% 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{mt43.tex}
\versiondate{21.8.15}
\copyrightdate{1997}

\def\chaptername{Topologies and measures II}
\def\sectionname{Introduction}

\newchapter{43}

The first chapter of this volume was `general' theory of topological
measure spaces;  I attempted to distinguish the most important
properties a topological measure can have -- inner regularity,
$\tau$-additivity -- and describe their interactions at an abstract
level.   I now turn to rather more specialized investigations, looking
for features which offer explanations of the behaviour of the most
important spaces, radiating outwards from Lebesgue measure.

In effect, this chapter consists of three
distinguishable parts and two appendices.   The first three sections are
based on ideas from descriptive set theory, in particular Souslin's
operation (\S431);  the properties of this operation are the foundation
for the theory of two classes of topological space of particular
importance in measure theory, the
K-analytic spaces (\S432) and the analytic spaces (\S433).
The second part of the chapter, \S\S434-435, collects
miscellaneous results on Borel and Baire measures, looking at the ways
in which topological properties of a space determine properties of the
measures it carries.   In \S436 I
present the most important theorems on the
representation of linear functionals by integrals;  if you like, this is
the inverse operation to the construction of integrals from measures in
\S122.   The ideas continue into \S437, where I discuss spaces of signed
measures representing the duals of spaces of continuous functions, and
topologies on spaces of measures.
The first appendix, \S438, looks at a
special topic:  the way in which
the patterns in \S\S434-435 are affected if we assume that our spaces
are not unreasonably complex in a rather
special sense defined in terms of measures on discrete spaces.
Finally, I end the chapter with a further collection of
examples, mostly to exhibit boundaries to the theorems of the chapter,
but also to show some of the variety of the structures we are dealing
with.

\discrpage


\versiondate{4.8.15}
\copyrightdate{2000}

\def\chaptername{Topological measure spaces II}
\def\sectionname{Souslin's operation}

\newsection{431}

I begin the chapter with a short section on Souslin's
operation\cmmnt{ (\S421)}.   The basic facts we need to know are that
(in a complete locally determined measure space) the family of
measurable sets is closed under Souslin's operation (431A), and that the
kernel of a Souslin scheme can be approximated from within in measure
(431D).   \cmmnt{As in \S421,}
I write\cmmnt{ $S$ for $\bigcup_{k\in\Bbb N}\BbbN^k$ and}
$S^*$ for $\bigcup_{k\ge 1}\BbbN^k$.

\leader{431A}{Theorem} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space.   Then $\Sigma$ is closed under Souslin's
operation.

\proof{ Let $\family{\sigma}{S^*}{E_{\sigma}}$ be a Souslin scheme
in $\Sigma$ with kernel $A$.
If $F\in\Sigma$ and $\mu F<\infty$, then $A\cap F\in\Sigma$.
\Prf\ For each $\sigma\in S$, set

\Centerline{$A_{\sigma}
=\bigcup_{\phi\in\BbbN^{\Bbb N},\phi\supseteq\sigma}
  \bigcap_{n\ge 1}E_{\phi\restr n}$,}

\noindent and let $G_{\sigma}$ be a measurable envelope of
$A_{\sigma}\cap F$.   Because $A_{\sigma}\subseteq E_{\sigma}$ (writing
$E_{\emptyset}=X$), we may suppose that $G_{\sigma}\subseteq
E_{\sigma}\cap F$.   Now, for any $\sigma\in S$,

\Centerline{$A_{\sigma}\cap F
=\bigcup_{i\in\BbbN}A_{\sigma^{\smallfrown}\fraction{i}}\cap F
\subseteq\bigcup_{i\in\BbbN}G_{\sigma^{\smallfrown}\fraction{i}}$,}

\noindent so

\Centerline{$H_{\sigma}
=G_{\sigma}\setminus\bigcup_{i\in\Bbb N}
   G_{\sigma^{\smallfrown}\fraction{i}}$}

\noindent is negligible.

Set $H=\bigcup_{\sigma\in S}H_{\sigma}$, so that $H$ is negligible.
Take any $x\in G_{\emptyset}\setminus H$.   Choose
$\sequence{i}{\phi(i)}$
inductively, as follows.   Given that
$\sigma=\langle\phi(i)\rangle_{i<k}$
has been chosen and $x\in G_{\sigma}$, then $x\notin H_{\sigma}$, so
there must be some $j\in\BbbN$ such that $x\in G_{\sigma^{\smallfrown}\fraction{j}}$;
set $\phi(k)=j$, and continue.   Now

\Centerline{$x\in\bigcap_{k\ge 1}G_{\phi\restr k}
\subseteq\bigcap_{k\ge 1}E_{\phi\restr k}
\subseteq A$.}

Thus we see that $G_{\emptyset}\setminus H\subseteq A$;  as
$G_{\emptyset}\subseteq F$, $G_{\emptyset}\setminus H\subseteq A\cap F$.
On the other hand, $A\cap F\subseteq G_{\emptyset}$.   Because $H$ is
negligible and $\mu$ is complete,
$A\cap F\in\Sigma$.\ \Qed

Because $\mu$ is locally determined, it follows that
$A\in\Sigma$.   As $\family{\sigma}{S^*}{E_{\sigma}}$ is arbitrary,
$\Sigma$ is closed under Souslin's operation.
}%end of proof of 431A

\leader{431B}{Corollary} If $(X,\frak T,\Sigma,\mu)$ is a complete
locally determined topological measure space, every Souslin-F set in
$X$\cmmnt{ (definition:  421K)} is measurable.

\leader{431C}{Corollary} Let $X$ be a set and $\theta$ an outer measure
on $X$.   Let $\mu$ be the measure defined by \Caratheodory's method,
and $\Sigma$ its domain.   Then $\Sigma$ is closed under Souslin's
operation.

\proof{ Let $\family{\sigma}{S^*}{E_{\sigma}}$ be a Souslin scheme in
$\Sigma$ with kernel $A$.   Take any $C\subseteq X$ such that
$\theta C<\infty$.   Then $\theta_C=\theta\restrp\Cal PC$ is an outer
measure on $C$;  let $\mu_C$ be the measure on $C$ defined from
$\theta_C$ by \Caratheodory's method, and $\Sigma_C$ its domain.   If
$\sigma\in S^*$ and $D\subseteq C$ then

$$\eqalign{\theta_C(D\cap C\cap E_{\sigma})
  +\theta_C(D\setminus(C\cap E_{\sigma}))
&=\theta(D\cap E_{\sigma})+\theta(D\setminus E_{\sigma})\cr
&=\theta D=\theta_CD;\cr}$$

\noindent as $D$ is arbitrary, $C\cap E_{\sigma}\in\Sigma_C$.   $\mu_C$
is a complete totally finite measure, so 431A tells us that the kernel
of the Souslin scheme $\family{\sigma}{S^*}{C\cap E_{\sigma}}$ belongs to
$\Sigma_C$.   But this is just $C\cap A$ (applying 421Cb to the identity
map from $C$ to $X$).   So

\Centerline{$\theta(C\cap A)+\theta(C\setminus A)
=\theta_C(C\cap A)+\theta_C(C\setminus A)
=\theta_CC=\theta C$.}

\noindent As $C$ is arbitrary, $A\in\Sigma$ (113D).   As
$\family{\sigma}{S^*}{E_{\sigma}}$ is arbitrary, we have the result.
}%end of proof of 431C

\leader{431D}{Theorem} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, and $\family{\sigma}{S^*}{E_{\sigma}}$ a Souslin
scheme in $\Sigma$ with kernel $A$.

(a)

$$\eqalign{\mu A
&=\sup\{\mu(\bigcup_{\phi\in K}\bigcap_{n\ge 1}E_{\phi\restr n}):
  K\subseteq\BbbN^{\BbbN}\text{ is compact}\}\cr
&=\sup\{\mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr n}):
  \psi\in\BbbN^{\BbbN}\},\cr}$$

\noindent writing $\phi\le\psi$ if $\phi(i)\le\psi(i)$ for every
$i\in\Bbb N$.

(b)\dvAformerly{4{}31Xc}
If $\family{\sigma}{S^*}{E_{\sigma}}$ is fully regular, then
$\mu A=\sup\{\mu(\bigcap_{n\ge 1}E_{\psi\restr n}):\psi\in\NN\}$, and if
in addition $\mu$ is totally finite,
$\mu A=\sup\{\inf_{n\ge 1}\mu E_{\psi\restr n}:\psi\in\NN\}$.

\proof{{\bf (a)(i)} By 431A, $A$ is measurable.   For
$K\subseteq\BbbN^{\Bbb N}$, set
$H_K=\bigcup_{\phi\in K}\bigcap_{n\ge 1}E_{\phi\restr n}$.   Of course
$H_K\subseteq A$, and we know from 421M (or otherwise) that
$H_K\in\Sigma$ if $K$ is compact.   So surely $\mu A\ge\mu H_K$ for
every compact $K\subseteq \BbbN^{\BbbN}$.
If $\psi\in\BbbN^{\BbbN}$, then
$\{\phi:\phi\le\psi\}=\prod_{i\in\Bbb N}(\psi(i)+1)$ is compact.   We
therefore have

$$\eqalignno{\mu A
&\ge\sup\{\mu(\bigcup_{\phi\in K}\bigcap_{n\ge 1}E_{\phi\restr
n}):K\subseteq\BbbN^{\BbbN}\text{ is compact}\}\cr
&\ge\sup\{\mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr
n}):\psi\in\BbbN^{\BbbN}\}.\cr}$$

\noindent So what I need to prove is that

\Centerline{$\mu A\le\sup\{\mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge
1}E_{\phi\restr n}):\psi\in\BbbN^{\BbbN}\}$.}

\medskip

\quad{\bf (ii)} Fix on a set $F\in\Sigma$ of finite measure.
For $\sigma\in S$ set

\Centerline{$A_{\sigma}
=\bigcup_{\phi\in\BbbN^{\Bbb N},\phi\supseteq\sigma}
  \bigcap_{n\ge 1}E_{\phi\restr n}$.}

\noindent We need to know that $A_{\sigma}$ belongs to $\Sigma$;  this
follows from 431A, because writing $E'_{\tau}=E_{\tau}$ if
$\tau\subseteq\sigma$ or $\sigma\subseteq\tau$, $\emptyset$ otherwise,

\Centerline{$A_{\sigma}
=\bigcup_{\phi\in\BbbN^{\BbbN}}\bigcap_{n\ge 1}E'_{\phi\restr n}
\in\Cal S(\Sigma)=\Sigma$,}

\noindent writing $\Cal S$ for Souslin's operation, as in \S421.

Let $\epsilon>0$, and take a family
$\family{\sigma}{S}{\epsilon_{\sigma}}$ of strictly positive real
numbers such that $\sum_{\sigma\in S}\epsilon_{\sigma}\le\epsilon$.
For each $\sigma\in S$ we have $A_{\sigma}=\bigcup_{i\in\Bbb
N}A_{\sigma^{\smallfrown}\fraction{i}}$, so there is an $m_{\sigma}\in\BbbN$ such
that

\Centerline{$\mu(F\cap A_{\sigma}\setminus\bigcup_{i\le
m_{\sigma}}A_{\sigma^{\smallfrown}\fraction{i}})\le\epsilon_{\sigma}$.}

\noindent Define $\psi\in\BbbN^{\BbbN}$ by saying that

\Centerline{$\psi(k)=\max\{m_{\sigma}:\sigma\in\Bbb
N^k,\,\sigma(i)\le\psi(i)$ for every $i<k\}$}

\noindent for each $k\in\BbbN$.  Set

\Centerline{$H=\bigcup_{\phi\in\BbbN^{\BbbN},\phi\le\psi}\bigcap_{n\ge
1}E_{\phi\restr n}$.}

\medskip

\quad{\bf (iii)} Set

\Centerline{$G=\bigcup_{\sigma\in S}F\cap
A_{\sigma}\setminus\bigcup_{i\le m_{\sigma}}A_{\sigma^{\smallfrown}\fraction{i}}$,}

\noindent so that $\mu G\le\epsilon$, by the choice of the
$\epsilon_{\sigma}$ and the $m_{\sigma}$.   Then
$F\cap A\setminus G\subseteq H$.
\Prf\ If $x\in F\cap A\setminus G$, choose
$\sequence{i}{\phi(i)}$ inductively, as follows.   Given that
$\phi(i)\le\psi(i)$ for $i<k$ and $x\in A_{\sigma}$, where
$\sigma=\langle\phi(i)\rangle_{i<k}$, then $x\notin
A_{\sigma}\setminus\bigcup_{j\le m_{\sigma}}A_{\sigma^{\smallfrown}\fraction{j}}$,
so there must be some $j\le m_{\sigma}$ such that
$x\in A_{\sigma^{\smallfrown}\fraction{j}}$;  set $\phi(k)=j$;  because
$\sigma\in\prod_{i<k}(\psi(i)+1)$, $j\le m_{\sigma}\le\psi(k)$, and the
induction continues.   At the end of the induction, $\phi\le\psi$ and

\Centerline{$x\in\bigcap_{n\ge 1}A_{\phi\restr n}\subseteq\bigcap_{n\ge
1}E_{\phi\restr n}\subseteq H$.   \Qed}

\medskip

\quad{\bf (iv)} It follows that

\Centerline{$\mu(A\cap F)\le\mu G+\mu H\le\epsilon+\mu H$.}

\noindent As $F$ and $\epsilon$ are arbitrary, and $\mu$ is semi-finite,

\Centerline{$\mu A\le\sup\{\mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge
1}E_{\phi\restr n}):\psi\in\BbbN^{\BbbN}\}$,}

\noindent and (a) is true.

\medskip

{\bf (b)} If $\family{\sigma}{S^*}{E_{\sigma}}$ is fully regular, and
$\phi\le\psi$, then $\phi\restr n\le\psi\restr n$ and
$E_{\phi\restr n}\subseteq E_{\psi\restr n}$ for every $n\ge 1$;
consequently $\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr n}
\subseteq\bigcap_{n\in\Bbb N}E_{\psi\restr n}$ for every $\psi\in\NN$, and

$$\eqalign{\mu A
&=\sup\{\mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr n}):
  \psi\in\BbbN^{\BbbN}\}
\cr&=\sup\{\mu(\bigcap_{n\ge 1}E_{\psi\restr n}):
  \psi\in\BbbN^{\BbbN}\}.
\cr}$$

\noindent Moreover, $E_{\psi\restr n}\subseteq E_{\psi\restr m}$ whenever
$1\le m\le n$, so if $\mu$ is totally finite,
$\mu(\bigcap_{n\ge 1}E_{\psi\restr n})=\inf_{n\ge 1}\mu E_{\psi\restr n}$
for every $\psi$, and
$\mu A=\sup_{\psi\in\NN}\inf_{n\ge 1}\mu E_{\psi\restr n}$.
}%end of proof of 431D

\leader{431E}{Corollary} If $(X,\frak T,\Sigma,\mu)$ is a topological
measure space and $E\subseteq X$ is a Souslin-F set with finite outer
measure, then $\mu^*E=\sup\{\mu F:F\subseteq E$ is closed$\}$.

\proof{ Let $\tilde\mu$ be the c.l.d.\ version of
$\mu$\cmmnt{ (213E)}.   Let $\family{\sigma}{S^*}{E_{\sigma}}$ be a
Souslin scheme of closed sets with kernel $E$.   Then 213Fb and 431D
tell us that

\Centerline{$\mu^*E=\tilde\mu E
=\sup_{K\subseteq\NN\text{ is compact}}\mu F_K$,}

\ifdim\pagewidth>467pt\fontdimen3\tenrm=2.5pt\fi
\ifdim\pagewidth>467pt\fontdimen4\tenrm=1.67pt\fi
\noindent where
$F_K=\bigcup_{\phi\in K}\bigcap_{n\ge 1}E_{\phi\restr n}$ for
$K\subseteq\NN$.   But every $F_K$ is closed, by 421M.   So
$\mu^*E\penalty-100\le\penalty-100
\sup_{F\subseteq E\text{ is closed}}\mu F$;  as the reverse
inequality is trivial, we have the result.
\fontdimen3\tenrm=1.67pt
\fontdimen4\tenrm=1.11pt
}%end of proof of 431E

\leader{*431F}{}\cmmnt{ Two further versions of the ideas in
431A will be useful.   The first is topological.

\medskip

\noindent}{\bf Theorem} Let $X$ be any topological space, and
$\widehat{\Cal B}$ its Baire-property algebra.

(a) For any $A\subseteq X$, there is a Baire-property envelope of $A$,
that is, a set $E\in\widehat{\Cal B}$ such that
$A\subseteq E$ and $E\setminus F$ is meager whenever
$A\subseteq F\in\widehat{\Cal B}$.

(b) $\widehat{\Cal B}$ is closed under Souslin's operation.

\proof{{\bf (a)} By 4A3Sa, there is an open set $H\subseteq X$ such that
$A\setminus H$ is meager and $H\cap G$ is empty whenever $G\subseteq X$
is open and $A\cap G$ is meager.   Set $E=A\cup H$;  then $E\supseteq A$
and $E\symmdiff H=A\setminus H$ is meager, so $E\in\widehat{\Cal B}$.
If $A\subseteq F\in\widehat{\Cal B}$, let $G$ be an open set such that
$G\symmdiff(X\setminus F)$ is meager.   Then $G\cap A\subseteq G\cap F$
is meager, so $G\cap H$ is empty and
$E\setminus F\subseteq(E\symmdiff H)\cup(G\symmdiff(X\setminus F))$ is
meager.  Thus $E$ is a Baire-property envelope of $A$.

\medskip

{\bf (b)} Let $\family{\sigma}{S^*}{E_{\sigma}}$ be a Souslin scheme
in $\widehat{\Cal B}$ with kernel $A$.   For each $\sigma\in S$, set

\Centerline{$A_{\sigma}
=\bigcup_{\phi\in\BbbN^{\Bbb N},\phi\supseteq\sigma}
  \bigcap_{n\ge 1}E_{\phi\restr n}$,}

\noindent and let $G_{\sigma}$ be a Baire-property envelope of
$A_{\sigma}$ as described in (a).   Because
$A_{\sigma}\subseteq E_{\sigma}$ (writing
$E_{\emptyset}=X$), we may suppose that
$G_{\sigma}\subseteq E_{\sigma}$.   Now, for any $\sigma\in S$,

\Centerline{$A_{\sigma}
=\bigcup_{i\in\BbbN}A_{\sigma^{\smallfrown}\fraction{i}}
\subseteq\bigcup_{i\in\BbbN}G_{\sigma^{\smallfrown}\fraction{i}}$,}

\noindent so

\Centerline{$H_{\sigma}
=G_{\sigma}\setminus\bigcup_{i\in\Bbb N}G_{\sigma^{\smallfrown}\fraction{i}}$}

\noindent is meager.

Set $H=\bigcup_{\sigma\in S}H_{\sigma}$, so that $H$ is meager.
Take any $x\in G_{\emptyset}\setminus H$.   Choose
$\sequence{i}{\phi(i)}$
inductively, as follows.   Given that
$\sigma=\langle\phi(i)\rangle_{i<k}$
has been chosen and $x\in G_{\sigma}$, then $x\notin H_{\sigma}$, so
there must be some $j\in\BbbN$ such that
$x\in G_{\sigma^{\smallfrown}\fraction{j}}$;   set $\phi(k)=j$, and continue.   Now

\Centerline{$x\in\bigcap_{k\ge 1}G_{\phi\restr k}
\subseteq\bigcap_{k\ge 1}E_{\phi\restr k}
\subseteq A$.}

Thus we see that $G_{\emptyset}\setminus H\subseteq A$.
On the other hand, $A\subseteq G_{\emptyset}$, so
$G_{\emptyset}\symmdiff A$ is meager and $A\in\widehat{\Cal B}$.   As
$\family{\sigma}{S^*}{E_{\sigma}}$ is arbitrary,
$\widehat{\Cal B}$ is closed under Souslin's operation.
}%end of proof of 431F

\vleader{48pt}{*431G}{}\cmmnt{ The second relies on a countable chain condition
to give the same envelope property.

\medskip

\noindent}{\bf Theorem} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$ and $\Cal I\subseteq\Sigma$ a $\sigma$-ideal of subsets
of $X$.   If $\Sigma/\Cal I$ is ccc then $\Sigma$ is closed under
Souslin's operation.

\proof{{\bf (a)} As before, the essential fact is that for every
$A\subseteq X$
there is an $E\in\Sigma$ such that $A\subseteq E$ and $F\in\Cal I$
whenever $F\in\Sigma$ and
$F\subseteq E\setminus A$.   \Prf\ Let $\Cal E$ be a maximal disjoint
family of members of $\Sigma\setminus\Cal I$ disjoint from $A$.
Because $\Sigma/\Cal I$ is
ccc, $\Cal E$ is countable (316C), so $E=X\setminus\bigcup\Cal E$
belongs to $\Sigma$;  now it is easy to see that this $E$ serves.\ \Qed

In this case I will call $E$ a `measurable envelope' of $A$.

\medskip

{\bf (b)} Now we can argue as in 431A or 431F.   Let
$\family{\sigma}{S^*}{E_{\sigma}}$ be a Souslin scheme
in $\Sigma$ with kernel $A$;  for $\sigma\in S$, set

\Centerline{$A_{\sigma}
=\bigcup_{\phi\in\BbbN^{\Bbb N},\sigma\subseteq\phi}
  \bigcap_{n\ge 1}E_{\phi\restr n}$,}

\noindent and let $G_{\sigma}\subseteq E_{\sigma}$
be a measurable envelope of $A_{\sigma}$.   Setting

\Centerline{$H=\bigcup_{\sigma\in S}
  (G_{\sigma}\setminus\bigcup_{i\in\Bbb N}
    G_{\sigma^{\smallfrown}\fraction{i}})$,}

\noindent $H\in\Cal I$ and $G_{\emptyset}\symmdiff A\subseteq H$, so
$A\in\Sigma$.
As $\family{\sigma}{S^*}{E_{\sigma}}$ is arbitrary, $\Sigma$ is closed under
Souslin's operation.
}%end of proof of 431G

\exercises{
\leader{431X}{Basic exercises (a)}
%\spheader 431Xa
Let $(X,\frak T,\Sigma,\mu)$ be a complete locally
determined topological measure space, $Y$ a topological space,
and $f:X\to Y$ a measurable function.   Let $\Cal B(Y)$ be the Borel
algebra of $Y$.   Show that $f^{-1}[B]\in\Sigma$ for every
$B\in\Cal S(\Cal B(Y))$.
%431A

\spheader 431Xb  Let $X$ be a topological space and $\mu$ a semi-finite
topological measure on $X$ which is inner regular with respect to the
Souslin-F sets.   Show that $\mu$ is inner regular with respect to the
closed sets.
%431D

\sqheader 431Xc Let $(X,\Sigma,\mu)$ be a measure space with locally
determined negligible sets (definition:  213I), and
$\family{\sigma}{S^*}{E_{\sigma}}$ a Souslin scheme in $\Sigma$
with kernel $A$.   Show that

\Centerline{$\mu^*A
=\sup_{\psi\in\BbbN^{\Bbb N}}
  \mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr n})$.}
%431D

\sqheader 431Xd Let $(X,\Sigma,\mu)$ be a semi-finite measure space, and
$\family{\sigma}{S^*}{E_{\sigma}}$ a Souslin scheme in $\Sigma$
with kernel $A$.   Show that

\Centerline{$\mu_*A
=\sup_{\psi\in\BbbN^{\Bbb N}}
  \mu(\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr n})$.}
%431D

\leader{431Y}{Further exercises (a)}
%\spheader 431Ya
Let $(X,\Sigma,\mu)$ be a complete measure space with
the measurable envelope property.   Show that $\Sigma$ is closed
under Souslin's operation.
%431A

\spheader 431Yb\dvAnew{2014} Let $(X,\Sigma,\mu)$ be a complete
totally finite
measure space, $Y$ a set and $\Tau$ a $\sigma$-algebra of subsets of $Y$.
Suppose that $A\in\Cal S(\Sigma\tensorhat\Tau)$.   Show that
$\{y:\mu A^{-1}[\{y\}]>\alpha\}$ belongs to $\Cal S(\Tau)$ for every
$\alpha\in\Bbb R$.
%431Db 421Cb 252P

\spheader 431Yc Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets
of $X$, and $\Cal I$ a $\sigma$-ideal of subsets of $X$ such that
$\Cal I\subseteq\Sigma$.   Suppose that for every $A\subseteq X$ there
is an $F\in\Sigma$ such that
$A\subseteq F$ and $F\setminus E\in\Cal I$ whenever
$A\subseteq E\in\Sigma$.   Show that $\Sigma$ is closed under Souslin's
operation.
%431F

\spheader 431Yd\dvAnew{2015}
Let $r\ge 1$ be an integer, $D\subseteq\BbbR^r$ a Borel set
and $f:D\to\Bbb R$ a Borel measurable function.   Show that the domain of
its first partial derivative $\pd{f}{\xi_1}$ is coanalytic, therefore
Lebesgue measurable, but may fail to be Borel.
%423T 431A out of order query

}%end of exercises

\endnotes{\Notesheader{431}
From the point of view of measure theory, the most important property of
Souslin's operation, after its idempotence, is the fact that (for many
measure spaces) the family of measurable sets is closed under the
operation (431A).   The proof I give here is based on the concept of
measurable envelope, which can be used in other cases of great interest
(431F, 431G, 431Yc).   But for some applications it is also very important to
know that if $A$ is the kernel of a Souslin scheme
$\family{\sigma}{S^*}{E_{\sigma}}$,
then $A$ can be approximated from inside by sets of the form
$H=\bigcup_{\phi\le\psi}\bigcap_{n\ge 1}E_{\phi\restr n}$ (431D, 431Xc),
which belong to the $\sigma$-algebra generated by the $E_{\sigma}$
(421M).   A typical application of this idea is when every $E_{\sigma}$
is a Borel subset of $\Bbb R$;  then we find not only that $A$ is
Lebesgue measurable
(indeed, measured by every Radon measure on $\Bbb R$) but that (for
any given Radon measure $\mu$) the Souslin scheme itself provides
Borel subsets $H$ of $A$ of measure approximating the measure of $A$.
A similar result, based on rather different hypotheses, is in 432K.

Let me repeat that the essence of descriptive set theory is that we are
not satisfied merely to know that a set of a certain type exists.   We
want also to know how to build it, because we expect that an explicit
construction will be valuable later on.   For instance, the construction
given in 431D shows that if the Souslin scheme consists of closed
compact sets, the sets $H$ will be compact (421Xn).

I mention 431B as a typical application of 431A, even though it is both
obvious and obviously less than what can be said.   The algebras
$\Sigma$ of this section are algebras closed under Souslin's operation.
In a complete locally determined topological measure space, the algebra
$\Sigma$ of measurable sets includes the open sets (by definition),
therefore the Borel algebra $\Cal B$, therefore $\Cal S(\Cal B)$;  but
now we can take the algebra $\Cal A_1$ generated by $\Cal S(\Cal B)$,
and $\Cal A_1$ and $\Cal S(\Cal A_1)$ will also be included in $\Sigma$,
so that $\Sigma$ will included the algebra $\Cal A_2$ generated by
$\Cal S(\Cal A_1)$, and so on.   (Note that $\Cal S(\Cal A_1)$ includes
the $\sigma$-algebra generated by $\Cal A_1$, by 421F, so I do not need
to mention that separately.)   We have to run through all the countable
ordinals before we can be sure of getting to the smallest algebra
$\Cal A_{\omega_1}=\bigcup_{\xi<\omega_1}\Cal A_{\xi}$ which contains
every open set and is closed under Souslin's operation, and we shall
then have $\Cal A_{\omega_1}\subseteq\Sigma$.
}%end of notes

\discrpage

\frfilename{mt432.tex}
\versiondate{2.10.13}
\copyrightdate{2008}

\def\NN{\BbbN^{\Bbb N}}

\def\chaptername{Topologies and measures II}
\def\sectionname{K-analytic spaces}

\newsection{432}

I describe the basic measure-theoretic properties of K-analytic
spaces\cmmnt{ (\S422)}.   I start with `elementary' results
(432A-432C), assembling ideas from \S\S421, 422 and 431.   The main
theorem of the section is 432D, one of the leading cases of the general
extension theorem 416P.   An important corollary (432G) gives a
sufficient condition for the existence of pull-back measures.   I
briefly mention `capacities' (432J-432L).

\leader{432A}{Proposition} Let $(X,\frak T,\Sigma,\mu)$ be a complete
locally determined Hausdorff topological measure space.   Then every
K-analytic subset of $X$ is measurable.

\proof{ If $A\subseteq X$ is K-analytic, it is Souslin-F (422Ha),
therefore measurable (431B).
}%end of proof of 432A

\leader{432B}{Theorem} Let $X$ be a K-analytic Hausdorff space, and
$\mu$ a semi-finite topological measure on $X$.   Then

\Centerline{$\mu X=\sup\{\mu K:K\subseteq X$ is compact$\}$.}

\proof{ If $\gamma<\mu X$, there is an $E\in\dom\mu$ such that
$\gamma<\mu E<\infty$;   set $\nu F=\mu(E\cap F)$ for every Borel set
$F\subseteq X$, so that $\nu$ is a totally finite Borel measure on $X$,
and $\nu X>\gamma$.   Let $\hat\nu$ be the completion of $\nu$.   Let
$R\subseteq\NN\times X$ be an usco-compact relation such
that $R[\NN]=X$.   Set $F_{\sigma}=\overline{R[I_{\sigma}]}$ for
$\sigma\in S^*=\bigcup_{k\ge 1}\BbbN^k$, where
$I_{\sigma}=\{\phi:\sigma\subseteq\phi\in\NN\}$.   Because $R$ is closed
in $\NN\times X$ (422Da), $X$ is the kernel of the Souslin scheme
$\family{\sigma}{S^*}{F_{\sigma}}$ (421I).   By 431D, there is a compact
$L\subseteq\NN$ such that
$\hat\nu(\bigcup_{\phi\in L}\bigcap_{n\in\Bbb N}F_{\phi\restr n})
\ge\gamma$.
But, by 421I, this is just $\hat\nu(R[L])$;  and $R[L]$ is compact, by
422D(e-i).   So $\mu R[L]$ is defined, with
$\mu R[L]\ge\nu R[L]=\hat\nu R[L]$, and we have a compact subset of $X$ of measure at least
$\gamma$.   As $\gamma$ is arbitrary, the theorem is proved.
}%end of proof of 432B

\leader{432C}{Proposition} Let $X$ be a Hausdorff space such that all
its open sets are K-analytic, and $\mu$ a Borel measure on $X$.

(a) If $\mu$ is semi-finite, it is tight.

(b) If $\mu$ is locally finite, its completion is a Radon measure on
$X$.

\proof{{\bf (a)} By 422Hb, every open subset of $X$ is Souslin-F.
Applying 421F to the family $\Cal E$ of closed
subsets of $X$, we see that every Borel subset of $X$ is Souslin-F,
therefore K-analytic (422Ha).
Now suppose that $E\subseteq X$ is a Borel set.   Then the subspace
measure $\mu_E$ is a semi-finite Borel measure on the
K-analytic space $E$, so by 432B
$\mu E=\sup_{K\subseteq E\text{ is compact}}\mu K$.   As $E$ is
arbitrary, $\mu$ is inner regular with respect to the compact sets;  but we are supposing that $X$ is Hausdorff, so these are all closed, and $\mu$ is tight.

\medskip

{\bf (b)} Because $X$ is Lindel\"of (422Gg), $\mu$ is $\sigma$-finite
(411Ge), therefore semi-finite.   So (a) tells us that $\mu$ is tight.
By 416F, its c.l.d.\ version
is a Radon measure.   But (because $\mu$ is $\sigma$-finite) this is
just its completion (213Ha).
}%end of proof of 432C

\leader{432D}{Theorem}\cmmnt{ ({\smc Aldaz \& Render 00})} Let
$X$ be a K-analytic Hausdorff space and $\mu$ a locally finite measure
on $X$ which is inner regular with respect to the closed sets.   Then
$\mu$ has an extension to a Radon measure on $X$.   In particular, $\mu$
is $\tau$-additive.

\proof{ The point is that if $\mu E>0$ then there is a compact
$K\subseteq E$ such that $\mu^*K>0$.   \Prf\ Write $\Sigma$ for the
domain of $\mu$.   Take $\gamma<\mu E$.   Because $X$ is Lindel\"of
(422Gg again), $\mu$ is $\sigma$-finite (411Ge), therefore semi-finite.   Let
$E'\subseteq E$ be such that
$\gamma<\mu E'<\infty$.   Because $\mu$ is inner regular with respect to
the closed sets, there is a closed set $F\subseteq E$ such that
$\mu F>\gamma$.   $F$ is K-analytic (422Gf);  let
$R\subseteq\NN\times F$ be an
usco-compact relation such that $R[\NN]=F$.  For
$\sigma\in S=\bigcup_{n\in\Bbb N}\BbbN^n$ set

\Centerline{$A_{\sigma}=\{x:(\phi,x)\in R$ for some $\phi\in\NN$ such
that $\phi(i)\le\sigma(i)$ for every $i<\#(\sigma)\}$.}

\noindent Then $\sequence{i}{A_{\sigma^{\smallfrown}\fraction{i}}}$ is a
non-decreasing sequence with union $A_{\sigma}$, so

\Centerline{$\mu^*A_{\sigma}
=\sup_{i\in\Bbb N}\mu^*A_{\sigma^{\smallfrown}\fraction{i}}$}

\noindent for every $\sigma\in S$ (132Ae).
We can therefore find a sequence $\psi\in\NN$ such that

\Centerline{$\mu^*A_{\psi\restr n}>\gamma$}

\noindent for every $n\in\Bbb N$.   Set

\Centerline{$K
=\{\phi:\phi\in\NN,\,\phi(i)\le\psi(i)$ for every $i\in\Bbb N\}$;}

\noindent then $K=\prod_{n\in\Bbb N}(\psi(n)+1)$ is compact, so $R[K]$
is compact (422D(e-i) again).

\Quer\ Suppose, if possible, that $\mu^*R[K]<\gamma$.   Then there is an
$H\in\Sigma$ such that $R[K]\subseteq H\subseteq F$ and
$\mu(F\setminus H)>\mu F-\gamma$.   Because $\mu$ is inner regular with
respect to the closed sets, there is a closed set $F'\in\Sigma$ such that
$F'\subseteq F\setminus H$ and $\mu F'>\mu F-\gamma$.    Since
$R[K]\cap F'=\emptyset$, $K\cap R^{-1}[F']=\emptyset$.   $R^{-1}[F']$
is closed, because $R$ is usco-compact, so there is some $n$ such that

\Centerline{$L=\{\phi:\phi\in\NN$, $\phi\restr n=\phi'\restr n$ for some
$\phi'\in K\}$}

\noindent does not meet $R^{-1}[F']$ (4A2F(h-vi)), and
$R[L]\cap F'=\emptyset$.
But $L$ is just $\{\phi:\phi(i)\le\psi(i)$ for every $i<n\}$, so
$R[L]=A_{\psi\restr n}$, and

\Centerline{$\gamma<\mu^*A_{\psi\restr n}\le\mu(F\setminus F')<\gamma$,}

\noindent which is absurd.\ \Bang

Thus $\mu^*R[K]\ge\gamma$.   As $\gamma>0$, we have the
result.\ \Qed

Now the theorem follows at once from 416P(ii)$\Rightarrow$(i).
}%end of proof of 432D

\leader{432E}{Corollary} Let $X$ be a K-analytic Hausdorff space, and
$\mu$ a locally finite quasi-Radon measure on $X$.   Then $\mu$ is a
Radon measure.

\proof{ By 432D, $\mu$ has an extension to a Radon measure $\mu'$.   But
of course $\mu$ and $\mu'$ must coincide, by 415H or otherwise.
}%end of proof of 432E

\leader{432F}{Corollary} Let $X$ be a K-analytic Hausdorff space,
and $\nu$ a locally finite Baire measure on $X$.   Then $\nu$ has an
extension to a Radon measure on $X$;  in particular, it is
$\tau$-additive.   If the topology of $X$ is
regular, the extension is unique.

\proof{ Because $X$ is Lindel\"of (422Gg once more), $\nu$ is $\sigma$-finite,
therefore semi-finite;  by 412D, it is inner regular with
respect to the closed sets.   So 432D tells us that it has an extension
to a Radon measure on $X$.   Since the extension is $\tau$-additive, so
is $\nu$.

If $X$ is regular, then it must be completely regular (4A2H(b-i)), and
the family $\Cal G$ of cozero sets is a base for the topology closed
under finite unions.   If $\mu$, $\mu'$ are Radon measures extending
$\nu$, they agree on $\Cal G$, and must be equal, by 415H(iv).
}%end of proof of 432F

\leader{432G}{Corollary} Let $X$ be a K-analytic Hausdorff space, $Y$ a
Hausdorff space and $\nu$ a locally finite measure on $Y$ which is inner
regular with respect to the closed sets.   Let $f:X\to Y$ be a
continuous function such that $f[X]$ has full outer measure in $Y$.
Then there is a Radon
measure $\mu$ on $X$ such that $f$ is \imp\ for $\mu$ and $\nu$.   If
$\nu$ is Radon, it is precisely the image measure $\mu f^{-1}$.

\proof{{\bf (a)} Write $\Tau$ for the domain of $\nu$, and set
$\Sigma_0=\{f^{-1}[F]:f\in\Tau\}$, so that $\Sigma_0$ is a
$\sigma$-algebra of subsets of $X$, and we have a measure $\mu_0$ on $X$
defined by setting $\mu_0f^{-1}[F]=\nu F$ whenever $F\in\Tau$
(234F).

\medskip

{\bf (b)} If $E\in\Sigma_0$ and $\gamma<\mu_0E$, there is an $F\in\Tau$
such that $E=f^{-1}[F]$.   Now there is a closed set $F'\subseteq F$
such that $\nu F'\ge\gamma$.   Because $f$ is continuous, $f^{-1}[F']$
is closed, and we have $f^{-1}[F']\subseteq E$ and
$\mu_0f^{-1}[F']\ge\gamma$.   As $E$ and $\gamma$ are arbitrary, $\mu_0$
is inner regular with respect to the closed sets.

If $x\in X$, then (because $\nu$ is locally finite) there is an open
set $H\subseteq Y$ such that $f(x)\in H$ and
$\nu^*H<\infty$;  as $f$ is of course \imp\
for $\mu_0$ and $\nu$, $\mu_0^*f^{-1}[H]\le\nu^*H$ (234B(f-i)) is finite,
while $f^{-1}[H]$ is an open set containing $x$.   Thus $\mu_0$ is locally
finite.

\medskip

{\bf (c)} By 432D, there is a Radon measure $\mu$ on $X$ extending
$\mu_0$.   Because $f$ is \imp\ for $\mu_0$ and $\nu$, it is surely
\imp\ for $\mu$ and $\nu$.

The image measure $\mu f^{-1}$ extends $\nu$, so must be locally finite;
it is therefore a Radon measure (418I).   So if $\nu$ itself is a Radon
measure, it must be identical with $\mu f^{-1}$, by 416Eb.
}%end of proof of 432G

\leader{432H}{Corollary} Suppose that $X$ is a set and that $\frak S$,
$\frak T$ are Hausdorff topologies on $X$ such that $(X,\frak T)$ is
K-analytic and $\frak S\subseteq\frak T$.   Then
the totally finite Radon measures on $X$ are the same for $\frak S$ and
$\frak T$.

\proof{ Write $f$ for the identity function on $X$ regarded as a
continuous function from $(X,\frak T)$ to $(X,\frak S)$.   If $\mu$ is a
totally finite $\frak T$-Radon measure on $X$, then $\mu=\mu f^{-1}$ is
$\frak S$-Radon, by 418I again.
If $\nu$ is a totally finite $\frak S$-Radon
measure on $X$, then 432G tells us that it is of the form
$\mu=\mu f^{-1}$ for some $\frak T$-Radon measure $\mu$, that is, is
itself $\frak T$-Radon.
}%end of proof of 432H

\leader{432I}{Corollary}\dvAnew{2010}
Let $X$ be a K-analytic Hausdorff space, and
$\Cal U$ a subbase for the topology of $X$.   Let $(Y,\Tau,\nu)$ be a
complete totally finite measure space and $\phi:Y\to X$ a function such
that $\phi^{-1}[U]\in\Tau$ for every $U\in\Cal U$.   Then there is a Radon
measure $\mu$ on $X$ such that $\int fd\mu=\int f\phi\,d\nu$ for every
bounded continuous $f:X\to\Bbb R$.

\proof{{\bf (a)} Let $\nu\phi^{-1}$ be the image measure on $X$, and
$\Sigma_0$ its domain.   Then $\Sigma_0$ is a
$\sigma$-algebra of subsets of $X$ including $\Cal U$.   So if $x$, $y$ are
distinct points of $X$, there are disjoint open sets $U$, $V\in\Sigma_0$
containing $x$, $y$ respectively.   \Prf\ Because $X$ is Hausdorff, there
are disjoint open sets $U_0$ and $V_0$ such that $x\in U_0$ and $y\in V_0$.
Because $\Sigma_0$ is closed under finite intersections and includes the
subbase $\Cal U$, it includes a base for the topology of $X$ (4A2B(a-i)),
and there are open sets $U$, $V\in\Sigma_0$ such that
$x\in U\subseteq U_0$ and $y\in V\subseteq V_0$.\ \Qed

Every
cozero subset of $X$ belongs to $\Sigma_0$.   \Prf\ If $G\subseteq X$ is a
cozero set, there is a sequence $\sequencen{F_n}$ of closed subsets of $X$
with union $G$.   For each $n$, $F_n$ and $X\setminus G$ are disjoint
K-analytic subsets of $X$ (422Gf), so there is an $E_n\in\Sigma_0$ such that
$F_n\subseteq E_n\subseteq G$ (422I).   Now $G=\bigcup_{n\in\Bbb N}E_n$
belongs to $\Sigma_0$.\ \Qed

\medskip

{\bf (b)} It follows that the Baire $\sigma$-algebra $\CalBa(X)$ of $X$ is
included in $\Sigma_0$.   So $\mu_0=\nu\phi^{-1}\restr\CalBa(X)$
is a Baire
measure on $X$.   By 432F, $\mu_0$ has an extension to a Radon measure
$\mu$ on $X$.

If $f\in C_b(X)$, then $f$ is $\mu_0$-integrable;  since $\phi$ is \imp\
for $\nu$ and $\mu_0$, $\int f\phi\,d\nu$ is defined and equal to
$\int fd\mu_0$ (235G).   Similarly $\int fd\mu=\int fd\mu_0$.
So $\int fd\mu=\int f\phi\,d\nu$, as required.
}%end of proof of 432I

\leader{432J}{\dvrocolon{Capacitability}}\cmmnt{ The next theorem is
not exactly measure theory as studied in most of this treatise;
but it is clearly very close to the other ideas of this section, and it has
important applications to measure theory in the narrow sense.

\medskip

\noindent}{\bf Definitions}\dvAformerly{4{}32I}
Let $(X,\frak T)$ be a topological space.

\spheader 432Ja A {\bf Choquet capacity} on $X$ is a function
$c:\Cal PX\to[0,\infty]$ such that

\quad(i) $c(A)\le c(B)$ whenever $A\subseteq B\subseteq X$;

\quad(ii) $\lim_{n\to\infty}c(A_n)=c(A)$ whenever $\sequencen{A_n}$ is a
non-decreasing sequence of subsets of $X$ with union $A$;

\quad(iii) $c(K)=\inf\{c(G):G\supseteq K$ is open$\}$ for every compact set
$K\subseteq X$.
%Kechris demands c(K)<\infty for compact K

\spheader 432Jb\dvAnew{2008}
A Choquet capacity $c$ on $X$ is {\bf outer regular} if
$c(A)=\inf\{c(G):G\supseteq A$ is open$\}$ for every $A\subseteq X$.

\leader{432K}{Theorem}\cmmnt{ ({\smc Choquet 55})} Let $X$ be a
Hausdorff space and $c$ a Choquet capacity on $X$.   If $A\subseteq X$ is
K-analytic, then $c(A)=\sup\{c(K):K\subseteq A$ is compact$\}$.

\proof{ Take $\gamma<c(A)$.   Let $R\subseteq\NN\times X$ be an
usco-compact relation such that $R[\NN]=A$;  for
$\sigma\in S=\bigcup_{n\in\Bbb N}\BbbN^n$ set

\Centerline{$A_{\sigma}=\{x:(\phi,x)\in R$ for some $\phi\in\NN$ such
that $\phi(i)\le\sigma(i)$ for every $i<\#(\sigma)\}$.}

\noindent Then $\sequence{i}{A_{\sigma^{\smallfrown}\fraction{i}}}$ is a
non-decreasing sequence with union $A_{\sigma}$, so

\Centerline{$c(A_{\sigma})=\sup_{i\in\Bbb
N}c(A_{\sigma^{\smallfrown}\fraction{i}})$}

\noindent for every $\sigma\in S$.   We can therefore find a sequence
$\psi\in\NN$ such that $c(A_{\psi\restr n})>\gamma$ for every
$n\in\Bbb N$.   Set

\Centerline{$K=\{\phi:\phi\in\NN,\,\phi(i)\le\psi(i)$ for every
$i\in\Bbb N\}$;}

\noindent then $K=\prod_{n\in\Bbb N}(\psi(n)+1)$ is compact, so $R[K]$
is compact (422D(e-i) once more).

\Quer\ Suppose, if possible, that $c(R[K])<\gamma$.   Then, by (iii) of
432J, there is an open set $G\supseteq R[K]$ such that $c(G)<\gamma$.
Set $F=X\setminus G$, so that $F$ is closed and
$K\cap R^{-1}[F]=\emptyset$.   $R^{-1}[F]$ is closed, because $R$ is
usco-compact, so there is some $n$ such that

\Centerline{$L=\{\phi:\phi\in\NN$, $\phi\restr n=\phi'\restr n$ for some
$\phi'\in K\}$}

\noindent does not meet $R^{-1}[F]$ (4A2F(h-vi) again), and
$R[L]\cap F=\emptyset$, that is, $R[L]\subseteq G$.   But $L$ is just
$\{\phi:\phi(i)\le\psi(i)$ for every $i<n\}$, so
$R[L]=A_{\psi\restr n}$, and

\Centerline{$\gamma<c(A_{\psi\restr n})\le c(G)<\gamma$,}

\noindent which is absurd.\ \Bang

Thus $c(R[K])\ge\gamma$.   As $\gamma$ is arbitrary and $R[K]$ is
compact, we have the result.
}%end of proof of 432K

\leader{432L}{Proposition}\dvAnew{2008}
Let $(X,\frak T)$ be a topological space.

(a) Let $c_0:\frak T\to[0,\infty]$ be a functional such that

\inset{$c_0(G)\le c_0(H)$ whenever $G$, $H\in\frak T$ and $G\subseteq H$;

$c_0$ is submodular\cmmnt{ (definition: 413Qb)}

$c_0(\bigcup_{n\in\Bbb N}G_n)=\lim_{n\to\infty}c_0(G_n)$ for every
non-decreasing sequence $\sequencen{G_n}$ in $\frak T$.}

\noindent Then $c_0$ has a unique extension to an outer regular
Choquet capacity $c$ on $X$, and $c$ is submodular.

(b) Suppose that $X$ is regular.   Let $\Cal K$ be the family of compact
subsets of $X$, and $c_1:\Cal K\to[0,\infty]$ a functional such that

\inset{$c_1$ is submodular;

$c_1(K)
=\inf_{G\in\frak T,G\supseteq K}\sup_{L\in\Cal K,L\subseteq G}c_1(L)$
for every $K\in\Cal K$.}

\noindent Then $c_1$ has a unique extension to an outer regular
Choquet capacity $c$ on $X$ such that

\Centerline{$c(G)=\sup\{c(K):K\subseteq G$ is compact$\}$
for every open $G\subseteq X$,}

\noindent and $c$ is submodular.

\proof{{\bf (a)} For $A\subseteq X$, set
$c(A)=\inf\{c_0(G):A\subseteq G\in\frak T\}$.   Then
$c:\Cal PX\to[0,\infty]$ extends $c_0$ because $c_0$ is order-preserving.
Conditions (i) and (iii) of
432J are obviously satisfied.   As for (ii), let $\sequencen{A_n}$ be a
non-decreasing sequence of subsets of $X$ with union $A$.   Then
$\lim_{n\to\infty}c(A_n)$ is defined and not greater than $c(A)$.
If the limit is infinite, then certainly it is equal to
$c(A)$.   Otherwise, take $\epsilon>0$.   For each $n\in\Bbb N$, choose
an open set $G_n\supseteq A_n$ such that
$c_0(G_n)\le c(A_n)+2^{-n}\epsilon$.   Set $H_n=\bigcup_{i\le n}G_i$
for $n\in\Bbb N$.   Then $c_0(H_n)\le c(A_n)+2\epsilon-2^{-n}\epsilon$ for
every $n$.   \Prf\ Induce on $n$.   If $n=0$ then $H_0=G_0$ and the
result is immediate.   For the inductive step to $n+1$, we have

$$\eqalignno{c_0(H_{n+1})+c_0(H_n\cap G_{n+1})
&\le c_0(H_n)+c_0(G_{n+1})\cr
\displaycause{because $c_0$ is submodular}
&\le c(A_n)+2\epsilon-2^{-n}\epsilon+c(A_{n+1})+2^{-n-1}\epsilon\cr
\displaycause{using the inductive hypothesis}
&\le c_0(H_n\cap G_{n+1})+2\epsilon-2^{-n-1}\epsilon+c(A_{n+1});\cr}$$

\noindent as $c_0(H_n\cap G_{n+1})\le c_0(G_{n+1})$ is finite,
$c_0(H_{n+1})\le 2\epsilon-2^{-n-1}\epsilon+c(A_{n+1})$, as required.\ \Qed

Set $H=\bigcup_{n\in\Bbb N}H_n$.   As $\sequencen{H_n}$ is non-decreasing,

\Centerline{$c(A)\le c_0(H)=\lim_{n\to\infty}c_0(H_n)
\le\lim_{n\to\infty}c(A_n)+2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $c(A)\le\lim_{n\to\infty}c(A_n)$ and
the final condition of 432J is satisfied.

Of course $c$ is the only Choquet capacity extending $c_0$ and outer
regular with respect to the open sets.

As for the submodularity of $c$, if $A$, $B\subseteq X$ and $\epsilon>0$,
there are open sets $G\supseteq A$ and $H\supseteq B$ such that
$c_0(G)+c_0(H)\le c(G)+c(H)+\epsilon$;  so that

$$\eqalign{c(A\cup B)+c(A\cap B)
&\le c_0(G\cup H)+c_0(G\cap H)\cr
&\le c_0(G)+c_0(H)
\le c(A)+c(B)+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$c(A\cup B)+c(A\cap B)\le c(A)+c(B)$, as required.

\medskip

{\bf (b)(i)} The key fact is this:  if $G$, $H\in\frak T$,
$K$, $L\in\Cal K$, $K\subseteq G\cup H$ and $L\subseteq G\cap H$, then
there are $K_1$, $L_1\in\Cal K$ such that $K_1\subseteq G$,
$L_1\subseteq H$, $K\subseteq K_1\cup L_1$ and $L\subseteq K_1\cap L_1$.
\Prf\ Because $(X,\frak T)$ is regular, there is an open set
$G_1$ such that $K\setminus H\subseteq G_1$ and
$\overline{G}_1\subseteq G$ (4A2F(h-ii)).   Set
$K_1=(K\cap\overline{G}_1)\cup L$,
$L_1=(K\setminus G_1)\cup L$;  these work.\ \Qed

\medskip

\quad{\bf (ii)} Define $c_0:\frak T\to[0,\infty]$ by setting
$c_0(G)=\sup_{K\in\Cal K,K\subseteq G}c_1(K)$ for open $G\subseteq X$.
Then $c_0$ satisfies the conditions of (a).   \Prf\ The first and third are
elementary.   As for the second, if $G$, $H\in\frak T$ and
$\gamma<c_0(G\cup H)+c_0(G\cap H)$, there are $K$, $L\in\Cal K$ such that
$K\subseteq G\cup H$, $L\subseteq G\cap H$ and
$\gamma\le c_1(K)+c_2(L)$.   Now (i) tells us that there are compact sets
$K_1\subseteq G$ and $L_1\subseteq H$ such that $K\subseteq K_1\cup L_1$
and $L\subseteq K_1\cap L_1$, in which case

$$\eqalign{c_0(G)+c_0(H)
&\ge c_1(K_1)+c_1(L_1)
\ge c_1(K_1\cup L_1)+c_1(K_1\cap L_1)\cr
&\ge c_1(K)+c_1(L)
\ge\gamma.\cr}$$

\noindent As $\gamma$ is arbitrary,
$c_0(G)+c_0(H)\ge c_0(G\cup H)+c_0(G\cap H)$, as required.\ \Qed

\medskip

\quad{\bf (iii)} We therefore have a submodular outer regular
Choquet capacity $c:\Cal PX\to[0,\infty]$
defined by setting $c(A)=\inf_{G\in\frak T,A\subseteq G}c_0(G)$ for every
$A\subseteq X$.   From the second condition on $c_1$, we see that $c$
extends $c_1$.   Clearly $c$ satisfies the two regularity conditions, and
is the only extension of $c_1$ which does so.
}%end of proof of 432L

\exercises{
\leader{432X}{Basic exercises (a)}
%\spheader 432Xa
Put 422Xe, 431Xb and 432D together to prove 432C.
%432C, 432D
%422Xe > Borel sets K-analytic [therefore Souslin-F].
%431Xb > \mu is i.r. wrt closed sets.

\spheader 432Xb Let $X$ be a K-analytic Hausdorff space, and $\mu$ a
measure on $X$ which is outer regular with respect to the open sets.
Show that $\mu X=\sup_{K\subseteq X\text{ is compact}}\mu^*K$.
\Hint{see the proof of 432D.}
%432D

\sqheader 432Xc Let $X$ be a K-analytic Hausdorff space, and $\mu$ a
semi-finite topological measure on $X$.   Show that if {\it either}
$\mu$ is inner regular with respect to the closed sets %432D
{\it or} $X$ is
regular and $\mu$ is a $\tau$-additive Borel measure, then $\mu$ is
tight. %414M
%432E

\spheader 432Xd Use 422Gf, 432B and 416C to prove 432E.
%432E
%422Gf > closed sset of K-analytic is K-analytic
%416C > quasi-Radon + \sup_K K^{\ssbullet}=1 => Radon

\sqheader 432Xe Suppose that $X$ is a set and that $\frak S$, $\frak T$
are Hausdorff topologies on $X$ such that $(X,\frak T)$ is K-analytic
and $\frak S\subseteq\frak T$.   Let $(Z,\frak U,\Tau,\nu)$ be a Radon
measure space and $f:Z\to X$ a function which is almost continuous for
$\frak U$ and $\frak S$.   Show that $f$ is almost continuous for
$\frak U$ and $\frak T$.   \Hint{it is enough to consider totally finite
$\nu$;  show that $\nu f^{-1}$ is $\frak T$-Radon, so is inner regular
for $\{K:\frak T_K=\frak S_K\}$, writing $\frak T_K$ for the subspace
topology induced by $\frak T$ on $K$.}
%432H

\spheader 432Xf Let $X$ be a topological space and $\mu$ a locally
finite measure on $X$ which is inner regular with respect to the closed
sets.   Show that $\mu^*$ is a submodular Choquet capacity.
%432J

\spheader 432Xg Let $X$ be a topological space and $F$ a closed subset
of $X$.   Define $c:\Cal PX\to\{0,1\}$ by setting $c(A)=1$ if $A$ meets
$F$, $0$ otherwise.   Show that $c$ is a submodular
Choquet capacity on $X$.
%432J

\spheader 432Xh Let $X$ and $Y$ be Hausdorff spaces, and
$R\subseteq X\times Y$ an usco-compact relation.   Show that if $c$ is a
Choquet capacity on $Y$, then $A\mapsto c(R[A])$ is a Choquet
capacity on $X$, which is submodular if $c$ is.
%432J

\spheader 432Xi Use 432K and 432Xf to shorten the proof of 432D.
%432K

\spheader 432Xj\dvArevised{2014}
Let $P$ be a lattice, and $c:P\to\Bbb R$ an
order-preserving functional.    Show that the following are equiveridical:
(i) $c$ is submodular;  (ii) $(p,q)\mapsto 2c(p\vee q)-c(p)-c(q)$ is a
pseudometric on $P$;  (iii) setting $c_r(p)=c(p\vee r)-c(r)$,
$c_r(p\vee q)\le c_r(p)+c_r(q)$ for all $p$, $q$, $r\in P$.
%432J out of order query

\spheader 432Xk\dvAnew{2014} Let $X$ be a topological space,
$c:X\to[0,\infty]$ a Choquet capacity, and $f:[0,\infty]\to[0,\infty]$ a
non-decreasing function.   (i) Show that if $f$ is continuous then
$fc$ is a Choquet capacity.   (ii) Show that if $f\restr\coint{0,\infty}$
is concave and $c$ is submodular, then $fc$ is submodular.
%432J

\spheader 432Xl\dvAnew{2016} Let $X$ be a Hausdorff space, $c$ a Choquet
capacity on $X$, and $\Cal K$ a non-empty
downwards-directed family of compact subsets of $X$.   Show that
$c(\bigcap\Cal K)=\inf_{K\in\Cal K}c(K)$.
%432J

\leader{432Y}{Further exercises (a)}
%\spheader 432Ya
Show that there are a K-analytic Hausdorff space $X$ and a probability
measure $\mu$ on $X$ such that (i) $\mu$ is inner regular with respect
to the Borel sets (ii) the domain of $\mu$ includes a base for the
topology of $X$ (iii) every compact subset of $X$ is negligible.   Show
that there is no extension of $\mu$ to a topological measure on $X$.
%432B
%432D %mt43bits

\spheader 432Yb\dvAnew{2011}
Let $X$, $Y$ be Hausdorff spaces, $R\subseteq X\times Y$ an usco-compact
relation and $\mu$ a Radon probability measure on $X$ such that
$\mu_*R^{-1}[Y]=1$.   Show that there is a Radon probability measure on $Y$
such that $\nu_*R[A]\ge\mu_*A$ for every $A\subseteq X$.
%418L applied to projection from  R  onto  R^{-1}[Y]

\leaveitout{\spheader 432Y? Let $X$ be a set with its discrete topology,
and $\kappa$ a cardinal of uncountable cofinality.   Define
$c:\Cal PX\to\{0,1\}$ by setting $c(A)=0$ if $\#(A)\le\kappa$, $1$
otherwise.   Show that $c$ is a Choquet capacity on $X$.
%432J
}%end of leaveitout


}%end of exercises

\leaveitout{\leader{432Z}{Problem} Let $X$ be a K-analytic Hausdorff
space and $\mu$ a probability measure on $X$ such that (i) the domain of
$\mu$ includes a base for the topology of $X$ (ii) $\mu$ is inner
regular with respect to the Borel sets.   Must there be a non-negligible
compact set?

$\mu$ measures every Baire set and
there is a unique
Radon measure agreeing with $\mu$ on the Baire sets, I think.
}%end of leaveitout

\cmmnt{\Notesheader{432}
The measure-theoretic properties of K-analytic spaces can largely be
summarised in the slogan `K-analytic spaces have lots of compact sets'.
I said above that it is sometimes helpful to think of K-analytic spaces
as an amalgam of compact Hausdorff spaces and Souslin-F subsets of
$\Bbb R$.   For the former, it is obvious that they have many compact
subsets;
for the latter, it is not obvious, but is of course one of their
fundamental properties, deducible from 422De.   432B and the proof of 432D (repeated in 432K) are typical manifestations of the phenomenon.
The real point of these theorems is that we can extend a Borel or Baire
measure to a Radon measure with no prior assumption of $\tau$-additivity
(432F).   A Radon measure must be $\tau$-additive just because it is
tight.   A (locally finite)
Borel or Baire measure must be $\tau$-additive whenever the measurable
open sets are K-analytic.

The condition `every open set is K-analytic' in 432C is of course a very
strong one in the context of compact Hausdorff spaces (422Xd).   But for
analytic spaces it is automatically satisfied (423Eb), and
that is the side on which the principal applications of 432C appear.

The results which I call corollaries of 432D can mostly be proved by
more direct methods (see 432Xd), but the line I choose here seems to be
the most powerful technique.   Indeed it can be used to deal with 432C
as well (432Xa).

In \S434 I will discuss `universally measurable' sets in topological
spaces.   In
fact K-analytic sets are universally measurable in a particularly strong
sense (432A).   The point here is that K-analyticity is intrinsic;  a
K-analytic space is measurable whenever embedded as a subspace of a
(complete locally determined) Hausdorff topological measure space.

The theorems here touch on two phenomena of particular importance.
First, in 432G we have an example of `pulling back' a measure, that
is, we have a measure $\nu$ on a set $Y$ and a function $f:X\to Y$ and
seek a Radon measure $\mu$ on $X$ such that $f$ is \imp, or, even
better, such
that $\nu=\mu f^{-1}$.   There was a similar result in 418L.   In both
cases we have to suppose that $f$ is continuous and (in effect) that
$\nu$ is a Radon measure.   (This is not part of the hypotheses of 432G,
but of course it is an easy consequence of them, using 432B.)   In 418L,
we need a special hypothesis to ensure that there are enough compact
subsets of $X$ to carry an appropriate Radon measure;  in 432G, this is
an automatic result of assuming that $X$ is K-analytic.   Both 418L and
432G can be regarded as consequences of Henry's theorem (416N).   The
difficulty arises from the requirement that $\mu$ should be a Radon
measure;  if we do not insist on this there is a much simpler solution,
since we need suppose only that $f[X]$ has full outer measure (234F).

The next theme I wish to mention is a related one, the investigation of
comparable topologies.   If $\frak S$ and $\frak T$ are (Hausdorff)
topologies on a set $X$, and $\frak S$ is coarser than $\frak T$ (so
that $(X,\frak S)$ is a continuous image of $(X,\frak T)$), then
418I tells us that any totally finite $\frak T$-Radon
measure is $\frak S$-Radon.   We very much want to know when the reverse
is true, so that the (totally finite) Radon measures for the two
topologies are the same.   432H provides one of the important cases in
which this occurs.   The hypothesis `$(X,\frak T)$ is K-analytic'
generalizes the alternative \lq$(X,\frak T)$ is compact';  in the latter
case, $\frak S=\frak T$, so that the result is, from our point of view
here, trivial.   (But from the point of view of elementary general
topology, of course, it is one of the pivots of the theory of compact
Hausdorff spaces.)   In a similar vein we have a variety of important
topological consequences of the same hypotheses (422Yb, 423Fb).

The paragraphs 432J-432L %432J 432K 432L
may appear to be no more that a minor
extension of ideas already set out.   I ought therefore to say plainly
that the topological and measure theory of K-analytic spaces have
co-evolved with the notion of capacity, and that 432K (`K-analytic
spaces are capacitable') is one of the cornerstones of a theory of which
I am giving only a minuscule part.   For a idea of the vitality and
scope of this theory, see {\smc Dellacherie 80}.
}%end of notes

\discrpage

\frfilename{mt433.tex}
\versiondate{27.6.10}
\copyrightdate{2003}
\def\NN{\BbbN^{\Bbb N}}
\def\tildeTau{\tilde{\text{T}}}

\def\chaptername{Topologies and measures II}
\def\sectionname{Analytic spaces}

\newsection{433}

We come now to the special properties of measures on `analytic'
spaces\cmmnt{,
that is, continuous images of $\NN$, as described in \S423}.   I start
with a couple of facts about spaces with countable networks.

\leader{433A}{Proposition} Let $(X,\frak T)$ be a topological space with
a countable network, and $\mu$ a localizable topological measure on $X$
which is inner regular with respect to the Borel sets.   Then $\mu$ has
countable Maharam type.

\proof{ Let $\tilde\mu$ be the c.l.d.\ version of $\mu$ (213E).   Then
the measure algebra $\frak A$ of $\tilde\mu$ can be identified with the
measure algebra of $\mu$ (322D(b-iii)).
Also $\tilde\mu$ is complete, locally determined and localizable, so
every subset of $X$ has a measurable envelope with respect to
$\tilde\mu$ (213J, 213L).    Let $\tilde\Sigma$ be the domain of
$\tilde\mu$, and $\Cal E$ a countable network for $\frak T$.   For each
$E\in\Cal E$, let $F_E\in\tilde\Sigma$ be a measurable envelope of $E$.

Let $\frak B$ be the order-closed subalgebra of $\frak A$ generated by
$\{F_E^{\ssbullet}:E\in\Cal E\}$, and set
$\Tau=\{F:F\in\tilde\Sigma,\,F^{\ssbullet}\in\frak B\}$.   Because
$\frak B$ is an order-closed subalgebra of $\frak A$, $\Tau$ is a
$\sigma$-subalgebra of $\tilde\Sigma$.   Now $\frak T\subseteq\Tau$.
\Prf\ If $G\in\frak T$, set $\Cal E_0=\{E:E\in\Cal E,\,E\subseteq G\}$.
Set $F=\bigcup_{E\in\Cal E_0}F_E$, so that $F\in\Tau$ and
$G\subseteq F$.   For each
$E\in\Cal E_0$, $F_E\setminus G$ is negligible, so $F\setminus G$ is
negligible, and $G^{\ssbullet}=F^{\ssbullet}\in\frak B$, so $G\in\Tau$.\
\Qed

It follows that $\Tau$ includes the Borel $\sigma$-algebra of $X$.
Because $\mu$ is inner regular with respect to the Borel sets, $\frak B$
is order-dense in $\frak A$, and $\frak B=\frak A$.   Thus the countable
set $\{F_E^{\ssbullet}:E\in\Cal E\}\,\,\tau$-generates $\frak A$, and
the Maharam
type of $\frak A$, which is the Maharam type of $\mu$, is countable.
}%end of proof of 433A

\leader{433B}{Lemma} If $(X,\frak T)$ is a Hausdorff space with a
countable network, then
any topological measure on $X$ is countably separated\cmmnt{ in the
sense of 343D}.

\proof{ By 4A2Nf, there is a countable family of open sets separating
the points of $X$.
}%end of proof of 433B used 4{}52, 4{}53, 5{}22

\vleader{48pt}{433C}{Theorem} Let $X$ be an analytic Hausdorff space,
and $\mu$ a Borel measure on $X$.

(a) If $\mu$ is semi-finite, it is tight.

(b) If $\mu$ is locally finite, its completion is a Radon measure on
$X$.

\proof{ $X$ is K-analytic (423C);  moreover, every open subset of $X$ is
again analytic (423Eb).   So 432C gives the result at once.
}%end of proof of 433C

\cmmnt{\medskip

\noindent{\bf Remark} Compare 256C.
}%end of comment

\leader{433D}{Theorem} Let $X$ and $Y$ be analytic Hausdorff spaces,
$\nu$ a totally finite Radon measure on $Y$ and $f:X\to Y$ a Borel
measurable function such that $f[X]$ has full outer measure for $\nu$.
Then there is a Radon measure $\mu$ on $X$ such that $\nu=\mu f^{-1}$.

\proof{ By 423Ga, the graph $R$ of $f$ is an analytic set in
$X\times Y$, therefore K-analytic.   Set $\pi_1(x,y)=x$, $\pi_2(x,y)=y$
for $(x,y)\in R$, so that $\pi_1$ and $\pi_2$ are continuous.   Now
$\pi_2[R]=f[X]$ has full outer measure, so by 432G there is a Radon
measure $\lambda$ on $R$
such that $\nu=\lambda\pi_2^{-1}$.   Next, because $\pi_1$ is
continuous, the image $\mu=\lambda\pi_1^{-1}$ is a Radon measure on $X$,
by 418I.   But $\pi_2=f\pi_1$, so

\Centerline{$\mu f^{-1}=(\lambda\pi_1^{-1})f^{-1}
=\lambda(f\pi_1)^{-1}=\lambda\pi_2^{-1}=\nu$,}

\noindent as required.
}%end of proof of 433D

\leader{433E}{Proposition} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space and $\frak T$ a topology on $X$ such that $\mu$ is inner regular
with respect to the closed sets.   Let $(Y,\frak S)$ be an analytic
Hausdorff space and $f:X\to Y$ a measurable function.   Then $f$ is almost
continuous.

\proof{ Take $E\in\Sigma$ and $\gamma<\mu E$.   Then there is an
$F\in\Sigma$ such that $F\subseteq E$ and $\gamma<\mu F<\infty$.   For
Borel sets $H\subseteq Y$, set $\nu H=\mu(F\cap f^{-1}[H])$.   Then
$\nu$ is a totally finite Borel measure on $Y$, so is tight (433C);
let $K\subseteq Y$ be a compact set
such that $\nu K>\gamma$, so that $\mu(F\cap f^{-1}[K])>\gamma$.   The
subspace measure on $L=F\cap f^{-1}[K]$ is still inner regular with
respect to the (relatively) closed sets (412Pc), and $f\restr L$ is
still measurable;
but $f\restr L$ is a function from $L$ to $K$, and $K$ is metrizable, by
423Dc.   So $f\restr L$ is almost continuous, by 418J, and there is
a set $F'\subseteq L$, of measure at least $\gamma$, such that
$f\restr F'$ is continuous.

As $E$ and $\gamma$ are arbitrary, $f$ is almost continuous.
}%end of proof of 433E

\cmmnt{\medskip

\noindent{\bf Remark} Compare 418Yf.}

\leader{433F}{}\cmmnt{ I give some simple corollaries of the von
Neumann-Jankow selection theorem (423N-423Q).  %423N 423O 423P 423Q

\medskip

\noindent}{\bf Proposition} Let $(X,\frak T)$ and $(Y,\frak S)$ be
analytic Hausdorff spaces, and $f:X\to Y$ a Borel measurable surjection.
Let $\nu$ be a complete locally determined topological measure on $Y$,
and $\Tau$ its domain.   Then there is a $\Tau$-measurable function
$g:Y\to X$ such that $gf$ is the identity on $X$.

\proof{ By 423Q we know that there is a function $g:Y\to X$ such that
$fg$ is the identity and $g$ is $\Tau_1$-measurable, where $\Tau_1$ is
the $\sigma$-algebra generated by the Souslin-F subsets of $Y$.   But
$\Tau$ contains every Souslin-F subset of $Y$, by 431B, therefore
includes $\Tau_1$, and $g$ is actually $\Tau$-measurable.
}%end of proof of 433F

\vleader{72pt}{433G}{Proposition} Let $(X,\frak T)$ be an analytic
Hausdorff
space, $(Y,\Tau,\nu)$ a complete locally determined measure
space, and $f:X\to Y$ a surjection.   Suppose that there is some
countable family $\Cal F\subseteq\Tau$ such that $\Cal F$ separates the
points of $Y$\cmmnt{ (that is, whenever $y$, $y'$ are distinct points
of $Y$ there is a member of $\Cal F$ containing one and not the other)}
and $f^{-1}[F]$ is a Borel subset of $X$ for every $F\in\Cal F$.   Then
there is a $\Tau$-measurable function $g:Y\to X$ such that $fg$ is the
identity on $Y$.

\proof{ Set $\Cal A=\Cal F\cup\{Y\setminus F:F\in\Cal F\}$.   The
topology $\frak T_1$ on $X$ generated by
$\frak T\cup\{f^{-1}[A]:A\in\Cal A\}$
is still analytic (423H).   If we take $\frak S$ to be the
topology on $Y$ generated by $\Cal A$, then $\frak S$ is Hausdorff and
$f$ is $(\frak T_1,\frak S)$-continuous, so $\frak S$ is analytic
(423Bb).

Because $\frak S$ is generated by a countable subset $\Cal A$ of $\Tau$,
it is second-countable, and
$\frak S\subseteq\Tau$ (4A3Da/4A3E).   So $\nu$ is a topological
measure with respect to $\frak S$.   By
433F, there is a function $g:Y\to X$, measurable for  $\Tau$ and the
topology $\frak T_1$, such that $gf$ is the identity on $X$;  and of
course $g$ is still measurable for $\Tau$ and the coarser original
topology $\frak T$ on $X$.
}%end of proof of 433G

\leader{433H}{Proposition}\dvAnew{2011}
Let $X$ be an analytic Hausdorff space, and $(Y,\Tau,\nu)$ a complete
locally determined measure space.   Suppose that
$W\subseteq X\times Y$ belongs to $\Cal S(\Cal B(X)\tensorhat\Tau)$,
where $\Cal B(X)$ is the Borel $\sigma$-algebra of $X$.   Then
$W[X]\in\Tau$ and there is a $\Tau$-measurable function
$g:W[X]\to X$ such that $(g(y),y)\in W$ for every $y\in W[X]$.

\proof{ Set
$\Cal V=\Cal S(\{E\times F:E\subseteq X$ is closed, $F\in\Tau\})$.   Then
$\Cal V$ contains $H\times Y$ for every Souslin-F
subset $H$ of $X$ (421Cb), and therefore for every $H\in\Cal B(X)$ (423Eb);
by 421F, it follows that $\Cal V$ includes $\Cal B(Y)\tensorhat\Tau$ and
therefore $W\in\Cal S(\Cal V)=\Cal V$ (421D).   By 423N,
$W[X]\in\Cal S(\Tau)$, which by 431A is just $\Tau$, and there is
a $\Tau$-measurable function which is a selector for $W^{-1}$.
}%end of proof of 433H

\leader{433I}{}\cmmnt{ Because analytic spaces have countable networks
(423C), and their compact subsets are therefore metrizable
(423Dc), their measure theory is very close to that of $\Bbb R$ or
$[0,1]$ or $\{0,1\}^{\Bbb N}$.   I give some simple manifestations of
this principle.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}33H}
Let $\familyiI{X_i}$ be a family of analytic Hausdorff spaces, and for each
$i\in I$ let $\mu_i$ be a Radon probability measure on $X_i$.   Let $\lambda$
be the ordinary product measure on $X=\prod_{i\in I}X_i$\cmmnt{, as
defined in \S254}.

(a) If $I$ is countable then $\lambda$ is a Radon measure.

(b) If every $\mu_i$ is strictly positive, then $\lambda$ is a
quasi-Radon measure.

\proof{{\bf (a)} In this case, $X$ is analytic (423Bc), therefore
hereditarily Lindel\"of (423Da).   Let $\Lambda$ be the domain of
$\lambda$ and
$\frak T$ the topology of $X$.  Then $\Lambda\cap\frak T$ is a base for
$\frak T$;  by 4A3Da, $\frak T\subseteq\Lambda$.   By 417Sb, $\lambda$
is the $\tau$-additive product measure on $X$;  by 417Q, this is a Radon
measure.

\medskip

{\bf (b)} By (a), the ordinary product measure on $\prod_{i\in J}X_i$ is
a Radon measure for every finite set $J\subseteq I$.   So 417Sc tells us
that $\lambda$ is the $\tau$-additive product measure on $X$;
by 417O, this is a quasi-Radon measure.
}%end of proof of 433I

\leader{433J}{Proposition}\dvAformerly{4{}33I}
Let $X$ be an analytic Hausdorff space, and
$\Tau$ a countably generated $\sigma$-subalgebra of the Borel
$\sigma$-algebra $\Cal B(X)$ of $X$.   Then
any locally finite measure with domain $\Tau$ has an extension
to a Radon measure on $X$.

\proof{ Let $\mu_0$ be a locally finite measure with domain $\Tau$.

\medskip

{\bf (a)} Consider first the case in which $\mu_0$ is totally
finite.   Let $\sequencen{F_n}$ be a sequence in $\Tau$ generating
$\Tau$ as $\sigma$-algebra.   Define $f:X\to\{0,1\}^{\Bbb N}$ by setting

\Centerline{$f(x)(n)=\chi F_n(x)$ for $n\in\Bbb N$, $x\in X$.}

\noindent Then $f$ is $\Tau$-measurable
(use 418Bd), so we have a Borel measure $\nu_0$
on $\{0,1\}^{\Bbb N}$ defined by setting $\nu_0E=\mu_0f^{-1}[E]$ for
every Borel set $E\subseteq\{0,1\}^{\Bbb N}$.   Now the completion $\nu$
of $\nu_0$ is a Radon
measure (433C).   Also $f[X]$ must be analytic, by 423Gb, because $f$ is
$\Cal B(X)$-measurable.    So $\nu$ measures $f[X]$ (432A), and

\Centerline{$\nu f[X]=\nu_0^*f[X]=\nu_0\{0,1\}^{\Bbb N}$,}

\noindent that is, $f[X]$ is $\nu$-conegligible.   By 433D, there is a
Radon measure $\mu$ on $X$ such that $\nu=\mu f^{-1}$.

Because every $F_n$ is expressible as $f^{-1}[E]$ for some
$E\in\Cal B(\{0,1\}^{\Bbb N})$,
so is every member of $\Tau$.   If $F\in\Tau$, take
$H\in\Cal B(\{0,1\}^{\Bbb N})$ such
that $F=f^{-1}[H]$;  then

\Centerline{$\mu F=\nu H=\nu_0H=\mu_0F$.}

\noindent Thus $\mu$ extends $\mu_0$ and $\mu\restr\Sigma$ will serve.

\medskip

{\bf (b)} In general, because $X$ is Lindel\"of and $\mu_0$ is locally
finite, $\mu_0$ is $\sigma$-finite.   Let $\sequencen{X_n}$ be a
partition of $X$ into members of $\Tau$ such that $\mu_0X_n$ is
finite for every $n$, and set $\mu^{(n)}_0F=\mu_0(F\cap X_n)$ for every
$n$ and every $F\in\Tau$;  then every $\mu^{(n)}_0$
has an extension to a Radon measure $\mu^{(n)}$.
Let $\mu$ be the sum $\sum_{n=0}^{\infty}\mu^{(n)}$
(234G\formerly{1{}12Ya}).
Of course $\mu$ extends $\mu_0=\sum_{n=0}^{\infty}\mu^{(n)}_0$.
Because $\mu_0$ is locally finite, so is $\mu$, and $\mu$ is a Radon
measure (416De).
}%end of proof of 433K

\leader{433K}{}\cmmnt{ I turn now to a brief mention of `standard
Borel spaces'.   From the point of view of this chapter, it is natural
to regard the following results as simple corollaries of theorems about
Polish spaces.   But, as remarked in \S424, there are cases in which a
standard Borel space is presented without any specific topology being
attached;  and in any case it is interesting to look at the ways in
which we can express these ideas as theorems about $\sigma$-algebras
rather than about topological spaces.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}33J}
Let $(X,\Sigma)$ be a standard Borel space
and $\Tau$ a countably generated $\sigma$-subalgebra of $\Sigma$.   Then
any $\sigma$-finite measure with domain $\Tau$ has an extension
to $\Sigma$.

\proof{ Let $\mu_0$ be a $\sigma$-finite measure with domain $\Tau$.

\medskip

{\bf (a)} If $\mu_0$ is totally finite, give $X$ a Polish topology for
which $\Sigma$ is the Borel $\sigma$-algebra of $X$, and use 433J.

\medskip

{\bf (b)} In general, let $\sequencen{X_n}$ be a partition of $X$
into members of $\Tau$ such that $\mu_0X_n<\infty$ for every $n$, and
set $\mu^{(n)}_0F=\mu_0(F\cap X_n)$ for every $n$ and every $F\in\Tau$;
then every $\mu^{(n)}_0$
has an extension to a measure $\mu^{(n)}$ with domain $\Sigma$, and we
can set $\mu=\sum_{n=0}^{\infty}\mu^{(n)}$.
}%end of proof of 433K

\leader{433L}{Proposition}\dvAformerly{4{}33K}
Let $\sequencen{(X_n,\Sigma_n,\mu_n)}$ be a
sequence of probability spaces such that $(X_n,\Sigma_n)$ is a standard
Borel space for every $n$.   Suppose that for each $n\in\Bbb N$ we are
given an \imp\ function $f_n:X_{n+1}\to X_n$.   Then we can find a
standard Borel space $(X,\Sigma)$, a probability measure $\mu$ with
domain $\Sigma$, and \imp\ functions $g_n:X\to X_n$ such that
$f_ng_{n+1}=g_n$ for every $n$.

\proof{ For each $n$, choose a Polish topology $\frak T_n$ on $X_n$ such
that $\Sigma_n$ is the algebra of $\frak T_n$-Borel sets.  Let
$\hat\mu_n$ be the completion of $\mu_n$;  then $\hat\mu_n$ is a Radon
measure (433C).   Every $f_n$ is \imp\ for $\hat\mu_{n+1}$ and
$\hat\mu_n$, by 234Ba\formerly{2{}35Hc}, and almost continuous, by 418J.

By 418Q, we have a Radon measure $\hat\mu$ on

\Centerline{$X=\{x:x\in\prod_{n\in\Bbb N}X_n,\,f_n(x(n+1))=x(n)$ for
every $n\in\Bbb N\}$}

\noindent such that the continuous maps $x\mapsto x(n)=g_n(x):X\to X_n$
are \imp\ for every $n$.   Now $X$ is a Borel subset of
$Z=\prod_{n\in\Bbb N}X_n$.   \Prf\  For each $n\in\Bbb N$, let
$\Cal U_n$ be a countable base for $\frak T_n$.   Then

\Centerline{$Z\setminus X
=\bigcup_{n\in\Bbb N}\bigcup_{U,V\in\Cal U_n,U\cap V=\emptyset}
  \{z:z(n)\in U,\,f_n(z(n+1))\in V\}$}

\noindent is a countable union of Borel sets because $\{z:z(n)\in U\}$
is open and $\{z:z(n+1)\in f_n^{-1}[V]\}$ is Borel whenever $n\in\Bbb N$
and $U$, $V\in\Cal U_n$.   So $Z\setminus X$ and $X$ are Borel sets.\
\Qed

Accordingly $(X,\Sigma)$ is a standard Borel space, where $\Sigma$ is
the Borel $\sigma$-algebra of $X$ (424G).   So if we take
$\mu=\hat\mu\restr\Sigma$, we shall have a suitable measure on $X$.
}%end of proof of 433L

\exercises{
\leader{433X}{Basic exercises (a)}
%\spheader 433Xa
Let $(X,\frak T)$ be a topological space with a
countable network, and $\mu$ a topological measure on $X$ which is inner
regular with respect to the Borel sets and has the measurable envelope
property (213Xl).   Show that $\mu$ has countable Maharam type.
%433A

\spheader 433Xb Show that an effectively locally finite measure on a
hereditarily Lindel\"of space (in particular, on any analytic
Hausdorff space) is $\sigma$-finite.
%433B

\spheader 433Xc Let $X\subseteq[0,1]$ be a set of Lebesgue outer measure
$1$ and inner measure $0$.  Show that the subspace measure on $X$ is a
totally finite Borel measure which is not tight.
%433C

\spheader 433Xd Let $X$ be a Hausdorff space and $\mu$ a locally finite
measure on $X$, inner regular with respect to the Borel sets, such that
$\dom\mu$ includes a base for the topology of $X$.   Suppose that
$Y\subseteq X$ is an analytic set of full outer measure.   Show that
$\mu$ has a unique extension to a Radon measure $\tilde\mu$ on $X$, and
that $Y$ is $\tilde\mu$-conegligible.
%433C

\spheader 433Xe Let $(X,\Sigma)$ be a standard Borel space.   (i) Show
that any semi-finite measure with domain $\Sigma$ is a compact measure
(definition: 342Ac, or 451Ab below), therefore perfect.
\Hint{if $X$ is given a
suitable topology, the measure is tight.}
(ii) Show that any measure with domain
including $\Sigma$ is countably separated.
%433C

\sqheader 433Xf (i) Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be atomless
probability spaces such that $(X,\Sigma)$ and $(Y,\Tau)$ are standard
Borel spaces.   Show that $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are
isomorphic.   \Hint{by 344I, their completions are isomorphic;  by 344H,
they have negligible sets with cardinal $\frak c$;  show that any
isomorphism between the completions is $(\Sigma,\Tau)$-measurable on a
conegligible set;  use 424Da to match residual negligible sets.}
(ii) Let $X$ be a Polish space and $\mu$ an atomless Radon measure on $X$.
Show that there is a Borel isomorphism between $X$ and $[0,1]$ which
matches $\mu$ to Lebesgue measure on $[0,1]$.
%433Xe, 433C

\spheader 433Xg Let $X$ be $[0,1]\times\{0,1\}$, with its usual
topology, and $I^{\|}$ the split interval (419L);  define
$f:X\to I^{\|}$ by
setting $f(t,0)=t^-$, $f(t,1)=t^+$ for $t\in[0,1]$.   (i) Give $I^{\|}$
its usual Radon measure $\nu$ (343J, 419Lc).   Show that there is no
Radon measure
$\lambda$ on $X$ such that $\nu=\lambda f^{-1}$.   (ii) Let $\mu$ be the
product Radon probability measure on $X$, starting from Lebesgue measure
on $[0,1]$ and the usual fair-coin measure on $\{0,1\}$.   Show that $f$
is \imp\ for $\mu$ and
$\nu$.   Show that $f$ is not almost continuous.
%433E

\leader{433Y}{Further exercises (a)}
%\spheader 433Ya
Find a Hausdorff topological space $X$ with a countable network and a
semi-finite Borel measure on $X$ which does not have countable Maharam
type.
%433A %mt43bits

\spheader 433Yb
Let $X$ be an analytic Hausdorff space and $\mu$ an atomless
Radon measure on $X$.   Show that $(X,\mu)$ is isomorphic to Lebesgue
measure on some measurable subset of $\Bbb R$.   \Hint{344I.}
%

\spheader 433Yc Let $(X,\frak T)$ be a Polish space without isolated
points, and $\mu$ a $\sigma$-finite topological measure on $X$.
Show that there is a conegligible meager set.   \Hint{Show that every
non-empty open set is uncountable.   Find a countable dense negligible
set and a negligible G$_{\delta}$ set including it.}
%/

\spheader 433Yd Let $\sequencen{X_n}$ be a sequence of analytic
Hausdorff spaces
and for each $n\in\Bbb N$ let $\mu_n$ be a Borel probability measure on
$X_n$.   Suppose that for each $n\in\Bbb N$ we are
given an \imp\ function $f_n:X_{n+1}\to X_n$.   Show that we can find a
standard Borel space $(X,\Sigma)$, a probability measure $\mu$ with
domain $\Sigma$, and \imp\ functions $g_n:X\to X_n$ such that
$f_ng_{n+1}=g_n$ for every $n$.
%433L
}%end of exercises

\cmmnt{\Notesheader{433}
The measure-theoretic results of 433C-433E %433C 433D 433E
are of much the same
type as those in \S432.   A characteristic difference is that Borel
measurable functions between analytic spaces behave in many ways like
continuous functions.   (Compare 433D and 432G.)   You may feel that
423Yc offers some explanation of this.   For
any question which refers to the Borel algebra of an analytic space $X$,
or to the class of its analytic subsets, we can expect to be able to
suppose that $X$ is separable and metrizable (see 423Xd), and that any
single Borel measurable function appearing is continuous.   (424H is a
particularly remarkable instance of this principle.)

433I here amounts to spelling out a special case of ideas already
treated in 417S.   As this territory is relatively unfamiliar, I give
detailed examples
(423Xi, 433Xc, 433Xg, 439A, 439K) to show that the theorems of this
section are not generally valid for compact Hausdorff spaces (the
archetype of
K-analytic spaces which need not be analytic), nor for separable metric
spaces (the archetypical spaces with countable network).   They really
do depend on the particular combination of properties possessed by
analytic spaces.

For large parts of probability theory, standard Borel spaces provide an
adequate framework, and have a number of advantages;  some of the
technical problems concerning measurability which loom rather large in
this treatise disappear in such contexts.   Many authors accordingly
give them great prominence.   I myself believe that the simplifications
are an entrapment rather than a liberation, that sooner or later
everyone has to leave the comfortable environment of Borel algebras on
Polish spaces, and that it is better to be properly equipped with a
suitable general theory when one does.   But it is surely important to
know what the simplifications are, and the results in 433K-433L will I
hope show at least that there are wonderful ideas here, even if my own
presentation tends to leave them on one side.

}%end of notes

\discrpage

\frfilename{mt434.tex}
\versiondate{18.1.14}
\copyrightdate{2000}

\def\Yprod#1{\prod_{i\in#1}Y_i}
\hyphenation{Pfef-fer}

\def\chaptername{Topologies and measures II}
\def\sectionname{Borel measures}

\newsection{434}

What one might call the fundamental question of topological measure
theory is the following.

\Centerline{\it What kinds of measures can arise on what kinds of
topological space?}

\noindent Of course this question has inexhaustible ramifications,
corresponding to all imaginable properties of measures and topologies
and connexions between them.   The challenge I face here is that of
identifying particular ideas as being more important than others, and
the chief difficulty lies in the bewildering variety of topological
properties which have been studied, any of which may have implications
for the measure theory of the spaces involved.   In this section and the
next I give a sample of what is known, necessarily biased and
incomplete.   I try however to include the results which are most often
applied and enough others for the proofs to contain, between them, most
of the non-trivial arguments which have been found effective in this
area.

In 434A I set out a crude classification of Borel measures on
topological spaces.   For compact Hausdorff spaces, at least, the first
question is whether they carry Borel measures which are not, in effect,
Radon measures;  this leads us to the definition of `Radon' space (434C)
which is also of interest in the context of general Hausdorff spaces.
I give a brief account of the properties of Radon spaces (434F, 434Nd).
I look also at two special topics:
`quasi-dyadic' spaces (434O-434Q) and a
construction of Borel product measures by integration of sections
(434R).

In the study of Radon spaces we find ourselves looking at `universally
measurable' subsets of topological spaces (434D-434E).   These are
interesting in themselves, and also interact with constructions
from earlier parts of this treatise (434S-434T).
Three further classes of topological space, defined in terms of the
types of topological measure which they carry, are the
`Borel-measure-compact', `Borel-measure-complete' and `pre-Radon'
spaces;  I discuss them briefly in 434G-434J.   They provide useful
methods for deciding whether Hausdorff spaces are Radon (434K).

\cmmnt{
\vleader{72pt}{434A}{Types of Borel measures} In \S411 I introduced
the following properties which a Borel measure may or may not have:

\qquad(i) inner regularity with respect to closed sets;

\qquad(ii) inner regularity with respect to zero sets;

\qquad(iii) tightness\cmmnt{ (that is, inner regularity with respect
to closed compact sets)};

\qquad(iv) $\tau$-additivity.

\noindent These are of course interrelated.   (ii)$\Rightarrow$(i) just
because zero sets are closed, and (iii)$\Rightarrow$(iv) by 411E;
in a Hausdorff space, (iii)$\Rightarrow$(i);  and for an effectively
locally finite measure on a regular topological space,
(iv)$\Rightarrow$(i) (414Mb).

On a regular Hausdorff space, therefore, we can divide totally finite
Borel measures into four classes:

\inset{(A) measures which are not inner regular with respect to the
closed sets,}

\inset{(B) measures which are inner regular with respect to the closed
sets, but not $\tau$-additive nor tight,}

\inset{(C) measures which are $\tau$-additive and inner regular with
respect to the closed sets, but not inner regular with respect to the
compact sets,}

\inset{(D) tight measures;}

\noindent and each of the classes (B)-(D) can be further subdivided into
those which are completion regular (B$_1$, C$_1$, D$_1$) and those which
are not (B$_0$, C$_0$, D$_0$).   Examples may be found in 434Xg (type
A), 411Q and 439Yf (type B$_0$), 439J (type B$_1$), 415Xc and 434Xa (type
C$_1$) and 434Xb (type D$_0$), while Lebesgue measure itself is of type
D$_1$, and any direct sum of spaces of types D$_0$ and C$_1$ will have
type C$_0$.   (The space in 439J depends for its construction on
supposing that there is a cardinal which is not measure-free.   It
seems that
no convincing example of a space of class B$_1$, that is, a completion
regular, non-$\tau$-additive Borel probability measure on a completely
regular Hausdorff space, is known which does not depend on some special
axiom beyond ordinary ZFC.   For one of the obstacles to finding such a
space, see 434Q.)

Note that a totally finite Borel measure $\mu$ on a regular Hausdorff
space can be extended to a quasi-Radon measure iff $\mu$ is of class C
or D (415M), and that in this case the quasi-Radon measure must be
just the completion $\hat\mu$ of $\mu$.   $\hat\mu$ will be of the same
type, on the classification here, as $\mu$;  in particular, $\hat\mu$
will be a Radon measure iff $\mu$ is of class D (416F).
}%end of comment

\cmmnt{
\leader{434B}{Compact, analytic and K-analytic spaces} For any
class of topological spaces, we can enquire which of the seven types of
measure described above can be realized by measures on spaces of that
class.   The enquiry is limited only by our enterprise and diligence in
seeking out new classes of topological space.   For the spaces studied
in \S\S432-433, however, we have something worth repeating here.   On a
K-analytic Hausdorff space, a semi-finite Borel measure which is inner
regular with respect to the closed sets is tight (432B, 432D);
consequently classes B and C of 434A
cannot appear, and we are left with only the types A, D$_0$ and D$_1$,
all of which appear on compact Hausdorff spaces (434Xb, 434Xg).   On an
analytic Hausdorff space
we have further simplifications:  every semi-finite Borel measure is
tight (433Ca), and (if $X$ is
regular) every closed set is a zero set (423Db).   Thus on an analytic
regular Hausdorff space only type D$_1$, of the seven types in 434A, can
appear.   (If the topology is not regular, we may also get measures of
type D$_0$;  see 434Ya.)
}%end of comment

\leader{434C}{Radon spaces:  Definition}\cmmnt{ For K-analytic
Hausdorff spaces, therefore, we have a large gap between the `bad'
measures of class A and the `good' measures of class D;  furthermore, we
have an important class of spaces in which type A cannot appear.   It is
natural to enquire further into the spaces in which every (totally
finite) Borel measure is of class D, and (given that no exact
description can be found) we are led, as usual, to a definition.}   A
Hausdorff space $X$ is {\bf Radon} if every totally finite Borel measure
on $X$ is tight.

\leader{434D}{Universally measurable sets}\dvArevised{2014}\cmmnt{ Before
going farther
with the study of Radon spaces it will be useful to spend a couple of
paragraphs on the following concept.}   Let $X$ be a topological space.

\spheader 434Da I will say that a subset $E$
of $X$ is {\bf universally measurable} (in $X$) if it is measured by the
completion of every Borel probability measure on $X$\cmmnt{;  that
is, for every Borel probability measure $\mu$ on $X$ there is a Borel
set $F\subseteq X$ such that $E\symmdiff F$ is $\mu$-negligible}.

\spheader 434Db
A subset of $X$ is universally measurable iff it is measured by every
complete locally determined topological measure on $X$.
\prooflet{\Prf\ (i) Suppose that $A\subseteq X$ is universally
measurable, and that
$\mu$ is a complete locally determined topological measure on $X$.
Take any $F\subseteq X$ be such that $\mu F$ is defined and
finite.   If $F$ is negligible then $F\cap A$ is negligible and
$\mu(F\cap A)$ is defined.   Otherwise,
we have a Borel probability measure $\nu$ on $X$
defined by setting $\nu E=\Bover1{\mu F}\mu(F\cap E)$ for every Borel set
$E\subseteq X$.   Now there are Borel sets $E$, $B\subseteq X$ such that
$A\symmdiff E\subseteq B$ and $\nu B=0$.   In this case,
$(A\cap F)\symmdiff(E\cap F)\subseteq B\cap F$ and $\mu(B\cap F)=0$, so
that $A\cap F$ is measured by $\mu$.
Because $F$ is arbitrary and $\mu$ is locally determined, $A$ is
measured by $\mu$.   (ii) Suppose that $A\subseteq X$ is measured by
every complete locally determined topological measure on $X$.   Then, in
particular, it is measured by the completion of any Borel probability
measure, so is universally measurable.\ \Qed}

\spheader 434Dc The family
$\Sigma_{\text{um}}$ of universally measurable subsets of $X$ is a
$\sigma$-algebra closed under Souslin's operation and including the
Borel $\sigma$-algebra.   \prooflet{(For it is the intersection of the
domains of a family of complete totally finite measures, and these
are all $\sigma$-algebras including the Borel $\sigma$-algebra and closed
under Souslin's operation, by 431A.)}   In particular, Souslin-F sets
are universally measurable, so (if $X$ is Hausdorff) K-analytic and
analytic sets are\cmmnt{ (422Ha, 423C)}.

\spheader 434Dd Note that a function $f:X\to\Bbb R$ is
$\Sigma_{\text{um}}$-measurable iff it is
$\mu$-virtually measurable\cmmnt{ (definition:  122Q)} for
every totally finite Borel measure $\mu$ on $X$\cmmnt{ (122Q, 212Fa)}.
Generally, if $Y$ is another topological space, I will say that
$f:X\to Y$ is {\bf universally measurable} if
$f^{-1}[H]\in\Sigma_{\text{um}}$
for every open set $H\subseteq Y$\cmmnt{;  that is, if $f$ is
$(\Sigma_{\text{um}},\Cal B(Y))$-measurable, where $\Cal B(Y)$ is the
Borel $\sigma$-algebra of $Y$}.   Continuous functions are universally
measurable\cmmnt{, of course}.

\spheader 434De In fact, if $f:X\to Y$ is universally measurable, then
it is $(\Sigma_{\text{um}},\Sigma^{(Y)}_{\text{um}})$-measurable, where
$\Sigma^{(Y)}_{\text{um}}$ is the algebra of universally measurable
subsets of $Y$.   \prooflet{\Prf\ Take $F\in\Sigma^{(Y)}_{\text{um}}$
and a totally finite Borel measure $\mu$ on $X$.   If $\hat\mu$ is the
completion of $\mu$, then the image measure $\nu=\hat\mu f^{-1}$ is a
complete totally finite topological measure on $Y$, so measures $F$,
and $f^{-1}[F]\in\dom\hat\mu$.   As $\mu$ is arbitrary,
$f^{-1}[F]\in\Sigma_{\text{um}}$;  as $F$ is arbitrary, $f$
is $(\Sigma_{\text{um}},\Sigma^{(Y)}_{\text{um}})$-measurable.\ \Qed}

\spheader 434Df It follows that if $Z$ is a third topological space
and $f:X\to Y$, $g:Y\to Z$ are universally measurable, then $gf:X\to Z$
is universally measurable.

\leader{434E}{Universally Radon-measurable sets}\cmmnt{ A companion
idea is the following.} Let $X$ be a Hausdorff space.

\spheader 434Ea I will say that a
subset $E$ of $X$ is {\bf universally Radon-measurable} if it is
measured by every Radon measure on $X$.

\spheader 434Eb The family
$\Sigma_{\text{uRm}}$ of universally Radon-measurable subsets of $X$ is
a $\sigma$-algebra closed under Souslin's operation and including the
algebra\cmmnt{ $\Sigma_{\text{um}}$}
of universally measurable subsets of $X$\cmmnt{ (and, {\it a fortiori},
including the Borel $\sigma$-algebra)}.   \prooflet{(Use 434Db and the
idea of 434Dc.)}

\spheader 434Ec If $Y$ is another topological space, I will say that a
function $f:X\to Y$ is {\bf universally
Radon-measurable} if $f^{-1}[H]\in\Sigma_{\text{uRm}}$ for every open set
$H\subseteq Y$.   A function $f:X\to\Bbb R$ is universally Radon-measurable
iff it is $\Sigma_{\text{uRm}}$-measurable iff it is
$\mu$-virtually measurable\cmmnt{ (that is, $\mu$-almost continuous,
see 418J)} for every totally finite tight Borel measure $\mu$ on $X$.
\cmmnt{(Compare 434Dd.)}   A universally measurable function is universally
Radon-measurable.

\leader{434F}{Elementary properties of Radon spaces:  Proposition}
Let $X$ be a Hausdorff space.

(a) The following are equiveridical:

\quad(i) $X$ is a Radon space;

\quad(ii) every semi-finite Borel measure on $X$ is tight;

\quad(iii) if $\mu$ is a locally finite Borel measure on $X$, its
c.l.d.\ version $\tilde\mu$ is a Radon measure;

\quad(iv) whenever $\mu$ is a totally finite Borel measure on $X$, and
$G\subseteq X$ is an open set with $\mu G>0$, then there is a compact
set $K\subseteq G$ such that $\mu K>0$;

\quad(v) whenever $\mu$ is a non-zero totally finite Borel measure on
$X$, there is a Radon subspace $Y$ of $X$ such that $\mu^*Y>0$.

(b) If $Y\subseteq X$ is a subspace which is a Radon space in its
induced topology, then $Y$ is universally measurable in $X$.

(c) If $X$ is a Radon space and $Y\subseteq X$,
then $Y$ is Radon iff it is universally measurable in $X$ iff it is
universally Radon-measurable in $X$.
In particular, all Borel subsets and all Souslin-F subsets of $X$ are
Radon spaces.

(d) The family of Radon subspaces of $X$ is closed under Souslin's
operation and set difference.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Let $\mu$ be a semi-finite Borel
measure on $X$, $E\subseteq X$ a Borel set and $\gamma<\mu E$.   Because
$\mu$ is semi-finite, there is a Borel set $H$ of finite measure such
that $\mu(E\cap H)>\gamma$.   Set $\nu F=\mu(F\cap H)$ for every Borel
set $F\subseteq X$;  then $\nu$ is a totally finite Borel measure on
$X$, and $\nu E>\gamma$.   Because $X$ is a Radon space, there is a
compact set $K\subseteq E$ such that $\nu K\ge\gamma$, and now $\mu
K\ge\gamma$.   As $\mu$, $E$ and $\gamma$ are arbitrary, (ii) is true.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} and {\bf (i)$\Rightarrow$(v)} are
trivial.

\medskip

\quad{\bf (v)$\Rightarrow$(iv)} Assume (v), and let $\mu$ be a totally
finite Radon measure on $X$ and $G$ a non-negligible open set.   Set
$\nu E=\mu(E\cap G)$ for every Borel set $E\subseteq X$.   Then $\nu$ is
a non-zero totally finite Borel measure on $X$, so there is a Radon
subspace $Y$ of $X$ such that $\nu^*Y>0$.   The subspace measure $\nu_Y$
on $Y$ is a Borel measure on $Y$, so is tight.   Since
$\nu_Y(Y\setminus G)=\nu(X\setminus G)=0$, $\nu_Y(Y\cap G)>0$ and there
is a compact set $K\subseteq Y\cap G$ such that $\nu_YK>0$.   Now $\mu
K>0$.   As $\mu$ and $G$ are arbitrary, (iv) is true.

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(iv)} If $X$ is not Radon, there is a
totally finite Borel measure $\mu$ on $X$ which is not tight.   By
416F(iii), there is an open set $G\subseteq X$ such that

\Centerline{$\mu G>\sup_{K\subseteq G\text{ is compact}}\mu K=\gamma$}

\noindent say.   Let $\Cal K$ be the family of compact subsets of $G$.
By 215B(v), there is a non-decreasing sequence $\sequencen{K_n}$ in
$\Cal K$ such that $\mu(K\setminus F)=0$ for every $K\in\Cal K$, where
$F=\bigcup_{n\in\Bbb N}K_n$.   Observe that

\Centerline{$\mu F=\lim_{n\to\infty}\mu K_n\le\gamma<\mu G$.}

\noindent Now set
$\nu E=\mu(E\cap G\setminus F)$ for every Borel set $E\subseteq X$.
Then $\nu$ is a
Borel measure on $X$, and $\nu G>0$.   If $K\subseteq G$ is compact,
then $\nu K=\mu(K\setminus F)=0$.   So $\nu$ and $G$ witness that (iv)
is false.

\medskip

\quad{\bf (i)$\Rightarrow$(iii)} The point is that $\tilde\mu$ is tight.
\Prf\ If
$\tilde\mu E>\gamma$, then, because $\tilde\mu$ is semi-finite, there is
a set $E'\subseteq E$ such that $\gamma<\tilde\mu E'<\infty$;  now there
is a Borel set $H\subseteq E'$ such that $\mu H=\tilde\mu E'$ (213Fc).
Setting $\nu F=\mu(H\cap F)$ for every Borel set $F$, $\nu$ is a totally
finite Borel measure on $X$ and $\nu H>\gamma$, so there is a compact
set $K\subseteq H$ such that $\nu K\ge\gamma$.   Since
$\mu K<\infty$, $\tilde\mu K=\mu K\ge\gamma$ (213Fa), while
$K\subseteq E$.   As $E$ and $\gamma$ are arbitrary, $\tilde\mu$ is tight.\ \Qed

On the other hand, every point of $X$ belongs to an open set of finite
measure for $\mu$, which is still of finite measure for $\tilde\mu$
(213Fa again).   So $\tilde\mu$ is locally finite;  since it is surely
complete and locally determined, it is a Radon measure.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} Assume (iii), and let $\mu$ be a totally
finite Borel measure on $X$.   Then its c.l.d.\ version $\tilde\mu$ is
tight.   But $\tilde\mu$
extends $\mu$ (213Hc), so $\mu$ also is tight.   As $\mu$ is arbitrary,
$X$ is a Radon space.

\medskip

{\bf (b)} Let $\mu$ be a totally finite Borel measure on $X$, and
$\hat\mu$ its completion;  let $\epsilon>0$.   Let $\mu_Y$ be the
subspace measure on $Y$, so that $\mu_Y$ is a totally finite Borel
measure on $Y$, and is tight.
There is a compact set $K\subseteq Y$ such that
$\mu_YK\ge\mu_YY-\epsilon$.   But this means that

\Centerline{$\mu^*Y=\mu_YY\le\mu_YK+\epsilon
=\mu^*(K\cap Y)+\epsilon=\mu K+\epsilon\le\mu_*Y+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\mu_*Y=\mu^*Y$, and $Y$ is
measured by $\hat\mu$ (413Ef);  as $\mu$ is
arbitrary, $Y$ is universally measurable.

\medskip

{\bf (c)}(i) If $Y$ is Radon, it is universally measurable, by
(b).   (ii) If $Y$ is universally measurable, it is universally
Radon-measurable, by 434Eb.  (iii) Suppose that $Y$ is universally
Radon-measurable, and that $\nu$
is a totally finite Borel measure on $Y$.   For Borel sets
$E\subseteq X$, set $\mu E=\nu(E\cap Y)$.   Then $\mu$ is a totally
finite Borel
measure on $X$, so its c.l.d.\ version $\tilde\mu$ is a Radon measure on
$X$, by (a-iii).
We are supposing that $Y$ is universally Radon-measurable, so, in
particular, it must be measured by $\tilde\mu$.   We have

$$\eqalignno{\tilde\mu(X\setminus Y)
&=\sup_{K\subseteq X\setminus Y\text{ is compact}}\tilde\mu K
=\sup_{K\subseteq X\setminus Y\text{ is compact}}\mu K\cr
\displaycause{213Ha, because $\mu$ is totally finite}
&=\sup_{K\subseteq X\setminus Y\text{ is compact}}\nu(K\cap Y)
=0,\cr}$$

\noindent and $Y$ is $\tilde\mu$-conegligible.

Now suppose that $E\subseteq Y$ is a (relatively) Borel subset of $Y$.
Then $E$ is of the form $F\cap Y$ where $F$ is a Borel subset of $X$, so
that

$$\eqalign{\nu E
&=\mu F
=\tilde\mu F
=\tilde\mu(Y\cap F)
=\tilde\mu E\cr
&=\sup_{K\subseteq E\text{ is compact}}\mu K
=\sup_{K\subseteq E\text{ is compact}}\nu K.\cr}$$

\noindent As $E$ is arbitrary, $\nu$ is tight;  as $\nu$ is arbitrary,
$Y$ is a Radon space.

By 434Dc, it follows that all Borel subsets and all Souslin-F subsets of
$X$ are Radon spaces.

\medskip

{\bf (d)} The first step is to note that if $\sequencen{E_n}$ is a
sequence of Radon subspaces of $X$ with union $E$, then $E$ is Radon;
this is immediate from (a-v) above.

Now let $\family{\sigma}{S}{E_{\sigma}}$ be a Souslin
scheme, consisting of Radon subsets of $X$, with kernel $A$.   We know
that $E=\bigcup_{\sigma\in S}E_{\sigma}$ is a Radon space.   Every
$E_{\sigma}$ is universally measurable in $E$, by (b), so $A$ also is
(434Dc), and must be Radon, by (c).   Thus the family of Radon subspaces
of $X$ is closed under Souslin's operation.

If $E$ and $F$ are Radon subsets of $X$, then $E\cup F$ is Radon, and,
just as above, $F$ is universally measurable in $E\cup F$.
But this means that $E\setminus F=(E\cup F)\setminus F$ is universally
measurable in $E\cup F$, so that $E\setminus F$ is Radon.
}%end of proof of 434F

\leader{434G}{}\cmmnt{ Just as we can address the question `when can
we be sure that every Borel measure is of class D?' in terms of the
definition of `Radon' space (434C), we can form other classes of
topological space by declaring that the Borel measures they support must
be of certain kinds.   Three definitions which lead to interesting
patterns of ideas are the following.

\medskip

\noindent}{\bf Definitions (a)} A topological space $X$ is {\bf
Borel-measure-compact}\cmmnt{ ({\smc Gardner \& Pfeffer 84})}
if every totally finite Borel measure on
$X$ which is inner regular with respect to the closed sets is
$\tau$-additive\cmmnt{, that is, $X$ carries no measure of class B in
the classification of 434A}.

\spheader 434Gb A topological space $X$ is
{\bf Borel-measure-complete}\cmmnt{ ({\smc Gardner \& Pfeffer 84})} if
every totally finite Borel measure on
$X$ is $\tau$-additive.   \cmmnt{(If $X$ is regular and Hausdorff,
this amounts to saying that $X$ carries no measures of classes A or B in
the classification of 434A.)}

\spheader 434Gc A Hausdorff space $X$ is {\bf pre-Radon}\cmmnt{ (also
called
`{\bf hypo-radonian}', `{\bf semi-radonian}')} if every $\tau$-additive
totally finite Borel measure on $X$ is tight.   \cmmnt{(If $X$ is
regular, this amounts to saying that
$X$ carries no measure of class C in the classification of 434A.)}

\leader{434H}{Proposition} Let $X$ be a topological space and $\Cal B$
its Borel $\sigma$-algebra.

(a) The following are equiveridical:

\quad(i) $X$ is Borel-measure-compact;

\quad(ii) every semi-finite Borel measure on $X$ which is inner regular
with respect to the closed sets is $\tau$-additive;

\quad(iii) every effectively locally
finite Borel measure on $X$ which is inner regular with respect to the
closed sets has an extension to a quasi-Radon measure;

\quad(iv) every totally finite Borel measure on $X$ which is inner
regular with respect to the closed sets has a support;

\quad(v) if
$\mu$ is a non-zero totally finite Borel measure on $X$, inner regular
with respect to the closed sets, and $\Cal G$ is an open cover of $X$,
then there is some $G\in\Cal G$ such that $\mu G>0$.

(b) If $X$ is Lindel\"of\cmmnt{ (in particular, if $X$ is a
K-analytic Hausdorff space)}, it is Borel-measure-compact.

(c) If $X$ is Borel-measure-compact and $A\subseteq X$ is a Souslin-F
set, then $A$ is Borel-measure-compact in its subspace topology.
In particular, any Baire subset of $X$ is Borel-measure-compact.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Assume (i), and let $\mu$ be a
semi-finite Borel measure on $X$ which is inner regular with respect to
the closed sets.   Let $\Cal G$ be an
upwards-directed family of open sets with union $G^*$, and
$\gamma<\mu G^*$.   Because $\mu$ is semi-finite, there is an
$H\in\Cal B$ such that $\mu H<\infty$ and $\mu(H\cap G^*)\ge\gamma$.
Set $\nu E=\mu(E\cap H)$ for every $E\in\Cal B$;  then $\nu$ is a
totally finite Borel measure on $X$.   For any $E\in\Cal B$,

\Centerline{$\nu E=\mu(E\cap H)
=\sup\{\mu F:F\subseteq E\cap H$ is closed$\}
\le\sup\{\nu F:F\subseteq E$ is closed$\}$,}

\noindent so $\nu$ is inner regular with respect to the closed sets, and
must be $\tau$-additive.   Now

\Centerline{$\gamma
\le\nu G^*
=\sup_{G\in\Cal G}\nu G
\le\sup_{G\in\Cal G}\mu G$.}

\noindent As $\gamma$ and $\Cal G$ is arbitrary, $\mu$ is
$\tau$-additive.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} Assume (ii), and let $\mu$ be an
effectively locally finite Borel measure on $X$ which is inner regular
with respect to the closed sets.   Then it is semi-finite (411Gd),
therefore $\tau$-additive.   By 415L, it has an extension to a
quasi-Radon measure on $X$.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} If (iii) is true and $\mu$ is a totally
finite Borel measure on $X$ which is inner regular with respect to the
closed sets, then $\mu$ has an extension to a
quasi-Radon measure, which is $\tau$-additive, so $\mu$ also is
$\tau$-additive (411C).

\medskip

\quad{\bf (ii)$\Rightarrow$(iv)} Use 411Nd.

\medskip

\quad{\bf (iv)$\Rightarrow$(v)} Suppose that (iv) is true, that $\mu$ is
a non-zero totally finite Borel measure on $X$ which is inner regular
with respect to the closed sets, and that $\Cal G$ is an open cover of
$X$.   If $F$ is the support of $\mu$, then $\mu F>0$ so
$F\ne\emptyset$;  there must be some $G\in\Cal G$ meeting $F$, and now
$\mu G>0$.

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(v)} Suppose that there is a totally
finite Borel measure $\mu$ on $X$, inner regular with respect to the
closed sets, which is not $\tau$-additive.   Let $\Cal G$ be an
upwards-directed family of open sets such that $\mu G^*>\gamma$, where
$G^*=\bigcup\Cal G$ and $\gamma=\sup_{G\in\Cal G}\mu G$.   Let
$\sequencen{G_n}$ be a non-decreasing sequence in $\Cal G$ such that
$\mu(G\setminus G^*_0)$ for every $G\in\Cal G$, where
$G^*_0=\bigcup_{n\in\Bbb N}G_n$ (215B(v) again).
Then $\mu G^*_0\le\gamma$,
so there is a closed set $F\subseteq G^*\setminus G^*_0$ such that
$\mu F>0$.

Let $\nu$ be the Borel measure on $X$ defined by setting
$\nu E=\mu(E\cap F)$ for every $E\in\Cal B$.   As in the argument for
(i)$\Rightarrow$(ii), $\nu$ is inner regular with respect to the closed
sets.   Consider $\Cal H=\Cal G\cup\{X\setminus F\}$;  this is an open
cover of $X$.   If $G\in\Cal G$ then $\nu G\le\mu(G\setminus G^*_0)=0$,
so $\nu H=0$ for every $H\in\Cal H$;  thus $\nu$ and $\Cal H$ witness
that (v) is false.

\medskip

{\bf (b)} Use (a-v) with 422Gg.

\medskip

{\bf (c)} Let $\mu$ be a Borel measure on $A$ which is inner regular
with respect to the closed sets, that is to say, the relatively closed
sets in $A$.   Let $\nu$ be the corresponding Borel measure on $X$,
defined by setting $\nu E=\mu(A\cap E)$ for every $E\in\Cal B$.   Let
$\hat\nu$ be the completion of $\nu$.   Putting 431D and 421M together,
we see that $\hat\nu A=\sup\{\hat\nu F:F\subseteq A$ is closed in $X\}$,
that is, $\nu X=\sup\{\mu F:F\subseteq A$ is closed in $X\}$.   But this
means that if $E\in\Cal B$ and $\gamma<\nu E$, there is a closed set $F$
in $X$ such that $F\subseteq A$ and $\mu(E\cap F)>\gamma$;  now
there is a relatively closed set $F'\subseteq A$ such that
$F'\subseteq E\cap F$ and $\mu F'\ge\gamma$, and as $F'$ must be
relatively closed in $F$ it is closed in $X$, while $\nu F'\ge\gamma$.
Since $E$ and $\gamma$ are arbitrary, $\nu$ is inner regular with
respect to the closed sets, and will be $\tau$-additive.

Now suppose that $\Cal G$ is an upwards-directed family of relatively
open subsets of $A$.   Set
$\Cal H=\{H:H\subseteq X$ is open, $H\cap A\in\Cal G\}$.   Then $\Cal H$
is upwards-directed, so

\Centerline{$\mu(\bigcup\Cal G)
=\nu(\bigcup\Cal H)
=\sup_{H\in\Cal H}\nu H
=\sup_{G\in\Cal G}\mu G$.}

\noindent As $\mu$ and $\Cal G$ are arbitrary, $A$ is
Borel-measure-compact.

By 421L, it follows that any Baire subset of $X$ is
Borel-measure-compact.
}%end of proof of 434H

\leader{434I}{Proposition} Let $X$ be a topological space.

(a) The following are equiveridical:

\quad(i) $X$ is Borel-measure-complete;

\quad(ii) every semi-finite Borel measure on $X$ is $\tau$-additive;

\quad(iii) every totally finite Borel measure on $X$ has
a support;

\quad(iv) whenever $\mu$ is a totally finite Borel measure on $X$ there
is a base $\Cal U$ for the topology of $X$ such that
$\mu(\bigcup\{U:U\in\Cal U,\,\mu U=0\})=0$.

(b) If $X$ is regular, it is Borel-measure-complete
iff every effectively locally finite Borel measure on $X$ has an
extension to a quasi-Radon measure.

(c) If $X$ is Borel-measure-complete, it is Borel-measure-compact.

(d) If $X$ is Borel-measure-complete, so is every subspace of $X$.

(e) If $X$ is hereditarily Lindel\"of\cmmnt{ (for instance, if $X$ is
separable and metrizable, see 4A2P(a-iii))}, it is
Borel-measure-complete\cmmnt{, therefore Borel-measure-compact}.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Use the argument of
(i)$\Rightarrow$(ii) of 434Ha;  this case is simpler, because we do not
need to check that the auxiliary measure $\nu$ is inner regular.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} is trivial.

\medskip

\quad{\bf (i)$\Rightarrow$(iv)} If $X$ is Borel-measure-complete and
$\mu$ is a totally finite Borel measure on $X$, take $\Cal U$ to be the
family of all open subsets of $X$.   This is surely a base for the
topology, and setting $\Cal U_0=\{U:U\in\Cal U,\,\mu U=0\}$, $\Cal U_0$
is upwards-directed so
$\mu(\bigcup\Cal U_0)=\sup_{U\in\Cal U_0}\mu U=0$, as required.

\medskip

\quad{\bf (iv)$\Rightarrow$(iii)} Assume (iv), and  let $\mu$ be a
totally finite Borel measure on $X$.   Take a base $\Cal U$ as in (iv),
so that $\mu(\bigcup\Cal U_0)=0$, where $\Cal U_0$ is the family of
negligible members of $\Cal U$.   Set $F=X\setminus\bigcup\Cal U_0$, so
that $F$ is a conegligible closed set.   If $G\subseteq X$ is an open
set meeting $F$, there is a member $U$ of $\Cal U$ such that $U\subseteq
G$ and $U\cap F\ne\emptyset$;  now $U\notin\Cal U$ so

\Centerline{$\mu(G\cap F)=\mu G\ge\mu U>0$.}

\noindent As $G$ is arbitrary, $F$ is self-supporting and is the support
of $\mu$.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} Assume (iii), and let $\mu$ be a
totally finite Borel measure on $X$.   Let $\Cal G$ be an
upwards-directed family of open sets with union $G^*$.   Set
$\gamma=\sup_{G\in\Cal G}\mu G$.   Let $\sequencen{G_n}$ be a
non-decreasing sequence in $\Cal G$ such that
$\mu(G\setminus G^*_0)$ for every $G\in\Cal G$, where
$G^*_0=\bigcup_{n\in\Bbb N}G_n$ (215B(v) once more).
Then $\mu G^*_0\le\gamma$.
Let $\nu$ be the Borel measure on $X$ defined by setting
$\mu E=\mu(E\cap G^*\setminus G_0^*)$ for every Borel set $E\subseteq X$.
Then
$\nu$ has a support $F$ say.   Now $\nu G=0$ for every $G\in\Cal G$, so
$F\cap G=\emptyset$ for every $G\in\Cal G$, and $F\cap G^*=\emptyset$;
but this means that

\Centerline{$\mu(G^*\setminus G_0^*)=\nu X=\nu F
=\mu(F\cap G^*\setminus G_0^*)=0$.}

\noindent Accordingly $\mu G^*=\gamma$.   As $\mu$ and $\Cal G$ are
arbitrary, $X$ is Borel-measure-complete.

\medskip

{\bf (b)} If $X$ is
Borel-measure-complete and $\mu$ is an effectively locally finite Borel
measure on $X$, then $\mu$ is $\tau$-additive, by (a-ii), so extends to
a quasi-Radon measure on $X$, by 415Cb.   If effectively locally finite
Borel measures on $X$ extend to quasi-Radon measures, then any totally
finite Borel measure is $\tau$-additive, by 411C, and $X$ is
Borel-measure-complete.

\medskip

{\bf (c)} Immediate from the definitions.

\medskip

{\bf (d)} If $Y\subseteq X$ and $\mu$ is a totally finite Borel measure
on $Y$, let $\nu$ be the Borel measure on $X$ defined by setting
$\nu E=\mu(E\cap Y)$ for every Borel set
$E\subseteq X$.   Then $\nu$ is $\tau$-additive.
So if $\Cal G$ is an upwards-directed family of relatively open subsets
of $Y$, and we set $\Cal H=\{H:H\subseteq X$ is open, $H\cap Y\in\Cal
G\}$, we shall get

\Centerline{$\mu(\bigcup\Cal G)=\nu(\bigcup\Cal H)
=\sup_{H\in\Cal H}\nu H=\sup_{G\in\Cal G}\mu G$.}

\noindent As $\mu$ and $\Cal G$ are arbitrary, $Y$ is
Borel-measure-complete.

\medskip

{\bf (e)} If $\mu$ is a totally finite Borel measure on $X$ and $\Cal G$
is a non-empty upwards-directed family of open subsets of $X$ with union
$G^*$, then there is a sequence $\sequencen{G_n}$ in $\Cal G$ with union
$G^*$, by 4A2H(c-i).   Because $\Cal G$ is upwards-directed, there is a
non-decreasing sequence $\sequencen{G'_n}$ in $\Cal G$ such that
$G'_n\supseteq G_n$ for every $n\in\Bbb N$, so that

\Centerline{$\mu G^*=\lim_{n\to\infty}\mu G'_n
\le\sup_{G\in\Cal G}\mu G$.}

\noindent As $\mu$ and $\Cal G$ are arbitrary, $X$ is
Borel-measure-complete.
}%end of proof of 434I

\vleader{72pt}{434J}{Proposition} Let $X$ be a Hausdorff space.

(a) The following are equiveridical:

\quad(i) $X$ is pre-Radon;

\quad(ii) every effectively locally finite $\tau$-additive Borel measure
on $X$ is tight;

\quad(iii) whenever $\mu$ is a non-zero totally finite
$\tau$-additive Borel measure on $X$, there is a compact set
$K\subseteq X$ such that $\mu K>0$;

\quad(iv) whenever $\mu$ is a totally finite $\tau$-additive Borel
measure on $X$, $\mu X=
\ifdim\pagewidth=390pt\break\fi
\sup_{K\subseteq X\text{ is compact}}\mu K$;

\quad(v) whenever $\mu$ is a locally finite effectively locally finite
$\tau$-additive Borel measure on $X$, the c.l.d.\ version of $\mu$ is a
Radon measure on $X$.

(b) If $X$ is pre-Radon, then every locally finite
quasi-Radon measure on $X$ is a Radon measure.

(c) If $X$ is regular and every totally finite quasi-Radon measure on
$X$ is a Radon measure, then $X$ is pre-Radon.

(d) If $X$ is pre-Radon, then any universally
Radon-measurable subspace\cmmnt{ (in particular, any Borel subset or
Souslin-F subset)} of $X$ is pre-Radon.

(e) If $A\subseteq X$ is pre-Radon in its subspace topology, it is
universally Radon-measurable in $X$.

(f) If $X$ is K-analytic\cmmnt{ (for instance, if it is compact)}, it
is pre-Radon.

(g) If $X$ is completely regular and \v{C}ech-complete\cmmnt{ (for
instance, if it is locally compact (4A2Gk), or metrizable and complete
under a metric inducing its topology (4A2Md))}, it is pre-Radon.

(h) If $X=\prod_{i\in I}X_i$ where $\familyiI{X_i}$ is a countable
family of pre-Radon Hausdorff spaces, then $X$ is pre-Radon.

(i) If every point of $X$ belongs to a pre-Radon open subset of $X$,
then $X$ is pre-Radon.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Suppose that $X$ is pre-Radon, that
$\mu$ is an effectively locally finite $\tau$-additive Borel measure on
$X$, that $E\subseteq X$ is Borel, and that
$\gamma<\mu E$.   Because $\mu$ is semi-finite, there is a Borel set
$H\subseteq E$ of finite measure such that $\mu(H\cap E)>\gamma$.   Set
$\nu F=\mu(F\cap H)$ for every Borel set $F\subseteq X$;  then $\nu$ is
a totally finite Borel measure on $X$, and is $\tau$-additive by 414Ea.
Now $\nu E>\gamma$, so there is a compact set $K\subseteq E$ such that
$\gamma\le\nu K\le\mu K$.   As $E$ is arbitrary, $\mu$ is tight.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} is trivial.

\medskip

\quad{\bf (iii)$\Rightarrow$(iv)} Assume (iii), and let $\mu$ be a
totally finite $\tau$-additive Borel measure on $X$.   Let $\Cal K$ be
the family of compact subsets of $X$ and set
$\alpha=\sup_{K\in\Cal K}\mu K$.
\Quer\ Suppose, if possible, that $\mu X>\alpha$.   Let
$\sequencen{K_n}$ be a sequence in $\Cal K$ such that
$\sup_{n\in\Bbb N}\mu K_n=\alpha$, and set $L=\bigcup_{n\in\Bbb N}K_n$;
then

\Centerline{$\mu L=\lim_{n\to\infty}\mu(\bigcup_{i\le n}K_i)=\alpha$.}

\noindent Set $\nu E=\mu(E\setminus L)$ for every Borel set
$E\subseteq X$.   Then $\nu$ is a non-zero totally finite Borel measure
on $X$, and is $\tau$-additive, by 414Ea again.   So there is a
$K\in\Cal K$ such that $\nu K>0$.   But now there is an $n\in\Bbb N$
such that
$\nu K+\mu K_n>\alpha$, and in this case $K\cup K_n\in\Cal K$ and

\Centerline{$\mu(K\cup K_n)=\mu(K\setminus K_n)+\mu K_n
\ge\nu K+\mu K_n>\alpha$,}

\noindent which is impossible.\ \BanG\
So $\mu X=\alpha$, as required.

\medskip

\quad{\bf (iv)$\Rightarrow$(i)} Assume (iv), and let $\mu$ be a totally
finite $\tau$-additive Borel measure on $X$.   Suppose that
$E\subseteq X$ is Borel and that $\gamma<\mu E$.   By (iv), there is a
compact set $K\subseteq X$ such that $\mu K>\mu X-\mu E+\gamma$, so that
$\mu(E\cap K)>\gamma$.   Consider the subspace measure $\mu_K$ on the
compact Hausdorff space $K$.   By 414K, $\mu_K$ is $\tau$-additive,
so is inner regular with respect to the
closed subsets of $K$ (414Mb).   There is therefore a relatively
closed subset $F$ of $K$ such that $F\subseteq K\cap E$ and
$\mu_KF\ge\gamma$;  but now $F$ is a compact subset of $E$ and $\mu
F\ge\gamma$.   As $E$ and $\gamma$ are arbitrary, $\mu$ is tight.   As
$\mu$ is arbitrary, $X$ is
pre-Radon.

\medskip

\quad{\bf (ii)$\Rightarrow$(v)} Assume (ii), and let $\mu$ be a locally
finite effectively locally finite $\tau$-additive Borel measure on $X$.
Then $\mu$ is tight, so by
416F(ii) its c.l.d.\ version is a Radon measure.

\medskip

\quad{\bf (v)$\Rightarrow$(iv)} Assume (v), and let $\mu$ be a totally
finite $\tau$-additive Borel measure on $X$.   Then the c.l.d.\ version
$\tilde\mu$ of $\mu$ is a Radon measure;  but $\tilde\mu$ extends $\mu$
(213Hc), so

\Centerline{$\sup_{K\subseteq X\text{ is compact}}\mu K
=\sup_{K\subseteq X\text{ is compact}}\tilde\mu K
=\tilde\mu X=\mu X$.}

\medskip

{\bf (b)} Let $\mu$ be a locally finite quasi-Radon measure on $X$.   By
(a-ii), applied to the restriction of $\mu$ to the Borel $\sigma$-algebra
of $X$, $\mu$ is tight;  by 416C, $\mu$ is a Radon measure.

\medskip

{\bf (c)} Let $\mu$ be a totally finite $\tau$-additive Borel measure on
$X$.   Because $X$ is regular, the c.l.d.\ version $\tilde\mu$ of $\mu$ is a
quasi-Radon measure (415Cb again),
therefore a Radon measure;  but $\tilde\mu$ extends $\mu$
(213Hc once more), so $\mu$, like $\tilde\mu$, must be tight.   As $\mu$ is
arbitrary, $X$ is pre-Radon.

\medskip

{\bf (d)} Let $A$ be a universally Radon-measurable subset of $X$, and $\mu$
a totally finite $\tau$-additive Borel measure on
$A$.   Set
$\nu E=\mu(E\cap A)$ for every Borel set $E\subseteq X$;  then $\nu$ is
a totally finite $\tau$-additive Borel measure on $X$.   So its c.l.d.\
version (that is, its completion $\hat\nu$, by 213Ha) is a Radon measure
on $X$, by (a-v).   Now $\hat\nu$ measures $A$, so

\Centerline{$\mu A
=\nu^*A
=\hat\nu A
=\sup\{\hat\nu K:K\subseteq A$ is compact$\}
=\sup\{\mu K:K\subseteq A$ is compact$\}$.}

\noindent By (a-iv), $A$ is pre-Radon.

\medskip

{\bf (e)} Let $\mu$ be a totally finite Radon measure on $X$.   Then the
subspace measure $\mu_A$ is $\tau$-additive (414K), so its restriction
$\nu$ to the Borel $\sigma$-algebra of $A$ is still $\tau$-additive.
Because $A$ is pre-Radon,

$$\eqalign{\mu^*A
=\mu_AA
&=\nu A
=\sup\{\nu K:K\subseteq A\text{ is compact}\}\cr
&=\sup\{\mu K:K\subseteq A\text{ is compact}\}
=\mu_*A,\cr}$$

\noindent and $\mu$ measures $A$ (413Ef again).
As $\mu$ is arbitrary, $A$ is universally Radon-measurable.

\medskip

{\bf (f)} Put 432B and (a-iv) together.

\medskip

{\bf (g)} If we identify $X$ with a G$_{\delta}$ set in a compact
Hausdorff space $Z$, then $Z$ is pre-Radon, by (f), so $X$ is pre-Radon,
by (d).

\medskip

{\bf (h)} Let $\mu$ be a totally finite $\tau$-additive Borel measure on
$X$, and $\epsilon>0$.   Let $\familyiI{\epsilon_i}$ be a family of
strictly positive real numbers such that
$\sum_{i\in I}\epsilon_i\le\epsilon$ (4A1P).   For each $i\in I$ and Borel
set $F\subseteq X_i$, set $\mu_iF=\mu\pi_i^{-1}[F]$, where
$\pi_i(x)=x(i)$ for $x\in X$;  because $\pi_i:X\to X_i$ is continuous,
$\mu_i$ is a totally finite $\tau$-additive Borel measure on $X_i$.
Because $X_i$ is pre-Radon, we can find a compact set $K_i\subseteq X_i$
such that $\mu_i(X_i\setminus K_i)\le\epsilon_i$, by (a-iv).   Now
$K=\prod_{i\in I}K_i$ is compact (3A3J), and
$X\setminus K\subseteq\bigcup_{i\in I}\pi_i^{-1}[X_i\setminus K_i]$, so

\Centerline{$\mu(X\setminus K)\le\sum_{i\in I}\mu_i(X_i\setminus K_i)
\le\sum_{i\in I}\epsilon_i\le\epsilon$.}

As $\epsilon$ and $\mu$ are arbitrary, $X$ satisfies the condition of
(a-iv), and is pre-Radon.

\medskip

{\bf (i)} Let $\Cal G$ be a cover of $X$ by pre-Radon open sets.   Let
$\mu$ be a non-zero totally finite $\tau$-additive Borel measure on $X$.
Then $\mu X=\sup\{\mu(\bigcup\Cal G_0):\Cal G_0\subseteq\Cal G$ is
finite$\}$, so there is some $G\in\Cal G$ such that $\mu G>0$.   Now the
subspace measure $\mu_G$ is a non-zero totally finite $\tau$-additive
Borel measure on $G$, so there is a compact set $K\subseteq G$ such that
$\mu_GK>0$, in which case $\mu K>0$.   As $\mu$ is arbitrary, $X$ is
pre-Radon, by (a-iii).
}%end of proof of 434J

\leader{434K}{}\cmmnt{ I return to criteria for deciding whether
Hausdorff spaces are Radon.

\wheader{434K}{6}{2}{2}{72pt}

\noindent}{\bf Proposition} (a) A Hausdorff space is Radon iff it is
Borel-measure-complete and pre-Radon.

(b) An analytic Hausdorff space is Radon.
\cmmnt{In particular, any compact metrizable space is Radon and any
Polish space is Radon.}

(c) $\omega_1$ and $\omega_1+1$, with their order topologies, are not
Radon.

(d) For a set $I$, $[0,1]^I$ is Radon iff $\{0,1\}^I$ is Radon iff $I$
is countable.

(e) A hereditarily Lindel\"of K-analytic Hausdorff space is Radon;  in
particular, the split interval\cmmnt{ (343J, 419L)} is Radon.

\proof{{\bf (a)} Put the definitions 434C, 434Gb and 434Gc together,
recalling that a tight measure is necessarily $\tau$-additive (411E).

\medskip

{\bf (b)} 433Cb.

\medskip

{\bf (c)} Dieudonn\'e's measure (411Q) is a Borel measure on $\omega_1$
which is not tight,
so $\omega_1$ is certainly not a Radon space;  as it is an open set in
$\omega_1+1$, and the subspace topology inherited from $\omega_1+1$ is
the order topology of $\omega_1$ (4A2S(a-iii)), $\omega_1+1$ cannot be
Radon (434Fc).

\medskip

{\bf (d)} If $I$ is countable, then $\{0,1\}^I$ and $[0,1]^I$ are
compact metrizable spaces, so are Radon.   If $I$ is uncountable, then
$\omega_1+1$, with its order topology, is homeomorphic to a closed
subset of $\{0,1\}^I$.   \Prf\ Set $\kappa=\#(I)$.   For
$\xi\le\omega_1$, $\eta<\kappa$ set $x_{\xi}(\eta)=1$ if $\eta<\xi$, $0$
if $\xi\le\eta$.   The map $\xi\mapsto
x_{\xi}:\omega_1+1\to\{0,1\}^{\kappa}$ is injective
because $\kappa\ge\omega_1$, and is continuous because all the sets
$\{\xi:x_{\xi}(\eta)=0\}=(\omega_1+1)\cap(\eta+1)$ are open-and-closed
in $\omega_1+1$.   Since $\omega_1+1$ is compact in its order topology
(4A2S(a-i)), it is homeomorphic to its image in
$\{0,1\}^{\kappa}\cong\{0,1\}^I$.\ \Qed

By 434Fc, $\{0,1\}^I$ cannot
be a Radon space.   Since $\{0,1\}^I$ is a closed subset of $[0,1]^I$,
$[0,1]^I$ also is not a Radon space.

\medskip

{\bf (e)} Suppose that $X$ is a hereditarily Lindel\"of K-analytic
Hausdorff space.   Then it is Borel-measure-complete by 434Ie and
pre-Radon by 434Jf, so by (a) here it is Radon.

Since the split interval is compact and Hausdorff and hereditarily
Lindel\"of (419La), it is a Radon space.
}%end of proof of 434K

\leader{434L}{}\cmmnt{ It is worth noting an elementary special
property of metric spaces.

\medskip

\noindent}{\bf Proposition} If $(X,\rho)$ is a metric space, then any
quasi-Radon measure on $X$ is inner regular with respect to the
totally bounded subsets of $X$.

\proof{ Let $\mu$ be a quasi-Radon measure on $X$ and $\Sigma$
its domain.   Suppose that $E\in\Sigma$ and $\gamma<\mu E$.   Then there
is an open set $G$ of finite measure such that $\mu(E\cap G)>\gamma$;
set $\delta=\mu(E\cap G)-\gamma$.   For $n\in\Bbb N$, $I\subseteq X$ set
$H(n,I)=\bigcup_{x\in I}\{y:\rho(y,x)<2^{-n}\}$.   Then
$\{H(n,I):I\in[X]^{<\omega}\}$ is an
upwards-directed family of open sets covering $X$.   Because $\mu$ is
$\tau$-additive, there is a finite set $I_n\subseteq X$ such that
$\mu(G\setminus H(n,I_n))\le 2^{-n-1}\delta$.   Consider
$F=\bigcap_{n\in\Bbb N}H(n,I_n)$.   This is
totally bounded and $\mu(G\setminus F)\le\delta$, so $E\cap F$ is
totally bounded and $\mu(E\cap F)\ge\gamma$.   As $E$ and $\gamma$ are
arbitrary,
$\mu$ is inner regular with respect to the totally bounded sets.
}%end of proof of 434L

\leader{434M}{}\cmmnt{ I turn next to a couple of ideas depending on
countable compactness.

\medskip

\noindent}{\bf Lemma} Let $X$ be a countably compact topological space
and $\Cal E$ a non-empty family of closed subsets of $X$ with the finite
intersection
property.   Then there is a Borel probability measure $\mu$ on $X$,
inner regular
with respect to the closed sets, such that $\mu F=1$ for every
$F\in\Cal E$.

\proof{{\bf (a)} By Zorn's lemma, $\Cal E$ is included in a maximal
family $\Cal E^*$ of
closed subsets of $X$ with the finite intersection property.

\quad(i) If $F\subseteq X$ is closed and $F\cap F_0\cap\ldots\cap
F_n\ne\emptyset$ for every $F_0,\ldots,F_n\in\Cal E^*$, then $\Cal
E^*\cup\{F\}$ has the finite intersection property, so $F\in\Cal E^*$.

\quad(ii) If $F$, $F'\in\Cal E^*$, then $F\cap F'\cap F_0\cap\ldots\cap
F_n\ne\emptyset$ for all $F_0,\ldots,F_n\in\Cal E^*$, so $F\cap
F'\in\Cal E^*$.

\quad(iii) If $F\subseteq X$ is closed and $F\cap F'\in\Cal E^*$ for every
$F'\in\Cal E^*$, then $F\cap
F_0\cap\ldots\cap F_n\in\Cal E^*$ for every $F_0,\ldots,F_n\in\Cal E^*$
(because $F_0\cap\ldots\cap F_n\in\Cal E^*$, by (ii)), so
$F\in\Cal E^*$.

\quad(iv) If $\sequencen{F_n}$ is a sequence in $\Cal E^*$,
with intersection
$F$, and $F'\in\Cal E^*$, then $F'\cap\bigcap_{i\le n}F_i\ne\emptyset$
for every $n\in\Bbb N$.   Because $X$ is countably compact,
$F'\cap F\ne\emptyset$ (4A2G(f-ii)).   As $F'$ is arbitrary,
$F\in\Cal E^*$, by (iii).
Thus $\Cal E^*$ is closed under countable intersections.

\medskip

{\bf (b)} Set

\Centerline{$\Sigma=\{E:E\subseteq X$, there is an $F\in\Cal E^*$ such
that either $F\subseteq E$ or $F\cap E=\emptyset\}$,}

\noindent and define $\hat\mu:\Sigma\to\{0,1\}$ by saying that
$\hat\mu E=1$ if there is some $F\in\Cal E^*$ such that $F\subseteq E$,
$0$ otherwise.   Then $\hat\mu$ is a probability measure on $X$.
\Prf\

\quad(i) $\emptyset\in\Sigma$ because $\Cal E^*\supseteq\Cal E$ is not empty.

\quad(ii) $X\setminus E\in\Sigma$ whenever $E\in\Sigma$ because the
definition of $\Sigma$ is symmetric between $E$ and $X\setminus E$.

\quad(iii) If $\sequencen{E_n}$ is any sequence in $\Sigma$ with union $E$,
then either there are $n\in\Bbb N$ and $F\in\Cal E^*$ such that
$F\subseteq E_n\subseteq E$ and $E\in\Sigma$, or for every $n\in\Bbb N$ there is an
$F_n\in\Cal E^*$ such that $F_n\cap E_n=\emptyset$.   In this case
$F=\bigcap_{n\in\Bbb N}F_n\in\Cal E^*$, by (a-iv), and $E\cap
F=\emptyset$, so again $E\in\Sigma$.   Thus $\Sigma$ is a
$\sigma$-algebra of subsets of $X$.

\quad(iv) $\hat\mu\emptyset=0$ because $\emptyset$ cannot belong to $\Cal
E^*$.

\quad(v) If $\sequencen{E_n}$ is any disjoint sequence in $\Sigma$ with union
$E$, then either there is some $n$ such that $\hat\mu E_n=1$, in which
case $\hat\mu E_i=0$ for every $i\ne n$ (because any two members of
$\Cal E^*$ must meet) and $\hat\mu E=1=\sum_{i=0}^{\infty}\hat\mu E_i$,
or $\hat\mu E_i=0$ for every $i$, in which case, just as in (iii),
$\hat\mu E=0=\sum_{i=0}^{\infty}\hat\mu E_i$.   Thus $\hat\mu$ is a
measure.

\quad(vi) Because $\Cal E^*\ne\emptyset$, $\hat\mu X=1$.   Thus $\hat\mu$ is
a probability measure.\ \Qed

\medskip

{\bf (c)} If $F\subseteq X$ is a closed set, then either $F$ itself
belongs to $\Cal E^*$, so $F\in\Sigma$, or there is some $F'\in\Cal E^*$
such that $F\cap F'=\emptyset$, in which case again $F\in\Sigma$.   So
$\Sigma$ contains every closed set, therefore every Borel set, and
$\hat\mu$ is a topological measure.   By construction, $\hat\mu$ is
inner regular with respect to $\Cal E^*$ and therefore with respect to
the closed sets.   Finally, if $F\in\Cal E$ then $F\in\Cal E^*$, so
$\hat\mu F=1$.   We may therefore take $\mu$ to be the restriction of
$\hat\mu$ to
the Borel $\sigma$-algebra of $X$, and $\mu$ will be a Borel measure on
$X$,
inner regular with respect to the closed sets, such that $\mu E=1$ for
every $E\in\Cal E$.
}%end of proof of 434M

\leader{434N}{Proposition} (a) Let $X$ be a Borel-measure-compact
topological space.   Then closed countably compact subsets of $X$ are
compact.

(b) Let $X$ be a Borel-measure-complete topological space.
Then countably compact subsets of $X$ are compact.

(c) Let $X$ be a Hausdorff Borel-measure-complete topological space.
Then compact subsets of $X$ are countably tight.

(d) In particular, any Radon compact Hausdorff space is countably tight.

\proof{{\bf (a)} Let $C$ be a closed countably compact subset of $X$.
Let $\Cal F$ be an ultrafilter on $C$.   Let $\Cal E$ be the
family of closed subsets of $C$ belonging to $\Cal F$.
Then $\Cal E$ has the finite intersection property, so by 434M there is
a Borel probability measure $\mu$ on $C$, inner regular with respect to
the closed sets, such that $\mu E=1$ for every $E\in\Cal E$.
Let $\nu$ be the Borel measure on $X$ defined by setting
$\nu H=\mu(C\cap H)$ for every Borel set $H\subseteq X$.   Then $\nu$ is
also inner regular with respect to the closed sets (of either $C$ or
$X$);  because $X$ is Borel-measure-compact, $\nu$ has a support $F$
(434H(a-iv)).   Since
$\nu F=\nu X=1$, $F\cap C\ne\emptyset$;  take $x\in F\cap C$.   If $G$
is any open
set (in $X$) containing $x$, then $\mu(C\setminus G)=\nu(X\setminus G)<1$, so
$C\setminus G\notin\Cal F$ and $C\cap G\in\Cal F$.   As $G$ is
arbitrary, $\Cal F\to x$;  as $\Cal F$ is arbitrary, $C$ is compact.

\medskip

{\bf (b)} Repeat the argument of (a).   Let $C$ be a countably compact
subset of $X$ and $\Cal F$ an ultrafilter on $C$.   Let $\Cal E$ be
the family of relatively closed subsets of $C$ belonging to $\Cal F$.
Then there is
a Borel probability measure $\mu$ on $C$ such that $\mu E=1$ for every
$E\in\Cal E$.
Let $\nu$ be the Borel measure on $X$ defined by setting
$\nu H=\mu(C\cap H)$ for every Borel set $H\subseteq X$.   Because $X$
is Borel-measure-complete, $\nu$ has a support $F$ (434I(a-iii)).
Since $\nu F=\nu X=1$, $F\cap C\ne\emptyset$;  take $x\in F\cap C$.   If
$G$ is any open
set containing $x$, then $\nu(X\setminus G)<1$, so
$C\setminus G\notin\Cal F$ and $C\cap G\in\Cal F$.   As $G$ is
arbitrary, $\Cal F\to x$;  as $\Cal F$ is arbitrary, $C$ is compact.

\medskip

{\bf (c)} Again let $C$ be a (countably) compact subset of $X$.   Take
$A\subseteq C$, and set
$C_0=\bigcup\{\overline{B}:B\in[A]^{\le\omega}\}$.   Then $C_0$ is
countably compact.   \Prf\ If $\sequencen{y_n}$ is any sequence in
$C_0$, it has a cluster point $y\in C$.   For
each $n\in\Bbb N$ there is a countable set $B_n\subseteq A$ such that
$y_n\in\overline{B_n}$.   Now $B=\bigcup_{n\in\Bbb N}B_n$ is a countable
subset of $A$, and $y\in\overline{B}\subseteq C_0$, so $y$ is a cluster
point of $\sequencen{y_n}$ in $C_0$.   As $\sequencen{y_n}$ is
arbitrary, $C_0$ is countably compact.\ \Qed

By (b), $C_0$ is compact, therefore closed, and must include
$\overline{A}$.   Thus every point of $\overline{A}$ is in the closure
of some countable subset of $A$.   As $A$ is arbitrary, $C$ is countably
tight.

\medskip

{\bf (d)} Finally, a compact Radon Hausdorff space is
Borel-measure-complete (434Ka) and countably compact, therefore
countably tight.
}%end of proof of 434N

\leader{434O}{Quasi-dyadic \dvrocolon{spaces}}\cmmnt{ I wish now to
present a result in an entirely different direction.   Measures of type
B$_1$ in the classification of 434A (completion regular, but not
$\tau$-additive) seem to be hard to come by.   The next theorem shows
that on a substantial class of spaces they cannot appear.   First, we
need a definition.

\medskip

\noindent}{\bf Definition} A topological space $X$ is {\bf quasi-dyadic}
if it is expressible as a continuous image of a product of separable
metrizable spaces.

\cmmnt{I give some elementary results to indicate what kind of spaces
we have here.}

\leader{434P}{Proposition} (a) A continuous image of a quasi-dyadic
space is quasi-dyadic.

(b) Any product of quasi-dyadic spaces is quasi-dyadic.

(c) A space with a countable network is quasi-dyadic.

(d) A Baire subset of a quasi-dyadic space is quasi-dyadic.

(e) If $X$ is any topological space, a countable union of quasi-dyadic
subspaces of $X$ is quasi-dyadic.

(f) A quasi-dyadic space is ccc.

\proof{{\bf (a)} Immediate from the definition.

\medskip

{\bf (b)} Again immediate;  if $X_i$ is a continuous image of
$\prod_{j\in J_i}Y_{ij}$, where $Y_{ij}$ is a separable metrizable space for
every $i\in I$ and $j\in J_i$, then $\prod_{i\in I}X_i$ is a continuous
image of $\prod_{i\in I,j\in J_i}Y_{ij}$.

\medskip

{\bf (c)} Let $\Cal E$ be a countable network for the
topology of $X$.   On $X$ let $\sim$ be the equivalence relation in
which $x\sim y$ if they
belong to just the same members of $\Cal E$;  let $Y$ be the space
$X/\backstep2\sim$ of equivalence classes, and $\phi:X\to Y$ the
canonical map.   $Y$ has a separable metrizable topology with base
$\{\phi[E]:E\in\Cal
E\}\cup\{\phi[X\setminus E]:E\in\Cal E\}$.   Let $I$ be any set such
that $\#(\{0,1\}^I)\ge\#(X)$, and for each $y\in Y$ let
$f_y:\{0,1\}^I\to y$ be a surjection.   Then we have a
continuous surjection $f:Y\times\{0,1\}^I\to X$ given by saying that
$f(y,z)=f_y(z)$ for $y\in Y$ and $z\in\{0,1\}^I$.

\medskip

{\bf (d)} Let $\familyiI{Y_i}$ be a
family of separable metrizable spaces with product $Y$ and $f:Y\to X$ a
continuous surjection.   If $W\subseteq Y$ is a Baire set, it is
determined by coordinates in a countable subset of $I$ (4A3Nb),
so can be regarded as $W'\times\prod_{i\in I\setminus J}Y_i$, where
$J\subseteq I$ is countable and $W'\subseteq\prod_{i\in J}Y_i$;  as
$\prod_{i\in J}Y_i$ and $W'$ are separable metrizable spaces
(4A2Pa), $W$ can be
thought of as a product of separable metrizable spaces.   Now the set
$\{E:E\subseteq X,\,f^{-1}[E]$ is a Baire set in $Y\}$ is a
$\sigma$-algebra containing every zero set in $X$, so contains every
Baire set.   Thus every Baire subset of $X$ is a continuous image of a
Baire subset of $Y$, and is therefore quasi-dyadic.

\medskip

{\bf (e)} If $E_n\subseteq X$ is quasi-dyadic for each $n\in\Bbb N$,
then $Z=\Bbb N\times\prod_{n\in\Bbb N}E_n$ is quasi-dyadic, and
$f:Z\to\bigcup_{n\in\Bbb N}E_n$ is a continuous surjection, where
$f(n,\langle x_i\rangle_{{i\in\Bbb N}})=x_n$.   So
$\bigcup_{n\in\Bbb N}E_n$ is quasi-dyadic.

\medskip

{\bf (f)} Use 4A2E(a-iii) and (a-iv).
}%end of proof of 434P

\leader{434Q}{Theorem}\cmmnt{ ({\smc Fremlin \& Grekas 95})} A
semi-finite completion regular topological measure on a quasi-dyadic
space is $\tau$-additive.

\proof{\Quer\ Suppose, if possible, otherwise.

\medskip

{\bf (a)} The first step is the standard reduction to the case in which
$\mu X=1$ and $X$ is covered by open sets of zero measure.   In detail:
suppose that $X$ is a quasi-dyadic space and $\mu_0$ is a semi-finite
completion regular topological measure on $X$ which is not
$\tau$-additive.   Let $\Cal G$ be an upwards-directed family of open sets in $X$ such that $\mu_0(\bigcup\Cal G)$ is strictly greater than $\sup_{G\in\Cal G}\mu_0G=\gamma$ say.   Let $\sequencen{G_n}$ be a
non-decreasing
sequence in $\Cal G$ such that $\lim_{n\to\infty}\mu_0G_n=\gamma$, and
set $H_0=\bigcup_{n\in\Bbb N}G_n$, so that $\mu_0H_0=\gamma$;  take a
closed set $Z\subseteq\bigcup\Cal G$ such that $\gamma<\mu_0Z<\infty$.
Set $\mu_1E=\mu_0(E\cap Z\setminus H_0)$ for every Borel set
$E\subseteq X$.   Then $\mu_1$ is a non-zero totally finite Borel
measure on $X$, and is
completion regular.   \Prf\ If $E\subseteq X$ is a Borel set and
$\epsilon>0$, there is
a zero set $F\subseteq E\cap Z\setminus H_0$ such that
$\mu_0F\ge\mu_0(E\cap X\setminus H_0)-\epsilon$, and now
$\mu_1F\ge\mu_1E-\epsilon$.\ \QeD\
Note that $\mu_1(X\setminus Z)=\mu_1G=0$ for every $G\in\Cal G$.

For Borel sets $E\subseteq X$,
set $\mu E=\mu_1E/\mu_1X$;  then $\mu$ is a completion regular Borel
probability measure on $X$, and $\Cal G\cup\{X\setminus Z\}$ is a cover
of $X$ by open negligible sets.

\medskip

{\bf (b)} Now let $\langle Y_i\rangle_{i\in I}$ be a family
of separable metrizable spaces such that there is a continuous surjection
$f:Y\to X$, where $Y=\Yprod{I}$.   For each $i\in I$ let
$\Cal B_i$ be a countable base for the topology of $Y_i$.   For
$J\subseteq I$ let $\Cal C(J)$ be the family of all non-empty open cylinders in
$Y$ expressible in the form

\Centerline{$\{s:s(i)\in B_i\Forall i\in K\}$}

\noindent where $K$ is a finite subset of $J$ and
$B_i\in\Cal B_i$ for each $i\in K$;  thus $\Cal C(I)$ is a base for the
topology of $Y$.   Set $\Cal C_0(J)=\{U:U\in\Cal C(J),\,\mu^*f[U]=0\}$
for each $J\subseteq I$.   Note that (because every $\Cal B_i$ is
countable) $\Cal C(J)$ and $\Cal C_0(J)$ are countable for every
countable subset $J$ of $I$.   It is easy to see that
$\Cal C(J)\cap\Cal C(K)=\Cal C(J\cap K)$ for all $J$, $K\subseteq I$,
because if $U\in\Cal C(I)$ is non-empty it belongs to $\Cal C(J)$ iff
its projection onto $X_i$ is the whole of $X_i$ for every $i\notin J$.

For each negligible set $E\subseteq X$, let
$\langle F_n(E)\rangle_{n\in\Bbb N}$ be a family of zero sets, subsets
of $X\setminus E$, such that $\sup_{n\in\Bbb N}\mu F_n(E)=1$.   Then each
$f^{-1}[F_n(E)]$ is a zero set in $Y$, so there is a countable set
$M(E)\subseteq I$ such that all the sets $f^{-1}[F_n(E)]$ are determined
by coordinates in $M(E)$ (4A3Nc).   Let $\Cal J$ be the family of
countable subsets $J$ of
$I$ such that $M(f[U])\subseteq J$ for every $U\in\Cal C_0(J)$;  then
$\Cal J$ is cofinal with $[I]^{\le\omega}$, that is, every countable
subset of $I$ is included in some member of $\Cal J$.   \Prf\ If we
start from any countable subset $J_0$ of $I$ and set

\Centerline{$J_{n+1}=J_n\cup\bigcup\{M(f[U]):U\in\Cal C_0(J_n)\}$}

\noindent for each $n\in\Bbb N$, then every $J_n$ is countable, and
$\bigcup_{n\in\Bbb N}J_n\in\Cal
J$, because $\sequencen{J_n}$ is non-decreasing, so $\Cal
C_0(\bigcup_{n\in\Bbb N}J_n)=\bigcup_{n\in\Bbb N}\Cal C_0(J_n)$.\ \Qed

\medskip

{\bf (c)} For each $J\in\Cal J$, set

\Centerline{$Q_J=\bigcap\{\bigcup_{n\in\Bbb N}F_n(f[U]):U\in\Cal
C_0(J)\}$.}

\noindent Then $\mu Q_J=1$ and $f^{-1}[Q_J]$ is determined by
coordinates
in $J$, while $f^{-1}[Q_J]\cap U=\emptyset$ whenever $U\in\Cal C_0(J)$.

If $G\subseteq X$ is an open set, then $G\cap Q_J=\emptyset$
whenever $J\in\Cal J$ and there is a negligible Baire set $Q\supseteq G$
such that $f^{-1}[Q]$ is determined by coordinates in $J$.   \Prf\  Set
$H=\pi_J^{-1}[\pi_J[f^{-1}[G]]]$, where $\pi_J:Y\to\Yprod{J}$ is the
canonical map;  then $H$ is a union of members of $\Cal C(J)$, because
$f^{-1}[G]$ is open in $Y$ and $\pi_J[f^{-1}[G]]$ is open in
$\Yprod{J}$.   Also, because $f^{-1}[Q]$ is determined by coordinates in
$J$, $H\subseteq f^{-1}[Q]$, so $f[H]\subseteq Q$ and $\mu^*f[H]=0$;
thus all the members of $\Cal C(J)$ included in $H$ actually belong to
$\Cal C_0(J)$, and $H\cap f^{-1}[Q_J]=\emptyset$.
But this means that $f^{-1}[G]\cap f^{-1}[Q_J]=\emptyset$ and (because
$f$ is a surjection) $G\cap Q_J=\emptyset$, as
claimed.\ \QeD\   In particular, if $G$ is a negligible open set in $X$,
then $G\cap
Q_J=\emptyset$ whenever $J\in\Cal J$ and $J\supseteq M(G)$.

\medskip

{\bf (d)} If $J\in\Cal J$, there are $s$, $s'\in f^{-1}[Q_J]$ such that
$s\restr J=s'\restr J$ and $f(s)$, $f(s')$ can be separated by open sets
in $X$.   \Prf\ Start from any $x\in Q_J$ and take a negligible
open set $G$ containing $x$ (recall that our hypothesis is that $X$ is
covered by negligible open sets).   For each $n\in\Bbb N$ let
$h_n:X\to\Bbb R$ be a continuous function such that
$F_n(G)=h_n^{-1}[\{0\}]$.   We know that $G\cap Q_J\ne\emptyset$, while
$G\subseteq X\setminus(\bigcup_{n\in\Bbb N}F_n(G)\cap Q_J)$, which is a
negligible Baire set;  by (c), $f^{-1}[X\setminus(\bigcup_{n\in\Bbb
N}F_n(G)\cap Q_J)]$ is not determined by coordinates in $J$, and there
must be some $n$ such that $f^{-1}[F_n(G)\cap Q_J]$ is not determined by
coordinates in $J$.   Accordingly
there must be $s$, $s'\in Y$ such that $s\restr J=s'\restr J$, $s\in
f^{-1}[F_n(G)\cap Q_J]$ and $s'\notin f^{-1}[F_n(G)\cap Q_J]$.   Now
$s\in f^{-1}[Q_J]$, which is determined by coordinates in $J$;  since
$s\restr J=s'\restr J$, $s'\in f^{-1}[Q_J]$ and
$s'\notin f^{-1}[F_n(G)]$.
Accordingly $h_n(f(s))=0\ne h_n(f(s'))$ and $f(s)$, $f(s')$ can be
separated by open sets.\ \Qed

\medskip

{\bf (e)} We are now ready to embark on the central construction of the
argument.   We can choose inductively, for ordinals $\xi<\omega_1$, sets
$J_{\xi}\in\Cal J$, negligible open sets $G_{\xi}$,
$G'_{\xi}\subseteq X$, points $s_{\xi}$, $s'_{\xi}\in Y$ and sets
$U_{\xi}$, $V_{\xi}$, $V'_{\xi}\in\Cal C(I)$ such that

\inset{$J_{\eta}\subseteq J_{\xi}$, $U_{\eta}$, $V_{\eta}$, $V'_{\eta}$
all belong to $\Cal C(J_{\xi})$ and $G_{\eta}\cap Q_{J_{\xi}}=\emptyset$
whenever $\eta<\xi<\omega_1$ (using the results of (b) and (c) to choose
$J_{\xi}$);}

\inset{$s_{\xi}\restr J_{\xi}=s'_{\xi}\restr J_{\xi}$, $s_{\xi}\in
f^{-1}[Q_{J_{\xi}}]$ and $f(s_{\xi})$ and $f(s'_{\xi})$ can be separated
by open sets in $X$ (using (d) to choose $s_{\xi}$, $s'_{\xi}$);}

\inset{$G_{\xi}$, $G'_{\xi}$ are disjoint negligible open sets
containing
$f(s_{\xi})$, $f(s'_{\xi})$ respectively (choosing $G_{\xi}$,
$G'_{\xi}$);}

\inset{$U_{\xi}\in\Cal C(J_{\xi})$, $V_{\xi}$, $V'_{\xi}\in\Cal
C(I\setminus J_{\xi})$, $s_{\xi}\in U_{\xi}\cap V_{\xi}\subseteq
f^{-1}[G_{\xi}]$, $s'_{\xi}\in U_{\xi}\cap V'_{\xi}
\subseteq f^{-1}[G'_{\xi}]$ (choosing
$U_{\xi}$, $V_{\xi}$, $V'_{\xi}$, using the fact that $s_{\xi}\restr
J_{\xi}=s'_{\xi}\restr J_{\xi}$).}

\noindent On completing this construction, take for each $\xi<\omega_1$
a finite set $K_{\xi}\subseteq J_{\xi+1}$ such that $U_{\xi}$, $V_{\xi}$
and $V'_{\xi}$ all belong to $\Cal C(K_{\xi})$.   By the $\Delta$-system
Lemma (4A1Db), there is an uncountable $A\subseteq\omega_1$ such
that $\langle K_{\xi}\rangle_{\xi\in A}$ is a $\Delta$-system with root
$K$ say.    For $\xi\in A$, express $U_{\xi}$ as
$\tilde U_{\xi}\cap U'_{\xi}$ where $\tilde U_{\xi}\in\Cal C(K)$ and
$U'_{\xi}\in\Cal C(K_{\xi}\setminus K)$.   Then there are only countably many
possibilities for $\tilde U_{\xi}$, so there is an uncountable
$B\subseteq A$ such that $\tilde U_{\xi}$ is constant for $\xi\in B$;
write $\tilde U$ for the constant value.   Let $C\subseteq B$ be an
uncountable set, not containing $\min A$, such that $K_{\xi}\setminus K$
does not meet $J_{\eta}$ whenever $\xi$, $\eta\in C$ and $\eta<\xi$
(4A1Eb).   Let $D\subseteq C$ be such that $D$ and $C\setminus D$
are both uncountable.

Note that $K\subseteq K_{\eta}\subseteq J_{\xi}$ whenever $\eta$,
$\xi\in A$ and $\eta<\xi$, so that $K\subseteq J_{\xi}$ for every
$\xi\in C$.   Consequently $U'_{\xi}$, $V_{\xi}$ and $V'_{\xi}$ all
belong to $\Cal C(K_{\xi}\setminus K)$ for every $\xi\in C$.

\medskip

{\bf (f)} Consider the open set

\Centerline{$G=\bigcup_{\xi\in D}G_{\xi}\subseteq X$.}

\noindent At this point the argument divides.

\medskip

\quad{\bf case 1} Suppose $\mu^*(G\cap f[\tilde U])>0$.   Then there is
a Baire set $Q\subseteq G$ such that $\mu^*(Q\cap f[\tilde U])>0$.   Let
$J\subseteq I$ be a countable set such that $f^{-1}[Q]$ is determined by
coordinates in $J$.
Let $\gamma\in C\setminus D$ be so large that $K_{\xi}\setminus K$ does
not meet $J$ for any $\xi\in A$ with $\xi\ge\gamma$.   Then
$Q\cap Q_{J_{\gamma}}\cap f[\tilde U]$ is not empty;  take
$s\in \tilde U\cap f^{-1}[Q\cap Q_{J_{\gamma}}]$.   Because the
$K_{\xi}\setminus K$ are disjoint from each other and from
$J\cup J_{\gamma}$ for $\xi\in C\setminus(\gamma+1)$,
we can modify $s$ to form $s'$ such that
$s'\restr J\cup J_{\gamma}=s\restr J\cup J_{\gamma}$ and
$s'\in U'_{\xi}\cap V'_{\xi}$ whenever $\xi\in C$ and $\xi>\gamma$;  now
$s'\in\tilde U$ (because $K\subseteq J_{\gamma}$), so
$s'\in\tilde U\cap U'_{\xi}\cap V'_{\xi}\subseteq f^{-1}[G'_{\xi}]$ and
$f(s')\notin G_{\xi}$ whenever $\xi\in C$ and $\xi>\gamma$.
On the other hand, if $\xi\in D$ and
$\xi<\gamma$,
$G_{\xi}\cap Q_{J_{\gamma}}=\emptyset$, while $s'\in f^{-1}[Q_{J_{\gamma}}]$ (because
$f^{-1}[Q_{J_{\gamma}}]$  is determined by coordinates in $J_{\gamma}$),
so again $f(s')\notin G_{\xi}$.

Thus $f(s')\notin G$.   But $s'\restr J=s\restr J$ so
$f(s)\in Q\subseteq G$;  which is impossible.

This contradiction disposes of the possibility that
$\mu^*(G\cap f[\tilde U])>0$.

\medskip

\quad{\bf case 2} Suppose that $\mu^*(G\cap f[\tilde U])=0$.   In this
case there is a negligible Baire set $Q\supseteq G\cap f[\tilde U]$.
Let $J\subseteq I$ be a countable set such that $f^{-1}[Q]$ is
determined
by coordinates in $J$.   Let $\gamma<\omega_1$ be such that
$J\cap J_{\gamma}=J\cap\bigcup_{\xi<\omega_1}J_{\xi}$ and
$J\cap K_{\xi}\setminus K=\emptyset$ for every $\xi\in A\setminus\gamma$.
Take $\xi\in D$ such that $\xi\ge\gamma$.   Then

\Centerline{$\tilde U\cap U'_{\xi}\cap V_{\xi}
\subseteq f^{-1}[G_{\xi}]\cap\tilde U
\subseteq f^{-1}[G\cap f[\tilde U]]
\subseteq f^{-1}[Q]$,}

\noindent so $\tilde U\subseteq f^{-1}[Q]$, because $U'_{\xi}\cap V_{\xi}$ is a
non-empty member of $\Cal C(I\setminus J)$.   But this
means that $\mu^*f[\tilde U]=0$ and $\mu^*f[U_{\xi}]=0$.   On the other
hand, we have $s_{\xi}\in U_{\xi}\cap f^{-1}[Q_{J_{\xi}}]$, so
$U_{\xi}\notin\Cal C_0(J_{\xi})$ and $\mu^*f[U_{\xi}]>0$.\ \Bang

Thus this route also is blocked and we must abandon the original
hypothesis that there is a quasi-dyadic space with a semi-finite
completion regular topological measure which is not $\tau$-additive.
}%end of proof of 434Q

\leader{434R}{}\cmmnt{ There is a useful construction of Borel product
measures which can be fitted in here.

\medskip

\noindent}{\bf Proposition} Let $X$ and $Y$ be topological spaces with
Borel measures $\mu$ and $\nu$;  write $\Cal B(X)$, $\Cal B(Y)$ for the
Borel $\sigma$-algebras of $X$ and $Y$ respectively.   If {\it either}
$X$ is first-countable {\it or} $\nu$ is $\tau$-additive and effectively
locally finite, there is a Borel measure $\lambda_B$ on $X\times Y$
defined by the formula

\Centerline{$\lambda_BW
=\sup_{F\in\Cal B(Y),\nu F<\infty}\int\nu(W[\{x\}]\cap F)\mu(dx)$}

\noindent for every Borel set $W\subseteq X\times Y$.   Moreover

\quad(i) if $\mu$ is semi-finite, then $\lambda_B$ agrees with the
c.l.d.\ product measure $\lambda$ on $\Cal B(X)\tensorhat\Cal B(Y)$, and
the c.l.d.\ version $\tilde\lambda_B$ of $\lambda_B$ extends $\lambda$;

\quad(ii) if $\nu$ is $\sigma$-finite, then
$\lambda_BW=\int\nu W[\{x\}]\mu(dx)$ for every Borel set
$W\subseteq X\times Y$;

\quad(iii) if both $\mu$ and $\nu$ are $\tau$-additive and effectively
locally finite, then $\lambda_B$ is just the restriction of the
$\tau$-additive product measure $\tilde\lambda$\cmmnt{ (417D, 417F)}
to the Borel $\sigma$-algebra of $X\times Y$;  in particular,
$\lambda_B$ is $\tau$-additive.

%when is $\lambda_B$ inner regular wrt closed sets?

\proof{{\bf (a)} The point is that $x\mapsto\nu(W[\{x\}]\cap F)$ is
lower semi-continuous whenever $W\subseteq X\times Y$ is open and
$\nu F<\infty$.   \Prf\ Of course $W[\{x\}]$ is always open, so $\nu$
always measures $W[\{x\}]\cap F$.   Take any $\alpha\in\Bbb R$ and set
$G=\{x:x\in X,\,\nu(W[\{x\}]\cap F)>\alpha\}$;  let $x_0\in G$.

\medskip

\quad\grheada\ Suppose that $X$ is first-countable.   Let
$\sequencen{U_n}$ be a
non-increasing sequence running over a base of open neighbourhoods of
$x_0$.   For each $n\in\Bbb N$, set

\Centerline{$V_n=\bigcup\{V:V\subseteq Y$ is open,
$U_n\times V\subseteq W\}$.}

\noindent Then $\sequencen{V_n}$ is a non-decreasing sequence with union
$W[\{x_0\}]$, so there is an $n\in\Bbb N$ such that
$\nu(V_n\cap F)>\alpha$.
Now $V_n\subseteq W[\{x\}]$ for every $x\in U_n$, so $U_n\subseteq G$.

\medskip

\quad\grheadb\ Suppose that $\nu$ is $\tau$-additive and effectively
locally finite.   Set

\Centerline{$\Cal V=\{V:V\subseteq Y$ is open, $U\times V\subseteq W$
for some open set $U$ containing $x_0\}$.}

\noindent Then $\Cal V$ is an upwards-directed family of open sets with
union $W[\{x_0\}]$, so there is a $V\in\Cal V$ such that
$\nu(V\cap F)>\alpha$ (414Ea).   Let $U$ be an open set containing $x_0$
such that $U\times V\subseteq W$;  then $V\subseteq W[\{x\}]$ for every
$x\in U$, so $U\subseteq G$.

\medskip

\quad\grheadc\ Thus in either case we have an open set containing $x_0$
and included in $G$.   As $x_0$ is arbitrary, $G$ is open;  as $\alpha$
is arbitrary, $x\mapsto\nu(W[\{x\}]\cap F)$ is lower semi-continuous.\
\Qed

\medskip

{\bf (b)} It follows that $x\mapsto\nu(W[\{x\}]\cap F)$ is Borel
measurable whenever $W\subseteq X\times Y$ is a Borel set and
$\nu F<\infty$.
\Prf\ Let $\Cal W$ be the family of sets $W\subseteq X\times Y$ such
that $W[\{x\}]$ is
a Borel set for every $x\in X$ and $x\mapsto\nu(W[\{x\}]\cap F)$ is
Borel measurable.   Then every open subset of $X\times Y$ belongs to
$\Cal W$ (by (a) above), $W\setminus W'\in\Cal W$ 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 $\Cal W$.   By the
Monotone Class Theorem (136B), $\Cal W$ includes the $\sigma$-algebra
generated by the open sets, that is, the Borel $\sigma$-algebra of
$X\times Y$.\
\Qed

\medskip

{\bf (c)} It is now easy to check that
$W\mapsto\int\nu(W[\{x\}]\cap F)\mu(dx)$ is a Borel measure on
$X\times Y$ whenever $\nu F<\infty$, and therefore that $\lambda_B$, as
defined here, is a Borel measure.

\wheader{434R}{4}{2}{2}{108pt}

{\bf (d)} Now suppose that $\mu$ is semi-finite, and that
$W\in\Cal B(X)\tensorhat\Cal B(Y)$.   Then

$$\eqalignno{\lambda W
&=\sup_{\mu E<\infty,\nu F<\infty}\lambda(W\cap(E\times F))\cr
\displaycause{by the definition of `c.l.d.\ product measure', 251F}
&=\sup_{\mu E<\infty,\nu F<\infty}\int_E\nu(W[\{x\}]\cap F)\mu(dx)\cr
\displaycause{by Fubini's theorem, 252C, applied to the product of the
subspace measures $\mu_E$ and $\nu_F$}
&=\sup_{\nu F<\infty}\int\nu(W[\{x\}]\cap F)\mu(dx)\cr
\displaycause{by 213B, because $\mu$ is semi-finite}
&=\lambda_BW.\cr}$$

\medskip

{\bf (e)} If, on the other hand, $\nu$ is $\sigma$-finite, let
$\sequencen{F_n}$ be a non-decreasing sequence of sets of finite measure
covering $Y$;  then

$$\eqalign{\lambda_BW
&=\sup_{\nu F<\infty}\int\nu(W[\{x\}]\cap F)\mu(dx)
\ge\sup_{n\in\Bbb N}\int\nu(W[\{x\}]\cap F_n)\mu(dx)\cr
&=\int\sup_{n\in\Bbb N}\nu(W[\{x\}]\cap F_n)\mu(dx)
=\int\nu W[\{x\}]\mu(dx)
\ge\lambda_BW\cr}$$

\noindent for any Borel set $W\subseteq X$.

\medskip

{\bf (f)} If both $\mu$ and $\nu$ are $\tau$-additive and effectively
locally finite, so that we have a $\tau$-additive product measure
$\tilde\lambda$, then Fubini's theorem for such measures (417G) tells us
that $\lambda_BW=\tilde\lambda W$ at least when $W\subseteq X\times Y$
is a Borel set and $\tilde\lambda W$ is finite.   If $W$ is any Borel
subset of $X\times Y$, then, as in (d),

$$\eqalign{\lambda_BW
&=\sup_{\mu E<\infty,\nu F<\infty}\int_E\nu(W[\{x\}]\cap F)\mu(dx)\cr
&=\sup_{\mu E<\infty,\nu F<\infty}\tilde\lambda(W\cap(E\times F))
=\tilde\lambda W\cr}$$

\noindent by 417C(b-iii).
}%end of proof of 434R

\cmmnt{\medskip

\noindent{\bf Remark} The case in which $X$ is first-countable is due to
{\smc Johnson 82}.}

\vleader{60pt}{*434S}{}\cmmnt{ The concept of `universally measurable'
set
enables us to extend a number of ideas from earlier sections.   First,
recall a problem from the very beginning of measure theory on the real
line:  the composition of Lebesgue measurable functions need not be
Lebesgue measurable (134Ib), while the composition of a Borel measurable
function with a Lebesgue measurable function is measurable (121Eg).   In
fact we can replace `Borel measurable' by `universally measurable', as
follows.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, $Y$ and $Z$ topological spaces, $f:X\to Y$ a
measurable function and $g:Y\to Z$ a universally measurable function.
Then $gf:X\to Z$ is measurable.   In particular, $f^{-1}[F]\in\Sigma$
for every universally measurable set $F\subseteq Y$.

\proof{ Let $H\subseteq Z$ be an open set and $E\in\Sigma$ a set of
finite measure.   Let $\mu_E$ be the subspace measure on $E$.   Then the
image measure $\nu=\mu_E(f\restr E)^{-1}$ is a complete totally finite
topological measure on $Y$, so its domain contains $g^{-1}[H]$, and

\Centerline{$E\cap(gf)^{-1}[H]
=(f\restr E)^{-1}[g^{-1}[H]]
\in\dom\mu_E\subseteq\Sigma$.}

\noindent As $E$ is arbitrary and $\mu$ is locally determined,
$(gf)^{-1}[H]\in\Sigma$;  as $H$ is arbitrary, $gf$ is measurable.

Applying this to $g=\chi F$, we see that $f^{-1}[F]\in\Sigma$ for every
universally measurable $F\subseteq Y$.
}%end of proof of 434S

\leader{*434T}{}\cmmnt{ The next remark concerns the concept
$\Bvalue{u\in E}$ of \S364.

\medskip

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ be a localizable
measure algebra.   Write $\Sigma_{\text{um}}$ for the algebra of
universally measurable subsets of $\Bbb R$.

(a) For any $u\in L^0=L^0(\frak A)$, we have a sequentially
order-continuous Boolean homomorphism
$E\mapsto\Bvalue{u\in E}:\Sigma_{\text{um}}\to\frak A$ defined by saying
that

$$\eqalign{\Bvalue{u\in E}
&=\sup\{\Bvalue{u\in F}:F\subseteq E\text{ is Borel}\}
=\sup\{\Bvalue{u\in K}:K\subseteq E\text{ is compact}\}\cr
&=\inf\{\Bvalue{u\in F}:F\supseteq E\text{ is Borel}\}
=\inf\{\Bvalue{u\in G}:G\supseteq E\text{ is open}\}\cr}$$

\noindent for every $E\in\Sigma_{\text{um}}$.

(b) For any $u\in L^0$ and universally measurable function
$h:\Bbb R\to\Bbb R$ we have a corresponding element $\bar h(u)$ of $L^0$
defined by the formula

\Centerline{$\Bvalue{\bar h(u)\in E}=\Bvalue{u\in h^{-1}[E]}$
for every $E\in\Sigma_{\text{um}}$, $u\in L^0$.}

\proof{ We can regard $(\frak A,\bar\mu)$ as the measure algebra of a
complete strictly localizable measure space $(X,\Sigma,\mu)$ (322O), in
which case $L^0$ can be identified with $L^0(\mu)$
(364Ic).   Write
$\Cal B$ for the Borel $\sigma$-algebra of $\Bbb R$.

\medskip

{\bf (a)} Let $f:X\to\Bbb R$ be a $\Sigma$-measurable
function representing $u$.   Then $f^{-1}[E]\in\Sigma$ for every
$E\in\Sigma_{\text{um}}$, by 434S.   Setting
$\phi E=(f^{-1}[E])^{\ssbullet}$, $\phi:\Sigma_{\text{um}}\to\frak A$ is
a sequentially order-continuous Boolean homomorphism.

We find that

\Centerline{$\phi E=\sup\{\Bvalue{u\in K}:
  K\subseteq E\text{ is compact}\}$}

\noindent for every $E\in\Sigma_{\text{um}}$.   \Prf\ If $H\in\Sigma$
and $\mu H<\infty$, then (writing $\mu_H$ for the subspace measure on
$H$) the image measure $\mu_H(f\restr H)^{-1}$ is a complete topological
measure, and its restriction $\nu$ to the Borel $\sigma$-algebra
$\Cal B$ of
$\Bbb R$ is a totally finite Borel measure.   Now $E$ is measured by the
completion $\hat\nu$ of $\nu$, which is a Radon measure (256C), so for any
$\epsilon>0$ there are a compact $K\subseteq E$ and a Borel $F\supseteq E$
such that $\nu F=\hat\nu E\le\nu K+\epsilon$.   In this
case,

\Centerline{$\Bvalue{u\in K}=(f^{-1}[K])^{\ssbullet}
\Bsubseteq\phi E\Bsubseteq(f^{-1}[F])^{\ssbullet}=\Bvalue{u\in F}$,}

\noindent using the formula of 364Ib, while

$$\eqalign{\bar\mu(H^{\ssbullet}\Bcap\phi E)
&\le\bar\mu(H^{\ssbullet}\Bcap\Bvalue{u\in F})
=\mu(H\cap f^{-1}[F])\cr
&=\nu F
\le\nu K+\epsilon
=\bar\mu(H^{\ssbullet}\Bcap\Bvalue{u\in K})+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$H^{\ssbullet}\Bcap\phi E
=\sup\{H^{\ssbullet}\Bcap\Bvalue{u\in K}:
  K\subseteq E\text{ is compact}\}$;}

\noindent as $H$ is arbitrary, and $(\frak A,\bar\mu)$ is semi-finite,
$\phi E=\sup\{\Bvalue{u\in K}:K\subseteq E\text{ is compact}\}$.
\QeD

Applying this to $\Bbb R\setminus E$, we see that
$\phi E=\inf\{\Bvalue{u\in G}:G\supseteq E\text{ is open}\}$.
Of course it follows at once that

\Centerline{$\Bvalue{u\in E}
=\sup\{\Bvalue{u\in F}:F\subseteq E\text{ is Borel}\}
=\inf\{\Bvalue{u\in F}:F\supseteq E\text{ is Borel}\}$.}

\noindent We can therefore identify the sequentially
order-continuous Boolean homomorphism $\phi$ with
$E\mapsto\Bvalue{u\in E}$, as described.

\medskip

{\bf (b)} Once again identifying $u$ with $f^{\ssbullet}$ where
$f$ is $\Sigma$-measurable, we see that $hf$ is $\Sigma$-measurable (by
434S), so we have a corresponding element $(hf)^{\ssbullet}$ of $L^0$.
If $E\in\Sigma_{\text{um}}$, then

\Centerline{$\Bvalue{(hf)^{\ssbullet}\in E}
=((hf)^{-1}[E])^{\ssbullet}
=(f^{-1}[h^{-1}[E]])^{\ssbullet}
=\Bvalue{u\in h^{-1}[E]}$,}

\noindent using 434De to check that $h^{-1}[E]\in\Sigma_{\text{um}}$, so
that we can identify $(hf)^{\ssbullet}$ with $\bar h(u)$, as described.
}%end of proof of 434T

\leader{434U}{}\cmmnt{ I give an elementary remark
on images of completion regular measures.
%434U

\medskip

\noindent}{\bf Proposition} Let $X$ and $Y$ be compact Hausdorff spaces and
$f:X\to Y$ a continuous open map.   If $\mu$ is a completion regular topological
measure on $X$, then the image measure $\mu f^{-1}$ on $Y$
is completion regular.

\proof{ If $F\subseteq Y$ is measured by $\nu=\mu f^{-1}$ and
$\gamma<\nu F=\mu f^{-1}[F]$, there is a zero set
$Z\subseteq f^{-1}[F]$ such that $\mu Z\ge\gamma$.   Now
$f[Z]\subseteq F$ is a zero set in $Y$ (4A2G(c-ii)) and

\Centerline{$\nu f[Z]=\mu f^{-1}[f[Z]]\ge\mu Z\ge\gamma$.}

\noindent As $F$ and $\gamma$ are arbitrary, $\nu$ is inner regular with
respect to the zero sets, so is completion regular.
}%end of proof of 434U

\exercises{
\leader{434X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 434Xa
Let $A\subseteq[0,1]$ be any non-measurable set.   Show
that the subspace measure on $A$ is completion regular and
$\tau$-additive but not tight.
%434A

\sqheader 434Xb Let $X$ be any Hausdorff space with a point $x$ such
that $\{x\}$ is not a G$_{\delta}$ set;  for instance, $X=\omega_1+1$
and $x=\omega_1$, or $X=\{0,1\}^I$ for any uncountable set $I$ and $x$
any point of $X$.   Show that setting $\mu E=\chi E(x)$ we get a tight
Borel measure on $X$ which is not completion regular.
%434A

\sqheader 434Xc Let $X$ be a topological space.
(i) Show that if $A\subseteq X$ is universally measurable in $X$, then
$A\cap Y$ is universally measurable in $Y$ for any set $Y\subseteq X$.
(ii) Show that if $Y\subseteq X$ is universally measurable in $X$, and
$A\subseteq Y$ is universally measurable in $Y$, then $A$ is universally
measurable in $X$.   (iii) Suppose that $X$ is the product of a
countable family $\familyiI{X_i}$ of
topological spaces, and $E_i\subseteq X_i$ is a universally measurable
set for each $i\in I$.   Show that $\prod_{i\in I}E_i$ is universally
measurable in $X$.
%434D

\spheader 434Xd Let $X$ be an analytic
Hausdorff space.   (i) Suppose that $Y$ is a
topological space and $W$ is a Borel subset of $X\times Y$.   Show that
$W[X]$ is a universally measurable subset of $Y$.   \Hint{423P.}
(ii) Let $A$ be a subset of $X$.   Show that the following are
equiveridical:  ($\alpha$) $A$ is universally measurable in $X$;  ($\beta$)
$f^{-1}[A]$ is Lebesgue measurable for every Borel measurable function
$f:[0,1]\to X$;  ($\gamma$) $f^{-1}[A]$ is measured by the usual measure on
$\{0,1\}^{\Bbb N}$ for every continuous function $f:\{0,1\}^{\Bbb N}\to X$.
%434D

\spheader 434Xe Let $\Sigma_{\text{um}}$
be the algebra of universally measurable
subsets of $\Bbb R$, and $\mu$ the restriction of Lebesgue measure to
$\Sigma_{\text{um}}$.
Show that $\mu$ is translation-invariant, but has no
translation-invariant lifting.   \Hint{345F.}
%434D query out of order

\spheader 434Xf Let $X$ be a Hausdorff space.
(i) Show that, for $A\subseteq X$, the following are equiveridical:
($\alpha$) $A$ is universally Radon-measurable in $X$;  ($\beta$) $A$ is
measured by every atomless Radon probability measure on $X$;  ($\gamma$)
$A\cap K$ is universally Radon-measurable in $K$ for every compact
$K\subseteq X$.
(ii) Show that if
$A\subseteq X$ is universally Radon-measurable in $X$, then $A\cap Y$ is
universally Radon-measurable in $Y$ for any set $Y\subseteq X$.
(iii) Show that if $Y\subseteq X$ is universally Radon-measurable in
$X$, and $A\subseteq Y$ is universally Radon-measurable in $Y$, then $A$
is universally Radon-measurable in $X$.
(iv) Show that if $\Cal G$ is an open cover of $X$, and $A\subseteq X$
is such that $A\cap G$ is universally Radon-measurable (in $G$ or in
$X$) for every $G\in\Cal G$, then $A$ is universally Radon-measurable in
$X$.
(v) Show that if $Y$ is another Hausdorff space, and
$\Sigma^{(X)}_{\text{uRm}}$, $\Sigma^{(Y)}_{\text{uRm}}$ are the
algebras of universally
Radon-measurable subsets of $X$, $Y$ respectively, then every continuous
function from $X$ to $Y$ is
$(\Sigma^{(X)}_{\text{uRm}},\Sigma^{(Y)}_{\text{uRm}})$-measurable.
(vi) Suppose that $X$ is the product of a countable family
$\familyiI{X_i}$ of
topological spaces, and $E_i\subseteq X_i$ is a universally
Radon-measurable set for each $i\in I$.   Show that $\prod_{i\in I}E_i$
is universally Radon-measurable in $X$.
%434E

\sqheader 434Xg(i) Let $\mu_0$ be Dieudonn\'e's measure on $\omega_1$.
Give $\omega_1+1=\omega_1\cup\{\omega_1\}$ its compact Hausdorff order
topology, and define a Borel measure $\mu$ on $\omega_1+1$ by setting
$\mu E=\mu_0(E\cap\omega_1)$ for every Borel set $E\subseteq\omega_1+1$.
Show that $\mu$ is a complete probability measure and is neither
$\tau$-additive nor inner regular with respect to the closed sets.
(ii) Show that the universally measurable subsets of $\omega_1+1$ are
just its Borel sets.
\Hint{4A3J, 411Q.}   (iii) Show that every totally finite
$\tau$-additive topological measure on $\omega_1+1$ has a countable
support.   (iv) Show that every subset of $\omega_1+1$ is universally
Radon-measurable.
%434E

\spheader 434Xh(i) Show that there is a set $X\subseteq[0,1]$ such that
$K\cap X$ and $K\setminus X$ both have cardinal $\frak c$ for every
uncountable compact set $K\subseteq[0,1]$.   \Hint{4A3Fa, 423L.}
(ii) Show that if we give $X$ its subspace topology, then every subset
of $X$ is universally Radon-measurable, but not every subset is
universally measurable.   \Hint{every compact subset of $X$ is
countable, so every Radon measure on $X$ is purely atomic, but $X$ has
full outer Lebesgue measure in $[0,1]$.}
%434E

\spheader 434Xi Show that a Hausdorff space $X$ is Radon iff ($\alpha$)
every compact subset of $X$ is Radon ($\beta$) for every non-zero
totally finite Borel measure $\mu$ on $X$ there is a compact subset $K$
of $X$ such that $\mu K>0$.   \Hint{434F(a-v).}
%434F

\sqheader 434Xj\dvAformerly{4{}34Xi}(i)
Let $X$ and $Y$ be K-analytic Hausdorff spaces and
$f:X\to Y$ a continuous surjection.   Suppose that $F\subseteq Y$ and
that $f^{-1}[F]$ is universally Radon-measurable in $X$.   Show that $F$
is universally Radon-measurable in $Y$.   \Hint{432G.}
(ii) Let $X$ and $Y$ be analytic Hausdorff spaces and $f:X\to Y$ a Borel
measurable surjection.
Suppose that $F\subseteq Y$ and that $f^{-1}[F]$ is universally
Radon-measurable in $X$.   Show that $F$ is universally Radon-measurable
in $Y$.   \Hint{433D.}
%434F

\spheader 434Xk Show that if $X$ is a perfectly normal space then it is
Borel-measure-compact iff it is Borel-measure-complete.
%434I

\spheader 434Xl Let $X$ be a Radon Hausdorff space.   (i) Show that $X\times Y$ is
Borel-measure-compact whenever $Y$ is Borel-measure-compact.   (ii) Show that
$X\times Y$ is Borel-measure-complete whenever $Y$ is Borel-measure-complete.
%434I

\spheader 434Xm Show that if we give $\omega_1+1$ its order topology, it
is Borel-measure-compact but not Borel-measure-complete or pre-Radon,
and its open subset $\omega_1$ is not Borel-measure-compact.
%434J

\spheader 434Xn Show that the Sorgenfrey line (415Xc, 439Q)
is Borel-measure-complete and Borel-measure-compact, but not
Radon or pre-Radon.
%434J

\spheader 434Xo Let $X$ be a topological space.   (i) Show that the
family of Borel-measure-complete subsets of $X$ is closed under
Souslin's operation.
(ii) Show that the union of a sequence of Borel-measure-compact subsets
of $X$ is Borel-measure-compact.
(iii) Show that if $X$ is Hausdorff then the family of pre-Radon subsets
of $X$ is closed under Souslin's operation.   \Hint{in (i) and (iii),
start by showing that the family under consideration is closed under
countable unions.}
%434J

\spheader 434Xp Show that $\ooint{0,1}^{\omega_1}$ is not pre-Radon.
%434J

\spheader 434Xq Let $X$ be a separable metrizable space.   Show that the following
are equiveridical:  (i) $X$ is a Radon space;  (ii) $X$ is a pre-Radon space;
(iii) there is a metric on $X$, defining the topology of $X$, such that $X$ is
universally Radon-measurable in its completion;  (iv) whenever $Y$ is a separable
metrizable space and $X'$ is a subset of $Y$ such that there is a Borel isomorphism
between $X$ and $X'$, then $X'$ is universally measurable in $Y$;  (v) $X$ is a Radon
space under any separable metrizable topology giving rise to the same Borel sets as
the original topology.
%434J

\sqheader 434Xr Show that a K-analytic Hausdorff space is Radon iff all
its compact subsets are Radon.   \Hint{432B, 434Xi.}
%434K

\spheader 434Xs Suppose that $X$ is a K-analytic Hausdorff space such
that every Radon measure on $X$ is completion regular.   Show that $X$
is a Radon space.
%434K 434Xr

\spheader 434Xt Let $X$ and $Y$ be topological spaces, and suppose that
$Y$ has a countable network.   (i) Show that if $X$ is
Borel-measure-complete, then $X\times Y$ is Borel-measure-complete.
(ii) Show that if $X$ and $Y$ are Radon Hausdorff spaces, then
$X\times Y$ is Radon.
%434K

\spheader 434Xu Let $\sequencen{X_n}$ be a sequence of topological
spaces;  write $X=\prod_{n\in\Bbb N}X_n$ and $Z_n=\prod_{i\le n}X_i$ for
each $n$.   (i) Show that if every $Z_n$ is Borel-measure-complete, so
is $X$.  (ii) Show that if every $Z_n$ is Hausdorff and pre-Radon, so is
$X$.   (iii) Show that if every $Z_n$ is Hausdorff and Radon, so is $X$.
%434K

\spheader 434Xv(i) Let $\sequencen{X_n}$ be a sequence of Radon
Hausdorff
spaces such that $\prod_{i\le n}K_i$ is Radon whenever $n\in\Bbb N$ and
$K_i\subseteq X_i$ is compact for every $i\le n$.   Show that
$X=\prod_{n\in\Bbb N}X_n$ is Radon.
(ii) Let $\sequencen{X_n}$ be a sequence of Radon Hausdorff
spaces with countable networks.   Show that $\prod_{n\in\Bbb N}X_n$ is
Radon.
%434K

\spheader 434Xw Show that if, in 434R, $\nu$ is $\sigma$-finite, then
$\int g\,d\lambda_B=\iint g(x,y)\nu(dy)\mu(dx)$ for every
$\lambda_B$-integrable function $g:X\times Y\to\Bbb R$.
%434R

\spheader 434Xx Show that the product measure construction of 434R is
`associative' and `distributive' in the sense that (under appropriate
hypotheses) the product measures on $(X\times Y)\times Z$ and
$X\times(Y\times Z)$ agree, and those on
$\bigcup_{i,j\in\Bbb N}(X_i\times Y_j)$ and
$(\bigcup_{i\in\Bbb N}X_i)\times(\bigcup_{j\in\Bbb N}Y_j)$ agree.
%434R

\sqheader 434Xy Show that the product measure construction of 434R is
not `commutative';  indeed, taking $\mu=\nu$ to be Dieudonn\'e's measure
on $\omega_1$, show that the Borel measures $\lambda_1$, $\lambda_2$ on
$\omega_1^2$ defined by setting

\Centerline{$\lambda_1 W=\biggerint\nu W[\{\xi\}]\mu(d\xi)$,
\quad$\lambda_2 W=\biggerint\mu W^{-1}[\{\eta\}]\nu(d\eta)$}

\noindent are different.
%434R

\spheader 434Xz Read through \S271, looking for ways to apply the
concept `$\Pr(\pmb{X}\in E)$' for random variables $\pmb{X}$ and
universally measurable sets $E$.
%434T

\leader{434Y}{Further exercises (a)}
%\spheader 434Ya
Set $X=\Bbb N\setminus\{0,1\}$.   For $m$, $p\in X$ set
$U_{mp}=m+p\Bbb N$;  show that $\{U_{mp}:m$, $p\in X$ are coprime$\}$ is
a base for a connected Hausdorff topology on $X$.
\Hint{$pq\in\overline{U}_{mp}$ for every $q\ge 1$.   See
{\smc Steen \& Seebach 78}, ex.\ 60.}   Show that $X$ is a second-countable analytic
Hausdorff space and carries a Radon measure which is not completion
regular.
%434B

\spheader 434Yb Let $(X,\Sigma,\mu)$ be a semi-finite measure space and
$\frak T$ a topology on $X$ such that $\mu$ is inner regular with
respect to the closed sets.   Suppose that $Y$ is a topological space
with a countable network consisting of universally measurable sets, and
that $f:X\to Y$ is measurable.   Show that $f$ is almost continuous.
%4{}18J 4{}18Yg 434D

\spheader 434Yc
If $X$ is a topological space, a set $A\subseteq X$ is
{\bf universally capacitable} if
$c(A)=\sup\{c(K):K\subseteq A$ is compact$\}$ for every Choquet capacity
$c$ on $X$.   (i) Show that if $X$ is a Hausdorff space and $\pi_1$,
$\pi_2:X\times X\to X$ are the coordinate maps, then we have a
Choquet capacity $c$
on $X\times X$ defined by saying that $c(A)=0$ if $A\subseteq X\times X$
and there is a Borel set $E\subseteq X$ including $\pi_1[A]$ and disjoint
from $\pi_2[A]$, and $c(A)=1$ for other $A\subseteq X\times X$.
(ii) Show that there is a universally measurable subset of $\Bbb R$ which
is not universally capacitable.   \Hint{423M.}
%434D out of order query
%take  A\subseteq\BbbR  analytic non-Borel, set
%X=A\times(\BbbR\setminus A);   then  X  is universally measurable not
%universally capacitable.

\spheader 434Yd Let $X$ be a Hausdorff space.   Let $\Sigma$ be the
family of those subsets $E$ of $X$ such that $f^{-1}[E]$ has the Baire
property in $Z$ whenever $Z$ is a compact Hausdorff space and $f:Z\to X$
is continuous.   Show that $\Sigma$ is a $\sigma$-algebra of subsets of
$X$ closed under Souslin's operation.   Show that every member of
$\Sigma$ is universally Radon-measurable.
%434E

\spheader 434Ye
Let $X$ be a Hausdorff space such that there is a countable algebra
$\Cal A$ of universally Radon-measurable subsets of $X$
which separates the
points of $X$ in the sense that whenever $I\in[X]^2$ there is an
$A\in\Cal A$ such that $\#(I\cap A)=1$.   Show that two
Radon probability measures on $X$ which agree on $\Cal A$ are identical.
%416Yj 434E out of order query

\spheader 434Yf Let $X$ be a completely regular Hausdorff space.   Show
that the following are equiveridical:  ($\alpha$) $X$ is pre-Radon;
($\beta$) $X$ is a universally Radon-measurable subset of its
Stone-\v{C}ech compactification;  ($\gamma$) whenever $Y$ is a Hausdorff
space and $X'$ is a subspace of $Y$ which is homeomorphic to $X$, then
$X'$ is universally Radon-measurable in $Y$.
%434J

\spheader 434Yg Set $X=\omega_1+1$, with its order topology, and let
$\Sigma$ be the $\sigma$-algebra of subsets of $X$ generated by the
countable sets and the set $\Omega$ of limit ordinals in $X$.   Show
that there is a unique probability measure $\mu$ on $X$ with domain
$\Sigma$ such that $\mu\xi=\mu\Omega=0$ for every $\xi<\omega_1$.   Show
that $\mu$ is inner regular with respect to the Borel sets, is defined
on a base for the topology of the compact Hausdorff space $X$, but has
no extension to a topological measure on $X$.
%434Xm 434J

\spheader 434Yh Let $X$ be a metrizable space without isolated points,
and $\mu$ a $\sigma$-finite Borel measure on $X$.   Show
that there is a conegligible meager set.   \Hint{there is a dense set
$D\subseteq X$ such that $\{\{d\}:d\in D\}$ is
$\sigma$-metrically-discrete.}
%434L

\spheader 434Yi Give an example of a Hausdorff uniform space $(X,\Cal W)$
with a quasi-Radon probability measure which is not inner regular with
respect to the totally bounded sets.
%434L \BbbR^{\omega_1}

\spheader 434Yj Show that $\beta\Bbb N$ is not countably tight,
therefore not Borel-measure-complete.
%434N

\spheader 434Yk(i) Show that the split interval is not quasi-dyadic.
(ii) Show that the Sorgenfrey line is
not quasi-dyadic.   (iii) Show that $\omega_1$ and $\omega_1+1$, with
their order topologies, are not quasi-dyadic.
%434P, 4{}23X?  mt43bits

\spheader 434Yl Show that a perfectly normal quasi-dyadic space is
Borel-measure-compact.
%434Q

\spheader 434Ym Let $\sequencen{X_n}$ be a sequence of first-countable
spaces, and $\mu_n$ a Borel probability measure on $X_n$ for each $n$.
For $n\in\Bbb N$ set $Z_n=\prod_{i<n}X_i$, and let $\lambda_n$ be the
product Borel measure on $Z_n$ constructed by repeatedly using the
method of 434R (cf.\ 434Xx).
(i) Show that there is a unique Borel measure
$\lambda$ on $Z=\prod_{n\in\Bbb N}X_n$ such that all the canonical maps
from $Z$ to $Z_n$ are \imp.   (ii) Show that, for any $n$, $\lambda$ can be
identified with the product of $\lambda_n$ and a suitable product
measure on $\prod_{i\ge n}X_i$.
%434R mt43bits

\spheader 434Yn ({\smc Aldaz 97}) A topological space $X$ is {\bf
countably metacompact} if whenever $\Cal G$ is a countable open cover of
$X$ then there is a point-finite open cover $\Cal H$ of $X$ refining
$G$.
(i) Show that $X$ is countably metacompact iff whenever
$\sequencen{F_n}$ is a
non-increasing sequence of closed sets with empty intersection in $X$
then there is a sequence $\sequencen{G_n}$ of open sets, with empty
intersection, such that $F_n\subseteq G_n$ for every $n$.
(ii) Let $X$
be any topological space and $\nu:\Cal PX\to[0,1]$ a finitely additive
functional such that $\nu X=1$.   Show that there is a finitely additive
$\nuprime:\Cal PX\to[0,1]$ such that
$\nu F\le\nuprime F=\inf\{\nuprime G:G\supseteq F$ is open$\}$ for every
closed $F\subseteq X$.   \Hint{413S.}
(iii) Show that if $X$ is countably metacompact and $\mu$ is any Borel
probability measure on $X$, there is a Borel probability measure $\mu'$
on $X$, inner regular with respect to the closed sets, such that
$\mu F\le\mu' F$ for every closed set $F\subseteq X$;  so that $\mu$ and
$\mu'$ agree on the Baire $\sigma$-algebra of $X$.
%+

\spheader 434Yo Let $X$ be a totally ordered set with its order
topology.   Show that any $\tau$-additive Borel probability measure on
$X$ has countable Maharam type.
\Hint{$\{\ocint{-\infty,x}^{\ssbullet}:x\in X\}$ generates the measure
algebra.}
%+

\spheader 434Yp If $X$ is a topological space and $\rho$ is a metric on
$X$, $X$ is {\bf$\sigma$-fragmented} by $\rho$ if for every $\epsilon>0$
there is a countable cover $\Cal A$ of $X$ such that whenever
$\emptyset\ne B\subseteq A\in\Cal A$ there is a non-empty relatively open
subset of $B$ of $\rho$-diameter at most $\epsilon$.   Now suppose that $X$
is a Hausdorff space which is $\sigma$-fragmented by a metric $\rho$ such
that (i) $X$ is complete under $\rho$ (ii) the topology generated by $\rho$
is finer than the given topology on $X$.   Show that $X$ is a pre-Radon
space.
%434+ mt43bits

\spheader 434Yq ({\smc Oxtoby 70}) Let $\mu$ be an atomless strictly
positive
Radon probability measure on $\BbbN^{\Bbb N}$.   (i) Show that if
$\sequencen{\alpha_n}$ is any sequence in $[0,1]$ such that
$\sum_{n=0}^{\infty}\alpha_n=1$, then there is a partition $\sequencen{U_n}$ of
$\BbbN^{\Bbb N}$ into open sets such that
$\mu U_n=\alpha_n$ for every $n$.   (ii) Show that if $\nu$ is any other
atomless strictly positive Radon probability measure on
$\BbbN^{\Bbb N}$, there is a homeomorphism
$f:\BbbN^{\Bbb N}\to\Bbb N^{\Bbb N}$ such that $\nu=\mu f^{-1}$.
%434+

\spheader 434Yr Let $X$ be a Hausdorff space and $\mu$ an atomless strictly
localizable tight Borel measure on $X$.
Show that $\mu$ is $\sigma$-finite.   \Hint{{\smc Fremlin n05}.}
%434+ n05426 Prop 9 uses Borel classes

}%end of exercises

\leader{434Z}{Problems (a)} Must every Radon compact Hausdorff space be
sequentially compact?

\spheader 434Zb Must a Hausdorff continuous image of a Radon compact
Hausdorff space be Radon?

\endnotes{\Notesheader{434} I said that the fundamental question of
topological measure theory is `which measures can appear on which
topological spaces'?   In this section I have concentrated on Borel
measures, classified according to the scheme laid out in \S411.   (Of
course there are other kinds of classification.   One of the most
interesting is the Maharam classification of Chapter 33:  we can ask
what measure algebras can appear from topological measures on a given
topological space.   I will return to this idea in \S531 of Volume 5.
For the moment I pass it by, with only 434Yo to give a taste.)
We can ask
this question from either of two directions.   The obvious approach is
to ask, for a given class of topological spaces, which types of measure
can appear.   But having discovered that (for instance) there are
several types of topological space on which all (totally finite) Borel
measures are tight, we can use
this as a definition of a class of topological spaces, and ask the
ordinary questions about this class.   Thus we have `Radon',
`Borel-measure-complete', `Borel-measure-compact' and `pre-Radon' spaces
(434C, 434G).   I have given precedence to the first partly
to honour the influence of {\smc Schwartz 73} and partly because a
compact Hausdorff space is always Borel-measure-compact and pre-Radon
(434Hb, 434Jf) and is Borel-measure-complete iff it is Radon (434Ka).
In effect, `Borel-measure-complete' means `Borel measures are
quasi-Radon'
(434Ib), `pre-Radon' means `quasi-Radon measures are Radon' (434Jb), and
`Radon' means `Borel measures are Radon' (434F(a-iii)).   These slogans
have to be interpreted with care;  but it is true that a Hausdorff space
is Radon iff it is both Borel-measure-complete and pre-Radon (434Ka).

The concept of `Radon' space is in fact one of the important
contributions of measure
theory to general topology, offering a variety of challenging
questions.   One which has attracted some attention is the problem of
determining when products of Radon spaces are Radon.   Uncountable
products hardly ever are (434Kd);  for countable products it is enough
to understand products of finitely many compact spaces (434Xv);  but the
product of two compact spaces already seems to lead us into undecidable
questions (438Xq, {\smc Wage 80}).   Two more very natural questions are
in 434Z.   One of the obstacles to the investigation is the rather small
number of Radon compact Hausdorff spaces which are known.   I should
remark that if the continuum hypothesis (for instance) is true, then
every compact Hausdorff space in which countably compact sets are closed
is sequentially compact ({\smc Ismail \& Nyikos 80}, or
{\smc Fremlin 84}, 24Nc), so that in this case we have a quick answer to
434Za, using 434Nb.

You will recognise the construction of 434M as a universal version of
Dieudonn\'e's measure (411Q).   `Tightness' is of great
interest for other reasons ({\smc Engelking 89}), and here is
very helpful when showing that spaces are not Radon (434Yj).

A large proportion of the definitions in general topology can be
regarded as different abstractions from the concept of metrizability.
Countable tightness is an obvious example;  so is `first-countability'
(434R).   In quite a different direction we have `metacompactness'
(438J, 434Yn).   The construction of the product measure in 434R is an
obvious idea, as soon as you have seen Fubini's theorem, but it is not
obvious just when it will work.

`Quasi-dyadic' spaces are a relatively recent invention;  I introduce
them here only as a vehicle for the argument of 434Q.   Of course a
dyadic space is
quasi-dyadic;   for basic facts on dyadic spaces, see 4A2D, 4A5T and
{\smc Engelking 89}, \S3.12 and 4.5.9-4.5.11.
}%end of notes

\discrpage

\frfilename{mt435.tex}
\versiondate{16.8.08}
\copyrightdate{1999}

\def\chaptername{Topologies and measures II}
\def\sectionname{Baire measures}

\newsection{435}

Imitating the programme of \S434, I apply a similar analysis to Baire
measures, starting with a simple-minded classification\cmmnt{ (435A)}.   This time the central section (435D-435H) is devoted to
`measure-compact'
spaces, those on which all (totally finite) Baire measures are
$\tau$-additive.

\cmmnt{
\leader{435A}{Types of Baire measures} In 434A I looked at a
list of four
properties which a Borel measure may or may not possess:  inner
regularity with respect to closed sets, inner regularity with respect to
zero sets, tightness\cmmnt{ (that is, inner regularity with respect to closed compact sets)}, and
$\tau$-additivity.   Since every (semi-finite) Baire measure is inner
regular with respect to the zero sets (412D), only two of the four are
important considerations for Baire measures:  tightness and
$\tau$-additivity.   On the other hand,
there is a new question we can ask.   Given a Baire measure on a
topological space, when can it be extended to a Borel measure?   And in
the case of a positive answer, we can ask whether the extension is
unique, and whether we can find extensions to Borel measures satisfying
the properties considered in 434A.

We already have some information on this.   If $X$ is a completely
regular space, and $\mu$ is a $\tau$-additive effectively locally finite
Baire measure on $X$, then $\mu$ has a (unique) extension to a
$\tau$-additive Borel measure (415N).
While if $\mu$ is tight, the extension will also be tight (cf.\ 416C).   
Perhaps I should
remark immediately that while there can be only one $\tau$-additive
Borel measure extending $\mu$, there might be another Borel measure, not
$\tau$-additive, also extending $\mu$;  see 435Xa.   Of course if there
is any completion regular Borel measure extending $\mu$, there is only
one;  moreover, if $\mu$ is $\sigma$-finite, and there is a completion
regular Borel measure extending $\mu$, this is the only Borel measure
extending $\mu$.   (For every Borel set will be measured by the
completion of $\mu$.)

A possible division of Baire measures is therefore into classes

\inset{(E) measures which are not $\tau$-additive,}

\inset{(F) measures which are $\tau$-additive, but not tight,}

\inset{(G) tight measures,}

\noindent and within these classes we can distinguish measures with no
extension to a Borel measure (type E$_0$), measures with more than one
extension to a Borel measure (types E$_1$, F$_1$ and G$_1$), measures
with exactly one extension to a Borel measure which is not completion
regular (types E$_2$, F$_2$ and G$_2$) and measures with an extension to
a completion regular Borel measure (types E$_3$, F$_3$ and G$_3$).   For
examples, see 439M and 439O (E$_0$), 439N (E$_2$), 439J
(E$_3$), 435Xc (F$_1$), 435Xd (F$_2$), 415Xc and 434Xa (F$_3$), 435Xa
(G$_1$), 435Xb (G$_2$) and the
restriction of Lebesgue measure to the Baire subsets of $\Bbb R$
(G$_3$);   other
examples may be constructed as direct sums of these.

A separate question we can ask of a Baire measure is whether it can be
extended to a Radon measure.   For this there is a straightforward
criterion (435B), which shows that (at least for totally finite measures
on completely regular spaces) only the types F$_1$ and F$_2$ are divided
by this question.   (If a Baire measure $\mu$ can be extended to a Radon
measure, it is surely
$\tau$-additive.   If $\mu$ is tight, it satisfies the criteria of 435B,
so has an extension to a Radon
measure.   If $\mu$ has an extension to a completion regular Borel
measure $\mu_1$ {\it and} has an extension to a Radon measure $\mu_2$,
then the completion $\hat\mu$ of $\mu$ extends $\mu_1$, while $\mu_2$
extends $\hat\mu$;   so $\mu_1$ is the restriction of $\mu_2$ to the
Borel sets and $\mu_2=\hat\mu_1=\hat\mu$ and $\mu$, like $\mu_2$, is
tight, by 412Hb or otherwise.
Thus no measure of type F$_3$ can be extended to a Radon measure.)

As with the classification of Borel measures that I offered in \S434,
any restriction on the topology of the underlying space may eliminate
some of these possibilities.   For instance, because a semi-finite Baire
measure is inner regular with respect to the closed sets, we can have no
(semi-finite) measure of classes E or F on a compact Hausdorff space.
On a locally compact Hausdorff space we can have no effectively locally
finite Baire measure of class F (435Xe), while on a
K-analytic Hausdorff space we can have no locally finite Baire
measure of class E (432F).   In a metrizable space, or a regular
space with a countable network (e.g., a regular analytic Hausdorff space),
the Baire and Borel $\sigma$-algebras coincide
(4A3Kb), so we can have no measures of type E$_0$, E$_1$,
F$_1$ or G$_1$.
}%end of comment

\leader{435B}{Theorem} Let $X$ be a Hausdorff space and $\mu$ a
locally finite Baire measure on $X$.   Then the following are
equiveridical:

\inset{(i) $\mu$ has an extension to a Radon measure on $X$;

(ii) for every non-negligible Baire set $E\subseteq X$ there is a
compact set $K\subseteq E$ such that $\mu^*K>0$.}

\noindent If $\mu$ is totally finite, we can add

\inset{(iii) $\sup\{\mu^*K:K\subseteq X$ is compact$\}=\mu X$.}

\proof{ Because $\mu$ is inner regular with respect to the closed sets
(412D), this is just a special case of 416P.
}%end of proof of 435B

\leader{435C}{Theorem}\cmmnt{ ({\smc \Marik\ 57})} Let $X$ be a
normal countably paracompact space.   Then any semi-finite Baire measure
on $X$ has an extension to a semi-finite Borel measure which is inner
regular with respect to the closed sets.

\proof{{\bf (a)} Let $\nu$ be a semi-finite Baire measure on $X$.   Let
$\Cal K$ be the family of those closed subsets of
$X$ which are included in zero sets of finite measure, and set
$\phi_0K=\nu^*K$ for $K\in\Cal K$.   Then $\Cal K$ and $\phi_0$ satisfy
the conditions of 413J, that is,

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

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

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

\inset{($\alpha$)
$\phi_0K=\phi_0L+\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$
whenever $K$, $L\in\Cal K$ and $L\subseteq K$,}

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

\Prf\ The first three are trivial.

\medskip

\quad\grheada\ Take $K$, $L\in\Cal K$ with $L\subseteq K$, and set
$\gamma=\sup\{\phi_0 K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$.   (i)
If $K'\subseteq K\setminus L$ is closed, then (because $X$ is normal)
there is a zero set $F$ including $K'$ and disjoint from $L$
(4A2F(d-iv)), so

\Centerline{$\phi_0K'+\phi_0L
=\nu^*((K'\cup L)\cap F)+\nu^*((K'\cup L)\setminus F)
=\nu^*(K'\cup L)
\le\nu^*K$.}

\noindent As $K'$ is arbitrary, $\gamma+\phi_0L\le\phi_0K$.
(ii) Let $\epsilon>0$.   Let $F_0$ be a zero set of finite measure
including $K$.   Because $\nu$ is inner regular with respect to the zero
sets (412D), there is a zero set $F\subseteq F_0\setminus L$ such that
$\nu F\ge\nu_*(F_0\setminus L)-\epsilon$ (413Ee), so that
$\nu(F_0\setminus F)\le\nu^*L+\epsilon$ (413Ec).   Set $K'=K\cap F$.
Then

\Centerline{$\nu^*K
=\nu^*(K\setminus F)+\nu^*(K\cap F)
\le\nu(F_0\setminus F)+\nu^*K'
\le \nu^*L+\epsilon+\gamma$.}

\noindent As $\epsilon$ is arbitrary, $\nu^*K\le\nu^*L+\gamma$.

\medskip

\quad\grheadb\ If $\sequencen{K_n}$ is a non-increasing sequence in
$\Cal K$ with empty intersection, then (because $X$ is countably
paracompact)
there is a sequence $\sequencen{G_n}$ of open sets such that
$K_n\subseteq G_n$ for every $n$ and $\bigcap_{n\in\Bbb N}G_n=\emptyset$
(4A2Ff).   Because $X$ is normal, there are zero sets $F_n$ such
that $K_n\subseteq F_n\subseteq G_n$ for each $n$ (4A2F(d-iv) again),
so that
$\bigcap_{n\in\Bbb N}F_n=\emptyset$.   We may suppose that $F_0$ has
finite measure.   In this case,

\Centerline{$\lim_{n\to\infty}\nu^*K_n
\le\lim_{n\to\infty}\nu(\bigcap_{i\le n}F_i)
=0$.}

\noindent Thus $\Cal K$ and $\phi_0$ satisfy the conditions ($\alpha$)
and ($\beta$) as well.\ \Qed

\medskip

{\bf (b)} By 413J, there is a complete locally determined measure $\mu$
on $X$, extending $\phi_0$ and inner regular with respect to $\Cal K$.
If $F\subseteq X$ is closed, then $F\cap K\in\Cal K$ for every
$K\in\Cal K$, so $F\in\dom\mu$ (413F(ii));  accordingly $\mu$ is a
topological measure, and because $\nu$ also is inner regular with
respect to $\Cal K$, $\mu$ must extend $\nu$.   So the restriction of
$\mu$ to the Borel
sets is a Borel extension of $\nu$ which is inner regular with respect
to the closed sets.
}%end of proof of 435C

\cmmnt{\medskip

\noindent{\bf Remark} If $X$ is normal, but not countably paracompact,
the result may fail;  see 439O.   I have stated the result in terms of
`countable paracompactness', but the formally distinct `countable
metacompactness' is also sufficient (435Ya).   If we are told that the
Baire measure is $\tau$-additive and effectively locally finite, we have
a much stronger result (415M).
}%end of comment

\leader{435D}{}\cmmnt{ Just as with the `Radon' spaces of \S434, we
can look at classes of topological spaces defined by the behaviour of
the Baire measures they carry.   The class which has aroused most
interest is the following.

\medskip

\noindent}{\bf Definition} A completely regular topological space $X$ is
{\bf measure-compact}\cmmnt{ (sometimes called {\bf almost Lindel\"of})} if
every totally finite Baire measure on $X$ is
$\tau$-additive\cmmnt{, that is, has an extension to a quasi-Radon
measure on $X$ (415N)}.

\leader{435E}{}\cmmnt{ The following lemma will make our path easier.

\medskip

\noindent}{\bf Lemma} Let $X$ be a completely regular topological space
and $\nu$ a totally finite Baire measure on $X$.   Suppose that
$\sup_{G\in\Cal G}\nu G=\nu X$ whenever $\Cal G$ is an upwards-directed
family of cozero sets with union $X$.   Then $\nu$ is $\tau$-additive.

\proof{ Let $\Cal G$ be an upwards-directed family of open Baire sets
such that $G^*=\bigcup\Cal G$ also is a Baire set, and $\epsilon>0$.
Because $\nu$ is inner regular with respect to the zero sets, there is a
zero set $F\subseteq G^*$ such that $\nu F\ge\nu G^*-\epsilon$.   Let
$\Cal G'$ be the family of cozero sets included in members of $\Cal G$;
because $X$ is completely regular, so that the cozero sets are a base
for its topology, $\bigcup\Cal G'=G^*$, and of course $\Cal G'$ is
upwards-directed.   Now

\Centerline{$\Cal H=\{G\cup(X\setminus F):G\in\Cal G'\}$}

\noindent is an upwards-directed family of cozero sets with union $X$,
so there is a $G_0\in\Cal G'$ such that
$\nu(G_0\cup(X\setminus F))\ge\nu X-\epsilon$.   In this case

\Centerline{$\sup_{G\in\Cal G}\nu G
\ge\nu G_0\ge\nu X-\epsilon-\nu(X\setminus F)=\nu F-\epsilon
\ge\nu G^*-2\epsilon$.}

\noindent As $\Cal G$ and $\epsilon$ are arbitrary, $\nu$ is
$\tau$-additive.
}%end of proof of 435E

\leader{435F}{Elementary facts (a)} If $X$ is a completely regular space
which is not measure-compact, there are a Baire probability measure
$\mu$ on $X$ and a
cover of $X$ by $\mu$-negligible cozero sets.   \prooflet{\Prf\ There is
a totally finite Baire measure $\nu$ on $X$ which is not
$\tau$-additive. By 435E, there is an upwards-directed family $\Cal G$
of cozero sets, covering $X$, such that $\sup_{G\in\Cal G}\nu G<\nu X$.
Let $\sequencen{G_n}$ be a sequence in $\Cal G$ such that
$\sup_{n\in\Bbb N}\nu G_n=\sup_{G\in\Cal G}\nu G$.   Then
$\gamma=\nu(X\setminus\bigcup_{n\in\Bbb N}G_n)>0$.   Set

\Centerline{$\mu H=\Bover1{\gamma}\nu(H\setminus\bigcup_{n\in\Bbb
N}G_n)$}

\noindent for Baire sets $H\subseteq X$;  then $\mu$ is a Baire
probability measure and $\Cal G$ is a cover of $X$ by $\mu$-negligible
cozero sets.\ \Qed
}%end of prooflet

\spheader 435Fb Regular Lindel\"of spaces are measure-compact.
\prooflet{(For if a Lindel\"of space can be covered by negligible open
sets, it can be covered by countably many negligible open sets, so is
itself negligible.)}   \cmmnt{In particular, compact Hausdorff spaces,
indeed all regular K-analytic
Hausdorff spaces (422Gg), are measure-compact.}

\cmmnt{Note that regular Lindel\"of spaces are normal and paracompact
(4A2H(b-i)), so their measure-{\vthsp}compactness is also a consequence
of 435C and 434Hb.}

\spheader 435Fc An open subset of a measure-compact space need not be
measure-compact\cmmnt{ (435Xi(i))}.   A continuous image of a
measure-compact space need not be measure-compact\cmmnt{ (435Xi(ii))}.
$\BbbN^{\frakc}$ is not measure-compact\cmmnt{ (439P)}.   The
product of two measure-compact spaces need not
be measure-compact\cmmnt{ (439Q)}.

\spheader 435Fd If $X$ is a measure-compact completely regular space it
is Borel-measure-compact.   \prooflet{\Prf\ Let $\mu$ be a non-zero
totally finite Borel measure on $X$ and $\Cal G$ an open cover of $X$.
Let $\nu$ be the restriction of $\mu$ to the Baire $\sigma$-algebra of
$X$, so that $\nu$ is $\tau$-additive.   Let $\Cal U$ be the set of
cozero sets $U\subseteq X$ included in members of $\Cal G$;  because the
family of cozero sets is a base for the topology of $X$,
$\bigcup\Cal U=X$, and there is some $U\in\Cal U$ such that $\nu U>0$.
This means that there is some $G\in\Cal G$ such that $\mu G>0$.   By
434H(a-v), $X$ is Borel-measure-compact.\ \Qed}

\leader{435G}{Proposition} A Souslin-F subset of a measure-compact
completely regular space is measure-compact.

\proof{{\bf (a)} Let $X$ be a measure-compact completely regular space,
$\family{\sigma}{S}{F_{\sigma}}$ a Souslin scheme consisting of closed
subsets of $X$ with kernel $A$, $\nu$ a totally finite Baire measure on
$A$, and $\Cal G$ an upwards-directed family of (relatively) cozero
subsets of $A$ covering $A$.   Let $\nu_1$ be the Baire measure on $X$
defined by setting $\nu_1H=\nu(A\cap H)$ for every Baire subset $H$ of
$X$.   Because $X$ is measure-compact, $\nu_1$ has an extension to a
quasi-Radon measure $\mu$ on $X$.   Let $\mu_A$ be the subspace measure
on $A$.

\medskip

{\bf (b)} By 431B, $A$ is measured by $\mu$.   In fact
$\mu A=\nu A$.  \Prf\ The construction of $\mu$ given in
415K-415N %415K 415L 415M 415N
ensures that
$\mu F=\nu_1^*F$ for every closed set $F$, and this is in any case a
consequence of the facts that $\mu$ is $\tau$-additive and $\dom\nu_1$
includes a base for the topology.   For each $\sigma\in S$, in
particular, $\mu F_{\sigma}=\nu_1^*F_{\sigma}$;  let
$F'_{\sigma}\supseteq F_{\sigma}$ be a Baire set such that
$\nu_1F'_{\sigma}=\nu_1^*F_{\sigma}$.   Then

\Centerline{$\mu F'_{\sigma}=\nu_1F'_{\sigma}=\nu_1^*F_{\sigma}
=\mu F_{\sigma}$}

\noindent and $\mu(F'_{\sigma}\setminus F_{\sigma})=0$ for every
$\sigma\in S$.   Let $A'$ be the kernel of the Souslin scheme
$\family{\sigma}{S}{F'_{\sigma}}$.   Then $A\subseteq A'$ and

\Centerline{$\mu(A'\setminus A)
\le\sum_{\sigma\in S}\mu(F'_{\sigma}\setminus F_{\sigma})=0$,}

\noindent so $\mu A=\mu A'$.   On the other hand, writing $\hat\nu_1$
for the completion of $\nu_1$, $A'$ is measured by $\hat\nu_1$, by
431A, so that (because $\mu$ extends $\nu_1$)

\Centerline{$\mu A=\mu A'=\mu^*A'\le\nu_1^*A'
=(\nu_1)_*A'\le\mu_*A'=\mu A'$.}

\noindent Thus $\mu A=\nu_1^*A'$.   But of course

\Centerline{$\nu A=\nu_1X=\nu_1^*A=\nu_1^*A'$,}

\noindent so that $\mu A=\nu A$.\ \Qed

Since we surely have

\Centerline{$\mu X=\nu_1X=\nu A$,}

\noindent we see that $\mu(X\setminus A)=0$.

\medskip

{\bf (c)} It follows that $\mu F=\nu F$ for every (relatively) zero set
$F\subseteq A$.   \Prf\ There is a closed set $F'\subseteq X$ such that
$F=A\cap F'$.   Now if $H\subseteq X$ is a Baire set including $F'$,
$H\cap A$ is a (relatively) Baire set including $F$, so $\nu
F\le\nu(H\cap A)=\nu_1H$;  as $H$ is arbitrary, $\nu F\le\nu_1^*F'$.
But $\nu_1^*F'=\mu F'$, as remarked in (b) above,
and $\mu(X\setminus A)=0$, so

\Centerline{$\mu F=\mu F'=\nu_1^*F'\ge\nu F$.}

\noindent On the other hand, $A\setminus F$ is (relatively) cozero, so
there is a non-decreasing sequence $\sequencen{F_n}$ of (relatively)
zero
subsets of $A$ with union $A\setminus F$, and

\Centerline{$\mu(A\setminus F)=\lim_{n\to\infty}\mu
F_n\ge\lim_{n\to\infty}\nu F_n=\nu(A\setminus F)$.}

\noindent Since we already know that $\mu A=\nu A$, it follows that

\Centerline{$\mu F=\mu A-\mu(A\setminus F)\le\nu A-\nu(A\setminus F)
=\nu F$,}

\noindent and $\mu F=\nu F$.\ \Qed

\medskip

{\bf (d)} The set

\Centerline{$\{E:E\subseteq A$ is a (relative) Baire set,
$\mu E=\nu E\}$}

\noindent therefore contains every (relatively) zero set, and by the
Monotone Class Theorem (136C) contains every (relatively) Baire set.
What this means is that $\mu$ actually extends $\nu$;  so the subspace
measure $\mu_A=\mu\restrp\Cal PA$ also extends $\nu$.   But $\mu_A$ is a
quasi-Radon measure (415B), therefore $\tau$-additive, and $\nu$ must
also be $\tau$-additive.
}%end of proof of 435G

\leader{435H}{Corollary} A Baire subset of a measure-compact completely
regular space is measure-compact.

\proof{ Put 435G and 421L together.
}%end of proof of 435H

\exercises{
\leader{435X}{Basic exercises $\pmb{>}$(a)}
%\spheader 435Xa
Give $\omega_1+1$ its order topology.   (i) Show that its  Baire
$\sigma$-algebra $\Sigma$ is just the family of sets
$E\subseteq\omega_1+1$ such that either $E$ or its complement is a
countable subset of $\omega_1$.   (ii) Show that there is a unique Baire
probability measure $\nu$ on $\omega_1+1$ such that $\nu\{\xi\}=0$
for every $\xi<\omega_1$.   (iii) Show that $\nu$ is $\tau$-additive.
(iv) Show that there is exactly one Radon measure on $\omega_1+1$
extending $\nu$, but that the measure $\mu$ of 434Xg is another Borel
measure also extending $\nu$.
%435A

\sqheader 435Xb Let $I$ be a set with cardinal $\omega_1$, endowed with
its discrete topology, and $X=I\cup\{\infty\}$ its one-point
compactification (3A3O).   Let $\mu$ be the Dirac measure on $X$
concentrated at $\infty$.   (i) Show that every subset of $X$ is
a Borel set.   (ii) Show that $\{\infty\}$ is not a zero set.   (iii)
Let $\nu$ be the restriction of $\mu$ to the Baire $\sigma$-algebra of
$X$.   Show that $\nu$ is tight.
Show that $\mu$ is the unique Borel measure extending $\nu$ ({\it
hint\/}:  you will need 419G), but is not completion regular.   (iv)
Show that the subspace measure $\nu_I$ on $I$ is the
countable-cocountable measure on $I$, and is not a Baire measure, nor
has any extension to a Baire measure on $I$.   (v) Show that $X$ is
measure-compact.
%435A

\spheader 435Xc On $\BbbR^{\omega_1}$ let $\mu$ be the Baire measure
defined by saying that $\mu E=1$ if $\chi\omega_1\in E$, $0$ otherwise.
(i) Show that $\mu$ is $\tau$-additive, but not tight.   \Hint{4A3P.}
(ii) Show that the
map $\xi\mapsto\chi\xi:\omega_1+1\to\BbbR^{\omega_1}$ is continuous, so
that $\mu$ has more than one extension to a Borel measure.   (iii) Show
that $\mu$ has an extension to a Radon measure.
%435A

\spheader 435Xd Set $X=\omega_1+1$ with the topology $\Cal
P\omega_1\cup\{X\setminus A:A\subseteq\omega_1$ is countable$\}$.   Let
$\mu$ be the Baire measure on $X$ defined by saying that, for Baire sets
$E\subseteq X$, $\mu E=1$ if $\omega_1\in E$, $0$ otherwise.   (i) Show
that a function $f:X\to\Bbb R$ is continuous iff $\{\xi:\xi\in
X,\,f(\xi)\ne f(\omega_1)\}$ is countable;  show that $X$ is completely
regular and Hausdorff.   (ii) Show that $\mu$ is
$\tau$-additive.   (iii) Show that every subset of $X$ is Borel.   (iv)
Show that the only Borel measure extending $\mu$ is the Dirac measure
concentrated at $\omega_1$, and that this is a Radon measure.   (v)
Show that all compact subsets of $X$ are finite, so that $\mu$ is not
tight.   *(vi) Show that $X$ is Lindel\"of.
%435A

\spheader 435Xe Let $X$ be a locally compact Hausdorff space and $\mu$
an effectively locally finite $\tau$-additive Baire measure on $X$.
Show that $\mu$ is tight.
\Hint{the relatively compact cozero sets cover $X$;  use 414Ea and
412D.}
%435A

\sqheader 435Xf Let $X$ be a completely regular space and $\mu$ a
totally finite $\tau$-additive Borel measure on $X$.   Let $\mu_0$ be
the restriction of $\mu$ to the Baire $\sigma$-algebra of $X$.   Show
that $\mu F=\mu_0^*F$ for every closed set $F\subseteq X$.
%435C

\spheader 435Xg Show that if a semi-finite Baire measure $\nu$ on a
normal countably paracompact space is extended to a Borel measure $\mu$
by the construction in 435C, then the measure algebra of $\nu$ becomes
embedded as an order-dense subalgebra of the measure algebra of $\mu$,
so that $L^1(\mu)$ can be identified with $L^1(\nu)$.
%435C

\spheader 435Xh Show
that a Borel-measure-compact normal countably paracompact
space is measure-compact.
%435F, 435C

\spheader 435Xi(i) Show that $\omega_1+1$ is measure-compact, in its
order topology, but that its open subset $\omega_1$ is not (cf.\ 434Xm).
(ii) Show
that a discrete space with cardinal $\omega_1$ is measure-compact, but
that it has a continuous image which is not measure-compact.
%435G

\spheader 435Xj Let $X$ be a metacompact completely regular space and
$\nu$ a totally finite strictly positive Baire measure on $X$.   Show
that $X$ is Lindel\"of, so that $\nu$ has an extension to a quasi-Radon
measure on $X$.   \Hint{if $\Cal H$ is a point-finite open cover of $X$,
not containing $\emptyset$, then for each $H\in\Cal H$ choose a
non-empty cozero set $G_H\subseteq H$;  show that
$\{H:\nu G_H\ge\delta\}$ is finite for every $\delta>0$.}
%435H

\spheader 435Xk A completely regular space $X$ is {\bf strongly
measure-compact} ({\smc Moran 69}) if
$\mu X=\sup\{\mu^*K:K\subseteq X$ is compact$\}$ for every totally
finite Baire measure $\mu$ on $X$.   (i) Show that a
completely regular Hausdorff space $X$ is strongly measure-compact iff
every totally finite Baire measure on $X$ has an extension to a Radon
measure iff $X$ is measure-compact and pre-Radon.   (ii) Show that a Souslin-F subset of a
strongly measure-compact completely regular space is strongly
measure-compact.   (iii) Show that a discrete space with cardinal
$\omega_1$ is strongly measure-compact.   (iv) Show that a countable
product of strongly measure-compact completely regular spaces is
strongly measure-compact.   (v) Show that $\BbbN^{\omega_1}$ is not
strongly measure-compact.   \Hint{take a non-trivial probability measure
on $\Bbb N$ and consider its power on $\BbbN^{\omega_1}$.}  (vi) Show
that if $X$ and $Y$ are completely regular
spaces, $X$ is measure-compact and $Y$ is strongly measure-compact then
$X\times Y$ is measure-compact.
%435+

\spheader 435Xl (T.D.Austin) Let $X$ be a topological space, $\mu$ an
atomless Baire probability measure on $X$ and $\hat\mu$ its completion.
Show that there is a continuous function $f:X\to[0,1]$ which is \imp\ for
$\hat\mu$ and Lebesgue measure on $[0,1]$.
\Hint{Check the case $X=[0,1]$ first.   For the general case,
let $Z$ be the set of continuous functions from $X$ to
$[0,1]$ with the complete metric induced by $\|\,\|_{\infty}$, and set
$\alpha(f)=\max\{\mu f^{-1}[\{t\}]:t\in[0,1]\}$ for $f\in Z$.
Show that $\interior\{f:\alpha(f)\le\epsilon\}$ is dense in
$Z$ for every $\epsilon>0$, so that there is an $f\in Z$ such that
$\mu f^{-1}$ is atomless.}
%435+

\spheader 435Xm Let $X$ be a normal space and $\mu$ a
complete $\sigma$-finite
topological probability measure on $X$ which is inner regular
with respect to the closed sets.
(i) Let $\nu$ be the restriction of $\mu$ to the Baire $\sigma$-algebra
of $X$.    Show that $\mu$ and $\nu$ have isomorphic measure algebras.
(ii) Show that if $\mu$ is an atomless probability measure there is a
continuous $f:X\to[0,1]$ which is \imp\ for $\mu$ and Lebesgue measure.
%435Xl

\spheader 435Xn Let $X$ be a topological space and $\Cal G$ the family of
cozero sets in $X$.
Show that a functional $\psi:\Cal G\to\coint{0,\infty}$ can
be extended to a Baire measure on $X$ iff $\psi$ is modular (definition:
413Qc) and
$\lim_{n\to\infty}\psi G_n=0$ whenever $\sequencen{G_n}$ is a
non-increasing sequence in $\Cal G$ with empty intersection.   \Hint{if
$\psi$ satisfies the conditions, first check that $\psi\emptyset=0$ and
that $\psi G\le\psi H$ whenever $G\subseteq H$;  now apply 413J with
$\phi K=\inf\{\psi G:K\subseteq G\in\Cal G\}$ for zero sets $K$.}
%435+

\spheader 435Xo Let $X$ be a countably compact topological space and $\mu$
a totally finite Baire measure on $X$.   Show that $\mu$ has an extension
to a Borel measure which is inner regular with respect to the closed sets.
\Hint{413P.}
%435F

\leader{435Y}{Further exercises (a)}
%\spheader 435Ya
Show that a normal countably metacompact space (434Yn)
is countably paracompact.
%435C

\spheader 435Yb
Let $X$ be a completely regular Hausdorff space and $\beta X$ its
Stone-\v Cech
compactification.   Show that $X$ is measure-compact iff whenever $\nu$
is a Radon measure on $\beta X$ such that $\nu X=0$, there is a
$\nu$-negligible Baire subset of $\beta X$ including $X$.
%435D

\leaveitout{Let $X$ be a completely regular Hausdorff space and
$\beta X$ its Stone-\v Cech
compactification.   Show that $X$ is measure-compact iff whenever
$K\subseteq\beta X\setminus X$ is a compact set which is the support of
a Radon measure, then there is a zero set $H$ of $\beta X$ such
that $K\subseteq H\subseteq\beta X\setminus X$.
}%is this true?

}%end of exercises

\endnotes{
\Notesheader{435} The principal reason for studying Baire measures is
actually outside the main line of this chapter.   For a completely
regular Hausdorff space $X$, write $C_b(X)$ for the $M$-space of bounded
continuous real-valued functions on $X$.   Then
$C_b(X)^*=C_b(X)^{\sim}$ is an $L$-space\cmmnt{ (356N)}, and inside
$C_b(X)^*$ we have the bands generated by the tight, smooth and
sequentially smooth
functionals (see 437A and 437F below), all identifiable, if we choose,
with spaces
of `signed Baire measures'.   {\smc Wheeler 83} argues convincingly that
for the questions a functional analyst naturally asks, these Baire
measures are often an effective aid.

From the point of view of the arguments in this section, the most
fundamental difference between `Baire' and `Borel'
measures lies in their action on subspaces.   If $X$ is a topological
space and $A$ is a subset of $X$, then any Borel or Baire measure $\mu$
on $A$ provides us with a measure $\mu_1$ of the same type on $X$,
setting $\mu_1E=\mu(A\cap E)$ for the appropriate sets $E$.   In the
other direction, if $\mu$ is a Borel measure on $X$, then the subspace
measure $\mu_A$ is a Borel measure on $A$, because the Borel
$\sigma$-algebra of
$A$ is just the subspace $\sigma$-algebra derived from the Borel algebra
of $X$ (4A3Ca).   But if $\mu$ is a Baire measure on $X$, it does
not follow that $\mu_A$ is a Baire measure on $A$;  this is because (in
general) not every continuous function $f:A\to[0,1]$ has a continuous
extension to $X$, so that not every zero set in $A$ is the intersection
of $A$ with a zero set in $X$ (see 435Xb).   The analysis of those pairs
$(X,A)$ for which the Baire $\sigma$-algebra of $A$ is just the subspace
algebra derived from the Baire sets in $X$ is a challenging problem in
general topology which I pass by here.   For the moment I note only that
avoiding it is the principal technical problem in the proof of 435G.

I do not know if I ought to apologise for
`countably tight' spaces (434N),
`first-countable' spaces (434R),
`metacompact' spaces (438J),
`normal countably paracompact' spaces (435C),
`quasi-dyadic' spaces (434O) and
`sequential' spaces (436F).
General topology is notorious for invoking
arcane terminology to stretch arguments to their utmost limit of
generality, and even specialists may find their patience tried by
definitions which seem to have only one theorem each.   In 438J, for
instance, it is obvious that the original result concerned metrizable
spaces (438H), and you may well feel at first that the extension is a
baroque over-elaboration.   On the other hand, there
are (if you look for them) some very interesting metacompact spaces
({\smc Engelking 89}, \S5.3), and metacompactness has taken its
place in the standard lists.   In this book I try to follow a rule of
introducing a class of topological spaces only when it is both genuinely
interesting, from the point of view of general topology, and
also a support for an idea which
is interesting from the point of view of measure theory.
}%end of notes

\discrpage

\frfilename{mt436.tex}
\versiondate{9.5.11}
\copyrightdate{2002}

\def\chaptername{Topologies and measures II}
\def\sectionname{Representation of linear functionals}

\newsection{436}

I began this treatise with the three steps which make measure theory, as
we know it, possible:  a construction of Lebesgue measure, a
definition of an integral from a measure, and a proof of the
convergence theorems.   I used what I am sure is the best route:
Lebesgue measure from Lebesgue
outer measure, and integrable functions from simple functions.   But of
course there are many other paths to the same ends, and some of them
show us slightly different aspects of the subject.   In this section I
come -- rather later than many authors would -- to an account of a
procedure for constructing measures from integrals.

I start with three fundamental theorems, the first and third being the
most important.   A positive linear functional on a truncated Riesz
space of functions is an integral iff it is sequentially smooth (436D);
a smooth linear functional corresponds to a quasi-Radon measure (436H);
and if $X$ is a compact Hausdorff space, any positive linear functional
on $C(X)$ corresponds to a Radon measure (436J-436K).

\leader{436A}{Definition} Let $X$ be a set, $U$ a Riesz subspace of
$\Bbb R^X$, and $f:U\to\Bbb R$ a positive linear functional.   I say
that $f$ is {\bf sequentially smooth} if whenever $\sequencen{u_n}$ is a
non-increasing sequence in $U$ such that $\lim_{n\to\infty}u_n(x)=0$ for
every $x\in X$, then $\lim_{n\to\infty}f(u_n)=0$.

If $(X,\Sigma,\mu)$ is a measure space and $U$ is a Riesz subspace of
the space of real-valued $\mu$-integrable functions defined everywhere
on $X$, then $\int\,d\mu:U\to\Bbb R$ is sequentially smooth\cmmnt{, by
Fatou's Lemma or Lebesgue's Dominated Convergence Theorem}.

\cmmnt{\medskip

\noindent{\bf Remark} It is essential to distinguish between
`sequentially smooth', as defined here, and `sequentially
order-continuous', as in 313Hb or 355G.   In the context here, a
positive linear operator $f:U\to\Bbb R$ is sequentially order-continuous
if $\lim_{n\to\infty}f(u_n)=0$ whenever $\sequencen{u_n}$ is a
non-increasing sequence in $U$ such that $0$ is the greatest lower bound
for $\{u_n:n\in\Bbb N\}$ in $U$;  while $f$ is sequentially smooth if
$\lim_{n\to\infty}f(u_n)=0$ whenever $\sequencen{u_n}$ is a
non-increasing sequence in $U$ such that $0$ is the greatest lower bound
for $\{u_n:n\in\Bbb N\}$ in $\Bbb R^X$.   So there can be sequentially
smooth functionals which are not sequentially order-continuous, as in
436Xi.   A sequentially order-continuous positive linear functional is
of course sequentially smooth.
}%end of comment

\leader{436B}{Definition} Let $X$ be a set.   I will say that a Riesz
subspace $U$ of $\Bbb R^X$ is {\bf truncated} (or satisfies {\bf Stone's
condition}) if $u\wedge\chi X\in U$ for every $u\in U$.

In this case, $u\wedge\gamma\chi X\in U$ for every $\gamma\ge 0$ and
$u\in U$\cmmnt{ (being $-u^-$ if $\gamma=0$,
$\gamma(\gamma^{-1}u\wedge\chi X)$ otherwise)}.

\leader{436C}{Lemma} Let $X$ be a set and $U$ a truncated Riesz subspace
of $\Bbb R^X$.   Write $\Cal K$ for the family of sets of the form
$\{x:x\in X,\,u(x)\ge 1\}$ as $u$ runs over $U$.   Let $f:U\to\Bbb R$ be
a sequentially smooth positive linear functional, and $\mu$ a measure on
$X$ such
that $\mu K$ is defined and equal to $\inf\{f(u):\chi K\le u\in U\}$ for
every $K\in\Cal K$.   Then $\int u\,d\mu$ exists and is equal to $f(u)$
for every $u\in U$.

\proof{ It is enough to deal with the case $u\ge 0$, since $U=U^+-U^+$
and both $f$ and $\int$ are linear.   Note that if $v\in U$,
$K\in\Cal K$ and $v\le\chi K$, then $v\le w$ whenever
$\chi K\le w\in U$, so $f(v)\le\mu K$.   For $k$, $n\in\Bbb N$ set

\Centerline{$K_{nk}=\{x:u(x)\ge  2^{-n}k\}$,
\quad$u_{nk}=u\wedge 2^{-n}k\chi X$.}

\noindent Then, for $k\ge 1$,

\Centerline{$K_{nk}=\{x:\Bover{2^n}{k}u\ge 1\}\in\Cal K$,}

\Centerline{$2^n(u_{n,k+1}-u_{nk})\le\chi K_{nk}
\le 2^n(u_{nk}-u_{n,k-1})$.}

\noindent So

\Centerline{$2^nf(u_{n,k+1}-u_{nk})\le\mu K_{nk}
\le 2^nf(u_{nk}-u_{n,k-1})$,}

\noindent and

\Centerline{$f(u_{n,4^n+1}-u_{n1})\le\sum_{k=1}^{4^n}2^{-n}\mu K_{nk}
\le f(u_{n,4^n}\le f(u)$.}

\noindent But setting $w_n=u_{n,4^n+1}-u_{n1}$, $\sequencen{w_n}$ is a
non-decreasing sequence of functions in $U$ and $\sup_{n\in\Bbb
N}w_n(x)=u(x)$ for every $x$, so
$\lim_{n\to\infty}f(u-w_n)=0$ and $\lim_{n\to\infty}f(w_n)=f(u)$.
Also, setting $v_n=\sum_{k=1}^{4^n}2^{-n}\chi K_{nk}$, we have
$w_n\le v_n\le u$ and $f(w_n)\le\int v_n\le f(u)$ for each $n$, so

\Centerline{$\biggerint u=\lim_{n\to\infty}\int v_n=f(u)$}

\noindent by B.Levi's theorem.
}%end of proof of 436C

\leader{436D}{Theorem} Let $X$ be a set and $U$ a truncated Riesz
subspace of $\Bbb R^X$.   Let $f:U\to\Bbb R$ be a positive linear
functional.   Then the following are equiveridical:

(i) $f$ is sequentially smooth;

(ii) there is a measure $\mu$ on $X$ such that $\int u\,d\mu$ is defined
and equal to $f(u)$ for every $u\in U$.

\proof{ I remarked in 436A that (ii)$\Rightarrow$(i) is a
consequence of Fatou's Lemma.   So the argument here is devoted to
proving that (i)$\Rightarrow$(ii).

\medskip

{\bf (a)} Let $\Cal K$ be the family of sets $K\subseteq X$ such that
$\chi K=\inf_{n\in\Bbb N}u_n$ for some sequence $\sequencen{u_n}$ in
$U$, taking the infimum in $\Bbb R^X$, so that
$(\inf_{n\in\Bbb N}u_n)(x)=\inf_{n\in\Bbb N}u_n(x)$ for every $x\in X$.
Then $\Cal K$
is closed under finite unions and countable intersections.   \Prf\ (i)
If $K$, $K'\in\Cal K$ take sequences $\sequencen{u_n}$,
$\sequencen{u'_n}$ in $U$ such that $\chi K=\inf_{n\in\Bbb N}u_n$ and
$\chi K'=\inf_{n\in\Bbb N}u'_n$; then
$\chi(K\cup K')=\inf_{m,n\in\Bbb N}u_m\vee u'_n$, so
$K\cup K'\in\Cal K$.   (ii) If $\sequencen{K_n}$ is a sequence in
$\Cal K$, then for each $n\in\Bbb N$ we can choose a sequence
$\sequence{i}{u_{ni}}$ in $U$ such
that $\chi K_n=\inf_{i\in\Bbb N}u_{ni}$;  now
$\chi(\bigcap_{n\in\Bbb N}K_n)=\inf_{n,i\in\Bbb N}u_{ni}$, so
$\bigcap_{n\in\Bbb N}K_n\in\Cal K$.\ \Qed

Note that $\emptyset\in\Cal K$ because $0\in U$.

\medskip
{\bf (b)} We need to know that if $u\in U$ then $K=\{x:u(x)\ge 1\}$
belongs to $\Cal K$.   \Prf\ Set

\Centerline{$u_n=2^n((u\wedge\chi X)-(u\wedge(1-2^{-n})\chi X))$.}

\noindent Because $U$ is truncated, every $u_n$ belongs to $U$, and it
is easy to
check that $\inf_{n\in\Bbb N}u_n=\chi K$.\ \QeD\  It follows that

\Centerline{$\{x:u(x)\ge\alpha\}=\{x:\Bover1\alpha u(x)\ge 1\}
\in\Cal K$}

\noindent whenever $u\in U$ and $\alpha>0$.

\medskip

{\bf (c)} For $K\in\Cal K$, set
$\phi_0K=\inf\{f(u):u\in U,\,u\ge\chi K\}$.   Then $\phi_0$ satisfies
the conditions of 413J.   \Prf\  I have
already checked ($\dagger$) and ($\ddagger$) of 413J.

\medskip

\quad\grheada\ Fix $K$, $L\in\Cal K$ with $L\subseteq K$.   Set
$\gamma=\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$.

\medskip

\qquad{\bf (i)} Suppose that $K'\in\Cal K$ is included in $K\setminus L$,
and $\epsilon>0$.   Let $\sequencen{u_n}$, $\sequencen{u'_n}$ be
sequences in $U$ such that $\chi L=\inf_{n\in\Bbb N}u_n$ and
$\chi K'=\inf_{n\in\Bbb N}u'_n$, and let $u\in U$ be such that $u\ge\chi K$
and $f(u)\le\phi_0K+\epsilon$.   Set $v_n=u\wedge\inf_{i\le n}u_i$,
$v'_n=u\wedge\inf_{i\le n}u'_i$ for each $n$.   Then
$\sequencen{v_n\wedge v'_n}$ is a non-increasing sequence in $U$ with
infimum $\chi L\wedge\chi K'=0$, so there is an $n$ such that
$f(v_n\wedge v'_n)\le\epsilon$.   In this case

$$\eqalign{\phi_0L+\phi_0K'
&\le f(v_n)+f(v'_n)
=f(v_n+v'_n)\cr
&=f(v_n\vee v'_n)+f(v_n\wedge v'_n)
\le f(u)+\epsilon
\le\phi_0K+2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\phi_0L+\phi_0K'\le \phi_0K$.
As $K'$ is arbitrary, $\phi_0L+\gamma\le\phi_0K$.

\medskip

\qquad{\bf (ii)} Next, given $\epsilon\in\ooint{0,1}$, there are $u$,
$v\in U$ such that $u\ge\chi K$, $v\ge\chi L$ and
$f(v)\le\phi_0L+\epsilon$.   Consider

\Centerline{$K'=\{x:x\in K,\,\min(1,u(x))-v(x)\ge\epsilon\}
\subseteq K\setminus L$.}

\noindent By (b), $K'\in\Cal K$.   If $w\in U$ and
$w\ge\chi K'$, then $v(x)+w(x)\ge 1-\epsilon$ for every $x\in K$, so

\Centerline{$\phi_0K
\le\Bover1{1-\epsilon}f(v+w)
\le\Bover1{1-\epsilon}(\phi_0L+\epsilon+f(w))$.}

\noindent As $w$ is arbitrary,

\Centerline{$(1-\epsilon)\phi_0K\le\phi_0
L+\epsilon+\phi_0K'\le\phi_0L+\epsilon+\gamma$.}

\noindent As $\epsilon$ is arbitrary, $\phi_0K\le\phi_0L+\gamma$ and we
have equality, as required by ($\alpha$) in 413J.

\medskip

\quad\grheadb\ Now suppose that $\sequencen{K_n}$ is a non-increasing
sequence in $\Cal K$ with empty intersection.   For each $n\in\Bbb N$
let $\sequence{i}{u_{ni}}$ be a sequence in $U$ with infimum $\chi K_n$
in $\Bbb R^X$.   Set $v_n=\inf_{i,j\le n}u_{ji}$ for each $n$;  then
$\sequencen{v_n}$ is a non-increasing sequence in $U$ with infimum
$\inf_{n\in\Bbb N}\chi K_n=0$, so $\inf_{n\in\Bbb N}f(v_n)=0$.   But

\Centerline{$v_n\ge\inf_{j\le n}\chi K_j=\chi K_n$,
\quad$\phi_0K_n\le f(v_n)$}

\noindent for every $n$, so $\inf_{n\in\Bbb N}\phi_0K_n=0$, as required
by ($\beta$) of 413J.\ \Qed

\medskip

{\bf (d)} By 413J, there is a complete locally determined measure $\mu$
on $X$, inner regular with respect to $\Cal K$, extending $\phi_0$.
By 436C, $f(u)=\int u\,d\mu$ for every $u\in U$, as required.
}%end of proof of 436D

\leader{436E}{Proposition} Let $X$ be any topological space,
and $C_b=C_b(X)$ the space of bounded continuous
real-valued functions on
$X$.   Then there is a one-to-one correspondence between totally finite
Baire measures $\mu$ on $X$ and sequentially smooth positive linear
functionals $f:C_b\to\Bbb R$, given by the formulae

\Centerline{$f(u)=\biggerint u\,d\mu$ for every $u\in C_b$,}

\Centerline{$\mu Z=\inf\{f(u):\chi Z\le u\in C_b\}$ for every zero
set
$Z\subseteq X$.}

\proof{{\bf (a)} If $\mu$ is a totally finite Baire measure on $X$, then
every continuous bounded real-valued function is integrable, and
$f=\int d\mu$ is a sequentially smooth positive linear operator on
$C_b$, by Fatou's Lemma, as usual.

\medskip

{\bf (b)} If $f:C_b\to\Bbb R$ is a sequentially smooth positive
linear
operator, then 436D tells us that there is a measure $\mu_0$ on $X$ such
that $\int u\,d\mu_0$ is defined and equal to $f(u)$ for every $u\in
C_b$.   By the construction in 436D, or otherwise, we may suppose
that $\mu_0$ is complete, so that every $u\in C_b$ is
$\Sigma$-measurable, where $\Sigma$ is the domain of $\mu_0$.   It
follows by the definition of the Baire $\sigma$-algebra $\CalBa$ of
$X$ (4A3K) that $\CalBa\subseteq\Sigma$, so that
$\mu=\mu_0\restr\Sigma$ is a Baire measure;  of course we still have
$f(u)=\int u\,d\mu$ for every $u\in C_b$.   Also, if $Z\subseteq X$
is a
zero set, $\mu Z=\inf\{f(u):\chi Z\le u\in C_b\}$.   \Prf\  Express
$Z$
as $\{x:v(x)=0\}$ where $v:X\to[0,1]$ is continuous.   Set

\Centerline{$u_n=(\chi X-2^nv)^+$}

\noindent for $n\in\Bbb N$;  then $\sequencen{u_n}$ is a non-increasing
sequence in $C_b$ and $\sequencen{u_n(x)}\to(\chi Z)(x)$ for every
$x\in X$, so

$$\eqalign{\mu Z
&\le\inf\{\int u\,d\mu:\chi Z\le u\in C_b\}
=\inf\{f(u):\chi Z\le u\in C_b\}\cr
&\le\inf_{n\in\Bbb N}f(u_n)
=\lim_{n\to\infty}\int u_nd\mu
=\mu Z.\text{ \Qed}\cr}$$

\medskip

{\bf (c)} The argument of (b) shows that if two totally finite Baire
measures give the same integrals to every member of $C_b$, then they
must agree on all zero sets.   By the Monotone Class Theorem (136C) they
agree on the $\sigma$-algebra generated by the zero sets, that is,
$\CalBa$, and are therefore equal.   Thus the operator
$\mu\mapsto\int d\mu$ from the set of totally finite Baire measures on
$X$ to the set of sequentially smooth positive linear operators on
$C_b$
is a bijection, and if $f=\int d\mu$ then $\mu Z=\inf\{f(u):\chi Z\le
u\in C_b\}$ for every zero set $Z$, as required.
}%end of proof of 436E

\leader{436F}{}\cmmnt{ Corresponding to 434R, we have the
following construction for product Baire measures, applicable to a
slightly larger class of spaces.

\medskip

\noindent}{\bf Proposition} Let $X$ be a sequential space, $Y$ a
topological space, and $\mu$, $\nu$ totally finite Baire measures on
$X$, $Y$ respectively.   Then there is a Baire measure $\lambda$ on
$X\times Y$ such that

\Centerline{$\lambda W=\biggerint\nu W[\{x\}]\mu(dx)$,
\quad$\int fd\lambda=\biggeriint f(x,y)\nu(dy)\mu(dx)$}

\noindent for every Baire set $W\subseteq X\times Y$ and every bounded
continuous function $f:X\times Y\to\Bbb R$.

\proof{{\bf (a)} $\phi(f)=\iint f(x,y)dydx$ is defined in $\Bbb R$ for
every bounded continuous function $f:X\times Y\to\Bbb R$.   \Prf\ For
each $x\in X$, $g(x)=\int f(x,y)dy$ is defined because $y\mapsto f(x,y)$
is continuous.    If $\sequencen{x_n}$ is any sequence in $X$ converging
to $x\in X$, then

\Centerline{$g(x)=\biggerint f(x,y)dy=\int\lim_{n\to\infty}f(x_n,y)dy
=\lim_{n\to\infty}\int f(x_n,y)dy=\lim_{n\to\infty}g(x_n)$}

\noindent by Lebesgue's Dominated Convergence Theorem.   So $g$ is
sequentially continuous;  because $X$ is sequential, $g$ is continuous
(4A2Kd).   So $\iint f(x,y)dydx=\int g(x)dx$ is defined in
$\Bbb R$.\ \Qed

\medskip

{\bf (b)} Of course $\phi$ is a positive linear functional on
$C_b(X\times Y)$, and B.Levi's theorem shows that it is sequentially
smooth.   By 436E, there is a Baire measure $\lambda$ on
$X\times Y$ such that $\int fd\lambda=\phi(f)$ for every
$f\in C_b(X\times Y)$.

\medskip

{\bf (c)} If $W\subseteq X\times Y$ is a zero set, there is a
non-increasing sequence $\sequencen{f_n}$ in $C_b(X\times Y)$ such that
$\chi W=\inf_{n\in\Bbb N}f_n$.   So B.Levi's theorem tells us that

\Centerline{$\biggerint\nu W[\{x\}]dx
=\lim_{n\to\infty}\int f_n(x,y)dydx
=\lim_{n\to\infty}\int f_nd\lambda
=\lambda W$.}

\noindent Now the Monotone Class Theorem (136B) tells us that

\Centerline{$\{W:W\subseteq X\times Y$ is Baire, $\int\nu W[\{x\}]dx$
exists $=\lambda W\}$}

\noindent includes the $\sigma$-algebra generated by the zero sets, that
is, contains every Baire set in $X\times Y$.   So $\lambda$ has the
required properties.
}%end of proof of 436F

\leader{436G}{Definition} Let $X$ be a set, $U$ a Riesz subspace of
$\Bbb R^X$, and $f:U\to\Bbb R$ a positive linear functional.   I say
that $f$ is {\bf smooth} if whenever $A$ is a non-empty
downwards-directed family in
$U$ such that $\inf_{u\in A}u(x)=0$ for every $x\in X$, then
$\inf_{u\in A}f(u)=0$.

Of course a smooth functional is sequentially smooth.   If
$(X,\frak T,\Sigma,\mu)$ is an effectively locally finite
$\tau$-additive
topological measure space and $U$ is a Riesz subspace of $\Bbb R^X$
consisting of integrable continuous functions, then
$\int\,d\mu:U\to\Bbb R$ is
smooth\cmmnt{, by 414Bb}.   \cmmnt{Corresponding to the remark in
436A, note that an order-continuous positive linear functional must
be smooth, but that a smooth positive linear functional need not be
order-continuous.}

\leader{436H}{Theorem} Let $X$ be a set and $U$ a truncated Riesz
subspace of $\Bbb R^X$.   Let $f:U\to\Bbb R$ be a positive linear
functional.   Then the following are equiveridical:

(i) $f$ is smooth;

(ii) there are a topology $\frak T$ and a measure $\mu$ on $X$ such that
$\mu$ is a quasi-Radon measure with respect to $\frak T$,
$U\subseteq C(X)$ and $\int u\,d\mu$ is defined and equal to $f(u)$ for
every $u\in U$;

(iii) writing $\frak S$ for the coarsest topology on $X$ for which every
member of $U$ is continuous, there is a
measure $\mu$ on $X$ such that $\mu$ is a quasi-Radon measure with
respect to $\frak S$,
and $\int u\,d\mu$ is defined and equal to $f(u)$ for every $u\in U$.

\proof{ As remarked in 436G, in a fractionally more general context,
(ii)$\Rightarrow$(i) is a
consequence of 414B.   Of course (iii)$\Rightarrow$(ii).   So the
argument here is devoted to proving that
(i)$\Rightarrow$(iii).   Except for part (b) it is a simple adaptation
of
the method of 436D.

\medskip

{\bf (a)} Let $\Cal K$ be the family of sets $K\subseteq X$ such that
$\chi K=\inf A$ in $\Bbb R^X$ for some non-empty set $A\subseteq U$.
Then $\Cal K$ is closed under finite unions.
\Prf\ If $K$, $K'\in\Cal K$ take $A$, $A'\subseteq U$ such that
$\chi K=\inf A$, $\chi K'=A'$;  then
$\chi(K\cup K')=\inf\{u\vee u':u\in A,\,u'\in A'\}$, so
$K\cup K'\in\Cal K$.\ \Qed

Note that $\emptyset\in\Cal K$ because $0\in U$.

As in part (b) of the proof of 436D, $\{x:u(x)\ge\alpha\}\in\Cal K$
whenever $\alpha>0$ and $u\in U$.

\medskip

{\bf (b)} Every member of
$\Cal K$ is closed for $\frak S$, being of the form
$\{x:u(x)\ge 1$ for every $u\in A\}$ for some $A\subseteq U$.   We need
to know that if $K\in\Cal K$ and $G\in\frak S$, then
$K\setminus G\in\Cal K$.   \Prf\ Take a
non-empty set $B\subseteq U$ such that $\chi K=\inf B$.
Because $\Cal K$ is closed under
finite unions and arbitrary intersections,
$\frak S_K=\{G:G\subseteq X,\,K\setminus G\in\Cal K\}$ is a topology on
$X$.  (i) If $u\in U$ and $G=\{x:u(x)>0\}$, then
$\chi(K\setminus G)=\inf\{(v-ku)^+:v\in B,\,k\in\Bbb N\}$ so
$K\setminus G\in\Cal K$ and $G\in\frak S_K$.
(ii) If $u\in U$ and $\alpha>0$, then
$\{x:u(x)>\alpha\}=\{x:(u-u\wedge\alpha\chi X)(x)>0\}$ belongs to
$\frak S_K$, by (i). (iii) If $u\in U$ and $\alpha>0$, set
$G=\{x:u(x)<\alpha\}$.   Then

\Centerline{$K\setminus G
=K\cap\{x:u(x)\ge\alpha\}\in\Cal K$,}

\noindent so $G\in\frak S_K$.
(iv) Thus every member of $U^+$ is $\frak S_K$-continuous (2A3Bc), so
every member of $U$ is $\frak S_K$-continuous (2A3Be), and
$\frak S\subseteq\frak S_K$, that is,
$K\setminus G\in\Cal K$ for every $G\in\frak S$.\ \Qed

\medskip

{\bf (c)} For $K\in\Cal K$, set
$\phi_0K=\inf\{f(u):u\in U,\,u\ge\chi K\}$.   Then $\phi_0$ satisfies
the conditions of 415K.   \Prf\  I have
already checked ($\dagger$) and ($\ddagger$) of 415K.

\medskip

\quad\grheada\ Fix $K$, $L\in\Cal K$ with $L\subseteq K$.   Set
$\gamma=\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq K\setminus L\}$.

\medskip

\qquad{\bf (i)} Suppose that $K'\in\Cal K$ is included in $K\setminus L$
and $\epsilon>0$.   Set $A=\{u:\chi L\le u\in U\}$, $A'=\{u:\chi K'\le
u\in
U\}$, so that $\chi L=\inf A$ and $\chi K'=\inf A'$, and let $v\in U$ be
such that $v\ge\chi K$ and $f(v)\le\phi_0K+\epsilon$.   Then $\{u\wedge
u':u\in A,\,u'\in A'\}$ is a downwards-directed family with infimum $0$
in
$\Bbb R^X$, so (because $f$ is smooth) there are $u\in A$, $u'\in A'$
such
that $f(u\wedge u')\le\epsilon$.   In this case

\Centerline{$\phi_0L+\phi_0K'
\le f(v\wedge u)+f(v\wedge u')
=f(v\wedge(u\vee u'))+f(v\wedge u\wedge u')
\le\phi_0K+2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\phi_0L+\phi_0K'\le \phi_0K$.
As
$K'$ is arbitrary, $\phi_0L+\gamma\le\phi_0K$.

\medskip

\qquad{\bf (ii)} Next, given $\epsilon\in\ooint{0,1}$, there are $u$,
$v\in
U$ such that $u\ge\chi K$, $v\ge\chi L$ and $f(v)\le\phi_0L+\epsilon$.
Consider

\Centerline{$K'=\{x:x\in K,\,\min(1,u(x))-v(x)\ge\epsilon\}$.}

\noindent By the last remark in (a), $K'\in\Cal K$.   If $w\in U$ and
$w\ge\chi K'$, then $v(x)+w(x)\ge 1-\epsilon$ for every $x\in K$, so

\Centerline{$\phi_0K
\le\Bover1{1-\epsilon}f(v+w)
\le\Bover1{1-\epsilon}(\phi_0L+\epsilon+f(w))$.}

\noindent As $w$ is arbitrary,

\Centerline{$(1-\epsilon)\phi_0K\le\phi_0
L+\epsilon+\phi_0K'\le\phi_0L+\epsilon+\gamma$.}

\noindent As $\epsilon$ is arbitrary, $\phi_0K\le\phi_0L+\gamma$ and we
have equality, as required by ($\alpha$) in 415K.

\medskip

\quad\grheadb\ Now suppose that $\Cal K'$ is a non-empty
downwards-directed subset of $\Cal K$ with empty intersection.   Set

\Centerline{$A=\bigcup_{K\in\Cal K'}\{u:\chi K\le u\in U\}$.}

\noindent Then $A$ is a downwards-directed subset of $U$ and $\inf A=0$
in $\Bbb R^X$.   Because $f$ is smooth,

\Centerline{$0=\inf_{u\in A}f(u)=\inf_{K\in\Cal K'}\phi_0K$.}

\noindent Thus ($\beta$) of 415K is satisfied.

\medskip

\quad\grheadc\ If $K\in\Cal K$ and $\phi_0K>0$, take $u\in U$ such that
$u\ge\chi K$, and consider $G=\{x:u(x)>\bover12\}$.   Then
$K\subseteq G$, while $G\subseteq\{x:2u(x)\ge 1\}$, so

\Centerline{$\sup\{\phi_0K':K'\in\Cal K,\,K'\subseteq G\}
\le 2f(u)<\infty$.}

\noindent Thus $\phi_0$ satisfies ($\gamma$) of 415K.\ \Qed

\medskip

{\bf (d)} By 415K, there is a quasi-Radon measure $\mu$ on $X$ extending
$\phi_0$.   By 436C, $f(u)=\int u\,d\mu$ for every $u\in U$.
}%end of proof of 436H

\medskip

\noindent{\bf Remark}\cmmnt{ It is worth noting explicitly that} $\mu$, as
constructed here, is inner regular with respect to the family $\Cal K$
of sets
$K\subseteq X$ such that $\chi K=\inf A$ for some set $A\subseteq U$.

\leader{436I}{Lemma} Let $X$ be a topological space.   Let $C_0=C_0(X)$
be the space of continuous functions $u:X\to\Bbb R$ which `vanish at
infinity' in the sense that $\{x:|u(x)|\ge\epsilon\}$ is compact for
every $\epsilon>0$.

(a) $C_0$ is a norm-closed solid linear subspace of $C_b=C_b(X)$, so is
a Banach lattice in its own right.

(b) $C_0^*=C_0^{\sim}$ is an $L$-space\cmmnt{ (definition: 354M)}.

(c) If $A\subseteq C_0$ is a non-empty downwards-directed set such
that $\inf_{u\in A}u(x)=0$ for every $x\in X$, then
$\inf_{u\in A}\|u\|_{\infty}=0$.

\proof{{\bf (a)(i)} If $u\in C_0$, then $K=\{x:|u(x)|\ge 1\}$ is
compact, so $\|u\|_{\infty}\le\sup(\{1\}\cup\{|u(x)|:x\in K\})$ is
finite, and $u\in C_b$.

\medskip

\quad{\bf (ii)} If $u$, $v\in C_0$ and $\alpha\in\Bbb R$ and
$w\in C_b$ and $|w|\le|u|$, then for any $\epsilon>0$

\Centerline{$\{x:|u(x)+v(x)|\ge\epsilon\}
\subseteq\{x:|u(x)|\ge\Bover12\epsilon\}
\cup\{x:|v(x)|\ge\Bover12\epsilon\}$,}

\Centerline{$\{x:|\alpha u(x)|\ge\epsilon\}
\subseteq\{x:|u(x)|\ge\Bover{\epsilon}{1+|\alpha|}\}$,}

\Centerline{$\{x:|w(x)|\ge\epsilon\}
\subseteq\{x:|u(x)|\ge\epsilon\}$}

\noindent are closed relatively compact sets, so are compact, and $u+v$,
$\alpha u$, $w$ belong to $C_0$.   Thus $C_0$ is a solid linear
subspace
of $C_b$.

\medskip

\quad{\bf (iii)} If $\sequencen{u_n}$ is a sequence in $C_0$ which
$\|\,\|_{\infty}$-converges to $u\in C_b$, then for any $\epsilon>0$
there is an $n\in\Bbb N$ such that
$\|u-u_n\|_{\infty}\le\bover12\epsilon$, so that

\Centerline{$\{x:|u(x)|\ge\epsilon\}
\subseteq\{x:|u_n(x)|\ge\Bover12\epsilon\}$}

\noindent is compact, and $u\in C_0$.   Thus $C_0$ is norm-closed
in
$C_b$.

\medskip

\quad{\bf (iv)} Being a norm-closed Riesz subspace of a Banach lattice,
$C_0$ is itself a Banach lattice.

\medskip

{\bf (b)} By 356Dc, $C_0^*=C_0^{\sim}$ is a Banach lattice.   Now
$\|f+g\|=\|f\|+\|g\|$ for all non-negative $f$, $g\in C_0^*$.   \Prf\
Of course $\|f+g\|\le\|f\|+\|g\|$.   On the other hand, for any
$\epsilon>0$ there are $u$, $v\in C_0$ such that $\|u\|_{\infty}\le
1$,
$\|v\|_{\infty}\le 1$ and
$|f(u)|\ge\|f\|-\epsilon$, $|g(v)|\ge\|g\|-\epsilon$.   Set
$w=|u|\vee|v|$;  then $w\in C_0$ and

\Centerline{$\|w\|_{\infty}
=\max(\|u\|_{\infty},\|v\|_{\infty})\le 1$.}

\noindent So

\Centerline{$\|f+g\|\ge(f+g)(w)
\ge f(|u|)+g(|v|)\ge|f(u)|+|g(v)|\ge\|f\|+\|g\|-2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\|f+g\|\ge\|f\|+\|g\|$.\ \Qed

So $C_0^*$ is an $L$-space.

\medskip

{\bf (c)} Let $\epsilon>0$.   For $u\in A$ set
$K_u=\{x:u(x)\ge\epsilon\}$.   Then $\{K_u:u\in A\}$ is a
downwards-directed family of closed compact sets with empty
intersection, so there must be some $u\in A$ such that $K_u=\emptyset$,
and $\|u\|_{\infty}\le\epsilon$.   As $\epsilon$ is arbitrary, we have
the result.
}%end of proof of 436I

\cmmnt{\medskip

\noindent{\bf Remark} (c) is a version of {\bf Dini's theorem}.
}

\vleader{48pt}{436J}{Riesz Representation Theorem (first form)} Let
$(X,\frak T)$ be a locally compact Hausdorff space, and $C_k=C_k(X)$ the
space of continuous real-valued functions on $X$ with compact
support.   If $f:C_k\to\Bbb R$ is any positive linear functional,
there is a unique Radon measure $\mu$ on $X$ such that
$f(u)=\int u\,d\mu$ for every $u\in C_k$.

\proof{{\bf (a)} The point is that $f$ is smooth.   \Prf\ Suppose that
$A\subseteq C_k$ is non-empty and downwards-directed and that
$\inf A=0$
in $\Bbb R^X$.   Fix $u_0\in A$ and set $K=\overline{\{x:u_0(x)>0\}}$,
so that $K$ is compact.   Because $X$ is locally compact, there is an
open relatively compact set $G\supseteq K$.   Now there is a continuous
function $u_1:X\to[0,1]$ such that $u_1(x)=1$ for $x\in K$ and
$u_1(x)=0$ for $x\in X\setminus G$ (4A2F(h-iii)).   Because $G$ is
relatively compact, $u_1\in C_k$.

Take any $\epsilon>0$.   By 436Ic,
there is a $v\in A$ such that $\|v\|_{\infty}\le\epsilon$.   Now there
is a $v'\in A$ such that $v'\le v\wedge u_0$, so that $v'(x)\le\epsilon$
for every $x\in K$ and $v'(x)=0$ for $x\notin K$.   In this case
$v'\le\epsilon u_1$, and

\Centerline{$\inf_{u\in A}f(u)\le f(v')\le\epsilon f(u_1)$.}

\noindent As $\epsilon$ is arbitrary, $\inf_{u\in A}f(u)=0$;  as $A$ is
arbitrary, $f$ is smooth.\ \Qed

\medskip

{\bf (b)} Note that because $\frak T$ is locally compact, it is the
coarsest topology on $X$ for which every function in $C_k$ is
continuous (4A2G(e-ii)).   Also $C_k$ is a truncated Riesz subspace
of $\BbbR^X$.
So 436H tells us that there is a quasi-Radon measure $\mu$
on $X$ such that $f(u)=\int u\,d\mu$ for every $u\in C_k$.
And $\mu$ is locally finite.   \Prf\ If $x_0\in X$, then (as in (a)
above) there is a $u_1\in C_k^+$ such that $u_1(x_0)=1$;  now
$G=\{x:u_1(x)>\bover12\}$ is an open set containing $x_0$, and
$\mu G\le 2f(u_1)$ is finite.\ \Qed

By 416G, or otherwise, $\mu$ is a Radon measure.

\medskip

{\bf (c)} By 416E(b-v), $\mu$ is unique.
}%end of proof of 436J

\leader{436K}{Riesz Representation Theorem (second form)} Let
$(X,\frak T)$ be a locally compact Hausdorff space.   If
$f:C_0(X)\to\Bbb R$ is any positive linear functional,
there is a unique totally finite Radon measure $\mu$ on $X$ such that
$f(u)=\int u\,d\mu$ for every $u\in C_0=C_0(X)$.

\proof{{\bf (a)} As noted in 436Ib, $C_0^*=C_0^{\sim}$, so $f$ is
$\|\,\|_{\infty}$-continuous.   $C_k(X)$ is a linear subspace of
$C_0$, and $f\restr C_k(X)$ is a positive linear functional;  so by
436J
there is a unique Radon measure $\mu$ on $X$ such that
$f(u)=\int u\,d\mu$ for every $u\in C_k(X)$.   Now $\mu$ is totally
finite.   \Prf\ By 414Ab,

\Centerline{$\mu X=\sup\{f(u):u\in C_k(X),\,0\le u\le\chi X\}
\le\|f\|<\infty$.  \Qed}

\medskip

{\bf (b)} Accordingly $\int u\,d\mu$ is defined for every $u\in C_b(X)$,
and in particular for every $u\in C_0$.   Next, $C_k=C_k(X)$ is
norm-dense in
$C_0$.   \Prf\ If $u\in C_0^+$, then $u_n=(u-2^{-n}\chi X)^+$
belongs to
$C_k$ and $\|u-u_n\|_{\infty}\le 2^{-n}$ for every $n\in\Bbb N$, so
$u\in\overline{C}_k$;  accordingly $C_0=C_0^+-C_0^+$ is
included in
$\overline{C}_k$.\ \QeD\   Since $\int\,d\mu$, regarded as a linear
functional on $C_0$, is positive, therefore continuous, and agrees
with $f$ on $C_k$, it must be identical to $f$.   Thus
$f(u)=\int u\,d\mu$ for every $u\in C_0$.

\medskip

{\bf (c)} Because there is only one Radon measure giving the right
integrals to members of $C_k$ (436J), $\mu$ is unique.
}%end of proof of 436K

\leader{*436L}{}\cmmnt{ The results here, by opening a path between
measure theory and the study of linear functionals on spaces of
continuous functions, provide an enormously powerful tool for the
analysis of dual spaces $C(X)^*$ and their relatives.   I will explore
some of these ideas in the next section.   Here I will give only a
sample pair of facts to show how measure theory can tell us things about
Banach lattices which seem difficult to reach by other methods.

\medskip

\noindent}{\bf Proposition} Let $X$ be a topological space;  write
$C_b$ for $C_b(X)$.   Suppose that $U$ is a
norm-closed linear subspace of $C_b^*$ such that the functional
$u\mapsto f(u\times v):C_b\to\Bbb R$ belongs to $U$ whenever $f\in U$
and $v\in C_b$.   Then $U$ is a band in the $L$-space $C_b^*$.

\proof{{\bf (a)} Let $e=\chi X$ be the standard order unit of
$C_b$, and if
$f\in C_b^*$ and $u$, $v\in C_b$ write $f_v(u)$ for
$f(u\times v)$.   By 356Na, $C_b^*=C_b^{\sim}$ is an $L$-space.

\medskip

{\bf (b)} I show first that $U$ is a Riesz subspace of $C_b^{\sim}$.
\Prf\ If $f\in U$ and $\epsilon>0$, there is a $v\in C_b$ such that
$|v|\le e$ and $f(v)\ge|f|(e)-\epsilon$ (356B).   Now $f_v\le|f|$ and

\Centerline{$\||f|-f_v\|=(|f|-f_v)(e)\le\epsilon$}

\noindent (356Nb), while
$f_v\in U$.   As $\epsilon$ is arbitrary, $|f|\in\overline{U}=U$;  as
$f$ is arbitrary, $U$ is a Riesz subspace of $C_b^*$ (352Ic).\ \Qed

\medskip

{\bf (c)} Now suppose that $X$ is a compact Hausdorff space.   Then $U$
is a solid linear subspace of $C_b^{\sim}=C(X)^{\sim}$.   \Prf\
Suppose that $f\in U$ and that $0\le g\le f$.   Let $\epsilon>0$.   By
either 436J or 436K, there are Radon measures $\mu$, $\nu$ on $X$ such
that $f(u)=\int u\,d\mu$ and $g(u)=\int u\,d\nu$ for every
$u\in C(X)$.   By 416Ea, $\nu\le\mu$ in the sense of 234P,
so $\nu$ is an indefinite-integral measure over
$\mu$ (415Oa, or otherwise);  let $w:X\to[0,1]$ be such that
$\int_Ew\,d\mu=\nu E$ for every $E\in\dom\nu$.    There is a continuous
function $v:X\to\Bbb R$ such that $\int|w-v|d\mu\le\epsilon$
(416I), and now

$$\eqalignno{|g(u)-f_v(u)|
&=|\int u\,d\nu-\int u\times v\,d\mu|
=|\int u\times(w-v)d\mu|\cr
\displaycause{235K}
&\le\|u\|_{\infty}\int|w-v|d\mu
\le\epsilon\|u\|_{\infty}\cr}$$

\noindent for every $u\in C(X)$, so $\|g-f_v\|\le\epsilon$, while
$f_v\in U$.   As $\epsilon$ is arbitrary, $g\in U$;  as $f$ and $g$ are
arbitrary (and $U$ is a Riesz subspace of $C(X)^*$), $U$ is a solid
linear subspace of $C(X)^*$.\ \Qed

Since every norm-closed solid linear subspace of an $L$-space is a band
(354Eg), it follows that (provided $X$ is a compact Hausdorff space)
$U$ is actually a band.

\medskip

{\bf (d)} For the general case, let $Z$ be the set of all Riesz
homomorphisms $z:C_b\to\Bbb R$ such that $z(e)=1$.   Then $Z$ is a
weak*-closed subset of the unit ball of $C_b^*$ so is a compact
Hausdorff space.   We have a
Banach lattice isomorphism $T:C_b\to C(Z)$ given by the formula
$(Tu)(z)=z(u)$ for $u\in C_b$, $z\in Z$ (see the proofs of 
353N\formerly{3{}53M} and
354K).   But note also that $T$ is multiplicative (353Qd\formerly{3{}53Pd}), 
and $T':C(Z)^*\to C_b^*$ is a Banach lattice isomorphism.   Let
$V$ be $(T')^{-1}[U]\subseteq C(Z)^*$;
then $V$ is a closed linear subspace
of $C(Z)^*$.   If $g\in V$ and $v$, $w\in C(Z)$, then

\Centerline{$g(v\times w)=(T'g)(T^{-1}v\times T^{-1}w)$,}

\noindent so $g_w$, defined in $C(Z)^*$ by the convention of
(a) above, is just $(T')^{-1}((T'g)_{T^{-1}w})$, and belongs to $V$.
By (b), $V$ is a band in $C(Z)^*$ so $U$ is a band in $C_b^*$, as
required.
}%end of proof of 436L

\leader{*436M}{Corollary} Let $\frak A$ be a Boolean algebra, and
$M(\frak A)$ the $L$-space of bounded finitely additive functionals on
$\frak A$\cmmnt{ (362B)}.   Let $U\subseteq M(\frak A)$ be a norm-closed linear
subspace such that $a\mapsto\nu(a\Bcap b)$ belongs to $U$ whenever
$\nu\in U$ and $b\in\frak A$.   Then $U$ is a band in $M(\frak A)$.

\proof{{\bf (a)}
Let $Z$ be the Stone space of $\frak A$, so that $C(Z)$ is the
$M$-space $L^{\infty}(\frak A)$ (363A), and we have an $L$-space
isomorphism $T:M(\frak A)\to C(Z)^*$ defined by saying that
$(T\nu)(\chi\widehat{a})=\nu a$ whenever $\nu\in M(\frak A)$, $a\in\frak A$
and $\widehat{a}$ is the open-and-closed subset of $Z$ corresponding to
$a$ (363K).   Now $V=T[U]$ is a norm-closed linear subspace of $C(Z)^*$.

\medskip

{\bf (b)} $V$ satisfies the condition of 436L.   \Prf\ Suppose that
$f\in V$ and $v\in C(Z)$;  set $f_v(u)=f(u\times v)$ for $u\in C(Z)$, and
$\nu=T^{-1}f\in U$.
(i) If $v$ is of the form $\chi\widehat{b}$,
where $b\in\frak A$, then $\nu_b\in U$, where
$\nu a=\nu(a\Bcap b)$ for $a\in\frak A$.   Now $T\nu_b\in V$ and

\Centerline{$(T\nu_b)(\chi\widehat{a})=\nu_ba=\nu(a\Bcap b)
=f(\chi(\widehat{a\Bcap b}))=f(\chi\widehat{a}\times\chi\widehat{b})
=f_v(\chi\widehat{a})$}

\noindent for every $a\in\frak A$, so $f_v=T\nu_b$ belongs to $U$.
(ii) If $v$ is of the form $\sum_{i=0}^n\alpha_i\chi\widehat{b}_i$, where
$b_0,\ldots,b_n\in\frak A$ and $\alpha_0,\ldots,\alpha_n\in\Bbb R$, then
$f_v=\sum_{i=0}^n\alpha_if_{v_i}$, where $v_i=\chi\widehat{b}_i$ for each
$i$;   as $f_{v_i}\in V$ for each $i$, $f_v\in V$, because $V$ is a linear
subspace.  (iii) In general, given
$v\in C(Z)=L^{\infty}(\frak A)$ and $\epsilon>0$, there is a
$w\in C(Z)$, expressible in the form of (ii), such that
$\|v-w\|_{\infty}\le\epsilon$ (363C).   In this case $f_w\in V$, by (ii),
while

\Centerline{$|f_v(u)-f_w(u)|
=|f(u\times(v-w))|\le\|f\|\|u\|_{\infty}\epsilon$}

\noindent for every $u\in C(Z)$, and $\|f_v-f_w\|\le\epsilon\|f\|$.   As
$\epsilon$ is arbitrary and $V$ is closed, $f_v\in V$.\ \Qed

\medskip

{\bf (c)} By 436L, $V$ is a band in $C(Z)^*$;  as $T$ is a Riesz space
isomorphism, $U$ is a band in $M(\frak A)$.
}%end of proof of 436M

\exercises{\leader{436X}{Basic exercises $\pmb{>}$(a)}
%\spheader 436Xa
Let $(X,\Sigma,\mu_0)$ be a measure space, and $U$ the set of
$\mu_0$-integrable $\Sigma$-measurable real-valued functions defined
everywhere on $X$.   For $u\in U$ set $f(u)=\int u\,d\mu_0$.   Show that
$U$ and $f$ satisfy the conditions of 436D, and that the measure $\mu$
constructed from $f$ by the procedure there is just the c.l.d.\ version
of $\mu_0$.
%436D

\spheader 436Xb Let $\mu$ and $\nu$ be two complete locally determined
measures on a set $X$, and suppose that $\int f\,d\mu=\int f\,d\nu$ for
every function $f:X\to\Bbb R$ for which either integral is defined in
$\Bbb R$.   Show that $\mu=\nu$.
%436D

\sqheader 436Xc Let $X$ be a set, $U$ a truncated Riesz subspace of
$\Bbb R^X$, and $f:U\to\Bbb R$ a sequentially smooth positive linear
functional.   For $A\subseteq X$ set

$$\eqalign{\theta A
&=\inf\{\sup_{n\in\Bbb N}f(u_n):
  \sequencen{u_n}\text{ is a non-decreasing sequence in }U^+,\cr
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad
  \lim_{n\to\infty}u_n(x)=1\text{ for every }x\in A\},\cr}$$

\noindent taking $\inf\emptyset=\infty$ if need be.   Show that $\theta$
is an outer measure on $X$.   Let $\mu_0$ be the measure defined from
$\theta$ by \Caratheodory's method.   Show that $f(u)=\int u\,d\mu_0$
for every $u\in U$.   Show that the measure $\mu$ constructed in 436D is
the c.l.d.\ version of $\mu_0$.
%436D

\spheader 436Xd Let $X$ be a set and $U$ a truncated Riesz subspace of
$\Bbb R^X$.   Let $\tau:U\to\coint{0,\infty}$ be a seminorm such that
(i) $\tau(u)\le\tau(v)$ whenever $|u|\le|v|$ (ii)
$\lim_{n\to\infty}\tau(u_n)=0$ whenever $\sequencen{u_n}$ is a
non-increasing sequence in $U$ and $\lim_{n\to\infty}u_n(x)=0$ for every
$x\in X$.   Show that for any $u_0\in U^+$ there is a measure $\mu$ on
$x$ such that $\int u\,d\mu$ is defined, and less than or equal to
$\tau(u)$, for every $u\in U$, and $\int u_0d\mu=\tau(u_0)$.   \Hint{put
the Hahn-Banach theorem together with 436D.}
%436D

\spheader 436Xe\dvArevised{2010}(i)
Let $X$ be any topological space.   Show that every
positive linear functional on $C(X)$ is sequentially smooth (compare
375A), so corresponds to a totally finite Baire measure on $X$.
(ii) Let $X$ be a regular Lindel\"of space.   Show that every positive
linear functional on $C(X)$ is smooth, so corresponds to a totally finite
quasi-Radon measure on $X$.   (iii) Let $X$ be a K-analytic
Hausdorff space.   Show that every positive linear functional on $C(X)$
corresponds to at least one totally finite Radon measure on $X$.
%436E

\spheader 436Xf Let $X$ be a completely regular space.   Show that it is
measure-compact iff every sequentially smooth positive linear functional
on $C_b(X)$ is smooth.   \Hint{436Xj.}
%4{}35D 436E

\spheader 436Xg A completely regular topological space
$X$ is called {\bf realcompact}
if every Riesz homomorphism from $C(X)$ to $\Bbb R$ is of the form
$u\mapsto\alpha u(x)$ for some $x\in X$ and $\alpha\ge 0$.
(i) Show that, for any topological space $X$, any Riesz
homomorphism from $C(X)$ to $\Bbb R$ is representable by a Baire measure
on $X$ which takes at most two values.  (ii) Show that
a completely regular space $X$ is realcompact iff every $\{0,1\}$-valued
Baire measure on $X$ is of the form $E\mapsto\chi E(x)$.   (iii) Show
that a completely regular space $X$ is realcompact iff every purely
atomic totally finite Baire measure on $X$ is $\tau$-additive.   (iii)
Show that a measure-compact completely regular space is realcompact.
(iv) Show that the discrete topology on $[0,1]$ is realcompact.
\Hint{if $\nu$ is a Baire measure taking only the values $0$ and $1$,
set $x_0=\sup\{x:\nu[x,1]=1\}$.}   (v) Show that any product of
realcompact completely regular spaces is realcompact.   (vi) Show that
a Souslin-F subset of a realcompact completely regular space is
realcompact.   (For realcompact spaces which are not measure-compact,
see 439Xp.)
%4{}35D 436E

\spheader 436Xh In 436F, suppose that $\mu$ and $\nu$ are
$\tau$-additive.   Let $\tilde\mu$ and $\tilde\nu$ be the corresponding
quasi-Radon measures (415N), and $\tilde\lambda$ the quasi-Radon product
of $\tilde\mu$ and $\tilde\nu$ (417N).   Show that $\lambda$ is the
restriction of $\tilde\lambda$ to the Baire $\sigma$-algebra of
$X\times Y$.
%436F

\sqheader 436Xi For $u\in C([0,1])$ let $f(u)$ be the Lebesgue integral
of $u$.   Show that $f$ is smooth (therefore sequentially smooth) but
not sequentially order-continuous (therefore not order-continuous).
\Hint{enumerate $\Bbb Q\cap[0,1]$ as $\sequencen{q_n}$, and set
$u_n(x)=\min_{i\le n}2^{i+2}|x-q_i|$ for $n\in\Bbb N$, $x\in[0,1]$;
show that $\inf_{n\in\Bbb N}u_n=0$ in $C([0,1])$ but
$\lim_{n\to\infty}f(u_n)>0$.}
%436G

\spheader 436Xj In 436E, show that $\mu$ is $\tau$-additive iff $f$ is
smooth.
%436G

\spheader 436Xk Suppose that $X$ is a set, $U$ is a truncated Riesz
subspace of $\Bbb R^X$ and $f:U\to\Bbb R$ is a smooth positive linear
functional.   For $A\subseteq X$ set

$$\eqalign{\theta A=\inf\{\sup_{u\in B}f(u):B\text{ is}
&\text{ an upwards-directed family in }U^+\cr
&\text{ such that }\sup_{u\in B}u(x)=1\text{ for every }x\in A\},\cr}$$

\noindent taking $\inf\emptyset=\infty$ if need be.   Show that $\theta$
is an outer measure on $X$.   Let $\mu_0$ be the measure defined from
$\theta$ by \Caratheodory's method.   Show that $f(u)=\int u\,d\mu_0$
for every $u\in U$.   Show that the measure $\mu$ constructed in 436H is
the c.l.d.\ version of $\mu_0$.
%436H

\spheader 436Xl Let $X$ be a completely regular topological space and
$f$ a smooth positive linear functional on $C_b(X)$.   Show that there
is a unique totally finite quasi-Radon measure $\mu$ on $X$ such that
$f(u)=\int u\,d\mu$ for every $u\in C_b(X)$.
%436H

\sqheader 436Xm For $u\in C([0,1])$ let $f(u)$ be the Riemann integral
of $u$ (134K).   Show that the Radon measure on $[0,1]$ constructed by
the method of 436J is just Lebesgue measure on $[0,1]$.   Explain how to
construct Lebesgue measure on $\Bbb R$ from an appropriate version of
the Riemann integral on $\Bbb R$.
%436J

\spheader 436Xn Let $X$ be a topological space.   Let
$f:C_b(X)\to\Bbb R$ be a linear functional.   (i) Show that the
following are equiveridical:  ($\alpha$) $f$ is {\bf tight},
that is, for every $\epsilon>0$ there is a closed compact
$K\subseteq X$ such that $|f(u)|\le\epsilon$ whenever
$0\le u\le\chi(X\setminus K)$ ($\beta$) there is a tight totally finite
Borel measure
$\mu$ on $X$ such that $|f(u)|\le\int|u|\,d\mu$ for every $u\in C_b(X)$.
\Hint{show that a positive tight functional is smooth.}
(ii)\dvAnew{2009}
Show that the set of tight functionals on $C_b(X)$ is a band in
$C_b(X)^{\sim}$.
%436J

\sqheader 436Xo Let $(X,\frak T,\Sigma,\mu)$ and
$(Y,\frak S,\Tau,\nu)$ be locally compact Radon measure spaces.   (i)
Show that the
function $x\mapsto\int w(x,y)\nu(dy)$ belongs to $C_k(X)$ for every
$w\in C_k(X\times Y)$, so we have a
positive linear functional $h:C_k(X\times Y)\to\Bbb R$ defined by
setting

\Centerline{$h(w)=\biggeriint w(x,y)\nu(dy)\mu(dx)$}

\noindent for $w\in C_k(X\times Y)$.   (ii) Show that the corresponding
Radon measure on $X\times Y$ is just the product Radon
measure as defined in 417P/417R.

\sqheader 436Xp Let $\frak A$ be a Boolean algebra and $Z$ its Stone
space;  identify $L^{\infty}(\frak A)$ with $C(Z)$, as in 363A.   Let
$\nu$ be a non-negative finitely additive functional on $\frak A$,  $f$
the corresponding positive linear functional on $L^{\infty}(\frak A)$
(363K), and $\mu$ the corresponding Radon measure on $Z$ (416Qb).   Show
that $f(u)=\int u\,d\mu$ for every $u\in L^{\infty}(\frak A)$.
%436J/436K

\spheader 436Xq Let $X$ be a locally compact Hausdorff space.   Show
that a sequence $\sequencen{u_n}$ in $C_0(X)$ converges to $0$ for the
weak topology on $C_0(X)$ iff $\sup_{n\in\Bbb N}\|u_n\|_{\infty}$ is
finite and $\lim_{n\to\infty}u_n(x)=0$ for every $x\in X$.
%436K

\spheader 436Xr Let $X$ be a non-empty compact Hausdorff space, and
$\phi:X\to X$ a continuous function.   Show that there is a Radon
probability measure $\mu$ on $X$ such that
$\phi$ is \imp\ for $\mu$.   \Hint{let $\Cal F$ be a non-principal
ultrafilter on $X$, $x_0$ any point of $X$, and define $\mu$ by the
formula
$\int ud\mu=\lim_{n\to\Cal F}\Bover1{n+1}\sum_{k=0}^nu(\phi^k(x_0))$ for
every $u\in C(X)$.   Use 416E(b-v) to show that $\mu\phi^{-1}=\mu$.}

\spheader 436Xs Let $X$ be a locally compact Hausdorff space.
(i) Write $M^{\infty+}_{\text{R}}$ for the set of all Radon measures on
$X$.   For $\mu\in M^{\infty+}_{\text{R}}$, let $S\mu$ be the
corresponding
functional on $C_k(X)$, defined by setting $(S\mu)(u)=\int u\,d\mu$ for
every $u\in C_k(X)$.   Show that
$S(\mu+\nu)=S\mu+S\nu$ and $S(\alpha\mu)=\alpha S\mu$
whenever $\mu$, $\nu\in M^{\infty+}_{\text{R}}$ and
$\alpha\ge 0$, where addition and scalar multiplication of measures
are defined as in 234G and 234Xf.
(ii) Write $M^+_{\text{R}}$ for the set of totally finite Radon measures on
$X$.   For $\mu\in M^+_{\text{R}}$, let $T\mu$ be the
corresponding
functional on $C_b(X)$, defined by setting $(T\mu)(u)=\int u\,d\mu$ for
every $u\in C_b(X)$.   Show that
$T(\mu+\nu)=T\mu+T\nu$ and $T(\alpha\mu)=\alpha T\mu$
whenever $\mu$, $\nu\in M^+_{\text{R}}$ and $\alpha\ge 0$.
%436I 436J out of order query

\leader{436Y}{Further exercises (a)}
%\spheader 436Ya
Let $X$ be a set, $U$ a Riesz subspace of $\Bbb R^X$, and $f:U\to\Bbb R$
a sequentially smooth positive linear functional.   (i) Write
$U_{\sigma}$ for the set of functions from $X$ to $[0,\infty]$
expressible as the
supremum of a non-decreasing sequence $\sequencen{u_n}$ in $U^+$ such
that $\sup_{n\in\Bbb N}f(u_n)<\infty$.   Show that there is a functional
$f_{\sigma}:U_{\sigma}\to\coint{0,\infty}$ such that
$f_{\sigma}(u)=\sup_{n\in\Bbb N}f(u_n)$ whenever $\sequencen{u_n}$ is a
non-decreasing sequence in $U^+$ with supremum $u\in U_{\sigma}$.
(Compare 122I.)   (ii) Show that $u+v\in U_{\sigma}$ and
$f_{\sigma}(u+v)=f_{\sigma}(u)+f_{\sigma}(v)$ for all $u$, $v\in
U_{\sigma}$.   (iii) Suppose that $u$, $v\in U_{\sigma}$, $u\le v$ and
$u(x)=v(x)$ whenever $v(x)$ is finite.   Show that
$f_{\sigma}(u)=f_{\sigma}(v)$.   \Hint{take non-decreasing sequences
$\sequencen{u_n}$, $\sequencen{v_n}$ with suprema $u$, $v$.   Consider
$\sequencen{f(v_k-u_n-\delta v_n)^+}$ where $k\in\Bbb N$, $\delta>0$.}
(iv) Let $V$ be the set of functions $v:X\to\Bbb R$ such that there are
$u_1$, $u_2\in U_{\sigma}$ such that $v(x)=u_1(x)-u_2(x)$ whenever
$u_1(x)$, $u_2(x)$ are both finite.   Show that $V$ is a linear subspace
of $\Bbb R^X$ and that there is a linear functional $g:V\to\Bbb R$
defined by setting $g(v)=f_{\sigma}(u_1)-f_{\sigma}(u_2)$ whenever
$v=u_1-u_2$ in the sense of the last sentence.   (v) Show that $V$ is a
Riesz subspace of $\Bbb R^X$.   (vi) Show that if $\sequencen{v_n}$ is a
non-decreasing sequence in $V$ and $\gamma=\sup_{n\in\Bbb N}g(v_n)$ is
finite, then there is a $v\in V$ such that $g(v)=\sup_{n\in\Bbb
N}g(v\wedge v_n)=\gamma$.   (This is a version of
the {\bf Daniell integral}.)
%436D

\spheader 436Yb Develop further the theory of 436Ya, finding a version
of Lebesgue's Dominated Convergence Theorem, a concept of `negligible'
subset of $X$, and an $L$-space of equivalence classes of `integrable'
functions.
%436Ya, 436D

\spheader 436Yc Let $X$ be a countably compact topological space.
Show that every positive linear functional on $C_b(X)$ is sequentially
smooth, so corresponds to a totally finite Baire measure on $X$.
%436Xe 436E

\spheader 436Yd Let $X$ be a sequential space, $Y$ a topological space,
$\mu$ a semi-finite Baire measure on $X$ and $\nu$ a $\sigma$-finite
Baire measure on $Y$.   Let $\tilde\mu$ be the c.l.d.\ version of $\mu$.
Show that there is a semi-finite Baire measure $\lambda$ on $X\times Y$
such that

\Centerline{$\lambda W=\biggerint\nu W[\{x\}]\tilde\mu(dx)$,
\quad$\int fd\lambda=\biggeriint f(x,y)\nu(dy)\tilde\mu(dx)$}

\noindent for every Baire set $W\subseteq X\times Y$ and every
non-negative continuous function $f:X\times Y\to\Bbb R$.   Show that the
c.l.d.\ version of $\lambda$ extends the c.l.d.\ product measure of
$\mu$ and $\nu$.
%436F

\spheader 436Ye Let $X_0,\ldots,X_n$ be sequential spaces and $\mu_i$ a
totally finite Baire measure on $X_i$ for each $i$.   (i) Show that if
$f:X_0\times\ldots\times X_n\to\Bbb R$ is a bounded separately
continuous function, then

\Centerline{$\phi(f)=\biggerint\ldots\int
f(x_0,\ldots,x_n)\mu_n(dx_n)\ldots\mu_0(dx_0)$}

\noindent is defined, so that we have a corresponding Baire product
measure on $X_0\times\ldots\times X_n$.   (ii) Show that this product is
associative.
%436F

\spheader 436Yf Let $X$ and $Y$ be compact Hausdorff spaces.   (i) Show
that there is a unique bilinear operator
$\phi:C(X)^*\times C(Y)^*\to C(X\times Y)^*$ which is separately
continuous for the weak* topologies and such that
$\phi(\delta_x,\delta_y)=\delta_{(x,y)}$ for all $x\in X$ and $y\in Y$,
setting $\delta_x(f)=f(x)$ for $f\in C(X)$ and $x\in X$.   (ii) Show
that if $\mu$ and $\nu$ are Radon measures on $X$, $Y$ respectively with
Radon measure product $\lambda$, then
$\phi(\int d\mu,\int d\nu)=\int d\lambda$.   (iii) Show that
$\|\phi\|\le 1$ (definition:  253Ab).
%436Xo, 436J

\spheader 436Yg Let $X$ be any topological space.   (i) Let $C_k$ be the
set of continuous functions $u:X\to\Bbb R$ such that
$\overline{\{x:u(x)\ne 0\}}$ is compact, and $f:C_k\to\Bbb R$ a positive
linear functional.   Show that there is a tight quasi-Radon measure $\mu$
on $X$ such that $f(u)=\int u\,d\mu$ for every $u\in C_k$.
(ii) Let $\tilde C_k$ be the
set of continuous functions $u:X\to\Bbb R$ such that
$\{x:u(x)\ne 0\}$ is relatively compact, and $f:\tilde C_k\to\Bbb R$ a
positive
linear functional.   Show that there is a tight quasi-Radon measure $\mu$
on $X$ such that $f(u)=\int u\,d\mu$ for every $u\in C_k$.
%436J
}%end of exercises

\endnotes{
\Notesheader{436}
From the beginning, integration has been at the centre of measure
theory.   My own view -- implicit in the arrangement of this treatise,
from Chapter
11 onward -- is that `measure' and `integration' are not quite the same
thing.   I freely acknowledge that my treatment of `integration' is
distorted by my presentation of it as part of measure theory;  on the
other side of the argument, I hold that regarding `measure' as a concept
subsidiary to `integral', as many authors do, seriously interferes with
the development of truly penetrating intuitions for the former.   But it
is undoubtedly true that every complete locally determined measure can
be derived from its associated integral (436Xb).   Moreover, it is
clearly of the highest importance that we should be able to recognise
integrals when we see them;  I mean, given a linear functional on a
linear space of functions, then if it can be expressed as an integral
with respect to a
measure this is something we need to know at once.   And thirdly,
investigation of linear functionals frequently leads us to measures of
great importance and interest.

Concerning the conditions in 436D, an integral must surely be `positive'
(because measures in this treatise are always non-negative) and
`sequentially smooth' (because measures are supposed to be countably
additive).   But it is not clear that we are forced to restrict our
attention to Riesz subspaces of $\Bbb R^X$, and even less clear that
they have to be `truncated'.   In 439I below I give an example to show
that this last condition is essential for 436D and 436H as stated.
However it is not necessary for large parts of the theory.   In many
cases, if $U\subseteq\Bbb R^X$ is a Riesz subspace which is not
truncated, we can take an element $e\in U^+$ and look at
$Y=\{x:e(x)>0\}$,
$V_e=\{u:u(x)=0$ for every $x\in X\setminus Y\}\cong W_e$, where
$W_e=\{u/e:u\in V_e\}$ is a truncated Riesz subspace of $\BbbR^Y$.
But there are applications in
which this approach is unsatisfactory and a more radical revision of the
basic theory of integration, as in 436Ya, is useful.

I have based the arguments of this section on the inner measure
constructions of \S413.   Of course it is also possible to approach them
by means of outer measures (436Xc, 436Xk).

I emphasize the exercise 436Xo because it is prominent in `Bourbakist'
versions of the
theory of Radon measures, in which (following {\smc Bourbaki 65} rather
than {\smc Bourbaki 69}) Radon measures are regarded as linear
functionals on spaces of continuous functions.   By 436J, this is a
reasonably effective approach as long as we are interested only in
locally compact spaces, and there are parts of the theory of topological
groups (notably the duality theory of \S445 below) in which it even has
advantages.   The construction of 436Xo shows that we can find a direct
approach to the tensor product of linear functionals which does not
require any attempt to measure sets rather than integrate functions.
I trust that the prejudices I am
expressing will not be taken as too sweeping a disparagement of
such methods.   Practically all correct arguments in mathematics (and
not a few incorrect ones) are valuable in some way, suggesting new
possibilities for investigation.   In particular, one of the challenges
of measure theory (not faced in this treatise) is that of devising
effective theories of vector-valued measures.   Typically this is much
easier with Riemann-type integrals, and any techniques for working
directly with these should be noted.

436L revisits ideas from Chapter 35, and the result would be easier to
find if it were in \S356.   I include it here as an example of the way
in which familiar material from measure theory (in particular, the
Radon-Nikod\'ym theorem) can be drafted to serve functional analysis.
I should perhaps remark that there are alternative routes which do not
use measure theory explicitly, and while longer are (in my view) more
illuminating.
}%end of notes

\discrpage


\frfilename{mt437.tex}
\versiondate{5.11.12}
\copyrightdate{2011}

\def\frakTKR{\frak T_{\text{KR}}}
\def\MqR{M^+_{\text{qR}}}
\def\MR{M^+_{\text{R}}}
\def\PqR{P_{\text{qR}}}
\def\PR{P_{\text{R}}}
\def\sigmaKR{\sigma_{\text{KR}}}

\def\chaptername{Topologies and measures II}
\def\sectionname{Spaces of measures}

\newsection{437}

Once we have started to take the correspondence between measures and
integrals as something which operates in both directions, we can go a
very long way.   While `measures', as dealt with in this treatise, are
essentially positive, an `integral' can be thought of as a member of a
linear space, dual in some sense to a space of functions.   Since the
principal spaces of functions are Riesz spaces, we find ourselves
looking at dual Riesz spaces as discussed in \S356;  while the
corresponding spaces of measures are close to those of \S362.   Here I
try to draw these ideas together with an
examination of spaces $U^{\sim}_{\sigma}$ and $U^{\sim}_{\tau}$ of
sequentially smooth and smooth functionals, and the matching spaces
$M_{\sigma}$ and $M_{\tau}$ of countably additive and $\tau$-additive
measures (437A-437I).   Because a (sequentially) smooth functional
corresponds to a countably additive measure, which can be expected to
integrate many more functions than those in the original Riesz space
(typically, a space of continuous functions), we find that relatively
large spaces of bounded measurable functions can be canonically embedded
into the biduals $(U^{\sim}_{\sigma})^*$ and $(U^{\sim}_{\tau})^*$
(437C, 437H, 437I).

The guiding principles of functional analysis encourage us not only to
form linear spaces, but also to examine linear space topologies,
starting with norm and weak topologies.   The theory of Banach lattices
described in \S354, particularly the theory of $M$- and $L$-spaces, is
an important part of the structure here.   In addition, our spaces
$U^{\sim}_{\sigma}$ have natural weak* topologies which can be
regarded as topologies on spaces of measures;  these are the `vague'
topologies of 437J, which have already been considered, in a
special case, in \S285.

It turns out that on the positive cone of $M_{\tau}$, at least, the
vague topology may have an alternative description directly in terms of
the behaviour of the measures on
open sets (437L).   This leads us to a parallel idea, the `narrow'
topology on non-negative additive functionals (437Jd).   The second half
of the section
is devoted to the elementary properties of narrow topologies
(437K-437N), %437K 437L 437M 437P 437R 437N
with especial reference to compact sets in these topologies (437P, 437Rf,
437T).
Seeking to identify narrowly compact sets, we come to the concept of
`uniform tightness' (437O).
Bounded uniformly tight sets are narrowly relatively compact (437P);  in
`Prokhorov spaces' (437U) the converse is true.  I end the section with
a list of the best-known Prokhorov spaces (437V).

\leader{437A}{Smooth and sequentially smooth duals} Let $X$ be a
set, and $U$ a Riesz subspace of $\Bbb R^X$.   \cmmnt{Recall that
$U^{\sim}$ is the Dedekind complete Riesz space of order-bounded linear
functionals on $U$, that $U^{\sim}_c$ is the band of differences of
sequentially order-continuous positive linear functionals, and that
$U^{\times}$ is the band of differences of
order-continuous positive linear functionals (356A).   A functional
$f\in(U^{\sim})^+$ is `sequentially smooth' if
$\inf_{n\in\Bbb N}f(u_n)=0$ whenever $\sequencen{u_n}$
is a non-increasing sequence in $U$ and $\lim_{n\to\infty}u_n(x)=0$ for
every $x\in X$, and `smooth' if $\inf_{u\in A}f(u)=0$ whenever
$A\subseteq U$ is a non-empty
downwards-directed set and $\inf_{u\in A}u(x)=0$ for every $x\in X$
(436A, 436G).}

\spheader 437Aa Set $U^{\sim}_{\sigma}=\{f:f\in U^{\sim}$, $|f|$ is
sequentially smooth$\}$, the {\bf sequentially smooth dual} of $U$.
Then $U^{\sim}_{\sigma}$ is a band in $U^{\sim}$.   \prooflet{\Prf\ (i)
If $f\in U^{\sim}_{\sigma}$, $g\in U^{\sim}$ and $|g|\le|f|$, then

\Centerline{$|g|(u_n)\le|f|(u_n)\to 0$}

\noindent whenever $\sequencen{u_n}$ is a non-increasing sequence in $U$
and $\lim_{n\to\infty}u_n(x)=0$ for every $x$, so $|g|$ is sequentially
smooth and $g\in U^{\sim}_{\sigma}$.   Thus $U^{\sim}_{\sigma}$ is a
solid subset of $U^{\sim}$.   (ii) If $f$, $g\in U^{\sim}_{\sigma}$ and
$\alpha\in\Bbb R$, then

\Centerline{$|f+g|(u_n)\le|f|(u_n)+|g|(u_n)\to 0$,
\quad$|\alpha f|(u_n)=|\alpha||f|(u_n)\to 0$}

\noindent whenever $\sequencen{u_n}$ is a non-increasing sequence in $U$
and $\lim_{n\to\infty}u_n(x)=0$ for every $x$, so $|f+g|$ and
$|\alpha f|$ are sequentially smooth.   Thus $U^{\sim}_{\sigma}$ is a
Riesz subspace of $U^{\sim}$.   (iii) Now suppose that
$B\subseteq(U^{\sim}_{\sigma})^+$ is an upwards-directed set with
supremum $g\in U^{\sim}$, and that $\sequencen{u_n}$ is a non-increasing
sequence in
$U$ such that $\lim_{n\to\infty}u_n(x)=0$ for every $x\in X$.   Then,
given $\epsilon>0$, there is an $f\in B$ such that
$(g-f)(u_0)\le\epsilon$ (355Ed), so that $g(u_n)\le f(u_n)+\epsilon$
for every $n$, and

\Centerline{$\limsup_{n\to\infty}g(u_n)
\le\epsilon+\lim_{n\to\infty}f(u_n)\le\epsilon$.}

\noindent As $\epsilon$ and $\sequencen{u_n}$ are arbitrary,
$g\in U^{\sim}_{\sigma}$;  as $B$ is arbitrary, $U^{\sim}_{\sigma}$ is a
band (352Ob).\ \Qed
}%end of prooflet

\cmmnt{As remarked in 436A, sequentially order-continuous positive
linear functionals are sequentially smooth,
so} $U^{\sim}_c\subseteq U^{\sim}_{\sigma}$.

\spheader 437Ab Set $U^{\sim}_{\tau}=\{f:f\in U^{\sim}$, $|f|$ is
smooth$\}$, the {\bf smooth dual} of $U$.   Then $U^{\sim}_{\tau}$ is a
band in $U^{\sim}$.   \prooflet{\Prf\ (i) Suppose that $f$,
$g\in U^{\sim}_{\tau}$, $\alpha\in\Bbb R$, $h\in U^{\sim}$ and
$|h|\le|f|$.   If $A\subseteq U$ is a non-empty downwards-directed set
and $\inf_{u\in A}u(x)=0$ for every $x\in X$, and $\epsilon>0$, then
there are $u_0$,
$u_1\in A$ such that $|f|(u_0)\le\epsilon$ and $|g|(u_1)\le\epsilon$,
and a $u\in A$ such that $u\le u_0\wedge u_1$.   In this case

\Centerline{$|h|(u)\le|f|(u)\le\epsilon$,}

\Centerline{$|f+g|(u)\le|f|(u)+|g|(u)\le 2\epsilon$,}

\Centerline{$|\alpha f|(u)=|\alpha||f|(u)\le|\alpha|\epsilon$.}

\noindent As $A$ and $\epsilon$ are arbitrary, $h$, $f+g$ and $\alpha f$
all belong to $U^{\sim}_{\tau}$;  so that $U^{\sim}_{\tau}$ is a solid
Riesz subspace of $U^{\sim}$.   (ii) Now suppose that $B\subseteq
(U^{\sim}_{\tau})^+$ is an upwards-directed set with supremum $g\in
U^{\sim}$, and that $A\subseteq U$ is a non-empty downwards-directed set
such that $\inf_{u\in A}u(x)=0$ for every $x\in X$.   Fix any $u_0\in
A$.   Then, given $\epsilon>0$, there is an $f\in B$ such that
$(g-f)(u_0)\le\epsilon$,
so that $g(u)\le f(u)+\epsilon$ whenever $u\in A$
and $u\le u_0$.   But $A_0=\{u:u\in A,\,u\le u_0\}$ is also a
downwards-directed set,
and $\inf_{u\in A_0}u(x)=0$ for every $x\in X$, so

\Centerline{$\inf_{u\in A}g(u)
\le\epsilon+\inf_{u\in A_0}f(u)\le\epsilon$.}

\noindent As $\epsilon$ and $A$ are arbitrary, $g\in U^{\sim}_{\tau}$;
as $B$ is arbitrary, $U^{\sim}_{\tau}$ is a band.\ \Qed
}%end of prooflet

\cmmnt{Just as $U^{\sim}_c\subseteq U^{\sim}_{\sigma}$,}
$U^{\times}\subseteq U^{\sim}_{\tau}$.

\leader{437B}{Signed measures}\cmmnt{ Collecting these ideas together
with those of \S\S362-363, we are ready to approach `signed measures'.}
Recall that if $X$ is a set and $\Sigma$ is a $\sigma$-algebra of
subsets of
$X$, we can identify $L^{\infty}=L^{\infty}(\Sigma)$\cmmnt{, as
defined in \S363,} with the space
$\eusm L^{\infty}=\eusm L^{\infty}(\Sigma)$ of bounded
$\Sigma$-measurable real-valued functions\cmmnt{ (363H)}.
\cmmnt{Now, because $\eusm L^{\infty}$ is sequentially order-closed in
$\Bbb R^X$, sequentially smooth functionals on $\eusm L^{\infty}$ are
actually sequentially order-continuous,
so }$(\eusm L^{\infty})^{\sim}_{\sigma}=(\eusm L^{\infty})^{\sim}_c$.
Next, we can
identify $(L^{\infty})^{\sim}_c$ with the space $M_{\sigma}$ of
countably additive functionals\cmmnt{, or `signed measures',} on
$\Sigma$\cmmnt{ (363K)};  if $\nu\in M_{\sigma}$, the corresponding
member of $(L^{\infty})^{\sim}_c$ is the unique
order-bounded\cmmnt{ (or norm-continuous)} linear functional $f$ on
$L^{\infty}$ such that
$f(\chi E)=\nu E$ for every $E\in\Sigma$.   \cmmnt{If $\nu\ge 0$, so
that $\nu$
is a totally finite measure with domain $\Sigma$, then of course
$f$, when interpreted as a functional on $\eusm L^{\infty}$, must be
just integration with respect to $\nu$.}

The identification between $(L^{\infty})^{\sim}_c$ and
$M_{\sigma}$\cmmnt{ described in 363K} is an $L$-space isomorphism.
So\cmmnt{ it tells us, for instance, that if we are willing to use the
symbol $\dashint$ for the duality between $L^{\infty}$ and the space of
bounded finitely additive functionals on $\Sigma$, as in 363L, then we
can write}

\Centerline{$\dashint u\,d(\mu+\nu)=\dashint u\,d\mu+\dashint u\,d\nu$}

\noindent for every $u\in\eusm L^{\infty}$ and all $\mu$,
$\nu\in M_{\sigma}$.

\leader{437C}{Theorem} Let $X$ be a set and $U$ a Riesz subspace of
$\ell^{\infty}(X)$\cmmnt{, the $M$-space of bounded real-valued
functions on $X$,} containing the constant functions.

(a) Let $\Sigma$ be the smallest $\sigma$-algebra of subsets of $X$ with
respect to which every member of $U$ is measurable.   Let
$M_{\sigma}=M_{\sigma}(\Sigma)$ be the $L$-space of countably additive
functionals on $\Sigma$\cmmnt{ (326I\formerly{3{}26E}, 362B)}.
Then there is a Banach lattice
isomorphism $T:M_{\sigma}\to U^{\sim}_{\sigma}$ defined by saying that
$(T\mu)(u)=\int u\,d\mu$ whenever $\mu\in M_{\sigma}^+$ and $u\in U$.

(b) We\cmmnt{ now} have
a sequentially order-continuous norm-preserving Riesz
homomorphism $S$, embedding the
$M$-space $\eusm L^{\infty}=\eusm L^{\infty}(\Sigma)$ of bounded
real-valued
$\Sigma$-measurable functions on $X$\cmmnt{ (363Hb)} into the
$M$-space $(U^{\sim}_{\sigma})^{\sim}=(U^{\sim}_{\sigma})^*
=(U^{\sim}_{\sigma})^{\times}$, defined by saying that
$(Sv)(T\mu)=\int v\,d\mu$ whenever $v\in\eusm L^{\infty}$ and
$\mu\in M_{\sigma}^+$.   If
$u\in U$, then $(Su)(f)=f(u)$ for every $f\in U^{\sim}_{\sigma}$.

\proof{{\bf (a)(i)} The norm $\|\,\|_{\infty}$ is an order-unit norm on
$U$ (354Ga), so $U^*=U^{\sim}$ is an $L$-space (356N), and the band
$U^{\sim}_{\sigma}$ (437Aa) is an $L$-space in its own right (354O).

\medskip

\quad{\bf (ii)} As noted in 437B, we have a Banach lattice isomorphism
$T_0:M_{\sigma}\to(\eusm L^{\infty})^{\sim}_c$ defined by saying that
$(T_0\mu)(u)=\int u\,d\mu$ whenever $u\in\eusm L^{\infty}$ and
$\mu\in M_{\sigma}^+$.   If we set $T\mu=T_0\mu\restr U$, then $T$
is a positive linear operator from $M_{\sigma}$ to $U^{\sim}$, just
because $U$ is a linear subspace of $\eusm L^{\infty}$;  and since
$T\mu\in U^{\sim}_{\sigma}$ for every $\mu\in M_{\sigma}^+$, $T$ is an
operator from $M_{\sigma}$ to $U^{\sim}_{\sigma}$.   Now every
$f\in(U^{\sim}_{\sigma})^+$ is of the form $T\mu$ for some
$\mu\in M_{\sigma}^+$.   \Prf\ By 436D, there is some measure $\lambda$
such that $\int u\,d\lambda=f(u)$ for every $u\in U$.   Completing
$\lambda$ if necessary, we see that we may suppose that every member of
$U$ is $(\dom\lambda)$-measurable, that is, that
$\Sigma\subseteq\dom\lambda$;  take $\mu=\lambda\restr\Sigma$.\ \QeD\
So $T$ is surjective.

\medskip

\quad{\bf (iii)} Write $\Cal K$ for the family of sets $K\subseteq X$
such that $\chi K=\inf_{n\in\Bbb N}u_n$ for some sequence
$\sequencen{u_n}$ in $U$.   (See the proof of 436D.)   We need to know
the following.
($\alpha$) $\Cal K\subseteq\Sigma$.   ($\beta$) If $K\in\Cal K$, then
there is a non-increasing sequence $\sequencen{u_n}$ in $U$ such that
$\chi K=\inf_{n\in\Bbb N}u_n$.   (For if $\sequencen{u'_n}$ is any
sequence in $U$ such that $\chi K=\inf_{n\in\Bbb N}u'_n$, we can set
$u_n=\inf_{i\le n}u'_i$ for each $i$.)   ($\gamma$) The $\sigma$-algebra
of subsets of $X$ generated by $\Cal K$ is $\Sigma$.   \Prf\ Let $\Tau$
be the $\sigma$-algebra of subsets of $X$ generated by $\Cal K$.
$\Tau\subseteq\Sigma$ because $\Cal K\subseteq\Sigma$.   If $u\in U$ and
$\alpha>0$ then $\{x:u(x)\ge\alpha\}\in\Cal K$ (see part (b) of the
proof of 436D).   So every member of $U^+$, therefore every member of
$U$, is $\Tau$-measurable, and $\Sigma\subseteq\Tau$.\ \Qed

\medskip

\quad{\bf (iv)} $T$ is injective.   \Prf\ If $\mu_1$,
$\mu_2\in M_{\sigma}$ and $T\mu_1=T\mu_2$, set
$\nu_i=\mu_i+\mu_1^-+\mu_2^-$ for
each $i$, so that $\nu_i$ is non-negative and $T\nu_1=T\nu_2$.   If
$K\in\Cal K$ then there is a non-increasing sequence $\sequencen{u_n}$
in $U$ such that $\chi K=\inf_{n\in\Bbb N}u_n$ in $\Bbb R^X$, so

\Centerline{$\nu_1K=\inf_{n\in\Bbb N}\biggerint u_nd\nu_1
=\inf_{n\in\Bbb N}\int u_nd\nu_2
=\nu_2K$.}

\noindent Now $\Cal K$ contains $X$ and is closed under finite
intersections and $\nu_1$ and $\nu_2$ agree on $\Cal K$.   By the
Monotone Class Theorem (136C), $\nu_1$ and $\nu_2$ agree on the
$\sigma$-algebra generated by $\Cal K$, which is $\Sigma$;  so
$\nu_1=\nu_2$ and $\mu_1=\mu_2$.\ \Qed

Thus $T$ is a linear space isomorphism between $M_{\sigma}$ and
$U^{\sim}_{\sigma}$.

\medskip

\quad{\bf (v)} As noted in (ii),
$T[M_{\sigma}^+]=(U^{\sim}_{\sigma})^+$;  so $T$ is a Riesz space
isomorphism.

\medskip

\quad{\bf (vi)} Now if $\mu\in M_{\sigma}$,

$$\eqalignno{\|T\mu\|
&=|T\mu|(\chi X)\cr
\noalign{\noindent (356Nb)}
&=(T|\mu|)(\chi X)\cr
\noalign{\noindent (because $T$ is a Riesz homomorphism)}
&=|\mu|(X)
=\|\mu\|\cr}$$

\noindent (362Ba).   So $T$ is norm-preserving and is an $L$-space
isomorphism, as claimed.

\medskip

{\bf (b)(i)} By 356Pb, $(U^{\sim}_{\sigma})^*=(U^{\sim}_{\sigma})^{\sim}
=(U^{\sim}_{\sigma})^{\times}$ is an $M$-space.

\medskip

\quad{\bf (ii)} We have a canonical map
$S_0:\eusm L^{\infty}\to((\eusm L^{\infty})^{\sim}_c)^{\times}$ defined
by saying that $(S_0v)(h)=h(v)$
for every $v\in\eusm L^{\infty}$ and $h\in(\eusm L^{\infty})^{\sim}_c$;
and by 356F, $S_0$ is a Riesz homomorphism.   If $\sequencen{v_n}$ is a
non-increasing sequence in $\eusm L^{\infty}$ with infimum $0$, then
$\inf_{n\in\Bbb N}(S_0v_n)(h)=\inf_{n\in\Bbb N}h(v_n)=0$ for every
$h\in((\eusm L^{\infty})^{\sim}_c)^+$, so $\inf_{n\in\Bbb N}S_0v_n=0$
(355Ee);  as $\sequencen{v_n}$ is arbitrary, $S_0$ is sequentially
order-continuous (351Gb).

Also $S_0$ is norm-preserving.   \Prf\ ($\alpha$) If
$h\in(\eusm L^{\infty})^{\sim}_c$ and $v\in\eusm L^{\infty}$, then

\Centerline{$|(S_0v)(h)|=|h(v)|\le\|h\|\|v\|_{\infty}$,}

\noindent so $\|S_0v\|\le\|v\|_{\infty}$.   ($\beta$)
If $v\in\eusm L^{\infty}$ and $0\le\gamma<\|v\|_{\infty}$, take $x\in X$
such that $|v(x)|>\gamma$, and define $h_x\in(\eusm
L^{\infty})^{\sim}_c$ by setting $h_x(w)=w(x)$ for every
$w\in\eusm L^{\infty}$;  then $\|h_x\|=1$ and

\Centerline{$|(S_0v)(h_x)|=|h_x(v)|=|v(x)|\ge\gamma$,}

\noindent so $\|S_0v\|\ge\gamma$.   As $\gamma$ is arbitrary,
$\|S_0v\|\ge\|v\|_{\infty}$ and $\|S_0v\|=\|v\|_{\infty}$.\ \Qed

\medskip

\quad{\bf (iii)} Now $T_0:M_{\sigma}\to(\eusm L^{\infty})^{\sim}_c$ and
$T:M_{\sigma}\to U^{\sim}_{\sigma}$ are both norm-preserving Riesz space
isomorphisms, so $T_0T^{-1}:U^{\sim}_{\sigma}\to(\eusm
L^{\infty})^{\sim}_c$ is another, and its adjoint $S_1:((\eusm
L^{\infty})^{\sim}_c)^*\to(U^{\sim}_{\sigma})^*$ must also be a
norm-preserving Riesz space isomorphism.   So if we set $S=S_1S_0$, $S$
will be a norm-preserving sequentially order-continuous Riesz
homomorphism from $\eusm L^{\infty}$ to
$(U^{\sim}_{\sigma})^{\times}=(U^{\sim}_{\sigma})^*$.

\medskip

\quad{\bf (iv)} Setting the construction out in this way tells us a lot
about the properties of the operator $S$, but undeniably leaves it
somewhat obscure.   So let us start again from $v\in\eusm L^{\infty}$
and $\mu\in M_{\sigma}^+$, and seek to calculate
$(Sv)(T\mu)$.   We have

$$\eqalignno{(Sv)(T\mu)
&=(S_1S_0v)(T\mu)
=(S_0v)(T_0T^{-1}T\mu)\cr
\noalign{\noindent (because $S_1$ is the adjoint of $T_0T^{-1}$)}
&=(S_0v)(T_0\mu)
=(T_0\mu)(v)
=\int v\,d\mu,\cr}$$

\noindent as claimed.

If $u\in U$, then $(T\mu)(u)=(T_0\mu)(u)$ for every $\mu\in M_{\sigma}$,
so if $f\in U^{\sim}_{\sigma}$ then

\Centerline{$(Su)(f)
=(S_1S_0u)(f)
=(S_0u)(T_0T^{-1}f)
=(T_0T^{-1}f)(u)
=(TT^{-1}f)(u)
=f(u)$.}

\noindent This completes the proof.
}%end of proof of 437C

\cmmnt{
\leader{437D}{Remarks} What is happening here is that the
canonical Riesz homomorphism $u\mapsto\hat u$ from $U$ to
$(U^{\sim}_{\sigma})^*$ (356F) has a natural extension to
$\eusm L^{\infty}(\Sigma)$.   The original homomorphism $u\mapsto\hat u$
is not, as a rule, sequentially order-continuous, just because
$U^{\sim}_{\sigma}$ is generally larger than $U^{\sim}_c$;  but the
extension to $\eusm L^{\infty}$ is sequentially order-continuous.   If
you like, it is sequential smoothness which is carried over to the
extension, and because the embedding of $\eusm L^{\infty}$ in $\Bbb R^X$
is sequentially order-continuous, a sequentially smooth operator on
$\eusm L^{\infty}$ is sequentially order-continuous.

\cmmnt{In the statement of 437C, I have used the formulae
$(T\mu)(u)=\int u\,d\mu$ and $(Sv)(T\mu)=\int v\,d\mu$ on the assumption
that $\mu\in M_{\sigma}^+$, so that $\mu$ is actually a measure on the
definition used in this treatise, and $\int\,d\mu$ is the ordinary
integral as constructed in \S122.   Since the functions $u$ and $v$ are
bounded, measurable and defined everywhere, we can choose to extend the
notion of integration to signed measures, as in 363L, in which case the
formulae $(T\mu)(u)=\dashint u\,d\mu$ and $(Sv)(T\mu)=\dashint v\,d\mu$
become meaningful, and true, for all $\mu\in M_{\sigma}$, $u\in U$ and
$v\in\eusm L^{\infty}$.}

In fact the ideas here can be pushed farther, as in 437Ib, 437Xf and
437Yd.
}%end of comment

\leader{437E}{Corollary} Let $X$ be a completely regular Hausdorff
space, and $\CalBa\cmmnt{\mskip5mu=\CalBa(X)}$ its Baire $\sigma$-algebra.
Then we can identify
$C_b(X)^{\sim}_{\sigma}$ with the $L$-space $M_{\sigma}(\CalBa)$ of
countably additive functionals on $\CalBa$, and we have a
norm-preserving sequentially order-continuous Riesz homomorphism $S$ from
$\eusm L^{\infty}(\CalBa)$ to $(C_b(X)^{\sim}_{\sigma})^*$ defined by
setting $(Sv)(f)=\int v\,d\mu_f$ for every $v\in\eusm L^{\infty}$ and
$f\in(C_b(X)^{\sim}_{\sigma})^+$, where $\mu_f$ is the Baire measure
associated with $f$.

\proof{ Apply 437C with $U=C_b(X)$ (cf.\ 436E).}

\leader{437F}{Proposition} Let $X$ be a topological space and
$\Cal B\cmmnt{\mskip5mu=\Cal B(X)}$ its
Borel $\sigma$-algebra.   Let $M_{\sigma}$ be the $L$-space of countably
additive functionals on $\Cal B$.

(a) Write $M_{\tau}\subseteq M_{\sigma}$ for the set of differences of
$\tau$-additive totally finite Borel measures.   Then $M_{\tau}$ is a
band in $M_{\sigma}$, so is an $L$-space in its own right.

(b) Write $M_t\subseteq M_{\tau}$ for the set of
differences of totally finite Borel measures which are
tight\cmmnt{ (that is, inner regular with respect to the
closed compact sets)}.   Then $M_t$ is a band in $M_{\sigma}$, so is an
$L$-space in its own right.

\proof{{\bf (a)(i)} Let $\mu_1$, $\mu_2$ be totally finite
$\tau$-additive
Borel measures on $X$, $\alpha\ge 0$, and $\mu\in M_{\sigma}$ such that
$0\le\mu\le\mu_1$.   Then $\mu_1+\mu_2$, $\alpha\mu_1$ and $\mu$ are
totally finite $\tau$-additive Borel measures.   \Prf\ They all belong
to $M_{\sigma}$, that is, are totally finite Borel measures.
Now let $\Cal G$ be a
non-empty upwards-directed family of open sets in $X$ with union $H$,
and $\epsilon>0$.   Then there are $G_1$, $G_2\in\Cal G$ such that
$\mu_1G_1\ge\mu_1H-\epsilon$ and $\mu_2G_2\ge\mu_2H-\epsilon$, and a
$G\in\Cal G$ such that $G\supseteq G_1\cup G_2$.   In this case,

\Centerline{$(\mu_1+\mu_2)(G)\ge(\mu_1+\mu_2)(H)-2\epsilon$,}

\Centerline{$(\alpha\mu_1)(G)\ge(\alpha\mu_1)(H)-\alpha\epsilon$}

\noindent and

\Centerline{$\mu G=\mu H-\mu(H\setminus G)\ge\mu H-\mu_1(H\setminus G)
\ge\mu H-\epsilon$.}

\noindent As $\Cal G$ and $\epsilon$ are arbitrary, $\mu_1+\mu_2$,
$\alpha\mu_1$ and $\mu$ are all $\tau$-additive.\ \Qed

It follows that $M_{\tau}$ is a solid linear subspace of $M_{\sigma}$.

\medskip

\quad{\bf (ii)} Now suppose that $B\subseteq M_{\tau}^+$ is non-empty
and upwards-directed and has a supremum $\nu$ in $M_{\sigma}$.   Then
$\nu\in M_{\tau}$.   \Prf\  If $\Cal G$ is a non-empty
upwards-directed family of open sets with union $H$, then

$$\eqalignno{\nu H
&=\sup_{\mu\in B}\mu H\cr
\noalign{\noindent (362Be)}
&=\sup_{\mu\in B,G\in\Cal G}\mu G
=\sup_{G\in\Cal G}\nu G;\cr}$$

\noindent as $\Cal G$ is arbitrary, $\nu$ is $\tau$-additive and belongs
to $M_{\tau}$.\ \Qed

As $B$ is arbitrary, $M_{\tau}$ is a band in $M_{\sigma}$.   By 354O, it
is itself an $L$-space.

\medskip

{\bf (b)} We can use the same arguments.   Suppose that $\mu_1$,
$\mu_2\in M_{\sigma}^+$ are tight,
$\alpha\ge 0$, and $\mu\in M_{\sigma}$ is such that
$0\le\mu\le\mu_1$.   If $E\in\Cal B$ and $\epsilon>0$, there are
closed compact sets $K_1$, $K_2\subseteq E$ such that
$\mu_1K_1\ge\mu_1E-\epsilon$ and $\mu_2K_2\ge\mu_2E-\epsilon$.   Set
$K=K_1\cup K_2$, so that $K$ also is a closed compact subset of $E$.
Then

\Centerline{$(\mu_1+\mu_2)(K)\ge(\mu_1+\mu_2)(E)-2\epsilon$,}

\Centerline{$(\alpha\mu_1)(K)\ge(\alpha\mu_1)(E)-\alpha\epsilon$}

\noindent and

\Centerline{$\mu K=\mu E-\mu(E\setminus K)\ge\mu E-\mu_1(E\setminus K)
\ge\mu E-\epsilon$.}

\noindent As $\Cal G$ and $\epsilon$ are arbitrary, $\mu_1+\mu_2$,
$\alpha\mu_1$ and $\mu$ are all tight;  as $\mu_1$, $\mu_2$, $\mu$ and
$\alpha$ are arbitrary, $M_t$ is a solid linear subspace of $M_{\sigma}$.

Now suppose that $B\subseteq M_t^+$ is non-empty and
upwards-directed and has a supremum $\nu$ in $M_{\sigma}$.   Take any
$E\in\Cal B$ and $\epsilon>0$.   Then there is a $\mu\in B$ such that
$\mu E\ge\nu E-\epsilon$;  there is a closed compact set $K\subseteq E$
such that $\mu K\ge\mu E-\epsilon$;  and now $\nu K\ge\nu E-2\epsilon$.
As $E$ and $\epsilon$ are arbitrary, $\nu$ is tight;  as $B$ is
arbitrary, $M_t$ is a band in $M_{\sigma}$, and is in itself an
$L$-space.
}%end of proof of 437F

\leader{437G}{Definitions} Let $X$ be a topological space.   A {\bf
signed Baire measure} on $X$ will be a countably additive functional on
the Baire
$\sigma$-algebra $\CalBa(X)$\cmmnt{, which by the Jordan decomposition
theorem (231F) is expressible as the difference of two totally finite
Baire measures};
a {\bf signed Borel measure} will be a countably additive functional on
the Borel $\sigma$-algebra $\Cal B(X)$\cmmnt{, that is, the difference
of two totally finite
Borel measures};  a {\bf signed $\tau$-additive Borel measure} will
be\cmmnt{ a member of the $L$-space $M_{\tau}$ as described in 437F,
that is,} the difference of two $\tau$-additive totally finite
Borel measures;  and a {\bf signed
tight Borel measure} will be\cmmnt{ a member of the $L$-space $M_t$ as
described in 437F, that is,}
the difference of two tight totally finite Borel measures.

\leader{437H}{Theorem} Let $X$ be a set and $U$ a Riesz subspace of
$\ell^{\infty}(X)$ containing the constant functions.   Let $\frak T$ be
the coarsest topology on $X$ rendering every member of $U$ continuous,
and $\Cal B$ the corresponding Borel $\sigma$-algebra.

(a) Let $M_{\tau}$ be the $L$-space of signed
$\tau$-additive Borel measures on $X$.   Then we have a Banach
lattice isomorphism $T:M_{\tau}\to U^{\sim}_{\tau}$ defined by saying
that $(T\mu)(u)=\int u\,d\mu$ whenever $\mu\in M_{\tau}^+$ and $u\in U$.

(b) We\cmmnt{ now} have
a sequentially order-continuous norm-preserving Riesz
homomorphism $S$, embedding\cmmnt{ the
$M$-space} $\cmmnt{\eusm L^{\infty}=\mskip5mu}\eusm L^{\infty}(\Cal
B)$\cmmnt{ of bounded Borel
measurable functions on $X$} into
$(U^{\sim}_{\tau})^{\sim}=(U^{\sim}_{\tau})^*
=(U^{\sim}_{\tau})^{\times}$, defined by saying that
$(Sv)(T\mu)=\int v\,d\mu$ whenever $v\in\eusm L^{\infty}$ and
$\mu\in M_{\tau}^+$.   If
$u\in U$, then $(Su)(f)=f(u)$ for every $f\in U^{\sim}_{\tau}$.

\proof{ The proof follows the same lines as that of 437C.

\medskip

{\bf (a)(i)} As before, the norm $\|\,\|_{\infty}$ is an order-unit norm
on $U$, so $U^*=U^{\sim}$ is an $L$-space, and the band
$U^{\sim}_{\tau}$ (437Ab) is an $L$-space in its own right, like
$M_{\tau}$ (437F).

\medskip

\quad{\bf (ii)} Let $M_{\sigma}$ be the $L$-space of all countably
additive functionals on $\Cal B$, so that $M_{\tau}$ is a band in
$M_{\sigma}$.   Let $T_0:M_{\sigma}\to(\eusm L^{\infty})^{\sim}_c$ be
the canonical Banach lattice isomorphism defined by saying that
$(T_0\mu)(u)=\int u\,d\mu$ whenever $u\in\eusm L^{\infty}$ and
$\mu\in M_{\sigma}^+$.   If we set
$T\mu=T_0\mu\restr U$ for $\mu\in M_{\tau}$,
then $T$ is a positive linear operator from $M_{\tau}$ to $U^{\sim}$,
just because $U$ is a Riesz subspace of $\eusm L^{\infty}$;  and since
$T\mu\in U^{\sim}_{\tau}$ for every $\mu\in M_{\tau}^+$ (436H), $T$ is
an operator from $M_{\tau}$ to $U^{\sim}_{\tau}$.   Now every
$f\in(U^{\sim}_{\tau})^+$ is of the form $T\mu$ for some
$\mu\in M_{\tau}^+$.
\Prf\ By 436H, there is a quasi-Radon measure $\lambda$
such that $\int u\,d\lambda=f(u)$ for every $u\in U$;  set
$\mu=\lambda\restr\Cal B$.\ \QeD\  So $T$ is surjective.

\medskip

\quad{\bf (iii)} Let $\Cal K$ be the family of subsets $K$ of $X$ such
that $\chi K=\inf A$ in $\Bbb R^X$ for some non-empty subset $A$ of $U$.
Then $\Cal K$ is just the family of closed sets for $\frak T$.   \Prf\
As noted in part (b) of the proof of 436H, every member of $\Cal K$ is
closed, and $K\setminus G\in\Cal K$ whenever $K\in\Cal K$ and
$G\in\frak T$;  but as, in the present case, $X\in\Cal K$, every closed
set belongs to $\Cal K$.\ \Qed

\medskip

\quad{\bf (iv)} $T$ is injective.   \Prf\ If $\mu_1$, $\mu_2\in
M_{\tau}$ and $T\mu_1=T\mu_2$, set $\nu_i=\mu_i+\mu_1^-+\mu_2^-$ for
each $i$, so that $\nu_i$ is non-negative and $T\nu_1=T\nu_2$.   If
$K\in\Cal K$, set $A=\{u:u\in U,\,u\ge\chi K\}$, so that $\chi
K=\inf A$ in $\Bbb R^X$, and $A$ is downwards-directed.   By 414Bb,

\Centerline{$\nu_1K=\inf_{u\in A}\int u\,d\nu_1
=\inf_{u\in A}\int u\,d\nu_2=\nu_2K$.}

\noindent Now $\Cal K$ contains $X$ and is closed under finite
intersections and $\nu_1$ and $\nu_2$ agree on $\Cal K$.   By the
Monotone Class Theorem, $\nu_1$ and $\nu_2$ agree on the
$\sigma$-algebra generated by $\Cal K$, which is $\Cal B$;  so
$\nu_1=\nu_2$ and $\mu_1=\mu_2$.\ \Qed

Thus $T$ is a linear space isomorphism between $M_{\tau}$ and
$U^{\sim}_{\tau}$.

\medskip

\quad{\bf (v)} As noted in (ii), $T[M_{\tau}^+]=(U^{\sim}_{\tau})^+$;
so $T$ is a Riesz space isomorphism.

\medskip

\quad{\bf (vi)} Now if $\mu\in M_{\tau}$,

\Centerline{$\|T\mu\|
=|T\mu|(\chi X)
=(T|\mu|)(\chi X)
=|\mu|(X)
=\|\mu\|$.}

\noindent So $T$ is norm-preserving and is an $L$-space isomorphism, as
claimed.

\medskip

{\bf (b)(i)} By 356Pb, $(U^{\sim}_{\tau})^*=(U^{\sim}_{\tau})^{\sim}
=(U^{\sim}_{\tau})^{\times}$ is an $M$-space.

\medskip

\quad{\bf (ii)} Because $T_0:M_{\sigma}\to(\eusm L^{\infty})^{\sim}_c$
is a Banach lattice isomorphism, and $M_{\tau}$ is a band in
$M_{\sigma}$, $W=T_0[M_{\tau}]$ is a band in
$(\eusm L^{\infty})^{\sim}_c$.   We therefore have a Riesz homomorphism
$S_0:\eusm L^{\infty}\to W^{\times}$ defined by writing $(S_0v)(h)=h(v)$
for $v\in\eusm L^{\infty}$, $h\in W$ (356F).   Just as in (b-ii) of the
proof of 437C, $S_0$ is sequentially order-continuous and
norm-preserving.   (We need to observe that $h_x$ in the second half of
the argument there always belongs to $W$;  this is because
$h_x=T_0(\delta_x)$, where $\delta_x\in M_{\tau}$ is defined by setting
$\delta_x(E)=\chi E(x)$ for every Borel set $E$.)

\medskip

\quad{\bf (iii)} Now $T_0:M_{\sigma}\to(\eusm L^{\infty})^{\sim}_c$ and
$T:M_{\tau}\to U^{\sim}_{\tau}$ are both norm-preserving Riesz space
isomorphisms, so $T_0T^{-1}:U^{\sim}_{\tau}\to W$ is another, and its
adjoint $S_1:W^*\to(U^{\sim}_{\tau})^*$ must also be a
norm-preserving Riesz space isomorphism.   So if we set $S=S_1S_0$, $S$
will be a norm-preserving sequentially order-continuous Riesz
homomorphism from $\eusm L^{\infty}$ to
$(U^{\sim}_{\tau})^{\times}=(U^{\sim}_{\tau})^*$.

\medskip

\quad{\bf (iv)} If $v\in\eusm L^{\infty}$ and $\mu\in M_{\tau}^+$,

$$\eqalignno{(Sv)(T\mu)
&=(S_1S_0v)(T\mu)
=(S_0v)(T_0T^{-1}T\mu)\cr
&=(S_0v)(T_0\mu)
=(T_0\mu)(v)
=\int v\,d\mu;\cr}$$

\noindent if $u\in U$ and $f\in U^{\sim}_{\tau}$, then
$(T\mu)(u)=(T_0\mu)(u)$ for every $\mu\in M_{\tau}$, so

\Centerline{$(Su)(f)=(S_1S_0u)(f)=(S_0u)(T_0T^{-1}f)
=(T_0T^{-1}f)(u)=(TT^{-1}f)(u)=f(u)$.}
}%end of proof of 437H

\leader{437I}{Proposition} Let $X$ be a locally compact Hausdorff space,
$\Cal B\cmmnt{\mskip5mu=\Cal B(X)}$
its Borel $\sigma$-algebra, and $\eusm L^{\infty}(\Cal B)$
the $M$-space of  bounded Borel measurable real-valued functions on $X$.

(a) Let $M_t$ be the $L$-space of signed tight Borel measures on $X$.
Then we have
a Banach lattice isomorphism $T:M_t\to C_0(X)^*$ defined by saying that
$(T\mu)(u)=\int u\,d\mu$ whenever $\mu\in M_t^+$ and $u\in C_0(X)$.

(b) Let $\Sigma_{\text{uRm}}$ be the algebra of universally
Radon-measurable subsets of $X$\cmmnt{ (definition: 434E)}, and
$\eusm L^{\infty}(\Sigma_{\text{uRm}})$ the $M$-space of bounded
$\Sigma_{\text{uRm}}$-measurable real-valued functions on $X$.   Then we
have a norm-preserving sequentially order-continuous Riesz
homomorphism $S:\eusm L^{\infty}(\Sigma_{\text{uRm}})\to C_0(X)^{**}$
defined by saying that $(Sv)(T\mu)=\int v\,d\mu$ whenever
$v\in\eusm L^{\infty}(\Sigma_{\text{uRm}})$ and
$\mu\in M_t^+$;  and $(Su)(f)=f(u)$ for every $u\in C_0(X)$,
$f\in C_0(X)^*$.

\proof{{\bf (a)} The point is just that in this context $M_t$ is equal
to $M_{\tau}$, as defined in 437F-437H (416H), while
$C_0(X)^*=C_0(X)^{\sim}_{\tau}$ (see part (a) of the proof of 436J), and
the topology of $X$ is completely regular, so we just have a special
case of 437Ha.

\medskip

{\bf (b)(i)} As in 437Hb, we have a
sequentially order-continuous Riesz
homomorphism $S_0:\eusm L^{\infty}(\Cal B)\to C_0(X)^{**}$ defined by
saying that $(S_0v)(T\nu)=\int v\,d\nu$ whenever
$v\in\eusm L^{\infty}(\Cal B)$ and $\nu\in M_t^+$.

\medskip

\quad{\bf (ii)} If $v\in\eusm L^{\infty}(\Sigma_{\text{uRm}})$, then

\Centerline{$\sup\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\le v\}
=\inf\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\ge v\}$}

\noindent in $C_0(X)^{**}$.   \Prf\ Set

\Centerline{$A=\{w:w\in\eusm L^{\infty}(\Cal B),\,w\le v\}$,
\quad$B=\{w:w\in\eusm L^{\infty}(\Cal B),\,w\ge v\}$.}

\noindent Because the constant functions belong to
$\eusm L^{\infty}(\Cal B)$, $A$ and $B$ are both non-empty;   of course
$w\le w'$ and $S_0w\le S_0w'$ for every $w\in A$ and $w'\in B$;  because
$C_0(X)^{**}$ is Dedekind complete, $\phi=\sup S_0[A]$ and
$\psi=\inf S_0[B]$ are both defined in $C_0(X)^{**}$, and $\phi\le\psi$.
If $f\ge 0$ in $C_0(X)^*$, then there is a $\nu\in M_t^+$ such that
$T\nu=f$.
Since $v$ is $\nu$-virtually measurable (see 434Ec), there are (bounded)
Borel measurable functions $w$, $w'$ such that $w\le v\le w'$ and
$w=w'\,\,\nu$-a.e., that is, $w\in A$, $w'\in B$ and

\Centerline{$(S_0w)(f)=\biggerint w\,d\nu=\int w'd\nu=(S_0w')(f)$.}

\noindent But as

\Centerline{$(S_0w)(f)\le\phi(f)\le\psi(f)\le(S_0w')(f)$,}

\noindent $\phi(f)=\psi(f)$;  as $f$ is arbitrary, $\phi=\psi$.\ \Qed

\medskip

\quad{\bf (iii)} We can therefore define
$S:\eusm L^{\infty}(\Sigma_{\text{uRm}})\to C_0(X)^{**}$ by setting

\Centerline{$Sv=\sup\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\le v\}
=\inf\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\ge v\}$}

\noindent for every $v\in\eusm L^{\infty}$.   The argument in (ii) tells
us also that $(Sv)(T\nu)=\int v\,d\nu$ for every $\nu\in M_t^+$;  that
is, that $(Sv)(T\mu)=\int v\,d\mu$ for every Radon measure $\mu$ on
$X$.

\medskip

\quad{\bf (iv)} Now $S$ is a norm-preserving
sequentially order-continuous Riesz homomorphism.   \Prf\ (Compare
355F.)   ($\alpha$) The
non-trivial part of this is actually the check that $S$ is additive.
But the formula

\Centerline{$Sv=\sup\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\le v\}$}

\noindent ensures that $Sv_1+Sv_2\le S(v_1+v_2)$ for all $v_1$,
$v_2\in\eusm L^{\infty}$, while the formula

\Centerline{$Sv=\inf\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\ge v\}$}

\noindent ensures that $Sv_1+Sv_2\ge S(v_1+v_2)$ for all $v_1$, $v_2$.
($\beta$) It is easy to check that $S(\alpha v)=\alpha Sv$ whenever
$v\in\eusm L^{\infty}(\Sigma_{\text{uRm}})$ and $\alpha>0$, so that $S$
is linear.   ($\gamma$) If $v_1\wedge v_2=0$ in
$\eusm L^{\infty}(\Sigma_{\text{uRm}})$,

$$\eqalignno{Sv_1\wedge Sv_2
&=\sup\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\le v_1\}
    \wedge\sup\{S_0w:w\in\eusm L^{\infty}(\Cal B),\,w\le v_2\}\cr
&=\sup\{S_0w_1\wedge S_0w_2:w_1,\,w_2\in\eusm L^{\infty}(\Cal B),\,
   w_1\le v_1,\, w_2\le v_2\}\cr
\displaycause{352Ea}
&=\sup\{S_0(w_1\wedge w_2):w_1,\,w_2\in\eusm L^{\infty}(\Cal B),\,
   w_1\le v_1,\, w_2\le v_2\}\cr
\displaycause{because $S_0$ is a Riesz homomorphism}
&=0.\cr}$$

\noindent So $S$ is a Riesz homomorphism (352G(iv)).   ($\delta$) Now
suppose that $\sequencen{v_n}$ is a non-increasing sequence in
$\eusm L^{\infty}(\Sigma_{\text{uRm}})$ with infimum $0$ in
$\eusm L^{\infty}(\Sigma_{\text{uRm}})$.   Then
$\inf_{n\in\Bbb N}v_n(x)=0$ for every $x\in X$, so
$\inf_{n\in\Bbb N}\int v_nd\nu=0$ for every $\nu\in M_t^+$ and
$\inf_{n\in\Bbb N}(Sv_n)(f)=0$ for every $f\in(C_0(X)^*)^+$.   So
$\inf_{n\in\Bbb N}Sv_n=0$;  as $\sequencen{v_n}$ is arbitrary, $S$ is
sequentially
order-continuous.   ($\epsilon$)
If $v\in\eusm L^{\infty}(\Sigma_{\text{uRm}})$,
then $|v|\le\|v\|_{\infty}\chi X$, so

\Centerline{$\|Sv\|\le\|v\|_{\infty}\|S(\chi X)\|=\|v\|_{\infty}$.}

\noindent On the other hand, for any $x\in X$ we have the Dirac measure
$\delta_x$ on $X$ concentrated at $x$, with the matching
functional $h_x\in C_0(X)^*$, and

\Centerline{$\|Sv\|=\||Sv|\|=\|S|v|\|\ge(S|v|)(h_x)
=\biggerint|v|d\delta_x=|v(x)|$,}

\noindent so $\|Sv\|\ge\|v\|_{\infty}$;  thus $S$ is
norm-preserving.\ \Qed
}%end of proof of 437I

\cmmnt{\medskip

\noindent{\bf Remark} As in 437D, we can write
$(Sv)(T\mu)=\dashint v\,d\mu$ whenever $v\in\eusm L^{\infty}(\Cal B)$
and $u\in U$ and $\mu\in M_{\tau}$ (in 437H) or
$v\in\eusm L^{\infty}(\Sigma_{\text{uRm}})$ and $u\in C_0(X)$ and
$\mu\in M_t$ (in 437I).
}%end of comment

\leader{437J}{Vague and narrow topologies}\cmmnt{ We are ready for
another look
at `vague' topologies on spaces of measures.}   Let $X$ be a topological
space.

\spheader 437Ja Let $\Sigma$ be an algebra of subsets of $X$.   I will
say that $\Sigma$ {\bf separates zero sets} if whenever $F$,
$F'\subseteq X$ are disjoint zero sets then there is an $E\in\Sigma$
such that $F\subseteq E$ and $E\cap F'=\emptyset$.

\spheader 437Jb If $\Sigma$ is any algebra of subsets of $X$,
we can identify the Banach algebra and Banach lattice
$L^{\infty}(\Sigma)$\cmmnt{, as defined in \S363,} with the
$\|\,\|_{\infty}$-closed linear subspace of
$\ell^{\infty}(X)$ generated by $\{\chi E:E\in\Sigma\}$\cmmnt{ (363C,
363Ha)}.   If we do this, then $C_b(X)\subseteq L^{\infty}(\Sigma)$ iff
$\Sigma$ separates zero sets.   \prooflet{\Prf\ (i) Suppose that
$C_b(X)\subseteq L^{\infty}(\Sigma)$ and that $F_1$, $F_2\subseteq X$
are disjoint zero sets.   Let
$u_1$, $u_2:X\to\Bbb R$ be continuous functions such that
$F_i=u_i^{-1}[\{0\}]$ for both $i$;  then $|u_1(x)|+|u_2(x)|>0$ for
every $x$;  set
$v=\Bover{|u_1|}{|u_1|+|u_2|}$, so that $v:X\to[0,1]$ is continuous,
$v(x)=0$ for $x\in F_1$ and $v(x)=1$ for $x\in F_2$.   Now
$v\in C_b(X)\subseteq L^{\infty}(\Sigma)$,
so there is a $w\in S(\Sigma)$, the linear subspace of
$L^{\infty}(\Sigma)$ generated by $\{\chi E:E\in\Sigma\}$, such that
$\|v-w\|_{\infty}<\bover12$ (363C).
Set $E=\{x:w(x)\le\bover12\}$;  then $E\in\Sigma$ and
$F_1\subseteq E\subseteq X\setminus F_2$.   As $F_1$ and $F_2$ are
arbitrary, $\Sigma$ separates zero sets.

(ii) Now suppose that $\Sigma$ separates zero sets, that $u:X\to[0,1]$
is continuous, and that $n\ge 1$ is an integer.   For $i\le n$, set
$F_i=\{x:x\in X$, $u(x)\le\bover{i}{n}\}$, $F'_i=\{x:x\in X$,
$u(x)\ge\bover{i+1}{n}\}$.   Then $F_i$ and $F'_i$ are disjoint zero
sets so there is an
$E_i\in\Sigma$ such that $F'_i\subseteq E_i\subseteq X\setminus F_i$.
Set $w=\bover1n\sum_{i=1}^n\chi E_i\in S(\Sigma)$.   If $x\in X$, let
$j\le n$ be such that
$\bover{j}{n}\le u(x)<\bover{j+1}{n}$;  then for $i\le n$

\Centerline{$i<j\Longrightarrow x\in F'_i\Longrightarrow x\in E_i\Longrightarrow
x\notin F_i\Longrightarrow i\le j$,}

\noindent and $w(x)=\bover1n\#(\{i:i\le n$, $x\in E_i\})$ is either
$\bover{j}n$ or $\bover{j+1}n$.   Thus $|w(x)-u(x)|\le\bover1n$.  As $x$
is arbitrary,
$\|u-w\|_{\infty}\le\bover1n$;  as $n$ is arbitrary, $u\in
L^{\infty}(\Sigma)$.   As $L^{\infty}(\Sigma)$ is a linear subspace of
$\ell^{\infty}(X)$,
this is enough to show that $C_b(X)\subseteq L^{\infty}(\Sigma)$.\
\Qed}%end of prooflet

\spheader 437Jc It follows that if $\Sigma$ is an algebra of subsets of
$X$ separating the zero sets, and $\nu:\Sigma\to\Bbb R$ is a bounded
additive functional, we can speak of $\dashint u\,d\nu$
for any $u\in C_b(X)$.   The map $\nu\mapsto\dashint\,d\nu$
is a Banach lattice isomorphism from the $L$-space $M(\Sigma)$ of
bounded additive functionals on $\Sigma$ to
$L^{\infty}(\Sigma)^*=L^{\infty}(\Sigma)^{\sim}$\cmmnt{ (363K)}.
We therefore have a positive linear operator $T:M(\Sigma)\to C_b(X)^*$
defined by setting $(T\nu)(u)=\dashint u\,d\nu$ for every
$\nu\in M(\Sigma)$ and $u\in C_b(X)$.
Except in the trivial case $X=\emptyset$, $\|T\|=1$\prooflet{ (if
$x\in X$, we have $\delta_x\in M(\Sigma)$ defined by setting
$\delta_x(E)=\chi E(x)$ for $E\in\Sigma$, and $\|T(\delta_x)\|=1$)}.

The {\bf vague topology} on $M(\Sigma)$ is\cmmnt{ now} the
topology generated by
the functionals $\nu\mapsto\dashint u\,d\nu$ as $u$ runs over $C_b(X)$;
that is, the coarsest topology on $M(\Sigma)$ such that the canonical
map $T:M(\Sigma)\to C_b(X)^*$ is
continuous for the weak* topology of $C_b(X)^*$.   Because the
functionals $\nu\mapsto|\dashint u\,d\nu|$ are seminorms on $M(\Sigma)$,
the vague topology is a locally convex linear space topology.

\spheader 437Jd \cmmnt{There is a variant of the vague topology which
can be applied directly to spaces of (non-negative) totally finite
measures.}
Let $\tilde M^+$ be the set of all non-negative real-valued additive
functionals defined on algebras of subsets of $X$ which contain every
open set.
The {\bf narrow topology} on $\tilde M^+$ is that generated by sets of
the form

\Centerline{$\{\nu:\nu\in\tilde M^+$, $\nu G>\alpha\}$,
\quad$\{\nu:\nu\in\tilde M^+$, $\nu X<\alpha\}$}

\noindent for open sets $G\subseteq X$ and real numbers $\alpha$.
\cmmnt{(See {\smc Tops{\o}e 70b}, 8.1.)}

Observe that $\nu\mapsto\nu X:\tilde M^+\to\coint{0,\infty}$ is
continuous for the narrow topology, and if
$G\subseteq X$ is open then $\nu\mapsto\nu G$ is lower semi-continuous
for the narrow topology.
Writing $P_{\text{top}}$ for the set of topological probability
measures on $X$, then the narrow topology on $P_{\text{top}}$ is
generated by sets of the form
$\{\mu:\mu\in P_{\text{top}},\,\mu G>\alpha\}$ for real numbers $\alpha$ and
open sets
$G\subseteq X$.   Writing $\delta_x$ for the Dirac measure on $X$ concentrated
at $x$, $x\mapsto\delta_x:X\to P_{\text{top}}$ is a homeomorphism between $X$
and $\{\delta_x:x\in X\}$\cmmnt{, since $\{x:\delta_xG>\alpha\}$ is $X$ if
$\alpha<0$, $G$ if $0\le\alpha<1$ and $\emptyset$ if
$\alpha\ge 1$}\dvAnew{2023}.

Writing $\tilde M^+_{\sigma}$ for the set of
totally finite topological measures on $X$, then
$\nu\mapsto\nu E:\tilde M^+_{\sigma}\to\coint{0,\infty}$ is Borel
measurable, for the narrow topology on $\tilde M^+_{\sigma}$, for every
Borel set $E\subseteq X$\prooflet{ (because the family of
sets $E$ for which
$\nu\mapsto\nu E$ is Borel measurable is a Dynkin class containing the
open sets)}.   Similarly,
$\nu\mapsto\int u\,d\nu:\tilde M^+_{\sigma}\to\Bbb R$ is Borel measurable
for every bounded Borel measurable function $u:X\to\Bbb R$\cmmnt{,
being the limit of a sequence
of linear combinations of Borel measurable functions}.

\spheader 437Je Vague topologies, being linear space
topologies, are necessarily\cmmnt{ associated with uniformities (3A4Ad),
therefore} completely regular\cmmnt{ (4A2Ja)}.
In the very general context of (c) here\cmmnt{, in which
we have a space $M(\Sigma)$ of all finitely additive functionals on an
algebra $\Sigma$}, we do not expect the vague topology to be Hausdorff.
\cmmnt{But if we look
at particular subspaces, such as the space $M_{\sigma}(\CalBa(X))$ of
signed Baire measures, or the space
$M_{\tau}$ of signed $\tau$-additive Borel measures on a
completely regular space $X$, we may well have a Hausdorff vague
topology (437Xg).}

Similarly, the narrow topology on $\tilde M^+$ is rarely Hausdorff.
But on important subspaces we can get Hausdorff topologies.
In particular, if $X$ is Hausdorff, then the
narrow topology on the space $\MR$ of totally finite Radon
measures on $X$ is Hausdorff\cmmnt{ (437R(a-ii))}.

\spheader 437Jf\dvAnew{2007}
It will be useful to know that if $u:X\to\Bbb R$ is bounded
and lower semi-continuous, then
$\nu\mapsto\dashint u\,d\nu:\tilde M^+\to\Bbb R$ is lower semi-continuous
for the narrow topology.
\prooflet{\Prf\ (i) Perhaps I should start by explaining why
$\dashint u\,d\nu$ is always defined;  this is because the algebra $\Tau$
generated by the open sets is always a subalgebra of $\dom\nu$, and
$\{x:u(x)>\alpha\}\in\Tau$ for every $\alpha$, so
$u\in L^{\infty}(\Tau)$ (363Ha).
(ii) Now suppose for a moment that $u\ge 0$.   If
$\nu_0\in\tilde M^+$ and $\gamma<\dashint u\,d\nu_0$,
let $\epsilon>0$ be such
that $\gamma+\epsilon(1+\nu_0X)<\dashint u\,d\nu_0$,
let $n\ge 1$ be such that
$\|u\|_{\infty}\le n\epsilon$, and for $i\le n$ set
$G_i=\{x:u(x)>i\epsilon\}$.   Then

\Centerline{$\epsilon\sum_{i=1}^n\chi G_i
\le u\le\epsilon(\chi X+\sum_{i=1}^n\chi G_i)$,}

\Centerline{$\dashint u\,d\nu_0
\le\epsilon(\nu_0X+\sum_{i=1}^n\nu_0G_i)$,}

\Centerline{$V=\{\nu:\nu\in\tilde M^+$,
$\sum_{i=0}^n\nu G_i>\sum_{i=0}^n\nu_0G_i-1\}$}

\noindent is a neighbourhood of $\nu_0$ in $\tilde M^+$, and

\Centerline{$\dashint u\,d\nu
\ge\epsilon\sum_{i=1}^n\nu G_i
>\gamma$}

\noindent for every $\nu\in V$.   As $\nu_0$ and $\gamma$ are
arbitrary, $\nu\mapsto\dashint u\,d\nu$ is lower semi-continuous.    (iii)
In general, $u$ is expressible as the sum of a constant function and a
non-negative lower semi-continuous function;  as $\nu\mapsto\nu X$ is
continuous, $\nu\mapsto\dashint u\,d\nu$ is the sum of two lower
semi-continuous functions and is lower semi-continuous.\ \Qed}

Of course it follows at once that if $u:X\to\Bbb R$ is bounded and
continuous, then $\nu\mapsto\dashint u\,d\nu$ is continuous for the narrow
topology;  that is, the vague topology is coarser than the narrow topology
in contexts in which both make sense.

\spheader 437Jg\cmmnt{ With the more liberal definitions I use when
considering
integrals with respect to $\sigma$-additive measures, we have another
version of the same idea.}   If $u:X\to[0,\infty]$ is a
lower semi-continuous function, then
$\nu\mapsto\int u\,d\nu:\tilde M^+_{\sigma}\to[0,\infty]$ is lower
semi-continuous for the narrow topology.
\prooflet{\Prf\ It is the supremum of the lower
semi-continuous functions $\nu\mapsto\int(u\wedge n\chi X)d\nu$.\ \Qed}

\spheader 437Jh\dvAnew{2010}
Let $X$ and $Y$ be topological spaces, $\phi:X\to Y$ a
continuous function, and $\tilde M^+(X)$, $\tilde M^+(Y)$ the spaces of
functionals described in (d).   For a functional $\nu$ defined on a subset
of $\Cal PX$, define
$\nu\phi^{-1}$ by saying that $(\nu\phi^{-1})(F)=\nu(\phi^{-1}[F])$
whenever $F\subseteq Y$ and $\phi^{-1}[F]\in\dom\nu$.   Then
$\nu\phi^{-1}\in\tilde M^+(Y)$ whenever $\nu\in\tilde M^+(X)$, and
the map $\nu\mapsto\nu\phi^{-1}:\tilde M^+(X)\to\tilde M^+(Y)$ is
continuous for the narrow topologies\prooflet{ (use 4A2B(a-ii))}.

\spheader 437Ji I am trying to
maintain the formal distinctions between `quasi-Radon measure' and
`$\tau$-additive effectively locally finite Borel measure inner regular
with respect to the closed sets', and between `Radon measure' and
`tight locally finite Borel measure'.   \cmmnt{There are obvious problems
in interpreting the sum and
difference of measures with different domains, which are readily soluble
(see 234G and 416De) but in the context of this section
are unilluminating.}   If\cmmnt{, however,} we
take $\MqR$ to be the set of totally
finite quasi-Radon measures on $X$, and $X$ is completely regular,
we have a canonical embedding of $\MqR$ into a cone in the
$L$-space $C_b(X)^*$;
\cmmnt{more generally,} even if\cmmnt{ our space} $X$ is
not completely regular, the map
$\mu\mapsto\mu\restr\Cal B(X):\MqR\to M_{\sigma}(\Cal B(X))$
is still injective,
and we can identify $\MqR$ with a cone in the $L$-space
$M_{\tau}$ of
signed $\tau$-additive Borel measures\cmmnt{ (often the whole positive
cone of $M_{\tau}$, as in 415M)}.
Similarly, when $X$ is Hausdorff, we can identify totally finite Radon
measures with tight totally finite Borel measures\cmmnt{ (416F)}.   The
definition in 437Jd makes it plain that these
identifications are homeomorphisms for the narrow topology,

\cmmnt{It is even possible to extend these ideas to measures which are not
totally finite (437Yi), though there may be new difficulties (415Ya).}

\cmmnt{\spheader 437Jj For a different kind of narrow topology, adapted
to the space of all Radon measures on a Hausdorff space, see 495R below.
}%end of comment

\leader{437K}{Proposition} Let $X$ be a topological space, and
$\tilde M^+$ the set of all non-negative real-valued additive
functionals defined on algebras of subsets of $X$ containing every open
set.

(a) We have a function $T:\tilde M^+\to C_b(X)^*$ defined by the formula
$(T\nu)(u)=\dashint u\,d\nu$ whenever $\nu\in\tilde M^+$ and
$u\in C_b(X)$.

(b) $T$ is continuous for the narrow topology
$\frak S$ on $\tilde M^+$ and the weak* topology on $C_b(X)^*$.

(c) Suppose now that $X$ is completely regular, and that
$W\subseteq\tilde M^+$ is a family of $\tau$-additive totally finite
topological measures such that
two members of $W$ which agree on the Borel $\sigma$-algebra are equal.
Then $T\restr W$ is a homeomorphism between $W$, with the narrow
topology, and $T[W]$, with the weak* topology.

\proof{{\bf (a)} We have only to assemble the operators of 437Jc, noting
that if an algebra of subsets of $X$ contains every open set then it
certainly separates the
zero sets (indeed, it actually contains every zero set).

\medskip

{\bf (b)} As already noted in 437Jf,
$\nu\mapsto (T\nu)(u)=\dashint u\,d\nu$ is
$\frak S$-continuous for every $u\in C_b(X)$.
Since the weak* topology on $C_b(X)^*$ is the coarsest
topology on $C_b(X)^*$
for which all the functionals $f\mapsto f(u)$ are continuous,
$T$ is continuous.

\medskip

{\bf (c)(i)}
Write $\frak T$ for the topology on $W$ induced by $T$, that is, the
family of sets of the form $W\cap T^{-1}[V]$ where $V\subseteq C_b(X)^*$
is weak*-open.   If $G\subseteq X$ is open, then
$A=\{u:u\in C_b(X),\,0\le u\le\chi G\}$ is upwards-directed and has
supremum $\chi G$, so $\mu G=\sup_{u\in A}\int u\,d\mu$ for every
$\mu\in W$ (414Ba).   Accordingly $\{\mu:\mu\in W$,
$\mu G>\alpha\}=\bigcup_{u\in A}\{\mu:(T\mu)(u)>\alpha\}$ belongs
to $\frak T$ for every $\alpha\in\Bbb R$.   Also, of course,
$\{\mu:\mu X<\alpha\}=\{\mu:(T\mu)(\chi X)<\alpha\}\in\frak T$ for
every $\alpha$.   So if $\frak S'$ is the narrow topology on $W$,
$\frak S'\subseteq\frak T$.
Putting this together with (b), we see that $\frak S'=\frak T$.

\medskip

\quad{\bf (ii)} Now the same formulae show that $T\restr W$ is
injective.   \Prf\ Suppose that $\mu_1$, $\mu_2\in W$ and that
$T\mu_1=T\mu_2$.   Then $\mu_1G=\mu_2G$ for every open set
$G\subseteq X$.   By the Monotone Class Theorem, $\mu_1$ and
$\mu_2$ agree on all Borel sets;
but our hypothesis is that this is enough to ensure that $\mu_1=\mu_2$.\
\Qed

Since $T:W\to T[W]$ is continuous and open, it is a homeomorphism.
}%end of proof of 437K

\leader{437L}{Corollary} Let $X$ be a completely regular topological
space, and $M_{\tau}$ the space of signed $\tau$-additive Borel
measures on $X$.   Then the narrow and vague topologies on $M_{\tau}^+$
coincide.   In particular, the narrow topology on $M_{\tau}^+$ is
completely regular.

\proof{437Kc, 3A4Ad, 4A2Ja.}
%end of proof of 437L

\leader{437M}{Theorem}\discrversionA{\footnote{Revised
2009.}}{}\cmmnt{ ({\smc Ressel 77})} For a topological space
$X$, write $\MqR(X)$ for the space of totally finite quasi-Radon
measures on $X$, $\PqR(X)$ for the space of quasi-Radon probability
measures on $X$, and $M_{\tau}(X)$ for the
$L$-space of signed $\tau$-additive Borel measures on $X$.

(a) Let $X$ and $Y$ be topological spaces.   If
$\mu\in\MqR(X)$ and $\nu\in\MqR(Y)$, write
$\mu\times\nu$ for their $\tau$-additive
product measure on $X\times Y$\cmmnt{ (417C, 417F)}.
Then $(\mu,\nu)\mapsto\mu\times\nu$ is continuous for the narrow
topologies on $\MqR(X)$, $\MqR(Y)$ and $\MqR(X\times Y)$.

(b) Let $\familyiI{X_i}$ be a family of topological
spaces, with product $X$.   If $\familyiI{\mu_i}$ is a family of
probability measures such that $\mu_i\in\PqR(X_i)$ for each
$i$, write $\prod_{i\in I}\mu_i$ for its
$\tau$-additive product on $X$.   Then
$\familyiI{\mu_i}\mapsto\prod_{i\in I}\mu_i$ is continuous for the
narrow topology on $\PqR(X)$ and the product of the narrow
topologies on $\prod_{i\in I}\PqR(X_i)$.

(c) Let $X$ and $Y$ be topological spaces.

\quad(i) We have a unique bilinear operator
$\psi:M_{\tau}(X)\times M_{\tau}(Y)\to M_{\tau}(X\times Y)$
such that $\psi(\mu,\nu)$ is the restriction of the $\tau$-additive product
of $\mu$ and $\nu$ to the
Borel $\sigma$-algebra of $X\times Y$ whenever $\mu$, $\nu$ are totally
finite Borel measures on $X$, $Y$ respectively.

\quad(ii) $\|\psi\|\le 1$\cmmnt{ (definition:  253Ab)}.

\quad(iii) $\psi$ is separately continuous for the vague topologies on
$M_{\tau}(X)$, $M_{\tau}(Y)$ and $M_{\tau}(X\times Y)$.

(d) In (c), suppose that $X$ and $Y$ are compact and Hausdorff.
If $B\subseteq M_{\tau}(X)$ and $B'\subseteq M_{\tau}(Y)$ are
norm-bounded, then $\psi\restr B\times B'$ is continuous for the vague
topologies.

\proof{{\bf (a)(i)} I ought to note that we need 417N to assure us that
$\mu\times\nu\in\MqR(X\times Y)$ whenever
$\mu\in\MqR(X)$ and $\nu\in\MqR(Y)$.

\medskip

\quad{\bf (ii)} If $W\subseteq X\times Y$ is open and
$\alpha\in\Bbb R$, then $Q=\{(\mu,\nu):(\mu\times\nu)(W)>\alpha\}$ is
open in $\MqR(X)\times\MqR(Y)$.   \Prf\
Suppose that $(\mu_0,\nu_0)\in Q$.   Because $\mu_0\times\nu_0$ is
$\tau$-additive, there is a subset $W'\subseteq W$, expressible in the
form $\bigcup_{i\le n}G_i\times H_i$ where $G_i\subseteq X$ and
$H_i\subseteq Y$ are open for every $i$, such that

\Centerline{$\alpha<(\mu_0\times\nu_0)(W')
=\int\nu_0 W'[\{x\}]\mu_0(dx)$}

\noindent (417C(b-v-$\beta$).   Set $u(x)=\nu_0W'[\{x\}]$ for $x\in X$, so that
$u$ is lower semi-continuous (417Ba).   Let $\eta>0$ be such that
$\int u\,d\mu_0>\alpha+(1+2\mu_0X)\eta$, and set
$E_i=\{x:u(x)>\eta i\}$ for $i\in\Bbb N$, so that
$\eta\sum_{i=1}^{\infty}\mu_0E_i>\int u\,d\mu_0-\eta\mu_0X$.   Because
every $E_i$ is open, there is a neighbourhood $U$ of $\mu_0$ in
$\MqR(X)$ such that

\Centerline{$\int u\,d\mu_0-\eta\mu_0X\le\eta\sum_{i=1}^{\infty}\mu E_i
\le\int u\,d\mu=(\mu\times\nu_0)(W')$}

\noindent for every $\mu\in U$;  shrinking $U$ if necessary, we can
arrange at the same time that $\mu X<\mu_0X+1$ for every $\mu\in U$.
Next, observe that $\Cal H=\{W'[\{x\}]:x\in X\}
\subseteq\{\bigcup_{i\in I}H_i:I\subseteq\{0,\ldots,n\}\}$ is finite, so
there is a neighbourhood $V$ of $\nu_0$ in $\MqR(Y)$ such
that $\nu H\ge\nu_0 H-\eta$ for every $H\in\Cal H$ and $\nu\in V$.   If
$\mu\in U$ and $\nu\in V$, we have

$$\eqalign{(\mu\times\nu)(W)
&\ge(\mu\times\nu)(W')
=\int\nu W'[\{x\}]\mu(dx)
\ge\int u(x)-\eta\,\mu(dx)\cr
&=\int u\,d\mu-\eta\mu X
\ge\int u\,d\mu_0-\eta\mu_0X-\eta(1+\mu_0X)
>\alpha.\cr}$$

\noindent As $\mu_0$ and $\nu_0$ are arbitrary, $Q$ is open.\ \Qed

\medskip

\quad{\bf (iii)} Since $(\mu\times\nu)(X\times Y)=\mu X\cdot\nu Y$, the
sets $\{(\mu,\nu):(\mu\times\nu)(X\times Y)<\alpha\}$ are also open for
every $\alpha\in\Bbb R$.   So $(\mu,\nu)\mapsto\mu\times\nu$ is
continuous (4A2B(a-ii) again).

\medskip

{\bf (b)} Similarly, we can refer to 417O to check that
$\prod_{i\in I}\mu_i\in\PqR(X)$ whenever $\mu_i\in\PqR(X_i)$ for each
$i$.   For finite sets $I$, the result is a simple induction on $\#(I)$,
using 417Db and part (a) just above.
For infinite $I$, let $W\subseteq X$ be an open set and
$\alpha\in\Bbb R$, and consider

\Centerline{$Q
=\{\familyiI{\mu_i}:\mu_i\in\PqR(X_i)$ for each $i$,
$(\prod_{i\in I}\mu_i)(W)>\alpha\}$.}

\noindent If $\familyiI{\mu_i}\in Q$, then
there is an open set $W'\subseteq W$, determined by coordinates in a
finite set $J\subseteq I$, such that $(\prod_{i\in I}\mu_i)(W')>\alpha$.
Setting $V=\{x\restr J:x\in W'\}$, we have
$(\prod_{i\in J}\mu_i)(V)>\alpha$.   Now we can find open sets $U_i$
in $\PqR(X_i)$, for $i\in J$, such that
$(\prod_{i\in J}\nu_i)(V)>\alpha$ whenever $\nu_i\in U_i$ for $i\in J$.
If now
$\familyiI{\nu_i}\in\prod_{i\in I}\PqR(X_i)$ is such that
$\nu_i\in U_i$ for every $i\in J$,

\Centerline{$(\prod_{i\in I}\nu_i)(W)\ge(\prod_{i\in I}\nu_i)(W')
=(\prod_{i\in J}\nu_i)(V)>\alpha$,}

\noindent so $\prod_{i\in I}\nu_i\in Q$.   As $\familyiI{\mu_i}$ is
arbitrary, $Q$ is open.

As $W$ and $\alpha$ are arbitrary,
$\familyiI{\mu_i}\mapsto\prod_{i\in I}\mu_i$ is continuous.

\medskip

{\bf (c)(i)} Start by writing
$\psi(\mu,\nu)=(\mu\times\nu)\restr\Cal B(X\times Y)$ for
$\mu\in M^+_{\tau}(X)$ and $\nu\in M^+_{\tau}(Y)$,
where $\Cal B(X\times Y)$ is the Borel $\sigma$-algebra of $X\times Y$.
If $\mu$, $\mu_1$, $\mu_2\in M^+_{\tau}(X)$ and
$\nu$, $\nu_1$, $\nu_2\in M^+_{\tau}(Y)$ and $\alpha\ge 0$, then

\Centerline{$\psi(\mu_1+\mu_2,\nu)
=\psi(\mu_1,\nu)+\psi(\mu_2,\nu)$.}

\noindent\Prf\ On each side of the equation we have a $\tau$-additive
Borel measure, and the two measures agree on the standard base $\Cal W$
for the topology of $X\times Y$ consisting of products of open sets;
since $\Cal W$ is closed under finite intersections, they agree on the
algebra generated by $\Cal W$ and therefore on all open sets and
therefore (using the Monotone Class Theorem yet again) on all Borel
sets.\ \QeD\   Similarly,

\Centerline{$\psi(\mu,\nu_1+\nu_2)
=\psi(\mu,\nu_1)+\psi(\mu,\nu_2)$,
\quad$\psi(\alpha\mu,\nu)=\psi(\mu,\alpha\nu)
=\alpha\psi(\mu,\nu)$}

\noindent whenever $\mu\in M^+_{\tau}(X)$, $\nu$, $\nu_1$,
$\nu_2\in M^+_{\tau}(Y)$ and $\alpha\in\Bbb R$.
Now if $\mu'_1$, $\mu'_2\in M^+_{\tau}(X)$ and $\nu'_1$,
$\nu'_2\in M^+_{\tau}(Y)$ are such that $\mu_1-\mu_2=\mu'_1-\mu'_2$ and
$\nu_1-\nu_2=\nu'_1-\nu'_2$, we shall have

$$\eqalign{\psi(\mu_1,\nu_1)-\psi(\mu_1,&\nu_2)
  -\psi(\mu_2,\nu_1)+\psi(\mu_2,\nu_2)\cr
&=\psi(\mu_1,\nu_1+\nu'_2)-\psi(\mu_1+\mu'_2,\nu'_2)
        +\psi(\mu'_2,\nu'_2)\cr
&\mskip75mu
   -\psi(\mu_1,\nu'_1+\nu_2)+\psi(\mu_1+\mu'_2,\nu'_1)
   -\psi(\mu'_2,\nu'_1)\cr
&\mskip75mu
   -\psi(\mu_2,\nu_1+\nu'_2)+\psi(\mu_2+\mu'_1,\nu'_2)
     -\psi(\mu'_1,\nu'_2)\cr
&\mskip75mu
   +\psi(\mu_2,\nu'_1+\nu_2)-\psi(\mu_2+\mu'_1,\nu'_1)
     +\psi(\mu'_1,\nu'_1)\cr
&=\psi(\mu'_2,\nu'_2)
 -\psi(\mu'_2,\nu'_1)
 -\psi(\mu'_1,\nu'_2)
 +\psi(\mu'_1,\nu'_1).\cr}$$

\noindent We can therefore extend $\psi$ to an operator on
$M_{\tau}(X)\times M_{\tau}(Y)$ by setting

\Centerline{$\psi(\mu_1-\mu_2,\nu_1-\nu_2)
  =\psi(\mu_1,\nu_1)
  -\psi(\mu_1,\nu_2)
  -\psi(\mu_2,\nu_1)
  +\psi(\mu_2,\nu_2)$}

\noindent whenever $\mu_1$, $\mu_2\in M^+_{\tau}(X)$ and $\nu_1$,
$\nu_2\in M^+_{\tau}(Y)$, and it is straightforward to check that $\psi$
is bilinear.

\medskip

\quad{\bf (ii)} If $\mu\in M_{\tau}(X)$, then
$\|\mu\|=\mu^+(X)+\mu^-(X)$, where $\mu^+$ and $\mu^-$ are evaluated in
the Riesz space $M_{\tau}(X)$.   Now if $\nu\in M_{\tau}(Y)$,

$$\eqalign{|\psi(\mu,\nu)|
&=|\psi(\mu^+,\nu^+)-\psi(\mu^+,\nu^-)
  -\psi(\mu^-,\nu^+)+\psi(\mu^-,\nu^-)|\cr
&\le\psi(\mu^+,\nu^+)+\psi(\mu^+,\nu^-)
  +\psi(\mu^-,\nu^+)+\psi(\mu^-,\nu^-),\cr}$$

\noindent so

$$\eqalign{\|\psi(\mu,\nu)\|
&=|\psi(\mu,\nu)|(X\times Y)\cr
&\le\psi(\mu^+,\nu^+)(X\times Y)+\psi(\mu^+,\nu^-)(X\times Y)\cr
&\mskip100mu
  +\psi(\mu^-,\nu^+)(X\times Y)+\psi(\mu^-,\nu^-)(X\times Y)\cr
&=\mu^+(X)\cdot\nu^+(Y)+\mu^+(X)\cdot\nu^-(Y)
  +\mu^-(X)\cdot\nu^+(Y)+\mu^-(X)\cdot\nu^-(Y)\cr
&=\|\mu\|\|\nu\|.\cr}$$

\noindent As $\mu$ and $\nu$ are arbitrary, $\|\psi\|\le 1$.

\medskip

\quad{\bf (iii)} Fix $\nu\in M^+_{\tau}(Y)$ and $w\in C_b(X\times Y)^+$,
and consider the map
$\mu\mapsto\dashint w\,d\psi(\mu,\nu):M_{\tau}(X)\to\Bbb R$.   Note first
that if $\mu\in M^+_{\tau}(X)$,

\Centerline{$\dashint w\,d\psi(\mu,\nu)=\int w\,d(\mu\times\nu)
=\iint w(x,y)\nu(dy)\mu(dx)=\dashiint w(x,y)\nu(dy)\mu(dx)$}

\noindent (417G).   Since both sides of this equation are linear in
$\mu$, we have

\Centerline{$\dashint w\,d\psi(\mu,\nu)=\dashiint w(x,y)\nu(dy)\mu(dx)$}

\noindent for every $\mu\in M_{\tau}(X)$.   Now
$x\mapsto\int w(x,y)\nu(dy)$ is
continuous.   \Prf\ By 417Bc, it is lower semi-continuous;  but if
$\alpha\ge\|w\|_{\infty}$ and $w'=\alpha\chi(X\times Y)-w$, then
$x\mapsto\int w'(x,y)\nu(dy)$ is lower semi-continuous, so

\Centerline{$x\mapsto\alpha\nu Y-\biggerint w'(x,y)\nu(dy)
=\int w(x,y)\nu(dy)$}

\noindent is also upper semi-continuous, therefore continuous.\ \QeD\
It follows at once that $\mu\mapsto\dashiint w(x,y)\nu(dy)\mu(dx)$ is
continuous for the vague topology on $M_{\tau}(X)$.   The argument has
supposed that $w$ and $\nu$ are positive;  but taking positive and
negative parts as usual, we see that $\mu\mapsto\dashint w\,d\psi(\mu,\nu)$
is vaguely continuous for every $w\in C_b(X\times Y)$ and
$\nu\in M_{\tau}(Y)$.   As $w$ is arbitrary, $\mu\mapsto\psi(\mu,\nu)$
is vaguely continuous, for every $\nu$.   Similarly,
$\nu\mapsto\psi(\mu,\nu)$ is vaguely continuous for every $\mu$, and
$\psi$ is separately continuous.

\medskip

{\bf (d)} Now suppose that $X$ and $Y$ are compact.   Let $W$ be the
linear subspace of $C(X\times Y)$ generated by
$\{u\otimes v:u\in C(X)$, $v\in C(Y)\}$, writing
$(u\otimes v)(x,y)=u(x)v(y)$ as in 253B.   Then $W$ is a subalgebra of
$C(X\times Y)$ separating the points of
$X\times Y$ and containing the constant functions, so is
$\|\,\|_{\infty}$-dense in $C(X\times Y)$ (281E).   Now

\Centerline{$(\mu,\nu)\mapsto\dashint u\otimes v\,d\psi(\mu,\nu)
=\dashint u\,d\mu\cdot\dashint v\,d\nu$}

\noindent is continuous whenever $u\in C(X)$ and $v\in C(Y)$, so

\Centerline{$(\mu,\nu)\mapsto\dashint w\,d\psi(\mu,\nu)$}

\noindent is continuous whenever $w\in W$.

Next suppose that $B\subseteq M_{\tau}(X)$ and $B'\subseteq M_{\tau}(Y)$
are bounded.   Let $\gamma\ge 0$ be such that $\|\mu\|\le\gamma$ for
every $\mu\in B$ and $\|\nu\|\le\gamma$ for every $\nu\in B$.   If
$w\in C(X\times Y)$ and $\epsilon>0$, there is a $w'\in W$ such that
$\|w-w'\|_{\infty}\le\epsilon$.   In this case

$$\eqalign{|\dashint w\,d\psi(\mu,\nu)-\dashint w'\,d\psi(\mu,\nu)|
&\le\|w-w'\|_{\infty}\|\psi(\mu,\nu)\|\cr
&\le\epsilon\|\mu\|\|\nu\|
\le\gamma^2\epsilon\cr}$$

\noindent whenever $\mu\in B$ and $\nu\in B'$.   As $\epsilon$ is
arbitrary, the function $(\mu,\nu)\mapsto\dashint w\,d\psi(\mu,\nu)$ is
uniformly approximated on $B\times B'$ by vaguely continuous functions,
and is therefore itself vaguely continuous on $B\times B'$.
}%end of proof of 437M

\leader{437N}{}\cmmnt{ One of the standard constructions of Radon measures
is as image measures.   It leads naturally to maps between spaces of
Radon measures, and of course we wish to know whether they are continuous.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}37Q} (a) Let $X$ and $Y$ be
Hausdorff spaces, and $\phi:X\to Y$ a continuous function.   Let $\MR(X)$,
$\MR(Y)$ be the spaces
of totally finite Radon measures on $X$ and $Y$ respectively.   Write
$\tilde\phi(\mu)$ for the image measure $\mu\phi^{-1}$ for
$\mu\in\MR(X)$.

\quad(i) $\tilde\phi:\MR(X)\to\MR(Y)$ is continuous for
the narrow topologies on $\MR(X)$ and $\MR(Y)$.

\quad(ii) $\tilde\phi(\mu+\nu)=\tilde\phi(\mu)+\tilde\phi(\nu)$ and
$\tilde\phi(\alpha\mu)=\alpha\tilde\phi(\mu)$ for all $\mu$, $\nu\in\MR(X)$
and $\alpha\ge 0$.

(b) If $Y$ is a Hausdorff space, $X$ a subset of $Y$, and $\phi:X\to Y$
the identity map, then $\tilde\phi$ is a homeomorphism between
$\MR(X)$ and
$\{\nu:\nu\in\MR(Y)$, $\nu(Y\setminus X)=0\}$.

\proof{{\bf (a)(i)} All we have to do is to recall from 418I that
$\mu\phi^{-1}\in\MR(Y)$ for every
$\mu\in\MR(X)$, and observe that

\Centerline{$\{\mu:(\mu\phi^{-1})(H)>\alpha\}
=\{\mu:\mu\phi^{-1}[H]>\alpha\}$,
\quad$\{\mu:(\mu\phi^{-1})(Y)<\alpha\}=\{\mu:\mu X<\alpha\}$}

\noindent are narrowly open in $\MR(X)$ for every open set
$H\subseteq Y$ and $\alpha\in\Bbb R$.

\medskip

\quad{\bf (ii)} As usual, since all the measures here are Radon measures,
it is enough to check that
$\tilde\phi(\mu+\nu)(E)=\tilde\phi(\mu)(E)+\tilde\phi(\nu)(E)$ and
$\tilde\phi(\alpha\mu)(E)=\alpha\tilde\phi(\mu)(E)$ for every Borel set
$E\subseteq X$, and this is easy.

\medskip

{\bf (b)} First note that if $\mu\in\MR(X)$, then certainly
$\tilde\phi(\mu)(Y\setminus X)=0$;  while if $\nu\in\MR(Y)$
and $\nu(Y\setminus X)=0$,
then $\mu=\nu\restr\Cal PX$ is a Radon measure on $X$ (416Rb) and
$\nu=\tilde\phi(\mu)$.   Thus $\tilde\phi$ is a continuous bijection
from $\MR(X)$ to
$\{\nu:\nu\in\MR(Y)$, $\nu(Y\setminus X)=0\}$.   Now if
$G\subseteq X$ is relatively open and $\alpha\in\Bbb R$, there is an
open set $H\subseteq Y$ such that $G=H\cap X$, so that

\Centerline{$\{\mu:\mu\in\MR(X)$,
$\mu G>\alpha\}=\{\mu:\tilde\phi(\mu)(H)>\alpha\}$}

\noindent is the inverse image of a narrowly open set in
$\MR(Y)$;  and of course

\Centerline{$\{\mu:\mu\in\MR(X)$,
$\mu X<\alpha\}=\{\mu:\tilde\phi(\mu)(Y)<\alpha\}$}

\noindent is also the inverse image of an open set.   So $\tilde\phi$ is
a homeomorphism between $\MR(X)$ and
$\{\nu:\nu\in\MR(Y)$, $\nu(Y\setminus X)=0\}$.
}%end of proof of 437N

\leader{437O}{Uniform tightness}\dvAformerly{4{}37T}
Let $X$ be a topological space.   If $\nu$ is a bounded additive functional
on an algebra of subsets of $X$,\cmmnt{ I say that} it is {\bf tight} if

\Centerline{$\nu E
\in\overline{\{\nu K:K\subseteq E,\,K\in\dom\nu,\,
  K\text{ is closed and compact}\}}$}

\noindent for every $E\in\dom\nu$, and that a set $A$ of tight functionals
is {\bf uniformly tight} if every member of $A$ is tight
and for every $\epsilon>0$ there is a
closed compact set $K\subseteq X$ such that $\nu K$ is defined and
$|\nu E|\le\epsilon$ whenever $\nu\in A$ and $E\in\dom\nu$ is disjoint from
$K$.

\leader{437P}{Proposition}\discrversionA{\footnote{Formerly 4{}37U;
revised 2009.}}{} Let $X$ be a topological space.

(a) Let $\MqR$ be the
set of totally finite quasi-Radon measures on $X$.   Suppose that
$A\subseteq\MqR$ is uniformly totally finite\cmmnt{ (that is,
$\{\mu X:\mu\in A\}$ has a finite upper bound)} and for every
$\epsilon>0$ there is a
closed compact $K\subseteq X$ such that $\mu(X\setminus K)\le\epsilon$ for
every $\mu\in A$.   Then $A$ is relatively compact in $\MqR$ for the narrow
topology.

(b) Suppose now that $X$ is Hausdorff, and that $\MR$ is the set of Radon
measures on $X$.   If $A\subseteq\MR$ is uniformly totally finite and
uniformly tight, then
it is relatively compact in $\MR$ for the narrow topology.

\proof{{\bf (a)(i)} I show first that the closure $\overline{A}$ of $A$ in
$\MqR$ has the same two properties.   \Prf\ Because $\mu\mapsto\mu X$ is
continuous for the narrow topology,
$\{\mu X:\mu\in\overline{A}\}\subseteq\overline{\{\mu X:\mu\in A\}}$ is
bounded.   If $\epsilon>0$, there is a closed compact set $K\subseteq X$
such that $\mu(X\setminus K)\le\epsilon$ for every $\mu\in A$.
In this case $\{\mu:\mu\in\MqR$, $\mu(X\setminus K)\le\epsilon\}
=\MqR\setminus\{\mu:\mu(X\setminus K)>\epsilon\}$
is closed in $\MqR$, so includes $\overline{A}$.   As $\epsilon$ is
arbitrary, we have the result.\ \Qed

\medskip

\quad{\bf (ii)} Now let $\Cal F$ be an ultrafilter on $\MqR$ containing
$\overline{A}$.

\medskip

\qquad\grheada\ For Borel sets
$E\subseteq X$, set $\theta E=\lim_{\nu\to\Cal F}\nu E$;  this is defined
in $\Bbb R$ because $\sup_{\nu\in\overline{A}}\nu X$ is finite.
$\theta$ is
a non-negative additive functional on the Borel $\sigma$-algebra of $X$.
The family $\Cal K$ of closed compact subsets of $X$ is a compact
class containing $\emptyset$
and closed under finite unions and countable intersections,
so 413Ub tells us that there is a complete measure $\mu$ on $X$
such that $\mu X\le\theta X$, $\Cal K\subseteq\dom\mu$, $\mu K\ge\theta K$
for every $K\in\Cal K$, and $\mu$ is inner regular with respect to
$\Cal K$.

If $F\subseteq X$ is closed, then $F\cap K\in\Cal K$ for every
$K\in\Cal K$, so $\mu$ measures $F$ (412Ja).   Thus $\mu$ is a topological
measure.   Because $\mu$ is tight, it is $\tau$-additive (411E).
So $\mu$ is a complete totally finite $\tau$-additive measure which is
inner regular with respect to the closed sets,
and is a quasi-Radon measure.

\medskip

\qquad\grheadb\ Given $\epsilon>0$, there is a $K\in\Cal K$ such that
$\nu(X\setminus K)\le\epsilon$ for every $\nu\in\overline{A}$, so
$\theta(X\setminus K)\le\epsilon$ and

\Centerline{$\theta X-\epsilon\le\theta K\le\mu K\le\mu X\le\theta X$.}

\noindent As $\epsilon$ is arbitrary,
$\mu X=\theta X=\lim_{\nu\to\Cal F}\nu X$.

\medskip

\qquad\grheadc\
If $G\subseteq X$ is open and $\epsilon>0$, there is a $K\in\Cal K$ such
that $\nu(X\setminus K)\le\epsilon$ for every $\nu\in\overline{A}$, so
$\theta(X\setminus K)\le\epsilon$ and

$$\eqalignno{\mu G
&=\mu X-\mu(X\setminus G)
\le\theta X-\mu(K\setminus G)
\le\theta X-\theta(K\setminus G)\cr
\displaycause{because $K\setminus G\in\Cal K$}
&\le\theta X-\theta(X\setminus G)+\theta(X\setminus K)
\le\theta G+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\mu G\le\theta G=\lim_{\nu\to\Cal F}\nu G$.}

\medskip

\qquad\grheadd\ Putting ($\beta$) and ($\gamma$) together, we see that
$\Cal F\to\mu$ for the narrow topology on
$\MqR$;  it follows that $\mu\in\overline{A}$.   Thus every ultrafilter on
$\MqR$ containing $\overline{A}$ has a limit in $\overline{A}$, and
$\overline{A}$ is compact.   Accordingly $A$ is relatively compact in
$\MqR$, as claimed.

\medskip

{\bf (b)} We know from the proof of (a) that the closure $\overline{A}$ of
$A$ in $\MqR$ is compact, so it will be enough to show that
$\overline{A}\subseteq\MR$.   If $\mu\in\overline{A}$ and $E\in\dom\mu$,
then for every
$\epsilon>0$ there is a compact set $K\subseteq X$ such that
$\mu(X\setminus K)\le\epsilon$, by (a-i) above;  also there is a closed set
$F\subseteq E$ such that $\mu(E\setminus F)\le\epsilon$.   But now
$F\cap K$ is compact and $\mu(E\setminus(F\cap K))\le 2\epsilon$.   As $E$
and $\epsilon$ are arbitrary, $\mu$ is inner regular with respect to the
compact sets, so is a Radon measure.
}%end of proof of 437P

\leader{437Q}{Two metrics}\dvAnew{2012}
\cmmnt{So far, as elsewhere in this volume, I have been writing about
topologies with as few restrictions on their nature as possible.   Of
course the repeated invocation of $L$-spaces in the first part of the
section indicates that there are norms and their associated metrics about,
and when the underlying set $X$ is metrizable we rather hope that the
constructions of 437J will lead to metrizable topologies on the spaces of
measures considered there.   I offer two definitions which
seem to give us interesting paths to explore.

\medskip

}{\bf (a)(i)} If $X$ is a set and $\mu$, $\nu$
are bounded additive functionals defined on algebras of subsets of $X$,
then $\mu-\nu:\dom\mu\cap\dom\nu\to\Bbb R$ is bounded and
additive, and we can set

\Centerline{$\rhotv(\mu,\nu)=|\mu-\nu|(X)
=\sup_{E,F\in\dom\mu\cap\dom\nu}(\mu-\nu)(E)-(\mu-\nu)(F)$.}

\noindent In this generality, $\rhotv$
is not even a pseudometric, but if we
have a class $M$ of totally finite measures on $X$ all of which are inner
regular with respect to a subset $\Cal K$ of $\bigcap_{\mu\in M}\dom\mu$,
then we have

\Centerline{$\rhotv(\mu,\nu)
=\sup_{K,L\in\Cal K}(\mu K-\mu L)-(\nu K-\nu L)$}

\noindent for all $\mu$, $\nu\in M$, and $\rhotv\restr M\times M$ is a
pseudometric on $M$.   If moreover $M$ is such that distinct members of $M$
differ on $\Cal K$ (as when $\Cal K$ is the family of
closed sets in a topological space $X$ and $M=\MqR(X)$, or when
$\Cal K$ the family of compact sets in a Hausdorff space $X$ and
$M=\MR(X)$), then $\rhotv$
gives us a metric on $M$.   In such a case I will call
$\rhotv\restr M\times M$ the {\bf total variation metric} on $M$.
\cmmnt{(Compare the `total variation norms' of 362B.)}

\medskip

\quad{\bf (ii)} Note that if $\Sigma\subseteq\dom\mu\cap\dom\nu$
is a $\sigma$-algebra then

\Centerline{$|\int u\,d\mu-\int u\,d\nu|\le\|u\|_{\infty}\rhotv(\mu,\nu)$}

\noindent whenever $u\in\eusm L^{\infty}(\Sigma)$.   So if, for instance,
$X$ is a topological space and $M\subseteq M^+_{\text{qR}}(X)$, then
$u\mapsto\int u\,d\mu$ will be continuous for the total variation metric on
$M$ whenever $u:X\to\Bbb R$ is a bounded universally measurable function.

\medskip

\quad{\bf (iii)}\cmmnt{ It is of course worth knowing when to expect a
complete
metric.}   When our set $M$ can be identified with the positive cone
of a band in some $L$-space
$M_{\sigma}$ of countably additive functions\cmmnt{, as in 437F}, then
we\cmmnt{ naturally}
have a complete metric\cmmnt{, because bands in $L$-spaces are closed
subspaces
(354Bd)}.   In particular, for any Hausdorff space $X$,
$M^+_{\text{R}}(X)$ can be identified with the
positive cone of the $L$-space of tight Borel measures on $X$, so is
complete.    \cmmnt{See also 437Xo.}

\cmmnt{\medskip

\quad{\bf (iv)} There is an obvious variation on $\rhotv$ as defined here:
the function

\Centerline{$(\mu,\nu)\mapsto\sup_{E\in\dom\mu\cap\dom\nu}|\mu E-\nu E|$,}

\noindent which will be a metric on nearly all occasions when $\rhotv$ is a
metric, and will then be uniformly equivalent to $\rhotv$.   But the more
complex formulation gives a better match to the Riesz norm metric of the
leading examples.
}%end of comment

\spheader 437Qb Suppose that $(X,\rho)$ is a metric space.   Write $\MqR$
for the set of totally finite quasi-Radon measures on $X$.
For $\mu$, $\nu\in\MqR$ set

\Centerline{$\rhoKR(\mu,\nu)
=\sup\{|\int u\,d\mu-\int u\,d\nu|:
   u:X\to[-1,1]$ is $1$-Lipschitz$\}$.}

\noindent Then $\rhoKR$ is a metric on $\MqR$.  \prooflet{\Prf\ It is
immediate from the form of the definition that $\rhoKR$ is a pseudometric.
If $\mu$, $\nu\in\MqR$ are different, there is an open set
$G$ such that $\mu G\ne\nu G$ (415G(iii));  suppose that $\mu G<\nu G$.
Set $u(x)=\rho(x,X\setminus G)$ for $x\in X$.
There must be an $n\in\Bbb N$ such that $\mu G<\nu F_n$
where $F_n=\{x:u(x)\ge 2^{-n}\}$.   In this case, setting
$u'=u\wedge 2^{-n}\chi X$,

\Centerline{$\int u'd\mu\le 2^{-n}\mu G<2^{-n}\nu F_n\le\int u'd\nu$,}

\noindent so $\rhoKR(\mu,\nu)\ge 2^{-n}(\nu F_n-\mu G)>0$.
As $\mu$ and $\nu$ are arbitrary, $\rhoKR$ is a metric.\ \Qed}

\cmmnt{\medskip

\noindent{\bf Remark} $\rhoKR$ here is taken from {\smc Bogachev 07},
\S8.3, where it is called the `Kantorovich-Rubinshtein metric'.
For its principal properties, see 437R(g)-(h) below.   A variation of
this construction will be used in 457L;  see also 437Xs.}
%Hutchinson J.E.\ [81] `Fractals and self similarity',
%Indiana Univ.\ Math.\ J.\ 30 (1981) 713-747.
%thanks to J.Pachl for reference.

\leader{437R}{Theorem}\dvAformerly{4{}37Je, 4{}37O and 4{}37R;
revised and expanded 2009-2012} Let $X$ be a topological space;  write
$\MqR=\MqR(X)$ for the set of totally finite
quasi-Radon measures on $X$, and if $X$ is Hausdorff write $\MR=\MR(X)$
for the set of totally finite Radon measures on $X$, both endowed with
their narrow topologies.

(a)(i) If $X$ is regular then $\MqR$ is Hausdorff.

\quad(ii) If $X$ is Hausdorff then $\MR$ is Hausdorff.

(b)\dvAnew{2011} If $X$ has a countable network then $\MqR$ has a countable
network.

(c) Suppose that $X$ is separable.

\quad(i) If $X$ is a T$_1$ space, then $\MqR$ is separable.

\quad(ii) If $X$ is Hausdorff, $\MR$ is separable.

(d)\dvAnew{2011} If $X$ is a K-analytic Hausdorff space, so is $\MqR=\MR$.

(e)\dvAnew{2011} If $X$ is an analytic Hausdorff space, so is $\MqR=\MR$.

(f)(i) If $X$ is compact, then for any real $\gamma\ge 0$ the sets
$\{\mu:\mu\in\MqR$, $\mu X\le\gamma\}$ and $\{\mu:\mu\in\MqR$,
$\mu X=\gamma\}$ are compact.

\quad(ii) If $X$ is compact and Hausdorff, then for any real
$\gamma\ge 0$ the sets
$\{\mu:\mu\in\MR$, $\mu X\le\gamma\}$ and $\{\mu:\mu\in\MR$,
$\mu X=\gamma\}$ are compact.  In particular, the set $\PR$ of
Radon probability measures on $X$ is compact.

(g) Suppose that $X$ is metrizable and
$\rho$ is a metric on $X$ inducing its topology.

\quad(i) The metric $\rhoKR$ on $\MqR$\cmmnt{ (437Qb)}
induces the narrow topology on $\MqR$.

\quad(ii) If $(X,\rho)$ is complete then $\MqR=\MR$ is complete
under $\rhoKR$.

(h) If $X$ is Polish, so is $\MqR=\MR$.

\proof{{\bf (a)(i)} (Cf.\ 437Qb.)   Take distinct
$\mu_0$, $\mu_1\in\MqR$.   If $\mu_0X\ne\mu_1X$ then they
can be separated by open sets of the form $\{\mu:\mu X<\alpha\}$,
$\{\mu:\mu X>\alpha\}$.   Otherwise, set $\gamma=\mu_0X=\mu_1X$.   There
is certainly an open set $G$ such that $\mu_0G\ne\mu_1G$ (415H);
suppose that $\mu_0G<\mu_1G$.   Because $\mu_1$ is inner regular with
respect to the closed sets, there is a closed set $F\subseteq G$ such
that $\mu_0G<\mu_1F$.   Now consider

\Centerline{$\Cal H
=\{H:H$ is open, $\overline{H}\subseteq G\}$.}

\noindent Then $\Cal H$ is upwards-directed;  because $X$ is regular,
$\bigcup\Cal H=G$;  because $\mu_1$ is quasi-Radon,

\Centerline{$\sup_{H\in\Cal H}\mu_1H
\ge\sup_{H\in\Cal H}\mu_1(H\cap F)=\mu_1F>\mu_0G$}

\noindent and there is an $H\in\Cal H$ such that $\mu_1H>\mu_0G$.   Now

\Centerline{$\mu_1H+\mu_0(X\setminus\overline{H})
\ge\mu_1H+\gamma-\mu_0G>\gamma$.}

\noindent Let $\alpha$, $\beta$ be such that $\mu_1H>\alpha$,
$\mu_0(X\setminus\overline{H})>\beta$ and $\alpha+\beta>\gamma$.   Then

\Centerline{$\{\mu:\mu\in\MqR$, $\mu(X\setminus\overline{H})>\beta\}$,
\quad$\{\mu:\mu\in\MqR$, $\mu H>\alpha$, $\mu X<\alpha+\beta\}$}

\noindent are disjoint open sets containing $\mu_0$, $\mu_1$ respectively,
so again we have separation.

\medskip

\quad{\bf (ii)} We can use the same ideas.   Take distinct
$\mu_0$, $\mu_1\in M^+_{\text{R}}$.   If $\mu_0X\ne\mu_1X$ then $\mu_0$ and
$\mu_1$ can be separated by open sets of the form $\{\mu:\mu X<\alpha\}$,
$\{\mu:\mu X>\alpha\}$.   Otherwise, set $\gamma=\mu_0X=\mu_1X$, and take
an open set $G$ such that $\mu_0G\ne\mu_1G$;
suppose that $\mu_0G<\mu_1G$.   Then $\mu_0(X\setminus G)+\mu_1G>\gamma$.
Because $\mu_0$ and $\mu_1$ are inner regular with respect to the compact
sets, there are compact sets
$K_0\subseteq X\setminus G$, $K_1\subseteq G$ such
that $\mu_0K_0+\mu_1K_1>\gamma$.   Now there are disjoint open sets
$H_0$, $H_1$ such that $K_i\subseteq H_i$ for both $i$ (4A2F(h-i)),
in which case $\mu_0H_0+\mu_1H_1>\gamma$.   Take $\alpha_0<\mu_0H_0$ and
$\alpha_1<\mu_1H_1$
such that $\alpha_0+\alpha_1>\gamma$.   In this case,
$\{\mu:\mu H_0>\alpha_0\}$ and
$\{\mu:\mu H_1>\alpha_1$, $\mu X<\alpha_0+\alpha_1\}$
are disjoint open sets containing $\mu_0$, $\mu_1$ respectively.

\medskip

{\bf (b)} Let $\Cal A$ be a countable network for the topology of
$X$;  replacing $\Cal A$ by
$\{\bigcup\Cal A_0:\Cal A_0\in[\Cal A]^{<\omega}\}$ if necessary,
we can suppose that $\Cal A$ is closed under finite unions.   Let $\Cal D$
be the family of sets of the form

\Centerline{$\{\mu:\mu\in\MqR$, $\mu X<\gamma$, $\mu^*A_i>\gamma_i$ for
$i\le n\}$}

\noindent where $n\in\Bbb N$, $A_0,\ldots,A_n\in\Cal A$ and $\gamma$,
$\gamma_0,\ldots,\gamma_n\in\Bbb Q$.   Then $\Cal D$ is countable.   If
$V\subseteq\MqR$ is an open set and $\mu_0\in V$, there must be
open sets $G_0,\ldots,G_n\subseteq X$ and $\gamma$,
$\gamma_0,\ldots,\gamma_n\in\Bbb Q$ such that

\Centerline{$\mu_0\in\{\mu:\mu X<\gamma$,
$\mu G_i>\gamma_i$ for every $i\le n\}
\subseteq V$.}

\noindent For each $i\le n$, $\{A:A\in\Cal A$, $A\subseteq G_i\}$ is a
countable upwards-directed set with union $G_i$, so there is a
non-decreasing sequence $\sequence{j}{A_{ij}}$ in $\Cal A$
with union $G_i$, and there
must be a $j_i\in\Bbb N$ such that $\mu_0^*A_{ij_i}>\gamma_i$ (132Ae).
Now

\Centerline{$\{\mu:\mu X<\gamma$,
$\mu^*A_{ij_i}>\gamma_i$ for every $i\le n\}$}

\noindent belongs to $\Cal D$, contains $\mu$ and is included in $V$.   As
$\mu$ and $V$ are arbitrary, $\Cal D$ is a countable network for the
topology of $\MqR$.

\medskip

{\bf (c)(i)} If $X$ is empty, then $\MqR$ is a
singleton, and
we can stop.   Otherwise, let $D$ be a countable dense subset of $X$.   Set
$D'=\{\sum_{i=0}^n\alpha_i\delta_{x_i}:x_0,\ldots,x_n\in D$,
$\alpha_0,\ldots,\alpha_n\in\Bbb Q\cap\coint{0,\infty}\}$, writing
$\delta_x$ for the Dirac measure concentrated at $x$ for each $x\in X$.
Because $X$ is T$_1$, $D'\subseteq\MqR$.   In fact $D'$ is dense in $\MqR$.
\Prf\ Take any $\mu\in\MqR$, a finite family $\Cal G$
of open subsets of $X$, and $\epsilon>0$.   Let $\Cal E$ be the
algebra of subsets of $X$ generated by $\Cal G$, and $\Cal A$ the set of
atoms of $\Cal E$.   For each $E\in\Cal A$ choose
$x_E\in D\cap\bigcap\{G:E\subseteq G\in\Cal G\}$ and
$\alpha_E\in\Bbb Q\cap\coint{0,\infty}$ such that
$|\alpha_E-\mu E|\le\Bover{\epsilon}{\#(\Cal A)}$.   Try
$\nu=\sum_{E\in\Cal A}\alpha_E\delta_{x_E}\in D'$.   If $G\in\Cal G$, then

$$\eqalign{\mu G
&=\sum_{E\in\Cal A,E\subseteq G}\mu E
\le\sum_{E\in\Cal A,x_E\in G}\mu E\cr
&\le\epsilon+\sum_{E\in\Cal A,x_E\in G}\alpha_E
=\epsilon+\nu G;\cr}$$

\noindent while

\Centerline{$\nu X
=\sum_{E\in\Cal A}\alpha_E
\le\epsilon+\sum_{E\in\Cal A}\mu E
=\epsilon+\mu X$.}

\noindent As $\mu$, $\Cal G$ and $\epsilon$ are arbitrary, $D'$ is dense in
$\MqR$.\ \QeD\   So $\MqR$ is separable.

\medskip

\quad{\bf (ii)} If $X$ is Hausdorff, use the same construction;  in this
case $D'\subseteq\MR$, so $\MR$ also is separable.

\medskip

{\bf (d)} Most of the argument will be devoted to proving
that the set $\PR$ of Radon probability measures on $X$ is K-analytic in
its narrow topology.

\medskip

\quad{\bf (i)} We are supposing that there is an usco-compact relation
$R\subseteq\NN\times X$ such that $R[\NN]=X$ (422F).   Set
$R_1=\{(\alpha,x):$ there is a $\beta\le\alpha$
such that $(\beta,\alpha)\in R\}$;  then $R_1$ also is usco-compact
(422Dh).

Set

$$\eqalign{\tilde R
&=\{(\pmb{\alpha},\mu):\pmb{\alpha}\in(\NN)^{\Bbb N},\,\mu\in\PR,\,
\mu R_1[\{\pmb{\alpha}(n)\}]\ge 1-2^{-n}\text{ for every }n\in\Bbb N\}\cr
&\subseteq(\NN)^{\Bbb N}\times\PR.\cr}$$

\noindent(Of course $R_1[\{\pmb{\alpha}(n)\}]$ is compact, therefore
universally measurable, whenever $\pmb{\alpha}\in(\NN)^{\Bbb N}$ and
$n\in\Bbb N$.)

\medskip

\quad{\bf (ii)} $\tilde R[(\NN)^{\Bbb N}]=\PR$.   \Prf\ $\NN\times X$ is
K-analytic (422Ge), while $R$ is a closed subset of $\NN\times X$ (422Da),
so is itself K-analytic (422Gf).   Let $\pi_1:R\to\NN$ and
$\pi_2:R\to X$ be the coordinate maps.   If $\mu\in\PR$, there is a Radon
probability measure $\lambda$ on $R$ such that $\mu=\lambda\pi_2^{-1}$
(432G).   For each $n\in\Bbb N$ let $L_n\subseteq R$ be a compact set such
that $\lambda L_n>1-2^{-n}$;  then $\pi_1[L_n]$ is a non-empty
compact subset of $\NN$.   Define $\pmb{\alpha}$ by setting

\Centerline{$\pmb{\alpha}(n)(m)=\sup\{\beta(m):\beta\in\pi_1[L_n]\}$}

\noindent for $m$, $n\in\Bbb N$.   Then

\Centerline{$\mu R_1[\{\pmb{\alpha}(n)\}]
\ge\mu R[\pi_1[L_n]]
\ge\mu\pi_2[L_n]
\ge\lambda L_n
\ge 1-2^{-n}$}

\noindent for every $n\in\Bbb N$, so $(\pmb{\alpha},\mu)\in\tilde R$ and
$\mu\in\tilde R[(\NN)^{\Bbb N}]$.\ \Qed

\medskip

\quad{\bf (iii)} $\tilde R[\{\pmb{\alpha}\}]$ is a compact subset of $\PR$
for every $\pmb{\alpha}\in(\NN)^{\Bbb N}$.   \Prf\ Since
$R_1[\{\pmb{\alpha}(n)\}]$ is compact for every $n$,
$\tilde R[\{\pmb{\alpha}\}]$ is uniformly tight, therefore
relatively compact in $\MR$, by
437Pb.   On the other hand, $\{\mu:\mu\in\MR$, $\mu X=1\}$
and $\{\mu:\mu\in\MR$,
$\mu(X\setminus R_1[\{\pmb{\alpha}(n)\}])\le 1-2^{-n}\}$ are closed for
every $n$, so $\tilde R[\{\pmb{\alpha}\}]$ is closed, therefore compact.\
\Qed

\medskip

\quad{\bf (iv)} If $F\subseteq\PR$ is closed, then $\tilde R^{-1}[F]$ is
closed in $(\NN)^{\Bbb N}$.   \Prf\ Let $\sequence{k}{\pmb{\alpha}_k}$ be a
sequence in $\tilde R^{-1}[F]$ converging to $\pmb{\alpha}$ in
$(\NN)^{\Bbb N}$.   For each $k\in\Bbb N$ choose $\mu_k\in F$ such that
$(\pmb{\alpha}_k,\mu_k)\in\tilde R$.   For $n$, $k\in\Bbb N$ set
$L_{nk}=\{\pmb{\alpha}(n)\}\cup\{\pmb{\alpha}_l(n):l\ge k\}$.   Then
$L_{nk}$ is a compact subset of $\NN$, so $R_1[L_{nk}]$ is a compact subset
of $X$.   \Quer\ If
$x\in\bigcap_{k\in\Bbb N}R_1[L_{nk}]\setminus R_1[\{\pmb{\alpha}(n)\}]$, then
for every $k\in\Bbb N$ there is an $l_k\ge k$ such that
$(\pmb{\alpha}_{l_k}(n),x)\in R_1$;  but $R_1^{-1}[\{x\}]$ is closed in
$\NN$, so contains
$\lim_{k\to\infty}\pmb{\alpha}_{l_k}(n)=\pmb{\alpha}(n)$, and
$x\in R_1[\{\pmb{\alpha}(n)\}]$.\ \BanG\   Thus
$\bigcap_{k\in\Bbb N}R_1[L_{nk}]=R_1[\{\pmb{\alpha}(n)\}]$.

For any $n$ and $k$,
$\mu_lR_1[L_{nk}]\ge\mu_lR_1[\{\pmb{\alpha}_l(n)\}]\ge 1-2^{-n}$ for every
$l\ge k$.   In the first place, taking $k=0$,
$\{\mu_l:l\in\Bbb N\}$ is uniformly tight, therefore relatively compact and
$\sequence{l}{\mu_l}$ has
a cluster point $\mu$ say, which must belong to $F$.   Now, for any $n$,

$$\eqalignno{\mu R_1[\{\pmb{\alpha}(n)\}]
&=\inf_{k\in\Bbb N}\mu R_1[L_{nk}]
\ge\inf_{k\in\Bbb N,l\ge k}\mu_lR_1[L_{nk}]\cr
\displaycause{because $R_1[L_{nk}]$ is compact, therefore closed, and
$\mu\in\overline{\{\mu_l:l\ge k\}}$, for each $k$}
&\ge\inf_{k\in\Bbb N,l\ge k}\mu_lR_1[\{\pmb{\alpha}_l(n)\}]
\ge 1-2^{-n}.\cr}$$

\noindent So $(\pmb{\alpha},\mu)\in\tilde R$ and
$\pmb{\alpha}\in\tilde R^{-1}[F]$.   As $\sequence{k}{\pmb{\alpha}_k}$ is
arbitrary, $\tilde R^{-1}[F]$ is closed.\ \Qed

\medskip

\quad{\bf (v)} Thus $\tilde R\subseteq(\NN)^{\Bbb N}\times\PR$ is
usco-compact.   Since $(\NN)^{\Bbb N}$, like $\NN$, is Polish (4A2Ub,
4A2Qc), $\tilde R[(\NN)^{\Bbb N}]$ is K-analytic (422Gd).   But we saw in
(ii) that $\tilde R[(\NN)^{\Bbb N}]=\PR$.   So $\PR$ is K-analytic.

\medskip

\quad{\bf (vi)} Now observe that
$(\alpha,\mu)\mapsto\alpha\mu:\coint{0,\infty}\times\PR\to\MR$ is
continuous.   \Prf\ We have only to note that

\Centerline{$(\alpha,\mu)\mapsto(\alpha\mu)(X)=\alpha$}

\noindent is continuous, and that

\Centerline{$\{(\alpha,\mu):(\alpha\mu)(G)>\gamma\}
=\bigcup_{\beta>0}\{(\alpha,\mu):\alpha>\beta$,
$\mu G>\Bover{\gamma}{\beta}\}$}

\noindent is open for every open $G\subseteq X$ and $\gamma\ge 0$.\ \QeD\
Since $\coint{0,\infty}$ and $\PR$ are K-analytic, and (except in the
trivial case $X=\emptyset$) every member of
$\MR$ is expressible as a non-negative multiple of a probability measure,
$\MR$ is K-analytic (using 422Ge and 422Gd again).

\medskip

\quad{\bf (vii)} Finally, $\MqR=\MR$ by 432E.

\medskip

{\bf (e)} Put (d), (b) and 423C together.

\medskip

{\bf (f)(i)} Because $X$ is compact, every quasi-Radon measure on $X$ is
tight, and $\MqR$ itself is uniformly tight;
by 437Pa, $\{\mu:\mu\in\MqR$, $\mu X\le\gamma\}$
is relatively compact in $\MqR$.   But as
it is also closed in $\MqR$, it is actually compact.   The same argument
applies to $\{\mu:\mu\in\MqR$, $\mu X=\gamma\}$.

\medskip

\quad{\bf (ii)} Use the same idea, but with 437Pb in place of 437Pa.

\medskip

{\bf (g)(i)} Write $\frakTKR$ for the topology generated by
$\rhoKR$.

\medskip

\qquad\grheada\ If $\mu\in\MqR$ and $\mu X>\alpha$, then $\nu X>\alpha$
whenever $\nu\in\MqR$ and $\rhoKR(\mu,\nu)<\mu X-\alpha$, just because
$\chi X$ is a $1$-Lipschitz function;  so
$\{\mu:\mu\in\MqR$, $\mu X>\alpha\}\in\frakTKR$ for every
$\alpha\in\Bbb R$.

If $G\subseteq X$ is open, $\alpha\ge 0$ and
$\mu\in\MqR$ is such that $\mu G>\alpha$, there is a
$\delta\in\ocint{0,1}$ such that $\mu F>\alpha+\delta$, where
$F=\{x:\rho(x,X\setminus G)\ge\delta\}$.   Let $u$ be a $1$-Lipschitz
function such that $\delta\chi F\le u\le\delta\chi G$.
If $\nu\in\MqR$ and $\rhoKR(\mu,\nu)\le\delta^2$, then

\Centerline{$\delta\nu G
\ge\int u\,d\nu
\ge\int u\,d\mu-\delta^2
\ge\delta\mu F-\delta^2
>\delta\alpha$}

\noindent and $\nu G>\alpha$.   This shows that
$\{\mu:\mu\in\MqR$, $\mu G>\alpha\}\in\frakTKR$.   As $G$ and $\alpha$ are
arbitrary, $\frakTKR$ is finer than the narrow topology.

\medskip

\qquad\grheadb\ Suppose that $\mu\in\MqR$ and
$\epsilon>0$;  let $\delta>0$ be such that
$\delta(3\delta+6\mu X+7)\le\epsilon$.   Then there is a
totally bounded closed set $F\subseteq X$ such that
$\mu(X\setminus F)\le\delta$ (434L).   Set $G=\{x:\rho(x,F)<\delta\}$.
Let $x_0,\ldots,x_n\in X$ be such
that $F\subseteq\bigcup_{i\le n}B(x_i,\delta)$;
then $G\subseteq\bigcup_{i\le n}B(x_i,2\delta)$ and
there are $v_0,\ldots,v_n\in C_b(X)^+$
such that $\chi G\le\sum_{i=0}^nv_i(x)\le\chi X$ and
$\{x:v_i(x)>0\}\subseteq B(x_i,3\delta)$ for every $i\le n$.   Let
$w\in C_b(X)$ be such that $\chi(X\setminus G)\le w\le\chi(X\setminus F)$.
By the choice of $F$, $\int w\,d\mu\le\delta$.

If $u:X\to[-1,1]$ is $1$-Lipschitz then

\Centerline{$|u-\sum_{i=0}^nu(x_i)v_i|\le 3\delta\chi X+2w$.}

\noindent\Prf\ If $x\in G$, then

$$\eqalignno{|u(x)-\sum_{i=0}^nu(x_i)v_i(x)|
&=|\sum_{i=0}^n(u(x)-u(x_i))v_i(x)|\cr
\displaycause{because $\sum_{i=0}^nv_i(x)=1$}
&\le\sum_{i=0}^n|u(x)-u(x_i)|v_i(x)
\le\sum_{i=0}^n3\delta v_i(x)\cr
\displaycause{because whenever $v_i(x)>0$,
$|u(x)-u(x_i)|\le\rho(x,x_i)\le 3\delta$}
&\le 3\delta.\cr}$$

\noindent If $x\in X\setminus G$, then

\Centerline{$|u(x)-\sum_{i=0}^nu(x_i)v_i(x)|
\le|u(x)|+\sum_{i=0}^n|u(x_i)|v_i(x)
\le 2=2w(x)$.  \Qed}

\noindent So if $\nu\in\MqR$,

\Centerline{$|\int u\,d\nu-\sum_{i=0}^nu(x_i)\int v_id\nu|
\le 3\delta\nu X+2\int w\,d\nu$.}

By 437Jf or 437L,
there is a neighbourhood $V$ of $\mu$ for the narrow topology
in $\MqR$ such that if $\nu\in V$ then $\nu X\le\mu X+\delta$,
$\int w\,d\nu\le 2\delta$ and
$|\int v_id\mu-\int v_id\nu|\le\Bover{\delta}{n+1}$ for every
$i\le n$.   So if $\nu\in V$ and
$u:X\to[-1,1]$ is $1$-Lipschitz, we shall have

$$\eqalign{\bigl|\int u\,d\nu-\int u\,d\mu\bigr|
&\le 3\delta(\nu X+\mu X)+2(\int w\,d\nu+\int w\,d\mu)\cr
&\mskip180mu
   +\sum_{i=0}^n\bigl|\int v_id\nu-\int v_id\mu\bigr|\cr
&\le 3\delta(2\mu X+\delta)+6\delta
   +\sum_{i=0}^n\Bover{\delta}{n+1}
\le\delta(6\mu X+3\delta+7)\le\epsilon.\cr}$$

\noindent Thus $\{\nu:\rhoKR(\nu,\mu)\le\epsilon\}\supseteq V$
is a neighbourhood
of $\mu$ for the narrow topology;  as $\mu$ and $\epsilon$ are arbitrary,
the narrow topology is finer than $\frakTKR$, and the two topologies are
equal.

\medskip

\quad{\bf (ii)}
If $X$ is $\rho$-complete then $\MqR=\MR$ by 434Jg and 434Jb.
Now suppose that $\sequencen{\mu_n}$ is a $\rhoKR$-Cauchy sequence
in $\MR$.

\medskip

\qquad\grheada\ For every $\epsilon\in\ocint{0,1}$ there is a compact
$K\subseteq X$
such that $\mu_n(X\setminus U(K,\epsilon))\le\epsilon$ for every
$n\in\Bbb N$, where
$U(K,\epsilon)=\{x:\rho(x,y)<\epsilon$ for some $y\in K\}$.   \Prf\
Take $m\in\Bbb N$ such that $\rhoKR(\mu_m,\mu_n)\le\bover12\epsilon^2$
for every $n\ge m$.   Let $K\subseteq X$ be a compact set such that
$\mu_n(X\setminus K)\le\bover12\epsilon$ for every $n\le m$.
Set $G=U(K,\epsilon)$.   There is a
$1$-Lipschitz function $u:X\to[0,\epsilon]$ such that
$\epsilon\chi(X\setminus G)\le u\le\chi(X\setminus K)$.   If $n\le m$, then
of course $\mu_n(X\setminus G)\le\epsilon$.   If $n\ge m$, then

$$\eqalign{\mu_n(X\setminus G)
&\le\Bover1{\epsilon}\int u\,d\mu_n
\le\Bover1{\epsilon}\bigl(\rhoKR(\mu_n,\mu_m)+\int u\,d\mu_m\bigr)\cr
&\le\Bover1{\epsilon}
  \bigl(\Bover{\epsilon^2}2+\epsilon\mu_m(X\setminus K)\bigr)
\le\Bover{\epsilon}2+\mu_m(X\setminus K)
\le\epsilon.\cr}$$

\noindent So we have an appropriate $K$.\ \Qed

\medskip

\qquad\grheadb\ $\{\mu_n:n\in\Bbb N\}$ is uniformly totally finite and
uniformly tight.   \Prf\ Since $|\mu_mX-\mu_nX|\le\rhoKR(\mu_m,\mu_n)$
for all $m$, $n\in\Bbb N$, $\{\mu_nX:n\in\Bbb N\}$ is bounded.
Of course all the $\mu_n$ are tight.   Now take any
$\epsilon\in\ocint{0,1}$.   For each $m\in\Bbb N$, (i) tells us that
there is a compact set $K_m\subseteq X$ such that
$\mu_n(X\setminus U(K_m,2^{-m}\epsilon))\le 2^{-m}\epsilon$ for every
$n\in\Bbb N$.   Set $E=\bigcap_{m\in\Bbb N}U(K_m,2^{-m}\epsilon)$,
$K=\overline{E}$.   Then $E$ and $K$ are totally bounded;  because
$(X,\rho)$ is complete, $K$ is compact.   And

\Centerline{$\mu_n(X\setminus K)
\le\sum_{m=0}^{\infty}\mu_n(X\setminus U(K_m,2^{-m}\epsilon))
\le 2\epsilon$}

\noindent for every $n\in\Bbb N$.   As $\epsilon$ is arbitrary,
$\{\mu_n:n\in\Bbb N\}$ is uniformly tight.\ \Qed

\medskip

\qquad\grheadc\ By 437Pb, $\sequencen{\mu_n}$ has a cluster point $\mu$
in $\MR$ for the narrow topology.   Now, for $m\in\Bbb N$,

$$\eqalignno{\rhoKR(\mu,\mu_m)
&=\sup\{|\int u\,d\mu-\int u\,d\mu_m|:
  u:X\to[-1,1]\text{ is }1\text{-Lipschitz}\}\cr
&\le\sup\{|\int u\,d\mu_n-\int u\,d\mu_m|:
  u:X\to[-1,1]\text{ is }1\text{-Lipschitz},\,n\ge m\}\cr
\displaycause{437Jf again}
&\le\sup_{n\ge m}\rhoKR(\mu_n,\mu_m),\cr}$$

\noindent and $\lim_{n\to\infty}\rhoKR(\mu,\mu_m)=0$.

\medskip

\qquad\grheadd\ Thus every $\rhoKR$-Cauchy sequence in $\MR$ has a
limit in $\MR$, and $\MR$ is complete.

\medskip

{\bf (h)} Put (g-ii) and (c-ii) together.
}%end of proof of 437R

\leader{437S}{}\cmmnt{ The sets of measures we have been considering have
generally been convex, if addition and multiplication by non-negative
scalars are defined as in 234G and 234Xf.   We can therefore look for
extreme points, in the hope that they will have straightforward
characterizations, as in the following.

\medskip

\noindent}{\bf Proposition}\dvAformerly{4{}37P} Let $X$ be a Hausdorff
space, and $\PR$ the
set of Radon probability measures on $X$.   Then the extreme points
of $\PR$ are just the Dirac measures on $X$.

\proof{{\bf (a)} Suppose that $x\in X$, and that $\delta_x$ is the Dirac
measure on $X$ concentrated at $x$.   If $\mu_1$,
$\mu_2\in\PR$ are
such that $\delta_x=\bover12(\mu_1+\mu_2)$, then we must have
$\mu_1E\le 2\mu E$ for every Borel set $E$;
in particular, $\mu_1(X\setminus\{x\})=0$ and $\mu_1\{x\}=1$, that is,
$\mu_1=\delta_x$.   Similarly, $\mu_2=\delta_x$;  as $\mu_1$ and $\mu_2$
are arbitrary, $\delta_x$ is an extreme point of $\PR$.

\medskip

{\bf (b)} Suppose that $\mu$ is an extreme point of $\PR$.   Let $K$ be
the support of $\mu$.   \Quer\ If $K$ has more than one point, take
distinct $x$, $y\in K$.
As $X$ is Hausdorff, there are disjoint open sets $G$, $H$ such that
$x\in G$ and $y\in H$.   Set $E=G\cap K$, $\alpha=\mu E$.   Because $K$
is the support of $\mu$,
$\alpha>0$.   But similarly $\mu(H\cap K)>0$ and $\alpha<1$.   Let
$\mu_1$, $\mu_2$ be the indefinite-integral measures defined over $\mu$
by $\Bover1{\alpha}\chi E$ and $\Bover1{\beta}\chi(X\setminus E)$
respectively.   Then both are Radon probability measures on $X$ (416S),
so belong to $\PR$.
Now $\mu F=\alpha\mu_1F+(1-\alpha)\mu_2F$ for every Borel set $F$;  as
$\mu$ and $\alpha\mu_1+(1-\alpha)\mu_2$ are both Radon measures, they
coincide;  as
neither $\mu_1$ nor $\mu_2$ is equal to $\mu$, $\mu$ is not extreme in
$\PR$.\ \Bang

Thus $K=\{x\}$ for some $x\in X$.   But this means that $\mu\{x\}=1$ and
$\mu(X\setminus\{x\})=0$, so $\mu=\delta_x$ is of the declared form.
}%end of proof of 437S

%437xT
\leader{437T}{}\cmmnt{ We now have a language in which to express a
fundamental result in the theory of dynamical systems.

\medskip

\noindent}{\bf Theorem}\dvAformerly{4{}37S} Let $X$ be a non-empty
compact Hausdorff space, and $\phi:X\to X$ a continuous function.
Write $Q_{\phi}$ for the set of Radon probability measures on $X$ for
which $\phi$ is \imp.   Then $Q_{\phi}$ is convex and not empty, and is
compact for the narrow topology.

\proof{{\bf (a)} Write $\MR$ for the set of totally finite Radon
measures on $X$, and let $\tilde\phi:\MR\to\MR$ be the function
corresponding to $\phi:X\to X$ as described in  437N.
Now, for $\mu\in\MR$, $\mu\in Q_{\phi}$ iff $\mu X=1$ and
$\mu(\phi^{-1}[E])=\mu E$ whenever $\mu$ measures $E$, that is, iff the
image measure $\mu\phi^{-1}=\tilde\phi(\mu)$ extends $\mu$.   But as
$\tilde\phi(\mu)$ and $\mu$ are Radon measures, $\mu\in Q_{\phi}$ iff
$\mu X=1$ and $\tilde\phi(\mu)=\mu$.

Since $\tilde\phi$ is continuous (and $\MR$ is Hausdorff, see 437Ra),
$Q_{\phi}$ is closed for the narrow topology.   By 437Pb/437R(f-ii),
it is compact.   Because $\tilde\phi$ respects addition and scalar
multiplication, $Q_{\phi}$ is convex.

\medskip

{\bf (b)} To see that $Q_{\phi}$ is not empty,
take any $x_0\in X$ and a non-principal ultrafilter $\Cal F$ on
$\Bbb N$.   Define $f:C(X)\to\Bbb R$ by setting
$f(u)=\lim_{n\to\Cal F}\bover1{n+1}\sum_{i=0}^nu(\phi^i(x_0))$ for every
$u\in C(X)$.   Then $f$ is a positive linear functional and
$f(\chi X)=1$.
So there is a $\mu\in\MR$ such that $f(u)=\int u\,d\mu$ for every
$u\in C(X)$.

If $u\in C(X)$, then $f(u)=f(u\phi)$.   \Prf\

$$\eqalign{|f(u\phi)-f(u)|
&=|\lim_{n\to\Cal F}\Bover1{n+1}
  \sum_{i=0}^nu(\phi^{i+1}(x_0))-u(\phi^i(x_0))|\cr
&=|\lim_{n\to\Cal F}\Bover1{n+1}u(\phi^{n+1}(x_0))-u(x_0)|\cr
&\le\lim_{n\to\Cal F}\Bover1{n+1}|u(\phi^{n+1}(x_0))-u(x_0)|
\le\lim_{n\to\Cal F}\Bover{2\|u\|_{\infty}}{n+1}
=0.  \text{ \Qed}\cr}$$

On the other hand,

$$\eqalignno{\int u\,d(\tilde\phi(\mu))
&=\int u\phi\,d\mu\cr
\displaycause{235G}
&=f(u\phi)
=f(u)
=\int u\,d\mu\cr}$$

\noindent for every $u\in C(X)$.   By the uniqueness of the representation
of $f$ as an integral, $\mu=\tilde\phi(\mu)$.   Of course
$\mu X=f(\chi X)=1$ so $\mu\in Q_{\phi}$, as required.
}%end of proof of 437T

\leader{437U}{}\cmmnt{ In important cases, the narrowly compact
subsets of $M^+_{\text{R}}(X)$ are exactly the bounded uniformly tight
sets.   Once again, it is worth introducing a word to describe when this
happens.

\medskip

\noindent}{\bf Definition}\dvAformerly{4{}37V} Let $X$ be a Hausdorff
space and $P_{\text{R}}(X)$ the set
of Radon probability measures on $X$.   $X$ is a {\bf Prokhorov space}
if every subset of $P_{\text{R}}(X)$ which is compact for the narrow
topology is uniformly tight.

\vleader{60pt}{437V}{Theorem}\dvAformerly{4{}37W}
(a) Compact Hausdorff spaces are Prokhorov spaces.

(b) A closed subspace of a Prokhorov Hausdorff space is a Prokhorov
space.

(c) An open subspace of a Prokhorov Hausdorff space is a Prokhorov
space.

(d) The product of a countable family of Prokhorov Hausdorff spaces is a
Prokhorov space.

(e) Any G$_{\delta}$ subset of a Prokhorov Hausdorff space is a
Prokhorov space.

(f) \v{C}ech-complete spaces are Prokhorov spaces.

(g) Polish spaces are Prokhorov spaces.

\proof{{\bf (a)} This is trivial;  on a compact Hausdorff space the set
of all Radon probability measures is uniformly tight.

\medskip

{\bf (b)} Let $X$ be a Prokhorov Hausdorff space, $Y$ a closed subset of
$X$, and $A\subseteq P_{\text{R}}(Y)$ a narrowly compact set.   Taking
$\phi$ to be the identity map from $Y$ to $X$, and defining
$\tilde\phi:M^+_{\text{R}}(Y)\to M^+_{\text{R}}(X)$ as in 437N,
$\tilde\phi[A]$ is narrowly compact in
$P_{\text{R}}(X)$, so is uniformly tight.   For any $\epsilon>0$, there
is a compact set $K\subseteq X$ such that
$\tilde\phi(\mu)(X\setminus K)\le\epsilon$
for every $\mu\in A$.   Now $K\cap Y$ is a compact subset of $Y$ and
$\mu(Y\setminus(K\cap Y))\le\epsilon$ for every $\mu\in A$.   As
$\epsilon$ is arbitrary,
$A$ is uniformly tight in $P_{\text{R}}(Y)$.

\medskip

{\bf (c)} Let $X$ be a Prokhorov Hausdorff space, $Y$ an open subset of
$X$, and $A\subseteq P_{\text{R}}(Y)$ a narrowly compact set.   Once
again, take $\phi$ to be the
identity map from $Y$ to $X$, so that
$\tilde\phi[A]\subseteq P_{\text{R}}(X)$ is narrowly compact and
uniformly tight in $P_{\text{R}}(X)$.

\Quer\ Suppose, if possible, that $A$ is not uniformly tight in
$P_{\text{R}}(Y)$.   Then there is an $\epsilon>0$ such that
$A_K=\{\mu:\mu\in A$, $\mu(Y\setminus K)\ge 5\epsilon\}$ is non-empty
for every compact set $K\subseteq Y$.   Note that $A_K\subseteq A_{K'}$
whenever
$K\supseteq K'$, so $\{A_K:K\subseteq Y$ is compact$\}$ has the finite
intersection property, and there is an ultrafilter $\Cal F$ on
$P_{\text{R}}(Y)$
containing every $A_K$.   Because $A$ is narrowly compact, there is a
$\lambda\in P_{\text{R}}(Y)$ such that $\Cal F\to\lambda$.   Let
$K^*\subseteq Y$ be
a compact set such that $\lambda(Y\setminus K^*)\le\epsilon$.

As $\tilde\phi[A]$ is uniformly tight, there is a
compact set $L\subseteq X$ such that
$\mu(Y\setminus L)=\tilde\phi(\mu)(X\setminus L)\le\epsilon$
for every $\mu\in A$.   Now $K^*$ and $L\setminus Y$ are disjoint
compact sets in the Hausdorff space $X$, so there are disjoint open sets
$G$, $H\subseteq X$
such that $K^*\subseteq G$ and $L\setminus Y\subseteq H$ (4A2F(h-i) again).   Set
$K=L\setminus H\supseteq L\cap G$;  then
$K$ is a compact subset of $Y$.   As $A_K\in\Cal F$, there must be a
$\mu\in A_K$ such that $\mu Y\le\lambda Y+\epsilon$ and
$\mu(G\cap Y)\ge\lambda(G\cap Y)-\epsilon$.   Accordingly

$$\eqalign{\mu(Y\setminus L)
&\le\epsilon,\cr
\mu(Y\setminus G)
&=\mu Y-\mu(G\cap Y)
\le\lambda Y+\epsilon-\lambda(G\cap Y)+\epsilon\cr
&=\lambda(Y\setminus G)+2\epsilon
\le\lambda(Y\setminus K^*)+2\epsilon\le 3\epsilon,\cr
\mu((Y\setminus L)\cup(Y\setminus G))
&=\mu(Y\setminus(L\cap G))\ge\mu(Y\setminus K)\ge 5\epsilon,\cr}$$

\noindent which is impossible.\ \Bang

Thus $A$ is uniformly tight.   As $A$ is arbitrary, $Y$ is a Prokhorov
space.

\medskip

{\bf (d)} Let $\sequencen{X_n}$ be a sequence of Prokhorov Hausdorff
spaces with product $X$.   Let $A\subseteq P_{\text{R}}(X)$ be a
narrowly compact set.
Let $\epsilon>0$.   For each $n\in\Bbb N$ let $\pi_n:X\to X_n$ be the
canonical map and
$\tilde\pi_n:M^+_{\text{R}}(X)\to M^+_{\text{R}}(X_n)$ the
associated function.
Then $\tilde\pi_n[A]$ is narrowly compact in $P_{\text{R}}(X_n)$,
therefore uniformly tight, and there is a compact set $K_n\subseteq X_n$
such that
$(\tilde\pi_n\mu)(X_n\setminus K_n)\le 2^{-n-1}\epsilon$ for every
$\mu\in A$.   Set
$K=\prod_{n\in\Bbb N}K_n$, so that $K$ is a compact subset of $X$ and
$X\setminus K=\bigcup_{n\in\Bbb N}\pi_n^{-1}[X_n\setminus K_n]$.   If
$\mu\in A$, then

\Centerline{$\mu(X\setminus K)
\le\sum_{n=0}^{\infty}\mu\pi_n^{-1}[X_n\setminus K_n]
\le\sum_{n=0}^{\infty}2^{-n-1}\epsilon=\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $A$ is uniformly tight;  as $A$ is
arbitrary, $X$ is a Prokhorov space.

\medskip

{\bf (e)} Let $X$ be a Prokhorov Hausdorff space and $Y$ a G$_{\delta}$
subset of $X$.   Express $Y$ as $\bigcap_{n\in\Bbb N}Y_n$ where every
$Y_n\subseteq X$ is open.
Set $Z=\{z:z\in\prod_{n\in\Bbb N}Y_n$, $z(m)=z(n)$ for all $m$,
$n\in\Bbb N\}$.
Because $X$ is Hausdorff, $Z$ is a closed subspace of
$\prod_{n\in\Bbb N}Y_n$ homeomorphic to $Y$.
Putting (c), (d) and (b) together, $Z$ and $Y$ are Prokhorov spaces.

\medskip

{\bf (f)} Put (a), (e) and the definition of `\v{C}ech-complete'
together.

\medskip

{\bf (g)} This is a special case of (f) (4A2Md).
}%end of proof of 437V

\exercises{
\leader{437X}{Basic exercises (a)}
%\spheader 437Xa
Let $X$ be a set, $U$ a Riesz subspace of $\Bbb R^X$ and
$f\in U^{\sim}$.   (i) Show that $f\in U^{\sim}_{\sigma}$ iff
$\lim_{n\to\infty}f(u_n)=0$ whenever $\sequencen{u_n}$ is a
non-increasing sequence in $U$ such that $\lim_{n\to\infty}u_n(x)=0$ for
every $x\in X$.   \Hint{show that in this case, if $0\le v_n\le u_n$, we
can find $k(n)$ such that $|f(v_n\vee u_{k(n)})-f(v_n)|\le 2^{-n}$.}
(ii) Show that $f\in U^{\sim}_{\tau}$ iff $\inf_{u\in A}|f(u)|=0$
whenever $A\subseteq U$ is a non-empty downwards-directed set and
$\inf_{u\in A}u(x)=0$ for every $x\in X$.   \Hint{given $\epsilon>0$,
set $B=\{v:f(v)\ge\inf_{u\in A}f^+(u)-\epsilon,\,\exists\,w\in A,\,v\ge
w\}$ and show that $B$ is a downwards-directed set with infimum $0$ in
$\Bbb R^X$.}
%437A

\spheader 437Xb Let $X$ be a set, $U$ a Riesz subspace of
$\ell^{\infty}(X)$ containing the constant functions, and $\Sigma$ the
smallest $\sigma$-algebra of subsets of $X$ with respect to which every
member of $U$ is measurable.   Let $\mu$ and $\nu$ be two totally finite
measures on $X$ with domain $\Sigma$, and $f$, $g$ the corresponding
linear functionals on $U$.   Show that $f\wedge g=0$ in $U^{\sim}$ iff
there is an $E\in\Sigma$ such that $\mu E=\nu(X\setminus E)=0$.
\Hint{326M\formerly{3{}26I}.}
%437C

\sqheader 437Xc\dvAnew{2012} Let $I\subseteq\Bbb R$ be a
interval with at least two points.   (i) Show that if $g:I\to\Bbb R$ is of
bounded variation on every compact subinterval of $I$,
there is a unique signed
tight Borel measure $\mu_g$ on $I$ such that
$\mu_g[a,b]=\lim_{x\downarrow b}g(x)-\lim_{x\uparrow a}g(x)$ whenever
$a\le b$ in $I$, counting $\lim_{x\uparrow a}g(x)$ as $g(a)$ if
$a=\min I$, and $\lim_{x\downarrow b}g(x)$ as $g(b)$ if $b=\max I$.
(ii) Show that if $h:I\to\Bbb R$ is another
function of bounded variation on every compact subinterval, then
$\mu_h=\mu_g$ iff $\{x:h(x)\ne g(x)\}$ is countable iff
$\{x:h(x)=g(x)\}$ is dense in $I$.
(iii) Show that if $\nu$ is any signed Baire
measure on $I$ there is a $g$ of bounded variation on every compact
subinterval such that $\nu=\mu_g$.
%437G

\spheader 437Xd(i) Show that $S$, in 437C, is the unique sequentially
order-continuous positive linear operator from $\eusm L^{\infty}$ to
$(U^{\sim}_{\sigma})^*$ which extends the canonical embedding of $U$ in
$(U^{\sim}_{\sigma})^*$.   (ii) Show that $S$, in 437H, is the unique
sequentially order-continuous positive linear operator from $\eusm
L^{\infty}$ to $(U^{\sim}_{\tau})^*$ which extends the canonical
embedding of $U$ in $(U^{\sim}_{\tau})^*$ and is `$\tau$-additive' in
the sense that whenever $\Cal G$ is a non-empty upwards-directed family
of open sets with union $H$ then $S(\chi H)=\sup_{G\in\Cal G}S(\chi G)$
in $(U^{\sim}_{\tau})^*$.
%437H

\spheader 437Xe Let $X$ and $Y$ be completely regular
topological spaces and $\phi:X\to Y$
a continuous function.   Define $T:C_b(Y)\to C_b(X)$ by setting
$T(v)=v\phi$ for every $v\in C_b(Y)$, and let $T':C_b(X)^*\to C_b(Y)^*$
be its adjoint.   (i) Show that $T'$ is a norm-preserving Riesz
homomorphism.   (ii) Show that
$T'[C_b(X)^{\sim}_{\sigma}]\subseteq C_b(Y)^{\sim}_{\sigma}$, and that
if $f\in C_b(X)^{\sim}_{\sigma}$ corresponds to a Baire measure $\mu$ on
$X$, then $T'f$ corresponds to the Baire measure
$\mu\phi^{-1}\restr\CalBa(Y)$.   (iii) Show that
$T'[C_b(X)^{\sim}_{\tau}]\subseteq C_b(Y)^{\sim}_{\tau}$, and that if
$f\in C_b(X)^{\sim}_{\tau}$ corresponds to a Borel measure $\mu$ on $X$,
then $T'f$ corresponds to the Borel measure
$\mu\phi^{-1}\restr\Cal B(Y)$.   (iv) Write $\Cal L^{\infty}_X$ and
$\Cal L^{\infty}_Y$ for the
$M$-spaces of bounded real-valued Borel measurable functions on $X$, $Y$
respectively, and $S_X:\eusm L^{\infty}_X\to(C_b(X)^{\sim}_{\tau})^*$,
$S_Y:\eusm L^{\infty}_Y\to(C_b(Y)^{\sim}_{\tau})^*$ for the canonical
Riesz homomorphisms as constructed in 437Hb.   Show that if
$T'':(C_b(Y)^{\sim}_{\tau})^*\to(C_b(X)^{\sim}_{\tau})^*$ is the adjoint
of $T'\restr C_b(X)^{\sim}_{\tau}$, then $T''S_Y(v)=S_X(v\phi)$ for
every $v\in\eusm L^{\infty}_Y$.
%437H

\spheader 437Xf Let $X$ be a topological space,
$\eusm L^{\infty}(\Sigma_{\text{um}})$ the
space of bounded universally measurable real-valued functions on $X$,
and $M_{\sigma}$ the space of countably additive functionals on the
Borel $\sigma$-algebra of $X$.
Show that we have a sequentially order-continuous
Riesz homomorphism
$S:\eusm L^{\infty}(\Sigma_{\text{um}})\to M_{\sigma}^*$ defined by the
formula $(Sv)(\mu)=\int v\,d\mu$ whenever
$v\in\eusm L^{\infty}(\Sigma_{\text{um}})$ and $\mu\in M_{\sigma}^+$.
%437I

\spheader 437Xg Let $X$ be a completely regular topological space.
Show that the vague topology on the space $M_{\tau}$ of differences of
$\tau$-additive totally finite Borel measures on $X$ is Hausdorff.
%437J

\sqheader 437Xh Let $X$ and $Y$ be topological spaces, and $\phi:X\to Y$
a continuous function.   Write $M_{\#}(X)$ for any of
$M(\CalBa(X))$, $M_{\sigma}(\CalBa(X))$, $M(\Cal B(X))$,
$M_{\sigma}(\Cal B(X))$, $M_{\tau}(X)$ or $M_t(X)$, where
$M_{\tau}(X)\subseteq M_{\sigma}(\Cal B(X))$
is the space of signed $\tau$-additive Borel measures and
$M_t(X)\subseteq M_{\tau}(X)$ is the space of signed tight
Borel measures;  and $M_{\#}(Y)$ for the corresponding space based on
$Y$.   Show that there is a positive linear operator
$\tilde\phi:M_{\#}(X)\to M_{\#}(Y)$ defined by saying that
$\tilde\phi(\mu)(E)=\mu\phi^{-1}[E]$ whenever $\mu\in M_{\#}(X)$
and $E$ belongs to $\CalBa(Y)$ or $\Cal B(Y)$, as appropriate, and that
$\tilde\phi$ is continuous for the vague topologies on $M_{\#}(X)$ and
$M_{\#}(Y)$.
%437J

\sqheader 437Xi Let $X$ be a zero-dimensional compact Hausdorff
space and $\Cal E$ the algebra of open-and-closed subsets of $X$.   (i)
Show that $\Cal E$ separates
zero sets.   (ii) Show that the vague topology on $M(\Cal E)$ is just
the topology of pointwise convergence induced by the usual topology of
$\BbbR^{\Cal E}$.
(iii) Writing $M_t$ for the space of signed tight Borel measures on $X$,
show that $\mu\mapsto\mu\restr\Cal E:M_t\to M(\Cal E)$ is a Banach
lattice isomorphism
between the $L$-spaces $M_t$ and $M(\Cal E)$, and is also a
homeomorphism when $M_t$ and $M(\Cal E)$ are given their vague
topologies.
%437J

\spheader 437Xj\dvAnew{2012}
(i) Let $X$ be a topological space, and $\Sigma$ an algebra of
subsets of $X$ containing every open set;  let $M(\Sigma)^+$ be the set of
non-negative real-valued additive functionals on $\Sigma$, endowed with its
narrow topology, $E$ a member
of $\Sigma$, and $\partial E$ its boundary.
Show that $\nu\mapsto\nu E:M(\Sigma)^+\to\coint{0,\infty}$
is continuous at $\nu_0\in M(\Sigma)^+$ iff $\nu_0(\partial E)=0$.
%Mushtari 96, p 2
(ii) Let $X$ be a completely regular topological space, and $\Sigma$ a
$\sigma$-algebra of subsets of $X$ including the Baire $\sigma$-algebra.
Write $M_{\sigma}$ for the $L$-space of countably additive functionals on
$\Sigma$.   Let $\Cal F$ be a filter on the positive cone $M_{\sigma}^+$
and $\mu$ a member of $M_{\sigma}^+$.
Show that $\Cal F\to\mu$ for the vague topology on $M_{\sigma}$ iff
$\mu E=\lim_{\nu\to\Cal F}\nu E$ whenever $E\in\Sigma$ and
$\mu(\partial E)=0$.
%437J

\spheader 437Xk\dvAnew{2009}
Let $X$ be a compact Hausdorff space, $M^+_{\text{R}}$
the set of Radon measures on $X$ and $P_{\text{R}}$ the set of Radon
probability measures on $X$.
(i) Show that $M^+_{\text{R}}$, with its narrow
topology and its natural convex structure,
can be identified with the positive cone of $C(X)^*$ with its
weak* topology.
(ii) Show that $P_{\text{R}}$, with its narrow
topology and its natural convex structure, can be identified with
$\{f:f\in C(X)^*$, $f\ge 0$, $f(\chi X)=1\}$ with its weak* topology.
%437L

\spheader 437Xl In 437Mc, show that $|\psi(\mu,\nu)|=\psi(|\mu|,|\nu|)$
for every $\mu\in M_{\tau}(X)$ and $\nu\in M_{\tau}(Y)$.
%437M

\spheader 437Xm\dvAnew{2009} Let $X$ be a topological space, $Y$ a regular
topological space and $\MqR(X)$, $\MqR(Y)$ the spaces of totally finite
quasi-Radon measures on $X$, $Y$ respectively.   For a continuous
$\phi:X\to Y$ define $\tilde\phi:\MqR(X)\to\MqR(Y)$ by saying that $\phi$
is \imp\ for $\mu$ and $\tilde\phi(\mu)$ for every $\mu\in\MqR(X)$ (418Hb).
Show that $\tilde\phi$ is continuous for the narrow topologies.
%437N

\sqheader 437Xn\dvAnew{2010} Let
$\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a countable family of Radon
probability spaces, and $Q$ the set of Radon probability measures $\mu$ on
$X=\prod_{i\in I}X_i$ such that the image of $\mu$ under the map
$x\mapsto x(i)$ is
$\mu_i$ for every $i\in I$.   Show that $Q$ is uniformly tight and is
compact for the narrow topology on the set of
totally finite topological measures on $X$.
%what if I uncountable? fails on  [0,1[^{\non\Cal N}  I think
%437N 4{}57L

\spheader 437Xo\dvAnew{2009}
Let $X$ be any topological space, and $M^+_{\text{qR}}$ the
space of totally finite quasi-Radon measures on $X$.   Show that
$M^+_{\text{qR}}$ is complete in the total variation metric.
%437Qa

\spheader 437Xp\dvAnew{2009}
Let $X$ and $Y$ be topological spaces, and $\rhotv^{(X)}$,
$\rhotv^{(Y)}$, $\rhotv^{(X\times Y)}$
the total variation metrics on the spaces
$M^+_{\text{qR}}(X)$, $M^+_{\text{qR}}(Y)$ and $M^+_{\text{qR}}(X\times Y)$
of quasi-Radon measures.   Let $\mu_1$, $\mu_2$ be totally finite
quasi-Radon measures
on $X$, $\nu_1$, $\nu_2$ totally finite quasi-Radon measures on $Y$, and
$\mu_1\times\nu_1$, $\mu_2\times\nu_2$ the quasi-Radon product measures.
Show that

\Centerline{$\rhotv^{(X\times Y)}(\mu_1\times\nu_1,\mu_2\times\nu_2)
\le\rhotv^{(X)}(\mu_1,\mu_2)\cdot\nu_2Y
  +\mu_1X\cdot\rhotv^{(Y)}(\nu_1,\nu_2)$.}
%437Qa

\spheader 437Xq(i)\dvAformerly{4{}37Xi} Show that the set
$M^+_{\sigma}(\Cal B(X))$ of totally finite
Borel probability measures on $X$ is T$_0$ in its narrow topology for any
topological space $X$.
(ii)\dvAformerly{4{}37Xk} Give $X=\omega_1+1$ its order topology.
Show that the narrow topology on $M^+_{\sigma}(\Cal B(X))$ is not T$_1$.
\Hint{consider interpretations of
Dieudonn\'e's measure on $\omega_1$ and the Dirac measure concentrated at
$\omega_1$ as Borel measures on $X$.}
%437R

\spheader 437Xr Let $X$ be any topological space and $\tilde M^+$ the
set of non-negative additive functionals defined on subalgebras of
$\Cal PX$ containing every open set.   For $\mu$,
$\nu\in\tilde M^+$ define $\mu+\nu\in\tilde M^+$ by setting
$(\mu+\nu)(E)=\mu E+\nu E$ for $E\in\dom\mu\cap\dom\nu$.   (i) Show that
addition on $\tilde M^+$ is continuous for the narrow topology.   (ii)
Show that $(\alpha,\mu)\mapsto\alpha\mu:
\coint{0,\infty}\times\tilde M^+\to \tilde M^+$ is continuous for the
narrow topology on $\tilde M^+$.   (iii) Writing $\tilde P$ for
$\{\mu:\mu\in\tilde M^+$, $\mu X=1\}$, and $\delta_x$ for the Dirac measure
concentrated at $x$ for each $x\in X$, show that the convex hull of
$\{\delta_x:x\in X\}$ is dense in $\tilde P$ for the narrow topology.
(iv) Suppose that $A$ and $B$
are uniformly tight subsets of $\tilde M^+$ and $\gamma\ge 0$.
Show that $A\cup B$, $A+B=\{\mu+\nu:\mu\in A$, $\nu\in B\}$ and
$\{\alpha\mu:\mu\in A$, $0\le\alpha\le\gamma\}$ are uniformly tight.
%437R

\spheader 437Xs\dvAnew{2012} Let $(X,\rho)$ be a metric space, $\Sigma$ a
subalgebra of $\Cal PX$ containing all the open sets, and $M=M(\Sigma)$ the
set of bounded finitely additive functionals on
$\Sigma$.   For $\mu$, $\nu\in M$ set

\def\rhoLP{\rho_{\text{LP}}}

\Centerline{$\rhoKR(\mu,\nu)=\sup\{|\dashint u\,d\mu-\dashint u\,d\nu|:
   u:X\to[-1,1]$ is $1$-Lipschitz$\}$,}

$$\eqalign{\rhoLP(\mu,\nu)
&=\inf\{\epsilon:\epsilon>0,\,
   \nu F-\mu U(F;\epsilon)\le\epsilon
   \text{ and }\mu F-\nu U(F;\epsilon)\le\epsilon\cr
&\mskip200mu\text{ for every non-empty closed }F\subseteq X\},\cr}$$

\noindent where $U(F;\epsilon)=\{x:\rho(x,F)<\epsilon\}$ for non-empty
subsets $F$ of $X$.   (i) Show that $\rhoKR$ and $\rhoLP$ are pseudometrics
on $M$.
(ii) Show that if $\mu$, $\nu\in M$ and $\rhoLP(\mu,\nu)=\delta$ then
$\min(1,\delta^2)\le\rhoKR(\mu,\nu)\le 2\delta(1+\delta+|\nu|X)$.
(iii) Show that $\rhoKR$ and $\rhoLP$ induce the
same topology on $M$ and the same uniformity on
$\{\mu:\mu\in M$, $|\mu|X\le\gamma\}$, for any $\gamma\ge 0$.
\Hint{{\smc Bogachev 07}, 8.10.43.
$\rhoLP$ is the {\bf L\'evy-Prokhorov} pseudometric.}
%437J  437R  but does either  \rhoKR  or  \rhoLP  induce vague topy?

%    "Fortet-Mourier metric of order p "  ask Kawabe

\spheader 437Xt\dvArevised{2014} Let $X$ be a Hausdorff space and
$P_{\text{R}}'$ the set of Radon measures $\mu$ on $X$ such that
$\mu X\le 1$, with its narrow topology.
(i) Show that the extreme points of $P_{\text{R}}'$ are the
Dirac measures on $X$, as in 437S, together with the zero measure.
(ii) For $x\in X$ let
$\delta_x$ be the Dirac measure on $X$ concentrated at $x$.
Show that
$x\mapsto\delta_x$ is a homeomorphism between $X$ and its image in
$P_{\text{R}}'$.
%437S

\spheader 437Xu\dvAformerly{4{}37Yw}
Let $X$ be a non-empty compact metrizable
space, and $\phi:X\to X$ a continuous function.   Show that there is an
$x\in X$ such that $\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nu(\phi^i(x))$
is defined for every $u\in C(X)$.   \Hint{372H\formerly{3{}72I}, 4A2Pe.}
%437T;  372H Ergodic Theorem, 4A2Pe C(X) separable
%fails if X the Stone space of Lebesgue prob algebra

\spheader 437Xv\dvAnew{2009} Let $X$ be a T$_0$ topological space, and
$\PqR$ the
set of quasi-Radon probability measures on $X$.   Show that
the extreme points of $\PqR$ are just the Dirac measures on $X$
concentrated on points $x$ such that $\{x\}$ is closed.
%437S

\sqheader 437Xw Let $X$ be a Prokhorov Hausdorff space, and $A$ a set of
totally finite
Radon measures on $X$ which is compact for the narrow topology.   Show
that $A$ is uniformly tight.   \Hint{(i) $\gamma=\sup_{\mu\in A}\mu X$
is finite;  (ii) for any $\epsilon>0$ the set
$\{\bover1{\mu X}\mu:\mu\in A$, $\mu X\ge\epsilon\}$ is narrowly
compact, therefore uniformly tight;  for any $\epsilon>0$ the set
$\{\mu:\mu\in A$, $\mu X\ge\epsilon\}$ is uniformly tight.}
%437U 437Xr(iv)

\spheader 437Xx Give $\omega_1$ its order topology, and let $M_t$ be the
$L$-space of signed tight Borel measures on $\omega_1$.
(i) Show that $\omega_1$ is a Prokhorov space.   (ii) For
$\xi<\omega_1$, define $\mu_{\xi}\in M_t$
by setting $\mu_{\xi}(E)=\chi E(\xi)-\chi E(\xi+1)$ for every Borel set
$E\subseteq\omega_1$.   Show that $\{\mu_{\xi}:\xi<\omega_1\}$ is
relatively compact in
$M_t$ for the vague topology, but is not uniformly tight.   (Compare 437Yy
below.)
%437V

\spheader 437Xy\dvAnew{2014}
Let $X$ and $Y$ be analytic spaces, and $P=\PR(X)$ the
space of Radon probability measures on $X$ with its narrow topology.
(i) Let $V$ be an analytic subset of $(P\times Y)\times X$.   Show that
$\{(\mu,y):\mu\in P$, $y\in Y$, $\mu V[\{(\mu,y)\}]>\alpha\}$ is analytic
for every $\alpha\in\Bbb R$.   \Hint{start with $V=C\times F\times E$ where
$C$, $F$ and $E$ are closed, recalling 437Re and 437Jd,
and apply 431Db;  compare 431Yb.}
(ii) Let $W$ be a coanalytic subset of $(P\times Y)\times X$.   Show that
$\{(\mu,y):\mu\in P$, $y\in Y$, $W[\{(\mu,y)\}]$ is not
$\mu$-negligible$\}$ is coanalytic.
%437R out of order query

\spheader 437Xz\dvAnew{2014}
Let $X$ be a second-countable topological space, $P=P_{\sigma}(X)$
the space of topological probability measures on $X$ with its narrow
topology and $\Cal C$ the space of closed subsets of $X$ with the Fell
topology.   For $\mu\in P$ write $\supp\mu$ for the support of $\mu$.
(i) Show that $\{(x,C):C\in\Cal C$, $x\in C\}$ is a Borel subset of
$X\times\Cal C$.
(ii) Show that $\{(\mu,x):\mu\in P$, $x\in\supp\mu\}$ is a Borel
subset of $P\times X$.
(iii) Show that
$\mu\mapsto\supp\mu:P\to\Cal C$ is Borel measurable.   (Cf.\ 424Ya.)

\leader{437Y}{Further exercises (a)}
%\spheader 437Ya
Let $X$ be a set and $U$ a Riesz subspace of $\Bbb R^X$.
Give formulae for the components of a given element of $U^{\sim}$ in the
bands $U^{\sim}_{\sigma}$, $(U^{\sim}_{\sigma})^{\perp}$,
$U^{\sim}_{\tau}$ and $(U^{\sim}_{\tau})^{\perp}$.   \Hint{356Yb.}
%437A

\spheader 437Yb Let $X$ be a compact Hausdorff space.   Show that the
dual $C(X;\Bbb C)^*$ of the
complex linear space of continuous functions from $X$ to $\Bbb C$ can be
identified with the space of `complex tight Borel measures' on $X$, that
is, the space of functionals $\mu:\Cal B(X)\to\Bbb C$ expressible as a
complex linear
combination of tight totally finite Borel measures;  explain how this
may be identified,
as Banach space, with the complexification of the $L$-space $M_t$
of signed tight Borel measures
as described in 354Yl.   Show that the complex Banach space
$\eusm L^{\infty}_{\Bbb C}(\Cal B(X))$ is canonically embedded in
$C(X;\Bbb C)^{**}$.
%437I

\spheader 437Yc Write $\mu_c$ for counting measure on $[0,1]$, and
$\mu_L$ for Lebesgue measure;  write $\mu_c\times\mu_L$ for the product
measure on $[0,1]^2$, and $\mu$ for the direct sum of $\mu_c$ and
$\mu_c\times\mu_L$.   Show that the $L$-space $C([0,1])^{\sim}$ is
isomorphic, as $L$-space, to $L^1(\mu)$.   \Hint{every Radon measure on
$[0,1]$ has countable Maharam type.}
%437I

\spheader 437Yd Let $X$ be a set, $U$ a Riesz subspace of
$\ell^{\infty}(X)$ containing the constant functions, and $\Sigma$ the
smallest $\sigma$-algebra of subsets of $X$ with respect to which every
member of $U$ is measurable.   Write $\tilde\Sigma$ for the intersection
of the domains of the completions of the totally finite measures with
domain $\Sigma$.   Show that there is a unique sequentially
order-continuous
norm-preserving Riesz homomorphism from $\eusm L^{\infty}(\tilde\Sigma)$
to $(U^{\sim}_{\sigma})^*\cong M_{\sigma}^*$ such that $(Su)(f)=f(u)$
whenever $u\in U$ and $f\in U^{\sim}_{\sigma}$.
%437I 437Xf

\spheader 437Ye\dvAnew{2010}
Show that in 437Ib the operator
$S:\eusm L^{\infty}(\Sigma_{\text{uRm}})\to C_0(X)^{**}$ is
multiplicative if $C_0(X)^{**}$ is given the Arens multiplication
described in 4A6O based on the ordinary multiplication
$(u,v)\mapsto u\times v$ on $C_0(X)$.
%437I out of order query

\spheader 437Yf Explain how to express the proof of
285L(iii)$\Rightarrow$(ii) as ($\alpha$) a proof that if the
characteristic functions of a sequence $\sequencen{\nu_n}$ of Radon
probability measures on $\BbbR^r$ converge pointwise to a characteristic
function, then $\{\nu_n:n\in\Bbb N\}$ is uniformly tight ($\beta$) the
observation that any subalgebra of $C_b(\BbbR^r)$ which separates the
points of $\BbbR^r$ and contains the constant functions will define the
vague topology on any vaguely compact set of measures.
%437J, 437Yf

\spheader 437Yg Let $X$ be a topological space.   (i)
Let $\Cal Ba$ be the Baire $\sigma$-algebra of $X$,
$M_{\sigma}(\Cal Ba)$ the space of signed Baire measures on
$X$, and $u:X\to\Bbb R$ a bounded Baire measurable function.   Show that
we have a linear functional
$\mu\mapsto\dashint u\,d\mu:M_{\sigma}(\CalBa)\to\Bbb R$ agreeing with
ordinary integration with respect to non-negative
measures.  Show that this functional is Baire
measurable with respect to the vague topology on $M_{\sigma}(\CalBa)$.
(ii) Let $M_{\tau}$ be the space of signed $\tau$-additive Borel
measures on $X$, and $u:X\to\Bbb R$ a bounded Borel measurable function.
Show that
we have a linear functional $\mu\mapsto\dashint u\,d\mu:M_{\tau}\to\Bbb R$
agreeing with ordinary integration with respect to non-negative
measures.  Show that this functional is Borel
measurable with respect to the vague topology on $M_{\tau}$.
%437J

\spheader 437Yh\dvAnew{2009}
Let $X$ be a topological space, $\mu_0$ a totally finite
$\tau$-additive topological measure on $X$, and $u:X\to\Bbb R$ a bounded
function which is continuous $\mu_0$-a.e.   Let $\tilde M_{\sigma}^+$
be the set of totally finite topological measures on $X$, with its narrow
topology.   Show that
$\nu\mapsto\overline{\int}u\,d\nu:\tilde M^+_{\sigma}\to\Bbb R$ is
continuous at $\mu_0$.
%437J

\spheader 437Yi\dvAnew{2015}
Let $X$ be a Hausdorff space, and
$M^{\infty+}_{\text{R}}$ the set of all Radon measures on $X$.   Define
addition and scalar multiplication (by positive scalars) on
$M^{\infty+}_{\text{R}}$ as in 234G,
234Xf and 416De and $\le$ by the formulae of 234P or 416Ea.
(i) Show that $M^{\infty+}_{\text{R}}$ is a Dedekind complete
lattice.
(ii) Show that if $A\subseteq M^{\infty+}_{\text{R}}$ is upwards-directed
and non-empty, it is bounded above iff $\{G:G\subseteq X$ is open,
$\sup_{\nu\in A}\nu G<\infty\}$ covers $X$, and in this case
$\dom(\sup A)=\bigcap_{\nu\in A}\dom\nu$ and
$(\sup A)(E)=\sup_{\nu\in A}\nu E$ for every $E\in\dom(\sup A)$.
(iii) Show that if $\mu$, $\nu\in M^{\infty+}_{\text{R}}$ then
$\nu=\sup_{n\in\Bbb N}\nu\wedge n\mu$ iff
every $\mu$-negligible set is $\nu$-negligible.
(iv) Show that if $\mu$, $\nu\in M^{\infty+}_{\text{R}}$ then
$\nu$ is uniquely expressible as $\nu_s+\nu_{ac}$ where
$\nu_s$, $\nu_{ac}\in\in M^{\infty+}_{\text{R}}$, $\mu\wedge\nu_s=0$ and
$\nu_{ac}=\sup_{n\in\Bbb N}\nu_{ac}\wedge n\mu$.
(v) Show that if $\mu$, $\nu\in M^{\infty+}_{\text{R}}$ then
$\dom(\mu\vee\nu)=\dom\mu\cap\dom\nu$ and $(\mu\vee\nu)(E)
=\sup\{\mu F+\nu(E\setminus F):F\in\dom\mu\cap\dom\nu$, $F\subseteq E\}$
for every $E\in\dom(\mu\vee\nu)$.
(vi) Show that if $\mu$, $\nu\in M^{\infty+}_{\text{R}}$ then
$\dom(\mu\wedge\nu)=\{E\cup F:E\in\dom\mu$, $F\in\dom\nu\}$.
(vii) Show that if $\mu$, $\nu\in M^{\infty+}_{\text{R}}$ then
$\mu\wedge\nu=0$ iff there is a set $E\subseteq X$ which is
$\mu$-negligible and $\nu$-conegligible.
(viii) Show that there is a Dedekind complete Riesz space $V$ such
that the positive cone of $V$ is isomorphic to
$M^{\infty+}_{\text{R}}$.
%437Ji

\spheader 437Yj For a topological space $X$ let $M_{\tau}(X)$ be the
$L$-space of signed $\tau$-additive Borel measures on $X$, and
$\psi:M_{\tau}(X)\times M_{\tau}(X)\to M_{\tau}(X\times X)$ the
canonical bilinear operator (437M);  give $M_{\tau}(X)$ and
$M_{\tau}(X\times X)$ their vague topologies.   (i) Show that if
$X=[0,1]$ then $\psi$ is not continuous.   (ii) Show that $X=\Bbb N$ and
$B$ is the unit ball of $M_{\tau}(X)$ then $\psi\restr B\times B$ is not
continuous.
%437M

\spheader 437Yk Let $X$ and $Y$ be topological spaces, and
$\psi:M_{\tau}(X)\times M_{\tau}(Y)\to M_{\tau}(X\times Y)$ the bilinear
map of 437Mc.   Write $M_t(X)$, etc., for the spaces of signed tight
Borel measures.   (i) Show that $\psi(\mu,\nu)\in M_t(X\times Y)$ for
every $\mu\in M_t(X)$, $\nu\in M_t(Y)$.   (ii) Show that if
$B\subseteq M_t(X)$, $B'\subseteq M_t(Y)$ are norm-bounded and uniformly
tight, then $\psi\restr B\times B'$ is continuous for the vague
topologies.
%437O 437M

\spheader 437Yl\dvAnew{2009} Let $X$ be a topological space, and
$\tilde M$ the
space of bounded additive functionals defined on subalgebras of $\Cal PX$
containing every open set.   For $\nu\in\tilde M$, say that
$|\nu|(E)=\sup\{\nu F-\nu(E\setminus F):F\in\dom\nu$, $F\subseteq E\}$ for
$E\in\dom\nu$.   Show that a set $A\subseteq\tilde M$ is uniformly tight
in the sense of 437O iff every member of $A$ is tight and
$\{|\nu|:\nu\in A\}$ is uniformly tight.
%437O

\spheader 437Ym Let $X$ be a completely regular space and
$\PqR$ the space of quasi-Radon probability measures on $X$.   Let
$B\subseteq\PqR$ be a non-empty set.   Show that the following are
equiveridical:  (i) $B$ is relatively compact in $\PqR$ for the
narrow topology;  (ii) whenever $A\subseteq C_b(X)$ is non-empty and
downwards-directed and $\inf_{u\in A}u(x)=0$ for every $x\in A$, then
$\inf_{u\in A}\sup_{\mu\in B}\int u\,d\mu=0$;  (iii) whenever $\Cal G$
is an upwards-directed family of open sets with union $X$, then
$\sup_{G\in\Cal G}\inf_{\mu\in B}\mu G=1$.
%437P

\spheader 437Yn\dvAnew{2010} (i) Let $X$ be a regular topological space,
and
$\MqR$ the space of totally finite quasi-Radon measures on $X$, with its
narrow topology.    Show that $\MqR$ is regular.   (ii)
Find a second-countable Hausdorff space $X$ such that the space
$P_{\text{qR}}$ of quasi-Radon probability measures on $X$ is not
Hausdorff in its narrow topology.
%437J 437Ra

\spheader 437Yo\dvAnew{2009}
Let $(X,\rho)$ and $(Y,\sigma)$ be metric
spaces, and give $\MqR(X)$ and $\MqR(Y)$ the corresponding metrics
$\rhoKR$, $\sigmaKR$ as in 437Rg.   For a continuous function
$\phi:X\to Y$, let $\tilde\phi:\MqR(X)\to\MqR(Y)$ be the map
described in 437Xm.   (i) Show that if $\phi$ is $\gamma$-Lipschitz,
where $\gamma\ge 1$, then $\tilde\phi$ is $\gamma$-Lipschitz.   (ii)
(J.Pachl) Show
that if $\phi$ is uniformly continuous, then $\tilde\phi$ is uniformly
continuous on any uniformly totally finite subset of $\MqR(X)$.
(iii) Show that if $(X,\rho)$ is $\Bbb R$ with its usual
metric, then $\rhoKR$ is not uniformly equivalent to L\'evy's metric as
described in 274Yc\formerly{2{}74Ya}.
(For a discussion of various metrics related to
$\rhoKR$, see {\smc Bogachev 07}, 8.10.43-8.10.48.)
%437Rg 437Xm 43bits

\spheader 437Yp\dvAnew{2009}
Let $(X,\rho)$ be a metric space.   For $f\in C_b(X)^{\sim}_{\sigma}$, set
$\|f\|_{\text{KR}}
=\sup\{|f(u)|:u\in C_b(X),\,\|u\|_{\infty}\le 1,\,u$ is $1$-Lipschitz$\}$.
(i) Show that $\|\,\|_{\text{KR}}$ is a
norm on $C_b(X)^{\sim}_{\sigma}$.
%Bogachev \S8.3 calls it `Kantorovich-Rubinstein norm'
(ii) Let $(X',\rho')$ and $(X'',\rho'')$ be metric spaces, and $\rho$ the
$\ell^1$-product metric on $X=X'\times X''$ defined by saying that
$\rho((x',x''),(y',y''))=\rho'(x',y')+\rho''(x'',y'')$.   Identifying the
spaces
$M_{\tau}(X')$, $M_{\tau}(X'')$ and $M_{\tau}(X)$ of signed $\tau$-additive
Borel measures with subspaces of
$C_b(X')^{\sim}_{\sigma}$, $C_b(X'')^{\sim}_{\sigma}$ and
$C_b(X')^{\sim}_{\sigma}$, as in 437E-437H, show that the bilinear map
$\psi:M_{\tau}(X')\times M_{\tau}(X'')\to M_{\tau}(X)$ described in
437Mc has norm $1$ when $M_{\tau}(X')$, $M_{\tau}(X'')$ and $M_{\tau}(X)$
are given the appropriate norms $\|\,\|_{\text{KR}}$.
%437Rg 437Yo 437E 437H 437M

\spheader 437Yq\dvAnew{2009}
Let $X$ be a topological space and $\tilde M^+$
the set of non-negative real-valued additive functionals defined on
algebras of subsets of $X$ containing every open set, endowed with its
narrow topology.   Show that the weight $w(\tilde M^+)$ of
$\tilde M^+$ is at most $\max(\omega,w(X))$.
%437Rb

\spheader 437Yr Let $X$ be a \v{C}ech-complete completely regular
Hausdorff space
and $P_{\text{R}}$ the set of Radon probability measures on $X$, with
its narrow topology.
Show that $P_{\text{R}}$ is \v{C}ech-complete.
%437Rh

\spheader 437Ys\dvAnew{2008}
Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a non-empty family of probability
algebras, and $\Cal F$ an ultrafilter on $I$.   Let
$(\frak A,\bar\mu)=\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$
be the reduced product as defined in 328C.   For each $i\in I$, let
$(Z_i,\nu_i)$ be the Stone space of $(\frak A_i,\bar\mu_i)$;
give $W=\{(z,i):i\in I$, $z\in Z_i\}$ its disjoint union topology, and let
$\beta W$ be the Stone-\v{C}ech compactification of $W$.   For each
$i\in I$, define $\phi_i:Z_i\to W\subseteq\beta W$ by setting
$\phi_i(z)=(z,i)$ for $z\in Z_i$, and let $\nu_i\phi_i^{-1}$
be the image measure on $\beta W$.
Let $\nu$ be the limit $\lim_{i\to\Cal F}\nu_i\phi_i^{-1}$ for the narrow
topology on the space of Radon probability measures on $\beta W$,
and $Z$ its support.   Show that
$(Z,\nu)$ can be identified with the Stone space of $(\frak A,\bar\mu)$.
%437R

\spheader 437Yt\dvAnew{2009} (i) Show that there are a continuous
$\phi:\{0,1\}^{\Bbb N}\to[0,1]$ and a positive linear operator
$T:C(\{0,1\}^{\Bbb N})\to C([0,1])$ such that $T(u\phi)=u$ for every
$u\in C([0,1])$.   \Hint{if
$I_{\sigma}=\{x:\sigma\subseteq x\in\{0,1\}^{\Bbb N}$ for
$\sigma\in\bigcup_{n\in\Bbb N}\{0,1\}^n$, arrange
that $\{t:T(\chi I_{\sigma})(t)>0\}$ is always an interval of length
$(\bover23)^{\#(\sigma)}$.}   (ii) Show that there are a continuous
$\tilde\phi:(\{0,1\}^{\Bbb N})^{\Bbb N}\to[0,1]^{\Bbb N}$ and a positive
linear operator
$\tilde T:C((\{0,1\}^{\Bbb N})^{\Bbb N})\to C([0,1]^{\Bbb N})$ such that
$\tilde T(h\tilde\phi)=h$ for every
$h\in C([0,1]^{\Bbb N})$.   \Hint{if, in (i), $(Tg)(t)=\int g\,d\nu_t$ for
$t\in[0,1]$ and $g\in C(\{0,1\}^{\Bbb N})$, take $\nu_{\pmb{t}}$ to be the
product measure $\prod_{n\in\Bbb N}\nu_{t_n}$ for
$\pmb{t}=\sequencen{t_n}\in[0,1]^{\Bbb N}$.}
%for 437Yu (437R)

\spheader 437Yu\dvAnew{2009} Let $X$ be a separable metrizable space
and $P=\PR(X)$ the set of Radon probability measures on $X$, with its
narrow topology.
Show that there is a family $\family{\mu}{P}{f_{\mu}}$ of
functions from
$[0,1]$ to $X$ such that (i) $(\mu,t)\mapsto f_{\mu}(t)$ is Borel
measurable (ii) writing $\mu_L$ for Lebesgue measure on
$[0,1]$, $\mu=\mu_Lf_{\mu}^{-1}$ for every $\mu\in P$ (iii) whenever
$\sequencen{\mu_n}$ is a sequence in $P$ converging
to $\mu\in P$, there is a countable set $A\subseteq[0,1]$ such that
$f_{\mu}(t)=\lim_{n\to\infty}f_{\mu_n}(t)$ for every
$t\in[0,1]\setminus A$.   \Hint{first consider
the cases $X=[0,1]$ and $X=\{0,1\}^{\Bbb N}$,
then use 437Yt to deal with $[0,1]^{\Bbb N}$ and its subspaces.
See {\smc Bogachev 07}, \S8.5.}
%437Yt 437R+

\spheader 437Yv\dvAnew{2012} Let $(\frak A,\bar\mu)$ be a measure algebra, and
$\frak A^f$ the ideal of elements of $\frak A$ of finite measure.   For
$a\in\frak A^f$ and $u\in L^0=L^0(\frak A)$, let $\nu_{au}$ be the totally
finite Radon measure on $\Bbb R$ defined by saying that
$\nu_{au}(E)=\bar\mu(a\Bcap\Bvalue{u\in E})$ (definition:  364G, 434T) for Borel
sets $E\subseteq\Bbb R$.   For $a\in\frak A^f$ and $u$, $v\in L^0$ set
$\bar\rho_a(u,v)=\rho_{KR}(\nu_{au},\nu_{av})$, where $\rho_{KR}$ is the metric
on $\MR=\MR(\Bbb R)$ defined from the usual metric on $\Bbb R$.   (i) Show that
the family $\Rho=\{\bar\rho_a:a\in\frak A^f\}$ of pseudometrics
defines the topology of convergence in measure on $L^0$ (definition:  367L).
(ii) Show that if $(\frak A,\bar\mu)$ is
semi-finite then the uniformity $\Cal U$
defined from $\Rho$ is metrizable iff $(\frak A,\bar\mu)$ is $\sigma$-finite and
$\frak A$ has countable Maharam type.   (iii) Show that if
$(\frak A,\bar\mu)$
is semi-finite then $L^0$ is complete under $\Cal U$ (definition:  3A4F)
iff $\frak A$ is purely
atomic.   \Hint{if $(\frak A,\bar\mu)$ is an atomless probability algebra
and $\sequencen{c_n}$ is an independent sequence of elements of $\frak A$
of measure $\bover12$, show that $\sequencen{\nu_{a,\chi c_n}}$ is
convergent in $\MR$ for every $a\in\frak A$,
so $\sequencen{\chi c_n}$ is $\Cal U$-Cauchy.}
%437R
%\query check Dudley

\spheader 437Yx(i) Let $X$ be a metrizable space, and $A$ a narrowly
compact subset of the set of Radon probability measures on $X$.   Show
that there is a separable subset
$Y$ of $X$ which is conegligible for every measure in $A$.   (ii) Show
that a metrizable space is Prokhorov iff all its closed separable
subspaces are Prokhorov.
%437V

\spheader 437Yy I say that a completely regular Hausdorff space
$X$ is {\bf strongly Prokhorov}
if every vaguely compact subset of the space $M_t(X)$ of signed tight
Borel measures
on $X$ is uniformly tight.   (i) Check that a strongly Prokhorov
completely regular Hausdorff space is Prokhorov.  (ii) Show that a
closed subspace of a strongly Prokhorov
completely regular Hausdorff space is strongly Prokhorov.   (iii) Show
that the product of a countable family of strongly Prokhorov completely
regular Hausdorff spaces is
strongly Prokhorov.   (iv) Show that a G$_{\delta}$ subset of a strongly
Prokhorov metrizable space is strongly Prokhorov.   (v) Show that if
$(X,\rho)$ is a complete metric space then $X$ is strongly Prokhorov.
%437V 437Xx mt43bits

%query: is every metrizable Prokhorov space strongly Prokhorov?

\spheader 437Yz\dvAnew{2010}
Let $X$ be a regular Hausdorff topological space and $C$ a non-empty
narrowly compact
set of totally finite topological measures on $X$, all inner regular with
respect to the closed sets.   Set $c(A)=\sup_{\mu\in C}\mu^*A$ for
$A\subseteq X$.   Show that $c:\Cal PX\to\coint{0,\infty}$
is a Choquet capacity.
%437V
}%end of exercises

\endnotes{\Notesheader{437}
The ramifications of the results here are enormous.   For completely
regular topological spaces $X$, the theorems of \S436 give effective
descriptions of the totally finite Baire, quasi-Radon and Radon measures
on $X$ as linear functionals on $C_b(X)$ (436E, 436Xl, 436Xn).   This
makes it possible, and natural, to integrate the topological measure
theory of $X$ into functional analysis, through the theory of
$C_b(X)^*$.   (See {\smc Wheeler 83} for an extensive discussion of this
approach.)   For the rest of this volume we shall never be far away from
such considerations.   In 437C-437I I give only a sample of the results,
heavily slanted towards the abstract theory of Riesz spaces in Chapter
35 and the first part of Chapter 36.

Note that while the constructions of the dual spaces $U^{\sim}$,
$U^{\sim}_c$ and $U^{\times}$ are `intrinsic' to a Riesz space $U$, in
that we can identify these functions as soon as we know the linear and
order structure of $U$, the spaces
$U^{\sim}_{\sigma}$ and $U^{\sim}_{\tau}$ are
definable only when $U$ is presented as a Riesz subspace of $\Bbb R^X$.
In the same way, while the space $M_{\sigma}(\Sigma)$ of countably
additive
functionals on a $\sigma$-algebra $\Sigma$ depends only on the Boolean
algebra structure, the spaces $M_{\tau}$ here (not to be confused with
the space of completely additive functionals considered in 362B) depend
on the topology as well as the Borel algebra.   (For an example in which
radically different topologies give rise to the same Borel algebra, see
{\smc Juh\'asz Kunen \& Rudin 76}.)

You may have been puzzled by the shift from `quasi-Radon' measures in
436H to `$\tau$-additive' measures in 437H;  somewhere the requirement
of inner regularity has got lost.   The point is that the topologies
being considered here, being defined by declaring certain families of
functions continuous, are (completely) regular;  so that $\tau$-additive
measures are necessarily inner regular with respect to the closed sets
(414Mb).

The theory of `vague' and `narrow' topologies in 437J-437V here hardly
impinges on
the questions considered in \S\S274 and 285, where vague topologies
first appeared.   This is because the earlier investigation was
dominated by the very special position of the functions
$x\mapsto e^{iy\dotproduct x}$ (what we shall in \S445 come to call the
`characters' of the additive groups of $\Bbb R$ or $\BbbR^r$).   One
idea which does appear essentially in the proof of 285L, and has a
natural interpretation in the general theory, is that of a `uniformly
tight' family of Radon measures (437O).   In 445Yh below I
set out a generalization of 285L to abelian locally compact groups.

In \S461 I will return to the general theory of extreme points in
compact convex sets.   Here I remark only that it is never surprising
that extreme points should be
special in some way, as in 437S and 461Q-461R;
but the precise ways in which they are special are often unexpected.
A good deal of work has been done on relationships between the
topological properties of a topological space $X$ and the space
$P_{\text{R}}$ of Radon probability measures on $X$ with the
narrow topology.   Here I give only a sample of basic facts in 437R
and 437Yq-437Yr.   Having observed that $\MqR(X)$ is metrizable
whenever $X$ is (437Rg), it is natural to seek ways of defining a metric on
$\MqR(X)$ from a metric on $X$.   The Kantorovich-Rubinstein metric
$\rhoKR$ I have
chosen here is only one of many possibilities;  compare the L\'evy metric
of 274Yc and the metric $\rho_{\text{W}}$ of 457K below.   Note that it can
make a difference whether we look at quasi-Radon (or $\tau$-additive)
probability measures, or at general Borel measures (438Yl).

The terms `vague' and `narrow' both appear in the literature on this
topic, and I take the opportunity to use them both, meaning slightly
different things.   Vague topologies,
in my usage, are linear space topologies on linear spaces of
functionals;  narrow topologies are topologies on spaces of (finitely
additive) measures, which are not linear spaces, though we can
in some cases define addition and multiplication by non-negative
scalars (437N, 437Xr, 437Yi).
I must warn you that this distinction is not standard.   I see that the
word `narrow' appears above
a good deal oftener than the word `vague', which
is in part a reflection of
a simple prejudice against signed measures;  but from the point of view
of this treatise as a whole, it is more natural to work with a concept
well adapted to measures with
variable domains, even if we are considering questions (like compactness
of sets of measures) which originate in linear analysis.
I should mention also that the definition in 437Jc includes a choice.
The duality considered there uses the space $C_b(X)$;  for locally
compact $X$, we have the rival spaces $C_0(X)$ and $C_k(X)$ (see 436J
and 436K), and there are occasions when one of these gives a more
suitable topology on a space of measures (as in 495Xl below).

The elementary theory of uniform tightness and Prokhorov spaces
(437O-437V) %437O 437P 437U 437V
is both pretty and useful.   The emphasis I give it here, however, is
partly because it provides the background to a remarkable construction
by D.Preiss (439S below), showing that $\Bbb Q$ is not a Prokhorov
space.
}%end of notes

\discrpage

\frfilename{mt438.tex}
\versiondate{13.12.06/10.10.07}
\copyrightdate{2007}

\def\Marik{Ma\v r\'\i k}

\def\chaptername{Topologies and measures II}
\def\sectionname{Measure-free cardinals}

\newsection{438}

At several points in \S418, and again in \S434, we had theorems about
separable metrizable spaces in which the proofs undoubtedly needed some
special property of these spaces (e.g., the fact that they are
Lindel\"of), but left it unclear whether something more general could be
said.   When we come to investigate further, asking (for instance)
whether complete metric spaces in general are Radon (438H), we find
ourselves once again approaching the
Banach-Ulam problem, already mentioned at several points in previous
volumes\cmmnt{, and in particular in 363S}.
It seems to be undecidable, in
ordinary set theory with the axiom of choice, whether or not every
discrete space is Radon in the sense of 434C.
On the other hand it
is known that discrete spaces with cardinal at most $\omega_{\omega_1}$ (for
instance) are indeed always Radon.   While as a rule I am deferring
questions of this type to Volume 5, this particular phenomenon is so
pervasive that I think it is worth taking a section now to clarify it.

The central definition is that of `measure-free cardinal' (438A), and
the basic results are 438B-438D.   In particular, `small' infinite
cardinals are measure-free (438C).   From the point of view of
measure theory, a metrizable space whose weight is measure-free is
almost separable, and most of the results in \S418 concerning separable
metrizable spaces can be extended (438E-438G).   In fact `measure-free
weight' exactly determines whether a metrizable space is
measure-compact (438J\cmmnt{, 438Xm}) and whether a complete metric
space is Radon (438H).   If $\frak c$ is measure-free, some
interesting spaces of functions are Radon (438T).   I approach these
last spaces through the concept of `hereditary weak
$\theta$-refinability' (438K), which enables us to do most of the work
without invoking any special axiom.

\leader{438A}{Measure-free cardinals:  Definition} A cardinal
$\kappa$ is {\bf measure-free} or {\bf of measure zero} if whenever
$\mu$ is a
probability measure with domain $\Cal P\kappa$ then there is a
$\xi<\kappa$ such that $\mu\{\xi\}>0$.
\cmmnt{In 363S I discussed some statements equiveridical with the
assertion `every cardinal is measure-free'.}%end of comment

\leader{438B}{}\cmmnt{ It is worth getting some basic facts out
into the open immediately.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space and $\familyiI{E_i}$ a point-finite family of subsets of $X$ such
that $\#(I)$ is measure-free and
$\bigcup_{i\in J}E_i\in\Sigma$ for every $J\subseteq I$.   Set
$E=\bigcup_{i\in I}E_i$.

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

(b) If $\familyiI{E_i}$ is disjoint, then
$\mu E=\sum_{i\in I}\mu E_i$.   In particular, if $\Sigma=\Cal PX$ and
$A\subseteq X$ has measure-free cardinal, then
$\mu A=\sum_{x\in A}\mu\{x\}$.

(c) If $\mu$ is $\sigma$-finite, then $L=\{i:i\in I,\,\mu E_i>0\}$ is
countable and $\bigcup_{i\in I\setminus L}E_i$ is negligible.

\proof{{\bf (a)(i)} The first step is to show, by induction on $n$, that
the result is true if $\mu X<\infty$ and every $E_i$ is negligible and
$\#(\{i:i\in I,\,x\in E_i\})\le n$ for every $x\in X$.   If $n=0$ this
is trivial, since every $E_i$ must be empty.   For the inductive
step to $n\ge 1$, define $\nu:\Cal PI\to\coint{0,\infty}$ by setting
$\nu J=\mu(\bigcup_{i\in J}E_i)$ for every $J\subseteq I$.   Then $\nu$
is a measure on $I$.   \Prf\ Write $F_J=\bigcup_{i\in J}E_j$ for
$J\subseteq I$.  ($\alpha$) If $J$, $K\subseteq I$ are disjoint, then
for $i\in I$ set $E'_i=E_i\cap F_K$ for $i\in J$, $\emptyset$ for
$i\in I\setminus J$.   In this case,
$\family{i}{I}{E'_i}$ is a family of negligible subsets of $X$,
$\bigcup_{i\in J'}E'_i=F_{J'\cap J}\cap F_K$ is measurable for every
$J'\subseteq I$, and $\#(\{i:x\in E'_i\})\le n-1$ for every $x\in X$;
so the inductive hypothesis tells us that

\Centerline{$\mu(\bigcup_{i\in I}E'_i)
=\sup_{J'\subseteq I\text{ is finite}}\mu(\bigcup_{i\in J'}\mu E'_i)
=0$,}

\noindent that is, $F_J\cap F_K$ is negligible.   But this means that

\Centerline{$\nu(J\cup K)=\mu F_{J\cup K}=\mu(F_J\cup F_K)
=\mu F_J+\mu F_K=\nu J+\nu K$.}

\noindent As $J$ and $K$ are arbitrary, $\nu$ is additive.   ($\beta$)
If $\sequencen{J_n}$ is a disjoint sequence in $\Cal PI$, then

$$\eqalign{\nu(\bigcup_{n\in\Bbb N}J_n)
&=\mu(\bigcup_{n\in\Bbb N}F_{J_n})
=\lim_{n\to\infty}\mu(\bigcup_{m\le n}F_{J_m})\cr
&=\lim_{n\to\infty}\sum_{m=0}^n\nu J_m
=\sum_{n=0}^{\infty}\nu J_n,\cr}$$

\noindent so $\nu$ is countably additive and is a measure.\ \Qed

At the same time, $\nu\{i\}=\mu E_i=0$ for every $i$.   Because $\#(I)$
is measure-free, $\nu I=0$.   \Prf\Quer\ Otherwise, let
$f:I\to\kappa=\#(I)$ be any bijection and set
$\lambda A=\Bover1{\nu I}\nu f^{-1}[A]$ for every $A\subseteq\kappa$;
then $\lambda$ is a
probability measure with domain $\Cal P\kappa$ which is zero on
singletons, and $\kappa$ is not measure-free.\ \Bang\QeD\   But this
means just that $\mu(\bigcup_{i\in I}E_i)=0$.
Thus the induction proceeds.

\medskip

\quad{\bf (ii)} \Quer\ Now suppose, if possible, that the general result
is false.   For finite sets $J\subseteq I$ set
$F_J=\bigcup_{i\in J}E_i$, as before, and consider
$\Cal E=\{F_J:J\in[\kappa]^{<\omega}\}$
(see 3A1J for this notation).   Then $\Cal E$ is closed under finite unions
and $\gamma=\sup_{H\in\Cal E}\mu H$ is finite, because it is less than
$\mu E$;  let $\sequencen{H_n}$ be a non-decreasing sequence in
$\Cal E$ such that $\mu(H\setminus H^*)=0$ for every $H\in\Cal E$, where
$H^*=\bigcup_{n\in\Bbb N}H_n$ and $\mu H^*=\gamma$ (215Ab).

Because $\mu$ is semi-finite, there is an $F\in\Sigma$ such that
$F\subseteq E$ and $\gamma<\mu F<\infty$.   For each $n\in\Bbb N$, set

\Centerline{$Y_n=\{x:x\in F\setminus H^*,\,\#(\{i:x\in E_i\})\le n\}$.}

\noindent Then there is some $n\in\Bbb N$ such that $\mu^*Y_n>0$.   Let
$\nu$ be the subspace measure on $Y_n$, so that $\nu$ is non-zero and
totally finite.   Now
$\familyiI{E_i\cap Y_n}$ is a family of negligible subsets of $Y_n$,
$\bigcup_{i\in J}E_i\cap Y_n=Y_n\cap\bigcup_{i\in J}E_i$ is measured by
$\nu$ for every $J\subseteq I$, and $\#(\{i:x\in E_i\cap Y_n\})\le n$
for every $x\in Y_n$.   But this contradicts (i) above.\ \Bang

This proves (a).

\medskip

{\bf (b)} If $\familyiI{E_i}$ is disjoint, then

\Centerline{$\sup_{J\in[I]^{<\omega}}\mu(\bigcup_{i\in J}E_i)
=\sup_{J\in[I]^{<\omega}}\sum_{i\in J}\mu E_i
=\sum_{i\in I}\mu E_i$.}

\noindent Setting $I=A$, $E_x=\{x\}$ for $x\in A$, we get the special
case.

\medskip

{\bf (c)} Let
$\sequencen{X_n}$ be a non-decreasing sequence of measurable sets of
finite measure covering $X$.  For each $n$, set
$L_n=\{i:i\in I,\,\mu(E_i\cap X_n)\ge 2^{-n}\}$.   \Quer\ If $L_n$ is
infinite, take a
sequence $\sequence{k}{i_k}$ of distinct elements in $L_n$, and consider
$G_m=\bigcup_{k\ge m}E_{i_k}$ for $m\in\Bbb N$;  then every $G_m$ has
measure at least $2^{-n}$, $G_0$ has finite measure, $\sequence{m}{G_m}$
is non-increasing, and $\bigcap_{m\in\Bbb N}G_m$ is empty, because
$\sequence{k}{E_{i_k}}$ is point-finite.   But this is impossible.\
\Bang

Thus every $L_n$ is finite and $L=\bigcup_{n\in\Bbb N}L_n$ is
countable.

Now (a), applied to $\familyiI{E'_i}$ where $E'_i=E_i$ if
$i\in I\setminus L$, $\emptyset$ if $i\in L$, tells us that
$\bigcup_{i\in I\setminus L}E_i$ is negligible.
}%end of proof of 438B

\leader{438C}{}\cmmnt{ I do not think we are ready for the most
interesting set-theoretic results concerning measure-free cardinals.
But the following facts may help to make sense of the general pattern.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Ulam 1930})} (a) $\omega$ is
measure-free.

(b) If $\kappa$ is a measure-free cardinal and $\kappa'\le\kappa$ is
a smaller cardinal, then $\kappa'$ is measure-free.

(c) If $\ofamily{\xi}{\lambda}{\kappa_{\xi}}$ is a family of
measure-free cardinals, and $\lambda$ also is measure-free, then
$\kappa=\sup_{\xi<\lambda}\kappa_{\xi}$ is measure-free.

(d) If $\kappa$ is a measure-free cardinal so is $\kappa^+$.

(e) The following are equiveridical:

\quad(i) $\frak c$ is not measure-free;

\quad(ii) there is a semi-finite measure space $(X,\Cal PX,\mu)$ which
is not purely atomic;

\quad(iii) there is a measure $\mu$ on $[0,1]$ extending Lebesgue
measure and measuring every subset of $[0,1]$.

(f) If $\kappa\ge\frak c$ is a measure-free cardinal then
$2^{\kappa}$ is measure-free.

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

\medskip

{\bf (b)} If $\mu$ is a probability measure with domain $\Cal P\kappa'$,
set $\nu A=\mu(\kappa'\cap A)$ for every $A\subseteq\kappa$.   Then
$\nu$ is a probability measure with domain $\Cal P\kappa$, so there is a
$\xi<\kappa$ such that $\nu\{\xi\}>0$;  evidently $\xi<\kappa'$ and
$\mu\{\xi\}>0$.

\medskip

{\bf (c)} Let $\mu$ be a probability measure on $\kappa$ with domain
$\Cal P\kappa$.   Define $f:\kappa\to\lambda$ by setting
$f(\alpha)=\min\{\xi:\alpha<\kappa_{\xi}\}$ for $\alpha<\kappa$.
Then the image measure $\mu f^{-1}$ is a probability measure on
$\lambda$ with domain $\Cal P\lambda$, so there is a $\xi<\lambda$ such
that $\mu f^{-1}[\{\xi\}]>0$.   Now $\mu\kappa_{\xi}>0$.   Applying
438Bb to $A=\kappa_{\xi}$, we see that there is an
$\alpha<\kappa_{\xi}$ such that $\mu\{\alpha\}>0$.   As $\mu$ is
arbitrary, $\kappa$ is measure-free.

\medskip

{\bf (d)} By (a) and (b), we need consider only the case
$\kappa\ge\omega$.   \Quer\
Suppose, if possible, that $\mu$ is a probability measure with domain
$\Cal P\kappa^+$ such that $\mu\{\alpha\}=0$ for every
$\alpha<\kappa^+$.   For each $\alpha<\kappa^+$, choose an injection
$f_{\alpha}:\alpha\to\kappa$.   For $\beta<\kappa^+$,
$\xi<\kappa$ set
$A(\beta,\xi)=\{\alpha:\beta<\alpha<\kappa^+,\,f_{\alpha}(\beta)=\xi\}$.
Then $\kappa^+\setminus\bigcup_{\xi<\kappa}A(\beta,\xi)=\beta+1$ has
cardinal at most $\kappa$, which is measure-free, so $\mu(\beta+1)=0$
and $\mu(\bigcup_{\xi<\kappa}A(\beta,\xi))>0$.   Also
$\ofamily{\xi}{\kappa}{A(\beta,\xi)}$ is disjoint.   There is therefore
a $\xi_{\beta}<\kappa$ such that $\mu A(\beta,\xi_{\beta})>0$, by
438Bb.   Now $\kappa^+>\max(\omega,\kappa)$, so there must be an
$\eta<\kappa$ such that $B=\{\beta:\xi_{\beta}=\eta\}$ is uncountable.
In this case, however, $\family{\beta}{B}{A(\beta,\eta)}$ is an
uncountable family of sets of measure greater than zero, and cannot be
disjoint, because $\mu$ is totally finite (215B(iii));  but if
$\alpha\in A(\beta,\eta)\cap A(\beta',\eta)$, where $\beta\ne\beta'$,
then $f_{\alpha}(\beta)=f_{\alpha}(\beta')=\eta$, which is impossible,
because $f_{\alpha}$ is supposed to be injective.\ \Bang

So there is no such measure $\mu$, and $\kappa^+$ is measure-free.

\medskip

{\bf (e)(i)$\Rightarrow$(ii)} Suppose that $\frak c$ is not
measure-free;  let $\mu$ be a probability measure with domain
$\Cal P\frak c$
such that $\mu\{\xi\}=0$ for every $\xi<\frak c$.   Then $\mu$ is
atomless.   \Prf\Quer\ Suppose, if possible, that $A\subseteq\frak c$ is
an atom for $\mu$.   Let $f:\frak c\to\Cal P\Bbb N$ be a bijection.
For each $n\in\Bbb N$, set $E_n=\{\xi:n\in f(\xi)\}$.   Set
$D=\{n:\mu(A\cap E_n)=\mu A\}$.   Because $A$ is an atom,
$\mu(A\cap E_n)=0$ for every $n\in\Bbb N\setminus D$.   This means
that $B=\bigcap_{n\in D}E_n\setminus\bigcup_{n\in\Bbb N\setminus D}E_n$
has measure $\mu A>0$;  but $f(\xi)=D$ for every $\xi\in B$, so
$\#(B)\le 1$, and $\mu\{\xi\}>0$ for some $\xi$, contrary to
hypothesis.\ \Bang\Qed

So (ii) is true.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} Suppose that there is a semi-finite
measure space $(X,\Cal PX,\mu)$ which is not purely atomic.   Then there
is a non-negligible set $E\subseteq X$ which does not include any atom;
let $F\subseteq E$ be a set of non-zero finite measure.   If we take
$\nu$ to be $\Bover1{\mu F}\mu_F$, where $\mu_F$ is the subspace measure
on $F$, then $\nu$ is an atomless probability measure with domain
$\Cal PF$.
Consequently there is a function $g:F\to[0,1]$ which is \imp\ for $\nu$
and Lebesgue measure (343Cb).   But this means that the image measure
$\nu g^{-1}$ is a measure defined on every subset of $[0,1]$ which
extends Lebesgue measure.

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(iii)} Conversely, if $\frak c$ is
measure-free, then any probability measure on $[0,1]$ measuring every
subset must give positive measure to some singleton, and cannot extend
Lebesgue measure.

\medskip

{\bf (f)} We are supposing that $\kappa\ge\frak c$ is measure-free,
so, in particular, $\frak c$ is measure-free.
Let $\mu$ be a probability measure with domain
$\Cal P(2^{\kappa})$.   By (e), it cannot be atomless;  let
$E\subseteq 2^{\kappa}$ be an atom.   Let $f:2^{\kappa}\to\Cal P\kappa$
be a bijection, and for $\xi<\kappa$ set
$E_{\xi}=\{\alpha:\alpha<2^{\kappa},\,\xi\in f(\alpha)\}$;  set
$D=\{\xi:\xi<\kappa,\,\mu(E\cap E_{\xi})=\mu E\}$.   Note that
$\mu(E\cap E_{\xi})$ must be zero for every $\xi\in\kappa\setminus D$,
so that $E\cap\{\alpha:\xi\in D\symmdiff f(\alpha)\}$ is always
negligible.   Consider

\Centerline{$A_{\xi}
=\{\alpha:\alpha\in E,\,\xi=\min(D\symmdiff f(\alpha))\}$}

\noindent for $\xi<\kappa$.   Then $\ofamily{\xi}{\kappa}{A_{\xi}}$ is a
disjoint family of negligible sets, so its union $A$ is negligible, by
438Bb, because $\kappa$ is measure-free.   But
$E\setminus A\subseteq f^{-1}[\{D\}]$ has at most one element, and is
not negligible;  so $\mu\{\alpha\}>0$ for some $\alpha$.   As $\mu$ is
arbitrary, $2^{\kappa}$ is measure-free.
}%end of proof of 438C

\cmmnt{\medskip

\noindent{\bf Remark} This extends the result of 419G, which used a
different approach to show that $\omega_1$ is measure-free.

We see from (d) above that $\omega_2$, $\omega_3,\ldots$ are all
measure-free;  so, by (c), $\omega_{\omega}$ also is;  generally, if
$\kappa$ is any measure-free cardinal, so is $\omega_{\kappa}$
(438Xa).   I ought to point out that there are more powerful
arguments showing that any cardinal which is not measure-free must be
enormous (see 541L in Volume 5).   In this context,
however, $\frak c=2^{\omega}$ can be `large', at least in the absence of
an axiom like the continuum hypothesis to locate it in the hierarchy
$\family{\xi}{\On}{\omega_{\xi}}$;  it is generally believed that it is
consistent to suppose that $\frak c$ is not measure-free.
}%end of comment

\leader{438D}{}\cmmnt{ I turn now to the contexts in which
measure-free cardinals behave as if they were `small'.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a $\sigma$-finite
measure space, $Y$ a metrizable space with measure-free weight, and
$f:X\to Y$ a measurable function.   Then there is a closed separable set
$Y_0\subseteq Y$ such that
$f^{-1}[Y_0]$ is conegligible;  that is, there is a conegligible
measurable set $X_0\subseteq X$ such that $f[X_0]$ is separable.

\proof{ Let $\Cal U$ be a $\sigma$-disjoint base for the topology of $Y$
(4A2L(g-ii));  express it as $\bigcup_{n\in\Bbb N}\Cal U_n$ where each
$\Cal U_n$ is a disjoint family of open sets.   If $n\in\Bbb N$,
$\#(\Cal U_n)\le w(Y)$ (4A2Db) is a measure-free cardinal (438Cb),
and $\family{U}{U_n}{f^{-1}[U]}$ is a disjoint family in $\Sigma$ such
that $\bigcup_{u\in\Cal V}f^{-1}[U]=f^{-1}[\bigcup\Cal V]$ is measurable
for every $\Cal V\subseteq\Cal U_n$;  so 438Bc tells us that there is a
countable set $\Cal V_n\subseteq\Cal U_n$ such that

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

\noindent is negligible.    Set

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

\noindent Then $f^{-1}[Y\setminus Y_0]
=\bigcup_{n\in\Bbb N}f^{-1}[\bigcup(\Cal U_n\setminus\Cal V_n)]$ is
negligible.   On the other hand,

\Centerline{$\{U\cap Y_0:U\in\Cal U\}
\subseteq\{\emptyset\}
  \cup\{V\cap Y_0:V\in\bigcup_{n\in\Bbb N}\Cal V_n\}$}

\noindent is countable, and is a base for the subspace topology of $Y_0$
(4A2B(a-vi));  so $Y_0$ is second-countable and must be separable
(4A2Oc).

Thus we have an appropriate $Y_0$.   Now $X_0=f^{-1}[Y_0]$ is
conegligible and measurable and $f[X_0]\subseteq Y_0$ is separable
(4A2P(a-iv)).
}%end of proof of 438D

\vleader{36pt}{438E}{Proposition}\cmmnt{ (cf.\ 418B)}
Let $(X,\Sigma,\mu)$ be a complete locally determined measure space.

(a) If $Y$ is a topological space, $Z$ is a
metrizable space, $w(Z)$ is measure-free, and $f:X\to Y$, $g:X\to Z$
are measurable functions, then $x\mapsto (f(x),g(x)):X\to Y\times Z$ is
measurable.

(b) If $\sequencen{Y_n}$ is a sequence of metrizable
spaces, with product $Y$, $w(Y_n)$ is measure-free for every
$n\in\Bbb N$, and $f_n:X\to Y_n$ is measurable for every $n\in\Bbb N$,
then $x\mapsto f(x)=\sequencen{f_n(x)}:X\to\prod_{n\in\Bbb N}Y_n$ is
measurable.

\proof{{\bf (a)(i)} Consider first the case in which $\mu$ is totally
finite.   Then there is a conegligible set $X_0\subseteq X$ such that
$g[X_0]$ is separable (438D).   Applying 418Bb to $f\restr X_0$ and
$g\restr X_0$, we see that $x\mapsto(f(x),g(x)):X_0\to Y\times g[Z_0]$
is measurable.   As $\mu$ is complete, it follows that
$x\mapsto(f(x),g(x)):X\to Y\times Z$ is measurable.

\medskip

\quad{\bf (ii)} In the general case, take any open set
$W\subseteq Y\times Z$ and any measurable set $F\subseteq X$ of finite
measure.
Set $Q=\{x:(f(x),g(x))\in W\}$.   By (i), applied to $f\restr F$ and
$g\restr F$, $F\cap Q\in\Sigma$;  as $F$ is arbitrary and $\mu$ is
locally determined, $Q\in\Sigma$;  as $W$ is arbitrary,
$x\mapsto(f(x),g(x))$ is measurable.

\medskip

{\bf (b)} As in (a), it is enough to consider the case in which $\mu$ is
totally finite.    In this case, we have for each $n\in\Bbb N$ a
conegligible set $X_n$ such that $f_n[X_n]$ is separable.   Set
$X'=\bigcap_{n\in\Bbb N}X_n$;  then 418Bd tells us that $f\restr X'$ is
measurable, so that $f$ is measurable.
}%end of proof of 438E

\leader{438F}{Proposition}\cmmnt{ (cf.\ 418J)} Let $(X,\Sigma,\mu)$ be
a semi-finite measure
space and $\frak T$ a topology on $X$ such that $\mu$ is inner regular
with respect to the closed sets.   Suppose that $Y$ is a
metrizable space, $w(Y)$ is measure-free and $f:X\to Y$ is
measurable.   Then $f$ is almost continuous.

\proof{ Take $E\in\Sigma$ and $\gamma<\mu E$.   Then there is a
measurable set $F\subseteq E$ such that $\gamma<\mu F<\infty$.   Let
$F_0\subseteq F$ be a measurable set such that $F\setminus F_0$ is
negligible and
$f[F_0]$ is separable (438D).   By 412Pc, the subspace measure on $F_0$
is still inner regular with respect to the closed sets, so $f\restr F_0$
is almost continuous (418J), and there is a measurable set
$H\subseteq F_0$, of measure at least $\gamma$, such that $f\restr H$ is
continuous.   As $E$ and $\gamma$ are arbitrary, $f$ is almost
continuous.
}%end of proof of 438F

\leader{438G}{Corollary}\cmmnt{ (cf.\ 418K)} Let
$(X,\frak T,\Sigma,\mu)$ be a quasi-Radon
measure space and $Y$ a metrizable space such that $w(Y)$ is
measure-free.   Then a function
$f:X\to Y$ is measurable iff it is almost continuous.

\leaveitout{438F}

\leader{438H}{}\cmmnt{ Now let us turn to questions which arose in
\S434.

\medskip

\noindent}{\bf Proposition} A complete metric space is Radon iff its
weight is measure-free.

\proof{ Let $(X,\rho)$ be a complete metric space, and $\kappa=w(X)$ its
weight.

\medskip

{\bf (a)} If $\kappa$ is measure-free, let $\mu$ be any totally
finite Borel measure on $X$.   Applying 438D to the identity map from
$X$ to itself, we see that there is a closed separable conegligible
subspace $X_0$.   Now $X_0$ is complete, so is a Polish space, and by
434Kb is a Radon space.   The subspace measure $\mu_{X_0}$ is
therefore tight (that is, inner regular with respect to the compact sets);  as $X_0$ is
conegligible, it follows at once that $\mu$ also is.   As $\mu$ is
arbitrary, $X$ is a Radon space.

\medskip

{\bf (b)} If $\kappa$ is not measure-free, take any $\sigma$-disjoint
base $\Cal U$ for the topology of $X$.   Express $\Cal U$ as
$\bigcup_{n\in\Bbb N}\Cal U_n$ where every $\Cal U_n$ is disjoint.
Then $\kappa\le\#(\Cal U)$
and there is a probability measure $\nu$ on $\Cal U$, with domain
$\Cal P\Cal U$, such that $\nu\{U\}=0$ for every $U\in\Cal U$.
Let $n\in\Bbb N$ be such that $\nu\Cal U_n>0$.
For each $U\in\Cal U_n$ choose
$x_U\in U$. For Borel sets $E\subseteq X$ set
$\mu E=\nu\{U:U\in\Cal U_n,\,x_U\in E\}$;
then $\mu$ is a Borel measure on $X$ and
$\mu(\bigcup\Cal U_n)=\nu\Cal U_n>0$, while
$\mu(\bigcup\Cal V)=\nu\Cal V=0$ for every finite
$\Cal V\subseteq\Cal U_n$.
Thus $\mu$ is not $\tau$-additive and cannot be tight,
and $X$ is not a Radon space.
}%end of proof of 438H

\leader{438I}{Proposition} Let $X$ be a metrizable space and
$\ofamily{\xi}{\kappa}{F_{\xi}}$ a non-decreasing family of closed
subsets of $X$, where $\kappa$ is a measure-free cardinal.   Then

\Centerline{$\mu(\bigcup_{\xi<\kappa}F_{\xi})
=\sum_{\xi<\kappa}\mu(F_{\xi}\setminus\bigcup_{\eta<\xi}F_{\eta})$}

\noindent for every semi-finite Borel measure $\mu$ on $X$.

\proof{{\bf (a)} I had better begin by remarking that
$H_{\xi}=\bigcup_{\eta<\xi}F_{\eta}$ is an F$_{\sigma}$ set for every
ordinal $\xi\le\kappa$, by 4A2Ld and 4A2Ka.   So, setting
$E_{\xi}=F_{\xi}\setminus H_{\xi}$, $\sum_{\xi<\kappa}\mu E_{\xi}$ is
defined.

\medskip

{\bf (b)} I show by induction on $\zeta$ that
$\mu H_{\zeta}=\sum_{\xi<\zeta}\mu E_{\xi}$ for every $\zeta\le\kappa$.
The induction starts trivially with $\mu H_0=0$.   The inductive step to
a successor ordinal $\zeta+1$ is also immediate, as
$H_{\zeta+1}=H_{\zeta}\cup E_{\zeta}$.   For the inductive step to a
limit ordinal $\zeta$ of countable cofinality, let $\sequencen{\zeta_n}$
be a non-decreasing sequence in $\zeta$ with supremum $\zeta$;  then

\Centerline{$\mu H_{\zeta}
=\sup_{n\in\Bbb N}\mu H_{\zeta_n}
=\sup_{n\in\Bbb N}\sum_{\xi<\zeta_n}\mu E_{\xi}
=\sum_{\xi<\zeta}\mu E_{\xi}$,}

\noindent as required.

\medskip

{\bf (c)} So we are left with the case in which $\zeta$ is a limit
ordinal of uncountable cofinality.   In this case,
$\mu(E\cap H_{\zeta})\le\sum_{\xi<\zeta}\mu E_{\xi}$ whenever $\mu E$ is
finite.   \Prf\ Let $\Cal U$ be a $\sigma$-disjoint base for the
topology of $X$ (4A2L(g-ii)), and express $\Cal U$ as
$\bigcup_{n\in\Bbb N}\Cal U_n$ where each $\Cal U_n$ is disjoint.   For
$n\in\Bbb N$, $\xi\le\zeta$ set

\Centerline{$\Cal V_{n\xi}
=\{U:U\in\Cal U_n,\,U\cap H_{\xi}\ne\emptyset\}$,
\quad$V_{n\xi}=\bigcup\Cal V_{n\xi}$.}

\noindent Define $\phi_n:V_{n\zeta}\to\zeta$ by saying that
$\phi_n(x)=\min(\{\xi:\xi<\zeta,\,x\in V_{n\xi}\})$.   Then, for any
$D\subseteq\zeta$,

\Centerline{$\phi_n^{-1}[D]
=\bigcup_{\xi\in D}
  \bigcup(\Cal V_{n\xi}\setminus\bigcup_{\eta<\xi}\Cal V_{n\eta})$}

\noindent is a union of members of $\Cal U_n$, so is open.   We
therefore have a measure $\nu_n$ on $\Cal P\zeta$ defined by saying that
$\nu_nD=\mu(E\cap\phi_n^{-1}[D])$ for every $D\subseteq\zeta$.   At this
point, recall that we are supposing that $\kappa$ is measure-free, so
$\#(\zeta)$ also is measure-free (438Cb) and
$\nu_n\zeta=\sum_{\xi<\zeta}\nu_n\{\xi\}=\sup_{\xi<\zeta}\nu_n\xi$
(438Bb).   Interpreting this in $X$, we have
$\mu(E\cap V_{n\zeta})=\sup_{\xi<\zeta}\mu(E\cap V_{n\xi})$.

This is true for every $n\in\Bbb N$.   So there is a countable set
$C\subseteq\zeta$ such that
$\mu(E\cap V_{n\zeta})=\sup_{\xi\in C}\mu(E\cap V_{n\xi})$ for every
$n\in\Bbb N$.   Because $\cf\zeta>\omega$, there is an $\alpha<\zeta$
such that $C\subseteq\alpha$, and
$\mu(E\cap V_{n\zeta})=\mu(E\cap V_{n\alpha})$, that is,
$E\cap V_{n\zeta}\setminus V_{n\alpha}$ is
negligible, for every $n\in\Bbb N$.

Now note that $F_{\alpha}$ is closed.   So

$$\eqalign{H_{\zeta}\setminus F_{\alpha}
&\subseteq\bigcup\{U:U\in\Cal U,\,H_{\zeta}\cap U\ne\emptyset,\,
   U\cap F_{\alpha}=\emptyset\}\cr
&=\bigcup_{n\in\Bbb N}V_{n\zeta}\setminus V_{n,\alpha+1},\cr}$$

\noindent and $E\cap H_{\zeta}\setminus F_{\alpha}$ is negligible.
Accordingly, using the inductive hypothesis,

\Centerline{$\mu(E\cap H_{\zeta})
\le\mu F_{\alpha}
=\mu H_{\alpha+1}\le\sum_{\xi\le\alpha}\mu E_{\xi}
\le\sum_{\xi<\zeta}\mu E_{\xi}$,}

\noindent as claimed.\ \Qed

Because $\mu$ is semi-finite, and $E$ is arbitrary,
$\mu H_{\zeta}\le\sum_{\xi<\zeta}E_{\xi}$;  but the reverse inequality
is trivial, so we have equality, and the induction proceeds in this case
also.

\medskip

{\bf (d)} At the end of the induction we have
$\mu H_{\kappa}=\sum_{\xi<\kappa}\mu E_{\xi}$, as stated.
}%end of proof of 438I

\leader{438J}{}\cmmnt{ So far we have been looking at metrizable
spaces, the obvious first step.   But it turns out that the concept of
`metacompactness' leads to generalizations of some of the results above.

\medskip

\noindent}{\bf Proposition}\cmmnt{ ({\smc Moran 70}, {\smc Haydon 74})}
Let $X$ be a metacompact space with measure-free weight.

(a) $X$ is Borel-measure-compact.

(b) If $X$ is normal, it is measure-compact.

(c) If $X$ is perfectly normal\cmmnt{ (for instance, if it is
metrizable)}, it is Borel-measure-complete.

\proof{{\bf (a)} \Quer\ If $X$ is not Borel-measure-compact, there are
a non-zero totally finite Borel measure $\mu$ on $X$ and a cover
$\Cal G$ of $X$ by negligible open sets (434H(a-v)).   Let $\Cal H$ be
a point-finite open cover of $X$ refining $\Cal G$.   By 4A2Dc,
$\#(\Cal H)$ is at most $\max(\omega,w(X))$, so is measure-free, by 438C.
Because $\mu$ is a Borel measure,
$\bigcup\Cal H'$ is
measurable for every $\Cal H'\subseteq\Cal H$;  $\mu H=0$ for every
$H\in\Cal H$;  while $\mu(\bigcup\Cal H)=\mu X>0$.   But this
contradicts 438Ba.\ \Bang

\medskip

{\bf (b)} Now suppose that $X$ is normal, and that $\mu$ is a totally
finite Baire measure on $X$.   Because a
normal metacompact space is countably paracompact (4A2F(g-iii)), $\mu$
has an extension to a Borel measure $\mu_1$ which is inner regular with
respect to the closed sets, by \Marik's theorem (435C).   Now $\mu_1$ is
$\tau$-additive, by (a) above, so $\mu$ also is (411C).   As $\mu$ is
arbitrary, $X$ is measure-compact.

\medskip

{\bf (c)} Since on a perfectly normal space the Baire and Borel measures are the same, $X$ is Borel-measure-complete iff it is measure-compact, and we can use (b).
}%end of proof of 438J

\cmmnt{\medskip

\noindent{\bf Remark} The arguments here can be adapted in various
ways, and in particular the hypotheses can be weakened;  see
438Yd-438Yf. %438Yd 438Ye 438Yf
}%end of comment

\leader{438K}{Hereditarily weakly $\theta$-refinable spaces} A
topological space $X$ is {\bf \hwtr}\cmmnt{ (also called {\bf
hereditarily $\sigma$-relatively metacompact},
{\bf hereditarily weakly submetacompact})} if for every family $\Cal G$
of open subsets of $X$ there is a
$\sigma$-isolated family $\Cal A$ of subsets of $X$, refining
$\Cal G$, such that $\bigcup\Cal A=\bigcup\Cal G$.

\vleader{48pt}{438L}{Lemma} (a) Any subspace of a \hwtr\ topological
space is \hwtr.

(b) A hereditarily metacompact
space\cmmnt{ (e.g., any metrizable space, see 4A2Lb)} is \hwtr.

(c) A hereditarily Lindel\"of space is \hwtr.

(d) A topological space with a $\sigma$-isolated network is
\hwtr.
%in particular, any space with a development is \hwtr

\proof{{\bf (a)} If $X$ is \hwtr, $Y$ is a subspace of $X$, and $\Cal H$
is a family of open subsets of $Y$, set
$\Cal G=\{G:G\subseteq X$ is open, $G\cap Y\in\Cal H\}$.   Then there is
a $\sigma$-isolated family $\Cal A$, refining $\Cal G$, with union
$\bigcup\Cal G$;  and $\{A\cap Y:A\in\Cal A\}$ is $\sigma$-isolated
(4A2B(a-viii)), refines $\Cal H$, and has union $\bigcup\Cal H$.   As
$\Cal H$ is arbitrary, $Y$ is \hwtr.

\medskip

{\bf (b)} If $X$ is hereditarily metacompact, and $\Cal G$ is any
family of open sets in $X$ with union $W$, then $\Cal G$ is an open
cover of the metacompact space $W$, so has a point-finite open
refinement $\Cal H$ with the same union.   For each $x\in W$, set
$\Cal H_x=\{H:x\in H\in\Cal H\}$, $V_x=\bigcap\Cal H_x$, so that
$\Cal H_x$ is
a non-empty finite set and $V_x$ is an open set containing $x$.   For
$n\ge 1$, set
$A_n=\{x:x\in W,\,\#(\Cal H_x)=n\}$;  then for any distinct $x$,
$y\in A_n$, either $\Cal H_x=\Cal H_y$ and $V_x=V_y$, or
$\#(\Cal H_x\cup\Cal H_y)>n$ and $V_x\cap V_y\cap A_n=\emptyset$.   This
means that $\Cal A_n=\{V_x\cap A_n:x\in A_n\}$ is a partition of
$A_n$ into relatively open sets, and is an isolated family.   Also,
$\Cal A_n$ is a refinement of $\Cal H$ and therefore of $\Cal G$;  so
$\bigcup_{n\ge 1}\Cal A_n$ is a $\sigma$-isolated refinement of
$\Cal G$, and its union is $\bigcup_{n\ge 1}A_n=W$.   As $\Cal G$ is
arbitrary, $X$ is \hwtr.

\medskip

{\bf (c)} If $X$ is hereditarily Lindel\"of and $\Cal G$ is a
family of open subsets of $X$, there is a countable
$\Cal G_0\subseteq\Cal G$ with the same union;  now $\Cal G_0$, being
countable, is $\sigma$-isolated.   As $\Cal G$ is arbitrary, $X$
is \hwtr.

\medskip

{\bf (d)} If $X$ has a $\sigma$-isolated network $\Cal A$, and
$\Cal G$ is any family of open subsets of $X$, then

\Centerline{$\Cal E=\{A:A\in\Cal A$, $A$ is included in some member of
$\Cal G\}$}

\noindent is a $\sigma$-isolated family (4A2B(a-viii) again), refining
$\Cal G$, with union $\bigcup\Cal G$.
}%end of proof of 438L

\leader{438M}{Proposition}\cmmnt{ ({\smc Gardner 75})}
If $X$ is a \hwtr\ topological space with measure-free weight, it is
Borel-measure-complete.

%Gardner & Pfeffer 10.2;  Gardner 75

\proof{ Let $\mu$ be a Borel probability measure on $X$, and $\Cal G$
the family of $\mu$-negligible open sets.   Let $\Cal A$ be a
$\sigma$-isolated family refining $\Cal G$ with union $\bigcup\Cal G$.
Express $\Cal A$ as $\bigcup_{n\in\Bbb N}\Cal A_n$ where each $\Cal A_n$
is an isolated family;  for each $n\in\Bbb N$, set
$X_n=\bigcup\Cal A_n$ and let
$\mu_n$ be the subspace measure on $X_n$.   Then $\Cal A_n$ is a
disjoint family of relatively open $\mu_n$-negligible sets;  as
$\#(\Cal A_n)\le w(X_n)\le w(X)$ (4A2D) is measure-free, and $\mu_n$
is a totally finite Borel measure on $X_n$,

\Centerline{$\mu^*X_n=\mu_nX_n=\mu_n(\bigcup\Cal A_n)=0$,}

\noindent by 438Bb.   Now
$\mu(\bigcup\Cal G)=\mu^*(\bigcup_{n\in\Bbb N}X_n)=0$.   As $\mu$ is
arbitrary, $X$ is Borel-measure-complete (434I(a-iv)).
}%end of proof of 438M

%[is a normal weakly $\theta$-refinable space countably paracompact?]

\leader{438N}{}\cmmnt{ For the next few paragraphs, I will use the
following notation.}   Let
$X$ be a topological space and $\Cal G$ a family of subsets of $X$.
Then $\Cal J(\Cal G)$ will be the family of subsets of $X$ expressible
as $\bigcup\Cal A$ for some $\sigma$-isolated family $\Cal A$
refining $\Cal G$.   \cmmnt{Observe that} $X$ is \hwtr\ iff
$\bigcup\Cal G$
belongs to $\Cal J(\Cal G)$ for every family $\Cal G$ of open subsets of
$X$.

\spheader 438Na $\Cal J(\Cal G)$ is always a $\sigma$-ideal of subsets
of $X$.
\prooflet{\Prf\ If $\Cal A$ is a $\sigma$-isolated family of
subsets of $X$, refining $\Cal G$, and $B$ is any set, then
$\{B\cap A:A\in\Cal A\}$ is still
$\sigma$-isolated and still refines $\Cal G$;  so any subset of a
member of $\Cal J(\Cal G)$ belongs to $\Cal J(\Cal G)$.   If
$\sequencen{\Cal A_n}$ is a sequence of $\sigma$-isolated
families refining $\Cal G$, then $\bigcup_{n\in\Bbb N}\Cal A_n$ is
$\sigma$-isolated and refines $\Cal G$;  so the union of
any sequence in $\Cal J(\Cal G)$ belongs to $\Cal J(\Cal G)$.\ \Qed}

\spheader 438Nb If $\Cal H$ refines $\Cal G$, then
$\Cal J(\Cal H)\subseteq\Cal J(\Cal G)$.  \prooflet{\Prf\ All we need to
remember is that any family refining $\Cal H$ also refines $\Cal G$.\
\Qed}

\spheader 438Nc If $X$ and $Y$ are topological spaces, $A\subseteq X$,
$f:A\to Y$ is
continuous, and $\Cal H$ is a family of subsets of $Y$, set
$\Cal G=\{f^{-1}[H]:H\in\Cal H\}$.   Then
$\Cal J(\Cal G)\supseteq\{f^{-1}[B]:B\in\Cal J(\Cal H)\}$.
\prooflet{\Prf\ If $B\in\Cal J(\Cal H)$, there is a $\sigma$-isolated
family $\Cal D$ of subsets of $Y$, refining $\Cal H$, and with union
$B$.   Now
$\Cal A=\{f^{-1}[D]:D\in\Cal D\}$ refines $\Cal G$ and has union
$f^{-1}[B]$.   We can express $\Cal D$ as
$\bigcup_{n\in\Bbb N}\Cal D_n$, where each $\Cal D_n$ is an isolated
family;  set
$\Cal A_n=\{f^{-1}[D]:D\in\Cal D_n\}$, so that
$\Cal A=\bigcup_{n\in\Bbb N}\Cal A_n$.   For each $n$, $\Cal A_n$ is
disjoint, because $\Cal D_n$ is.   Moreover, if $D\in\Cal D_n$, then
$D=H\cap\bigcup\Cal D_n$ for some open set $H\subseteq Y$, so that
$f^{-1}[D]=f^{-1}[H]\cap\bigcup\Cal A_n$ is relatively open in
$\bigcup\Cal A_n$;  this shows that $\Cal A_n$ is an isolated
family.   Accordingly $\Cal A$ is $\sigma$-isolated and witnesses
that $f^{-1}[B]\in\Cal J(\Cal G)$.   As $B$ is arbitrary, we have the
result.\ \Qed}

\spheader 438Nd If $X$ is a topological space, $\Cal G$ is a family of
subsets of $X$,
and $\familyiI{D_i}$ is an isolated family in $\Cal J(\Cal G)$,
then $\bigcup_{i\in I}D_i\in\Cal J(\Cal G)$.   \prooflet{\Prf\ For each
$i\in I$, let $\sequencen{\Cal A_{ni}}$ be a sequence of isolated
families, all refining $\Cal G$, such that
$D_i=\bigcup_{n\in\Bbb N}\bigcup\Cal A_{in}$.   Set
$\Cal A_n=\bigcup_{i\in I}\Cal A_{in}$ for each $n$.   Then $\Cal A_n$
refines $\Cal G$, and
$\bigcup_{n\in\Bbb N}\bigcup\Cal A_n=\bigcup_{i\in I}D_i$.   It is easy
to check that every $\Cal A_n$ is isolated, so that
$\bigcup_{n\in\Bbb N}\Cal A_n$ witnesses that $\bigcup_{i\in I}D_i$
belongs to $\Cal J(\Cal G)$.\ \Qed}

\leader{438O}{Lemma} Give
$\Bbb R$ the topology $\frak S$ generated by the closed intervals
$\ocint{-\infty,t}$ for $t\in\Bbb R$, and let $r\ge 1$.   Then
$\BbbR^r$, with the product topology corresponding to $\frak S$, is
\hwtr.

\proof{ Induce on $r$.   Write $\le$ for the usual partial order of
$\BbbR^r$, and $\ocint{-\infty,x}$ for $\{y:y\le x\}$;  set
$V_A=\bigcup_{x\in A}\ocint{-\infty,x}$ for $A\subseteq\BbbR^r$.   The
sets $\ocint{-\infty,x}$, as $x$ runs over $\BbbR^r$, form a base for
the topology of $\BbbR^r$.

The induction starts easily because $\frak S$ itself is hereditarily
Lindel\"of.   \Prf\ If $\Cal G\subseteq\frak S$, set

\Centerline{$A
=\{x:x\in\Bbb R$, there is some $G\in\Cal G$ such that
$\ocint{-\infty,x}\subseteq G\}$.}

\noindent Then $A$ has a
countable cofinal set $D$, so that there is a corresponding countable
subset of $\Cal G$ with the same union as $\Cal G$.\ \QeD\  By
438Lc, $\frak S$ is \hwtr.

For the inductive step to $r+1$, where $r\ge 1$, let $\Cal G$ be a
family of open subsets of $\BbbR^{r+1}$, and set

\Centerline{$A
=\{x:x\in\BbbR^r$, there is some $G\in\Cal G$ such that
$\ocint{-\infty,x}\subseteq G\}$.}

\noindent For each $k\le r$, $q\in\Bbb Q$ set
$K_k=(r+1)\setminus\{k\}$ and let $B_{kq}\subseteq\BbbR^{K_k}$ be the
set $\{z:z^{\smallfrown}\fraction{q}\in V_A\}$,
writing $z^{\smallfrown}\fraction{q}$ for that
member $x$ of $\BbbR^{r+1}$ such that $x\restr K_k=z$ and $x(k)=q$.   (I
am thinking of members of $\BbbR^{r+1}$ as functions from
$r+1=\{0,\ldots,r\}$ to $\Bbb R$.)   Set
$A_{kq}=\{x:x\in V_A,\,x(k)=q\}$,
$\Cal G_{kq}=\{\ocint{-\infty,x}:x\in A_{kq}\}$.   Then, in the notation
of 438N, $V_{A_{kq}}\in\Cal J(\Cal G)$.   \Prf\ Set $f(x)=x\restr K_k$
for each $x\in V_{A_{kq}}$, so that $f:V_{A_{kq}}\to\BbbR^{K_k}$ is
continuous.   For $x\in A_{kq}$,
$\ocint{-\infty,x}=f^{-1}[\,\ocint{-\infty,f(x)}\,]$.   Now
$\BbbR^{K_k}$ is \hwtr, by the inductive hypothesis, so if we set
$\Cal H_{kq}=\{\ocint{-\infty,f(x)}:x\in A_{kq}\}$,
$\bigcup\Cal H_{kq}\in\Cal J(\Cal H_{kq})$ and (by 438Nc)

\Centerline{$V_{A_{kq}}=f^{-1}[\bigcup\Cal H_{kq}]
\in\Cal J(\Cal G_{kq})\subseteq\Cal J(\Cal G)$.  \Qed}

\noindent Accordingly $W\in\Cal J(\Cal G)$, where
$W=\bigcup_{k\le r,q\in\Bbb Q}V_{A_{kq}}$, by 438Na.

Now consider $V_A\setminus W$.   If $x$, $x'\in V_A\setminus W$ and
$x\le x'$ then $x=x'$.   \Prf\Quer\ Otherwise, there are a $k\le n$ and
a $q\in\Bbb Q$ such that $x(k)\le q\le x'(k)$.   In this case, setting
$y\restr K_k=x\restr K_k$ and $y(k)=q$, we have $y\in A_{kq}$ and
$x\in V_{A_{kq}}$.\ \Bang\Qed

But this means that the subspace topology of $V_A\setminus W$ is
discrete, so that $\{\{x\}:x\in V_A\setminus W\}$ is an isolated
family covering $V_A\setminus W$ and refining $\Cal G$;  thus
$V_A\setminus W\in\Cal J(\Cal G)$ and $\bigcup\Cal G=V_A$ belongs to
$\Cal J(\Cal G)$.   As $\Cal G$ is arbitrary, $\BbbR^{r+1}$ is \hwtr\
and the induction proceeds.
}%end of proof of 438O

\leader{438P}{Lemma} Let $X$ be a Polish space, and
$\tildeClll=\tildeClll(X)$ the family of functions
$\omega:\Bbb R\to X$ such that $\lim_{s\uparrow t}\omega(s)$ and
$\lim_{s\downarrow t}\omega(s)$ are defined in $X$ for every
$t\in\Bbb R$.

(a) For $A\subseteq B\subseteq\Bbb R$ and $f\in X^B$, set

$$\eqalign{\jump_A(f,\epsilon)
&=\sup\{n:\text{ there is an }I\in[A]^n
\text{ such that }\rho(f(s),f(t))>\epsilon\cr
&\mskip180mu
\text{ whenever }s<t\text{ are successive elements of }I\}.\cr}$$

\noindent Now a function $\omega\in X^{\Bbb R}$ belongs to $\tildeClll$ iff
$\jump_{[-n,n]}(\omega,\epsilon)$ is finite for every $n\in\Bbb N$ and
$\epsilon>0$.

(b) If $\omega\in\tildeClll$ then $\omega$ is continuous at all but
countably many points of $\Bbb R$.

(c) If $\omega\in\tildeClll$ then $\omega[\,[-n,n]\,]$ is relatively
compact in $X$ for every $n\in\Bbb N$.

\proof{Fix on a complete metric $\rho$ inducing the topology of $X$.

\medskip

{\bf (a)(i)} Suppose that $\omega\in\tildeClll$, $n\in\Bbb N$ and
$\epsilon>0$.
For every $t\in[-n,n]$, there is a $\delta_t>0$ such that
$\rho(\omega(s),\omega(s'))\le\epsilon$ whenever either
$t<s\le s'\le t+\delta_t$
or $t-\delta_t\le s\le s'<t$.   Now there is an $m\ge 1$ such that
whenever $s$, $s'\in[-n,n]$ and $|s-s'|\le\Bover{2n}{m}$ there is a
$t\in[-n,n]$ such that both $s$ and $s'$ belong to
$[t-\delta_t,t+\delta_t]$.   Suppose now that
$-n\le t_0<t_1<\ldots<t_{3m}\le n$.   Then there must be
an $i<m$ such that $t_{3i+3}-t_{3i}\le\Bover{2n}m$.    Let $t$ be such
that both $t_{3i}$ and $t_{3i+3}$ belong to $[t-\delta_t,t+\delta_t]$.
There is at least one $j$ such that $3i\le j\le 3i+2$ and
$t\notin[t_j,t_{j+1}]$;  in which case
$\rho(\omega(t_j),\omega(t_{j+1}))\le\epsilon$.
So $\jump_{[-n,n]}(\omega,\epsilon)\le 3m$.   As $n$ and $\epsilon$ are
arbitrary, the condition is satisfied.

\medskip

\quad{\bf (ii)} Suppose that $\omega$ satisfies the condition.
If $t\in\Bbb R$ and $\epsilon>0$, take $n\ge|t|+1$ and
$m=\jump_{[-n,n]}(\omega,\epsilon)$;  then there must be a $\delta>0$
such that $\diam\{\omega(s):t<s\le t+\delta\}\le 2\epsilon$, since
otherwise we should be able to find $t_0>t_1>\ldots>t_m>t$ such that
$t_0=t+1$ and
$\rho(\omega(t_{i+1}),\omega(t_i))>\epsilon$ for $i<m$.   Because $X$ is
$\rho$-complete, $\lim_{s\downarrow t}\omega(s)$ is defined.   Similarly,
$\lim_{s\uparrow t}\omega(s)$ is defined;  as $t$ is arbitrary,
$\omega\in\tildeClll$.

\medskip

{\bf (b)} For $k\in\Bbb N$, set
set $A_k=\{t:t\in\Bbb R$,
$\limsup_{s\to t}\rho(\omega(s),\omega(t))>2^{-k+1}\}$.
Then $\#(A_k\cap\coint{-n,n})\le\jump_{[-n,n]}(\omega,2^{-k})$ for every
$n\in\Bbb N$.   \Prf\ If $t_0,\ldots,t_m\in A_k$ and
$-n\le t_0<\ldots<t_m<n$, then we can choose
$s_0,\ldots,s_m$ such that $s_0=t_0$,

\Centerline{$s_{i-1}<s_i<t_{i+1}$,
\quad $\rho(\omega(s_i),\omega(s_{i-1}))>2^{-k}$}

\noindent whenever $1\le i\le m$, interpreting $t_{m+1}$ as $n$.
Now $\{s_i:i\le m\}$ witnesses that
$\jump_{[-n,n]}(\omega,2^{-k})>m$.\ \Qed

By (a), $A_k\cap\coint{-n,n}$ is finite for every $n$,
and

\Centerline{$\{t:\omega$ is discontinuous at $t\}=\bigcup_{k\in\Bbb N}A_k$}

\noindent is countable.

\medskip

{\bf (c)} If $\sequence{k}{t_k}$ is any monotonic sequence in $\Bbb R$ with
limit $t$,
$\sequence{k}{\omega(t_k)}$ is convergent to one of
$\lim_{s\uparrow t}\omega(s)$, $\lim_{s\downarrow t}\omega(s)$.
But this means that if $\sequence{k}{t_k}$ is any sequence in
$[-n,n]$, $\sequence{k}{\omega(t_k)}$ has a subsequence which is convergent
in $X$;  by 4A2Le, $\omega[\,[-n,n]\,]$ is relatively compact in $X$.
}%end of proof of 438P

\leader{438Q}{Theorem} Let $X$ be a Polish space, and
$\tildeClll=\tildeClll(X)$ the family of functions
$\omega:\Bbb R\to X$ such that $\lim_{s\uparrow t}\omega(s)$ and
$\lim_{s\downarrow t}\omega(s)$ are defined in $X$ for every
$t\in\Bbb R$.

(a) $\tildeClll$,
with its topology of pointwise convergence
inherited from the product topology of $X^{\Bbb R}$, is K-analytic.

(b) $\tildeClll$ is \hwtr.

\proof{ Fix on a complete metric $\rho$ inducing the topology of $X$.

\medskip

{\bf (a)} By 4A2Qg, $X$ can be regarded as a G$_{\delta}$ set in a compact
metrizable space $Z$.

\medskip

\quad{\bf (i)}
Give the space $\Cal C=\Cal C(Z)$ of closed subsets of $Z$ its
Fell topology;  then $\Cal C$ is compact and metrizable (4A2T(b-iii),
4A2Tf).    Let $\Cal K$ be the family of compact subsets of $X$, that is,
the set of those $K\in\Cal C$ which are included in $X$.   Then $\Cal K$ is
a G$_{\delta}$ set in $\Cal C$.   \Prf\ $Z\setminus X$ is an F$_{\sigma}$
set in $Z$, so is expressible as the union of a sequence $\sequencen{L_n}$
of compact sets;  now
$\Cal K=\bigcap_{n\in\Bbb N}\{K:K\in\Cal C$, $K\cap L_n=\emptyset\}$ is
G$_{\delta}$, by the definition of the Fell topology (4A2T(a-ii)).\ \Qed

\medskip

\quad{\bf (ii)} For $n\in\Bbb N$, set

\Centerline{$Q_n
=\{\omega:\omega\in Z^{\Bbb R}$, $\omega[\,[-n,n]\,]$ is a
relatively compact subset of $X\}$.}

\noindent Then $Q_n$ is K-analytic.  \Prf\ Set

\Centerline{$R_n
=\{(K,\omega):K\in\Cal K$, $\omega\in Z^{\Bbb R}$,
$\omega[\,[-n,n]\,]\subseteq K\}$.}

\noindent Then

\Centerline{$R_n
=\bigcap_{t\in[-n,n]}\{(K,\omega):K\in\Cal K$, $\omega\in Z^{\Bbb R}$,
$\omega(t)\in K\}$}

\noindent is a closed set in $\Cal K\times Z^{\Bbb R}$, by 4A2T(e-i).
Since $\Cal K$ is Polish and $Z^{\Bbb R}$ is compact,
$R_n$ is K-analytic (423Ba, 423C, 422Ge, 422Gf).   But $Q_n$ is the
projection of $R_n$ onto the second coordinate, so it too is K-analytic
(422Gd).\ \Qed

\medskip

\quad{\bf (iii)} Next, for $m$, $k\in\Bbb N$, defining the function
$\jump_{[-n,n]}$ from $\rho$ as in 438P,

\Centerline{$V_{mk}=\{\omega:\omega\in Q_n$,
$\jump_{[-n,n]}(\omega,2^{-k})\le m\}$}

\noindent is relatively closed in $Q_n$, therefore K-analytic, and

\Centerline{$Q'_n=Q_n\cap\bigcap_{k\in\Bbb N}\bigcup_{m\in\Bbb N}V_{mk}$}

\noindent is K-analytic (422Hc).

\medskip

\quad{\bf (iv)} Consequently $Q=\bigcap_{n\in\Bbb N}Q'_n$ is K-analytic.
But $Q=\tildeClll$.
\Prf\ For $n$, $k\in\Bbb N$, $\tildeClll\subseteq Q_n$ by 438Pc,
and $\tildeClll\subseteq\bigcup_{m\in\Bbb N}V_{mk}$ by 438Pa.
So $\tildeClll\subseteq Q$.
Conversely, if $\omega\in Q$, then surely
$\omega(t)\in X$ for every $t\in\Bbb R$, and
$\jump_{[-n,n]}(\omega,2^{-k})$ is finite for all $n$, $k\in\Bbb N$;  so
$\omega\in\tildeClll$ by 438Pa in the other direction.\ \Qed

Accordingly $\tildeClll$ is K-analytic, as claimed.

\medskip

{\bf (b)} Let $\Cal G$ be a family of open sets in $\tildeClll$.

\medskip

\quad{\bf (i)} In the
notation of 438N, I seek to show that $\bigcup\Cal G$ belongs to
$\Cal J(\Cal G)$.   Of course it will be enough to consider the case in
which $\bigcup\Cal G$ is non-empty.   The following elementary remarks
will be useful.

\medskip

\qquad\grheada\
If $D\subseteq\bigcup\Cal G$, and $\Cal E$ is a partition of
$D$ into relatively open sets such that $\Cal E$ refines $\Cal G$, then
$D\in\Cal J(\Cal G)$.

\medskip

\qquad\grheadb\ If $\familyiI{D_i}$ is any family in $\Cal J(\Cal G)$, and
$\familyiI{H_i}$ is a  family of open sets, and
$D\subseteq\{\omega:\#(\{i:\omega\in H_i\})=1\}$, then
$D\cap\bigcup_{i\in I}H_i\cap D_i$ belongs to
$\Cal J(\Cal G)$.   \Prf\ $\familyiI{D\cap H_i\cap D_i}$ is an isolated
family in $\Cal J(\Cal G)$;  use 438Nd.\ \Qed

\medskip

\quad{\bf (ii)} Let $\Cal I$ be the family of non-empty open intervals in
$\Bbb R$ with rational endpoints, and $\Cal U$ a countable base for the
topology of $X$.   Write $Q$ for the family of all
finite sequences

\Centerline{$\pmb{q}=
(I_0,U_0,V_0,W_0,I_1,U_1,V_1,W_1,\ldots,I_n,U_n,V_n,W_n)$}

\noindent where $I_0,I_1,\ldots,I_n$ are disjoint members of $\Cal I$,
all the $U_i$, $V_i$, $W_i$ belong to $\Cal U$, and, for each $i\le n$,
any pair of $U_i$, $V_i$, $W_i$ are either equal or disjoint.   Fix
$\pmb{q}=(I_0,\ldots,W_n)\in Q$ for the moment.

\medskip

\quad{\bf (iii)} Set $T_{\pmb{q}}=\prod_{i\le n}I_i$.   For
$\tau\in T_{\pmb{q}}$, write $F_{\pmb{q}\tau}$ for the set of those
$\omega\in\tildeClll$ such that, for every $i\le n$,
$\omega(s)\in U_i$ for $s\in I_i\cap\ooint{-\infty,\tau(i)}$,
$\omega(\tau(i))\in V_i$ and $\omega(s)\in W_i$ for
$s\in I_i\cap\ooint{\tau(i),\infty}$.   Set
$\Omega_{\pmb{q}}=\bigcup\{F_{\pmb{q}\tau}:\tau\in T_{\pmb{q}}\}$,
and for $\tau\in T_{\pmb{q}}$ set
$H_{\pmb{q}\tau}
=\{\omega:\omega\in \Omega_{\pmb{q}},\,\omega(\tau(i))\in V_i$
for every $i\le n\}$.
Finally, set

\Centerline{$S_{\pmb{q}}
=\{\tau:\tau\in T_{\pmb{q}}$ and $H_{\pmb{q}\tau}$ is included in some
member of $\Cal G\}$.}

\medskip

\quad{\bf (iv)} If $T\subseteq S_{\pmb{q}}$ then
$H=\bigcup_{\tau\in T}H_{\pmb{q}\tau}$ belongs to $\Cal J(\Cal G)$.
\Prf\ Induce on $\#(L(T))$, where

\Centerline{$L(T)=\{i:i\le n$, there are $\tau$, $\tau'\in T$ such that
$\tau(i)\ne\tau'(i)\}$.}

\noindent If $L(T)=\emptyset$, then $\#(T)\le 1$, so $H$ is either empty
or included in some member of $\Cal G$, and the induction
starts.   For the inductive step to $\#(L(T))=k\ge 1$, consider three
cases.

\medskip

\qquad{\bf case 1} Suppose there is a $j\in L(T)$ such that
$U_j=V_j=W_j$.   Then $\omega(t)\in V_j$ whenever
$\omega\in\Omega_{\pmb{q}}$ and
$t\in I_j$.   Fix any $t^*\in I_j$ and for $\tau\in T_{\pmb{q}}$ define
$\tau^*\in T_{\pmb{q}}$ by setting $\tau^*(j)=t^*$, $\tau^*(i)=\tau(i)$
for $i\ne j$;  then $H_{\pmb{q}\tau^*}=H_{\pmb{q}\tau}$.   Set
$T^*=\{\tau^*:\tau\in T\}$;  then $L(T^*)=L(T)\setminus\{j\}$, so
$\#(L(T^*))<\#(L(T))$, while $T^*\subseteq S_{\pmb{q}}$.   By the
inductive hypothesis, $H=\bigcup_{\tau\in T^*}H_{\pmb{q}\tau}$ belongs to
$\Cal J(\Cal G)$.

\medskip

\qquad{\bf case 2} Suppose there is a $j\in L(T)$ such that $U_j\ne V_j$
and $V_j\ne W_j$.   Then $V_j\cap(U_j\cup W_j)=\emptyset$.   For
$s\in I_j$ set $T^*_s=\{\tau:\tau\in T,\,\tau(j)=s\}$.   Then
$\#(L(T^*_s))<\#(L(T))$ so, by the inductive hypothesis,
$H^*_s\in\Cal J(\Cal G)$,
where $H^*_s=\bigcup_{\tau\in T^*_s}H_{\pmb{q}\tau}$.   But, for
$\tau\in T$ and $\omega\in H_{\pmb{q}\tau}$, $\omega(s)\in V_j$ iff
$s=\tau(j)$;
so $H^*_s=\{\omega:\omega\in H,\,\omega(s)\in V_j\}$ and
$\family{s}{I_j}{H^*_s}$ is a
partition of $H$ into relatively open sets.   By (i-$\beta$),
$H\in\Cal J(\Cal G)$.

\medskip

\qquad{\bf case 3} Otherwise, $L(T)=J\cup J'$ where
$J=\{i:i\in L(T),\,U_i=V_i\}$ and $J'=\{i:i\in L(T),\,V_i=W_i\}$ are
disjoint.   For $\omega\in H$ and $i\in J$ we see that there is a
largest $t\in I_i$ such that $\omega(t)\in V_i$;  set $\phi_i(\omega)=-t$.
Similarly, if $\omega\in H$, $i\in J'$ there is a smallest
$t\in I_i$ such that $\omega(t)\in V_i$;  in this case, set
$\phi_i(\omega)=t$.   Observe that, for $i\in J$ and $s\in\Bbb R$,

$$\eqalign{\{\omega:\omega\in H,\,\phi_i(\omega)\le s\}
&=\emptyset\text{ if }s<-t\text{ for every }t\in I_i,\cr
&=H\text{ if }-t<s\text{ for every }t\in I_i,\cr
&=\{\omega:\omega\in H,\,\omega(-s)\in V_i\}\text{ if }-s\in I_i,\cr}$$

\noindent so is always relatively open in $H$, and $\phi_i:H\to\Bbb R$ is
continuous if $\Bbb R$ is given the left-facing
topology $\frak S$ of Lemma 438O.   Similarly, for $i\in J'$, $s\in\Bbb R$,

$$\eqalign{\{\omega:\omega\in H,\,\phi_i(\omega)\le s\}
&=\emptyset\text{ if }s<t\text{ for every }t\in I_i,\cr
&=H\text{ if }t<s\text{ for every }t\in I_i,\cr
&=\{\omega:\omega\in H,\,\omega(s)\in V_i\}\text{ if }s\in I_i.\cr}$$

\noindent So in this case also $\phi_i$ is continuous.

Accordingly, giving $\BbbR^{L(T)}$ the product topology corresponding to
$\frak S$, we have a continuous map $\phi:H\to\BbbR^{L(T)}$
defined by setting $\phi(\omega)=\family{i}{L(T)}{\phi_i(\omega)}$ for
$\omega\in H$.   For $\tau\in T$, set $\tilde\tau(i)=-\tau(i)$ if $i\in J$,
$\tau(i)$ if $i\in J'$, and
$\tilde H_{\tau}=\ocint{-\infty,\tilde\tau}\subseteq\BbbR^{L(T)}$.  Then

$$\eqalignno{H_{\pmb{q}\tau}
&=\{\omega:\omega\in H,\,\omega(\tau(i))\in V_i
  \text{ for every }i\le n\}\cr
&=\{\omega:\omega\in H,\,\omega(\tau(i))\in V_i
  \text{ for every }i\in L(T)\}\cr
\displaycause{because if $\omega\in H$, $i\le n$ and $i\notin L(T)$ then
there is some $\tau'\in T$ such that $\omega\in H_{\pmb{q}\tau'}$, so that
$\omega(\tau'(i))\in V_i$ and therefore $\omega(\tau(i))\in V_i$}
&=\{\omega:\omega\in H,\,\phi_i(\omega)\le\tilde\tau(i)
  \text{ for every }i\in L(T)\}
=\phi^{-1}[\tilde H_{\tau}].\cr}$$

Set $\tilde\Cal G=\{\tilde H_{\tau}:\tau\in T\}$.   Because
$\BbbR^{L(T)}$ is \hwtr\ (438O), and $\tilde\Cal G$ is a family of open
subsets of $\BbbR^{L(T)}$,
$\bigcup\tilde\Cal G\in\Cal J(\tilde\Cal G)$.   By 438Nc,
$H=\bigcup_{\tau\in T}H_{\pmb{q}\tau}=\phi^{-1}[\bigcup\tilde\Cal G]$
belongs to $\Cal J(\{H_{\pmb{q}\tau}:\tau\in T\}$ and therefore to
$\Cal J(\Cal G)$, by 438Nb.

Thus in all three cases the induction proceeds.\ \Qed

\medskip

\quad{\bf (v)} This means that, for any $\pmb{q}\in Q$,
$Y_{\pmb{q}}=\bigcup\{H_{\pmb{q}\tau}:\tau\in S_{\pmb{q}}\}$ belongs to
$\Cal J(\Cal G)$.   Since $Q$ is countable, $Y\in\Cal J(\Cal G)$, where
$Y=\bigcup_{\pmb{q}\in Q}Y_{\pmb{q}}$.   But $\bigcup\Cal G\subseteq Y$.
\Prf\ If $\omega\in G\in\Cal G$, there are $t_0<\ldots<t_n$ and
$V'_i\in\Cal I$, for $i\le n$, such that

\Centerline{$\omega
\in\{\omega':\omega'\in\tildeClll,\,\omega'(t_i)\in V'_i$ for every
  $i\le n\}\subseteq G$.}

\noindent Set $x_i=\omega(t_i)$, $x^-_i=\lim_{s\uparrow t_i}\omega(s)$,
$x^+_i=\lim_{s\downarrow t_i}\omega(s)$ for $i\le n$;  let $U_i$, $V_i$,
$W_i\in\Cal U$ be such that $x^-_i\in U_i$, $x_i\in V_i\subseteq V'_i$,
$x^+_i\in W_i$ and any pair of $U_i$, $V_i$, $W_i$ are either equal or
disjoint;  and let $I_0,\ldots,I_n\in\Cal I$ be disjoint and such that
$t_i\in I_i$, $\omega(s)\in U_i$ for $s\in I_i\cap\ooint{-\infty,t_i}$ and
$\omega(s)\in W_i$ for $s\in I_i\cap\ooint{t_i,\infty}$ for each $i\le n$.
Then, setting $\pmb{q}=(I_0,\ldots,W_n)$ and $\tau(i)=t_i$ for $i\le n$,

\Centerline{$\omega\in F_{\pmb{q}\tau}
\subseteq H_{\pmb{q}\tau}\subseteq G$,}

\noindent so that $\tau\in S_{\pmb{q}}$ and
$\omega\in Y_{\pmb{q}}\subseteq Y$.\ \Qed

As $\Cal G$ is arbitrary, $\tildeClll $ is \hwtr.
}%end of proof of 438Q

\leader{438R}{Corollary} (a) Let $I^{\|}$ be the split
interval\cmmnt{ (419L)}.   Then any countable power of $I^{\|}$ is a
\hwtr\ compact Hausdorff space.

(b) Let $Y$ be the `Helly space', the space of non-decreasing functions
from $[0,1]$ to itself with the topology of pointwise convergence
inherited from the product topology on
$[0,1]^{[0,1]}$\cmmnt{ ({\smc Kelley 55}, Ex.\ 5M)}.
Then $Y$ is a \hwtr\ compact Hausdorff space.

\medskip

\proof{ These are both (homeomorphic to) subspaces of the
space $\tildeClll$ of Proposition 438Q, if we take $X$ there to be
$\Bbb R$.   To see this, argue as follows.   For
(a), observe that we have a function $f:I^{\|}\to\tildeClll$
defined by setting
$f(t^-)(s)=f(t^+)(s)=1$ if $s<t$, $f(t^-)(s)=f(t^+)(s)=0$ if $s>t$, and
$f(t^-)(t)=0$, $f(t^+)(t)=1$, and that $f$ is a homeomorphism between
$I^{\|}$ and its image.   Next, for any $L\subseteq\Bbb N$, we can
define $g:(I^{\|})^L\to\tildeClll$ by setting

$$\eqalign{g(\tbf{t})(s)
&=f(t_n)(s-2n)\text{ if }n\in L\text{ and }2n\le s\le 2n+1,\cr
&=0\text{ if }s\in\Bbb R\setminus\bigcup_{n\in L}[2n,2n+1]\cr}$$

\noindent for $\tbf{t}=\family{n}{L}{t_n}\in (I^{\|})^L$;  it is
easy to check that $g$ is a homeomorphism between $(I^{\|})^L$
and its image in $\tildeClll$.
As for (b), if we take $g(y)$ to be the extension of the function
$y:[0,1]\to[0,1]$ to the function which is constant on each of the
intervals $\ocint{-\infty,0}$ and $\coint{1,\infty}$, then
$g:Y\to\tildeClll$ is
a homeomorphism between $Y$ and its image $g[Y]$.

Since both $(I^{\|})^L$ and $Y$ are compact, they are
homeomorphic to closed subsets of $\tildeClll$, and are \hwtr\ (438La).
}%end of proof of 438R

\leader{*438S}{C\`all\`al \dvrocolon{functions}}\cmmnt{ To support some
of the theory of L\'evy processes which I will present in \S455, I give a
further consequence of 438Q.

\medskip

\noindent}{\bf Proposition} Let $X$ be a
Polish space.   Let $\Clll=\Clll(X)$ be the set of \callal\
functions\cmmnt{ (definition:  4A2A)} from $\coint{0,\infty}$ to $X$,
with its topology of
pointwise convergence inherited from the product topology of
$X^{\coint{0,\infty}}$.

(a)(i) If $\omega\in\Clll$, then $\omega$ is continuous at all but
countably many points of $\coint{0,\infty}$.

\quad(ii) If $\omega$, $\omega'\in\Clll$, $D$ is a dense subset of
$\coint{0,\infty}$ containing every point at which $\omega$ is
discontinuous, and $\omega'\restr D=\omega\restr D$, then $\omega'=\omega$.

(b) $\Clll$ is \hwtr.

(c) $\Clll$ is K-analytic.

\proof{ Fix a complete metric $\rho$ on $X$ defining its topology.
Let $\tildeClll\subseteq X^{\Bbb R}$ be the space of 438P-438Q.

\medskip

{\bf (a)} If $X=\emptyset$ the results are trivial.
Otherwise, fix $x_0\in X$, and for $\omega\in X^{\coint{0,\infty}}$ define
$\tilde\omega\in X^{\Bbb R}$ to be that extension of $\omega$
which takes the value $x_0$ everywhere on $\ooint{-\infty,0}$.

\medskip

\quad{\bf (i)}
If $\omega\in\Clll$, then $\tilde\omega\in\tildeClll$;
so the result follows from 438Pb.

\medskip

\quad{\bf (ii)} If $t\in D$, $\omega'(t)$ is certainly equal to
$\omega(t)$.   Next,

\Centerline{$\omega'(0)
=\lim_{s\downarrow 0}\omega'(s)
=\lim_{s\in D,s\downarrow 0}\omega'(s)
=\lim_{s\in D,s\downarrow 0}\omega(s)
=\omega(0)$.}

\noindent If $t\in\ooint{0,\infty}\setminus D$, then $\omega$ is continuous
at $t$, so

\Centerline{$\lim_{s\uparrow t}\omega'(s)
=\lim_{s\in D,s\uparrow t}\omega'(s)
=\lim_{s\in D,s\uparrow 0}\omega(s)
=\omega(t)$,}

\Centerline{$\lim_{s\downarrow t}\omega'(s)
=\lim_{s\in D,s\downarrow t}\omega'(s)
=\lim_{s\in D,s\downarrow 0}\omega(s)
=\omega(t)$.}

\noindent Since $\omega'(t)$ must be either
$\lim_{s\uparrow t}\omega'(s)$ or $\lim_{s\downarrow t}\omega'(s)$, it is
again equal to $\omega(t)$.   So $\omega'=\omega$.

\medskip

{\bf (b)} Since $\Clll$ is homeomorphic to a subspace of $\tildeClll$, it
is \hwtr\ (438Qb, 438La).

\medskip

{\bf (c)} Set $\tilde Q=\{\tilde\omega:\omega\in\Clll\}$;  then
$\tilde Q\subseteq\tildeClll$ is homeomorphic to $\Clll$.   But
$\tilde Q$ is a Souslin-F set in $\tildeClll$.   \Prf\ If $\omega\in\Clll$
then it belongs to $\tilde Q$ iff $\omega(t)=x_0$ for every $t<0$.
$\lim_{t\downarrow 0}\omega(t)=\omega(0)$, and
$\omega(t)\in\{\lim_{s\uparrow t}\omega(s),\lim_{s\downarrow t}\omega(s)\}$
for every $t>0$.   Now

\Centerline{$\{\omega:\omega(t)=x_0$ for every $t<0\}$}

\noindent is closed, while

$$\eqalignno{\{\omega:\omega\in\tildeClll,\,
  \omega(0)=\lim_{t\downarrow 0}\omega(t)\}
&=\{\omega:\omega\in\tildeClll,\,
  \omega(0)=\lim_{i\to\infty}\omega(2^{-i})\}\cr
\displaycause{because $\lim_{t\downarrow 0}\omega(t)$ is defined for every
$\omega\in\tildeClll$}
&=\bigcap_{k\in\Bbb N}\bigcup_{m\in\Bbb N}\bigcap_{i\ge m}
 \{\omega:\omega\in\tildeClll,\,\rho(\omega(2^{-i}),\omega(0))\le 2^{-k}\}
\cr}$$

\noindent is Souslin-F.   As for the other condition, note that for $t>0$
and $\omega\in\tildeClll$,
$\omega(t)$ belongs to
$\{\lim_{s\uparrow t}\omega(s),\lim_{s\downarrow t}\omega(s)\}$
iff for every $\epsilon>0$ there are distinct rational
numbers $q$, $q'\in[t-\epsilon,t+\epsilon]$ such that
$\omega(q)$ and $\omega(q')$ belong to $B(\omega(t),\epsilon)$.
Let $\Cal U$ be a countable base for the topology of $X$ and
$\Cal I$ a countable base for the topology of $\ooint{0,\infty}$ not
containing $\emptyset$;  then for $\omega\in\tildeClll$,
$\omega(t)\in\{\lim_{s\uparrow t}\omega(s),\lim_{s\downarrow t}\omega(s)\}$
for every $t>0$ if and only if

\inset{\noindent for every $U\in\Cal U$ and $I\in\Cal I$
either $I\cap\omega^{-1}[U]=\emptyset$ or
$I\cap\Bbb Q\cap\omega^{-1}[\overline{U}]$ has at least two members.}

\noindent Since, for $U\in\Cal U$ and $I\in\Cal I$,

\Centerline{$\{\omega:\omega\in\tildeClll$,
$I\cap\omega^{-1}[U]=\emptyset\}
=\bigcap_{t\in I}\{\omega:\omega\in\tildeClll$, $\omega(t)\notin U\}$}

\noindent is closed in $\tildeClll$, and

$$\eqalign{\{\omega:\omega\in\tildeClll,\,
I\cap\Bbb Q\cap\omega^{-1}&[\overline{U}]\text{ has at least two members}\}
\cr
&=\bigcup_{\Atop{q,q'\in I\cap\Bbb Q}{q<q'}}
   \{\omega:\omega(q),\,\omega(q')\in\overline{U}\}\cr}$$

\noindent is F$_{\sigma}$ in $\tildeClll$, while $\Cal I$ and $\Cal U$ are
countable,

\Centerline{$\{\omega:\omega\in\tildeClll$,
$\omega(t)\in\{\lim_{s\uparrow t}\omega(s),\lim_{s\downarrow t}\omega(s)\}$
for every $t>0\}$}

\noindent is Souslin-F in $\tildeClll$.   Taking the intersection, we see
that $\tilde Q$ is Souslin-F.\  \Qed

Accordingly $\tilde Q$ and $\Clll$ are K-analytic (422Hb).
}%end of proof of 438S

\leader{438T}{Proposition} Assume that $\frak c$ is measure-free.
Then $(I^{\|})^{\Bbb N}$, the Helly
space\cmmnt{ (438Rb)} and the spaces $\tildeClll(X)$, $\Clll(X)$ of 438Q
and 438S, for any Polish space $X$, are all Radon spaces.

\proof{ By 438Q-438S, %438Q 438R 438S
they are K-analytic and \hwtr, also they have weight at most
$w(X^{\Bbb R})\le\frak c$.   They are therefore
pre-Radon (434Jf), Borel-measure-complete (438M) and Radon
(434Ka).
}%end of proof of 438T

\leader{438U}{}\cmmnt{ In 434R I described a construction of
product measures.   In accordance with my general practice of
examining the measure algebra of any new measure, I give the
following result.

\medskip

\noindent}{\bf Proposition} Let $X$ and $Y$ be topological spaces with
$\sigma$-finite Borel measures $\mu$, $\nu$ respectively.   Suppose that
{\it either} $X$ is first-countable {\it or} $\nu$ is $\tau$-additive
and effectively locally finite.   Write $\lambda$ for the Borel measure
on $X\times Y$ defined by the formula

\Centerline{$\lambda W=\int\nu W[\{x\}]\mu(dx)$
for every Borel set $W\subseteq X\times Y$}

\noindent as in 434R(ii).   If {\it either} the weight of $X$ {\it or}
the Maharam type of $\nu$ is a measure-free cardinal, then for every
Borel set $W\subseteq X\times Y$ there is a set
$W'\in\Cal B(X)\tensorhat\Cal B(Y)$ such that
$\lambda(W\symmdiff W')=0$;  consequently, the measure algebra of
$\lambda$ can be identified with
the localizable measure algebra free product of the measure algebras of
$\mu$ and $\nu$.

\proof{{\bf (a)} Write $(\frak B,\bar\nu)$ for the measure algebra of
$\nu$.   With its measure-algebra topology, this is metrizable (323Gb).
Let $\sequencen{Y_n}$ be
a non-decreasing sequence of Borel sets of finite measure in $Y$ with
union $Y$.

\medskip

{\bf (b)} For the moment (down to the end of (e) below) fix on an open
set $W\subseteq X\times Y$.   For $x\in X$, set
$f(x)=W[\{x\}]^{\ssbullet}$ in $\frak B$.   Then $f:X\to\frak B$ is
Borel measurable.

\medskip

\Prf\ {\bf (i)} Let $H\subseteq\frak B$ be an open set.   For $k$,
$n\in\Bbb N$ set

\Centerline{$E_{nk}
=\{x:x\in X,\,2^{-n}k\le\nu(Y_n\cap W[\{x\}])<2^{-n}(k+1)\}$.}

\noindent Just as in part (a) of the proof of 434R, the function
$x\mapsto\nu(Y_n\cap W[\{x\}])$ is lower semi-continuous, so $E_{nk}$ is
a Borel set.   Set

\Centerline{$G_{nk}
=\bigcup\{G:G\subseteq X$ is open, $G\cap E_{nk}\subseteq f^{-1}[H]\}$;}

\noindent then $E=\bigcup_{n,k\in\Bbb N}(G_{nk}\cap E_{nk})$ is a Borel
set included in $f^{-1}[H]$.

\medskip

\quad{\bf (ii)} The point is that $E=f^{-1}[H]$.   To see this, take any
$x$ such that $f(x)\in H$.   Then there are $b\in\frak B$, $\epsilon>0$
such that $\bar\nu b<\infty$ and $c\in H$ whenever $c\in\frak B$ and
$\bar\nu(b\Bcap(c\Bsymmdiff f(x)))\le 5\epsilon$.   Since
$b=\sup_{n\in\Bbb N}b\Bcap Y_n^{\ssbullet}$, there is an $n\in\Bbb N$
such that $\bar\nu(b\Bsetminus Y_n^{\ssbullet})\le\epsilon$ and
$2^{-n}\le\epsilon$.   In this case $c\in H$ whenever $c\in\frak B$ and
$\bar\nu(Y_n^{\ssbullet}\Bcap(c\Bsymmdiff f(x)))\le 4\epsilon$;  thus

\Centerline{$\{x':\nu(Y_n\cap(W[\{x'\}]\symmdiff W[\{x\}]))\le
4\epsilon\}\subseteq f^{-1}[H]$.}

Let $k\in\Bbb N$ be such that
$2^{-n}k\le\nu(Y_n\cap W[\{x\}])<2^{-n}(k+1)$, that is, $x\in E_{nk}$.
Again using the ideas of part (a) of the proof of 434R, there are an
open set $G$ containing $x$ and an open set $V\subseteq Y$ such that
$G\times V\subseteq W$ and $\nu(Y_n\cap V)\ge 2^{-n}(k-1)$.   Now if
$x'\in G\cap E_{nk}$, $V\subseteq W[\{x'\}]\cap W[\{x\}]$, so

$$\eqalign{\nu(Y_n\cap(W[\{x'\}]\symmdiff W[\{x\}]))
&\le\nu(Y_n\cap W[\{x'\}])+\nu(Y_n\cap W[\{x\}])-2\nu(Y_n\cap V)\cr
&\le 2^{-n}((k+1)+(k+1)-2(k-1))
=4\cdot 2^{-n}\le 4\epsilon.\cr}$$

\noindent But this means that $G\cap E_{nk}\subseteq f^{-1}[H]$, so
$G\subseteq G_{nk}$ and $x\in G\cap E_{nk}\subseteq E$.   As $x$ is
arbitrary, $f^{-1}[H]\subseteq E$ and $E=f^{-1}[H]$.

\medskip

\quad{\bf (iii)} Thus $f^{-1}[H]$ is a Borel set.   As $H$ is arbitrary,
$f$ is Borel measurable.\ \Qed

\medskip

{\bf (c)} We need to know also that if
$\Cal H$ is a disjoint family of open subsets of $\frak B$ all meeting
$f[X]$, then

\Centerline{$\#(\Cal H)\le\max(\omega,\min(w(X),\tau(\frak B)))$}

\noindent\Prf\ Repeat the ideas of (b) above, setting

\Centerline{$G^{(H)}_{nk}=\bigcup\{G:G\subseteq X$ is open, $G\cap
E_{nk}\subseteq f^{-1}[H]\}$}

\noindent for $H\in\Cal H$ and $k$, $n\in\Bbb N$, so that
$f^{-1}[H]=\bigcup_{n,k\in\Bbb N}G^{(H)}_{nk}\cap E_{nk}$.   For fixed
$n$ and $k$ the family $\family{H}{\Cal H}{G^{(H)}_{nk}\cap E_{nk}}$ is
disjoint, so can have at most $w(E_{nk})\le w(X)$ non-empty members
(4A2D again).   But this means that

\Centerline{$\Cal H
=\bigcup_{n,k\in\Bbb N}\{H:G^{(H)}_{nk}\cap E_{nk}\ne\emptyset\}$}

\noindent has cardinal at most $\max(\omega,w(X))$.

On the other hand, there is a set $B\subseteq\frak B$, with cardinal
$\tau(\frak B)$, which $\tau$-generates $\frak B$.   The algebra
$\frak B_0$ generated by $B$ has cardinal at most $\max(\omega,\#(B))$
(331Gc), and $\frak B_0$ is topologically dense in $\frak B$ (323J), so
every member of $\Cal H$ meets $\frak B_0$, and

\Centerline{$\#(\Cal H)\le\#(\frak B_0)\le\max(\omega,\tau(\frak B))$.}

\noindent Putting these together, we have the result.\ \Qed

In particular, under the hypotheses above, $\#(\Cal H)$ is measure-free
whenever $\Cal H$ is a disjoint family of open subsets of $\frak B$
all meeting $f[X]$.

\medskip

{\bf (d)} The next step is to observe that there is a conegligible Borel
set $Z\subseteq X$ such that $f[Z]$ is separable.   \Prf\ Let $\Cal H$
be a $\sigma$-disjoint base for the topology of $\frak B$;
express it as $\bigcup_{n\in\Bbb N}\Cal H_n$ where each $\Cal H_n$ is
disjoint.   Let $\sequence{m}{X_m}$ be a cover of $X$ by Borel sets of
finite measure.   For $n\in\Bbb N$ consider
$\Cal H'_n=\{H:H\in\Cal H_n,\,H\cap f[X]\ne\emptyset\}$.   For
$m\in\Bbb N$, we have a totally
finite measure $\nu_{nm}$ with domain $\Cal P\Cal H'_n$ defined by
saying

\Centerline{$\nu_{nm}\Cal E=\mu(X_m\cap f^{-1}(\bigcup\Cal E))$}

\noindent for every $\Cal E\subseteq\Cal H'_n$.   Since $\Cal H'_n$ has
measure-free cardinal, by (c), there must be a countable set
$\Cal E_{nm}\subseteq\Cal H'_n$ such that
$\nu_{nm}(\Cal H'_n\setminus\Cal E_{nm})=0$.   Set

\Centerline{$Z=X\setminus\bigcup_{m,n\in\Bbb N}
  (X_m\cap f^{-1}[\bigcup(\Cal H'_n\setminus\Cal E_{nm})])$;}

\noindent then $Z$ is conegligible.   If $x\in Z$ and
$f(x)\in H\in\Cal H_n$, then there is some $m\in\Bbb N$ such that
$x\in X_m$, while
$H\in\Cal H'_n$, so $H$ must belong to $\Cal E_{nm}$.   But this means
that $\{f[Z]\cap H:H\in\Cal H\}$, which is a base for the topology of
$f[Z]$, is just $\{f[Z]\cap H:H\in\bigcup_{m,n\in\Bbb N}\Cal E_{mn}\}$,
and is countable.   So $f[Z]$ is separable (4A2Oc), as required.\
\Qed

\medskip

{\bf (e)} 418T(a-ii) now tells us that there is a set
$W'\in\Cal B(X)\tensorhat\Cal B(Y)$ such that
$f(x)=W'[\{x\}]^{\ssbullet}$ for every $x\in Z$, so that
$\nu(W[\{x\}]\symmdiff W'[\{x\}])=0$ for almost every $x$, that is,
$\lambda(W\symmdiff W')=0$.
And this is true for every open set $W\subseteq X\times Y$.

\medskip

{\bf (f)} Now let $\Cal W$ be the family of those Borel sets
$W\subseteq X\times Y$ for which there is a
$W'\in\Cal B(X)\tensorhat\Cal B(Y)$ such
that $\lambda(W\symmdiff W')=0$.   This is a $\sigma$-algebra containing
every open set, so is the whole Borel $\sigma$-algebra, as required.

Since the c.l.d.\ product measure $\lambda_0$ on $X\times Y$ is just the
completion of its restriction to $\Cal B(X)\tensorhat\Cal B(Y)$
(251K), and $\lambda_0$ and $\lambda$ agree on
$\Cal B(X)\tensorhat\Cal B(Y)$ (by Fubini's theorem), the embedding
$\Cal B(X)\tensorhat\Cal B(Y)\embedsinto\Cal B(X\times Y)$ induces an
isomorphism
between the measure algebras of $\lambda$ and $\lambda_0$.   As remarked
in 325Eb, because $\mu$ and $\nu$ are strictly localizable, the latter
may be
identified with `the' localizable measure algebra free product of the
measure algebras of $\mu$ and $\nu$.
}%end of proof of 438U

\cmmnt{\medskip

\noindent{\bf Remark} The hypothesis on the weight of $X$ can be
slightly weakened;  see 438Yg.   439L below shows that some restriction on
$(X,\mu)$ and $(Y,\nu)$ is necessary.
}%end of comment

\exercises{\leader{438X}{Basic exercises (a)}
%\spheader 438Xa
Show that a cardinal $\kappa$ is measure-free iff
$M_{\sigma}=M_{\tau}$, where $M_{\sigma}$, $M_{\tau}$ are the spaces of
countably additive and completely additive functionals on the algebra
$\Cal P\kappa$ (362B).
%438A

\spheader 438Xb\dvArevised{2017}
 Let $(X,\Sigma,\mu)$ be a localizable measure space.   Show
that the following are equiveridical:  ($\alpha$) the magnitude of
$\mu$ (definition:  332Ga) is either finite or a measure-free
cardinal ($\beta$) every absolutely continuous countably additive
functional $\nu:\Sigma\to\Bbb R$ is truly continuous.   \Hint{363S.}
%438A, 363S

\spheader 438Xc Let $\frak A$ be a Dedekind complete Boolean algebra
with measure-free cellularity.   Show that any countably additive
functional $\nu:\frak A\to\Bbb R$ is completely additive.
%438A, 363S

\spheader 438Xd Let $U$ be a Dedekind complete Riesz space such that any
disjoint order-bounded family in $U^+$ has measure-free cardinal.
Show that $U^{\sim}_c=U^{\times}$.
%438A, 363S

\spheader 438Xe Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, $(Y,\Tau,\nu)$ a strictly localizable measure space, and
$f:X\to Y$ an \imp\ function.   Suppose that the magnitude of $\nu$ is
either finite or a measure-free cardinal.   Show that $\mu$ is strictly
localizable.
%438B query out of order

\spheader 438Xf Let $(X_1,\Sigma_1,\mu_1)$, $(X_2,\Sigma_2,\mu_2)$,
$(Y_1,\Tau_1,\nu_1)$ and $(Y_2,\Tau_2,\nu_2)$ be measure spaces, and
$\lambda_1$, $\lambda_2$ the c.l.d.\ product measures on $X_1\times Y_1$,
$X_2\times Y_2$ respectively;   suppose that $f:X_1\to X_2$ and
$g:Y_1\to Y_2$ are \imp\ functions, and that
$h(x,y)=(f(x),g(y))$ for $x\in X_1$, $y\in Y_1$.   Show that if $\mu_2$ and
$\nu_2$ are both strictly localizable, with
magnitudes which are either finite or measure-free
cardinals, then $h$ is \imp.   (Compare 251L.)
%438Xe 438B

\sqheader 438Xg Show that if $\kappa$ is a measure-free cardinal, so
is $\omega_{\kappa}$.   \Hint{show by induction on ordinals $\xi$ that
if $\#(\xi)$ is measure-free, then so is $\omega_{\xi}$.}
%438C

\sqheader 438Xh
Let $(X,\Sigma,\mu)$ be a complete locally determined measure space,
$(Y,\rho)$ a complete metric space with measure-free weight, and
$\sequencen{f_n}$ a sequence of
measurable functions from $X$ to $Y$.   Show that
$\{x:\lim_{n\to\infty}f_n(x)$ is defined in $Y\}$ is measurable.
(Cf.\ 418C.)
%438E+

\spheader 438Xi Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces with c.l.d.\
product\discretionary{}{}{}
$(X\times Y,\discretionary{}{}{}\Lambda,\lambda)$.   Give $L^0(\nu)$ the
topology of convergence in measure.   Suppose that $f:X\to L^0(\nu)$ is
measurable and there is a conegligible set
$X_0\subseteq X$ such that $w(f[X_0])$ is measure-free.   Show that
there is an $h\in\eusm L^0(\lambda)$ such that $f(x)=h_x^{\ssbullet}$
for every $x\in X$, where $h_x(y)=h(x,y)$ for $(x,y)\in\dom h$.
(Cf.\ 418S.)
%438G+

\spheader 438Xj Let $(Y,\Tau,\nu)$ be a
$\sigma$-finite measure space, and $(\frak B,\bar\nu)$ its measure
algebra,  with its usual topology;  assume that the Maharam type of
$\frak B$ is measure-free.
Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space and
$\Lambda$ the domain of the c.l.d.\ product measure $\lambda$ on
$X\times Y$.   Show that if $f:X\to\frak B$ is measurable, then there is
a $W\in\Lambda$ such that $f(x)=W[\{x\}]^{\ssbullet}$ for every
$x\in X$.   (Cf.\ 418T(b-ii).)
%438G+

\spheader 438Xk Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $V$ a normed space such that $w(V)$ is measure-free.
(i) Show that the space $\eusm L$ of measurable functions from
$X$ to $V$ is a linear space, setting $(f+g)(x)=f(x)+g(x)$, etc.   (ii)
Show that if $V$ is a Riesz space with a Riesz norm then $\eusm L$ is a
Riesz space under the natural operations.
%438G+

\sqheader 438Xl Let $X$ be a topological space and $\Cal G$ a
point-finite open cover of $X$ such that $\#(\Cal G)$ is measure-free.
Suppose that $E\subseteq X$ is such that $E\cap G$ is
universally measurable for every $G\in\Cal G$.   Show that $E$ is
universally measurable.    (Compare 434Xf(iv).)
%438J-
%query:  do we need to know that \#(\Cal G) is measure-free?
% or that \Cal G is point-finite?  for the latter, yes - try
% X=\omega_1

\sqheader 438Xm Show
that for a metrizable space $X$, the following are equiveridical:
(i) $X$ is Borel-measure-compact;  (ii) $X$ is
Borel-measure-complete;  (iii) $X$ is measure-compact;
(iv) $w(X)$ is measure-free.
%438J

\spheader 438Xn Let $X$ be a topological space and $\Cal G$ a family of
open subsets of $X$.   Show that the following are equiveridical:  (i)
there is a $\sigma$-isolated family $\Cal A$ of sets, refining $\Cal G$,
such that $\bigcup\Cal A=\bigcup\Cal G$;  (ii) there is a sequence
$\Cal H_n$ of families of open sets, all refining $\Cal G$,
such that for every $x\in\bigcup\Cal G$ there is an $n\in\Bbb N$ such
that $\{H:x\in H\in\Cal H_n\}$ is finite and not empty.
%438K

\spheader 438Xo Let $Y$ be the Helly space.  (i) Show that $Y$ is a compact convex subset of $\BbbR^{[0,1]}$ with its usual topology.   (ii) Show that there is
a natural one-to-one correspondence between the split interval $I^{\|}$ and the set of extreme points of $Y$, matching $t^-\in I^{\|}$ with the function
$\chi\coint{0,t}$ and $t^+$ with $\chi[0,t]$.
(iii) Let $P_{\text{R}}$ be the set of Radon probability measures on
$I^{\|}$ with its narrow topology (437J).   Show that
there is a natural homeomorphism $\phi:P_{\text{R}}\to Y$
defined by setting $\phi(\mu)(t)=\mu[0^-,t^-]$ for
$\mu\in P_{\text{R}}$, $t\in[0,1]$.
%438R

\spheader 438Xp Show that any countable power of the Sorgenfrey line
(415Xc, 439Q) is \hwtr.
%438R

\sqheader 438Xq Let $I^{\|}$ be the split interval.   Show that
$I^{\|}\times I^{\|}$ is a Radon space iff $\frak c$ is measure-free.
\Hint{$\{(\alpha^+,(1-\alpha)^+):\alpha\in[0,1]\}$ is a discrete Borel
subset with cardinal $\frak c$.}
%438T

\spheader 438Xr Give
$\Bbb R$ the right-facing Sorgenfrey topology (415Xc).   Show that the
following are equiveridical:  (i) $\frak c$ is measure-free;  (ii)
$\BbbR^{\Bbb N}$, with the corresponding product topology, is
Borel-measure-complete;  (iii) $\BbbR^2$, with the product topology, is
Borel-measure-compact.   (Compare 439Q.)
%438T, 438Xq

\spheader 438Xs Suppose that $\frak c$ is measure-free.   Let
$X\subseteq\BbbR^{\Bbb R}$ be the set of functions of bounded variation
on $\Bbb R$, with the topology of pointwise convergence inherited from
the product
topology of $\BbbR^{\Bbb R}$.   Show that $X$ is a Radon space.
\Hint{$X$ is an F$_{\sigma}$ subset of the space $\tildeClll$ of 438Q.}
%438T

\leader{438Y}{Further exercises (a)}
%\spheader 438Ya
Let $(X,\Sigma,\mu)$ be a probability space and $\familyiI{E_i}$ a
point-finite family of measurable sets such that
$\nu J=\mu(\bigcup_{i\in J}E_i)$ is defined for every $J\subseteq I$.
Show directly that $\nu$ is a uniformly exhaustive Maharam submeasure on
$\Cal PI$, and use the Kalton-Roberts theorem to prove 438Ba.
%438B

\spheader 438Yb Suppose that $\frak c$ is measure-free, but that
$\kappa>\frak c$ is not measure-free.   Show that there is a
non-principal $\omega_1$-complete ultrafilter on $\kappa$.   \Hint{part
(b) of the proof of 451Q.}

\spheader 438Yc Show that if $X$ is a metrizable space and
$\min(\frak c,w(X))$ is measure-free, then every $\sigma$-finite
Borel measure on $X$ has countable Maharam type.
%438H

\spheader 438Yd Let $X$ be a metacompact T$_1$ space.   Show that $X$ is
Borel-measure-compact iff every closed discrete
subspace has measure-free cardinal.
%438J

\spheader 438Ye Let $X$ be a topological space such that every subspace
of $X$ is metacompact and has measure-free cellularity.   Show that
$X$ is Borel-measure-complete.
%438J

\spheader 438Yf Let $X$ be a normal metacompact Hausdorff space.   Show
that it is measure-compact iff every closed discrete subspace has
measure-free cardinal.
%438J

\spheader 438Yg In 438U, show that it would be enough to suppose that
every discrete subset of $X$ has measure-free cardinal.
%438U

\spheader 438Yh Suppose that $\frak c$ is measure-free.   Let $D$ be
any subset of $\Bbb R$ and $X\subseteq\BbbR^D$ the set of functions of
bounded variation on $D$, with the topology of pointwise convergence
inherited from the
product topology of $\BbbR^D$.   Show that $X$ is a Radon space.
%438T, 438Xs

\spheader 438Yi Let $X$ be a totally ordered set with its order
topology.   Show that if $c(X)$ is measure-free then every
$\sigma$-finite Borel measure on $X$ has countable Maharam type.
(Cf.\ 434Yo.)
%+

\spheader 438Yj Suppose that $X$ is a normal metacompact Hausdorff space
which is not realcompact.   Show that there are a closed discrete subset
$D$ of $X$ and a non-principal $\omega_1$-complete ultrafilter on $D$.
\Hint{in 435C, if we start with a $\{0,1\}$-valued Baire measure we obtain
a $\{0,1\}$-valued Borel measure;  in the proof of 438Ba,
if $\mu$ is $\{0,1\}$-valued then $\nu$ is $\{0,1\}$-valued.}

\spheader 438Yk Let $Z$ be a regular
Hausdorff space, $T$ a Dedekind complete totally
ordered space with least and greatest elements $a$, $b$,
and $x:T\to Z$ a function such that
$\lim_{s\uparrow t}x(s)$ and $\lim_{s\downarrow t}x(s)$ are defined in
$Z$ for every $t\in T$ (except $t=a$ in the first case and $t=b$ in the
second).   Show that $x[T]$ is relatively compact in $Z$.
%438Q out of order query

\spheader 438Yl\dvAnew{2009} Let $(X,\rho)$ be a metric space, and
$P_{\text{Bor}}$ the set of Borel probability measures on $X$.
For $\mu$, $\nu\in P_{\text{Bor}}$ set
$\bar\rho_{\text{KR}}(\mu,\nu)
=\sup\{|\int u\,d\mu-\int u\,d\nu|:u:X\to[-1,1]$ is $1$-Lipschitz$\}$.
(i) Show that $\bar\rho_{\text{KR}}$ is a metric on
$P_{\text{Bor}}$.   (ii) Let $\frak T_{\text{KR}}$ be the topology it
induces on $P_{\text{Bor}}$.   Show that $\frak T_{\text{KR}}$ is finer
than the narrow topology on $P_{\text{Bor}}$.   (iii) Show that the
following are equiveridical:  ($\alpha$) the narrow topology on
$P_{\text{Bor}}$ is metrizable;  ($\beta$) $\frak T_{\text{KR}}$ is the
narrow topology on $P_{\text{Bor}}$;  ($\gamma$) $w(X)$ is measure-free.
(Cf.\ 437Rg, 437Yp.)

\spheader 438Ym\dvAnew{2016} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, and $\theta=\bover12(\mu^*+\mu_*)$ the outer measure
described in 413Xd.   
Show that if the measure $\mu_{\theta}$ defined by \Caratheodory's
method is not equal to $\mu$, then there is a set $A\subseteq X$ such that
$0<\mu^*A<\infty$ and the subspace measure on $A$ induced by $\mu$ measures
every subset of $A$.
%413Xd mt43bits
}%end of exercises

\endnotes{
\Notesheader{438} Since the axiom `every cardinal is measure-free' is
admissible -- that is, will not lead to a paradox unless one is already
latent in the Zermelo-Fraenkel axioms for set theory -- it is tempting,
in the context of this section, to assume it;  so that `every complete
metric space is Radon' becomes a theorem, along with `every measurable
function from a quasi-Radon measure space to a metrizable space is
almost continuous' (438G), `$U^{\sim}_c=U^{\times}$ for every Dedekind
complete Riesz space $U$' (438Xd), `metacompact spaces are
Borel-measure-compact' (438J), `the sum of two measurable functions
from a complete probability space to a normed space is measurable'
(438Xk) and `the Helly space is Radon' (438T).   Undoubtedly the
consequent mathematical universe is tidier.   In my view, the tidiness
is the tidiness of poverty.   Apart
from anything else, it leads us to neglect such questions as `is every
measurable function from a Radon measure space to a metrizable space
almost continuous?', which have answers in ZFC (451T).

From the point of view of measure theory, the really interesting
question is whether $\frak c$ is measure-free.
It is not quite clear from the results above why this should be so;
438T is a very small part of the story.   There is a larger hint
in 438Ce-438Cf:  if $\frak c$ is measure-free, but $\kappa>\frak c$
is not measure-free, then the witnessing measures will be purely
atomic.   I will return to this point in \S543 of Volume 5.   For a general
exploration of universes in which $\frak c$ is {\it not} measure-free,
see \S544 and {\smc Fremlin 93}.   For fragments of what happens if we
suppose that we have an atom for a measure which witnesses that $\kappa$
is not measure-free, see 438Yb and the notes on normal filters in
4A1I-4A1L.  %4A1I 4A1J 4A1K 4A1L

There are many further applications of 438Q besides those in 438R and
438Xp-438Xs.  %438Xp 438Xr 438Xs
But the most obvious candidate, the space $C(\Bbb R)$ of
continuous real-valued functions on $\Bbb R$, although indeed it is a
Borel subset of the potentially Radon space of 438Q, is in fact Radon
whether or not $\frak c$ is measure-free (454Sa).   As soon as we
start using any such special axiom as `$\frak c=\omega_1$' or `$\frak c$
is measure-free', we must make a determined effort to check, through
such examples as 438Xq, that our new theorems do indeed depend on
something more than ZFC.
}%end of notes

\discrpage

\frfilename{mt439.tex}
\versiondate{7.7.14}
\copyrightdate{2002}

\def\chaptername{Topologies and measures II}
\def\sectionname{Examples}
\def\NN{\BbbN^{\Bbb N}}

\newsection{439}

As in Chapter 41, I end this chapter with a number of examples,
exhibiting some of the boundaries around the results in the rest of the
chapter, and filling in a gap with basic facts about Lebesgue measure
(439E).   The first three examples (439A) are measures defined on
$\sigma$-subalgebras of the Borel $\sigma$-algebra of $[0,1]$ which have
no extensions to the whole Borel algebra.   The next part of the section
(439B-439G) deals with `universally negligible' sets;  I use properties
of these to show that Hausdorff measures are generally not semi-finite
(439H), closing some unfinished business from \S264, and that
smooth linear functionals may fail to be representable by integrals in
the absence of Stone's condition (439I).   In 439J-439R I set out some
examples relevant to \S\S434-435, filling out the classification schemes
of 434A and 435A, with spaces which just miss being Radon (439K) or
measure-compact (439N, 439P, 439Q).   In 439S I present the canonical
example of a non-Prokhorov topological space, answering an obvious
question from \S437.

\leader{439A}{Example} Let $\Cal B$ be the Borel $\sigma$-algebra of
$[0,1]$.
There is a probability measure $\nu$ defined on a $\sigma$-subalgebra
$\Tau$ of $\Cal B$ which has no extension to a measure on $\Cal B$.

\prooflet{
\medskip

\noindent{\bf first construction}
Let $A\subseteq[0,1]$ be an analytic set which is not Borel (423M).
Let $\Cal I$ be the family of sets of the form $E\cup F$ where $E$, $F$
are Borel sets, $E\subseteq A$ and $F\subseteq[0,1]\setminus A$.   Then
$\Cal I$ is a $\sigma$-ideal of $\Cal B$ not containing $[0,1]$.   Set
$\Tau=\Cal I\cup\{[0,1]\setminus H:H\in\Cal I\}$, and define
$\nu:\Tau\to\{0,1\}$ by setting $\nu H=0$, $\nu([0,1]\setminus H)=1$ for
every $H\in\Cal I$;  then $\nu$ is a probability measure (cf.\
countable-cocountable measures (211R) or Dieudonn\'e's measure (411Q)).

\Quer\ If $\mu:\Cal B\to[0,1]$ is a measure extending $\nu$, then its
completion $\hat\mu$ measures $A$ (432A).   Also $\hat\mu$ is a Radon
measure (433Cb).   Now every compact subset of $A$ belongs to $\Cal I$, so

\Centerline{$\hat\mu A
=\sup\{\hat\mu K:K\subseteq A$ is compact$\}
=\sup\{\nu K:K\subseteq A$ is compact$\}
=0$.}

\noindent Similarly $\hat\mu([0,1]\setminus A)=0$, which is absurd.\
\Bang

\medskip

\noindent{\bf second construction} This time, let $\Cal I$ be the family
of meager Borel sets in $[0,1]$.   As before, let $\Tau$ be $\Cal
I\cup\{[0,1]\setminus E:E\in\Cal I\}$, and set $\nu E=0$,
$\nu([0,1]\setminus E)=1$ for $E\in\Cal I$.   \Quer\ If $\mu$ is a Borel
measure extending $\nu$, then $\mu([0,1]\setminus\Bbb Q)=1$, and $\mu$
is tight (that is, inner regular with respect to the compact sets), so there is a closed subset $F$
of $[0,1]\setminus\Bbb Q$ such that $\mu F>0$.   But $F$ is nowhere
dense, so $\nu F=0$.\ \Bang

\medskip

\noindent{\bf third construction}\cmmnt{\footnote{I am grateful to
M.Laczkovich and D.Preiss for showing this to me.}{}}
There is a function
$f:[0,1]\to\{0,1\}^{\frak c}$ which is $(\Cal B,\CalBa)$-measurable,
where $\CalBa$ is the Baire $\sigma$-algebra of $\{0,1\}^{\frak c}$, and
such that $f[\,[0,1]\,]$ meets every non-empty member of
$\CalBa$.
\Prf\ Set $X=C([0,1])^{\Bbb N}$ with the product of the norm
topologies,
so that $X$ is an uncountable Polish space (4A2Pe, 4A2Qc), and
$([0,1],\Cal B)$ is isomorphic to $(X,\Cal B(X))$, where $\Cal B(X)$ is
the Borel $\sigma$-algebra of $X$ (424Da).   Define
$g:X\to\{0,1\}^{[0,1]}$ by saying that $g(\sequence{i}{u_i})(t)=1$ iff
$\lim_{i\to\infty}u_i(t)=1$.   For each $t\in[0,1]$,

$$\{\sequence{i}{u_i}:\lim_{i\to\infty}u_i(t)=1\}
=\bigcap_{m\in\Bbb N}\bigcup_{n\in\Bbb N}\{\sequence{i}{u_i}:
|u_i(t)-1|\le 2^{-m}\text{ for every }i\ge n\}$$

\noindent is a Borel subset of $X$, so $g$ is
$(\Cal B(X),\CalBa(\{0,1\}^{[0,1]}))$-measurable, where
$\CalBa(\{0,1\}^{[0,1]})$ is the Baire $\sigma$-algebra of
$\{0,1\}^{[0,1]}$ (4A3Ne).   If $E\in\CalBa(\{0,1\}^{[0,1]})$ is
non-empty, there is a countable set $I\subseteq[0,1]$ such that $E$ is
determined by coordinates in $I$ (4A3Nb), so that
$E\supseteq\{w:w\restr I=z\}$ for some $z\in\{0,1\}^I$.   Now we can
find a sequence $\sequence{i}{u_i}$ in $C([0,1])$ such that
$\lim_{i\to\infty}u_i(t)=z(t)$ for every $t\in I$ (if
$I\subseteq\{t_j:j\in\Bbb N\}$, take $u_i$ such that
$|u_i(t_j)-z(t_j)|\le 2^{-i}$ whenever $j\le i$), and in this case
$g(\sequence{i}{u_i})\in E$.

Because $(X,\Cal B(X))\cong([0,1],\Cal B)$ and
$(\{0,1\}^{[0,1]},\CalBa(\{0,1\}^{[0,1]}))
\cong(\{0,1\}^{\frak c},\CalBa)$, we can copy $g$ to a function $f$ with
the required properties.\ \QeD\

In particular, $f[\,[0,1]\,]$ has full outer measure for the usual measure
$\nu_{\frakc}$ on $\{0,1\}^{\frak c}$,
because $\nu_{\frakc}$ is completion regular (415E).   Setting
$\Tau=\{f^{-1}[H]:H\in\CalBa\}$, we have a measure $\nu$ with domain $\Tau$
such that $f$ is \imp\ for $\nu$ and $\nu_{\frak c}$ (234F).   The map
$H^{\ssbullet}\mapsto f^{-1}[H]^{\ssbullet}$ from the measure algebra of
$\nu_{\frakc}$ to the measure algebra of $\nu$ is measure-preserving;
since
it is surely surjective, the measure algebras are isomorphic, and $\nu$
has Maharam type $\frak c$.

However, any probability measure on the whole algebra $\Cal B$ has
countable Maharam type (433A), so cannot extend $\nu$.
}%end of prooflet

\cmmnt{\medskip

\noindent{\bf Remark} Compare 433J-433K.
}%end of comment

\leader{439B}{Definition} Let $X$ be a Hausdorff space.   I will call
$X$ {\bf universally negligible} if there is no Borel probability
measure $\mu$ defined on $X$ such that $\mu\{x\}=0$ for every $x\in X$.
A subset of $X$ will be `universally negligible' if it is universally
negligible in its subspace topology.

\vleader{48pt}{439C}{Proposition} Let $X$ be a Hausdorff space.

(a) If $A$ is a subset of $X$, the following are equiveridical:

\inset{(i) $A$ is universally negligible;

(ii) $\mu^*A=0$ whenever $\mu$ is a Borel probability measure on
$X$ such that $\mu\{x\}=0$ for every $x\in X$;

(iii) $\mu^*A=0$ whenever $\mu$ is a $\sigma$-finite topological
measure on $X$ such that $\mu\{x\}=0$ for every $x\in A$;

(iv) for every $\sigma$-finite topological measure $\mu$ on $X$ there is
a countable set $B\subseteq A$ such that $\mu^*A=\mu B$;

(v) $A$ is a Radon space and every compact subset of $A$ is scattered.}

\noindent In particular, countable subsets of $X$ are universally
negligible.

(b) The family of universally
negligible subsets of $X$ is a $\sigma$-ideal.

(c) Suppose that $Y$ is a universally
negligible Hausdorff space and that $f:X\to Y$ is
a Borel measurable function
such that $f^{-1}[\{y\}]$ is universally negligible for every $y\in Y$.
Then $X$ is universally negligible.

(d) If the topology on $X$ is discrete, $X$ is universally negligible
iff $\#(X)$ is measure-free.

\proof{{\bf (a)(i)$\Rightarrow$(iii)} If $A$ is universally
negligible and $\mu$ is a $\sigma$-finite topological measure on $X$
such that $\mu\{x\}=0$ for every $x\in A$, let $\mu_A$ be the subspace
measure on $A$.
\Quer\ If $\mu^*A=\alpha>0$, then (because $\mu$ is $\sigma$-finite)
there is a measurable set $E\subseteq X$ such that
$\gamma=\mu^*(E\cap A)$ is finite and non-zero.   The subspace measure
$\mu_{E\cap A}$ is a topological measure on $E\cap A$;  set
$\nu F=\gamma^{-1}\mu_{E\cap A}(E\cap F)$ for relatively Borel sets
$F\subseteq A$;  then $\nu$ is a
Borel probability measure on $A$ which is zero on singletons.\ \BanG\
So $\mu^*A=0$.

\medskip

\quad{\bf (iii)$\Rightarrow$(iv)} If (iii) is true and $\mu$ is a
$\sigma$-finite topological measure on $X$, set
$B=\{x:x\in X,\,\mu\{x\}>0\}$.   Because $\mu$ is $\sigma$-finite, $B$
must be countable, therefore measurable, and if we set
$\nu E=\mu(E\setminus B)$ for every Borel set $E\subseteq X$, $\nu$ is a
$\sigma$-finite Borel measure on $X$ and $\nu\{x\}=0$ for every
$x\in X$.   By (iii), $\nu^*A=0$, that is, there is a Borel set
$E\supseteq A$ such that $\mu(E\setminus B)=0$;  in which case

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

\noindent so $\mu^*A=\mu(A\cap B)$.   As $\mu$ is
arbitrary, (iv) is true.

\medskip

\quad{\bf (iv)$\Rightarrow$(ii)} is trivial.

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(ii)} If $A$
is not universally negligible, let $\mu$ be a Borel probability measure
on $A$ which is zero on singletons.   Set $\nu E=\mu(E\cap A)$ for any
Borel set $E\subseteq X$;  then $\nu$ is a Borel probability measure on
$X$ which is zero on singletons, and $\nu^*A=1$.

\medskip

\quad{\bf (i)$\Rightarrow$(v)} Suppose that $A$ is universally
negligible.   Let $\mu$ be a totally finite Borel measure on $A$.
Applying (i)$\Rightarrow$(iv) with $X=A$, we see that there is a
countable set $B\subseteq A$ such that $\mu B=\mu A$;  but this means
that $\mu$ is inner regular with respect to the finite subsets of $B$,
which of course are compact.   As $\mu$ is arbitrary, $A$ is a Radon
space.

\Quer\ Suppose, if possible, that $A$ has a compact set $K$ which is not
scattered.   In this case there is a continuous surjection $f:K\to[0,1]$
(4A2G(j-iv)).   Now there is a Radon probability measure $\nu$ on $K$
such that $f$ is \imp\ for $\nu$ and Lebesgue measure on $[0,1]$ and
induces an isomorphism of the measure algebras, so that $\nu$ is
atomless (418L).   Accordingly we have a Borel probability measure $\mu$
on $A$
defined by setting $\mu E=\nu(K\cap E)$ for every relatively Borel set
$E\subseteq A$, and $\mu\{x\}=0$ for every $x\in A$, so $A$ is not
universally negligible.\ \BanG\  Thus all compact subsets of $A$ are
scattered, and (v) is true.

\medskip

\quad{\bf (v)$\Rightarrow$(i)} Now suppose that (v) is true and that
that $\mu$ is a Borel probability measure on $A$.   Then $\mu$ has an
extension to a Radon measure $\tilde\mu$ (434F(a-iii)).   Let
$K\subseteq A$ be a non-empty compact set which is self-supporting for
$\tilde\mu$ (416Dc).   $K$ is scattered, so has an isolated point
$\{x\}$;  because $K$ is self-supporting, $\mu\{x\}=\tilde\mu\{x\}>0$.
As $\mu$ is arbitrary, $A$ is universally negligible.

\medskip

{\bf (b)} This is immediate from (a-ii).

\medskip

{\bf (c)} Let $\mu$ be a Borel probability measure on $X$.   Then
$F\mapsto\nu f^{-1}[F]$ is a Borel probability measure on $Y$.   Because
$Y$ is universally negligible, there must be a $y\in Y$ such that
$\mu f^{-1}[\{y\}]>0$.   Set $E=f^{-1}[\{y\}]$ and let $\mu_E$ be the
subspace measure on $E$.   Then $\mu_E$ is a non-zero totally finite
Borel measure on $E$.   Since $E$ is supposed to be universally
negligible, there must be some $x\in E$ such that
$0<\mu_E\{x\}=\mu\{x\}$.

\medskip

{\bf (d)} This is just a re-phrasing of the definition in 438A.
}%end of proof of 439C


\leader{439D}{Remark\dvro{}{s (a)}}\cmmnt{ The following will be useful when
interpreting the definition in 439B.}   Let $X$ be a hereditarily
Lindel\"of Hausdorff space and $\mu$ a topological probability measure
on $X$ such that $\mu\{x\}=0$ for every $x\in X$.   Then $\mu$ is
atomless.

\prooflet{\Prf\ Suppose that $\mu H>0$.   Write

\Centerline{$\Cal G=\{G:G\subseteq X$ is open, $\mu(G\cap H)=0\}$.}

\noindent Then there is a countable $\Cal G_0\subseteq\Cal G$ such that
$\bigcup\Cal G_0=\bigcup\Cal G$ (4A2H(c-i)), so

\Centerline{$\mu(H\cap\bigcup\Cal G)=\mu(H\cap\bigcup\Cal G_0)=0$,}

\noindent and $\mu(H\setminus\bigcup\Cal G)>0$.   Because $\mu$ is zero
on singletons, $H\setminus\bigcup\Cal G$ has at least two points $x$,
$y$ say.   Now there are disjoint open sets $G_0$, $G_1$ containing $x$,
$y$ respectively, and neither belongs to $\Cal G$, so $H\cap G_0$,
$H\cap G_1$ are disjoint subsets of $H$ of positive measure.   Thus $H$
is not an atom.   As $H$ is arbitrary, $\mu$ is atomless.\ \Qed
}%end of prooflet

\cmmnt{\spheader 439Db The obvious applications of (a) are when $X$
is separable and metrizable;  but, more generally, we can use it on any
Hausdorff space with a countable network, e.g., on any analytic space.
}%end of comment

\leader{439E}{Lemma} (a) Let $E$, $B\subseteq\Bbb R$ be such that $E$ is
measurable and $\mu_LE$, $\mu_L^*B$ are both greater than $0$, where
$\mu_L$ is Lebesgue measure.   Then $E-B=\{x-y:x\in E,\,y\in B\}$
includes a non-trivial interval.

(b) If $A\subseteq\Bbb R$ and $\mu_L^*A>0$, then $A+\Bbb Q$ is
of full outer measure in $\Bbb R$.

\proof{{\bf (a)} By 223B or 261Da, there are $a\in E$, $b\in B$ such that

$$\eqalign{\lim_{\delta\downarrow 0}\Bover1{2\delta}
  \mu(E\cap[a-\delta,a+\delta])
=\lim_{\delta\downarrow 0}\Bover1{2\delta}
  \mu^*(B\cap[b-\delta,b+\delta])
=1.\cr}$$

\noindent Let $\gamma>0$ be such that

\Centerline{$\mu_L(E\cap[a-\delta,a+\delta])>\Bover32\delta$,
\quad$\mu_L^*(B\cap[b-\delta,b+\delta])>\Bover32\delta$}

\noindent whenever $0<\delta\le\gamma$.   Now suppose that
$0<\delta\le\gamma$.   Then

$$\eqalign{\mu_L((E+b)\cap[a+b,a+b+\delta])
&=\mu_L(E\cap[a,a+\delta])\cr
&\ge\mu_L(E\cap[a-\delta,a+\delta])-\delta
>\Bover12\delta,\cr}$$

\noindent and similarly

$$\eqalign{\mu_L^*((B+a+\delta)\cap[a+b,a+b+\delta])
&=\mu_L^*(B\cap[b-\delta,b])\cr
&\ge\mu_L^*(B\cap[b-\delta,b+\delta])-\delta
>\Bover12\delta.\cr}$$

\noindent But this means that
$(E+b)\cap(B+a+\delta)$ cannot be empty.   If
$u\in(E+b)\cap(B+a+\delta)$, then $u-b\in E$ and $u-a-\delta\in B$ so

\Centerline{$a-b+\delta=(u-b)-(u-a-\delta)\in E-B$.}

As $\delta$ is arbitrary, $E-B$ includes the interval
$\ocint{a-b,a-b+\gamma}$.

\medskip

{\bf (b)} \Quer\ Suppose, if possible, otherwise;  that there is a
measurable set $E\subseteq\Bbb R$ such that $\mu_LE>0$ and
$E\cap(A+\Bbb Q)=\emptyset$.   Then $E-A$ does not meet $\Bbb Q$ and
cannot include any non-trivial interval.\ \Bang
}%end of proof of 439E

\cmmnt{\medskip

\noindent{\bf Remark} There will be a dramatic generalization of (a) in
443Db.
}%end of comment

\leader{439F}{Proposition} Let $\kappa$ be the least cardinal of any set of
non-zero Lebesgue outer measure in $\Bbb R$.

(a) There is a set $X\subseteq[0,1]$ with cardinal $\kappa$ and full outer
Lebesgue measure.

(b) If $(Z,\Tau,\nu)$ is any atomless complete locally determined
measure space and $A\subseteq Z$ has cardinal less than $\kappa$, then
$\nu^*A=0$.

(c)\cmmnt{ ({\smc Grzegorek 81})} There is a universally negligible
set $Y\subseteq[0,1]$ with cardinal $\kappa$.

\proof{{\bf (a)} Take any set $A\subseteq\Bbb R$ such that
$\#(A)=\kappa$ and $\mu_L^*A>0$, where $\mu_L$ is Lebesgue measure.
Set $B=A+\Bbb Q$.   Then $(\mu_L)_*(\Bbb R\setminus B)=0$, by 439Eb.
Set $X=[0,1]\cap B$;  then $\mu_L^*X=1$ while $\#(X)\le\#(B)=\kappa$.
By the definition of $\kappa$, $\#(X)$ must be exactly $\kappa$.

\medskip

{\bf (b)} \Quer\ Otherwise, by 412Jc, there is a set $F\subseteq Z$ such
that $\nu F<\infty$ and $\nu^*(F\cap A)>0$.   By 343Cc, there is
a function $f:F\to[0,\nu F]$ which is \imp\ for
the subspace measure $\nu_F$ and Lebesgue measure on
$[0,\nu F]$.   But $f[A\cap F]$ has cardinal less than $\kappa$, so
$\mu_Lf[A\cap F]=0$ and

\Centerline{$0<\nu^*(A\cap F)\le\nu f^{-1}[f[A\cap F]]=0$,}

\noindent which is absurd.\ \Bang

\medskip

{\bf (c)(i)} Enumerate $X$ as $\langle x_{\xi}\rangle_{\xi<\kappa}$.
For each $\xi<\kappa$, $A_{\xi}=\{x_{\eta}:\eta\le\xi\}$ has cardinal
less than $\kappa$, so is Lebesgue negligible;  let
$\sequencen{I_{\xi n}}$ be a sequence of intervals covering $A_{\xi}$ with
$\sum_{n=0}^{\infty}\mu_LI_{\xi n}<\bover12$.   Enlarging the intervals
slightly if necessary, we may suppose that every $I_{\xi n}$ has
rational endpoints;  let $\sequence{m}{J_m}$ enumerate the family of
intervals in $\Bbb R$ with rational endpoints.

Set

\Centerline{$C_{mn}=\{\xi:\xi<\kappa,\,I_{\xi n}=J_m\}$}

\noindent for each $m$, $n\in\Bbb N$.

\medskip

\quad{\bf (ii)} If $\nu$ is an atomless totally finite measure on
$\kappa$ which measures every $C_{mn}$, then $\nu\kappa=0$.   \Prf\ Note
first that (by (b), applied to the completion of $\nu$) $\nu^*\xi=0$ for
every $\xi<\kappa$.   Let $\lambda$
be the (c.l.d.) product of $\mu_X$, the subspace measure on $X$,
with $\nu$.   Set

\Centerline{$B=\bigcup_{m,n\in\Bbb N}((X\cap J_m)\times C_{mn})
\subseteq X\times\kappa$.}

\noindent Then $B$ is measured by $\lambda$, so, by Fubini's theorem,

\Centerline{$\int\nu B[\{x\}]\mu_X(dx)
=\int\mu_X B^{-1}[\{\xi\}]\nu(d\xi)$}

\noindent (252D).

Now look at the sectional measures $\nu B[\{x\}]$,
$\mu_XB^{-1}[\{\xi\}]$.   (Because $B$ is actually a countable union of
measurable rectangles, these are always defined.)   For any $x\in X$,
there is an $\eta<\kappa$ such that $x=x_{\eta}$, and now

$$\eqalign{B[\{x\}]
&=\{\xi:\text{ there are }m,\,n\in\Bbb N
  \text{ such that }x\in J_m\text{ and }\xi\in C_{mn}\}\cr
&=\{\xi:\text{ there are }m,\,n\in\Bbb N
  \text{ such that }x_{\eta}\in J_m\text{ and }I_{\xi n}=J_m\}\cr
&=\{\xi:\text{ there is an }n\in\Bbb N
  \text{ such that }x_{\eta}\in I_{\xi n}\}
\supseteq\kappa\setminus\eta\cr}$$

\noindent by the choice of the $I_{\xi n}$.   But as $\nu^*\eta=0$, this
means that $\nu B[\{x\}]=\nu\kappa$.

On the other hand, if $\xi<\kappa$, then

$$\eqalign{B^{-1}[\{\xi\}]
&=\{x:\text{ there are }m,\,n\in\Bbb N\text{ such that }x\in J_m
  \text{ and }\xi\in C_{mn}\}\cr
&=\{x:\text{ there are }m,\,n\in\Bbb N\text{ such that }x\in J_m
  \text{ and }I_{\xi n}=J_m\}\cr
&=\{x:\text{ there is an }n\in\Bbb N
  \text{ such that }x\in I_{\xi n}\}
=X\cap\bigcup_{n\in\Bbb N}I_{\xi n},\cr}$$

\noindent so that

\Centerline{$\mu_XB^{-1}[\{\xi\}]
\le\sum_{n=0}^{\infty}\mu_LI_{\xi n}\le\Bover12$.}

Returning to the integrals, we have

\Centerline{$\nu\kappa
=\int\nu B[\{x\}]\mu_X(dx)
=\int\mu_X B^{-1}[\{\xi\}]\nu(d\xi)
\le\Bover12\nu\kappa$,}

\noindent so that $\nu\kappa$ must be $0$, as claimed.\ \Qed

\medskip

\quad{\bf (iii)} Now there is an injective function $g:\kappa\to[0,1]$
such that $g[C_{mn}]$ is relatively Borel in $g[\kappa]$ for every $m$,
$n\in\Bbb N$.   \Prf\ Define
$h:\kappa\to\{0,1\}^{\Bbb N\times\Bbb N\times\Bbb N}$ by setting

$$\eqalign{h(\xi)(m,n,k)&=1\text{ if }\xi\in C_{mn}
  \text{ and }x_{\xi}\in J_k,\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then $h$ is injective (because if $\xi\ne\eta$ then
$x_{\xi}\ne x_{\eta}$, so there is some $k$ such that $x_{\xi}\in J_k$
and $x_{\eta}\notin J_k$), and

\Centerline{$h[C_{mn}]=h[\kappa]\cap\{w:\text{ there is some }k
  \text{ such that }w(m,n,k)=1\}$}

\noindent is relatively Borel in $h[\kappa]$ for every $m$,
$n\in\Bbb N$.   But now recall that
$\{0,1\}^{\Bbb N\times\Bbb N\times\Bbb N}\cong\{0,1\}^{\Bbb N}$ is
homeomorphic to the Cantor set $C\subseteq[0,1]$ (4A2Uc).   If
$\phi:\{0,1\}^{\Bbb N\times\Bbb N\times\Bbb N}\to C$ is any
homeomorphism,
then $\phi h$ has the required properties.\ \Qed

\medskip

\quad{\bf (iv)} Set $Y=g[\kappa]$.   Because $g$ is injective,
$\#(Y)=\kappa$.   Also $Y$ is universally negligible.   \Prf\ Suppose
that $\tilde\nu$ is a Borel measure on $Y$ which is zero on singletons.
Then it is atomless, because $Y$ is separable and metrizable (439D).
So its copy $\nu=\tilde\nu(g^{-1})^{-1}$ on $\kappa$ is atomless.
Because $g[C_{mn}]$ is a Borel subset of $Y$, $\nu$ measures $C_{mn}$
for all $m$, $n\in\Bbb N$, so $\tilde\nu Y=\nu\kappa=0$, by (ii)
above.\ \Qed
}%end of proof of 439F

\leader{439G}{Corollary} A metrizable continuous image of a universally
negligible metrizable space need not be universally negligible.

\proof{ Take $X$ and $Y$ from 439Fa and 439Fc above, and let $f:X\to Y$
be any bijection.   Let $\Gamma$ be the graph of $f$.   The projection
map $(x,y)\mapsto y:\Gamma\to Y$ is continuous and injective, so
$\Gamma$ is universally negligible, by 439Cc.   On the other hand, the
projection map $(x,y)\mapsto x:\Gamma\to X$ is continuous and
surjective, and $X$ is
surely not universally negligible, since it is not Lebesgue negligible.
}%end of proof of 439G

\leader{439H}{Corollary} One-dimensional Hausdorff measure on $\BbbR^2$
is not semi-finite.

\proof{ Let $\mu_{H1}$ be one-dimensional Hausdorff measure on
$\BbbR^2$.    Let $X$, $\Gamma$ be the sets described in 439F and the
proof of 439G.

\medskip

{\bf (a)} $\mu_{H1}^*\Gamma>0$.   \Prf\ The first-coordinate map
$\pi_1:\BbbR^2\to\Bbb R$ is 1-Lipschitz, so, writing $\mu_L$ for
Lebesgue measure on $\Bbb R$,

\Centerline{$1=\mu_L^*X=\mu_L^*\pi_1[\Gamma]\le\mu_{H1}^*\Gamma$}

\noindent by 264G/471J and 264I.\ \Qed

\medskip

{\bf (b)} If $E\subseteq\BbbR^2$ and $\mu_{H1}E<\infty$, then
$\mu_{H1}(E\cap\Gamma)=0$, because $\Gamma$ is universally negligible,
so $E\cap\Gamma$ is universally
negligible (439Cb), and $\mu_{H1}$ is a topological measure (264E/471Da)
which is zero on singletons.

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that $\Gamma$ is not measured by
$\mu_{H1}$.   Then there is a set $A\subseteq\BbbR^2$ such that
$\mu^*_{H1}A<\mu^*_{H1}(A\cap\Gamma)+\mu^*_{H1}(A\setminus\Gamma)$
(264Fb/471Dc).   Let $E$ be a Borel set including $A$ such that
$\mu_{H1}E=\mu_{H1}^*A$ (264Fa/471Db);  then $\mu_{H1}(E\cap\Gamma)=0$, so

\Centerline{$\mu^*_{H1}(A\cap\Gamma)+\mu^*_{H1}(A\setminus\Gamma)
\le\mu_{H1}(E\cap\Gamma)+\mu_{H1}^*A=\mu^*_{H1}A$.  \Bang}

\medskip

{\bf (d)} Since $\Gamma$ is measurable, not negligible, and meets every
measurable set of finite measure in a negligible set, it is purely
infinite, and $\mu_{H1}$ is not semi-finite.
}%end of proof of 439H

\cmmnt{\medskip

\noindent{\bf Remark} Compare 471S below.
}

\leader{439I}{Example} There are a set $X$, a Riesz subspace $U$ of
$\Bbb R^X$ and a smooth positive linear functional $h:U\to\Bbb R$ which
is not expressible as an integral.

\proof{ By 439F, we have a non-negligible $X$ and a universally negligible
set $Y$, both subsets of $[0,1]$, of the same cardinality.
Replacing $Y$ by $Y\setminus\{0\}$
if need be, we may suppose that $0\notin Y$.   Let $f:X\to Y$ be any
bijection.

Let $U$ be the Riesz subspace
$\{u\times f:u\in C_b\}\subseteq\Bbb R^X$, where $C_b$ is the space of
bounded continuous functions from $X$ to $\Bbb R$.   Because $f$ is
strictly positive, $u\mapsto u\times f:C_b\to U$ is a bijection,
therefore a Riesz space isomorphism;
moreover, for a non-empty set $A\subseteq C_b$, $\inf_{u\in A}u(x)=0$
for every $x\in X$ iff $\inf_{u\in A}u(x)f(x)=0$ for every $x\in X$.
We therefore have a smooth linear functional $h:U\to\Bbb R$ defined by
setting $h(u\times f)=\int u\,d\mu_X$ for every $u\in C_b$, where
$\mu_X$ is the subspace measure on $X$ induced by Lebesgue measure.
(By 415B, $\mu_X$ is quasi-Radon, so the integral it defines on $C_b$ is
smooth, as noted in 436H.)

\Quer\ But suppose, if possible, that $h$ is the integral with respect
to some measure $\nu$ on $X$.   Since $f\in U$, it must be
$\Tau$-measurable, where $\Tau$ is the domain of the completion
$\hat\nu$ of $\nu$.   Note that $\hat\nu\{x\}=0$ for every $x\in X$.
\Prf\ Set $u_n(y)=\max(0,1-2^n|y-x|)$ for $y\in X$.   Then

$$\eqalign{f(x)\hat\nu\{x\}
&=\lim_{n\to\infty}\int u_n\times f\,d\nu
=\lim_{n\to\infty}h(u_n\times f)\cr
&=\lim_{n\to\infty}\int u_n\,d\mu_X
=\mu_X\{x\}
=0,\cr}$$

\noindent so $\hat\nu\{x\}=0$.\ \Qed

For Borel sets $E\subseteq[0,1]$ set
$\lambda E=\hat\nu f^{-1}[E]$.   Then the completion $\hat\lambda$ of
$\lambda$ is a Radon measure on $[0,1]$ (433Cb or 256C).   If $t\in[0,1]$
then $f^{-1}[\{t\}]$ contains at most one point, so
$\hat\lambda\{t\}=\lambda\{t\}=0$.   But $Y$ is supposed to be
universally negligible, so $\lambda^*Y=\hat\lambda^*Y=0$ (439Ca), that
is, there is a Borel set $E\supseteq Y$ with $\lambda E=0$;  in which
case $\nu X=\hat\nu f^{-1}[E]=0$, which is impossible.\ \Bang

Thus $h$ is not an integral, despite being a smooth linear functional on
a Riesz subspace of $\Bbb R^X$.
}%end of proof of 439I

\cmmnt{\medskip

\noindent{\bf Remark} Compare 436H.   This example is adapted from
{\smc Fremlin \& Talagrand 78}.
}%end of comment

\leader{439J}{Example} Assume that there is some cardinal $\kappa$ which
is not measure-free.   Give $\kappa$ its discrete topology, and let
$\mu$ be a probability measure with domain $\Cal P\kappa$ such that
$\mu\{\xi\}=0$ for every $\xi<\kappa$.   Now every subset of $\kappa$ is
open-and-closed, so $\mu$ is simultaneously a Baire probability measure
and a completion regular Borel probability measure.    Of course it is
not $\tau$-additive.   \cmmnt{In the classification schemes of 434A
and 435A, we have a measure which is of type B$_1$ as a Borel measure
and type E$_3$ as a Baire measure.}
%4{}35A (type E_3) %434A (type B_1)

\leader{439K}{Example} There is a first-countable compact Hausdorff
space which is not Radon.

\proof{ The construction starts from a compact metrizable space
$(Z,\frak S)$ with an atomless Radon probability measure $\mu$.   The
obvious candidate is $[0,1]$ with Lebesgue measure;  but for technical
convenience in a later application I will instead use
$Z=\{0,1\}^{\Bbb N}$ with its usual product topology and measure (254J).

\medskip

{\bf (a)} There is a topology $\frak T_{\frak c}$ on $Z$ such that

\quad($\alpha$) $\frak S\subseteq\frak T_{\frak c}$;

\quad($\beta$) every point of $Z$ belongs to a countable set which is
compact and open for $\frak T_{\frak c}$;

\quad($\gamma$) if $\sequencen{F_n}$ is a sequence of
$\frak T_{\frak c}$-closed sets with empty intersection, then
$\bigcap_{n\in\Bbb N}\overline{F}_n^{\frak S}$ is countable, where I
write $\overline{F}^{\frak S}$ for the $\frak S$-closure of $F$.

\medskip

\Prf{\bf (i)} $\frak T_{\frak c}$ will be the last in a family
$\langle\frak T_{\xi}\rangle_{\xi\le\frak c}$ of topologies.   We must
begin by enumerating $Z$ as
$\langle z_{\xi}\rangle_{\xi<\frak c}$
and taking a family $\langle\sequencen{I_{\xi n}}\rangle_{\xi<\frak c}$
running over $([Z]^{\le\omega})^{\Bbb N}$ with cofinal repetitions.
(This can be done because $\#([Z]^{\le\omega})=\frak c$, by 2A1Hb.)
Together with $\langle\frak T_{\xi}\rangle_{\xi\le\frak c}$ we choose
simultaneously families $\langle x_{\xi}\rangle_{\xi<\frak c}$,
$\langle y_{\xi}\rangle_{\xi<\frak c}$ of points in $Z$, and the
inductive hypothesis will be

\inset{$\frak T_{\xi}$ is a topology on
$X_{\xi}=\{x_{\eta}:\eta<\xi\}\cup\{y_{\eta}:\eta<\xi\}$ finer than the
topology on $X_{\xi}$ induced by $\frak S$;}

\inset{if $\eta<\xi\le\frak c$, then $X_{\eta}\in\frak T_{\xi}$ and
$\frak T_{\eta}$ is the subspace topology on $X_{\eta}$ induced by
$\frak T_{\xi}$;}

\inset{every point of $X_{\xi}$ belongs to a countable set which is
compact and open for $\frak T_{\xi}$.}

\noindent The induction starts with $X_0=\emptyset$,
$\frak T_0=\{\emptyset\}$.

\medskip

\quad{\bf (ii)} {\it Inductive step to a successor ordinal} Suppose that
we have found $X_{\xi}$ and $\frak T_{\xi}$ where $\xi<\frak c$.

\medskip

\qquad\grheada\ Start by picking $y_{\xi}\in Z\setminus X_{\xi}$ such
that $y_{\xi}=z_{\xi}$ if $z_{\xi}\notin X_{\xi}$.
Examine the sequence $\sequencen{I_{\xi n}}$.   If either
$\bigcup_{n\in\Bbb N}I_{\xi n}\not\subseteq X_{\xi}$ or
$\bigcap_{n\in\Bbb N}\overline{I}_{\xi n}^{\frak S}$ is countable, take
$x_{\xi}$ to be any point of $Z\setminus(X_{\xi}\cup\{y_{\xi}\})$
and set $K_m=\emptyset$ for every $m$ before proceeding to ($\gamma$)
below.

\medskip

\qquad\grheadb\ If $I_{\xi n}\subseteq X_{\xi}$ for every $n$ and
$\bigcap_{n\in\Bbb N}\overline{I}_{\xi n}^{\frak S}$ is uncountable, it
must have cardinal $\frak c$, by 423L, so cannot be included in
$X_{\xi}\cup\{y_{\xi}\}$.    Take any
$x_{\xi}\in\bigcap_{n\in\Bbb N}\overline{I}_{\xi n}^{\frak S}
\setminus(X_{\xi}\cup\{y_{\xi}\})$.   Let $\sequence{m}{t_m}$ be a
sequence in $Z$ such that $t_m\restr m=x_{\xi}\restr m$ for every
$m\in\Bbb N$ and $t_m\in I_{\xi n}$ whenever $r\in\Bbb N$, $n\le 2r$ and
$m=r^2+n$.   (Thus, for each $n$, $t_m\in I_{\xi n}$ for infinitely many
$m$, while $\sequence{m}{t_m}\to x_{\xi}$ in the ordinary sense.)   By
the inductive hypothesis, we can find countable sets
$K_m\subseteq X_{\xi}$,
compact and open for $\frak T_{\xi}$, such that $t_m\in K_m$ for each
$m$.   Because $\{t:t\in X_{\xi},\,t\restr m=x_{\xi}\restr m\}$ is
open-and-closed for $\frak T_{\xi}$ and contains $t_m$, we may suppose
that $t\restr m=x_{\xi}\restr m$ for every $t\in K_m$.

\medskip

\qquad\grheadc\ Let $\frak T_{\xi+1}$ be the topology on
$X_{\xi+1}=X_{\xi}\cup\{x_{\xi},y_{\xi}\}$ generated by

\Centerline{$\frak T_{\xi}\cup\{\{y_{\xi}\}\}\cup\{L_n:n\in\Bbb N\}$,}

\noindent where $L_n=\{x_{\xi}\}\cup\bigcup_{m\ge n}K_m$ for each $n$.

\medskip

\qquad\grheadd\ Because $\frak T_{\xi}\subseteq\frak T_{\xi+1}$,
$X_{\xi}$ will be open in $X_{\xi+1}$.   Because the $K_m$ are always
$\frak T_{\xi}$-open, and $x_{\xi}$, $y_{\xi}$ are distinct points of
$Z\setminus X_{\xi}$, the topology on $X_{\xi}$ induced by
$\frak T_{\xi+1}$ is just $\frak T_{\xi}$.   Consequently (by the
inductive
hypothesis) the topology on $X_{\eta}$ induced by $\frak T_{\xi+1}$ is
$\frak T_{\eta}$ for every $\eta\le\xi$.   We have
$t\restr n=x_{\xi}\restr n$ for every $t\in L_n$, so $\frak T_{\xi+1}$
is finer than the usual topology on $X_{\xi+1}$.

If $x\in X_{\xi}$, then there is a countable $\frak T_{\xi}$-open
$\frak T_{\xi}$-compact set containing $x$, which is still
$\frak T_{\xi+1}$-open and $\frak T_{\xi+1}$-compact.   Of course
$\{y_{\xi}\}$
is a countable $\frak T_{\xi+1}$-open $\frak T_{\xi+1}$-compact set
containing $y_{\xi}$.   As for $x_{\xi}$, $L_0$ is surely countable and
$\frak T_{\xi+1}$-open.   To see that it is $\frak T_{\xi+1}$-compact,
observe that any ultrafilter containing $L_0$ either contains every
$L_n$, and converges to $x_{\xi}$, or contains some $K_m$ and converges
to a point of $K_m$.

Thus the induction proceeds at successor stages.

\medskip

\quad{\bf (iii)} {\it Inductive step to a limit ordinal} If
$\xi\le\frak c$ is a non-zero limit ordinal, then we have
$X_{\xi}=\bigcup_{\eta<\xi}X_{\eta}$, and can take $\frak T_{\xi}$ to be
the topology generated by $\bigcup_{\eta<\xi}\frak T_{\eta}$.   It is
easy to check that this works (because the topologies $\frak T_{\eta}$
are consistent with each other).

\medskip

\quad{\bf (iv)} At the end of the induction, we have $X_{\frak c}=Z$
because $z_{\xi}\in X_{\xi+1}\subseteq X_{\frak c}$ for every $\xi$.
The final topology $\frak T_{\frak c}$ on $Z$ will have the properties
($\alpha$) and ($\beta$) required.   \Quer\ Now suppose, if possible,
that $\sequencen{F_n}$ is a sequence of $\frak T_{\frak c}$-closed sets
with empty intersection, and that
$\bigcap_{n\in\Bbb N}\overline{F}^{\frak S}_n$ is uncountable.   For
each $n\in\Bbb N$, let
$J_n\subseteq F_n$ be a countable $\frak S$-dense set.   Then there is
some $\zeta<\frak c$ such that
$\bigcup_{n\in\Bbb N}J_n\subseteq X_{\zeta}$ (because
$\cf\frak c>\omega$, see 4A1A(c-iii)).
Let $\xi\ge\zeta$ be such that $J_n=I_{\xi n}$ for every $n\in\Bbb N$.
Then in the construction of $\frak T_{\xi+1}$ we must be in case
($\beta$) of (ii) above.   Taking
$\sequence{m}{t_m}$ as described there, we have
$\sequence{m}{t_m}\to x_{\xi}$ for $\frak T_{\xi+1}$, and therefore for
$\frak T_{\frak c}$.
But for any $n\in\Bbb N$, $t_m\in J_n\subseteq F_n$ for infinitely many
$m$, so $x_{\xi}\in F_n$.   Thus $x_{\xi}\in\bigcap_{n\in\Bbb N}F_n$;
but this is impossible.\ \Bang

So we have a topology of the type required.\ \Qed

\medskip

{\bf (b)} There is a probability measure $\nu$ on $Z$, extending the
usual measure $\mu$, such that with respect to
$\frak T_{\frak c}\,\,\nu$ is a topological measure inner regular with
respect to the closed sets, but is not $\tau$-additive.

\medskip

\Prf\ Let $\Cal K$ be the family of $\frak T_{\frak c}$-closed
subsets of $Z$.   For $F\in\Cal K$, set
%$\phi F=\mu^*F$.
$\phi F=\mu\overline{F}^{\frak S}$.

\medskip

\quad{\bf (i)} If $E$, $F$ are disjoint $\frak T_{\frak c}$-closed
sets, then $\overline{E}^{\frak S}\cap\overline{F}^{\frak S}$ must be
countable (take $F_{2n}=E$, $F_{2n+1}=F$ in (a-$\gamma$)).   So

$$\eqalign{\phi(E\cup F)
%=\mu^*(E\cup F)
%&=\mu^*((E\cup F)\cap\overline{E}^{\frak S})
%  +\mu^*((E\cup F)\setminus\overline{E}^{\frak S})\cr
%&=\mu^*E+\mu^*F
&=\mu(\overline{E\cup F}^{\frak S})
=\mu\overline{E}^{\frak S}+\mu\overline{F}^{\frak S}
   -\mu(\overline{E}^{\frak S}\cap\overline{F}^{\frak S})\cr
&=\mu\overline{E}^{\frak S}+\mu\overline{F}^{\frak S}
=\phi E+\phi F.\cr}$$

\medskip

\quad{\bf (ii)} If $E$, $F\in\Cal K$, $E\subseteq F$ and $\epsilon>0$,
there is an $\frak S$-open set
%$G\supseteq E$
$G\supseteq\overline{E}^{\frak S}$
such that
%$\mu G\le\mu^*E+\epsilon$.
$\mu G\le\mu\overline{E}^{\frak S}+\epsilon$.
Now $F\setminus G\in\Cal K$ and

\Centerline{$\phi F
%=\mu^*(F\cap G)+\mu^*(F\setminus G)
=\mu(\overline{F}^{\frak S}\cap G)+\mu(\overline{F}^{\frak S}\setminus G)
\le\mu G+\phi(F\setminus G)
\le\phi E+\phi(F\setminus G)+\epsilon$.}

\noindent Putting this together with (i), we see that

\Centerline{$\phi F
=\phi E+\sup\{\phi E':E'\in\Cal K,\,E'\subseteq F\setminus E\}$.}

\medskip

\quad{\bf (iii)} If $\sequencen{F_n}$ is a non-increasing sequence in
$\Cal K$ with empty intersection, then

\Centerline{$\lim_{n\to\infty}\phi F_n
%\le\lim_{n\to\infty}\mu\overline{F}_n^{\frak S}
=\lim_{n\to\infty}\mu\overline{F}_n^{\frak S}
=\mu(\bigcap_{n\in\Bbb N}\overline{F}_n^{\frak S})
=0$.}

\medskip

\quad{\bf (iv)} Thus $\Cal K$ and $\phi$ satisfy all the conditions of
413J, and there is a measure $\nu$, extending $\phi$, which is defined
on every member of $\Cal K$ and inner regular with respect to $\Cal K$,
and therefore is (for $\frak T_{\frak c}$) a topological measure inner
regular with respect to the closed sets.

If we write $\Cal V$ for the family of $\frak T_{\frak c}$-compact
$\frak T_{\frak c}$-open countable subsets of $Z$, then for any
$K\in\Cal V$

\Centerline{$\nu K=\phi K=\mu K=0$,}

\noindent while $\Cal V$ is upwards-directed and has union $Z$;  so that
$\nu$ is not $\tau$-additive.\ \Qed

\medskip

{\bf (c)} So far we seem to have very little more than is provided by
$\omega_1$ with the order topology and Dieudonn\'e's measure.   The
point of doing all this work is the next step.   Set
$X=Z\times\{0,1\}$ and give $X$ the topology $\frak T$ generated by

\Centerline{$\{G\times\{0,1\}:G\in\frak S\}
\cup\{H\times\{1\}:H\in\frak T_{\frak c}\}
\cup\{X\setminus(K\times\{1\}):K$ is $\frak T_{\frak c}$-compact$\}$.}

\medskip

\quad{\bf (i)} $\frak T$ is Hausdorff.   \Prf\ If $w$, $z$ are distinct
points of $X$, then either their first coordinates differ and they are
separated by sets of the form $G_0\times\{0,1\}$, $G_1\times\{0,1\}$
where $G_0$, $G_1$ belong to $\frak S$, or they are of the form $(x,1)$,
$(x,0)$ and are separated by open sets of the form $K\times\{1\}$,
$X\setminus(K\times\{1\})$ for some set $K$ which is compact and open
for $\frak T_{\frak c}$.\ \Qed

\medskip

\quad{\bf (ii)} $\frak T$ is compact.   \Prf\ Let $\Cal F$ be an
ultrafilter on $X$.   Writing $\pi_1(x,0)=\pi_1(x,1)=x$ for $x\in Z$,
$\pi_1[[\Cal F]]$ is $\frak S$-convergent, to $x_0$ say.   If
$K\times\{1\}\in\Cal F$ for some $\frak T_{\frak c}$-compact set $K$,
then $\Cal F$ is $\frak T$-convergent to $(x,1)$;  otherwise, it is
$\frak T$-convergent to $(x,0)$ (using 4A2B(a-iv)).\ \Qed
%characterizing convergent filters given a base

\medskip

\quad{\bf (iii)} $\frak T$ is first-countable.   \Prf\ If $x\in Z$,
then $\{(x,0),(x,1)\}=\pi_1^{-1}[\{x\}]$ is a G$_{\delta}$ set in $X$
because $\{x\}$ is a G$_{\delta}$ set in $Z$ and $\pi_1$ is
continuous (4A2C(a-iii)).   Now $\{(x,0)\}$ and $\{(x,1)\}$ are
relatively open in $\{(x,0),(x,1)\}$, so are G$_{\delta}$ sets in $X$
(4A2C(a-iv)).   Thus singletons are G$_\delta$ sets.   Because $\frak T$
is compact and Hausdorff, it is first-countable (4A2Kf).\ \Qed

\medskip

\quad{\bf (iv)} $(X,\frak T)$ is not a Radon space.   \Prf\
$Z\times\{1\}$ is an open subset of $X$, homeomorphic to $Z$
with the topology $\frak T_{\frak c}$.   But the measure $\nu$ of (b)
above (or, if you prefer, its restriction to the
$\frak T_{\frak c}$-Borel
algebra) witnesses that $\frak T_{\frak c}$ is not a Radon topology, so
$\frak T$ also cannot be a Radon topology, by 434Fc.\ \Qed
}%end of proof of 439K

\cmmnt{\medskip

\noindent{\bf Remark} Aficionados will recognise $\frak T_{\frak c}$ as
a kind of `JKR-space', derived from the construction in
{\smc Juh\'asz Kunen \& Rudin 76}.
}%end of comment

\leader{439L}{Example} Suppose that $\kappa$ is a cardinal which is not
measure-free;  let $\mu$ be a probability measure with domain
$\Cal P\kappa$ which is zero on singletons.   Give $\kappa$ its discrete
topology\cmmnt{, so that $\mu$ is a Borel measure and $\kappa$ is
first-countable}.   Let $\nu$ be the restriction of the usual measure on
$Y=\{0,1\}^{\kappa}$ to the algebra $\Cal B$ of Borel subsets of $Y$, so
that $\nu$ is a $\tau$-additive probability measure,
and $\lambda$ the product measure on $\kappa\times Y$ constructed by the
method of 434R.   Then

\Centerline{$W=\{(\xi,y):\xi<\kappa$, $y(\xi)=1\}
=\bigcup_{\xi<\kappa}\{\xi\}\times\{y:y(\xi)=1\}$}

\noindent is open in $\kappa\times Y$.

If $W'\in\Cal P\kappa\tensorhat\Cal B$ then
$\lambda(W\symmdiff W')=\bover12$.   \prooflet{\Prf\  There is
a countable set $\Cal E\subseteq\Cal B$ such that $W'$ belongs to the
$\sigma$-algebra
generated by $\{A\times E:A\subseteq\kappa$, $E\in\Cal E\}$ (331Gd).
For $J\subseteq\kappa$, write $\pi_J(y)=y\restr J$ for $y\in Y$, let
$\nu_J$ be the usual measure on $\{0,1\}^J$ and $\Tau_J$ its domain,
and let $\Tau'_J$ be the family of sets
$E\subseteq Y$ such that there are $H$, $H'\in\Tau_J$ such that
$\pi_J^{-1}[H]\subseteq E\subseteq\pi_J^{-1}[H']$ and
$\nu_J(H'\setminus H)=0$.   Then $\Tau'_J\subseteq\Tau'_K$
whenever
$J\subseteq K\subseteq\kappa$, and every set measured by $\nu$ belongs
to $\Tau'_J$ for some countable $J$ (254Oc).   There is therefore a
countable set $J\subseteq\kappa$ such that $\Cal E\subseteq\Tau'_J$.
Also, of course, $\Tau'_J$ is a $\sigma$-algebra of subsets of $Y$.

The set

\Centerline{$\{V:V\subseteq\kappa\times Y,\,V[\{\xi\}]\in\Tau'_J$ for
every $\xi<\kappa\}$}

\noindent is a $\sigma$-algebra of subsets of $\kappa\times Y$
containing $A\times E$ whenever $A\subseteq\kappa$ and $E\in\Cal E$, so
contains $W'$.   But this means that if $\xi\in\kappa\setminus J$,
$W[\{\xi\}]$ and $W'[\{\xi\}]$ are stochastically independent, and
$\nu(W[\{\xi\}]\symmdiff W'[\{\xi\}])=\bover12$.   Since
$\mu(\kappa\setminus J)=1$,

\Centerline{$\lambda(W\symmdiff W')
=\int\nu(W[\{\xi\}]\symmdiff W'[\{\xi\}])\mu(d\xi)
=\Bover12$,}

\noindent as claimed.\ \Qed
}%end of prooflet

In particular, $W^{\ssbullet}$ in the the measure algebra of $\lambda$
cannot be represented by a member of $\Cal P\kappa\tensorhat\Cal B$.

\leader{439M}{Example} There is a first-countable locally compact
Hausdorff space $X$ with a
Baire probability measure $\mu$ which is not $\tau$-additive and
has no extension to a Borel
measure.   \cmmnt{In the classification of 435A, $\mu$ is of type
E$_0$.}

\proof{ Let $\Omega$ be the set of non-zero countable limit ordinals,
and for each $\xi\in\Omega$ let $\sequence{i}{\theta_{\xi}(i)}$ be a
strictly increasing sequence of ordinals with supremum $\xi$.   Set
$X=\omega_1\times(\omega+1)$, and define a topology $\frak T$ on $X$ by
saying that $G\subseteq X$ is open iff

\inset{$\{\xi:(\xi,n)\in G\}$ is open in the order topology of
$\omega_1$ for every $n<\omega$,}

\inset{whenever $\xi\in\Omega$ and $(\xi,\omega)\in G$ then there is
some $n<\omega$ such that $(\eta,i)\in G$ whenever $n\le i<\omega$ and
$\theta_{\xi}(i)<\eta\le\xi$.}

\noindent This is finer than the product of the order topologies, so is
Hausdorff.   For every $\xi<\omega_1$ and $n<\omega$,
$(\xi+1)\times\{n\}$ is a countable compact open set containing
$(\xi,n)$;  for
every $\xi\in\omega_1\setminus\Omega$, $\{(\xi,\omega)\}$ is a countable
compact open set containg $(\xi,\omega)$;  and for every $\xi\in\Omega$,

\Centerline{$\{(\xi,\omega)\}
\cup\{(\eta,i):i<\omega,\,\theta_{\xi}(i)<\eta\le\xi\}$}

\noindent is a countable compact open subset of $X$ containing
$(\xi,\omega)$.   Thus $\frak T$ is locally compact, and every singleton
subset of $X$ is G$_{\delta}$, so $\frak T$ is first-countable
(4A2Kf again).

If $f:X\to\Bbb R$ is continuous, then for every $n<\omega$ there is a
$\zeta_n<\omega_1$ such that $f$ is constant on
$\{(\xi,n):\zeta_n\le\xi<\omega_1\}$ (4A2S(b-iii)).   Setting
$\zeta=\sup_{n<\omega}\zeta_n$, $f$ must be constant on
$\{(\xi,\omega):\xi\in\Omega,\,\xi>\zeta\}$.   \Prf\ If $\xi$,
$\eta\in\Omega\setminus(\zeta+1)$, then
$f(\xi,\omega)=\lim_{i\to\infty}f(\theta_{\xi}(i)+1,i)$ and
$f(\eta,\omega)=\lim_{i\to\infty}f(\theta_{\eta}(i)+1,i)$.   But there
is some $n$ such that both $\theta_{\xi}(i)$ and $\theta_{\eta}(i)$ are
greater than $\zeta$ for every $i\ge n$, so that
$f(\theta_{\xi}(i)+1,i)=f(\theta_{\eta}(i)+1,i)$ for every $i\ge n$ and
$f(\xi,\omega)=f(\eta,\omega)$.\ \Qed

Writing $\Sigma$ for the family of subsets $E$ of $X$ such that
$\{\xi:\xi\in\Omega,\,(\xi,\omega)\in E\}$ is either countable or
cocountable in $\Omega$, $\Sigma$ is a $\sigma$-algebra of subsets of
$X$ such that every continuous function is $\Sigma$-measurable, so every
Baire set belongs to $\Sigma$.   We therefore have a Baire measure
$\mu_0$ on $X$ defined by saying that $\mu_0E=0$ if
$E\cap(\Omega\times\{\omega\})$ is countable, $1$ otherwise.
$\{(\xi+1)\times(\omega+1):\xi<\omega_1\}$ is a cover of $X$ by
negligible open-and-closed sets, so $\mu_0$ is not $\tau$-additive.

\Quer\ Suppose, if possible, that $\mu$ were a Borel measure on $X$
extending $\mu_0$.   Then we must have
$\mu(\omega_1\times\{n\})=\mu_0(\omega_1\times\{n\})=0$ for every
$n<\omega$, so $\mu(\omega_1\times\{\omega\})=1$.   Let $\lambda$ be
the subspace measure on $\omega_1\times\{\omega\}$ induced by $\mu$.
If $A\subseteq\omega_1$, $(A\times\{\omega\})\cup(\omega_1\times\omega)$
is an open set, so $\lambda$ is defined on every subset of
$\omega_1\times\{\omega\}$;  and if $\xi<\omega_1$, then
$\mu_0((\xi+1)\times(\omega+1))=0$, so $\lambda$ is zero on singletons.
And this contradicts Ulam's theorem (419G, 438Cd).\ \Bang
}%end of proof of 439M
%4{}35A (type E_0)

\leader{439N}{Example} Give $\omega_1$ its order topology.

(i) $\omega_1$ is a normal Hausdorff space which is not measure-compact.

(ii) There is a Baire probability measure $\mu_0$ on $\omega_1$ which is
not $\tau$-additive and has a unique extension to a Borel measure, which
is not completion regular\cmmnt{;  that is, $\mu_0$ is of type E$_2$
in the classification of 435A}.

\proof{{\bf (a)} As noted in 4A2Rc, order topologies are always normal
and Hausdorff.

\medskip

{\bf (b)} Let $\mu$ be Dieudonn\'e's measure on $\omega_1$, and
$\mu_0$ its restriction to the Baire $\sigma$-algebra, which is also the
countable-cocountable algebra (4A3P), so that $\mu_0$ is the
countable-cocountable measure.   Then $\mu$ is
the only Borel measure extending $\mu_0$.   \Prf\ Let $\nu$ be any Borel
measure extending $\mu_0$.   Every set $[0,\xi]=\coint{0,\xi+1}$, where
$\xi<\omega_1$, is open-and-closed, so

\Centerline{$\nu[0,\xi]=\mu_0[0,\xi]=\mu[0,\xi]=0$;}

\noindent also, of course, $\nu\omega_1=1$.   Let $F\subseteq\omega_1$
be any closed
set.   If $F$ is countable, then it is included in some initial segment
$[0,\xi]$, so $\nu F=\mu F=0$.   Now suppose that $F$ is uncountable.
Set $G=\omega_1\setminus F$.   For each $\xi\in F$, set
$\zeta_{\xi}=\min\{\eta:\xi<\eta\in F\}$ and
$G_{\xi}=\ooint{\xi,\zeta_{\xi}}$.   Then
$\langle G_{\xi}\rangle_{\xi\in F}$ is a disjoint family of open sets.
By 438Bb and 419G/438Cd,

\Centerline{$\nu(\bigcup_{\xi\in F}G_{\xi})
=\sum_{\xi\in F}\nu G_{\xi}=0$.}

\noindent But now

\Centerline{$1=\nu\omega_1
=\nu F+\nu\coint{0,\min F}+\nu(\bigcup_{\xi\in F}G_{\xi})
=\nu F=\mu F$.}

Thus $\mu$ and $\nu$ agree on the family $\Cal E$ of closed sets.   By
the Monotone Class Theorem (136C), they agree on the $\sigma$-algebra
generated by $\Cal E$, which is their common domain;  so they are
equal.\ \Qed

\medskip

{\bf (c)} I have already remarked in 411Q-411R that $\mu$ and $\mu_0$
are not $\tau$-additive and $\mu$ is not completion regular.   So of
course $\omega_1$ is not measure-compact.
}%end of proof of 439N
%4{}35A (type E_2)

\leader{439O}{}\cmmnt{ In 439M I described a Baire measure with no
extension to a Borel measure.   In view of \Marik's theorem
(435C), it is natural to ask whether this can be done with a normal
space.   This leads us into relatively deep water, and the only examples
known need special assumptions.

\medskip

\noindent}{\bf Example} Assume Ostaszewski's $\clubsuit$.
Then there is a normal
Hausdorff space with a Baire probability measure $\mu$ which is not
$\tau$-additive and not extendable to
a Borel measure.   \cmmnt{(In the classification of 435A, $\mu$ is of
type E$_0$.)}

\proof{{\bf (a)} $\clubsuit$ implies that there is a family
$\langle C_{\xi}\rangle_{\xi<\omega_1}$ of sets such that (i)
$C_{\xi}\subseteq\xi$ for every $\xi<\omega_1$ (ii) $C_{\xi}\cap\eta$ is
finite whenever $\eta<\xi<\omega_1$ (iii) for any uncountable sets $A$,
$B\subseteq\omega_1$ there is a $\xi<\omega_1$ such that $A\cap C_{\xi}$
and $B\cap C_{\xi}$ are both infinite (4A1N).   For
$A\subseteq\omega_1$, set $A'=\{\xi:\xi<\omega_1$, $A\cap C_{\xi}$ is
infinite$\}$;  then $A'\cap B'$ is non-empty whenever $A$,
$B\subseteq\omega_1$ are uncountable.   But this means that $A'\cap B'$
is actually uncountable for uncountable $A$, $B$, since $A'\cap
B'\setminus\gamma\supseteq(A\setminus\gamma)'\cap(B\setminus\gamma)'$ is
non-empty for every $\gamma<\omega_1$.

Set $X=\omega_1\times\Bbb N$.   For
$x=(\xi,n)\in X$, say that

$$\eqalign{I_x&=C_{\xi}\times\{n-1\}\text{ if }n\ge 1,\cr
&=\emptyset\text{ otherwise }.\cr}$$

\medskip

{\bf (b)} Define a topology $\frak
T$ on $X$ by saying  that a set $G\subseteq X$ is open iff $I_x\setminus
G$ is finite for every $x\in G$.

The form of the construction ensures that $\frak T$ is T$_1$.   In fact,
$I_x\cap I_y$ is finite whenever $x\ne y$ in $X$.   \Prf\
Express $x$ as $(\xi,m)$ and $y$ as
$(\eta,n)$ where $\eta\le\xi$.   If either $m=0$ or $n=0$ or $m\ne n$,
$I_x\cap I_y=\emptyset$.   If
$n\ge 1$ and $\eta<\xi$, then

\Centerline{$I_x\cap I_y
\subseteq(C_{\xi}\cap\eta)\times\{n-1\}$}

\noindent is finite.   Similarly, $I_x\cap I_y$ is finite if $m\ge 1$
and $\xi<\eta$.\ \QeD\   Consequently $\{x\}\cup J$
is closed whenever $x\in X$ and $J\subseteq I_x$.

Observe that $(\xi+1)\times\Bbb N$ is open and closed for every
$\xi<\omega_1$, again because $C_{\eta}\cap(\xi+1)$ is finite
whenever $\eta\in\Omega$ and $\eta>\xi$, while $C_{\xi}\subseteq\xi$ for
every $\xi$.

\medskip

{\bf (c)} The next step is to understand the uncountable closed
subsets of $X$.   First, if $F\subseteq X$ is closed and $n\in\Bbb N$,
then
$F^{-1}[\{n\}]'$, as defined in (a), is a subset of $F^{-1}[\{n+1\}]$,
since if
$\xi\in F^{-1}[\{n\}]'$ then $I_{(\xi,n+1)}\cap F$ is infinite.   If $F$
is uncountable, there is some $n\in\Bbb N$ such that $F^{-1}[\{n\}]$ is
uncountable, so that (inducing on $m$) $F^{-1}[\{m\}]$ is uncountable
for every $m\ge n$.   Finally, this means that if $E$, $F\subseteq X$
are uncountable closed sets, there is an $m\in\Bbb N$ such that
$E^{-1}[\{m\}]$ and $F^{-1}[\{m\}]$ are both uncountable, so that
$E^{-1}[\{m\}]'\cap F^{-1}[\{m\}]'$ is non-empty and $E\cap F$ is
non-empty.

\medskip

{\bf (d)} It follows that $X$ is normal.   \Prf\ Let $E$ and $F$ be
disjoint closed sets in $X$.   By (c), at least one of them is
countable;  let us take it that $E\subseteq\zeta\times\Bbb N$ where
$\zeta<\omega_1$.   Enumerate the open-and-closed set
$W=(\zeta+1)\times\Bbb N$ as $\sequencen{x_n}$.   Choose
$\sequencen{U_n}$, $\sequencen{V_n}$ inductively, as follows.   $U_0=E$,
$V_0=F\cap W$.   If $x_n\in U_n$, then $U_{n+1}=U_n\cup
(I_{x_n}\setminus V_n)$ and $V_{n+1}=V_n$;  if $x_n\notin U_n$, then
$U_{n+1}=U_n$ and $V_{n+1}=V_n\cup\{x_n\}\cup(I_{x_n}\setminus U_n)$.
An easy induction shows that, for every $n$, ($\alpha$) $U_n\cap
V_n=\emptyset$ ($\beta$) $U_n\cup V_n\subseteq W$ ($\gamma$)
$I_x\cap(U_n\cup V_n)$ is finite for every $x\in X\setminus(U_n\cup
V_n)$ ($\delta$) $I_x\cap V_n$ is finite for every $x\in U_n$
($\epsilon$) $I_x\cap U_n$ is finite for every $x\in V_n$.

At the end of the induction, set $G=\bigcup_{n\in\Bbb N}U_n$,
$H=\bigcup_{n\in\Bbb N}V_n\cup(X\setminus W)$.   Then $E\subseteq G$,
$F\subseteq H$ and $G\cap H=\emptyset$.   If $x\in G$, it is of the form
$x_n$ for some $n$, in which case $x_n\in U_n$ (because $x_n\notin
V_{n+1}$) and $I_x\setminus U_{n+1}=I_x\cap V_n$ is finite;  thus $G$ is
open.   If $x\in H\cap W$, again it is of the form $x_n$ where this time
$x_n\notin U_n$, so that $I_x\setminus V_{n+1}=I_x\cap U_n$ is finite;
so $H$ is open.

Thus $E$ and $F$ are separated by open sets;  since $E$ and $F$ are
arbitrary, $X$ is normal.\ \Qed

Being T$_1$ (see (b)), $X$ is also Hausdorff.

\medskip

{\bf (e)} Because disjoint closed sets in $X$ cannot both be uncountable
((c) above), any bounded continuous function on $X$ must be constant
on a cocountable set.   (Compare 4A2S(b-iii).)
%same for usual topology on $\omega_1$.
The countable-cocountable measure $\mu_0$ is therefore a Baire measure
on $X$ (cf.\ 411R).   But it has no extension to a Borel measure.
\Prf\ The point is that if $A$ is any subset of $\omega_1$, and
$n\in\Bbb N$, then

\Centerline{$(A\times\{n\})\cup(\omega_1\times n)$,
\quad$\omega_1\times n$}

\noindent are both open, so $A\times\{n\}$ is Borel;  accordingly every
subset of $X$ is a Borel set.   But $\omega_1$ is measure-free
(419G, 438Cd), so there can be no Borel probability measure on $X$ which
is zero on singletons.\ \Qed

Of course $\mu_0$ is not $\tau$-additive, because $\{(\xi+1)\times\Bbb
N:\xi<\omega_1\}$ is a cover of $X$ by open-and-closed negligible sets.
}%end of proof of 439O

\cmmnt{\medskip

\noindent{\bf Remark} Thus in Ma\v r\'\i k's theorem we really do
need `countably paracompact' as well as `normal', at least if we want a
theorem valid in ZFC.

Observe that any example of this phenomenon must involve a {\bf Dowker
space}, that is, a normal Hausdorff space which is not countably
paracompact.   The one here
is based on {\smc de Caux 76}.   Such spaces are hard to come by in ZFC
if we do not allow
ourselves to use special principles like $\clubsuit$.   `Real' Dowker
spaces have been
described by {\smc Rudin 71} and {\smc Balogh 96};  for a survey, see
{\smc Rudin 84}.   I do not know if either of these can be adapted to
provide a ZFC example to replace the one above.
}%end of comment
\leaveitout{In {\smc Simon 71} it is noted that Rudin's Dowker space
gives a Baire probability measure $\mu$ which has no extension to a
Borel measure which is inner regular with respect to the closed sets.
}

\allowmorestretch{468}{
\leader{439P}{Example}\cmmnt{ (cf.\ {\smc Moran 68})}
$\BbbN^{\frak c}$ is
not Borel-measure-compact, therefore not Borel-measure-complete,
measure-compact or Radon.
}

\proof{ Consider the topology $\frak T_{\frak c}$ on
$Z=\{0,1\}^{\Bbb N}$, as constructed in 439K.   Then
$(Z,\frak T_{\frak c})$ is homeomorphic to a closed subset of
$\BbbN^Z\times\{0,1\}^{\Bbb N}$, where in this product the second factor
$\{0,1\}^{\Bbb N}$ is given
its usual topology $\frak S$.   \Prf\ For each
$x\in Z$, let  $L_x$ be a
$\frak T_{\frak c}$-open $\frak T_{\frak c}$-compact subset of $Z$.
The first thing to observe is that if $x\in Z$, and we write
$V_{xm}=\{y:y\in Z,\,y\restr m=x\restr m\}$ for each $m\in\Bbb N$, then
$\Cal U_x=\{L_x\cap V_{xm}:m\in\Bbb N\}$ is a downwards-directed family
of compact open neighbourhoods of $x$ with intersection $\{x\}$, so is a
base of neighbourhoods of $x$ (4A2Gd);  thus
$\Cal U=\{L_x:x\in Z\}\cup\frak S$ generates $\frak T_{\frak c}$.   Now,
for $x\in Z$, define $\phi_x:Z\to\Bbb N$ by setting

$$\eqalign{\phi_{x}(y)&=0\text{ if }y\in L_x,\cr
&=m+1\text{ if }y\in V_{xm}\setminus(L_x\cup V_{x,m+1}).\cr}$$

\noindent Then every $\phi_x$ is $\frak T_{\frak c}$-continuous, so we
have a $\frak T_{\frak c}$-continuous function
$\phi:Z\to\BbbN^Z\times\{0,1\}^{\Bbb N}$ defined by setting
$\phi(y)=(\family{z}{Z}{\phi_z(y)},y)$ for $y\in Z$.   Because every
element of $\Cal U$ is of the form $\phi^{-1}[H]$ for some open set
$H\subseteq\BbbN^Z\times\{0,1\}^{\Bbb N}$, $Z$ is homeomorphic to its
image $\phi[Z]$.

Now suppose that $(w,z)\in\overline{\phi[Z]}$.   In this case, there is
a filter $\Cal G$ containing $\phi[Z]$ which converges to $(w,z)$ (4A2Bc).
Let $\Cal F$ be an ultrafilter on $Z$ including
$\{\phi^{-1}[A]:A\in\Cal G\}$;  then $\phi[[\Cal F]]$ includes $\Cal G$ so
converges to $(w,z)$, and $\Cal F\to z$ for $\frak S$.
\Quer\ If $z$ is not the
$\frak T_{\frak c}$-limit of $\Cal F$, then $\Cal F$ can have no
$\frak T_{\frak c}$-limit, and can contain no
$\frak T_{\frak c}$-compact set (2A3R).   In particular,
$L_z\notin\Cal F$;  but in this case $V_{zm}\setminus L_z\in\Cal F$ for
every $m$, so that $\{(v,y):v(x)>m\}\in\phi[[\Cal F]]$ for every $m$,
and $w(x)>m$ for every $m$, which is impossible.\ \BanG\   Thus
$\Cal F\to z$, and (as $\phi$ is continuous) $(w,z)=\phi(z)$.

This shows that $\phi[Z]$ is closed, so we have the required
homeomorphism between $Z$ and a closed subset of
$\BbbN^Z\times\{0,1\}^{\Bbb N}$.\ \Qed

Of course $\BbbN^Z\times\{0,1\}^{\Bbb N}$ is a closed subset of
$\BbbN^Z\times\BbbN^{\Bbb N}\cong\BbbN^{\frak c}$.   So $Z$ is
homeomorphic to a closed subset of $\BbbN^{\frak c}$.  But $Z$, with
$\frak T_c$, carries a Borel probability measure $\nu$ which is inner
regular with respect to the closed sets and is not $\tau$-additive
(439Kb).   So $(Z,\frak T_{\frak c})$ is not Borel-measure-compact.
By 434Hc, $\BbbN^{\frak c}$ is not Borel-measure-compact.   By 434Ic,
$\BbbN^{\frak c}$ is not
Borel-measure-complete;  by 434Ka, it is not Radon;  by 435Fd, it is not
measure-compact.
}%end of proof of 439P

\leader{439Q}{Example} Let $X$ be the Sorgenfrey line\cmmnt{ (415Xc)}.
Then $X$ is measure-compact but $X^2$ is not.

\proof{{\bf (a)} Note that every set $\coint{a,b}$ is open-and-closed in
$X$, so that the topology is zero-dimensional, therefore completely
regular;  and it is finer than the usual topology of $\Bbb R$, so is
Hausdorff.

$X$ is Lindel\"of.   \Prf\ Let $\Cal G$ be an open cover of $X$.
For each $q\in\Bbb Q$, set

\Centerline{$A_q=\{a:a\in\ooint{-\infty,q}$ and there is some
$G\in\Cal G$ such that $\coint{a,q}\subseteq G\}$.}

\noindent Then $\bigcup_{q\in\Bbb Q}A_q=\Bbb R$.   For each
$q\in\Bbb Q$, there is a countable set $A'_q\subseteq A_q$ such that
$\inf A'_q=\inf A_q$ in $[-\infty,\infty]$ and $A'_q$ contains
$\min A_q$ if $A_q$ has a least element.   Now, for each pair $(a,q)$
where $q\in\Bbb Q$ and $a\in A'_q$, choose $G_{aq}\in\Cal G$ such that
$\coint{a,q}\subseteq G_{aq}$.   It is easy to see that
$\bigcup\{G_{aq}:a\in A'_q\}\supseteq A_q$, so that the countable family
$\{G_{aq}:q\in\Bbb Q,\,a\in A'_q\}$ covers $X$.   As $\Cal G$ is
arbitrary, $X$ is Lindel\"of.\ \Qed

It follows that $X$ is measure-compact (435Fb).

\medskip

{\bf (b)} Let $\frak S$ be the usual topology on $\BbbR^2$, and $\frak
T$ the product topology on $X^2$.

\medskip

\quad{\bf (i)} Whenever $G$, $H$ are disjoint $\frak T$-open sets, there
is an $\frak S$-Borel set $E$ such that $G\subseteq E\subseteq
X^2\setminus H$.   \Prf\ For $n\in\Bbb N$, set

\Centerline{$A_n
=\{(a,b):\coint{a,a+2^{-n}}\times\coint{b,b+2^{-n}}\subseteq G\}$.}

\noindent\Quer\ Suppose, if possible, that there is a point
$(x,y)\in\overline{A}_n^{\frak S}\cap H$, where I write
$\overline{\phantom{A}}^{\frak S}$ to denote closure for the topology
$\frak
S$.   Let $\delta>0$ be such that
$\coint{x,x+2\delta}\times\coint{y,y+2\delta}\subseteq H$ and
$2\delta<2^{-n}$.   Then there must be $(a,b)\in A_n$ such that
$|a-x|\le\delta$ and
$|b-y|\le\delta$.   In this case, $a\le x+\delta<a+2^{-n}$ and $b\le
y+\delta<b+2^{-n}$, so $(x+\delta,y+\delta)\in G$;  while $\delta$ was
chosen so that $(x+\delta,y+\delta)$ would belong to $H$.\ \Bang

Accordingly $E=\bigcup_{n\in\Bbb N}\overline{A}_n^{\frak S}$ is an
$\frak S$-Borel set disjoint from $H$.   But $G=\bigcup_{n\in\Bbb
N}A_n$, so $G\subseteq E$.\ \Qed

\medskip

\quad{\bf (ii)} Consequently every $\frak T$-continuous real-valued
function is $\frak S$-Borel measurable.   \Prf\ If $f:X^2\to\Bbb R$ is
$\frak T$-continuous and $\alpha\in\Bbb R$, then there is an
$\frak S$-Borel set $E_{\alpha}$ such that

\Centerline{$\{(x,y):f(x,y)<\alpha\}
\subseteq E_{\alpha}\subseteq\{(x,y):f(x,y)\le\alpha\}$.}

\noindent But this means that
$\{(x,y):f(x,y)<\alpha\}=\bigcup_{n\in\BbbN}E_{\alpha-2^{-n}}$ is
$\frak S$-Borel.\ \Qed

\medskip

\quad{\bf (iii)} It follows that every $\frak T$-Baire set is
$\frak S$-Borel.   We therefore have a $\frak T$-Baire probability
measure $\nu$ on $X^2$ defined by setting

\Centerline{$\nu E=\mu_L\{t:t\in[0,1],\,(t,1-t)\in E\}$}

\noindent for every $\frak T$-Baire subset of $X^2$, where $\mu_L$ is
Lebesgue measure on $\Bbb R$.   In this case
every point $(x,y)$ of $X^2$ belongs to a $\frak T$-open set of zero
measure for $\nu$.   \Prf\ Set $K=\{(t,1-t):t\in[0,1]\}$.   Then $K$ is
$\frak S$-closed, therefore $\frak T$-closed, and $\nu(X^2\setminus
K)=0$, so if $(x,y)\notin K$ then we can stop.   If $(x,y)\in K$, then
$\coint{x,x+1}\times\coint{y,y+1}$ is a $\frak T$-open $\frak T$-closed
set meeting $K$
in the single point $(x,y)$, so is a negligible $\frak T$-neighbourhood
of $(x,y)$.\ \Qed

Thus $\nu$ is not $\tau$-additive and $X^2$ is not measure-compact.
}%end of proof of 439Q

\cmmnt{\medskip

\noindent{\bf Remark} Contrast this with 438Xr.
}%end of comment

\leader{439R}{Example} There are first-countable completely regular
Hausdorff spaces $X$, $Y$ with Baire probability measures $\mu$, $\nu$
such that the Baire measures $\lambda$, $\lambda'$ on $X\times Y$
defined by the formulae

\Centerline{$\int fd\lambda=\biggeriint f(x,y)\nu(dy)\mu(dx)$,
\quad$\int fd\lambda'=\biggeriint f(x,y)\mu(dx)\nu(dy)$}

\noindent\cmmnt{(436F) }are different.

\proof{ Let $X$, $Y$ be disjoint stationary subsets of $\omega_1$
(4A1Cd).   Give each the topology induced by the order topology of
$\omega_1$.   Let $\tilde\mu$ be Dieudonn\'e's measure on $\omega_1$,
and $\tilde\mu_X$, $\tilde\mu_Y$ the subspace measures
induced on $X$ and $Y$ by $\tilde\mu$;  let $\mu$ and $\nu$ be the
restrictions of $\tilde\mu_X$, $\tilde\mu_Y$ to the Baire
$\sigma$-algebras of $X$, $Y$ respectively.   Then

\Centerline{$\mu X=\tilde\mu_XX=\tilde\mu^*X=1$}

\noindent because $X$ meets every cofinal closed set in $\omega_1$;
similarly,
$\nu Y=1$.

Set

\Centerline{$W=\{(x,y):x\in X,\,y\in Y,\,x<y\}
=\{(x,y):x\in X,\,y\in Y,\,x\le y\}$.}

\noindent Then $W$ is open-and-closed in $X\times Y$ (use 4A2Rl),
so that $f=\chi W$ is continuous.   But

\Centerline{$\biggeriint f(x,y)\nu(dy)\mu(dx)
=\int\nu\{y:y\in Y,\,x<y\}\mu(dx)
=1$,}

\Centerline{$\biggeriint f(x,y)\mu(dx)\nu(dy)
=\int\mu\{x:x\in X,\,x<y\}\nu(dy)
=0$.}
}%end of proof of 439R

\cmmnt{\medskip

\noindent{\bf Remark} Contrast this with 434Xy and 439Yi.
}%end of comment

\leader{439S}{}\cmmnt{ The results of 437V leave open the question of
which familiar spaces, beyond \v{C}ech-complete spaces, can be
Prokhorov.   In fact rather
few are.   The basis of any further investigation must be the following
result.

\medskip

\noindent}{\bf Theorem}\cmmnt{ ({\smc Preiss 73})} $\Bbb Q$ is not a
Prokhorov space.

\proof{{\bf (a)} There is a non-decreasing sequence $\sequence{k}{X_k}$
of non-empty compact subsets of $X=\Bbb Q\cap[0,1]$, with union $X$, such
that whenever $k\in\Bbb N$, $x\in X_k$
and $\delta>0$, then $X_{k+1}\cap[x-\delta,x+\delta]$ is infinite.
\Prf\ Start by enumerating $X$ as $\sequence{k}{q_k}$.   Set
$X_0=\{q_0\}$.   Given
that $X_k\subseteq X$ is compact, then for each $m\in\Bbb N$ let
$\Cal E_m$ be a finite cover of $X_k$ by open intervals of length at most
$2^{-m}$ all meeting
$X_k$, and
let $I_{km}$ be a finite subset of $X\setminus X_k$ meeting every member
of $\Cal E_m$;  set
$X_{k+1}=X_k\cup\{q_{k+1}\}\cup\bigcup_{m\in\Bbb N}I_{km}$.
If $\Cal H$ is any cover of $X_{k+1}$ by open sets in $\Bbb R$, then
there is a finite $\Cal H_0\subseteq\Cal H$ covering $X_k$.   There must
be an $m\in\Bbb N$ such that
$[x-2^{-m},x+2^{-m}]\subseteq\bigcup\Cal H_0$ for every $x\in X_k$
(2A2Ed), so that $I_{kl}\subseteq\bigcup\Cal H_0$ for every
$l\ge m$, and $X_{k+1}\setminus\bigcup\Cal H_0$ is finite;  accordingly
there is a finite $\Cal H_1\subseteq\Cal H$ covering $X_{k+1}$.   As
$\Cal H$ is arbitrary,
$X_{k+1}$ is compact, and the induction can proceed.   If $x\in X_k$ and
$\delta>0$, then for every $m\in\Bbb N$ there is an $x'\in
X_{k+1}\setminus X_k$ such that
$|x'-x|\le 2^{-m}$, so that $[x-\delta,x+\delta]\cap X_{k+1}$ must be
infinite.\ \Qed

\medskip

{\bf (b)} If $\sequence{k}{\epsilon_k}$ is any sequence in
$\ooint{0,\infty}$, and $F\subseteq[0,1]$ is a countable closed set,
then there is an $x^*\in X\setminus F$ such that
$\rho(x^*,X_k)<\epsilon_k$ for every $k\in\Bbb N$.   \Prf\ We can
suppose that $\lim_{k\to\infty}\epsilon_k=0$.
Define $\sequence{k}{H_k}$ inductively, as follows.   $H_0=\Bbb R$.
Given $H_k$, set
$H_{k+1}=H_k\cap\{x:\rho(x,X_k\cap H_k)<\epsilon_k\}$, where
$\rho(x,A)=\inf_{y\in A}|x-y|$ when $x\in\Bbb R$ and $A\subseteq\Bbb R$
is non-empty.   Observe that every $H_k$ is an
open subset of $\Bbb R$ and that $X_k\cap H_k\subseteq H_{k+1}\subseteq
H_k$ for every $k$;  consequently, setting $E=\bigcap_{k\in\Bbb N}H_k$,
$E$ is a
G$_{\delta}$ subset of $\Bbb R$ and $X_k\cap H_k\subseteq E$ for every
$k$.   In particular, $E\cap X$ contains $q_0$ and is not empty.   Next,
for each $k$,
$\rho(x,E\cap X_k)<\epsilon_k$ for every
$x\in H_{k+1}$ and therefore for every $x\in E$;  accordingly $E\cap X$
is dense in $E$.   Moreover, if $x\in E\cap X$, there is
a $k\in\Bbb N$ such that $x\in X_k$;  we must have $x\in H_k$, and in
this case $H_{k+1}$ is a neighbourhood of $x$.   So every neighbourhood
of $x$ contains
infinitely many points of $H_{k+1}\cap X_{k+1}\subseteq E\cap X$.   Thus
$E\cap X$ has no isolated points;  it follows that $E$ has no isolated
points.   By 4A2Mc and 4A2Me, $E$ is uncountable.

There is therefore a point $z\in E\setminus F$.   Let $m\in\Bbb N$ be
such that $\rho(z,F)\ge\epsilon_m$.   As $z\in H_{m+1}$, there is an
$x^*\in H_m\cap X_m$ such that
$|z-x^*|<\epsilon_m$ and $x^*\notin F$.   Let $k\in\Bbb N$.   If
$k\ge m$ then certainly $\rho(x^*,X_k)=0<\epsilon_k$.
If $k<m$ then $x^*\in H_{k+1}$ so
$\rho(x^*,X_k)\le\rho(x^*,H_k\cap X_k)<\epsilon_k$.   So we have a
suitable $x^*$.\ \Qed

\medskip

{\bf (c)} For $n$, $k\in\Bbb N$ set

\Centerline{$G_{kn}=\{x:x\in\Bbb R\setminus X_k$,
$\rho(x,X_n)>2^{-k}\}$.}

\noindent Then $G_{kn}$ is an open subset of $\Bbb R$.   Let $A$ be the
set of Radon probability
measures $\mu$ on $X$ such that $\mu(G_{kn}\cap X)\le 2^{-n}$ for all
$n$,
$k\in\Bbb N$.

\medskip

{\bf (d)} Write $\tilde A$ for the set of Radon probability measures
$\mu$ on $[0,1]$ such that $\mu(G_{kn}\cap[0,1])\le 2^{-n}$ for all
$k$, $n\in\Bbb N$.   Then $\tilde A$ is a narrowly closed subset of the
set of Radon probability measures on $[0,1]$, which is itself narrowly
compact (437R(f-ii)).
Also $\mu([0,1]\setminus X)=0$ for every $\mu\in\tilde A$.   \Prf\ Let
$K\subseteq [0,1]\setminus X$ be compact, and $n\in\Bbb N$.   Then
$K$ and
$X_n$ are disjoint compact sets, so there is some $k\in\Bbb N$ such that
$|x-y|>2^{-k}$ for every $x\in X_n$ and $y\in K$.   In this case
$K\subseteq G_{kn}$ so
$\mu K\le 2^{-n}$.   As $n$ is arbitrary, $\mu K=0$;  as $K$ is
arbitrary, $\mu([0,1]\setminus X)=0$.\ \Qed

$A$ is compact in the narrow topology.   \Prf\
The identity map $\phi:X\to[0,1]$ induces a map
$\tilde\phi:M^+_{\text{R}}(X)\to M^+_{\text{R}}([0,1])$ which is a
homeomorphism between $M^+_{\text{R}}(X)$ and
$\{\mu:\mu\in M^+_{\text{R}}([0,1])$, $\mu([0,1]\setminus X)=0\}$ (437Nb).
The definition of $A$ makes it plain that it is
$\tilde\phi^{-1}[\tilde A]$;  since
$\tilde A\subseteq\{\mu:\mu\in M^+_{\text{R}}([0,1])$,
$\mu([0,1]\setminus X)=0\}$, $\tilde\phi\restr A$ is a homeomorphism
between $A$ and $\tilde A$, and $A$ is compact.\ \Qed

\medskip

{\bf (e)} $A$, regarded as a subset of $M^+_{\text{R}}(X)$, is not
uniformly tight.
\Prf\ Let $K\subseteq X$ be compact.   Consider the set
$C$ of those $w\in[0,1]^X$ such that $w(x)=0$ for every $x\in K$,
$\sum_{x\in X}w(x)\le 1$ and $\sum_{x\in G_{kn}\cap X}w(x)\le 2^{-n}$
for all $k$, $n\in\Bbb N$.
Then $C$ is a compact subset of $[0,1]^X$.   If $D\subseteq C$ is any
non-empty upwards-directed set, then $\sup D$, taken in $[0,1]^X$,
belongs to $C$.   By Zorn's
Lemma, $C$ has a maximal member $w$ say.   \Quer\ Suppose, if possible,
that $\sum_{x\in X}w(x)=\gamma<1$.
For each $n\in\Bbb N$, let $L_n\subseteq X$ be a finite set such that
$\sum_{x\in L_n}w(x)\ge\gamma-2^{-n-1}$, and $m_n\in\Bbb N$ such that
$L_n\subseteq X_{m_n}$.
By (b), there is an $x^*\in X\setminus K$
such that $\rho(x^*,X_n)<2^{-m_n}$ for every $n\in\Bbb N$.   Let
$r\in\Bbb N$ be such that $x^*\in X_r$ and $\gamma+2^{-r}\le 1$, and set
$w'(x^*)=w(x^*)+2^{-r}$,
$w'(x)=w(x)$ for every $x\in X\setminus\{x^*\}$.   Then certainly
$w'\in[0,1]^X$ and $\sum_{x\in X}w'(x)\le 1$.   If $k$, $n\in\Bbb N$ and
$x^*\notin G_{kn}$, then $\sum_{x\in G_{kn}\cap X}w'(x)=\sum_{x\in
G_{kn}\cap X}w(x)\le 2^{-n}$.   If $x^*\in G_{kn}$, then $n<r$ and
$2^{-k}<\rho(x^*,X_n)<2^{-m_n}$, so $m_n<k$ and $L_n\subseteq X_k$ and

\Centerline{$\sum_{x\in G_{kn}\cap X}w(x)\le\sum_{x\in X\setminus
X_k}w(x)\le\sum_{x\in X\setminus L_n}w(x)\le 2^{-n-1}$,}

\Centerline{$\sum_{x\in G_{kn}\cap X}w'(x)\le 2^{-n-1}+2^{-r}
\le 2^{-n}$.}

\noindent Thus $w'\in C$ and $w$ was not maximal.\ \Bang

Accordingly $\sum_{x\in X}w(x)=1$ and the point-supported measure $\mu$
defined by $w$ is a probability measure on $X$.   By the definition of
$C$, $\mu\in A$ and
$\mu(X\setminus K)=1$.   As $K$ is arbitrary, $A$ cannot be uniformly
tight.\ \Qed

\medskip

{\bf (f)} Thus $A$ witnesses that $X=\Bbb Q\cap[0,1]$ is not a Prokhorov
space.   Since $X$ is a closed subset of $\Bbb Q$, 437Vb tells us that
$\Bbb Q$ is not a Prokhorov space.
}%end of proof of 439S

\exercises{
\leader{439X}{Basic exercises (a)}%
%\spheader 439Xa
(i) Show that there is a set $A\subseteq[0,1]$ such that $\mu_L^*A=1$,
where $\mu_L$ is Lebesgue measure, and every member of $[0,1]$ is
uniquely expressible as $a+q$ where $a\in A$ and $q\in\Bbb Q$.
\Hint{134B.}
(ii) Define $f:[0,1]\to A$ by setting $f(x)=a$
when $x\in a+\Bbb Q$.   Show that the image measure
$\mu_Lf^{-1}$ takes only the values $0$ and $1$.   ({\smc Aldaz 95}.
Compare 342Xg.)
%439A

\spheader 439Xb
Let $X$ be a Radon Hausdorff space and $A$ a subset of
$X$.   Show that $A$ is universally negligible iff $\mu A=0$ for every
atomless Radon measure on $X$.
%439C

\sqheader 439Xc Let $X$ be a Hausdorff space.   Show that a set
$A\subseteq X$ is universally negligible iff $\mu A=0$ whenever $\mu$ is
a topological measure on $X$ with locally determined negligible sets
such that $\mu\{x\}=0$ for every $x\in X$.
%439C

\spheader 439Xd Let $X$ be a Hausdorff space.   Show that any
universally negligible subset of $X$ is universally measurable in the
sense of 434D.
%439C

\spheader 439Xe(i) Show that there is an analytic set
$A\subseteq\Bbb R$ such that for any Borel subset $E$ of
$\Bbb R\setminus A$ there is an uncountable Borel subset of
$\Bbb R\setminus(A\cup E)$.   \Hint{423Sb.}
(ii) Show that $A$ is universally measurable, but there is no
Borel set $E$ such that $A\symmdiff E$ is universally negligible.
%439C

\spheader 439Xf Show that a first-countable compact Hausdorff space is
universally negligible iff it is scattered iff it is countable.
%439C

\spheader 439Xg Show that the product of two universally negligible
Hausdorff spaces is universally negligible.
%439C

\spheader 439Xh Let us say that a Hausdorff space $X$ is {\bf
universally $\tau$-negligible} if there is no $\tau$-additive Borel
probability measure on $X$ which is zero on singletons.   (i) Show that
if $X$ is a Hausdorff space and $A\subseteq X$, then $A$ is universally
$\tau$-negligible iff $\mu^*A=0$ for every $\tau$-additive Borel
probability measure on $X$ such that $\mu\{x\}=0$ for every $x\in X$.
(ii) Show that if $X$ is a regular Hausdorff space, then a subset $A$ of
$X$ is universally $\tau$-negligible iff $\mu A=0$ for every atomless
quasi-Radon measure on $X$.   (iii) Show that if $X$ is a completely
regular Hausdorff space, it is universally $\tau$-negligible iff
whenever $\mu$ is an atomless Radon measure on a space $Z$, and
$X'\subseteq Z$ is homeomorphic to $X$, then $\mu X'=0$.   (iv) Show
that a Hausdorff space $X$ is universally negligible iff it is
Borel-measure-complete and universally $\tau$-negligible.   (v) Show
that if $X$ is a Hausdorff space, $Y$ is a universally $\tau$-negligible
Hausdorff space, and $f:X\to Y$ is a continuous function such that
$f^{-1}[\{y\}]$ is universally $\tau$-negligible for every $y\in Y$,
then $X$ is universally $\tau$-negligible.   (vi) Show that the product
of two universally $\tau$-negligible Hausdorff spaces is universally
$\tau$-negligible.   (vi) Show that a scattered Hausdorff space (in
particular, any discrete space) is universally $\tau$-negligible.
(vii) Show that a compact
Hausdorff space is universally $\tau$-negligible iff it is scattered.
%439C

\spheader 439Xi Let $X$ be a Polish space, $A\subseteq X$
an analytic set which is not
Borel (423Sb, 423Ye), and $\ofamily{\xi}{\omega_1}{E_{\xi}}$ a
family of Borel constituents of $X\setminus A$ (423R).   Suppose that
$x_{\xi}\in E_{\xi}\setminus\bigcup_{\eta<\xi}E_{\eta}$ for every
$\xi<\omega_1$.   Show that $\{x_{\xi}:\xi<\omega_1\}$ is universally
negligible.   Hence show that any probability measure with domain
$\Cal P\omega_1$ is point-supported.
%439C

\spheader 439Xj Let $(X,\le)$ be any well-ordered set and $\mu$ a
non-zero $\sigma$-finite measure on $X$ such that every
singleton is negligible.   Show that $\{(x,y):x\le y\}$ is not
measured by the (c.l.d.) product measure on $X\times X$.
\Hint{Reduce to the case in which $\mu$ is complete and
totally finite, $X=\zeta$ is an ordinal and $\mu\xi=0$ for every
$\xi<\zeta$.   You will probably need 251Q.}
%439F

\sqheader 439Xk Show that 439Fc, or any of the examples of 439A, can be
regarded as an example of a probability space $(X,\mu)$ and
a function $f:X\to[0,1]$ such that there is no extension of $\mu$ to a
measure $\nu$ such that $f$ is $\dom\nu$-measurable;  and accordingly can
provide an example of a probability space $(X,\mu)$ with a countable
totally ordered family $\Cal A$ of subsets of $X$ such that there is no
extension of $\mu$ to a measure measuring every member of $\Cal A$.
Contrast with 214P, 214Xm-214Xn and 214Yb.
%439F 439A

\spheader 439Xl Show that $1$-dimensional Hausdorff measure on $\BbbR^2$
is not inner regular with respect to the closed sets.   \Hint{439H,
439C(a-v);  see also 471S.}
%439H

\spheader 439Xm Show that the one-point compactification of the space
$(Z,\frak T_{\frak c})$ described in 439K is a scattered compact Hausdorff
space with an atomless Borel probability measure.
%439K

\spheader 439Xn\dvAformerly{4{}39Yf}
Show that a semi-finite Borel measure
on $\omega_1$, with its order topology, must be purely atomic.
%439N

\spheader 439Xo Show that $\BbbN^I$ is not pre-Radon for any uncountable
set $I$.   \Hint{417Xq.}
%439P

\spheader 439Xp\dvArevised{2014}(i)
Suppose that $X$ is a completely regular
space and there is a continuous function $f$ from $X$ to a
realcompact completely regular space $Z$ such that
$f^{-1}[\{z\}]$ is realcompact for every
$z\in Z$.   Show that $X$ is realcompact (definition:  436Xg).   (ii)
Show that the spaces $X$ of 439K and $X^2$ of 439Q are realcompact.
%439Q

\leader{439Y}{Further exercises (a)}
%\spheader 439Ya
Show that a subset $A$ of $\Bbb R$ is universally
negligible iff $f[A]$ is Lebesgue negligible for every continuous
injective function $f:\Bbb R\to\Bbb R$.   \Hint{if $\nu$ is an atomless
Borel probability measure on $\Bbb R$, set
$f(x)=x+\nu[0,x]$ for $x\ge 0$, and show that $\mu_Lf[E]=\mu_LE+\nu E$ for
every Borel set $E\subseteq\coint{0,\infty}$.}
%439C

\spheader 439Yb For this exercise only, let us say that a `universally
negligible measurable space' is a pair $(X,\Sigma)$ where $X$ is a set
and $\Sigma$ a $\sigma$-algebra of subsets of $X$ containing every
countable subset of $X$ such that there is no probability measure $\mu$
with domain $\Sigma$ such that $\mu\{x\}=0$ for every $x\in X$.
(i) Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets of $X$
containing all countable subsets of $X$, $A\subseteq X$ and $\Sigma_A$
the subspace $\sigma$-algebra.   Show that $(A,\Sigma_A)$ is universally
negligible iff $\mu^*A=0$ whenever $\mu$ is a probability measure with
domain $\Sigma$ which is zero on singletons.   Show that if $(X,\Sigma)$
is universally negligible so is $(A,\Sigma_A)$.   (ii) Let $X$ and $Y$
be sets, $\Sigma$ and $\Tau\,\,\sigma$-algebras of subsets of $X$ and
$Y$ containing all appropriate countable sets, and $f:X\to Y$ a
$(\Sigma,\Tau)$-measurable function.   Suppose that $(Y,\Tau)$ and
$(f^{-1}[\{y\}],\Sigma_{f^{-1}[\{y\}]})$ are universally negligible for
every $y\in Y$.   Show that $(X,\Sigma)$ is universally negligible.
(iii) Let $X$ be a set and $\Sigma$ a $\sigma$-algebra of subsets of $X$
containing all countable subsets of $X$.   Show that the set of those
$A\subseteq X$ such that $(A,\Sigma_A)$ is universally negligible is a
$\sigma$-ideal of subsets of $X$.
%439C

\spheader 439Yc Let $X$ be an analytic Hausdorff
space and $A$ an analytic subset
of $X$.   Show that $X\setminus A$ is universally negligible iff all the
constituents of $X\setminus A$ (for any Souslin scheme defining $A$) are
countable.
%439C

\spheader 439Yd(i) A pair
$(\ofamily{\xi}{\omega_1}{a_{\xi}},\ofamily{\xi}{\omega_1}{b_{\xi}})$ of
families of subsets
of $\Bbb N$ is a {\bf Hausdorff gap} if $a_{\xi}\setminus a_{\eta}$,
$a_{\xi}\setminus b_{\xi}$ and $b_{\eta}\setminus b_{\xi}$ are finite whenever
$\xi\le\eta<\omega_1$, $a_{\eta}\setminus a_{\xi}$ and
$b_{\xi}\setminus b_{\eta}$ are infinite whenever
$\xi<\eta<\omega_1$, and moreover
$\{\xi:\xi<\eta$, $a_{\xi}\subseteq b_{\eta}\cup n\}$ is finite for every
$\eta<\omega_1$.   (For a construction of a Hausdorff gap, see
{\smc Fremlin 84}, 21L.)   Show that in this case there is no $c\subseteq\Bbb N$
such that $a_{\xi}\setminus c$ and $c\setminus b_{\xi}$ are finite for every
$\xi<\omega_1$, and that
$\{a_{\xi}:\xi<\omega_1\}\cup\{b_{\xi}:\xi<\omega_1\}$ is universally
negligible in $\Cal P\Bbb N$.   (ii) Let
$\phi:(\Cal P\Bbb N)^{\Bbb N}\to\Cal P\Bbb N$ be a homeomorphism.
For $0<\xi<\omega_1$ let
$\sequencen{\theta(\xi,n)}$ be a sequence running over $\xi$.   Set
$a_0=\emptyset$ and for $0<\xi<\omega_1$ set
$a_{\xi}=\phi(\sequencen{a_{\theta(\xi,n)}})$.   Show that
$\{a_{\xi}:\xi<\omega_1\}$ is universally negligible.
%439F mt43bits

\spheader 439Ye(i) Let $X$ be a metrizable space such that $f[X]$ is
Lebesgue negligible for every continuous function $f:X\to\Bbb R$.   Show
that $X$ is universally negligible.   (ii) Let $X$ be a completely
regular Hausdorff space such that $f[X]$ is Lebesgue negligible for
every continuous function $f:X\to\Bbb R$.   Show that $X$ is universally
$\tau$-negligible.
%439G mt43bits

\spheader 439Yf Let $\frak T_{\frak c}$ be the topology on
$\{0,1\}^{\Bbb N}$
constructed in the proof of 439K.   (i) Show that it is normal and
countably paracompact.   (ii)
Show that any $\frak T_{\frak c}$-zero set is an $\frak S$-Borel set.
(iii) Show that the measure $\nu$ of part (b) of the proof of 439K is not
completion regular, so that its restriction to the
$\frak T_{\frak c}$-Borel $\sigma$-algebra is of type B$_0$ in the
classification of 434A.
%439K mt43bits

\spheader 439Yg Show that the space of 439O is locally compact and locally
countable, therefore first-countable.
%439O

\spheader 439Yh Show that the Sorgenfrey line
is hereditarily Lindel\"of, but that its square is not Lindel\"of.
%439Q

\spheader 439Yi Show that if $\omega_1$ is given its order topology, and
$f:\omega_1^2\to\Bbb R$ is continuous, then there is a $\zeta<\omega_1$
such that $f$ is constant on $(\omega_1\setminus\zeta)^2$.
\Hint{4A2S(b-iii).}   Show that if $\mu$ and $\nu$ are Baire probability
measures on $\omega_1$, then the Baire probability measures
$\mu\times\nu$, $\nu\times\mu$ on $\omega_1^2$ defined by the formulae
of 436F coincide.
%439R

\spheader 439Yj(i) Let $X\subseteq[0,1]$ be a dense set with no
uncountable compact subset.   Show that $X$ is not a Prokhorov space.
(ii)\dvAnew{2015} Show that $\BbbR^{\frak c}$ is not a Prokhorov space.
%439S mt43bits

\spheader 439Yk\dvAnew{2015}
Show that the Sorgenfrey line is not a Prokhorov space.
\Hint{{\smc Fremlin n15}.}
%439S


}%end of exercises

\cmmnt{\Notesheader{439} I give three separate constructions in 439A
because the phenomenon here is particularly important.   For two
chapters I have, piecemeal, been offering theorems on the extension of
measures.   The principal ones so far seem to be 413P, 415L, 416N,
417C, 417E and 435C, and I have used methods reflecting my belief
that the essential feature on which each such theorem depends is inner
regularity of an appropriate kind.   I think we should simultaneously
seek to develop an intuition for measures which do {\it not} extend, and
those in 439A are especially significant because they refer to the Borel
algebra of the unit interval, which in so many other contexts is
comfortably clear of the obstacles which beset more exotic structures.

Note that because the Borel $\sigma$-algebra of $\Bbb R$ is countably
generated, the examples here are examples of measures which cannot be
extended to measure every member of a countable family of sets.
Recall that in 214P I showed that measures can be extended to
measure the sets in arbitrary {\it well-ordered} families.

Outside the context of Polish spaces, the terms `universally measurable'
and `universally negligible' are not properly settled.   I have tried to
select definitions which lead to a reasonable pattern.   At least
a universally negligible subset of a Hausdorff space is
universally measurable (439Xd), and both concepts can be expressed in
terms of sets with $\sigma$-algebras, as in 439Yb.   It is important to
notice, in 439B,
that I write `$\mu\{x\}=0$ for every $x\in X$', not `$\mu$ is atomless'.
For instance, Dieudonn\'e's measure shows that $\omega_1$, with
its order topology, is not universally negligible on the definition
here;  but it is easy to show that there is no atomless Borel
probability measure on $\omega_1$ (439Xn).   In many cases, of course,
we do not need to make this distinction (439D).

The cardinal $\kappa$ of 439F (the `uniformity' of the Lebesgue null
ideal) is one of a large family of cardinals which will be
examined in Chapter 52 in the next volume.

% 439J rvmc  439O cofinal closed
In some of the arguments above (439J, 439L, 439O) I appeal to
(different) principles (`there is a cardinal which is not measure-free',
$\clubsuit$) which are not
theorems according to the rules I follow in this book.   Such examples
would in some ways fit better into Volume 5, where I mean to investigate
such principles properly.   I include the examples here because they do
at least exhibit bounds on what can be proved in ZFC.   I should not
want anyone to waste her time trying to show, for instance, that all
completion regular Borel measures are $\tau$-additive.   Nevertheless,
the absence of a `real' counter-example (obviously we want a probability
measure on a completely regular Hausdorff space) remains in my view a
significant gap.   It remains conceivable that there is a mathematical
world in which no such space exists.   Clearly the discovery of such a
world is likely to require familiarity with the many worlds already
known, and I am not going to embark on any such exploration in this
volume.   On the other hand, it is also very possible that all we need
is a bit of extra ingenuity to construct a counter-example in ZFC.   In
this section we have two examples of successes of this kind.   In
439F-439H, for instance, we have results which were long known as
consequences of the continuum hypothesis;  the particular insight of
{\smc Grzegorek 81} was the observation that they depended on
determinate properties of the cardinal $\kappa$ of 439F, and that its
indeterminate position between $\omega_1$ and $\frak c$ was unimportant.
In 439K I show how a
re-working of ideas in {\smc Juh\'asz Kunen \& Rudin 76}, where a
similar example was constructed (for an entirely different purpose)
assuming the continuum hypothesis, provides us with an interesting space
(a first-countable non-Radon compact Hausdorff space) in ZFC.   Let me
emphasize that these ideas were originally set out in a framework
supported by an extra axiom, where some technical details were easier
and the prize aimed at (a non-Lindel\"of hereditarily separable space)
more important.

The examples in 439K-439R are mostly based on constructions more or less
familiar from general topology.   I have already mentioned the origins
of 439K.   439M is related to the Tychonoff and Dieudonn\'e planks
({\smc Steen \& Seebach 78}, \S\S86-89).   439N and 439Q revisit yet
again $\omega_1$ and the Sorgenfrey line.   439O is adapted from one of
the standard constructions of Dowker spaces.   Products of disjoint
stationary sets (439R) have also been used elsewhere.
}%end of notes

\frnewpage

\frfilename{mt43conc.tex}
\versiondate{21.11.10}
\copyrightdate{2009}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my
knowledge) appeared in print in references to this chapter, and which
have since been changed.

\medskip

\leader{4{}32I}{Capacitability} 4{}32I, referred to in the 2008 edition of
Volume 5, is now 432J.

\leader{4{}34S-4{}34T}{Vague topologies} The material on vague topologies,
referred to in the 2001 edition of Volume 2, has been moved to \S437.

\leaveitout{\leader{4{}37C}{Daniell representation} %437C
{\smc Pachl 13}, P.24
}

\leaveitout{\leader{4{}37H}{Representation of smooth functionals} %437H
{\smc Pachl 13}, P.28
}

\leaveitout{\leader{4{}37Xa}{Daniell representation} %437Xa
{\smc Pachl 13}, P.24 \& P.28
}

\leaveitout{\leader{4{}38A}{Measure-free cardinals} %438A
{\smc Pachl 13}, P.32
}

\leaveitout{\leader{4{}38C}{Measure-free cardinals} %438C
{\smc Pachl 13}, P.33
}

\leaveitout{\leader{4{}38D}{Metrizable spaces} %438D
{\smc Pachl 13}, P.34
}

\leader{4{}39H}{$\tau$-smooth functionals} The example of a $\tau$-smooth
functional which is not representable as an integral, referred to in
{\smc Bogachev 07}, is now 439I.

\leader{4{}39J}{A non-Radon space} The example of a first-countable compact
Hausdorff space which is not Radon,
referred to in {\smc Bogachev 07}, is now 439K.

\leader{4{}39N}{Baire measure} The example of a Baire probability measure
with no extension to a Borel measure, 
referred to in {\smc Bogachev 07}, is now 439M.

\bigskip

\noindent{\bf References}

\medskip

Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

%Pachl J.\ [13] {\it Uniform Spaces and Measures.}   Springer, 2013.

\frnewpage
%\fi


\references{
\frfilename{mt4r.tex}
\versiondate{9.2.13}
\copyrightdate{2002}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\Loadcyrillic

\Centerline{\bf References for Volume 4}


\medskip%A

Aldaz J.M.\ [95] `On compactness and Loeb measures', Proc.\ Amer.\
Math.\ Soc.\ 123 (1995) 173-175.
\cmmnt{[439Xa.]}

Aldaz J.M.\ [97] `Borel extensions of Baire measures', Fund.\ Math.\ 154
(1997) 275-293.
\cmmnt{[434Yn.]}

Aldaz J.M.\ \& Render H.\ [00] `Borel measure extensions of measures
defined on sub-$\sigma$-algebras', Advances in Math.\ 150 (2000)
233-263.
\cmmnt{[432D.]}

Alexandroff P.S.\ \& Urysohn P.S.\ [1929] {\it  M\'emoire sur les espaces
topologiques compacts},
Verhandelingen Akad.\ Wetensch.\ Amsterdam 14 (1929) 1-96.
\cmmnt{[419L.]}

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P.,
1987.
\cmmnt{[4A4N.]}

Andretta A.\ \& Camerlo R.\ [13] `The descriptive set theory of the
Lebesgue density theorem', Advances in Math.\ 234 (2013).
%{\tt http://arxiv.org/abs/1105.3355}.
\cmmnt{[475Yg.]}

Arkhangel'skii A.V.\ [92] {\it Topological Function Spaces.}  Kluwer,
1992.
\cmmnt{[\S462 {\it intro.}, \S467 {\it notes\/}.]}

\leaveitout{
Armitage D.H.\ \& Gardiner S.J.\ [01] {\it Classical Potential Theory.}
Springer, 2001.
478Yd}

Aronov B., Basu S., Pach J.\ \& Sharir M.\ [03] (eds.)
{\it Discrete and Computational Geometry;  the Goodman-Pollack
Festschrift,}  Springer, 2003.

Asanov M.O.\ \& Velichko N.V.\ [81] `Compact sets in $C_p(X)$', Comm.\
Math.\ Helv.\ 22 (1981) 255-266.
\cmmnt{[462Ya.]}

Austin T.\ [10a] `On the norm convergence of nonconventional ergodic
averages', Ergodic Theory and Dynamical Systems 30 (2010) 321-338.
%({\tt http://{\penalty-100}arxiv.org/abs/{\penalty-100}0805.0320}).
\cmmnt{[497M.]}

Austin T.\ [10b] `Deducing the multidimensional Szemer\'edi theorem from
the infinitary hypergraph removal lemma',
J.\ d'Analyse Math.\ 111 (2010) 131-150.
%({\tt http://arxiv.org{\bsp}abs/0808.2267}).
\cmmnt{[497M.]}

\leaveitout{
Avil\'es A., Plebanek G.\ \& Rodr\'{\i}guez J.\ [p11] `Measurability in
$C(2^{\kappa})$ and Kunen cardinals', 2011.
%material on  C(\{0,1\}^{\kappa}) ;  Baire and Borel \sigma-algebras
%in weak and pointwise topologies
}

\leaveitout{Axler S. \& Ramey W.\ [95]
`Harmonic polynomials and Dirichlet-type problems',
Proc.\ Amer.\  Math.\ Soc.\ 123 (1995) 3765-3773.
http://www.axler.net/HarmonicPolynomials.html
478Yd}

\medskip%B

Balogh Z.\ [96] `A small Dowker space in ZFC', Proc.\ Amer.\ Math.\
Soc.\ 124 (1996) 2555-2560.
\cmmnt{[439O.]}

Banach S.\ [1923] `Sur le probl\`eme de la mesure',
Fundamenta Math.\ 4 (1923) 7-33.
\cmmnt{[449O.]}

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996 (London Math.\ Soc.\
Lecture Note Series 232).  \cmmnt{[424H, 448P, \S448 {\it notes\/}.]}

Bellow A.\ {\it see} Ionescu Tulcea A.

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).
%4r

Bergelson V.\ [85] `Sets of recurrence of $\Bbb Z^m$-actions and properties of
sets of differences in $\Bbb Z^m$',
J.\ London Math.\ Soc.\ (2) 31 (1985) 295-304.
%http://people.dm.unipi.it/dinasso/ULTRA2015/BUMN/Bergelson%20-%20Sets%20of%20recurrence%20of%20Zm-actions%20and%20properties%20of%20sets%20of%20differences%20in%20Zm%20(1985).pdf
\cmmnt{[491Yt.]}

Bergman G.M.\ [06] `Generating infinite symmetric groups', Bull.\ London
Math.\ Soc.\ 38 (2006) 429-440.
%http://math.berkeley.edu/~gbergman/papers/
%({\tt http:/{\bsp}arxiv.org{\bsp}abs{\bsp}math{\bsp}0401304v2}).
\cmmnt{[\S494 {\it notes}.]}

Billingsley P.\ [99] {\it Convergence of Probability Measures.}  Wiley,
1999.
\cmmnt{[\S4A3 {\it notes}.]}
%second edition ISBN 0-471-19745-9

Blackwell D.\ [56] `On a class of probability spaces',
pp.\ 1-6 in {\smc Neyman 56}.
\cmmnt{[419K, 452P.]}

Bochner S.\ [1933] `Monotone Funktionen, Stieltjessche Integrale, und
harmonische Analyse', Math.\ Ann.\ 108 (1933) 378-410.
\cmmnt{[445N.]}

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.
\cmmnt{[437Q, 437Xs, 437Yo, 437Yu, 457K.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[4A1G, 4A4N.]}

Bongiorno B., Piazza L.di \ \& Preiss D.\ [00]
`A constructive minimal integral which includes Lebesgue integrable
functions and derivatives', J.\ London Math.\ Soc.\ (2) 62 (2000) 117-126.
\cmmnt{[483Yi.]}

Borodulin-Nadzieja P.\ \& Plebanek G.\ [05] `On compactness of measures on
Polish spaces', Illinois J.\ Math.\ 49 (2005) 531-545.
\cmmnt{[451L.]}

Bourbaki N.\ [65] {\it Int\'egration,} chaps.~1-4.   Hermann, 1965
(Actualit\'es Scientifiques et Industrielles 1175).
\cmmnt{[\S416 {\it notes\/}, \S436 {\it notes\/}.]}

Bourbaki N.\ [66] {\it Espaces Vectoriels Topologiques}, chaps.~1-2,
$2^{\text{e}}$ \'ed.
Hermann, 1966 (Actualit\'es Scientifiques et Industrielles 1189).
\cmmnt{[4A2J.]}
%Store A QA 37.B6 no 7792 0070 5

Bourbaki N.\ [69] {\it Int\'egration,} chap.~9.   Hermann, 1969
(Actualit\'es Scientifiques et Industrielles 1343).
\cmmnt{[\S436 {\it notes.}]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S461 {\it notes\/}, \S4A4.]}

\indexheader{Bourgain}
Bourgain J., Fremlin D.H.\ \& Talagrand M.\ [78] `Pointwise compact
sets of Baire-measurable functions', Amer.\ J.\ Math.\ 100 (1978)
845-886.
\cmmnt{[462Ye.]}

\indexheader{Brodski\v{\i}}
Brodski\v{\i} M.L.\ [1949] `On some properties of sets of positive measure',
Uspehi Matem.\ Nauk (N.S.) 4 (1949) 136-138.
\cmmnt{[498B.]}

Brook R.B.\ [70] `A construction of the greatest ambit',
Math.\ Systems Theory 4 (1970) 243-248.
\cmmnt{[449D.]}

Burke D.K.\ \& Pol R.\ [05] `Note on separate continuity and the Namioka
property', Top.\ Appl.\ 152 (2005) 258-268.
\cmmnt{[\S463 {\it notes.}]}

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p21] `The strong marginal lifting
problem for hyperstonian spaces', preprint (2021).
%bms5s.pdf

\medskip%C

Carrington D.C.\ [72] `The generalised Riemann-complete integral', PhD
thesis, Cambridge, 1972.
\cmmnt{[481L.]}

\v{C}ech E.\ [66] {\it Topological Spaces.}  Wiley, 1966.
\cmmnt{[\S4A2, 4A3S.]}

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[494F.]}

\indexheader{Choquet}
Choquet G.\ [55] `Theory of capacities', Ann.\ Inst.\ Fourier (Grenoble)
5 (1955) 131-295.
\cmmnt{[432K.]}

\indexheader{Chung}
Chung K.L.\ [95] {\it Green, Brown and Probability.}   World Scientific,
1995.
% 978-9810225339  HA33.4.C5  479Xt

\indexheader{Ciesielski}
Ciesielski K.\ \& Pawlikowski J.\ [03]
`Covering Property Axiom CPA$_{\text{cube}}$ and its consequences',
Fundamenta Math. 176 (2003) 63-75.
\cmmnt{[498C.]}

Cs\'asz\'ar \'A.\ [78] {\it General Topology.}  Adam Hilger, 1978.
\cmmnt{[\S4A2, \S4A5.]}

\medskip%D

Davies R.O.\ [70] `Increasing sequences of sets and Hausdorff measure', Proc.\ London Math.\ Soc.\ (3) 20 (1970) 222-236.
\cmmnt{[471G.]}

Davies R.O.\ [71] `Measures not approximable or not specifiable by means
of balls', Mathematika 18 (1971) 157-160.
\cmmnt{[\S466 {\it notes\/}.]}

Davis W.J., Ghoussoub N.\ \& Lindenstrauss J.\ [81] `A lattice-renorming
theorem and applications to vector-valued processes', Trans.\ Amer.\
Math.\ Soc.\ 263 (1981) 531-540.
\cmmnt{[\S467 {\it notes\/}.]}

\indexheader{de Caux}
de Caux P.\ [76] `A collectionwise normal, weakly $\theta$-refinable
Dowker space which is neither irreducible nor realcompact', Topology
Proc.\ 1 (1976) 67-77.
\cmmnt{[439O.]}

Dellacherie C.\ [80] `Un cours sur les ensembles analytiques', pp.\
183-316 in {\smc Rogers 80}.
\cmmnt{[\S432 {\it notes\/}.]}

de Maria J.L.\ \& Rodriguez-Salinas B.\ [91] `The space
$(\ell_{\infty}/c_0)$,weak) is not a Radon space', Proc.\ Amer.\ Math.\
Soc.\ 112 (1991) 1095-1100.
\cmmnt{[466I.]}

Deville R., Godefroy G.\ \& Zizler V.\ [93] {\it Smoothness and
Renormings in Banach Spaces.}  Pitman, 1993.
\cmmnt{[\S467 {\it intro.}, 467Ye, \S467 {\it notes\/}.]}

Diaconis, P.\ \& Freedman, D.\ [80] `Finite exchangeable sequences.'
Ann. of Probability 8 (1980) 745-764.
\cmmnt{[459Xd.] }

Droste M., Holland C.\ \& Ulbrich G.\ [08]
`On full groups of measure preserving and ergodic transformations with
uncountable cofinalities',
Bull.\ London Math. Soc.\ 40 (2008) 463-472.
%{\tt http://web.mac.com/chollan1{\bsp}iWeb/Site{\bsp}publications.html}.
\cmmnt{[494Q.]}

Dubins L.E.\ \& Freedman D.A.\ [79]
`Exchangeable processes need not be mixtures of independent,
identically distributed random variables',
Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 48 (1979) 115-132.
\cmmnt{[459K.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S4A4.]}

Du Plessis, N.\ [70] {\it Introduction to Potential Theory.}  Oliver \&
Boyd, 1970.
%used for \S478

\medskip%E

Eggleston H.G.\ [54] `Two measure properties of Cartesian product sets',
Quarterly J.\ Math.\ (2) 5 (1954) 108-115.
\cmmnt{[498B.]}

Emerson W.\ \& Greenleaf F.\ [67] `Covering properties and F{\o}lner
conditions for locally compact groups',
Math.\ Zeitschrift 102 (1967) 370-384.
\cmmnt{[449J.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[4A1A.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[Vol.\ 4 {\it intro.}, 415Yb, \S434 {\it notes\/},
\S435 {\it notes\/}, 462L, \S4A2, 4A4B.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[Chap.\ 47 {\it intro.}, \S473 {\it intro.}, \S473 {\it notes\/},
\S474 {\it notes\/}, \S475 {\it notes\/}.]}

\medskip%F

Falconer K.J.\ [90] {\it Fractal Geometry.}  Wiley, 1990.
\cmmnt{[\S471 {\it intro.}]}

Federer H.\ [69] {\it Geometric Measure Theory.}
Springer, 1969 (reprinted 1996).
\cmmnt{[\S441 {\it notes\/}, 442Ya,
\S443 {\it notes\/}, Chap.\ 47 {\it intro.}, 471Yk,
\S471 {\it notes\/}, \S472 {\it notes\/}, \S475 {\it notes\/}.]}

Fernique X.\ [97] {\it Fonctions al\'eatoires gaussiennes,
vecteurs al\'eatoires gaussiens}, CRM Publications, 1997.
\cmmnt{[\S456 {\it notes\/}.]}

Folland G.B.\ [95] {\it A Course in Abstract Harmonic Analysis.}   CRC
Press, 1995
\cmmnt{[\S4A5, \S4A6.]}

\indexheader{F{\o}lner}
F{\o}lner E.\ [55] `On groups with full Banach mean value',
Math.\ Scand.\ 3 (1955) 243-254.
\cmmnt{[\S449 {\it notes\/}.]}

\indexheader{Ford}
Ford L.R.\ \& Fulkerson D.R.\ [56] `Maximal flow through a network',
Canadian J.\ Math.\ 8 (1956) 399-404.
\cmmnt{[4A4N.]}

Frankl P.\ \& R\"odl V.\ [02] `Extremal problems on set systems',
Random Structures Algorithms 20 (2002) 131-164.
\cmmnt{[497L.]}

\indexheader{Fremlin}
Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S462 {\it notes\/}.]}

Fremlin D.H.\ [75a] `Pointwise compact sets of measurable functions',
Manuscripta Math.\ 15 (1975) 219-242.
\cmmnt{[463K.]}

Fremlin D.H.\ [75b] `Topological measure spaces:  two counter-examples',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 78 (1975) 95-106.
\cmmnt{[419C, 419D.]}

Fremlin D.H.\ [76] `Products of Radon measures:  a counterexample',
Canadian Math.\ Bull.\ 19 (1976) 285-289.
\cmmnt{[419E.]}

Fremlin D.H.\ [81] `Measurable functions and almost continuous
functions', Manuscripta Math.\ 33 (1981) 387-405.
\cmmnt{[451T.]}

Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S434 {\it notes\/}, 439Yd, 4A2E.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S438 {\it notes\/}.]}

Fremlin D.H.\ [95] `The generalized McShane integral',
Illinois J.\ Math.\ 39 (1995) 39-67.
\cmmnt{[481N.]}

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
Fundamenta Math.\ 165 (2000) 67-94.
\cmmnt{[451V.]}

Fremlin D.H.\ [n05] `Strictly localizable Borel measures', note of 7.5.05
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[434Yr.]}

Fremlin D.H.\ [n15] `The Sorgenfrey line is not a Prokhorov space',
note of 5.8.15
({\tt http:/{\bsp}www.\penalty-100essex.ac.uk{\bsp}maths/people{\bsp}fremlin/preprints.htm}).
\cmmnt{[439Yk.]}

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fund.\ Math.\ 147 (1995) 27-37.
\cmmnt{[434Q.]}

Fremlin D.H.\ \& Talagrand M.\ [78] `On the representation of increasing
linear functionals on Riesz spaces by measures',
Mathematika 25 (1978) 213-215.
\cmmnt{[439I.]}

Fremlin D.H.\ \& Talagrand M.\ [79] `A decomposition theorem for
additive set functions and applications to Pettis integral and ergodic
means', Math.\ Zeitschrift 168 (1979) 117-142.
\cmmnt{[\S464.]}  %{\it 464C}, 464I, 464M, 464N, \S464 {\it notes\/}

Fristedt B.\ \& Gray L.\ [97] {\it A Modern Approach to Probability
Theory.}  Birkh\"auser, 1997.
\cmmnt{[\S455 {\it notes\/}, \S495 {\it notes\/}.]}
%HA33.4.F7

Frol{\'\i}k Z.\ [61] `On analytic spaces', Bull.\ Acad.\ Polon.\ Sci.\ 
%8 (1961) 747-750.
9 (1961) 721-726.   %Paulo Monteiro, email 8.9.22
\cmmnt{[422F.]}

Frol\'\i k Z.\ [88] (ed) {\it General Topology and its Relations to
Modern Analysis and Algebra VI} (Proc.\ Sixth Prague Topological
Symposium), Heldermann, 1988.

Furstenberg H.\ [77]
`Ergodic behavior of diagonal measures and a theorem of Szemer\'edi
on arithmetic progressions', J. d'Analyse Math.\ 31 (1977), 204-256.
\cmmnt{[\S497 {\it intro.}, \S497 {\it notes\/}.]}

Furstenberg H.\ [81]
{\it Recurrence in Ergodic Theory and Combinatorial Number Theory.}
Princeton U.P., 1981.
\cmmnt{[\S497 {\it intro.}, 497N.]}
% QA614

Furstenberg H.\ \& Katznelson Y.\ [85]
`An ergodic Szemer\'edi theorem for IP-systems and combinatorial theory',
J.\ d'Analyse Math.\ 45 (1985) 117-168.
\cmmnt{[\S497 {\it intro.}]}

\medskip%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S4A2.]}

Gardner R.J.\ [75] `The regularity of Borel measures and Borel
measure-compactness', Proc.\ London Math.\ Soc.\ (3) 30 (1975) 95-113.
\cmmnt{[438M.]}

Gardner R.J.\ \& Pfeffer W.F.\ [84] `Borel measures', pp. 961-1043 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[434G.]}

Giordano T.\ \& Pestov V.\ [02] `Some extremely amenable groups',
C.R.A.S.\ Paris (S\'er.\ I) 334 (2002) 273-278.
\cmmnt{[494I.]}

Glasner S.\ [98] `On minimal actions of Polish groups', Topology and its Applications 85 (1998) 119-125.
\cmmnt{[\S493 {\it notes\/}.]}

Gordon R.A.\ [94] {\it The Integrals of Lebesgue, Denjoy, Perron and
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4).
\cmmnt{[481Q, \S483 {\it notes\/}.]}

Greenleaf F.P.\ [69] {\it Invariant Means on Topological Groups and Their
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\cmmnt{[\S449 {\it notes.}]}

Grothendieck A. [92] {\it Topological Vector Spaces.}  Gordon \& Breach,
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\cmmnt{[\S462 {\it notes\/}.]}

\indexheader{Grzegorek}
Grzegorek E.\ [81] `On some results of Darst and Sierpi\'nski concerning
universal null and universally negligible sets', Bull.\ Acad.\ Polon.\
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\cmmnt{[439F, \S439 {\it notes\/}.]}

\medskip%H

\indexheader{Halmos}
Halmos P.R.\ [1944] `In general a measure-preserving transformation is
mixing', Annals of Math.\ 45 (1944) 786-792.
\cmmnt{[494E.]}

Halmos P.R.\ [50] {\it Measure Theory.}   Springer, 1974.
\cmmnt{[\S441 {\it notes\/}, \S442 {\it notes\/}, 443M, \S443 {\it notes\/}.]}

Halmos P.R.\ [56] {\it Lectures on ergodic theory.}  Chelsea, 1956.
\cmmnt{[494A.]}

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Banach spaces', Serdica Math.\ J.\ 27 (2001) 1-66.
\cmmnt{[466D.]}

Hart J.E.\ \& Kunen K.\ [99] `Orthogonal continuous functions',
Real Analysis Exchange 25 (1999) 653-659.
\cmmnt{[416Yh.]}

Hartman S.\ \& Mycielski J.\ [58] `On the imbedding of topological groups into connected topological groups', Colloq.\ Math.\ 5 (1958) 167-169.
\cmmnt{[493Ya.]}

Haydon R.\ [74] `On compactness in spaces of measures and
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\cmmnt{[438J.]}

Henry J.P.\ [69] `Prolongement des mesures de Radon', Ann.\ Inst.\
Fourier 19 (1969) 237-247.
\cmmnt{[416N.]}

Henstock R.\ [63] {\it Theory of Integration.}  Butterworths, 1963.
\cmmnt{[Chap.\ 48 {\it intro.}, 481J.]}

Henstock R.\ [91] {\it The General Theory of Integration.}   Oxford
U.P., 1991.
\cmmnt{[\S481 {\it notes\/}.]}

Herer W.\ \& Christensen J.P.R.\ [75] `On the existence of pathological submeasures and the construction of exotic topological groups', Math.\ Ann.\ 213 (1975) 203-210.
\cmmnt{[\S493 {\it notes\/}.]}

Herglotz G.\ [1911] `\"Uber Potenzreihen mit positivem reellen Teil im
Einheitskreis', Leipziger Berichte 63 (1911) 501-511.
\cmmnt{[445N.]}

Hewitt E.\ \& Ross K.A.\ [63] {\it Abstract Harmonic Analysis I.}
Springer, 1963 and 1979.
\cmmnt{[\S442 {\it notes\/}, \S444 {\it intro.}, \S444 {\it notes\/},
\S4A5, \S4A6.]}

Hewitt E.\ \& Savage L.J.\ [55]
`Symmetric measures on Cartesian products',
Trans.\ Amer.\ Math.\ Soc.\ 80 (1955) 470-501.
\cmmnt{[459Xe.]}

%Holick\'y P.\ \& Spurn\'y J.\ [03] `Perfect images of absolute Souslin
%and absolute Borel Tychonoff spaces', Topology and its Appl.\ 131 (2003)
%281-294.
%\cmmnt{[434Yr.]}

\indexheader{Howroyd}
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\cmmnt{[471R, 471S.]}

\medskip%I

\indexheader{Ionescu Tulcea}
Ionescu Tulcea A.\ [73] `On pointwise convergence, compactness and
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\cmmnt{[463C.]}

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equicontinuity II', Advances in Math.\ 12 (1974) 171-177.
\cmmnt{[463G.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [67] `On the existence of a
lifting commuting with the left translations of an arbitrary locally
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\cmmnt{[\S447 {\it intro.}, 447I, \S447 {\it notes\/}.]}

Ionescu Tulcea A.\ \& Ionescu Tulcea C.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.
\cmmnt{[\S453 {\it intro.}]}

\indexheader{Ismail}
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\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.

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\cmmnt{[\S4A4.]}

\indexheader{Jayne}
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23 (1976) 208-211.
\cmmnt{[422Yc.]}

Jayne J.E.\ \& Rogers C.A.\ [95] `Radon measures on Banach spaces with
their weak topologies', Serdica Math.\ J.\ 21 (1995) 283-334.
\cmmnt{[466H, \S466 {\it notes\/}.]}

\indexheader{Jech}
Jech T.\ [78] {\it Set Theory.}  Academic, 1978.
\cmmnt{[423S, \S4A1.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003.
%ISBN 3540440852
\cmmnt{[\S4A1.]}

Johnson R.A.\ [82] `Products of two Borel measures', Trans.\ Amer.\
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\cmmnt{[434R.]}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
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Juh\'asz I., Kunen K.\ \& Rudin M.E.\ [76] `Two more hereditarily
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\cmmnt{[\S437 {\it notes\/}, 439K, \S439 {\it notes\/}.]}

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}  Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[4A1A.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}  Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[\S4A1.]}

\medskip%K

Kampen, E.R.van [1935] `Locally bicompact abelian groups and their character groups',
Ann.\ of Math.\ (2) 36 (1935) 448-463.
\cmmnt{[445U.]}

\indexheader{Kaplansky}
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\cmmnt{[\S446 {\it notes.}]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
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\cmmnt{[Chap.\ 42 {\it intro.}, 423T, \S423 {\it notes\/},
\S448 {\it notes\/}, \S4A2, 4A3S, 4A3T.]}

%Kechris A.S.\ [p09] {\it Global Aspects of Ergodic Group Actions.}
%Amer.\ Math.\ Soc.\

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\cmmnt{[457M.]}

Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[438R.]}

Kelley J.L.\ \& Srinivasan T.P.\ [71] `Pre-measures on lattices of sets',
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\cmmnt{[413Xr.]}

Kittrell J.\ \& Tsankov T.\ [09] `Topological properties of full groups',
Ergodic Theory and Dynamical Systems 2009 (doi:10.1017/S0143385709000078).
\cmmnt{[494O, \S494 {\it notes}.]}

Kolmogorov A.N.\ [1933] {\it Grundbegriffe der
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\cmmnt{[454D.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[\S462 {\it notes\/}, \S4A4, 4A6C.]}

Koumoullis G.\ \& Prikry K.\ [83] `The Ramsey property and measurable
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\cmmnt{[451T.]}

Kuipers L.\ \& Niederreiter H.\ [74] {\it Uniform Distribution of
Sequences.}   Wiley, 1974.
\cmmnt{[491Z.]}

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[\S4A1, 4A2E.]}

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.
%4r

Kuratowski K.\  [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[423T, 462Ye, \S4A2, 4A3S.]}

Kurzweil J.\ [57] `Generalized ordinary differential equations and
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\cmmnt{[Chap.\ 48 {\it intro.}]}

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%{\tt http://www.math.uiuc.edu/\penalty-100$^{\sim}$ssolecki/papers.html}.
\cmmnt{[\S448 {\it notes\/}.]}

\medskip%L

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\cmmnt{[\S449 {\it notes\/}.]}

LeCam L.M.\ \& Neyman J.\ [67] (eds.) {\it Proc.\ Fifth Berkeley
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\cmmnt{[\S4A1.]}

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\cmmnt{[\S495 {\it notes\/}.]}

\indexheader{Losert}
Losert V.\ [79] `A measure space without the strong lifting property',
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\cmmnt{[453N, \S453 {\it notes\/}.]}

\indexheader{Luke\v{s}}
Luke\v{s} J., Mal\'y J.\ \& Zaj\'\i\v{c}ek L.\ [86]
{\it Fine topology methods
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\cmmnt{[414Ye.]}

\medskip%M

\indexheader{Mackey}
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\cmmnt{[448S.]}

Marczewski E.\ [53] `On compact measures', Fundamenta Math.\ 40 (1953)
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\cmmnt{[413T, Chap.\ 45 {\it intro.}, \S451 {\it intro.}, 451J.]}

Marczewski E.\ \& Ryll-Nardzewski C.\ [53] `Remarks on the compactness
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\cmmnt{[454C.]}

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\cmmnt{[435C.]}

Mattila P.\ [95] {\it Geometry of Sets and Measures in Euclidean
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\cmmnt{[Chap.\ 47 {\it intro.}, \S471 {\it intro.}, \S472 {\it notes\/},
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%QA311.M2 9520 6456 1

Mauldin R.D. \& Stone A.H.\ [81] `Realization of maps', Michigan Math.\
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\cmmnt{[418T.]}

Maurey B.\ [79] `Construction des suites sym\'etriques', C.R.A.S.\ (Paris) 288 (1979) 679-681.
\cmmnt{[492H.]}
  % one m in sym\'etrique

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Monthly 80 (1973) 349-359.
\cmmnt{[481M.]}

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\cmmnt{[\S495 {\it notes\/}.]}

Milman V.D.\ \& Schechtman G.\ [86]
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\cmmnt{[492G, \S492 {\it notes\/}.]}

\leaveitout{Mohammed D.\ \& Saadi K.\ [p09#]
`Probl\`emes de mesurabilit\'es sur
l'espace des fonctions continues sur un compact', 2009.
\cmmnt{[466Y?.]}
%refereeing for Canad Math Bull;  rejected
}

Molt\'o A., Orihuela J., Troyanski S.\ \& Valdivia M.\ [09] {\it A
Non-linear Transfer Technique for Renorming.}   Springer, 2009 (Lecture
Notes in Mathematics 1951).
\cmmnt{[\S467 {\it notes.}]}

Montgomery D.\ \& Zippin L.\ [55] {\it Topological Transformation
Groups.}   Interscience, 1955.
\cmmnt{[\S446 {\it notes.}]}

\indexheader{Moran}
Moran W.\ [68] `The additivity of measures on completely regular
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\cmmnt{[439P.]}

Moran W.\ [69] `Measures and mappings on topological spaces', Proc.\
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\cmmnt{[435Xk.]}

Moran W.\ [70] `Measures on metacompact spaces', Proc.\ London Math.\
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\cmmnt{[438J.]}

\indexheader{Moschovakis}
Moschovakis Y.N.\ [80] {\it Descriptive Set Theory.}   Springer, 1980.
\cmmnt{[Chap.\ 42 {\it intro.}, \S423 {\it notes\/}.]}

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\cmmnt{[481P, \S482 {\it notes\/}.]}

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%ISBN 5-900975-01-0

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\cmmnt{[451U.]}

\medskip%N

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%Naimark M.A. [70] {\it Normed Rings.}  Wolters-Noordhoff, 1970.


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\medskip%O
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\cmmnt{[434Yq.]}

\medskip%P

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\cmmnt{[452I, 452S, 452Ye, \S452 {\it notes\/}.]}

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\cmmnt{[449K, \S449 {\it notes\/}.]}
% QA 403.P2 store A

Perron O.\ [1914] `\"Uber den Integralbegriff', Sitzungsberichte der
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\cmmnt{[\S493 {\it notes\/}.]}

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\indexheader{Plewik}
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\indexheader{Pontryagin}
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\indexheader{Preiss}
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Z.\ Wahrscheinlichkeitstheorie und verw.\ Gebiete 27 (1973) 109-116.
\cmmnt{[439S.]}
%ILL 31.7.15

Preiss D.\ \& Ti\v{s}er J.\ [91]
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\cmmnt{[\S466 {\it notes\/}.]}

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\cmmnt{[462B, 462C.]}

\medskip%Q\medskip%R

Radon J.\ [1913] `Theorie und Anwendungen der absolut additivien
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\cmmnt{[\S416 {\it notes\/}.]}

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\cmmnt{[\S451 {\it notes\/}.]}

\indexheader{Rao B.V.}
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\cmmnt{[419F.]}

Rao B.V. \& Srivastava S.M.\ [94] `Borel isomorphism theorem',
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\cmmnt{[424C.]}

\indexheader{Rao M.}
Rao M.\ [77] {\it Brownian motion and classical potential theory.}
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{\tt http://www1.essex.ac.uk/maths/people{\bsp}fremlin{\bsp}rao.htm}.
%used for \S477

\indexheader{Ressel}
Ressel P.\ [77] `Some continuity and measurability results on spaces of
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\cmmnt{[417C, 437M.]}

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\cmmnt{[\S494 {\it notes\/}.]}

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\cmmnt{[\S471 {\it notes\/}.]}

Rogers C.A.\ [80] (ed.) {\it Analytic Sets.}   Academic, 1980.
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Rogers L.C.J.\ \& Williams D.\ [94] {\it Diffusions, Markov Processes
and Martingales,} vol.\ I.   Wiley, 1994.
\cmmnt{[\S455 {\it notes\/}.]}
%HA33.4.W5

Rokhlin V.A.\ [1948] `A ``general'' measure-preserving transformation is not
mixing', Doklady Akademii Nauk SSSR 60 (1948) 349-351.
\cmmnt{[494E.]}

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`The generic isometry and measure preserving homeomorphism are conjugate
to their powers',  Fundamenta Math.\ 205 (2009) 1-27.
%({\tt http://www.math.uic.edu{\bsp}$\sptilde$rosendal/}).
\cmmnt{[494Ye.]}

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fixed points on metric compacta', Israel J.\ Math.\ 162 (2007) 349-371.
\cmmnt{[494Z, \S494 {\it notes}.]}

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normal', Fund.\ Math.\ 73 (1971) 179-186.
\cmmnt{[439O.]}

Rudin M.E.\ [84] `Dowker spaces', pp.\ 761-780 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[439O.]}

Rudin W.\ [67] {\it Fourier Analysis on Groups.} Wiley, 1967 and 1990.
\cmmnt{[\S445 {\it intro\/}.]}

Rudin W.\ [91] {\it Functional Analysis.}  McGraw-Hill, 1991.
\cmmnt{[\S4A4, \S4A6.]}

Ryll-Nardzewski C.\ [53] `On quasi-compact measures', Fundamenta Math.\
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\cmmnt{[451C.]}

\medskip%S

Sazonov V.V.\ [66] `On perfect measures', A.M.S.\ Translations (2) 48
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\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',
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\cmmnt{[444Ye.]}

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\cmmnt{[449Yi.]}

\indexheader{Szemer\'edi}
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\cmmnt{[497L.]}
\indexheader{Talagrand} %? before \medskip to act correctly

\medskip%T

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\cmmnt{[467M.]}

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\cmmnt{[463Ya, 463Yd, 463Za, \S465 {\it intro.}, 465B, 465N,
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466I, \S466 {\it notes\/}.]}

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%465Xm

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%used in chap 41

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\cmmnt{[425D, 425Ya.]}

\medskip%U

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\medskip%V

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% `Ta{sh}ken{t}skogo instituta' taken from ref in Fund Math

\medskip%W

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\indexheader{Wheeler}
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  % printer's error "2" for "1" at head of paper refers to part, not
%volume
\cmmnt{[\S435 {\it notes\/}, \S437 {\it notes\/}.]}

%\medskip%X\medskip%Y\medskip%Z

%Zindulka O.\ [p98] `Killing residual measures', preprint, 1998.

\frnewpage
}%end of references


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