% 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{mt32.tex} 
\versiondate{21.7.11} 
\copyrightdate{2008} 
 
\def\chaptername{Measure algebras} 
\def\sectionname{Introduction} 
 
\newchapter{32} 
 
I now come to the real work of this volume, the study of the Boolean 
algebras of equivalence classes of measurable sets.   In this chapter I 
work through the `elementary' theory, defining this to consist of the 
parts which do not depend on Maharam's theorem or the lifting theorem or 
non-trivial set theory. 
 
\S321 gives the definition of `measure algebra', and relates this idea 
to its origin as the quotient of a $\sigma$-algebra of measurable sets 
by a $\sigma$-ideal of negligible sets, both in its elementary 
properties (following those of measure spaces treated in \S112) and in 
an appropriate version of the Stone representation. 
\S322 deals with the classification of measure algebras according to the 
scheme already developed in \S211 for measure spaces.   \S323 discusses 
the standard topology and uniformity of a measure algebra.   \S324 
contains results 
concerning Boolean homomorphisms between measure algebras, with the 
relationships between topological continuity, order-continuity and 
preservation of measure.   \S325 is devoted to the measure algebras of 
product measures, and their abstract characterization as completed free
products.    
\S\S326-327 address the properties of additive functionals on Boolean 
algebras, generalizing the ideas of Chapter 23.   Finally, \S328 looks at 
`reduced products' of probability algebras and some related constructions, 
including inductive limits. 
 
\discrpage 
 
 
\frfilename{mt321.tex} 
\versiondate{3.1.11} 
\copyrightdate{1995} 
 
\def\chaptername{Measure algebras} 
\def\sectionname{Measure algebras} 
 
\newsection{321} 
 
I begin by defining `measure algebra' and relating this concept to the 
work of Chapter 31 and to the elementary properties of measure 
spaces. 
 
\leader{321A}{Definition} A {\bf measure algebra} is a pair  
$(\frak A,\bar\mu)$, where $\frak A$ is a Dedekind 
$\sigma$-complete Boolean algebra and $\bar\mu:\frak A\to[0,\infty]$ is 
a function such that 
 
\qquad $\bar\mu 0=0$; 
 
\qquad whenever $\sequencen{a_n}$ is a disjoint sequence in $\frak A$, 
$\bar\mu(\sup_{n\in\Bbb N}a_n)=\sum_{n=0}^{\infty}\bar\mu a_n$; 
 
\qquad $\bar\mu a>0$ whenever $a\in\frak A$ and $a\ne 0$. 
 
\leader{321B}{Elementary properties of measure algebras} 
\cmmnt{Corresponding 
to the most elementary properties of measure spaces (112C in Volume 1), 
we have the following basic properties of measure algebras.} 
Let $(\frak A,\bar\mu)$ be a measure algebra. 
 
\header{321Ba}{\bf (a)} If $a$, $b\in\frak A$ and $a\Bcap b=0$ then 
$\bar\mu(a\Bcup b)=\bar\mu a+\bar\mu b$.   \prooflet{\Prf\ Set $a_0=a$, 
$a_1=b$, 
$a_n=0$ for $n\ge 2$;  then 
 
\Centerline{$\bar\mu(a\Bcup b)=\bar\mu(\sup_{n\in\Bbb N}a_n) 
=\sum_{n=0}^{\infty}\bar\mu a_n=\bar\mu a+\bar\mu b$.   \Qed}} 
 
\header{321Bb}{\bf (b)} If $a$, $b\in\frak A$ and $a\Bsubseteq b$ then 
$\bar\mu a\le\bar\mu b$.   \prooflet{\Prf\ 
 
\Centerline{$\bar\mu a\le\bar\mu a+\bar\mu(b\Bsetminus a)=\bar\mu b$. 
\Qed}} 
 
\header{321Bc}{\bf (c)} For any $a$, $b\in\frak A$,  
$\bar\mu(a\Bcup b)\le\bar\mu a+\bar\mu b$.   \prooflet{\Prf\ 
 
\Centerline{$\bar\mu(a\Bcup b)=\bar\mu a+\bar\mu(b\Bsetminus 
a)\le\bar\mu a+\bar\mu b$. 
\Qed}} 
 
\header{321Bd}{\bf (d)} If $\sequencen{a_n}$ is any sequence in  
$\frak A$, then  
$\bar\mu(\sup_{n\in\Bbb N}a_n)\le\sum_{n=0}^{\infty}\bar\mu a_n$. 
\prooflet{\Prf\ For each $n$, set $b_n=a_n\Bsetminus\sup_{i<n}a_i$. 
Inducing on $n$, we see that $\sup_{i\le n}a_i=\sup_{i\le n}b_i$ for 
each $n$, so $\sup_{n\in\Bbb N}a_n=\sup_{n\in\Bbb N}b_n$ and 
 
\Centerline{$\bar\mu(\sup_{n\in\Bbb N}a_n)=\bar\mu(\sup_{n\in\Bbb N}b_n) 
=\sum_{n=0}^{\infty}\bar\mu b_n\le\sum_{n=0}^{\infty}\bar\mu a_n$} 
 
\noindent because $\sequencen{b_n}$ is disjoint.   \Qed} 
 
\header{321Be}{\bf (e)} If $\sequencen{a_n}$ is a non-decreasing 
sequence in $\frak A$, then $\bar\mu(\sup_{n\in\Bbb 
N}a_n)=\lim_{n\to\infty}\bar\mu a_n$.   \prooflet{\Prf\ Set $b_0=a_0$, 
$b_n=a_n\Bsetminus a_{n-1}$ for $n\ge 1$.   Then 
 
$$\eqalign{\bar\mu(\sup_{n\in\Bbb N}a_n) 
&=\bar\mu(\sup_{n\in\Bbb N}b_n) 
=\sum_{n=0}^{\infty}\bar\mu b_n\cr 
&=\lim_{n\to\infty}\sum_{i=0}^n\bar\mu b_i 
=\lim_{n\to\infty}\bar\mu(\sup_{i\le n}b_i) 
=\lim_{n\to\infty}\bar\mu a_n.\text{  \Qed}\cr}$$} 
 
\spheader 321Bf If $\sequencen{a_n}$ is a non-increasing sequence in 
$\frak A$ and $\inf_{n\in\Bbb N}\bar\mu a_n<\infty$, then  
$\bar\mu(\inf_{n\in\Bbb N}a_n)=\lim_{n\to\infty}\bar\mu a_n$.   
\prooflet{\Prf\ (Cf.\  
112Cf.)   Set $a=\inf_{n\in\Bbb N}a_n$.   Take $k\in\Bbb N$ such that  
$\bar\mu a_k<\infty$.   Set $b_n=a_k\Bsetminus a_n$ for $n\in\Bbb N$;  then  
$\sequencen{b_n}$ is  
non-decreasing and $\sup_{n\in\Bbb N}b_n=a_k\Bsetminus a$ (313Ab).   Because  
$\bar\mu a_k$ is finite, 
 
$$\eqalignno{\bar\mu a 
&=\bar\mu a_k-\bar\mu(a_k\Bsetminus a) 
=\bar\mu a_k-\lim_{n\to\infty}\bar\mu b_n\cr 
\displaycause{by (e) above} 
&=\lim_{n\to\infty}\bar\mu(a_k\Bsetminus b_n) 
=\lim_{n\to\infty}\bar\mu a_n.  \text{ \Qed}\cr}$$
} 
 
\leader{321C}{Proposition} Let $(\frak A,\bar\mu)$ be a measure algebra, 
and $A\subseteq\frak A$ a non-empty upwards-directed set.   If 
$\sup_{a\in A}\bar\mu a<\infty$, then $\sup A$ is defined in $\frak A$ 
and $\bar\mu(\sup A)=\sup_{a\in A}\bar\mu a$. 
 
\proof{ (Compare 215A.)   Set $\gamma=\sup_{a\in A}\bar\mu a$, and for 
each $n\in\Bbb N$ 
choose $a_n\in A$ such that $\bar\mu a_n\ge\gamma-2^{-n}$.   Next, 
choose $\sequencen{b_n}$ in $A$ such that 
$b_{n+1}\Bsupseteq b_n\Bcup a_n$ for 
each $n$, and set $b=\sup_{n\in\Bbb N}b_n$.   Then 
 
\Centerline{$\bar\mu b=\lim_{n\to\infty}\bar\mu b_n\le\gamma$, 
\quad$\bar\mu a_n\le\bar\mu b$ for every $n\in\Bbb N$,} 
 
\noindent so $\bar\mu b=\gamma$. 
 
If $a\in A$, then for every $n\in\Bbb N$ there is an $a_n'\in A$ such 
that $a\Bcup a_n\Bsubseteq a_n'$, so that 
 
\Centerline{$\bar\mu(a\Bsetminus b)\le\bar\mu(a\Bsetminus a_n) 
\le\bar\mu(a_n'\Bsetminus a_n) 
=\bar\mu a'_n-\bar\mu a_n\le\gamma-\bar\mu a_n\le 2^{-n}$.} 
 
\noindent This means that $\bar\mu(a\setminus b)=0$, so $a\Bsetminus 
b=0$ 
and $a\Bsubseteq b$.   Accordingly $b$ is an upper bound of $A$, and is 
therefore $\sup A$;  since we already know that $\bar\mu b=\gamma$, the 
proof is complete. 
}%end of proof of 321C 
 
\leader{321D}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra 
and $A\subseteq\frak A$ a non-empty upwards-directed set.   If $\sup A$ 
is defined in $\frak A$, then $\bar\mu(\sup A)=\sup_{a\in A}\bar\mu a$. 
 
\proof{ If $\sup_{a\in A}\bar\mu a=\infty$, this is trivial;  otherwise 
it follows from 321C. 
}%end of proof of 321D 
 
\leader{321E}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra 
and $A\subseteq\frak A$ a disjoint set.   If $\sup A$ is defined in 
$\frak A$, then $\bar\mu(\sup A)=\sum_{a\in A}\bar\mu a$. 
 
\proof{ If $A=\emptyset$ then $\sup A=0$ and the result is trivial. 
Otherwise, set $B=\{a_0\Bcup\ldots\Bcup a_n:a_0,\ldots,a_n\in A$ are 
distinct$\}$.   Then $B$ is upwards-directed, and 
$\sup_{b\in B}\bar\mu b=\sum_{a\in A}\bar\mu a$ because $A$ is disjoint. 
Also $B$ has the same upper bounds as $A$, so $\sup B=\sup A$ and 
 
\Centerline{$\bar\mu(\sup A)=\bar\mu(\sup B) 
=\sup_{b\in B}\bar\mu b=\sum_{a\in A}\bar\mu a$.} 
 
}%end of proof of 321E 
 
\leader{321F}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra 
and $A\subseteq\frak A$ a non-empty downwards-directed set.   If 
$\inf_{a\in A}\bar\mu a<\infty$, then $\inf A$ is defined in $\frak A$ 
and $\bar\mu(\inf A)=\inf_{a\in A}\bar\mu a$. 
 
\proof{ Take $a_0\in A$ with $\bar\mu a_0<\infty$, and set 
$B=\{a_0\Bsetminus a:a\in A\}$.   Then $B$ is upwards-directed, and 
$\sup_{b\in B}\bar\mu b\le\bar\mu a_0<\infty$, so $\sup B$ is defined. 
Accordingly $\inf A=a_0\setminus\sup B$ is defined (313Aa), and 
 
$$\eqalign{\bar\mu(\inf A) 
&=\bar\mu a_0-\bar\mu(\sup B) 
=\bar\mu a_0-\sup_{b\in B}\bar\mu b\cr 
&=\inf_{b\in B}\bar\mu(a_0\Bsetminus b) 
=\inf_{a\in A}\bar\mu(a_0\Bcap a) 
=\inf_{a\in A}\bar\mu a.\cr}$$ 
}%end of proof of 321F 
 
\leader{321G}{Subalgebras} If $(\frak A,\bar\mu)$ is a measure algebra, 
and $\frak B$ is a $\sigma$-subalgebra of $\frak A$, then 
$(\frak B,\bar\mu\restrp\frak B)$ is a measure algebra. 
\prooflet{\Prf\ 
As remarked in 314Eb, $\frak B$ is Dedekind $\sigma$-complete.   If 
$\sequencen{b_n}$ is a disjoint sequence in $\frak B$, then the supremum 
$b=\sup_{n\in\Bbb N}b_n$ is the same whether taken in $\frak B$ or 
$\frak A$, so that we have $\bar\mu b=\sum_{n=0}^{\infty}\bar\mu b_n$. 
\Qed} 
 
\leader{321H}{The measure algebra of a measure 
\dvrocolon{space}}\cmmnt{ I introduce the 
abstract notion of `measure algebra' because I believe that this is the 
right language in which to formulate the questions addressed in this 
volume.   However it is very directly linked with the idea of  
`measure space', as the next two results show. 
 
\wheader{321H}{4}{2}{2}{72pt} 
 
\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a measure space, and 
$\Cal N$ the null ideal of $\mu$.   Let $\frak A$ be the 
Boolean algebra quotient $\Sigma/\Sigma\cap\Cal N$.   Then we have a 
functional $\bar\mu:\frak A\to[0,\infty]$ defined by setting 
 
\Centerline{$\bar\mu E^{\ssbullet}=\mu E$ for every $E\in\Sigma$,} 
 
\noindent and $(\frak A,\bar\mu)$ is a measure algebra.  The canonical 
map $E\mapsto E^{\ssbullet}:\Sigma\to\frak A$ is sequentially 
order-continuous. 
 
\proof{{\bf (a)} By 314C, $\frak A$ is a Dedekind $\sigma$-complete 
Boolean algebra.   By 313Qb, $E\mapsto E^{\ssbullet}$ is sequentially 
order-continuous, because $\Sigma\cap\Cal N$ is a $\sigma$-ideal of 
$\Sigma$. 
 
\medskip 
 
{\bf (b)} If $E$, $F\in\Sigma$ and $E^{\ssbullet}=F^{\ssbullet}$ in 
$\frak A$, then $E\symmdiff F\in\Cal N$, so 
 
\Centerline{$\mu E\le\mu F+\mu(E\setminus F)=\mu F 
\le\mu E+\mu(F\setminus E)=\mu E$} 
 
\noindent and $\mu E=\mu F$.   Accordingly the given formula does indeed 
define a function $\bar\mu:\frak A\to[0,\infty]$. 
 
\medskip 
 
{\bf (c)} Now 
 
\Centerline{$\bar\mu 0=\bar\mu\emptyset^{\ssbullet}=\mu\emptyset=0$.} 
 
\noindent If $\sequencen{a_n}$ is a disjoint sequence in $\frak A$, 
choose for each $n\in\Bbb N$ an $E_n\in\Sigma$ such that 
$E_n^{\ssbullet}=a_n$.   Set $F_n=E_n\setminus\bigcup_{i<n}E_i$;  then 
 
\Centerline{$F_n^{\ssbullet} 
=E_n^{\ssbullet}\Bsetminus\sup_{i<n}E_i^{\ssbullet} 
=a_n\Bsetminus\sup_{i<n}a_i=a_n$} 
 
\noindent for each $n$, so $\bar\mu a_n=\mu F_n$ for each $n$. 
Now set $E=\bigcup_{n\in\Bbb N}E_n=\bigcup_{n\in\Bbb N}F_n$;  then 
$E^{\ssbullet}=\sup_{n\in\Bbb N}F_n^{\ssbullet}=\sup_{n\in\Bbb N}a_n$. 
So 
 
\Centerline{$\bar\mu(\sup_{n\in\Bbb N}a_n) 
=\mu E 
=\sum_{n=0}^{\infty}\mu F_n 
=\sum_{n=0}^{\infty}\bar\mu a_n$.} 
 
\noindent Finally, if $a\ne 0$, then there is an $E\in\Sigma$ such that 
$E^{\ssbullet}=a$, and $E\notin\Cal N$, so $\bar\mu a=\mu E>0$. 
Thus $(\frak A,\bar\mu)$ is a measure algebra. 
}%end of proof of 321H 
 
\leader{321I}{Definition} For any measure space $(X,\Sigma,\mu)$ I will 
call $(\frak A,\bar\mu)$, as constructed above, the {\bf measure 
algebra of} $(X,\Sigma,\mu)$. 
 
\leader{321J}{The Stone representation of a 
measure \dvrocolon{algebra}}\cmmnt{ Just as 
with Dedekind $\sigma$-complete Boolean algebras (314N), every measure 
algebra is obtainable from the construction above. 
 
\medskip 
 
\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be any measure algebra. 
Then it is isomorphic, as measure algebra, to the measure algebra of 
some measure space. 
 
\proof{{\bf (a)} 
We know from 314M that $\frak A$ is isomorphic, as Boolean algebra, to a 
quotient algebra $\Sigma/\Cal M$ where $\Sigma$ is a 
$\sigma$-algebra of subsets of the Stone space $Z$ of $\frak A$, and 
$\Cal M$ is the ideal of meager subsets of $Z$.   Let 
$\pi:\Sigma/\Cal M\to\frak A$ be the canonical isomorphism, and set 
$\theta E=\pi E^{\ssbullet}$ for each $E\in\Sigma$;  then 
$\theta:\Sigma\to\frak A$ is a sequentially 
order-continuous surjective Boolean homomorphism with kernel $\Cal M$. 
 
\medskip 
 
{\bf (b)} For $E\in\Sigma$, set 
 
\Centerline{$\nu E=\bar\mu(\theta E)$.} 
 
\noindent Then $(Z,\Sigma,\nu)$ is a measure space.   \Prf\ (i) We know 
already that $\Sigma$ is a $\sigma$-algebra of subsets of $Z$.   (ii) 
 
\Centerline{$\nu\emptyset=\bar\mu(\theta\emptyset)=\bar\mu 0=0$.} 
 
\noindent (iii) If $\sequencen{E_n}$ is a disjoint sequence in $\Sigma$, 
then (because $\theta$ is a Boolean homomorphism) 
$\sequencen{\theta E_n}$ 
is a disjoint sequence in $\frak A$ and (because $\theta$ is 
sequentially order-continuous) 
$\theta(\bigcup_{n\in\Bbb N}E_n)=\sup_{n\in\Bbb N}\theta E_n$;  so 
 
\Centerline{$\nu(\bigcup_{n\in\Bbb N}E_n) 
=\bar\mu(\sup_{n\in\Bbb N}\theta E_n) 
=\sum_{n=0}^{\infty}\bar\mu(\theta E_n) 
=\sum_{n=0}^{\infty}\nu E_n$.  \Qed} 
 
\medskip 
 
{\bf (c)} For $E\in\Sigma$, 
 
\Centerline{$\nu E=0 
\iff\bar\mu(\theta E)=0 
\iff\theta E=0 
\iff E\in\Cal M$.} 
 
\noindent So the measure algebra of $(Z,\Sigma,\nu)$ is just 
$\Sigma/\Cal M$, with 
 
\Centerline{$\bar\nu E^{\ssbullet}=\nu E 
=\bar\mu(\theta E)=\bar\mu(\pi E^{\ssbullet})$} 
 
\noindent for every $E\in\Sigma$.   Thus the Boolean algebra isomorphism 
$\pi$ is also an isomorphism between the measure algebras 
$(\Sigma/\Cal M,\bar\nu)$ and $(\frak A,\bar\mu)$, and 
$(\frak A,\bar\mu)$ is represented in the required form. 
}%end of proof of 321J 
 
\leader{321K}{Definition} I will call the measure space $(Z,\Sigma,\nu)$ 
constructed in the proof of 321J 
the {\bf Stone space} of the measure algebra $(\frak A,\bar\mu)$. 
 
\cmmnt{For later reference, I repeat the description of this space as 
developed in 311E, 311I, 314M and 321J.}   $Z$ 
is a compact Hausdorff space, being the Stone space of $\frak A$. 
$\frak A$ can be identified with the algebra of open-and-closed sets in 
$Z$.   The null ideal of $\nu$ coincides with the ideal of 
meager subsets of $Z$;  \cmmnt{in particular,} $\nu$ is complete. 
The measurable sets are precisely those 
expressible in the form $E=\widehat{a}\symmdiff M$ where $a\in\frak A$, 
$\widehat{a}\subseteq Z$ is the corresponding open-and-closed set, and 
$M$ is meager;  in this case $\nu E=\bar\mu a$ and 
$a\cmmnt{\mskip5mu=\theta E}$ is the 
member of $\frak A$ corresponding to $E$. 
 
\cmmnt{For the most important classes of measure algebras, more can be 
said;  see 322O {\it et seq.} below.} 
 
\exercises{ 
\leader{321X}{Basic exercises $\pmb{>}$(a)} 
Let $(\frak A,\bar\mu)$ be a measure 
algebra, and $a\in\frak A$;  write $\frak A_a$ for the principal ideal of 
$\frak A$ generated by $a$.   Show that 
$(\frak A_a,\bar\mu\restrp\frak A_a)$ is a measure algebra. 
 
\spheader 321Xb Let $(X,\Sigma,\bar\mu)$ be a measure space, and 
$\frak A$ its measure algebra.   (i) Show that if $\Tau$ is a 
$\sigma$-subalgebra of $\Sigma$, then $\{E^{\ssbullet}:E\in\Tau\}$ is a 
$\sigma$-subalgebra of $\frak A$.   (ii) Show that if 
$\frak B$ is a $\sigma$-subalgebra of $\frak A$ then 
$\{E:E\in\Sigma,\,E^{\ssbullet}\in\frak B\}$ is a 
$\sigma$-subalgebra of $\Sigma$. 
 
\leader{321Y}{Further exercises (a)}  
%\spheader 321Ya 
Let $(\frak A,\bar\mu)$ be a 
measure algebra, and $I\normalsubgroup\frak A$ a $\sigma$-ideal.   For 
$u\in\frak A/I$ set  
$\bar\nu u=\inf\{\bar\mu a:a\in\frak A,\,a^{\ssbullet}=u\}$.   
(i) Show that the infimum is always attained.
(ii) Show that $(\frak A/I,\bar\nu)$ is a measure algebra. 
}%end of exercises 
 
\cmmnt{ 
\Notesheader{321} The idea behind taking the quotient $\Sigma/\Cal N$, 
where $\Sigma$ is the algebra of measurable sets and $\Cal N$ is the null ideal, is just that if negligible sets can be ignored 
-- as is the case for a very large proportion of the results of measure 
theory -- then two measurable sets can be counted as virtually the same 
if they differ by a negligible set, that is, if they represent the same 
member of the measure algebra.   The definition in 321A is designed to 
be 
an exact characterization of these quotient algebras, taking into 
account the measures with which they are endowed.   In the course of the 
present chapter I will work through many of the basic ideas dealt with 
in Volumes 1 and 2 to show how they can be translated into theorems 
about measure algebras, as I have done in 321B-321F.   It is worth 
checking these correspondences carefully, because some of the ideas 
mutate significantly in translation.   In measure algebras, it becomes 
sensible to take seriously the suprema and infima of uncountable sets 
(see 321C-321F). 
 
I should perhaps remark that while the Stone representation (321J-321K) 
is significant, it is not the most important method of representing 
measure algebras, which is surely Maharam's theorem, to be dealt with in 
the next chapter.   Nevertheless, the Stone representation is a 
canonical one, and will 
appear at each point that we meet a new construction involving measure 
algebras, just as the ordinary Stone representation of Boolean algebras 
can be expected to throw 
light on any aspect of Boolean algebra. 
}%end of notes 
 
\discrpage 
 
\frfilename{mt322.tex}
\versiondate{24.4.06}
\copyrightdate{1995}

\def\chaptername{Measure algebras}
\def\sectionname{Taxonomy of measure algebras}

\newsection{322}

Before going farther with the general theory of measure algebras, I run
through those parts of the classification of measure spaces in \S211
which have expressions in terms of measure algebras.   The most
important concepts at this stage are those of `semi-finite',
`localizable' and
`$\sigma$-finite' measure algebra (322Ac-322Ae);  these correspond
exactly to the same terms applied to measure spaces (322B).   I briefly
investigate the Boolean-algebra properties of
semi-finite and $\sigma$-finite measure algebras (322F, 322G), with
mentions of completions and c.l.d.\ versions (322D), subspace measures
(322I-322J), indefinite-integral measures (322K),
direct sums of measure spaces (322L, 322M) and
subalgebras of measure algebras (322N).   It turns out that
localizability
of a measure algebra is connected in striking ways to the properties of
the canonical measure on its Stone space (322O).   I end the section
with a description of the `localization' of a
semi-finite measure algebra (322P-322Q) and with some further properties
of Stone spaces (322R).

\vleader{72pt}{322A}{Definitions} Let $(\frak A,\bar\mu)$ be a measure
algebra.

\header{322Aa}{\bf (a)}\cmmnt{ I will say that} $(\frak A,\bar\mu)$ is
a {\bf probability algebra} if $\bar\mu 1=1$.

\header{322Ab}{\bf (b)} $(\frak A,\bar\mu)$ is {\bf
totally finite} if $\bar\mu 1<\infty$.

\header{322Ac}{\bf (c)} $(\frak A,\bar\mu)$ is
{\bf $\sigma$-finite} if there is a sequence $\sequencen{a_n}$ in
$\frak A$ such that $\bar\mu a_n<\infty$ for every $n\in\Bbb N$ and
$\sup_{n\in\Bbb N}a_n=1$.   Note that in this case $\sequencen{a_n}$ can be taken {\it either} to be non-decreasing\prooflet{ (consider $a'_n=\sup_{i<n}a_i$)}
{\it or} to be disjoint\prooflet{ (consider
$a''_n=a_n\Bsetminus a'_n$)}.

\header{322Ad}{\bf (d)} $(\frak A,\bar\mu)$ is
{\bf semi-finite} if whenever $a\in\frak A$ and $\bar\mu a=\infty$ there
is a non-zero $b\Bsubseteq a$ such that $\bar\mu b<\infty$.

\header{322Ae}{\bf (e)} $(\frak A,\bar\mu)$ is {\bf
localizable} if it is semi-finite and the Boolean
algebra $\frak A$ is Dedekind complete.

\leader{322B}{}\cmmnt{ The first step is to relate these concepts to
the corresponding ones for measure spaces.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a measure space, and
$(\frak A,\bar\mu)$ its measure algebra.   Then

(a) $(X,\Sigma,\mu)$ is a probability space iff $(\frak A,\bar\mu)$ is
a probability algebra;

(b) $(X,\Sigma,\mu)$ is totally finite iff $(\frak A,\bar\mu)$ is;

(c) $(X,\Sigma,\mu)$ is $\sigma$-finite iff $(\frak A,\bar\mu)$ is;

(d) $(X,\Sigma,\mu)$ is semi-finite iff $(\frak A,\bar\mu)$ is;

(e) $(X,\Sigma,\mu)$ is localizable iff $(\frak A,\bar\mu)$ is;

(f) if $E\in\Sigma$, then $E$ is an atom for $\mu$ iff $E^{\ssbullet}$
is an atom in $\frak A$;

(g) $(X,\Sigma,\mu)$ is atomless iff $\frak A$ is;

(h) $(X,\Sigma,\mu)$ is purely atomic iff $\frak A$ is.

\proof{{\bf (a), (b)} are trivial, since $\bar\mu 1=\mu X$.

\medskip

{\bf (c)(i)} If $\mu$ is $\sigma$-finite, let $\sequencen{E_n}$ be a
sequence of sets of finite measure covering $X$;  then $\bar\mu
E_n^{\ssbullet}<\infty$ for every $n$, and

\Centerline{$\sup_{n\in\Bbb N}E_n^{\ssbullet}=(\bigcup_{n\in\Bbb
N}E_n)^{\ssbullet}=1$,}

\noindent so $(\frak A,\bar\mu)$ is $\sigma$-finite.

\medskip

\quad{\bf (ii)} If $(\frak A,\bar\mu)$ is $\sigma$-finite, let
$\sequencen{a_n}$ be a sequence in $\frak A$ such that
$\bar\mu a_n<\infty$ for every $n$ and $\sup_{n\in\Bbb N}a_n=1$.   For each $n$, choose $E_n\in\Sigma$ such that $E_n^{\ssbullet}=a_n$.   Set
$E=\bigcup_{n\in\Bbb N}E_n$;  then
$E^{\ssbullet}=\sup_{n\in\Bbb N}a_n=1$, so $E$ is conegligible.   
Now $(X\setminus E,E_0,E_1,\ldots)$
is a sequence of sets of finite measure covering $X$, so $\mu$ is
$\sigma$-finite.

\medskip

{\bf (d)(i)} Suppose that $\mu$ is semi-finite and that $a\in\frak A$,
$\bar\mu a=\infty$.   Then there is an $E\in\Sigma$ such that
$E^{\ssbullet}=a$, so that $\mu E=\bar\mu a=\infty$.   As $\mu$ is
semi-finite, there is an $F\in\Sigma$ such that $F\subseteq E$ and
$0<\mu F<\infty$.   Set $b=F^{\ssbullet}$;  then $b\Bsubseteq a$ and
$0<\bar\mu b<\infty$.

\medskip

\quad{\bf (ii)} Suppose that $(\frak A,\bar\mu)$ is semi-finite and
that $E\in\Sigma$, $\mu E=\infty$.   Then $\bar\mu
E^{\ssbullet}=\infty$, so there is a $b\Bsubseteq
E^{\ssbullet}$ such that $0<\bar\mu b<\infty$.   Let $F\in\Sigma$ be
such that $F^{\ssbullet}=b$.   Then $F\cap
E\in\Sigma$, $F\cap E\subseteq E$ and $(F\cap
E)^{\ssbullet}=E^{\ssbullet}\Bcap b=b$, so that $\mu(F\cap E)=\bar\mu
b\in\ooint{0,\infty}$.

\medskip

{\bf (e)(i)} Note first that if $\Cal E\subseteq\Sigma$ and
$F\in\Sigma$, then

$$\eqalign{E\setminus F&\text{ is negligible for every }E\in\Cal E\cr
&\iff E^{\ssbullet}\Bsetminus F^{\ssbullet}=0
    \text{ for every }E\in\Cal E\cr
&\iff F^{\ssbullet}\text{ is an upper bound for }
    \{E^{\ssbullet}:E\in\Cal E\}.\cr}$$

\noindent So if $\Cal E\subseteq\Sigma$ and $H\in\Sigma$, then $H$ is an
essential supremum of $\Cal E$ in $\Sigma$, in the sense of 211G, iff
$H^{\ssbullet}$ is the
supremum of $A=\{E^{\ssbullet}:E\in\Cal E\}$ in $\frak A$.   \Prf\
Writing $\Cal F$ for

\Centerline{$\{F:F\in\Sigma,\,E\setminus F$ is negligible
for every $E\in\Cal E\}$,}

\noindent we see that $B=\{F^{\ssbullet}:F\in\Cal F\}$ is just the set
of upper bounds of $A$, and that $H$ is an essential supremum of $\Cal
E$ iff $H\in\Cal F$ and $H^{\ssbullet}$ is a lower bound for $B$;  that
is, iff $H^{\ssbullet}=\sup A$.\ \Qed

\medskip

\quad{\bf (ii)} Thus $\frak A$ is Dedekind complete iff every family in
$\Sigma$ has an essential supremum in $\Sigma$.   Since we already know
that $(\frak A,\bar\mu)$ is semi-finite iff $\mu$ is, we see that
$(\frak A,\bar\mu)$ is localizable iff $\mu$ is.

\medskip

{\bf (f)} This is immediate from the definitions in 211I and 316K, if
we remember always that $\{b:b\Bsubseteq E^{\ssbullet}\}
=\{F^{\ssbullet}:F\in\Sigma,\,F\subseteq E\}$ (312Lb).

\medskip

{\bf (g), (h)} follow at once from (f).
}%end of proof of 322B

\leader{322C}{}\cmmnt{ I copy out the relevant parts of Theorem 211L
in the new context.

\medskip

\noindent}{\bf Theorem} (a) A probability algebra is totally finite.

(b) A totally finite measure algebra is $\sigma$-finite.

(c) A $\sigma$-finite measure algebra is localizable.

(d) A localizable measure algebra is semi-finite.

\proof{ All except (c) are trivial;  and (c) may be deduced from
211Lc-211Ld, 322Bc, 322Be and 321J, or from 316Fa and 322G below.
}%end of proof of 322C

\leader{322D}{}\cmmnt{ Of course not all the definitions in \S211 are
directly relevant to measure algebras.   The concepts of `complete',
`locally determined' and `strictly localizable' measure space do not
correspond in any direct way to properties of the measure algebras.
Indeed, completeness is just irrelevant, as the next proposition shows.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space,
with completion $(X,\hat\Sigma,\hat\mu)$ and c.l.d.\ version
$(X,\tilde\Sigma,\tilde\mu)$\cmmnt{ (213E)}.   Write
$(\frak A,\bar\mu)$, $(\frak A_1,\bar\mu_1)$ and $(\frak A_2,\bar\mu_2)$
for the measure algebras of $\mu$, $\hat\mu$ and $\tilde\mu$
respectively.

(a) The embedding
$\Sigma\embedsinto\hat\Sigma$ corresponds to an isomorphism between
$(\frak A,\bar\mu)$ and $(\frak A_1,\bar\mu_1)$.

(b)(i) The embedding $\Sigma\embedsinto\tilde\Sigma$ defines an
order-continuous Boolean
homomorphism $\pi:\frak A\to\frak A_2$.   Setting
$\frak A^f=\{a:a\in\frak A,\,\bar\mu a<\infty\}$, $\pi\restrp\frak A^f$
is a measure-preserving bijection between $\frak A^f$ and
$\frak A_2^f=\{c:c\in\frak A_2,\,\bar\mu_2c<\infty\}$.

\quad(ii) $\pi$ is injective iff $\mu$ is semi-finite, and in this case
$\bar\mu_2(\pi a)=\bar\mu a$ for every $a\in\frak A$.

\quad(iii) If $\mu$ is localizable, $\pi$ is a bijection.

\proof{ For $E\in\Sigma$, I write $E^{\smallcirc}$ for its image in
$\frak A$;  for $F\in\hat\Sigma$, I write $F^*$ for its image in
$\frak A_1$;  and for $G\in\tilde\Sigma$, I write $G^{\ssbullet}$ for its image in $\frak A_2$.

\medskip

{\bf (a)} This is nearly trivial.   The map $E\mapsto
E^*:\Sigma\to\frak A_1$ is a
Boolean homomorphism, being the composition of the Boolean homomorphisms
$E\mapsto E:\Sigma\to\hat\Sigma$ and $F\mapsto
F^*:\hat\Sigma\to\frak A_1$.   Its kernel is
$\{E:E\in\Sigma,\,\hat\mu E=0\}=\{E:E\in\Sigma,\,\mu E=0\}$, so it
induces an injective Boolean homomorphism $\phi:\frak A\to\frak A_1$
given by the formula $\phi(E^{\smallcirc})=E^*$ for every
$E\in\Sigma$ (312F, 3A2G).   To see that $\phi$ is surjective, take any
$b\in\frak A_1$.   There is an $F\in\hat\Sigma$ such that
$F^*=b$, and there is an $E\in\Sigma$ such that $E\subseteq F$
and $\hat\mu(F\setminus E)=0$, so that

\Centerline{$\pi(E^{\smallcirc})=E^*=F^*=b$.}

\noindent Thus $\pi$ is a Boolean algebra isomorphism.   It is a measure
algebra isomorphism because for any $E\in\Sigma$

\Centerline{$\bar\mu_1\phi(E^{\smallcirc})
=\bar\mu_1E^*
=\hat\mu E
=\mu E
=\bar\mu E^{\smallcirc}$.}

\medskip

{\bf (b)(i)} The map $E\mapsto E^{\ssbullet}:\Sigma\to\frak A_2$ is a
Boolean homomorphism with kernel
$\{E:E\in\Sigma,\,\tilde\mu E=0\}\supseteq\{E:E\in\Sigma,\,\mu E=0\}$,
so induces a Boolean homomorphism
$\pi:\frak A\to\frak A_2$, defined by saying that
$\pi E^{\smallcirc}=E^{\ssbullet}$ for every $E\in\Sigma$.

If $a\in\frak A^f$, it is expressible as $E^{\smallcirc}$ where
$\mu E<\infty$.   Then $\tilde\mu E=\mu E$ (213Fa), so
$\pi a=E^{\ssbullet}$ belongs to $\frak A_2^f$, and
$\bar\mu_2(\pi a)=\bar\mu a$.   If $a$, $a'$
are distinct members of $\frak A^f$, then

\Centerline{$\bar\mu_2(\pi a\Bsymmdiff\pi a')
=\bar\mu_2\pi(a\Bsymmdiff a')
=\bar\mu(a\Bsymmdiff a')>0$,}

\noindent so $\pi a\ne\pi a'$;  thus $\pi\restrp\frak A^f$ is an
injective
map from $\frak A^f$ to $\frak A_2^f$.   If $c\in\frak A_2^f$, then
$c=G^{\ssbullet}$ where $\tilde\mu G<\infty$;   by 213Fc, there is an
$E\in\Sigma$ such that $E\subseteq G$, $\mu E=\tilde\mu G$ and
$\tilde\mu(G\setminus E)=0$, so that $E^{\smallcirc}\in\frak A^f$ and

\Centerline{$\pi E^{\smallcirc}=E^{\ssbullet}=G^{\ssbullet}=c$.}

\noindent As $c$ is arbitrary, $\phi[\frak A^f]=\frak A_2^f$.

Finally, $\pi$ is order-continuous.   \Prf\  Let $A\subseteq\frak A$ be
a non-empty downwards-directed set with infimum $0$, and $b\in\frak A_2$
a lower bound for $\pi[A]$.   \Quer\ If $b\ne 0$, then (because $(\frak
A_2,\bar\mu_2)$ is semi-finite) there is a $b_0\in\frak A_2^f$ such that
$0\ne b_0\Bsubseteq b$.   Let $a_0\in\frak A$ be such that $\pi
a_0=b_0$.   Then $a_0\ne 0$, so there is an $a\in A$ such that
$a\notBsupseteq a_0$,
that is, $a\Bcap a_0\ne a_0$.   But now, because $\pi\restrp\frak A^f$
is injective,

\Centerline{$b_0=\pi a_0\ne\pi(a\Bcap a_0)=\pi a\Bcap\pi a_0
=\pi a\Bcap b_0$,}

\noindent and $b_0\notBsubseteq\pi a$, which is
impossible.\ \BanG\ Thus $b=0$, and $0$ is the only lower bound of
$\pi[A]$.   As $A$ is arbitrary, $\pi$ is order-continuous
(313L(b-ii)).\ \Qed

\medskip

\quad{\bf (ii)} ($\alpha$) If $\mu$ is semi-finite, then
$\tilde\mu E=\mu E$ for every $E\in\Sigma$ (213Hc), so

\Centerline{$\bar\mu_2(\pi E^{\smallcirc})
=\bar\mu_2E^{\ssbullet}
=\tilde\mu E
=\mu E
=\bar\mu E^{\smallcirc}$}

\noindent for every $E\in\Sigma$.   In particular,

\Centerline{$\pi a=0\Longrightarrow 0=\bar\mu_2(\pi a)=\bar\mu a
\Longrightarrow a=0$,}

\noindent so $\pi$ is injective.   ($\beta$) If $\mu$ is not
semi-finite, there is an $E\in\Sigma$ such that $\mu E=\infty$ but $\mu
H=0$
whenever $H\in\Sigma$, $H\subseteq E$ and $\mu H<\infty$;  so that
$\tilde\mu E=0$ and

\Centerline{$E^{\smallcirc}\ne 0$,
\quad$\pi E^{\smallcirc}=E^{\ssbullet}=0$.}

\noindent So in this case $\pi$ is not injective.

\medskip

\quad{\bf (iii)} Now suppose that $\mu$ is localizable.   Then for every
$G\in\tilde\Sigma$ there is an $E\in\Sigma$ such that
$\tilde\mu(E\symmdiff G)=0$, by 213Hb;  accordingly
$\pi E^{\smallcirc}=E^{\ssbullet}=G^{\ssbullet}$.   As $G$ is arbitrary,
$\pi$ is surjective;  and we know from (ii) that $\pi$ is injective, so
it is a bijection, as claimed.
}%end of proof of 322D

\leader{322E}{Proposition} Let $(\frak A,\bar\mu)$ be a measure algebra.

(a) $(\frak A,\bar\mu)$ is semi-finite iff it has a partition of unity
consisting of elements of finite measure.

(b) If $(\frak A,\bar\mu)$ is semi-finite,
$a=\sup\{b:b\Bsubseteq a,\,\bar\mu b<\infty\}$ and
$\bar\mu a=\sup\{\bar\mu b:b\Bsubseteq a,\,\bar\mu b<\infty\}$
for every $a\in\frak A$.

\proof{ Set $\frak A^f=\{b:b\in\frak A,\,\bar\mu b<\infty\}$.

\medskip

{\bf (a)(i)} If $(\frak A,\bar\mu)$ is semi-finite, then $\frak A^f$ is
order-dense in $\frak A$, so there is a partition of unity consisting of
members of $\frak A^f$ (313K).
\medskip

\quad{\bf (ii)}  If there is a partition of unity $C\subseteq\frak A^f$,
and $\bar\mu a=\infty$, then there is a $c\in C$ such that
$a\Bcap c\ne 0$, and now $a\Bcap c\Bsubseteq a$ and
$0<\bar\mu(a\Bcap c)<\infty$;  as
$a$ is arbitrary, $(\frak A,\bar\mu)$ is semi-finite.

\medskip

{\bf (b)} Of course $\frak A^f$ is upwards-directed, by 321Bc, and we
are supposing that its supremum is $1$.   If $a\in\frak A$, then

\Centerline{$B=\{b:b\in\frak A^f,\,b\Bsubseteq a\}
=\{a\Bcap b:b\in\frak A^f\}$}

\noindent is upwards-directed and has supremum $a$ (313Ba), so
$\bar\mu a=\sup_{b\in B}\bar\mu b$, by 321D.
}%end of proof of 322E

\cmmnt{\medskip

\noindent{\bf Remark} Compare 213A.}

\leader{322F}{Proposition} If $(\frak A,\bar\mu)$ is a semi-finite
measure algebra, then $\frak A$ is a \wsid\ Boolean algebra.

\proof{ Let $\sequencen{A_n}$ be a sequence of non-empty
downwards-directed subsets of $\frak A$, all with infimum $0$.   Set

\Centerline{$B
=\{b:$ for every $n\in\Bbb N$ there is an $a\in A_n$ such that
$b\Bsupseteq a\}$.}

\noindent If $c\in\frak A\setminus\{0\}$, let $c'\Bsubseteq c$ be such
that $0<\bar\mu c'<\infty$.   For each $n\in\Bbb N$,
$\inf_{a\in A_n}\bar\mu(c'\Bcap a)=0$, by 321F;  so we may choose
$a_n\in A_n$ such
that $\bar\mu(c'\Bcap a_n)\le 2^{-n-2}\bar\mu b$.   Set
$b=\sup_{n\in\Bbb N}a_n\in B$.   Then

\Centerline{$\bar\mu(c'\Bcap b)
\le\sum_{n=0}^{\infty}\bar\mu(c'\Bcap a_n)<\bar\mu c'$,}

\noindent so $c'\notBsubseteq b$ and $c\notBsubseteq b$.   As $c$ is
arbitrary, $\inf B=0$;  as $\sequencen{A_n}$ is arbitrary, $\frak A$ is
weakly $(\sigma,\infty)$-distributive (316G).
}%end of proof of 322F

\vleader{72pt}{322G}{}\cmmnt{ Corresponding to 215B, we have the
following description of $\sigma$-finite algebras.

\medskip

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra.   Then the following are equiveridical:

(i) $(\frak A,\bar\mu)$ is $\sigma$-finite;

(ii) $\frak A$ is ccc;

(iii) {\it either} $\frak A=\{0\}$ {\it or} there is a functional
$\bar\nu:\frak A\to[0,1]$ such
that $(\frak A,\bar\nu)$ is a probability algebra.

\proof{{\bf (i)$\Leftrightarrow$(ii)} By 321J,
it is enough to consider the case in
which $(\frak A,\bar\mu)$ is the measure algebra of a measure space
$(X,\Sigma,\mu)$, and $\mu$ is semi-finite, by 322Bd.   We know that
$\frak A$ is ccc iff there is no uncountable disjoint set in
$\Sigma\setminus\Cal N$, where $\Cal N$ is the null ideal of $\mu$
(316D).   But 215B(iii)
shows that this is equivalent to $\mu$ being $\sigma$-finite, which is
equivalent to $(\frak A,\bar\mu)$ being $\sigma$-finite, by 322Bc.

\medskip

{\bf (i)$\Rightarrow$(iii)} If $(\frak A,\bar\mu)$ is $\sigma$-finite,
and $\frak A\ne\{0\}$, let $\sequencen{a_n}$ be a disjoint sequence in
$\frak A$ such that $\bar\mu a_n<\infty$ for every $n$ and
$\sup_{n\in\Bbb N}a_n=1$.
Then $\bar\mu a_n>0$ for some $n$, so there are $\gamma_n>0$ such that
$\sum_{n=0}^{\infty}\gamma_n\bar\mu a_n=1$.
(Set $\gamma'_n=2^{-n}/(1+\bar\mu a_n)$,
$\gamma_n=\gamma'_n/(\sum_{i=0}^{\infty}\gamma'_i\bar\mu a_i)$.)   Set
$\bar\nu a=\sum_{n=0}^{\infty}\gamma_n\bar\mu(a\Bcap a_n)$ for every
$a\in\frak A$;
it is easy to check that $(\frak A,\bar\nu)$ is a probability algebra.

\medskip

{\bf (iii)$\Rightarrow$(i)} is a consequence of (i)$\Leftrightarrow$(ii).
}%end of proof of 322G

\leader{322H}{Principal ideals} If $(\frak A,\bar\mu)$ is a measure
algebra and $a\in\frak A$, then\cmmnt{ it is easy to see (using
314Eb) that} $(\frak A_a,\bar\mu\restrp\frak{A}_a)$
is a measure algebra, where $\frak A_a$ is the
principal ideal of $\frak A$ generated by $a$.

\leader{322I}{Subspace \dvrocolon{measures}}\cmmnt{ General
subspace measures give rise to
complications in the measure algebra (see 322Xf, 322Yd).   But subspaces
with measurable envelopes (132D, 213L) are manageable.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$A\subseteq X$ a set with a measurable envelope $E$.   Let $\mu_A$ be
the subspace measure on $A$, and $\Sigma_A$ its domain;  let
$(\frak A,\bar\mu)$ be the measure algebra of $(X,\Sigma,\mu)$ and
$(\frak A_A,\bar\mu_A)$ the measure algebra of $(A,\Sigma_A,\mu_A)$.
Set $a=E^{\ssbullet}$ and let $\frak A_a$ be the principal ideal of
$\frak A$ generated by $a$.   Then we
have an isomorphism between $(\frak A_a,\bar\mu\restrp\frak{A}_a)$ and
$(\frak A_A,\bar\mu_A)$ given by the formula

\Centerline{$F^{\ssbullet}\mapsto (F\cap A)^{\smallcirc}$}

\noindent whenever $F\in\Sigma$ and $F\subseteq E$, writing
$F^{\ssbullet}$ for the equivalence class of $F$ in $\frak A$ and
$(F\cap A)^{\smallcirc}$
for the equivalence class of $F\cap A$ in $\frak A_A$.

\proof{ Set $\Sigma_E=\{E\cap F:F\in\Sigma\}$.   For $F$,
$G\in\Sigma_E$,

\Centerline{$F^{\ssbullet}=G^{\ssbullet}
\iff \mu(F\symmdiff G)=0
\iff \mu_A(A\cap(F\symmdiff G))=0
\iff (F\cap A)^{\smallcirc}=(G\cap A)^{\smallcirc}$,}

\noindent because $E$ is a measurable envelope of $A$.   Accordingly the
given formula defines an injective function from the image
$\{F^{\ssbullet}:F\in\Sigma_E\}$ of $\Sigma_E$ in $\frak A$ to
$\frak A_A$;  but this image is just the principal ideal $\frak A_a$.   It is
easy to check that the map is a Boolean homomorphism from $\frak A_a$ to
$\frak A_A$, and it is a Boolean isomorphism because
$\Sigma_A=\{F\cap A:F\in\Sigma_E\}$.   Finally, it is measure-preserving
because

\Centerline{$\bar\mu F^{\ssbullet}=\mu F=\mu^*(F\cap A)
=\mu_A(F\cap A)=\bar\mu_A(F\cap A)^{\smallcirc}$}

\noindent for every $F\in\Sigma_E$, again using the fact that $E$ is a
measurable envelope of $A$.
}%end of proof of 322I

\leader{322J}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space, with
measure algebra $(\frak A,\bar\mu)$.

(a) If $E\in\Sigma$, then the measure algebra of the subspace measure
$\mu_E$ can be identified with the principal ideal
$\frak A_{E^{\ssbullet}}$ of $\frak A$.

(b) If $A\subseteq X$ is a set of full outer measure (in
particular, if $\mu^*A=\mu X<\infty$), then the
measure algebra of the subspace measure $\mu_A$ can be identified with
$\frak A$.

\leader{322K}{Indefinite-integral measures:  Proposition}\dvAnew{20??}
Let $(X,\Sigma,\mu)$ be a measure space and $\nu$ an
indefinite-integral measure over $\mu$\cmmnt{ (234J)}.
Then the measure
algebra of $\nu$ can be identified, as Boolean algebra, with a principal
ideal of the measure algebra of $\mu$.

\proof{ Taking $(X,\hat\Sigma,\hat\mu)$ to be the completion of
$(X,\Sigma,\mu)$, then we can identify the measure algebras of $\mu$ and
$\hat\mu$, by 322Da;  and $\nu$ is still an indefinite-integral measure
over $\hat\mu$, just because $\mu$ and $\hat\mu$ give rise to the same
theory of integration (212Fb).   Now there is a $G\in\hat\Sigma$
such that the
domain $\Tau$ of $\nu$ is $\{E:E\subseteq X$, $E\cap G\in\hat\Sigma\}$
and the null ideal $\Cal N_{\nu}$ of $\nu$ is
$\{A:A\subseteq X$, $A\cap G\in\Cal N_{\mu}\}$, where $\Cal N_{\mu}$ is the
null ideal of $\mu$ or $\hat\mu$ (234Lc\footnote{Formerly 2{}34D.}, 212Eb).
Writing $\frak A$ for the
measure algebra of $\hat\mu$, $c=G^{\ssbullet}\in\frak A$,
and $\frak A_c$ for
the principal ideal of $\frak A$ generated by $c$, we have a Boolean
homomorphism $E\mapsto(E\cap G)^{\ssbullet}:\Tau\to\frak A_c$ with kernel
$\Cal N_{\nu}$.   So, writing $E^{\smallcirc}\in\frak B$ for the
equivalence class of $E\in\Tau$, we have an injective Boolean homomorphism
$\pi:\frak B\to\frak A_c$ defined by setting
$\pi E^{\smallcirc}=(E\cap G)^{\ssbullet}$ for every $E\in\Tau$.   Of
course

\Centerline{$\pi[\frak B]\supseteq\{(E\cap G)^{\ssbullet}:E\in\hat\Sigma\}
=\{a\Bcap c:a\in\frak A\}=\frak A_c$,}

\noindent so $\pi$ is actually an isomorphism, as required.
}%end of proof of 322K

\leader{322L}{Simple products}\dvAformerly{3{}22K}
{\bf (a)} Let $\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$ be an
indexed family of measure
algebras.   Let $\frak A$ be the simple product Boolean algebra
$\prod_{i\in I}\frak A_i$\cmmnt{ (315A)}, and for $a\in\frak A$ set
$\bar\mu a=\sum_{i\in I}\bar\mu_ia(i)$. Then\cmmnt{ it is easy to check
(using 315D(e-ii)) that} $(\frak A,\bar\mu)$ is a
measure algebra; I will call it the {\bf simple product} of the family
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$.   Each of the $\frak A_i$
corresponds to a principal ideal $\frak A_{e_i}$ say in $\frak A$,
where $e_i\in\frak A$ corresponds to
$1_{\frak A_i}\in\frak A_i$\cmmnt{ (315E)},
and the Boolean isomorphism between $\frak A_i$ and $\frak A_{e_i}$ is a
measure algebra isomorphism between $(\frak A_i,\bar\mu_i)$ and
$(\frak A_{e_i},\bar\mu\restrp\frak{A}_{e_i})$.

\header{322Lb}{\bf (b)} If $\familyiI{(X_i,\Sigma_i,\mu_i)}$ is a family
of measure spaces, with direct sum
$(X,\Sigma,\mu)$\cmmnt{ (214L)}, then the measure algebra
$(\frak A,\bar\mu)$ of
$(X,\Sigma,\mu)$ can be identified with the simple product of the
measure algebras $(\frak A_i,\bar\mu_i)$ of the
$(X_i,\Sigma_i,\mu_i)$.   \prooflet{\Prf\ If, as in 214L, we set
$X=\{(x,i):i\in I,\,x\in X_i\}$, and for $E\subseteq X$, $i\in I$ we set
$E_i=\{x:(x,i)\in E\}$, then the Boolean isomorphism
$E\mapsto\familyiI{E_i}:\Sigma\to\prod_{i\in I}\Sigma_i$ induces a
Boolean isomorphism from $\frak A$ to $\prod_{i\in I}\frak A_i$, which
is also a measure algebra isomorphism, because

\Centerline{$\bar\mu E^{\ssbullet}=\mu E=\sum_{i\in I}\mu_iE_i
=\sum_{i\in I}\bar\mu_iE_i^{\ssbullet}$}

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

\spheader 322Lc A simple product of measure algebras is semi-finite, or
localizable, or atomless, or purely atomic, iff every factor is.
\cmmnt{(Compare 214Kb.)}

\spheader 322Ld\dvAnew{2014} Let $(\frak A,\bar\mu)$ be a measure
algebra, and $\familyiI{e_i}$ a countable partition of unity in $\frak A$.
Then $(\frak A,\bar\mu)$ is isomorphic to the
product $\prod_{i\in I}(\frak A_{e_i},\bar\mu\restrp\frak{A}_{e_i})$ of
the corresponding principal ideals.   \prooflet{\Prf\ By 315F(ii),
the map $a\mapsto\langle a\Bcap e_i\rangle_{i\in I}$ is a Boolean
isomorphism between $\frak A$ and $\prod_{i\in I}\frak A_i$.   Because
$\langle e_i\rangle_{i\in I}$ is disjoint and
$a=\sup_{i\in I}a\Bcap e_i$, $\bar\mu a=\sum_{i\in I}\bar\mu(a\cap e_i)$
for every $a\in\frak A$ (321E, or otherwise).   So
$a\mapsto\langle a\Bcap e_i\rangle_{i\in I}$ is a measure
algebra isomorphism between $(\frak A,\bar\mu)$ and
$\prod_{i\in I}(\frak A_i,\bar\mu\restrp\frak{A}_{e_i})$.\ \Qed}

\spheader 322Le Let $(\frak A,\bar\mu)$ be a localizable measure
algebra.

\medskip

\quad{\bf (i)} If  $\langle e_i\rangle_{i\in I}$ is any partition of
unity in $\frak A$, then $(\frak A,\bar\mu)$ is isomorphic to the
product $\prod_{i\in I}(\frak A_{e_i},\bar\mu\restrp\frak{A}_{e_i})$ of
the corresponding principal ideals.   \prooflet{\Prf\ By 315F(iii),
the map $a\mapsto\langle a\Bcap e_i\rangle_{i\in I}$ is a Boolean
isomorphism between $\frak A$ and $\prod_{i\in I}\frak A_i$.   Because
$\langle e_i\rangle_{i\in I}$ is disjoint and
$a=\sup_{i\in I}a\Bcap e_i$, $\bar\mu a=\sum_{i\in I}\bar\mu(a\cap e_i)$
(321E, in its full strength), for every $a\in\frak A$.   So
$a\mapsto\langle a\Bcap e_i\rangle_{i\in I}$ is a measure
algebra isomorphism between $(\frak A,\bar\mu)$ and
$\prod_{i\in I}(\frak A_i,\bar\mu\restrp\frak{A}_{e_i})$.\ \Qed}

\medskip

\quad{\bf (ii)}\cmmnt{ In particular, since $\frak A$ has a partition
of unity consisting of elements of finite measure (322Ea),
$(\frak A,\bar\mu)$
is isomorphic to a simple product of totally finite measure algebras.
Each of these is isomorphic to the measure algebra of a totally finite
measure space, so} $(\frak A,\bar\mu)$ is isomorphic to the measure
algebra of a
direct sum of totally finite measure spaces, which is strictly
localizable.

\cmmnt{Thus every localizable measure algebra is isomorphic to the
measure algebra of a strictly localizable measure space.   (See also
322O below.)}

\leader{*322M}{Strictly
localizable\dvrocolon{ spaces}}\dvAformerly{3{}22L}\cmmnt{ The
following fact is occasionally useful.

\medskip

\noindent}{\bf Proposition}  Let $(X,\Sigma,\mu)$ be a strictly
localizable measure space with $\mu X>0$, and $(\frak A,\bar\mu)$ its
measure algebra.   If $\familyiI{a_i}$ is a partition of unity in
$\frak A$, there is a partition $\familyiI{X_i}$ of $X$ into members of
$\Sigma$ such that $X_i^{\ssbullet}=a_i$ for every $i\in I$ and

\Centerline{$\Sigma
=\{E:E\subseteq X,\,E\cap X_i\in\Sigma\Forall i\in I\}$,}

\Centerline{$\mu E=\sum_{i\in I}\mu(E\cap X_i)$ for every $E\in\Sigma$;}

\noindent that is, the isomorphism between $\frak A$ and the simple
product $\prod_{i\in I}\frak A_{a_i}$ of its principal
ideals\cmmnt{ (315F)} corresponds to an isomorphism between
$(X,\Sigma,\mu)$ and the direct sum of the subspace measures on $X_i$.

\proof{{\bf (a)} Suppose to begin with that $\mu X<\infty$.   In this
case $J=\{i:a_i\ne 0\}$ must be countable (322G).   For each $i\in J$,
choose $E_i\in\Sigma$ such that $E_i^{\ssbullet}=a_i$, and set
$F_i=E_i\setminus\bigcup_{j\in J,j\ne i}E_j$;  then
$F_i^{\ssbullet}=a_i$ for each $i\in J$, and $\family{i}{J}{F_i}$ is
disjoint.   Because $\mu X>0$, $J$ is non-empty;  fix some $j_0\in J$
and set

$$\eqalign{X_i
&=F_{j_0}\cup(X\setminus\bigcup_{j\in J}F_j)\text{ if }i=j_0,\cr
&=F_i\text{ for }i\in J\setminus\{j_0\},\cr
&=\emptyset\text{ for }i\in I\setminus J.\cr}$$

\noindent Then $\familyiI{X_i}$ is a disjoint family in $\Sigma$,
$\bigcup_{i\in I}X_i=X$ and $X_i^{\ssbullet}=a_i$ for every $i$.
Moreover, because only countably many of the $X_i$ are non-empty, we
certainly have

\Centerline{$\Sigma
=\{E:E\subseteq X,\,E\cap X_i\in\Sigma\Forall i\in I\}$,}

\Centerline{$\mu E=\sum_{i\in I}\mu(E\cap X_i)$ for every $E\in\Sigma$.}

\medskip

{\bf (b)} For the general case, start by taking a decomposition
$\family{j}{J}{Y_j}$ of $X$.   We can suppose that no $Y_j$ is
negligible, because there is certainly some $j_0$ such that
$\mu Y_{j_0}>0$, and we can if necessary replace $Y_{j_0}$ by
$Y_{j_0}\cup\bigcup\{Y_j:\mu Y_j=0\}$.   For each $j$, we can identify
the measure algebra of the subspace measure on $Y_j$ with the principal
ideal $\frak A_{b_j}$ generated by $b_j=Y_j^{\ssbullet}$ (322I).   Now
$\familyiI{a_i\Bcap b_j}$ is a partition of unity in $\frak A_{b_j}$, so
by (a) just above we can find a disjoint family $\familyiI{X_{ji}}$ in
$\Sigma$ such that $\bigcup_{i\in I}X_{ji}=Y_j$,
$X_{ji}^{\ssbullet}=a_i\Bcap b_j$ for every $i$ and

\Centerline{$\Sigma\cap\Cal PY_j
=\{E:E\subseteq Y_j,\,E\cap X_{ji}\in\Sigma\Forall i\in I\}$,}

\Centerline{$\mu E=\sum_{i\in I}\mu(E\cap X_{ji})$ for every
$E\in\Sigma\cap\Cal PY_j$.}

Set $X_i=\bigcup_{j\in I}X_{ji}$ for every $i\in I$.   Then
$\familyiI{X_i}$ is a partition of $X$.   Because
$X_i\cap Y_j=X_{ji}$ is measurable for every $j$, $X_i\in\Sigma$.
Because $X_i^{\ssbullet}\Bsupseteq a_i\Bcap b_j$ for every $j$, and
$\family{j}{J}{b_j}$ is a partition of unity in $\frak A$ (322Lb),
$X_i^{\ssbullet}\Bsupseteq a_i$ for each $i$;  because
$\familyiI{X_i^{\ssbullet}}$ is disjoint and $\sup_{i\in I}a_i=1$,
$X_i^{\ssbullet}=a_i$ for every $i$.   If $E\subseteq X$ is such that
$E\cap X_i\in\Sigma$ for every $i$, then $E\cap X_{ji}\in\Sigma$ for all
$i\in I$ and $j\in J$, so $E\cap Y_j\in\Sigma$ for every $j\in J$ and
$E\in\Sigma$.   If $E\in\Sigma$, then

$$\eqalign{\mu E
&=\sum_{j\in J}\mu(E\cap Y_j)
=\sum_{j\in J}\sum_{i\in I}\mu(E\cap X_{ji})\cr
&=\sum_{i\in I}\sum_{j\in J}\mu(E\cap X_i\cap Y_j)
=\sum_{i\in I}\mu(E\cap X_i).\cr}$$

\noindent Thus $\familyiI{X_i}$ is a suitable family.
}%end of proof of 322M

\leader{322N}{Subalgebras:  Proposition}\dvAformerly{322M} Let
$(\frak A,\bar\mu)$ be a
measure algebra, and $\frak B$ a $\sigma$-subalgebra of $\frak A$.  Set
$\bar\nu=\bar\mu\restrp\frak{B}$.

(a) $(\frak B,\bar\nu)$ is a measure algebra.

(b) If $(\frak A,\bar\mu)$ is totally finite, or a probability algebra,
so is $(\frak B,\bar\nu)$.

(c) If $(\frak A,\bar\mu)$ is $\sigma$-finite and $(\frak B,\bar\nu)$ is
semi-finite, then $(\frak B,\bar\nu)$ is $\sigma$-finite.

(d) If $(\frak A,\bar\mu)$ is localizable and $\frak B$ is
order-closed and $(\frak B,\bar\nu)$ is semi-finite, then
$(\frak B,\bar\nu)$ is localizable.

(e) If $(\frak B,\bar\nu)$ is a probability algebra, or totally finite,
or $\sigma$-finite, so is $(\frak A,\bar\mu)$.

\proof{{\bf (a)} By 314Eb, $\frak B$ is Dedekind $\sigma$-complete, and
the identity map $\pi:\frak B\to\frak A$ is sequentially
order-continuous;  so that $\bar\nu=\bar\mu\pi$ will be countably
additive and $(\frak B,\bar\nu)$ will be a measure algebra.

\medskip

{\bf (b)} This is trivial.

\medskip

{\bf (c)} Use 322G.   Every disjoint subset of $\frak B$ is disjoint in
$\frak A$, therefore countable, because $\frak A$ is ccc;  so $\frak B$
also is ccc and $(\frak B,\bar\nu)$ (being semi-finite) is
$\sigma$-finite.

\medskip

{\bf (d)} By 314Ea, $\frak B$ is Dedekind complete;  we are supposing
that $(\frak B,\bar\nu)$ is semi-finite, so it is localizable.

\medskip

{\bf (e)} This is elementary.
}%end of proof of 322N

\leader{322O}{The Stone space of a localizable measure
\dvrocolon{algebra}}\dvAformerly{3{}22N}\cmmnt{ I
said above that the concepts
`strictly localizable' and `locally determined' measure space have no
equivalents in the theory of measure algebras.   But when we look at the
canonical measure on the Stone space of a measure algebra, we can of
course hope that properties of the measure algebra will be reflected in
the properties of this measure, as happens in the next theorem.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra,
$Z$ the Stone space of $\frak A$, and $\nu$ the standard measure on
$Z$\cmmnt{ constructed by
the method of 321J-321K}.   Then the following are equiveridical:

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

(ii) $\nu$ is localizable;

(iii) $\nu$ is locally determined;

(iv) $\nu$ is strictly localizable.

\proof{ Write $\Sigma$ for the domain of $\nu$, that is,

\Centerline{$\{E\symmdiff A:E\subseteq Z$ is open-and-closed,
$A\subseteq Z$ is meager$\}$,}

\noindent and $\Cal M$ for the ideal of meager subsets of $Z$, that is,
the null ideal of $\nu$ (314M, 321K).   Then
$a\mapsto\widehat{a}^{\ssbullet}:\frak A\to\Sigma/\Cal M$ is an
isomorphism
between $(\frak A,\bar\mu)$ and the measure algebra of $(Z,\Sigma,\nu)$
(314M).   Note that because any subset of a meager set is meager, $\nu$
is surely complete.

\medskip

{\bf (a)(i)\hbox{$\Leftrightarrow$}(ii)} is a consequence of 322Be.

\medskip

{\bf (b)(ii)$\Rightarrow$(iii)} Suppose that $\nu$ is localizable.   Of
course it is semi-finite.   Let $V\subseteq Z$ be a set such that
$V\cap E\in\Sigma$ whenever $E\in\Sigma$ and $\nu E<\infty$.   Because
$\nu$ is
localizable, there is a $W\in\Sigma$ which is an essential supremum in
$\Sigma$ of $\{V\cap E:E\in\Sigma,\,\nu E<\infty\}$, that is,
$W^{\ssbullet}=\sup\{(V\cap E)^{\ssbullet}:\nu E<\infty\}$ in
$\Sigma/\Cal M$.    I claim that $W\symmdiff V$ is nowhere dense.  \Prf\
Let $G\subseteq Z$ be a
non-empty open set.   Then there is a non-zero $a\in\frak A$ such that
$\widehat{a}\subseteq G$.   Because $(\frak A,\bar\mu)$ is semi-finite,
we may suppose that $\bar\mu a<\infty$.   Now

\Centerline{$(W\cap\widehat{a})^{\ssbullet}
=W^{\ssbullet}\Bcap\widehat{a}^{\ssbullet}
=\sup_{\nu E<\infty}(V\cap E)^{\ssbullet}\Bcap\widehat{a}^{\ssbullet}
=\sup_{\nu E<\infty}(V\cap E\cap\widehat{a})^{\ssbullet}
=(V\cap\widehat{a})^{\ssbullet}$,}

\noindent so $(W\symmdiff V)\cap\widehat{a}$ is negligible, therefore
meager.   But we know that $\frak A$ is weakly
$(\sigma,\infty)$-distributive (322F),
so that meager sets in $Z$ are nowhere dense (316I), and there is a
non-empty open set $H\subseteq\widehat{a}\setminus(W\symmdiff V)$.
Now $H\subseteq G\setminus\overline{W\symmdiff V}$.   As $G$ is
arbitrary,
$\interior\overline{W\symmdiff V}=\emptyset$ and $W\symmdiff V$ is
nowhere dense.\ \Qed

But this means that $W\symmdiff V\in\Cal M\subseteq\Sigma$ and
$V=W\symmdiff(W\symmdiff V)\in\Sigma$.   As $V$ is arbitrary, $\nu$ is
locally determined.

\medskip

{\bf (c)(iii)$\Rightarrow$(iv)} Assume that $\nu$ is locally determined.
Because $(\frak A,\bar\mu)$ is semi-finite, there is a partition of
unity $C\subseteq\frak A$ consisting of elements of finite measure
(322Ea).   Set $\Cal C=\{\widehat{c}:c\in C\}$.   This is a disjoint
family of sets of finite measure for $\nu$.   Now suppose that
$F\in\Sigma$ and $\nu F>0$.   Then there is an open-and-closed set
$E\subseteq Z$ such that $F\symmdiff E$
is meager, and $E$ is of the form $\widehat{a}$ for some $a\in\frak A$.
Since

\Centerline{$\bar\mu a=\nu\widehat{a}=\nu F>0$,}

\noindent there is some $c\in C$ such that $a\Bcap c\ne 0$, and now

\Centerline{$\nu(F\cap\widehat{c})=\bar\mu(a\Bcap c)>0$.}

\noindent This means that $\nu$ satisfies the conditions of 213Oa and
must be strictly localizable.

\medskip

{\bf (d)(iv)$\Rightarrow$(ii)} This is just 211Ld.
}%end of proof of 322O

\leader{322P}{Theorem}\dvAformerly{3{}22O} Let
$(\frak A,\bar\mu)$ be a semi-finite
measure algebra, and let $\widehat{\frak A}$ be the Dedekind completion
of $\frak A$\cmmnt{ (314U)}.   Then there is a unique extension of
$\bar\mu$ to
a functional $\tilde\mu$ on $\widehat{\frak A}$ such that
$(\widehat{\frak A},\tilde\mu)$ is a localizable measure algebra.
The embedding $\frak A\embedsinto\widehat{\frak A}$ identifies the ideals
$\{a:a\in\frak A,\,\bar\mu a<\infty\}$ and
$\{a:a\in\widehat{\frak A},\,\tilde\mu a<\infty\}$.

\proof{ (I write the argument out as if $\frak A$ were actually a
subalgebra of $\widehat{\frak A}$.)   For $c\in\widehat{\frak A}$, set

\Centerline{$\tilde\mu c=\sup\{\bar\mu a:a\in\frak A,\,a\Bsubseteq c\}$.}

\noindent Evidently $\tilde\mu$ is a function from $\widehat{\frak A}$
to $[0,\infty]$ extending $\bar\mu$, so $\tilde\mu 0=0$.   Because
$\frak A$ is
order-dense in $\widehat{\frak A}$, $\tilde\mu c>0$ whenever $c\ne 0$,
because any such $c$ includes a non-zero member of $\frak A$.   If
$\sequencen{c_n}$ is a disjoint sequence in $\widehat{\frak A}$ with
supremum $c$, then $\tilde\mu c=\sum_{n=0}^{\infty}\tilde\mu c_n$.
\Prf\ Let $A$ be the set of all members of $\frak A$ expressible as
$a=\sup_{n\in\Bbb N}a_n$ where $a_n\in\frak A$ and $a_n\Bsubseteq c_n$ for
every $n\in\Bbb N$.   Now

$$\eqalign{\sup_{a\in A}\bar\mu a
&=\sup\{\sum_{n=0}^{\infty}\bar\mu a_n:a_n\in\frak A,\,a_n\Bsubseteq
c_n\text{ for every }n\in\Bbb N\}\cr
&=\sum_{n=0}^{\infty}\sup\{\bar\mu a_n:a_n\Bsubseteq c_n\}
=\sum_{n=0}^{\infty}\tilde\mu c_n.\cr}$$

\noindent Also, because $\frak A$ is order-dense in $\widehat{\frak A}$,
$c_n=\sup\{a:a\in\frak A,\,a\Bsubseteq c_n\}$ for each $n$, and
$\sup A$, taken in $\widehat{\frak A}$, must be $c$.
But this means that if
$a'\in\frak A$ and $a'\Bsubseteq c$ then $a'=\sup_{a\in A}a'\Bcap a$ in
$\widehat{\frak A}$ and therefore also in $\frak A$;  so that

\Centerline{$\bar\mu a'=\sup_{a\in A}\bar\mu(a'\Bcap a)
\le\sup_{a\in A}\bar\mu a$.}

\noindent   Accordingly

\Centerline{$\tilde\mu c=\sup_{a\in A}\bar\mu a
=\sum_{n=0}^{\infty}\tilde\mu c_n$.   \Qed}

This shows that $(\widehat{\frak A},\tilde\mu)$ is a measure algebra.
It is semi-finite because $(\frak A,\bar\mu)$ is and every non-zero
element of
$\widehat{\frak A}$ includes a non-zero element of $\frak A$, which in
turn includes a non-zero element of finite measure.    Since
$\widehat{\frak A}$ is Dedekind complete, $(\widehat{\frak A},\bar\mu)$
is localizable.

If $\bar\mu a$ is finite, then surely $\tilde\mu a=\bar\mu a$ is finite.
If $\tilde\mu c$ is finite, then $A=\{a:a\in\frak A,\,a\Bsubseteq c\}$ is
upwards-directed and $\sup_{a\in A}\bar\mu a=\tilde\mu c$ is finite, so
$b=\sup A$ is defined in $\frak A$ and $\bar\mu b=\tilde\mu c$.
Because $\frak A$ is order-dense in $\widehat{\frak A}$, $b=c$ (313K,
313O) and $c\in\frak A$, with $\bar\mu c=\tilde\mu c$.
}%end of proof of 322P

\leader{322Q}{Definition}\dvAformerly{3{}22P} Let
$(\frak A,\bar\mu)$ be any semi-finite
measure algebra.   I will call $(\widehat{\frak A},\tilde\mu)$, as
constructed
above, the {\bf localization} of $(\frak A,\bar\mu)$.   \cmmnt{Of
course it is unique just in so far as the Dedekind completion of
$\frak A$ is.}

\leader{322R}{Further properties of Stone
spaces:  Proposition}\dvAformerly{3{}22Q} Let
$(\frak A,\bar\mu)$ be a semi-finite measure algebra and
$(Z,\Sigma,\nu)$ its Stone space.

(a) Meager sets in $Z$ are nowhere dense;  every $E\in\Sigma$ is
uniquely expressible as $G\symmdiff M$ where $G\subseteq Z$ is
open-and-closed and $M$ is nowhere dense, and
$\nu E=\sup\{\nu H:H\subseteq E$ is open-and-closed$\}$.

(b) The c.l.d.\ version $\tilde\nu$ of $\nu$ is strictly localizable,
and has the same negligible sets as $\nu$.

(c) If $(\frak A,\bar\mu)$ is totally finite then
$\nu E=\inf\{\nu H:H\supseteq E$ is open-and-closed$\}$ for every
$E\in\Sigma$.

\proof{{\bf (a)} I have already remarked (in the proof of 322O) that
$\frak A$ is weakly $(\sigma,\infty)$-distributive, so that meager sets
in $Z$ are nowhere dense.   But we know that every member of $\Sigma$ is
expressible as $G\symmdiff M$ where $G$ is open-and-closed and $M$ is
meager, therefore nowhere dense.   Moreover, the expression is unique,
because if $G\symmdiff M=G'\symmdiff M'$ then
$G\symmdiff G'\subseteq M\cup M'$ is open and nowhere dense, therefore
empty, so $G=G'$ and $M=M'$.

Now let $a\in\frak A$ be such that $\widehat{a}=G$, and consider
$B=\{b:b\in\frak A,\,\widehat{b}\subseteq E\}$.   Then $\sup B=a$ in
$\frak A$.   \Prf\ If $b\in B$, then
$\widehat{b}\setminus\widehat{a}\subseteq M$
is nowhere dense, therefore empty;  so $a$ is an upper bound for $B$.
\Quer\ If $a$ is not the supremum of $B$, then there is a non-zero
$c\Bsubseteq a$ such that $b\Bsubseteq a\Bsetminus c$ for every
$b\in B$.   But now $\widehat{c}$ cannot be empty, so
$\widehat{c}\setminus\overline{M}$
is non-empty, and there is a non-zero $d\in\frak A$ such that
$\widehat{d}\subseteq\widehat{c}\setminus\overline{M}$.   In this case
$d\in B$ and $d\notBsubseteq a\Bsetminus c$.\ \BanG\  Thus $a=\sup B$.\
\Qed

It follows that

$$\eqalign{\nu E
&=\nu G
=\bar\mu a
=\sup_{b\in B}\bar\mu b\cr
&=\sup_{b\in B}\nu\widehat{b}
\le\sup\{\nu H:H\subseteq E\text{ is open-and-closed}\}
\le\nu E\cr}$$

\noindent and $\nu E=\sup\{\nu H:H\subseteq E$ is open-and-closed$\}$.

\medskip

{\bf (b)} This is the same as part (c) of the proof of 322O.   We have a
disjoint family $\Cal C$ of sets of finite measure for $\nu$ such that
whenever $E\in\Sigma$ and $\nu E>0$ there is a $C\in\Cal C$ such that
$\mu(C\cap E)>0$.   Now if $\tilde\nu F$ is defined and not $0$, there
is an $E\in\Sigma$ such that $E\subseteq F$ and $\nu E>0$ (213Fc), so
that there is a $C\in\Cal C$ such that $\nu(E\cap C)>0$;  since
$\nu C<\infty$, we have

\Centerline{$\tilde\nu(F\cap C)\ge\tilde\nu(E\cap C)=\nu(E\cap C)>0$.}

\noindent And of course $\tilde\nu C<\infty$ for every $C\in\Cal C$.
This means that $\Cal C$ witnesses that $\tilde\nu$ satisfies the
conditions of 213Oa, so that $\tilde\nu$ is strictly localizable.

Any $\nu$-negligible set is surely $\tilde\nu$-negligible.   If $M$ is
$\tilde\nu$-negligible then it is nowhere dense.   \Prf\ If
$G\subseteq Z$ is open and not empty then there is a non-empty
open-and-closed set
$H_1\subseteq G$, and now $H_1\in\Sigma$, so there is a non-empty
open-and-closed set $H\subseteq H_1$ such that $\nu H$ is finite
(because $\nu$ is semi-finite).   In this case $H\cap M$ is
$\nu$-negligible, therefore nowhere dense, and
$H\not\subseteq\overline{M}$.   But this means that
$G\not\subseteq\overline{M}$;  as $G$ is arbitrary, $M$ is nowhere
dense.\ \QeD\  Accordingly $M\in\Cal M$ and is $\nu$-negligible.

Thus $\nu$ and $\tilde\nu$ have the same negligible sets.

\medskip

{\bf (c)} Because $\nu Z<\infty$,

$$\eqalign{\nu E
&=\nu Z-\nu(Z\setminus E)
=\nu Z-\sup\{\nu H:H\subseteq Z\setminus E
  \text{ is open-and-closed}\}\cr
&=\inf\{\nu(Z\setminus H):H\subseteq Z\setminus E
  \text{ is open-and-closed}\}\cr
&=\inf\{\nu H:H\supseteq E\text{ is open-and-closed}\}.\cr}$$
}%end of proof of 322R

\exercises{
\leader{322X}{Basic exercises $\pmb{>}$(a)}
%\spheader 322Xa
Let $(\frak A,\bar\mu)$ be a measure
algebra.   Let $I_{\infty}$ be the set of those $a\in \frak A$ which are
either $0$ or
`purely infinite', that is, $\bar\mu b=\infty$ for every non-zero
$b\Bsubseteq a$.   Show that $I_{\infty}$ is a $\sigma$-ideal of
$\frak A$.   Show that there is a function
$\bar\mu_{\text{sf}}:\frak A/I_{\infty}\to[0,\infty]$ defined by setting
$\bar\mu_{\text{sf}}a^{\ssbullet}
=\sup\{\bar\mu b:b\Bsubseteq a,\,\bar\mu b<\infty\}$ for
every $a\in\frak A$.   Show that $(\frak A/I_{\infty},\bar\mu_{\text{sf}})$ is
a semi-finite measure algebra.

\header{322Xb}{\bf (b)} Let $(X,\Sigma,\mu)$ be a measure space and let
$\mu_{\text{sf}}$ be the `semi-finite version' of $\mu$, as defined in
213Xc.
Let $(\frak A,\bar\mu)$ be the measure algebra of $(X,\Sigma,\mu)$.
Show that the measure algebra of $(X,\Sigma,\mu_{\text{sf}})$ is isomorphic to
the measure algebra $(\frak A/I_{\infty},\bar\mu_{\text{sf}})$ of (a) above.

\spheader 322Xc Let $(X,\Sigma,\mu)$ be a measure space and
$(X,\tilde\Sigma,\tilde\mu)$ its c.l.d.\ version.   Let
$(\frak A,\bar\mu)$
and $(\frak A_2,\bar\mu_2)$ be the corresponding measure algebras, and
$\pi:\frak A\to\frak A_2$ the canonical homomorphism, as in 322Db.
Show that the kernel of $\pi$ is the ideal $I_{\infty}$, as described in
322Xa, so that $\frak A/I_{\infty}$ is isomorphic, as Boolean algebra,
to $\pi[\frak A]\subseteq\frak A_2$.   Show that this isomorphism
identifies $\bar\mu_{\text{sf}}$, as described in 322Xa, with
$\bar\mu_2\restr\pi[\frak A]$.
%322Xa, 322D

\spheader 322Xd Give a direct proof of 322G, not relying on 215B
and 321J.
%322G

\sqheader 322Xe Let $(\frak A,\bar\mu)$ be any measure algebra, $A$ a
non-empty subset of $\frak A$, and $c\in\frak A$ such that
$\bar\mu c<\infty$.   Show that (i) $c_0=\sup\{a\Bcap c:a\in A\}$ is
defined in
$\frak A$ (ii) there is a countable set $B\subseteq A$ such that
$c_0=\sup\{a\Bcap c:a\in B\}$.
%322G

\spheader 322Xf Let $(X,\Sigma,\mu)$ be a measure space and $A$ any
subset of $X$;  let $\mu_A$ be the subspace measure on $A$ and
$\Sigma_A$ its domain.   Write $(\frak A,\bar\mu)$ for the measure
algebra of $(X,\Sigma,\mu)$ and $(\frak A_A,\bar\mu_A)$ for the
measure algebra of $(A,\Sigma_A,\mu_A)$.   Show that the formula
$F^{\ssbullet}\mapsto (F\cap A)^{\ssbullet}$ defines a
sequentially order-continuous Boolean homomorphism
$\pi:\frak A\to\frak A_A$ which has kernel
$I=\{F^{\ssbullet}:F\in\Sigma,\,F\cap A=\emptyset\}$.
Show that for any $a\in\frak A$,
$\bar\mu_A(\pi a)=\min\{\bar\mu b:b\in\frak A,\,a\Bsetminus b\in I\}$.
%322I

\spheader 322Xg Let $(\frak A,\bar\mu)$ be a measure algebra and
$\frak B$ a regularly embedded $\sigma$-subalgebra of $\frak A$.
Suppose that
$(\frak B,\bar\mu\restrp\frak{B})$ is semi-finite.   Show that
$(\frak A,\bar\mu)$ is semi-finite.

\spheader 322Xh Let $(\frak A,\bar\mu)$ be any measure algebra and
$(Z,\Sigma,\nu)$ its Stone space.   Show that the c.l.d.\ version of
$\nu$ is strictly localizable.
%322O

\leader{322Y}{Further exercises (a)} Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, and $\Cal I$ a $\sigma$-ideal of
$\Sigma$.   Set $\Cal N=\{N:\,\exists\,F\in\Cal I,\,N\subseteq F\}$.
Show that $\Cal N$ is a $\sigma$-ideal of subsets of $X$.   Set
$\hat\Sigma=\{E\symmdiff N:E\in\Sigma,\,N\in\Cal N\}$.   Show that
$\hat\Sigma$ is a $\sigma$-algebra of subsets of $X$ and that
$\hat\Sigma/\Cal N$ is isomorphic to $\Sigma/\Cal I$.

\header{322Yb}{\bf (b)} Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $(Z,\Sigma,\nu)$ its Stone space.   Let $\tilde\nu$ be the
c.l.d.\ version of $\nu$, and $\tilde\Sigma$ its domain.   Show that
$\tilde\Sigma$ is precisely the Baire-property algebra
$\{G\symmdiff A:G\subseteq Z$ is open, $A\subseteq Z$ is meager$\}$, so
that $\tilde\Sigma/\Cal M$ can be
identified with the regular open algebra of $Z$ (314Yd) and the measure
algebra of $\tilde\nu$ can be identified with the localization of
$\frak A$.

\spheader 322Yc Give an example of a localizable measure algebra
$(\frak A,\bar\mu)$ with a $\sigma$-subalgebra $\frak B$ such that
$(\frak B,\bar\mu\restrp\frak{B})$ is semi-finite and atomless, but
$\frak A$ has an atom.

\spheader 322Yd Let $(X,\Sigma,\mu)$ be a measure space and
$A\subseteq X$ a subset;  let $\mu_A$ be the subspace measure on $A$,
$\frak A$ and $\frak A_A$ the measure algebras of $\mu$ and $\mu_A$, and
$\pi:\frak A\to\frak A_A$ the canonical homomorphism, as described in
322Xf.   (i) Show that if
$\mu_A$ is semi-finite, then $\pi$ is order-continuous.   (ii) Show that
if $\mu$ is semi-finite but $\mu_A$ is not, then $\pi$ is not
order-continuous.
%322Xf, 322I

\spheader 322Ye Show that if $(\frak A,\bar\mu)$ is a semi-finite measure algebra,
with Stone space $(Z,\Sigma,\nu)$, then $\nu$ has locally determined
negligible sets in the sense of 213I.

\spheader 322Yf Let $(\frak A,\bar\mu)$ be a localizable measure algebra
and $(Z,\Sigma,\nu)$ its Stone space.   (i) Show that a function
$f:Z\to\Bbb R$
is $\Sigma$-measurable iff there is a conegligible set $G\subseteq X$
such that $f\restr G$ is continuous.   \Hint{316Yi.}
(ii) Show that $f:Z\to[0,1]$ is $\Sigma$-measurable iff there is a
continuous function $g:Z\to[0,1]$ such that $f=g\,\,\nu$-a.e.
}%end of exercises

\cmmnt{
\Notesheader{322} I have taken this leisurely tour through the concepts
of Chapter 21 partly to recall them (or persuade you to look them up)
and partly to give you practice in the elementary manipulations of
measure algebras.   The really vital result here is the correspondence
between `localizability' in measure spaces and measure algebras.   Part
of the object of this volume (particularly in Chapter 36) is to try to
make sense of the properties of localizable measure spaces, as discussed
in Chapter 24 and elsewhere, in terms of their measure algebras.   I
hope that 322Be has already persuaded you that the concept really
belongs to measure algebras, and that the formulation in terms of
`essential suprema' is a dispensable expedient.

I have given proofs of 322C and 322G depending on the realization of an
arbitrary measure algebra as the measure algebra of a measure space, and
the corresponding theorems for measure spaces, because this seems the
natural approach from where we presently stand;  but I am sympathetic to
the view that such proofs must be inappropriate, and that it is in some
sense better
style to look for arguments which speak only of measure algebras
(322Xd).

For any measure algebra $(\frak A,\bar\mu)$, the set $\frak A^f$ of
elements of finite measure is an ideal of $\frak A$;  consequently it is
order-dense iff it includes a partition of unity (322E).   In 322F we
have something deeper:  any semi-finite measure algebra must be weakly
$(\sigma,\infty)$-distributive when regarded as a Boolean algebra, and
this has significant consequences in its Stone space, which are used in
the proofs of 322O and 322R.   Of course a result of this kind must
depend on
the semi-finiteness of the measure algebra, since any Dedekind
$\sigma$-complete Boolean algebra becomes a measure algebra if we give
every non-zero element the measure $\infty$.   It is natural to look for
algebraic conditions on a Boolean algebra sufficient to make it
`measurable', in the sense that it should carry a semi-finite measure;
this is an unresolved
problem to which I will return in Chapter 39.

Subspace measures, indefinite-integral measures,
simple products, direct sums, principal ideals and
order-closed subalgebras give no real surprises;  I spell out the
details in 322H-322N %322H 322I 322J 322L 322M 322N
and 322Xf-322Xg.   It is worth noting that completing a
measure space has no effect on its measure algebra (322D, 322Ya).   We
see also that
from the point of view of measure algebras there is no distinction to be
made between `localizable' and `strictly localizable', since every
localizable measure algebra is representable as the measure algebra of a
strictly localizable measure space (322Le).   (But strict localizability
does have implications for some processes starting in the measure
algebra;  see 322M.)   It is nevertheless remarkable that the
canonical measure on the Stone space of a semi-finite measure algebra is
localizable iff it is strictly localizable (322O).   This canonical
measure has many other interesting properties, which I skim over in
322R, 322Xh, 322Yb and 322Yf.
In Chapter 21 I discussed a number of methods of improving measure
spaces, notably `completions' (212C) and `c.l.d.\ versions' (213E).
Neither of
these is applicable in any general way to measure algebras.   But in
fact we have a more effective construction, at least for
semi-finite measure algebras, that of `localization'
(322P-322Q);  I say that it is more effective just because
localizability is more important than completeness or local
determinedness, being of vital importance in the behaviour of function
spaces (241Gb, 243Gb, 245Ec, 363M,
364M, 365L, 367M, 369A, 369C).   Note that the localization of a
semi-finite
measure algebra does in fact correspond to the c.l.d.\ version of a
certain measure (322Yb).   But of course $\frak A$ and
$\widehat{\frak A}$ do
{\it not} have the same Stone spaces, even when $\widehat{\frak A}$ can
be effectively represented as the measure algebra of a measure on the
Stone space of $\frak A$.   What is happening in 322Yb is that we are
using all
the open sets of $Z$ to represent members of $\widehat{\frak A}$, not
just the open-and-closed sets, which correspond to members of $\frak A$.
}%end of notes

\discrpage

\frfilename{mt323.tex}
\versiondate{20.7.06}
\copyrightdate{1999}

\def\chaptername{Measure algebras}
\def\sectionname{The topology of a measure algebra}

\newsection{323}

I take a short section to discuss one of the fundamental tools for
studying totally finite measure algebras, the natural metric that each
carries.   The same ideas, suitably adapted, can be applied to an
arbitrary
measure algebra, where we have a topology corresponding closely to the
topology of convergence in measure on the function space $L^0$.   Most
of the section consists of an analysis of the relations between this
topology and the order structure of the measure algebra.

\leader{323A}{The pseudometrics $\rho_a$ (a)} Let $(\frak A,\bar\mu)$ be
a measure algebra.   Write $\frak A^f=\{a:a\in\frak A,\,\bar\mu
a<\infty\}$.   For $a\in\frak A^f$ and $b$, $c\in\frak A$, write
$\rho_a(b,c)=\bar\mu(a\Bcap(b\Bsymmdiff c))$.
\cmmnt{Then} $\rho_a$ is a pseudometric on $\frak A$.
\prooflet{\Prf\ (i)
Because $\bar\mu a<\infty$, $\rho_a$ takes values in $\coint{0,\infty}$.
(ii) If $b$, $c$, $d\in\frak A$ then
$b\Bsymmdiff d\Bsubseteq(b\Bsymmdiff c)\Bcup(c\Bsymmdiff d)$, so

$$\eqalign{\rho_a(b,d)
&=\bar\mu(a\Bcap(b\Bsymmdiff d))
\le\bar\mu((a\Bcap(b\Bsymmdiff c))\Bcup(a\Bcap(c\Bsymmdiff d)))\cr
&\le\bar\mu(a\Bcap(b\Bsymmdiff c))+\bar\mu(a\Bcap(c\Bsymmdiff d))
=\rho_a(b,c)+\rho_a(c,d).\cr}$$

\noindent (iii) If $b$, $c\in\frak A$ then

\Centerline{$\rho_a(b,c)
=\bar\mu(a\Bcap(b\Bsymmdiff c))
=\bar\mu(a\Bcap(c\Bsymmdiff b))
=\rho_a(c,b)$.  \Qed}
}%end of prooflet

\header{323Ab}{\bf (b)} Now the {\bf measure-algebra topology} of the
measure algebra $(\frak A,\bar\mu)$ is that generated by the family
$\Rho=\{\rho_a:a\in\frak A^f\}$ of pseudometrics on $\frak A$.
Similarly the {\bf measure-algebra uniformity} on $\frak A$ is that
generated by $\Rho$.   \cmmnt{For the rest of this section I will take
it that every measure algebra is endowed with its measure-algebra topology
and uniformity.}

\cmmnt{(For a general discussion of topologies defined by pseudometrics,
see 2A3F {\it et seq.}   For the associated uniformities see \S3A4.)}

\header{323Ac}{\bf (c)}\cmmnt{ Note that} $\Rho$ is
upwards-directed\cmmnt{, since
$\rho_{a\Bcup a'}\ge\max(\rho_a,\rho_{a'})$ for all $a$,
$a'\in\frak A^f$}.

\spheader 323Ad\discrversionA{\footnote{Revised 2006.}}{} 
On the ideal $\frak A^f$ we have an actual metric $\rho$
defined by saying that $\rho(a,b)=\bar\mu(a\Bsymmdiff b)$ for $a$,
$b\in\frak A^f$\prooflet{ (to see that $\rho$ is a metric, repeat the
formulae of (a) above)};  this is the {\bf measure metric}\cmmnt{ or
{\bf Fr\'echet-Nikod\'ym metric}}. %Plebanek 97
I will call the topology it generates the {\bf strong measure-algebra
topology} on $\frak A^f$.

When $\bar\mu$ is totally finite\cmmnt{, that is, $\frak A^f=\frak A$},
$\rho=\rho_1$ defines the measure-algebra topology and uniformity of
$\frak A$.

\leader{323B}{Proposition} Let $(\frak A,\bar\mu)$ be any measure
algebra, and give $\frak A$ its measure-algebra topology.

(a) The operations $\Bcup$, $\Bcap$, $\Bsetminus$ and $\Bsymmdiff$ are
all uniformly continuous.

(b)\dvAnew{2009} $\frak A^f$ is dense in $\frak A$.

\proof{{\bf (a)} 
The point is that for any $b$, $c$, $b'$, $c'\in\frak A$ we have

\Centerline{$(b*c)\Bsymmdiff(b'*c')\Bsubseteq(b\Bsymmdiff
b')\Bcup(c\Bsymmdiff c')$}

\noindent for any of the operations $*=\Bcup$, $\Bcap$ etc.;  so that if
$a\in\frak A^f$ then

\Centerline{$\rho_a(b*c,b'*c')\le\rho_a(b,b')+\rho_a(c,c')$.}

\noindent Consequently the operation $*$ must be uniformly continuous.

\medskip

{\bf (b)} Given $b\in\frak A$, $a\in\frak A^f$ and
$\epsilon>0$, then $a\Bcap b\in\frak A^f$ and $\rho_a(b,a\Bcap b)=0$.
Because the family $\{\rho_a:a\in\frak A^f\}$ is upwards-directed, this is
enough to show that every neighbourhood of $b$ meets $\frak A^f$;  as $b$
is arbitrary, $\frak A^f$ is dense.
}%end of proof of 323B

\leader{323C}{Proposition} (a) Let $(\frak A,\bar\mu)$ be a totally
finite measure algebra.   Then $\bar\mu:\frak A\to\coint{0,\infty}$ is
uniformly continuous.

(b) Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.   Then
$\bar\mu:\frak A\to[0,\infty]$ is lower semi-continuous.

(c) Let $(\frak A,\bar\mu)$ be any measure algebra.   If $a\in\frak A$
and $\bar\mu a<\infty$, then
$b\mapsto\bar\mu(b\Bcap a):\frak A\to\Bbb R$ is uniformly continuous.

\proof{{\bf (a)} For any $a$, $b\in\frak A$,

\Centerline{$|\bar\mu a-\bar\mu b|
\le\bar\mu(a\Bsymmdiff b)=\rho_1(a,b)$.}

\medskip

{\bf (b)} Suppose that $b\in\frak A$ and $\bar\mu b>\alpha\in\Bbb R$.
Then there is an $a\Bsubseteq b$ such that $\alpha<\bar\mu a<\infty$
(322Eb).   If $c\in\frak A$ is such that $\rho_a(b,c)<\bar\mu a-\alpha$,
then

\Centerline{$\bar\mu c\ge\bar\mu(a\Bcap c)
=\bar\mu a-\bar\mu(a\Bcap(b\Bsetminus c))>\alpha$.}

\noindent Thus $\{b:\bar\mu b>\alpha\}$ is open;  as $\alpha$ is
arbitrary, $\bar\mu$ is lower semi-continuous.

\medskip

{\bf (c)}
$|\bar\mu(a\Bcap b)-\bar\mu(a\Bcap c)|\le\rho_a(b,c)$ for all
$b$, $c\in\frak A$.
}%end of proof of 323C

\vleader{108pt}{323D}{}\cmmnt{ The following facts are basic to any
understanding
of the relationship between the order structure and topology of a
measure algebra.

\medskip

\noindent}{\bf Lemma} Let $(\frak A,\bar\mu)$ be a measure algebra.

(a) Let $B\subseteq\frak A$ be a non-empty upwards-directed set.   For
$b\in B$ set $F_b=\{c:b\Bsubseteq c\in B\}$.

\quad (i) $\{F_b:b\in B\}$ generates a Cauchy filter
$\Cal F(B\closeuparrow)$ on $\frak A$.

\quad (ii) If $\sup B$ is defined in $\frak A$, then it is a
topological limit of $\Cal F(B\closeuparrow)$;  \cmmnt{in particular,}
it belongs to the topological closure of $B$.

(b) Let $B\subseteq\frak A$ be a non-empty downwards-directed set.   For
$b\in B$ set $F'_b=\{c:b\Bsupseteq c\in B\}$.

\quad (i) $\{F'_b:b\in B\}$ generates a Cauchy filter
$\Cal F(B\closedownarrow)$ on $\frak A$.

\quad (ii) If $\inf B$ is defined in $\frak A$, then it is a
topological limit of $\Cal F(B\closedownarrow)$;  \cmmnt{in
particular,} it
belongs to the topological closure of $B$.

(c)(i) Closed subsets of $\frak A$ are order-closed\cmmnt{ in the
sense of 313Da}.

\quad (ii) An order-dense subalgebra of $\frak A$ must be dense in the
topological sense.

(d) Now suppose that $(\frak A,\bar\mu)$ is semi-finite.

\quad (i) The sets $\{b:b\Bsubseteq c\}$, $\{b:b\Bsupseteq c\}$ are
closed for every $c\in\frak A$.

\quad (ii) If $B\subseteq\frak A$ is non-empty and upwards-directed and
$e$ is a cluster point of $\Cal F(B\closeuparrow)$, then $e=\sup B$.

\quad (iii) If $B\subseteq\frak A$ is non-empty and downwards-directed
and $e$ is a cluster point of $\Cal F(B\closedownarrow)$, then 
$e=\inf B$.

\proof{ I use the notations $\frak A^f$, $\rho_a$ from 323A.

\medskip

{\bf (a)(i)} ($\alpha$) If $b$, $c\in B$ then there is a $d\in B$ such
that $b\Bcup c\Bsubseteq d$, so that $F_d\subseteq F_b\cap F_c$;
consequently

\Centerline{$\Cal F(B\closeuparrow)
=\{F:F\subseteq \frak A,\,\exists\,b\in B,\,F_b\subseteq F\}$}

\noindent is a filter on $\frak A$.  ($\beta$) Let $a\in\frak A^f$,
$\epsilon>0$.   Then there is a $b\in B$ such that
$\bar\mu(a\Bcap c)\le\bar\mu(a\Bcap b)+\bover12\epsilon$ for every
$c\in B$, and $F_b\in\Cal F(B\closeuparrow)$.   If now $c$, $c'\in F_b$,
$c\Bsymmdiff c'\Bsubseteq(c\Bsetminus b)\Bcup(c'\Bsetminus b)$, so

\Centerline{$\rho_a(c,c')
\le\bar\mu(a\Bcap c\Bsetminus b)+\bar\mu(a\Bcap c'\Bsetminus b)
=\bar\mu(a\Bcap c)+\bar\mu(a\Bcap c')-2\bar\mu(a\Bcap b)
\le\epsilon$.}

\noindent As $a$ and $\epsilon$ are arbitrary, $\Cal F(B\closeuparrow)$
is Cauchy.

\medskip

\quad{\bf (ii)} Suppose that $e=\sup B$ is defined in $\frak A$.
Let $a\in\frak A^f$, $\epsilon>0$.   By 313Ba,
$a\Bcap e=\sup_{b\in B}a\Bcap b$;
but $\{a\Bcap b:b\in B\}$ is upwards-directed,
so  $\bar\mu(a\Bcap e)=\sup_{b\in B}\bar\mu(a\Bcap b)$, by 321D.
Let $b\in B$ be such that
$\bar\mu(a\Bcap b)\ge\bar\mu(a\Bcap e)-\epsilon$.
Then for any $c\in F_b$, $e\Bsymmdiff c\Bsubseteq e\Bsetminus b$, so

\Centerline{$\rho_a(e,c)
=\bar\mu(a\Bcap(e\Bsymmdiff c))
\le\bar\mu(a\Bcap(e\Bsetminus b))
=\bar\mu(a\Bcap e)-\bar\mu(a\Bcap b)
\le\epsilon$.}

\noindent As $a$ and $\epsilon$ are arbitrary,
$\Cal F(B\closeuparrow)\to e$.

Because $B\in\Cal F(B\closeuparrow)$, $e$ surely belongs to the
topological closure of $B$.

\medskip

{\bf (b)} Either repeat the arguments above, with appropriate
inversions, using 321F in place of 321D, or apply (a) to the set
$\{1\Bsetminus b:b\in B\}$.

\medskip

{\bf (c)(i)} This follows at once from (a) and (b) and the definition in
313Da.

\medskip

\quad{\bf (ii)} If $\frak B\subseteq\frak A$ is an order-dense
subalgebra
and $a\in\frak A$, then $B=\{b:b\in\frak B,\,b\Bsubseteq a\}$ is
upwards-directed and has supremum $a$ (313K);  by (a-ii),
$a\in\overline{B}\subseteq\overline{\frak B}$.   As $a$ is arbitrary,
$\frak B$ is topologically dense.

\medskip

{\bf (d)(i)} Set $F=\{b:b\Bsubseteq c\}$.   If $d\in\frak A\setminus F$,
then (because $(\frak A,\bar\mu)$ is semi-finite) there is an $a\in\frak
A^f$ such that $\delta=\bar\mu(a\Bcap d\Bsetminus c)>0$;  now if $b\in
F$,

\Centerline{$\rho_a(d,b)\ge\bar\mu(a\Bcap d\Bsetminus b)\ge\delta$,}

\noindent so that $d$ cannot belong to the closure of $F$.   As $d$ is
arbitrary, $F$ is closed.   Similarly, $\{b:b\Bsupseteq c\}$ is closed.

\medskip

\quad{\bf (ii)} ($\alpha$) If $b\in B$, then $e\in\overline{F_b}$,
because $F_b\in\Cal F(B\closeuparrow)$;  but
$\{c:b\Bsubseteq c\}$ is a closed set including $F_b$, so
contains $e$, and $b\Bsubseteq e$.   As $b$ is arbitrary, $e$ is an
upper bound for $B$.   ($\beta$)
If $d$ is an upper bound of $B$, then $\{c:c\Bsubseteq d\}$
is a closed set belonging to $\Cal F(B\closeuparrow)$, so
contains $e$.   As $d$ is arbitrary, this shows that $e$ is the
supremum of $B$, as claimed.

\medskip

\quad{\bf (iii)} Use the same arguments as in (ii), but inverted.
}%end of proof of 323D

\leader{323E}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra.

(a) If $\sequencen{b_n}$ is a non-decreasing sequence in $\frak A$ with
supremum $b$, then $\sequencen{b_n}$ converges topologically to $b$.

(b) If $\sequencen{b_n}$ is a non-increasing sequence in $\frak A$ with
infimum $b$, then $\sequencen{b_n}$ converges topologically to $b$.

\proof{ I call this a `corollary' because it is the special case of
323Da-323Db in which $B$ is the set of terms of a monotonic sequence;
but it is probably easier to work directly from the definition in 323A,
and use 321Be or 321Bf to see that $\lim_{n\to\infty}\rho_a(b_n,b)=0$
whenever $\bar\mu a<\infty$.
}%end of proof of 323E


\leader{323F}{}\cmmnt{ The following is a useful calculation.

\medskip

\noindent}{\bf Lemma} Let $(\frak A,\bar\mu)$ be a measure algebra and
$\sequencen{c_n}$ a sequence in $\frak A$ such that \dvro{}{the sum}
$\sum_{n=0}^{\infty}\bar\mu(c_n\Bsymmdiff c_{n+1})$ is finite.   Set
$d_0=\sup_{n\in\Bbb N}\inf_{m\ge n}c_m$, $d_1=\inf_{n\in\Bbb
N}\sup_{m\ge n}c_m$.   Then $d_0=d_1$ and, writing $d$ for their common
value, $\lim_{n\to\infty}\bar\mu (c_n\Bsymmdiff d)=0$.

\proof{ Write $\alpha_n=\bar\mu(c_n\Bsymmdiff c_{n+1})$,
$\beta_n=\sum_{k=n}^{\infty}\alpha_k$ for $n\in\Bbb N$;  we are
supposing that $\lim_{n\to\infty}\beta_n=0$.   Set $b_n=\sup_{m\ge
n}c_m\Bsymmdiff c_{m+1}$;  then

\Centerline{$\bar\mu b_n\le\sum_{m=n}^{\infty}\bar\mu(c_m\Bsymmdiff
c_{m+1})=\beta_n$}

\noindent for each $n$.   If $m\ge n$, then

\Centerline{$c_m\Bsymmdiff c_n\Bsubseteq\sup_{n\le k<m}c_k\Bsymmdiff
c_{k+1}\Bsubseteq b_n$,}

\noindent so

\Centerline{$c_n\Bsetminus b_n\Bsubseteq c_m\Bsubseteq c_n\Bcup b_n$.}
\noindent Consequently

\Centerline{$c_n\Bsetminus b_n\Bsubseteq\inf_{k\ge m}c_k\Bsubseteq
\sup_{k\ge m}c_k\Bsubseteq c_n\Bcup b_n$}

\noindent for every $m\ge n$, and

\Centerline{$c_n\Bsetminus b_n\Bsubseteq d_0\Bsubseteq d_1\Bsubseteq
c_n\Bcup b_n$,}

\noindent so that

\Centerline{$c_n\Bsymmdiff d_0\Bsubseteq b_n$,
\quad$c_n\Bsymmdiff d_1\Bsubseteq b_n$,
\quad$d_1\Bsetminus d_0\Bsubseteq b_n$.}

\noindent As this is true for every $n$,

\Centerline{$\lim_{n\to\infty}\bar\mu(c_n\Bsymmdiff d_i)
\le\lim_{n\to\infty}\bar\mu b_n=0$}

\noindent for both $i$, and

\Centerline{$\bar\mu(d_1\Bsymmdiff d_0)\le\inf_{n\in\Bbb N}\bar\mu
b_n=0$,}

\noindent so that $d_1=d_0$.
}%end of proof of 323F

\leader{323G}{The classification of measure algebras:  Theorem} Let
$(\frak A,\bar\mu)$ be a measure algebra, $\frak T$ its measure-algebra
topology and $\Cal U$ its measure-algebra uniformity.

(a) $(\frak A,\bar\mu)$ is semi-finite iff $\frak T$ is Hausdorff.

(b) $(\frak A,\bar\mu)$ is $\sigma$-finite iff $\frak T$ is metrizable,
and in this case $\Cal U$ also is metrizable.

(c) $(\frak A,\bar\mu)$ is localizable iff $\frak T$ is Hausdorff and
$\frak A$ is complete under $\Cal U$.

\proof{ I use the notations $\frak A^f$, $\rho_a$ from 323A.

\medskip

{\bf (a)(i)} Suppose that $(\frak A,\bar\mu)$ is semi-finite and that
$b$, $c$ are distinct members of $\frak A$.   Then there is an
$a\Bsubseteq
b\Bsymmdiff c$ such that $0<\bar\mu a<\infty$, and now $\rho_a(b,c)>0$.
As $b$ and $c$ are arbitrary, $\frak T$ is Hausdorff (2A3L).

\medskip

\quad{\bf (ii)} Suppose that $\frak T$ is Hausdorff and that $b\in\frak
A$ has $\bar\mu b=\infty$.   Then $b\ne 0$ so there must be an
$a\in\frak
A^f$ such that $\bar\mu(a\Bcap b)=\rho_a(0,b)>0$;   in which case
$a\Bcap b\Bsubseteq b$ and $0<\bar\mu(a\Bcap b)<\infty$.   As $b$ is
arbitrary, $\bar\mu$ is semi-finite.

\medskip

{\bf (b)(i)} Suppose that $\bar\mu$ is $\sigma$-finite.   Let
$\sequencen{a_n}$ be a non-decreasing sequence in $\frak A^f$ with
supremum $1$.   Set

$$\eqalign{\rho(b,c)
=\sum_{n=0}^{\infty}\bover{\rho_{a_n}(b,c)}{1+2^n\bar\mu a_n}}$$

\noindent for $b$, $c\in\frak A$.   Then $\rho$ is a metric on
$\frak A$, because if $\rho(b,c)=0$ then $a_n\Bcap(b\Bsymmdiff c)=0$ for
every $n$, so $b\Bsymmdiff c=0$ and $b=c$.

If $a\in\frak A^f$ and $\epsilon>0$, take $n$ such that
$\bar\mu(a\Bsetminus
a_n)\le\bover12\epsilon$.   If $b$, $c\in\frak A$ and
$\rho(b,c)\le\epsilon/2(1+2^n\bar\mu a_n)$, then

$$\eqalign{\rho_a(b,c)
&=\rho_{a\Bsetminus a_n}(b,c)+\rho_{a\Bcap a_n}(b,c)
\le\bar\mu(a\Bsetminus a_n)+\rho_{a_n}(b,c)\cr
&\le\Bover12\epsilon+(1+2^n\bar\mu a_n)\rho(b,c)
\le\epsilon.\cr}$$

In the other direction, given $\epsilon>0$, take $n\in\Bbb N$ such that
$2^{-n}\le\bover12\epsilon$;   then $\rho(b,c)\le\epsilon$ whenever
$\rho_{a_n}(b,c)\le\epsilon/2(n+1)$.

This shows that $\Cal U$ is the same as the metrizable uniformity
defined by $\{\rho\}$;  accordingly $\frak T$ also is defined by $\rho$.

\medskip

\quad{\bf (ii)} Now suppose that $\frak T$ is metrizable, and let $\rho$
be a metric defining $\frak T$.   For each $n\in\Bbb N$ there must be
$a_{n0},\ldots,a_{nk_n}\in\frak A^f$ and $\delta_n>0$ such that

\Centerline{$\rho_{a_{ni}}(b,1)\le\delta_n$ for every $i\le
k_n\Longrightarrow\rho(b,1)\le 2^{-n}$.}

\noindent Set $d=\sup_{n\in\Bbb N,i\le k_n}a_{ni}$.   Then
$\rho_{a_{ni}}(d,1)=0$ for every $n$ and $i$, so $\rho(d,1)\le 2^{-n}$ for
every $n$ and $d=1$.   Thus $1$ is the supremum of countably
many elements of finite measure and $(\frak A,\bar\mu)$ is
$\sigma$-finite.

\medskip

{\bf (c)(i)} Suppose that $(\frak A,\bar\mu)$ is localizable.   Then
$\frak T$ is Hausdorff, by (a).   Let $\Cal F$ be a Cauchy filter on
$\frak A$.   For each $a\in\frak A^f$, choose a sequence
$\sequencen{F_n(a)}$ in $\Cal F$ such that
$\rho_a(b,c)\le 2^{-n}$ whenever $b$, $c\in F_n(a)$ and $n\in\Bbb N$.
Choose $c_{an}\in\bigcap_{k\le n}F_k(a)$ for each $n$;  then
$\rho_a(c_{an},c_{a,n+1})\le 2^{-n}$ for each $n$.   Set
$d_a=\sup_{n\in\Bbb N}\inf_{k\ge n}a\Bcap c_{ak}$.   Then

\Centerline{$\lim_{n\to\infty}\rho_a(d_a,c_{an})
=\lim_{n\to\infty}\bar\mu(d_a\Bsymmdiff(a\Bcap c_{an}))=0$,}

\noindent by 323F.

If $a$, $b\in\frak A^f$ and $a\Bsubseteq b$, then $d_a=a\Bcap d_b$.
\Prf\ For each $n\in\Bbb N$, $F_n(a)$ and $F_n(b)$ both belong to
$\Cal F$, so must have a point $e$ in common;  now

$$\eqalign{\rho_a(d_a,d_b)
&\le\rho_a(d_a,c_{an})+\rho_a(c_{an},e)
  +\rho_a(e,c_{bn})+\rho_a(c_{bn},d_b)\cr
&\le\rho_a(d_a,c_{an})+\rho_a(c_{an},e)
  +\rho_b(e,c_{bn})+\rho_b(c_{bn},d_b)\cr
&\le\rho_a(d_a,c_{an})+2^{-n}+2^{-n}+\rho_b(c_{bn},d_b)\cr
&\to 0\text{ as }n\to\infty.\cr}$$

\noindent Consequently $\rho_a(d_a,d_b)=0$, that is,

\Centerline{$d_a=a\Bcap d_a=a\Bcap d_b$.   \Qed}

Set $d=\sup\{d_b:b\in\frak A^f\}$;  this is defined because $\frak A$ is
Dedekind complete.   Then $\Cal F\to d$.   \Prf\  If
$a\in\frak A^f$ and $\epsilon>0$, then

\Centerline{$a\Bcap d
=\sup_{b\in\frak A^f}a\Bcap d_b
=\sup_{b\in\frak A^f}a\Bcap b\Bcap d_{a\Bcup b}
=\sup_{b\in\frak A^f}a\Bcap b\Bcap d_{a}
=a\Bcap d_a$.}

\noindent So if we choose $n\in\Bbb N$ such that
$2^{-n}+\rho_a(c_{an},d_a)\le\epsilon$, then for any $e\in F_n(a)$ we
shall have

\Centerline{$\rho_a(e,d)
\le\rho_a(e,c_{an})+\rho_a(c_{an},d)
\le 2^{-n}+\rho_a(c_{an},d_a)
\le\epsilon$.}

\noindent   Thus

\Centerline{$\{e:\rho_a(d,e)\le \epsilon\}\supseteq F_n(a)\in\Cal F$.}

\noindent As $a$, $\epsilon$ are arbitrary, $\Cal F$ converges to $d$.\
\QeD\  As $\Cal F$ is arbitrary, $\frak A$ is complete.

\medskip

\quad{\bf (ii)} Now suppose that $\frak T$ is Hausdorff and that 
$\frak A$ is complete under $\Cal U$.   By (a), $(\frak A,\bar\mu)$ is
semi-finite.
Let $B$ be any non-empty subset of $\frak A$, and set
$B'=\{b_0\Bcup\ldots\Bcup b_n:b_0,\ldots,b_n\in B\}$, so that $B'$ is
upwards-directed and has the same upper bounds as $B$.   By 323Da, we
have a Cauchy filter $\Cal F(B'\backstep3\uparrow)$;  because $\frak A$
is complete, this is convergent;  and because $(\frak A,\bar\mu)$ is
semi-finite, its limit must be $\sup B'=\sup B$, by 323Dd.
As $B$ is arbitrary, $\frak A$ is Dedekind complete, so
$(\frak A,\bar\mu)$ is localizable.
}%end of proof of 323G

\leader{323H}{Closed\dvrocolon{ subalgebras}}\cmmnt{ The ideas used in
the proof of
(c) above have many other applications, of which one of the most
important is the following.   You may find it helpful to read the next
theorem first on the assumption that $(\frak A,\bar\mu)$ is a
probability algebra.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a localizable measure
algebra, and $\frak B$ a subalgebra of $\frak A$.   Then it is
topologically closed iff it is order-closed.

\proof{{\bf (a)} If $\frak B$ is closed, it must be order-closed, by
323Dc.

\medskip

{\bf (b)} Now suppose that $\frak B$ is order-closed.   I repeat the
ideas of part (c-i) of the proof of 323G.   Let
$e$ be any member of the closure of $\frak B$ in $\frak A$.   For each
$a\in\frak A^f$ and $n\in\Bbb N$ choose $c_{an}\in \frak B$ such that
$\rho_a(c_{an},e)\le 2^{-n}$.   Then

$$\eqalign{\sum_{n=0}^{\infty}\bar\mu((a\Bcap c_{an})\Bsymmdiff(a\Bcap
c_{a,n+1}))
&=\sum_{n=0}^{\infty}\rho_a(c_{an},c_{a,n+1})\cr
&\le\sum_{n=0}^{\infty}\rho_a(c_{an},e)+\rho_a(e,c_{a,n+1})
<\infty.\cr}$$

\noindent So if we set $e_a=\sup_{n\in\Bbb N}\inf_{k\ge n}c_{ak}$, then

\Centerline{$\rho_a(e_a,c_{an})=\rho_a(a\Bcap e_a,a\Bcap c_{an})\to 0$}

\noindent as $n\to\infty$, by 323F, and $\rho_a(e,e_a)=0$, that is,
$a\Bcap e_a=a\Bcap e$.   Also, because $\frak B$ is order-closed,
$\inf_{k\ge n}c_{ak}\in\frak B$ for every $n$, and $e_a\in\frak B$.

Because $\frak A$ is Dedekind complete, we can set

\Centerline{$e'_a=\inf\{e_b:b\in\frak A^f,\,a\Bsubseteq b\}$;}

\noindent then $e'_a\in\frak B$ and

\Centerline{$e'_a\Bcap a
=\inf_{b\Bsupseteq a}e_b\Bcap a
=\inf_{b\Bsupseteq a}e_b\Bcap b\Bcap a
=\inf_{b\Bsupseteq a}e\Bcap b\Bcap a
=e\Bcap a$.}

\noindent Now $e'_a\Bsubseteq e'_b$ whenever $a\Bsubseteq b$, so
$B=\{e'_a:a\in\frak A^f\}$ is upwards-directed, and

\Centerline{$\sup B=\sup\{e'_a\Bcap a:a\in\frak A^f\}=\sup\{e\Bcap
a:a\in\frak A^f\}=e$}

\noindent because $(\frak A,\bar\mu)$ is semi-finite.   Accordingly
$e\in\frak B$.   As $e$ is arbitrary, $\frak B$ is closed, as claimed.
}%end of proof of 323H

\leader{323I}{Notation} In the context of 323H, I will
say\cmmnt{ simply} that
$\frak B$ is a {\bf closed subalgebra} of $\frak A$.

\leader{323J}{Proposition} If $(\frak A,\bar\mu)$ is a localizable
measure algebra and $\frak B$ is a subalgebra of $\frak A$, then the
topological
closure $\overline{\frak B}$ of $\frak B$ in $\frak A$ is precisely the
order-closed subalgebra of $\frak A$ generated by $\frak B$.

\proof{ Write $\frak B_{\tau}$ for the smallest order-closed subset of
$\frak A$ including $\frak B$.   By 313Gc, $\frak B_{\tau}$ is a
subalgebra of $\frak A$, and is the order-closed subalgebra of $\frak A$
generated by $\frak B$.   Being an order-closed subalgebra of $\frak A$,
it is topologically closed, by 323H, and must include
$\overline{\frak B}$.   On the other hand, $\overline{\frak B}$, being
topologically
closed, is order-closed (323D(c-i)), so includes $\frak B_{\tau}$.
Thus $\overline{\frak B}=\frak B_{\tau}$ is the order-closed subalgebra
of $\frak A$ generated by $\frak B$.
}%end of proof of 323J

\leader{323K}{}\cmmnt{ I note some simple results for future
reference.

\medskip

\noindent}{\bf Lemma} If $(\frak A,\bar\mu)$ is a localizable measure
algebra and $\frak B$ is a closed subalgebra of $\frak A$, then for any
$a\in\frak A$ the subalgebra\cmmnt{ $\frak C$} of $\frak A$ generated
by $\frak B\cup\{a\}$ is closed.

\proof{ By 314Ja, $\frak C$ is order-closed.
}%end of proof of 323K

\leader{323L}{Proposition} Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a
family of measure algebras with simple product
$(\frak A,\bar\mu)$\cmmnt{ (322K)}.
Then the measure-algebra topology on
$\frak A=\prod_{i\in I}\frak A_i$ defined by $\bar\mu$ is\cmmnt{ just}
the product of the measure-algebra topologies of the $\frak A_i$.

\proof{ I use the notations $\frak A^f$, $\rho_a$ from 323A.   Write
$\frak T$ for the topology of $\frak A$ and $\frak S$
for the product topology.
For $i\in I$ and $d\in\frak A_i^f$ define a pseudometric $\tilde\rho_{di}$
on $\frak A$ by setting

\Centerline{$\tilde\rho_{di}(b,c)=\rho_{d}(b(i),c(i))$}

\noindent whenever $b$, $c\in\frak A$;  then $\frak S$ is defined by
$\Rho=\{\tilde\rho_{di}:i\in I,\,a\in\frak A_i^f\}$ (3A3Ig).   Now
each $\tilde\rho_{di}$ is one of the defining pseudometrics for
$\frak T$, since

\Centerline{$\tilde\rho_{di}(b,c)=\bar\mu(\tilde d\Bcap(b\symmdiff c))$}

\noindent where $\tilde d(i)=d$, $\tilde d(j)=0$ for $j\ne i$.   So
$\frak S\subseteq\frak T$.

Now suppose that $a\in\frak A^f$ and $\epsilon>0$.   Then $\sum_{i\in
I}\bar\mu_ia(i)=\bar\mu a$ is finite, so there is a finite set
$J\subseteq I$ such that $\sum_{i\in I\setminus
J}\bar\mu_ia(i)\le\bover12\epsilon$.   For each $j\in J$,
$\tau_j=\tilde\rho_{a(j),j}$ belongs to $\Rho$, and

$$\eqalign{\rho_a(b,c)
&=\sum_{i\in I}\bar\mu_i(a(i)\Bcap(b(i)\Bsymmdiff c(i)))\cr
&\le\sum_{j\in J}\bar\mu_j(a(j)\Bcap(b(j)\Bsymmdiff c(j)))
   +\Bover12\epsilon
=\sum_{j\in J}\tau_j(b,c)+\Bover12\epsilon
\le\epsilon\cr}$$

\noindent whenever $b$, $c$ are such that
$\tau_j(b,c)\le\epsilon/(1+2\#(J))$ for every $j\in J$.   By 2A3H,
the identity map from $(\frak A,\frak S)$ to $(\frak A,\frak T)$ is
continuous, that is, $\frak T\subseteq\frak S$.

Putting these together, we see that $\frak S=\frak T$, as claimed.
}%end of proof of 323L

\leader{*323M}{}\dvAformerly{3{}23Xg}\cmmnt{ In this volume we shall have 
little need to
consider the measure metric on $\frak A^f$, but the following facts are
sometimes useful.

\medskip

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ be a measure algebra,
and give $\frak A^f$ its measure metric.   

(a) The Boolean operations $\Bsymmdiff$, $\Bcap$, $\Bcup$ and $\Bsetminus$
on $\frak A^f$ are uniformly continuous.

(b) $\bar\mu\restrp\frak A^f:\frak A^f\to\coint{0,\infty}$ is
$1$-Lipschitz, therefore uniformly continuous.

(c) $\frak A^f$ is complete.

\proof{{\bf (a)} Writing $\rho$ for the measure metric on $\frak A^f$,
then, just as in the proof of 323Ba,

\Centerline{$\rho(b*c,b'*c')\le\rho(b,b')+\rho(c,c')$}

\noindent for all $b$, $c$, $b'$, $c'\in\frak A^f$ and any of the Boolean
operations $*=\Bsymmdiff$, $\Bcap$, $\Bcup$ and $\Bsetminus$.

\medskip

{\bf (b)} If $a$, $b\in\frak A^f$ then 

\Centerline{$|\bar\mu a-\bar\mu b|
\le|\bar\mu a-\bar\mu(a\Bcap b)|+|\bar\mu b-\bar\mu(a\Bcap b)|
=\bar\mu(a\Bsetminus b)+\bar\mu(b\Bsetminus a)
=\rho(a,b)$.}

\medskip

{\bf (c)} If $\sequencen{a_n}$ is a sequence in $\frak A^f$ such that
$\sum_{n=0}^{\infty}\rho(a_n,a_{n+1})<\infty$, set 
$d=\sup_{n\in\Bbb N}\inf_{m\ge n}a_m$.   By 323F, 
$\lim_{n\to\infty}\bar\mu(d\Bsymmdiff a_n)=0$.   In particular, there is
some $n\in\Bbb N$ such that $\bar\mu(d\Bsetminus a_n)$ is finite, so
$d\in\frak A^f$ and $\lim_{n\to\infty}\rho(d,a_n)=0$.   As in 2A4E, this is
enough to show that $\frak A^f$ is complete.
}%end of proof of 323M

\exercises{
\leader{323X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 323Xa
Let $(X,\Sigma,\mu)$ be a measure space, and
$(\frak A,\bar\mu)$ its measure algebra.   (i) Show that we have an
injection $\chi:\frak A\to L^0(\mu)$ (see \S241) given by setting
$\chi(E^{\ssbullet})=(\chi E)^{\ssbullet}$ for every $E\in\Sigma$.
(ii) Show that $\chi$ is a homeomorphism between $\frak A$ and its image
if $\frak A$ is given its measure-algebra topology and $L^0(\mu)$ is
given its topology of convergence in measure (245A).
%323A

\sqheader 323Xb\dvAformerly{3{}23Xg} 
Let $(\frak A,\bar\mu)$ be a measure algebra and $\rho$
the measure metric on the ideal $\frak A^f$ of elements of finite
measure.   (i)
Show that the embedding $\frak A^f\subseteq\frak A$ is
uniformly continuous for
the measure-algebra uniformity on $\frak A$.   (ii) In the context of
323Xa, show that $\chi:\frak A^f\to L^0(\mu)$ is an isometry between
$\frak A^f$ and a subset of $L^1(\mu)$.
%323Xa, 323B

\spheader 323Xc
Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra.   Show
that the set $\{(a,b):a\Bsubseteq b\}$ is a closed set in
$\frak A\times\frak A$.
%323B

\sqheader 323Xd Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space
and $(\frak A,\bar\mu)$ its measure algebra.   (i) Show that if $\Tau$
is a $\sigma$-subalgebra of $\Sigma$, then $\{F^{\ssbullet}:F\in\Tau\}$
is a closed subalgebra of $\frak A$.   (ii) Show that if $\frak B$ is a
closed subalgebra of $\frak A$, then
$\{F:F\in\Sigma,\,F^{\ssbullet}\in\frak B\}$ is a $\sigma$-subalgebra of
$\Sigma$.
%323H 

\spheader 323Xe Let $(\frak A,\bar\mu)$ be a localizable measure
algebra, and $C\subseteq\frak A$ a set such that $\sup A$, $\inf A$
belong to $C$ for all non-empty subsets $A$ of $C$.   Show that $C$ is
topologically closed.
%323H

\spheader 323Xf Show that if $(\frak A,\bar\mu)$ is any measure
algebra and $\frak B$ is a subalgebra of $\frak A$, then its topological
closure $\overline{\frak B}$ is again a subalgebra.
%323J

\spheader 323Xg Let $(\frak A,\bar\mu)$ be a measure algebra, and
$e\in\frak A$;  let $\frak A_e$ be the principal ideal of $\frak A$
generated by $e$, and $\bar\mu_e$ its measure (322H).  (i) Show that the
measure-algebra 
topology on $\frak A_e$ defined by $\bar\mu_e$ is just the subspace
topology induced by the measure-algebra topology of $\frak A$.
(ii)\dvAnew{2011} Show that the
measure-algebra uniformity on $\frak A_e$ is the subspace
uniformity induced by the measure-algebra uniformity of $\frak A$.
(iii) Show that the strong
measure-algebra topology on $\frak A_e^f$ is the subspace
topology induced by the strong measure-algebra topology of $\frak A^f$.
%323L

\spheader 323Xh Let $(\frak A,\bar\mu)$ be a measure algebra.  Show that
its localization (322P) can be identified with its completion under its
measure-algebra uniformity.
%323C out of order query

\leader{323Y}{Further exercises (a)}
%\spheader 323Ya
Let $(\frak A,\bar\mu)$ be a $\sigma$-finite
measure algebra.   Show that a set $F\subseteq\frak A$ is
topologically closed iff $e\in F$ whenever there are non-empty sets
$B$, $C\subseteq \frak A$ such that $B$ is upwards-directed,
$C$ is downwards-directed, $\sup B=\inf C=e$ and
$[b,c]\cap F\ne\emptyset$ for every $b\in B$, $c\in C$, writing
$[b,c]=\{d:b\Bsubseteq d\Bsubseteq c\}$.
%323H

\spheader 323Yb Give an example to show that (a) is false for
general localizable measure algebras.   %mt32bits
%323Ya, 323H

\spheader 323Yc Give an example of a semi-finite measure algebra
$(\frak A,\bar\mu)$ with an order-closed subalgebra which is not
topologically closed.   %mt32bits
%323H

\spheader 323Yd Let $(\frak A,\bar\mu)$ be a probability algebra and
write $\Bbb B$ for the family of closed subalgebras of $\frak A$.   For
$\frak B$, $\frak C\in\Bbb B$ set
$\rho(\frak B,\frak C)
=\max(\sup_{b\in\frak B}\inf_{c\in\frak C}\bar\mu(b\Bsymmdiff c),
\sup_{c\in\frak C}\inf_{b\in\frak B}\bar\mu(b\Bsymmdiff c))$.   Show
that $(\Bbb B,\rho)$ is a complete metric space.   (Cf.\ 246Yb, 4A2T.)
%323+

\spheader 323Ye Let $(\frak A,\bar\mu)$ be the measure algebra of
Lebesgue measure on $\Bbb R$.   Show that it is separable in its
measure-algebra topology.   \Hint{245Yj.}

}%end of exercises

\cmmnt{
\Notesheader{323} The message of this section is that the topology
of a measure algebra is essentially defined by its order and algebraic
structure;  see also 324F-324H %324F 324G 324H
below.   Of course the results are really
about semi-finite measure algebras, and indeed this whole volume, like
the rest of measure theory, has little of interest to say about others;
they are included only because they arise occasionally and it is not
absolutely essential to exclude them.   We therefore expect to be able
to describe such things as closed subalgebras and continuous
homomorphisms in terms of the ordering, as in 323H and 324G.
For $\sigma$-finite algebras, indeed, there is an easy description of
the topology in terms of the order (323Ya).   I think the result of this
section on which I shall most often depend is 323H:  in most
contexts, there is no need to distinguish between `topologically
closed subalgebra' and `order-closed subalgebra'.
%323H not especially often quoted by name;  but phrase `closed subalgebra'
%very common.
However a $\sigma$-subalgebra of
a localizable measure algebra need not be topologically sequentially
closed;  there is an example in {\smc Fremlin Pagter \& Ricker 05}.

It is also the case that the topology of a measure algebra corresponds
very closely indeed to the topology of convergence in measure.   A
description of this correspondence is in 323Xa.  Indeed all the results
of this section have analogues in the theory of topological Riesz
spaces.   I will enlarge on the
idea here in \S367.   For the moment, however, if you look back to
Chapter 24, you will see that 323B and 323G are closely paralleled by
245D and 245E, while 323Ya is related to 245L.

It is I think natural to ask whether there are any other topological
Boolean algebras with the properties 323B-323D.   %323B 323C 323D
In fact there are;  see 393G and 393Xf below.
}%end of notes

\discrpage

\frfilename{mt324.tex} 
\versiondate{29.11.17} 
\copyrightdate{1998} 
      
\def\chaptername{Measure algebras} 
\def\sectionname{Homomorphisms} 
      
\newsection{324} 
      
In the course of Volume 2, I had occasion to remark that elementary 
measure theory is unusual among abstract topics in pure mathematics in 
not being dominated by any particular class of structure-preserving 
operators.   We now come to what I think is one of the reasons for the 
gap:  the most important operators of the theory are not between measure 
spaces at all, but between their measure algebras.   In this section I 
run through the most elementary facts about Boolean homomorphisms 
between measure algebras.   I start with results on the construction of 
such homomorphisms from functions between measure spaces 
(324A-324E), %324A 324B 324C 324D 324E
then investigate continuity and order-continuity of homomorphisms 
(324F-324H) %324F 324G 324H
before turning to measure-preserving homomorphisms 
(324I-324P). %324I 324J 324K 324L 324M 324N 324O 324P
      
\leader{324A}{Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ 
be measure spaces, and $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ their 
measure algebras.   Write $\hat\Sigma$ for the domain of the completion 
$\hat\mu$ of $\mu$.   Let $D\subseteq X$ be a set of full outer 
measure\cmmnt{ (definition: 132F)}, and $\hat\Sigma_D$ the 
subspace $\sigma$-algebra on $D$ induced by $\hat\Sigma$\cmmnt{ (121A)}.  Let 
$\phi:D\to Y$ be a function such that 
$\phi^{-1}[F]\in\hat\Sigma_D$ for every $F\in\Tau$ and 
$\phi^{-1}[F]$ is $\mu$-negligible whenever $\nu F=0$.   Then 
there is a sequentially order-continuous Boolean homomorphism  
$\pi:\frak B\to\frak A$ defined by the formula 
      
\Centerline{$\pi F^{\ssbullet}=E^{\ssbullet}$ whenever $F\in\Tau$, 
$E\in\Sigma$ and $(E\cap D)\symmdiff\phi^{-1}[F]$ is negligible.} 
      
\proof{ Let $F\in\Tau$.  Then there is an $H\in\hat\Sigma$ such that 
$H\cap D=\phi^{-1}[F]$;  now there is an $E\in\Sigma$ such that 
$E\symmdiff H$ is negligible, so that $(E\cap D)\symmdiff\phi^{-1}[F]$ 
is negligible.   If $E_1$ is another member of $\Sigma$ such that 
$(E_1\cap D)\symmdiff\phi^{-1}[F]$ is negligible, then $(E\symmdiff 
E_1)\cap D$ is negligible, so is included in a negligible member $G$ of 
$\Sigma$.   Since $(E\symmdiff E_1)\setminus G$ belongs to $\Sigma$ and 
is disjoint from $D$, it is negligible;  accordingly $E\symmdiff E_1$ is 
negligible and $E^{\ssbullet}=E_1^{\ssbullet}$ in $\frak A$. 
      
What this means is that the formula offered defines a map $\pi:\frak 
B\to\frak A$.   It is now easy to check that $\pi$ is a Boolean 
homomorphism, because if 
      
\Centerline{$(E\cap D)\symmdiff\phi^{-1}[F]$, \quad$(E'\cap 
D)\symmdiff\phi^{-1}[F']$} 
      
\noindent are negligible, so are 
      
\Centerline{$((E\cap E')\cap D)\symmdiff\phi^{-1}[F\cap F']$, 
\quad$((X\setminus E)\cap D)\symmdiff\phi^{-1}[Y\setminus F]$,} 

\noindent and we can apply 312H.
      
To see that $\pi$ is sequentially order-continuous, let 
$\sequencen{b_n}$ be a sequence in $\frak B$.  For each $n$ we may 
choose an $F_n\in\Tau$ such that $F_n^{\ssbullet}=b_n$, and 
$E_n\in\Sigma$ such that $(E_n\cap D)\symmdiff\phi^{-1}[F_n]$ is 
negligible;  now, setting $F=\bigcup_{n\in\Bbb N}F_n$, 
$E=\bigcup_{n\in\Bbb N}E_n$, 
      
\Centerline{$(E\cap D)\symmdiff\phi^{-1}[F] 
\subseteq\bigcup_{n\in\Bbb N}(E_n\cap D)\symmdiff\phi^{-1}[F_n]$} 
      
\noindent is negligible, so 
      
\Centerline{$\pi(\sup_{n\in\Bbb N}b_n)=\pi(F^{\ssbullet}) 
=E^{\ssbullet}=\sup_{n\in\Bbb N}E_n^{\ssbullet} 
=\sup_{n\in\Bbb N}\pi b_n$.} 
      
\noindent (Recall that the maps $E\mapsto E^{\ssbullet}$,  
$F\mapsto F^{\ssbullet}$ are sequentially order-continuous, by 321H.)    
So $\pi$ is sequentially order-continuous (313L(c-iii)). 
}%end of proof of 324A 
      
      
\leader{324B}{Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ 
be measure spaces, and $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ their 
measure algebras.   Let $\phi:X\to Y$ be a function such that 
$\phi^{-1}[F]\in\Sigma$ for every $F\in\Tau$ and $\mu\phi^{-1}[F]=0$ 
whenever $\nu F=0$.   Then there is a sequentially order-continuous 
Boolean homomorphism $\pi:\frak B\to\frak A$ defined by the formula 
      
\Centerline{$\pi F^{\ssbullet}=(\phi^{-1}[F])^{\ssbullet}$ for every 
$F\in\Tau$.} 
      
      
\cmmnt{ 
\leader{324C}{Remarks (a)} In \S235 and elsewhere in Volume 2 I spent 
a good deal of time on functions between measure spaces which satisfy 
the conditions of 324A.   Indeed, I take the trouble to spell 324A out 
in such generality just in order to catch these applications.   Some of 
the results of the present chapter (322D, 322Jb) can also be regarded as 
special cases of 324A. 
      
\spheader 324Cb The question of which homomorphisms between the measure 
algebras of 
measure spaces $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ can be realized 
by functions between $X$ and $Y$ is important and deep;  I will return 
to it in \S\S343-344. 
      
\spheader 324Cc In the simplified context of 324B, I have actually 
defined a contravariant functor;  the relevant facts are the following. 
}%end of comment 
      
\leader{324D}{Proposition} Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ 
and $(Z,\Lambda,\lambda)$ be measure spaces, with measure algebras 
$(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$, $(\frak C,\bar\lambda)$. 
Suppose that $\phi:X\to Y$ and $\psi:Y\to Z$ satisfy the conditions of 
324B, that is, 
      
\Centerline{$\phi^{-1}[F]\in\Sigma$ if $F\in\Tau$, 
\quad$\mu\phi^{-1}[F]=0$ if $\nu F=0$,} 
      
\Centerline{$\psi^{-1}[G]\in\Tau$ if $G\in\Lambda$, 
\quad$\mu\psi^{-1}[G]=0$ if $\lambda G=0$.} 
      
\noindent Let $\pi_{\phi}:\frak B\to\frak A$, $\pi_{\psi}:\frak 
C\to\frak B$ be the corresponding homomorphisms.   Then $\psi\phi:X\to 
Z$ is another map of the same type, and 
$\pi_{\psi\phi}=\pi_{\phi}\pi_{\psi}:\frak C\to\frak A$. 
      
\proof{ The necessary checks are all elementary. 
}%end of proof of 324D 
      
\leader{324E}{Stone \dvrocolon{spaces}}\cmmnt{ While in the context of 
general measure 
spaces the question of realizing homomorphisms is difficult, in the case 
of the Stone representation it is relatively straightforward. 
      
\medskip 
      
\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ and 
$(\frak B,\bar\nu)$ be measure algebras, with Stone spaces $Z$ and $W$;   
let $\mu$, $\nu$ be the corresponding measures on $Z$ and 
$W$\cmmnt{, as described in 321J-321K}, and $\Sigma$, $\Tau$ their domains.   If 
$\pi:\frak B\to\frak A$ is any order-continuous Boolean homomorphism, 
let $\phi:Z\to W$ be 
the corresponding continuous function\cmmnt{, as described in 312Q}. 
Then $\phi^{-1}[F]\in\Sigma$ for every $F\in\Tau$, $\mu\phi^{-1}[F]=0$ 
whenever $\nu F=0$, and (writing $E^*$ for the member of $\frak A$ 
corresponding to $E\in\Sigma$) $\pi F^*=(\phi^{-1}[F])^*$ for every 
$F\in\Tau$. 
      
\proof{ Recall that $E^*=a$ iff 
$E\symmdiff\widehat{a}$ is meager, where $\widehat{a}$ is the 
open-and-closed subset of $Z$ corresponding to $a\in\frak A$.   In 
particular, $\mu E=0$ 
iff $E$ is meager.   Now the point is that $\phi^{-1}[F]$ is nowhere 
dense in $Z$ whenever $F$ is a nowhere dense subset of $W$, by 313R. 
Consequently $\phi^{-1}[F]$ is meager whenever $F$ is meager in $W$, 
since $F$ is then just a countable union of nowhere dense sets. 
Thus we see already that $\mu\phi^{-1}[F]=0$ whenever $\nu F=0$. 
If $F$ is any member of $\Tau$, there is an open-and-closed set $F_0$ 
such that $F\symmdiff F_0$ is meager;  now $\phi^{-1}[F_0]$ is 
open-and-closed, so 
$\phi^{-1}[F]=\phi^{-1}[F_0]\symmdiff\phi^{-1}[F\symmdiff F_0]$ belongs 
to $\Sigma$.   Moreover, if $b\in \frak B$ is such that 
$\widehat{b}=F_0$, 
and $a=\pi b$, then $\widehat{a}=\phi^{-1}[F_0]$, so 
      
\Centerline{$\pi F^*=\pi b=a=(\phi^{-1}[F_0])^*=(\phi^{-1}[F])^*$,} 
      
\noindent as required. 
}%end of proof of 324E 
      
\leader{324F}{}\cmmnt{ I turn now to the behaviour of 
order-continuous 
homomorphisms between measure algebras. 
      
\medskip 
      
\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ 
be measure 
algebras and $\pi:\frak A\to\frak B$ a Boolean homomorphism. 
Give $\frak A$ and $\frak B$ their measure-algebra topologies and 
uniformities\cmmnt{ (323Ab)}. 
      
(a) $\pi$ is continuous iff it is continuous at $0$ iff it is uniformly 
continuous. 
      
(b) If $(\frak B,\bar\nu)$ is semi-finite and $\pi$ is continuous, then 
it is order-continuous. 
      
(c) If $(\frak A,\bar\mu)$ is semi-finite and $\pi$ is order-continuous, 
then it is continuous. 
      
\proof{ I use the notations $\frak A^f$, $\rho_a$ from 323A. 
      
\medskip 
      
{\bf (a)} Suppose that $\pi$ is continuous at $0$;   I seek to show that 
it is uniformly continuous.  Take $b\in\frak B^f$ and $\epsilon>0$. 
Then there are $a_0,\ldots,a_n\in\frak A^f$ and $\delta>0$ such that 
      
\Centerline{$\bar\nu(b\Bcap \pi c)=\rho_b(\pi c,0)\le\epsilon$ 
whenever $\max_{i\le n}\rho_{a_i}(c,0)\le\delta$;} 
      
\noindent setting $a=\sup_{i\le n}a_i$, 
      
\Centerline{$\bar\nu(b\Bcap\pi c)\le\epsilon$ whenever $\bar\mu(a\Bcap 
c)\le\delta$.} 
      
Now suppose that $\rho_a(c,c')\le\delta$.   Then 
$\bar\mu(a\Bcap(c\Bsymmdiff c'))\le\delta$, so 
      
\Centerline{$\rho_b(\pi c,\pi c') 
=\bar\nu(b\Bcap(\pi c\Bsymmdiff\pi c')) 
=\bar\nu(b\Bcap\pi(c\Bsymmdiff c')) 
\le\epsilon$.} 
      
\noindent As $b$ and $\epsilon$ are arbitrary, $\pi$ is uniformly 
continuous.   The rest of the implications are elementary. 
      
\medskip 
      
{\bf (b)} Let $A$ be a non-empty downwards-directed set in $\frak A$ 
with infimum $0$.   Then $0\in\overline{A}$ (323D(b-ii));  because $\pi$ 
is continuous, $0\in\overline{\pi[A]}$.   \Quer\ If $b$ is a non-zero 
lower bound for $\pi[A]$ in $\frak B$, then (because $(\frak B,\bar\nu)$ 
is semi-finite) there is a $c\Bsubseteq b$ with $0<\bar\nu c<\infty$; 
now 
      
\Centerline{$\rho_c(\pi a,0)=\bar\nu(c\Bcap\pi a)=\bar\nu c>0$} 
      
\noindent for every $a\in A$, so $0\notin\overline{\pi[A]}$.\ \Bang 
      
Thus $\inf\pi[A]=0$ in $\frak B$;  as $A$ is arbitrary, $\pi$ is 
order-continuous (313L(b-ii)). 
      
\medskip 
      
{\bf (c)} By (a), it will be enough to show that $\pi$ is continuous at 
$0$.   Take $b\in\frak B^f$ and $\epsilon>0$.   \Quer\ Suppose, if possible, 
that whenever $a\in\frak A^f$ and $\delta>0$ there is a $c\in\frak A$ such 
that $\bar\mu(a\Bcap c)\le\delta$ but $\bar\nu(b\Bcap\pi c)\ge\epsilon$. 
For 
each $a\in\frak A^f$, $n\in\Bbb N$ choose $c_{an}$ such that 
$\bar\mu(a\Bcap 
c_{an})\le2^{-n}$ but $\bar\nu(b\Bcap\pi c_{an})\ge\epsilon$.   Set 
$c_a=\inf_{n\in\Bbb N}\sup_{m\ge n}c_{am}$;  then 
      
\Centerline{$\bar\mu(a\Bcap c_a)\le\inf_{n\in\Bbb 
N}\sum_{m=n}^{\infty}\bar\mu(a\Bcap c_{an})=0$,} 
      
\noindent so $c_a\Bcap a=0$.   On the other hand, because $\pi$ is 
order-continuous, $\pi c_a=\inf_{n\in\Bbb N}\sup_{m\ge n}\pi c_{am}$, 
so that 
      
\Centerline{$\bar\nu(b\Bcap\pi 
c_a)=\lim_{n\to\infty}\bar\nu(b\Bcap\sup_{m\ge 
n}\pi c_{am})\ge\epsilon$.} 
      
\noindent This shows that 
      
\Centerline{$\rho_b(\pi(1\Bsetminus a),0) 
=\bar\nu(b\Bcap\pi(1\Bsetminus a)) 
\ge\bar\nu(b\Bcap\pi c_a)\ge\epsilon$.} 
      
\noindent But now observe that $A=\{1\Bsetminus a:a\in\frak A^f\}$ is a 
downwards-directed subset of $\frak A$ with infimum $0$, because  
$(\frak A,\bar\mu)$ is semi-finite.   So $\pi[A]$ is downwards-directed 
and has infimum $0$,  
and $0$ must be in the closure of $\pi[A]$, by 323D(b-ii) again; 
while we have just seen that $\rho_b(d,0)\ge\epsilon$ for every 
$d\in\pi[A]$.   \Bang 
      
Thus there must be $a\in\frak A^f$, $\delta>0$ such that 
      
\Centerline{$\rho_b(\pi c,0)=\bar\nu(b\Bcap\pi c)\le\epsilon$} 
      
\noindent whenever 
      
\Centerline{$\rho_a(c,0)=\bar\mu(a\Bcap c)\le\delta$.} 
      
\noindent As $b$ and $\epsilon$ are arbitrary, $\pi$ is continuous at $0$ 
and therefore continuous. 
}%end of proof of 324F 
      
\leader{324G}{Corollary} If $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ 
are semi-finite measure algebras, a Boolean homomorphism  
$\pi:\frak A\to\frak B$ is continuous iff it is order-continuous. 
      
\leader{324H}{Corollary} If $\frak A$ is a Boolean algebra and 
$\bar\mu$, 
$\bar\nu$ are two measures both rendering $\frak A$ a semi-finite 
measure 
algebra, then they endow $\frak A$ with the same uniformity 
(and\cmmnt{, of 
course,} the same topology). 
      
\proof{ By 324G, the identity map from $\frak A$ to itself is continuous 
whichever of the topologies we place on $\frak A$;  and by 324Fa it is 
therefore uniformly continuous. 
}%end of proof of 324H 
      
\leader{324I}{Definition} Let $(\frak A,\bar\mu)$ and 
$(\frak B,\bar\nu)$ be 
measure algebras.   A Boolean homomorphism $\pi:\frak A\to\frak B$ is 
{\bf measure-preserving} if $\bar\nu(\pi a)=\bar\mu a$ for every 
$a\in\frak A$. 
      
\leader{324J}{Proposition} Let $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ 
and $(\frak C,\bar\lambda)$ be measure algebras, and 
$\pi:\frak A\to\frak B$, 
$\theta:\frak B\to\frak C$ measure-preserving Boolean homomorphisms. 
Then $\theta\pi:\frak A\to\frak C$ is a measure-preserving Boolean 
homomorphism. 
      
\proof{ Elementary. 
}%end of proof of 324J 
      
\vleader{108pt}{324K}{Proposition} Let $(\frak A,\bar\mu)$ and 
$(\frak B,\bar\nu)$ be measure algebras, and $\pi:\frak A\to\frak B$ a 
measure-preserving Boolean homomorphism. 
      
(a) $\pi$ is injective. 
      
(b) $(\frak A,\bar\mu)$ is totally finite iff $(\frak B,\bar\nu)$ is, 
and in this case $\pi$ is 
order-continuous, therefore continuous, and $\pi[\frak A]$ is a closed 
subalgebra of $\frak B$. 
      
(c) If $(\frak A,\bar\mu)$ is semi-finite and $(\frak B,\bar\nu)$ is 
$\sigma$-finite, then $(\frak A,\bar\mu)$ is $\sigma$-finite. 
      
(d) If $(\frak A,\bar\mu)$ is $\sigma$-finite and $\pi$ is sequentially 
order-continuous, then $(\frak B,\bar\nu)$ is $\sigma$-finite. 
      
(e) If $(\frak A,\bar\mu)$ is semi-finite and $\pi$ is 
order-continuous, then $(\frak B,\bar\nu)$ is semi-finite. 
      
(f) If $(\frak A,\bar\mu)$ is atomless and semi-finite, and $\pi$ is 
order-continuous, then $\frak B$ is atomless. 
      
(g) If $\frak B$ is purely atomic and $(\frak A,\bar\mu)$ is 
semi-finite, then $\frak A$ is purely atomic. 
      
\proof{{\bf (a)} If $a\ne 0$ in $\frak A$, then  
$\bar\nu\pi a=\bar\mu a>0$ so $\pi a\ne 0$.   By 3A2Db, $\pi$ is injective. 
      
\medskip 
      
{\bf (b)} Because 
      
\Centerline{$\bar\nu 1_{\frak B}=\bar\nu\pi 1_{\frak A} 
=\bar\mu 1_{\frak A}$,} 
      
\noindent $(\frak A,\bar\mu)$ is totally finite iff $(\frak B,\bar\nu)$ 
is.   Now suppose that $A\subseteq\frak A$ is downwards-directed and 
non-empty and that $\inf A=0$.   Then 
      
\Centerline{$\inf_{a\in A}\bar\nu\pi a=\inf_{a\in A}\bar\mu a=0$} 
      
\noindent by 321F.   So $\bar\nu b=0$ for any lower bound $b$ of 
$\pi[A]$, and $\inf\pi[A]=0$.   As $A$ is arbitrary, $\pi$ is 
order-continuous, by 313Lb again. 
      
By 324Fc, $\pi$ is continuous.   By 314Fa, $\pi[\frak A]$ is 
order-closed in $\frak B$, that is, `closed' in the sense of 323I. 
      
\medskip 
      
{\bf (c)} I appeal to 322G.   If $C$ is a disjoint family in  
$\frak A\setminus\{0\}$, then  
$\langle\pi c\rangle_{c\in C}$ is a disjoint family in  
$\frak B\setminus\{0\}$, 
so is countable, and $C$ must be countable, because $\pi$ is injective. 
Thus $\frak A$ is ccc and 
(being semi-finite) $(\frak A,\bar\mu)$ is $\sigma$-finite. 
      
\medskip 
      
{\bf (d)} Let $\sequencen{a_n}$ be a sequence in $\frak A$ such that 
$\bar\mu a_n<\infty$ for every $n$ and $\sup_{n\in\Bbb N}a_n=1$.   Then 
$\bar\nu\pi a_n<\infty$ for every $n$ and (because $\pi$ is sequentially 
order-continuous) $\sup_{n\in\Bbb N}\pi a_n=1$, so $(\frak B,\bar\nu)$ 
is $\sigma$-finite. 
      
\medskip 
      
{\bf (e)} Setting $\frak A^f=\{a:\bar\mu a<\infty\}$, $\sup \frak 
A^f=1$; 
because $\pi$ is order-continuous, $\sup\pi[\frak A^f]=1$ in $\frak B$. 
So if $\bar\nu b=\infty$, there is an $a\in\frak A^f$ such that $\pi 
a\Bcap 
b\ne 0$, and now $0<\bar\nu(b\Bcap\pi a)<\infty$. 
      
\medskip 
      
{\bf (f)} Take any non-zero $b\in\frak B$.   As in (e), there is an 
$a\in\frak A$ such that $\bar\mu a<\infty$ and $\pi a\Bcap b\ne 0$.   If 
$\pi a\Bcap b\ne b$, then surely $b$ is not an atom.   Otherwise, set 
      
\Centerline{$C=\{c:c\in\frak A,\,c\Bsubseteq a,\,b\Bsubseteq\pi c\}$.} 
      
\noindent Then $C$ is downwards-directed and contains $a$, so 
$c_0=\inf C$ is defined in $\frak A$ (321F again), and 
      
\Centerline{$\bar\mu c_0=\inf_{c\in C}\bar\mu c\ge\bar\nu b>0$,} 
      
\noindent so $c_0\ne 0$.   Because $\frak A$ is atomless, there is a 
$d\Bsubseteq c_0$ such that neither $d$ nor $c_0\Bsetminus d$ is zero, 
so that neither $c_0\Bsetminus d$ nor $d$ can belong to $C$.  But this 
means that $b\Bcap\pi d$ and $b\Bcap\pi(c_0\Bsetminus d)$ are both 
non-zero, so that again $b$ is not an atom. 
As $b$ is arbitrary, $\frak B$ is atomless. 
      
\medskip 
      
{\bf (g)} Take any non-zero $a\in\frak A$.   Then there is an 
$a'\Bsubseteq a$ such that $0<\bar\mu a'<\infty$.   Because $\frak B$ is 
purely atomic, there is an atom $b$ of $\frak B$ with $b\Bsubseteq\pi 
a'$.   Set 
      
\Centerline{$C=\{c:c\in\frak A,\,c\Bsubseteq a',\,b\Bsubseteq\pi c\}$.} 
      
\noindent Then $C$ is downwards-directed and contains $a'$, so $c_0=\inf 
C$ is defined in $\frak A$, and 
      
\Centerline{$\bar\mu c_0=\inf_{c\in C}\bar\mu c\ge\bar\nu b>0$,} 
      
\noindent so $c_0\ne 0$.  If $d\Bsubseteq c_0$, then $b\Bcap \pi d$ must 
be either $b$ or $0$.   If $b\Bcap\pi d=b$, then $d\in C$ and $d=c_0$. 
If $b\Bcap\pi d=0$, then $c_0\Bsetminus d\in C$ and $d=0$.   Thus $c_0$ 
is an atom in $\frak A$.   As $a$ is arbitrary, $\frak A$ is purely 
atomic. 
}%end of proof of 324K 
      
\leader{324L}{Corollary} Let $(\frak A,\bar\mu)$ be a totally finite 
measure algebra, $(\frak B,\bar\nu)$ a measure algebra, and 
$\pi:\frak A\to\frak B$ 
a measure-preserving homomorphism.   If $C\subseteq\frak A$ and 
$\frak C$ is the closed subalgebra of $\frak A$ generated by $C$, then 
$\pi[\frak C]$ is the closed subalgebra of $\frak B$ generated by 
$\pi[C]$. 
      
\proof{ By 324Ka, $\pi$ is order-continuous, so we can apply 314H. 
}%end of proof of 324L 
      
\leader{324M}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be 
measure spaces, with measure algebras $(\frak A,\bar\mu)$ and 
$(\frak B,\bar\nu)$.   Let $\phi:X\to Y$ be \imp.    
Then we have a sequentially order-continuous measure-preserving Boolean 
homomorphism $\pi:\frak B\to\frak A$ defined by setting 
$\pi F^{\ssbullet}=\phi^{-1}[F]^{\ssbullet}$ for every $F\in\Tau$. 
      
\proof{ This is immediate from 324B. 
}%end of proof of 324M 
      
\leader{324N}{Proposition} Let $(\frak A,\bar\mu)$ and 
$(\frak B,\bar\nu)$ be measure algebras, with Stone spaces $Z$ and $W$;    
let $\mu$, $\nu$ be the corresponding measures on $Z$ and $W$.   If 
$\pi:\frak B\to\frak A$ is an order-continuous 
measure-preserving Boolean homomorphism, and $\phi:Z\to W$ the 
corresponding continuous function, then $\phi$ is \imp. 
      
\proof{ Use 324E.   In the notation there, if $F\in\Tau$, then 
      
\Centerline{$\nu F=\bar\nu F^*=\bar\mu\pi F^*=\bar\mu\phi^{-1}[F]^* 
=\mu\phi^{-1}[F]$.} 
}%end of proof of 324N 
      
\leader{324O}{Proposition} Let $(\frak A,\bar\mu)$ and 
$(\frak B,\bar\nu)$ be 
totally finite measure algebras, $\frak A_0$ a topologically dense 
subalgebra of $\frak A$, and $\pi:\frak A_0\to\frak B$ a Boolean 
homomorphism such that $\bar\nu\pi a=\bar\mu a$ for every 
$a\in\frak A_0$. 
Then $\pi$ has a unique extension to a measure-preserving homomorphism 
from $\frak A$ to $\frak B$. 
      
\proof{ Let $\rho$, $\sigma$ be the measure metrics on $\frak A$, 
$\frak B$ respectively, as in 323Ad.   Then for any $a$,  
$a'\in\frak A_0$ 
      
\Centerline{$\sigma(\pi a,\pi a')=\bar\nu(\pi a\symmdiff\pi a') 
=\bar\nu\pi(a\symmdiff a')=\bar\mu(a\symmdiff a')=\rho(a,a')$;} 
      
\noindent that is, $\pi:\frak A_0\to\frak B$ is an isometry.   Because 
$\frak A_0$ is dense in the metric space $(\frak A,\rho)$, while  
$\frak B$ is complete under $\sigma$ (323Gc), there is a unique continuous 
function $\hat\pi:\frak A\to\frak B$ extending $\pi$ (3A4G).   Now 
the operations 
      
\Centerline{$(a,a')\mapsto \hat\pi(a\Bcap a')$, 
\quad$(a,a')\mapsto \hat\pi 
a\Bcap\hat\pi a':\frak A\times\frak A\to\frak B$,} 
      
\noindent are continuous and agree on the dense subset  
$\frak A_0\times\frak A_0$ of $\frak A\times\frak A$;  because the topology of 
$\frak B$ is Hausdorff, they agree on $\frak A\times\frak A$, that is, 
$\hat\pi(a\Bcap a')=\hat\pi a\Bcap\hat\pi a'$ for all $a$,  
$a'\in\frak A$ (2A3Uc).   Similarly, the operations 
      
\Centerline{$a\mapsto\hat\pi(1\Bsetminus a)$, 
\quad$a\mapsto 1\Bsetminus\hat\pi a:\frak A\to\frak B$} 
      
\noindent are continuous and agree on the dense subset $\frak A_0$ of 
$\frak A$, so they agree on $\frak A$, that is, 
$\hat\pi(1\Bsetminus a)=1\Bsetminus a$ for every $a\in\frak A$.   
Thus $\hat\pi$ is a Boolean 
homomorphism (312H again).   To see that it is measure-preserving, observe that 
      
\Centerline{$a\mapsto\bar\mu a=\rho(a,0)$, 
\quad$a\mapsto\bar\nu(\hat\pi 
a)=\sigma(\hat\pi a,0):\frak A\to\Bbb R$} 
      
\noindent are continuous and agree on $\frak A_0$, so agree on $\frak A$. 
Finally, $\hat\pi$ is the only measure-preserving Boolean 
homomorphism extending $\pi$, because any such map must be continuous 
(324Kb), and $\hat\pi$ is the only continuous extension of $\pi$. 
}%end of proof of 324O 
      
\leader{*324P}{}\cmmnt{ The following fact will be applied in \S387, 
by which time it will seem perfectly elementary;  for the moment, it may 
be a useful exercise. 
      
\medskip 
      
\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ and  
$(\frak B,\bar\nu)$ be totally finite measure algebras such that  
$\bar\mu 1=\bar\nu 1$.   Suppose that $A\subseteq\frak A$ and  
$\phi:A\to\frak B$ 
are such that $\bar\nu(\inf_{i\le n}\phi a_i)=\bar\mu(\inf_{i\le n}a_i)$ 
for all $a_0,\ldots,a_n\in A$.   Let $\frak C$ be the smallest closed 
subalgebra of $\frak A$ including $A$.   Then $\phi$ has a unique 
extension to a measure-preserving Boolean homomorphism from $\frak C$ to 
$\frak B$. 
      
\proof{{\bf (a)} Let $\Psi$ be the family of all functions $\psi$ 
extending $\phi$ and having the same properties;  that is, $\psi$ is a 
function from a subset of $\frak A$ to $\frak B$, and 
$\bar\nu(\inf_{i\le n}\psi a_i)=\bar\mu(\inf_{i\le n}a_i)$ for all 
$a_0,\ldots,a_n\in\dom\psi$.   By Zorn's Lemma, $\Psi$ has a maximal 
member $\theta$.   Write $D$ for the domain of $\theta$. 
      
\medskip 
      
{\bf (b)(i)} If $c$, $d\in D$ then $c\Bcap d\in D$.   \Prf\Quer\ 
Otherwise, set $D'=D\cup\{c\Bcap d\}$ and extend $\theta$ to 
$\theta':D'\to\frak B$ by writing  
$\theta'(c\Bcap d)=\theta c\Bcap\theta d$.    
It is easy to check that $\theta'\in\Psi$, which is supposed to be 
impossible.\ \Bang\Qed 
      
Now 
      
\Centerline{$\bar\nu(\theta c\Bcap\theta d\Bcap\theta(c\Bcap d)) 
=\bar\mu(c\Bcap d) 
=\bar\nu(\theta c\Bcap\theta d) 
=\bar\nu\theta(c\Bcap d)$,} 
      
\noindent so $\theta(c\Bcap d)=\theta c\Bcap\theta d$. 
      
\medskip 
      
\quad{\bf (ii)} If $d\in D$ then $1\Bsetminus d\in D$.   \Prf\Quer\ 
Otherwise, set $D'=D\cup\{1\Bsetminus d\}$ and extend $\theta$ to $D'$ 
by writing $\theta'(1\Bsetminus d)=1\Bsetminus\theta d$.   Once again, 
it is easy to check that $\theta'\in\Psi$, which is 
impossible.\ \Bang\Qed 
      
Consequently (since $D$ is certainly not empty, even if $A$ is), $D$ is 
a subalgebra of $\frak A$ (312B(iii)). 
      
\medskip 
      
\quad{\bf (iii)} Since 
      
\Centerline{$\bar\nu\theta 1=\bar\mu 1=\bar\nu 1$,} 
      
\noindent $\theta 1=1$.   If $d\in D$ then 
      
\Centerline{$\bar\nu\theta(1\Bsetminus d) 
=\bar\mu(1\Bsetminus d) 
=\bar\mu 1-\bar\mu d 
=\bar\nu 1-\bar\nu\theta d 
=\bar\nu(1\Bsetminus\theta d)$,} 
      
\noindent while 
      
\Centerline{$\bar\nu(\theta d\Bcap\theta(1\Bsetminus d)) 
=\bar\mu(d\Bcap(1\Bsetminus d))=0$,} 
      
\noindent so $\theta d\Bcap\theta(1\Bsetminus d))=0$, 
$\theta(1\Bsetminus d)\Bsubseteq 1\Bsetminus\theta d$ and 
$\theta(1\Bsetminus d)$ must be equal to $1\Bsetminus\theta d$. 
      
By 312H(ii), $\theta:D\to\frak B$ is a Boolean homomorphism. 
      
\medskip 
      
\quad{\bf (iv)} Let $\frak D$ be the topological closure of $D$ in 
$\frak A$.   Then it is an order-closed subalgebra of $\frak A$ (323J), 
so, with $\bar\mu\restrp\frak D$, is a totally finite measure algebra in 
which $D$ is a topologically dense subalgebra.   By 324O, there is an 
extension of $\theta$ to a measure-preserving Boolean homomorphism from 
$\frak D$ to $\frak B$;  of course this extension belongs to $\Psi$, so 
in fact $D=\frak D$ is a closed subalgebra of $\frak A$. 
      
\medskip 
      
{\bf (c)} Since $A\subseteq D$, $\frak C\subseteq\frak D$ and 
$\phi_1=\theta\restrp\frak C$ is a suitable extension of $\phi$. 
      
To see that $\phi_1$ is unique, let $\phi_2:\frak C\to\frak B$ be any 
other measure-preserving Boolean homomorphism extending $\phi$.   Set 
$C=\{a:\phi_1a=\phi_2a\}$;  then $C$ is a topologically closed 
subalgebra of $\frak A$ including $A$, so is the whole of $\frak C$, and 
$\phi_2=\phi_1$. 
}%end of proof of 324P 
      
\exercises{ 
\leader{324X}{Basic exercises (a)} 
%\spheader 324Xa 
Let $\frak A$ and $\frak B$ be Boolean algebras, 
of which $\frak A$ is Dedekind $\sigma$-complete, and 
$\phi:\frak A\to\frak B$ a sequentially order-continuous Boolean homomorphism.   Let 
$I$ be an ideal of $\frak A$ included in the kernel of $\phi$.   Show 
that we have a sequentially order-continuous Boolean homomorphism 
$\pi:\frak A/I\to \frak B$ given by setting $\pi(a^{\ssbullet})=\phi a$ 
for every $a\in\frak A$. 
%324B 
      
\spheader 324Xb Let $(\frak A,\bar\mu)$ be a measure algebra, and 
$\frak B$ an order-closed subalgebra of $\frak A$ such that
$(\frak B,\bar\mu\restrp\frak B)$ is semi-finite.   Show that
the topology on $\frak B$ induced by $\bar\mu\restrp\frak B$ is just 
the subspace topology induced by the topology of $\frak A$.   
%324F 
      
\spheader 324Xc Let $(X,\Sigma,\mu)$ be a measure space and 
$(X,\tilde\Sigma,\tilde\mu)$ its c.l.d.\ version.   Let $\frak A$, 
$\frak A_2$ be the corresponding measure algebras and  
$\pi:\frak A\to\frak A_2$ the canonical homomorphism (see 322Db).   Show that $\pi$ is topologically continuous. 
%324F 
      
\spheader 324Xd Let $(\frak A,\bar\mu)$ and  
$(\frak B,\bar\nu)$ be measure algebras, and $\pi:\frak A\to\frak B$ a  
bijective measure-preserving Boolean homomorphism.   Show that 
$\pi^{-1}:\frak B\to\frak A$ is a measure-preserving homomorphism. 
%324I 
      
\spheader 324Xe Let $\bar\mu$ be counting measure on $\Cal P\Bbb N$. 
Show that $(\Cal P\Bbb N,\bar\mu)$ is a $\sigma$-finite measure algebra. 
Find a measure-preserving Boolean homomorphism from $\Cal P\Bbb N$ to 
itself which is not sequentially order-continuous. 
%324K 
      
\leader{324Y}{Further exercises (a)} Let $\frak A$ and $\frak B$ be 
Boolean algebras, of which $\frak A$ is Dedekind complete, and 
$\phi:\frak A\to\frak B$ an order-continuous Boolean homomorphism.   Let 
$I$ be an ideal of $\frak A$ included in the kernel of $\phi$.   Show 
that we have an order-continuous Boolean homomorphism  
$\pi:\frak A/I\to \frak B$ given by setting $\pi(a^{\ssbullet})=\phi a$ for every 
$a\in\frak A$. 
%324B, 324Xa 
      
\header{324Yb}{\bf (b)} Let $\frak A$ be a Dedekind 
$\sigma$-complete Boolean algebra, and $Z$ its Stone space.   Write 
$\Cal E$ for the algebra of open-and-closed subsets of $Z$, and $\Cal Z$ 
for the family of nowhere dense zero sets of $Z$;  let $\Cal Z_{\sigma}$ 
be the $\sigma$-ideal of subsets of $Z$ generated by $\Cal Z$.   Show 
that $\Sigma=\{E\symmdiff U:E\in\Cal E,\,U\in\Cal Z_{\sigma}\}$ is a 
$\sigma$-algebra of subsets of $Z$, and describe a canonical isomorphism 
between $\Sigma/\Cal Z_{\sigma}$ and $\frak A$. 
%324E 
      
\header{324Yc}{\bf (c)} Let $\frak A$ and $\frak B$ be Dedekind 
$\sigma$-complete Boolean algebras, with Stone spaces $Z$ and $W$. 
Construct $\Cal Z_{\sigma}\subseteq\Sigma\subseteq\Cal PZ$ as in 324Yb, 
and let $\Cal W_{\sigma}\subseteq\Tau\subseteq\Cal PW$ be the 
corresponding structure defined from $\frak B$.   Let 
$\pi:\frak B\to\frak A$ be a sequentially order-continuous  
Boolean homomorphism, and $\phi:Z\to W$ the corresponding continuous map. 
Show that if $E^*\in\frak A$ corresponds to $E\in\Sigma$, then 
$\pi F^*=\phi^{-1}[F]^*$ for every $F\in\Tau$.   
%324E, 324Yb 313Ye
      
\spheader 324Yd Let $\frak A$ be a Boolean algebra, $\frak B$ a ccc 
Boolean algebra and $\pi:\frak A\to\frak B$ an injective Boolean 
homomorphism.   Show that $\frak A$ is ccc. 
%324K 
      
\spheader 324Ye Let $\frak A$ be a Dedekind complete Boolean 
algebra, $\frak B$ a Boolean algebra, and $\pi:\frak A\to\frak B$ an 
order-continuous Boolean homomorphism.   Show that for every atom 
$b\in\frak B$ there is an atom $a\in\frak A$ such that  
$\pi a\Bsupseteq b$.    
Hence show that if $\frak A$ is atomless so is $\frak B$, and that 
if $\frak B$ is purely atomic and $\pi$ is injective then $\frak A$ is 
purely atomic. 
%324K 
      
\spheader 324Yf Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
localizable measure algebras and $\frak A_0$ an order-dense subalgebra 
of $\frak A$. 
Suppose that $\pi:\frak A_0\to\frak B$ is an order-continuous Boolean 
homomorphism such that $\bar\nu\pi a=\bar\mu a$ for every 
$a\in\frak A_0$.  Show that $\pi$ has a unique extension to a 
measure-preserving Boolean homomorphism from $\frak A$ to $\frak B$. 
%324O 
      
\spheader 324Yg Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be  
probability algebras, and $f:\frak A\to\frak B$ an isometry for the  
measure metrics.   Show that $a\mapsto f(a)\Bsymmdiff f(0)$ is a  
measure-preserving Boolean homomorphism. 
%324O 
% reduce to case  f(0)=0,  show  f(1)=1  , 
% f(1\Bsetminus a)=1\Bsetminus f(a) ,  a\Bsubseteq b iff f(a)\Bsubseteq f(b) ,
% f(a\Bcup b)\Bsupseteq f(a)\Bcup f(b)\Bsupseteq f(a)\Bcap f(b)
%   \Bsupseteq f(a\Bcap b)
}%end of exercises 
      
\endnotes{ 
\Notesheader{324} If you examine the arguments of this section 
carefully, you will see that rather little depends on the measures 
named.   Really this material deals with structures $(X,\Sigma,\Cal I)$ 
where $X$ is a set, $\Sigma$ is a $\sigma$-ideal of subsets of $X$, and 
$\Cal I$ is a $\sigma$-ideal of $\Sigma$, corresponding to the family of 
measurable negligible sets. 
In this abstract form it is natural to think in terms of 
sequentially order-continuous homomorphisms, as in 324Yc. 
I have stated 324E in terms of order-continuous homomorphisms just for a 
slight gain in simplicity.   But in fact, when there is a difference, it 
is likely that order-continuity, rather than sequential 
order-continuity, will be the more significant condition.   Note that 
when the domain algebra is $\sigma$-finite, it is ccc (322G), 
so the two concepts coincide (316Fd). 
      
Of course I need to refer to measures when looking at such concepts as 
$\sigma$-finite measure algebra or measure-preserving homomorphism, but 
even here the real ideas involved are such notions as 
order-continuity and the countable chain condition, as you will see if 
you work through 324K.   It is instructive to look at the translations 
of these facts into the context of inverse-measure-preserving functions; 
see 234B. 
      
324H shows that we may speak of `the' topology and uniformity of a 
Dedekind $\sigma$-complete Boolean algebra which carries any semi-finite 
measure;  the topology of such an algebra is determined by its algebraic 
structure.   Contrast this with the theory of normed spaces:  two Banach 
spaces (e.g., $\ell^1$ and $\ell^2$) can be isomorphic as linear spaces, 
both being of algebraic dimension $\frak c$, while they are not 
isomorphic as topological linear spaces.   When we come to the theory of 
ordered linear topological spaces, however, we shall again find 
ourselves with operators whose algebraic properties guarantee continuity 
(355C, 367O). 
}%end of notes 
      
\discrpage 
  
\frfilename{mt325.tex}
\versiondate{30.8.06}
\copyrightdate{1999}

\def\chaptername{Measure algebras}
\def\sectionname{Free products and product measures}

\newsection{325}

In this section I aim to describe the measure algebras of product
measures as defined in Chapter 25.   This will involve the concept of
`free product' set out in \S315.   It turns out that we cannot determine
the measure algebra of a product measure from the measure algebras of
the factors (325B), unless we are told that
the product measure is localizable;  but that
there is nevertheless a general construction of `localizable measure
algebra free product', applicable to any pair of semi-finite measure
algebras (325D), which represents the measure algebra of the product
measure in the most important cases (325Eb).   In the second part of the
section (325I-325M) I deal with
measure algebra free products of probability algebras, corresponding to
the products of probability spaces treated in \S254.

\leader{325A}{Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with measure algebras $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$.   Let $\lambda$ be the c.l.d.\ product measure on
$X\times Y$, and $\Lambda$ its domain;  let $(\frak C,\bar\lambda)$ be
the corresponding measure algebra.

(a)(i) The map $E\mapsto E\times Y:\Sigma\to\Lambda$ induces an
order-continuous Boolean homomorphism from $\frak A$ to $\frak C$.

\quad(ii) The map $F\mapsto X\times F:\Tau\to\Lambda$ induces an
order-continuous Boolean homomorphism from $\frak B$ to $\frak C$.

(b) The map $(E,F)\mapsto E\times F:\Sigma\times\Tau\to\Lambda$ induces
a Boolean homomorphism $\psi:\frak A\otimes\frak B\to\frak C$.

(c) $\psi[\frak A\otimes\frak B]$ is topologically dense in $\frak C$ for
the measure-algebra topology of $\frak C$.

(d) For every $c\in\frak C$,

\Centerline{$\bar\lambda c
=\sup\{\bar\lambda(c\Bcap\psi(a\otimes b)):
a\in\frak A,\,b\in\frak B,\,\bar\mu a<\infty,\,\bar\nu b<\infty\}$.}

(e) If $\mu$ and $\nu$ are semi-finite, $\psi$ is injective and
$\bar\lambda\psi(a\otimes b)=\bar\mu a\cdot\bar\mu b$ for every
$a\in\frak A$, $b\in\frak B$.

\proof{{\bf (a)} $E\times Y\in\Lambda$ for every $E\in\Sigma$ (251E),
and $\lambda(E\times Y)=0$ whenever $\mu E=0$ (251Ia).   Thus
$E\mapsto(E\times Y)^{\ssbullet}:\Sigma\to\frak C$ is a Boolean
homomorphism with kernel including $\{E:\mu E=0\}$, so descends to a
Boolean homomorphism $\varepsilon_1:\frak A\to\frak C$.

To see that $\varepsilon_1$ is order-continuous,
let $A\subseteq\frak A$ be a non-empty downwards-directed set with
infimum $0$.   \Quer\ If there is a non-zero lower bound $c$ of
$\varepsilon_1[A]$, express $c$ as $W^{\ssbullet}$ where $W\in\Lambda$.
We have $\lambda(W)>0$;  by the definition of $\lambda$ (251F), there
are $G\in\Sigma$, $H\in\Tau$ such that $\mu G<\infty$, $\nu H<\infty$
and $\lambda(W\cap(G\times H))>0$.   Of course
$\inf_{a\in A}a\Bcap G^{\ssbullet}=0$ in $\frak A$, so
$\inf_{a\in A}\bar\mu(a\Bcap G^{\ssbullet})=0$, by 321F;
let $a\in A$ be such that
$\bar\mu(a\Bcap G^{\ssbullet})\cdot\nu H<\lambda(W\cap(G\times H))$.
Express $a$ as $E^{\ssbullet}$, where $E\in\Sigma$.   Then
$\lambda(W\setminus(E\times Y))=0$.   But this means that

\Centerline{$\lambda(W\cap(G\times H))
\le\lambda((E\cap G)\times H)
=\mu(E\cap G)\cdot\nu H
=\bar\mu(a\Bcap G^{\ssbullet})\cdot\nu H$,}

\noindent contradicting the choice of $a$.\ \Bang\   Thus
$\inf\varepsilon_1[A]=0$ in $\frak C$;  as $A$ is arbitrary,
$\varepsilon_1$ is order-continuous.

Similarly $\varepsilon_2:\frak B\to\frak C$, induced by $F\mapsto
X\times F:\Tau\to\Lambda$, is order-continuous.

\medskip

{\bf (b)} Now there must be a corresponding Boolean homomorphism
$\psi:\frak A\otimes\frak B\to\frak C$ such that
$\psi(a\otimes b)=\varepsilon_1a\Bcap\varepsilon_2b$ for every
$a\in\frak A$ and $b\in\frak B$, that is,

\Centerline{$\psi(E^{\ssbullet}\otimes F^{\ssbullet})
=(E\times Y)^{\ssbullet}\Bcap(X\times F)^{\ssbullet}
=(E\times F)^{\ssbullet}$}

\noindent for every $E\in\Sigma$, $F\in\Tau$ (315Jb).

\medskip

{\bf (c)} Suppose that $c$, $e\in\frak C$, $\bar\lambda e<\infty$
and $\epsilon>0$.   Express $c$, $e$ as $U^{\ssbullet}$, $W^{\ssbullet}$
where $U$, $W\in\Lambda$.   By 251Ie, there are
$E_0,\ldots,E_n\in\Sigma$, $F_0,\ldots,F_n\in\Tau$, all of finite
measure, such that
$\lambda((U\cap W)\symmdiff\bigcup_{i\le n}E_i\times F_i)\le\epsilon$.
Set

\Centerline{$c_1=(\bigcup_{i\le n}E_i\times
F_i)^{\ssbullet}\in\psi[\frak A\otimes\frak B]$;}

\noindent then

\Centerline{$\bar\lambda(e\Bcap(c\Bsymmdiff c_1))
=\lambda(W\cap(U\symmdiff\bigcup_{i\le n}E_i\times F_i))
\le\epsilon$.}

\noindent As $c$, $e$ and $\epsilon$ are arbitrary,
$\psi[\frak A\otimes\frak B]$ is topologically dense in $\frak C$.

\medskip

{\bf (d)} By the definition of $\lambda$, we have

\Centerline{$\lambda W=\sup\{\lambda(W\cap(E\times F)):
E\in\Sigma,\,F\in\Tau,\,\mu E<\infty,\,\nu F<\infty\}$}

\noindent for every $W\in\Lambda$;   so all we have to do is express $c$
as $W^{\ssbullet}$.

\medskip

{\bf (e)} Now suppose that $\mu$ and $\nu$ are semi-finite.   Then
$\lambda(E\times F)=\mu E\cdot\nu F$
for any $E\in\Sigma$, $F\in\Tau$ (251J), so
$\bar\lambda\psi(a\otimes b)=\bar\mu a\cdot\bar\nu b$ for every
$a\in\frak A$ and $b\in\frak B$.

To see that $\psi$ is injective, take any non-zero
$c\in\frak A\otimes\frak B$;  then there must be non-zero $a\in\frak A$,
$b\in\frak B$ such that $a\otimes b\Bsubseteq c$ (315Kb), so that

\Centerline{$\bar\lambda\psi c\ge\bar\lambda\psi(a\otimes b)
=\bar\mu a\cdot\bar\nu b>0$}

\noindent and $\psi c\ne 0$.
}%end of proof of 325A

\leader{325B}{Characterizing the measure algebra of a
product \dvrocolon{space}}\cmmnt{ A
very natural question to ask is, whether it is possible to define a
`measure algebra free product' of two abstract measure algebras in a way
which will correspond to one of the constructions above.  I give an
example to show the difficulties involved.

\medskip

\noindent}{\bf Example} There are complete locally determined
localizable measure spaces $(X,\mu)$, $(X',\mu')$, with isomorphic
measure algebras, and a probability space $(Y,\nu)$ such that the
measure algebras of the c.l.d.\ product measures on $X\times Y$,
$X'\times Y$ are not isomorphic.

\proof{ Let $(X,\Sigma,\mu)$ be the complete
locally determined localizable not-strictly-localizable measure space
described in 216E.   Recall that, for $E\in\Sigma$,
$\mu E=\#(\{\gamma:\gamma\in C,\,f_{\gamma}\in E\})$ if this is finite,
$\infty$ otherwise (216Eb), where $C$ is a set with cardinal greater than
$\frak c$.   The map
$E\mapsto\{\gamma:f_{\gamma}\in E\}:\Sigma\to\Cal PC$ is surjective
(216Ec), so descends to an isomorphism between
$\frak A$, the measure algebra of $\mu$, and $\Cal PC$.   Let
$(X',\Sigma',\mu')$ be $C$ with counting measure, so that its measure
algebra $(\frak A',\bar\mu')$ is isomorphic to $(\frak A,\bar\mu)$,
while $\mu'$ is of course strictly localizable.

Let $(Y,\Tau,\nu)$ be $\{0,1\}^C$ with its usual measure.   Let
$\lambda$, $\lambda'$ be the c.l.d.\ product measures on $X\times Y$,
$X'\times Y$ respectively, and $(\frak C,\bar\lambda)$,
$(\frak C',\bar\lambda')$ the corresponding measure algebras.   Then
$\lambda$ is not localizable (254U), so $(\frak C,\bar\lambda)$ is not
localizable (322Be).   On the other hand, $\lambda'$, being the c.l.d.\
product of strictly localizable measures, is strictly localizable
(251O), therefore localizable, so $(\frak C',\bar\lambda')$ is
localizable, and is not isomorphic to $(\frak C,\bar\lambda)$.
}%end of proof of 325Bb

\leader{325C}{}\cmmnt{ Thus there can be no universally applicable
method of identifying the measure algebra of a product measure from the
measure algebras of
the factors.   However, you have no doubt observed that the example
above involves non-$\sigma$-finite spaces, and conjectured that this is
not an accident.   In contexts in which we know that the algebras
involved are localizable, there are positive results available, such as
the following.

\medskip

\noindent}{\bf Theorem} Let $(X_1,\Sigma_1,\mu_1)$ and
$(X_2,\Sigma_2,\mu_2)$ be semi-finite measure spaces, with measure
algebras $(\frak A_1,\bar\mu_1)$ and $(\frak A_2,\bar\mu_2)$.   Let
$\lambda$ be the c.l.d.\ product measure on $X_1\times X_2$, and
$(\frak C,\bar\lambda)$ the corresponding measure algebra.   Let
$(\frak B,\bar\nu)$ be a localizable measure algebra, and
$\phi_1:\frak A_1\to\frak B$, $\phi_2:\frak A_2\to\frak B$
order-continuous Boolean homomorphisms such that
$\bar\nu(\phi_1(a_1)\Bcap\phi_2(a_2))=\bar\mu_1a_1\cdot\bar\mu_2a_2$ for
all $a_1\in\frak A_1$, $a_2\in\frak A_2$.   Then there is a unique
order-continuous measure-preserving Boolean homomorphism
$\phi:\frak C\to\frak B$ such that
$\phi(\psi(a_1\otimes a_2))=\phi_1(a_1)\Bcap\phi_2(a_2)$ for all
$a_1\in\frak A_1$,
$a_2\in\frak A_2$, writing $\psi:\frak A_1\otimes\frak A_2\to\frak C$
for the canonical map\cmmnt{ described in 325A}.

\proof{{\bf (a)} Because $\psi$ is injective, it is an isomorphism
between $\frak A_1\otimes\frak A_2$ and its image in $\frak C$.  I trust
it will cause no confusion if I abuse notation slightly and treat
$\frak A_1\otimes\frak A_2$ as actually a subalgebra of $\frak C$.
Now the
Boolean homomorphisms $\phi_1$, $\phi_2$ correspond to a Boolean
homomorphism $\theta:\frak A_1\otimes\frak A_2\to\frak B$.   The point
is that $\bar\nu\theta c=\bar\lambda c$ for every
$c\in\frak A_1\otimes\frak A_2$.
\Prf\ By 315Kb, every member of $\frak A_1\otimes\frak A_2$ is
expressible as $\sup_{i\le n}a_i\otimes a'_i$, where $a_i\in\frak A_1$,
$a'_i\in\frak A_2$ for each $i$
and $\langle a_i\otimes a'_i\rangle_{i\le n}$ is
disjoint.   Now for each $i$ we have

\Centerline{$\bar\nu\theta(a_i\otimes a'_i)
=\bar\nu(\phi_1(a_i)\Bcap\phi_2(a'_i))
=\bar\mu_1a_i\cdot\bar\mu_2a'_i
=\bar\lambda(a_i\otimes a'_i)$,}

\noindent by 325Ae.   So

\Centerline{$\bar\nu\theta(c)
=\sum_{i=0}^n\bar\nu\theta(a_i\otimes a'_i)
=\sum_{i=0}^n\bar\lambda(a_i\otimes a'_i)
=\bar\lambda c$.   \Qed}

\medskip

{\bf (b)} The following fact will underlie many of the arguments below.
If $e\in\frak B$, $\bar\nu e<\infty$ and $\epsilon>0$, there are
$e_1\in\frak A_1^f$, $e_2\in\frak A_2^f$ such that
$\bar\nu(e\Bsetminus\theta(e_1\otimes e_2))\le\epsilon$,
writing $\frak A_i^f$ for $\{a:\bar\mu_i a<\infty\}$.
\Prf\ Because $(\frak A_1,\bar\mu_1)$ is semi-finite, $\frak A_1^f$
has supremum $1$ in
$\frak A_1$;  because $\phi_1$ is order-continuous,
$\sup\{\phi_1(a):a\in\frak A_1^f\}=1$ in $\frak B$, and
$\inf\{e\Bsetminus\phi_1(a):a\in\frak A_1^f\}=0$ (313Aa).   Because
$\frak A_1^f$ is upwards-directed,
$\{e\Bsetminus\phi_1(a):a\in\frak A_1^f\}$ is downwards-directed, so
$\inf\{\bar\nu(e\Bsetminus\phi(a)):a\in\frak A_1^f\}=0$ (321F again).   Let
$e_1\in\frak A_1^f$ be such that
$\bar\nu(e\Bsetminus\phi_1(e_1))\le\bover12\epsilon$.

In the same way, there is an $e_2\in\frak A_2^f$ such that
$\bar\nu(e\Bsetminus\phi_2(e_2))\le\bover12\epsilon$.   Consider
$e'=e_1\otimes e_2\in\frak C$.   Then

\Centerline{$\bar\nu(e\Bsetminus\theta e')
=\bar\nu(e\Bsetminus(\phi_1(e_1)\Bcap\phi_2(e_2)))
\le\bar\nu(e\Bsetminus\phi_1(e_1))+\bar\nu(e\Bsetminus\phi_2(e_2))
\le\epsilon$.  \Qed}

\medskip

{\bf (c)} The next step is to check that $\theta$ is uniformly
continuous for the measure-algebra uniformities defined by $\bar\nu$
and $\bar\lambda$.   \Prf\ Take any $e\in\frak B^f$ and $\epsilon>0$.
By (b), there are $e_1$, $e_2$ such that $\bar\lambda(e_1\otimes
e_2)<\infty$ and $\bar\nu(e\Bsetminus\theta(e_1\otimes
e_2))\le\bover12\epsilon$.   Set $e'=e_1\otimes e_2$.
Now suppose that $c$, $c'\in\frak A_1\otimes\frak A_2$ and
$\bar\lambda((c\Bsymmdiff c')\Bcap e')\le\bover12\epsilon$.   Then

\Centerline{$\bar\nu((\theta(c)\Bsymmdiff\theta(c'))\Bcap e)
\le\bar\nu\theta((c\Bsymmdiff c')\Bcap e')+\bar\nu(e\Bsetminus\theta e')
\le\bar\lambda((c\Bsymmdiff c')\Bcap e')+\Bover12\epsilon
\le\epsilon$.}

\noindent By 3A4Cc, $\theta$ is uniformly continuous for the
subspace uniformity on $\frak A_1\otimes\frak A_2$.\ \Qed

\medskip

{\bf (d)} Recall that $\frak A_1\otimes\frak A_2$ is topologically dense
in $\frak C$ (325Ac), while $\frak B$ is complete for its uniformity
(323Gc).   So there is a uniformly continuous function
$\phi:\frak C\to\frak B$ extending $\theta$ (3A4G).

\medskip

{\bf (e)} Because $\theta$ is a Boolean homomorphism, so is $\phi$.
\Prf\ (i) The functions $c\mapsto\phi(1\Bsetminus c)$,
$c\mapsto1\Bsetminus\phi(c)$ are continuous and the topology of
$\frak B$ is Hausdorff, so
$\{c:\phi(1\Bsetminus c)=1\Bsetminus\phi(c)\}$ is
closed;   as it includes $\frak A_1\otimes\frak A_2$, it must be the
whole of $\frak C$.   (ii) The functions $(c,c')\mapsto\phi(c\Bcup c')$,
$(c,c')\mapsto\phi(c)\Bcup\phi(c')$ are continuous, so
$\{(c,c'):\phi(c\Bcup c')=\phi(c)\Bcup\phi(c')\}$ is closed in
$\frak C\times\frak C$;  as it includes
$(\frak A_1\otimes\frak A_2)\times(\frak A_1\otimes\frak A_2)$, it must be the whole of $\frak C\times\frak C$.\ \Qed

\medskip

{\bf (f)} Because $\theta$ is measure-preserving, so is $\phi$.   \Prf\
Take any $e_1\in\frak A_1^f$, $e_2\in\frak A_2^f$.   Then the functions
$c\mapsto\bar\lambda(c\Bcap(e_1\otimes e_2))$,
$c\mapsto\bar\nu\phi(c\Bcap(e_1\otimes e_2))$ are continuous and equal
on $\frak A_1\otimes\frak A_2$, so are equal on $\frak C$.   The
argument of (b) shows that for any $b\in\frak B$,

$$\eqalign{\bar\nu b
&=\sup\{\bar\nu(b\Bcap e):e\in\frak B^f\}\cr
&=\sup\{\bar\nu(b\Bcap\phi(e_1\otimes e_2)):e_1\in\frak
A_1^f,\,e_2\in\frak
A_2^f\},\cr}$$

\noindent so that

$$\eqalign{\bar\nu\phi(c)
&=\sup\{\bar\nu\phi(c\Bcap(e_1\otimes e_2)):e_1\in\frak
A_1^f,\,e_2\in\frak
A_2^f\}\cr
&=\sup\{\bar\lambda(c\Bcap(e_1\otimes e_2)):e_1\in\frak
A_1^f,\,e_2\in\frak A_2^f\}
=\bar\lambda c\cr}$$

\noindent for every $c\in\frak C$.\ \Qed

\medskip

{\bf (g)} To see that $\phi$ is order-continuous, take any non-empty
downwards-directed set $C\subseteq\frak C$ with infimum $0$.   \Quer\ If
$\phi[C]$ has a non-zero lower bound $b$ in $\frak B$, let
$e\Bsubseteq b$ be such that $0<\bar\nu e<\infty$.   Let $e'\in\frak C$
be such that
$\bar\lambda e'<\infty$ and $\bar\nu(e\Bsetminus\phi(e'))<\bar\nu e$, as
in (b) above, so that $\bar\nu(e\Bcap\phi(e'))>0$.   Now, because
$\inf C=0$, there is a $c\in C$ such that
$\bar\lambda(c\Bcap e')<\bar\nu(e\Bcap\phi(e'))$.   But this means that

\Centerline{$\bar\nu(b\Bcap\phi(e'))\le\bar\nu\phi(c\Bcap e')
=\bar\lambda(c\Bcap e')<\bar\nu(e\Bcap\phi(e'))
\le\bar\nu(b\Bcap\phi(e'))$,}

\noindent which is absurd.\ \BanG   Thus $\inf\phi[C]=0$ in $\frak B$.
As $C$ is arbitrary, $\phi$ is order-continuous.

\medskip

{\bf (h)} Finally, to see that $\phi$ is unique, observe that any
order-continuous Boolean homomorphism from $\frak C$ to $\frak B$ must
be continuous (324Fc);  so that if it agrees with $\phi$ on
$\frak A_1\otimes\frak A_2$ it must agree with $\phi$ on $\frak C$.
}%end of proof of 325C

\vleader{72pt}{325D}{Theorem} Let $(\frak A_1,\bar\mu_1)$ and
$(\frak A_2,\bar\mu_2)$ be semi-finite measure algebras.

(a) There is a localizable measure algebra $(\frak C,\bar\lambda)$,
together with order-continuous Boolean homomorphisms
$\varepsilon_1:\frak A_1\to\frak C$ and $\varepsilon_2:\frak A_2\to\frak C$, such that
whenever $(\frak B,\bar\nu)$ is a localizable measure algebra, and
$\phi_1:\frak A_1\to\frak B$, $\phi_2:\frak A_2\to\frak B$ are
order-continuous Boolean homomorphisms and
$\bar\nu(\phi_1(a_1)\Bcap\phi_2(a_2))=\bar\mu_1a_1\cdot\bar\mu_2a_2$ for
all $a_1\in\frak A_1$, $a_2\in\frak A_2$, then there is a unique
order-continuous measure-preserving Boolean homomorphism
$\phi:\frak C\to\frak B$ such that $\phi\varepsilon_j=\phi_j$ for both $j$.

(b) The structure $(\frak C,\bar\lambda,\varepsilon_1,\varepsilon_2)$ is determined
up to isomorphism by this property.

(c)(i) The Boolean homomorphism
$\psi:\frak A_1\otimes\frak A_2\to\frak C$ defined from $\varepsilon_1$ and
$\varepsilon_2$ is injective, and $\psi[\frak A_1\otimes\frak A_2]$ is
topologically dense in $\frak C$.

\quad(ii) The closed subalgebra of $\frak C$ generated by
$\psi[\frak A_1\otimes\frak A_2]$ is the whole of $\frak C$.

(d) If $j\in\{1,2\}$ and $(\frak A_j,\bar\mu_j)$ is localizable, then
$\varepsilon_j[\frak A_j]$ is a closed subalgebra of $(\frak C,\bar\lambda)$.

\proof{{\bf (a)(i)} We may regard $(\frak A_1,\bar\mu_1)$ as the
measure algebra of $(Z_1,\Sigma_1,\mu_1)$ where $Z_1$ is the Stone space
of $\frak A_1$, $\Sigma_1$ is the algebra of subsets of $Z_1$ differing
from an open-and-closed set by a meager set, and $\mu_1$ is an
appropriate measure (321K).   Note that in this representation,
each $a\in\frak A_1$ becomes identified with $\widehat{a}^{\ssbullet}$,
where $\widehat{a}$ is the open-and-closed subset of $Z_1$ corresponding
to $a$.   Similarly, we may think of $(\frak A_2,\bar\mu_2)$ as the measure algebra of $(Z_2,\Sigma_2,\mu_2)$,
where $Z_2$ is the Stone space of $\frak A_2$.

\medskip

\quad{\bf (ii)} Let $\lambda$ be the c.l.d.\ product measure on
$Z_1\times Z_2$.   The point is that $\lambda$ is strictly localizable.
\Prf\ By 322Ea, both $\frak A_1$ and $\frak A_2$ have partitions of unity
consisting of elements of finite measure;  let
$\langle c_i\rangle_{i\in I}$, $\langle d_j\rangle_{j\in J}$ be such partitions.
Then $\langle\widehat{c}_i\times\widehat{d}_j\rangle_{i\in I,j\in J}$ is
a disjoint family of sets of finite measure in $Z_1\times Z_2$.   If
$W\subseteq Z_1\times Z_2$ is such that $\lambda W>0$, there must be sets $E_1$, $E_2$ of finite measure such that
$\lambda(W\cap(E_1\times E_2))>0$.   Because
$E_1^{\ssbullet}=\sup_{i\in I}E_1^{\ssbullet}\Bcap c_i$, we must have

\Centerline{$\mu_1E_1=\bar\mu_1E_1^{\ssbullet}
=\sum_{i\in I}\bar\mu_1(E_1^{\ssbullet}\Bcap c_i)
=\sum_{i\in I}\mu_1(E_1\cap\widehat{c}_i)$.}

\noindent Similarly,
$\mu_2E_2=\sum_{i\in J}\mu_2(E_2\cap\widehat{d}_j)$.
But this means that there must be finite $I'\subseteq I$,
$J'\subseteq J$ such that

\Centerline{$\sum_{i\in I',j\in J'}
  \mu_1(E_1\cap\widehat{c}_i)\mu_2(E_2\cap\widehat{d}_j)
>\mu_1E_1\cdot\mu_2E_2-\lambda(W\cap(E_1\times E_2))$,}

\noindent so that there have to be $i\in I'$, $j\in J'$ such that
$\lambda(W\cap(\widehat{c}_i\times\widehat{d}_j))>0$.

Now this means that
$\langle\widehat{c}_i\times\widehat{d}_j\rangle_{i\in I,j\in J}$
satisfies the conditions of 213O.   Because $\lambda$ is surely
complete and locally determined, it is strictly localizable.\ \Qed

\medskip

\quad{\bf (iii)} We may therefore take $(\frak C,\bar\lambda)$ to be
just the measure algebra of $\lambda$.   The maps $\varepsilon_1$, $\varepsilon_2$
will be the canonical maps described in 325Aa, inducing the map
$\psi:\frak A_1\otimes\frak A_2\to\frak C$ referred to in 325C;  and
325C now gives the result.

\medskip

{\bf (b)} This is nearly obvious.   Suppose we had an alternative
structure $(\frak C',\bar\lambda',\varepsilon'_1,\varepsilon'_2)$
with the same
property.   Then we must have an order-continuous measure-preserving
Boolean homomorphism $\phi:\frak C\to\frak C'$ such that
$\phi\varepsilon_j=\varepsilon'_j$ for both $j$;  and similarly we have an
order-continuous measure-preserving Boolean homomorphism
$\phi':\frak C'\to\frak C$ such that $\phi'\varepsilon'_j=\varepsilon_j$
for both $j$.   Now $\phi'\phi:\frak C\to\frak C$ is an
order-continuous measure-preserving Boolean homomorphism such that
$\phi'\varepsilon_j=\varepsilon_j$ for both $j$.   By the uniqueness assertion in
(a), applied with $\frak B=\frak C$, $\phi'\phi$ must be the
identity on $\frak C$.   In the same way, $\phi\phi'$ is the
identity on $\frak C'$.   So $\phi$ and $\phi'$ are the two halves
of the required isomorphism.

\medskip

{\bf (c)} In view of the construction for $\frak C$ offered in part (a)
of the proof, (i) is just a consequence of 325Ac and 325Ae.   Now (ii)
follows by 323J.

\medskip

{\bf (d)} If $\frak A_j$ is Dedekind complete then $\varepsilon_j[\frak A_j]$
is order-closed in $\frak C$ because $\varepsilon_j$ is order-continuous
(314F(a-i)).
}%end of proof of 325D

\leader{325E}{Remarks}\cmmnt{ \bf (a)}  We could say that a
measure algebra $(\frak C,\bar\lambda)$, together with embeddings
$\varepsilon_1$ and $\varepsilon_2$, as
described in 325D, is a {\bf localizable measure algebra free product}
of $(\frak A_1,\bar\mu_1)$ and $(\frak A_2,\bar\mu_2)$\cmmnt{;  and
its uniqueness up to isomorphism makes it safe, most of the time, to
call it `the' localizable measure algebra free product.   Observe that
it can equally well be regarded as the uniform space completion of the
algebraic free product;  see 325Yc}.

\cmmnt{\spheader 325Eb As the example in
325B shows, the localizable
measure algebra free product of the measure algebras of given measure
spaces need not appear directly as the measure algebra of their product.
But there is one context in which it must so appear:  if
the product measure is localizable, 325C tells us at once
that it has the right measure algebra.   For $\sigma$-finite
measure algebras, of course, any corresponding measure spaces have to
be strictly localizable, so again we can use the product measure
directly.
}%end of comment

\leader{325F}{}\cmmnt{ I ought not to proceed to the next topic
without giving
another pair of examples to show the subtlety of the concept of `measure
algebra free product'.

\medskip

\noindent}{\bf Example} Let $(\frak A,\bar\mu)$ be the measure algebra
of Lebesgue measure $\mu$ on $[0,1]$, and $(\frak C,\bar\lambda)$ the
measure algebra of Lebesgue measure $\lambda$ on $[0,1]^2$.   Then
$(\frak C,\bar\lambda)$ can be regarded as the localizable measure
algebra free
product of $(\frak A,\bar\mu)$ with itself\cmmnt{, by 251N and 325Eb}.
Let $\psi:\frak A\otimes\frak A\to\frak C$ be the canonical map\cmmnt{, as described
in 325A}.   Then $\psi[\frak A\otimes\frak A]$ is not order-dense
in $\frak C$, and $\psi$ is not order-continuous.

\proof{{\bf (a)} Let $\sequencen{\epsilon_n}$ be a sequence in $[0,1]$
such that $\sum_{n=0}^{\infty}\epsilon_n=\infty$, but
$\sum_{n=0}^{\infty}\epsilon_n^2<1$;  for instance, we could take
$\epsilon_n=\bover1{n+2}$.   Let $\sequencen{E_n}$ be a stochastically
independent sequence of measurable subsets of $[0,1]$ such that
$\mu E_n=\epsilon_n$ for each $n$.   In $\frak A$ set
$a_n=E_n^{\ssbullet}$,
and consider $c_n=\sup_{i\le n}a_i\otimes a_i\in\frak A\otimes\frak A$
for each $n$.

\medskip

{\bf (b)} We have $\sup_{n\in\Bbb N}c_n=1$ in $\frak A\otimes\frak A$.
\Prf\Quer\ Otherwise, there is a non-zero $a\in\frak A\otimes\frak A$
such that $a\Bcap(a_n\otimes a_n)=0$ for every $n$, and now there are
non-zero $b$, $b'\in\frak A$ such that $b\otimes b'\Bsubseteq a$.   Set
$I=\{n:a_n\Bcap b=0\}$, $J=\{n:a_n\Bcap b'\}=0$.   Then
$\langle E_n\rangle_{n\in I}$ is an independent family and
$\mu(\bigcup_{n\in I}E_i)\le 1-\bar\mu b<1$, so
$\sum_{n\in I}\mu E_n<\infty$, by the
Borel-Cantelli lemma (273K).   Similarly $\sum_{n\in J}\mu E_n<\infty$.
Because $\sum_{n\in\Bbb N}\mu E_n=\infty$, there must be some
$n\in\Bbb N\setminus(I\cup J)$.   Now $a_n\Bcap b$ and $a_n\Bcap b'$ are
both non-zero, so

\Centerline{$0\ne(a_n\Bcap b)\otimes(a_n\Bcap b')
=(a_n\otimes a_n)\Bcap(b\otimes b')=0$,}

\noindent which is absurd.\ \Bang\Qed

\medskip

{\bf (c)} On the other hand,

\Centerline{$\sum_{n=0}^{\infty}\bar\lambda\psi(c_n)
\le\sum_{n=0}^{\infty}(\bar\mu a_n)^2
=\sum_{n=0}^{\infty}\epsilon_n^2<1$,}

\noindent by the choice of the $\epsilon_n$.   So
$\sup_{n\in\Bbb N}\psi(c_n)$ cannot be $1$ in $\frak C$.

Thus $\psi$ is not order-continuous.

\medskip

{\bf (d)} By 313P(a-ii) and 313O, $\psi[\frak A\otimes\frak A]$ cannot
be order-dense in $\frak C$;  alternatively, (b) shows that there can be
no non-zero member of $\psi[\frak A\otimes\frak A]$ included in
$1\Bsetminus\sup_{n\in\Bbb N}\psi(c_n)$.   (Both these arguments rely
tacitly on the fact that $\psi$ is injective, as noted in 325Ae.)
}%end of proof of 325F

\leader{325G}{}\cmmnt{ Since 325F shows that the free product
and the localizable measure algebra free product are very different
constructions, I had better repeat an idea from \S315 in the new
context.

\medskip

\noindent}{\bf Example} Again, let $(\frak A,\bar\mu)$ be the measure
algebra of Lebesgue measure on $[0,1]$, and $(\frak C,\bar\lambda)$
the measure algebra of Lebesgue measure on $[0,1]^2$.   Then there is no
order-continuous Boolean homomorphism $\phi:\frak C\to\frak A$ such that
$\phi(a\otimes b)=a\Bcap b$ for all $a$, $b\in\frak A$.
\prooflet{\Prf\ Let $\phi:\frak C\to\frak A$ be a Boolean homomorphism
such that $\phi(a\otimes b)=a\Bcap b$ for all $a$, $b\in \frak A$.   For
$i<2^n$ let $a_{ni}$ be the equivalence class in $\frak A$ of the
interval $[2^{-n}i,2^{-n}(i+1)]$, and set $c_n=\sup_{i<2^n}a_{ni}\otimes
a_{ni}$.   Then $\phi c_n=1$ for every $n$, but
$\bar\lambda c_n=2^{-n}$ for each $n$, so $\inf_{n\in\Bbb N}c_n=0$ in
$\frak C$;  thus $\phi$ cannot be order-continuous.\ \Qed}
\cmmnt{(Compare 315Q.)}

\leader{*325H}{Products of more than two factors} \cmmnt{We can of
course extend the ideas of 325A, 325C and 325D to products of any finite
number of factors.   No new ideas are needed, so I spell the results out
without proofs.

\medskip

}{\bf (a)} Let $\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$ be a
non-empty finite family of semi-finite measure algebras.
Then there is a localizable measure algebra
$(\frak C,\bar\lambda)$, together with order-continuous Boolean
homomorphisms $\varepsilon_i:\frak A_i\to\frak C$ for $i\in I$, such that
whenever $(\frak B,\bar\nu)$ is a localizable measure algebra, and
$\phi_i:\frak A_i\to\frak B$ are
order-continuous Boolean homomorphisms such that
$\bar\nu(\inf_{i\in I}\phi_i(a_i))=\prod_{i\in I}\bar\mu_ia_i$ whenever
$a_i\in\frak A_i$ for each $i$, then there is a unique
order-continuous measure-preserving Boolean homomorphism
$\phi:\frak C\to\frak B$ such that $\phi\varepsilon_i=\phi_i$ for every $i$.

\spheader 325Hb The structure
$(\frak C,\bar\lambda,\langle\varepsilon_i\rangle_{i\in I})$ is determined up to
isomorphism by this property.

\spheader 325Hc The Boolean homomorphism
$\psi:\bigotimes_{i\in I}\frak A_i\to\frak C$ defined from the
$\varepsilon_i$
is injective, and $\psi[\bigotimes_{i\in I}\frak A_i]$ is topologically
dense in $\frak C$.

\spheader 325Hd Write $\Locmafp_{i\in I}(\frak A_i,\bar\mu_i)$
for (a particular version of) the localizable measure algebra free
product described in (a).   If
$\langle\frak(A_i,\bar\mu_i)\rangle_{i\in I}$ is a
finite family of semi-finite measure algebras and
$\langle I_k\rangle_{k\in K}$ is a partition of $I$ into non-empty sets,
then $\Locmafp_{i\in I}(\frak A_i,\bar\mu_i)$ is isomorphic,
in a canonical way, to
$\Locmafp_{k\in K}\bigl(\Locmafp_{i\in I_k}(\frak A_i,\bar\mu_i)\bigr)$.

\spheader 325He Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a finite family of semi-finite measure spaces, and write
$(\frak A_i,\bar\mu_i)$ for the measure algebra of $(X_i,\Sigma_i,\mu_i)$.
Let $\lambda$ be the c.l.d.\ product measure on
$\prod_{i\in I}X_i$\cmmnt{ (251W)}, and
$(\frak C,\bar\lambda)$ the corresponding measure algebra.   Then there
is a canonical order-continuous measure-preserving embedding of
$(\frak C,\bar\lambda)$ into the localizable measure algebra free
product of the $(\frak A_i,\bar\mu_i)$.   If each $\mu_i$ is strictly
localizable, this embedding is an isomorphism.

\leader{325I}{Infinite \dvrocolon{products}}\cmmnt{ Just as in \S254,
we can now turn to products of infinite families of probability
algebras.

\wheader{325I}{4}{2}{2}{96pt}

\noindent}{\bf Theorem}  Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any family of probability spaces, with measure algebras
$(\frak A_i,\bar\mu_i)$.   Let $\lambda$ be the product measure on
$X=\prod_{i\in I}X_i$, and $(\frak C,\bar\lambda)$ the corresponding
measure algebra.   For each $i\in I$, we have a measure-preserving
homomorphism $\varepsilon_i:\frak A_i\to\frak C$ corresponding to the
inverse-measure-preserving function $x\mapsto x(i):X\to X_i$.   Let
$(\frak B,\bar\nu)$ be a probability algebra, and
$\phi_i:\frak A_i\to\frak B$ Boolean homomorphisms such that $\bar\nu(\inf_{i\in J}\phi_i(a_i))=\prod_{i\in J}\bar\mu_ia_i$ whenever $J\subseteq I$ is
a finite set and $a_i\in\frak A_i$ for every $i$.   Then there is a
unique measure-preserving Boolean homomorphism $\phi:\frak C\to\frak B$
such that $\phi\varepsilon_i=\phi_i$ for every $i\in I$.

\proof{{\bf (a)} As remarked in 254Fb, all the maps $x\mapsto x(i)$ are
inverse-measure-preserving, so correspond to measure-preserving
homomorphisms $\varepsilon_i:\frak A_i\to\frak C$ (324M).   It will be helpful
to use some notation from \S254.   Write $\Cal C$ for the family of
measurable cylinders in $X$ expressible in the form

\Centerline{$E=\{x:x\in X,\,x(i)\in E_i$ for every $i\in J\}$,}

\noindent where $J\subseteq I$ is finite and $E_i\in\Sigma_i$ for every
$i\in J$.   Note that in this case

\Centerline{$E^{\ssbullet}=\inf_{i\in J}\varepsilon_i(E_i^{\ssbullet})$.}

\noindent Set

\Centerline{$C=\{E^{\ssbullet}:E\in\Cal C\}\subseteq\frak C$,}

\noindent so that $C$ is precisely the family of elements of $\frak C$
expressible in the form $\inf_{i\in J}\phi_i(a_i)$ where $J\subseteq I$
is finite and $a_i\in \frak A_i$ for each $i$.

The homomorphisms $\varepsilon_i:\frak A_i\to\frak C$ define a Boolean
homomorphism $\psi:\bigotimes_{i\in I}\frak A_i\to\frak C$ (315J),
which is injective.   \Prf\ If
$c\in\bigotimes_{i\in I}\frak A_i$ is
non-zero, there must be a finite set $J\subseteq I$ and a family
$\langle a_i\rangle_{i\in J}$ such that $a_i\in\frak A_i\setminus\{0\}$
for each $i$ and $c\Bsupseteq\inf_{i\in J}\tilde\varepsilon_i(a_i)$, where
for the moment I write $\tilde\varepsilon_i$ for the canonical map from
$\frak A_i$ to $\bigotimes_{i\in I}\frak A_i$ (315Kb).
Express each $a_i$ as $E_i^{\ssbullet}$, where $E_i\in\Sigma_i$.   Then

\Centerline{$E=\{x:x\in X,\,x(i)\in E_i$ for each $i\in J\}$}

\noindent has measure

\Centerline{$\lambda E=\prod_{i\in J}\mu E_i
=\prod_{i\in J}\bar\mu a_i\ne 0$,}

\noindent while

\Centerline{$E^{\ssbullet}
=\psi(\inf_{i\in J}\tilde\varepsilon_i(a_i))\Bsubseteq\psi(c)$,}

\noindent so $\psi(c)\ne 0$.   As $c$ is arbitrary, $\psi$ is
injective.\ \Qed

\medskip

{\bf (b)} Because $\psi$ is injective, it is an isomorphism between
$\bigotimes_{i\in I}\frak A_i$ and its image in $\frak C$.  I trust it
will cause no confusion if I abuse notation slightly and treat
$\bigotimes_{i\in I}\frak A_i$ as actually a subalgebra of $\frak C$, so
that $\varepsilon_j:\frak A_j\to\frak C$ becomes identified with
$\tilde\varepsilon_j:\frak A_j\to\bigotimes_{i\in I}\frak A_i$.   Now the
Boolean homomorphisms $\phi_i:\frak A_i\to\frak B$ correspond to a
Boolean homomorphism $\theta:\bigotimes_{i\in I}\frak A_i\to\frak B$.
The point is that $\bar\nu\theta(c)=\bar\lambda c$ for every
$c\in\bigotimes_{i\in I}\frak A_i$.   \Prf\ Suppose to begin with that
$c\in C$.   Then we have $c=E^{\ssbullet}$, where $E=\{x:x(i)\in
E_i\Forall i\in J\}$ and $E_i\in\Sigma_i$ for each $i\in J$.   So

$$\eqalign{\bar\lambda c
&=\lambda E
=\prod_{i\in J}\mu E_i
=\prod_{i\in J}\bar\mu_iE_i^{\ssbullet}
=\bar\nu(\inf_{i\in J}\phi a_i)\cr
&=\bar\nu(\inf_{i\in J}\theta\varepsilon_i(a_i))
=\bar\nu\theta(\inf_{i\in J}\varepsilon_i(a_i))
=\bar\nu\theta(c).\cr}$$

\noindent Next, any $c\in\bigotimes_{i\in I}\frak A_i$
is expressible as the supremum of a
finite disjoint family $\langle c_k\rangle_{k\in K}$ in $C$ (315Kb),  so

\Centerline{$\bar\nu\theta(c)=\sum_{k\in K}\bar\nu\theta(c_k)
=\sum_{k\in K}\bar\lambda(c_k)=\bar\lambda c$.   \Qed}


\medskip

{\bf (c)} It follows that $\theta$ is uniformly continuous for the
measure metrics defined by $\bar\nu$ and $\bar\lambda$, since

\Centerline{$\bar\nu(\theta(c)\Bsymmdiff\theta(c'))
=\bar\nu\theta(c\Bsymmdiff c')
=\bar\lambda(c\Bsymmdiff c')$}

\noindent for all $c$, $c'\in\bigotimes_{i\in I}\frak A_i$.

\medskip

{\bf (d)} Next, $\bigotimes_{i\in I}\frak A_i$ is topologically dense in
$\frak C$.   \Prf\ Let $c\in\frak C$, $\epsilon>0$.   Express $c$ as
$W^{\ssbullet}$.  Then by 254Fe there are $H_0,\ldots,H_k\in\Cal C$ such
that $\lambda(W\symmdiff\bigcup_{j\le k}H_j)\le\epsilon$.   Now
$c_j=H_j^{\ssbullet}\in C$ for each $j$, so

\Centerline{$c'=\sup_{j\le k}c_j=(\bigcup_{j\le
k}H_j)^{\ssbullet}\in\bigotimes_{i\in I}\frak A_i$,}

\noindent and $\bar\lambda(c\Bsymmdiff c')\le\epsilon$.\ \Qed

Since $\frak B$ is complete for its uniformity (323Gc), there is a
uniformly continuous function $\phi:\frak C\to\frak B$ extending
$\theta$ (3A4G).

\medskip

{\bf (e)} Because $\theta$ is a Boolean homomorphism, so is $\phi$.
\Prf\ (i) The functions $c\mapsto\phi(1\Bsetminus c)$,
$1\Bsetminus\phi(c)$ are continuous and the topology of $\frak B$ is
Hausdorff, so  $\{c:\phi(1\Bsetminus c)=1\Bsetminus\phi(c)\}$ is closed;
as it includes $\bigotimes_{i\in I}\frak A_i$, it must be the whole of
$\frak C$.   (ii) The functions $(c,c')\mapsto\phi(c\Bcup c')$,
$(c,c')\mapsto\phi(c)\Bcup\phi(c')$ are continuous, so
$\{(c,c'):\phi(c\Bcup c')=\phi(c)\Bcup\phi(c')\}$ is closed in
$\frak C\times\frak C$;  as it includes
$\bigotimes_{i\in I}\frak A_I\times\bigotimes_{i\in I}\frak A_i$,
it must be the whole of $\frak C\times\frak C$.\ \Qed

\medskip

{\bf (f)} Because $\theta$ is measure-preserving, so is $\phi$.   \Prf\
The functions $c\mapsto\bar\lambda c$, $c\mapsto\bar\nu\phi(c)$ are
continuous and equal on $\bigotimes_{i\in I}\frak A_i$, so are equal on
$\frak C$.\ \Qed

\medskip

{\bf (g)} Finally, to see that $\phi$ is unique, observe that any
measure-preserving Boolean homomorphism from $\frak C$ to $\frak B$ must
be continuous,  so that if it agrees with $\phi$ on
$\bigotimes_{i\in I}\frak A_i$ it must agree with $\phi$ on $\frak C$.
}%end of proof of 325I

\leader{325J}{}\cmmnt{ Of course this leads at once to a result
corresponding to 325D.

\medskip

\noindent}{\bf Theorem} Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a
family of probability algebras.

(a) There is a probability algebra $(\frak C,\bar\lambda)$, together
with measure-preserving Boolean homomorphisms
$\varepsilon_i:\frak A_i\to\frak C$ for $i\in I$,
such that whenever $(\frak B,\bar\nu)$ is a probability algebra, and
$\phi_i:\frak A_i\to\frak B$ are
Boolean homomorphisms such that
$\bar\nu(\inf_{i\in J}\phi_i(a_i))=\prod_{i\in J}\bar\mu_ia_i$ whenever
$J\subseteq I$ is
finite and $a_i\in\frak A_i$ for each $i\in J$, then there is a unique
measure-preserving Boolean homomorphism $\phi:\frak C\to\frak B$ such
that $\phi\varepsilon_i=\phi_i$ for every $i\in I$.

(b) The structure $(\frak C,\bar\lambda,\familyiI{\varepsilon_i})$ is
determined up to isomorphism by this property.

(c) The Boolean homomorphism $\psi:\bigotimes_{i\in I}\frak A_i\to\frak C$
defined from the $\varepsilon_i$ is injective, and
$\psi[\bigotimes_{i\in I}\frak A_i]$ is topologically dense in $\frak C$.

\proof{ For (a) and (c), all we have to do is represent each $(\frak
A_i,\bar\mu_i)$ as the measure algebra of a probability space, and
apply 325I.   The uniqueness of $\frak C$ and the $\varepsilon_i$ follows from
the uniqueness of the homomorphisms $\phi$, as in 325Db.
}%end of proof of 325J

\leader{325K}{Definition} As in 325Ea, we can say that
$(\frak C,\bar\lambda,\langle\varepsilon_i\rangle_{i\in I})$
is\dvro{ the}{ a, or the,} {\bf probability algebra free product} of
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$.

\leader{325L}{Independent subalgebras} If $(\frak A,\bar\mu)$ is a
probability algebra, we say that a family
$\langle\frak B_i\rangle_{i\in I}$ of subalgebras of $\frak A$ is
{\bf stochastically independent} if
$\bar\mu(\inf_{i\in J}b_i)=\prod_{i\in J}\bar\mu b_i$ whenever
$J\subseteq I$ is finite and $b_i\in\frak B_i$ for each $i$.
\cmmnt{(Compare
272Ab.)}   If every $\frak B_i$ is closed, so that
$(\frak B_i,\bar\mu\restrp\frak B_i)$ is a probability algebra,
the identity maps $\iota_i:\frak B_i\to\frak A$
satisfy the conditions of the universal mapping theorem 325Ja,
so we have a probability algebra free product
$(\frak C,\bar\mu\restrp\frak C,\familyiI{\iota_i})$ of
$\langle(\frak B_i,\bar\mu\restrp\frak B_i)\rangle_{i\in I}$, where
$\frak C=\bigvee_{i\in I}\frak B_i$ is the closed subalgebra
of $\frak A$ generated by $\bigcup_{i\in I}\frak B_i$.

Conversely, if $\familyiI{(\frak A_i,\bar\mu_i)}$ is any family of
probability algebras with probability algebra free product
$(\frak C,\bar\lambda,\familyiI{\varepsilon_i})$, then
$\familyiI{\varepsilon_i[\frak A_i]}$ is an independent family of
closed subalgebras of $\frak C$.   \cmmnt{(Compare 272J, 315Xp.)}


\leader{325M}{}\cmmnt{ We can now make a general trawl
through Chapters 25 and 27 seeking results which can be expressed in
the language of this section.   I give some in
325Xf-325Xi.   %325Xf 325Xg 325Xh 325Xi
Some ideas from \S254 which are thrown into sharper relief by a
reformulation are in the following theorem.

\medskip

\noindent}{\bf Theorem} Let
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$ be a family of
probability algebras and
$(\frak C,\bar\lambda,\langle\varepsilon_i\rangle_{i\in I})$ their probability
algebra free product.   For
$J\subseteq I$ let $\frak C_J=\bigvee_{i\in J}\varepsilon_i[\frak A_i]$
be the closed subalgebra of $\frak C$
generated by $\bigcup_{i\in J}\varepsilon_i[\frak A_i]$.

(a) For any $J\subseteq I$,
$(\frak C_J,\bar\lambda\restrp\frak C_J,
   \langle\varepsilon_i\rangle_{i\in J})$
is a probability algebra free product of
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in J}$.

(b)(i)  For any $c\in\frak C$, there is a unique smallest $J_c\subseteq I$
such that $c\in\frak C_{J_c}$, and this $J_c$ is countable.

\quad(ii)\dvAnew{2008}
If $c$, $d\in\frak C$ and $c\Bsubseteq d$, then there is an
$e\in\frak C_{J_c\cap J_d}$ such that $c\Bsubseteq e\Bsubseteq d$.

(c) For any non-empty family $\Cal J\subseteq\Cal PI$,
$\bigcap_{J\in\Cal J}\frak C_J=\frak C_{\bigcap\Cal J}$.

\proof{{\bf (a)} If $(\frak B,\bar\nu,\langle\phi_i\rangle_{i\in J})$ is
any probability algebra free product of
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in J}$,
then we have a measure-preserving
homomorphism $\psi:\frak B\to\frak C$ such that $\psi\phi_i=\varepsilon_i$ for
every $i\in J$.   Because the subalgebra $\frak B_0$ of $\frak B$
generated by $\bigcup_{i\in J}\phi_i[\frak A_i]$ is topologically dense
in $\frak B$ (325Jc), and $\psi$ is continuous (324Kb),
$\bigcup_{i\in J}\varepsilon_i[\frak A_i]$ is topologically dense in
$\psi[\frak B]$;  also
$\psi[\frak B]$ is closed in $\frak C$ (324Kb again).   But this means
that $\psi[\frak B]$ is just the topological closure of
$\bigcup_{i\in I}\varepsilon_i[\frak A_i]$ and must be $\frak C_J$.
Thus $\psi$ is an isomorphism, and

\Centerline{$(\frak C_J,\bar\lambda\restrp\frak C_J,
   \family{i}{J}{\varepsilon_i})
=(\psi[\frak B],\bar\nu\psi^{-1},\family{i}{J}{\psi\phi_i})$}

\noindent also is a probability algebra free product of
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in J}$.

\medskip

{\bf (b)} As in 325J, we may suppose that each $(\frak A_i,\bar\mu_i)$
is the measure algebra of a probability space $(X_i,\Sigma_i,\mu_i)$,
and that $\frak C$ is the measure algebra of their product
$(X,\Lambda,\lambda)$.   For $J\subseteq I$ let $\Lambda_J$ be
the set of members of $\Lambda$ which are determined by coordinates
in $J$.   Then
$\{x:x(i)\in E\}\in\Lambda_J$ for every $i\in J$ and $E\in\Sigma_i$;  so
$\{U^{\ssbullet}:U\in\Lambda_J\}$ is a closed subalgebra of $\frak C$
including $\varepsilon_i[\frak A_i]$ for every $i\in J$, and therefore
including $\frak C_J$.   On the other hand, as observed in 254Ob,
any member of $\Lambda_J$ is approximated, in measure, by sets in the
$\sigma$-algebra $\Tau_J$ generated by sets of the form
$\{x:x(i)\in E\}$ where $i\in J$ and $E\in\Sigma_i$.   Of course
$\Tau_J\subseteq\Lambda_J$, so
$\{W^{\ssbullet}:W\in\Lambda_J\}=\{W^{\ssbullet}:W\in\Tau_J\}$ is the
closed subalgebra of $\frak C$ generated by
$\bigcup_{i\in J}\varepsilon_i[\frak A_i]$, which is $\frak C_J$.

\medskip

\quad{\bf (i)} Let $W\in\Lambda$ be such that $c=W^{\ssbullet}$.
By 254Rd, there is a smallest $J_c\subseteq I$ such that
$W\symmdiff U$ is negligible for some $U\in\Lambda_{J_c}$,
and $J_c$ is countable.   By the remarks above, $J_c$ is also the
unique smallest subset of $I$ such that $c\in\frak C_{J_c}$.

\medskip

\quad{\bf (ii)} Let $U\in\Lambda_{J_c}$, $V\in\Lambda_{J_d}$ be such that
$c=U^{\ssbullet}$ and $d=V^{\ssbullet}$.   We can think of
$\lambda$ as a product $\lambda'\times\lambda''$ where $\lambda'$ is the
product measure on $X'=\prod_{i\in J_d}X_i$ and $\lambda''$ is the product
measure on $X''=\prod_{i\in I\setminus J_d}X_i$ (254N).   Express $V$ as
$V_0\times X''$ where $V_0\subseteq X'$ belongs to the domain of
$\lambda'$ (254Ob).   Consider

\Centerline{$W_0=\{z:z\in X'$, $\{w:w\in X''$, $(z,w)\in U\}$ is not
$\lambda''$-negligible$\}$;}

\noindent then $W_0$ is measured by $\lambda'$, by Fubini's theorem
(252B or 252D).
Because $c\Bsubseteq d$, $U\setminus V$ is $\lambda$-negligible
and $W_0\setminus V_0$ is $\lambda'$-negligible, while $W_0$ is determined
by coordinates in $J_c\cap J_d$.   So $W=W_0\times X''$ also is determined
by coordinates in $J_c\cap J_d$, while $U\setminus W$ and $W\setminus V$
are $\lambda$-negligible.   We can therefore take $e=W^{\ssbullet}$.


\medskip

{\bf (c)} Of course $\frak C_K\subseteq\frak C_J$ whenever $K\subseteq
J\subseteq I$, so
$\bigcap_{J\in\Cal J}\frak C_J\supseteq\frak C_{\bigcap\Cal J}$.   On
the other hand, suppose that $c\in\bigcap_{J\in\Cal J}\frak C_J$;  then
by (b-i) there is some $K\subseteq\bigcap\Cal J$ such that $c\in\frak
C_K\subseteq\frak C_{\bigcap\Cal J}$.   As $c$ is arbitrary,
$\bigcap_{J\in\Cal J}\frak C_J=\frak C_{\bigcap\Cal J}$.
}%end of proof of 325M

\leader{*325N}{Notation} In this context, I will say that an element
$c$ of $\frak C$ is {\bf determined by coordinates in} $J$ if
$c\in\frak C_J$.

\exercises{
\leader{325X}{Basic exercises (a)}
%\spheader 325Xa
Let $(\frak A_1,\bar\mu_1)$, $(\frak A_2,\bar\mu_2)$ be two semi-finite
measure algebras, and suppose that
for each $j$ we are given a closed subalgebra $\frak B_j$ of $\frak A_j$
such that $(\frak B_j,\bar\nu_j)$ also is semi-finite, where
$\bar\nu_j=\bar\mu_j\restrp\frak B_j$.   Show that the localizable
measure algebra free product
$(\frak B_1,\bar\nu_1)\locmafp(\frak B_2,\bar\nu_2)$ can be thought of as
a closed subalgebra of
$(\frak A_1,\bar\mu_1)\locmafp(\frak A_2,\bar\mu_2)$.
%325E

\spheader 325Xb Let $(\frak A_1,\bar\mu_1)$ and
$(\frak A_2,\bar\mu_2)$ be two semi-finite measure algebras,
and suppose that
for each $j$ we are given a principal ideal $\frak B_j$ of $\frak A_j$.
Set $\bar\nu_j=\bar\mu_j\restrp\frak B_j$.   Show that the
localizable measure algebra free product
$(\frak B_1,\bar\nu_1)\locmafp(\frak B_2,\bar\nu_2)$ can be thought of as
a principal ideal of $(\frak A_1,\bar\mu_1)\locmafp(\frak A_2,\bar\mu_2)$.
%325E

\spheader 325Xc Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure algebras with localizations
$(\widehat{\frak A},\tilde\mu)$ and $(\widehat{\frak B},\tilde\nu)$.
Show that the localizable measure algebra free products
$(\frak A,\bar\mu)\locmafp(\frak B,\bar\nu)$ and
$(\widehat{\frak A},\tilde\mu)\locmafp(\widehat{\frak B},\tilde\nu)$ are
isomorphic.
%325E

\sqheader 325Xd Let $\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$ and
$\langle(\frak B_j,\bar\nu_j)\rangle_{j\in J}$ be families of
semi-finite measure algebras, with simple products $(\frak A,\bar\mu)$
and $(\frak B,\bar\nu)$ (322L).   Show that the localizable measure
algebra free product $(\frak A,\bar\mu)\locmafp(\frak B,\bar\nu)$ can be
identified with the simple product of the family
$\langle(\frak A_i,\bar\mu_i)\locmafp(\frak B_j,\bar\nu_j)
 \rangle_{i\in I,j\in J}$.
%325E, 325H

\sqheader 325Xe Let $\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$
and $\familyiI{(\frak A_i',\bar\mu'_i)}$ be two
families of probability algebras, and
$(\frak C,\bar\lambda,\langle\varepsilon_i\rangle_{i\in I})$,
$(\frak C',\bar\lambda',\langle\varepsilon'_i\rangle_{i\in I})$ their
probability algebra free products.   Suppose that
for each $i\in I$ we are given a measure-preserving Boolean homomorphism
$\pi_i:\frak A_i\to\frak A'_i$.   Show that there is a unique
measure-preserving Boolean homomorphism $\pi:\frak C\to\frak C'$ such
that $\pi\varepsilon_i=\varepsilon'_i\pi_i$ for every $i\in I$.
%325K

\sqheader 325Xf Let $(\frak A,\bar\mu)$ be a probability
algebra.   We say that a family $\langle a_i\rangle_{i\in I}$ in
$\frak A$ is {\bf stochastically independent} if
$\bar\mu(\inf_{i\in J}a_i)=\prod_{i\in J}\bar\mu a_i$ for every non-empty
finite $J\subseteq I$.   Show that this is so iff
$\langle\frak A_i\rangle_{i\in I}$ is stochastically
independent, where $\frak A_i=\{0,a_i,1\Bsetminus a_i,1\}$ for each $i$.
(Compare 272F.)
%325L

\sqheader 325Xg Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\langle\frak A_i\rangle_{i\in I}$ a stochastically
independent family of closed subalgebras of $\frak A$.   Let
$\langle J(k)\rangle_{k\in K}$ be a disjoint family of subsets of $I$,
and for each $k\in K$ let $\frak B_k=\bigvee_{i\in J(k)}\frak A_i$
be the closed subalgebra of $\frak A$
generated by $\bigcup_{i\in J(k)}\frak A_i$.   Show that
$\langle\frak B_k\rangle_{k\in K}$ is
stochastically independent.   (Compare 272K.)
%325L

\spheader 325Xh Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\langle\frak A_i\rangle_{i\in I}$ a stochastically
independent family of closed subalgebras of $\frak A$.   For
$J\subseteq I$ set $\frak B_J=\bigvee_{i\in J}\frak A_i$.
Show that
$\bigcap\{\frak B_{I\setminus J}:J$ is a finite subset of $I\}=\{0,1\}$.
\Hint{For
$J\subseteq I$, show that $\bar\mu(b\Bcap c)=\bar\mu b\cdot\bar\mu c$
for every $b\in\frak B_{I\setminus J}$ and $c\in \frak B_J$.   Compare
272O, 325M.}
%325L

\spheader 325Xi Let
$\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$ be a family of
probability algebras with probability algebra free
product $(\frak C,\bar\lambda,\langle\varepsilon_i\rangle_{i\in I})$.   For
$J\subseteq I$ set $\frak C_J=\bigvee_{i\in J}\varepsilon_i[\frak A_i]$.
Show that for any $J$, $K\subseteq I$ and $c\in\frak C$,
$\frak C_J\cap\frak C_K=\frak C_{J\cap K}$ and the upper envelope
$\upr(c,\frak C_{J\cap K})$ is equal to
$\upr(\upr(c,\frak C_J),\frak C_K)$.
%325M

\leader{325Y}{Further exercises (a)}
%\spheader 325Ya
Let $\mu$ be counting measure on $X=\{0\}$, $\mu'$ the
countable-cocountable measure on $X'=\omega_1$, and $\nu$ counting
measure on $Y=\omega_1$.   Show that the measure algebras of the
primitive product measures on $X\times Y$, $X'\times Y$ are not
isomorphic.
%325B

\spheader 325Yb Let $(\frak A_1,\bar\mu_1)$, $(\frak A_2,\bar\mu_2)$,
$(\frak A'_1,\bar\mu'_1)$ and $(\frak A'_2,\bar\mu'_2)$ be semi-finite
measure algebras with localizable measure algebra free products
$(\frak C,\bar\lambda,\varepsilon_1,\varepsilon_2)$ and
$(\frak C',\bar\lambda',\varepsilon'_1,\varepsilon'_2)$.   Suppose that
$\pi_1:\frak A_1\to\frak A'_1$ and $\pi_2:\frak A_2\to\frak A'_2$ are
measure-preserving Boolean homomorphisms.   Show that there is a
measure-preserving Boolean homomorphism $\pi:\frak C\to\frak C'$ such that
$\pi\varepsilon_i=\varepsilon'_i\pi_i$ for both $i$, but that $\pi$ is not necessarily
unique.
%325C

\spheader 325Yc
Let $\frak A$ be a Boolean algebra, and
$\mu:\frak A\to[0,\infty]$ a functional such that $\mu 0=0$, $\mu a>0$ for
every $a\ne 0$, and
$\mu(a\Bcup b)=\mu a+\mu b$ whenever $a$, $b\in \frak A$ and
$a\Bcap b=0$;   suppose that $\frak A^f=\{a:\mu a<\infty\}$ is
order-dense in $\frak A$.  For $e\in\frak A^f$, $a$, $b\in\frak A$ set
$\rho_e(a,b)=\mu(e\Bcap(a\Bsymmdiff b))$.   Give $\frak A$
the uniformity defined by $\{\rho_e:\mu e<\infty\}$.   (i) Show that the
completion $\widehat{\frak A}$ of $\frak A$ under this uniformity has a
measure $\hat\mu$, extending
$\mu$, under which it is a localizable measure algebra.   (ii) Show that
if $a\in\widehat{\frak A}$, $\hat\mu a<\infty$ and $\epsilon>0$, there
is a $b\in\frak A$ such that $\hat\mu(a\Bsymmdiff b)\le\epsilon$.
(iii) Show that for every $a\in\widehat{\frak A}$ there is a sequence
$\sequencen{a_n}$ in $\frak A$ such that
$a\Bsupseteq\sup_{n\in\Bbb N}\inf_{m\ge n}a_m$ and
$\hat\mu a=\hat\mu(\sup_{n\in\Bbb N}\inf_{m\ge n}a_m)$.
(iv) In particular, the set of infima in $\widehat{\frak A}$
of sequences in $\frak A$ is
order-dense in $\widehat{\frak A}$.   (v) Explain the relevance of this
construction to the embedding
$\frak A_1\otimes\frak A_2\embedsinto\frak C$ in 325D.
%325D

\spheader 325Yd In 325F, set $W=\bigcup_{n\in\Bbb N}E_n\times E_n$.   Show that if $A$, $B$ are any non-negligible subsets
of $[0,1]$, then $W\cap(A\times B)$ is not negligible.
%325F

\spheader 325Ye Let $(\frak A,\bar\mu)$ be the measure algebra of
Lebesgue measure on $[0,1]$.   Show that $\frak A\otimes\frak A$ is ccc
but not \wsid.   \Hint{(i) $\frak A\otimes\frak A$ is embeddable as a
subalgebra of a probability algebra (ii) in the notation of 325F, look
at $c_{mn}=\sup_{m\le i\le n}e_i\otimes e_i$.}
%325F

\spheader 325Yf Repeat 325F-325G and 325Yd-325Ye with an arbitrary
atomless probability space in place of $[0,1]$.
%325G, 325Yd

\spheader 325Yg Let $(\frak A,\bar\mu)$ be a probability algebra and
$\langle a_i\rangle_{i\in I}$ a stochastically independent family in
$\frak A$.   Show that for any $a\in\frak A$ and $\epsilon>0$ the set
$\{i:i\in I,\,
  |\bar\mu(a\Bcap a_i)-\bar\mu a\cdot\bar\mu a_i|\ge\epsilon\}$ is finite,
so that
$\{i:\bar\mu(a\Bcap a_i)\ne\bar\mu a\cdot\bar\mu a_i\}$ is countable.
\Hint{272Ye\formerly{2{}72Yd}.}
%325M
}%end of exercises

\endnotes{
\Notesheader{325} 325B shows that the measure algebra of a product
measure may be irregular if we have factor measures which are not
strictly localizable.   But two facts lead the way
to the `localizable measure algebra free product' in 325D-325E.   The
first is that every semi-finite measure algebra is embeddable, in a
canonical way, in a localizable measure algebra (322P);  and the second
is that the Stone representation of a localizable measure algebra is
strictly localizable (322O).   It is a happy coincidence that we can
collapse these two facts together in the construction of 325D.   Another
way of looking at the localizable measure algebra free product of two
localizable measure algebras is to express it as the simple product of
measure algebra free products of totally finite measure algebras, using
325Xd and the fact that for $\sigma$-finite measure algebras there is
only one reasonable measure algebra free product, being that provided
by any representation of them as measure algebras of measure spaces
(325Eb).

Yet a third way of approaching measure algebra free products is as the
uniform space completions of algebraic free products, using 325Yc.
This gives the same result as the construction of 325D because the
algebraic free product appears as a topologically dense subalgebra of
the localizable measure algebra free product, which is complete
as uniform space (325Dc).
(I have to repeat such phrases as `topologically dense' because the
algebraic free product is emphatically {\it not} order-dense in the
measure algebra free product (325F).)   The results in 251I on
approximating measurable sets for a c.l.d.\ product measure by
combinations of measurable rectangles correspond to general facts about
completions of finitely-additive measures (325Yc(ii), 325Yc(iii)).
It is worth noting that the completion process can be regarded as made
up of two steps;  first take infima of sequences of sets of finite
measure, and then take arbitrary suprema (325Yc(iv)).

The idea of 325F appears in many guises, and this is only the first time
that I shall wish to call on it.   The point of the set
$W=\bigcup_{n\in\Bbb N}E_n\times E_n$ is that it is a measurable subset
of the square (indeed, by taking the $E_n$ to be open sets we can
arrange that $W$ should be open), of measure strictly less than $1$ (in
fact, as small as we wish), such that its complement does not include
any non-negligible `measurable rectangle' $G\times H$;   indeed,
$W\cap(A\times B)$ is non-negligible for any non-negligible sets $A$,
$B\subseteq[0,1]$ (325Yd).   I believe that the first published example
of such a set was by {\smc Erd\H{o}s \& Oxtoby 55} 
(a version of which is in
532N in Volume 5);  I learnt the method of 325F from R.O.Davies.

I include 325G as a kind of guard-rail.   The relationship between
preservation of measure and order-continuity is a subtle one, as I have
already tried to show in 324K, and it is often worth considering the
possibility that a result involving order-continuous measure-preserving
homomorphisms has a form applying to all order-continuous homomorphisms.
However, there is no simple expression of such an idea in the present
context.

In the context of infinite free products of probability algebras, there
is a degree of simplification, since there is only one algebra which can
plausibly be called the probability algebra free product, and this is
produced by any realization of the algebras as measure algebras of
probability spaces (325I-325K).   The examples 325F-325G apply equally,
of course, to this context.   At this point I mention the concept of
`stochastically independent' family (325L, 325Xf) because we have the
machinery to translate several results from \S272 into the language of
measure algebras (325Xf-325Xh).   I feel that I have to use the phrase
`stochastically independent' here because there is the much weaker
alternative concept of `Boolean independence' (315Xp) also present.
But I leave most of this as exercises, because the language of measure
algebras offers few ideas to the probability theory already covered in
Chapter 27.   All it can do is formalise the ever-present principle that
negligible sets often can and should be ignored.
}%end of notes

\discrpage

\frfilename{mt326.tex}
\versiondate{21.5.11}
\copyrightdate{2001}

\def\chaptername{Measure algebras}
\def\sectionname{Additive functionals on Boolean algebras}

\newsection{326}

I devote two sections to the general theory of additive functionals on
measure algebras.   As many readers will rightly be in a hurry to get on
to the next two chapters, I remark that the only significant result
needed for \S\S331-332 is the Hahn decomposition of a countably
additive functional (326M), and that this is no more than a translation
into the language of measure algebras of a theorem already given in
Chapter 23.   The concept of `standard extension' of a
countably additive functional from a subalgebra (327F-327G) will be used
for a theorem in \S333, and as preparation for Chapter 36.

I begin with notes on the space of additive functionals on an arbitrary
Boolean algebra (326A-326D), corresponding to 231A-231B, but adding a
more general form of the Jordan decomposition of a bounded additive
functional into positive and negative parts (326D).
The next four paragraphs are starred, because they will not be needed in
this volume;  326E is essential if you want to look at additive functionals
on free products, 326F is a basic classification criterion, and 326H is an
important extension of a fundamental fact about atomless measures noted in
215D, but all can be passed over on first reading.
The next
subsection (326I-326M) deals with countably additive functionals,
corresponding to 231C-231F.   In 326N-326T I develop a new idea, that of
`completely additive' functional, which does not match anything in
the previous treatment.

\leader{326A}{Additive functionals:  Definition}  Let $\frak A$ be a
Boolean algebra.   A functional $\nu:\frak A\to\Bbb R$ is {\bf
finitely additive}, or just {\bf additive}, if
$\nu(a\Bcup b)=\nu a+\nu b$ whenever $a$, $b\in\frak A$ and
$a\Bcap b=0$.

\cmmnt{
A non-negative additive functional is sometimes called a {\bf finitely
additive measure} or {\bf charge}.}

\leader{326B}{Elementary facts} Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to\Bbb R$ a finitely additive functional.   \cmmnt{The
following
will I hope be obvious.}

\header{326Ba}{\bf (a)} $\nu 0=0$\cmmnt{ (because $\nu 0=\nu 0+\nu
0$)}.

\header{326Bb}{\bf (b)} If $c\in\frak A$, then $a\mapsto\nu(a\Bcap c)$
is additive\cmmnt{ (because $(a\Bcap c)\Bcup(b\Bcap c)=(a\Bcup
b)\Bcap c$)}.

\header{326Bc}{\bf (c)} $\alpha\nu$ is an additive functional
for any
$\alpha\in\Bbb R$.   If $\nuprime$ is another finitely additive functional
on $\frak
A$, then $\nu+\nuprime$ is  additive.

\header{326Bd}{\bf (d)} If $\langle\nu_i\rangle_{i\in I}$ is any family
of finitely additive functionals such that
$\nuprime a=\sum_{i\in I}\nu_i a$ is defined in $\Bbb R$ for every
$a\in \frak A$, then $\nuprime$ is additive.

\header{326Be}{\bf (e)} If $\frak B$ is another Boolean algebra and
$\pi:\frak B\to\frak A$ is a Boolean homomorphism, then $\nu\pi:\frak
B\to\Bbb R$ is additive.   In particular, if $\frak B$ is a subalgebra
of $\frak A$, then
$\nu\restrp\frak B:\frak B\to\Bbb R$ is  additive.

\header{326Bf}{\bf (f)} $\nu$ is non-negative iff it is
order-preserving -- that is,

\Centerline{$\nu a\ge 0\text{ for every }a\in\frak A
\iff \nu b\le\nu c\text{ whenever }b\Bsubseteq c$\dvro{.}{}}

\prooflet{\noindent (because $\nu c=\nu b+\nu(c\Bsetminus b)$ if
$b\Bsubseteq c$).}

\leader{326C}{The space of additive functionals} Let $\frak A$ be any
Boolean algebra.   From 326Bc we see that the set $M$ of all finitely
additive real-valued functionals on $\frak A$ is a linear
space\cmmnt{ (a
linear subspace of $\BbbR^{\frak A}$)}.   We give it the ordering
induced by that of $\BbbR^{\frak A}$\cmmnt{, so that $\nu\le\nuprime$ iff
$\nu a\le\nuprime a$ for every $a\in\frak A$}.   This renders it a partially
ordered linear space\cmmnt{ (because $\BbbR^{\frak A}$ is)}.

\leader{326D}{The Jordan decomposition (I):  Proposition} Let $\frak A$
be a Boolean algebra, and $\nu$ a finitely additive real-valued
functional on $\frak A$.   Then the following are equiveridical:

(i) $\nu$ is bounded;

(ii) $\sup_{n\in\Bbb N}|\nu a_n|<\infty$ for every disjoint sequence
$\sequencen{a_n}$ in $\frak A$;

(iii) $\lim_{n\to\infty}|\nu a_n|=0$ for every disjoint sequence
$\sequencen{a_n}$ in $\frak A$;

(iv) $\sum_{n=0}^{\infty}|\nu a_n|<\infty$ for every disjoint sequence
$\sequencen{a_n}$ in $\frak A$;

(v) $\nu$ is expressible as the difference of two non-negative additive
functionals.

\proof{{\bf (a)(i)$\Rightarrow$(v)} Assume that $\nu$ is bounded.   For
each
$a\in\frak A$, set

\Centerline{$\nu^+a=\sup\{\nu b:b\Bsubseteq a\}$.}

\noindent Because $\nu$ is bounded, $\nu^+$ is real-valued.   Now
$\nu^+$ is  additive.   \Prf\ If $a$, $b\in\frak A$ and $a\Bcap b=0$,
then

$$\eqalignno{\nu^+(a\Bcup b)
&=\sup_{c\Bsubseteq a\Bcup b}\nu c
=\sup_{d\Bsubseteq a,e\Bsubseteq b}\nu(d\Bcup e)
=\sup_{d\Bsubseteq a,e\Bsubseteq b}\nu d+\nu e\cr
\noalign{\noindent (because $d\Bcap e\Bsubseteq a\Bcap b=0$ whenever
$d\Bsubseteq a$, $e\Bsubseteq b$)}
&=\sup_{d\Bsubseteq a}\nu d+\sup_{e\Bsubseteq b}\nu e
=\nu^+a+\nu^+b.\text{ \Qed}\cr}$$

\noindent Consequently $\nu^-=\nu^+-\nu$ also is additive (326Bc).

Since

\Centerline{$0=\nu 0\le\nu^+a$,\quad $\nu a\le\nu^+a$}

\noindent for every $a\in\frak A$, $\nu^+\ge 0$ and $\nu^-\ge 0$.   Thus
$\nu=\nu^+-\nu^-$ is the difference of two non-negative  additive
functionals.

\medskip

{\bf (b)(v)$\Rightarrow$(iv)} If $\nu$ is expressible as
$\nu_1-\nu_2$, where $\nu_1$ and $\nu_2$ are non-negative  additive
functionals, and $\sequencen{a_n}$ is disjoint, then

\Centerline{$\sum_{i=0}^n\nu_ja_i=\nu_j(\sup_{i\le n}a_i)\le\nu_j1$}

\noindent for every $n$, both $j$, so that

\Centerline{$\sum_{i=0}^{\infty}|\nu a_i|
\le\sum_{i=0}^{\infty}\nu_1a_i+\sum_{i=0}^{\infty}\nu_2a_i
\le\nu_11+\nu_21<\infty$.}

\medskip

{\bf (c)(iv)$\Rightarrow$(iii)$\Rightarrow$(ii)} are trivial.

\medskip

{\bf (d) not-(i)$\Rightarrow$not-(ii)} Suppose that $\nu$ is unbounded.
Choose sequences $\sequencen{a_n}$, $\sequencen{b_n}$ inductively, as
follows.   $b_0=1$.   Given that $\sup_{a\Bsubseteq b_n}|\nu a|=\infty$,
choose $c_n\Bsubseteq b_n$ such that $|\nu c_n|\ge|\nu b_n|+n$;  then
$|\nu c_n|\ge n$ and

\Centerline{$|\nu(b_n\Bsetminus c_n)|=|\nu b_n-\nu c_n|\ge|\nu c_n|-|\nu
b_n|\ge n$.}

\noindent We have

$$\eqalign{\infty
&=\sup_{a\Bsubseteq b_n}|\nu a|
=\sup_{a\Bsubseteq b_n}|\nu (a\Bcap c_n)+\nu(a\Bsetminus c_n)|\cr
&\le\sup_{a\Bsubseteq b_n}|\nu (a\Bcap c_n)|+|\nu(a\Bsetminus c_n)|
\le\sup_{a\Bsubseteq b_n\cap c_n}|\nu a|+
   \sup_{a\Bsubseteq b_n\Bsetminus c_n}|\nu a|,\cr}$$

\noindent so at least one of
$\sup_{a\Bsubseteq b_n\cap c_n}|\nu a|$, $\sup_{a\Bsubseteq
b_n\Bsetminus
c_n}|\nu a|$  must be infinite;  take $b_{n+1}$ to be one of $c_n$,
$b_n\Bsetminus c_n$ such that $\sup_{a\Bsubseteq b_{n+1}}|\nu
a|=\infty$, and set $a_n=b_n\Bsetminus b_{n+1}$, so that $|\nu a_n|\ge
n$.   Continue.

On completing the induction, we have a disjoint sequence
$\sequencen{a_n}$ such that $|\nu a_n|\ge n$ for every $n$, so that (ii)
is false.
}%end of proof of 326D

\cmmnt{\medskip

\noindent{\bf Remark} I hope that this reminds you of the decomposition
of a function of bounded variation as the difference of monotonic
functions (224D).
}

\leader{*326E}{Additive functionals on free
\dvrocolon{products}} \cmmnt{In Volume 4, when we
return to the
construction of measures on product spaces, the following fundamental
fact will be useful.

\medskip

\noindent}{\bf Theorem} Let $\familyiI{\frak A_i}$ be a non-empty family
of Boolean algebras, with free product $\frak A$;  write
$\varepsilon_i:\frak A_i\to\frak A$ for the canonical maps, and

\Centerline{$C=\{\inf_{j\in J}\varepsilon_j(a_j):J\subseteq I$ is
finite, $a_j\in\frak A_j$ for every $j\in J\}$.}

\noindent Suppose that $\theta:C\to\Bbb R$ is such that

\Centerline{$\theta c
=\theta(c\Bcap\varepsilon_i(a))
  +\theta(c\Bcap\varepsilon_i(1\Bsetminus a))$}

\noindent whenever $c\in C$, $i\in I$ and $a\in\frak A_i$.   Then there
is a unique finitely additive functional $\nu:\frak A\to\Bbb R$ extending
$\theta$.

\proof{{\bf (a)} It will help if I note at once that $\theta 0=0$.
\Prf

\Centerline{$\theta
0=\theta(0\Bcap\varepsilon_i(0))+\theta(0\Bcap\varepsilon_i(1))
=2\theta 0$}

\noindent for any $i\in I$.\ \Qed

\medskip

{\bf (b)} The key is of course the following fact:  if
$\langle c_r\rangle_{r\le m}$ and $\langle d_s\rangle_{s\le n}$ are two
disjoint families in $C$ with the same supremum in $\frak A$, then
$\sum_{r=0}^m\theta c_r=\sum_{s=0}^n\theta d_s$.   \Prf\ Let
$J\subseteq I$ be a finite set and $\frak B_i\subseteq\frak A_i$ a
finite subalgebra, for each $i\in J$, such that every $c_r$ and every
$d_s$ belongs to the subalgebra $\frak A_0$ of $\frak A$ generated by
$\{\varepsilon_j(b):j\in J,\,b\in\frak B_j\}$.   Next, if $j\in J$ and
$b\in\frak B_j$, then

\Centerline{$\sum_{r=0}^m\theta c_r
=\sum_{r=0}^m\theta(c_r\Bcap\varepsilon_j(b))
+\sum_{r=0}^m\theta(c_r\Bsetminus\varepsilon_j(b))$.}

\noindent We can therefore find a disjoint family
$\langle c'_r\rangle_{r\le m'}$ in $C\cap\frak A_0$ such that

\Centerline{$\sup_{r\le m'}c'_r=\sup_{r\le m}c_r$,
\quad$\sum_{r=0}^{m'}\theta c'_r=\sum_{r=0}^m\theta c_r$,}

\noindent and whenever $r\le m'$, $j\in J$ and $b\in\frak B_j$ then
either $c'_r\Bsubseteq\varepsilon_j(b)$ or $c'_r\Bcap\varepsilon_j(b)=0$;
that is, every $c'_r$ is either $0$ or of the form
$\inf_{j\in J}\varepsilon_j(b_j)$ where $b_j$ is an atom of $\frak B_j$
for every $j$.   Similarly, we can find $\langle d'_s\rangle_{s\le n'}$
such that

\Centerline{$\sup_{s\le n'}d'_s=\sup_{s\le n}d_s$,
\quad$\sum_{s=0}^{n'}\theta d'_s=\sum_{s=0}^n\theta d_s$,}

\noindent and whenever $s\le n'$ and $j\in J$ then
$d'_s$ is either $0$ or of the form
$\inf_{j\in J}\varepsilon_j(b_j)$ where $b_j$ is an atom of $\frak B_j$
for every $j$.   But we now have $\sup_{r\le m'}c'_r=\sup_{s\le n'}d'_s$
while for any $r\le m'$, $s\le n'$ either $c'_r=d'_s$ or
$c'_r\Bcap d'_s=0$.   It follows that the non-zero terms in the finite
sequence $\langle c'_r\rangle_{r\le m'}$ are just a rearrangement of the
non-zero terms in $\langle d'_s\rangle_{s\le n'}$, so that

\Centerline{$\sum_{r=0}^m\theta c_r
=\sum_{r=0}^{m'}\theta c'_r=\sum_{s=0}^{n'}\theta d'_s
=\sum_{s=0}^n\theta d_s$,}

\noindent as required.\ \Qed

\medskip

{\bf (c)} By 315Kb, this means that we have a functional
$\nu:\frak A\to\Bbb R$ such that
$\nu(\sup_{r\le m}c_r)=\sum_{r=0}^m\theta c_r$ whenever
$\langle c_r\rangle_{r\le m}$ is a disjoint family in $C$.   It is now
elementary to check that $\nu$ is additive, and it is clearly the only
additive functional on $\frak A$ extending $\theta$.
}%end of proof of 326E

\leader{*326F}{}\cmmnt{ I give a couple of pages to an interesting
property of
additive functionals on Dedekind $\sigma$-complete Boolean
algebras.   I do not think it will be used in this book, and it really
belongs to the theory of vector measures, which is hardly considered here,
but the ideas are important, and the following definition has other
uses.

\medskip

\noindent}{\bf Definition} Let $\frak A$ be a Boolean algebra, and $\nu$ a
finitely additive functional on $\frak A$.
\cmmnt{I will say that} $\nu$ is
{\bf properly atomless} if for every $\epsilon>0$ there is a finite
partition
$\familyiI{a_i}$ of unity in $\frak A$ such that $|\nu a|\le\epsilon$
whenever $i\in I$ and $a\Bsubseteq a_i$.

\leader{*326G}{Lemma} Let $\frak A$ be a Boolean algebra.

(a)(i) If $\nu$, $\nu':\frak A\to\Bbb R$ are properly atomless
finitely additive functionals and $\alpha\in\Bbb R$, then $\alpha\nu$
and $\nu+\nu'$ are properly atomless additive functionals.

\quad(ii) If $\nu:\frak A\to\Bbb R$ is a properly atomless
finitely additive
functional, then $\nu$ is bounded and $\nu$ can be expressed as the
difference of two non-negative properly atomless additive
functionals.

(b) Suppose that $\frak A$ is Dedekind $\sigma$-complete and that
$\familyiI{\nu_i}$ is a family of non-negative additive functionals on
$\frak A$ such that for every $a\in\frak A$ there are an
$\alpha\in[\bover13,\bover23]$ and an $a'\Bsubseteq a$ such that
$\nu_ia'=\alpha\nu_ia$ for every $i\in I$.
Then for any $a\in\frak A$ there is a
non-decreasing family $\family{t}{[0,1]}{a_t}$ in $\frak A$ such that
$a_0=0$, $a_1=a$ and $\nu_ia_t=t\nu_ia$ for every $t\in[0,1]$ and $i\in I$.

(c) Suppose that $\frak A$ is Dedekind $\sigma$-complete and that
$\nu_0,\ldots,\nu_n:\frak A\to\coint{0,\infty}$ are
properly atomless additive functionals such that $\nu_ia\le\nu_0a$ for
every $i\le n$ and $a\in\frak A$.   Then for any $a\in\frak A$ there is a
non-decreasing family $\family{t}{[0,1]}{a_t}$ in $\frak A$ such that
$a_0=0$, $a_1=a$ and $\nu_ia_t=t\nu_ia$ for every $t\in[0,1]$ and $i\le n$.

\proof{{\bf (a)(i)} Let $\epsilon>0$.   Then there are finite partitions
$\familyiI{a_i}$, $\family{j}{J}{b_j}$ of unity in $\frak A$ such that
$|\nu a|\le\Bover{\epsilon}{2+|\alpha}$ whenever $i\in I$ and
$a\Bsubseteq a_i$, while $|\nu'a|\le\Bover{\epsilon}2$ whenever $j\in J$
and $a\Bsubseteq b_j$.   Now $|(\alpha\nu)(a)|\le\epsilon$ whenever
$i\in I$ and $a\Bsubseteq a_i$.   Moreover,
$\family{(i,j)}{I\times J}{a_i\Bcap b_j}$ is a finite partition of unity in
$\frak A$, and $|(\nu+\nu')(a)|\le\epsilon$ whenever $i\in I$, $j\in J$ and
$a\Bsubseteq a_i\Bcap b_j$.

\medskip

\quad{\bf (ii)}\grheada\ There is a finite partition
$\family{j}{J}{c_j}$ of unity in $\frak A$ such that $|\nu a|\le 1$
whenever $i\in J$ and $a\Bsubseteq c_j$;  now
$|\nu a|\le\sum_{j\in J}|\nu(a\Bcap c_j)|\le\#(J)$ for every $a\in\frak A$,
so $\nu$ is bounded.

\medskip

\qquad\grheadb\ Define $\nu^+$ as in part (a) of the proof of 326D, so that
$\nu^+:\frak A\to\coint{0,\infty}$ is additive.    Now $\nu^+$ is properly
atomless.   \Prf\ Given $\epsilon>0$, there is a finite partition
$\familyiI{a_i}$ of unity in $\frak A$ such that $|\nu a|\le\epsilon$
whenever $i\in I$ and $a\Bsubseteq a_i$;  in which case
$\nu^+a=\sup_{b\Bsubseteq a}\nu b\le\epsilon$ whenever $i\in I$ and
$a\Bsubseteq a_i$.\ \QeD\   As in 326D, $\nu^-=\nu^+-\nu$ is
non-negative, and by
(i) just above (or otherwise)
it is properly atomless, so $\nu=\nu^+-\nu^-$ is the
difference of non-negative properly atomless functionals.

\medskip

{\bf (b)} If $\nu_ia=0$ for every $i\in I$, we can take $a_t=0$ for
$0\le t<1$ and $a_1=a$.   So suppose that $k\in I$ is such that
$\nu_ka>0$.   For $i\in I$, set $\gamma_i=\Bover{\nu_ia}{\nu_ka}$.   Choose
$\sequencen{D_n}$ inductively, as follows.   $D_0=\{0,a\}$.
Given that $D_n$ is a finite totally ordered subset of
$\{b:b\Bsubseteq a\}$ containing $0$ and $a$
and $\nu_id=\gamma_i\nu_kd$ for every $d\in D_n$ and $i\in I$,
then for each
$d\in D_n\setminus\{a\}$ let $d'$ be the next member of $D_n$ strictly
including $d$, and take $b_d\Bsubseteq d'\Bsetminus d$,
$\alpha_d\in[\bover13,\bover23]$ such that
$\nu_ib_d=\alpha_d\nu_i(d'\Bsetminus d)$ for every $i\in I$.
Then

\Centerline{$\nu_i(d\Bcup b_d)
=(1-\alpha_d)\nu_id+\alpha_d\nu_id'
=\gamma_i((1-\alpha_d)\nu_kd+\alpha_d\nu_kd')
=\gamma_i\nu_k(d\Bcup b_d)$}

\noindent for every $i$.   Set $D_{n+1}=D_n\cup\{d\Bcup b_d:d\in D_n\}$;
observe that $D_{n+1}$ is still totally ordered, and continue.
At the end of the induction,
it is easy to see that $\nu_k(d'\Bsetminus d)\le(\bover23)^n\nu_ia$
whenever $n\in\Bbb N$ and $d\Bsubset d'$ are successive members
of $D_n$.

Set $D=\bigcup_{n\in\Bbb N}D_n$.   Then $D$ is a
countable totally ordered set with
least element $0$ and greatest element $a$, and $\{\nu_kd:d\in D\}$ is
dense in $[0,\nu_ka]$.   For $t\in\ocint{0,1}$, set
$a_t=\sup\{d:d\in D$, $\nu_kd\le t\nu_ka\}$;  this is where we need to know
that $\frak A$ is Dedekind $\sigma$-complete.   Set $a_0=0$.
Then $\langle a_t\rangle_{t\in[0,1]}$ is a non-decreasing family with
$a_0=0$ and $a_1=a$.   If $0<t<1$, $i\in I$ and $\epsilon>0$, there are
$d$, $d'\in D$ such that

\Centerline{$t\nu_ka-\epsilon\le\nu_k d\le t\nu_ka<\nu_kd'
\le t\nu_ka+\epsilon$,}

\Centerline{$t\nu_ia-\gamma_i\epsilon
\le\nu_i d\le t\nu_ia<\nu_id'
\le t\nu_ia+\gamma_i\epsilon$;}

\noindent in this case $d\Bsubseteq a_t\Bsubseteq d'$, so

\Centerline{$t\nu_ia-\gamma_i\epsilon\le\nu_ia_t
\le t\nu_ia+\gamma_i\epsilon$;}

\noindent as $\epsilon$ is arbitrary, $\nu_ia_t=t\nu_ia$.   Thus we have a
suitable family $\langle a_t\rangle_{t\ge 0}$.

\medskip

{\bf (c)} Induce on $n$.

\medskip

\quad{\bf (i)} The induction starts with a single non-negative
properly atomless functional $\nu_0$.    Now for any $a\in\frak A$ there is
an $a'\Bsubseteq a$ such that $\bover13\nu_0a\le\nu_0a'\le\bover23\nu_0a$.
\Prf\ This is trivial if $\nu_0a=0$.   Otherwise, let $C$ be a
finite partition of unity in $\frak A$ such that $\nu_0c\le\bover13\nu_0a$
for every $c\in C$.   Enumerate $C$ as $\langle c_i\rangle_{i<m}$ and for
$i\le m$ set $b_i=a\Bcap\sup_{j<i}c_j$.   Then $b_0=0$, $b_m=a$ and
$\nu_0b_{i+1}-\nu_0b_i\le\nu_0c_i\le\bover13\nu_0a$ for each $i$.
So there
must be an $i\le m$ such that $\bover13\nu_0a\le\nu_0b_i\le\bover23\nu_0a$,
and we can set $a'=b_i$.\ \Qed

Now (b), with $I=\{0\}$, gives the result.

\medskip

\quad{\bf (ii)} For the inductive step to $n\ge 1$, I show first that if
$a\in\frak A$ there is an $a'\Bsubseteq a$ such that
$\nu_ia'=\bover12\nu_ia$ for every $i\le n$.   \Prf\ By the inductive
hypothesis, we have a non-decreasing family $\family{t}{[0,1]}{a_t}$ such
that $a_0=0$, $a_1=a$ and $\nu_ia_t=t\nu_ia$ whenever $t\in[0,1]$ and
$i<n$.   Now observe that for $0\le s\le t\le 1$,

\Centerline{$|\nu_na_t-\nu_na_s|=\nu_n(a_t\Bsetminus a_s)
\le\nu_0(a_t\Bsetminus a_s)=(t-s)\nu_0a$.}

\noindent So the functions $t\mapsto\nu_na_t:[0,1]\to\coint{0,\infty}$ and
$f:[0,\bover12]\to\coint{0,\infty}$ are continuous, where
$f(t)=\nu_na_{t+\bover12}-\nu_na_t$ for $0\le t\le\bover12$.   However,
$f(0)+f(\bover12)=\nu_na$, so $\bover12\nu_na$ lies between $f(0)$ and
$f(\bover12)$ and there is a $t\in[0,\bover12]$ such that
$f(t)=\bover12\nu_na$.   Set $a'=a_{t+\bover12}\Bsetminus a_t$;  then
$\nu_ia'=\bover12\nu_ia$ for every $i\le n$, as required.\ \Qed

Once again (b), with $I=\{0,\ldots,n\}$, shows that for any $a\in\frak A$
we have a non-decreasing family $\family{t}{[0,1]}{a_t}$ such that
$a_0=0$, $a_1=1$ and $\nu_ia_t=t\nu_ia$ whenever $t\in[0,1]$ and $i\le n$.
}%end of proof of 326G

\leader{*326H}{Liapounoff's convexity
theorem}\cmmnt{ ({\smc Liapounoff 1940})}
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra, and $r\ge 1$ an integer.   Suppose that $\nu:\frak A\to\BbbR^r$
is additive in the sense that $\nu(a\Bcup b)=\nu a+\nu b$ whenever
$a\Bcap b=0$\cmmnt{ (see 361B)}, and
properly atomless in the sense that for every $\epsilon>0$ there is a
finite partition $\langle a_j\rangle_{j\in J}$ of unity in $\frak A$ such
that $\|\nu a\|\le\epsilon$ whenever $j\in J$ and $a\Bsubseteq a_j$.
Then $\{\nu a:a\in\frak A\}$ is a convex set in
$\BbbR^r$.\cmmnt{(\footnote{I
learnt this version of the theorem from K.P.S.Bhaskara Rao.})}

\proof{ For $1\le i\le r$, let $\nu_i$ be the $i$th component of $\nu$, so
that $\nu a=\langle\nu_ia\rangle_{1\le i\le r}$ for each $a\in\frak A$.
Then every $\nu_i$ is additive.   Moroever, it is properly atomless.
\Prf\ Given $\epsilon>0$, there is a finite partition $\family{j}{J}{a_j}$
of unity in $\frak A$ such that $|\nu_ia|\le\|\nu a\|\le\epsilon$ whenever
$j\in J$ and $a\Bsubseteq a_j$.\ \QeD\   So we can express $\nu_i$ as
$\nu_i^+-\nu_i^-1$ where $\nu_i^+$ and $\nu_i^{-1}$ are non-negative
properly atomless non-negative functionals (326G(a-ii)).
Set $\tilde\nu a=\sum_{i=1}^r\nu_i^+a+\nu_i^-a$ for $a\in\frak A$.   Then
$\tilde\nu$ is again properly atomless (326G(a-i)).

Suppose that $x$, $y\in\nu[\frak A]$ and $\alpha\in[0,1]$.   Let $a$,
$b\in\frak A$ be such that $\nu a=x$ and $\nu b=y$.   By 326Gc, applied to
$\tilde\nu,\nu_1^+,\nu_1^-,\ldots,\nu_r^+,\nu_r^-$, there is an
$c\Bsubseteq a\Bsetminus b$ such that

\Centerline{$\nu_i^+c=\alpha\nu_i^+(a\Bsetminus b)$,
\quad$\nu_i^-c=\alpha\nu_i^-(a\Bsetminus b)$,}

\noindent for every $i\le r$, so that $\nu_ic=\alpha\nu_i(a\Bsetminus b)$
for every $i\le r$.   Similarly, there is a $d\Bsubseteq b\Bsetminus a$
such that $\nu d=(1-\alpha)\nu(b\Bsetminus d)$.   Now
$e=c\Bcup(a\Bcap b)\Bcup d$,

$$\eqalign{\alpha x+(1-\alpha)y
&=\alpha\nu a+(1-\alpha)\nu b\cr
&=\alpha\nu(a\Bsetminus b)+\alpha\nu(a\Bcap b)+(1-\alpha)\nu(a\Bcap b)
   +(1-\alpha)\nu(b\Bsetminus a)\cr
&=\nu c+\nu(a\Bcap b)+\nu d
=\nu(c\Bcup(a\Bcap b)\Bcup d)
\in\nu[\frak A].\cr}$$

\noindent As $x$, $y$ and $\alpha$ are arbitrary, $\nu[\frak A]$ is convex.
}%end of proof of 326H

\leader{326I}{Countably additive functionals:  Definition}
Let $\frak A$ be a Boolean algebra.   A functional
$\nu:\frak A\to\Bbb R$ is {\bf countably additive} or
{\bf $\sigma$-additive} if $\sum_{n=0}^{\infty}\nu a_n$ is defined and
equal to $\nu(\sup_{n\in\Bbb N}a_n)$ whenever
$\sequencen{a_n}$ is a disjoint sequence in $\frak A$ and
$\sup_{n\in\Bbb N}a_n$ is defined in $\frak A$.

\cmmnt{A warning is perhaps in order.   It can happen that $\frak A$
is presented to us as a subalgebra of a larger algebra $\frak B$;  for
instance, $\frak A$ might be an algebra of sets, a subalgebra of some
$\sigma$-algebra $\Sigma\subseteq\Cal PX$.   In this case, there may be
sequences in $\frak A$ which
have a supremum in $\frak A$ which is not a supremum in $\frak B$
(indeed, this will happen just when the embedding is not sequentially
order-continuous).   So we can have a countably additive functional
$\nu:\frak B\to\Bbb R$ such that $\nu\restrp\frak A$ is not countably
additive in the sense used here.   A similar phenomenon will arise when
we come to the Daniell integral in Volume 4 (\S436).
}

\leader{326J}{Elementary facts} Let $\frak A$ be a Boolean algebra
and $\nu:\frak A\to\Bbb R$ a countably additive functional.

\header{326Ja}{\bf (a)} $\nu$ is finitely additive.   \prooflet{(Setting
$a_n=0$ for every $n$, we see from the definition in 326I that
$\nu 0=0$.   Now, given $a\Bcap b=0$, set $a_0=a$, $a_1=b$, $a_n=0$ for
$n\ge 2$ to see that $\nu(a\Bcup b)=\nu a+\nu b$.)}

\header{326Jb}{\bf (b)} If $\sequencen{a_n}$ is a non-decreasing
sequence in $\frak A$ with a supremum $a\in\frak A$, then

\Centerline{$\nu a\cmmnt{\mskip5mu =\nu
a_0+\sumop_{n=0}^{\infty}\nu(a_{n+1}\Bsetminus a_n)
}=\lim_{n\to\infty}\nu a_n$.}

\header{326Jc}{\bf (c)} If $\sequencen{a_n}$ is a non-increasing
sequence in $\frak A$ with an infimum $a\in\frak A$,
then\cmmnt{ $\sequencen{a_0\Bsetminus a_n}$ is a non-decreasing sequence
with supremum $a_0\Bsetminus a$, so}

\Centerline{$\nu a\cmmnt{\mskip5mu =\nu a_0-\nu(a_0\Bsetminus a)
=\nu a_0-\lim_{n\to\infty}\nu(a_0\Bsetminus a_n)
}=\lim_{n\to\infty}\nu a_n$.}

\header{326Jd}{\bf (d)} If $c\in\frak A$, then $a\mapsto\nu(a\Bcap c)$
is countably additive.   \prooflet{(For
$\sup_{n\in\Bbb N}a_n\Bcap c=c\Bcap\sup_{n\in\Bbb N}a_n$ whenever the
right-hand-side is defined, by 313Ba.)}

\header{326Je}{\bf (e)} $\alpha\nu$ is a countably additive functional
for any
$\alpha\in\Bbb R$.   If $\nuprime$ is another countably additive functional
on $\frak A$, then $\nu+\nuprime$ is countably additive.

\header{326Jf}{\bf (f)} If $\frak B$ is another Boolean algebra and
$\pi:\frak B\to\frak A$ is a sequentially order-continuous Boolean
homomorphism, then $\nu\pi$ is a countably additive functional on $\frak
B$.   \prooflet{(For if $\sequencen{b_n}$ is a disjoint sequence in
$\frak B$ with supremum $b$, then $\sequencen{\pi b_n}$ is a disjoint
sequence with supremum $\pi b$.)}

\header{326Jg}{\bf (g)} If $\frak A$ is Dedekind $\sigma$-complete and
$\frak B$ is a $\sigma$-subalgebra of $\frak A$, then
$\nu\restrp\frak B:\frak B\to\Bbb R$ is countably additive.
\prooflet{(For the identity map from $\frak B$ to $\frak A$ is
sequentially order-continuous, by 314Gb.)}


\leader{326K}{Corollary} Let $\frak A$ be a Boolean algebra and $\nu$ a
finitely additive real-valued functional on $\frak A$.

(a) $\nu$ is countably additive iff $\lim_{n\to\infty}\nu a_n=0$
whenever $\sequencen{a_n}$ is a non-increasing sequence in $\frak A$
with infimum $0$ in $\frak A$.

(b) If $\nuprime$ is an additive functional on $\frak A$ and
$|\nuprime a|\le\nu a$ for every $a\in\frak A$, and $\nu$ is countably
additive, then $\nuprime$ is countably additive.

(c) If $\nu$ is non-negative, then $\nu$ is countably additive iff it is
sequentially order-continuous.

\proof{{\bf (a)(i)} If $\nu$ is countably additive and $\sequencen{a_n}$
is a non-increasing sequence in $\frak A$ with infimum $0$, then
$\lim_{n\to\infty}\nu a_n\penalty-100=0$ by 326Jc.
{\bf (ii)} If $\nu$ satisfies
the condition, and $\sequencen{a_n}$ is a disjoint sequence in $\frak A$
with supremum $a$, set $b_n=a\Bsetminus\sup_{i\le n}a_i$ for each
$n\in\Bbb N$;  then $\sequencen{b_n}$ is non-increasing and has infimum
$0$, so

\Centerline{$\nu a-\sum_{i=0}^n\nu a_i
=\nu a-\nu(\sup_{i\le n}a_i)
=\nu b_n\to 0$}

\noindent as $n\to\infty$, and $\nu a=\sum_{n=0}^{\infty}\nu a_n$;  thus
$\nu$ is countably additive.

\medskip

{\bf (b)} If $\sequencen{a_n}$ is a disjoint sequence in $\frak A$ with
supremum $a$, set $b_n=\sup_{i\le n}a_i$ for each $n$;  then $\nu
a=\lim_{n\to\infty}\nu b_n$, so

\Centerline{$\lim_{n\to\infty}|\nuprime a-\nuprime b_n|
=\lim_{n\to\infty}|\nuprime(a\Bsetminus b_n)|
\le\lim_{n\to\infty}\nu(a\Bsetminus b_n)
=0$,}

\noindent and

\Centerline{$\sum_{n=0}^{\infty}\nuprime a_n=\lim_{n\to\infty}\nuprime b_n
=\nuprime a$.}

\medskip

{\bf (c)}  If $\nu$ is countably additive, then it is sequentially
order-continuous by 326Jb-326Jc.   If $\nu$ is sequentially
order-continuous, then of course it satisfies the condition of (a), so
is countably additive.
}%end of proof of 326K

\leader{326L}{The Jordan decomposition (II):  Proposition} Let $\frak A$
be a Boolean algebra and $\nu$ a bounded countably additive
real-valued functional on $\frak A$.   Then $\nu$ is expressible as the
difference of two non-negative countably additive functionals.

\proof{ Consider the functional $\nu^+a=\sup_{b\Bsubseteq a}\nu b$
defined in the proof of 326D.   If $\sequencen{a_n}$ is a disjoint
sequence in $\frak A$ with supremum $a$, and $b\Bsubseteq a$, then

\Centerline{$\nu b=\sum_{n=0}^{\infty}\nu(b\Bcap
a_n)\le\sum_{n=0}^{\infty}\nu^+a_n$.}

\noindent As $b$ is arbitrary, $\nu^+a\le\sum_{n=0}^{\infty}\nu^+a_n$.
But of course

\Centerline{$\nu^+a\ge\nu^+(\sup_{i\le n}a_i)=\sum_{i=0}^n\nu^+a_i$}

\noindent for every $n\in\Bbb N$, so
$\nu^+a=\sum_{n=0}^{\infty}\nu^+a_n$.   As $\sequencen{a_n}$ is
arbitrary, $\nu^+$ is countably additive.

Now $\nu^-=\nu^+-\nu$ also is countably additive, and $\nu=\nu^+-\nu^-$
is the difference of non-negative countably additive functionals.
}%end of proof of 326L

\leader{326M}{The Hahn decomposition:  Theorem} Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $\nu:\frak A\to\Bbb R$ a
countably additive functional.   Then $\nu$ is bounded and there is a
$c\in\frak A$ such that $\nu a\ge 0$ whenever $a\Bsubseteq c$, while
$\nu a\le 0$ whenever $a\Bcap c=0$.

\ifwithproofs
\medskip

\noindent{\bf first proof} By 314M, there are a set $X$ and a
$\sigma$-algebra $\Sigma$ of subsets of $X$ and a sequentially
order-continuous Boolean homomorphism $\pi$ from $\Sigma$ onto
$\frak A$.   Set $\nu_1=\nu\pi:\Sigma\to\Bbb R$.   Then $\nu_1$ is
countably additive (326Jf).   So $\nu_1$ is bounded and there is a set
$H\in\Sigma$
such that $\nu_1F\ge 0$ whenever $F\in\Sigma$ and $F\subseteq H$ and
$\nu_1F\le 0$ whenever $F\in\Sigma$ and $F\cap H=\emptyset$ (231Eb).
Set $c=\pi H\in\frak A$.   If $a\Bsubseteq c$, then there is an
$F\in\Sigma$ such that $\pi F=a$;  now $\pi(F\cap H)=a\Bcap c=a$, so
$\nu a=\nu_1(F\cap H)\ge 0$.   If $a\Bcap c=0$, then there is an
$F\in\Sigma$
such that $\pi F=a$;  now $\pi(F\setminus H)=a\Bsetminus c=a$, so
$\nu a=\nu_1(F\setminus H)\le 0$.

\medskip

\noindent{\bf second proof (a)} Note first that $\nu$ is bounded.
\Prf\ If $\sequencen{a_n}$ is a disjoint sequence in $\frak A$, then
$\sum_{n=0}^{\infty}\nu a_n$ must exist and be equal to
$\nu(\sup_{n\in\Bbb N}a_n)$;  in particular,
$\lim_{n\to\infty}\nu a_n=0$.   By 326D, $\nu$ is bounded.
\Qed

\medskip

{\bf (b)(i)} We know that $\gamma=\sup\{\nu a:a\in\frak A\}<\infty$.
Choose
a sequence $\sequencen{a_n}$ in $\frak A$ such that
$\nu a_n\ge\gamma-2^{-n}$ for every $n\in\Bbb N$.   For $m\le n\in\Bbb
N$, set $b_{mn}=\inf_{m\le i\le n}a_i$.   Then
$\nu b_{mn}\ge\gamma-2\cdot 2^{-m}+2^{-n}$ for every $n\ge m$.   \Prf\
Induce
on $n$.   For $n=m$, this is due to the choice of $a_m=b_{mm}$.   For
the inductive step, we have $b_{m,n+1}=b_{mn}\Bcap a_{n+1}$, while
surely
$\gamma\ge\nu(a_{n+1}\Bcup b_{mn})$, so

$$\eqalignno{\gamma+\nu b_{m,n+1}
&\ge\nu(a_{n+1}\Bcup b_{mn})+\nu(a_{n+1}\Bcap b_{mn})\cr
&=\nu a_{n+1}+\nu b_{mn}
\ge\gamma-2^{-n-1}+\gamma-2\cdot 2^{-m}+2^{-n}\cr
\noalign{\noindent (by the choice of $a_{n+1}$ and the inductive
hypothesis)}
&=2\gamma-2\cdot 2^{-m}+2^{-n-1}.\cr}$$

\noindent Subtracting $\gamma$ from both sides, $\nu b_{m,n+1}\ge\gamma
-2\cdot 2^{-m}+2^{-n-1}$ and the induction proceeds.\ \Qed

\medskip

\quad{\bf (ii)} Set

\Centerline{$b_m=\inf_{n\ge m}b_{mn}=\inf_{n\ge m}a_n$.}

\noindent Then

\Centerline{$\nu b_m=\lim_{n\to\infty}\nu b_{mn}
\ge\gamma-2\cdot 2^{-m}$,}

\noindent by 326Jc.
Next, $\sequencen{b_n}$ is non-decreasing, so setting $c=\sup_{n\in\Bbb
N}b_n$ we have

\Centerline{$\nu c=\lim_{n\to\infty}\nu b_n\ge\gamma$;}

\noindent  since $\nu c$
is surely less than or equal to $\gamma$, $\nu c=\gamma$.

If $b\in\frak A$ and $b\Bsubseteq c$, then

\Centerline{$\nu c-\nu b=\nu(c\Bsetminus
b)\le\gamma=\nu c$,}

\noindent so $\nu b\ge 0$.   If $b\in\frak A$ and $b\Bcap
c=0$ then

\Centerline{$\nu c+\nu b=\nu(c\Bcup b)\le\gamma=\nu c$}

\noindent so $\nu b\le
0$.   This completes the proof.
\else\fi %end of proof of 326M

\leader{326N}{Completely additive functionals:  Definition} Let $\frak
A$ be a Boolean algebra.   A functional $\nu:\frak A\to\Bbb R$ is {\bf
completely additive} or {\bf $\tau$-additive} if it is finitely additive
and $\inf_{a\in A}|\nu a|=0$ whenever $A$ is a non-empty
downwards-directed set in $\frak A$ with infimum $0$.

\leader{326O}{Basic facts} Let $\frak A$ be a Boolean algebra and $\nu$
a completely additive real-valued functional on $\frak A$.

\header{326Oa}{\bf (a)} $\nu$ is countably additive.   \prooflet{\Prf\
If $\sequencen{a_n}$ is a non-increasing sequence in $\frak A$ with
infimum $0$, then for any infinite $I\subseteq\Bbb N$ the set
$\{a_i:i\in I\}$ is downwards-directed and has infimum $0$, so
$\inf_{i\in I}|\nu a_i|=0$;  which means that $\lim_{n\to\infty}\nu a_n$
must be zero.   By 326Ka, $\nu$ is countably additive.\ \Qed}

\header{326Ob}{\bf (b)} Let $A$ be a non-empty downwards-directed set in
$\frak A$ with infimum $0$.   Then for every $\epsilon>0$ there is an
$a\in A$ such that $|\nu b|\le\epsilon$ whenever $b\Bsubseteq a$.
\prooflet{\Prf\Quer\ Suppose, if possible, otherwise.   Set

\Centerline{$B=\{b:|\nu b|\ge\epsilon,\,\exists\,a\in A,\,b\Bsupseteq
a\}$.}

\noindent If $a\in A$ there is a $b'\Bsubseteq a$ such that $|\nu
b'|>\epsilon$.   Now $\{a'\Bsetminus b':a'\in A,\,a'\Bsubseteq a\}$ is
downwards-directed and has infimum $0$, so there is an $a'\in A$ such
that $a'\Bsubseteq a$ and $|\nu(a'\Bsetminus b')|\le|\nu b'|-\epsilon$.
Set $b=b'\Bcup a'$;  then $a'\Bsubseteq b$ and

\Centerline{$|\nu b|=|\nu b'+\nu(a'\Bsetminus b')|
\ge|\nu b'|-|\nu(a'\Bsetminus b')|\ge\epsilon$,}

\noindent so $b\in B$.   But also $b\Bsubseteq a$.   Thus every member
of $A$ includes some member of $B$.   Since every member of $B$ includes
a member of $A$, $B$ is downwards-directed and has infimum $0$;  but
this is impossible, since $\inf_{b\in B}|\nu b|\ge\epsilon$.\ \Bang\Qed}

\header{326Oc}{\bf (c)} If $\nu$ is non-negative, it is
order-continuous.   \prooflet{\Prf\ (i) If $A$ is a non-empty
upwards-directed set with supremum $a_0$, then $\{a_0\Bsetminus a:a\in
A\}$ is a non-empty downwards-directed set with infimum $0$, so

\Centerline{$\sup_{a\in A}\nu a
=\nu a_0-\inf_{a\in A}\nu(a_0\Bsetminus a)
=\nu a_0$.}

\noindent (ii) If $A$ is a non-empty
downwards-directed set with infimum $a_0$, then $\{a\Bsetminus a_0:a\in
A\}$ is a non-empty downwards-directed set with infimum $0$, so

\Centerline{$\inf_{a\in A}\nu a
=\nu a_0+\inf_{a\in A}\nu(a\Bsetminus a_0)
=\nu a_0$. \Qed}
}

\header{326Od}{\bf (d)} If $c\in\frak A$, then $a\mapsto\nu(a\Bcap c)$
is completely additive.   \prooflet{\Prf\ If $A$ is a non-empty
downwards-directed set with infimum $0$, so is $\{a\Bcap c:a\in A\}$, and
$\inf_{a\in A}|\nu(a\Bcap c)|=0$.\ \Qed}

\header{326Oe}{\bf (e)} $\alpha\nu$ is a completely additive functional
for any
$\alpha\in\Bbb R$.   If $\nuprime$ is another completely additive functional
on $\frak A$, then $\nu+\nuprime$ is completely additive.   \prooflet{\Prf\
We know from 326Bc that $\nu+\nuprime$ is additive.   Let $A$ be a
non-empty
downwards-directed set with infimum $0$.   For any $\epsilon>0$, (b)
tells us that there are $a$, $a'\in A$ such that $|\nu b|\le\epsilon$
whenever $b\Bsubseteq a$ and $|\nuprime b|\le\epsilon$ whenever $b\Bsubseteq
a'$.   But now, because $A$ is downwards-directed, there is a $b\in A$
such that $b\Bsubseteq a\Bcap a'$, which means that $|\nu
b+\nuprime b|\le|\nu b|+|\nuprime b|$ is at most $2\epsilon$.   As $\epsilon$ is
arbitrary, $\inf_{a\in A}|(\nu+\nuprime)(a)|=0$, and $\nu+\nuprime$ is
completely additive.\ \Qed}

\header{326Of}{\bf (f)} If $\frak B$ is another Boolean algebra and
$\pi:\frak B\to\frak A$ is an order-continuous Boolean homomorphism,
then $\nu\pi$ is a completely additive functional on $\frak B$.
\prooflet{\Prf\ By
326Be, $\nu\pi$ is additive.   If $B\subseteq\frak B$ is a non-empty
downwards-directed set with infimum $0$ in $\frak B$, then $\pi[B]$ is a
non-empty downwards-directed set with infimum $0$ in $\frak A$, because
$\pi$ is order-continuous, so $\inf_{b\in B}|\nu\pi b|=0$.\ \Qed}
In particular, if $\frak B$ is a regularly embedded subalgebra of
$\frak A$, then $\nu\restrp\frak B$ is completely additive.

\header{326Og}{\bf (g)} If $\nuprime$ is another additive functional on
$\frak A$ and $|\nuprime a|\le\nu a$ for every $a\in\frak A$, then $\nuprime$ is
completely additive.   \prooflet{\Prf\ If $A\subseteq\frak A$ is
non-empty and downwards-directed and $\inf A=0$, then
$\inf_{a\in A}|\nuprime a|\le\inf_{a\in A}\nu a=0$.\ \Qed}

\leader{326P}{}\cmmnt{ I squeeze a useful fact in here.

\medskip

\noindent}{\bf Proposition} If $\frak A$ is a ccc Boolean algebra, a
functional $\nu:\frak A\to\Bbb R$ is countably additive iff it is
completely additive.

\proof{ If $\nu$ is completely additive it is countably additive, by
326Oa.   If $\nu$ is countably additive and $A$ is a
non-empty downwards-directed set in $\frak A$ with infimum $0$, then
there is a (non-empty) countable subset $B$ of $A$ also with infimum $0$
(316E).   Let $\sequencen{b_n}$ be a sequence running over $B$, and
choose $\sequencen{a_n}$ in $A$ such that $a_0=b_0$, $a_{n+1}\Bsubseteq
a_n\Bcap b_n$ for every $n\in\Bbb N$.   Then $\sequencen{a_n}$ is a
non-increasing sequence with infimum $0$, so
$\lim_{n\to\infty}\nu a_n=0$
(326Jc) and $\inf_{a\in A}|\nu a|=0$.   As $A$ is arbitrary, $\nu$ is
completely additive.
}%end of proof of 326P

\leader{326Q}{The Jordan decomposition (III):  Proposition} Let
$\frak A$ be a Boolean algebra and $\nu$ a completely additive
real-valued functional on $\frak A$.   Then $\nu$ is bounded and
expressible as the difference
of two non-negative completely additive functionals.

\proof{{\bf (a)} I must first check that $\nu$ is bounded.   \Prf\ Let
$\sequencen{a_n}$ be a disjoint sequence in $\frak A$.   Set

\Centerline{$A=\{a:a\in\frak A,$ there is an $n\in\Bbb N$ such that
$a_i\Bsubseteq a$ for every $i\ge n\}$.}

\noindent Then $A$ is closed under $\Bcap$, and if $b$ is any lower
bound for $A$ then $b\Bsubseteq 1\Bsetminus a_n\in A$, so
$b\Bcap a_n=0$,
for every $n\in\Bbb N$;  but this means that $1\Bsetminus b\in A$, so
that $b\Bsubseteq 1\Bsetminus b$ and $b=0$.   Thus $\inf A=0$.   By
326Ob, there is an $a\in A$ such that $|\nu b|\le 1$ whenever
$b\Bsubseteq a$.   By the
definition of $A$, there must be an $n\in\Bbb N$ such that
$|\nu a_i|\le 1$ for every $i\ge n$.   But this means that
$\sup_{n\in\Bbb N}|\nu a_n|$ is finite.   As $\sequencen{a_n}$ is
arbitrary, $\nu$ is bounded, by 326D(ii).\ \Qed

\medskip

{\bf (b)} As in 326D and 326L, set $\nu^+a=\sup_{b\Bsubseteq a}\nu b$
for every $a\in\frak A$.   Then $\nu^+$ is completely additive.   \Prf\
We know that $\nu^+$ is additive.   If $A$ is a non-empty
downwards-directed subset of $\frak A$ with infimum $0$, then for every
$\epsilon>0$ there is an $a\in A$ such that $|\nu b|\le\epsilon$
whenever $b\Bsubseteq a$;  in particular, $\nu^+a\le\epsilon$.   As
$\epsilon$ is arbitrary, $\inf_{a\in A}\nu^+a=0$;  as $A$ is arbitrary,
$\nu^+$ is completely additive.\ \Qed

Consequently $\nu^-=\nu^+-\nu$ is completely additive (326Oe) and
$\nu=\nu^+-\nu^-$ is the difference of non-negative completely additive
functionals.
}%end of proof of 326Q

\leader{326R}{}\cmmnt{ I give an alternative definition of
`completely additive' which you may feel clarifies the concept.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra, and
$\nu:\frak A\to\Bbb R$ a function.   Then the following are
equiveridical:

(i) $\nu$ is completely additive;

(ii) $\nu 1=\sum_{i\in I}\nu a_i$ whenever $\langle a_i\rangle_{i\in I}$
is a partition of unity in $\frak A$;

(iii) $\nu a=\sum_{i\in I}\nu a_i$ whenever
$\langle a_i\rangle_{i\in I}$ is a disjoint family in $\frak A$ with
supremum $a$.

\proof{ (For notes on sums $\sum_{i\in I}$, see 226A.)

\medskip

{\bf (a)(i)$\Rightarrow$(ii)} If $\nu$ is completely additive and
$\langle a_i\rangle_{i\in I}$ is a partition of unity in $A$, then
(inducing on
$\#(J)$) $\nu(\sup_{i\in J}a_i)=\sum_{i\in J}\nu a_i$ for every finite
$J\subseteq I$.   Consider

\Centerline{$A=\{1\Bsetminus\sup_{i\in J}a_i:J\subseteq I$ is
finite$\}$.}

\noindent Then $A$ is non-empty and downwards-directed and has infimum
$0$, so for every
$\epsilon>0$ there is an $a\in A$ such that $|\nu b|\le\epsilon$
whenever $b\Bsubseteq a$ (326Ob again).
Express $a$ as $1\Bsetminus\sup_{i\in J}a_i$
where $J\subseteq I$ is finite.   If now $K$ is another finite subset of
$I$ including $J$,

\Centerline{$|\nu 1-\sum_{i\in K}a_i|
=|\nu(1\Bsetminus\sup_{i\in K}a_i)|\le\epsilon$.}

\noindent As remarked in 226Ad, this means that
$\nu 1=\sum_{i\in I}\nu a_i$, as claimed.

\medskip

{\bf (b)(ii)$\Rightarrow$(iii)} Suppose that $\nu$ satisfies the
condition (ii), and that $\langle a_i\rangle_{i\in I}$ is a disjoint
family with supremum $a$.   Take any $j\notin I$, set $J=I\cup\{j\}$ and
$a_j=1\Bsetminus a$;  then $\langle a_i\rangle_{i\in J}$,
$(a,1\Bsetminus a)$ are both partitions of unity, so

\Centerline{$\nu(1\Bsetminus a)+\nu a=\nu 1=\sum_{i\in J}\nu a_i
=\nu(1\Bsetminus a)+\sum_{i\in I}\nu a_i$,}

\noindent and $\nu a=\sum_{i\in I}\nu a_i$.

\medskip

{\bf (c)(iii)$\Rightarrow$(i)} Suppose that $\nu$ satisfies (iii).
Then $\nu$ is additive.

\medskip

\quad\grheada\ $\nu$ is bounded.   \Prf\ Let $\sequencen{a_n}$ be a
disjoint sequence in $\frak A$.   Applying Zorn's Lemma to the set $\Cal
C$ of all disjoint families $C\subseteq\frak A$ including $\{a_n:n\in
\Bbb N\}$, we find a partition of unity $C\supseteq\{a_n:n\in\Bbb N\}$.
Now $\sum_{c\in C}\nu c$ is defined in $\Bbb R$, so $\sup_{n\in\Bbb
N}|\nu a_n|\le\sup_{c\in C}|\nu c|$ is finite.   By 326D, $\nu$ is
bounded.\ \Qed

\medskip

\quad\grheadb\ Define $\nu^+$ from $\nu$ as in 326D.  Then $\nu^+$
satisfies the same condition as $\nu$.   \Prf\ Let $\langle
a_i\rangle_{i\in I}$ be a disjoint family in $\frak A$ with supremum
$a$.   Then
for any $b\subseteq a$, we have $b=\sup_{i\in I}b\Bcap a_i$, so

\Centerline{$\nu b=\sum_{i\in I}\nu(b\Bcap a_i)
\le\sum_{i\in I}\nu^+a_i$.}

\noindent Thus $\nu^+a\le\sum_{i\in I}\nu^+a_i$.   But of course

$$\eqalign{\sum_{i\in I}\nu^+a_i
&=\sup\{\sum_{i\in J}\nu^+a_i:J\subseteq I\text{ is finite}\}\cr
&=\sup\{\nu^+(\sup_{i\in J}a_i):J\subseteq I\text{ is finite}\}
\le\nu^+a,\cr}$$

\noindent so $\nu^+a=\sum_{i\in I}\nu^+a_i$.\ \Qed

\medskip

\quad\grheadc\ It follows that $\nu^+$ is completely additive.   \Prf\
If $A$ is a non-empty downwards-directed set with infimum $0$, then
$B=\{b:\,\Exists a\in A,\,b\Bcap a=0\}$ is order-dense in $\frak A$, so
there is a
partition of unity $\langle b_i\rangle_{i\in I}$ lying in $B$ (313K).
Now if $J\subseteq I$ is finite, there is an $a\in A$ such that
$a\Bcap\sup_{i\in J}b_i=0$ (because $A$ is downwards-directed), and

\Centerline{$\nu^+a+\sum_{i\in J}\nu^+b_i\le\nu^+1$.}

\noindent Since $\nu^+1=\sup_{J\subseteq I\text{ is finite}}\sum_{i\in
J}\nu^+b_i$, $\inf_{a\in A}\nu^+a=0$.   As $A$ is arbitrary, $\nu^+$ is
completely additive.\ \Qed

\medskip

\quad\grheadd\ Now consider $\nu^-=\nu^+-\nu$.   Of course

\Centerline{$\nu^-a
=\nu^+a-\nu a
=\sum_{i\in I}\nu^+a_i-\sum_{i\in I}\nu a_i
=\sum_{i\in I}\nu^-a_i$}

\noindent whenever $\langle a_i\rangle_{i\in I}$ is a disjoint family in
$\frak A$ with supremum $a$.   Because $\nu^-$ is non-negative, the
argument of ($\gamma$)
shows that $\nu^-=(\nu^-)^+$ is completely additive.   So
$\nu=\nu^+-\nu^-$ is completely additive, as required.
}%end of proof of 326R

\leader{326S}{}\cmmnt{ For completely additive functionals, we have a
useful refinement of the Hahn decomposition.   I give it in a form
adapted to the applications I have in mind.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\nu:\frak A\to\Bbb R$ a
completely additive functional.   Then there is a unique element of
$\frak A$, which I will denote $\Bvalue{\nu>0}$,\cmmnt{ `the region
where $\nu>0$',} such that $\nu a>0$ whenever
$0\ne a\Bsubseteq\Bvalue{\nu>0}$, while $\nu a\le 0$ whenever
$a\Bcap\Bvalue{\nu>0}=0$.

\proof{ Set

\Centerline{$C_1=\{c:c\in\frak A\setminus\{0\},\,\nu a>0$ whenever
$0\ne a\Bsubseteq c\}$,}

\Centerline{$C_2=\{c:c\in\frak A,\,\nu a\le 0$ whenever $a\Bsubseteq c\}$.}

\noindent Then $C_1\cup C_2$ is order-dense in $\frak A$.   \Prf\ There
is a $c_0\in\frak A$ such that $\nu a\ge 0$ for every $a\Bsubseteq c_0$ and
$\nu a\le 0$ whenever $a\Bcap c_0=0$ (326M).   Given
$b\in\frak A\setminus\{0\}$, then $b\Bsetminus c_0\in C_2$, so if $b\Bsetminus c_0\ne 0$ we can stop.   Otherwise, $b\Bsubseteq c_0$.   If $b\in C_1$
we can stop.   Otherwise, there is a non-zero $c\Bsubseteq b$ such that
$\nu c\le 0$;  but in this case $\nu a\ge 0$ and $\nu(c\Bsetminus a)\ge 0$
so $\nu a=0$ for every $a\Bsubseteq c$, and $c\in C_2$.\ \Qed

There is therefore a partition of unity $D\subseteq C_1\cup C_2$.   Now
$D\cap C_1$ is countable.   \Prf\ If $d\in D\cap C_1$, $\nu d>0$.   Also

\Centerline{$\#(\{d:d\in D,\,\nu d\ge 2^{-n}\})\le 2^n\sup_{a\in\frak
A}\nu a$}

\noindent is finite for each $n$, so $D\cap C_1$ is the union of a
sequence of finite sets, and is countable.\ \Qed

Accordingly $D\cap C_1$ has a supremum $e$.   If $0\ne a\Bsubseteq e$
then

\Centerline{$\nu a=\sum_{c\in D}\nu(a\Bcap c)=\sum_{c\in D\cap
C_1}\nu(a\Bcap c)\ge 0$}

\noindent by 326R.   Also there must be some $c\in D\cap C_1$ such that
$a\Bcap c\ne 0$, in which case $\nu(a\Bcap c)>0$, so that $\nu a>0$.
If $a\Bcap e=0$, then

\Centerline{$\nu a=\sum_{c\in D}\nu(a\Bcap c)=\sum_{c\in D\cap
C_2}\nu(a\Bcap c)\le 0$.}

Thus $e$ has the properties demanded of $\Bvalue{\nu>0}$.   To see that
$e$ is unique, we need observe only that if $e'$ has the same properties
then $\nu(e\Bsetminus e')\le 0$ (because $(e\Bsetminus e')\Bcap e'=0$),
so $e\Bsetminus e'=0$ (because $e\Bsetminus e'\Bsubseteq e$).
Similarly, $e'\Bsetminus e=0$ and $e=e'$.   Thus we may properly denote
$e$ by the formula $\Bvalue{\nu>0}$.
}%end of proof of 326S

\leader{326T}{Corollary} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $\mu$, $\nu$ two completely additive functionals on
$\frak A$.   Then there is a unique element of $\frak A$, which I will
denote $\Bvalue{\mu>\nu}$,\cmmnt{ `the region where $\mu>\nu$',} such
that

\Centerline{$\mu a>\nu a$ whenever $0\ne a\Bsubseteq\Bvalue{\mu>\nu}$,}

\Centerline{$\mu a\le\nu a$ whenever $a\Bcap\Bvalue{\mu>\nu}=0$.}

\proof{ Apply 326S to the functional $\mu-\nu$, and set
$\Bvalue{\mu>\nu}=\Bvalue{\mu-\nu>0}$.
}%end of proof of 326T

\exercises{
\leader{326X}{Basic exercises (a)}
%\spheader 326Xa
Let $\frak A$ be a Boolean algebra
and $\nu:\frak A\to\Bbb R$ a finitely additive functional.  Show that
(i) $\nu(a\Bcup b)=\nu a+\nu b-\nu(a\Bcap b)$ (ii)
$\nu(a\Bcup b\Bcup c)=\nu a+\nu
b+\nu c-\nu(a\Bcap b)-\nu(a\Bcap c)-\nu(b\Bcap c)+\nu(a\Bcap b\Bcap c)$
for all $a$, $b$, $c\in\frak A$.
Generalize
these results to longer sequences in $\frak A$.
%326B

\spheader 326Xb Let $\frak A$ be a Boolean algebra.
(i) Show that a finitely additive functional $\nu$ is properly atomless iff
there is a properly atomless additive functional $\nu'$ such
that $|\nu a|\le \nu'a$ for every $a\in\frak A$.   (ii) Show that
a non-negative finitely additive functional $\nu$ on $\frak A$
is properly atomless iff whenever $\nuprime$ is a
non-zero finitely additive functional such that
$0\le\nuprime a\le\nu a$ for every $a\in\frak A$ there
is an $a\in\frak A$ such that $\nuprime a$ and $\nuprime(1\Bsetminus a)$
are both non-zero.
%326F

\spheader 326Xc(i) Suppose that $\frak A$ is a Dedekind $\sigma$-complete
Boolean algebra
and $\nu:\frak A\to\Bbb R$ is countably additive.
Show that $\Cal I=\{a:\nu b=0$ for every $b\Bsubseteq a\}$ is an ideal of
$\frak A$.   Show that the following
are equiveridical:  ($\alpha$) $\nu$ is
properly atomless;  ($\beta$) whenever $\nu a\ne 0$ there is a
$b\Bsubseteq a$ such that $\nu b\notin\{0,\nu a\}$;  ($\gamma$) the
quotient algebra $\frak A/\Cal I$ is atomless.
(ii) Find an atomless Dedekind complete Boolean algebra $\frak A$ and a
finitely additive
$\nu:\frak A\to[0,1]$ such that $\nu a>0$ for every non-zero $a\in\frak A$
but $\nu$ is not properly atomless.
%326J

\spheader 326Xd Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to \Bbb R$ a finitely additive functional.
Show that the following are equiveridical:
(i) $\nu$ is countably additive;
(ii) $\lim_{n\to\infty}\nu a_n=\nu a$ whenever $\sequencen{a_n}$
is a non-decreasing sequence in $\frak A$ with supremum $a$.
%326K

\spheader 326Xe Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\nu:\frak A\to \Bbb R$ a
finitely additive functional.
Show that the following are equiveridical:
(i) $\nu$ is countably additive;
(ii) $\lim_{n\to\infty}\nu a_n=0$ whenever $\sequencen{a_n}$ is a
sequence in $\frak A$ and $\inf_{n\in\Bbb N}\sup_{m\ge
n}a_m=0$;
(iii) $\lim_{n\to\infty}\nu a_n=\nu a$ whenever $\sequencen{a_n}$
is a sequence in $\frak A$ and
$a=\inf_{n\in\Bbb N}\sup_{m\ge n}a_m=\sup_{n\in\Bbb N}\inf_{m\ge n}a_m$.
\noindent\Hint{for (i)$\Rightarrow$(iii), consider
non-negative $\nu$ first.}
%326K

\spheader 326Xf Let $X$ be an uncountable set, and $J$ an
infinite subset of $X$.   Let $\frak A$ be the
finite-cofinite algebra of $X$ (316Yl), and for $a\in A$ set
$\nu a=\#(a\cap J)$ if $a$ is finite,
$-\#(J\setminus a)$ if $a$ is cofinite.
Show that $\nu$ is countably additive and unbounded.
%326M

\sqheader 326Xg Let $\frak A$ be the algebra of subsets of $[0,1]$
generated by the family of (closed) intervals.   Show that there is a
unique additive functional $\nu:\frak A\to\Bbb R$ such that
$\nu[\alpha,\beta]=\beta-\alpha$ whenever $0\le \alpha\le\beta\le 1$.
Show that $\nu$ is  countably additive but not completely additive.
%326N

\spheader 326Xh(i) Let $(X,\Sigma,\mu)$ be any atomless probability
space.   Show that $\mu:\Sigma\to\Bbb R$ is a countably additive
functional which is not completely additive.   (ii) Let $X$ be any
uncountable set and $\mu$ the countable-cocountable measure on $X$
(211R).   Show that $\mu$ is countably additive but not completely
additive.
%326N

\spheader 326Xi Let $\frak A$ be an atomless Boolean algebra.   Show that
every completely additive functional on $\frak A$ is properly atomless.
%326O

\spheader 326Xj Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to\Bbb R$ a function.   (i) Show that $\nu$ is finitely
additive iff $\sum_{i\in I}\nu a_i=\nu 1$ for every finite partition of
unity $\langle a_i\rangle_{i\in I}$.   (ii) Show that $\nu$ is countably
additive iff $\sum_{i\in I}\nu a_i=\nu 1$ for every countable partition
of unity $\langle a_i\rangle_{i\in I}$.
%326R

\spheader 326Xk Show that 326S can fail if $\nu$ is only countably
additive, rather than completely additive.   \Hint{326Xh.}
%326S

\spheader 326Xl Let $\frak A$ be a Boolean algebra and $\nu$ a finitely
additive real-valued functional on $\frak A$.   Let us say that
$a\in\frak A$ is
a {\bf support} of $\nu$ if ($\alpha$) $\nu b=0$ whenever $b\Bcap a=0$
($\beta$) for every non-zero $b\Bsubseteq a$ there is a $c\Bsubseteq b$
such that $\nu c\ne 0$.   (i) Check that $\nu$ can have at most one
support.   (ii) Show that if $a$ is a support for $\nu$ and $\nu$ is
bounded, then the principal ideal $\frak A_a$ generated by $a$ is ccc.
(iii) Show that if $\frak A$ is Dedekind $\sigma$-complete and $\nu$ is
countably additive, then $\nu$ is completely additive iff it has a
support, and that in the language of 326S this is
$\Bvalue{\nu>0}\Bcup\Bvalue{-\nu>0}$.   (iv) Taking $J=X$ in 326Xf, show
that $X$ is the support of the functional $\nu$ there.
%326S
}

\exercises{
\leader{326Y}{Further exercises (a)}
%\spheader 326Ya
Show that there is a finitely additive functional
$\nu:\Cal P\Bbb N\to\Bbb R$ such that $\nu\{n\}=1$ for every
$n\in\Bbb N$, so that $\nu$ is not bounded.
({\it Hint\/}:  Use Zorn's Lemma to
construct a maximal linearly independent subset of $\ell^{\infty}$
including $\{\chi\{n\}:n\in\Bbb N\}$, and hence to construct a linear
map $f:\ell^{\infty}\to\Bbb R$ such that $f(\chi\{n\})=1$ for every
$n$.)
%326D Hamel basis would be useful?

\spheader 326Yb Let $\frak A$ be any infinite Boolean algebra.
Show that there is an unbounded finitely additive functional
$\nu:\frak A\to\Bbb R$.   ({\it Hint\/}: let $\sequencen{t_n}$ be a
sequence of distinct points in the Stone space of $\frak A$, and set
$\nu a=\nuprime\{n:t_n\in\widehat a\}$ for a suitable $\nuprime$.)
%326D, 326Ya

\spheader 326Yc Let $\frak A$ be a Boolean algebra, and give
$\Bbb R^{\frak A}$ its product topology.   Show that the space of
finitely additive functionals on $\frak A$ is a closed subset of
$\BbbR^{\frak A}$, but that the space of bounded finitely additive
functionals is closed only when $\frak A$ is finite.
%326D

\spheader 326Yd Let $\frak A$ be a Boolean algebra, and $M$ the
linear space of all bounded finitely additive real-valued functionals on
$\frak A$.   For $\nu$, $\nuprime\in M$ say that $\nu\le\nuprime$ if
$\nu a\le\nuprime a$ for every $a\in\frak A$.   Show that

\quad (i) $\nu^+$, as defined in the proof of 326D, is just
$\sup\{0,\nu\}$ in $M$;

\quad(ii) $M$ is a Dedekind complete Riesz space (241E-241F, 353H);

\quad(iii) for $\nu$, $\nuprime\in M$, $|\nu|=\nu\vee(-\nu)$,
$\nu\vee\nuprime$ and
$\nu\wedge\nuprime$ are given by the formulae

\Centerline{$|\nu|(a)=\sup_{b\Bsubseteq a}\nu b-\nu(a\Bsetminus b)$,
\quad$(\nu\vee\nuprime)(a)=\sup_{b\Bsubseteq a}\nu b+\nuprime(a\Bsetminus b)$,}

\Centerline{$(\nu\wedge\nuprime)(a)
=\inf_{b\Bsubseteq a}\nu b+\nuprime(a\Bsetminus b)$;}

\quad(iv) for any non-empty $A\subseteq M$, $A$ is bounded above in $M$
iff

\Centerline{$\sup\{\sum_{i=0}^n\nu_ia_i:\nu_i\in A$ for each $i\le n$,
$\langle a_i\rangle_{i\le n}$ is disjoint$\}$}

\noindent is finite, and then $\sup A$ is defined by the formula

\Centerline{$(\sup A)(a)=\sup\{\sum_{i=0}^n\nu_ia_i:\nu_i\in A$ for each
$i\le n$, $\langle a_i\rangle_{i\le n}$ is disjoint,
$\sup_{i\le n}a_i=a\}$}

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

\quad(v) setting $\|\nu\|=|\nu|(1)$, $\|\,\|$ is an order-continuous
norm (definition:  354Dc) on $M$ under which $M$ is a Banach lattice.
%326D

\spheader 326Ye Let $\frak A$ be a Boolean algebra.   A functional
$\nu:\frak A\to\Bbb C$ is {\bf finitely additive} if its real and
imaginary parts are.
Show that the space of bounded finitely additive functionals from
$\frak A$ to $\Bbb C$ is a Banach space under the total variation norm
$\|\nu\|=\sup\{\sum_{i=0}^n|\nu a_i|:\langle a_i\rangle_{i\le n}$
is a partition of unity in $\frak A\}$.
%326D, 326Yd

\spheader 326Yf Let $\frak A$ and $\frak B$ be Boolean algebras and
$\mu$, $\nu$ finitely additive functionals on $\frak A$, $\frak B$
respectively.   Show that there is a unique finitely
additive functional $\lambda$ on the free product
$\frak A\otimes\frak B\to\Bbb R$ such that
$\lambda(a\otimes b)=\mu a\cdot\nu b$ for all $a\in\frak A$,
$b\in\frak B$.
%326E

\spheader 326Yg Let $\langle\frak A_i\rangle_{i\in I}$ be a family of
Boolean algebras, with free product
$(\bigotimes_{i\in I}\frak A_i,\langle\varepsilon_i\rangle_{i\in I})$,
and for each $i\in I$ let
$\nu_i$ be a finitely additive functional on $\frak A_i$ such that
$\nu_i1=1$.   Show that there is a unique finitely additive functional
$\nu:\bigotimes_{i\in I}\frak A_i\to\Bbb R$ such that
$\nu(\inf_{i\in J}\varepsilon_i(a_i))=\prod_{i\in J}\nu_ia_i$ whenever
$J\subseteq I$ is
non-empty and finite and $a_i\in\frak A_i$ for each $i\in J$.
%326E

\spheader 326Yh Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\nu:\frak A\to\coint{0,\infty}$ a countably additive
functional.   Show that $\nu$ is
properly atomless iff whenever $a\in\frak A$ and
$\nu a\ne 0$ there is a $b\Bsubseteq a$ such that $0<\nu b<\nu a$.
%326F out of order query

\spheader 326Yi Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\nu:\frak A\to\Bbb R$ a countably additive functional.   Show
that $\nu[\frak A]$ is a compact subset of $\Bbb R$.
%taking  a_n  such that  \nu a_n\to\alpha , subsequence deciding each atom,
%reduce to atomless case in which we have a convex set and need only look
%at the endpts
%326H 326M

\spheader 326Yj Let $\frak G$ be the regular open algebra of $\Bbb R$
(314P).   Find a properly atomless finitely additive
$\nu:\frak G\to\Bbb R$ such that $\nu[\frak G]$ is not compact.
%take \nu^+ , \nu^-  corresponding to mutually singular prob measures with
% support \Bbb R , so that  \nu  cannot take value  1
%326Yi 326M

\spheader 326Yk ({\smc Halmos 1948})
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $r\ge 1$ an integer.   (i) Let $C\subseteq\BbbR^r$ be a
non-empty bounded convex set, and for $z\in\BbbR^r$ set
$H_z=\{x:\varinnerprod{x}{z}=\sup_{y\in C}\varinnerprod{y}{z}\}$.
Suppose that $H_z\cap\overline{C}\subseteq C$ for every
$z\in\BbbR^r\setminus\{0\}$.   Show that $C$ is closed.
(ii) Suppose that $\nu:\frak A\to\BbbR^r$ is
countably additive in the sense that all its coordinates are countably
additive functionals.   Show that $\nu[\frak A]$ is compact.
% Dinculeanu 67?  Aliprantis & Border 13.33 atomless case only
%326Yi 326H 326M

\spheader 326Yl Let $\frak A$ be a Boolean algebra, and give it
the topology $\frak T_{\sigma}$ for which the closed sets are the
sequentially order-closed sets.   Show that a finitely additive
functional $\nu:\frak A\to\Bbb R$ is countably additive iff it is
continuous for $\frak T_{\sigma}$.
%326K

\spheader 326Ym Let $\frak A$ be a Boolean algebra, and
$M_{\sigma}$ the set of all bounded countably additive real-valued
functionals on $\frak A$.   Show that $M_{\sigma}$ is a closed
and order-closed linear subspace of the normed space $M$ of all additive
functionals on $\frak A$ (326Yd), and that $|\nu|\in M_{\sigma}$
whenever $\nu\in M_{\sigma}$.
%326L

\spheader 326Yn Let $\frak A$ be a Boolean algebra and $\nu$ a
non-negative finitely additive functional on $\frak A$.   Set

\Centerline{$\nu_{\sigma}a
=\inf\{\sup_{n\in\Bbb N}\nu a_n:\sequencen{a_n}$ is a non-decreasing
sequence with supremum $a\}$}

\noindent for every $a\in\frak A$.   Show that $\nu_{\sigma}$ is
countably additive, and is $\sup\{\nuprime:\nuprime\le\nu$ is countably
additive$\}$.
%326L

\spheader 326Yo Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\sequencen{\nu_n}$ a sequence of
countably additive real-valued functionals on $\frak A$ such that
$\nu a=\lim_{n\to\infty}\nu_na$ is defined in $\Bbb R$ for every
$a\in\frak A$.   Show that $\nu$ is countably additive.   ({\it Hint\/}:
use arguments from part (a) of the proof of 247C to see that
$\lim_{n\to\infty}\sup_{k\in\Bbb N}|\nu_ka_n|=0$ for every disjoint
sequence $\sequencen{a_n}$ in $\frak A$, and therefore that
$\lim_{n\to\infty}\sup_{k\in\Bbb N}|\nu_ka_n|=0$ whenever
$\sequencen{a_n}$ is a non-increasing sequence with infimum $0$.)
%326M

\spheader 326Yp Let $\frak A$ be a Boolean algebra, and
$M_{\tau}$ the set of all completely additive real-valued functionals on
$\frak A$.   Show that $M_{\tau}$ is a closed and order-closed linear
subspace of the normed space $M$ of all additive functionals, and that
$|\nu|\in M_{\tau}$ whenever $\nu\in M_{\tau}$.
%326O

\spheader 326Yq Let $\frak A$ be a Boolean algebra and $\nu$ a
non-negative finitely additive functional on $\frak A$.   Set

\Centerline{$\nu_{\tau}b=\inf\{\sup_{a\in A}\nu a:A$ is a non-empty
upwards-directed  set with supremum $b\}$}

\noindent for every $b\in\frak A$.   Show that $\nu_{\tau}$ is
completely additive, and is $\sup\{\nuprime:\nuprime\le\nu$ is
completely additive$\}$.
%326O

\spheader 326Yr Let $\frak A$ be a Boolean algebra, and give it
the topology $\frak T$ for which the closed sets are the order-closed
sets (313Xb).   Show that a finitely additive functional
$\nu:\frak A\to\Bbb R$ is completely additive iff it is continuous for
$\frak T$.
%326O

\spheader 326Ys Let $X$ be a set, $\Sigma$ any $\sigma$-algebra
of subsets of $X$, and $\nu:\Sigma\to\Bbb R$ a functional.   Show that
$\nu$ is completely additive iff there are sequences $\sequencen{x_n}$,
$\sequencen{\alpha_n}$ such that $\sum_{n=0}^{\infty}|\alpha_n|<\infty$
and $\nu E=\sum_{n=0}^{\infty}\alpha_n\chi E(x_n)$ for every
$E\in\Sigma$.
%326O

}%end of exercises

\cmmnt{
\Notesheader{326}
I have not mentioned the phrase `measure algebra' anywhere in this
section, and in principle this material could have been part of Chapter
31;  but
countably additive functionals are kissing cousins of measures, and most
of the ideas here surely belong to `measure theory' rather than to
`Boolean algebra', in so far as such divisions are meaningful at all.
I have given as much as possible of the theory in a
general form because the simplifications which are possible when we look
only at measure algebras are seriously confusing if they are allowed too
much prominence.   In particular, it is important to understand that the
principal properties of
completely additive functionals do not depend on Dedekind completeness
of the algebra, provided we take care over the definitions.
Similarly, the definition of `countably additive' functional for
algebras which are not Dedekind $\sigma$-complete needs a moment's
attention to the phrase `and $\sup_{n\in\Bbb N}a_n$ is defined in
$\frak A$'.    It can happen that a functional is countably additive
mostly because there are too few such sequences (326Xf).

The formulations I have chosen as principal definitions (326A, 326I,
326N) are those which I find closest to my own intuitions of the
concepts, but you may feel that 326K(i),
326Xe(iii) and 326R, or 326Yl and 326Yr, provide useful alternative
patterns.   The point is that countable additivity corresponds to
sequential order-continuity
(326Jb, 326Jc, 326Jf), while  complete additivity corresponds to
order-continuity (326Oc, 326Of);  the difficulty is that we must
consider functionals which are not order-preserving, so that the simple
definitions in 313H cannot be applied directly.   It is fair to say that
all the additive functionals $\nu$ we need to understand are bounded,
and therefore may be studied in terms of their positive and negative
parts $\nu^+$, $\nu^-$, which are order-preserving (326Bf);  but many of
the most important applications of these ideas depend precisely on using
facts about $\nu$ to deduce facts about $\nu^+$ and $\nu^-$.

It is in 326D that we seem to start getting more out of the theory than
we have put in.   The ideas here have vast ramifications.   What it
amounts to is that we can discover much more than we might expect by
looking at disjoint sequences.   To begin with, the conditions here lead
directly to 326M and 326Q:  every completely additive functional is
bounded, and every countably additive functional on a Dedekind
$\sigma$-complete Boolean algebra is bounded.   (But note 326Ya-326Yb.)

I have expressed 326H in terms of an additive function from a Boolean
algebra to a finite-dimensional space (it is already non-trivial in the
two-dimensional case, which would correspond to an additive complex-valued
functional, as in 326Ye).
It is usually regarded as a theorem about countably additive
functions, or `vector measures' (see 394O below), but rather remarkably we
do not in fact need countable additivity.   Of course it can also be
regarded as a kind of ham-sandwich theorem for measures;  we can
simultaneously bisect an element of a Dedekind $\sigma$-complete Boolean
algebra with respect to finitely many additive functionals.   If you like,
the dimensionality requirement of the ordinary ham-sandwich theorems of
topology is met by the requirement of atomlessness here.   A companion
result, also due to Liapounoff,
which requires countable additivity but allows atoms, is in 362Yx.

Naturally enough, the theory of countably additive functionals on
general Boolean algebras corresponds closely to the special case of
countably additive functionals on $\sigma$-algebras of sets, already
treated in \S\S231-232 for the sake of the Radon-Nikod\'ym theorem.
This should make 326I-326M very straightforward.
When we come to completely additive functionals, however, there is room
for many surprises.   The natural map from a $\sigma$-algebra of
measurable sets to the corresponding measure algebra is sequentially
order-continuous but rarely order-continuous, so that there can be
completely additive functionals on the measure algebra which do not
correspond to completely additive functionals on the $\sigma$-algebra.
Indeed there are very few completely additive functionals on
$\sigma$-algebras of sets (326Ys).   Of course these surprises can arise
only when there is a difference between completely additive and
countably additive functionals, that is, when the algebra involved is
not ccc (326P).   But I think that neither 326Q nor 326R is obvious.

I find myself generally using the phrase `countably additive' in
preference to `completely additive' in the context of ccc algebras,
where there is no difference between them.   This is an attempt at
user-friendliness;  the phrase `countably additive' is the commoner one
in ordinary use.   But I must say that my personal inclination is to the
other side.   The reason why so many theorems apply to countably
additive functionals in these contexts is just that they are
completely additive.

I have given two proofs of 326M.   I certainly assume
that if you have got this far you are acquainted with the
Radon-Nikod\'ym theorem and the associated basic facts about countably
additive functionals on $\sigma$-algebras of sets;  so that the `first
proof' should be easy and natural.   On the other hand, there are purist
objections on two fronts.   First, it relies on the Stone
representation, which involves a much stronger form of the axiom of
choice than is actually necessary.   Second, the classical Hahn
decomposition in 231E is evidently a special case of 326M, and if we
need both (as we certainly do) then one expects the ideas to stand out
more clearly if they are applied directly to the general case.   In fact
the two versions of the argument are so nearly identical that (as you
will observe, if you have Volume 2 to hand) they can share nearly every
word.   You can take the `second proof', therefore, as a worked example
in the translation of ideas from the context of $\sigma$-algebras of
sets to the context of Dedekind $\sigma$-complete Boolean algebras.
What makes it possible is the fact that the only limit operations
referred to involve countable families.

Arguments not involving limit operations can generally, of course, be
applied to all Boolean algebras;  I have lifted some exercises
(326Yd, 326Yn) from \S231 %231Yh, 231Yf
to give you some practice in such generalizations.

Almost any non-trivial measure provides an example of a countably
additive functional on a Dedekind $\sigma$-complete algebra which is not
completely additive (326Xh).   The question of whether such a functional
can exist on a Dedekind complete algebra is the `Banach-Ulam
problem', to which I will return in 363S.

In this section I have looked only at questions which can be adequately
treated in terms of the underlying algebras $\frak A$, without using any
auxiliary structure.   To go much farther we shall need to study the
`function spaces' $S(\frak A)$ and $L^{\infty}(\frak A)$ of Chapter 36.
In particular, the ideas of 326Ya, 326Yd-326Ye and
326Ym-326Yq %326Ym 326Yn 326Yo 326Yp 326Yq
will make better sense when redeveloped in \S362.
}%end of notes

\discrpage

\frfilename{mt327.tex}
\versiondate{13.7.11}
\copyrightdate{1995}

\def\chaptername{Measure algebras}
\def\sectionname{Additive functionals on measure algebras}

\newsection{327}

When we turn to measure algebras, we have a simplification, relative to
the general context of \S326, because the algebras are always
Dedekind $\sigma$-complete;  but there are also elaborations, because we
can ask how the additive functionals we examine are related to the
measure.   In 327A-327C I work through the relationships between the
concepts of `absolute continuity', `(true) continuity' and
`countable additivity', following \S232, and adding
`complete additivity' from \S326.   These ideas provide a new
interpretation of the Radon-Nikod\'ym theorem (327D).
I then use this theorem to develop some machinery (the `standard
extension' of an additive functional from a closed subalgebra to the
whole algebra, 327F-327G) which will be used in \S333.

\leader{327A}{}\cmmnt{ I start with the following definition and
theorem corresponding to 232A-232B.

\medskip

\noindent}{\bf Definition} Let $(\frak A,\bar\mu)$ be a measure algebra
and $\nu:\frak A\to\Bbb R$ a finitely additive functional.   Then $\nu$
is {\bf absolutely continuous} with respect to $\bar\mu$ if for every
$\epsilon>0$ there is a $\delta>0$ such that $|\nu a|\le\epsilon$
whenever $\bar\mu a\le\delta$.

\vleader{108pt}{327B}{Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra, and
$\nu:\frak A\to\Bbb R$ a finitely additive functional.   Give $\frak A$
its measure-algebra topology and uniformity\cmmnt{ (\S323)}.

(a) If $\nu$ is continuous at $0$, it is completely additive.

(b) If $\nu$ is countably additive, it is absolutely continuous with
respect to $\bar\mu$.

(c) The following are equiveridical:

\quad(i) $\nu$ is continuous at $0$;

\quad(ii) $\nu$ is countably additive and whenever $a\in\frak A$ and
$\nu a\ne 0$ there is a $b\in\frak A$ such that $\bar\mu b<\infty$ and
$\nu(a\Bcap b)\ne 0$;

\quad(iii) $\nu$ is continuous everywhere on $\frak A$;

\quad(iv) $\nu$ is uniformly continuous.

(d) If $(\frak A,\bar\mu)$ is semi-finite, then $\nu$ is continuous iff
it is completely additive.

(e) If $(\frak A,\bar\mu)$ is $\sigma$-finite, then $\nu$ is continuous
iff it is countably additive iff it is completely additive.

(f) If $(\frak A,\bar\mu)$ is totally finite, then $\nu$ is continuous
iff it is absolutely continuous with
respect to $\bar\mu$ iff it is countably additive iff it is
completely additive.

\proof{{\bf (a)} If $\nu$ is continuous, and $A\subseteq\frak A$ is
non-empty, downwards-directed and has infimum $0$, then
$0\in\overline{A}$ (323D(b-ii)), so $\inf_{a\in A}|\nu a|=0$.

\medskip

{\bf (b)}  \Quer\ Suppose, if possible, that $\nu$ is countably additive
but not absolutely continuous.   Then there is an
$\epsilon>0$ such that for every $\delta>0$ there is an $a\in\frak A$
such that $\bar\mu a\le\delta$ but
$|\nu a|\ge\epsilon$.   For each $n\in\Bbb N$ we may choose a
$b_n\in\frak A$ such that $\bar\mu b_n\le 2^{-n}$ and
$|\nu b_n|\ge\epsilon$.   Consider $b^*_n=\sup_{k\ge n}b_k$,
$b=\inf_{n\in\Bbb N}b^*_n$.   Then we have

\Centerline{$\bar\mu b\le\inf_{n\in\Bbb N}\bar\mu(\sup_{k\ge n}b_k)
\le\inf_{n\in\Bbb N}\sum_{k=n}^{\infty}2^{-k}=0$,}

\noindent so $\bar\mu b=0$ and $b=0$.   On the other hand, $\nu$ is
expressible as a difference $\nu^+-\nu^-$ of non-negative countably
additive functionals (326L), each of which is sequentially
order-continuous (326Kc), and

\Centerline{$0=\lim_{n\to\infty}(\nu^++\nu^-)b^*_n
\ge\inf_{n\in\Bbb N}(\nu^++\nu^-)b_n
\ge\inf_{n\in\Bbb N}|\nu b_n|
\ge\epsilon$,}

\noindent which is absurd.\ \Bang

\medskip

{\bf (c)(i)$\Rightarrow$(ii)} Suppose that $\nu$ is continuous
at $0$.   Then it is
completely additive, by (a), therefore countably additive.   If
$\nu a\ne 0$, there
must be a $b$ of finite measure such that $|\nu d|<|\nu a|$ whenever
$d\Bcap b=0$, so that $|\nu(a\Bsetminus b)|<|\nu a|$ and
$\nu(a\Bcap
b)\ne 0$.   Thus the conditions are satisfied.

\medskip

\quad{\bf (ii)$\Rightarrow$(iv)} Now suppose that $\nu$ satisfies the
two conditions in (ii).   Because $\frak A$ is Dedekind
$\sigma$-complete, $\nu$ must be bounded (326M),
therefore expressible as the
difference $\nu^+-\nu^-$ of countably additive functionals.   Set
$\nu_1=\nu^++\nu^-$.   Set

\Centerline{$\gamma=\sup\{\nu_1b:b\in\frak A,\,\bar\mu b<\infty\}$,}

\noindent and
choose a sequence $\sequencen{b_n}$ of elements of $\frak A$ of finite
measure such that
$\lim_{n\to\infty}\nu_1b_n=\gamma$;  set $b^*=\sup_{n\in\Bbb N}b_n$.
If $d\in\frak A$ and $d\Bcap b^*=0$ then $\nu d=0$.   \Prf\  If
$b\in\frak A$ and $\bar\mu b<\infty$, then

\Centerline{$|\nu(d\Bcap b)|\le\nu_1(d\Bcap b)\le\nu_1(b\Bsetminus b_n)
=\nu_1(b\Bcup b_n)-\nu_1b_n\le\gamma-\nu_1b_n$}

\noindent for every $n\in\Bbb N$, so $\nu(d\Bcap b)=0$.   As $b$ is
arbitrary, the second condition in (ii) tells us that $\nu d=0$.\ \Qed

Setting $b_n^*=\sup_{k\le n}b_k$ for each $n$, we have
$\lim_{n\to\infty}\nu_1(b^*\Bsetminus b^*_n)=0$.
Take any $\epsilon>0$, and (using (b) above) let $\delta>0$
be such that $|\nu a|\le\epsilon$ whenever $\bar\mu a\le\delta$.   Let
$n$ be such that
$\nu_1(b^*\Bsetminus b_n^*)\le\epsilon$.   Then

$$\eqalign{|\nu a|
&\le|\nu(a\Bcap b^*_n)|+|\nu(a\Bcap(b^*\Bsetminus b^*_n))|
  +|\nu(a\Bsetminus b^*)|\cr
&\le|\nu(a\Bcap b^*_n)|+\nu_1(b^*\Bsetminus b^*_n)
\le|\nu(a\Bcap b^*_n)|+\epsilon\cr}$$

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

   Now if $b$, $c\in\frak A$
and $\bar\mu((b\Bsymmdiff c)\Bcap b^*_n)\le\delta$ then


$$\eqalign{|\nu b-\nu c|
&\le|\nu(b\Bsetminus c)|+|\nu(c\Bsetminus b)|\cr
&\le|\nu((b\Bsetminus c)\Bcap b^*)|
  +|\nu((c\Bsetminus b)\Bcap b^*)|+2\epsilon
\le\epsilon+\epsilon+2\epsilon
=4\epsilon\cr}$$

\noindent because $\bar\mu((b\Bsetminus c)\Bcap b^*_n)$,
$\bar\mu((c\Bsetminus b)\Bcap b^*_n)$ are both less than or equal to
$\delta$.   As $\epsilon$ is arbitrary, $\nu$ is uniformly continuous.

\medskip

\quad{\bf (iv)$\Rightarrow$(iii)$\Rightarrow$(i)} are trivial.

\medskip

{\bf (d)} One implication is covered by (a).   For the other, suppose
that $\nu$ is completely additive.   Then it is countably additive.   On
the other hand, if $\nu a\ne 0$, consider
$B=\{b:b\Bsubseteq a,\,\bar\mu b<\infty\}$.   Then $B$ is
upwards-directed and $\sup B=a$, because
$\bar\mu$ is semi-finite (322Eb), so $\{a\Bsetminus b:b\in B\}$ is
downwards-directed and has infimum $0$.   Accordingly
$\inf_{b\in B}|\nu(a\Bsetminus b)|=0$, and there must be a $b\in B$ such
that $\nu b\ne 0$.   But this means that condition (ii) of (c) is
satisfied, so that $\nu$ is continuous.

\medskip

{\bf (e)} Now suppose that $(\frak A,\bar\mu)$ is $\sigma$-finite.   In
this case $\frak A$ is ccc (322G) so  complete additivity and countable
additivity are the same (326P) and we have a special case of (d).

\medskip

{\bf (f)} Finally, suppose that $\bar\mu 1<\infty$ and that $\nu$ is
absolutely continuous with respect to $\bar\mu$.   If
$A\subseteq\frak A$ is non-empty and downwards-directed and has infimum
$0$, then
$\inf_{a\in A}\bar\mu a=0$ (321F), so $\inf_{a\in A}|\nu a|$ must be
$0$;  thus $\nu$ is completely additive.   With (b) and (e) this shows
that all four conditions are equiveridical.
}%end of proof of 327B

\leader{327C}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space
and $(\frak A,\bar\mu)$ its measure algebra.

(a) There is a one-to-one correspondence between finitely additive
functionals $\bar\nu$ on $\frak A$ and finitely additive functionals
$\nu$ on $\Sigma$ such that $\nu E=0$ whenever $\mu E=0$,
given by the formula $\bar\nu E^{\ssbullet}=\nu E$ for every
$E\in\Sigma$.

(b) In (a), $\bar\nu$ is absolutely continuous with respect to $\bar\mu$
iff $\nu$ is absolutely continuous with respect to $\mu$.

(c) In (a), $\bar\nu$ is countably additive iff $\nu$ is countably
additive;  so that we have a one-to-one correspondence between the
countably additive functionals on $\frak A$ and the absolutely
continuous countably additive functionals on $\Sigma$.

(d) In (a), $\bar\nu$ is continuous for the measure-algebra topology on
$\frak A$ iff $\nu$ is truly continuous in the sense of 232Ab.

(e) Suppose that $\mu$ is semi-finite.   Then, in (a), $\bar\nu$ is
completely additive iff $\nu$ is truly continuous.

\proof{{\bf (a)} This should be nearly obvious.   If
$\bar\nu:\frak A\to\Bbb R$ is additive, then the formula defines a
functional $\nu:\Sigma\to\Bbb R$ which is additive by 326Be.
Also, of course,

\Centerline{$\mu E=0\,\Longrightarrow\,E^{\ssbullet}=0
  \,\Longrightarrow\,\nu E=0$.}

\noindent On the other hand, if $\nu$ is an additive functional on
$\Sigma$ which is zero on negligible sets, then, for $E$, $F\in\Sigma$,

$$\eqalign{E^{\ssbullet}=F^{\ssbullet}
&\Longrightarrow\,\mu(E\setminus F)=\mu(F\setminus E)=0\cr
&\Longrightarrow\,\nu(E\setminus F)=\nu(F\setminus E)=0\cr
&\Longrightarrow\,\nu F=\nu E-\nu(E\setminus F)+\nu(F\setminus E)=\nu
E,\cr}$$

\noindent so we have a function $\bar\nu:\frak A\to\Bbb R$ defined by
the
given formula.   If $E$, $F\in\Sigma$ and $E^{\ssbullet}\Bcap
F^{\ssbullet}=0$, then

$$\eqalign{\bar\nu(E^{\ssbullet}\Bcup F^{\ssbullet})
&=\bar\nu(E\cup F)^{\ssbullet}
=\nu(E\cup F)\cr
&=\nu(E\setminus F)+\nu F
=\bar\nu E^{\ssbullet}+\bar\nu F^{\ssbullet}\cr}$$

\noindent because
$(E\setminus F)^{\ssbullet}=E^{\ssbullet}\Bsetminus F^{\ssbullet}
=E^{\ssbullet}$.   Thus $\bar\nu$ is additive, and the
correspondence is complete.

\medskip

{\bf (b)} This is immediate from the definitions.

\medskip

{\bf (c)(i)} If $\nu$ is countably additive, and $\sequencen{a_n}$
is a disjoint sequence in $\frak A$, we can express it as
$\sequencen{E_n}$ where $\sequencen{E_n}$ is a sequence in $\Sigma$.
Setting $F_n=E_n\setminus\bigcup_{i<n}E_i$, $\sequencen{F_n}$ is a
disjoint sequence in $\Sigma$ and

\Centerline{$F_n^{\ssbullet}=a_n\Bsetminus\sup_{i<n}a_i=a_n$}

\noindent for each $n$.   So

\Centerline{$\bar\nu(\sup_{n\in\Bbb N}a_n)
=\nu(\bigcup_{n\in\Bbb N}F_n)
=\sum_{n=0}^{\infty}\nu F_n
=\sum_{n=0}^{\infty}\bar\nu a_n$.}

\noindent As $\sequencen{a_n}$ is arbitrary, $\bar\nu$ is countably
additive.

\medskip

\quad{\bf (ii)} If $\bar\nu$ is countably additive, then $\nu$ is
countably additive by 326Jf.

\medskip

\quad{\bf (iii)} For the last remark, note that by 232Ba a countably
additive functional on $\Sigma$ is absolutely continuous with respect to
$\mu$ iff it is zero on the $\mu$-negligible sets.

\medskip

{\bf (d)} The definition of `truly continuous' functional translates
directly to continuity at $0$ in the measure algebra.   But by 327Bc
this is the same thing as continuity.

\medskip

{\bf (e)} Put (d) and 327Bd together.
}%end of proof of 327C

\leader{327D}{The Radon-Nikod\'ym theorem}\cmmnt{ We are now ready for
another look at this theorem.

\medskip

\noindent{\bf Theorem}} Let $(X,\Sigma,\mu)$ be a semi-finite measure
space, with measure algebra $(\frak A,\bar\mu)$.   Let $L^1$ be the
space of equivalence classes of real-valued integrable functions on
$X$\cmmnt{ (\S242)}, and write $M_{\tau}$ for the set of
completely additive
real-valued functionals on $\frak A$.   Then there is an ordered linear
space bijection between $M_{\tau}$ and $L^1$ defined by saying that
$\bar\nu\in M_{\tau}$ corresponds to $u\in L^1$ if

\Centerline{$\bar\nu a=\int_Ef$ whenever $a=E^{\ssbullet}$ in $\frak A$
and $f^{\ssbullet}=u$ in $L^1$.}

\proof{{\bf (a)} Given $\bar\nu\in M_{\tau}$, we have a truly continuous
$\nu:\Sigma\to\Bbb R$ given by setting $\nu E=\bar\nu E^{\ssbullet}$
for every $E\in\Sigma$ (327Ce).   Now there is an integrable function
$f$ such that $\nu E=\int_Ef$ for every $E\in\Sigma$ (232E).   There
is likely to be more than one such function, but any two must be equal
almost everywhere (232Hd), so the corresponding equivalence class
$u_{\bar\nu}=f^{\ssbullet}$ is uniquely defined.

\medskip

{\bf (b)} Conversely, given $u\in L^1$, we have a well-defined
functional $\nu_u$ on $\Sigma$ given by setting

\Centerline{$\nu_u E=\int_Eu=\int_Ef$ whenever $f^{\ssbullet}=u$}

\noindent for every $E\in\Sigma$ (242Ac).   By 232D, $\nu_u$ is
additive and truly continuous, and of course it is zero when $\mu$
is zero, so corresponds to a completely additive functional $\bar\nu_u$
on $\frak A$ (327Ce).

\medskip

{\bf (c)} Clearly the maps $u\mapsto\bar\nu_u$ and
$\bar\nu\mapsto u_{\bar\nu}$ are
now the two halves of a one-to-one correspondence.   To see that it is
linear, we need note only that

\Centerline{$(\bar\nu_u+\bar\nu_v)E^{\ssbullet}
=\bar\nu_uE^{\ssbullet}+\bar\nu_vE^{\ssbullet}
=\int_Eu+\int_Ev
=\int_Eu+v
=\bar\nu_{u+v}E^{\ssbullet}$}

\noindent for every $E\in\Sigma$, so $\bar\nu_u+\bar\nu_v=\bar\nu_{u+v}$
for all
$u$, $v\in L^1$;  and similarly $\bar\nu_{\alpha u}=\alpha\bar\nu_u$
for $u\in L^1$ and $\alpha\in\Bbb R$.    As for the ordering, given $u$ and
$v\in L^1$, take integrable $f$, $g$ such that $u=f^{\ssbullet}$ and
$v=g^{\ssbullet}$;  then

$$\eqalign{\bar\nu_u\le\bar\nu_v
&\iff\bar\nu_uE^{\ssbullet}\le\bar\nu_vE^{\ssbullet}
   \text{ for every }E\in\Sigma\cr
&\iff\int_Eu\le\int_Ev
   \text{ for every }E\in\Sigma\cr
&\iff\int_Ef\le\int_Eg
   \text{ for every }E\in\Sigma\cr
&\iff f\leae g
\iff u\le v,\cr}$$

\noindent using 131Ha.
}%end of proof of 327D

\leader{327E}{}\cmmnt{ I slip in an elementary fact.

\medskip

\noindent}{\bf Proposition} If $(\frak A,\bar\mu)$ is a measure algebra,
then the functional $a\mapsto\mu_ca=\bar\mu(a\Bcap c)$ is
completely additive whenever $c\in\frak A$ and $\bar\mu c<\infty$.

\proof{ $\mu_c$ is additive because $\bar\mu$ is additive, and by 321F
again
$\inf_{a\in A}\mu_ca=0$ whenever $A$ is non-empty, downwards-directed
and has infimum $0$.
}%end of proof of 327E

\leader{327F}{Standard \dvrocolon{extensions}}\cmmnt{ The machinery of
327D provides the
basis of a canonical method for extending countably additive functionals
from closed subalgebras, which we shall need in \S333.

\wheader{327F}{4}{2}{2}{96pt}

\noindent}{\bf Lemma} Let $(\frak A,\bar\mu)$ be a totally finite
measure algebra and $\frak C\subseteq\frak A$ a closed subalgebra.
Write $M_{\sigma}(\frak A)$, $M_{\sigma}(\frak C)$ for the spaces of
countably additive real-valued functionals on $\frak A$, $\frak C$
respectively.

(a) There is an operator $R:M_{\sigma}(\frak C)\to M_{\sigma}(\frak A)$
defined by saying that, for every $\nu\in M_{\sigma}(\frak C)$, $R\nu$
is the unique member of $M_{\sigma}(\frak A)$ such that
$\Bvalue{R\nu>\alpha\bar\mu}=\Bvalue{\nu>\alpha\bar\mu\restrp\frak C}$
for every $\alpha\in\Bbb R$.

(b)(i) $R\nu$ extends $\nu$ for every $\nu\in M_{\sigma}(\frak C)$.

\quad(ii) $R$ is linear and order-preserving.

\quad(iii) $R(\bar\mu\restrp\frak C)=\bar\mu$.

\quad(iv) If $\sequencen{\nu_n}$ is a sequence of non-negative
functionals in $M_{\sigma}(\frak C)$ such that
$\sum_{n=0}^{\infty}\nu_nc=\bar\mu c$ for every $c\in\frak C$, then
$\sum_{n=0}^{\infty}(R\nu_n)(a)=\bar\mu a$ for every $a\in\frak A$.

\cmmnt{\medskip

\noindent{\bf Remarks} When saying that $\frak C$ is `closed', I mean,
indifferently, `topologically closed' or `order-closed';  see 323H-323I.

For the notation `$\Bvalue{\nu>\alpha\bar\mu}$' see 326S-326T.
}%end of comment

\proof{{\bf (a)(i)} By 321J-321K, we may represent $(\frak A,\bar\mu)$
as the measure algebra of a measure space $(X,\Sigma,\mu)$;  write $\pi$
for the canonical map from $\Sigma$ to $\frak A$.   Write $\Tau$ for
$\{E:E\in\Sigma,\,\pi E\in\frak C\}$.   Because $\frak C$ is a
$\sigma$-subalgebra of $\frak C$ and $\pi$ is a sequentially
order-continuous Boolean homomorphism, $\Tau$ is a $\sigma$-subalgebra
of $\Sigma$.

\medskip

\quad{\bf (ii)} For each $\nu\in M_{\sigma}(\frak C)$,
$\nu\pi:\Tau\to\Bbb R$ is countably additive and zero on
$\{F:F\in\Tau,\,\mu F=0\}$, so we can choose a $\Tau$-measurable
function $f_{\nu}:X\to\Bbb R$ such that
$\int_Ff_{\nu}d(\mu\restrp\Tau)=\nu\pi F$ for every $F\in\Tau$.   Of
course we can now think of $f_{\nu}$ as a
$\mu$-integrable function (233B), so we get a corresponding countably
additive functional $R\nu:\frak A\to\Bbb R$ defined by setting
$(R\nu)(\pi E)=\int_Ef_{\nu}$ for every $E\in\Sigma$ (327D).   (In this
context, of course, countably additive functionals are completely
additive, by 327Bf.)    Note that if $c\in\frak C$ there is an $F\in\Tau$
such that $F^{\ssbullet}=c$, so that

\Centerline{$(R\nu)(c)=\int_Ff_{\nu}=\nu c$.}

For $\alpha\in\Bbb R$, set $H_{\alpha}=\{x:f_{\nu}(x)>\alpha\}\in\Tau$.
Then for any $E\in\Sigma$,

\Centerline{$E\subseteq H_{\alpha},\,\mu E>0
\Longrightarrow\int_Ef_{\nu}>\alpha\mu E$,}

\Centerline{$E\cap H_{\alpha}=\emptyset
\Longrightarrow\int_Ef_{\nu}\le\alpha\mu E$.}

\noindent Translating into terms of elements of $\frak A$, and setting
$c_{\alpha}=\pi H_{\alpha}\in\frak C$, we have

\Centerline{$0\ne a\Bsubseteq c_{\alpha}
\Longrightarrow(R\nu)(a)>\alpha\bar\mu a$,}

\Centerline{$a\Bcap c_{\alpha}=0
\Longrightarrow(R\nu)(a)\le\alpha\bar\mu a$.}

\noindent So $\Bvalue{R\nu>\alpha\bar\mu}=c_{\alpha}\in\frak C$.   Of
course we now have

\Centerline{$\nu c=(R\nu)(c)>\alpha\bar\mu c$ when $c\in\frak C$,
$0\ne c\Bsubseteq c_{\alpha}$,}

\Centerline{$\nu c\le\alpha\bar\mu c$ when $c\in\frak C$,
$c\Bcap c_{\alpha}=0$,}

\noindent so that $c_{\alpha}$ is also equal to
$\Bvalue{\nu>\alpha\bar\mu\restrp\frak C}$.

Thus the functional $R\nu$ satisfies the declared formula.

\medskip

\quad{\bf (iii)} To see that $R\nu$ is uniquely defined, observe that if
$\lambda\in M_{\sigma}(\frak A)$ and
$\Bvalue{\lambda>\alpha\bar\mu}=\Bvalue{R\nu>\alpha\bar\mu}$ for every
$\alpha$, then there is a $\Sigma$-measurable function $g:X\to\Bbb R$
such that $\int_Eg\,d\mu=\lambda\pi E$ for every $E\in\Sigma$;  but in
this case (just as in (ii)) $\Bvalue{\lambda>\alpha\bar\mu}=\pi
G_{\alpha}$, where $G_{\alpha}=\{x:g(x)>\alpha\}$, for each $\alpha$.
So we must have $\pi G_{\alpha}=\pi H_{\alpha}$, that is,
$\mu(G_{\alpha}\symmdiff H_{\alpha})=0$, for every $\alpha$.
Accordingly

\Centerline{$\{x:f_{\nu}(x)\ne g(x)\}=\bigcup_{q\in\Bbb Q}G_q\symmdiff
H_q$}

\noindent is negligible;  $f_{\nu}\eae g$,
$\int_Ef_{\nu}d\mu=\int_Eg\,d\mu$ for every $E\in\Sigma$ and
$\lambda=R\nu$.

\medskip

{\bf (b)(i)} I have already noted that $(R\nu)c=\nu c$ for every
$\nu\in M_{\sigma}(\frak C)$ and $c\in\frak C$.

\medskip

\quad{\bf (ii)} If $\nu=\nu_1+\nu_2$, we must have, in the language of (a)
above,

\Centerline{$\int_Ff_{\nu}=\nu\pi F=\nu_1\pi F+\nu_2\pi F
=\int_Ff_{\nu_1}+\int_Ff_{\nu_2}=\int_Ff_{\nu_1}+f_{\nu_2}$}

\noindent for every $F\in\Tau$, so $f_{\nu}\eae f_{\nu_1}+f_{\nu_2}$,
and we can repeat the formulae

\Centerline{$(R\nu)(\pi E)=\int_Ef_{\nu}=\int_Ef_{\nu_1}+f_{\nu_2}
=\int_Ef_{\nu_1}+\int_Ef_{\nu_2}=(R\nu_1)(\pi E)+(R\nu_2)(\pi E)$,}

\noindent in a different order, for every $E\in\Sigma$, to see that
$R\nu=R\nu_1+R\nu_2$.   Similarly, if $\nu\in M_{\sigma}(\frak C)$ and
$\gamma\in\Bbb R$, $f_{\gamma\nu}\eae\gamma f_{\nu}$ and
$R(\gamma\nu)=\gamma R\nu$.   If $\nu_1\le\nu_2$ in
$M_{\sigma}(\frak C)$, then

\Centerline{$\int_Ff_{\nu_1}=\nu_1\pi F\le\nu_2\pi F=\int_Ff_{\nu_2}$}

\noindent for every $F\in\Tau$, so $f_{\nu_1}\leae f_{\nu_2}$
(131Ha again), and $R\nu_1\le R\nu_2$.

Thus $R$ is linear and order-preserving.

\medskip

\quad{\bf (iii)} If $\nu=\bar\mu\restrp\frak C$ then

\Centerline{$\int_Ff_{\nu}=\nu\pi F=\mu F=\int_F\chi X$}

\noindent for every $F\in\Tau$, so $f_{\nu}\eae\chi X$ and
$R\nu=\bar\mu$.

\medskip

\quad{\bf (iv)} Now suppose that $\sequencen{\nu_n}$ is a sequence in
$M_{\sigma}(\frak C)$ such that, for every $c\in\frak C$, $\nu_nc\ge 0$
for every $n$ and $\sum_{n=0}^{\infty}\nu_nc=\bar\mu c$.  Set
$g_n=\sum_{i=0}^nf_{\nu_i}$ for each $n$;  then
$0\leae g_n\leae g_{n+1}\leae\chi X$ for every $n$, and

\Centerline{$\lim_{n\to\infty}\int g_n
=\lim_{n\to\infty}$$\sum_{i=0}^n\nu_i1=\bar\mu 1$.}

\noindent But this means that, setting $g=\lim_{n\to\infty}g_n$,
$g\leae\chi X$ and $\int g=\int\chi X$, so that $g\eae\chi X$ and

\Centerline{$\sum_{n=0}^{\infty}(R\nu_i)(\pi E)
=\lim_{n\to\infty}$$\int_Eg_n=\mu E$}

\noindent for every $E\in\Sigma$.   Thus
$\sum_{n=0}^{\infty}(R\nu_i)(a)=\bar\mu a$ for every $a\in\frak A$.
}%end of proof of 327F

\leader{327G}{Definition} In the context of 327F, I will call $R\nu$
the {\bf standard extension} of $\nu$ to $\frak A$.

\cmmnt{\medskip

\noindent{\bf Remark} The point of my insistence on the
uniqueness of $R$, and on the formula in 327Fa, is that $R\nu$ really is
defined by the abstract structure $(\frak A,\bar\mu,\frak C,\nu)$, even
though I have used a proof which runs through the representation of
$(\frak A,\bar\mu)$ as the measure algebra of a measure space
$(X,\Sigma,\mu)$.
}%end of comment

\exercises{\leader{327X}{Basic exercises (a)}
%\spheader 327Xa  
Let $(X,\Sigma,\mu)$ be a probability space,
and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Let $(\frak A,\bar\mu)$
be the measure algebra of $(X,\Sigma,\mu)$.   Show that $\frak
C=\{F^{\ssbullet}:F\in\Tau\}$ is a closed subalgebra of $\frak A$.
Identify the spaces $M_{\sigma}(\frak A)$, $M_{\sigma}(\frak C)$ of
countably additive functionals with $L^1(\mu)$,
$L^1(\mu\restrp\Tau)$, as in 327D.   Show that the
conditional expectation operator
$P:L^1(\mu)\to L^1(\mu\restrp\Tau)$ (242Jd) corresponds to the map
$\nu\mapsto\nu\restrp\frak C:
M_{\sigma}(\frak A)\to M_{\sigma}(\frak C)$.
%327D

\spheader 327Xb Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra and $\nu:\frak A\to\Bbb R$ a countably additive functional.
Show that, for any $a\in\frak A$,

\Centerline{$\nu a
=\int_0^{\infty}\bar\mu(a\Bcap\Bvalue{\nu>\alpha\bar\mu})d\alpha
-\int_{-\infty}^0
\bar\mu(a\Bsetminus\Bvalue{\nu>\alpha\bar\mu})d\alpha$,}

\noindent the integrals being taken with respect to Lebesgue measure.
\Hint{take $(\frak A,\bar\mu)$ to be the measure algebra of
$(X,\Sigma,\mu)$;  represent $\nu$ by a $\mu$-integrable function $f$;
apply Fubini's theorem to the sets $\{(x,t):x\in E,\,0\le t<f(x)\}$,
$\{(x,t):x\in E,\,f(x)\le t\le 0\}$ in $X\times\Bbb R$, where
$a=E^{\ssbullet}$.}
%327F

\spheader 327Xc
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\mu')$ be
totally finite measure algebras, and $\pi:\frak A\to\frak B$ a
measure-preserving Boolean homomorphism.   Let $\frak C$ be a closed
subalgebra of $\frak A$, and $\nu$ a countably additive functional on
the closed subalgebra $\pi[\frak C]$ of $\frak B$ (324L).   (i) Show that
$\nu\pi$ is a countably additive functional on $\frak C$.   (ii) Show
that if $\tilde\nu$ is the standard extension of $\nu$ to $\frak B$,
then $\tilde\nu\pi$ is the standard extension of $\nu\pi$ to $\frak A$.
\Hint{take $\alpha\in\Bbb R$ and set
$e_0=\Bvalue{\tilde\nu>\alpha\bar\mu'}
=\Bvalue{\nu>\alpha\bar\mu'\restr\pi[\frak C]}$;  there is a
$c_0\in\frak C$ such that $\pi c_0=e_0$;  check that
$c_0=\Bvalue{\tilde\nu\pi>\alpha\bar\mu}
=\Bvalue{\nu\pi>\alpha\bar\mu\restrp\frak C}$.}
%327G

\spheader 327Xd Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra, $\frak C$ a closed subalgebra of $\frak A$ and $\nu:\frak
C\to\Bbb R$ a countably additive functional with standard extension
$\tilde\nu:\frak A\to\Bbb R$.   Show that, for any $a\in\frak A$,

\Centerline{$\tilde\nu a
=\int_0^{\infty}
   \bar\mu(a\Bcap\Bvalue{\nu>\alpha\bar\mu\restrp\frak C})d\alpha
 -\int_{-\infty}^0
   \bar\mu(a\Bsetminus\Bvalue{\nu>\alpha\bar\mu\restrp\frak
C})d\alpha$.}
%327G

\spheader 327Xe Let $(\frak A,\bar\mu)$ be a probability algebra, and
$\frak B$, $\frak C$ stochastically independent closed subalgebras of
$\frak A$ (definition: 325L).   Let $\nu$ be a countably additive
functional on $\frak C$, and $\tilde\nu$ its standard extension to
$\frak A$.   Show that $\tilde\nu(b\Bcap c)=\bar\mu b\cdot\nu c$ for
every $b\in\frak B$, $c\in\frak C$.
%327G

\spheader 327Xf Let $(X,\Sigma,\mu)$ be a probability space, and $\Tau$
a $\sigma$-subalgebra of $\Sigma$.   Let $\nu$ be a probability measure
with domain $\Tau$ such that $\nu E=0$ whenever $E\in\Tau$ and
$\mu E=0$.   Show that there is a probability measure $\lambda$ with
domain $\Sigma$ which extends $\nu$.

\leader{327Y}{Further exercises (a)}
%\spheader 327Ya
Let $(\frak A_1,\bar\mu_1)$ and $(\frak A_2,\bar\mu_2)$ be localizable
measure algebras with localizable measure algebra free product
$(\frak C,\bar\lambda)$.   Show that if $\nu_1$, $\nu_2$ are completely
additive
functionals on $\frak A_1$, $\frak A_2$ respectively, there is a unique
completely additive functional $\nu:\frak C\to\Bbb R$ such that
$\nu(a_1\otimes a_2)=\nu_1a_1\cdot\nu_2a_2$ for every $a_1\in\frak A_1$,
$a_2\in\frak A_2$.   \Hint{253D.}
%327D

\spheader 327Yb Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra and $\frak C$ a closed subalgebra;  let
$R:M_{\sigma}(\frak C)\to M_{\sigma}(\frak A)$ be the standard extension
operator (327G).
Show (i) that $R$ is order-continuous (ii) that $R(\nu^+)=(R\nu)^+$,
$\|R\nu\|=\|\nu\|$ for every $\nu\in M_{\sigma}(\frak C)$, defining
$\nu^+$ and $\|\nu\|$ as in 326Yd.
%327G

\spheader 327Yc Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra and $\frak C$ a closed subalgebra of $\frak A$.   For a
countably additive functional $\nu$ on $\frak C$ write $\tilde\nu$ for
its standard extension to $\frak A$.   Show that if $\nu$,
$\sequencen{\nu_n}$ are countably additive functionals on $\frak C$ and
$\lim_{n\to\infty}\nu_nc=\nu c$ for every $c\in\frak C$, then
$\lim_{n\to\infty}\tilde\nu_na=\tilde\nu a$ for every $a\in\frak A$.
({\it Hint\/}:  use ideas from \S\S246-247, as well as from 327F and
326Yo.)
%327G
}%end of exercises

\cmmnt{
\Notesheader{327}
When we come to measure algebras, it is the completely additive
functionals which fit most naturally into the topological theory
(327Bd);   they correspond to the `truly
continuous' functionals which I discussed in \S232 (327Cd), and
therefore to the Radon-Nikod\'ym theorem (327D).   I will return to some
of these questions in Chapter 36.    I myself regard the form
here as the best expression of the essence of the Radon-Nikod\'ym
theorem, if not the one most commonly applied.

The concept of `standard extension' of a countably additive functional
(or, as we could equally well say, of a completely additive functional,
since in the context of 327F the two coincide) is in a sense dual to the
concept of `conditional expectation'.   If
$(X,\Sigma,\mu)$ is a probability space and $\Tau$ is a
$\sigma$-subalgebra of $\Sigma$, then we have a corresponding closed
subalgebra $\frak C$ of the measure algebra $(\frak A,\bar\mu)$ of
$\mu$, and identifications between the spaces $M_{\sigma}(\frak A)$,
$M_{\sigma}(\frak C)$ of countably additive functionals and the spaces
$L^1(\mu)$, $L^1(\mu\restrp\Tau)$.   Now we have a natural
embedding $S$ of
$L^1(\mu\restrp\Tau)$ as a subspace of $L^1(\mu)$ (242Jb), and a
natural restriction map from $M_{\sigma}(\frak A)$ to
$M_{\sigma}(\frak C)$.   These give rise to corresponding operators
between the opposite
members of each pair;  the standard extension operator $R$ of 327F-327G,
and the conditional expectation operator $P$ of 242Jd.    (See 327Xa.)
The fundamental fact

\Centerline{$PSv=v$ for every $v\in L^1(\mu\restrp\Tau)$}

\noindent (242Jg) is matched by the fact that

\Centerline{$R\nu\restrp\frak C=\nu$ for every
$\nu\in M_{\sigma}(\frak C)$.}

\noindent The further identification of $R\nu$ in terms of integrals
$\int\bar\mu(a\Bcap\Bvalue{\nu>\alpha\bar\mu})d\alpha$ (327Xd) is
relatively inessential, but is
striking, and perhaps makes it easier to believe that $R$ is truly
`standard' in the abstract contexts which will arise in \S333 below.
It is also useful in such calculations as 327Xe.

The isomorphisms between $M_{\tau}$ spaces and $L^1$ spaces described
here mean that any of the concepts involving $L^1$ spaces discussed in
Chapter 24 can be applied to $M_{\tau}$ spaces, at least in the case of
measure algebras.   In fact, as I will show in Chapter 36, there is much
more to be said here;  the space of bounded additive functionals on a
Boolean algebra is already an $L^1$ space in an abstract sense, and
ideas such as `uniform integrability' are relevant and significant
there, as well as in the spaces of countably additive and
completely additive functionals.   I
hope that 326Yd, 326Ym-326Yn, 326Yp-326Yq and 327Yb will provide some
hints to be going on with for the moment.
}%end of comment

\frnewpage

\frfilename{mt328.tex}
\versiondate{2.6.09}
\copyrightdate{2008}

\def\chaptername{Measure algebras}
\def\sectionname{Reduced products and other constructions}

\newsection{*328\dvAnew{2008}}

I devote a section to some related constructions.   At the
end of \S315 %315R 315S
I mentioned projective and inductive limits of systems of
Boolean algebras with linking homomorphisms.   In the context of the
present chapter, we naturally ask whether similar constructions can be
found for probability algebras.   For projective limits there is no
difficulty (328I).   For inductive limits the situation is more complex
(328H).   Some ideas in Volume 5 will
depend on what I call `reduced products'
(328A-328F), %328A 328B 328C 328D 328E 328F
which also provide a route to 328H.   The same methods give a route to
a useful result relating measure-preserving Boolean homomorphisms on a
probability algebra to measure-preserving automorphisms on a larger
probability algebra (328J).

\leader{328A}{Construction}
Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a non-empty family of probability
algebras, and $\Cal F$ an ultrafilter on $I$.

\spheader 328Aa Set

\Centerline{$\Cal J
=\{\familyiI{a_i}:\familyiI{a_i}\in\prod_{i\in I}\frak A_i$,
  $\lim_{i\to\Cal F}\bar\mu_ia_i=0\}$.}

\noindent Then $\Cal J$ is an ideal in the simple
product Boolean algebra $\prod_{i\in I}\frak A_i$.
\prooflet{\Prf\ If $\familyiI{a_i}$ and $\familyiI{b_i}$ belong to
$\Cal J$, and $\familyiI{c_i}\in\prod_{i\in I}\frak A_i$ is such that
$\familyiI{c_i}\Bsubseteq\familyiI{a_i}\Bcup\familyiI{b_i}$, then
$c_i\Bsubseteq a_i\Bcup b_i$ for every $i$, so

\Centerline{$\lim_{i\to\Cal F}\bar\mu_ic_i
\le\lim_{i\to\Cal F}\bar\mu_ia_i+\bar\mu_ib_i
=\lim_{i\to\Cal F}\bar\mu_ia_i+\lim_{i\to\Cal F}\bar\mu_ib_i
=0$}

\noindent and $\familyiI{c_i}\in\Cal J$.   Of course
$\familyiI{0_{\frak A_i}}$ belongs to $\Cal J$, so
$\Cal J\normalsubgroup\prod_{i\in I}\frak A_i$.\ \Qed}

\spheader 328Ab Let $\frak A$ be the
quotient Boolean algebra $\prod_{i\in I}\frak A_i/\Cal J$.   Then we have a
functional $\bar\mu:\frak A\to[0,1]$ defined by saying that

\Centerline{$\bar\mu(\familyiI{a_i}^{\ssbullet})
=\lim_{i\to\Cal F}\bar\mu_ia_i$}

\noindent whenever $\familyiI{a_i}\in\prod_{i\in I}\frak A_i$.
\prooflet{\Prf\ If $\familyiI{a_i}$,
$\familyiI{b_i}\in\prod_{i\in I}\frak A_i$ and
$\familyiI{a_i}^{\ssbullet}=\familyiI{b_i}^{\ssbullet}$, then
$\familyiI{a_i\Bsymmdiff b_i}\in\Cal J$, so

\Centerline{$|\lim_{i\to\Cal F}\bar\mu_ia_i-\lim_{i\to\Cal F}\bar\mu_ib_i|
=\lim_{i\to\Cal F}|\bar\mu_ia_i-\bar\mu_ib_i|
\le\lim_{i\to\Cal F}\bar\mu_i(a_i\Bsymmdiff b_i)
=0$.  \Qed}}

\leader{328B}{Proposition}
Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a non-empty family of probability
algebras and $\Cal F$ an ultrafilter on $I$, and construct $\frak A$ and
$\bar\mu$ as in 328A.   Then $(\frak A,\bar\mu)$ is a probability algebra.

\proof{{\bf (a)} If $\familyiI{a_i}$,
$\familyiI{b_i}\in\prod_{i\in I}\frak A_i$ and
$\familyiI{a_i}^{\ssbullet}\Bcap\familyiI{b_i}^{\ssbullet}=0$, then
$\familyiI{a_i\cap b_i}\in\Cal J$, so

$$\eqalign{
\bar\mu(\familyiI{a_i}^{\ssbullet}\Bcup\familyiI{b_i}^{\ssbullet})
&=\bar\mu(\familyiI{a_i\Bcup b_i}^{\ssbullet})
=\lim_{i\to\Cal F}\bar\mu_i(a_i\Bcup b_i)\cr
&=\lim_{i\to\Cal F}\bar\mu_ia_i+\bar\mu_ib_i-\bar\mu_i(a_i\Bcap b_i)\cr
&=\lim_{i\to\Cal F}\bar\mu_ia_i
  +\lim_{i\to\Cal F}\bar\mu_ib_i
  -\lim_{i\to\Cal F}\bar\mu_i(a_i\Bcap b_i)\cr
&=\lim_{i\to\Cal F}\bar\mu_ia_i
  +\lim_{i\to\Cal F}\bar\mu_ib_i
=\bar\mu(\familyiI{a_i}^{\ssbullet})+\bar\mu(\familyiI{b_i}^{\ssbullet}).
\cr}$$

\noindent So $\bar\mu$ is additive.

\medskip

{\bf (b)}
$1_{\frak A}=\familyiI{1_{\frak A_i}}^{\ssbullet}$ so

\Centerline{$\bar\mu 1_{\frak A}
=\lim_{i\to\Cal F}\bar\mu_i1_{\frak A_i}=1$.}

\medskip

{\bf (c)} If $\familyiI{a_i}\in\prod_{i\in I}\frak A_i$ and
$\bar\mu(\familyiI{a_i}^{\ssbullet})=0$, then
$\familyiI{a_i}\in\Cal J$ and $\familyiI{a_i}^{\ssbullet}=0$;  thus
$\bar\mu$ is strictly positive.

\medskip

{\bf (d)} Suppose that $\sequencen{\tilde a_n}$ is a disjoint sequence in
$\frak A$.   Express each $\tilde a_n$ as $\familyiI{a_{ni}}^{\ssbullet}$
where
$a_{ni}\in\frak A_i$ for each $i$.   Set $b_{ni}=\sup_{m\le n}a_{mi}$ for
$n\in\Bbb N$ and $i\in I$;  then
$\familyiI{b_{ni}}^{\ssbullet}=\sup_{m\le n}\tilde a_m$ in $\frak A$.   Set

\Centerline{$\gamma=\sum_{n=0}^{\infty}\bar\mu\tilde a_n
=\sup_{n\in\Bbb N}\bar\mu(\familyiI{b_{ni}}^{\ssbullet})
=\sup_{n\in\Bbb N}\lim_{i\to\Cal F}\bar\mu_ib_{ni}$.}

\noindent Set $A_n=\{i:i\in I$, $\bar\mu_ib_{ni}\le\gamma+2^{-n}\}$ for
each $n\in\Bbb N$;  then $\sequencen{A_n}$ is a non-increasing sequence in
$\Cal F$, and $A_0=I$.   For $i\in I$ set

$$\eqalign{b_i
&=b_{ni}\text{ if }i\in A_n\setminus A_{n+1},\cr
&=\sup_{n\in\Bbb N}b_{ni}\text{ if }i\in\bigcap_{n\in\Bbb N}A_n.\cr}$$

\noindent Consider $\tilde b=\familyiI{b_i}^{\ssbullet}\in\frak A$.
For each
$n\in\Bbb N$, $\{i:a_{ni}\Bsubseteq b_i$, $\bar\mu b_i\le\gamma+2^{-n}\}$
includes $A_n\in\Cal F$, so
$\tilde a_n\Bsubseteq\tilde b$ for every $n$ and
$\bar\mu\tilde b\le\gamma$.

If $\tilde c\in\frak A$ is another upper bound for
$\{\tilde a_n:n\in\Bbb N\}$, then, using (a),

\Centerline{$\gamma=\sup_{n\in\Bbb N}\bar\mu(\sup_{m\le n}\tilde a_m)
\le\bar\mu(\tilde b\Bcap\tilde c)\le\bar\mu\tilde b\le\gamma$;}

\noindent so $\bar\mu(\tilde b\Bsetminus\tilde c)=0$ and
$\tilde b\Bsetminus\tilde c=0$, by (c).
Thus $\tilde b=\sup_{n\in\Bbb N}\tilde a_n$ in $\frak A$, while
$\bar\mu\tilde b=\sum_{n=0}^{\infty}\bar\mu\tilde a_n$.

\medskip

{\bf (e)} If $\sequencen{\tilde a_n}$ is any sequence in
$\frak A$, then (iv) tells us that
$\{\tilde a_n\Bsetminus\sup_{m<n}\tilde a_m:n\in\Bbb N\}$
has a supremum in $\frak A$, which is also the supremum of
$\{\tilde a_n:n\in\Bbb N\}$.   So $\frak A$ is Dedekind $\sigma$-complete.
Now (d) tells us also that $\bar\mu$ is countably additive, so that
$(\frak A,\bar\mu)$ is a probability algebra.
}%end of proof of 328B

\leader{328C}{Definition} In the context of 328A/328B,
I will call $(\frak A,\bar\mu)$ the {\bf probability
algebra reduced product} of $\familyiI{(\frak A_i,\bar\mu_i)}$
modulo $\Cal F$;  I will
sometimes write it as $\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$.
\cmmnt{(There are dangers in this notation.   In 351M I will speak of
`reduced powers' $\BbbR^I|\Cal F$, and the rules will be significantly
different there.)}

If all the $(\frak A_i,\bar\mu_i)$ are the same, with common value
$(\frak B,\bar\nu)$, I will write
$(\frak B,\bar\nu)^I|\Cal F$ for
$\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$, and call it the
{\bf probability algebra reduced power}.

\vleader{108pt}{328D}{Proposition} Let $I$ be a set,
$\familyiI{(\frak A_i,\bar\mu_i)}$,
$\familyiI{(\frak B_i,\bar\nu_i)}$ and
$\familyiI{(\frak C_i,\bar\lambda_i)}$ three families of probability
algebras, and $\Cal F$ an ultrafilter on $I$;  let
$(\frak A,\bar\mu)=\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$,
$(\frak B,\bar\nu)=\prod_{i\in I}(\frak B_i,\bar\nu_i)|\Cal F$ and
$(\frak C,\bar\lambda)=\prod_{i\in I}(\frak C_i,\bar\lambda_i)|\Cal F$
be the corresponding reduced products.

(a) If $\pi_i:\frak A_i\to\frak B_i$ is a measure-preserving Boolean
homomorphism for each $i\in I$, we have a measure-preserving Boolean
homomorphism $\pi:\frak A\to\frak B$ given by saying that

\Centerline{$\pi(\familyiI{a_i}^{\ssbullet})
=\familyiI{\pi_ia_i}^{\ssbullet}$}

\noindent whenever $a_i\in\frak A_i$ for every $i\in I$.

(b) If, in addition, $\phi_i:\frak B_i\to\frak C_i$ is a measure-preserving
Boolean homomorphism for each $i\in I$, and $\phi:\frak B\to\frak C$ is
constructed as in (a), then $\phi\pi:\frak A\to\frak C$ corresponds to the
family $\familyiI{\phi_i\pi_i}$.

\proof{{\bf (a)} Following through the construction in 328A, we have ideals

\Centerline{$\Cal J=\{\familyiI{a_i}:\lim_{i\to\Cal F}\bar\mu_ia_i=0\}
\normalsubgroup\prod_{i\in I}\frak A_i$,}

\Centerline{$\Cal K=\{\familyiI{b_i}:\lim_{i\to\Cal F}\bar\nu_ib_i=0\}
\normalsubgroup\prod_{i\in I}\frak B_i$,}

\noindent and a Boolean homomorphism
$\hat\pi:\prod_{i\in I}\frak A_i\to\prod_{i\in I}\frak B_i$ given by
the formula $\hat\pi\familyiI{a_i}=\familyiI{\pi_ia_i}$ (use 315Bb).
Because the
homomorphisms $\pi_i$ are measure-preserving, $\hat\pi\pmb{a}\in\Cal K$
whenever $\pmb{a}\in\Cal J$.   Consequently we have a Boolean homomorphism
$\pi:\prod_{i\in I}\frak A_i/\Cal J\to\prod_{i\in I}\frak B_i/\Cal K$ given
by setting $\pi\pmb{a}^{\ssbullet}=(\hat\pi\pmb{a})^{\ssbullet}$ whenever
$\pmb{a}\in\prod_{i\in I}\frak A_i$ (3A2G).   And

\Centerline{$\bar\nu\pi(\familyiI{a_i}^{\ssbullet})
=\bar\nu(\familyiI{\pi_ia_i}^{\ssbullet})
=\lim_{i\to\Cal F}\bar\nu_i\pi_ia_i
=\lim_{i\to\Cal F}\bar\mu_ia_i
=\bar\mu(\familyiI{a_i}^{\ssbullet})$}

\noindent whenever $\familyiI{a_i}\in\prod_{i\in I}\frak A_i$, so
$\pi$ is measure-preserving.

\medskip

{\bf (b)} is now just a matter of writing the defining formulae out.
}%end of proof of 328D

\leader{328E}{Proposition}\dvAformerly{3{}28D}
Let $I$ be a non-empty set, $\le$ a reflexive
transitive relation on $I$, and $\Cal F$ an ultrafilter on $I$ such that
$\{j:j\in I$, $j\ge i\}$ belongs to $\Cal F$ for every $i\in I$.
Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a family of probability algebras,
and suppose
that we are given a family $\langle\pi_{ji}\rangle_{i\le j}$  such that

\inset{$\pi_{ji}$ is a measure-preserving Boolean
homomorphism from $\frak A_i$ to $\frak A_j$ whenever $i\le j$ in $I$,

$\pi_{ki}=\pi_{kj}\pi_{ji}$ whenever $i\le j\le k$ in $I$.}

\noindent Let $(\frak A,\bar\mu)$ be the probability algebra reduced
product $\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$.

(a) For each $i\in I$ we have a measure-preserving Boolean
homomorphism $\pi_i:\frak A_i\to\frak A$ defined by saying that
$\pi_ia=\family{j}{I}{a_j}^{\ssbullet}$ whenever $a_j=\pi_{ji}a$ for every
$j\ge i$, and $\pi_i=\pi_j\pi_{ji}$ whenever $i\le j$ in $I$.

(b) $\familyiI{a_i}^{\ssbullet}\Bsubseteq\sup_{j\in A}\pi_ja_j$
whenever $\familyiI{a_i}\in\prod_{i\in I}\frak A_i$ and $A\in\Cal F$.

\proof{{\bf (a)} $\pi_i$ is well-defined because $\{j:j\ge i\}\in\Cal F$.
It is a measure-preserving Boolean homomorphism because every $\pi_{ji}$
is.   If $i\le j$ in $I$, $a\in\frak A_i$ and $a_k=\pi_{ki}a$ for every
$k\ge i$, then $a_k=\pi_{kj}\pi_{ji}a$ for every $k\ge j$, so
$\pi_j\pi_{ji}a=\family{k}{I}{a_k}^{\ssbullet}=\pi_ia$;  as $a$ is
arbitrary, $\pi_j\pi_{ji}=\pi_i$.

\medskip

{\bf (b)} Set $c=\sup_{j\in A}\pi_ja_j$ in $\frak A$.
For any $\epsilon>0$, there is a finite $K\subseteq A$ such that
$\bar\mu c\le\epsilon+\bar\mu(\sup_{j\in K}\pi_ja_j)$ (321C).
The set $B=\{k:k\in I$, $j\le k$ for every $j\in K\}$ belongs to $\Cal F$,
so is not empty;  fix $k\in B$, and set
$b=\sup_{j\in K}\pi_{kj}a_j\in\frak A_k$,

$$\eqalign{b_i
&=\pi_{ik}b\text{ if }i\ge k,\cr
&=0\text{ for other }i\in I.\cr}$$

\noindent Then

\Centerline{$\familyiI{b_i}^{\ssbullet}
=\pi_kb
=\pi_k(\sup_{j\in K}\pi_{kj}a_j)
=\sup_{j\in K}\pi_k\pi_{kj}a_j
=\sup_{j\in K}\pi_ja_j\Bsubseteq c$.}

\noindent If $i\in A$ and $i\ge k$, then

$$\eqalign{\bar\mu_i(a_i\Bsetminus b_i)
&=\bar\mu(\pi_ia_i\Bsetminus\pi_ib_i)
=\bar\mu(\pi_ia_i\Bsetminus\pi_i\pi_{ik}b)\cr
&=\bar\mu(\pi_ia_i\Bsetminus\pi_kb)
=\bar\mu(\pi_ia_i\Bsetminus\sup_{j\in K}\pi_ja_j)
\le\bar\mu(c\Bsetminus\sup_{j\in K}\pi_ja_j)
\le\epsilon\cr}$$

\noindent by the choice of $K$.   So

$$\eqalign{\bar\mu(\familyiI{a_i}^{\ssbullet}\Bsetminus c)
&\le\bar\mu(\familyiI{a_i}^{\ssbullet}\Bsetminus\familyiI{b_i}^{\ssbullet})
=\bar\mu(\familyiI{a_i\Bsetminus b_i}^{\ssbullet})\cr
&=\lim_{i\to\Cal F}\bar\mu_i(a_i\Bsetminus b_i)
\le\sup_{i\in A,i\ge k}\bar\mu_i(a_i\Bsetminus b_i)
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\familyiI{a_i}^{\ssbullet}\Bsubseteq c$.
}%end of proof of 328E

\leader{328F}{Corollary} Suppose that $\sequencen{(\frak A_n,\bar\mu_n)}$
is a sequence of
probability algebras, $\phi_n:\frak A_n\to\frak A_{n+1}$ is a
measure-preserving Boolean homomorphism for each $n$ and $\Cal F$ is a
non-principal ultrafilter on $\Bbb N$.   Let
$(\frak A,\bar\mu)$ be the probability algebra reduced product
$\prod_{n\in\Bbb N}(\frak A_n,\bar\mu_n)|\Cal F$.   Then we have canonical
measure-preserving Boolean homomorphisms $\pi_n:\frak A_n\to\frak A$ such
that
$\sequencen{a_n}^{\ssbullet}\Bsubseteq\sup_{n\in A}\pi_na_n$ whenever
$\sequencen{a_n}\in\prod_{n\in\Bbb N}\frak A_n$ and $A\in\Cal F$,
and $\pi_{n+1}\phi_n=\pi_n$ for every $n\in\Bbb N$.

\proof{ Apply 328E with $\pi_{ji}=\phi_{j-1}\ldots\phi_{i+1}\phi_i$
whenever $i<j$.
}%end of proof of 328F

\leader{328G}{Corollary}
Let $(\frak B,\bar\nu)$ be a probability algebra, $I$ a non-empty set, and
$\Cal F$ an ultrafilter on $I$.   Let $(\frak A,\bar\mu)$ be the
probability algebra reduced power $(\frak B,\bar\nu)^I|\Cal F$.

(a) We have a measure-preserving Boolean homomorphism
$\pi:\frak B\to\frak A$ defined by saying that
$\pi b=\familyiI{b}^{\ssbullet}$ for $b\in\frak B$.

(b)

\Centerline{$\familyiI{b_i}^{\ssbullet}
\Bsubseteq\sup_{j\in A}\pi b_j=\pi(\sup_{j\in A}b_j)$}

\noindent whenever $A\in\Cal F$ and $\familyiI{b_i}\in\frak B^I$.

\proof{ Apply 328E with $\le\mskip5mu=I\times I$ and
$\pi_{ji}$ the identity
operator on $\frak B$ for all $i$, $j\in I$.
}%end of proof of 328G

\leader{328H}{Proposition} Let $(I,\le)$ be an upwards-directed partially
ordered set, and $\familyiI{(\frak A_i,\bar\mu_i)}$
a family of probability algebras;  suppose
that $\pi_{ji}:\frak A_i\to\frak A_j$ is a measure-preserving
Boolean homomorphism whenever $i\le j$, and that
$\pi_{ki}=\pi_{kj}\pi_{ji}$ whenever $i\le j\le k$.
Then there are a probability algebra $(\frak C,\bar\lambda)$ and a family
$\familyiI{\pi_i}$ such that

\inset{$\pi_i:\frak A_i\to\frak C$ is a measure-preserving
Boolean homomorphism for each $i\in I$,

$\pi_i=\pi_j\pi_{ji}$ whenever $i\le j$,

$\{0,1\}\cup\bigcup_{i\in I}\pi_i[\frak A_i]$ is topologically dense in
$\frak C$,}

\noindent and whenever $(\frak B,\bar\nu)$, $\familyiI{\phi_i}$ are such
that

\inset{$(\frak B,\bar\nu)$ is a probability algebra,

$\phi_i:\frak A_i\to\frak B$ is a measure-preserving
Boolean homomorphism for each $i\in I$,

$\phi_i=\phi_j\pi_{ji}$ whenever $i\le j$,}

\noindent then there is a unique measure-preserving Boolean homomorphism
$\phi:\frak C\to\frak B$ such that $\phi\pi_i=\phi_i$ for every $i\in I$.

\proof{{\bf (a)}
If $I$ is empty the result is trivial (take $\frak C=\{0,1\}$);  so
let us suppose henceforth that $I\ne\emptyset$.   In this case,

\Centerline{$\{A:A\subseteq I$, there is some $i\in I$ such that $j\in A$
whenever $i\le j\}$}

\noindent is a filter on $I$, and is included in an ultrafilter $\Cal F$
say (2A1O).   Let $(\frak A,\bar\mu)$ be the reduced product
$\prod_{i\in I}(\frak A_i,\bar\mu_i)|\Cal F$.   Then we have for each
$i\in I$ a measure-preserving Boolean homomorphism
$\pi_i:\frak A_i\to\frak A$ such that
$\pi_i=\pi_j\pi_{ji}$ whenever $i\le j$ (328E).
If $i\le j$ in $I$, then
$\pi_i[\frak A_i]\subseteq\pi_j[\frak A_j]$;  because $(I,\le)$
is upwards-directed,
$\familyiI{\pi_i[\frak A_i]}$ is an upwards-directed family of
subalgebras of $\frak A$, and
$\frak D=\bigcup_{i\in I}\pi_i[\frak A_i]$ is a subalgebra of
$\frak A$;  let $\frak C$ be its closure (323J).   Set
$\bar\lambda=\bar\mu\restrp\frak C$, so that $(\frak C,\bar\lambda)$ is a
probability algebra, and $\pi_i:\frak A_i\to\frak C$ is a
measure-preserving Boolean homomorphism for each $i\in I$, with
$\pi_i=\pi_j\pi_{ji}$ whenever $i\le j$.

\medskip

{\bf (b)}
Now suppose that $\frak B$ and $\familyiI{\phi_i}$ are as declared.

\medskip

\quad{\bf (i)} Set

\Centerline{$\phi'=\{(\pi_ia,\phi_ia):i\in I$, $a\in\frak A_i\}
\subseteq\frak D\times\frak B$.}

\noindent Then $\phi'$ is (the graph of) a function from $\frak D$ to
$\frak B$.
\Prf\ If $c\in\frak D$, there is surely an $i\in I$ such that
$c\in\pi_i[\frak A_i]$, so that $(c,\phi_ia)\in\phi'$ for some
$a\in\frak A_i$.
If $(c,b)$ and $(c,b')$ belong to $\phi'$, there are $i$, $j\in I$ and
$a\in\frak A_i$, $a'\in\frak A_j$ such that

\Centerline{$\pi_ia=\pi_ja'=c$,
\quad$\phi_ia=b$,
\quad$\phi_ja'=b'$.}

\noindent Let $k\in I$ be such that $i\le k$ and $j\le k$;  then

\Centerline{$\pi_k\pi_{ki}a=\pi_ia=c=\pi_ja'=\pi_k\pi_{kj}a'$.}

\noindent As $\pi_k$ is measure-preserving, therefore injective,
$\pi_{ki}a=\pi_{kj}a'$, and

\Centerline{$b=\phi_ia=\phi_k\pi_{ki}a=\phi_k\pi_{kj}a'=\phi_ja'=b'$.}

\noindent So each element of $\frak D$ is the first member of exactly one
element of $\phi'$, and $\phi'$ is the graph of a function.\ \QeD\   Of
course the defining formula for $\phi'$ guarantees that
$\phi'\pi_i=\phi_i:\frak A_i\to\frak B$ for every $i\in I$.

\medskip

\quad{\bf (ii)} Next, $\phi':\frak D\to\frak B$
is a measure-preserving Boolean homomorphism.   \Prf\ If
$c$, $c'\in\frak D$ then there are $i$, $j\in I$ and $a\in\frak A_i$,
$a'\in\frak A_j$ such that $c=\pi_ia$ and $c'=\pi_ja'$.   Again take
$k\in I$ such that $i\le k$ and $j\le k$;  then

\Centerline{$c=\pi_k\pi_{ki}a$,
\quad$c'=\pi_k\pi_{kj}a'$,
\quad$\phi'c=\phi_k\pi_{ki}a$,
\quad$\phi'c'=\phi_k\pi_{kj}a'$.}

\noindent In this case, for either of the Boolean operations
$\star=\Bsymmdiff$ or $\star=\Bcap$, we have

$$\eqalign{\phi'c\star\phi'c'
&=\phi_k\pi_{ki}a\star\phi_k\pi_{kj}a'
=\phi_k(\pi_{ki}a\star\pi_{kj}a')\cr
&=\phi'\pi_k(\pi_{ki}a\star\pi_{kj}a')
=\phi'(\pi_k\pi_{ki}a\star\pi_k\pi_{kj}a')
=\phi'(c\star c').\cr}$$

\noindent As $c$, $c'$ and $\star$ are arbitrary,
$\phi'$ is a ring homomorphism.   Moreover, in the same context,

\Centerline{$\bar\nu\phi'c=\bar\nu\phi_ia=\bar\mu_ia
=\bar\mu\pi_ia=\bar\lambda c$,}

\noindent so $\phi'$ is measure-preserving.
It follows that $\phi'1_{\frak C}=1_{\frak B}$,
and $\phi'$ is a Boolean homomorphism.\ \Qed

\medskip

\quad{\bf (iii)}
By 324O, there is a unique extension of $\phi'$ to a measure-preserving
Boolean homomorphism $\phi:\frak C\to\frak B$;  and of course we still have
$\phi\pi_i=\phi_i$ for every $i\in I$.

\medskip

\quad{\bf (iv)}
To see that $\phi$ is unique, take any measure-preserving Boolean
homomorphism $\tilde\phi:\frak C\to\frak B$ such that
$\tilde\phi\pi_i=\phi_i$ for every $i$.   Then $\tilde\phi$ must agree with
$\phi$ on $\pi_i[\frak A_i]$ for every $i$, so
$\tilde\phi\restrp\frak D=\phi\restrp\frak D$;  as $\frak D$ is
topologically dense in $\frak C$, $\tilde\phi=\phi$ (324O again).
}%end of proof of 328H

\leader{328I}{}\cmmnt{ For completeness, I spell out the relatively
elementary construction for projective limits.

\medskip

\noindent}{\bf Proposition} Let $(I,\le)$ be a non-empty
upwards-directed set, and $\familyiI{(\frak A_i,\bar\mu_i)}$
a family of probability algebras;
suppose that $\pi_{ij}:\frak A_j\to\frak A_i$ is a measure-preserving
Boolean homomorphism for $i\le j$ in $I$, and that
$\pi_{ij}\pi_{jk}=\pi_{ik}$ whenever $i\le j\le k$.
Then
there are a probability algebra $(\frak C,\bar\lambda)$ and a family
$\familyiI{\pi_i}$ such that

\inset{$\pi_i:\frak C\to\frak A_i$ is a measure-preserving
Boolean homomorphism for each $i\in I$,

$\pi_i=\pi_{ij}\pi_j$ whenever $i\le j$,}

\noindent and whenever $(\frak B,\bar\nu)$, $\familyiI{\phi_i}$
are such that

\inset{$(\frak B,\bar\nu)$ is a probability algebra,

$\phi_i:\frak B\to\frak A_i$ is a measure-preserving
Boolean homomorphism for each $i\in I$,

$\phi_i=\pi_{ij}\phi_j$ whenever $i\le j$,}

\noindent then there is a unique measure-preserving Boolean homomorphism
$\phi:\frak B\to\frak C$ such that $\pi_i\phi=\phi_i$ for every $i\in I$.

\proof{{\bf (a)} Let $\frak C\subseteq\prod_{i\in I}\frak A_i$ be the set

\Centerline{$\{\familyiI{a_i}:\pi_{ij}a(j)=a(i)$ whenever $i\le j$ in
$I\}$.}

\noindent Because every $\pi_{ij}$ is a Boolean homomorphism, $\frak C$ is
a subalgebra of $\prod_{i\in I}\frak A_i$;  taking
$\pi_j(\familyiI{a_i})=a_j$ whenever $\familyiI{a_i}\in\frak C$,
$\pi_j:\frak C\to\frak A_j$ is a Boolean homomorphism for every $j\in I$,
and $\pi_i=\pi_{ij}\pi_j$ whenever $i\le j$.

Because every $\pi_{ij}$ is order-continuous, $\frak C$ is an order-closed
subalgebra of $\prod_{i\in I}\frak A_i$, so is Dedekind complete.

\medskip

{\bf (b)} If $c=\familyiI{a_i}\in\frak C$, then

\Centerline{$\bar\mu_i\pi_ic=\bar\mu_ia_i=\bar\mu_i\pi_{ij}a_j
=\bar\mu_ja_j=\bar\mu_j\pi_jc$}

\noindent whenever $i\le j$ in $I$;  because $I$ is upwards-directed,
$\bar\mu_i\pi_ic=\bar\mu_j\pi_jc$ for all $i$, $j\in I$.   So we have a
functional $\bar\lambda:\frak C\to[0,1]$ defined by setting
$\bar\lambda c=\bar\mu_i\pi_ic$ whenever $c\in\frak C$ and $i\in I$.
Note that $1_{\frak C}=\familyiI{1_{\frak A_i}}$, so
$\bar\lambda 1_{\frak C}=\bar\mu_i1_{\frak A_i}=1$, for any $i\in I$.

If $\sequencen{c_n}$ is a disjoint sequence in $\frak C$ with supremum $c$,
then express each $c_n$ as $\familyiI{a_{ni}}$;  we must have
$c=\familyiI{\sup_{n\in\Bbb N}a_{ni}}$, so

\Centerline{$\bar\lambda c
=\bar\mu_i(\sup_{n\in\Bbb N}a_{ni})
=\sum_{n=0}^{\infty}\bar\mu_ia_{ni}
=\sum_{n=0}^{\infty}\bar\lambda c_n$}

\noindent for any $i\in I$.   Thus $\bar\lambda$ is countably additive.
If $c\in\frak C$ is non-zero, express it as $\familyiI{a_i}$;  there must
be an $i\in I$ such that $a_i\ne 0$, so that
$\bar\lambda c=\bar\mu_ia_i>0$.   Thus $\bar\lambda$ is strictly positive,
and $(\frak C,\bar\lambda)$ is a probability algebra.

\medskip

{\bf (c)} If $(\frak B,\bar\nu)$ is a probability algebra and
$\familyiI{\phi_i}$ is a family such that $\phi_i:\frak B\to\frak A_i$ is a
measure-preserving Boolean homomorphism and $\phi_i=\pi_{ij}\phi_j$
whenever $i\le j$ in $I$, set $\phi b=\familyiI{\phi_ib}$ for
$b\in\frak B$.   Then $\phi:\frak B\to\prod_{i\in I}\frak A_i$ is a Boolean
homomorphism;  also

\Centerline{$\pi_{ij}(\phi b)(j)=\pi_{ij}\phi_jb=\phi_ib=(\phi b)(i)$}

\noindent whenever $i\le j$ and $b\in\frak B$, so
$\phi[\frak B]\subseteq\frak C$, while $\pi_i\phi=\phi_i$ for every
$i\in I$.   And of course this uniquely determines $\phi$.   To see that
$\phi$ is measure-preserving, we have only to check that

\Centerline{$\bar\lambda\phi b
=\bar\mu_i\pi_i\phi b=\bar\mu_i\phi_ib=\bar\nu b$}

\noindent whenever $b\in\frak B$ and $i\in I$.
}%end of proof of 328I

\leader{328J}{}\cmmnt{ A different application of the method in 328A yields
the following result on commuting families of Boolean homomorphisms.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\Phi$
a family of measure-preserving Boolean homomorphisms from $\frak A$ to
itself such that $\phi\psi=\psi\phi$ for all $\phi$, $\psi\in\Phi$.   Then
there are a probability algebra $(\frak C,\bar\lambda)$, a
measure-preserving Boolean homomorphism $\pi:\frak A\to\frak C$
and a family $\family{\phi}{\Phi}{\tilde\phi}$ such that

(i) $\tilde\phi:\frak C\to\frak C$ is a measure-preserving Boolean
automorphism and $\tilde\phi\pi=\pi\phi$ for every $\phi\in\Phi$;

(ii) $(\phi\psi)\ssptilde=\tilde\phi\tilde\psi$ for all $\phi$,
$\psi\in\Phi$.

\proof{{\bf (a)} Let $\Psi$ be the set of all products
$\phi_0\phi_1\ldots\phi_n$ where $\phi_i\in\Phi\cup\{\iota\}$ for every
$i\le n$, $\iota$ here being the identity map from $\frak A$ to itself.
Then $\Psi$ is a family of measure-preserving Boolean homomorphisms from
$\frak A$ to itself, and $\phi\psi=\psi\phi\in\Psi$
for all $\phi$, $\psi\in\Psi$.

\medskip

{\bf (b)} For $\phi$, $\psi\in\Psi$, say that $\phi\le\psi$ if there is a
$\theta\in\Psi$ such that $\phi\theta=\psi$.   Then $\le$ is a reflexive
transitive relation on $\Psi$.   Note that if $\phi\le\psi$ in $\Psi$
then there is exactly one $\theta\in\Psi$ such that $\phi\theta=\psi$,
because $\phi$ is injective.   So we may define $\pi_{\psi,\phi}\in\Psi$
by saying that $\phi\pi_{\psi,\phi}=\psi$ whenever $\phi\le\psi$ in
$\Psi$;  that is,
$\pi_{\phi\psi,\phi}=\psi$ whenever $\phi$, $\psi\in\Psi$.
Observe that if $\phi\le\psi\le\theta$ in $\Psi$, then

\Centerline{$\phi\pi_{\psi,\phi}\pi_{\theta,\psi}
=\psi\pi_{\theta,\psi}=\theta=\phi\pi_{\theta,\phi}$,}

\noindent so

\Centerline{$\pi_{\theta,\phi}=\pi_{\psi,\phi}\pi_{\theta,\psi}
=\pi_{\theta,\psi}\pi_{\psi,\phi}$.}

\noindent Of course $\iota\le\phi$ for every
$\phi\in\Psi$.

\medskip

{\bf (c)} If $\phi_1$, $\phi_2\in\Psi$ then $\phi_1\le\phi_1\phi_2$ and
$\phi_2\le\phi_2\phi_1=\phi_1\phi_2$;  generally, if $D\subseteq\Psi$ is
finite, there is a $\psi\in\Psi$ such that $\phi\le\psi$ for every
$\phi\in D$.   Consequently

\Centerline{$\{A:A\subseteq\Psi$, there is some $\phi\in\Psi$
such that $\psi\in A$ whenever $\phi\le\psi\}$}

\noindent is a filter on $\Psi$, and is included in an ultrafilter
$\Cal F$ say.   Let $(\frak C_0,\bar\lambda_0)$ be the
probability algebra reduced power $(\frak A,\bar\mu)^{\Psi}|\Cal F$.
By 328E, we have for each $\phi\in\Psi$ a measure-preserving Boolean
homomorphism $\pi_{\phi}:\frak A\to\frak C_0$ defined by saying that
$\pi_{\phi}a=\family{\psi}{\Psi}{a_{\psi}}^{\ssbullet}$ if
$a_{\psi}=\pi_{\psi,\phi}a$ whenever $\phi\le\psi$ in $\Psi$, and
$\pi_{\phi}=\pi_{\psi}\pi_{\psi,\phi}$ whenever $\phi\le\psi$.
Re-interpreting this in terms of the definitions of $\le$ and
$\pi_{\psi,\phi}$, we have $\pi_{\phi}=\pi_{\phi\psi}\psi$ whenever
$\phi$, $\psi\in\Psi$.

\medskip

{\bf (d)} If $\phi$, $\psi$ in $\Psi$, then

\Centerline{$\pi_{\phi}[\frak A]\cup\pi_{\psi}[\frak A]
=\pi_{\phi\psi}[\psi[\frak A]]\cup\pi_{\psi\phi}[\phi[\frak A]]
\subseteq\pi_{\phi\psi}[\frak A]\cup\pi_{\psi\phi}[\frak A]
=\pi_{\phi\psi}[\frak A]$,}

\noindent which is a subalgebra of $\frak C_0$.
So $\frak D=\bigcup_{\phi\in\Psi}\pi_{\phi}[\frak A]$ is a
subalgebra of
$\frak C_0$, and its closure $\frak C$ is a closed subalgebra of
$\frak C_0$;  set $\bar\lambda=\bar\lambda_0\restrp\frak C$.   Then
$\pi=\pi_{\iota}:\frak A\to\frak C$ is a measure-preserving Boolean
homomorphism.

\medskip

{\bf (e)} If $\theta\in\Psi$, we have a measure-preserving Boolean
homomorphism $\hat\theta:\frak C\to\frak C$ defined by the formula

\Centerline{$\hat\theta(\family{\psi}{\Psi}{a_{\psi}}^{\ssbullet})
=\family{\psi}{\Psi}{\theta a_{\psi}}^{\ssbullet}$}

\noindent for every family
$\family{\psi}{\Psi}{a_{\psi}}$ in $\frak A$ (328Da);  and
%$(\theta\phi)\sphat=\hat\theta\hat\phi$
%$(\theta\phi)\mskip-3mu\hat{\phantom{x}}=\hat\theta\hat\phi$
$\widehat{\theta\phi}=\hat\theta\hat\phi$
for all $\theta$, $\phi\in\Psi$
(328Db).   Also $\hat\theta\pi_{\phi}=\pi_{\phi}\theta$ for every $\phi$,
$\theta\in\Psi$.   \Prf\ Let $a\in\frak A$.   Define
$\family{\psi}{\Psi}{a_{\psi}}$,
$\family{\psi}{\Psi}{a'_{\psi}}$ by setting

$$\eqalign{a_{\psi}
&=\pi_{\psi,\phi}a\text{ when }\phi\le\psi,\cr
&=0\text{ otherwise},\cr
a'_{\psi}&=\pi_{\psi,\phi}\theta a=\theta\pi_{\psi,\phi}a
   \text{ when }\phi\le\psi,\cr
&=0\text{ otherwise}.}$$

\noindent Then

\Centerline{$\pi_{\phi}a=\family{\psi}{\Psi}{a_{\psi}}^{\ssbullet}$,}

\Centerline{$\hat\theta\pi_{\phi}a
=\family{\psi}{\Psi}{\theta a_{\psi}}^{\ssbullet}
=\family{\psi}{\Psi}{a'_{\psi}}^{\ssbullet}
=\pi_{\phi}\theta a$.   \Qed}

\medskip

{\bf (f)} It follows that, for $\theta\in\Psi$,

\Centerline{$\hat\theta[\frak D]
=\bigcup_{\phi\in\Psi}\hat\theta[\pi_{\phi}[\frak A]]
=\bigcup_{\phi\in\Psi}\pi_{\phi}[\theta[\frak A]]
\subseteq\frak D$.}

\noindent But in fact $\hat\theta[\frak D]=\frak D$.   \Prf\ If
$d\in\frak D$, there are $\phi\in\Psi$ and $a\in\frak A$ such that
$\pi_{\phi}a=d$.   Now define

$$\eqalign{a_{\psi}
&=\pi_{\psi,\phi}a\text{ if }\phi\le\psi,\cr
&=0\text{ for other }\psi\in\Psi,\cr
a'_{\psi}
&=\pi_{\psi,\phi\theta}a\text{ if }\phi\theta\le\psi,\cr
&=0\text{ for other }\psi\in\Psi,\cr
d'
&=\pi_{\phi\theta}a=\family{\psi}{\Psi}{a'_{\psi}}^{\ssbullet}.\cr
}$$

\noindent In this case, if $\phi\theta\le\psi$,

\Centerline{$\phi\theta a'_{\psi}=\psi a$,
\quad$\phi a_{\psi}=\psi a$}

\noindent so $\theta a'_{\psi}=a_{\psi}$.   Consequently

$$\eqalignno{\hat\theta d'
&=\hat\theta(\family{\psi}{\Psi}{a'_{\psi}}^{\ssbullet})
=\family{\psi}{\Psi}{\theta a'_{\psi}}^{\ssbullet}
=\family{\psi}{\Psi}{a_{\psi}}^{\ssbullet}\cr
\displaycause{because $\{\psi:\phi\theta\le\psi\}\in\Cal F$}
&=d,\cr}$$

\noindent and $d=\hat\theta\pi_{\phi\theta}a\in\hat\theta[\frak D]$.\ \Qed

\medskip

{\bf (g)} Since $\hat\theta[\frak C]$ is a closed subalgebra of $\frak C_0$
(324Kb) in which $\hat\theta[\frak D]=\frak D$ is topologically dense
(3A3Eb), $\hat\theta[\frak C]=\frak C$.   Setting
$\tilde\theta=\hat\theta\restrp\frak C$, we see that
$\tilde\theta:\frak C\to\frak C$ is a surjective measure-preserving Boolean
homomorphism, so is a Boolean automorphism.   Since
%$(\phi\theta)\sphat=\hat\phi\hat\theta$,
%$(\phi\theta)\mskip-3mu\hat{\phantom{x}}=\hat\phi\hat\theta$,
$\widehat{\phi\theta}=\hat\phi\hat\theta$,
we have $(\phi\theta)\ssptilde=\tilde\phi\tilde\theta$ for all $\phi$,
$\theta\in\Psi$.

\medskip

{\bf (h)} Finally, as observed at the beginning of (e),

\Centerline{$\tilde\theta\pi
=\tilde\theta\pi_{\iota}
=\hat\theta\pi_{\iota}
=\pi_{\iota}\theta
=\pi\theta$}

\noindent for every $\theta\in\Psi$.   So
$(\frak C,\bar\lambda,\pi,\family{\theta}{\Phi}{\tilde\theta})$
has the required properties.
}%end of proof of 328J

\exercises{\leader{328X}{Basic exercises (a)}
%\spheader 328Xa
Write out a version of the proof of 328J adapted to the
case in which $\Phi=\{\phi\}$ is a singleton.   (This is an
abstract version of a construction known as the `natural extension' of
an \imp\ function;  see {\smc Petersen 83}, 1.3G.)
%328J

\spheader 328Xb Let $\nu_{\Bbb N}$ be the usual measure on
$X=\{0,1\}^{\Bbb N}$, and $(\frak B_{\Bbb N},\bar\nu_{\Bbb N})$
its measure algebra.
(i) Find \imp\ functions $f$, $g:X\to X$ such that $gf=g$ but
$f(x)\ne x$ for every $x\in X$.  \Hint{try
$g(x)(n)=x(n+1)$.}   (ii) Find measure-preserving Boolean homomorphisms
$\phi$, $\psi:\frak B_{\Bbb N}\to\frak B_{\Bbb N}$
such that $\phi\psi=\psi$ but $\phi$ is
not the identity.   (iii) In 328J, show that the hypothesis that members of
$\Phi$ commute cannot be omitted.
%328J

\spheader 328Xc\dvAnew{2014}
Let $(\frak A,\bar\mu)$ be a purely atomic probability
algebra, $I$ a non-empty set and $\Cal F$ an ultrafilter on $I$.   Show
that $(\frak A,\bar\mu)^I|\Cal F$ is isomorphic to $(\frak A,\bar\mu)$.
%328C out of order query

%\leader{328Y}{Further exercises (a)}
}%end of exercises

\endnotes{
\Notesheader{328} I have starred this section because it is far from the
main line of argument of the volume, and most readers should be moving on
to Maharam's theorem and the Lifting Theorem.   However the results here,
while natural enough, have some features which demand a little attention,
and it will be useful to be able to call on exact formulations of the
ideas.

The proof of 328H begins by taking an ultrafilter on $I$.   This ought to
ring bells.   It should be clear from the statement of the proposition that
$(\frak C,\bar\lambda,\familyiI{\pi_i})$ is determined up to isomorphism by
the properties declared here.   It cannot therefore depend on
which ultrafilter we pick, and there ought to be a construction not relying
on this approach (and, we can hope, not demanding any
application of the axiom of choice).   This is indeed the case, and
in 392Yd below I will sketch a method which can be adapted to give such a
proof.   Yet another proof of 328H is proposed in 418Yn\formerly{4{}18Yp} 
in Volume 4.

The same remarks apply to the proof of 328J.   In the result as stated, I
have not imposed conditions on the structure
$(\frak C,\bar\lambda,\pi,\family{\phi}{\Phi}{\tilde\phi})$
sufficient to define it uniquely, but once again
it is not necessary to employ an ultrafilter, and in fact the filter

\Centerline{$\{A:A\subseteq\Psi$, there is some $\phi\in\Psi$
such that $\psi\in A$ whenever $\phi\le\psi\}$}

\noindent is already enough, if we take the trouble to move to the right
subalgebra of $\frak A^{\Psi}$ before taking the quotient algebra.
}%end of notes

\discrpage
%\ifresultsonly\else
\frfilename{mt32conc.tex}
\versiondate{24.4.06}
\copyrightdate{2006}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{Measure algebras}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my
knowledge) appeared in print in references to this chapter, and which
have since been changed.

\medskip

\leader{3{}22K}{} Paragraphs 3{}22K (simple products of measure algebras), 
3{}22N (the Stone space of a measure algebra) and 3{}22Q (further
properties of Stone spaces), referred to in
the 2003 and 2006 editions of Volume 4, are now 322L, 322O and 322R.

\leader{3{}26E}{Countably additive functionals} Definition 3{}26E, 
referred to in the 2003 and 2006 
editions of Volume 4 and the 2008 edition of Volume 5, is now 326I.
%4{}13K 4{}37C 5{}63A

\leader{3{}26G}{} Corollary 3{}26G, referred to in the 2008 edition of Volume 5, 
is now 326K.
%5{}45F

\leader{3{}26I}{Hahn decomposition} Theorem 3{}26I, referred to in the 2003 and 2006 editions of Volume 4, 
is now 326M.
%4{}37Xb 461

\leader{3{}26K}{Completely additive functionals} The notes in 3{}26K, 
referred to in the 2003 and 2006 editions of Volume 4, 
have been moved to 326O.
%4{}14D

\leader{3{}26Q}{Finitely additive functionals on free products} Theorem 3{}26Q, 
referred to in the 2003 and 2006 editions of Volume 4 and the 2008 edition 
of Volume 5, is now 326E.
%4{}54E 4{}58 564O

\leaveitout{\leader{3{}26Yo}{Convergent sequences of measures} Exercise 3{}26Yo, 
referred to in the 2003 and 2006 editions of
Volume 4, is now 326Yo.
%4{}91Yg
}
 
\leader{3{}28D}{Reduced products of probability algebras} Paragraph 3{}28D,
referred to in the 2008 edition of Volume 5, is now 328E.
 
\frnewpage
%\fi

\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.]}

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Proc.\ Amer.\ Math.\ Soc.\ 136 (2008) 1285-1294.
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  % 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
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%Wendt S IN5 EL24 T IT

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\medskip %L

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\medskip %M

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\medskip %N

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\medskip %O

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\medskip%P

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\medskip %Q%R

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\medskip %S

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%Store A	QA 611.S3    6640 0159 9

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\medskip %T

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\medskip %U%V

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\medskip %W

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\medskip %X%Y%Z

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\frnewpage

}%end of references


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