% 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\Sous{\Cal S\text{ous}}  %used in vol 6
\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\tinyplus{{\vrule height 1.6pt depth -1.3pt width 2.5pt}
    \hskip-1.3pt{\vrule height 2.6pt depth -0.1pt width 0.3pt}}
\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{mt56.tex} 
\versiondate{3.1.15} 
\copyrightdate{2006} 
 
\def\chaptername{Choice and determinacy} 
 
\newchapter{56} 
 
Nearly everyone reading this 
book will have been taking the axiom of choice for granted nearly all the 
time.   This is the home territory of twentieth-century abstract analysis, 
and the one in which the great majority of the results 
have been developed. 
But I hope that everyone is aware that there are other ways of doing 
things.   In this chapter I want to explore what seem to me to be the most 
interesting alternatives.   In one sense they are minor variations on the 
standard approach, since I keep strictly to ideas expressible  
within the framework of Zermelo-Fraenkel set theory;  but in other ways 
they are dramatic enough to rearrange our prejudices. 
The arguments I will present in this chapter are mostly not  
especially difficult by the 
standards of this volume, but they do depend on intuitions for 
which familiar results which are likely to remain valid under the new 
rules being considered.    
 
Let me say straight away that the real aim of the chapter is \S567, on 
the axiom of determinacy. 
The significance of this axiom is that it is (so far) the most striking 
rival to the axiom of choice, in that it leads us quickly to a large number 
of propositions directly contradicting familiar theorems;  for instance, 
every subset of the real line is now Lebesgue measurable (567G).   But we 
need also to know which theorems are still true, and the first six sections 
of the chapter are devoted to a discussion of what can be done in ZF alone 
(\S\S561-565) and with countable or dependent choice (\S566).   Actually 
\S\S562-565 are rather off the straight line to \S567, because they examine 
parts of real analysis in which the standard proofs depend only on 
countable choice or less;  but a great deal more can be done than most of 
us would expect, and the methods are instructive. 
 
Going into details, \S561 looks at basic facts from real analysis, 
functional analysis and general topology which can be proved in ZF.   \S562 
deals with `codable' Borel sets and functions, using Borel codes to keep 
track of constructions for objects, so that if we know a sequence of  
codes we can avoid having to make a sequence of choices.   A `Borel-coded 
measure' (\S563) is now one which behaves well with  
respect to codable sequences of 
measurable sets;  for such a measure we have an integral with 
versions of the convergence theorems (\S564), and Lebesgue measure fits 
naturally into the structure (\S565).   In \S566, with ZF + AC($\omega$), 
we are back in familiar territory, and most of the results of Volumes 1 and 
2 can be proved if we are willing to re-examine some definitions and 
hypotheses.   Finally, in \S567, I look at infinite games and half a dozen 
of the consequences of AD, with a postscript on determinacy in the context 
of ZF + AC. 
 
\discrpage 
 
\frfilename{mt561.tex}
\versiondate{8.9.13}
\copyrightdate{2006}

\def\chaptername{Choice and determinacy}
\def\sectionname{Analysis without choice}

\newsection{561}

Elementary courses in analysis
are often casual about uses of weak forms of choice;  a typical argument
runs `for every $\epsilon>0$ there is an $a\in A$ such that
$|a-x|\le\epsilon$, so there is a sequence in $A$ converging to $x$'.
This is a direct call on the countable axiom of choice:  setting
$A_n=\{a:a\in A$, $|a-x|\le 2^{-n}\}$, we are told that every $A_n$ is
non-empty, and conclude that $\prod_{n\in\Bbb N}A_n$ is non-empty.   In the
present section I will abjure such methods and investigate what can
still be done with the ideas important in measure theory.
We have useful partial versions of Tychonoff's theorem (561D), Baire's
theorem (561E), Stone's theorem (561F) and Kakutani's theorem on the
representation of $L$-spaces (561H);  moreover, there is a direct
construction of Haar measures, regarded as linear functionals (561G).

{\it Unless explicitly stated otherwise}, throughout this section (and the
next four) I am working entirely without any form of the axiom of choice.

\leader{561A}{Set theory without choice}\cmmnt{ In \S\S1A1 and 2A1
I tried to lay out some of the basic ideas of set theory without
appealing to the axiom of choice except when this was clearly necessary.}
The most obvious point is that in the absence of choice

\Centerline{\bf the union of a sequence of countable sets need not be
countable\dvro{.}{}}

\noindent\cmmnt{(see the note in 1A1G).
In fact {\smc Feferman \& Levy 63} (see
{\smc Jech 73}, 10.6) described a
model of set theory in which $\Bbb R$ is the union of a sequence of
countable sets.   But not all is lost.   }The elementary arguments of 1A1E
still give

\Centerline{$\Bbb N\simeq\Bbb Z\simeq\Bbb N\times\Bbb N\simeq\Bbb Q$;}

\cmmnt{\noindent there is no difficulty in extending them to show such
things as}

\Centerline{$\Bbb N\simeq[\Bbb N]^{<\omega}
\simeq\bigcup_{n\ge 1}\BbbN^n\simeq\BbbQ^r\times\BbbQ^r$}

\noindent for every integer $r\ge 1$.   The Schr\"oder-Bernstein theorem
survives\cmmnt{ (the method in 344D is easily translated back into its
original form as a proof of the ordinary Schr\"oder-Bernstein theorem)}.
Consequently\cmmnt{ we still have enough bijections to establish}

\Centerline{$\Bbb R\simeq\Cal P\Bbb N\simeq\{0,1\}^{\Bbb N}
\simeq\Cal P(\Bbb N\times\Bbb N)
\simeq(\Cal P\Bbb N)^{\Bbb N}\simeq\BbbR^{\Bbb N}\simeq\NN$.}

\noindent\cmmnt{Cantor's theorem
that }$X\not\simeq\Cal PX$\cmmnt{ is unaffected}, so\cmmnt{ we
still know that} $\Bbb R$ is not countable.

We can still use transfinite recursion\cmmnt{;  see 2A1B}.
We still have a class
$\On$ of von Neumann ordinals such that every well-ordered set is
isomorphic to exactly one
ordinal\cmmnt{ (2A1Dg)} and\cmmnt{ therefore} equipollent with
exactly one initial ordinal\cmmnt{ (2A1Fb)}.
I will say that a set $X$ is {\bf well-orderable} if there is a
well-ordering of $X$.   The standard
arguments for Zermelo's Well-Ordering Theorem\cmmnt{ (2A1K)} now
tell us that for any set $X$ the following are equiveridical:

\inset{(i) $X$ is well-orderable;

(ii) $X$ is equipollent with some ordinal;

(iii) there is an injective function from $X$ into a well-orderable set;

(iv) there is a choice function for $\Cal PX\setminus\{\emptyset\}$}

\noindent (that is, a function $f$ such that
$f(A)\in A$ for every non-empty $A\subseteq X$).
What this means is that if we are given a family $\familyiI{A_i}$
of non-empty sets, and $X=\bigcup_{i\in I}A_i$ is
well-orderable\cmmnt{ (e.g., because it is countable)}, then
$\prod_{i\in I}A_i$ is not empty\cmmnt{
(it contains $\familyiI{f(A_i)}$ where $f$ is a function as in (iv)
above)}.

Note\cmmnt{ also} that while we still have a first uncountable ordinal
$\omega_1$\cmmnt{ (the set of
countable ordinals)}, it can have countable cofinality\cmmnt{ (561Ya)}.
The union of a sequence of finite sets need not be
countable\cmmnt{ ({\smc Jech 73}, \S5.4)};   but the union
of a sequence of finite subsets of a given totally ordered set {\it is}
countable\cmmnt{, because we can use the total ordering to simultaneously
enumerate each of the finite sets in ascending order}.   Consequently, if
$\gamma:\omega_1\to\Bbb R$ is a monotonic function there is a
$\xi<\omega_1$ such that $\gamma(\xi+1)=\gamma(\xi)$.
\prooflet{\Prf\ It is enough to consider the case in which $\gamma$ is
non-decreasing.   Set

\Centerline{$A_n=\{\xi:\gamma(\xi)+2^{-n}\le\gamma(\xi+1)\le n\}$.}

\noindent Then $A_n$ has at most $2^n\max(0,n-\gamma(0))$ members, so is
finite;  consequently $\bigcup_{n\in\Bbb N}A_n$ is countable, and there is
a $\xi\in\omega_1\setminus\bigcup_{n\in\Bbb N}A_n$.   Of course we now find
that $\gamma(\xi+1)=\gamma(\xi)$.\ \Qed}

\cmmnt{
\leader{561B}{Real analysis without choice}
In fact all the standard theorems of elementary
real and complex analysis are
essentially unchanged.   The kind of tightening required in some proofs, to
avoid explicit dependence on the existence of sequences, is similar to the
adaptations needed when we move to general topological spaces.   For
instance, we must define `compactness' in terms of open covers;
compactness and sequential compactness, even for subsets of $\Bbb R$, may
no longer coincide (561Xc).
But we do still have the Heine-Borel theorem in the form `a subset of
$\BbbR^r$ is compact iff it is closed and bounded' (provided, of course,
that we understand that `closed' is not the same thing as `sequentially
closed');  see the proof in 2A2F.
}%end of comment

\leader{561C}{}\cmmnt{ Some new difficulties arise when we move away
from `concrete' questions like the Prime Number Theorem and start looking
at general metric spaces, or even general subsets of $\Bbb R$.   For
instance, a subset of $\Bbb R$, regarded as a topological space, must be
second-countable but need not be separable.   However we can go a long way
if we take care.   The following is an elementary example which will be
useful below.

\medskip

\noindent}{\bf Lemma}\dvArevised{2013}
Let $\Cal E$ be the set of non-empty closed subsets
of $\NN$.   Then there is a family $\family{F}{\Cal E}{f_F}$ such that,
for each $F\in\Cal E$, $f_F$ is a continuous function from $\NN$ to $F$
and $f_F(\alpha)=\alpha$ for every $\alpha\in F$.

\proof{ For $F\in\Cal E$,
set $T_F=\{\alpha\restr n:\alpha\in F$, $n\in\Bbb N\}$.
If $\alpha\in\NN\setminus F$ then, because $F$ is closed,
there is some $n\in\Bbb N$ such that $\beta\restr n\ne\alpha\restr n$ for
any $\beta\in F$, that is, $\alpha\restr n\notin T_F$.   For
$\sigma\in T_F$ define $\beta_{F\sigma}\in\NN$ inductively by saying that

$$\eqalign{\beta_{F\sigma}(n)&=\sigma(n)\text{ if }n<\#(\sigma),\cr
&=\inf\{i:\text{ there is some }\alpha\text{ such that }
\beta_{F\sigma}\restr n\subseteq\alpha\in F\text{ and }\alpha(n)=i\}
\text{ otherwise},\cr}$$

\noindent counting $\inf\emptyset$ as $0$ if necessary.
We see that in fact $\beta_{F\sigma}\restr n\in T_F$ for
every $n\in\Bbb N$, so that $\beta_{F\sigma}\in F$.

We can therefore define $f_F:\NN\to\NN$ by setting

$$\eqalign{f_F(\alpha)&=\alpha\text{ if }\alpha\in F,\cr
&=\beta_{F,\alpha\restr n}\text{ for the largest }n
\text{ such that }\alpha\restr n\in T_F\text{ otherwise}.\cr}$$

\noindent (Because $F$ is not empty, the empty sequence $\alpha\restr 0$
belongs to $T_F$ for every $\alpha\in\NN$.)   We see that
$f_F(\alpha)\in F$ for all $F$ and $\alpha$, and $f_F(\alpha)=\alpha$ if
$\alpha\in F$.   To see that $f_F$ is always
continuous, note that in fact
if $\alpha\in\NN\setminus F$, and $n$ is the largest integer such that
$\alpha\restr n$ belongs to $T$, then $f_F(\beta)=f_F(\alpha)$ whenever
$\beta\restr n+1=\alpha\restr n+1$, so $f_F$ is continuous at $\alpha$.
While if $\alpha\in F$, $n\in\Bbb N$ and $\beta\restr n=\alpha\restr n$,
then either $\beta\in F$ so $f_F(\beta)\restr n=f_F(\alpha)\restr n$, or
$f_F(\beta)=\beta_{F\sigma}$ where
$\alpha\restr n\subseteq\sigma\subseteq\beta_{F\sigma}$, and again
$f_F(\beta)\restr n=\beta_{\sigma}\restr n=\alpha\restr n$.   So we have a
suitable family of functions.
}%end of proof of 561C

\leader{561D}{Tychonoff's theorem}\cmmnt{ It is a classic
result ({\smc Kelley 50}) that Tychonoff's theorem, in a general form, is
actually equivalent to the axiom of choice.   But nevertheless we have
useful partial results which do not depend on the axiom of choice.   The
following will help in the proofs of 561F and 563I.

\medskip

\noindent{\bf Theorem}} Let $\familyiI{X_i}$ be a family of compact
topological spaces such that $I$ is well-orderable.   For each $i\in I$ let
$\Cal E_i$ be the family of non-empty closed subsets of $X_i$, and suppose
that there is a choice function for $\bigcup_{i\in I}\Cal E_i$.   Then
$X=\prod_{i\in I}X_i$ is compact.

\proof{ Since $I$ is well-orderable, we may suppose that $I=\kappa$ for
some initial ordinal $\kappa$.   Fix a choice function $\psi$ for
$\bigcup_{\xi<\kappa}\Cal E_i$.
For $\xi<\kappa$ write $\pi_{\xi}:X\to X_{\xi}$ for the coordinate map.
If $X$ is empty
the result is trivial.   Otherwise, let $\Cal F$ be any family of closed
subsets of $X$ with the finite intersection property.   I
seek to define a non-decreasing family
$\langle\Cal F_{\xi}\rangle_{\xi\le\kappa}$ of filters on $X$ such that
the image filter $\pi_{\xi}[[\Cal F_{\xi+1}]]$ (2A1Ib) is convergent for
each $\xi<\kappa$.   Start with $\Cal F_0$ the filter generated by
$\Cal F$.   Given $\Cal F_{\xi}$,
let $F_{\xi}$ be the set of cluster points of $\pi_{\xi}[[\Cal F_{\xi}]]$;
because $X_{\xi}$ is compact, this is a non-empty closed subset of
$X_{\xi}$, and $x_{\xi}=\psi(F_{\xi})$ is defined.   Let
$\Cal F_{\xi+1}$ be the filter on $X$ generated by

\Centerline{$\Cal F_{\xi}
\cup\{\pi_{\xi}^{-1}[U]:U$ is a neighbourhood of $x_{\xi}$ in $X_{\xi}\}$.}

\noindent For limit ordinals $\xi\le\kappa$, let $\Cal F_{\xi}$ be the
filter on $X$ generated by $\bigcup_{\eta<\xi}\Cal F_{\eta}$.

Now $\Cal F_{\kappa}$ is a filter including $\Cal F$ converging to
$x=\ofamily{\xi}{\kappa}{x_{\xi}}$, and $x$ must belong to $\bigcap\Cal F$.
As $\Cal F$ is arbitrary, $X$ is compact.
}%end of proof of 561D

\cmmnt{\medskip

\noindent{\bf Remark} The point of the condition
`there is a choice function for $\bigcup_{i\in I}\Cal E_i$'
is that it is
satisfied if every $X_i$ is the unit interval $[0,1]$, for instance;
we could take
$\psi(E)=\min E$ for non-empty closed sets $E\subseteq[0,1]$.
You will have no difficulty in devising other examples, using the
technique of the proof above, or otherwise.   Note that 561C shows that
there is a choice function for the family $\Cal E$
of non-empty closed subsets of
$\NN$, since we can use the function $F\mapsto f_F(\tbf{0})$ where
$\family{F}{\Cal E}{f_F}$ is the family of functions defined there.
}

\leader{561E}{Baire's theorem} (a) Let $(X,\rho)$ be a complete metric
space with a well-orderable dense subset.   Then $X$ is a Baire space.

(b) Let $X$ be a compact Hausdorff space
with a well-orderable $\pi$-base.   Then $X$ is a Baire space.

\proof{{\bf (a)} Let $D$ be a dense subset of $X$ with a well-ordering
$\preccurlyeq$.   If $\sequencen{G_n}$ is a sequence of dense
open subsets of $X$, and $G$ is a non-empty open set, define
$\sequencen{H_n}$, $\sequencen{x_n}$ and $\sequencen{\epsilon_n}$
inductively, as follows.   $H_0=G$.   Given $H_n$,
$x_n$ is to be the $\preccurlyeq$-first point of $H_n$.
Given $x_n$ and $H_n$, $\epsilon_n$ is to be the first rational number
in $\ocint{0,2^{-n}}$ such that $B(x_n,\epsilon_n)\subseteq H_n$.
(I leave it to you to decide which rational numbers come first.)
Now set $H_{n+1}=\{y:y\in G_n$, $\rho(y,x_n)<\epsilon_n\}$;  continue.

At the end of the induction, $\sequencen{x_n}$ is a Cauchy sequence so has
a limit $x$ in $X$.   Since $x_n\in H_n\subseteq H_m$ whenever $m\le n$,
$x\in\overline{H}_{n+2}\subseteq H_{n+1}\subseteq G_n$ for every $n$, and
$x$ witnesses that $G\cap\bigcap_{n\in\Bbb N}G_n$ is non-empty.
As $G$ and $\sequencen{G_n}$ are arbitrary, $X$ is a Baire space.

\medskip

{\bf (b)} Let $\Cal U$ be a $\pi$-base for the topology of $X$, not
containing $\emptyset$, with a
well-ordering $\preccurlyeq$.   If $\sequencen{G_n}$ is a sequence of dense
open subsets of $X$, and $G$ is a non-empty open set, define
$\sequencen{U_n}$ in $\Cal U$ inductively by saying that

\inset{$U_0$ is the
$\preccurlyeq$-first member of $\Cal U$ included in $G$,

$U_{n+1}$ is the $\preccurlyeq$-first member of $\Cal U$ such that
$\overline{U}_{n+1}\subseteq U_n\cap G_n$}

\noindent for each $n$.   Then $\bigcap_{n\in\Bbb N}\overline{U}_n$ is
non-empty and included in $G\cap\bigcap_{n\in\Bbb N}G_n$.
}%end of proof of 561E

\leader{561F}{Stone's Theorem} Let $\frak A$ be a well-orderable Boolean
algebra.   Then there is a compact Hausdorff
Baire space $Z$ such that $\frak A$
is isomorphic to the algebra of open-and-closed subsets of $Z$.

\proof{ As in 311E, let $Z$ be the set of ring homomorphisms from $\frak A$
onto $\Bbb Z_2$.   Writing $\Bbb B$ for the set of finite subalgebras of
$\frak A$, $Z=\bigcap_{\frak B\in\Bbb B}Z_{\frak B}$ where

\Centerline{$Z_{\frak B}
=\{z:z\in\Bbb Z_2^{\frak A}$, $z\restrp\frak B$ is a Boolean
homomorphism$\}$.}

\noindent So $Z$ is a closed subset of the compact Hausdorff
space $\{0,1\}^{\frak A}$, and is compact.   Setting
$\widehat{a}=\{z:z\in Z$, $z(a)=1\}$, the map $a\mapsto\widehat{a}$ is a
Boolean homomorphism from $\frak A$ to the algebra $\frak E$ of
open-and-closed subsets of $Z$.
If $a\in\frak A\setminus\{0\}$ and $\frak B$ is a finite
subalgebra of $\frak A$, then the subalgebra $\frak C$ generated by
$\{a\}\cup\frak B$ is still finite, and there is a Boolean homomorphism
$w:\frak C\to\Bbb Z_2$ such that $w(a)=1$;  extending $w$ arbitrarily to a
member of $\{0,1\}^{\frak A}$, we obtain a $z\in Z_{\frak B}$ such that
$z(a)=1$;  as $\frak B$ is arbitrary, there is a $z\in Z$ such that
$z(a)=1$.   So the map $a\mapsto\widehat{a}$ is injective.   If
$G\subseteq Z$ is open and $z\in G$, there must be a finite set
$A\subseteq\frak A$ such that $G$ includes
$\{z':z'\in Z$, $z'\restr A=z\restr A\}$;  in this case, setting
$c=\inf\{a:a\in A$, $z(a)=1\}\Bsetminus\sup\{a:a\in A$, $z(a)=0\}$,
$z\in\widehat{c}\subseteq G$.   It follows that any member of $\frak E$
is of the form $\widehat{a}$ for some $a\in\frak A$, so that
$a\mapsto\widehat{a}$ is an isomorphism between $\frak A$ and $\frak E$.

Because $\frak A$ is well-orderable, $\Bbb Z_2^{\frak A}$ and $Z$ have
well-orderable bases, and $Z$ is a Baire space, by 561E.
}%end of proof of 561F

\leader{561G}{Haar\dvrocolon{ measure}}\cmmnt{ Now I come to something
which demands a rather less sketchy treatment.

\medskip

\noindent}{\bf Theorem} Let $X$ be a completely regular
locally compact Hausdorff topological group.

(i) There is a non-zero
left-translation-invariant positive linear functional on $C_k(X)$.

(ii) If $\phi$, $\phi'$ are non-zero
left-translation-invariant positive linear functionals on $C_k(X)$ then
each is a scalar multiple of the other.
%maybe drop "Hausdorff"

\proof{ %{\smc Naimark 70}, VI.27.5 p 361
{\bf (a)} Write $\Phi$ for $\{g:g\in C_k(X)^+$, $g(e)=\|g\|_{\infty}=1\}$
where $e$ is the identity of $X$.  For $f\in C_k(X)^+$ and $g\in\Phi$, set

$$\eqalign{\high{f:g}
&=\inf\{\sum_{i=0}^n\alpha_i:\alpha_0,\ldots,\alpha_n\ge 0\cr
&\mskip100mu \text{ and there are }a_0,\ldots,a_n\in X\text{ such that }
f\le\sum_{i=0}^n\alpha_ia_i\action_l g\},\cr}$$

\noindent writing $(a\action_lg)(x)=g(a^{-1}x)$ as in
4A5Cc\dvAformerly{4{}41Ac}.
We have to confirm that this infimum is always defined in
$\coint{0,\infty}$.   \Prf\ Set $K=\overline{\{x:f(x)>0\}}$ and
$U=\{x:g(x)>\bover12\}$, so that $K$ is compact,
$U$ is open and $U\ne\emptyset$.
Then $K\subseteq\bigcup_{a\in X}aU$, so there are $a_0,\ldots,a_n\in X$
such that $K\subseteq\bigcup_{i\le n}a_iU$.   In this case

\Centerline{$f
\le\sum_{i=0}^n2\|f\|_{\infty}a_i\action_l g$}

\noindent and $\high{f:g}\le 2(n+1)\|f\|_{\infty}$.\ \QeD\

It is now easy to check that

\Centerline{$\high{a\action_l f:g}=\high{f:g}$,
\quad$\high{f_1+f_2:g}\le\high{f_1:g}+\high{f_2:g}$,}

\Centerline{$\high{\alpha f:g}=\alpha\high{f:g}$,
\quad$\|f\|_{\infty}\le\high{f:g}$,
\quad$\high{f:h}\le\high{f:g}\high{g:h}$}

\noindent whenever $f$, $f_1$, $f_2\in C_k(X)^+$,
$g$, $h\in\Phi$, $a\in X$
and $\alpha\in\coint{0,\infty}$.  (Compare part (c) of the proof of 441C.)

\medskip

{\bf (b)} Fix $g_0\in\Phi$, and for $g\in\Phi$ set

$$\psi_g(f)=\bover{\high{f:g}}{\high{g_0:g}}$$

\noindent for $f\in C_k(X)$.   Then

\Centerline{$\psi_g(a\action_l f)=\psi_g(f)$,
\quad$\psi_g(f_1+f_2)\le\psi_g(f_1)+\psi_g(f_2)$,}

\Centerline{$\psi_g(\alpha f)=\alpha\psi_g(f),
\quad\psi_g(f)\le\high{f:g_0}$,}

\Centerline{$\psi_g(f)\le\psi_g(h)\high{f:h}$,
\quad$1\le\psi_g(h)\high{g_0:h}$}

\noindent whenever $f$, $f_1$, $f_2\in C_k(X)^+$, $h\in\Phi$,
$a\in X$ and $\alpha\ge 0$.   For a neighbourhood $U$ of the identity $e$
of $X$, write $\Phi_U$ for the set of those $g\in\Phi$ such that
$g(x)=0$ for every $x\in X\setminus U$;  because $X$ is locally compact and
completely regular, $\Phi_U\ne\emptyset$.

\medskip

{\bf (c)(i)} If $f_0,\ldots,f_m\in C_k(X)^+$ and $\epsilon>0$, there
is a neighbourhood $U$ of $e$ such that

\Centerline{$\sum_{j=0}^m\psi_g(f_j)
\le\psi_g(\sum_{j=0}^mf_j)+\epsilon$}

\noindent whenever $g\in\Phi_U$.
\Prf\ Set $f=\sum_{j=0}^mf_j$.   Let $K$ be the compact set
$\overline{\{x:f(x)\ne 0\}}$, and let $\hat f\in C_k(X)$ be
such that $\chi K\le\hat f$.   Let $\eta>0$ be such that

\Centerline{$(1+(m+1)\eta)(\psi_g(f)+\eta\high{\hat f:g_0})
\le\psi_g(f)+\epsilon$,}

\noindent and set $f^*=f+\eta\hat f$.
Then we can express each $f_j$ as
$f^*\times h_j$ where $h_j\in C_k(X)^+$ and $\sum_{j=0}^mh_j\le\chi X$.
Let $U$ be a neighbourhood of $e$ such that $|h_j(x)-h_j(y)|\le\eta$
whenever $x^{-1}y\in U$ and $j\le m$ (compare 4A5Pa).

Take $g\in\Phi_U$.   Let $\alpha_0,\ldots,\alpha_n\ge 0$ and
$a_0,\ldots,a_n\in X$ be such that $f^*\le\sum_{i=0}^n\alpha_ia_i\action_l g$
and $\sum_{i=0}^n\alpha_i\le\high{f^*:g}+\eta$.
Then, for any $x\in X$ and $j\le m$,

\Centerline{$f_j(x)=f^*(x)h_j(x)
\le\sum_{i=0}^n\alpha_ig(a_i^{-1}x)h_j(x)
\le\sum_{i=0}^n\alpha_ig(a_i^{-1}x)(h_j(a_i)+\eta)$}

\noindent because if $i$ is such that $g(a_i^{-1}x)\ne 0$ then
$a_i^{-1}x\in U$ and $h_j(x)\le h_j(a_i)+\eta$.   So
$\high{f_j:g}\le\sum_{i=0}^n\alpha_i(h_j(a_i)+\eta)$.   Summing over $j$,

\Centerline{$\sum_{j=0}^m\high{f_j:g}
\le\sum_{i=0}^n\alpha_i(1+(m+1)\eta)$}

\noindent because $\sum_{j=0}^mh_j(a_i)\le 1$ for every $i$.
As $\alpha_0,\ldots,\alpha_n$ and $a_0,\ldots,a_n$ are arbitrary,

\Centerline{$\sum_{j=0}^m\high{f_j:g}
\le(1+(m+1)\eta)\high{f^*:g}
\le(1+(m+1)\eta)(\high{f:g}+\eta\high{\hat f:g})$,}

\noindent and

$$\eqalign{\sum_{j=0}^m\psi_g(f_i)
&\le(1+(m+1)\eta)(\psi_g(f)+\eta\psi_g(\hat f))\cr
&\le(1+(m+1)\eta)(\psi_g(f)+\eta\high{\hat f:g_0})
\le\psi_g(f)+\epsilon\cr}$$

\noindent as required.\ \Qed

\medskip

\quad{\bf (ii)} If $f_0,\ldots,f_m\in C_k(X)^+$, $M\ge 0$ and $\epsilon>0$,
there is a neighbourhood $U$ of $e$ such that

\Centerline{$\sum_{j=0}^m\psi_g(\gamma_jf_j)
\le\psi_g(\sum_{j=0}^m\gamma_jf_j)+\epsilon$}

\noindent whenever $g\in\Phi_U$ and $\gamma_0,\ldots,\gamma_m\in[0,M]$.
\Prf\ Let $\eta>0$ be such that
$\eta(1+\sum_{j=0}^m\high{f_j:\nobreak g_0})\le\epsilon$.
By (i), applied finitely often,
there is a neighbourhood $U$ of $e$ such that

\Centerline{$\sum_{j=0}^m\psi_g(\gamma_jf_j)
\le\psi_g(\sum_{j=0}^m\gamma_jf_j)+\eta$}

\noindent whenever $g\in\Phi_U$ and $\gamma_0,\ldots,\gamma_m\in[0,M]$ are
multiples of $\eta$.   Now, given arbitrary
$\gamma_0,\ldots,\gamma_m\in[0,M]$ and $g\in\Phi_U$,
let $\gamma_0',\ldots,\gamma_m'$ be
multiples of $\eta$ such that $\gamma_j'\le\gamma_j<\gamma_j'+\eta$ for
each $j$.   Then

$$\eqalign{\sum_{j=0}^m\psi_g(\gamma_jf_j)
&\le\sum_{j=0}^m\psi_g(\gamma'_jf_j)+\eta\psi_g(f_j)\cr
&\le\psi_g(\sum_{j=0}^m\gamma'_jf_j)+\eta+\eta\sum_{j=0}^m\psi_g(f_j)
\le\psi_g(\sum_{j=0}^m\gamma_jf_j)+\epsilon\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (d)} (We are coming to the magic bit.)
Suppose that $f\in C_k(X)^+$, $\epsilon>0$ and that $U$ is a
neighbourhood of $e$ such that $|f(x)-f(y)|\le\epsilon$ whenever
$x^{-1}y\in U$.   Then if $g\in\Phi_U$ and $\gamma>\epsilon$ there are
$\alpha_0,\ldots,\alpha_n\ge 0$ and $a_0,\ldots,a_n\in X$ such that
$\|f-\sum_{i=0}^n\alpha_ia_i\action_l g\|_{\infty}\le\gamma$.
\Prf\ For all $x$, $y\in X$ we have

\Centerline{$(f(x)-\epsilon)g(x^{-1}y)
\le f(y)g(x^{-1}y)\le(f(x)+\epsilon)g(x^{-1}y)$.}

\noindent Let $\eta>0$ be such that
$\eta(1+\high{f:\Reverse{g}})\le\gamma-\epsilon$,
where $\Reverse{g}(x)=g(x^{-1})$ for $x\in X$.
Let $V$ be an open neighbourhood of $e$ such that $|g(x)-g(y)|\le\eta$
whenever $xy^{-1}\in V$.   Then we have $a_0,\ldots,a_n$ such that
$\bigcup_{i\le n}a_iV\supseteq\overline{\{x:f(x)\ne 0\}}$, and
$h_0,\ldots,h_n\in C_k(X)^+$ such that $\sum_{i=0}^nh_i(x)=1$ whenever
$f(x)>0$, while $h_i(x)=0$ if $i\le n$ and $x\notin a_iV$.
By (c-ii), there is an $h\in\Phi$ such that

\Centerline{$\sum_{i=0}^n\psi_h(\gamma_if\times h_i)
\le\psi_h(\sum_{i=0}^n\gamma_if\times h_i)+\eta$}

\noindent whenever $0\le\gamma_i\le\high{g_0:\Reverse{g}}$ for each $i$.

Now, for $i\le n$ and $x$, $y\in X$,

\Centerline{$h_i(y)f(y)(g(a_i^{-1}x)-\eta)
\le h_i(y)f(y)g(y^{-1}x)
\le h_i(y)f(y)(g(a_i^{-1}x)+\eta)$.}

\noindent Accordingly

$$\eqalign{(f(x)-\epsilon)(x\action_l\Reverse{g})(y)
&=(f(x)-\epsilon)g(y^{-1}x)
\le f(y)g(y^{-1}x)
=\sum_{i=0}^nh_i(y)f(y)g(y^{-1}x)\cr
&\le\sum_{i=0}^nh_i(y)f(y)(g(a_i^{-1}x)+\eta)
=\eta f(y)+\sum_{i=0}^nh_i(y)f(y)g(a_i^{-1}x);\cr}$$

\noindent similarly,

\Centerline{$(f(x)+\epsilon)(x\action_l\Reverse{g})(y)
\ge\sum_{i=0}^nh_i(y)f(y)g(a_i^{-1}x)-\eta f(y)$.}

Fixing $x$ for the moment, and applying the functional $\psi_h$ to the
expressions here (regarded as functions of $y$), we get

\Centerline{$(f(x)-\epsilon)\psi_h(\Reverse{g})
\le\eta\psi_h(f)+\psi_h(\sum_{i=0}^ng(a_i^{-1}x)f\times h_i)$}

\noindent so

$$\eqalign{f(x)-\gamma
&\le f(x)-\epsilon-\eta\high{f:\Reverse{g}}
\le f(x)-\epsilon-\eta\Bover{\psi_h(f)}{\psi_h(\Reverse{g})}\cr
&\le\psi_h(\sum_{i=0}^n
  \Bover{g(a_i^{-1}x)}{\psi_h(\Reverse{g})}f\times h_i)
\le\sum_{i=0}^n\Bover{g(a_i^{-1}x)}{\psi_h(\Reverse{g})}\psi_h(f\times h_i)
=\sum_{i=0}^n\alpha_ig(a_i^{-1}x)\cr}$$

\noindent where $\alpha_i=\Bover{\psi_h(f\times h_i)}{\psi_h(\Reverse{g})}$.
On the other side,

\Centerline{$(f(x)+\epsilon)\psi_h(\Reverse{g})+\eta\psi_h(f)
\ge\psi_h(\sum_{i=0}^ng(a_i^{-1}x)f\times h_i)$,}

\noindent so

$$\eqalignno{f(x)+\gamma
&\ge f(x)+\epsilon+\eta\Bover{\psi_h(f)}{\psi_h(\Reverse{g})}+\eta\cr
&\ge\psi_h(\sum_{i=0}^n\Bover{g(a_i^{-1}x)}{\psi_h(\Reverse{g})}f\times h_i)
  +\eta
\ge\sum_{i=0}^n\Bover{g(a_i^{-1}x)}{\psi_h(\Reverse{g})}\psi_h(f\times h_i)\cr
\displaycause{because
$\Bover{g(a_i^{-1}x)}{\psi_h(\Reverse{g})}\le\high{g_0:\Reverse{g}}$ for every $i$}
&=\sum_{i=0}^n\alpha_ig(a_i^{-1}x).}$$

\noindent All this is valid for every $x\in X$;  so

\Centerline{$\|f-\sum_{i=0}^n\alpha_ia_i\action_l g\|_{\infty}\le\gamma$.
\Qed}

\medskip

{\bf (e)} For any $f\in C_k(X)^+$ and $\epsilon>0$ there are a
$\gamma\ge 0$ and a neighbourhood $U$ of $e$ such that
$|\psi_h(f)-\gamma|\le\epsilon$ for every $h\in\Phi_U$.   \Prf\ Let $V$ be
a compact neighbourhood of $0$ and $K=\overline{\{x:f(x)+g_0(x)>0\}}$;
let $f^*\in C_k(X)$ be such that $\chi(KV^{-1}V)\le f^*\le\chi X$.
Let $\delta$, $\eta>0$ be such that

\Centerline{$\delta(1+2(\delta+\high{f:g_0}))\le\epsilon$,
\quad$\delta\le\bover12$,
\quad$\eta(1+\high{f^*:g_0})\le\delta$.}

\noindent By (d), there are $g\in\Phi_V$,
$\alpha_0,\ldots,\alpha_n,\beta_0,\ldots,\beta_m\ge 0$ and
$a_0,\ldots,a_n,b_0,\ldots,b_n\in X$ such that

\Centerline{$\|f-\sum_{i=0}^n\alpha_ia_i\action_l g\|_{\infty}\le\eta$,
\quad$\|g_0-\sum_{j=0}^m\beta_jb_j\action_l g\|_{\infty}\le\eta$.}

\noindent
We can suppose that all the $a_i$, $b_j$ belong to $KV^{-1}$, since
$g(a^{-1}x)=0$ if $x\in K$ and $a\notin KV^{-1}$;  consequently

\Centerline{$|f-\sum_{i=0}^n\alpha_ia_i\action_l g|\le\eta f^*$,
\quad$|g_0-\sum_{j=0}^m\beta_jb_j\action_l g|\le\eta f^*$.}

\noindent Set $\alpha=\sum_{i=0}^n\alpha_i$, $\beta=\sum_{j=0}^m\beta_j$
and $\gamma=\Bover{\alpha}{\beta}$.   ($\beta$ is non-zero because
$\|g_0\|_{\infty}=1$ and $\eta\|f^*\|_{\infty}\le\bover12$.)

Let $U\subseteq V$ be a neighbourhood of $e$ such that

\Centerline{$\sum_{i=0}^n\alpha_i\psi_h(a_i\action_l g)
\le\psi_h(\sum_{i=0}^n\alpha_ia_i\action_l g)+\eta$,}

\Centerline{$\sum_{j=0}^m\beta_j\psi_h(b_j\action_l g)
\le\psi_h(\sum_{j=0}^m\beta_jb_j\action_l g)+\eta$}

\noindent for every $h\in\Phi_U$  ((c) above).   Take any $h\in\Phi_U$.
Then

$$\eqalign{|\psi_h(f)-\alpha\psi_h(g)|
&=|\psi_h(f)-\sum_{i=0}^n\alpha_i\psi_h(a_i\action_l g)|
\le|\psi_h(f)-\psi_h(\sum_{i=0}^n\alpha_ia_i\action_l g)|+\eta\cr
&\le\eta\psi_h(f^*)+\eta
\le\eta\high{f^*:g_0}+\eta;\cr}$$

\noindent similarly,

\Centerline{$|1-\beta\psi_h(g)|
=|\psi_h(g_0)-\beta\psi_h(g)|\le\eta(\high{f^*:g_0}+1)$.}

\noindent But this means that

$$\eqalign{|\psi_h(f)-\gamma|
&\le\eta(1+\high{f^*:g_0})+|\alpha\psi_h(g)-\gamma|\cr
&\le\delta+\gamma|\beta\psi_h(g)-1|
\le\delta(1+\gamma).\cr}$$

\noindent Consequently

\Centerline{$\gamma\le\Bover{\psi_h(f)+\delta}{1-\delta}
\le 2(\delta+\high{f:g_0})$,}

\Centerline{$|\psi_h(f)-\gamma|\le\delta(1+2(\delta+\high{f:g_0}))
\le\epsilon$,}

\noindent as required.\ \Qed

\medskip

{\bf (f)} We are nearly home.   Let $\Cal F$ be the filter on $\Phi$
generated by $\{\Phi_U:U$ is a neighbourhood of $e\}$.   By (e),
$\phi(f)=\lim_{h\to\Cal F}\psi_h(f)$ is defined for every $f\in C_k(X)^+$.
From the formulae in (b) we have

\Centerline{$\phi(a\action_l f)=\phi(f)$,
\quad$\phi(f_1+f_2)\le\phi(f_1)+\phi(f_2)$,
\quad$\phi(\alpha f)=\alpha\phi(f)$}

\noindent whenever $f$,
$f_1$, $f_2\in C_k(X)^+$, $a\in X$ and $\alpha\ge 0$.
By (c-i), we have $\phi(f_1)+\phi(f_2)\le\phi(f_1+f_2)$ for all $f_1$,
$f_2\in C_k(X)^+$.   So $\phi$ is additive and extends to an invariant
positive linear functional on $C_k(X)$ which is non-zero because
$\phi(g_0)=1$.

\medskip

{\bf (g)} As for uniqueness, we can repeat the arguments in (e).
Suppose that $\phi'$ is another
left-translation-invariant positive linear functional on $C_k(X)$ such that
$\phi'(g_0)=1$, and $f\in C_k(X)^+$.   Let
$K$ be the closure of ${\{x:f(x)+g_0(x)>0\}}$ and $V$ a compact
neighbourhood
of $e$;  let $f^*\in C_k(X^*)$ be such that $\chi(KV^{-1}V)\le f^*$.
Take $\epsilon>0$.   Let $\delta$, $\eta>0$ be such that

\Centerline{$\delta\le\Bover12$,
\quad$\delta(1+2(\phi(f)+1))\le\epsilon$,
\quad$\delta(1+2(\phi'(f)+1))\le\epsilon$,}

\Centerline{$\eta\phi(f^*)\le\delta$,
\quad$\eta\phi'(f^*)\le\delta$.}

\noindent Then there
is a neighbourhood $U$ of $e$, included in $V$, such that
$|f(x)-f(y)|\le\eta$
and $|g_0(x)-g_0(y)|\le\eta$ whenever $x^{-1}y\in U$.   By (d),
there are $g\in\Phi_V$,
$\alpha_0,\ldots,\alpha_n,\beta_0,\ldots,\beta_m\ge 0$ and
$a_0,\ldots,a_n,b_0,\ldots,b_m\in X$ such that

\Centerline{$|f(x)-\sum_{i=0}^n\alpha_i(a_i\action_l g)(x)|\le\eta$,
\quad$|g_0(x)-\sum_{j=0}^m\beta_j(b_j\action_l g)(x)|\le\eta$}

\noindent for every $x\in X$;  as in (e), we may suppose that every $a_i$,
$b_j$ belongs to $KV^{-1}$ so that

\Centerline{$|f-\sum_{i=0}^n\alpha_ia_i\action_l g|\le\eta f^*$,
\quad$|g_0-\sum_{j=0}^m\beta_jb_j\action_l g|\le\eta f^*$.}

\noindent Consequently, setting $\alpha=\sum_{i=0}^n\alpha_i$,
$\beta=\sum_{i=0}^n\beta_i$ and $\gamma=\alpha/\beta$,

\Centerline{$|\phi(f)-\alpha\phi(g)|
=|\phi(f)-\sum_{i=0}^n\alpha_i\phi(a_i\action_l g)|
\le\eta\phi(f^*)\le\delta$,}

\Centerline{$|1-\beta\phi(g)|
=|\phi(g_0)-\sum_{j=0}^m\beta_j\phi(b_j\action_l g)|
\le\eta\phi(f^*)\le\delta$.}

\noindent So

\Centerline{$|\phi(f)-\gamma|
\le\eta\phi(f^*)+\gamma|\beta\phi(g)-1|
\le\eta\phi(f^*)(1+\gamma)
\le\delta(1+\gamma)$}

\noindent and

\Centerline{$\gamma\le\Bover{\phi(f)+\delta}{1-\delta}\le 2(\phi(f)+1)$,}

\Centerline{$|\phi(f)-\gamma|\le\delta(1+2(\phi(f)+\delta))
\le\epsilon$.}

\noindent Similarly, $|\phi'(f)-\gamma|\le\epsilon$ and
$|\phi(f)-\phi'(f)|\le 2\epsilon$.   As $\epsilon$ and $f$ are arbitrary,
$\phi=\phi'$.
}%end of proof of 561G

\leader{561H}{Kakutani's theorem} (a) Let $U$ be an Archimedean Riesz space
with a weak order unit.
Then there are a Dedekind complete Boolean algebra
$\frak A$ and an order-dense Riesz subspace of $L^0(\frak A)$, containing
$\chi 1$, which is isomorphic to $U$.

(b) Let $U$ be an $L$-space with a weak order unit $e$.   Then there is a
totally finite measure algebra $(\frak A,\bar\mu)$ such that $U$ is
isomorphic, as normed Riesz space, to $L^1(\frak A,\bar\mu)$, and we can
choose the isomorphism to match $e$ with $\chi 1$.

\proof{ All the required ideas are in Volume 3;  but we have quite a lot of
checking to do.

\medskip

{\bf (a)(i)} The first step is to observe that, for any Dedekind
$\sigma$-complete Boolean algebra $\frak A$, the definition of
$L^0=L^0(\frak A)$ in 364A gives no difficulties, and that the
formulae of 364D
can be used to define a Riesz space structure on $L^0$.   \Prf\
I recall the formulae in question:

\Centerline{$\Bvalue{u>\alpha}=\sup_{\beta>\alpha}\Bvalue{u>\beta}$
for every $\alpha\in\Bbb R$,}

\Centerline{$\inf_{\alpha\in\Bbb R}\Bvalue{u>\alpha}=0$,
\quad $\sup_{\alpha\in\Bbb R}\Bvalue{u>\alpha}=1$,}

\Centerline{$\Bvalue{u+v>\alpha}
=\sup_{q\in\Bbb Q}\Bvalue{u>q}\Bcap\Bvalue{v>\alpha-q}$,}

\noindent whenever $u$, $v\in L^0$ and $\alpha\in\Bbb R$,

\Centerline{$\Bvalue{\gamma u>\alpha}=\Bvalue{u>\bover{\alpha}{\gamma}}$}

\noindent whenever $u\in L^0$, $\gamma\in\ooint{0,\infty}$ and
$\alpha\in\Bbb R$.
The distributive laws in 313A-313B are enough to ensure that $u+v$ and
$\gamma u$, so defined, belong to $L^0$, and also that
$u+v=v+u$, $u+(v+w)=(u+v)+w$, $\gamma(u+v)=\gamma u+\gamma v$ for $u$, $v$,
$w\in L^0$ and $\gamma>0$.   Defining $\tbf{0}\in L^0$ by saying that

\Centerline{$\Bvalue{\tbf{0}>\alpha}=1$ if $\alpha<0$, $0$ if
$\alpha\ge 0$,}

\noindent we can check that $u+0=u$ for every $u$.   Defining $-u\in L^0$
by saying that

\Centerline{$\Bvalue{-u>\alpha}
=\sup_{q\in\Bbb Q,q>\alpha}1\Bsetminus\Bvalue{u>-q}$}

\noindent for $u\in L^0$ and $\alpha\in\Bbb R$, we find (again using the
distributive laws, of course) that $u+(-u)=\tbf{0}$;  we can now define
$\gamma u$, for $\gamma\le 0$, by saying that $0\cdot u=\tbf{0}$ and
$\gamma u=(-\gamma)(-u)$ if $\gamma<0$, and we shall have a linear space.
Turning to the ordering, it is nearly trivial to check that the definition

\Centerline{$u\le v\iff\Bvalue{u>\alpha}\Bsubseteq\Bvalue{v>\alpha}$ for
every $\alpha\in\Bbb R$}

\noindent gives us a partially ordered linear space.   It is a Riesz space
because the formula

\Centerline{$\Bvalue{u\vee v>\alpha}
=\Bvalue{u>\alpha}\Bcup\Bvalue{v>\alpha}$}

\noindent defines a member of $L^0$ which must be $\sup\{u,v\}$ in $L^0$.\
\Qed

We need to know that if $\frak A$ is Dedekind
complete, so is $L^0$;  the argument of 364M still applies.
Note also that $a\mapsto\chi a:\frak A\to L^0$ is
order-continuous, as in 364Jc.

\medskip

\quad{\bf (ii)} Now suppose that $U$ is an Archimedean Riesz space with an
order unit $e$.   Let $\frak A$ be the band algebra of $U$ (353B).
Then we can argue as in 368E, but with the simplification
that the maximal disjoint set $C$ in $U^+\setminus\{0\}$ is just $\{e\}$,
to see that we have an injective Riesz homomorphism $T:U\to L^0(\frak A)$
defined by taking $\Bvalue{Tu>\alpha}$ to be the band generated by
$e\wedge(u-\alpha e)^+$ (or, if you prefer, by $(u-\alpha e)^+$, since it
comes to the same thing).   We shall have $T[U]$ order-dense, as before,
with $Te=\chi 1$.

\medskip

{\bf (b)(i)} Again, the bit we have to concentrate on is the check that,
starting from a totally finite measure algebra $(\frak A,\bar\mu)$, we can
define $L^1(\frak A,\bar\mu)$ as in 365A.
We have to be a bit careful, because already in Proposition 321C there is
an appeal to AC$(\omega)$ (see 561Yi(vi));
but I think we need to know very little about
measure algebras to get through the arguments here.   Of course another
difficulty arises at once in 365A, because I write

\Centerline{$\|u\|_1
=\int_0^{\infty}\bar\mu\Bvalue{|u|>\alpha}\,d\alpha$,}

\noindent and say that the integration is with respect to `Lebesgue
measure', which won't do, at least until I redefine Lebesgue integration as
in \S565.   But we are integrating a monotonic
function, so the integral can be thought of as an improper Riemann
integral;  if you like,

\Centerline{$\|u\|_1
=\lim_{n\to\infty}2^{-n}\sum_{i=1}^{4^n}\bar\mu\Bvalue{|u|>2^{-n}i}
=\sup_{n\in\Bbb N}2^{-n}\sum_{i=1}^{4^n}\bar\mu\Bvalue{|u|>2^{-n}i}$.}

\noindent Next, we can't use the Loomis-Sikorski theorem to prove 365C, and
have to go back to first principles.   To see that $\|\,\|_1$ is
subadditive, and additive on $(L^0)^+$, look first at `simple' non-negative
$u$, expressed as $u=\sum_{i=0}^n\alpha_i\chi a_i$, and check that
$\int u=\|u\|_1=\sum_{i=0}^n\alpha_i\bar\mu a_i$;  now confirm that
every element of $(L^0)^+$ is expressible as the supremum of a
non-decreasing sequence of such elements, and that $\|\,\|_1$ is
sequentially order-continuous on the left on $(L^0)^+$.
(We need 321Be.)
This is enough to show that $L^1$ is a solid linear subspace of $L^0$ with
a Riesz norm and a sequentially order-continuous integral.
(I do {\it not} claim, yet, that $L^1$ is an $L$-space, because
I do not know, in the absence of countable choice, that every Cauchy
filter on $L^1$ converges.)

\medskip

\quad{\bf (ii)} Now let $U$ be an $L$-space with a weak order unit $e$.
As in (a), let $\frak A$ be the band algebra of $U$ and $T:U\to L^0$ an
injective Riesz homomorphism onto an order-dense Riesz subspace of $L^0$
with $Te=\chi 1$.   Now $U$ is Dedekind complete (354N, 354Ee).
Consequently $T[U]$ must be solid in $L^0$ (353L\formerly{3{}53K}).

\medskip

\quad{\bf (iii)} For $a\in\frak A$, set $\bar\mu a=\|T^{-1}(\chi a)\|$.
Because the map $a\mapsto T^{-1}\chi a$ is additive and order-continuous
and injective, $(\frak A,\bar\mu)$ is a measure algebra;  indeed,
$\bar\mu$ is actually order-continuous.   So we have a space
$L^1=L^1(\frak A,\bar\mu)$.   Because $\bar\mu$ is order-continuous,
364L(a-ii) tells us that $\|w\|_1=\sup_{v\in B}\|v\|_1$ whenever
$B\subseteq L^0$ is a non-empty upwards-directed set in $L^0$
with supremum $w$ in $L^0$.

Writing $S\subseteq L^0$ for
the linear span of $\{\chi a:a\in\frak A\}$, we see that
$\|w\|_1=\|T^{-1}w\|$ for every $w\in S$.   Since $S$ is order-dense in
$L^0$ it is order-dense in $L^1$, and $T^{-1}[S]$ is order-dense in $U$,
therefore norm-dense (354Ef).

\medskip

\quad{\bf (iv)}
$Tu\in L^1$ for every $u\in U^+$.   \Prf\ For $n\in\Bbb N$ set
$a_n=\Bvalue{Tu>2^n}\Bsetminus\Bvalue{Tu>2^{n+1}}$, $u_n=T^{-1}(\chi a_n)$.
Set $w_n=\sum_{i=0}^n2^i\chi a_i$ for $n\in\Bbb N$.   Then $w_n\le Tu$
and $\|w_n\|_1=\|T^{-1}w_n\|\le\|u\|$ for every $n$.   By 364L(a-i),
$w=\sup_{n\in\Bbb N}w_n$ is defined in $L^0$, and
$\|w\|_1=\sup_{n\in\Bbb N}\|w_n\|_1$ is finite.   But $Tu\le 2w+\chi 1$, so
$Tu\in L^1$.\ \QeD\

\medskip

\quad{\bf (v)} If $w\in(L^1)^+$ there is a $v\in U^+$ such that $w=Tv$ and
$\|v\|=\|w\|_1$.   \Prf\ Consider
$A=\{u:u\in T^{-1}[S]$, $Tu\le w\}$.   This is upwards-directed and
norm-bounded, so has a supremum $v$ in $U$ (354N again), and
$Tv\ge w'$ whenever $w'\in S$ and $w'\le w$.   But $S$ is order-dense in
$L^0$ so $Tv\ge w$.   Because $T$ is order-continuous, (iii) tells us that

\Centerline{$\|Tv\|_1=\sup_{u\in A}\|Tu\|_1=\sup_{u\in A}\|u\|=\|v\|$,}

\noindent while surely $\|w\|_1\ge\sup_{u\in A}\|Tu\|_1$.   So $Tv=w$.\
\Qed

\medskip

\quad{\bf (vi)} Putting (iv) and (v) together, we see that $T[U]=L^1$ and
that $T$ is a normed Riesz space isomorphism, as required.
}%end of proof of 561H

\vleader{72pt}{561I}{Hilbert spaces:  Proposition}
Let $U$ be a Hilbert space.

(a) If $C\subseteq U$ is a non-empty closed convex set then for
any $u\in U$ there is a unique $v\in C$ such that
$\|u-v\|=\min\{\|u-w\|:w\in C\}$.

(b) Every closed linear subspace of $U$ is the image of an orthogonal
projection, that is, has an orthogonal complement.

(c) Every member of $U^*$ is of the form
$u\mapsto\innerprod{u}{v}$ for some $v\in U$.

(d) $U$ is reflexive.

(e) If $C\subseteq U$ is a norm-closed convex set then it is
weakly closed.

\proof{{\bf (a)} Set $\gamma=\inf\{\|u-w\|:w\in C\}$ and let $\Cal F$ be
the filter on $U$ generated by sets of the form
$F_{\epsilon}=\{w:w\in C$, $\|u-w\|\le\gamma+\epsilon\}$ for $\epsilon>0$.
Then $\Cal F$ is Cauchy.   \Prf\ Suppose that $\epsilon>0$ and $w_1$,
$w_2\in F_{\epsilon}$.   Then

$$\eqalignno{\|w_1-w_2\|^2
&=2\|u-w_1\|^2+2\|u-w_2\|^2-\|2u-w_1-w_2\|^2
\le 4(\gamma+\epsilon)^2-4\gamma^2\cr
\displaycause{because $\Bover12(w_1+w_2)\in C$}
&=8\gamma\epsilon+4\epsilon^2.\cr}$$

\noindent So

\Centerline{$\inf_{F\in\Cal F}\diam F=\inf_{\epsilon>0}\diam F_{\epsilon}
=0$.   \Qed}

We therefore have a limit $v$ of $\Cal F$, which is in $C$ because $C$ is
closed, and $\|u-v\|=\lim_{w\to\Cal F}\|u-w\|=\gamma$.
If now $w$ is any other member of $C$, $\|u-\Bover12(v+w)\|\ge\gamma$ so
$\|u-w\|>\gamma$.

\medskip

{\bf (b)} Let $V$ be a closed linear subspace of $U$.   By (a), we have a
function $P:U\to V$ such that $Pu$ is the unique closest element of $V$ to
$u$, that is, $\|u-Pu\|\le\|u-Pu+\alpha v\|$ for every $v\in V$ and
$\alpha\in\Bbb R$.   It follows that $\innerprod{u-Pu}{v}=0$ for every
$v\in V$, that is, that $u-Pu\in V^{\perp}$.   As $u$ is arbitrary,
$U=V+V^{\perp}$;  as $V\cap V^{\perp}=\{0\}$, $P$ must be the projection
onto $V$ with kernel $V^{\perp}$, and is an orthogonal projection.

\medskip

{\bf (c)} Take $f\in U^*$.   If $f=0$ then $f(u)=\innerprod{u}{0}$ for
every $u$.   Otherwise, set $C=\{w:f(w)=0\}$.   Then $C$ is a proper closed
linear subspace of $U$.   Take any $u_0\in U\setminus C$.
Let $v_0$ be the point of $C$ nearest to $u_0$, and consider $u_1=u_0-v_0$.
Then $0$ is the point of $C$ nearest to $u_1$, so that
$\innerprod{u}{u_1}=0$ for every $u\in C$.   Set
$v=\Bover{f(u_1)}{\|u_1\|^2}u_1$;  then $\innerprod{u}{v}=0$ for every
$u\in C$, while $f(v)=\innerprod{v}{v}$.   So $f(u)=\innerprod{u}{v}$ for
every $u\in U$.

\medskip

{\bf (d)} From (c) it follows that we can identify $U^*$ with $U$ and
therefore $U^{**}$ also becomes identified with $U$.

\medskip

{\bf (e)} If $C$ is empty this is trivial.   Otherwise, take any
$u\in U\setminus C$.   Let $v$ be the point of $C$ nearest to $U$, and set
$f(w)=\innerprod{w}{u-v}$ for $w\in U$.   Then $f(w)\le f(v)<f(u)$ for
every $w\in C$.   So $u$ does not belong to the weak closure of $C$;  as
$u$ is arbitrary, $C$ is weakly closed.
}%end of proof of 561I

\exercises{\leader{561X}{Basic exercises (a)}
%\spheader 561Xa
Let $X$ be any set.   (i) Show
that $\ell^p(X)$, for $1\le p\le\infty$, is a Banach space.
(ii) Show that $\ell^1(X)^*$ can be identified with $\ell^{\infty}(X)$.
(iii) Show that if $1<p<\infty$ and $\bover1p+\bover1q=1$ then
$\ell^p(X)^*$ can be identified with $\ell^q(X)$.
%561B

\spheader 561Xb Let $X$ be any topological space.   Show that $C_b(X)$,
with $\|\,\|_{\infty}$, is a Banach space.
%561B

\spheader 561Xc Suppose that there is
an infinite subset $X$ of $\Bbb R$ with no infinite countable subset
({\smc Jech 73}, \S10.1).   Show that $X$ is sequentially closed
but not closed, second-countable but not separable,
sequentially compact but not compact, sequentially complete
(that is, every Cauchy sequence converges)
but not complete.   Show that the topology of $\Bbb R$ is not countably
tight.
%561B

\sqheader 561Xd (i) Let $C$ be the set of those
$R\subseteq\Bbb N\times\Bbb N$ which are total orderings of subsets of
$\Bbb N$.   Show that $C$ is a closed subset of
$\Cal P(\Bbb N\times\Bbb N)$ with its usual topology.
(ii) For $\xi<\omega_1$, let $C_{\xi}$ be the set of those
$R\in C$ which are well-orderings of order type $\xi$ of subsets of
$\Bbb N$.   Show that $C_{\xi}$ is a Borel subset of
$\Cal P(\Bbb N\times\Bbb N)$.   \Hint{induce on $\xi$.}
(iii) Show that there is an injective function from $\omega_1$ to the
Borel $\sigma$-algebra $\Cal B(\Bbb R)$ of $\Bbb R$.
%561B

\spheader 561Xe(i) Show that every non-empty closed subset of $\NN$ has a
lexicographically-first member.   (ii) Show that if a T$_1$
topological space $X$ is a continuous image
of $\NN$, then there is an injection from $X$ to $\Cal P\Bbb N$.
%561C

\spheader 561Xf Let $X$ be a topological space.
(i) Show that if $X$ is separable, then
$X^{\Bbb R}$ is separable.  (ii) Show that if $X$ has a countable network,
then $X^{\Bbb R}$ has a countable network.
%561C

\spheader 561Xg(i) Show that a locally compact Hausdorff space is regular.
(ii) Show that a compact regular space is normal.
%561D

\spheader 561Xh Let $U$ be a normed space with a well-orderable subset $D$
such that the linear span of $D$ is dense in $U$.
(i) Show that if $V$ is a
linear subspace of $U$ and $f\in V^*$, there is a $g\in U^*$, extending
$f$, with the same norm as $f$.   (ii) Show that the unit ball
$B$ of $U^*$ is
weak*-compact and has a well-orderable base for its topology.   (iii) Show
that if $K\subseteq B$ is weak*-closed then $K$ has an extreme point.
%561D

\spheader 561Xi Let $(X,\rho)$ be a separable
compact metric space, and $G$ the
isometry group of $X$ with its topology of pointwise convergence (441G).
Show that $G$ is compact.   \Hint{$X^{\Bbb N}$ is compact.}
%561D

\spheader 561Xj Let $X$ be a regular topological space and $A$ a subset of
$X$.   Show that the following are equiveridical:  (i) $A$ is relatively
compact in $X$;  (ii) $\overline{A}$ is compact;  (iii) every filter on $X$
containing $A$ has a cluster point.
%561D

\sqheader 561Xk Let $(X,\frak T)$ be a regular second-countable
topological space, and write $\frak S$ for the usual topology on
$\BbbR^{\Bbb N}$.   (i) Show that there are a continuous function
$f:X\to\BbbR^{\Bbb N}$ and a function $\phi:\frak T\to\frak S$ such that
$G=f^{-1}[\phi(G)]$ for every $G\in\frak T$.   (ii) Show that if $X$ is
Hausdorff it is metrizable.
%561E 561Xl

\sqheader 561Xl Let $X$ be a regular second-countable topological space,
$\Cal C$ the family of closed subsets of $X$, and $\Cal D$ the set of
disjoint pairs
$(F_0,F_1)\in\Cal C\times\Cal C$.   (i) Show that $X$ is
normal, and that there
is a function $\psi:\Cal D\to\Cal C$ such that
$F_0\subseteq\interior\psi(F_0,F_1)$ and $F_1\cap\psi(F_0,F_1)=\emptyset$
whenever $(F_0,F_1)\in\Cal D$.
(ii) Show that there is a function $\phi:\Cal D\to C(X)$ such
$\phi(F_0,F_1)(x)=0$, $\phi(F_0,F_1)(y)=1$ whenever $(F_0,F_1)\in\Cal D$,
$x\in F_0$ and $y\in F_1$.
%561E

\spheader 561Xm Let $X$ be a well-orderable discrete abelian group.   Show
that its dual group, as defined in 445A, is a completely regular compact
Hausdorff group.
%561G

\spheader 561Xn Let $U$ be a Riesz space with a Riesz norm.   Let
$\Delta:U^+\to\coint{0,\infty}$ be such that
($\alpha$) $\Delta$ is non-decreasing,
($\beta$) $\Delta(\alpha u)=\alpha\Delta(u)$ whenever
$u\in U^+$ and $\alpha\ge 0$,
($\gamma$) $\Delta(u+v)=\Delta(u)+\Delta(v)$ whenever $u\wedge v=0$
($\delta$) $|\Delta(u)-\Delta(v)|\le\|u-v\|$ for all $u$, $v\in U^+$.
Show that $\Delta$ has an  to a member of $U^*$.
%561H

\spheader 561Xo Let $U$ be an $L$-space.
Show that $\|u\|=\sup\{f(u):f\in U^*$, $\|f\|\le 1\}$ for every $u\in U$.
%561H

\spheader 561Xp Let $(\frak A,\bar\mu)$ be a measure algebra.   Show that
$L^1(\frak A,\bar\mu)$ is a Dedekind $\sigma$-complete Riesz space and a
sequentially complete normed space.
%561H

\spheader 561Xq Let $\frak A$ be a Boolean algebra.   Show that there are a
set $X$, an algebra $\Cal E$ of subsets of $X$ and a surjective Boolean
homomorphism from $\Cal E$ onto $\frak A$.   \Hint{566L.}
%561H

\spheader 561Xr Let $U$ be a Hilbert space.
(i) Show that a bounded sequence
$\sequencen{u_n}$ in $U$ is weakly convergent in $U$ iff
$\lim_{n\to\infty}\innerprod{u_n}{u_m}$ is defined for every
$m\in\Bbb N$.
(ii) Show that the unit ball
of $U$ is sequentially compact for the weak topology.   (iii)
Show that if
$T:U\to U$ is a self-adjoint compact linear operator, then $T[U]$ is
included in the closed linear span of $\{Tv:v$ is an eigenvector of $T\}$.
\Hint{reduce to the case in which $U$ is separable, and show that there is
then a sequence $\sequencen{u_n}$ in the unit ball $B$ of $U$ such that
$\lim_{n\to\infty}\innerprod{Tu_n}{u_n}=\sup_{u\in B}\innerprod{Tu}{u}$.}
%561I

\spheader 561Xs\dvAnew{2013}
In 561C, show that $(F,\alpha)\mapsto f_F(\alpha):\Cal E\times\NN\to\NN$ is
continuous if $\Cal E$ is given its Vietoris topology (4A2T) and
$\NN$ its usual topology.
%561C out of order query

\leader{561Y}{Further exercises (a)}
%\spheader 561Ya
Suppose that there is a sequence $\sequencen{A_n}$ of
countable sets such that $\bigcup_{n\in\Bbb N}A_n=\Bbb R$.   Show that
$\cf\omega_1=\omega$.
%561A

\spheader 561Yb(i) Show that there is a bijection between $\omega_1$ and
$\Bbb N\times\omega_1$.   (ii) Show that $\omega_2$ is not expressible as the
union of a sequence of
countable sets.   (iii) Show that $\Cal P\omega_1$ is not expressible as the
union of a sequence of
countable sets.   (iv) Show that $\Cal P(\Cal P\Bbb N)$ is not
expressible as the union of a sequence of countable sets.
%561Ya 561A

\spheader 561Yc Suppose there is a countable family of doubleton
sets with no choice function ({\smc Jech 73}, \S5.5).   Show that
(i) there is
a set $I$ such that $\{0,1\}^I$ has an open-and-closed set which is not
determined by coordinates in any countable subset of $I$ (ii) there is a
compact metrizable space which is not ccc, therefore not second-countable
(iii) there is a complete totally bounded metric space which is neither ccc
nor compact (iv) there is a probability algebra which is not ccc.
%561D mt56bits

\spheader 561Yd Let $X$ be a metrizable space.   Show that it
is second-countable iff it has a countable $\pi$-base
iff it has a countable network.
%561E

\spheader 561Ye(i) Let $(X,\rho)$ be a complete metric space.
Show that $X$ has a well-orderable dense subset iff it has a well-orderable
base iff it has a well-orderable $\pi$-base iff it has a well-orderable
network iff there is a choice function
for the family of its non-empty closed subsets.
(ii) Let $X$ be a locally compact Hausdorff space.
($\alpha$) Show that if it
has a well-orderable $\pi$-base then it has a well-orderable dense subset.
($\beta$) Show that if it has a well-orderable base then
it is completely regular and there is a choice
function for the family of its non-empty closed subsets.
%561E 561D

\spheader 561Yf Let $X$ be a metrizable space.
Show that every continuous real-valued
function defined on a closed subset of $X$ has an extension to
a continuous real-valued function on $X$.
%561E mt56bits

\spheader 561Yg(i) Show that if $\frak A$ is a Boolean algebra,
there is an essentially unique Dedekind complete Boolean algebra
$\widehat{\frak A}$ in which $\frak A$ can
be embedded as an order-dense subalgebra.   (ii) Show that if $\frak A$ and
$\frak B$ are two Boolean algebras and $\pi:\frak A\to\frak B$ is an
order-continuous Boolean homomorphism, $\pi$ has a unique extension to an
order-continuous Boolean homomorphism from $\widehat{\frak A}$ to
$\widehat{\frak B}$.   \Hint{take $\widehat{\frak A}$ to be the set of
pairs $(A,A')$ of subsets of $\frak A$ such that $A$ is the set of lower
bounds of $A'$ and $A'$ is the set of upper bounds of $A$.}
%3{}14T 561H

\spheader 561Yh Let $\familyiI{\frak A_i}$ be a family of Boolean algebras.
(i) Show that there is an essentially unique structure
$(\frak A,\familyiI{\varepsilon_i})$ such that ($\alpha$) $\frak A$ is a
Boolean algebra ($\beta$) $\varepsilon_i:\frak A_i\to\frak A$ is a Boolean
homomorphism for every $i$ ($\gamma$) whenever $\frak B$ is a Boolean
algebra
and $\phi_i:\frak A_i\to\frak B$ is a Boolean homomorphism for every $i$,
there is a unique Boolean homomorphism $\pi:\frak A\to\frak B$ such that
$\pi\varepsilon_i=\phi_i$ for every $i$.   (ii) Show that if
$\nu_i:\frak A_i\to\Bbb R$ is additive, with $\nu_i1=1$,
for every $i\in I$, there is a
unique additive $\nu:\frak A\to\Bbb R$ such that
$\nu(\inf_{i\in J}\varepsilon_i(a_i))=\prod_{i\in J}\nu_ia_i$ whenever
$J\subseteq I$ is finite and $a_i\in\frak A_i$ for $i\in J$.
%3{}15 561H

\spheader 561Yi Suppose that there is an infinite totally ordered set $I$
with no countably infinite subset ({\smc Jech 73}, \S10.1).   Let $\Cal E$
be the algebra of subsets of $\{0,1\}^I$ determined by coordinates in
finite sets.   (i) Show that the union of any countable family of finite
subsets of $I$ is finite.
(ii) Show that $\Cal E$ has no countably infinite subset, so
that every finitely additive real-valued functional on $\Cal E$ is
countably additive.   (iii) Show that there is no infinite disjoint
family in $\Cal E$.
(iv) Show that $\Cal E$ is Dedekind complete.   (v) Show that there
is a functional $\mu_1$ such that $(\Cal E,\mu_1)$
is a probability algebra and
$\mu_1$ is order-continuous.   (vi) Show that there is a functional
$\mu_2$ such that $(\Cal E,\mu_2)$ is a probability algebra and $\mu_2$ is
not order-continuous.
%3{}91  56bits

\spheader 561Yj Let $(X,\rho)$ be a complete metric space with a
well-orderable base.   Show that a subset of $X$ is compact iff it is
sequentially compact iff it is closed and totally bounded.
%561Ye 563Ff
}%end of exercises

\endnotes{
\Notesheader{561}
The arguments of this section will I hope give an idea of
the kind of discipline which will be imposed for the rest of the chapter.
Apart from anything else, we have to fix on the correct definitions.
Typically, when defining something like `compactness' or `completeness',
the definition to use is that which is most useful in the most general
context;  so that even in metrizable spaces we should prefer filters to
sequences (cf.\ 561Xc).

We can distinguish two themes in the methods I have used here.   First, in
the presence of a well-ordering we can hope to adapt the standard attack on
a problem;  see 561D-561F.  %561D, 561E, 561F.
Second, if (in the presence of the
axiom of choice) there is a {\it unique} solution to a problem, then we can
hope that it is still a unique solution without choice.   This is what
happens in 561G and also in 561Ia-561Ic.   In 561I we just go through the
usual arguments with a little more care.   In 561G (taken from
{\smc Naimark 70}) we need new ideas.   But in the key step, part (d) of
the proof, the two variables $x$ and $y$ reflect an adaptation of a
repeated-integration argument as in \S442.   Note that the scope of 561G
may be limited if we have fewer locally compact groups than we expect.

A regular second-countable Hausdorff space is
metrizable (561Yf).   But it may not be separable (561Xc).
We do not have Urysohn's Lemma in its usual form, so cannot be sure that
a locally compact Hausdorff space is completely regular;  a topological
group has left, right and bilateral uniformities, but a uniformity need not
be defined by pseudometrics and a uniform space need not be completely
regular.   So in such results as 561G we may need an extra `completely
regular' in the hypotheses.

I give a version of Kakutani's theorem (561H) to show that some of the
familiar patterns are distorted in possibly unexpected ways, and that
occasionally it is the more abstract parts of the theory which survive
best.
I suppose I ought to remark explicitly that I define `measure algebra'
exactly as in 321A:  a Dedekind $\sigma$-complete Boolean algebra with a
strictly positive countably additive $[0,\infty]$-valued functional.
I do not claim that every $\sigma$-finite measure algebra is
either localizable or ccc (561Yc),
nor that every measure algebra can be represented in terms of a measure
space.   I set up a construction of a normed Riesz space
$L^1(\frak A,\bar\mu)$, but do not claim that this is an $L$-space.
However, if we start from an $L$-space $U$
with a weak order unit, we can build
a measure on its band algebra and proceed to an $L^1(\frak A,\bar\mu)$
which is isomorphic to $U$ (and is therefore an $L$-space).
}%end of notes


\discrpage

\frfilename{mt562.tex}
\versiondate{20.10.13}
\copyrightdate{2006}

\def\chaptername{Choice and determinacy}
\def\sectionname{Borel codes}

\newsection{562}

The concept of `Borel set', either in
the real line or in general topological spaces, has been fundamental in
measure theory since before the modern subject existed.   It is at this
point that the character of the subject changes if we do not allow
ourselves even the countable axiom of choice.   I have already mentioned
the Feferman-L\'evy model in which
$\Bbb R$ is a countable union of countable sets;
immediately, every subset of $\Bbb R$ is a countable union of countable
sets and is `Borel' on the definition of 111G.   In these circumstances
that definition becomes unhelpful.

An alternative which leads to a non-trivial theory, coinciding
with the usual theory in the presence of AC, is the
algebra of `codable Borel sets' (562B).   This is not necessarily a
$\sigma$-algebra, but is closed under unions and intersections of `codable
sequences' (562K).   When we come to look for measurable functions, the
corresponding concept is that of `codable Borel function' (562L);  again,
we do not expect the limit of an arbitrary sequence of codable Borel
functions to be measurable in any useful sense, but the limit of a codable
sequence of codable Borel functions is again a codable Borel function
(562Ne).
The same ideas can be used to give a theory of `codable Baire sets' in
any topological space (562T).

\vleader{60pt}{562A}{Trees} \cmmnt{I review some ideas from \S421.

\medskip

}{\bf (a)} Set $S^*=\bigcup_{n\ge 1}\BbbN^n$.
For $\sigma\in S^*$ and $T\subseteq S^*$, write $T_{\sigma}$
for $\{\tau:\tau\in S^*$, $\sigma^{\smallfrown}\tau\in T\}$ (notation:
5A1C).

%maybe switch \Cal T and \Cal T_0 ?  see \S421
% but  \Cal T  is much commoner here!

\spheader 562Ab Let $\Cal T_0$ be the family of sets
$T\subseteq S^*$ such that
$\sigma\restr n\in T$ whenever $\sigma\in T$ and $n\ge 1$.
Recall from 421N\footnote{Early editions of Volume 4
used a slightly different
definition of iterated derivations, so that the `rank' of a tree was not
quite the same.} that we have a derivation $\partial:\Cal T_0\to\Cal T_0$
defined by setting

\Centerline{$\partial T
=\{\sigma:\sigma\in S^*$, $T_{\sigma}\ne\emptyset\}$,}

\noindent with iterates $\partial^{\xi}$, for $\xi<\omega_1$, defined by
setting

\Centerline{$\partial^0T=T$,
\quad$\partial^{\xi}T=\bigcap_{\eta<\xi}\partial(\partial^{\eta}T)$ for
$\xi\ge 1$.}

\noindent Now for any $T\in\Cal T_0$ there is a $\xi<\omega_1$ such that
$\partial^{\xi}T=\partial^{\eta}T$ whenever $\xi\le\eta<\omega_1$.
\prooflet{\Prf\
The argument in 421Nd assumed that $\omega_1$ has uncountable cofinality,
but we can avoid this assumption, as follows.   Let
$\family{\sigma}{S^*}{\epsilon_{\sigma}}$ be a summable family of strictly
positive real numbers, and set
$\gamma_T(\xi)=\sum_{\sigma\in\partial^{\xi}T}\epsilon_{\sigma}$;  then
$\gamma_T:\omega_1\to\coint{0,\infty}$ is non-increasing, so 561A tells us
that there is a $\xi<\omega_1$ such that $\gamma_T(\xi+1)=\gamma_T(\xi)$,
that is, $\partial^{\xi+1}T=\partial^{\xi}T$.   Of course we now have
$\partial^{\eta}T=\partial^{\xi}T$ for every $\eta\ge\xi$.\ \Qed}

\spheader 562Ac We therefore still have a rank function
$r:\Cal T_0\to\omega_1$ defined by saying that $r(T)$ is the least ordinal
such that $\partial^{r(T)}T=\partial^{r(T)+1}T$.   Now $\partial^{r(T)}T$
is empty iff there is no $\alpha\in\NN$ such that $\alpha\restr n\in T$ for
every $n\ge 1$.   \prooflet{\Prf\ The argument in 421Nf used the word
`choose';  but we can avoid this by being more specific.   If
$\sigma\in\partial^{r(T)}T$, then we can define a sequence
$\sequencen{\sigma_n}$ by saying that $\sigma_0=\sigma$ and, given
$\sigma_n\in\partial^{r(T)}T$,
$\sigma_{n+1}={\sigma_n}^{\smallfrown}\fraction{i}$ for the least $i$ such
that ${\sigma_n}^{\smallfrown}\fraction{i}\in\partial^{r(T)}T$;
$\alpha=\bigcup_{n\in\Bbb N}\sigma_n$ will now have $\alpha\restr n\in T$
for every $n\ge 1$.   The argument in the other direction is unchanged.\
\Qed}

Let $\Cal T$ be the set of those $T\in\Cal T_0$
with no infinite branch, that is, such that $\partial^{r(T)}T=\emptyset$.
\cmmnt{Note that if $T\in\Cal T$ then $r(T)=0$ iff $T=\emptyset$, while
$r(T)=1$ iff there is a non-empty set $A\subseteq\Bbb N$ such that
$T=\{\fraction{i}:i\in A\}$.}

\spheader 562Ad For $T\in\Cal T$, set $A_T=\{i:\fraction{i}\in T\}$.
We need a fact not covered in \S421:  for any $T\in\Cal T$,
$r(T)=\sup\{r(T_{\fraction{i}})+1:i\in A_T\}$.   \prooflet{\Prf\ An easy
induction on $\xi$ shows that
$\partial^{\xi}(T_{\sigma})=(\partial^{\xi}T)_{\sigma}$ for any
$\xi<\omega_1$, $T\in\Cal T_0$ and $\sigma\in S^*$.   So, for $T\in\Cal T$
and $\xi<\omega_1$,

$$\eqalign{r(T)>\xi
&\Longrightarrow\partial^{\xi}T\ne\emptyset\cr
&\Longrightarrow\,\Exists i,\,\fraction{i}\in\partial^{\xi}T
  =\bigcap_{\eta<\xi}\partial^{\eta+1}T\cr
&\Longrightarrow\,\Exists i\in A_T,\,
  \partial^{\eta}(T_{\fraction{i}})
  =(\partial^{\eta}T)_{\fraction{i}}\ne\emptyset\Forall\eta<\xi\cr
&\Longrightarrow\,\Exists i\in A_T,\,
  r(T_{\fraction{i}})>\eta\Forall\eta<\xi\cr
&\Longrightarrow\,\Exists i\in A_T,\,
  r(T_{\fraction{i}})\ge\xi;\cr}$$

\noindent thus $r(T)\le\sup\{r(T_{\fraction{i}})+1:i\in A_T\}$.   In the
other direction, if $i\in A_T$ and $\eta<\xi=r(T_{\fraction{i}})$, then

\Centerline{$(\partial^{\eta}T)_{\fraction{i}}
=\partial^{\eta}(T_{\fraction{i}})\ne\emptyset$,}

\noindent so $\fraction{i}\in\partial^{\eta+1}T$;  as $\eta$ is arbitrary,
$\fraction{i}\in\partial^{\xi}T$ and $\xi<r(T)$;  as $i$ is arbitrary,
$r(T)\ge\sup\{r(T_{\fraction{i}})+1:i\in A_T\}$.\ \Qed}

\leader{562B}{Coding sets with trees (a)}
Let $X$ be a set and $\sequencen{E_n}$ a sequence of
subsets of $X$.   Define $\phi:\Cal T\to\Cal PX$ inductively by saying that

$$\eqalign{\phi(T)&=\bigcup_{i\in A_T}E_i\text{ if }r(T)\le 1,\cr
&=\bigcup_{i\in A_T}X\setminus\phi(T_{\fraction{i}})
   \text{ if }r(T)>1.\cr}$$

\cmmnt{\noindent By 562Ad, this definition is sound.}
I will call $\phi$ the {\bf interpretation of Borel codes} defined by
$X$ and $\sequencen{E_n}$.

\spheader 562Bb Of course $\phi(\emptyset)=\emptyset$.
If we set

\Centerline{$T^*
=\{\fraction{0},\fraction{0}^{\smallfrown}\fraction{0},
\fraction{0}^{\smallfrown}\fraction{0}^{\smallfrown}\fraction{0},
\fraction{1},\fraction{1}^{\smallfrown}\fraction{0}\}$}

\noindent and

\Centerline{$T
=\{\fraction{0}\}\cup\{\fraction{0}^{\smallfrown}\sigma:\sigma\in T^*\}$,}

\noindent then

\ifwithproofs
$$\eqalign{\phi(T^*)
&=(X\setminus\phi(T^*_{\fraction{0}}))
  \cup(X\setminus\phi(T^*_{\fraction{1}}))\cr
&=(X\setminus\phi(\{\fraction{0},\fraction{0}^{\smallfrown}\fraction{0}\}))
  \cup(X\setminus\phi(\{\fraction{0}\}))\cr
&=(X\setminus(X\setminus\phi(\{\fraction{0}\})))
  \cup(X\setminus\phi(\{\fraction{0}\}))
=X,\cr
\phi(T)
&=X\setminus\phi(T^*)
=\emptyset,\cr}$$
\else
\Centerline{$\phi(T^*)=X$,\quad$\phi(T)=\emptyset$,}
\fi

\noindent while $T\ne\emptyset$\cmmnt{, which it will be useful to know}.

\spheader 562Bc Now suppose that $X$ is a second-countable topological
space and that $\sequencen{U_n}$, $\sequencen{V_n}$ are two sequences
running over bases for the topology of $X$.   Let $\phi:\Cal T\to\Cal PX$
and $\phi':\Cal T\to\Cal PX$ be the interpretations of Borel codes defined
by $\sequencen{U_n}$, $\sequencen{V_n}$ respectively.   Then
there is a function $\Theta:\Cal T\to\Cal T\setminus\{\emptyset\}$
such that $\phi'\Theta=\phi$.
\prooflet{\Prf\ Define $\Theta$ inductively, as
follows.   If $r(T)\le 1$, then
$\phi(T)=\bigcup_{i\in A_T}U_i$ is open.   If $\phi(T)\ne\emptyset$,
set $\Theta(T)=\{\fraction{j}:j\in\Bbb N$, $V_j\subseteq\phi(T)\}$;  then

\Centerline{$\phi'(\Theta(T))
=\bigcup\{V_j:j\in\Bbb N$, $V_j\subseteq\phi(T)\}=\phi(T)$.}

\noindent If $\phi(T)=\emptyset$, take
$\Theta(T)$ to be any non-empty member of $\Cal T$ such that
$\phi'(\Theta(T))=\emptyset$;  e.g., that presented in (b) just above.

For the inductive step to $r(T)>1$, set

\Centerline{$\Theta(T)=\{\fraction{i}:i\in A_T\}
  \cup\{\fraction{i}^{\smallfrown}\sigma:
i\in A_T$, $\sigma\in\Theta(T_{\fraction{i}})\}$;}

\noindent then $r(\Theta(T))>1$ and

$$\eqalign{\phi'(\Theta(T))
&=\bigcup_{i\in A_{\Theta(T)}}X\setminus\phi'(\Theta(T)_{\fraction{i}})
=\bigcup_{i\in A_T}X\setminus\phi'(\Theta(T_{\fraction{i}}))\cr
&=\bigcup_{i\in A_T}X\setminus\phi(T_{\fraction{i}})
=\phi(T),\cr}$$

\noindent so the induction continues.\ \Qed}

\cmmnt{(There will be a substantial strengthening of this idea in
562Ma.)}

\spheader 562Bd Now say that a {\bf codable Borel set}
in $X$ is one expressible as
$\phi(T)$ for some $T\in\Cal T$, starting from some sequencen running over
a base for the topology of $X$\cmmnt{;  in view of (c),
we can restrict our calculations to a
fixed enumeration of a fixed base if we wish}.   I will write $\Cal B_c(X)$
for the family of codable Borel sets of $X$.

The definition of `interpretation of Borel codes'
makes it plain that any $\sigma$-algebra of
subsets of $X$ containing every open set will also contain every codable
Borel set;\cmmnt{ so} every codable Borel set is\cmmnt{ indeed} a
`Borel set' on the definition of 111G or 4A3A.

\cmmnt{As in the argument for (c) just above,
it will sometimes be useful to know that every element of $\Cal B_c(X)$ can
be coded by a non-empty member of $\Cal T$;  we
have only to check the case
of the empty set, which is dealt with in the formula in (b).}

\leader{562C}{}\cmmnt{
The point of these codings is that we can define explicit
functions on $\Cal T$ which will have appropriate reflections in the
coded sets.

\medskip

}{\bf (a)} For instance, there are functions
$\Theta_0:\Cal T\to\Cal T$, $\Theta_1:\Cal T\times\Cal T\to\Cal T$,
$\Theta_2:\Cal T\times\Cal T\to\Cal T$,
$\Theta_3:\Cal T\times\Cal T\to\Cal T$ such that, for any interpretation
$\phi$ of Borel codes,

\Centerline{$\phi(\Theta_0(T))=X\setminus\phi(T)$,
\quad$\phi(\Theta_1(T,T'))=\phi(T)\cup\phi(T')$,}

\Centerline{$\phi(\Theta_2(T,T'))=\phi(T)\cap\phi(T')$,
\quad$\phi(\Theta_3(T,T'))=\phi(T)\setminus\phi(T')$}

\noindent for all $T$, $T'\in\Cal T$.
\prooflet{\Prf\ Let $T^*$ be the tree described in 562Bb, so that
$\phi(T^*)=X$.   Set

$$\eqalign{\Theta_0(T)
&=T^*\text{ if }T=\emptyset,\cr
&=\{\fraction{0}\}\cup\{\fraction{0}^{\smallfrown}\sigma:\sigma\in T\}
\text{ otherwise};\cr}$$

\noindent then

$$\eqalign{\phi(\Theta_0(T))
&=\phi(T^*)=X=X\setminus\phi(T)\text{ if }T=\emptyset,\cr
&=X\setminus\phi(\Theta_0(T)_{\fraction{0}})
=X\setminus\phi(T)\text{ otherwise}.\cr}$$

\noindent Now set

\Centerline{$\Theta_1(T,T')
=\{\fraction{0},\fraction{1}\}
   \cup\{\fraction{0}^{\smallfrown}\sigma:\sigma\in\Theta_0(T)\}
   \cup\{\fraction{1}^{\smallfrown}\sigma:\sigma\in\Theta_0(T')\}$,}

\noindent so that

$$\eqalign{\phi(\Theta_1(T,T'))
&=(X\setminus\phi(\Theta_1(T,T'))_{\fraction{0}})
  \cup(X\setminus\phi(\Theta_1(T,T'))_{\fraction{1}})\cr
&=(X\setminus\phi(\Theta_0(T)))
  \cup(X\setminus\phi(\Theta_0(T')))\cr
&=(X\setminus(X\setminus\phi(T)))
  \cup(X\setminus(X\setminus\phi(T')))
=\phi(T)\cup\phi(T').\cr}$$

\noindent So we can take

\Centerline{$\Theta_2(T,T')
=\Theta_0(\Theta_1(\Theta_0(T),\Theta_0(T')))$,
\quad$\Theta_3(T,T')
=\Theta_2(T,\Theta_0(T'))$}

\noindent and get

\Centerline{$\phi(\Theta_2(T,T'))
=X\setminus((X\setminus\phi(T))\cup(X\setminus\phi(T')))
=\phi(T)\cap\phi(T')$,}

\Centerline{$\phi(\Theta_3(T,T'))
=\phi(T)\cap(X\setminus\phi(T'))
=\phi(T)\setminus\phi(T')$.  \Qed}}

\spheader 562Cb\cmmnt{ We can find codes for unions and intersections of
sequences, provided the sequences are presented in the right way;  I give
a general formulation of the process.}   For any countable set $K$ we have
functions $\tilde\Theta_1$,
$\tilde\Theta_2:\bigcup_{J\subseteq K}\Cal T^J\to\Cal T$ such that whenever
$X$ is a set, $\sequencen{E_n}$ is a sequence of subsets
of $X$ and $\phi$ is the corresponding interpretation of Borel codes, then
$\phi(\tilde\Theta_1(\tau))=\bigcup_{j\in J}\phi(\tau(j))$ and
$\phi(\tilde\Theta_2(\tau))=X\cap\bigcap_{j\in J}\phi(\tau(j))$ whenever
$J\subseteq K$ and $\tau\in\Cal T^J$.
\prooflet{\Prf\ Let $\sequencen{k_n}$ be a
sequence running over $K\cup\{\emptyset\}$.
For $J\subseteq K$ and $\tau\in\Cal T^J$, set
$A=\{n:k_n\in J$, $\tau(k_n)\ne\emptyset\}$ and

\Centerline{$\tilde\Theta_1(\tau)
=\{\fraction{n}:n\in A\}
\cup\{\fraction{n}^{\smallfrown}\sigma:
  n\in A,\,\sigma\in\Theta_0(\tau(k_n))\}$,}

\Centerline{$\tilde\Theta_2(\tau)
=\Theta_0(\tilde\Theta_1(\family{j}{J}{\Theta_0(\tau(j))}))$.}

\noindent Then

$$\eqalign{\phi(\tilde\Theta_1(\tau))
&=\emptyset=\bigcup_{j\in J}\phi(\tau(j))\text{ if }A=\emptyset,\cr
&=\bigcup_{n\in A}X\setminus\phi(\Theta_0(\tau(k_n)))
=\bigcup_{n\in A}\phi(\tau(k_n))
=\bigcup_{j\in J}\phi(\tau_j)\text{ otherwise},\cr
\phi(\tilde\Theta_2(\tau))
&=X\setminus\bigcup_{j\in J}(X\setminus\phi(\tau_j))
=X\cap\bigcap_{j\in J}\phi(\tau_j).  \text{\Qed}\cr}$$
}

\spheader 562Cc\cmmnt{ A
more sophisticated version of two of the codings in (a)
will be useful in \S564.}
%Fubini's theorem
Let $X$ be a regular second-countable space, $\sequencen{U_n}$ a sequence
running over a base for the topology of $X$ containing $\emptyset$, and
$\phi:\Cal T\to\Cal PX$ the associated interpretation of Borel codes.
Then there are functions
$\Theta'_1$, $\Theta'_2:\Cal T\times\Cal T\to\Cal T$ such that

\Centerline{$\phi(\Theta'_1(T,T'))=\phi(T)\cup\phi(T')$,
\quad$\phi(\Theta'_2(T,T'))=\phi(T)\cap\phi(T')$,}

\Centerline{$r(\Theta'_1(T,T'))=r(\Theta'_2(T,T'))=\max(r(T),r(T'))$}

\noindent for all $T$, $T'\in\Cal T$.   \prooflet{\Prf\ The point is
just that open sets in a regular second-countable space
are F$_{\sigma}$.   Because of the slightly awkward
form taken by the definition of $\phi$, we need to start with an
auxiliary function.   Define $T\mapsto\tilde T:\Cal T\to\Cal T$ by saying
that

$$\eqalign{\tilde T
&=\{\fraction{n}:\overline{U}_n\subseteq\phi(T)\}
\cup\{\fraction{n}^{\smallfrown}\fraction{i}:
  \overline{U}_n\subseteq\phi(T),\,U_i\cap U_n=\emptyset\}\cr
&\mskip300mu \text{ if }r(T)\le 1,\cr
&=T\text{ otherwise }.\cr}$$

\noindent Then $\phi(\tilde T)=\phi(T)$ and $r(\tilde T)=\max(2,r(T))$
for every $T$ (because if $r(T)\le 1$ there is some $n$ such that
$U_n=\emptyset$ and
$\fraction{n}^{\smallfrown}\fraction{n}\in\tilde T$).
Note that $\Tilde{\Tilde T}=\tilde T$.
We also need to fix a bijection $n\mapsto(i_n,j_n)$ between $\Bbb N$ and
$\Bbb N\times\Bbb N$.

Now define $\Theta'_1$ by saying that

\inset{----- if $\max(r(T),r(T'))\le 1$,
$\Theta'_1(T,T')=\Theta'_1(T)\cup\Theta'_1(T')$;

----- if $\max(r(T),r(T'))>1$, then

$$\eqalign{\Theta'_1(T,T')
&=\{\fraction{2n}:n\in A_{\tilde T}\}
\cup\{\fraction{2n}^{\smallfrown}\sigma:\sigma\in\tilde T_{\fraction{n}}\}
\cr
&\mskip100mu\cup\{\fraction{2n+1}:n\in A_{\tilde T'}\}
\cup\{\fraction{2n+1}^{\smallfrown}\sigma:
  \sigma\in\tilde T'_{\fraction{n}}\}.\cr}$$
}

\noindent For $\Theta'_2$ induce on $\max(r(T),r(T'))$:

\inset{----- if $T=T'=\emptyset$, $\Theta'_2(T,T')=\emptyset$;

----- if $\max(r(T),r(T'))=1$,

\Centerline{$\Theta'_2(T,T')
=\{\fraction{n}:U_n\subseteq\phi(T)\cap\phi(T')\}$;}

----- if $\max(r(T),r(T'))>1$, set $A=\{n:i_n\in A_{\tilde T}$,
$j_n\in A_{\tilde T'}\}$ and

\Centerline{$\Theta'_2(T,T')
=\{\fraction{n}:n\in A\}
\cup\{\fraction{n}^{\smallfrown}\sigma:n\in A$, $\sigma
  \in\Theta'_1(\tilde T_{\fraction{i_n}},\tilde T'_{\fraction{j_n}})\}$}

\noindent(interpreting $\tilde T'_{\fraction{j_n}}$ as
$((T')\sptilde)_{\fraction{j_n}}$).
}

\noindent These formulae work.   I run through the calculations for
$\Theta'_2(T,T')$ when $\max(r(T),r(T'))>1$.    We have
$r(\tilde T)\ge 2$ and $r(\tilde T')\ge 2$, so $A_{\tilde T}$,
$A_{\tilde T'}$ and $A$ are non-empty,

$$\eqalign{r(\Theta'_2(T,T'))
&=\sup_{n\in A}
   r(\Theta'_1(\tilde T_{\fraction{i_n}},\tilde T'_{\fraction{j_n}}))+1
=\sup_{i\in A_{\tilde T},j\in A_{\tilde T'}}
   r(\Theta'_1(\tilde T_{\fraction{i}},\tilde T'_{\fraction{j}}))+1\cr
&=\sup_{i\in A_{\tilde T},j\in A_{\tilde T'}}
   \max(r(\tilde T_{\fraction{i}}),r(\tilde T'_{\fraction{j}}))+1\cr
&=\max(\sup_{i\in A_{\tilde T}}r(\tilde T_{\fraction{i}})+1,
   \sup_{j\in A_{\tilde T'}}r(\tilde T'_{\fraction{j}})+1)\cr
&=\max(r(\tilde T),r(\tilde T'))
=\max(2,r(T),r(T'))
=\max(r(T),r(T'))\cr}$$

\noindent and

$$\eqalign{\phi(\Theta'_2(T,T'))
&=\bigcup_{n\in A}X\setminus
  \phi(\Theta'_1(\tilde T_{\fraction{i_n}},\tilde T'_{\fraction{j_n}}))\cr
&=\bigcup_{n\in A}X\setminus
  \phi(\Theta'_1(\Tilde{\Tilde T}_{\fraction{i_n}},
    \Tilde{\Tilde T}'_{\fraction{j_n}}))
=\bigcup_{n\in A}X\setminus
  (\phi(\tilde T_{\fraction{i_n}})\cup\phi(\tilde T'_{\fraction{j_n}}))\cr
&=\bigcup_{n\in A}(X\setminus\phi(\tilde T_{\fraction{i_n}}))
  \cap(X\setminus\phi(\tilde T'_{\fraction{j_n}}))\cr
&=\bigcup_{i\in A_{\tilde T},j\in A_{\tilde T'}}
   (X\setminus\phi(\tilde T_{\fraction{i_n}}))
     \cap(X\setminus\phi(\tilde T'_{\fraction{j_n}}))\cr
&=(\bigcup_{i\in A_{\tilde T}}X\setminus\phi(\tilde T_{\fraction{i_n}}))
   \cap(\bigcup_{j\in A_{\tilde T'}}
          X\setminus\phi(\tilde T'_{\fraction{j_n}}))\cr
&=\phi(\tilde T)\cap\phi(\tilde T')
=\phi(T)\cap\phi(T'). \text{ \Qed}\cr}$$
}

\leader{562D}{Proposition} (a) If $X$ is a second-countable space, then the
family of codable Borel subsets of $X$ is an algebra of subsets
of $X$ containing every G$_{\delta}$ set and every F$_{\sigma}$ set.

(b) [AC$(\omega)$] Every Borel set is a codable Borel set.

\proof{{\bf (a)} Let $\sequencen{U_n}$ be a sequence running over a base
for the topology of $X$, and
$\phi:\Cal T\to\Cal B_c(X)$ the corresponding surjection.
From 562Ca we see that $X\setminus E$ and $E\cup E'$ belong to
$\Cal B_c(X)$ for all $E$, $E'\in\Cal B_c(X)$;
since $\emptyset=\phi(\emptyset)$ belongs
to $\Cal B_c(X)$, $\Cal B_c(X)$ is an algebra of subsets of $X$.

If $E\subseteq X$ is an F$_{\sigma}$ set, there is a sequence
$\sequencen{F_n}$ of closed sets with union $E$.   Set

\Centerline{$T=\{\fraction{n}:n\in\Bbb N\}
  \cup\{\fraction{n}^{\smallfrown}\fraction{i}:
    n$, $i\in\Bbb N$, $U_i\subseteq X\setminus F_n\}$.}

\noindent Then $r(T)=2$, $\phi(T_{\fraction{n}})=X\setminus F_n$ for every
$n$ and $\phi(T)=E$.

Thus every F$_{\sigma}$ set belongs to $\Cal B_c(X)$;  it follows at once
that every G$_{\delta}$ set is also a codable Borel set.

\medskip

{\bf (b)} We can repeat the argument in (a), but this time in a more
general form.   If $\sequencen{E_n}$ is any sequence in $\Cal B_c(X)$, then
for each $n\in\Bbb N$ {\bf choose}
$T^{(n)}\in\Cal T\setminus\{\emptyset\}$ such that $\phi(T^{(n)})=E_n$;  set

\Centerline{$T=\{\fraction{n}:n\in\Bbb N\}
\cup\{\fraction{n}^{\smallfrown}\sigma:n\in\Bbb N$,
$\sigma\in T^{(n)}\}$;}

\noindent then $\bigcup_{n\in\Bbb N}X\setminus E_n=\phi(T)$ is a
codable Borel set.   Because $\Cal B_c(X)$ is an algebra, this is enough to
show that it is a $\sigma$-algebra and therefore equal to the
$\sigma$-algebra $\Cal B(X)$.
}%end of proof of 562D

\leader{562E}{Proposition} Let $X$ be a second-countable space and
$Y\subseteq X$ a subspace of $X$.   Then
$\Cal B_c(Y)=\{Y\cap E:E\in\Cal B_c(X)\}$.

\proof{ Let $\sequencen{U_n}$ be a sequence running over a base for the
topology of $X$, and set $V_n=Y\cap U_n$ for each
$n$;  let $\phi_X:\Cal T\to\Cal B_c(X)$ and $\phi_Y:\Cal T\to\Cal B_c(Y)$
be the interpretations of Borel codes corresponding to $\sequencen{U_n}$,
$\sequencen{V_n}$ respectively.   Then an easy induction on the rank of $T$
shows that $\phi_Y(T)=Y\cap\phi_X(T)$ for every $T\in\Cal T$.   So

\Centerline{$\Cal B_c(Y)=\phi_Y[\Cal T]=\{Y\cap\phi_X(T):T\in\Cal T\}
=\{Y\cap E:E\in\Cal B_c(X)\}$.}
}%end of proof of 562E

\leader{*562F}{}\cmmnt{ I do not expect to rely on the next
result, but it is
interesting that two of the basic facts of descriptive set theory have
versions in the new context.

\medskip

\noindent}{\bf Theorem}
(a) If $X$ is a Hausdorff second-countable space and $A$, $B$ are disjoint
analytic subsets of $X$, there is a codable Borel set $E\subseteq X$ such
that $A\subseteq E$ and $B\cap E=\emptyset$.

(b) Let $X$ be a Polish space.   Then a subset $E$ of $X$ is a codable
Borel set iff $E$ and $X\setminus E$ are analytic.

\proof{{\bf (a)(i)} If either $A$ or $B$ is empty, this is trivial, just
because $\emptyset$ and $X$ are codable Borel sets;  so suppose otherwise.
Let $f:\NN\to X$ and $g:\NN\to X$ be continuous functions such that
$f[\NN]=A$ and $g[\NN]=B$.   Fix an enumeration $\sequencen{(j_n,k_n)}$ of
$\Bbb N\times\Bbb N$, and a sequence $\sequencen{U_n}$ running over a base
for the topology of $X$;  let $\phi$ be the interpretation of Borel codes
defined by $\sequencen{U_n}$.   For
$\sigma\in S=\bigcup_{n\in\Bbb N}\BbbN^n$ set

\Centerline{$A_{\sigma}
  =\{f(\alpha):\alpha\in\NN$, $\alpha_i=j_{\sigma(i)}$
     for $i<\#(\sigma)\}$,}

\Centerline{$B_{\sigma}
  =\{g(\alpha):\alpha\in\NN$, $\alpha_i=k_{\sigma(i)}$
     for $i<\#(\sigma)\}$.}

\noindent Then $A_{\emptyset}=A$ and
$A_{\sigma}=\bigcup_{i\in\Bbb N}A_{\sigma^{\smallfrown}\fraction{i}}$ for
every $\sigma$, and similarly for $B$.

\medskip

\quad{\bf (ii)} Still in the setting-up stage, we need general union and
intersection operators on $\Cal T$.   As in 562Cb, let
$\tilde\Theta_1$,
$\tilde\Theta_2:\bigcup_{J\subseteq\Bbb N}\Cal T^J\to\Cal T$ be such that
$\phi(\tilde\Theta_1(\tau))=\bigcup_{j\in J}\phi(\tau(j))$ and
$\phi(\tilde\Theta_2(\tau))=X\cap\bigcap_{j\in J}\phi(\tau(j))$
whenever $J\subseteq\Bbb N$ and $\tau\in\Cal T^J$.

\medskip

\quad{\bf (iii)} Set

$$\eqalign{T
&=\{\sigma:\sigma\in S^*,\text{ there are no }i,\,n\in\Bbb N
\text{ such that }n<\#(\sigma)\cr
&\mskip200mu \text{and }A_{\sigma\restr n}\subseteq U_i
\subseteq X\setminus B_{\sigma\restr n}\}.\cr}$$

\noindent If $\sigma\in S$ and $n\in\Bbb N$ then
$A_{\sigma\restr n}\supseteq A_{\sigma}$ and
$B_{\sigma\restr n}\supseteq B_{\sigma}$, so $\sigma\restr m\in T$ whenever
$\sigma\in T$ and $m\ge 1$;
thus $T$ belongs to $\Cal T_0$ as defined in 562Ab.   In fact
$T\in\Cal T$.   \Prf\Quer\ Otherwise, by 562Ac, there is a $\gamma\in\NN$
such that $\gamma\restr n\in T$ for every $n\ge 1$.   Set
$\alpha=\sequencen{\gamma_{j_n}}$, $\beta=\sequencen{\gamma_{k_n}}$;
then $f(\alpha)\in A$ and $g(\beta)\in B$, so $f(\alpha)\ne g(\beta)$.
Because $X$ is Hausdorff, there are $i$, $j\in\Bbb N$ such that
$f(\alpha)\in U_i$, $g(\beta)\in U_j$ and $U_i\cap U_j=\emptyset$.
Because $f$ and $g$ are continuous, there is an $n\ge 1$ such that
$f(\alpha')\in U_i$ and $g(\beta')\in U_j$ whenever
$\alpha'$, $\beta'\in\NN$, $\alpha'\restr n=\alpha\restr n$ and
$\beta'\restr n=\beta\restr n$;  that is, such that
$A_{\gamma\restr n}\subseteq U_i$
and $B_{\gamma\restr n}\subseteq U_j$.   But this means that
$\gamma\restr n+1\notin T$.\ \Bang\Qed

\medskip

\quad{\bf (iv)} We know that $\langle\partial^{\xi}T\rangle_{\xi\le r(T)}$
is a non-increasing family in $\Cal T$ with last member $\emptyset$, and
moreover that $\partial^{\xi}T=\bigcap_{\eta<\xi}\partial^{\eta}T$ for
non-zero limit ordinals $\xi\le r(T)$.   So for $\sigma\in T$ there is
a unique $h(\sigma)<r(T)$ such that
$\sigma\in\partial^{h(\sigma)}T\setminus\partial(\partial^{h(\sigma)}T)$.
I seek to define $T^{(\sigma)}\in\Cal T$, for $\sigma\in T$, such that
$A_{\sigma}\subseteq\phi(T^{(\sigma)})\subseteq X\setminus B_{\sigma}$
for every $\sigma$.   I do this inductively.

\medskip

\quad{\bf (v)} If $h(\sigma)=0$, that is, $\sigma\in T\setminus\partial T$,
then $\sigma^{\smallfrown}0\notin T$.
So there is a first $i\in\Bbb N$ such that
$A_{\sigma}\subseteq U_i\subseteq X\setminus B_{\sigma}$.
Set $T^{(\sigma)}=\{\fraction{i}\}$,
so that $\phi(T^{(\sigma)})=U_i$ includes $A_{\sigma}$ and is
disjoint from $B_{\sigma}$.

\medskip

\quad{\bf (vi)} Now suppose that we have $\xi\le r(t)$ such that
$\xi\ge 1$ and
$T^{(\sigma)}$ has been defined for every $\sigma\in T$ with
$h(\sigma)<\xi$.
Take $\sigma\in T$ such that $h(\sigma)=\xi$;   then
$\sigma^{\smallfrown}\fraction{n}\in T$ for some, therefore every,
$n\in\Bbb N$, while
$h(\sigma^{\smallfrown}\fraction{n})<\xi$ and $T^{(\sigma^{\smallfrown}\fraction{n})}$ is defined
for every $n\in\Bbb N$.   Now, for each $n$, we have
$A_{\sigma^{\smallfrown}\fraction{n}}
\subseteq\phi(T^{(\sigma^{\smallfrown}\fraction{n})})$.   But of course
$A_{\sigma^{\smallfrown}\fraction{n}}=A_{\sigma^{\smallfrown}\fraction{m}}$
whenever $j_m=j_n$.   So we have
$A_{\sigma^{\smallfrown}\fraction{n}}
\subseteq\bigcap_{j_m=j_n}\phi(T^{(\sigma^{\smallfrown}\fraction{m})})$
for each $n$, and

\Centerline{$A_{\sigma}
\subseteq\bigcup_{n\in\Bbb N}\bigcap_{j_m=j_n}
   \phi(T^{(\sigma^{\smallfrown}\fraction{m})})
=\bigcup_{j\in\Bbb N}\bigcap_{j_m=j}
   \phi(T^{(\sigma^{\smallfrown}\fraction{m})})$.}

\noindent Similarly,
$B_{\sigma^{\smallfrown}\fraction{n}}=B_{\sigma^{\smallfrown}\fraction{m}}$
is disjoint from
$\phi(T^{(\sigma^{\smallfrown}\fraction{m})})$ whenever
$k_m=k_n$, so

\Centerline{$B_{\sigma}
=\bigcup_{k\in\Bbb N}\bigcap_{k_m=k}B_{\sigma^{\smallfrown}\fraction{m}}$}

\noindent is disjoint from
$\bigcap_{k\in\Bbb N}\bigcup_{k_m=k}
  \phi(T^{(\sigma^{\smallfrown}\fraction{m})})$.

On the other hand, for any $j$, $k\in\Bbb N$, there is a $p\in\Bbb N$ such
that $j_p=j$ and $k_p=k$, so that

\Centerline{$\bigcap_{j_m=j}
   \phi(T^{(\sigma^{\smallfrown}\fraction{m})})
\subseteq\phi(T^{(\sigma^{\smallfrown}\fraction{p})})
\subseteq\bigcup_{k_m=k}
  \phi(T^{(\sigma^{\smallfrown}\fraction{m})})$.}

\noindent But this means that
$\bigcup_{j\in\Bbb N}\bigcap_{j_m=j}
   \phi(T^{(\sigma^{\smallfrown}\fraction{m})})
\subseteq\bigcap_{k\in\Bbb N}\bigcup_{k_m=k}
  \phi(T^{(\sigma^{\smallfrown}\fraction{m})})$
is disjoint from $B_{\sigma}$.

If therefore we set

\Centerline{$T^{(\sigma)}
=\tilde\Theta_1(\sequence{j}{\tilde\Theta_2(
   \langle T^{(\sigma^{\smallfrown}\fraction{m})}\rangle_{j_m=j})})$}

\noindent we shall have
$A_{\sigma}\subseteq\phi(T^{(\sigma)})\subseteq X\setminus B_{\sigma}$, and
we have a formula defining a suitable tree
$T^{(\sigma)}$ whenever $\sigma\in T$ and
$h(\sigma)=\xi$, so we can continue the induction.

\medskip

\quad{\bf (vii)} This gives us a family
$\family{\sigma}{T}{T^{(\sigma)}}$ in $\Cal T$.
Of course what we are really
looking for is a tree $T^{(\emptyset)}$.   But if $T$ is empty, this is
because there is an $i\in\Bbb N$ such that
$A\subseteq U_i\subseteq X\setminus B$;  in which case $U_i$ is a
codable Borel set separating $A$ from $B$.   While if $T$ is not empty,
$\fraction{n}\in T$ for every $n\in\Bbb N$, and just as in (vi) we can set

\Centerline{$T^{(\emptyset)}
=\tilde\Theta_1(\sequence{j}
  {\tilde\Theta_2(\langle T^{(\fraction{m})}\rangle_{j_m=j})})$}

\noindent to obtain a codable Borel set $E=\phi(T^{\emptyset})$ such that
$A\subseteq E$ and $B\cap E=\emptyset$.

\medskip

{\bf (b)} If $X$ is empty, this is trivial;  suppose henceforth that
$X$ is not empty.

\medskip

\quad{\bf (i)} If $E$ and $X\setminus E$ are analytic, then (a) tells us
that there is a codable Borel set $F$ including $E$ and disjoint from
$X\setminus E$, so that $E=F$ is a codable Borel set.   So the rest of this
part of the proof will be devoted to the converse.

Let $\rho$ be a complete metric on $X$ inducing its
topology, $\sequencen{x_n}$ a sequence running over a dense subset of
$X$, and $\sequencen{U_n}$ a sequence running over a
base for the topology of $X$;  let $\phi$ be the interpretation of Borel
codes defined from $\sequencen{U_n}$.

\medskip

\quad{\bf (ii)} We need to fix on a continuous surjection from a
closed subset of $\NN$ onto $X$;  a convenient one is the following.   Set

\Centerline{$F
=\{\alpha:\alpha\in\NN$, $\rho(x_{\alpha(n+1)},x_{\alpha(n)})\le 2^{-n}$
for every $n\in\Bbb N$\};}

\noindent then $F\subseteq\NN$ is closed.   Define $f:F\to X$ by saying
that $f(\alpha)=\lim_{n\to\infty}x_{\alpha(n)}$ for every $\alpha\in\NN$.
If $\alpha$, $\beta\in\NN$ and $\alpha\restr n=\beta\restr n$ where
$n\ge 1$, then $\rho(f(\alpha),f(\beta))\le 2^{-n+2}$, so $f$ is
continuous.   If $x\in X$, we can define $\alpha\in\NN$ by saying that
$\alpha(n)$ is to be the least $i$ such that $\rho(x,x_i)\le 2^{-n-1}$;
then $\rho(x_{\alpha(n)},x_{\alpha(n+1)})\le 2^{-n-1}+2^{-n-2}\le 2^{-n}$
for every $n$, so $\alpha\in F$, and of course $f(\alpha)=x$.   So $f$ is
surjective.

The next thing we need is a choice function for the set $\Cal F$ of
non-empty closed subsets of $\NN$;  I described one in 561D.   Let
$g:\Cal F\to\NN$ be such that $g(F)\in F$ for every $F\in\Cal F$.

\medskip

\quad{\bf (iii)}
There is a family $\family{T}{\Cal T}{(F_T,f_T,F'_T,f'_T)}$
such that

\Centerline{$F_T$, $F'_T$ are closed subsets of $\NN$,}

\Centerline{$f_T:F_T\to\phi(T)$, $f'_T:F'_T\to X\setminus\phi(T)$ are
continuous surjections}

\noindent for each $T\in\Cal T$.   \Prf\ Start by fixing a homeomorphism
$\alpha\mapsto\sequence{i}{h_i(\alpha)}:\NN\to(\NN)^{\Bbb N}$.
Define the quadruples
$(F_T,f_T,F'_T,f'_T)$ inductively on the rank of $T$.

If $r(T)\le 1$ then $\phi(T)$ is open.   Set

\Centerline{$F_T
=\{\alpha:\alpha\in F$, $x_{\alpha(0)}\in\phi(T)$,
$\rho(x_{\alpha(n)},x_{\alpha(0)})
\le\Bover12\rho(x_{\alpha(0)},X\setminus\phi(T))$ for every $n\ge 1\}$}

\noindent (interpreting $\rho(x,\emptyset)$ as $\infty$ if necessary), and
$f_T=f\restr F_T$.   Then $F_T$ is a closed subset of $\NN$ and
$\rho(f(\alpha),x_{\alpha(0)})
\le\Bover12\rho(x_{\alpha(0)},X\setminus\phi(T))$, so
$f(\alpha)\in\phi(T)$, for every $\alpha\in F_T$.   If $x\in\phi(T)$ then
we can define $\alpha\in\NN$ by taking

\Centerline{$\alpha(n)
=\min\{i:\rho(x_i,x)\le\min(2^{-n-1},\Bover15\rho(x,X\setminus\phi(T)))\}$}

\noindent for every $n$, and now we find that $\alpha\in F_T$ and
$f_T(\alpha)=x$.
As for $F'_T$ and $f'_T$, just set
$F'_T=f^{-1}[X\setminus\phi(T)]$ and $f'_T=f\restr F'_T$.

For the inductive step to $r(T)>1$, set
$A_T=\{i:\fraction{i}\in T\}$, as in 562Ad.
We have $\phi(T)=\bigcup_{i\in A_T}X\setminus\phi(T_{\fraction{i}})$ and
$X\setminus\phi(T)=\bigcap_{i\in A_T}\phi(T_{\fraction{i}})$, while
$r(T_{\fraction{i}})<r(T)$ for every $i\in A_T$.   Set

\Centerline{$F_T=\bigcup_{i\in A_T}
\{\fraction{i}^{\smallfrown}\alpha:\alpha\in F'_{T_{\fraction{i}}}\}$,}

$$\eqalign{F'_T
&=\{\alpha:h_i(\alpha)\in F_{T_{\fraction{i}}}
  \text{ for every }i\in A_T,\cr
&\mskip100mu
f_{T_{\fraction{i}}}(h_i(\alpha))=f_{T_{\fraction{j}}}(h_j(\alpha))
  \text{ for all }i,\,j\in A_T\},\cr}$$

\Centerline{$f_T(\fraction{i}^{\smallfrown}\alpha)
=f'_{T_{\fraction{i}}}(\alpha)$ whenever
$i\in A_T$ and $\alpha\in F'_{T_{\fraction{i}}}$,}

\Centerline{$f'_T(\alpha)=f_{T_{\fraction{i}}}(h_i(\alpha))$ whenever
$i\in A_T$ and $\alpha\in F'_T$.}

\noindent It is straightforward to confirm that $F_T$ and $F'_T$ are
closed, $f_T:F_T\to\phi(T)$ and $f'_T:F'_T\to X\setminus\phi(T)$
are continuous and $f_T[F_T]=\phi(T)$.  To see that
$f'_T[F'_T]=X\setminus\phi(T)$, take any
$x\in X\setminus\phi(T)=\bigcap_{i\in A_T}\phi(T_{\fraction{i}})$.
Then $f_{T_{\fraction{i}}}^{-1}[\{x\}]$ is a non-empty closed subset
of $F_{T_{\fraction{i}}}$ for each $i\in A_T$.
Set $\alpha_i=g(f_{T_{\fraction{i}}}^{-1}[\{x\}])$, so that
$f_{T_{\fraction{i}}}(\alpha_i)=x$ for each $i\in A_T$.   For
$i\in\Bbb N\setminus A_T$, take $\alpha_i=\tbf{0}$.   Now
$\alpha=h^{-1}(\sequence{i}{\alpha_i})$ belongs to $F'_T$ and
$f'_T(\alpha)=x$.   Thus $f'_T[F'_T]=X\setminus\phi(T)$ and the induction
continues.\ \Qed

\medskip

\quad{\bf (iv)}
In particular, $\phi(T)=f_T[F_T]$ is a continuous image of a
closed subset of $\NN$ for every $T\in\Cal T$.

\medskip

\quad{\bf (v)} The definition of `analytic set' in 423A refers to
continuous images of $\NN$, so there is a final step to make.
If $E\subseteq X$ is a non-empty codable Borel set, it is a continuous
image of a closed subset $F_T$ of $\NN$;  but 561C tells us that $F_T$ is a
continuous image of $\NN$, so $E$ also is, and $E$ is analytic.
}%end of proof of 562F

\leader{562G}{Resolvable \dvrocolon{sets}}\cmmnt{ The essence of the
concept of `codable Borel set' is that it is not enough to know, in the
abstract, that a set is `Borel';  we need to know its pedigree.   For a
significant number of elementary sets, however,
starting with open sets and closed sets, we can determine codes from the
sets themselves.

\medskip

\noindent}{\bf Definition}\cmmnt{ (see {\smc Kuratowski 66}, \S12)}
I will say that a subset $E$ of a
topological space $X$ is {\bf resolvable} if there is no non-empty
set $A\subseteq X$ such that
$A\subseteq\overline{A\cap E}\cap\overline{A\setminus E}$.

\leader{562H}{Proposition} Let $X$ be a topological space, and $\Cal E$ the
set of resolvable subsets of $X$.   Then $\Cal E$ is an algebra of sets
containing every open subset of $X$.

\proof{{\bf (a)} If $G\subseteq X$ is open and $A\subseteq X$ is
non-empty, then either $A$ meets $G$ and
$A\not\subseteq\overline{A\setminus G}$, or $A\cap G=\emptyset$ then
$A\not\subseteq\overline{A\cap G}$.   So every open set is resolvable.

\medskip

{\bf (b)}
If $E$ is resolvable and $A\subseteq X$ is not empty, there is an open set
$G$ such that $A\cap G$ is not empty but one of $A\cap G\cap E$,
$A\cap G\setminus E$ is empty.   \Prf\ If
$A\not\subseteq\overline{A\cap E}$ take
$G=X\setminus\overline{A\cap E}$;  otherwise
take $G=X\setminus\overline{A\setminus E}$.\ \Qed

\medskip

{\bf (c)} If $E$, $E'\subseteq X$ are resolvable, so is $E\cup E'$.
\Prf\ Suppose that $A\subseteq X$ is non-empty.   Then there is an open set
$G$ such that $A\cap G$ is non-empty and disjoint from one of $E$,
$X\setminus E$.   Now there is an open set $H$ such that $A\cap G\cap H$ is
non-empty and disjoint from one of $E'$, $X\setminus E'$.   Consequently
one of $A\cap G\cap H\cap(E\cup E')$, $A\cap G\cap H\setminus(E\cup E')$ is
empty, and the open set $G\cap H$ is disjoint from
$\overline{A\cap(E\cup E')}\cap\overline{A\setminus(E\cup E')}$;  in which
case $A$ cannot be included in
$\overline{A\cap(E\cup E')}\cap\overline{A\setminus(E\cup E')}$.   As
$A$ is arbitrary, $E\cup E'$ is resolvable.\ \Qed

\medskip

{\bf (d)} Immediately from the definition in 562G we see that
the complement of a resolvable set is resolvable, so $\Cal E$ is
an algebra of subsets of $X$.
}

\leader{562I}{Theorem} Let $X$ be a second-countable space,
$\sequencen{U_n}$ a sequence running over a base for the topology of $X$,
and $\phi:\Cal T\to\Cal B_c(X)$ the associated
interpretation of Borel codes.
Let $\Cal E$ be the algebra of resolvable subsets of $X$.
Then there is a function $\psi:\Cal E\to\Cal T$ such that
$\phi(\psi(E))=E$ for every $E\in\Cal E$.

\proof{ We need to start by settling on functions

\Centerline{$\Theta'_1:\Cal T\times\Bbb N\to\Cal T$,
\quad$\Theta'_2:\Cal T\times\Cal T\times\Bbb N\to\Cal T$,
\quad$\tilde\Theta'_1:\Cal T^{\Bbb N}\to\Cal T$,
\quad$\tilde\Theta'_2:\Cal T^{\Bbb N}\to\Cal T$}

\noindent such that

\Centerline{$\phi(\Theta'_1(T,n))=\phi(T)\setminus U_n$,
\quad$\phi(\Theta'_2(T,T',n))=\phi(T)\cup(\phi(T')\cap U_n)$,}

\Centerline{$\phi(\tilde\Theta'_1(\tau))=\bigcup_{i\in\Bbb N}\phi(\tau(i))$,
\quad$\phi(\tilde\Theta'_2(\tau))=\bigcap_{i\in\Bbb N}\phi(\tau(i))$}

\noindent for $T\in\Cal T$, $n\in\Bbb N$ and $\tau\in\Cal T^{\Bbb N}$.
(We can take $\tilde\Theta_1$ and $\tilde\Theta_2$ directly from
562Cb, and

\Centerline{$\Theta'_1(T,n)=\Theta_3(T,\{\fraction{n}\}),
\Theta'_2(T,T',n)=\Theta_1(T,\Theta_2(T,\{\fraction{n}\}))$}

\noindent where $\Theta_1$, $\Theta_2$ and $\Theta_3$ are the functions of
562Ca.)

Now, given $E\in\Cal E$, define
$\ofamily{\xi}{\omega_1}
{(F_{\xi},\tilde T^{(\xi)},T^{(\xi)},n_{\xi})}$ inductively,
as follows.   The inductive
hypothesis will be that
$F_{\xi}\subseteq X$ is closed, $F_{\xi}\subseteq F_{\eta}$ for every
$\eta\le\xi$,
$\tilde T^{(\xi)}$, $T^{(\xi)}\in\Cal T$,
$\phi(\tilde T^{(\xi)})=F_{\xi}$ and
$\phi(T^{(\xi)})=E\setminus F_{\xi}$.
Start with $F_0=X$, $T^{(0)}=\emptyset$,
$\tilde T^{(0)}=\{\fraction{n}:n\in\Bbb N\}$.   For the
inductive step to $\xi+1$,

\inset{----- if $F_{\xi}=\emptyset$, set $n_{\xi}=0$ and
$(F_{\xi+1},\tilde T^{(\xi+1)},T^{(\xi+1)})
=(F_{\xi},\tilde T^{(\xi)},T^{(\xi)})$;

----- if there is an $n$ such that
$\emptyset\ne F_{\xi}\cap U_n\subseteq E$, let $n_{\xi}$ be the least
such, and set

\Centerline{$F_{\xi+1}=F_{\xi}\setminus U_{n_{\xi}}$,
\quad$\tilde T^{(\xi+1)}
=\Theta'_1(\tilde T^{(\xi)},n_{\xi})$,
\quad$T^{(\xi+1)}=\Theta'_2(T^{(\xi)},\tilde T^{(\xi)},n_{\xi})$;}

----- otherwise, $n_{\xi}$ is to be the least $n$ such that
$\emptyset\ne F_{\xi}\cap U_n\subseteq X\setminus E$, and

\Centerline{$F_{\xi+1}=F_{\xi}\setminus U_{n_{\xi}}$,
\quad$\tilde T^{(\xi+1)}
=\Theta'_1(\tilde T^{(\xi)},n_{\xi})$
\quad$T^{(\xi+1)}=T^{(\xi)}$.}
}%end of inset

\noindent (Because $E$ is resolvable, these three cases exhaust the
possibilities.)   It is easy to check that the inductive hypothesis
remains valid at level $\xi+1$.

For the inductive step to a non-zero
limit ordinal $\xi$, then if there is an $\eta<\xi$ such that
$F_{\eta}=\emptyset$, take the first such $\eta$ and set $n_{\xi}=0$ and
$(F_{\xi},T^{(\xi)},\tilde T^{(\xi)})
=(F_{\eta},T^{(\eta)},\tilde T^{(\eta)})$.   Otherwise,
we must have
$F_{\zeta}\subseteq F_{\eta}\setminus U_{n_{\eta}}\subset F_{\eta}$
whenever $\eta<\zeta<\xi$, so that $\eta\mapsto n_{\eta}:\xi\to\Bbb N$ is
injective.   Set

$$\eqalign{\tilde\tau(i)&=\tilde T^{(\eta)}
  \text{ if }\eta<\xi\text{ and }i=n_{\eta},\cr
&=\tilde T^{(0)}\text{ if there is no such }\eta,\cr
\tau(i)&=T^{(\eta)}\text{ if }\eta<\xi\text{ and }i=n_{\eta},\cr
&=\emptyset\text{ if there is no such }\eta;\cr}$$

\noindent now set

\Centerline{$F_{\xi}=\bigcap_{\eta<\xi}F_{\eta}$,
\quad$\tilde T^{(\xi)}=\tilde\Theta'_2(\tilde\tau)$,
\quad$T^{(\xi)}=\tilde\Theta'_1(\tau)$.}

\noindent Again, it is easy to check that the induction proceeds.

Now, with the family
$\ofamily{\xi}{\omega_1}
{(F_{\xi},\tilde T^{(\xi)},T^{(\xi)},n_{\xi})}$ complete, observe that
$\ofamily{\xi}{\omega_1}{n_{\xi}}$ cannot be injective.   There is
therefore a first $\xi=\xi_E$ for which $F_{\xi_E}$ is empty.   Set
$\psi(E)=T^{(\xi_E)}$;  then $\phi(\psi(E))=E\setminus F_{\xi_E}=E$, as
required.
}%end of proof of 562I

\leader{562J}{Codable families of sets}
Let $X$ be a second-countable space
and $\Cal B_c(X)$ the algebra of codable Borel subsets of $X$.   Let
$\sequencen{U_n}$, $\sequencen{V_n}$ be sequences running over bases for
the topology of $X$, and
$\phi:\Cal T\to\Cal B_c(X)$,
$\phi':\Cal T\to\Cal B_c(X)$ the corresponding interpretations of Borel
codes.   Let us say that
a family $\familyiI{E_i}$ is {\bf$\phi$-codable} if there is a family
$\familyiI{T^{(i)}}$ in $\Cal T$ such that $\phi(T^{(i)})=E_i$ for every
$i\in I$.   Then\cmmnt{ 562Bc tells us that} $\familyiI{E_i}$ is
$\phi$-codable iff it is $\phi'$-codable.

We may therefore say that a family $\familyiI{E_i}$ in $\Cal B_c(X)$ is
{\bf codable} if it is $\phi$-codable for some\cmmnt{, therefore any,}
interpretation of Borel codes defined by the procedure of 562B from a
sequence running over a base for the topology of $X$.

Note that any finite family in $\Cal B_c(X)$ is codable, and that any
family of resolvable sets is codable\cmmnt{, because
we can use 562I to provide codes};  also any subfamily of a codable family
is codable.   Slightly more generally, if
$\familyiI{E_i}$ is a codable family in $\Cal B_c(X)$, $J$ is a set, and
$f:J\to I$ is a function, then $\family{j}{J}{E_{f(j)}}$ is codable.
If $\familyiI{E_i}$ and
$\familyiI{F_i}$ are codable families in $\Cal B_c(X)$, then so are
$\familyiI{X\setminus E_i}$,
$\familyiI{E_i\cup F_i}$, $\familyiI{E_i\cap F_i}$ and
$\familyiI{E_i\setminus F_i}$\prooflet{,
since we have formulae to transform codes for $E$, $F$ into codes for
$X\setminus E$, $E\cup F$, $E\cap F$ and $E\setminus F$}.

\leader{562K}{Proposition} Let $X$ be a second-countable space and
$\sequencen{E_n}$ a codable sequence in $\Cal B_c(X)$.   Then

(a) $\bigcup_{n\in\Bbb N}E_n$, $\bigcap_{n\in\Bbb N}E_n$ belong to
$\Cal B_c(X)$;

(b) $\sequencen{\bigcup_{i<n}E_i}$ is a codable family in $\Cal B_c(X)$;

(c) $\sequencen{E_n\setminus\bigcup_{i<n}E_i}$ is a codable family in
$\Cal B_c(X)$.
%for 563B

\proof{ Let $\phi:\Cal T\to\Cal B_c(X)$ be an interpretation
of Borel codes defined from a sequence running over a base for the topology
of $X$;  then we have a sequence $\sequencen{T^{(n)}}$ in $\Cal T$
such that $\phi(T^{(n)})=E_n$ for every $n$, and using 562Bb we
can arrange that $T^{(n)}\ne\emptyset$ for every $n$.

\medskip

{\bf (a)} Setting

\Centerline{$T=\{\fraction{n}:n\in\Bbb N\}
\cup\{\fraction{n}^{\smallfrown}\fraction{0}:n\in\Bbb N\}
\cup\{\fraction{n}^{\smallfrown}\fraction{0}^{\smallfrown}\sigma:
n\in\Bbb N$, $\sigma\in T^{(n)}\}$,}

\Centerline{$T'=\{\fraction{0}\}
\cup\{\fraction{0}^{\smallfrown}\fraction{n}:n\in\Bbb N\}
\cup\{\fraction{0}^{\smallfrown}\fraction{n}^{\smallfrown}\sigma:
n\in\Bbb N$, $\sigma\in T^{(n)}\}$,}

\noindent we have $\phi(T)=\bigcup_{n\in\Bbb N}E_n$ and
$\phi(T')=\bigcap_{n\in\Bbb N}E_n$.

\medskip

{\bf (b)} Setting

\Centerline{$\hat T^{(n)}=\{\fraction{i}:i<n\}
\cup\{\fraction{i}^{\smallfrown}\fraction{0}:i<n\}
\cup\{\fraction{i}^{\smallfrown}\fraction{0}^{\smallfrown}\sigma:
i<n$, $\sigma\in T^{(i)}\}$,}

\noindent $\phi(\hat T^{(n)})=\bigcup_{i<n}E_i$ for every $n$.

\medskip

{\bf (c)} Setting

$$\eqalign{T''
&=\{\fraction{n}:n\in\Bbb N\}
  \cup\{\fraction{n}^{\smallfrown}\fraction{0}:n\in\Bbb N\}
\cup\{\fraction{n}^{\smallfrown}\fraction{0}^{\smallfrown}\sigma:
   \sigma\in T^{(n)}\}\cr
&\mskip100mu
  \cup\{\fraction{n}^{\smallfrown}\fraction{1}:n\in\Bbb N\}
  \cup\{\fraction{n}^{\smallfrown}\fraction{1}^{\smallfrown}\fraction{0}:
    n\in\Bbb N\}\cr
&\mskip100mu
\cup\{\fraction{n}^{\smallfrown}\fraction{1}^{\smallfrown}
  \fraction{0}^{\smallfrown}\sigma:
  \sigma\in \hat T^{(n)}\},\cr}$$

\noindent $\phi(T'')=\bigcup_{n\in\Bbb N}(E_n\setminus\bigcup_{i<n}E_i)$.
}%end of proof of 562K

\leader{562L}{Codable Borel functions} Let $X$ and
$Y$ be second-countable
spaces.   A function $f:X\to Y$ is a {\bf codable Borel function} if
$\langle f^{-1}[H]\rangle_{H\subseteq Y\text {is open}}$
is a codable family in $\Cal B_c(X)$.

\leader{562M}{Theorem} Let $X$ be a second-countable space,
$\sequencen{U_n}$ a sequence running over a base for the topology of $X$,
and $\phi:\Cal T\to\Cal B_c(X)$ the corresponding interpretation of Borel
codes.

(a) If $Y$ is another second-countable space, $\sequencen{V_n}$ a
sequence running over a base for the topology of $Y$ containing
$\emptyset$, $\phi_Y:\Cal T\to\Cal B_c(Y)$ the corresponding interpretation
of Borel codes, and
$f:X\to Y$ is a function, then the following are equiveridical:

\quad(i) $f$ is a codable Borel function;

\quad(ii) $\sequencen{f^{-1}[V_n]}$ is a codable sequence in $\Cal B_c(X)$;

\quad(iii) there is a function $\Theta:\Cal T\to\Cal T$ such that
$\phi(\Theta(T))=f^{-1}[\phi_Y(T)]$ for every $T\in\Cal T$.

(b) If $Y$ and $Z$ are second-countable spaces and $f:X\to Y$, $g:Y\to Z$
are codable Borel functions then $gf:X\to Z$ is a codable Borel function.

(c) If $Y$ and $Z$ are second-countable spaces and $f:X\to Y$, $g:X\to Z$
are codable Borel functions then $x\mapsto(f(x),g(x))$ is a codable Borel
function from $X$ to $Y\times Z$.

(d) If $Y$ is a second-countable space then any continuous function
from $X$ to $Y$ is a codable Borel function.

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

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} This is really a full-strength version
of 562Bc.    Because $\sequencen{f^{-1}[V_n]}$ is codable, we have a
sequence $\sequencen{T^{(n)}}$ in $\Cal T$ such that
$\phi(T^{(n)})=f^{-1}[V_n]$ for every $n$.
As in 562C, let $\Theta_0:\Cal T\to\Cal T$
and $\tilde\theta_1:\bigcup_{I\subseteq\Bbb N}\Cal T^I\to\Cal T$
be such that
$\phi(\Theta_0(T))=X\setminus\phi(T)$ for every $T\in\Cal T$ and
$\phi(\tilde\Theta_1(\tau))=\bigcup_{i\in I}\phi(\tau(i))$ whenever
$I\subseteq\Bbb N$ and $\tau\in\Cal T^I$.
Define $\Theta:\Cal T\to\Cal T$
inductively, as follows.   Given $T\in\Cal T$,
set $A_T=\{n:\fraction{n}\in T\}$.
If $r(T)=0$ set $\Theta(T)=T=\emptyset$.   If $r(T)=1$ set
$\Theta(T)=\tilde\Theta_1(\family{n}{A_T}{T^{(n)}})$, so that

\Centerline{$\phi(\Theta(T))=\bigcup_{n\in A_T}\phi(T^{(n)})
=\bigcup_{n\in A_T}f^{-1}[V_n]=f^{-1}[\phi_Y(T)]$.}

\noindent If $r(T)>1$ set

\Centerline{$\Theta(T)=\tilde\Theta_1(\family{n}{A_T}
   {\Theta_0(\Theta(T_{\fraction{n}}))})$}

\noindent so that

$$\eqalign{\phi(\Theta(T))
&=\bigcup_{n\in A_T}X\setminus\phi(\Theta(T_{\fraction{n}}))
=\bigcup_{n\in A_T}X\setminus f^{-1}[\phi_Y(T_{\fraction{n}})]\cr
&=f^{-1}[\bigcup_{n\in A_T}Y\setminus\phi_Y(T_{\fraction{n}})]
=f^{-1}[\phi_Y(T)]\cr}$$

\noindent and the induction continues.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} For open $H\subseteq Y$ set
$\psi_Y(H)=\{\fraction{n}:V_n\subseteq H\}$.   Taking $\Theta$ as above,

\Centerline{$\phi(\Theta(\psi_Y(H))=f^{-1}[\phi_Y(\psi_Y(H))]=f^{-1}[H]$}

\noindent for every $H$, so
$\langle\phi(\Theta(\psi_Y(H)))\rangle_{H\subseteq Y\text{ is open}}$ is a
family of codes for
$\langle f^{-1}[H]\rangle_{H\subseteq Y\text {is open}}$.

\medskip

{\bf (b)} Take $\sequencen{V_n}$, $\phi_Y$ and $\Theta:\Cal T\to\Cal T$
as in (a).   Write $\frak U$ for the
topology of $Z$;  then we have a function $\theta:\frak U\to\Cal T$ such
that $\phi_Y(\theta(H))=g^{-1}[H]$ for every $H\in\frak U$.
Now $\family{H}{\frak U}{\Theta(\theta(H))}$ is a coding for
$\family{H}{\frak U}{(gf)^{-1}[H]}$, so $gf$ is codable.

\medskip

{\bf (c)} Let $\sequencen{V_n}$, $\sequencen{W_n}$ be sequences running
over bases for the topologies of $Y$ and $Z$, and
$\sequencen{(i_n,j_n)}$ an enumeration of $\Bbb N\times\Bbb N$.
Set $H_n=V_{i_n}\times W_{j_n}$;  then $\sequencen{H_n}$ is a base for the
topology of $Y\times Z$.
Let $\Theta_2:\Cal T\times\Cal T\to\Cal T$ be such that
$\phi(\Theta_2(T,T'))=\phi(T)\cap\phi(T')$ for all $T$, $T'\in\Cal T$
(562Ca).   Let $\sequencen{T^{(n)}}$, $\sequencen{\hat T^{(n)}}$ be codings for
$\sequencen{f^{-1}[V_n]}$, $\sequencen{g^{-1}[W_n]}$.   Then
$\sequencen{\Theta_2(T^{(i_n)},\hat T^{(j_n)})}$ is a coding for
$\sequencen{h^{-1}[H_n]}$, where $h(x)=(f(x),g(x))$ for $x\in X$.
So $h$ is a codable Borel function.

\medskip

{\bf (d)} If $f:X\to Y$ is continuous, then
$\langle f^{-1}[H]\rangle_{H\subseteq Y\text{ is open}}$ is a family of
resolvable sets, therefore codable, as noted in 562J.
}%end of proof of 562M

\cmmnt{\medskip

\noindent{\bf Remark} Note in part
(a)(ii)$\Rightarrow$(iii) of the proof the function $\Theta$ is
constructed by a definite process from $\sequencen{T^{(n)}}$;  so we shall be
able to uniformize the process to define families $\familyiI{\Theta_i}$
from families $\familyiI{f_i}$, at least if we can reach a family
$\langle T^{(i,n)}\rangle_{i\in I,n\in\Bbb N}$ such that
$T^{(n,i)}$ codes $f_i^{-1}[V_n]$ for all $i\in I$ and $n\in\Bbb N$.
}

\leader{562N}{Proposition} Let $X$ be a second-countable space,
and $\phi:\Cal T\to\Cal B_c(X)$ the interpretation of Borel codes
associated with some sequence running over a base for the topology of $X$.

(a) If $f:X\to\Bbb R$ is a function, the following are equiveridical:

\quad(i) $f$ is a codable Borel function;

\quad(ii) the family
$\family{\alpha}{\Bbb R}{\{x:f(x)>\alpha\}}$ is codable;

\quad(iii)
$\family{q}{\Bbb Q}{\{x:f(x)>q\}}$ is codable.

(b) Write $\tilde{\Cal T}$ for the set of
functions $\tau:\Bbb R\to\Cal T$ such that

\Centerline{$\phi(\tau(\alpha))
=\bigcup_{\beta>\alpha}\phi(\tau(\beta))$ for every $\alpha\in\Bbb R$,}

\Centerline{$\bigcap_{n\in\Bbb N}\phi(\tau(n))=\emptyset$,
\quad$\bigcup_{n\in\Bbb N}\phi(\tau(-n))=X$.}

\noindent Then

\quad(i) for every $\tau\in\tilde{\Cal T}$ there is a unique codable Borel
function $\tilde\phi(\tau):X\to\Bbb R$ such that
$\phi(\tau(\alpha))=\{x:\tilde\phi(\tau)(x)>\alpha\}$ for every
$\alpha\in\Bbb R$;

\quad(ii) every codable Borel function from $X$ to $\Bbb R$ is expressible
as $\tilde\phi(\tau)$ for some $\tau\in\tilde{\Cal T}$.

(c) If $\sequencen{\tau_n}$ is a sequence in $\tilde{\Cal T}$ such
that $f(x)=\sup_{n\in\Bbb N}\tilde\phi(\tau_n)(x)$ is finite for every
$x\in X$, then $f$ is a codable Borel function.

(d) If $f$, $g:X\to\Bbb R$ are codable Borel functions and
$\gamma\in\Bbb R$, then
$f+g$, $\gamma f$, $|f|$ and $f\times g$ are codable Borel functions.

(e) If $\sequencen{\tau_n}$ is a sequence in $\tilde{\Cal T}$, then there
is a codable Borel function $f$ such
that $\liminf_{n\to\infty}\tilde\phi(\tau_n)(x)=f(x)$ whenever the lim inf
is finite.

(f) A subset $E$ of $X$ belongs to $\Cal B_c(X)$ iff $\chi E:X\to\Bbb R$ is
a codable Borel function.

\proof{{\bf (a)(i)$\Rightarrow$(ii)}
If $f:X\to\Bbb R$ is codable then of course
$\family{\alpha}{\Bbb R}{\{x:f(x)>\alpha\}}$ is codable, because it is
a subfamily of
$\langle f^{-1}[H]\rangle_{H\subseteq\Bbb R\text{ is open}}$.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)} Similarly, if
$\family{\alpha}{\Bbb R}{\{x:f(x)>\alpha\}}$ is codable, its subfamily
$\family{q}{\Bbb Q}{\{x:f(x)>q\}}$ is codable.

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} If $\family{q}{\Bbb Q}{\{x:f(x)>q\}}$
is codable, we have a family
$\family{q}{\Bbb Q}{T^{(q)}}$ in $\Cal T$ coding it.
Let $\sequencen{(q_n,q'_n)}$ be an
enumeration of $\{(q,q'):q$, $q'\in\Bbb Q$, $q<q'\}$.
As in 562C, we have functions

\Centerline{$\Theta_3:\Cal T\times\Cal T\to\Cal T$,
\quad$\tilde\Theta_1:\bigcup_{I\subseteq\Bbb Q}\Cal T^I\to\Cal T$,
\quad$\tilde\Theta_2:\bigcup_{J\subseteq\Bbb N}\Cal T^J\to\Cal T$}

\noindent such that

\Centerline{$\phi(\Theta_3(T,T'))=\phi(T)\setminus\phi(T')$,
\quad$\phi(\tilde\Theta_1(\tau))=\bigcup_{q\in I}\phi(\tau(q)))$,}

\Centerline{$\phi(\tilde\Theta_2(\tau))
=X\cap\bigcap_{q\in I}\phi(\tau(j))$}

\noindent for $T$, $T'\in\Cal T$, $I\subseteq\Bbb Q$ and
$\tau\in\Cal T^I$.   Now for $n\in\Bbb N$ consider

\Centerline{$\hat T^{(n)}
=\tilde\Theta_1(\langle\Theta_3(T^{(q_n)},T^{(r)})\rangle_{r\in\Bbb Q,
  r<q'_n})$,}

\noindent so that
$\phi(\hat T^{(n)})
%=\bigcup_{q\in\Bbb Q,r<q'_n}\phi(T^{(q_n)},T^{(r)})
%=\bigcup_{q\in\Bbb Q,r<q'_n}\phi(T^{(q_n)})\setminus\phi(T^{(r)})
%=\bigcup_{q\in\Bbb Q,r<q'_n}\{x:q_n<f(x)\le r\}
%=\{x:q_n<f(x)<q'_n\}
=f^{-1}[\,\ooint{q_n,q'_n}\,]$ for every $n$, and
$\sequencen{f^{-1}[\,\ooint{q_n,q'_n}\,]}$ is codable;  by
562Ma, $f$ is a codable Borel function.

\medskip

{\bf (b)} This is elementary;  given $\tau\in\tilde{\Cal T}$ we can, and
must, set $\tilde\phi(\tau)(x)=\sup\{\alpha:x\in\phi(\tau(\alpha))\}$
for every $x\in X$;  and given $f$ we have a coding $\tau$ for
$\family{\alpha}{\Bbb R}{\{x:f(x)>\alpha\}}$ which must belong to
$\tilde{\Cal T}$ and be such that $\tilde\phi(\tau)=f$.

\medskip

{\bf (c)} Given $\sequencen{\tau_n}$ as described, and taking
$\tilde\Theta_1$ as in (a)(iii)$\Rightarrow$(i) above,

\Centerline{$\alpha\mapsto\tilde\Theta_1(\sequencen{\tau_n(\alpha)})$}

\noindent will be a Borel code for $f$.

\medskip

{\bf (d)} Use 562M(b)-(d).

\medskip

{\bf (e)} Let

\Centerline{$\Theta_0:\Cal T\to\Cal T$,
\quad$\Theta_1:\Cal T\times\Cal T\to\Cal T$}

\noindent be such that

\Centerline{$\phi(\Theta_0(T))=X\setminus\phi(T)$,
\quad$\phi(\Theta_1(T,T'))=\phi(T)\cup\phi(T')$}

\noindent for every $T$, $T'\in\Cal T$.
Now, given $\sequencen{\tau_n}$ as described, set

\Centerline{$\tau(\alpha)
=\tilde\Theta_1(\sequencen{\tilde\Theta_1(\langle\tilde\Theta_2(\langle
  \tau_m(q)\rangle_{m\ge n})\rangle_{q\in\Bbb Q,q>\alpha})})$}

\noindent for $\alpha\in\Bbb R$.   Then

\Centerline{$\phi(\tau(\alpha))
=\bigcup_{q>\alpha,n\in\Bbb N}\bigcap_{m\ge n}\{x:f_m(x)>q\}
=\{x:\liminf_{n\to\infty}f_n(x)>\alpha\}$}

\noindent for each $\alpha$.   We don't yet have a code for a
real-value function defined everywhere in $X$.   But if we set

\Centerline{$T=\tilde\Theta_1(\sequencen{\Theta_3(\tau(-n),\tau(n))})$,}

\noindent then

\Centerline{$\phi(T)
=\bigcup_{n\in\Bbb N}\phi(\tau(-n))\setminus\phi(\tau(n))
=\{x:\liminf_{n\to\infty}f_n(x)\text{ is finite}\}$.}

\noindent So take

$$\eqalign{\tau'(\alpha)
&=\Theta_3(\tau(\alpha),\Theta_0(T))\text{ if }\alpha\ge 0,\cr
&=\Theta_1(\tau(\alpha),\Theta_0(T))\text{ if }\alpha<0;\cr}$$

\noindent this will get $\tau'\in\tilde{\Cal T}$ such that

$$\eqalign{\tilde\phi(\tau')(x)
&=\liminf_{n\to\infty}f_n(x)\text{ if this is finite},\cr
&=0\text{ otherwise}.\cr}$$

\medskip

{\bf (f)} Elementary.
}%end of proof of 562N

\leader{562O}{Remarks (a)} For some purposes there are advantages
in coding real-valued functions by functions from $\Bbb Q$ to $\Cal T$
rather than by functions
from $\Bbb R$ to $\Cal T$\cmmnt{;  see 364Af and 556A}.

\spheader 562Ob As in 562C,
\cmmnt{it will be useful to observe that} the constructions here
are largely determinate.    For instance, the function $\Theta$ of
562M(a-iii) can be built by a definite rule from the sequence
$\sequencen{T^{(n)}}$ provided by the hypothesis (a-ii) there.
What this means is that if we have a family
$\familyiI{(Y_i,\sequencen{V_{in}},f_i)}$ such that $Y_i$ is a
second-countable space, $\sequencen{V_{in}}$ is a sequence running over a
base for the topology of $Y_i$, and $f_i:X\to Y_i$ is a continuous function
for each $i\in I$, then there will be a function
$\tilde\Theta:\Cal T\times I\to\Cal T$ such that
$\phi(\tilde\Theta(T,i))=f_i^{-1}[\phi_i(T)]$ for every $i\in I$ and
$T\in\Cal T$, where $\phi_i:\Cal T\to\Cal B_c(Y_i)$ is the interpretation
of Borel codes
corresponding to the sequence $\sequencen{V_{in}}$.   \cmmnt{(Start from

\Centerline{$T^{(i,n)}
=\{\fraction{j}:U_j\subseteq f_i^{-1}[V_{in}]\}$}

\noindent for $i\in I$ and $n\in\Bbb N$, and build $\tilde\Theta(T,i)$ as
562M.)}

\spheader 562Oc Similarly\cmmnt{, when we look at 562N(d)-(e),
we have something
better than just existence proofs for codes for $f+g$ and
$\liminf_{n\to\infty}f_n$.   For instance}, we have a function
$\tilde\Theta_1:\tilde{\Cal T}\times\tilde{\Cal T}\to\tilde{\Cal T}$
such that
$\tilde\phi(\tilde\Theta_1(\tau,\tau'))$ will always be
$\tilde\phi(\tau)-\tilde\phi(\tau')$
for $\tau$, $\tau'\in\tilde{\Cal T}$.
\prooflet{\Prf\ We need to have

\Centerline{$\phi(\tilde\Theta(\tau,\tau')(\alpha))
=\bigcup_{q\in\Bbb Q}\phi(\tau(q))\setminus\phi(\tau'(q-\alpha))$}

\noindent for every $\alpha$, and this is easy to build from a
set-difference operator, as in 562Ca,
and a general countable-union operator as built in 562Cb.\ \QeD}
Equally, we have a function
$\tilde\Theta_1^*:\tilde{\Cal T}^{\Bbb N}\to\tilde{\Cal T}^{\Bbb N}$ such
that

\Centerline{$\tilde\phi(\tilde\Theta_1^*(\sequencen{\tau_n})(m))
=\inf_{n\ge m}\tilde\phi(\tau_n)$}

\noindent for every $m$
whenever $\sequence{n}{\tau_n}$ is a sequence in $\tilde{\Cal T}$
such that $\inf_{n\in\Bbb N}\tilde\phi(\tau_n)$ is defined as a real-valued
function on $X$.   \prooflet{\Prf\ This time we need

\Centerline{$\phi(\tilde\Theta_1(\sequencen{\tau_n})(m)(\alpha))
=\bigcup_{q\in\Bbb Q,q>\alpha}
  \bigl(X\setminus\bigcup_{n\ge m}(X\setminus\phi(\tau_n(q)))\bigr)$}

\noindent for all $m$ and $\alpha$,
and once again a complementation operator and a general
countable-union operator will do the trick.
\Qed}

\leader{562P}{Codable Borel equivalence (a)} If $X$ is a set, we can say
that two second-countable topologies $\frak S$, $\frak T$ on $X$ are
{\bf codably Borel equivalent} if the identity functions
$(X,\frak S)\to(X,\frak T)$ and $(X,\frak T)\to(X,\frak S)$ are
codable Borel functions.   In this case, $\frak S$ and $\frak T$ give the
same families of codable Borel
functions and the same algebra $\Cal B_c(X)$\cmmnt{ (562Mb, 562Nf)}.

\spheader 562Pb If $(X,\frak T)$ is a second-countable space and
$\sequencen{E_n}$ is any codable sequence in $\Cal B_c(X)$, there is a
topology $\frak S$ on $X$, generated by a countable algebra of subsets of
$X$, such that
$\frak S$ and $\frak T$ are codably Borel equivalent and every $E_n$
belongs to $\frak S$.   \prooflet{\Prf\ Since there is certainly a codable
sequence running over a base for the topology of $X$, we can suppose that
such a sequence has been amalgamated with $\sequencen{E_n}$, so that
$\{E_n:n\in\Bbb N\}$ includes a base for $\frak T$.
Let $\Cal E$ be the algebra of subsets of $X$
generated by $\{E_n:n\in\Bbb N\}$ and
$\frak S$ the topology generated by $\Cal E$.
As $\Cal E$ is an algebra, $\frak S$ is zero-dimensional;  as
$\Cal E$ is countable, $\frak S$ is second-countable.

The identity map $(X,\frak S)\to(X,\frak T)$ is continuous, therefore a
codable Borel function (562Md).
In the reverse direction, we have a sequence
$\sequencen{T^{(n)}}$ of codes for $\sequencen{E_n}$.   From these we can
build, using our standard operations, codes $T_I$, for
$I\in[\Bbb N]^{<\omega}$, $T'_{IJ}$, for $I$, $J\in[\Bbb N]^{<\omega}$,
and $T''_{\Cal K}$, for
$\Cal K\in[[\Bbb N]^{<\omega}\times[\Bbb N]^{<\omega}]^{<\omega}$, such that

\inset{$T_I$ codes $\bigcup_{i\in I}E_i$,

$T'_{IJ}$ codes $\bigcup_{i\in I}E_i\setminus\bigcup_{i\in J}E_i$,

$T''_{\Cal K}$ codes
$\bigcup_{(I,J)\in\Cal K}
  (\bigcup_{i\in I}E_i\setminus\bigcup_{i\in J}E_i)$.}

\noindent But of course
$[[\Bbb N]^{<\omega}\times[\Bbb N]^{<\omega}]^{<\omega}$ is countable and
the $T''_{\Cal K}$ can be enumerated as a sequence $\sequencen{T^*_n}$
coding a sequence $\sequencen{V_n}$ running over $\Cal E$.
By 562Ma,
the identity map $(X,\frak T)\to(X,\frak S)$ is a codable Borel function.\
\Qed}

\cmmnt{Note that $\frak S$ here is necessarily regular;
this will be useful at
more than one point in the next couple of sections.}

\leader{562Q}{Resolvable functions} Let $X$ be a topological space.
I will say that a function $f:X\to[-\infty,\infty]$ is
{\bf resolvable} if whenever
$\alpha<\beta$ in $\Bbb R$ and $A\subseteq X$ is a non-empty set,
then\cmmnt{ at least} one of $\{x:x\in A$, $f(x)\le\alpha\}$,
$\{x:x\in A$, $f(x)\ge\beta\}$ is not dense in $A$.

\medskip

\noindent{\bf Examples (a)} Any semi-continuous
function from $X$ to $[-\infty,\infty]$ is resolvable.
\prooflet{\Prf\ If $f:X\to[-\infty,\infty]$ is
lower semi-continuous, $A\subseteq X$ is non-empty, and $\alpha<\beta$ in
$\Bbb R$, then $U=\{x:f(x)>\alpha\}$ is open;  if $A\cap U\ne\emptyset$
then $\{x:x\in A$, $f(x)\le\alpha\}$ is not dense in $A$;  otherwise
$\{x:x\in A$, $f(x)\ge\beta\}$ is not dense in $A$.\ \Qed}

\spheader 562Qb
If $f:X\to\Bbb R$ is such that $\{x:f(x)>\alpha\}$ is resolvable for
every $\alpha$, then $f$ is resolvable.   \prooflet{\Prf\ Suppose that
$A\subseteq X$ is non-empty and $\alpha<\beta$ in $\Bbb R$.   Set
$E=\{x:f(x)>\alpha\}$.   If $A\not\subseteq\overline{A\cap E}$, then
$\{x:x\in A$, $f(x)\ge\beta\}$ is not dense in $A$.
Otherwise $\{x:x\in A$, $f(x)\le\alpha\}=A\setminus E$ is not dense
in $A$.\ \Qed

}In particular, the indicator
function of a resolvable set is resolvable.

\spheader 562Qc
A function $f:\Bbb R\to\Bbb R$ which has bounded variation on every
bounded set is resolvable.  \prooflet{\Prf\ If $A\subseteq\Bbb R$ is
non-empty and $\alpha<\beta$ in $\Bbb R$, take $y\in A$.   If $y$ is
isolated in $A$, then we have an open set $U$ such that $U\cap A=\{y\}$,
so that one of $\{x:x\in A$, $f(x)\le\alpha\}$,
$\{x:x\in A$, $f(x)\ge\beta\}$ does not contain $y$ and is
not dense in $A$.   Otherwise, $y$ is in the
closure of one of $A\cap\ooint{y,\infty}$, $A\cap\ooint{-\infty,y}$;
suppose the former.   For each $n\in\Bbb N$ set
$I_n=[y+2^{-n-1},y+2^{-n}]$,
$\delta_n=\Var_{I_n}(f)$.   We have

\Centerline{$\infty>\Var_{\ocint{y,y+1}}(f)
=\sum_{n=0}^{\infty}\delta_n$,}

\noindent so there is an $n\in\Bbb N$ such that
$\delta_m\le\bover14(\beta-\alpha)$ for $m\ge n$.   Take $m>n$ such that
$I_m\cap A\ne\emptyset$, and consider
$U=\interior(I_{m-1}\cup I_m\cup I_{m+1})$.   Then
$\Var_U(f)\le\bover34(\beta-\alpha)$ so $U$ cannot meet both
$\{x:x\in A$, $f(x)\ge\beta\}$ and $\{x:x\in A$, $f(x)\ge\alpha\}$, and one
of these is not dense in $A$.\ \Qed}

\leader{562R}{Theorem} Let $X$ be a second-countable space,
$\sequencen{U_n}$ a sequence running over a base for the topology of $X$,
and $\phi:\Cal T\to\Cal B_c(X)$ the associated
interpretation of Borel codes.   Let $\Cal R$
be the family of resolvable real-valued functions on $X$.   Then
there is a function $\tilde\psi:\Cal R\to\Cal T^{\Bbb R}$ such that

\Centerline{$\phi(\tilde\psi(f)(\alpha))=\{x:f(x)>\alpha\}$}

\noindent for every $f\in\Cal R$ and $\alpha\in\Bbb R$.

\proof{{\bf (a)} Start by fixing a bijection

\Centerline{$k\mapsto(n_k,q_k,q'_k):
\Bbb N\to\Bbb N\times\{(q,q'):q$, $q'\in\Bbb Q$, $q<q'\}$.}

\noindent Next, fix a function $\Theta^*_1:\Cal T^3\times\Bbb N\to\Cal T$
such that

\Centerline{$\phi(\Theta^*_1(T,T',T'',n))
=\phi(T)\cup(U_n\setminus(\phi(T')\cup\phi(T'')))$}

\noindent for $T$, $T'$, $T''\in\Cal T$ and $n\in\Bbb N$, and a function
$\tilde\Theta^*_1:\bigcup_{J\subseteq\Bbb N}\Cal T^J\to\Cal T$ such that
$\phi(\tilde\Theta^*_1(\tau))=\bigcup_{i\in J}\phi(\tau(i))$ whenever
$J\subseteq\Bbb N$ and $\tau\in\Cal T^J$.   (See 562Ca.)

\medskip

{\bf (b)} Given $f\in\Cal R$, define $\zeta<\omega_1$ and a family
$\langle(\tau_{\xi},\tau'_{\xi},k_{\xi})\rangle_{\xi\le\zeta}$ in
$\Cal T^{\Bbb R}\times\Cal T^{\Bbb R}\times\Bbb N$
inductively, as follows.   The inductive hypothesis will be that
$k_{\eta}\ne k_{\xi}$ whenever $\eta<\xi<\zeta$.   Start with
$\tau_0(\alpha)=\tau_0'(\alpha)=\emptyset$ for every $\alpha\in\Bbb R$.

\medskip

{\it Inductive step to a successor ordinal $\xi+1$} Given $\tau_{\xi}$ and
$\tau'_{\xi}$ in $\Cal T^{\Bbb R}$, then for $q<q'$ in $\Bbb Q$
set
$F_{\xi}(q,q')=X\setminus(\phi(\tau_{\xi}(q))\cup\phi(\tau'_{\xi}(q')))$.
Now

\inset{----- if there is a $k\in\Bbb N\setminus\{k_{\eta}:\eta<\xi\}$
such that
$U_{n_k}\cap F_{\xi}(q_k,q'_k)\ne\emptyset$ and $f(x)\ge q_k$ for every
$x\in U_{n_k}\cap F_{\xi}(q_k,q'_k)$, take the first such $k$, and set

$$\eqalign{\tau_{\xi+1}(\alpha)&=\tau_{\xi}(\alpha)
   \text{ for every }\alpha\in\Bbb R,\cr
\tau'_{\xi+1}(\alpha)
&=\Theta^*_1(\tau'_{\xi}(\alpha),\tau_{\xi}(q_k),\tau'_{\xi}(q'_k),n_k)
  \text{ if }\alpha\le q_k,\cr
&=\tau'_{\xi}(\alpha)\text{ if }\alpha>q_k,\cr
k_{\xi}&=k;\cr}$$

----- if this is not so, but there is a
$k\in\Bbb N\setminus\{k_{\eta}:\eta<\xi\}$ such that
$U_{n_k}\cap F_{\xi}(q_k,q'_k)\ne\emptyset$ and $f(x)\le q'_k$ for every
$x\in U_{n_k}\cap F_{\xi}(q_k,q'_k)$, take the first such $k$, and set

$$\eqalign{\tau_{\xi+1}(\alpha)
&=\Theta^*_1(\tau_{\xi}(\alpha),\tau_{\xi}(q_k),\tau'_{\xi}(q'_k),n_k)
  \text{ if }\alpha\ge q'_k,\cr
&=\tau_{\xi}(\alpha)\text{ if }\alpha<q'_k,\cr
\tau'_{\xi+1}(\alpha)
&=\tau'_{\xi}(\alpha)\text{ for every }\alpha\in\Bbb R,\cr
k_{\xi}&=k;\cr}$$

----- and if that doesn't happen either, set $\zeta=\xi$
and stop.}

\medskip

{\it Inductive step to a countable limit ordinal $\xi$} Given
$\ofamily{\eta}{\xi}{(\tau_{\eta},\tau'_{\eta},k_{\eta})}$, set
$I=\{k_{\eta}:\eta<\xi\}$ and define $g:I\to\xi$ by setting
$g(i)=\eta$ whenever $\eta<\xi$ and $k_{\eta}=i$.
Now set

\Centerline{$\tau_{\xi}(\alpha)
=\tilde\Theta^*_1(\familyiI{\tau_{g(i)}(\alpha)})$,
\quad$\tau'_{\xi}(\alpha)
=\tilde\Theta^*_1(\familyiI{\tau'_{g(i)}(\alpha)})$}

\noindent for every $\alpha\in\Bbb R$.

\medskip

{\bf (c)} Now an induction on $\xi$ shows that

\Centerline{$\phi(\tau_{\eta}(\alpha))\subseteq\phi(\tau_{\xi}(\alpha))$,
\quad$\phi(\tau'_{\eta}(\alpha))\subseteq\phi(\tau'_{\xi}(\alpha))$,}

\Centerline{$\phi(\tau_{\xi}(\alpha))\subseteq\{x:f(x)\le\alpha\}$,
\quad$\phi(\tau'_{\xi}(\alpha))\subseteq\{x:f(x)\ge\alpha\}$}

\noindent whenever $\eta\le\xi$, $\alpha\in\Bbb R$ and the codes here are
defined.   Next, if $k_{\eta}=k$ is defined, we must have
$U_{n_k}\cap F_{\eta}(q_k,q'_k)\ne\emptyset$ and

\inset{----- either $f(x)\ge q_k$ for every
$x\in U_{n_k}\cap F_{\eta}(q_k,q'_k)$ and
$\phi(\tau'_{\eta+1}(q_k))
=\phi(\tau'_{\eta}(q_k))\cup(U_{n_k}\cap F_{\eta}(q_k,q_k'))$

----- or $f(x)\le q'_k$ for every
$x\in U_{n_k}\cap F_{\eta}(q_k,q'_k)$ and
$\phi(\tau_{\eta+1}(q_k))
=\phi(\tau_{\eta}(q_k))\cup(U_{n_k}\cap F_{\eta}(q_k,q_k'))$.}

\noindent In either case,
$U_{n_k}\cap F_{\eta}(q_k,q_k')$ must be disjoint from $F_{\xi}(q_k,q'_k)$
for every $\xi>\eta$ for which $F_{\xi}$ is defined;
consequently we cannot have $k_{\xi}=k$ for any
$\xi>\eta$.   The induction must therefore stop.

$F_{\zeta}(q,q')=\emptyset$ whenever $q$, $q'\in\Bbb Q$ and $q<q'$.
\Prf\Quer\ Otherwise, because $f$ is resolvable, there is an
$n\in\Bbb N$ such that $V=U_n\cap F_{\zeta}(q,q')$ is non-empty and
either $f(x)\ge q$ for every $x\in V$ or $f(x)\le q'$ for every $x\in V$.
Let $k\in\Bbb N$ be such that $n_k=n$, $q_k=q$ and $q'_k=q'$;  then
$U_{n_k}$ meets $F_{\zeta}(q_k,q'_k)$ so $k\ne k_{\eta}$ for any
$\eta<\zeta$.   But this means that we ought to have proceeded according to
one of the first two alternatives in the single-step inductive stage,
and ought not to have stopped at $\zeta$.\ \Bang\Qed

\medskip

{\bf (d)} Now set

\Centerline{$\tau(\alpha)
=\tilde\Theta^*_1(\langle\tau'_{\zeta}(q_n)\rangle_{n\in\Bbb N,q_n>\alpha})$}

\noindent for $\alpha\in\Bbb R$.   Then

$$\eqalign{\phi(\tau(\alpha))
&=\bigcup_{n\in\Bbb N,q_n>\alpha}\phi(\tau'_{\zeta}(q_n))\cr
&\subseteq\bigcup_{n\in\Bbb N,q_n>\alpha}\{x:f(x)\ge q_n\}
\subseteq\{x:f(x)>\alpha\}\cr}$$

\noindent for every $\alpha$.   \Quer\ If $\alpha$ is such that
$\phi(\tau(\alpha))\subset\{x:f(x)>\alpha\}$, let $x\in X$ and
$n\in\Bbb N$ be such that $f(x)>q'_n>q_n>\alpha$ and
$x\notin\phi(\tau(\alpha))$.   Then $x\notin\phi(\tau'_{\zeta}(q_n))$;
but also $f(y)\le q'_n$ for every $y\in\phi(\tau_{\zeta}(q'_n))$, so
$x\notin\phi(\tau_{\zeta}(q'_n))$ and $x\in F_{\zeta}(q_n,q'_n)$, which is
supposed to be impossible.\ \Bang

So we can set $\tilde\psi(f)=\tau$.
}%end of proof of 562R

\leader{562S}{Codable families of codable functions (a)} If $X$ and $Y$
are second-countable spaces, a family $\familyiI{f_i}$ of functions from
$X$ to $Y$ is a {\bf codable family of codable Borel functions} if
$\langle f_i^{-1}[H]\rangle_{i\in I,H\subseteq Y\text{ is open}}$ is a
codable family in $\Cal B_c(X)$.

\spheader 562Sb Uniformizing the arguments of 562N, it is easy to check
that a family $\familyiI{f_i}$ of real-valued functions on $X$ is a codable
family of codable Borel functions iff there is a family
$\familyiI{\tau_i}$ in $\tilde{\Cal T}$ such that\cmmnt{, in
the language there,} $f_i=\tilde\phi(\tau_i)$ for every $i$.

\spheader 562Sc \cmmnt{In this language,} 562Ne can be rephrased as

\inset{if $\sequencen{f_n}$ is a codable sequence of real-valued
codable Borel functions on $X$, there
is a codable Borel function $f$ such
that $f(x)=\liminf_{n\to\infty}f_n(x)$ whenever the lim inf is finite,}

\noindent and 562R implies that

\inset{the family of resolvable real-valued functions on $X$ is a codable
family of codable Borel functions.}

\spheader 562Sd If $X$, $Y$ and $Z$ are second-countable spaces,
$\familyiI{f_i}$ is a codable family of codable Borel functions from
$X$ to $Y$, and $\familyiI{g_i}$ is a codable family of codable
Borel functions from $Y$ to $Z$, then $\familyiI{g_if_i}$ is a codable
family of codable functions from $X$ to $Z$\prooflet{;  this
is because the proof of 562Mb
gives a recipe for calculating a code for the composition of codable
functions, which can be performed simultaneously on the compositions
$g_if_i$ if we are given codes for the functions $g_i$ and $f_i$}.

\cmmnt{\spheader 562Se Extending the remarks in 562O(b)-(c), we see that
(for instance) we
can define a sequence $\sequencen{\Phi_n}$ such that
$\Phi_n$ is a function from $\tilde{\Cal T}^{n+1}$ to $\tilde{\Cal T}$
for every $n$, and
$\tilde\phi(\Phi_n(\langle\tau_i\rangle_{i\le n}))
=\sum_{i=0}^n\tilde\phi(\tau_i)$ whenever
$\tau_0,\ldots,\tau_n\in\tilde{\Cal T}$;  so that if
$\sequencen{f_n}$ is a codable sequence of codable Borel functions, then
$\sequencen{\sum_{i=0}^nf_i}$ is codable.}

\leader{562T}{Codable Baire sets}\cmmnt{ The
ideas here can be adapted to give a
theory of Baire algebras in general topological spaces.}
Start by settling on a sequence running over a base for the topology
of $\BbbR^{\Bbb N}$, with the associated interpretation
$\phi:\Cal T\to\Cal B_c(\BbbR^{\Bbb N})$  of Borel codes.
Let $X$ be a topological space.

\spheader 562Ta A subset $E$ of $X$ is a {\bf codable Baire set} if it
is of the form $f^{-1}[F]$ for some continuous $f:X\to\BbbR^{\Bbb N}$ and
$F\in\Cal B_c(\BbbR^{\Bbb N})$;  write $\CalBa_c(X)$ for the family of such
sets.   If $E\in\CalBa_c(X)$, then a {\bf code} for $E$ will be a pair
$(f,T)$ where $f:X\to\BbbR^{\Bbb N}$ is continuous, $T\in\Cal T$ and
$E=f^{-1}[\phi(T)]$.   A family $\familyiI{E_i}$ in $\CalBa_c(X)$
is\cmmnt{ now} a {\bf codable} family if there is a family
$\familyiI{(f_i,T^{(i)})}$ such that $(f_i,T^{(i)})$ codes $E_i$ for every $i$.

\spheader 562Tb{\bf (i)} Suppose that $\familyiI{f_i}$ is a countable
family of
continuous functions from $X$ to $\BbbR^{\Bbb N}$, and
$\familyiI{T^{(i)}}$ a
family in $\Cal T$.   Then there are a continuous function
$f:X\to\BbbR^{\Bbb N}$ and a sequence $\sequence{i}{\hat T^{(i)}}$ in $\Cal T$ such
that $(f,\hat T^{(i)})$ codes the same Baire set as $(f_i,T^{(i)})$ for every
$i\in I$.   \prooflet{\Prf\ If $I$ is empty, this is trivial.   Otherwise,
$I\times\Bbb N$ is countably infinite, so
$(\BbbR^{\Bbb N})^I\cong\BbbR^{I\times\Bbb N}$ is homeomorphic to
$\BbbR^{\Bbb N}$;  let
$h:\BbbR^{\Bbb N}\to(\BbbR^{\Bbb N})^I$ be a homeomorphism, and
set $f(x)=h^{-1}(\familyiI{f_i(x)})$ for each $x\in\Bbb N$.
Then $f_i=\pi_ihf$ for each $i$, where $\pi_i(z)=z(i)$ for
$z\in(\BbbR^{\Bbb N})^I$.   Now
$\langle(\pi_ih)^{-1}[V_n]\rangle_{i\in I,n\in\Bbb N}$
is a family of open sets
in $\BbbR^{\Bbb N}$, so is codable (562I, or otherwise);  let
$\langle T^{(i,n)}\rangle_{i\in I,n\in\Bbb N}$ be a family in $\Cal T$
such that $\phi(T^{(i,n)})=(\pi_ih)^{-1}[V_n]$ whenever $n\in\Bbb N$
and $i\in I$.
The construction of part (a)(ii)$\Rightarrow$(iii) in the proof of 562M
gives us a family $\familyiI{\Theta_i}$ of functions from $\Cal T$ to
$\Cal T$ such that $(\pi_ih)^{-1}[\phi(T)]=\phi(\Theta_i(T))$ whenever
$i\in I$ and $T\in\Cal T$.   So we can take $\hat T^{(i)}=\Theta_i(T^{(i)})$,
and we shall have

\Centerline{$f_i^{-1}[\phi(T^{(i)})]
=f^{-1}[(\pi_ih)^{-1}[\phi(T^{(i)})]]
=f^{-1}[\phi(\Theta_i(T^{(i)}))]
=f^{-1}[\phi(\hat T^{(i)})]$}

\noindent for every $i$, as required.\ \Qed}

\medskip

\quad{\bf (ii)} It follows that if $\sequence{i}{E_i}$ is a codable
sequence in $\CalBa_c(X)$ then $\bigcup_{i\in\Bbb N}E_i$ and
$\bigcap_{i\in\Bbb N}E_i$ belong to $\CalBa_c(X)$.   \prooflet{\Prf\
By (i), we have a continuous $f:X\to\BbbR^{\Bbb N}$ and a sequence
$\sequence{i}{\hat T^{(i)}}$ in $\Cal T$ such that $E_i=f^{-1}[\phi(\hat T^{(i)})]$ for
every $i\in\Bbb N$.   Now 562K tells us that
$F=\bigcup_{i\in\Bbb N}\phi(\hat T^{(i)})$ and
$F'=\bigcap_{i\in\Bbb N}\phi(\hat T^{(i)})$ are codable, so
$f^{-1}[F]=\bigcup_{i\in\Bbb N}E_i$ and
$f^{-1}[F']=\bigcap_{i\in\Bbb N}E_i$ belong to $\CalBa_c(X)$.\ \Qed}

\medskip

\quad{\bf (iii)}\cmmnt{ In particular,} $\CalBa_c(X)$ is\cmmnt{ closed
under finite
intersections; as it is certainly closed under complementation,
it is} an algebra of subsets of $X$.   Every zero set belongs to
$\CalBa_c(X)$.   \prooflet{\Prf\ If $g:X\to\Bbb R$ is continuous, set
$f(x)(i)=g(x)$ for $x\in X$, $i\in\Bbb N$;  then
$H=\{z:z\in\BbbR^{\Bbb N}$, $z(0)=0\}$ is closed, therefore a codable Borel
set, and $g^{-1}[\{0\}]=f^{-1}[H]$ is a codable Baire set.\ \Qed}

\medskip

\quad{\bf (iv)} If $Y$ is another topological space and
$g:X\to Y$ is continuous,
then $\familyiI{g^{-1}[F_i]}$ is a codable family
in $\CalBa_c(X)$ for every codable family $\familyiI{F_i}$ in
$\CalBa_c(Y)$.   \prooflet{\Prf\ If $\familyiI{(f_i,T^{(i)})}$ codes
$\familyiI{F_i}$, then $\familyiI{(f_ig,T^{(i)})}$ codes
$\familyiI{g^{-1}[F_i]}$.\ \Qed}

\spheader 562Tc{\bf (i)}
A function $f:X\to\Bbb R$ is a {\bf codable Baire
function} if there are a continuous $g:X\to\BbbR^{\Bbb N}$ and a codable
Borel function $h:\BbbR^{\Bbb N}\to\Bbb R$ such that $f=hg$.
A family $\familyiI{f_i}$ of
codable Baire functions is a {\bf codable} family if there is a family
$\familyiI{(g_i,h_i)}$ such that $g_i:X\to\BbbR^{\Bbb N}$ is a continuous
function for every $i\in I$ and $\familyiI{h_i}$ is a codable family of
codable Borel functions from $\BbbR^{\Bbb N}$ to $\Bbb R$.

\medskip

\quad{\bf (ii)} Suppose that $\sequencen{f_n}$ is a codable sequence of
codable Baire functions from $X$ to $\Bbb R$.   Then there are a continuous
function $g:X\to\BbbR^{\Bbb N}$ and a codable sequence $\sequencen{h_n}$ of
codable Borel functions from $\BbbR^{\Bbb N}$ to $\Bbb R$ such that
$f_n=h_ng$ for every $n\in\Bbb N$.   \prooflet{\Prf\ Let
$\sequencen{(g_n,h'_n)}$ be such that $g_n:X\to\BbbR^{\Bbb N}$ is continuous
for every $n$, $\sequencen{h'_n}$ is a codable sequence of codable Borel
functions from $\BbbR^{\Bbb N}\to\Bbb R$, and $f_n=h'_ng_n$ for each $n$.
Now $\langle\{x:h'_n(x)>q\}\rangle_{n\in\Bbb N,q\in\Bbb Q}$ is a codable
family in $\Cal B_c(\BbbR^{\Bbb N})$;  let
$\langle T'_{nq}\rangle_{n\in\Bbb N,q\in\Bbb Q}$ be a family in $\Cal T$
coding it.   By (b-i) above, there are a continuous function
$g:X\to\BbbR^{\Bbb N}$ and a family
$\langle T^{(n,q)}\rangle_{n\in\Bbb N,q\in\Bbb N}$ in $\Cal T$ such that

\Centerline{$g^{-1}[\phi(T^{(n,q)})]=g_n^{-1}[\phi(T'_{nq})]
=\{x:f_n(x)>q\}$}

\noindent for every $n\in\Bbb N$ and $q\in\Bbb Q$.

To convert $\langle T^{(n,q)}\rangle_{n\in\Bbb N,q\in\Bbb Q}$ into a code for
a sequence of real-valued functions on $\BbbR^{\Bbb N}$, I copy ideas from
the proof of 562N.   Let

\Centerline{$\Theta_0:\Cal T\to\Cal T$,
\quad$\Theta_1:\Cal T\times\Cal T\to\Cal T$,}

\Centerline{$\Theta_3:\Cal T\times\Cal T\to\Cal T$,
\quad$\tilde\Theta_1:\bigcup_{I\subseteq\Bbb Q}\Cal T^I\to\Cal T$}

\noindent be such that

\Centerline{$\phi(\Theta_0(T))=X\setminus\phi(T)$,
\quad$\phi(\Theta_1(T,T'))=\phi(T)\cup\phi(T'),$}

\Centerline{$\phi(\Theta_3(T,T'))=\phi(T)\setminus\phi(T')$,
\quad$\phi(\tilde\Theta_1(\tau))=\bigcup_{q\in I}\phi(\tau(q)))$}

\noindent for $T$, $T'\in\Cal T$, $I\subseteq\Bbb Q$ and $\tau\in\Cal T^I$.
Now, for $\alpha\in\Bbb R$ and $n\in\Bbb N$, set

\Centerline{$\tau'_n(\alpha)
=\tilde\Theta_1(\langle T^{(n,q)}\rangle_{q\in\Bbb Q,q\ge\alpha})$,}

\Centerline{$T^{(n)}
=\tilde\Theta_1(\sequence{k}{\Theta_3(\tau'_n(-k),\tau'_n(k))})$,}

$$\eqalign{\tau_n(\alpha)
&=\Theta_3(\tau'_n(\alpha),\Theta_0(T^{(n)}))\text{ if }\alpha\ge 0,\cr
&=\Theta_1(\tau'_n(\alpha),\Theta_0(T^{(n)}))\text{ if }\alpha<0.\cr}$$

\noindent We now have a sequence $\sequencen{\tau_n}$ in $\tilde{\Cal T}$
(as defined in 562N) coding a sequence $\sequencen{h_n}$ of Borel functions
from $\BbbR^{\Bbb N}$ to $\Bbb R$
such that $f_n=h_ng$ for every $n$ (see 562Sb).\ \Qed}

\medskip

\quad{\bf (iii)}
If $\sequencen{f_n}$ is a codable sequence of codable Baire
functions, there is a codable Baire function $f$ such that
$f(x)=\liminf_{n\to\infty}f_n(x)$ whenever the lim inf is finite.
\prooflet{\Prf\ Take $g$ and $\sequencen{h_n}$ as in (i);  by
562Ne, there is a codable Borel function $h$ such that
$h(z)=\liminf_{n\to\infty}h_n(z)$ whenever $z\in\BbbR^{\Bbb N}$ is such
that the $\liminf$ is finite, and $f=hg:X\to\Bbb R$ will serve.\ \Qed}

\medskip

\quad{\bf (iv)}
The family of codable Baire functions is a Riesz subspace of
$\BbbR^X$ containing all continuous functions and closed under
multiplication.   \prooflet{(This time, use (i) and 562Nd.)}

\medskip

\quad{\bf (v)} The family of continuous real-valued
functions on $X$ is a codable family of codable Baire functions.
\prooflet{(For $f\in C(X)$, define $g_f\in C(X;\BbbR^{\Bbb N})$ by setting
$g_f(x)(n)=f(x)$ for every $x\in X$ and $n\in\Bbb N$;  setting
$\pi_0(z)=z(0)$ for $z\in\BbbR^{\Bbb N}$, $\family{f}{C(X)}{(g_f,\pi_0)}$
is a family of codes for $C(X)$.)}

\medskip

\quad{\bf (vi)}\dvAnew{2014} If $E\subseteq X$, then $E\in\CalBa_c(X)$ iff
$\chi E:X\to\Bbb R$ is a codable Baire function.   \prooflet{\Prf\

$$\eqalignno{E\in\CalBa_c(X)
&\iff\text{ there are a continuous }g:X\to\BbbR^{\Bbb N}
\cr&\mskip100mu\text{ and an }
F\in\Cal B_c(\BbbR^{\Bbb N})\text{ such that }E=g^{-1}[F]
\cr&\iff\text{ there are a continuous }g:X\to\BbbR^{\Bbb N}
\cr&\mskip100mu\text{ and an }
F\in\Cal B_c(\BbbR^{\Bbb N})\text{ such that }\chi E=(\chi F)g
\cr&\iff\text{ there are a continuous }g:X\to\BbbR^{\Bbb N}
\cr&\mskip100mu\text{ and a codable
Borel function }h:\BbbR^{\Bbb N}\text{ such that }\chi E=hg\cr
\displaycause{562Nf, because if $\chi E=hg$ then $\chi E=(\chi F)g$ where
$F=\{y:h(y)>0\}$}
&\iff\chi E\text{ is a codable Baire function}.  \text{ \Qed}\cr}$$
}

\spheader 562Td\dvAnew{2014}
If $\sequencen{f_n}$ is a codable sequence of codable Baire
functions from $X$ to $\Bbb R$, then
$\langle f_n^{-1}[H]\rangle_{n\in\Bbb N,H\subseteq\Bbb R\text{ is open}}$
is codable.   \prooflet{\Prf\ By (c-i), we have a continuous
$g:X\to\BbbR^{\Bbb N}$ and a codable sequence $\sequencen{h_n}$ of
codable Borel functions from $\BbbR^{\Bbb N}$ to $\Bbb R$ such that
$f_n=h_ng$ for every $n$.   Let $\psi:\Cal T\to\Cal B_c(\Bbb R)$
be an interpretation of Borel codes corresponding to some enumeration of a
base for the topology of $\Bbb R$.
By 562Md, $g$ is codable;  by 562M(a-iii),
there is a function $\Theta:\Cal T\to\Cal T$ such that
$g^{-1}[\psi(T)]=\phi(\Theta(T))$ for every $T\in\Cal T$.   Now
$\langle h_n^{-1}[H]\rangle_{n\in\Bbb N,H\subseteq\Bbb R\text{ is open}}$
is codable, by the definition in 562S;  that is, there is a family
$\langle T_{nH}\rangle_{n\in\Bbb N,H\subseteq\Bbb R\text{ is open}}$ in
$\Cal T$ such that $\phi(T_{nH})=h_n^{-1}[H]$ for all $n$ and $H$.   Now

\Centerline{$f_n^{-1}[H]=g^{-1}[h_n^{-1}[H]]
=g^{-1}[\phi(T_{nH})]=\psi(\Theta(T_{nH}))$}

\noindent for all $n$ and $H$, so we have a coding of
$\langle f_n^{-1}[H]\rangle_{n\in\Bbb N,H\subseteq\Bbb R\text{ is open}}$.\
\Qed}

\leader{562U}{Proposition}
Let $(X,\frak T)$ be a second-countable space.   Then there is a
second-countable topology $\frak S$ on $X$, codably Borel equivalent to
$\frak T$, such that
$\Cal B_c(X)=\CalBa_c(X,\frak S)$ and the codable families in
$\Cal B_c(X)$ are exactly the codable families in $\CalBa_c(X,\frak S)$.

\proof{{\bf (a)} By 562Pb there is a topology $\frak S$ on $X$,
finer than $\frak T$, generated by a countable algebra $\Cal E$ of subsets
of $X$, which is codably Borel equivalent to $\frak T$.
Let $\sequencen{U_n}$ be a sequence running over $\Cal E$.
Define $g_0:X\to\BbbR^{\Bbb N}$ by setting
$g_0(x)=\sequencen{\chi U_n(x)}$ for each $x\in X$.
Then $g_0$ is continuous.
Set $W_n=\{z:z\in\BbbR^{\Bbb N}$, $z(n)>0\}$ for each $n$, so that
$W_n\subseteq\BbbR^{\Bbb N}$ is open and $U_n=g_0^{-1}[W_n]$;  let
$\sequencen{V_n}$ be a sequence running over a base for the topology of
$\BbbR^{\Bbb N}$ and such that $V_{2n}=W_n$ for
every $n$.   Let $\phi:\Cal T\to\Cal B_c(X)$,
$\phi':\Cal T\to\Cal B_c(\BbbR^{\Bbb N})$ be the interpretations of Borel
codes corresponding to $\sequencen{U_n}$, $\sequencen{V_n}$ respectively.

\medskip

{\bf (b)} We have a function $\Theta:\Cal T\to\Cal T\setminus\{\emptyset\}$
such that $\phi(T)=g^{-1}[\phi'(\Theta(T))]$ for every $T\in\Cal T$.
\Prf\ Induce on
$r(T)$.   As usual, set $A_T=\{n:\fraction{n}\in T\}$.
If $r(T)=0$, take $\Theta(T)\in\Cal T\setminus\{\emptyset\}$ such
that $\phi'(\Theta(T))=\emptyset$.   If $r(T)=1$,
set $\Theta(T)=\{\fraction{2n}:n\in A_T\}$;   then

\Centerline{$\phi'(\Theta(T))=\bigcup\{V_{2n}:n\in A_T\}$,
\quad$g_0^{-1}[\phi'(\Theta(T))]
=\bigcup\{U_n:n\in A_T\}=\phi(T)$.}

\noindent If $r(T)>1$, set

\Centerline{$\Theta(T)=\{\fraction{i}:i\in A_T\}
\cup\{\fraction{i}^{\smallfrown}\sigma:i\in A_T$,
  $\sigma\in\Theta(T_{\fraction{i}})\}$.  \Qed}

This means that if we have any codable family in $\Cal B_c(X)$, coded by a
family $\familyiI{T^{(i)}}$ in $\Cal T$, $\familyiI{(g_0,\Theta(T^{(i)}))}$ will
code the same family in $\CalBa_c(X,\frak S)$.

\medskip

{\bf (c)} Next, there is a function
$\Phi:C((X,\frak S);\BbbR^{\Bbb N})\times\Cal T
\to\Cal T\setminus\{\emptyset\}$ such that
$g^{-1}[\phi'(T))]=\phi(\Phi(g,T))$ for every $\frak S$-continuous
$g:X\to\BbbR^{\Bbb N}$ and $T\in\Cal T$.   \Prf\ If $r(T)\le 1$
and $g^{-1}[\phi'(T)]$ is
empty take $\Phi(g,T)\in\Cal T\setminus\{\emptyset\}$ such that
$\phi(\Phi(g,T))=\emptyset$.   If $r(T)=1$ and $g^{-1}[\phi'(T)]$ is
not empty set

\Centerline{$\Phi(g,T)
=\{\fraction{n}:U_n\subseteq g^{-1}[\phi'(T)]\}$.}

\noindent If $r(T)>1$ set

\Centerline{$\Phi(g,T)
=\{\fraction{i}:i\in A_T\}\cup\{\fraction{i}^{\smallfrown}\sigma:
  i\in A_T$, $\sigma\in\Phi(g,T_{\fraction{i}})\}$.   \Qed}

\noindent So given any codable family in $\CalBa_c(X,\frak S)$, coded by a
family $\familyiI{(g_i,T^{(i)})}$ in
$C((X,\frak S);\BbbR^{\Bbb N})\times\Cal T$, $\familyiI{\Phi(g_i,T^{(i)})}$
will code it in $\Cal B_c(X)$.
}%end of proof of 562U

\leader{562V}{}\cmmnt{ A
different use of Borel codes will appear when we come to re-examine
a result in Volume 3.   I will defer the application to 566O, but
the first part of the argument fits naturally into the ideas of this
section.

\medskip

\noindent}{\bf Theorem} (a) Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, and $\sequencen{a_n}$ a sequence in $\frak A$.   Then we
have an interpretation $\phi:\Cal T\to\frak A$ of Borel codes such that

$$\eqalign{\phi(T)&=\sup_{i\in A_T}a_i\text{ if }r(T)\le 1,\cr
&=\sup_{i\in A_T}1\Bsetminus\phi(T_{\fraction{i}})\text{ if }r(T)>1,\cr}$$

\noindent where $A_T=\{i:\fraction{i}\in T\}$ as usual.

(b) For $n\in\Bbb N$, set $E_n=\{x:x\in\{0,1\}^{\Bbb N}$, $x(n)=1\}$.
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra, and $\sequencen{a_n}$ a sequence in $\frak A$.   Let
$\phi:\Cal T\to\frak A$ and $\psi:\Cal T\to\Cal P(\{0,1\}^{\Bbb N})$
be the interpretations of Borel codes
corresponding to $\sequencen{a_n}$ and $\sequencen{E_n}$.
If $T$, $T'\in\Cal T$ are such that $\phi(T)\notBsubseteq\phi(T')$, then
$\psi(T)\not\subseteq\psi(T')$.

\proof{{\bf (a)} Define $\phi(T)$ inductively on the rank of $T$, as in
562Ba.

\medskip

{\bf (b)} Let $\sequencen{T^{(n)}}$ be a sequence running over
$\{T,T'\}\cup\{T_{\sigma}:\sigma\in S^*\}
\cup\{T'_{\sigma}:\sigma\in S^*\}$.
Define $\sequencen{c_n}$ inductively, as follows.
$c_0=\phi(T)\Bsetminus\phi(T')$.   Given that
$c_n\in\frak A\setminus\{0\}$, then

\inset{----- if $r(T^{(n)})\le 1$ and there
is an $i\in A_{T^{(n)}}$ such that $c_n\Bcap a_i\ne 0$, take the first such
$i$ and set $c_{n+1}=c_n\Bcap a_i$;

----- if $r(T^{(n)})>1$ and there is an $i\in A_{T^{(n)}}$ such that
$c_n\Bsetminus\phi(T^{(n)}_{\fraction{i}})\ne 0$, take the first such $i$
and set $c_{n+1}=c_n\Bsetminus\phi(T^{(n)}_{\fraction{i}})$;

----- otherwise, set $c_{n+1}=c_n$.}

\noindent Continue.

At the end of the induction, define $x\in\{0,1\}^{\Bbb N}$ by saying that
$x(i)=1$ iff there is an $n\in\Bbb N$ such that $c_n\Bsubseteq a_i$.
Now we find that, for every $m\in\Bbb N$,

\inset{----- if $x\in\psi(T^{(m)})$
there is an $n\in\Bbb N$ such that $c_n\Bsubseteq\phi(T^{(m)})$,

----- if $x\notin\psi(T^{(m)})$ there is an $n\in\Bbb N$ such that
$c_n\Bcap\phi(T^{(m)})=0$.}

\Prf\ Induce on $r(T^{(m)})$.   If $r(T^{(m)})\le 1$ then

$$\eqalign{x\in\psi(T^{(m)})
&\Longrightarrow\text{ there is an }i\in A_{T^{(m)}}\text{ such that }
  x\in E_i\cr
&\Longrightarrow\text{ there are }i\in A_{T^{(m)}},\,n\in\Bbb N
  \text{ such that }c_n\Bsubseteq a_i\cr
&\Longrightarrow\text{ there is an }n\in\Bbb N\text{ such that }
  c_n\Bsubseteq\phi(T^{(m)}),\cr}$$

$$\eqalign{x\notin\psi(T^{(m)})
&\Longrightarrow x\notin E_i\text{ for every }i\in A_{T^{(m)}}\cr
&\Longrightarrow c_{m+1}\notBsubseteq a_i
  \text{ for every }i\in A_{T^{(m)}}\cr
&\Longrightarrow c_m\Bcap a_i=0
  \text{ for every }i\in A_{T^{(m)}}
\Longrightarrow c_m\Bcap\phi(T^{(m)})=0.\cr}$$

\noindent If $r(T^{(m)})>1$ then

$$\eqalignno{x\in\psi(T^{(m)})
&\Longrightarrow\text{ there is an }i\in A_{T^{(m)}}\text{ such that }
  x\notin\psi(T^{(m)}_{\fraction{i}})\cr
&\Longrightarrow\text{ there are }i\in A_{T^{(m)}},\,n\in\Bbb N
  \text{ such that }c_n\Bcap\phi(T^{(m)}_{\fraction{i}})=0\cr
\displaycause{by the inductive hypothesis, because $T^{(m)}_{\fraction{i}}$
is always equal to $T^{(k)}$ for some $k$, and
$r(T^{(m)}_{\fraction{i}})<r(T^{(m)})$}
&\Longrightarrow\text{ there is an }n\in\Bbb N
  \text{ such that }c_n\Bsubseteq\phi(T^{(m)}),\cr}$$

$$\eqalign{x\notin\psi(T^{(m)})
&\Longrightarrow x\in\psi(T^{(m)}_{\fraction{i}})
  \text{ for every }i\in A_{T^{(m)}}\cr
&\Longrightarrow\text{ for every }i\in A_{T^{(m)}}
  \text{ there is an }n\in\Bbb N\text{ such that }
  c_n\Bsubseteq\phi(T^{(m)}_{\fraction{i}})\cr
&\Longrightarrow
  c_{m+1}\notBsubseteq 1\Bsetminus\phi(T^{(m)}_{\fraction{i}})
  \text{ for every }i\in A_{T^{(m)}}\cr
&\Longrightarrow c_m\Bsetminus\phi(T^{(m)}_{\fraction{i}})=0
  \text{ for every }i\in A_{T^{(m)}}\cr
&\Longrightarrow c_m\Bcap\phi(T^{(m)})=0.  \text{ \Qed}\cr}$$

In particular, since both $T$ and $T'$ appear in the list
$\sequence{m}{T^{(m)}}$, $c_n\Bcap\phi(T)\ne 0$ and $c_n\Bcap\phi(T')=0$
for every $n$, $x\in\psi(T)\setminus\psi(T')$ and
$\psi(T)\not\subseteq\psi(T')$.
}%end of proof of 562V

\exercises{\leader{562X}{Basic exercises (a)}
%\spheader 562Xa
Let $X$ be a regular second-countable space.   Show that a
resolvable subset of $X$ is F$_{\sigma}$.   \Hint{in the proof of 562I,
show that $\phi(T^{(\xi)})$ is always F$_{\sigma}$.}
%562I

\spheader 562Xb Let $X$ be a second-countable space and
$\langle E_{ni}\rangle_{n,i\in\Bbb N}$ a family of resolvable subsets of
$X$.   Show that $\bigcap_{n\in\Bbb N}\bigcup_{i\in\Bbb N}E_{ni}$ is
a codable Borel set.
%562K

\spheader 562Xc Let $X$ be a second-countable space and
$\familyiI{E_i}$ a codable family in $\Cal B_c(X)$.   (i) Show that
$\family{i}{J}{E_i}$ is codable for every $J\subseteq I$.
(ii) Show that if $I$ is countable and not empty then
$\bigcup_{i\in I}E_i$ and $\bigcap_{i\in I}E_i$ are codable Borel sets.
(iii) Show that if $h:J\times\Bbb N\to I$ is a function, where $J$ is any
other set, then $\family{j}{J}{\bigcup_{n\in\Bbb N}E_{h(j,n)}}$ is a
codable family.   (iv) Show that if $\familyiI{F_i}$ is another codable
family in $\Cal B_c(X)$ then $\familyiI{E_i\cap F_i}$ and
$\familyiI{E_i\symmdiff F_i}$ are codable families.
%562K

\spheader 562Xd Let $X$ and $Y$ be
second-countable spaces and $f:X\to Y$ a function.   Suppose that
$\{F:F\subseteq Y$, $f^{-1}[F]$ is resolvable$\}$ includes a countable
network for the topology of $Y$.   Show that $f$ is a codable Borel
function.
%562K

\spheader 562Xe Let $X$ be a second-countable space and $\familyiI{E_i}$
a family in $\Cal B_c(X)$.   (i) Show that $\{J:J\subseteq I$,
$\family{i}{J}{E_i}$ is codable$\}$ is an ideal of subsets of $I$.
(ii) Show that if every $E_i$ is resolvable then $\familyiI{E_i}$ is
codable.
%562K

\spheader 562Xf Let $X$ be a second-countable space and $f:X\to\Bbb R$ a
function.   Show that $f$ is a codable Borel function iff
$\{(x,\alpha):x\in X$, $\alpha<f(x)\}$ is a codable Borel subset of
$X\times\Bbb R$.
%562N

\spheader 562Xg Let $X$ be a topological space and $f$, $g:X\to\Bbb R$
resolvable functions.   (i) Show that $f\vee g$ and
$\alpha f$ are resolvable
for any $\alpha\in\Bbb R$.   (ii) Show that if $f$ is bounded
then $f+g$ is resolvable.   (iii) Show that if $f$ and $g$ are bounded,
$f\times g$ is resolvable.   (iv) Show that if $f$ and $g$
are non-negative, then $f+g$ is resolvable.   (v) Show that if
$h:\Bbb R\to\Bbb R$ is continuous and $h^{-1}[\{\alpha\}]$ is finite for
every $\alpha\in\Bbb R$, then $hf$ is resolvable.
%562Q

\spheader 562Xh Let $f:\Bbb R\to\Bbb R$ be such that
$\lim_{t\downarrow x}f(t)$ is defined in $[-\infty,\infty]$
for every $x\in\Bbb R$.   Show that $f$ is resolvable.
%562Q

\spheader 562Xi Let $X$ be a second-countable space and
$\sequencen{f_n}$ a sequence of resolvable real-valued functions on $X$.
Show that there is a codable Borel function $g$ such that
$g(x)=\lim_{n\to\infty}f_n(x)$ for any $x$ such that the limit is defined
in $\Bbb R$.
%562R

\spheader 562Xj Let $X$ be a second-countable space and $Y$ a subspace of
$X$.   Show that a family $\familyiI{g_i}$ in $\BbbR^Y$ is a codable family
of codable Borel functions iff there is a codable family $\familyiI{f_i}$
of real-valued codable Borel functions on $X$ such that $g_i=f_i\restr Y$
for every $i\in I$.
%562S 562E

\sqheader 562Xk Let $X$ be a second-countable space.   Show that every
codable family of codable Baire subsets of $X$ is a codable family of
codable Borel subsets of $X$.
%562T

\sqheader 562Xl Let $X$ be a regular second-countable space.
Show that every
codable family of codable Borel subsets of $X$ is a codable family of
codable Baire subsets of $X$.   \Hint{561Xk.}
%562T

\leader{562Y}{Further exercises (a)}
%\spheader 562Ya
Let $X$ be a second-countable space, $Y$ a T$_0$
second-countable space and $f:X\to Y$ a function with graph
$\Gamma\subseteq X\times Y$.   (i) Show that if $f$ is a codable Borel
function, then $\Gamma$ is a codable Borel subset of $X\times Y$.
(ii) Show that if $X$ and $Y$ are Polish and $\Gamma$ is a codable Borel
subset of $X\times Y$, then $f$ is a codable Borel function.
%562F  562L

\spheader 562Yb
Show that there is an analytic subset of $\NN$ which is not
a codable Borel set.   \Hint{423M.}
%*562F out of order query

\spheader 562Yc Show
that if $X$ is a Polish space then a subset of $X$ is resolvable iff it is
both F$_{\sigma}$ and G$_{\delta}$.
%562I

\spheader 562Yd Let $X$ be a Polish space.   Show that a function
$f:X\to\Bbb R$ is resolvable iff $\{x:\alpha<f(x)<\beta\}$ is
F$_{\sigma}$ for all $\alpha$, $\beta\in\Bbb R$.
%562Q

\spheader 562Ye Let $X$ be a topological space.   Let $\Phi$ be the set of
functions $f:X\to\Cal P\Bbb N$ such that $\{x:n\in f(x)\}$ is open for
every $n\in\Bbb N$.   Write $\Cal B'_c(X)$ for
$\{f^{-1}[F]:f\in\Phi$,  $F\in\Cal B_c(\Cal P\Bbb N)\}$;  say a family
$\familyiI{E_i}$ in $\Cal B'_c(X)$ is codable if there is a family
$\familyiI{(f_i,F_i)}$ in $\Phi\times\Cal B_c(\Cal P\Bbb N)$ such that
$\familyiI{F_i}$ is codable and $E_i=f_i^{-1}[F_i]$ for every $i$.   (i)
Show that if $X$ is second-countable then $\Cal B'_c(X)=\Cal B_c(X)$ and
the codable families on the definition here coincide with the codable
families of 562J.   (ii) Develop a theory of codable Borel sets and
functions corresponding to that in 562T.
%562T
}%end of exercises


\endnotes{
\Notesheader{562} The idea of `Borel code' is of great importance in
mathematical logic, for reasons quite separate from the questions
addressed here;
see {\smc Jech 78}, {\smc Jech 03} or {\smc Kunen 80}.   (Of course it is
not a coincidence that an approach which is effective in the absence of the
axiom of choice should also be relevant to absoluteness in the presence of
choice.)   Every author has his favoured formula corresponding to that in
562Ba.   The particular one
I have chosen is intended to be economical and direct, but is
slightly awkward at the initial stages, and some proofs demand an extra
moment's attention to the special case of trees of rank $1$.   The real
motivation for the calculations here will have to wait for \S565;  Lebesgue
measure can be defined in such a way that it is countably additive with
respect to {\it codable} sequences of Borel sets, and there are enough of
these to make the theory non-trivial.

Borel codes are wildly non-unique, which is why the concept of codable
family is worth defining.   But it is also important that certain sets,
starting with the open sets, are self-coding in the sense that from the set
we can pick out an appropriate code.   `Resolvable' sets and functions
(562G, 562Q) are
common enough to be very useful, and for these we can work with the objects
themselves, just as we always have, and leave the coding until we need it.

The Borel codes described here can be used only in second-countable
spaces.   It is easy enough to find variations of the concept
which can be applied
in more general contexts (562Ye), though it is not obvious that there are
useful
theorems to be got in such a way.   More relevant to the work of the next
few sections is the idea of `codable Baire set' (562T).   Because any
codable sequence of codable Baire sets can be factored through a single
continuous function to $\BbbR^{\Bbb N}$ (562T(b-i)),
we have easy paths to the elementary results given here.

}%end of notes

\discrpage

\frfilename{mt563.tex}
\versiondate{3.12.13}
\copyrightdate{2008}

\def\chaptername{Choice and determinacy}
\def\sectionname{Borel measures without choice}

\newsection{563}

Having decided that a `Borel set' is to be one obtainable by
a series of operations described by a Borel code, it is a natural step to
say that a `Borel measure' should be one which respects these operations
(563A).   In regular spaces, such measures have strong inner and outer
regularity properties also based on the Borel coding (563D-563F), and we
have effective methods of constructing such measures (563H).   Analytic
sets are universally measurable (563I).   We can use similar ideas to give
a theory of Baire measures on general topological spaces (563J-563K).
In the basic case, of a second-countable space with a codably
$\sigma$-finite measure, we have a measure algebra with many of the same
properties as in the standard theory (563M-563N).

The theory would not be very significant if there were no interesting
Borel-coded measures, so you may wish to glance ahead to \S565 to confirm
that Lebesgue measure can be brought into the framework developed here.

\leader{563A}{Definitions (a)}\cmmnt{ ({\smc Foreman \& Wehrung 91})}
Let $X$ be a second-countable space and
$\Cal B_c(X)$ the algebra of codable Borel subsets of $X$.
I will say that a {\bf Borel-coded measure} on $X$ is a functional
$\mu:\Cal B_c(X)\to[0,\infty]$ such that
$\mu\emptyset=0$ and
$\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$ whenever
$\sequencen{E_n}$ is a disjoint codable family in $\Cal B_c(X)$.

\cmmnt{I will try to remember to say `Borel-coded measure' everywhere
in this section, because these are dangerously different from the `Borel
measures' of \S434.   Their domains are not necessarily $\sigma$-algebras
and while they are finitely additive they need not be countably additive
even in the sense of 326I.}

\spheader 563Ab
\cmmnt{As usual, }I
will say that a subset of $X$ is {\bf negligible} if it is
included in a set of measure $0$\cmmnt{, which here must be a codable
Borel set;
the terms `conegligible', `almost everywhere', `null ideal'
will take their meanings from
this}.   We can now define the {\bf completion} of $\mu$ to be the
natural extension of $\mu$ to the algebra
$\{E\symmdiff A:E\in\Cal B_c(X)$, $A$ is $\mu$-negligible$\}$.

\spheader 563Ac\cmmnt{ Some of the other definitions from the ordinary
theory can be transferred without difficulty (e.g., `totally finite',
`probability'), but we may need to make some
finer distinctions.   For instance,} I will say that a Borel-coded measure
$\mu$ is {\bf semi-finite} if
$\sup\{\mu F:F\subseteq E$, $\mu F<\infty\}=\infty$ whenever
$\mu E=\infty$\cmmnt{;  we no longer have the ordinary
principle of exhaustion (215A), and the definition in 211F, taken
literally, may be too weak.   For `locally finite', however, 411Fa can be
taken just as it is, since all open sets are measurable}.

\spheader 563Ad \dvro{A}{For `$\sigma$-finite' we again have to make a
choice.
The definition in 211C calls only for `a sequence of measurable sets of
finite measure'.   Here the following will be more useful:  a} Borel-coded
measure on $X$ is {\bf codably $\sigma$-finite} if there is a codable
sequence
$\sequencen{E_n}$ in $\Cal B_c(X)$ such that $X=\bigcup_{n\in\Bbb N}E_n$
and $\mu E_n$ is finite for every $n$.

\leader{563B}{Proposition} Let $(X,\frak T)$
be a second-countable space and $\mu$ a Borel-coded measure on $X$.

(a) Let $\sequencen{E_n}$ be a codable sequence in $\Cal B_c(X)$.

\quad(i) $\mu(\bigcup_{n\in\Bbb N}E_n)\le\sum_{n=0}^{\infty}\mu E_n$.

\quad(ii) If $\sequencen{E_n}$ is non-decreasing,
$\mu(\bigcup_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu E_n$.

\quad(iii) If $\sequencen{E_n}$ is non-increasing and $\mu E_0$ is finite,
then $\mu(\bigcap_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu E_n$.

(b) $\mu$ is $\tau$-additive.

(c)\dvArevised{2013} Suppose that $\frak T$ is T$_1$.
If $\Cal E$ is the algebra of resolvable subsets of $X$\cmmnt{ (562H)},
then $\mu\restr\Cal E$ is countably additive in the sense that
$\mu E=\sum_{n=0}^{\infty}\mu E_n$ for any disjoint
family $\sequencen{E_n}$ in $\Cal E$ such that
$E=\sup_{n\in\Bbb N}E_n$ is defined in $\Cal E$.

\proof{{\bf (a)(i)} Set $F_n=E_n\setminus\bigcup_{i<n}E_i$ for
$n\in\Bbb N$;  then $\sequencen{F_n}$ is codable (562Kc), so

\Centerline{$\mu(\bigcup_{n\in\Bbb N}E_n)
=\mu(\bigcup_{n\in\Bbb N}F_n)
=\sum_{n=0}^{\infty}\mu F_n
\le\sum_{n=0}^{\infty}\mu E_n$.}

\medskip

\quad{\bf (ii)} If $\sequencen{E_n}$ is non-decreasing then, in the
language of (i), $E_n=\bigcup_{i\le n}F_i$ for each $n$, so

\Centerline{$\lim_{n\to\infty}\mu E_n
=\lim_{n\to\infty}\sum_{i=0}^n\mu F_i
=\mu(\bigcup_{n\in\Bbb N}E_n)$.}

\medskip

\quad{\bf (iii)} Apply (ii) to $\sequencen{E_0\setminus E_n}$.
(As remarked in 562J, this will be a codable sequence.)

\medskip

{\bf (b)} Suppose that $\Cal G$ is an upwards-directed family of
open sets with union $H$.   Set $\gamma=\sup_{G\in\Cal G}\mu G$.   Let
$\sequencen{U_n}$ be a sequence running over a base for the topology of
$X$, and for $n\in\Bbb N$ set

\Centerline{$V_n
=\bigcup\{U_i:i\le n$, $U_i\subseteq G$ for some $G\in\Cal G\}$.}

\noindent Then $\sequencen{V_n}$ is a non-decreasing sequence of open sets
with union $H$.   As every $V_n$ is resolvable, $\sequencen{V_n}$ is
codable and

\Centerline{$\mu H=\lim_{n\to\infty}\mu V_n
\le\sup_{G\in\Cal G}\mu G\le\mu H$}

\noindent by (a-ii).

\medskip

{\bf (c)} If $\sequencen{E_n}$ is a disjoint sequence in $\Cal E$ with a
supremum $E$ in $\Cal E$, then $E\supseteq\bigcup_{n\in\Bbb N}E_n$.
If $x\in E$ then $\{x\}$ is closed,
because $\frak T$ is T$_1$, so $\{x\}$ is resolvable (562H) and
$E\setminus\{x\}\in\Cal E$;  as $E\setminus\{x\}$ is not an upper bound of
$\{E_n:n\in\Bbb N\}$, $x\in\bigcup_{n\in\Bbb N}E_n$.   So
$E=\bigcup_{n\in\Bbb N}E_n$.
Now $\sequencen{E_n}$ is codable, as noted in
562J, so $\mu E=\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$.
}%end of proof of 563B

\leader{563C}{Corollary} Let $X$ be a second-countable space, $\mu$ a
Borel-coded measure on $X$ and
$\sequencen{E_n}$ a sequence of resolvable sets in $X$.

(a)(i) $\bigcup_{n\in\Bbb N}E_n$ is measurable;

\quad(ii) $\mu(\bigcup_{n\in\Bbb N}E_n)\le\sum_{n=0}^{\infty}\mu E_n$;

\quad(iii) if $\sequencen{E_n}$ is disjoint,
$\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$;

\quad(iv) if $\sequencen{E_n}$ is non-decreasing,
$\mu(\bigcup_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu E_n$.

(b)(i) $\bigcap_{n\in\Bbb N}E_n$ is measurable;

\quad(ii) if $\sequencen{E_n}$ is non-increasing and
$\inf_{n\in\Bbb N}\mu E_n$ is finite, then
$\mu(\bigcap_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu E_n$.

\proof{{\bf (a)}
Use 562I to find a sequence of codes for $\sequencen{E_n}$,
and apply 563B.

\medskip

{\bf (b)} follows, because $X\setminus E_n$ is resolvable for each $n$.
}%end of proof of 563C

\leader{563D}{}\cmmnt{ The next lemma is primarily intended as a basis for
Theorem 563H, but it will be useful in 563F.

\medskip

\noindent}{\bf Lemma} Let $(X,\frak T)$
be a regular second-countable space and
$\mu:\frak T\to[0,\infty]$ a functional such that

\inset{$\mu\emptyset=0$,

$\mu G\le\mu H$ if $G\subseteq H$,

$\mu$ is modular\cmmnt{ (definition:  413Qc)},

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

$\bigcup\{G:G\in\frak T$, $\mu G<\infty\}=X$.}

(a) $\mu(\bigcup_{i\in I}G_i)\le\sum_{i\in I}\mu G_i$ for every countable
family $\familyiI{G_i}$ in $\frak T$.

(b) There is a function $\pi^*:\frak T\times\Bbb N\to\frak T$ such that

\Centerline{$X\setminus G\subseteq\pi^*(G,k)$,
\quad$\mu(G\cap\pi^*(G,k))\le 2^{-k}$}

\noindent whenever $G\in\frak T$ and $k\in\Bbb N$.

(c) Let $\phi:\Cal T\to\Cal B_c(X)$ be an interpretation of Borel codes
defined from a sequence running over $\frak T$\cmmnt{, where
$\Cal T$ is the set of subtrees of $\bigcup_{n\in\Bbb N}\BbbN^n$ without
infinite branches (562A, 562B)}.   Then there are functions $\pi$,
$\pi':\Cal T\times\Bbb N\to\frak T$ such that

\Centerline{$\phi(T)\subseteq\pi(T,n)$,
\quad$X\setminus\phi(T)\subseteq\pi'(T,n)$,
\quad$\mu(\pi(T,n)\cap\pi'(T,n))\le 2^{-n}$}

\noindent for every $T\in\Cal T$ and $n\in\Bbb N$.

\proof{{\bf (a)} This is elementary.   First, $\mu(G\cup H)\le\mu G+\mu H$
for all open sets $G$ and $H$, because $\mu(G\cap H)\ge 0$.   Next,
if $\sequencen{G_n}$ is a sequence of open sets
with union $G$, then

\Centerline{$\mu G=\lim_{n\to\infty}\mu(\bigcup_{i\le n}G_i)
\le\lim_{n\to\infty}\sum_{i=0}^n\mu G_i
=\sum_{n=0}^{\infty}\mu G_n$.}

\noindent Now the step to general countable $I$ is immediate.

\medskip

{\bf (b)} Set $I=\{n:n\in\Bbb N$, $\mu U_n<\infty\}$;  because
$\mu$ is locally finite, $\{U_n:n\in I\}$ is a base for $\frak T$.
Given $G\in\frak T$ and $k\in\Bbb N$,
then for $n\in I$ and $m\in\Bbb N$ set

\Centerline{$W_{nm}=\bigcup\{U_i:i\le m$,
$\overline{U}_i\subseteq U_n\cap G\}$.}

\noindent Because $\frak T$ is regular,
$\bigcup_{m\in\Bbb N}W_{nm}=U_n\cap G$ and
$\mu(U_n\cap G)=\lim_{m\to\infty}\mu W_{nm}$.
Let $m_n$ be the least integer such that
$\mu W_{nm_n}\ge\mu(U_n\cap G)-2^{-k-n-1}$.   Set

\Centerline{$\pi^*(G,k)
=\bigcup_{n\in I}U_n\setminus\overline{W}_{nm_n}\in\frak T$.}

\noindent Because $\overline{W}_{nm}\subseteq U_n\cap G$ for all $m$ and
$n$, while $\bigcup_{n\in I}U_n=X$,
$\pi^*(G,k)\supseteq X\setminus G$.   Now

\Centerline{$\mu(G\cap\pi^*(G,k))
\le\sum_{n\in I}\mu(G\cap U_n\setminus\overline{W}_{nm_n})
\le\sum_{n=0}^{\infty}2^{-k-n-1}
=2^{-k}$,}

\noindent as required.

\medskip

{\bf (c)} Define $\pi(T)$ and $\pi'(T)$ inductively on the rank $r(T)$
of $T$.

\medskip

\quad{\bf (i)} If $r(T)=0$,
set $\pi(T,n)=\emptyset$ and $\pi'(T,n)=X$ for every $n$.
If $r(T)=1$ then $G=\phi(T)$ is open;  set $\pi(T,n)=G$ and
$\pi'(T,n)=\pi^*(G,n)$ for each $n$.

\medskip

\quad{\bf (ii)} For the inductive step to $r(T)>1$,
set $A_T=\{i:\fraction{i}\in T\}$
and
$T_{\fraction i}=\{\sigma:\fraction{i}^{\smallfrown}\sigma\in T\}$ for
$i\in\Bbb N$, as in 562A.   Set

\Centerline{$\pi(T,n)=\bigcup_{i\in A_T}\pi'(T_{\fraction{i}},n+i+2)$,}

\Centerline{$\pi'(T,n)
=\bigcup_{i\in A_T}
(\pi(T_{\fraction{i}},n+i+2)\cap\pi'(T_{\fraction{i}},n+i+2))
  \cup\pi^*(\pi(T,n),n+1)$.}

\noindent Then

\Centerline{$\phi(T)
=\bigcup_{i\in A_T}X\setminus T_{\fraction{i}}
\subseteq\bigcup_{i\in A_T}\pi'(T_{\fraction{i}},n+i+2)
=\pi(T,n)$,}

$$\eqalign{X\setminus\phi(T)
&=\bigcap_{i\in A_T}\phi(T_{\fraction{i}})
\subseteq(\pi(T,n)\cap\bigcap_{i\in A_T}\phi(T_{\fraction{i}}))
   \cup\pi^*(\pi(T,n),n+1)\cr
&\subseteq\bigl(\bigcup_{i\in A_T}\pi'(T_{\fraction{i}},n+i+2)
     \cap\bigcap_{i\in A_T}\pi(T_{\fraction{i}},n+i+2)\bigr)
   \cup\pi^*(\pi(T,n),n+1)\cr
&\subseteq\bigcup_{i\in A_T}\bigl(\pi'(T_{\fraction{i}},n+i+2)
                                 \cap\pi(T_{\fraction{i}},n+i+2)\bigr)
   \cup\pi^*(\pi(T,n),n+1)\cr
&=\pi'(T,n),\cr}$$

$$\eqalignno{\mu(\pi(T,n)\cap\pi'(T,n))
&\le\sum_{i\in A_T}
  \mu(\pi(T_{\fraction{i}},n+i+2)\cap\pi'(T_{\fraction{i}},n+i+2))\cr
&\mskip200mu +\mu(\pi(T,n)\cap\pi^*(\pi(T,n),n+1))\cr
&\le\sum_{i\in A_T}2^{-n-i-2}+2^{-n-1}
\le 2^{-n}\cr}$$

\noindent for every $n$, so the induction continues.
}%end of proof of 563D

\leader{563E}{Lemma} Let $X$ be a second-countable space and $\Mu$ a
non-empty
upwards-directed set of Borel-coded measures on $X$.   For each codable
Borel set $E\subseteq X$, set $\nu E=\sup_{\mu\in\Mu}\mu E$.   Then
$\nu$ is a Borel-coded measure on $X$.

\proof{ Immediate from the definition in 563Aa.
}%end of proof of 563E

\leader{563F}{Proposition} Let $(X,\frak T)$ be a second-countable
space and $\mu$ a Borel-coded measure on $X$.

(a) For any $F\in\Cal B_c(X)$, we have a Borel-coded measure $\mu_F$ on $X$
defined by saying that $\mu_FE=\mu(E\cap F)$ for every $E\in\Cal B_c(X)$.

(b) We have a semi-finite Borel-coded measure $\mu_{\text{sf}}$ defined by saying
that

\Centerline{$\mu_{\text{sf}}(E)
=\sup\{\mu F:F\in\Cal B_c(X)$, $F\subseteq E$, $\mu F<\infty\}$}

\noindent for every $E\in\Cal B_c(X)$.

(c)(i) If $\mu$ is locally finite it is codably $\sigma$-finite.

\quad(ii) If $\mu$ is codably $\sigma$-finite, it is semi-finite and
there is a totally finite
Borel-coded measure $\nu$ on $X$ with the same null ideal as $\mu$.

\quad(iii) If $\mu$ is codably $\sigma$-finite, there is a non-decreasing
codable sequence of codable Borel sets of finite measure which covers $X$.

(d) If $X$ is regular then the following are equiveridical:

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

\quad(ii) $\mu$ is semi-finite,
outer regular with respect to the open sets and inner regular with
respect to the closed sets;

\quad(iii) $\mu$ is semi-finite
and outer regular with respect to the open sets.

(e) If $X$ is regular and $\mu$ is semi-finite, then $\mu$
is inner regular with
respect to the closed sets of finite measure.

(f) If $X$ is Polish and $\mu$ is semi-finite, then $\mu$ is
inner regular with respect to the compact sets.

(g) If $\mu$ is locally finite, and $\nu$ is another Borel-coded measure on
$X$ agreeing with $\mu$ on the open sets, then $\nu=\mu$.

\proof{ Fix a sequence $\sequencen{U_n}$ running over a base for the
topology of $X$.

\medskip

{\bf (a)} The point is just that $\sequencen{E_n\cap F}$ is a
codable family whenever $\sequencen{E_n}$ is a codable family in
$\Cal B_c(X)$ and $F$ is codable.   (562J again.)

\medskip

{\bf (b)} Writing $\mu_F$ for the Borel-coded measure corresponding to a
set $F$ of finite measure, as in (a), we have an upwards-directed family of
measures;  by 563E, its supremum $\mu_{\text{sf}}$ is a Borel-coded measure.
If $E\subseteq X$ is a codable Borel set and $\gamma<\mu_{\text{sf}}E$, then
there is a set $F$ of finite measure such that $\mu(E\cap F)\ge\gamma$;
now

\Centerline{$\gamma\le\mu_{\text{sf}}(E\cap F)=\mu(E\cap F)<\infty$.}

\medskip

{\bf (c)(i)} Set $I=\{i:i\in\Bbb N$, $\mu U_i<\infty\}$;  then
$\familyiI{U_i}$ is a codable family of sets of finite measure covering
$X$.

\medskip

\quad{\bf (ii)} Let $\sequencen{H_n}$ be a codable sequence of sets of
finite measure covering $X$.

\medskip

\qquad\grheada\ If $E\in\Cal B_c(X)$, then
$\sequencen{E\cap\bigcup_{i\le n}H_i}$ is a non-decreasing codable sequence
with union $E$, so

\Centerline{$\mu E
=\sup_{n\in\Bbb N}\mu(E\cap\bigcup_{i\le n}H_i)
\le\sup\{\mu F:F\subseteq E$, $\mu F<\infty\}\le\mu E$.}

\noindent As $E$ is arbitrary, $\mu$ is semi-finite.

\medskip

\qquad\grheadb\ Let $\sequencen{\epsilon_n}$ be a sequence
of strictly positive real numbers such that
$\sum_{n=0}^{\infty}\epsilon_n\mu H_n$ is finite.   Set
$\nu E=\sum_{n=0}^{\infty}\epsilon_n\mu(E\cap H_n)$ for $E\in\Cal B_c(X)$.
Of course $\nu\emptyset=0$ and $\nu X<\infty$.
If $\sequence{k}{E_k}$ is a disjoint codable sequence in $\Cal B_c(X)$,
then $\sequence{k}{E_k\cap H_n}$ is codable for every $n$, so

$$\eqalign{\nu(\bigcup_{k\in\Bbb N}E_k)
&=\sum_{n=0}^{\infty}\epsilon_n\mu(\bigcup_{k\in\Bbb N}E_k\cap H_n)
=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\epsilon_n\mu(E_k\cap H_n)\cr
&=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\epsilon_n\mu(E_k\cap H_n)
=\sum_{k=0}^{\infty}\nu E_k.\cr}$$

\noindent So $\nu$ is a Borel-coded measure.

If $E\in\Cal B_c(X)$ and $\mu E=0$, then of course
$\nu E=\sum_{n=0}^{\infty}\epsilon_n\mu(E\cap H_n)=0$.   Conversely, if
$\nu E=0$, then $\mu(E\cap H_n)=0$ for every $n$;  but
$\sequencen{E\cap H_n}$, like $\sequencen{H_n}$, is codable, so
$\mu E=\mu(\bigcup_{n\in\Bbb N}E\cap H_n)=0$.   Thus $\mu$ and $\nu$ have
the same sets of zero measure;  it follows at once that they have the
same null ideals.

\medskip

\quad{\bf (iii)} All we have to note is that if $\sequencen{E_n}$ is
a codable sequence of sets of finite measure covering $X$, then
$\sequencen{\bigcup_{i\le n}E_i}$ is codable (562Kb), so gives the
required non-decreasing witness.

\medskip

{\bf (d)}{\bf (i)$\Rightarrow$(ii)}\grheada\  Observe first that
$\mu\restr\frak T$ satisfies the conditions of 563D.
\Prf\ The first three are consequences of the fact that
$\mu:\Cal B_c(X)\to[0,\infty]$ is additive.   If $\sequencen{G_n}$ is a
non-decreasing sequence in $\frak T$, it is a codable sequence of codable
Borel sets, by 562I as usual;  so
$\mu(\bigcup_{n\in\Bbb N}G_n)=\lim_{n\to\infty}\mu G_n$ by 563B(a-ii).
Finally, we are assuming that $\mu$ is locally finite, so the last
condition is satisfied.\ \Qed

Take an interpretation $\phi$ of Borel codes and
functions $\pi$, $\pi':\Cal T\times\Bbb N\to\frak T$ as in 563Dc.

\medskip

\quad\grheadb\ If $E\in\Cal B_c(X)$ and
$\mu E<\gamma$, take $n$ such that
$2^{-n}\le\gamma-\mu E$.   There is a $T\in\Cal T$ such that
$\phi(T)=E$, and now $G=\pi(T,n)$ is
open, $E\subseteq G$ and

\Centerline{$\mu(G\setminus E)\le\mu(G\cap\pi'(T,n))\le 2^{-n}$,}

\noindent so $\mu G\le\gamma$.

\medskip

\quad\grheadc\ If $E\in\Cal B_c(X)$ and $\gamma<\mu E$, take $T\in\Cal T$
such that $\phi(T)=E$ and $n\in\Bbb N$ such that $2^{-n}<\mu E-\gamma$;
now $F=X\setminus\pi'(T,n)$ is closed, $F\subseteq E$ and
$\mu(E\setminus F)\le 2^{-n}$, so $\mu F>\gamma$.   Next, if we set

\Centerline{$F_m
=F\cap\bigcup\{\overline{U_i}:i\le m$, $\mu\overline{U}_i<\infty\}$,}

\noindent $\sequence{m}{F_m}$ will be a non-decreasing sequence of closed
sets of finite measure with union $F$.   The sets $F_m$ are all resolvable,
so $\mu F=\lim_{m\to\infty}\mu F_m$ and there is an $m$ such that
$\mu F_m\ge\gamma$, while $F_m\subseteq E$ is a set of finite
measure.   As $E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with
respect to the closed sets and also semi-finite.

\medskip

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

\medskip

\quad{\bf(iii)$\Rightarrow$(i)}\grheada\ If $x\in X$ then $x$ belongs to
some set of finite measure.   \Prf\ Set

\Centerline{$F
=\bigcap\{U_n:n\in\Bbb N$, $x\in U_n\}\setminus\bigcup\{U_n:n\in\Bbb N$,
$x\notin U_n\}$.}

\noindent Then $F$ is a codable Borel set, being the difference of
G$_{\delta}$ sets (562Da),
and the subspace topology on $F$ is indiscrete.   If $\mu F=0$
we can stop.   Otherwise, there must be an $F'\subseteq F$ such that
$0<\mu F'<\infty$;  but $F'\in\Cal B_c(F)=\{\emptyset,F\}$ (562E),
so $F'=F$ and
again $F$ has finite measure.\ \Qed

\medskip

\qquad\grheadb\ Now as $\mu$ is outer regular with respect to the open
sets, every set of finite measure is included in an open set of finite
measure.   So $\mu$ must be locally finite.

\medskip

{\bf (e)} Suppose that $E\in\Cal B_c(X)$ and
$\gamma<\mu E$.   Then there is an $H\in\Cal B_c(X)$ such that
$H\subseteq E$ and $\gamma<\mu H<\infty$.   Consider the Borel-coded
measure
$\mu_H$ defined from $\mu$ and $H$ as in (a).   This is totally finite,
so (d) tells us that it is outer regular with respect to the open sets and
therefore inner regular with respect to the closed sets,
and there is a closed set $F\subseteq H$ such that $\mu F=\mu_HF\ge\gamma$.
As $E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with respect to
the closed sets of finite measure.

\medskip

{\bf (f)} Let $\rho$ be a complete
metric on $X$ inducing its topology.
If $E\in\Cal B_c(X)$ and $\gamma<\mu E$, let $F\subseteq E$ be a
closed set such that $F\subseteq E$ and $\gamma<\mu F<\infty$.
For each $n\in\Bbb N$ set $J_n=\{i:\diam U_i\le 2^{-n}\}$.
Define $\sequencen{k_n}$, $\sequencen{F_n}$ inductively by saying that
$F_0=F$ and

\Centerline{$k_n=\min\{k:\mu(F_n\cap\bigcup_{i\in J_n\cap k}U_i)>\gamma\}$,
\quad$F_{n+1}=F_n\cap\bigcup_{i\in J_n\cap k_n}\overline{U}_i$}

\noindent for each $n$;  set $K=\bigcap_{n\in\Bbb N}F_n\subseteq E$.   Then
$\mu K=\lim_{n\to\infty}\mu F_n\ge\gamma$.   The point is that $K$ is
compact.   \Prf\ Set $L=\prod_{n\in\Bbb N}J_n\cap k_n\subseteq\NN$.
Then $L$ is compact (561D).   Set
$L'=\{\alpha:\alpha\in L$,
$F\cap\bigcap_{i\le n}\overline{U}_{\alpha(i)}\ne\emptyset$
for every $n\}$;
then $L'$ is a closed subset of $L$ so is compact.   For $\alpha\in L'$,
$\{F\cap\overline{U}_{\alpha(i)}:i\in\Bbb N\}$ generates a filter
$\Cal F_{\alpha}$ on $X$ which
is a Cauchy filter because $\diam\overline{U}_{\alpha(i)}
=\diam U_{\alpha(i)}\le 2^{-i}$ for every $i$;  because $X$ is
$\rho$-complete, $f(\alpha)=\lim\Cal F_{\alpha}$ is defined, and belongs to
$F\cap\bigcap_{i\in\Bbb N}\overline{U}_{\alpha(i)}\subseteq K$.   If
$\alpha$, $\beta\in L'$ and $\alpha(i)=\beta(i)$, then
$f(\alpha)$, $f(\beta)$ both belong to $\overline{U}_{\alpha(i)}$ so
$\rho(f(\alpha),f(\beta))\le 2^{-i}$;  thus $f$ is continuous and
$f[L']$ is a compact subset of $K$.
On the other hand, given $x\in K$, we can set
$\alpha(n)=\min\{i:i\in J_n\cap k_n$, $x\in\overline{U}_i\}$ for each
$n$, and now $\alpha\in L'$ and $f(\alpha)=x$.   So $K=f[L']$ is compact.
(See 561Yj.)\ \Qed

As $E$ and $\gamma$ are arbitrary, $\mu$ is inner regular with respect to
the compact sets.

\medskip

{\bf (g)(i)} Consider first the case in which $X$ is regular.   In this
case both $\mu$ and $\nu$ must be outer regular with respect to the open
sets, by (d);  as they agree on the open sets they must be equal.

\medskip

\quad{\bf (ii)} Next, suppose that $\mu X=\nu X$ is finite.
Let $\Cal E$ be the algebra of subsets of $X$
generated by $\{U_n:n\in\Bbb N\}$, and $\frak S$ the topology generated by
$\Cal E$.   As noted in the argument of 562Pb, $\frak S$ is codably Borel
equivalent to the original topology of $X$, so $\mu$ and $\nu$ are still
Borel-coded measures with respect to $\frak S$, and are still locally
finite, because $\frak S$ is finer than $\frak T$;  while $\frak S$ is
regular.   Now any member of $\Cal E$ is expressible in the form
$E=\bigcup_{i\le n}G_i\setminus H_i$ where the $G_i$, $H_i$ are open and
$\langle G_i\setminus H_i\rangle_{i\le n}$ is disjoint.   So

\Centerline{$\mu E=\sum_{i=0}^n\mu G_i-\mu(G_i\cap H_i)=\nu E$.}

\noindent More generally, if $H\in\frak S$, there is a non-decreasing
sequence $\sequencen{E_n}$
in $\Cal E$ with union $H$;  as all the sets in $\Cal E$ are resolvable,
$\sequencen{E_n}$ is codable and

\Centerline{$\mu H=\sup_{n\in\Bbb N}\mu E_n=\nu H$.}

\noindent Thus $\mu$ and $\nu$ agree on $\frak S$;  by (i), they are
equal.

\medskip

\quad{\bf (iii)} Finally, for the general case, set
$V_n=\bigcup\{U_i:i\le n$, $\mu U_i<\infty\}$ for each $n$.   Because $\mu$
is locally finite, $\bigcup_{n\in\Bbb N}V_n=X$.   For each $n\in\Bbb N$ let
$\mu_{V_n}$, $\nu_{V_n}$ be the Borel-coded measures defined from $V_n$ as
in (a).   Then $\mu_{V_n}$ and $\nu_{V_n}$ are totally finite and agree on
the open sets, so are equal.   Now $\sequencen{V_n}$, being a sequence of
open sets, is codable;  so if $E\in\Cal B_c(X)$ the sequence
$\sequencen{E\cap V_n}$ is codable, and

\Centerline{$\mu E=\lim_{n\to\infty}\mu(E\cap V_n)
=\lim_{n\to\infty}\mu_{V_n}E=\nu E$.}

\noindent So in this case also we have $\mu=\nu$.
}%end of proof of 563F

\vleader{72pt}{563G}{Proposition} Let $X$ be a set and
$\theta:\Cal PX\to[0,\infty]$ a submeasure.

(a)

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

\noindent is an algebra of subsets of $X$, and $\theta\restr\Sigma$ is
additive in the sense that $\theta(E\cup F)=\theta E+\theta F$ in
$[0,\infty]$ whenever $E$, $F\in\Sigma$ are disjoint.

(b) If $E\subseteq X$ and for
every $\epsilon>0$ there is an $F\in\Sigma$ such that $E\subseteq F$ and
$\theta(F\setminus E)\le\epsilon$, then $E\in\Sigma$.

\proof{{\bf (a)} Parts (a)-(c) of the proof of 113C apply unchanged.

\wheader{563G}{6}{2}{2}{36pt}
{\bf (b)} Take any $A\subseteq X$ and $\epsilon>0$.   Let $F\in\Sigma$ be
such that $E\subseteq F$ and $\theta(F\setminus E)\le\epsilon$.   Then

\Centerline{$\theta A\le\theta(A\cap E)+\theta(A\setminus E)
\le\theta(A\cap F)+\theta(A\setminus F)+\theta(F\setminus E)
\le\theta A+\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\theta A=\theta(A\cap E)+\theta(A\setminus E)$;  as $A$ is arbitrary,
$E\in\Sigma$.
}%end of proof of 563G

\leader{563H}{Theorem} Let $(X,\frak T)$
be a regular second-countable space and
$\mu:\frak T\to[0,\infty]$ a functional such that

\inset{$\mu\emptyset=0$,

$\mu G\le\mu H$ if $G\subseteq H$,

$mu$ is modular,

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

$\bigcup\{G:G\in\frak T$, $\mu G<\infty\}=X$.}

\noindent Then $\mu$ has a unique extension to a Borel-coded measure on
$X$.

\proof{{\bf (a)} For $A\subseteq X$ set
$\theta A=\inf\{\mu G:A\subseteq G\in\frak T\}$.   Then $\theta$ is a
submeasure on $\Cal PX$ (because $\mu(G\cup H)\le\mu G+\mu H$ for all $G$,
$H\in\frak T$), extending $\mu$ (because $\mu G\le\mu H$ if
$G\subseteq H$).   Set

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

\noindent and $\nu=\theta\restr\Sigma$, as in 563G.
Let $\phi:\Cal T\to\Cal B_c(X)$ be an interpretation of Borel codes and
$\pi$, $\pi':\Cal T\times\Bbb N\to\frak T$ corresponding
functions as in 563Dc.
Now $\Cal B_c(X)\subseteq\Sigma$.   \Prf\
Given $T\in\Cal T$, $A\subseteq X$ and $n\in\Bbb N$,
let $G\in\frak T$ be such that $A\subseteq G$ and
$\mu G\le\theta A+2^{-n}$.   Then

$$\eqalign{\theta A
&\le\theta(A\cap\phi(T))+\theta(A\setminus\phi(T))
\le\theta(A\cap\pi(T,n))+\theta(A\cap\pi'(T,n))\cr
&\le\mu(G\cap\pi(T,n))+\mu(G\cap\pi'(T,n))\cr
&=\mu(G\cap(\pi(T,n)\cup\pi'(T,n))+\mu(G\cap\pi(T,n)\cap\pi'(T,n))\cr
&\le\mu G+\mu(\pi(T,n)\cap\pi'(T,n))
\le\theta A+2^{-n+1}.\cr}$$

\noindent As $A$ and $n$ are arbitrary, $\phi(T)\in\Sigma$.\ \Qed

\medskip

{\bf (b)} Let $\sequencen{T_n}$ be a sequence in $\Cal T$ such that
$\sequencen{E_n}$ is disjoint, where $E_n=\phi(T_n)$ for each $n$;  set
$E=\bigcup_{n\in\Bbb N}E_n$.
Then, for any $k\in\Bbb N$,
$E\subseteq\bigcup_{n\in\Bbb N}\pi(T_n,k+n)$, so

$$\eqalignno{\sum_{n=0}^{\infty}\nu E_n
&=\lim_{n\to\infty}\nu(\bigcup_{i\le n}E_i)
\le\nu E
\le\nu(\bigcup_{n\in\Bbb N}\pi(T_n,k+n))\cr
\displaycause{563Da}
&=\sum_{n=0}^{\infty}\nu E_n+\nu(\pi(T_n,k+n)\setminus E_n)\cr
\displaycause{563Ga}
&\le\sum_{n=0}^{\infty}\nu E_n+\mu(\pi(T_n,k+n)\cap\pi'(T_n,k+n))\cr
&\le\sum_{n=0}^{\infty}\nu E_n+2^{-k-n}
=2^{-k+1}+\sum_{n=0}^{\infty}\nu E_n;\cr}$$

\noindent as $k$ is arbitrary, $\sum_{n=0}^{\infty}\nu E_n=\nu E$;  as
$\sequencen{T_n}$ is arbitrary, $\nu\restr\Cal B_c(X)$ is a Borel-coded
measure extending $\mu$.

\medskip

{\bf (c)} At the same time we see that if $\lambda$ is any other
Borel-coded
measure extending $\mu$, we must have $\lambda E\le\theta E=\nu E$ for
every $E\in\Cal B_c(X)$.   In the other direction,

$$\eqalign{\lambda(\phi(T))
&\ge\lambda(\pi(T,n))-\lambda(\pi(T,n)\cap\pi'(T,n))\cr
&=\mu(\pi(T,n))-\mu(\pi(T,n)\cap\pi'(T,n))
\ge\nu(\phi(T))-2^{-n}\cr}$$

\noindent for every $T\in\Cal T$ and $n\in\Bbb N$, so $\lambda E\ge\nu E$
for every $E\in\Cal B_c(X)$.   Thus $\nu\restr\Cal B_c(X)$ is the only
Borel-coded measure extending $\mu$.
}%end of proof of 563H

\leader{563I}{Theorem} Let $X$ be a Hausdorff second-countable space,
$\mu$ a codably $\sigma$-finite Borel-coded measure on $X$, and
$A\subseteq X$ an analytic set.   Then there are a codable Borel set
$E\supseteq A$ and a sequence $\sequencen{K_n}$ of compact subsets
of $A$ such that $E\setminus\bigcup_{n\in\Bbb N}K_n$ is negligible.
Consequently $A$ is measured by the completion of $\mu$.

\proof{{\bf (a)} By 563F(c-ii),
there is a totally finite Borel-coded measure on
$X$ with the same negligible sets as $\mu$;  so it will be enough to
consider the case in which $\mu$ itself is totally finite.

If $A$ is empty, the result is trivial.   So we may suppose that
there is a continuous surjection $f:\NN\to A$.   For
$\sigma\in S=\bigcup_{n\in\Bbb N}\BbbN^n$ set
$I_{\sigma}=\{\alpha:\sigma\subseteq\alpha\in\NN\}$.
Fix on a sequence running over a base for the topology of $X$ and the
corresponding interpretation
$\phi:\Cal T\to\Cal B_c(X)$ of Borel codes.

\medskip

{\bf (b)} For $\sigma\in S$ and $\xi<\omega_1$ define $E_{\sigma\xi}$ by
saying that

\Centerline{$E_{\sigma 0}=\overline{f[I_{\sigma}]}$,}

\Centerline{$E_{\sigma,\xi+1}
=\bigcup_{i\in\Bbb N}E_{\sigma^{\smallfrown}\fraction{i},\xi}$,}

\Centerline{$E_{\sigma\xi}=\bigcap_{\eta<\xi}E_{\sigma\eta}$ if $\xi>0$ is
a countable limit ordinal.}

\noindent Then $\ofamily{\xi}{\omega_1}{E_{\sigma\xi}}$
is a non-increasing family of sets including $f[I_{\sigma}]$.

\medskip

{\bf (c)} For every $\xi<\omega_1$,
$\langle E_{\sigma\eta}\rangle_{\sigma\in S,\eta\le\xi}$ is a codable
family of codable Borel sets.   \Prf\ It is enough to consider the case
$\xi\ge\omega$.   Because $\xi$ is countable, we have a function
$\tilde\Theta_1:\bigcup_{J\subseteq\xi}\Cal T^J\to\Cal T$
such that $\phi(\tilde\Theta_1(\family{\eta}{J}{T_{\eta}}))
=\bigcup_{\eta\in J}\phi(T_{\eta})$ for every $J\subseteq\xi$ (562Cb).
Also, of course, we have a function $\Theta_0:\Cal T\to\Cal T$
such that $\phi(\Theta_0(T))=X\setminus\phi(T)$ for every $T\in\Cal T$.
Next, all the sets $E_{\sigma 0}$ are closed, therefore
resolvable.   So we have a family $\family{\sigma}{S}{T_{\sigma 0}}$ in
$\Cal T$ such that $\phi(T_{\sigma 0})=E_{\sigma 0}$ for every $\sigma$.
Now we can set

\Centerline{$T_{\sigma,\eta+1}
=\tilde\Theta_1(\sequence{i}{T_{\sigma^{\smallfrown}\fraction{i},\eta}})$}

\noindent if $\eta<\xi$,

\Centerline{$T_{\sigma\eta}
=\Theta_0(\tilde\Theta_1(\ofamily{\zeta}{\eta}
   {\Theta_0(T_{\sigma\zeta})}))$}

\noindent if $\eta\le\xi$ is a non-zero limit ordinal, and
$\phi(T_{\sigma\eta})$ will be equal to $E_{\sigma\eta}$ as required.\ \Qed

\medskip

{\bf (d)} Let
$\langle\epsilon_{\sigma}\rangle_{\sigma\in S}$ be a
summable family of strictly positive real numbers, and for $\xi<\omega_1$
set

\Centerline{$\gamma(\xi)
=\sum_{\sigma\in S}
  \epsilon_{\sigma}\mu(E_{\sigma\xi})$.}

\noindent Then $\gamma:\omega_1\to\Bbb R$ is non-increasing.   There is
therefore a $\xi<\omega_1$ such that $\gamma(\xi+1)=\gamma(\xi)$ (561A),
that is,
$\mu(E_{\sigma,\xi+1})=\mu(E_{\sigma\xi})$ for every $\sigma\in S$.

\medskip

{\bf (e)(i)} Set $E=E_{\emptyset\xi}$.
Of course $A=f[I_{\emptyset}]\subseteq E$.
Now define $\alpha_n\in\NN$, for $n\in\Bbb N$, as follows.   Given
$\ofamily{i}{m}{\alpha_n(i)}$, set

\Centerline{$G_{nm}
=\bigcup\{E_{\sigma\xi}:\sigma\in\BbbN^m$, $\sigma(i)\le\alpha_n(i)$
for every $i<m\}$,}

\Centerline{$G_{nmk}
=\bigcup\{E_{\sigma^{\smallfrown}\fraction{j},\xi}:
\sigma\in\BbbN^m$, $j\le k$, $\sigma(i)\le\alpha_n(i)$
for every $i<m\}$,}

\noindent Then $\langle G_{nm}\rangle_{n,m\in\Bbb N}$ is codable,
and $\lim_{k\to\infty}\mu G_{nmk}=\mu G_{nm}$ for all $m$,
$n\in\Bbb N$.   \Prf\ By (c), there is a family
$\langle T_{\sigma\eta}\rangle_{\sigma\in S,\eta\le\xi+1}$
in $\Cal T$ such
that $\phi(T_{\sigma\eta})=E_{\sigma\eta}$ whenever $\sigma\in S$ and
$\eta\le\xi+1$.
This time, we need a function
$\tilde\Theta_1:\bigcup_{J\subseteq S}\Cal T^J\to\Cal T$
such that $\phi(\tilde\Theta_1(\family{\sigma}{J}{T_{\sigma}}))
=\bigcup_{\sigma\in J}\phi(T_{\sigma})$ whenever
$J\subseteq S$ and $\family{\sigma}{J}{T_{\sigma}}$ is a family
in $\Cal T$, and a function $\Theta_3:\Cal T\times\Cal T\to\Cal T$ such
that $\phi(\Theta_3(T,T'))=\phi(T)\setminus\phi(T')$ for all $T$,
$T'\in\Cal T$.   Setting

\Centerline{$T'_{nm}
=\tilde\Theta_3(\langle T_{\sigma\xi}
  \rangle_{\sigma\in\BbbN^m,\,\sigma(i)\le\alpha_n(i)\Forall i<m})$,}

\Centerline{$T'_{nmk}
=\tilde\Theta_3(\langle T_{\sigma^{\smallfrown}\fraction{j},\xi}
  \rangle_{\sigma\in\BbbN^m,\,j\le k,\,\sigma(i)\le\alpha_n(i)
  \Forall i<m})$,}

\noindent we have $\phi(T'_{nm})=G_{nm}$ and $\phi(T'_{nmk})=G_{nmk}$ for
all $m$, $n$, $k\in\Bbb N$.   In particular, all the $G_{nm}$ and $G_{nmk}$
are codable Borel sets, and $\langle G_{nm}\rangle_{n,m\in\Bbb N}$ is
codable.   Moreover, for any particular pair $m$ and $n$,
$\sequence{k}{G_{nmk}}$ is a codable sequence;
we therefore have $\lim_{k\to\infty}\mu G_{nmk}=\mu G$, where
$G=\bigcup_{k\in\Bbb N}G_{nmk}$.   Next,

\Centerline{$G=\bigcup\{E_{\sigma,\xi+1}:\sigma\in\BbbN^m$,
$\sigma(i)\le\alpha_n(i)$ for every $i<m\}$,}

\noindent so

\Centerline{$G\symmdiff G_{nm}
\subseteq\bigcup\{E_{\sigma\xi}\setminus E_{\sigma,\xi+1}:
\sigma\in S\}$.}

\noindent Since
$\langle E_{\sigma\xi}\setminus E_{\sigma,\xi+1}
\rangle_{\sigma\in S}$ is a countable family of negligible sets coded by
$\langle\Theta_3(T_{\sigma\xi},T_{\sigma,\xi+1})\rangle_{\sigma\in S}$,
$G\symmdiff G_{nm}$ also is negligible and

\Centerline{$\mu G_{nm}=\mu G=\lim_{k\to\infty}\mu G_{nmk}$.\ \Qed}

\noindent Take the least $\alpha_n(m)\in\Bbb N$ such that
$\mu G_{n,m,\alpha_n(m)}\ge\mu G_{nm}-2^{-n-m}$, and continue.

\medskip

\quad{\bf (ii)} Set

\Centerline{$L_n
=\{\alpha:\alpha\in\NN$, $\alpha(i)\le\alpha_n(i)$ for every
$i\in\Bbb N\}$.}

\noindent Then $L_n$ is compact (561D), and $f[L_n]\subseteq A$.
Also $f[L_n]\supseteq\bigcap_{m\in\Bbb N}G_{nm}$.   \Prf\ If
$x\in\bigcap_{m\in\Bbb N}G_{nm}$, then for each $m\in\Bbb N$ let $\sigma_m$
be the lexicographically first member of
$\{\sigma:\sigma\in\Bbb N^m$, $\sigma(i)\le\alpha_n(i)$ for every $i<m\}$
such that $x\in E_{\sigma_m,\xi}$, and let $\beta_m\in\NN$ be such
that $\sigma_m\subseteq\beta_m$ and $\beta_m(i)=0$ for $i\ge m$.   Then
$\beta_m\in L_n$ for every $m$, so $\sequence{m}{\beta_m}$ has a cluster
point $\alpha\in L_n$.   \Quer\ If $f(\alpha)\ne x$, we have an open
neighbourhood $U$ of $f(\alpha)$ such that $x\notin\overline{U}$.
Let $m\in\Bbb N$ be such that $I_{\alpha\restr m}\subseteq f^{-1}[U]$;
then there is a $k\ge m$ such that
$\alpha\restr m=\beta_k\restr m=\sigma_k\restr m$.   Now

\Centerline{$x\in E_{\sigma_k,\xi}
\subseteq E_{\sigma_k,0}
\subseteq\overline{f[I_{\sigma_k}]}
\subseteq\overline{f[I_{\alpha\restr m}]}\subseteq\overline{U}$.
\Bang}

\noindent So $x=f(\alpha)\in f[L_n]$.\ \Qed

But $\sequence{m}{G_{nm}}$ is codable, and $G_{n0}=E$,
so we must have

$$\eqalign{\mu(E\setminus f[L_n])
&\le\mu(E\setminus G_{n0})
  +\sum_{m=0}^{\infty}\mu(G_{nm}\setminus G_{n,m+1})\cr
&=\sum_{m=0}^{\infty}\mu(G_{nm}\setminus G_{n,m,\alpha_n(m)})
\le\sum_{m=0}^{\infty}2^{-n-m}
=2^{-n+1}.\cr}$$

\noindent(Of course $f[L_n]$ is compact, therefore closed, therefore
measurable.)

\medskip

{\bf (f)} Set $K_n=f[L_n]$ for each $n$.   Then $\sequencen{K_n}$ is a
sequence of compact subsets of $A$;  because the $K_n$ are resolvable,
$F=\bigcup_{n\in\Bbb N}K_n$ is a codable Borel set.   For each $n$,

\Centerline{$\mu(E\setminus F)\le\mu(E\setminus K_n)
\le 2^{-n+1}$;}

\noindent so $E\setminus F$ is negligible.
Thus $E$ and $\sequencen{K_n}$ have the required properties.

Of course it now follows that $E\setminus A\subseteq E\setminus F$ is
negligible, so that the completion of $\mu$ measures $A$.
}%end of proof of 563I

\leader{563J}{Baire-coded measures} \cmmnt{Working from 562T,
we can develop a
theory of Baire measures on general topological spaces, as follows.}
If $X$ is a topological space, and $\CalBa_c(X)$ its algebra of
codable Baire sets,
a {\bf Baire-coded measure} on $X$ will be a
function $\mu:\CalBa_c(X)\to[0,\infty]$ such that $\mu\emptyset=0$ and
$\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$ for every
disjoint codable sequence $\sequencen{E_n}$ in $\CalBa_c(X)$.

\vleader{72pt}{563K}{Proposition} (a)
If $X$ and $Y$ are topological spaces, $f:X\to Y$ is a
continuous function and $\mu$ is a Baire-coded measure on $X$, then
$F\mapsto\mu f^{-1}[F]:\CalBa_c(Y)\to[0,\infty]$ is a Baire-coded measure
on $Y$.

(b) Suppose that $\mu$ is a Baire-coded measure on a
topological space $X$, and $\sequencen{E_n}$ is a codable family in
$\CalBa_c(X)$.   Then

\quad(i) $\mu(\bigcup_{n\in\Bbb N}E_n)\le\sum_{n=0}^{\infty}\mu E_n$;

\quad(ii) If $\sequencen{E_n}$ is non-decreasing,
$\mu(\bigcup_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu E_n$;

\quad(iii) If $\sequencen{E_n}$ is non-increasing and $\mu E_0$ is finite,
then $\mu(\bigcap_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu E_n$.

(c) Let $X$ be a topological space and $\Mu$ a non-empty
upwards-directed family of Baire-coded measures on $X$.   Set
$\nu E=\sup_{\mu\in\Mu}\mu E$ for every codable Baire set $E\subseteq X$.
Then $\nu$ is a Baire-coded measure on $X$.

\proof{{\bf (a)} Use 562T(b-iv).

\medskip

{\bf (b)} Recall that, by 562T(b-i),
there must be a continuous function $f:X\to\BbbR^{\Bbb N}$ and a
codable sequence $\sequencen{F_n}$ in $\Cal B_c(\BbbR^{\Bbb N})$ such that
$E_n=f^{-1}[F_n]$ for every $n$.
By (a), $F\mapsto\mu f^{-1}[F]:\Cal B_c(\BbbR^{\Bbb N})\to[0,\infty]$ is a
Borel-coded measure on $\BbbR^{\Bbb N}$.   Applying 563Ba to
$\sequencen{F_n}$, we get the result here.

\medskip

{\bf (c)} As 563E.
}%end of proof of 563K

\leader{563L}{Proposition}
Suppose that $X$ is a topological space;  write $\Cal G$
for the lattice of cozero subsets of $X$.   Let
$\mu:\Cal G\to[0,\infty]$ be such that

\inset{$\mu\emptyset=0$,

$\mu G\le\mu H$ if $G\subseteq H$,

$\mu$ is modular,

$\mu(\bigcup_{n\in\Bbb N}G_n)=\lim_{n\to\infty}\mu G_n$ whenever
$\sequencen{G_n}$ is a
non-decreasing sequence in $\Cal G$ and there is a sequence
$\sequencen{f_n}$ of continuous functions from $X$ to $\Bbb R$ such that
$G_n=\{x:f_n(x)\ne 0\}$ for every
$n$,\cmmnt{\footnote{\smallerfonts Observe that
$\bigcup_{n\in\Bbb N}G_n$ is a cozero set, defined by
$f:X\to\eightBbb{R}$ where
$f(x)=\sup_{n\in\Bbb N}\min(2^{-n},|f_n(x)|)$ for each $x$.}}

$\mu G=\sup\{\mu H:H\in\Cal G$, $H\subseteq G$, $\mu H<\infty\}$ for every
$G\in\Cal G$.
}

\noindent Then there is a Baire-coded measure on $X$ extending
$\mu$;  if $\mu X$ is finite, then the extension is unique.

\proof{{\bf (a)} Suppose to begin with that $\mu X$ is finite.

\medskip

\quad{\bf (i)} For
each continuous $f:X\to\BbbR^{\Bbb N}$, consider the
functional $G\mapsto\mu f^{-1}[G]$ for open $G\subseteq\BbbR^{\Bbb N}$.
This satisfies the conditions of 563H.   \Prf\ Only the fourth requires
attention.    Fix a metric $\rho$ defining the topology of
$\BbbR^{\Bbb N}$.   If $\sequencen{H_n}$ is a non-decreasing sequence
of open sets in $\BbbR^{\Bbb N}$ with union $H$, set

\Centerline{$h_n(z)=\min(1,\rho(z,\BbbR^{\Bbb N}\setminus H_n))$}

\noindent for $n\in\Bbb N$ and $z\in\BbbR^{\Bbb N}$, counting
$\rho(z,\emptyset)$ as $\infty$ if necessary.   In this case,
setting $G_n=f^{-1}[H_n]$, $G_n=\{x:h_nf(x)>0\}$ for each $n$;  so

\Centerline{$\mu f^{-1}[H]=\mu(\bigcup_{n\in\Bbb N}G_n)
=\lim_{n\to\infty}\mu G_n=\lim_{n\to\infty}\mu f^{-1}[H_n]$.  \Qed}

\noindent There is therefore a unique
Borel-coded measure $\nu_f$ on $\BbbR^{\Bbb N}$ such that
$\nu_fH=\mu f^{-1}[H]$ for every open set $G\subseteq\BbbR^{\Bbb N}$.

\medskip

\quad{\bf (ii)} If $f:X\to\BbbR^{\Bbb N}$ is continuous and
$F\in\Cal B_c(\BbbR^{\Bbb N})$, then
$\nu_fF=\inf\{\mu G:f^{-1}[F]\subseteq G\in\Cal G\}$.   \Prf\
By 563Fd, $\nu_f$ is outer regular with respect to the open sets, so

$$\eqalign{\nu_fF
&=\inf\{\nu_fH:H\subseteq\BbbR^{\Bbb N}\text{ is open}, F\subseteq H\}\cr
&=\inf\{\mu f^{-1}[H]:H\subseteq\BbbR^{\Bbb N}\text{ is open},
   F\subseteq H\}
\ge\inf\{\mu G:f^{-1}[F]\subseteq G\in\Cal G\}.\cr}$$

\noindent In the other direction, if $G\in\Cal G$ and
$f^{-1}[F]\subseteq G$, take any $\epsilon>0$.
There is an open set $H\subseteq\BbbR^{\Bbb N}$ such that
$\BbbR^{\Bbb N}\setminus H\subseteq F$ and
$\nu_f(F\cap H)\le\epsilon$.
But this means $G\cup f^{-1}[H]=X$ and

\Centerline{$\mu G\ge\mu X-\mu f^{-1}[H]
=\nu_f\BbbR^{\Bbb N}-\nu_fH\ge\nu_fF-\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu_fF\le G$;  as $G$ is arbitrary,
$\nu_fF\le\inf\{\mu G:f^{-1}[F]\subseteq G\in\Cal G\}$.\ \Qed

\medskip

\quad{\bf (iii)}
This means that if we set $\nu E=\inf\{\mu G:E\subseteq G\in\Cal G\}$ for
$E\in\CalBa_c(X)$, we shall have $\nu f^{-1}[F]=\nu_fF$ whenever
$f:X\to\BbbR^{\Bbb N}$ is continuous and $F\in\Cal B_c(X)$.
It follows that $\nu$ is a Baire-coded measure on $X$.   \Prf\ Of course
$\nu\emptyset=0$.
If $\sequencen{E_n}$ is a disjoint
codable sequence in $\CalBa_c(X)$, there are a continuous
$f:X\to\BbbR^{\Bbb N}$ and a codable sequence $\sequencen{F_n}$ of coded
Borel sets in $\BbbR^{\Bbb N}$ such that $E_n=f^{-1}[F_n]$ for every $n$,
by 562T(b-i).   Set $F'_n=F_n\setminus\bigcup_{i<n}F_i$ for $n\in\Bbb N$;
then $\sequencen{F'_n}$ is a codable sequence (562Kc), so

\Centerline{$\nu(\bigcup_{n\in\Bbb N}E_n)
=\nu_f(\bigcup_{n\in\Bbb N}F'_n)
=\sum_{n=0}^{\infty}\nu_fF'_n
=\sum_{n=0}^{\infty}\nu E_n$.  \Qed}

\noindent Of course $\nu$ extends $\mu$.

\medskip

\quad{\bf (iv)}
As for uniqueness, if $\nu'$ is any other Baire-coded measure on $X$
extending $\mu$, and $f:X\to\BbbR^{\Bbb N}$ is a continuous function, then
$F\mapsto\nu'f^{-1}[F]$ is a Borel-coded measure on
$\BbbR^{\Bbb N}$ which agrees
with $\nu_f$ on open sets and is therefore equal to $\nu_f$ (563Fg);
it follows at once that $\nu'=\nu$.

\medskip

{\bf (b)} For the general case, let $\Cal G^f$ be
$\{H:H\in\Cal G$, $\mu H<\infty\}$, and for $H\in\Cal G^f$ define
$\mu_H:\Cal G\to\coint{0,\infty}$ by setting $\mu_HG=\mu(G\cap H)$ for
every $G\in\Cal G$.   Then $\mu_H$ satisfies all the conditions of the
proposition.   \Prf\ Everything is elementary;  for the hypothesis on
non-decreasing sequences in $\Cal G$, note that there is a continuous
function $f:X\to\Bbb R$ such that $H=\{x:f(x)\ne 0\}$, so that
if $\sequencen{f_n}$ is a
sequence of real-valued continuous function defining a sequence
$\sequencen{G_n}$ in $\Cal G$, then $\sequencen{f_n\times f}$ defines
$\sequencen{G_n\cap H}$.\ \Qed

There is therefore a unique Baire-coded measure $\nu_H$ on $X$ extending
$\mu_H$.   Now if $H$, $H'\in\Cal G^f$ and $H\subseteq H'$,
$\nu_HE=\nu_{H'}(E\cap H)$ for every $E\in\CalBa_c(X)$.   \Prf\ The
functional $E\mapsto\nu_{H'}(E\cap H)$ is a Baire-coded measure on $X$
extending $\mu_H$, so must be equal to $\nu_H$.\ \QeD\   In particular,
$\nu_HE\le\nu_{H'}E$ for every codable Baire set $E\subseteq X$.

Now set $\nu E=\sup\{\nu_HE:H\in\Cal G^f\}$ for $E\in\CalBa_c(X)$.
By 563Kc, $\nu$ is a Baire-coded measure on $X$;  and by the final
hypothesis of this proposition, $\nu$ extends $\mu$.
}%end of proof of 563L

\leader{563M}{Measure algebras} If $\mu$ is either a Borel-coded measure or
a Baire-coded measure, we can form the quotient Boolean algebra
$\frak A=\dom\mu/\{E:\mu E=0\}$ and the functional
$\bar\mu:\frak A\to[0,\infty]$ defined by setting
$\bar\mu E^{\ssbullet}=\mu E$ for every $E\in\dom\mu$;\cmmnt{ as in 321H,}
$\bar\mu$ is a strictly positive additive functional from $\frak A$ to
$[0,\infty]$.   As in \S323, we have a topology and
uniformity on $\frak A$ defined by the pseudometrics
$(a,b)\mapsto\bar\mu(c\Bcap(a\Bsymmdiff b))$ for $c\in\frak A$ of
finite measure;  if $\mu$ is semi-finite, the topology is Hausdorff.

\leader{563N}{Theorem} Let $X$ be a second-countable space, and $\mu$ a
codably $\sigma$-finite Borel-coded measure on $X$.   Let $\frak A$ and
$\bar\mu$ be as in 563M.   Then $\frak A$ is complete for its
measure-algebra uniformity, therefore Dedekind complete.

\proof{{\bf (a)} There is a codable sequence of sets of finite measure
covering $X$.   By 562Pb, we can find a codably Borel equivalent
second-countable topology $\frak S$
on $X$, generated by a countable algebra $\Cal E$ of subsets of $X$,
for which all these sets are open, so that $\mu$ becomes locally finite,
while $\Cal S$ is regular and second-countable.
Let $\sequencen{H_n}$ be a sequence running over $\Cal E$;  note that
$\sequencen{H_n}$ is codable.

\medskip

{\bf (b)} $\{H^{\ssbullet}:H\in\Cal E\}$ is dense in $\frak A$ for the
measure-algebra topology.   \Prf\ Suppose that $a$, $c\in\frak A$,
$\epsilon>0$ and $\bar\mu c<\infty$.   Express $a$ as $E^{\ssbullet}$
and $c$ as $F^{\ssbullet}$
where $E$, $F\in\Cal B_c(X)$.   By 563Fd, there is a $G\in\frak S$ such
that $E\cap F\subseteq G$ and $\mu(G\setminus(E\cap F))\le\epsilon$.
Setting
$G_n=\bigcup\{H_i:i\le n$, $H_i\subseteq G\}$,  $\sequence{n}{G_n\cap F}$
is a non-decreasing codable sequence with union $G\cap F$, so there is an
$n\in\Bbb N$ such that $\mu((G\setminus G_n)\cap F)\le\epsilon$.   In this
case

\Centerline{$\bar\mu(c\Bcap(a\Bsymmdiff G_n^{\ssbullet}))
=\mu(F\cap(E\symmdiff G_n))
\le\mu(F\cap(G\setminus G_n))+\mu(G\setminus(E\cap F))
\le 2\epsilon$,}

\noindent while $G_n\in\Cal E$.   As $a$, $c$ and $\epsilon$ are arbitrary,
we have the result.\ \Qed

\medskip

{\bf (c)} $\frak A$ is complete for the measure-algebra uniformity.   \Prf\
Set $\tilde H_n=\bigcup\{H_i:i\le n$, $\mu H_i<\infty\}$,
$c_n=\tilde H_n^{\ssbullet}$ for each $n$.
Let $\Cal F$ be a Cauchy filter on $\frak A$ for the measure-algebra
uniformity.   For each $n\in\Bbb N$, there is an $A\in\Cal F$ such that
$\bar\mu(c_n\Bcap(a\Bsymmdiff b))\le 2^{-n}$ for all
$a$, $b\in A$;  there is a $b_0\in A$;  and there is an $m\in\Bbb N$ such
that $\bar\mu(c_n\Bcap(b_0\Bsymmdiff H_m^{\ssbullet}))\le 2^{-n}$, so
that

\Centerline{\hfill
$\{a:\bar\mu(c_n\Bcap(a\Bsymmdiff H_m^{\ssbullet}))\le 2^{-n+1}\}
\in\Cal F$.\hfill(*)}

\noindent Let $m_n$ be the first $m$ for which (*) is true, and set $d_n=H_{m_n}^{\ssbullet}$.   Note that

\Centerline{$\bar\mu(c_i\Bcap(d_{i+1}\Bsymmdiff d_i))\le 3\cdot 2^{-i}$}

\noindent for each $i$, because there must be an $a\in\frak A$ such that
$\bar\mu(c_i\Bcap(a\Bsymmdiff d_i))\le 2^{-i+1}$
and $\bar\mu(c_{i+1}\Bcap(a\Bsymmdiff d_{i+1}))\le 2^{-i}$.

Set $E=\bigcap_{n\in\Bbb N}\bigcup_{i\ge n}H_{m_i}$;  because
$\sequencen{\bigcup_{i\ge n}H_{m_i}}$ is codable, $E\in\Cal B_c(X)$.   Set
$d=E^{\ssbullet}$.   If $n\in\Bbb N$, then

\Centerline{$E\symmdiff H_{m_n}
\subseteq\bigcup_{i\ge n}H_{m_{i+1}}\symmdiff H_{m_i}$}

\noindent and $\sequence{i}{\tilde H_n\cap(H_{m_{i+1}}\symmdiff H_{m_i})}$
is codable, so

$$\eqalign{\bar\mu(c_n\Bcap(d\Bsymmdiff d_n))
&=\mu(\tilde H_n\cap(E\symmdiff H_{m_n}))
\le\sum_{i=n}^{\infty}\mu(\tilde H_n\cap(H_{m_{i+1}}\symmdiff H_{m_i}))\cr
&\le\sum_{i=n}^{\infty}3\cdot 2^{-i}
=6\cdot 2^{-n}.\cr}$$

Take any $c\in\frak A$ such that $\bar\mu c$ is finite, and $\epsilon>0$.
Express $c$ as $F^{\ssbullet}$, where $\mu F<\infty$.   Then
$\sequencen{F\cap\tilde H_n}$ is a non-decreasing codable sequence with
union $F$, so there is an $n\in\Bbb N$ such that
$\mu(F\setminus\tilde H_n)\le\epsilon$ and $2^{-n}\le\epsilon$.   Now

$$\eqalign{
\{a:\bar\mu(c\Bcap(a\Bsymmdiff d))\le 9\epsilon\}
&\supseteq\{a:\bar\mu(c_n\Bcap(a\Bsymmdiff d))\le 8\epsilon\}\cr
&\supseteq\{a:\bar\mu(c_n\Bcap(a\Bsymmdiff d_n))\le 2\epsilon\}
\in\Cal F.\cr}$$

\noindent As $c$ and $\epsilon$ are arbitrary, $\Cal F\to d$
for the measure-algebra topology;  as $\Cal F$ is arbitrary, $\frak A$ is
complete.\ \Qed

\medskip

{\bf (d)} Now suppose that $A\subseteq\frak A$ is a non-empty set,
and $B$ the family of its upper bounds, so that $B$ is downwards-directed.
As in 323D, the filter $\Cal F(B\closedownarrow)$ generated by
$\{B\cap[0,b]:b\in B\}$ is Cauchy for the measure-algebra uniformity,
so has a limit, which is
$\inf B=\sup A$.   As $A$ is arbitrary, $\frak A$ is Dedekind complete.
}%end of proof of 563N

\exercises{\leader{563X}{Basic exercises (a)}
%\spheader 563Xa
Let $X$, $Y$ be second-countable spaces, $\mu$ a
Borel-coded measure on $X$, and $f:X\to Y$ a codable Borel function.
Show that $F\mapsto\mu f^{-1}[F]:\Cal B_c(Y)\to[0,\infty]$ is a Borel-coded
measure on $Y$.
%563A

\spheader 563Xb Let $X$ be a regular second-countable space and $\mu$ a
locally finite Borel-coded measure on $X$.   Show that for every
$E\in\Cal B_c(X)$ there are an F$_{\sigma}$ set $F\subseteq E$ and a
G$_{\delta}$ set $H\supseteq E$ such that $\mu(H\setminus F)=0$.
%563F

\spheader 563Xc Let $X$ be a regular second-countable space.   Show that a
function $\mu$ is a codable Borel measure on $X$ iff it is a codable Baire
measure on $X$.   \Hint{562Xk, 562Xl.}
%563J

\spheader 563Xd Let $X$ be a topological space.   Show that any semi-finite
Baire-coded measure on $X$ is inner regular with respect to the zero
sets.
%563K

\spheader 563Xe\dvAnew{2014}
Let $X$ be a zero-dimensional compact Hausdorff space, $\Cal E$ the
algebra of open-and-closed subsets of $X$ and
$\mu_0:\Cal E\to\coint{0,\infty}$ an additive functional.
Show that there is a unique Baire-coded measure on $X$ extending $\mu_0$.
%563L

%\leader{563Y}{Further exercises (a)}
}%end of exercises

\leader{563Z}{Problem} Suppose we define `probability space' in the
conventional way, following literally the formulations in 111A, 112A and
211B.   Is it relatively consistent with ZF to suppose that every
probability space is purely atomic in the sense of 211K?

\endnotes{
\Notesheader{563} The arguments above are generally
drawn from those used earlier in this treatise;  the new discipline
required
is just to systematically respect the self-denying ordinance renouncing the
axiom of choice, as in part (f) of the proof of 563F.
This does involve us in deeper analyses at a number of
points.   In 563Dc, for instance, we need functions $\pi$,
$\pi'$ defined on $\Cal T\times\Bbb N$, not $\Cal B_c(X)\times\Bbb N$,
because the rank function of $\Cal T$ gives us a foundation for induction.
(In 563Db we can use a function $\pi^*$ defined on $\frak T\times\Bbb N$,
but this is because we have canonical codes for open sets.)
In 563I we can no longer assume the existence of measurable envelopes, let
alone a whole family of them as used in the standard proof in 431A, and
have to find another construction, watching carefully to make sure that we
get not only a countable ordinal $\xi$ but a codable family of sets
$E_{\sigma\eta}$ leading to the measurable envelopes $E_{\sigma\xi}$;
back in 561A, there was a moment when we needed to resist the temptation to
suppose that a sequence in $\omega_1$ must have a supremum in $\omega_1$.

Note that we have to distinguish between `negligible' and `outer measure
zero'.   The natural meaning of the latter is
`for every $\epsilon>0$ there is a measurable set
$E\supseteq A$ with $\mu E\le\epsilon$'.   Even for outer regular
measures, when a set of outer measure zero must be included in open sets of
small measure, we cannot be sure that there is a sequence of such sets from
which we can define a set of measure zero including $A$ (565Xb).

In 563K I have kept the proofs short by quoting results from earlier in the
section.   But you may find it illuminating to look for a list of
properties of codable families of codable Baire sets which would support
formally independent proofs.

In 563M-563N I am taking care to avoid the phrase `measure algebra' in
the formal exposition.   The reason is that the definition in \S321
demands a Dedekind $\sigma$-complete algebra, and in the generality of
563M there is no reason to suppose that this will be satisfied.   In the
special context of 563N, of course, there is no difficulty.

There is something I ought to point out here.   The problem is not that the
principal arguments of \S\S111-113 and \S\S121-123 depend on the axiom of
choice.   If you wish, you can continue to define `$\sigma$-algebra',
`measure', `outer measure', `measurable function' and `integral' with the
same forms of words as used in Volume 1, and the basic theorems, up to and
including the convergence theorems, will still be true.   The problem is
that on these definitions the formulae of \S\S114-115 may not give an outer
measure, and we may have nothing corresponding to Lebesgue measure.   It
does not quite follow that every probability space is purely atomic
(there is a question here:  see
563Z), but clearly we are not going to get a theory which can respond to
any of the basic challenges dealt with in Volume 2
(Fundamental Theorem of Calculus, geometric measure
theory, probability distributions, Fourier series), and I think it more
useful to develop a new structure which can carry an effective
version of the Lebesgue theory (see \S565).
}%end of notes

\discrpage

\frfilename{mt564.tex}
\versiondate{9.2.14}
\copyrightdate{2006}

\def\chaptername{Choice and determinacy}
\def\sectionname{Integration without choice}

\newsection{564}

I come now to the problem
of defining an integral with respect to a Borel- or Baire-coded measure.
Since a Borel-coded measure can be regarded as a Baire-coded measure on a
second-countable space\cmmnt{ (562U)},
I will give the basic results in terms of the wider class.
I seek to follow the general plan of Chapter 12, starting from simple
functions and taking integrable functions to be almost-everywhere
limits of sequences of simple functions (564A);  the concept of
`virtually measurable' function has to be re-negotiated (564Ab).   The
basic convergence theorems from \S123 are restricted but recognisable
(564F).   We also have versions of two of the
representation theorems from \S436 (564H, 564I).

There is a significant change when we come to the completeness of
$L^p$ spaces (564K) and the Radon-Nikod\'ym theorem (564L),
where it becomes necessary to
choose sequences, and we need a well-orderable dense set of functions to
pick from.   Subject to this, we have workable notions of conditional
expectation operator (564Mc) and product measures (564N, 564O).

\leader{564A}{Definitions (a)} Given a topological space $X$ and a
Baire-coded measure $\mu$ on $X$\cmmnt{ (563J)}, I will write
$\CalBa_c(X)^f$ for
the ring of codable Baire sets of finite measure;  $S=S(\CalBa_c(X)^f)$
will be the linear subspace of $\BbbR^X$ generated by
$\{\chi E:E\in\CalBa_c(X)^f\}$\cmmnt{ (see 122Ab,
361D\footnote{\S\S361-362
are written on a general assumption of AC.   The only essential use of it
to begin with, however, is in asserting that an arbitrary Boolean ring
can be faithfully represented as a ring of sets;  and even that can be
dispensed with for a while if we work a little harder, as in 361Ya.})}.
\cmmnt{Then }$S$ is a Riesz subspace of $\BbbR^X$, and also an
$f$-algebra\cmmnt{ in the sense of 352W}.

\spheader 564Ab I will write $\eusm L^0$
for the space of real-valued functions $f$ defined almost
everywhere in $X$ such that there is a codable Baire function
$g:X\to\Bbb R$ such that $f\eae g$.

\spheader 564Ac Let $\int:S\to\Bbb R$ be the
positive linear functional defined by
saying that $\int\chi E=\mu E$ for every $E\in\CalBa_c(X)^f$.
\cmmnt{(The arguments of 361E-361G still apply, so there is such a
functional.)}

\spheader 564Ad $\eusm L^1$ will be the set of those real-valued
functions $f$ defined almost everywhere in $X$ for
which there is a codable sequence $\sequencen{h_n}$ in
$S$ converging to $f$ almost everywhere and such that
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|<\infty$;  I will call such functions
{\bf integrable}.

\leader{564B}{Lemma} Let $X$ be a topological space and $\mu$ a
Baire-coded measure on $X$.

(a) $\eusm L^1\subseteq\eusm L^0$.

(b) If $\sequencen{h_n}$ is a non-increasing codable sequence in
$S=S(\CalBa_c(X)^f)$ and
$\lim_{n\to\infty}h_n(x)=0$ for almost every $x$, then
$\lim_{n\to\infty}\int h_n=0$.

(c) If $\sequencen{h_n}$ and $\sequencen{h'_n}$ are two codable
sequences in
$S$ such that $\lim_{n\to\infty}h_n$ and $\lim_{n\to\infty}h'_n$
are defined and equal almost everywhere, and
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$
and $\sum_{n=0}^{\infty}\int|h'_{n+1}-h'_n|$ are both finite, then
$\lim_{n\to\infty}\int h_n$ and $\lim_{n\to\infty}\int h'_n$ are defined
and equal.

(d) If $\sequencen{h_n}$ is a codable sequence in $S$ and
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$ is finite, then
$\sequencen{h_n}$ converges almost everywhere.   In particular, if
$\sequencen{h_n}$ is a non-decreasing codable sequence in $S$ and
$\sup_{n\in\Bbb N}\int h_n$ is finite, $\sequencen{h_n}$ converges a.e.

(e) If $\sequencen{h_n}$ is a codable sequence in $S^+$ and
$\liminf_{n\to\infty}\int h_n=0$, then $\liminf_{n\to\infty}h_n=0$ a.e.

\proof{{\bf (a)} If $f\in\eusm L^1$, there is a codable sequence
$\sequencen{h_n}$ in $S$ converging almost everywhere to $f$.
Now 562T(c-iii) tells us that there is a codable Baire function $g$ equal
to
$\lim_{n\to\infty}h_n$ wherever this is defined as a real number, so that
$f\eae g$ and $f\in\eusm L^0$.

\medskip

{\bf (b)} Set $E=\{x:h_0(x)>0\}$.   Take any $\epsilon>0$.
For each
$n\in\Bbb N$ set $E_n=\{x:h_n(x)>\epsilon\}$.   Then $\sequencen{E_n}$
is a non-increasing codable sequence in $\CalBa_c(X)^f$ (562Td), and
$\bigcap_{n\in\Bbb N}E_n\subseteq\{x:\lim_{n\to\infty}h_n(x)\ne 0\}$ is
negligible;  also $E_0$ has finite measure.
Accordingly $\lim_{n\to\infty}\mu E_n=0$ (563K(b-iii)).   But

\Centerline{$h_n\le\|h_0\|_{\infty}\chi E_n+\epsilon\chi E$,
\quad$\int h_n\le\|h_0\|_{\infty}\mu E_n+\epsilon\mu E$}

\noindent for every $n$, so
$\limsup_{n\to\infty}\int h_n\le\epsilon\mu E$.
As $\epsilon$ is arbitrary, $\lim_{n\to\infty}\int h_n=0$.

\medskip

{\bf (c)} Since $\int$ is a positive linear functional on the Riesz space
$S$,

\Centerline{$\sum_{n=0}^{\infty}|\int h_{n+1}-\int h_n|
\le\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$}

\noindent is finite, and
the limit $\lim_{n\to\infty}\int h_n$ is defined in $\Bbb R$.   Similarly,
$\lim_{n\to\infty}\int h'_n$ is defined.   To see that the limits are
equal, set $g_n=h_n-h'_n$ for each $n$, so that $\lim_{n\to\infty}g_n=0$
a.e.\ and $\sum_{n=0}^{\infty}\int|g_{n+1}-g_n|<\infty$.   Then
$\int|g_n|\le\sum_{m=n}^{\infty}\int|g_{m+1}-g_m|$ for every $n$.   \Prf\
For $k\ge n$, set $f_k=(|g_n|-\sum_{m=n}^k|g_{m+1}-g_m|)^+$.   Then
$0\le f_k\le|g_{k+1}|$ for each $k$, so
$\langle f_k\rangle_{k\ge n}$ is a non-increasing codable
sequence in $S$ converging to
$0$ almost everywhere.   (To check that $\langle f_k\rangle_{k\ge n}$
is codable, use 562T(c-ii) and the idea of 562Se.)
By (b), $\lim_{n\to\infty}\int f_k=0$;  but
$\int f_k\ge\int|g_n|-\sum_{m=n}^k\int|g_{m+1}-g_m|$ for every $k$.\ \Qed

Consequently

$$\eqalign{|\lim_{n\to\infty}\int h_n-\lim_{n\to\infty}\int h'_n|
&=\lim_{n\to\infty}|\int h_n-\int h'_n|
\le\lim_{n\to\infty}\int|g_n|
\cr&\le\lim_{n\to\infty}\sum_{m=n}^{\infty}\int|g_{m+1}-g_m|
=0\cr}$$

\noindent as required.

\medskip

{\bf (d)} For $k\in\Bbb N$ let $n_k\in\Bbb N$ be the least integer
such that
$\sum_{i=n_k}^{\infty}\int|h_{i+1}-h_i|\le 2^{-k}$, and for $m\ge n_k$ set

\Centerline{$G_{km}=\{x:\sum_{i=n_k}^m|h_{i+1}(x)-h_i(x)|\ge 1\}$.}

\noindent Then $\mu G_{km}\le 2^{-k}$, because
$\chi G_{km}\le\sum_{i=n_k}^m|h_{i+1}-h_i|$, so
$\mu G_k\le 2^{-k}$, where $G_k=\bigcup_{m\ge n_k}G_{km}$, by 563K(b-i).
(Of course we have to check that all the sequences of sets and functions
involved here are codable.)   Accordingly,
setting $E=\bigcap_{k\in\Bbb N}G_k$, $\mu E=0$.   But observe
that if $x\in X\setminus E$ there is a $k\in\Bbb N$ such that
$x\notin G_k$ and $\sum_{i=n_k}^{\infty}|h_{i+1}(x)-h_i(x)|\le 1$;
in which case $\lim_{n\to\infty}h_n(x)$ is defined.


\medskip

{\bf (e)} For $k\in\Bbb N$ let $n_k$ be the least integer such that
$n_k>n_i$ for $i<k$ and $\int h_{n_k}\le 4^{-n_k}$.   Set
$G_k=\{x:h_{n_k}(x)\ge 2^{-k}\}$;  then $\mu G_k\le 2^{-k}$ and
$\sequence{k}{G_k}$ is codable.   So
$\mu(\bigcup_{k\ge n}G_k)\le 2^{-n+1}$ for every $n$ and
$E=\bigcap_{n\in\Bbb N}\bigcup_{k\ge n}G_k$ is negligible.   But
$E\supseteq\{x:\liminf_{n\to\infty}h_n(x)>0\}$.
}%end of proof of 564B

\leader{564C}{Definition} Let $X$ be a topological space and $\mu$ a
Baire-coded measure on $X$.   For $f\in\eusm L^1$, define its integral
$\int f$ by saying that $\int f=\lim_{n\to\infty}\int h_n$ whenever
$\sequencen{h_n}$ is a codable sequence in $S=S(\CalBa_c(X)^f)$ converging to $f$ almost
everywhere and $\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$ is finite.
\cmmnt{By
564Bc, this definition is sound;  and clearly it is consistent with the
previous definition of the integral on $S$.}

\leader{564D}{Lemma} Let $X$ be a topological space and $\sequencen{f_n}$ a
codable sequence of codable Baire functions on $X$.
Let $\sequence{i}{q_i}$ be an enumeration of $\Bbb Q\cap\coint{0,\infty}$,
starting with $q_0=0$.   Set

\Centerline{$f'_n(x)=\max\{q_i:i\le n$, $q_i\le\max(0,f_n(x))\}$}

\noindent for $n\in\Bbb N$ and $x\in X$.   Then $\sequencen{f'_n}$ is a
codable sequence of codable Baire functions.

\proof{ Take a sequence running over a base for the topology of
$\BbbR^{\Bbb N}$ and the corresponding interpretation
$\phi:\Cal T\to\Cal B_c(\BbbR^{\Bbb N})$ of Borel codes, as in 562B, and
let $\tilde\phi:\tilde T\to\BbbR^X$ be the corresponding interpretation of codes for
real-valued codable Borel functions, as in 562N.   By
562T(c-ii), there are a continuous function $g:X\to\BbbR^{\Bbb N}$ and
a sequence $\sequencen{\tau_n}$ of codes such that
$f_n=\tilde\phi(\tau_n)\frsmallcirc g$ for every $n$.   We need
a sequence $\sequencen{\tau'_n}$ of codes such that

$$\eqalign{\phi(\tau'_n(\alpha))
&=\bigcup_{\Atop{i\le n}{q_i>\alpha}}
  \bigcap_{\Atop{j\in\Bbb N}{q_j<q_i}}\phi(\tau_n(q_j))
\text{ if }\alpha\ge 0,\cr
&=X\text{ if }\alpha<0;\cr}$$

\noindent and this is easy to build using complementation and
general union operators as in
562C.   Now take $f'_n=\tilde\phi(\tau'_n)\frsmallcirc g$ for each $n$.
}%end of proof of 564D

\leader{564E}{Theorem} Let $X$ be a
topological space and $\mu$ a Baire-coded measure on $X$.

(a)(i) If $f$, $g\in\eusm L^0$ and $\alpha\in\Bbb R$,
then $f+g$, $\alpha f$, $|f|$ and $f\times g$ belong to $\eusm L^0$.

\quad(ii) If $h:\Bbb R\to\Bbb R$ is a codable Borel function,
$hf\in\eusm L^0$ for every $f\in\eusm L^0$.

(b) If $f$, $g\in\eusm L^1$ and $\alpha\in\Bbb R$,
then

\quad(i) $f+g$, $\alpha f$ and $|f|$ belong to $\eusm L^1$;

\quad(ii) $\int f+g=\int f+\int g$, $\int\alpha f=\alpha\int f$;

\quad(iii) if $f\leae g$ then $\int f\le\int g$.

(c)(i) If $f\in\eusm L^0$, $g\in\eusm L^1$ and $|f|\leae g$, then
$f\in\eusm L^1$.

\quad(ii) If $E\in\CalBa_c(X)$ and $\chi E\in\eusm L^1$ then $\mu E$ is
finite.

\proof{{\bf (a)(i)} Use 562T(c-iv).

\medskip

\quad{\bf (ii)} We know that $f$ is equal almost everywhere to a product
$f'g$ where
$g:X\to\BbbR^{\Bbb N}$ is continuous and $f':\BbbR^{\Bbb N}\to\Bbb R$ is a
codable Borel function.   Now $hf'$ is a codable Borel function, by
562Mb, so $hf'g$ is a codable Baire function and $hf\eae hf'g$ belongs to
$\eusm L^0$.

\medskip

{\bf (b)(i)-(ii)} These proceed by the same arguments as in (a-i).
To deal with $|f|$, we need to note that if $\sequencen{h_n}$ is any
codable sequence in $S=S(\CalBa_c(X)^f)$ then $\sum_{n=0}^{\infty}\int||h_{n+1}|-|h_n||
\le\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$.

\medskip

\quad{\bf (iii)} If $f\leae g$, let $\sequencen{f_n}$, $\sequencen{g_n}$ be
codable sequences in $S$ converging almost everywhere to $f$, $g$
respectively, and such that $\sum_{n=0}^{\infty}\int|f_{n+1}-f_n|$ and
$\sum_{n=0}^{\infty}\int|g_{n+1}-g_n|$ are finite.   Set
$h_n=f_n\wedge g_n$ for each $n$.   Then $\sequencen{h_n}$ is codable,
$f\eae\lim_{n\to\infty}h_n$,
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|<\infty$ and

\Centerline{$\int f=\lim_{n\to\infty}\int h_n\le\lim_{n\to\infty}\int g_n
=\int g$.}

\medskip

{\bf (c)(i)} Let $\sequencen{h_n}$ be a codable sequence in $S$ such that
$g\eae\lim_{n\to\infty}h_n$ and $\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$ is
finite.   Set $h'_n=\sup_{i\le n}h_i^+$ for each $n$;  then
$\sequencen{h'_n}$ is a non-decreasing codable sequence in $S$ and
$\sup_{n\in\Bbb N}\int h'_n$ is finite, while
$|f|\leae g\leae\sup_{n\in\Bbb N}h'_n$.   There is a codable
Baire function $\tilde f$ such that
$f\eae\tilde f$.   Now $\tilde f^+$ is a codable Baire function, so
$\sequencen{\tilde f^+\wedge h'_n}$ is a codable sequence of
non-negative codable Baire functions.

For each $n\in\Bbb N$ consider $h''_n$ where

\Centerline{$h''_n(x)
  =\max\{q_i:i\le n$, $q_i\le\max(0,(\tilde f^+\wedge h'_n)(x)\}$}

\noindent for $x\in X$.
By 564D, $\sequencen{h''_n}$ is a codable
sequence;  it is non-decreasing and converges a.e.\ to
$\tilde f^+\eae f^+$.   Because $0\le h''_n\le h'_n$,
$h''_n\in S$ for each $n$, and
$\sup_{n\in\Bbb N}\int h''_n\le\sup_{n\in\Bbb N}\int h'_n$ is finite;  so
564Bd tells us that $f^+$ is integrable.

Similarly, $f^-$ is integrable, so $f$ is integrable.

\medskip

\quad{\bf (ii)} Let $\sequencen{h_n}$ be a codable sequence in $S$ such
that $\chi E\eae\lim_{n\to\infty}h_n$ and
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|$ is finite.   Set
$h'_n=\sup_{i\le n}h_i$ for each $n$;  then $\sequencen{h'_n}$ is a codable
sequence in $S$ and
$\sup_{n\in\Bbb N}\int h'_n$ is finite.   Set
$E_n=\{x:h'_n(x)>\bover12\}$;  then $\sequencen{E_n}$ is a non-decreasing
codable sequence in $\CalBa_c(X)^f$.   Also
$E\setminus\bigcup_{n\in\Bbb N}E_n$ is negligible, so

\Centerline{$\mu E\le\mu(\bigcup_{n\in\Bbb N}E_n)
=\sup_{n\in\Bbb N}\mu E_n=\sup_{n\in\Bbb N}\int\chi E_n
\le 2\sup_{n\in\Bbb N}\int h'_n$}

\noindent is finite.
}%end of proof of 564E

\leader{564F}{}\cmmnt{ I come now to versions of the fundamental
convergence theorems.

\medskip

\noindent}{\bf Theorem} Let $X$ be a topological space and
$\mu$ a Baire-coded measure on $X$.   Suppose that $\sequencen{f_n}$ is a
codable sequence of integrable codable Baire functions on $X$.

(a) If $\sequencen{f_n}$ is non-decreasing and
$\gamma=\sup_{n\in\Bbb N}\int f_n$ is finite, then
$f=\lim_{n\to\infty}f_n$ is defined a.e.\ and is integrable, and
$\int f=\gamma$.

(b) If every $f_n$ is non-negative and $\liminf_{n\to\infty}\int f_n$
is finite, then $f=\liminf_{n\to\infty}f_n$ is defined a.e.\ and is
integrable, and $\int f\le\liminf_{n\to\infty}\int f_n$.

(c) Suppose that
there is a $g\in\eusm L^1$ such that $|f_n|\leae g$ for every $n$,
and $f=\lim_{n\to\infty}f_n$ is defined a.e.   Then $\int f$ and
$\lim_{n\to\infty}\int f_n$ are defined and equal.

(d) If $\sum_{n=0}^{\infty}\int|f_{n+1}-f_n|$
is finite, then $f=\lim_{n\to\infty}f_n$ is defined a.e., and $\int f$ and
$\lim_{n\to\infty}\int f_n$ are defined and equal.

(e) If $\sum_{n=0}^{\infty}\int|f_n|$ is
finite, then $f=\sum_{n=0}^{\infty}f_n$ is defined a.e., and $\int f$ and
$\sum_{n=0}^{\infty}\int f_n$ are defined and equal.

\proof{{\bf (a)} Replacing $f_n$ by $f_n-f_0$ for each $n$, we may suppose
that $f_n\ge 0$ for each $n$.
Let $\sequence{i}{q_i}$ be an enumeration of $\Bbb Q\cap\coint{0,\infty}$
and set

\Centerline{$h_n(x)
  =\max\{q_i:i\le n$, $q_i\le\max(0,f_n(x))\}$}

\noindent for each $x\in X$.   Then $\sequencen{h_n}$ is a codable sequence
of codable Baire functions (use 564D again).   Moreover, $h_n$ takes only
finitely many values, all non-negative, and for $\alpha>0$ the set
$E_{n\alpha}=\{x:h_n(x)>\alpha\}$ is a codable Baire set such that
$\chi E_{n\alpha}\leae\bover1{\alpha}f_n$;  by 564Ec, $E_{n\alpha}$ has
finite measure;  as $\alpha$ is arbitrary, $h_n\in S$.

Now $\sequencen{h_n}$ is non-decreasing, and $\int h_n\le\int f_n\le\gamma$
for every $n$;   so by 564Bd $\sequencen{h_n}$ converges almost
everywhere to an integrable function $f_1$, with $\int f_1\le\gamma$.
Of course $f_1\eae\lim_{n\to\infty}f_n=f$;  as $f\geae f_n$ for every
$n$, $\int f=\int f_1=\gamma$ exactly.

\medskip

{\bf (b)} By 562T(c-ii) and 562Oc,
$\sequencen{f'_n}$ is codable, where $f'_n=\inf_{m\ge n}f_m$ for
every $n$, and of course

\Centerline{$\int f'_n\le\inf_{m\ge n}\int f_m
\le\liminf_{m\to\infty}\int f_m$}

\noindent for every $n$.   Now $\sequencen{f'_n}$ is non-decreasing,
so (a) tells us that
$\int\liminf_{n\to\infty}f_n=\int\lim_{n\to\infty}f'_n$ is defined and
equal to $\lim_{n\to\infty}\int f'_n\le\liminf_{n\to\infty}\int f_n$.

\medskip

{\bf (c)} Let $g'$ be a codable Baire function such that $g'\eae g$, and
set $f'_n=\med(-g',f_n,g')$ for each $n$;  once again, 562T(c-ii) and the
ideas of 562Oc show that $\sequencen{g'+f'_n}$ is codable.
So we can use (b) to see that
$\int\liminf_{n\to\infty}g'+f'_n$ is defined and is at most
$\liminf_{n\to\infty}\int g'+f'_n=\int g'+\liminf_{n\to\infty}\int f_n$.
Subtracting $g'$, we get
$\int\liminf_{n\to\infty}f'_n\le\liminf_{n\to\infty}\int f_n$.
Similarly, $\int\limsup_{n\to\infty}f'_n\ge\limsup_{n\to\infty}\int f_n$.

Once again, the sequences
$\sequencen{f_n}$, $\sequencen{f'_n}$, $\sequencen{|f'_n-f_n|}$ and
$\sequencen{\{x:f'_n(x)\ne f_n(x)\}}$ are all codable.
Since all the sets
$\{x:f'_n(x)\ne f_n(x)\}$ are negligible, so is their union;  but this
means that $\lim_{n\to\infty}f'_n\eae\lim_{n\to\infty}f_n$ is defined
almost everywhere.   So (just as in 123C) the integrals are sandwiched, and
$\int\lim_{n\to\infty}f_n=\lim_{n\to\infty}\int f_n$.

\medskip

{\bf (d)} Of course $\sum_{n=0}^{\infty}|\int f_{n+1}-\int f_n|$ is finite,
so $\gamma=\lim_{n\to\infty}\int f_n$ is defined.
Next, (a) tells us that $g=|f_0|+\sum_{n=0}^{\infty}|f_{n+1}-f_n|$ is
defined a.e.\ and is integrable (of course this depends on our having a
procedure -- induction is allowed -- for building a sequence of Baire codes
representing $\sequencen{|f_0|+\sum_{i=0}^n|f_{i+1}-f_i|}$ out of a
sequence of codes representing $\sequencen{f_n}$).
Since $\lim_{n\to\infty}f_n(x)$ is
defined whenever $g(x)$ is defined and finite, which is almost everywhere,
and $|f_n|\leae g$ for every $n$, (c) gives the result we're looking for.

\medskip

{\bf (e)} Similarly, $\sequencen{\sum_{i=0}^nf_i}$ is
codable and we can apply (d).
}%end of proof of 564F

\vleader{72pt}{564G}{Integration over subsets:  Proposition} Let $X$ be a
topological space and $\mu$ a Baire-coded measure on $X$.

(a) If $f\in\eusm L^1$, the functional
$E\mapsto\int f\times\chi E:\CalBa_c(X)\to\Bbb R$ is additive and truly
continuous with respect to $\mu$.\cmmnt{\footnote{The
definition of `truly continuous' in 232Ab assumed that $\mu$ was defined on
a $\sigma$-algebra.   I hope it is obvious that the same formulation makes
sense when the domain of $\mu$ is any Boolean algebra.}}

(c) If $f$, $g\in\eusm L^1$, then $f\leae g$ iff
$\int f\times\chi E\le\int g\times\chi E$ for every $E\in\CalBa_c(X)$.
\cmmnt{So }$f\eae g$ iff
$\int f\times\chi E=\int g\times\chi E$ for every $E\in\CalBa_c(X)$.

\proof{{\bf (a)} If $E\in\CalBa_c(X)$ then $\chi E$ is a codable Baire
function (use 562Nf), so that $f\times\chi E$ is
integrable (564E(a-i), 564E(c-i)).
Because $\chi:\CalBa_c(X)\to\eusm L^0$ is additive,
$E\mapsto\int f\times\chi E$ is additive.   To see that it is truly
continuous, take $\epsilon>0$.   There is a codable
sequence $\sequencen{h_n}$ in $S=S(\CalBa_c(X)^f)$ such that
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|<\infty$ and
$f\eae\lim_{n\to\infty}h_n$.   For each $n$,

\Centerline{$\int|f-h_n|=\lim_{m\to\infty}\int|h_m-h_n|
\le\sum_{m=n}^{\infty}\int|h_{m+1}-h_m|$,}

\noindent so there is an $n$ such that $\int|f-h_n|\le\bover12\epsilon$.
Set $E=\{x:h_n(x)\ne 0\}$ and $\delta=\epsilon/(1+2\|h_n\|_{\infty})$.
Then $E$ has finite measure.   If $F\in\Cal B_c(X)$ and
$\mu(E\cap F)\le\delta$, then

\Centerline{$|\int f\times\chi F|
\le\int|f-h_n|+\int|h_n|\times\chi F
\le\Bover12\epsilon+\|h_n\|_{\infty}\mu(E\cap F)
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, the functional is truly continuous.

\medskip

{\bf (b)(i)} If $f\leae g$ and $E\in\Cal B_c(X)$, then
$f\times\chi E\leae g\times\chi E$ so
$\int f\times\chi E\le\int g\times\chi E$.

\medskip

\quad{\bf (ii)} If $\int f\times\chi E\le\int g\times\chi E$ for every
$E\in\Cal B_c(X)$, let $\sequencen{h_n}$ be a codable sequence in $S$ such
that $f-g\eae\lim_{n\to\infty}h_n$ and
$\sum_{n=0}^{\infty}\int|h_{n+1}-h_n|<\infty$.   For $k\in\Bbb N$ let
$n_k$ be the least integer such that
$\sum_{m=n_k}^{\infty}\int|h_{m+1}-h_m|\le 2^{-k}$.   For $m$, $k\in\Bbb N$
set $E_{mk}=\{x:h_{n_k}(x)\ge 2^{-m}\}$.   Then

$$\eqalign{\mu E_{mk}
&\le 2^m\int h_{n_k}\times\chi E_{mk}
\le 2^m\int(h_{n_k}-f+g)\times\chi E_{mk}\cr
&=2^m\lim_{i\to\infty}\int(h_{n_k}-h_i)\times\chi E_{mk}
\le 2^m\lim_{i\to\infty}\int|h_{n_k}-h_i|
\le 2^{m-k}.\cr}$$

\noindent Also
$\langle E_{mk}\rangle_{m,k\in\Bbb N}$ is a codable family in
$\CalBa_c(X)$, so $\mu(\bigcup_{k\ge 2m}E_{mk})\le 2^{-m+1}$ for every
$m$ and $\mu E=0$, where

\Centerline{$E
=\bigcup_{l\in\Bbb N}\bigcap_{m\ge l}\bigcup_{k\ge 2m}E_{mk}$.}

\noindent But for $x\in X\setminus E$,
$\limsup_{k\to\infty}h_{n_k}(x)\le 0$.   Since
$f-g\eae\lim_{k\to\infty}h_{n_k}$, $f\leae g$.
}%end of proof of 564G

\leader{564H}{Theorem} Let $X$ be a topological space, and
$f:C_b(X)\to\Bbb R$ a sequentially smooth
positive linear functional\cmmnt{, where $C_b(X)$ is the space of
bounded continuous real-valued functions on $X$}.
Then there is a totally finite Baire-coded measure $\mu$ on
$X$ such that $f(u)=\int u\,d\mu$ for every $u\in C_b(X)$.

\proof{{\bf (a)} For cozero sets $G\subseteq X$ set
$\mu_0 G=\sup\{f(u):u\in C_b(X)$, $0\le u\le\chi G\}$.   Then
$\mu_0 G=\lim_{n\to\infty}f(u_n)$ whenever $G\subseteq X$ is a cozero set
and $\sequencen{u_n}$ is a
non-decreasing sequence in $C_b(X)^+$ with supremum $\chi G$ in $\BbbR^X$.
\Prf\ Setting
$\gamma=\sup_{n\in\Bbb N}f(u_n)$, then of course

\Centerline{$\mu_0 G\ge\gamma=\lim_{n\to\infty}f(u_n)$.}

\noindent On the other hand, if $v\in C_b(X)$ and $0\le v\le\chi G$,
$\sequencen{(v-u_n)^+}$ is a non-increasing sequence
converging to $\tbf{0}$ pointwise, so

\Centerline{$f(v)\le f(u_n)+f(v-u_n)^+\le\gamma+f(v-u_n)^+\to\gamma$}

\noindent as $n\to\infty$.   As $v$ is arbitrary, $\mu_0 G\le\gamma$.\ \Qed

\medskip

{\bf (b)} It follows that $\mu_0$ satisfies the conditions of 563L.
\Prf\ Of course $\mu_0\emptyset=0$ and $\mu_0$ is monotonic.
If $G$, $H\subseteq X$
are cozero sets, express them as $\{x:u(x)>0\}$ and $\{x:v(x)>0\}$ where
$u$, $v\in C_b(X)^+$.   Set $u_n=nu\wedge\chi X$, $v_n=nv\wedge\chi X$ for
each $n$;  then $\sequencen{u_n}$, $\sequencen{v_n}$ are non-decreasing
sequences in $C_b(X)^+$ converging pointwise to $\chi G$, $\chi H$
respectively.   Now $\sequencen{u_n\wedge v_n}$ and
$\sequencen{u_n\vee v_n}$ are also non-decreasing sequences in $C_b(X)^+$
converging to $\chi(G\cap H)$, $\chi(G\cup H)$;  so (a) tells us that

$$\eqalign{\mu_0(G\cup H)+\mu_0(G\cap H)
&=\lim_{n\to\infty}f(u_n\wedge v_n)+\lim_{n\to\infty}f(u_n\vee v_n)\cr
&=\lim_{n\to\infty}f(u_n\wedge v_n+u_n\vee v_n)\cr
&=\lim_{n\to\infty}f(u_n+v_n)
=\mu_0 G+\mu_0 H.\cr}$$

\noindent As for the penultimate condition in 563L,
let $\sequencen{G_n}$ be a
non-decreasing sequence of cozero sets such that there is a sequence
$\sequencen{v_n}$ in $C(X)$ such that $G_n=\{x:v_n(x)\ne 0\}$ for each $n$.
Set $u_n=\chi X\wedge n\sup_{i\le n}|v_i|$ for each $n$, and
$G=\bigcup_{n\in\Bbb N}G_n$;  then
$\sequencen{u_n}\uparrow\chi G$, so

\Centerline{$\mu_0 G
=\lim_{n\to\infty}f(u_n)\le\lim_{n\to\infty}\mu_0 G_n\le\mu_0G$,}

\noindent as required.\ \Qed

\medskip

{\bf (c)} We therefore have a Baire-coded measure $\mu$ on $X$ extending
$\mu_0$.   Now take any $u\in C_b(X)$ such that $0\le u\le\chi X$, and
$n\ge 1$.   For each $i<n$ set $G_i=\{x:u(x)>\bover{i}{n}\}$;  then

\Centerline{$\Bover1n\sum_{i=0}^{n-1}\chi G_i\le u+\Bover1n\chi X$,}

\noindent so

\Centerline{$\Bover1n\sum_{i=0}^{n-1}\mu G_i
\le\int u\,d\mu+\Bover1n\mu X$.}

\noindent Next, setting

\Centerline{$v_i=u\wedge\Bover{i+1}n\chi X-u\wedge\Bover{i}n\chi X$}

\noindent for $i<n$, $u=\sum_{i=0}^{n-1}v_i$ and $nv_i\le\chi G_i$ for
each $i$, so

\Centerline{$f(u)=\sum_{i=0}^{n-1}f(v_i)\le\Bover1n\sum_{i=0}^{n-1}\mu G_i
\le\int u\,d\mu+\Bover1n\mu X$.}

\noindent As $n$ is arbitrary, $f(u)\le\int u\,d\mu$.   On the other hand,
$f(\chi X)=\mu X=\int\chi X\,d\mu$ and $f(\chi X-u)\le\int(\chi X-u)d\mu$;
so in fact $f(u)=\int u\,d\mu$.

\medskip

{\bf (d)} It follows at once that $f(u)=\int u\,d\mu$ for every
$u\in C_b(X)^+$ and therefore for every $u\in C_b(X)$, as required.
}%end of proof of 564H

\leader{564I}{Riesz Representation Theorem} Let $X$ be a completely
regular locally compact space, and $f:C_k(X)\to\Bbb R$ a positive
linear functional\cmmnt{, where $C_k(X)$ is the space of continuous
real-valued functions with compact support}.
Then there is a Baire-coded measure $\mu$ on $X$ such
that $\int u\,d\mu$ is defined and equal to $f(u)$ for every $u\in C_k(X)$.

\proof{ We can follow the plan of 564H, with minor modifications.

\medskip

{\bf (a)} For open sets $G\subseteq X$ write
$D_G=\{u:u\in C_k(X)$, $0\le u\le\chi X$, $\supp u\subseteq G\}$, where
$\supp u=\overline{\{x:u(x)\ne 0\}}$.   We need to know that if $G$,
$H\subseteq X$ are open and $K\subseteq G\cup H$, $K'\subseteq G\cap H$
are compact, there are
$u\in D_G$, $v\in D_H$ such that $\chi K\le u\vee v$ and
$\chi K'\le u\wedge v$.   \Prf\ Because
$X$ is completely regular, the family
$\{\interior\{x:u(x)=1\}:u\in D_G\cup D_H\}$ is an open cover of $G\cup H$
and has a finite subfamily covering $K$;  because $D_G$ and $D_H$ are
upwards-directed, we can reduce this finite subfamily to two terms, one
corresponding to $u_1\in D_G$ and the other to $v_1\in D_H$, so that
$\chi K\le u_1\vee v_1$.
Next, $\{\interior\{x:u(x)=1\}:u\in D_G\}$ is an open cover of
$G\supseteq K'$, so we can find a $u_2\in D_G$ such that $\chi K'\le u_2$;
similarly, there is a $v_2\in D_H$ such that $\chi K'\le v_2$;  set
$u=u_1\vee u_2$ and $v=v_1\wedge v_2$.\ \Qed

\medskip

{\bf (b)} For cozero $G\subseteq X$, set
$\mu_0G=\sup\{f(u):u\in D_G\}$.   If $G$, $H\subseteq X$ are cozero sets,
$u\in D_G$ and $v\in D_H$, then $u\vee v\in D_{G\cup H}$ and
$u\wedge v\in D_{G\cap H}$;  this is enough to show that
$\mu_0G+\mu_0H\le\mu_0(G\cup H)+\mu_0(G\cap H)$.   If $w\in D_{G\cup H}$
and $w'\in D_{G\cap H}$, (a) tells us that
there are $u\in D_G$ and $v\in D_H$ such that
$u\vee v\ge\chi(\supp w)\ge w$
and $u\wedge v\ge\chi(\supp w')\ge w'$;  this is what we need to show that so that
$\mu_0G+\mu_0H\ge\mu_0(G\cup H)+\mu_0(G\cap H)$.

If $\sequencen{G_n}$ is a non-decreasing sequence of cozero sets, defined
from a sequence of continuous functions so that
$G=\bigcup_{n\in\Bbb N}G_n$ is a cozero set, then
$D_G=\bigcup_{n\in\Bbb N}D_{G_n}$ so that
$\mu_0G=\sup_{n\in\Bbb N}\mu_0G_n$.

If $G$ is a relatively compact cozero set then $\mu_0G<\infty$.
\Prf\ There is a $w\in C_k(X)$ such that $\chi\overline{G}\le w$,
so that $\mu G\le f(w)$.\ \QeD\   If $G$ is a cozero set and
$\gamma<\mu_0G$, there is a $u\in D_G$ such that $f(u)\ge\gamma$.
Now there is a $v\in D_G$ such that $\chi(\supp(u))\le v$, so that
$\mu_0H\ge\gamma$, where $H=\{x:v(x)>0\}$;  as $H$ is relatively compact,
$\mu_0H$ is finite.   Thus $\mu_0G=\sup\{\mu_0H:H\subseteq G$ is a cozero
set, $\mu_0H<\infty\}$.

The other hypotheses of 563L are elementary, so we have a Baire-coded
measure on $X$ extending $\mu_0$.

\medskip

{\bf (c)} If $u\in C_k(X)$ and $0\le u\le\chi X$ and $\epsilon>0$,
let $G$ be a relatively
compact cozero set including $\supp u$, and $v\in D_G$ such that
$\chi(\supp u)\le v$ and $f(v)\ge\mu G-\epsilon$.   The argument of
part (c) of the proof of 564H, with $v$ in place of $\chi X$, shows that
$f(u)\le\int u\,d\mu+\bover1n\int v\,d\mu$ for every $n$, so that
$f(u)\le\int u\,d\mu$.   On the other hand,

$$\eqalign{f(u)
&=f(v)-f(v-u)
\ge\mu G-\epsilon-\int v-u\,d\mu\cr
&=\mu G-\int v\,d\mu+\int u\,d\mu-\epsilon
\ge\int u\,d\mu-\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $f(u)=\int u\,d\mu$.   Of course it
follows at once that $f$ agrees with $\int\,d\mu$ on the whole of $C_k(X)$.
}%end of proof of 564I

\leader{564J}{The space $L^1$} Let $X$ be a topological space and
$\mu$ a Baire-coded measure on $X$.

\spheader 564Ja If $f$, $g\in\eusm L^1$ then $f\eae g$ iff $\int|f-g|=0$.
\prooflet{\Prf\
If $f\eae g$ then $|f-g|=0$ a.e.\ and $\int|f-g|=0$ by the definition
in 564Ad.   If $\int|f-g|=0$, let $f_1$, $g_1$ be codable Baire functions
such that $f\eae f_1$ and $g\eae g_1$ (564Ba);  then $|f_1-g_1|$ is
codable.
For each $n\in\Bbb N$, set $E_n=\{x:|f_1(x)-g_1(x)|\ge 2^{-n}\}$.
Then $E_n\in\CalBa_c(X)$ and
$|f_1-g_1|\ge 2^{-n}\chi E_n$ so $\mu E_n=\int\chi E_n=0$.
But $\sequencen{E_n}$ is a
codable sequence so $\bigcup_{n\in\Bbb N}E_n=\{x:f_1(x)\ne g_1(x)\}$ is
negligible and

\Centerline{$f\eae f_1\eae g_1\eae g$.  \Qed}}

\spheader 564Jb As in \S242, we have an equivalence
relation $\sim$ on $\eusm L^1$ defined by saying that $f\sim g$ if
$f\eae g$.   The set $L^1$ of equivalence classes has a Riesz space
structure and a Riesz norm inherited from the addition, scalar
multiplication, ordering and integral on $\eusm L^1$.

\spheader 564Jc\cmmnt{ As in \S242,} I\cmmnt{ will} define
$\int:L^1\to\Bbb R$ by saying
that $\int f^{\ssbullet}=\int f$ for every $f\in\eusm L^1$.   Similarly, we
can define $\int_Eu$, for $u\in L^1$ and $E\in\CalBa_c(X)$, by saying that
$\int_Ef^{\ssbullet}=\int f\times\chi E$ for $f\in\eusm L^1$.

\leader{564K}{}\cmmnt{ In order to prove that an $L^1$-space is
norm-complete, it seems that we need extra conditions.

\medskip

\noindent}{\bf Theorem} Let $X$ be a second-countable space and $\mu$ a
codably $\sigma$-finite Borel-coded measure on $X$.   Then $L^1(\mu)$ is a
separable $L$-space.

\proof{ (Compare 563N.)

\medskip

{\bf (a)} There is a codable sequence of sets of finite measure
covering $X$.   By 562Pb, we can find a codably Borel equivalent
zero-dimensional second-countable topology on $X$ for which all the
these sets
are open, so that $\mu$ becomes locally finite.   Since this procedure does
not change $\eusm L^1$ and $L^1$, we may suppose from the beginning that
$X$ is regular and $\mu$ is locally finite.   Let $\sequencen{U_n}$ run
over a countable base for the topology of $X$ containing $\emptyset$
and closed under finite unions.

\medskip

{\bf (b)} If $E\in\Cal B_c(X)^f$ and $\epsilon>0$, there is an open
$G\subseteq X$ such that $E\subseteq G$ and $\mu(G\setminus E)\le\epsilon$,
by 563Fd.
Next, $G=\bigcup\{U_n:n\in\Bbb N$, $U_n\subseteq G\}$, so there is a finite
set $I\subseteq\Bbb N$ such that $G'=\bigcup_{n\in I}U_n\subseteq G$ and
$\mu(G\setminus G')\le\epsilon$;  now $\mu(G'\symmdiff E)\le 2\epsilon$ and
$G'=U_m$ for some $m$.

\medskip

{\bf (c)} If $f\in\eusm L^1$ and $\epsilon>0$, there is an
$h\in S(\Cal B_c(X)^f)$ such that $\int|f-h|\le\epsilon$;  now there must
be an $n\in\Bbb N$ and a family $\langle q_i\rangle_{i\le n}$
in $\Bbb Q$ such that
$\int|h-\sum_{i=0}^nq_i\chi U_i|\le\epsilon$.   The set $D$ of such
rational linear combinations of the $\chi U_i$ is countable;  enumerate it
as $\sequencen{h_n}$.   All the $h_n$ are differences of semi-continuous
functions, therefore resolvable, so $\sequencen{h_n}$ is a codable
sequence;  and for any $u\in L^1$ and $\epsilon>0$ there is an $n$ such
that $\|u-h_n^{\ssbullet}\|_1\le 2\epsilon$.

\medskip

{\bf (d)} This shows that $L^1$ is separable.   To see that it is complete,
take a Cauchy filter $\Cal F$ on $L^1$.   For each
$k\in\Bbb N$ we can take the first $n_k\in\Bbb N$ such that
$\{u:\|u-h_{n_k}^{\ssbullet}\|_1\le 2^{-k}\}$ belongs to $\Cal F$.   Now
$\int|h_{n_k}-h_{n_{k+1}}|\le 2^{-k}+2^{-k-1}$ for every $k$, so the
codable sequence $\sequence{k}{h_{n_k}}$ converges a.e.\ to some
$f\in\eusm L^1$ (564Fd),
and $\int|f-h_{n_k}|\le 3\cdot 2^{-k}$ for every $k$.   So

\Centerline{$f^{\ssbullet}
=\lim_{k\to\infty}h_{n_k}^{\ssbullet}=\lim\Cal F$.}

\medskip

{\bf (e)} Thus $L^1$ is norm-complete.   We know it is a
Riesz space with a Riesz norm, so it is a Banach lattice.   As for the
additivity of the norm on the positive cone, we have only to observe that
if $f$, $g\in\eusm L^1$ and $f^{\ssbullet}$, $g^{\ssbullet}$ are
non-negative, then

$$\eqalign{\|f^{\ssbullet}+g^{\ssbullet}\|_1
&=\||f|^{\ssbullet}+|g|^{\ssbullet}\|_1
=\|(|f|+|g|)^{\ssbullet}\|_1\cr
&=\int|f|+|g|
=\int|f|+\int|g|
=\|f^{\ssbullet}\|_1+\|g^{\ssbullet}\|_1.\cr}$$
}%end of proof of 564K

\leader{564L}{Radon-Nikod\'ym theorem} Let $X$ be a second-countable space
with a codably $\sigma$-finite
Borel-coded measure $\mu$.   Let $\nu:\Cal B_c(X)\to\Bbb R$ be a
truly continuous additive functional.   Then there is an
$f\in\eusm L^1(\mu)$
such that $\nu E=\int f\times\chi E$ for every $E\in\Cal B_c(X)$.

\proof{{\bf (a)} Let $M$ be the space of bounded additive functionals on
$\Cal B_c(X)$;  as in 362B, $M$ is an $L$-space.
I will write $\Cal L^1$ for the Riesz space of integrable real-valued
codable Borel functions on $X$.   For $f\in\Cal L^1$ and $E\in\Cal B_c(X)$,
set $\nu_fE=\int f\times\chi E$;  this is defined by 564Ea and
564E(c-i).   The map $f\mapsto\nu_f:\Cal L^1\to M$ is a Riesz
homomorphism, and
norm-preserving in the sense that $\|\nu_f\|=\int|f|$ for every
$f\in\Cal L^1$.   Accordingly $M_1=\{\nu_f:f\in\Cal L^1\}$
is a Riesz subspace
of $M$ isomorphic, as normed Riesz space, to $L^1$;  in particular, it is
norm-complete, by 564K, therefore norm-closed.

\medskip

{\bf (b)} If $\nu\in M^+$ is truly continuous and $\epsilon>0$, there are
an $E\in\Cal B_c(X)^f$ and a $\gamma>0$ such that
$\|(\nu-\gamma\nu_{\chi E})^+\|\le\epsilon$.   \Prf\ There
are $E\in\Cal B_c(X)$ and $\delta>0$ such that $\mu E<\infty$ and
$\nu F\le\epsilon$ whenever $\mu(E\cap F)\le\delta$.   Set
$\gamma=\Bover{\|\nu\|}{\delta}$.   Then

$$\eqalign{(\nu-\gamma\nu_{\chi E})(F)
&=\nu F-\Bover{\|\nu\|}{\delta}\mu(F\cap E)\cr
&\le 0\text{ if }\mu(F\cap E)\ge\delta,\cr
&\le\epsilon\text{ otherwise}.\cr}$$

\noindent So $\|(\nu-\gamma\nu_{\chi E})^+\|\le\epsilon$.\ \Qed

\medskip

{\bf (c)} Suppose that $\nu\in M$, $E\in\Cal B_c(X)^f$ and $\gamma>0$
are such that $0\le\nu\le\gamma\nu_{\chi E}$.   Let $\epsilon>0$.   Then
there are an $f\in\Cal L^1$ and a $\nu'\in M^+$ such that
$\|\nu-\nu_f-\nu'\|\le\epsilon$ and $\nu'\le\bover12\gamma\nu_{\chi E}$.
\Prf\
Set $\alpha=\sup_{F\in\Cal B_c(X)}\nu F-\bover12\gamma\mu F$;
let $H\in\Cal B_c(X)$ be such that
$\nu H-\bover12\gamma\mu H\ge\alpha-\bover13\epsilon$;  set
$f=\bover12\gamma\chi(H\cap E)$ and
$\nu'=(\nu-\nu_f)^+\wedge\bover12\gamma\nu_{\chi E}$.

If $F\in\Cal B_c(X)$ then

$$\eqalignno{(\nu_f-\nu)(F)
&=\Bover12\gamma\mu(F\cap H\cap E)-\nu F
\le\Bover12\gamma\mu(F\cap H)-\nu(F\cap H)\cr
&=\Bover12\gamma\mu H-\nu H-\Bover12\gamma\mu(H\setminus F)
  +\nu(H\setminus F)\cr
\displaycause{of course $\mu H$ must be finite, as
$\nu H-\bover12\gamma\mu H$ is finite}
&\le-\alpha+\Bover13\epsilon+\alpha
=\Bover13\epsilon,\cr
(\nu-\nu_f-\Bover12\gamma\nu_{\chi E})(F)
&=\nu F-\Bover12\gamma\mu(F\cap E\cap H)
  -\Bover12\gamma\mu(F\cap E)\cr
&=\nu(F\cap E\setminus H)-\Bover12\gamma\mu(F\cap E\setminus H)\cr
&\mskip100mu
  +\nu(F\setminus E)+\nu(F\cap E\cap H)-\gamma\mu(F\cap E\cap H)\cr
&\le\nu(F\cap E\setminus H)-\Bover12\gamma\mu(F\cap E\setminus H)\cr
\displaycause{because $\nu\le\gamma\nu_{\chi E}$}
&=\nu((F\cap E)\cup H)-\Bover12\gamma\mu((F\cap E)\cup H)
  -\nu H+\Bover12\gamma\mu H\cr
&\le\alpha-(\alpha-\Bover13\epsilon)
=\Bover13\epsilon.\cr}$$

\noindent So $\|(\nu_f-\nu)^+\|\le\bover13\epsilon$ and
$\|(\nu-\nu_f-\bover12\gamma\nu_{\chi E})^+\|\le\bover13\epsilon$.
But this means that

$$\eqalign{\|\nu-\nu_f-\nu'\|
&=\|\nu-\nu_f-(\nu-\nu_f)^++((\nu-\nu_f)^+-\Bover12\gamma\nu_{\chi E})^+\|
  \cr
&\le 2\|\nu-\nu_f-(\nu-\nu_f)^+\|
   +\|(\nu-\nu_f-\Bover12\gamma\nu_{\chi E})^+\|\cr
&\le 2\|(\nu_f-\nu)^+\|+\Bover13\epsilon
\le\epsilon,\cr}$$

\noindent as required.\ \Qed

\medskip

{\bf (d)} Again suppose that $\nu\in M$, $E\in\Cal B_c(X)^f$, $\gamma>0$
and $\epsilon>0$ are such that $0\le\nu\le\gamma\nu_{\chi E}$.
Then for any $n\in\Bbb N$
there are an $f\in\Cal L^1$ and a $\nu'\in M^+$ such that
$\|\nu-\nu_f-\nu'\|\le\epsilon$ and $\nu'\le 2^{-n}\gamma\nu_{\chi E}$.
\Prf\ Induce on $n$.\ \Qed

\medskip

{\bf (e)} If $\nu\in M^+$ is truly continuous and $\epsilon>0$, there is
an $f\in\Cal L^1$ such that $\|\nu-\nu_f\|\le\epsilon$.   \Prf\ By
(b), there are an $E\in\Cal B_c(X)^f$ and a $\gamma>0$ such that
$\|(\nu-\gamma\nu_{\chi E})^+\|\le\bover13\epsilon$.   Let $n\in\Bbb N$ be
such that $2^{-n}\gamma\mu E\le\bover13\epsilon$.   By (d), we have an
$f\in\Cal L^1$ and a $\nu'\in M$ such that
$\|(\nu\wedge\gamma\nu_{\chi E})-\nu_f-\nu'\|\le\bover13\epsilon$ and
$0\le\nu'\le 2^{-n}\gamma\nu_{\chi E}$.   But this means that

$$\eqalign{\|\nu-\nu_f\|
&\le\|(\nu-\gamma\nu_{\chi E})^+\|+\|(\nu\wedge\gamma\nu_{\chi E})-\nu_f\|
   \cr
&\le\Bover13\epsilon+\Bover13\epsilon+\|\nu'\|
\le\Bover23\epsilon+2^{-n}\gamma\mu E
\le\epsilon. \text{ \Qed}\cr}$$

\medskip

{\bf (f)} Since any truly continuous $\nu\in M$ has truly continuous
positive and negative parts, the space
$M_{tc}$ of truly continuous functionals is included in the closure of
$M_1=\{\nu_f:f\in\Cal L^1\}$.   But I noted in (a)
that $M_1$ is norm-isomorphic to $L^1$, so is complete, therefore closed,
and must include $M_{tc}$.
}%end of proof of 564L

\leader{564M}{Inverse-measure-preserving functions (a)} Let $X$ and $Y$ be
second-countable spaces, with Borel-coded measures $\mu$ and $\nu$.
Suppose that $\varphi:X\to Y$ is a codable Borel function such that
$\mu\varphi^{-1}[F]=\nu F$ for every $F\in\Cal B_c(Y)$.   Then
$h\varphi\in S_X$ and $\int h\varphi\,d\mu=\int h\,d\nu$ for every
$h\in S_Y$, writing $S_X=S(\Cal B_c(X)^f)$, $S_Y$ for the spaces of
simple functions.
\cmmnt{By 562Mb,} $f\varphi\in\eusm L^0(\mu)$ for every
$f\in\eusm L^0(\nu)$.   \cmmnt{By 562Sd,}
$\sequencen{h_n\varphi}$ is a codable sequence in $S_X$ whenever
$\sequencen{h_n}$ is a codable sequence in $S_Y$;  \cmmnt{consequently}
$f\varphi\in\eusm L^1(\mu)$ whenever $f\in\eusm L^1(\nu)$, and
we have a norm-preserving Riesz homomorphism $T:L^1(\nu)\to L^1(\mu)$
defined by setting
$Tf^{\ssbullet}=(f\varphi)^{\ssbullet}$ for $f\in\eusm L^1(\mu)$.

\spheader 564Mb If $\nu$ is codably $\sigma$-finite, we have a conditional
expectation operator in the reverse direction\cmmnt{, as follows}.   For
any $f\in\eusm L^1(\mu)$, consider the functional
$\lambda_f$ defined by setting
$\lambda_fF=\int f\times\chi(\varphi^{-1}[F])$ for $F\in\Cal B_c(Y)$.
This is additive and truly continuous.   \prooflet{\Prf\ Let $\epsilon>0$.
By 564Ga, there are an $E_0\in\Cal B_c(X)$ and a $\delta>0$ such that
$\mu E_0<\infty$ and $\int|f|\times\chi E\le\epsilon$ whenever
$E\in\Cal B_c(X)$ and
$\mu(E\cap E_0)\le 2\delta$.   Next, there is a non-decreasing
codable sequence $\sequencen{F_n}$ in $\Cal B_c(Y)$ such that
$\nu F_n<\infty$ for every $n$ and $Y=\bigcup_{n\in\Bbb N}F_n$.   In this
case, $\sequencen{\varphi^{-1}[F_n]}$ is a non-decreasing codable sequence
in $\Cal B_c(X)$ with union $X$, so there is an $n$ such that
$\mu(E_0\setminus\varphi^{-1}[F_n])\le\delta$.   Now suppose that
$F\in\Cal B_c(Y)$ and $\nu(F\cap F_n)\le\delta$.   In this case,

\Centerline{$\mu(E_0\cap\varphi^{-1}[F])
\le\mu(E_0\setminus\varphi^{-1}[F_n])+\mu(\varphi^{-1}[F_n\cap F])
\le 2\delta$,}

\noindent so

\Centerline{$|\lambda_fF|\le\int|f|\times\chi(\varphi^{-1}[F])
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lambda_f$ is truly continuous.\
\Qed}

There is\cmmnt{ therefore} a unique $v_f\in L^1(\nu)$ such that
$\int_Fv_f=\lambda_fF$ for every $F\in\Cal B_c(Y)$.  \prooflet{\Prf\
By 564L, there is a $g\in\eusm L^1(\nu)$ such that
$\lambda_fF=\int g\times\chi F$ for every $F\in\Cal B_c(Y)$.   By 564Gb,
any two such functions are equal almost everywhere, so have the same
equivalence class in $L^1$, which we may call $v_f$.\ \Qed}

We may call $v_f$ the {\bf conditional expectation} of $f$ with respect to
the \imp\ function $\varphi$.

\spheader 564Mc Still supposing that $\nu$ is codably $\sigma$-finite,
\cmmnt{we see that}
$\lambda_f=\lambda_{f'}$ whenever $f$, $f'\in\eusm L^1(\mu)$ are
equal almost everywhere, so that we have an operator
$P:L^1(\mu)\to L^1(\nu)$ defined by saying that $Pf^{\ssbullet}=v_f$ for
every $f\in\eusm L^1(\mu)$;  \cmmnt{that is, that}
$\int_FPu=\int_{\varphi^{-1}[F]}u$ for every $u\in L^1(\mu)$ and
$F\in\Cal B_c(Y)$.   \cmmnt{Because this defines each $Pu$ uniquely,}
$P$ is
linear.   It is positive\dvro{. }{ because if $f^{\ssbullet}\ge 0$ then
$\lambda_f\ge 0$;  if now $g\in\eusm L^1(\nu)$ is such that
$\int g\times\chi F=\lambda_fF\ge 0$ for every $F\in\Cal B_c(X)$,
$g\ge 0$ a.e., by 564Gb, and

\Centerline{$Pf^{\ssbullet}=u_f=g^{\ssbullet}\ge 0$.}

\noindent}%end of dvro
It is elementary to check that if $T$ is the operator of (a)
above then $PT$ is the identity operator on $L^1(\nu)$.

\spheader 564Md Now consider the special case in which $Y=X$,
the topology of $Y$ is the topology generated by a codable sequence
$\sequencen{V_n}$ in $\Cal B_c(X)^f$, $\nu=\mu\restr\Cal B_c(Y)$ and
$\varphi$ is the identity function.   \cmmnt{(Of course this can be
done only when
$\mu$ is codably $\sigma$-finite.)}   In this case, we can identify
$L^1(\nu)$ with its image in $L^1(\mu)$ under $T$, and $P$ becomes a
conditional expectation operator of the kind examined in 242J.

\leader{564N}{Product measures:  Theorem} Let $X$ and $Y$ be
second-countable spaces, and $\mu$, $\nu$ semi-finite
Borel-coded measures on $X$, $Y$ respectively.

(a) There is a Borel-coded measure $\lambda$
on $X\times Y$ such that
$\lambda(E\times F)=\mu E\cdot\nu F$ for all
$E\in\Cal B_c(X)$ and $F\in\Cal B_c(Y)$.

(b) If $\nu$ is codably $\sigma$-finite then we can arrange that
$\iint f(x,y)\nu(dy)\mu(dx)$ is defined and equal to $\int fd\lambda$
for every $\lambda$-integrable real-valued function $f$.

(c) If $\mu$ and $\nu$ are both codably $\sigma$-finite then
$\lambda$ is uniquely defined by the formula in (a).

\proof{{\bf (a)(i)} Start by fixing sequences $\sequencen{U_n}$,
$\sequencen{V_n}$ running over bases for the topologies of $X$, $Y$
respectively containing $\emptyset$, and a bijection
$n\mapsto(i_n,j_n):\Bbb N\to\Bbb N$;  then
$\sequencen{U_{i_n}\times V_{j_n}}$ runs over a base for the topology of
$X\times Y$ containing $\emptyset$.   Let

\Centerline{$\phi_X:\Cal T\to\Cal B_c(X)$,
\quad$\tilde{\Cal T}_X\subseteq\Cal T^{\Bbb R}$,
\quad$\tilde\phi_X:\tilde{\Cal T}_X\to\BbbR^X$,}

\Centerline{$\phi_Y:\Cal T\to\Cal B_c(Y)$,}

\Centerline{$\phi:\Cal T\to\Cal B_c(X\times Y)$,
\quad$\tilde{\Cal T}\subseteq\Cal T^{\Bbb R}$,
\quad$\tilde\phi:\tilde{\Cal T}\to\BbbR^{X\times Y}$}

\noindent be the interpretations of codes
associated with the sequences $\sequencen{U_n}$,
$\sequencen{V_n}$ and $\sequencen{U_{i_n}\times V_{j_n}}$,
as described in 562B and 562N.
Let $\Cal R_X$ be the space of resolvable real-valued functions on $X$,
and $\tilde\psi_X:\Cal R_X\to\tilde{\Cal T}_X$ a function such that
$\tilde\phi_X(\tilde\psi_X(f))=f$ for ever $f\in\Cal R_X$,
as in 562R.
The argument will depend on the existence of a number of
further functions;  it may help if I lay them out explicitly.
Fix a member $\tau_0$ of $\tilde{\Cal T}_X$.

\medskip

\qquad\grheada\
Let $\Theta_2':\Cal T\times\Cal T\to\Cal T$ be such that

\Centerline{$\phi(\Theta_2'(T,T'))=\phi(T)\cap\phi(T')$,
\quad$r(\Theta_2'(T,T'))=\max(r(T),r(T'))$}

\noindent for all $T$, $T'\in\Cal T$ (562Cc);  now define
$\Theta_2^*:\bigcup_{I\in[\Bbb N]^{<\omega}}\Cal T^I\to\Cal T$ by setting

$$\eqalign{\Theta_2^*(\familyiI{T_i})
&=\{\emptyset\}\cup\{\fraction{n}:n\in\Bbb N\}\text{ if }I=\emptyset,\cr
&=\Theta_2'(\Theta_2^*(\family{i}{I\cap n}{T_i}),T_n)
  \text{ if }n=\max I.\cr}$$

\noindent Then

\Centerline{$\phi(\Theta_2^*(\familyiI{T_i}))
=(X\times Y)\cap\bigcap_{i\in I}\phi(T_i)$,
\quad$r(\Theta_2^*(\familyiI{T_i}))=\max(1,\sup_{i\in I}r(T_i))$}

\noindent whenever $I\subseteq\Bbb N$ is finite and $T_i\in\Cal T$ for
$i\in I$.

\medskip

\qquad\grheadb\ Let $\tilde\Theta_1:\Cal T^{\Bbb N}\to\Cal T$ be such that

\Centerline{$\phi_X(\tilde\Theta_1(\sequencen{T_n}))
=\bigcup_{n\in\Bbb N}\phi_X(T_n)$}

\noindent for every sequence $\sequencen{T_n}$ in $\Cal T$ (526Cb).

\medskip

\qquad\grheadc\ There is a function
$\Theta_2:\tilde{\Cal T}_X\times\Cal T\to\tilde{\Cal T}_X$ such that

\Centerline{$\tilde\phi_X(\Theta_2(\tau,T'))
=(\nu\phi_Y(T'))\chi X-\tilde\phi_X(\tau)$}

\noindent whenever $\tau\in\tilde{\Cal T}_X$ and $T'\in\Cal T$ is such that
$\nu(\phi_Y(T'))$ is finite.   \Prf\ Taking $\Theta_0:\Cal T\to\Cal T$
such that $\phi_X(\Theta_0(T))=X\setminus\phi_X(T)$ for every $T\in\Cal T$
(562Ca), set

\Centerline{$\hat\Theta(\tau,\beta)
=\tilde\Theta_1(\sequencen{\Theta_0(\tau(\beta-2^{-n}))})$}

\noindent for $\tau\in\tilde{\Cal T}_X$ and $\beta\in\Bbb R$, so that
$\hat\Theta$ is a function from $\tilde{\Cal T}_X\times\Bbb R$ to
$\Cal T$ and

\Centerline{$\phi_X(\hat\Theta(\tau,\beta))
=\bigcup_{n\in\Bbb N}X\setminus\phi_X(\tau(\beta-2^{-n}))
=\{x:\tilde\phi_X(\tau)(x)<\beta\}$}

\noindent for $\tau\in\tilde{\Cal T}_X$ and $\beta\in\Bbb R$.   If
$\nu(\phi_Y(T'))=\infty$, take $\Theta_2(\tau,T')=\tau_0$
for every $\tau\in\tilde{\Cal T}_X$;  otherwise set

\Centerline{$\Theta_2(\tau,T')(\alpha)
=\hat\Theta(\tau,\nu\phi_Y(T')-\alpha)$}

\noindent for $\tau\in\tilde{\Cal T}_X$, $T'\in\Cal T$ and
$\alpha\in\Bbb R$, so that

$$\eqalign{\phi_X(\Theta_2(\tau,T')(\alpha))
&=\{x:\tilde\phi_X(\tau)(x)<\nu\phi_Y(T')-\alpha\}
\cr&=\{x:\nu\phi_Y(T')-\tilde\phi_X(\tau)(x)>\alpha\}\cr}$$

\noindent for every $\alpha$,
$\Theta_2(\tau,T')\in\tilde{\Cal T}_X$ and
$\tilde\phi_X(\Theta_2(\tau,T'))(x)=\nu\phi_Y(T')-\tilde\phi_X(\tau)(x)$
for every $x\in X$.\ \Qed

\medskip

\qquad\grheadd\ Define
$\Theta_1^*:\tilde{\Cal T}_X^{\Bbb N}\to\Cal T^{\Bbb R}$ by saying that

\Centerline{$\Theta_1^*(\sequencen{\tau_n})(\alpha)
=\tilde\Theta_1(\sequencen{\tau_n(\alpha)})$}

\noindent for every sequence $\sequencen{\tau_n}$ in $\tilde{\Cal T}_X$, so
that $\Theta_1^*(\sequencen{\tau_n})\in\tilde{\Cal T}_X$ and

\Centerline{$\tilde\phi_X(\tilde\Theta_1(\sequencen{\tau_n}))
=\sup_{n\in\Bbb N}\tilde\phi_X(\tau_n)$}

\noindent whenever $\sequencen{\tau_n}$ is a sequence in
$\tilde{\Cal T}_X$ such that $\sup_{n\in\Bbb N}\tilde\phi_X(\tau_n)$ is
defined in $\BbbR^X$.

\medskip

\qquad\grheade\ As in 562Ob,
we can find a function $\Theta^*:\Cal T\times X\to\Cal T$ such that

\Centerline{$\phi_Y(\Theta^*(T,x))=\{y:(x,y)\in\phi(T)\}$}

\noindent for $T\in\Cal T$ and $x\in X$.

\medskip

\quad{\bf (ii)} If $W\subseteq X\times Y$ is open and $F\in\Cal B_c(Y)$,
$x\mapsto\nu(F\cap W[\{x\}]):X\to[0,\infty]$ is lower semi-continuous.
\Prf\ Take $\gamma\in\Bbb R$ and consider
$G=\{x:\nu(F\cap W[\{x\}])>\gamma\}$.
Given $x\in G$ let $K$ be $\{(m,n):x\in U_m$, $U_m\times V_n\subseteq W\}$;
then $W[\{x\}]=\bigcup_{(m,n)\in K}V_n$.   Now $\family{(m,n)}{K}{V_n}$ and
$\family{(m,n)}{K}{F\cap V_n}$ are codable families (562J),
so there is a finite set
$L\subseteq K$ such that $\nu(\bigcup_{(m,n)\in L}F\cap V_n)>\gamma$
(563B(a-ii)).   In this case,
$H=X\cap\bigcap_{(m,n)\in L}U_m$ is an open neighbourhood of $x$ included
in $G$, and $\nu(F\cap W[\{x'\}])>\gamma$ for every $x'\in H$.
As $x$ is arbitrary, $G$ is open;  as $\gamma$ is arbitrary, the
function is lower semi-continuous.\ \Qed

\medskip

\quad{\bf (iii)} For $T$, $T'\in\Cal T$ and $x\in X$, set

\Centerline{$h_{TT'}(x)=\nu\{y:y\in\phi_Y(T')$, $(x,y)\in\phi(T)\}
=\nu(\phi_Y(T')\cap\phi_Y(\Theta^*(T,x)))$.}

\noindent Then there is a function
$\Theta:\Cal T\times\Cal T\to\tilde{\Cal T}_X$ such that

\Centerline{$\tilde\phi_X(\Theta(T,T'))=h_{TT'}$}

\noindent whenever $T$, $T'\in\Cal T$ are such that
$\nu\phi_Y(T')$ is finite.   \Prf\
If $\nu\phi_Y(T')=\infty$ set $\Theta(T,T')=\tau_0$.
For other $T'$, build $\Theta$ by
induction on the rank of $T$, as usual.
If $r(T)\le 1$, then $\phi(T)$ is open;  by (ii), $h_{TT'}$ is
lower semi-continuous, therefore resolvable (562Qa).   So we can set
$\Theta(T,T')=\tilde\psi_X(h_{TT'})$.

For the inductive step to $r(T)\ge 2$, set
$A_T=\{n:\fraction{n}\in T\}$, so that

$$\eqalign{\phi(T)
&=\bigcup_{n\in A_T}(X\times Y)\setminus\phi(T_{\fraction{n}})\cr
&=\bigcup_{m\in\Bbb N}(X\times Y)
  \setminus((X\times Y)\cap\bigcap_{n\in A_T\cap m}
     \phi(T_{\fraction{n}}))\cr
&=\bigcup_{m\in\Bbb N}(X\times Y)
    \setminus\phi(\Theta_2^*(\family{n}{A_T\cap m}{T_{\fraction{n}}}))
\cr}$$

\noindent and

\Centerline{$h_{TT'}(x)
=\lim_{m\to\infty}\nu\phi_Y(T')-h_{T^{(m)}T'}(x)
=\sup_{m\in\Bbb N}\nu\phi_Y(T')-h_{T^{(m)}T'}(x)$}

\noindent for every $x$, where

\Centerline{$T^{(m)}=\Theta_2^*(\family{n}{A_T\cap m}{T_{\fraction{n}}})$,
\quad$\phi(T^{(m)})
=(X\times Y)\cap\bigcap_{n\in A_T\cap m}\phi(T_{\fraction{n}})$}

\noindent for $m\in\Bbb N$.   Now $r(T^{(m)})<r(T)$ for every $m$,
so each $\Theta(T^{(m)},T')$ has been defined, and we can speak of
$\Theta_2(\Theta(T^{(m)},T'),\penalty-100T')$ for each $m$;  we shall have

$$\eqalign{\tilde\phi_X(\Theta_2(\Theta(T^{(m)},T'),T'))(x)
&=\nu\phi_Y(T')-\tilde\phi_X(\Theta(T^{(m)},T'))(x)
=\nu\phi_Y(T')-h_{T^{(m)}T'}(x)\cr
&=\nu\{y:y\in\phi_Y(T'),\,(x,y)\in\bigcup_{n\in A_T\cap m}
(X\times Y)\setminus\phi(T_{\fraction{n}})\}\cr}$$

\noindent for $m\in\Bbb N$ and $x\in X$.   So if we set

\Centerline{$\Theta(T,T')
=\Theta_1^*(\sequence{m}{\Theta_2(\Theta(T^{(m)},T'),T')})$,}

\noindent we shall have

$$\eqalign{\tilde\phi_X(\Theta(T,T'))
&=\sup_{m\in\Bbb N}\tilde\phi_X(\Theta_2(\Theta(T^{(m)},T'),T'))\cr
&=\sup_{m\in\Bbb N}(\nu\phi_Y(T'))\chi X
   -\tilde\phi_X(\Theta(T^{(m)},T'))\cr
&=\sup_{m\in\Bbb N}(\nu\phi_Y(T'))\chi X-h_{T^{(m)}T'}
=h_{TT'},\cr}$$

\noindent as required for the induction to proceed.\ \Qed

\medskip

\quad{\bf (iv)} Thus we see that $h_{TT'}\in\eusm L^0(\mu)$ whenever
$T$, $T'\in\Cal T$ and $\nu\phi_Y(T')$ is finite.

\medskip

\quad{\bf (v)}
Let $\Cal B_c(Y)^f$ be the ring of subsets of $Y$ of finite measure.
For $F\in\Cal B_c(Y)^f$ and $W\in\Cal B_c(X\times Y)$ we have $T$,
$T'\in\Cal T$ such that $\phi(T)=W$ and $\phi_Y(T')=F$, and now
$\nu(F\cap W[\{x\}])=h_{TT'}(x)$ for every $x\in X$.   So
we have a functional
$\lambda_F:\Cal B_c(X\times Y)\to[0,\infty]$ defined by saying that

$$\eqalign{\lambda_FW&=\int\nu(F\cap W[\{x\}])\mu(dx)
  \text{ if the integral is defined in }\Bbb R,\cr
&=\infty\text{ otherwise}.\cr}$$

\noindent Of course $\lambda_F$ is additive.   If $E\in\Cal B_c(X)$ and
$F'\in\Cal B_c(Y)$, then

$$\eqalign{\lambda_F(E\times F')
&=0=\mu E\cdot\nu(F\cap F')\text{ if }\nu(F\cap F')=0,\cr
&=\int\nu(F\cap F')\chi E\,d\mu=\mu E\cdot\nu(F\cap F')
   \text{ if }\mu E<\infty,\cr
&=\infty=\mu E\cdot\nu(F\cap F')\text{ if }\mu E=\infty
   \text{ and }\nu(F\cap F')>0.\cr}$$

\noindent (To see that
$E\times F'\in\Cal B_c(X\times Y)$, use 562Mc.)

\medskip

\quad{\bf (vi)} Now suppose that $\sequencen{W_n}$ is a codable disjoint
sequence in $\Cal B_c(X\times Y)$ with union $W$, and that
$F\in\Cal B_c(Y)^f$.   We surely have
$\lambda_FW\ge\sum_{n=0}^{\infty}\lambda_FW_n$.   If
$\sum_{n=0}^{\infty}\lambda_FW_n$ is finite, let $\sequencen{T_n}$ be a
sequence in $\Cal T$ such that $\phi(T_n)=W_n$ for each $n$, and take
$T'\in\Cal T$ such that $\phi_Y(T')=F$.   Then
$\sequencen{h_{T_nT'}}=\sequencen{\tilde\phi_X(\Theta(T_n,T'))}$
is a codable sequence of integrable Borel functions,
so 564Fe tells us that the sum of the
integrals is the integral of the sum;  but

$$\eqalign{\sum_{n=0}^{\infty}h_{T_nT'}(x)
&=\sum_{n=0}^{\infty}\nu(\phi_Y(T')\cap\phi_Y(\Theta^*(T_n,x)))
=\nu(\bigcup_{n\in\Bbb N}\phi_Y(T')\cap\phi_Y(\Theta^*(T_n,x)))\cr
&=\nu(\bigcup_{n\in\Bbb N}F\cap W_n[\{x\}])
=\nu(F\cap W[\{x\}])\cr}$$

\noindent for each $x$, so we have

$$\eqalign{\lambda_FW
&=\int\nu(F\cap W[\{x\}])\mu(dx)
=\int\sum_{n=0}^{\infty}h_{T_nT'}d\mu\cr
&=\sum_{n=0}^{\infty}\int h_{T_nT'}d\mu
=\sum_{n=0}^{\infty}\lambda_FW_n.\cr}$$

\noindent As $\sequencen{W_n}$ is arbitrary, $\lambda_F$ is a Borel-coded
measure.

\medskip

\quad{\bf (vii)} If $W\in\Cal B_c(X\times Y)$
and $F\subseteq F'$ in $\Cal B_c(Y)^f$, then

$$\eqalignno{\lambda_FW
&=\int\nu(F\cap W[\{x\}])\mu(dx)\cr
\displaycause{counting $\int h\,d\mu$ as $\infty$ for a non-negative
function $h\in\eusm L^0(\mu)\setminus\eusm L^1(\mu)$}
&=\int\nu(F'\cap(W\cap(X\times F))[\{x\}])\mu(dx)
=\lambda_{F'}(W\cap(X\times F))
\le\lambda_{F'}W.\cr}$$

\noindent Thus $\langle\lambda_F(W)\rangle_{F\in\Cal B_c(Y)^f}$ is an
upwards-directed family for each $W\in\Cal B_c(X\times Y)$;  let
$\lambda W$ be its supremum.   Then $\lambda$ is a Borel-coded measure on
$X\times Y$ (563E).   Also

\Centerline{$\lambda(E\times F)
=\lambda_F(E\times F)=\mu E\cdot\nu_FF=\mu E\cdot\nu F$}

\noindent whenever $E\in\Cal B_c(X)$ and $F\in\Cal B_c(Y)^f$
have finite measure.
For other measurable $E$ and $F$, if either is negligible then
$\lambda(E\times F)=0$, while if one has infinite measure and the other has
non-zero measure then $\lambda(E\times F)=\infty$ because $\mu$ and $\nu$
are both semi-finite.

Observe that the construction ensures that if $\lambda W<\infty$ and
$W\subseteq X\times F$ for some $F\in\Cal B_c(Y)^f$, then
$\lambda W=\int\nu W[\{x\}]\mu(dx)$.

\medskip

{\bf (b)} Now suppose that $\nu$ is codably $\sigma$-finite.

\medskip

\quad{\bf (i)} Let
$\sequencen{F_n}$ be a codable sequence in $\Cal B_c(Y)^f$ covering $Y$;
since $\sequencen{\bigcup_{i\le n}F_i}$ also is codable, we can suppose
that $\sequencen{F_n}$ is non-decreasing.   Let $\sequencen{T'_n}$ be
a sequence in $\Cal T$ such that $\phi_Y(T'_n)=F_n$ for each $n$.   By
562Mc, as usual, $\sequencen{X\times F_n}$ is a codable sequence in
$\Cal B_c(X\times Y)$, so
$\lambda W=\sup_{n\in\Bbb N}\lambda(W\cap(X\times F_n))$ whenever
$\lambda$ measures $W$.

\medskip

\quad{\bf (ii)} Let
$f:X\times Y\to\coint{0,\infty}$ be an integrable codable Borel function.
Then $\iint f(x,y)\nu(dy)\mu(dx)$ is defined and equal to
$\int fd\lambda$.

\quad\Prf\grheada\ For $n$, $k\in\Bbb N$ set

\Centerline{$W_{nk}=\{(x,y):y\in F_n$, $f(x,y)\ge 2^{-n}k\}$;}

\noindent then $\langle W_{nk}\rangle_{n,k\in\Bbb N}$ is a codable family
in $\Cal B_c(X\times Y)$.   Let $\langle T_{nk}\rangle_{n,k\in\Bbb N}$ be a
family in $\Cal T$ such that $W_{nk}=\phi(T_{nk})$ for $n$, $k\in\Bbb N$.
For $n\in\Bbb N$, define $v_n:X\times Y\to\Bbb R$ by setting

\Centerline{$v_n=2^{-n}\sum_{k=1}^{4^n}\chi W_{nk}$;}

\noindent then $\sequencen{v_n}$ is a codable sequence of codable Borel
functions on $X\times Y$.   Moreover, setting $v_{nx}(y)=v_n(x,y)$,
$\sequencen{v_{nx}}$ is a codable sequence of codable Borel functions on
$Y$, for each $x\in X$.   Now set

\Centerline{$u_{nk}(x)=\nu W_{nk}[\{x\}]$,
\quad$u_n(x)=\int v_n(x,y)\nu(dy)$}

\noindent for $x\in X$ and $n$, $k\in\Bbb N$.   Then, in the language of
part (a) of this proof,

\Centerline{$u_{nk}=\tilde\phi_X(\Theta(T_{nk},T'_n))$}

\noindent for all $n$ and $k$, so $\langle u_{nk}\rangle_{n,k\in\Bbb N}$ is
a codable family of codable Borel functions on $X$.   Since

\Centerline{$u_n=2^{-n}\sum_{k=1}^{4^n}u_{nk}$}

\noindent for each $n$, $\sequencen{u_n}$ is a codable sequence of codable
Borel functions on $X$.

\medskip

\qquad\grheadb\ Next, for each $n\in\Bbb N$,

$$\eqalignno{\int u_nd\mu
&=2^{-n}\sum_{k=1}^{4^n}\int\nu W_{nk}[\{x\}]\mu(dx)
=2^{-n}\sum_{k=1}^{4^n}\lambda W_{nk}\cr
\displaycause{by the final remark in part (a) of the proof}
&=\int v_nd\lambda.\cr}$$

\noindent At this point, observe that $\sequencen{v_n}$ is a non-decreasing
codable sequence with limit $f$.   So

\Centerline{$\lim_{n\to\infty}\int u_nd\mu
=\lim_{n\to\infty}\int v_nd\lambda=\int fd\lambda$}

\noindent is finite;  since $\sequencen{u_n}$ also is non-decreasing,
$u(x)=\lim_{n\to\infty}u_n(x)$ is finite for $\mu$-almost all $x$, and

\Centerline{$\int u\,d\mu
=\lim_{n\to\infty}\int u_nd\mu=\int fd\lambda$}

\noindent (564Fa).   On the other hand, for each $x\in X$,
$\sequencen{v_{nx}}$ is a non-decreasing codable sequence with limit
$f_x$, where $f_x(y)=f(x,y)$ for $y\in Y$;  so

\Centerline{$u(x)=\lim_{n\to\infty}\int v_{nx}d\nu=\int f_xd\nu$}

\noindent for almost all $x$, and

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)
=\iint f_xd\nu\mu(dx)=\int u\,d\mu=\int fd\lambda$.  \Qed}

\medskip

\quad{\bf (iii)} It follows at once, taking the difference of positive and
negative parts, that

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)=\int fd\lambda$}

\noindent for every $\lambda$-integrable codable Borel function $f$.

\medskip

\quad{\bf (iv)} In particular (or more directly), if
$W\in\Cal B_c(X\times Y)$ is $\lambda$-negligible,
then $\mu$-almost every vertical section of $W$ is $\nu$-negligible.   So
starting from a general $\lambda$-integrable function $f$, we move to a
codable Borel function $g$ such that $f\eae g$;  now
$\int f(x,y)\nu(dy)$ must be defined and equal to $\int g(x,y)\nu(dy)$ for
almost every $x$, and

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)
=\iint g(x,y)\nu(dy)\mu(dx)
=\int g\,d\lambda=\int fd\lambda$.}

\noindent This completes the proof of (b).

\medskip

{\bf (c)} Let $\sequencen{E_n}$, $\sequencen{F_n}$ be codable sequences of
sets of finite measure covering $X$, $Y$ respectively;  we may suppose that
both sequences are non-decreasing.   Then
$\sequencen{E_n\times F_n}=\sequencen{(E_n\times Y)\cap(X\times F_n)}$ is a
codable sequence (562Mc).   Suppose that
$\lambda$, $\lambda'$ are two Borel-coded measures on $X\times Y$ agreeing
on measurable rectangles.
For each $n\in\Bbb N$ let $\lambda_n$, $\lambda'_n$ be
the totally finite measures defined by setting

\Centerline{$\lambda_nW=\lambda(W\cap(E_n\times F_n))$,
\quad$\lambda'_nW=\lambda'(W\cap(E_n\times F_n))$}

\noindent for $W\in\Cal B_c(X\times Y)$.   Now, given $n$, set
$\Cal W_n=\{W:W\in\Cal B_c(X\times Y)$, $\lambda_nW=\lambda'_nW\}$.
Then $W\cup W'\in\Cal W_n$ whenever $W$, $W'\in\Cal W_n$ are disjoint,
and $E\times F\in\Cal W_n$ whenever $E\in\Cal B_c(X)$ and
$F\in\Cal B_c(Y)$.   So $\Cal W_n$ includes the algebra of subsets of
$X\times Y$ generated by
$\{E\times F:E\in\Cal B_c(X)$, $F\in\Cal B_c(Y)\}$.
In particular, $\Cal W_n$ includes any
set of the form $\bigcup_{(i,j)\in K}U_i\times V_j$ where
$K\subseteq\Bbb N\times\Bbb N$ is finite.   But any open subset of
$X\times Y$ is expressible as the union of a non-decreasing
codable sequence of such sets,
so also belongs to $\Cal W_n$.   By 563Fg, $\lambda_n=\lambda'_n$.

This is true for every $n\in\Bbb N$.   Since

\Centerline{$\lambda W=\sup_{n\in\Bbb N}\lambda_nW$,
\quad$\lambda'W=\sup_{n\in\Bbb N}\lambda'_nW$}

\noindent for every $W\in\Cal B_c(X\times Y)$, $\lambda=\lambda'$, as
claimed.
}%end of proof of 564N

\leader{564O}{Theorem} Let $\sequencen{(X_k,\rho_k)}$ be a sequence of
complete metric spaces, and suppose that we have a double sequence
$\langle U_{ki}\rangle_{k,i\in\Bbb N}$ such that $\{U_{ki}:i\in\Bbb N\}$ is
a base for the topology of $X_k$
for each $k$.   Let $\sequencen{\mu_k}$ be a sequence such
that $\mu_k$ is a Borel-coded probability measure on $X_k$ for each $k$.
Set $X=\prod_{k\in\Bbb N}X_k$.   Then $X$ is a Polish space and
there is a Borel-coded probability measure $\lambda$ on $X$
such that $\lambda(\prod_{k\in\Bbb N}E_k)=\prod_{k\in\Bbb N}\mu_kE_k$
whenever $\sequence{k}{E_k}\in\prod_{k\in\Bbb N}\Cal B_c(X_k)$ and
$\{k:E_k\ne X_k\}$ is finite.

\proof{{\bf (a)(i)}
Of course $X$ is Polish;  we have a complete metric $\rho$
on $X$ defined by saying that
$\rho(x,y)=\sup_{k\in\Bbb N}\min(2^{-k},\penalty-100\rho_k(x(k),y(k)))$
for $x$, $y\in X$, and a countable base generated by sets of the form
$\{x:x(k)\in U_{ki}\}$.

\medskip

\quad{\bf (ii)} Writing $\Cal F_k$ for the family of closed subsets of
$X_k$ for $k\in\Bbb N$, we have a choice function $\zeta$ on
$\bigcup_{k\in\Bbb N}\Cal F_k\setminus\{\emptyset\}$.   \Prf\ Given
a non-empty
$F\in\bigcup_{k\in\Bbb N}\Cal F_k$, take the first $k$ such that
$F\in\Cal F_k$, and define $\sequence{m}{F_m}$, $\sequence{m}{i_m}$ by
saying that

\inset{$F_0=F$,

$i_m=\min\{i:i\in\Bbb N$, $U_{ki}\cap F_m\ne\emptyset$,
$\diam U_{ki}\le 2^{-m}\}$}

\noindent (taking the diameter as measured by $\rho_k$, of course),

\inset{$F_{m+1}=\overline{F_m\cap U_{ki_m}}$}

\noindent for each $m$.   Now $\sequence{m}{F_m}$ generates a Cauchy
filter in $X_k$ which must have a unique limit;  take this limit for
$\zeta(F)$.\ \Qed

\medskip

{\bf (b)(i)} Let
$\Tau=\bigotimes_{k\in\Bbb N}\Cal B_c(X_k)$ be the algebra of subsets of
$X$ generated by $\{\{x:x(k)\in E\}:k\in\Bbb N$, $E\in\Cal B_c(X_k)\}$.
Note that all these sets belong to $\Cal B_c(X)$, by 562Md, so
$\Tau\subseteq\Cal B_c(X)$.   Set

\Centerline{$\Cal C
=\{\prod_{k\in\Bbb N}E_k:E_k\in\Cal B_c(X_k)$ for every $k\in\Bbb N$,
$\{k:E_k\ne X_k\}$ is finite$\}$,}

\Centerline{$\Cal C_o
=\{\prod_{k\in\Bbb N}G_k:G_k\subseteq X_k$ is open for every
$k\in\Bbb N$, $\{k:G_k\ne X_k\}$ is finite$\}$,}

\Centerline{$\Cal C_c
=\{\prod_{k\in\Bbb N}F_k:F_k\subseteq X_k$ is closed
for every $k\in\Bbb N$, $\{k:F_k\ne X_k\}$ is finite$\}$.}

\noindent Then every member of $\Tau$ can be
expressed as the union of a finite disjoint family in $\Cal C$.
$\overline{C}\in\Cal C_c$ for every $C\in\Cal C$, so the closure of any
member of $\Tau$ can be expressed as the union
of finitely many members of $\Cal C_c$ and belongs to $\Tau$.
The complement of a member of $\Cal C_c$
can be expressed as the union of finitely many members of $\Cal C_o$, so
any open set belonging to $\Tau$ can be expressed as the union of
finitely many members of $\Cal C_o$.

\medskip

\quad{\bf (ii)} For $m\in\Bbb N$ write
$\Tau_m=\bigotimes_{k\ge m}\Cal B_c(X_k)$ for the algebra of
subsets of $\prod_{k\ge m}X_k$ generated by sets of the form
$\{x:x(k)\in E_k\}$ for $k\ge m$ and $E_k\in\Cal B_c(X_k)$.   Then we have an
additive functional $\nu_m:\Tau_m\to[0,1]$ defined by saying that
$\nu_m(\prod_{k\ge m}E_k)=\prod_{k=m}^{\infty}\mu_kE_k$ whenever
$E_k\in\Cal B_c(X_k)$ for every $k\ge m$ and $\{k:E_k\ne X_k\}$ is finite
(326E).   Now if $m\in\Bbb N$ and $W\in\Tau_m$ then

\Centerline{$\nu_mW
=\int\nu_{m+1}\{v:\fraction{t}^{\smallfrown}v\in W\}\mu_m(dt)$}

\noindent (notation:  5A1C).   \Prf\ This is elementary for cylinder sets
$W=\prod_{k\ge m}E_k$;  now any other member of $\Tau_m$ is
expressible as a finite disjoint union of such sets.\ \Qed

\medskip

{\bf (c)(i)} For open sets $W\subseteq X$ define

\Centerline{$\lambda_0 W
=\sup\{\nu_0V:V\in\Tau$, $\overline{V}\subseteq W\}$.}

\noindent Then if $\sequencen{W_n}$ is a non-decreasing sequence of open
sets with union $X$, $\lim_{n\to\infty}\lambda_0 W_n=1$.   \Prf\ Starting
from the double sequence $\langle U_{ki}\rangle_{k,i\in\Bbb N}$, it is easy
to build a sequence $\sequencen{U_n}$ in $\Cal C_o$ which runs over a base
for the topology of $X$.   Set
$W'_n=\bigcup\{U_i:i\le n$, $\overline{U}_i\subseteq W_n\}$ for each $n$;
then $\sequencen{W'_n}$ is a non-decreasing sequence of open sets belonging
to $\Tau=\Tau_0$, and $\bigcup_{n\in\Bbb N}W'_n=X$.   \Quer\
Suppose, if possible, that $\lim_{n\to\infty}\nu_0W'_n\le 1-2^{-l}$ for
some $l\in\Bbb N$.   Then we can
define $\sequence{k}{t_k}$ inductively, as follows.   The inductive
hypothesis will be that $\nu_mV_{mn}\le 1-2^{-l-m}$ for every $n$, where

\Centerline{$V_{mn}=\{v:v\in\prod_{k\ge m}X_k$,
$\ofamily{k}{m}{t_k}\cup v\in W'_n\}$.}

\noindent In this case, define $f_{mn}:X_m\to[0,1]$ by setting

\Centerline{$f_{mn}(t)
=\nu_{m+1}\{w:w\in\prod_{k\ge m+1}X_k$,
$\fraction{t}^{\smallfrown}w\in V_{mn}\}$.}

\noindent By (b-ii),
$\nu_mV_{mn}=\int f_{mn}d\mu_m$ for every $m$, while $\sequencen{f_{mn}}$
is non-decreasing.

Because every $W'_n$ is a finite union of open cylinder sets, so
is $V_{mn}$, and $f_{mn}$ is lower semi-continuous, therefore resolvable;
so

\Centerline{$\int\sup_{n\in\Bbb N}f_{mn}d\mu_m
=\sup_{n\in\Bbb N}\int f_{mn}d\mu_m
=\lim_{n\to\infty}\nu_mV_{mn}\le 1-2^{-l-m}$.}

\noindent The set
$F=\{t:\sup_{n\in\Bbb N}f_{mn}(t)\le 1-2^{-l-m-1}\}$ must be closed and
non-empty, and we can
set $t_m=\zeta(F)$, where $\zeta$ is the choice function of (a-ii).
In this case,

\Centerline{$V_{m+1,n}=\{w:\fraction{t_m}^{\smallfrown}w\in V_{mn}\}$,
\quad$\nu_{m+1}V_{m+1,n}=f_{mn}(t_m)\le 1-2^{-l-m-1}$}

\noindent for every $n$, and the induction continues.

At the end of the induction, however, $x=\sequence{k}{t_k}$ belongs to $X$,
so belongs to $W'_n$ for some $n$.   There must be an $m$ such that $W'_n$
is determined by coordinates less than $m$, and now
$V_{mn}=\prod_{k\ge m}X_k$, so $\nu_mV_{mn}=1$;  which is supposed to be
impossible.\ \Bang

We conclude that

\Centerline{$1=\lim_{n\to\infty}\nu_0W'_n
=\lim_{n\to\infty}\nu_0\overline{W'_n}
=\lim_{n\to\infty}\lambda_0 W_n$}

\noindent because $\overline{W'_n}$ is a closed member of $\Tau$ included
in $W_n$ for each $n$.\ \Qed

\medskip

\quad{\bf (ii)} $\lambda_0$ satisfies the conditions of 563H.   \Prf\

\medskip

\qquad\grheada\ Of course $\lambda_0\emptyset=\emptyset$ and
$\lambda_0W\le\lambda_0W'$ whenever $W\subseteq W'$;  also $\lambda_0X=1$
is finite.

\medskip

\qquad\grheadb\ Suppose that $\sequencen{W_n}$ is a non-decreasing sequence
of open sets in $X$ with union $W$, and $\epsilon>0$.   Then
there is a closed
$V\in\Tau$ such that $V\subseteq W$ and $\nu_0V\ge\lambda_0W-\epsilon$.
Set $W'_n=(X\setminus V)\cup W_n$ for each $n$;  then $\sequencen{W'_n}$
is a non-decreasing sequence of open sets with union $X$, so by (i)
there are an
$n\in\Bbb N$ such that $\lambda_0W'_n\ge 1-\epsilon$, and a closed
$V'\in\Tau$ such that $V'\subseteq W'_n$ and $\nu_0V'\ge 1-2\epsilon$.
Now $V\cap V'$ is a closed member of $\Tau$ included in $W_n$ and

\Centerline{$\lambda_0W_n
\ge\nu_0(V\cap V')\ge\nu_0V-2\epsilon\ge\lambda_0W-3\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lambda_0W\le\lim_{n\to\infty}\lambda_0W_n$;  the reverse inequality is
trivial, so we have equality.

\medskip

\qquad\grheadc\ Let
$W$, $W'\subseteq X$ be open sets.   As in (i), we have
non-decreasing sequences
$\sequencen{W_n}$, $\sequencen{W'_n}$ of open members of $\Tau$ such that

\Centerline{$W=\bigcup_{n\in\Bbb N}W_n=\bigcup_{n\in\Bbb N}\overline{W}_n$,
\quad$W'=\bigcup_{n\in\Bbb N}W'_n=\bigcup_{n\in\Bbb N}\overline{W'_n}$.}

\noindent In this case

\Centerline{$W\cap W'
=\bigcup_{n\in\Bbb N}W_n\cap W'_n
=\bigcup_{n\in\Bbb N}\overline{W_n\cap W'_n}$,}

\Centerline{$W\cup W'=\bigcup_{n\in\Bbb N}W_n\cup W'_n
=\bigcup_{n\in\Bbb N}\overline{W_n\cup W'_n}$.}

\noindent Also

\Centerline{$\lambda_0W=\lim_{n\to\infty}\lambda_0W_n
\le\lim_{n\to\infty}\nu_0W_n
\le\lim_{n\to\infty}\nu_0\overline{W}_n\le\lambda_0W$,}

\noindent so these are all equal;  the same applies to the sequences
converging to $W'$, $W\cap W'$ and $W\cup W'$, so

$$\eqalign{\lambda_0W+\lambda_0W'
&=\lim_{n\to\infty}\nu_0W_n+\nu_0W'_n\cr
&=\lim_{n\to\infty}\nu_0(W_n\cap W'_n)+\nu_0(W_n\cup W'_n)\cr
&=\lambda_0(W\cap W')+\lambda_0(W\cup W'). \text{ \Qed}\cr}$$

\medskip

{\bf (d)} By 563H, we have a Borel-coded measure $\lambda$ on $X$
extending $\lambda_0$.   Now $\lambda$ extends $\nu_0$.
\Prf\ If $C\in\Cal C$ and $\epsilon>0$, express $C$ as
$\prod_{k\in\Bbb N}E_k$ where $E_k\in\Cal B_c(X_k)$ for every $k$ and there is
an $m$ such that $E_k=X_k$ for $k>m$.   For each $k\le m$, there is an open
set $G_k\supseteq E_k$ such that
$\mu_kG_k\le\mu_kE_k+\bover{\epsilon}{m+1}$ (563Fd again);
setting $G_k=X_k$ for $k>m$
and $W=\prod_{k\in\Bbb N}G_k$, $C\subseteq W$ and

\Centerline{$\lambda C\le\lambda W
=\lambda_0W\le\nu_0W=\prod_{k=0}^m\mu_kG_k\le\nu_0C+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lambda C\le\nu_0C$.   This is true
for every $C\in\Cal C$.   As both $\lambda$ and $\nu_0$ are additive,
$\lambda W\le\nu_0W$ for every $W\in\Tau$;  as $\lambda X=\nu_0X=1$,
$\lambda$ agrees with $\nu_0$ on $\Tau$.\ \Qed

In particular, $\lambda$ agrees with $\nu_0$ on $\Cal C$, as required.
}%end of proof of 564O

\exercises{\leader{564X}{Basic exercises (a)}
%\spheader 564Xa
Let $X$ be a
second-countable space and $\mu$ a Borel-coded measure on $X$.
Let $E\in\Cal B_c(X)$ and let $\mu_E$ be the
Borel-coded measure on $X$ defined as in 563Fa.   Show that
$\int fd\mu_E$ is defined and equal to $\int f\times\chi E\,d\mu$ for
every $f\in\eusm L^1(\mu)$.
%564G

\spheader 564Xb\dvAnew{2014} Let $X$ be a topological space, $\mu$ a
Baire-coded measure on $X$, and $f$ a non-negative
integrable real-valued function
defined almost everywhere in $X$.   Set $\nu E=\int f\times\chi E$ for
$E\in\CalBa_c(X)$.   Show that $\nu$ is a Baire-coded measure, and that
$\int g\,d\nu=\int g\times f\,d\mu$ for every $\nu$-integrable $g$, if we
interpret $(g\times f)(x)$ as $0$ when $f(x)=0$ and $g(x)$ is undefined.
(Compare 235K.)
%564G

\spheader 564Xc Let $X$ be a countably compact topological space.   (i)
Show that $C(X)=C_b(X)$.   (ii) Show that every positive linear functional
$f:C(X)\to\Bbb R$ is sequentially smooth.   (iii) Show that a
norm-bounded sequence
$\sequencen{u_n}$ in the normed space $C(X)$ is weakly convergent to $0$
iff it is pointwise convergent to $0$.
(iv) Prove this
without using measure theory.   \Hint{{\smc Fremlin 74}, A2F.   Also see
564Ya.}
%564H

\spheader 564Xd Let $X$ be a topological space and $\mu$ a Baire-coded
measure on $X$.
(i) Describe constructions for normed Riesz spaces $L^p(\mu)$
for $1<p\le\infty$.
(ii) Show that if $X$ is second-countable,
$\mu$ is codably $\sigma$-finite and $1<p<\infty$ then
$L^p(\mu)$ is a Dedekind complete Banach lattice with an order-continuous
norm, while $L^2(\mu)$ is a Hilbert space.
%564K

\spheader 564Xe In 564O, show that $\lambda$ is uniquely defined.   Hence
show that we have commutative and associative laws for the product measure
construction.
%564O

\leader{564Y}{Further exercises (a)}\dvAnew{2014}
%\spheader 564Ya
Let $X$ be a topological space and $\sequencen{f_n}$ a
codable sequence of bounded codable Baire real-valued functions on $X$
such that
$\{\int f_nd\mu:n\in\Bbb N\}$ is bounded for every totally finite
Baire-coded measure $\mu$ on $X$.   (i) Show that if $\sequencen{E_n}$ is a
disjoint codable sequence in $\CalBa_c(X)$ and $\mu$ is a Baire-coded
measure on $X$, then $\lim_{n\to\infty}\int f_n\times\chi E_nd\mu=0$.
(ii) Now suppose in addition that
$\lim_{n\to\infty}f_n(x)=0$ for every $x\in X$.
Show that if $\mu$ is a Baire-coded
measure on $X$, then $\lim_{n\to\infty}\int f_nd\mu=0$.
(iii) Use this result to strengthen (iii) of 564Xc to `a sequence
$\sequencen{u_n}$ in $C(X)$ is weakly convergent to $0$ iff it is bounded
for the weak topology and pointwise convergent to $0$'.
%564Xc 564H 56bits 564Xb

\spheader 564Yb
Let $X$ be a locally compact completely regular topological group.
Show that there is a non-zero \lti\ Baire-coded measure on $X$.
%564I

\spheader 564Yc
Let $I$ be a set and $X=\{0,1\}^I$.   Write $Z$ for
$\{0,1\}^{\Bbb N}$.   For
$\theta:\Bbb N\to I$ define $g_{\theta}:X\to Z$ by setting
$g_{\theta}(x)=x\theta$ for $x\in X$.   Let
$\phi:\Cal T\to\Cal B_c(Z)$ be an interpretation of Borel codes for subsets
of $Z$ defined from a sequence running over a base for the topology of $Z$.
Let $\Sigma$ be the family of
subsets of $X$ of the form $\phi'(\theta,T)=g_{\theta}^{-1}[\phi(T)]$ where
$\theta\in I^{\Bbb N}$ and $T\in\Cal T$;  say that a codable family in
$\Sigma$ is one of the form $\familyiI{\phi'(\theta_i,T_i)}$.
Show that there is a functional $\mu:\Sigma\to[0,1]$ such that
$\mu\emptyset=0$, $\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$
whenever $\sequencen{E_n}$ is a disjoint codable sequence in $\Sigma$, and
$\mu\{x:x\restr J=w\}=2^{-\#(J)}$ whenever $J\subseteq I$ is finite and
$w\in\{0,1\}^J$.
%564N

\spheader 564Yd
Suppose there is a disjoint sequence $\sequencen{I_n}$ of doubleton sets
such that for every function $f$ with domain $\Bbb N$ the set
$\{n:f(n)\in I_n\}$ is finite ({\smc Jech 73}, 4.4).   Set
$I=\bigcup_{n\in\Bbb N}I_n$ and let $\Sigma$ be the algebra of subsets of
$\{0,1\}^I$ determined by coordinates in finite sets.   Let
$\lambda:\Sigma\to[0,1]$ be the additive functional such that
$\lambda\{x:z\subseteq x\}=2^{-k}$ whenever $J\in[I]^k$ and
$z\in\{0,1\}^J$.   Show that there is a sequence $\sequencen{E_n}$ in
$\Sigma$, covering $\{0,1\}^I$, such that
$\sum_{n=0}^{\infty}\lambda E_n<1$.
% E_n=\{x:x\restr I_m$ constant for $n\le m\le 4^{n+1}\}$.
%564N
}%end of exercises

\endnotes{
\Notesheader{564}
In the definitions of 564A, I follow the principles of earlier volumes in
allowing virtually measurable functions with conegligible domains to be
counted as integrable.   But you will see that in
564F and elsewhere I work with real-valued Baire measurable
functions defined everywhere.   The point is that while, if you wish
to work through the basic theorems of Fourier analysis under the new rules,
you will certainly need to deal with functions which are not defined
everywhere, all the main theorems will depend on establishing that you have
sequences of sets and functions which are codable in appropriate senses.
There is no way of coding members of $\eusm L^0$ or $\eusm L^1$ as I have
defined them in 564A.   What you will need to do is to build parallel
structures, so that associated with each almost-everywhere-summable Fourier
series $f(x)=\bover12a_0+\sum_{k=1}^{\infty}a_k\cos kx+b_k\sin kx$ you
have in hand a code $\tau$ for a codable Borel function $\tilde f$ equal
almost everywhere to $f$, together with a code $T$ for a conegligible
codable Borel set $E$ included in $\{x:x\in\dom f$, $f(x)=\tilde f(x)\}$.
Provided that associated with every relevant sequence $\sequencen{f_n}$ you
can define appropriate sequences $\sequencen{\tau_n}$ and
$\sequencen{T_n}$, you can hope to deduce the reqired properties of
$\sequencen{f_n}$ by applying 564F to the sequence coded by
$\sequencen{\tau_n}$.

Of course there are further significant technical differences between the
treatment here and the more orthodox one I have employed elsewhere.
In the ordinary theory, using the axiom of choice whenever convenient,
a measure $\mu$, thought of as a function defined on a $\sigma$-algebra of
sets, carries in itself all the information needed to describe the space
$\eusm L^0(\mu)$.   In the present context, we are dealing with functions
$\mu$ defined on algebras which do not directly code the topologies on
which the definition relies.   So it would be safer to write
$\eusm L^0(\frak T,\mu)$.   But of course what really matters is the
collection of codable families of codable sets,
and perhaps we should be thinking of a
different level of abstraction.   In the proof of 564N I have tried to cast
the proof in a language which might be adaptable to other ways of coding
sets and functions.

From 564K on, most of
the results seem to depend on second-countability;  it may be
that something can be done with spaces which have well-orderable bases.

In the shift from 564Xc(iii) to 564Ya(iii) I find myself asking for a
reason why a weakly bounded sequence in $C(X)$ should be norm-bounded.
As far as I know, there is no useful general result in ZF in this
direction.   But in 564Ya I have suggested a method which will serve in
this special context.

I offer 564Yc and 564Yd as positive and negative examples.   The point is
that in 564Yc there may be few sequences of
functions from $\Bbb N$ to $I$, so that we get few codable sequences of
sets.   Of course, if $I$ is well-orderable then $\{0,1\}^I$ is compact
(561D) and we can use 564H.   For well-orderable $I$,
any continuous
real-valued function on $\{0,1\}^I$ is determined by coordinates in some
countable set, so that the methods of 564H and 564Yc will give the same
measure.
}%end of notes

\discrpage

\frfilename{mt565.tex}
\versiondate{25.4.14}
\copyrightdate{2006}

\def\chaptername{Choice and determinacy}
\def\sectionname{Lebesgue measure without choice}

\newsection{565}

I come now to the construction of specific
non-trivial Borel-coded measures.
Primary among them is of course Lebesgue measure on $\BbbR^r$;  we also
have Hausdorff measures (565N-565O).   For Lebesgue
measure I begin, as in \S115, with
half-open intervals.   The corresponding `outer measure' may no longer be
countably subadditive, so I call it `Lebesgue submeasure'.
Carath\'eodory's method no longer seems quite appropriate, as it smudges
the distinction between `negligible' and `outer measure zero', so I use
563H to show that there is a Borel-coded measure agreeing with Lebesgue
submeasure on open sets (565C-565D);  it is the completion of this
Borel-coded measure which I will call Lebesgue measure.
We have a version of Vitali's theorem for well-orderable families
(in particular, for countable families)  of balls (565F).   From this we
can prove the Fundamental Theorem of Calculus in essentially its standard
form (565M).

\leader{565A}{Definitions} Throughout this section, except when otherwise
stated, $r\ge 1$ will be a
fixed integer.   \cmmnt{As in \S115, }I will say that a 
{\bf half-open interval} in $\BbbR^r$ is a set of the form

\Centerline{$\coint{a,b}=\{x:x\in\BbbR^r$, $a(i)\le x(i)<b(i)$ for
$i<r\}$}

\noindent where $a$, $b\in\BbbR^r$.
For a half-open interval $I$, set $\lambda I=0$ if $I=\emptyset$
and otherwise $\lambda I=\prod_{i=0}^{r-1}b(i)-a(i)$ where $I=\coint{a,b}$.
Now for $A\subseteq\BbbR^r$ set

\Centerline{$\theta A=\inf\{\sum_{j=0}^{\infty}\lambda I_j:
\sequence{j}{I_j}
\text{ is a sequence of half-open intervals covering }A\}$.}

\leader{565B}{Proposition} In the notation of 565A,

(a) the function
$\theta:\Cal P\BbbR^r\to[0,\infty]$ is a submeasure,

(b) $\theta I=\lambda I$ for every half-open interval $I\subseteq\BbbR^r$.

\proof{{\bf (a)} As in parts (a-i) to (a-iii) of the proof of 115D,
$\theta\emptyset=0$ and $\theta A\le\theta B$ whenever $A\subseteq B$.
If $A$, $B\subseteq\BbbR^r$ and $\epsilon>0$, we have sequences
$\sequencen{I_n}$ and $\sequencen{J_n}$ of half-open intervals such that

\Centerline{$A\subseteq\bigcup_{n\in\Bbb N}I_n$,
\quad$B\subseteq\bigcup_{n\in\Bbb N}J_n$,}

\Centerline{$\sum_{n=0}^{\infty}\lambda I_n\le\theta A+\epsilon$,
\quad$\sum_{n=0}^{\infty}\lambda J_n\le\theta B+\epsilon$.}

\noindent Set $K_{2n}=I_n$, $K_{2n+1}=J_n$ for $n\in\Bbb N$;  then
$A\cup B\subseteq\bigcup_{n\in\Bbb N}K_n$ so

\Centerline{$\theta(A\cup B)
\le\sum_{n=0}^{\infty}\lambda K_n
=\sum_{n=0}^{\infty}\lambda I_n+\sum_{n=0}^{\infty}\lambda J_n
\le\theta A+\theta B+2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\theta(A\cup B)\le\theta A+\theta B$.

\medskip

{\bf (b)} The arguments of 114B/115B/115Db nowhere called on 
any form of the axiom of choice, so can be used unchanged.
}%end of proof of 565B

\medskip

\noindent{\bf Definition} I will call the submeasure $\theta$ {\bf Lebesgue
submeasure} on $\BbbR^r$.

\leader{565C}{Lemma} Let $\Cal I$ be the family of half-open intervals in
$\BbbR^r$;  let $\theta$ be Lebesgue submeasure, and set

\Centerline{$\Sigma
=\{E:E\subseteq X$, $\theta A=\theta(A\cap E)+\theta(A\setminus E)$
for every $A\subseteq X\}$,
\quad$\nu=\theta\restr\Sigma$\dvro{.}{}}

\cmmnt{\noindent (563G).}

(a) Let $\sequencen{I_n}$ be a disjoint sequence in $\Cal I$.   Then
$E=\bigcup_{n\in\Bbb N}I_n$ belongs to $\Sigma$
and $\nu E=\sum_{n=0}^{\infty}\nu I_n$.

(b) Every open set in $\BbbR^r$ belongs to $\Sigma$.

(c) If $G$, $H\subseteq\BbbR^r$ are open, then
$\nu G+\nu H=\nu(G\cap H)+\nu(G\cup H)$.
%or, $\nu\restr\frak T$ is modular?

(d) If $\sequencen{G_n}$ is a non-decreasing
sequence of open sets then
$\nu(\bigcup_{n\in\Bbb N}G_n)=\lim_{n\to\infty}G_n$.

\proof{{\bf (a)(i)} If $i<r$ and $\alpha\in\Bbb R$ then
$\{x:x\in\BbbR^r$, $x(i)<\alpha\}\in\Sigma$, as in 115F.   So every
half-open interval belongs to $\Sigma$.   By 565Bb,
$\nu I=\theta I=\lambda I$ for every $I\in\Cal I$.

\medskip

\quad{\bf (ii)}\grheada\ $\theta E=\sum_{n=0}^{\infty}\nu I_n$.
\Prf\ Because
$\sequencen{I_n}$ is a sequence in $\Cal I$ covering $E$,
$\theta E$ is at most
$\sum_{n=0}^{\infty}\lambda I_n\le\sum_{n=0}^{\infty}\nu I_n$.
In the other direction,

\Centerline{$\theta E
\ge\sup_{n\in\Bbb N}\theta(\bigcup_{i\le n}I_i)
=\sup_{n\in\Bbb N}\nu(\bigcup_{i\le n}I_i)
=\sup_{n\in\Bbb N}\sum_{i\le n}\nu I_i
=\sum_{n=0}^{\infty}\nu I_n$.  \Qed}

\medskip

\qquad\grheadb\ $E\in\Sigma$.   \Prf\ Let
$A\subseteq\BbbR^r$ be such that $\theta A$ is finite, and $\epsilon>0$.
We have a sequence $\sequence{m}{J_m}$ in $\Cal I$ such
that $A\subseteq\bigcup_{m\in\Bbb N}J_m$ and
$\sum_{m=0}^{\infty}\lambda J_m\le\theta A+\epsilon$ is finite.
Let $m$ be such that $\sum_{j=m+1}^{\infty}\lambda J_j\le\epsilon$,
and set $K=\bigcup_{j\le m}J_j$;   then

\Centerline{$A\cap E\subseteq(K\cap E)\cup\bigcup_{j>m}J_j$,}

\noindent so

\Centerline{$\theta(A\cap E)
\le\theta(K\cap E)+\theta(\bigcup_{j>m}J_j)
\le\theta(K\cap E)+\sum_{j=m+1}^{\infty}\lambda J_j
\le\theta(K\cap E)+\epsilon$.}

\noindent Similarly,
$\theta(A\setminus E)\le\theta(K\setminus E)+\epsilon$.   Next, by
($\alpha$) applied to $\sequence{i}{J_j\cap I_i}$ or
otherwise, $\sum_{i=0}^{\infty}\nu(J_j\cap I_i)$ is finite for every $j$,
so there is an $n\in\Bbb N$ such that
$\sum_{j=0}^m\sum_{i=n+1}^{\infty}\nu(J_j\cap I_i)\le\epsilon$.   Set
$L=\bigcup_{i\le n}I_i$;  then

\Centerline{$K\cap E
\subseteq(K\cap L)\cup\bigcup_{j\le m,i>n}J_j\cap I_i$,}

$$\eqalign{\theta(K\cap E)
&\le\theta(K\cap L)+\theta(\bigcup_{j\le m,i>n}J_j\cap I_i)\cr
&\le\theta(K\cap L)+\sum_{j=0}^m\sum_{i=n+1}^{\infty}\nu(J_j\cap I_i)
\le\theta(K\cap L)+\epsilon.\cr}$$

\noindent Assembling these,

$$\eqalignno{\theta A
&\le\theta(A\cap E)+\theta(A\setminus E)
\le\theta(K\cap E)+\theta(K\setminus E)+2\epsilon\cr
&\le\theta(K\cap L)+\theta(K\setminus L)+3\epsilon
=\theta K+3\epsilon\cr
\displaycause{because we know that $L\in\Sigma$}
&\le\theta A+3\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta A=\theta(A\cap E)+\theta(A\setminus E)$.   This was on the
assumption that $\theta A$ was finite;  but of course it is also true if
$\theta A=\infty$.   As $A$ is arbitrary, $E\in\Sigma$.\ \Qed

\medskip

\qquad\grheadc\ Accordingly $\nu E=\theta E=\sum_{n=0}^{\infty}\nu I_n$.

\medskip

{\bf (b)} Let $\Cal I_0$ be the family of dyadic half-open intervals in
$\BbbR^r$ of the form $\coint{2^{-k}z,2^{-k}(z+\tbf{1})}$ where
$k\in\Bbb N$, $z\in\BbbZ^r$ and $\tbf{1}=(1,\ldots,1)$.   Note that
$\Cal I_0$ is countable and that if $I$, $J\in\Cal I_0$ then either
$I\subseteq J$ or $J\subseteq I$ or $I\cap J=\emptyset$.   Also any
non-empty subset of $\Cal I_0$ has a maximal element.

If $G\subseteq\BbbR^r$ is open, set
$\Cal J=\{I:I\in\Cal I_0$, $I\subseteq G\}$ and let $\Cal J'$ be the set of
maximal elements of $\Cal J$.   Then $\Cal J'$ is disjoint and countable,
so by (a-ii) $G=\bigcup\Cal J=\bigcup\Cal J'$ belongs to $\Sigma$.

\medskip

{\bf (c)} Because $\nu$ is additive on $\Sigma$,

\Centerline{$\nu(G\cup H)+\nu(G\cap H)
=\nu G+\nu(H\setminus G)+\nu(H\cap G)
=\nu G+\nu H$}

\noindent for all open sets $G$, $H\subseteq\BbbR^r$.

\medskip

{\bf (d)} This time, let $\Cal J$ be
$\bigcup_{n\in\Bbb N}\{I:I\in\Cal I_0$, $I\subseteq G_n\}$;  again, let
$\Cal J'$ be the set of maximal elements of $\Cal J$.   Then
$G=\bigcup\Cal J=\bigcup\Cal J'$, so

\Centerline{$\nu G=\sum_{J\in\Cal J'}\nu J
=\sup_{\Cal K\subseteq\Cal J'\text{ is finite}}\sum_{J\in\Cal K}\nu J
\le\sup_{n\in\Bbb N}\nu G_n=\lim_{n\to\infty}\nu G_n\le\nu G$}

\noindent because $\sequencen{G_n}$ is non-decreasing.
}%end of proof of 565C

\leader{565D}{Definition} Let $\theta$ and $\nu$ be as in 565C.
By 563H, there is a unique Borel-coded measure
$\mu$ on $\BbbR^r$ such that $\mu G=\nu G=\theta G$ for every open set
$G\subseteq\BbbR^r$.   I will say that {\bf Lebesgue measure} on $\BbbR^r$
is the completion $\mu_L$ of $\mu$;
the sets it measures will be {\bf Lebesgue measurable}.

\leader{565E}{Proposition} Let
$\Cal I$, $\theta$, $\Sigma$, $\nu$, $\mu$ and $\mu_L$ be as in
565A-565D.

(a) $\mu$ is the restriction of $\theta$ to the
algebra $\Cal B_c(\BbbR^r)$ of codable Borel sets.

(b) For every $A\subseteq\BbbR^r$,

\Centerline{$\theta A
=\inf\{\mu_L E:E\supseteq A$ is Lebesgue measurable$\}
=\inf\{\mu G:G\supseteq A$ is open$\}$.}

(c) $E\in\Sigma$ and $\mu_L E=\nu E=\theta E$
whenever $E$ is Lebesgue measurable.

(d) $\mu_L$ is inner regular with respect to the compact sets and outer
regular with respect to the open sets.

\proof{{\bf (a)} If $E\in\Cal B_c(\BbbR^r)$, then

$$\eqalignno{\mu E
&=\inf\{\mu G:G\supseteq E\text{ is open}\}\cr
\displaycause{563Fd}
&=\inf\{\theta G:G\supseteq E\text{ is open}\}
\ge\theta E.\cr}$$

\noindent Next, if $I\subseteq\BbbR^r$ is a half-open interval, it is a
codable Borel set and

$$\eqalign{\lambda I
&=\inf\{\lambda J:J\in\Cal I,\,I\subseteq\interior J\}
\ge\inf\{\theta(\interior J):J\in\Cal I,\,I\subseteq\interior J\}\cr
&\ge\inf\{\theta G:G\supseteq I\text{ is open}\}
=\mu I
\ge\theta I
=\lambda I.\cr}$$

\noindent So $\mu$ and $\lambda$ agree on $\Cal I$.   If now
$E$ is a codable Borel set and $\epsilon>0$, there is a sequence
$\sequencen{I_n}$ in $\Cal I$ such that $E\subseteq\bigcup_{n\in\Bbb N}I_n$
and $\sum_{n=0}^{\infty}\lambda I_n\le\theta E+\epsilon$.   But every $I_n$
is resolvable (because it belongs to the algebra of sets generated by the
open sets), so $\sequencen{I_n}$ is a codable sequence (562J) and

\Centerline{$\mu E
\le\mu(\bigcup_{n\in\Bbb N}I_n)
\le\sum_{n=0}^{\infty}\mu I_n
=\sum_{n=0}^{\infty}\lambda I_n
\le\theta E+\epsilon$.}

\noindent As $E$ and $\epsilon$ are arbitrary,
$\mu=\theta\restr\Cal B_c(\BbbR^r)$.

\medskip

{\bf (b)} Suppose that $A\subseteq\BbbR^r$.   If $E\supseteq A$ is
Lebesgue measurable, there are $F$, $H\in\Cal B_c(\BbbR^r)$ such that
$E\symmdiff F\subseteq H$ and $\mu H=0$, so that
$E\subseteq F\cup H$ and

\Centerline{$\theta A\le\theta(F\cup H)=\mu(F\cup H)
=\mu_L E$.}

\noindent So we have

\Centerline{$\theta A
\le\inf\{\mu_L E:E\supseteq A$ is Lebesgue measurable$\}
\le\inf\{\mu G:G\supseteq A$ is open$\}$.}

\noindent In the other direction, given $\epsilon>0$ there is a sequence
$\sequencen{I_n}$ in $\Cal I$, covering $A$, such that
$\sum_{n=0}^{\infty}\lambda I_n\le\theta A+\epsilon$.   As in (a) just
above, $E=\bigcup_{n\in\Bbb N}I_n$ is a codable Borel set and
$\mu E\le\sum_{n=0}^{\infty}\lambda I_n$;  now there is an open
$G\supseteq E$ such that $\mu G\le\mu E+\epsilon\le\theta A+2\epsilon$.
As $\epsilon$ is arbitrary,

\Centerline{$\inf\{\mu G:G\supseteq A$ is open$\}\le\theta A$}

\noindent and we have the equalities.

\medskip

{\bf (c)} Suppose that $E$ is Lebesgue measurable, $A\subseteq\BbbR^r$ and
$\epsilon>0$.   By (b), there is an open set $G\supseteq A$ such that
$\mu G\le\theta A+\epsilon$.   Now

$$\eqalignno{\theta(A\cap E)+\theta(A\setminus E)
&\le\theta(G\cap E)+\theta(G\setminus E)
\le\mu_L(G\cap E)+\mu_L(G\setminus E)\cr
\displaycause{by (b)}
&=\mu_L G
=\mu G
\le\theta A+\epsilon.\cr}$$

\noindent As usual, this is enough to ensure that $E\in\Sigma$.   Now (b)
again tells us that $\mu_L E=\theta E=\nu E$.

\medskip

{\bf (d)} Of course $\mu$ is locally finite, while $\BbbR^r$
is a regular topological space.   So 563F(d-ii) tells us that $\mu$ is
inner regular with respect to the closed sets and outer regular with
respect to the open sets;  it follows that $\mu_L$ also is.
Next, every closed set is K$_{\sigma}$, while compact sets are
resolvable and all sequences of compact sets are codable, so
$\mu F=\sup\{\mu K:K\subseteq F$ is compact$\}$ for every closed set
$F\subseteq\BbbR^r$;  consequently
 $\mu_L$ is inner regular with respect to the compact sets.
}%end of proof of 565E

\leader{565F}{Vitali's Theorem} Let $\Cal C$ be a well-orderable
family of non-singleton closed balls
in $\BbbR^r$.   For $\Cal I\subseteq\Cal C$ set

\Centerline{$A_{\Cal I}
=\bigcap_{\delta>0}\bigcup\{C:C\in\Cal I$, $\diam C\le\delta\}$.}

\noindent Let $\frak T$ be the family of open subsets of $\BbbR^r$.
Then there are functions
$\Psi:\Cal P\Cal C\to\Cal P\Cal C$ and
$\Theta:\Cal P\Cal C\times\Bbb N\to\frak T$ such that
$\Psi(\Cal I)\subseteq\Cal I$, $\Psi(\Cal I)$ is disjoint and countable,
$\mu_L(\Theta(\Cal I,k))\le 2^{-k}$ and
$A_{\Cal I}\subseteq\bigcup\Psi(\Cal I)\cup\Theta(\Cal I,k)$ whenever
$\Cal I\subseteq\Cal C$ and $k\in\Bbb N$.   In particular,

\Centerline{$A_{\Cal I}\setminus\bigcup\Psi(\Cal I)
\subseteq\bigcap_{k\in\Bbb N}\Theta(\Cal I,k)$}

\noindent is negligible.

\proof{ We use the greedy algorithm of 221A/261B, but watching more
carefully.   Start by fixing on a well-ordering $\preccurlyeq$ of
$\Cal C\cup\{\emptyset\}$.
Next, for each $n\in\Bbb N$, set $U_n=\{x:x\in\BbbR^r$, $n<\|x\|<n+1\}$,
where $\|\,\|$ is the Euclidean norm on $\BbbR^r$.
It will be convenient to fix at this point on a family
$\langle G_{kn}\rangle_{k,n\in\Bbb N}$ of open sets such that
$\mu_L G_{kn}\le 2^{-k-n-2}$ and $\{x:\|x\|=n\}\subseteq G_{kn}$ for all $k$
and $n$;  for instance, $G_{kn}$ could be an open shell with rational inner
and outer radii (except for $G_{k0}$, which should be an open
ball).\footnote{Of course we can still use the similarity argument
from part (g) of the proof of 261B to check that thin shells have small
measure.}

\woddheader{565F}{0}{0}{0}{60pt}

Now define $C_{\Cal Inm}$, for $\Cal I\subseteq\Cal C$ and $m$,
$n\in\Bbb N$, by saying that

\inset{\noindent given $\ofamily{i}{m}{C_{\Cal Ini}}$, $C_{\Cal Inm}$ is to
be the $\preccurlyeq$-first member of
$\{\emptyset\}\cup(\Cal I\cap\Cal PU_n)$ which is disjoint from
$\bigcup_{i<m}C_{\Cal Ini}$ and has diameter at least

\Centerline{$\Bover12\sup\{\diam C:
C\in\{\emptyset\}\cup(\Cal I\cap\Cal PU_n)$ is disjoint from
$\bigcup_{i<m}C_{\Cal Ini}\}$.}}

\noindent (I take the diameter of the empty set to be $0$, as usual.)
Set

\Centerline{$\Psi(\Cal I)
=\{C_{\Cal Inm}:n$, $m\in\Bbb N\}\setminus\{\emptyset\}$.}

Because the $U_n$ are disjoint, $\Psi(\Cal I)$ is a disjoint subfamily of
$\Cal I$, and of course it is countable.
Just as in 261B, we find that for each $\Cal I\subseteq\Cal C$ and
$n\in\Bbb N$ we have

\Centerline{$A_{\Cal I}\cap U_n
\subseteq\bigcup_{i<m}C_{\Cal Ini}\cup\bigcup_{i\ge m}C'_{\Cal Ini}$,}

\noindent where for $C\in\Cal C$ I write $C'$ for the {\it open} ball
with the same centre and {\it six} times the radius;
$\emptyset'$ will be $\emptyset$.   Just as in 261B,
$\sum_{m=0}^{\infty}\mu_L C'_{\Cal Inm}\le 6^r\mu_L B(\tbf{0},n+1)$
is finite.
So, for each $k$ and $n$, we can take the first $m_{kn}$ such that
$\sum_{i=m_{kn}}^{\infty}\mu_L C'_{\Cal Ini}\le 2^{-n-k-2}$.   Now set

\Centerline{$\Theta(\Cal I,k)
=\bigcup_{n\in\Bbb N}G_{kn}
  \cup\bigcup_{n\in\Bbb N,i\ge m_{kn}}C'_{\Cal Ini}$;}

\noindent we shall have
$A\setminus\bigcup\Psi(\Cal I)\subseteq\Theta(\Cal I,k)$ and

\Centerline{$\mu_L\Theta(\Cal I,k)\le\sum_{n=0}^{\infty}\mu_L G_{kn}
  +\sum_{n=0}^{\infty}\sum_{i=m_{kn}}^{\infty}
     \mu_L C'_{\Cal Ini}\le 2^{-k}$,}

\noindent as required.
}%end of proof of 565F

\leader{565G}{Proposition} Let $A\subseteq\BbbR^r$ be any set.   Then its
Lebesgue submeasure is

\Centerline{$\theta A
=\inf\{\sum_{n=0}^{\infty}\mu_L B_n:\sequencen{B_n}$ is a sequence of
closed balls covering $A\}$.}

\proof{ Let $\epsilon>0$.   Then there is a (non-empty) open set
$G\supseteq A$ with
$\mu_L G\le\theta A+\epsilon$.   Use Vitali's theorem, with $\Cal C$
the family of closed balls with rational centres and non-zero
rational radii, to see that there is a
disjoint sequence $\sequencen{C_n}$ of balls included in $G$ such that
$\mu_L(A\setminus\bigcup_{n\in\Bbb N}C_n)=0$ and
$\sum_{n\in\Bbb N}\mu_L C_n\le\theta A+\epsilon$.   Next, cover
$A\setminus\bigcup_{n\in\Bbb N}C_n$ by a sequence of half-open intervals
with measures summing to not more than $\epsilon$, and expand these to
balls with measures summing to not more than $\epsilon r^{r/2}$.   
Interleaving this sequence with the $C_n$, we get a
sequence $\sequencen{B_n}$ of balls, covering $A$, with
$\sum_{n=0}^{\infty}\mu_L B_n\le\theta A+(1+r^{r/2})\epsilon$.   So

\Centerline{$\theta A
\ge\inf\{\sum_{n=0}^{\infty}\mu_L B_n:\sequencen{B_n}$ is a sequence of
closed balls covering $A\}$.}

\noindent The reverse inequality is elementary (563C(a-ii)).
}%end of proof of 565G

\leader{565H}{Corollary} Lebesgue measure is invariant under isometries.

\proof{ We can see from its definition that Lebesgue submeasure is
translation-invariant, so Lebesgue measure also is.   Consequently two
balls with the same radii have the same measure.
Isometries of $\BbbR^r$ take closed balls to closed balls with the
same radii, so 565G gives the result.
}%end of proof of 565H

\leader{565I}{Lemma} (a) Writing $C_k(\BbbR^r)$ for the space of
continuous real-valued functions on $\BbbR^r$ with compact support,
$C_k(\BbbR^r)\subseteq\eusm L^1(\mu_L)$.

(b) There is a countable set $D\subseteq C_k(\BbbR^r)$ such that
$\{g^{\ssbullet}:g\in D\}$ is norm-dense in $L^1(\mu_L)$.

\proof{{\bf (a)} This is elementary;  every continuous function is
resolvable, therefore a codable Borel function and belongs to $\eusm L^0$;
if in addition it has compact support it is dominated by an integrable
function and is integrable, by 564E(c-i).

\medskip

{\bf (b)(i)} Let $\Cal U$ be a countable base for the topology of
$\BbbR^r$, consisting of bounded sets and closed under finite unions.
Let $D_0$ be the set of functions of the form
$x\mapsto\max(0,1-2^k\rho(x,\BbbR^r\setminus U))$ for $U\in\Cal U$ and
$k\in\Bbb N$, where $\rho$ is the Euclidean metric on $\BbbR^r$,
and $D$ the set of rational linear combinations of members of
$D_0$;  then $D$ is a countable subset of $C_k(\BbbR^r)$.

\medskip

\quad{\bf (ii)}
If $E\subseteq\BbbR^r$ is a codable Borel set of finite
measure, and $\epsilon>0$, then by 565Ed
there are a compact set $K\subseteq E$ and
an open set $G\supseteq E$ such that $\mu_L(G\setminus K)\le\epsilon$.
Now there are a $U\in\Cal U$ such that $K\subseteq U\subseteq G$ and
a $g\in D_0$ such that $\chi K\le g\le\chi U$, so that
$\int|g-\chi E|\le\epsilon$.

\medskip

\quad{\bf (iii)} It follows that whenever $f$ is a simple
codable Borel function, in the sense of 564Aa, and $\epsilon>0$ there is
a $g\in D$ such that $\int|f-g|\le\epsilon$.

\medskip

\quad{\bf (iv)} If $f\in\eusm L^1$ and $\epsilon>0$ there are a simple
codable Borel function $g$ and an $h\in D$ such that 
$\int|f-g|\le\bover12\epsilon$ such that $\int|g-h|\le\bover12\epsilon$, so that
$\int|f-h|\le\epsilon$.
}%end of proof of 565I

\leader{565J}{Lemma} Suppose that $f$ is an integrable function on
$\BbbR^r$, and that $\int_If\ge 0$ for every
half-open interval $I\subseteq\BbbR^r$.   Then $f(x)\ge 0$ for almost every
$x\in\BbbR^r$.

\proof{{\bf (a)} Note first that any finite union $E$
of half-open intervals is
expressible as a finite disjoint union of half-open intervals.
So $\int_Ef\ge 0$.

\medskip

{\bf (b)} Suppose that $g$ is a simple codable Borel function such that
$\int_Eg\le\epsilon$ whenever $E$ is a finite union of
half-open intervals.   Then $\int g^+\le\epsilon$.   \Prf\ Set
$F=\{x:g(x)>0\}$, and take any $\eta>0$.   Then there are a compact
$K\subseteq F$ and an open $G\supseteq F$ such that
$\mu_L(G\setminus K)\le\eta$.   There is a set $E$,
a finite union of half-open intervals, such that $K\subseteq E\subseteq G$.
In this case,

\Centerline{$\int g^+-\int_Eg\le\int|g\times\chi(E\symmdiff F)|
\le\|g\|_{\infty}\mu_L(G\setminus K)$,
\quad$\int g^+\le\epsilon+\eta\|g\|_{\infty}$;}

\noindent as $\eta$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (c)} We know that there is a codable
sequence $\sequencen{g_n}$ of simple codable
Borel functions such that $f\eae\lim_{n\to\infty}g_n$ and
\ifnum\stylenumber=12 the sum \else\fi
$\sum_{n=0}^{\infty}\int|g_{n+1}-g_n|$ is finite.   Set
$\epsilon_n=\sum_{i=n}^{\infty}\int|g_{i+1}-g_i|$ for each $n$;  then
$\int|f-g_n|\le\epsilon_n$, because $\int|g_m-g_n|\le\epsilon_n$ for every
$m\ge n$.   So if $E$ is a finite union of half-open intervals,

\Centerline{$\int_Eg_n
=\int g_n\times\chi E\ge\int f\times\chi E-\int|f-g_n|
\ge-\epsilon_n$;}

\noindent by (a), applied to $-g_n$,
$\int g_n^-\le\epsilon_n$.   By 564Be,

\Centerline{$f^-\eae\lim_{n\to\infty}g_n^-\eae
\liminf_{n\to\infty}g_n^-=0$}

\noindent almost everywhere, as required.
}%end of proof of 565J

\leader{565K}{Theorem} A monotonic function $f:\Bbb R\to\Bbb R$ is
differentiable almost everywhere.

\cmmnt{\medskip

\noindent{\bf Remark} Of course `almost everywhere'
here is with respect to Lebesgue
measure on $\Bbb R$;  in this result and the next two I am taking $r=1$.}

\proof{ We can use the ideas in 222A if we refine them using 565F.   First,
$\Cal C$ will be the set of closed non-trivial intervals with rational
endpoints;  take $\Psi$ and $\Theta$ as in 565F.   It will be enough to
deal with the case of non-decreasing $f$.   For $a<b$ in $\Bbb R$, set
$f^*([a,b])=[f(a),f(b)]$.   I shall repeatedly use the fact that if
$\Cal I\subseteq\Cal C$ is disjoint, then

\Centerline{$\mu_L(\bigcup_{C\in\Cal I}f^*(C))
=\sum_{C\in\Cal I}\mu_L f^*(C)$,}

\noindent because $\Cal I$ is countable and
$f^*(C)\cap f^*(C')$ contains at most one point for any distinct
$C$, $C'\in\Cal I$, and we can use 563C(a-iv).

\wheader{565K}{6}{2}{2}{72pt}
{\bf (a)} Again set

\Centerline{$(\DiniD f)(x)=\limsup_{h\to 0}{1\over h}(f(x+h)-f(x))$,}

\Centerline{$(\Dinid f)(x)=\liminf_{h\to 0}{1\over h}(f(x+h)-f(x))$}
     
\noindent for $x\in\Bbb R$.   To see that $\DiniD f<\infty$ a.e., set
$E_m=\{x:|x|<m$, $(\DiniD f)(x)>2^m(1+f(m)-f(-m))\}$ and

$$\eqalign{\Cal I_m
&=\{[\alpha,\beta]:\alpha,\,\beta\in\Bbb Q,\,-m<\alpha<\beta<m,\cr
&\mskip 200mu f(\beta)-f(\alpha)>2^m(1+f(m)-f(-m))(\beta-\alpha)\}\cr
&=\{C:C\in\Cal C,\,C\subseteq\ooint{-m,m},
\mu_L f^*(C)>2^m(1+f(m)-f(-m))\mu_L C\}\cr}$$

\noindent for each $m$.   Then, in the language of 565F,
$E_m\subseteq A_{\Cal I_m}$.   (If $x\in E_m$ and $\delta>0$, then
$x$ is an endpoint of a non-trivial
closed interval $[\alpha,\beta]\subseteq\ooint{-m,m}$, of length less than
$\delta$, such that
$f(\beta)-f(\alpha)>2^m(1+f(m)-f(-m))(\beta-\alpha)$.
Now we can expand $[\alpha,\beta]$
slightly to get an interval $[\alpha',\beta']\in\Cal I_m$ of length at most
$\delta$.)   So $E_m\subseteq\bigcup\Psi(\Cal I_m)\cup\Theta(\Cal I_m,k)$
for each $k$.   $\Psi(\Cal I_m)$ is a countable family of closed
sets, and

$$\eqalign{2^m(1+f(m)-f(-m))\sum_{C\in\Psi(\Cal I_m)}\mu_L C
&\le\sum_{C\in\Psi(\Cal I_m)}\mu_L f^*(C)\cr
&=\mu_L\bigl(\bigcup_{C\in\Psi(\Cal I_m)}f^*(C)\bigr)
\le f(m)-f(-m).\cr}$$

\noindent So $\sum_{C\in\Psi(\Cal I_m)}\mu_L C\le 2^{-m}$.
Setting 
$H_m=\Theta(\Cal I_m,m)\cup\bigcup\{\interior C:C\in\Psi(\Cal I_m)\}$,
$H_m$ is open, $\mu H_m\le 2^{-m+1}$ and $E_m\setminus\Bbb Q\subseteq H_m$.

Set $E=\{x:(\DiniD f)(x)=\infty\}$, and take any $n\in\Bbb N$.
Then

\Centerline{$E\setminus\Bbb Q
\subseteq\bigcup_{m\ge n}E_m\setminus\Bbb Q
\subseteq\bigcup_{m\ge n}H_m$.}

\noindent Now 563C(a-ii) tells us that

\Centerline{$\mu_L(\bigcup_{m\ge n}H_m)
\le\sum_{m=n}^{\infty}\mu H_m\le 2^{-n+1}$}

\noindent for each $n$, so that
$E\setminus\Bbb Q$ is included in a negligible G$_{\delta}$ set and
$\mu_L E=\mu_L(E\setminus\Bbb Q)=0$.   Thus $\DiniD f$ is finite a.e.

\medskip

{\bf (b)} To see that $\DiniD f\leae\Dinid f$, we use similar ideas, but with
an extra layer of complexity, corresponding to the double use of
Vitali's theorem.   Set $F=\{x:(\Dinid f)(x)<(\DiniD f)(x)\}$.   Take any
$\epsilon>0$;  because $\Bbb Q$ is countable, there is a family
$\langle\epsilon_{mqq'}\rangle_{m\in\Bbb N,q,q'\in\Bbb Q}$
of strictly positive numbers such that
$\sum_{m\in\Bbb N,q,q'\in\Bbb Q}\epsilon_{mqq'}\le\bover12\epsilon$.
For $q$, $q'\in\Bbb Q$ and $m$, $k\in\Bbb N$ let $\Cal I_{mqk}$,
$\Cal J_{mqk}$ be

\Centerline{$\{C:C\in\Cal C$, $C\subseteq\ooint{-m,m}$,
$\mu_L C\le 2^{-k}$, $\mu_L f^*(C)\ge q\mu_L C\}$,}

\Centerline{$\{C:C\in\Cal C$, $C\subseteq\ooint{-m,m}$,
$\mu_L C\le 2^{-k}$, $\mu_L f^*(C)\le q\mu_L C\}$}

\noindent respectively.  For $m$, $k\in\Bbb N$ and $q$, $q'\in\Bbb Q$ set

\Centerline{$G_{mqq'k}=\bigcup\{\interior C:C\in\Cal I_{mq'k}\}
   \cap\bigcup\{\interior C:C\in\Cal J_{mqk}$\};}

\noindent then $\sequence{k}{G_{mqq'k}}$ is a
non-increasing sequence of open sets of finite measure.
So, setting $F_{mqq'}=\bigcap_{k\in\Bbb N}G_{mqq'k}$, we can find a family
$\langle k(m,q,q')\rangle_{m\in\Bbb N,q,q'\in\Bbb Q}$ in $\Bbb N$ such that

\Centerline{$\mu_L(G_{m,q,q',k(m,q,q')}\setminus F_{mqq'})
\le\min(1,\Bover{q'-q}{q})\epsilon_{mqq'}$}

\noindent whenever $m\in\Bbb N$, $q$,
$q'\in\Bbb Q$ and $0<q<q'$ (563C(b-ii)).   Write $H_{mqq'}$ for
$G_{m,q,q',k(m,q,q')}$.

If $m\in\Bbb N$ and $0<q<q'$ in $\Bbb Q$, set

\Centerline{$\Cal J'_{mqq'}=\{C:C\in\Cal C$,
$C\subseteq H_{mqq'}$, $\mu_L f^*(C)\le q\mu_L C\}$.}

\noindent Then $F_{mqq'}\subseteq H_{mqq'}$,
so every point of $F_{mqq'}$ belongs to the interiors of arbitrarily small
intervals belonging to $\Cal J'_{mqq'}$;  accordingly
$F_{mqq'}\setminus\bigcup\Psi(\Cal J'_{mqq'})$ is negligible.

Now let $\Cal I'_{mqq'}$ be the set

\Centerline{$\{C:C\in\Cal C$, $C\subseteq C'$ for some
$C'\in\Psi(\Cal J'_{mqq'})$, $\mu_L f^*(C)\ge q'\mu_L C\}$.}

\noindent Then every point of
$F_{mqq'}\cap\bigcup\Psi(\Cal J'_{mqq'})\setminus\Bbb Q$ belongs to
arbitrarily small members of $\Cal I'_{mqq'}$, so
$F_{mqq'}\setminus\bigcup\Psi(\Cal I'_{mqq'})$ is negligible.

Now we come to the calculation at the heart of the proof.   If $m\in\Bbb N$
and $0<q<q'$ in $\Bbb Q$,

$$\eqalignno{q'\mu_L F_{mqq'}
&\le q'\mu_L(\bigcup\Psi(\Cal I'_{mqq'}))
=q'\sum_{C\in\Psi(\Cal I'_{mqq'})}\mu_L C\cr
&\le\sum_{C\in\Psi(\Cal I'_{mqq'})}\mu_L f^*(C)
=\mu_L\bigl(\bigcup_{C\in\Psi(\Cal I'_{mqq'})}f^*(C)\bigr)
\le\mu_L\bigl(\bigcup_{C\in\Psi(\Cal J'_{mqq'})}f^*(C)\bigr)\cr
\displaycause{because every member of $\Cal I'_{mqq'}$ is included in a
member of $\Psi(\Cal J'_{mqq'})$}
&=\sum_{C\in\Psi(\Cal J'_{mqq'})}\mu_L f^*(C)
\le q\sum_{C\in\Psi(\Cal J'_{mqq'})}\mu_L C\cr
&=q\mu_L(\bigcup\Psi(\Cal J'_{mqq'})
\le q\mu_L H_{mqq'}
\le q\mu_L F_{mqq'}+(q'-q)\epsilon_{mqq'},\cr}$$

\noindent and $\mu_L F_{mqq'}\le\epsilon_{mqq'}$,
$\mu_L H_{mqq'}\le 2\epsilon_{mqq'}$.   But this means that

\Centerline{$F\setminus\Bbb Q
\subseteq\bigcup_{m\in\Bbb N,q,q'\in\Bbb Q,0<q<q'}F_{mqq'}
\subseteq\bigcup_{m\in\Bbb N,q,q'\in\Bbb Q,0<q<q'}H_{mqq'}$,}

\noindent which has measure at most

\Centerline{$\sum_{m\in\Bbb N,q,q'\in\Bbb Q,0<q<q'}\mu_L H_{mqq'}
\le 2\sum_{m\in\Bbb N,q,q'\in\Bbb Q,0<q<q'}\epsilon_{mqq'}
\le\epsilon$.}

The process described here gives a recipe, starting from $\epsilon>0$, for
finding an open set of measure at most $\epsilon$ including
$F\setminus\Bbb Q$.   So we can repeat this for each term of a sequence
converging to $0$ to define a negligible G$_{\delta}$ set including
$F\setminus\Bbb Q$, and $F$ must be negligible, as required.
}%end of proof of 565K

\leader{565L}{Lemma} Suppose that $F:\Bbb R\to\Bbb R$ is a bounded
non-decreasing function.   Then $\int F'$ is defined and is at most
$\lim_{x\to\infty}F(x)-\lim_{x\to-\infty}F(x)$.

\wheader{565L}{0}{0}{0}{48pt}

\proof{ I copy the ideas of 222C.
For each $n\in\Bbb N$, define $g_n:\Bbb R\to\Bbb R$ by setting
$a_{nk}=2^{-n+1}k(n+1)-n$ for $k\le 2^n$,

$$\eqalign{g_n(x)
&=\Bover{2^{n-1}}{n+1}(F(a_{n,k+1})-F(a_{nk}))\text{ if }k<2^n
  \text{ and }a_{nk}\le x<a_{n,k+1},\cr
&=0\text{ if }x<-n\text{ or }x\ge n+2.\cr}$$

\noindent Then $g_n$ is a simple Borel measurable function
and $F'(x)=\lim_{n\to\infty}g_n(x)$ whenever
$F'(x)$ is defined, which is almost everywhere, by 565K.
Also $\int g_n=F(n+2)-F(-n)$.   Because the $g_n$ are resolvable,
$\sequencen{g_n}$ is codable;  by Fatou's Lemma (564Fb),

\Centerline{$\int F'
\le\liminf_{n\to\infty}\int g_n
=\lim_{x\to\infty}F(x)-\lim_{x\to-\infty}F(x)$.}
}%end of proof of 565L

\leader{565M}{Theorem} Let $F:\Bbb R\to\Bbb R$ be a function.   Then the
following are equiveridical:

\inset{(i) there is an integrable function $f$ such that
$F(x)=\int_{\ooint{-\infty,x}}f$ for every $x\in\Bbb R$,

(ii) $F$ is of bounded variation, absolutely continuous on every bounded
interval, and
\ifdim\pagewidth>467pt\break\fi
$\lim_{x\to-\infty}F(x)=0$,}

\noindent and in this case $F'\eae f$.

\proof{{\bf (a)} If $f$ is integrable and
$F(x)=\int_{\ooint{-\infty,x}}f$ for every $x\in\Bbb R$, take any
$\epsilon>0$.   Then there is a $g\in C_k(\Bbb R)$ such that
$\int|f-g|\le\epsilon$ (565Ib).
Let $x_0$ be such that $g(x)=0$ for $x\le x_0$;  then

\Centerline{$|F(x)|\le\int|f-g|\le\epsilon$}

\noindent whenever $x\le x_0$.   Set
$\delta=\Bover{\epsilon}{1+\|g\|_{\infty}}$.   If
$a_0\le b_0\le\ldots\le a_n\le b_n$ and $\sum_{i=0}^nb_i-a_i\le\delta$,
then

$$\eqalign{\sum_{i=0}^n|F(b_i)-F(a_i)|
&=\sum_{i=0}^n|\int_{\coint{a_i,b_i}}f|
\le\sum_{i=0}^n\int_{\coint{a_i,b_i}}|g|+\int_{\coint{a_i,b_i}}|f-g|\cr
&\le\sum_{i=0}^n(b_i-a_i)\|g\|_{\infty}+\int|f-g|
\le\delta\|g\|_{\infty}+\epsilon
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\lim_{x\to-\infty}F(x)=0$ and $F$ is
absolutely continuous on every bounded interval.   As for the variation of
$F$, if $a_0\le a_1\le\ldots\le a_n$ then

$$\eqalign{\sum_{i=1}^n|F(a_i)-F(a_{i-1})|
&=\sum_{i=1}^n|\int_{\coint{a_{i-1},a_i}}f|
\le\sum_{i=1}^n\int_{\coint{a_{i-1},a_i}}|f|\cr
&=\int_{\coint{a_0,a_n}}|f|
\le\int|f|,\cr}$$

\noindent so $\Var(F)\le\int|f|$ is finite.

Thus (i)$\Rightarrow$(ii).

\medskip

{\bf (b)} Moreover, under the conditions of (a),
$F'\eae f$.   \Prf\ Because $f$ is the difference of two
non-negative integrable functions, it is enough to consider the case
$f\ge 0$ a.e., so that $F$ is non-decreasing.
In this case, applying 565L to the function $x\mapsto\med(F(a),F(x),F(b))$,
we see that $\int_{\coint{a,b}}F'\le\int_{\coint{a,b}}f$ whenever $a\le b$
in $\Bbb R$;  also, applying 565L to $F$ itself, $F'$ is integrable.
Applying 565J to $f-F'$, we see that $F'\leae f$.

Recall that there is a countable subset $D$ of $C_k(\Bbb R)$
approximating all integrable functions in mean (565Ib).
So there is a sequence  $\sequencen{g_n}$ in $D$ such that
$\sum_{n=0}^{\infty}\int|g_n-f|$ is finite.   Set
$\tilde g_n=\sup_{i\le n}g_i^+$ for $n\in\Bbb N$;  then all the
$\tilde g_n$ are continuous, therefore resolvable, and
$\sequencen{\tilde g_n}$ is a codable sequence of integrable functions.
By 564Fa, $g=\lim_{n\to\infty}\tilde g_n$ is defined a.e.\ and integrable.
Let $G_n$, $G$ be the indefinite integrals of $\tilde g_n$, $g$
respectively.   Then the arguments just used show that $G'\leae g$.
But note that each $G_n$, being the indefinite integral of a continuous
function, has $G'_n=\tilde g_n$ exactly,
while $G'_n\le G'$ whenever $G'$ is defined.   So

\Centerline{$g\eae\lim_{n\to\infty}\tilde g_n=\lim_{n\to\infty}G'_n
\leae G'$,}

\noindent and $g\eae G'$.

At this point observe that $\int\liminf_{n\to\infty}|g_n-f|=0$, by 564Fb,
so $f\leae g$, while $G-F$ is the indefinite
integral of the essentially non-negative integrable function $g-f$.
So $G'-F'\leae g-f\eae G'-f$ and $f\leae F'$.   So actually $f\eae F'$,
as hoped for.\ \Qed

\medskip

{\bf (c)} Now suppose that $F:\Bbb R\to\Bbb R$
is of bounded variation and absolutely continuous on every bounded
interval, and that $\lim_{x\to-\infty}F(x)=0$.   By 224D and 565L,
$F'$ is integrable;  set $G(x)=\int_{\ooint{-\infty,x}}F'$
and $H(x)=F(x)-G(x)$ for $x\in\Bbb R$.   By (b), $H'=F'-G'$ is zero a.e.,
while $H$, like $F$ and $G$, is absolutely continuous on every bounded
interval.   But this means that $H$ is constant.   \Prf\ Suppose that
$a<b$ in $\Bbb R$ and $\epsilon>0$.   Let $\delta\in\ooint{0,b-a}$
be such that $\sum_{i=0}^n|H(b_i)-H(a_i)|\le\epsilon$ whenever
$a\le a_0\le b_0\le a_1\le b_1\le\ldots\le a_n\le b_n\le b$ and
$\sum_{i=0}^nb_i-a_i\le\delta$   Set
$E=\{x:x\in\ooint{a,b}$, $H'(x)=0\}$.   Let $\Cal C$ be the family of
non-trivial closed subintervals $[c,d]$ of $\ooint{a,b}$ with rational
endpoints such that $|H(d)-H(c)|\le\epsilon(d-c)$;  then every point of $E$
belongs to arbitrarily small members of $\Cal C$.   By Vitali's theorem
(565F) there is a disjoint countable
family $\Cal I\subseteq\Cal C$ such that
$E\setminus\bigcup\Cal I$ is negligible, so that

\Centerline{$\sum_{I\in\Cal I}\mu_LI=\mu_L(\bigcup\Cal I)=b-a$.}

\noindent Let $\Cal J\subseteq\Cal I$ be a finite subset such that
$\sum_{I\in\Cal J}\mu_LI\ge b-a-\delta$;  express $\Cal J$ as
$\ofamily{i}{n}{[b_i,a_{i+1}]}$ where $\ofamily{i}{n}{b_i}$ is strictly
increasing.   Setting $a_0=a$ and $b_n=b$, we have
$a=a_0\le b_0\le\ldots\le a_n\le b_n=b$ and $\sum_{i=0}^nb_i-a_i\le\delta$.
So

$$\eqalign{|H(b)-H(a)|
&\le\sum_{i=0}^n|H(b_i)-H(a_i)|+\sum_{i=0}^{n-1}|H(a_{i+1})-H(b_i)|\cr
&\le\epsilon+\epsilon\sum_{i=0}^{n-1}(a_{i+1}-b_i)
\le\epsilon(1+b-a).\cr}$$

\noindent As $a$, $b$ and $\epsilon$ are arbitrary, $H$ is constant.\
\Qed

So $F-G$ is constant.   As both $F$ and $G$ tend to $0$ at $-\infty$,
they are equal.   Thus $F(x)=\int_{\ooint{-\infty,x}}F'$ for every $x$, and
$F$ is an indefinite integral.
}%end of proof of 565M

\leader{565N}{Hausdorff measures} Let $(X,\rho)$ be a metric space and
$s\in\ooint{0,\infty}$.   As in \S471, we can define {\bf Hausdorff
$s$-dimensional submeasure} $\theta_s:\Cal PX\to[0,\infty]$ by
writing

$$\eqalign{\theta_sA
=\sup_{\delta>0}\inf\{\sum_{n=0}^{\infty}(\diam D_n)^s:
  \sequencen{D_n}
&\text{ is a sequence of subsets of }X\text{ covering }A,\cr
&\qquad\qquad\qquad\quad
  \diam D_n\le\delta\text{ for every }n\in\Bbb N\},\cr}$$

\noindent counting $\diam\emptyset$ as $0$ and
$\inf\emptyset$ as $\infty$.
\cmmnt{As with Lebesgue submeasure,} $\theta_s$ is a submeasure.

\leader{565O}{Theorem} Let $(X,\rho)$ be a second-countable metric
space, and $s>0$.   Then there is a Borel-coded measure $\mu$ on $X$ such
that $\mu K=\theta_sK$ whenever $K\subseteq X$ is compact and $\theta_sK$
is finite.

%If $\theta_sX$ is finite, then there is a Borel-coded
%measure $\mu$ on $X$ such that $\mu G=\theta_sG$ for every open set
%$G\subseteq X$.

\proof{{\bf (a)} To begin with, suppose that $X$ is compact and
$\theta_sX$ is finite.

\medskip

\quad{\bf (i)} Let $\Cal U$ be a countable base for the topology of $X$
closed under finite unions;  let
$\preccurlyeq$ be a well-ordering of $\Cal U$.
Then for any compact $K\subseteq X$,
$\delta>0$ and $\epsilon>0$, there are $U_0,\ldots,U_n\in\Cal U$ such that
$K\subseteq\bigcup_{i\le n}U_i$, $\diam U_i\le\delta$ for every $i$ and
$\sum_{i=0}^n(\diam U_i)^s\le\theta_sK+\epsilon$.   \Prf\ There is a sequence
$\sequencen{A_n}$ of subsets of $X$ such that
$K\subseteq\bigcup_{n\in\Bbb N}A_n$, $\diam A_n\le\bover12\delta$ for every
$n$ and $\sum_{n=0}^{\infty}(\diam A_n)^s\le\theta_sK+\bover12\epsilon$.
Let $\sequencen{\eta_n}$ be a sequence in $\ooint{0,\bover14\delta}$ such
that $\sum_{n=0}^{\infty}(2\eta_n+\diam A_n)^s\le\theta_sK+\epsilon$.

For each $n\in\Bbb N$, set $G_n
=\{x:\rho(x,A_n)<\eta_n\}$.   Then $\overline{A}_n$ is a compact subset 
of $G_n$ so there is a $\preccurlyeq$-first $U_n\in\Cal U$ such that
$\overline{A}_n\subseteq U_n\subseteq G_n$.   Now
$\diam U_n\le\min(\delta,\diam A_n+2^{-n-2}\epsilon)$ for each $n$ so
$\sum_{n=0}^{\infty}(\diam U_n)^s\le\theta_sK+\epsilon$.   But as $K$ is
compact there is an $n$ such that $K\subseteq\bigcup_{i\le n}U_i$.\ \Qed

\medskip

\quad{\bf (ii)} As in the proof of 471Da, $\theta_s$ is a `metric
submeasure', that is, $\theta_s(A\cup B)=\theta_sA+\theta_sB$ whenever $A$,
$B\subseteq X$ and $\rho(A,B)>0$.   (It will be convenient here to say
that $\rho(A,B)=\infty$ if either $A$ or $B$ is empty.)   It follows that
$\theta_s(\bigcup_{n\in\Bbb N}K_n)=\sum_{n=0}^{\infty}\theta_sK_n$ whenever
$\sequencen{K_n}$ is a disjoint sequence of compact subsets of $X$.   \Prf\
Recall that $\rho(K,K')>0$ whenever $K$, $K'$ are
disjoint compact subsets of $X$;  this is because $K\times K'$ is compact
and $\rho:X\times X\to\Bbb R$ is continuous.   So

\Centerline{$\theta_s(\bigcup_{n\in\Bbb N}K_n)
\ge\theta_s(\bigcup_{i\le n}K_i)
=\sum_{i=0}^n\theta_sK_i$}

\noindent for every $n\in\Bbb N$, and
$\theta_s(\bigcup_{n\in\Bbb N}K_n)\ge\sum_{n=0}^{\infty}\theta_sK_n$.   In
the other direction, let $\epsilon>0$.   Let $\preccurlyeq'$ be a
well-ordering of $\bigcup_{n\in\Bbb N}\Cal U^n$.   Then for each
$n\in\Bbb N$ there is a $\preccurlyeq'$-first finite sequence
$U_{n0},\ldots,U_{nm_n}$ in $\Cal U$ such that
$K_n\subseteq\bigcup_{i\le m_n}U_{ni}$, $\diam U_{ni}\le\epsilon$ for every
$i$ and $\sum_{i=0}^{m_n}(\diam U_{ni})^s\le\theta_sK_n+2^{-n}\epsilon$.
Now $\langle U_{ni}\rangle_{n\in\Bbb N,i\le m_n}$ witnesses that

$$\eqalign{\sum_{n=0}^{\infty}\theta_sK_n+2\epsilon
&\ge\inf\{\sum_{j=0}^{\infty}(\diam D_j)^s:
  \sequence{j}{D_j}\text{ is a sequence of subsets of }X\cr
&\mskip150mu\text{ covering }\bigcup_{n\in\Bbb N}K_n,\,
  \diam D_j\le\epsilon\text{ for every }j\in\Bbb N\}.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta_s(\bigcup_{n\in\Bbb N}K_n)\le\sum_{n=0}^{\infty}\theta_s(K_n)$.\
\Qed

\medskip

\quad{\bf (iii)} If $G\subseteq X$ is open and $\epsilon>0$, there is a compact
set $K\subseteq G$ such that $\theta_s(G\setminus K)\le\epsilon$.   \Prf\
Set $K_0=\{x:\rho(x,X\setminus G)\ge 1\}$ and for $n\ge 1$ set

\Centerline{$K_n=\{x:2^{-n}\le\rho(x,X\setminus G)\le 2^{n+1}\}$.}

\noindent Then $\sum_{n=0}^{\infty}\theta_sK_{2n}$ and
$\sum_{n=0}^{\infty}\theta_sK_{2n+1}$ are both bounded by
$\theta_sX<\infty$, so there is an $n\in\Bbb N$ such that
$\sum_{i=n}^{\infty}\theta_sK_n\le\epsilon$.   But this means that
$\theta_s(\bigcup_{i\ge n}K_n)\le\epsilon$ (apply (ii) to the odd and even
terms separately).   Set $K=\bigcup_{i\le n}K_i$;  this works.\ \Qed

\medskip

\quad{\bf (iv)} Writing $\frak T$ for the topology of $X$,
$\theta_s\restrp\frak T$ satisfies the conditions of 563H.   \Prf\ Of course
it is zero at $\emptyset$, monotonic and locally finite.   If $G$,
$H\in\frak T$ and $\epsilon>0$, let $K\subseteq G$, $L\subseteq H$ be
compact sets such
that $\theta_s(G\setminus K)+\theta_s(H\setminus L)\le\epsilon$.   Then
$K\setminus H$, $K\cap L$ and $L\setminus G$ are disjoint compact sets and

\Centerline{$(G\cup H)
\setminus((K\setminus H)\cup(K\cap L)\cup(L\setminus G))$,
\quad$(G\cap H)\setminus(K\cap L)$,}

\Centerline{$G\setminus((K\setminus H)\cup(K\cap L))$,
\quad$H\setminus((L\setminus G)\cup(K\cap L))$}

\noindent are all included in $(G\setminus K)\cup(H\setminus L)$, so all
have submeasure at most $\epsilon$.   But this means that
$\theta_s(G\cup H)+\theta_s(G\cap H)$ and $\theta_sG+\theta_sH$ both differ
from $\theta_s(K\setminus H)+2\theta_s(K\cap L)+\theta_s(L\setminus G)$ by
at most $2\epsilon$ (upwards)
and differ from each other by at most $2\epsilon$ also.
As $\epsilon$ is arbitrary, we have the modularity condition.

As for the sequential order-continuity, this is elementary;  if
$\sequencen{G_n}$ is a non-decreasing sequence with union $G$, and
$\epsilon>0$, there is a compact $K\subseteq G$ such that
$\theta_s(G\setminus K)\le\epsilon$;  now $K\subseteq G_n$ for some $n$,
and $\theta_sG\le\theta_sG_n+\epsilon$.\ \Qed

\medskip

\quad{\bf (v)} So 563H tells us that there is a Borel-coded
measure $\mu$ on $X$
extending $\theta_s\restrp\frak T$.   Now $\mu K=\theta_sK$ for every
compact $K\subseteq X$.   \Prf\

\Centerline{$\mu K+\mu(X\setminus K)=\mu X
=\theta_sX\le\theta_sK+\theta_s(X\setminus K)
=\theta_sK+\mu(X\setminus K)$,}

\noindent so $\mu K\le\theta_sK$.   On the other hand, given $\epsilon>0$,
there is a compact $L\subseteq X\setminus K$ such that
$\theta_sL\ge\theta_s(X\setminus K)-\epsilon$, and now

\Centerline{$\theta_sK=\theta_s(K\cup L)-\theta_sL
\le\theta_sX-\theta_s(X\setminus K)+\epsilon
=\mu K+\epsilon$;}

\noindent as $\epsilon$ is arbitrary, $\mu K=\theta_sK$.\ \Qed

\medskip

\quad{\bf (vi)} Note that 563H tells us that $\mu$ is the only Borel-coded
measure extending $\theta_s\restrp\frak T$, and must therefore be the only
Borel-coded measure agreeing with $\theta_s$ on the compact sets.

\medskip

{\bf (b)} For the general case, let $\Cal K$ be
$\{K:K\subseteq X$ is compact, $\theta_sK<\infty\}$.
Then (a) tells us that for every $K\in\Cal K$ there is a unique Borel-coded measure
$\mu_K$ on $K$ agreeing with $\theta_s$ on the compact subsets of $K$.
If $K$, $L\in\Cal K$ and $K\subseteq L$, $\mu_L\restr\Cal B_c(K)$ is a
Borel-coded measure on $K$ (563Fa)
agreeing with $\theta_s$ on the compact subsets of $K$, so $\mu_L$ extends
$\mu_K$.   We therefore have a Borel-coded measure $\mu$ on $X$
defined by setting $\mu E=\sup_{K\in\Cal K}\mu_K(E\cap K)$ for every
$E\in\Cal B_c(X)$ (cf.\ 563E),
and $\mu$ agrees with $\theta_s$ on $\Cal K$, as required.
}%end of proof of 565O

%any hope of Howroyd's theorem?

\exercises{\leader{565X}{Basic exercises (a)}%
%\spheader 565Xa
(i) Show that Lebesgue submeasure $\theta$ and Lebesgue measure are
translation-invariant.   (ii) Show that if $A\subseteq\BbbR^r$ and
$\alpha\ge 0$ then $\theta(\alpha A)=\alpha^r\theta A$.  (iii)
Show that if $E\subseteq\BbbR^r$ is measurable and
$\alpha\in\Bbb R$ then $\alpha E$ is measurable.
%565C

\spheader 565Xb Suppose that there is a sequence $\sequencen{A_n}$ of
countable subsets of $[0,1]$ with union $[0,1]$.   (i) Set
$A=\bigcup_{m\le n}A_m+n$.    Show that $A$ belongs to the algebra $\Sigma$
of 565C, that the Lebesgue submeasure of $A$ is $\infty$,
but that $A\cap[0,n]$ is Lebesgue negligible for every $n$.
(ii) Set $B=\{2^{-n}x:n\in\Bbb N$, $x\in A_n\}$.   Show that $B$ has
Lebesgue submeasure $0$, but is not Lebesgue negligible.
%565D

\spheader 565Xc Let $g:\Bbb R\to\Bbb R$ be a non-decreasing function.   For
half-open intervals $I\subseteq\Bbb R$ define $\lambda_gI$ by setting

\Centerline{$\lambda_g\emptyset=0$,
\quad$\lambda_g\coint{a,b}
=\lim_{x\uparrow b}g(x)-\lim_{x\uparrow a}g(x)$}

\noindent if $a<b$.   For any set $A\subseteq\Bbb R$ set

\Centerline{$\theta_gA=\inf\{\sum_{j=0}^{\infty}\lambda_gI_j:
\sequence{j}{I_j}
\text{ is a sequence of }\text{half-open intervals covering }A$.}

\noindent Show that $\theta_g$ is a submeasure on $\Cal P\Bbb R$.
Show that there is a Borel-coded measure
$\mu_g$ on $\Bbb R$ agreeing with $\theta_g$ on open sets.
%565D

\spheader 565Xd Apply 564N to relate Lebesgue measure on $\BbbR^2$ to
Lebesgue measure on $\Bbb R$.
%565D

\spheader 565Xe Suppose that there is a sequence $\sequencen{A_n}$ of
countable sets with union $[0,1]$.   Show that there is a set
$A\subseteq[0,1]^2$, with two-dimensional Lebesgue submeasure zero, such
that all the vertical sections $A[\{x\}]$, for $x\in[0,1]$, have non-zero
one-dimensional Lebesgue measure.
%565L

\spheader 565Xf Confirm that the principal
results of \S281 can be proved without the axiom of choice.

\leader{565Y}{Further exercises (a)}
%\spheader 565Ya
Show that if $X$ is a second-countable space and $\mu$ is a
codably $\sigma$-finite Borel-coded
measure on $X$, then there is a non-decreasing function
$g:\Bbb R\to\Bbb R$ such that the Lebesgue-Stieltjes measure $\mu_g$
of 565Xc has measure algebra isomorphic to that of $\mu$.
%565D 565Xc
}%end of exercises


\endnotes{
\Notesheader{565} In these five sections I have tried to indicate,
without succumbing to the temptation to re-write the whole treatise, a
possible version of Lebesgue's theory which can be used in plain ZF.
With the Fundamental Theorem of Calculus (565M), the
Radon-Nikod\'ym theorem (564L), Fubini's theorem (564N) and at least some
infinite product measures (564O), it is clear that most of the ideas of
Volume 2 should be expressible in forms not relying on the axiom of choice.
We must expect restrictions of the type already found in the convergence
theorems (564F);  for versions of the Central Limit Theorem or the strong
law of large numbers or Koml\'os's theorem, for instance, we should
certainly start by changing any hypothesis `let $\sequencen{f_n}$ be a
sequence of random variables' into `let $\sequencen{f_n}$ be a codable
sequence of codable Borel functions'.   I am not sure how to approach
martingales, but the best chance of positive results will be with `codable
martingales' in which we have a full set of Borel codes for countable sets
generating each of the $\sigma$-algebras involved, along the lines of
564Md.   If you glance at the formulae of
Chapter 28, you will see that while there are many appeals to the
convergence theorems, they are generally applied to sequences of the form
$\sequencen{f\times g_n}$ where $f$ is integrable and the $g_n$ are
continuous;  but this means that $\sequencen{g_n}$ is necessarily codable
(562Qa, 562Sc) 
so that $\sequencen{f\times g_n}$ will be a codable sequence if $f$
itself is a codable function.

In Volumes 3 and 4 we encounter much more solid obstacles, and I see no way
in which Maharam's theorem, or the Lifting Theorem, can be made to work
without something approaching the full axiom of choice, or a strong
hypothesis declaring the existence of a well-orderable set at a crucial
point.   I give an example of such a hypothesis in the statement of
Vitali's theorem (565F).   But in the applications of Vitali's theorem
later in this section, we can always work with a countable family of balls,
for which well-orderability is not an issue.   Separability and
second-countability hypotheses can be expected to act in similar ways;  so
that, for instance, we have 565Ya, which is a kind of primitive case of
Maharam's theorem.


}%end of notes

\discrpage


\frfilename{mt566.tex}
\versiondate{22.8.14}
\copyrightdate{2006}

\def\chaptername{Choice and determinacy}
\def\sectionname{Countable choice}

\def\spwidehat{\widehat{\phantom{m}}}

\newsection{566}

With AC($\omega$) measure theory becomes recognisable.   The
definition of Lebesgue measure used in Volume 1 gives us a true
countably additive Radon measure;  the most important divergence from the
standard theory is the possibility that every subset of $\Bbb R$ is
Lebesgue measurable\cmmnt{ (see 567G below)}.
With occasional exceptions (most
notably, in the theory of infinite products) we can use the work of Volume
2.   In Volume 3, we lose the two best theorems in the abstract
theory of measure algebras, Maharam's theorem and the Lifting Theorem;
but function spaces and ergodic theory are relatively unaffected.
Even in Volume 4, a good proportion of the ideas can be applied in some
form.

\cmmnt{\leader{566A}{} Nearly all
mathematicians working on the topics of this treatise
spend most of their time thinking in the framework of ZFC.   When we move
to weaker theories, we have a number of alternative strategies available.

\spheader 566Aa Some of the time, all we have to do is to check that our
previous arguments remain valid.   In the present context, moving from full
ZF + AC to ZF + AC($\omega$), this is true of
most of Volumes 1 and 2 and useful fragments thereafter.   In particular,
for most of the basic theory of the Lebesgue integral countable choice is
adequate.
Sometimes, of course, we have to trim our theorems back a bit\cmmnt{,
as in 566E, 566I, 566M, 566N, 566R and 566Xc}.

\spheader 566Ab Some results have to be dropped altogether.   For
instance, we no longer have a construction of a non-Lebesgue-measurable
subset of $\Bbb R$, and the Lifting Theorem disappears.

\spheader 566Ac Some results become so much weaker that they change their
character entirely.   For instance, the Hahn-Banach theorem, Baire's
theorem, Stone's theorem and Maharam's theorem survive only in sharply
restricted forms\cmmnt{ (561Xh, 561E, 561F, 566Nb)}.

\spheader 566Ad Sometimes we find that while proofs rely on the axiom of
choice, the results can be proved without it, or with something much
weaker.   Of course this is often a reason to regard the original
proof as inappropriate.
\cmmnt{Some of the ultrafilters in Volume 4 are there
just to save a couple of lines of argument, and renouncing them actually
brings ideas into clearer focus.}   But there are occasions when
the less scrupulous approach makes it a good deal easier for us to develop
appropriate intuitions.
\cmmnt{There is an example in the theory of the spaces
$S(\frak A)$ and $L^{\infty}(\frak A)$ in Chapter 36.   If we think of
$S(\frak A)$ as a quotient of a free linear space\cmmnt{ (361Ya)} and of
$L^{\infty}(\frak A)$ as the $\|\,\|_{\infty}$-completion of $S(\frak A)$,
we can prove all the basic results which come from their identification
with spaces of functions on the Stone space of $\frak A$;  but for most of
us such an approach would seriously complicate the process of understanding
the nature of the objects being constructed.   I used the representation
theorems in the theory of free products\cmmnt{ (\S315, \S325)} for
the same reason.}

On other occasions, we may need new ideas\cmmnt{,
as in 566F-566H, %566F 566G 566H
566L and 566P-566Q.   A deeper example is
in \cmmnt{562V/}566O\cmmnt{, where I set out
alternative routes to the results of 364F and 434T}.
Here we have quite a lot of extra distance to travel,
but at the same time we see some new territory}.

\spheader 566Ae More subtly, it may be useful to re-consider some
definitions\cmmnt{;  e.g., the distinction between `ccc' and `countable sup
property' for Boolean algebras\cmmnt{ (566Xd)}}.
I have made an effort in this book to use definitions which
will be appropriate in the absence of the axiom of choice, but in a number
of places this would lead to a division of a concept in potentially
confusing ways.

The ordinary theory of cardinals
depends so essentially on the existence of well-orderings that
it is often unclear what we can do without them.
However some theorems, which appear to involve the
theory of infinite cardinals, can be rescued if we re-interpret the
statements.
\cmmnt{Sometimes the cardinal $\frak c$ can be simply replaced by
$\Bbb R$ or $\Cal P\Bbb N$\cmmnt{ (343I, 491G)}.   Sometimes a statement
`$\#(X)\ge\frak c$' can be replaced by `there is
an injection from $\Cal P\Bbb N$ into $X$' or
`there is a surjection from $X$ onto
$\Cal P\Bbb N$'\cmmnt{ (344H, 4A2G(j-ii))};  %mt56bits
similarly, `$\#(X)\le\frak c$' might mean
`there is an injection from $X$ into $\Cal P\Bbb N$'
or `there is a surjection from $\Cal P\Bbb N$ onto
$X\cup\{\emptyset\}$'\cmmnt{ (4A1O, 4A3Fa)}.
Of course `$\#(X)=\frak c$' usually
becomes `there is a bijection between $X$ and
$\Cal P\Bbb N$'\cmmnt{ (423L)};  but
it might mean `there are an injection from $\Cal P\Bbb N$ into $X$ and a
surjection from $\Cal P\Bbb N$ onto $X$'\cmmnt{ (4A3Fb)},
or the other way round, or just two surjections.}

When dealing with a property which is invariant under
equipollence, it may be right to drop the concept of `cardinal' altogether,
and re-phrase a definition in more primitive terms\cmmnt{, as in 566Xl}.

Elsewhere\cmmnt{, as in 2A1Fd and 4A1E,}
we have results which refer to initial ordinals and hence to
well-orderable sets.   But the theory of cardinal functions is so bound up
with the idea that cardinal numbers form a well-ordered class that much
greater adjustments are necessary.
\cmmnt{I offer the following idea for
consideration.   For a metric space $(X,\rho)$ and a dense set
$D\subseteq X$, set

\Centerline{$\Cal U(X,\rho,D)=\{\{y:y\in X$, $\rho(x,y)<2^{-n}\}:
x\in D$, $n\in\Bbb N\}$,}

\noindent so that $\Cal U(X,\rho,D)$ is a base for the topology of $X$.
The existence (in ZF) of this function $\Cal U$
corresponds to the ZFC result that
`$w(X)$ is at most the cardinal product $\omega\times d(X)$ for every
metrizable space $X$'.}

\medskip

\spheader 566Af Another way to preserve the ideas of a theorem in the new
environment is to make some small variation in its hypotheses.
\cmmnt{For
instance, Urysohn's Lemma, in its usual form, demands DC.   So if we are
working with AC($\omega$) alone, we cannot be sure that compact Hausdorff
spaces are completely regular;  similarly, there may be uniformities
not definable from pseudometrics.   For a general topologist, this is
important.   But a measure theorist may be happy to simply add `completely
regular' to the hypotheses of a theorem\cmmnt{, as in 561G and 566Xk}.
In \S\S412-413 I repeatedly mention families $\Cal K$ which are closed
under disjoint finite unions.   Results starting from this hypothesis tend
to depend on DC;  but if we take $\Cal K$ to be closed under $\cup$,
AC($\omega$) may well be enough\cmmnt{ (566D)}.
A more dramatic change, but one which still leads to interesting results,
is in 566I.}
}%end of comment
%so assume all uniform sps, in particular all top gps, are c reg

\cmmnt{\leader{566B}{Volume 1} With countable choice,
Lebesgue outer measure becomes an outer measure in the usual sense, so we
can use \Caratheodory's method to define a measure space in the sense of
112A.   No further
difficulties arise in the work of Chapters 11 and 12, and we
can proceed exactly as before to the convergence theorems.
Indeed all the theorems of Volume 1 are available, with a single
exceptional feature:
the construction of non-measurable sets\cmmnt{ in 134B and 134D,} and a
non-measurable function\cmmnt{ in 134Ib}.   \cmmnt{(I will return
to this point in \S567.)}   In particular, the union of countably many
countable sets is countable.
}%end of comment

\cmmnt{\leader{566C}{Volume 2}
In Volume 2 also we find that arguments using more than countable choice
are the exception rather than the rule.   \cmmnt{Naturally,
they appear oftener in the more abstract topics of Chapter 21.}
One is in 211L;  we can no longer
be sure that a strictly localizable space is localizable, though a
$\sigma$-finite measure space does have to be localizable\cmmnt{,
since the choice demanded in the proof of 211Ld can then be performed
over a countable index set}.   There is a similar problem in 213J;
a strictly localizable space might fail to have locally determined
negligible sets, and might have a subset without a measurable
envelope.   Again\cmmnt{, in 214Ia}, it is not clear that a subspace of a
strictly localizable space must be strictly localizable.
\cmmnt{In 211P I ask for a non-Borel subset of
$\Bbb R$, and give an answer involving a non-measurable set;  but with
AC($\omega$) we have a non-Borel analytic set\cmmnt{ as in 423M}.
\cmmnt{(See also 566Xb.)}}

A more important gap arises in the theory of infinite products of
probability spaces.   The first problem is that if we have an
uncountable family $\familyiI{(X_i,\Sigma_i,\mu_i)}$ of probability spaces,
there is no assurance that $\prod_{i\in I}X_i$ is non-empty.
\cmmnt{In concrete
cases, this is not usually a serious worry.}   But there is another one.
\cmmnt{The proof of 254F makes an appeal to DC.}
I do not think that there can be
a construction of a product measure on even a sequence of arbitrary
probability spaces which does not use some form of dependent choice.
However a partial version, adequate for many purposes\cmmnt{ (including
the essential needs of Chapter 27)}, can be done with countable choice
alone\cmmnt{ (566I)}.
We can now continue through \S254 with the proviso that
every infinite family of probability spaces for which we consider a
product measure
should be a family of perfect probability spaces with non-empty
product.   There will be a difficulty in 254L, concerning the product of
subspaces of full outer measure\cmmnt{, where the modification
essentially confines it to non-empty products of conegligible sets}.
%so no 464C I think
%but maybe under AD, all semi-finite measure spaces perfect?
%495D might be fun to investigate, see 495G
\cmmnt{For 254N, it will be helpful to know that (under the conditions
of 566I)
the product of perfect spaces is again perfect.   \prooflet{The
proof of this fact
(451Jc) is scattered through Volume 4, but (given that we have a product
probability measure) needs only countably many choices at each step.}}

When we come to products of probability spaces in Chapter 27, we shall
again have to restrict the applications of the results, but at each point
only sufficiently to ensure that we have the product probability measures
discussed.
}%end of comment

\leader{566D}{\dvrocolon{Exhaustion}}\cmmnt{ The
versions of the principle of exhaustion in 215A all seem to require
DC rather than AC($\omega$).   For many applications, however, we can make
do with a weaker result, as follows.   I
include some corollaries showing that
in many familiar cases we can continue to use the intuitions developed in
the main text.

\medskip

\noindent}{\bf Proposition} [AC($\omega$)]
(a) Let $P$ be a partially ordered set such that
$p\vee q=\sup\{p,q\}$ is defined for all $p$, $q\in P$, and
$f:P\to\Bbb R$ an order-preserving function.    Then there is a
non-decreasing sequence $\sequencen{p_n}$ in $P$ such that
$\lim_{n\to\infty}f(p_n)=\sup_{p\in P}f(p)$.

(b) Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal E\subseteq\Sigma$ a non-empty set such that
$\sup_{E\in\Cal E}\mu E$ is
finite and $E\cup F\in\Cal E$ for every $E$, $F\in\Cal E$.
Then there is a non-decreasing
sequence $\sequencen{F_n}$ in $\Cal E$ such that, setting
$F=\bigcup_{n\in\Bbb N}F_n$, $\mu F=\sup_{E\in\Cal E}\mu E$ and
$E\setminus F$ is negligible for every $E\in\Cal E$.

(c) Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K$ a family of sets such that

\inset{($\alpha$) $K\cup K'\in\Cal K$ for all $K$, $K'\in\Cal K$,

($\beta$) whenever $E\in\Sigma$ is non-negligible there is a
non-negligible $K\in\Cal K\cap\Sigma$ such that $K\subseteq E$.}

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

(d)(i) Let $(X,\Sigma,\mu)$ be a semi-finite measure space.
Then $\mu$ is inner regular with respect to the family of sets of finite
measure.

\quad(ii) Let $(X,\Sigma,\mu)$ be a perfect measure space.   Then whenever
$E\in\Sigma$, $f:X\to\Bbb R$ is measurable and $\gamma<\mu E$, there is a
compact set $K\subseteq f[E]$ such that $\mu f^{-1}[K]\ge\gamma$.

\proof{{\bf (a)} For each $n\in\Bbb N$, set
$\gamma_n=\sup_{p\in P}\min(n,f(p)-2^{-n})$.   Then there is a sequence
$\sequencen{q_n}$ in $P$ such that $f(q_n)\ge\gamma_n$ for each $n$;  set
$p_n=\sup_{i\le n}q_i$ for each $n$.

\medskip

{\bf (b)} By (a) there is a non-decreasing
sequence $\sequencen{F_n}$ in $\Cal E$ such that
$\sup_{n\in\Bbb N}\mu F_n=\sup_{E\in\Cal E}\mu E$;  set
$F=\bigcup_{n\in\Bbb N}F_n$.

\medskip

{\bf (c)} Because $\mu$ is inner regular with respect to $\Cal K$ iff it is
inner regular with respect to $\Cal K\cup\{\emptyset\}$, we may suppose
that $\emptyset\in\Cal K$.   Take $F\in\Sigma$, and consider
$\Cal E=\{K:K\in\Cal K\cap\Sigma$, $K\subseteq F\}$.   \Quer\ If
$\sup_{E\in\Cal E}\mu E<\mu F$, let $\sequencen{E_n}$ be a non-decreasing
sequence in $\Cal E$ such that $\mu(E\setminus\bigcup_{n\in\Bbb N}E_n)=0$
for every $E\in\Cal E$ ((b) above).
Set $G=\bigcup_{n\in\Bbb N}E_n$;  then
$\mu G=\sup_{n\in\Bbb N}\mu E_n<\mu F$, so $\mu(F\setminus G)>0$.
But now there ought to be a non-negligible $K\in\Cal K\cap\Sigma$ such that
$K\subseteq F\setminus G$, in which case $K\in\Cal E$ and
$\mu(K\setminus G)>0$.\ \Bang

\medskip

{\bf (d)(i)} Apply (c) with $\Cal K$ the family of sets of finite measure.

\medskip

\quad{\bf (ii)} Apply (c) to the subspace measure $\mu_E$ and
$\Cal K=\{f^{-1}[K]:K\subseteq f[E]$ is compact$\}$.
}%end of proof of 566D

\leader{566E}{}\cmmnt{ The problem recurs in parts of 215B, where
I list characterizations of $\sigma$-finiteness, and in 215C.   It seems
equally
that a ccc semi-finite measure algebra may fail to be $\sigma$-finite,
though a $\sigma$-finite measure algebra has to be ccc.
%322G
We have a stripped-down version of 215B, with one of its fragments used
in \S235, as follows:

\medskip

\noindent}{\bf Proposition} [AC($\omega$)]
Let $(X,\Sigma,\mu)$ be a semi-finite
measure space.   Write $\Cal N$ for the $\sigma$-ideal
of $\mu$-negligible sets.

(a) The following are equiveridical:

\quad(i) $\mu$ is $\sigma$-finite;

\quad(ii) either $\mu X=0$ or there is a probability measure
$\nu$ on $X$ with the same domain and the same negligible sets as $\mu$;

\quad(iii) there is a measurable integrable function $f:X\to\ocint{0,1}$;

\quad(iv) either $\mu X=0$ or there is a measurable function
$f:X\to\ooint{0,\infty}$ such that $\int fd\mu=1$.

(b) If $\mu$ is $\sigma$-finite, then

\quad(i) every disjoint family in $\Sigma\setminus\Cal N$ is countable;

\quad(ii) for every $\Cal E\subseteq\Sigma$ there is a countable
$\Cal E_0\subseteq\Cal E$ such that $E\setminus\bigcup\Cal E_0$ is
negligible for every $E\in\Cal E$.

(c) Suppose that $\mu$ is $\sigma$-finite, $(Y,\Tau,\nu)$ is a
semi-finite measure space, and $\phi:X\to Y$ is a
$(\Sigma,\Tau)$-measurable function such that $\mu\phi^{-1}[F]>0$ whenever
$\nu F>0$.   Then $\nu$ is $\sigma$-finite.

\proof{{\bf (a)} Use the methods of 215B.

\medskip

{\bf (b)} By (a-ii), we may suppose that $\mu$ is totally finite.

\medskip

\quad{\bf (i)} If $\Cal E\subseteq\Sigma\setminus\Cal N$ is disjoint, then
$\Cal E_n=\{E:E\in\Cal E$, $\mu E\ge 2^{-n}\}$ is finite for every $n$, so
$\Cal E=\bigcup_{n\in\Bbb N}\Cal E_n$ is countable.

\medskip

\quad{\bf (ii)} Let $\Cal H$ be the set of finite unions of members of
$\Cal E$.   By 566Db, there is a sequence $\sequencen{H_n}$ in $\Cal H$
such that $\mu(H\setminus\bigcup_{n\in\Bbb N}H_n)=0$ for every
$H\in\Cal H$.   For each $n\in\Bbb N$, choose a finite set
$\Cal H_n\subseteq\Cal E$ such that $H_n=\bigcup\Cal H_n$;  then
$\Cal E_0=\bigcup_{n\in\Bbb N}\Cal H_n$ has the required properties.

\medskip

{\bf (c)} Again, we may suppose that $\mu$ is totally finite.
For each $m\in\Bbb N$ let $\Cal H_m$ be the set of those $F\in\Tau$ such
that $\nu F<\infty$ and $\mu\phi^{-1}[F]\ge 2^{-m}$.   Then any disjoint
family in $\Cal H_m$ has at most $\lfloor 2^m\mu X\rfloor$ members,
so each $\Cal H_m$ has a maximal disjoint subset;  choose a sequence
$\sequence{m}{\Cal E_m}$ such that $\Cal E_m$ is a maximal disjoint subset
of $\Cal H_m$ for each $m$.   Then $\Cal E=\bigcup_{m\in\Bbb N}\Cal E_m$ is
a countable family of sets of finite measure in $Y$.   Now
$Z=Y\setminus\bigcup\Cal E$ is negligible.   \Prf\Quer\ Otherwise, there is a
non-negligible set $F$ of finite measure disjoint from $\bigcup\Cal E$;
now there is an $m$ such that $F\in\Cal H_m$, so that $\Cal E_m$ was not
maximal.\ \Bang\QeD\   So $\Cal E\cup\{Z\}$ witnesses that $\nu$ is
$\sigma$-finite.
}%end of proof of 566E

\leader{566F}{Atomless \dvrocolon{algebras}}\cmmnt{ To
make atomless measure spaces and
measure algebras recognisable, we need a more penetrating
argument than that previously used in 215D.

\medskip

\noindent}{\bf Lemma}\dvArevised{2014} [AC($\omega$)]
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra, and
$\mu$ a positive countably additive functional on $\frak A$ such that
$\mu 1=1$.   Suppose that whenever $a\in\frak A$ and $\mu a>0$ there is a
$b\Bsubseteq a$ such that $0<\mu b<\mu a$.
Then there is a function $f:\frak A\times[0,1]\to\frak A$ such that
$f(a,\alpha)\Bsubseteq a$ and $\bar\mu f(a,\alpha)=\min(\alpha,\bar\mu a)$
for $a\in\frak A$ and $\alpha\in[0,1]$, and
$\alpha\mapsto f(a,\alpha)$ is non-decreasing for every $a\in\frak A$.

\proof{{\bf (a)} Just as in part (a) of the proof of
215D, we see by induction on $n$
that for every $b\in\frak A$ such that $\mu b>0$ and every
$n\in\Bbb N$, there is a
$c\Bsubseteq b$ such that $0<\mu c\le 2^{-n}\mu b$.

\medskip

{\bf (b)} If $b\in\frak A$ and $\mu b>0$,
there is a $c\subseteq b$ such that
$\bover13\mu b<\mu c\le\bover23\mu b$.  \Prf\Quer\ Otherwise,
set $\gamma=\sup\{\mu c:c\subseteq b$, $\mu c\le\bover23\mu b\}$
and let
$\sequencen{c_n}$ be a sequence in $\frak A$ such that
$c_n\subseteq b$ and $\gamma-2^{-n}\le\mu c_n\le\gamma$ for every $n$.
Set $d_n=\sup_{i\le n}c_i$ for each $n$, and
$d=\sup_{n\in\Bbb N}d_n$.
Inducing on $n$, we see that $\mu d_n\le\bover23\mu b$ so
$\mu d_n\le\bover13\mu b$ for each $n$, and $\mu d\le\bover13\mu b$.
Now by (a) there is an $e\subseteq b\setminus d$ such that
$0<\mu e\le\bover13\mu b$.   In this case,
$\mu(d\cup e)\le\bover23\mu b$, so

\Centerline{$\gamma
\ge\mu(d\cup e)\ge\mu e+\sup_{n\in\Bbb N}\mu c_n>\gamma$.
\Bang\Qed}

\medskip

{\bf (c)} For each $n\in\Bbb N$
there is a finite partition of unity into
elements of measure at most $(\bover23)^n$.
\Prf\ Induce on $n$, using (b) for the inductive step.\ \Qed

\medskip

{\bf (d)} Choose a sequence $\sequence{k}{C_k}$ of finite partitions of
unity such that $\mu c\le 2^{-k}$ for every
$k\in\Bbb N$ and $c\in C_n$.   Set $C=\bigcup_{k\in\Bbb N}C_k$;  then
$C$ is countable.   Moreover, whenever $a\in\frak A$ and
$\beta>0$, there must be a $c\in C$ such that
$a\Bcap c\ne 0$ and $\mu c\le\beta$.
Let $\sequencen{c_n}$ be a sequence running over $C$.

\medskip

{\bf (e)} Define
$\sequencen{f_n}$, $\sequencen{g_n}$ inductively by saying that,
for $a\in\frak A$ and $\alpha\in[0,1]$,

\Centerline{$f_0(a,\alpha)=0$, \quad$g_0(a,\alpha)=a$}

$$\eqalign{f_{n+1}(a,\alpha)
&=f_n(a,\alpha)\Bcup(c_n\Bcap g_n(a,\alpha))
   \text{ if }\mu(f_n(a,\alpha)\Bcup(c_n\Bcap g_n(a,\alpha)))\le\alpha,\cr
&=f_n(a,\alpha)\text{ otherwise},\cr
g_{n+1}(a,\alpha)
&=g_n(a,\alpha)
   \text{ if }\mu(f_n(a,\alpha)\Bcup(c_n\Bcap g_n(a,\alpha)))\le\alpha,\cr
&=f_n(a,\alpha)\Bcup(c_n\Bcap g_n(a,\alpha))
   \text{ otherwise}.\cr}$$

\noindent Then

\Centerline{$f_n(a,\alpha)\Bsubseteq f_{n+1}(a,\alpha)
\Bsubseteq g_{n+1}(a,\alpha)\Bsubseteq g_n(a,\alpha)\Bsubseteq a$,}

\Centerline{$\mu f_n(a,\alpha)\le\alpha$,
\quad$\mu g_n(a,\alpha)\ge\min(\mu a,\alpha)$}

\noindent for every $n\in\Bbb N$.   Set
$f(a,\alpha)=\sup_{n\in\Bbb N}f_n(a,\alpha)$.   Then
$f(a,\alpha)\Bsubseteq a$ and $\mu f(a,\alpha)\le\alpha$ whenever
$a\in\frak A$ and $\alpha\in[0,1]$.

\medskip

{\bf (f)(i)} \Quer\ If $a\in\frak A$ and $\alpha\in[0,1]$ are such that
$\mu f(a,\alpha)<\min(\mu a,\alpha)$, set
$b=\inf_{n\in\Bbb N}g_n(a,\alpha)$.   Then $f(a,\alpha)\Bsubseteq b$ and
$\mu b\ge\min(\mu a,\alpha)$.   By (d), there is an $n\in\Bbb N$ such that

\Centerline{$c_n\Bcap b\Bsetminus f(a,\alpha)\ne 0$,
\quad$\mu c_n\le\min(\mu a,\alpha)-\mu f(a,\alpha)$.}

\noindent In this case,
$\mu(f_n(a,\alpha)\Bcup(c_n\Bcap g_n(a,\alpha)))\le\min(\mu a,\alpha)$
so

\Centerline{$f(a,\alpha)\Bsupseteq f_{n+1}(a,\alpha)
\Bsupseteq c_n\Bcap g_n(a,\alpha))
\Bsupseteq c_n\Bcap b$,}

\noindent which is impossible.\ \Bang

So $\mu f(a,\alpha)=\min(\mu a,\alpha)$ for all $a\in\frak A$ and
$\alpha\in[0,1]$.

\medskip

\quad{\bf (ii)} If $a\in\frak A$ and $0\le\alpha\le\beta\le 1$, then for
every $n\in\Bbb N$

\inset{{\it either} $f_n(a,\alpha)=f_n(a,\beta)$ and
$g_n(a,\alpha)=g_n(a,\beta)$

{\it or} $g_n(a,\alpha)\Bsubseteq f_n(a,\beta)$.}

\noindent(Induce on $n$.)   So $f(a,\alpha)\Bsubseteq f(a,\beta)$.

Thus we have a suitable function $f$.
}%end of proof of 566F

%216E looks problematic

\cmmnt{
\leader{566G}{Vitali's theorem}\cmmnt{ The arguments I presented
for Vitali's theorem
in 221A/261B and 471N-471O, and for the similar result in 472B, involve
the inductive construction of a sequence, which
ordinarily is a signal that DC is being used.   In 565F I suggested a
weaker form of Vitali's theorem which is adequate for its most important
applications in measure theory.}   With AC($\omega$)\cmmnt{, however,} we
can get most of the results as previously stated\cmmnt{, if
we refine our methods slightly}.

\cmmnt{
\spheader 566Ga In 261B, we have a family $\Cal I$ of closed balls
in $\BbbR^r$ and
we wish to choose inductively a disjoint sequence $\sequencen{I_n}$ in
$\Cal I$ such that

\Centerline{$\diam I_n
\ge\Bover12\sup\{\diam I:I\in\Cal I$, $I\cap\bigcup_{i<n}I_i=\emptyset\}$}

\noindent for every $n$.   We have already reduced the problem to the case
in which
$\sup_{I\in\Cal I}\diam I$ is finite and for any finite disjoint
subset of $\Cal I$ there is a member of $\Cal I$ disjoint from all of them.
Let $\sequence{m}{G_m}$ run over the family of all open balls with
centres in $\BbbQ^r$ and rational radii.   For $m\in\Bbb N$ set
$\Cal K_m=\Cal I\cap\Cal PG_m$, and let $\Cal I'\subseteq\Cal I$ be a
countable set such that
$\sup_{I\in\Cal I'\cap\Cal K_m}\diam I=\sup_{I\in\Cal K_m}\diam I$ for
every $m\in\Bbb N$ such that $\Cal K_m$ is non-empty;  this can be found
with countably many choices.

Now, when we come to choose $I_n$, we can always pick a member of
$\Cal I'$.   \Prf\ If
$\Cal I_n=\{I:I\in\Cal I$, $I\cap\bigcup_{i<n}I_i=\emptyset\}$,
$\gamma_n=\sup_{I\in\Cal I_n}\diam I$
and $I\in\Cal I_n$ is such that $\diam I>\bover12\gamma_n$, there is an
$m\in\Bbb N$ such that
$I\subseteq G_m\subseteq\BbbR^r\setminus\bigcup_{i<n}I_i$, in which case
there is an $I'\in\Cal I'\cap\Cal K_m$ such that
$\diam I'\ge\bover12\gamma_n$, and $I'$ is eligible to be $I_n$.\
\QeD\  Because $\Cal I'$ is well-orderable, we can set out a rule
for making these choices, and the argument can proceed as written, without
recourse to the devices of \S565.

\spheader 566Gb A similar trick can be used in 472B.   Here, given a
family $\Cal I$ of closed balls, we wish to choose a sequence
$\sequencen{B_n}$ in $\Cal I$ such that the centre of $B_n$ does not belong
to $\bigcup_{i<n}B_i$ and, subject to this, the diameter of $B_n$ is
nearly as large as it
could be.   This time, take $\Cal K_m$ to be the set of members of $\Cal I$
with centres in $G_m$, use countably many choices to find a countable set
$\Cal I'\subseteq\Cal I$ with adequately large intersections with every
$\Cal K_m$, and choose $\sequencen{B_n}$ from $\Cal I'$.

At the next step, in 472C, we have to do this repeatedly, but the same
method works;  in fact, we can work inside a fixed family $\Cal I'$
chosen as above.   (See 472Yd.)

\spheader 566Gc The version in 471N-471O is not manageable in quite the
same way.   If, however, we assume that the metric spaces there are
locally compact and separable, we can use the same idea as in (a) above
to limit our search to countable subfamilies of the given family $\Cal F$.
}%end of comment
}%end of comment

\leader{566H}{Bounded additive \dvrocolon{functionals}}\cmmnt{ We
come to another obstacle in the proof of 231E.   The
argument given there
relies on DC to show that a countably additive functional
is bounded.   But we can avoid this, at the cost of an
extra manoeuvre, as follows.

\medskip

\noindent}{\bf Lemma} [AC($\omega$)]
Let $\frak A$ be a Boolean algebra and $\nu:\frak A\to\Bbb R$ an additive
functional such that $\{\nu a_n:n\in\Bbb N\}$ is bounded
for every disjoint
sequence $\sequencen{a_n}$ in $\frak A$.   Then $\nu$ is bounded.

\proof{ \Quer\ Suppose, if possible, otherwise.   Then there is a
sequence $\sequencen{b_n}$ in $\Sigma$ such that $|\nu b_n|\ge 2^nn$ for
every $n\in\Bbb N$.   For each $n\in\Bbb N$, let $\frak B_n$ be the
subalgebra of $\frak A$
generated by $\{b_i:i<n\}$;  then $\frak B_n$ has
at most $2^n$ atoms, so there must be an atom $a$ of $\frak B_n$ such that
$|\nu(a\Bcap b_n)|\ge n$.   Choose a sequence $\sequencen{c_n}$ such that
$c_n$ is an atom of $\frak B_n$ and $|\nu d_n|\ge n$ for every $n$, where
$d_n=c_n\Bcap b_n$;  note that $d_n$ is an atom of $\frak B_{n+1}$,
so that if $n<m$ then
either $d_m\Bsubseteq d_n$ or $d_m\Bcap d_n=0$.   By Ramsey's
theorem (4A1G), there is an infinite $I\subseteq\Bbb N$ such that

\inset{{\it either} $\family{n}{I}{d_n}$ is disjoint

{\it or} $d_m\subseteq d_n$ whenever $m$, $n\in I$ and
$n<m$.}

\noindent Now the first alternative is certainly impossible, because
$\{\nu d_n:n\in I\}$ is unbounded.
So we have the second.   But in this case we can define a strictly
increasing sequence $\sequence{k}{n_k}$ in $I$ such that
$n_{k+1}\ge k+|\nu d_{n_k}|$ for each $k$.   Set
$a_k=d_{n_k}\Bsetminus d_{n_{k+1}}$ for each $k$;  then $\sequence{k}{a_k}$
is disjoint and $|\nu a_k|\ge k$ for each $k$, so again we have a
contradiction.\ \Bang
}%end of proof of 566H

\leader{566I}{Infinite products:  Theorem} [AC($\omega$)] Let
$\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of perfect probability
spaces such that $X=\prod_{i\in I}X_i$ is non-empty.
Then there is a complete probability measure $\lambda$ on $X$ such that

(i) if $E_i\in\Sigma_i$ for every $i\in I$, and $\{i:E_i\ne X_i\}$ is
countable, then $\lambda(\prod_{i\in I}E_i)$ is defined and equal to
$\prod_{i\in I}\mu_iE_i$;

(ii) $\lambda$ is inner regular with respect to
$\Tensorhat_{i\in I}\Sigma_i$.

\proof{ The only point at which the construction in 254A-254F needs
re-examination is in the proof that the standard outer measure on $X$ gives
it outer measure $1$.

\medskip

{\bf (a)} I recall the definitions.   For a cylinder
$C=\prod_{i\in I}C_i$, set $\theta_0C=\prod_{i\in I}\mu_iC_i$;  for
$A\subseteq X$, set

\Centerline{$\theta A
=\inf\{\sum_{n=0}^{\infty}\theta_0C_n:C_n\in\Cal C
  \text{ for every }n\in\Bbb N$, $A\subseteq\bigcup_{n\in\Bbb N}C_n\}$;}

\noindent $\lambda$ will be the measure defined from $\theta$ by
\Caratheodory's method.

\Quer\ Suppose, if possible, that $\theta X<1$.
Then we have a sequence
$\sequencen{C_n}$ of cylinder sets, covering $X$, with
$\sum_{n=0}^{\infty}\theta C_n=1-2\epsilon$ where $\epsilon>0$.
Express each $C_n$ as
$\prod_{i\in I}E_{ni}$ where
$J_n=\{i:E_{ni}\ne X_i\}$ is finite;  let $J$ be the countable set
$\bigcup_{n\in\Bbb N}J_n$;
take $K=\#(J)$ (identifying $\Bbb N$ with $\omega$),
and a bijection $k\mapsto i_k:K\to J$.

For each $k\in K$ and $n\in\Bbb N$, set $L_k=\{i_j:j<k\}\subseteq J$ and
$\alpha_{nk}=\prod_{i\in I\setminus L_k}\mu_iE_{ni}$.
If $J$ is finite, $L_{\#(J)}=J$ and $\alpha_{n,\#(J)}=1$ for every
$n$.   We have $\alpha_{n0}=\theta_0C_n$ for each $n$, so
$\sum_{n=0}^{\infty}\alpha_{n0}=1-2\epsilon$.   For $n\in\Bbb N$,
$k\in K$ and $t\in X_{i_k}$ set

$$\eqalign{f_{nk}(t)&=\alpha_{n,k+1}\text{ if }t\in E_{n,i_k},\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then

\Centerline{$\int f_{nk}d\mu_{i_k}
=\alpha_{n,k+1}\mu_{i_k}E_{ni_k}=\alpha_{nk}$.}

\medskip

{\bf (b)} For each $k\in K$, let $h_k:X_{i_k}\to\Bbb R$ be the Marczewski
functional defined by setting

\Centerline{$h_k(t)=\sum_{n=0}^{\infty}3^{-n}\chi E_{ni_k}(t)$}

\allowmorestretch{468}{
\noindent for $t\in X_k$.   Because $\mu_k$ is perfect, there is
for each $k\in K$ a compact set $Q\subseteq h_k[X_{i_k}]$
such that $\mu_{i_k}h_k^{-1}[Q]\ge 1-2^{-k}\epsilon$.
Choose $\family{k}{K}{Q_k}$ such that $Q_k\subseteq h_k[X_{i_k}]$ is
compact and
$\mu_{i_k}h_k^{-1}[Q_k]\ge 1-2^{-k}\epsilon$ for every $k\in K$.
Observe that if $k\in K$ and $n\in\Bbb N$ then
$f_{nk}=\alpha_{nk}\chi E_{ni_k}$ is of the form $\alpha_{nk}g_nh_k$ where
$g_n:\Bbb R\to[0,1]$ is continuous.
}

\medskip

{\bf (c)} Define non-empty sets
$F_k\subseteq H_k\subseteq X_{i_k}$ inductively, for $k\in K$,
as follows.   The inductive hypothesis will be that
$\sum_{n\in M_k}\alpha_{nk}\le 1-2^{-k+1}\epsilon$, where
$M_k=\{n:n\in\Bbb N$, $F_j\subseteq E_{ni_j}$ whenever $j<k\}$;
of course $M_0=\Bbb N$, so the induction starts.
Given that $k\in K$ and that

\Centerline{$1-2^{-k+1}\epsilon
\ge\sum_{n\in M_k}\alpha_{nk}
=\sum_{n\in M_k}\int f_{nk}d\mu_{i_k}
=\int(\sum_{n\in M_k}f_{nk})d\mu_{i_k}$,}

\noindent the set

\Centerline{$H_k=\{t:t\in X_{i_k}$,
  $\sum_{n\in M_k}f_{nk}(t)\le 1-2^{-k}\epsilon\}$}

\allowmorestretch{468}{
\noindent must have measure greater than $2^{-k}\epsilon$ and meets
$h_k^{-1}[Q_k]$.   But observe
that $\sum_{n\in M_k}f_{nk}=g'_kh_k$ where
$g'_k=\penalty-100\sum_{n\in M_k}\alpha_{nk}g_n$ is lower semi-continuous, so that
$H_k=h_k^{-1}[G_k]$ where
$G_k=\{\alpha:g'_k(\alpha)\le 1-2^{-k}\epsilon\}$ is closed.
Since $H_k$ meets $h_k^{-1}[Q_k]$, $Q_k\cap G_k$ is non-empty and has a
least member $\beta_k$;  set $F_k=h_k^{-1}[\{\beta_k\}]$.
Because $Q_k\subseteq h_k[X_{i_k}]$, $F_k$ is non-empty.
}

Examine

\Centerline{$M_{k+1}=\{n:n\in M_k$, $F_k\subseteq E_{ni_k}\}$.}

\noindent There certainly is some $t^*\in F_k$, and because $h_k\restr F_k$
is constant, $M_{k+1}=\{n:n\in M_k$, $t^*\in E_{ni_k}\}$.   In this case

\Centerline{$\sum_{n\in M_{k+1}}\alpha_{n,k+1}
=\sum_{n\in M_k}f_{nk}(t^*)\le 1-2^{-k}\epsilon$}

\noindent and the induction proceeds.

\medskip

{\bf (d)} At the end of the induction, either finite or infinite,
choose $t_k\in F_k$ for $k\in K$.   We are supposing that $X$ has a member
$x^*$;  define $x\in X$ by setting $x(i_k)=t_k$ for $k\in K$ and
$x(i)=x^*(i)$ for $i\in I\setminus J$.   Then there is supposed to be an
$m\in\Bbb N$ such that $x\in C_m$, so that $m\in M_k$ for every $k$.
But at some stage we shall have
$J_m\subseteq L_k$ (allowing $k=\#(K)$ if $K$ is finite) and
$\alpha_{mk}=1$, which is impossible.\ \Bang
}%end of proof of 566I

\leader{566J}{}\cmmnt{ In
particular, 566I applies to all products $\{0,1\}^I$ and
$[0,1]^I$ with their usual measures.   For these we have Kakutani's theorem
that the usual measures are topological measures (415E),
which turns out to be valid with countable choice alone.

\medskip

\noindent}{\bf Theorem} [AC($\omega$)]
(a) Let $\familyiI{(X_i,\frak T_i,\Sigma_i,\mu_i)}$ be a
family of metrizable Radon probability spaces such that every $\mu_i$ is
strictly positive and $X=\prod_{i\in I}X_i$ is non-empty.
Then the product measure on $X$ is a quasi-Radon measure.

(b) If $I$ is well-orderable then the product measure on $\{0,1\}^I$ is a
completion regular Radon measure.

\proof{{\bf (a)(i)} We had better check immediately that every $X_i$ is
separable.   The point is that because $\mu_i$ is a
totally finite measure inner regular with respect to the compact sets,
there is a conegligible K$_{\sigma}$ set;  because $\mu_i$ is strictly
positive, this is dense;  and countable choice is enough to ensure that a
compact metrizable space is second-countable, therefore separable.
It follows that $\prod_{i\in J}X_i$ is separable, therefore
second-countable, for every countable $J\subseteq I$.

\medskip

\quad{\bf (ii)} Because every $\mu_i$ is a Radon measure it is perfect,
so we
have a product probability measure on $X$.   Now we can repeat the argument
of 416Ua.

\medskip

{\bf (b)} Put (a), 561D and G together.
}%end of proof of 566J

\cmmnt{\leader{566K}{Volume 3} Turning to the concerns of Volume 3,
the elementary theory
of measure algebras is not radically changed.   But Lemma 311D is
hopelessly lost;  we no longer have Stone
spaces, and need to re-examine any proof which appears to rely on them.
Another result which changes is 313K;  order-dense sets\cmmnt{, as
defined in
313J,} need no longer give rise to partitions of unity.   So a localizable
measure algebra does not need to be isomorphic to a simple product of
totally finite measure algebras.   Similarly,
condition (ii) of 316H is no longer sufficient to prove weak
$(\sigma,\infty)$-distributivity.   However some of the
constructions which I described in terms of Stone spaces, in particular,
the Loomis-Sikorski theorem, the
Dedekind completion of a Boolean algebra, the localization of a
semi-finite measure algebra, free products and measure-algebra free
products, can be done by other methods which remain effective with
AC($\omega$) at most\cmmnt{;  see 566L, 561Yg, 323Xh and 325Yc}.

The theory of ccc algebras is rather different\cmmnt{ (566M, 566Xd)}.
Maharam's theorem\cmmnt{ (331I, 332B)} is surely unprovable
without something like the full axiom of choice;  and the Lifting
Theorem\cmmnt{ (341K)} is
equally inaccessible under the rules of this section.    We do
however have useful special cases of results in Chapters 33 and
34\cmmnt{ (566N)}.
%any hope for 332Q?

A good start can be made on the elementary
theory of Riesz spaces without any form of the axiom of
choice\cmmnt{ (see 561H)}, and with
AC($\omega$) we can go a long way\cmmnt{, as in 566Q}.   What is
missing is the Hahn-Banach theorem (for non-separable spaces)
and many representation theorems.   Similarly, the function spaces of
Chapter 36 are recognisable, provided that (for general Boolean algebras
$\frak A$)
we think of $S(\frak A)$ as a quotient space of the free linear space
generated by $\frak A$, and of
$L^{\infty}(\frak A)$ as the $\|\,\|_{\infty}$-completion of
$S(\frak A)$.   \cmmnt{Of
course we have to take care at every point to avoid the
use of Stone spaces.}   \cmmnt{One
place at which this involves us in a new
argument is in 566O.}   Most of the arguments of Chapter 24 remain valid,
so the basic theory of $L^p$ spaces in \S\S365-366 survives.
What is perhaps
surprising is that if we take the trouble we can still reach the most
important results on weak compactness (566P, 566Q).

In the ergodic theory of Chapter 38, a good proportion of the classical
results survive.   There are difficulties with some of the extensions
of the classical theory in \S\S381-382.   For instance, the definition of
`full subgroup' of the group of automorphisms of a Boolean
algebra\cmmnt{ in 381Be} assumes that order-dense sets include
partitions of unity.   \cmmnt{If not,
this definition may fail to be equivalent to
the formulation in 381Ia.   The latter would seem to be the more natural
one to use.}   However, the definition as given seems to work for the
principal needs of Chapter 38\cmmnt{ (see 381I)}.

Frol\'\i k's theorem\cmmnt{ in the generality 382D-382E} needs
something approaching AC, and with AC($\omega$) alone
there seems no hope of getting results for
general Dedekind complete algebras along the lines of the main theorems of
\S382.   For measurable algebras, however, we do have a version of
382Eb\cmmnt{ (566R)}.

Many of the later results of Chapter 38 are equally robust,
at least in their leading applications to measure algebras.   We
have to remember that we do not know that measurable algebras have many
involutions\cmmnt{, and
even among those which do there is no assurance that 382Q
will be true}.   So in \S\S383-384 we find ourselves restricted rather
further, to those measurable algebras in which every
non-zero element is the support of an involution;  but these include the
standard examples (566N).
}%end of comment

\leader{566L}{The Loomis-Sikorski theorem} [AC($\omega$)] (a) Let $\frak A$
be a Dedekind $\sigma$-complete Boolean algebra.   Then there are a set
$X$, a $\sigma$-algebra $\Sigma$ of subsets of $X$ and a $\sigma$-ideal
$\Cal I$ of $\Sigma$ such that $\frak A\cong\Sigma/\Cal I$.

(b) Let $(\frak A,\bar\mu)$ be a measure algebra.   Then it is isomorphic
to the measure algebra of a measure space.

\proof{{\bf (a)(i)} Set $X=\{0,1\}^{\frak A}$, and
$\Sigma=\Tensorhat_{\frak A}\Cal P(\{0,1\})$.   For $a\in\frak A$ set
$\widehat{a}=\{x:x\in X$, $x(a)=1\}\in\Sigma$.   Let $\Cal I$ be the
$\sigma$-ideal of $\Sigma$ generated by sets of the form

\Centerline{$\widehat{a\Bsymmdiff b}\symmdiff\widehat{a}
  \symmdiff\widehat{b}$,
\quad$(\inf_{n\in\Bbb N}a_n)\spwidehat
\symmdiff\bigcap_{n\in\Bbb N}\widehat{a_n}$}

\noindent for $a$, $b\in\frak A$ and sequences $\sequencen{a_n}$ in
$\frak A$, together with the set $\{x:x(1)=0\}$.

\medskip

\quad{\bf (ii)} (The key.) $\widehat{a}\notin\Cal I$ for any
$a\in\frak A\setminus\{0\}$.   \Prf\ If $E\in\Cal I$ then (using
AC($\omega$)) we can find sequences $\sequencen{a_n}$, $\sequencen{b_n}$ in
$\frak A$, together with a double sequence
$\langle c_{ni}\rangle_{n,i\in\Bbb N}$, such that, setting
$c_n=\inf_{i\in\Bbb N}c_{ni}$ for each $n$,

\Centerline{$F
=\{x:x(1)=0\}
 \cup\bigcup_{n\in\Bbb N}\bigl(\widehat{a_n\Bsymmdiff b_n}
   \symmdiff\widehat{a_n}\symmdiff\widehat{b_n}\bigr)
 \cup\bigcup_{n\in\Bbb N}\bigl(\widehat{c_n}
   \symmdiff\bigcap_{i\in\Bbb N}\widehat{c_{ni}}\bigr)$}

\noindent includes $E$.   Let $\frak B$ be the subalgebra of $\frak A$
generated by

\Centerline{$\{a\}\cup\{a_n:n\in\Bbb N\}\cup\{b_n:n\in\Bbb N\}
  \cup\{c_n:n\in\Bbb N\}\cup\{c_{ni}:n,i\in\Bbb N\}$.}

\noindent Then $\frak B$ is countable, so we can choose inductively a
sequence $\sequencen{d_n}$ in $\frak B\setminus\{0\}$ such that $d_0=a$
and, for each $n\in\Bbb N$,

\inset{$d_{n+1}\Bsubseteq d_n$,

either $d_{n+1}\Bsubseteq a_n$ or $d_{n+1}\Bcap a_n=0$,

either $d_{n+1}\Bsubseteq b_n$ or $d_{n+1}\Bcap b_n=0$,

either $d_{n+1}\Bsubseteq c_n$ or there is an $i\in\Bbb N$ such that
$d_{n+1}\Bcap c_{ni}=0$.}

\noindent Define $x\in X$ by saying that

$$\eqalign{x(d)
&=1\text{ if }d\Bsupseteq d_n\text{ for some }n\in\Bbb N,\cr
&=0\text{ otherwise.}\cr}$$

\noindent Then $x\in\widehat{a}\setminus F$ and
$\widehat{a}\not\subseteq E$;  as $E$ is arbitrary,
$\widehat{a}\notin\Cal I$.\ \Qed

\medskip

\quad{\bf (iii)} Set

\Centerline{$\Sigma_0=\{E:E\in\Sigma$, $E\symmdiff\widehat{a}\in\Cal I$ for
some $a\in\frak A\}$.}

\noindent Then $\Sigma_0$ is closed under symmetric difference and
countable intersections and contains $X$ (because
$X\symmdiff\widehat{1}\in\Cal I$).   So $\Sigma_0$ is a $\sigma$-algebra of
sets;  as it contains $\widehat{a}$ for every $a\in\frak A$, it is equal to
$\Sigma$.

\medskip

\quad{\bf (iv)} From (ii) we see that
$\widehat{a\Bsymmdiff b}$, and therefore $\widehat{a}\symmdiff\widehat{b}$,
do not belong to $\Cal I$ for any distinct $a$, $b\in\frak A$.   With
(iii), this tells us that we have a function $\pi:\Sigma\to\frak A$ defined
by setting $\pi E=a$ whenever $E\symmdiff\widehat{a}\in\Cal I$.   Now
$\pi X=1$ and $\pi$ preserves symmetric difference and countable infima, so
is a sequentially order-continuous Boolean homomorphism;  its kernel is
$\Cal I$, so $\frak A\cong\Sigma/\Cal I$, as required.

\medskip

{\bf (b)} This is now easy;  we can use the familiar argument of 321J.
}%end of proof of 566L

\leader{566M}{Measure algebras:  Proposition} [AC($\omega$)]
(a) Let $\frak A$ be a measurable algebra.

\quad(i) For any $A\subseteq\frak A$ there is a countable $B\subseteq A$
with the same upper bounds as $A$.

\quad(ii) $\frak A$ is Dedekind complete.

\quad(iii) If $D\subseteq\frak A$ is order-dense and $c\in D$ whenever
$c\Bsubseteq d\in D$, there is a partition of unity included in $D$.

(b) Let $(\frak A,\bar\mu)$ be a $\sigma$-finite measure algebra and
$\frak B$ a subalgebra of $\frak A$ such that
$(\frak B,\bar\mu\restrp\frak B)$ is a semi-finite measure algebra.
Then $(\frak B,\bar\mu\restrp\frak B)$ is a $\sigma$-finite
measure algebra.

\proof{{\bf (a)} (Cf.\ 322G, 316E, 322Cc.)
Let $\bar\mu$ be such that $(\frak A,\bar\mu)$
is a totally finite measure algebra.

\medskip

\quad{\bf (i)}
Let $A^*$ be the set of suprema of finite
subsets of $A$, and set $\gamma=\sup_{a\in A^*}\bar\mu a$.
There is a sequence $\sequencen{a_n}$ in $A^*$ such that
$\sup_{n\in\Bbb N}\bar\nu a_n=\gamma$;  let
$B\subseteq A$ be a countable set such that every $a_n$ is the supremum of
a finite subset of $B$.   Then any upper bound $c$ of $B$ is an upper bound
of $A$.   \Prf\ Take $d\in A$.   Then $a\Bcup a_n\in A^*$, so

\Centerline{$\bar\mu(a\Bsetminus c)\le\bar\mu(a\Bsetminus a_n)
=\bar\mu(a\Bcup a_n)-\bar\mu a_n\le\gamma-\bar\mu a_n$}

\noindent for every $n$, and $\bar\mu(a\Bsetminus c)=0$, that is,
$a\Bsubseteq c$.\ \Qed

\medskip

\quad{\bf (ii)} follows at once from (i), since $\frak A$ is Dedekind
$\sigma$-complete.

\medskip

\quad{\bf (iii)} By (i), there is a sequence $\sequencen{d_n}$ in $D$ with
supremum $1$;  now $\sequencen{d_n\Bsetminus\sup_{i<n}d_i}$ is a partition
of unity included in $D$.

\medskip

{\bf (b)} (Cf.\ 322Nc.)   Write $\frak B^f$ for the ring
$\{b:b\in\frak B$, $\bar\mu b<\infty\}$.   Let $\sequencen{a_n}$ be a
non-decreasing sequence in $\frak A$, with supremum $1$, such that
$\bar\mu a_n<\infty$ for every $n$.    For each $n\in\Bbb N$, set
$\alpha_n=\sup\{\bar\mu(b\Bcap a_n):b\in\frak B^f\}$;  choose a sequence
$\sequencen{b_n}$ in $\frak B^f$ such that
$\bar\mu(b_n\Bcap a_n)\ge\alpha_n-2^{-n}$ for every $n$.   \Quer\ If
$1$ is not the supremum of $\{b_n:n\in\Bbb N\}$ in $\frak B$, let
$b\in\frak B\setminus\{0\}$ be such that $b\Bcap b_n=0$ for every $n$.
Because $\bar\mu\restrp\frak B$ is semi-finite, there is a non-zero
$b'\in\frak B^f$ included in $b$.   But now
$0<\bar\mu b'=\sup_{n\in\Bbb N}\bar\mu(b'\Bcap a_n)$, so there is an
$n\in\Bbb N$ such that $\bar\mu(b'\Bcap a_n)>2^{-n}$;  in which case
$b'\Bcup b_n\in\frak B^f$ and $\bar\mu((b'\Bcup b_n)\Bcap a_n)>\alpha_n$,
which is impossible.\ \Bang
}%end of proof of 566M

\leader{566N}{Characterizing the usual measure on $\{0,1\}^{\Bbb N}$:
Theorem} [AC($\omega$)]
(a) Let $(X,\Sigma,\mu)$ be an atomless, perfect,
complete, countably separated probability space.   Then it is isomorphic to
$\{0,1\}^{\Bbb N}$ with its usual measure.

(b) Let $(\frak A,\bar\mu)$ be an atomless
probability algebra of countable Maharam type.   Then it
is isomorphic to the measure algebra of the usual measure on
$\{0,1\}^{\Bbb N}$.

(c) An atomless measurable algebra of countable Maharam type is
homogeneous.

(d) For any infinite set $I$, the measure algebra of the usual measure on
$\{0,1\}^I$ is homogeneous.

\proof{{\bf (a)} (Cf.\ 344I.)   Write $\nu$ for the usual measure on
$Y=\{0,1\}^{\Bbb N}$, and $\Tau$ for its domain.

\medskip

\quad{\bf (i)} Let $\Cal H$ be a countable subset of $\Sigma$ separating
the points of $X$, and $\sequencen{E_n}$ a sequence running over
$\Cal H$ with cofinal repetitions.   Let $f:\Sigma\times[0,1]\to\Sigma$ and
be a function as in 566F.
Define $g:\Sigma\times\Bbb N\to\frak A$ by setting

$$\eqalign{g(E,n)&=f(E\Bcap E_n,\Bover12\bar\mu E)
  \text{ if }\bar\mu(E\Bcap E_n)\ge\Bover12\bar\mu E,\cr
&=f(E\setminus E_n,\Bover12\bar\mu E)\text{ otherwise }.\cr}$$

\noindent
Define $\sequencen{\Cal E_n}$ inductively by saying that
$\Cal E_0=\{X\}$ and

\Centerline{$\Cal E_{n+1}
=\{g(E,n):E\in\Cal E_n\}\cup\{E\setminus g(E,n):E\in\Cal E_n\}$}

\noindent for each $n$.   Then each $\Cal E_n$ is a partition of unity
consisting of $2^n$ elements of measure $2^{-n}$.   Set
$G_n=\bigcup_{E\in\Cal E_n}g(E,n)$ for each $n$, and let
$\Sigma_0$ be the $\sigma$-subalgebra of $\Sigma$ generated by
$\{G_n:n\in\Bbb N\}$.

For $H\in\Sigma$ and $n\in\Bbb N$, set

\Centerline{$\gamma_n(H)
=2^{-n}\#(\{E:E\in\Cal E_n$, $E\cap H\ne\emptyset$ and
$E\not\subseteq H\}$.}

\noindent Then $\gamma_{n+1}(H)\le\gamma_n(H)$ for every $n$, and
$\gamma_{n+1}(E_n)\le\bover12\gamma_n(E_n)$.   Since every member of
$\Cal H$
appears infinitely often as an $E_n$, $\lim_{n\to\infty}\gamma_n(H)=0$ for
every $H\in\Cal H$.   But this means that if $H\in\Cal H$ and we set
$H'=\bigcup\{E:E\in\bigcup_{n\in\Bbb N}\Cal E_n$, $E\subseteq H\}$ and
$H''=X\setminus\bigcup\{E:E\in\bigcup_{n\in\Bbb N}\Cal E_n$,
$H\cap E=\emptyset\}$, then $H'$ and $H''$ both belong to $\Sigma_0$,
$H'\subseteq H\subseteq H''$ and $H''\setminus H'$ is negligible.

\medskip

\quad{\bf (ii)} Define $\phi_0:X\to Y$ by setting
$\phi_0(x)=\sequencen{\chi G_n(x)}$ for $x\in Y$.   Then $\phi_0$ is
$\Sigma_0$-measurable.   Consider the image measure $\mu\phi_0^{-1}$.
This is a topological measure, and because $\mu$ is perfect and complete
(and $Y$ is homeomorphic to a subset of $\Bbb R$)
$\mu\phi_0^{-1}$ is a Radon measure.   If $n\in\Bbb N$ and $z\in\{0,1\}^n$
then $\phi_0^{-1}\{y:z\subseteq y\in Y\}$ belongs to
$\Cal E_n$ and has measure $2^{-n}$, so $\mu\phi_0^{-1}$ and
$\nu$ agree on such sets;  both being Radon measures, they must be equal.

\medskip

\quad{\bf (iii)} Now observe that
$\Sigma_0=\{\phi_0^{-1}[F]:F\subseteq Y$ is Borel$\}$.   We have seen that
if $H\in\Cal H$ there are $H'$, $H''\in\Sigma_0$ such that
$H'\subseteq H\subseteq H''$ and $H''\setminus H'$ is negligible.
Set $X_1=X\setminus\bigcup_{H\in\Cal H}H\setminus H'$, so that
$X_1\subseteq X$ is $\mu$-conegligible.   Now $\phi_0\restr X_1$ is
injective.
\Prf\ If $x$, $x'$ are distinct members of $X_0$, there is an
$H\in\Cal H$ containing one and not the other;  as neither belongs to
$H\setminus H'$, $H'$ contains one and not the other;  as
$H'=\phi_0^{-1}[F]$ for some $F\subseteq Y$, $\phi_0(x)\ne\phi_0(x')$.\
\QeD\   We also find that $\phi_0[X_1]$ is $\nu$-conegligible.   \Prf\
Because $\nu=\mu\phi_0^{-1}$, $\phi_0[X]$ is $\nu$-conegligible.   For each
$H\in\Cal H$,

$$\eqalignno{\nu\phi_0[H\setminus H']
&=\mu\phi_0^{-1}[\phi_0[H\setminus H']]
\le\mu\phi_0^{-1}[\phi_0[H''\setminus H']]
=\mu(H''\setminus H')
\Displaycause{because $H''\setminus H'\in\Sigma_0$ so is the inverse image
of a subset of $Y$}
=0.
\cr}$$

\noindent So $\phi_0[\bigcup_{H\in\Cal H}H\setminus H']$ is
$\nu$-negligible and $\phi_0[X_1]$ is $\nu$-conegligible.\ \Qed

\medskip

\quad{\bf (iv)}
It follows that if we set $\phi_1=\phi_0\restr X_1$ then the subspace
measure $\nu_{Y_1}$ is just the image measure $\mu_{X_1}\phi_1^{-1}$.
\Prf\ If $F\subseteq Y_1$ then

\Centerline{$\mu_{X_1}\phi_1^{-1}[F]=\mu(X_1\cap\phi_0^{-1}[F])
=\mu\phi_0^{-1}[F]=\nu F=\nu_{Y_1}F$}

\noindent if any of these is defined.\ \QeD\  But as $\phi_1$ is a
bijection, this means that it is an isomorphism between $(X_1,\mu_{X_1})$
and $(Y_1,\nu_{Y_1})$.

\medskip

\quad{\bf (v)} There is no reason to suppose that $X\setminus X_1$ and
$Y\setminus Y_1$ are equipollent, so $\phi_1$ may not be directly
extendable to an isomorphism between $X$ and $Y$.   However, there is a
negligible subset $D$ of $Y_1$ which is equipollent with
$\Bbb R$.
\Prf\ Let $K\subseteq Y_1$ be a non-negligible compact set.   Set
$S_2=\bigcup_{n\in\Bbb N}\{0,1\}^n$ and define $\family{z}{S_2}{K_z}$
inductively, as follows.   $K_{\emptyset}=K$.   Given that $z\in\{0,1\}^n$
and that $K_z$ is a non-negligible compact set, take the first $m\ge 2n+2$
such that $J=\{w:w\in\{0,1\}^m$, $\nu\{y:w\subseteq y\in K_z\}>0\}$ has
more than one member, let $w$, $w'$ be the lexicographically two first
members of $J$, and set

\Centerline{$K_{z^{\smallfrown}0}=\{y:w\subseteq y\in K_z\}$,
\quad$K_{z^{\smallfrown}1}=\{y:w'\subseteq y\in K_z\}$;}

\noindent continue.   This will ensure that
$0<\nu K_z\le 4^{-n}$ for every
$z\in\{0,1\}^n$.   Set $D=\bigcap_{n\in\Bbb N}\bigcup_{z\in\{0,1\}^n}K_z$;
then $D$ is negligible and
equipollent with $\{0,1\}^{\Bbb N}$ and $\Bbb R$.\ \Qed

Now set $X_2=X_1\setminus\phi_1^{-1}[D]$ and $Y_2=Y_1\setminus D$.
$\phi_2=\phi_1\restr X_2$ is an isomorphism between the conegligible sets
$X_2$ and $Y_2$ with their subspace measures.   Since $\Cal H$
separates the points of $X$, we surely have an injective function from
$X\setminus X_2$ to $\Bbb R$, while we also have an injective function from
$\Bbb R$ to $\phi_1^{-1}[D]\subseteq X\setminus X_2$.   So $X\setminus X_2$
is equipollent with $\Bbb R$.   Similarly, $Y\setminus Y_2$ is equipollent
with $\Bbb R$.   So $\phi_2:X_2\to Y_2$ can be extended to a bijection
$\phi:X\to Y$, which will be the required isomorphism between
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$.

\medskip

{\bf (b)} (Cf.\ 331I.)   We can use the same idea as in (a).
Let $\sequencen{a_n}$ be a sequence running over a
$\tau$-generating set $A\subseteq\frak A$ with cofinal repetitions.
Let $f:\frak A\times[0,1]\to\frak A$ be a function as in 566F.
Define $g:\frak A\times\Bbb N\to\frak A$ by setting

$$\eqalign{g(a,n)&=f(a\Bcap a_n,\Bover12\bar\mu a)
  \text{ if }\bar\mu(a\Bcap a_n)\ge\Bover12\bar\mu a,\cr
&=f(a\Bsetminus a_n,\Bover12\bar\mu a)\text{ otherwise }.\cr}$$

\noindent
Define $\sequencen{B_n}$ inductively by saying that
$B_0=\{1\}$ and

\Centerline{$B_{n+1}
=\{g(b,n):b\in B_n\}\cup\{b\Bsetminus g(b,n):b\in B_n\}$}

\noindent for each $n$.   Then each $B_n$ is a partition of unity
consisting of $2^n$ elements of measure $2^{-n}$.   Let $\frak B$ be the
closed subalgebra of $\frak A$ generated by $\bigcup_{n\in\Bbb N}B_n$;
then $\frak B$ is isomorphic to the measure algebra of the usual measure on
$\{0,1\}^{\Bbb N}$.

For $a\in\frak A$ and $n\in\Bbb N$, set

\Centerline{$\gamma_n(a)
=2^{-n}\#(\{b:b\in B_n$, $a\cap b\notin\{0,b\}\})$.}

\noindent Then $\gamma_{n+1}(a)\le\gamma_n(a)$ for every $n$, and
$\gamma_{n+1}(a_n)\le\bover12\gamma_n(a_n)$.   Since every member of $A$
appears infinitely often as an $a_n$, $\lim_{n\to\infty}\gamma_n(a)=0$ for
every $a\in A$.   But this means that $A\subseteq\frak B$ and
$\frak B=\frak A$.   So we have the required isomorphism.

\medskip

{\bf (c)} (Cf.\ 331N.)   If $\frak A$ is such an algebra,
any non-zero principal ideal of $\frak A$ is atomless and of
countable Maharam type and supports a probability measure,
so must be isomorphic to the measure algebra of the usual measure on
$\{0,1\}^{\Bbb N}$ and to $\frak A$.

\medskip

{\bf (d)} For $J\subseteq I$,
write $\nu_J$ for the usual measure on $\{0,1\}^J$, $\Tau_J$ for its
domain and $(\frak B_J,\bar\nu_J)$
for its measure algebra.   If $a\in\frak B_I$ is
non-zero, then it is of the form $E^{\ssbullet}$ for some $E\in\Tau_I$
determined by coordinates in a countable subset $J$ of $I$.   Identifying
$\{0,1\}^I$ with $\{0,1\}^J\times\{0,1\}^{I\setminus J}$, we have an
$F\in\Tau_J$ such that $E=F\times\{0,1\}^{I\setminus J}$.   Let
$b\in\frak B_J$ be the equivalence class of $F$.   Now we can
think of the probability algebra free product
$\frak B_J\tensorhat\frak B_{I\setminus J}$ as the metric completion of the
algebraic free product $\frak B_J\otimes\frak B_{I\setminus J}$, and as
such isomorphic to $\frak B_I$ under an isomorphism which identifies the
principal ideal $(\frak B_I)_a$ with
$(\frak B_J)_b\tensorhat\frak B_{I\setminus J}$.   By (b),
$((\frak B_J)_b,\bar\nu_J\restr(\frak B_J)_b)$ is isomorphic, up to a
scalar multiple of the measure, to $(\frak B_J,\bar\nu_J)$;  so we have

\Centerline{$(\frak B_I)_a
\cong(\frak B_J)_b\tensorhat\frak B_{I\setminus J}
\cong\frak B_J\tensorhat\frak B_{I\setminus J}
\cong\frak B_I$.}

\noindent As $a$ is arbitrary, $\frak B_I$ is homogeneous.
}%end of proof of 566N

\leader{566O}{Boolean values:  Proposition} [AC($\omega$)]
(a) Let $\frak B$ be the algebra
of open-and-closed subsets of $\{0,1\}^{\Bbb N}$, and
$\Cal B(\{0,1\}^{\Bbb N})$ the Borel $\sigma$-algebra.   If $\frak A$ is a
Dedekind $\sigma$-complete Boolean algebra and $\pi:\frak B\to\frak A$ is
a Boolean homomorphism, $\pi$ has a unique extension to a sequentially
order-continuous Boolean homomorphism from $\Cal B(\{0,1\}^{\Bbb N})$ to
$\frak A$.

(b) Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra.   Then there is a bijection between
$L^0=L^0(\frak A)$ and the set $\Phi$ of sequentially order-continuous
Boolean homomorphisms from the algebra $\Cal B(\Bbb R)$ of Borel subsets of
$\Bbb R$ to $\frak A$, defined by saying that $u\in L^0$ corresponds to
$\phi\in\Phi$ iff $\Bvalue{u>\alpha}=\phi(\ooint{\alpha,\infty})$ for
every $\alpha\in\Bbb R$.

(c) Let $(\frak A,\bar\mu)$ be a localizable
measure algebra.   Write $\Sigma_{\text{um}}$ for the algebra of
universally measurable subsets of $\Bbb R$.
Then for any $u\in\cmmnt{ L^0=\mskip5mu}L^0(\frak A)$, we have a sequentially
order-continuous Boolean homomorphism
$E\mapsto\Bvalue{u\in E}:\Sigma_{\text{um}}\to\frak A$ such 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 $E\in\Sigma_{\text{um}}$, while

\Centerline{$\Bvalue{u\in\ooint{\alpha,\infty}\,}=\Bvalue{u>\alpha}$}

\noindent for every $\alpha\in\Bbb R$.

\proof{{\bf (a)}
As in \S562, let $\Cal T$ be the set of trees without
infinite branches in $S^*=\bigcup_{n\ge 1}\BbbN^n$.   For $n\in\Bbb N$
set $E_n=\{x:x\in\{0,1\}^{\Bbb N}$, $x(n)=1\}\in\frak B$ and
$a_n=\pi E_n\in\frak A$.   Let $\phi:\Cal T\to\frak A$ and
$\psi:\Cal T\to\Cal B(\{0,1\}^{\Bbb N})$ be the corresponding
interpretations of Borel codes, as in 562V.
Then $\phi(T)=\phi(T')$ whenever $\psi(T)=\psi(T')$ (562V), and (using
AC($\omega$)) it is easy to check that
$\psi[\Cal T]=\Cal B(\{0,1\}^{\Bbb N})$ (cf.\ 562Db), so
we have a function $\tilde\pi:\Cal B(\{0,1\}^{\Bbb N})\to\frak A$ defined
by saying that $\tilde\pi(\psi(T))=\phi(T)$ for every $T\in\Cal T$.
Now if $\sequencen{F_n}$ is any sequence of Borel subsets of
$\{0,1\}^{\Bbb N}$, we have a
$T\in\Cal T$ such that $F_n=\psi(T_{\fraction{n}})$ for every $n$
and no $T_{\fraction{n}}$ is empty (see 562Bb).   In this case

$$\eqalign{\tilde\pi(\bigcup_{n\in\Bbb N}\{0,1\}^{\Bbb N}\setminus F_n)
&=\tilde\pi(\psi(T))
=\phi(T)\cr
&=\sup_{n\in\Bbb N}1\Bsetminus\phi(T_{\fraction{n}})
=\sup_{n\in\Bbb N}1\Bsetminus\tilde\pi F_n.\cr}$$

\noindent So $\tilde\pi$ is a sequentially order-continuous Boolean
homomorphism.   Since it agrees with $\pi$ on $\{E_n:n\in\Bbb N\}$ it must
agree with $\pi$ on $\frak B$.

Of course the extension is unique because if
$\tilde\pi':\Cal B(\{0,1\}^{\Bbb N}\to\frak A$ is any
sequentially order-continuous Boolean homomorphism extending $\pi$ then
$\{E:\tilde\pi'E=\tilde\pi E\}$ is a $\sigma$-algebra of sets including
$\frak B$ and therefore containing every open set.

\medskip

{\bf (b)} (Cf.\ 364F.)  Let $\Cal E$ be the algebra of subsets of $\Bbb R$
generated by sets of the form $\ooint{q,\infty}$ for $q\in\Bbb Q$.   Then
$\Cal E$ is an atomless countable Boolean algebra, so is isomorphic to the
algebra $\frak B$;  let $\theta:\frak B\to\Cal E$ be an isomorphism.
Define $f:\{0,1\}^{\Bbb N}\to[-\infty,\infty]$ by setting
$f(x)=\sup\{q:q\in\Bbb Q$, $x\in\theta^{-1}\ooint{q,\infty}\}$.   Then $f$
is Borel measurable.

Take any $u$ in $L^0$.   It is easy to check that we
have a Boolean homomorphism
$\pi:\Cal E\to\frak A$ defined by saying that
$\pi\ooint{q,\infty}=\Bvalue{u>q}$ for every $q\in\Bbb Q$.
By (a), there is a sequentially order-continuous Boolean homomorphism
$\tilde\pi:\Cal B(\{0,1\}^{\Bbb N})\to\frak A$ extending
$\pi\theta:\frak B\to\frak A$.
Set $\phi E=\tilde\pi(f^{-1}[E])$ for
$E\in\Cal B(\Bbb R)$.

If $\alpha\in\Bbb R$ then

$$\eqalign{\phi(\ooint{\alpha,\infty})
&=\tilde\pi\{x:f(x)>\alpha\}
=\tilde\pi(\bigcup\{\theta^{-1}\ooint{q,\infty}:q\in\Bbb Q,\,q>\alpha\})
  \cr
&=\sup\{\tilde\pi(\theta^{-1}\ooint{q,\infty}):q\in\Bbb Q,\,q>\alpha\}\cr
&=\sup\{\pi\ooint{q,\infty}:q\in\Bbb Q,\,q>\alpha\}\cr
&=\sup\{\Bvalue{u>q}:q\in\Bbb Q,\,q>\alpha\}
=\Bvalue{u>\alpha}.\cr}$$

\noindent It follows that

$$\eqalign{\phi\Bbb R
&=\sup_{n\in\Bbb N}\tilde\pi(f^{-1}\ooint{-n,\infty})
  \Bsetminus\inf_{n\in\Bbb N}\tilde\pi(f^{-1}\ooint{n,\infty})\cr
&=\sup_{n\in\Bbb N}\Bvalue{u>-n}\Bsetminus\inf_{n\in\Bbb N}\Bvalue{u>n}
=1,\cr}$$

\noindent and therefore that $\phi\in\Phi$.   For the rest of the argument
we can follow the method of 364F.

\medskip

{\bf (c)} (Cf.\ 434T.)

\medskip

\quad{\bf (i)}
To begin with, consider the case in which $\bar\mu$ is
totally finite.   In this case, we have a non-decreasing function
$g:\Bbb R\to\coint{0,\infty}$ defined by saying that
$g(\alpha)=\bar\mu 1-\bar\mu\Bvalue{u>\alpha})$ for
$\alpha\in\Bbb R$.   Let $\nu_g$
be the corresponding Lebesgue-Stieltjes measure (114Xa), and
$(\frak C,\bar\nu_g)$ its measure algebra.   Note that $g$ is continuous on
the right, so that
$\nu_g\ocint{\alpha,\beta}
=\bar\mu\Bvalue{u>\alpha}-\bar\mu\Bvalue{u>\beta}$
whenever $\alpha\le\beta$ in $\Bbb R$.   Let $\frak D$ be the
subalgebra of $\frak C$ generated by
$\{\ocint{-\infty,\alpha}^{\ssbullet}:\alpha\in\Bbb R\}$.   Then we have a
measure-preserving Boolean
homomorphism $\pi:\frak D\to\frak A$ defined uniquely by saying
that $\pi(\ooint{\alpha,\infty}^{\ssbullet})=\Bvalue{u>\alpha}$ for
$\alpha\in\Bbb R$.   Because $\frak D$ is dense in $\frak C$ for the
measure-algebra topology, $\pi$ has a unique extension to a
measure-preserving Boolean homomorphism $\tilde\pi:\frak C\to\frak A$.

Because $\Sigma_{\text{um}}\subseteq\dom\nu_g$, we can define
$\Bvalue{u\in E}$ to be $\tilde\pi E^{\ssbullet}$ for
$E\in\Sigma_{\text{um}}$, and this will give us a sequentially
order-continuous Boolean homomorphism from $\Sigma_{\text{um}}$ to
$\frak A$ such that
$\Bvalue{u\in\ooint{\alpha,\infty}\,}=\Bvalue{u>\alpha}$
for every $\alpha$.   As for the other formulae, they are immediate from
the facts that $\nu_g$ is inner regular with respect to the compact sets
and outer regular with respect to the open sets.

\medskip

\quad{\bf (ii)} We need to observe that these properties uniquely define
$\Bvalue{u\in E}$.   \Prf\ Let $\Cal E$ be the algebra of subsets of
$\Bbb R$ generated by $\{\ooint{\alpha,\infty}:\alpha\in\Bbb R\}$.
The requirement
$\Bvalue{u\in\ooint{\alpha,\infty}\,}=\Bvalue{u>\alpha}$
determines the values of $\Bvalue{u\in E}$ for $E\in\Cal E$.
Next, if $G\subseteq\Bbb R$ is open and
$K\subseteq G$ is compact there is an $E\in\Cal E$ such that
$K\subseteq E\subseteq G$.   Consequently
$\Bvalue{u\in K}=\inf\{\Bvalue{u\in E}:E\in\Cal E$, $E\supseteq K\}$
is fixed for every compact $K\subseteq\Bbb R$.   Finally, the inner
regularity condition
$\Bvalue{u\in E}=\sup\{\Bvalue{u\in K}:K\subseteq E$ is compact$\}$
determines $\Bvalue{u\in E}$ for other $E\in\Sigma_{\text{um}}$.\ \Qed

\medskip

\quad{\bf (iii)} Now turn to the general case of a localizable measure
algebra $(\frak A,\bar\mu)$ and $u\in L^0(\frak A)$.   Let $\frak A^f$ be
the ideal of elements of finite measure.   Then for each $a\in\frak A^f$ we
have a corresponding homomorphism
$E\mapsto\Bvalue{u\in E}_a$ from $\Sigma_{\text{um}}$ to the principal
ideal $\frak A_a$.   If $a\Bsubseteq b\in\frak A^f$, we can use the
uniqueness described in (ii) to see that
$\Bvalue{u\in E}_a=a\Bcap\Bvalue{u\in E}_b$ for every $E$.   So if we set
$\Bvalue{u\in E}=\sup_{a\in\frak A^f}\Bvalue{u\in E}_a$, we shall have
$\Bvalue{u\in E}_a=a\Bcap\Bvalue{u\in E}$ whenever $a\in\frak A^f$ and
$E\in\Sigma_{\text{um}}$.   It is now easy to check that
$E\mapsto\Bvalue{u\in E}$ has the required properties.
}%end of proof of 566O

\leader{566P}{Weak \dvrocolon{compactness}}\cmmnt{ In the
absence of
Tychonoff's theorem, the theory of weak compactness in normed spaces
becomes uncertain.   However AC($\omega$) is enough to give a couple of the
principal results involving classical Banach spaces, starting with
Hilbert space.

\medskip

\noindent}{\bf Theorem} [AC($\omega$)] Let $U$ be a Hilbert space.   Then
bounded sets in $U$ are relatively weakly compact.

\proof{ If $U$ is finite-dimensional, this is trivial;  so let us suppose
that $U$ is infinite-dimensional.   Let $A\subseteq U$ be a bounded set,
and $\overline{A}$ its closure for the weak topology;  let $\Cal F_0$ be a
family of weakly closed subsets of $\overline{A}$
with the finite intersection
property, and $\Cal F$ the filter on $U$ generated by $\Cal F_0$.

\medskip

{\bf (a)} For closed subspaces $V$ of $U$, let
$P_V:U\to V$ be the orthogonal projection from $U$ onto $V$ (561Ib), and
set $\gamma_V=\liminf_{u\to\Cal F}\|P_Vu\|^2$.
Because $\Cal F$ contains a
bounded set, $\gamma_V\le\gamma_U<\infty$ for every $V$.
If $V_0$, $V_1$ are
orthogonal subspaces of $U$, then
$\|P_{V_0+V_1}u\|^2=\|P_{V_0}u\|^2+\|P_{V_1}u\|^2$ for every $u\in U$, so
$\gamma_{V_0+V_1}\ge\gamma_{V_0}+\gamma_{V_1}$.

\medskip

{\bf (b)} Set
$\gamma=\sup\{\gamma_V:V$ is a finite-dimensional linear subspace of $U\}$,
and choose a sequence $\sequencen{V_n}$ of finite-dimensional subspaces of
$U$ such that $\gamma=\sup_{n\in\Bbb N}\gamma_{V_n}$;  because $U$ is
infinite-dimensional, we can do this in such a way that $\dim V_n\ge n$ for
each $n$.  Let $W$ be the
closed linear span of $\bigcup_{n\in\Bbb N}V_n$.   If $V$ is a
finite-dimensional linear subspace of $W^{\perp}$, then

\Centerline{$\gamma\ge\gamma_{V+V_n}\ge\gamma_V+\gamma_{V_n}$}

\noindent for every $n$, so $\gamma_V=0$.

\medskip

{\bf (c)} If $F\in\Cal F$, $V\subseteq W^{\perp}$ is a finite-dimensional
linear subspace, and $\epsilon>0$, then $F\cap\{u:\|P_Vu\|\le\epsilon\}$ is
non-empty.   We can therefore extend $\Cal F$ to the filter $\Cal G$
generated by sets of this type, and $\lim_{u\to\Cal G}\innerprod{u}{w}=0$
for every $w\in W^{\perp}$.

\medskip

{\bf (d)} Let $\sequencen{e_n}$ be an orthonormal basis for $W$.   Define
$\sequencen{\Cal G_n}$ as follows.   $\Cal G_0=\Cal G$.   Given that
$\Cal G_n$ is a filter on $U$ containing a bounded set, set
$\alpha_n=\liminf_{u\to\Cal G_n}\innerprod{u}{e_n}$, and let $\Cal G_{n+1}$
be the filter generated by
$\Cal G_n\cup\{\{u:\innerprod{u}{e_n}<\alpha\}:\alpha>\alpha_n\}$;  then
$\alpha_n=\lim_{u\to\Cal G_{n+1}}\innerprod{u}{e_n}$.   Set
$\Cal H=\bigcup_{n\in\Bbb N}\Cal G_n$;  then
$\alpha_n=\lim_{u\to\Cal H}\innerprod{u}{e_n}$ for each $n$.

For any $n\in\Bbb N$,

\Centerline{$\sum_{i=0}^n\alpha_i^2
=\sum_{i=0}^n\lim_{u\to\Cal H}\innerprod{u}{e_i}^2
=\lim_{u\to\Cal H}\sum_{i=0}^n\innerprod{u}{e_i}^2
\le\limsup_{u\to\Cal H}\|u\|^2
<\infty$}

\noindent because $\Cal H$ contains a bounded set.   So
$\sum_{n=0}^{\infty}\alpha_n^2$ is finite and
$v=\sum_{n=0}^{\infty}\alpha_ne_n$ is defined in $U$.

\medskip

{\bf (e)} Now

\Centerline{$\innerprod{v}{e_n}
=\alpha_n=\lim_{u\to\Cal H}\innerprod{u}{e_n}$}

\noindent for every $n$;  again because $\Cal H$ contains a bounded set,
$\innerprod{v}{w}=\lim_{u\to\Cal H}\innerprod{u}{w}$ for every $w\in W$.
On the other hand, if $w\in W^{\perp}$,

\Centerline{$\lim_{u\to\Cal H}\innerprod{u}{w}
=\lim_{u\to\Cal G}\innerprod{u}{w}=0=\innerprod{v}{w}$.}

\noindent Since $W+W^{\perp}=U$,
$\lim_{u\to\Cal H}\innerprod{u}{w}=\innerprod{v}{w}$ for every $w\in U$.
By 561Ic, $v$ is the limit of $\Cal H$ for the weak topology on $U$,
and must belong to every member of $\Cal F_0$.

As $\Cal F_0$ is arbitrary, $\overline{A}$ is weakly
compact and $A$ is relatively weakly compact.
}%end of proof of 566P

\leader{566Q}{\bf Theorem} [AC($\omega$)]
Let $U$ be an $L$-space.   Then a subset of $U$ is weakly relatively
compact iff it is uniformly integrable.

\proof{{\bf (a)(i)}\grheada\ Recall that $U$ is a Banach lattice with an
order-continuous norm (354N), so is Dedekind complete (354Ee) and all its
bands are projection bands (353J\formerly{3{}53I});
for a band $V$ in $U$, let $P_V:U\to V$ be the band projection onto $V$.

\medskip

\qquad\grheadb\ If $u\in U$ there is an $f\in U^*$ such that
$\|f\|\le 1$ and
$f(u)=\|u\|$.   \Prf\ Let $V$ be the band generated by $u^+$ and
$W=V^{\perp}$ its band complement.
Set $f(v)=\int P_Vv-\int P_Wv$ for $v\in U$.
Since $\|v\|=\|P_Vv\|+\|P_Wv\|$ for
every $v\in U$, $\|f\|\le 1$.   Also $P_Vu=u^+$ and $P_Wu=-u^-$
so $f(u)=\int|u|=\|u\|$.\ \Qed

\medskip

\qquad\grheadc\ If $A\subseteq U$ is weakly bounded it is norm-bounded.
\Prf\Quer\ Otherwise, choose for each $n\in\Bbb N$ a $u_n\in A$ and
$f_n\in U^*$ such that
$\|u_n\|\ge n$, $\|f_n\|=1$ and $f_n(u_n)=\|u_n\|\ge n$.
For $f\in U^*$ set $\rho_A(f)=\sup_{u\in A}|f(u)|$.
Define $\sequence{k}{n_k}$ by setting
$n_k=\lceil 2\cdot 3^k(k+\sum_{i=0}^{k-1}3^{k-i}\rho_A(f_{n_i}))\rceil$
for each $k$.
Set $f=\sum_{i=0}^{\infty}3^{-i}f_{n_i}$.   Then for any $k\in\Bbb N$ we
have

$$\eqalign{\rho_A(f)
&\ge f(u_{n_k})
=\sum_{i=0}^{\infty}3^{-i}f_{n_i}(u_{n_k})\cr
&\ge 3^{-k}f_{n_k}(u_{n_k})-\sum_{i=0}^{k-1}3^{-i}\rho_A(f_{n_i})
     -\sum_{i=k+1}^{\infty}3^{-i}\|u_{n_k}\|\cr
&=\Bover1{2\cdot 3^k}\|u_{n_k}\|-\sum_{i=0}^{k-1}3^{-i}\rho_A(f_{n_i})
\ge k.  \text{ \Bang\Qed}\cr}$$

\medskip

\quad{\bf (ii)}
Now let $K\subseteq U$ be a weakly relatively countably compact
set.   Let $\frak A$ be the band algebra of $U$.   For $V\in\frak A$
set $\nu V=\sup_{u\in K}\|P_Vu\|$ (counting $\sup\emptyset$ as $0$).
Then $\nu$ is a submeasure on $\frak A$.   By (i-$\gamma$),
$K$ is norm-bounded and $\nu$ is
finite-valued;  set $\alpha=\nu U=\sup_{u\in K}\|u\|$.

$\nu$ is exhaustive.
\Prf\Quer\ Otherwise, let $\sequencen{V_n}$ be a disjoint sequence in
$\frak A$ such that
$\epsilon=\bover16\inf_{n\in\Bbb N}\nu V_n$ is greater than $0$.
For each $n\in\Bbb N$ choose $u_n\in K$ and $f_n\in U^*$
such that $\|f_n\|\le 1$ and
$f_n(P_nu_n)=\|P_nu_n\|\ge 5\epsilon$, where here I write $P_n$
for $P_{V_n}$.   Let $v_0$
be a cluster point of $\sequencen{u_n}$ in $U$ for the weak topology of
$U$.   Note that $\sum_{n=0}^{\infty}\|P_nu\|\le\|u\|$ for any $u\in U$;
let $m\in\Bbb N$ be such that $\sum_{n=m}^{\infty}\|P_nv_0\|\le\epsilon$.
For $n\in\Bbb N$, set $g_n(u)=f_n(P_nu)$ for $u\in U$.

We can now build a strictly increasing sequence $\sequence{k}{n_k}$ such
that

\Centerline{$n_0\ge m$,}

\Centerline{$\sum_{i=0}^{k-1}|g_{n_i}(u_{n_k})|
\le\epsilon+\sum_{i=0}^{k-1}|g_{n_i}(v_0)|$,}

\Centerline{$\|P_{n_k}u_{n_i}\|\le 2^{-k}\epsilon$ whenever $i<k$}

\noindent for every $k\in\Bbb N$.   Let $v_1$ be a weak cluster point of
$\sequence{k}{u_{n_k}}$, and $l\in\Bbb N$ such that
$\sum_{k=l}^{\infty}\|P_{n_k}v_1\|\le\epsilon$.   Set
$g=\sum_{k=l}^{\infty}g_{n_k}$;  this is defined in $U^*$ because

\Centerline{$\sum_{k=l}^{\infty}|g_{n_k}(u)|
\le\sum_{k=l}^{\infty}\|P_{n_k}u\|\le\|u\|$}

\noindent for every $u\in U$.   Of course $|g(v_1)|\le\epsilon$.   On the
other hand, for any $k\ge l$,

$$\eqalign{g(u_{n_k})
&=g_{n_k}(u_{n_k})+\sum_{i=l}^{k-1}g_{n_i}(u_{n_k})
   +\sum_{i=k+1}^{\infty}g_{n_i}(u_{n_k})\cr
&\ge 5\epsilon-\sum_{i=0}^{k-1}|g_{n_i}(u_{n_k})|
   -\sum_{i=k+1}^{\infty}\|P_{n_i}u_{n_k}\|   \cr
&\ge 5\epsilon-\sum_{i=0}^{k-1}|g_{n_i}(v_0)|-\epsilon
   -\sum_{i=k+1}^{\infty}2^{-i}\epsilon\cr
&\ge 4\epsilon-\sum_{n=m}^{\infty}\|P_nv_0\|
   -2^{-k}\epsilon
\ge 2\epsilon\cr}$$

\noindent and $v_1$ cannot be a weak cluster point of
$\sequence{k}{u_{n_k}}$.\ \Bang\Qed

\medskip

\quad{\bf (iii)} In fact $\nu$ is uniformly exhaustive.
\Prf\Quer\ Otherwise, let $\epsilon>0$ be such that there are arbitrarily
long disjoint strings in $\frak A$ of elements of submeasure greater than
$2\epsilon$.   Set
$q(n)=\lceil\Bover{2^nn\alpha}{\epsilon}\rceil$ for each $n$,
and choose a family
$\langle V_{ni}\rangle_{n\in\Bbb N,i\le q(n)}$ such that
$\langle V_{ni}\rangle_{i\le q(n)}$ is a disjoint family in $\frak A$
for each $n$ and $\nu V_{ni}>2\epsilon$ for all $n$ and $i$;
adapting the temporary notation of (ii), I set $P_{ni}=P_{V_{ni}}$ for
$i\le q(n)$.   Now choose
$u_{ni}\in K$ such that $\|P_{ni}u_{ni}\|\ge 2\epsilon$ for all $i$ and
$n$.   Because $\sum_{i=0}^{q(n)}\|P_{ni}u\|\le\|u\|$ for every $u\in U$
and $n\in\Bbb N$, we can define inductively a sequence $\sequencen{i_n}$
such that $i_n\le q(n)$ and
$\|P_{ni_n}u_{mi_m}\|\le 2^{-n}\epsilon$ whenever $m<n$.

Now set

\Centerline{$W_{mn}=V_{mi_m}\cap\bigcap_{m<k\le n}V_{ki_k}^{\perp}$,
\quad$Q_{mn}=P_{W_{mn}}$}

\noindent for $m\le n$,

\Centerline{$W_m=\bigcap_{n\ge m}W_{mn}$,
\quad$Q_m=P_{W_m}$}

\noindent for $m\in\Bbb N$.
For any $u\in U$ and $m\le n$,

\Centerline{$|P_{mi_m}u|
=P_{mi_m}|u|\le Q_{mn}|u|+\sum_{k=m+1}^nP_{ki_k}|u|$,}

\Centerline{$\|P_{mi_m}u\|
\le\|Q_{mn}u\|+\sum_{k=m+1}^n\|P_{ki_k}u\|$,}

\noindent so

$$\eqalign{\|Q_{mn}u_{mi_m}\|
&\ge\|P_{mi_m}u_{mi_m}\|-\sum_{k=m+1}^n\|P_{ki_k}u_{mi_m}\|\cr
&\ge 2\epsilon-\sum_{k=m+1}^{\infty}2^{-k}\epsilon
\ge\epsilon.\cr}$$

\noindent Next, if $u\ge 0$, $\langle Q_{mn}u\rangle_{n\ge m}$ is a
non-increasing sequence, and its infimum belongs to
$\bigcap_{n\ge m}W_{mn}$, so must be equal to $Q_mu$;  accordingly
$Q_mu$ is the norm-limit of $\langle Q_{mn}u\rangle_{n\ge m}$.   The same
is therefore true for every $u\in U$, and in particular

\Centerline{$\|Q_mu_{mi_m}\|=\lim_{n\to\infty}\|Q_{mn}u_{mi_m}\|
\ge\epsilon$.}

\noindent Consequently $\nu W_m\ge\epsilon$.   But
$W_m\cap W_n\subseteq V_{ni_n}^{\perp}\cap V_{ni_n}=\{0\}$ whenever $n>m$,
so this contradicts (ii).\ \Bang\Qed

\medskip

\quad{\bf (iv)} Now take any $\epsilon>0$.
Then there is a $u^*\in U^+$ such
that $\int(|u|-u^*)^+\le\epsilon$ for every $u\in K$.   \Prf\
If $\alpha\le\epsilon$ we can take $u^*=0$ and stop.   Otherwise, there
is a largest $n\in\Bbb N$ such that there are disjoint
$V_0,\ldots,V_n\in\frak A$ such that $\nu V_i>\epsilon$ for every $n$.
Take $u_0,\ldots,u_n\in K$ such that $\|P_{V_i}u_i\|>\epsilon$ for each
$i$.   Let $\gamma>0$ be such that
$\|P_{V_i}u_i\|-\Bover{\alpha}{\gamma}>\epsilon$ for every $i\le n$, and
set $u^*=\gamma\sum_{i=0}^n|u_i|$.
\Quer\ Suppose that $u\in K$ is such that $\int(|u|-u^*)^+>\epsilon$.
Let $W$ be the band generated by $(|u|-u^*)^+$, so that
$\nu W\ge\|P_Wu\|>\epsilon$.   For each $i\le n$, set
$W_i=V_i\cap W^{\perp}$;  then

\Centerline{$|P_Wu_i|\le\Bover1{\gamma}P_Wu^*\le\Bover1{\gamma}|u|$,
\quad$|P_{W_i}u_i|\ge|P_{V_i}u_i|-\Bover1{\gamma}|u|$,}

\Centerline{$\nu W_i\ge\|P_{W_i}u_i\|
\ge\|P_{V_i}u_i\|-\Bover{\alpha}{\gamma}>\epsilon$.}

\noindent
But now $W_0,\ldots,W_n,W$ witnesses that $n$ was not maximal.\ \Bang\
So $\sup_{u\in K}\int(|u|-u^*)^+\le\epsilon$, as required.\ \Qed

As $\epsilon$ is arbitrary, $K$ is uniformly integrable.   Thus every
relatively weakly compact subset of $U$ is uniformly integrable.

\medskip

{\bf (b)(i)} In the reverse direction, suppose to begin with
that $(\frak A,\bar\mu)$ is a totally finite measure algebra, and
that $A\subseteq L^1=L^1(\frak A,\bar\mu)$ is uniformly integrable;  let
$\Cal F$ be a filter on $L^1$ containing $A$.    Write $\Cal V$ for the set
of neighbourhoods of $0$ for the weak topology
$\frak T_s(L^1,(L^1)^*)$.\footnote{\smallerfonts
Of course $(L^1)^*$ can be identified
with $L^{\infty}(\frak A)$, but if
you don't wish to trace through the arguments for this, and confirm that
they can be carried out without appealing to anything more than
AC($\omega$), you can defer the exercise for the time being.}

\medskip

\qquad\grheada\ For each $n\in\Bbb N$ let $M_n\ge 0$ be such that
$\|(|u|-M_n\chi 1)^+\|_1\le 2^{-n}$ for every $u\in A$, and
define sets $K_n\subseteq[-M_n\chi 1,M_n\chi 1]$ and filters $\Cal F_n$
as follows.   $\Cal F_0=\Cal F$.   Given that $\Cal F_n$ contains $A$,
define $\phi_n:L^1\to L^2=L^2(\frak A,\bar\mu)$ by setting
$\phi_n(u)=\med(-M_n\chi 1,u,M_n\chi 1)$ for each $u\in L^1$, and consider
the filter $\phi_n[[\Cal F_n]]$.   This is a filter on the Hilbert space
$L^2$ containing the $\|\,\|_2$-bounded set $[-M_n\chi 1,M_n\chi 1]$, so
the set $K_n^*$ of its $\frak T_s(L^2,L^2)$-cluster points is non-empty, by
566P;  as $K_n^*$ is $\frak T_s(L^2,L^2)$-closed, it is
$\frak T_s(L^2,L^2)$-compact.
As $[-M_n\chi 1,M_n\chi 1]$ is $\|\,\|_2$-closed and convex, it is
$\frak T_s(L^2,L^2)$-closed (561Ie) and includes $K_n^*$.
Set $\gamma_n=\inf\{\|u\|_2:u\in K_n^*\}$.   As all the sets
$\{u:\|u\|_2\le\alpha\}$, for $\alpha>\gamma_n$, are
$\frak T_s(L^2,L^2)$-closed and meet $K_n^*$,
$K_n=\{u:u\in K_n^*$, $\|u\|_2\le\gamma_n\}$ is non-empty.

Suppose that $G\in\Cal V$.
Because the embedding $L^2\embedsinto L^1$ is norm-continuous, it is weakly
continuous, and $G\cap L^2$ is a $\frak T_s(L^2,L^2)$-neighbourhood of $0$.
It follows that
$x+G$ meets every member of $\phi_n[[\Cal F_n]]$ for every $x\in K^*_n$;
so $K_n+G$ meets every member of $\phi_n[[\Cal F_n]]$.   We can therefore
extend $\Cal F_n$ to the filter $\Cal F_{n+1}$ generated by

\Centerline{$\Cal F_n\cup\{\phi_n^{-1}[K_n+G]:G\in\Cal V\}$}

\noindent and continue.

\medskip

\qquad\grheadb\
Set $\Cal G=\bigcup_{n\in\Bbb N}\Cal F_n$ and
$B=\{u:u\in L^1$, $\|u\|_1\le 1\}$.   Then for each $n\in\Bbb N$
there is a finite set $J\subseteq L^1$ such that
$J+G+2^{-n+1}B\in\Cal G$ for every $G\in\Cal V$.  \Prf\ $K_n$ is a
$\frak T_s(L^2,L^2)$-closed subset of $K_n^*$, so is
$\frak T_s(L^2,L^2)$-compact;  also it is included in the sphere
$S=\{u:\|u\|_2=\gamma_n\}$.   Because $\|\,\|_2$ is locally uniformly
rotund, it is a Kadec norm (467B)
and the norm and weak topologies on $S$ coincide;  consequently $K_n$ is
$\|\,\|_2$-compact.   Since $\|\,\|_1$ and $\|\,\|_2$ give rise to the same
topology on any $\|\,\|_{\infty}$-bounded set, $K_n$ is $\|\,\|_1$-compact.
There is therefore a finite set $J\subseteq K_n$ such that
$K_n\subseteq J+2^{-n}B$.

Take any $G\in\Cal V$.   Then
$\|u-\phi_n(u)\|=\|(|u|-M_n\chi 1)^+\|\le 2^{-n}$ for
every $u\in A$, so

\Centerline{$J+G+2^{-n+1}B\supseteq(K_n+G)+2^{-n}B
\supseteq A\cap\phi_n^{-1}[K_n+G]\in\Cal F_{n+1}\subseteq\Cal G$.\ \Qed}

\medskip

\qquad\grheadc\ For each $n\in\Bbb N$ choose a minimal
finite set $J_n\subseteq L^1$ such that $J_n+G+2^{-n+1}B\in\Cal G$
for every
$G\in\Cal V$.   Note that $(x+G+2^{-n+1}B)\cap D$ must be non-empty
whenever $n\in\Bbb N$, $x\in J_n$, $G\in\Cal V$ and $D\in\Cal G$.
\Prf\Quer\ Otherwise,

\Centerline{$(J_n\setminus\{x\})+G'+2^{-n+1}B
\supseteq(J_n+(G\cap G')+2^{-n+1}B)\cap D$}

\noindent belongs to $\Cal G$ for every $G'\in\Cal V$, and $J_n$ was not
minimal.\ \Bang\Qed

\medskip

\qquad\grheadd\ For any $n\in\Bbb N$ and $u\in J_n$ there is a
$v\in J_{n+1}$ such that $\|u-v\|_1\le 2^{-n+1}+2^{-n}$.   \Prf\Quer\
Otherwise, by (a-i-$\beta$) above, we can
choose for each $v\in J_{n+1}$ an $f_v\in(L^1)^*$ such that
$\|f_v\|=1$ and
$f_v(v-u)=\|v-u\|=2^{-n+1}+2^{-n}+\delta_v$ where $\delta_v>0$;  set

\Centerline{$G
=\{w:|f_v(w)|<\bover12\delta_v$ for every $v\in J_{n+1}\}\in\Cal V$.}

\noindent Then $u+G+2^{-n+1}B$ does not meet $J_{n+1}+G+2^{-n}B$,
contradicting ($\gamma$) here.\ \Bang\Qed

\medskip

\qquad\grheade\ Because $\bigcup_{n\in\Bbb N}J_n$ is countable, therefore
well-orderable, we can define inductively a sequence $\sequencen{u_n}$ such
that $u_n\in J_n$ and $\|u_n-u_{n+1}\|\le 2^{-n+1}+2^{-n}$ for every $n$.
Now $\sequencen{u_n}$ is Cauchy, so has a limit $u$ in $L^1$.
If $G\in\Cal V$ there is an $n\in\Bbb N$ such that
$u+G\supseteq u_n+\bover12G+2^{-n}B$, so $u+G$ meets every member of
$\Cal G$;  thus $u$ is a weak cluster point of $\Cal G$ and of $\Cal F$.
As $\Cal F$ is arbitrary, $A$ is relatively weakly compact.

\medskip

\quad{\bf (ii)} Now suppose that $U$ is an arbitrary $L$-space and
$A\subseteq U$ is a uniformly integrable set.   Then we can choose a
sequence $\sequencen{e_n}$ in $U^+$ such that
$\|(|u|-e_n)^+\|\le 2^{-n}$ for every $n\in\Bbb N$ and $u\in A$.   Set
$e=\sum_{n=0}^{\infty}\Bover1{1+2^n\|e_n\|}e_n$ in $U$, and let $V$ be the
band in $U$ generated by $e$.   Then $A\subseteq V$, and of course $A$ is
uniformly integrable in $V$.   By 561Hb, we have a
totally finite measure algebra $(\frak A,\bar\mu)$ and a normed Riesz space
isomorphism $T:V\to L^1(\frak A,\bar\mu)$;   now $T[A]$ is uniformly
integrable in $L^1(\frak A,\bar\mu)$, therefore relatively weakly
compact, by (i).   But this means that $A$ is relatively weakly compact in
$V$;  as the embedding $V\embedsinto U$ is weakly continuous, $A$ is
relatively weakly compact in $U$.

This completes the proof.
}%end of proof of 566Q

\vleader{48pt}{566R}{Automorphisms of measurable algebras:
Theorem} [AC($\omega$)] Let $\frak A$ be a measurable algebra.

(a) Every automorphism of $\frak A$ has a separator.

(b) Every $\pi\in\Aut\frak A$ is a product of at most three exchanging
involutions belonging to the full subgroup of $\Aut\frak A$ generated by
$\pi$.

\proof{{\bf (a)}  (Cf.\ 382Eb.)   Take $\pi\in\Aut\frak A$.
Let $\bar\mu$ be such that $(\frak A,\bar\mu)$ is a totally finite
measure algebra.   For $a\in\Bbb N$ set $\psi(a)=\sup_{n\in\Bbb Z}\pi^na$,
so that $\pi(\psi(a))=\psi(a)$.   Note that if $a\Bcap\psi(b)=0$ then
$\psi(a)\Bcap\psi(b)=0$.   Set
$A=\{a:a\in\frak A$, $a\Bcap\pi a=0\}$ and choose a sequence
$\sequencen{a_n}$ in $A$ such that

\Centerline{$\sup_{n\in\Bbb N}\bar\mu(\psi(a_n))
=\sup_{a\in A}\bar\mu(\psi(a))$.}

\noindent Define $\sequencen{b_n}$, $\sequencen{c_n}$ by saying that

\Centerline{$c_0=0$,
\quad$b_n=a_n\setminus\psi(c_n)$,
\quad$c_{n+1}=b_n\Bcup c_n$}

\noindent for each $n$.   Inducing on $n$ we see that $b_n$ and $c_n$
belong to $A$ and that $\psi(c_{n+1})\Bsupseteq\psi(a_n)$ for every $n$.
Set $c=\sup_{n\in\Bbb N}c_n$;  then $c\in A$ and
$\psi(c)\Bsupseteq\psi(a_n)$ for every $n$.

Now $c$ is a separator for $\pi$.   \Prf\Quer\ Otherwise, there is a
non-zero $d\Bsubseteq 1\Bsetminus\psi(c)$ such that $d\Bcap\pi d=0$
(381Ei).   In this case $d\Bcup c\in A$ and

\Centerline{$\bar\mu(\psi(d\Bcup c))>\bar\mu(\psi(c))
=\sup_{n\in\Bbb N}\bar\mu(\psi(a_n))=\sup_{a\in A}\bar\mu(\psi(a))$,}

\noindent which is impossible.\ \Bang\Qed

\medskip

{\bf (b)} We can now work through the proofs of 382A-382M to confirm that
there is no essential use of anything beyond countable choice there, so
long as we suppose that we are working with measurable algebras.
(There is an inductive construction in the proof of 382J.   To do this with
AC($\omega$) rather than DC, we need to check that every element of the
construction can be made determinate following an initial countable set of
choices;  in the case there, we need to
check that the existence assertions of 382D
and 382I can be represented as functions, as in
566Xh and 566Xj.) %566Xh 566Xi 566Xj
Since the proof of 382K speaks of the Stone representation
theorem, there seems to be a difficulty here, unless we take the
alternative route suggested in 382Yb.
But note that while the general Stone theorem
has a strength little short of full AC, the representation of a
{\it countable} Boolean algebra $\frak B$
as the algebra of open-and-closed subsets
of a compact Hausdorff Baire space can be done in ZF alone (561F).
In part (f) of the proof of 382K, therefore,
take $\frak B$ to be a countable subalgebra of $\frak A$ such that

\inset{$e_n$, $u'_n$, $u''_n$, $v'_l$, $v''_l$, $d_{lj}$, $d'_{lj}$,
$\supp(\pi\phi)^k$, $\supp(\pi\phi_1)^k\in\frak B$
whenever $n\in\Bbb N$ and $j$, $k$, $l\ge 1$,

$c_0$, $c_1$, $\supp\phi_2\in\frak B$,

$\frak B$ is closed under the functions $\pi$, $\phi_1$, $\phi_2$, $\phi$
and $\tilde\pi_n$ for $n\in\Bbb N$,}

\noindent and let $Z$ be the Stone space of $\frak B$.   Now we can perform
the arguments of the rest of the proof in $Z$ to show that
$c_0=\inf_{n\ge 1}\supp(\pi\phi)^n$ is zero, as required.
}%end of proof of 566R

\cmmnt{\leader{566S}{Volume 4} In Volume 4, naturally,
a rather larger proportion of the
ideas become inaccessible without strong forms of the axiom of choice.
Since we are missing the most useful representation theorems, many
results have to be abandoned altogether.   More subtly, we seem to lose the
result that Radon measures are localizable\cmmnt{ (416B)}.
Nevertheless, a good deal can still be done\cmmnt{, if
we follow the principles
set out in 566Ae-566Af}.   Most notably, we have a workable theory of Haar
measure on completely regular locally compact
topological groups\cmmnt{, because the Riesz representation theorems of
\S436 are still available, and we can use 561G instead of 441C}.
I should remark, however, that in the absence of Tychonoff's theorem we may
have fewer compact groups than we expect.
And the theory of dual groups in \S445 depends heavily on AC.

The descriptive set theory of Chapter 42 is hardly touched, and
enough of the rest of the volume survives to make it worth
checking any point of particular interest.   Most of Chapter 46 depends
heavily on the Hahn-Banach theorem and therefore becomes limited to
cases in which we have a good grasp of dual spaces\cmmnt{, as in 561Xh}.
There are some difficulties in the geometric measure theory of arbitrary
metric spaces in \S471, but the rest of the chapter seems to stand up.
The abstract theory of
gauge integrals in \S482 is expressed in forms which need DC at least, but
I think that the basic facts about the Henstock
integral\cmmnt{ (\S483)} are
unaffected.   There are some interesting challenges in Chapter 49, but
there the eclectic nature of the arguments
means that we cannot expect much of the theory to keep its shape.
}%end of comment

\leader{566T}{}\cmmnt{ I give one result which may not be obvious and
helps to keep things in order.

\medskip

\noindent}{\bf Proposition} [AC($\omega$)] Let $I$ be any set, and $X$ a
separable metrizable space.
Then the Baire $\sigma$-algebra $\CalBa(X^I)$ of
$X^I$ is equal to the $\sigma$-algebra $\Tensorhat_I\Cal B(X)$ generated by
sets of the form $\{x:x(i)\in E\}$ for $i\in I$ and Borel sets
$E\subseteq X$.

\proof{{\bf (a)} Every open set in $X$ is a cozero set, so $\Cal B(X)=\CalBa(X)$
and $\{x:x(i)\in E\}\in\CalBa(X^I)$ whenever $i\in I$ and $E\in\Cal B(X)$;
accordingly $\Tensorhat_I\Cal B(X)\subseteq\CalBa(X^I)$.

\medskip

{\bf (b)} Fix a sequence $\sequencen{U_n}$ running
over a base for the topology of $X$.   For
$\sigma\in S=\bigcup_{J\in[I]^{<\omega}}\Bbb N^J$ set

\Centerline{$C_{\sigma}
=\{x:x\in X$, $x(i)\in U_{\sigma(i)}$ for every $i\in\dom\sigma\}
\in\Tensorhat_I\Cal B(X)$.}

\noindent Then $\{C_{\sigma}:\sigma\in S\}$ is a base for the topology of
$X^I$.   If $W\subseteq X^I$ is a regular open set, there is a countable
set $R\subseteq S$ such that $W=\bigcup_{\sigma\in R}C_{\sigma}$.
\Prf\ Let $R^*$ be the set of those $\sigma\in S$ such that
$C_{\sigma}\subseteq W$, and $R$ the set of minimal members of $R^*$
(ordering $S$ by extension of functions).   Then every member of $R^*$
extends some member of $R$, so

\Centerline{$\bigcup_{\sigma\in R}C_{\sigma}
=\bigcup_{\sigma\in R^*}C_{\sigma}=W$.}

\noindent For $n\in\Bbb N$ set
$R_n=\{\sigma:\sigma\in R$, $\#(\sigma)=n$, $\sigma(i)<n$ for every
$i\in\dom\sigma\}$.

\Quer\ Suppose, if possible, that $n\in\Bbb N$ and $R_n$ is infinite.
Then there is a sequence $\sequence{k}{\sigma_k}$ of distinct elements of
$R_n$;  set $J_k=\dom\sigma_k$ for each $k$.   Let $M\subseteq\Bbb N$ be an
infinite set such that $\family{k}{M}{J_k}$ is a $\Delta$-system with root
$J$ say.   Then there is a $\sigma\in n^J$ such that
$M'=\{k:k\in M$, $\sigma_k\restr J=\sigma\}$ is infinite.

In this case, however,

\Centerline{$C_{\sigma}
\subseteq\interior\overline{\bigcupop_{k\in M'}C_{\sigma_k}}
\subseteq\interior\overline{W}=W$}

\noindent and $\sigma\in R^*$, so that $\sigma_k\notin R$ for $k\in M'$;
which is impossible.\ \Bang

Thus every $R_n$ is countable and $R=\bigcup_{n\in\Bbb N}R_n$ is
countable.\  \Qed

\medskip

{\bf (c)} This shows that every regular open subset of $X^I$ is a countable
union of open cylinder sets and belongs to
$\Tensorhat_I\Cal B(X)$.   Consequently every cozero set belongs to
$\Tensorhat_I\Cal B(X)$.   \Prf\ If $f:X^I\to\Bbb R$ is continuous, then
for each rational $q>0$ set $W_q=\interior\{x:|f(x)|\ge q\}$.   Then
$W_q$ is a regular open set so belongs to
$\Tensorhat_I\Cal B(X)$.   But now
$\{x:f(x)\ne 0\}=\bigcup_{q\in\Bbb Q,q>0}W_q$ is the union of countably
many sets in $\Tensorhat_I\Cal B(X)$ and itself belongs to
$\Tensorhat_I\Cal B(X)$.\ \Qed

So $\Tensorhat_I\Cal B(X)\supseteq\CalBa(X^I)$ and the two are equal.
}%end of proof of 566T

\cmmnt{
\leader{566U}{Dependent choice} If we allow ourselves to use the stronger
principle DC rather than AC($\omega$) alone, we get some useful
simplifications.   The difficulties with the principle of exhaustion in
\S215 and 566D above
disappear, and there is no longer any obstacle to the
construction of product measures in 254F, provided only that we know we
have a non-empty product space.   So a typical theorem on product measures
will now begin `let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces such that $X=\prod_{i\in I}X_i$ is non-empty'.   Later,
we now have Baire's theorem (both for complete metric spaces and for
locally compact Hausdorff spaces) and Urysohn's Lemma (so we can drop the
formulation `completely regular locally compact topological group').
The most substantial gap in Volume 4
which is now filled seems to be in the abstract
theory of gauge integrals in \S482.
But I cannot point to a result which is essential to the structure of
this treatise and can be proved in ZF + DC but not in ZF + AC($\omega$).
}%end of comment

\exercises{\leader{566X}{Basic exercises (a)}
%\spheader 566Xa
[AC($\omega$)] Let $(X,\rho)$ be a metric space.
(i) Show that $X$
is compact iff it is sequentially compact iff it is countably compact
iff it is complete and totally bounded.  (ii) Show that if $X$ is separable
then every subspace of $X$ is separable.

\spheader 566Xb [AC($\omega$)] Show that there is a surjection from
$\Bbb R$ onto its Borel $\sigma$-algebra,
so that there must be a non-Borel subset of $\Bbb R$.
%566C

\spheader 566Xc\dvAnew{2014}(i) (Cf.\ 313K)
Let $\frak A$ be a Boolean algebra, and
$D\subseteq\frak A$ an order-dense set.   Show that
$a=\sup\{d:d\in D$, $d\Bsubseteq a\}$ for every $a\in\frak A$.
(ii) (Cf.\ 322Eb)
Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.   Show that
$a=\sup\{b:b\Bsubseteq a$, $\bar\mu b<\infty\}$ for every $a\in\frak A$.
%566M

\spheader 566Xd
Let us say that a Boolean algebra $\frak A$ has the
{\bf countable sup property} if for every $A\subseteq\frak A$ there is a
countable $B\subseteq A$ with the same upper bounds as $A$.
(i) Show that a Dedekind $\sigma$-complete Boolean
algebra with the countable sup property is Dedekind complete.
(ii) Show that a countably additive functional on a
Boolean algebra with the countable sup property is completely additive.
%566M

\spheader 566Xe [AC($\omega$)] Show that if there is a
translation-invariant lifting for Lebesgue measure then there is a subset
of $\Bbb R$ which is not Lebesgue measurable.   \Hint{345F.}
%345F

\spheader 566Xf [AC($\omega$)] Show that if $1<p<\infty$ and
$(\frak A,\bar\mu)$ is a
measure algebra, the unit ball of $L^p(\frak A,\bar\mu)$ (\S366) is weakly
compact.   \Hint{part (b) of the proof of 566Q.}
%566Q

\spheader 566Xg [AC($\omega$)] (i) Let $\frak A$ be a measurable algebra.
Show that the
unit ball of $L^{\infty}=L^{\infty}(\frak A)$ is compact for
$\frak T_s(L^{\infty},(L^{\infty})^{\times})$ (definition: 3A5Ea).
(ii) Let $U$ be an $L$-space with a weak order unit.   Show
that the unit ball of $U^*$ is weak*-compact.   \Hint{561Hb.}
%566Q

\spheader 566Xh Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   Show that there is a function
$f:\Aut\frak A\times\frak A\to\frak A$ such that if $\pi\in\Aut\frak A$ and
$a$ is a separator for $\pi$ then $a\Bcap f(\pi,a)=0$ and
$f(\pi,a)\Bcup\pi f(\pi,a)\Bcup\pi^2f(\pi,a)$ is the support of $\pi$.
\Hint{382D.}
%566R

\spheader 566Xi Let $\frak A$ be a Dedekind complete Boolean
algebra and $G$ a well-orderable subgroup of $\Aut\frak A$.
Let $G^*$ be the full subgroup of $\Aut\frak A$ generated by $G$.
Show that there is a function
$f:G^*\times G\to\frak A$ such that $\family{\phi}{G}{f(\pi,\phi)}$ is a
partition of unity for each $\pi\in G^*$ and $\pi a=\phi a$ whenever
$\pi\in G^*$, $\phi\in G$ and $a\Bsubseteq f(\pi,\phi)$.   \Hint{381I.}
%566R

\spheader 566Xj [AC($\omega$)]
Let $\frak A$ be a Dedekind complete Boolean algebra and
$G$ a countable subgroup of $\Aut\frak A$ such that every
member of $G$ has a separator.   Let $G^*$ be the full subgroup
of $\Aut\frak A$ generated by $G$.    Show that there is a
function
$g:G^*\to\frak A$ such that $g(\pi)$ is a separator for $\pi$ for every
$\pi\in G^*$.   \Hint{566Xi, 382Id.}
%566R

\spheader 566Xk [AC($\omega$)] Let $(X,\frak T)$ be a completely regular
locally compact Hausdorff space, and $f:C_k(X)\to\Bbb R$ a positive
linear functional.   Show that
there is a unique Radon measure $\mu$ on $X$ such that
$f(u)=\int u\,d\mu$ for every $u\in C_k(X)$.
%436J

\spheader 566Xl [AC($\omega$)]
Say that a set $X$ is {\bf measure-free} if whenever $\mu$
is a probability measure with domain $\Cal PX$ there is an $x\in X$ such
that $\mu\{x\}>0$.   (i) Show that
the following are equiveridical:
($\alpha$) $\Bbb R$ is not measure-free;
($\beta$) there is a semi-finite measure space $(X,\Cal PX,\mu)$ which
is not purely atomic;
($\gamma$) there is a measure $\mu$ on $[0,1]$ extending Lebesgue
measure and measuring every subset of $[0,1]$.
(ii) Prove 438B for point-finite families $\familyiI{E_i}$ such that
the index set $I$ is measure-free.
%438B

\leader{566Y}{Further exercises (a)}
%\spheader 566Ya
[AC($\omega$)] Show that if $U$ is an $L$-space, and
$\sequencen{u_n}$ is a bounded sequence in $U$, then there are a
subsequence $\sequencen{v_n}$ of $\sequencen{u_n}$ and a $w\in U$ such that
$\sequencen{\bover1{n+1}\sum_{i=0}^nw_i}$ is order*-convergent to $w$ for
every subsequence $\sequencen{w_n}$ of $\sequencen{v_n}$.   \Hint{in the
proof of 276H, show that we can find a countably-generated filter to
replace the ultrafilter $\Cal F$.}
%276H

\spheader 566Yb [AC($\omega$)] Let
$X$ be a completely regular compact Hausdorff topological group and $\mu$ a
left Haar measure on $X$.   Show that if $w\in L^2(\mu)$ then
$u\mapsto u*w:L^2(\mu)\to C(X)$ is a
compact linear operator.   \Hint{444V.}
%444

\spheader 566Yc [AC($\omega$)]
Let $\familyiI{(X_i,\sequencen{U_{in}})}$ be a family such
that $X_i$ is a separable metrizable space and $\sequencen{U_{in}}$ is a
base for the topology of $X_i$ for each $i\in I$.   Show that
$\CalBa(\prod_{i\in I}X_i)=\Tensorhat_{i\in I}\Cal B(X_i)$.
%566S

\spheader 566Yd
[DC] Let $U$ be an inner product space and $K\subseteq U$
a convex weakly compact set.   Show that $K$ has an extreme point.
}%end of exercises

\leader{566Z}{Problem} Is it relatively consistent with ZF + AC($\omega$)
to suppose that there is a non-zero atomless rigid measurable algebra?

\endnotes{
\Notesheader{566} In this section I have taken a lightning tour through the
material of Volumes 1 to 4, pausing over a rather odd selection of results,
mostly chosen to exhibit the alternative arguments which are available.
In the first place, I am trying to suggest something of the quality of the
world of measure theory, and of analysis in general, under this particular
set of rules.   Perhaps I should say that my real objective is the next
section, with DC rather than AC($\omega$), because DC is believed to be
compatible with the axiom of determinacy, and ZF + DC + AD is not a poor
relation of ZFC, as ZF + AC($\omega$) sometimes seems to be, but a
potential rival.

I have a second reason for taking all this trouble,
which is a variation on one of the reasons
for `generalization' as found in
twentieth-century pure mathematics.   When we `generalize' an argument,
moving (for example) from metric spaces to topological spaces, or from
Lebesgue measure to abstract measures, we are usually stimulated by some
particular question which demands the new framework.   But the process
frequently has a lasting value which is quite independent of its
motivation.
It forces us to re-examine the nature of the proofs we are using,
discarding or adapting those steps which depend on the original context,
and isolating those which belong in some other class of ideas.   In the
same way, renouncing the use of AC forces us to look more closely at
critical points, and decide which of them correspond to some deeper
principle.

Something I have not attempted to do is to look for models in which my
favourite theorems are actually false.
An interesting class of problems is concerned with
`exact engineering', that is, finding combinatorial propositions which will
be equivalent, in ZF, to given results which are not provable in ZF.   For
instance, Baire's theorem for complete metric spaces is actually equivalent
to DC ({\smc Blair 77}), while Baire's theorem for compact Hausdorff spaces
may be weaker ({\smc Fossy \& Morillon 98}).   I am not presenting any such
results here.   However, if we take Maharam's theorem as an example of a
central result of measure theory with ZFC which is surely unprovable
without a strong form of AC, we can ask just how false it can be;  and I
offer 566Z as a sample target.
}%end of notes

\discrpage

\frfilename{mt567.tex}
\versiondate{31.10.14}
\copyrightdate{2006}

\def\Game{\mathop{\text{Game}}\nolimits}
\def\StrI{\mathop{\text{Str}}\nolimits_{\text{I}}}
\def\StrII{\mathop{\text{Str}}\nolimits_{\text{II}}}
\def\WO{\mathop{\text{WO}}}

\def\chaptername{Choice and determinacy}
\def\sectionname{Determinacy}

\newsection{567}

So far, this chapter has been looking at set theories which are weaker than
the standard theory ZFC, and checking which of the principal results of
measure theory can still be proved.   I now turn to an axiom which directly
contradicts the axiom of choice, and leads to a very different world.
This is AD, the `axiom of determinacy', defined in terms of strategies for
infinite games (567A-567C).  %567A 567B 567C
The first step is to confirm that we automatically have a weak version of
countable choice which is enough to make Lebesgue measure well-behaved
(567D-567E).   Next, in separable metrizable spaces all
subsets are universally measurable and have the Baire property (567G).
Consequently (at least when we can use AC($\omega$)) linear operators
between Banach spaces are bounded (567H), additive functionals on
$\sigma$-complete Boolean algebras are countably additive (567J), and
many $L$-spaces are reflexive (567K).   In a different direction, we
find that $\omega_1$ is \2vm\ (567L) and that there are many surjections
from $\Bbb R$ onto ordinals (567M).

At the end of the section I include two celebrated results in ZFC
(567N, 567O) which depend on some of the same ideas.

\leader{567A}{Infinite games} \cmmnt{ I return to
an idea introduced in \S451. %451V

\medskip

}{\bf (a)} Let $X$ be a non-empty set and $A$ a subset of
$X^{\Bbb N}$.   In the corresponding infinite game $\Game(X,A)$, players I
and II choose members of $X$ alternately, so that I chooses
$x(0),x(2),\ldots$ and II chooses $x(1),x(3),\ldots$;  a {\bf play} of the
game is an element of $X^{\Bbb N}$;  player I wins the play $x$ if
$x\in A$, otherwise II wins.   A {\bf strategy} for I is a function
$\sigma:\bigcup_{n\in\Bbb N}X^n\to X$;  a play $x\in X^{\Bbb N}$ is {\bf
consistent} with $\sigma$ if
$x(2n)=\sigma(\ofamily{i}{n}{x(2i+1)})$ for every $n$\cmmnt{, that is,
if I uses the function $\sigma$ to decide his move from the previous
moves by his opponent};
$\sigma$ is a {\bf winning strategy} if every play consistent with $\sigma$
belongs to $A$\cmmnt{, that is, if I wins whenever he follows the
strategy $\sigma$}.   Similarly, a strategy for II is a function
$\tau:\bigcup_{n\ge 1}X^n\to X$;  a play $x$ is consistent with $\tau$ if
$x(2n+1)=\tau(\langle x(2i)\rangle_{i\le n})$ for every $n$;  and $\tau$
is a winning strategy for II if
$x\notin A$ whenever $x\in X^{\Bbb N}$ and $x$ is consistent with $\tau$.

\spheader 567Ab A set $A\subseteq X^{\Bbb N}$ is {\bf determined} if
either I or II has a winning strategy in $\Game(X,A)$.
\cmmnt{Note that we need to know the set $X$ as well as the set $A$ to
specify the game in question.}

\spheader 567Ac It will sometimes be convenient to describe games with
`rules', so that the players are required to choose points in subsets
of $X$ (determined by the moves so far) at each move.   Such a description
can be regarded as specifying $A$ in the form $(A'\cup G)\setminus H$,
where $G$ is the set of plays in which II is the first to break a rule, $H$
is the set of plays in which I is the first to break a rule, and $A'$ is
the set of plays in which both obey the rules and I wins.

\spheader 567Ad Not infrequently the `rules' will specify different sets
for the moves of the two players, so that I always chooses a point in
$X_1$ and II always chooses a point in $X_2$;
setting $X=X_1\cup X_2$ we can reduce this
to the formalization above.

\leader{567B}{Theorem} Let $X$ be a non-empty well-orderable set.
Give $X$ its discrete topology and $X^{\Bbb N}$ the product topology.
If $F\subseteq X^{\Bbb N}$ is closed then $\Game(X,F)$ is determined.

\proof{{\bf (a)} Fix a well-ordering $\preccurlyeq$ of $X$.
Define $\family{\xi}{\On}{W_{\xi}}$ by setting

\Centerline{$W_0
=\{w:w\in\bigcup_{n\in\Bbb N}X^{2n+1}$, $w\not\subseteq x$ for any
$x\in F\}$,}

$$\eqalign{W_{\xi}
&=\{w:w\in\bigcup_{n\in\Bbb N}X^{2n+1}
  \text{, there is some }t\in X\text{ such that}\cr
&\mskip200mu w^{\smallfrown}\fraction{t}^{\smallfrown}\fraction{u}
   \in\bigcup_{\eta<\xi}W_{\eta}
\text{ for  every }u\in X\}\cr}$$

\noindent if $\xi>0$.  (See 5A1C for the notation here.)
If $w\in W_0$ then of course
$w^{\smallfrown}\fraction{t}^{\smallfrown}\fraction{u}\in W_0$ for
all $t$, $u\in X$;  so $W_0\subseteq W_1$, and of course
that $W_{\xi}\subseteq W_{\xi'}$ whenever $1\le\xi\le\xi'$ in $\On$.
There is therefore an ordinal $\zeta$ such that $W_{\zeta+1}=W_{\zeta}$;
write $W$ for $W_{\zeta}$.

For $w\in W$, let $r(w)\le\zeta$ be the least ordinal such that
$w\in W_{r(w)}$.   If $r(w)>0$ then there is some $t\in X$ such that
such that
$w^{\smallfrown}\fraction{t}^{\smallfrown}\fraction{u}
\in\bigcup_{\eta<r(w)}W_{\eta}$, that
is, $r(w^{\smallfrown}\fraction{t}^{\smallfrown}\fraction{u})<r(w)$, for
every $u\in X$.

Let
$V$ be the set of those $v\in\bigcup_{n\in\Bbb N}X^{2n}$ such that there is
a $u\in X$ such that $v^{\smallfrown}\fraction{u}\notin W$.
Observe that if $w\in\bigcup_{n\in\Bbb N}X^{2n+1}\setminus W$ then
$w\notin W_{\zeta+1}$ so $w^{\smallfrown}\fraction{t}\in V$ for every
$t\in X$.

\medskip

{\bf (b)} Suppose that $\emptyset\in V$.   Define
$\sigma:\bigcup_{n\in\Bbb N}X^n\to X$ inductively by saying that

\inset{$\sigma(\emptyset)$ is the $\preccurlyeq$-least member
$t$ of $X$ such
that the one-element sequence $\emptyset^{\smallfrown}\fraction{t}$
does not belong to $W$,

if $v\in X^{n+1}$ and
$w=(\sigma(v\restr 0),v(0),\sigma(v\restr 1),v(1),\ldots,
\sigma(v\restr n),v(n))\in V$,
take $\sigma(v)$ to be the $\preccurlyeq$-least member $t$ of $X$ such that
$w^{\smallfrown}\fraction{t}\notin W$,

for other $v\in X^{n+1}$ take $\sigma(v)$ to be the $\preccurlyeq$-least
member of $X$.}

\noindent Then $\sigma$ is a winning strategy for I.   \Prf\ If
$x$ is a play consistent with $\sigma$, then an induction on $n$ shows that
$x\restr 2n\in V$ and $x\restr 2n+1\notin W$ for every $n$.   In
particular, $x\restr 2n+1\notin W_0$, that is, there is a member of $F$
extending $x\restr 2n+1$, for every $n$.   As $F$ is closed, $x\in F$ and I
wins the play $x$.\ \Qed

\medskip

{\bf (c)} Suppose that $\emptyset\notin V$, that is, $w\in W$ for every
$w\in X^1$.   Define
$\tau:\bigcup_{n\ge 1}X^n\to X$ inductively by saying

\inset{if $v\in X^n$ and
$w=(v(0),\tau(v\restr 1),v(1),\tau(v\restr 2),\ldots,v(n-1))$ belongs to
$W\setminus W_0$,
then $\tau(v)$ is the $\preccurlyeq$-least $t\in X$ such that
$r(w^{\smallfrown}\fraction{t}^{\smallfrown}\fraction{u})<r(w)$
for every $u\in X$,

for other $v\in X^n$, $\tau(v)$ is the $\preccurlyeq$-least member of $X$.}

\noindent Then $\tau$ is a winning strategy for II.   \Prf\ Let $x$ be a
play consistent with $\tau$.   Then an induction on $n$ tells us that

\Centerline{$x\restr 2n+1\in W$,
\quad if $x\restr 2n+1\notin W_0$ then $r(x\restr 2n+3)<r(x\restr 2n+1)$}

\noindent for every $n\in\Bbb N$.  Since $\sequencen{r(x\restr 2n+1)}$
cannot be strictly decreasing, there is some $n\in\Bbb N$ such that
$x\restr 2n+1\in W_0$ and $x\notin F$.   Thus II wins the play $x$.\ \Qed

\medskip

{\bf (d)} Putting (b) and (c) together we see that $F$ is determined.
}%end of proof of 567B

\vleader{36pt}{567C}{The axiom of determinacy (a)}
The standard `axiom of determinacy' is the statement

\Centerline{(AD)\quad Every subset of $\BbbN^{\Bbb N}$ is determined.}

\noindent Evidently it will follow that every subset of $X^{\Bbb N}$ is
determined for any countable set $X$.  \cmmnt{(If $X\subseteq\Bbb N$,
a game on $X$ can be regarded as a game on $\Bbb N$ in which there is a
rule that the players must always choose points in $X$.   See also 567Xc.)}

\spheader 567Cb At the same time, it will be useful to consider a weak form
of the axiom of countable choice:   for any set $X$, write
AC($X;\omega$) for the statement

\Centerline{$\prod_{n\in\Bbb N}A_n\ne\emptyset$ whenever $\sequencen{A_n}$
is a sequence of non-empty subsets of $X$.}

\leader{567D}{Theorem}\cmmnt{ ({\smc Mycielski 64})}
AD implies AC($\Bbb R;\omega$).

\proof{ Since we know that $\Bbb R$ is equipollent with $\NN$, we can look
at AC($\NN;\omega$).
Let $\sequencen{A_n}$ be a sequence of non-empty subsets of $\NN$.   Set

\Centerline{$A
=\{x:x\in\BbbN^{\Bbb N}$, $\sequencen{x(2n+1)}\notin A_{x(0)}\}$.}

\noindent Then I has no winning strategy in $\Game(\Bbb N,A)$,
because if $\sigma$
is a strategy for I in $\Game(\Bbb N,A)$ set $k=\sigma(\emptyset)$;
there is a
point $y\in A_k$, and II need only play $x(2n+1)=y(n)$ for each $n$.

So II has a winning strategy $\tau$ say.   Define
$g:\Bbb N\to\BbbN^{\Bbb N}$ by saying that
$g(n)(i)=\tau(e_{ni})$ for $n$, $i\in\Bbb N$,
where $e_{ni}\in\BbbN^{i+1}$, $e_{ni}(0)=n$,
$e_{ni}(j)=0$ for $1\le j\le i$.   If now $n\in\Bbb N$,
I plays $(n,0,0,\ldots)$ and
II follows the strategy $\tau$, the resulting play
$(n,g(n)(0),0,g(n)(1),0,\ldots)$ must not belong to $A$ so $g(n)\in A_n$.
}%end of proof of 567D

\leader{567E}{Consequences of AC($\Bbb R;\omega$)} Suppose that
AC($\Bbb R;\omega$) is true.

\spheader 567Ea If a set $X$ is the image of a subset $Y$ of $\Bbb R$
under a function $f$, then AC($X;\omega$) is true.   \prooflet{\Prf\ If
$\sequencen{A_n}$ is a sequence of non-empty subsets of $X$, then
there is an $x\in\prod_{n\in\Bbb N}f^{-1}[A_n]$, and
$\sequencen{f(x(n))}\in\prod_{n\in\Bbb N}A_n$.\ \Qed}

\spheader 567Eb In particular, taking
$S^*=\bigcup_{n\ge 1}\BbbN^n$\cmmnt{ as
in \S562}, AC($\Cal PS^*;\omega$) is true.   It follows that (in any
second-countable space $X$) every sequence of
codable Borel sets is codable and the family of codable Borel sets is a
$\sigma$-algebra, coinciding with the Borel $\sigma$-algebra $\Cal B(X)$
on its
ordinary definition.   Moreover, since $\Cal B(X)$ is an image of
$\Cal PS^*$, we have AC($\Cal B(X);\omega$)\cmmnt{,
countable choice for collections
of Borel sets}.  Similarly, the family of codable Borel functions
becomes the ordinary family of Borel-measurable functions, and we have
countable choice for sets of Borel real-valued functions on $X$.

\spheader 567Ec Consequently the results of \S562-565 give us
large parts of the elementary theory of Borel measures on
second-countable spaces.   At the same time, if $X$ is second-countable,
the union of a sequence of meager subsets of $X$ is
meager\cmmnt{ (because we have
countable choice for sequences of nowhere dense closed sets)}, so the
Baire-property algebra of $X$ is a $\sigma$-algebra.

\spheader 567Ed We also find that the supremum of a sequence
of countable
ordinals is again countable.   \prooflet{\Prf\ Let $\sequencen{\xi_n}$
be a sequence in $\omega_1$.   Using
AC($\Bbb R;\omega)$), we can choose for each $n\in\Bbb N$ a subset
$\preccurlyeq_n$ of $\Bbb N\times\Bbb N$ which is a well-ordering
of $\Bbb N$ with order type $\max(\omega,\xi_n)$.   Now we have a
well-ordering $\preccurlyeq$ of $\Bbb N^2$ defined by saying that
$(i,j)\preccurlyeq(i',j')$ if $i<i'$ or $i=i'$ and $j\preccurlyeq_ij'$.
In this case, the order type $\xi$ of $\preccurlyeq$ will be greater than
or equal to
every $\xi_n$, so that $\sup_{n\in\Bbb N}\xi_n\le\xi$ is
countable.\ \Qed}

\leader{567F}{Lemma}\cmmnt{ (see {\smc Mycielski \& \'Swierczkowski 64})}
[AC($\Bbb R;\omega$)] Suppose that
$A\subseteq\{0,1\}^{\Bbb N}$ is a continuous image of a subset $B$ of
$\{0,1\}^{\Bbb N}$ such that $(h^{-1}[B]\cap F)\cup H\subseteq\NN$
is determined
whenever $h:\BbbN^{\Bbb N}\to\{0,1\}^{\Bbb N}$ is continuous,
$F\subseteq\BbbN^{\Bbb N}$ is closed and $H\subseteq\BbbN^{\Bbb N}$ is
open.

(a) $A$ is universally measurable.

(b) $A$ has the Baire property in $\{0,1\}^{\Bbb N}$.

\proof{ Fix a continuous surjection $f:B\to A$.   Let $\Cal E$ be the
countable algebra
of subsets of $\{0,1\}^{\Bbb N}$ determined by coordinates in finite
sets, that is to say, the algebra of open-and-closed subsets of
$\{0,1\}^{\Bbb N}$ (311Xh).

\medskip

{\bf (a)(i)} Let $\mu$ be a Borel probability measure on
$\{0,1\}^{\Bbb N}$ and $\hat\mu$ its completion.
If $Z\subseteq\{0,1\}^{\Bbb N}$ is closed and not
negligible, then at least one of $Z\cap A$, $Z\setminus A$
has non-zero inner measure.

\Prf\ Let $\sequencen{E_n}$ enumerate
$\Cal E$.   Set $\epsilon_n=2^{-2n-2}\mu Z$ for $n\in\Bbb N$.
In $(\{0,1\}\times\Cal E)^{\Bbb N}$ consider the
game in which the players choose $(k_0,K_0),(k_1,K_1),\ldots$ such that
$K_0=Z$ and for each $n\in\Bbb N$

\Centerline{$k_n\in\{0,1\}$,
\quad$K_n\in\Cal E$,
\quad$\mu K_{2n+1}\le\epsilon_n$.}

\noindent I wins if $y=\sequencen{k_{2n}}$ belongs to $B$ and
$f(y)\notin\bigcup_{n\in\Bbb N}K_{2n+1}$.   Observe that when $y\in B$,
$f(y)\in\bigcup_{n\in\Bbb N}K_{2n+1}$ iff there is an $m\in\Bbb N$
such that $f(w)\in\bigcup_{i<m}K_{2i+1}$ whenever $w\in B$ and
$w\restr m=y\restr m$;  so  I
wins iff $y\in B$ and at every stage $((k_0,K_0),\ldots,(k_{2m},K_{2m}))$
there is a $w\in B$ such that $w(i)=k_{2i}$ for $i<m$ and
$f(w)\notin\bigcup_{i<m}K_{2i+1}$.   So the payoff set $D$
of plays $\sequencen{(k_n,K_n)}$ won by I is of the form
$(h^{-1}[B]\cap F)\cup H$ where
$h:(\{0,1\}\times\Cal E)^{\Bbb N}\to\{0,1\}^{\Bbb N}$ is
continuous, $F\subseteq (\{0,1\}\times\Cal E)^{\Bbb N}$ is closed and
$H\subseteq(\{0,1\}\times\Cal E)^{\Bbb N}$ is open.
(Here $H$ is
the set of plays which are won because II is the first to break a rule.)
Consequently $D$ is determined.

\medskip

\qquad{\bf case 1} Suppose that I has a winning strategy $\sigma$.
For each play $\sequencen{(k_n,K_n)}$ consistent with $\sigma$,
$f(\sequencen{k_{2n}})$ is defined and belongs to $A$.
Since the set of plays consistent with $\sigma$ is a closed subset of
$(\{0,1\}\times\Cal E)^{\Bbb N}$, the set $C$ of points obtainable in this
way is an analytic subset of $Z$, therefore measured by $\hat\mu$ (563I).
\Quer\ If $\hat\mu C=0$, then there is an open set $G\supseteq C$ such that
$\mu G<\epsilon_0$ (563Fd).   In this case, II can play in such a way that

\Centerline{$K_{2n+1}\subseteq G$,
\quad$\mu(G\setminus\bigcup_{i\le n}K_{2i+1})<\epsilon_{n+1}$,}

\Centerline{if
$E_n\subseteq G$ then $E_n\subseteq\bigcup_{i\le n}K_{2i+1}$}

\noindent for every $n$.   But now, taking I's responses under $\sigma$,
we have a play of $\Game(\{0,1\}\times\Cal E,D)$
in which $\bigcup_{n\in\Bbb N}K_{2n+1}=G$
includes $C$, so contains $f(\sequencen{k_n})$, and is won by II;  which is
supposed to be impossible.\ \Bang

So in this case $\mu_*A\ge\mu C>0$.

\medskip

\qquad{\bf case 2} Suppose that II has a winning strategy $\tau$.
For each $n\in\Bbb N$ and $u\in\{0,1\}^n$, let $L(u)$ be the second
component of
$\tau(\ofamily{i}{n}{(u(i),\emptyset)})$;  set
$G=\bigcup_{n\in\Bbb N}\bigcup_{u\in\{0,1\}^n}L(u)$, so that
$\mu G\le\sum_{n=0}^{\infty}2^n\epsilon_n<\mu Z$.   If we take any
$y\in B$, then we have a play $\sequencen{(k_n,K_n)}$
of $\Game(\{0,1\}\times\Cal E,D)$,
consistent with $\tau$, in which $k_{2n}=y(n)$ and $K_{2n}=\emptyset$ for
each $n$.   Since II wins this play, $f(y)$ must belong to

\Centerline{$\bigcup_{n\in\Bbb N}K_{2n+1}
=\bigcup_{n\in\Bbb N}L(y\restr n)\subseteq G$.}

\noindent As $y$ is arbitrary, $A\subseteq G$
and $\mu_*(Z\setminus A)\ge\mu(Z\setminus G)>0$.\ \Qed

\medskip

\quad{\bf (ii)} Write $\Cal K$ for the family of compact sets
$K\subseteq\{0,1\}^{\Bbb N}$ such that $A\cap K$ is Borel.
If $E\subseteq\{0,1\}^{\Bbb N}$ and $\mu E>0$, there is a
$K\in\Cal K$ such that $K\subseteq E$ and $\mu K>0$.   \Prf\ There is a
closed $Z\subseteq E$ such that $\mu Z>0$ (563Fd again).   By (i),
at least one of $\mu_*(Z\cap A)$,
$\mu_*(Z\setminus A)$ is non-zero, and there is a compact
set $K$ of non-zero measure which is included in one of $Z\cap A$,
$Z\setminus A$.   But now $K\in\Cal K$.\ \Qed

Now (because we have countable choice for subsets of $\Cal K$)
there is a
sequence $\sequencen{K_n}$ in $\Cal K$ such that
$\sup_{n\in\Bbb N}\mu K_n=\sup_{K\in\Cal K}\mu K$;  setting
$E=\{0,1\}^{\Bbb N}\setminus\bigcup_{n\in\Bbb N}K_n$, $E$ must be
negligible, while $A\setminus E$ is a Borel set;  so $A$ is measured by
$\hat\mu$.   As $\mu$ is arbitrary, $A$ is universally measurable.

\medskip

{\bf (b)(i)} If $V\in\Cal E\setminus\{\emptyset\}$ then
either $V\cap A$ is meager or there is a
$V'\in\Cal E\setminus\{\emptyset\}$ such that $V'\subseteq V$
and $V'\setminus A$ is meager.   \Prf\ Set
$\Cal U=\{E:E\in\Cal E\setminus\{\emptyset\}$, $E\subseteq V\}$
and let $\preccurlyeq$ be a
well-ordering of $\Cal U$ (in order type $\omega$, if you like).
Consider the game on $\{0,1\}\times\Cal U$
in which the players choose $(k_0,U_0),(k_1,U_1),\ldots$ such that,
for each $n\in\Bbb N$,

\Centerline{$k_n\in\{0,1\}$,
\quad$U_n\in\Cal U$,
\quad$U_{n+1}\subseteq U_n$.}

\noindent I wins if $y=\sequencen{k_{2n}}$ belongs to $B$ and
$f(y)\in\bigcap_{n\in\Bbb N}U_n$.   Because the $U_n$ are all
open-and-closed,
this game is determined for the same reasons as the game of (a).

\medskip

\qquad{\bf case 1} Suppose that I has a winning strategy $\sigma$;
say that $\sigma_2(w)\in\Cal U$ is the second component of $\sigma(w)$ for each
$w\in\bigcup_{n\in\Bbb N}(\{0,1\}\times\Cal U)^n$.   For each $n\in\Bbb N$
let $\Cal U_n$ be the set of those $U\in\Cal U$ such that
$v\restr n=v'\restr n$ for all $v$, $v'\in U$.   Let
$(k',V')=\sigma(\emptyset)$ be
I's first move when following $\sigma$.   Let $Q$ be the set of
positions in the game consistent with $\sigma$ and with II to move,
that is, finite sequences

\Centerline{$q=\langle(k_i,U_i)\rangle_{i\le 2n}
\in(\{0,1\}\times\Cal U)^{2n+1}$}

\noindent such that
$(k_{2m},U_{2m})=\sigma(\langle(k_{2i+1},U_{2i+1})\rangle_{i<m})$
for every $m\le n$ and $\langle U_i\rangle_{i\le 2n}$ is non-increasing.
For such a $q$, set $V_q=U_{2n}$ and

\Centerline{$W_q
=\bigcup\{\sigma_2({\langle(k_{2i+1},U_{2i+1})\rangle_{i<n}}
   ^{\smallfrown}\fraction{(k,U)}):
k\in\{0,1\}$, $U\in\Cal U_n$, $U\subseteq V_q\}$.}

\noindent Then $W_q$ is an open subset of $V_q$;  but also it is dense in
$V_q$, because if $W\subseteq V_q$ is open and not empty there is a
$U\in\Cal U_n$ included in $W$ and
$\sigma_2({\langle(k_{2i+1},U_{2i+1})\rangle_{i<n}}^{\smallfrown}
  \fraction{(k,U)})$
is a non-empty subset of $U$.    $Q$ is countable, so
$E=\bigcap_{q\in Q}W_q\cup(\{0,1\}^{\Bbb N}\setminus\overline{V}_q)$ is
comeager in $\{0,1\}^{\Bbb N}$.

\Quer\ If $V'\setminus A$ is not meager, there is an
$x\in E\cap V'\setminus A$.
Define $\sequencen{(k_n,U_n)}$ inductively,
as follows.   $(k_0,U_0)=(k',V')$.   Given that
$q=\langle(k_i,U_i)\rangle_{i\le 2n}$ belongs to $Q$ and $x\in V_q$,
then $x\in W_q$ so there are $k\in\{0,1\}$, $U\in\Cal U_n$ such that $x
\in\sigma_2({\langle(k_{2i+1},U_{2i+1})\rangle_{i<n}}^{\smallfrown}
  \fraction{(k,U)})$;
take the
lexicographically first such pair $(k,U)$ for $(k_{2n+1},U_{2n+1})$,
and set
$(k_{2n+2},U_{2n+2})=\sigma(\langle(k_{2i+1},U_{2i+1})\rangle_{i\le n})$.
Then $q'=\langle(k_i,U_i)\rangle_{i\le 2n+2}$ belongs to $Q$
and $V_{q'}=U_{2n+2}=\sigma_2(\langle(k_{2i+1},U_{2i+1})\rangle_{i\le n})$
contains $x$, so the induction can continue.

At the end of this induction, $\sequencen{(k_n,U_n)}$ will be a play of the
game consistent with $\sigma$ in which the only point of
$\bigcap_{n\in\Bbb N}U_n$ is $x$ and does not belong to $A$.
So either $y=\sequencen{k_{2n}}$ does not belong to $B$ or
$f(y)\notin\bigcap_{n\in\Bbb N}U_n$;  in either case, II wins the play;
which is supposed to be impossible.\ \Bang

So in this case $V'\setminus A$ is meager.

\medskip

\qquad{\bf case 2} Suppose that II has a winning strategy $\tau$;
say that $\tau_2(w)$ is the second component of $\tau(w)$ for each
$w\in\bigcup_{n\ge 1}(\{0,1\}\times\Cal U)^n$.
Let $Q$ be the set of objects

\Centerline{$q=(\ofamily{i}{2n}{(k_i,U_i)},k)$}

\noindent such that $\ofamily{i}{2n}{(k_i,U_i)}$ is a finite sequence in
$\{0,1\}\times\Cal U$ consistent with $\tau$ (allowing the empty string
when $n=0$) and $k\in\{0,1\}$.   For such a $q$, set $V_q=U_{2n-1}$ (if
$n>0$) or $V_q=V$ (if $n=0$);  set

\Centerline{$W_q
=\bigcup\{\tau_2({\ofamily{i}{n}{(k_{2i},U_{2i})}}^{\smallfrown}
  \fraction{(k,U)}):
U\in\Cal U$, $U\subseteq V_q\}$,}

\noindent so that $W_q$ is a dense subset of $V_q$.   $Q$ is countable, so
$E=\bigcap_{q\in Q}W_q\cup(\{0,1\}^{\Bbb N}\setminus\overline{V}_q)$ is
comeager.

\Quer\ If there is an $x$ in $A\cap V\cap E$, let
$y\in B$ be such that $f(y)=x$, and define
$\sequencen{(k_n,U_n)}$ as follows.   Given that
$q=(\ofamily{i}{2n}{(k_i,U_i)},y(n))$ belongs to $Q$ and $x\in V_q$,
then $x\in W_q$ so there is a $U\in\Cal U$ such that
$x\in\tau_2({\ofamily{i}{n}{(k_{2i},U_{2i})}}^{\smallfrown}
  \penalty-100\fraction{(y(n),U)})$;
take the
$\preccurlyeq$-first such $U$ for $U_{2n}$, set $k_{2n}=y(n)$ and
$(k_{2n+1},U_{2n+1})=\tau(\langle(k_{2i},U_{2i})\rangle_{i\le n})$,
so that $q'=(\langle(k_i,U_i)\rangle_{i\le 2n+1},\penalty-100y(n+1))$
belongs to $Q$ and
$V_{q'}=U_{2n+1}=\tau_2(\langle(k_{2i},U_{2i})\rangle_{i\le n})$
contains $x$.

At the end of this induction, $\sequencen{(k_n,U_n)}$ will be a play of the
game consistent with $\tau$ in which
$f(\sequencen{k_{2n}})=x\in\bigcap_{n\in\Bbb N}U_n$, so that I wins, which
is supposed to be impossible.\ \Bang

Thus in this case $A\cap V$ must be meager.\ \Qed

\medskip

\quad{\bf (ii)} Now let $G$ be the union of those $V\in\Cal E$
such that $V\setminus A$ is meager;  then
$G\setminus A$ is meager.   (This is where we need AC($\Bbb R;\omega$).)
If $V\in\Cal E$ and $V\subseteq\{0,1\}^{\Bbb N}\setminus G$,
then $V'\setminus A$ is non-meager for every whenever
$V\in\Cal E\setminus\{\emptyset\}$ and $V'\subseteq V$,
so $V\cap A$ is meager;  accordingly $G'\cap A$ is meager, where
$G'=\{0,1\}^{\Bbb N}\setminus\overline{G}$.   But this means that
$G\symmdiff A
\subseteq(G\setminus A)\cup(G'\cap A)\cup(\overline{G}\setminus G)$ is
meager and $A$ has the Baire property.
}%end of proof of 567F

\leader{567G}{Theorem} [AD] In any Hausdorff
second-countable space, every subset is
universally measurable and has the Baire property.

\proof{
Let $X$ be a Hausdorff second-countable space, $\sequencen{U_n}$ a sequence
running over a base for the topology of $X$, and $A\subseteq X$.

\medskip

{\bf (a)} Define
$g:X\to\{0,1\}^{\Bbb N}$ by setting $g(x)=\sequencen{\chi U_n(x)}$ for
$x\in X$;  then $g$ is injective and Borel measurable.
If $\mu$ is a Borel probability measure on $X$, we have a Borel
probability measure $\nu=\mu g^{-1}\restr\Cal B(\{0,1\}^{\Bbb N})$
on $\{0,1\}^{\Bbb N}$.   By 567Fa, $g[A]$ is measured by the completion
$\hat\nu$ of $\nu$;   let $F$, $H\subseteq\{0,1\}^{\Bbb N}$ be Borel sets
such that $\nu H=0$ and $g[A]\symmdiff F\subseteq H$;  then
$A\symmdiff g^{-1}[F]\subseteq g^{-1}[H]$ is $\mu$-negligible, so $A$ is
measured by $\hat\mu$.   As $\mu$ is arbitrary, $A$ is universally
measurable.

\medskip

{\bf (b)} Set
$G=\bigcup\{U_n:n\in\Bbb N$, $U_n\cap A$ has the Baire property$\}$,
so that $G\cap A$ has the Baire property.   (Remember that as we have a
bijection between $X$ and a subset of $\Bbb R$, we have
countable choice for subsets of $X$, so that the ideal of meager subsets
of $X$
is a $\sigma$-ideal and the Baire-property algebra is a $\sigma$-algebra.)
Set
$V=X\setminus(\bigcup_{n\in\Bbb N}\partial U_n\cup\overline{G})$;  then
$G\cup V$ is comeager in $X$, and $A\setminus V$ has the Baire
property.   If $V$ is empty, we can stop.   Otherwise, let $\Cal V$
be the countable algebra of subsets of $V$ generated by
$\{V\cap U_n:n\in\Bbb N\}$.   Since $A\cap U$ does not have the Baire
property (in $X$) for any non-empty relatively open subset $U$ of $V$,
$V$ has no isolated points and $\Cal V$ is atomless.   So $\Cal V$ is
isomorphic to the algebra of open-and-closed subsets of $\{0,1\}^{\Bbb N}$
(316M) and there is a Boolean-independent sequence
$\sequencen{V_n}$ in $\Cal V$ generating $\Cal V$.   Define
$h:V\to\{0,1\}^{\Bbb N}$ by setting $h(x)=\sequencen{\chi V_n(x)}$ for
$x\in V$.   Then $h[V]$ is dense in $\{0,1\}^{\Bbb N}$ and $h^{-1}[H]$ is
dense in $V$ for every dense open set $H\subseteq\{0,1\}^{\Bbb N}$;
consequently $h^{-1}[M]$ is meager in $V$ and in $X$ whenever
$M\subseteq\{0,1\}^{\Bbb N}$ is either nowhere dense or meager.
By 567Fb, $h[A]$ has the Baire property in $\{0,1\}^{\Bbb N}$;  express it
as $H\symmdiff M$ where $H$ is open and $M$ is meager;  then
$A\cap V=h^{-1}[h[A]]=h^{-1}[H]\symmdiff h^{-1}[M]$ has the
Baire property in $X$, so $A$ has the Baire property in $X$, as required.
}%end of proof of 567G

\leader{567H}{Theorem} (a) [AD] Let $X$ be a Polish group and $Y$ a
topological group which is either separable or Lindel\"of.
Then every group homomorphism from $X$ to $Y$ is continuous.

(b) [AD+AC($\omega$)] Let $X$ be an abelian
topological group which is complete under a metric defining its topology,
and $Y$ a topological group which is either separable or Lindel\"of.
Then every group homomorphism from $X$ to $Y$ is continuous.

(c) [AD+AC($\omega$)]
Let $X$ be a complete metrizable linear topological space,
$Y$ a linear topological space and $T:X\to Y$ a linear operator.
Then $T$ is continuous.   In particular, every linear operator between
Banach spaces is a bounded operator.

\proof{{\bf (a)(i)} Let $f:X\to Y$ be a homomorphism, and
$V$ a neighbourhood of the identity in $Y$.
Let $W$ be an open neighbourhood of the identity in $Y$
such that $W^{-1}W\subseteq V$.   Then there is countable family $\Cal H$
of left translates of $W$ which covers $Y$.   \Prf\ If $Y$ is
separable, let $D$ be a countable dense subset of $Y$, and set
$\Cal H=\{yW:y\in D\}$.   If $Y$ is Lindel\"of, we have only to note that
$\{yW:y\in Y\}$ is an open cover of $Y$, so has a countable subcover.\ \Qed

\medskip

\quad{\bf (ii)} Since $X$ is a Baire space (561Ea),
and the ideal of meager subsets of $X$ is a $\sigma$-ideal (see
part (b) of the proof of 567G), and $\{f^{-1}[H]:H\in\Cal H\}$ is a
countable cover of $X$, there is an $H\in\Cal H$ such that
$E=f^{-1}[H]$ is non-meager.
Now $E^{-1}E$ is a neighbourhood of the
identity in $X$.   \Prf\ By 567G, $E$ has the Baire property;  let
$G$ be a non-empty open set in $X$ such that $G\setminus E$ is meager.
Set $U=\{x:Gx\cap G\ne\emptyset\}$;  then $U$ is a neighbourhood of the
identity in $X$.   If $x\in U$, then

\Centerline{$Gx\cap G
\subseteq(Ex\cap E)\cup(Gx\setminus Ex)\cup(G\setminus E)
=(Ex\cap E)\cup(G\setminus E)x\cup(G\setminus E)$.}

\noindent Since $Gx\cap G$ is non-meager, while $G\setminus E$ and
$(G\setminus E)x$ are meager, $Ex\cap E\ne\emptyset$ and $x\in E^{-1}E$.
Thus $E^{-1}E\supseteq U$ is a neighbourhood of the identity.\ \Qed

\medskip

\quad{\bf (iii)} Let $y\in Y$ be such that $H=yW$.
If $x$, $z\in E$, $y^{-1}f(x)$ and $y^{-1}f(z)$ both belong to
$W$, so

\Centerline{$f(x^{-1}z)=f(x)^{-1}f(z)\in W^{-1}yy^{-1}W=W^{-1}W
\subseteq V$.}

\noindent Thus $f^{-1}[V]\supseteq E^{-1}E$ is a neighbourhood of the
identity in $X$.   As $V$ is arbitrary, $f$ is continuous at the identity,
therefore continuous.

\medskip

{\bf (b)} \Quer\ Otherwise, there is a neighbourhood $V$ of the identity
$e_Y$ of $Y$ such that $f^{-1}[V]$ is not a neighbourhood of the identity
$e_X$ of $X$.   Let $\rho$ be a metric on $X$, defining its topology, under
which $X$ is complete.   Then for each $n\in\Bbb N$ we can choose an
$x_n\in X$ such that $\rho(x_n,e_X)\le 2^{-n}$ and $f(x_n)\notin V$.
(This is where we need AC($\omega$).)   For finite $J\subseteq\Bbb N$
set $u_J=\prod_{n\in J}x_n$, starting from $u_{\emptyset}=e_X$.
We can define an infinite $I\subseteq\Bbb N$ inductively by saying that

\Centerline{$I=\{n:$ whenever $J\subseteq I\cap n$
then $\rho(u_J,u_Jx_n)\le 2^{-\#(I\cap n)}\}$.}

\noindent This will ensure
that $v_K=\lim_{n\to\infty}u_{K\cap n}$ is defined for every
$K\subseteq I$.   Note that $v_{K\cup\{m\}}=v_Kx_m$ whenever $m\in I$ and
$K\subseteq I\setminus\{m\}$
(this is where we need to know that $X$ is abelian).

Give $\Cal PI$ its usual topology.
Let $W$ be a neighbourhood of $e_Y$ such that $W^{-1}W\subseteq V$.
By the argument of (a) above, applied to the map
$K\mapsto f(v_K):\Cal PI\to Y$,
there is a $y\in Y$ such that $E=\{K:K\subseteq I$, $f(v_K)\in yW\}$
is non-meager in $\Cal PI$.   Looking at the topological group
$(\Cal PI,\symmdiff)$, we see that there is a neighbourhood $U$ of
$\emptyset$ in $\Cal PI$ included in $\{K\symmdiff L:K$, $L\in E\}$.
Taking any sufficiently large $n\in I$, we have $\{n\}\in U$, so there must
be a $K\in E$ such that $n\notin K$ and $K\cup\{n\}\in E$.
In this case $f(v_K)\in yW$, $f(v_{K\cup\{n\}})\in yW$ and

\Centerline{$f(x_n)=f(v_K^{-1}v_{K\cup\{n\}})
=f(v_K)^{-1}f(v_{K\cup\{n\}})\in W^{-1}W\subseteq V$,}

\noindent which is impossible.\ \Bang

\medskip

{\bf (c)} \Quer\ Otherwise, there is a neighbourhood $V$ of $0$ in $Y$ such
that $T^{-1}[V]$ is not a neighbourhood of $0$ in $X$;  we can suppose that
$\alpha y\in V$ whenever $y\in V$ and $|\alpha|\le 1$.
Let $\rho$ be a metric on $X$, defining its topology, under
which $X$ is complete.   Let $W$ be a neighbourhood of $0$ in $Y$ such that
$W-W\in V$.   Then for each $n\in\Bbb N$ we can choose an
$x_n\in X\setminus nT^{-1}[V]$ such that
$\rho(x_n,0)\le 2^{-n}$.   Define $I\in[\Bbb N]^{\omega}$ and
$\langle v_K\rangle_{K\subseteq I}$ as in (b), but using additive notation
rather than multiplicative.   This time we are not
supposing that $Y$ is separable.   However, there must be an $m\in\Bbb N$
such that $E=\{K:T(v_K)\in mW\}$ is non-meager.   As before, we can find
$n\in I\setminus m$ and $K\in E$ such that
$n\notin L$ and $K\cup\{n\}\in E$.   So the calculation gives

\Centerline{$Tx_n=Tv_{K\cup\{n\}}-Tv_K\in mW-mW
\subseteq mV\subseteq nV$,}

\noindent again contrary to the choice of $x_n$.\ \Bang
}%end of proof of 567H

\leader{567I}{Proposition} [AC($\Bbb R;\omega$)] Let $\widehat{\Cal B}$ be
the Baire-property algebra of $\Cal P\Bbb N$.   Then every
$\widehat{\Cal B}$-measurable real-valued additive functional on
$\Cal P\Bbb N$ is of the form $a\mapsto\sum_{n\in a}\gamma_n$ for some
$\sequencen{\gamma_n}\in\ell^1$.

\proof{ As noted in 567Ec, $\widehat{\Cal B}$ is a $\sigma$-algebra of
subsets of $\Cal P\Bbb N$.

\medskip

{\bf (a)(i)} If $G\subseteq\Cal P\Bbb N$ is a dense open set and
$m\in\Bbb N$, there are an $m'>m$ and an $L\subseteq m'\setminus m$ such
that $\{a:a\subseteq\Bbb N$, $a\cap m'\setminus m=L\}\subseteq G$.   \Prf\
The set $H=\{b:b\subseteq\Bbb N\setminus m$, $I\cup b\in G$ for every
$I\subseteq m\}$ is a dense open subset of $\Cal P(\Bbb N\setminus m)$, so
there are an $m'>m$ and an $L\subseteq m'\setminus m$ such that
$H\supseteq\{b:b\subseteq\Bbb N\setminus m$, $b\cap m'\setminus m=L\}$;
this pair $m'$, $L$ works.\ \Qed

\medskip

\quad{\bf (ii)} If $G\subseteq\Cal P\Bbb N$ is comeager, there are a
strictly increasing sequence $\sequencen{m_n}$ in $\Bbb N$ and sets
$L_n\subseteq m_{n+1}\setminus m_n$, for $n\in\Bbb N$, such that

\Centerline{$G
\supseteq\{a:a\subseteq\Bbb N$, $a\cap m_{n+1}\setminus m_n=L_n$ for
infinitely many $n\}$.}

\noindent\Prf\ Let $\sequencen{G_n}$ be a non-increasing sequence of dense
open sets such that $G\supseteq\bigcap_{n\in\Bbb N}G_n$, and choose
$\sequencen{m_n}$, $\sequencen{L_n}$ inductively such that $m_n<m_{n+1}$,
$L_n\subseteq m_{n+1}\setminus m_n$ and
$\{a:a\subseteq\Bbb N$, $a\cap m_{n+1}\setminus m_n=L_n\}\subseteq G_n$ for
every $n$.\ \Qed

\medskip

\quad{\bf (iii)} If $G\subseteq\Cal P\Bbb N$ is comeager, and
$a\subseteq\Bbb N$, then there are $b_0$, $b'_0$, $b_1$, $b'_1\in G$ such
that

\Centerline{$b_0\subseteq b'_0$,
\quad$b_1\subseteq b'_1$,
\quad$(b'_0\setminus b_0)\cap(b'_1\setminus b_1)=\emptyset$,
\quad$(b'_0\setminus b_0)\cup(b'_1\setminus b_1)=a$.}

\noindent\Prf\ Let $\sequencen{m_n}$ and $\sequencen{L_n}$ be as in (ii).
Set

\Centerline{$b_0=\bigcup_{n\in\Bbb N}L_{2n}$,
\quad
$b'_0=b_0\cup(a\cap m_0)
\cup\bigcup_{n\in\Bbb N}a\cap m_{2n+2}\setminus m_{2n+1}$,}

\Centerline{$b_1=\bigcup_{n\in\Bbb N}L_{2n+1}$,
\quad$b'_1=b_1\cup\bigcup_{n\in\Bbb N}a\cap m_{2n+1}\setminus m_{2n}$.\
\Qed}

\noindent So if $\nu:\Cal P\Bbb N\to\Bbb R$ is additive,
$\sup_{a\subseteq\Bbb N}|\nu a|\le 4\sup_{b\in G}|\nu b|$.

\medskip

\quad{\bf (iv)} If $G\subseteq\Cal P\Bbb N$ is comeager, there is a
disjoint sequence $\sequencen{a_n}$ in $G$.   \Prf\ Take
$\sequencen{m_n}$ and $\sequencen{L_n}$ as in (ii), and set
$a_n=\bigcup_{i\in\Bbb N}L_{2^n(2i+1)}$ for each $n$.\ \Qed

\medskip

{\bf (b)} If $\nu:\Cal P\Bbb N\to\Bbb R$ is additive and
$\widehat{\Cal B}$-measurable,
it is bounded.   \Prf\ Let $M\in\Bbb N$ be such that
$E=\{a:|\nu a|\le M\}$ is non-meager.   Then there are an $m\in\Bbb N$ and
$J\subseteq m$ such that $V_{mJ}\setminus E$ is meager, where
$V_{mJ}=\{a:a\cap m=J\}$.   For $K\subseteq m$, $a\subseteq\Bbb N$
set $\phi_K(a)=a\symmdiff K$;  then $\phi_K$ is an autohomeomorphism of
$\Cal P\Bbb N$, so $\phi_K[V_{mJ}\setminus E]$ is meager.   Let $G$ be the
comeager set
$\Cal P\Bbb N\setminus\bigcup_{K\subseteq m}\phi_K[V_{mJ}\setminus E]$.
Set $\delta=\sum_{i<m}|\nu\{i\}|$;  then
$|\nu\phi_K(a)-\nu a|\le\delta$ whenever $K\subseteq m$
and $a\subseteq\Bbb N$.   If $b\in G$, set $K=(b\cap m)\symmdiff J$;  then
$\phi_K(b)\in V_{mJ}\setminus(V_{mJ}\setminus E)\subseteq E$, so
$|\nu b|\le M+\delta$.   So (a-iii) tells us that $|\nu a|\le 4(M+\delta)$
for every $a\subseteq\Bbb N$, and $\nu$ is bounded.\ \Qed

\medskip

{\bf (c)} If $\nu:\Cal P\Bbb N\to\Bbb R$ is
additive and $\widehat{\Cal B}$-measurable
and $\nu\{n\}=0$ for every $n\in\Bbb N$, then $E=\{a:\nu a\ge\epsilon\}$ is
meager for every $\epsilon>0$.
\Prf\Quer\ Otherwise, let $m\in\Bbb N$ and $J\subseteq m$ be such that
$V_{mJ}\setminus E$ is meager.   Let $G$ be the comeager set
$\Cal P\Bbb N\setminus\bigcup_{K\subseteq m}\phi_K[V_{mJ}\setminus E]$, as
in (b).   This time, $\nu a=\nu\phi_K(a)$ whenever $K\subseteq m$ and
$a\subseteq\Bbb N$, so $\nu a\ge\epsilon$ for every $a\in G$.   But
(a-iv) tells us that there is a disjoint sequence
$\sequencen{a_n}$ in $G$, and now
$\sup_{n\in\Bbb N}\nu(\bigcup_{i\le n}a_i)=\infty$, contradicting (b).\
\Bang\Qed

\medskip

{\bf (d)} If $\nu:\Cal P\Bbb N\to\Bbb R$ is additive and
$\widehat{\Cal B}$-measurable
and $\nu\{n\}=0$ for every $n\in\Bbb N$, then $\nu=0$.   \Prf\ By (c),
applied to $\nu$ and $-\nu$, $G=\{a:\nu a=0\}$ is comeager.   By (a-iii),
$\nu$ must be identically zero.\ \Qed

\medskip

{\bf (e)} Now suppose that $\nu$ is any additive
$\widehat{\Cal B}$-measurable functional.   Set $\gamma_n=\nu\{n\}$ for
each $n$.   By (b), $\sequencen{\gamma_n}\in\ell^1$.   Setting
$\nu'a=\nu a-\sum_{n\in a}\gamma_n$ for $a\subseteq\Bbb N$, $\nu'$ is still
additive and $\widehat{\Cal B}$-measurable, and $\nu'\{n\}=0$ for every
$n$, so (d) tells us that $\nu'=0$ and
$\nu a=\sum_{n\in a}\gamma_n$ for every $a$, as required.
}%end of proof of 567I

\leader{567J}{Proposition} [AD] A finitely additive functional on a
Dedekind $\sigma$-complete Boolean algebra is countably additive.

\proof{ Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra,
$\nu$ a finitely additive functional on $\frak A$ and $\sequencen{a_n}$ a
disjoint sequence in $\frak A$ with supremum $a$.   Set
$\lambda c=\nu(\sup_{n\in c}a_n)$ for $c\subseteq\Bbb N$.
Then $\lambda$ is an
additive functional on $\Cal P\Bbb N$.   By 567G, it is
$\widehat{\Cal B}(\Cal P\Bbb N)$-measurable;  by 567I,

\Centerline{$\nu a=\lambda\Bbb N=\sum_{n=0}^{\infty}\lambda\{n\}
=\sum_{n=0}^{\infty}\nu a_n$.}
}%end of proof of 567J

\leader{567K}{Theorem} [AD+AC($\omega$)] If $U$ is an $L$-space with a weak
order unit, it is reflexive.

\proof{ By 561Hb, $U$ is isomorphic to $L^1(\frak A,\bar\mu)$ for
some totally finite measure algebra $(\frak A,\bar\mu)$;  now $U^*$ can be
identified with $L^{\infty}(\frak A)$.   Next, $L^{\infty}(\frak A)^*$ can
be identified with the space of bounded finitely additive functionals on
$\frak A$, as in 363K;  by 567J, these are all countably additive.
Because we have countable choice, $\frak A$ is ccc (566M),
so countably additive functionals are completely additive and correspond
to members of $L^1$, as in 365Ea.   Thus the canonical embedding of $U$ in
$U^{**}$ is surjective.
}%end of proof of 567K

\leader{567L}{Theorem}\cmmnt{ (R.M.Solovay)} [AD] $\omega_1$ is \2vm.

\medskip

\noindent{\bf Remark}\cmmnt{ The definition in 541M speaks of
`regular uncountable
cardinals'.}   In the present context I will use the formulation
`an initial ordinal $\kappa$ is
\2vm\ if there is a proper $\kappa$-additive
$2$-saturated ideal $\Cal I$
of $\Cal P\kappa$ containing singletons', where here
`$\kappa$-additive' means that $\bigcup_{\eta<\xi}J_{\eta}\in\Cal I$
whenever $\xi<\kappa$ and $\ofamily{\eta}{\xi}{J_{\eta}}$ is a family in
$\Cal I$.

\proof{{\bf (a)} Let $\StrI$ be the set of strategies for player I in games
of the form $\Gamma(\Bbb N,.)$, that is, $\StrI$ is the set of functions
from $\bigcup_{n\in\Bbb N}\BbbN^n$ to $\Bbb N$;  for $\sigma\in\StrI$ and
$x\in\NN$, let $\sigma*x\in\NN$ be the play in which I follows the strategy
$\sigma$ and II plays the sequence  $x$, that is,

\Centerline{$(\sigma*x)(2n)=\sigma(x\restr n)$,
\quad$(\sigma*x)(2n+1)=x(n)$}

\noindent  for $n\in\Bbb N$.
Similarly, let $\StrII$ be the set of functions from
$\bigcup_{n\ge 1}\BbbN^n$ to $\Bbb N$
and for $\tau\in\StrII$, $x\in\NN$, $n\in\Bbb N$ set

\Centerline{$(\tau*x)(2n)=x(n)$,
\quad$(\tau*x)(2n+1)=\tau(x\restr(n+1))$.}

We can find bijections $g:\NN\to\StrI\cup\StrII$ and
$h:\NN\to\WO(\Bbb N)$, where $\WO(\Bbb N)\subseteq\Cal P(\BbbN^2)$ is the
set of well-orderings of $\Bbb N$.   \Prf\ Since
$S=\bigcup_{n\in\Bbb N}\BbbN^n$ and $S^*=\bigcup_{n\ge 1}\BbbN^n$ are
countably infinite, $\StrI=\BbbN^{S}$ and $\StrII=\BbbN^{S^*}$ are
equipollent with $\NN$.    As $\Cal P(\BbbN^2)\sim\Cal P\Bbb N\sim\NN$,
there is an injection from $\WO(\Bbb N)$ to $\NN$.   In the reverse
direction, there are an injection from $\NN$ to the set $F$ of permutations
of $\Bbb N$, and an injection from $F$ to $\WO(\Bbb N)$;  so
the Schroeder-Bernstein theorem tells us that $\WO(\Bbb N)\sim\NN$.\ \Qed

Define $f:\WO(\Bbb N)\to\omega_1$ by
saying that $f(\preccurlyeq)=\otp(\Bbb N,\preccurlyeq)$ for
$\preccurlyeq\mskip5mu\in\WO(\Bbb N)$.

\medskip

{\bf (b)} For $x\in\NN$ let $L_x\subseteq\NN$ be the smallest set such that

\inset{$x\in L_x$,

whenever $y$, $z\in L_x$ then $g(y)*z\in L_x$,

whenever $y\in L_x$ then $\sequencen{y(2n)}$ and
$\sequencen{y(2^k(2n+1))}$ belong to $L_x$ for every $k\in\Bbb N$.}

\noindent Observe that $L_x$ is countable and that $L_y\subseteq L_x$
whenever $y\in L_x$.   For $x\in\NN$, set $C_x=\{y:y\in\NN$, $x\in L_y\}$.

\medskip

{\bf (c)} For any sequence $\sequencen{x_n}$ in $\NN$ there is an
$x\in\NN$ such that $C_x\subseteq\bigcap_{n\in\Bbb N}C_{x_n}$.   \Prf\ Set
$x(0)=0$ and $x(2^k(2n+1))=x_k(n)$ for $k$, $n\in\Bbb N$.
Then $x_k\in L_x$
for every $k$.   So if $y\in C_x$ and $n\in\Bbb N$, we have
$x_n\in L_x\subseteq L_y$ and $y\in C_{x_n}$.\ \Qed

Let $\Cal F$ be the filter on $\NN$ generated by $\{C_x:x\in\NN\}$;  then
(because AC($\Bbb R;\omega$) is true)
$\Cal F$ is closed under countable intersections.

\medskip

{\bf (d)} Suppose that $A\subseteq\NN$ is such that whenever $x\in A$ and
$L_y=L_x$ then $y\in A$.

\medskip

\quad{\bf (i)} If I has a winning strategy in $\Game(\Bbb N,A)$ then
$A\in\Cal F$.   \Prf\ Let $\sigma\in\StrI$ be a winning strategy for I, and
consider $x=g^{-1}(\sigma)\in\NN$.   Suppose that $y\in\NN$ and
$x\in L_y$, and consider $z=\sigma*y\in A$.   As $z=g(x)*y$ belongs to
$L_y$, $L_z\subseteq L_y$;  on the other hand,
$y(n)=z(2n+1)$ for every $n$, so $y\in L_z$ and $L_y\subseteq L_z$.
So $L_y=L_z$ and $y\in A$.
As $y$ is arbitrary, $C_x\subseteq A$ and $A\in\Cal F$.\ \Qed

\medskip

\quad{\bf (ii)} If II has a winning strategy in $\Game(\Bbb N,A)$ then
$\NN\setminus A\in\Cal F$.   \Prf\ Let $\tau\in\StrII$ be a winning
strategy for II, and
consider $x=g^{-1}(\tau)\in\NN$.   Suppose that $y\in\NN$ and
$x\in L_y$, and consider $z=\tau*y\in\NN\setminus A$.   As before,
$L_z\subseteq L_y$;  this time,
$y(n)=z(2n)$ for every $n$ so $y\in L_z$ and $L_y\subseteq L_z$.
So $y\notin A$.   As $y$ is arbitrary, $C_x\subseteq\NN\setminus A$ and
$\NN\setminus A\in\Cal F$.\ \Qed

\medskip

{\bf (e)} For $x\in\NN$ set $\phi(x)=\sup_{y\in L_x}f(h(y))$;  because
$L_x$ is countable, $\phi(x)<\omega_1$ (567Ed).   Let $\Cal G$ be the image
filter $\phi[[\Cal F]]$.   Because $\Cal F$ is closed under countable
intersections, so is $\Cal G$.   If $B\subseteq\omega_1$ then $\phi^{-1}[B]$
satisfies the condition of (d), so that one of $\phi^{-1}[B]$,
$\NN\setminus\phi^{-1}[B]$ belongs to $\Cal F$ and one of $B$,
$\omega_1\setminus B$ belongs to $\Cal G$;  as $B$ is arbitrary, $\Cal G$
is an ultrafilter.

\medskip

{\bf (f)} Finally, $\Cal G$ does not contain any singletons.   \Prf\ If
$\xi<\omega_1$, there is an $x\in\NN$ such that $f(h(x))=\xi+1$.   Now
$C_x\in\Cal F$ so $\phi[C_x]\in\Cal G$.   If $y\in C_x$ then $x\in L_y$ so
$\xi+1\le\phi(y)$;  accordingly $\xi\notin\phi[C_x]$ and
$\{\xi\}\notin\Cal G$.\ \QeD\   So $\Cal G$ (or, if you like, the ideal
$\{\omega_1\setminus B:B\in\Cal G\}$) witnesses that $\omega_1$ is \2vm.
}%end of proof of 567L

\leader{567M}{Theorem}\cmmnt{ ({\smc Moschovakis 70})}
[AD] Let $\alpha$ be an
ordinal such that there is a surjection from $\Cal P\Bbb N$ onto $\alpha$.
Then there is a surjection from $\Cal P\Bbb N$ onto $\Cal P\alpha$.

\proof{ The formulae will run slightly more smoothly if we work with
surjections
from $\NN$ rather than from $\Cal P\Bbb N$;  of course this makes no
difference to the result.

\medskip

{\bf (a)}  We may suppose that $\alpha$ is uncountable.
Let $f:\NN\to\alpha$ be a surjection.   I seek to define
inductively a family $\langle g_{\xi}\rangle_{\xi\le\alpha}$ such that
$g_{\xi}$ is a surjection from $\NN$ onto $\Cal P\xi$ for every
$\xi\le\alpha$.
As in the proof of 567L, let
$\StrI$ be the set of functions from $\bigcup_{n\in\Bbb N}\BbbN^n$ to
$\Bbb N$, and
$\StrII$ the set of functions from $\bigcup_{n\ge 1}\BbbN^n$ to
$\Bbb N$;  fix a surjection $h:\NN\to\StrI\cup\StrII$.
For $\sigma\in\StrI$, $\tau\in\StrII$ and $x\in\NN$ let $\sigma*x$,
$\tau*x$ be the plays in games on $\Bbb N$ as described in the proof of
567L.

\medskip

{\bf (b)} Start by setting $g_n(x)=n\cap x[\Bbb N]$ for $x\in\NN$ and
$n\in\Bbb N$.

\medskip

{\bf (c)} For the inductive step to a non-zero limit
ordinal $\xi\le\alpha$, given $\ofamily{\eta}{\xi}{g_{\eta}}$, then for
$x\in\NN$ set

\Centerline{$\eta_x=f(\sequencen{x(4n)})$,
\quad$\zeta_x=f(\sequencen{x(4n+1)})$,}

$$\eqalign{E_x
&=g_{\eta_x}(\sequencen{x(4n+2)})\text{ if }\eta_x<\xi,\cr
&=\emptyset\text{ otherwise},\cr
F_x
&=g_{\zeta_x}(\sequencen{x(4n+3)})\text{ if }\zeta_x<\xi,\cr
&=\emptyset\text{ otherwise}.\cr}$$

\noindent Next, for $D\subseteq\xi$, set

$$\eqalign{A_D
&=\{x:x\in\NN,\,\eta_x<\xi,\,E_x=D\cap\eta_x\cr
&\mskip70mu
\text{ and either }\zeta_x\le\eta_x\text{ or }\zeta_x\ge\xi
\text{ or }F_x\ne D\cap\zeta_x\}.\cr}$$

\noindent (The idea is that the players are competing to see who can
capture the largest initial segment of $D$ with the pair $(\eta_x,E_x)$
determined by I's moves or the pair $(\zeta_x,F_x)$ determined by
II's moves;  for definiteness, if neither correctly defines an
initial segment, then II wins, while if they seize the same
segment $(\eta_x,E_x)=(\zeta_x,F_x)$, then I wins.)   Finally, define
$g:\StrI\cup\StrII\to\Cal P\xi$ by setting

\Centerline{$g(\sigma)=\bigcup\{D:\sigma$ is a winning strategy for I in
$\Game(\Bbb N,A_D)\}$ if $\sigma\in\StrI$,}

\Centerline{$g(\tau)=\bigcup\{D:\tau$ is a winning strategy for II in
$\Game(\Bbb N,A_D)\}$ if $\tau\in\StrII$.}

We find that $g$ is a surjection onto $\Cal P\xi$.   \Prf\ Take any
$D\subseteq\xi$.

\medskip

\quad{\bf case 1}
Suppose that I has a winning strategy $\sigma$ in $\Game(\Bbb N,A_D)$.
Then $D\subseteq g(\sigma)$.   \Quer\ If $D\ne g(\sigma)$, there is a $D'$,
distinct from $D$, such that $\sigma$ is a winning strategy for I in
$\Game(\Bbb N,D')$.   Let $\zeta<\xi$ be such that
$D\cap\zeta\ne D'\cap\zeta$.   Then there is
a $z\in\NN$ such that $f(\sequencen{z(2n)})=\zeta$ and
$g_{\zeta}(\sequencen{z(2n+1)})=D\cap\zeta$.   In this case, taking
$x=\sigma*z$, we have $x(4n+1)=z(2n)$ and $x(4n+3)=z(2n+1)$ for every $n$,
so $\zeta_x=\zeta$ and $F_x=D\cap\zeta$.   Since $x\in A_D$,
we have $\eta_x<\xi$, $E_x=D\cap\eta_x$ and $\zeta\le\eta_x$.
But also $x\in A_{D'}$, so $E_x=D'\cap\eta_x$ and $D\cap\zeta=D'\cap\zeta$,
contrary to the choice of $\zeta$.\ \BanG\   Thus $D=g(\sigma)$.

\medskip

\quad{\bf case 2}
Suppose that II has a winning strategy $\tau$ in $\Game(\Bbb N,A_D)$.
Then $D\subseteq g(\tau)$.   \Quer\ If $D\ne g(\tau)$, there is a $D'$,
distinct from $D$, such that $\tau$ is a winning strategy for II in
$\Game(\Bbb N,D')$.   Let $\zeta<\xi$ be such that
$D\cap\zeta\ne D'\cap\zeta$.   Again, there is
a $z\in\NN$ such that $f(\sequencen{z(2n)})=\zeta$ and
$g_{\zeta}(\sequencen{z(2n+1)})=D\cap\zeta$.   This time, taking
$x=\tau*z$, we have $x(4n)=z(2n)$ and $x(4n+2)=z(2n+1)$ for every $n$,
so $\eta_x=\zeta$ and $E_x=D\cap\eta_x$.   Since $x\notin A_D$, we must
have $\eta_x<\zeta_x<\xi$ and $F_x=D\cap\zeta_x$;  since also
$x\notin A_{D'}$, $F_x=D'\cap\zeta_x$;  so that
$D\cap\zeta=F_x\cap\zeta=D'\cap\zeta$, which is impossible.\ \BanG\
Thus $D=g(\tau)$.

\medskip

Thus in either case $D\in g[\StrI\cup\StrII]$.   As $D$ is arbitrary,
$g[\StrI\cup\StrII]=\Cal P\xi$.\ \Qed\

Setting $g_{\xi}=gh$, the induction proceeds.

\medskip

{\bf (d)} For the inductive step to $\xi+1$ where $\omega\le\xi<\alpha$,
set

\Centerline{$h_{\xi}(0)=\xi$,
$h_{\xi}(n)=n-1$ for $n\in\omega\setminus\{0\}$,
$h_{\xi}(\eta)=\eta$ if $\omega\le\eta<\xi$,}

\Centerline{$g_{\xi+1}(x)=h_{\xi}[g_{\xi}(x)]$ for $x\in\NN$.}

\medskip

{\bf (e)} At the end of the induction, $g_{\alpha}$
is the required surjection onto $\Cal P\alpha$.
%Kanamori 28.15
}%end of proof of 567M

\leader{567N}{Theorem}\cmmnt{ ({\smc Martin 70})} [AC]
Assume that there is a \2vm\
cardinal.   Then every coanalytic subset of $\BbbN^{\Bbb N}$ is determined.

\proof{ Let $A\subseteq\NN$ be a coanalytic set.

\medskip

{\bf (a)} Set $S^*=\bigcup_{n\ge 1}\BbbN^n$.   For $\upsilon$,
$\upsilon'\in S^*$ say that $\upsilon\preccurlyeq\upsilon'$ if either
$\upsilon$ extends $\upsilon'$ or there is an
$n<\min(\#(\upsilon),\#(\upsilon'))$ such that
$\upsilon\restr n=\upsilon'\restr n$ and $\upsilon(n)<\upsilon'(n)$.   Then
$\preccurlyeq$ is a total order, and its restriction to $\BbbN^n$ is the
lexicographic well-ordering for each $n\ge 1$.

For $w\in\bigcup_{n\in\Bbb N}\BbbN^n$, set
$I_w=\{x:w\subseteq x\in\NN\}$.   Fix an enumeration
$\sequence{i}{\upsilon_i}$ of $S^*$ such that
$\#(\upsilon_i)\le i+1$ for every $i\in\Bbb N$.

\medskip

{\bf (b)} $A'=\NN\setminus A$ is Souslin-F (423Eb);  express it as

\Centerline{$A'=\bigcup_{y\in\NN}\bigcap_{n\ge 1}F_{y\restr n}$}

\noindent where $F_{\upsilon}$ is closed for every $\upsilon\in S^*$.
Replacing $F_{\upsilon}$ by
$\bigcap_{1\le i\le\#(\upsilon)}F_{\upsilon\restr i}$ if necessary, we may
suppose that $F_{\upsilon}\subseteq F_{\upsilon'}$ whenever
$\upsilon\supseteq\upsilon'$, as in 421Cf\Latereditions.

For $x\in\NN$, set

\Centerline{$T_x=\{\upsilon:\upsilon\in S^*$,
$I_{x\restr\#(\upsilon)}\cap F_{\upsilon}\ne 0\}$,}

\noindent and define a relation $\preccurlyeq_x$ on $S^*$ by saying that

$$\eqalign{\upsilon\preccurlyeq_x\upsilon'\iff
&\text{ either }\upsilon,\,\upsilon'\in T_x
   \text{ and }\upsilon\preccurlyeq\upsilon'\cr
&\text{ or }\upsilon\in T_x\text{ and }\upsilon'\notin T_x\cr
&\text{ or }\upsilon,\,\upsilon'\notin T_x\text{ and }i\le j
  \text{ where }\upsilon=\upsilon_i,\,\upsilon'=\upsilon_j.\cr}$$

\noindent Then $\preccurlyeq_x$ is a total ordering, since it copies the
total ordering $\preccurlyeq$ on $T_x$ and the
well-ordering induced by the
enumeration $\sequence{i}{\upsilon_i}$ on $S^*\setminus T_x$, and puts one
below the other.

Note that if $n\in\Bbb N$, $x$, $y\in\NN$ are such that
$x\restr n=y\restr n$, and $i<n$,
then $x\restr\#(\upsilon_i)=y\restr\#(\upsilon_i)$,
so $\upsilon_i\in T_x$ iff $\upsilon_i\in T_y$.   Consequently, for $i$,
$j<n$, $\upsilon_i\preccurlyeq_x\upsilon_j$ iff
$\upsilon_i\preccurlyeq_y\upsilon_j$.   It follows that for every
$w\in\BbbN^n$ we have a total ordering $\preccurlyeq'_w$ of $n$
defined by saying that $i\preccurlyeq'_wj$ iff
$\upsilon_i\preccurlyeq_x\upsilon_j$ whenever $x\in I_w$.

\medskip

{\bf (c)} If $x\in\NN$ and $\preccurlyeq_x$ is not a well-ordering, then
$x\notin A$.   \Prf\ Let $D\subseteq S^*$ be a non-empty set with no
$\preccurlyeq_x$-least member.   Then $D\cap T_x$ is an initial segment of
$D$.   Since $S^*\setminus T_x$ is certainly
well-ordered by $\preccurlyeq_x$, $D\cap T_x\ne\emptyset$.
Define $\sequencen{D_n}$, $\sequencen{y(n)}$ as follows.   $D_0=D\cap T_x$.
Given that $D_n$ is a non-empty
initial segment of $D$ and that $\upsilon\supseteq y\restr n$ for every
$\upsilon\in D_n$, then $y\restr n$ cannot be the least member of $D$, so
$D_n\ne\{y\restr n\}$;  set
$y(n)=\min\{\upsilon(n):\upsilon\in D_n\setminus\{y\restr n\}\}$,

\Centerline{$D_{n+1}
=\{\upsilon:\upsilon\in D_n\setminus\{y\restr n\}$, $\upsilon(n)=y(n)\}$.}

\noindent Because
$\preccurlyeq_x$ agrees with $\preccurlyeq$ on $T_x$, $D_{n+1}$ is a
non-empty initial segment of $D$, and the induction continues.

If $m$, $n\in\Bbb N$, then there is an $\upsilon\in T_x$ such that
$\upsilon\supseteq y\restr\max(m,n)$, and

\Centerline{$I_{x\restr m}\cap F_{y\restr n}
\supseteq I_{x\restr\#(\upsilon)}\cap F_{\upsilon}\ne\emptyset$.}

\noindent As $m$ is arbitrary and
$F_{y\restr n}$ is closed, $x\in F_{y\restr n}$;  as
$n$ is arbitrary, $x\in A'$ and $x\notin A$.\ \Qed

\medskip

{\bf (d)} Let $\kappa$ be a \2vm\ cardinal, and give $\Bbb N\times\kappa$
its discrete topology.
In $(\Bbb N\times\kappa)^{\Bbb N}$ consider the set
$F$ of sequences $\sequencen{(x(n),\xi(n))}$ such that

\Centerline{whenever $i$, $j\in\Bbb N$,
$\upsilon_i\subset\upsilon_j$ and $x\in F_{\upsilon_j}$, then
$\xi(2j)<\xi(2i)$.}

\noindent Then $F$ is closed
for the product topology;  by 567B, $F$ is determined.

\medskip

{\bf (e)} Suppose I has a winning strategy $\sigma$ in the game
$\Game(\Bbb N\times\kappa,F)$.
Then I has a winning strategy in $\Game(\Bbb N,A)$.   \Prf\ For
$\ofamily{i}{n}{k_i}\in\Bbb N^n$ take
$\sigma'(\ofamily{i}{n}{k_i})$ to be the first component of
$\sigma(\ofamily{i}{n}{(k_i,0)})$.   If $x$ is any play of
$\Game(\Bbb N,A)$ consistent with $\sigma'$, then for each $n$ set
$\xi(2n+1)=0$ and let $\xi(2n)$ be the second component of
$\sigma(\ofamily{i}{n}{(x(2i+1),0)})$.   Then
$\sequencen{(x(n),\xi(n))}$ is a play of $\Game(\Bbb N\times\kappa,F)$
consistent with $\sigma$, so is won by I.   \Quer\ If $x\notin A$,
let $y\in\NN$ be such that $x\in F_{y\restr n}$ for every $n\in\Bbb N$.
Set $I=\{i:i\in\Bbb N$, $y\supseteq\upsilon_i\}$;  then $I$ is infinite,
and there is an infinite $J\subseteq I$ such that
$\upsilon_i\subset\upsilon_j$ whenever $i$, $j\in J$ and $i<j$, while
$x\in F_{\upsilon_j}$ for every $j\in J$.
But now we see that $\xi(2j)<\xi(2i)$ whenever $i<j$ in
$J$, which is impossible.\ \Bang

Thus $x\in A$;  as $x$ was arbitrary, $\sigma'$ is a winning strategy for I
in $\Game(\Bbb N,A)$.\ \Qed

\medskip

{\bf (f)} Suppose II has a winning strategy $\tau$ in
$\Game(\Bbb N\times\kappa,F)$.   Then II has a winning strategy in
$\Game(\Bbb N,A)$.   \Prf\ Fix a normal $\kappa$-additive ultrafilter
$\Cal F$
on $\kappa$ (541Ma).   For $w=(k_0,\ldots,k_{2n})\in\Bbb N^{2n+1}$
consider the
function $f_w:[\kappa]^{n+1}\to\Bbb N$ defined by saying that
$f_w(J)$ is to be the first component of
$\tau(\langle(k_{2i},\xi_i)\rangle_{i\le n})$
where $(\xi_0,\ldots,\xi_n)$ is
that enumeration of $J$ such that, for $i$, $j\le n$, $\xi_i\le\xi_j$ iff
$i\preccurlyeq'_wj$.   Then for each $m\in\Bbb N$ there is a
$C_{wm}\in\Cal F$ such that either $f_w(J)\le m$
for every $J\in[C_{wm}]^{n+1}$
or $f_w(J)>m$ for every $J\in[C_{wm}]^{n+1}$ (4A1L).  Setting
$C=\bigcap_{m,n\in\Bbb N}\bigcap_{w\in\BbbN^{2n+1}}C_{ w m}$,
$C\in\Cal F$ and every $f_w$ is constant on $[C]^{n+1}$.
Let $\rho(w)$ be the constant value of $f_w\restr[C]^{n+1}$.

Define $\tau':\bigcup_{n\ge 1}\BbbN^n\to\Bbb N$ inductively, saying that
$\tau'(k_0,\ldots,k_n)=\rho(w)$
whenever $ w(2i)=k_i$ for $i\le n$ and
$ w(2i+1)=\tau'(k_0,\ldots,k_i)$ for $i<n$.
Suppose that $x$ is a play of $\Game(\Bbb N,A)$ consistent with
$\tau'$.   \Quer\ If $x\in A$, then
$\preccurlyeq_x$ is a well-ordering, by (c).   The order type of
$(S^*,\preccurlyeq_x)$ is countable, so is surely less than
$\otp(C)=\kappa$, and we have a function $\theta:S^*\to C$ such that
$\theta(\upsilon)\le\theta(\upsilon')$ iff
$\upsilon\preccurlyeq_x\upsilon'$.

Define $\sequencen{\xi(n)}$ by saying that

$$\eqalign{\xi(n)&=\theta(\upsilon_j)\text{ if }n=2j
  \text{ is even },\cr
&=\text{ the second component of }
  \tau((x(0),\xi(0)),(x(2),\xi(2)),\ldots,(x(2j),\xi(2j)))\cr
&\mskip200mu\text{ if }n=2j+1\text{ is odd}.\cr}$$

\noindent For $i$, $j\le n\in\Bbb N$, setting
$ w=x\restr 2n+1$,
we have $\xi(2i)$, $\xi(2j)\in C$ and

\Centerline{$\xi(2i)\le\xi(2j)
\iff\theta(\upsilon_i)\le\theta(\upsilon_j)
\iff\upsilon_i\preccurlyeq_x\upsilon_j
\iff i\preccurlyeq'_wj$.}

\noindent So

\Centerline{$x(2n+1)=\rho(w)=f_w(\{\xi(2i):i\le n\})$}

\noindent is the first component of
$\tau((x(0),\xi(0)),\ldots,\penalty-100(x(2n),\xi(2n)))$;  thus
$\sequencen{(x(n),\xi(n))}$ is a play of $\Game(\Bbb N\times\kappa,F)$
consistent with $\tau$, and is won by II.   There must therefore be
$i$, $j\in\Bbb N$ such that $\upsilon_i\subset\upsilon_j$,
$x\in F_{\upsilon_j}$ and and
$\xi(2i)\le\xi(2j)$.   Now $\upsilon_j\in T_x$ and
$\upsilon_i\preccurlyeq_x\upsilon_j$, so
$\upsilon_i\preccurlyeq\upsilon_j$;  which is impossible.\ \Bang

So $x\notin A$;  as $x$ is arbitrary, $\tau'$ is a winning strategy for
II in $\Game(\Bbb N,A)$.\ \Qed

\medskip

{\bf (g)} Putting (d), (e) and (f) together, we see that $A$ is determined.
}%end of proof of 567N

\leader{567O}{Corollary}\dvArevised{2010} [AC]
If there is a \2vm\ cardinal, then every
PCA\cmmnt{ ($=\pmb{\Sigma}^1_2$)} subset of any Polish space is
universally measurable.

\proof{{\bf (a)} Let $A\subseteq\{0,1\}^{\Bbb N}$ be PCA.   Then
there is a coanalytic subset $B$ of $\NN\times\{0,1\}^{\Bbb N}$
such that $A$ is the projection of $B$.   Of course this means that there
is a coanalytic subset $B'$ of $\{0,1\}^{\Bbb N}$ such that
$A$ is a continuous image of
$B'$, since $\NN\times\{0,1\}^{\Bbb N}$ is homeomorphic to a G$_{\delta}$
subset of $\{0,1\}^{\Bbb N}$, and any homeomorphism must carry $B$ to a
coanalytic subset of $\{0,1\}^{\Bbb N}$, by 423Tc.
Now $(h^{-1}[B']\cap F)\cup H$ is coanalytic whenever
$h:\NN\to\{0,1\}^{\Bbb N}$ is continuous, $F\subseteq\NN$ is closed and
$H\subseteq\NN$ is open;  by 567N, $(\Bbb N,(h^{-1}[B']\cap F)\cup H)$ is
always determined;  by 567F, $A$ is measured by the usual measure $\nu$ on
$\{0,1\}^{\Bbb N}$.

\medskip

{\bf (b)} Now suppose that $X$ is a Polish space,
$A\subseteq X$ is a PCA set and $\mu$ is a
Borel probability measure on $X$ with completion $\hat\mu$.
Then there is a Borel measurable function $f:\{0,1\}^{\Bbb N}\to X$
such that $\hat\mu$ is the image measure $\nu f^{-1}$.
\Prf\ Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be the measure
algebras of $\mu$, $\nu$ respectively.   Then $\frak A$ has Maharam type at
most $w(X)=\omega$ (531Aa), so there is a measure-preserving Boolean
homomorphism $\pi:\frak A\to\frak B$ (332N).   Now $\hat\mu$ is
a Radon measure (433Cb), so there is a function
$f_0:\{0,1\}^{\Bbb N}\to X$ such that $f_0^{-1}[E]$ is measured by
$\nu$, and $\nu f_0^{-1}[E]=\hat\mu E$, whenever $E$ is measured by
$\hat\mu$
(416Wb).   In this case, $f_0$ is almost continuous (433E) and there is a
sequence $\sequencen{K_n}$ of compact subsets of $\{0,1\}^{\Bbb N}$ such
that $\lim_{n\to\infty}\nu K_n=1$ and $f_0\restr K_n$ is continuous for
every $n$.   Fix any $x_0\in X$ and set $f(z)=f_0(z)$ for
$z\in\bigcup_{n\in\Bbb N}K_n$, $x$ for
other $z\in\{0,1\}^{\Bbb N}$;  then $f$ is Borel measurable and equal
$\nu$-a.e.\ to $f_0$, so $f$ also is \imp\ for $\nu$ and $\hat\mu$.
Finally, because $f$ likewise is almost continuous, the image measure
$\nu f^{-1}$ on $X$ is a Radon measure (418I), and must be exactly
$\hat\mu$ (416Eb).\ \Qed

Since $f^{-1}[A]$ is PCA (423Td), $\nu$ measures $f^{-1}[A]$ and $\hat\mu$
measures $A$.   As $\mu$ is arbitrary, $A$ is universally measurable.
}%end of proof of 567O

\exercises{\leader{567X}{Basic exercises (a)}
%\spheader 567Xa
Let $X$ be a non-empty well-orderable set, with its discrete topology,
and $G\subseteq X^{\Bbb N}$ an open set.   Show that $G$ is determined.
%567B

\spheader 567Xb [AC($\Bbb R;\omega$)] Let $A\subseteq\NN$ be such that
$\{x:\fraction{n}^{\smallfrown}x\in A\}$ is determined for every
$n\in\Bbb N$.   Show that $\NN\setminus A$ is determined.
%567B

\spheader 567Xc Show that AD is true iff every subset of $\{0,1\}^{\Bbb N}$
is determined.   \Hint{For $x\in\{0,1\}^{\Bbb N}$ set
$I_x=\{n:x(2n)=1\}$, $J_x=\{n:x(2n+1)=1\}$;  set
$C_{\text{I}}=\{x:\sup I_x>\sup J_x\}$,
$D=\{x:I_x$ and $J_x$ are both infinite$\}$.   Define $f:D\to\NN$ by
setting $f(x)(0)=\min I_x$, $f(x)(2n+1)=\min\{k:f(x)(2n)+k\in J_x\}$,
$f(x)(2n+2)=\min\{k:f(x)(2n+1)+k+1\in I_x\}$.   Show that if
$A\subseteq\NN$ and $C_{\text{I}}\cup f^{-1}[A]$ is determined, then
$A$ is determined.}
%567C

\spheader 567Xd [AC($\Bbb R;\omega$)] (i) Show that the intersection of a
sequence of closed cofinal subsets of $\omega_1$ is cofinal.   (ii) Show
that we have a unique topological probability measure on $\omega_1$ which
is zero on singletons and inner regular with respect to the closed sets.
%567E

\spheader 567Xe [AD] Show that if $f:[0,1]^2\to\Bbb R$ is a bounded
function, then $\iint f(x,y)dxdy$ and $\iint f(x,y)dydx$ are defined and
equal, where the integrations are with respect to Lebesgue measure on
$[0,1]$.
%567G

\spheader 567Xf [AD] Let $\mu$ be a Radon measure on a Polish space $X$,
and $\Cal E$ a well-ordered family of subsets of $X$.   Show that
$\mu(\bigcup\Cal E)=\sup_{E\in\Cal E}\mu E$.   \Hint{567Xe.}
%567Xe 567G

\spheader 567Xg [AD+AC($\omega$)]
Show that there are no interesting Sierpi\'nski sets, in the
sense that every atomless probability space has an uncountable negligible
subset.
%567G

\spheader 567Xh [AD] Show that every semi-finite measure space is
perfect.
%567G

\spheader 567Xi [AD] Show that if $X$ is a separable Banach space and $Y$
is a normed space then every linear operator from $X$ to $Y$ is bounded.
%567H

\spheader 567Xj\dvAnew{2011}
[DC] Let $I$ be a set, and $\widehat{\Cal B}$ the Baire-property algebra of
$\Cal PI$ with its usual topology.   Show that every
$\widehat{\Cal B}$-measurable real-valued
finitely additive functional on $\Cal PI$ is completely additive.
\Hint{remember to prove that $\Cal PI$ is a Baire space.}
%567I out of order query

\spheader 567Xk [AD] (i) Show that there is no non-principal ultrafilter on
$\Bbb N$.   (ii) Show that $\{0,1\}^{\Bbb R}$ is not compact.
%567J

\spheader 567Xl [AD] Show that there is no linear
lifting for Lebesgue measure on $\Bbb R$.   \Hint{567J.}
%567J

\spheader 567Xm [AD] (i) Show that $\ell^1(\Bbb R)$ is not reflexive.
(ii) Show that $\ell^1(\omega_1)$ is not reflexive.
%567G 567L

\spheader 567Xn [AD] (i) Show that there is no injective function from
$\omega_1$ to $\Bbb R$.   (ii) Show that there is no family
$\ofamily{\xi}{\omega_1}{f_{\xi}}$ such that $f_{\xi}$ is an injective
function from $\xi$ to $\Bbb N$ for every $\xi<\omega_1$.   (iii) Show that
there is no function $f:\omega_1\times\Bbb N\to\omega_1$ such that
$\{f(\xi,n):n\in\Bbb N\}$ is a cofinal subset
of $\xi$ for every non-zero limit ordinal $\xi<\omega_1$.   \Hint{567L.}
%567L

\spheader 567Xo (i) Show that there is a set $A\subseteq\omega_1^{\Bbb N}$
such that $\Game(\omega_1,A)$ is not determined.   \Hint{Set II the task of
enumerating $x(0)$;  see 567D and 567Xn.}
(ii) Show that there is a set $A\subseteq(\Cal P\Bbb R)^{\Bbb N}$
such that $\Game(\Cal P\Bbb R,A)$ is not determined.
%567Xn 567L Mycielski 64? Kanamori 03, 27.12

\spheader 567Xp [AD] Show that there is a surjective function from
$\Bbb R$ to $\Cal B(\Bbb R)$, but no injective function from
$\Cal B(\Bbb R)$ to $\Bbb R$.   \Hint{567E, 561Xd.}
%567Xn, 567L

\spheader 567Xq [AC] Show that if there is a \2vm\ cardinal and
$A\subseteq\NN$ is analytic then $A$ is determined.
%567N

\spheader 567Xr [AC] Suppose that there is a \2vm\ cardinal.   Show
that every PCA subset of $\Bbb R$ has the Baire property.
%567O

\leader{567Y}{Further exercises (a)}
%\spheader 567Ya
Let $X$ be a non-empty
set and $A\subseteq X^{\Bbb N}$.   A
{\bf quasi-strategy for I} in $\Game(X,A)$ is a function
$\sigma:\bigcup_{n\in\Bbb N}X^n\to\Cal PX\setminus\{\emptyset\}$;  it
is a winning quasi-strategy if $x\in A$ whenever $x\in X^{\Bbb N}$ and
$x(2n)\in\sigma(\ofamily{i}{n}{x(2i+1)})$ for every $n$.   Similarly, a
winning quasi-strategy for II is a function
$\tau:\bigcup_{n\ge 1}X^n\to\Cal PX\setminus\{\emptyset\}$ such that
$x\notin A$ whenever $x\in X^{\Bbb N}$ and
$x(2n+1)\in\tau(\ofamily{i}{n}{x(2i)})$ for every $n$.
(i) Show that if $X$ is any non-empty discrete space
and $F\subseteq X^{\Bbb N}$ is closed then at least one player has a
winning quasi-strategy in $\Game(X,F)$.
(ii) Show that DC is true iff there is no game $\Game(X,A)$ such that
both players have winning quasi-strategies.
%567B mt56bits

\spheader 567Yb [AD] Show that every uncountable subset of $\Bbb R$ has a
non-empty perfect subset.
({\it Hint\/}:  Let $A\subseteq\{0,1\}^{\Bbb N}$.
Enumerate $\bigcup_{n\in\Bbb N}\{0,1\}^n$ as
$\sequence{j}{\upsilon_j}$.   For $x\in\NN$ set

\Centerline{$f(x)
={\upsilon_{x(0)}}^{\smallfrown}\fraction{\min(1,x(1))}^{\smallfrown}
  {\upsilon_{x(2)}}^{\smallfrown}\fraction{\min(1,x(3))}^{\smallfrown}
  \ldots$.}

\noindent Consider $\Game(\Bbb N,f^{-1}[A])$.)
%567G Kanamori 03, 27.5

\spheader 567Yc\dvAnew{2010} [AD]
Let $X$ be an analytic Hausdorff space, and
$c:\Cal PX\to[0,\infty]$ a submodular Choquet capacity.   Show that
$c(A)=\sup\{c(K):K\subseteq A$ is compact$\}$  for every
$A\subseteq X$.  (Cf.\ 479Yj.)
%567G out of order query
%mt56bits

\spheader 567Yd\dvAnew{2011}
[AC($\Bbb R;\omega$)] Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra, and $\nu:\frak A\to\Bbb R$ an additive functional which is Borel
measurable for the order-sequential topology on $\frak A$.   Show that
$\nu$ is countably additive.
%567J out of order query

\spheader 567Ye Let $\Theta$ be the least ordinal such that there
is no surjection from $\Cal P\Bbb N$ onto $\Theta$.
(i) [AC($\omega$)] Show that $\cf\Theta>\omega$.
(ii) [AD] Show that $\Theta=\omega_{\Theta}$.
%use method of 567M to build a family
%$\ofamily{\xi}{\Theta}{f_{\xi}}$ such that
%$f_{\xi}:\Cal P\Bbb N\to\omega_{\xi}$ is surjective for every
%$\xi<\theta$.
%any chance of  \cf\Theta>\omega  with  AD  bu not  AC(\omega) ? query
%567M

\spheader 567Yf
[AC] Suppose that there is a \2vm\ cardinal.   Show that
every uncountable PCA subset of $\Bbb R$ has a non-empty perfect subset.
%567O 567Yb Kanamori 27.14
}%end of exercises

\endnotes{
\Notesheader{567} The consequences of the axiom of determinacy are so
striking that the question of its consistency is particularly pressing.
In fact W.H.Woodin has determined
its consistency strength, in terms of large cardinals
({\smc Kanamori 03}, 32.16, or {\smc Jech 03}, 33.27), and
this is less than that of the existence of a supercompact cardinal;  so it
seems safe enough.

%\query what about AD+DC?

In ZFC, 567B is most naturally thought of as a basic special case of
Martin's theorem that
every Borel subset of $X^{\Bbb N}$, for any discrete space $X$,
is determined ({\smc Martin 75}, or {\smc Kechris 95}, 20.5).
The idea of the proof is that if II has no winning strategy, then all I has
to do is to avoid positions from which II can win.   But for a proof in ZF
we need more than this.   It would not be enough to show that for every
first move by I, there is a winning strategy for II from the resulting
position;  we should
need to show that these can be pieced together as a single
function $\tau:\bigcup_{n\ge 1}X^n\to X$.   Turning this round, AD must
imply a weak form of the axiom of choice (567D;  see also 567Xo).
In the particular case
of 567B, we have a basic set $W_0$ of winning positions for II with a
trivial family $\family{w}{W_0}{\tau_w}$ of strategies.   (Starting
from a position in
$W_0$, II can simply play the $\preccurlyeq$-least point of $X$ to get
a position from which I cannot avoid $W_0$.)
From these we can work backwards to construct a
family $\family{w}{W}{\tau_w}$ of
strategies, where $W=\bigcup_{\xi\in\On}W_{\xi}$;  so that if
$\fraction{u}\in W$ for every $u\in X$, we can assemble these into a
winning strategy for II in $\Game(X,F)$.

The central result of the section is I suppose 567G.   From the point of
view of a real analyst like myself, as opposed to a logician or set
theorist, this is the door into a different world, explored in
567H-567K, %567H 567J 567K
567Xe-567Xm %567Xe 567Xf 567Xg 567Xi 567Xl 567Xk 567Xm
and 567Yb.
In 567I we have a result which is already interesting in ZFC.   Recall that
in ZFC there are non-trivial
additive functionals on $\Cal P\Bbb N$ which are measurable
in the sense of \S464 (464Jb);
none of them can be Baire-property-measurable.

I have not talked about `automatic continuity' in this book.   If you have
seen anything of this subject you will recognise the three parts of
567H as versions of standard results on homomorphisms which are
measurable in some sense.   I do not know whether the hypothesis `abelian'
is necessary in 567Hb.
If you like, 567J can also be thought of as an
automatic-continuity result.

You will see that 567H-567J %567H 567I 567J
depend on 567Gb rather than on 567Ga;  that is, on category rather than on
measure.   It is not clear how much can be proved if we assume, as an
axiom, that every subset of $\Bbb R$ is Lebesgue measurable (together with
AC($\omega$) at least, of course), rather than that every subset of
$\Bbb R$ has the Baire property.

In 567L far more is true, at least with AD+DC;  $\omega_2$, as
well as $\omega_1$,
is \2vm, and the filter on $\omega_1$ generated by the closed
cofinal sets is an ultrafilter ({\smc Kanamori 03}, \S28, or
{\smc Jech 03}, Theorem 33.12).
I am not sure what we should think of as a `\rvm\ cardinal' in this
context.   In the language of 566Xl, AD implies that $\Bbb R$ is not
measure-free, and Lebesgue measure is $\kappa$-additive for every initial
ordinal $\kappa$ (567Xf).
For further combinatorial consequences of AD, see {\smc Kanamori 03}.
Note that AD implies CH in the form `every uncountable subset of $\Bbb R$
is equipollent with $\Bbb R$' (567Yb).   But the relationship of $\Bbb R$
with $\omega_1$ is quite different.   ZF is enough to build a surjection
from $\Bbb R$ onto $\omega_1$.   AD implies that there is no injection
from $\omega_1$ into $\Bbb R$ (567Xn) but that
there are surjections from $\Bbb R$ onto much larger initial ordinals
(567M, 567Ye).

In 567N-567O I return to the world of ZFC;  they are in this section
because they depend on 567B and 567F.   Once again, much more is known
about determinacy compatible with AC, and
may be found in {\smc Kanamori 03} or {\smc Jech 03}.

}%end of notes

\discrpage
\references{
\frfilename{mt5r.tex}
\versiondate{28.4.23}
\copyrightdate{2008}

% 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}

\centerline{\bf References for Volume 5}

%use subparagraph details (5A1Aa) only for X. Y. Z.
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\medskip%A

%Antonovskii M.Ja.\ \& Chudnovsky D.V.\ [76] `Some questions of general
%topology and Tikhonov semifields II', Russian Math.\ Surveys 31 (1976)
%69-128.

Argyros S.A.\ \& Kalamidas N.D.\ [82] `The $K_{\alpha n}$ property on
spaces with strictly positive measure', Canadian J.\ Math.\ 34 (1982)
1047-1058.
\cmmnt{[525T.]}

Argyros S.A.\ \& Tsarpalias A.\ [82] `Calibers of compact spaces',
Trans.\ Amer.\ Math.\ Soc.\ 270 (1982) 149-162.
\cmmnt{[525N.]}

\medskip%B

Babiker A.G.\ [76] `On uniformly regular topological measure spaces',
Duke Math.\ J.\ 43 (1976) 775-789.
\cmmnt{[533Xd.]}

Balcar B.\ \& Fran\v{e}k F.\ [82] `Independent families in complete Boolean
algebras', Trans.\ Amer.\ Math.\ Soc.\ 274 (1982) 607-618.
\cmmnt{[515H.]}

Balcar B., Jech T.\ \& Paz\'ak T.\ [05]
`Complete ccc Boolean algebras, the order sequential topology,
and a problem of von Neumann',
Bull.\ London Math.\ Soc.\ 37 (2005) 885-898.
\cmmnt{[\S539 {\it intro.}, 539E, 539L, 539N, \S539 {\it notes\/}, 555K.]}
 % arXiv.math.LO/0312473

Balcar B., Jech T.\ \& Zapletal J.\ [97] `Semi-Cohen Boolean algebras',
Ann.\ Pure \& Applied Logic 87 (1997) 187-208.
\cmmnt{[\S515 {\it notes\/}, \S547 {\it notes\/}.]}

Balcar B.\ \& Vojt\'a\v{s} P.\ [77] `Refining systems on Boolean algebras',
pp.\ 45-58 in {\smc Lachlan Srebny \& Zarach 77}.
\cmmnt{[515E.]}

Bar-Hillel Y.\ [70] {\it Mathematical Logic and Foundations of Set Theory.}
North-Holland, 1970.

Bartoszy\'nski T.\ [84] `Additivity of measure implies additivity of
category', Trans.\ Amer.\ Math.\ Soc.\ 281 (1984) 209-213.
\cmmnt{[Chap.\ 52 {\it intro.}, 522Q, \S524 {\it notes\/}.]}

Bartoszy\'nski T.\ [87] `Combinatorial aspects of measure and category',
Fund.\ Math.\ 127 (1987) 225-239.
\cmmnt{[522S.]}
% http://matwbn.icm.edu.pl/ksiazki/fm/fm127/fm127118.pdf

% Bartoszy\'nski T.\ [88] `On covering of real line by null sets',
% Pacific J.\ Math.\ 131 (1988) 1-12.

% Bartoszy\'nski T.\ [p97] `Invariants of measure and category', in
% preparation for {\smc Foreman Kanamori \& Magidor ??}

% Bartoszy\'nski T.\ [00] `A note on duality between measure and category',
% Proc.\ Amer.\ Math.\ Soc.\ 128 (2000) 2745-2748.
   % (2^N,+,meager) not iso to (2^N,+,negligible)

Bartoszy\'nski T.\ \& Judah H.\ [89] `On the cofinality of the smallest
covering of the real line by meager sets', J.\ Symbolic Logic 54 (1989) 828-832.
\cmmnt{[522V.]}

Bartoszy\'nski T.\ \& Judah H.\ [95] {\it Set Theory:  on the structure
of the real line}.  A.K.Peters, 1995.
\cmmnt{[522R, \S522 {\it notes\/},
526M, \S528 {\it notes\/}, 534Q, \S534 {\it notes\/}, 546I,
553C, 554Ya, 5A1Q.]}

%Bartoszy\'nski T., Judah H.\ \& Shelah S.\ [93] `The Cicho\'n diagram',
%J.\ Symbolic Logic 58 (1993) 401-423.

%Bartoszy\'nski T.\ \& Scheepers M.\ [96] (eds.) {\it Set Theory (BEST, 1992-1994).}
%Amer.\ Math.\ Soc., 1996 (Contemporary Mathematics 192).
% ISBN 0821803069

Bartoszy\'nski T.\ \& Shelah S.\ [92] `Closed measure zero sets', Ann.\
Pure \& Applied Logic 58 (1992) 93-110.
\cmmnt{[526M.]}

Baumgartner J.E., Taylor A.D.\ \& Wagon S.\ [77] `On splitting
stationary subsets of large cardinals',
J.\ Symbolic Logic 42 (1977) 203-214.
\cmmnt{[\S541 {\it notes\/}, \S544 {\it notes\/}.]}

Bell M.G.\ [81] `On the combinatorial principle $P(c)$',
Fundamenta Math.\ 114 (1981) 149-157.
\cmmnt{[\S517 {\it notes\/}.]}

Benedikt M.\ [98] `Ultrafilters which extend measures',
J.\ Symbolic Logic 63 (1998) 638-662.
\cmmnt{[538M, 538Xm, \S538 {\it notes\/}.]}

Benedikt M.\ [99] `Hierarchies of measure-theoretic ultrafilters',
Annals of Pure and Applied Logic 97 (1999) 203-219.
\cmmnt{[538H.]}

Blair C.E.\ [77] `The Baire category theorem implies the principle of
dependent choice', Bull.\ Acad.\ Polon.\ Sci.\ (Math.\ Astron.\ Phys.) 25
(1977) 933-934.
\cmmnt{[\S566 {\it notes\/}.]}

%Blass A.\ [88] `Selective ultrafilters and homogeneity', Ann.\ Pure and
%Applied Logic 38 (1988) 252-285.

%Blass A.\ [p01b] `Combinatorial cardinal characteristics of the continuum',
%pp. ?? in {\smc Foreman Kanamori \& Magidor ??}.

Blass A.\ \& Laflamme C.\ [89] `Consistency results about filters and the
number of inequivalent growth types', J.\ Symbolic Logic 54 (1989) 50-56.
% u<g  filter dichotomy
\cmmnt{[5A6Ib, 5A6J.]}

Booth D.\ [70] `Ultrafilters on a countable set',
Ann.\ Math.\ Logic 2 (1970) 1-24.
\cmmnt{[517R.]}

Borel E.\ [1919] `Sur la classification des ensembles de mesure nulle', Bull.\ Soc.\ Math.\ France 47 (1919) 97-125.
\cmmnt{[\S534 {\it intro.}]}

Brendle J.\ [00] {\it Inner model theory and large cardinals.}
Research report, March 2000.
\cmmnt{[528N, 529F-529H.]} %529F 529G 529H

%Brendle J.\ [01] `Cardinal invariants of the continuum --- a survey',
%J.\ Research Institute for Mathematical Sciences, Kyoto University
%(JRIMS Kokyuroku) 1202 (2001) 7-32.

Brendle J.\ [06] `Cardinal invariants of the continuum and combinatorics
on uncountable cardinals',
Annals of Pure and Applied Logic 144 (2006) 43-72.
\cmmnt{[529F, 529H.]}

Burke M.R.\ [93] `Liftings for Lebesgue measure', pp.\ 119-150 in
{\smc Judah 93}.
\cmmnt{[\S535 {\it notes\/}.]}

%Burke M.R.\ [n95] `Consistent liftings', privately circulated, 1995.

%Burke M.R.\ [n96] `A theorem of Gitik and Shelah on disjoint refinements
%of sequences of subsets of the real line', note of 23.8.96.
%\cmmnt{[5x46J, \S547 {\it intro.}]}

Burke M.R.\ [n05] `Non-null sets of minimal cofinality', note of 2.3.05.
\cmmnt{[523K.]}

Burke M.R.\ \& Magidor M.\ [90] `Shelah's pcf theory and
its applications', Ann.\ Pure and Applied Logic 50 (1990) 207-254.
\cmmnt{[513J, \S5A2 {\it intro.}.]}

Burke M.R.\ \& Shelah S.\ [92]
`Linear liftings for non-complete probability spaces',
Israel J.\ Math.\ 79 (1992) 289-296.
\cmmnt{[\S535 {\it notes\/}.]}

\medskip%C

Carlson T.J.\ [84] `Extending Lebesgue measure to infinitely many sets',
Pacific J.\ Math.\ 115 (1984) 33-45.
\cmmnt{[552N, \S552 {\it notes.}]}

Carlson T.J.\ [93] `Strong measure zero and strongly meager sets',
Proc.\ Amer.\ Math.\ Soc.\ 118 (1993) 577-586.
\cmmnt{[534R.]}

Carlson T., Frankiewicz R.\ \& Zbierski P.\ [94]
`Borel liftings of the measure
algebra and the failure of the continuum hypothesis', Proc.\ Amer.\ Math.\ Soc.\
120 (1994) 1247-1250.
\cmmnt{[\S535 {\it notes\/}, 554I.]}

Choksi J.R.\ \& Fremlin D.H.\ [79] `Completion regular measures on
product spaces', Math.\ Ann.\ 241 (1979) 113-128.
\cmmnt{[532I.]}

%Cicho\'n J., Kamburelis T.\ \& Pawlikowski J.\ [85] `On dense subsets of
%the measure algebra', Proc.\ Amer.\ Math.\ Soc.\ 94 (1985) 142-146.

Cicho\'n J.\ \& Pawlikowski J.\ [86]
`On ideals of subsets of the plane and on Cohen
reals', J.\ Symbolic Logic 51 (1986) 560-569.
\cmmnt{[527F.]}

%Cohen P.\ [63] `The independence of the continuum hypothesis',
% Proc.\ Nat.\ Acad.\ Sci.\ U.S.A.\ 50 (1963) 1143-1148.

Comfort W.W.\ \& Negrepontis S.\ [74] {\it The Theory of Ultrafilters.}
Springer, 1974.
\cmmnt{[538F, 538Yb.]}

Comfort W.W.\ \& Negrepontis S.\ [82] {\it Chain Conditions in
Topology.}   Cambridge U.P., 1982.
\cmmnt{[516 {\it notes\/}.]}

Cummings J.\ [92] `A model in which GCH holds at successors but fails at
limits',
Trans.\ Amer.\ Math.\ Soc.\ 329 (1992) 115-142.
\cmmnt{[525Z.]}

\medskip%D

Dellacherie C., Meyer P.A.\ \& Weil M.\ (eds.) [73]
{\it S\'eminaire de Probabilit\'es VII.}   Springer, 1973
(Springer Lecture Notes in Mathematics 321).

Devlin K.J.\ [84] {\it Constructibility.} Springer, 1984.
\cmmnt{[5A6D.]}

%Dow A.\ \& Hart K.P.\ [p98] `The measure algebra does not always embed',
%preprint, 1998

Dow A.\ \& Stepr\=ans J.\ [94] `The $\sigma$-linkedness of the measure
algebra', Canad.\ Math.\ Bulletin 37 (1994) 42-45.
\cmmnt{[524L.]}

%Drake F.\ [74] {\it Set Theory.}  North-Holland, 1974.

%Drake F.R.\ \& Singh D.\ [96] {\it Intermediate Set Theory.}
%Wiley, 1996.  \cmmnt{[\S5A1.]}
% ISBN 0 471 96496 4

D\v{z}amonja M.\ \& Kunen K.\ [93] `Measures on compact HS spaces',
Fundamenta Math.\ 143 (1993) 41-54.
\cmmnt{[\S531 {\it notes\/}.]}

%D\v{z}amonja M.\ \& Kunen K.\ [95] `Properties of the class of
%measure-separable compact spaces', Fundamenta Math.\ 147 (1995) 261-277.

D\v{z}amonja M.\ \& Plebanek G.\ [04]
`Precalibre pairs of measure algebras',
Topology Appl.\ 144 (2004) 67-94.
\cmmnt{[525L.]}

\medskip%E

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).
\cmmnt{[\S5A4.]}

%Engelking R.\ \& Kar{\l}owicz M.\ [65]
%`Some theorems of set theory and their topological consequences',
% Fund.\ Math.\ 57 (1965) 275-285.

Erd\H{o}s P., Hajnal A., M\'at\'e A.\ \& Rado R.\ [84] {\it Combinatorial
Set Theory:  Partition Relations for Cardinals.}   Akad\'emiai Kiad\'o,
1984 (Disquisitiones Math.\ Hung.\ 13).
\cmmnt{[5A1F, 5A1H, 5A6F.]}

\medskip%F

Farah I.\ [00] {\it Analytic Quotients.}
Mem.\ Amer.\ Math.\ Soc.\ 148 (2000).
\cmmnt{[5A6H.]}

Farah I.\ [03] `How many algebras $\Cal P(\Bbb N)/\Cal I$ are there?',
Illinois J.\ Math.\ 46 (2003) 999-1033.
\cmmnt{[\S556 {\it notes\/}.]}

Farah I.\ [06] `Analytic Hausdorff gaps II: the density zero ideal',
Israel J.\ Math.\ 154 (2006) 235-246.
\cmmnt{[556S.]}

Farah I.\ \& Veli\v{c}kovi\'c B.\ [06]
`Von Neumann's problem and large cardinals',
Bull.\ London Math.\ Soc.\ 38 (2006) 907-912.
\cmmnt{[539Q.]}

Feferman S. \& L\'evy A. [63] `Independence results in set theory by Cohen's
method', Notices Amer.\ Math.\ Soc.\ 10 (1963) 53.
\cmmnt{[561A.]}

Fleissner W.G.\ [91]
`Normal measure axiom and Balogh's theorems', Topology and its Appl.\
39 (1991) 123-143.
\cmmnt{[555N.]}

% Fleissner W.G., Hansell R.W.\ \& Junnila H.J.K.\ [82] `PMEA implies
% Proposition P', Topology and its Appl.\ 13 (1982) 255-262.

% Foreman M., Kanamori A.\ \& Magidor M.\ [??] {\it Handbook of Set
% Theory}.   Kluwer, to appear.

%Foreman M., Magidor M.\ \& Shelah S.\ [88] `Martin's maximum, saturated ideals, and non-regular ultrafilters', Annals of Math.\ (2) 127 (1988) 1-47.

Foreman M.\ [10] `Ideals and generic elementary embeddings',  pp.\ 885-1147
in {\smc Foreman \& Kanamori 10}, vol.\ 2.
\cmmnt{[547Z.]}

Foreman M.\ \& Kanamori A.\  [10] {\it Handbook of Set Theory.}
Springer, 2010.

Foreman M.\ \& Wehrung F.\ [91] `The Hahn-Banach theorem implies the
existence of a non-Lebesgue measurable set', Fundamenta Math.\ 138
(1991) 13-19.
\cmmnt{[563A.]}

Foreman M.\ \& Woodin W.\ [91] `The generalized continuum hypothesis can
fail everywhere', Annals of Math.\ 133 (1991) 1-35.
\cmmnt{[525Z.]}

Fossy J.\ \& Morillon M.\ [98] `The Baire category property and some
notions of compactness', J.\ London Math.\ Soc. (2) 57 (1998) 1-19.
\cmmnt{[\S566 {\it notes\/}.]}

% Frankiewicz R.\ \& Plebanek G.\ [94] `Nonaccessible filters in
% measure algebras and functionals on $L^{\infty}(\lambda)^*$', Studia
% Math.\ 108 (1994) 191-200.

%Frankiewicz R.\ \& Plebanek G.\ [96] `On asymptotic density and
%uniformly distributed sequences', Studia Math.\ 119 (1996) 17-26.

Freese R. \& Nation J.B.\ [78] `Projective lattices',
Pacific J.\ Math.\ 75 (1978) 93-106.
\cmmnt{[\S518 {\it notes\/}.]}

Freiling C.\ [86] `Axioms of symmetry;  throwing darts at the real number
line', J.\ Symbolic Logic 51 (1986) 190-200.
\cmmnt{[537K.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and Measure Theory.}
Cambridge U.P., 1974.
\cmmnt{[564Xc.]}

Fremlin D.H.\ [77]
`Uncountable powers of {\bf R} can be almost Lindel\"of',
Manuscripta Math.\ 22 (1977) 77-85.
\cmmnt{[533J.]}

Fremlin D.H.\ [84a] {\it Consequences of Martin's Axiom.}
Cambridge U.P., 1984.
\cmmnt{[\S511 {\it notes\/}, \S517 {\it notes\/}, \S531 {\it notes\/}.]}

Fremlin D.H.\ [84b] `On the additivity and cofinality of Radon measures',
Mathematika 31 (1984) 323-335.
\cmmnt{[\S524 {\it notes\/}.]}

Fremlin D.H.\ [87] {\it Measure-additive Coverings and
Measurable Selectors.} Dissertationes Math.\ 260 (1987).
\cmmnt{[551A.]}

Fremlin D.H.\ [88] `Large correlated families of positive
random variables',
Math.\ Proc.\ Cambridge Phil.\ Soc.\ 103 (1988) 147-162.
\cmmnt{[525S.]}

Fremlin D.H.\ [91] `The partially ordered sets of measure
theory and Tukey's ordering,' Note di Matematica  11 (1991) 177-214.
[\S524 {\it notes\/}, 526B, 526I, 527J, 529C, 529D, 534L.]

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304
in {\smc Judah 93}.
\cmmnt{[\S537 {\it notes\/}, \S541 {\it notes\/}, \S544 {\it notes\/},
\S545 {\it notes\/}, \S5A2 {\it intro.}.]}

Fremlin D.H.\ [97] `On compact spaces carrying Radon measures of
uncountable Maharam type', Fundamenta Math.\  154 (1997) 295-304.
\cmmnt{[531T.]}

%Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces',
%Fundamenta Math.\ 165 (2000) 67-94.

Fremlin D.H.\ [03] `Skew products of ideals',
J.\ Applied Analysis 9 (2003) 1-18.
\cmmnt{[\S527 {\it notes\/}.]}
%n01910.tex

%Fremlin D.H.\ [n97] `A theorem of Gitik and Shelah on disjoint
%refinements of sequences of subsets of the real line', note of 4.4.97.
%n96j01.tex

%Fremlin D.H.\ [n99] `Embedding the Lebesgue measure algebra into
%$\Cal P\Bbb N/\Cal Z$', note of 25.5.99.
%n99525.tex

%Fremlin D.H.\ [n99] `Notes on {\smc Farah p99}', note of 26.7.99.
%n99612.tex

%Fremlin D.H.\ [n99] `Notes on {\smc Solecki p96}', note of 17.6.99.
%n99617.tex

%Fremlin D.H.\ [n02] `Games and the Tukey classification', note of 29.9.02.
%n02929.tex

Fremlin D.H.\ \& Grekas S.\ [95] `Products of completion regular
measures', Fundamenta Math.\ 147 (1995) 27-37.
\cmmnt{[532D.]}

Fremlin D.H.\ \& Miller A.W.\ [88]
`On some properties of Hurewicz, Menger and Rothberger',
Fund.\ Math.\ 129 (1988) 17-33.
\cmmnt{[534Q.]}

Fremlin D.H., Natkaniec T.\ \& Rec{\l}aw I.\ [00]
`Universally Kuratowski-Ulam spaces', Fundamenta Math.\ 165 (2000) 239-247.
\cmmnt{[\S527 {\it notes\/}.]}

%Friedman H.\ [80]
%`A consistent Fubini-Tonelli theorem for nonmeasurable functions',
% Illinois J.\ Math.\ 24 (1980) 390-395.

Friedman S.D. \& Koepke P.\ [97] `An elementary approach to the fine
structure of L', Bull.\ Symbolic Logic 3 (1997) 453-468.
\cmmnt{[5A6D.]}

Fuchino S.,  Geschke S., Shelah S.\ \& Soukup L.\ [01]
`On the weak Freese-Nation property of complete Boolean algebras',
Ann.\ Pure Appl.\ Logic 110 (2001) 89-105.
\cmmnt{[518K.]}

Fuchino S., Geschke S.\ \& Soukup L.\ [01] `On the weak Freese-Nation
property of $\Cal P(\omega)$', Arch.\ Math.\ Logic 40 (2001) 425-435.
\cmmnt{[522U, \S522 {\it notes\/}.]}

Fuchino S., Koppelberg S.\ \& Shelah S.\ [96] `Partial orderings with
the weak Freese-Nation property',
Ann.\ Pure and Applied Logic 80 (1996) 35-54.
\cmmnt{[518A, 518G, 518Yb, \S518 {\it notes\/}, 522U.]}

Fuchino S.\ \& Soukup L.\ [97] `More set theory around the weak
Freese-Nation property', Fundamenta Math.\ 154 (1997) 159-176.
\cmmnt{[518I, 518K.]}

\medskip%G

Galvin F.\ [80] `Chain conditions and products',
Fundamenta Math.\ 108 (1980) 33-48.
\cmmnt{[\S553 {\it notes\/}.]}

Galvin F., Mycielski J.\ \& Solovay R.M.\ [79] `Strong measure zero sets',
Notices Amer.\ Math.\ Soc.\ 26 (1979) A280.
%(abstract 79T-E25);  often assigned to year 1973
\cmmnt{[534K.]}

Gandy R.O.\ \& Hyland J.M.E.\ [77](eds.) {\it Logic Colloquium '76.}
North-Holland, 1977.

Geschke S.\ [02] `On tightly $\kappa$-filtered Boolean algebras',
Algebra Universalis 47 (2002) 69-93.
\cmmnt{[518P, 518S, \S535 {\it notes\/}.]}

Gitik M.\ \& Shelah S.\ [89] `Forcings with ideals and
simple forcing notions', Israel J.\ Math.\ 68 (1989) 129-160.
\cmmnt{[543E, 547F.]}

Gitik M.\ \& Shelah S.\ [93]
`More on simple forcing notions and forcings with ideals',
Annals of Pure and Applied Logic 59 (1993) 219-238.
\cmmnt{[542E, 543E, 547F.]}

Gitik M.\ \& Shelah S.\ [01]
`More on real-valued measurable cardinals and forcing with ideals',
Israel J.\ Math. 124 (2001) 221-242.
\cmmnt{[548E.]}
%GiSh582

G\l\'owczy\'nski W.\ [91] `Measures on Boolean algebras',
Proc.\ Amer.\ Math.\ Soc.\ 111 (1991) 845-849.
\cmmnt{[555K.]}

G\l\'owczy\'nski W.\ [08] `Outer measure on Boolean algebras', Acta Univ.\
Carolinae (Math.\ et Phys.) 49 (2008) 3-8.
\cmmnt{[555K.]}

Goldstern M., Judah H.\ \& Shelah S.\ [93] `Strong measure zero sets
without Cohen reals',
J.\ Symbolic Logic 58 (1993) 1323-1341.
\cmmnt{[\S534 {\it notes\/}.]}

\medskip%H

% Hart J.\ \& Kunen K.\ [98] `Weak Measure Extension Axioms',
% Topology and its Applications 85 (1998) 219-246.
% http://www.math.wisc.edu/~kunen/me.ps

Haydon R.G.\ [77] `On Banach spaces which contain $\ell^1(\tau)$ and
types of measures on compact spaces', Israel J.\ Math.\ 28 (1977) 313-324.
\cmmnt{[531E, 531L.]}

%Haydon R.G.\ [78] `On dual $L^1$-spaces and injective bidual Banach
%spaces,' Israel J.\ Math.\ 31 (1978) 142-152.

Hodges W., Hyland M., Steinhorn C.\ \& Truss J.\ [96] (eds.)
{\it Logic: from Foundations to Applications,
European Logic Colloquium, 1993.}
Clarendon, 1996.
%[ISBN 0-19-853862-6]

Humke P.D.\ \& Laczkovich M.\ [05] `Symmetrically approximately
continuous functions, consistent density theorems, and Fubini type
inequalities', Trans.\ Amer.\ Math.\ Soc.\ 357 (2005) 31-44.
\cmmnt{[537Q, \S537 {\it notes\/}.]}

\medskip%I

Ihoda J.I.\ [88] `Strong measure zero sets and rapid filters',
J.\ Symbolic Logic 53 (1988) 393-402.
%model of Borel conjecture + Ramsey filter
\cmmnt{[534Q.]}

\medskip%J

Jech T.\ [73] {\it The Axiom of Choice.}  North-Holland, 1973.
\cmmnt{[561A, 561Xc, 561Yc, 561Yi, 564Yd.]}

Jech T.\ [78] {\it Set Theory.}  Academic, 1978 (ISBN 0123819504).
\cmmnt{[{\it Intro.}, 553H,
555O, \S555 {\it notes\/}, \S562 {\it notes\/}, \S5A1, 5A6B.]}
%5A1B 5A1D 5A1E 5A1F B
%ASIN B0095LMZ9O

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2003
(ISBN 3540440852).
\cmmnt{[\S521 {\it notes\/}, \S555 {\it notes\/},
\S562 {\it notes\/}, \S567 {\it notes\/}, \S5A1, 5A3N, 5A3P, 5A6B.]}
%5A1B 5A1D 5A1E 5A1F B

%Jin R.\ \& Shelah S.\ [p96] `Compactness of Loeb spaces', preprint, 1996.

%Jovanovi\'c A.\ [p01]
%`Real valued measurability, some set-theoretic aspects',
%pp.\ ?? in {\smc Pap 0?}

Judah H.\ [93] (ed.) {\it Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

Judah H.\ \& Repick\'y M.\ [95] `Amoeba reals', J.\ Symbolic Logic 60 (1995) 1168-1185.
\cmmnt{[528N.]}

%Juh\'asz I.\ [71]  {\it Cardinal Functions in Topology.}  North-Holland,
%1975 (Math.\ Centre Tracts 34).

%Juh\'asz I.\ [80]  {\it Cardinal Functions in Topology -- Ten Years Later.}  Mathematisch Centrum, Amsterdam, 1980 (Math.\ Centre Tracts 123).   ISBN 9061961963.

Just W.\ [92] `A modification of Shelah's oracle-c.c., with
applications', Trans.\ Amer.\ Math.\ Soc.\ 329 (1992) 325-356.
\cmmnt{[527M.]}

%Just W.\ \& Krawczyk A.\ [84] `On certain Boolean algebras
%$\Cal P(\omega)/I$', Trans.\ Amer.\ Math.\ Soc.\ 285 (1984) 411-429.

Just W.\ \& Weese M.\ [96] {\it Discovering Modern Set Theory I.}
Amer.\ Math.\ Soc., 1996 (Graduate Studies in Mathematics 8).
\cmmnt{[{\it Intro.}.]}

Just W.\ \& Weese M.\ [97] {\it Discovering Modern Set Theory II.}
Amer.\ Math.\ Soc., 1997 (Graduate Studies in Mathematics 18).
\cmmnt{[5A1B, 5A1F, 5A1H.]}

\medskip%K

%Kada M. \& Yuasa Y.\ [96] `Cardinal invariants about shrinkability of unbounded
%sets', Topology and its Applications 74 (1996) 215-223.

%Kamburelis A.\ [89] `Iterations of Boolean algebras with measure',
%Arch.\ Math.\ Logic 29 (1989) 21-28.
% chargeable algebras

%Kamburelis A.\ [90] `A new proof of the Gitik-Shelah theorem', Israel
%J.\ Math.\ 72 (1990) 373-380.

Kanamori A.\ [03] {\it The Higher Infinite.}   Springer, 2003.
\cmmnt{[\S541 {\it notes\/}, \S555 {\it notes\/}, \S567 {\it notes\/},
5A1H, 5A6B, 5A6F.]}

Kechris A.S.\ [95] {\it Classical Descriptive Set Theory.}  Springer,
1995.
\cmmnt{[5A1D, \S567 {\it notes\/}.]}
%QA248.K4

Keisler H.J.\ \& Tarski A.\ [64] `From accessible to
inaccessible cardinals', Fundamenta Math.\ 53 (1964) 225-308;  errata
Fundamenta Math.\ 57 (1965) 119.
\cmmnt{[\S541 {\it intro.\/}, \S541 {\it notes\/}.]}

Kelley J.L.\ [50] `The Tychonoff product theorem implies the axiom of
choice', Fundamenta Math.\ 37 (1950) 75-76.
\cmmnt{[561D.]}

Koppelberg S.\ [75]
`Homomorphic images of $\sigma$-complete Boolean algebras',
Proc.\ Amer.\ Math.\ Soc.\ 51 (1975) 171-175.
\cmmnt{[515L.]}

Koppelberg S.\ [89] {\it General Theory of Boolean
Algebras\/}, vol.\ 1 of {\smc Monk 89}.
\cmmnt{[\S515 {\it notes\/}.]}

Koppelberg S.\ \& Shelah S.\ [96]
`Subalgebras of Cohen algebras need not be Cohen',
pp.\ 261-275 in {\smc Hodges Hyland Steinhorn \& Truss 96}.
\cmmnt{[\S515 {\it notes\/}, \S547 {\it notes\/}.]}

Kraszewski J.\ [01] `Properties of ideals on the generalized Cantor spaces',
J.\ Symbolic Logic 66 (2001) 1303-1320.
\cmmnt{[523G, 523H, 523J, \S523 {\it notes\/}.]}

%Krawczyk A.\ [n85] `Dominating reals add random reals', 1985.

Kumar A.\ [13] `Avoiding rational distances', Real Analysis Exchange 38
(2012/13), 493-498.
%{\tt http://www.math.wisc.edu/\~{}akumar/ARD.pdf}.
%{\tt
%https:/{\bsp}projecteuclid.org/journals/real-analysis-exchange/volume-38%
%/issue-2/Avoiding-Rational-{\vthsp}Distances/rae/1403894908.full}
\cmmnt{[548Xb.]}

%Kumar A.\ [n16] `On a theorem of Gitik and Shelah', note of January 2016
%({\tt http:/{\bsp}www.math.huji.ac.il/\~{}akumar/otgs.pdf}).

Kumar A.\ \& Shelah S.\ [17] `A transversal of full outer measure',
Advances in Math.\ 321 (2017) 475-485.
\cmmnt{[547P, 547Q, \S548 {\it intro.\/}, 548C.]}
%http://home.iitk.ac.in/~krashu/tfom.pdf;  printed version not seen

%Kunen K.\ [75] `Some points in $\beta N$',
% Math.\ Proc.\ Cambridge Phil.\ Soc.\ 80 (1975) 385-398.

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[{\it Intro.}, 511A, \S521 {\it notes\/}, \S522 {\it notes\/},
\S541 {\it notes\/}, 551Q, \S551 {\it notes\/}, 556F,
\S562 {\it notes}, Appendix {\it intro.}, \S5A1, \S5A3, 5A6B.]}

Kunen K.\ [81] `A compact $L$-space under CH', Topology and its Appl.\
12 (1981) 283-287.
\cmmnt{[\S531 {\it notes\/}.]}

Kunen K.\ [84] `Random and Cohen reals', pp.\ 887-911 in
{\smc Kunen \& Vaughan 84}.
\cmmnt{[\S554 {\it notes}.]}

Kunen K.\ [n70] `$\Pi^1_1$ reflection at the continuum', note of January
1970.
\cmmnt{[543C, 544C, 544E, 544F.]}

Kunen K.\ \& Mill J.van [95] `Measures on Corson compact spaces',
Fundamenta Math.\ 147 (1995) 61-72.
\cmmnt{[531O.]}
%http://www.math.wisc.edu/~kunen/corson.ps

Kunen K.\ \& Vaughan J.E.\ [84] (eds.) {\it Handbook of Set-Theoretic
Topology.}   North-Holland, 1984.

Kuratowski K.\ [66] {\it Topology,} vol.\ I.   Academic, 1966.
\cmmnt{[562G.]}

\medskip%L

Lachlan A., Srebny M.\ \& Zarach A.\ [77] (eds.) {\it Set Theory and Hierarchy Theory.}  Springer, 1977 (Lecture Notes in Math.\ 619).

%Laczkovich M.\ \& Miller A.W.\ [95] `Measurability of functions with
%approximately continuous vertical sections and measurable horizontal
%sections', Colloquium Math.\ 69 (1995) 299-308.

Larson P.\ [09] `The filter dichotomy and medial limits',
J.\ Math.\ Logic 9 (2009) 159-165.
\cmmnt{[538S.]}

Larson P., Neeman I.\ \& Shelah S.\ [10] `Universally measurable sets in
generic extensions', Fundamenta Math.\ 208 (2010) 173-192.
\cmmnt{[553O.]}
	
%Larson P.\ \& Shelah S.\ [p09]
%`There may be just continuum many universally measurable sets', 2009.
%LrSh:947

Laver R.\ [76] `On the consistency of Borel's conjecture', Acta Math.\
137 (1976) 151-169.
\cmmnt{[534Q.]}

Laver R.\ [87] `Random reals and Souslin trees',
Proc.\ Amer.\ Math.\ Soc.\ 100 (1987) 531-534.
\cmmnt{[553M.]}

Levinski J.-P., Magidor M.\ \& Shelah S.\ [90] `Chang's conjecture for
$\aleph_{\omega}$', Israel J.\ Math.\ 69 (1990) 161-172.
\cmmnt{[5A6F.]}

Levy A.\ [71] `The sizes of the indescribable cardinals',
pp.\ 205-218 in {\smc Scott 71}.
\cmmnt{[\S541 {\it notes\/}, \S544 {\it notes\/}.]}

Lipecki Z.\ [09]
`Semivariations of an additive function on a Boolean ring',
Colloquium Math.\ 117 (2009) 267-279.
\cmmnt{[552Xe.]}

%Loeb P.A.\ \& Talvila E.\ [p00]
%`Covering theorems and Lebesgue integration', preprint, June 2000.

%Losada M. \& Todor\v{c}evi\'c S. [p98] `Chains and antichains in Boolean
%algebras', note of 5.12.98.

%Louveau A.\ \& Veli\v{c}kovi\'c B.\ [99] `Analytic ideals and cofinal
%types', Ann.\ Pure and Applied Logic 99 (1999) 171-195.
%\ell^1\not\prT\CalNwd

\medskip%M

Maharam D.\ [1947] `An algebraic characterization of measure algebras',
Ann.\ Math.\ 48 (1947) 154-167.
\cmmnt{[539P.]}

Makowsky J.A.\ \& Ravve E.V.\ (eds.) [98]
{\it Logic Colloquium '95.}  Springer, 1998 (Lecture Notes in Logic 11).
%ISBN 3-540-63994-2

%Malliaris M.\ \& Shelah S.\ [16] `Cofinality spectrum theorems in
%model theory, set theory and general topology',
%J.\ Amer.\ Math.\ Soc.\ 29 (2016) 237-297.

Martin D.A.\ [70] `Measurable cardinals and analytic games',
Fundamenta Math.\ 66 (1970) 287-291.
\cmmnt{[567N.]}

Martin D.A.\ [75] `Borel determinacy', Ann.\ of Math.\ (2) 102 (1975)
363-371.
\cmmnt{[\S567 {\it notes\/}.]}

Martin D.A.\ \& Solovay R.M.\ [70] `Internal Cohen
extensions', Ann.\ Math.\ Logic 2 (1970) 143-178.
\cmmnt{[Chap.\ 52 {\it intro.}, \S528 {\it intro.}, \S528 {\it notes\/}.]}

%Mathias A.R.D.\ [78] `0\# and the $p$-point problem', pp.\ 375-384 in
%{\smc M\"uller \& Scott 78}.

M\'atrai T.\ [p09] `More cofinal types of definable directed orders', 2009
({\tt https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.414.3938}).
\cmmnt{[526L, \S526 {\it notes\/}.]}
% books/matrai.p09.pdf

%Mercourakis S.\ [96] `Some remarks on countably determined measures and
%uniform distribution of sequences', Monatschefte f\"ur Math.\ 121 (1996)
%79-101.

Meyer P.-A.\ [73] `Limites m\'ediales d'apr\`es Mokobodzki',
pp.\ 198-204 in {\smc Dellacherie Meyer \& Weil 73}.
\cmmnt{[538Yl.]}

Meyer P.A.\ [75] (ed.) {\it S\'eminaire de Probabilit\'es IX.}
Springer, 1975 (Lecture Notes in Mathematics 465).

Miller A.W.\ [80] `There are no Q-points in Laver's model for the Borel
conjecture', Proc.\ Amer.\ Math.\ Soc.\ 78 (1980) 103-106.
\cmmnt{[\S538 {\it notes\/}.]}
%actually, no rapid ultrafilters, therefore no rapid filters

Miller A.W.\ [81] `Some properties of measure and category',
Trans.\ Amer.\ Math.\ Soc.\ 266 (1981) 93-114.
\cmmnt{[522J.]}

Miller A.W.\ [82] `The Baire category theorem and cardinals of countable
cofinality', J.\ Symbolic Logic 47 (1982) 275-288.
\cmmnt{[552G, 552Xb.]}

%Miller A.W.\ [83] `Mapping a set of reals onto the reals',
%J.\ Symbolic Logic 48 (1983) 575-584.

Mokobodzki G.\ [75] `R\'el\`evement bor\'elien compatible avec une
classe d'ensembles n\'egligeables.  Application \grv{a} la
d\'esint\'egration des mesures', pp.\ 437-442 in {\smc Meyer 75}.
\cmmnt{[535I.]}

Mokobodzki G.\ [7?] `D\'esintegration des mesures et rel\`evements
Boreliens de sous-espaces de $L^{\infty}(X,\Cal B,\mu)$',
Lab.\ d'Analyse Fonctionnelle, Univ.\ Paris VI (?).
\cmmnt{[535E, \S535 {\it notes\/}.]\footnote{I have not been able to
locate this paper;  I believe it was a seminar report.   I took notes from
it in 1977.}}
% "On two theorems of Mokobodzki", 23.6.77.

Monk J.D.\ [89] (ed.) {\it Handbook of Boolean Algebra.}
North-Holland, 1989.

%Monk J.D.\ [96] {\it Cardinal Invariants on Boolean Algebras.}
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%Pm%Pn%Po

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%Pp%Pq%Pr

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\medskip%R

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\medskip%S

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% Sh400a
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%Sh 430

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%http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pl/1235419814

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% Sh594;  arXiv:math.LO/9611221.

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 % Sh620

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% Sh592

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% Sh619

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% Con(\non\Cal N(\mu_H)<\non\Cal N)

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\medskip%T\medskip

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%U
\medskip

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%V
\medskip

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%W
\medskip

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\medskip%X\medskip%Y\medskip%Z

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\discrpage

}%end of references

%---------------------------------------------------------------%

\end



