% The following TeXfile comprises part of

%     MEASURE THEORY, by D.H.Fremlin
%          (University of Essex, Colchester, England;
%                   e-mail  fremdh@essex.ac.uk)

% Please contact me with any difficulties not covered by the remarks
% below.

% The file is set up for processing by plain TeX in the presence of
% files  amstex.tex ,  amssym.tex  and  amsppt.sty  suitably supported by the
% appropriate fonts.   If these are unavailable, a readable version
% can be obtained by removing the comment symbol  %  from the line

%        %   \amstexsupportedfalse

% below.   Some systems do not allow access to files  amstex.tex  and
% amsppt.sty , but instead employ a command "amstex <filename.tex>";
% for such systems it is sometimes right to delete or disable the
% commands

%        \input amstex
%        \documentstyle{amsppt}

% (I have little experience with this and should welcome comments.)

% Various formats of output are obtainable.   The principal variations
% are

%    1. "Standard" output uses the command
%               \smallprint
% and gives output adapted to A4 paper (21 cm x 29.7 cm).

%    2.  Using the command
%               \USsmallprint
% gives output adapted to 8.5" x 11" .

%    3.  Using the command
%               \noproofs
% produces an abridged version, in which proofs are omitted.

%    4.   Using the command
%               \resultsonly
% produces a super-abridged version, omitting comments and exercises
% as well as proofs.

% These commands are grouped together at the end of the introductory
% definitions;  they may be activated by deleting the initial comment
% symbols  % .

% Any diagrams included will be called for by inputting postscript files.
% (You can look for these by hunting for   \picture  in the code below.)
% To get these you will need the relevant  .ps  files with the expected
% names  ( \picture{mt133g}{...}  demands the file  mt133g.ps , for instance).


%(If you can read TeX fluently, you may wish to fast-forward to a line
%beginning   %file starts here   or   \frfilename
%if one of these is present.)

%TeXfiles with this heading are designed for processing with AmSTeX,
%accessed through TeXfiles  amstex.tex (with amssym.tex  and
% amsppt.sty  and possibly  amsppt.sti)  during compilation.   
%If these are unavailable, a readable
%version can (I hope) be produced by deleting the  %  before
% \amstexsupportedfalse  below.

%Some special difficulties arise with the importation of  .ps  files
%in diagrams.   Look for  \picture  below for notes on these.

%The following \count registers are used for special purposes:
%   \count1=\stylenumber  to specify style
%   \count2=\footnotenumber  to count footnotes

%AmSTeX declaration
\newif\ifamstexsupported
     \amstexsupportedtrue
%     \amstexsupportedfalse               %use when AMSTeX unavailable
\newif\ifDHFamstexloaded        %for detection of a non-standard AMSTeX
     \DHFamstexloadedfalse    
\newif\ifDHFamspptloaded        
     \DHFamspptloadedfalse    
\newif\iffourteensavailable
     \fourteensavailabletrue
    \fourteensavailablefalse
\newif\ifathome
     \athometrue     %to cope with a quirk     of my son's Ghostscript
     \athomefalse
\newif\iffirstpage
     \firstpagetrue

%Style assignments
\countdef\stylenumber=1
\countdef\footnotenumber=2
\def\fullsize{\magnification=\magstep1
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=1}
\def\smallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=2}
\def\printersize{\magnification=\magstep1
     \baselineskip24pt
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{8.5truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=3}
\def\refsize{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=4}
\def\rrversion{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=5}
\def\bigsize{\magnification=\magstep2
     \overfullrule0pt
     \def\pagewidth{325pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=6}
\def\bbigsize{\magnification=\magstep3
     \overfullrule0pt
     \def\pagewidth{271pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=7}
\def\USsmallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{482pt}
     \def\pageheight{8.7truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=8}
\def\halfmagsize{\magnification=\magstephalf
     \def\pagewidth{427pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=9}
           % register  stylenumber  is used to control the output format

%addresses, letterheads
\def\deptaddress{\hskip 3truein Department of Mathematical Sciences\par
   \noindent e-mail: {\tt fremdh}\hskip 2.9truein\today\par
   \medskip}
\long\def\univaddress#1{\hskip 3truein Department of Mathematical Sciences\par
   \hskip 3truein University of Essex\par
   \hskip 3truein Colchester CO4 3SQ\par
   \hskip 3truein England\par
   %\hskip 3truein van Vleck Hall\par
   %\hskip 3truein 480 Lincoln Drive\par
   %\hskip 3truein Madison\par
   %\hskip 3truein WI 53706-1388, U.S.A.\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
       %{\tt fremlin\@math.wisc.edu}
   \bigskip}
\long\def\homeaddress#1{\hskip 3truein 25 Ireton Road\par
   \hskip 3truein Colchester CO3 3AT\par
   \hskip 3truein England\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
   \bigskip}
\long\def\letterhead#1#2#3{\univaddress{#1}
   {\parindent=0pt
    #2
   }
   \medskip

   \noindent Dear #3,
     }
\def\shortaddress{\noindent Departmentof Mathematical Sciences, University of Essex,
   Colchester CO4 3SQ, England\par
   \noindent e-mail: david\@fremlin.org}
\def\valediction#1#2{\bigskip
   \hskip 3truein #1,
   \vskip 1truein
   \hskip 3.3truein #2
   \bigskip}

%picture importation
   %It is clear that there is no general standard.   The syntax I
   %use when calling for a  .ps  file is
   %    \def\Caption{<...>}
   %    \picture{<filename>}{<dim>}
   %which is supposed to import  <filename>.ps  at height  <dim> ,
   %centered, and adding a one-line centered caption  <...> .
   %The basic macro listed in  \def\picture  is that which
   %used to work at Essex.   I now generally use the macro in
   % \def\picturewithpsfig , which requires a previous input
   % psfig.tex  or  psfig.sty ;
   %if you interpolate  \atUEssex  early in your file it will
   %give you the latter.
   %   I find that  psfig  is erratic in its centering, especially
   %under magnification, so I have written a routine
   % \centerpicturewithpsfig  (which may be called from
   % \centerpicture ) with the syntax
   %    \centerpicture{<filename>}{<dim>}{<dim>}
   %the first dimension being width and the second height.

   % A second package is called for by  \usegraphicx .   This requires
   % miniltx.tex  and  graphicx.sty ;  miniltx.tex  is currently (April 2016)
   % available from
   % http://mirror.ox.ac.uk/sites/ctan.org/macros/plain/graphics/miniltx.tex
   % The advantage of this is that it seems to allow manipulation of the
   % dvi>pdf stage using  dvipdfmx .

\def\picture#1#2{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at height #2
         with dvitops}
  \vbox{\vskip #2
    \special{dvitops: import #1.ps \pagewidth #2}
    \Centerline{\Caption}
    }
   \medskip
   \def\Caption{}}
%an alternative version, without caption
\def\centerpicture#1#2#3{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with dvitops}
  \vbox{\vskip #3
    \special{dvitops: import #1.ps \pagewidth #3}
  }}
%to put pictures on same line
\def\sideshiftedpicture#1#2#3#4{\bigskip
   \bigskip
   \immediate\write0{importing diagram #1.ps at height #4
        and width #3 with dvitops}
   \vbox{\special{dvitops:  import #1.ps #3 #4}}
   \hskip #2 plus 0pt minus 0pt
   }
\def\startsideshiftedpicture{}
\newcount\scaleup
\newcount\scaledown
\scaleup=1
\scaledown=1
\def\scalesideshiftedpicture#1#2#3#4{
  \dimen4=#4
  \dimen3=#3
  \multiply\dimen4 by\scaleup
  \divide\dimen4 by \scaledown
  \multiply\dimen3 by\scaleup
  \divide\dimen3 by \scaledown
  \sideshiftedpicture{#1}{#2}{\dimen3}{\dimen4}}

\def\picturewithpsfig#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with psfig}
   \vbox{\hsize\pagewidth
   \centerline{\psfig{figure=#1.ps,height=#2}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithpsfig#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with psfig}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\psfig{file=#1.ps,width=\dimen3,height=\dimen4}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithpsfig#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with psfig}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\psfig{figure=#1.ps,height=\dimen4,width=\dimen3}}
   \psfig{figure=#1.ps,height=\dimen4,width=\dimen3}
   \hskip #2 plus 0pt minus 0pt}

\def\picturewithgraphicx#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with graphicx}
   \vbox{\hsize\pagewidth
   \centerline{\includegraphics[height=#2]{#1.ps}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithgraphicx#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with graphicx}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\includegraphics[width=\dimen3,height=\dimen4]{#1.ps}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithgraphicx#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with graphicx}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\includegraphics[height=\dimen4,width=\dimen3]{#1.ps}}
   \includegraphics[height=\dimen4,width=\dimen3]{#1.ps}
   \hskip #2 plus 0pt minus 0pt}

%alternative picture importation code;  I've been told about this
  %but have not used it successfully myself
%\def\Caption{}
%\input epsf
%\epsfverbosetrue
%\def\picture#1#2{\bigskip
%    \bigskip
%    \immediate\write0{importing diagram #1.ps at height #2 with epsf}
%    \epsfysize=#2
%    \centerline{\epsffile{#1.ps}}
%    \medskip
%    \Centerline{\Caption}
%    \medskip
%    }

%in addition it may be necessary to amend the imported  .ps  file
%by replacing the   %%BoundingBox   header by  %%BoundingBox:

\def\atUEssex{
   \input psfig.sty
   \def\picture##1##2{\picturewithpsfig{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithpsfig{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithpsfig{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
   }
\def\usegraphicx{
   \input miniltx
   \def\Gin@driver{dvips.def}
   \input graphicx.sty
   \def\picture##1##2{\picturewithgraphicx{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithgraphicx{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithgraphicx{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
}


%special fonts
\def\biggerscriptfonts{\scriptfont0=\eightrm
     \scriptfont1=\eighti
     \scriptfont2=\eightsy
     \scriptfont3=\eightex
     \scriptfont\bffam=\eightbf
     \scriptfont\msbfam=\eightmsb
     \scriptfont\eufmfam=\eighteufm
     }
\def\Loadcyrillic{\ifamstexsupported
      \input cyracc.def
      \font\tencyr=wncyr10
      \def\cyr{\tencyr\cyracc} \def\itcyr{\tenitcyr\cyracc}
    \else\def\cyr{\smc}\def\cdprime{u\i}\def\Cdprime{UI}\fi}
%http://wkweb3.cableinet.co.uk/linux/LDP/HOWTO/Cyrillic-HOWTO-7.html
%cyrillic.tex

\def\Loadfourteens{\iffourteensavailable
       \font\fourteenrm=cmr14
       \font\fourteenbf=cmbx14
       \font\fourteenit=cmsl14
   \else\font\fourteenrm=cmr10 scaled \magstep2
      \font\fourteenbf=cmbx10 scaled \magstep2
      \font\fourteenit=cmsl10 scaled \magstep2
   \fi}
\def\smallerfonts{\eightpoint
   \textfont\msbfam=\eightmsb
   \textfont\eufmfam=\eighteufm
   \textfont\bffam=\eightbf
   \textfont\eurmfam=\eighteurm
   \textfont\eusmfam=\eighteusm
   }
\font\twelveex=cmex10 scaled \magstep1


%Copyright declarations
\newcount\oldcopyrightdate
\newcount\newcopyrightdate
\oldcopyrightdate=0
\def\copyrightdate#1{\newcopyrightdate=#1
       \ifnum\newcopyrightdate=\oldcopyrightdate\else
       \oldcopyrightdate=#1
       \oldfootnote{}{\eightrm \copyright\ #1 D.~H.~Fremlin}
       \immediate\write0{Copyright #1}
       \fi}

%hyphenation
\hyphenation{equi-ver-id-ical}

%miscellaneous definitions
\def\Aut{\mathop{\text{Aut}}}
\def\backstep#1{\hskip -0.#1em plus 0 pt minus 0 pt}
\def\Bang{\vthsp\hbox{\bf X\hskip0.07em\llap X\hskip0.07em\llap X\rm}}
\def\BanG{\Bang{\hskip 1.111pt plus 3.333pt minus 0pt}}
%\def\BbbN{\mathchoice{\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb N$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb N$\hskip0.04em}}}
%\def\BbbQ{\mathchoice{\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Q$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Q$\hskip0.04em}}}
%\def\BbbR{\mathchoice{\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb R$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb R$\hskip0.04em}}}
%  %do not use with $\Bbb R^X$
%\def\BbbZ{\mathchoice{\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Z$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Z$\hskip0.04em}}}
\def\BbbN{\openBbb{N}}
\def\BbbQ{\openBbb{Q}}
\def\BbbR{\openBbb{R}}
  %do not use with $\Bbb R^X$  do not use in $_{\Bbb R^r}$
\def\BbbZ{\openBbb{Z}}
\def\Bbbone{\hbox{1\hskip0.13em\llap{1}}}
\def\biggerint{\textfont3=\twelveex\intop\nolimits}
  % you can get this size into a displayed formula with
  %  $$\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}$$
  % but it may then seem cramped
  % or try \hbox{$\biggerint_{\Cal S}\,$}
\def\biggeriint{\textfont3=\twelveex\iint}
  % remarkable fact:  this seems to infect all \int within a given
  % $..$ ; but  $\biggerint$$\int$ seems to leave the $\int$ normal
  % really  \textfont3=\twelveex  is more reliable, since  \biggerint
  % as written gets turned off rather easily (e.g. by  \comment,
  % or by an extra  {...})
\def\bover#1#2{{{#1}\over{#2}}}
\def\Bover#1#2{\backstep4\hbox{
     \biggerscriptfonts${\shortstrut#1}\over{\shortstrut#2}$}}
  % when there are superscripts in the denominator, as in \Bover1{2^m}
  % or  \Bover1{\sqrt{2\pi}^r} , these can sometimes come out with better
  % horizontal alignment if you use  \Bover1{{\sshortstrut 2}^m}  or
  % \Bover1{{\sqrt{2\pi}}^r}
\def\bsover#1#2{\textstyle\bover{#1}{#2}}
\def\bsp{/\penalty-100}
  % \penalty seems to behave oddly in a \tt string in \fullsize mode
\def\Caption{}
\def\Center#1{

   \Centerline{#1}

   \noindent}
\def\Centerline#1{\vskip 5pt
     {\def\int{\text{}$$\textfont3=\twelveex\intop\nolimits}
      \def\dashint{\text{}$$\textfont3=\twelveex
         \mskip2mu\vrule height2.65pt width4.5pt depth-2.35pt\mskip-11mu
	 \intop\nolimits\mskip-2mu}
      \def\idashint{\text{}$$\int\mskip-7mu\dashint}	 
      \def\dashiint{\text{}$$\dashint\mskip-6mu\int}	 
      \def\dashidashint{\text{}$$\dashint\mskip-6mu\dashint}	 
      \def\underlineint{
           \text{}$$\textfont3=\twelveex\underline{\intop\nolimits}}
      \def\overlineint{
           \text{}$$\textfont3=\twelveex\overline{\intop\nolimits}}
      \def\iint{\text{}$$\textfont3=\twelveex
           \intop\nolimits\hskip-0.5em\intop\nolimits}
      \def\sum{\text{}$$\sumop}
      \def\prod{\text{}$$\prodop}
      \def\bigcap{\text{}$$\bigcapop}
      \def\bigcup{\text{}$$\bigcupop}
   \centerline{#1}}\vskip 5pt}
   \mathchardef\bigcapop="1354
   \mathchardef\bigcupop="1353
   \mathchardef\prodop="1351
   \mathchardef\sumop="1350
   %in  \Centerline{$\overline{}$, $\cmmnt{}$} must use \sumop etc
\def\cf{\mathop{\text{cf}}}
\def\circumflex#1{{\accent"5E #1}}
\def\closeplus{\hbox{+}}
\long\def\cmmnt#1{\ifresultsonly\else{#1}\fi}
\def\coint#1{\left[#1\right[}
\def\curl{\mathop{\text{curl}}}
\def\dashint{\mathchoice{-\mskip-20mu\intop\nolimits}
   {{\vrule height2.75pt width4.5pt depth-2.45pt\mskip-10mu\intop\nolimits}}
      {-\mskip-16.5mu\intop\nolimits} %not yet used
      {-\mskip-16.5mu\intop\nolimits}} %not yet used
\def\idashint{\mathchoice{\int\mskip-6mu\dashint}
      {\int\mskip-5mu\dashint}
      {\int\mskip-5mu\dashint} %not yet used
      {\int\mskip-5mu\dashint}} %not yet used
\def\dashiint{\mathchoice{\dashint\mskip-10mu\int}
      {\dashint\mskip-6mu\int}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashidashint{\mathchoice{\dashint\mskip-10mu\dashint}
      {\dashint\mskip-6mu\dashint}
      {\dashint\mskip-6mu\dashint} %not yet used
      {\dashint\mskip-6mu\dashint}} %not yet used
\def\dashidashiint{\mathchoice{\dashint\mskip-8mu\dashint\mskip-10mu\int}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashiidashint{\mathchoice{\dashint\mskip-10mu\int\mskip-8mu\dashint}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\idashidashint{\mathchoice{\int\mskip-8mu\dashint\mskip-8mu\dashint}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\diam{\mathop{\text{diam}}}
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
           \else\ifnum\stylenumber=3{\frnewpage}
           \else\ifnum\stylenumber=5{\frnewpage}
           \else\bigskip\fi\fi\fi}
\long\def\discrversionA#1#2{\ifnum\stylenumber=1#1\else#2\fi}
\long\def\discrversionB#1#2{\ifnum\stylenumber=2#1\else#2\fi}
\long\def\discrversionC#1#2{\ifnum\stylenumber=3#1\else#2\fi}
\long\def\discrversionD#1#2{\ifnum\stylenumber=4#1\else#2\fi}
\long\def\discrversionE#1#2{\ifnum\stylenumber=5#1\else#2\fi}
\long\def\discrversionF#1#2{\ifnum\stylenumber=6#1\else#2\fi}
\long\def\discrversionG#1#2{\ifnum\stylenumber=7#1\else#2\fi}
\long\def\discrversionH#1#2{\ifnum\stylenumber=8#1\else#2\fi}
           % facility for special interpolations in particular output
           % versions
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\diverg{\mathop{\text{div}}}
\def\dom{\mathop{\text{dom}}}
\def\eae{=_{\text{a.e.}}}

\def\esssup{\mathop{\text{ess sup}}}
\mathchardef\Exists="3239

\def\family#1#2#3{\langle#3\rangle_{#1\in#2}}
\def\familyi#1#2{\family{i}{#1}{#2}}
\def\familyiI#1{\familyi{I}{#1}}

\def\frfilename#1{\ifnum\stylenumber=1{\noindent filename #1}
               \else\ifnum\stylenumber=6{\noindent{\fiverm filename #1}}
               \else\ifnum\stylenumber=7{\noindent{\fiverm filename #1}}
               \else\fi\fi\fi}
\def\Footnote#1#2{\footnote"#1"{#2}}
\mathchardef\Forall="3238
\def\frakc{\frak c\phantom{.}}
\def\frnewpage{\vskip \pageheight plus 0pt minus\pageheight
           \eject
           \gdef\topparagraph{}
           \gdef\bottomparagraph{}
           }
\def\frsmallcirc{{\hskip 1pt plus 0pt minus 0pt}
     {\raise 1pt\hbox{$\smallcirc$}}{\hskip 1pt plus 0pt minus 0pt}}

\def\geae{\ge_{\text{a.e.}}}
\def\grad{\mathop{\text{grad}}}
\def\grheada{{\bf ($\pmb{\alpha}$)}}
\def\grheadb{{\bf ($\pmb{\beta}$)}}
\def\grheadc{{\bf ($\pmb{\gamma}$)}}
\def\grheadd{{\bf ($\pmb{\delta}$)}}
\def\grheade{{\bf ($\pmb{\epsilon}$)}}
\def\grheadz{{\bf ($\pmb{\zeta}$)}}
\def\grv#1{{\accent"12 #1}}

\def\hfillall{\hskip\pagewidth plus0pt minus\pagewidth}

\def\Imag{\mathop{\Cal I\text{m}}}
\long\def\inset#1{{\narrower\narrower{\vskip 1pt plus 1pt minus 0pt}
     #1
     {\vskip 1pt plus 1pt minus 0pt}}}
\def\Int#1#2{\displaystyle{\int_{#1}^{#2}}}
\def\interior{\mathop{\text{int}}}

\def\leae{\le_{\text{a.e.}}}
\long\def\leaveitout#1{}
\def\LLcorner{\mathchoice{\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height5pt width0.3pt depth0pt
       \vrule height0.3pt width4pt depth0pt}\,}
   {\,\hbox{\vrule height4pt width0.25pt depth0pt
       \vrule height0.25pt width3pt depth0pt}\,}
  }
\def\Ln{\mathop{\text{Ln}}}
\def\Loadfont#1{\font#1}
\def\loibr{\hbox{\hskip0.1em]\hskip0.05em}}
\def\Matrix#1{\pmatrix#1\endpmatrix}
\def\Mu{\text{M}}
\def\noamslogo{\def\amslogo{}}
\def\normalsubgroup{\vartriangleleft}
\def\Nu{\text{N}}
\def\ocint#1{\left]#1\right]}
\def\ofamily#1#2#3{\langle#3\rangle_{#1<#2}}
\let\oldcases=\cases
\let\oldfootnote=\footnote
\def\On{\mathop{\text{On}}}
\def\ooint#1{\left]#1\right[}
\def\openBbb#1{\Bbb#1\mskip1mu}
\def\pd#1#2{\bover{\partial#1}{\partial#2}}
\def\Pd#1#2{\Bover{\partial#1}{\partial#2}}
\def\pdd#1#2{{{\partial^2#1}\over{\partial#2^2}}}
\def\Pdd#1#2{\Bover{\partial^2#1}{\partial#2^2}}
\def\pde#1#2#3{{{\partial^2#1}\over{\partial#2\partial#3}}}
\def\Pde#1#2#3{\Bover{\partial^2#1}{\partial#2\partial#3}}
\def\Prf{\vthsp\hbox{\bf P\hskip0.07em\llap P\hskip0.07em\llap P\rm}}
\def\pushbottom#1{\hbox{\vrule height0pt depth#1 width0pt}}
\def\Qed{\vthsp\hbox{\bf Q\hskip0.07em\llap Q\hskip0.07em\llap Q\rm}}
\def\QeD{\Qed{\hskip 1.111pt plus 3.333pt minus 0pt}}
\def\Quer{\vthsp\hbox{\bf ?\hskip0.05em\llap ?\hskip0.05em\llap
   ?\thinspace\rm}}
\def\query{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{query}}{}}
\def\Real{\mathop{\Cal R\text{e}}}
\def\restr{\mathchoice
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\scriptstyle\restriction$}\mskip 1mu}
  {\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
\def\restrp{\mathchoice
  {\hbox{$\restriction$}\mskip 2mu}
  {\hbox{$\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptstyle\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
    %use in \restrp V \restrp W \restrp Y
    % \restrp\Cal P \restrp\Bbb R \restrp\frak
    % \restrp\Omega \restrp\Tau \restrp\Upsilon
    % \restrp[ \restrp\widehat{}
\def\Rho{\text{P}}
\def\roibr{\hbox{\hskip0.05em[\hskip0.1em}}
\def\RoverC{\hbox{\biggerscriptfonts${{\Bbb R}\atop{\Bbb C}}$}}
\def\sequence#1#2{\family{#1}{\Bbb N}{#2}}
\def\sequencen#1{\sequence{n}{#1}}
\def\setover#1#2{\overset{#1}\to{#2}}
\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}
\def\sshortstrut{\hbox{\vrule height5.8pt depth2.5pt width0pt}}
\def\smallcirc{{\scriptstyle\circ}}
\def\ssbullet{{\scriptscriptstyle\bullet}}
\def\symmdiff{\triangle}
\def\thicken#1{#1\hskip0.03em\llap #1\hskip0.03em\llap
   #1\hskip0.03em\llap #1}
\def\Tau{{\text{T}}}
\def\tbf#1{\text{\bf #1}}
\def\terror#1#2#3{#1:  `#2' for `#3'}
\def\today{\number\day.\number\month.\number\year}
\def\UEMDRR{University of Essex Mathematics Department Research Report}
\def\UEMDWP{University of Essex Mathematics Department Working Paper}
\def\versiondate#1{
    \immediate\write0{Version of #1}
    \def\vsdate{#1}
               \ifnum\stylenumber=1\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=2\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=6\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=7\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=8\hfill Version of #1
                              \medskip
               \else\fi\fi\fi\fi\fi}
\def\versionhead{\ifnum\stylenumber=3 \hfill PRINTER'S VERSION

                        \bigskip
                   \else\ifnum\stylenumber=4 \hfill REFEREE'S VERSION

                        \bigskip
                   \else \fi\fi}
\def\versionref{\ifnum\stylenumber=1\workingref
           \else\ifnum\stylenumber=2\workingref
           \else\ifnum\stylenumber=8\workingref
           \else\printerref\fi\fi\fi}
\def\vthsp{\hskip0.001em}

%Missing AmSTeX can be coped with by using the following lines:

\ifamstexsupported\else
   \immediate\write0{AmSTeX-unsupported version of  fremtex.tex }
   \fourteensavailablefalse
   \def\@{@}
   \def\\{\cr}
   \def\AmSTeX{AmS\TeX}
   \def\Bbb#1{\underline{\hbox{\bf #1}}}
   \def\biggerscriptfonts{}
   \def\bumpeq{=_{\text{approx}}}
   \def\Cal#1{{\cal#1}}
   \def\curlyeqprec{\preceq}
   \def\eightit{\it}
   \def\eightrm{\rm}
   \def\eightsmc{\smc}
   \def\eurm#1{\underline{\cal#1}}
   \def\Footnote#1#2{\footnote#1#2}
   \def\frac#1#2{{#1}\over{#2}}
   \def\frak#1{{\hbox{\bf #1\hskip0.1em\llap #1}}}
   \def\frnewpage{\vfill\eject}
   \def\iint{\int\!\!\!\int}
   \def\iiint{\int\!\!\!\int\!\!\!\int}
   \def\iiiint{\int\!\!\!\int\!\!\!\int\!\!\!\int}
   \def\llcorner{\lfloor}
   \def\loadeurm{}
   \def\loadeusm{}
%   \def\loadeusmx#1{\def\eusm#1{\underline{\cal#1}}}
   \def\ltimes{|\hskip-0.5em\times}
   \def\eusm#1{\underline{\cal#1}}
   \def\ltimes{|\times}
   \def\Loadfont#1{}
   \def\Matrix#1{\pmatrix{#1}}
   \def\noamslogo{}
   \def\nrightarrow{\not\rightarrow}
   \def\preccurlyeq{\preceq}
   \def\query{\discrversionA{\immediate\write0{query}
                             \footnote{$^*$}{query}}{}}
   \def\restriction{|}
   \def\rtimes{\times|}
   \def\setover#1#2{#2^{#1}}
   \def\shortstrut{\strut}
   \def\smallerfonts{\eightrm}
   \def\smallfrown{\wedge}
   \def\smc{}
   \def\spcheck{\check{\phantom{x}}}
   \def\sphat{\hat{\phantom{x}}}
   \def\square{\mathchoice\sqr64\sqr64\sqr43\sqr33}
   \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
          \hbox{\vrule width.#2pt height#1pt \kern#1pt
                  \vrule width.#2pt}
          \hrule height.#2pt}}}}
   \def\pmb#1{\underline{#1}}
   \def\text#1{\hbox{#1}}
   \def\therefore{\ssbullet\raise.9ex\hbox{$\ssbullet$}\ssbullet}
   \def\twelvebf{\bf}
   \def\vartriangleleft{\triangleleft}
   \def\Vdash{|\vdash}
   \def\vDash{|=}
   \fi

%End of lines for missing AmSTeX

\ifamstexsupported\input amstex
%              \documentstyle{amsppt}
               \input amsppt.sty
               \PSAMSFonts
	       \tenpoint
               %\font\eightmsb=msym8
               \font\eightmsb=msbm10
               \font\eighteufm=eufm10 at 8 pt
               \font\eighteurm=eurm8
               \font\eighteusm=eusm8
               \font\varseveneufm=eufm10 at 7 pt
               \scriptfont\eufmfam=\varseveneufm
   \def\.#1{{\accent"5F #1}}
   \def\=#1{{\accent"16 #1}}
   \gdef\nopagenumbers{\NoPageNumbers}
   \else\fi

\newcount\footmarkcount
\ifDHFamstexloaded
    \loadmsBm
    \def\eightBbb#1{{\fam\msBfam#1}}
    \loadmsFm
    \def\eightfrak#1{{\fam\msFfam#1}}
  \else
    \def\eightBbb#1{{\scriptstyle\Bbb#1}}
    \def\eightfrak#1{{\scriptstyle\frak#1}}
  \fi
\ifathome\athome\fi

%interword glue (\tenrm standard):
%  (space) 3.333 + 1.667 - 1.111
%  (,)     3.333 + 2.083 - 0.889
%  (.)     4.444 + 5.    - 0.370

%very odd fact:  \discrversionA{2.}{}  doesn't work,
%   \discrversionA{ 2.}{} does

%The false line feed introduced by my Fortran programs, leading to
%a surplus  ae  at the top of the next page, seems to be associated
%with  Fk(\032)3081  or something similar in the PostScript file.


\immediate\write0{Extract from MEASURE THEORY,
   by D.H.Fremlin, University of Essex, Colchester.
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   http://dsl.org/copyleft/dsl.txt.   This is a development version and the
   source files are not permanently archived.
   For further information contact david\@fremlin.org.}

%hyphenations
\hyphenation{Archi-me-dean}

%new style
\def\smallprintersize{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \nologo
     \def\copyrightdate##1{}
     \def\folio{\ifnum\pageno<0{}\else\number\pageno\fi}
     \def\mtfootline{}
     \stylenumber=10
     \pageno=-1000}

%With-proofs, results-only versions
\newif\ifwithproofs
\withproofstrue
\newif\ifresultsonly
\resultsonlyfalse
\def\noproofs{\withproofsfalse}
\def\resultsonly{\resultsonlytrue\withproofsfalse}
\long\def\proof#1{\ifwithproofs{\medskip

     \noindent{\bf proof} #1}\else\fi}
\long\def\prooflet#1{\ifwithproofs{#1}\else\fi}
%\long\def\Prooflet#1{\prooflet{\Prf\ #1 \Qed\ }}
\long\def\endnotes#1{\ifresultsonly\else{#1}\fi}
\long\def\exercises#1{\ifresultsonly\else{#1}\fi}
\long\def\references#1{\ifresultsonly\else{#1}\fi}
\long\def\dvro#1#2{\ifresultsonly{#1}\else{#2}\fi}

%Headline assignments

\def\topparagraph{}
\def\bottomparagraph{}
\def\newparagraph{}
\newif\ifpageturned

\newdimen\rhoffset
\newdimen\lhoffset
\lhoffset=0cm
\rhoffset=0cm

\newcount\volumeno
\volumeno=5
\newcount\luluvolumeno
\luluvolumeno=0

\let\plainpagebody=\pagebody
\def\pagebody{\ifodd\pageno\global\hoffset=\rhoffset
         \else\global\hoffset=\lhoffset\fi
     \plainpagebody
     \turnpage}
\def\turnpage{
     \iffirstpage\else{\baselineskip=24pt\line{\mtfootline}}\fi
%     \immediate\write0{turning page \folio,
%           top paragraph \topparagraph,   bottom \bottomparagraph,
%           new \newparagraph, pagetotal \the\pagetotal}
     \global\pageturnedtrue
     \global\firstpagefalse
     \global\let\topparagraph=\bottomparagraph
     \global\let\bottomparagraph=\newparagraph
     }

\newdimen\pagetotalplus
\newdimen\pagetotalminus

%wheader parameters:  #1 new paragraph name
 % #2#3#4  \vskip parameters if page break not forced (no. of points)
 % #5  spare height demanded if page break not to be forced
\long\def\wheader#1#2#3#4#5{\gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #5
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
%     \immediate\write0{forcing page break}
     \eject
     \else\vskip #2pt plus #3pt minus #4pt\fi
     \else\vskip #2pt plus #3pt minus #4pt\fi
  \ifpageturned\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
     \fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
%   \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  }
\def\header#1{
     \wheader{#1}{10}{4}{4}{24pt}}
\def\leader#1#2{\immediate\write0{#1 #2}
     \wheader{#1}{14}{7}{5}{24pt}
     {\bf #1 #2}}
\def\vleader#1#2#3{\immediate\write0{#2 #3}
     \wheader{#2}{14}{7}{5}{#1}
     {\bf #2 #3}}
\def\Notesheader#1{\immediate\write0{#1 Notes and comments}
     \wheader{#1 Notes}{16}{8}{5}{48pt}
     \noindent{\bf #1 Notes and comments}}
\def\vNotesheader#1#2{\immediate\write0{#2 Notes and comments}
     \wheader{#2 Notes}{16}{8}{5}{#1}
     \noindent{\bf #2 Notes and comments}}
\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\def\spstheader#1#2#3#4#5{\header{*#1#2#3#4#5}{\bf (#5)}}
\def\vspheader#1#2#3#4#5#6{
     \wheader{#2#3#4#5#6}{10}{4}{4}{#1}{\bf (#6)}}
\def\sqheader#1#2#3#4#5{\header{#1#2#3#4#5}{$\pmb{>}${\bf (#5)}}}
\def\errheader#1#2#3{
     \wheader{#3}{6}{4}{2}{24pt}
     {\bf p #1 l #2}}
\long\def\woddheader#1#2#3#4#5{
   \ifodd\pageno\wheader{#1}{#2}{#3}{#4}{#5}
   \else\vskip #2pt plus #3pt minus #4pt\fi}

\def\indexmedskip{
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by 18 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
      \else\medskip\fi\else\medskip\fi}

\def\indexheader#1{\immediate\write0{#1}\vindexheader{#1}{24}}
\def\vindexheader#1#2{
  \gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #2 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
%  \ifdim\pagetotal>\pagegoal
%      \immediate\write0{warning: pagetotal greater than pagegoal}
%      \immediate\write0{...   last paragraph reached is #1}
%      \immediate\write0{...   check for underfull vbox here}
%      \else\fi
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal{\eject}
      \fi\fi
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
  \ifpageturned{\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
      }\else\fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%    \bottomparagraph}
  }

\def\leftheadline{\rlap{\folio}\hfill{\eightit\chaptername}\hfill
       \llap{\eightbf\topparagraph}}
\def\rightheadline{\rlap{\eightbf\bottomparagraph}\hfill
       {\eightit\headlinesectionname}\hfill\llap{\folio}}

%Footline assignments
\def\mtfootline{\ifodd\pageno\rightfootline\else\leftfootline\fi}
\def\rightfootline{\hfill\eightsmc D.H.Fremlin}
\def\leftfootline{\ifwithproofs{\eightsmc Measure Theory\hfill}
    \else{\eightsmc Measure Theory \eightit (abridged version)\hfill}
    \fi}

%Style assignments
\def\newchapter#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Chap.\ #1 {\it intro.}}
   \gdef\newparagraph{Chap.\ #1 {\it intro.}}
   \footmarkcount=0
   \gdef\sectionname{Introduction}
   \gdef\headlinesectionname{Introduction}
   \centerline{\bf Chapter #1}
   \medskip
   \centerline{\bf \chaptername}
   \medskip
   }
\def\newsection#1{\gdef\bottomparagraph{\S#1 {\it intro.}}
   \noindent{\bf #1 \sectionname}
   \smallskip
   \let\headlinesectionname=\sectionname}
\def\newvolume#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Vol.\ #1 {\it intro.}}
   \Loadfont{\twelvebf=cmbx12}
   \def\chaptername{Volume #1}
   \def\sectionname{\volumename}
   \def\headlinesectionname{\volumename}
   \centerline{\twelvebf Volume #1:  \volumename}
   \bigskip
   }

%font variations
\newif\ifvolwp
\volwpfalse
\def\tenrmstretch#1{\fontdimen3\tenrm=#1}
\font\elevenex=cmex10 at 11pt

%contents and introductions
\newif\iflargelogo
\def\vtmpb#1{}
\def\newvolume#1{}
\def\pagereference#1#2{\hfill\hbox{\ifresultsonly#2\qquad \hskip 1truein
                                                 \else#1\qquad \fi}}
\long\def\section#1#2#3#4#5#6{\ifresultsonly
    \qquad #1 #2\pagereference{#4}{#5}
  \else\vskip 2pt plus 1pt minus 0pt
    \vbox{\qquad #1 #2\pagereference{#4}{#5}\par
    \abstr{#6}
    }\vskip -1pt plus 1pt minus 0pt
  \fi}
\long\def\chapintrosection#1#2#3{\ifresultsonly
    \qquad Introduction\pagereference{#2}{#3}
  \else\vskip 1pt plus 1pt minus 0pt
    \qquad Introduction\pagereference{#2}{#3}
    \vskip 1pt plus 1pt minus 0pt
  \fi}
\def\abstr#1{\ifresultsonly{}
   \else{\advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \vskip1pt plus 0pt minus0pt
     \parindent=0pt
     \smallerfonts #1%
     \hbox{\vrule height0pt depth2pt width0pt}
     \vskip1pt plus 0pt minus0pt}\fi}
     
%language variations
\newif\ifUSEnglish
\USEnglishfalse
%\USEnglishtrue

%"Later editions only"
\newcount\latereditionpage
\newcount\latereditionmark
\latereditionpage=-1
\latereditionmark=0
\newcount\newpageno

%removing the \write0 instruction altogether disorganizes it!
\def\Latereditions{\newpageno=\pageno%
  \ifdim\pagetotal>\pagegoal\advance\newpageno by 1\fi%
     \divide\newpageno by 2%
  \immediate\write0{%
%     "Later editions only";
%      pagetotal \the\pagetotal, pagegoal \the\pagegoal,
%      pageno \number\pageno,\par newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage%
  }\ifnum\latereditionpage=\newpageno%
     \ifDHFamspptloaded$^{\number\latereditionmark}$%
     \else\footnote{Later editions only.}\fi%
  \else\footnote{Later editions only.}%
  \global\latereditionpage=\newpageno%
  \global\latereditionmark=\footmarkcount\fi}%
%  \immediate\write0{newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage}

   
%general definitions
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
                       {\lower-.3ex\hbox{$\ssbullet$}}
                       {\ssbullet}
                       {\ssbullet}}  %see also \varaction
\def\add{\mathop{\text{add}}\nolimits}
\long\def\allowmorestretch#1#2{
  \ifdim\pagewidth<#1pt\else
    \fontdimen3\tenrm=2pt\fontdimen4\tenrm=1.67pt\fi
  #2
    \fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt}
\def\am{atom\discretionary{-}{}{}lessly-measurable}
\def\AmuA{\Aut_{\bar\mu}\mskip-2mu\frak A}
\def\ASint{\mathchoice{\Bbb A\mskip-5mu}
                  {\Bbb A\mskip-3mu}
                  {\Bbb A\mskip-2mu}
                  {\Bbb A\mskip-2mu}\Sint}
\def\Asisfour{$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$}
\def\Asissix{$(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Asisfourt{$(\frak A,\bar\mu,\coint{0,\infty},
   \langle\frak A_t\rangle_{t\ge 0})$}
\def\Asissixt{$(\frak A,\bar\mu,\coint{0,\infty},
   \penalty-100\langle\frak A_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Atop#1#2{{\scriptstyle #1}\atop{\scriptstyle #2}}
\def\Aut{\mathop{\text{Aut}}}

\def\Bcap{\Booleanop{\cap}}
\def\Bcup{\Booleanop{\cup}}
\def\Booleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
 %   {{\mskip2mu\scriptstyle#1\mskip2mu}}  %until 15.10.17
    {{\scriptscriptstyle#1}}  %from 15.10.17
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminus{\Booleanop{\setminus}}
\def\Bsubseteq{\hskip.1em\Booleanop{\subseteq}\hskip.1em}
\def\Bsubset{\Booleanop{\subset}}
\def\Bsupseteq{\Booleanop{\supseteq}}
\def\Bsupset{\Booleanop{\supset}}
\def\Bsymmdiff{\Booleanop{\symmdiff}}
\def\Bcapshort{\Booleanopshort{\cap}}
\def\Bcupshort{\Booleanopshort{\cup}}
\def\Booleanopshort#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hskip-1pt\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminusshort{\Booleanopshort{\setminus}}
\def\BSint{\mathchoice{\Bbb B\mskip-5mu}
                  {\Bbb B\mskip-3mu}
                  {\Bbb B\mskip-2mu}
                  {\Bbb B\mskip-2mu}\Sint}
%\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}\hskip.1em}
\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}}
\def\Bsubsetshort{\Booleanopshort{\subset}}
\def\Bsupseteqshort{\Booleanopshort{\supseteq}}
\def\Bsymmdiffshort{\Booleanopshort{\symmdiff}}

\def\closeBooleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\closeBcup{\closeBooleanop{\cup}}
\def\closeBcap{\closeBooleanop{\cap}}
\def\closeBsetminus{\closeBooleanop{\setminus}}
\def\closeBsymmdiff{\closeBooleanop{\symmdiff}}

\def\biggerHeint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 2.6pt{\sevenrm H}\hss}
   \intop\nolimits}
\def\biggerPfint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}
   \intop\nolimits\hskip 1.2pt}
\def\biggerSint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 3.6pt{\sevenrm S}\hss}
   \intop\nolimits}
\def\bu{\mathop{\text{bu}}}
\def\Bvalue#1{\mathchoice
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$\scriptstyle[\![#1]\!]$}}
   {\hbox{$\scriptscriptstyle[\![#1]\!]$}}}
\def\cadlag{c\`adl\`ag}
\def\CalBa{\Cal{B}\eurm{a}}
   %note: neither  \cal  nor  \eusm  has lower case
\def\CalFr{\Cal F_{\text{Fr}}}
\def\callal{c\`all\`al}
\def\CalNwd{\Cal{N}\hskip-1.2pt\eurm{w}\hskip-.8pt\eurm{d}}

\def\CalRbg{\Cal{R}\mskip-1mu\eurm{b}\mskip-1mu\eurm{g}}
\def\CalSmz{\Cal{S}\mskip-2mu\eurm{mz}}
\def\CalUn{\Cal{U}\eurm{n}}
\def\capacity{\mathop{\text{cap}}\nolimits}
\def\Caratheodory{Cara\-th\'eo\-dory}
\def\Cdlg{C_{\text{dlg}}}
\def\cff{\cf\mskip1mu}
\def\checkpmb#1{\check{\pmb #1}}
\def\ci{\mathop{\text{ci}}}
\def\cl{\mathop{\text{cl}}}
\def\Clll{\hbox{$C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\clMsimp{\overline{M}_{\text{simp}}}
\def\closeuparrow{\mathchar"0222}
\def\closedownarrow{\mathchar"0223}
\def\clstar{\text{cl*}}
\def\cov{\mathop{\text{cov}}}
\def\covar#1#2{\mathchoice
  {[#1{\lower4pt\hbox{${\scriptstyle *}\atop{\vrule height 18pt depth -10pt}$}}#2]}
  {[#1{*\atop{\vrule height 8pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}}
\def\covSh{\mathop{\text{cov}}\nolimits_{\text{Sh}}}
\def\cdownarrow{c\mskip2mu^{\downarrow}}
\def\cycle#1{(\overleftarrow{#1})}
\def\ddaggerloc{$(\ddagger)_{\text{loc}}$}
\def\ddownarrow{d\mskip2mu^{\downarrow}}
%\def\DiniD{\setover{
%  \hbox{\hskip.1em\vrule height-1pt width.5em depth1.4pt\hfil}}D}
%\def\Dinid{\hskip.1em
%  \hbox{\hskip.05em\vrule height-1.3pt width.5em depth1.7pt}\hskip-.65em D}
\def\Dinid{\hbox{$D$\hskip-8.3pt\vrule height-1pt width7pt depth1.85pt}
   \hskip1pt}
\def\DiniD{\hbox{$D$\hskip-8.3pt\vrule height9.2pt width7ptdepth-8.35pt}
   \hskip1pt}

\def\discrcenter#1#2{\ifdim\pagewidth>#1{#2}\else{

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt34.tex}
\versiondate{28.5.07}
\copyrightdate{1996}

\def\chaptername{The lifting theorem}

\newchapter{34}

Whenever we have a surjective homomorphism $\phi:P\to Q$, where $P$ and
$Q$ are mathematical structures, we can ask whether there is a right
inverse of $\phi$, a homomorphism $\psi:Q\to P$ such that $\phi\psi$ is
the identity on $Q$.   As a general rule, we expect a negative answer;
those categories in which epimorphisms always have right inverses (e.g.,
the category of linear spaces) are rather special, and elsewhere the
phenomenon is relatively rare and almost always important.   So it is
notable that we have a case of this at the very heart of the theory of
measure algebras:  for any complete probability space $(X,\Sigma,\mu)$
(in fact, for any complete strictly localizable space of non-zero
measure) the canonical homomorphism from $\Sigma$ to the measure algebra
of $\mu$ has a right inverse (341K).   This is the von Neumann-Maharam
lifting theorem.   Its proof, together with some essentially elementary
remarks, takes up the whole of of \S341.

As a first application of the theorem (there will be others in Volume
4) I apply it to one of the central problems of measure theory:  under
what circumstances will a homomorphism between measure algebras be
representable by a function between measure spaces?   Variations on this
question are addressed in \S343.   For a reasonably large proportion of
the measure spaces arising naturally in analysis, homomorphisms are
representable (343B).   New difficulties arise if we ask for
isomorphisms of measure algebras to be representable by isomorphisms of
measure spaces, and here we have to work rather hard for rather narrowly
applicable results;  but in the case of Lebesgue measure and its
closest relatives, a good deal can be done, as in 
344I-344K.  %344I 344J 344K

Returning to liftings, there are many difficult questions concerning the
extent to which liftings can be required to have special properties,
reflecting the natural symmetries of the standard measure spaces.   For
instance, Lebesgue measure is translation-invariant;  if liftings were
in any sense canonical, they could be expected to be automatically
translation-invariant in some sense.   It seems sure that there is no
canonical lifting for Lebesgue measure -- all constructions of liftings
involve radical use of the axiom of choice -- but even so we do have
many translation-invariant liftings (\S345).   We have less luck with
product spaces;  here the construction of liftings which respect the
product structure is fraught with difficulties.   I give the currently
known results in \S346.


\discrpage


\frfilename{mt341.tex}
\versiondate{9.4.10}
\copyrightdate{2000}

\def\chaptername{Liftings}
\def\sectionname{The lifting theorem}

\newsection{341}

I embark directly on the principal theorem of this chapter (341K, `every
non-trivial complete strictly localizable measure space has a lifting'),
using the minimum of advance preparation.   341A-341B give the
definition
of `lifting';  the main argument is in 341F-341K, using the concept of
`lower density' (341C-341E) and a theorem on martingales from \S275.
In 341P I describe an alternative way of thinking about liftings in
terms of the Stone space of the measure algebra.

\leader{341A}{Definition} Let $(X,\Sigma,\mu)$ be a measure space,
and $\frak A$ its measure algebra.   By a {\bf lifting} for $\frak A$ (or
for $(X,\Sigma,\mu)$, or for $\mu$) I shall mean

{\it either} a Boolean homomorphism $\theta:\frak A\to\Sigma$ such that
$(\theta a)^{\ssbullet}=a$ for every $a\in\frak A$

{\it or} a Boolean homomorphism $\phi:\Sigma\to\Sigma$ such that (i)
$\phi E=\emptyset$ whenever $\mu E=0$ (ii) $\mu(E\symmdiff\phi E)=0$ for
every $E\in\Sigma$.

\cmmnt{
\leader{341B}{Remarks (a)} I trust that the ambiguities permitted by
this terminology will not cause any confusion.   The point is that there
is a natural one-to-one correspondence between liftings
$\theta:\frak A\to\Sigma$ and liftings $\phi:\Sigma\to\Sigma$ given by the formula

\Centerline{$\theta E^{\ssbullet}=\phi E$ for every $E\in\Sigma$.}

\noindent\prooflet{\Prf\ (i) Given a lifting $\theta:\frak A\to\Sigma$,
the formula defines a Boolean homomorphism $\phi:\Sigma\to\Sigma$ such
that

\Centerline{$\phi\emptyset=\theta 0=\emptyset$,
\quad$(E\Bsymmdiff\phi E)^{\ssbullet}
=E^{\ssbullet}\Bsymmdiff(\theta E^{\ssbullet})^{\ssbullet}
=0\,\Forall\,E\in\Sigma$,}

\noindent so that $\phi$ is a lifting.   (ii) Given a lifting
$\phi:\Sigma\to\Sigma$, the kernel of $\phi$ includes
$\{E:\mu E=0\}$, so there is a Boolean homomorphism
$\theta:\frak A\to\Sigma$
such that $\theta E^{\ssbullet}=\phi E$ for every $E$ (3A2G), and now

\Centerline{$(\theta E^{\ssbullet})^{\ssbullet}
=(\phi E)^{\ssbullet}=E^{\ssbullet}$}

\noindent for every $E\in\Sigma$, so $\theta$ is a lifting.\ \Qed}

I suppose that the word `lifting' applies most naturally to functions
from $\frak A$ to $\Sigma$;  but for applications in measure theory the
other type of lifting is used at least equally often.


\header{341Bb}{\bf (b)} Note that if $\phi:\Sigma\to\Sigma$ is a lifting
then $\phi^2=\phi$.   \prooflet{\Prf\ For any $E\in\Sigma$,

\Centerline{$\phi^2E\Bsymmdiff\phi E=\phi(E\Bsymmdiff\phi E)=\emptyset$.
\Qed}
}

\noindent If $\phi$ is associated with $\theta:\frak A\to\Sigma$, then
$\phi\theta a=\theta a$ for every $a\in\frak A$.   \prooflet{\Prf\
$\phi\theta a=\theta((\theta a)^{\ssbullet})=\theta a$.\ \Qed}

\spheader 341Bc In the theorems to follow, there will occasionally intrude
a hypothesis `$\mu X>0$'.   The point is that if we have a measure space
$(X,\Sigma,\mu)$ which is trivial in the sense that $\mu X=0$, then the
only candidate for a `lifting' $\phi:\Sigma\to\Sigma$ is the constant
function with value $\emptyset$;  and if $X\ne\emptyset$ this is not a
Boolean homomorphism in the sense of this book.   The simplest way of
dealing with these cases is to rule them out of the discussion.
}%end of comment

\leader{341C}{Definition} Let $(X,\Sigma,\mu)$ be a measure space,
and $\frak A$ its measure algebra.   By a {\bf lower density} for
$\frak A$ (or for $(X,\Sigma,\mu)$, or for $\mu$) I shall mean

{\it either} a function $\undtheta:\frak A\to\Sigma$ such that
(i) $(\undtheta a)^{\ssbullet}=a$ for every $a\in\frak A$
(ii) $\undtheta0=\emptyset$
(iii) $\undtheta(a\Bcap b)=\undtheta a\cap\undtheta b$ for all $a$,
$b\in\frak A$

{\it or} a function $\undphi:\Sigma\to\Sigma$ such that (i)
$\undphi E=\undphi F$ whenever $E$, $F\in\Sigma$ and
$\mu(E\symmdiff F)=0$ (ii) $\mu(E\symmdiff\undphi E)=0$
for every $E\in\Sigma$ (iii) $\undphi\emptyset=\emptyset$ (iv)
$\undphi(E\cap F)=\undphi E\Bcap\undphi F$ for all $E$, $F\in\Sigma$.

\leader{341D}{Remarks (a)} As in 341B, there is a natural one-to-one
correspondence between lower densities
$\undtheta:\frak A\to\Sigma$ and lower densities
$\undphi:\Sigma\to\Sigma$ given by the formula

\Centerline{$\undtheta E^{\ssbullet}=\undphi E$ for
every $E\in\Sigma$.}

\cmmnt{\noindent (For the requirement $\undphi E=\undphi F$
whenever $E^{\ssbullet}=F^{\ssbullet}$ in $\frak A$ means that every
$\undphi$ corresponds to a function $\undtheta$, and
the other clauses match each other directly.)}

\cmmnt{\header{341Db}{\bf (b)} As before, if $\undphi:\Sigma\to\Sigma$
is a lower density then $\undphi^2=\undphi$.   If
$\undphi$ is associated with $\undtheta:\frak
A\to\Sigma$, then $\undphi\undtheta
=\undtheta$.
}

\spheader 341Dc\cmmnt{ It will be convenient, in the course of the
proofs of 341F-341H below, to have the following concept available.}
If $(X,\Sigma,\mu)$ is a measure space with measure algebra $\frak A$,
a {\bf partial lower density} of $\frak A$ is a function
$\undtheta:\frak B\to\Sigma$ such that (i) the domain $\frak B$
of $\undtheta$ is a subalgebra of $\frak A$ (ii)
$(\undtheta b)^{\ssbullet}=b$ for every $b\in\frak B$ (iii)
$\undtheta0=\emptyset$ (iv)
$\undtheta(a\Bcap b)=\undtheta a\cap\undtheta b$ for all $a$,
$b\in\frak B$.

Similarly, if $\Tau$ is a subalgebra of $\Sigma$, a function
$\undphi:\Tau\to\Sigma$ is a {\bf partial lower density} if (i) $\undphi
E=\undphi F$ whenever $E$, $F\in\Tau$ and $\mu(E\symmdiff F)=0$ (ii)
$\mu(E\symmdiff\undphi E)=0$ for every $E\in\Tau$ (iii)
$\undphi\emptyset=\emptyset$
(iv) $\undphi(E\cap F)=\undphi E\cap\undphi F$ for all $E$, $F\in\Tau$.

\cmmnt{\spheader 341Dd Note that lower densities and partial lower
densities
are order-preserving;  if $a\Bsubseteq b$ in $\frak A$, and $\undtheta$
is a lower density for $\frak A$, then

\Centerline{$\undtheta a=\undtheta(a\Bcap b)=\undtheta a\Bcap\undtheta
b\Bsubseteq\undtheta b$.}
}

\cmmnt{\spheader 341De Of course a Boolean homomorphism from $\frak A$ to
$\Sigma$, or from $\Sigma$ to itself, is a lifting iff it is a lower
density.
}%end of comment

\leader{341E}{Example} Let $\mu$ be Lebesgue measure on $\BbbR^r$,
where $r\ge 1$, and $\Sigma$ its domain.   For $E\in\Sigma$ set

\Centerline{$\intstar E
=\{x:x\in\BbbR^r,\,\lim_{\delta\downarrow 0}
\Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}=1\}$.}

\noindent (Here $B(x,\delta)$ is the closed ball with centre $x$ and
radius $\delta$.)   Then $\intstar$ is a lower density for $\mu$;  we may
call it {\bf lower Lebesgue density}.   \prooflet{\Prf\ (You may prefer
at first to suppose that $r=1$, so that
$B(x,\delta)=[x-\delta,x+\delta]$
and $\mu B(x,\delta)=2\delta$.)   By 261Db (or 223B, for the
one-dimensional case) $E\symmdiff\intstar E$ is negligible for
every $E$;  in particular, $\intstar E\in\Sigma$ for every
$E\in\Sigma$.   If $E\symmdiff F$ is negligible, then
$\mu(E\cap B(x,\delta))=\mu(F\cap B(x,\delta))$ for every $x$ and
$\delta$, so $\intstar E=\intstar F$.   If $E\subseteq F$, then
$\mu(E\cap B(x,\delta))\le \mu(F\cap B(x,\delta))$ for every
$x$, $\delta$, so $\intstar E\subseteq\intstar F$;
consequently $\intstar(E\cap F)\subseteq\intstar E\cap\intstar F$ for all
$E$, $F\in\Sigma$. If $E$, $F\in\Sigma$ and
$x\in\intstar E\cap\intstar F$, then

$$\eqalign{\mu(E\cap F\cap B(x,\delta))
&=\mu(E\cap B(x,\delta))+\mu(F\cap B(x,\delta))
-\mu((E\cup F)\cap B(x,\delta))\cr
&\ge\mu(E\cap B(x,\delta))+\mu(F\cap B(x,\delta))
-\mu(B(x,\delta))\cr}$$

\noindent for every $\delta$, so

\Centerline{$\Bover{\mu(E\cap F\cap B(x,\delta))}{\mu B(x,\delta)}
\ge\Bover{\mu(E\cap B(x,\delta))}{\mu B(x,\delta)}
  +\Bover{\mu(F\cap B(x,\delta))}{\mu B(x,\delta)}-1
\to 1$}

\noindent as $\delta\downarrow 0$, and $x\in\intstar(E\cap F)$.
Thus $\intstar(E\cap F)=\intstar E\cap\intstar F$
for all $E$, $F\in\Sigma$, and $\intstar$ is a lower density.\ \Qed}

\cmmnt{\medskip

\noindent{\bf Remark} In Chapter 47 of Volume 4 I will return to the
operator $\intstar$ in a context in which an alternative name, `essential
interior', is more natural.
}

\leader{341F}{}\cmmnt{ The hard work of this section is in the proof
of 341H below.   To make it a little more digestible, I extract two
parts of the proof as separate lemmas.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a probability space and
$\frak A$ its measure algebra.   Let $\frak B$ be a closed subalgebra of
$\frak A$ and $\undtheta:\frak B\to\Sigma$ a partial lower
density.   Then for any $e\in\frak A$ there is a partial lower density
$\undtheta_1$, extending $\undtheta$, defined on the
subalgebra $\frak B_1$ of $\frak A$ generated by $\frak B\cup\{e\}$.

\proof{{\bf (a)} Because $\frak B$ is order-closed, therefore Dedekind
complete in itself (314Ea),

\Centerline{$v=\upr(e,\frak B)=\inf\{a:a\in\frak B,\,a\Bsupseteq e\}$,
\quad$w=\upr(1\Bsetminus e,\frak B)$}

\noindent are defined in $\frak B$.   Let $E\in\Sigma$ be such
that $E^{\ssbullet}=e$.

\medskip

{\bf (b)} We have a function $\undtheta_1:\frak B_1\to\Sigma$ defined by
writing

\Centerline{$\undtheta_1((a\Bcap e)\Bcup(b\Bsetminus e))
=\bigl(\undtheta((a\Bcap v)\Bcup(b\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(b\Bcap w))
     \setminus E\bigr)$}

\noindent for $a$, $b\in\frak B$.   \Prf\ By 312N, every element of
$\frak B_1$ is expressible as $(a\Bcap e)\Bcup(b\Bsetminus e)$ for some
$a$, $b\in\frak B$.   If $a$, $a'$, $b$, $b'\in\frak B$ are such that
$(a\Bcap e)\Bcup(b\Bsetminus e)=(a'\Bcap e)\Bcup(b'\Bsetminus e)$, then
$a\Bcap e=a'\Bcap e$ and $b\Bsetminus e=b'\Bsetminus e$, that is,

\Centerline{$a\Bsymmdiff a'\Bsubseteq 1\Bsetminus e\Bsubseteq w$,
\quad$b\Bsymmdiff b'\Bsubseteq e\Bsubseteq v$.}

\noindent This means that
$e\Bsubseteq 1\Bsetminus(a\Bsymmdiff a')\in\frak B$ and
$1\Bsetminus e\Bsubseteq 1\Bsetminus(b\Bsymmdiff b')\in\frak B$.   So we
also have $v\Bsubseteq 1\Bsetminus(a\Bsymmdiff a')$ and
$w\Bsubseteq 1\Bsetminus(b\Bsymmdiff b')$.   Accordingly

\Centerline{$a\Bcap v=a'\Bcap v$,
\quad$b\Bcap w=b'\Bcap w$.
\quad$a\Bsetminus w=a'\Bsetminus w$,
\quad$b\Bsetminus v=b'\Bsetminus v$.}

\noindent But this means that

$$\eqalign{\bigl(\undtheta((a\Bcap v)&\Bcup(b\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(b\Bcap w))
      \setminus E\bigr)\cr
&=\bigl(\undtheta((a'\Bcap v)\Bcup(b'\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((a'\Bsetminus w)\Bcup(b'\Bcap w))
      \setminus E\bigr).\cr}$$

\noindent Thus the formula given defines $\undtheta_1$ uniquely.\ \Qed

\medskip

{\bf (c)} Now $\undtheta_1$ is a partial lower density.

\medskip

\Prf{\bf (i)} If $a$, $b\in\frak B$,

$$\eqalign{(\undtheta_1((a\Bcap e)\Bcup(b\Bsetminus e)))^{\ssbullet}
&=\bigl(\bigl(\undtheta((a\Bcap v)\Bcup(b\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(b\Bcap w))\setminus E\bigr)
   \bigr)^{\ssbullet}\cr
&=\bigl(((a\Bcap v)\Bcup(b\Bsetminus v))\Bcap e\bigr)
   \cup\bigl(((a\Bsetminus w)\Bcup(b\Bcap w))\Bsetminus e\bigr)\cr
&=(a\Bcap e)
   \cup(b\Bsetminus e).\cr}$$

\noindent So $(\undtheta_1c)^{\ssbullet}=c$ for every $c\in\frak B_1$.

\medskip

\quad{\bf (ii)}

\Centerline{$\undtheta_1(0)
=\bigl(\undtheta((0\Bcap v)\Bcup(0\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((0\Bsetminus w)\Bcup(0\Bcap w))\setminus E\bigr)
=\emptyset$.}

\medskip

\quad{\bf (iii)} If $a$, $a'$, $b$, $b'\in\frak B$, then

$$\eqalign{\undtheta_1(((a\Bcap e)&\Bcup(b\Bsetminus e))
  \Bcap((a'\Bcap e)\Bcup(b'\Bsetminus e)))\cr
&=\undtheta_1((a\Bcap a'\Bcap e)\Bcup(b\cap b'\Bsetminus e))\cr
&=\bigl(\undtheta((a\Bcap a'\Bcap v)\Bcup(b\Bcap b'\Bsetminus v))
   \cap E\bigr)
   \cup\bigl(\undtheta((a\Bcap a'\Bsetminus w)\Bcup(b\Bcap b'\Bcap w))
   \setminus E\bigr)\cr
&=\bigl(\undtheta(((a\Bcap v)\Bcup(b\Bsetminus v))
   \Bcap((a'\Bcap v)\Bcup(b'\Bsetminus v)))
   \cap E\bigr)\cr
&\qquad\qquad \cup\bigl(\undtheta(((a\Bsetminus w)\Bcup(b\Bcap w))
   \Bcap((a'\Bsetminus w)\Bcup(b'\Bcap w)))
   \setminus E\bigr)\cr
&=\bigl(\undtheta((a\Bcap v)\Bcup(b\Bsetminus v))
   \cap\undtheta((a'\Bcap v)\Bcup(b'\Bsetminus v))
   \cap E\bigr)\cr
&\qquad\qquad \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(b\Bcap w))
   \cap\undtheta((a'\Bsetminus w)\Bcup(b'\Bcap w))
   \setminus E\bigr)\cr
&=\bigl(\bigl(\undtheta((a\Bcap v)\Bcup(b\Bsetminus v))
   \cap E\bigr)
  \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(b\Bcap w))
   \setminus E\bigr)\bigr)\cr
&\qquad\qquad
  \cap\bigl(\bigl(\undtheta((a'\Bcap v)\Bcup(b'\Bsetminus v))
   \cap E\bigr)
  \cup\bigl(\undtheta((a'\Bsetminus w)\Bcup(b'\Bcap w))
   \setminus E\bigr)\bigr)\cr
&=\undtheta_1((a\Bcap e)\Bcup(b\Bsetminus e))
  \cap\undtheta_1((a'\Bcap e)\Bcup(b'\Bsetminus e)).\cr}$$

\noindent So $\undtheta_1(c\Bcap c')=\undtheta_1(c)\Bcap\undtheta_1(c')$
for all $c$, $c'\in\frak B_1$.\ \Qed

\wheader{341F}{6}{2}{2}{72pt}

{\bf (d)} If $a\in\frak B$, then

$$\eqalign{\undtheta_1(a)
&=\undtheta_1((a\Bcap e)\Bcup(a\Bsetminus e))\cr
&=\bigl(\undtheta((a\Bcap v)\Bcup(a\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(a\Bcap w))
     \setminus E\bigr)\cr
&=\bigl(\undtheta(a)\cap E\bigr)
   \cup\bigl(\undtheta(a)\setminus E\bigr)
=\undtheta a.\cr}$$

\noindent Thus $\undtheta_1$ extends $\undtheta$, as required.
}%end of proof of 341F

\leader{341G}{Lemma} Let $(X,\Sigma,\mu)$ be a probability space and
$(\frak A,\bar\mu)$ its measure algebra.   Suppose we have a sequence
$\sequencen{\undtheta_n}$ of partial lower densities such that, for each
$n$, (i) the domain $\frak B_n$ of $\undtheta_n$ is a closed subalgebra
of $\frak A$ (ii) $\frak B_n\subseteq\frak B_{n+1}$ and
$\undtheta_{n+1}$ extends $\undtheta_n$.   Let $\frak B$ be the closed
subalgebra of $\frak A$ generated by $\bigcup_{n\in\Bbb N}\frak B_n$.
Then there is a partial lower density $\undtheta$, with domain $\frak
B$, extending every $\undtheta_n$.

\proof{{\bf (a)} For each $n$, set

\Centerline{$\Sigma_n=\{E:E\in\Sigma,\,E^{\ssbullet}\in\frak B_n\}$,}

\noindent and set

\Centerline{$\Sigma_{\infty}=\{E:E\in\Sigma,\,E^{\ssbullet}\in\frak
B\}$.}

\noindent Then (because all the $\frak B_n$, $\frak B$ are
$\sigma$-subalgebras of $\frak A$, and $E\mapsto E^{\ssbullet}$ is
sequentially order-continuous) all the $\Sigma_n$, $\Sigma_{\infty}$ are
$\sigma$-subalgebras of $\Sigma$.
We need to know that $\Sigma_{\infty}$ is just the
$\sigma$-algebra $\Sigma_{\infty}^*$ of subsets of $X$ generated by
$\bigcup_{n\in\Bbb N}\Sigma_n$.   \Prf\ Because $\Sigma_{\infty}$ is a
$\sigma$-algebra including $\bigcup_{n\in\Bbb N}\Sigma_n$,
$\Sigma_{\infty}^*\subseteq\Sigma_{\infty}$.   On the other hand,
$\frak B^*=\{E^{\ssbullet}:E\in\Sigma^*_{\infty}\}$ is a
$\sigma$-subalgebra of $\frak A$ including $\frak B_n$ for every
$n\in\Bbb N$.   Because $\frak A$ is ccc, $\frak B^*$ is
(order-\nobreak)closed (316Fb), so includes $\frak B$.
This means that if $E\in\Sigma_{\infty}$ there must be an
$F\in\Sigma_{\infty}^*$ such that $E^{\ssbullet}=F^{\ssbullet}$.   But
now $(E\symmdiff F)^{\ssbullet}=0\in\frak B_0$, so
$E\symmdiff F\in\Sigma_0\subseteq \Sigma_{\infty}^*$, and $E$ also
belongs to $\Sigma^*_{\infty}$.   This shows that
$\Sigma_{\infty}\subseteq\Sigma_{\infty}^*$ and the two algebras are
equal.\ \Qed

\medskip

{\bf (b)} For each $n\in\Bbb N$, we have the partial lower density
$\undtheta_n:\frak B_n\to\Sigma$.   Since
$(\undtheta_na)^{\ssbullet}=a\in\frak B_n$ for every
$a\in\frak B_n$, $\undtheta_n$ takes all its values in $\Sigma_n$.
For $n\in\Bbb N$, let $\undphi_n:\Sigma_n\to\Sigma_n$ be the
lower density corresponding to $\undtheta_n$ (341Ba), that is,
$\undphi_nE=\undtheta_nE^{\ssbullet}$ for every
$E\in\Sigma_n$.

\medskip

{\bf (c)} For $a\in\frak A$, $n\in\Bbb N$ choose $G_a\in\Sigma$,
$g_{an}$ such that $G_a^{\ssbullet}=a$ and $g_{an}$ is a conditional
expectation of $\chi G_a$ on $\Sigma_n$;  that is,

\Centerline{$\int_Eg_{an}=\int_E\chi G_a=\mu(E\cap
G_a)=\bar\mu(E^{\ssbullet}\Bcap a)$}

\noindent for every $E\in\Sigma_n$.   As remarked in 233Db, such a
function $g_{an}$ can always be found, and moreover we may take it to be
$\Sigma_n$-measurable and defined everywhere on $X$.
Now if $a\in\frak B$, $\lim_{n\to\infty}g_{an}(x)$ exists and is equal
to $\chi G_a(x)$ for almost every $x$.   \Prf\ By L\'evy's martingale
theorem (275I), $\lim_{n\to\infty}g_{an}$ is defined almost everywhere
and is a conditional expectation of $\chi G_a$ on the $\sigma$-algebra
generated by $\bigcup_{n\in\Bbb N}\Sigma_n$.   As observed in (a), this
is just $\Sigma_{\infty}$;  and as $\chi G_a$ is itself
$\Sigma_{\infty}$-measurable, it is also a conditional expectation of
itself on $\Sigma_{\infty}$, and must be equal almost everywhere to
$\lim_{n\to\infty}g_{an}$.\ \Qed

\medskip

{\bf (d)} For $a\in\frak B$, $k\ge 1$, $n\in\Bbb N$ set

\Centerline{$H_{kn}(a)
=\{x:x\in X,\,g_{an}(x)\ge 1-2^{-k}\}\in\Sigma_n$,
\quad $\tilde H_{kn}(a)=\undphi_n(H_{kn}(a))$,}

\Centerline{$\undtheta a
=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}\tilde H_{km}(a)$.}

\noindent The rest of the proof is devoted to showing that
$\undtheta:\frak B\to\Sigma$ has the required properties.

\medskip

{\bf (e)} $G_0$ is negligible, so every $g_{0n}$ is zero almost
everywhere, every $H_{kn}(0)$ is negligible and every $\tilde H_{kn}(0)$
is empty;  so $\undtheta 0=\emptyset$.

\medskip

{\bf (f)} If $a\Bsubseteq b$ in $\frak B$, then $\undtheta a\subseteq
\undtheta b$.  \Prf\  $G_a\setminus G_b$ is negligible, $g_{an}\le
g_{bn}$ almost everywhere for every $n$, every $H_{kn}(a)\setminus
H_{kn}(b)$ is negligible, $\tilde H_{kn}(a)\subseteq\tilde H_{kn}(b)$
for every $n$ and $k$, and $\undtheta a\subseteq \undtheta b$.\ \Qed

\medskip

{\bf (g)}  If $a$, $b\in\frak B$ then $\undtheta(a\Bcap b)=\undtheta
a\cap\undtheta b$.
\Prf\ $\chi G_{a\Bcap b}\geae\chi G_a+\chi G_b-1$ so
$g_{a\Bcap b,n}\geae g_{an}+g_{bn}-1$ for every $n$.   Accordingly

\Centerline{$H_{k+1,n}(a)\cap H_{k+1,n}(b)\setminus H_{kn}(a\Bcap b)$}

\noindent is negligible, and (because $\undphi_n$ is a lower density)

\Centerline{$\tilde H_{kn}(a\Bcap b)\supseteq
\undphi_n(H_{k+1,n}(a)\cap H_{k+1,n}(b))
=\tilde H_{k+1,n}(a)\cap\tilde H_{k+1,n}(b)$}

\noindent for all $k\ge 1$, $n\in\Bbb N$.   Now, if $x\in\undtheta
a\cap\undtheta b$, then, for any $k\ge 1$, there are $n_1$, $n_2\in\Bbb
N$ such that

\Centerline{$x\in\bigcap_{m\ge n_1}\tilde H_{k+1,m}(a)$,
\quad$x\in\bigcap_{m\ge n_2}\tilde H_{k+1,m}(b)$.}

\noindent But this means that

\Centerline{$x\in\bigcap_{m\ge\max(n_1,n_2)}\tilde H_{km}(a\Bcap b)$.}

\noindent As $k$ is arbitrary, $x\in\undtheta (a\Bcap b)$;  as $x$ is
arbitrary, $\undtheta a\cap\undtheta b\subseteq\undtheta(a\Bcap b)$.
We know already from (f) that $\undtheta(a\Bcap b)\subseteq\undtheta
a\cap\undtheta b$, so
$\undtheta(a\Bcap b)=\undtheta a\cap\undtheta b$.\ \Qed

\medskip

{\bf (h)}
If $a\in\frak B$, then $\undtheta a^{\ssbullet}=a$.   \Prf\
$\sequencen{g_{an}}\to\chi G_a$ a.e., so setting

\Centerline{$V_a
=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}H_{km}(a)
=\{x:\liminf_{n\to\infty}g_{an}(x)\ge 1\}$,}

\noindent $V_a\symmdiff G_a$ is negligible, and
$V_a^{\ssbullet}=a$;  but

\Centerline{$\undtheta a\symmdiff V_a
\subseteq\bigcup_{k\ge 1,n\in\Bbb N}H_{kn}(a)\symmdiff\tilde H_{kn}(a)$}

\noindent is negligible, so $\undtheta a^{\ssbullet}$ is also equal
to $a$.\ \QeD\   Thus $\undtheta$ is a partial lower density with domain
$\frak B$.

\medskip

{\bf (i)} Finally, $\undtheta$ extends $\undtheta_n$ for every
$n\in\Bbb N$.   \Prf\ If $a\in\frak B_n$, then $G_a\in\Sigma_m$ for every $m\ge n$, so $g_{am}\eae\chi G_a$ for every $m\ge n$;  $H_{km}(a)\symmdiff G_a$ is negligible for $k\ge 1$, $m\ge n$;

\Centerline{$\tilde H_{km}=\undphi_mG_a=\undtheta_m a=\undtheta_na$}

\noindent for $k\ge 1$, $m\ge n$ (this is where I use the hypothesis
that $\undtheta_{m+1}$ extends $\undtheta_m$ for every $m$); and

$$\eqalign{\undtheta a
&=\bigcap_{k\ge 1}\bigcup_{r\in\Bbb N}
  \bigcap_{m\ge r}\tilde H_{km}(a)\cr
&=\bigcap_{k\ge 1}\bigcup_{r\ge n}\bigcap_{m\ge r}\tilde H_{km}(a)
=\bigcap_{k\ge 1}\bigcup_{r\ge n}\undtheta_na
=\undtheta_n a.\text{ \Qed}\cr}$$

The proof is complete.
}%end of proof of 341G


\leader{341H}{}\cmmnt{ Now for the first main theorem.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be any strictly
localizable measure space.   Then it has a lower density
$\undphi:\Sigma\to\Sigma$.   If $\mu X>0$ we can take $\undphi X=X$.

\proof{{\bf: Part A} I deal first with the case of probability spaces.
Let $(X,\Sigma,\mu)$ be a probability space, and $(\frak A,\bar\mu)$ its
measure algebra.

\medskip

{\bf (a)} Set $\kappa=\#(\frak A)$ and enumerate $\frak A$ as $\langle
a_{\xi}\rangle_{\xi<\kappa}$.   For $\xi\le\kappa$ let $\frak A_{\xi}$
be the closed subalgebra of $\frak A$ generated by
$\{a_{\eta}:\eta<\xi\}$.
I seek to define a lower density $\undtheta:\frak A\to\Sigma$ as the
last of a family $\langle\undtheta_{\xi}\rangle_{\xi\le\kappa}$, where
$\undtheta_{\xi}:\frak A_{\xi}\to\Sigma$ is a partial lower
density for each $\xi$.
The inductive hypothesis will be that $\undtheta_{\xi}$ extends
$\undtheta_{\eta}$ whenever $\eta\le\xi\le\kappa$.

To start the induction, we have $\frak A_0=\{0,1\}$,
$\undtheta_00=\emptyset$, $\undtheta_01=X$.
\medskip

{\bf (b)} {\it Inductive step to a successor ordinal $\xi$}   Given a
successor ordinal $\xi\le\kappa$, express it as $\zeta+1$;  we are
supposing that $\undtheta_{\zeta}:\frak A_{\zeta}\to\Sigma$ has
been defined.   Now $\frak A_{\xi}$ is the subalgebra of $\frak A$
generated by $\frak A_{\zeta}\cup\{a_{\zeta}\}$ (because this is a
closed subalgebra, by 323K).   So 341F tells us that $\undtheta_{\zeta}$
can be extended to a partial lower density $\undtheta_{\xi}$ with domain
$\frak A_{\xi}$.

\medskip

{\bf (c)} {\it Inductive step to a non-zero limit ordinal $\xi$ of
countable cofinality} In this case, there is a strictly increasing
sequence $\sequencen{\zeta(n)}$ with supremum $\xi$.   Applying 341G
with $\frak B_n=\frak A_{\zeta(n)}$, we see that there is a partial
lower density $\undtheta_{\xi}$, with domain the closed subalgebra
$\frak B$ generated by $\bigcup_{n\in\Bbb N}\frak A_{\zeta(n)}$,
extending every $\undtheta_{\zeta(n)}$.   Now
$\frak A_{\zeta(n)}\subseteq\frak A_{\xi}$ for every $n$, so
$\frak B\subseteq\frak A_{\xi}$;  but also, if $\eta<\xi$, there is an
$n\in\Bbb N$ such that $\eta<\zeta(n)$, so that
$a_{\eta}\in\frak A_{\zeta(n)}\subseteq\frak B$;  as $\eta$ is arbitrary,
$\frak A_{\xi}\subseteq\frak B$ and $\frak A_{\xi}=\frak B$.   Again, if
$\eta<\xi$, there is an $n$ such that $\eta\le\zeta(n)$, so that
$\undtheta_{\zeta(n)}$ extends $\undtheta_{\eta}$ and $\undtheta_{\xi}$
extends $\undtheta_{\eta}$.   Thus the induction continues.

\medskip


{\bf (d)} {\it Inductive step to a limit ordinal $\xi$ of uncountable
cofinality} In this case,
$\frak A_{\xi}=\bigcup_{\eta<\xi}\frak A_{\eta}$.
\Prf\ Because $\frak A$ is ccc, every member $a$ of
$\frak A_{\xi}$ must be in the closed subalgebra of $\frak A$ generated
by some countable subset $A$ of $\{a_{\eta}:\eta<\xi\}$ (331Gd-331Ge).
Now $A$ can be expressed as $\{a_{\eta}:\eta\in I\}$ for some countable
$I\subseteq\xi$.   As $I$ cannot be cofinal with $\xi$, there is a
$\zeta<\xi$ such that $\eta<\zeta$ for every $\eta\in I$, so that
$A\subseteq\frak A_{\zeta}$ and $a\in\frak A_{\zeta}$.\ \Qed

But now, because $\undtheta_{\zeta}$ extends
$\undtheta_{\eta}$ whenever $\eta\le\zeta<\xi$, we have a
function $\undtheta_{\xi}:\frak A_{\xi}\to\Sigma$ defined by
writing $\undtheta_{\xi}a=\undtheta_{\eta}a$ whenever
$\eta<\xi$ and $a\in\frak A_{\eta}$.   Because the family $\{\frak
A_{\eta}:\eta<\xi\}$ is totally ordered and every
$\undtheta_{\eta}$ is a partial lower density,
$\undtheta_{\xi}$ is a partial lower density.

Thus the induction proceeds when $\xi$ is a limit ordinal of uncountable
cofinality.

\medskip

{\bf (e)} The induction stops when we reach
$\undtheta_{\kappa}:\frak A\to\Sigma$, which is a lower
density such that $\undtheta_{\kappa}1=X$.   Setting
$\undphi E=\undtheta_{\kappa}E^{\ssbullet}$,
$\undphi$ is a lower density such that $\undphi X=X$.

\medskip

{\bf Part B} The general case of a strictly localizable
measure space
follows easily.   First, if $\mu X=0$, then $\frak A=\{0\}$ and we
can set $\undphi 0=\emptyset$.   Second, if $\mu$ is totally finite
but not zero, we can replace it by $\nu$, where $\nu E=\mu
E/\mu X$ for every $E\in\Sigma$;  a lower density for $\nu$ is
also a lower density for $\mu$.   Third, if $\mu$ is not totally finite,
let $\langle X_i\rangle_{i\in I}$ be a decomposition of $X$ (211E).
There is surely some $j$ such that $\mu X_j>0$;  replacing $X_j$ by
$X_j\cup\bigcup\{X_i:i\in I,\,\mu X_i=0\}$, we may assume that $\mu
X_i>0$ for every $i\in I$.   For each
$i\in I$, let $\undphi_i:\Sigma_i\to\Sigma_i$ be a
lower density for $\mu_i$, where $\Sigma_i=\Sigma\cap\Cal PX_i$ and
$\mu_i=\mu\restr\Sigma_i$, such that $\undphi_iX_i=X_i$.   Then it is
easy to check that we
have a lower density $\undphi:\Sigma\to\Sigma$ given by setting

\Centerline{$\undphi E=\bigcup_{i\in I}\undphi_i(E\cap X_i)$}

\noindent for every $E\in\Sigma$, and that $\undphi X=X$.
}%end of proof of 341H

\leader{341I}{}\cmmnt{ The next step is to give a method of moving
from lower densities to liftings.   I start with an elementary remark on
lower densities on complete measure spaces.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a complete measure space
with
measure algebra $\frak A$.

(a) Suppose that $\undtheta:\frak A\to\Sigma$ is a lower density and
$\undtheta_1:\frak A\to\Cal PX$ is a function such that
$\undtheta_10=\emptyset$,
$\undtheta_1(a\Bcap b)=\undtheta_1a\cap\undtheta_1b$ for all
$a$, $b\in\frak A$ and $\undtheta_1a\supseteq\undtheta a$ for all
$a\in\frak A$.   Then $\undtheta_1$ is a lower density.   If
$\undtheta_1$
is a Boolean homomorphism, it is a lifting.

(b) Suppose that $\undphi:\Sigma\to\Sigma$ is a lower density and
$\undphi_1:\Sigma\to\Cal PX$ is a function such that
$\undphi_1E=\undphi_1F$ whenever $E\symmdiff F$ is negligible,
$\undphi_1\emptyset=\emptyset$,
$\undphi_1(E\Bcap F)=\undphi_1E\cap\undphi_1F$ for all
$E$, $F\in\Sigma$ and $\undphi_1E\supseteq\undphi E$ for all
$E\in\Sigma$.
Then $\undphi_1$ is a lower density.   If $\undphi_1$ is a Boolean
homomorphism, it is a lifting.

\proof{{\bf (a)} All I have to check is that $\undtheta_1a\in\Sigma$ and
$(\undtheta_1a)^{\ssbullet}=a$ for every $a\in\frak A$.   But

\Centerline{$\undtheta a\subseteq\undtheta_1a$,
\quad$\undtheta(1\Bsetminus a)\subseteq\undtheta_1(1\Bsetminus a)$,
\quad$\undtheta_1a\cap\undtheta_1(1\Bsetminus a)
=\undtheta_10=\emptyset$.}

\noindent So

\Centerline{$\undtheta a\subseteq\undtheta_1a\subseteq
X\setminus\undtheta(1\Bsetminus a)$.}

\noindent Since

\Centerline{$(\undtheta a)^{\ssbullet}=a
=(X\setminus\undtheta(1\Bsetminus a))^{\ssbullet}$,}

\noindent and $\mu$ is complete, $\undtheta_1$ is a lower density.   If
it is a Boolean homomorphism, then it is also a lifting (341De).

\medskip

{\bf (b)} This follows by the same argument, or by looking at the
functions from $\frak A$ to $\Sigma$ defined by $\undphi$ and
$\undphi_1$ and using (a).
}%end of proof of 341I

\leader{341J}{Proposition} Let $(X,\Sigma,\mu)$ be a complete
measure space such that $\mu X>0$, and $\frak A$ its measure algebra.

(a) If $\undtheta:\frak A\to\Sigma$ is any lower density, there is a
lifting $\theta:\frak A\to\Sigma$ such that $\theta a\supseteq\undtheta
a$
for every $a\in\frak A$.

(b) If $\undphi:\Sigma\to\Sigma$ is any lower density, there is a
lifting $\phi:\Sigma\to\Sigma$ such that $\phi E\supseteq\undphi E$ for
every $E\in\Sigma$.

\proof{{\bf (a)}  For each $x\in\undtheta 1$, set

\Centerline{$I_x=\{a:a\in\frak A,\,x\in\undtheta(1\Bsetminus a)\}$.}

\noindent Then $I_x$ is a proper ideal of $\frak A$.   \Prf\ We have

\inset{$0\in I_x$, because $x\in\undtheta 1$,}

\inset{if $b\Bsubseteq a\in I_x$ then $b\in I_x$, because
$x\in\undtheta(1\Bsetminus a)\subseteq\undtheta(1\Bsetminus b)$,}

\inset{if $a$, $b\in I_x$ then $a\Bcup b\in I_x$, because
$x\in\undtheta(1\Bsetminus a)\cap\undtheta(1\Bsetminus
b)=\undtheta(1\Bsetminus(a\Bcup b))$,}

\inset{$1\notin I_x$ because $x\notin\emptyset=\undtheta 0$.\ \Qed}

\noindent For $x\in X\setminus\undtheta 1$, set $I_x=\{0\}$;  this is
also a proper ideal of $\frak A$, because $\frak A\ne\{0\}$.   By 311D,
there is a surjective Boolean homomorphism $\pi_x:\frak A\to\{0,1\}$
such that $\pi_xd=0$ for every $d\in I_x$.

Define $\theta:\frak A\to\Cal PX$ by setting

\Centerline{$\theta a=\{x:x\in X,\,\pi_x(a)=1\}$}

\noindent for every $a\in\frak A$.   It is easy to check that, because
every $\pi_x$ is a surjective Boolean homomorphism, $\theta$ is a
Boolean homomorphism.   Now for any $a\in\frak A$, $x\in X$,

\Centerline{$x\in\undtheta a\Longrightarrow 1\Bsetminus a\in I_x
\Longrightarrow \pi_x(1\Bsetminus a)=0\Longrightarrow \pi_xa=1
\Longrightarrow x\in\theta a$.}

\noindent Thus $\theta a\supseteq\undtheta a$ for every $a\in\frak A$.
By 341I, $\theta$ is a lifting, as required.

\medskip

{\bf (b)} Repeat the argument above, or apply it, defining $\undtheta$
by setting $\undtheta(E^{\ssbullet})=\undphi E$ for every $E\in\Sigma$,
and $\phi$ by setting $\phi E=\theta(E^{\ssbullet})$ for every $E$.
}%end of proof of 341J

\leader{341K}{The Lifting Theorem} Every complete strictly localizable
measure space of non-zero measure has a lifting.

\proof{ By 341H, it has a lower density, so by 341J it has a lifting.}

\cmmnt{
\leader{341L}{Remarks} If we count 341F-341K as a single argument, it
may be the longest proof, after Carleson's theorem (\S286), which I have
yet presented in this treatise, and
perhaps it will be helpful if I suggest ways of looking at its
components.

\header{341La}{\bf (a)} The first point is that the theorem should be
thought of as one about probability spaces.   The shift to general
strictly localizable spaces (Part B of the proof of 341H) is purely a
matter of technique.   I would not have presented it if I did not think
that it's worth doing, for a variety of reasons, but there is no
significant idea needed, and if -- for instance -- the result were valid
only for $\sigma$-finite spaces, it would still be one of the great
theorems of mathematics.   So the rest of these remarks will be directed
to the ideas needed in probability spaces.

\header{341Lb}{\bf (b)} All the proofs I know of the theorem depend in
one way or another on an inductive construction.   We do not, of course,
need a transfinite induction written out in the way I have presented it
in 341H above.   Essentially the same proof can be presented as an
application of Zorn's Lemma;  if we take $P$ to be the set of partial
lower densities, then the arguments of 341G and part (A-d) of the proof
of 341H can be adapted to prove that any totally ordered subset of $P$
has an upper bound in $P$, while the argument of 341F shows that any
maximal element of $P$ must have domain $\frak A$.   I think it is
purely a matter of taste which form one prefers.   I suppose I have used
the ordinal-indexed form largely because that seemed appropriate for
Maharam's theorem in the last chapter.

\header{341Lc}{\bf (c)} There are then three types of inductive step to
examine, corresponding to 341F, 341G and (A-d) in 341H.   The first and
last are easier than the second.   Seeking the one-step extension of
$\undtheta:\frak B\to\Sigma$ to $\undtheta_1:\frak B_{1}\to\Sigma$,
the natural model to use is the one-step extension of a
Boolean homomorphism presented in 312O.    The situation here is
rather more complicated, as $\undtheta_1$ is not fully specified by the
value of $\undtheta_1e$, and we do in fact have more freedom at this
point than is entirely welcome.   The formula used in the proof of 341F
is derived from {\smc Graf \& Weizs\"acker 76}.

\header{341Ld}{\bf (d)} At this point I must call attention to the way
in which the whole proof is dominated by the choice of {\it closed}
subalgebras as the domains of our partial liftings.   This is what makes
the inductive step to a limit ordinal $\xi$ of countable cofinality
difficult, because $\frak A_{\xi}$ will ordinarily be larger than
$\bigcup_{\eta<\xi}\frak A_{\eta}$.   But it is absolutely essential in
the one-step extensions as treated here.   (I will return to this point in
\S535 of Volume 5.   See also 341Ye.)

Because we are dealing with a ccc algebra $\frak A$, the requirement
that the $\frak A_{\xi}$ should be closed is not a problem when
$\cf\xi$ is uncountable, since in this case
$\bigcup_{\eta<\xi}\frak A_{\eta}$ is already a closed subalgebra;
this is the only idea needed in (A-d) of 341H.

\header{341Le}{\bf (e)} So we are left with the inductive step to $\xi$
when $\cf\xi=\omega$, which is 341G.   Here we actually need some
measure theory, and a particularly striking bit.   (You will see that
the {\it measure} $\mu$, as opposed to the algebras $\Sigma$ and
$\frak A$ and the homomorphism $E\mapsto E^{\ssbullet}$ and the ideal of
negligible sets, is simply not mentioned anywhere else in the whole
argument.)

\medskip

\quad{\bf (i)}
The central idea is to use the fact that bounded martingales converge to
define $\undtheta a$ in terms of a sequence of conditional
expectations.   Because I have chosen a fairly direct assault on the
problem, some of the surrounding facts are not perhaps so clearly
visible as they might have been if I had used a more leisurely route.
For each $a\in\frak A$, I start by choosing a representative
$G_a\in\Sigma$;  let me emphasize that this is a crude application of
the axiom of choice, and that the different sets $G_a$ are in no way
coordinated.   (The theorem we are proving is that
they {\it can} be coordinated, but we have not reached that point yet.)
Next, I choose, arbitrarily, a conditional expectation $g_{an}$ of
$\chi G_a$ on each
$\Sigma_n$.   Once again, the choices are not coordinated;
but the martingale theorem assures us that $g_a=\lim_{n\to\infty}g_{an}$
is defined almost everywhere, and is equal almost everywhere to
$\chi G_a$ if $a\in\frak B$.   Of course I could have gone to the $g_{an}$
directly, without mentioning the $G_a$;  $g_{an}$ is a Radon-Nikod\'ym
derivative of the countably additive functional
$E\mapsto\bar\mu(E^{\ssbullet}\Bcap a):\Sigma_n\to\Bbb R$.   Now the
$g_{an}$, like the $G_a$, are not uniquely defined.   But they are
defined `up to a negligible set', so that any alternative functions
$g'_{an}$ would have $g'_{an}\eae g_{an}$.   This means that the sets
$H_{kn}(a)=\{x:g_{an}(x)\ge 1-2^{-k}\}$ are also defined `up to a
negligible set', and consequently the sets
$\tilde H_{kn}(a)=\undphi_{n}(H_{kn}(a))$ are uniquely defined.   I point
this out to show that it is not a complete miracle that we have formulae

\Centerline{$\tilde H_{kn}(a)\subseteq\tilde H_{kn}(b)$ if
$a\Bsubseteq b$,}

\Centerline{$\tilde H_{kn}(a\Bcap b)
\supseteq\tilde H_{k+1,n}(a)\cap\tilde H_{k+1,n}(b)$
for all $a$, $b\in\frak A$}

\noindent which do not ask us to turn a blind eye to any negligible
sets.   I note in passing that I could have defined the
$\tilde H_{kn}(a)$ without mentioning the $g_{an}$;  in fact

\Centerline{$\tilde H_{kn}(a)
=\undtheta_{n}(\sup\{c:c\in\frak B_{n},\,
\bar\mu(a\Bcap d)\ge 1-2^{-k}\bar\mu d$ whenever $d\in\frak B_{n}$
and $d\Bsubseteq c\})$.}

\medskip

\quad{\bf (ii)} Now, with the sets $\tilde H_{kn}(a)$ in hand, we can
look at

\Centerline{$\tilde V_a=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb
N}\bigcap_{m\ge n}\tilde H_{kn}(a)$;}

\noindent because $g_{an}\to \chi G_a$ a.e., $\tilde V_a\symmdiff G_a$
is negligible and $\tilde V_a^{\ssbullet}=a$ for every
$a\in\frak A_{\xi}$.   The rest of the argument amounts to checking that
$a\mapsto\tilde V_a$ will serve for $\undtheta$.

\spheader 341Lf The arguments above apply to all probability spaces, and
show that every probability space has a lower density.   The next step
is to convert a lower density into a lifting.   It is here that we need
to assume completeness.   The point is that we can find a Boolean
homomorphism $\theta:\frak A\to\Cal PX$ such that $\undtheta
a\subseteq\theta a$ for every $a$;   this corresponds just to extending
the ideals $I_x=\{a:x\in\undtheta(1\Bsetminus a)\}$ to maximal ideals
(and giving a moment's thought to $x\in X\setminus\undtheta 1$).   In
order to ensure that $\theta a\in\Sigma$ and $(\theta a)^{\ssbullet}=a$,
we have to observe that $\theta a$ is sandwiched between $\undtheta a$
and $X\setminus\undtheta(1\Bsetminus a)$, which differ by a negligible
set;  so that if $\mu$ is complete all will be well.

\spheader 341Lg The fact that completeness is needed at only one point
in the argument makes it natural to wonder whether the theorem might be
true for probability spaces in general.   (I will come later, in
341M, to non-strictly-localizable spaces.)   There is as yet no
satisfactory answer to this.   For Borel measure on $\Bbb R$, the
question is known to be undecidable from the ordinary axioms of set
theory (including the axiom of choice, but not the continuum hypothesis,
as usual);  I will give the easy part of the argument in \S535;
see {\smc Burke 93} for the rest.
But I conjecture that there is a counter-example under the ordinary
axioms (see 341Z below).

\spheader 341Lh Quite apart from whether completeness is needed in the
argument, it is not absolutely clear why measure theory is required.
The general question of whether a lifting exists can be formulated for
any triple $(X,\Sigma,\Cal I)$ where $X$ is a set, $\Sigma$ is a
$\sigma$-algebra of subsets of $X$, and $\Cal I$ is a $\sigma$-ideal of
$\Sigma$.   (See 341Ya below.)   S.Shelah has given an example of such a
triple without a lifting in which two of the basic properties of the
measure-theoretic case are satisfied:  $(X,\Sigma,\Cal I)$ is
`complete' in the sense that every subset of any member of $\Cal I$
belongs to $\Sigma$ (and therefore to $\Cal I$), and $\Cal I$ is
$\omega_1$-saturated in $\Sigma$ in the sense of 316C (see 
{\smc Shelah 98}).   
But many other cases are known (e.g., 341Yb) in which liftings
do exist.

\spheader 341Li It is of course possible to prove 341K without
mentioning `lower densities', and there are even some advantages in
doing so.   The idea is to follow the lines of 341H, but with `liftings'
instead of `lower densities' throughout.   The inductive step to a
successor ordinal is actually easier, because we have a Boolean
homomorphism $\theta$ in 341F to extend, and we can use 312O as it
stands if we can choose the pair $E$, $F=X\setminus E$ correctly.   The
inductive step to an ordinal of uncountable cofinality remains
straightforward.   But in the inductive step to an ordinal of countable
cofinality, we find that in 341G we get no help from assuming that the
$\undtheta_n$ are actually liftings;  we are still led to a lower
density $\undtheta$.   So at this point we have to interpolate the
argument of 341J to convert this lower density into a lifting.

I have chosen the more leisurely exposition, with the extra concept,
partly in order to get as far as possible without assuming completeness
of the measure and partly because lower densities are an important tool
for further work (see \S\S345-346).

\spheader 341Lj For more light on the argument of 341G see also 363Xe
and 363Yf below.
}%end of comment 341L

\leader{341M}{}\cmmnt{ I remarked above that the shift from
probability spaces to general strictly localizable spaces was simply a
matter of technique.   The question of which spaces have liftings is
also primarily a matter concerning probability spaces, as the next
result shows.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined space with $\mu X>0$.   Then it has a lifting iff it has a
lower density iff it is strictly localizable.

\proof{ If $(X,\Sigma,\mu)$ is strictly localizable then it has a
lifting, by 341K.   A lifting is already a lower density, and if
$(X,\Sigma,\mu)$ has a lower density it has a lifting, by 341J.   So we
have only to prove that if it has a lifting then it is strictly
localizable.

Let $\theta:\frak A\to\Sigma$ be a lifting, where $\frak A$ is the
measure algebra of $(X,\Sigma,\mu)$.   Let $C$ be a partition of unity
in $\frak A$ consisting of elements of finite measure (322Ea).   Set
$\Cal A=\{\theta c:c\in C\}$.   Because $C$ is disjoint, so is
$\Cal A$.   Because $\sup C=1$ in $\frak A$, every set of positive
measure meets some member of $\Cal A$ in a set of positive measure.   So
the conditions of 213Oa are satisfied, and $(X,\Sigma,\mu)$ is strictly
localizable.
}%end of proof of 341M

\leader{341N}{Extension of partial \dvrocolon{liftings}}\cmmnt{ The
following facts are obvious from the proof of 341H, but it will be
useful to have them out in the open.

\wheader{341N}{6}{2}{2}{48pt}

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a probability space
and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.

(a) Any partial lower density $\undphi_0:\Tau\to\Sigma$ has an extension
to a lower density $\undphi:\Sigma\to\Sigma$.

(b) Suppose now that $\mu$ is complete.   If $\undphi_0$ is a Boolean
homomorphism, it has an extension to a lifting $\phi$ for $\mu$.

%query:  example of partial lower density on a subalgebra which has
%no extension?

\proof{{\bf (a)} In Part A of the proof of 341H, let $\frak A_{\xi}$ be
the closed subalgebra of $\frak A$ generated by
$\{E^{\ssbullet}:E\in\Tau\}\cup\{a_{\eta}:\eta<\xi\}$, and set
$\undtheta_0E^{\ssbullet}=\undphi_0E$ for every $E\in\Tau$.   Proceed
with the induction as before.   The only difference is that we no longer
have a guarantee that $\undphi X=X$.

\medskip

{\bf (b)} Suppose now that $\undphi_0$ is a Boolean homomorphism and $\mu$
is complete.   341J
tells us that there is a lifting $\phi:\Sigma\to\Sigma$ such that $\phi
E\supseteq\undphi E$ for every $E\in\Sigma$.   But if $E\in\Tau$ we must
have $\phi E\supseteq\undphi_0E$,

\Centerline{$\phi E\setminus\undphi_0E
=\phi E\cap\undphi_0(X\setminus E)
\subseteq\phi E\cap\phi(X\setminus E)=\emptyset$,}

\noindent so that $\phi E=\undphi_0E$, and $\phi$ extends $\undphi_0$.
}%end of proof of 341N


\leader{341O}{Liftings and Stone spaces}\cmmnt{ The arguments of this
section so far involve repeated use of the axiom of choice, and offer no
suggestion that any liftings (or lower densities) are in any sense
`canonical'.   There is however one context in which we have a
distinguished
lifting.}   Suppose that we have the Stone space $(Z,\Tau,\nu)$ of a
measure algebra $(\frak A,\bar\mu)$;  \cmmnt{as in 311E,} I think of
$Z$ as being the set of surjective Boolean homomorphisms from $\frak A$
to $\Bbb Z_2$, so that each $a\in\frak A$ corresponds to the
open-and-closed set $\widehat{a}=\{z:z(a)=1\}$.   Then we have a lifting
$\theta:\frak A\to\Tau$ defined by setting $\theta a=\widehat{a}$ for
each $a\in\frak A$.   \cmmnt{(I am identifying
$\frak A$ with the measure algebra of $\nu$, as in 321J.)}   The
corresponding
lifting $\phi:\Tau\to\Tau$ is defined by taking $\phi E$ to be that
unique open-and-closed set such that $E\symmdiff\phi E$ is
negligible\cmmnt{ (or, if you prefer, meager)}.

\cmmnt{Generally, liftings can be described in terms of Stone spaces,
as follows.}

\leader{341P}{Proposition} Let
$(X,\Sigma,\mu)$ be a measure space, $(\frak A,\bar\mu)$ its measure
algebra, and $(Z,\Tau,\nu)$ the Stone space of $(\frak A,\bar\mu)$ with
its canonical measure.

(a) There is a one-to-one correspondence between liftings
$\theta:\frak A\to\Sigma$ and functions $f:X\to Z$ such that
$f^{-1}[\widehat{a}]\in\Sigma$
and $(f^{-1}[\widehat{a}])^{\ssbullet}=a$ for every $a\in\frak A$,
defined by the formula

\Centerline{$\theta a=f^{-1}[\widehat{a}]$ for every $a\in\frak A$.}

(b) If $(X,\Sigma,\mu)$ is complete and locally determined, then a
function $f:X\to Z$ satisfies the conditions of (a) iff ($\alpha$) it is
inverse-measure-preserving ($\beta$) the homomorphism it induces between
the measure algebras of $\mu$ and $\nu$ is the canonical
isomorphism defined by the construction of $Z$.

\proof{ Recall that $\Tau$ is just the set
$\{\widehat{a}\symmdiff M:a\in\frak A$, $M\subseteq Z$ is meager$\}$,
and that $\nu(\widehat{a}\symmdiff M)=\bar\mu a$ for all such $a$, $M$;
while the canonical isomorphism $\pi$ between $\frak A$ and the measure
algebra of $\nu$ is defined by the formula

\Centerline{$\pi F^{\ssbullet}=a$ whenever $F\in\Tau$, $a\in\frak A$ and
$F\symmdiff\widehat{a}$ is meager}

\noindent (341K).

\medskip

{\bf (a)} If $\theta:\frak A\to\Sigma$ is any Boolean homomorphism, then
for every $x\in X$ we have a surjective Boolean homomorphism
$f_{\theta}(x):\frak A\to\Bbb Z_2$ defined by saying that
$f_{\theta}(x)(a)=1$ if $x\in\theta a$, $0$ otherwise.   $f_{\theta}$ is
a function from $X$ to $Z$.   We can recover
$\theta$ from $f_{\theta}$ by the formula

\Centerline{$\theta
a=\{x:f_{\theta}(x)(a)=1\}=\{x:f_{\theta}(x)\in\widehat{a}\}
=f_{\theta}^{-1}[\widehat{a}]$.}

\noindent So $f_{\theta}^{-1}[\widehat{a}]\in\Sigma$ and, if $\theta$ is
a lifting,


\Centerline{$(f_{\theta}^{-1}[\widehat{a}])^{\ssbullet}=(\theta
a)^{\ssbullet}=a$.}

\noindent for every $a\in\frak A$.

Similarly, given a function $f:X\to Z$ with this property, then we can
set $\theta a=f^{-1}[\widehat{a}]$ for every $a\in\frak A$ to obtain a
lifting $\theta:\frak A\to\Sigma$;  and of course we now have

\Centerline{$f(x)(a) = 1\iff f(x)\in\widehat{a}\iff x\in\theta a$,}

\noindent so $f_{\theta}=f$.

\medskip

{\bf (b)} Assume now that $(X,\Sigma,\mu)$ is complete and locally
determined.

\medskip

\quad{\bf (i)} Let $f:X\to Z$ be the function associated with a lifting
$\theta$, as in (a).   I show first that $f$ is \imp.
\Prf\ If $F\in\Tau$, express it as $\widehat{a}\symmdiff M$, where
$a\in\frak A$ and $M\subseteq Z$ is meager.   By 322F, $\frak A$ is
\wsid, so $M$ is nowhere dense (316I).
Consider $f^{-1}[M]$.  If $E\subseteq X$ is measurable and of finite
measure, then $E\cap f^{-1}[M]$ has a measurable envelope $H$ (132Ee).
\Quer\ If $\mu H>0$, then $b=H^{\ssbullet}\ne 0$ and $\widehat{b}$ is a
non-empty open set in $Z$.   Because $M$ is nowhere dense, there is a
non-zero $a\in\frak A$ such that
$\widehat{a}\subseteq\widehat{b}\setminus
M$.   Now $\mu(f^{-1}[\widehat{b}]\symmdiff H)=0$, so
$f^{-1}[\widehat{a}]\setminus
H$ is negligible, and $f^{-1}[\widehat{a}]\cap H$ is a non-negligible
measurable set disjoint from $E\cap f^{-1}[M]$ and included in $H$;
which is
impossible.\ \BanG\   Thus $H$ and $E\cap f^{-1}[M]$ are negligible.
This is true for every measurable set $E$ of finite measure.  Because
$\mu$ is complete and locally determined, $f^{-1}[M]\in\Sigma$ and
$\mu f^{-1}[M]=0$.
So $f^{-1}[F]=f^{-1}[\widehat{a}]\symmdiff f^{-1}[M]$ is measurable, and

\Centerline{$\mu f^{-1}[F]=\mu f^{-1}[\widehat{a}]=\mu\theta a=\bar\mu
a=\nu\widehat{a}=\nu F$.}

\noindent As $F$ is arbitrary, $f$ is \imp.\ \Qed

It follows at once that for any $F\in\Tau$,

\Centerline{$f^{-1}[F]^{\ssbullet}=a=\pi F^{\ssbullet}$}
\noindent where $a$ is that element of $\frak A$ such that $M=F\symmdiff
a$ is meager, because in this case $f^{-1}[\widehat{a}]^{\ssbullet}=a$,
by (a), while $f^{-1}[M]$ is negligible.   So $\pi$ is the
homomorphism induced by $f$.

\medskip

\quad{\bf (ii)} Now suppose that $f:X\to Z$ is an
inverse-measure-preserving function such that $f^{-1}[F]^{\ssbullet}=\pi
F^{\ssbullet}$ for every $F\in\Tau$.   Then, in particular,

\Centerline{$f^{-1}[\widehat{a}]^{\ssbullet}=\pi
\widehat{a}^{\ssbullet}=a$}

\noindent for every $a\in\frak A$, so that $f$ satisfies the conditions
of (a).
}%end of proof of 341P

\leader{341Q}{Corollary} Let $(X,\Sigma,\mu)$ be a strictly
localizable measure space, $(\frak A,\bar\mu)$ its measure algebra, and
$Z$ the Stone space of $\frak A$;  suppose that $\mu X>0$.   For
$E\in\Sigma$ write $E^*$ for the
open-and-closed subset of $Z$ corresponding to
$E^{\ssbullet}\in\frak A$.   Then there is a function $f:X\to Z$ such
that $E\symmdiff f^{-1}[E^*]$ is negligible for every $E\in\Sigma$.
If $\mu$ is complete, then $f$ is \imp.

\proof{ Let $\hat\mu$ be the completion of $\mu$, and $\hat\Sigma$ its
domain.   Then we can identify $(\frak A,\bar\mu)$ with the measure
algebra of $\hat\mu$ (322Da).   Let $\theta:\frak A\to\hat\Sigma$ be a
lifting, and $f:X\to Z$ the
corresponding function.   If $E\in\Sigma$ then $E^*=\widehat{a}$ where
$a=E^{\ssbullet}$, so
$E\symmdiff f^{-1}[E^*]=E\symmdiff\theta E^{\ssbullet}$ is negligible.
If $\mu$ is itself complete, so that $\hat\Sigma=\Sigma$, then $f$ is
\imp, by 341Pb.
}%end of proof of 341Q

\exercises{
\leader{341X}{Basic exercises (a)} Let $(X,\Sigma,\mu)$ be a measure
space and $\phi:\Sigma\to\Sigma$ a function.   Show that $\phi$ is a
lifting iff it is a lower density and $\phi E\cup\phi(X\setminus E)=X$
for every $E\in\Sigma$.
%341A

\sqheader 341Xb Let $\nu_{\Bbb N}$ be the usual measure on
$X=\{0,1\}^{\Bbb N}$, and $\Tau_{\Bbb N}$ its domain.
For $x\in X$ and $n\in\Bbb N$ set
$U_n(x)=\{y:y\in X,\,y\restr n=x\restr n\}$.   For $E\in\Tau_{\Bbb N}$ set
$\undphi E=\{x:\lim_{n\to\infty}2^n\mu(E\cap U_n(x))=1\}$.   Show
that $\undphi$ is a lower density for $\nu_{\Bbb N}$.
%341E

\sqheader 341Xc Let $\frak A$ be a Boolean algebra, $I$ an ideal of
$\frak A$, and $\frak B$ a countable subalgebra of the quotient algebra
$\frak A/I$.
Show that there is a Boolean homomorphism $\theta:\frak B\to\frak A$
such that
$(\theta b)^{\ssbullet}=b$ for every $b\in\frak B$.
\Hint{let $\sequencen{b_n}$ run over $\frak B$;  let $\frak B_n$ be the
subalgebra of $\frak B$ generated
by $\{b_i:i<n\}$;  given $\theta\restr\frak B_n$, show that there is an
$a_n\in\frak A$ such that $a_n^{\ssbullet}=b_n$ and
$\theta b'\Bsubseteq a_n\Bsubseteq \theta b''$
whenever $b'$, $b''\in\frak B_n$ and $b'\Bsubseteq b_n\Bsubseteq b''$.}
%new 2003;  341F

\sqheader 341Xd Let $P$ be the set of all lower densities of a complete
measure space $(X,\Sigma,\mu)$, with measure algebra $\frak A$, ordered
by saying that $\undtheta\le\undtheta'$ if
$\undtheta a\subseteq\undtheta'a$
for every $a\in\frak A$.   Show that any non-empty totally ordered
subset of $P$ has an upper bound
in $P$.   Show that if $\undtheta\in P$, $a\in\frak A\setminus\{0\}$ and
$x\in X\setminus(\undtheta a\cup\undtheta(1\Bsetminus a))$, then
$\undtheta':\frak A\to\Sigma$ is a lower density, where
$\undtheta'b=\undtheta b\cup\{x\}$ if either $a\Bsubseteq b$ or there is
a $c\in\frak A$ such that
$x\in\undtheta c$ and $a\Bcap c\Bsubseteq b$, and
$\undtheta'b=\undtheta b$ otherwise.   Hence prove 341J.
%341J

\spheader 341Xe Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces and suppose that there is an inverse-measure-preserving
function $f:X\to Y$ such that the associated homomorphism from the
measure algebra of $\nu$ to that of $\mu$ (324M) is an isomorphism.
Show that for every lifting $\phi$ for $(Y,\Tau,\nu)$ we have a
corresponding lifting $\psi$ of $(X,\Sigma,\mu)$ defined uniquely by the
formula

\Centerline{$\psi(f^{-1}[F])=f^{-1}[\phi F]$ for every $F\in\Tau$.}
%341P

\spheader 341Xf Let $(X,\Sigma,\mu)$ be a measure space, and write
$\eusm L^{\infty}(\Sigma)$ for the linear space of all bounded
$\Sigma$-measurable functions from $X$ to $\Bbb R$.   Show that for any
lifting $\phi:\Sigma\to\Sigma$ of $\mu$ there is a unique linear
operator $T:L^{\infty}(\mu)\to\eusm L^{\infty}(\Sigma)$ such that
$T(\chi E)^{\ssbullet}=\chi(\phi E)$ for every $E\in\Sigma$ and
$Tu\ge 0$ in $\eusm L^{\infty}(\Sigma)$ whenever $u\ge 0$ in
$L^{\infty}(\mu)$.   Show that (i) $(Tu)^{\ssbullet}=u$ and
$\sup_{x\in X}|(Tu)(x)|=\|u\|_{\infty}$ for every $u\in L^{\infty}(\mu)$
(ii) $T(u\times v)=Tu\times Tv$ for all $u$, $v\in L^{\infty}(\mu)$.

\spheader 341Xg Let $\mu$ be Lebesgue measure on $[0,1]$.
Write $\eusm L^1_{\Sigma_L}$ for the
linear space of integrable functions $f:[0,1]\to\Bbb R$.   Show that there
is no operator $T:L^1(\mu)\to\eusm L^1_{\Sigma_L}$ such that (i)
$(Tu)^{\ssbullet}=u$ for every $u\in L^1(\mu)$ (ii) $Tu\ge Tv$ whenever
$u\ge v$ in $L^1(\mu)$.
\Hint{Let $F\subseteq\eusm L^1_{\Sigma_L}$ be the countable set
$\{n\chi[2^{-n}k,2^{-n}(k+1)]:n\in\Bbb N$, $k<2^n\}$.
Show that if $T$
satisfies (i) then there is an $x\in\{0,1\}^{\Bbb N}$ such that
$T(f^{\ssbullet})(x)=f(x)$ for every $f\in F$;  find a sequence
$\sequencen{f_n}$ in $F$ such that $\{f_n^{\ssbullet}:n\in\Bbb N\}$ is
bounded above in $L^1(\mu)$ but $\sup_{n\in\Bbb N}f_n(x)=\infty$.}
%341Xf

\leader{341Y}{Further exercises (a)} Let $X$ be a set, $\Sigma$ an
algebra of subsets of $X$ and $\Cal I$ an ideal of $\Sigma$;  let
$\frak A$ be the quotient Boolean algebra $\Sigma/\Cal I$.   We say that
a function $\theta:\frak A\to\Sigma$ is a {\bf lifting} if it is a
Boolean homomorphism and $(\theta a)^{\ssbullet}=a$ for every
$a\in\frak A$, and that $\undtheta:\frak A\to\Sigma$ is a {\bf lower density} if $\undtheta 0=\emptyset$,
$\undtheta(a\Bcap b)=\undtheta a\cap\undtheta b$ for all $a$,
$b\in\frak A$, and $(\undtheta a)^{\ssbullet}=a$ for
every $a\in\frak A$.

Show that if $(X,\Sigma,\Cal I)$ is `complete' in the sense that
$F\in\Sigma$ whenever $F\subseteq E\in\Cal I$, and if $X\notin\Cal I$,
and $\undtheta:\frak A\to\Sigma$ is a lower density, then there is a
lifting $\theta:\frak A\to\Sigma$ such that
$\undtheta a\subseteq\theta a$ for every $a\in\frak A$.
%341J

\spheader 341Yb Let $X$ be a Baire space, $\widehat{\Cal B}$ the
Baire-property algebra of $X$ (314Yd)
and $\Cal M$ the ideal of meager subsets of $X$.   Show that there is a
lifting $\theta$ from $\widehat{\Cal B}/\Cal M$ to $\widehat{\Cal B}$
such that $\theta G^{\ssbullet}\supseteq G$ for every open
$G\subseteq X$.   \Hint{in
341Ya, set $\undtheta(G^{\ssbullet})=G$ for every regular open set
$G$.}
%341Ya 341J

\spheader 341Yc Let $(X,\Sigma,\mu)$ be a \Mth\
probability space with Maharam type $\kappa\ge\omega$.    Let
$\CalBa_{\kappa}$
be the Baire $\sigma$-algebra of $Y=\{0,1\}^{\kappa}$, that is, the
$\sigma$-algebra of subsets of $Y$ generated by the family
$\{\{x:x(\xi)=1\}:\xi<\kappa\}$, and let $\nu$ be the restriction to
$\CalBa_{\kappa}$ of the usual
measure on $\{0,1\}^{\kappa}$.  Show that there is an
inverse-measure-preserving function $f:X\to Y$ which induces an
isomorphism between the measure algebras of $\mu$ and $\nu$.
%341Yf

\spheader 341Yd Let $(X,\Sigma,\mu)$ be a complete \Mth\ probability
space with Maharam type $\kappa\ge\omega$, and give
$Y=\{0,1\}^{\kappa}$ its usual measure $\nu_{\kappa}$.  Show that
there is an
inverse-measure-preserving function $f:X\to Y$ which induces an
isomorphism between the measure algebras of $\mu$ and $\nu_{\kappa}$.
%341Yc

\header{341Ye}{\bf *(e)} Give an example of a
complete probability space $(X,\Sigma,\mu)$,
a subalgebra $\Tau$ of $\Sigma$, and a partial lower density
$\undphi:\Tau\to\Sigma$ which has no extension to a lower density for
$\mu$.   \Hint{There is a subset of $\{0,1\}^{\frakc}$, with cardinal
$\frak c$, which is non-negligible for the usual measure on
$\{0,1\}^{\frakc}$.}
%341N mt34bits

\spheader 341Yf\dvAnew{2014} 
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\familyiI{a_i}$ a family in $\frak A$.   Let $\CalBa_I$
be the Baire $\sigma$-algebra of $Y=\{0,1\}^I$, that is, the
$\sigma$-algebra of subsets of $Y$ generated by the family
$\{E_i:i\in I\}$ where $E_i=\{y:y\in Y$, $y(i)=1\}$ for $i\in I$.
Show that there is a unique sequentially order-continuous Boolean
homomorphism $\phi:\CalBa_I\to\frak A$ such that $\phi E_i=a_i$ for every
$i\in I$, and that $\phi[\CalBa_I]$ is the $\sigma$-subalgebra of $\frak A$
generated by $\{a_i:i\in I\}$.
%out of order query
}%end of exercises


\leader{341Z}{Problems (a)} Can we construct, using the ordinary axioms
of mathematics\cmmnt{ (including the axiom of choice, but not the
continuum
hypothesis)}, a probability space $(X,\Sigma,\mu)$ with no lifting?

\header{341Zb}{\bf (b)} Set $\kappa=\omega_3$.  \cmmnt{(There is a
reason for taking $\omega_3$ here;  see 535E in Volume 5.)}
Let $\CalBa_{\kappa}$ be the Baire $\sigma$-algebra of
$\{0,1\}^{\kappa}$\cmmnt{ (as in 341Yc)}, and $\mu$ the
restriction to $\CalBa_{\kappa}$ of the usual measure on
$\{0,1\}^{\kappa}$.   Can we show that $\mu$ has no lifting?

\cmmnt{
\Notesheader{341} Innumerable variations of the proof of 341K have been
devised, as each author has struggled with the technical complications.
I have discussed the reasons for my own choices in 341L.

The theorem has a curious history.   It was originally announced by von
Neumann, but he seems never to have written his proof down, and the
first published proof is that of {\smc Maharam 58}.   That argument is
based on Maharam's theorem, 341Xe and 341Yd, which show that it is
enough to find liftings for every $\{0,1\}^{\kappa}$;  this requires
most of the ideas presented above, but feels more concrete, and some of
the details are slightly simpler.   The argument as I have written it
owes a great deal to {\smc Ionescu Tulcea \& Ionescu Tulcea 69}.

The lifting theorem and Maharam's theorem are the twin pillars of modern
abstract measure theory.   But there remains a degree of mystery about
the lifting theorem which is absent from the other.   The first point is
that there is nothing canonical about the liftings we can construct,
except in the quite exceptional case of Stone spaces (341O).   Even
when there is a more or less canonical lower density present (341E,
341Xb), the conversion of this into a lifting requires arbitrary
choices, as in 341J.   While we can distinguish some liftings as being
somewhat more regular than others, I know of no criterion which marks
out any particular lifting for Lebesgue measure, for instance, among the
rest.   Perhaps associated with this arbitrariness is the extreme
difficulty of deciding whether liftings of any given type exist.
Neither positive nor negative results are easily come by (I will present
a few in the later sections of this chapter), and the nature of the
obstacles remains quite unclear.
}

\discrpage
\frfilename{mt342.tex}
\versiondate{9.7.10}
\copyrightdate{2001}

\def\chaptername{The lifting theorem}
\def\sectionname{Compact measure spaces}

\newsection{342}

The next three sections amount to an extended parenthesis, showing how
the Lifting Theorem can be used to attack one of the fundamental
problems of
measure theory:  the representation of Boolean homomorphisms between
measure algebras by functions between appropriate measure spaces.   This
section prepares for the main idea by introducing the class of `locally
compact' measures (342Ad), with the associated concepts of `compact' and
`perfect' measures (342Ac, 342K).   These depend on the notions of
`inner regularity'
(342Aa, 342B) and `compact class' (342Ab, 342D).   I list the basic
permanence properties for compact and locally compact measures
(342G-342I)
and mention some of the compact measures which we have already seen
(342J). Concerning perfect measures, I content myself with the proof
that a locally
compact measure is perfect (342L).   I end the section with two examples
(342M, 342N).

\leader{342A}{Definitions (a)} Let $(X,\Sigma,\mu)$ be a measure space.
If $\Cal K\subseteq\Cal PX$, I will say that $\mu$ is 
{\bf inner regular} with respect to $\Cal K$ if

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

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

\cmmnt{Of course $\mu$ is inner regular with respect to $\Cal K$ iff
it is inner regular with respect to $\Cal K\cap\Sigma$.
It is convenient in this context to interpret $\sup\emptyset$ as $0$, so
that we have to check the definition only when $\mu E>0$, and need not
insist that $\emptyset\in\Cal K$.
}

\spheader 342Ab A family $\Cal K$ of sets is a {\bf compact class} if
$\bigcap\Cal K'\ne\emptyset$ whenever $\Cal K'\subseteq\Cal K$ has the
finite intersection property.

Note that any subset of a compact class is again a compact class.
\cmmnt{(In particular, it is convenient to allow the empty set as a
compact class.)}

\spheader 342Ac A measure space $(X,\Sigma,\mu)$, or a measure $\mu$, is
{\bf compact} if $\mu$ is inner regular with respect to some compact
class of subsets of $X$.

\cmmnt{Allowing $\emptyset$ as a compact class, and interpreting
$\sup\emptyset$ as $0$ in (a) above,} $\mu$ is a compact
measure whenever $\mu X=0$.

\spheader 342Ad A measure space $(X,\Sigma,\mu)$, or a measure $\mu$, is
{\bf locally compact} if the subspace measure $\mu_E$ is compact
whenever $E\in\Sigma$ and $\mu E<\infty$.

\cmmnt{\medskip

\noindent{\bf Remark} I ought to point out that the original
definitions of `compact class' and `compact measure' ({\smc Marczewski 53}) 
correspond to what I will call `countably compact class' and
`countably
compact measure' in Volume 4.   For another variation on the concept
of `compact class' see condition ($\beta$) in 343B(ii)-(iii).

For examples of compact measure spaces see 342J and 342Xf.
}%end of comment

\leader{342B}{}\cmmnt{ I prepare the ground with some straightforward
lemmas.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a measure space, and
$\Cal K\subseteq\Sigma$ a set such that whenever $E\in\Sigma$ and
$\mu E>0$ there is a $K\in\Cal K$ such that $K\subseteq E$ and
$\mu K>0$.   Let $E\in\Sigma$.

(a) There is a countable disjoint set $\Cal K_1\subseteq\Cal K$ such
that $K\subseteq E$ for every $K\in\Cal K_1$ and
$\mu(\bigcup\Cal K_1)=\mu E$.

(b) If $\mu E<\infty$ then $\mu(E\setminus\bigcup\Cal K_1)=0$.

(c) In any case, there is for any $\gamma<\mu E$ a finite
disjoint $\Cal K_0\subseteq\Cal K$ such that $K\subseteq E$ for every
$K\in\Cal K_0$ and $\mu(\bigcup\Cal K_0)\ge\gamma$.

\proof{ Set
$\Cal K'=\{K:K\in\Cal K,\,K\subseteq E,\,\mu K>0\}$.   Let
$\Cal K^*$ be a maximal disjoint subfamily of $\Cal K'$.   If $\Cal K^*$
is uncountable, then there is some $n\in\Bbb N$ such that
$\{K:K\in\Cal K^*,\,\mu K\ge 2^{-n}\}$ is infinite, so that there is a
countable $\Cal K_1\subseteq\Cal K^*$ such that
$\mu(\bigcup\Cal K_1)=\infty=\mu E$.

If $\Cal K^*$ is countable, set $\Cal K_1=\Cal K^*$.   Then
$F=\bigcup\Cal K_1$ is measurable, and
$F\subseteq E$.   Moreover, there is no member of $\Cal K'$ disjoint
from $F$;  but this means that $E\setminus F$ must be negligible.   So
$\mu F=\mu E$, and (a) is true.   Now (b) and (c) follow at once,
because

\Centerline{$\mu(\bigcup\Cal K_1)
=\sup\{\mu(\bigcup\Cal K_0):\Cal K_0\subseteq\Cal K_1$ is finite$\}$.}
}%end of proof of 342B

\cmmnt{\medskip

\noindent{\bf Remark} This lemma can be thought of as more versions
of the principle of exhaustion;  compare 215A.
}

\leader{342C}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space and
$\Cal K\subseteq\Cal PX$ a family of sets such that ($\alpha$)
$K\cup K'\in\Cal K$
whenever $K$, $K'\in\Cal K$ and $K\cap K'=\emptyset$ ($\beta$) whenever
$E\in\Sigma$ and $\mu E>0$, there is a $K\in\Cal K\cap\Sigma$ such that
$K\subseteq E$ and $\mu K>0$.   Then $\mu$ is inner regular with respect
to $\Cal K$.

\proof{ Apply 342Bc to $\Cal K\cap\Sigma$.
}

\vleader{60pt}{342D}{Lemma} Let $X$ be a set and $\Cal K$ a family of subsets
of $X$.

(a) The following are equiveridical:

\quad (i) $\Cal K$ is a compact class;

\quad (ii) there is a topology $\frak T$ on $X$ such that $X$ is compact
and every member of $\Cal K$ is a closed set for $\frak T$.

(b) If $\Cal K$ is a compact class, so are the families
$\Cal K_1=\{K_0\cup\ldots\cup K_n:K_0,\ldots,K_n\in\Cal K\}$ and
$\Cal K_2=\{\bigcap\Cal K':\emptyset\ne\Cal K'\subseteq\Cal K\}$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Let $\frak T$ be the topology
generated
by $\{X\setminus K:K\in\Cal K\}$.   Then of course every member of
$\Cal K$ is closed for $\frak T$.   Let $\Cal F$ be an ultrafilter on
$X$.   Then $\Cal K\cap\Cal F$ has the finite intersection property;
because $\Cal K$ is a compact class, it has
non-empty intersection;  take $x\in X\cap\bigcap(\Cal K\cap\Cal F)$.   The
family

\Centerline{$\{G:G\subseteq X$, either $G\in\Cal F$ or $x\notin G\}$}

\noindent is easily seen to be a topology on $X$, and contains
$X\setminus K$ for every $K\in\Cal K$ (because if
$X\setminus K\notin\Cal F$ then
$K\in\Cal F$ and $x\in K$), so includes $\frak T$;  but this just means
that every $\frak T$-open set containing $x$ belongs to $\Cal F$, that
is, that $\Cal F\to x$.   As $\Cal F$ is arbitrary, $X$ is compact for
$\frak T$ (2A3R).

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} Use 3A3Da.

\medskip

{\bf (b)} Let $\frak T$ be a topology on $X$ such that $X$ is compact
and every member of $\Cal K$ is closed for $\frak T$;  then the same is
true of every member of $\Cal K_1$ or $\Cal K_2$.
}%end of proof of 342D

\leader{342E}{Corollary} Suppose that $(X,\Sigma,\mu)$ is a measure
space and that $\Cal K$ is a compact class such that whenever
$E\in\Sigma$ and
$\mu E>0$ there is a $K\in\Cal K\cap\Sigma$ such that $K\subseteq E$ and
$\mu K>0$.   Then $\mu$ is compact.

\proof{ Set
$\Cal K_1=\{K_0\cup\ldots\cup K_n:K_0,\ldots,K_n\in\Cal K\}$.
By 342Db, $\Cal K_1$ is a compact class, and by 342C $\mu$ is inner
regular with respect to $\Cal K_1$.
}%end of proof of 342E

\leader{342F}{Corollary} A measure space $(X,\Sigma,\mu)$ is compact iff
there is a topology on $X$ such that $X$ is compact and $\mu$ is inner
regular with respect to the closed sets.

\proof{{\bf (a)} If $\mu$ is inner regular with respect to a compact
class $\Cal K$, then there is a compact topology on $X$ such that every
member of $\Cal K$ is closed (342Da);
now the family $\Cal F$ of closed sets includes
$\Cal K$, so $\mu$ is also inner regular with respect to $\Cal F$.

\medskip

{\bf (b)} If there is a compact topology on $X$ such that $\mu$ is inner
regular with respect to the family $\Cal K$ of closed sets, then this is
a compact class, so $\mu$ is a compact measure.
}%end of proof of 342F

\vleader{72pt}{342G}{}\cmmnt{ Now I look at the standard questions
concerning
preservation of the properties of `compactness' or `local compactness'
under the usual manipulations.

\medskip

\noindent}{\bf Proposition}
(a) Any measurable subspace of a compact measure space is compact.

(b) The completion and c.l.d.\ version of a compact measure space are
compact.

(c) A semi-finite measure space is compact iff its completion is compact
iff its c.l.d.\ version is compact.

(d) The direct sum of a family of compact measure spaces is compact.

(e) The c.l.d.\ product of two compact measure spaces is compact.

(f) The product of any family of compact probability spaces is compact.

\proof{{\bf (a)} Let $(X,\Sigma,\mu)$ be a compact measure space, and
$E\in\Sigma$.   If $\Cal K$ is a compact class such that $\mu$ is inner
regular with respect to $\Cal K$, then $\Cal K_E=\Cal K\cap\Cal PE$ is a
compact class (just because it is a subset of $\Cal K$) and the subspace
measure $\mu_E$ is inner regular with respect to $\Cal K_E$.

\medskip

{\bf (b)} Let $(X,\Sigma,\mu)$ be a compact measure space.   Write
$(X,\check\Sigma,\check\mu)$ for {\it either} the completion {\it or}
the c.l.d.\ version of $(X,\Sigma,\mu)$.   Let $\Cal K\subseteq\Cal PX$
be a compact class such that $\mu$ is inner regular with respect to
$\Cal K$. Then $\check\mu$ also is inner regular with respect to
$\Cal K$.   \Prf\ If
$E\in\check\Sigma$ and $\gamma<\check\mu E$ there is an $E'\in\Sigma$
such that $E'\subseteq E$ and $\mu E'>\gamma$;  if $\check\mu$ is the
c.l.d.\
version of $\mu$, we may take $\mu E'$ to be finite.   There is a
$K\in\Cal K\cap\Sigma$ such that $K\subseteq E'$ and $\mu K\ge\gamma$.
Now $\check\mu K=\mu K\ge\gamma$ and $K\subseteq E$ and
$K\in\Cal K\cap\check\Sigma$.\ \Qed

\medskip

{\bf (c)} Now suppose that $(X,\Sigma,\mu)$ is semi-finite;  again write
$(X,\check\Sigma,\check\mu)$ for {\it either} its completion {\it or}
its c.l.d.\ version.   We already know that if $\mu$ is compact, so is
$\check\mu$.   If $\check\mu$ is compact, let $\Cal K\subseteq\Cal PX$
be a compact class such that $\check\mu$ is inner regular with respect
to $\Cal K$.   Set
$\Cal K^*=\{\bigcap\Cal K':\emptyset\ne\Cal K'\subseteq\Cal K\}$;
then $\Cal K^*$ is a compact class (342Db).   Now $\mu$ is inner regular
with respect to $\Cal K^*$.   \Prf\ Take $E\in\Sigma$ and
$\gamma<\mu E$.
Choose $\sequencen{E_n}$, $\sequencen{K_n}$ as follows.   Because $\mu$
is semi-finite, there is an $E_0\subseteq E$ such that $E_0\in\Sigma$
and $\gamma<\mu E_0<\infty$.   Given $E_n\in\Sigma$ such that
$\mu E_n>\gamma$,
there is a $K_n\in\Cal K\cap\check\Sigma$ such that $K_n\subseteq E_n$
and $\check\mu K_n>\gamma$.   Now there is an $E_{n+1}\in\Sigma$ such
that $E_{n+1}\subseteq K_n$ and $\mu E_{n+1}>\gamma$.   Continue.   On
completing the induction, set
$K=\bigcap_{n\in\Bbb N}K_n=\bigcap_{n\in\Bbb N}E_n$, so that
$K\in\Cal K^*\cap\Sigma$ and $K\subseteq E$ and
$\mu K=\lim_{n\to\infty}\mu E_n\ge\gamma$.   As $E$ and $\gamma$ are
arbitrary, $\mu$ is inner regular with
respect to $\Cal K^*$.\ \QeD\   As $\Cal K^*$ is a compact class, $\mu$
is a compact measure.

\medskip

{\bf (d)} Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family
of compact measure spaces, with direct sum $(X,\Sigma,\mu)$.   We may
suppose that each $X_i$ is actually a subset of $X$, with $\mu_i$ the
subspace measure.   For each $i\in I$ let $\Cal K_i\subseteq\Cal PX_i$
be a compact class such that $\mu_i$ is inner regular with respect to
$\Cal K_i$.   Then $\Cal K=\bigcup_{i\in I}\Cal K_i$ is a compact class,
for if $\Cal K'\subseteq\Cal K$ has the finite intersection property,
then $\Cal K'\subseteq\Cal K_i$ for some $i$, so has non-empty
intersection.   Now if
$E\in\Sigma$ and $\mu E>0$ there is some $i\in I$ such that
$\mu_i(E\cap X_i)>0$, and we can find a
$K\in\Cal K_i\cap\Sigma_i\subseteq\Cal K\cap\Sigma$ such that
$K\subseteq E\cap X_i$ and $\mu_iK>0$, in which
case $\mu K>0$.   By 342E, $\mu$ is compact.

\medskip

{\bf (e)} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two compact measure
spaces, with c.l.d.\ product measure $(X\times Y,\Lambda,\lambda)$.
Let $\frak T$, $\frak S$ be topologies on $X$, $Y$ respectively such
that $X$
and $Y$ are compact spaces and $\mu$, $\nu$ are inner regular with
respect to the closed sets.   Then the product topology on $X\times Y$
is compact (3A3J).

The point is that $\lambda$ is inner regular with respect to the family
$\Cal K$ of closed subsets of $X\times Y$.   \Prf\ Suppose that
$W\in\Lambda$ and $\lambda W>\gamma$.   Then there are $E\in\Sigma$,
$F\in\Tau$ such that $\mu E<\infty$, $\nu F<\infty$ and
$\lambda(W\cap(E\times F))>\gamma$ (251F).   Now
there are sequences $\sequencen{E_n}$, $\sequencen{F_n}$ in $\Sigma$,
$\Tau$ respectively such that

\Centerline{$(E\times F)\setminus W
\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$,}

\Centerline{$\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
<\lambda((E\times F)\setminus W)
  +\lambda((E\times F)\cap W)-\gamma
=\lambda(E\times F)-\gamma$}

\noindent (251C).   Set

\Centerline{$W'
=(E\times F)\setminus\bigcup_{n\in\Bbb N}E_n\times F_n
=\bigcap_{n\in\Bbb N}((E\times(F\setminus F_n))
  \cup((E\setminus E_n)\times F))$.}

\noindent Then $W'\subseteq W$, and

\Centerline{$\lambda((E\times F)\setminus W')
\le\lambda(\bigcup_{n\in\Bbb N}E_n\times F_n)
\le\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
<\lambda(E\times F)-\gamma$,}

\noindent so $\lambda W'>\gamma$.

Set $\epsilon=\bover14(\lambda W'-\gamma)/(1+\mu E+\mu F)$.   For each
$n$, we can find
closed measurable sets $K_n$, $K'_n\subseteq X$ and $L_n$,
$L'_n\subseteq Y$ such that

\Centerline{$K_n\subseteq E$,
\quad$\mu(E\setminus K_n)\le 2^{-n}\epsilon$,}

\Centerline{$L'_n\subseteq F\setminus F_n$,
\quad$\nu((F\setminus F_n)\setminus L'_n)\le 2^{-n}\epsilon$,}

\Centerline{$K'_n\subseteq E\setminus E_n$,
\quad$\mu((E\setminus E_n)\setminus K'_n)\le 2^{-n}\epsilon$,}

\Centerline{$L_n\subseteq F$,
\quad$\nu(F\setminus L_n)\le 2^{-n}\epsilon$.}

\noindent Set

\Centerline{$V
=\bigcap_{n\in\Bbb N}(K_n\times L'_n)\cup(K'_n\times L_n)
\subseteq W'\subseteq W$.}

\noindent   Now

$$\eqalign{W'\setminus V
&\subseteq\bigcup_{n\in\Bbb N}((E\setminus K_n)\times F)
   \cup(E\times((F\setminus F_n)\setminus L'_n))\cr
&\qquad\qquad\cup(((E\setminus E_n)\setminus K'_n)\times F)
   \cup(E\times(F\setminus L_n)),\cr}$$

\noindent so

$$\eqalign{\lambda(W'\setminus V)
&\le\sum_{n=0}^{\infty}\mu(E\setminus K_n)\cdot\nu F
   +\mu E\cdot\nu((F\setminus F_n)\setminus L'_n)\cr
&\qquad\qquad+\mu((E\setminus E_n)\setminus K'_n)\cdot\nu F
   +\mu E\cdot\nu(F\setminus L_n)\cr
&\le\sum_{n=0}^{\infty}2^{-n}\epsilon(2\mu E+2\mu F)
\le\lambda W'-\gamma,\cr}$$

\noindent and $\lambda V\ge\gamma$.   But $V$ is a countable
intersection of
finite unions of products of closed measurable sets, so is itself a
closed measurable set, and belongs to $\Cal K\cap\Lambda$.\ \Qed

Accordingly the product topology on $X\times Y$ witnesses that $\lambda$
is a compact measure.

\medskip

{\bf (f)} The same method works.   In detail:  let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of compact
probability spaces, with product $(X,\Lambda,\lambda)$.   For each $i$,
let $\frak T_i$ be a topology on $X_i$ such that $X_i$ is compact and
$\mu_i$ is inner regular with respect to the closed sets.   Give $X$ the
product topology;  this is compact.   If $W\in\Lambda$ and $\epsilon>0$,
let $\sequencen{C_n}$ be a sequence of measurable cylinders (in the
sense of 254A) such that $X\setminus W\subseteq\bigcup_{n\in\Bbb N}C_n$
and $\sum_{n=0}^{\infty}\lambda C_n\le\lambda(X\setminus W)+\epsilon$.
Express each $C_n$ as $\prod_{i\in I}E_{ni}$ where $E_{ni}\in\Sigma_i$
for each $i$
and $J_n=\{i:E_{ni}\ne X_i\}$ is finite.   For $n\in\Bbb N$
set $\epsilon_n=2^{-n}\epsilon/(1+\#(J_n))$.   Choose closed
measurable sets $K_{ni}\subseteq X_i\setminus E_{ni}$ such that
$\mu_i((X_i\setminus E_{ni})\setminus K_{ni})\le\epsilon_n$ whenever
$n\in\Bbb N$ and $i\in J_n$.   For each $n\in\Bbb N$, set

\Centerline{$V_n=\bigcup_{i\in J_n}\{x:x\in X,\,x(i)\in K_{ni}\}$,}

\noindent so that $V_n$ is a closed measurable subset of $X$.   Observe
that

\Centerline{$X\setminus V_n=\{x:x(i)\in X\setminus K_{ni}$ for
$i\in J_n\}$}

\noindent includes $C_n$, and that

\Centerline{$\lambda(X\setminus(V_n\cup C_n))
\le\sum_{i\in J_n}\lambda\{x:x(i)\in X_i\setminus(K_{ni}\cup E_{ni})\}
\le\sum_{i\in J_n}\epsilon_n
\le 2^{-n}\epsilon$.}

Now set $V=\bigcap_{n\in\Bbb N}V_n$;  then $V$ is again a closed
measurable set, and

\Centerline{$X\setminus V
\subseteq\bigcup_{n\in\Bbb N}C_n\cup(X\setminus(C_n\cup V_n))$}

\noindent has measure at most

\Centerline{$\sum_{n=0}^{\infty}\lambda C_n+2^{-n}\epsilon
\le 1-\lambda W+\epsilon+2\epsilon$,}

\noindent so $\lambda V\ge\lambda W-3\epsilon$.
As $W$ and $\epsilon$ are arbitrary, $\lambda$ is inner regular with
respect to the closed sets, and is a compact measure.
}%end of proof of 342G

\leader{342H}{Proposition} (a) A compact measure space is locally
compact.

(b) A strictly localizable locally compact measure space is compact.

(c) Let $(X,\Sigma,\mu)$ be a measure space.   Suppose that whenever
$E\in\Sigma$ and $\mu E>0$ there is an $F\in\Sigma$ such that
$F\subseteq E$,
$\mu F>0$ and the subspace measure on $F$ is compact.   Then $\mu$ is
locally compact.

\proof{{\bf (a)} This is immediate from 342Ga and the definition of
`locally compact' measure space.

\medskip

{\bf (b)} Suppose that $(X,\Sigma,\mu)$ is a strictly localizable
locally compact measure space.   Let $\langle X_i\rangle_{i\in I}$ be a
decomposition of $X$, and for each $i\in I$ let $\mu_i$ be the subspace
measure on $X_i$.   Then $\mu_i$ is compact.   Now $\mu$ can be
identified
with the direct sum of the $\mu_i$, so itself is compact, by 342Gd.

\medskip

{\bf (c)} Write $\Cal F$ for the set of measurable sets $F\subseteq X$
such that the subspace measures $\mu_F$ are compact.   Take $E\in\Sigma$
with $\mu E<\infty$.   By 342Bb, there is a countable disjoint family
$\langle F_i\rangle_{i\in I}$ in $\Cal F$ such that $F_i\subseteq E$ for
each $i$,
and $F'=E\setminus\bigcup_{i\in I}F_i$ is negligible;  now this means
that $F'\in\Cal F$ (342Ac), so we may take it that
$E=\bigcup_{i\in I}F_i$.   In this case $\mu_E$ is isomorphic to the
direct sum of the measures
$\mu_{F_i}$ and is compact.   As $E$ is arbitrary, $\mu$ is locally
compact.
}%end of proof of 342H

\leader{342I}{Proposition} (a) Any measurable subspace of a locally
compact measure space is locally compact.

(b) A measure space is locally compact iff its completion is locally
compact iff its c.l.d.\ version is locally compact.

(c) The direct sum of a family of locally compact measure spaces is
locally compact.

(d) The c.l.d.\ product of two locally compact measure spaces is locally
compact.

\proof{{\bf (a)} Trivial:  if $(X,\Sigma,\mu)$ is locally compact, and
$E\in\Sigma$, and $F\subseteq E$ is a measurable set of finite measure
for the subspace measure on $E$, then $F\in\Sigma$ and $\mu F<\infty$,
so the subspace measure on $F$ is compact.

\medskip

{\bf (b)} Let $(X,\Sigma,\mu)$ be a measure space, and write
$(X,\check\Sigma,\check\mu)$ for {\it either} its completion {\it or}
its c.l.d.\ version.

\medskip

\quad{\bf (i)} Suppose that $\mu$ is locally compact, and that
$\check\mu F<\infty$.   Then there is an $E\in\Sigma$ such that
$E\subseteq F$ and $\mu E=\check\mu F$.   Let $\mu_E$ be the subspace
measure on $E$ induced by the
measure $\mu$;  then we are assuming that $\mu_E$ is compact.   Let
$\Cal K\subseteq\Cal PE$ be a compact class such that $\mu_E$ is inner
regular with respect to $\Cal K$.   Then, as in the proof of 342Gb, the
subspace measure $\check\mu_F$ on $F$ induced by $\check\mu$ is also
inner regular
with respect to $\Cal K$, so $\check\mu_F$ is compact;  as $F$ is
arbitrary, $\check\mu$ is locally compact.

\medskip

\quad{\bf (ii)} Now suppose that $\check\mu$ is locally compact, and
that $\mu E<\infty$.   Then the subspace measure $\check\mu_E$ is
compact.   But
this is just the completion of the subspace measure $\mu_E$, so $\mu_E$
is compact, by 342Gc;  as $E$ is arbitrary, $\mu$ is locally compact.

\medskip

{\bf (c)} Put (a) and 342Hc together.

\medskip

{\bf (d)} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be locally compact
measure spaces, with product $(X\times Y,\Lambda,\lambda)$.   If
$W\in\Lambda$ and
$\lambda W>0$, there are $E\in\Sigma$, $F\in\Tau$ such that
$\mu E<\infty$,
$\nu F<\infty$ and $\lambda(W\cap(E\times F))>0$.   Now the subspace
measure $\lambda_{E\times F}$ induced by $\lambda$ on $E\times F$ is
just the product of the subspace measures (251Q(ii-$\alpha$)), so is
compact, and the subspace measure $\lambda_{W\cap(E\times F)}$ is
therefore again compact, by
342Ga.   By 342Hc, this is enough to show that $\lambda$ is locally
compact.
}%end of proof of 342I

\leader{342J}{Examples} \cmmnt{It is time I listed some examples of
compact measure spaces.

\medskip

}{\bf (a)} Lebesgue measure on $\BbbR^r$ is
compact.   \prooflet{(Let $\Cal K$ be the family of subsets of $\BbbR^r$
which are compact for the usual topology.   By 134Fb, Lebesgue measure
is inner regular with respect to $\Cal K$.)}

\spheader 342Jb Similarly, any Radon measure on
$\BbbR^r$\cmmnt{ (256A)} is compact.

\spheader 342Jc If $(\frak A,\bar\mu)$ is any semi-finite measure
algebra, the standard measure\cmmnt{ $\nu$} on its Stone space $Z$ is 
compact.   \prooflet{(By
322Ra, $\nu$ is inner regular with respect to the family of
open-and-closed subsets
of $Z$, which are all compact for the standard topology of $Z$, so form
a compact class.)}

\spheader 342Jd The usual measure on $\{0,1\}^I$ is compact, for any set
$I$.   \prooflet{(It is obvious that the usual measure on $\{0,1\}$ is
compact;  now use 342Gf.)}

\cmmnt{\medskip

\noindent{\bf Remark} Actually all these measures are `Radon' in the sense
of Volume 4.}

\leader{342K}{}\cmmnt{ One of the most important properties of
(locally) compact measure spaces has been studied under the following
name.

\medskip

\noindent}{\bf Definition} Let $(X,\Sigma,\mu)$ be a measure space.
Then $(X,\Sigma,\mu)$, or $\mu$, is {\bf perfect} if whenever
$f:X\to\Bbb R$ is measurable, $E\in\Sigma$ and $\mu E>0$, then there
is a compact set $K\subseteq f[E]$ such that $\mu f^{-1}[K]>0$.

\leader{342L}{Theorem} A semi-finite locally compact measure space is
perfect.

\proof{ Let $(X,\Sigma,\mu)$ be a semi-finite locally compact measure
space, $f:X\to\Bbb R$ a measurable function, and $E\in\Sigma$ a set of
non-zero measure.   Because $\mu$ is semi-finite, there is an
$F\in\Sigma$ such that $F\subseteq E$ and $0<\mu F<\infty$.   Now the
subspace measure $\mu_F$ is compact;  let $\frak T$ be a topology on $F$
such that $F$ is compact and $\mu_F$ is inner regular with respect to
the family $\Cal K$ of closed sets for $\frak T$.

Let $\langle\epsilon_q\rangle_{q\in\Bbb Q}$ be a family of strictly
positive real numbers such that
$\sum_{q\in\Bbb Q}\epsilon_q<\bover12\mu F$.   (For
instance, you could set $\epsilon_{q(n)}=2^{-n-3}\mu F$ where
$\sequencen{q(n)}$ is an enumeration of $\Bbb Q$.)   For each
$q\in\Bbb Q$,
set $E_q=\{x:x\in F,\,f(x)\le q\}$, $E'_q=\{x:x\in F,\,f(x)>q\}$, and
choose $K_q$, $K'_q\in\Cal K\cap\Sigma$ such that $K_q\subseteq E_q$,
$K'_q\subseteq E'_q$, $\mu(E_q\setminus K_q)\le\epsilon_q$ and
$\mu(E'_q\setminus K'_q)\le\epsilon_q$.   Then
$K=\bigcap_{q\in\Bbb Q}(K_q\cup K'_q)\in\Cal K\cap\Sigma$,
$K\subseteq F$ and

\Centerline{$\mu(F\setminus K)
\le\sum_{q\in\Bbb Q}\mu(E_q\setminus K_q)+\mu(E'_q\setminus K'_q)
<\mu F$,}

\noindent so $\mu K>0$.

The point is that $f\restr K$ is continuous.   \Prf\ For any
$q\in\Bbb Q$, $\{x:x\in K,\,f(x)\le q\}=K\cap K_q$ and
$\{x:x\in K,\,f(x)>q\}=K\cap K'_q$.
If $H\subseteq\Bbb R$ is open and $x\in K\cap f^{-1}[H]$, take $q$,
$q'\in\Bbb Q$ such that $f(x)\in\ocint{q,q'}\subseteq H$;  then
$G=K\setminus(K_q\cup K'_{q'})$ is a relatively open subset of $K$
containing $x$ and included in $f^{-1}[H]$.   Thus $K\cap f^{-1}[H]$ is
relatively open in $K$;  as $H$ is arbitrary, $f\restr K$ is continuous.\
\Qed

Accordingly $f[K]$ is a continuous image of a compact set, therefore
compact;   it is
a subset of $f[E]$, and $\mu f^{-1}[f[K]]\ge\mu K>0$.   As $f$ and $E$
are arbitrary, $\mu$ is perfect.
}%end of proof of 342L

\leader{342M}{}\cmmnt{ I ought to give examples to distinguish between
the concepts introduced here, partly on general principles, but also
because it
is not obvious that the concept of `locally compact' measure space is
worth spending time on at all.   It is easy to distinguish between
`perfect' and
`(locally) compact';  `locally compact' and `compact' are harder to
separate.

\medskip

\noindent}{\bf Example} Let $X$ be an uncountable set and $\mu$ the
countable-cocountable measure on $X$\cmmnt{ (211R)}.   Then $\mu$ is
perfect but not compact or locally compact.

\proof{{\bf (a)} If $f:X\to\Bbb R$ is measurable and $E\subseteq X$ is
measurable, with measure greater than $0$, set
$A=\{\alpha:\alpha\in\Bbb R,\,\{x:x\in X,\,f(x)\le\alpha\}$ is
negligible$\}$.   Then $\alpha\in A$
whenever $\alpha\le\beta\in A$.   Since
$X=\bigcup_{n\in\Bbb N}\{x:f(x)\le n\}$, there is some $n$ such that
$n\notin A$, in which case $A$ is bounded
above by $n$.   Also there is some $m\in\Bbb N$ such that
$\{x:f(x)>-m\}$ is non-negligible, in which case it must be
conegligible, and $-m\in A$, so $A$ is non-empty.   Accordingly
$\gamma=\sup A$ is defined in $\Bbb R$.   Now for any $k\in\Bbb N$,
$\{x:f(x)\le \gamma-2^{-k}\}$ is negligible, so $\{x:f(x)<\gamma\}$ is
negligible.   Also, for any $k$, $\{x:f(x)\le\gamma+2^{-k}\}$ is
non-negligible, so $\{x:f(x)>\gamma+2^{-k}\}$ must be negligible;
accordingly, $\{x:f(x)>\gamma\}$ is negligible.   But this means that
$\{x:f(x)=\gamma\}$ is conegligible and has measure $1$.   Thus we have
a compact set
$K=\{\gamma\}$ such that $\mu f^{-1}[K]=1$, and $\gamma$ must belong to
$f[E]$.   As $f$ and $E$ are arbitrary, $\mu$ is perfect.

\medskip

{\bf (b)} $\mu$ is not compact.   \Prf\Quer\ Suppose, if possible, that
$\Cal K\subseteq\Cal PX$ is a compact class such that $\mu$ is inner
regular with respect to $\Cal K$.   Then for every $x\in X$ there is a
measurable set $K_x\in\Cal K$ such that $K_x\subseteq X\setminus\{x\}$
and $\mu K_x>0$, that is, $K_x$ is conegligible.   But this means that
$\{K_x:x\in X\}$ must have the finite intersection property;  as it also
has empty intersection, $\Cal K$ cannot be a compact class.\ \Bang\Qed

\medskip

{\bf (c)} Because $\mu$ is totally finite, it cannot be locally compact.
}%end of proof of 342M

\cmmnt{\medskip

\noindent{\bf Remark} See also 342X(n-viii).}

\leader{*342N}{Example} There is a complete locally determined
localizable locally compact measure space which is not compact.

\proof{{\bf (a)} I refer to the example of 216E.   In that construction,
we have a set $I$ and a family $\langle x_{\gamma}\rangle_{\gamma\in C}$
in $X=\{0,1\}^I$ such that for every $D\subseteq C$ there is an $i\in I$
such that $D=\{\gamma:x_{\gamma}(i)=1\}$;  moreover, $\#(C)>\frak c$.
The $\sigma$-algebra $\Sigma$ is the family of sets $E\subseteq X$ such
that for
every $\gamma$ there is a countable set $J\subseteq I$ such that
$\{x:x\restr J=x_{\gamma}\restr J\}$ is a subset of either $E$ or
$X\setminus E$;  and for $E\in\Sigma$, $\mu E$ is
$\#(\{\gamma:x_{\gamma}\in E\})$ if this is finite, $\infty$ otherwise.
Note that any subset of
$X$ determined by coordinates in a countable set belongs to $\Sigma$.

For each $\gamma\in C$, let $i_{\gamma}\in I$ be such that
$x_{\gamma}(i_{\gamma})=1$, $x_{\delta}(i_{\gamma})=0$ for
$\delta\ne\gamma$.   (In 216E I took $I$ to be $\Cal PC$, and
$i_{\gamma}$ would be $\{\gamma\}$.)   Set

\Centerline{$Y=\{x:x\in X,\,\{\gamma:\gamma\in C,\,x(i_{\gamma})=1\}$ is
finite$\}$.}

\noindent Give $Y$ its subspace measure $\mu_Y$ with domain $\Sigma_Y$.
Then $\mu_Y$ is complete, locally determined and localizable (214Ie).
Note that $x_{\gamma}\in Y$ for every $\gamma\in C$.

\medskip

{\bf (b)} $\mu_Y$ is locally compact.   \Prf\ Suppose that
$F\in\Sigma_Y$
and $\mu_YF<\infty$.   If $\mu_YF=0$ then surely the subspace measure
$\mu_F$ is compact.   Otherwise, we can express $F$ as $E\cap Y$ where
$E\in\Sigma$ and $\mu E=\mu_YF$.   Then
$D=\{\gamma:x_{\gamma}\in E\}=\{\gamma:x_{\gamma}\in F\}$ is finite.
For $\gamma\in D$ set

\Centerline{$G'_{\gamma}=\{x:x\in X,\,x(i_{\gamma})=1,\,x(i_{\delta})=0$
for
every $\delta\in D\setminus\{\gamma\}\}\in\Sigma$,}

\Centerline{$\Cal K_{\gamma}
=\{K:x_{\gamma}\in K\subseteq F\cap G'_{\gamma}\}$.}

\noindent Then each $\Cal K_{\gamma}$ is a compact class, and members of
different $\Cal K_{\gamma}$'s are disjoint, so
$\Cal K=\bigcup_{\gamma\in D}\Cal K_{\gamma}$ is a compact class.

Now suppose that $H$ belongs to the subpsace $\sigma$-algebra $\Sigma_F$
and $\mu_FH>0$.   Then there is a $\gamma\in D$ such that
$x_{\gamma}\in H$, so
that $H\cap G'_{\gamma}\in\Cal K\cap\Sigma_F$ and
$\mu_F(H\cap G'_{\gamma})>0$.   By 342E, this is enough to show that
$\mu_F$ is compact.
As $F$ is arbitrary, $\mu_Y$ is locally compact.\ \Qed

\medskip

{\bf (c)} $\mu_Y$ is not compact.   \Prf\Quer\ Suppose, if possible,
that $\mu_Y$ is inner regular with respect to a compact class
$\Cal K\subseteq\Cal PY$.   For each $\gamma\in C$ set
$G_{\gamma}=\{x:x\in X,\,x(i_{\gamma})=1\}$, so that
$x_{\gamma}\in G_{\gamma}\in\Sigma$ and
$\mu_Y(G_{\gamma}\cap Y)=1$.   There must
therefore be a $K_{\gamma}\in\Cal K$ such that
$K_{\gamma}\subseteq G_{\gamma}\cap Y$ and $\mu_YK_{\gamma}=1$ (since
$\mu_Y$ takes no value in
$\ooint{0,1}$).   Express $K_{\gamma}$ as $Y\cap E_{\gamma}$, where
$E_{\gamma}\in\Sigma$, and let $J_{\gamma}\subseteq I$ be a countable
set such that

\Centerline{$E_{\gamma}
\supseteq\{x:x\in X,\,x\restr J_{\gamma}=x_{\gamma}\restr J_{\gamma}\}$.}

At this point I call on the full strength of 2A1P.   There is a set
$B\subseteq C$, with cardinal greater than $\frak c$, such that
$x_{\gamma}\restr J_{\gamma}\cap J_{\delta}
=x_{\delta}\restr J_{\gamma}\cap J_{\delta}$ for all $\gamma$,
$\delta\in B$.   But this means that, for any
finite set $D\subseteq B$, we can define $x\in X$ by setting

$$\eqalign{x(i)
&=x_{\alpha}(i)\text{ if }\alpha\in D,\,i\in J_{\alpha},\cr
&=0\text{ if }i\in I\setminus\bigcup_{\alpha\in D}J_{\alpha}.\cr}$$

\noindent It is easy to check that
$\{\gamma:\gamma\in C,\,x(i_{\gamma})=1\}=D$, so that $x\in Y$;  but now

\Centerline{$x\in Y\cap\bigcap_{\alpha\in D}E_{\alpha}
=\bigcap_{\alpha\in D}K_{\alpha}$.}

What this shows is that $\{K_{\alpha}:\alpha\in B\}$ has the finite
intersection property.   It must therefore have non-empty intersection;
say

\Centerline{$y\in\bigcap_{\alpha\in B}K_{\alpha}
\subseteq\bigcap_{\alpha\in B}G_{\alpha}$.}

\noindent But now we have a member $y$ of $Y$ such that
$\{\gamma:y(i_{\gamma})=1\}\supseteq B$ is infinite, contrary to the
definition of $Y$.\ \Bang\Qed
}%end of proof of 342N

\exercises{
\leader{342X}{Basic exercises $\pmb{>}$(a)}
%\spheader 342Xa
Show that a measure space $(X,\Sigma,\mu)$ is semi-finite
iff $\mu$ is inner regular with respect to $\{E:\mu E<\infty\}$.
%342A

\spheader 342Xb Find a proof of 342B based on 215A.
%342B

\spheader 342Xc Let $(X,\Sigma,\mu)$ be a locally compact semi-finite
measure space in which all singleton sets are negligible.   Show that it
is atomless.
%342D

\spheader 342Xd Let $(X,\Sigma,\mu)$ be a measure
space, and $\nu$ an indefinite-integral measure over $\mu$
(234J\footnote{Formerly 2{}34B.}).
Show that $\nu$ is compact, or locally compact, if $\mu$ is.   \Hint{if
$\Cal K$ satisfies the conditions of 342E with respect to $\mu$, then it
satisfies them for $\nu$.}
%342E

\spheader 342Xe Let $f:\Bbb R\to\Bbb R$ be any non-decreasing function,
and $\nu_f$ the corresponding Lebesgue-Stieltjes measure.   Show
that $\nu_f$ is compact.   \Hint{256Xg.}
%342J

\spheader 342Xf Let $\mu$ be Lebesgue measure on $[0,1]$, $\nu$ the
countable-cocountable measure on $[0,1]$, and $\lambda$ their c.l.d.\
product.   Show that $\lambda$ is a compact measure.   \Hint{let
$\Cal K$ be the family of sets $K\times A$ where $A\subseteq[0,1]$ is
cocountable and $K\subseteq A$ is compact.}
%342J

\spheader 342Xg(i) Give an example of a compact probability space
$(X,\Sigma,\mu)$, a set $Y$ and a function $f:X\to Y$ such that the
image
measure $\mu f^{-1}$ is not compact.  (ii) Give an example of a compact
probability space $(X,\Sigma,\mu)$ and a $\sigma$-subalgebra $\Tau$ of
$\Sigma$ such that $(X,\Tau,\mu\restrp\Tau)$ is not compact.
\Hint{342Xf.}
%342Xf, 342J

\spheader 342Xh Let $(X,\Sigma,\mu)$ be a perfect measure space, and
$f:X\to\Bbb R$ a measurable function.   Show that the image measure
$\mu f^{-1}$ is inner regular with respect to the compact subsets of
$\Bbb R$, so is a compact measure.
%342K

\spheader 342Xi Let $(X,\Sigma,\mu)$ be a $\sigma$-finite
measure space.   Show that it is perfect iff for every measurable
$f:X\to\Bbb R$ there is a Borel set $H\subseteq f[X]$ such that
$f^{-1}[H]$ is conegligible in $X$.   \Hint{342Xh for `only if', 256C
for `if'.}
%342K, 342Xh

\spheader 342Xj Let $(X,\Sigma,\mu)$ be a complete totally finite
perfect measure space and $f:X\to\Bbb R$ a measurable function.   Show
that the image measure $\mu f^{-1}$ is a Radon measure, and is the only
Radon measure on $\Bbb R$ for which $f$ is \imp.   \Hint{256G.}
%342K

\spheader 342Xk Suppose that $(X,\Sigma,\mu)$ is a perfect measure
space.   (i) Show that if $(Y,\Tau,\nu)$ is a measure space, and $f:X\to
Y$ is a function such that $f^{-1}[F]\in\Sigma$ for every $F\in\Tau$ and
$f^{-1}[F]$ is $\mu$-negligible for every $\nu$-negligible set
$F$, then $(Y,\Tau,\nu)$ is perfect.   (ii) Show that if $\Tau$ is a
$\sigma$-subalgebra of $\Sigma$ then $(X,\Tau,\mu\restrp\Tau)$ is
perfect.
%342K

\spheader 342Xl Let $(X,\Sigma,\mu)$ be a perfect measure space such
that $\Sigma$ is the $\sigma$-algebra generated by a sequence of sets.
Show that $\mu$ is compact.   \Hint{if $\Sigma$ is generated by
$\{E_n:n\in\Bbb
N\}$, set $f=\sum_{n=0}^{\infty}3^{-n}\chi E_n$ and consider
$\{f^{-1}[K]:K\subseteq f[X]$ is compact$\}$.}
%342L

\spheader 342Xm Let $(X,\Sigma,\mu)$ be a
semi-finite measure space.   Show that $\mu$
is perfect iff $\mu\restrp\Tau$ is compact for every countably generated
$\sigma$-subalgebra $\Tau$ of $\Sigma$.
%342L

\spheader 342Xn Show that (i) a measurable subspace of a perfect measure
space is perfect (ii) a semi-finite measure space is perfect iff all its
totally finite subspaces are perfect (iii) the direct sum of any family
of perfect measure spaces is perfect (iv) the c.l.d.\ product of two
perfect measure spaces is perfect ({\it hint\/}: put 342Xm and 342Ge
together) (v) the product of any family of
perfect probability spaces is perfect (vi) a measure space is perfect
iff its completion is perfect (vii) the c.l.d.\ version of a perfect
measure space is perfect (viii) any purely atomic measure space is
perfect (ix) an indefinite-integral measure over a perfect measure is
perfect (x)\dvAnew{2010} a sum (234G\Latereditions) of perfect measures
is perfect.
%342L

\spheader 342Xo Let $\mu$ be Lebesgue measure on $\Bbb R$, $A$ a
subset of $\Bbb R$, and $\mu_A$ the subspace measure on $A$.   Show that
$\mu_A$ is compact iff it is perfect iff $A$ is Lebesgue measurable.
\Hint{if $\mu_A$ is perfect, consider the image measure $\mu_Ah^{-1}$ on
$\Bbb R$, where $h(x)=x$ for $x\in A$.}
%342L

\leader{342Y}{Further exercises (a)}
%\spheader 342Ya
Let $U$ be a Banach space such that there is a linear
operator $T:U^{**}\to U$, of norm at most $1$, such that $T\hat u=u$ for
every $u\in U$, writing $\hat u$ for the member of $U^{**}$
corresponding to $u$.   Show that the family of closed balls in $U$ is a
compact class.
%342D

\spheader 342Yb\dvAnew{2010} Give an example of a compact class
$\Cal K$ of subsets of $\Bbb N$
such that there is no compact Hausdorff topology on $\Bbb N$ for which
every member of $\Cal K$ is closed.
%mt34bits

\spheader 342Yc
Show that the space $(X,\Sigma,\mu)$ of 216E and 342N is a
compact measure space.   \Hint{use the usual topology on $X=\{0,1\}^I$.}
%342N

\spheader 342Yd Give an example of a compact complete locally
determined measure space which is not localizable.   \Hint{in 216D, add
a point to each horizontal and vertical section of $X$, so that all the
sections become compact measure spaces.}
%342N
}%end of exercises

\cmmnt{\Notesheader{342} The terminology I find myself using in this
section -- `compact', `locally compact', `perfect'
-- is not entirely satisfactory, in that it risks collision with the
same words applied to
topological spaces.   For the moment, this is not a serious problem;
but when in Volume 4 we come to the systematic analysis of spaces which
have both topologies and measures present, it will be necessary to watch
our language carefully.   Of course there are cases in which a `compact
class' of the sort discussed here can be taken to be the family of
compact sets for
some familiar topology, as in 342Ja-342Jd, but in others this is not so
(see 342Xf);  and even when we have a familiar compact class, the
topology constructed from it by the method of 342Da need not be one we
might expect.
(Consider, for instance, the topology on $\Bbb R$ for which the closed
sets are just the sets which are compact for the usual topology, together
with the set $\Bbb R$ itself.)

I suppose that `compact' and `perfect' measure spaces look reasonably
natural objects to study;  they offer to illuminate one of the basic
properties of Radon measures, the fact that (at least for totally finite
Radon measures on Euclidean space) the image measure of a Radon measure
under a measurable function is again Radon (256G, 342Xj).   Indeed this
was the original impetus for the
study of perfect measures ({\smc Gnedenko \& Kolmogorov 54}, 
{\smc Sazonov 66}).   It is not obvious that there is any need to examine
`locally compact' measure spaces, but actually they are the chief
purpose of this
section, since the main theorem of the next section is an alternative
characterization of
semi-finite locally compact measure spaces (343B).   Of course you may
feel that the fact that `locally compact' and `compact' coincide for
strictly localizable spaces (342Hb) excuses you from troubling about the
distinction at first reading.

As with any new classification of measure spaces, it is worth finding
out how the classes of `compact' and `perfect' measure spaces
behave with respect to the standard constructions.   I run through the
basic facts in 342G-342I, 342Xd, 342Xk and 342Xn.   We can also look for
relationships between the new properties and those already studied.
Here, in fact, there is not much to be said;  342N and 342Yd show that
`compactness' is largely independent of the classification in \S211.
However there are
interactions with the concept of `atom' (342Xc, 342Xn(viii)).

I give examples to show that perfect measure spaces need not be locally
compact, and that locally compact measure spaces need not be compact
(342M, 342N).   The standard examples of measure spaces which are not
perfect are non-measurable subspaces (342Xo);  I will return to these in
the next section (343L-343M).

Something which is not important to us at the moment, but is perhaps
worth taking note of, is the following observation.   To determine
whether a
measure space $(X,\Sigma,\mu)$ is compact, we need only the structure
$(X,\Sigma,\Cal N)$, where $\Cal N$ is the $\sigma$-ideal of
negligible sets, since that is all that is referred to in the criterion
of 342E.   The same is true of local compactness, by 342Hc, and of
perfectness,
by the definition in 342K.   Compare 342Xd, 342Xk and 342Xn(ix).

Much of the material of this section will be repeated in Volume 4 as
part of a more systematic analysis of inner regularity.
}%end of notes

\discrpage

\frfilename{mt343.tex}
\versiondate{17.11.10}
\copyrightdate{2001}

\def\chaptername{The lifting theorem}
\def\sectionname{Realization of homomorphisms}

\newsection{343}

We are now in a position to make progress in one of the basic questions
of abstract measure theory.   In \S324 I have already described the
way in which a function between two measure spaces can give rise to a
homomorphism between their measure algebras.   In this section I discuss
some conditions under which we can be sure that a
homomorphism can be represented by a function.

The principal theorem of the section is 343B.  If a measure space
$(X,\Sigma,\mu)$ is locally compact, then many homomorphisms from the
measure algebra of $\mu$ to other measure algebras will be representable
by functions into $X$;  moreover, this characterizes locally compact
spaces.   In general, a homomorphism between measure
algebras can be represented by widely different functions (343I,
343J).  But in some of the most important cases (e.g., Lebesgue
measure) representing functions are `almost' uniquely defined;  I
introduce the concept of `countably separated' measure space to describe these (343D-343H).

\leader{343A}{Preliminary remarks} \cmmnt{It will be helpful to
establish some vocabulary and a couple of elementary facts.

\medskip

}{\bf (a)} If $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are
measure spaces, with measure algebras $\frak A$ and $\frak B$, I will
say that a function $f:X\to Y$ {\bf represents} a homomorphism
$\pi:\frak B\to\frak A$ if $f^{-1}[F]\in\Sigma$ and
$(f^{-1}[F])^{\ssbullet}=\pi(F^{\ssbullet})$ for every $F\in\Tau$.

\cmmnt{(Perhaps I should emphasize here that some homomorphisms are
representable in this sense, and some are not;  see 343M below for
examples of non-representable homomorphisms.)}

\spheader 343Ab If $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are
measure spaces, with measure algebras $\frak A$ and $\frak B$,
$f:X\to Y$ is a function, and $\pi:\frak B\to\frak A$ is a
sequentially order-continuous Boolean homomorphism, then

\Centerline{$\{F:F\in\Tau,\,f^{-1}[F]\in\Sigma$ and
$f^{-1}[F]^{\ssbullet}=\pi F^{\ssbullet}\}$}

\noindent is a $\sigma$-subalgebra of $\Tau$.   \cmmnt{(The
verification is elementary.)}

\spheader 343Ac Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with measure algebras $\frak A$ and $\frak B$, and
$\pi:\frak B\to\frak A$ a Boolean homomorphism which is represented by
a function $f:X\to Y$.   Let $(X,\hat\Sigma,\hat\mu)$,
$(Y,\hat\Tau,\hat\nu)$ be the completions of $(X,\Sigma,\mu)$,
$(Y,\Tau,\nu)$;  then $\frak A$ and $\frak B$ can be identified with
the measure algebras of $\hat\mu$ and $\hat\nu$\cmmnt{ (322Da)}.   Now
$f$ still represents $\pi$ when regarded as a function from
$(X,\hat\Sigma,\hat\mu)$ to $(Y,\hat\Tau,\hat\nu)$.   \prooflet{\Prf\
If $G$ is $\nu$-negligible, there is a negligible $F\in\Tau$ such
that $G\subseteq F$;  since

\Centerline{$f^{-1}[F]^{\ssbullet}=\pi F^{\ssbullet}=0$,}

\noindent $f^{-1}[F]$ is $\mu$-negligible, so $f^{-1}[G]$ is
negligible, therefore belongs to $\hat\Sigma$.   If $G$ is any element
of $\hat\Tau$, there is an $F\in\Tau$ such that $G\symmdiff F$ is
negligible, so that

\Centerline{$f^{-1}[G]
=f^{-1}[F]\symmdiff f^{-1}[G\symmdiff F]
\in\hat\Sigma$,}

\noindent and

\Centerline{$f^{-1}[G]^{\ssbullet}=f^{-1}[F]^{\ssbullet}
=\pi F^{\ssbullet}=\pi G^{\ssbullet}$.   \Qed}
}%end of prooflet

\leader{343B}{Theorem} Let $(X,\Sigma,\mu)$ be a non-empty
semi-finite measure space, and $(\frak A,\bar\mu)$ its measure algebra.
Let $(Z,\Lambda,\lambda)$ be the Stone space of $(\frak A,\bar\mu)$;
for $E\in\Sigma$ write $E^*$ for the
open-and-closed subset of $Z$ corresponding to the image $E^{\ssbullet}$
of $E$ in $\frak A$.   Then the following are equiveridical.

(i) $(X,\Sigma,\mu)$ is locally compact\cmmnt{ in the sense of 342Ad}.

(ii) There is a family $\Cal K\subseteq\Sigma$ such that ($\alpha$)
whenever $E\in\Sigma$ and $\mu E>0$ there is a $K\in\Cal K$ such that
$K\subseteq E$
and $\mu K>0$ ($\beta$) whenever $\Cal K'\subseteq\Cal K$ is such that
$\mu(\bigcap\Cal K_0)>0$ for every non-empty finite set
$\Cal K_0\subseteq\Cal K'$, then $\bigcap\Cal K'\ne\emptyset$.

(iii) There is a family $\Cal K\subseteq\Sigma$ such that ($\alpha$)$'$
$\mu$ is inner regular with respect to $\Cal K$ ($\beta$) whenever
$\Cal K'\subseteq\Cal K$ is such that $\mu(\bigcap\Cal K_0)>0$ for every
non-empty finite set $\Cal K_0\subseteq\Cal K'$, then
$\bigcap\Cal K'\ne\emptyset$.

(iv) There is a function $f:Z\to X$ such that $f^{-1}[E]\symmdiff E^*$
is negligible for every $E\in\Sigma$.

(v) Whenever $(Y,\Tau,\nu)$ is a complete strictly localizable
measure space, with measure algebra $\frak B$, and
$\pi:\frak A\to\frak B$ is an order-continuous Boolean homomorphism,
then there is a $g:Y\to X$ representing $\pi$.

(vi) Whenever $(Y,\Tau,\nu)$ is a complete strictly localizable
measure space, with measure algebra $\frak B$, and
$\pi:\frak A\to\frak B$ is an order-continuous measure-preserving Boolean
homomorphism, then there is a $g:Y\to X$ representing $\pi$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Because $\mu$ is semi-finite, there
is a partition of unity $\langle a_i\rangle_{i\in I}$ in $\frak A$ such
that $\bar\mu a_i<\infty$ for each $i$.   For each $i\in I$, let
$E_i\in\Sigma$
be such that $E_i^{\ssbullet}=a_i$.   Then the subspace measure $\mu_{E_i}$
on $E_i$ is compact;  let $\Cal K_i\subseteq\Cal PE_i$ be a compact class
such that $\mu_{E_i}$ is inner regular with respect to $\Cal K_i$.   Set
$\Cal K=\bigcup_{i\in I}\Cal K_i$.   If $\Cal K'\subseteq\Cal K$ and
$\mu(\bigcap\Cal K_0)>0$ for every non-empty finite
$\Cal K_0\subseteq\Cal K$, then $\Cal K'\subseteq\Cal K_i$ for some $i$,
and surely has the
finite intersection property, so $\bigcap\Cal K'\ne\emptyset$;  thus
$\Cal K'$ satisfies ($\beta$) of condition (ii).   And if $E\in\Sigma$,
$\mu E>0$ then
there must be some $i\in I$ such that $E_i^{\ssbullet}\Bcap a_i\ne 0$,
that is, $\mu(E\cap E_i)>0$, in which case there is a
$K\in\Cal K_i\subseteq\Cal K$
such that $K\subseteq E\cap E_i$ and $\mu K>0$;  so that $\Cal K$
satisfies condition ($\alpha$).

\medskip

{\bf (b)(ii)$\Rightarrow$(iii)} Suppose that $\Cal K\subseteq\Sigma$
witnesses that (ii) is true.   If $\mu X=0$ then $\Cal K$ already
witnesses that (iii) is true, so we need consider only the case
$\mu X>0$.   Set
$\Cal L=\{K_0\cup\ldots\cup K_n:K_0,\ldots,K_n\in\Cal K\}$.   Then
$\Cal L$
witnesses that (iii) is true.   \Prf\ By 342Ba, $\mu$ is inner regular
with respect to $\Cal L$.
Let $\Cal L'\subseteq\Cal L$ be such that $\mu(\bigcap\Cal L_0)>0$ for
every non-empty finite $\Cal L_0\subseteq\Cal L'$.   Then

\Centerline{$\Cal F_0=\{A:A\subseteq X$, there is a finite $\Cal
L_0\subseteq\Cal L'$ such that $X\cap\bigcap\Cal L_0\setminus A$ is
negligible$\}$}

\noindent is a filter on $X$, so there is an ultrafilter $\Cal F$ on $X$
including $\Cal F_0$.   Note that every conegligible set belongs to
$\Cal F_0$, so no negligible set can belong to $\Cal F$.   Set $\Cal
K'=\Cal K\cap\Cal F$;  then $\bigcap\Cal K_0$ belongs to $\Cal F$, so is
not negligible, for every non-empty finite $\Cal K_0\subseteq\Cal K'$.
Accordingly there is some $x\in\bigcap\Cal K'$.   But any member of
$\Cal L'$ is of the form $L=K_0\cup\ldots\cup K_n$ where each
$K_i\in\Cal K$;
because $\Cal F$ is an ultrafilter and $L\in\Cal F$, there must be some
$i\le n$ such that $K_i\in\Cal F$, in which case $x\in K_i\subseteq L$.
Thus $x\in\bigcap\Cal L'$.   As $\Cal L'$ is arbitrary, $\Cal L$
satisfies the condition ($\beta$).\ \Qed

\medskip

{\bf (c)(iii)$\Rightarrow$(iv)}
Let $\Cal K\subseteq\Sigma$ witness that (iii) is true.   For any $z\in
Z$, set $\Cal
K_z=\{K:K\in\Cal K,\,z\in K^*\}$.    If $K_0,\ldots,K_n\in\Cal K_z$,
then $z\in\bigcap_{i\le n}K^*_i=(\bigcap_{i\le n}K_i)^*$, so
$(\bigcap_{i\le n}K_i)^*\ne\emptyset$ and $\mu(\bigcap_{i\le n}K_i)>0$.
By ($\beta$) of condition (iii), $\bigcap\Cal
K_z\ne\emptyset$;  and even if $\Cal K_z=\emptyset$, $X\cap\bigcap\Cal
K_z\ne\emptyset$ because $X$ is non-empty.   So we may choose $f(z)\in
X\cap\bigcap\Cal K_z$.   This defines a function $f:Z\to X$.   Observe
that, for $K\in\Cal K$ and $z\in Z$,

\Centerline{$z\in K^*\Longrightarrow K\in\Cal K_z\Longrightarrow f(z)\in
K\Longrightarrow z\in f^{-1}[K]$,}

\noindent so that $K^*\subseteq f^{-1}[K]$.

Now take any $E\in\Sigma$.   Consider

\Centerline{$U_1=\bigcup\{K^*:K\in\Cal K,\,K\subseteq E\}
\subseteq\bigcup\{E^*\cap f^{-1}[K]:K\in\Cal K,\,K\subseteq E\}
\subseteq E^*\cap f^{-1}[E]$,}

\Centerline{$U_2=\bigcup\{K^*:K\in\Cal K,\,K\subseteq X\setminus E\}
\subseteq (X\setminus E)^*\cap f^{-1}[X\setminus E]
=Z\setminus(f^{-1}[E]\cup E^*)$,}

\noindent so that $f^{-1}[E]\symmdiff E^*\subseteq Z\setminus(U_1\cup
U_2)$.   Now $U_1$ and $U_2$ are open subsets of $Z$, so
$M=Z\setminus(U_1\cup U_2)$ is closed, and in fact $M$ is nowhere dense.
\Prf\Quer\ Otherwise, there is a non-zero $a\in\frak A$ such that the
corresponding open-and-closed set
$\widehat{a}$ is included in $M$, and an $F\in\Sigma$ of non-zero
measure such that
$a=F^{\ssbullet}$.   At least one of $F\cap E$, $F\setminus E$ is
non-negligible and therefore includes a non-negligible member $K$ of
$\Cal K$.   But in this case $K^*$ is a non-empty open subset of $M$
which is included in either $U_1$ or $U_2$, which is
impossible.\ \Bang\Qed

By the definition of $\lambda$ (321J-321K), $M$ is
$\lambda$-negligible, so $f^{-1}[E]\symmdiff E^*\subseteq M$ is
negligible, as required.

\medskip

{\bf (d)(iv)$\Rightarrow$(v)} Now assume that $f:Z\to X$ witnesses
(iv), and let $(Y,\Tau,\nu)$ be a complete strictly localizable measure
space, with measure algebra $\frak B$, and $\pi:\frak A\to\frak B$ an
order-continuous Boolean homomorphism.   If $\nu Y=0$ then any function
from $Y$ to $X$ will represent $\pi$, so we may suppose that $\nu Y>0$.
Write $W$ for the Stone space
of $\frak B$.   Then we have a continuous function $\phi:W\to Z$ such
that $\phi^{-1}[\widehat{a}]=\widehat{\pi a}$ for every $a\in\frak A$
(312Q), and
$\phi^{-1}[M]$ is nowhere dense in $W$ for every nowhere dense
$M\subseteq Z$ (313R).   It follows that $\phi^{-1}[M]$ is meager for
every meager $M\subseteq Z$, that is, $\phi^{-1}[M]$ is negligible in
$W$ for every negligible $M\subseteq Z$.   By 341Q, there is an \imp\
function $h:Y\to W$ such that
$h^{-1}[\vthsp\widehat{b}\vthsp]^{\ssbullet}=b$ for
every $b\in\frak B$.   Consider $g=f\phi h:Y\to X$.

If $E\in\Sigma$, set $a=E^{\ssbullet}\in\frak A$, so that
$E^*=\widehat{a}\subseteq Z$, and $M=f^{-1}[E]\symmdiff E^*$ is
$\lambda$-negligible;  consequently $\phi^{-1}[M]$ is negligible in $W$.
Because $h$ is \imp,

\Centerline{$g^{-1}[E]\symmdiff h^{-1}[\phi^{-1}[E^*]]
=h^{-1}[\phi^{-1}[f^{-1}[E]]]\symmdiff h^{-1}[\phi^{-1}[E^*]]
=h^{-1}[\phi^{-1}[M]]$}

\noindent is negligible.   But $\phi^{-1}[E^*]=\widehat{\pi a}$, so

\Centerline{$g^{-1}[E]^{\ssbullet}
=h^{-1}[\phi^{-1}[E^*]]^{\ssbullet}=\pi a$.}
\noindent As $E$ is arbitrary, $g$ induces the homomorphism $\pi$.

\medskip

{\bf (e)(v)$\Rightarrow$(vi)} is trivial.

\medskip

{\bf (f)(vi)$\Rightarrow$(iv)} Assume (vi).   Let $\nu$ be the
c.l.d.\ version of $\lambda$, $\Tau$ its domain, and $\frak B$ its
measure algebra;  then $\nu$ is strictly localizable (322Rb).   The
embedding $\Lambda\embedsinto\Tau$ corresponds to
an order-continuous measure-preserving Boolean homomorphism from
$\frak A$ to $\frak B$ (322Db).
By (vi), there is a function $f:Z\to X$ such that $f^{-1}[E]\in\Tau$ and
$f^{-1}[E]^{\ssbullet}=(E^*)^{\ssbullet}$ in $\frak B$ for
every $E\in\Sigma$.   But as $\nu$ and $\lambda$ have the same
negligible sets (322Rb), $f^{-1}[E]\symmdiff E^*$ is
$\lambda$-negligible for every $E\in\Sigma$, as required by (iv).

\medskip

{\bf (g)(iv)$\Rightarrow$(i)}\grheada\ To begin with (down to the end of
($\gamma$) below) I suppose that $\mu$ is totally finite.   In this case
we have a function $g:X\to Z$ such that $E\symmdiff g^{-1}[E^*]$ is
negligible
for every $E\in\Sigma$ (341Q again).    We are supposing also that
there is a function $f:Z\to X$ such that $f^{-1}[E]\symmdiff E^*$
is negligible for every $E\in\Sigma$.   Write $\Cal K$ for the family of
sets $K\subseteq X$ such that $K\in\Sigma$ and there is a compact set
$L\subseteq Z$ such that $f[L]\subseteq K\subseteq g^{-1}[L]$.

\medskip

\quad\grheadb\ $\mu$ is inner regular with respect to $\Cal K$.   \Prf\
Take $F\in\Sigma$ and $\gamma<\mu F$.   Choose $\sequencen{V_n}$,
$\sequencen{F_n}$
as follows.  $F_0=F$.   Given that $\mu F_n>\gamma$, then

\Centerline{$\lambda(f^{-1}[F_n]\cap F_n^*)=\lambda F_n^*
=\mu F_n>\gamma$,}

\noindent so there is an open-and-closed set
$V_n\subseteq f^{-1}[F_n]\cap F_n^*$ with $\lambda V_n>\gamma$.
Express $V_n$ as $F_{n+1}^*$ where
$F_{n+1}\in\Sigma$;  since $F_n\symmdiff g^{-1}[F_n^*]$ is negligible,
and $V_n\subseteq F_n^*$, we may take it that
$F_{n+1}\subseteq g^{-1}[F_n^*]$.   Continue.

At the end of the induction, set $K=\bigcap_{n\in\Bbb N}F_n\in\Sigma$
and $L=\bigcap_{n\in\Bbb N}F_n^*$.   Because
$F_{n+1}\setminus F_n\subseteq g^{-1}[F_n^*]\setminus F_n$ is negligible
for each $n$,
$\mu K=\lim_{n\to\infty}\mu F_n\ge\gamma$, while $K\subseteq F$ and $L$
is surely compact.   We have

\Centerline{$L\subseteq\bigcap_{n\in\Bbb N}V_n
\subseteq\bigcap_{n\in\Bbb N}f^{-1}[F_n]=f^{-1}[K]$,}

\noindent so $f[L]\subseteq K$.   Also

\Centerline{$K\subseteq\bigcap_{n\in\Bbb N}F_{n+1}
\subseteq\bigcap_{n\in\Bbb N}g^{-1}[F_n^*]=g^{-1}[L]$.}

\noindent So $K\in\Cal K$.   As $F$ and $\gamma$ are arbitrary, $\mu$ is
inner regular with respect to $\Cal K$.\ \Qed

\medskip

\quad\grheadc\ Next, $\Cal K$ is a compact class.   \Prf\ Suppose that
$\Cal K'\subseteq\Cal K$ has the finite intersection property.   If
$\Cal K'=\emptyset$, of course $\bigcap\Cal K'\ne\emptyset$;  suppose
that $\Cal K'$ is non-empty.   Let $\Cal L$ be the family of closed sets
$L\subseteq Z$
such that $g^{-1}[L]$ includes some member of $\Cal K'$.   Then $\Cal L$
has the finite intersection property, and $Z$ is compact, so there is
some $z\in\bigcap\Cal L$;  also $Z\in\Cal L$, so $z\in Z$.   For any
$K\in\Cal K'$, there is some closed set $L\subseteq Z$ such that
$f[L]\subseteq
K\subseteq g^{-1}[L]$, so that $L\in\Cal L$ and $z\in L$ and $f(z)\in
K$.   Thus $f(z)\in\bigcap\Cal K'$.   As $\Cal K'$ is arbitrary, $\Cal
K$ is a compact class.\ \Qed

So $\Cal K$ witnesses that $\mu$ is a compact measure.

\medskip

\quad\grheadd\ Now consider the general case.
Take any $E\in\Sigma$ of finite measure.   If $E=\emptyset$ then surely
the subspace measure $\mu_E$ is compact.   Otherwise, we can identify
the measure algebra of $\mu_E$ with the principal ideal
$\frak A_{E^{\ssbullet}}$ of
$\frak A$ generated by $E^{\ssbullet}$ (322Ja), and $E^*\subseteq Z$
with the Stone space of $\frak A_{E^{\ssbullet}}$ (312T).   Take any
$x_0\in E$ and define
$\tilde f:E^*\to E$ by setting $\tilde f(z)=f(z)$ if
$z\in E^*\cap f^{-1}[E]$, $x_0$ if $z\in E^*\setminus f^{-1}[E]$.   Then
$f$ and $\tilde f$ agree almost everywhere in $E^*$, so
$\tilde f^{-1}[F]\symmdiff F^*$ is negligible for every $F\in\Sigma_E$,
that is, $\tilde f$ represents the canonical isomorphism between the
measure algebras of $\mu_E$ and the subspace measure $\lambda_{E^*}$ on
$E^*$.   But
this means that condition (iv) is true of $\mu_E$, so $\mu_E$ is
compact, by
($\alpha$)-($\gamma$) above.   As $E$ is arbitrary, $\mu$ is locally
compact.

This completes the proof.
}%end of proof of 343B

\leader{343C}{Examples (a)} Let $I$ be any set.
\dvro{The}{We know that the} usual measure $\nu_I$ on
$\{0,1\}^I$ is
compact\cmmnt{ (342Jd)}.   \dvro{If}{It follows that if}
$(X,\Sigma,\mu)$
is any complete probability space such that the measure algebra
$\frak B_I$ of $\nu_I$ can be embedded as a subalgebra of
the measure algebra $\frak A$ of $\mu$, there is an \imp\ function from
$X$ to $\{0,1\}^I$.   \cmmnt{For infinite $I$, this is so iff every
non-zero principal ideal of $\frak A$ has Maharam type at least
$\kappa$, by 332P.   Of
course this does not depend in any way on the results of the present
chapter.   If
$\frak B_{\kappa}$ can be embedded in $\frak A$, there must be a
stochastically independent family $\langle E_{\xi}\rangle_{\xi<\kappa}$
of sets of measure $\bover12$;  now we get a map
$h:X\to\{0,1\}^{\kappa}$ by
saying that $h(x)(\xi)=1$ iff $x\in E_{\xi}$, which by 254G is \imp.}

\spheader 343Cb In particular, if $\mu$ is atomless, there is an \imp\
function from $X$ to $\{0,1\}^{\Bbb N}$;  since this is
isomorphic\cmmnt{, as measure space,} to $[0,1]$ with Lebesgue
measure\cmmnt{ (254K)}, there is an \imp\ function from $X$ to
$[0,1]$.

\spheader 343Cc More generally, if $(X,\Sigma,\mu)$ is any complete
atomless totally finite measure space, there is an \imp\ function from
$X$ to the interval $[0,\mu X]$ endowed with Lebesgue measure.
\cmmnt{(If $\mu X>0$, apply (b) to the normalized measure
$(\mu X)^{-1}\mu$;  or argue directly from 343B, using the fact that
Lebesgue measure on $[0,\mu X]$ is compact;  or use the idea suggested
in 343Xd.)}

\spheader 343Cd\dvAnew{2008}
In the other direction, if $(X,\Sigma,\mu)$ is a compact
probability space with Maharam type at most $\kappa\ge\omega$,
then there is an
\imp\ function from $\{0,1\}^{\kappa}$ to $X$.   \prooflet{\Prf\
By 332N, there is a measure-preserving homomorphism from the measure
algebra of $\mu$ to the measure algebra of $\nu_{\kappa}$;  by 343B, this
is represented by an \imp\ function from $\{0,1\}^{\kappa}$ to $X$.\ \Qed}

\cmmnt{\spheader 343Ce Throughout the work above -- in \S254 as well
as in 343B -- I have taken the measures involved to be complete.   It
does occasionally happen, in this context, that this restriction is
inconvenient.   Typical results not depending on completeness in the
domain space $X$ are in
343Xc-343Xd.   Of course these depend not only on the very special
nature of the codomain spaces $\{0,1\}^I$ or $[0,1]$, but also on the
measures on these spaces being taken to be incomplete.
}%end of comment

\leader{343D}{Uniqueness of \dvrocolon{realizations}}\cmmnt{ The
results of 342E-342J, %342E 342F 342G 342H 342I 342J
together with 343B, give a respectable number of contexts in
which homomorphisms
between measure algebras can be represented by functions between measure
spaces.   They say nothing about whether such functions are unique, or
whether we can distinguish, among the possible representations of a
homomorphism, any canonical one.   In fact the proof of 343B, using the
Lifting Theorem as it does, strongly suggests that this is like looking
for a canonical lifting, and I am sure that (outside a handful of very
special cases) any such search is vain.
Nevertheless, we do have a weak kind of uniqueness theorem, valid in a
useful number of spaces, as follows.

\medskip

\noindent}{\bf Definition} A measure space $(X,\Sigma,\mu)$ is
{\bf countably separated} if there is a countable set
$\Cal A\subseteq\Sigma$
separating the points of $X$ in the sense that for any distinct $x$,
$y\in X$ there is an $E\in\Cal A$ containing one but not the other.
\cmmnt{(Of course this is a property of the structure $(X,\Sigma)$
rather than of $(X,\Sigma,\mu)$.)}

\leader{343E}{Lemma} A measure space $(X,\Sigma,\mu)$ is countably
separated iff there is an injective measurable function from $X$ to
$\Bbb R$.

\proof{ If $(X,\Sigma,\mu)$ is countably separated, let $\Cal
A\subseteq\Sigma$ be a countable set separating the points of $X$.   Let
$\sequencen{E_n}$ be a sequence running over $\Cal A\cup\{\emptyset\}$.
Set

\Centerline{$f=\sum_{n=0}^{\infty}3^{-n}\chi E_n:X\to\Bbb R$.}

\noindent Then $f$ is measurable (because every $E_n$ is measurable) and
injective (because if $x\ne y$ in $X$ and
$n=\min\{i:\#(E_i\cap\{x,y\})=1\}$ and $x\in E_n$, then

\Centerline{$f(x)\ge 3^{-n}+\sum_{i<n}3^{-i}\chi E_i(x)
>\sum_{i>n}3^{-i}+\sum_{i<n}3^{-i}\chi E_i(y)\ge f(y)$.)}

On the other hand, if $f:X\to\Bbb R$ is measurable and injective, then
$\Cal A=\{f^{-1}[\,\ocint{-\infty,q}\,]:q\in\Bbb Q\}$ is a countable
subset of $\Sigma$ separating the points of $X$, so $(X,\Sigma,\mu)$ is
countably separated.
}%end of proof of 343E

\cmmnt{\medskip

\noindent{\bf Remark} The construction of the function $f$ from the
sequence $\sequencen{E_n}$ in the proof above is a standard trick;  such
$f$ are sometimes called {\bf Marczewski functionals}.
}%end of comment

\leader{343F}{Proposition} Let $(X,\Sigma,\mu)$ be a countably
separated measure space and $(Y,\Tau,\nu)$ any measure space.  Let
$f$, $g:Y\to X$ be two functions such that $f^{-1}[E]$ and
$g^{-1}[E]$ both belong to $\Tau$, and $f^{-1}[E]\symmdiff g^{-1}[E]$ is
$\nu$-negligible, for every $E\in\Sigma$.   Then
$f=g\,\,\nu$-almost everywhere, and $\{y:y\in Y,\,f(y)\ne g(y)\}$ is
measurable as well as negligible.

\proof{ Let $\Cal A\subseteq\Sigma$ be a countable set separating the
points of  $X$.  Then

\Centerline{$\{y:f(y)\ne g(y)\}=\bigcup_{E\in\Cal A}f^{-1}[E]\symmdiff
g^{-1}[E]$}

\noindent is measurable and negligible.
}%end of proof of 343F

\leader{343G}{Corollary} If, in 343B, $(X,\Sigma,\mu)$ is countably
separated, then the functions $g:Y\to X$ of 343B(v)-(vi) are almost
uniquely defined in the sense that if $f$, $g$ both represent the same
homomorphism from $\frak A$ to $\frak B$ then $f\eae g$.

\leader{343H}{Examples} Leading examples of countably separated measure
spaces are

(i) $\Bbb R$ \cmmnt{(take $\Cal A=\{\ocint{-\infty,q}:q\in\Bbb Q\}$)};

(ii) $\{0,1\}^{\Bbb N}$ \cmmnt{(take $\Cal A=\{E_n:n\in\Bbb N\}$,
where $E_n=\{x:x(n)=1\}$)};

(iii) subspaces (measurable or not) of countably separated spaces;

(iv) finite products of countably separated spaces;

(v) countable products of countably separated probability spaces;

(vi) completions and c.l.d.\ versions of countably separated spaces.

\cmmnt{As soon as we move away from these elementary ideas, however,
some interesting difficulties arise.}

\leader{343I}{Example} Let $\nu_{\frakc}$ be the usual measure on
$X=\{0,1\}^{\frakc}$\cmmnt{, where $\frak c=\#(\Bbb R)$}, and
$\Tau_{\frakc}$ its domain.   Then there is a function $f:X\to X$
such that $f(x)\ne x$ for every $x\in X$, but $E\symmdiff f^{-1}[E]$ is
negligible for every $E\in\Tau_{\frakc}$.   \prooflet{\Prf\ The
set $\frak c\setminus\omega$ is still with cardinal $\frak c$, so there is
an injection $h:\{0,1\}^{\omega}\to\frak c\setminus\omega$.   (As
usual, I am identifying the cardinal number $\frak c$ with the
corresponding initial ordinal.    But if you prefer to argue without the
full axiom of choice, you can express all the same ideas with $\Bbb R$
in the place of $\frak c$ and $\Bbb N$ in the place of $\omega$.)    For
$x\in X$, set

$$\eqalign{f(x)(\xi)&=1-x(\xi)\text{ if }\xi=h(x\restr\omega),\cr
&=x(\xi)\text{ otherwise }.\cr}$$

\noindent Evidently $f(x)\ne x$ for every $x$.   If $E\subseteq X$ is
measurable, then we can find a countable set $J\subseteq\frak c$ and
sets $E'$, $E''$, both determined by coordinates in $J$, such that
$E'\subseteq E\subseteq E''$ and $E''\setminus E'$ is negligible
(254Oc).   Now for any particular $\xi\in\frak c\setminus\omega$,
$\{x:h(x\restr\omega)=\xi\}$ is negligible, being either empty or of the
form $\{x:x(n)=z(n)$ for every $n<\omega\}$ for some
$z\in\{0,1\}^{\omega}$.   So $H=\{x:h(x\restr\omega)\in J\}$ is
negligible.   Now we see that for $x\in X\setminus H$,
$f(x)\restr J=x\restr J$, so for
$x\in X\setminus(H\cup(E''\setminus E'))$,

\Centerline{$x\in E\Longrightarrow x\in E'\Longrightarrow f(x)\in E'\Longrightarrow
f(x)\in E$,}

\Centerline{$x\notin E\Longrightarrow x\notin E''\Longrightarrow f(x)\notin
E''\Longrightarrow f(x)\notin E$.}

\noindent Thus $E\symmdiff f^{-1}[E]\subseteq H\cup(E''\setminus E')$ is
negligible.\ \Qed}

\leader{343J}{The split interval} \cmmnt{I introduce a construction
which here will
seem essentially elementary, but in other contexts is of great
interest, as will appear in Volume 4.

\medskip

}{\bf (a)} Take $I^{\|}$ to consist of two
copies of each point of the unit interval, so that
$I^{\|}=\{t^+:t\in[0,1]\}\cup\{t^-:t\in[0,1]\}$.   For
$A\subseteq I^{\|}$ write
$A_l=\{t:t^-\in A\}$, $A_r=\{t:t^+\in A\}$.   Let $\Sigma$ be the set

\Centerline{$\{E:E\subseteq I^{\|},\,E_l$ and $E_r$ are Lebesgue
measurable and $E_l\symmdiff E_r$ is Lebesgue negligible$\}$.}

\noindent For $E\in\Sigma$, set

\Centerline{$\mu E=\mu_LE_l=\mu_LE_r$}

\noindent where $\mu_L$ is Lebesgue measure on $[0,1]$.
\cmmnt{It is easy to check that} $(I^{\|},\Sigma,\mu)$ is a complete
probability space\cmmnt{ (cf.\ 234F, 234Ye)}.
Also it is compact.   \prooflet{\Prf\ Take $\Cal K$
to be the family of sets $K\subseteq I^{\|}$ such that $K_l=K_r$ is a
compact subset of $[0,1]$, and check that $\Cal K$ is a compact class
and that $\mu$
is inner regular with respect to $\Cal K$;  or use 343Xa below.\ \Qed\ }
The sets $\{t^-:t\in[0,1]\}$
and $\{t^+:t\in[0,1]\}$ are non-measurable subsets of $I^{\|}$;  on both
of them
the subspace measures correspond exactly to $\mu_L$.   We have a
canonical \imp\ function $h:I^{\|}\to[0,1]$ given by
setting $h(t^+)=h(t^-)=t$ for every $t\in[0,1]$;  $h$ induces an
isomorphism between the measure algebras of $\mu$ and $\mu_L$.

$I^{\|}$ is called the {\bf split interval}\cmmnt{ or (especially when
given its
standard topology, as in 343Yc below) the {\bf double arrow space} or
{\bf two arrows space}}.

Now\cmmnt{ the relevance to the present discussion is this:}  we have
a map $f:I^{\|}\to I^{\|}$ given by setting

\Centerline{$f(t^+)=t^-$, $f(t^-)=t^+$ for every $t\in[0,1]$}

\noindent such that $f(x)\ne x$ for every $x$, but
$E\symmdiff f^{-1}[E]$ is negligible for every $E\in\Sigma$, so that $f$
represents the identity homomorphism on the measure algebra of $\mu$.
The\cmmnt{ function $h:I^{\|}\to[0,1]$ is canonical enough, but is
two-to-one, and the} canonical map from the measure algebra of $\mu$ to
the measure
algebra of $\mu_L$ is represented equally by the functions
$t\mapsto t^-$ and $t\mapsto t^+$, which are nowhere equal.

\spheader 343Jb Consider the direct sum
$(Y,\nu)$ of $(I^{\|},\mu)$ and $([0,1],\mu_L)$;\cmmnt{ for
definiteness,} take $Y$ to be
$(I^{\|}\times\{0\})\cup([0,1]\times\{1\})$.
Setting

\Centerline{$h_1(t^+,0)=h_1(t^-,0)=(t,1)$,\quad $h_1(t,1)=(t^+,0)$,}

\noindent\cmmnt{ we see that }$h_1:Y\to Y$ induces a measure-preserving
involution of the measure algebra $\frak B$ of $\nu$\cmmnt{, corresponding
to its expression as a simple product of the isomorphic measure algebras
of $\mu$ and $\mu_L$}.   But\cmmnt{ $h_1$ is not invertible, and
indeed} there is no invertible function from $Y$ to itself which induces
this involution of $\frak B$.   \prooflet{\Prf\Quer\ Suppose, if
possible,
that $g:Y\to Y$ were such a function.   Looking at the sets

\Centerline{$E_q=[0,q]\times\{1\}$,\quad
$F_q=\{(t^+,0):t\in[0,q]\}\cup\{(t^-,0):t\in[0,q]\}$}

\noindent  for $q\in \Bbb Q$,
we must have $g^{-1}[E_q]\symmdiff F_q$ negligible for every
$q$, so that we must have $g(t^+,0)=g(t^-,0)=(t,1)$ for almost every
$t\in[0,1]$, and $g$ cannot be injective.\ \Bang\Qed}

\spheader 343Jc Thus even with a compact probability space, and an
automorphism $\phi$ of its measure algebra, we cannot be sure of
representing $\phi$ and $\phi^{-1}$ by functions which will be inverses
of each other.

\leader{343K}{}\cmmnt{ 342L has a partial converse.

\medskip

\noindent}{\bf Proposition} If $(X,\Sigma,\mu)$ is a semi-finite
countably separated measure space, it is compact iff it is locally
compact iff it is perfect.

\proof{ We already know that compact measure spaces are locally compact
and locally compact semi-finite measure spaces are perfect (342Ha,
342L).   So suppose that $(X,\Sigma,\mu)$ is a perfect semi-finite
countably
separated measure space.   Let $f:X\to\Bbb R$ be an injective measurable
function (343E).    Consider

\Centerline{$\Cal K
=\{f^{-1}[L]:L\subseteq f[X],\,L$ is compact in $\Bbb R\}$.}

\noindent The definition of `perfect' measure space states exactly that
whenever $E\in\Sigma$ and $\mu E>0$ there is a $K\in\Cal K$ such
that $K\subseteq E$ and $\mu K>0$.   And $\Cal K$ is a compact class.
\Prf\ If $\Cal K'\subseteq\Cal K$ has
the finite intersection property, $\Cal L=\{f[K]:K\in\Cal K'\}$ is
a family of compact sets in $\Bbb R$ with the finite intersection
property, and has non-empty intersection;   so that $\bigcap\Cal K'$ is
also non-empty, because $f$ is injective.\ \QeD\   By 342E,
$(X,\Sigma,\mu)$ is compact.
}%end of proof of 343K

\leader{343L}{}\cmmnt{ The time has come to give examples of spaces
which are {\it not} locally compact, so that we can expect to have
measure-preserving homomorphisms not representable by \imp\ functions.
The most commonly arising ones are covered by the following result.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a complete locally
determined countably separated measure space, and $A\subseteq X$ a set
such that the subspace measure $\mu_A$ is perfect.    Then $A$ is
measurable.

\proof{\Quer\ Otherwise, there is a set $E\in\Sigma$ such that
$\mu E<\infty$ and $B=A\cap E\notin\Sigma$.   Let $f:X\to\Bbb R$ be an
injective measurable function (343E again).   Then
$f\restr B$
is $\Sigma_B$-measurable, where $\Sigma_B$ is the domain of the subspace
measure $\mu_B$ on $B$.   Set

\Centerline{$\Cal K=\{f^{-1}[L]:L\subseteq f[B],\,L$ is compact in
$\Bbb R\}$.}

\noindent Just as in the proof of 343K, $\Cal K$ is a compact class and
$\mu_B$ is inner regular with respect to $\Cal K$.
By 342Bb, there is a sequence $\sequencen{K_n}$ in $\Cal K$ such that
$\mu_B(B\setminus\bigcup_{n\in\Bbb N}K_n)=0$.   But of course
$\Cal K\subseteq\Sigma$, because $f$ is $\Sigma$-measurable, so
$\bigcup_{n\in\Bbb N}K_n\in\Sigma$.   Because $\mu$ is complete,
$B\setminus\bigcup_{n\in\Bbb N}K_n\in\Sigma$ and $B\in\Sigma$.\ \Bang
}%end of proof of 343L

\leader{343M}{Example} 343L tells us that any non-measurable set
$X$ of $\BbbR^r$, or of $\{0,1\}^{\Bbb N}$, with their usual measures,
is not perfect, therefore not (locally) compact, when given its subspace
measure.

\cmmnt{To find a non-representable homomorphism, we do not need to go
through the whole apparatus of 343B.   Take $Y$ to be a measurable
envelope of $X$ (132Ee).   Then the identity function from $X$ to $Y$
induces an isomorphism of their measure algebras.   But there is no
function from $Y$ to $X$ inducing the same isomorphism.
\prooflet{\Prf\Quer\ Writing $Z$ for $\BbbR^r$ or $\{0,1\}^{\Bbb N}$
and $\mu$ for its measure, $Z$ is countably separated;  suppose
$\sequencen{E_n}$ is a sequence of measurable sets in $Z$ separating its
points.   For each $n$, $(Y\cap E_n)^{\ssbullet}$ in the measure algebra
of $\mu_Y$ corresponds to $(X\cap E_n)^{\ssbullet}$ in the measure
algebra of $\mu_X$.   So if $f:Y\to X$ were a function representing the
isomorphism of the measure algebras, $(Y\cap E_n)\symmdiff f^{-1}[E_n]$
would have to be negligible for each $n$, and
$A=\bigcup_{n\in\Bbb N}(Y\cap E_n)\symmdiff f^{-1}[E_n]$ would be
negligible.   But for $y\in Y\setminus A$, $f(y)$ belongs to just the
same $E_n$ as $y$ does, so must be equal to $y$.   Accordingly
$X\supseteq Y\setminus A$ and $X$ is measurable.\ \Bang\Qed}
}%end of comment

\exercises{
\leader{343X}{Basic exercises (a)}
%\spheader 343Xa
Let $(X,\Sigma,\mu)$ be a
semi-finite measure space.   (i) Suppose that there is a set
$A\subseteq X$, of full outer measure, such that the
subspace measure on $A$ is compact.   Show that $\mu$ is locally
compact.   \Hint{show that $\mu$ satisfies (ii) or (v) of 343B.}  (ii)
Suppose that for every non-negligible $E\in\Sigma$ there is a
non-negligible set $A\subseteq E$ such that
the subspace measure on $A$ is compact.   Show that $\mu$ is locally
compact.
%343B

\spheader 343Xb Let $\langle X_i\rangle_{i\in I}$ be a
family of non-empty sets, with product $X$;  write $\pi_i:X\to X_i$ for
the coordinate map.   Suppose we are given a $\sigma$-algebra
$\Sigma_i$ of subsets of $X_i$ for each $i$;  let
$\Sigma=\Tensorhat_{i\in I}\Sigma_i$ be the corresponding
$\sigma$-algebra of subsets of $X$
generated by $\{\pi_i^{-1}[E]:i\in I,\,E\in\Sigma_i\}$.
Let $\mu$ be a totally finite measure with domain $\Sigma$, and for
$i\in I$ let $\mu_i$ be the image measure $\mu\pi_i^{-1}$.   Check that
the domain of $\mu_i$ is $\Sigma_i$.   Show that if
every $(X_i,\Sigma_i,\mu_i)$ is compact,
then so is $(X,\Sigma,\mu)$.   \Hint{{\it either} show that $\mu$
satisfies (v) of 343B {\it or} adapt the method of 342Gf.}
%343B

\spheader 343Xc Let  $I$ be any set.   Let $\CalBa$ be the
$\sigma$-algebra of
subsets of $\{0,1\}^I$ generated by the sets $F_i=\{z:z(i)=1\}$ for
$i\in I$, and $\nu$ any probability measure with domain $\CalBa$;  let
$\frak B$ be
the measure algebra of $\nu$.   Let $(X,\Sigma,\mu)$ be a measure space
with measure algebra $\frak A$, and $\phi:\frak B\to\frak A$ an
order-continuous
Boolean homomorphism.   Show that there is an \imp\ function
$f:X\to\{0,1\}^I$ representing $\phi$.   \Hint{for each $i\in I$, take
$E_i\in\Sigma$ such that $E_i^{\ssbullet}=\phi F_i^{\ssbullet}$;  set
$f(x)(i)=1$ if $x\in E_i$, and use 343Ab.}
%343C

\spheader 343Xd Let $(X,\Sigma,\mu)$ be an atomless probability space.
Let $\mu_{\Cal B}$ be the restriction of Lebesgue measure to the
$\sigma$-algebra of Borel subsets of $[0,1]$.   Show that there is a
function $g:X\to[0,1]$ which is \imp\ for $\mu$ and $\mu_{\Cal B}$.
\Hint{find an $f:X\to\{0,1\}^{\Bbb N}$ as in 343Xc, and set $g=hf$ where
$h(z)=\sum_{n=0}^{\infty}2^{-n-1}g(n)$, as in 254K;  or choose
$E_q\in\Sigma$
such that $\mu E_q=q$, $E_q\subseteq E_{q'}$ whenever $q\le q'$ in
$[0,1]\cap\Bbb Q$, and set $f(x)=\inf\{q:x\in E_q\}$ for $x\in E_1$.}
%343D

\spheader 343Xe Let $(X,\Sigma,\mu)$ be a countably separated measure
space, with measure algebra $\frak A$.   (i) Show that $\{x\}\in\Sigma$
for every $x\in X$.   (ii) Show that every atom of $\frak A$ is of the
form $\{x\}^{\ssbullet}$ for some $x\in X$.
%343D

\spheader 343Xf Let $(X,\Sigma,\mu)$ be a semi-finite countably separated
measure space.
(i) Show that $\mu$ is point-supported iff it is complete,
strictly localizable
and purely atomic.   (ii) Show that $\mu$ is atomless iff $\mu\{x\}=0$
for every $x\in X$.
%343D

\spheader 343Xg Let $I^{\|}$ be the split interval, with its usual
measure
$\mu$ described in 343J, and $h:I^{\|}\to[0,1]$ the canonical
surjection.   Show that the canonical isomorphism between the measure
algebras of $\mu$ and Lebesgue measure on $[0,1]$ is given by the
formula `$E^{\ssbullet}\mapsto h[E]^{\ssbullet}$ for every measurable
$E\subseteq I^{\|}$'.
%343J

\spheader 343Xh Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces with measure algebras $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$.
Suppose that $X\cap Y=\emptyset$ and that we have a measure-preserving
isomorphism $\pi:\frak A\to\frak B$.   Set

\Centerline{$\Lambda=\{W:W\subseteq X\cup Y,\,W\cap X\in\Sigma,\,W\cap
Y\in\Tau,\,\pi(W\cap X)^{\ssbullet}=(W\cap Y)^{\ssbullet}\}$,}

\noindent and for $W\in\Lambda$ set $\lambda W=\mu(W\cap X)=\nu(W\cap
Y)$.   Show that $(X\cup Y,\Lambda,\lambda)$ is a measure space which is
locally compact, or perfect, if $(X,\Sigma,\mu)$ is.
%343J

\sqheader 343Xi Let $(X,\Sigma,\mu)$ be a complete perfect totally
finite measure space, $(Y,\Tau,\nu)$ a complete countably separated
measure space, and $f:X\to Y$ an inverse-measure-preserving function.
Show that $\Tau=\{F:F\subseteq Y,\,f^{-1}[F]\in\Sigma\}$, so that a
function $h:Y\to\Bbb R$ is $\nu$-integrable iff $hf$ is
$\mu$-integrable.
\Hint{if $A\subseteq Y$ and $E=f^{-1}[A]\in\Sigma$, $f\restr E$ is \imp\
for the subspace measures $\mu_E$, $\nu_A$;  by 342Xk, $\nu_A$ is
perfect, so by 343L $A\in\Tau$.   Now use 235J.}
%343L

\spheader 343Xj Let $(X,\Sigma,\mu)$ be a complete compact measure space,
$Y$ a set and $f:Y\to X$ a surjection;  set

\Centerline{$\Tau=\{F:F\subseteq Y$, $f[F]\in\Sigma$,
$\mu(f[F]\cap f[Y\setminus F])=0\}$,
\quad$\nu F=\mu f[F]$ for $F\in\Tau$,}

\noindent so that $\nu$ is a measure on $Y$ and $f$ is \imp\
(234Ye).   Show that $\nu$ is a compact measure.
%343J

\leader{343Y}{Further exercises (a)}
%\spheader 343Ya
Let $(X,\Sigma,\mu)$ be a semi-finite measure space, and suppose that
there is a compact class $\Cal K\subseteq\Cal PX$ such that ($\alpha$)
whenever $E\in\Sigma$ and $\mu E>0$ there is a non-negligible
$K\in\Cal K$ such that
$K\subseteq E$ ($\beta$) whenever $K_0,\ldots,K_n\in\Cal K$ and
$\bigcap_{i\le n}K_i=\emptyset$ then there are measurable sets
$E_0,\ldots,E_n$ such that $E_i\supseteq K_i$ for every $i$ and
$\bigcap_{i\le n}E_i$ is negligible.   Show that $\mu$ is locally
compact.
%343B

\spheader 343Yb(i)\dvArevised{2010}
Show that a countably separated semi-finite measure
space has magnitude and Maharam type at most
$2^{\frak c}$.   (ii) Show that the direct sum of $\frak c$ or fewer
countably separated measure spaces is countably separated.
(iii)\dvAnew{2011} Show that a countably separated perfect measure space
has countable Maharam type.
%343E

\spheader 343Yc Let
$I^{\|}=\{t^+:t\in[0,1]\}\cup\{t^-:t\in[0,1]\}$ be the split interval
(343J).   (i) Show that the rules

\Centerline{$s^-\le t^-\iff s^+\le t^+\iff s\le t$,
\quad $s^+\le t^-\iff s<t$,}

\Centerline{$t^-\le t^+$ for all $t\in[0,1]$}

\noindent define a Dedekind complete total order on $I^{\|}$ with
greatest and least elements.   (ii) Show that the intervals $[0^-,t^-]$,
$[t^+,1^+]$,
interpreted for this ordering, generate a compact Hausdorff topology on
$I^{\|}$ for which the map $h:I^{\|}\to[0,1]$ of 343J is continuous.
(iii) Show
that a subset $E$ of $I^{\|}$ is Borel for this topology iff the sets
$E_r$, $E_l\subseteq[0,1]$, as described in 343Ja, are Borel and
$E_r\symmdiff E_l$ is countable.    (iv) Show that if $f:[0,1]\to\Bbb R$
is of bounded
variation then there is a continuous $g:I^{\|}\to\Bbb R$ such that
$g=fh$ except perhaps at countably many points.
(v) Show that the measure $\mu$ of 343J is inner regular with respect to
the compact subsets of $I^{\|}$.   (vi) Show that we have a lower
density $\phi$ for $\mu$ defined by setting

$$\eqalign{\phi E=\{t^-:\,&0<t\le 1,\,\lim_{\delta\downarrow 0}
  \Bover1{\delta}\mu(E\cap[(t-\delta)^+,t^-])=1\}\cr
&\cup\{t^+:0\le t<1,\,\lim_{\delta\downarrow 0}
  \Bover1{\delta}\mu(E\cap[t^+,(t+\delta)^-])=1\}\cr}$$

\noindent for measurable sets $E\subseteq I^{\|}$.
%343H

\spheader 343Yd Set $X=\{0,1\}^{\frak c}$, with its usual measure
$\nu_{\frakc}$.   Show that there is an inverse-measure-preserving function
$f:X\to X$ such that $f[X]$ is non-measurable but $f$ induces the
identity automorphism of the measure algebra of $\nu_{\frakc}$.
\Hint{use the
idea of 343I.}   Show that under these conditions $f[X]$, with its
subspace measure, must be compact.   \Hint{use 343B(iv).}
%343H

\spheader 343Ye Let $\mu_{Hr}$ be $r$-dimensional Hausdorff measure on
$\BbbR^s$, where $s\ge 1$ is an integer and $r\ge 0$ (\S264).   (i)
Show that $\mu_{Hr}$ is countably separated.   (ii) Show that the
c.l.d.\ version of $\mu_{Hr}$ is compact.   \Hint{264Yi.}
%343K

\spheader 343Yf Give an example of a countably separated probability
space $(X,\Sigma,\mu)$ and a function $f$ from $X$ to a set $Y$ such
that the image measure $\mu f^{-1}$ is not countably separated.
\Hint{use 223B to show that if $E\subseteq\Bbb R$ is Lebesgue measurable
and not negligible, then $E+\Bbb Q$ is conegligible;  or use the
zero-one law to show that if $E\subseteq\Cal P\Bbb N$ is measurable and
not negligible for the usual measure on $\Cal P\Bbb N$, then
$\{a\symmdiff b:a\in E,\,b\in[\Bbb N]^{<\omega}\}$ is conegligible.}
%343M

}%end of exercises

\cmmnt{\Notesheader{343} The points at which the Lifting Theorem
impinges on the work of this section are in the proofs of
(iv)$\Rightarrow$(i) and (iv)$\Rightarrow$(v) in Theorem 343B.
In fact the ideas can be rearranged to give a proof of 343B which does
not rely on the Lifting Theorem;  I give a hint in Volume 4 (413Ye).

I suppose the significant new ideas of this section are in 343B and
343K.   The rest is mostly a matter of being thorough and careful.
But I take this material at a slow pace because there are some
potentially confusing features, and the underlying question is of the
greatest importance:  when, given a Boolean homomorphism from one
measure algebra to another, can we be sure of representing it by a
measurable function between measure spaces?   The concept of `compact'
measure puts the burden firmly on the measure space corresponding to the
{\it domain} of the Boolean homomorphism, which will be the {\it
codomain} of the measurable function.   So the first step is to try to
understand properly which measures are compact, and what other properties
they can be expected to have;  which accounts for much of the length of
\S342.   But having understood that many of our favourite measures
are compact, we have to come to terms with the fact that we still cannot
count on a measure algebra isomorphism corresponding to a measure space
isomorphism.   I introduce the split interval (343J, 343Xg,
343Yc) as a close
approximation to Lebesgue measure on $[0,1]$ which is not isomorphic to
it.   Of course we have already seen a more dramatic example:  the Stone
space of the Lebesgue measure algebra also has the same measure algebra
as Lebesgue measure, while being in almost every other way very much
more complex, as will appear in Volumes 4 and 5.

As 343C suggests, elementary cases in which 343B can be applied are
often amenable to more primitive methods, avoiding not only the concept
of `compact' measure, but also Stone spaces and the Lifting Theorem.
For substantial examples in which we can prove that a measure space
$(X,\mu)$ is
compact, without simultaneously finding direct constructions for \imp\
functions into $X$ (as in 343Xc-343Xd), I think we shall have to wait
until Volume 4.

The concept of `countably separated' measure space does not involve the
measure at all, nor even the null ideal;  it belongs to
the theory of $\sigma$-algebras of sets.   Some simple
permanence properties are in 343H and 343Yb(ii).
Let us note in passing that 343Xi describes some more situations in
which the `image measure catastrophe', described
in 235H, cannot arise.

I include the variants 343B(ii), 343B(iii) and 343Ya of the notion of
`local compactness' because they are not obvious and may illuminate it.
}%end of notes

\discrpage
\frfilename{mt344.tex}
\versiondate{22.3.06}

\def\chaptername{The lifting theorem}
\def\sectionname{Realization of automorphisms}

\newsection{344}

In 343Jb, I gave an example of a `good' (compact, complete) probability
space $X$ with an automorphism $\phi$ of its measure algebra such that
both $\phi$ and $\phi^{-1}$ are representable by functions from $X$ to
itself, but there is no such representation in which the two
functions are
inverses of each other.   The present section is an attempt to describe
the further refinements necessary to ensure that automorphisms of
measure algebras can be represented by automorphisms of the measure
spaces.   It turns out that in the most important contexts in which this
can be done, a little extra work yields a significant generalization:
the simultaneous realization of countably many homomorphisms by a
consistent family of functions.

I will describe three cases in which such simultaneous realizations can
be achieved:  Stone spaces (344A), perfect complete countably separated
spaces (344C) and suitable measures on $\{0,1\}^I$ (344E-344G).   The
arguments for 344C, suitably refined, give a complete description of
perfect complete countably separated strictly localizable spaces which
are not purely atomic (344I\cmmnt{, 344Xc}).   At the same time we
find that Lebesgue measure, and the usual measure on $\{0,1\}^I$, are
`homogeneous' in the strong sense that two measurable subspaces (of
non-zero measure) are isomorphic iff they have the same measure (344J,
344L).

\leader{344A}{Stone spaces}\cmmnt{ The first case is
immediate from the work of \S\S312, 313 and 321, as collected in
324E.} If $(Z,\Sigma,\mu)$ is\cmmnt{ actually} the Stone space of a
measure algebra $(\frak A,\bar\mu)$, then every order-continuous Boolean
homomorphism
$\phi:\frak A\to\frak A$ corresponds to a unique continuous function
$f_{\phi}:Z\to Z$\cmmnt{ (312Q)} which represents
$\phi$\cmmnt{ (324E)}.
\cmmnt{The uniqueness of $f_{\phi}$
means that we can be sure that} $f_{\phi\psi}=f_{\psi}f_{\phi}$ for all
order-continuous homomorphisms $\phi$ and $\psi$;
and\cmmnt{ of course} $f_{\iota}$ is the identity map on $Z$, so that
$f_{\phi^{-1}}$ will
have to be $f_{\phi}^{-1}$ whenever $\phi$ is invertible.
Thus in this\cmmnt{ special} case we can consistently, and
canonically,
represent all
order-continuous Boolean homomorphisms from $\frak A$ to itself.

\cmmnt{Now for two cases where we have to work for the results.}

\leader{344B}{Theorem} Let $(X,\Sigma,\mu)$ be a countably separated
measure space with measure algebra $\frak A$, and $G$ a countable
semigroup of Boolean homomorphisms from $\frak A$ to itself such that
every member of $G$ can be represented by some function from $X$ to
itself.   Then a family $\langle f_{\phi}\rangle_{\phi\in G}$ of such
representatives can be chosen in such a way that
$f_{\phi\psi}=f_{\psi}f_{\phi}$ for all $\phi$, $\psi\in G$;  and if the
identity automorphism $\iota$ belongs to $G$, then we may arrange that
$f_{\iota}$ is the identity function on $X$.

\proof{{\bf (a)} Because $G\cup\{\iota\}$ satisfies the same conditions
as $G$, we may suppose from the beginning that $\iota$ belongs to $G$
itself.   Let $\Cal A\subseteq\Sigma$ be a countable set separating the
points of $X$.   For each $\phi\in G$ take some representing
function $g_{\phi}:X\to X$;  take $g_{\iota}$ to be the identity
function.   If $\phi$, $\psi\in G$, then of course

\Centerline{$((g_{\phi}g_{\psi})^{-1}[E])^{\ssbullet}
=(g_{\psi}^{-1}[g_{\phi}^{-1}[E]])^{\ssbullet}
=\psi(g_{\phi}^{-1}[E])^{\ssbullet}
=\psi\phi E^{\ssbullet}
=(g_{\psi\phi}^{-1}[E])^{\ssbullet}$}

\noindent for every $E\in\Sigma$.   By 343F, the set

\Centerline{$H_{\phi\psi}=\{x:g_{\psi\phi}(x)\ne g_{\phi}g_{\psi}(x)\}$}

\noindent is negligible and belongs to $\Sigma$.

\medskip

{\bf (b)} Set

\Centerline{$H=\bigcup_{\phi,\psi\in G}
H_{\phi\psi}$;}

\noindent because $G$ is countable, $H$ also is measurable and
negligible.   Try defining $f_{\phi}:X\to X$ by setting
$f_{\phi}(x)=g_{\phi}(x)$ if $x\in X\setminus H$, $f_{\phi}(x)=x$ if
$x\in H$.   Because $H$ is measurable, $f_{\phi}^{-1}[E]\in\Sigma$ for
every $E\in\Sigma$;  because $H$ is negligible,

\Centerline{$(f_{\phi}^{-1}[E])^{\ssbullet}
=(g_{\phi}^{-1}[E])^{\ssbullet}=\phi E^{\ssbullet}$}

\noindent for every $E\in\Sigma$, and $f_{\phi}$ represents $\phi$, for
every $\phi\in G$.   Of course $f_{\iota}=g_{\iota}$ is the identity
function on $X$.

\medskip

{\bf (c)} If $\theta\in G$ then $f_{\theta}^{-1}[H]=H$.   \Prf\ (i) If
$x\in
H$ then $f_{\theta}(x)=x\in H$.   (ii) If $f_{\theta}(x)\in H$ and
$f_{\theta}(x)=x$ then of course $x\in H$.   (iii) If
$f_{\theta}(x)=g_{\theta}(x)\in H$ then there are $\phi$, $\psi\in
G$ such that $g_{\phi}g_{\psi}g_{\theta}(x)\ne
g_{\psi\phi}g_{\theta}(x)$.   So either

\Centerline{$g_{\psi}g_{\theta}(x)\ne g_{\theta\psi}(x)$,}

\noindent or

\Centerline{$g_{\phi}g_{\theta\psi}(x)\ne g_{\theta\psi\phi}(x)$}

\noindent or

\Centerline{$g_{\theta\psi\phi}(x)\ne g_{\psi\phi}g_{\theta}(x)$,}

\noindent and in any case $x\in H$.\ \Qed

\medskip

{\bf (d)} It follows that $f_{\phi}f_{\psi}=f_{\psi\phi}$ for every
$\phi$, $\psi\in G$.   \Prf\ (i) If $x\in H$ then

\Centerline{$f_{\phi}f_{\psi}(x)=x=f_{\psi\phi}(x)$.}

\noindent (ii) If $x\in X\setminus H$ then $f_{\psi}(x)\notin H$, by
(c), so

\Centerline{$f_{\phi}f_{\psi}(x)=g_{\phi}g_{\psi}(x)=g_{\psi\phi}(x)
=f_{\psi\phi}(x)$.   \Qed}
}%end of proof of 344B

\leader{344C}{Corollary} Let $(X,\Sigma,\mu)$ be a countably
separated perfect complete strictly localizable measure space with
measure algebra $\frak A$, and $G$ a countable semigroup of
order-continuous Boolean homomorphisms from $\frak A$ to itself.  Then
we can choose simultaneously, for each $\phi\in G$, a function
$f_{\phi}:X\to X$ representing
$\phi$, in such a way that $f_{\phi\psi}=f_{\psi}f_{\phi}$ for all
$\phi$, $\psi\in G$;  and if the identity automorphism $\iota$ belongs
to $G$, then we may arrange that $f_{\iota}$ is the identity function on
$X$.   In particular, if $\phi\in G$ is invertible, and
$\phi^{-1}\in G$, we shall have $f_{\phi^{-1}}=f_{\phi}^{-1}$;  so that
if moreover
$\phi$ and $\phi^{-1}$ are measure-preserving, $f_{\phi}$ will be an
automorphism of the measure space $(X,\Sigma,\mu)$.

\proof{ By 343K, $(X,\Sigma,\mu)$ is compact.   So 343B(v) tells us that
every member of $G$ is representable, and we can apply 344B.
}%end of proof of 344C

\cmmnt{\medskip

\noindent{\bf Reminder}  Spaces satisfying the conditions of this
corollary include Lebesgue measure on $\BbbR^r$, the usual measure on
$\{0,1\}^{\Bbb N}$, and their measurable subspaces;  see also 342J,
342Xe, 343H and 343Ye.}

\leader{344D}{}\cmmnt{ The third case I wish to present requires a
more elaborate argument.   I start with a kind of Schr\"oder-Bernstein
theorem for measurable spaces.

\medskip

\noindent}{\bf Lemma} Let $X$ and $Y$ be sets, and 
$\Sigma\subseteq\Cal PX$, $\Tau\subseteq\Cal PY\,\,\sigma$-algebras.   
Suppose that there are
$f:X\to Y$, $g:Y\to X$ such that $F=f[X]\in\Tau$, $E=g[Y]\in\Sigma$, $f$
is an isomorphism between $(X,\Sigma)$ and $(F,\Tau_F)$ and $g$ is an
isomorphism between $(Y,\Tau)$ and $(E,\Sigma_E)$, writing $\Sigma_E$,
$\Tau_F$ for the subspace $\sigma$-algebras\cmmnt{ (see 121A)}.   Then
$(X,\Sigma)$ and $(Y,\Tau)$ are isomorphic, and there is an isomorphism
$h:X\to Y$ which is covered by $f$ and $g$ in the sense that

\Centerline{$\{(x,h(x)):x\in X\}\subseteq\{(x,f(x)):x\in X\}
\cup\{(g(y),y):y\in Y\}$.}

\proof{ Set $X_0=X$, $Y_0=Y$, $X_{n+1}=g[Y_n]$ and $Y_{n+1}=f[X_n]$ for
each $n\in\Bbb N$;  then $\sequencen{X_n}$ is a non-increasing sequence
in $\Sigma$ and $\sequencen{Y_n}$ is a non-increasing sequence in
$\Tau$.   Set $X_{\infty}=\bigcap_{n\in\Bbb N}X_n$,
$Y_{\infty}=\bigcap_{n\in\Bbb N}Y_n$.   Then $f\restr X_{2k}\setminus
X_{2k+1}$ is an isomorphism between $X_{2k}\setminus X_{2k+1}$ and
$Y_{2k+1}\setminus Y_{2k+2}$, while
$g\restrp Y_{2k}\setminus Y_{2k+1}$ is an isomorphism between
$Y_{2k}\setminus Y_{2k+1}$ and $X_{2k+1}\setminus X_{2k+2}$;  and
$g\restrp Y_{\infty}$ is an isomorphism between $Y_{\infty}$ and
$X_{\infty}$.    So the formula

$$\eqalign{h(x)
&=f(x)\text{ if }x\in\bigcup_{k\in\Bbb N}X_{2k}\setminus X_{2k+1},\cr
&=g^{-1}(x)\text{ for other }x\in X\cr}$$

\noindent gives the required isomorphism between $X$ and $Y$.
}%end of proof of 344D

\cmmnt{\medskip

\noindent{\bf Remark} You will recognise the ordinary
Schr\"oder-Bernstein theorem (2A1G) as the case $\Sigma=\Cal PX$,
$\Tau=\Cal PY$.}

\leader{344E}{Theorem} Let $I$ be any set, and let $\mu$ be a
$\sigma$-finite measure on $X=\{0,1\}^I$ with domain the
$\sigma$-algebra $\CalBa_I$ generated by the sets $\{x:x(i)=1\}$ as $i$
runs over $I$;  write $\frak A$ for the measure algebra of $\mu$.
Let $G$ be a countable semigroup of order-continuous Boolean
homomorphisms from $\frak A$ to itself.   Then we can choose
simultaneously, for each $\phi\in G$, a function $f_{\phi}:X\to X$
representing $\phi$, in such a way that $f_{\phi\psi}=f_{\psi}f_{\phi}$
for all $\phi$, $\psi\in G$;  and if the identity automorphism $\iota$
belongs to $G$, then we may arrange that $f_{\iota}$ is the identity
function on $X$.   In particular, if $\phi\in G$ is invertible and
$\phi^{-1}\in G$, we shall have $f_{\phi^{-1}}=f_{\phi}^{-1}$;  so that
if moreover $\phi$ is measure-preserving, $f_{\phi}$ will be an
automorphism of the measure space $(X,\CalBa_I,\mu)$.

\proof{{\bf (a)} As in 344C, we may as well suppose from the beginning
that $\iota\in G$.   The case of finite $I$ is trivial, so I will
suppose that $I$ is infinite.      For $i\in I$, set
$E_i=\{x:x(i)=1\}$;  for $J\subseteq I$, let $\Cal B_J$ be the
$\sigma$-subalgebra of $\CalBa_I$ generated by $\{E_i:I\in J\}$.    For
$i\in I$, $\phi\in G$ choose
$F_{\phi i}\in\CalBa_I$ such that
$F_{\phi i}^{\ssbullet}=\phi E_i^{\ssbullet}$.    (Of course we take
$F_{\iota i}=E_i$ for every $i$.)
Let $\Cal J$ be the family of those subsets $J$ of $I$ such that
$F_{\phi i}\in\Cal B_J$ for every $i\in J$ and $\phi\in G$.

\medskip

{\bf (b)} For the purposes of this proof, I will say that a pair
$(J,\langle g_{\phi}\rangle_{\phi\in
G})$ is {\bf consistent} if $J\in\Cal J$ and, for each $\phi\in G$,
$g_{\phi}$ is a function from $X$ to itself such that

\inset{$g_{\phi}^{-1}[E_i]\in\Cal B_J$ and
$(g_{\phi}^{-1}[E_i])^{\ssbullet}=\phi E_i^{\ssbullet}$ whenever $i\in
J$, $\phi\in G$,}


\inset{$g_{\phi}^{-1}[E_i]=E_i$ whenever $i\in I\setminus J$, $\phi\in
G$,}

\inset{$g_{\phi}g_{\psi}=g_{\psi\phi}$ whenever $\phi$, $\psi\in G$,}

\inset{$g_{\iota}(x)=x$ for every $x\in X$.}

\noindent Now the key to the proof is the following fact:  if
$(J,\langle g_{\phi}\rangle_{\phi\in G})$ is consistent, and $\tilde J$
is a member of $\Cal J$ such that $\tilde J\setminus J$ is countably
infinite, then there is a family
$\family{\phi}{G}{tilde g_{\phi}}$ such that
$(\tilde J,\langle\tilde g_{\phi}\rangle_{\phi\in G})$ is
consistent and $\tilde g_{\phi}^{-1}[E_i]=g_{\phi}^{-1}[E_i]$ whenever
$i\in J$ and $\phi\in G$, that is,
$\tilde g_{\phi}(x)\restr J=g_{\phi}(x)\restr J$ whenever $\phi\in G$ and
$x\in X$.   The construction is as follows.

\medskip

\quad{\bf (i)} Start by fixing on any infinite set
$K\subseteq\tilde J\setminus J$ such that
$(\tilde J\setminus J)\setminus K$ also is infinite.   For
$z\in\{0,1\}^K$, set $V_z=\{x:x\in X,\,x\restr K=z\}$;  then
$V_z\in\Cal B_{\tilde J}$.   All the sets $V_z$, as $z$ runs over the
uncountable
set $\{0,1\}^K$, are disjoint, so they cannot all have non-zero measure
(because $\mu$ is $\sigma$-finite), and we can choose $z$
such that $V_z$ is $\mu$-negligible.

\medskip

\quad{\bf (ii)} Define $h_{\phi}:X\to X$, for $\phi\in G$, by setting

$$\eqalign{h_{\phi}(x)(i)&=g_{\phi}(x)(i)\text{ if }i\in J,\cr
&=x(i)\text{ if }i\in I\setminus\tilde J,\cr
&=x(i)\text{ if }i\in \tilde J\setminus J\text{ and }x\in V_z,\cr
&=1\text{ if }i\in\tilde J\setminus J\text{ and }
  x\in F_{\phi i}\setminus V_z,\cr
&=0\text{ if }i\in\tilde J\setminus J\text{ and }
  x\notin F_{\phi i}\cup V_z.\cr}$$

\noindent Because $V_z\in\Cal B_{\tilde J}$ and $\mu V_z=0$, we see
that

\inset{($\alpha$) $h_{\phi}^{-1}[E_i]=g_{\phi}^{-1}[E_i]\in\Cal B_J$ if
$i\in J$,}

\inset{($\beta$) $h_{\phi}^{-1}[E_i]\in\Cal B_{\tilde J}$ and
$h_{\phi}^{-1}[E_i]\symmdiff F_{\phi i}$ is negligible
if $i\in\tilde J\setminus J$,}

\noindent and consequently

\inset{($\gamma$) $(h_{\phi}^{-1}[E_i])^{\ssbullet}=\phi E_i^{\ssbullet}$
for every $i\in\tilde J$,}

\inset{($\delta$) $(h_{\phi}^{-1}[E])^{\ssbullet}=\phi E^{\ssbullet}$
for every $E\in\Cal B_{\tilde J}$}

\noindent (by 343Ab);  moreover,

\inset{($\epsilon$) $h_{\phi}^{-1}[E]=g_{\phi}^{-1}[E]$ for every
$E\in\Cal B_J$,}

\inset{($\zeta$) $h_{\phi}^{-1}[E]\in\Cal B_{\tilde J}$ for every
$E\in\Cal B_{\tilde J}$,}

\inset{($\eta$) $h_{\phi}^{-1}[E_i]=E_i$ if $i\in I\setminus\tilde J$,}

\noindent so that

\inset{($\theta$) $h_{\phi}^{-1}[E]\in\CalBa_I$ for every $E\in\CalBa_I$;}

\noindent finally, because $F_{\iota i}=E_i$,

\inset{($\iota$) $h_{\iota}(x)=x$ for every $x\in X$.}

\medskip

\quad{\bf (iii)} The next step is to note that if $\phi$, $\psi\in G$
then

\Centerline{$H_{\phi,\psi}=\{x:x\in X,\,h_{\phi}h_{\psi}(x)\ne
h_{\psi\phi}(x)\}$}

\noindent belongs to $\Cal B_{\tilde J}$ and is negligible.
\Prf

\Centerline{$H_{\phi,\psi}
=\bigcup_{i\in I}h_{\psi}^{-1}[h_{\phi}^{-1}[E_i]]\symmdiff
h_{\psi\phi}^{-1}[E_i]$.}

\noindent Now if $i\in J$, then $h_{\phi}^{-1}[E_i]
=g_{\phi}^{-1}[E_i]\in\Cal B_J$, so

\Centerline{$h_{\psi}^{-1}[h_{\phi}^{-1}[E_i]]
=h_{\psi}^{-1}[g_{\phi}^{-1}[E_i]]
=g_{\psi}^{-1}[g_{\phi}^{-1}[E_i]]
=g_{\psi\phi}^{-1}[E_i]
=h_{\psi\phi}^{-1}[E_i]$.}

\noindent Next, for $i\in I\setminus\tilde J$,

\Centerline{$h_{\psi}^{-1}[h_{\phi}^{-1}[E_i]]
=h_{\psi}^{-1}[E_i]
=E_i
=h_{\psi\phi}^{-1}[E_i]$.}

\noindent So

\Centerline{$H_{\phi,\psi}
=\bigcup_{i\in\tilde J\setminus J}
h_{\psi}^{-1}[h_{\phi}^{-1}[E_i]]\symmdiff h_{\psi\phi}^{-1}[E_i]$.}

\noindent But for any particular $i\in\tilde J\setminus J$,
$E_i$ and $h_{\phi}^{-1}[E_i]$ belong to $\Cal B_{\tilde J}$, so

\Centerline{$(h_{\psi}^{-1}[h_{\phi}^{-1}[E_i]])^{\ssbullet}
=\psi(h_{\phi}^{-1}[E_i])^{\ssbullet}
=\psi\phi E_i^{\ssbullet}
=(h_{\psi\phi}^{-1}[E_i])^{\ssbullet}$,}

\noindent and $h_{\psi}^{-1}[h_{\phi}^{-1}[E_i]]\symmdiff
h_{\psi\phi}^{-1}[E_i]$ is a negligible set, which by (ii-$\zeta$)
belongs to $\Cal B_{\tilde J}$.   So
$H_{\phi,\psi}$ is a countable union of sets of measure $0$ in
$\Cal B_{\tilde J}$ and is itself a negligible member of $\Cal B_{\tilde
J}$, as claimed.\ \Qed

\medskip

\quad{\bf (iv)} Set

\Centerline{$H=\bigcup_{\phi,\psi\in G}H_{\phi,\psi}\cup\bigcup_{\phi\in
G}h_{\phi}^{-1}[V_z]$.}

\noindent Then $H\in\Cal B_{\tilde J}$ and $\mu H=0$.   \Prf\  We
know that every $H_{\phi,\psi}$ is negligible and belongs to
$\Cal B_{\tilde J}$ ((iii) above), that every $h_{\phi}^{-1}[V_z]$
belongs to $\Cal B_{\tilde J}$ (by (ii-$\zeta$), and that
$(h_{\phi}^{-1}[V_z])^{\ssbullet}=\phi V_z^{\ssbullet}=0$, so that
$h_{\phi}^{-1}[V_z]$ is negligible, for every $\phi\in G$
(by (ii-$\delta$)).   Consequently $H$ is negligible and belongs to
$\Cal B_{\tilde J}$.\ \Qed\
Also, of course, $V_z=h_{\iota}^{-1}[V_z]\subseteq H$.

Next, $h_{\phi}(x)\notin H$ whenever $x\in X\setminus H$ and $\phi\in
G$.   \Prf\ If $\psi$, $\theta\in G$ then

\Centerline{$h_{\theta\psi}h_{\phi}(x)
=h_{\phi\theta\psi}(x)
=h_{\psi}h_{\phi\theta}(x)=h_{\psi}h_{\theta}h_{\phi}(x)$,}

\Centerline{$h_{\psi}h_{\phi}(x)=h_{\phi\psi}(x)\notin V_z$}

\noindent because

\Centerline{$x\notin H_{\theta\psi,\phi}\cup H_{\psi,\phi\theta}\cup
H_{\theta,\phi}\cup H_{\psi,\phi}\cup h_{\phi\psi}^{-1}[V_z]$;}

\noindent thus
$h_{\phi}(x)\notin H_{\psi,\theta}\cup h_{\psi}^{-1}[V_z]$;  as $\psi$
and $\theta$ are arbitrary, $h_{\phi}(x)\notin H$.\ \Qed

\medskip

\quad{\bf (v)} The next fact we need is that there is a bijection
$q:X\to H$ such that ($\alpha$) for $E\subseteq H$,
$E\in\Cal B_{\tilde J}$ iff
$q^{-1}[E]\in\Cal B_{\tilde J}$ ($\beta$) $q(x)(i)=x(i)$ for every
$i\in I\setminus(\tilde J\setminus J)$ and $x\in X$.
\Prf\ Fix any bijection
$r:\tilde J\setminus J\to\tilde J\setminus(J\cup K)$.   Consider the
maps $p_1:X\to H$, $p_2:H\to X$ given by

$$\eqalign{p_1(x)(i)&=x(r^{-1}(i))\text{ if }
  i\in\tilde J\setminus (J\cup K),\cr
&=z(i)\text{ if }i\in K,\cr
&=x(i)\text{ if }i\in X\setminus(\tilde J\setminus J),\cr
p_2(y)&=y\cr}$$

\noindent for $x\in X$, $y\in H$.   Then $p_1$ is actually an
isomorphism between $(X,\Cal B_{\tilde J})$ and 
$(V_z,\Cal B_{\tilde J}\cap\Cal PV_z)$.   So $p_1$, $p_2$ are 
isomorphisms between
$(X,\Cal B_{\tilde J})$, $(H,\Cal B_{\tilde J}\cap\Cal PH)$ and
measurable subspaces of $H$, $X$ respectively.   By 344D, there is an
isomorphism $q$ between $X$ and $H$ such that, for every $x\in X$,
either $q(x)=p_1(x)$ or $p_2(q(x))=x$.   Since $p_1(x)\restr
I\setminus(\tilde J\setminus J)=x\restr I\setminus(\tilde J\setminus J)$
for every $x\in X$, and $p_2(y)\restr I\setminus(\tilde J\setminus
J)=y\restr I\setminus(\tilde J\setminus J)$ for every $y\in H$,
$q(x)\restr I\setminus(\tilde J\setminus J)=x\restr I\setminus(\tilde
J\setminus J)$ for every $x\in X$.\ \Qed

\medskip

\quad{\bf (vi)} An incidental fact which will be used below is the
following:  if $i\in \tilde J$ and $\phi\in G$ then $g_{\phi}^{-1}[E_i]$
belongs to $\Cal B_{\tilde J}$, because it belongs to $\Cal B_J$ if
$i\in J$, and otherwise is equal to $E_i$.   Consequently
$g_{\phi}^{-1}[E]\in\Cal B_{\tilde J}$ for every $E\in\Cal B_{\tilde
J}$.

\medskip

\quad{\bf (vii)} I am at last ready to give a formula for $\tilde
g_{\phi}$.   For $\phi\in G$ set

$$\eqalign{\tilde g_{\phi}(x)&=h_{\phi}(x)
  \text{ if }x\in X\setminus H,\cr
&=qg_{\phi}q^{-1}(x)\text{ if }x\in H.\cr}$$

\noindent Now ($\tilde J,\langle \tilde g_{\phi}\rangle_{\phi\in G})$ is
consistent.   \Prf\

\qquad($\alpha$) If $i\in\tilde J$ and $\phi\in G$,

\Centerline{$\tilde g_{\phi}^{-1}[E_i]=(h_{\phi}^{-1}[E_i]\setminus
H)\cup q[g_{\phi}^{-1}[q^{-1}[E_i\cap H]]]\in \tilde\Cal B_J$}

\noindent because $H\in\Cal B_{\tilde J}$ and
$h_{\phi}^{-1}[E]$, $q^{-1}[H\cap E]$, $g_{\phi}^{-1}[E]$ and $q[E]$ all
belong to $\Cal B_{\tilde J}$ for every $E\in\Cal B_{\tilde J}$.   At
the same time, because $\tilde g_{\phi}$ agrees with $h_{\phi}$ on the
conegligible set $X\setminus H$,

\Centerline{$(\tilde g_{\phi}^{-1}[E_i])^{\ssbullet}
=(h_{\phi}^{-1}[E_i])^{\ssbullet}
=\phi E_i^{\ssbullet}$.}

\qquad($\beta$) If $i\in I\setminus\tilde J$, $\phi\in G$ and $x\in X$ then

\Centerline{$g_{\phi}(x)(i)=h_{\phi}(x)(i)=q(x)(i)=x(i)$,}

\noindent and if $x\in H$ then $q^{-1}(x)(i)$ also is equal to $x(i)$;
so $\tilde g_{\phi}(x)(i)=x(i)$.   But this means that 
$\tilde g_{\phi}^{-1}[E_i]=E_i$.

\qquad($\gamma$) If $\phi$, $\psi\in G$ and $x\in X\setminus H$, then

\Centerline{$\tilde g_{\psi}(x)=h_{\psi}(x)\in X\setminus H$}

\noindent by (iv) above.   So

\Centerline{$\tilde g_{\phi}\tilde g_{\psi}(x)
=h_{\phi}h_{\psi}(x)
=h_{\psi\phi}(x)
=\tilde g_{\psi\phi}(x)$}

\noindent because $x\notin H_{\phi,\psi}$.   While if $x\in H$, then

\Centerline{$\tilde g_{\psi}(x)=qg_{\psi}q^{-1}(x)\in H$,}

\noindent so

\Centerline{$\tilde g_{\phi}\tilde g_{\psi}(x)
=qg_{\phi}q^{-1}qg_{\psi}q^{-1}(x)
=qg_{\phi}g_{\psi}q^{-1}(x)
=qg_{\psi\phi}q^{-1}(x)
=\tilde g_{\psi\phi}(x)$.}

\noindent Thus $\tilde g_{\phi}\tilde g_{\psi}=\tilde g_{\psi\phi}$.

\qquad($\delta$) Because $g_{\iota}(x)=h_{\iota}(x)=x$ for every $x$,
$\tilde g_{\iota}(x)=x$ for every $x$.\ \Qed

\medskip

\quad{\bf (viii)} Finally, if $i\in J$ and $\phi\in G$,
$q^{-1}[E_i]=E_i$, so that $q[E_i]=E_i\cap H$.
Accordingly $q(x)\restr J=x\restr J$ for
every $x\in X$, while $q^{-1}(x)\restr J=x\restr J$ for $x\in H$.   So
$g_{\phi}q^{-1}(x)\restr J=g_{\phi}(x)\restr J$ for $x\in H$, and

$$\eqalign{\tilde g_{\phi}(x)\restr J
&=h_{\phi}(x)\restr J=g_{\phi}(x)\restr J\text{ if }x\in X\setminus H,\cr
&=qg_{\phi}q^{-1}(x)\restr J=g_{\phi}q^{-1}(x)\restr J=g_{\phi}(x)\restr J
  \text{ if }x\in H.\cr}$$

\noindent Thus $(\tilde J,\langle \tilde g_{\phi}\rangle_{\phi\in G})$
satisfies all the required conditions.

\medskip

{\bf (c)} The remaining idea we need is the following:  there is a
non-decreasing family $\langle J_{\xi}\rangle_{\xi\le\kappa}$ in $\Cal
J$, for some cardinal $\kappa$, such that $J_{\xi+1}\setminus J_{\xi}$
is countably infinite for every $\xi<\kappa$,
$J_{\xi}=\bigcup_{\eta<\xi}J_{\eta}$ for every limit ordinal
$\eta<\kappa$, and $J_{\kappa}=I$.   \Prf\ Recall that I am already
supposing that $I$ is infinite.   If $I$ is countable, set $\kappa=1$,
$J_0=\emptyset$, $J_1=I$.   Otherwise, set $\kappa=\#(I)$ and let
$\langle i_{\xi}\rangle_{\xi<\kappa}$ be an enumeration of $I$.   For
$i\in I$, $\phi\in G$ let $K_{\phi i}\subseteq I$ be a countable set
such that $F_{\phi i}\in\Cal B_{K_{\phi i}}$.   Choose the $J_{\xi}$
inductively, as follows.   The inductive hypothesis must include the
requirement that $\#(J_{\xi})\le\max(\omega,\#(\xi))$ for every $\xi$.
Start by setting $J_0=\emptyset$.   Given $\xi<\kappa$ and
$J_{\xi}\in\Cal J$ with $\#(J_{\xi})\le\max(\omega,\#(\xi))<\kappa$,
take an infinite set $L\subseteq\kappa\setminus J_{\xi}$ and set
$J_{\xi+1}=J_{\xi}\cup\bigcup_{n\in\Bbb N}L_n$, where

\Centerline{$L_0=L\cup\{i_{\xi}\}$,}

\Centerline{$L_{n+1}=\bigcup_{i\in L_n,\phi\in G}K_{\phi i}$}

\noindent for $n\in\Bbb N$, so that every $L_n$ is countable,

\Centerline{$F_{\phi i}\in\Cal B_{L_{n+1}}$ whenever $i\in L_n$,
$\phi\in G$}

\noindent  and $J_{\xi+1}\in\Cal J$;  since $L\subseteq
J_{\xi+1}\setminus J_{\xi}\subseteq\bigcup_{n\in\Bbb N}L_n$,
$J_{\xi+1}\setminus J_{\xi}$ is countably infinite, and

\Centerline{$\#(J_{\xi+1})=\max(\omega,\#(J_{\xi}))
\le\max(\omega,\#(\xi))=\max(\omega,\#(\xi+1))$.}

\noindent For non-zero limit ordinals $\xi<\kappa$, set
$J_{\xi}=\bigcup_{\eta<\xi}J_{\eta}$;  then

\Centerline{$\#(J_{\xi})
\le\max(\omega,\#(\xi),\sup_{\eta<\xi}\#(J_{\eta}))
\le\max(\omega,\#(\xi))$.}

\noindent Thus the induction proceeds.   Observing that the construction
puts $i_{\xi}$ into $J_{\xi+1}$ for every $\xi$, we see that
$J_{\kappa}$ will be the whole of $I$, as required.\ \Qed

\medskip

{\bf (d)} Now put (b) and (c) together, as follows.   Take
$\langle J_{\xi}\rangle_{\xi\le\kappa}$ from (c).   Set
$f_{\phi 0}(x)=x$ for every
$\phi\in G$, $x\in X$;  then, because $J_0=\emptyset$,
$(J_0,\langle f_{\phi 0}\rangle_{\phi\in G})$ is consistent in the sense
of (b).   Given that $(J_{\xi},\langle f_{\phi\xi}\rangle_{\phi\in G})$
is consistent, where $\xi<\kappa$, use the construction of (b) to find a
family $\langle f_{\phi,\xi+1}\rangle_{\phi\in G}$ such that
$(J_{\xi+1},\langle f_{\phi,\xi+1}\rangle_{\phi\in G})$ is consistent
and $f_{\phi,\xi+1}(x)(i)=f_{\phi\xi}(x)(i)$ for every $i\in J_{\xi}$ and
$x\in X$.   At a non-zero limit ordinal $\xi\le\kappa$, set

$$\eqalign{f_{\phi\xi}(x)(i)&=f_{\phi\eta}(x)(i)
  \text{ if }x\in X,\,\eta<\xi,\,i\in J_{\eta},\cr
&=x(i)\text{ if }i\in I\setminus J_{\xi}.\cr}$$

\noindent (The inductive hypothesis includes the requirement that
$f_{\phi\eta}(x)\restr J_{\zeta}=f_{\phi\zeta}(x)\restr J_{\zeta}$
whenever $\phi\in G$, $x\in X$ and $\zeta\le\eta<\xi$.)   To see that
$(J_{\xi},\langle f_{\phi\xi}\rangle_{\phi\in G})$
is consistent, the only non-trivial point to check is that

\Centerline{$f_{\phi,\xi}f_{\psi,\xi}=f_{\psi\phi,\xi}$}

\noindent for all $\phi$, $\psi\in G$.   But if $i\in J_{\xi}$ there is
some $\eta<\xi$ such that $i\in J_{\eta}$, and in this case

\Centerline{$f_{\psi,\xi}^{-1}[E_i]
=f_{\psi,\eta}^{-1}[E_i]\in\Cal B_{J_{\eta}}$}

\noindent is determined by coordinates in $J_{\eta}$, so that (because
$f_{\phi,\xi}(x)\restr J_{\eta}=f_{\phi,\eta}(x)\restr J_{\eta}$ for
every $x$)

\Centerline{$f_{\phi,\xi}^{-1}[f_{\psi,\xi}^{-1}[E_i]]
=f_{\phi,\eta}^{-1}[f_{\psi,\eta}^{-1}[E_i]]
=f_{\psi\phi,\eta}^{-1}[E_i]
=f_{\psi\phi,\xi}^{-1}[E_i]$;}

\noindent while if $i\in I\setminus J_{\xi}$ then

\Centerline{$f_{\psi\phi,\xi}^{-1}[E_i]=E_i
=f_{\phi,\xi}^{-1}[E_i]=f_{\psi,\xi}^{-1}[E_i]
=f_{\psi,\xi}^{-1}[f_{\phi,\xi}^{-1}[E_i]]$.}

\noindent Thus

\Centerline{$f_{\psi,\xi}^{-1}[f_{\phi,\xi}^{-1}[E_i]]
=f_{\psi\phi,\xi}^{-1}[E_i]$}

\noindent for every $i$, and
$f_{\phi,\xi}f_{\psi,\xi}=f_{\psi\phi,\xi}$.

On completing the induction, set $f_{\phi}=f_{\phi\kappa}$ for every
$\phi\in G$;  it is easy to see that
$\langle f_{\phi}\rangle_{\phi\in G}$ satisfies the conditions of the
theorem.
}%end of proof of 344E

\leader{344F}{Corollary} Let $I$ be any set, and let $\mu$ be a
$\sigma$-finite measure on $X=\{0,1\}^I$.   Suppose that $\mu$ is
the completion of its restriction to the
$\sigma$-algebra $\CalBa_I$ generated by the sets $\{x:x(i)=1\}$ as $i$
runs over $I$.  Write $\frak A$ for the measure algebra of $\mu$.
Let $G$ be a countable semigroup of order-continuous Boolean
homomorphisms from $\frak A$ to itself.   Then we can choose
simultaneously, for each $\phi\in G$, a function $f_{\phi}:X\to X$
representing $\phi$, in such a way that $f_{\phi\psi}=f_{\psi}f_{\phi}$
for all $\phi$, $\psi\in G$;  and if the identity automorphism $\iota$
belongs to $G$, then we may arrange that $f_{\iota}$ is the identity
function on $X$.   In particular, if $\phi\in G$ is invertible and
$\phi^{-1}\in G$, we shall have $f_{\phi^{-1}}=f_{\phi}^{-1}$;  so that
if moreover $\phi$ is measure-preserving, $f_{\phi}$ will be an
automorphism of the measure space $(X,\Sigma,\mu)$.

\proof{ Apply 344E to $\mu\restr\CalBa_I$;  of course $\frak A$ is
canonically isomorphic to the measure algebra of $\mu\restr\CalBa_I$
(322Da).   The functions $f_{\phi}$ provided by 344E still represent the
homomorphisms $\phi$ when re-interpreted as functions on the completed
measure space $(\{0,1\}^I,\mu)$, by 343Ac.
}%end of proof of 344F

\leader{344G}{Corollary} Let $I$ be any set, $\nu_I$ the usual measure
on $\{0,1\}^{I}$, and $\frak B_I$ its measure algebra.   Then any
measure-preserving automorphism of $\frak B_I$ is representable by a
measure space automorphism of $(\{0,1\}^I,\nu_I)$.

\leaveitout{344F}

\leader{344H}{Lemma}\dvArevised{2010} Let $(X,\Sigma,\mu)$ be a perfect
semi-finite measure space.   If $H\in\Sigma$ is a non-negligible set which
includes no atom, there is a
negligible subset of $H$ with cardinal $\frak c$.

\proof{{\bf (a)} Consider first the case in which $\mu$ is atomless,
compact and totally finite, and $H=X$.
Let $\Cal K\subseteq\Cal PX$ be a compact class such that 
$\mu$ is inner regular with respect to $\Cal K$.
Set $S=\bigcup_{n\in\Bbb N}\{0,1\}^n$, and
choose $\langle K_{\sigma}\rangle_{\sigma\in S}$ inductively, as follows.
$K_{\emptyset}$ is to be any non-negligible member of $\Cal K\cap\Sigma$.
Given that $\mu K_{\sigma}>0$, where
$\sigma\in\{0,1\}^n$, take $F_{\sigma}$, $F'_{\sigma}\subseteq K_{\sigma}$ 
to be disjoint non-negligible
measurable sets both of measure at most $3^{-n}$;  such exist because
$\mu$ is atomless (215D).   Choose
$K_{\sigma^{\smallfrown}\fraction{0}}\subseteq F_{\sigma}$,
$K_{\sigma^{\smallfrown}\fraction{1}}\subseteq F'_{\sigma}$ to be 
non-negligible members of $\Cal K\cap\Sigma$.

For each $w\in\{0,1\}^{\Bbb N}$, $\sequencen{K_{w\restr n}}$ is a
decreasing sequence of members of $\Cal K$ all of non-zero measure, so
has non-empty intersection;  choose a point
$x_w\in\bigcap_{n\in\Bbb N}K_{w\restr n}$.   Since
$K_{\sigma^{\smallfrown}\fraction{0}}
\cap K_{\sigma^{\smallfrown}\fraction{1}}=\emptyset$ for every 
$\sigma\in S$, all the $x_w$ are
distinct, and $A=\{x_w:w\in\{0,1\}^{\Bbb N}\}$ has cardinal $\frak c$.
Also

\Centerline{$A\subseteq\bigcup_{\sigma\in\{0,1\}^{n}}K_{\sigma}$}

\noindent which has measure at most $2^n3^{-(n-1)}$ for every $n\ge 1$,
so $\mu^*A=0$ and $A$ is negligible.

\medskip

{\bf (b)} Now consider the case in which $\mu$ is atomless and totally
finite and perfect, but not necessarily compact, while again $H=X$.   
In this case, by 215D, 
we can choose $\sequencen{E_n}$ inductively so that 
$\mu(E_n\cap E)=\bover12\mu E$ whenever $n\in\Bbb N$ and $E$ is an atom of
the subalgebra of $\Cal PX$ generated by $\{E_i:i<n\}$.
Now define $f:X\to\{0,1\}^{\Bbb N}$ by setting
$f(x)=\sequencen{\chi E_n(x)}$ for $x\in X$.   Consider the image measure
$\nu=\mu f^{-1}$ on $Y=f[X]\subseteq\{0,1\}^{\Bbb N}$.   
This is perfect.   \Prf\ If
$g:Y\to\Bbb R$ is $\Tau$-measurable, where $\Tau=\dom\nu$,
and $\nu F>0$, then
$gf:X\to\Bbb R$ is $\Sigma$-measurable and $\mu f^{-1}[F]>0$.   There is
therefore a compact set $K\subseteq gf[f^{-1}[F]]$ such that
$\mu(gf)^{-1}[K]>0$.   In this case, $K\subseteq g[F]$ and 
$\nu g^{-1}[K]>0$.\ \Qed

Next, for every $n\in\Bbb N$ and $\sigma\in\{0,1\}^n$,

\Centerline{$\nu\{y:y\in Y,\,y\restr n=\sigma\}
=\mu\{x:\,\Forall i<n,\,x\in E_i\iff \sigma(i)=1\}=2^{-n}\mu X$.}

\noindent So $\nu$ can have no atom of measure greater than $2^{-n}\mu X$; 
as $n$ is arbitrary, $\nu$ is atomless.   Thirdly, $(Y,\Tau,\nu)$ is
countably separated, because $\sequencen{\{y:y\in Y$, $y(n)=1\}}$ is a
sequence of measurable sets separating the points of $Y$.   By 343K, $\nu$
is compact;  by (a) here, there is a $\nu$-negligible set $B\subseteq Y$ of
cardinal $\frak c$.   Now $f^{-1}[B]$ is $\mu$-negligible, and because
$B\subseteq f[X]$, $\#(f^{-1}[B])\ge\#(B)=\frak c$.   We therefore have a
set $A\subseteq f^{-1}[B]$ with cardinal $\frak c$, and $A$ is
$\mu$-negligible.

\medskip

{\bf (c)} Finally, for the general case in which $\mu$ is just semi-finite
and perfect, and $H$ is a non-negligible subset of $X$ not including an
atom, let $E\subseteq H$ be a set of non-zero finite
measure.   Then the subspace measure $\mu_E$ is
atomless.   Also $\mu_E$ is perfect.   \Prf\ Let $f:E\to\Bbb R$ be a
measurable function.   Define $g:X\to\Bbb R$ by setting 

$$\eqalign{g(x)
&=e^{f(x)}\text{ if }x\in E,\cr
&=0\text{ if }x\in X\setminus E.\cr}$$

\noindent Then $g$ is measurable.   There is therefore a compact set
$K\subseteq g[E]$ such that $\mu g^{-1}[K]>0$.   Now 
$\ln[K]\subseteq f[E]$ is compact and 
$\mu_Ef^{-1}[\ln[K]]=\mu g^{-1}[K]>0$.\ \Qed

By (b), there is a $\mu_E$-negligible
set $A\subseteq E$ with cardinal $\frak c$, and of course $A$ is also a
$\mu$-negligible subset of $H$.
}%end of proof of 344H

\cmmnt{\medskip

\noindent{\bf Remark} I see that in this proof I have slipped into a
notation which is a touch more sophisticated than what I have used so
far.   See 3A1H for a note on the interpretations of `$\{0,1\}^n$',
`$\{0,1\}^{\Bbb N}$' which make sense of the formulae here.
}

\leader{344I}{Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$
be atomless, perfect, complete, strictly localizable, countably
separated measure spaces of the same non-zero magnitude.   Then they are
isomorphic.

\proof{{\bf (a)} The point is that the measure
algebra $(\frak A,\bar\mu)$ of $\mu$ has Maharam type $\omega$.  \Prf\
Let $\sequencen{E_n}$ be a sequence in $\Sigma$ separating the points of
$X$.   Let $\Sigma_0$ be
the $\sigma$-subalgebra of $\Sigma$ generated by $\{E_n:n\in\Bbb N\}$,
and $\frak A_0$ the order-closed subalgebra of $\frak A$ generated by
$\{E_n^{\ssbullet}:n\in\Bbb N\}$;  then $E^{\ssbullet}\in\frak A_0$ for
every $E\in\Sigma_0$, and $(X,\Sigma_0,\mu\restr\Sigma_0)$ is countably
separated.   Let $f:X\to\Bbb R$ be $\Sigma_0$-measurable and injective
(343E).   Of course $f$ is also $\Sigma$-measurable.   If
$a\in\frak A\setminus\{0\}$, express $a$ as $E^{\ssbullet}$ where
$E\in\Sigma$.   Because $(X,\Sigma,\mu)$ is perfect, there is a compact
$K\subseteq\Bbb R$ such that $K\subseteq f[E]$ and
$\mu f^{-1}[K]>0$.  $K$ is surely a Borel set, so
$f^{-1}[K]\in\Sigma_0$ and

\Centerline{$b=f^{-1}[K]^{\ssbullet}\in\frak A_0\setminus\{0\}$.}

\noindent   But because $f$ is injective, we also have
$f^{-1}[K]\subseteq E$ and $b\Bsubseteq a$.   As $a$ is arbitrary,
$\frak A_0$ is order-dense in $\frak A$;   but $\frak A_0$ is
order-closed, so must be the whole of $\frak A$.   Thus $\frak A$ is
$\tau$-generated by the countable set $\{E_n^{\ssbullet}:n\in\Bbb N\}$,
and $\tau(\frak A)\le\omega$.\ \Qed

On the other hand, because $\frak A$ is atomless, and not $\{0\}$, none
of its principal ideals can have finite Maharam type, and it is \Mth,
with type $\omega$.

\medskip

{\bf (b)} Writing $(\frak B,\bar\nu)$ for the measure algebra of $\nu$,
we see that the argument of (a) applies equally to $(\frak B,\bar\nu)$,
so that $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ are atomless
localizable
measure algebras, with Maharam type $\omega$ and the same magnitude.
Consequently they are isomorphic as measure algebras, by 332J.   Let
$\phi:\frak A\to\frak B$ be a measure-preserving isomorphism.

By 343K, both $\mu$ and $\nu$ are (locally) compact.   As they are also
complete and strictly localizable, 343B tells us that
there are functions $g:Y\to X$ and $f:X\to Y$ representing
$\phi$ and $\phi^{-1}$.   Now $fg:Y\to Y$ and $gf:X\to X$
represent the identity automorphisms on $\frak B$, $\frak A$, so by 343F
are equal almost everywhere to the identity functions on $Y$, $X$
respectively.   Set

\Centerline{$E=\{x:x\in X,\,gf(x)=x\}$,
\quad$F=\{y:y\in Y,\,fg(y)=y\}$;}

\noindent then both $E$ and $F$ are conegligible.   Of course
$f[E]\subseteq F$ (since $fgf(x)=f(x)$ for every $x\in E$), and
similarly $g[F]\subseteq E$;  consequently $f\restr E$, $g\restr F$ are
the two halves of a one-to-one correspondence between $E$ and $F$.
Because $\phi$ is measure-preserving, $\mu f^{-1}[H]=\nu H$ and
$\nu g^{-1}[G]=\mu G$ for every $G\in\Sigma$, $H\in\Tau$;
accordingly $f\restr E$ is an isomorphism between the subspace measures
on $E$ and $F$.

\medskip

{\bf (c)} By 344H, there is a
negligible set $A\subseteq E$ with cardinal $\frak c$.   Now $X$ and $Y$,
being countably separated, both have cardinal at most $\frak c$.
(There are injective functions from $X$ and $Y$ to $\Bbb R$.)   Set

\Centerline{$B=A\cup(X\setminus E)$,\quad $C=f[A]\cup(Y\setminus F)$.}

\noindent Then $B$ and $C$ are negligible subsets of $X$, $Y$
respectively, and both have cardinal $\frak c$ precisely, so there is a
bijection $h:B\to C$.   Set

$$\eqalign{f_1(x)&=f(x)\text{ if }x\in X\setminus B=E\setminus A,\cr
&=h(x)\text{ if }x\in B.\cr}$$

\noindent Then, because $\mu$ and $\nu$ are complete, $f_1$ is an
isomorphism between the measure spaces
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$, as required.
}%end of proof of 344I

\leader{344J}{Corollary} Suppose that $E$, $F$ are two Lebesgue
measurable subsets of $\BbbR^r$ of the same non-zero measure.   Then
the subspace measures on $E$ and $F$ are isomorphic.

\leader{344K}{Corollary} (a) A measure space is isomorphic to Lebesgue
measure on $[0,1]$ iff it is an atomless countably separated compact (or
perfect) complete probability space;  in this case it is also isomorphic
to the usual measure on $\{0,1\}^{\Bbb N}$.

(b) A measure space is isomorphic to Lebesgue measure on $\Bbb R$ iff it
is an atomless countably separated compact (or perfect) $\sigma$-finite
measure space which is not totally finite;  in this case it is also
isomorphic to Lebesgue measure on any Euclidean space $\BbbR^r$.

(c) Let $\mu$ be Lebesgue measure on $\Bbb R$.   If $0<\mu E<\infty$ and
we set $\nu F=\Bover1{\mu E}\mu F$ for every measurable $F\subseteq E$,
then $(E,\nu)$ is isomorphic to Lebesgue measure on $[0,1]$.

\leader{344L}{}\cmmnt{ The homogeneity property of Lebesgue
measure described in 344J is repeated in $\{0,1\}^I$ for any infinite $I$.

\medskip

\noindent}{\bf Theorem}\discrversionA{\footnote{Revised 2006.}}{} 
Let $I$ be an infinite set, and $\nu_I$ the usual
measure on $\{0,1\}^I$.   If $E\subseteq\{0,1\}^I$ is a
measurable set of non-zero measure, the subspace
measure on $E$ is isomorphic to $(\nu_IE)\nu_I$.

\proof{ For $J\subseteq I$ let $\nu_J$ be the usual measure on
$X_J=\{0,1\}^J$.

\medskip

{\bf (a)} If $I$ is countably infinite,
then the subspace measure on $E$ is perfect and complete and countably
separated, so is isomorphic to Lebesgue measure on the interval
$[0,\nu_IE]$, by 344I.
But by 344Kc, or otherwise, this is isomorphic,
up to a scalar multiple of the measure, to Lebesgue measure on $[0,1]$,
which is in turn isomorphic to $\nu_I$.

So henceforth we can suppose that $I$ is uncountable.

\medskip

{\bf (b)} By 254Oc there are a countable set $J\subseteq I$ and a set
$E'\subseteq E$, determined by coordinates in $J$, such that
$E\setminus E'$ is
negligible.   Identifying $X_I$ with $X_J\times X_{I\setminus J}$ (254N),
we can think of $E'$ as $V\times X_{I\setminus J}$ where $V$ is
measured by $\nu_J$ (see 254O).   Take $v_0\in V$ and set

\Centerline{$V'=V\setminus\{v_0\}$,
\quad$W'=X_J\setminus\{v_0\}$,
\quad$E''=V'\times X_{I\setminus J}$,
\quad$F''=W'\times X_{I\setminus J}$.}

\noindent Then by (a), applied to $V'$ and $W'$ in turn,
we have a bijection
$g:V'\to W'$ which, up to a scalar multiple of the measure, is an
isomorphism between the subspace measures.   Now the subspace measure on
$V'\times X_{I\setminus J}$ is just the product of the subspace
measure on $V'$ with $\nu_{I\setminus J}$
(251Q(ii)), so if we set $f_0(x,z)=(g(x),z)$ for $x\in V'$ and
$z\in X_{I\setminus J}$, then $f_0:E''\to F''$ is an isomorphism of
the subspace measures on $E''$ and
$F''$, up to a scalar multiple of the
measures as always.   On the other hand, $E\setminus E''$ and
$X_I\setminus F''$ are negligible and both have cardinal
$\#(X_{I\setminus J})=\#(X_I)$, so we have a bijection
$f_1:E\setminus E''\to X_I\setminus F''$.
Putting $f_0$ and $f_1$ together,
we have a bijection $f:E\to X_I$ which, up to a scalar multiple of the
measure, is an isomorphism of the subspace measure on $E$ with $\nu_I$.
}%end of proof of 344L

\exercises{
\leader{344X}{Basic exercises (a)}
%\spheader 344Xa
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces, and suppose
that there are $E\in\Sigma$, $F\in\Tau$ such that $(X,\Sigma,\mu)$ is
isomorphic to the subspace $(F,\Tau_F,\nu_F)$, while $(Y,\Tau,\nu)$ is
isomorphic to $(E,\Sigma_E,\mu_E)$.   Show that $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ are isomorphic.
%344D

\spheader 344Xb Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be perfect
countably separated complete strictly localizable measure spaces with
isomorphic measure algebras.   Show that there are conegligible subsets
$X'\subseteq X$, $Y'\subseteq Y$ such that $X'$ and $Y'$, with the
subspace measures, are isomorphic.
%344I

\spheader 344Xc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be perfect
countably separated complete strictly localizable measure spaces with
isomorphic measure algebras.   Suppose that they are not purely atomic.
Show that they are isomorphic.
%344I

\spheader 344Xd Give an example of two perfect countably separated
complete probability spaces, with isomorphic measure algebras, which are
not isomorphic.
%344I

\spheader 344Xe Let $(Z,\Sigma,\mu)$ be the Stone space of a homogeneous
measure algebra.   Show that if $E$, $F\in\Sigma$ have the same non-zero
finite measure, then the subspace measures on $E$ and $F$ are
isomorphic.
%344J maybe should be in 344Y

\spheader 344Xf Let $(I^{\|},\Sigma,\mu)$ be the split interval with its
usual measure (343J), and $\frak A$ its measure algebra.  (i) Show that
every measure-preserving
automorphism of $\frak A$ is represented by a measure space
automorphism of $I^{\|}$.   (ii) Show that if $E$, $F\in\Sigma$ and
$\mu E=\mu F>0$ then the subspace measures on $E$ and $F$ are
isomorphic.
%344L

\vleader{72pt}{344Y}{Further exercises (a)}
%\spheader 344Ya
Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$, $\Cal I$ a $\sigma$-ideal of $\Sigma$, and $\frak A$ the
quotient $\Sigma/\Cal I$.   Suppose that there is a countable set
$\Cal A\subseteq\Sigma$ separating the points of $X$.   Let $G$ be a
countable semigroup of Boolean homomorphisms from $\frak A$ to itself
such that every member of $G$ can be represented by some function from
$X$ to itself.   Show that a family
$\langle f_{\phi}\rangle_{\phi\in G}$ of such representatives can be
chosen in such a way that $f_{\phi\psi}=f_{\psi}f_{\phi}$ for all
$\phi$, $\psi\in G$;  and if the identity automorphism $\iota$ belongs
to $G$, then we may arrange that $f_{\iota}$ is the identity function on
$X$.
%344B

\spheader 344Yb Let $\frak A$, $\frak B$ be Dedekind $\sigma$-complete
Boolean algebras.   Suppose that each is isomorphic to a principal ideal
of the other.   Show that they are isomorphic.
%344D

\spheader 344Yc Let $I$ be an infinite set, and
write $\CalBa_I$ for the $\sigma$-algebra of subsets of $X=\{0,1\}^I$
generated by the sets $\{x:x(i)=1\}$ as $i$ runs over $I$.   Let
$\mu$ and $\nu$ be $\sigma$-finite measures on $X$, both with domain
$\CalBa_I$, and with measure algebras $(\frak A,\bar\mu)$,
$(\frak B,\bar\nu)$.   Show that any Boolean isomorphism
$\phi:\frak A\to\frak B$ is represented by a permutation $f:X\to X$ such
that $f^{-1}$ represents $\phi^{-1}:\frak B\to\frak A$, and hence that
$(\frak A,\bar\mu)$ is isomorphic to $(\frak B,\bar\nu)$ iff
$(X,\CalBa_I,\mu)$ is isomorphic to $(X,\CalBa_I,\nu)$.
%344E

\spheader 344Yd Let $I$ be any set, and write $\CalBa_I$ for the
$\sigma$-algebra of subsets of $X=\{0,1\}^I$ generated by the sets
$\{x:x(i)=1\}$ as $i$
runs over $I$.   Let $\Cal I$ be an $\omega_1$-saturated $\sigma$-ideal of
$\CalBa_I$,
and write $\frak A$ for the quotient Boolean algebra
$\Cal B/\Cal I$.   Let $G$ be a countable semigroup of order-continuous
Boolean homomorphisms from $\frak A$ to itself.   Show that we can
choose simultaneously, for each $\phi\in G$, a function $f_{\phi}:X\to
X$ representing $\phi$, in such a way that
$f_{\phi\psi}=f_{\psi}f_{\phi}$
for all $\phi$, $\psi\in G$;  and if the identity automorphism $\iota$
belongs to $G$, then we may arrange that $f_{\iota}$ is the identity
function on $X$.   In particular, if $\phi\in G$ is invertible and
$\phi^{-1}\in G$, $f_{\phi}$ will be an automorphism of the structure
$(X,\CalBa_I,\Cal I)$.
%344E

\spheader 344Ye Let $I$ be any set, and write $\CalBa_I$ for the
$\sigma$-algebra of subsets of $X=\{0,1\}^I$ generated by the sets
$\{x:x(i)=1\}$ as $i$
runs over $I$.   Let $\Cal I$, $\Cal J$ be $\omega_1$-saturated
$\sigma$-ideals of $\CalBa_I$.
Show that if the Boolean algebras $\CalBa_I/\Cal I$
and $\CalBa_I/\Cal J$ are isomorphic, so are the structures
$(X,\CalBa_I,\Cal I)$ and $(X,\CalBa_I,\Cal J)$.
%344E
}%end of exercises

\cmmnt{\Notesheader{344} In this section and the last, I have allowed
myself to drift some distance from the avowed subject of this chapter;
but it seemed a suitable place for this material, which is fundamental
to abstract measure theory.   We find that the concepts of \S\S342-343
are just what is needed to characterize Lebesgue measure (344K), and the
characterization shows that among non-negligible measurable subspaces of
$\BbbR^r$ the isomorphism classes are determined by a single parameter,
the measure of the subspace.   Of course a very large number of other
spaces -- indeed, most of those appearing in ordinary applications of
measure theory to other topics -- are perfect and countably separated
(for example, those of 342Xe and 343Ye), and therefore covered by this
classification.   I note that it includes, as a special case, the
isomorphism between Lebesgue measure on $[0,1]$ and the usual measure on
$\{0,1\}^{\Bbb N}$ already described in 254K.

In 344I, the first part of the proof is devoted to showing that a
perfect countably separated measure space has countable Maharam type;  I
ought perhaps to note here that we must resist the temptation to suppose
that
all countably separated measure spaces have countable Maharam type.   In
fact there are countably separated probability spaces with Maharam type
as high as $2^{\frak c}$.   The arguments are elementary but seem to fit
better into \S521 of Volume 5 than here.

I have offered three contexts in which automorphisms of measure algebras
are represented by automorphisms of measure spaces (344A, 344C, 344E).
In the first case, every automorphism can be represented simultaneously
in a consistent way.   In the other two cases, there is, I am sure, no
such consistent family of representations which can be constructed
within ZFC;  but the theorems I give offer consistent simultaneous
representations of countably many homomorphisms.   The question
arises, whether `countably many' is the true natural limit of
the arguments.   In fact it is possible to extend both results to
families of at most $\omega_1$ automorphisms.
%I hope to return to this in Volume 5.

Having successfully characterized Lebesgue measure -- or, what is very
nearly the same thing, the usual measure on $\{0,1\}^{\Bbb N}$ -- it is
natural to seek similar characterizations of the usual measures on
$\{0,1\}^{\kappa}$ for uncountable cardinals $\kappa$.   This seems to
be hard.   A variety of examples (some
touched on in the exercises to \S521)
show that none of the most natural conjectures can be provable in ZFC.

In fact the principal new ideas of this section do not belong
specifically to measure theory;  rather, they belong to the general
theory of $\sigma$-algebras and $\sigma$-ideals of sets.   In the case
of the Schr\"oder-Bernstein-type theorem 344D, this is obvious from the
formulation I give.   (See also 344Yb.)   In the case of 344B and 344E,
I offer generalizations in 344Ya-344Ye.   Of course the applications of
344B here, in 344C and its corollaries, depend on Maharam's theorem and
the concept of `compact' measure space.   The former has no
generalization to the wider context, and the value of the latter is
based on the
equivalences in Theorem 343B, which also do not have simple
generalizations.

The property described in 344J -- a measure space
$(X,\Sigma,\mu)$ in which any two measurable subsets of the same
non-zero measure are isomorphic -- seems to be a natural concept of
`homogeneity' for measure spaces;  it seems unreasonable to ask for all
sets of zero measure to be isomorphic, since finite sets of different
cardinalities can be expected to be of zero measure.   An extra
property, shared by Lebesgue measure and the usual measure on
$\{0,1\}^I$ and by the measure on the split interval (344Kc, 344L, 344Xf)
but not by
counting measure, would be the requirement that
measurable sets of different non-zero finite measures should be
isomorphic up
to a scalar multiple of the measure.   All these examples have the
further property, that all automorphisms of their measure algebras
correspond to automorphisms of the measure spaces.
}%end of notes


\discrpage

\frfilename{mt345.tex}
\versiondate{27.6.06}
\copyrightdate{1995}

\def\upr{\mathop{\text{upr}}}

\def\chaptername{Liftings}
\def\sectionname{Translation-invariant liftings}

\newsection{345}

In this section and the next I complement the work of \S341 by
describing some
important special properties which can, in appropriate circumstances, be
engineered into our liftings.   I begin with some remarks on
translation-invariance.   I restrict
my attention to measure spaces which we have already seen, delaying a
general discussion of translation-invariant measures on groups until
Volume 4.

\vleader{60pt}{345A}{Translation-invariant liftings} I shall
consider two forms of translation-invariance, as follows.

\spheader 345Aa Let $\mu$ be Lebesgue measure on $\BbbR^r$, and
$\Sigma$ its domain.   A lifting $\phi:\Sigma\to\Sigma$ is {\bf
translation-invariant} if $\phi(E+x)=\phi E+x$ for every $E\in\Sigma$,
$x\in\BbbR^r$.   \cmmnt{(Recall from 134A that $E+x=\{y+x:y\in E\}$
belongs to $\Sigma$ for every $E\in\Sigma$, $x\in\BbbR^r$.)}

Similarly, writing $\frak A$ for the measure algebra of $\mu$, a
lifting $\theta:\frak A\to\Sigma$ is {\bf translation-invariant} if
$\theta(E+x)^{\ssbullet}=\theta E^{\ssbullet}+x$ for every $E\in\Sigma$,
$x\in\BbbR^r$.

\cmmnt{It is easy to see that if $\theta$ and $\phi$ correspond to
each other in the manner of 341B, then one is translation-invariant if
and only if the other is.}

\spheader 345Ab Now let $I$ be any set, and let $\nu_I$ be the usual
measure on $X=\{0,1\}^I$, with $\Tau_I$ its domain and $\frak B_I$ its
measure algebra.   For $x$, $y\in X$, define $x+y\in X$ by setting
$(x+y)(i)=x(i)+_2y(i)$ for every $i\in I$\cmmnt{;  that is, give $X$
the group structure of the product group $\Bbb Z_2^I$}.
\cmmnt{This makes $X$ an abelian
group (isomorphic to the additive group $(\Cal PI,\symmdiff)$ of the
Boolean algebra $\Cal PI$, if we match $x\in X$ with
$\{i:x(i)=1\}\subseteq I$).

\cmmnt{Recall that the measure $\nu_I$ is a product measure (254J),
being the product of copies of the fair-coin probability measure on the
two-element set
$\{0,1\}$.}   If $x\in X$,\cmmnt{ then for each $i\in I$ the map
$\epsilon\mapsto\epsilon+_2x(i):\{0,1\}\to\{0,1\}$ is a measure space
automorphism of $\{0,1\}$, since the two singleton sets $\{0\}$ and
$\{1\}$ have the same measure $\bover12$.   It follows at once that} the
map $y\mapsto y+x:X\to X$ is a measure space automorphism.
}

\ifresultsonly{We }\else{Accordingly we can again }\fi
say that a lifting $\theta:\frak B_I\to\Tau_I$,
or $\phi:\Tau_I\to\Tau_I$, is {\bf translation-invariant} if

\Centerline{$\theta(E+x)^{\ssbullet}=\theta E^{\ssbullet}+x$,
\quad$\phi(E+x)=\phi E+x$}

\noindent whenever $E\in\Sigma$ and $x\in X$.

\leader{345B}{Theorem} For any $r\ge 1$, there is a
translation-invariant lifting for Lebesgue measure on $\BbbR^r$.

\proof{{\bf (a)} Write $\mu$ for Lebesgue measure on $\BbbR^r$,
$\Sigma$ for its domain.   Let $\undphi:\Sigma\to\Sigma$ be lower
Lebesgue density (341E).   Then $\undphi$ is translation-invariant in
the sense that $\undphi(E+x)=\undphi E+x$ for every $E\in\Sigma$,
$x\in\BbbR^r$.   \Prf\

$$\eqalignno{\undphi(E+x)
&=\{y:y\in\BbbR^r,\,\lim_{\delta\downarrow 0}
\Bover{\mu(E+x)\cap B(y,\delta)}{\mu(B(y,\delta))}=1\}\cr
&=\{y:y\in\BbbR^r,\,\lim_{\delta\downarrow 0}
\Bover{\mu(E\cap B(y-x,\delta))}{\mu(B(y-x,\delta))}=1\}\cr
\noalign{\noindent (because $\mu$ is translation-invariant)}
&=\{y+x:y\in\BbbR^r,\,\lim_{\delta\downarrow 0}
\Bover{\mu(E\cap B(y,\delta))}{\mu(B(y,\delta))}=1\}\cr
&=\undphi E+x.   \text{  \Qed}\cr}$$

\medskip

{\bf (b)} Let $\phi_0$ be any lifting for $\mu$ such that
$\phi_0E\supseteq\undphi E$ for every $E\in\Sigma$ (341Jb).    Consider

\Centerline{$\phi E=\{y:\tbf{0}\in\phi_0(E-y)\}$}

\noindent for $E\in\Sigma$.
It is easy to check that $\phi:\Sigma\to\Sigma$ is a Boolean
homomorphism because $\phi_0$ is, so that, for instance,

$$\eqalign{y\in\phi E\symmdiff\phi F
&\iff \tbf{0}\in\phi_0(E-y)\symmdiff\phi_0(F-y)\cr
&\iff \tbf{0}\in\phi_0((E-y)\symmdiff(F-y))=\phi_0((E\symmdiff F)-y)\cr
&\iff y\in\phi(E\symmdiff F).\cr}$$

\medskip

{\bf (c)} If $\mu E=0$, then $E-y$ is negligible for every $y\in\Bbb
R^r$, so $\phi_0(E-y)$ is always empty and $\phi E=\emptyset$.

\medskip

{\bf (d)} Next, $\undphi E\subseteq\phi E$ for every $E\in\Sigma$.
\Prf\ If $y\in\undphi E$, then

\Centerline{$\tbf{0}=y-y\in\undphi E-y
=\undphi(E-y)\subseteq\phi_0(E-y)$,}

\noindent so $y\in\phi E$.\ \QeD\  By 341Ib, $\phi$ is a lifting for
$\mu$.

\medskip

{\bf (e)} Finally, $\phi$ is translation-invariant, because if
$E\in\Sigma$ and $x$, $y\in\BbbR^r$ then

$$\eqalign{y\in\phi(E+x)
&\iff\tbf{0}\in\phi_0(E+x-y)=\phi_0(E-(y-x))\cr
&\iff y-x\in\phi E\cr
&\iff y\in\phi E+x.\cr}$$
}%end of proof of 345B

\leader{345C}{Theorem} For any set $I$, there is a translation-invariant
lifting for the usual measure on $\{0,1\}^I$.

\proof{ I base the argument on the same programme as in 345B.   This
time we have to work rather harder, as we have no simple formula for a
translation-invariant lower density.   However, the ideas already used
in 341F-341H are in fact adequate, if we take care, to produce one.

\medskip

{\bf (a)} Since there is certainly a bijection between
$I$ and its cardinal $\kappa=\#(I)$, it is enough to consider the case
$I=\kappa$.   Write $\nu_{\kappa}$ for the usual measure on
$X=\{0,1\}^I=\{0,1\}^{\kappa}$ and $\Tau_{\kappa}$ for its domain.   For each
$\xi<\kappa$ set $E_{\xi}=\{x:x\in X,\,x(\xi)=1\}$, and let
$\Sigma_{\xi}$ be the $\sigma$-algebra generated by
$\{E_{\eta}:\eta<\xi\}$.   Because $x+E_{\eta}$ is either $E_{\eta}$ or
$X\setminus E_{\eta}$, and in either case belongs to $\Sigma_{\xi}$, for
every $\eta<\xi$ and $x\in X$, $\Sigma_{\xi}$ is translation-invariant.
(Consider
the algebra

\Centerline{$\Sigma_{\xi}'=\{E:E+x\in\Sigma_{\xi}$ for every $x\in
X\}$;}

\noindent this must be $\Sigma_{\xi}$.)   Let $\Phi_{\xi}$ be the set of
partial lower densities $\undphi:\Sigma_{\xi}\to\Tau_{\kappa}$ which are
translation-invariant in the sense that $\undphi(E+x)=\undphi E+x$ for
any $E\in\Sigma_{\xi}$, $x\in X$.


\medskip

{\bf (b)(i)} For $\xi<\kappa$,
$\Sigma_{\xi+1}$ is just the algebra of subsets of $X$ generated by
$\Sigma_{\xi}\cup\{E_{\xi}\}$, that is, sets of the form
$(F\cap E_{\xi})\cup(G\setminus E_{\xi})$ where $F$, $G\in\Sigma_{\xi}$
(312N).
Moreover, the expression is unique.   \Prf\ Define $x_{\xi}\in X$ by
setting $x_{\xi}(\xi)=1$, $x_{\xi}(\eta)=0$ if $\eta\ne\xi$.   Then
$x_{\xi}+E_{\eta}=E_{\eta}$ for every $\eta<\xi$, so $x_{\xi}+F=F$ for
every $F\in\Sigma_{\xi}$.   If
$H=(F\cap E_{\xi})\cup(G\setminus E_{\xi})$ where $F$,
$G\in\Sigma_{\xi}$, then

\Centerline{$x_{\xi}+H
=((x_{\xi}+F)\cap(x_{\xi}+E_{\xi}))\cup((x_{\xi}+G)\setminus
(x_{\xi}+E_{\xi}))
=(F\setminus E_{\xi})\cup(G\cap E_{\xi})$,}

\noindent so

\Centerline{$F=(H\cap E_{\xi})\cup((x_{\xi}+H)\setminus E_{\xi})=F_H$,}

\Centerline{$G=(H\setminus E_{\xi})\cup((x_{\xi}+H)\cap E_{\xi})=G_H$}

\noindent are determined by $H$.\ \Qed

\medskip

\quad{\bf (ii)} The functions $H\mapsto F_H$,
$H\mapsto G_H:\Sigma_{\xi+1}\to\Sigma_{\xi}$ defined above are clearly
Boolean homomorphisms;  moreover, if $H$, $H'\in\Sigma_{\xi+1}$ and
$H\symmdiff H'$ is negligible, then

\Centerline{$(F_H\symmdiff F_{H'})\cup(G_H\symmdiff G_{H'})
\subseteq(H\symmdiff H')\cup(x_{\xi}+(H\symmdiff H'))$}

\noindent is negligible.   It follows at once that if
$\xi<\kappa$ and $\undphi\in\Phi_{\xi}$, we can define
$\undphi_1:\Sigma_{\xi+1}\to\Tau_{\kappa}$ by setting

\Centerline{$\undphi_1H
=(\undphi F_H\cap E_{\xi})\cup(\undphi G_H\setminus E_{\xi})$,}

\noindent and $\undphi_1$ will be a lower density.   If
$H\in\Sigma_{\xi}$ then $F_H=G_H=H$, so $\undphi_1H=\undphi H$.
Generally, if $H$, $H'\in\Sigma_{\xi}$ then

\Centerline{$\undphi_1((H\cap E_{\xi})\cup(H'\setminus E_{\xi}))
=(\undphi F_H\cap E_{\xi})\cup(\undphi G_{H'}\setminus E_{\xi})
=(\undphi H\cap E_{\xi})\cup(\undphi H'\setminus E_{\xi})$.}

\medskip

\quad{\bf (iii)} To see that $\undphi_1$ is translation-invariant,
observe that if $x\in X$ and $x(\xi)=0$ then $x+E_{\xi}=E_{\xi}$, so,
for any $F$, $G\in\Sigma_{\xi}$,

$$\eqalign{\undphi_1(x+((F\cap E_{\xi})\cup(G\setminus E_{\xi})))
&=\undphi_1(((F+x)\cap E_{\xi})\cup((G+x)\setminus E_{\xi}))\cr
&=(\undphi(F+x)\cap E_{\xi})\cup(\undphi(G+x)\setminus E_{\xi})\cr
&=((\undphi F+x)\cap E_{\xi})\cup((\undphi G+x)\setminus E_{\xi})\cr
&=x+(\undphi F\cap E_{\xi})\cup(\undphi G\setminus E_{\xi})\cr
&=x+\undphi_1((F\cap E_{\xi})\cup(G\setminus E_{\xi})).\cr}$$

\noindent While if $x(\xi)=1$ then $x+E_{\xi}=X\setminus E_{\xi}$, so

$$\eqalign{\undphi_1(x+((F\cap E_{\xi})\cup(G\setminus E_{\xi})))
&=\undphi_1(((F+x)\setminus E_{\xi})\cup((G+x)\cap E_{\xi}))\cr
&=(\undphi(F+x)\setminus E_{\xi})\cup(\undphi(G+x)\cap E_{\xi})\cr
&=((\undphi F+x)\setminus E_{\xi})\cup((\undphi G+x)\cap E_{\xi})\cr
&=x+(\undphi F\cap E_{\xi})\cup(\undphi G\setminus E_{\xi})\cr
&=x+\undphi_1((F\cap E_{\xi})\cup(G\setminus E_{\xi})).\cr}$$

\noindent So $\undphi_1\in\Phi_{\xi+1}$.

\medskip

\quad{\bf (iv)} Thus every member of $\Phi_{\xi}$ has an extension to a
member of $\Phi_{\xi+1}$.

\medskip

{\bf (c)} Now suppose that $\sequencen{\zeta(n)}$ is a non-decreasing
sequence in $\kappa$ with supremum $\xi<\kappa$.   Then $\Sigma_{\xi}$
is just the $\sigma$-algebra generated by
$\bigcup_{n\in\Bbb N}\Sigma_{\zeta(n)}$.   If we have a sequence
$\sequencen{\undphi_n}$
such that $\undphi_n\in\Phi_{\zeta(n)}$ and $\undphi_{n+1}$ extends
$\undphi_n$ for every $n$, then there is a $\undphi\in\Phi_{\xi}$
extending every $\undphi_n$.   \Prf\ I repeat the ideas of 341G.

\medskip

\quad{\bf (i)} For $E\in\Sigma_{\xi}$, $n\in\Bbb N$ choose
$g_{En}$ such that $g_{En}$ is a conditional
expectation of $\chi E$ on $\Sigma_{\zeta(n)}$;  that is,

\Centerline{$\int_Fg_{En}=\int_F\chi E=\nu_{\kappa}(F\cap E)$}

\noindent for every $E\in\Sigma_{\zeta(n)}$.   Moreover, make these
choices in such a way that ($\alpha$) every $g_{En}$ is
$\Sigma_{\zeta(n)}$-measurable and defined everywhere on $X$ ($\beta$)
$g_{En}=g_{E'n}$ for every $n$ if $E\symmdiff E'$ is negligible.
Now $\lim_{n\to\infty}g_{En}$ exists and is equal
to $\chi E$ almost everywhere, by L\'evy's martingale
theorem (275I).

\medskip

\quad{\bf (ii)} For $E\in\Sigma_{\xi}$, $k\ge 1$, $n\in\Bbb N$ set

\Centerline{$H_{kn}(E)
=\{x:x\in X,\,g_{En}(x)\ge 1-2^{-k}\}\in\Sigma_{\zeta(n)}$,
\quad $\tilde H_{kn}(E)=\undphi_n(H_{kn}(E))$,}

\Centerline{$\undphi E
=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}\tilde H_{km}(E)$.}

\medskip

\quad{\bf (iii)} Every $g_{\emptyset n}$ is zero almost
everywhere, every $H_{kn}(\emptyset)$ is negligible and every
$\tilde H_{kn}(\emptyset)$
is empty;  so $\undphi\emptyset=\emptyset$.   If $E$,
$E'\in\Sigma_{\xi}$ and $E\symmdiff E'$ is negligible, $g_{En}=g_{E'n}$
for every $n$, $H_{nk}(E)=H_{nk}(E')$ and
$\tilde H_{nk}(E)=\tilde H_{nk}(E')$ for all $n$, $k$,
and $\undphi E=\undphi E'$.

\medskip

\quad{\bf (iv)} If $E\subseteq F$ in $\Sigma_{\xi}$, then $g_{En}\le
g_{Fn}$ almost everywhere for every $n$, every $H_{kn}(E)\setminus
H_{kn}(F)$ is negligible, $\tilde H_{kn}(E)\subseteq\tilde H_{kn}(F)$
for every $n$, $k$, and $\undphi E\subseteq \undphi F$.

\medskip

\quad{\bf (v)} If $E$, $F\in\Sigma_{\xi}$ then
$\chi(E\cap F)\geae\chi E+\chi F-1$ so
$g_{E\cap F,n}\geae g_{En}+g_{Fn}-1$ for every $n$.   Accordingly

\Centerline{$H_{k+1,n}(E)\cap H_{k+1,n}(F)\setminus H_{kn}(E\cap F)$}

\noindent is negligible, and (because $\undphi_n$ is a lower density)

\Centerline{$\tilde H_{kn}(E\Bcap F)\supseteq
\undphi_n(H_{k+1,n}(E)\cap H_{k+1,n}(F))
=\tilde H_{k+1,n}(E)\cap\tilde H_{k+1,n}(F)$}

\noindent for all $k\ge 1$, $n\in\Bbb N$.   Now, if $x\in\undphi
E\cap\undphi F$, then, for any $k\ge 1$, there are $n_1$, $n_2\in\Bbb
N$ such that

\Centerline{$x\in\bigcap_{m\ge n_1}\tilde H_{k+1,m}(E)$,
\quad$x\in\bigcap_{m\ge n_2}\tilde H_{k+1,m}(F)$.}

\noindent But this means that

\Centerline{$x\in\bigcap_{m\ge\max(n_1,n_2)}\tilde H_{km}(E\cap F)$.}

\noindent As $k$ is arbitrary, $x\in\undphi (E\cap F)$;  as $x$ is
arbitrary, $\undphi E\cap\undphi F\subseteq\undphi(E\cap F)$.
We know already from (iv) that $\undphi(E\cap F)\subseteq\undphi
E\cap\undphi F$, so $\undphi(E\cap F)=\undphi E\cap\undphi F$.

\medskip

\quad{\bf (vi)} If $E\in\Sigma_{\xi}$, then $g_{En}\to\chi E$ a.e., so
setting

\Centerline{$V=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge
n}H_{km}(E)=\{x:\limsup_{n\to\infty}g_{En}(x)\ge 1\}$,}

\noindent $V\symmdiff E$ is negligible;  but

\Centerline{$\undphi E\symmdiff V
\subseteq\bigcup_{k\ge 1,n\in\Bbb N}H_{kn}(E)\symmdiff\tilde H_{kn}(E)$}

\noindent is also negligible, so $\undphi E\symmdiff E$ is negligible.
Thus $\undphi$ is a partial lower density with domain $\Sigma_{\xi}$.

\medskip

\quad{\bf (vii)}  If $E\in\Sigma_{\zeta(n)}$, then
$E\in\Sigma_{\zeta(m)}$ for every $m\ge
n$, so $g_{Em}\eae\chi E$ for every $m\ge n$;  $H_{km}(E)\symmdiff E$ is negligible for $k\ge 1$, $m\ge n$;

\Centerline{$\tilde H_{km}(E)=\undphi_mE=\undphi_nE$}

\noindent for $k\ge 1$, $m\ge n$; and
$\undphi E=\undphi_nE$.    Thus $\undphi$ extends every $\undphi_n$.

\medskip

\quad{\bf (viii)} I have still to check the translation-invariance of
$\undphi$.   If $E\in\Sigma_{\xi}$ and $x\in X$, consider $g'_n$,
defined by setting

\Centerline{$g'_n(y)=g_{En}(y-x)$}

\noindent for every $y\in X$, $n\in\Bbb N$;  that is, $g'_n$ is the
composition $g_{En}\psi$, where $\psi(y)=y-x$ for $y\in X$.   (I am not
sure whether it is more, or less, confusing to distinguish between the
operations of addition and subtraction in $X$.   Of course
$y-x=y+(-x)=y+x$ for every $y$.)   Because $\psi$ is a measure space
automorphism, and in particular is \imp, we have

\Centerline{$\int_{F+x}g'_n
=\int_{\psi^{-1}[F]}g'_n=\int_Fg_{En}=\nu_{\kappa}(E\cap F)$}

\noindent whenever $F\in\Sigma_{\zeta(n)}$ (235Gc\formerly{2{}35I}).
But because
$\Sigma_{\zeta(n)}$ is itself translation-invariant, we can apply this
to $F-x$ to get

\Centerline{$\int_Fg'_n=\nu_{\kappa}(E\cap(F-x))=\nu_{\kappa}((E+x)\cap F)$}

\noindent for every $F\in\Sigma_{\zeta(n)}$.   Moreover, for any
$\alpha\in\Bbb R$,

\Centerline{$\{y:g'_n(y)\ge\alpha\}=\{y:g_{En}(y)\ge\alpha\}+x
\in\Sigma_{\zeta(n)}$}

\noindent for every $\alpha$, and $g'_n$ is
$\Sigma_{\zeta(n)}$-measurable.   So $g'_n$ is a conditional expectation
of $\chi(E+x)$ on $\Sigma_{\zeta(n)}$, and must be equal almost
everywhere to $g_{E+x,n}$.

This means that if we set

\Centerline{$H'_{kn}=\{y:g'_n(y)\ge 1-2^{-k}\}=H_{kn}(E)+x$}

\noindent for $k$, $n\in\Bbb N$, we shall have
$H'_{kn}\in\Sigma_{\zeta(n)}$ and $H'_{kn}\symmdiff H_{kn}(E+x)$ will be
negligible, so

$$\eqalign{\tilde H_{kn}(E+x)
&=\undphi_n(H_{kn}(E+x))
=\undphi_n(H'_{kn})\cr
&=\undphi_n(H_{kn}(E)+x)
=\undphi_n(H_{kn}(E)) + x
=\tilde H_{kn}(E)+x.\cr}$$

\noindent Consequently

$$\eqalign{\undphi(E+x)
&=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
\tilde H_{kn}(E+x)\cr
&=\bigcap_{k\ge 1}\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
\tilde H_{kn}(E)+x
=\undphi E+x.\cr}$$

\noindent As $E$ and $x$ are arbitrary, $\undphi$ is
translation-invariant and belongs to $\Phi_{\xi}$.\ \Qed

\medskip

{\bf (d)} We are now ready for the proof that there is a
translation-invariant lower density for $\nu_{\kappa}$.   \Prf\ Build inductively a
family $\langle\undphi_{\xi}\rangle_{\xi\le\kappa}$ such that ($\alpha$)
$\undphi_{\xi}\in\Phi_{\xi}$ for each $\xi$ ($\beta$) $\undphi_{\xi}$
extends
$\undphi_{\eta}$ whenever $\eta\le\xi\le\kappa$.   The induction starts
with $\Sigma_0=\{\emptyset,X\}$, $\undphi_0\emptyset=\emptyset$,
$\undphi_0X=X$.   The inductive step to a successor ordinal is dealt
with in (b), and the inductive step to a non-zero ordinal of countable
cofinality is dealt with in (c).   If $\xi\le\kappa$ has uncountable
cofinality, then $\Sigma_{\xi}=\bigcup_{\eta<\xi}\Sigma_{\eta}$, so we
can (and must) take $\undphi_{\xi}$ to be the unique common extension of
all the previous $\undphi_{\eta}$.

The induction ends with
$\undphi_{\kappa}:\Sigma_{\kappa}\to\Tau_{\kappa}$.   Note that
$\Sigma_{\kappa}$ is not in general the whole of $\Tau_{\kappa}$.   But for
every $E\in\Tau_{\kappa}$ there is an $F\in\Sigma_{\kappa}$ such that
$E\symmdiff F$ is negligible (254Ff).   So we can extend
$\undphi_{\kappa}$ to a function $\undphi$ defined on the whole of
$\Tau_{\kappa}$ by setting

\Centerline{$\undphi E=\undphi_{\kappa}F$ whenever $E\in\Tau_{\kappa}$,
$F\in\Sigma_{\kappa}$ and $\nu_{\kappa}(E\symmdiff F)=0$}

\noindent (the point being that $\undphi_{\kappa}F=\undphi_{\kappa}F'$
if $F$, $F'\in\Sigma_{\kappa}$ and $\nu_{\kappa}(E\symmdiff
F)=\nu_{\kappa}(E\symmdiff F')=0$).   It is easy to check that $\undphi$ is a
lower density, and it is translation-invariant because if $E\in\Tau_{\kappa}$,
$x\in X$, $F\in\Sigma_{\kappa}$ and $E\symmdiff F$ is negligible, then
$(E+x)\symmdiff(F+x)=(E\symmdiff F)+x$ is negligible, so

\Centerline{$\undphi(E+x)=\undphi_{\kappa}(F+x)
=\undphi_{\kappa}F+x=\undphi E+x$.  \Qed}

\medskip

{\bf (e)} The rest of the argument is exactly that of parts (b)-(e) of
the proof of 345B;  you have to change $\BbbR^r$ into $X$ wherever it
appears, but otherwise you can use it word for word, interpreting
`$\tbf{0}$' as the identity of the group $X$, that is, the constant
function with value $0$.
}%end of proof of 345C

\leader{345D}{}\cmmnt{ Translation-invariant liftings are of great
importance, and I will return to them in \S447 with a theorem
dramatically generalizing the results above.   Here I shall content
myself with giving one of their basic properties, set out for the two
kinds of translation-invariant lifting we have seen.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be {\it either}
Lebesgue measure on $\BbbR^r$ {\it or} the usual measure on $\{0,1\}^I$
for some set $I$, and let $\phi:\Sigma\to\Sigma$ be a
translation-invariant lifting.   Then for any open set $G\subseteq X$ we
must have $G\subseteq\phi G\subseteq\overline{G}$, and for any closed
set $F$ we must have $\interior F\subseteq\phi F\subseteq F$.

\proof{{\bf (a)} Suppose that $G\subseteq X$ is open and that $x\in G$.
Then there is an open set $U$ such that $\tbf{0}\in U$ and
$x+U-U=\{x+y-z:y,\,z\in U\}\subseteq G$.   \Prf\ ($\alpha$) If
$X=\Bbb R^r$, take $\delta>0$ such that
$\{y:\|y-x\|\le\delta\}\subseteq G$, and
set $U=\{y:\|y-x\|<\bover12\delta\}$.   ($\beta$) If $X=\{0,1\}^I$, then
there is a finite set $K\subseteq I$ such that
$\{y:y\restr K=x\restr K\}\subseteq G$ (3A3K);
set $U=\{y:y(i)=0$ for every $i\in K\}$.\ \Qed

It follows that $x\in\phi G$.   \Prf\ Consider $H=x+U$.   Then
$\mu H=\mu U>0$ so $H\cap\phi H\ne\emptyset$.   Let $y\in U$ be such that
$x+y\in\phi H$.   Then

\Centerline{$x=(x+y)-y\in\phi(H-y)\subseteq\phi G$}

\noindent because

\Centerline{$H-y\subseteq x+U-U\subseteq G$.  \Qed}

\medskip

{\bf (b)} Thus $G\subseteq\phi G$ for every open set $G\subseteq X$.
But it follows at once that if $G$ is open and $F$ is closed,

\Centerline{$\interior F\subseteq\phi(\interior F)\subseteq\phi F$,}

\Centerline{$\overline{G}=X\setminus\interior(X\setminus G)
\supseteq X\setminus\phi(X\setminus G)
=\phi G$,}

\Centerline{$F=X\setminus(X\setminus F)\supseteq
X\setminus\phi(X\setminus F)=\phi F$.}
}%end of proof of 345D

\leader{345E}{}\cmmnt{ I remarked in 341Lg that it is 
undecidable in
ordinary set theory whether there is a lifting for Borel measure on
$\Bbb R$.   It is however known that there can be no
translation-invariant Borel lifting.   The argument depends on the
following fact about measurable sets in $\{0,1\}^{\Bbb N}$.

\medskip

\noindent}{\bf Lemma}\dvArevised{2011}
Give $X=\{0,1\}^{\Bbb N}$ its usual measure
$\nu_{\Bbb N}$, and let $E\subseteq X$ be any non-negligible measurable
set.   Then there is an $n\in\Bbb N$ such that for every $k\ge n$ there are
$x$, $x'\in E$ which differ at $k$ and nowhere else.

\proof{ By 254Fe, there is a set $F$, determined by coordinates in a
finite set, such that
$\nu_{\Bbb N}(E\symmdiff F)\le\bover14\nu_{\Bbb N}E$;  we have
$\nu_{\Bbb N}F\ge\bover34\nu_{\Bbb N}E$, so
$\nu_{\Bbb N}(E\symmdiff F)\le\bover13\nu_{\Bbb N}F$.
Let $n\in\Bbb N$ be such that 
$F$ is determined by coordinates in $\{0,\ldots,n-1\}$.   Take any
$k\ge n$.   Then the map
$\psi:X\to X$, defined by setting $(\psi x)(k)=1-x(k)$,
$(\psi x)(i)=x(i)$ for $i\ne k$, is a measure space automorphism, and

\Centerline{$\nu_{\Bbb N}(\psi^{-1}[E\symmdiff F]\cup(E\symmdiff F))
\le 2\nu_{\Bbb N}(E\symmdiff F)<\nu_{\Bbb N}F$.}

\noindent Take any
$x\in F\setminus((E\symmdiff F)\cup\psi^{-1}[E\symmdiff F])$.   Then
$x'=\psi x$ differs from $x$ at $k$, and only there;  but also
$x'\in F$, by the choice of $n$, so both $x$ and $x'$ belong to $E$.
}%end of proof of 345E

\leader{345F}{\bf Proposition} Let $\mu$ be the restriction of
Lebesgue measure to the algebra $\Cal B$ of Borel subsets
of $\Bbb R$.   Then $\mu$ is translation-invariant, but has no
translation-invariant lifting.

\proof{{\bf (a)} To see that $\mu$ is translation-invariant all we
have to know is that $\Cal B$ is translation-invariant and that Lebesgue
measure is translation-invariant.   I have already cited 134A for the
proof that Lebesgue measure is invariant, and $\Cal B$ is invariant
because $G+x$ is open for every open set $G$ and every $x\in\Bbb R$.

\medskip

{\bf (b)} The argument below is most easily expressed in terms of the
geometry of the Cantor set $C$.
Recall that $C$ is defined as the intersection $\bigcap_{n\in\Bbb N}C_n$
of a sequence of closed subsets of $[0,1]$;  each $C_n$ consists of
$2^n$ closed intervals of length $3^{-n}$;  $C_{n+1}$ is obtained from
$C_n$ by deleting the middle third of each interval of $C_n$.   Any
point of $C$ is uniquely expressible as
$f(e)=\bover23\sum_{n=0}^{\infty}3^{-n}e(n)$ for some
$e\in\{0,1\}^{\Bbb N}$.   (See 134Gb.)
Let $\nu_{\Bbb N}$ be the usual measure of $\{0,1\}^{\Bbb N}$.
Because the map
$e\mapsto e(n):\{0,1\}^{\Bbb N}\to\{0,1\}$ is
measurable for each $n$, $f:\{0,1\}^{\Bbb N}\to\Bbb R$ is
measurable.

We can label the closed intervals constituting $C_n$ as
$\langle J_z\rangle_{z\in\{0,1\}^n}$, taking $J_{\emptyset}$ to be the
unit interval $[0,1]$ and, for $z\in\{0,1\}^n$, taking
$J_{z^{\smallfrown}\fraction{0}}$
to be the left-hand third of $J_z$ and
$J_{z^{\smallfrown}\fraction{1}}$ to be the
right-hand third of $J_z$.   (If the notation here seems odd to you,
there is an explanation in 3A1H.)

For $n\in\Bbb N$ and $z\in\{0,1\}^n$, let $J'_z$ be the open interval with
the same centre as $J_z$ and twice the length.   Then $J'_z\setminus
J_z$ consists of two open intervals of length $3^{-n}/2$ on either side
of $J_z$;  call the left-hand one $V_z$ and the right-hand one $W_z$.
Thus $V_{z^{\smallfrown}\fraction{1}}$ is the right-hand half of the middle third
of
$J_z$, and $W_{z^{\smallfrown}\fraction{0}}$ is the left-hand half of the middle
third of $J_z$.

Construct sets $G$, $H\subseteq\Bbb R$ as follows.

\inset{$G$ is to be the union of
the intervals $V_z$ where $z$ takes the value $1$ an even number of
times, together with the intervals $W_z$ where $z$ takes the value $0$
an odd number of times;}

\inset{$H$ is to be the union of the intervals $V_z$ where $z$ takes
the value $1$ an odd number of times, together with the intervals $W_z$
where $z$ takes the value $0$ an even number of times.}

\noindent $G$ and $H$ are open sets.   The intervals $V_z$, $W_z$
between them cover the whole of the interval
$\ooint{-\bover12,\bover32}$ with the exception of the set $C$ and the
countable set of midpoints of the intervals $J_z$;  so that
$\ooint{-\bover12,\bover32}\setminus(G\cup H)$ is negligible.   We have
to observe that $G\cap H=\emptyset$.   \Prf\ For each $z$,
$J'_{z^{\smallfrown}\fraction{0}}$ and $J'_{z^{\smallfrown}\fraction{1}}$ are disjoint subsets
of $J'_z$.   Consequently $J'_z\cap J'_{w}$ is non-empty just when one
of $z$, $w$ extends the other, and we need consider only the
intersections of the four sets $V_z$, $W_z$, $V_{w}$, $W_{w}$ when
$w$ is a proper extension of $z$;  say $w\in\{0,1\}^n$ and $z= w\restr
m$, where $m<n$.  ($\alpha$)
If in the extension $(w(m),\ldots,w(n-1))$ both values $0$ and $1$
appear, $J'_{w}$ will be a subset of $J_z$, and certainly the four sets
will all be disjoint.   ($\beta$) If $w(i)=0$ for $m\le i<n$, then
$W_{w}\subseteq J_z$ is disjoint from the rest, while $V_{w}\subseteq
V_z$;  but $z$ and $w$ take the value $1$ the same number of times, so
$V_{w}$ is assigned to $G$ iff $V_z$ is, and otherwise both are
assigned to $H$.   ($\gamma$) Similarly, if $w(i)=1$ for $m\le i<n$,
$V_{w}\subseteq J_z$, $W_{w}\subseteq W_z$ and $z$, $w$ take the
value $0$ the same number of times, so $W_z$ and $W_{w}$ are assigned
to the same set.\ \Qed

The following diagram may help you to see what is supposed to be
happening:

\def\Caption{}
\ifdim\pagewidth>467pt\picture{mt345f}{125pt}
\else\picture{mt345f}{104pt}\fi
%122pt made diagram a little indented on the right in \fullsize

\noindent The assignment rule can be restated as follows:

\inset{$V=V_{\emptyset}$ is assigned to $G$, $W=W_{\emptyset}$ is assigned to $H$;

$V_{z^{\smallfrown}\fraction{0}}$ is assigned to the same set as $V_z$, and
$V_{z^{\smallfrown}\fraction{1}}$ to the other;

$W_{z^{\smallfrown}\fraction{1}}$ is assigned to the same set as $W_z$, and
$W_{z^{\smallfrown}\fraction{0}}$ to the other.}

\medskip

{\bf (c)} Now take any $n\in\Bbb N$ and $z\in\{0,1\}^n$.   Consider the
two open intervals $I_0=J'_{z^{\smallfrown}\fraction{0}}$,
$I_1=J'_{z^{\smallfrown}\fraction{1}}$.   These are both of length
$\gamma=2\cdot 3^{-n-1}$ and abut at the centre of $J_z$, so $I_1$ is
just
the translate
$I_0+\gamma$.   I claim that $I_1\cap H=(I_0\cap G)+\gamma$.   \Prf\ Let
$A$ be the set

\Centerline{$\bigcup_{m>n}\{w:w\in\{0,1\}^m,\,w$ extends
$z^{\smallfrown}\fraction{0}\}$,}

\noindent and for $w\in A$ let $w'$ be the finite sequence obtained from
$w$ by changing $w(n)=0$ into $w'(n)=1$ but leaving the other values of
$w$ unaltered.   Then
$V_{w'}=V_w+\gamma$ and $W_{w'}=W_w+\gamma$ for every $w\in A$.   Now

$$\eqalign{I_0\cap G
&=\bigcup\{V_w:w\in A,\,
w\text{ takes the value }1\text{ an even number of times}\}\cr
&\qquad\cup\bigcup\{W_w:w\in A,\,
w\text{ takes the value }0\text{ an odd number of times}\},\cr}$$

\noindent so

$$\eqalign{(I_0\cap G)+\gamma
&=\bigcup\{V_{w'}:w\in A,\,
w\text{ takes the value }1\text{ an even number of times}\}\cr
&\qquad\cup\bigcup\{W_{w'}:w\in A,\,
w\text{ takes the value }0\text{ an odd number of times}\}\cr
&=\bigcup\{V_{w'}:w\in A,\,
w'\text{ takes the value }1\text{ an odd number of times}\}\cr
&\qquad\cup\bigcup\{W_{w'}:w\in A,\,
w'\text{ takes the value }0\text{ an even number of times}\}\cr
&=I_1\cap H.  \text{  \Qed}\cr}$$

\medskip

{\bf (d)} \Quer\ Now suppose, if possible, that $\phi:\Cal B\to\Cal B$
is a translation-invariant lifting.   Note first that $U\subseteq\phi U$
for every open $U\subseteq\Bbb R$.   \Prf\  The argument is exactly that
of 345D as applied to $\Bbb R=\BbbR^1$.\ \QeD\ Consequently

\Centerline{$J'_{\emptyset}=\ooint{-\bover12,\bover32}\subseteq\phi
J'_{\emptyset}$.}

\noindent But as $J'_{\emptyset}\setminus(G\cup H)$ is negligible,

\Centerline{$C\subseteq\ooint{-\bover12,\bover32}\subseteq\phi G\cup\phi
H$.}

\noindent Consider the sets $E=f^{-1}[\phi G]$, $F=\{0,1\}^{\Bbb
N}\setminus E=f^{-1}[\phi H]$.   Because $f$ is measurable and
$\phi G$, $\phi H$ are Borel sets, $E$ and $F$ are measurable subsets of
$\{0,1\}^{\Bbb N}$,
and at least one of them has positive measure for $\nu_{\Bbb N}$.
There must therefore be
$e$, $e'\in\{0,1\}^{\Bbb N}$, differing at exactly one coordinate, such
that either both belong to $E$ or both belong to $F$ (345E).   Let us
suppose that $n$ is such that $e(n)=0$, $e'(n)=1$ and $e(i)=e'(i)$ for
$i\ne n$.   Set $z=e\restr n=e'\restr n$.
Then $f(e)$ belongs to the open interval
$I_0=J'_{z^{\smallfrown}\fraction{0}}$, so
$f(e)\in\phi I_0$ and $f(e)\in\phi G$ iff $f(e)\in\phi(I_0\cap G)$.
But now

\Centerline{$f(e')=f(e)+2\cdot 3^{-n-1}
\in I_1=J'_{z^{\smallfrown}\fraction{1}}$,}

\noindent so

$$\eqalignno{e\in E
&\iff f(e)\in\phi G
\iff f(e)\in\phi(I_0\cap G)\cr
&\iff f(e')\in\phi((I_0\cap G)+2\cdot 3^{-n-1})\cr
\noalign{\noindent (because $\phi$ is translation-invariant)}
&\iff f(e')\in\phi(I_1\cap H)\cr
\noalign{\noindent (by (c) above)}
&\iff f(e')\in\phi H\cr
\noalign{\noindent (because $f(e')\in I_1\subseteq\phi I_1$)}
&\iff e'\in F.\cr}$$

\noindent But this contradicts the choice of $e$.   \Bang

Thus there is no translation-invariant lifting for $\mu$.
}%end of proof of 345F

\cmmnt{\medskip

\noindent{\bf Remark} This result is due to {\smc Johnson 80};  the
proof here follows {\smc Talagrand 82b}.   For references to various
generalizations see {\smc Burke 93}, $\S$3.
}

\exercises{
\leader{345X}{Basic exercises (a)} In 345Ab I wrote `It follows at once
that the map $y\mapsto y+x:X\to X$ is a measure space automorphism'.
Write the details out in full, using 254G or otherwise.
%345A

\spheader 345Xb Let $S^1$ be the unit circle in $\BbbR^2$, and let
$\mu$ be one-dimensional Hausdorff measure on $S^1$ (\S\S264-265).
Show that $\mu$ is translation-invariant, if $S^1$ is given its
usual group operation corresponding to complex multiplication (255M),
and that it has a translation-invariant lifting $\phi$.
\Hint{Identifying $S^1$ with $\ocint{-\pi,\pi}$ with the group operation
$+_{2\pi}$, show that we can set
$\phi E=\ocint{-\pi,\pi}\cap\phi'(\bigcup_{n\in\Bbb Z}E+2\pi n)$, where
$\phi'$ is any translation-invariant lifting for Lebesgue measure.}
%345B

\sqheader 345Xc   Show that there is no lifting $\phi$ of Lebesgue
measure on $\Bbb R$ which is `symmetric' in the sense that
$\phi(-E)=-\phi E$ for every measurable set $E$, writing
$-E=\{-x:x\in E\}$.   \Hint{can $0$ belong to $\phi(\coint{0,\infty})$?}
%345B

\sqheader 345Xd Let $\mu$ be Lebesgue measure on
$X=\Bbb R\setminus\{0\}$.   Show that there is a lifting $\phi$ of $\mu$
such that $\phi(xE)=x\phi E$ for every $x\in X$ and every measurable
$E\subseteq X$, writing $xE=\{xy:y\in E\}$.
%345B

\spheader 345Xe Let $\nu_I$ be the usual measure on $X=\{0,1\}^I$, for
some set $I$, $\Tau_I$ its domain, and $(\frak B_I,\bar\nu_I)$ its measure
algebra.   (i) Show that we can define $\pi_x(a)=a+x$, for $a\in\frak B_I$
and $x\in X$, by the formula $E^{\ssbullet}+x=(E+x)^{\ssbullet}$;  and
that $x\mapsto\pi_x$ is a group homomorphism from $X$ to the group of
measure-preserving automorphisms of $\frak A$.   (ii) Define
$\Sigma_{\xi}$ as in the proof of 345C, and set
$\frak A_{\xi}=\{E^{\ssbullet}:E\in\Sigma_{\xi}\}$.   Say that a partial
lifting $\undtheta:\frak A_{\xi}\to\Tau_I$ is
translation-invariant if $\undtheta(a+x)=\undtheta a+x$ for every
$a\in\frak A_{\xi}$ and $x\in X$.   Show that any such partial lifting
can be extended to a translation-invariant partial lifting on
$\frak A_{\xi+1}$.   (iii) Write out a proof of 345C in the language of
341F-341H.  %341F 341G 341H
%345C

\sqheader 345Xf Let $\undphi$ be a lower density for Lebesgue
measure on $\BbbR^r$ which is translation-invariant in the sense that
$\undphi(E+x)=\undphi E+x$ for every $x\in\BbbR^r$ and every measurable
set $E$.   Show that $\undphi G\supseteq G$ for every open set
$G\subseteq\BbbR^r$.
%345D

\spheader 345Xg Let $\mu$ be 1-dimensional Hausdorff measure on $S^1$, as in 345Xb.   Show that there is no translation-invariant lifting $\phi$ of $\mu$ such that $\phi E$ is a Borel set for every $E\in\dom\mu$.
%345F

\leader{345Y}{Further exercises (a)}  Let $(X,\Sigma,\mu)$ be a
complete measure space, and suppose that $X$ has a group operation
$(x,y)\mapsto xy$ (not necessarily abelian!) such that $\mu$ is
left-translation-invariant, in the sense that
$xE=\{xy:y\in E\}\in\Sigma$ and $\mu(xE)=\mu E$ whenever $E\in\Sigma$
and $x\in X$.   Suppose that $\undphi:\Sigma\to\Sigma$ is a lower
density which is
left-translation-invariant in the sense that $\undphi(xE)=x(\undphi E)$
for every $E\in\Sigma$ and $x\in X$.   Show that there is a
left-translation-invariant lifting $\phi:\Sigma\to\Sigma$ such that
$\undphi E\subseteq\phi E$ for every $E\in\Sigma$.
%345C

\spheader 345Yb Write $\Sigma$ for the $\sigma$-algebra of Lebesgue
measurable subsets of $\Bbb R$, and $\eusm L^0(\Sigma)$ for the linear
space of $\Sigma$-measurable functions from $\Bbb R$ to itself.   Show
that there is a linear operator
$T:L^0(\mu)\to\eusm L^0(\Sigma)$ such that ($\alpha$)
$(Tu)^{\ssbullet}=u$ for every $u\in L^0(\mu)$ ($\beta$)
$\sup_{x\in\Bbb R}|(Tu)(x)|=\|u\|_{\infty}$ for every
$u\in L^{\infty}(\mu)$ ($\gamma$) $Tu\ge 0$ whenever
$u\in L^{\infty}(\mu)$ and $u\ge 0$ ($\delta$) $T$ is
translation-invariant in the sense that
$T(S_xf)^{\ssbullet}=S_xTf^{\ssbullet}$ for every $x\in\Bbb R$ and
$f\in\eusm L^0(\Sigma)$, where $(S_xf)(y)=f(x+y)$ for
$f\in\eusm L^0(\Sigma)$ and $x$, $y\in\Bbb R$ ($\epsilon$) $T$ is
reflection-invariant in the sense that
$T(Rf)^{\ssbullet}=RTf^{\ssbullet}$ for every $f\in\eusm L^0(\Sigma)$,
where $(Rf)(x)=f(-x)$ for $f\in\eusm L^0(\Sigma)$ and $x\in\Bbb R$.
({\it Hint\/}:  for $f\in\eusm L^0(\Sigma)$, set

\Centerline{$p(f^{\ssbullet})
=\inf\{\alpha:\alpha\in[0,\infty],\,\lim_{\delta\downarrow 0}
  \Bover1{2\delta}\mu\{x:|x|\le\delta,\,|f(x)|>\alpha\}=0\}$.}

\noindent Set $V=\{u:u\in L^0(\mu),\,p(u)<\infty\}$ and show that $V$ is
a linear subspace of $L^0(\mu)$ and that $p\restr V$ is a seminorm.
Let $h_0:V\to\Bbb R$ be a linear functional such that
$h_0(\chi\Bbb R)^{\ssbullet}=1$ and $h_0(u)\le p(u)$ for every $u\in V$.
Extend $h_0$ arbitrarily to a linear functional $h_1:L^0(\mu)\to\Bbb R$;
set $h(f^{\ssbullet})=\bover12(h_1(f^{\ssbullet})+h_1(Rf)^{\ssbullet})$.
Set $(Tf^{\ssbullet})(x)=h(S_{-x}f)^{\ssbullet}$.   You will need 223C.)
Show that
there must be a $u\in L^1(\mu)$ such that $u\ge 0$ but $Tu\not\ge 0$.
%341Xg

\spheader 345Yc Show that there is no translation-invariant lifting
$\phi$ of the usual measure on $\{0,1\}^{\Bbb N}$ such that $\phi E$ is a
Borel set for every measurable set $E$.
%345F  mt34bits
}%end of exercises

\cmmnt{
\Notesheader{345} I have taken a great deal of care over the concept of
`translation-invariance'.   I hope that you are already a little
impatient with some of the details as I have written them out;  but
while it is very easy to guess at the structure of such arguments as
part (e) of
the proof of 345B, or (b-iii) and (c-viii) in the proof of 345C, I am
not sure that one can always be certain of guessing correctly.   A fair
test of your intuition will be how quickly you can generate the formulae
appropriate to a non-abelian group operation, as in 345Ya.

Part (b) of the proof of 345C is based on the same idea as the proof
of 341F.   There is a useful simplification because the set $E_{\xi}$ in
345C, corresponding to the set $E$ of the proof of 341F, is independent
of the algebra $\Sigma_{\xi}$ in a very strong sense, so that the
expression of an element of $\Sigma_{\xi+1}$ in the form
$(F\cap E_{\xi})\cup(G\setminus E_{\xi})$ is unique.   Interpreted in
the terms of 341F, we have $w=v=1$, so that the formula

\Centerline{$\undtheta_1((a\Bcap e)\Bcup(b\Bsetminus e))
=\bigl(\undtheta((a\Bcap v)\Bcup(b\Bsetminus v))\cap E\bigr)
   \cup\bigl(\undtheta((a\Bsetminus w)\Bcup(b\Bcap w))
     \setminus E\bigr)$}

\noindent used there becomes

\Centerline{$\undtheta_1((a\Bcap e)\Bcup(b\Bsetminus e))
=(\undtheta a\cap E)\cup(\undtheta b\setminus E)$,}

\noindent matching the formula for $\undphi_1$ in the proof of 345C.

The results of this section are satisfying and natural;  they have
obvious generalizations, many of which are true.   The most important
measure spaces come equipped with a variety of automorphisms, and we can
always ask which of these can be preserved by a lifting.   The answers
are not always obvious;  I offer 345Xc and 346Xc as warnings, and 345Xd
as an encouragement.   345Yb is striking (I have made it as striking as
I can), but slightly off the most natural target;  the sting is in the
last sentence (see 341Xg).
}%end of comment

\discrpage

\frfilename{mt346.tex}
\versiondate{17.12.10}
\copyrightdate{1995}

\def\undtheta{\underline{\theta}}
\def\undphi{{\underline{\phi}}}
\def\undpsi{\underline{\psi}}
\def\upr{\mathop{\text{upr}}}
\def\tmputp{\underline{\tilde\phi}}

\def\chaptername{Liftings}
\def\sectionname{Consistent liftings}

\newsection{346}

I turn now to a different type of condition
which we should naturally prefer our liftings to satisfy.   If
we have a product measure $\mu$ on a product $X=\prod_{i\in I}X_i$ of
probability spaces, then we can look for liftings $\phi$ which `respect
coordinates', that is, are
compatible with the product structure in the sense that they factor
through subproducts (346A).   There seem to be obstacles in the way of
the natural conjecture (346Za), and I give the partial results which are
known.   For \Mth\ spaces $X_i$, there is always a
lifting which respects coordinates (346E), and indeed the
translation-invariant liftings of \S345
on $\{0,1\}^I$ already have this property (346C).   There is always a
lower density for the product measure
which respects coordinates, and we can ask
for a little more (346G);  using the full strength of 346G, we can
enlarge this lower density to a lifting which respects single
coordinates and initial segments of a well-ordered product (346H).   In
the case in which all the factors are copies of each other, we can
arrange for the induced liftings on the factors to be copies also (346I,
346J\cmmnt{, 346Ye}).   I end the section with an important fact about
Stone spaces which is relevant here (346K-346L).

\leader{346A}{Definition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
probability spaces, with product $(X,\Sigma,\mu)$.   I will say that
a lifting $\phi:\Sigma\to\Sigma$ {\bf respects coordinates} if $\phi E$
is determined by coordinates in $J$ whenever $E\in\Sigma$ is determined
by coordinates in $J\subseteq I$.

\cmmnt{
\medskip

\noindent{\bf Remark} Recall that a set $E\subseteq X$ is
`determined by coordinates in $J$' if $x'\in E$ whenever $x\in E$,
$x'\in X$ and $x'\restr J=x\restr J$;  that is, if $E$ is expressible as
$\pi_J^{-1}[F]$ for some $F\subseteq\prod_{i\in J}X_i$, where
$\pi_J(x)=x\restr J$ for every $x\in X$;  that is, if
$E=\pi_J^{-1}[\pi_J[E]]$.   See 254M.   Recall also that in this case,
if $E$ is measured by the product measure on $X$, then $\pi_J[E]$ is
measured by the product measure on $\prod_{i\in J}X_i$ (254Ob).
}%end of comment

\leader{346B}{Lemma} (a)\dvAnew{2010} Let $(X,\Sigma,\mu)$ be a measure
space with a lifting $\phi:\Sigma\to\Sigma$.   Suppose that $Y$ is a set
and $f:X\to Y$ a 
surjective function such that whenever $E\in\Sigma$ is such that
$f^{-1}[f[E]]=E$, then $f^{-1}[f[\phi E]]=\phi E$.
Then we have a lifting 
$\psi$ for the image measure $\mu f^{-1}$ defined by the formula

\Centerline{$f^{-1}[\psi F]=\phi(f^{-1}[F])$ whenever $F\subseteq Y$ and
$f^{-1}[F]\in\Sigma$.}

(b) Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
probability spaces, with product $(Z,\Lambda,\lambda)$.   For
$J\subseteq I$ let $(Z_J,\Lambda_J,\lambda_J)$ be the product of
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in J}$, and $\pi_J:Z\to Z_J$
the canonical map.   Let $\phi:\Lambda\to\Lambda$ be a lifting.   If
$J\subseteq I$ is such that $\phi W$ is determined by coordinates in $J$
whenever $W\in\Lambda$ is determined by coordinates in $J$, then $\phi$
induces a lifting $\phi_J:\Lambda_J\to\Lambda_J$ defined by the formula

\Centerline{$\pi_J^{-1}[\phi_J E]
=\phi(\pi_J^{-1}[E])$ for every $E\in\Lambda_J$.}

\proof{{\bf (a)} Set $\psi F=f[\phi(f^{-1}[F])]$ for
$F\in\dom(\mu f^{-1})$.   Because $f$ is surjective, $\psi Y=Y$,
and it is now elementary to check that
$\psi$ is a lifting for $\mu f^{-1}$.

\medskip

{\bf (b)} By 254Oa, $\lambda_J$ is the image measure $\lambda\pi_J^{-1}$,
so we can use (a).
}%end of proof of 346B

\cmmnt{\medskip

\noindent{\bf Remark} Of course we frequently wish to use part (b) here
with a singleton set $J=\{j\}$.   In this case we must remember that
$(Z_J,\Sigma_J,\lambda_J)$ corresponds to the {\it completion} of
the probability space $(X_j,\Sigma_j,\mu_j)$.
}%end of comment

\leader{346C}{Theorem} Let $I$ be any set, and $\nu_I$ the usual
measure on $\cmmnt{X=\mskip5mu}\{0,1\}^I$.   
Then any translation-invariant lifting for $\nu_I$ respects coordinates.

\proof{ Suppose that $E\subseteq X$ is a measurable set determined by
coordinates in $J\subseteq I$;  take $x\in\phi E$ and $x'\in X$ such
that $x'\restr J=x\restr J$.   Set $y=x'-x$;  then $y(i)=0$ for
$i\in J$, so that $E+y=y$.   Now

\Centerline{$x'=x+y\in\phi E + y=\phi(E+y)=\phi E$}

\noindent because $\phi$ is translation-invariant.   As $x$, $x'$ are
arbitrary, $\phi E$ is determined by coordinates in $J$.   As $E$ and
$J$ are arbitrary, $\phi$ respects coordinates.
}%end of proof of 346C

\leader{346D}{}\cmmnt{ I describe a standard method of constructing
liftings from other liftings.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with measure algebras $\frak A$, $\frak B$;  suppose
that $f:X\to Y$ represents an isomorphism
$F^{\ssbullet}\mapsto f^{-1}[F]^{\ssbullet}:\frak B\to\frak A$.   Then
if $\phi:\Tau\to\Tau$ is a lifting for $\nu$, there is a
corresponding lifting $\phi':\Sigma\to\Sigma$ given by the formula

\Centerline{$\phi'E=f^{-1}[\phi F]$ whenever
$\mu(E\symmdiff f^{-1}[F])=0$.}

\proof{ If we say that $\pi:\frak B\to\frak A$ is the isomorphism
induced by $f$, then

\Centerline{$\phi'E=f^{-1}[\theta(\pi^{-1} E^{\ssbullet})]$,}

\noindent where $\theta:\frak B\to\Tau$ is the lifting corresponding to
$\phi:\Tau\to\Tau$.   Since $\theta$, $\pi^{-1}$ and
$F\mapsto f^{-1}[F]$ are all Boolean homomorphisms, so is $\phi'$, and
it is easy to check that $(\phi'E)^{\ssbullet}=E^{\ssbullet}$ for every
$E\in\Sigma$ and that $\phi'E=\emptyset$ if $\mu E=0$.
}%end of proof of 346D

\cmmnt{\medskip

\noindent{\bf Remark} Compare the construction in 341P.
}

\leader{346E}{Theorem} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of \Mth\
probability spaces, with product
$(X,\Sigma,\mu)$.   Then there is a lifting for $\mu$ which
respects coordinates.

\proof{{\bf (a)} Replacing each $\mu_i$ by its completion does not
change $\mu$ (254I), so we may suppose that all the $\mu_i$ are
complete.   In this case there is for each $i$ an isomorphism between
the measure algebra $(\frak A_i,\bar\mu_i)$ of $\mu_i$ and the measure
algebra $(\frak B_{J_i},\bar\nu_{J_i})$ of some $\{0,1\}^{J_i}$ with its
usual
measure $\nu_{J_i}$ (331L).   We may suppose that the sets $J_i$ are
disjoint.   Each $\nu_{J_i}$ is compact (342Jd), so the isomorphisms are
represented by \imp\ functions $f_i:X_i\to\{0,1\}^{J_i}$ (343Ca).

Set $K=\bigcup_{i\in I}J_i$, and let $\nu_K$ be the usual measure on
$Y=\{0,1\}^K$, $\Tau_K$ its domain.  We have a natural bijection between
$\prod_{i\in I}\{0,1\}^{J_i}$ and $Y$, so we obtain a function
$f:X\to Y$;  literally speaking,

\Centerline{$f(x)(j)=f_i(x(i))(j)$}

\noindent for $i\in I$, $j\in J_i$ and $x\in X$.

\medskip

{\bf (b)} Now $f$ is
\imp\ and induces an isomorphism between the measure algebras $\frak A$,
$\frak B_K$ of $\mu$, $\nu_K$.

\medskip

\Prf{\bf (i)} If $L\subseteq K$ is finite and $z\in\{0,1\}^L$, then,
setting $L_i=L\cap J_i$ for $i\in I$,

$$\eqalignno{\mu\{x:x\in X,\,f(x)\restr L=z\}
&=\mu(\prod_{i\in I}\{w:w\in X_i,\,
  f_i(w)\restr L_i=z\restr L_i\})\cr
&=\prod_{i\in I}\mu_i\{w:w\in X_i,\,
  f_i(w)\restr L_i=z\restr L_i\}\cr
&=\prod_{i\in I}\nu_{J_i}\{v:v\in\{0,1\}^{J_i},\,
  v\restr L_i=z\restr L_i\}\cr
\noalign{\noindent (because every $f_i$ is \imp)}
&=\prod_{i\in I}2^{-\#(L_i)}
=2^{-\#(L)}
=\nu_K\{y:y\in Y,\,y\restr L=z\}.\cr}$$

\noindent So $\mu f^{-1}[C]=\nu_KC$ for every basic cylinder set
$C\subseteq Y$.   By 254G, $f$ is \imp.

\medskip

\quad{\bf (ii)} Accordingly $f$ induces a measure-preserving
homomorphism
$\pi:\frak B_K\to\frak A$.   To see that $\pi$ is surjective, consider

\Centerline{$\Lambda'=\{E:E$ is $\Sigma$-measurable,
$E^{\ssbullet}\in\pi[\frak B_K]\}$.}

\noindent Because $\pi[\frak B_K]$ is a closed subalgebra of $\frak A$
(324Kb), $\Lambda'$ is a $\sigma$-subalgebra of the domain $\Lambda$ of
$\mu$, and of course it contains all $\mu$-negligible sets.   If
$i\in J$ and $G\in\Sigma_i$, then there is an $H\subseteq\{0,1\}^{J_i}$
such that $G\symmdiff f_i^{-1}[H]$ is $\mu_i$-negligible.   Now if
$E=\{x:x\in X,\,x(i)\in G\}$ and $F=\{y:y\in Y,\,y\restr J_i\in H\}$,

\Centerline{$E\symmdiff f^{-1}[F]
=\{x:x(i)\in G\symmdiff f_i^{-1}[H]\}$}

\noindent is $\mu$-negligible, and $E\in\Lambda'$.   But this means
that $\Lambda'\supseteq\Tensorhat_{i\in I}\Sigma_i$, and must therefore
be the whole of $\Lambda$ (254Ff).
\Qed

\medskip

{\bf (c)} By 345C, there is a translation-invariant lifting $\phi$ for
$\nu_K$;  by 346C, this respects coordinates.   By 346D, we have a
corresponding lifting $\phi'$ for $\mu$ such that

\Centerline{$\phi'f^{-1}[F]=f^{-1}[\phi F]$}

\noindent for every $F\in\Tau_K$.   Now suppose that
$E\in\Lambda$ is determined by coordinates in $L\subseteq I$.   Then
there is an $E'$ belonging to the $\sigma$-algebra $\Lambda'_L$
generated by

\Centerline{$\{\{x:x(i)\in G\}:i\in L,\,G\in\Sigma_i\}$}

\noindent such that $\mu(E\symmdiff E')=0$ (254Ob).   Write
$\Tau_L$ for the family of sets in $\Tau_K$ determined by coordinates in
$\bigcup_{i\in L}J_i$.   Then, just as in (b-ii), every member of
$\Lambda'_L$ differs by a negligible set from some set of the form
$f^{-1}[F]$ with $F\in\Tau_L$.   So there is an $F\in\Tau_L$ such that
$E\symmdiff f^{-1}[F]$ is $\mu$-negligible.   Consequently

\Centerline{$\phi'E=\phi'f^{-1}[F]=f^{-1}[\phi F]$.}

\noindent But $\phi$ respects coordinates, so $\phi F$ is determined by
coordinates in $\bigcup_{i\in L}J_i$.   It follows at once that
$f^{-1}[\phi F]$ is determined by coordinates in $L$;  that is, that
$\phi'E$ is determined by coordinates in $L$.   As $E$ and $L$ are
arbitrary, $\phi'$ respects coordinates, and witnesses the truth of the
theorem.
}%end of proof of 346E

\leader{346F}{}\cmmnt{ It seems to be unknown whether 346E is true of
arbitrary probability spaces (346Za);  I give some partial results in
this direction.   The following general method of constructing lower
densities will be useful.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
complete probability spaces, with product $(X\times
Y,\Lambda,\lambda)$.   If $\undphi:\Lambda\to\Lambda$ is a lower
density, then we have a lower density $\undphi_1:\Sigma\to\Sigma$
defined by saying that

\Centerline{$\undphi_1E=\{x:x\in X,\,\{y:(x,y)\in\undphi(E\times Y)\}$
is conegligible in $Y\}$}

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

\proof{ For $E\in\Sigma$, $(E\times Y)\symmdiff\undphi(E\times Y)$ is
negligible, so that

\Centerline{$H_x
=\{y:(x,y)\in (E\times Y)\symmdiff\undphi(E\times Y)\}$}

\noindent is $\nu$-negligible for almost
every $x\in X$ (252D).   Now $E\symmdiff\undphi_1E=\{x:H_x$ is not
negligible$\}$ is negligible, so $\undphi_1E\in\Sigma$.
If $E$, $F\in\Sigma$, then

\Centerline{$\undphi((E\cap F)\times Y)=\undphi((E\times Y)\cap(F\times
Y))=\undphi(E\times Y)\cap\undphi(F\times Y)$,}

\noindent so that

\Centerline{$\{y:(x,y)\in\undphi((E\cap F)\times Y)\}
=\{y:(x,y)\in\undphi(E\times Y)\}\cap\{y:(x,y)\in\undphi(F\times Y)\}$}

\noindent is conegligible iff both
$\{y:(x,y)\in\undphi(E\times Y)\}$ and $\{y:(x,y)\in\undphi(F\times
Y)\}$ are conegligible, and $\undphi_1(E\cap
F)=\undphi_1E\cap\undphi_1F$.

The rest is easy.   Of course $\undphi(\emptyset\times Y)=\emptyset$ so
$\undphi_1\emptyset=\emptyset$.   If $E$, $F\in\Sigma$ and $E\symmdiff
F$ is negligible, then $(E\times Y)\symmdiff(F\times Y)$ is negligible,
$\undphi(E\times Y)=\undphi(F\times Y)$ and $\undphi_1E=\undphi_1F$.
So $\undphi_1$ is a lower density, as claimed.
}%end of proof of 346F

\leader{346G}{Theorem} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces with product
$(X,\Sigma,\mu)$.   For $J\subseteq I$ let $\Sigma_J$ be the
set of members of $\Sigma$ which are determined by coordinates in $J$.
Then there is a lower density $\undphi:\Sigma\to\Sigma$ such that

(i) whenever $J\subseteq I$ and $E\in\Sigma_J$ then
$\undphi E\in\Sigma_J$,

(ii) whenever $J$, $K\subseteq I$ are disjoint, $E\in\Sigma_J$ and
$F\in\Sigma_K$ then $\undphi(E\cup F)=\undphi E\cup\undphi F$.

\proof{ For each $i\in I$, set $Y_i=X_i^{\Bbb N}$, with the product
measure $\nu_i$;  set $Y=\prod_{i\in I}Y_i$, with its product
measure $\nu$;  set $Z_i=X_i\times Y_i$, with its product measure
$\lambda_i$, and $Z=\prod_{i\in I}Z_i$, with its product measure
$\lambda$.   Then the natural identification of
$Z=\prod_{i\in I}X_i\times Y_i$ with
$\prod_{i\in I}X_i\times\prod_{i\in I}Y_i=X\times Y$ makes $\lambda$
correspond to the product of $\mu$ and $\nu$ (254N).

Each $(Z_i,\lambda_i)$ can be identified with an infinite power of
$(X_i,\mu_i)$, and is therefore \Mth\ (334E).
Consequently there is a lifting $\phi:\Lambda\to\Lambda$ which
respects coordinates (346E).   Regarding $(Z,\lambda)$ as the
product of $(X,\mu)$ and $(Y,\nu)$, we see that $\phi$ induces
a lower density $\undphi:\Sigma\to\Sigma$ by the formula of 346F.

If $J\subseteq I$ and $E\in\Sigma$ is determined by coordinates in $J$,
then $E\times Y$ (regarded as a subset of $\prod_{i\in I}Z_i$) is
determined by coordinates in $J$, so $\phi(E\times Y)$ also is.   Now
suppose that $x\in\undphi E$, $x'\in X$ and $x\restr J=x'\restr J$.
Then for any $y\in Y$, $(x\restr J,y\restr J)=(x'\restr J,y\restr J)$,
so $(x,y)\in\phi(E\times Y)$ iff $(x',y)\in\phi(E\times Y)$.   Thus

\Centerline{$\{y:(x',y)\in\phi(E\times
Y)\}=\{y:(x,y)\in\phi(E\times Y)\}$}

\noindent is conegligible in $Y$, and $x'\in\undphi E$.   This shows
that $\undphi E$ is determined by coordinates in $J$.

Now suppose that $J$ and $K$ are disjoint subsets of $I$, that $E$,
$F\in\Sigma$ are determined by coordinates in $J$, $K$ respectively, and
that $x\notin\undphi E\cup\undphi F$.   Then
$A=\{y:(x,y)\notin\phi(E\times Y)\}$ and
$B=\{y:(x,y)\notin\phi(F\times Y)\}$ are non-negligible.   As noted
just above, $\phi(E\times Y)$ is determined by coordinates in $J$, so
$A$ is determined by coordinates in $J$, and can be expressed as
$\{y:y\restr J\in A'\}$, where $A'\subseteq Y_J=\prod_{i\in J}Y_i$.
Because $y\mapsto y\restr J:Y\to Y_J$ is \imp, $A'$ cannot be negligible
in $Y_J$.    Similarly, $B$ can be expressed as $\{y:y\restr K\in B'\}$
for some non-negligible $B'\subseteq Y_K$.

By 251S/251Wm, $A'\times B'\times Y_{I\setminus(J\cup K)}$, regarded as
a subset of $Y$, is non-negligible, that is,

\Centerline{$C=\{y:y\in Y,\,y\restr J\in A',\,y\restr K\in B'\}$}

\noindent is non-negligible.   But

\Centerline{$C=A\cap B
=\{y:(x,y)\notin\phi(E\times Y)\cup\phi(F\times Y)\}
=\{y:(x,y)\notin\phi((E\cup F)\times Y\}$.}

\noindent So $x\notin\undphi(E\cup F)$.   As $x$ is arbitrary,
$\undphi(E\cup F)\subseteq\undphi E\cup\undphi F$;  but of course
$\undphi E\cup\undphi F\subseteq\undphi(E\cup F)$, because $\undphi$ is
a lower density, so that $\undphi(E\cup F)=\undphi E\cup\undphi F$, as
required.
}%end of proof of 346G

\cmmnt{\medskip

\noindent{\bf Remark} See {\smc Macheras Musia\l\ \& Strauss 99} for an
alternative proof.
}%end of comment

\leader{346H}{Theorem} Let $\zeta$ be an ordinal, and
$\langle(X_{\xi},\Sigma_{\xi},\mu_{\xi})\rangle_{\xi<\zeta}$ a family
of probability spaces, with product $(Z,\Lambda,\lambda)$.
For $J\subseteq\zeta$ let $\Lambda_J$ be the set of those $W\in\Lambda$
which are determined by coordinates in $J$.   Then there is a lifting
$\phi:\Lambda\to\Lambda$ such that $\phi W\in\Lambda_J$ whenever
$W\in\Lambda_J$ and $J$ is {\it either} a singleton subset of $\zeta$
{\it or} an initial segment of $\zeta$.

\proof{{\bf (a)} Let $P$ be the set of all lower densities
$\undphi:\Lambda\to\Lambda$ such that, for every $\xi<\zeta$, (i)
whenever $E\in\Lambda_{\xi}$ then $\undphi
E\in\Lambda_{\xi}$ (ii) whenever $E\in\Lambda_{\{\xi\}}$ then $\undphi
E\in\Lambda_{\{\xi\}}$  (iii) whenever $E\in\Lambda_{\xi}$ and
$F\in\Lambda_{\zeta\setminus\xi}$ then $\undphi(E\cup F)=\undphi
E\cup\undphi F$.   By 346G, $P$ is not empty.   Order $P$ by saying that
$\undphi\le\undphi'$ if $\undphi E\subseteq\undphi'E$ for every
$E\in\Lambda$;  then $P$ is a partially ordered set.   Note that if
$\undphi\in P$ then $\undphi Z=Z$ (because $\Lambda_0=\{\emptyset,Z\}$).

\medskip

{\bf (b)} Any non-empty totally ordered subset $Q$ of $P$ has an upper
bound in $P$.   \Prf\ Define $\undphi^*:\Lambda\to\Cal PX$ by setting
$\undphi^*E=\bigcup_{\undphi\in Q}\undphi E$ for every $E\in\Lambda$.
(i)

\Centerline{$\undphi^*\emptyset=\bigcup_{\undphi\in
Q}\emptyset=\emptyset$.}

\noindent (ii) If $E$, $F\in\Lambda$ and $\lambda(E\symmdiff F)=0$
then $\undphi E=\undphi F$ for every $\undphi\in Q$ so
$\undphi^*E=\undphi^*F$.   (iii) If $E$, $F\in\Lambda$ and $E\subseteq
F$ then $\undphi E\subseteq\undphi F$ for every $\undphi\in Q$ so
$\undphi^*E\subseteq\undphi^*F$.   (iv) If $E$, $F\in\Lambda$ and
$x\in\undphi^*E\cap\undphi^*F$, then there are $\undphi_1$,
$\undphi_2\in Q$ such that $x\in\undphi_1E\cap\undphi_2F$;   now either
$\undphi_1\le\undphi_2$ or $\undphi_2\le\undphi_1$, so that

\Centerline{$x\in(\undphi_1E\cap\undphi_1F)
\cup(\undphi_2E\cap\undphi_2F)=\undphi_1(E\cap F)\cup\undphi_2(E\cap F)
\subseteq\undphi^*(E\cap F)$.}

\noindent Accordingly
$\undphi^*E\cap\undphi^*F\subseteq\undphi^*(E\cap F)$ and
$\undphi^*E\cap\undphi^*F=\undphi^*(E\cap F)$.   (v) Taking any
$\undphi_0\in Q$, we have $\undphi_0E\subseteq\undphi^*E$ for every
$E\in\Lambda$, so (because $\lambda$ is complete) $\undphi^*$ is a
lower density, by 341Ib.   (vi) Now suppose that $J\subseteq I$ is
either a singleton $\{\xi\}$ or an initial segment $\xi$, and that
$E\in\Lambda_J$.   Then $\undphi E$ is determined by coordinates in $J$
for every $\undphi\in Q$, so $\undphi^*E$ is determined by coordinates
in $J$.   (vii) Finally, suppose that $\xi<\zeta$ and that
$E\in\Lambda_{\xi}$, $F\in\Lambda_{\zeta\setminus\xi}$.   If
$x\in\undphi^*(E\cup F)$ then there is a $\undphi\in Q$ such that

\Centerline{$x\in\undphi(E\cup F)=\undphi E\cup\undphi F
\subseteq\undphi^* E\cup\undphi^* F$.}

\noindent So $\undphi^*(E\cup F)\subseteq\undphi^*E\cup\undphi^*F$ and
(using (iii) again) $\undphi^*(E\cup F)=\undphi^*E\cup\undphi^*F$.
Thus
$\undphi^*$ belongs to $P$ and is an upper bound for $Q$ in $P$.\ \Qed

By Zorn's Lemma, $P$ has a maximal element $\tmputp$.

\medskip


{\bf (c)} For any $H\in\Lambda$ we may define a function $\undphi_H$ as
follows.   Set $A_H=Z\setminus(\tmputp H\cup\tmputp(Z\setminus H))$,

\Centerline{$\undphi_H E=\tmputp E\cup(A_H\cap \tmputp(H\cup E))$}

\noindent for $E\in\Lambda$.   Then $\undphi_H$ is a lower density.
\Prf\ (i) Because $H\symmdiff\tmputp H$ and $(Z\setminus
H)\symmdiff\tmputp(Z\setminus H)$ are both negligible, $A_H$ is
negligible and $\undphi_HE$ is measurable and
$(\undphi_H E)^{\ssbullet}=(\tmputp E)^{\ssbullet}=E^{\ssbullet}$ for
every $E\in\Lambda$.
(ii) Because $A_H\cap\tmputp H=\emptyset$,
$\undphi_H\emptyset=\emptyset$.
(iii) If $E$, $F\in\Lambda$ and $\lambda(E\symmdiff F)=0$ then
$\tmputp E=\tmputp F$ and $\tmputp(E\cup H)=\tmputp(F\cup H)$, so
$\undphi_H E=\undphi_H F$.
(iv) If $E$, $F\in\Lambda$ and $E\subseteq F$ then $\tmputp
E\subseteq\tmputp F$ and $\tmputp(E\cup H)\subseteq\tmputp(F\cup H)$, so
$\undphi_H E\subseteq\undphi_H F$.
(v) If $E$, $F\in\Lambda$ and $x\in\undphi_H E\cap\undphi_H F$, then

\qquad($\alpha$) if $x\notin A_H$,

\Centerline{$x\in\tmputp E\cap\tmputp
F=\tmputp(E\cap F)\subseteq\undphi_H(E\cap F)$,}

\qquad($\beta$) if $x\in A_H$,

\Centerline{$x\in\tmputp(E\cup H)\cap\tmputp(F\cup
H)=\tmputp((E\cap F)\cup H)\subseteq\undphi_H(E\cap F)$.}

\noindent Thus $\undphi_H E\cap\undphi_H F\subseteq\undphi_H(E\cap F)$
and $\undphi_H E\cap\undphi_H F=\undphi_H(E\cap F)$.\ \Qed

\medskip

{\bf (d)}
It is worth noting the following.

\medskip

\quad{\bf (i)} If $E$, $H\in\Lambda$ and $\tmputp(E\cup H)=\tmputp
E\cup\tmputp H$ then $\undphi_HE=\tmputp E$.   \Prf\ We have

\Centerline{$\undphi_H E
=\tmputp E\cup(A_H\cap\tmputp(E\cup H))
=\tmputp E\cup(A_H\cap\tmputp E)\cup(A_H\cap\tmputp H)
=\tmputp E$}

\noindent because $A_H\cap\tmputp H=\emptyset$.\ \Qed

\wheader{346H}{6}{2}{2}{36pt}

\quad{\bf (ii)} If $H\in\Lambda$ and $\undphi_H\in P$ then $\tmputp
H\cup\tmputp(Z\setminus H)=Z$.   \Prf\ By the maximality of $\tmputp$,
we must have $\undphi_H=\tmputp$.   But

\Centerline{$A_H=\undphi_H(Z\setminus H)\setminus\tmputp(Z\setminus
H)$,}

\noindent so $A_H=\emptyset$, that is, $\tmputp H\cup\tmputp(Z\setminus
H)=Z$.\ \Qed

\medskip

\quad{\bf (iii)} If $E$, $F\in\Lambda$ and
$\tmputp E\cup\tmputp(Z\setminus E)=Z$, then
$\tmputp(E\cup F)=\tmputp E\cup\tmputp F$.   \Prf\

\Centerline{$\tmputp(E\cup F)\setminus\tmputp E
=\tmputp(E\cup F)\cap\tmputp(Z\setminus E)
=\tmputp((E\cup F)\cap(Z\setminus E))
=\tmputp(F\setminus E)
\subseteq\tmputp F$,}

\noindent so  $\tmputp(E\cup F)\subseteq\tmputp E\cup\tmputp F$;  as the
reverse inclusion is true for all $E$ and $F$, we have the result.
\Qed

\medskip

{\bf (e)} If $\xi<\zeta$ and $H\in\Lambda_{\{\xi\}}$, then
$\undphi_H\in P$.

\medskip

\Prf{\bf (i)} If $J\subseteq I$ is either a singleton or an inital
segment, and $E\in\Lambda_J$, then

\qquad($\alpha$) if $\xi\in J$, $E\cup H$ and $\tmputp E$ and
$\tmputp(E\cup H)$ and $A_H$ all belong to $\Lambda_J$, so
$\undphi_HE\in\Lambda_J$.

\qquad($\beta$) If $\xi\notin J$,
$\tmputp(E\cup H)=\tmputp E\cup\tmputp H$, because there is some $\eta$
such that $J\subseteq\eta$ and $\{\xi\}\subseteq\zeta\setminus\eta$;
so $\undphi_HE=\tmputp E\in\Lambda_J$ by (d-i).

\medskip

\quad{\bf (ii)} If $\eta<\zeta$, $E\in\Lambda_{\eta}$ and
$F\in\Lambda_{\zeta\setminus\eta}$, then

\inset{if $\xi<\eta$, $E\cup H\in\Lambda_{\eta}$ so $\tmputp(E\cup F\cup
H)=\tmputp(E\cup H)\cup\tmputp F$, and

$$\eqalign{\undphi_H(E\cup F)
&=\tmputp(E\cup F)\cup(A_H\cap\tmputp(E\cup F\cup H))\cr
&=\tmputp E\cup\tmputp F\cup(A_H\cap\tmputp(E\cup H))
   \cup(A_H\cap\tmputp F)
\subseteq\undphi_H E\cup\undphi_H F;\cr}$$
}

\inset{if $\eta\le\xi$, $F\cup H\in\Lambda_{\zeta\setminus\eta}$ so
$\tmputp(E\cup F\cup H)=\tmputp(E)\cup\tmputp(F\cup H)$, and

$$\eqalign{\undphi_H(E\cup F)
&=\tmputp(E\cup F)\cup(A_H\cap\tmputp(E\cup F\cup H))\cr
&=\tmputp E\cup\tmputp F\cup(A_H\cap\tmputp E))
   \cup(A_H\cap\tmputp(F\cup H))
\subseteq\undphi_H E\cup\undphi_H F;\cr}$$
}

\noindent accordingly $\undphi_H(E\cup F)=\undphi_H E\cup\undphi_H F$.
\Qed


By (d-ii) we have

\Centerline{$\tmputp H\cup\tmputp(Z\setminus H)=Z$}

\noindent whenever $\xi<\zeta$ and $H\in\Lambda_{\{\xi\}}$.

\medskip

{\bf (f)} If $\xi\le\zeta$ and $H\in\Lambda_{\xi}$, then
$\undphi_H\in P$.   \Prf\ Induce on $\xi$.   For $\xi=0$,
$H\in\Lambda_0=\{\emptyset,Z\}$ so $\tmputp H$ is either $\emptyset$ or
$Z$, $A_H=\emptyset$ and $\undphi_H=\tmputp$ belongs to $P$.
For the inductive step to
$\xi\le\zeta$, we have the following.

\medskip

\quad{\bf (i)} If $\eta<\zeta$ and $E\in\Lambda_{\eta}$, then

\qquad($\alpha$) if $\xi\le\eta$, $E\cup H$ and $\tmputp E$ and
$\tmputp(E\cup H)$ and $A_H$ all belong to $\Lambda_{\eta}$, so
$\undphi_HE\in\Lambda_{\eta}$.

\qquad($\beta$) if $\eta<\xi$, then, by the inductive hypothesis,
$\undphi_E\in P$, $\tmputp E=Z\setminus\tmputp(Z\setminus E)$ and
$\tmputp(E\cup H)=\tmputp E\cup\tmputp H$, by (d-ii) and (d-iii) above;
so $\undphi_HE=\tmputp E\in\Lambda_{\eta}$ by (d-i).

\medskip

\quad{\bf (ii)} If $\eta<\zeta$ and $E\in\Lambda_{\{\eta\}}$, then,
by (e), $\tmputp E\cup\tmputp(Z\setminus E)=Z$, so that $\tmputp(E\cup
H)=\tmputp E\cup\tmputp H$, by (d-iii), and $\undphi_HE=\tmputp
E\in\Lambda_{\{\eta\}}$, by (d-i).

\medskip

\quad{\bf (iii)} If $\eta<\zeta$, $E\in\Lambda_{\eta}$ and
$F\in\Lambda_{\zeta\setminus\eta}$, then

\qquad($\alpha$) if $\xi\le\eta$, then $E\cup H\in\Lambda_{\eta}$ and
$F\in\Lambda_{\zeta\setminus\eta}$, so that
$\tmputp(E\cup F\cup H)=\tmputp(E\cup H)\cup\tmputp F$, and

$$\eqalign{\undphi_H(E\cup F)
&=\tmputp(E\cup F)\cup(A_H\cap\tmputp(E\cup F\cup H))\cr
&=\tmputp E\cup\tmputp F\cup(A_H\cap\tmputp(E\cup H))
   \cup(A_H\cap\tmputp F)
\subseteq\undphi_H E\cup\undphi_H F,\cr}$$


\noindent as in (e-ii) above, and accordingly
$\undphi_H(E\cup F)=\undphi_H E\cup\undphi_H F$.

\qquad($\beta$) If $\eta<\xi$ then, as in (ii), using the inductive
hypothesis, we have
$\tmputp(E\cup F\cup H)=\tmputp E\cup\tmputp(F\cup H)$, and (just as in
($\alpha$)) we get $\undphi_H(E\cup F)=\undphi_H E\cup\undphi_H F$.

Thus $\undphi_H\in P$ and the induction continues.
\Qed

\medskip

{\bf (g)} But the case $\xi=\zeta$ of (f) just tells us that

\Centerline{$\tmputp H\cup\tmputp(Z\setminus H)=Z$}

\noindent for every $H\in\Lambda$.   This means that $\tmputp$ is
actually a lifting (since it preserves intersections and complements).
And the definition of $P$ is just what is needed to ensure that it is a
lifting of the right type.
}%end of proof of 346H

\cmmnt{\medskip

\noindent{\bf Remark} This result is due to {\smc Macheras \& Strauss 96b}.
}%end of comment

\leader{346I}{Theorem} Let $(X,\Sigma,\mu)$ be a complete
probability space.   For any set $I$, write $\lambda_I$ for the
product measure on $X^I$, $\Lambda_I$ for its domain and
$\pi_{Ii}(x)=x(i)$ for $x\in X^I$, $i\in I$.   Then there is a lifting
$\psi:\Sigma\to\Sigma$ such that for every set $I$ there
is a lifting $\phi:\Lambda_I\to\Lambda_I$ such that
$\phi(\pi_{Ii}^{-1}[E])=\pi_{Ii}^{-1}[\psi E]$ whenever $E\in\Sigma$ and
$i\in I$.

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

Let $\Psi$ be the set of all liftings for $\mu$.   We are supposing
that for every $\psi\in\Psi$ there is a set $I_{\psi}$ for which there
is no lifting for $\lambda_{I_{\psi}}$ consistent with $\psi$ in the
sense above.   Let $\kappa$ be a cardinal greater than
$\max(\omega,\#(\Psi),\sup_{\psi\in\Psi}\#(I_{\psi}))$.   Let
$\phi_0:\Lambda_{\kappa}\to\Lambda_{\kappa}$ be a lifting satisfying the
conditions of 346H.   346Bb tells us that for every $\xi<\kappa$ we have
a lifting $\psi$ for $\mu$ defined by the formula
$\pi_{\kappa\xi}^{-1}[\psi E]=\phi_0(\pi_{\kappa\xi}^{-1}[E])$.
For $\psi\in\Psi$ set

\Centerline{$K_{\psi}
=\{\xi:\xi<\kappa,\,\phi_0(\pi_{\kappa\xi}^{-1}[E])
  =\pi_{\kappa\xi}^{-1}[\psi E]$ for every $E\in\Sigma\}$.}

\noindent Then $\bigcup_{\psi\in\Psi}K_{\psi}=\kappa$, so
$\kappa\le\max(\omega,\#(\Psi),\sup_{\psi\in\Psi}\#(K_{\psi}))$ and
there is some $\psi\in\Psi$ such that $\#(K_{\psi})>\#(I_{\psi})$.
Take $I\subseteq K_{\psi}$ such that $\#(I)=\#(I_{\psi})$.

We may regard $X^{\kappa}$ as $X^I\times X^{\kappa\setminus I}$, and in
this form we can use the method of 346F to obtain a lower density
$\undphi:\Lambda_I\to\Lambda_I$ from
$\phi_0:\Lambda_{\kappa}\to\Lambda_{\kappa}$.   Now

\Centerline{$\undphi(\pi_{I\xi}^{-1}[E])=\pi_{I\xi}^{-1}[\psi E]$ for
every $E\in\Sigma$, $\xi\in I$.}

\noindent\Prf\ The point is that
$\pi_{I\xi}^{-1}[E]\times X^{\kappa\setminus I}$ corresponds to
$\pi_{\kappa\xi}^{-1}[E]\subseteq X^{\kappa}$, while
$\phi_0(\pi_{\kappa\xi}^{-1}[E])=\pi_{\kappa\xi}^{-1}[\psi E]$ can be
identified with
$\pi_{I\xi}^{-1}[\psi E]\times X^{\kappa\setminus I}$.   Now the
construction
of 346F obviously makes $\undphi(\pi_{I\xi}^{-1}[E])$ equal to
$\pi_{I\xi}^{-1}[\psi E]$.\ \Qed

By 341Jb, there is a lifting $\phi:\Lambda_I\to\Lambda_I$ such that
$\phi W\supseteq\undphi W$ for every $W\in\Lambda_I$.   But now we must
have

$$\eqalign{\pi_{I\xi}^{-1}[\psi E]
&=\undphi(\pi_{I\xi}^{-1}[E])
\subseteq\phi(\pi_{I\xi}^{-1}[E])\cr
&=X^I\setminus\phi(\pi_{I\xi}^{-1}[X\setminus E])
\subseteq X^I\setminus\undphi(\pi_{I\xi}^{-1}[X\setminus E])\cr
&=X^I\setminus\pi_{I\xi}^{-1}[\psi(X\setminus E)]
=X^I\setminus\pi_{I\xi}^{-1}[X\setminus\psi E]
=\pi_{I\xi}^{-1}[\psi E]\cr}$$

\noindent and $\phi(\pi_{I\xi}^{-1}[E])=\pi_{I\xi}^{-1}[\psi E]$ for
every $E\in\Sigma$ and $\xi\in I$.   But since $\#(I)=\#(I_{\psi})$, this
must be impossible, by the choice of $I_{\psi}$.\ \Bang

This contradiction proves the theorem.
}%end of proof of 346I

\leader{346J}{Consistent liftings} Let $(X,\Sigma,\mu)$ be a measure
space.   A lifting $\psi:\Sigma\to\Sigma$ is {\bf consistent} if for
every $n\ge 1$ there is a lifting $\phi_n$ of the product measure on
$X^n$ such that $\phi_n(E_1\times\ldots\times E_n)
=\psi E_1\times\ldots\times\psi E_n$ for all $E_1,\ldots,E_n\in\Sigma$.
Thus\cmmnt{ 346I tells us, in part, that every complete probability
space has a consistent lifting;  it follows that} every non-trivial
complete totally finite measure space has a consistent lifting.

\cmmnt{I do not suppose you will be surprised to be told that not all
liftings on probability spaces are consistent.   What may be surprising
is the fact that one of the standard liftings already introduced is not
consistent.   This depends on a general fact about Stone spaces of
measure algebras which has further important applications, so I present
it as a lemma.}

\leader{346K}{Lemma} Let $(Z,\Tau,\nu)$ be the Stone space of the
measure algebra of
Lebesgue measure on $[0,1]$, and let $\lambda$ be the product
measure on $Z\times Z$, with $\Lambda$ its domain.   Then there is a set
$W\in\Lambda$, with $\lambda W<1$, such that $\lambda^*\tilde W=1$,
where

\Centerline{$\tilde W
=\bigcup\{G\times H:G,\,H\subseteq Z$ are open-and-closed,
  $(G\times H)\setminus W$ is negligible$\}$.}

\cmmnt{\medskip

\noindent{\bf Remark} For the sake of anybody who has already become
acquainted with the alternative measures which can be put on the product
of topological measure spaces, I ought to insist that the `product
measure' $\lambda$ here is, as always in this volume, the ordinary
completed product measure as defined in Chapter 25.
}

\proof{{\bf (a)} Let $\sequencen{E_n}$ be a sequence of measurable
subsets of $[0,1]$, stochastically independent for Lebesgue measure
$\mu$ on $[0,1]$, such that $\mu E_n=\bover1{n+2}$ for each $n$.
Set $a_n=E_n^{\ssbullet}$ in the measure of algebra of $\mu$, and
$E_n^*=\widehat{a}_n$ the corresponding compact open subset of $Z$.
Set $W=\bigcup_{n\in\Bbb N}E_n^*\times E_n^*$.   Then

\Centerline{$\lambda W\le\sum_{n=0}^{\infty}(\nu
E_n)^2=\sum_{n=2}^{\infty}\Bover1{n^2}<1$.}

\Quer\ Suppose, if possible, that $\lambda^*\tilde W<1$.   Then there
are sequences $\sequencen{G_n}$, $\sequencen{H_n}$ in $\Tau$ such that
$\tilde W\subseteq\bigcup_{n\in\Bbb N}G_n\times H_n$ and
$\lambda(\bigcup_{n\in\Bbb N}G_n\times H_n)<1$.   Recall from
322Rc that

\Centerline{$\nu F
=\inf\{\nu G:G$ is compact and open, $F\subseteq G\}$}

\noindent for every $F\in\Tau$.   Accordingly we can find compact open
sets $\tilde G_n$, $\tilde H_n$ such that $G_n\subseteq \tilde G_n$,
$H_n\subseteq
\tilde H_n$ for every $n\in\Bbb N$ and

\Centerline{$\sum_{n=0}^{\infty}\nu(\tilde G_n\setminus
G_n)+\sum_{n=0}^{\infty}\nu(\tilde H_n\setminus H_n)
<1-\lambda(\bigcup_{n\in\Bbb N}G_n\times H_n)$,}

\noindent so that
$\lambda(\bigcup_{n\in\Bbb N}\tilde G_n\times \tilde H_n)<1$.

Let $\Cal U_0$ be the family

\Centerline{$\{Z\}\cup\{E_n^*:n\in\Bbb N\}\cup\{Z\setminus
\tilde G_n:n\in\Bbb N\}\cup\{Z\setminus \tilde H_n:n\in\Bbb N\}$,}

\noindent so that $\Cal U_0$ is a countable subset of $\Tau$.   Let
$\Cal U$ be the set of finite intersections $U_0\cap U_1\cap\ldots\cap U_n$
where $U_0,\ldots,U_n\in\Cal U_0$, so that $\Cal U$ also is a
countable subset of $\Tau$, and $\Cal U$ is closed under $\cap$.

\medskip

{\bf (b)} For $U\in\Cal U$, define $Q(U)$ as follows.   If
$\nu U=0$, then $Q(U)=U$.   Otherwise,

\Centerline{$Q(U)
=Z\setminus\bigcup\{E_n^*:n\in\Bbb N,\,\nu(E_n^*\cap U)>0\}$.}

\noindent Then $\nu Q(U)$ is always $0$.   \Prf\ Of course this is
true if $\nu U=0$, so suppose that $\nu U>0$.   Set
$I=\{n:\nu(E_n^*\cap U)=0\}$.   Then we have
$\nu U'>0$, where $U'=U\setminus\bigcup_{n\in I}E_n^*$, and
$Z\setminus E_n^*\supseteq U'$ for every $n\in I$.   Because
$\sequencen{E_n}$ is stochastically independent for $\mu$,
$\sequencen{E_n^*}$ is stochastically independent for $\nu$, while

\Centerline{$\nu(\bigcup_{n\in I}E_n^*)\le 1-\nu U'<1$.}

\noindent By the
Borel-Cantelli lemma (273K), $\sum_{n\in I}\nu E_n^*<\infty$.
Consequently
$\sum_{n\in\Bbb N\setminus I}\nu E_n^*=\infty$, because $\sum_{n=0}^{\infty}\bover1{n+2}$ is infinite, so

\Centerline{$\nu(Z\setminus Q(U))
=\nu(\bigcup_{n\in\Bbb N\setminus I}E_n^*)=1$,}

\noindent and $\nu Q(U)=0$.\ \Qed

\medskip

{\bf (c)} Set $Q_0=\bigcup_{U\in\Cal U}Q(U)$;  because $\Cal U$ is
countable, $Q_0$ is negligible.   Accordingly $(Z\setminus Q_0)^2$ has
measure $1$ and cannot be included in $\bigcup_{n\in\Bbb N}\tilde
G_n\times
\tilde H_n$;  take $(w,z)\in(Z\setminus Q_0)^2\setminus\bigcup_{n\in\Bbb
N}\tilde G_n\times \tilde H_n$.

\medskip

{\bf (d)} We can find sequences $\sequencen{C_n}$, $\sequencen{D_n}$,
$\sequencen{U_n}$ and $\sequencen{V_n}$ in $\Cal U$ such that

\inset{$w\in U_{n+1}\subseteq U_n$, $z\in V_{n+1}\subseteq V_n$,
$(U_{n+1}\times V_{n+1})\cap(\tilde G_n\times \tilde H_n)=\emptyset$,}

\inset{$\nu C_n>0$, $\nu D_n>0$,}

\inset{$C_n\subseteq U_{n}$, $D_n\subseteq V_{n+1}$,}

\inset{$C_n\times V_{n+1}\subseteq W$, $U_{n+1}\times D_n\subseteq W$}

\noindent for every $n\in\Bbb N$.   \Prf\ Build the sequences
inductively, as follows.   Start with $U_0=V_0=Z$.   Given that
$w\in U_n\in\Cal U$ and $z\in V_n\in\Cal U$,
then we know that $(w,z)\notin
\tilde G_n\times \tilde H_n$.   If $w\notin \tilde G_n$, set
$U'_n=U_n\setminus \tilde G_n$,
$V'_n=V_n$;  otherwise set $U'_n=U_n$, $V'_n=V_n\setminus \tilde H_n$.
In
either case, we have $w\in U'_n\in\Cal U$, $z\in V'_n\in\Cal U$ and
$(U'_n\times V'_n)\cap(\tilde G_n\times \tilde H_n)=\emptyset$.

Because $U'_n\in\Cal U$, $w\notin Q(U'_n)$.   But $w\in U'_n$, so this
must be because $\nu U'_n>0$.    Now $z\notin Q(U'_n)$, so
$z\in\bigcup\{E^*_k:k\in\Bbb N,\,\nu(E^*_k\cap U'_n)>0\}$.   Take
some $k\in\Bbb N$ such that $z\in E^*_k$ and $\nu(E^*_k\cap
U'_n)>0$, and set

\Centerline{$V_{n+1}=V'_n\cap E^*_k$,\quad $C_n=E^*_k\cap U'_n$,}

\noindent so that

\Centerline{$z\in V_{n+1}\in\Cal U$,\quad $C_n\subseteq U_n$, \quad
$C_n\times V_{n+1}\subseteq E^*_k\times E^*_k\subseteq W$,\quad $\nu
C_n>0$.}

\noindent Next, $z\notin Q(V_{n+1})$ and $\nu V_{n+1}>0$;  also
$w\notin Q(V_{n+1})$, so there is an $l$ such that $w\in E^*_l$ and
$\nu(E^*_l\cap V_{n+1})>0$.   Set

\Centerline{$U_{n+1}=U'_n\cap E^*_l$,\quad $D_n=E^*_l\cap V_{n+1}$,}

\noindent so that

\Centerline{$w\in U_{n+1}\in\Cal U$,\quad $D_n\subseteq V_{n+1}$, \quad
$U_{n+1}\times D_{n}\subseteq E^*_l\times E^*_l\subseteq W$,\quad
$\nu D_n>0$,}

\Centerline{$(U_{n+1}\times V_{n+1})\cap(\tilde G_n\times \tilde
H_n)\subseteq
(U'_n\times V'_n)\cap(\tilde G_n\times \tilde H_n)=\emptyset$,}

\noindent and continue the process.
\Qed

\medskip

{\bf (e)} Setting $C=\bigcup_{n\in\Bbb N}C_n$
and $D=\bigcup_{n\in\Bbb N}D_n$ we see that $C\times D\subseteq W$.
\Prf\ If $m\le n$,
$D_n\subseteq V_{n+1}\subseteq V_{m+1}$, so $C_m\times D_n\subseteq W$.
If $m>n$, $C_m\subseteq U_m\subseteq U_{n+1}$, so $C_m\times
D_n\subseteq W$.\ \Qed

Recall from 321K that the measurable sets of $Z$ are precisely those of
the form $G\symmdiff M$ where $M$ is nowhere dense and negligible and
$G$ is compact and open.   There must therefore be compact open sets
$G$, $H\subseteq Z$ such that $G\symmdiff C$ and $H\symmdiff D$ are
negligible.   Consequently

\Centerline{$(G\times H)\setminus W\subseteq ((G\setminus C)\times
Z)\cup(Z\times(H\setminus D))$}

\noindent is negligible, and

\Centerline{$G\times H\subseteq\tilde W\subseteq\bigcup_{n\in\Bbb
N}\tilde G_n\times \tilde H_n$.}

\noindent But because $G\times H$ is compact (3A3J), and all the
$\tilde G_n\times \tilde H_n$ are open, there must be some $n$ such that
$G\times H\subseteq\bigcup_{k\le n}\tilde G_k\times\tilde H_k=S$ say.
Now $(U_{k+1}\times V_{k+1})\cap(\tilde G_k\times \tilde H_k)=\emptyset$
for every $k$, so

\Centerline{$(C_{n+2}\times D_{n+2})\cap(G\times H)
\subseteq(U_{n+1}\times V_{n+1})\cap S
=\emptyset$,}

\noindent and either $C_{n+2}\cap G=\emptyset$ or
$D_{n+2}\cap H=\emptyset$.   Since

\Centerline{$C_{n+2}\setminus G\subseteq C\setminus G$,
\quad$D_{n+2}\setminus H\subseteq D\setminus H$}

\noindent are both negligible, one of $C_{n+2}$, $D_{n+2}$ is
negligible.   But the construction took care to ensure that all the
$C_k$, $D_k$ were non-negligible.\ \Bang

\medskip

{\bf (f)} Thus $\lambda^*\tilde W=1$, as required.
}%end of proof of 346K

\leader{346L}{Proposition} Let $(Z,\Tau,\nu)$ be the Stone space of
the measure algebra of Lebesgue measure on $[0,1]$.   Let
$\psi:\Tau\to\Tau$ be the canonical lifting, defined by setting
$\psi E=G$ whenever $E\in\Tau$, $G$ is open-and-closed and
$E\symmdiff G$ is negligible\cmmnt{ (341O)}.   Then $\psi$ is not
consistent.

\proof{ \Quer\ Suppose, if possible, that $\phi$ is a lifting on
$Z\times Z$ such that $\phi(E\times F)=\psi E\times\psi F$ for every
$E$, $F\in\Tau$.   Let $W\subseteq Z\times Z$ be a set as in 346K, and
consider $\phi W$.   If $G$, $H\subseteq Z$ are open-and-closed and
$(G\times H)\setminus W$ is negligible, then

\Centerline{$G\times H=\psi G\times\psi H=\phi(G\times H)\subseteq\phi
W$;}

\noindent that is, in the language of 346K, we must have    $\tilde
W\subseteq\phi W$.   But this means that

\Centerline{$\lambda(\phi W)\ge\lambda^*\tilde W=1>\lambda
W$,}

\noindent which is impossible.\ \Bang

Thus $\psi$ fails the first test and cannot be consistent.
}%end of proof of 346L

\exercises{\leader{346X}{Basic exercises (a)}
%\spheader 346Xa
Let $(X,\Sigma,\mu)$ be a measure space and
$\sequencen{\undphi_n}$ a sequence of lower densities for $\mu$.
(i) Show that $E\mapsto\bigcap_{n\in\Bbb N}\undphi_nE$ and
$E\mapsto\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}\undphi_mE$ are also lower
densities for $\mu$.   (ii) Show that if $\mu$ is complete and $\Cal F$
is any filter on $\Bbb N$, then
$E\mapsto\bigcup_{F\in\Cal F}\bigcap_{n\in F}\undphi_nE$ is a lower
density for $\mu$.

\spheader 346Xb Let $(X,\Sigma,\mu)$ be a strictly localizable
measure space, and $G$ a countable group of measure space automorphisms
from $X$ to itself.   Show that there is a lower density
$\undphi:\Sigma\to\Sigma$ which is $G$-invariant in the sense that
$\undphi(g^{-1}[E])=g^{-1}[\undphi E]$ for every $E\in\Sigma$
and $g\in G$.
\Hint{set $\undphi E=\bigcap_{g\in G}g[\undphi_0(g^{-1}[E])]$.}
%346Xa

\sqheader 346Xc   Show that there is no lifting $\phi$ of Lebesgue
measure on $[0,1]^2$ which is `symmetric' in the sense that
$\phi(E^{-1})=(\phi E)^{-1}$ for every measurable set $E$, writing
$E^{-1}=\{(y,x):(x,y)\in E\}$.   \Hint{345Xc.}
%346E

\spheader 346Xd Let $(X,\Sigma,\mu)$
be a measure space and $\undphi$ a lower density for $\mu$.  Take
$H\in\Sigma$ and set $A=X\setminus(\undphi H\cup\undphi(Z\setminus H))$,
$\undphi' E=\undphi E\cup(A\cap \undphi(H\cup E))$ for $E\in\Sigma$.
Show that $\undphi'$ is a lower density.
%346H

\spheader 346Xe\dvAnew{2010}
Describe the connections between 346B, 346D and 346F.
%346F

\sqheader 346Xf Suppose, in 341H, that $(X,\Sigma,\mu)$ is a product of
probability spaces, and that in the proof, instead of taking
$\langle a_{\xi}\rangle_{\xi<\kappa}$ to run over the whole measure
algebra $\frak A$, we take it to run over the elements of $\frak A$
expressible as $E^{\ssbullet}$ where $E\in\Sigma$ is determined by a
single coordinate.   Show that the resulting lower density $\undtheta$
respects coordinates in the sense that $\undtheta E^{\ssbullet}$ is
determined by
coordinates in $J$ whenever $E\in\Sigma$ is determined by coordinates in
$J$.   (Compare {\smc Macheras \& Strauss 95}, Theorem 2.)
%346G

\sqheader 346Xg\dvAformerly{3{}46Xe;  revised 2010}
Let $\undphi$ be lower Lebesgue density on $\Bbb R$, and
$\phi$ a translation-invariant lifting for Lebesgue measure on
$\Bbb R$ such that $\phi E\supseteq\undphi E$ for every measurable set
$E$.   Show that $\phi$ is consistent.  ({\it Hint\/}:  given $n\ge 1$,
let $\undphi_n$ be lower Lebesgue density on $\BbbR^n$.
Let $\Cal I$ be the ideal generated by

\Centerline{$\{W:\tbf{0}\in\undphi_n(\BbbR^n\setminus W)\}
  \cup\bigcup_{i<n}\{\pi_i^{-1}[E]:0\in\phi(\Bbb R\setminus E)\}$;}

\noindent show that $\BbbR^n\notin\Cal I$, so that we can use the
method of 345B to construct a lifting for Lebesgue measure on $\BbbR^n$.)
%346J   do we need \phi to be t-i?

\spheader 346Xh Show that Lemma 346K is valid for any $(Z,\Tau,\nu)$
which is the Stone space of an atomless probability space.
%346K

\leader{346Y}{Further exercises (a)}
%\spheader 346Ya 
Let $(X_1,\Sigma_1,\mu_1),\ldots,(X_n,\Sigma_n,\mu_n)$
be probability spaces with product $(X,\Sigma,\mu)$.   Show that there
is a lifting for $\mu$ which respects coordinates.   ({\smc Burke n95}.)
%346E decompose each factor into M-homog pieces

\spheader 346Yb Let $(X,\Sigma,\mu)$ be a
probability space, $I$ any set, and $\lambda$ the product measure on
$X^I$.   Show that there is a lower density for $\lambda$ which is
invariant under transpositions of pairs of coordinates.
%mt34bits 346Zb 346Xc 346E

\spheader 346Yc Suppose that $(X,\Sigma,\mu)$
and $(Y,\Tau,\nu)$ are complete probability spaces with product
$(X\times Y,\Lambda,\lambda)$.   Show that for any lifting
$\psi_1:\Sigma\to\Sigma$ there are liftings $\psi_2:\Tau\to\Tau$ and
$\phi:\Lambda\to\Lambda$ such that
$\phi(E\times F)=\psi_1E\times\psi_2F$ for all $E\in\Sigma$, $F\in\Tau$.
\Hint{use the methods of \S341.   In the inductive construction of
341H, start with $\undphi_0(E\times Y)=(\psi_1E)\times Y$ for every
$E\in\Sigma$.   Extend each lower density $\undphi_{\xi}$ to the algebra
generated by $\dom(\undphi_{\xi})\cup\{X\times F_{\xi}\}$ for some
$F_{\xi}\in\Tau$.   Make sure that $\undphi_{\xi}(X\times F)$ is always
of the form $X\times F'$, and that
$\undphi_{\xi}((E\times Y)\cup(X\times F))
=\undphi_{\xi}(E\times Y)\cup\undphi_{\xi}(X\times F)$;  adapt the
construction of
341G to maintain this.   Use the method of 346H to generate a lifting
from the final lower density $\undphi$.   See 
{\smc Macheras \& Strauss 96a}, Theorem 4.}
%346H

\spheader 346Yd Use 346Yc and induction on $\zeta$ to prove 346H.
({\smc Macheras \& Strauss 96b}.)
%346Yc 346H

\spheader 346Ye Let $(X,\Sigma,\mu)$ be a complete probability space.
Show that there is a lifting $\psi:\Sigma\to\Sigma$ such that whenever
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ is a family of probability
spaces, with product measure $\lambda$, there is a lifting $\phi$ for
$\lambda$ such that
$\phi(\pi_i^{-1}[E])=\pi_i^{-1}[\psi E]$ whenever $E\in\Sigma$ and
$i\in I$ is such that $(X_i,\Sigma_i,\mu_i)=(X,\Sigma,\mu)$, writing
$\pi_i(x)=x(i)$ for $x\in\prod_{i\in I}X_i$.
%346I
}%end of exercises

\leader{346Z}{Problems (a)} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of
probability spaces, with product $(X,\Sigma,\mu)$.   Is there always
a lifting for $\mu$ which respects coordinates in the sense of 346A?

\spheader 346Zb Is there a lower density $\undphi$ for the usual measure
on $\{0,1\}^{\Bbb N}$ which is invariant under all permutations of
coordinates?

\endnotes{
\Notesheader{346} I ought to say at once that in writing this section I
have been greatly assisted by M.R.Burke.

The theorem that every complete probability space has a consistent
lifting (346J) is due to {\smc Talagrand 82a};  it is the inspiration
for the whole of the section.   `Consistent' liftings were devised in
response to some very interesting questions (see {\smc Talagrand 84},
\S6) which I do not discuss here;
one will be mentioned in Theorem 465P in Volume 4.   My aim
here is rather to suggest further ways in which a lifting on a product
space can
be consistent with the product structure.   The labour is substantial
and the results achieved are curiously partial.   I offer 346Za as the
easiest natural question which does not appear amenable to the methods I
describe.

The arguments I use are based on the fact that the translation-invariant
measures of 345C already respect coordinates
(346C).   Maharam's theorem now
makes it easy to show that any product of \Mth\
probability spaces has a lifting which respects coordinates
(346E).   A kind of
projection argument (346F) makes it possible to obtain a lower density
which respects coordinates on any product of probability spaces (346G).
In fact the methods of \S341, very slightly refined, automatically
produce such lower densities (346Xf).   But the extra power of 346G lies
in the condition (ii):  if $E$ and $F$ are `fully
independent' in the sense of being determined by coordinates in disjoint sets, then $\undphi(E\cup F)=\undphi E\cup\undphi F$, that is,
$\undphi$ is making a tentative step towards being a lifting.
(Remember that the difference between a lifting and a lower density is
mostly that a lifting preserves finite unions as well as finite
intersections;  see 341Xa.)   This can also be achieved by a
modification of the previous method, but we have to work harder at one
point in the proof.

The next step is to move to liftings which continue, as far as possible,
to respect coordinates.   Here there seem to be quite new obstacles, and
346H is the best result I know;  the lifting respects {\it individual}
coordinates, and also, for a given well-ordering of the index set,
initial segments of the coordinates.   The treatment
of initial segments makes essential use of the well-ordering, which is
what leaves 346Za open.

Finally, if all the factors are identical, we can seek lower densities
and liftings which are invariant under permutation of coordinates.
I give 345Xc and 346Xc as examples to show that we must not just assume
that a symmetry in the underlying measure space can be reflected in a
symmetry of a lifting.   The problems there concern liftings themselves,
not lower densities, since we can frequently find lower densities which
share symmetries (346Xb, 346Yb).   (Even for lower densities
there seem to be difficulties if we are more ambitious (346Zb).)
However a very simple argument (346I) shows that at least we can make
each individual coordinate look more or less the same, as long as we do
not investigate its relations with others.

Still on the question of whether, and when, liftings can be `good', note
346L/346Xh and 346Xg.   The most natural liftings for Lebesgue
measure are necessarily consistent;  but the only example we have of a
truly canonical lifting is not consistent in any non-trivial context.

I have deliberately used a variety of techniques here, even though 346H
(for instance) has an alternative proof based on the ideas of \S341
(346Yc-346Yd).   In particular, I give some of the standard methods of
constructing liftings and lower densities (346B, 346D, 346F, 346Xd,
346Xa).   In fact 346D was one of the elements of Maharam's original
proof of the lifting theorem ({\smc Maharam 58}).
}%end of comment

\frnewpage


\frfilename{mt34conc.tex}
\versiondate{10.4.10}
\copyrightdate{2010}

\def\chaptername{Concordance}
\def\sectionname{Concordance}

\centerline{\bf Concordance for Volume 3}

\bigskip

I list here the section and paragraph numbers which have 
(to my knowledge) appeared in print in references to this volume, and which have since been changed.

\medskip

\leaveitout{3{}41P 341P Burke Macheras & Strauss p22}


\leader{}{3{}41X} Exercises 3{}41Xd and 3{}41Xf, referred to in the 2003
and 2006 editions of Volume 4, are now 341Xc and 341Xe.

\leaveitout{{\smc Fremlin p99} \S342  compact measure, perfect measure}

\leaveitout{342D {\smc Fremlin 00} 342{}D A family $\Cal K$ of subsets of 
a set $X$ is a compact class
iff there is a compact topology (not necessarily Hausdorff) on $X$ such
that every member of $\Cal K$ is closed 
}

\leaveitout{342G {\smc Fremlin 00} 342{}G The product of compact 
probability spaces is compact 
}

\leaveitout{343K {\smc Fremlin 00} 343{}K If $(X,\Sigma,\mu)$ is 
`countably separated', in the sense that there is a sequence of 
measurable sets separating the points of $X$, then $\mu$ is compact 
iff it is perfect 
}

\leaveitout{
\leader{3{}44I}{Isomorphisms} The isomorphism theorem for atomless, perfect,
complete countably separated spaces,
referred to in {\smc Bogachev 07}, is now 
344I.
}

\leaveitout{\leader{3{}45B}{Translation-invariant liftings} The description
of a translation-invariant lifting for Lebesgue measure, referred to in 
{\smc Burke Macheras \& Strauss p10}, is now in 345B.
}

\leaveitout{\bigskip

\noindent{\bf References}

\medskip

Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p10] `Densities and liftings for
derived algebras', 9.10.10.

Burke M.R., Macheras N.D.\ \& Strauss W.\ [p22] `Density topologies and the marginal strong 
lifting problem for hyperstonian spaces', preprint, 2022.
%BMS.2022.pdf

Fremlin D.H.\ [00] `Weakly $\alpha$-favourable measure spaces', Fundamenta Math.\ 165 (2000) 67-94.
  %n99821.tex
}%end of leaveitout

\discrpage
\references{
\frfilename{mt3r.tex}
\versiondate{25.8.17}
\copyrightdate{1999}

% 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 3}

\medskip

%A

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P., 1987.
\cmmnt{[332Xk, 3A1K.]}

% Anzai H.\ [51] `Ergodic skew product transformations of the torus',
% % Osaka Math.\ J.\ 3 (1951) 83-99.

Anzai H.\ [51] `On an example of a measure-preserving transformation
which is not conjugate to its inverse',
Proc.\ Japanese Acad.\ Sci.\ 27 (1951) 517-522.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip
%B

Balcar B., G\l\'owczy\'nski W. \& Jech T.\ [98] `The sequential
topology on complete Boolean algebras', Fundamenta Math.\ 155 (1998)
59-78.
\cmmnt{[393L, 393Q.]}

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{[393L, 393Q.]}
% arXiv.math.LO/0312473

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996
(London Math.\ Soc.\ Lecture Note Series 232).
\cmmnt{[\S395 {\it notes\/}.]}

Bekkali M.\ \& Bonnet R.\ [89] `Rigid Boolean algebras', pp.\ 637-678 in
{\smc Monk 89}.
\cmmnt{[384L.]}

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).

Bhaskara Rao, K.P.S.\ \& Bhaskara Rao, M.\ {\it Theory of Charges.}
Academic, 1983.
%326H

Billingsley P.\ [65] {\it Ergodic Theory and Information.}  Wiley, 1965.
\cmmnt{[\S372 {\it notes\/}.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[332Xk, 3A1K.]}

Bose R.C.\ \& Manvel B.\ [84] {\it Introduction to Combinatorial Theory.}   Wiley, 1984.
\cmmnt{[3A1K.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[\S3A3, \S3A4.]}

Bourbaki N. [68] {\it Theory of Sets.}   Hermann/Addison-Wesley, 1968.
\cmmnt{[\S315 {\it notes\/}.]}

Bourbaki N. [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S3A5.]}

Bukhvalov A.V.\ [95] `Optimization without compactness, and its
applications', pp.\ 95-112 in
{\smc Huijsmans Kaashoek Luxemburg \& Pagter 95}.
\cmmnt{[367U.]}

Burke M.R.\ [93] `Liftings for Lebesgue measure', pp.\ 119-150 in
{\smc Judah 93}.   \cmmnt{[341L, 345F.]}

Burke M.R.\ [n95] `Consistent liftings', privately circulated, 1995.
\cmmnt{[346Ya.]}

Burnside W.\ [1911] {\it Theory of Groups of Finite Order.}   Cambridge
U.P., 1911 (reprinted by Dover, 1955).
\cmmnt{[\S384 {\it notes\/}.]}

\medskip
%C

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[372R.]}

Chacon R.V.\ \& Krengel U.\ [64] `Linear modulus of a linear
operator', Proc.\ Amer.\ Math.\ Soc.\ 15 (1964) 553-559.
\cmmnt{[\S371 {\it notes\/}.]}

Choksi J.R.\ \& Prasad V.S.\ [82] `Ergodic theory of homogeneous measure
algebras', pp.\ 367-408 of {\smc K\"olzow \& Maharam-Stone 82}.
\cmmnt{[383I.]}

%Christenson C.O.\ \& Voxmann L.\ [77] {\it Aspects of Topology.}
%Dekker, 1977.

Cohn H.\ [06] `A short proof of the simple continued fraction expansion of
$e$', Amer.\ Math.\ Monthly 113 (2006) 57-62;
{\tt arXiv:math.NT/0601660}.
\cmmnt{[372L.]}

Coleman A.J.\ \& Ribenboim P.\ [67] (eds.) {\it Proceedings of the Symposium in Analysis, Queen's University, June 1967.}  Queen's University, Kingston, Ontario, 1967.

Comfort W.W.\ \& Negrepontis S.\ [82] {\it Chain Conditions in
Topology.}   Cambridge U.P., 1982.   \cmmnt{[\S391 {\it notes\/}.]}

\leaveitout{
Cornfeld I.P., Fomin S.V.\ \& Sina\v\i Ya.G.\ [82] {\it Ergodic Theory.}
Springer, 1982 (Grundlehren der mathematischen Wissenschaften 245)
}

Czisz\'ar I.\ [67] `Information-type measures of difference of
probability distributions and indirect observations',
Studia Scientiarum Math.\ Hungarica 2 (1967) 299-318.   \cmmnt{[386G.]}

\medskip
%D

Davey B.A. \& Priestley H.A. [90] `Introduction to Lattices and Order',
Cambridge U.P., 1990.  \cmmnt{[3A6C.]}

%Dow A.\ \& Stepr\=ans J.\ [93] `The $\sigma$-linkedness of the measure algebra', Canad.\ Math.\ Bulletin 37 (1993) 42-45.

Dugundji J.\ [66] {\it Topology.}  Allyn \& Bacon, 1966.
\cmmnt{[\S3A3, 3A4Bb.]}

Dunford N.\ [1936] `Integration and linear operators',
Trans.\ Amer.\ Math.\ Soc.\ 40 (1936) 474-494.
\cmmnt{[\S376 {\it notes\/}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S356 {\it notes\/}, \S371 {\it notes\/}, \S376 {\it notes\/},
\S3A5.]}

Dye H.A.\ [59] `On groups of measure preserving transformations I',
Amer.\ J.\ Math.\ 81 (1959) 119-159.   \cmmnt{[\S388 {\it notes\/}.]}

\medskip
%E

Eigen S.J.\ [82] `The group of measure-preserving transformations of
$[0,1]$ has no outer automorphisms', Math.\ Ann.\ 259 (1982) 259-270.
\cmmnt{[384D.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).  \cmmnt{[\S3A3, \S3A4.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S3A1.]}
%QA248.E6  7850 2996 6

Erd\H{o}s P.\ \& Oxtoby J.C.\ [55] `Partitions of the plane into sets
having positive measure in every non-null product set', Trans.\ Amer.\
Math.\ Soc.\ 79 (1955) 91-102.   \cmmnt{[\S325 {\it notes\/}.]}

\medskip
%F

%Farah I.\ [04] `Examples of $\epsilon$-exhaustive pathological
%submeasures', Fundamenta Math.\ 181 (2004) 257-272.

Fathi A.\ [78] `Le groupe des transformations de $[0,1]$ qui
pr\'eservent la mesure de Lebesgue est un groupe simple', Israel J.\
Math.\ 29 (1978) 302-308.  \cmmnt{[\S382 {\it notes\/}, 383I.]}

Fremlin D.H.\ [74a] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[Chap.\ 35 {\it intro\/}., 354Yb, \S355 {\it notes\/},
\S356 {\it notes\/}, \S363 {\it notes\/}, \S365 {\it notes\/},
\S371 {\it notes\/}, \S376 {\it notes\/}.]}

Fremlin D.H.\ [74b] `A characterization of $L$-spaces', Indag.\ Math.\
36 (1974) 270-275.  \cmmnt{[\S371 {\it notes\/}.]}

%Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
%Cambridge U.P., 1984.

\leaveitout{Fremlin D.H.\ [89] `Measure algebras', pp.\ 876-980 in
{\smc Monk 89}.}

Fremlin D.H., de Pagter B.\ \&  Ricker W.J.\ [05]
`Sequential closedness of Boolean algebras of projections in Banach
spaces', Studia Math.\ 167 (2005) 45-62.
\cmmnt{[\S323 {\it notes\/}.]}

Frol\'\i k Z.\ [68] `Fixed points of maps of extremally disconnected
spaces and complete Boolean algebras', Bull.\ Acad.\ Polon.\ Sci.\ 16
(1968) 269-275.  \cmmnt{[382E.]}

\medskip
%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S3A3, \S3A4.]}

Gaifman H.\ [64] `Concerning measures on Boolean algebras', Pacific J.\
Math.\ 14 (1964) 61-73.  \cmmnt{[\S391 {\it notes\/}.]}

Gale D. [60] {\it The theory of linear economic models.}  McGraw-Hill,
1960.  \cmmnt{[3A5D.]}

Gnedenko B.V.\ \& Kolmogorov A.N.\ [54] {\it Limit Distributions for
Sums of Independent Random Variables.}  Addison-Wesley, 1954.
\cmmnt{[\S342 {\it notes\/}.]}

Graf S.\ \& Weizs\"acker H.von [76] `On the existence of lower densities in non-complete measure spaces',
pp.\ 155-158 in {\smc Bellow \& K\"olzow 76}.  \cmmnt{[341L.]}

\medskip%H

Hajian A.\ \& Ito Y.\ [69]
`Weakly wandering sets and invariant measures for a group of
transformations', J.\ of Math.\ and Mech.\ 18 (1969) 1203-1216.
\cmmnt{[396B.]}

Hajian A., Ito Y.\ \& Kakutani S.\ [75] `Full groups and a theorem of
Dye', Advances in Math.\ 17 (1975) 48-59.
\cmmnt{[\S388 {\it notes\/}.]}

Halmos P.R.\ [1948] `The range of a vector measure', Bull.\ Amer.\ Math.\
Soc.\ 54 (1948) 416-421.
\cmmnt{[326Yk.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[3A1D.]}

\leaveitout{
Hopf E.\ [1932] `Theory of measures and invariant integrals', Trans.\
Amer.\ Math.\ Soc.\ 34 (1932) 372-393.
}

Huijsmans C.B., Kaashoek M.A., Luxemburg W.A.J.\ \& de Pagter B.\ [95]
(eds.) {\it Operator Theory in Function Spaces and Banach Lattices.}
Birkh\"auser, 1995.

\medskip %I

Ionescu Tulcea C.\ \& Ionescu Tulcea A.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.  \cmmnt{[\S341 {\it notes\/}.]}

\medskip %J

James I.M.\ [87] {\it Topological and Uniform Spaces.}  Springer, 1987.
\cmmnt{[\S3A3, \S3A4.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2002.
\cmmnt{[\S3A1.]}

Jech T.\ [08] `Algebraic characterizations of measure algebras',
Proc.\ Amer.\ Math.\ Soc.\ 136 (2008) 1285-1294.
%{\tt http://www.math.psu.edu/jech/preprints/papers.html}
\cmmnt{[393Xj.]}

Johnson R.A.\ [80] `Strong liftings which are not Borel liftings',
Proc.\ Amer.\ Math.\ Soc.\ 80 (1980) 234-236.   \cmmnt{[345F.]}
%Russell A. Johnson

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.

\medskip %K

Kakutani S.\ [1941] `Concrete representation of abstract $L$-spaces
and the mean ergodic theorem', Annals of Math. 42 (1941) 523-537.
\cmmnt{[369E.]}

Kalton N.J., Peck N.T.\ \& Roberts J.W.\ [84] `An F-space sampler',
Cambridge U.P., 1984.  \cmmnt{[\S375 {\it notes\/}.]}

Kalton N.J.\ \& Roberts J.W.\ [83] `Uniformly exhaustive submeasures and
nearly additive set functions', Trans.\ Amer.\ Math.\ Soc.\ 278 (1983)
803-816.  \cmmnt{[392D, \S392 {\it notes\/}.]}

Kantorovich L.V., Vulikh B.Z. \& Pinsker A.G.\ [50]
{\it Functional Analysis in
Partially Ordered Spaces}, Gostekhizdat, 1950.
\cmmnt{[391D.]}

Kawada Y.\ [1944] `\"Uber die Existenz der invarianten Integrale', Jap.\
J.\ Math.\ 19 (1944) 81-95.  \cmmnt{[\S395 {\it notes\/}.]}

Kelley J.L.\ [59] `Measures on Boolean algebras', Pacific J.\ Math.\ 9
(1959) 1165-1177.  \cmmnt{[\S391 {\it notes\/}.]}

\leaveitout{
Kelley J.L. \& Namioka I.\ [76] {\it Linear Topological Spaces.}
Springer, 1976.
}

\leaveitout{
King J.L.F.\ [00] `The generic transformation has roots of all orders',
Colloquium Mathematicum 84/85  (2000) 521-547.
  % if  \frak A  is the Lebesgue measure algebra
  % a comeager set in the Polish group
  % \Aut_{\bar\mu}\frak A   has roots of all orders
}

Kolmogorov A.N. [58] `New metric invariants of transitive dynamical
systems and automorphisms of Lebesgue spaces', Dokl.\ Akad.\ Nauk SSSR
119 (1958) 861-864.  \cmmnt{[385P.]}

K\"olzow D.\ \& Maharam-Stone D.\ [82] (eds.) {\it Measure Theory
Oberwolfach 1981.}   Springer, 1982 (Lecture Notes in Math.\ 945).

Koppelberg S.\ [89] {\it General Theory of Boolean
Algebras\/}, vol.\ 1 of {\smc Monk 89}.
\cmmnt{[Chap.\ 31 {\it intro\/}., \S332 {\it notes\/}.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[\S356 {\it notes\/}, 3A5N.]}

Kranz P.\ \& Labuda I.\ [93] (eds.) {\it Proceedings of the Orlicz
Memorial Conference, 1991\/}, unpublished manuscript, available from
first editor ({\tt mmkranz{\@}olemiss.edu}).

Krengel, U.\ [63] `"Uber den Absolutbetrag stetiger linearer
Operatoren und seine Anwendung auf ergodische Zerlegungen', Math.\
Scand.\ 13 (1963) 151-187.   \cmmnt{[371Xb.]}

Krieger W.\ [76] `On ergodic flows and the isomorphism of factors',
Math.\ Ann.\ 223 (1976) 19-70.   [\S388 {\it notes}.]
   %Dye's theorem in general form, von-Neumann-algebra version

Krivine J.-L.\ [71] {\it Introduction to Axiomatic Set Theory.}
D.Reidel, 1971.  \cmmnt{[\S3A1.]}

Kullback S.\ [67]
`A lower bound for discrimination information in terms of variation',
IEEE Trans.\ on Information Theory 13 (1967) 126-127.
\cmmnt{[386G.]}
%Wendt S IN5 EL24 T IT

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[\S3A1.]}

Kwapien S.\ [73] `On the form of a linear operator on the space of
all measurable functions', Bull.\ Acad.\ Polon.\ Sci.\ 21 (1973)
951-954.  \cmmnt{[\S375 {\it notes\/}.]}

\medskip %L

Lang S.\ [93] {\it Real and Functional Analysis.}  Springer, 1993.
\cmmnt{[\S3A5.]}

Liapounoff A.A.\ [1940] `Sur les fonctions-vecteurs compl\`etement
additives', Bull.\ Acad.\ Sci.\ URSS (Izvestia Akad.\ Nauk SSSR) 4 (1940)
465-478.
\cmmnt{[326H.]}

Lindenstrauss J.\ \& Tzafriri L.\ [79] {\it Classical Banach Spaces II.}
Springer, 1979, reprinted in {\smc Lindenstrauss \& Tzafriri 96}.
\cmmnt{[\S354 {\it notes\/}, 374Xj.]}

Lindenstrauss J.\ \& Tzafriri L.\ [96] {\it Classical Banach Spaces I \& II.}  Springer, 1996.

Lipschutz S.\ [64] {\it Set Theory and Related Topics.}  McGraw-Hill,
1964 (Schaum's Outline Series).  \cmmnt{[3A1D.]}

Luxemburg W.A.J.\ [67a] `Is every integral normal?',
Bull.\ Amer.\ Math.\ Soc.\ 73 (1967) 685-688.   \cmmnt{[363S.]}

Luxemburg W.A.J.\ [67b] `Rearrangement-invariant Banach function spaces',
pp.\ 83-144 in {\smc Coleman \& Ribenboim 67}.
\cmmnt{[\S374 {\it notes}.]}

Luxemburg W.A.J.\ \& Zaanen A.C.\ [71] {\it Riesz Spaces
I.} North-Holland, 1971.  \cmmnt{[Chap.\ 35 {\it intro\/}.]}

\medskip %M

Macheras N.D., Musia\l\ K.\ \& Strauss W.\ [99] `On products of
admissible liftings and densities',
J.\ for Analysis and its Applications 18 (1999) 651-668. \cmmnt{[346G.]}

Macheras N.D.\ \& Strauss W.\ [95] `Products of lower densities', J.\
for Analysis and its Applications 14 (1995) 25-32.  \cmmnt{[346Xf.]}

Macheras N.D.\  \& Strauss W.\ [96a] `On products of almost strong
liftings', J.\ Australian Math.\ Soc.\ (A) 60 (1996) 1-23.
\cmmnt{[346Yc.]}

Macheras N.D.\ \& Strauss W.\ [96b] `The product lifting for arbitrary products of complete
probability spaces', Atti Sem.\ Math.\ Fis.\ Univ.\ Modena 44 (1996) 485-496.
\cmmnt{[346H, 346Yd.]}

Maharam D.\ [1942] `On homogeneous measure algebras', Proc.\ Nat.\ Acad.\
Sci.\ U.S.A.\ 28 (1942) 108-111.  \cmmnt{[331F, 332B.]}

Maharam D.\ [1947] `An algebraic characterization of measure algebras',
Ann.\ Math.\ 48 (1947) 154-167.
\cmmnt{[393J.]}

Maharam D.\ [58] `On a theorem of von Neumann', Proc.\ Amer.\ Math.\
Soc.\ 9 (1958) 987-994.
\cmmnt{[\S341 {\it notes\/}, \S346 {\it notes\/}.]}

Marczewski E.\ [53] `On compact measures',
Fund.\ Math.\ 40 (1953) 113-124.  \cmmnt{[342A.]}

McCune W.\ [97] `Solution of the Robbins problem',
J.\ Automated Reasoning 19 (1997) 263-276.
\cmmnt{[311Yc.]}

Miller B.D.\ [04] PhD Thesis, University of California, Berkeley, 2004.
%\query
\cmmnt{[382Xc, \S382 {\it notes\/}.]}

Monk J.D.\ [89] (ed.) {\it Handbook of Boolean Algebra.}
North-Holland, 1989.

\medskip %N

Nadkarni M.G.\ [90] `On the existence of a finite invariant measure',
Proc.\ Indian Acad.\ Sci., Math.\ Sci.\ 100 (1990) 203-220.
\cmmnt{[\S395 {\it notes\/}.]}

\medskip %O

Ornstein D.S.\ [74] {\it Ergodic Theory, Randomness and Dynamical Systems.}  Yale U.P., 1974.
\cmmnt{[\S387 {\it notes\/}.]}

Ornstein D.S.\ \& Shields P.C.\ [73] `An uncountable family of
$K$-automorphisms', Advances in Math.\ 10 (1973) 63-88.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip%P

%Pedersen G.K.\ [89] {\it Analysis Now.}  Springer, 1989.

Perovi\'c \v{Z}.\ \& Veli\v{c}kovi\'c B.\ [18] `Ranks of Maharam algebras',
Advances in Math.\ 330 (2018) 253-279.
\cmmnt{[394A.]}

Petersen K.\ [83] {\it Ergodic Theory.}   Cambridge U.P., 1983.
\cmmnt{[328Xa, 385C, \S385 {\it notes\/}, 386E.]}

\medskip %Q%R

Roberts J.W.\ [93] `Maharam's problem', in {\smc Kranz \& Labuda 93}.
\cmmnt{[\S394 {\it notes\/}.]}

Rotman J.J.\ [84] `An Introduction to the Theory of Groups', Allyn \&
Bacon, 1984.
\cmmnt{[\S384 {\it notes\/}, 3A6B.]}

Rudin W. [91] {\it Functional Analysis.}  McGraw-Hill, 1991.
\cmmnt{[\S3A5.]}

Ryzhikov V.V.\ [93] `Factorization of an automorphism of a complete Boolean algebra into a product of three involutions', Mat.\ Zametki (=Math.\ Notes of Russian Acad.\ Sci.) 54 (1993) 79-84.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip %S

Sazonov V.V.\ [66] `On perfect measures', A.M.S.\ Translations (2) 48
(1966) 229-254.   \cmmnt{[\S342 {\it notes\/}.]}

Schaefer H.H.\ [66] {\it Topological Vector Spaces.}  MacMillan, 1966;  reprinted with corrections Springer, 1971.
\cmmnt{[3A4A, 3A5J, 3A5N.]}
%Store A	QA 611.S3    6640 0159 9

Schaefer H.H.\ [74] {\it Banach Lattices and Positive Operators.}
Springer, 1974.   \cmmnt{[Chap.\ 35 {\it intro\/}., \S354 {\it notes\/}.]}

Schubert H.\ [68] {\it Topology.}  Allyn \& Bacon, 1968.
\cmmnt{[\S3A3, \S3A4.]}

Shelah S.\ [98] `The lifting problem with the full ideal',
J.\ Applied Analysis 4 (1998) 1-17.   \cmmnt{[341L.]}

Sikorski R.\ [64] {\it Boolean Algebras.}  Springer, 1964.
\cmmnt{[Chap.\ 31 {\it intro\/}.]}

Sina\v\i\ Ya.G.\ [59] `The notion of entropy of a dynamical system',
Dokl.\ Akad.\ Nauk SSSR 125 (1959) 768-771.  \cmmnt{[385P.]}

Sina\v\i\ Ya.G.\ [62] `Weak isomorphism of transformations with an invariant measure', Soviet Math.\ 3 (1962) 1725-1729.
\cmmnt{[387E.]}
% ({\cyr Dokl.\ Akad.\ Nauk SSSR 147 (1962) 797-800}).

Smorodinsky M.\ [71] {\it Ergodic Theory, Entropy.}
Springer, 1971 (Lecture Notes in Math., 214).
\cmmnt{[\S387 {\it notes\/}.]}

\v St\v ep\'anek P.\ \& Rubin M.\ [89] `Homogeneous Boolean algebras',
pp.\ 679-715 in {\smc Monk 89}.   \cmmnt{[382S, \S382 {\it notes\/}.]}

\medskip %T

Talagrand M.\ [82a] `Closed convex hull of set of measurable functions,
Riemann-\vthsp measurable functions and measurability of translations',
Ann.\ Institut Fourier (Grenoble) 32 (1982) 39-69.
\cmmnt{[\S346 {\it notes\/}.]}

Talagrand M.\ [82b] `La pathologie des rel\grv{e}vements invariants',
Proc.\ Amer.\ Math.\ Soc.\ 84 (1982) 379-382.
\cmmnt{[345F.]}

Talagrand M.\ [84] {\it Pettis integral and measure theory.}
Mem.\ Amer.\ Math.\ Soc.\  307 (1984).
\cmmnt{[\S346 {\it notes\/}.]}

%Talagrand M.\ [06] `Maharam's problem', C.R.Acad.Sci.\ Paris 342 (2006)
%501-503.

Talagrand M.\ [08] `Maharam's problem', Annals of Math.\ 168 (2008)
981-1009.
\cmmnt{[\S394.]}

Taylor A.E.\ [64] {\it Introduction to Functional Analysis.}
Wiley, 1964.
\cmmnt{[\S3A5.]}

Todor\v{c}evi\'c S.\ [04]
`A problem of von Neumann and Maharam about algebras supporting
continuous submeasures',
Fund.\ Math.\ 183 (2004) 169-183.
\cmmnt{[393S.]}

Truss J.K.\ [89] `Infinite permutation groups I:  products of conjugacy
classes',
J.\ Algebra 120 (1989) 454-493.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip %U%V

Vladimirov D.A.\ [02] {\it Boolean algebras in analysis.}
Kluwer, 2002 (Math.\ and its Appl.\ 540).
\cmmnt{[367L.]}

Vulikh B.C.\ [67] {\it Introduction to the Theory of Partially Ordered
Vector Spaces.}   Wolters-Noordhoff, 1967.
\cmmnt{[\S364 {\it notes\/}.]}

\medskip %W

Wagon S. [85] {\it The Banach-Tarski Paradox.}   Cambridge U.P., 1985.
\cmmnt{[\S395 {\it notes\/}.]}

Wilansky A. [64] {\it Functional Analysis.}  Blaisdell, 1964.
\cmmnt{[\S3A5.]}

\medskip %X%Y%Z

Zaanen A.C.\ [83] {\it Riesz Spaces II.}  North-Holland, 1983.
\cmmnt{[Chap.\ 35 {\it intro.\/}, 376K, \S376 {\it notes\/}.]}


\frnewpage

}%end of references


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