g(x)\}>0$;  because $\mu$ is semi-finite, this
happens iff $\mu\{x:f(x)\ne g(x)\}>0$.

\medskip

{\bf (c)} If $\frak T_p^{(K)}=\frak T_m^{(K)}$ then of course $\frak T_m^{(K)}$
is Hausdorff, because $\frak T_p^{(K)}$ is.   If $\frak T_m^{(K)}$ is
Hausdorff then the identity map $(K,\frak T_p^{(K)})\to(K,\frak T_m^{(K)})$ is
an injective function from a compact space to a Hausdorff space and (by
(a)) is continuous, therefore a homeomorphism, so the two topologies are
equal.

\medskip

{\bf (d)} If $\frak T_p^{(K)}=\frak T_m^{(K)}$ then $\frak T_m^{(K)}$ must be
Hausdorff, just as in (c).   So let us suppose that $\frak T_m^{(K)}$ is
Hausdorff.   Note that, by 245Eb, the topology of convergence in measure
on $L^0$ is metrizable;  in terms of $\eusm L^0$, this says just that
the topology of convergence in measure on $\eusm L^0$ is
pseudometrizable.   So $\frak T_m^{(K)}$ is Hausdorff and pseudometrizable,
therefore metrizable (4A2La).

We are told that any sequence in $K$ has a $\frak T_p^{(K)}$-convergent
subsequence.   But this subsequence is now $\frak T_m^{(K)}$-convergent
(463B), so $\frak T_m^{(K)}$ is sequentially compact;  being metrizable, it
is compact (4A2Lf).   Moreover, the same is true of any
$\frak T_p^{(K)}$-closed subset of $K$, so every $\frak T_p^{(K)}$-closed set
is $\frak T_m^{(K)}$-compact, therefore $\frak T_m^{(K)}$-closed.   Thus the
identity map from $(K,\frak T_m^{(K)})$ to $(K,\frak T_p^{(K)})$ is
continuous.   Since $\frak T_m^{(K)}$ is compact and $\frak T_p^{(K)}$ is
Hausdorff, the two topologies are equal;  and, in particular, 
$\frak T_p^{(K)}$ is compact and metrizable.

\medskip

{\bf (e)} Let $\rho$ be a metric on $K$ inducing the topology 
$\frak T_p^{(K)}$.   Let $D\subseteq K$ be a countable dense set.   For each
$n\in\Bbb N$, set

\Centerline{$G_n=\{x:|f(x)-g(x)|\le\Bover13\epsilon$ whenever $f$, $g\in
D$ and $\rho(f,g)\le 2^{-n}\}$.}

\noindent Because $D$ is countable, $G_n$ is measurable.   Now
$\bigcup_{n\in\Bbb N}G_n=X$.   \Prf\Quer\ If $x\in
X\setminus\bigcup_{n\in\Bbb N}G_n$, then for each $n\in\Bbb N$ we can
find $f_n$, $g_n\in D$ such that $\rho(f_n,g_n)\le 2^{-n}$ and
$|f_n(x)-g_n(x)|\ge\bover13\epsilon$.   Because $K$ is compact, there is
a strictly increasing sequence $\sequence{k}{n_k}$ such that
$\sequence{k}{f_{n_k}}$ and $\sequence{k}{g_{n_k}}$ are both convergent
to $f$, $g$ say.   Now

\Centerline{$\rho(f,g)=\lim_{k\to\infty}\rho(f_{n_k},g_{n_k})=0$,
\quad$|f(x)-g(x)|
=\lim_{k\to\infty}|f_{n_k}(x)-g_{n_k}(x)|\ge\Bover13\epsilon$,}

\noindent so $f=g$ while $f(x)\ne g(x)$, which is impossible.\ \Bang\Qed

There is therefore some $n\in\Bbb N$ such that $\mu(E\cap G_n)>0$.
Since $K$, being compact, is totally bounded for $\rho$, there is a
finite set $D'\subseteq D$ such that every member of $D$ is within a
distance of $2^{-n}$ of some member of $D'$.   Now there is a measurable
set $F\subseteq E\cap G_n$ such that $\mu F>0$ and
$|g(x)-g(y)|\le\bover13\epsilon$ whenever $g\in D'$ and $x$, $y\in F$.
So $|f(x)-f(y)|\le\epsilon$ whenever $f\in D$ and $x$, $y\in F$.   But
as $D$ is dense in $K$, $|f(x)-f(y)|\le\epsilon$ whenever $f\in K$ and
$x$, $y\in F$, as required.
}%end of proof of 463C

\leader{463D}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space, and
$\eusm L^0$ the space of $\Sigma$-measurable real-valued functions on
$X$.   Write $\frak T_p$ for the topology of pointwise
convergence on $\eusm L^0$.   Suppose that
$K\subseteq\eusm L^0$ is $\frak T_p$-compact and that there is no
$\frak T_p$-continuous surjection from any closed subset of $K$ onto
$\{0,1\}^{\omega_1}$.   If $E\in\Sigma$ has finite measure, then every sequence in $K$ has a subsequence which is convergent almost everywhere in $E$.

\proof{{\bf (a)} Let $\sequencen{f_n}$ be a sequence in $K$.   Set
$q(x)=\max(1,2\sup_{n\in\Bbb N}|f_n(x)|)$ for each $x\in X$.
For any infinite $I\subseteq\Bbb N$, set

\Centerline{$g_I=\liminf_{i\to I}f_i
=\sup_{n\in\Bbb N}\inf_{i\in I,i\ge n}f_i$,}

\Centerline{$h_I=\limsup_{i\to I}f_i
=\inf_{n\in\Bbb N}\sup_{i\in I,i\ge n}f_i$;}

\noindent because $\sup_{f\in K}|f(x)|$ is surely finite for each
$x\in X$, $g_I$ and $h_I$ are defined in $\eusm L^0$, and $g_I\le h_I$.
For $f\in\eusm L^0$ set $\tau'(f)=\int_E\min(1,|f|/q)$, and for
$I\in[\Bbb N]^{\omega}$ (the set of infinite subsets of $\Bbb N$) set
$\Delta(I)=\tau'(h_I-g_I)$.   Since $h_I-g_I\le q$,
$\Delta(I)=\int_E(h_I-g_I)/q$.   If $I$, $J\in[\Bbb N]^{\omega}$ and
$J\setminus I$ is finite, then $g_I\le g_J\le h_J\le h_I$, so
$\Delta(J)\le\Delta(I)$, with equality only when $g_I=g_J$ a.e.\ on $E$
and $h_I=h_J$ a.e.\ on $E$.

\medskip

{\bf (b)} There is a $J\in[\Bbb N]^{\omega}$ such that
$\Delta(I)=\Delta(J)$ for every $I\in[J]^{\omega}$.   \Prf\ For
$J\in[\Bbb N]^{\omega}$, set
$\underline{\Delta}(J)=\inf\{\Delta(I):I\in[J]^{\omega}\}$.   Choose
$\sequencen{I_n}$ in $[\Bbb N]^{\omega}$ inductively in such a way that
$I_{n+1}\subseteq I_n$ and
$\Delta(I_{n+1})\le\underline{\Delta}(I_n)+2^{-n}$ for every $n$.   If
we now set

\Centerline{$J=\{\min\{i:i\in I_n,\,i\ge n\}:n\in\Bbb N\}$,}

\noindent $J\subseteq\Bbb N$ will be an infinite set and
$J\setminus I_n$ is finite for every $n$.   If $I\in[J]^{\omega}$ then,
for every $n$,

\Centerline{$\Delta(J)\le\Delta(I_{n+1})
\le\underline{\Delta}(I_n)+2^{-n}
\le\Delta(I_n\cap I)+2^{-n}=\Delta(I)+2^{-n}$;}

\noindent as $n$ and $I$ are
arbitrary, $\underline{\Delta}(J)=\Delta(J)$, as required.\ \Qed

Now for any $I\in[J]^{\omega}$ we have 
$g_I=g_J$ a.e.\ on $E$ and $h_I=h_J$ a.e.\ on $E$.

\medskip

{\bf (c)} $\Delta(J)=0$.   \Prf\Quer\ Otherwise,
$F=\{x:x\in E,\,g_J(x)<h_J(x)\}$ has positive measure.   Write $K_0$ for
\penalty-100$\bigcap_{n\in\Bbb N}\overline{\{f_i:i\in J,\,i\ge n\}}$, the
closure being taken for $\frak T_p$, so that $K_0$ is
$\frak T_p$-compact.   Let
$\Cal A$ be the family of sets $A\subseteq F$ such that whenever $L$,
$M\subseteq A$ are finite and disjoint there is an $f\in K_0$ such that
$f(x)=g_J(x)$ for $x\in L$ and $f(x)=h_J(x)$ for $x\in M$.   Then
$\Cal A$ has a maximal member $A_0$ say.   If we define
$\phi:\eusm L^0\to[0,1]^{A_0}$ by setting
$\phi(f)(x)=\Bover{\med(0,f(x)-g_J(x),h_J(x)-g_J(x))}{h_J(x)-g_J(x)}$ for
$x\in A_0$ and $f\in\eusm L^0$,
$\phi[K_0]$ is a compact subset of $[0,1]^{A_0}$, and whenever $L$,
$M\subseteq A_0$ are finite there is a $g\in\phi[K_0]$ such that
$g(x)=0$ for $x\in L$ and $g(x)=1$ for $x\in M$.   This means that
$\phi[K_0]\cap\{0,1\}^{A_0}$ is dense in $\{0,1\}^{A_0}$ and must
therefore be the whole of $\{0,1\}^{A_0}$.   So $\{0,1\}^{A_0}$ is a
continuous image of a closed subset of $K$.

Since $\{0,1\}^{\omega_1}$ is not a continuous image of a closed subset of
$K$, it is not a continuous image of $\{0,1\}^{A_0}$, and cannot be
homeomorphic to $\{0,1\}^A$ for any $A\subseteq A_0$.   Thus no subset of
$A_0$ can have cardinal $\omega_1$ and $A_0$ is countable.

For each pair $L$, $M$ of disjoint finite
subsets of $A_0$, we have a cluster point $f_{LM}$ of
$\family{j}{J}{f_j}$ such that $f_{LM}(x)=g_J(x)$ for $x\in L$ and
$f_{LM}(x)=h_J(x)$ for $x\in M$.   Let $I(L,M)$ be an infinite subset of
$J$ such that
$\lim_{i\to\infty,i\in I(L,M)}f_i(x)=f_{LM}(x)$ for every $x\in A_0$.
Then $g_{I(L,M)}=g_J$ and $h_{I(L,M)}=h_J$ almost everywhere in $E$.
Because $\mu F>0$ and $A_0$ has only countably many finite subsets,
there is a $y\in F$ such that $g_{I(L,M)}(y)=g_J(y)$ and
$h_{I(L,M)}(y)=h_J(y)$ whenever $L$ and $M$ are disjoint finite subsets
of $A_0$.

What this means is that if $L$ and $M$ are disjoint finite subsets of
$A_0$, then there are infinite sets $I'$, $I''\subseteq I(L,M)$ such
that $\lim_{i\to\infty,i\in I'}f_i(y)=g_J(y)$ and
$\lim_{i\to\infty,i\in I''}f_i(y)=h_J(y)$;  so that there are $f'$,
$f''\in K_0$ such that

\Centerline{$f'(x)=g_J(x)$ for $x\in L\cup\{y\}$,
\quad$f'(x)=h_J(x)$ for $x\in M$,}

\Centerline{$f''(x)=g_J(x)$ for $x\in L$,
\quad$f''(x)=h_J(x)$ for $x\in M\cup\{y\}$.}

\noindent But this means that $A_0\cup\{y\}\in\Cal A$, and also that $y\notin A_0$;
and $A_0$ was chosen to be maximal.\ \Bang\Qed

\medskip

{\bf (d)} So $\int_E(h_J-g_J)/q=0$ and $g_J=h_J$ almost everywhere in
$E$.   But if we enumerate $J$ in ascending order as
$\sequence{i}{n_i}$,
$g_J=\liminf_{i\to\infty}f_{n_i}$ and $h_J=\limsup_{i\to\infty}f_{n_i}$,
so $\sequence{i}{f_{n_i}}$ converges almost everywhere in $E$.
}%end of proof of 463D

\leader{463E}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$\eusm L^0$ the space of $\Sigma$-measurable real-valued functions on
$X$.   Write $\frak T_p$, $\frak T_m$ for the topologies of pointwise
convergence and convergence in measure on $\eusm L^0$.   Suppose that
$K\subseteq\eusm L^0$ is $\frak T_p$-compact and that there is no
$\frak T_p$-continuous surjection from any closed subset of $K$ onto
$\omega_1+1$ with its order topology.   Then the identity map from
$(K,\frak T_p)$ to $(K,\frak T_m)$ is continuous.

\proof{{\bf (a)} It is worth noting straight away that
$\xi\mapsto\chi\xi:\omega_1+1\to\{0,1\}^{\omega_1}$ is a homeomorphism
between $\omega_1+1$ and a subspace of $\{0,1\}^{\omega_1}$.   So our
hypothesis tells us that there is no continuous surjection from any
closed subset of $K$ onto $\{0,1\}^{\omega_1}$, and therefore none onto
$\{0,1\}^A$ for any uncountable $A$.

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that the identity map from
$(K,\frak T_p)$ to $(K,\frak T_m)$ is not continuous at $f_0\in K$.
Then there are an $E\in\Sigma$, of finite measure, and an $\epsilon>0$
such that $C=\{f:f\in K,\,\tau_E(f-f_0)\ge\epsilon\}$ meets every
$\frak T_p$-neighbourhood of $f$, where $\tau_E(f)=\int_E\min(1,|f|)$
for every $f\in\eusm L^0$, and there is an ultrafilter $\Cal F$ on
$\eusm L^0$ which contains $C$ and converges to $f_0$ for $\frak T_p$.
Consider the map $\psi:\eusm L^0\to L^0(\mu_E)$, where $\mu_E$ is the
subspace measure on $E$, defined by setting
$\psi(f)=(f\restr E)^{\ssbullet}$ for $f\in\eusm L^0$.
We know from 463D that every sequence in $K$ has a subsequence convergent
almost everywhere in $E$, so every sequence in $\psi[K]$ has a
subsequence which is convergent for the topology of convergence in
measure on $L^0(\mu_E)$.   Since this is metrizable, $\psi[K]$ is
relatively compact in $L^0(\mu_E)$ (4A2Le), and the image filter
$\psi[[\Cal F]]$ has a limit $v\in L^0(\mu_E)$.   Let $f_1\in\eusm L^0$
be such that $\psi(f_1)=v$.

For any countable set $A\subseteq X$ there is a $g\in C$ such that
$g\restr A=f_0\restr A$ and $g=f_1$ almost everywhere in $E$.   \Prf\ If
$X=\emptyset$ this is trivial, so we may, if necessary, enlarge $A$ by
one point so that it is not empty.      Let $\sequencen{x_n}$ be a
sequence running over $A$.   Then for each $n\in\Bbb N$ the set

\Centerline{$\{g:g\in C,\,|g(x_i)-f_0(x_i)|\le 2^{-n}$ for every
$i\le n$, $\tau_E(g,f_1)\le 2^{-n}\}$}

\noindent belongs to $\Cal F$, so is not empty;  take $g_n$ in this set.
Let $g\in K$ be any cluster point of $\sequencen{g_n}$.    Since
$\sequencen{g_n}$ converges to $f_1$ almost everywhere in $E$, $g=f_1$
a.e.\ on $E$ and $\sequencen{g_n}$ converges to $g$ almost everywhere in
$E$.   Consequently $\tau_E(g-f_0)=\lim_{n\to\infty}\tau_E(g_n-f_0)$, by
the dominated convergence theorem, and $g\in C$.   
Since $\sequencen{g_n(x_i)}$
converges to $f_0(x_i)$ for every $i$, $g\restr A=f_0\restr A$.   So we
have the result.\ \Qed

In particular, there is a $g\in C$ such that $g=f_1$ a.e.\ on $E$, so
$\tau_E(f_1-f_0)=\tau(g-f_0)\ge\epsilon$ and
$F=\{x:x\in E,\,f_0(x)\ne f_1(x)\}$ has non-zero measure.   Now choose
$\ofamily{\xi}{\omega_1}{g_{\xi}}$ in $K$ and
$\ofamily{\xi}{\omega_1}{x_{\xi}}$ in $F$ inductively so that

\Centerline{$g_{\xi}\in C$,
\quad $g_{\xi}=f_1$ almost everywhere in $E$,
\quad $g_{\xi}(x_{\eta})=f_0(x_{\eta})$ for $\eta<\xi$}

\noindent (choosing $g_{\xi}$),

\Centerline{$x_{\xi}\in F$,
\quad $g_{\eta}(x_{\xi})=f_1(x_{\xi})$ for $\eta\le\xi$}

\noindent (choosing $x_{\xi}$).
If we now set $A=\{x_{\xi}:\xi<\omega_1\}$,

\Centerline{$K_1=\bigcap_{\xi\le\eta<\omega_1}\{f:f\in K$, either
$f(x_{\xi})=f_0(x_{\xi})$ or $f(x_{\eta})=f_1(x_{\eta})\}$,}

\noindent then $K_1$ is a closed subset of $K$ containing every
$g_{\xi}$ and also $f_0$.   But if we look at $\{f\restr A:f\in K_1\}$,
this is homeomorphic to $\omega_1+1$;   which is supposed to be
impossible.\ \Bang

So we conclude that the identity map from $(K,\frak T_p)$ to
$(K,\frak T_m)$ is continuous.
}%end of proof of 463E

\leader{463F}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space, and
$\eusm L^0$ the space of $\Sigma$-measurable real-valued functions on
$X$.   Write $\frak T_p$, $\frak T_m$ for the topologies of pointwise
convergence and convergence in measure on $\eusm L^0$.   Suppose that
$K\subseteq\eusm L^0$ is compact
and countably tight for $\frak T_p$.   Then the identity map from
$(K,\frak T_p)$ to $(K,\frak T_m)$ is continuous.   If $\frak T_m$ is
Hausdorff on $K$, the two topologies coincide on $K$.

\proof{ Since $\omega_1+1$ is not countably tight (the top point
$\omega_1$ is not in the closure of any countable subset of $\omega_1$),
$\omega_1+1$ is not a continuous image of any closed subset of $K$
(4A2Kb), and we can apply 463E to see that the identity map is
continuous.   It follows at once that if $\frak T_m$ is Hausdorff on
$K$, then the topologies coincide.
}%end of proof of 463F

\vleader{48pt}{463G}{Theorem}\cmmnt{ ({\smc Ionescu Tulcea 74})} Let
$(X,\Sigma,\mu)$ be a $\sigma$-finite measure space, and $K$ a convex
set of measurable functions from $X$ to $\Bbb R$ such that (i) $K$ is
compact for the topology $\frak T_p$ of pointwise convergence (ii)
$\{x:f(x)\ne g(x)\}$ is not
negligible for any distinct $f$, $g\in K$.   Then $K$ is metrizable for
$\frak T_p$, which agrees with the topology of convergence in
measure on $K$.

\proof{{\bf (a)} Let $\sequencen{f_n}$ be a sequence in $K$.   Then it
has a pointwise convergent subsequence.   \Prf\ Because $K$ is compact,
we surely have $\sup_{f\in K}|f(x)|<\infty$ for every $x\in X$.
Let $\sequence{k}{X_k}$ be a sequence of measurable sets
of finite measure covering $X$, and set

\Centerline{$Y_k=\{x:x\in X_k$, $|f_n(x)|\le k$ for every $n\in\Bbb N\}$}

\noindent for $k\in\Bbb N$,

\Centerline{$q=\sum_{k=0}^{\infty}\Bover1{2^k(1+\mu Y_k)}\chi Y_k$,}

\noindent so that $q$ is a strictly positive measurable function and

\Centerline{$\|f_n\times q\|_2\le\sum_{k=0}^{\infty}\Bover{k}{2^k}=2$}

\noindent for every $n$.

Setting $g_n=f_n\times q$, the sequence $\sequencen{g_n^{\ssbullet}}$ of
equivalence
classes is a norm-bounded sequence in the Hilbert space $L^2=L^2(\mu)$.
It therefore has a weakly convergent subsequence
$\sequence{i}{g^{\ssbullet}_{n(i)}}$ say (4A4Kb), with limit $v$.

\Quer\ Suppose, if possible, that $\sequence{i}{f_{n(i)}}$ is not
pointwise convergent.   Then there must be some $x_0\in X$ such that
$\liminf_{i\to\infty}f_{n(i)}(x_0)<\limsup_{i\to\infty}f_{n(i)}(x_0)$;
let $\alpha<\beta$ in $\Bbb R$ be such that
$I=\{i:f_{n(i)}(x_0)\le\alpha\}$, $I'=\{i:f_{n(i)}(x_0)\ge\beta\}$ are
both infinite.   In this case
$v$ belongs to the weak closures of both
$D=\{g_{n(i)}^{\ssbullet}:i\in I\}$ and
$D'=\{g_{n(i)}^{\ssbullet}:i\in I'\}$.   It must therefore
belong to the {\it norm} closures of their convex hulls $\Gamma(D)$,
$\Gamma(D')$ (4A4Ed).   Accordingly we can find $v_n\in\Gamma(D)$,
$v'_n\in\Gamma(D')$ such that $\|v-v_n\|_2\le 3^{-n}$,
$\|v-v'_n\|_2\le 3^{-n}$ for every $n\in\Bbb N$.

Setting $A=\{f_{n(i)}:i\in I\}$, $A'=\{f_{n(i)}:i\in I'\}$, we see that
there must be $h_n\in\Gamma(A)$, $h'_n\in\Gamma(A')$ such that
$v_n=(h_n\times q)^{\ssbullet}$, $v'_n=(h'_n\times
q)^{\ssbullet}$ for every $n\in\Bbb N$.   Now if $g:X\to\Bbb R$ is
a measurable function such that $g^{\ssbullet}=v$, and 
$\tilde h=g/q$, we have

$$\eqalign{\mu\{x:|\tilde h(x)-h_n(x)|\ge\Bover1{2^nq(x)}\}
&=\mu\{x:|g(x)-(h_n\times q)(x)|\ge 2^{-n}\}\cr
&\le 4^n\|v-v_n\|_2^2
\le 2^{-n}\cr}$$

\noindent for every $n\in\Bbb N$, and $h_n\to\tilde h$ a.e.   Similarly,
$h'_n\to\tilde h$ a.e.

At this point, recall that $K$ is supposed to be convex, so all the
$h_n$, $h'_n$ belong to $K$.   Let $\Cal F$ be any non-principal
ultrafilter on $\Bbb N$.   Because $K$ is pointwise compact,
$h=\lim_{n\to\Cal F}h_n$ and $h'=\lim_{n\to\Cal F}h'_n$ are both defined
in $K$ for the topology of pointwise convergence.   For any $x$ such
that
$\lim_{n\to\infty}h_n(x)=\tilde h(x)$, we surely have $h(x)=\tilde
h(x)$;  so $h\eae\tilde h$.   Similarly, $h'\eae\tilde h$, and
$h\eae h'$.

Now at last we apply the hypothesis that distinct members of $K$ are not
equal almost everywhere, to see that $h=h'$.   But if we look at what
happens at the distinguished point $x_0$ above, we see that
$f(x_0)\le\alpha$ for every $f\in A$, so that $f(x_0)\le\alpha$ for
every $f\in\Gamma(A)$, $h_n(x_0)\le\alpha$ for every $n\in\Bbb N$, and
$h(x_0)\le\alpha$;  and similarly $h'(x_0)\ge\beta$.   So $h\ne h'$,
which is absurd.\ \Bang

This contradiction shows that $\sequence{i}{f_{n(i)}}$ is pointwise
convergent, and is an appropriate subsequence.\ \Qed

\medskip

{\bf (b)} Now 463Cd tells us that $K$ is metrizable for $\frak T_p$, and
that $\frak T_p$ agrees on $K$ with the topology of convergence in
measure.
}%end of proof of 463G

\leader{463H}{Corollary} Let $(X,\frak T,\Sigma,\mu)$ be a
$\sigma$-finite topological measure space in which $\mu$ is strictly
positive.   Suppose that

\inset{\noindent
whenever $h\in\BbbR^X$ is such that $h\restr Q$ is continuous
for every relatively countably compact $Q\subseteq X$, then $h$ is
continuous.}

\noindent If $K\subseteq C_b(X)$ is a norm-bounded $\frak T_p$-compact
set, then it is $\frak T_p$-metrizable.

\proof{ By 462L, the $\frak T_p$-closed convex hull
$\overline{\Gamma(K)}$ of $K$ in $C(X)$ is $\frak T_p$-compact.
Because $\mu$ is strictly positive, $\mu\{x:f(x)\ne g(x)\}>0$ whenever
$f$ and $g$ are distinct continuous real-valued functions on $X$.   So
the result is immediate from 463G.
}%end of proof of 463H

\vleader{60pt}{463I}{Lemma} Let $(X,\Sigma,\mu)$ be a perfect probability
space, and $\sequencen{E_n}$ a sequence in $\Sigma$.   Suppose that
there is an $\epsilon>0$ such that

\Centerline{$\epsilon\mu F\le\liminf_{n\to\infty}\mu(F\cap E_n)
\le\limsup_{n\to\infty}\mu(F\cap E_n)\le(1-\epsilon)\mu F$}

\noindent for
every $F\in\Sigma$.   Then $\sequencen{E_n}$ has a subsequence
$\sequence{k}{E_{n_k}}$ such that $\mu_*A=0$ and $\mu^*A=1$ for any
cluster point $A$ of $\sequence{k}{E_{n_k}}$ in $\Cal PX$;  in
particular, $\sequence{k}{E_{n_k}}$ has no measurable cluster point.

\proof{{\bf (a)} If we replace $\mu$ by its completion, we do not change
$\mu^*$ and $\mu_*$ (212Ea, 413Eg), so we may suppose that $\mu$ is
already complete.

\medskip

{\bf (b)} Suppose that $\sequencen{E_n}$ is actually stochastically
independent, with $\mu E_n=\bover1k$ for every $n$, where $k\ge 2$ is an
integer.   In this case $\mu^*A=1$ for any cluster point $A$ of
$\sequencen{E_n}$.

\medskip

\Prf\ {\bf (i)} There is a non-principal ultrafilter $\Cal F$ on
$\Bbb N$ such that $A=\lim_{n\to\Cal F}E_n$ in $\Cal PX$ (4A2F(a-ii));  
that
is, $\chi A(x)=\lim_{n\to\Cal F}\chi E_n(x)$ for every $x\in X$;  that
is, $A=\{x:x\in X,\,\{n:x\in E_n\}\in\Cal F\}$.

\medskip

\quad{\bf (ii)} We have a measurable function
$\phi:X\to Y=\{0,1\}^{\Bbb N}$ defined by setting
$\phi(x)(n)=(\chi E_n)(x)$ for every $x\in X$,
$n\in\Bbb N$.   Because $\mu$ is complete and perfect and totally
finite, and $\{0,1\}^{\Bbb N}$ is compact and metrizable, the image
measure $\nu=\mu\phi^{-1}$ is a Radon measure (451O).   
For any basic open set of the form $H=\{y:y(i)=\epsilon_i$ for
every $i\le n\}$, $\mu\phi^{-1}[H]=\tilde\nu H$, where $\tilde\nu$ is
the product measure corresponding to the measure $\nu_0$ on $\{0,1\}$
for which $\nu_0\{1\}=\bover1k$, $\nu_0\{0\}=\bover{k-1}{k}$.   Since
$\tilde\nu$ also is a Radon measure (416U), $\tilde\nu=\nu$ (415H(v)).

Set $B=\{y:y\in Y$, $\{n:y(n)=1\}\in\Cal F\}$, so that $\phi^{-1}[B]=A$
and $B$ is determined by coordinates in $\{n,n+1,\ldots\}$ for every
$n\in\Bbb N$.   By 451Pc, $\mu^*A=\nu^*B$;  by the zero-one law
(254Sa), $\nu^*B$ must be either 0 or 1.   So $\mu^*A$ is either 0 or 1.

\medskip

\quad{\bf (iii)} To see that $\mu^*A$ cannot be $0$, we repeat the
arguments of (ii) from the other side, as follows.   Let $\lambda_0$ be
the uniform probability measure on $\{0,1,\ldots,k-1\}$, giving measure
$\bover1k$ to each point;  let $\lambda$ be the corresponding product
measure on
$Z=\{0,\ldots,k-1\}^{\Bbb N}$.   Let $\psi:Z\to Y$ be defined by setting
$\psi(z)(n)=1$ if $z(n)=0$, $0$ otherwise;  then $\psi$ is \imp\ (254G).
Since $\lambda$ is a Radon measure and $\psi$ is continuous,
$\lambda\psi^{-1}$ is a Radon measure on $Y$ and must be equal to $\nu$.
Accordingly $\nu^*B=\lambda^*\psi^{-1}[B]$, by 451Pc again or
otherwise.

\medskip

\quad{\bf (iv)}
We have a measure space automorphism $\theta:Z\to Z$ defined by setting
$\theta(z)(n)=z(n)+_k1$ for every $z\in Z$, $n\in\Bbb N$, where $+_k$
is addition mod $k$.   So, writing $C=\psi^{-1}[B]$,
$\lambda^*C=\lambda^*\theta^i[C]$ for every $i\in\Bbb N$.   Now, for
$z\in Z$,

\Centerline{$\{n:z(n)=0\}\in\Cal F
\iff\{n:\psi(z)(n)=1\}\in\Cal F
\iff\psi(z)\in B
\iff z\in C$.}

\noindent But for any $z\in Z$, there is some $i<k$ such that
$\{n:z(n)=i\}\in\Cal F$, so that $\theta^{k-i}(z)\in C$.   Thus
$\bigcup_{i<k}\theta^i[C]=Z$ and
$\sum_{i=0}^{k-1}\lambda^*\theta^i[C]\ge 1$ and $\lambda^*C>0$.   But
this means that $\mu^*A=\nu^*B=\lambda^*C$ is non-zero, and $\mu^*A$
must be $1$.\ \Qed

\medskip

{\bf (c)} Now return to the general case considered in (a).   Note first
that $\mu$ is atomless, because if $\mu F>0$ there is some $n\in\Bbb N$
such that $0<\mu(F\cap E_n)<\mu F$.

Let $k\ge 2$ be such that $\bover1k<\epsilon$.   Then there are a
strictly increasing sequence $\sequence{i}{m(i)}$ in $\Bbb N$ and a
stochastically independent sequence $\sequence{i}{F_i}$ in $\Sigma$ such
that $F_i\subseteq E_{m(i)}$ and $\mu F_i=\bover1k$ for every
$i\in\Bbb N$.   \Prf\ Choose $\sequence{i}{m(i)}$, $F_i$ inductively, as
follows.   Let $\Sigma_i$ be the (finite) algebra generated by
$\{F_j:j<i\}$.   Choose $m(i)$ such that $m(i)>m(j)$ for any $j<i$ and
$\mu(F\cap E_{m(i)})\ge\bover1k\mu F$ for every $F\in\Sigma_i$.   List
the atoms of $\Sigma_i$ as $G_{i0},\ldots,G_{ip_i}$, and choose
$F_{ir}\subseteq E_{m(i)}\cap G_{ir}$ such that
$\mu F_{ir}=\bover1k\mu G_{ir}$, for each
$r\le p_i$;  215D tells us that this is possible.   Set
$F_i=\bigcup_{r\le p_i}F_{ir}$;  then $\mu(F_i\cap F)=\bover1k\mu F$ for
every $F\in\Sigma_i$, and $F_i\subseteq E_{m(i)}$.   Continue.
It is easy to check that $\mu(F_{i_1}\cap\ldots\cap F_{i_r})=1/k^r$
whenever $i_1<\ldots<i_r$, so that $\sequence{i}{F_i}$ is stochastically
independent.\ \Qed

If $A$ is a cluster point of $\sequence{i}{E_{m(i)}}$, then there is a
non-principal ultrafilter $\Cal F$ on $\Bbb N$ such that
$A=\lim_{i\to\Cal F}E_{m(i)}$ in $\Cal PX$.   In this case,
$A\supseteq A'$, where $A'=\lim_{i\to\Cal F}F_i$.
But (b) tells us that $\mu^*A'$
must be $1$, so $\mu^*A=1$.

\medskip

{\bf (d)} Thus we have a subsequence $\sequence{i}{E_{m(i)}}$ of
$\sequencen{E_n}$ such that any cluster point of
$\sequence{i}{E_{m(i)}}$ has outer measure $1$.   But the same argument
applies to $\sequence{i}{X\setminus E_{m(i)}}$ to show that there is a
strictly increasing sequence $\sequence{k}{i_k}$ such that every cluster
point of $\sequence{k}{X\setminus E_{m(i_k)}}$ has outer measure $1$.
Since complementation is a homeomorphism of $\Cal PX$, $\mu^*(X\setminus
A)=1$, that is, $\mu_*A=0$, for every cluster point $A$ of
$\sequence{k}{E_{m(i_k)}}$.   So if we set $n_k=m(i_k)$, any cluster
point of $\sequence{k}{E_{n_k}}$ will have inner measure 0 and outer
measure 1, as claimed.
}%end of proof of 463I

\leader{463J}{Lemma} Let $(X,\Sigma,\mu)$ be a perfect probability
space, and $\sequencen{E_n}$ a sequence in $\Sigma$.   Then

\inset{{\it either} $\sequencen{\chi E_n}$ has a subsequence which is
convergent almost everywhere}

\inset{{\it or} $\sequencen{E_n}$ has a subsequence with no measurable
cluster point in $\Cal PX$.}

\proof{ Consider the sequence $\sequencen{\chi E_n^{\ssbullet}}$ in the
Hilbert space $L^2=L^2(\mu)$.   This is a norm-bounded sequence, so has
a weakly convergent subsequence $\sequence{i}{\chi E_{n_i}^{\ssbullet}}$
with limit $v$ say (4A4Kb again).   Express $v$ as $g^{\ssbullet}$ where
$g:X\to\Bbb R$ is $\Sigma$-measurable.

\woddheader{463J}{4}{2}{2}{30pt}

\quad{\bf case 1} Suppose that $g(x)\in\{0,1\}$ for almost every $x\in
X$;  set $F=\{x:g(x)=1\}$.    Then

\Centerline{$\lim_{i\to\infty}\int_F\chi E_{n_i}=\int_Fg=\mu F$,
\quad$\lim_{i\to\infty}\int_{X\setminus F}\chi E_{n_i}
=\int_{X\setminus F}g=0$.}

\noindent So, replacing $\sequence{i}{\chi E_{n_i}}$ with a
sub-subsequence if necessary, we may suppose that

\Centerline{$|\mu F-\int_F\chi E_{n_i}|\le 2^{-i}$,
\quad$|\int_{X\setminus F}\chi E_{n_i}|\le 2^{-i}$}

\noindent for every $i$.   But as $0\le \chi E_{n_i}\le 1$ everywhere,
we have $\int|\chi F-\chi E_{n_i}|\le 2^{-i+1}$ for every $i$, so that $\chi E_{n_i}\to\chi F$ a.e., and we
have a subsequence of $\sequencen{\chi E_n}$ which is convergent almost
everywhere.

\medskip

\quad{\bf case 2} Suppose that $\{x:g(x)\notin\{0,1\}\}$ has positive
measure.   Note that because
$\int_Fg=\lim_{i\to\infty}\mu(F\cap E_{n_i})$ lies between $0$ and
$\mu F$ for every $F\in\Sigma$,
$0\le g\le 1$ a.e., and $\mu\{x:0<g(x)<1\}>0$.   There is therefore an
$\epsilon>0$ such that $\mu G>0$, where
$G=\{x:\epsilon\le g(x)\le 1-\epsilon\}$.

Write $\mu_G$ for the subspace measure on $G$, and $\Sigma_G$ for its
domain;  set $\nu=(\mu G)^{-1}\mu_G$, so that $\nu$ is a probability
measure.   We know that $\mu_G$ is perfect (451Dc), so $\nu$ also is
(see the definition in 451Ad).   Now if $F\in\Sigma_G$,

\Centerline{$\lim_{i\to\infty}\nu(F\cap E_{n_i})
=(\mu G)^{-1}\int_Fg$}

\noindent lies between $\epsilon\mu F/\mu G=\epsilon\nu F$ and
$(1-\epsilon)\nu F$.

By 463I, there is a strictly increasing sequence $\sequence{k}{i(k)}$
such that $B\notin\Sigma_G$ whenever $B$ is a cluster point of
$\sequence{k}{G\cap E_{n_{i(k)}}}$ in $\Cal PG$.
If $A$ is any cluster point of $\sequence{k}{E_{n_{i(k)}}}$ in
$\BbbR^X$, then $A\cap G$ is a cluster point of $\sequence{k}{G\cap
E_{n_{i(k)}}}$ in $\Cal PG$, so cannot belong to $\Sigma_G$.   Thus
$A\notin\Sigma$.

So in this case we have a subsequence $\sequence{k}{E_{n_{i(k)}}}$ of
$\sequencen{E_n}$ which has no measurable cluster point.
}%end of proof of 463J

\leader{463K}{Fremlin's Alternative}\cmmnt{ ({\smc Fremlin 75a})} Let
$(X,\Sigma,\mu)$ be a perfect $\sigma$-finite measure space, and
$\sequencen{f_n}$ a sequence of real-valued measurable functions on $X$.
Then

\inset{{\it either} $\sequencen{f_n}$ has a subsequence which is
convergent almost everywhere}

\inset{{\it or} $\sequencen{f_n}$ has a subsequence with no measurable
cluster point in $\BbbR^X$.}

\proof{{\bf (a)} If
$\mu X=0$ then of course $\sequencen{f_n}$ itself is convergent a.e., so
we may suppose that $\mu X>0$.   If there is any $x\in X$ such that
$\sup_{n\in\Bbb N}|f_n(x)|=\infty$, then $\sequencen{f_n}$ has a
subsequence with no cluster point in $\BbbR^X$, measurable or
otherwise;
so we may suppose that $\sequencen{f_n}$ is bounded at each point of
$X$.

\medskip

{\bf (b)} Let $\lambda$ be the c.l.d.\ product of $\mu$ with Lebesgue measure on
$\Bbb R$, and $\Lambda$ its domain.   Then $\lambda$ is perfect (451Ic)
and also $\sigma$-finite (251K).   There is therefore a probability
measure $\nu$ on $X\times\Bbb R$ with the same domain and the same
negligible sets
as $\lambda$ (215B(vii)), so that $\nu$ also is perfect.
For any function $h\in\BbbR^X$, write

\Centerline{$\Omega(h)=\{(x,\alpha):x\in X,\,\alpha\le h(x)\}
\subseteq X\times\Bbb R$}

\noindent (compare 252N).

\medskip

{\bf (c)} By 463J, applied to the measure space
$(X\times\Bbb R,\Lambda,\nu)$ and the sequence
$\sequencen{\chi\Omega(f_n)}$, we have
a strictly increasing sequence $\sequence{i}{n(i)}$ such that either
$\sequence{i}{\chi\Omega(f_{n(i)})}$ is convergent $\nu$-a.e.\ or
$\sequence{i}{\Omega(f_{n(i)})}$ has no cluster point in $\Lambda$.

\medskip

\quad{\bf case 1} Suppose that $\sequence{i}{\chi\Omega(f_{n(i)})}$ is
convergent $\nu$-a.e.   Set

\Centerline{$W=\{(x,\alpha):
\lim_{i\to\infty}\chi\Omega(f_{n(i)})(x,\alpha)$ is defined$\}$.}

\noindent Then $W$ is $\lambda$-conegligible, so
$W^{-1}[\{\alpha\}]=\{x:(x,\alpha)\in W\}$ is $\mu$-conegligible for
almost every $\alpha$ (apply 252D to the complement of $W$).   Set
$D=\{\alpha:W^{-1}[\{\alpha\}]$ is $\mu$-conegligible$\}$, and let
$Q\subseteq D$ be a countable dense set;  then
$G=\bigcap_{\alpha\in Q}W^{-1}[\{\alpha\}]$ is $\mu$-conegligible.   But if $x\in G$, then for
any $\alpha\in Q$ the set
$\{i:f_{n(i)}(x)\ge\alpha\}=\{i:\chi\Omega(f_{n(i)})(x,\alpha)=1\}$
is either finite or has finite complement in $\Bbb N$, so
$\sequence{i}{f_{n(i)}(x)}$ must be convergent in $[-\infty,\infty]$.
Since $\sequencen{f_n(x)}$ is supposed to be bounded,
$\sequence{i}{f_{n(i)}(x)}$ is convergent in $\Bbb R$.   Thus in this
case we have an almost-everywhere-convergent subsequence of
$\sequencen{f_n}$.

\medskip

\quad{\bf case 2} Suppose that $\sequence{i}{\Omega(f_{n(i)})}$ has no
cluster point in $\Lambda$.   Let $h$ be any cluster point of
$\sequence{i}{f_{n(i)}}$ in $\BbbR^X$.   Then there is a non-principal
ultrafilter $\Cal F$ on $\Bbb N$ such that $h=\lim_{i\to\Cal F}f_{n(i)}$
in $\BbbR^X$.   Set $A=\lim_{i\to\Cal F}\Omega(f_{n(i)})$, so that
$A\notin\Lambda$.   If $x\in X$ and $\alpha\in\Bbb R$, then

\Centerline{$\alpha<h(x)\Longrightarrow\{i:\alpha<f_{n(i)}(x)\}\in\Cal F
\Longrightarrow(x,\alpha)\in A$,}

\Centerline{$h(x)<\alpha\Longrightarrow\{i:\alpha<f_{n(i)}(x)\}\notin\Cal F
\Longrightarrow(x,\alpha)\notin A$.}

\noindent Thus $\Omega'(h)\subseteq A\subseteq\Omega(h)$, where
$\Omega'(h)=\{(x,\alpha):\alpha<h(x)\}$.

\Quer\ If $h$ is $\Sigma$-measurable, then

\Centerline{$\Omega'(h)
=\bigcup_{q\in\Bbb Q}\{x:h(x)>q\}\times\ocint{-\infty,q}$,}

\Centerline{$\Omega(h)
=(X\times\Bbb R)\setminus\bigcup_{q\in\Bbb Q}
   \{x:h(x)<q\}\times\coint{q,\infty}$}

\noindent belong to $\Lambda$, and
$\lambda(\Omega(h)\setminus\Omega'(h))=0$ (because every vertical
section of $\Omega(h)\setminus\Omega'(h)$ is negligible).   But as
$\Omega'(h)\subseteq A\subseteq\Omega(h)$, $A\in\Lambda$ (remember that
product measures in this book are complete), which is
impossible.\ \Bang

Thus $h$ is not $\Sigma$-measurable.   As $h$ is arbitrary,
$\sequence{i}{f_{n(i)}}$ has no measurable cluster point in $\BbbR^X$.

So at least one of the envisaged alternatives must be true.
}%end of proof of 463K

\leader{463L}{Corollary} Let $(X,\Sigma,\mu)$ be a perfect
$\sigma$-finite measure space.   Write $\eusm L^0\subseteq\BbbR^X$ for
the space of real-valued $\Sigma$-measurable functions on $X$.

(a) If $K\subseteq\eusm L^0$ is relatively countably compact for the
topology $\frak T_p$ of pointwise convergence on $\eusm L^0$, then every
sequence in
$K$ has a subsequence which is convergent almost everywhere.
Consequently $K$ is relatively compact in $\eusm L^0$ for the topology
$\frak T_m$ of convergence in measure.

(b) If $K\subseteq\eusm L^0$ is countably compact for $\frak T_p$, then
it is compact for $\frak T_m$.

(c) Suppose that $K\subseteq\eusm L^0$ is countably compact for
$\frak T_p$ and that $\mu\{x:f(x)\ne g(x)\}>0$ for any distinct $f$,
$g\in K$.   Then the topologies $\frak T_m$ and $\frak T_p$ agree on $K$,
so both are compact and metrizable.

\proof{{\bf (a)} Since every sequence in $K$ must have a
$\frak T_p$-cluster point in $\eusm L^0$, 463K tells us that every
sequence in $K$ has a subsequence which is convergent almost everywhere,
therefore $\frak T_m$-convergent.   Now $K$ is relatively sequentially
compact in the pseudometrizable space $(\eusm L^0,\frak T_m)$, therefore
relatively compact (4A2Le again).

\medskip

{\bf (b)} As in (a), every sequence $\sequencen{f_n}$ in $K$ has a
subsequence $\sequencen{g_n}$ which is convergent almost everywhere.
But $\sequencen{g_n}$ has a $\frak T_p$-cluster point $g$ in $K$, and
now $g(x)=\lim_{n\to\infty}g_n(x)$ for every $x$ for which the limit is
defined;  accordingly $g_n\to g$ a.e., and $g$ is a $\frak T_m$-limit of
$\sequencen{g_n}$ in $K$.   Thus every sequence in $K$ has a $\frak
T_m$-cluster point in
$K$, and (because $\frak T_m$ is pseudometrizable) $K$ is
$\frak T_m$-compact.

\medskip

{\bf (c)} The point is that $K$ is sequentially compact under
$\frak T_p$.   \Prf\ Note that as $K$ is countably compact,
$\sup_{f\in K}|f(x)|$ is finite for every $x\in K$.   (I am passing over
the trivial
case $K=\emptyset$.)   If $\sequencen{f_n}$ is a sequence in $K$, then,
by (a), it has a subsequence $\sequencen{g_n}$ which is convergent a.e.
\Quer\ If $\sequencen{g_n}$ is not $\frak T_p$-convergent, then
there are a point $x_0\in X$ and two further subsequences
$\sequencen{g'_n}$,
$\sequencen{g''_n}$ of $\sequencen{g_n}$ such that
$\lim_{n\to\infty}g'_n(x_0)$, $\lim_{n\to\infty}g''_n(x_0)$ exist and
are different.   Now $\sequencen{g'_n}$,
$\sequencen{g''_n}$ must have cluster points $g'$, $g''\in K$ with
$g'(x_0)\ne g''(x_0)$.

However,

\Centerline{$g'(x)=\lim_{n\to\infty}g_n(x)=g''(x)$}

\noindent whenever the limit is defined, which is almost everywhere;  so
$g'\eae g''$.   And this contradicts the hypothesis that if two elements
of $K$ are equal a.e., they are identical.\  \BanG\  Thus
$\sequencen{g_n}$ is a $\frak T_p$-convergent subsequence of
$\sequencen{f_n}$.   As $\sequencen{f_n}$ is arbitrary, $K$ is
$\frak T_p$-sequentially compact.\ \Qed

Now 463Cd gives the result.
}%end of proof of 463L

\leader{463M}{Proposition} Let $X_0,\ldots,X_n$ be countably
compact topological spaces, each carrying a $\sigma$-finite 
perfect strictly positive measure which measures every Baire set.
Let $X$ be their product and $\CalBa(X_i)$ the Baire $\sigma$-algebra of
$X_i$ for each $i$.   
Then any separately continuous function $f:X\to\Bbb R$ is measurable
with respect to the $\sigma$-algebra
$\Tensorhat_{i\le n}\CalBa(X_i)$ generated by 
$\{\prod_{i\le n}E_i:E_i\in\CalBa(X_i)$ for $i\le n\}$.

\proof{ For $i\le n$ let $\mu_i$ be a $\sigma$-finite perfect
strictly positive measure on $X_i$ such that 
$\CalBa(X_i)\subseteq\dom\mu_i$;
let $\mu$ be the product measure on $X$.

\medskip

{\bf (a)} The proof relies on the fact that

\inset{($*$) if $g$, $g':X\to\Bbb R$ are distinct separately continuous
functions, then $\mu\{x:g(x)\ne g'(x)\}>0$;}

\noindent I seek to prove this, together with the stated result, by
induction on $n$.   The induction starts easily with $n=0$, so that $X$
can be identified with $X_0$, a separately continuous function on $X$ is
just a continuous function on $X_0$, and ($*$) is true because
$\mu=\mu_0$ is strictly positive.

\medskip

{\bf (b)} For the inductive step to $n+1$, given a separately continuous
function $f:X_0\times\ldots\times X_{n+1}\to\Bbb R$, set $f_t(y)=f(y,t)$
for every $y\in Y=X_0\times\ldots\times X_n$ and $t\in X_{n+1}$,
and $K=\{f_t:t\in X_{n+1}\}$   Then
every $f_t$ is separately continuous, therefore
$\Tensorhat_{i\le n}\CalBa(X_i)$-measurable, by the inductive 
hypothesis.   So $K$ consists of 
$\Tensorhat_{i\le n}\CalBa(X_i)$-measurable functions.  Moreover, again
because $f$ is separately continuous, the function $t\mapsto f_t(y)$ is
continuous for every $y$, that is, $t\mapsto f_t:X_{n+1}\to\BbbR^Y$ is
continuous;  it follows that $K$ is countably compact (4A2G(f-vi)).
Finally, by the inductive hypothesis ($*$),
$\nu\{y:f_t(y)\ne f_{t'}(y)\}>0$ whenever $t$, $t'\in X_{n+1}$ and
$f_t\ne f_{t'}$, where $\nu$ is the product measure on $Y$.

Since $\nu$ is perfect (451Ic) and 
$\Tensorhat_{i\le n}\CalBa(X_i)\subseteq\dom\nu$,
we can apply 463Lc to see that $K$ is metrizable for the
topology of pointwise convergence.   Let $\rho$ be a metric on $K$
inducing its
topology, and $\sequence{i}{g_i}$ a sequence running over a dense subset
of $K$.   (I am passing over the trivial case $K=\emptyset=X_{n+1}$.)
For $m$, $i\in\Bbb N$ set $E_{mi}=\{t:\rho(f_t,g_i)\le 2^{-m}\}$.
Because $t\mapsto f_t$ and $t\mapsto\rho(f_t,t_i)$ are continuous,
$E_{mi}\in\CalBa(X_{n+1})$.   Set
$f^{(m)}(y,t)=g_i(y)$ for $t\in E_{mi}\setminus\bigcup_{j<i}E_{mj}$
for $m$, $i\in\Bbb N$, $y\in Y$ and $t\in X_{n+1}$.   
Then $f^{(m)}:X\to\Bbb R$ is
$\Tensorhat_{i\le n+1}\CalBa(X_i)$-measurable because every $g_i$ is
$\Tensorhat_{i\le n}\CalBa(X_i)$-measurable and
every $E_{mi}$ belongs to $\CalBa(X_{n+1})$.
Also $\sequence{m}{f^{(m)}}\to f$ at every point, because
$\rho(f^{(m)}_t,f_t)\le 2^{-m}$ for every $m\in\Bbb N$ and
$t\in X_{n+1}$.   So $f$ is $\Tensorhat_{i\le n+1}\CalBa(X_i)$-measurable.

\medskip

{\bf (c)} We still have to check that ($*$) is true at the new level.
But if $h$, $h':X\to\Bbb R$ are distinct separately continuous
functions, then there are $t_0\in X_{n+1}$, $y_0\in Y$ such that
$h(y_0,t_0)\ne h'(y_0,t_0)$.   Let $G$ be an open set containing $t_0$
such that $h(y_0,t)\ne h'(y_0,t)$ whenever $t\in G$.   Then
$\nu\{y:h(y,t)\ne h'(y,t)\}>0$ for every $t\in G$, by the inductive
hypothesis, so

\Centerline{$\mu\{(y,t):h(y,t)\ne h'(y,t)\}
=\int\nu\{y:h(y,t)\ne h'(y,t)\}\mu_{n+1}(dt)>0$}

\noindent because $\mu_{n+1}$ is strictly positive.   Thus the induction
continues.
}%end of proof of 463M

\leader{463N}{Corollary} Let $X_0,\ldots,X_n$ be Hausdorff spaces with
product $X$.   Then every separately continuous function
$f:X\to\Bbb R$ is universally Radon-measurable\cmmnt{ in the sense of
434Ec}.

\proof{ Let $\mu$ be a Radon measure on $X$ and $\Sigma$ its
domain.

\medskip

{\bf (a)} Suppose first that the support $C$ of $\mu$ is compact.
For each $i\le n$, let $\pi_i:X\to X_i$ be the coordinate
projection, and $\mu_i=\mu\pi_i^{-1}$ the image Radon measure;  let
$Z_i$ be the support of $\mu_i$ and $Z=\prod_{i\le n}Z_i$.   Note that
$\pi_i[C]$ is compact and $\mu_i$-conegligible, so that 
$Z_i\subseteq\pi_i[C]$ is compact, for each $i$.   At the same time,
$\pi^{-1}[Z_i]$ is $\mu$-conegligible for each $i$, so that
$Z$ is $\mu$-conegligible.

By 463M, $f\restr Z$ is $\Tensorhat_{i\le n}\CalBa(Z_i)$-measurable;  
because $Z$ is conegligible, $f$ is $\Sigma$-measurable.

\medskip

{\bf (b)} In general, if $C\subseteq X$ is compact, then we can apply
(a) to the measure $\mu\LLcorner C$ (234M) to see that $f\restr C$ is
$\Sigma$-measurable.   As $\mu$ is complete and locally determined and inner
regular with respect to the compact sets, $f$ is $\Sigma$-measurable 
(see 412Ja).   

As $\mu$ is arbitrary, $f$ is universally Radon-measurable.
}%end of proof of 463N

\exercises{\leader{463X}{Basic exercises $\pmb{>}$(a)}
%\spheader 463Xa
Let $(X,\Sigma,\mu)$ be a measure space, $\eusm L^0$ the
space of $\Sigma$-measurable real-valued functions on $X$, $\frak T_p$
the topology of pointwise convergence on $\eusm L^0$ and $\frak T_m$ the
topology of convergence in measure on $\eusm L^0$.   (i) Show that $\frak T_m\subseteq\frak T_p$ iff for every measurable set $E$ of finite
measure there is a countable set $D\subseteq E$ such that
$\mu^*D=\mu E$.  (ii) Show that $\frak T_p\subseteq\frak T_m$ iff
$0<\mu^*\{x\}<\infty$ for every $x\in X$.
%463A

\spheader 463Xb Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space,
and $K\subseteq\eusm L^0$ a $\frak T_p$-countably compact set.
Show that the following are equiveridical:  (i) every sequence in $K$
has a subsequence which converges almost everywhere;  (ii) $K$ is
$\frak T_m$-compact;   (iii) $K$ is totally bounded for the uniformity
associated with the linear space topology $\frak T_m$.   Show that if
moreover the topology on $K$ induced by $\frak T_m$ is Hausdorff, then $K$ is $\frak T_p$-metrizable.
%463C

\spheader 463Xc(i) Show that there is a set of Borel measurable
functions on $[0,1]$ which is countably tight, compact and
non-metrizable
for the topology of pointwise convergence.   (ii) Show that there is a
strictly localizable measure space $(X,\Sigma,\mu)$ with a set $K$ of
measurable functions which is countably tight, compact, Hausdorff and
non-metrizable for both the topology of pointwise convergence and the
topology of convergence in measure.   \Hint{the one-point
compactification of any discrete space is countably tight.}
%463E

\spheader 463Xd Let $X$ be a topological space and
$K\subseteq C(X)$ a convex $\frak T_p$-compact set.   Show that if there is a strictly positive $\sigma$-finite topological measure on $X$, then $K$ is $\frak T_p$-metrizable.
%463G

\spheader 463Xe Use Koml\'os's theorem (276H) to shorten the proof of
463G.
%463G

\spheader 463Xf Let $(X,\Sigma,\mu)$ be any complete $\sigma$-finite
measure space.   Show that if $A\subseteq\eusm L^0$ is
$\frak T_m$-relatively compact, and $\sup_{f\in A}|f(x)|$ is finite for
every $x\in X$, then $A$ is $\frak T_p$-relatively countably compact in
$\eusm L^0$.
%463K

\spheader 463Xg Let $K$ be the set of non-decreasing functions from
$[0,1]$ to $\{0,1\}$.   Show that $K$, with its topology of pointwise
convergence, is homeomorphic to the split interval (419L).   Show that
(for any Radon measure $\mu$ on $[0,1]$) the identity map from
$(K,\frak T_p)$ to $(K,\frak T_m)$ is continuous.
%463L

\sqheader 463Xh Let $K$ be the set of non-decreasing functions from
$\omega_1$ to $\{0,1\}$.
%Show that $K$, with its topology of pointwise
%convergence, is homeomorphic to $\omega_1+1$.
Show that if $\mu$ is
the countable-cocountable measure on $\omega_1$ then $K$ is a
$\frak T_p$-compact set of measurable functions and is also
$\frak T_m$-compact, but the identity map from $(K,\frak T_p)$ to
$(K,\frak T_m)$ is not continuous.
%463L, 463Xg, 463F

\sqheader 463Xi Let $K$ be the set of functions $f:[0,1]\to\Bbb R$ such
that $\max(\|f\|_{\infty},\Var_{[0,1]}f)\le 1$, where $\Var_{[0,1]}f$ is
the variation
of $f$ (224A).   Show that $K$ is $\frak T_p$-compact and that (for any
Radon measure on $[0,1]$) the identity map from $(K,\frak T_p)$  to
$(K,\frak T_m)$ is continuous.
%463L, 463Xg

\sqheader 463Xj Let $A$ be the set of functions $f:[0,1]\to[0,1]$ such
that $\overline{\int}|f|d\mu\cdot\Var_{[0,1]}f\le 1$, where $\mu$ is
Lebesgue measure.   Show that every member of $A$ is measurable and that
every sequence in $A$ has a subsequence
which converges almost everywhere to a member of $A$, but that
$\int:A\to[0,1]$ is not $\frak T_p$-continuous, while $A$ is
$\frak T_p$-dense in $[0,1]^{[0,1]}$.
%463L, 463Xi

\spheader 463Xk Let $X$ be a Hausdorff space and $K\subseteq C(X)$ a
$\frak T_p$-compact set.   Show that if there is a strictly positive
$\sigma$-finite Radon measure on $X$ then $K$ is $\frak T_p$-metrizable.
%463L

\spheader 463Xl Let $(X,\Sigma,\mu)$ be a localizable measure space and
$K\subseteq\eusm L^0$ a non-empty $\frak T_p$-compact set.
Show that $\sup\{f^{\ssbullet}:f\in K\}$ is defined in $L^0(\mu)$.
%463L

\leader{463Y}{Further exercises (a)}
%\spheader 463Ya
Let $(X,\Sigma,\mu)$ be a probability space and $V$ a Banach space.   A
function $\phi:X\to V$ is {\bf scalarly measurable} (often called {\bf
weakly measurable}) if
$h\phi:X\to\Bbb R$ is $\Sigma$-measurable for every $h\in V^*$.
$\phi$ is {\bf Pettis integrable}, with {\bf indefinite
Pettis integral} $\theta:\Sigma\to V$, if $\int_Eh\phi\,d\mu$ is defined
and equal to $h(\theta E)$ for every $E\in\Sigma$ and every $h\in V^*$.
(i) Show that if $\phi$ is
scalarly measurable, then $K=\{h\phi:h\in V^*$, $\|h\|\le 1\}$ is a
$\frak T_p$-compact subset of $\eusm L^0$.   (ii) Show that if
$\phi$ is scalarly measurable, then it is Pettis integrable iff every
function in $K$ is integrable and $f\mapsto\int_Ef:K\to\Bbb R$ is
$\frak T_p$-continuous for every $E\in\Sigma$.   \Hint{4A4Cg.}   (iii)
In particular, if $\phi$ is bounded and scalarly measurable and the
identity map from $(K,\frak T_p)$ to $(K,\frak T_m)$ is continuous, then
$\phi$ is Pettis integrable.   (See {\smc Talagrand 84}, chap.\ 4.)

\spheader 463Yb Show that any Bochner integrable function (253Yf) is
Pettis integrable.
%463Ya

\spheader 463Yc Let $\mu$ be Lebesgue measure on $[0,1]$, and define
$\phi:[0,1]\to L^{\infty}(\mu)$ by setting
$\phi(t)=\chi[0,t]^{\ssbullet}$ for every $t\in[0,1]$.   (i) Show that
if $h\in L^{\infty}(\mu)^*$ and $\|h\|\le 1$, then $h\phi$ has variation
at most 1.  (ii) Show that $K=\{h\phi:h\in L^{\infty}(\mu)^*,\,\|h\|\le
1\}$ is a $\frak T_p$-compact set of Lebesgue measurable functions, and
that the identity map from $(K,\frak T_p)$ to $(K,\frak T_m)$ is
continuous, so that $\phi$ is Pettis integrable.   (iii) Show that
$\phi$ is not Bochner integrable.
%463Xi, 463Ya, 463Yb

\spheader 463Yd Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space
and suppose that $\mu$ is inner regular with respect to some family
$\Cal E\subseteq\Sigma$ with cardinal at most $\omega_1$.   (Subject to
the continuum hypothesis, this is true for any subset of $\Bbb R$, for
instance.)   Show that if $K\subseteq\eusm L^0$ is
$\frak T_p$-compact then it is $\frak T_m$-compact.   (See
536C, or {\smc Talagrand 84}, 9-3-3.)
%463L

\spheader 463Ye Assume that the continuum hypothesis is true;  let
$\preccurlyeq$ be a well-ordering of $[0,1]$ with order type $\omega_1$
(4A1Ad).   Let $(Z,\nu)$ be the Stone space of the measure algebra of
Lebesgue measure on $[0,1]$, and $q:Z\to[0,1]$ the canonical \imp\ map
(416V). Let $g:[0,1]\to\coint{0,\infty}$ be any function.   Show that
there is a
function $f:[0,1]\times Z\to\coint{0,\infty}$ such that ($\alpha$) $f$
is continuous in the second variable ($\beta$) $f(t,z)=0$ whenever
$q(z)\preccurlyeq t$ ($\gamma$) $\int f(t,z)\nu(dz)=g(t)$ for every
$t\in[0,1]$.   Show that $f$ is universally measurable in the first
variable, but need not be $\tilde\Lambda$-measurable, where $\tilde\Lambda$
is the domain of the product Radon measure on
$[0,1]\times Z$.   Setting $f_z(t)=f(t,z)$, show that $K=\{f_z:z\in Z\}$
is a $\frak T_p$-compact set of Lebesgue measurable functions and that
$g$ belongs to the $\frak T_p$-closed convex hull of $K$ in
$\BbbR^{[0,1]}$.
%463N
}%end of exercises

\leader{463Z}{Problems} {\bf (a) A.Bellow's problem}
%463Za
Let $(X,\Sigma,\mu)$ be a probability space, and
$K\subseteq\eusm L^0$ a $\frak T_p$-compact set such that
$\{x:f(x)\ne g(x)\}$ is
non-negligible for any distinct functions $f$, $g\in K$, as in 463G and
463Lc.   Does it follow that $K$ is metrizable for $\frak T_p$?

\cmmnt{A positive answer would displace several of the arguments of
this section, and have other consequences (see 462Z, for instance).   It
is known that under any of a variety of special axioms (starting with
the continuum hypothesis) there is indeed a positive answer;
see \S536 in Volume 5, or {\smc Talagrand 84}, chap.\ 12.}

\spheader 463Zb Let $X\subseteq[0,1]$ be a set of outer Lebesgue measure
$1$, and $\mu$ the subspace measure on $X$, with $\Sigma$ its domain.
Let $K$ be a $\frak T_p$-compact subset of $\eusm L^0$.   Must
$K$ be $\frak T_m$-compact?

\spheader 463Zc Let $X_0,\ldots,X_n$ be compact Hausdorff spaces and
$f:X_0\times\ldots\times X_n\to\Bbb R$ a separately continuous function.
Must $f$ be universally measurable?

\endnotes{
\Notesheader{463} The relationship between the topologies $\frak T_p$
and $\frak T_m$ is complex, and I do not think that the results here are
complete;  in particular, we have a remarkable outstanding problem in
463Za.   Much of the work presented here has been stimulated by problems
concerning the integration of vector-valued functions.   I am keeping
this theory firmly in the `further exercises' (463Ya-463Yc), but it is
certainly the most important source of examples of pointwise compact
sets of measurable functions.   In particular, since the set
$\{h\phi:h\in V^*,\,\|h\|\le 1\}$ is necessarily convex whenever $V$ is
a Banach space and $\phi:X\to V$ is a function, we are led to look at
the special properties of convex sets, as in 463G.   There are obvious
connexions with the theory of measures on linear topological spaces,
which I will come to in \S466.

The dichotomy in 463K shows that sets of measurable functions on perfect
measure spaces are either `good' (relatively countably compact for
$\frak T_p$, relatively compact for $\frak T_m$) or `bad' (with neither
property).   It is known that the result is not true for arbitrary
$\sigma$-finite measure spaces (see \S464 below), but it is not clear
whether there are important non-perfect spaces in which it still applies
in some form;  see 463Zb.

Just as in \S462, many questions concerning the topology $\frak T_p$ on
$\BbbR^X$ can be re-phrased as questions about real-valued functions on
products $X\times K$ which are continuous in the second variable.   For
the topology of pointwise convergence on sets of measurable functions,
we find ourselves looking at functions which are measurable in the first
variable.   In this way we are led to such results as 463M-463N and
463Ye.   Concerning 463M and 463Zc, it is the case that
if $X$ and $Y$ are {\it any} compact Hausdorff spaces, and
$f:X\times Y\to\Bbb R$ is separately continuous, then $f$ is Borel
measurable\cmmnt{ ({\smc Burke \& Pol 05}, 5.2)}.

A substantial proportion of the questions which arise naturally in this
topic are known to be undecidable without using special axioms.   I am
avoiding such questions in this volume, but it is worth noting that the
continuum hypothesis, in particular, has many striking consequences
here, of which 463Ye is a sample.   It also decides 463Za and 463Zb (see
463Yd).

}%end of notes

\discrpage

\frfilename{mt464.tex}
\versiondate{25.5.13}
\copyrightdate{1999}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Talagrand's measure}

\newsection{464}

An obvious question arising from 463I and its corollaries is, do we
really need the
hypothesis that the measure involved is perfect?   A very remarkable
construction by M.Talagrand (464D) shows that these results are
certainly not true of all probability spaces (464E).   Investigating the
properties of this measure we are led to some surprising facts about
additive functionals on algebras $\Cal PI$ and the duals of
$\ell^{\infty}$ spaces (464M, 464R).

\leader{464A}{The usual measure on $\Cal PI$} Recall\cmmnt{ from
254J and 416U} that for any set $I$ we have a standard measure $\nu$, a
Radon measure for the usual topology on $\Cal PI$\cmmnt{, defined by
saying that $\nu\{a:a\subseteq I,\,a\cap J=c\}=2^{-\#(J)}$ whenever
$J\subseteq I$ is a finite set and $c\subseteq J$, or by copying from
the usual product measure on $\{0,1\}^I$ by means of the bijection
$a\mapsto\chi a:\Cal PI\to\{0,1\}^I$}.   \cmmnt{We shall need a couple
of simple facts about these measures.}

\spheader 464Aa If $\family{j}{J}{I_j}$ is any partition of $I$, then
$\nu$ can be identified with the product of the family
$\family{j}{J}{\nu_j}$,
where $\nu_j$ is the usual measure on $\Cal PI_j$\cmmnt{, and we
identify $\Cal PI$ with $\prod_{j\in J}\Cal PI_j$ by matching
$a\subseteq I$ with
$\family{j}{J}{a\cap I_j}$;  this is the `associative law' 254N}.   It
follows that if we have any family $\family{j}{J}{A_j}$ of subsets of
$\Cal PI$, and if for each $j$ the set $A_j$ is `determined by
coordinates in $I_j$' in the sense that, for $a\subseteq I$, $a\in A_j$
iff $a\cap I_j\in A_j$, then
$\nu^*(\bigcap_{j\in J}A_j)=\prod_{j\in J}\nu^*A_j$\cmmnt{ (use
254Lb)}.

\spheader 464Ab Similarly, if $f_1$, $f_2$ are non-negative real-valued
functions on $\Cal PI$, and if there are disjoint sets $I_1$,
$I_2\subseteq I$ such that $f_j(a)=f_j(a\cap I_j)$ for every
$a\subseteq I$ and both $j$, then the upper integral
$\overline{\int}f_1+f_2\,d\nu$ is
$\overline{\int}f_1d\nu+\overline{\int}f_2d\nu$.   \prooflet{\Prf\ We
may suppose that $I_2=I\setminus I_1$.   For
each $j$, define $g_j:\Cal PI_j\to\coint{0,\infty}$ by setting
$g_j=f_j\restrp\Cal PI_j$, so that $f_j(a)=g_j(a\cap I_j)$ for every
$a\subseteq I$.   Let $\nu_j$ be the usual measure on $\Cal PI_j$, so
that we can identify $\nu$ with the product measure $\nu_1\times\nu_2$,
if we identify $\Cal PI$ with $\Cal PI_1\times\Cal PI_2$;  that is, we
think of a subset of $I$ as a pair $(a_1,a_2)$ where $a_j\subseteq I_j$
for both $j$.

Now we have

$$\eqalignno{\overline{\int}f_1+f_2\,d\nu
&=\overline{\int}g_1\,d\nu_1
   +\overline{\int}g_2\,d\nu_2\cr
\noalign{\noindent (253K)}
&=\overline{\int}g_1\,d\nu_1\cdot\overline{\int}\chi(\Cal PI_2)d\nu_2
   +\overline{\int}\chi(\Cal PI_1)d\nu_1\cdot\overline{\int}g_2\,d\nu_2\cr
&=\overline{\int}f_1\,d\nu+\overline{\int}f_2\,d\nu\cr}$$

\noindent by 253J, because we can think of $f_1(a_1,a_2)$ as
$g_1(a_1)\cdot(\chi\Cal PI_2)(a_2)$ for all $a_1$, $a_2$.\ \Qed}

\spheader 464Ac If $A\subseteq\Cal PI$ is such that $b\in A$ whenever
$a\in A$, $b\subseteq I$ and $a\symmdiff b$ is finite, then $\nu^*A$
must be either $0$ or $1$\cmmnt{;  this is the zero-one law 254Sa,
applied to the set
$\{\chi a:a\in A\}\subseteq\{0,1\}^I$ and the usual measure on
$\{0,1\}^I$}.

\vleader{72pt}{464B}{Lemma} Let $I$ be any set, and $\nu$ the usual measure on
$\Cal PI$.

(a)(i) There is a sequence $\sequencen{m(n)}$ in $\Bbb N$ such that
$\prod_{n=0}^{\infty}1-2^{-m(n)}=\bover12$.

\quad(ii) Given such a sequence, write $X$ for
$\prod_{n\in\Bbb N}(\Cal PI)^{m(n)}$, and let $\lambda$ be the product
measure on $X$.   We have a function
$\phi:X\to\Cal PI$ defined by setting

\Centerline{$\phi(\sequencen{\langle a_{ni}\rangle_{i<m(n)}})
=\bigcup_{n\in\Bbb N}\bigcap_{i<m(n)}a_{ni}$}

\noindent whenever $\sequencen{\langle a_{ni}\rangle_{i<m(n)}}\in X$.
Now $\phi$ is \imp\ for $\lambda$ and $\nu$.

(b) The map

\Centerline{$(a,b,c)\mapsto(a\cap b)\cup(a\cap c)\cup(b\cap c):
(\Cal PI)^3\to\Cal PI$}

\noindent is \imp\ for the product measure on $(\Cal PI)^3$.

\proof{{\bf (a)(i)} Choose $m(n)$ inductively so that, for each $n$ in
turn, $m(n)$ is minimal subject to the requirement
$\prod_{k=0}^n1-2^{-m(k)}>\bover12$.

\medskip

\quad{\bf (ii)} If $t\in I$, then

\Centerline{$\{\frak x:\frak x\in X,\,t\notin\phi(\frak x)\}
=\{\sequencen{\langle a_{ni}\rangle_{i<m(n)}}:
  t\notin\bigcup_{n\in\Bbb N}\bigcap_{i<m(n)}a_{ni}\}$}

\noindent has measure

\Centerline{$\prod_{n=0}^{\infty}
\lambda_n\{\ofamily{i}{m(n)}{a_i}:t\notin\bigcap_{i<m(n)}a_i\}$,}

\noindent where $\lambda_n$ is the product measure on $(\Cal PI)^{m(n)}$
for each $n$.   But this is just

\Centerline{$\prod_{n=0}^{\infty}1-2^{-m(n)}=\Bover12$,}

\noindent by the choice of $\sequencen{m(n)}$.   Accordingly
$\lambda\{\frak x:t\in\phi(\frak x)\}=\bover12$ for every $t\in I$.
Next, if we identify $X$ with
$\Cal P(\{(t,n,i):t\in I$, $n\in\Bbb N$, $i<m(n)\})$, each set
$E_t=\{\frak x:t\in\phi(\frak x)\}$ is determined by
coordinates in $J_t=\{(t,n,i):n\in\Bbb N,\,i<m(n)\}$.   Since the sets
$J_t$ are disjoint, the sets $E_t$, for different $t$, are
stochastically independent (464Aa), so if $J\subseteq I$ is finite,

\Centerline{$\lambda\{\frak x:J\subseteq\phi(\frak x)\}
=\prod_{t\in J}\lambda E_t=2^{-\#(J)}=\nu\{a:J\subseteq a\}$.}

This shows that $\lambda\phi^{-1}[F]=\nu F$ whenever $F$ is of the form
$\{a:J\subseteq a\}$ for some finite $J\subseteq I$.   By the Monotone
Class Theorem (136C), $\lambda\phi^{-1}[F]=\nu F$ for every $F$
belonging to the $\sigma$-algebra generated by sets of this form.   But
this $\sigma$-algebra certainly contains all sets of the form
$\{a:a\cap J=c\}$ where $J\subseteq I$ is finite and $c\subseteq J$,
which are the
sets corresponding to the basic cylinder sets in the product
$\{0,1\}^I$.   By 254G, $\phi$ is \imp.

\medskip

{\bf (b)} This uses the same idea as (a-ii).   Writing $\nu^3$ for the
product measure on $(\Cal PI)^3$, then, for any $t\in I$,

$$\eqalign{\nu^3\{(a,b,&c):t\in(a\cap b)\cup(a\cap c)\cup(b\cap c)\}\cr
&=\nu^3\{(a,b,c):t\in a\cap b\}+\nu^3\{(a,b,c):t\in a\cap c\}\cr
&\qquad\qquad\qquad+\nu^3\{(a,b,c):t\in b\cap c\}
  -2\nu^3\{(a,b,c):t\in a\cap b\cap c\}\cr
&=3\cdot\Bover14-2\cdot\Bover18=\Bover12.\cr}$$

\noindent Once again, these sets are independent for different $t$, and
this is all we need to know in order to be sure that the map is \imp.
}%end of proof of 464B

\leader{464C}{Lemma} Let $I$ be any set, and let $\nu$ be the usual
measure on $\Cal PI$.

(a)\cmmnt{ (see {\smc Sierpi\'nski 1945})} If $\Cal F\subseteq\Cal PI$
is any filter containing every cofinite
set, then $\nu_*\Cal F=0$ and $\nu^*\Cal F$ is either $0$ or $1$.   If
$\Cal F$ is a non-principal ultrafilter then $\nu^*\Cal F=1$.

(b)\cmmnt{ ({\smc Talagrand 80}
\footnote{The date of this paper is misleading, as there was an unusual
backlog in the journal;  in reality it preceded
{\smc Fremlin \& Talagrand 79}.})}
If $\sequencen{\Cal F_n}$ is a sequence of filters on $I$, all of
outer measure $1$, then $\bigcap_{n\in\Bbb N}\Cal F_n$ also has outer
measure $1$.

\proof{{\bf (a)} That $\nu^*\Cal F\in\{0,1\}$ is immediate from 464Ac.
If $\Cal F$ is an ultrafilter, then
$\Cal PI=\Cal F\cup\{I\setminus a:a\in\Cal F\}$;  but as
$a\mapsto I\setminus a$ is a measure space
automorphism of $(\Cal PI,\nu)$,
$\nu^*\{I\setminus a:a\in\Cal F\}=\nu^*\Cal F$, and both must be at
least $\bover12$, so $\nu^*\Cal F=1$.   Equally,
$\nu^*(\Cal PI\setminus\Cal F)=\nu^*\{I\setminus a:a\in\Cal F\}=1$, so
$\nu_*\Cal F=0$.   Returning to a general filter containing every
cofinite set, this is included in a non-principal ultrafilter, so also
has inner measure $0$.

\medskip

{\bf (b)} Let $\sequencen{m(n)}$ be a sequence in $\Bbb N$ such that
$\prod_{n=0}^{\infty}1-2^{-m(n)}=\bover12$, and let $X$, $\lambda$ and
$\phi:X\to\Cal PI$ be as in 464Ba.  Consider the set
$D=\prod_{n\in\Bbb N}\Cal F_n^{m(n)}$ as
a subset of $X$.   By 254Lb, $\lambda^*D=1$.   If we set
$\Cal F=\bigcap_{n\in\Bbb N}\Cal F_n$, then we see that whenever
$\frak x=\sequencen{\langle a_{ni}\rangle_{i<m(n)}}$ belongs to $D$,

\Centerline{$\phi(\frak x)\supseteq\bigcap_{i<m(n)}a_{ni}\in\Cal F_n$}

\noindent for every $n$, so that $\phi(\frak x)\in\Cal F$.   Thus
$D\subseteq\phi^{-1}[\Cal F]$ and

\Centerline{$\nu^*\Cal F\ge\lambda^*\phi^{-1}[\Cal F]\ge\lambda^*D=1$}

\noindent because $\phi$ is \imp\ (413Eh).

As $\sequencen{\Cal F_n}$ is arbitrary, we have the result.
}%end of proof of 464C

\leader{464D}{Construction}\cmmnt{ ({\smc Talagrand 80})} Let $I$ be
any set, and $\nu$ the usual
Radon measure on $\Cal PI$, with $\Tau$ its domain.   Let $\Sigma$ be
the set

$$\eqalign{\{E:E\subseteq\Cal PI,
&\text{ there are a set }F\in\Tau
   \text{ and a filter }\Cal F\text{ on }I\cr
&\hskip5em
 \text{ such that }\nu^*\Cal F=1
   \text{ and }E\cap\Cal F=F\cap\Cal F\}.\cr}$$

\noindent Then there is a unique extension of $\nu$ to a complete
probability measure $\mu$, with domain $\Sigma$, defined by saying that
$\mu E=\nu F$ whenever $E\in\Sigma$, $F\in\Tau$ and there is a filter
$\Cal F$ on $I$ such that $\nu^*\Cal F=1$ and $E\cap\Cal F=F\cap\Cal F$.
\prooflet{(By 464C, we can apply 417A with

\Centerline{$\Cal A=\{\Cal PI\setminus\Cal F:\Cal F$ is a filter on $I$
such that $\nu^*\Cal F=1\}$.)}
}%end of prooflet

\medskip

\noindent{\bf Definition} This measure $\mu$ is {\bf Talagrand's
measure} on $\Cal PI$.

\leader{464E}{Example} If $\mu$ is Talagrand's measure on
$X=\Cal P\Bbb N$, and $\Sigma$ its domain, then there is a set
$K\subseteq\Bbb R^X$, consisting of $\Sigma$-measurable functions and
separable and compact for the topology $\frak T_p$ of pointwise convergence,
such that $K$ is not compact for the topology of convergence in
measure\cmmnt{, that is, there is a sequence in $K$
with no subsequence which is convergent almost everywhere, even though
every cluster point is $\Sigma$-measurable}.

\proof{ Take $K$ to be
$\{\chi\Cal F:\Cal F$ is an ultrafilter on $\Bbb N\}$.   Then $K$ is
$\frak T_p$-compact (in fact, is precisely the set of Boolean
homomorphisms from $\Cal P\Bbb N$ to $\{0,1\}$, identified with the
Stone space of $\Cal P\Bbb N$ in 311E).   As such, it is separable, the
countable set of principal ultrafilters being dense.
By 464Ca, we see that any
non-principal ultrafilter $\Cal F$ on $\Bbb N$ belongs to $\Sigma$, and
$\mu\Cal F=1$.   On the other hand, all the principal ultrafilters
$\Cal F_n=\{a:n\in a\subseteq\Bbb N\}$ are
measured by $\nu$ and therefore by $\mu$, and form a stochastically
independent sequence of sets of measure $\bover12$.   So $K$ consists of
$\Sigma$-measurable functions;  but the sequence $\sequencen{\chi F_n}$
has no subsequence which is convergent almost everywhere, and $K$ is not
relatively compact for the topology of convergence in measure.
}%end of proof of 464E

\cmmnt{\medskip

\noindent{\bf Remark} In this example, a very large number of members of
$K$ are equal almost everywhere;  indeed, all non-principal ultrafilters
are equal a.e., and if we look at $\{f^{\ssbullet}:f\in K\}$ in
$L^0(\mu)$, it is a countable discrete set.   Given that $K$ is a set of
measurable functions homeomorphic to $\beta\Bbb N$, something like this
has to happen (see 536D\formerly{5{}36C}
in Volume 5).   Looking at this from a different angle, if
we wish to extend the usual measure $\nu$ on $\Cal P\Bbb N$ to measure
every ultrafilter, there will have to be two distinct ultrafilters
$\Cal F_1$, $\Cal F_2$ such that $\Cal F_1\symmdiff\Cal F_2$ is negligible for the extended measure.
}%end of comment

\leader{464F}{The $L$-space $\ell^{\infty}(I)^*$}\cmmnt{ For the next
step, we shall need to recall some facts from Volume 3.}
Let $I$ be any set.

\spheader 464Fa \cmmnt{The space} $\ell^{\infty}(I)$\cmmnt{ of
bounded real-valued
functions on $I$} is an $M$-space\cmmnt{ (354Ha)}, so\cmmnt{ its
dual} $\ell^{\infty}(I)^{\sim}=\ell^{\infty}(I)^*$ is an
$L$-space\cmmnt{ (356N),
that is, a Banach lattice such that $\|f+g\|=\|f\|+\|g\|$ for all
non-negative $f$, $g\in\ell^{\infty}(I)^*$}.
\dvro{We}{Since $\ell^{\infty}(I)$ can be identified with the space
$L^{\infty}(\Cal PI)$ as described in \S363 (see 363Ha), we} can
identify $\ell^{\infty}(I)^*$ with the $L$-space $M$ of bounded finitely
additive functionals on $\Cal PI$\cmmnt{ (363K)}, matching any
$f\in\ell^{\infty}(I)^*$ with
the functional $a\mapsto f(\chi a):\Cal PI\to\Bbb R$ in $M$.

\spheader 464Fb\dvro{ Write $M_{\tau}$ for the
band of completely additive functionals on $\Cal PI$.}{ In \S\S361-363 I
examined some of the bands in $M$.   The most
significant ones for our present purposes are the band $M_{\tau}$ of
completely additive functionals (362Bb) and its complement
$M_{\tau}^{\perp}$ (352P);  because $M$ is Dedekind complete, we have
$M=M_{\tau}\oplus M_{\tau}^{\perp}$ (353J\formerly{3{}53I}).}   \cmmnt{In
fact} $M_{\tau}$ is just the set of those $\theta\in M$ such that
$\theta a=\sum_{t\in a}\theta\{t\}$ for every $a\subseteq I$, while
$M_{\tau}^{\perp}$ is the set of those $\theta\in M$ such that
$\theta\{t\}=0$ for every $t\in I$.   \prooflet{\Prf\ For any $\theta\in
M$, we can set $\alpha_t=\theta\{t\}$ for each $t\in I$;  in this
case,

\Centerline{$|\sum_{t\in J}\alpha_t|=|\theta J|\le\|\theta\|$}

\noindent for every finite set $J\subseteq I$, so
$\sum_{t\in I}|\alpha_t|$ is finite, and we have a functional $\theta_1$
defined by setting
$\theta_1a=\sum_{t\in a}\alpha_t$ for every $a\subseteq I$.   It is easy
to check that $\theta_1\in M_{\tau}$.   If we write
$\theta_2=\theta-\theta_1$ then $\theta_2\{t\}=0$ for every $t\in
I$.   So if $\phi\in M_{\tau}$ and $0\le\phi\le|\theta_2|$, we still
have $\phi\{t\}=0$ for every $t\in I$, and $\phi a=0$ for every
finite $a\subseteq I$ and $\phi=0$;  thus $\theta_2\in
M_{\tau}^{\perp}$.   Now $\theta\in M_{\tau}$ iff $\theta_2=0$, that is,
iff $\theta=\theta_1$ and $\theta a=\sum_{t\in a}\theta\{t\}$ for every
$a\subseteq I$;  while $\theta\in M_{\tau}^{\perp}$ iff $\theta_1=0$,
that is, $\theta\{t\}=0$ for every $t\in I$.\ \Qed
}%end of prooflet

Observe that if $\theta\in M_{\tau}^{\perp}$ and $a$, $b\subseteq I$ are
such that $a\symmdiff b$ is finite, then $\theta a=\theta b$\cmmnt{,
because $\theta(a\setminus b)=\theta(b\setminus a)=0$}.

\spheader 464Fc \cmmnt{It will be useful to have an elementary fact
out in the open.}   If $\theta\in M^+\setminus\{0\}$,
then $\{a:a\subseteq I,\,\theta a=\theta I\}$ is a filter\cmmnt{;
this is because $\{a:\theta a=0\}$ is a proper ideal in $\Cal PI$}.

\leader{464G}{}\cmmnt{ We also need a new result not exactly covered
by those in Chapters 35 and 36.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be any Boolean algebra.   Write $M$
for the $L$-space of bounded additive functionals on $\frak A$, and
$M^+$ for its positive cone\cmmnt{, the set of non-negative additive
functionals}.   Suppose that $\Delta:M^+\to\coint{0,\infty}$ is a
functional such that

\inset{($\alpha$) $\Delta$ is non-decreasing,}

\inset{($\beta$) $\Delta(\alpha\theta)=\alpha\Delta(\theta)$ whenever
$\theta\in M^+$, $\alpha\ge 0$,}

\inset{($\gamma$)
$\Delta(\theta_1+\theta_2)=\Delta(\theta_1)+\Delta(\theta_2)$ whenever
$\theta_1$, $\theta_2\in M^+$ are such that, for some $e\subseteq I$,
$\theta_1(1\setminus e)=\theta_2e=0$,}

\inset{($\delta$) $|\Delta(\theta_1)-\Delta(\theta_2)|
\le\|\theta_1-\theta_2\|$ for all $\theta_1$, $\theta_2\in M^+$.}

\noindent Then there is a non-negative $h\in M^*$ extending $\Delta$.

\proof{{\bf (a)} If $\theta_1$, $\theta_2\in M^+$ are such that
$\theta_1\wedge\theta_2=0$ in $M^+$, then
$\Delta(\theta_1+\theta_2)=\Delta(\theta_1)+\Delta(\theta_2)$.
\Prf\ Let $\epsilon>0$.   Then there is an $e\in\frak A$ such that
$\theta_1(1\Bsetminus e)+\theta_2e\le\epsilon$ (362Ba).   Set
$\theta_1'a=\theta_1(a\Bcap e)$, $\theta_2'(a)=\theta_2(a\Bsetminus e)$
for $a\in\frak A$;  then $\theta_1'$ and $\theta_2'$ belong to $M^+$ and
$\theta_1'(1\setminus e)=\theta_2'e=0$, so
$\Delta(\theta_1'+\theta_2')=\Delta(\theta_1')+\Delta(\theta_2')$, by
hypothesis ($\gamma$).   On the other hand,

\Centerline{$0\le\theta_1a-\theta_1'a=\theta(a\Bsetminus
e)\le\theta(1\Bsetminus e)\le\epsilon$}

\noindent for every $a\in\frak A$, so
$\|\theta_1-\theta_1'\|\le\epsilon$.   Similarly,
$\|\theta_2-\theta_2'\|\le\epsilon$ and
$\|(\theta_1+\theta_2)-(\theta_1'-\theta_2')\|\le 2\epsilon$.   But this
means that we can use the hypothesis ($\delta$) to see that

\Centerline{$|\Delta(\theta_1+\theta_2)
  -\Delta(\theta_1)-\Delta(\theta_2)|
\le|\Delta(\theta'_1+\theta'_2)
  -\Delta(\theta'_1)-\Delta(\theta'_2)|+4\epsilon
=4\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\Delta(\theta_1+\theta_2)=\Delta(\theta_1)+\Delta(\theta_2)$.\ \Qed

\medskip

{\bf (b)} Now recall that $M$, being a Dedekind complete Riesz space,
can be identified with an order-dense solid linear subspace of
$L^0(\frak C)$ for some Boolean algebra $\frak C$ (368H).   Inside
$L^0(\frak C)$ we have the order-dense Riesz subspace $S(\frak C)$
(364Ja).   Write $S_1$ for $M\cap S(\frak C)$, so that $S_1$ is an
order-dense Riesz subspace of $M$ (352Nc, 353A).

If $\theta_1$, $\theta_2\in S_1^+=S_1\cap M^+$, then
$\Delta(\theta_1+\theta_2)=\Delta(\theta_1)+\Delta(\theta_2)$.   \Prf\
We can express $\theta_1$ as $\sum_{i=0}^m\alpha_i\chi c_i$, where
$c_0,\ldots,c_m$ are disjoint members of $\frak C$, and $\alpha_i\ge 0$
for each $i$ (361Ec);  adding a term $0\cdot\chi(1\Bsetminus\sup_{i\le
m}c_i)$ if necessary, we may suppose that $\sup_{i\le m}c_i=1$.
Similarly, we can express $\theta_2$ as $\sum_{j=0}^n\beta_j\chi d_j$
where $d_0,\ldots,d_n\in\frak C$ are disjoint, $\beta_j\ge 0$ for every
$j$ and $\sup_{j\le n}d_j=1$.   In this case,
$\theta_1=\sum_{i\le m,j\le n}\alpha_i\chi(c_i\Bcap d_j)$ and
$\theta_2=\sum_{i\le m,j\le n}\beta_j\chi(c_i\Bcap d_j)$.
Re-enumerating $\{c_i\Bcap d_j:i\le m,\,j\le n\}$ as $\langle e_i:i\le
k\}$ we have expressions of $\theta_1$, $\theta_2$ in the form
$\sum_{i\le k}\gamma_i\chi e_i$, $\sum_{i\le k}\delta_i\chi e_i$, while
$e_0,\ldots,e_k$ are disjoint.

Setting $\theta_1^{(r)}=\sum_{i=0}^r\gamma_i\chi e_i$ for $r\le k$, we
see that $\theta_1^{(r)}\wedge\gamma_{r+1}\chi e_{r+1}=0$ for $r<k$, so
(a) above, together with the hypothesis ($\beta$), tell us that

\Centerline{$\Delta(\theta^{(r+1)}_1)
=\Delta(\theta^{(r)}_1)+\Delta(\gamma_{r+1}\chi e_{r+1})
=\Delta(\theta^{(r)}_1)+\gamma_{r+1}\Delta(\chi e_{r+1})$}

\noindent for $r<k$.   Accordingly
$\Delta(\theta_1)=\sum_{i=0}^k\gamma_i\Delta(\chi e_i)$.   Similarly,
$\Delta(\theta_2)=\sum_{i=0}^k\delta_i\Delta(\chi e_i)$ and

\Centerline{$\Delta(\theta_1+\theta_2)
=\sum_{i=0}^k(\gamma_i+\delta_i)\Delta(\chi e_i)
=\Delta(\theta_1)+\Delta(\theta_2)$,}

\noindent as claimed.\ \Qed

\medskip

{\bf (c)} Consequently,
$\Delta(\theta_1+\theta_2)=\Delta(\theta_1)+\Delta(\theta_2)$ for all
$\theta_1$, $\theta_2\in M^+$.   \Prf\ Let $\epsilon>0$.   Because $S_1$
is order-dense in $M$, and the norm of $M$ is order-continuous (354N),
$S_1$ is norm-dense (354Ef), and there are $\theta'_1$,
$\theta'_2\in S_1^+$ such that $\|\theta_j-\theta'_j\|\le\epsilon$ for
both $j$ (354Be).   But now, just as in (a),

\Centerline{$|\Delta(\theta_1+\theta_2)
  -\Delta(\theta_1)-\Delta(\theta_2)|
\le|\Delta(\theta'_1+\theta'_2)
  -\Delta(\theta'_1)-\Delta(\theta'_2)|+4\epsilon
=4\epsilon$.}

\noindent As $\epsilon$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (d)} Now (c) and the hypothesis ($\beta$) are sufficient to ensure
that $\Delta$ has an extension to a positive linear functional (355D).
}%end of proof of 464G

\leader{464H}{}\cmmnt{ The next lemma contains the key ideas needed
for the rest of the section.

\medskip

\noindent}{\bf Lemma} Let $I$ be any set, and $M$ the $L$-space of
bounded additive functionals on $\Cal PI$;  let $\nu$ be the usual
measure on $\Cal PI$.   For $\theta\in M^+$, set

\Centerline{$
\Delta(\theta)=\overline{\intop}\theta\,d\nu$.}

(a) For every $\theta\in M^+$,
$\bover12\theta I\le\Delta(\theta)\le\theta I$.

(b) There is a non-negative $h\in M^*$ such that
$h(\theta)=\Delta(\theta)$ for every $\theta\in M^+$.

(c) If $\theta\in(M_{\tau}^{\perp})^+$, where $M_{\tau}\subseteq M$ is
the band of completely additive functionals, then
$\theta\le\Delta(\theta)\,\,\nu$-a.e., and
$\nu^*\{a:\alpha\le\theta a\le\Delta(\theta)\}=1$ for every
$\alpha<\Delta(\theta)$.

(d) Suppose that $\theta\in(M_{\tau}^{\perp})^+$ and $\beta$,
$\gamma\in[0,1]$ are such that $\theta I=1$ and
$\beta\theta'I\le\Delta(\theta')\le\gamma\theta'I$ whenever
$\theta'\le\theta$ in $M^+$.   Then, for any $\alpha<\beta$,

\quad(i) for any finite set $K\subseteq\Cal PI$, the set

\Centerline{$\{a:a\subseteq I,\,
\alpha\theta b\le\theta(a\cap b)\le\gamma\theta b$ for every $b\in K\}$}

\noindent has outer measure $1$ in $\Cal PI$;

\quad(ii) if $\alpha\ge\bover12$, the set

\Centerline{$R=\{(a,b,c):a,\,b,\,c\subseteq I,\,
\theta((a\cap b)\cup(a\cap c)\cup(b\cap c))
\ge 2\alpha^2+(1-2\alpha)\gamma^2\}$}

\noindent has outer measure $1$ in $(\Cal PI)^3$;

\quad(iii) if $\alpha\ge\bover12$, then
$2\alpha^2+(1-2\alpha)\gamma^2\le\gamma$.

(e) Any $\theta\in M^+$ can be expressed as $\theta_1+\theta_2$ where
$\Delta(\theta_1)=\bover12\theta_1I$ and $\Delta(\theta_2)=\theta_2I$.

(f) Suppose that $0\le\theta'\le\theta$ in $M$.

\quad(i) If $\Delta(\theta)=\bover12\theta I$, then
$\Delta(\theta')=\bover12\theta'I$.

\quad(ii) If $\Delta(\theta)=\theta I$, then $\Delta(\theta')=\theta'I$.

\proof{{\bf (a)} Of course

\Centerline{$\Delta(\theta)\le\biggerint\theta I\,d\nu=\theta I$.}

\noindent On the other hand, because
$a\mapsto I\setminus a:\Cal PI\to\Cal PI$ is an automorphism of the
measure space $(\Cal PI,\nu)$,
$\overline{\int}f(I\setminus a)\nu(da)=\overline{\int}f(a)\nu(da)$ for
any real-valued function $f$ (cf.\ 235Xn).   In particular,

\Centerline{$\Delta(\theta)
=\overline{\intop}\theta(a)\nu(da)
=\overline{\intop}\theta(I\setminus a)\nu(da)$.}

\noindent So

$$\eqalignno{2\Delta(\theta)
&=\overline{\int}\theta(a)\nu(da)
  +\overline{\int}\theta(I\setminus a)\nu(da)
\ge\overline{\int}\theta(a)+\theta(I\setminus a)\nu(da)\cr
\noalign{\noindent (133J(b-ii))}
&=\overline{\int}\theta I\,\nu(da)
=\theta I,\cr}$$

\noindent and $\Delta(\theta)\ge\bover12\theta I$.

\medskip

{\bf (b)} I use 464G with $\frak A=\Cal PI$.   Examine the conditions
($\alpha$)-($\delta$) there.

\medskip

\quad\grheada\ Of course $\Delta$ is non-decreasing (133Jc).

\medskip

\quad\grheadb\ $\Delta(\alpha\theta)=\alpha\Delta(\theta)$ for every
$\theta\in M^+$ and every $\alpha\ge 0$, by 133J(b-iii).

\medskip

\quad\grheadc\ If $\theta_1$, $\theta_2\in M^+$ then
$\Delta(\theta_1+\theta_2)\le\Delta(\theta_1)+\Delta(\theta_2)$ by
133J(b-ii), as in (a) above.
If $\theta_1$, $\theta_2\in M^+$
and $e\subseteq I$ are such that $\theta_1(I\setminus e)=\theta_2e=0$,
then $\theta_1a=\theta_1(a\cap e)$, $\theta_2(a)=\theta_2(a\setminus e)$
for every $a\subseteq I$.   So
$\Delta(\theta_1+\theta_2)=\Delta(\theta_1)+\Delta(\theta_2)$ by 464Ab.

\medskip

\quad\grheadd\ For every $a\subseteq I$,
$\theta_2a\le\theta_1a+\|\theta_2-\theta_1\|$, so
$\Delta(\theta_2)\le\Delta(\theta_1)+\|\theta_2-\theta_1\|$.
Similarly, $\Delta(\theta_1)
\le\Delta(\theta_2)+\|\theta_1-\theta_2\|$.   So
$|\Delta(\theta_1)-\Delta(\theta_2)|\le\|\theta_1-\theta_2\|$, and
$\Delta$ satisfies condition ($\delta$) of 464G.

\medskip

\quad\grheade\ Accordingly 464G tells us that $\Delta$ has an extension
to a member of $M^*$.

\medskip

{\bf (c)} If $\theta\in M_{\tau}^{\perp}$, then $\theta a=\theta b$
whenever $a\symmdiff b$ is finite (464Fb), so all the sets
$\{a:\theta a>\alpha\}$ have outer measure either $0$ or $1$, by 464Ac.
But this means that if $f$ is a measurable function
and $\theta\leae f$ and $\int f=\Delta(\theta)$, as in 133J(a-i),
$\{a:f(a)\le\Delta(\theta)\}$ has
positive measure and meets $A=\{a:\theta a>\Delta(\theta)\}$ in a
negligible set;  so $A$ cannot have full outer measure and is
negligible.

On the other hand, if $\alpha<\Delta(\theta)$, then $\theta$ cannot be
dominated a.e.\ by $\alpha\chi(\Cal PI)$, so
$\{a:\theta a>\alpha\}$ is not negligible and has outer measure $1$.
Consequently $\nu^*\{a:\alpha<\theta a\le\Delta(\theta)\}=1$.

\medskip

{\bf (d)(i)} The point is just that for any $b\subseteq I$, the
functional $\theta_b$, defined by saying that
$\theta_b(a)=\theta(a\cap b)$ for every $a\subseteq I$, belongs to $M^+$
and is dominated by
$\theta$, so that $\beta\theta_bI\le\Delta(\theta_b)\le\gamma\theta_bI$,
and $\{a:\alpha\theta_bI\le\theta_ba\le\gamma\theta_bI\}$ has outer
measure $1$, by (c).   (If $\alpha\theta_bI=\Delta(\theta_b)$, this is
because $\theta_bI=0$, and the result is trivial;  otherwise,
$\alpha\theta_bI<\Delta(\theta_b)\le\gamma\theta_bI$.)   But this just
says that, for any $b\subseteq I$,

\Centerline{$\{a:\alpha\theta b\le\theta(a\Bcap b)\le\gamma\theta b\}$}

\noindent has outer measure $1$.

Now, given any finite set $K\subseteq\Cal PI$, let $\Cal B$ be the
subalgebra of $\Cal PI$ generated by $K$, and $b_0,\ldots,b_n$ the atoms
of $\Cal B$.   Then all the sets

\Centerline{$A_i=\{a:\alpha\theta b_i\le\theta(a\cap b_i)
\le\gamma\theta b_i\}$}

\noindent have outer measure $1$;  because each $A_i$ is determined by
coordinates in $b_i$, and $b_0,\ldots,b_n$ are disjoint,
$A=\bigcap_{i\le n}A_i$ still has outer measure $1$, by 464Aa.  But if
$a\in A$ and $b\in K$, then $b=\bigcup_{i\in J}b_i$ for some
$J\subseteq\{0,\ldots,n\}$ and

$$\eqalign{\alpha\theta b
&=\sum_{i\in J}\alpha\theta b_i
\le\sum_{i\in J}\theta(a\cap b_i)\cr
&=\theta(a\cap b)
\le\sum_{i\in J}\gamma\theta b_i
=\gamma\theta b.\cr}$$

\noindent So $\{a:\alpha\theta b\le\theta(a\cap b)\le\gamma\theta b$ for
every $b\in K\}$ includes $A$ and has outer measure $1$.

\medskip

\quad{\bf (ii)} \Quer\ Suppose, if possible, otherwise;  then
$\nu^3_*(\Cal PI\setminus R)>0$, where $\nu^3$ is the product measure on
$(\Cal PI)^3$, and there is a measurable set
$W\subseteq(\Cal PI)^3\setminus R$ such that $\nu^3W>0$.   For
$a\subseteq I$, set
$W_a=\{(b,c):(a,b,c)\in W\}$;  then $\nu^3W=\int\nu^2W_a\nu(da)$, where
$\nu^2$ is the product measure on $(\Cal PI)^2$, by Fubini's theorem
(252D).   Set $E=\{a:\nu^2W_a$ is defined and not zero$\}$;  then
$\nu E>0$.   Since $\{a:\alpha\le\theta a\le\gamma\}$ has outer measure
$1$, there is an $a\in E$ such that $\alpha\le\theta a\le\gamma$.

For $b\subseteq I$, set $W_{ab}=\{c:(b,c)\in W_a\}=\{c:(a,b,c)\in W\}$.
Then $0<\nu^2W_a=\int\nu W_{ab}\nu(db)$, so $F=\{b:\nu W_{ab}$ is
defined and not $0\}$ has non-zero measure.   But also, by (i),

\Centerline{$\{b:\theta b\ge\alpha,\,\theta(a\cap b)\le\gamma\theta
a\}$}

\noindent has outer measure $1$, so we can find a $b\in F$ such that
$\theta b\ge\alpha$ and $\theta(a\cap b)\le\gamma\theta a$.

By (i) again,

\Centerline{$\{c:\theta(c\cap(a\symmdiff b))\ge\alpha\theta(a\symmdiff
b)\}$}

\noindent has outer measure $1$, so meets $W_{ab}$;  accordingly we have
a $c\subseteq I$ such that $(a,b,c)\in W$ while $\theta(c\cap(a\symmdiff
b))\ge\alpha\theta(a\symmdiff b)$.

Now calculate, for this triple $(a,b,c)$,

$$\eqalignno{\theta((a\cap b)\cup(a\cap c)\cup(b\cap c))
&=\theta(a\cap b)+\theta(c\cap(a\symmdiff b))
\ge\theta(a\cap b)+\alpha\theta(a\symmdiff b)\cr
\noalign{\noindent (by the choice of $c$)}
&=\alpha(\theta a+\theta b)+(1-2\alpha)\theta(a\cap b)\cr
&\ge\alpha(\theta a+\alpha)+(1-2\alpha)\gamma\theta a\cr
\noalign{\noindent (by the choice of $b$, recalling that
$1-2\alpha\le 0)$}
&\ge 2\alpha^2+(1-2\alpha)\gamma^2\cr}$$

\noindent by the choice of $a$.   But this means that $(a,b,c)\in W\cap
R$, which is supposed to be impossible.\ \Bang

\medskip

\quad{\bf (iii)} Now recall that the map
$(a,b,c)\mapsto(a\cap b)\cup(a\cap c)\cup(b\cap c)$ is \imp\ (464Bb).
Since $\theta a\le\Delta(\theta)\le\gamma$ for $\nu$-almost every $a$,
we must have
$\theta((a\cap b)\cup(a\cap c)\cup(b\cap c))\le\gamma$ for
$\nu^3$-almost every $(a,b,c)$.   But as $R$ is not negligible, there
must be some $(a,b,c)\in R$ such that
$\theta((a\cap b)\cup(a\cap c)\cup(b\cap c))\le\gamma$, and
$2\alpha^2+(1-2\alpha)\gamma^2\le\gamma$.

\medskip

{\bf (e)(i)} $M^*$, being the dual of an $L$-space, is an $M$-space
(356Pb), so can be represented as $C(Z)$ for some compact Hausdorff space
$Z$ (354L).   The functional $h$ of (b) above therefore corresponds to a
function $w\in C(Z)$.   Any $\theta\in M^+$ acts on $M^*$ as a positive
linear functional, so corresponds to a Radon measure $\mu_{\theta}$ on
$Z$ (436J/436K);  we have
$\Delta(\theta)=h(\theta)=\int w\,d\mu_{\theta}$.
The inequalities $\bover12\theta I\le\Delta(\theta)\le\theta I$ become
$\bover12\mu_{\theta}Z\le\int w\,d\mu_{\theta}\le\mu_{\theta}Z$, because
the constant function $\chi Z$ corresponds to the standard order unit
of $M^*$ (356Pb again), so that

\Centerline{$\mu_{\theta}Z=\biggerint\chi Z\,d\mu_{\theta}
=\|\theta\|=\theta I$}

\noindent for every $\theta\ge 0$.   Since $0\le h(\theta)\le\|\theta\|$
for every $\theta\ge 0$, $\|w\|_{\infty}=\|h\|\le 1$ and $0\le w\le\chi
Z$.

\medskip

\quad{\bf (ii)} Now suppose that $\beta<\gamma$ and that
$G=\{z:z\in Z,\,\beta<w(z)<\gamma\}$ is non-empty.   In this case there
is a non-zero $\theta_0\in M^+$ such that
$\beta\theta'I\le\Delta(\theta')\le\gamma\theta'I$ whenever
$0\le\theta'\le\theta_0$.

\Prf\grheada\ We have a
solid linear subspace $V=\{v:v\in C(Z),\,v(z)=0$ for every $z\in G\}$ of
$C(Z)$.   Consider $U=\{\theta:\theta\in M,\,\innerprod{\theta}{v}=0$
for every $v\in V\}$, where I write $\innerprod{\,}{\,}$ for the duality
between $M$ and $C(Z)$ corresponding to the identification of $C(Z)$
with $M^*$.

\medskip

\qquad\grheadb\ If $\theta\in U\cap M^+$, then
$\beta\theta I\le\Delta(\theta)\le\gamma\theta I$.   To see this,
observe that $\int v\,d\mu_{\theta}=\innerprod{\theta}{v}=0$ for every
$v\in V$, so

$$\eqalignno{\mu_{\theta}(Z\setminus\overline{G})
&=\sup\{\mu_{\theta}K:K\subseteq Z\setminus\overline{G}\text{ is
compact}\}\cr
&=\sup\{\int v\,d\mu_{\theta}:v\in C(Z),\,
  0\le v\le\chi(Z\setminus\overline{G})\}\cr
\displaycause{because whenever
$K\subseteq Z\setminus\overline{G}$ is compact there is a $v\in C(Z)$
such that $\chi K\le v\le\chi(Z\setminus\overline{G})$, by 4A2F(h-ii)}
&=0.\cr}$$

\noindent Because $w$ is continuous,
$\beta\le w(z)\le\gamma$ for every $z\in\overline{G}$;  thus
$\beta\le w\le\gamma\,\,\mu_{\theta}$-a.e.\ and
$\int w\,d\mu_{\theta}$ must belong to
$[\beta\mu_{\theta}Z,\gamma\mu_{\theta}Z]=[\beta\theta I,\gamma\theta I]$.

\medskip

\qquad\grheadc\ The dual $M^*=M^{\times}$ of $M$ is
perfect (356Lb), and $C(Z)$ is perfect;  moreover, $M$ is perfect
(356Pa), so the duality
$\innerprod{\,}{\,}$ identifies $M$ with $C(Z)^{\times}$.   Now
$V^{\perp}$, taken in $C(Z)$, contains any continuous function zero on
$Z\setminus G$, so is not $\{0\}$;  since $V^{\perp}$, like $C(Z)$, must
be perfect (356La), $(V^{\perp})^{\times}$ is non-trivial.   Take any
$\psi>0$ in $(V^{\perp})^{\times}$.   Being perfect, $C(Z)$ is Dedekind
complete (356K), so there is a band projection $P:C(Z)\to V^{\perp}$
(353J again).   Now $\psi P$ is a positive element of $C(Z)^{\times}$
which is zero on $V$, and must correspond to a non-zero element
$\theta_0$ of $U\cap M^+$.

\medskip

\qquad\grheadd\ If $0\le\theta'\le\theta_0$ in $M$,
then, for any $v\in V$,

\Centerline{$|\innerprod{\theta'}{v}|
\le\innerprod{\theta'}{|v|}
\le\innerprod{\theta_0}{|v|}
=0$,}

\noindent because $|v|\in V$.   So $\theta'\in U$ and
$\beta\theta'I\le\Delta(\theta')\le\gamma\theta'I$, by ($\beta$).   Thus
$\theta_0$ has the required property.\ \Qed

\medskip

\quad{\bf (iii)} It follows at once that $w(z)\ge\bover12$ for every
$z\in Z$.   \Prf\Quer\ If $w(z_0)<\bover12$, then we can apply (ii) with
$\beta=-1$, $\gamma\in\ooint{w(z_0),\bover12}$ to see that there is a
non-zero $\theta\in M^+$ such that $\Delta(\theta)\le\gamma\theta
I<\bover12\theta I$, which is impossible, by (a).\ \Bang\Qed

\medskip

\quad{\bf (iv)} But we find also that $w(z)\notin\ooint{\bover12,1}$ for
any $z\in Z$.  \Prf\Quer\ If $w(z_0)=\delta\in\ooint{\bover12,1}$, then
$2\delta^2+(1-2\delta)\delta^2>\delta$ (because
$\delta(2\delta-1)(1-\delta)>0$).   We can therefore find $\alpha$,
$\beta$ and $\gamma$ such that $\bover12\le\alpha<\beta<\delta<\gamma$
and $2\alpha^2+(1-2\alpha)\gamma^2>\gamma$.   But now
$\{z:\beta<w(z)<\gamma\}$ is non-empty, so by (ii) there is a non-zero
$\theta\in M^+$ such that
$\beta\theta'I\le\Delta(\theta')\le\gamma\theta'I$ whenever
$0\le\theta'\le\theta$.   Multiplying $\theta$ by a suitable scalar if
necessary, we can arrange that $\theta I$ should be $1$.   But this is
impossible, by (d-iii).\ \Bang\Qed

\medskip

\quad{\bf (v)} Thus $w$ takes only the values $\bover12$ and $1$;  let
$H_1$ and $H_2$ be the corresponding open-and-closed subsets of $Z$.

Take $\theta\in M^+$.   For $u\in C(Z)$, set
$\phi(u)=\int u\,d\mu_{\theta}$ and
$\phi_j(u)=\int_{H_j}u\,d\mu_{\theta}$ for each
$j$.   Then each $\phi_j$ is a positive linear functional on $C(Z)$ and
$\phi_j\le\phi$.   But $\phi$ is the image of $\theta$ under the
canonical isomorphism from $M$ to $C(Z)^{\times}\cong M^{\times\times}$,
and $C(Z)^{\times}$ is solid in $C(Z)^{\sim}$ (356B), so both $\phi_1$
and $\phi_2$ belong to the image of $M$, and correspond to $\theta_1$,
$\theta_2\in M$.   For any $u\in C(Z)\cong M^*$,

\Centerline{$\innerprod{\theta_1+\theta_2}{u}
=\phi_1(u)+\phi_2(u)
=\innerprod{\theta}{u}$,}

\noindent so $\theta=\theta_1+\theta_2$.   We have

$$\eqalign{\Delta(\theta_j)=\phi_j(w)&=\int_{H_j}w\,d\mu_{\theta}\cr
&=\Bover12\mu_{\theta}(H_1)=\Bover12\theta_1I\text{ if }j=1,\cr
&=\mu_{\theta}(H_2)=\theta_2I\text{ if }j=2.\cr}$$

\noindent So we have a suitable decomposition
$\theta=\theta_1+\theta_2$.

\medskip

{\bf (f)} This is easy.   Set $\theta''=\theta-\theta'$;  then

\Centerline{$\bover12\theta'I\le\Delta(\theta')\le\theta'I$,}

\Centerline{$\bover12\theta''I\le\Delta(\theta'')\le\theta''I$}

\noindent by (a), while
$\Delta(\theta')+\Delta(\theta'')=\Delta(\theta)$ by (b), and of course
$\theta'I+\theta''I=\theta I$.   But this means that

\Centerline{$\Delta(\theta')-\bover12\theta'I
\le\Delta(\theta)-\bover12\theta I$,
\quad$\theta'I-\Delta(\theta')\le\theta I-\Delta(\theta)$,}

\noindent and the results follow.
}%end of proof of 464H

\leader{464I}{Measurable and purely non-measurable functionals} As
before, let $I$ be any set, $\nu$ the usual measure on $\Cal PI$, $\Tau$
its domain, and
$M$ the $L$-space of bounded additive functionals on $\Cal PI$.
\cmmnt{Following {\smc Fremlin \& Talagrand 79},} I say that
$\theta\in M$ is {\bf measurable} if it is $\Tau$-measurable when
regarded as a
real-valued function on $\Cal PI$, and {\bf purely
non-measurable} if $\{a:a\subseteq I,\,|\theta|(a)=|\theta|(I)\}$ has
outer measure $1$.   \cmmnt{(Of course the zero functional is both
measurable and purely non-measurable.)}

\leader{464J}{Examples}\cmmnt{ Before going farther, I had better
offer some examples of measurable and purely non-measurable
functionals.}   Let $I$, $\nu$ and $M$ be as in 464I.

\spheader 464Ja Any $\theta\in M_{\tau}$ is measurable\cmmnt{, where
$M_{\tau}$ is the space of completely additive functionals on $\Cal
PI$}.   \prooflet{\Prf\ By 464Fb, $\theta$ can be expressed as a sum of
point masses;  say $\theta a=\sum_{t\in a}\alpha_t$ for some family
$\family{t}{I}{\alpha_t}$ in $\Bbb R$.   Since $\sum_{t\in I}|\alpha_t|$
must be finite, $\{t:\alpha_t\ne 0\}$ is countable, and we can express
$\theta$ as the limit of a sequence of finite sums $\sum_{t\in
K}\alpha_t\hat t$, where $\hat t(a)=1$ if $t\in a$, $0$ otherwise.   But
of course every $\hat t$ is a measurable function, so $\sum_{t\in
K}\alpha_t\hat t$ is measurable for every finite set $K$, and $\theta$
is measurable.\ \Qed}

\spheader 464Jb\cmmnt{ For a less elementary measurable functional,
consider the following construction.}   Let $\sequencen{t_n}$ be any
sequence of distinct points in $I$.
Then\cmmnt{ $\lim_{n\to\infty}\Bover1n\#(\{i:i<n,\,t_i\in a\})
=\Bover12$
for $\nu$-almost every $a\subseteq I$.   \prooflet{\Prf\ Set $f_n(a)=1$
if $t_n\in a$, $0$ otherwise.   Then $\sequencen{f_n}$ is an independent
sequence of random variables.   By any of the versions of the Strong Law
of Large Numbers in \S273 (273D, 273H, 273I),
$\lim_{n\to\infty}\Bover1n\sum_{i=0}^{n-1}f_i=\Bover12$ a.e., which is
what was claimed.\ \QeD\ }   So} if we take any non-principal
ultrafilter $\Cal F$ on $\Bbb N$, and set
$\theta a=\lim_{n\to\Cal F}\Bover1n\#(\{i:i<n,\,t_i\in a\})$ for
$a\subseteq I$, $\theta$ will be
constant $\nu$-almost everywhere, and measurable\cmmnt{;  and it is
easy to check that $\theta$ is additive}.   \cmmnt{Note that
$\theta\{t\}=0$
for every $t$, so} $\theta\in M_{\tau}^{\perp}$\cmmnt{, by 464Fb}.

\spheader 464Jc If $\Cal F$
is any non-principal ultrafilter on $I$, and we set $\theta a=1$
for $a\in\Cal F$, $0$ otherwise, then $\theta$ is an additive functional
which is purely non-measurable\cmmnt{, by 464Ca}.

\cmmnt{For further remarks on where to look for measurable and purely
non-measurable functionals, see 464P-464Q below.}

\leader{464K}{The space $M_{\text{m}}$:  Lemma} Let $I$ be any set,
$\nu$ the usual measure on $\Cal PI$, and $M$ the $L$-space of bounded
additive functionals on $\Cal PI$.
Write $M_{\text{m}}$ for the set of measurable $\theta\in M$,
$M_{\tau}$ for the space of completely additive
functionals on $\Cal PI$ and
$\Delta(\theta)=\overline{\int}\theta\,d\nu$ for
$\theta\in M^+$\cmmnt{, as in 464H}.

(a) If $\theta\in M_{\text{m}}\cap M_{\tau}^{\perp}$ and $b\subseteq I$,
then $\theta(a\cap b)=\bover12\theta b$ for $\nu$-almost every
$a\subseteq I$.

(b) $|\theta|\in M_{\text{m}}$ for every $\theta\in M_{\text{m}}$.

(c) A functional $\theta\in M^+$ is measurable iff
$\Delta(\theta)=\bover12\theta I$.

(d) $M_{\text{m}}$ is a solid linear subspace of $M$.

\proof{{\bf (a)(i)} $\theta=\bover12\theta I\,\,\nu$-a.e.   \Prf\ For any
$\alpha\in\Bbb R$, $A_{\alpha}=\{a:\theta a<\alpha\}$ is measurable;
but also $a'\in A$ whenever $a\in A$ and $a\symmdiff a'$ is finite, by
464Fb, so $\nu A$ must be either $1$ or $0$, by 464Ac.   Setting
$\delta=\sup\{\alpha:\nu A_{\alpha}=0\}$, we see that
$\nu A_{\delta}=0$,
$\nu A_{\delta+2^{-n}}=1$ for every $n\in\Bbb N$, so that
$\theta=\delta$ a.e.   Also, because $a\mapsto I\setminus a$ is a
measure space automorphism, $\theta(I\setminus a)=\delta$ for almost
every $a$, so there is some $a$ such that
$\theta a=\theta(I\setminus  a)=\delta$, and
$\delta=\bover12\theta I$.\ \Qed

\medskip

\quad{\bf (ii)} $\theta(a\cap b)=\bover12\theta b$ for almost every $a$.
\Prf\ We know that $\theta a=\bover12\theta I$ for almost every $a$.
But $a\mapsto a\symmdiff b:\Cal PI\to\Cal PI$ is \imp, so
$\theta(a\symmdiff b)=\bover12\theta I$
for almost every $a$.   This means that $\theta a=\theta(a\symmdiff b)$
for almost every $a$, and

\Centerline{$\theta(a\cap b)
=\Bover12(\theta b+\theta a-\theta(a\symmdiff b))=\Bover12\theta b$}

\noindent for almost every $a$.\ \Qed

\medskip

{\bf (b)(i)} If $\theta\in M_{\text{m}}\cap M_{\tau}^{\perp}$, then
$\theta^+$, taken in $M$, is measurable.  \Prf\ For any
$n\in\Bbb N$ we can find $b_n\subseteq I$ such that
$\theta^-b_n+\theta^+(I\setminus b_n)\le 2^{-n}$, so that

\Centerline{$|\theta^+a-\theta(a\cap b_n)|
=|\theta^+a-\theta^+(a\cap b_n)+\theta^-(a\cap b_n)|
\le\theta^+(I\setminus b_n)+\theta^-b_n
\le 2^{-n}$}

\noindent for every $a\subseteq I$.   But as $a\mapsto\theta(a\cap b_n)$ is
constant a.e.\ for every $n$, by (a), so is $\theta^+$, and
$\theta^+$ is measurable.\ \Qed

Consequently $|\theta|=2\theta^+-\theta$ is measurable.

\medskip

\quad{\bf (ii)} Now take an arbitrary $\theta\in M_{\text{m}}$.
Because $M$ is Dedekind complete (354N, 354Ee),
$M=M_{\tau}+M_{\tau}^{\perp}$ (J once more), and we can express $\theta$ as
$\theta_1+\theta_2$ where $\theta_1\in M_{\tau}$ and
$\theta_2\in M_{\tau}^{\perp}$;
moreover, $|\theta|=|\theta_1|+|\theta_2|$ (352Fb).
Now $\theta_1$ is measurable, by 464Ja, so $\theta_2=\theta-\theta_1$ is
measurable;  as $\theta_2\in M_{\tau}^{\perp}$, (i) tells us
that $|\theta_2|$ is measurable.   On the other hand, $|\theta_1|$
belongs to $M_{\tau}$ and is measurable, so
$|\theta|=|\theta_1|+|\theta_2|$ is measurable.

Thus (b) is true.

\medskip

{\bf (c)} Let $f$ be a $\nu$-integrable function such that
$\theta\leae f$ and $\int fd\nu=\Delta(\theta)$.   Then

\Centerline{$\theta I-f(a)\le\theta I-\theta a=\theta(I\setminus a)$}

\noindent for almost every $a$, so

\Centerline{$
\theta I-\Delta(\theta)
=\int\theta I-f(a)\nu(da)
\le\underline{\intop}\theta(I\setminus a)\nu(da)
=\underline{\intop}\theta(a)\nu(da)$}

\noindent because $a\mapsto I\setminus a$ is a measure space
automorphism, as in the proof of 464Ha.   So if
$\Delta(\theta)=\bover12\theta I$ then
$\underline{\int}\theta\,d\nu=\overline{\int}\theta\,d\nu$ and $\theta$
is $\nu$-integrable (133Jd), therefore (because $\nu$ is complete)
$(\dom\nu)$-measurable.   On the other hand, if $\theta$ is measurable,
then

\Centerline{$\Delta(\theta)=\biggerint\theta\,d\nu
=\int\theta(I\setminus a)\nu(da)
=\theta I-\int\theta\,d\nu=\theta I-\Delta(\theta)$,}

\noindent so surely $\Delta(\theta)=\bover12\theta I$.

\medskip

{\bf (d)} Of course $M_{\text{m}}$ is a linear subspace.   If
$\theta_0\in M_{\text{m}}$ and $|\theta|\le|\theta_0|$, then
$|\theta_0|\in M_{\text{m}}$, by (b), so
$\Delta(|\theta_0|)=\bover12|\theta_0|(I)$, by (c).   Because
$\theta^+\le|\theta|\le|\theta_0|$, $\Delta(\theta^+)=\bover12\theta^+I$
(464H(f-i)), and $\theta^+$ is measurable, by (c) in the reverse
direction.   Similarly, $\theta^-$ is measurable, and
$\theta=\theta^+-\theta^-$ is measurable.   As $\theta$ and $\theta_0$
are arbitrary, $M_{\text{m}}$ is solid.
}%end of proof of 464K

\leader{464L}{The space $M_{\text{pnm}}$:  Lemma} Let $I$ be any set,
$\nu$ the usual measure on $\Cal PI$, and $M$ the $L$-space of bounded
additive functionals on $\Cal PI$.   This time,
write $M_{\text{pnm}}$ for the set of those members of $M$ which are
purely non-measurable\cmmnt{ in the sense of 464I}.

(a) If $\theta\in M^+$, then $\theta$ is purely non-measurable iff
$\Delta(\theta)=\theta I$.

(b) $M_{\text{pnm}}$ is a solid linear subspace of $M$.

\proof{{\bf (a)(i)} If $\theta$ is purely non-measurable, and
$f\ge\theta$ is integrable, then $\{a:f(a)\ge\theta I\}$ is a measurable
set including $\{a:\theta a=\theta I\}$, so has measure $1$, and $\int
f\ge\theta I$;  as $f$ is arbitrary, $\Delta(\theta)=\theta I$.

\medskip

\quad{\bf (ii)} If $\Delta(\theta)=\theta I$, then
$\Delta(\theta')=\theta'I$ whenever $0\le\theta'\le\theta$, by 464Hf.
But this means that $\theta'$ cannot be measurable whenever
$0<\theta'\le\theta$, by 464Kc above, so that $\theta'\notin M_{\tau}$
whenever $0<\theta'\le\theta$, by 464Ja.   Thus $\theta\in
M_{\tau}^{\perp}$.

By 464Hc, $\nu^*\{a:\alpha\le\theta
a\le\Delta(\theta)\}=1$ for every $\alpha<\Delta(\theta)$.
Let $\sequencen{m(n)}$ be a sequence in $\Bbb N$ such that
$\prod_{n=0}^{\infty}1-2^{-m(n)}=\bover12$, and define $X$, $\lambda$
and $\phi$ as in 464Ba.   Set $\eta_n=2^{-n}/m(n)>0$ for each $n$.
Consider the sets
$A_n=\{a:\theta a\ge(1-\eta_n)\theta I\}$ for each $n\in\Bbb N$.   Then
$\nu^*A_n=1$ for each $n$, and
$\lambda^*(\prod_{n\in\Bbb N}A_n^{m(n)})=1$.   Because $\phi$ is \imp,
$\nu^*(\phi[\prod_{n\in\Bbb N}A_n^{m(n)}])=1$ (413Eh again).   But if
$\frak x=\sequencen{\langle a_{ni}\rangle_{i<m(n)}}$ belongs to
$\prod_{n\in\Bbb N}A_n^{m(n)}$, then

\Centerline{$\theta(\bigcap_{i<m(n)}a_{ni})
\ge\theta I-m(n)\eta_n\theta I=(1-2^{-n})\theta I$}

\noindent for each $n$, and

\Centerline{$\theta(\phi(\frak x))
\ge\sup_{n\in\Bbb N}\theta(\bigcap_{i<m(n)}a_{ni})=\theta I$.}

\noindent Thus the filter $\{a:\theta a=\theta I\}$ includes
$\phi[\prod_{n\in\Bbb N}A_n^{m(n)}]$ and has outer measure $1$, so that
$\theta$ is purely non-measurable.

\medskip

{\bf (b)(i)} If $\theta\in M_{\text{pnm}}$ and $|\theta'|\le|\theta|$,
then $\Delta(|\theta|)=|\theta|(I)$, by the definition in 464I, so
$\Delta(|\theta'|)=|\theta'|(I)$, by 464H(f-ii), and
$\theta'\in M_{\text{pnm}}$, by (a) above.   Thus $M_{\text{pnm}}$ is
solid.

\medskip

\quad{\bf (ii)} If $\theta_1$, $\theta_2\in M_{\text{pnm}}$, then

\Centerline{$\Delta(|\theta_1|+|\theta_2|)
=\Delta(|\theta_1|)+\Delta(|\theta_2|)
=|\theta_1|(I)+|\theta_2|(I)
=(|\theta_1|+|\theta_2|)(I)$}

\noindent (using 464Hb), and $|\theta_1|+|\theta_2|\in M_{\text{pnm}}$;
as $|\theta_1+\theta_2|\le|\theta_1|+|\theta_2|$, $\theta_1+\theta_2\in
M_{\text{pnm}}$.   Thus $M_{\text{pnm}}$ is closed under addition.

\medskip

\quad{\bf (iii)} It follows from (ii) that if $\theta\in M_{\text{pnm}}$
then $n\theta\in M_{\text{pnm}}$ for every integer $n\ge 1$, and then
from (i) that $\alpha\theta\in M_{\text{pnm}}$ for every $\alpha\in \Bbb
R$;  so that $M_{\text{pnm}}$ is closed under scalar multiplication, and
is a linear subspace.
}%end of proof of 464L

\leader{464M}{Theorem}\cmmnt{ ({\smc Fremlin \& Talagrand 79})} Let
$I$ be any set.   Write $M$ for the $L$-space of bounded finitely
additive functionals on $\Cal PI$,
and $M_{\text{m}}$, $M_{\text{pnm}}$ for the spaces of measurable and
purely non-measurable functionals\cmmnt{, as in 464K-464L}.   Then
$M_{\text{m}}$ and $M_{\text{pnm}}$ are complementary bands in $M$.

\proof{{\bf (a)} We know from 464K and 464L that these are both solid
linear subspaces of $M$.   Next, $M_{\text{m}}\cap M_{\text{pnm}}=\{0\}$.
\Prf\ If $\theta$ belongs to the intersection,
then $\Delta(|\theta|)=\bover12|\theta|(I)=|\theta|(I)$, by 464Kc and
464La;  so $\theta=0$.\ \Qed

\medskip

{\bf (b)} Now recall that every element of $M^+$ is expressible in the
form $\theta_1+\theta_2$ where $\theta_1\in M_{\text{m}}^+$ and
$\theta_2\in M_{\text{pnm}}^+$;  this is 464He, using 464Kc and 464La
again.   Because $M_{\text{m}}$ and $M_{\text{pnm}}$ are linear
subspaces, with intersection $\{0\}$,
$M=M_{\text{m}}\oplus M_{\text{pnm}}$.   Now
$M_{\text{pnm}}\subseteq M_{\text{m}}^{\perp}$,
so $M_{\text{m}}+M_{\text{m}}^{\perp}=M$ and
$M_{\text{m}}=M_{\text{m}}^{\perp\perp}$ is a complemented band (352Ra);
similarly, $M_{\text{pnm}}$ is a complemented band.   Since
$(M_{\text{m}}+M_{\text{pnm}})^{\perp}=\{0\}$, $M_{\text{m}}$ and
$M_{\text{pnm}}$ are complementary bands (see 352S).
}%end of proof of 464M

\leader{464N}{Corollary}\cmmnt{ ({\smc Fremlin \& Talagrand 79})} Let
$I$ be any set, and let $\mu$ be Talagrand's measure on $\Cal PI$;
write $\Sigma$ for its domain.   Then every bounded additive functional
on $\Cal PI$ is $\Sigma$-measurable.

\proof{ Defining $M$, $M_{\text{m}}$ and $M_{\text{pnm}}$ as in
464K-464M, %464K 464L 464M
we see that every functional in $M_{\text{m}}$ is
$\Sigma$-measurable because it is (by definition)
$(\dom\nu)$-measurable,
where $\nu$ is the usual measure on $\Cal PI$.   If
$\theta\in M_{\text{pnm}}^+$, then $\Cal F=\{a:\theta a=\theta I\}$ is a
non-measurable filter;  but this means that $\mu\Cal F=1$, by the
construction of $\mu$, so that $\theta=\theta I\,\,\mu$-a.e.   So if
$\theta$ is any member of $M_{\text{pnm}}$, both $\theta^+$ and
$\theta^-$ are $\Sigma$-measurable, and $\theta=\theta^+-\theta^-$ also
is.
}%end of proof of 464N

\leader{464O}{Remark}\cmmnt{ Note that we have a very simple
description of the behaviour of additive functionals as seen by the
measure $\mu$.}   Since $M_{\tau}\subseteq M_{\text{m}}$, we have a
three-part band decomposition
$M=M_{\tau}\oplus(M_{\text{m}}\cap M_{\tau}^{\perp})
\oplus M_{\text{pnm}}$.

(i) Functionals in $M_{\tau}$ are $\Tau$-measurable, where $\Tau$ is the
domain of $\nu$, therefore
$\Sigma$-measurable\cmmnt{, just because they can be built up from the
functionals $a\mapsto\chi a(t)$, as in 464Fb}.

(ii) A functional\cmmnt{ in $M_{\text{m}}$ is $\Tau$-measurable, by
definition;  but a functional} $\theta$ in
$M_{\text{m}}\cap M_{\tau}^{\perp}$ is\cmmnt{ actually} constant, with
value $\bover12\theta I$, $\nu$-almost everywhere\cmmnt{, by 464Ka}.
\cmmnt{Thus the almost-constant nature of the functionals described in
464Jb is typical of measurable functionals in $M_{\tau}^{\perp}$.}

(iii) Finally, a functional $\theta\in M_{\text{pnm}}$ is equal to
$\theta I\,\,\mu$-almost everywhere\cmmnt{;  once again, this follows
from the definition of `purely non-measurable' and the construction of
$\mu$ for $\theta\ge 0$, and from the fact that $M_{\text{pnm}}$ is
solid for other $\theta$}.

(iv) Thus\cmmnt{ we see that} any
$\theta\in M_{\tau}^{\perp}
=(M_{\text{m}}\cap M_{\tau}^{\perp})\oplus M_{\text{pnm}}$ is constant
$\mu$-a.e.   \cmmnt{We also have}
$\int\theta\,d\mu=\Delta(\theta)$ for every $\theta\ge 0$
\prooflet{(look at $\theta\in M_{\text{m}}$ and $\theta\in
M_{\text{pnm}}$ separately, using 464La for the latter)}, so that if
$h\in M^*$ is the linear functional of 464Hb, then
$\int\theta\,d\mu=h(\theta)$ for every $\theta\in M$.

\leader{464P}{More on purely non-measurable functionals (a)} We can
discuss non-negative additive functionals on $\Cal PI$ in terms of the
Stone-\v Cech compactification $\beta I$ of $I$\cmmnt{, as follows}.
For any set $A\subseteq\beta I$ set
$H_A=\{a:a\subseteq I,\,A\subseteq\widehat{a}\}$, where
$\widehat{a}\subseteq\beta I$ is the
open-and-closed set corresponding to $a\subseteq I$.   If
$A\ne\emptyset$, $H_A$ is a filter on $I$.   Write $\Cal A$ for the
family of those sets $A\subseteq\beta I$ such that $\nu^*H_A=1$,
where $\nu$ is the usual measure on $\Cal PI$.   Then $\Cal A$ is a
$\sigma$-ideal.
\prooflet{\Prf\ Of course $A\in\Cal A$ whenever $A\subseteq B\in\Cal A$,
since then $H_A\supseteq H_B$.   If $\sequencen{A_n}$ is a sequence in
$\Cal A$ with union $A$, then $\nu^*(\bigcap_{n\in\Bbb N}H_{A_n})=1$, by
464C;  but $H_A\supseteq\bigcap_{n\in\Bbb N}H_{A_n}$.\ \QeD\ }   Note
that if $A\in\Cal A$ then $\overline{A}\in\Cal A$\cmmnt{, because
$H_A=H_{\overline{A}}$}.   \cmmnt{We see also that} $\{z\}\in\Cal A$
for every $z\in\beta I\setminus I$\cmmnt{ (since $H_{\{z\}}$ is a
non-principal ultrafilter, as in 464Jc)}, while $\{t\}\notin\Cal A$ for
any $t\in I$\cmmnt{ (since $H_{\{t\}}=\{a:t\in a\}$ has measure
$\bover12$)}.

\dvro{We}{Because $\beta I$ can be identified with the Stone space of
$\Cal PI$ (4A2I(b-i)), we}
have a one-to-one correspondence between non-negative additive
functionals $\theta$ on $\Cal PI$ and Radon measures $\mu_{\theta}$ on
$\beta I$, defined by
writing $\mu_{\theta}(\widehat{a})=\theta a$ whenever $a\subseteq I$ and
$\theta\in M^+$\cmmnt{ (416Qb)}.
\cmmnt{(This is not the same as the measure $\mu_{\theta}$ of part (e)
of the proof of 464H, which is on a much larger space.)}
Now suppose that $\theta$ is a non-negative additive functional on
$\Cal PI$.   Then $\Cal F_{\theta}=\{a:\theta a=\theta I\}$ is either
$\Cal PI$ or a filter on $I$.   If we set
$F_{\theta}=\bigcap\{\widehat{a}:a\in\Cal F_{\theta}\}$, then
$\Cal F_{\theta}=H_{F_{\theta}}$\cmmnt{ (4A2I(b-iii))}.
\cmmnt{Since $a\in\Cal F_{\theta}$ iff $\theta a=\theta I$,

$$\eqalign{F_{\theta}
&=\bigcap\{\widehat{a}:a\in\Cal F_{\theta}\}
=\bigcap\{\widehat{a}:\theta a=\theta I\}\cr
&=\bigcap\{\widehat{a}:\mu_{\theta}\widehat{a}=\mu_{\theta}(\beta I)\}
=\beta I\setminus\bigcup\{\widehat{a}:\mu_{\theta}\widehat{a}=0\}.\cr}$$

\noindent But this is just the support of $\mu_{\theta}$ (411N),
because $\{\widehat{a}:a\subseteq I\}$ is a base for the topology of
$\beta I$.
}%end of comment

\cmmnt{Thus we see that} $\theta\in M^+$ is purely non-measurable iff
the support\cmmnt{ $F_{\theta}$} of the measure $\mu_{\theta}$ belongs to
$\Cal A$.   \cmmnt{If you like, $\theta$ is purely non-measurable iff the
support of $\mu_{\theta}$ is `small'.}

\spheader 464Pb\cmmnt{ Yet another corollary of 464C is the
following.}   Since $M$ is a set of real-valued functions on $\Cal PI$,
it has the corresponding topology $\frak T_p$ of pointwise convergence
as a subspace of
$\BbbR^{\Cal PI}$.   Now if $C\subseteq M_{\text{pnm}}$ is countable,
its $\frak T_p$-closure $\overline{C}$ is included in
$M_{\text{pnm}}$.   \prooflet{\Prf\ It is enough to consider the case in
which $C$ is non-empty and $0\notin C$.   For each $\theta\in C$,
$\Cal F_{|\theta|}=\{a:|\theta|(a)=|\theta|(I)\}$ is a filter with outer
measure $1$, so $\Cal F=\{a:|\theta|(a)=|\theta|(I)$ for every
$\theta\in C\}$ also has outer measure $1$, by 464Cb.   Now suppose that
$\theta_0\in\overline{C}$.   If $a\in\Cal F$, then
$|\theta|(I\setminus a)=0$ for every $\theta\in C$, that is,
$\theta b=0$ whenever $\theta\in C$ and $b\subseteq I\setminus a$
(362Ba).   But this means that
$\theta_0b=0$ whenever $b\subseteq I\setminus a$, so
$|\theta_0|(I\setminus a)=0$ and $|\theta_0|(a)=|\theta_0|(I)$.   Thus
$\Cal F_{|\theta_0|}$ includes $\Cal F$, so has outer measure $1$, and
$\theta_0$ also is purely non-measurable.\ \Qed}

\spheader 464Pc If $\theta\in M$ is such that $\theta a=0$ for every
countable set $a\subseteq I$, then $\theta\in M_{\text{pnm}}$.
\prooflet{\Prf\  $\nu$ is inner regular with respect to the family
$\Cal W$ of sets
which are determined by coordinates in countable subsets of $I$, by
254Ob.   But if $W\in\Cal W$ and $\nu W>0$, let $J\subseteq I$
be a countable set such that $W$ is determined by coordinates in $J$;
then $|\theta|(J)=0$, so if $a$ is any member of $W$ we shall have
$a\cup(I\setminus J)\in W\cap\Cal F_{|\theta|}$.   As $W$ is arbitrary,
$\nu^*\Cal F_{|\theta|}=1$ and $\theta$ is purely non-measurable.\ \Qed}

In particular, if $\theta\in M_{\sigma}\cap M_{\tau}^{\perp}$, where
$M_{\sigma}$ is the space of countably additive functionals on
$\Cal PI$\cmmnt{ (362B)}, then $\theta\in M_{\text{pnm}}$.
\cmmnt{(For `ordinary' sets $I$, $M_{\sigma}=M_{\tau}$;  see
438Xa.   But this observation is peripheral to the concerns of the
present section.)}

In the language of (a) above, we have a closed set in $\beta I$, being
$F=\beta I\setminus\bigcup\{\widehat{a}:a\in[I]^{\le\omega}\}$;  and if
$\theta$ is such that the support of $\mu_{\theta}$ is included in $F$,
then $\theta$ is purely non-measurable.

\leader{464Q}{More on measurable functionals (a)} We know that
$M_{\text{m}}$ is a band in $M$, and that it includes the band
$M_{\tau}$.   So it is natural to look at the band
$M_{\text{m}}\cap M_{\tau}^{\perp}$.

\spheader 464Qb If $\theta$ is any non-zero non-negative
functional in $M_{\text{m}}\cap M_{\tau}^{\perp}$, we can find a family
$\langle a_{\xi}\rangle_{\xi<\omega_1}$ in $\Cal PI$ which is
independent in the sense that $\theta(\bigcap_{\xi\in K}a_{\xi})
=2^{-\#(K)}\theta I$ for every non-empty finite $K\subseteq I$.
\prooflet{\Prf\ Choose
the $a_{\xi}$ inductively, observing that at the inductive step we have
to satisfy only countably many conditions of the form
$\theta(a_{\xi}\cap b)=\bover12\theta b$, where $b$ runs over the
subalgebra generated by $\{a_{\eta}:\eta<\xi\}$, and that each such
condition is satisfied $\nu$-a.e., by 464Ka;  so that $\nu$-almost
any $a$ will serve for $a_{\xi}$.\ \Qed}

In terms of the associated measure $\mu_{\theta}$ on $\beta I$, this
means that $\mu_{\theta}$ has Maharam type at least
$\omega_1$\cmmnt{ (use 331Ja)}.
If $\theta I=1$\cmmnt{, so that $\mu_{\theta}$ is a probability
measure}, then
$\langle(\widehat{a}_{\xi})^{\ssbullet}\rangle_{\xi<\omega_1}$ is an
uncountable stochastically
independent family in the measure algebra of
$\mu_{\theta}$\cmmnt{ (325Xf)},

Turning this round, we see that if $\lambda$ is a Radon measure on
$\beta I$, of countable Maharam type, and $\lambda I=0$, then the
corresponding functional on $\Cal PI$ is purely non-measurable.

\cmmnt{[For a stronger result in this direction, see 521T
in Volume 5.]}

\spheader 464Qc \cmmnt{Another striking property of measurable
additive functionals is the following.}   If
$\theta\in M_{\text{m}}\cap M_{\tau}^{\perp}$, and $n\in\Bbb N$, then $\theta(a_0\cap a_1\cap\ldots\cap a_n)=2^{-n-1}\theta I$ for $\nu^{n+1}$-almost every
$a_0,\ldots,a_n\subseteq I$, where $\nu^{n+1}$ is the product measure on
$(\Cal PI)^{n+1}$.   \prooflet{\Prf\ For $K\subseteq\{0,\ldots,n\}$
write $\psi_K(a_0,\ldots,a_n)=I\cap\bigcap_{i\in
K}a_i\setminus\bigcup_{i\le n,i\notin K}a_i$ for
$a_0,\ldots,a_n\subseteq I$;  for $\Cal S\subseteq\Cal P\{0,\ldots,n\}$
write $\phi_{\Cal S}(\frak x)=\bigcup_{K\in\Cal S}\psi_K(\frak x)$ for
$\frak x\in(\Cal PI)^{n+1}$.   Then, for any $t\in I$,

\Centerline{$\nu^{n+1}\{\frak x:t\in\phi_{\Cal S}(\frak x)\}
=\sum_{K\in\Cal S}\nu^{n+1}\{\frak x:t\in\phi_K(\frak x)\}
=2^{-n-1}\#(\Cal S)$.}

\noindent Moreover, for different $t$, these sets are independent.   So
if $\#(\Cal S)=2^n$, that is, if $\Cal S$ is just half of
$\Cal P\{0,\ldots,n\}$, then $\phi_{\Cal S}$ will be \imp, by the arguments in 464B.   (In fact 464Bb is the special case $n=2$,
$\Cal S=\{K:\#(K)\ge 2\}$.)

Accordingly we shall have
$\theta(\phi_{\Cal S}(\frak x))=\bover12\theta I$ for
$\nu^{n+1}$-almost every
$\frak x$ whenever $\Cal S\subseteq\Cal P\{0,\ldots,n\}$ has $2^n$ members,
as in 464Ka.   Since there are only finitely many
sets $\Cal S$, the set $E$ is $\nu^{n+1}$-conegligible, where

\Centerline{$E=\{\frak x:\frak x\in(\Cal PI)^{n+1},\,
  \theta(\phi_{\Cal S}(\frak x))=\Bover12\theta I$ whenever
  $\Cal S\subseteq\Cal P\{0,\ldots,n\}$ and $\#(\Cal S)=2^n\}$.}

\noindent But given $\frak x=(a_0,\ldots,a_n)\in E$, let
$K_1,\ldots,K_{2^{n+1}}$ be a listing of $\Cal P\{0,\ldots,n\}$ in such
an order that
$\theta(\psi_{K_i}(\frak x))\le\theta(\psi_{K_j}(\frak x))$ whenever $i\le j$, and consider $\Cal S=\{K_1,\ldots,K_{2^n}\}$; since

\Centerline{$\sum_{i=1}^{2^n}\theta(\psi_{K_i}(\frak x))
=\theta(\phi_{\Cal S}(\frak x))
=\Bover12\theta I
=\Bover12\sum_{i=0}^{2^{n+1}}\theta(\psi_{K_i}(\frak x))$,}

\noindent we must have $\theta(\psi_{K_i}(\frak x))=2^{-n-1}\theta I$
for every $i$.   In particular, $\theta(\psi_{\{0,\ldots,n\}}(\frak
x))=2^{-n-1}\theta I$, that is,
$\theta(\bigcap_{i\le n}a_i)=2^{-n-1}\theta I$.   As this is true
whenever $a_0,\ldots,a_n\in E$, we have the result.\ \Qed
}%end of prooflet

\leader{464R}{A note on $\ell^{\infty}(I)$}\cmmnt{ As already noted in
464F, we have, for any set $I$, a natural additive map
$\chi:\Cal PI\to\ell^{\infty}(I)\cong L^{\infty}(\Cal PI)$, giving rise
to an isomorphism between the $L$-space $M$ of bounded additive
functionals on
$\Cal PI$ and $\ell^{\infty}(I)^*$.}   If we write $\tilde\mu$ for the
image measure $\mu\chi^{-1}$ on $\ell^{\infty}(I)$, where $\mu$ is
Talagrand's measure on $\Cal PI$, and $\tilde\Sigma$ for the domain of
$\tilde\mu$, then\cmmnt{ every member of $\ell^{\infty}(I)^*$ is
$\tilde\Sigma$-measurable, by 464N.   Thus} $\tilde\Sigma$ includes the
cylindrical $\sigma$-algebra of $\ell^{\infty}(I)$\cmmnt{ (4A3U)}.
\cmmnt{We also have a band decomposition $\ell^{\infty}(I)^*
=\ell^{\infty}(I)^*_{\text{m}}\oplus\ell^{\infty}(I)^*_{\text{pnm}}$
corresponding to the decomposition
$M=M_{\text{m}}\oplus M_{\text{pnm}}$ (464M).}

\cmmnt{In this context,} $M_{\tau}$ corresponds to
$\ell^{\infty}(I)^{\times}$\cmmnt{ (363K, as before)}.
\cmmnt{Since we can
identify $\ell^{\infty}(I)$ with $\ell^1(I)^*=\ell^1(I)^{\times}$
(243Xl), and $\ell^1(I)$, like any
$L$-space, is perfect, $\ell^{\infty}(I)^{\times}$ is the canonical
image of $\ell^1(I)$ in $\ell^{\infty}(I)^*$.}  \dvro{Any}{Because any
functional in $M_{\tau}^{\perp}$ is $\mu$-almost constant (464O),
any} functional in $(\ell^{\infty}(I)^{\times})^{\perp}$ will be
$\tilde\mu$-almost constant.



\exercises{\leader{464X}{Basic exercises (a)}
%\spheader 464Xa
Let $I$ be any set and $\lambda$ a Radon measure on
$\beta I$.   Show that if the support of $\lambda$ is a separable subset
of $\beta I\setminus I$, then the corresponding additive functional on
$\Cal PI$ is purely non-measurable.
%464P

\spheader 464Xb Let $I$ be any set, and $\tilde\mu$ the image of
Talagrand's measure on $\ell^{\infty}(I)$, as in 464R.   Show that
$\tilde\mu$ has a barycenter in $\ell^{\infty}(I)$ iff $I$ is finite.
%464R

\leader{464Y}{Further exercises (a)}
%\spheader 464Ya
Show that there is a sequence $\sequencen{\Cal F_n}$ of distinct
non-principal ultrafilters on $\Bbb N$ with the following property:  if
we define $h(a)=\{n:a\in\Cal F_n\}$ for $a\subseteq\Bbb N$, then
$\{h(a):a\subseteq\Bbb N\}$ is negligible for the usual measure on
$\Cal P\Bbb N$.
%464E  %mt46bits

\spheader 464Yb
Let $\mu$ be Talagrand's measure on $\Cal P\Bbb N$, and
$\lambda$ the corresponding product measure on
$X=\Cal P\Bbb N\times\Cal P\Bbb N$.   Define $\Phi:X\to\ell^{\infty}$ by
setting $\Phi(a,b)=\chi a-\chi b$ for all $a$, $b\subseteq\Bbb N$.
Show that $\Phi$ is Pettis integrable (463Ya) with indefinite Pettis
integral $\Theta$ defined by setting $(\Theta E)(n)=\int_Ef_nd\lambda$,
where $f_n(a,b)=\chi a(n)-\chi b(n)$.   Show that
$K=\{h\Phi:h\in\ell^{\infty*},\,\|h\|\le 1\}$ contains every $f_n$, and
in particular is not $\frak T_m$-compact, so the identity map from
$(K,\frak T_p)$ to $(K,\frak T_m)$ is not continuous.
%464N

\spheader 464Yc Let $I$ be any set.   Write $\pmb{c}_0(I)$ for the closed
linear subspace of $\ell^{\infty}(I)$ consisting of those $x\in\Bbb R^I$
such that $\{t:t\in I$, $|x(t)|\ge\epsilon\}$ is finite for every
$\epsilon>0$;  that is, $C_0(I)$ if $I$ is given its discrete topology
(436I).   Show that, in 464R, $M_{\tau}^{\perp}$ can be identified as
Banach lattice with $(\ell^{\infty}(I)/\pmb{c}_0(I))^*$.
%464R

\spheader 464Yd\dvAnew{2011}(i) Let $\theta:\Cal P\Bbb N\to\Bbb R$
be an additive functional which is $\Tau$-measurable in the
sense of 464I.   Show that
$\{\theta\{n\}:n\in\Bbb N\}$ is bounded.
(ii) Let $\theta:\Cal P\Bbb N\to\Bbb R$ be an additive functional which is
universally measurable for the usual topology of $\Cal P\Bbb N$.
Show that $\theta$ is bounded.
(iii) Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $\theta:\frak A\to\Bbb R$ an
additive functional which is universally measurable for the
order-sequential topology on $\frak A$
(definition:  393L).   Show that $\theta$ is bounded
and $\theta^+$ is universally measurable.
%464J mt46bits

\spheader 464Ye\dvAnew{2011} Show that there is a $\Tau$-measurable
finitely additive
functional $\theta:\Cal P\Bbb N\to\Bbb R$ which is not bounded.
%464Jb 464Ye out of order query
%query:  Borel measurable => completely additive?
}%end of exercises

\leader{464Z}{Problem} Let $I$ be an infinite set, and $\tilde\mu$ the
image on $\ell^{\infty}(I)$ of Talagrand's measure\cmmnt{ (464R)}.
Is $\tilde\mu$ a topological measure for the
weak topology of $\ell^{\infty}(I)$?

\endnotes{
\Notesheader{464} The central idea of this section appears in 464B:  the
algebraic structure of $\Cal PI$ leads to a variety of \imp\ functions
$\phi$ from powers $(\Cal PI)^K$ to $\Cal PI$.   The simplest of these
is the measure space automorphism $a\mapsto I\setminus a$, as used in
the proofs of 464Ca, 464Ha, 464Ka and 464Kc.   Then we have the map
$(a,b,c)\mapsto(a\cap b)\cup(a\cap c)\cup(b\cap c)$, as in 464Bb, and
the generalization of this in the argument of 464Qc;  and, most
important of all, the map
$\sequencen{\langle a_{ni}\rangle_{i<m(n)}}
\mapsto\bigcup_{n\in\Bbb N}\bigcap_{i<m(n)}a_{ni}$ of 464Ba.   In each
case we can use
probabilistic intuitions to guide us to appropriate formulae, since the
events $t\in\phi(\frak x)$ are always independent, so if they have
probability $\bover12$ for every $t\in I$, the function $\phi$ will be
\imp.   Of course this depends on the analysis of product measures in
\S254.   It means also that we must use the `ordinary' product measure
defined there;  but happily this coincides with the `Radon' product
measure of \S417 (416U).

Talagrand devised his measure when seeking an example of a pointwise
compact set of measurable functions which is not compact for the
topology of convergence in measure, as in 464E.   The remarkable fact
that it is already, in effect, a measure on the cylindrical
$\sigma$-algebra of $\ell^{\infty}$ (464R) became apparent later, and
requires a much more detailed analysis.   An alternative argument not
explicitly involving the Riesz space structure of the space $M$ of
bounded additive functionals may be found in
{\smc Fremlin \& Talagrand 79}.   The proof I give here depends on the surprising fact that, for
non-negative additive functionals, the upper integral
$\overline{\int}\,d\nu$ is additive (464Hb), even though the functionals
may be very far from being measurable.   Once we know this, we can apply
the theory of Banach lattices to investigate the corresponding linear
functional on the $L$-space $M$.   There is a further key step in 464Hd.
We there have a non-negative $\theta\in M_{\tau}^{\perp}$ such that
$\theta I=1$ and
$\beta\theta'I\le\overline{\int}\theta'd\nu\le\gamma\theta'I$ whenever
$0\le\theta'\le\theta$.   It is easy to deduce that
$\{a:\alpha\theta b\le\theta(a\Bcap b)\le\gamma\theta b$
for every $b\in K\}$ has outer
measure $1$ for every finite $K\subseteq\Cal PI$ and $\alpha<\beta$.
If $\alpha$ and
$\gamma$ are very close, this means that there will be many families
$(a,b,c)$ in $\Cal PI$ which, as measured by $\theta$, look like
independent sets of measure $\gamma$, so that $\theta((a\cap
b)\cup(a\cap c)\cup(b\cap c))\bumpeq 3\gamma^2-2\gamma^3$.   But since
we must also have $\theta((a\cap b)\cup(a\cap c)\cup(b\cap c))\le\gamma$
almost everywhere, we can get information about the possible values of
$\gamma$.

Once having noted this remarkable dichotomy between `measurable' and
`purely non-measurable' functionals, it is natural to look for other
ways in which they differ.   The position seems to be that `simple'
Radon measures on $\beta I\setminus I$ (e.g., all measures with
separable support (464Xa) or countable Maharam type (464Qb)) have to
correspond to purely non-measurable functionals.   Of course the
simplest possible measures on $\beta I$ are those concentrated on $I$,
which give rise to functionals which belong to $M_{\tau}$ and are
therefore measurable;  it is functionals in
$M_{\text{m}}\cap M_{\tau}^{\perp}$ which have to give rise to
`complicated' Radon measures.

The fact that every element of $M_{\tau}^{\perp}$ is almost constant for
Talagrand's measure leads to an interesting Pettis integrable function
(464Yb).
The suggestion that Talagrand's measure on $\ell^{\infty}$ might be a
topological measure for the weak topology (464Z) is a bold one, but no
more outrageous than the suggestion that it might measure every
continuous linear functional once seemed.   Talagrand's measure does of
course measure every Baire set for the weak topology (4A3V).
I note here that the usual measure on $\{0,1\}^I$, when transferred to
$\ell^{\infty}(I)$, is actually a Radon measure for the weak*
topology $\frak T_s(\ell^{\infty},\ell^1)$, because on $\{0,1\}^I$ this
is just the ordinary topology.
}%end of notes

\discrpage


\frfilename{mt465.tex}
\versiondate{22.3.16}
\copyrightdate{1999}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Stable sets}

\newsection{465}

The structure of general pointwise compact sets of measurable functions
is complex and elusive.   One particular class of such sets, however, is
relatively easy to describe, and has a variety of remarkable properties,
some of them relevant to important questions arising in the theory of
empirical measures.   In this section I outline the theory of `stable'
sets of measurable functions from {\smc Talagrand 84} and
{\smc Talagrand 87}.

The first steps are straightforward enough.   The definition of stable
set (465B) is not obvious, but given this the basic properties of stable
sets listed in 465C are natural and easy to check, and we come quickly
to the fact that (for complete locally determined spaces) pointwise
bounded stable sets are relatively pointwise compact sets of measurable
functions (465D).   A less transparent, but still fairly elementary,
argument leads to the next reason for looking at stable sets:  the
topology of pointwise convergence on a stable set is finer than the
topology of convergence in measure (465G).

At this point we come to a remarkable fact:  a uniformly bounded set $A$
of functions on a complete probability space is stable if and only if
certain laws of large numbers apply `nearly uniformly' on $A$.   These
laws are expressed in conditions (ii), (iv) and (v) of 465M.   For
singleton sets $A$, they can be thought of as versions of the strong
law of large numbers described in \S273.   To get the full strength of
465M a further idea in this direction needs to be added,
described in 465H here.

The theory of stable sets applies in the first place to sets of true
functions.   There is however a corresponding notion applicable in
function spaces, which I explore briefly in 465O-465R.   Finally, I
mention the idea of `R-stable' set (465S-465U), obtained by using
$\tau$-additive product measures instead of c.l.d.\ product measures in
the definition.

\leader{465A}{Notation}\cmmnt{ Throughout this section, I will use
the following notation.

\medskip

} {\bf (a)} If $X$ is a set and $\Sigma$ a $\sigma$-algebra of subsets of
$X$, I will write $\eusm L^0(\Sigma)$ for the space of
$\Sigma$-measurable functions from $X$ to $\Bbb R$\cmmnt{, as in
\S463};  $\eusm L^{\infty}(\Sigma)$ will be the space of bounded
functions in $\eusm L^0(\Sigma)$.

\spheader 465Ab I will identify $\Bbb N$ with the set of finite
ordinals, so that\cmmnt{ each $n\in\Bbb N$ is the set of its
predecessors, and} a power $X^n$ becomes identified with the set of
functions from $\{0,\ldots,n-1\}$ to $X$.

\spheader 465Ac If $(X,\Sigma,\mu)$ is any measure space, then for
finite sets $I$\cmmnt{ (in particular, if $I=k\in\Bbb N$)} I write
$\mu^I$ for the\cmmnt{ c.l.d.} product measure on $X^I$\cmmnt{, as
defined in 251W}.   \cmmnt{(For definiteness, let us take $\mu^0$ to
be the unique probability measure on $X^0=\{\emptyset\}$.)}   If
$(X,\Sigma,\mu)$ is a probability space, then for any set $I\,\,\mu^I$
is to be the product probability measure on $X^I$\cmmnt{, as defined
in \S254}.

\spheader 465Ad\cmmnt{ We shall have occasion to look at free powers
of algebras of sets.}   If $X$ is a set and $\Sigma$ is an algebra of
subsets of $X$, then for any set $I$ write $\bigotimes_I\Sigma$ for the
algebra of subsets of $X^I$ generated by sets of the form
$\{w:w(i)\in E\}$ where $i\in I$ and $E\in\Sigma$, and
$\Tensorhat_I\Sigma$ for the
$\sigma$-algebra generated by $\bigotimes_I\Sigma$.

\spheader 465Ae\cmmnt{ Now for a new idea, which will be used in
almost every paragraph of the section.}   If $X$ is a set,
$A\subseteq\Bbb R^X$\cmmnt{ a set of real-valued functions defined on
$X$}, $E\subseteq X$, $\alpha<\beta$ in $\Bbb R$ and $k\ge 1$, write

$$\eqalign{D_k(A,E,\alpha,\beta)
=\bigcup_{f\in A}\{w:w\in E^{2k},\,
&f(w(2i))\le\alpha,\cr
&f(w(2i+1))\ge\beta\text{ for }i<k\}.\cr}$$

\spheader 465Af\cmmnt{ In this context it will be useful to have a
special
notation.}   If $X$ is a set, $k\ge 1$, $u\in X^k$ and $v\in X^k$, then I
will write $u\#v=(u(0),v(0),u(1),v(1),\ldots,u(k-1),v(k-1))\in X^{2k}$.
Note that if $(X,\Sigma,\mu)$ is a measure space then
$(u,v)\mapsto u\#v$ is an isomorphism between the c.l.d.\ product
$(X^k,\mu^k)\times(X^k,\mu^k)$ and
$(X^{2k},\mu^{2k})$\cmmnt{ (see 251Wh)}.

\cmmnt{We are now ready for the main definition.}

\leader{465B}{Definition} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space.   \cmmnt{Following {\smc Talagrand 84},} I say that a
set $A\subseteq\Bbb R^X$ is {\bf stable} if whenever $E\in\Sigma$,
$0<\mu
E<\infty$ and $\alpha<\beta$ in $\Bbb R$, there is some $k\ge 1$ such
that $(\mu^{2k})^*D_k(A,E,\alpha,\beta)<(\mu E)^{2k}$.

\cmmnt{\medskip

\noindent{\bf Remark} I hope that the next few results will show why
this concept is important.   It is worth noting at once that these sets
$D_k$ need not be measurable, and that some of the power of the
definition derives precisely from the fact that quite naturally arising
sets $A$ can give rise to non-measurable sets $D_k(A,E,\alpha,\beta)$.
If, however, the set $A$ is countable, then all the corresponding $D_k$
will be measurable;  this will be important in the results following
465R.
}%end of comment

\leader{465C}{}\cmmnt{ I start with a list of the `easy'
properties of stable sets, derivable more or less directly from the
definition.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a semi-finite
measure space.

(a) Let $A\subseteq\Bbb R^X$ be a stable set.

\quad(i) Any subset of $A$ is stable.  %4{}65Ca

\quad(ii) $\overline{A}$, the closure
of $A$ in $\Bbb R^X$ for the topology of pointwise convergence, is
stable.  %4{}65Cb

\quad(iii)
$\gamma A=\{\gamma f:f\in A\}$ is stable, for any $\gamma\in\Bbb R$.
%4{}65Ce

\quad(iv) If $g\in\eusm L^0=\eusm L^0(\Sigma)$, then
$A+g=\{f+g:f\in A\}$ is stable.  %4{}65Cf

\quad(v) If $g\in\eusm L^0$, then
$A\times g=\{f\times g:f\in A\}$ is stable.  %4{}65Ch

\quad(vi) Let $h:\Bbb R\to\Bbb R$ be a continuous non-decreasing function.
Then $\{hf:f\in A\}$ is stable.  %4{}65Ck

\ifdim\pagewidth>467pt\fontdimen3\tenrm=2pt\fi
\ifdim\pagewidth>467pt\fontdimen4\tenrm=1.67pt\fi
(b)(i) Suppose that $A\subseteq\Bbb R^X$, $E\in\Sigma$, $n\ge 1$ and
$\alpha<\beta$ are such that $0<\mu E<\infty$ and
$(\mu^{2n})^*D_n(A,E,\alpha,\beta)<(\mu E)^{2n}$.   Then
\fontdimen3\tenrm=1.67pt
\fontdimen4\tenrm=1.11pt

\Centerline{$\lim_{k\to\infty}
\Bover1{(\mu E)^{2k}}(\mu^{2k})^*D_k(A,E,\alpha,\beta)=0$.}
%4{}65Cc

\quad(ii) If $A$, $B\subseteq\Bbb R^X$ are stable, then $A\cup B$ is stable.
%4{}65Cd

\quad(iii) If $A\subseteq\eusm L^0$ is finite it is stable. %4{}65Cg

\quad(iv) If $A\subseteq\Bbb R^X$ is stable, so is
$\{f^+:f\in A\}\cup\{f^-:f\in A\}$. %4{}65Cl

(c) Let $A$ be a subset of $\BbbR^X$.

\quad(i) If $\hat\mu$, $\tilde\mu$ are the completion and c.l.d.\ version of
$\mu$, then $A$ is stable with respect to one of the
measures $\mu$, $\hat\mu$, $\tilde\mu$ iff it is stable with respect to
the others.   %4{}65Ci

\quad(ii) Let $\nu$ be an indefinite-integral measure over
$\mu$\cmmnt{ (234J)}.   If $A$ is stable with respect
to $\mu$, it is stable with respect to $\nu$ and with respect to
$\nu\restr\Sigma$.  %4{}65Cj

\quad(iii) If $A$ is stable, and $Y\subseteq X$ is such that the
subspace measure $\mu_Y$ is semi-finite, then $A_Y=\{f\restr Y:f\in A\}$ is
stable in $\BbbR^Y$ with respect to the measure $\mu_Y$.  %4{}65Cm

\quad(iv) $A$ is stable iff $A_E=\{f\restr E:f\in A\}$ is
stable in $\BbbR^E$ with respect to the subspace measure $\mu_E$ whenever
$E\in\Sigma$ has finite measure.  %4{}65Cn

\quad(v)\dvAnew{2016}
$A$ is stable iff $A_n=\{\med(-n\chi X,f,n\chi X):f\in A\}$ is stable
for every $n\in\Bbb N$.

(d)\dvAformerly{4{}65Xe}
Suppose that $\mu$ is $\sigma$-finite, $(Y,\Tau,\nu)$ is another measure
space and $\phi:Y\to X$ is \imp.   If $A\subseteq\BbbR^X$ is stable with
respect to $\mu$, then $B=\{f\phi:f\in A\}$ is stable with respect to $\nu$.

\proof{{\bf (a)(i)} This is immediate from the definition in 465B, since
$D_k(B,E,\alpha,\beta)\subseteq D_k(A,E,\alpha,\beta)$ for all $k$, $E$,
$\alpha$, $\beta$ and $B\subseteq A$.

\medskip

\quad{\bf (ii)} Given $E$ such that $0<\mu E<\infty$, and $\alpha<\beta$, take
$\alpha'$, $\beta'$ such that $\alpha<\alpha'<\beta'<\beta$.
Then it is easy to see that $D_k(\overline{A},E,\alpha,\beta)\subseteq
D_k(A,E,\alpha',\beta')$, so

\Centerline{$(\mu^{2k})^*D_k(\overline{A},E,\alpha,\beta)
\le(\mu^{2k})^*D_k(A,E,\alpha',\beta')
<(\mu E)^{2k}$}

\noindent for some $k\ge 1$.   As $E$,
$\alpha$ and $\beta$ are arbitrary, $\overline{A}$ is stable.

\medskip

\quad{\bf (iii)}\grheada\ If $\gamma>0$,
$D_k(\gamma A,E,\alpha,\beta)=D_k(A,E,\alpha/\gamma,\beta/\gamma)$ for
all $k$, $E$, $\alpha$ and $\beta$, so the result is elementary.
Similarly, if $\gamma=0$, then $D_k(\gamma A,E,\alpha,\beta)=\emptyset$
whenever $k\ge 1$, $E\in\Sigma$ and $\alpha<\beta$, so again we see that
$\gamma A$ is stable.

\medskip

\qquad\grheadb\ If
$\gamma=-1$ then, for any $k$, $E$, $\alpha$ and $\beta$,

$$\eqalign{D_k(-A,E,\alpha,\beta)
&=\bigcup_{f\in A}\{w:w\in E^{2k},\,f(w(2i))\ge-\alpha,\cr
&\mskip170mu f(w(2i+1))\le-\beta\Forall i<k\}\cr
&=\phi[D_k(A,E,-\beta,-\alpha)],\cr}$$

\noindent where $\phi:X^{2k}\to X^{2k}$ is the measure space
automorphism defined by setting

\Centerline{$\phi(w)=(w(1),w(0),w(3),w(2),\ldots,w(2k-1),w(2k-2))$}

\noindent for $w\in X^{2k}$.   So, given $E\in\Sigma$ and
$\alpha<\beta$, there is a $k\ge 1$ such that

\Centerline{$(\mu^{2k})^*D_k(-A,E,\alpha,\beta)
=(\mu^{2k})^*D_k(A,E,-\beta,-\alpha)
<(\mu E)^{2k}$.}

\noindent So $-A$ is stable.

\medskip

\qquad\grheadc\ Together with ($\alpha$) this shows that $\gamma A$ is stable
for every $\gamma\in\Bbb R$.

\medskip

\quad{\bf (iv)} Take $E$ such that $0<\mu E<\infty$, and $\alpha<\beta$.   Set
$\eta=\bover12(\beta-\alpha)>0$.   Then there is a $\gamma\in\Bbb R$
such that $F=\{x:x\in E,\,\gamma\le g(x)\le\gamma+\eta\}$ has
non-zero measure.   Set $\alpha'=\alpha-\gamma$,
$\beta'=\beta-\gamma-\eta$.   Then $D_k(A+g,F,\alpha,\beta)\subseteq
D_k(A,F,\alpha',\beta')$, while $\alpha'<\beta'$.   So if we take $k\ge
1$ such that
$(\mu^{2k})^*D_k(A,F,\alpha',\beta')<(\mu F)^{2k}$, then

\Centerline{$(\mu^{2k})^*D_k(A+g,E,\alpha,\beta)
\le(\mu^{2k})^*D_k(A+g,F,\alpha,\beta)+\mu^{2k}(E^{2k}\setminus F^{2k})
<(\mu E)^{2k}$.}

\medskip

\quad{\bf (v)} Let $E\in\Sigma$, $\alpha<\beta$ be such that
$0<\mu E<\infty$.   Set

\Centerline{$E_0=\{x:x\in E,\,g(x)=0\}$,
\quad$E_1=\{x:x\in E,\,g(x)>0\}$,}

\Centerline{$E_2=\{x:x\in E,\,g(x)<0\}$.}

\medskip

\qquad\grheada\ Suppose that $\mu E_0>0$.   Then
$D_1(A\times g,E,\alpha,\beta)$ does not meet $E_0^2$, so
$\mu^*D_1(A\times g,E,\alpha,\beta)<\mu E$.

\medskip

\qquad\grheadb\ Suppose that $\mu E_1>0$.   Let $\eta>0$ be such that

\Centerline{$\max(\alpha,\Bover{\alpha}{1+\eta})=\alpha'
<\beta'=\min(\beta,\Bover{\beta}{1+\eta})$.}

\noindent Let $\gamma>0$ be such that $\mu F>0$, where
$F=\{x:x\in E,\,\gamma\le g(x)\le\gamma(1+\eta)\}$.   If $x\in F$ then

\Centerline{$f(x)g(x)\le\alpha
\Longrightarrow f(x)\le\Bover{\alpha}{g(x)}\le\Bover{\alpha'}{\gamma}$,}

\Centerline{$f(x)g(x)\ge\beta
\Longrightarrow f(x)\ge\Bover{\beta}{g(x)}\ge\Bover{\beta'}{\gamma}$.}

\noindent So

\Centerline{$D_k(A\times g,F,\alpha,\beta)
\subseteq D_k(A,F,\Bover{\alpha'}{\gamma},\Bover{\beta'}{\gamma})$}

\noindent for every $k$.   Now, because $A$ is stable, there is some
$k\ge 1$ such that

\Centerline{$(\mu^{2k})^*D_k(A,F,\Bover{\alpha'}{\gamma},
\Bover{\beta'}{\gamma})<(\mu F)^{2k}$,}

\noindent and in this case
$(\mu^{2k})^*D_k(A\times g,E,\alpha,\beta)<(\mu E)^{2k}$, just as in the
argument for (a-iv) above.

\medskip

\qquad\grheadc\ If $\mu E_2>0$, then we know from (iii) above that $-A$ is
stable, so ($\beta$) here tells us that
there is a $k\ge 1$ such that
$(\mu^{2k})^*D_k((-A)\times(-g),E,\alpha,\beta)<(\mu E)^{2k}$.

Since one of these three cases must occur, and since $E$, $\alpha$ and
$\beta$ are arbitrary, $A\times g$ is stable.

\medskip

\quad{\bf (vi)} Write $B=\{hf:f\in A\}$.   Suppose that $0<\mu E<\infty$ and
$\alpha<\beta$.   If either $\alpha<h(\gamma)$ for every $\gamma\in\Bbb R$
or $h(\gamma)<\beta$ for every $\gamma\in\Bbb R$,

\Centerline{$\mu^*D_1(B,E,\alpha,\beta)=\mu\emptyset=0<\mu E$.}

\noindent Otherwise, because $h$ is continuous, the Intermediate Value
Theorem tells us that there are $\alpha'<\beta'$ such that
$\alpha<h(\alpha')<h(\beta')<\beta$.   In this case
$D_k(B,E,\alpha,\beta)\subseteq D_k(A,E,\alpha',\beta')$ for every $k$.
Because $A$ is stable, there is some $k\ge 1$ such that
$(\mu^{2k})^*D_k(A,E,\alpha',\beta')<(\mu E)^{2k}$, so that
$(\mu^{2k})^*D_k(B,E,\alpha,\beta)<(\mu E)^{2k}$.  As $E$, $\alpha$ and
$\beta$ are arbitrary, $B$ is stable.

\medskip

{\bf (b)(i)} For any $m\ge 1$ and $l<n$, if we identify $X^{2(mn+l)}$ with
$(X^{2n})^m\times X^{2l}$, we see that $D_{mn+l}(A,E,\alpha,\beta)$
becomes identified
with a subset of $D_n(A,E,\alpha,\beta)^m\times E^{2l}$.   (If
$w\in D_{mn+l}(A,E,\alpha,\beta)$, there is an $f\in A$ such that
$f(w(2i))\le\alpha$, $f(w(2i+1))\ge\beta$ for $i<mn+l$.   Now
$(w(2rn),w(2rn+1),\ldots,w(2rn+2n-1))\in D_n(A,E,\alpha,\beta)$ for
$r<m$.)   So

$$\eqalignno{\Bover1{(\mu E)^{2(mn+l)}}
   (\mu^{2(mn+l)})^*&D_{mn+l}(A,E,\alpha,\beta)\cr
&\le\Bover1{(\mu E)^{2(mn+l)}}
   \bigl((\mu^{2n})^*D_n(A,E,\alpha,\beta)\bigr)^m(\mu E)^{2l}\cr
\displaycause{251Wm}
&=\bigl(\Bover1{(\mu E)^{2n}}(\mu^{2n})^*D_n(A,E,\alpha,\beta)\bigr)^m
\to 0\cr}$$

\noindent as $m\to\infty$.

\medskip

\quad{\bf (ii)} Note that $D_k(A\cup B,E,\alpha,\beta)
=D_k(A,E,\alpha,\beta)\cup D_k(B,E,\alpha,\beta)$
for all $E$, $k$, $\alpha$ and $\beta$.  Now, given that
$0<\mu E<\infty$ and $\alpha<\beta$, there are $m$, $n\ge 1$ such that
$(\mu^{2m})^*D_m(A,E,\alpha,\beta)<(\mu E)^{2m}$ and
$(\mu^{2n})^*D_n(A,E,\alpha,\beta)<(\mu E)^{2n}$.   So, by (i) just above,

$$\eqalignno{\Bover1{(\mu E)^{2k}}(\mu^{2k})^*
  &D_k(A\cup B,E,\alpha,\beta)\cr
&\le\Bover1{(\mu E)^{2k}}\bigl((\mu^{2k})^*D_k(A,E,\alpha,\beta)
  +(\mu^{2k})^*D_k(B,E,\alpha,\beta)\bigr)
\to 0\cr}$$

\noindent as $k\to\infty$, and there is some $k$ such that
$(\mu^{2k})^*D_k(A\cup B,E,\alpha,\beta)<(\mu E)^{2k}$.   As $E$,
$\alpha$ and $\beta$ are arbitrary, $A\cup B$ is stable.

\medskip

\quad{\bf (iii)} If $A$ is empty, or contains only the constant function
$\tbf{0}$ with value $0$, this is trivial.   Now (a-iv) tells us that
$\{g\}=\{\tbf{0}\}+g$ is stable for every $g\in\eusm L^0$, and from (ii)
here it follows that any finite subset of $\eusm L^0$ is stable.

\medskip

\quad{\bf (iv)} From (a-vi) we see that $\{f^+:f\in A\}$ is stable;   now from
(a-iii) and (ii) here we see that $\{f^-:f\in A\}=\{f^+:f\in -A\}$ and
$\{f^+:f\in A\}\cup\{f^-:f\in A\}$ are stable.

\medskip

{\bf (c)(i)} The product measures $\mu^{2k}$, $\hat\mu^{2k}$ and
$\tilde\mu^{2k}$ are all the same (251Wn), so this follows immediately
from the definition in 465B.

\medskip

\quad{\bf (ii)} Let $h$ be a Radon-Nikod\'ym derivative of $\nu$ with respect
to $\mu$ (234J).
Suppose that $0<\nu E<\infty$ and $\alpha<\beta$.
Then there is an $F\in\Sigma$ such that $F\subseteq E\cap\dom h$,
$h(x)>0$ for every $x\in F$, and $0<\mu F<\infty$.   There is a $k\ge 1$
such that $(\mu^{2k})^*D_k(A,F,\alpha,\beta)<(\mu F)^{2k}$, that is,
there is a $W\subseteq F^{2k}\setminus D_k(A,F,\alpha,\beta)$ such that
$\mu^{2k}W>0$.   In this case, $\nu^{2k}W>0$.   \Prf\ Set
$\tilde h(w)=\prod_{i=0}^{2k-1}h(w(i))$ for $w\in(\dom h)^{2k}$.   Then
$\nu^{2k}$ is
the indefinite integral of $\tilde h$ with respect to $\mu^{2k}$
(253I, extended by induction to the product of more than two factors),
and $\tilde h(w)>0$ for every $w\in F^{2k}$.\ \QeD\
Since $W\subseteq E^{2k}\setminus D_k(A,E,\alpha,\beta)$,
$(\nu^{2k})^*D_k(A,E,\alpha,\beta)<(\nu E)^{2k}$;  as $E$, $\alpha$ and
$\beta$ are arbitrary, $A$ is stable with respect to $\nu$.

Since $\nu$ is the completion of its restriction $\nu\restr\Sigma$
(234Lb),
$A$ is also stable with respect to $\nu\restr\Sigma$, by (i) just above.

\medskip

\quad{\bf (iii)} Writing $\Sigma_Y$ for the subspace $\sigma$-algebra, take
$F\in\Sigma_Y$ such that $\mu_YF=\mu^*F$ is finite and non-zero, and
$\alpha<\beta$ in $\Bbb R$.   Let $E\in\Sigma$ be a measurable envelope of
$F$.   Then there is a $k\ge 1$ such that
$(\mu^{2k})^*D_k(A,E,\alpha,\beta)<(\mu E)^{2k}$.   Consider
$D_k(A_Y,F,\alpha,\beta)=F^{2k}\cap D_k(A,E,\alpha,\beta)$.   The identity map
from $F$ to $E$ is \imp\ for the subspace measures $\mu_F$ and $\mu_E$
(214Ce),
so the identity map from $F^{2k}$ to $E^{2k}$ is \imp\ for the product
measures $\mu_F^{2k}$ and $\mu_E^{2k}$ (apply 254H to
appropriate normalizations of
$\mu_E$, $\mu_F$).   Also  $\mu_E^{2k}$ is the subspace measure on $E^{2k}$
induced by $\mu^{2k}$ (251Wl), and similarly $\mu_F^{2k}$ is the subspace
measure on $F^{2k}$ induced by $\mu_Y^{2k}$, so

$$\eqalignno{(\mu_Y^{2k})^*D_k(A_Y,F,\alpha,\beta)
&=(\mu_F^{2k})^*D_k(A_Y,F,\alpha,\beta)
\le(\mu_E^{2k})^*D_k(A,E,\alpha,\beta)\cr
\displaycause{413Eh}
&=(\mu^{2k})^*D_k(A,E,\alpha,\beta)
<(\mu E)^{2k}
=(\mu_YF)^{2k}.\cr}$$

\noindent As $F$, $\alpha$ and $\beta$ are arbitrary, $A_Y$ is stable, as
claimed.

\medskip

\quad{\bf (iv)} If $A$ is stable, then (iii) just above
tells us that $A_E$ will be stable for
every $E\in\Sigma$.   Conversely, if $A_E$ is stable for every $E$ of
finite measure, take $E\in\Sigma$ such that $0<\mu E<\infty$ and
$\alpha<\beta$ in $\Bbb R$.   Then there is a $k\ge 1$ such that

\Centerline{$(\mu E)^{2k}
=(\mu_EE)^{2k}
>(\mu_E^{2k})^*D_k(A_E,E,\alpha,\beta)
=(\mu^{2k})^*D_k(A,E,\alpha,\beta)$.}

\noindent As $E$, $\alpha$ and $\beta$ are arbitrary, $A$ is stable.

\medskip

\quad{\bf (v)} If $A$ is stable, then every $A_n$ is stable, because
$A_n=\{h_nf:f\in A\}$ where $h_n(\alpha)=\med(-n,\alpha,n)$ and we can apply
(a-v).   Conversely, if every $A_n$ is stable, and we are given a set $E$ of
non-zero finite measure and $\alpha<\beta$, take $n\in\Bbb N$ such that
$-n\le\alpha<\beta\le n$.   Then $D_k(A,E,\alpha,\beta)=D_k(A_n,E,\alpha,\beta)$
for every $k$, and we have a $k\ge 1$ such that

\Centerline{$(\mu^{2k})^*D_k(A,E,\alpha,\beta)
=(\mu^{2k})^*D_k(A_n,E,\alpha,\beta)<(\mu E)^{2k}$.}

\noindent So $A$ is stable.

\medskip

{\bf (d)} Let $\sequencen{X_n}$ be a sequence of measurable sets of finite
measure covering $X$.   Take $F\in\Tau$ such that $\nu F>0$, and
$\alpha<\beta$ in $\Bbb R$.   Let $n\in\Bbb N$ be such that
$F_0=F\cap\phi^{-1}[X_n]$ has non-zero measure.
Set $\lambda E=\nu(F_0\cap\phi^{-1}[E])$ for $E\in\Sigma$;
then $\lambda$ is countably
additive and $\lambda\le\mu$.   By the Radon-Nikod\'ym theorem, there is a
$\Sigma$-measurable $h:X\to[0,1]$ such that $\int_Eh=\lambda E$ for every
$E\in\Sigma$.   Set $G=\{x:x\in X_n$, $h(x)>0\}$;  then
$\nu(F_0\setminus\phi^{-1}[G])=\int_{X_n\setminus G}h=0$.

Because $A$ is stable,
there is a $k\ge 1$ such that $(\mu^{2k})^*D_k(A,G,\alpha,\beta)<(\mu G)^{2k}$.
Let $V\in\Tensorhat_{2k}\Sigma$ be such that
$V\subseteq G^{2k}\setminus D_k(A,G,\alpha,\beta)$ and $\mu^{2k}V>0$.
Then $\lambda^{2k}V=\int_Vh^{\#}d\mu^{2k}$ where
$h^{\#}(w)=\prod_{i<2k}h(w(i))$ for $w\in X^{2k}$ (253I).   Because
$h^{\#}(w)>0$ for every $w\in G^{2k}$, $\lambda^{2k}V>0$.

Note next that if we define $\psi:Y^{2k}\to X^{2k}$ by setting
$\psi(\ofamily{i}{2k}{y_i})=\ofamily{i}{2k}{\phi(y_i)}$ when
$y_i\in Y$ for $i<2k$, then

\Centerline{$\{W:W\in\Tensorhat_{2k}\Sigma$,
$\psi^{-1}[W]\in\Tensorhat_{2k}\Tau$,
$\lambda^{2k}W=\nu^{2k}(F_0^{2k}\cap\psi^{-1}[W])\}$}

\noindent is a Dynkin class containing $\prod_{i<2k}E_i$ whenever
$E_i\in\Sigma$ for every $i<2k$, and therefore contains $V$, by the
Monotone Class Theorem (136B).   So

\Centerline{$\nu^{2k}(F^{2k}\cap\psi^{-1}[V])
\ge\nu^{2k}(F_0^{2k}\cap\psi^{-1}[V])>0$.}

\noindent On the other hand, if
$v\in D_k(B,F,\alpha,\beta)$ then there is an $f\in A$ such that
$f\phi(v(2i))\le\alpha$ and $f\phi(v(2i+1))\ge\beta$ for every
$i<k$, that is, $\psi(v)\in D_k(A,X,\alpha,\beta)$, and $\psi(v)\notin V$.
But this means that

\Centerline{$(\nu^{2k})^*D_k(B,F,\alpha,\beta)
\le\nu^{2k}(F^{2k}\setminus\psi^{-1}[V])
<\nu^{2k}F^{2k}=(\nu F)^{2k}$.}

\noindent As $F$, $\alpha$ and $\beta$ are arbitrary, $B$ is stable.
}%end of proof of 4{}65C

\leader{465D}{}\cmmnt{ Now for the first result connecting the notion
of `stable' set with the concerns of this chapter.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, and $A\subseteq\Bbb R^X$ a stable set.

(a) $A\subseteq\eusm L^0\cmmnt{\mskip5mu =\eusm L^0(\Sigma)}$\cmmnt{
(that is, every member of $A$ is $\Sigma$-measurable)}.

(b) If $\{f(x):f\in A\}$ is bounded for each $x\in X$, then $A$ is
relatively compact in $\eusm L^0$ for
the topology of pointwise convergence.

\proof{{\bf (a)} \Quer\ Suppose, if possible, that there is a
non-measurable $f\in A$.   By 413G, there are $\alpha<\beta$ and
$E\in\Sigma$ such that $0<\mu E<\infty$ and $\mu^*P=\mu^*Q=\mu E$ where
$P=\{x:x\in E,\,f(x)\le\alpha\}$ and $Q=\{x:x\in E,\,f(x)\ge\beta\}$.

Now suppose that $k\ge 1$.   Then
$D_k(\{f\},E,\alpha,\beta)\supseteq(P\times Q)^k$, so

\Centerline{$(\mu^{2k})^*D_k(\{f\},E,\alpha,\beta)=(\mu^*P\cdot\mu^*Q)^k
=(\mu E)^{2k}$}

\noindent (251Wm again).   Since this is true for every $k$,
$\{f\}$ is not stable, and
(by 465C(a-i)) $A$ cannot be stable;  which contradicts our hypothesis.\
\Bang

\medskip

{\bf (b)} Because $\{f(x):x\in A\}$ is bounded for each $x$,
$\overline{A}$, the closure of $A$ in $\Bbb R^X$, is compact for the
topology of pointwise convergence.   But $\overline{A}$ is stable, by
465C(a-ii), so is included in $\eusm L^0$, by (a).
}%end of proof of 465D

\leader{465E}{The topology
$\frak T_s(\eusm L^2,\eusm L^2)$}\cmmnt{ Some of the arguments below
will rely on ideas of compactness in function spaces.   There are of
course many ways of expressing the method, but a reasonably accessible
one uses the Hilbert space $L^2$, as follows.}   Let $(X,\Sigma,\mu)$ be
any measure space.   Then $L^2=L^2(\mu)$ is a Hilbert space with a
corresponding weak topology
$\frak T_s(L^2,L^2)$\cmmnt{ defined by the functionals
$u\mapsto\innerprod{u}{v}$ for $v\in L^2$}.   In the present section it
will be more convenient to regard this as a topology
$\frak T_s(\eusm L^2,\eusm L^2)$ on the space $\eusm L^2=\eusm L^2(\mu)$ of square-integrable real-valued functions\cmmnt{, defined
by the functionals $f\mapsto\int f\times g$ for $g\in\eusm L^2$}.   The
essential fact we need is that norm-bounded sets are relatively weakly
compact.   \prooflet{In $L^2$, this is because Hilbert spaces are
reflexive (4A4Ka).   In $\eusm L^2$, given an ultrafilter $\Cal F$
containing a $\|\,\|_2$-bounded set $B\subseteq\eusm L^2$,
$v=\lim_{f\to\Cal F}f^{\ssbullet}$ must be defined in $L^2$ for
$\frak T_s(L^2,L^2)$, and now there is a $g\in\eusm L^2$ such that
$g^{\ssbullet}=v$;  in which case

\Centerline{$\lim_{f\to\Cal F}\int f\times h
=\lim_{f\to\Cal F}\innerprod{f^{\ssbullet}}{h^{\ssbullet}}
=\innerprod{g^{\ssbullet}}{h^{\ssbullet}}
=\int g\times h$}

\noindent for every $h\in\eusm L^2$.   Note that we are free to take $g$
to be a $\Sigma$-measurable function with domain $X$ (241Bk).}

\leader{465F}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space, and
$B\subseteq\eusm L^2=\eusm L^2(\mu)$ a $\|\,\|_2$-bounded set.
Suppose that $h\in\eusm L^2$ belongs to the closure of $B$ for
$\frak T_s(\eusm L^2,\eusm L^2)$.   Then for any $\delta>0$ and $k\ge 1$
the set

$$\eqalign{W=\bigcup_{f\in B}\{w:w\in X^k,\,
&w(i)\in\dom f\cap\dom h\cr
&\text{and }f(w(i))\ge h(w(i))-\delta\text{ for every }i<k\}\cr}$$

\noindent is $\mu^k$-conegligible in $X^k$.

\proof{{\bf (a)} Since completing the measure $\mu$ does not change the
space $\eusm L^2$ (244Xa) nor the product measure $\mu^k$ (251Wn), we
may suppose that $\mu$ is complete.

\medskip

{\bf (b)} The first substantive fact to note is that there is a sequence
$\sequencen{f_n}$ in $B$ converging to $h$ for
$\frak T_s=\frak T_s(\eusm L^2,\eusm L^2)$.   \Prf\ Setting
$C=\{f^{\ssbullet}:f\in B\}$,
$C$ is a bounded set in $L^2=L^2(\mu)$ and $h^{\ssbullet}$ belongs to
the $\frak T_s(L^2,L^2)$-closure of $C$.   But $L^2$, being a normed
space, is angelic in its weak topology (462D), and $C$ is relatively
compact in $L^2$, so there is a sequence in $C$ converging to
$h^{\ssbullet}$.   We can represent this sequence as
$\sequencen{f_n^{\ssbullet}}$ where $f_n\in B$ for every $n$, and now
$\sequencen{f_n}\to h$ for $\frak T_s$.\ \Qed

\medskip

{\bf (c)} The second component of the proof is the following simple
idea.   Suppose that $\sequencen{E_n}$ is a sequence in $\Sigma$ such
that $\bigcap_{n\in I}E_n$ is negligible for every infinite set
$I\subseteq\Bbb N$.   For $m\ge 1$ and $I\subseteq\Bbb N$ set

\Centerline{$V_m(I)
=\bigcap_{n\in I}\{w:w\in X^m,\,\exists\,i<m,\,w(i)\in E_n\}$.}

\noindent Then $\mu^mV_m(I)=0$ for every infinite $I\subseteq\Bbb N$.
\Prf\ Induce on $m$.   For $m=1$ this is just the original hypothesis on
$\sequencen{E_n}$.   For the inductive step to $m+1$, identify
$\mu^{m+1}$ with the product of $\mu^m$ and $\mu$, and observe that

\Centerline{$V_{m+1}(I)^{-1}[\{x\}]
=\{w:(w,x)\in V_{m+1}(I)\}
=V_m(\{n:n\in I,\,x\notin E_n\})$}

\noindent for every $x\in X$.   Now, setting $I_x=\{n:n\in I,\,x\notin
E_n\}$, $F=\{x:I_x$ is finite$\}$, we have

\Centerline{$F=\bigcup_{r\in\Bbb N}\bigcap_{n\in I\setminus r}E_n$,}

\noindent so $F$ is negligible, while if $x\notin F$ then $V_m(I_x)$ is
negligible, by the inductive hypothesis.   But this means that almost
every horizontal section of $V_{m+1}(I)$ is negligible, and
$V_{m+1}(I)$, being measurable, must be negligible, by Fubini's theorem
(252F).   Thus the induction proceeds.\ \Qed

\medskip

{\bf (d)} Now let us return to the main line of the argument from (b).
For each $n\in\Bbb N$, set $E_n=\{x:x\in\dom f_n\cap\dom
h,\,f_n(x)<h(x)-\delta\}$.   If $I\subseteq\Bbb N$ is infinite, then
$\bigcap_{n\in I}E_n$ is negligible.   \Prf\ Setting
$G=\bigcap_{n\in I}E_n$, $\mu G$ is finite (because
$\delta\chi E_n\leae|h-f_n|$, so
$\mu E_n<\infty$ for every $n$) and $\int_Gf_n\le\int_Gh-\delta\mu G$
for every $n\in I$.   But $\lim_{n\to\infty}\int_Gf_n=\int_Gh$
and $I$ is infinite, so

\Centerline{$\delta\mu G\le\inf_{n\in I}|\int_Gh-\int_Gf_n|=0$.
\Qed}

By (c), it follows that

\Centerline{$V
=\bigcap_{n\in\Bbb N}\{w:w\in X^k,\,\exists\,i<k,\,w(i)\in E_n\}$}

\noindent is negligible.   But if we set
$Y=\bigcap_{n\in\Bbb N}\dom f_n\cap\dom h$, $Y$ is a conegligible subset
of $X$, $Y^k$ is a conegligible subset of $X^k$, and
$Y^k\setminus V\subseteq W$, so $W$ is conegligible, as required.
}%end of proof of 465F

\leader{465G}{Theorem} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space, and $A\subseteq\eusm L^0\cmmnt{\mskip5mu =\eusm L^0(\Sigma)}$ a
stable
set of measurable functions.   Let $\frak T_p$ and $\frak T_m$ be the
topologies of pointwise convergence and convergence in
measure\cmmnt{, as in \S463}.   Then the identity map from $A$ to
itself is $(\frak T_p,\frak T_m)$-continuous.

\proof{\Quer\ Suppose, if possible, otherwise.

\medskip

{\bf (a)} We must have an $f_0\in A$, a set $F\in\Sigma$ of finite
measure, and an $\epsilon>0$ such that for every
$\frak T_p$-neighbourhood $U$ of $f_0$ there is an $f\in A\cap U$ such
that $\int\chi F\wedge|f-f_0|d\mu\ge 2\epsilon$.   Set

\Centerline{$B=\{\chi F\wedge(f-f_0)^+:f\in A\}
  \cup\{\chi F\wedge(f-f_0)^-:f\in A\}$.}

\noindent Then

\Centerline{$B=\{\chi F-(\chi F-(f-f_0)^+)^+:f\in A\}
  \cup\{\chi F-(\chi F-(f-f_0)^-)^+:f\in A\}$}

\noindent is stable, by 465C(a-iv), 465C(b-iv) and 465C(a-iii), used repeatedly.
Setting $B'=\{f:f\in B,\,\int f\ge\epsilon\}$, $B'$ is again stable
(465C(a-i)).
Our hypothesis is that $f_0$ is in the $\frak T_p$-closure of

$$\eqalign{\{f:f\in A,\,\int\chi F\wedge|f-f_0|\ge 2\epsilon\}
&\subseteq\{f:f\in A,\,\int\chi F\wedge(f-f_0)^+\ge\epsilon\}\cr
&\mskip50mu\cup\{f:f\in A,\,\int\chi F\wedge(f-f_0)^-\ge\epsilon\};\cr}$$

\noindent since
$f\mapsto\chi F\wedge(f-f_0)^+$ and
$f\mapsto\chi F\wedge(f-f_0)^-$ are $\frak T_p$-continuous,
$0$ belongs to the $\frak T_p$-closure of $B'$.

\medskip

{\bf (b)} Let $\Cal F$ be an ultrafilter on $B'$ which
$\frak T_p$-converges to $0$.   Because $B'$ is $\|\,\|_2$-bounded
(since $0\le f\le\chi F$ for every $f\in B'$), $\Cal F$ also has a
$\frak T_s(\eusm L^2,\eusm L^2)$-limit $h$ say, as noted in 465E;  and
we can suppose that $h$ is measurable and defined everywhere.   We must
have

\Centerline{$\int_Fh=\lim_{f\to\Cal F}\int_Ff\ge\epsilon>0$.}

\noindent So there is a $\delta>0$ such that
$E=\{x:x\in F,\,h(x)\ge 3\delta\}$ has measure greater than $0$.

\medskip

{\bf (c)} Because $B'$ is stable, there must be some $k\ge 1$ such that
$(\mu^{2k})^*D_k(B',E,\delta,2\delta)<(\mu E)^{2k}$.   Let
$W\subseteq E^{2k}\setminus D_k(B',E,\delta,2\delta)$
be a measurable set of
positive measure.   By Fubini's theorem, there must be a $u\in X^k$
such that $\mu^kV$ is defined and greater than $0$, where

\Centerline{$V=\{v:v\in X^k,\,u\#v\in W\}$.}

\noindent Set $C=\{f:f\in B',\,f(u(i))\le\delta$ for every $i<k\}$;  then
$C\in\Cal F$, because $\Cal F\to 0$ for $\frak T_p$.   Accordingly $h$
belongs to the $\frak T_s(\eusm L^2,\eusm L^2)$-closure of $C$.   But
now 465F tells us that there must be a $v\in V$ and an $f\in C$ such
that $f(v(i))\ge h(v(i))-\delta$ for every $i<k$.

Consider $w=u\#v$.   We know that $w\in W$
(because $v\in V$), so, in particular, $w\in E^{2k}$ and
$h(w(i))\ge 3\delta$ for every $i<2k$;  accordingly

\Centerline{$f(w(2i+1))=f(v(i))\ge h(v(i))-\delta\ge2\delta$}

\noindent for every $i<k$.   On the other hand,
$f(w(2i))=f(u(i))\le\delta$ for every $i<k$, because $f\in C$.   But this
means that $f$ witnesses that $w\in D_k(B',E,\delta,2\delta)$, which is
supposed to be disjoint from $W$.\ \Bang

This contradiction shows that the theorem is true.
}%end of proof of 465G

\leader{465H}{}\cmmnt{ We shall need some interesting and important
general facts concerning powers of measures.   I start with a useful
elaboration of the strong law of large numbers.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be any probability space.
For $n\in\Bbb N$, write $\Lambda_n$ for the domain of the product
measure $\mu^n$.    For $w\in X^{\Bbb N}$, $k\ge 1$, $n\ge 1$ write
$\nu_{wk}$ for the probability measure with domain $\Cal PX$ defined by
writing

\Centerline{$\nu_{wk}(E)=\Bover1k\#(\{i:i<k,\,w(i)\in E\})$}

\noindent for $E\subseteq X$, and $\nu_{wk}^n$ for the corresponding
product measure on $X^n$.

Then whenever $n\ge 1$ and $f:X^n\to\Bbb R$ is bounded and
$\Lambda_n$-measurable, $\lim_{k\to\infty}\int fd\nu_{wk}^n$ exists, and
is equal to $\int fd\mu^n$, for $\mu^{\Bbb N}$-almost every
$w\in X^{\Bbb N}$.

\proof{{\bf (a)} For $k\ge n$ and $w\in X^{\Bbb N}$,  set

\Centerline{$h_k(w)=\Bover{(k-n)!}{k!}\sum_{\pi:n\to k\text{ is
injective}}f(w\pi)$.}

\noindent Setting $M=\sup_{v\in X^n}|f(v)|$, we have

$$\eqalign{\bigl|h_k(w)-\int fd\nu_{wk}^n\bigr|
&=\bigl|\Bover{(k-n)!}{k!}\sum_{\pi:n\to k\text{ is injective}}f(w\pi)
  -\Bover1{k^n}\sum_{\pi:n\to k}f(w\pi)\bigr|\cr&
\le\Bover{Mk!}{(k-n)!}\bigl(\Bover{(k-n)!}{k!}-\Bover1{k^n}\bigr)
  +\Bover{M}{k^n}\bigl(k^n-\Bover{k!}{(k-n)!}\bigr)\cr&
=2M\bigl(1-\Bover{k!}{k^n(k-n)!}\bigr)
\to 0\cr}$$

\noindent as $k\to\infty$, for every $w\in X^{\Bbb N}$.

\medskip

{\bf (b)} Write $\Lambda$ for the domain of $\mu^{\Bbb N}$, and for
$k\ge n$ set

\Centerline{$\Tau_k=\{W:W\in\Lambda,\,w\pi\in W$ whenever $w\in W$ and
$\pi\in S_k\}$,}

\noindent where $S_k$ is the set of permutations $\psi:\Bbb N\to\Bbb N$
such that $\psi(i)=i$ for $i\ge k$.   Then $\langle\Tau_k\rangle_{k\ge
n}$ is a non-increasing sequence of $\sigma$-subalgebras of $\Lambda$.
For any injective function $\pi:n\to k$ there are just $(k-n)!$
extensions of $\pi$ to a member of $S_k$.   So

\Centerline{$h_k(w)=\Bover1{k!}\sum_{\pi\in S_k}f(w\pi\restr n)$}

\noindent for every $w$.   Observe that $h_k(w\psi)=h_k(w)$ whenever
$\psi\in S_k$ and $w\in X^{\Bbb N}$, so $h_k$ is $\Tau_k$-measurable.   (Of
course we need to look back at the definition of $h_k$ to confirm that
it is $\Lambda$-measurable.)

\medskip

{\bf (c)} For any $k\ge n$, $h_k$ is a conditional expectation of $h_n$
on $\Tau_k$.   \Prf\ If $W\in\Tau_k$, then

\Centerline{$\int_Wh_k(w)dw
=\Bover1{k!}\sum_{\pi\in S_k}\int_Wf(w\pi\restr n)dw
=\Bover1{k!}\sum_{\pi\in S_k}\int_Wg(w\pi)dw$}

\noindent where $g(w)=f(w\restr n)$ for $w\in X^{\Bbb N}$.   Now
observe that for every $\pi\in S_k$ the map $w\mapsto w\pi$ is a measure
space automorphism of $X^{\Bbb N}$ which leaves $W$ unchanged, because
$W\in\Tau_k$;  so that $\int_Wg(w\pi)dw=\int_Wg(w)dw$, by 235Gc.   So
$\int_Wh_k=\int_Wg$.   But (since $W\in\Tau_k\subseteq\Tau_n$)
$\int_Wh_n$ is also equal to $\int_Wg$, and $\int_Wh_k=\int_Wh_n$.   As
$W$ is arbitrary, $h_k$ is a conditional expectation of $h_n$ on
$\Tau_k$.\ \Qed

\medskip

{\bf (d)} By the reverse martingale theorem (275K),
$h_{\infty}(w)=\lim_{k\to\infty}h_k(w)$ is defined for almost every
$w\in X^{\Bbb N}$.   Accordingly $\lim_{k\to\infty}\int fd\nu_{wk}^n$ is
defined for almost every $w$.

\medskip

{\bf (e)} To see that the limit is $\int fd\mu^n$, observe that if
$W\in\Tau_{\infty}=\bigcap_{k\in\Bbb N}\Tau_k$ then $\mu^{\Bbb N}W$ must
be either $0$ or $1$.   \Prf\ Set $\gamma=\mu^{\Bbb N}W$.   Let
$\epsilon>0$.   Then there is a $V\in\bigotimes_{\Bbb N}\Sigma$
(notation:  465Ad) such that
$\mu^{\Bbb N}(W\symmdiff V)\le\epsilon$ (254Fe).   There is some $k$
such that $V$
is determined by coordinates in $k$.   If we set $\pi(i)=2k-i$ for
$i<2k$, $i$ for $i\ge 2k$, then $V'=\{w\pi:w\in V\}$ is determined by
coordinates in $2k\setminus k$, so
$\mu^{\Bbb N}(V\cap V')=(\mu^{\Bbb N}V)^2$.    On the other hand,
because $W\in\Tau_{2k}$, the measure
space automorphism $w\mapsto w\pi$ does not move $W$, and
$\mu^{\Bbb N}(W\setminus V')=\mu^{\Bbb N}(W\setminus V)$.   Accordingly

$$\eqalign{\gamma
&=\mu^{\Bbb N}(W\cap W)
\le\mu^{\Bbb N}(V\cap V')+2\mu^{\Bbb N}(W\setminus V)
\le(\mu^{\Bbb N}V)^2+2\epsilon
\le(\gamma+\epsilon)^2+2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\gamma\le\gamma^2$ and
$\gamma\in\{0,1\}$.\ \Qed

\medskip

{\bf (e)} Now 275K tells us that $h_{\infty}$ is
$\Tau_{\infty}$-measurable, therefore essentially constant, and must be
equal to its expectation almost everywhere.   But, setting
$W=X^{\Bbb N}$ in the proof of (c), we see that

\Centerline{$\int h_k(w)dw=\int f(w\restr n)dw=\int fd\mu^n$}

\noindent for every $k\ge n$, so

$$\eqalign{\lim_{k\to\infty}\int fd\nu_{wk}^n
&=\lim_{k\to\infty}h_k(w)
=h_{\infty}(w)\cr&
=\int h_{\infty}(w)dw
=\lim_{k\to\infty}\int h_k(w)dw
=\int fd\mu^n\cr}$$

\noindent for almost every $w$, as claimed.
}%end of proof of 465H

\leader{465I}{}\cmmnt{ Now for a string of lemmas, working towards the
portmanteau Theorem 465M.   The first is elementary.

\medskip

\noindent}{\bf Lemma} Let $X$ be a set, and $\Sigma$ a $\sigma$-algebra
of subsets of $X$.    For $w\in X^{\Bbb N}$ and $k\ge 1$, write
$\nu_{wk}$ for the probability measure
with domain $\Cal PX$ defined by writing

\Centerline{$\nu_{wk}(E)=\Bover1k\#(\{i:i<k,\,w(i)\in E\})$}

\noindent for $E\subseteq X$.   Then for any $k\in\Bbb N$ and any set
$I$, $w\mapsto\nu_{wk}^I(W)$ is
$\Tensorhat_{\Bbb N}\Sigma$-measurable\cmmnt{ (notation:  465Ad)} for
every $W\in\Tensorhat_I\Sigma$.

\proof{ Write $\Cal W$ for the set of subsets $W$ of $X^I$ such that
$w\mapsto\nu_{wk}^I(W)$ is $\Tensorhat_{\Bbb N}\Sigma$-measurable.
Then $X^I\in\Cal W$, $W'\setminus W\in\Cal W$ 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 $\Cal W$.   Write
$\Cal V$ for the set of $\Sigma$-cylinders in $X^I$, that is, sets
expressible in the form $\{v:v(i)\in E_i$ for every $i\in J\}$, where
$J\subseteq I$ is finite and $E_i\in\Sigma$ for $i\in J$.   Then $\Cal
V\subseteq\Cal W$.   \Prf\ If $J\subseteq I$ is finite and
$E_i\in\Sigma$ for $i\in J$, then

\Centerline{$w\mapsto\nu_{wk}E_i=\Bover1k\sum_{j=0}^{k-1}\chi
E_i(w(j))$}

\noindent is $\Tensorhat_{\Bbb N}$-measurable for every $i\in J$.  So

\Centerline{$w\mapsto\nu_{wk}^I\{v:v(i)\in E_i\Forall i\in J\}
=\prod_{i\in J}\nu_{wk}E_i$}

\noindent is also measurable.\ \Qed

Because $V\cap V'\in\Cal V$ for all $V$, $V'\in\Cal V$, the Monotone
Class Theorem (136B) tells us that $\Cal W$ must include the
$\sigma$-algebra generated by $\Cal V$, which is $\Tensorhat_I\Sigma$.
}%end of proof of 465I

\leader{465J}{}\cmmnt{ The next three lemmas are specifically adapted
to the study of stable sets of functions.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a probability space.   For
any $n\in\Bbb N$ and $W\subseteq X^n$ I say that $W$ is {\bf symmetric} if
$w\pi\in W$ whenever $w\in W$ and $\pi:n\to n$ is a permutation.
For each $n$, write $\Lambda_n$ for the domain of the product measure $\mu^n$.

(a) Suppose that for each $n\ge 1$ we are given $W_n\in\Lambda_n$, and that
$W_{m+n}\subseteq W_m\times W_n$ for all $m$, $n\ge 1$, identifying
$X^{m+n}$ with $X^m\times X^n$.
Then $\lim_{n\to\infty}(\mu^nW_n)^{1/n}$ is defined and equal to
$\delta=\inf_{n\ge 1}(\mu^nW_n)^{1/n}$.

(b) Now suppose that
each $W_n$ is symmetric.   Then
there is an $E\in\Sigma$ such that $\mu E=\delta$ and $E^n\setminus W_n$
is negligible for every $n\in\Bbb N$.

(c) Next, let $\langle D_n\rangle_{n\ge 1}$ be a sequence of sets such
that

\inset{$D_n\subseteq X^n$ is symmetric for every $n\ge 1$,}

\inset{whenever $1\le m\le n$ and $v\in D_n$ then $v\restr m\in D_m$.}

\noindent Then $\delta=\lim_{n\to\infty}((\mu^n)^*D_n)^{1/n}$ is defined
and there is an $E\in\Sigma$ such that $\mu E=\delta$ and
$(\mu^n)^*(D_n\cap E^n)=(\mu E)^n$ for every $n\in\Bbb N$.

\proof{{\bf (a)} For any $\eta>0$, there is an $m\ge 1$ such that
$\mu^mW_m\le(\delta+\eta)^m$.   If $n=mk+i$, where $k\ge 1$ and $i<m$,
then (identifying $X^n$ with $(X^m)^k\times X^i$)
$W_n\subseteq(W_m)^k\times X^i$, so

\Centerline{$\mu^nW_n\le(\delta+\eta)^{mk}
\le\gamma(\delta+\eta)^{mk+i}=\gamma(\delta+\eta)^n$,}

\noindent where

\Centerline{$\gamma=\max_{i<m}(\Bover1{\delta+\eta})^i$.}

\noindent So

\Centerline{$\limsup_{n\to\infty}(\mu^nW_n)^{1/n}
\le(\delta+\eta)\limsup_{n\to\infty}\gamma^{1/n}=\delta+\eta$.}

\noindent As $\eta$ is arbitrary,

\Centerline{$\delta\le\liminf_{n\to\infty}(\mu^nW_n)^{1/n}\le
\limsup_{n\to\infty}(\mu^nW_n)^{1/n}\le\delta$,}

\noindent  and $\lim_{n\to\infty}(\mu^nW_n)^{1/n}=\delta$.

\medskip

{\bf (b)}  It is enough to consider the case $\delta>0$.

\medskip

\quad{\bf (i)} Consider the family $\Bbb V$ of sequences
$\langle V_n\rangle_{n\ge 1}$ such that

\inset{for each $n\ge 1$, $V_n\in\Lambda_n$ is symmetric and
$\mu^nV_n\ge\delta^n$,}

\inset{if $1\le m\le n$ then $v\restr m\in V_m$ for every $v\in V_n$.}

\noindent Observe that $\pmb{W}=\langle W_n\rangle_{n\ge 1}\in\Bbb V$.
Order $\Bbb V$ by saying that
$\langle V_n\rangle_{n\ge 1}\le\langle V'_n\rangle_{n\ge 1}$ if $V_n\subseteq V'_n$ for every $n$.   For
$\pmb{V}=\langle V_n\rangle_{n\ge 1}$ in $\Bbb V$, set
$\theta(\pmb{V})=\sum_{n=1}^{\infty}2^{-n}\mu^nV_n$.
Any non-increasing sequence
$\sequence{k}{\langle V_{kn}\rangle_{n\ge 1}}$ in $\Bbb V$ has a lower bound $\langle\bigcap_{k\in\Bbb N}V_{kn}\rangle_{n\ge 1}$
in $\Bbb V$, so there must be a
$\pmb{W}'=\langle W'_n\rangle_{n\ge 1}\in\Bbb V$ such that $\pmb{W}'\le\pmb{W}$ and
$\theta(\pmb{V})=\theta(\pmb{W}')$ whenever
$\pmb{V}\in\Bbb V$ and $\pmb{V}\le\pmb{W}'$;  that is, whenever
$\langle V_n\rangle_{n\ge 1}\in\Bbb V$ and $V_n\subseteq W'_n$ for every
$n\ge 1$, then $\mu^nV_n=\mu^nW'_n$ for every $n$.

\medskip

\quad{\bf (ii)} For $x\in X$, $n\ge 1$ set
$V^{(x)}_{n}=\{w:(x,w)\in W'_{n+1}\}$.   Then $V^{(x)}_{n}\in\Lambda_n$
for almost every $x$;
let $X_1\subseteq X$ be a conegligible set such that $V^{(x)}_n\in\Lambda_n$
for every $x\in X_1$ and every $n\ge 1$.   Every
$V^{(x)}_n$ is symmetric, and if $1\le m\le n$ and $v\in V^{(x)}_n$ then
$v\restr m\in V^{(x)}_m$.   It follows that if $m$, $n\ge 1$ then
$V^{(x)}_{m+n}$ becomes identified with a subset of $V^{(x)}_m\times
V^{(x)}_n$.

From (a) we see that $\delta_x=\lim_{n\to\infty}(\mu^nV_n^{(x)})^{1/n}$
is defined for every $x\in X_1$.   The map
$x\mapsto\mu^nV_n^{(x)}:X_1\to[0,1]$ is measurable for each
$n$, by Fubini's theorem (252D),
%do we need completeness?  \query
so $x\mapsto\delta_x$ is also measurable.   Since
$V_n^{(x)}\subseteq W'_n\subseteq W_n$ for every $x$ and $n$,
$\delta_x\le\delta$ for every $x\in X_1$.

\medskip

\quad{\bf (iii)} Set $E=\{x:x\in X_1,\,\delta_x=\delta\}$.   Then
$\mu E\ge\delta$.   \Prf\Quer\ Otherwise, there is some $\beta<\delta$
such that $\mu F<\beta$, where $F=\{x:x\in X_1,\,\delta_x\ge\beta\}$.
Now

$$\eqalign{X_1\setminus F
&=\{x:x\in X_1,\,\lim_{n\to\infty}(\mu^nV^{(x)}_n)^{1/n}<\beta\}\cr
&\subseteq\bigcup_{m\in\Bbb N}
  \bigcap_{n\ge m}\{x:x\in X_1,\,\mu^nV^{(x)}_n\le\beta^n\},\cr}$$

\noindent so there is some $m\in\Bbb N$ such that $\mu H\ge 1-\beta$,
where

\Centerline{$H
=\bigcap_{n\ge m}\{x:x\in X_1,\,\mu^nV^{(x)}_n\le\beta^n\}$.}

Set $\gamma_n=\mu^nW'_n\ge\delta^n$ for each $n$.   Then, for any
$n\ge m$,

$$\eqalign{\gamma_{n+1}
&=\mu^{n+1}W'_{n+1}
=\int_{X_1}\mu^nV^{(x)}_n\mu(dx)\cr
&=\int_H\mu^nV^{(x)}_n\mu(dx)
   +\int_{X_1\setminus H}\mu^nV^{(x)}_n\mu(dx)
\le\beta^n+\beta\gamma_n\cr}$$

\noindent because $V^{(x)}_n\subseteq W'_n$ for every $x$, and
$\mu(X_1\setminus H)\le\beta$.   An easy induction shows that
$\gamma_{m+k}\le k\beta^{m+k-1}+\beta^k\gamma_m$ for every $k\in\Bbb N$.
But this means that

\Centerline{$\delta^k=\delta^{-m}\delta^{m+k}\le\delta^{-m}\gamma_{m+k}
\le\beta^k\delta^{-m}(k\beta^{m-1}+\gamma_m)$}

\noindent for every $k$;  setting $\eta=(\delta-\beta)/\beta>0$,

\Centerline{$\Bover{k(k-1)}2\eta
\le(1+\eta)^k=\bigl(\Bover{\delta}{\beta}\bigr)^k
\le\delta^{-m}(k\beta^{m-1}+\gamma_m)$}

\noindent for every $k$, which is impossible.\ \Bang\Qed

\medskip

\quad{\bf (iv)} Next, for any $x\in E$,
$\pmb{V}^{(x)}=\langle V^{(x)}_n\rangle_{n\ge 1}\in\Bbb V$ and
$\pmb{V}^{(x)}\le\pmb{W}'$, so
$\mu^n(W'_n\setminus V^{(x)}_n)=0$ for every $n\ge 1$.   This means
that, given $n\ge 1$, every vertical section of
$(E\times W'_n)\setminus W'_{n+1}$ (regarded as a subset of
$X\times X^n$) is negligible;  so $(E\times W'_n)\setminus W'_{n+1}$ is
negligible.   We are assuming that $\delta>0$, so

\Centerline{$E\subseteq\{x:x\in X_1,\,V^{(x)}_1\ne\emptyset\}
\subseteq\{x:\{w:(x,w)\in W'_2\}\ne\emptyset\}
\subseteq W'_1$.}

\noindent Now a simple induction shows that $E^n\setminus W'_n$ is
negligible for every $n\ge 1$, so that $E^n\setminus W_n$ is negligible
for every $n$, and we have an appropriate $E$.   (Of
course $\mu E=\delta$ exactly, because $(\mu E)^n\le\mu W_n$ for every
$n$.)

\medskip

{\bf (c)} For each $n\in\Bbb N$ let $V_n\in\Lambda_n$
be a measurable envelope of
$D_n$ in $X^n$.   Define $\langle W_n\rangle_{n\ge 1}$ inductively by
saying

\Centerline{$W_1=V_1$,}

$$\eqalign{W_{n+1}=\{w:w\in X^{n+1},\,&w\pi\restr n\in W_n,\cr
&w\pi\in V_{n+1}\text{ for every permutation }\pi:n+1\to n+1\}}$$

\noindent for each $n\ge 1$.   Then an easy induction on $n$ shows
that $W_n$ is measurable and symmetric and that
$D_n\subseteq W_n\subseteq V_n$, so that $W_n$ is a measurable envelope of
$D_n$ and $\mu^nW_n=(\mu^n)^*D_n$.

Now $\langle W_n\rangle_{n\ge 1}$ satisfies the hypotheses of (a)-(b), so

\Centerline{$\delta=\lim_{n\to\infty}(\mu^nW_n)^{1/n}
=\lim_{n\to\infty}((\mu^n)^*D_n)^{1/n}$}

\noindent is defined and there is a set
$E\in\Sigma$, of measure $\delta$, such that $E^n\setminus W_n$ is
negligible for every $n$;  but this means that

\Centerline{$(\mu^n)^*(E^n\cap D_n)=\mu^n(E^n\cap W_n)=\delta^n$}

\noindent for every $n$, as required.
}%end of proof of 465J

\leader{465K}{Lemma} Let $(X,\Sigma,\mu)$ be a complete probability
space, and $A\subseteq[0,1]^X$ a stable set.   Suppose that $\epsilon>0$
is such that $\int
fd\mu\le\epsilon^2$ for every $f\in A$.   Then there are an $n\ge 1$ and
a $W\in\Tensorhat_n\Sigma$\cmmnt{ (notation: 465Ad)} and a
$\gamma>\mu^nW$ such that $\int fd\nu\le 3\epsilon$ whenever $f\in A$
and $\nu$ is a probability measure on $X$
with domain including $\Sigma$ such that $\nu^nW\le\gamma$.

\proof{{\bf (a)} For $n\in\Bbb N$ write
$\tilde C_n=\bigcup_{f\in A}\{x:f(x)\ge\epsilon\}^n$.   Then
$\sequencen{\tilde C_n}$ satisfies
the conditions of 465Jc, so
$\delta=\lim_{n\to\infty}((\mu^n)^*\tilde C_n)^{1/n}$ is defined, and
there is an $E\in\Sigma$ such that $\mu E=\delta$ and
$(\mu^n)^*(E^n\cap\tilde C_n)=\delta^n$ for every
$n\in\Bbb N$.   Now, for $B\subseteq[0,1]^X$ and $n\in\Bbb N$, write
$C_n(B)=\bigcup_{f\in B}\{x:x\in E,\,f(x)\ge\epsilon\}^n$, so that
$C_n(A)=E^n\cap\tilde C_n$, and $(\mu^n)^*C_n(A)=\delta^n$ for every
$n$.   For any $B\subseteq[0,1]^X$, $\sequencen{C_n(B)}$ also satisfies
the conditions of 465Jc, and
$\delta_B=\lim_{n\to\infty}((\mu^n)^*C_n(B))^{1/n}$ is defined;  we have
$\delta_A=\delta$.

\medskip

{\bf (b)} If $B$, $B'\subseteq[0,1]^X$, then

\Centerline{$(\mu^n)^*C_n(B)\le(\mu^n)^*C_n(B\cup B')
\le(\mu^n)^*C_n(B)+(\mu^n)^*C_n(B')$}

\noindent for every $n$, so
$\delta_B\le\delta_{B\cup B'}\le\max(\delta_B,\delta_{B'})$.   It
follows that if

\Centerline{$\Cal G=\{G:G\subseteq[0,1]^X$ is $\frak T_p$-open,
$\delta_{G\cap A}<\delta\}$,}

\noindent where $\frak T_p$ is the usual topology of $[0,1]^X$, no
finite subfamily of $\Cal G$ can cover $A$.   Accordingly, since the
$\frak T_p$-closure $\overline{A}$ of $A$ is $\frak T_p$-compact, there
is an $h\in\overline{A}$ such that $\delta_{G\cap A}=\delta$ for every
$\frak T_p$-open set $G$ containing $h$.

\medskip

{\bf (c)} At this point recall that every function in $\overline{A}$ is
measurable (465C(a-ii), 465Da) and that $\int:\overline{A}\to[0,1]$ is
$\frak T_p$-continuous (465G).   So $\int h\le\epsilon^2$ and
$\mu\{x:h(x)\ge\epsilon\}\le\epsilon$.

\Quer\ Suppose, if possible, that $\delta>\epsilon$.   Then there is
some $\eta>0$ such that $\mu F>0$, where
$F=\{x:x\in E,\,h(x)<\epsilon-\eta\}$.   For $k\in\Bbb N$ and $u\in F^k$
set $G_u=\{f:f\in[0,1]^X,\,f(u(i))<\epsilon-\eta$ for every $i<k\}$.
Then $G_u$ is an open neighbourhood of $h$, so
$\delta_{G_u\cap A}=\delta$ and $(\mu^k)^*(C_k(G_u\cap A))\ge\delta^k$.
But because
$F\subseteq E$ and $C_k(G_u\cap A)\subseteq E^k$ and $\mu E=\delta$,
this means that $(\mu^k)^*(F^k\cap C_k(G_u\cap A))=(\mu F)^k$.

In the notation of 465Af,

\Centerline{$C_k(G_u\cap A)\cap F^k
\subseteq\{v:u\#v\in D_k(A,F,\epsilon-\eta,\epsilon)\}$}

\noindent for any $u\in F^k$.   So
$(\mu^k)^*\{v:u\#v\in D_k(A,F,\epsilon-\eta,\epsilon)\}=(\mu F)^k$
for every $u\in F^k$.   But this means that
$(\mu^{2k})^*D_k(A,F,\epsilon-\eta,\epsilon)=(\mu F)^{2k}$.   Since this
is so for every $k\ge 1$, $A$ is not stable.\
\Bang

\medskip

{\bf (d)} Thus $\delta\le\epsilon$.   There is therefore some $n\ge 1$
such that $(\mu^n)^*\tilde C_n<(2\epsilon)^n$.   Let
$W\in\Tensorhat_n\Sigma$ be a measurable envelope of $\tilde C_n$, and
try $\gamma=(2\epsilon)^n$.   If $\nu$ is any probability measure on $X$ with domain including $\Sigma$ such that $\nu^nW\le\gamma$, then for any
$f\in A$ we have

\Centerline{$\{x:f(x)\ge\epsilon\}^n\subseteq\tilde C_n\subseteq W$,
\quad$(\nu\{x:f(x)\ge\epsilon\})^n\le\nu^nW\le(2\epsilon)^n$,}

\noindent so that $\nu\{x:f(x)\ge\epsilon\}\le 2\epsilon$.   As
$0\le f(x)\le 1$ for every $x\in X$, $\int fd\nu\le3\epsilon$, as
required.
}%end of proof of 465K

\leader{465L}{Lemma} ({\smc Talagrand 87}) Let $(X,\Sigma,\mu)$ be a
complete probability space, and
$A\subseteq[0,1]^X$ a set which is not stable.   Then there are
measurable functions $h_0$, $h_1:X\to[0,1]$ such that
$\int h_0\,d\mu<\int h_1\,d\mu$ and $(\mu^{2k})^*\tilde D_k=1$ for every
$k\ge 1$, where

$$\eqalignno{\tilde D_k
=\bigcup_{f\in A}\{w:w\in X^{2k},\,&f(w(2i))\le h_0(w(2i)),\cr
&f(w(2i+1))\ge h_1(w(2i+1))\text{ for every }i<k\}.
&\cmmnt{(*)}\cr}$$

\proof{ The proof divides into two cases.

\medskip

{\bf case 1} Suppose that there is an ultrafilter $\Cal F$ on $A$ such
that the $\frak T_p$-limit $g_0$ of $\Cal F$ is not measurable.   Let
$h'_0$, $h_1':X\to[0,1]$ be measurable functions such that
$h'_0\le g_0\le h'_1$
and $\int h'_0=\underline{\int}g_0$, $\int h'_1=\overline{\int}g_0$
(133Ja).   Then $\delta=\bover15\int h'_1-h'_0>0$ (133Jd).   Set
$h_0=h'_0+2\delta\chi X$, $h_1=h'_1-2\delta\chi X$, so that
$\int h_0<\int h_1$.

Set $Q_0=\{x:x\in X,\,g_0(x)\le h_0(x)-\delta\}$.  Then $\mu^*Q_0=1$, by
133J(a-i).   Similarly, $\mu^*Q_1=1$, where
$Q_1=\{x:g_0(x)\ge h_1(x)+\delta\}$.

If $k\ge 1$, $u\in Q_0^k$ and $v\in Q_1^k$,
then there is an $f\in A$ such that
$|f(u(i))-g_0(u(i))|\le\delta$,
$|f(v(i))-g_0(v(i))|\le\delta$ for every $i<k$.   But this means that
$f(u(i))\le h_0(u(i))$ and $f(v(i))\ge h_1(v(i))$ for every $i<k$, and
$u\#v\in\tilde D_k$.   Thus

\Centerline{$\tilde D_k
\supseteq Q_0\times Q_1\times Q_0\times Q_1
  \times\ldots\times Q_0\times Q_1$,}

\noindent which has full outer measure, by 254L.   As $k$ is arbitrary,
we have found appropriate $h_0$, $h_1$ in this case.

\medskip

{\bf case 2} Now suppose that for every ultrafilter $\Cal F$ on $A$,
the $\frak T_p$-limit of $\Cal F$ is measurable.

\medskip

\quad{\bf (i)} We are supposing that $A$ is not stable, so there are
$E\in\Sigma$ and $\alpha<\beta$ such that
$\mu E>0$ and $(\mu^{2k})^*D_k(A,E,\alpha,\beta)\penalty-100=(\mu E)^{2k}$
for every
$k\ge 1$.   The first thing to note is that if $\Cal I$ is the set
of those $B\subseteq A$ for which there is some $k\in\Bbb N$ such that
$(\mu^{2k})^*D_k(B,E,\alpha,\beta)<(\mu E)^{2k}$, then $\Cal I$ is an
ideal of subsets of $A$.   \Prf\ Of course $\emptyset\in\Cal I$ and
$B\in\Cal I$ whenever $B\subseteq B'\in\Cal I$.   Also 465C(b-i) tells us
that, if $B\in\Cal I$, then
$\lim_{k\to\infty}\Bover1{(\mu E)^{2k}}(\mu^{2k})^*D_k(B,E,\alpha,\beta)
=0$.   It follows easily (as in
the proof of 465C(b-ii)) that $B\cup B'\in\Cal I$ for all $B$,
$B'\in\Cal I$.\ \Qed

\medskip

\quad{\bf (ii)} $\Cal I$ is a proper ideal of subsets of $A$ (by the
choice of $E$, $k$, $\alpha$ and $\beta$), so there is an ultrafilter
$\Cal F$ on $A$ such that $\Cal F\cap\Cal I=\emptyset$.   Let $g_0$ be
the $\frak T_p$-limit of $\Cal F$.   Then $g_0$ is measurable.

Set $\delta=\bover13(\beta-\alpha)\mu E>0$, and define $h_0$, $h_1$ by
setting

$$\eqalign{h_0(x)&=\alpha\text{ if }x\in E,\cr
&=g_0(x)+\delta\text{ if }x\in X\setminus E,\cr
h_1(x)&=\beta\text{ if }x\in E,\cr
&=g_0(x)-\delta\text{ if }x\in X\setminus E.\cr}$$

\noindent Then

\Centerline{$\int h_1-\int h_0
\ge(\beta-\alpha)\mu E-2\delta\mu(X\setminus E)>0$.}

\medskip

\quad{\bf (iii)} We shall need to know a little more about sets of the
form $D_k(B,E,\alpha,\beta)$ for $B\in\Cal F$.   In fact, if
$B\subseteq A$ and $B\notin\Cal I$, then for any finite sets $I$ and $J$

\Centerline{$(\mu^I\times\mu^J)^*D_{I,J}(B,E,\alpha,\beta)
=(\mu E)^{\#(I)+\#(J)}$,}

\noindent where $D_{I,J}(B,E,\alpha,\beta)$ is

\Centerline{$\bigcup_{f\in B}\{(u,v):u\in E^I,\,v\in
E^J,\,f(u(i))\le\alpha$ for $i\in I$, $f(v(j))\ge\beta$ for $j\in J\}$.}

\noindent\Prf\ We may suppose that $I=k$ and $J=l$ where $k$,
$l\in\Bbb N$.   Take $m=\max(1,k,l)$.   Then we have an \imp\ map
$\phi:X^{2m}\to X^k\times X^l$ defined by saying that $\phi(w)=(u,v)$
where $u(i)=w(2i)$
for $i<k$ and $v(i)=w(2i+1)$ for $i<l$.   \Quer\ If
$(\mu^k\times\mu^l)^*D_{k,l}(B,E,\alpha,\beta)<(\mu E)^{k+l}$, there is
a non-negligible measurable set $V\subseteq (E^k\times E^l)\setminus
D_{k,l}(B,E,\alpha,\beta)$.   Now $\phi^{-1}[V]$ is
non-negligible and depends only on coordinates in
$\{2i:i<k\}\cup\{2i+1:i<l\}$, so

\Centerline{$\mu^{2m}(E^{2m}\cap\phi^{-1}[V])
=\mu^{2m}(\phi^{-1}[V])\cdot(\mu E)^{2m-k-l}>0$.}

\ifdim\pagewidth>467pt\fontdimen3\tenrm=2pt\fi
\ifdim\pagewidth>467pt\fontdimen4\tenrm=1.67pt\fi
\noindent But $\phi^{-1}[V]\cap D_m(B,E,\alpha,\beta)=\emptyset$, so
$(\mu^{2m})^*D_m(B,E,\alpha,\beta)<(\mu E)^{2m}$, and $B\in\Cal I$.\
\BanG\  So
$(\mu^k\times\mu^l)^*D_{k,l}(B,E,\alpha,\beta)
=(\mu E)^{k+l}$, as required.\ \Qed
\fontdimen3\tenrm=1.67pt
\fontdimen4\tenrm=1.11pt

\medskip

\quad{\bf (iv)} \Quer\ Suppose, if possible, that $k\ge 1$ is such that
$(\mu^{2k})^*\tilde D_k<1$, where $\tilde D_k$ is defined from $h_0$ and
$h_1$ by the formula (*) in the statement of the lemma.   Let
$W\subseteq X^{2k}$ be a
measurable set of positive measure disjoint from $\tilde D_k$.   For
$I$, $J\subseteq k$ write $W_{IJ}$ for

$$\eqalign{\{w:w\in W,\,
&w(2i)\in E\text{ for }i\in I,\,
  w(2i)\notin E\text{ for }i\in k\setminus I,\cr
&w(2i+1)\in E\text{ for }i\in J,\,
  w(2i+1)\notin E\text{ for }i\in k\setminus J\}.\cr}$$

\noindent Then there are $I$, $J\subseteq K$ such that
$\mu^{2k}W_{IJ}>0$.

We can identify $X^{2k}$ with
$X^I\times X^J\times X^{k\setminus I}\times X^{k\setminus J}$, matching
any $w\in X^{2k}$ with $(w_0,w_1,w_2,w_3)$ where

$$\eqalign{w_0(i)&=w(2i)\text{ for }i\in I,\cr
w_1(i)&=w(2i+1)\text{ for }i\in J,\cr
w_2(i)&=w(2i)\text{ for }i\in k\setminus I,\cr
w_3(i)&=w(2i+1)\text{ for }i\in k\setminus J.\cr}$$

\noindent Write $\tilde W$ for the image of $W_{IJ}$ under this
matching.
The condition $W_{IJ}\cap\tilde D_k=\emptyset$ translates into

\inset{($\dagger$) whenever $(w_0,w_1,w_2,w_3)\in\tilde W$, $f\in A$,

\quad either there is an $i\in I$ such that $f(w_0(i))>\alpha$

\quad or there is an $i\in J$ such that $f(w_1(i))<\beta$

\quad or there is an $i\in k\setminus I$ such that
$f(w_2(i))>g_0(w_2(i))+\delta$

\quad or there is an $i\in k\setminus J$ such that
$f(w_3(i))<g_0(w_3(i))-\delta$.}

\medskip

\quad{\bf (v)} By Fubini's theorem, applied to
$(X^I\times X^J)\times(X^{k\setminus I}\times X^{k\setminus J})$, we can
find $w_2\in X^{k\setminus I}$, $w_3\in X^{k\setminus J}$ such that
$(\mu^I\times\mu^J)(V)$ is defined and greater than $0$, where
$V=\{(w_0,w_1):(w_0,w_1,w_2,w_3)\in\tilde W\}$.   Set

$$\eqalign{B=\{f:f\in A,\,
&|f(w_2(i))-g_0(w_2(i))|\le\delta\text{ for }i\in k\setminus I,\cr
&|f(w_3(i))-g_0(w_3(i))|\le\delta\text{ for }i\in k\setminus J\}.\cr}$$

\noindent Then $B\in\Cal F$, because $\Cal F\to g_0$ for $\frak T_p$.
So $(\mu^I\times\mu^J)^*D_{I,J}(B,E,\alpha,\beta)
=(\mu E)^{\#(I)+\#(J)}$, by (iii) above.   Since
$W_{IJ}\subseteq W$, $V$ is included in $E^I\times E^J$
and meets
$D_{I,J}(B,E,\alpha,\beta)$;  that is, there are $f\in B$ and
$(w_0,w_1)\in V$ such that $f(w_0(i))\le\alpha$ for $i\in I$ and
$f(w_1(i))\ge\beta$ for $i\in J$.   But because $f\in B$ we also have
$f(w_2(i))\le g_0(w_2(i))+\delta$ for $i\in k\setminus I$ and
$f(w_3(i))\ge g_0(w_3(i))-\delta$ for $i\in k\setminus J$;  which
contradicts the list in ($\dagger$) above.\ \Bang

\medskip

\quad{\bf (vi)} Thus $(\mu^{2k})^*\tilde D_k=1$ for every $k$, and in
this case also we have an appropriate pair $h_0$, $h_1$.
}%end of proof of 465L

\vleader{108pt}{465M}{Theorem}\cmmnt{ ({\smc Talagrand 82},
{\smc Talagrand 87})} Let $(X,\Sigma,\mu)$ be a complete probability
space, and $A$ a non-empty uniformly bounded set of real-valued
functions defined on $X$.   Then the following are equiveridical.

(i) $A$ is stable.

(ii) Every function in $A$ is measurable, and
$\lim_{k\to\infty}\sup_{f\in A}
|\Bover1k\sum_{i=0}^{k-1}f(w(i))-\int f|=0$ for almost every
$w\in X^{\Bbb N}$.

(iii) Every function in $A$ is measurable, and for every $\epsilon>0$
there are a finite subalgebra $\Tau$ of $\Sigma$ in which every atom
is non-negligible and a sequence $\langle h_k\rangle_{k\ge 1}$ of
measurable functions on $X^{\Bbb N}$ such that

\Centerline{$h_k(w)\ge\sup_{f\in A}\Bover1k\sum_{i=0}^{k-1}
\bigl|f(w(i))-\Expn(f|\Tau)(w(i))\bigr|$}

\noindent for every $w\in X^{\Bbb N}$ and $k\ge 1$, and

\Centerline{$\limsup_{k\to\infty}h_k(w)\le\epsilon$}

\noindent for almost every $w\in X^{\Bbb N}$.   (Here
$\Expn(f|\Tau)$ is the\cmmnt{ (unique)} conditional expectation
of $f$ on $\Tau$.)

(iv) $\lim_{k,l\to\infty}\sup_{f\in A}
|\Bover1k\sum_{i=0}^{k-1}f(w(i))-\Bover1l\sum_{i=0}^{l-1}f(w(i))|=0$ for
almost every $w\in X^{\Bbb N}$.

(v) $\lim_{k,l\to\infty}\overline{\int}\sup_{f\in A}
|\Bover1k\sum_{i=0}^{k-1}f(w(i))-\Bover1l\sum_{i=0}^{l-1}f(w(i))|
 \mu^{\Bbb N}(dw)
=0$.

\proof{ All the statements (i)-(v) are unaffected by translations (by
constant functions) and non-zero
scalar multiplications of the set $A$, so it will be enough to consider
the case in which $A\subseteq[0,1]^X$.

As in 465H-465I, I write $\nu_{wk}(E)=\bover1k\#(\{i:i<k,\,w(i)\in E\})$
for $w\in X^{\Bbb N}$, $k\ge 1$ and $E\subseteq X$;  so for any function
$f:X\to\Bbb R$, $\int fd\nu_{wk}=\Bover1k\sum_{i=0}^{r-1}f(w(i))$.

\medskip

{\bf (a)(i)$\Rightarrow$(iii)}\grheada\ If $A$ is stable, then every
function in $A$ is measurable, by 465Da.   Let $\epsilon>0$.   Set
$\eta=\bover1{108}\epsilon^2>0$.   By 465C(a-ii) and 465Db, the
$\frak T_p$-closure
$\overline{A}$ of $A$ in $\Bbb R^X$ is a stable $\frak T_p$-compact set of
measurable functions, and by 465G it is $\frak T_m$-compact;  because
$A$ is uniformly bounded, it must be totally bounded for the
pseudometric induced
by $\|\,\|_1$.   So there are $f_0,\ldots,f_m\in A$ such that for every
$f\in A$ there is an $i\le m$ such that $\|f-f_i\|_1\le\eta$.   Let
$\Tau_0$ be the finite subalgebra of $\Sigma$ generated by the sets
$\{x:j\eta\le f_i(x)<(j+1)\eta\}$ for $i\le m$ and $j\le\Bover1{\eta}$.
Then $\Tau_0$ may have negligible atoms, but if we absorb these into
non-negligible atoms we get a finite subalgebra $\Tau$ of $\Sigma$ such
that $|f_i(x)-\Expn(f_i|\Tau)(x)|\le\eta$ for almost every $x\in X$,
every $i\le m$.   (Because $\Tau$ is a finite algebra with
non-negligible atoms, two $\Tau$-measurable functions which are equal
almost everywhere must be identical, and we have unique conditional
expectations with respect to $\Tau$.)   Since
$\|\Expn(f|\Tau)-\Expn(g|\Tau)\|_1\le\|f-g\|_1$ for all integrable
functions $f$ and $g$ (242Je), $\|f-\Expn(f|\Tau)\|_1\le 3\eta$ for
every $f\in A$.

\medskip

\quad\grheadb\ Set $A'=\{f-\Expn(f|\Tau):f\in A\}$.   Then $A'$ is
stable.   \Prf\ Suppose that $\mu E>0$ and $\alpha<\beta$.   The set
$B=\{\Expn(f|\Tau):f\in A\}$ is a uniformly bounded subset of a
finite-dimensional space of functions, so is $\|\,\|_{\infty}$-compact.
So there are $g_0,\ldots,g_r\in B$ such that
$B\subseteq\bigcup_{i\le r}\{g:\|g-g_i\|_{\infty}\le\bover13(\beta-\alpha)\}$.
By 465C(a-iv) and
465C(b-ii), $C=\bigcup_{i\le r}A-g_i$ is stable.   So there is a $k\ge 1$
such that

\Centerline{$(\mu^{2k})^*D_k(C,E,\bover23\alpha+\bover13\beta,
\bover13\alpha+\bover23\beta)<(\mu E)^{2k}$.}

\noindent But for every $g\in A'$ there is an $h\in C$ such that
$\|g-h\|_{\infty}\le\bover13(\beta-\alpha)$, so

\Centerline{$D_k(A',E,\alpha,\beta)\subseteq
D_k(C,E,\bover23\alpha+\bover13\beta,\bover13\alpha+\bover23\beta)$,
\quad$(\mu^{2k})^*D_k(A',E,\alpha,\beta)<(\mu E)^{2k}$.}

\noindent As $E$, $\alpha$ and $\beta$ are arbitrary, $A'$ is stable.\
\Qed

\medskip

\quad\grheadc\ By 465C(b-iv),

\Centerline{$A''=\{g^+:g\in A'\}\cup\{g^-:g\in A'\}$}

\noindent is stable.
By 465K there are an $n\ge 1$, a $W\in\Tensorhat_n\Sigma$
and a $\gamma>\mu^nW$ such that
$\int h\,d\nu\le 3\sqrt{3\eta}=\bover12\epsilon$
whenever $h\in A''$ and $\nu$ is a probability measure on $X$,
with domain including $\Sigma$, such that $\nu^nW\le\gamma$.   So
$\int|g|d\nu\le\epsilon$ whenever $g\in A'$ and $\nu$ is such a measure.
If $w\in X^{\Bbb N}$ and $k\in\Bbb N$,
$\nu_{wk}$ is a probability measure on $X$;  set
$q_k(w)=\nu_{wk}^n(W)$.
Applying 465H to $\chi W$, we see that
$\lim_{k\to\infty}q_k(w)=\mu^nW$ for almost every $w\in X^{\Bbb N}$.
Also, because $W\in\Tensorhat_n\Sigma$, every $q_k$ is measurable, by
465I.

Set $h_k(w)=1$ if $q_k(w)>\gamma$, $\epsilon$ if $q_k(w)\le\gamma$.
Then every $h_k$ is measurable and $\lim_{k\to\infty}h_k(w)=\epsilon$
for almost every $w$.

For any $w\in X^{\Bbb N}$ and any $f\in A$, $g=f-\Expn(f|\Tau)\in A'$
and $\|g\|_{\infty}\le 1$.   So either $h_k(w)=1$ and certainly
$\int|g|d\nu_{wk}\le h_k(w)$, or $h_k(w)=\epsilon$,
$\nu_{wk}^n(W)\le\gamma$ and $\int|g|d\nu_{wk}\le\epsilon$.   Thus we
have

\Centerline{$\Bover1k\sum_{i=0}^{k-1}\bigl|f(w(i))
  -\Expn(f|\Tau)(w(i))\bigr|
=\int\bigl|f-\Expn(f|\Tau)\bigr|d\nu_{wk}\le h_k(w)$}

\noindent for every $w\in X^{\Bbb N}$ and every $f\in A$, as required by
(iii).

\medskip

{\bf (b)(iii)\,$\Rightarrow$\,(ii)\&(v)} Set

\Centerline{$g_k(w)
=\sup_{f\in A}|\Bover1k\sum_{i=0}^{k-1}f(w(i))-\int fd\mu|$,}

\Centerline{$g'_{kl}(w)
=\sup_{f\in A}|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  -\Bover1l\sum_{i=0}^{l-1}f(w(i))|$,}

\noindent for $w\in X^{\Bbb N}$ and $k$, $l\ge 1$.
Let $\epsilon>0$.   Let $\Tau$ and $\langle h_k\rangle_{k\ge 1}$ be as in
(iii);  as $A\subseteq[0,1]^X$, we can suppose that $h_k(w)\le 2$ for every
$k$ and $w$.   Let $E_0,\ldots,E_r$ be the atoms of $\Tau$.   For
$w\in X^{\Bbb N}$, $k\ge 1$, $j\le r$ set
$q_{kj}(w)=|\mu E_j-\nu_{wk}E_j|$.   Then for any $f\in A$,
$\Expn(f|\Tau)$ is expressible as $\sum_{j=0}^r\alpha_j\chi E_j$ where
$\alpha_j\in[0,1]$ for every $j$ (remember that $A\subseteq[0,1]^X$), so

$$\eqalign{\bigl|\Bover1k\sum_{i=0}^{k-1}f(w(i))&-\int fd\mu\bigr|
=\bigl|\int fd\nu_{wk}-\int fd\mu\bigr|\cr
&\le\bigl|\int fd\nu_{wk}-\int\Expn(f|\Tau)d\nu_{wk}\bigr|
  +\bigl|\int\Expn(f|\Tau)d\nu_{wk}-\int\Expn(f|\Tau)d\mu\bigr|\cr
&\le\Bover1k\sum_{i=0}^{k-1}\bigl|f(w(i))-\Expn(f|\Tau)(w(i))\bigr|
  +\sum_{j=0}^r\alpha_j\bigl|\nu_{wk}E_j-\mu E_j\bigr|\cr
&\le h_k(w)+\sum_{j=0}^rq_{kj}(w),\cr
\bigl|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  &-\Bover1l\sum_{i=0}^{l-1}f(w(i))\bigr|\cr
&\le h_k(w)+\sum_{j=0}^rq_{kj}(w)+h_l(w)+\sum_{j=0}^rq_{lj}(w).\cr}$$

\noindent Taking the supremum over $f$, we have

\Centerline{$g_k(w)\le h_k(w)+\sum_{j=0}^rq_{kj}(w)$,}

\Centerline{$g'_{kl}(w)
\le h_k(w)+\sum_{j=0}^rq_{kj}(w)+\sum_{j=0}^rq_{lj}(w)+h_l(w)$.}

\noindent But, for each $j\le r$,
$\lim_{k\to\infty}q_{kj}(w)=0$ for almost every $w$, by 273J
applied to $\chi E_j$.   So

\Centerline{$\limsup_{k\to\infty}g_k(w)
\le\limsup_{k\to\infty}h_k(w)\le\epsilon$}

\noindent for almost every $w$.   At the same time,

\Centerline{$
\limsup_{k,l\to\infty}\overlineint g'_{kl}
\le\limsup_{k,l\to\infty}\int h_k+\sum_{j=0}^r\int q_{kj}
  +\sum_{j=0}^r\int q_{lj}+\int h_l
\le 2\epsilon$.}

As $\epsilon$ is arbitrary,
$\{w:\limsup_{k\to\infty}g_k(w)\ge 2^{-i}\}$ is negligible for every
$i\in\Bbb N$, and $\lim_{k\to\infty}g_k=0$ almost everywhere, as
required, while equally $\lim_{k,l\to\infty}\overline{\int}g'_{kl}=0$.
Thus (ii) and (v) are true.

\wheader{465M}{4}{2}{2}{36pt}

{\bf (c)(ii)$\Rightarrow$(iv)} is trivial, since

\Centerline{$\bigl|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  -\Bover1l\sum_{i=0}^{l-1}f(w(i))\bigr|
\le\bigl|\Bover1k\sum_{i=0}^{k-1}f(w(i))-\int f\bigr|
  +\bigl|\Bover1l\sum_{i=0}^{l-1}f(w(i))-\int f\bigr|$.}

\medskip

{\bf (d) not-(i)\,$\Rightarrow$\,not-(iv)\&not-(v)} Now suppose that $A$
is not stable.

\medskip

\quad\grheada\ In this case, by 465L, there are measurable functions
$h_0$, $h_1:X\to[0,1]$ such that $\int h_0d\mu<\int h_1d\mu$ and
$(\mu^{2k})^*\tilde D_k=1$ for every $k\in\Bbb N$, where

$$\eqalign{\tilde D_k
=\bigcup_{f\in A}\{w:w\in X^{2k},\,&f(w(2i))\le h_0(w(2i)),\cr
&f(w(2i+1))\ge h_1(w(2i+1))\text{ for every }i<k\}.\cr}$$

\medskip

\quad\grheadb\ Set $\delta=\bover14\int h_1-h_0>0$.   Let $k_0\ge 1$ be
so large that

\Centerline{$\mu^k\{w:w\in X^k,\,
|\int h_j-\Bover1k\sum_{i=0}^{k-1}h_j(w(i))|
\le\delta\}\ge\Bover12$}

\noindent for both $j$ and for every $k\ge k_0$ (273J or 465H.   The
point is that

$$\eqalign{\mu^k\{w:w\in X^k,\,
&|\int h_j-\Bover1k\sum_{i=0}^{k-1}h_j(w(i))|
\le\delta\}\cr
&=\mu^{\Bbb N}\{w:w\in X^{\Bbb N},\,
|\int h_j-\Bover1k\sum_{i=0}^{k-1}h_j(w(i))|
\le\delta\}
\to 1\text{ as }k\to\infty.)\cr}$$

\noindent Let $\langle k_n\rangle_{n\ge 1}$ be such that
$k_n\ge\Bover2{\delta}\sum_{i=0}^{n-1}k_i$ for every $n\ge 1$.   Since
$\delta\le\bover14$, every $k_n$ is at least as large as $k_0$.

\medskip

\quad\grheadc\ Set $m_n=2\sum_{i<n}k_i$ for each $n\in\Bbb N$.   Then we
have a measure space isomorphism
$\phi:\prod_{n\in\Bbb N}X^{2k_n}\to X^{\Bbb N}$ defined by setting

\Centerline{$\phi(w)(m_n+i)=w(n)(2i)$,
\quad$\phi(w)(m_n+k_n+i)=w(n)(2i+1)$}

\noindent for $n\in\Bbb N$ and $i<k_n$.   For each $n\in\Bbb N$,
$\tilde D_{k_n}$ has outer measure $1$ in $X^{2k_n}$, so
$\tilde D=\prod_{n\in\Bbb N}\tilde D_{k_n}$ has outer measure $1$ in
$\prod_{n\in\Bbb N}X^{2k_n}$, and $\phi[\tilde D]$ has outer measure $1$
in $X^{\Bbb N}$.   Note that $\phi[\tilde D]$ is just the set of
$w\in X^{\Bbb N}$ such that, for every $n\in\Bbb N$, there is an
$f\in A$ such that

$$\eqalign{f(w(i))&\le h_0(w(i))\text{ for }m_n\le i<m_n+k_n,\cr
f(w(i))&\ge h_1(w(i))\text{ for }m_n+k_n\le i<m_n+2k_n=m_{n+1}.\cr}$$

If we set

\Centerline{$V_{n0}=\{w:w\in X^{\Bbb N},\,
|\Bover1{k_n}\sum_{i=m_n}^{m_n+k_n-1}h_0(w(i))-\int h_0|
\le\delta\}$,}

\Centerline{$V_{n1}=\{w:w\in X^{\Bbb N},\,
|\Bover1{k_n}\sum_{i=m_n+k_n}^{m_n+2k_n-1}
h_1(w(i))-\int h_1|\le\delta\}$,}

\noindent every $V_{n0}$ and $V_{n1}$ has measure at least $\bover12$,
because $k_n\ge k_0$.

\medskip

\quad\grheadd\ Now suppose that $n\in\Bbb N$ and
$w\in V_{n0}\cap V_{n1}\cap\phi[\tilde D]$.   Then

$$\eqalign{\sup_{f\in A}\bigl|\Bover1{m_n+2k_n}
  \sum_{i=0}^{m_n+2k_n-1}&f(w(i))
  -\Bover1{m_n+k_n}\sum_{i=0}^{m_n+k_n-1}f(w(i))\bigr|
\ge\Bover{\delta}{(2+\delta)(1+\delta)}.\cr}$$

\noindent\Prf\ Since $w\in\phi[\tilde D]$, there must be an
$f\in A$ such that

$$\eqalign{f(w(i))&\le h_0(w(i))\text{ for }m_n\le i<m_n+k_n,\cr
f(w(i))&\ge h_1(w(i))\text{ for }m_n+k_n\le i<m_n+2k_n=m_{n+1}.\cr}$$

\noindent Set $s=\sum_{i=0}^{m_n-1}f(w(i))$,
$t=\sum_{i=m_n}^{m_n+k_n}f(w(i))$ and
$t'=\sum_{i=m_n+k_n}^{m_n+2k_n-1}f(w(i))$.   Then

$$\eqalign{\Bover1{m_n+2k_n}\sum_{i=0}^{m_n+2k_n-1}&f(w(i))
  -\Bover1{m_n+k_n}\sum_{i=0}^{m_n+k_n-1}f(w(i))\cr
&=\bover{s+t+t'}{m_n+2k_n}-\bover{s+t}{m_n+k_n}
=\bover{(t'-t-s)k_n+t'm_n}{(m_n+2k_n)(m_n+k_n)}\cr
&\ge\bover{(t'-t-m_n)k_n}{(m_n+2k_n)(m_n+k_n)}
\ge\bover{t'-t-\delta k_n}{(2+\delta)(1+\delta)k_n}\cr}$$

\noindent because $m_n\le\delta k_n$, by the choice of $k_n$.

To estimate $t$ and $t'$, we have

\Centerline{$t
=\sum_{i=m_n}^{m_n+k_n-1}f(w(i))
\le\sum_{i=m_n}^{m_n+k_n-1}h_0(w(i))
\le k_n(\int h_0+\delta)$}

\noindent because $w\in V_{n0}$.   On the other hand,

\Centerline{$t'
=\sum_{i=m_n+k_n}^{m_n+2k_n-1}f(w(i))
\ge\sum_{i=m_n+k_n}^{m_n+2k_n-1}h_1(w(i))
\ge k_n(\int h_1-\delta)$}

\noindent because $w\in V_{n1}$.   So

\Centerline{$t'-t\ge k_n(\int h_1-h_0-2\delta)=2k_n\delta$,}

\noindent and

$$\eqalign{\Bover1{m_n+2k_n}\sum_{i=0}^{m_n+2k_n-1}&f(w(i))
  -\Bover1{m_n+k_n}\sum_{i=0}^{m_n+k_n-1}f(w(i))\cr
&\ge\bover{2\delta k_n-\delta k_n}{(2+\delta)(1+\delta)k_n}
=\bover{\delta}{(2+\delta)(1+\delta)}.\text{ \Qed}\cr}$$

\medskip

\quad\grheade\ The $V_{n0}$ and $V_{n1}$ are all
independent.   So $\mu^{\Bbb N}(V_{n0}\cap V_{n1})\ge\bover14$ for every
$n$, and

\Centerline{$W=\{w:w\in X^{\Bbb N},\,w\in V_{n0}\cap V_{n1}$ for
infinitely many $n\}$}

\noindent has measure $1$ (by the Borel-Cantelli lemma, 273K, or
otherwise).
Accordingly $W\cap\phi[\tilde D]$ has outer measure $1$ in
$X^{\Bbb N}$.   But if $w\in W\cap\phi[\tilde D]$, then ($\delta$) tells
us that

$$\eqalign{\limsup_{k,l\to\infty}
\sup_{f\in A}\bigl|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  -\Bover1l\sum_{i=0}^{l-1}f(w(i))\bigr|
\ge\Bover{\delta}{(2+\delta)(1+\delta)},\cr}$$

\noindent because there are infinitely many $n$ such that
$w\in V_{n0}\cap V_{n1}\cap\phi[\tilde D]$.   So (iv) must be false.

\medskip

\quad\grheadz\ We see also that, for any $n\in\Bbb N$,

%$$\eqalign{\overline{\int}g'_{m_n+k_n,m_n+2k_n}
%&\ge\Bover{\delta}{(2+\delta)(1+\delta)}
%  \mu^{\Bbb N}(V_{n0}\cap V_{n1})\cr
%&\ge\Bover{\delta}{4(2+\delta)(1+\delta)},\cr}$$

\Centerline{$\overlineint g'_{m_n+k_n,m_n+2k_n}
\ge\Bover{\delta}{(2+\delta)(1+\delta)}
  \mu^{\Bbb N}(V_{n0}\cap V_{n1})
\ge\Bover{\delta}{4(2+\delta)(1+\delta)}$,}

\noindent writing

\Centerline{$g'_{kl}(w)=\sup_{f\in A}
 |\Bover1k\sum_{i=0}^{k-1}f(w(i))-\Bover1l\sum_{i=0}^{l-1}f(w(i))|$}

\noindent for $k$, $l\in\Bbb N$.   So
$\limsup_{k,l\to\infty}\overline{\int}g'_{kl}>0$, and (v) is false.

\medskip

{\bf (e)} Assembling these, we have

\Centerline{
(i)$\Rightarrow$(iii)$\Rightarrow$(ii)$\Rightarrow$(iv)$\Rightarrow$(i),
\quad(iii)$\Rightarrow$(v)$\Rightarrow$(i),}

\noindent so the five assertions are equiveridical.
}%end of proof of 465M

\cmmnt{\medskip

\noindent{\bf Remark} If $(X,\Sigma,\mu)$ is a probability space, a set
$A\subseteq\eusm L^1(\mu)$ is a {\bf Glivenko-Cantelli class} if
$\sup_{f\in A}|\Bover1k\sum_{i=0}^{k-1}\penalty-100f(w(i))-\int f|\to 0$ as
$k\to\infty$ for $\mu^{\Bbb N}$-almost every $w\in X^{\Bbb N}$.
Compare 273Xi.}

\leader{465N}{Theorem}
Let $(X,\Sigma,\mu)$ be a semi-finite measure space.

(a)\cmmnt{ ({\smc Talagrand 84})} %11-2-1
Let $A\subseteq\BbbR^X$ be a stable
set.   Suppose that there is a measurable function
$g:X\to\coint{0,\infty}$ such that $|f(x)|\le g(x)$ whenever $x\in X$ and
$f\in A$.   Then the convex hull $\Gamma(A)$ of $A$ in $\Bbb R^X$ is
stable.

(b) If $A\subseteq\BbbR^X$ is stable, then $|A|=\{|f|:f\in A\}$ is stable.

(c) Let $A$, $B\subseteq\Bbb R^X$ be two stable sets such that
$\{f(x):f\in A\cup B\}$ is bounded for every $x\in X$.   Then
$A+B=\{f_1+f_2:f_1\in A,\,f_2\in B\}$ is stable.

(d) Suppose that $\mu$ is complete and locally determined.
Let $A\subseteq\BbbR^X$ be a stable set such that
$\{f(x):f\in A\}$ is bounded for every $x\in X$.   Then $\Gamma(A)$ is
relatively compact in $\eusm L^0(\Sigma)$ for the topology of pointwise
convergence.

\proof{{\bf (a)(i)} Consider first the case in which $\mu X=1$ and
$A\subseteq[-1,1]^X$.   In this case,

\Centerline{$\sup_{f\in\Gamma(A)}|
  \Bover1k\sum_{i=0}^{k-1}f(w(i))-\int f|
=\sup_{f\in A}|\Bover1k\sum_{i=0}^{k-1}f(w(i))-\int f|$}

\noindent for every $k\ge 1$ and $w\in X^{\Bbb N}$.   So $\Gamma(A)$
satisfies condition (ii) of 465M whenever $A$ does, and we have the
result.

\medskip

\quad{\bf (ii)} Now suppose just that $\mu X=1$.   Set
$A'=\{\Bover{f}{g+\chi X}:f\in A\}$.   Then $A'$ is stable (465C(a-v)), so
$\Gamma(A')$ is stable, by (i), and
$\Gamma(A)=\{f\times(g+\chi X):f\in A'\}$ is stable.

\medskip

\quad{\bf (iii)} If $\mu X=0$, the result is trivial.   If $\mu X<\infty$,
apply (ii) to a multiple of the measure $\mu$.   For the general case,
write $A_E=\{f\restr E:f\in A\}$ for $E\subseteq X$.   Then $A_E$ is stable
for the subspace measure on $E$, by 4{}65Cm.
It follows that $\Gamma(A_E)$ is stable
whenever $\mu E<\infty$.   But $\Gamma(A_E)=(\Gamma(A))_E$, so 4{}65Cn tells
us that $\Gamma(A)$ is stable.

\medskip

{\bf (b)(i)} I begin with a basic special case of (c).   If $A$,
$B\subseteq\BbbR^X$ are stable and uniformly bounded, then $A+B$ is stable.
\Prf\ Putting 465C(b-ii), (a) of this theorem,
465C(a-iii) and 465C(a-i) together, we see
that $A\cup B$, $\Gamma(A\cup B)$ and $A+B\subseteq 2\Gamma(A\cup B)$ are
stable.\ \Qed

\medskip

\quad{\bf (ii)} Adding this to 465C(b-iv),
$|A|\subseteq\{f^+:f\in A\}+\{f^-:f\in A\}$ is stable
whenever $A\subseteq\BbbR^X$ is stable and uniformly bounded.

\medskip

\quad{\bf (iii)} For the general case, use 465C(c-v).   For $n\in\Bbb N$, set
$A_n=\{\med(-n\chi X,f,n\chi X):f\in A\}$,

\Centerline{$|A|_n=\{\med(-n\chi X,f,n\chi X):f\in|A|\}
=\{\med(-n\chi X,|f|,n\chi X):f\in A\}
=|A_n|$.}

\noindent Now 465C(c-v) and (i)-(ii) above tell us that
if $A$ is stable, so are every $A_n$ and every $|A_n|=|A|_n$, so that $A$ is
stable.

\medskip

{\bf (c)(i)} By 4{}65Cn, it is enough to consider the case of totally finite
$\mu$, so let us suppose from now on that $\mu X<\infty$.   We may also
suppose that neither $A$ nor $B$ is empty;  finally, by 465C(c-i),
we can suppose that $\mu$ is complete, so that
$A\cup B\subseteq\eusm L^0(\Sigma)$ (465Da).

I introduce some temporary notation:  if $E\subseteq X$, $k\ge 1$,
$\epsilon>0$ and $A\subseteq\BbbR^X$, set

\Centerline{$\tilde D_k(A,E,\epsilon)
=\bigcup_{f\in A}\{u:u\in E^k$, $|f(u(i))|\ge\epsilon$ for $i<k\}$.}

\medskip

\quad{\bf (ii)} We need to know that if $A\subseteq\BbbR^X$ and every
$f\in A$ is zero a.e., then $A$ is stable iff whenever $E\in\Sigma$,
$0<\mu E<\infty$ and $\epsilon>0$ there is a $k\ge 1$ such that
$(\mu^k)^*\tilde D_k(A,E,\epsilon)<(\mu E)^k$.

\medskip

\quad\Prf\grheada\ If $A$ is stable, then $|A|$ is stable,
by (b), so if $0<\mu E<\infty$ and $\epsilon>0$ there is a $k\ge 1$ such
that $(\mu^{2k})^*D_{2k}(|A|,E,0,\epsilon)<(\mu E)^{2k}$.   Let
$W\in\Tensorhat_{2k}\Sigma$ be such that
$D_{2k}(|A|,E,0,\epsilon)\subseteq W\subseteq E^{2k}$
and $\mu^{2k}W<(\mu E)^{2k}$.   Because $(u,v)\mapsto u\#v$ is a measure
space isomorphism,

\Centerline{$\mu^{2k}W=\int\mu^k\{u:u\#v\in W\}\mu^k(dv)$,}

\noindent so if we set
$V=\{v:v\in E^k$, $\mu^k\{u:u\#v\in W\}=(\mu E)^k\}$ we must have
$\mu^kV<(\mu E)^k$.   If
$v\in\tilde D_k(A,E,\epsilon)$, there is an $f\in A$ such that
$|f(v(i))|\ge\epsilon$ for every $i<k$;  now

\Centerline{$\{u:u\#v\in W\}
\supseteq\{u:u\in E^k$, $f(u(i))=0$ for every $i<k\}$}

\noindent has measure $(\mu E)^k$, because $f=0$ a.e.   So
$\tilde D_k(A,E,\epsilon)\subseteq V$ and
$(\mu^k)^*\tilde D_k(A,E,\epsilon)<(\mu E)^k$.
As $E$ and $\epsilon$ are arbitrary, $A$ satisfies the
condition.

\medskip

\qquad\grheadb\ Now suppose that $A$ satisfies the condition.
Take $E\in\Sigma$ such
that $\mu E>0$, and $\alpha<\beta$ in $\Bbb R$.   If $\beta>0$, set
$\epsilon=\beta$;  otherwise, set $\epsilon=-\alpha$.   Then there is a
$k\in\Bbb N$ such that $(\mu^k)^*\tilde D_k(A,E,\epsilon)<(\mu E)^k$.
As $D_k(A,E,\alpha,\beta)$ is included in
$\{(u,v):u\in E^k$, $v\in\tilde D_k(A,E,\epsilon)\}$ (if $\beta>0$) or
$\{(u,v):u\in\tilde D_k(A,E,\epsilon)$, $v\in E^k\}$ (if $\beta\le 0$),
$(\mu^{2k})^*D_k(A,E,\alpha,\beta)<(\mu E)^{2k}$.   As $E$, $\alpha$ and
$\beta$ are arbitrary, $A$ is stable.\ \Qed

\medskip

\quad{\bf (iii)} Suppose that $A$ and $B$ are stable sets such that
$f=0$ a.e.\ for every $f\in A\cup B$.   Then $A+B$ is stable.   \Prf\ Set
$A'=\{|f|\wedge\chi X:f\in A\}$, $B'=\{|f|\wedge\chi X:f\in B\}$.   Then
$A'$ and $B'$ are stable, so $A'+B'$ is stable, by (b)(i) above.
But now observe that if $u\in\tilde D_k(A+B,E,\epsilon)$, where
$E\subseteq X$, $\epsilon>0$ and $k\ge 1$, then there are $f_1\in A$,
$f_2\in B$ such that $|f_1(u(i))+f_2(u(i))|\ge\epsilon$ for every $i<k$.
In this case, setting $f_j'=|f_j|\wedge\chi X$ for both $j$, $g=f_1'+f_2'$
belongs to $A'+B'$ and $g(u(i))\ge\min(1,\epsilon)$ for every $i<k$.
This shows that
$\tilde D_k(A+B,E,\epsilon)\subseteq\tilde D_k(A'+B',E,\min(1,\epsilon))$.
Also every function in either $A+B$ or $A'+B'$ is zero a.e.   So
(ii) tells us that $A+B$ also is stable.\ \Qed

\medskip

\quad{\bf (iv)} Suppose that $A$, $B\subseteq\BbbR^X$ are stable, that
$|f|\le\chi X$ for every $f\in A$, and that $g=0$ a.e.\ for every $g\in B$.
Then $A+B$ is stable.   \Prf\ For $g\in B$ set
$g'(x)=\med(-2,g(x),2)$
for $x\in X$;  set $B'=\{g':g\in B\}$.   Then $B'$ is stable, by 465C(a-vi), and
both $A$ and $B'$ are uniformly bounded, so $A+B'$
is stable.   Take $E\in\Sigma$ such that $\mu E>0$, and
$\alpha<\beta$ in $\Bbb R$.

If $\beta>1$, then, by (ii), there is a $k\ge 1$ such that
$(\mu^k)^*\tilde D_k(B,E,\beta-1)<(\mu E)^k$.   Now if
$w\in D_k(A+B,E,\alpha,\beta)$ there are $f\in A$, $g\in B$ such that
$f(w(2i+1))+g(w(2i+1))\ge\beta$ for every $i<k$;  accordingly
$g(w(2i+1))\ge\beta-1$ for $i<k$ and $w=u\#v$ for some $u\in E^k$,
$v\in\tilde D_k(B,E,\beta-1)$.   So

\Centerline{$(\mu^{2k})^*D_k(A+B,E,\alpha,\beta)
\le(\mu E)^k\cdot(\mu^k)^*\tilde D(B,E,\beta-1)<(\mu E)^{2k}$.}

\noindent Similarly, if $\alpha<-1$, then

\Centerline{$(\mu^{2k})^*D_k(A+B,E,\alpha,\beta)
\le(\mu E)^k\cdot(\mu^k)^*\tilde D(B,E,-1-\alpha)<(\mu E)^{2k}$}

\noindent for some $k$.

On the other hand, if $-1\le\alpha<\beta\le 1$, there is a $k\ge 1$ such
that $(\mu^{2k})^*D_k(A+B',E,\alpha,\beta)<(\mu E)^{2k}$.   If now
$w\in D_k(A+B,E,\alpha,\beta)$, take $f\in A$ and $g\in B$ such that
$f(w(2i))+g(w(2i))\le\alpha$ and $f(w(2i+1))+g(w(2i+1))\ge\beta$ for $i<k$.
In this case, for each $i<k$,

\inset{----- either $g'(w(2i))\le g(w(2i))$ and
$f(w(2i))+g'(w(2i))\le\alpha$,
or $g'(w(2i))=-2$ and $f(w(2i))+g'(w(2i))\le-1\le\alpha$,

----- either $g'(w(2i+1))\ge g(w(2i+1))$ and
$f(w(2i+1))+g'(w(2i+1))\ge\beta$,
or $g'(w(2i+1))=2$ and $f(w(2i+1))+g'(w(2i+1))\ge 1\ge\beta$.}

\noindent So $w\in D_k(A+B',E,\alpha,\beta)$.   Accordingly
$(\mu^{2k})^*D_k(A+B,E,\alpha,\beta)<(\mu E)^{2k}$.

As $E$, $\alpha$ and $\beta$ are arbitrary, $A+B$ is stable.\ \Qed

\medskip

\quad{\bf (v)} Now suppose that $|f|\leae\chi X$ for every $f\in A\cup B$.
For $f\in A\cup B$ and $x\in X$, set $f_0(x)=\med(-1,f(x),1)$,
$f_1(x)=\max(0,f(x)-1)$ and $f_2(x)=\max(0,-1-f(x))$; then $f=f_0+f_1-f_2$,
$|f_0|\le\chi X$ and $f_1$, $f_2$ are zero a.e.   Also
$A_0=\{f_0:f\in A\}$, $A_1=\{f_1:f\in A\}$, $A_2=\{f_2:f\in A\}$,
$B_0=\{f_0:f\in B\}$, $B_1=\{f_1:f\in B\}$ and $B_2=\{f_2:f\in B\}$ are all
stable, by 465C(a-vi).   Accordingly $A_0+B_0$ is stable, by (ii);
by (iii), $A_1-A_2+B_1-B_2$ is stable;  by (iv),

\Centerline{$A+B\subseteq A_0+B_0+A_1-A_2+B_1-B_2$}

\noindent is stable.

\medskip

\quad{\bf (vi)} Finally, turn to the hypothesis stated in the proposition:
that $A$ and $B$ are stable and pointwise bounded.   Let
$h:X\to\coint{0,\infty}$ be such that $|f(x)|\le h(x)$ for every
$f\in A\cup B$ and $x\in X$;  note that I do {\it not} assume here that $h$
is measurable.   However, we are supposing that $\mu$ is totally finite, so
there must be a sequence $\sequencen{f_n}$ in $A\cup B$ such that
$|f|\leae\sup_{n\in\Bbb N}|f_n|$ for every $f\in A\cup B$.   \Prf\ For each
$q\in\Bbb Q$, choose a countable set $C_q\subseteq A\cup B$ such that
$\{x:|f(x)|\ge q\}\setminus\bigcup_{g\in C_q}\{x:|g(x)|\ge q\}$ is
negligible for any $f\in A\cup B$
(215B(iv));  let $\sequencen{f_n}$ run over
$\bigcup_{q\in\Bbb Q}C_q$.\ \QeD\   Set
$h_1=\chi X+\sup_{n\in\Bbb N}|f_n|$;
then $h_1$ is finite-valued, strictly positive and measurable, and
$|f|\leae h_1$ for every $f\in A\cup B$.   By 465C(a-v), $A_1=\{f/h_1:f\in A\}$
and $B_1=\{f/h_1:f\in B\}$ are stable;  by (v) here, $A_1+A_2$ is stable;
by 465C(a-v) again, $A+B=\{g\times h_1:g\in A_1+A_2\}$ is stable.   So we're
done.

\medskip

{\bf (d)} Since $A$ is pointwise
bounded, the closure $\overline{\Gamma(A)}$ of $\Gamma(A)$ in $\BbbR^X$
for the topology
of pointwise convergence is compact.   \Quer\ Suppose, if possible, that
there is a $g\in\overline{\Gamma(A)}\setminus\eusm L^0(\Sigma)$.
Then there must
be a measurable set $E$ of finite measure and $\alpha<\beta$ in $\Bbb R$
such that $\mu^*P=\mu^*Q=\mu E>0$, where

\Centerline{$P=\{x:x\in E$, $g(x)\le\alpha\}$,
\quad$Q=\{x:x\in E$, $g(x)\ge\beta\}$}

\noindent by 413G again.   Set
$Y_n=\{x:x\in E$, $|f(x)|\le n$ for every $f\in A\}$;  then
$\sequencen{Y_n}$ is a non-decreasing sequence with union $E$, so there is
an $n\in\Bbb N$ such that $\mu^*(P\cap Y_n)$ and $\mu^*(Q\cap Y_n)$ are
both at least $\bover23\mu E$.   Let $F'$, $F''$ be measurable envelopes of
$P\cap Y_n$ and $Q\cap Y_n$ respectively, and $Y=F'\cap F''\cap Y_n$;  then

$$\eqalign{0
&<\mu(F'\cap F'')
=\mu^*(F'\cap F''\cap P\cap Y_n)\cr
&=\mu^*(F'\cap F''\cap Q\cap Y_n)
=\mu^*Y
=\mu^*(P\cap Y)
=\mu^*(Q\cap Y).\cr}$$

Let $\mu_Y$ be the subspace measure on $Y$ and $\Sigma_Y$ its domain,
and consider the set
$A_Y=\{f\restr Y:f\in A\}$.   With respect to the measure $\mu_Y$,
this is stable (4{}65Cm).   Also it is uniformly bounded.   So $\Gamma(A_Y)$
is $\mu_Y$-stable, by (a) of this theorem.
As $\mu_Y$ is complete and totally finite, the closure
$\overline{\Gamma(A_Y)}$ for the topology of pointwise convergence in
$\BbbR^Y$ is included in $\eusm L^0(\Sigma_Y)$ (465Da).   Since
$f\mapsto f\restr Y:\BbbR^X\to\BbbR^Y$ is linear and continuous for the
topologies of pointwise convergence, $g\restr Y\in\overline{\Gamma(A_Y)}$
and $g\restr Y$ is $\Sigma_Y$-measurable.   But

\Centerline{$\mu_Y^*(P\cap Y)=\mu^*(P\cap Y)=\mu^*(Q\cap Y)
=\mu_Y^*(Q\cap Y)=\mu_YY\in\ooint{0,\infty}$}

\noindent (214Cd), so this is impossible.\ \Bang

Thus $\overline{\Gamma(A)}\subseteq\eusm L^0(\Sigma)$ and $\Gamma(A)$ is
relatively compact.
}%end of proof of 465N

\leader{465O}{Stable sets in $L^0$}\cmmnt{ The notion of `stability'
as defined in 465B is applicable only to true functions;  in such
examples as 465Xj, the irregularity of the set $A$ is erased entirely if
we look at its image in the space $L^0$ of equivalence classes of
measurable functions.   We do, however, have a corresponding concept for
subsets of function spaces, which can be expressed in the language of
\S325.}
If $(\frak A,\bar\mu)$ is a semi-finite measure algebra, and $k\ge 1$, I
write $(\Tensorhat_k\frak A,\bar\mu^k)$ for the localizable measure
algebra free product of $k$ copies of $(\frak A,\bar\mu)$\cmmnt{, as
described in 325H}.   If $Q\subseteq L^0(\frak A)$, $k\ge 1$,
$a\in\frak A$ has finite measure and $\alpha<\beta$ in $\Bbb R$, set

$$\eqalign{d_k(Q,a,\alpha,\beta)
=\sup_{v\in Q}\bigl((a\Bcap\Bvalue{v\le\alpha})
&\otimes(a\Bcap\Bvalue{v\ge\beta})
\otimes\ldots\cr
&\otimes(a\Bcap\Bvalue{v\le\alpha})\otimes(a\Bcap\Bvalue{v\ge\beta})
\bigr)\cr}
$$

\noindent in $\Tensorhat_{2k}\frak A$, taking $k$ repetitions of the
formula $(a\Bcap\Bvalue{v\le\alpha})
\otimes(a\Bcap\Bvalue{v\ge\beta})$\dvro{.}{ to match the corresponding
formula

\Centerline{$D_k(A,E,\alpha,\beta)
=\bigcup_{f\in A}
  ((E\cap\{x:f(x)\le\alpha\})\times(E\cap\{x:f(x)\ge\beta))^k$.}

\noindent (Note that the supremum $\sup_{v\in Q}\ldots$ is defined
because $a^{\otimes 2k}=a\otimes\ldots\otimes a$ has finite measure in
the measure
algebra $(\Tensorhat_{2k}\frak A,\bar\mu^{2k})$.   Of course I mean to
take $d_k(Q,a,\alpha,\beta)=0$ if $Q=\emptyset$.)
Now we can say that} $Q$ is {\bf stable} if whenever
$0<\bar\mu a<\infty$ and $\alpha<\beta$
there is a $k\ge 1$ such that
$\bar\mu^{2k}d_k(Q,a,\alpha,\beta)<(\bar\mu a)^{2k}$\cmmnt{;  that is,
$d_k(Q,a,\alpha,\beta)\ne a\otimes\ldots\otimes a$}.

\cmmnt{We have the following relationships between the two concepts of
stability.}

\vleader{48pt}{465P}{Theorem} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space, with measure algebra $(\frak A,\bar\mu)$.

(a) Suppose that $A\subseteq\eusm L^0(\Sigma)$ and that
$Q=\{f^{\ssbullet}:f\in A\}\subseteq L^0(\mu)$, identified with
$L^0=L^0(\frak A)$\cmmnt{ (364Ic)}.
Then $Q$ is stable\cmmnt{ in the sense of 465O} iff every countable
subset of $A$ is stable\cmmnt{ in the sense of 465B}.

(b) Suppose that $\mu$ is complete and
strictly localizable and $Q$ is a stable subset of $L^{\infty}(\mu)$, identified
with $L^{\infty}(\frak A)$ (363I).
Then there is a stable set $B\subseteq\eusm L^{\infty}(\Sigma)$
such that $Q=\{f^{\ssbullet}:f\in B\}$.

\proof{{\bf (a)(i)} Suppose that all countable subsets of $A$ are
stable, and take $a\in\frak A$ such that $0<\bar\mu a<\infty$ and
$\alpha<\beta$ in $\Bbb R$.   For each $k\in\Bbb N$ there is a countable
set $Q_k\subseteq Q$ such that
$d_k(Q_k,a,\alpha,\beta)=d_k(Q,a,\alpha,\beta)$, because
$a^{\otimes 2k}$ has finite measure in $\Tensorhat_{2k}\frak A$.
Now there is a
countable set $A'\subseteq A$ such that  $\{f^{\ssbullet}:f\in
A'\}=\bigcup_{k\in\Bbb N}Q_k$.   Let $E\in\Sigma$ be such that
$E^{\ssbullet}=a$.   As $\mu E=\bar\mu a\in\ooint{0,\infty}$ and $A'$ is
stable, there is a $k\ge 1$ such that
$(\mu^{2k})^*D_k(A',E,\alpha,\beta)<(\mu E)^{2k}$.   Because $A'$ is
countable, $D_k(A',E,\alpha,\beta)$ is measurable.   But 325He tells us
that we have an order-continuous
measure-preserving Boolean homomorphism $\pi$ from the measure algebra
of $\mu^{2k}$ to $\Tensorhat_{2k}\frak A$, such that
$\pi(\prod_{i<2k}F_i)^{\ssbullet}
=F_0^{\ssbullet}\otimes\ldots\otimes F_{2k-1}^{\ssbullet}$ for all
$F_0,\ldots,F_{2k-1}\in\Sigma$;  accordingly

$$\eqalign{\bar\mu^{2k}d_k(Q,a,\alpha,\beta)
&=\bar\mu^{2k}d_k(Q_k,a,\alpha,\beta)
\le\bar\mu^{2k}d_k(\bigcup_{i\in\Bbb N}Q_i,a,\alpha,\beta)\cr
&=\bar\mu^{2k}(\pi D_k(A',E,\alpha,\beta)^{\ssbullet})
=\mu^{2k}D_k(A',E,\alpha,\beta)\cr
&<(\mu E)^{2k}
=\bar\mu^{2k}a^{\otimes 2k}.\cr}$$

\noindent As $a$, $\alpha$ and $\beta$ are arbitrary, $Q$ is stable.

\medskip

\quad{\bf (ii)} Now suppose that $Q$ is stable and that $A'$ is any
countable subset of $A$.   Take $E\in\Sigma$ such that $0<\mu E<\infty$,
and $\alpha<\beta$ in $\Bbb R$.   Set $a=E^{\ssbullet}\in\frak A$.
This time, writing $Q'$ for $\{f^{\ssbullet}:f\in A'\}$, we have

\Centerline{$\pi D_k(A',E,\alpha,\beta)^{\ssbullet}
=d_k(Q',a,\alpha,\beta)
\Bsubseteq d_k(Q,a,\alpha,\beta)$}

\noindent for every $k\ge 1$.   There is some $k$ such that
$d_k(Q,a,\alpha,\beta)\ne a^{\otimes 2k}$, and in this case
$\mu^{2k}D_k(A',E,\alpha,\beta)<(\mu E)^{2k}$;  as $E$, $\alpha$ and
$\beta$ are arbitrary, $A'$ is stable.

\medskip

{\bf (b)(i)} If $\mu X=0$ this is trivial;  suppose that $\mu X>0$.
Let $\familyiI{E_i}$ be a decomposition of $X$ into sets of finite
measure.   Amalgamating any negligible $E_i$ into other non-negligible
ones, we may suppose that $\mu E_i>0$ for each $i$.   Writing $\mu_i$
for the subspace measure on $E_i$, we have a consistent lifting $\phi_i$
for $\mu_i$ (346J).   Set $\phi E=\bigcup_{i\in I}\phi_i(E\cap X_i)$
for $E\in\Sigma$;  then $\phi$ is a lifting for $\mu$.   Let $\theta$ be
the corresponding lifting from $\frak A$ to $\Sigma$ (341B) and
$T:L^{\infty}(\frak A)\to\eusm L^{\infty}(\Sigma)$ the associated
linear operator, defined by saying that
$T(\chi a)=\chi(\theta a)$ for
every $a\in\frak A$ (363F).   Since $\theta(a)^{\ssbullet}=a$ for every
$a\in\frak A$, $(Tv)^{\ssbullet}=v$ for every
$v\in L^{\infty}=L^{\infty}(\frak A)$, identified with $L^{\infty}(\mu)$
(cf.\ 363Xe).

\medskip

\quad{\bf (ii)} We need to know that if $v\in L^{\infty}$ and
$\alpha<\alpha'$, then
$\{x:(Tv)(x)\le\alpha\}\subseteq\phi(\{x:(Tv)(x)\le\alpha'\})$.   \Prf\
Let $v'\in S(\frak A)$ be such that
$\|v-v'\|_{\infty}\le\bover12(\alpha'-\alpha)$ (363C), and set
$\gamma=\bover12(\alpha+\alpha')$.  Express $v'$ as
$\sum_{i=0}^n\alpha_i\chi a_i$ where $a_0,\ldots,a_n\in\frak A$ are
disjoint.   Then $Tv'=\sum_{i=0}^n\alpha_i\chi(\theta a_i)$.   Now

\Centerline{$\|Tv-Tv'\|_{\infty}\le\|v-v'\|_{\infty}\le\gamma-\alpha
=\alpha'-\gamma$,}

\noindent so

$$\eqalignno{\{x:(Tv)(x)\le\alpha\}
&\subseteq\{x:(Tv')(x)\le\gamma\}\cr
&=\bigcup\{\theta a_i:i\le n,\,\alpha_i\le\gamma\}
=\phi(\bigcup\{\theta a_i:i\le n,\,\alpha_i\le\gamma\})\cr
\displaycause{because $\phi(\theta a)=\theta a$ for every $a\in\frak A$}
&=\phi(\{x:(Tv')(x)\le\gamma\})
\subseteq\phi(\{x:(Tv)(x)\le\alpha'\}),\cr}$$

\noindent as claimed.\ \Qed

Similarly, if $\beta'<\beta$ then
$\{x:(Tv)(x)\ge\beta\}\subseteq\phi(\{x:(Tv)(x)\ge\beta'\})$.

\medskip

\quad{\bf (iii)} Set $B=T[Q]$, so that $B\subseteq\eusm L^{\infty}(\Sigma)$
and $Q=\{f^{\ssbullet}:f\in B\}$.   Then $B$
is stable.   \Prf\ Let $E\in\Sigma$ be such that $0<\mu E<\infty$, and
$\alpha<\beta$.   Let $i\in I$ be such that $\mu(E\cap E_i)>0$, and
$\alpha'$, $\beta'\in\Bbb R$ such that $\alpha<\alpha'<\beta'<\beta$.
Setting $a=(E\cap E_i)^{\ssbullet}$, we have $0<\bar\mu a<\infty$, so
there is some $k\in\Bbb N$ such that
$d_k(Q,a,\alpha',\beta')\ne a^{\otimes 2k}$.
Let $\pi$ be the measure-preserving Boolean
homomorphism from the measure algebra of $\mu^{2k}$ to
$\Tensorhat_{2k}\frak A$ described in part (a) of this proof;  as noted
in 325He, the present context is enough to ensure that $\pi$ is an
isomorphism.   So there is a $W\in\dom\mu^{2k}$ such that
$\pi W^{\ssbullet}=d_k(Q,a,\alpha',\beta')$;  since

\Centerline{$d_k(Q,a,\alpha',\beta')\Bsubseteq a^{\otimes 2k}
\Bsubseteq(E_i^{2k})^{\ssbullet}$,}

\noindent we may suppose that $W\subseteq E_i^{2k}$.   If $f\in B$, then

\Centerline{$\pi D_k(\{f\},E,\alpha',\beta')^{\ssbullet}
=d_k(\{f^{\ssbullet}\},a,\alpha',\beta')
\Bsubseteq d_k(Q,a,\alpha',\beta')$,}

\noindent so $D_k(\{f\},E,\alpha',\beta')\setminus W$ is negligible.

At this point, recall that $\phi_i$ was supposed to be a consistent
lifting for $\mu_i$.   So we have a lifting $\phi'$ of $\mu_i^{2k}$ such
that $\phi'(\prod_{j<2k}F_j)=\prod_{j<2k}\phi_iF_j$ for all
$F_0,\ldots,F_{2k-1}\in\Sigma_i$.   In particular, if $f\in B$ and we
set

\Centerline{$F_{2j}=\{x:x\in E\cap E_i,\,f(x)\le\alpha\}$,
\quad$F_{2j+1}=\{x:x\in E\cap E_i,\,f(x)\ge\beta\}$,}

\Centerline{$F'_{2j}=\{x:x\in E\cap E_i,\,f(x)\le\alpha'\}$,
\quad$F'_{2j+1}=\{x:x\in E\cap E_i,\,f(x)\ge\beta'\}$,}

\noindent for $j<k$, we shall have

$$\eqalignno{\prod_{j<2k}F_j
&\subseteq\prod_{j<2k}\phi F'_j\cr
\displaycause{by (ii) above, because $f=Tv$ for some $v$}
&=\prod_{j<2k}\phi_iF'_j
=\phi'(\prod_{j<2k}F'_j)
\subseteq\phi'W\cr}$$

\noindent because $\prod_{j<2k}F'_j=D_k(\{f\},E,\alpha',\beta')$.
As $f$ is arbitrary, $D_k(B,E\cap E_i,\alpha,\beta)\subseteq\phi'W$.
But now

$$\eqalignno{(\mu^{2k})^*D_k(B,E\cap E_i,\alpha,\beta)
&\le\mu^{2k}(\phi'W)
=\mu_i^{2k}(\phi'W)\cr
\displaycause{251Wl}
&=\mu_i^{2k}W
=\mu^{2k}W
=\bar\mu^{2k}d_k(Q,a,\alpha',\beta')\cr
&<\bar\mu^{2k}(a^{\otimes 2k})
=\mu(E\cap E_i)^{2k}.\cr}$$

\noindent As usual, it follows that
$(\mu^{2k})^*D_k(B,E,\alpha,\beta)<(\mu E)^{2k}$;  as $E$, $\alpha$ and
$\beta$ are arbitrary, $B$ is stable, as claimed.\ \Qed
}%end of proof of 465P

\cmmnt{
\leader{465Q}{Remarks} Using 465Pa, we can work through the
first part
of this section to get a list of properties of stable subsets of $L^0$.
For instance,
the convex hull of an order-bounded stable set in $L^0$ is stable, as in
465Na.   It is harder to relate such results as 465M to the idea of
stability in $L^0$, but the argument of 465Pb gives a line to follow:
if $(X,\Sigma,\mu)$ is complete and strictly localizable, there is a
linear operator $T:L^{\infty}(\mu)\to\eusm L^{\infty}(\Sigma)$, defined
from a lifting, such that, for $Q\subseteq L^{\infty}$, $T[Q]$ is stable
iff $Q$ is stable.   So when $\mu$ is a complete probability measure, we
can look at the averages
$\psi_{wk}(v)=\bover1k\sum_{i=0}^{k-1}(Tv)(w(i))$ for $v\in Q$,
$w\in X^{\Bbb N}$ to devise criteria for stability of $Q$ in terms of
the linear functionals $\psi_{wk}$.

Working in $L^1$, however, we can seek results of a different type,
as follows.
}%end of comment

\leader{465R}{Theorem}\cmmnt{ ({\smc Talagrand 84})} % 11-4-1
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be measure algebras, and
$T:L^1(\frak A,\bar\mu)\to L^1(\frak B,\bar\nu)$ a bounded linear
operator.   If $Q$ is stable and order-bounded in
$L^1(\frak A,\bar\mu)$, then $T[Q]\subseteq L^1(\frak B,\bar\nu)$
is stable.

\proof{{\bf (a)} To begin with (down to (d) below) let us
suppose that

\inset{$(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ are the measure algebras of measure spaces
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$,}

\noindent so that we can identify $L^1_{\bar\mu}$,
$L^1_{\bar\nu}$ with $L^1(\mu)$ and $L^1(\nu)$ (365B),

\inset{$Q$ is countable,}

\noindent so that $Q$ can be expressed as $\{f^{\ssbullet}:f\in A\}$,
where $A\subseteq\eusm L^0(\Sigma)$ is countable and stable (465P),

\inset{$T$ is positive,

$\mu$ and $\nu$ are totally finite,

$T(\chi X^{\ssbullet})=\chi Y^{\ssbullet}$,

$Q\subseteq L^{\infty}(\frak A)$ is $\|\,\|_{\infty}$-bounded,}

\noindent so that we may take $A\subseteq\eusm L^{\infty}$ to be
$\|\,\|_{\infty}$-bounded,

\inset{$\nu Y=1$,

$\mu X=1$,

and that $\|T\|\le 1$.}

\medskip

{\bf (b)} The idea of the argument is that for any $n\ge 1$ we have a
positive linear
operator $U_n:L^1(\mu^n)\to L^1(\nu^n)$ defined as follows.

If $f_0,\ldots,f_{n-1}\in\eusm L^1(\mu)$, set
$(f_0\otimes\ldots\otimes f_{n-1})(w)=\prod_{i=0}^nf_i(w(i))$ whenever
$w\in\prod_{i<n}\dom f_i$.
Now we can define
$u_0\otimes u_1\otimes\ldots\otimes u_{n-1}\in L^1(\mu^n)$, for
$u_0,\ldots,u_{n-1}\in L^1(\mu)$, by saying that
$f_0^{\ssbullet}\otimes\ldots\otimes f_{n-1}^{\ssbullet}
=(f_0\otimes\ldots\otimes f_{n-1})^{\ssbullet}$ for all
$f_0,\ldots,f_{n-1}\in\eusm L^1(\mu)$, as in 253E.

Define the operators $U_n$ inductively.   $U_1=T$.   Given that
$U_n:L^1(\mu^n)\to
L^1(\nu^n)$ is a positive linear operator, then we have a bilinear
operator $\psi:L^1(\mu^n)\times L^1(\mu)\to L^1(\nu^{n+1})$ defined by
saying that $\psi(q,u)=U_nq\otimes Tu$ for $q\in L^1(\mu^n)$, $u\in
L^1(\mu)$, where
$\otimes:L^1(\nu^n)\times L^1(\nu)
\to L^1(\nu^n\times\nu)\cong L^1(\nu^{n+1})$ is the operator of 253E.   By
253F, there is a (unique) bounded linear operator
$U_{n+1}:L^1(\mu^{n+1})\to L^1(\nu^{n+1})$ such that
$U_{n+1}(q\otimes u)=\psi(q,u)$ for all $q\in L^1(\mu^n)$,
$u\in L^1(\mu)$.   To see that
$U_{n+1}$ is positive, use 253Gc.   (Remember that we are
supposing that $T$ is positive.)   Continue.

Now it is easy to check that

\Centerline{$U_n(u_0\otimes\ldots\otimes u_{n-1})
=Tu_0\otimes\ldots\otimes Tu_{n-1}$}

\noindent for all $u_0,\ldots,u_{n-1}\in L^1(\mu)$.   Moreover,
$\|U_{n+1}\|\le\|U_n\|\|T\|$ for every $n$ (see 253F), so $\|U_n\|\le 1$
for every $n$.

For $i<n\in\Bbb N$, we have a natural operator $R_{ni}:L^1(\mu)\to
L^1(\mu^n)$, defined by saying that
$R_{ni}f^{\ssbullet}=(f\pi_{ni})^{\ssbullet}$ for every
$f\in\eusm L^1(\mu)$, where $\pi_{ni}(w)=w(i)$ for $w\in X^n$.
Similarly, we have
an operator $S_{ni}:L^1(\nu)\to L^1(\nu^n)$.   Observe that

\Centerline{$R_{ni}u
=e\otimes\ldots\otimes e\otimes u\otimes e\otimes\ldots\otimes e$}

\noindent where $e=\chi X^{\ssbullet}$ and the $u$ is put in the
position corresponding to the coordinate $i$.   Since
$Te=(\chi Y)^{\ssbullet}=e'$ say,

\Centerline{$U_nR_{ni}u
=e'\otimes\ldots\otimes Tu\otimes\ldots\otimes e'=S_{ni}Tu$}

\noindent for every $u\in L^1(\mu)$.

\medskip

{\bf (c)} Let $B\subseteq\eusm L^{\infty}(\Tau)$ be a countable
$\|\,\|_{\infty}$-bounded set such that $T[Q]=\{g^{\ssbullet}:g\in B\}$.
($T[Q]$ is $\|\,\|_{\infty}$-bounded because $T$ is positive and
$T(\chi X)^{\ssbullet}=(\chi Y)^{\ssbullet}$.)   I seek to show that $B$
is stable by using the criterion 465M(v).   Let $\epsilon>0$.   Then
there is an $m\ge 1$ such that
$\int f_{kl}(w)\mu^{\Bbb N}(dw)\le\epsilon$ for any $k$, $l\ge m$,
writing

\Centerline{$f_{kl}(w)=\sup_{f\in A}|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  -\Bover1l\sum_{i=0}^{l-1}f(w(i))|$}

\noindent for $w\in X^{\Bbb N}$;  note that $f_{kl}$ is measurable
because $A$ is countable.

Take any $k$, $l\ge m$ and consider

\Centerline{$g_{kl}(z)=\sup_{g\in B}|\Bover1k\sum_{i=0}^{k-1}g(z(i))
  -\Bover1l\sum_{i=0}^{l-1}g(z(i))|$}

\noindent for $z\in Y^{\Bbb N}$.   I claim that
$\int g_{kl}d\nu^{\Bbb N}\le\epsilon$.   \Prf\ Set $n=\max(k,l)$.   Then
$\int g_{kl}d\nu^{\Bbb N}=\int\tilde g\,d\nu^n$, where

\Centerline{$\tilde g(z)=\sup_{g\in B}|\Bover1k\sum_{i=0}^{k-1}g(z(i))
  -\Bover1l\sum_{i=0}^{l-1}g(z(i))|$}

\noindent for $z\in Y^n$.   If we look at $\tilde g^{\ssbullet}$ in
$L^1(\nu^n)$, we see that it is

\Centerline{$\sup_{g\in B}|\Bover1k\sum_{i=0}^{k-1}S_{ni}g^{\ssbullet}
  -\Bover1l\sum_{i=0}^{l-1}S_{ni}g^{\ssbullet}|$}

\noindent where $S_{ni}:L^1(\nu)\to L^1(\nu^n)$ is defined in (b) above.
Thus

\Centerline{$\tilde g^{\ssbullet}
=\sup_{v\in T[Q]}|\Bover1k\sum_{i=0}^{k-1}S_{ni}v
  -\Bover1l\sum_{i=0}^{l-1}S_{ni}v|$.}

\noindent Similarly, setting

\Centerline{$\tilde f(w)=\sup_{f\in A}|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  -\Bover1l\sum_{i=0}^{l-1}f(w(i))|$}

\noindent for $w\in X^n$,

\Centerline{$\tilde f^{\ssbullet}
=\sup_{u\in Q}|\Bover1k\sum_{i=0}^{k-1}R_{ni}u
  -\Bover1l\sum_{i=0}^{l-1}R_{ni}u|$.}

\noindent Now consider $U_n\tilde f^{\ssbullet}$.   For any $v\in T[Q]$,
we can express $v$ as $Tu$ where $u\in Q$, so

$$\eqalignno{\bigl|\Bover1k\sum_{i=0}^{k-1}S_{ni}v
  -\Bover1l\sum_{i=0}^{l-1}S_{ni}v\bigr|
&=\bigl|\Bover1k\sum_{i=0}^{k-1}S_{ni}Tu
  -\Bover1l\sum_{i=0}^{l-1}S_{ni}Tu\bigr|\cr
&=\bigl|\Bover1k\sum_{i=0}^{k-1}U_nR_{ni}u
  -\Bover1l\sum_{i=0}^{l-1}U_nR_{ni}u\bigr|\cr
\displaycause{because $U_nR_{ni}=S_{ni}T$, as noted in (b) above}
&=\bigl|U_n(\Bover1k\sum_{i=0}^{k-1}R_{ni}u
  -\Bover1l\sum_{i=0}^{l-1}R_{ni}u)\bigr|\cr
&\le U_n(|\Bover1k\sum_{i=0}^{k-1}R_{ni}u
  -\Bover1l\sum_{i=0}^{l-1}R_{ni}u|)\cr
\displaycause{because $U_n$ is positive}
&\le U_n\tilde f^{\ssbullet}.\cr}$$

\noindent As $v$ is arbitrary, $\tilde g^{\ssbullet}\le U_n\tilde
f^{\ssbullet}$, and

$$\eqalignno{\int g_{kl}d\nu^{\Bbb N}
&=\int\tilde g\,d\nu^n
=\|\tilde g^{\ssbullet}\|_1
\le\|U_n\tilde f^{\ssbullet}\|_1
\le\|\tilde f^{\ssbullet}\|_1\cr
\displaycause{because $\|U_n\|\le 1$}
&\le\epsilon\cr}$$

\noindent because $k$, $l\ge m$.\ \Qed

\medskip

{\bf (d)} As $\epsilon$ is arbitrary, $B$ satisfies the criterion
465M(v), and is stable.   So $T[Q]$ is stable, by 465Pa in the other
direction.

\medskip

{\bf (e)} Now let us seek to unwind the list of special assumptions used
in the argument above.   Suppose we drop the last two, and assume only
that

\inset{$(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ are the measure algebras of measure spaces
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$,

$Q$ is countable,

$T$ is positive,

$\mu$ and $\nu$ are totally finite,

$T(\chi X^{\ssbullet})=\chi Y^{\ssbullet}$,

$Q\subseteq L^{\infty}(\frak A)$ is $\|\,\|_{\infty}$-bounded,

$\nu Y=1$.}

\noindent Then $T[Q]$ is stable.   \Prf\ Define
$\mu_1:\Sigma\to\coint{0,\infty}$ by setting
$\mu_1E=\int T(\chi E^{\ssbullet})$ for every $E\in\Sigma$.   Then
$\mu_1$ is countably
additive because $T$ is (sequentially) order-continuous (355Ka).   If
$\mu E=0$ then $\chi E^{\ssbullet}=0$ in $L^1(\mu)$ and $\mu_1E=0$, so
$\mu_1$ is truly continuous with respect to $\mu$ (232Bb) and has a
Radon-Nikod\'ym derivative (232E).   By 465C(c-ii), $A$ is stable with
respect to $\mu_1$, while $\mu_1X=\nu Y=1$, because
$T(\chi X)^{\ssbullet}=\chi Y^{\ssbullet}$.

Let $(\frak A_1,\bar\mu_1)$ be the measure algebra of $\mu_1$.   If
$E\in\Sigma$ and $\mu_1E=0$, then $T(\chi E^{\ssbullet})=0$.
Accordingly we can define an additive function
$\theta:\frak A_1\to L^1(\nu)$ by setting
$\theta E^{\ssbullet}=T(\chi E^{\ssbullet})$ for
every $E\in\Sigma$.   (Note that the two $^{\ssbullet}$s here must be
interpreted differently.   In the formula $\theta E^{\ssbullet}$, the
equivalence class $E^{\ssbullet}$ is to be taken in $\frak A_1$.   In
the formula $\chi E^{\ssbullet}=(\chi E)^{\ssbullet}$, the equivalence
class is to be taken in $L^0(\mu)$.   In the rest of this proof I will
pass over such points without comment;  I hope the context will always
make it clear how each $^{\ssbullet}$ is to be read.)
Because $T$ is positive, $\theta$ is non-negative, and by the definition
of $\mu_1$ we have $\|\theta a\|_1=\bar\mu_1a$ for every
$a\in\frak A_1$.   So we have a positive linear operator
$T_1:L^1(\frak A_1,\bar\mu_1)\to L^1(\nu)$ defined by setting

\Centerline{$T_1(\chi E^{\ssbullet})=\theta E^{\ssbullet}
=T(\chi E^{\ssbullet})$}

\noindent for every $E\in\Sigma$ (365J).   As
$\|T_1(\chi E^{\ssbullet})\|_1=\mu_1E$ for every $E\in\Sigma$,
$\|T_1\|=1$ (365I(ii-$\beta$)).

If $f:X\to\Bbb R$ is simple (that is, a linear combination of
indicator functions of sets in $\Sigma$), then
$T_1f^{\ssbullet}=Tf^{\ssbullet}$.   So this is also true for every
$f\in\eusm L^{\infty}(\Sigma)$;  in particular, it is true for every
$f\in A$, so that $T[Q]=T_1[Q_1]$, where
$Q_1=\{f^{\ssbullet}:f\in A\}\subseteq L^1(\mu_1)$.   But $\mu_1$, $Q_1$
and $T_1$ satisfy all the
conditions of (a), so (b)-(d) tell us that $T_1[Q_1]$ is stable, and
$T[Q]$ is stable, as required.\ \Qed

\medskip

{\bf (f)} The next step is to drop the condition `$\nu Y=1$'.   But this
is elementary, since we are still assuming that $\nu$ is totally finite,
and multiplying $\nu$ by a non-zero scalar doesn't change $L^0(\nu)$ or
the stability of any of its subsets, while the case $\nu Y=0$ is
trivial.   So we conclude that if

\inset{$(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ are the measure algebras of measure spaces
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$,

$Q$ is countable,

$T$ is positive,

$\mu$ and $\nu$ are totally finite,

$T(\chi X^{\ssbullet})=\chi Y^{\ssbullet}$,

$Q\subseteq L^{\infty}(\frak A)$ is $\|\,\|_{\infty}$-bounded,}

\noindent then $T[Q]$ is stable.

\medskip

{\bf (g)} We can now attack what remains.   We find that if

\inset{$(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ are the measure algebras of measure spaces
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$,

$Q$ is countable,

$T$ is positive,}

\noindent then $T[Q]$ is stable.   \Prf\ At this point recall that we
are supposing that $Q$ is order-bounded.   Let $u_0\in L^1(\mu)$ be such
that $|u|\le u_0$ for every $u\in Q$;  let $f_0\in\eusm L^0(\Sigma)^+$
be such that $f_0^{\ssbullet}=u_0$.   Setting
$A'=\{\med(-f_0,f,f_0):f\in A\}$, $A'$ is still stable, because
its image in $L^1(\mu)$ is still $Q$, or otherwise.
Set $v_0=Tu_0$, and let $g_0\in\eusm L^0(\Tau)^+$ be such that
$g_0^{\ssbullet}=v_0$.   Because $T$ is positive, $|Tu|\le T|u|\le v_0$
for every $u\in Q$.   So we can represent $T[Q]$ as
$\{g^{\ssbullet}:g\in B\}$, where $B\subseteq\eusm L^0(\Tau)$ is a
countable set and $|g|\le g_0$ for every $g\in B$.   Set
$F_0=\{y:g_0(y)\ne 0\}$.

Define measures $\mu_1$, $\nu_1$ by setting $\mu_1E=\int_Ef_0d\mu$ for
$E\in\Sigma$, $\nu_1F=\int_Fg_0d\nu$ for $F\in\Tau$.   Then both $\mu_1$
and $\nu_1$ are totally finite.   By 465C(c-ii), $A'$ is stable with respect
to $\mu_1$.   Set $A_1=\{\Bover{f}{f_0}:f\in A'\}$, interpreting
$\Bover{f}{f_0}(x)$ as $0$ if $f_0(x)=0$;  then $A_1$ is stable with
respect to $\mu_1$, by 465C(a-v), and $\|f\|_{\infty}\le 1$ for every
$f\in A_1$.   Take $Q_1=\{f^{\ssbullet}:f\in A_1\}\subseteq L^1(\mu_1)$,
so that $Q_1$ is stable.

We have a norm-preserving positive linear operator
$R:L^1(\mu_1)\to L^1(\mu)$ defined by setting
$Rf^{\ssbullet}=(f\times f_0)^{\ssbullet}$
for every $f\in\eusm L^1(\mu_1)$ (use 235A).   Observe that $R[Q_1]=Q$
and $R(\chi X)^{\ssbullet}=u_0$.   Similarly, we have a norm-preserving
positive linear operator $S:L^1(\nu_1)\to L^1(\nu)$ defined by setting
$Sg^{\ssbullet}=(g\times g_0)^{\ssbullet}$ for $g\in\eusm L^1(\nu_1)$.
The set of values of $S$ is just

\Centerline{$\{g^{\ssbullet}:g\in\eusm L^1(\nu),\,g(y)=0$ whenever
$g_0(y)=0\}$,}

\noindent which is the band in $L^1(\nu)$ generated by $v_0$.   So

\Centerline{$\{u:u\in L^1(\mu_1),\,TR|u|\in S[L^1(\nu_1)]\}$}

\noindent is a band in $L^1(\nu_1)$ containing $\chi X^{\ssbullet}$, and
must be the whole of $L^1(\mu_1)$.   Thus we have a positive linear
operator $T_1=S^{-1}TR:L^1(\mu_1)\to L^1(\nu_1)$, and $T_1(\chi
X)^{\ssbullet}=\chi Y^{\ssbullet}$ in $L^1(\nu_1)$.

By (f), $T_1[Q_1]$ is stable in $L^1(\nu_1)$.   Observe that
$T_1[Q_1]=\{S^{-1}g^{\ssbullet}:g\in B\}=\{g^{\ssbullet}:g\in B_1\}$,
where $B_1=\{\Bover{g}{g_0}:g\in B\}$, interpreting $\Bover{g}{g_0}(y)$
as $0$ if $y\in Y\setminus F_0$.   Consequently $B_1$ and
$B=\{g\times g_0:g\in B_1\}$ are stable with respect to $\nu_1$.
By 465C(c-ii), once more, $B$ is stable with respect to $\nu_0$, where

\Centerline{$\nu_0F=\int_F\Bover{1}{g_0}d\nu_1=\nu(F\cap F_0)$}

\noindent for any $F\in\Tau$.   But because $g(y)=0$ whenever $g\in B$
and $y\in Y\setminus F_0$,

\Centerline{$(\nu^{2k})^*D_k(B,F,\alpha,\beta)
=(\nu^{2k})^*D_k(B,F\cap F_0,\alpha,\beta)
=(\nu_0^{2k})^*D_k(B,F,\alpha,\beta)$}

\noindent whenever $F\in\Tau$, $\alpha<\beta$ and $k\ge 1$;  so $B$ is
also stable with respect to $\nu$, and $Q=\{g^{\ssbullet}:g\in B\}$ is
stable in $L^1(\nu)$.\ \Qed

\medskip

{\bf (h)} The worst is over.   If we are not told that $T$ is positive,
we know that it is expressible as the difference of positive linear
operators $T_1$ and $T_2$ (371D);  now $T_1[Q]$ and $T_2[Q]$ will be
stable, by the work above, so $T[Q]\subseteq T_1[Q]-T_2[Q]$ is stable,
by 465Nc.   If we are not told that $Q$ is countable, we refer to 465P
to see that we need only check that countable subsets of $T[Q]$ are
stable, and these are images of countable subsets of $Q$.   Finally, the
identification of the abstract measure algebras $(\frak A,\bar\mu)$
and $(\frak B,\bar\nu)$ with the measure algebras of measure spaces
is Theorem 321J.
}%end of proof of 465R

\leader{*465S}{R-stable sets}\cmmnt{ The theory above has been
developed in the
context of general measure (or probability) spaces and the `ordinary'
product measure of measure spaces.   For $\tau$-additive measures -- in
particular, for Radon measures -- we have an alternative product
measure, as described in \S417.}   If $(X,\frak T,\Sigma,\mu)$ is a
semi-finite $\tau$-additive topological measure space such that $\mu$ is
inner regular with respect to the Borel sets, write $\tilde\mu^I$ for
the $\tau$-additive product measure on $X^I$\cmmnt{, as described in
417C (for
the product of two spaces) and 417E (for the product of any family of
probability spaces);  we can extend the construction of
417C to arbitrary finite products (417D)}.   \cmmnt{Now say that}
$A\subseteq\Bbb R^X$ is
{\bf R-stable} if whenever $0<\mu E<\infty$ and $\alpha<\beta$ there is
a $k\ge 1$ such that
$(\tilde\mu^{2k})^*D_k(A,E,\alpha,\beta)<(\mu E)^{2k}$.
\cmmnt{Because we have a version of Fubini's theorem for the products
of $\tau$-additive topological measures (417G), all the
arguments of this section can be applied to R-stable sets, yielding
criteria for R-stability exactly like those in 465M.}

%417Xf for product of sets of full outer measure

Because the $\tau$-additive product measure extends the c.l.d.\ product
measure, stable sets are always R-stable.   \cmmnt{(We must have

\Centerline{$(\tilde\mu^{2k})^*D_k(A,E,\alpha,\beta)
\le(\mu^{2k})^*D_k(A,E,\alpha,\beta)$}

\noindent for all $k$, $A$, $E$, $\alpha$ and $\beta$.)   For an example
of an R-stable set which is not stable, see 465U.}

\cmmnt{The concept of `R-stability' is used in {\smc Talagrand 84}
in applications to the integration of vector-valued
functions.   I give one result, however, to show how it is relevant to a
question with a natural expression in the language of this chapter.}

\leader{*465T}{Proposition}\cmmnt{ ({\smc Talagrand 84})} % 9-4-2
Let $(X,\frak T,\Sigma,\mu)$ be a semi-finite
$\tau$-additive topological measure space such that $\mu$ is inner
regular with respect to the Borel sets.   If $A\subseteq C(X)$ is such
that every countable subset of $A$ is R-stable, then $A$ is R-stable.

\proof{ For any $\alpha<\beta$ and $k\ge 1$,

$$\eqalign{D'_k(A,X,\alpha,\beta)
=\bigcup_{f\in A}\{w:w\in
X^{2k},\,&f(w(2i))<\alpha,\cr
&f(w(2i+1))>\beta\text{ for }i<k\}\cr}$$

\noindent is open.   Suppose that $0<\mu E<\infty$.   Because all the
product measures $\tilde\mu^{2k}$ are $\tau$-additive, we can find a
countable set $A'\subseteq A$ such that

\Centerline{$\tilde\mu^{2k}D'_k(A,E,\alpha,\beta)
=\tilde\mu^{2k}D'_k(A',E,\alpha,\beta)$}

\noindent for every $k\ge 1$ and all rational $\alpha$, $\beta$.   Now,
if $\alpha<\beta$, there are rational $\alpha'$, $\beta'$ such that
$\alpha<\alpha'<\beta'<\beta$, and a $k\ge 1$ such that
$\tilde\mu^{2k}D_k(A',E,\alpha',\beta')<(\mu E)^{2k}$;  in which case

$$\eqalign{(\tilde\mu^{2k})^*D_k(A,E,\alpha,\beta)
&\le\tilde\mu^{2k}D'_k(A,E,\alpha',\beta')
=\tilde\mu^{2k}D'_k(A',E,\alpha',\beta')\cr
&\le\tilde\mu^{2k}D_k(A',E,\alpha',\beta')
<(\mu E)^{2k}.\cr}$$

\noindent As $E$, $\alpha$ and $\beta$ are arbitrary, $A$ is R-stable.
}%end of proof of 465T

\leader{*465U}{}\cmmnt{ I come now to the promised example of an
R-stable set which is not stable.   I follow the construction in
{\smc Talagrand 88}, which displays an interesting characteristic related
to 465O-465P above.

\medskip

\noindent}{\bf Example} There is a Radon probability space with an
R-stable set of continuous functions which is not stable.

\proof{{\bf (a)} Let $(X,\Sigma,\mu)$ be an atomless
probability space\cmmnt{ (e.g., the unit interval with Lebesgue
measure)}.   Define $\sequencen{r_n}$ by setting $r_0=1$, $r_1=2$,
$r_{n+1}=2^{nr_n}$ for $n\ge 1$;  then $2^n\le r_n<r_{n+1}$ for every
$n$.   Let $\sequencen{\Sigma_n}$ be an increasing sequence of finite
subalgebras of $\Sigma$, each $\Sigma_n$ having $r_n$ atoms of the same
size;  this
is possible because $r_{n+1}$ is always a multiple of $r_n$.   Write
$\Cal H_n$ for the set of atoms of $\Sigma_n$.   Next, for each
$n\in\Bbb N$, let $\family{H}{\Cal H_n}{G_H}$ be an independent family
in $\Sigma_{n+1}$ of sets of measure $2^{-n}$;  such a family exists
because $r_{n+1}$ is a multiple of $2^{nr_n}$.

Let $\Cal E$ be the family of all sets expressible in the form
$E=\bigcup_{i\in\Bbb N}H_i$ where, for some strictly increasing sequence
$\sequence{i}{n_i}$ in $\Bbb N$, $H_i\in\Cal H_{n_i}$ and
$H_j\subseteq G_{H_i}$ whenever $i<j$ in $\Bbb N$.   Set
$A=\{\chi E:E\in\Cal E\}\subseteq\eusm L^0(\Sigma)$.

\medskip

{\bf (b)} $A$ is stable.   \Prf\ Suppose that $F\in\Sigma$ and that
$\mu F>0$.   Take $n\in\Bbb N$ so large that
$3\cdot 2^{-n}<(\mu F)^2$.   Set
$\Cal H=\{H:H\in\Cal H_n,\,\mu(H\cap F)>0\}$;  enumerate $\Cal H$ as
$\ofamily{i}{m}{H_i}$;  set $F_1=F\cap\bigcup_{i<m}H_i$,
$V=\prod_{j<m}(H_j\cap F_1)\subseteq F_1^m$.   Because $\mu F_1=\mu F$,
$m\ge r_n\mu F\ge 3$.

Consider

\Centerline{$U_k=\bigcup_{H\in\Cal H_k}(H\times H)
\cup(H\times G_H)\cup(G_H\times H)\subseteq X^2$}

\noindent for $k\in\Bbb N$.   Then

\Centerline{$\mu^2U_k
\le r_k(\Bover1{r_k^2}+\Bover1{2^kr_k}+\Bover1{2^kr_k})
\le 3\cdot 2^{-k}$,}

\noindent so

\Centerline{$\mu^2(F_1^2\setminus\bigcup_{k>n}U_k)
\ge(\mu F)^2-3\sum_{k=n+1}^{\infty}2^{-k}>0$.}

\noindent Set

\Centerline{$V=\{w:w\in F_1^{2m},\,w(2i)\in H_i
\text{ for every }i<m,\,
(w(1),w(3))\notin\bigcup_{k>n}U_k\}$.}

\noindent Then

\Centerline{$\mu^{2m}(V)
\ge\prod_{i=0}^{m-1}\mu(F_1\cap H_i)
\cdot\mu^2(F_1^2\setminus\bigcup_{k>n}U_k)\cdot(\mu F_1)^{m-2}
>0$.}

\noindent\Quer\ Suppose, if possible, that there is a point
$w\in V\cap D_m(A,F,0,1)$.   Then there is an $E\in\Cal E$ such that
$w(2i)\notin E$, $w(2i+1)\in E$ for $i<m$.   Express $E$ as
$\bigcup_{j\in\Bbb N}E_j$
where $E_j\in\Cal H_{n_j}$ for every $j\in\Bbb N$, where
$\sequence{j}{n_j}$ is strictly increasing, and $E_k\subseteq G_{E_j}$
whenever $j<k$.   Because $w(2i)\in F_1\cap H_i\setminus E$, $E_j$
cannot include $H_i$ for any $i<m$, $j\in\Bbb N$;  so
$E_j\cap H_i=\emptyset$ whenever $i<m$ and $n_j\le n$.
Because $w(1)$, $w(3)\in E$, there are $j_0$, $j_1\in\Bbb N$ such that
$w(1)\in E_{j_0}$ and $w(3)\in E_{j_1}$.   Since both $w(1)$ and $w(3)$
belong to $F_1\subseteq\bigcup_{i<m}H_i$, $n_{j_0}$ and $n_{j_1}$ are
both greater than $n$.   But now

--- if $j_0=j_1$, then $(w(1),w(3))\in E_{j_0}^2$,

--- if $j_0<j_1$, then $(w(1),w(3))\in E_{j_0}\times E_{j_1}
\subseteq E_{j_0}\times G_{E_{j_0}}$,

--- if $j_1<j_0$, then $(w(1),w(3))\in E_{j_0}\times E_{j_1}
\subseteq G_{E_{j_1}}\times E_{j_1}$,

\noindent so in any case $(w(1),w(3))\in U_k$ where
$k=\min(n_{j_0},n_{j_1})>n$,
which is impossible.\ \Bang

Thus $D_m(A,F,0,1)$ does not meet $V$, and
$(\mu^{2m})^*D_m(A,F,0,1)<(\mu F)^{2m}$.

Now if $\alpha<\beta$, then
$D_m(A,F,\alpha,\beta)=D_m(A,F,0,1)$ if $0\le\alpha<\beta\le 1$,
$\emptyset$ otherwise;  so in all cases
$(\mu^{2m})^*D_m(A,F,\alpha,\beta)\penalty-100<(\mu F)^{2m}$.
As $F$ also is arbitrary, $A$ is stable, as claimed.\ \Qed

\medskip

{\bf (c)} Now let $(Z,\frak S,\Tau,\nu)$ be the Stone space of the
measure algebra of
$(X,\Sigma,\mu)$ (321K), so that $\nu$ is a Radon measure (411Pe).   For
$E\in\Sigma$, write $E^*$ for the
corresponding open-and-closed set in $Z$, so that $E\mapsto
E^*:\Sigma\to\Tau$ is a measure-preserving Boolean homomorphism.
Set $A^*=\{\chi E^*:E\in\Cal E\}\subseteq C(Z)$.   Write $\nu^m$ for the
c.l.d.\ product measure on $Z^m$ for $m\ge 1$.   We already know that
$\nu^2$ is not a topological measure (419E, 419Xc).

\medskip

{\bf (d)} The point is that $(\nu^{2m})^*D_m(A^*,Z,0,1)=1$ for every
$m\ge 1$.   \Prf\Quer\ Suppose, if possible, otherwise.   Then there is
a set $\tilde W\subseteq Z^{2m}$ such that $\nu^{2m}(\tilde W)>0$ and
$\tilde W\cap D_m(A^*,Z,0,1)\ne\emptyset$.   There
is an $\epsilon>0$ such that

\Centerline{$\tilde V=\{v:v\in Z^m,\,
\nu^m\{u:u\in Z^m,\,u\#v\in\tilde W\}$ is
defined and greater than $m\epsilon\}$}

\noindent has non-zero inner measure for $\nu^m$.   Now there are sets
$\tilde F_{ij}\in\Tau$, for $i\in\Bbb N$ and $j<m$, such that

\Centerline{$Z^m\setminus\tilde V
\subseteq\bigcup_{i\in\Bbb N}\prod_{j<m}\tilde F_{ij}$,
\quad$\sum_{i=0}^{\infty}\prod_{j=0}^m\nu\tilde F_{ij}<1$.}

\noindent Enlarging the $\tilde F_{ij}$ slightly if need be, we may
suppose that they are all open-and-closed (322Rc), therefore expressible
as $F_{ij}^*$ where $F_{ij}\in\Sigma$ for $i\in\Bbb N$, $j<m$.
Set $V=X^m\setminus\bigcup_{i\in\Bbb N}\prod_{j<m}F_{ij}$, so that

\Centerline{$\mu^mV
=\nu^m(Z^m\setminus\bigcup_{i\in\Bbb N}\prod_{j<m}F_{ij}^*)>0$.}

I seek to choose $\sequence{k}{n_k}$ in $\Bbb N$, $\sequence{k}{H_k}$
and $\langle E_{kj}\rangle_{k\in\Bbb N,j<m}$ in $\Sigma$ inductively, in
such a way that

\Centerline{$2^{-n_0}\le\bover12\epsilon$,}

\Centerline{$E_{kj}\subseteq E_{ij}\cap G_{H_i}$ for $i<k$ and $j<m$,}

\Centerline{$\prod_{j<m}E_{kj}\cap\prod_{j<m}F_{kj}=\emptyset$,}

\Centerline{$\mu^m(V\cap G_{H_k}^m\cap\prod_{j<m}E_{kj})>0$,}

\Centerline{$n_i<n_k$ for every $i<k$,}

\Centerline{$H_k\in\Cal H_{n_k}$,}

\Centerline{if $k=i_km+j_k$, where $i_k\in\Bbb N$ and $j_k<m$, then
$\mu(H_k\cap E_{kj_k})>0$,}

\noindent for every $k\in\Bbb N$.   The induction proceeds as follows.
Set $E'_{kj}=X$ if $k=0$, $G_{H_{k-1}}\cap E_{k-1,j}$ otherwise, so that
$\mu^m(V\cap\prod_{j<m}E'_{kj})>0$.   Because $V\cap\prod_{j<m}F_{kj}$ is
empty, we can find $E_{kj}\subseteq E'_{kj}$, for $j<m$, such that
$\mu^m(V\cap\prod_{j<m}E_{kj})>0$ and
$\prod_{j<m}E_{kj}\cap\prod_{j<m}F_{kj}=\emptyset$.   Set
$\eta=\mu E_{kj_k}$, $\delta=\mu^m(V\cap\prod_{j<m}E_{kj})$, so that
$\eta$ and $\delta$ are both strictly positive.

Now take $n_k$ so large that

\Centerline{$2^{-n_k}\le\bover12\epsilon$,
\quad$n_k>n_i$ for $i<k$,
\quad$(1-2^{-mn_k})^{\eta r_{n_k}}<\delta$.}

\noindent (This is possible because
$\lim_{n\to\infty}2^{-mn}r_n=\infty$.)
Set $\Cal H=\{H:H\in\Cal H_{n_k},\,\mu(H\cap E_{kj_k})>0\}$;  then
$\#(\Cal H)\ge\eta r_{n_k}$.   Consider the family
$\family{H}{\Cal H}{G_H^m}$.   These are stochastically independent sets
of measure $2^{-mn_k}$, so their union has measure
$1-(1-2^{-mn_k})^{\#(\Cal H)}>1-\delta$, and there is an $H_k\in\Cal H$
such that $\mu^m(V\cap G_{H_k}^m\cap\prod_{j<m}E_{kj})>0$.   Thus the
induction continues.

Look at the sequence $\sequence{k}{H_k}$ and its union $E$.   We have
$H_k\in\Cal H_{n_k}$ for every $k$;  moreover, if $i<k$, then
$\mu(H_k\cap E_{kj_k})>0$, while $E_{kj_k}\subseteq G_{H_i}$;  since
$H_k$ is an atom of $\Sigma_{n_k}$, while $G_{H_i}\in\Sigma_{n_k}$,
$H_k\subseteq G_{H_i}$.   Thus $E\in\Cal E$.   Next, whenever
$i\le k\in\Bbb N$,

\Centerline{$\prod_{j<m}E_{kj}\cap\prod_{j<m}F_{ij}
\subseteq\prod_{j<m}E_{ij}\cap\prod_{j<m}F_{ij}=\emptyset$,}

\noindent so
$\prod_{j<m}E_{kj}\cap\bigcup_{i\le k}\prod_{j<m}F_{ij}=\emptyset$.   At
the same time, we know that

\Centerline{$E^m\cap\prod_{j<m}E_{kj}
\supseteq\prod_{j<m}H_{km+j}\cap E_{kj}
\supseteq\prod_{j<m}H_{km+j}\cap E_{km+j,j}$}

\noindent has non-zero measure.   So
$\mu^m(E^m\setminus\bigcup_{i\le k}\prod_{j<m}F_{ij})>0$.

Moving back to $Z$, this translates into

\Centerline{$\nu^m((E^*)^m\setminus\bigcup_{i\le k}\prod_{j<m}F^*_{ij})
>0$.}

\noindent But this means that $(E^*)^m$ is not included in
$\bigcup_{i\le k}\prod_{j<m}F^*_{ij}$, for any $k\in\Bbb N$.   Because
$E^*$ is compact and every $F^*_{ij}$ is open,
$(E^*)^m$ is not included in $\bigcup_{i\in\Bbb N}\prod_{j<m}F^*_{ij}$,
and there is some $v\in(E^*)^m\cap\tilde V$.

By the definition of $\tilde V$,

$$\eqalign{\nu^m\{u:u\#v\in\tilde W\}
&>m\epsilon
\ge m\sum_{k=0}^{\infty}2^{-n_0-k}
\ge m\sum_{k=0}^{\infty}2^{-n_k}\cr
&=m\sum_{k=0}^{\infty}\mu H_k
\ge m\mu E
=m\nu E^*.\cr}$$

\noindent So there must be some $u$ such that $u\#v\in\tilde W$ and
$u(j)\notin E^*$ for every $j<m$.   But now, setting $w=u\#v$, we have
$w(2j)\notin E^*$, $w(2j+1)\in E^*$ for $j<m$, and $w\in
D_m(A^*,Z,0,1)\cap\tilde W$;  which is supposed to be impossible.\
\Bang\Qed

\medskip

{\bf (e)} This shows that $A^*$ is not stable.   It is, however,
R-stable.   \Prf\ We have a measure algebra isomorphism between the
measure algebras of $\mu$ and $\nu$ defined by the map $E\mapsto
E^*:\Sigma\to\Tau$.   The corresponding isomorphism between $L^0(\mu)$
and $L^0(\nu)$ takes $\{f^{\ssbullet}:f\in A\}$ to
$\{h^{\ssbullet}:h\in A^*\}$.   By 465Pa and (b) above,
$\{f^{\ssbullet}:f\in A\}$ is stable in $L^0(\mu)$, so
$\{h^{\ssbullet}:h\in A^*\}$ is stable in $L^0(\nu)$, and every
countable subset of $A^*$ is stable.
Since $A^*\subseteq C(X)$, it follows that $A^*$ is stable (465T).\ \Qed
}%end of proof of 465U

\cmmnt{
\leader{*465V}{Remark} This example is clearly related to
419E.   The argument here is significantly deeper, but it does have an
idea in common with that in 419E, besides the obvious point that both
involve the Stone spaces of atomless probability spaces.
Suppose that, in the context of 465U, we take $\Cal E_0$ to be the
family of sets $E$ expressible as the union of a {\it finite} chain
$H_0,\ldots,H_k$ where $H_i\in\Cal H_{n_i}$ for $i\le k$ and
$H_j\subseteq G_{H_i}$ for $i<j\le k$.   Then we find, on repeating the
argument of (b) in the proof of 465U, that the countable set
$A_0^*=\{\chi E^*:E\in\Cal E_0\}$ is stable, so that, setting
$W_m=D_m(A_0^*,Z,0,1)$, $\nu^{2m}W_m$ is small for large $m$.   On the
other hand, setting

\Centerline{$\tilde W_m=\bigcup\{V:V\subseteq Z^m$ is open,
$V\setminus W_m$ is negligible$\}$,}

\noindent we see that $D_m(A^*,Z,0,1)\subseteq\tilde W_m$, so that
$(\nu^{2m})^*\tilde W_m=1$ for all $m\ge 1$.   Of course, writing
$\tilde\nu^{2m}$ for the Radon product measure on $Z^{2m}$, we have
$\tilde\nu^{2m}(\tilde W_m)=\nu^{2m}W_m<1$ for large $m$, just as in
419E.

Both $A\subseteq\eusm L^0(\Sigma)$ and
$A^*\subseteq\eusm L^0(\Tau)$ are relatively pointwise compact.   Note
that while I took $(X,\Sigma,\mu)$ and $(Z,\Tau,\nu)$ to be quite
separate, it is entirely possible for them to be actually the same
space.   In this case it is natural to take every $\Sigma_n$ to consist
of open-and-closed sets, so that every member of $\Cal E$ is open, and
$E^*$ becomes identified with the closure of $E$ for $E\in\Cal E$.
}%end of comment

\exercises{\leader{465X}{Basic exercises (a)}
%\spheader 465Xa
Let $(X,\Sigma,\mu)$ be a semi-finite measure space, and
$\sequencen{f_n}$ a sequence of measurable real-valued functions on $X$
which converges a.e.   Show that $\{f_n:n\in\Bbb N\}$ is stable.
%465B

\spheader 465Xb Let $\Cal C$ be the family of convex sets in $\BbbR^r$, where
$r\ge 1$ is an integer.
Show that $\{\chi C:C\in\Cal C\}$ is stable with respect to Lebesgue
measure on $\BbbR^r$, but that if $r\ge 2$ there is a Radon probability
measure $\nu$ on $\BbbR^r$ such that
$\{\chi C:C\in\Cal C$, $C$ is closed$\}$ is not stable with respect to
$\nu$.
%465B

\spheader 465Xc Show that, for any $M\ge 0$, the set of functions
$f:\Bbb R\to\Bbb R$ of variation at most $M$ is stable with
respect to any Radon measure on $\Bbb R$.   \Hint{show that if $\mu$ is
a Radon measure on $\Bbb R$ and $E\in\dom\mu$ has non-zero finite
measure, and $(2k-1)(\beta-\alpha)>M$, then
$(\mu^{2k})^*D_k(A,E,\alpha,\beta)<(\mu E)^{2k}$.}
%465B

\spheader 465Xd Let $(X,\Sigma,\mu)$ be a
semi-finite measure space, and
$A\subseteq\Bbb R^X$.   Show that $A$ is stable iff $\{f^+:f\in A\}$ and
$\{f^-:f\in A\}$ are both stable.
%4{}65C

%gap here \query

\spheader 465Xf Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$A\subseteq\Bbb R^X$.   Suppose that $\mu$ is inner regular with
respect to the family
$\{F:F\in\Sigma,\,\{f\times\chi F:f\in A\}$ is stable$\}$.   Show that
$A$ is stable.
%4{}65C

\spheader 465Xg Let $(X,\Sigma,\mu)$ be a totally finite measure space
and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Let
$A\subseteq\eusm L^0(\Tau)$ be any set.   (i) Show that if $A$ is
$\mu\restrp\Tau$-stable
then it is $\mu$-stable.   (ii) Give an example to show that $A$ can be
$\mu$-stable and pointwise compact without being
$\mu\restrp\Tau$-stable.   \Hint{take $\mu$ to be Lebesgue measure on
$[0,1]$ and $\Tau$ the countable-cocountable algebra.}
%4{}65C

\spheader 465Xh\dvAnew{2016} Let $(X,\Sigma,\mu)$ be a semi-finite measure space
and $A$ a subset of $\BbbR^X$.   Suppose that for every $\epsilon>0$ there is a
stable set $B\subseteq\BbbR^X$ such that for every $f\in A$ there is a $g\in B$
such that $\|f-g\|_{\infty}\le\epsilon$.   Show that $A$ is stable.
%465C

\sqheader 465Xi\dvAnew{2016} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space and $\sequencen{E_n}$ a sequence in $\Sigma$.   (i) Suppose that
whenever $F\in\Sigma$ and $\mu F<\infty$ there is a $k\ge 1$ such that
$\sum_{n=0}^{\infty}(\mu(F\cap E_n)\,\mu(F\setminus E_n))^k$ is finite.
Show that $\sequencen{\chi E_n}$ is stable.   \Hint{465Cc.}
(ii) Suppose that $\mu X=1$,
that $\sequencen{E_n}$ is independent, and that
$\sum_{n=0}^{\infty}((1-\mu E_n)\mu E_n)^k=\infty$ for every $k\ge 1$.
Show that $\sequencen{\chi E_n}$ is not stable.   \Hint{273K.}
%465C

\sqheader 465Xj Show that there is a disjoint family $\Cal I$ of finite
subsets of $[0,1]$ such that $A=\{\chi I:I\in\Cal I\}$ is not stable with
respect to Lebesgue measure,
though $A$ is pointwise compact and the identity map on $A$ is
continuous for the topology of pointwise convergence and the topology of
convergence in measure.   \Hint{show by induction on $m$ that if $m\ge 1$ and
$K\subseteq[0,1]^m$ is a compact set with non-zero measure, then
$K\cap([0,1]\setminus D)^m\ne\emptyset$ whenever $\#(D)<\frak c$;  now put this
together with 4A3Fa.}
%465F

\spheader 465Xk Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$A\subseteq\Bbb R^X$ a stable set such that $\{f(x):f\in A\}$ is bounded for
every $x\in X$.   Let $\overline{A}$ be the closure
of $A$ for the topology of pointwise convergence.    Show that
$\{f^{\ssbullet}:f\in\overline{A}\}$ is just the closure of
$\{f^{\ssbullet}:f\in A\}\subseteq L^0(\mu)$ for the topology of
convergence in measure.
%465D 465G

\spheader 465Xl Let $(X,\Sigma,\mu)$ be a probability space.   Show that
$A\subseteq\Bbb R^X$ is stable iff

\Centerline{$\inf_{m\in\Bbb N}
\bigl((\mu^{2m})^*D_m(A,X,\alpha,\beta)\bigr)^{1/m}=0$}

\noindent whenever $\alpha<\beta$ in $\Bbb R$.
%465J

\spheader 465Xm Let $(X,\Sigma,\mu)$ be a
semi-finite measure space.   Show that a
countable set $A\subseteq\eusm L^0(\Sigma)$ is not stable iff there are
$E\in\Sigma$ and $\alpha<\beta$ such that $0<\mu E<\infty$ and
$\mu^m\tilde D_m(A,E,\alpha,\beta)=(\mu E)^m$ for every $m\ge 1$, where
$\tilde D_m(A,E,\alpha,\beta)$ is the set of those $w\in E^m$ such that
for every $I\subseteq m$ there is an $f\in A$ such that
$f(w(i))\le\alpha$
for $i\in I$, $f(w(i))\ge\beta$ for $i\in m\setminus I$.   \Hint{see
part (iii) of case 2 of the proof of 465L.}
%465L

\spheader 465Xn Let $(X,\Sigma,\mu)$ be a semi-finite measure space, and
$A\subseteq\eusm L^0(\Sigma)$ a set which is compact and metrizable for
the topology of pointwise convergence.   Show that $A$ is stable.
\Hint{otherwise, apply the
ideas of case 2 in the proof of 465L to a countable dense subset of $A$
to obtain a sequence which contradicts the conclusion of 465Xa.}
%465L

\spheader 465Xo Let $(X,\Sigma,\mu)$ be a probability space and
$A\subseteq\eusm L^0(\Sigma)$ a uniformly bounded set of functions.
Show that $A$ is stable iff

\Centerline{$
\lim_{n\to\infty}\overlineint
  \sup_{f\in A,k,l\ge n}
\bigl|\Bover1k\sum_{i=0}^{k-1}f(w(i))
  -\Bover1l\sum_{i=0}^{l-1}f(w(i))\bigr|\mu^{\Bbb N}(dw)=0$.}
%465M

\spheader 465Xp Show that there is a set $A\subseteq\BbbR^{[0,1]}$ such
that $\sup_{f\in A}|f(x)|$ is finite for every $x$, and $A$ is
stable with respect to Lebesgue measure, but its convex hull $\Gamma(A)$
is not.   \Hint{take $\Cal I$ as in 465Xj and
$A=\{\bover1{\#(I)}\chi\{x\}:x\in I\in\Cal I\setminus\{\emptyset\}\}$.}
%465N

\spheader 465Xq Show that there is a sequence $\sequencen{f_n}$ of
functions from $[0,1]$ to $\Bbb N$ such that $\{f_n:n\in\Bbb N\}$ is stable
for Lebesgue measure on $[0,1]$, but $\{f_m-f_n:m$, $n\in\Bbb N\}$ is not.
%465N 465Xi

\spheader 465Xr Let $(X,\Sigma,\mu)$ be a semi-finite measure
space and $Q\subseteq L^0(\mu)$ a set which is stable in the sense of
465O.  Show that the closure of $Q$ (for the topology of convergence in
measure) is stable.   \Hint{465Xk.}
%465P 465Xk

\spheader 465Xs Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a
$\sigma$-subalgebra of $\Sigma$, and $A\subseteq\Bbb R^X$ a countable
stable set of $\Sigma$-measurable functions such that
$\int\sup_{f\in A}|f(x)|\mu(dx)<\infty$.   Show that if for each
$f\in A$ we choose a conditional expectation $g_f$ of $f$ on $\Tau$
(requiring
each $g_f$ to be $\Tau$-measurable and defined everywhere on $X$), then
$\{g_f:f\in A\}$ is stable.
%465R

\spheader 465Xt Give an example of a probability algebra
$(\frak A,\bar\mu)$, a conditional expectation operator
$P:L^1_{\bar\mu}\to L^1_{\bar\mu}$ (365Q),
and a uniformly integrable
stable set $A\subseteq L^1_{\bar\mu}$ such that $P[A]$ is not stable.
\Hint{start by picking
a sequence $\sequencen{v_n}$ in $P[L^1_{\bar\mu}]$ which is
norm-convergent to $0$ but not stable, and express this as
$\sequencen{Pu_n}$ where $\bar\mu\Bvalue{u_n>0}\le 2^{-n}$ for every
$n$.}
%465R

\leader{465Y}{Further exercises (a)}
%\spheader 465Ya
Let $(X,\Sigma,\mu)$ be a
semi-finite measure space.   A set
$A\subseteq\BbbR^X$ is said to have the {\bf Bourgain property} if whenever
$E\in\Sigma$, $\mu E>0$ and $\epsilon>0$, there are
non-negligible measurable sets $F_0,\ldots,F_n\subseteq E$ such that for
every $f\in A$ there is an $i\le n$ such that the oscillation
$\sup_{x,y\in F_i}|f(x)-f(y)|$ of $f$ on $F_i$ is at most $\epsilon$.
Show that in this case $A$ is stable.
%4{}65C  or perhaps in 465Y

\spheader 465Yb Let $X$ be a topological space, and $\mu$ a
$\tau$-additive effectively locally finite topological measure on $X$.
Show that any equicontinuous subset of $C(X)$ has the Bourgain property, so
is stable.
%465Ya 4{}65C

\spheader 465Yc
Find an integrable continuous function
$f:\coint{0,1}^2\to\coint{0,\infty}$ such that, in the notation of
465H, $\limsup_{k\to\infty}\int fd\nu_{wk}^2=\infty$ for almost every
$w\in\coint{0,1}^{\Bbb N}$, if $\coint{0,1}^{\Bbb N}$ is given its usual
measure.
%\Hint{arrange
%for $f(x,x)$ to grow very fast as $x\uparrow 1$.}
%465H

\spheader 465Yd Show that in 465M we may replace the condition `$A$ is
uniformly bounded' by the condition `$|f|\le f_0$ for every $f\in A$,
where $f_0$ is integrable'.   \Hint{{\smc Talagrand 87}.}
%465M

\spheader 465Ye Let $(X,\Sigma,\mu)$ be a probability space, and
$A\subseteq\BbbR^X$ a uniformly bounded set.   Show that $A$ is stable iff
for every $\epsilon>0$ there are a stable set $B\subseteq\BbbR^X$, a
sequence $\sequence{k}{h_k}$ of measurable functions on $X^{\Bbb N}$, and a
family $\family{f}{A}{g_f}$ in $B$ such that

\Centerline{$h_k(w)\ge\Bover1k\sum_{i=0}^{k-1}|f(w(i))-g_f(w(i))|$
for every $w\in X^{\Bbb N}$, $k\ge 1$ and $f\in A$,}

\Centerline{$\limsup_{k\to\infty}h_k(w)\le\epsilon$ for almost every
$w\in X^{\Bbb N}$.}

%\noindent\Hint{465M(iii)$\Rightarrow$465M(v), with a little help from
%465Nc.}
%465N

\spheader 465Yf Let $(X,\Sigma,\mu)$ be a semi-finite measure space,
and $A$, $B\subseteq\BbbR^X$ pointwise bounded stable sets.   Show that
$\{f\times g:f\in A$, $g\in B\}$ is stable.
%\Hint{for the uniformly
%bounded case use 465M(iii) and 465Ye;  then extend as in 465Nc.}
%465Ye 465N

\spheader 465Yg Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be semi-finite
measure spaces, with c.l.d.\ product $(X\times Y,\Lambda,\lambda)$.
Suppose that $A\subseteq\BbbR^X$ and $B\subseteq\BbbR^Y$ are pointwise
bounded stable sets.   Show that $\{f\otimes g:f\in A$, $g\in B\}$ is
stable with respect to $\lambda$, where $(f\otimes g)(x,y)=f(x)g(y)$.
%465Yf 465N

\spheader 465Yh Let $(X,\Sigma,\mu)$ be a semi-finite measure space,
$A\subseteq\BbbR^X$ a uniformly bounded stable set and
$h:\Bbb R\to\Bbb R$ a continuous
function.   Show that $\{hf:f\in A\}$ is stable.
%\Hint{use 465Yf, 465Nc and the Stone-Weierstrass theorem.}
%465Yf 465N

\spheader 465Yi\dvArevised{2016} Let $(X,\frak T,\Sigma,\mu)$ be an
effectively locally finite $\tau$-additive
topological measure space.   Show that a countable R-stable subset of $\Bbb R^X$
is stable.
%\tau-additive product measure extends c.l.d. product measure
%465S

\spheader 465Yj Let $(X,\frak T,\Sigma,\mu)$ be a complete
$\tau$-additive topological probability
space such that $\mu$ is inner regular with respect to the Borel sets,
and $A\subseteq[0,1]^X$ an R-stable set.   Suppose that $\epsilon>0$
is such that $\int fd\mu\le\epsilon^2$ for every $f\in A$.   Show that
there are an $n\ge 1$ and a Borel set $W\subseteq X^n$ and a
$\gamma>\tilde\mu^nW$ (writing $\tilde\mu^n$ for the $\tau$-additive
product measure on $X^n$) such that $\int fd\nu\le 3\epsilon$ whenever
$f\in A$, $\nu:\Cal PX\to[0,1]$ is a point-supported
probability measure and $\nu^nW\le\gamma$.
%465S
%465K gives us a  W\in\Tensorhat_n\Sigma ;  it just has to be a measurable
%envelope so need to know that  \mu^n  is outer regular for the Borel sets
%to get Borel  W

\spheader 465Yk Set $X=\prod_{n=2}^{\infty}\Bbb Z_n$, where each $\Bbb Z_n$
is the cyclic group of order $n$ with its discrete topology;  let $\mu$ be
the Haar probability measure on $X$.   Let $\action_l$ be the left shift
action of $X$ on $\BbbR^X$, so that
$(a\action_lf)(x)=f(x-a)$ for $a$, $x\in X$ and $f\in\BbbR^X$.
(i) Show that for any $n\in\Bbb N$ there is a
compact negligible set $K_n\subseteq X$ such that for every $I\in[X]^n$
there are uncountably many $a\in X$ such that $I\subseteq K_n+a$.   (ii)
Show that there is a negligible set $E\subseteq X$ such that
$\{a\action_l\chi E:a\in X\}$ is dense in $\{0,1\}^X$ for the topology of
pointwise convergence.   (iii) Show that if $f\in C(X)$ then
$\{a\action_lf:a\in X\}$ is stable.   (iv) Show that there is a sequence
$\sequencen{f_n}$ in $C(X)$ such that $\{f_n:n\in\Bbb N\}$ is stable but
$\{a\action_lf_n):a\in X$, $n\in\Bbb N\}$ is not.
(v) Find expressions of these
results when $X$ is replaced by the circle group or by $\Bbb R$.
%mt46bits  465Yb
}%end of exercises

\endnotes{
\Notesheader{465}
The definition in 465B arose naturally when M.Talagrand and I were
studying pointwise compact sets of measurable functions;  we found that
in many cases a set of functions was relatively pointwise compact
because it was stable (465Db).   Only later did it appear that the
concept was connected with Glivenko-Cantelli classes in the theory of
empirical measures, as explained in {\smc Talagrand 87}.

It is not the case that all pointwise compact sets of
measurable functions are stable.   In fact I have already offered
examples in 463Xh
%countable-cocountable
and 464E above.
%Talagrand's measure
In both cases it is easy to check from the definition
in 465B that they are not stable, as can also be deduced from 465G.
Another example is in 465Xj.   You will observe however that all these
examples are `pathological' in the sense that either the measure space
is irregular (from some points of view, indeed, any measure space not
isomorphic to Lebesgue measure on the unit interval can be dismissed as
peripheral), or the set of functions is uninteresting.   It is clear
from 465R and 465T, for instance, that we should start with countable
sets.   So it is natural to ask:  if we have a {\it separable} pointwise
compact set of real-valued measurable functions on the unit interval,
must it be stable?   If the continuum hypothesis is true, yes
(536F\Latereditions), but the question is undecided by the ordinary axioms of
ZFC ({\smc Shelah \& Fremlin 93}).   (If we ask for `metrizable',
instead of `separable', we get a positive answer;  see 465Xn.)

At the very beginning of our work on this topic, one of
the leading questions was:  under what circumstances will an independent family
$\sequencen{E_n}$ of sets give rise to a relatively pointwise compact set
$\{\chi E_n:n\in\Bbb N\}$ in $\eusm L^0$?   There remain interesting open
questions here.   But for stability, rather than relative compactness, we have
a straightforward answer (465Xi).

The curious phrasing of the statement of 465M(iii), with the auxiliary
functions $h_k$, turns on the fact that all the expressions
`$\sup_{f\in A}\ldots$' here give rise to functions which need not be measurable.
Thus the simple pointwise convergence described in (ii) and (iv) is not
at all the same thing as the convergence in (v), which may be thought of
as a kind of $\|\,\|_1$-convergence if we write
$\|g\|_1=\overline{\int}|g|$ for arbitrary real-valued functions $g$.
(Since the sets $A$ here are uniformly bounded, it may equally be
thought of as convergence in measure.)   Similarly, 465M(iii) is saying
much more than just

\Centerline{$\lim_{k\to\infty}\sup_{f\in A}\Bover1k\sum_{i=0}^{k-1}
\bigl|f(w(i))-\Expn(f|\Tau)(w(i))\bigr|=0$ a.e.,}

\noindent though of course for countable sets $A$ these distinctions
disappear.   On the other hand, since the convergence is certainly not
monotonic in $k$, $\|\,\|_1$-convergence does not imply pointwise
convergence.   So we can look for something stronger than either (iv) or
(v) of 465M, as in 465Xo.   But this is still nowhere near the strength
of 465M(iii), in which $|\sum\ldots|$ is replaced by $\sum|\ldots|$.
For further variations, see {\smc Talagrand 87} and {\smc Talagrand 96}.

If we wish to adapt the ideas here to spaces of equivalence classes of
functions rather than spaces of true functions, we find that problems of
measurability evaporate, and that (because the definition of stability
looks only at sets of finite measure) all the relevant suprema can be
interpreted as suprema of countable sets.   Consequently a subset of
$L^0(\mu)$ or $L^0(\frak A)$ is stable if all its countable subsets are
stable (465Pa).   It is remarkable that, for strictly localizable
measures $\mu$, we can lift any stable set in $L^0$ to a stable set in
$\eusm L^0$ (465Pb, 465Q).   By moving to function spaces we get a
language in which to express a new kind of permanence property of stable
sets (465R, 465Xs).   See also {\smc Talagrand 89}.

The definition of `stable set' of functions seems to be utterly
dependent on the underlying measure space.   But 465R tells us that in
fact the property of being an order-bounded stable subset of
$L^1(\bar\mu)$ is invariant under normed space automorphisms.
(`Order-boundedness' is a normed space invariant in $L^1$ spaces by the
Chacon-Krengel theorem, 371D.)    Since stability can be defined in
terms of order-bounded sets (465Xh), we could, for instance, develop a
theory of stable sets in abstract $L$-spaces.

The theory of stable sets is of course bound intimately to the theory of
product measures, and such results as 465J have independent interest as
theorems about sets in product spaces.   So any new theory of product
measures will give rise to a new theory of stable sets.   In particular,
the $\tau$-additive product measures in \S417 lead to R-stable sets
(465S).   It is instructive to work through the details, observing how
the properties of the product are employed.   Primarily, of course, we
need `associative' and `commutative' laws, and Fubini's theorem;  but
some questions of measurability need to be re-examined, as in 465Yj.

I have starred 465U because it involves the notion of R-stability.   In
fact this appears only in the final stage, and the construction, as set
out in part (a) of the proof, is an instructive challenge to any
intuitive concept of what stable sets are like.
}%end of notes

\discrpage

\frfilename{mt466.tex}
\versiondate{2.8.13}
\copyrightdate{2000}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Measures on linear topological spaces}

\newsection{466}

In this section I collect a number of results on the special properties
of topological measures on linear topological spaces.   The most
important is surely Phillips' theorem (466A-466B):  on any Banach space,
the weak and norm topologies give rise to the same totally finite Radon
measures.   This is not because the weak and norm topologies have the
same Borel $\sigma$-algebras, though this does happen
in interesting cases (466C-466E, \S467).   When the Borel
$\sigma$-algebras are different, we can still
ask whether the Borel measures
are `essentially' the same, that is, whether every (totally finite)
Borel measure for the weak topology extends to a Borel measure for the
norm topology.   A construction due to M.Talagrand (466H, 466Ia) gives a
negative answer to the general question.

Just as in $\BbbR^r$, a totally finite quasi-Radon measure on a locally
convex linear topological space is determined by its characteristic
function (466K).   I end the section with a note on measurability
conditions sufficient to ensure that a linear operator between Banach
spaces is continuous (466L-466M), and with brief remarks on Gaussian
measures (466N-466O).

\leader{466A}{Theorem} Let $(X,\frak T)$ be a metrizable locally convex
linear topological space and $\mu$ a
$\sigma$-finite measure on $X$ which is quasi-Radon for the weak
topology $\frak T_s(X,X^*)$.   Then the support of $\mu$ is
separable, so $\mu$ is quasi-Radon for the original topology $\frak T$.
If $X$ is complete and $\mu$ is locally finite with respect to
$\frak T$, then $\mu$ is Radon for $\frak T$.

\proof{{\bf (a)} Let $\sequencen{V_n}$ be a sequence running over a base
of neighbourhoods of $0$ in $X$, and for $n\in\Bbb N$ set
$V_n^{\smallcirc}=\{f:f\in X^*$, $f(x)\le 1$ for every $x\in V_n\}$.
Then each $V_n^{\smallcirc}$ is a convex
$\frak T_s$-compact subset of $X^*$ (4A4Bf), and
$X^*=\bigcup_{n\in\Bbb N}V_n^{\smallcirc}$.   Let $Z\subseteq X$ be the
support of $\mu$, and for $n\in\Bbb N$ set
$K_n=\{f\restr Z:f\in V_n^{\smallcirc}\}$.   Since the map
$f\mapsto f\restr Z:X^*\to C(Z)$ is linear and continuous for
$\frak T_s$ and the topology $\frak T_p$ of pointwise convergence on
$Z$, each $K_n$ is convex and $\frak T_p$-compact.   Next, the subspace
measure $\mu_Z$ on $Z$ is $\sigma$-finite and strictly positive, so by
463G each $K_n$ is $\frak T_p$-metrizable.

Let $\Cal U$ be the base for $\frak T_p$ consisting of sets of the form

\Centerline{$U(I,h,\epsilon)
=\{f:f\in C(Z),\,|f(x)-h(x)|<\epsilon$ for every $x\in I\}$,}

\noindent where $I\subseteq Z$ is finite, $h\in\BbbR^I$ and
$\epsilon>0$.
For each $n\in\Bbb N$, write $\Cal V_n=\{K_n\cap U:U\in\Cal U\}$, so
that $\Cal V_n$ is a base for the subspace topology of $K_n$
(4A2B(a-vi)).   Since this is compact and metrizable, therefore
second-countable, there is a countable base $\Cal V'_n\subseteq\Cal V_n$
(4A2Ob).   Now there is a countable set $\Cal U'\subseteq\Cal U$
such that $\Cal V'_n\subseteq\{K_n\cap U:U\in\Cal U'\}$ for every
$n\in\Bbb N$, and a countable set $D\subseteq Z$ such that

\Centerline{$\Cal U'\subseteq\{U(I,h,\epsilon):I\subseteq D$ is finite,
$h\in\BbbR^I$, $\epsilon>0\}$.}

Let $D'\subseteq X$ be the linear span of $D$ over the rationals.   Then
$D'$ is again countable, and its $\frak T$-closure $Y$ is a linear
subspace of $X$ (4A4Bg).   \Quer\ If $Z\not\subseteq Y$, then
$\mu(X\setminus Y)>0$.   But, writing

$$\eqalign{Y^{\smallcirc}
&=\{f:f\in X^*,\,f(x)\le 1\text{ for every }x\in Y\} \cr
&=\{f:f\in X^*,\,f(x)=0\text{ for every }x\in Y\},\cr}$$

\noindent $X\setminus Y$ must be
$\bigcup_{f\in Y^{\smallcirc}}\{x:|f(x)|>0\}$, by 4A4Eb.
Because $\mu$ is $\tau$-additive, there must be an
$f\in Y^{\smallcirc}$ such that $\mu\{x:|f(x)|>0\}>0$.   Let
$n\in\Bbb N$ be such that $f\in V_n^{\smallcirc}$.   Since $f(x)=0$ for
every $x\in D$,
$f\restr Z$ belongs to the same members of $\Cal V'_n$ as the zero
function;  since $\Cal V'_n$ is a base for the Hausdorff subspace
topology of $K_n$, $f\restr Z$ actually is the zero function, and
$\{x:|f(x)|>0\}\subseteq X\setminus Z$, so $\mu Z<\mu X$, contrary to
the choice of $Z$.\ \Bang

Thus $Z\subseteq Y$.   Because $Y=\overline{D'}$ is $\frak T$-separable,
and $\frak T$ is metrizable, $Z$ also is $\frak T$-separable
(4A2P(a-iv)).

\medskip

{\bf (b)} The subspace topology $\frak T_Y$ induced on $Y$ by $\frak T$
is a separable metrizable locally convex linear space topology, so the
Borel $\sigma$-algebras of $\frak T_Y$ and the associated weak topology
$\frak T_s(Y,Y^*)$ are equal (4A3W).   But $\frak T_s(Y,Y^*)$ is
just the subspace topology induced on $Y$ by $\frak T_s$ (4A4Ea),
Accordingly the subspace measure $\mu_Y$ on $Y$ is
$\frak T_s(Y,Y^*)$-quasi-Radon (415B).   Since every $\frak T_Y$-open
set is $\frak T_s(Y,Y^*)$-Borel, $\mu_Y$ is a topological measure for
$\frak T_Y$.   Since $\frak T_Y$ is finer than $\frak T_s(Y,Y^*)$,
$\mu_Y$ is effectively locally finite for $\frak T_Y$ and inner regular
with respect to the $\frak T_Y$-closed sets, and therefore is a
quasi-Radon measure for $\frak T_Y$ (415D(i)).

By 415J, there is a measure $\tilde\mu$ on $X$, quasi-Radon for
$\frak T$, such that $\tilde\mu E=\mu_Y(E\cap Y)$ whenever $\tilde\mu$
measures $E$.   But as $Y$ is $\frak T$-closed and $\mu$-conegligible,
and $\mu$ is complete, we have $\mu=\tilde\mu$, and $\mu$ is quasi-Radon
for $\frak T$.

\medskip

{\bf (c)} If $X$ is complete and $\mu$ is locally finite with respect to
$\frak T$, then $(X,\frak T)$ is a pre-Radon space (434Jg), so $\mu$ is
a Radon measure for $\frak T$ (434Jb).
}%end of proof of 466A

\leader{466B}{Corollary}\cmmnt{ (Compare 462I.)}  If $X$ is a Banach
space and $\mu$ is a totally
finite measure on $X$ which is quasi-Radon for the weak topology of $X$,
it is a Radon measure for both the norm topology and the weak topology.

\proof{ By 466A, $\mu$ is a Radon measure for the norm topology;  by
418I, or otherwise, it is a Radon measure for the weak topology.
}%end of proof of 466B

\cmmnt{\medskip

\noindent{\bf Remark} Thus Banach spaces, with their weak topologies,
are pre-Radon.
}%end of comment

\leader{466C}{Definition} A normed space $X$ has a {\bf Kadec norm}
(also called {\bf Kadec-Klee norm}) if
the norm and weak topologies coincide on the sphere $\{x:\|x\|=1\}$.
Of course they will then also coincide on any sphere
$\{x:\|x-y\|=\alpha\}$.

\cmmnt{\medskip

\noindent{\bf Example} For any set $I$ and
any $p\in\coint{1,\infty}$, the
Banach space $\ell^p(I)$ has a Kadec norm.   \prooflet{\Prf\
Set $S=\{x:\|x\|_p=1\}$.   If $x\in S$ and $\epsilon>0$, take
$\eta\in\ocint{0,1}$ such that
$2\eta+(2p\eta)^{1/p}\le\epsilon$.   Let
$J\subseteq I$ be a finite set such that
$\sum_{i\in I\setminus J}|x(i)|^p\le\eta^p$.   Set
$H=\{y:y\in\ell^p(I),\,\sum_{i\in J}|y(i)-x(i)|^p<\eta^p\}$;
then $H$ is open for the weak topology of $\ell^p(I)$.
If $y\in H\cap S$, then, writing $x_J$ for
$x\times\chi J$, etc.,

\Centerline{$\|x_J\|_p\ge 1-\|x_{I\setminus J}\|_p
\ge 1-\eta$,}

\Centerline{$\|y_J\|_p
\ge\|x_J\|_p-\eta\ge 1-2\eta$,
\quad$\|y_J\|_p^p\ge 1-2p\eta$,}


$$\eqalign{\|y-x\|_p
&\le\|y_J-x_J\|_p+\|x_{I\setminus J}\|_p+\|y_{I\setminus J}\|_p\cr
&\le\eta+\eta+(1-\|y_J\|_p^p)^{1/p}
\le 2\eta+(2p\eta)^{1/p}
\le\epsilon.\cr}$$

\noindent Thus $\{y:y\in S$, $\|y-x\|_p\le\epsilon\}$ is a
neighbourhood of $x$ for the subspace weak topology on $S$;   as $x$ and
$\epsilon$ are arbitrary, the weak and norm topologies agree on $S$.\ \Qed}%end of prooflet

For further examples, see 467B {\it et seq}.}%end of comment

\leader{466D}{Proposition}\cmmnt{ ({\smc Hansell 01})} Let $X$ be a
normed space with a Kadec norm.   Then there is a network for the norm
topology on $X$ expressible in the form $\bigcup_{n\in\Bbb N}\Cal V_n$,
where for each $n\in\Bbb N\,\,\Cal V_n$ is an isolated family for
the weak topology and
$\bigcup\Cal V_n$ is the difference of two closed sets for the weak
topology.

\proof{ Let $\Cal U$ be a
$\sigma$-disjoint base for the norm topology of $X$ (4A2L(g-ii));
express it as
$\bigcup_{n\in\Bbb N}\Cal U_n$ where every $\Cal U_n$ is disjoint.   For
rational numbers $q$, $q'$ with $0<q<q'$, set
$S_{qq'}=\{x:q<\|x\|\le q'\}$, and for $A\subseteq X$ write $W(A,q,q')$
for the interior of $A\cap S_{qq'}$ taken in the subspace weak topology
of $S_{qq'}$.   Set
$\Cal V_{nqq'}=\{W(U,q,q'):U\in\Cal U_n\}$, so that $\Cal V_{nqq'}$ is a
disjoint family of relatively-weakly-open subsets of $S_{qq'}$, and is
an isolated family for the weak topology.   Now
$\bigcup_{n\in\Bbb N,q,q'\in\Bbb Q,0<q<q'}\Cal V_{nqq'}$ is a network
for the norm topology on $X\setminus\{0\}$.   \Prf\ If
$x\in X\setminus\{0\}$ and $\epsilon>0$, then take
$n\in\Bbb N$, $U\in\Cal U_n$ such that
$x\in U\subseteq\{y:\|y-x\|\le\epsilon\}$.  Let $\delta>0$ be such that
$\{y:\|y-x\|\le\delta\}\subseteq U$.   Next, because $\|\,\|$ is a Kadec
norm, there is a weak neighbourhood $V$ of $0$ such that
$\|y-x\|\le\bover12\delta$ whenever $y\in x-V$ and $\|y\|=\|x\|$.   Let
$V'$ be an open weak neighbourhood of $0$ such that $V'+V'\subseteq V$.
Let $\eta\in\ooint{0,1}$ be such that $\eta\|x\|\le\bover12\delta$ and
$y\in V'$ whenever $\|y\|\le\eta\|x\|$.
If $y\in x-V'$ and $(1-\eta)\|x\|\le\|y\|\le(1+\eta)\|x\|$, then

\Centerline{$\|y-\Bover{\|x\|}{\|y\|}y\|
=|1-\Bover{\|x\|}{\|y\|}|\,\|y\|
=|\Bover{\|y\|}{\|x\|}-1|\,\|x\|\le\eta\|x\|\le\Bover12\delta$,}

\Centerline{$x-\Bover{\|x\|}{\|y\|}y
=(x-y)+(y-\Bover{\|x\|}{\|y\|}y)\in V'+V'\subseteq V$,}

\noindent so $\|x-\Bover{\|x\|}{\|y\|}y\|\le\bover12\delta$ and
$\|x-y\|\le\delta$
and $y\in U$.   This means that if we take $q$, $q'\in\Bbb Q$ such that
$(1-\eta)\|x\|\le q\le\|x\|\le q'\le(1+\eta)\|x\|$, then
$(x-V')\cap S_{qq'}\subseteq U$ and $x\in W(U,q,q')\in\Cal V_{nqq'}$.
Since of course $W(U,q,q')\subseteq U\subseteq\{y:\|y-x\|\le\epsilon\}$,
and $x$ and $\epsilon$ are arbitrary, we have the result.\ \Qed

To get a $\sigma$-isolated family for the weak topology which is
a network for the norm topology on the whole of $X$, we just have to add
the singleton set $\{0\}$.   To see that the union of each of our
isolated families is the difference of two weakly open sets,
observe that $\bigcup\Cal V_{nqq'}$
is a relatively weakly open subset of $S_{qq'}$, which is the difference
of the weakly open sets $\{x:\|x\|>q\}$ and $\{x:\|x\|>q'\}$.
}%end of proof of 466D

\leader{466E}{Corollary} Let $X$ be a normed space with a Kadec norm.

(a) The norm and weak topologies give rise to the same Borel
$\sigma$-algebras.

(b) The weak topology has a $\sigma$-isolated network, so is
hereditarily weakly $\theta$-refinable.

\proof{{\bf (a)} Write $\Cal B_{\|\,\|}$, $\Cal B_{\frak T_s}$ for the
Borel $\sigma$-algebras for the weak and norm topologies.   Of course
$\Cal B_{\frak T_s}\subseteq\Cal B_{\|\,\|}$.   Let
$\sequencen{\Cal V_n}$ be a sequence covering a network for the norm
topology as in 466D.   Because
$\Cal V_n$ is (for the weak topology) isolated and its union
belongs to $\Cal B_{\frak T_s}$, $\bigcup\Cal W\in\Cal B_{\frak T_s}$
for every $n\in\Bbb N$ and $\Cal W\subseteq\Cal V_n$.   But this means that
$\bigcup\Cal W\in\Cal B_{\frak T_s}$ for every
$\Cal W\subseteq\Cal V=\bigcup_{n\in\Bbb N}\Cal V_n$;  and as
$\Cal V$ is a network for the norm topology, every
norm-open set belongs to $\Cal B_{\frak T_s}$, and
$\Cal B_{\|\,\|}\subseteq\Cal B_{\frak T_s}$.   Thus the two Borel
$\sigma$-algebras are equal.

\medskip

{\bf (b)} Of course $\Cal V$ is also a network for the weak topology, so
the weak topology has a $\sigma$-isolated network;  by
438Ld, it is hereditarily weakly $\theta$-refinable.
}%end of proof of 466E

\vleader{48pt}{466F}{Proposition}
Let $X$ be a Banach space with a Kadec norm.
Then the following are equiveridical:

(i) $X$ is a Radon space in its norm topology;

(ii) $X$ is a Radon space in its weak topology;

(iii) the weight of $X$ (for the norm topology) is
measure-free\cmmnt{ in the sense of \S438}.

\proof{{\bf (a)(i)$\Leftrightarrow$(ii)} By 466Ea, the norm and weak topologies
give rise to the same algebra $\Cal B$ of Borel sets.   If $X$ is a
Radon space in its norm topology, then any totally finite measure with
domain $\Cal B$ is inner regular with respect to the norm-compact sets,
therefore inner regular with respect to the weakly compact sets, and $X$
is Radon in its weak topology.   If $X$ is a Radon space in its weak
topology, then any totally finite measure $\mu$ with domain $\Cal B$ has
a completion $\hat\mu$ which is a Radon measure for the weak topology,
therefore also for the norm topology, by 466B;  as $\mu$ is arbitrary,
$X$ is a Radon space for the norm topology.

\medskip

{\bf (b)(i)$\Leftrightarrow$(iii)} is a special case of 438H.
}%end of proof of 466F

\leader{466G}{Definition} A partially ordered set $X$ has the
{\bf $\sigma$-interpolation property}\cmmnt{ or
{\bf countable separation
property}} if whenever $A$, $B$ are non-empty countable subsets of $X$ and
$x\le y$ for every $x\in A$, $y\in B$, then there is a $z\in X$ such
that $x\le z\le y$ for every $x\in A$ and $y\in B$.   \cmmnt{A Dedekind
$\sigma$-complete partially ordered set (314Ab) always has the
$\sigma$-interpolation property.}

\leader{466H}{Proposition}\cmmnt{ ({\smc Jayne \& Rogers 95})} Let $X$
be a Riesz space with a Riesz norm, given its
weak topology $\frak T_s=\frak T_s(X,X^*)$.   Suppose that ($\alpha$)
$X$ has the $\sigma$-interpolation property ($\beta$) there is a
strictly increasing
family $\langle p_{\xi}\rangle_{\xi<\omega_1}$ in $X$.   Then there is a
$\frak T_s$-Borel probability measure $\mu$ on $X$ such that

(i) $\mu$ is not inner regular with respect to the $\frak T_s$-closed
sets;

(ii) $\mu$ is not $\tau$-additive for the topology $\frak T_s$;

(iii) $\mu$ has no extension to a norm-Borel measure on $X$.

\noindent Accordingly $(X,\frak T_s)$ is not a Radon space (indeed, is
not Borel-measure-complete).

\proof{{\bf (a)} Let $K$ be the set

\Centerline{$\{f:f\in X^*,\,f\ge 0,\,\|f\|\le
1\}=\{f:\|f\|\le 1\}\cap\bigcap_{x\in X^+}\{f:f(x)\ge 0\}$,}

\noindent so that $K$ is a weak*-closed subset of the unit ball of
$X^*$ and is weak*-compact.   Because $X^*$ is a solid linear subspace
of the order-bounded dual $X^{\sim}$ of $X$ (356Da), every member of
$X^*$ is the difference of two non-negative members of $X^*$, and $K$
spans $X^*$.   We shall need to know that if $x<y$ in $X$, there is an
$f\in K$ such that $f(x)<f(y)$;  set $f=\Bover1{\|g\|}|g|$ where
$g\in X^*$ is such that $g(x)\ne g(y)$.   (Recall that the norm of $X^*$
is a Riesz norm, as also noted in 356Da.)

Set

\Centerline{$A=\bigcup_{\xi<\omega_1}\{x:x\in X,\,x\le p_{\xi}\}$;}

\noindent then every sequence in $A$ has an upper bound in $A$, but $A$
has no greatest member.   It follows that if $B\subseteq K$ is countable
there is an $x\in A$ such that $f(x)=\sup_{y\in A}f(y)$ for every
$f\in B$, so that $f(x)=f(y)$ whenever $x\le y\in A$ and $f\in B$.

Let $\Cal I$ be the family of those sets $E\subseteq X$ such
that $A\cap E$ is bounded above in $A$.   Then $\Cal I$ is a $\sigma$-ideal
of subsets of $X$.
\Prf\ Of course $\emptyset\in\Cal I$.   If $\sequencen{E_n}$ is a sequence
in $\Cal I$ and $E\subseteq\bigcup_{n\in\Bbb N}E_n$,
there is for each $n\in\Bbb N$ an $x_n\in A$ which is an upper bound for
$E_n\cap A$.   Let $x\in A$ be an upper bound for $\{x_n:n\in\Bbb N\}$;
then $x$ is an upper bound for
$E\cap A$ in $A$.   So $E\in\Cal I$.\ \Qed

\medskip

{\bf (b)} For $G\in\frak T_s$ and $k\in\Bbb N$, let $W(G,k)$ be the set of
those $x\in X$ for which there are $f_0,\ldots,f_k\in K$
such that $\{y:y\in X$, $|f_i(y)-f_i(x)|\le 2^{-k}$ for every $i\le k\}$
is included in $G$.   Because $K$ spans $X^*$,
$G=\bigcup_{k\in\Bbb N}W(G,k)$.   So if $G\in\frak T_s\setminus\Cal I$
there is a $k\in\Bbb N$ such that $W(G,k)\notin\Cal I$.

\medskip

{\bf (c)} (The key.)  If $\sequencen{G_n}$ is any sequence in
$\frak T_s\setminus\Cal I$, then
$\bigcap_{n\in\Bbb N}G_n\notin\Cal I$.   \Prf\ Start from any $z^*\in A$.
For each $n\in\Bbb N$, take $k_n\in\Bbb N$ such that
$W(G_n,k_n)\notin\Cal I$.   For each $z\in A$ and $n\in\Bbb N$, choose
$w_{zn}\in A\cap W(G_n,k_n)$ such that
$w_{zn}\ge z$;  now choose a family
$\langle f_{nzi}\rangle_{z\in A,i\le k_n}$ in $K$ such that

\Centerline{$\{y:|f_{nzi}(y-w_{zn})|\le 2^{-k_n}$ for every $i\le k_n\}
\subseteq G_n$.}

\noindent Let $\Cal F$ be any ultrafilter on $A$ containing
$\{x:x\in A$, $x\ge z\}$ for every $z\in A$, and write
$f_{ni}=\lim_{z\to\Cal F}f_{nzi}$ for $n\in\Bbb N$ and $i\le k_n$, the
limit being taken for the weak* topology on $K$.   Let $z_1^*>z^*$ be
such that $z_1^*\in A$ and $f_{ni}(x-z_1^*)=0$ whenever $x\in A$,
$x\ge z_1^*$, $n\in\Bbb N$ and $i\le k_n$.

Choose sequences $\sequencen{x_n}$, $\sequencen{y_n}$ and
$\sequencen{z_n}$ in $A$ inductively, as follows.   Set $y_0=z_1^*$.
Given that $y_n\ge z_1^*$, the set

\Centerline{$C_n=\{z:z\in A,\,z\ge y_n,\,
|(f_{nzi}-f_{ni})(y_n-z_1^*)|\le 2^{-k_n}$ for every $i\le k_n\}$}

\noindent belongs to $\Cal F$, so is not empty;  take $z_n\in C_n$.
Because $y_n\ge z_1^*$, we have $f_{ni}(y_n-z_1^*)=0$ and
$|f_{nz_ni}(y_n-z_1^*)|\le 2^{-k_n}$, for every $i\le k_n$.
Set $x_n=w_{z_nn}$, so that $x_n\in A$, $x_n\ge z_n$ and
$\{y:|f_{nz_ni}(y-x_n)|\le 2^{-k_n}$ for every $i\le k_n\}$ is included in
$G_n$.   Now let $y_{n+1}\in A$ be such that
$y_{n+1}\ge x_n$ and $f_{nz_ni}(y-y_{n+1})=0$ whenever $y\in A$ and
$y\ge y_{n+1}$.   Of course

\Centerline{$y_{n+1}\ge x_n\ge z_n\ge y_n\ge z_1^*$.}

\noindent Continue.

At the end of the induction, let $z_2^*$ be an upper bound for
$\{y_n:n\in\Bbb N\}$ in $A$.   For $n\in\Bbb N$, set

\Centerline{$u_n=z_1^*+\sum_{j=0}^nx_j-y_j$,
\quad$v_n=u_n+z_2^*-y_{n+1}$.}

\noindent Then $\sequencen{u_n}$ is non-decreasing and
$u_n\le v_n\le z_2^*$ for every $n\in\Bbb N$;  moreover,

\Centerline{$v_n-v_{n+1}=y_{n+1}-x_{n+1}-y_{n+1}+y_{n+2}\ge 0$}

\noindent for every $n$, so $\sequencen{v_n}$ is non-increasing, and
$u_m\le v_n$ for all $m$, $n\in\Bbb N$.   Because $X$ has the
$\sigma$-interpolation property, there is an $x\in X$ such that
$u_n\le x\le v_n$ for every $n\in\Bbb N$.   Since
$z_1^*\le x\le z_2^*$, $x\in A$ and $x>z^*$.

Take any $n\in\Bbb N$.   Then

\Centerline{$0\le f_{nz_ni}(x-u_n)\le f_{nz_ni}(v_n-u_n)
=f_{nz_ni}(z_2^*-y_{n+1})=0$}

\noindent for every $i\le k_n$.   On the other hand,

\Centerline{$x_n-u_n=(y_n-z_1^*)-\sum_{j=0}^{n-1}(x_j-y_j)$}

\noindent lies between $0$ and $y_n-z_1^*$, so

\Centerline{$0\le f_{nz_ni}(x_n-u_n)\le 2^{-k_n}$}

\noindent for every $i\le k_n$, and $|f_{nz_ni}(x-x_n)|\le 2^{-k_n}$ for
every $i$.   Thus $x\in G_n$.   As $n$ is arbitrary,
$x\in\bigcap_{n\in\Bbb N}G_n$.

As $x>z^*$, this shows that $z^*$ is not an upper bound of
$\bigcap_{n\in\Bbb N}G_n\cap A$.   As $z^*$ is arbitrary,
$\bigcap_{n\in\Bbb N}G_n\notin\Cal I$.\ \Qed

\medskip

{\bf (d)} Set

\Centerline{$\Cal K_0=\{G\setminus H:G,\,H\in\frak T_s$, $G\notin\Cal I$,
$H\in\Cal I\}$.}

\noindent Then $\bigcap_{n\in\Bbb N}E_n\ne\emptyset$ for any sequence
$\sequencen{E_n}$ in $\Cal K_0$.   \Prf\ Express each $E_n$ as
$G_n\setminus H_n$ where $G_n$, $H_n$ are $\frak T_s$-open,
$G_n\notin\Cal I$ and $H_n\in\Cal I$.   Then
$\bigcup_{n\in\Bbb N}H_n\in\Cal I$,
as noted in (a), while $\bigcap_{n\in\Bbb N}G_n\notin\Cal I$, by (c);  so
$\bigcap_{n\in\Bbb N}E_n
=\bigcap_{n\in\Bbb N}G_n\setminus\bigcup_{n\in\Bbb N}H_n$ is
non-empty.\ \Qed

It follows that $\Cal K=\Cal K_0\cup\{\emptyset\}$ is a countably
compact class in the sense of 413M.   Moreover, $E\cap E'\in\Cal K$ for
all $E$, $E'\in\Cal K$ (using (a) and (c) again), so if we define
$\phi_0:\Cal K\to\{0,1\}$ by
writing $\phi_0(E)=1$ for $E\in\Cal K_0$ and $\phi_0(\emptyset)=0$, then
$\Cal K$ and $\phi_0$ will satisfy all the conditions of 413N.   There
is therefore a measure $\hat\mu$ on $X$ extending $\phi_0$ and inner
regular with respect to $\Cal K_{\delta}$, the family of sets
expressible as intersections of sequences in $\Cal K$.   The domain of
$\hat\mu$ must include every member of $\Cal K$;  but if $G\in\frak T_s$
then either $G$ or $X\setminus G$ belongs to $\Cal K_0$, so is
measured by $\hat\mu$, and $\hat\mu$ is a topological measure.

We need to observe that, because $\phi_0$ takes only the values $0$ and
$1$, $\hat\mu E\le 1$ for every $E\in\Cal K_{\delta}$, and $\hat\mu X\le
1$;  since $\phi_0X=1$, $\hat\mu X=1$ and $\hat\mu$ is a probability
measure.

\medskip

{\bf (e)} We may therefore take $\mu$ to be the restriction of $\hat\mu$
to the algebra $\Cal B$ of $\frak T_s$-Borel sets,
and $\mu$ is a $\frak T_s$-Borel probability
measure.   Now $\mu$ is not inner regular with respect to the
$\frak T_s$-closed
sets.   \Prf\ For each $\xi<\omega_1$, $p_{\xi}<p_{\xi+1}<p_{\xi+2}$, so
there are $g_{\xi}$, $h_{\xi}\in K$ such that
$g_{\xi}(p_{\xi})<g_{\xi}(p_{\xi+1})$ and
$h_{\xi}(p_{\xi+1})<h_{\xi}(p_{\xi+2})$.   Let $D\subseteq\omega_1$ be
any set such that $D$ and $\omega_1\setminus D$ are both uncountable,
and set

\Centerline{$G=\bigcup_{\xi\in D}\{x:g_{\xi}(p_{\xi})<g_{\xi}(x)$,
$h_{\xi}(x)<h_{\xi}(p_{\xi+2})\}$.}

\noindent Then $G\in\frak T_s$, and
$p_{\xi+1}\in G$ for every $\xi\in D$,
so $G\notin\Cal I$ and $\mu G=\phi_0G=1$.   On the other hand, if
$\eta\in\omega_1\setminus D$, then for every $\xi\in D$ either
$\xi<\eta$ and $h_{\xi}(p_{\xi+2})\le h_{\xi}(p_{\eta+1})$, or
$\xi>\eta$ and $g_{\xi}(p_{\eta+1})\le g_{\xi}(p_{\xi})$;  thus
$p_{\eta+1}\notin G$ for any $\eta\in\omega_1\setminus D$, and
$X\setminus G\notin\Cal I$.   But this means that if $F\subseteq G$ is
closed then $X\setminus F\in\Cal K_0$ and $\mu F=0$.   Thus
$\mu G>\sup_{F\subseteq G\text{ is closed}}\mu F$.\ \Qed

\medskip

{\bf (f)} Because $\frak T_s$ is regular, $\mu$ cannot
be $\tau$-additive, by 414Mb.   It follows at once that $(X,\frak T_s)$
is not Borel-measure-complete, and in particular is not a Radon space.
To see that $\mu$ has no extension to a
norm-Borel measure, we need to look again at the set $A$.
For each $\xi<\omega_1$, set $F_{\xi}=\{x:x\le p_{\xi}\}$.   Then every
$F_{\xi}$ is norm-closed (354Bc) and $\ofamily{\xi}{\omega_1}{F_{\xi}}$
is an increasing family with union $A$.   Consequently, $A$ is
norm-closed (4A2Ld, 4A2Ka).   At the same time, every $F_{\xi}$ is
convex (cf.\ 351Ce), so $A$ is also convex.   It follows that $A$, like
every $F_{\xi}$, is
$\frak T_s$-closed (3A5Ee).   So $\mu$ measures $A$ and every $F_{\xi}$.
Because $X\setminus F_{\xi}$ is a $\frak T_s$-open set not belonging to
$\Cal I$, $\mu F_{\xi}=0$, for every $\xi<\omega_1$;  because
$X\setminus A$ is a $\frak T_s$-open set belonging to $\Cal I$,
$\mu A=1$.

But $\omega_1$ is a measure-free cardinal (438Cd), so 438I tells us
that $\lambda A=\sup_{\xi<\omega_1}\lambda F_{\xi}$ for any
semi-finite norm-Borel measure $\lambda$ on $X$.   Thus $\mu$ has no
extension to a norm-Borel measure, and the proof is complete.
}%end of proof of 466H

\leader{466I}{Examples} The following spaces satisfy the hypotheses of
466H.

\spheader 466Ia \cmmnt{ ({\smc Talagrand 78a}, or {\smc Talagrand 84},
16-1-2)} $X=\ell^{\infty}(I)$ or
$\{x:x\in\ell^{\infty}(I),\,\{i:x(i)\ne 0\}$ is countable$\}$, where $I$
is uncountable.   \prooflet{\Prf\ $X$
has the $\sigma$-interpolation property because it is Dedekind complete,
and if $\ofamily{\xi}{\omega_1}{i_{\xi}}$ is any family of distinct
elements of $I$, we can set $p_{\xi}(i_{\eta})=1$ for $\eta\le\xi$,
$p_{\xi}(i)=0$ for all other $i\in I$ to obtain a strictly increasing
family $\ofamily{\xi}{\omega_1}{p_{\xi}}$ in $X$.\ \Qed}

\spheader 466Ib \cmmnt{ ({\smc de Maria \& Rodriguez-Salinas 91})}
$X=\ell^{\infty}/\pmb{c}_0$\cmmnt{, where $\pmb{c}_0$ is the space of
real sequences converging to $0$}.

\prooflet{\Prf\ {\bf (i)} To see that $X$ has the
$\sigma$-interpolation property, let $A$, $B\subseteq X$ be non-empty
countable sets such that $u\le v$ for all $u\in A$, $v\in B$.   Let
$\sequencen{x_n}$, $\sequencen{y_n}$ be sequences in $\ell^{\infty}$
such that $A=\{x_n^{\ssbullet}:n\in\Bbb N\}$ and
$B=\{y_n^{\ssbullet}:n\in\Bbb N\}$.   Set $\tilde x_n=\sup_{i\le n}x_i$,
$\tilde y_n=\inf_{i\le n}y_i$ for $n\in\Bbb N$;  then
$\tilde x_n^{\ssbullet}\le\tilde y_n^{\ssbullet}$, so
$(\tilde x_n-\tilde y_n)^+\in\pmb{c}_0$.   Set

\Centerline{$k_n
=\max(\{n\}\cup\{i:\tilde x_n(i)\ge\tilde y_n(i)+2^{-n}\})$}

\noindent for $n\in\Bbb N$, and define $x\in\ell^{\infty}$ by writing

$$\eqalign{x(i)&=0\text{ if }i\le k_0,\cr
&=\tilde x_n(i)\text{ if }k_n<i\le k_{n+1}.\cr}$$

\noindent Then it is easy to check that $u\le x^{\ssbullet}\le v$ for
every $u\in A$, $v\in B$;  as $A$ and $B$ are arbitrary, $X$ has the
$\sigma$-interpolation property.

\medskip

\quad{\bf (ii)} To see that $X$ has the other property, recall that
there is a family $\ofamily{\xi}{\omega_1}{I_{\xi}}$ of infinite subsets
of $\Bbb N$ such that $I_{\xi}\setminus I_{\eta}$ is finite if
$\eta\le\xi$, infinite if $\xi<\eta$ (4A1Fa).   Setting
$p_{\xi}=\chi(\Bbb N\setminus I_{\xi})^{\ssbullet}$, we have a strictly
increasing family $\ofamily{\xi}{\omega_1}{p_{\xi}}$ in $X$.\ \Qed
}%end of prooflet

\leader{466J}{Theorem} Let $X$ be a linear topological space and
$\Sigma$ its cylindrical $\sigma$-algebra.   If $\mu$ and $\nu$ are
probability measures with domain $\Sigma$ such that
$\int e^{if(x)}\mu(dx)=\int e^{if(x)}\nu(dx)$ for every
$f\in X^*$, then $\mu=\nu$.

\proof{ Define $T:X\to\BbbR^{X^*}$ by setting $(Tx)(f)=f(x)$ for
$f\in X^*$, $x\in X$.   Then $T$ is linear and continuous for the weak
topology of $X$.   So if $F\subseteq\BbbR^{X^*}$ is a Baire set for the
product topology of $\BbbR^{X^*}$, $T^{-1}[F]$ is a Baire set for the
weak topology of $X$ (4A3Kc) and belongs to $\Sigma$ (4A3V).   We
therefore have Baire measures $\mu'$, $\nuprime$ on $\BbbR^{X^*}$ defined by
saying that $\mu'F=\mu T^{-1}[F]$ and $\nuprime F=\nu T^{-1}[F]$ for every
Baire set $F\subseteq\BbbR^{X^*}$.

If $h:\BbbR^{X^*}\to\Bbb R$ is a continuous linear functional, it can be
expressed in the form $h(z)=\sum_{i=0}^n\alpha_iz(f_i)$ where
$f_0,\ldots,f_n\in X^*$ and $\alpha_0,\ldots,\alpha_n\in\Bbb R$.   So

\Centerline{$h(Tx)=\sum_{i=0}^n\alpha_i(Tx)(f_i)
=\sum_{i=0}^n\alpha_if_i(x)=f(x)$}

\noindent for every $x\in X$, where $f=\sum_{i=0}^n\alpha_if_i$.   This
means that

\Centerline{$\int e^{ih(z)}\mu'(dz)
=\int e^{ih(Tx)}\mu(dx)
=\int e^{if(x)}\mu(dx)
=\int e^{if(x)}\nu(dx)
=\int e^{ih(z)}\nuprime(dz)$.}

\noindent As $h$ is arbitrary, $\mu'=\nuprime$ (454Pa).

Now let $\Sigma'$ be the family of subsets of $X$ of the form
$T^{-1}[F]$ where $F\subseteq\BbbR^{X^*}$ is a Baire set.   This is a
$\sigma$-algebra and contains all sets of the form $\{x:f(x)\ge\alpha\}$
where $f\in X^*$ and $\alpha\in\Bbb R$.   So every member of $X^*$ is
$\Sigma'$-measurable and $\Sigma'$ must include the cylindrical
$\sigma$-algebra of $X$.   Since $\mu$ and $\nu$ agree on $\Sigma'$ they
must be identical.
}%end of proof of 466J

\leader{466K}{Proposition} If $X$ is a locally convex linear topological
space and $\mu$, $\nu$ are quasi-Radon probability measures on $X$
such that $\int e^{if(x)}\mu(dx)=\int e^{if(x)}\nu(dx)$ for every
$f\in X^*$, then $\mu=\nu$.

\proof{ Write $\frak T$ for the given topology on $X$ and
$\frak T_s=\frak T_s(X,X^*)$ for the weak topology.   By 466J, $\mu$
and $\nu$ must agree on the cylindrical
$\sigma$-algebra $\Sigma$ of $X$.   Since $\Sigma$ includes a base for
$\frak T_s$, every weakly open set $G$ is the union of an
upwards-directed family of open sets belonging to $\Sigma$;  as $\mu$
and $\nu$ are $\tau$-additive, $\mu G=\nu G$.   Consequently $\mu$ and
$\nu$ agree on $\frak T_s$-closed sets, and therefore on
$\frak T$-closed convex sets (4A4Ed).   Write $\Cal H$ for the family of
$\frak T$-open sets which are expressible as the union of a
non-decreasing sequence of $\frak T$-closed convex sets;  then $\mu$ and
$\nu$ agree on $\Cal H$.   If $\tau$ is a $\frak T$-continuous seminorm
on $X$, $x_0\in X$ and $\alpha>0$, then
$\{x:\tau(x-x_0)<\alpha\}\in\Cal H$;  and sets of this kind constitute a
base for $\frak T$ (4A4Cb).   Also the intersection of two members of
$\Cal H$ belongs to $\Cal H$.   By 415H(v), $\mu=\nu$.
}%end of proof of 466K

\cmmnt{\medskip

\noindent{\bf Remark} This generalizes 285M and 454Xl, which are the
special cases $X=\BbbR^r$ (for finite $r$) and $X=\BbbR^I$;  see also
445Xq.
}%end of comment

\leader{466L}{Proposition} Suppose that $X$ and $Y$ are Banach spaces
and that $T:X\to Y$ is a linear operator
such that $gT:X\to\Bbb R$ is universally Radon-measurable\cmmnt{, in
the sense of 434Ec,} for every $g\in Y^*$.   Then $T$ is continuous.

\proof{\Quer\ Suppose, if possible, otherwise.   Then there is a
$g\in Y^*$ such that $gT$ is
not continuous (4A4Ib).   For each $n\in\Bbb N$, take $x_n\in X$ such
that $\|x_n\|=2^{-n}$ and $g(Tx_n)>2n$.
Define $h:\{0,1\}^{\Bbb N}\to X$ by setting
$h(t)=\sum_{n=0}^{\infty}t(n)x_n$ (4A4Ie).   Then $h$ is continuous,
because $\|h(t)-h(t')\|\le\sum_{n=0}^{\infty}2^{-n}|t(n)-t'(n)|$ for all
$t$, $t'\in\{0,1\}^{\Bbb N}$.   Let $\nu$ be the usual measure on
$\{0,1\}^{\Bbb N}$;  then the image measure $\mu=\nu h^{-1}$ is a Radon
measure on $X$ (418I), so $gT$ must be
$\dom\mu$-measurable, and $\phi=gTh$ is $\dom\nu$-measurable.   In this
case there is an $m\in\Bbb N$ such that
$E=\{t:|\phi(t)|\le m\}$ has measure greater than $\bover12$.   But as
$g(Tx_m)>2m$, we see that if $t\in E$ then $t'\notin E$, where $t'$
differs from $t$ at the $m$th coordinate only, so that
$|\phi(t)-\phi(t')|=g(Tx_m)$.   Since the map $t\mapsto t'$ is an
automorphism of the measure space $(\{0,1\}^{\Bbb N},\nu)$,
$\nu E\le\bover12$, which is impossible.\ \Bang
}%end of proof of 466L

\leader{466M}{Corollary} If $X$ is a Banach space, $Y$ is a separable
Banach space, and $T:X\to Y$ is a
linear operator such that the graph of $T$ is a Souslin-F set in
$X\times Y$, then $T$ is continuous.

\proof{ It will be enough to show that $T\restr Z$ is continuous for every
separable closed linear subspace $Z$ of $X$ (because then it must be
sequentially continuous, and we can use 4A2Ld).   Write
$\Gamma\subseteq X\times Y$ for the graph of $T$.   If $H\subseteq Y$
is open, then $\Gamma\cap(Z\times H)$ is a Souslin-F set in the
Polish space $Z\times Y$, so is analytic (423Eb), and its projection
$(T\restr Z)^{-1}[H]$ also is analytic (423Bb), therefore
universally measurable (434Dc).   Thus $T\restr Z:Z\to Y$ is a
universally measurable function, and $gT\restr Z$ must be universally
measurable for any $g\in Y^*$ (434Df).   By 466L, $T\restr Z$ is
continuous;  as $Z$ is arbitrary, $T$ is continuous.
}%end of proof of 466M.

%even if $T$ has closed graph, we do need both $X$ and $Y$ to be
%complete

\leader{466N}{Gaussian \dvrocolon{measures}}\dvAnew{2010}\cmmnt{ Some of
the ideas of \S456 can be adapted to the present context, as follows.

\medskip

\noindent}{\bf Definition} If $X$ is a linear topological space,
\cmmnt{I will say that}
a probability measure $\mu$ on $X$ is a {\bf centered Gaussian measure} if
its domain includes the cylindrical $\sigma$-algebra of $X$ and
every continuous linear functional on $X$
is either zero almost everywhere or a normal random
variable with zero expectation.   \cmmnt{(Thus
a `centered Gaussian distribution' on $\BbbR^I$, as defined in 456A,
is a distribution in the sense of 454K which is a centered Gaussian
measure when $\BbbR^I$ is thought of as a linear topological space.)}

\cmmnt{Warning!  many authors reserve the phrase `Gaussian measure' for
strictly positive measures.}

\leader{466O}{Proposition}\dvAnew{2010} Let $X$ be a separable Banach
space, and $\mu$ a
probability measure on $X$.   Suppose that there is a linear subspace $W$
of $X^*$, separating the points of $X$, such that every element of $W$ is
$\dom\mu$-measurable and either zero a.e.\ or a normal random
variable with zero expectation.
Then $\mu$ is a centered Gaussian measure with respect to
the norm topology of $X$.

\proof{{\bf (a)} As $W$ separates the points of $X$,
$X\setminus\{0\}=\bigcup_{f\in W}\{x:f(x)\ne 0\}$.
Because $X$ is Polish, therefore hereditarily Lindel\"of, there is a
countable set $I\subseteq W$ still separating the points of $X$.
Let $W_0$ be the linear subspace of $X^*$ generated by $I$.

Define $T:X\to\BbbR^I$ by setting $(Tx)(f)=f(x)$ for
$f\in I$.   Then $T$ is an injective linear operator, and is continuous for
$\frak T_s(X,W_0)$ and the usual topology of $\BbbR^I$.
Let $\lambda$ be the distribution of the family $\family{f}{I}{f}$;
$T$ is \imp\ for $\hat\mu$ and $\lambda$, where $\hat\mu$ is
the completion  of $\mu$ (454J(iv)).
If $g:\BbbR^I\to\Bbb R$ is a continuous linear functional, then
$gT\in W_0$ (use 4A4Be);  now
the distribution of $g$, with respect to the probability measure
$\lambda$, is
just the distribution of $gT$ with respect to $\hat\mu$ and $\mu$,
and is therefore
either normal or the Dirac measure concentrated at $0$.   So $\lambda$ is a
centered Gaussian distribution in the sense of 456Ab.   Because $I$ is
countable, $\lambda$ is a Radon measure (454J(iii)).

\medskip

{\bf (b)} If $\epsilon>0$, there is a norm-compact $K\subseteq X$ such that
$\hat\mu K$ is defined and is at least $1-\epsilon$.
\Prf\ As $X$ and $\BbbR^I$ are analytic (423B), there is a Radon
measure $\mu'$ on $X$ such that $\lambda=\mu'T^{-1}$ (432G).
Of course $\mu'X=\lambda\BbbR^I=1$.
There is a compact set $K\subseteq X$ such that
$\mu'K\ge 1-\epsilon$;  now $T[K]$ is compact and $K=T^{-1}[T[K]]$, so

\Centerline{$\hat\mu K=\mu T^{-1}[T[K]]=\lambda T[K]=\mu'K\ge 1-\epsilon$.
\Qed}

\medskip

{\bf (c)} Now suppose that $g\in X^*$.   Then there is a sequence
$\sequencen{g_n}$ in $W_0$ such that $\sequencen{g_n}\to g\,\,\mu$-a.e.
\Prf\ For each $n\in\Bbb N$, there is a compact set $K_n\subseteq X$ such
that $\hat\mu K_n\ge 1-2^{-n-1}$;  we can suppose that
$K_{n+1}\supseteq K_n$ for each $n$.
$W_0$ is dense in $X^*$ for the weak*-topology $\frak T_s(X^*,X)$ (4A4Eh);
being convex, it is dense for the Mackey topology $\frak T_k(X^*,X)$
(4A4F), and there is a $g_n\in W_0$ such that
$\sup_{x\in K_n}|g_n(x)-g(x)|\le 2^{-n}$.   Now
$g(x)=\lim_{n\to\infty}g_n(x)$ for every $x$ in the $\mu$-conegligible set
$\bigcup_{n\in\Bbb N}K_n$.\ \Qed

Set $\sigma_n=\sqrt{\Var(g_n)}$ for each $n$.   Then
$\{\sigma_n:n\in\Bbb N\}$ is bounded.   \Prf\
Set $M=\sup_{x\in K_0}|g(x)|$.   If $n\in\Bbb N$ and $\sigma_n\ne 0$,
$|g_n(x)|\le M+1$ for every $x\in K_0$, and

$$\eqalign{\Bover12
&\le\mu K_0
\le\Pr(|g_n|\le M+1)
=\Bover1{\sigma_n\sqrt{2\pi}}\int_{-M-1}^{M+1}e^{-t^2/2\sigma_n^2}dt
\le\Bover{2(M+1)}{\sigma_n\sqrt{2\pi}},\cr}$$

\noindent so $\sigma_n\le\Bover{4(M+1)}{\sqrt{2\pi}}$.\ \Qed

We therefore have a strictly increasing
sequence $\sequence{k}{n_k}$ such that
$\sigma=\lim_{k\to\infty}\sigma_{n_k}$
is defined in $\coint{0,\infty}$.   For each $k$, let $\nu_k$ be the
distribution of $g_{n_k}$ and $\varphi_k$ its characteristic function;
let $\nu$, $\varphi$ be the distribution and characteristic function of
$g$.   Since $\sequence{k}{g_{n_k}}\to g$ a.e.,

\Centerline{$\int h\,d\nu
=\int hg\,d\mu=\lim_{k\to\infty}\int hg_{n_k}d\mu
=\lim_{k\to\infty}\int h\,d\nu_k$}

\noindent for every bounded continuous function $h:\Bbb R\to\Bbb R$,
and $\varphi(t)=\lim_{k\to\infty}\varphi_k(t)$ for every $t\in\Bbb R$, by
285L.   But, for each $k$,

\Centerline{$\varphi_k(t)
=\exp(-\Bover12\sigma_{n_k}^2t^2)$}

\noindent by 285E if $\sigma_{n_k}>0$ and by direct calculation if
$\sigma_{n_k}=0$, as then $g_{n_k}=0$ almost everywhere.

Accordingly $\varphi(t)=\exp(-\Bover12\sigma^2t^2)$ for every $t$.   But this
means that $\nu$ is either the Dirac measure concentrated at $0$ (if
$\sigma=0$) or a normal distribution with zero expectation (if $\sigma>0$).

\medskip

{\bf (d)} As $g$ is arbitrary, $\mu$ is a centered Gaussian measure.
}%end of proof of 466O


\exercises{\leader{466X}{Basic exercises (a)}
%\spheader 466Xa
Let $(X,\frak T)$ be a metrizable locally convex linear topological
space and $\mu$ a totally finite measure on $X$ which is quasi-Radon for
the topology $\frak T$.   Show that $\mu$ is quasi-Radon for the weak
topology $\frak T_s(X,X^*)$.
%466A

\sqheader 466Xb Let $\sequencen{e_n}$ be the usual orthonormal basis of
$\ell^2$.   Give $\ell^2$ the Radon probability measure $\nu$ such that
$\nu\{e_n\}=2^{-n-1}$ for every $n$.   Let $I$ be an uncountable set,
and set $X=(\ell^2)^I$ with the product linear structure and the product
topology $\frak T$, each copy of $\ell^2$ being given its norm topology.
(i) Let $\lambda$ be the $\tau$-additive product of copies of
$\nu$ (417F).   Show that $\lambda$ is quasi-Radon for $\frak T$ but is
not inner regular with respect to the $\frak T_s$-closed sets.
(ii) Write $\frak T_s$ for the weak topology of $X$.
Let $\lambda_s$ be the $\tau$-additive product measure of copies of
$\nu$ when each copy of $\ell^2$ is given its weak topology instead of
its norm topology.   Show that $\lambda_s$ is quasi-Radon for $\frak T_s$
but does not measure every $\frak T$-Borel set.   \Hint{setting
$E=\{e_n:n\in\Bbb N\}$,
$\lambda(E^I)=1$ and $E^I$ is relatively $\frak T_s$-compact.}
%466A, 466Xa

\sqheader 466Xc Let $X$ be a metrizable locally convex linear
topological space and $\mu$ a $\tau$-additive totally finite measure on
$X$ with domain the cylindrical $\sigma$-algebra of $X$.   Show that
$\mu$ has an extension to a quasi-Radon measure on $X$.   \Hint{4A3V,
415N.}
%466A

\spheader 466Xd Let $X$ be a metrizable locally convex linear
topological space which is Lindel\"of in its weak topology, and $\Sigma$
the cylindrical $\sigma$-algebra of $X$.   Show that any totally finite
measure with domain $\Sigma$ has an extension to a quasi-Radon measure
on $X$.
%466A

\sqheader 466Xe Let $X$ be a separable Banach space.   Show that it is a
Radon space when given its weak topology.
%466A 466Xd
%is there a ZFC example of weight \omega_1?

\spheader 466Xf Let $K$ be a compact metrizable space, and $C(K)$ the
Banach space of continuous real-valued functions on $K$.   Show that the
$\sigma$-algebra of subsets of $C(K)$ generated by the functionals
$x\mapsto x(t):C(K)\to\Bbb R$, for $t\in K$, is just the cylindrical
$\sigma$-algebra of $C(K)$.   \Hint{4A2Pe.}   Examine the connexions
between this and 454Sa, 462Z and 466Xd.
%466E

\spheader 466Xg Let $K$ be a scattered compact Hausdorff space.   Show that
the weak topology and the topology of pointwise convergence on $C(K)$ have
the same Borel $\sigma$-algebras.
%466E 466Xf

\spheader 466Xh Re-write part (d) of the proof of 466H to avoid any
appeal to results from \S413.
%466H

\spheader 466Xi Let $X$ be a locally convex linear topological space and
$\mu$, $\nu$ two totally finite quasi-Radon measures on $X$.   Show that
if $\mu$ and $\nu$ give the same measure to every half-space
$\{x:f(x)\ge\alpha\}$, where $f\in X^*$ and $\alpha\in\Bbb R$, then
$\mu=\nu$.
%466K

\spheader 466Xj Let $X$ be a Hilbert space and $\mu$, $\nu$ two totally
finite Radon measures on $X$.   Show that if $\mu$ and $\nu$ give the
same measure to every ball $B(x,\delta)$, where $x\in X$ and
$\delta\ge 0$, then $\mu=\nu$.   \Hint{every open half-space is the
union of a non-decreasing sequence of balls.}
%466Xi, 466K

\spheader 466Xk Let $f:\Bbb R\to\Bbb R$ be a function such that $f(1)=1$
and $f(x+y)=f(x)+f(y)$ for all $x$, $y\in\Bbb R$.    Show that the
following are equiveridical:  (i) $f(x)=x$ for every $x\in\Bbb R$;  (ii)
$f$ is continuous at some point;  (iii) $f$ is bounded on some non-empty
open set;  (iv) $f$ is bounded on some Lebesgue measurable set of
non-zero measure;  (v) $f$ is Lebesgue measurable;  (vi) $f$ is Borel
measurable;  *(vii) $f$ is bounded on some non-meager G$_{\delta}$
set;  *(viii) $f$ is $\widehat{\Cal B}$-measurable, where
$\widehat{\Cal B}$ is the Baire-property algebra of $\Bbb R$.
\Hint{443Db.}
%466L

\spheader 466Xl Set $X=\{x:x\in\BbbR^{\Bbb N},\,\{n:x(n)\ne 0\}$ is
finite$\}$, and give $X$ any norm.   Show that any linear operator from
$X$ to any normed space is universally measurable.
%466L

\spheader 466Xm\dvAnew{2010}
Let $X$ be a linear topological space and $\mu$ a
centered Gaussian measure on $X$.
(i) Let $Y$ be another linear topological space and $T:X\to Y$ a
continuous linear operator.
Show that the image measure $\mu T^{-1}$ is a centered Gaussian measure on
$Y$.
(ii) Show that $X^*\subseteq\eusm L^2(\mu)$.
(iii) Let us say that the {\bf covariance matrix} of
$\mu$ is the family $\langle\sigma_{fg}\rangle_{f,g\in X^*}$, where
$\sigma_{fg}=\int f\times g\,d\mu$ for $f$, $g\in X^*$.   Suppose that
$\nu$ is
another centered Gaussian measure on $X$ with the same covariance matrix.
Show that $\mu$ and $\nu$ agree on the cylindrical $\sigma$-algebra of $X$.
%466N

\spheader 466Xn\dvAnew{2010}
Let $\familyiI{X_i}$ be a family of linear topological
spaces with product $X$.   Suppose that for each $i$ we have a centered
Gaussian measure $\mu_i$ on $X_i$.   Show that the product probability
measure $\prod_{i\in I}\mu_i$ is a centered Gaussian measure on $X$.
%466Xm 466N

\spheader 466Xo\dvAnew{2010}
Let $X$ be a linear topological space.   Show that the convolution of
two quasi-Radon centered Gaussian measures on $X$ is a centered Gaussian
measure.
%466Xn 466N

\spheader 466Xp\dvAnew{2010}
Let $X$ be a separable Banach space, and $\mu$ a
complete measure on $X$.   Show that the following are
equiveridical:  (i) $\mu$ is a centered Gaussian measure on $X$;
(ii) $\mu$ extends a centered Gaussian Radon measure on $X$;
(iii) there are a set $I$, an injective continuous linear operator
$T:X\to\BbbR^I$ and a centered Gaussian distribution $\lambda$ on
$\BbbR^I$ such that $T$ is \imp\ for $\mu$ and $\lambda$;
(iv) whenever $I$ is a set and $T:X\to\BbbR^I$ is a continuous linear
operator there is a centered Gaussian distribution $\lambda$ on
$\BbbR^I$ such that $T$ is \imp\ for $\mu$ and $\lambda$.
%466O

\spheader 466Xq Let $X$ be a Banach space, and $\mu$ a Radon measure on
$X$.   Show that, with respect to $\mu$, the unit ball of $X^*$ is a
stable set of functions in the sense of \S465.   \Hint{465Yb.}
%466Z out of order query

\sqheader 466Xr Let $I$ be an infinite set.
Show that Talagrand's measure, interpreted as a measure
on $\ell^{\infty}(I)$ (464R), is not $\tau$-additive for the weak
topology.
%466Z out of order query

\leader{466Y}{Further exercises (a)}
%\spheader 466Ya
Give an example of a Hausdorff locally convex linear topological space
$(X,\frak T)$ with a probability measure $\mu$ on $X$ which is a Radon
measure for the weak topology $\frak T_s(X,X^*)$ but not for the
topology $\frak T$.   \Hint{take $C=C([0,1])$ and $X=C^*$ with the
Mackey topology for the dual pair $(X,C)$, that is, the topology of
uniform convergence on weakly compact subsets of $C$.}
%466A

\spheader 466Yb\dvAnew{2011}
Let $X$ be a normed space and $\frak T$ a linear space topology on $X$
such that the unit ball of $X$ is $\frak T$-closed
and the topology on the unit sphere $S$ induced by $\frak T$ is finer
than the norm topology on $S$.
(i) Show that every norm-Borel subset of $X$
is $\frak T$-Borel.   (ii) Show that if $\frak T$ is coarser than the norm
topology, then it has a $\sigma$-isolated network.
%466E 466D

\spheader 466Yc\dvAnew{2011}
(i) Let $X$ be a Banach space.   Set
$S_2=\bigcup_{n\in\Bbb N}\{0,1\}^n$ and
suppose that $\family{\sigma}{S_2}{K_{\sigma}}$ is a family of
non-empty weakly compact
convex subsets of $X$ such that $K_{\sigma}\subseteq K_{\tau}$ whenever
$\sigma$, $\tau\in S_2$ and $\sigma$ extends $\tau$.   ($\alpha$) Show that
there is a weakly Radon
probability measure on $X$ giving measure at least $2^{-n}$ to $K_{\sigma}$
whenever $n\in\Bbb N$ and $\sigma\in\{0,1\}^n$.   ($\beta$) Show that
there are a $\sigma\in S_2$ and
$x\in K_{\sigma^{\smallfrown}\fraction{0}}$,
$y\in K_{\sigma^{\smallfrown}\fraction{1}}$ such that $\|x-y\|\le 1$.
(ii) Let $X$ be a locally convex Hausdorff linear topological space.
and $\family{\sigma}{S_2}{A_{\sigma}}$ a family of
non-empty relatively weakly
compact subsets of $X$ such that $A_{\sigma}\subseteq A_{\tau}$ whenever
$\sigma$, $\tau\in S_2$ and $\sigma$ extends $\tau$.
For $\sigma\in S_2$, set
$C_{\sigma}=A_{\sigma^{\smallfrown}\fraction{1}}
  -A_{\sigma^{\smallfrown}\fraction{0}}$.   Show that
$0\in\overline{\bigcup_{\sigma\in S_2}C_{\sigma}}$.
%466A 432Yb

\spheader 466Yd Find Banach spaces $X$ and $Y$ and a linear operator
from $X$ to $Y$ which is not continuous but whose graph is an
F$_{\sigma}$ set in $X\times Y$.
%466M  %mt46bits

\spheader 466Ye\dvAnew{2010}
Let $X$ be a complete Hausdorff locally convex linear topological
space and $\mu$ a Radon probability measure on $X$.   Suppose that there is
a linear subspace $W$ of $X^*$, separating the points of $X$, such that
every member of $W$ is either zero a.e.\ or a normal random variable with
zero expectation.   Show that $\mu$ is a centered Gaussian measure.
%466O mt46bits


\discrversionA{\inset{\spheader 466Y?\dvAnew{2009}
Let $I^{\|}$ be the split interval with its usual compact
Hausdorff topology.   Show that the weak topology and the topology of
pointwise convergence on $C(I^{\|})$ have the same Borel $\sigma$-algebras.
({\smc Mohammed \& Saadi p09}\#.)
%466Xg 466E Mohammed \& Saadi p09#
\leaveitout{there is a countable family $\Cal V$ of $\frak T_p$-closed
sets such
that $\{U\cap V:U$ is $\frak T_p$-open, $V\in\Cal V\}$ is a
$\frak T_s$-network.
}}}{}
}%end of exercises

\allowmorestretch{468}{
\leader{466Z}{Problems (a)} Does every probability measure defined on
the $\frak T_s(\ell^{\infty},(\ell^{\infty})^*)$-Borel sets of
$\ell^{\infty}\cmmnt{\mskip5mu =\ell^{\infty}(\Bbb N)}$ extend to a
measure defined on the $\|\,\|_{\infty}$-Borel sets?
}%end of allowmorestretch

\cmmnt{It is by no means obvious that the Borel sets of $\ell^{\infty}$ are
different for the weak and norm topologies;  for a proof see
{\smc Talagrand 78b}.}

\spheader 466Zb Assume that $\frak c$ is measure-free.   Does it
follow that $\ell^{\infty}$, with its weak topology, is a Radon space?

\cmmnt{Note that a positive answer to 464Z (with $I=\Bbb N$) would
settle this, since Talagrand's measure, when interpreted as a measure
on $\ell^{\infty}$, cannot agree on the weakly Borel sets with any
Radon measure on $\ell^{\infty}$ (466Xq, 466Xr).}

\endnotes{
\Notesheader{466}
I have given a proof of 466A using the machinery of \S463;  when the
measure $\mu$ is known to be a Radon measure for the weak topology,
rather than just a
quasi-Radon measure or a $\tau$-additive measure on the cylindrical
algebra (466Xc), the theorem is older than this, and for an instructive
alternative approach see {\smc Talagrand 84}, 12-1-4.   Another proof is
in {\smc Jayne \& Rogers 95}.

On any Banach space we have at least three important $\sigma$-algebras:
the norm-Borel $\sigma$-algebra (generated by the norm-open sets), the
weak-Borel
algebra (generated by the weakly open sets) and the cylindrical algebra
(generated by the continuous linear functionals).   (Note
that the Baire
$\sigma$-algebras corresponding to the norm and weak topologies are the
norm-Borel algebra (4A3Kb) and the cylindrical algebra (4A3V).)   If our
Banach space is naturally represented as a subspace of some $\BbbR^I$
(e.g., because it is a space of continuous functions), then we have in
addition the $\sigma$-algebra generated by the functionals
$x\mapsto x(i)$ for $i\in I$, and the Borel algebra for the topology of
pointwise convergence.
We correspondingly have natural questions concerning when these algebras
coincide, as in 466E and 4A3W and 466Xf, and when a measure on one of
the algebras leads to a measure on another, as in 466A-466B and 466Xc.

The question of which Banach spaces are Radon spaces in their norm
topologies is, if not exactly `solved', at least reducible to a
classical problem in set theory by the results in \S438.   It seems much
harder to decide which
non-separable Banach spaces are Radon spaces in their weak topologies.
We have a simple positive result for spaces with Kadec norms (466F), and
after some labour a negative result for a couple of standard examples
(466I), but no effective general criterion is known.
Even the case of $\ell^{\infty}$ seems still to be open in
`ordinary' set theories (466Zb).   $\ell^{\infty}$ is of particular
importance in this context because the dual of any separable Banach
space is isometrically isomorphic to a linear subspace of
$\ell^{\infty}$ (4A4Id).   So a positive answer to either question in
466Z would have very interesting consequences -- and would be
correspondingly surprising.

466L and 466M belong to a large family of results of the general form:
if, between spaces with both topological and algebraic structures, we
have a homomorphism (for the algebraic structures) which is not
continuous, then it is wildly irregular.   I hope to return to some of
these ideas in Volume 5.   For the moment I just give a version of the
classical result that an additive function $f:\Bbb R\to\Bbb R$ which is
Lebesgue measurable must be continuous (466Xk).   The definition of
`universally measurable' function which I gave in \S434 has a number of
paradoxical aspects.   I have already remarked that in some contexts we
might prefer to use the notion of `universally
Radon-measurable' function;  this is also appropriate for 466L.   But
when our space $X$, for any reason, has few Borel
measures, as in 466Xl, there are correspondingly many universally
measurable functions defined on $X$.   Of course 466M can also be
thought of as a generalization of the closed graph theorem;  but note
that, unlike the closed graph theorem, it needs a separable codomain
(466Yd).

The point of 466O is that the most familiar separable Banach spaces are
presented with continuous linear embeddings into $\BbbR^{\Bbb N}$, and of
course any separable Banach space $X$ has such a presentation.   We can now
describe the centered Gaussian Radon measures on $X$ in terms of
centered Gaussian distributions on $\BbbR^{\Bbb N}$, as in 466Xp.   But
perhaps the most important centered Gaussian measure is Wiener measure
(477Yj), which is not in fact on a Banach space.

A curious geometric question concerning measures on metric spaces is the
following.   If two totally finite Radon measures on a metric space
agree on balls, must they be identical?   It is known that (even for
compact spaces) the answer, in general, is `no' ({\smc Davies 71});
in Hilbert spaces the answer is `yes' (466Xj);   and in fact the same is
true in any normed space ({\smc Preiss \& Ti\v{s}er 91}).


}%end of notes

\discrpage

\frfilename{mt467.tex}
\versiondate{13.1.10}
\copyrightdate{2000}

\def\chaptername{Pointwise compact sets of measurable functions}
\def\sectionname{Locally uniformly rotund norms}

\def\normY#1{\|#1\|_{\lower1pt\hbox{\fivei Y}}}

%$\normY{y}$\quad  $q_{\normY{\,}}$ \quad  $q_{\normY{\,}}$

\newsection{*467}

In the last section I mentioned Kadec norms.   These are interesting in
themselves, but the reason for including them in this book is that in a
normed space with a Kadec norm the weak topology has the same Borel sets
as the norm topology\cmmnt{ (466Ea)}.
The same will evidently be true of any space which has an equivalent
Kadec norm.   Now Kadec norms
themselves are not uncommon, but equivalent Kadec norms appear in a
striking variety of cases.   Here I describe the principal class of spaces
(the `weakly K-countably determined' Banach spaces, 467H) which have
equivalent
Kadec norms.   In fact they have `locally uniformly rotund' norms, which
are much easier to do calculations with.

Almost everything here is pure functional analysis, mostly taken from
{\smc Deville Godefroy \& Zizler 93}, which is why I have starred the
section.   The word `measure' does not appear until 467P.   At that
point, however, we find ourselves with a striking result
(Schachermayer's theorem) which appears to need the structure theory of
weakly compactly generated Banach spaces developed in
467C-467M. %467C 467D 467E 467F 467G 467H 467I 467J 467K 467L 467M

\leader{467A}{Definition} Let $X$ be a linear space with a norm
$\|\,\|$.   $\|\,\|$ is {\bf locally uniformly rotund} or {\bf locally
uniformly convex} if whenever $\|x\|=1$ and $\epsilon>0$, there is a
$\delta>0$ such that
$\|x-y\|\le\epsilon$ whenever $\|y\|=1$ and $\|x+y\|\ge 2-\delta$.

If $X$ has a locally uniformly rotund norm, then every subspace of $X$
has a locally uniformly rotund norm.   \cmmnt{Of course any
uniformly convex norm (definition:  2A4K)
is locally uniformly rotund.}

\leader{467B}{Proposition} A locally uniformly rotund norm is a Kadec
norm.

\proof{ Let $X$ be a linear space with a locally uniformly rotund norm
$\|\,\|$.   Set $S_X=\{x:\|x\|=1\}$.   Suppose that $G$ is open for the
norm topology and that $x\in G\cap S_X$.   Then there is an $\epsilon>0$
such that $G\supseteq B(x,\epsilon)=\{y:\|y-x\|\le\epsilon\}$.   Let
$\delta>0$ be such that
$\|x-y\|\le\epsilon$ whenever $\|y\|=1$ and $\|x+y\|\ge 2-\delta$.
Now there is an
$f\in X^*$ such that $f(x)=\|f\|=1$ (3A5Ac).   So
$V=\{y:f(y)>1-\delta\}$ is open for the weak topology.   But if
$y\in V\cap S_X$, then
$\|x+y\|\ge f(x+y)\ge 2-\delta$, so $\|x-y\|\le\epsilon$ and $y\in G$.
As $x$ is arbitrary, $G\cap S_X$ is open for the weak topology on $S_X$;
as $G$ is arbitrary, the norm and weak topologies agree on $S_X$.
}%end of proof of 467B

\leader{467C}{A technical device (a)} I will use the following
notation for the rest of the section.   Let $X$ be a linear space and
$p:X\to\coint{0,\infty}$ a seminorm.   Define
$q_p:X\times X\to\coint{0,\infty}$ by setting

\Centerline{$q_p(x,y)=2p(x)^2+2p(y)^2-p(x+y)^2
=(p(x)-p(y))^2+(p(x)+p(y))^2-p(x+y)^2$}

\noindent for $x\in X$.

\spheader 467Cb A norm $\|\,\|$ on $X$ is locally uniformly rotund iff
whenever $x\in X$ and $\epsilon>0$ there is a $\delta>0$ such that
$\|x-y\|\le\epsilon$ whenever $q_{\|\,\|}(x,y)\le\delta$.

\prooflet{\medskip

\Prf{\bf (i)} Suppose that $\|\,\|$ is locally uniformly rotund,
$x\in X$ and $\epsilon>0$.   ($\alpha$) If $x=0$ then
$q_{\|\,\|}(x,y)=\|y\|^2=\|x-y\|^2$ for every $y$ so we can take
$\delta=\epsilon^2$.   ($\beta$) If $x\ne 0$ set $x'=\Bover1{\|x\|}x$.
Let $\eta>0$ be such that
$\|x'-y'\|\le\bover12\epsilon\|x\|$ whenever $\|y'\|=1$ and
$\|x'+y'\|\ge 2-\eta$.   Let $\delta>0$ be such that

\Centerline{$\delta+2\sqrt{\delta}\|x\|\le\eta\|x\|^2$,
\quad$\sqrt{\delta}<\|x\|$,
\quad$\sqrt{\delta}\le\Bover12\epsilon\|x\|^2$.}

\noindent Now if $q_{\|\,\|}(x,y)\le\delta$, we must have

\Centerline{$(\|x\|-\|y\|)^2\le\delta<\|x\|^2$,
\quad$(\|x\|+\|y\|)^2-\|x+y\|^2\le\delta$,}

\noindent so that $y\ne 0$ and

\Centerline{$\bigl|\Bover1{\|x\|}-\Bover1{\|y\|}\bigr|\|y\|
=\Bover{|\|y\|-\|x\||}{\|x\|}
\le\Bover{\sqrt{\delta}}{\|x\|}$,}

\Centerline{$\|x\|+\|y\|-\|x+y\|
\le\Bover{\delta}{\|x\|+\|y\|+\|x+y\|}\le\Bover{\delta}{\|x\|}$.}

\noindent Set $y'=\Bover1{\|y\|}y$, $y''=\Bover1{\|x\|}y$.   Then
$\|y'\|=1$, and

\Centerline{$\|y'-y''\|
=\big|\Bover1{\|y\|}-\Bover1{\|x\|}\bigr|\|y\|
\le\Bover{\sqrt{\delta}}{\|x\|}\le\Bover12\epsilon\|x\|$.}

\noindent Accordingly

$$\eqalign{\|x'+y'\|
&\ge\|x'+y''\|-\|y'-y''\|
\ge\Bover1{\|x\|}\|x+y\|-\Bover{\sqrt{\delta}}{\|x\|}\cr
&\ge\Bover1{\|x\|}(\|x\|+\|y\|-\Bover{\delta}{\|x\|})
   -\Bover{\sqrt{\delta}}{\|x\|}
=1+\Bover{\|y\|}{\|x\|}-\Bover{\delta}{\|x\|^2}
   -\Bover{\sqrt{\delta}}{\|x\|}\cr
&\ge 1+\Bover{\|x\|-\sqrt{\delta}}{\|x\|}-\Bover{\delta}{\|x\|^2}
   -\Bover{\sqrt{\delta}}{\|x\|}
=2-\Bover{\delta}{\|x\|^2}
   -\Bover{2\sqrt{\delta}}{\|x\|}
\ge 2-\eta.\cr}$$

\noindent But this means that $\|x'-y'\|\le\bover12\epsilon\|x\|$, so
that $\|x'-y''\|\le\epsilon\|x\|$ and $\|x-y\|\le\epsilon$.   As $x$ and
$\epsilon$ are arbitrary, the condition is satisfied.

\medskip

\quad{\bf (ii)} Suppose the condition is satisfied.   If $\|x\|=1$ and
$\epsilon>0$, take $\delta\in\ocint{0,2}$ such that $\|x-y\|\le\epsilon$
whenever $q(x,y)\le 4\delta$;  then if $\|y\|=1$ and
$\|x+y\|\ge 2-\delta$,
$q(x,y)=4-\|x+y\|^2\le 4\delta$ and $\|x-y\|\le\epsilon$.   As $x$ and
$\epsilon$ are arbitrary, $\|\,\|$ is locally uniformly rotund.\
\Qed}

\spheader 467Cc\cmmnt{ We have the following elementary facts.}   Let
$X$ be a linear space.

\medskip

\quad{\bf (i)} For any seminorm $p$ on $X$,
$q_p(x,y)\ge(p(x)-p(y))^2\ge 0$ for all $x$,
$y\in X$.   \prooflet{\Prf\ $(p(x)+p(y))^2-p(x+y)^2\ge 0$
because $p(x+y)\le p(x)+p(y)$.\ \Qed}

\medskip

\quad{\bf (ii)}  Suppose that $\familyiI{p_i}$ is a family of seminorms
on $X$ such that $\sum_{i\in I}p_i(x)^2$ is finite for every $x\in X$.
Set $p(x)=\sqrt{\sum_{i\in I}p_i(x)^2}$ for $x\in X$;  then $p$ is a
seminorm on $X$ and $q_p=\sum_{i\in I}q_{p_i}$.
\prooflet{\Prf\ Of course $p(\alpha x)=|\alpha|p(x)$ for
$\alpha\in\Bbb R$ and $x\in X$.   If $x\in X$, then
$p(x)=\|\phi(x)\|_2$, where $\phi(x)=\familyiI{p_i(x)}\in\ell^2(I)$.
Now for $x$, $y\in X$,

\Centerline{$0\le\phi(x+y)\le\phi(x)+\phi(y)$}

\noindent in $\ell^2(I)$, so

\Centerline{$p(x+y)=\|\phi(x+y)\|_2\le\|\phi(x)+\phi(y)\|_2
\le\|\phi(x)\|_2+\|\phi(y)\|_2=p(x)+p(y)$.}

\noindent Thus $p$ is a seminorm.   Now the calculation of
$q_p=\sum_{i\in I}q_{p_i}$ is elementary.\ \QeD}  \cmmnt{In
particular,} $q_p\ge q_{p_i}$ for every $i\in I$.

\medskip

\quad{\bf (iii)} If $\|\,\|$ is an inner product norm on $X$, then
$q_{\|\,\|}(x,y)=\|x-y\|^2$ for all $x$, $y\in X$.  \prooflet{\Prf\

$$\eqalign{2\|x\|^2+2\|y\|^2-\|x+y\|^2
&=2\innerprod{x}{x}+2\innerprod{y}{y}-\innerprod{x+y}{x+y}\cr
&=\innerprod{x}{x}+\innerprod{y}{y}-\innerprod{x}{y}-\innerprod{y}{x}
=\innerprod{x-y}{x-y}.  \text{ \Qed}\cr}$$
}%end of prooflet

\leader{467D}{Lemma} Let $(X,\|\,\|)$ be a normed space.   Suppose that
there are a space $Y$ with a locally uniformly rotund norm $\normY{\,}$
and a bounded
linear operator $T:Y\to X$ such that $T[Y]$ is dense in $X$ and, for
every $x\in X$ and $\gamma>0$, there is a $z\in Y$ such that
$\|x-Tz\|^2+\gamma\normY{z}^2=\inf_{y\in Y}\|x-Ty\|^2+\gamma\normY{y}^2$.
Then $X$ has an equivalent locally uniformly rotund norm.

\proof{{\bf (a)} For each $n\in\Bbb N$, $x\in X$ set

\Centerline{$p_n(x)
=\sqrt{\inf_{y\in Y}\|x-Ty\|^2+2^{-n}\normY{y}^2}$.}

\noindent Then $p_n:X\to\coint{0,\infty}$ is a norm on $X$, equivalent
to $\|\,\|$.   \Prf\ (i) The functionals $(x,y)\mapsto\|x-Ty\|$,
$(x,y)\mapsto 2^{-n/2}\normY{y}$ from $X\times Y$ to $\coint{0,\infty}$ are
both seminorms, so the
functional $(x,y)\mapsto\phi(x,y)=\sqrt{\|x-Ty\|^2+2^{-n}\normY{y}^2}$ also
is, by 467C(c-ii).   (ii) If $x\in X$ and $\alpha\in\Bbb R$, take
$z\in Y$ such that $p_n(x)=\phi(x,z)$;  then

\Centerline{$p_n(\alpha x)\le\phi(\alpha x,\alpha z)=|\alpha|\phi(x,z)
=|\alpha|p_n(x)$.}

\noindent If $\alpha\ne 0$, apply the same argument to see that
$p_n(x)\le|\alpha|^{-1}p_n(\alpha x)$, so that
$p_n(\alpha x)=|\alpha|p_n(x)$.
(iii) Now take any $x_1$, $x_2\in X$.   Let $z_1$, $z_2\in Y$ be such
that $p_n(x_i)=\phi(x_i,z_i)$ for both $i$.   Then

$$\eqalign{p_n(x_1+x_2)
&=2p_n(\Bover12x_1+\Bover12x_2)
\le 2\phi(\Bover12x_1+\Bover12x_2,\Bover12z_1+\Bover12z_2)\cr
&\le 2(\Bover12\phi(x_1,z_1)+\Bover12\phi(x_2,z_2))
=p_n(x_1)+p_n(x_2).\cr}$$

\noindent Thus $p_n$ is a seminorm.   (iv) $p_n(x)\le\phi(x,0)=\|x\|$
for every $x\in X$.   (vi) For any $x\in X$ and $y\in Y$, either
$\|Ty\|\ge\bover12\|x\|$ and

\Centerline{$\phi(x,y)\ge 2^{-n/2}\normY{y}
\ge\Bover1{2\cdot 2^{n/2}\|T\|}\|x\|$,}

\noindent or $\|Ty\|\le\bover12\|x\|$ and

\Centerline{$\phi(x,y)\ge\|x-Ty\|\ge\Bover12\|x\|$;}

\noindent this shows that
$p_n(x)\ge\min(\bover12,\bover122^{-n/2}\|T\|^{-1})\|x\|$.   (I am
passing over the trivial case $X=\{0\}$, $\|T\|=0$.)   In particular,
$p_n(x)=0$ only when $x=0$.   Thus $p_n$ is a norm equivalent to $p$.\
\Qed

\medskip

{\bf (b)} For any $x\in X$, $\lim_{n\to\infty}p_n(x)=0$.   \Prf\ Let
$\epsilon>0$.   Let $y\in Y$ be such that $\|x-Ty\|\le\epsilon$;  then

\Centerline{$\limsup_{n\to\infty}p_n(x)^2
\le\limsup_{n\to\infty}\|x-Ty\|^2+2^{-n}\normY{y}^2
\le\epsilon^2$.}

\noindent As $\epsilon$ is arbitrary, $\lim_{n\to\infty}p_n(x)=0$.\ \Qed

\medskip

{\bf (c)} Set $\|x\|'=\sqrt{\sum_{n=0}^{\infty}2^{-n}p_n(x)^2}$ for
$x\in X$.   The sum is always finite because $p_n(x)\le\|x\|$ for every
$n$, so $\|\,\|'\le\sqrt 2\|\,\|$ is a seminorm;  and it is a norm equivalent to
$\|\,\|$ because $p_0$ is.   Now $\|\,\|'$ is locally uniformly rotund.
\Prf\ Take $x\in X$ and $\epsilon>0$.   Let $n\in\Bbb N$ be such that
$p_n(x)\le\bover14\epsilon$.   Choose $y\in Y$ such that
$p_n(x)^2=\|x-Ty\|^2+2^{-n}\normY{y}^2$.   (This is where we really use the
hypothesis that the infimum in the definition of $p_n$ is attained.)
Let $\delta>0$ be such that $2^n\delta\le(\bover14\epsilon)^2$ and
$\|T\|\|y'-y\|\le\bover14\epsilon$
whenever $q_{\normY{\,}}(y',y)\le 2^{2n}\delta$.

If $q_{\|\,\|'}(x,x')\le\delta$, then $q_{2^{-n/2}p_n}(x,x')\le\delta$,
by 467C(c-ii), that is, $q_{p_n}(x,x')\le 2^n\delta$.   Let $y'\in Y$ be
such that $p_n(x')^2=\|x'-Ty'\|^2+2^{-n}\normY{y'}^2$.   Then

$$\eqalign{p_n(x+x')^2
&\le\|x+x'-Ty-Ty'\|^2+2^{-n}\normY{y+y'}^2\cr
&\le(\|x-Ty\|+\|x'-Ty'\|)^2+2^{-n}\normY{y+y'}^2,\cr}$$

\noindent so

$$\eqalign{q_{p_n}(x,x')
&=2p_n(x)^2+2p_n(x')^2-p_n(x+x')^2\cr
&\ge 2(\|x-Ty\|^2+2^{-n}\normY{y}^2)
  +2(\|x'-Ty'\|^2+2^{-n}\normY{y'}^2)\cr
&\qquad\qquad-(\|x-Ty\|+\|x'-Ty'\|)^2-2^{-n}\normY{y+y'}^2\cr
&=(\|x-Ty\|-\|x'-Ty'\|)^2
  +2^{-n}(2\normY{y}^2+2\normY{y'}^2-\normY{y+y'}^2).\cr}$$

\noindent This means that

\Centerline{$q_{\normY{\,}}(y,y')\le 2^nq_{p_n}(x,x')\le 2^{2n}\delta$,}

\noindent so $\|T\|\normY{y-y'}\le\bover14\epsilon$, while also

\Centerline{$\|x-Ty\|+\|x'-Ty'\|
\le 2\|x-Ty\|+\sqrt{q_{p_n}(x,x')}
\le 2p_n(x)+2^{n/2}\sqrt{\delta}
\le \Bover34\epsilon$.}

\noindent Finally

\Centerline{$\Bover1{\sqrt 2}\|x-x'\|'\le\|x-x'\|
\le\|T\|\normY{y-y'}+\|x-Ty\|+\|x'-Ty'\|
\le\Bover14\epsilon+\Bover34\epsilon
=\epsilon$.}

\noindent As $x$ and $\epsilon$ are arbitrary, this shows that $\|\,\|'$
is locally uniformly rotund.\ \Qed

This completes the proof.
}%end of proof of 467D

\leader{467E}{Theorem} Let $X$ be a separable normed space.   Then it
has an equivalent locally uniformly rotund norm.

\proof{{\bf (a)} It is enough to show that the completion of $X$ has an
equivalent locally uniformly rotund norm;  since the completion of $X$
is separable, we may suppose that $X$ itself is complete.   Let
$\sequence{i}{x_i}$ be a sequence in $X$ running
over a dense subset of $X$.   Define $T:\ell^2\to X$ by setting

\Centerline{$Ty=\sum_{i=0}^{\infty}\Bover{y(i)}{2^i(1+\|x_i\|)}x_i$}

\noindent for $y\in\ell^2=\ell^2(\Bbb N)$;  then $Ty$ is always defined
(4A4Ie);  $T$ is a linear operator and

\Centerline{$\|Ty\|
\le\sum_{i=0}^{\infty}2^{-i}|y(i)|
\le\sqrt{\sumop_{i=0}^{\infty}2^{-2i}}\,\|y\|_2$}

\noindent for every $y\in\ell^2$, by Cauchy's inequality (244Eb).  So
$T$ is a bounded linear operator.

\medskip

{\bf (b)} $T$ satisfies the conditions of 467D.   \Prf\ $T[\ell^2]$ is
dense because it contains every $x_i$.   Given $x\in X$, $\gamma>0$ and
$\alpha\ge 0$, the function
$y\mapsto\sqrt{\|x-Ty\|^2+\gamma\|y\|_2^2}$ is convex and
norm-continuous, so the set

\Centerline{$C_{\alpha}(x)
=\{y:y\in\ell^2,\,\|x-Ty\|^2+\gamma\|y\|_2^2\le\alpha^2\}$}

\noindent is convex and norm-closed.   Consequently,
$C_{\alpha}(x)$ is weakly closed (4A4Ed);  since
$\|y\|_2\le\gamma^{-1/2}\alpha$ for every $y\in C_{\alpha}(x)$,
$C_{\alpha}(x)$ is weakly compact (4A4Ka).   Set $\beta=\inf_{y\in\ell^2}\|x-Ty\|^2+\gamma\|y\|_2^2$.   Then
$\{C_{\alpha}(x):\alpha>\beta\}$ is a downwards-directed set of
non-empty weakly compact sets, so has non-empty intersection;  taking
any $z\in\bigcap_{\alpha>\beta}C_{\alpha}(x)$,
$\beta=\|x-Tz\|^2+\gamma\|z\|_2^2$.\ \Qed

\medskip

{\bf (c)} So 467D gives the result.
}%end of proof of 467E

\leader{467F}{Lemma} Let $(X,\|\,\|)$ be a Banach space, and
$\familyiI{T_i}$ a family of bounded linear operators from $X$ to itself
such that

\inset{(i) for each $i\in I$, the subspace $T_i[X]$ has an equivalent
locally uniformly rotund norm,

(ii) for each $x\in X$, $\epsilon>0$ there is a finite set
$J\subseteq I$ such that $\|x-\sum_{i\in J}T_ix\|\le\epsilon$,

(iii) for each $x\in X$, $\epsilon>0$  the set
$\{i:i\in I,\,\|T_ix\|\ge\epsilon\}$ is finite.}

\noindent  Then $X$ has an equivalent locally uniformly rotund norm.

\proof{{\bf (a)} Let  $\|\,\|_i$ be a locally uniformly rotund norm on
$X_i=T_i[X]$ equivalent to $\|\,\|$ on $X_i$.   Reducing $\|\,\|_i$ by a
scalar multiple if necessary, we may suppose that $\|T_ix\|_i\le\|x\|$
for every
$x\in X$ and $i\in I$.   By (iii), $\sup_{i\in I}\|T_ix\|$ is finite for
every $x\in X$;  by the Uniform Boundedness Theorem (3A5Ha),
$M=\sup_{i\in I}\sup_{\|x\|\le 1}\|T_ix\|$ is finite.   (This is where we
need to suppose that $X$ is complete.)   For finite sets
$J\subseteq I$ and $k\ge 1$, set

\Centerline{$p_{Jk}(x)=\sqrt{\sumop_{i\in J}\|T_ix\|_i^2
  +\bover1k\sumop_{K\subseteq J}\|x-\sumop_{i\in K}T_ix\|^2}$;}

\noindent for $n\in\Bbb N$ and $k\ge 1$ set

\Centerline{$p^{(n)}_k(x)=\sup\{p_{Jk}(x):J\subseteq I,\,\#(J)\le n\}$.}

\noindent By 467C(c-ii), as usual, all the $p_{Jk}$ are seminorms, and
it follows at once that the $p^{(n)}_k$ are seminorms.   Observe that if
$K\subseteq I$ is finite, then
$\|x-\sum_{i\in K}T_ix\|\le(1+M\#(K))\|x\|$ for every $x$, so if
$J\subseteq I$ is finite then

\Centerline{$p_{Jk}(x)\le\sqrt{\#(J)+2^{\#(J)}(1+M\#(J))}\|x\|$,}

\noindent and $p^{(n)}_k(x)\le\sqrt{n+2^n(1+Mn)}\|x\|$ whenever
$n\in\Bbb N$ and $k\ge 1$.   Setting $\beta_{nk}=2^{2n+k}$ for $n$,
$k\in\Bbb N$,

\Centerline{$\|x\|'
=\sqrt{\sumop_{n=0}^{\infty}\sumop_{k=1}^{\infty}
  \beta_{nk}^{-1}p^{(n)}_k(x)^2}$}

\noindent is finite for every $x\in X$, so that $\|\,\|'$ is a seminorm
on $X$;  moreover, $\|x\|'\le\beta\|x\|$ for every $x\in X$, where

\Centerline{$\beta
=\sqrt{\sumop_{n=0}^{\infty}\sumop_{k=1}^{\infty}
  \beta_{nk}^{-1}(n+2^n(1+Mn))}$}

\noindent is finite.   Since we also have

\Centerline{$\|x\|'\ge\Bover1{\sqrt{2}}p^{(0)}_1(x)
=\Bover1{\sqrt{2}}p_{\emptyset 1}(x)=\Bover1{\sqrt{2}}\|x\|$}

\noindent for every $x$, $\|\,\|'$ is a norm on $X$ equivalent to
$\|\,\|$.

\medskip

{\bf (b)} Now $\|\,\|'$ is locally uniformly rotund.   \Prf\ Take
$x\in X$ and $\epsilon>0$.   Let $K\subseteq I$ be a finite set such
that $\|x-\sum_{i\in K}T_ix\|\le\bover14\epsilon$;  we may suppose that
$T_ix\ne 0$ for every $i\in K$.   Set
$\alpha_1=\min_{i\in K}\|T_ix\|_i$,
$J=\{i:i\in I,\,\|T_ix\|_i\ge\alpha_1\}$ and
$\alpha_0=\sup_{i\in I\setminus J}\|T_ix\|_i$.   (For completeness, if
$K=\emptyset$, take $J=\emptyset$,  $\alpha_0=\sup_{i\in I}\|T_ix\|_i$
and $\alpha_1=\alpha_0+1$.)
Then $J$ is finite and $\alpha_0<\alpha_1$, by hypothesis (iii) of the
lemma.   Set $n=\#(J)$.   Let $k$ be so large that
$\Bover{2^n(Mn+1)}{k}\|x\|^2<\bover12(\alpha_1^2-\alpha_0^2)$.
Let $\eta>0$ be such that

\inset{$(\beta_{nk}+1)\eta
\le\min(\Bover{\epsilon^2}{16k},\alpha_1^2-\alpha_0^2)$,}

\inset{$\|T_ix-z\|\le\Bover{\epsilon}{1+4n}$ whenever $i\in K$, $z\in
X_i$ and $q_{\|\,\|_i}(T_ix,z)\le(\beta_{nk}+1)\eta$;}

\noindent this is where we use the hypothesis that every $\|\,\|_i$ is
locally uniformly rotund (and equivalent to $\|\,\|$ on $X_i$).

Now suppose that $y\in X$ and $q_{\|\,\|'}(x,y)\le\eta$.   Then
$q_{p^{(n)}_k}(x,y)\le\beta_{nk}\eta$, by 467C(c-ii).   Let $L\in[I]^n$
be such that $p^{(n)}_k(x+y)^2\le p_{Lk}(x+y)^2+\eta$.   Then

$$\eqalign{q_{p_{Lk}}(x,y)
&=2p_{Lk}(x)^2+2p_{Lk}(y)^2-p_{Lk}(x+y)^2\cr
&\le 2p^{(n)}_k(x)^2+2p^{(n)}_k(y)^2-p^{(n)}_k(x+y)^2+\eta
\le(\beta_{nk}+1)\eta.\cr}$$

\noindent We also have

$$\eqalign{2p_{Lk}(x)^2
&\ge 2p_{Lk}(x)^2+2p_{Lk}(y)^2-2p^{(n)}_k(y)^2
\ge p_{Lk}(x+y)^2-2p^{(n)}_k(y)^2\cr
&\ge p^{(n)}_k(x+y)^2-\eta-2p^{(n)}_k(y)^2
\ge 2p^{(n)}_k(x)^2-(\beta_{nk}+1)\eta,\cr}$$

\noindent so

$$\eqalign{\sum_{i\in J}\|T_ix\|_i^2
&\le p_{Jk}(x)^2
\le p_k^{(n)}(x)^2
\le p_{Lk}(x)^2+\Bover12(\beta_{nk}+1)\eta\cr
&\le\sum_{i\in L}\|T_ix\|_i^2+\Bover{2^n(Mn+1)}{k}\|x\|^2
  +\Bover12(\beta_{nk}+1)\eta\cr
&<\sum_{i\in L}\|T_ix\|_i^2+\alpha_1^2-\alpha_0^2.\cr}$$

\noindent Since $\#(L)=\#(J)$ and

\Centerline{$\|T_ix\|_i^2\le\alpha_0^2<\alpha_1^2\le\|T_jx\|_j^2$}

\noindent whenever $j\in J$ and $i\in I\setminus J$,  we must actually
have $L=J$.   In particular, $K\subseteq L$.   But this means that (by
467C(c-ii) again)

\Centerline{$q_{\|\,\|_i}(T_ix,T_iy)
\le q_{p_{Lk}}(x,y)\le(\beta_{nk}+1)\eta$}

\noindent and (by the choice of $\eta$)
$\|T_ix-T_iy\|\le\Bover{\epsilon}{1+4n}$ for every $i\in K$, so that
$\|\sum_{i\in K}T_ix-\sum_{i\in K}T_iy\|\le\bover14\epsilon$.

The last element we need is that, setting
$\tilde p(z)=\bover1{\sqrt{k}}\|z-\sum_{i\in K}T_iz\|$, $\tilde p$ is a
seminorm on $X$ and is one of the constituents of $p_{Lk}$;  so that

$$\eqalign{\Bover1k(\|x-\sum_{i\in K}T_ix\|-\|y-\sum_{i\in K}T_iy\|)^2
&\le q_{\tilde p}(x,y)
\le q_{p_{Lk}}(x,y)\cr
&\le(\beta_{nk}+1)\eta
\le\Bover{\epsilon^2}{16k},\cr}$$

\noindent and
$\bigl|\|x-\sum_{i\in K}T_ix\|-\|y-\sum_{i\in K}T_iy\|\bigr|
\le\bover14\epsilon$.   It follows that

\Centerline{$\|y-\sum_{i\in K}T_iy\|
\le\bover14\epsilon+\|x-\sum_{i\in K}T_ix\|\le\bover12\epsilon$.}

\noindent Putting these together,

\Centerline{$\|x-y\|
\le\|x-\sum_{i\in K}T_ix\|+\sum_{i\in K}\|T_ix-T_iy\|
  +\|y-\sum_{i\in K}T_iy\|
\le\epsilon$.}

\noindent And this is true whenever $q_{\|\,\|'}(x,y)\le\eta$.   As $x$
and $\epsilon$ are arbitrary, $\|\,\|'$ is locally uniformly rotund.\
\Qed
}%end of proof of 467F

\leader{467G}{Theorem} Let $X$ be a Banach space.   Suppose that there
are an ordinal $\zeta$ and a family
$\langle P_{\xi}\rangle_{\xi\le\zeta}$ of bounded linear operators from
$X$ to itself such that

\inset{(i) if $\xi\le\eta\le\zeta$ then
$P_{\xi}P_{\eta}=P_{\eta}P_{\xi}=P_{\xi}$;

(ii) $P_0(x)=0$ and $P_{\zeta}(x)=x$ for every $x\in X$;

(iii) if $\xi\le\zeta$ is a non-zero limit ordinal, then
$\lim_{\eta\uparrow\xi}P_{\eta}(x)=P_{\xi}(x)$ for every $x\in X$;

(iv) if $\xi<\zeta$ then $X_{\xi}=\{(P_{\xi+1}-P_{\xi})(x):x\in X\}$
has an equivalent locally uniformly rotund norm.}

\noindent Then $X$ has an equivalent locally uniformly rotund norm.

\medskip

\noindent{\bf Remark} A family $\langle P_{\xi}\rangle_{\xi\le\zeta}$ satisfying (i), (ii) and (iii) here is called a {\bf projectional resolution of the identity}.

\proof{ For $\xi<\zeta$ set $T_{\xi}=P_{\xi+1}-P_{\xi}$.   From
condition (i) we see easily that $T_{\xi}T_{\eta}=T_{\xi}$ if
$\xi=\eta$, $0$ otherwise;  and that $T_{\xi}P_{\eta}=T_{\xi}$ if
$\xi<\eta$, $0$ otherwise.

I seek to show that the conditions of 467F are satisfied by
$\ofamily{\xi}{\zeta}{T_{\xi}}$.   Condition (i) of 467F is just
condition (iv) here.   Let $Z$ be the set of those $x\in X$ for which
conditions (ii) and (iii) of 467F are satisfied;  that is,

\inset{for each $\epsilon>0$ there is a finite set $J\subseteq\zeta$
such that $\|x-\sum_{\xi\in J}T_{\xi}x\|\le\epsilon$, and
$\{\xi:\|T_{\xi}x\|\ge\epsilon\}$ is finite.}

\noindent Then $Z$ is a linear subspace of $X$.
For $\xi\le\zeta$, set $Y_{\xi}=P_{\xi}[X]$.   Then
$Y_{\xi}\subseteq Z$.   \Prf\ Induce on $\xi$.   Since $Y_0=\{0\}$, the
induction starts.   For the inductive step to a successor ordinal
$\xi+1\le\zeta$,
$Y_{\xi+1}=Y_{\xi}+X_{\xi}\subseteq Z$.   For the inductive step to a
non-zero limit ordinal $\xi\le\zeta$, given $x\in Y_{\xi}$ and
$\epsilon>0$, we know that there is a $\xi'<\xi$ such that
$\|P_{\eta}x-P_{\xi}x\|\le\bover13\epsilon$ whenever
$\xi'\le\eta\le\xi$.   So $\|T_{\eta}x\|\le\bover23\epsilon$ whenever
$\xi'\le\eta<\xi$, and

\Centerline{$\{\eta:\|T_{\eta}x\|\ge\epsilon\}
=\{\eta:\eta<\xi',\,\|T_{\eta}x\|\ge\epsilon\}
=\{\eta:\|T_{\eta}P_{\xi'}x\|\ge\epsilon\}$}

\noindent is finite, by the inductive hypothesis.   Moreover, there is a
finite set $J\subseteq\xi'$ such that
$\|P_{\xi'}x-\sum_{\eta\in J}T_{\eta}P_{\xi'}x\|\le\bover23\epsilon$,
and now $\|x-\sum_{\eta\in J}T_{\eta}x\|\le\epsilon$.   As $x$ and
$\epsilon$ are arbitrary, $Y_{\xi}\subseteq Z$.\ \Qed

In particular, $X=Y_{\zeta}\subseteq Z$ and conditions (ii) and (iii) of
467F are satisfied.   So 467F gives the result.
}%end of proof of 467G

\leader{467H}{Definitions (a)} A topological space $X$ is
{\bf K-countably determined} or a {\bf Lindel\"of-$\pmb{\Sigma}$} space if there are a subset $A$ of $\NN$ and an
usco-compact relation $R\subseteq A\times X$ such that $R[A]=X$.
\cmmnt{Observe that all K-analytic Hausdorff spaces (\S422) are
K-countably determined.}

\spheader 467Hb A normed space $X$ is {\bf weakly K-countably
determined} if it is K-countably determined in its weak topology.

\spheader 467Hc Let $X$ be a normed space and $Y$, $W$ closed linear
subspaces of $X$, $X^*$ respectively.   I will say that $(Y,W)$ is a
{\bf projection pair} if $X=Y\oplus W^{\smallcirc}$ and
$\|y+z\|\ge\|y\|$ for every
$y\in Y$, $z\in W^{\smallcirc}$\dvro{.}{, where

$$\eqalign{W^{\smallcirc}
&=\{z:z\in X,\,f(z)\le 1\text{ for every }f\in W\}\cr
&=\{z:z\in X,\,f(z)=0\text{ for every }f\in W\}.\cr}$$
}%end of dvro

\leader{467I}{Lemma} (a) If $X$ is a weakly K-countably determined
normed space, then any closed linear subspace of $X$ is weakly
K-countably determined.

(b) If $X$ is a weakly K-countably determined normed space, $Y$ is a
normed space, and $T:X\to Y$ is a continuous linear surjection, then $Y$
is weakly K-countably determined.

(c) If $X$ is a Banach space and $Y\subseteq X$ is a dense linear
subspace which is weakly K-countably determined, then $X$ is weakly
K-countably determined.

\proof{{\bf (a)} Let $A\subseteq\NN$, $R\subseteq A\times X$ be such
that $R$ is usco-compact (for the weak topology on $X$) and $R[A]=X$.
Let $Y$ be a (norm-\nobreak)closed linear subspace of $X$;
then $Y$ is closed
for the weak topology (3A5Ee).   Also the weak topology on $Y$ is just
the subspace topology induced by the weak topology of $X$ (4A4Ea).   Set
$R'=R\cap(A\times Y)$.   Then $R'$ is usco-compact whether regarded as a
subset of $A\times X$ or as a subset of $A\times Y$ (422Db,
422Dg).   Since $Y=R'[A]$, $Y$ is weakly K-countably determined.

\medskip

{\bf (b)} Let $A\subseteq\NN$, $R\subseteq A\times X$ be such that $R$
is usco-compact for the weak topology on $X$ and $R[A]=X$.   Because $T$
is continuous for the weak topologies on $X$ and $Y$ (3A5Ec),

\Centerline{$R_1=\{(\phi,y)$: there is some $x\in X$ such that
$(\phi,x)\in R$ and $Tx=y\}$}

\noindent is usco-compact in $A\times Y$ (422Dc, 422Df).   Also
$R_1[A]=T[R[A]]=Y$.   So $Y$ is weakly K-countably determined.

\medskip

{\bf (c)} Let $A\subseteq\NN$, $R\subseteq A\times Y$ be such that $R$
is usco-compact (for the weak topology on $Y$) and $R[A]=Y$.   Then, as
in (a), $R$ is usco-compact when regarded as a subset of $A\times X$.
By 422Dd, the set

\Centerline{$R_1
=\{(\sequencen{\phi_n},\sequencen{y_n}):(\phi_n,y_n)\in R$ for every
$n\in\Bbb N\}$}

\noindent is usco-compact in $A^{\Bbb N}\times Y^{\Bbb N}$.   Now
examine

\Centerline{$R_2=\{(\sequencen{y_n},x):x\in X$, $y_n\in Y$ and
$\|y_n-x\|\le 2^{-n}$ for every $n\in\Bbb N\}$.}

\noindent Then $R_2$ is usco-compact in $Y^{\Bbb N}\times X$.   \Prf\
(Remember that we are using weak topologies on $X$ and $Y$ throughout.)
If $\pmb{y}=\sequencen{y_n}$ is a sequence in $Y$ and
$(\pmb{y},x)\in R_2$, then $x=\lim_{n\to\infty}y_n$;  so $R_2[\{\pmb{y}\}]$
has at most one member and is certainly compact.   Let $F\subseteq X$ be
a weakly closed set and $\pmb{y}\in Y^{\Bbb N}\setminus R_2^{-1}[F]$.

{\bf case 1} If there are $m$, $n\in\Bbb N$ such that
$\|y_m-y_n\|>2^{-m}+2^{-n}$, let $f\in Y^*$ be such that $\|f\|\le 1$
and $f(y_m-y_n)>2^{-m}+2^{-n}$.   Then
$G=\{\pmb{z}:f(z_m-z_n)>2^{-m}+2^{-n}\}$ is an open set in $Y^{\Bbb N}$
containing $\pmb{y}$ and disjoint from $R_2^{-1}[F]$, because $R_2[G]$ is
empty.

{\bf case 2} Otherwise, $\pmb{y}$ is a Cauchy sequence and (because $X$
is a Banach space) has a limit $x\in X$, which does not belong to $F$.
Let $\delta>0$ and $f_0,\ldots,f_r\in X^*$ be such that
$\{w:|f_i(w)-f_i(x)|\le\delta$ for every $i\le r\}$ does not meet $F$.
Let $n\in\Bbb N$ be such that $2^{-n}\|f_i\|\le\bover13\delta$ for every
$i\le r$.   Then $G=\{\pmb{z}:|f_i(z_n)-f_i(y_n)|<\bover13\delta$ for
every $i\le r\}$ is an open set in $Y^{\Bbb N}$ containing $Y$.   If
$\pmb{z}\in G$ and $(\pmb{z},w)\in R_2$, then $\|z_n-w\|\le 2^{-n}$ so

$$\eqalign{|f_i(w)-f_i(x)|
&\le|f_i(w)-f_i(z_n)|+|f_i(z_n)-f_i(y_n)|+|f_i(y_n)-f_i(x)|\cr
&\le 2^{-n}\|f_i\|+\Bover13\delta+2^{-n}\|f_i\|
\le\delta\cr}$$

\noindent for every $i\le r$, and $w\notin F$.   Thus again
$G\cap R_2^{-1}[F]$ is empty.

This shows that there is always an open set containing $y$ and disjoint
from $R_2^{-1}[F]$.   As $y$ is arbitrary, $R_2^{-1}[F]$ is closed.   As $F$
is arbitrary, $R_2$ is usco-compact.\ \Qed

It follows that $R_2R_1\subseteq A^{\Bbb N}\times X$ is usco-compact
(422Df), while

\Centerline{$(R_2R_1)[A^{\Bbb N}]=R_2[R_1[A^{\Bbb N}]]=R_2[Y^{\Bbb N}]=X$}

\noindent because $Y$ is dense in $X$.   Finally, $A^{\Bbb N}$ is
homeomorphic to a subset of $\NN$ because it is a subspace of
$(\NN)^{\Bbb N}\cong\NN$.   So $X$ is weakly K-countably determined.
}%end of proof of 467I

\vleader{72pt}{467J}{Lemma}
Let $X$ be a weakly K-countably determined Banach
space.   Then there is a family $\Cal M$ of subsets of $X\cup X^*$ such
that

(i) whenever $B\subseteq X\cup X^*$ there is an $M\in\Cal M$ such that
$B\subseteq M$ and $\#(M)\le\max(\omega,\#(B))$;

(ii) whenever $\Cal M'\subseteq\Cal M$ is upwards-directed, then
$\bigcup\Cal M'\in\Cal M$;

(iii) whenever $M\in\Cal M$ then
$(\overline{M\cap X},\overline{M\cap X^*})$\cmmnt{ (where the closures
are taken for the norm topologies)} is a projection pair of subspaces of
$X$ and $X^*$.

\proof{{\bf (a)} Let $A\subseteq\NN$, $R\subseteq A\times X$ be such
that $R$ is usco-compact in $A\times X$ and $R[A]=X$.   Set
$S=\bigcup_{n\in\Bbb N}\BbbN^n$ and for $\sigma\in S$ set
$F_{\sigma}=R[I_{\sigma}]$, where
$I_{\sigma}=\{\phi:\sigma\subseteq\phi\in\NN\}$;  set
$S_0=\{\sigma:\sigma\in S,\,F_{\sigma}\ne\emptyset\}$.

Let $\Cal M$ be the family of those sets $M\subseteq X\cup X^*$ such
that ($\alpha$) whenever $x$, $y\in M\cap X$ and $q\in\Bbb Q$ then $x+y$
and $qx$ belong to $M$ ($\beta$) whenever $f$, $g\in M\cap X^*$ and
$q\in\Bbb Q$ then $f+g$ and $qf$ belong to $M$ ($\gamma$)
$\|x\|=\max\{f(x):f\in M\cap X^*,\|f\|\le 1\}$ for every $x\in M\cap X$
($\delta$)
$\sup_{x\in F_{\sigma}}f(x)=\sup_{x\in F_{\sigma}\cap M}f(x)$ for every
$f\in M\cap X^*$, $\sigma\in S_0$.

\medskip

{\bf (b)} For each $x\in X$, choose $h_x\in X^*$ such that
$\|h_x\|\le 1$ and $h_x(x)=\|x\|$;  for each $f\in X^*$ and
$\sigma\in S_0$ choose a countable set $C_{f\sigma}\subseteq F_{\sigma}$
such that $\sup\{f(x):x\in C_{f\sigma}\}=\sup\{f(x):x\in F_{\sigma}\}$.
Given $B\subseteq X$, define $\sequencen{B_n}$ by setting

$$\eqalign{B_{n+1}
&=B_n\cup\{x+y:x,\,y\in B_n\cap X\}
  \cup\{qx:q\in\Bbb Q,\,x\in B_n\cap X\}\cr
&\qquad\qquad\cup\{f+g:f,\,g\in B_n\cap X\}
  \cup\{qf:q\in\Bbb Q,\,f\in B_n\cap X\}\cr
&\qquad\qquad\cup\{h_x:x\in B_n\cap X\}
  \cup\bigcup\{C_{f\sigma}:f\in B_n\cap X^*,\,\sigma\in S_0\},\cr}$$

\noindent for each $n\in\Bbb N$.   Then $M=\bigcup_{n\in\Bbb N}B_n$
belongs to $\Cal M$ and has cardinal at most $\max(\omega,\#(B))$.

\medskip

{\bf (c)} The definition of $\Cal M$ makes it plain that if
$\Cal M'\subseteq\Cal M$ is upwards-directed then $\bigcup\Cal M'$
belongs to $\Cal M$.

\medskip

{\bf (d)} Now take $M\in\Cal M$ and set $Y=\overline{M\cap X}$,
$W=\overline{M\cap X^*}$.   These are linear subspaces (2A5Ec).   If
$y\in M\cap X$ and $z\in W^{\smallcirc}$, then there is an
$f\in M\cap X^*$ such that $\|f\|\le 1$ and $f(y)=\|y\|$, so that

\Centerline{$\|y+z\|\ge f(y+z)=f(y)=\|y\|$.}

\noindent Because the function $y\mapsto\|y+z\|-\|y\|$ is continuous,
$\|y+z\|\ge\|y\|$ for every $y\in Y$ and $z\in W^{\smallcirc}$.   In
particular, if $y\in Y\cap W^{\smallcirc}$, $\|y\|\le\|y-y\|=0$ and
$y=0$, so $Y+W^{\smallcirc}=Y\oplus W^{\smallcirc}$.   If
$x\in\overline{Y+W^{\smallcirc}}$, then there are sequences
$\sequencen{y_n}$ in $Y$ and $\sequencen{z_n}$ in $W^{\smallcirc}$ such
that $x=\lim_{n\to\infty}y_n+z_n$;  now
$\|y_m-y_n\|\le\|(y_m+z_m)-(y_n+z_n)\|\to 0$ as $n\to\infty$,  so
(because $X$ is a Banach space) $\sequencen{y_n}$ is convergent to $y$
say;  in this case, $y\in Y$ and $x-y=\lim_{n\to\infty}z_n$ belongs to
$W^{\smallcirc}$, so $x\in Y+W^{\smallcirc}$.   This shows that
$Y\oplus W^{\smallcirc}$ is a closed linear subspace of $X$.

\medskip

{\bf (e)} \Quer\ Suppose, if possible, that
$Y\oplus W^{\smallcirc}\ne X$.   Then there is an
$x_0\in X\setminus(Y\oplus W^{\smallcirc})$.   By 4A4Eb, there is an
$f\in X^*$ such that $f(x_0)>0$ and $f(y)=f(z)=0$ whenever $y\in Y$ and
$z\in W^{\smallcirc}$;  multiplying $f$ by a scalar if necessary, we can
arrange that $f(x_0)=1$.   By 4A4Eg, $f$ belongs to the weak*
closure of $W$ in $X^*$.

Let $\phi\in A$ be such that $(\phi,x_0)\in R$.   Then $K=R[\{\phi\}]$
is weakly compact.   Now the weak* closure of $W$ is also its closure
for the Mackey topology of uniform convergence on weakly compact subsets
of $X$ (4A4F).   So there is a $g\in W$ such that
$|g(x)-f(x)|\le\bover17$ for every $x\in K$.   Next, because $K$ is
bounded, and $g$ belongs to the norm closure of $M\cap X^*$, there is an
$h\in M\cap X^*$ such that $|h(x)-g(x)|\le\bover17$ for every $x\in K$.
This means that $|h(x)-f(x)|\le\bover27$ for every
$x\in K$, and $K$ is included in the weakly open set
$G=\{x:|h(x)-f(x)|<\bover37\}$, that is, $\phi$ does not belong to
$R^{-1}[X\setminus G]$, which is relatively closed in $A$, because $R$
is usco-compact regarded as a relation between $A$ and $X$ with the weak
topology.   There is therefore a $\sigma\in S$ such that
$\phi\in I_{\sigma}$ and
$I_{\sigma}\cap R^{-1}[X\setminus G]=\emptyset$, that is,
$F_{\sigma}\subseteq G$.   In this case, $x_0\in F_{\sigma}$, so
$\sigma\in S_0$, while $h(x)-f(x)<\bover37$ for every
$x\in F_{\sigma}$.   But, because $M\in\Cal M$, there is a
$y\in M\cap X\cap F_{\sigma}$ such that $h(y)\ge h(x_0)-\bover17$, and as $y\in Y$ we must now have

$$\eqalign{0
&=f(y)
=h(y)-(h(y)-f(y))
>h(x_0)-\Bover17-\Bover37\cr
&=f(x_0)-(f(x_0)-h(x_0))-\Bover47
\ge 1-\Bover37-\Bover47=0,\cr}$$

\noindent which is absurd.\ \Bang

Thus $X=Y\oplus W^{\smallcirc}$ and $(Y,W)$ is a projection pair.   This
completes the proof.
}%end of proof of 467J
%elementary submodel argument

\leader{467K}{Theorem} Let $X$ be a weakly K-countably determined Banach
space.   Then it has an equivalent locally uniformly rotund norm.

\proof{ Since the completion of $X$ is weakly K-countably determined
(467Ic), we may suppose that $X$ is complete.   The proof proceeds by
induction on the weight of $X$.

\medskip

{\bf (a)} The induction starts by observing that if $w(X)\le\omega$ then
$X$ is separable (4A2Li/4A2P(a-i))
so has an equivalent locally uniformly rotund norm by 467E.

\medskip

{\bf (b)} So let us suppose that $w(X)=\kappa>\omega$ and that any
weakly K-countably determined Banach space of weight less than $\kappa$
has an equivalent locally uniformly rotund norm.

Let $\Cal M$ be a family of subsets of $X\cup X^*$ as in 467J.   Then
there is a non-decreasing family
$\langle M_{\xi}\rangle_{\xi\le\kappa}$ in $\Cal M$ such that
$\#(M_{\xi})\le\max(\omega,\#(\xi))$ for every $\xi\le\kappa$,
$M_{\kappa}$ is dense in $X$, and $M_{\xi}=\bigcup_{\eta<\xi}M_{\eta}$
for every limit ordinal $\xi\le\kappa$.   \Prf\ By 4A2Li, there is a
dense subset of $X$ with cardinal $\kappa$;  enumerate it as
$\ofamily{\xi}{\kappa}{x_{\xi}}$.   Choose $M_{\xi}$ inductively, as
follows.    $M_0=\emptyset$.   GIven $M_{\xi}$ with
$\#(M_{\xi})\le\max(\omega,\#(\xi))$, then by 467J(i) there is an
$M_{\xi+1}\in\Cal M$ such that
$M_{\xi+1}\supseteq M_{\xi}\cup\{x_{\xi}\}$ and

\Centerline{$\#(M_{\xi+1})\le\max(\omega,\#(M_{\xi}\cup\{x_{\xi}\})
\le\max(\omega,\#(\xi+1))$.}

\noindent Given that $\ofamily{\eta}{\xi}{M_{\eta}}$ is a non-decreasing
family in $\Cal M$ with $\#(M_{\eta})\le\max(\omega,\#(\eta))$ for every
$\eta<\xi$, set $M_{\xi}=\bigcup_{\eta<\xi}M_{\eta}$;  then 467J(ii)
tells us that $M_{\xi}\in\Cal M$, while
$\#(M_{\xi})\le\max(\omega,\#(\xi))$, as required by the inductive
hypothesis.\ \Qed

At the end of the induction, $M_{\kappa}\supseteq\{x_{\xi}:\xi<\kappa\}$
will be dense in $X$.

\medskip

{\bf (c)} For each $\xi<\kappa$, set $Y_{\xi}=\overline{M_{\xi}\cap X}$,
$W_{\xi}=\overline{M_{\xi}\cap X^*}$.   Then $(Y_{\xi},W_{\xi})$ is a
projection pair, by 467J(iii).    Since
$X=Y_{\xi}\oplus W_{\xi}^{\smallcirc}$, we have a projection
$P_{\xi}:X\to Y_{\xi}$ defined by saying that $P_{\xi}(y+z)=y$ whenever
$y\in Y_{\xi}$ and $z\in W_{\xi}^{\smallcirc}$.   Now
$P_{\xi}P_{\eta}=P_{\eta}P_{\xi}=P_{\xi}$ whenever $\xi\le\eta$.   \Prf\
The point is that $Y_{\xi}\subseteq Y_{\eta}$ and
$W_{\xi}\subseteq W_{\eta}$, so
$W_{\eta}^{\smallcirc}\subseteq W_{\xi}^{\smallcirc}$.   If $x\in X$,
express it as $y+z_1$ where $y\in Y_{\xi}$ and
$z_1\in W_{\xi}^{\smallcirc}$, and express $z_1$ as $y'+z$ where
$y'\in Y_{\eta}$ and $z'\in W_{\eta}^{\smallcirc}$.   Then

\Centerline{$P_{\xi}x=y\in Y_{\xi}\subseteq Y_{\eta}$}

\noindent so $P_{\eta}P_{\xi}x=P_{\xi}x$.   On the other hand,
$x=y+y'+z$, $y+y'\in Y_{\eta}$ and $z\in W_{\eta}^{\smallcirc}$, so
$P_{\eta}x=y+y'$;  and as $y'=z_1-z$ belongs to $W_{\xi}^{\smallcirc}$,
$P_{\xi}(y+y')=y$, so $P_{\xi}P_{\eta}x=P_{\xi}x$.\ \Qed

Note that the condition

\Centerline{$\|y+z\|\ge\|y\|$ whenever $y\in Y_{\xi}$,
$z\in W_{\xi}^{\smallcirc}$}

\noindent ensures that $\|P_{\xi}\|\le 1$ for every $\xi$.

Next, if $\xi\le\kappa$ is a non-zero limit ordinal,
$P_{\xi}x=\lim_{\eta\uparrow\xi}P_{\eta}x$.   \Prf\ We know that

\Centerline{$P_{\xi}x\in Y_{\xi}
=\overline{M_{\xi}\cap X}
=\overline{\bigcupop_{\eta<\xi}M_{\eta}\cap X}$.}

\noindent So, given $\epsilon>0$, there is an
$x'\in\bigcup_{\eta<\xi}M_{\eta}$ such that
$\|P_{\xi}x-x'\|\le\bover12\epsilon$.   Let $\eta<\xi$ be such that
$x'\in M_{\eta}$.   If $\eta\le\eta'\le\xi$, then

$$\eqalignno{\|P_{\xi}x-P_{\eta'}x\|
&=\|P_{\xi}(P_{\xi}x-x')-P_{\eta'}(P_{\xi}x-x')\|\cr
\displaycause{because $x'\in Y_{\eta}$, so $P_{\xi}x'=P_{\eta'}x'=x'$}
&\le 2\|P_{\xi}x-x'\|
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$P_{\xi}x=\lim_{\eta\uparrow\xi}P_{\eta}x$.\ \Qed

\medskip

{\bf (d)} Now observe that every $Y_{\xi}$ is weakly K-countably
determined (467Ia), while $w(Y_{\xi})\le\max(\omega,\#(\xi))<\kappa$ for
every $\xi<\kappa$ (using 4A2Li, as usual).   So the inductive
hypothesis tells us that $Y_{\xi}=P_{\xi}[X]$ has an equivalent locally
uniformly rotund norm for every $\xi<\kappa$.   By 467G,
$X=P_{\kappa}[X]$ has an equivalent locally uniformly rotund norm.
Thus the induction proceeds.
}%end of proof of 467K

\leader{467L}{Weakly compactly generated Banach spaces}\cmmnt{ The
most important class of weakly K-{\vthsp}countably determined spaces is the following.}   A normed space $X$ is {\bf weakly compactly generated}
if there is a sequence $\sequencen{K_n}$ of weakly compact subsets of $X$ such that $\bigcup_{n\in\Bbb N}K_n$ is dense in $X$.

\allowmorestretch{468}{
\leader{467M}{Proposition}\cmmnt{ ({\smc Talagrand 75})} A weakly
compactly generated Banach space is weakly K-countably determined.
}%end of allowmorestretch

\proof{ Let $X$ be a Banach space with a sequence $\sequencen{K_n}$ of
weakly compact subsets of $X$ such that $\bigcup_{n\in\Bbb N}K_n$ is
dense in $X$.   Set

\Centerline{$L_n=\{\sum_{i=0}^n\alpha_ix_i:|\alpha_i|\le n,\,
x_i\in\bigcup_{j\le n}K_j$ for every $i\le n\}$}

\noindent for $n\in\Bbb N$.   Then every $L_n$ is weakly compact, and
$Y=\bigcup_{n\in\Bbb N}L_n$ is a linear subspace of $X$ including
$\bigcup_{n\in\Bbb N}K_n$, therefore dense.   Now $Y$ is a countable
union of weakly compact sets, therefore K-analytic for its weak topology
(422Gc, 422Hc);  in particular, it is weakly K-countably determined.
By 467Ic, $X$ is weakly K-countably determined.
}%end of proof of 467M

\leader{467N}{Theorem} Let $X$ be a Banach lattice with an
order-continuous norm\cmmnt{ (\S354)}.   Then it has an equivalent
locally uniformly rotund norm.

\proof{{\bf (a)} Consider first the case in which $X$ has a weak order
unit $e$.   Then the interval $[0,e]$ is weakly compact.   \Prf\ We have
$X^*=X^{\times}$ and $X^{**}=X^{\times\sim}$ (356D).   The canonical
identification of $X$ with its image in $X^{**}$ is an order-continuous
Riesz homomorphism from $X$ onto a solid order-dense Riesz subspace of
$X^{\times\times}$ (356I) and therefore of $X^{\times\sim}=X^{**}$.
In particular, $[0,e]\subseteq X$ is matched with an interval
$[0,\hat e]\subseteq X^{**}$.   But
$[0,\hat e]=\{\theta:\theta\in X^{**},\,0\le\theta(f)\le f(e)$ for every
$f\in(X^*)^+\}$ is weak*-closed and norm-bounded in $X^{**}$, therefore
weak*-compact;  as the weak* topology on $X^{**}$ corresponds to
the weak topology of $X$, $[0,e]$ is weakly compact.\ \Qed

Now, for each $n\in\Bbb N$, $K_n=[-ne,ne]=n[0,e]-n[0,e]$ is weakly
compact.   If $x\in X^+$, then $\sequencen{x\wedge ne}$ converges to
$X$, because the norm of $X$ is order-continuous;  so for any $x\in X$,
$\sequencen{x^+\wedge ne-x^-\wedge ne}$ converges to $x$.   Thus
$\bigcup_{n\in\Bbb N}K_n$ is dense in $X$ and $X$ is weakly compactly
generated.   By 467M and 467K, $X$ has an equivalent locally uniformly
rotund norm.

\medskip

{\bf (b)} For the general case,
let $\familyiI{x_i}$ be a maximal disjoint family in $X^+$.   For each
$i\in I$ let $X_i$ be the band in $X$ generated by $x_i$, and
$T_i:X\to X_i$ the band projection onto $X_i$ (354Ee, 353Ib\formerly{3{}53Hb}).   Then
$\familyiI{T_i}$ satisfies the conditions of 467F.   \Prf\ (i) Each
$T_i$ is a continuous linear operator because the given norm $\|\,\|$ of
$X$ is a Riesz norm.   Next, each $X_i$ has a weak order unit $x_i$, and
the norm on $X_i$ is order-continuous, so (a) tells us that there is an
equivalent locally uniformly rotund norm on $X_i=T_i[X]$.   (ii) If
$x\in X$, set $x'=\sup_{i\in I}T_i|x|$;  then $(|x|-x')\wedge x_i=0$ for
every $i$, so, by the maximality of $\familyiI{x_i}$, $x'=|x|$.   If
$\epsilon>0$ then, because the norm of $X$ is order-continuous, there is
a finite $J\subseteq I$ such that

\Centerline{$\|x-\sum_{j\in J}T_ix\|
=\|x'-\sup_{j\in J}T_i|x|\|\le\epsilon$.}

\noindent Moreover, if $i\in I\setminus J$, then

\Centerline{$\|T_ix\|\le\|x-\sum_{j\in J}T_jx\|\le\epsilon$.}

\noindent Thus conditions (ii) and (iii) of 467F are satisfied.\ \Qed

Accordingly 467F tells us that $X$ has an equivalent locally uniformly
rotund norm.
}%end of proof of 467N

\leader{467O}{Eberlein compacta:  Definition} A topological space $K$ is
an {\bf Eberlein compactum} if it is homeomorphic to a weakly compact
subset of a Banach space.

\leader{467P}{Proposition} Let $K$ be a compact Hausdorff space.

(a) The following are equiveridical:

\quad(i) $K$ is an Eberlein compactum;

\quad(ii) there is a set $L\subseteq C(K)$, separating the points of
$K$, which is compact for the topology of pointwise convergence.

(b) Suppose that $K$ is an Eberlein compactum.

\quad(i) $K$ has a $\sigma$-isolated network, so is hereditarily
weakly $\theta$-refinable.

\quad(ii)\cmmnt{ ({\smc Schachermayer 77})} If $w(K)$ is measure-free,
$K$ is a Radon space.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} If $K$ is an Eberlein compactum, we
may suppose that it is a weakly compact subset of a Banach space $X$.
Set $L=\{f\restr K:f\in X^*,\,\|f\|\le 1\}$;  since the map
$f\mapsto f\restr K:X^*\to C(K)\}$ is continuous for the weak* topology
of $X^*$ and the topology $\frak T_p$ of pointwise convergence on
$C(K)$, $L$ is $\frak T_p$-compact;  and $L$ separates the points of $K$
because $X^*$ separates the points of $X$.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} If $L\subseteq C(K)$ is
$\frak T_p$-compact and separates the points of $K$, set
$L_n=\{f:f\in L,\,\|f\|_{\infty}\le n\}$ for each $n\in\Bbb N$.   Then
$L_n$ is $\frak T_p$-compact for each $n$.   Set
$L'=\{0\}\cup\bigcup_{n\in\Bbb N}2^{-n}L_n$;  then $L'\subseteq C(K)$ is
norm-bounded and $\frak T_p$-compact and separates the points of $K$.
Now define $x\mapsto\hat x:K\to\BbbR^{L'}$ by setting $\hat x(f)=f(x)$
for $x\in K$ and $f\in L'$.   Then $\hat x\in C(L')$ for every $x$ and
$x\mapsto\hat x$ is continuous for the given topology of $K$ and the
topology of pointwise convergence on $C(L')$;  so the image
$\hat K=\{\hat x:x\in K\}$ is $\frak T_p$-compact.   Since it is also
bounded, it is weakly compact (462E).   But $x\mapsto\hat x$ is injective, because $L'$ separates
the points of $K$;  so $K$ is homeomorphic to $\hat K$, and is an
Eberlein compactum.

\medskip

{\bf (b)} Again suppose $K$ is actually a weakly compact subset of a
Banach space $X$.   As in 467M, set
$L_n=\{\sum_{i=0}^n\alpha_ix_i:|\alpha_i|\le n,\,x_i\in K$ for every
$i\le n\}$ for each $n\in\Bbb N$.   Then
$Y=\overline{\bigcup_{n\in\Bbb N}L_n}$ is a weakly compactly generated
Banach space.   (I am passing
over the trivial case $K=\emptyset$.)   So $Y$ has an equivalent locally
uniformly rotund norm (467M, 467K), which is a Kadec norm (467B), and
$Y$, with the weak topology, has a $\sigma$-isolated network
(466Eb).   It follows at once that $K$ has a $\sigma$-isolated
network (4A2B(a-ix)), so is hereditarily weakly $\theta$-refinable
(438Ld);  and if $w(K)$ is measure-free, $K$ is
Borel-measure-complete (438M), therefore Radon (434Jf, 434Ka).
}%end of proof of 467P

\exercises{\leader{467X}{Basic exercises (a)}%
%\spheader 467Xa
(i) Show that a continuous image of a K-countably
determined space is K-countably determined.   (ii) Show that the product
of a sequence of K-countably determined spaces is K-countably
determined.   (iii) Show that any K-countably determined topological
space is Lindel\"of.  \Hint{422De.}
(iv) Show that any Souslin-F subset of a K-countably determined
topological space is K-countably determined.   \Hint{422Hc.}
%467H

\spheader 467Xb Let $X$ be a $\sigma$-compact Hausdorff space.   Show
that a subspace $Y$ of $X$ is K-countably determined iff there is a
countable family $\Cal K$ of compact subsets of $X$ such that
$\bigcap\{K:y\in K\in\Cal K\}\subseteq Y$ for every $y\in Y$.
%467H mt46bits

\spheader 467Xc Show that if $X$ and $Y$ are weakly K-countably
determined normed spaces, then $X\times Y$, with an appropriate norm, is
weakly K-countably determined.
%467I

\spheader 467Xd Show that a normed space $X$ is weakly compactly
generated iff there is a weakly compact set $K\subseteq X$ such that the
linear subspace of $X$ generated by $K$ is dense in $X$.
%467L

\spheader 467Xe Show that any separable normed space is weakly compactly
generated.
%467Xd 467L

\spheader 467Xf Show that any reflexive Banach space is weakly compactly generated.
%467L

\sqheader 467Xg Show that if $X$ is a weakly compactly generated Banach
space, then it is K-analytic in its weak topology.   \Hint{in 467M, use
the proof of 467Ic.}
%467M

\spheader 467Xh Show that if $X$ is a Banach space and there is a set
$A\subseteq X$ such that $A$ is K-countably determined for the weak
topology and the linear subspace generated by $A$ is dense, then $X$ is
weakly K-countably determined.
%467M

\spheader 467Xi Show that the one-point compactification of any discrete
space is an Eberlein compactum.
%467P

\sqheader 467Xj Let $K$ be an Eberlein compactum,
and $\mu$ a Radon measure on $K$.
Show that $\mu$ is completion regular and inner regular with respect to
the compact metrizable subsets of $K$.   \Hint{466B.}
%467P

\leader{467Y}{Further exercises (a)}
%\spheader 467Ya
Let $\sequencen{X_n}$ be a sequence of weakly
K-countably determined normed spaces.   Investigate normed subspaces of
$\prod_{n\in\Bbb N}X_n$ which will be weakly K-countably determined.
%467I

\spheader 467Yb(i) Show that if $(\Omega,\Sigma,\mu)$ is a probability
space, then $L^1(\mu)$ has a locally uniformly rotund Riesz norm.
\Hint{apply the construction of 467D with $Y=L^2(\mu)$ and $T$ the
identity operator;  show that all the norms $p_n$ are Riesz norms.}
(ii) Show that if $X$ is any $L$-space then it has a locally uniformly
rotund Riesz norm.   \Hint{apply the construction of 467F/467N, noting
that if the $T_i$ in 467F are band projections and $\|\,\|$ and all the
$\|\,\|_i$ are Riesz norms, then all the norms in the proof of 467F are
Riesz norms.}
%467E

\spheader 467Yc\dvAnew{2011}
Let $(X,\|\,\|)$ be a Banach space, and $\frak T$ a linear
space topology on $X$ such that the unit ball of $X$ is $\frak T$-closed.
Suppose that
$\familyiI{T_i}$ is a family of bounded linear operators from $X$ to itself
such that

\inset{(i) for each $i\in I$, $T_i$ is $\frak T$-continuous as well as
norm-continuous,

(ii) for each $i\in I$, the subspace $T_i[X]$ has an equivalent
locally uniformly rotund norm for which the unit ball is closed for the
topology on $T_i[X]$ induced by $\frak T$,

(ii) for each $x\in X$, $\epsilon>0$ there is a finite set
$J\subseteq I$ such that $\|x-\sum_{i\in J}T_ix\|\le\epsilon$,

(iii) for each $x\in X$, $\epsilon>0$  the set
$\{i:i\in I,\,\|T_ix\|\ge\epsilon\}$ is finite.}

\noindent  Show that
$X$ has an equivalent locally uniformly rotund norm for which the unit ball
is $\frak T$-closed.
%467F

\spheader 467Yd\dvAnew{2011}
Let $X$ be a normed space with a locally uniformly rotund
norm, and $\frak T$ a linear space
topology on $X$ such that the unit ball of $X$ is $\frak T$-closed.
Show that every norm-Borel subset of $X$ is $\frak T$-Borel.
%466Yb

\spheader 467Ye\dvAnew{2011}
Let $\kappa$ be any cardinal, and $K$ a dyadic space.   (i) Show that
$C(K)$ has a locally uniformly rotund norm, equivalent to
the usual supremum norm $\|\,\|_{\infty}$, for which the unit ball is
closed for the topology $\frak T_p$ of pointwise convergence.
(See {\smc Deville Godefroy \& Zizler 93}, VII.1.10.)
(ii) Show that the norm topology on $C(K)$, the weak topology on $C(K)$ and
$\frak T_p$ give rise to the same Borel $\sigma$-algebras.
(iii) Show that $\frak T_p$ has a $\sigma$-isolated network.   (iv) Show
that if
$w(K)$ is measure-free, then $(C(K),\frak T_p)$ is Radon, and every
$\frak T_p$-Radon measure on $C(K)$ is norm-Radon.
%438H 438L 438M 466F 466Yb 467Yc 467Yd 467G 467P mt46bits

}%end of exercises

\endnotes{
\Notesheader{467} The purpose of this section has been to give an idea
of the scope of Proposition 466F.   `Local uniform rotundity' has an
important place in the geometrical theory of Banach spaces, but for the
many associated ideas I refer you to
{\smc Deville Godefroy \& Zizler 93}.   From our point of view, Theorem
467E is therefore purely accessory, since we know by different arguments
that on separable Banach spaces the weak and norm topologies have the
same Borel $\sigma$-algebras (4A3W).   We need it to provide the first
step in the inductive proof of 467K.

Since the concept of `K-analytic' space is one of the fundamental ideas
of Chapter 43, it is natural here to look at `K-countably determined'
spaces, especially as many of the ideas of \S422 are directly
applicable (467Xa).   But the goal of this part of the argument is
Schachermayer's theorem 467P(b-ii), which uses `weakly compactly
generated' spaces (467L).   `Eberlein compacta' are of great
interest in other ways;  they are studied at length in
{\smc Arkhangel'skii 92}.

I mention order-continuous norms here (467N) because they are prominent in the theory of Banach lattices in Volume 3.
Note that the methods here do {\it not} suffice in
general to arrange that the locally uniformly rotund norm found on $X$
is a Riesz norm;  though see 467Yb.   It is in fact the case that every Banach lattice with an order-continuous norm has an equivalent locally uniformly rotund Riesz norm,
but this requires further ideas (see
{\smc Davis Ghoussoub \& Lindenstrauss 81}).

The general question of identifying Banach spaces with equivalent Kadec
norms remains challenging.
For a recent survey see {\smc Molt\'o Orihuela Troyanski \& Valdivia 09}.
}%end of notes

\discrpage

\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



