% The following TeXfile comprises part of

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

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

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

%        %   \amstexsupportedfalse

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

%        \input amstex
%        \documentstyle{amsppt}

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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


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

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

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

   \Centerline{#1}

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

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

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

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

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

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

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

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

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

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

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

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

%End of lines for missing AmSTeX

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

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

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

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

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


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

%hyphenations
\hyphenation{Archi-me-dean}

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

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

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

%Headline assignments

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

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

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

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

\newdimen\pagetotalplus
\newdimen\pagetotalminus

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\Sous{\Cal S\text{ous}}  %used in vol 6
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\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{mt23.tex}
\versiondate{17.11.04}
\copyrightdate{1995}


\def\chaptername{The Radon-Nikod\'ym Theorem}
\newchapter{23}
\def\chaptername{The Radon-Nikod\'ym theorem}

In Chapter 22, I discussed the indefinite integrals of
integrable functions on $\Bbb R$, and gave what I hope you feel are
satisfying descriptions both of
the functions which are indefinite integrals (the absolutely
continuous functions) and of how to find which functions they are
indefinite integrals of (you differentiate them).   For general measure
spaces, we have no structure present which can give such simple
formulations;  but nevertheless the same questions can be asked and, up
to a point, answered.

The first section of this chapter introduces the basic machinery needed,
the concept of `countably additive' functional and its decomposition
into positive and negative parts.   The main theorem takes up the second
section:  indefinite integrals are the `truly continuous' additive
functionals;  on $\sigma$-finite spaces, these are the `absolutely
continuous' countably additive functionals.   In \S233 I discuss the
most important single application of the theorem, its use in providing a
concept of `conditional expectation'.   This is one of the central
concepts of probability theory -- as you very likely know;  but the form
here is a dramatic generalization of the elementary concept of the
conditional probability of one event given another, and needs the whole
strength of the general theory of measure and integration as developed
in Volume 1 and this chapter.   I include some notes on convex
functions, up to and including versions of Jensen's inequality
(233I-233J).

While we are in the area of `pure' measure theory, I take the opportunity 
to discuss some further topics.
I begin with some essentially elementary constructions,
image measures, sums of measures and indefinite-integral measures;
I think the details need a little attention,
and I work through them in \S234.
Rather deeper ideas are needed to deal with `measurable
transformations'.   In \S235 I set out the techniques necessary to provide an
abstract basis for a general method of integration-by-substitution, with
a detailed account of sufficient conditions for a formula of the type

\Centerline{$\int g(y)dy=\int g(\phi(x))J(x)dx$}

\noindent to be valid.

\discrpage
\frfilename{mt231.tex} 
\versiondate{25.8.15} 
\copyrightdate{1994} 
      
\def\chaptername{The Radon-Nikod\'ym theorem} 
\def\sectionname{Countably additive functionals} 
      
\newsection{231} 
      
I begin with an abstract description of the objects which will, in 
appropriate circumstances, correspond to the indefinite integrals of 
general integrable functions.   In this section I give those parts of 
the theory which do not involve a measure, but only a set with a 
distinguished $\sigma$-algebra of subsets.   The basic concepts are 
those of `finitely additive' and `countably additive' functional, and 
there is one substantial theorem, the `Hahn decomposition' (231E). 
      
\leader{231A}{Definition} Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$\cmmnt{ (136E)}.   A functional $\nu:\Sigma\to\Bbb R$ is 
{\bf finitely additive}, or just {\bf additive}, 
if $\nu(E\cup F)=\nu E+\nu F$ whenever $E$, 
$F\in\Sigma$ and $E\cap F=\emptyset$. 
      
\leader{231B}{Elementary facts} Let $X$ be a set, $\Sigma$ an algebra of 
subsets of $X$, and $\nu:\Sigma\to\Bbb R$ a finitely additive 
functional. 
      
\header{231Ba}{\bf (a)} $\nu\emptyset=0$.   \prooflet{(For 
$\nu\emptyset=\nu(\emptyset\cup\emptyset)=\nu\emptyset+\nu\emptyset$.) 
} 
      
\header{231Bb}{\bf (b)} If $E_0,\ldots,E_n$ are disjoint members of 
$\Sigma$ then $\nu(\bigcup_{i\le n}E_i)=\sum_{i=0}^n\nu E_i$. 
      
\header{231Bc}{\bf (c)} If $E$, $F\in\Sigma$ and $E\subseteq F$ then 
$\nu F=\nu E + \nu(F\setminus E)$.   More generally, for any $E$, 
$F\in\Sigma$, 
      
\Centerline{$\nu F=\nu(F\cap E)+\nu(F\setminus E)$,}

\Centerline{$\nu E+\nu F
\prooflet{\mskip5mu=\nu E+\nu(F\setminus E)+\nu(F\cap E)}
=\nu(E\cup F)+\nu(E\cap F)$,}

\Centerline{$\nu E-\nu F 
\prooflet{\mskip5mu = \nu(E\setminus F)+\nu(E\cap F) 
-\nu(E\cap F)-\nu(F\setminus E)} 
=\nu(E\setminus F)-\nu(F\setminus E)$.} 
      
\leader{231C}{Definition} Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$.   A function 
$\nu:\Sigma\to\Bbb R$ is {\bf countably additive} or 
{\bf $\sigma$-additive} if $\sum_{n=0}^{\infty}\nu E_n$ exists in $\Bbb 
R$ and is equal to $\nu(\bigcup_{n\in\Bbb N}E_n)$ for every 
disjoint sequence $\sequencen{E_n}$ in $\Sigma$ such that 
$\bigcup_{n\in\Bbb N}E_n\in\Sigma$. 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} Note that when I use the phrase `countably 
additive functional' I 
mean to exclude the possibility of $\infty$ as a value of the 
functional.   Thus a measure is a countably additive functional iff it 
is totally finite (211C). 
      
You will sometimes see the phrase `{\bf signed measure}' used to mean 
what I call a countably additive functional. 
} 
      
\leader{231D}{Elementary facts} Let $X$ be a set, $\Sigma$ a 
$\sigma$-algebra of subsets of $X$ and $\nu:\Sigma\to\Bbb R$ a countably 
additive functional. 
      
\header{231Da}{\bf (a)} $\nu$ is finitely additive.   \prooflet{\Prf\ 
(i) Setting $E_n=\emptyset$ for every $n\in\Bbb N$, 
$\sum_{n=0}^{\infty}\nu\emptyset$ must be defined in $\Bbb R$ so 
$\nu\emptyset$ must be $0$.   (ii) Now if $E$, $F\in\Sigma$ 
and $E\cap F=\emptyset$ we can set $E_0=E$, $E_1=F$, $E_n=\emptyset$ for 
$n\ge 2$ and get 
      
\Centerline{$\nu(E\cup F)=\nu(\bigcup_{n\in\Bbb N}E_n) 
=\sum_{n=0}^{\infty}\nu E_n=\nu E+\nu F$.  \Qed} 
} 
      
\header{231Db}{\bf (b)} If $\sequencen{E_n}$ is a non-decreasing 
sequence in $\Sigma$, with union $E\in\Sigma$, then 
      
\Centerline{$\nu E 
=\nu E_0+\sum_{n=0}^{\infty}\nu(E_{n+1}\setminus E_n) 
=\lim_{n\to\infty}\nu E_n$.} 
      
\header{231Dc}{\bf (c)} If $\sequencen{E_n}$ is a non-increasing 
sequence in $\Sigma$ with intersection $E\in\Sigma$, then 
      
\Centerline{$\nu E 
\prooflet{\mskip5mu =\nu E_0-\lim_{n\to\infty}\nu(E_0\setminus E_n)} 
=\lim_{n\to\infty}\nu E_n$.} 
      
\header{231Dd}{\bf (d)} If $\nuprime:\Sigma\to\Bbb R$ is another countably 
additive functional, and $c\in\Bbb R$, then $\nu+\nuprime:\Sigma\to\Bbb R$ 
and $c\nu:\Sigma\to\Bbb R$ are countably additive. 
      
\header{231De}{\bf (e)} If $H\in\Sigma$, then 
$\nu_H:\Sigma\to\Bbb R$ is countably additive, where 
$\nu_HE=\nu(E\cap H)$ for every 
$E\in\Sigma$.   \prooflet{\Prf\ If $\sequencen{E_n}$ is a disjoint 
sequence in 
$\Sigma$ with union $E\in\Sigma$ then $\sequencen{E_n\cap H}$ is 
disjoint, with union $E\cap H$, so 
      
\Centerline{$\nu_HE 
=\nu(H\cap E) 
=\nu(\bigcup_{n\in\Bbb N}(H\cap E_n)) 
=\sum_{n=0}^{\infty}\nu(H\cap E_n) 
=\sum_{n=0}^{\infty}\nu_HE_n$.  \Qed}} 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} For the time being, we shall be using the notion 
of `countably additive functional' only on $\sigma$-algebras $\Sigma$, 
in which case we can take it for granted that the unions and 
intersections above belong to $\Sigma$. 
}%end of comment 
      
\leader{231E}{}\cmmnt{ All the ideas above amount to minor 
modifications of 
ideas already needed at the very beginning of the theory of measure 
spaces.   We come now to something more substantial. 
      
\medskip 
      
\noindent}{\bf Theorem} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of 
subsets of $X$, and $\nu:\Sigma\to\Bbb R$ a countably additive 
functional.   Then 
      
(a) $\nu$ is bounded; 
      
(b) there is a set $H\in\Sigma$ such that 
      
\Centerline{$\nu F\ge 0$ whenever 
$F\in\Sigma$ and $F\subseteq H$,} 
      
\Centerline{$\nu F\le 0$ whenever $F\in\Sigma$ and 
$F\cap H=\emptyset$.} 
      
      
\proof{{\bf (a)} \Quer\ Suppose, if possible, otherwise.    For 
$E\in\Sigma$, set $M(E)=\sup\{|\nu F|:F\in\Sigma,\,F\subseteq E\}$; 
then $M(X)=\infty$.   Moreover, whenever $E_1$, $E_2$, $F\in\Sigma$ and 
$F\subseteq E_1\cup E_2$, then 
      
\Centerline{$|\nu F|=|\nu(F\cap E_1)+\nu(F\setminus E_1)| 
\le|\nu(F\cap E_1)|+|\nu(F\setminus E_1)|\le M(E_1)+M(E_2)$,} 
      
\noindent so $M(E_1\cup E_2)\le 
M(E_1)+M(E_2)$.   Choose a sequence $\sequencen{E_n}$ in $\Sigma$ as 
follows.   $E_0=X$.   Given that $M(E_n)=\infty$, where $n\in\Bbb N$, 
then surely there is an $F_n\subseteq E_n$ such that 
$|\nu F_n|\ge 1+|\nu E_n|$, in which case 
$|\nu(E_n\setminus F_n)|\ge 1$.   Now 
at least one of $M(F_n)$, $M(E_n\setminus F_n)$ is infinite;  if 
$M(F_n)=\infty$, set $E_{n+1}=F_n$;  otherwise, set 
$E_{n+1}=E_n\setminus F_n$;  in either case, note that 
$|\nu(E_n\setminus E_{n+1})|\ge 1$ and $M(E_{n+1})=\infty$, so that the 
induction will continue. 
      
On completing this induction, set $G_n=E_n\setminus E_{n+1}$ for 
$n\in\Bbb N$.   Then $\sequencen{G_n}$ is a disjoint sequence in 
$\Sigma$, so $\sum_{n=0}^{\infty}\nu G_n$ is defined in $\Bbb R$ and 
$\lim_{n\to\infty}\nu G_n=0$;  but $|\nu G_n|\ge 1$ for every $n$. 
\Bang 
      
\medskip 
      
{\bf (b)(i)} By (a), $\gamma=\sup\{\nu E:E\in\Sigma\}<\infty$.   Choose 
a sequence $\sequencen{E_n}$ in $\Sigma$ such that 
$\nu E_n\ge\gamma-2^{-n}$ for every $n\in\Bbb N$.   For 
$m\le n\in\Bbb N$, set $F_{mn}=\bigcap_{m\le i\le n}E_i$.   Then 
$\nu F_{mn}\ge\gamma-2\cdot 2^{-m}+2^{-n}$ for every $n\ge m$.   \Prf\ 
Induce on $n$.   For $n=m$, this is due to the choice of $E_m=F_{mm}$. 
For the inductive step, we have $F_{m,n+1}=F_{mn}\cap E_{n+1}$, while 
surely $\gamma\ge\nu(E_{n+1}\cup F_{mn})$, so 
      
$$\eqalignno{\gamma+\nu F_{m,n+1} 
&\ge\nu(E_{n+1}\cup F_{mn})+\nu(E_{n+1}\cap F_{mn})\cr&
=\nu E_{n+1}+\nu F_{mn}
\Displaycause{231Bc}
\ge\gamma-2^{-n-1}+\gamma-2\cdot 2^{-m}+2^{-n}
\Displaycause{by the choice of $E_{n+1}$ and the inductive 
hypothesis} 
=2\gamma-2\cdot 2^{-m}+2^{-n-1}.
\cr}$$ 
      
\noindent Subtracting $\gamma$ from both sides, 
$\nu F_{m,n+1}\ge\gamma-2\cdot 2^{-m}+2^{-n-1}$ and the induction 
proceeds. \Qed 
      
\medskip 
      
\quad{\bf (ii)} For $m\in\Bbb N$, set 
      
\Centerline{$F_m=\bigcap_{n\ge m}F_{mn}=\bigcap_{n\ge m}E_n$.} 
      
\noindent Then 
      
\Centerline{$\nu F_m 
=\lim_{n\to\infty}\nu F_{mn}\ge\gamma-2\cdot 2^{-m}$,} 
      
\noindent by 231Dc. 
Next, $\sequence{m}{F_m}$ is non-decreasing, so setting 
$H=\bigcup_{m\in\Bbb N}F_m$ we have 
      
\Centerline{$\nu H=\lim_{m\to\infty}\nu F_m\ge\gamma$;} 
      
\noindent  since $\nu H$ 
is surely less than or equal to $\gamma$, $\nu H=\gamma$. 
      
\medskip

\quad{\bf (iii)} If $F\in\Sigma$ and $F\subseteq H$, then 
      
\Centerline{$\nu H-\nu F=\nu(H\setminus 
F)\le\gamma=\nu H$,} 
      
\noindent so $\nu F\ge 0$.   If $F\in\Sigma$ and $F\cap 
H=\emptyset$ then 
      
\Centerline{$\nu H+\nu F=\nu(H\cup F)\le\gamma=\nu H$} 
      
\noindent so $\nu F\le 
0$.   This completes the proof. 
}%end of proof of 231E 
      
      
\leader{231F}{Corollary} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra 
of subsets of $X$, and $\nu:\Sigma\to\Bbb R$ a countably additive 
functional.   Then $\nu$ can be expressed as the difference of two 
totally finite measures with domain $\Sigma$. 
      
\proof{ Take $H\in\Sigma$ as described in 231Eb.   Set 
$\nu_1 E=\nu(E\cap H)$, $\nu_2 E=-\nu(E\setminus H)$ for $E\in\Sigma$. 
Then, as in 231Dd-e, both $\nu_1$ and $\nu_2$ are countably 
additive functionals on $\Sigma$, and of course $\nu=\nu_1-\nu_2$.   But 
also, by the choice of $H$, both $\nu_1$ and $\nu_2$ are non-negative, 
so are totally finite measures. 
}%end of proof of 231F 
      
\medskip 
      
\noindent{\bf Remark} This is called the `{\bf Jordan decomposition}' 
of 
$\nu$.   The expression of 231Eb is a `{\bf Hahn decomposition}'. 
      
      
      
\exercises{ 
\leader{231X}{Basic exercises (a)} 
%\spheader 231Xa 
Let $\Sigma$ be the family of subsets $A$ of 
$\Bbb N$ such that one of $A$, $\Bbb N\setminus A$ is finite.   Show 
that $\Sigma$ is an algebra of subsets of $\Bbb N$.   (This is the {\bf 
finite-cofinite algebra} of subsets of $\Bbb N$;  compare 211Ra.) 
%231A 
      
\spheader 231Xb Let $X$ be a set, $\Sigma$ an algebra of subsets 
of $X$ and $\nu:\Sigma\to\Bbb R$ a finitely additive functional.  Show 
that $\nu(E\cup F\cup G)+\nu(E\cap F)+\nu(E\cap G)+\nu(F\cap G) 
=\nu E+\nu F+\nu G+\nu(E\cap F\cap G)$ for all $E$, $F$, $G\in\Sigma$. 
Generalize this result to longer sequences of sets. 
%231A 
      
\sqheader 231Xc Let $\Sigma$ be the finite-cofinite algebra of 
subsets of $\Bbb N$, as in 231Xa.   Define $\nu:\Sigma\to\Bbb Z$ by 
setting 
      
\Centerline{$\nu E 
=\lim_{n\to\infty}\bigl(\#(\{i:i\le n,\,2i\in E\}) 
  -\#(\{i:i\le n,\,2i+1\in E\})\bigr)$} 
      
\noindent for every $E\in\Sigma$.   Show that $\nu$ is well-defined and 
finitely additive and unbounded. 
%231Xa, 231A 
      
\spheader 231Xd Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$.   (i) Show that if $\nu:\Sigma\to\Bbb R$ and 
$\nuprime:\Sigma\to\Bbb R$ are finitely additive, so are $\nu+\nuprime$ and 
$c\nu$ for 
any $c\in\Bbb R$.   (ii) Show that if $\nu:\Sigma\to\Bbb R$ is finitely 
additive and $H\in\Sigma$, then $\nu_H$ is finitely additive, where 
$\nu_H(E)=\nu(H\cap E)$ for every $E\in\Sigma$. 
%231A 
      
\spheader 231Xe Let $X$ be a set, $\Sigma$ an algebra of subsets 
of $X$ and $\nu:\Sigma\to\Bbb R$ a finitely additive functional.   Let 
$S$ be the linear space of those real-valued functions on $X$ 
expressible in the form $\sum_{i=0}^na_i\chi E_i$ where $E_i\in\Sigma$ 
for each $i$.   (i) Show that we have a linear functional 
$\dashint:S\to\Bbb R$ given by writing 
      
\Centerline{$\dashint\sum_{i=0}^na_i\chi E_i 
=\sum_{i=0}^na_i\nu E_i$} 
      
\noindent whenever $a_0,\ldots,a_n\in\Bbb R$ and 
$E_0,\ldots,E_n\in\Sigma$.   (ii) Show that if $\nu E\ge 0$ for every 
$E\in\Sigma$ then $\dashint f\ge 0$ whenever $f\in S$ and $f(x)\ge 0$ 
for every $x\in X$.   (iii) Show that if $\nu$ is bounded and 
$X\ne\emptyset$ then 
      
\Centerline{$\sup\{|\dashint f|:f\in S,\,\|f\|_{\infty}\le 1\} 
=\sup_{E,F\in\Sigma}|\nu E-\nu F|$,} 
      
\noindent writing $\|f\|_{\infty}=\sup_{x\in X}|f(x)|$. 
%231B 
      
\sqheader 231Xf Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of 
subsets of $X$ and $\nu:\Sigma\to \Bbb R$ a finitely additive 
functional.   Show that the following are equiveridical: 
      
\quad (i) $\nu$ is countably additive; 
      
\quad (ii) $\lim_{n\to\infty}\nu E_n=0$ whenever $\sequencen{E_n}$ is a 
non-increasing sequence in $\Sigma$ and 
$\bigcap_{n\in\Bbb N}E_n=\emptyset$; 
      
\quad (iii) $\lim_{n\to\infty}\nu E_n=0$ whenever $\sequencen{E_n}$ is a 
sequence in $\Sigma$ and 
$\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m=\emptyset$; 
      
\quad (iv) $\lim_{n\to\infty}\nu E_n=\nu E$ whenever $\sequencen{E_n}$ 
is a sequence in $\Sigma$ and 
      
\Centerline{$E=\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m 
=\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}E_m$.} 
      
\noindent({\it Hint}:  for (i)$\Rightarrow$(iv), consider non-negative 
$\nu$ first.) 
%231D 
      
\spheader 231Xg Let $X$ be a set and $\Sigma$ a $\sigma$-algebra 
of subsets of 
$X$, and let $\nu:\Sigma\to\coint{-\infty,\infty}$ be a function which 
is countably additive in the sense that $\nu\emptyset=0$ and whenever 
$\sequencen{E_n}$ is a disjoint sequence in $\Sigma$, 
$\sum_{n=0}^{\infty}\nu E_n=\lim_{n\to\infty}\sum_{i=0}^n\nu E_i$ is 
defined in $\coint{-\infty,\infty}$ and is 
equal to $\nu(\bigcup_{n\in\Bbb N}E_n)$.   Show that $\nu$ is bounded 
above and attains its upper bound (that is, there is an $H\in\Sigma$ 
such that $\nu H=\sup_{F\in\Sigma}\nu F$).   Hence, or otherwise, show 
that $\nu$ is expressible as the difference of a totally finite measure 
and a measure, both with domain $\Sigma$. 
%231F 
      
\leader{231Y}{Further exercises (a)} Let $X$ be a set, $\Sigma$ an 
algebra of subsets of $X$, and $\nu:\Sigma\to\Bbb R$ a bounded finitely 
additive 
functional.   Set 
      
\Centerline{$\nu^+E=\sup\{\nu F:F\in\Sigma,\,F\subseteq E\}$,} 
      
\Centerline{$\nu^-E=-\inf\{\nu F:F\in\Sigma,\,F\subseteq E\}$,} 
      
\Centerline{$|\nu|E=\sup\{\nu F_1-\nu 
F_2:F_1,\,F_2\in\Sigma,\,F_1,\,F_2\subseteq E\}$.} 
      
\noindent Show that $\nu^+$, $\nu^-$ and $|\nu|$ are all bounded 
finitely additive functionals on $\Sigma$ and that $\nu=\nu^+-\nu^-$, 
$|\nu|=\nu^++\nu^-$.   Show that if $\nu$ is countably additive so are 
$\nu^+$, $\nu^-$ and $|\nu|$.   ($|\nu|$ is sometimes called the {\bf 
variation} of $\nu$.) 
      
\header{231Yb}{\bf (b)} Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$.   Let $\nu_1$, $\nu_2$ be two bounded finitely additive 
functionals defined on $\Sigma$.   Set 
      
\Centerline{$(\nu_1\vee\nu_2)(E) 
=\sup\{\nu_1 F+\nu_2(E\setminus F):F\in\Sigma,\,F\subseteq E\}$,} 
      
\Centerline{$(\nu_1\wedge\nu_2)(E) 
=\inf\{\nu_1 F+\nu_2(E\setminus F):F\in\Sigma,\,F\subseteq E\}$.} 
      
\noindent Show that $\nu_1\vee\nu_2$ and $\nu_1\wedge\nu_2$ are finitely 
additive functionals, and that 
$\nu_1+\nu_2=\nu_1\vee\nu_2+\nu_1\wedge\nu_2$. 
Show that, in the language of 231Ya, 
      
\Centerline{$\nu^+=\nu\vee 0$, 
\quad$\nu^-=(-\nu)\vee 0=-(\nu\wedge 0)$, 
\quad$|\nu|=\nu\vee(-\nu)=\nu^+\vee\nu^-=\nu^++\nu^-$,} 
      
\Centerline{$\nu_1\vee\nu_2=\nu_1+(\nu_2-\nu_1)^+$, 
\quad$\nu_1\wedge\nu_2=\nu_1-(\nu_1-\nu_2)^+$,} 
      
\noindent so that $\nu_1\vee\nu_2$ and $\nu_1\wedge\nu_2$ are countably 
additive if $\nu_1$ and $\nu_2$ are. 
      
\header{231Yc}{\bf (c)} Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$.   Let $M$ be the set of all bounded finitely additive 
functionals 
from $\Sigma$ to $\Bbb R$.   Show that $M$ is a linear space under the 
natural definitions of addition and scalar multiplication.   Show that 
$M$ has a partial order $\le$ defined by saying that 
      
\Centerline{$\nu\le\nuprime$ iff $\nu E\le\nuprime E$ for every $E\in\Sigma$,} 
      
\noindent and that for this partial order $\nu_1\vee\nu_2$, 
$\nu_1\wedge\nu_2$, as defined in 231Yb, are $\sup\{\nu_1,\nu_2\}$, 
$\inf\{\nu_1,\nu_2\}$. 
      
\header{231Yd}{\bf (d)} Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$.   Let $\nu_0,\ldots,\nu_n$ be bounded finitely additive 
functionals on $\Sigma$ and set 
      
\Centerline{$\check\nu 
E=\sup\{\sum_{i=0}^n\nu_iF_i:F_0,\ldots,F_n\in\Sigma,\,\bigcup_{i\le 
n}F_i=E,\,F_i\cap F_j=\emptyset$ for $i\ne j\}$,} 
      
\Centerline{$\hat\nu 
E=\inf\{\sum_{i=0}^n\nu_iF_i:F_0,\ldots,F_n\in\Sigma,\,\bigcup_{i\le 
n}F_i=E,\,F_i\cap F_j=\emptyset$ for $i\ne j\}$} 
      
\noindent for $E\in\Sigma$.   Show that $\check\nu$ and $\hat\nu$ are 
finitely additive and are, respectively, $\sup\{\nu_0,\ldots,\nu_n\}$ 
and 
$\inf\{\nu_0,\ldots,\nu_n\}$ in the partially ordered set of finitely 
additive 
functionals on $\Sigma$. 
      
\header{231Ye}{\bf (e)} Let $X$ be a set and $\Sigma$ a $\sigma$-algebra 
of subsets of $X$;  let $M$ be the partially ordered set of all bounded 
finitely additive functionals from $\Sigma$ to $\Bbb R$.   (i) Show that 
if $A\subseteq M$ is non-empty and bounded above  in $M$, then $A$ has a 
supremum $\check\nu$ in $M$, given by the formula 
      
$$\eqalign{\check\nu E 
=\sup\{\sum_{i=0}^n\nu_iF_i:\nu_0,\ldots,\nu_n\in 
A,\,F_0,&\ldots,F_n\in\Sigma,\,\bigcup_{i\le n}F_i=E,\cr 
&\qquad\quad F_i\cap F_j=\emptyset\text{ for }i\ne j\}.\cr}$$ 
      
\noindent (ii) Show that if $A\subseteq M$ is non-empty and bounded 
below in $M$ then it has an infimum $\hat\nu\in M$, given by the formula 
      
$$\eqalign{\hat\nu E 
=\inf\{\sum_{i=0}^n\nu_iF_i:\nu_0,\ldots,\nu_n\in 
A,\,F_0,&\ldots,F_n\in\Sigma,\,\bigcup_{i\le n}F_i=E,\cr 
&\qquad\quad F_i\cap F_j=\emptyset\text{ for }i\ne j\}.\cr}$$ 
      
\header{231Yf}{\bf (f)} Let $X$ be a set, $\Sigma$ an algebra of subsets 
of $X$, and $\nu:\Sigma\to\Bbb R$ a non-negative finitely additive 
functional.   For $E\in\Sigma$ set 
      
\Centerline{$\nu_{ca}(E)=\inf\{\sup_{n\in\Bbb N}\nu 
F_n:\sequencen{F_n}$ is a non-decreasing sequence in $\Sigma$ with union 
$E\}$.} 
      
\noindent Show that $\nu_{ca}$ is a countably additive functional on 
$\Sigma$ and that if $\nuprime$ is any countably additive functional with 
$\nuprime\le\nu$ then $\nuprime\le\nu_{ca}$.   Show that 
$\nu_{ca}\wedge(\nu-\nu_{ca})=0$. 
      
\header{231Yg}{\bf (g)} Let $X$ be a set, $\Sigma$ an algebra of subsets 
of $X$, and $\nu:\Sigma\to\Bbb R$ a bounded finitely additive 
functional.   Show that $\nu$ is uniquely expressible as 
$\nu_{ca}+\nu_{pfa}$, where $\nu_{ca}$ is countably 
additive, $\nu_{pfa}$ is finitely additive and if 
$0\le\nuprime\le|\nu_{pfa}|$ and $\nuprime$ is countably additive then 
$\nuprime=0$. 
      
\header{231Yh}{\bf (h)} Let $X$ be a set and $\Sigma$ an algebra of 
subsets of $X$.   Let $M$ be the linear space of bounded finitely 
additive functionals on $\Sigma$, and for $\nu\in M$ set 
$\|\nu\|=|\nu|(X)$, defining $|\nu|$ as in 231Ya.   ($\|\nu\|$ is the 
{\bf total variation} of $\nu$.)   Show that $\|\,\|$ 
is a norm on $M$ under which 
$M$ is a Banach space.   Show that the space of bounded countably 
additive functionals on $\Sigma$ is a closed linear subspace of $M$. 
      
\header{231Yi}{\bf (i)} Repeat as many as possible of the results of 
this section for complex-valued functionals. 
}%end of exercises 
      
\endnotes{ 
\Notesheader{231} The real purpose of this section has been to describe 
the Hahn decomposition of a countably additive functional (231E).   The 
leisurely exposition in 231A-231D is intended as a review of the 
most elementary properties of measures, in the slightly more general 
context of `signed measures', with those properties corresponding to 
`additivity' alone separated from those which depend on `countable 
additivity'.   In 231Xf I set out necessary and sufficient conditions 
for a finitely additive functional on a $\sigma$-algebra to be countably 
additive, designed to suggest that a finitely additive functional is 
countably additive iff it is `sequentially order-continuous' in some 
sense.   The fact that a countably additive functional can be expressed 
as the difference of non-negative countably additive functionals (231F) 
has an important counterpart in the theory of finitely additive 
functionals:  a 
finitely additive functional can be expressed as the difference of 
non-negative finitely additive functionals if (and only if) it is 
bounded (231Ya).   But I do not think that this, or the further 
properties of bounded finitely additive functionals described in 231Xe 
and 231Y, will be important to us before Volume 3. 
}%end of notes 
      
      
\discrpage 
      
\frfilename{mt232.tex}
\versiondate{19.5.17}
\copyrightdate{1994}

\def\chaptername{The Radon-Nikod\'ym theorem}
\def\sectionname{The Radon-Nikod\'ym theorem}

\newsection{232}

I come now to the chief theorem of this chapter, one of the central
results of measure theory, relating countably additive functionals to
indefinite integrals.   The objective is to give a complete description
of the functionals which can arise as indefinite integrals of integrable
functions (232E).   These can be characterized as the `truly continuous'
additive functionals (232Ab).   A more commonly used concept, and one
adequate in many cases, is that of `absolutely continuous' additive
functional (232Aa);  I spend the first few paragraphs (232B-232D) on
elementary facts about truly continuous and absolutely continuous
functionals.   I end the section with a discussion of decompositions
of general countably additive functionals (232I).

\leader{232A}{Absolutely continuous functionals} Let $(X,\Sigma,\mu)$ be
a measure space and $\nu:\Sigma\to\Bbb R$ a finitely additive
functional.

\spheader 232Aa $\nu$ is {\bf absolutely continuous} with
respect to $\mu$ (sometimes written `$\nu\ll\mu$') if
for every $\epsilon>0$ there is a $\delta>0$ such that
$|\nu E|\le\epsilon$ whenever  $E\in\Sigma$ and $\mu E\le\delta$.


\spheader 232Ab $\nu$ is {\bf truly continuous} with respect to
$\mu$ if for
every $\epsilon>0$ there are $E\in\Sigma$ and $\delta>0$ such that
$\mu E<\infty$ and $|\nu F|\le\epsilon$ whenever $F\in\Sigma$ and
$\mu(E\cap F)\le\delta$.


\spheader 232Ac\cmmnt{ For reference, I add another definition
here.}   If $\nu$ is countably additive, it is
{\bf singular} with respect to $\mu$
if there is a set $F\in\Sigma$ such that $\mu F=0$ and
$\nu E=0$ whenever $E\in\Sigma$ and $E\subseteq X\setminus F$.


\leader{232B}{Proposition} Let $(X,\Sigma,\mu)$ be a measure
space and $\nu:\Sigma\to\Bbb R$ a finitely additive functional.

(a) If $\nu$ is countably additive, it is absolutely continuous with
respect to $\mu$ iff $\nu E=0$
whenever $\mu E=0$.

(b) $\nu$ is truly continuous with respect to $\mu$ iff ($\alpha$) it is
countably additive ($\beta$) it is absolutely continuous with respect to
$\mu$ ($\gamma$) whenever $E\in\Sigma$ and $\nu E\ne 0$
there is an $F\in\Sigma$ such that $\mu F<\infty$ and
$\nu(E\cap F)\ne 0$.

(c) If $(X,\Sigma,\mu)$ is $\sigma$-finite, then $\nu$ is truly
continuous with respect
to $\mu$ iff it is countably additive and absolutely
continuous with respect to $\mu$.

(d) If $(X,\Sigma,\mu)$ is totally finite, then $\nu$ is truly
continuous with respect to $\mu$ iff it is absolutely continuous with
respect to $\mu$.

\proof{{\bf (a)(i)}  If $\nu$ is absolutely continuous with
respect to $\mu$ and $\mu E=0$, then $\mu E\le\delta$ for every
$\delta>0$, so $|\nu E|\le\epsilon$ for every $\epsilon>0$ and $\nu
E=0$.

\medskip

\quad{\bf (ii)}  \Quer\ Suppose, if possible, that $\nu E=0$ whenever
$\mu E=0$, but $\nu$ is not absolutely continuous.   Then there is an
$\epsilon>0$ such that for every $\delta>0$ there is an $E\in\Sigma$
such that $\mu E\le\delta$ but
$|\nu E|\ge\epsilon$.   For each $n\in\Bbb
N$ we may choose an $F_n\in\Sigma$ such that $\mu F_n\le 2^{-n}$ and
$|\nu F_n|\ge\epsilon$.   Consider $F=\bigcap_{n\in\Bbb N}\bigcup_{k\ge
n}F_k$.   Then we have

\Centerline{$\mu F\le\inf_{n\in\Bbb N}\mu(\bigcup_{k\ge n}F_k)
\le\inf_{n\in\Bbb N}\sum_{k=n}^{\infty}2^{-k}=0$,}

\noindent so $\mu F=0$.

Now recall that by 231Eb there is an $H\in\Sigma$ such that $\nu G\ge 0$
when $G\in\Sigma$ and $G\subseteq H$, while $\nu G\le 0$ when $G\in\Sigma$
and $G\cap H=\emptyset$.   As in 231F, set $\nu_1 G=\nu(G\cap H)$,
$\nu_2G=-\nu(G\setminus H)$ for $G\in\Sigma$, so that $\nu_1$ and
$\nu_2$ are totally finite measures, and $\nu_1F=\nu_2F=0$ because
$\mu(F\cap H)=\mu(F\setminus H)=0$.   Consequently

\Centerline{$0=\nu_iF=\lim_{n\to\infty}\nu_i(\bigcup_{m\ge n}F_m)
\ge\limsup_{n\to\infty}\nu_iF_n\ge 0$}

\noindent for both $i$, and

\Centerline{$0=\lim_{n\to\infty}(\nu_1F_n+\nu_2F_n)
\ge\liminf_{n\to\infty}|\nu F_n|\ge\epsilon>0$,}

\noindent which is absurd.\ \Bang

\medskip

{\bf (b)(i)} Suppose that $\nu$ is truly continuous with respect to
$\mu$.   It is obvious from the definitions that $\nu$ is absolutely
continuous with respect to $\mu$.   If $\nu E\ne 0$, there
must be an $F$ of finite measure such that $|\nu G|<|\nu E|$ whenever
$G\cap F=\emptyset$, so that $|\nu(E\setminus F)|<|\nu E|$ and
$\nu(E\cap F)\ne 0$.   This deals with the conditions ($\beta$) and
($\gamma$).

To check that $\nu$ is countably additive, let $\sequencen{E_n}$ be a
disjoint sequence in $\Sigma$, with union $E$, and $\epsilon>0$.   Let
$\delta>0$ and $F\in\Sigma$ be such $\mu F<\infty$ and $|\nu G|\le\epsilon$
whenever $G\in\Sigma$ and $\mu(F\cap G)\le\delta$.   Then

\Centerline{$\sum_{n=0}^{\infty}\mu(E_n\cap F)\le\mu F<\infty$,}

\noindent so there is an $n\in\Bbb N$ such that
$\sum_{i=n}^{\infty}\mu(E_i\cap F)\le\delta$.   Take any $m\ge n$ and
consider $E^*_m=\bigcup_{i\le m}E_i$.   We have

%$$\eqalign{|\nu E-\sum_{i=0}^m\nu E_i|
%&=|\nu E-\nu E^*_m|\cr
%&=|\nu(E\setminus E^*_m)|\cr
%&\le\epsilon\cr}$$

\Centerline{$|\nu E-\sum_{i=0}^m\nu E_i|
=|\nu E-\nu E^*_m|
=|\nu(E\setminus E^*_m)|
\le\epsilon$,}


\noindent because

\Centerline{$\mu(F\cap E\setminus E^*_m)
=\sum_{i=m+1}^{\infty}\mu(F\cap E_i)
\le\delta$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\nu E=\sum_{i=0}^{\infty}\nu E_i$;}

\noindent as $\sequencen{E_n}$ is arbitrary, $\nu$ is countably
additive.

\medskip

\quad{\bf (ii)} Now suppose that $\nu$ satisfies the three conditions.
By 231F, $\nu$ can be expressed as the difference of two non-negative
countably additive functionals $\nu_1$, $\nu_2$;  set
$\nuprime=\nu_1+\nu_2$, so that $\nuprime$ is a non-negative countably
additive
functional and $|\nu F|\le\nuprime F$ for every $F\in\Sigma$.    Set

\Centerline{$\gamma=\sup\{\nuprime F:F\in\Sigma,\,\mu
F<\infty\}\le\nuprime
X<\infty$,}

\noindent and
choose a sequence $\sequencen{F_n}$ of sets of finite measure such that
$\lim_{n\to\infty}\nuprime F_n=\gamma$;  set $F^*=\bigcup_{n\in\Bbb
N}F_n$.
If $G\in\Sigma$ and $G\cap F^*=\emptyset$ then $\nu G=0$.   \Prf\Quer\
Otherwise, by condition ($\gamma$), there is an $F\in\Sigma$ such that
$\mu F<\infty$ and $\nu(G\cap F)\ne 0$.   It follows that

\Centerline{$\nuprime(F\setminus F^*)\ge\nuprime(F\cap G)\ge|\nu(F\cap
G)|>0$,}

\noindent and there must be an $n\in\Bbb N$ such that

\Centerline{$\gamma<\nuprime F_n+\nuprime(F\setminus
F^*)=\nuprime(F_n\cup(F\setminus
F^*))\le\nuprime(F\cup F_n)\le\gamma$}

\noindent because $\mu(F\cup F_n)<\infty$;  but this is impossible.
\Bang\Qed

Setting $F_n^*=\bigcup_{k\le n}F_k$ for each $n$, we have
$\lim_{n\to\infty}\nuprime(F^*\setminus F^*_n)=0$.
Take any $\epsilon>0$, and (using condition ($\beta$)) let $\delta>0$
be such that $|\nu E|\le{1\over 2}\epsilon$ whenever $\mu E\le\delta$.
Let $n$ be such that
$\nuprime(F^*\setminus F_n^*)\le\bover12\epsilon$.   Now if $F\in\Sigma$
and $\mu(F\cap F^*_n)\le\delta$ then

$$\eqalign{|\nu F|
&\le|\nu(F\cap F^*_n)|
   +|\nu(F\cap F^*\setminus F^*_n)|
   +|\nu(F\setminus F^*)|\cr
&\le\Bover12\epsilon+\nuprime(F\cap F^*\setminus F^*_n)+0\cr
&\le\Bover12\epsilon+\nuprime(F^*\setminus F^*_n)
\le\Bover12\epsilon+\Bover12\epsilon
=\epsilon.\cr}$$

\noindent And $\mu F^*_n<\infty$.   As $\epsilon$ is arbitrary, $\nu$ is
truly continuous.

\medskip

{\bf (c)} Now suppose that $(X,\Sigma,\mu)$ is $\sigma$-finite and that
$\nu$ is countably additive and absolutely continuous with respect to
$\mu$.   Let $\sequencen{X_n}$ be a non-decreasing sequence of sets
of finite measure covering $X$ (211D).   If $\nu E\ne 0$, then
$\lim_{n\to\infty}\nu(E\cap X_n)\ne 0$, so $\nu(E\cap X_n)\ne 0$ for
some $n$.   This shows that $\nu$ satisfies condition ($\gamma$) of (b),
so is truly continuous.

Of course the converse of this fact is already covered by (b).

\medskip

{\bf (d)} Finally, suppose that $\mu X<\infty$ and that $\nu$ is
absolutely continuous with respect to $\mu$.   Then it must be truly
continuous, because we can take $F=X$ in the definition 232Ab.
}%end of proof of 232B

\leader{232C}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space and $\nu$,
$\nuprime$ two countably additive functionals on $\Sigma$ which are
truly continuous with respect to $\mu$.   Take $c\in\Bbb R$ and
$H\in\Sigma$,
and set $\nu_HE=\nu(E\cap H)$ for $E\in\Sigma$\cmmnt{, as in 231De}.   Then
$\nu+\nuprime$, $c\nu$ and $\nu_H$ are
all truly continuous with respect to $\mu$, and $\nu$ is expressible as
the difference of non-negative countably additive functionals which are
truly continuous with respect to $\mu$.

\proof{ Let $\epsilon>0$.   Set $\eta=\epsilon/(2+\epsilon+|c|)>0$.
Then there are $\delta$, $\delta'>0$ and $E$, $E'\in\Sigma$ such that
$\mu E<\infty$, $\mu E'<\infty$ and $|\nu F|\le\eta$ whenever $\mu(F\cap
E)\le\delta$, $|\nuprime F|\le\eta$ whenever $\mu(F\cap E)\le\delta'$.
Set
$\delta^*=\min(\delta,\delta')>0$, $E^*=E\cup E'\in\Sigma$;  then

\Centerline{$\mu
E^*\le\mu E+\mu E'<\infty$.}

\noindent   Suppose that $F\in\Sigma$ and $\mu(F\cap
E^*)\le\delta^*$;  then

\Centerline{$\mu(F\cap H\cap E)\le\mu(F\cap E)\le\delta^*\le\delta$,
\quad $\mu(F\cap E')\le\delta^*\le\delta'$}

\noindent so

\Centerline{$|(\nu+\nuprime)F|\le|\nu
F|+|\nuprime F|\le\eta+\eta\le\epsilon$,}

\Centerline{$|(c\nu)F|=|c||\nu F|\le|c|\eta\le\epsilon$,}

\Centerline{$|\nu_HF|=|\nu(F\cap H)|\le\eta\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu+\nuprime$, $c\nu$ and $\nu_H$
are
all truly continuous.

Now, taking $H$ from 231Eb, we see that $\nu_1=\nu_H$ and
$\nu_2=-\nu_{X\setminus H}$ are truly continuous and non-negative, and
$\nu=\nu_1-\nu_2$ is the difference of truly continuous measures.
}%end of proof of 232C

\leader{232D}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$f$ a $\mu$-integrable real-valued function.   For $E\in\Sigma$ set
$\nu E=\int_Ef$.   Then $\nu:\Sigma\to\Bbb R$ is a countably additive
functional and is truly continuous with respect to $\mu$, therefore
absolutely continuous with respect to $\mu$.

\proof{ Recall that $\int_Ef=\int f\times\chi E$ is defined
for every $E\in\Sigma$ (131Fa).   So $\nu:\Sigma\to\Bbb R$ is
well-defined.   If $E$, $F\in\Sigma$ are disjoint then

$$\eqalign{\nu(E\cup F)
&=\int f\times\chi(E\cup F)
=\int (f\times\chi E)+(f\times\chi F)\cr
&=\int f\times\chi E+\int f\times\chi F
=\nu E+\nu F,\cr}$$

\noindent so $\nu$ is finitely additive.

Now 225A, without using the phrase `truly continuous', proved exactly
that $\nu$ is truly continuous with respect to $\mu$.   It follows from
232Bb that $\nu$ is countably additive and absolutely continuous.
}%end of proof of 232D

\medskip

\noindent{\bf Remark} The functional $E\mapsto\int_Ef$ is called the
{\bf indefinite integral} of $f$.


\leader{232E}{}\cmmnt{ We are now at last ready for the theorem.

\medskip

\noindent}{\bf The Radon-Nikod\'ym theorem}
Let $(X,\Sigma,\mu)$ be a measure space and $\nu:\Sigma\to\Bbb R$ a
function.   Then the following are equiveridical:

(i) there is a $\mu$-integrable function $f$
such that $\nu E=\int_Ef$ for every $E\in\Sigma$;

(ii) $\nu$ is finitely additive and truly continuous with respect to $\mu$.

\proof{{\bf (a)} If $f$ is a $\mu$-integrable real-valued
function and $\nu E=\int_Ef$ for
every $E\in\Sigma$, then 232D tells us that $\nu$ is finitely additive
and truly continuous.

\medskip

{\bf (b)} In the other direction, suppose that $\nu$ is
finitely additive and truly continuous;  note that (by 232B(a-b))
$\nu E=0$ whenever $\mu E=0$.   To begin with, suppose that $\nu$ is
non-negative and not zero.

In this case, there is a non-negative simple function $f$ such that
$\int f>0$ and $\int_Ef\le\nu E$ for every $E\in\Sigma$.   \Prf\ Let
$H\in\Sigma$ be such that $\nu H>0$;  set $\epsilon={1\over 3}\nu H>0$.
Let $E\in\Sigma$, $\delta>0$ be such that $\mu E<\infty$ and $\nu
F\le\epsilon$ whenever $F\in\Sigma$ and $\mu(F\cap E)\le\delta$;  then
$\nu(H\setminus E)\le\epsilon$ so $\nu E\ge\nu(H\cap E)\ge2\epsilon$ and
$\mu E\ge\mu(H\cap E)>0$.   Set
$\mu_EF=\mu(F\cap E)$ for every $F\in\Sigma$;  then $\mu_E$ is a
countably
additive functional on $\Sigma$.   Set $\nuprime=\nu-\alpha\mu_E$, where
$\alpha=\epsilon/\mu E$;  then $\nuprime$ is a countably additive
functional
and $\nuprime E>0$.   By 231Eb, as usual, there is a set $G\in\Sigma$
such that
$\nuprime F\ge 0$ if $F\in\Sigma$ and $F\subseteq G$, but $\nuprime F\le 0$
if $F\in\Sigma$ and $F\cap G=\emptyset$.
As $\nuprime(E\setminus G)\le 0$,

\Centerline{$0<\nuprime E\le\nuprime(E\cap G)\le\nu(E\cap G)$}

\noindent and $\mu(E\cap G)>0$.   Set
$f={\alpha}\chi(E\cap G)$;  then $f$ is a non-negative simple function
and $\int f={\alpha}\mu(E\cap G)>0$.

If $F\in\Sigma$ then $\nuprime(F\cap G)\ge 0$, that is,

\Centerline{$\nu(F\cap G)\ge\alpha\mu_E(F\cap G)
=\alpha\mu(F\cap E\cap G)=\int_Ff$.}

\noindent So

\Centerline{$\nu F\ge\nu(F\cap G)\ge\int_Ff$,}

\noindent as required.\ \Qed

\medskip

{\bf (c)} Still supposing that $\nu$ is a non-negative, truly continuous
additive functional,   let $\Phi$ be the set of non-negative
simple functions $f:X\to\Bbb R$ such that
$\int_Ef\le\nu E$ for every $E\in\Sigma$;  then
the constant function $\tbf{0}$ belongs to $\Phi$, so $\Phi$ is not
empty.

If $f$, $g\in\Phi$ then $f\vee g\in\Phi$, where $(f\vee
g)(x)=\max(f(x),g(x))$ for $x\in X$.   \Prf\    Set
$H=\{x:(f-g)(x)\ge 0\}\in\Sigma$;   then $f\vee g=(f\times\chi
H)+(g\times\chi(X\setminus H))$ is a non-negative simple function, and
for any $E\in\Sigma$,

\Centerline{$\int_Ef\vee g=\int_{E\cap H}f+\int_{E\setminus H}g
\le\nu(E\cap H)+\nu(E\setminus H)=\nu E$. \Qed}

Set

\Centerline{$\gamma=\sup\{\int f:f\in\Phi\}\le\nu X<\infty$.}

\noindent    Choose a
sequence $\sequencen{f_n}$ in $\Phi$  such that $\lim_{n\to\infty}\int
f_n=\gamma$.   For each $n$, set $g_n=f_0\vee f_1\vee\ldots\vee f_n$;
then $g_n\in\Phi$ and $\int f_n\le\int g_n\le\gamma$ for each $n$, so
$\lim_{n\to\infty}\int g_n=\gamma$.   By B.Levi's theorem,
$f=\lim_{n\to\infty}g_n$ is integrable and $\int f=\gamma$.   Note that
if $E\in\Sigma$ then

\Centerline{$\int_Ef=\lim_{n\to\infty}\int_Ef_n\le\nu E$.}


\vthsp\Quer\ Suppose, if possible, that there is an $H\in\Sigma$ such
that $\int_Hf\ne\nu H$.   Set

\Centerline{$\nu_1F=\nu F-\int_Ff\ge 0$}

\noindent for every
$F\in\Sigma$;  then by (a) of this proof and 232C, $\nu_1$ is a truly
continuous finitely additive functional, and we are supposing that
$\nu_1\ne 0$.   By (b) of this proof, there
is a non-negative simple function $g$ such that $\int_Fg\le\nu_1F$ for
every $F\in\Sigma$ and $\int g>0$.   Take $n\in\Bbb N$ such that
$\int f_n+\int g>\gamma$.   Then $f_n+g$ is a non-negative simple
function and

\Centerline{$\int_F(f_n+g)=\int_Ff_n+\int_Fg\le\int_Ff+\int_Fg
=\nu F-\nu_1F+\int_Fg\le\nu F$}

\noindent for any $F\in\Sigma$, so $f_n+g\in\Phi$, and

\Centerline{$\gamma<\int f_n+\int g=\int f_n+g\le\gamma$,}

\noindent which is absurd.\ \BanG\ Thus we have $\int_Hf=\nu H$ for
every $H\in\Sigma$.

\medskip

{\bf (d)} This proves the theorem for non-negative $\nu$.   For general
$\nu$, we need only observe that $\nu$ is expressible as $\nu_1-\nu_2$,
where $\nu_1$ and $\nu_2$ are non-negative truly continuous countably
additive functionals, by 232C;  so that there are integrable functions
$f_1$, $f_2$ such that $\nu_iF=\int_Ff_i$ for both $i$ and every
$F\in\Sigma$.
Of course $f=f_1-f_2$ is integrable and $\nu F=\int_Ff$ for every
$F\in\Sigma$.   This completes the proof.
}%end of proof of 232E

\leader{232F}{Corollary} Let $(X,\Sigma,\mu)$ be a $\sigma$-finite
measure
space and $\nu:\Sigma\to\Bbb R$ a function.   Then there is a
$\mu$-integrable function $f$ such that $\nu E=\int_Ef$ for every
$E\in\Sigma$ iff $\nu$ is countably additive and absolutely continuous
with respect to $\mu$.

\proof{ Put 232Bc and 232E together.
}%end of proof of 232F

\leader{232G}{Corollary} Let $(X,\Sigma,\mu)$ be a  totally finite
measure
space and $\nu:\Sigma\to\Bbb R$ a function.   Then there is a
$\mu$-integrable function $f$ on $X$ such that $\nu E=\int_Ef$ for every
$E\in\Sigma$ iff $\nu$ is finitely additive and absolutely continuous
with respect to $\mu$.

\proof{ Put 232Bd and 232E together.
}%end of proof of 232G

\leader{232H}{Remarks}\cmmnt{ {\bf (a)} Most authors are satisfied with
232F as the
`Radon-Nikod\'ym theorem'.   In my view the problem of identifying
indefinite integrals is of sufficient importance to justify an analysis
which applies to all measure spaces, even if it requires a new concept
(the notion of `truly continuous' functional).

\spheader 232Hb I ought to offer an example of an absolutely
continuous functional which is not truly continuous.   A simple one is
the following.   Let $X$ be any uncountable set.   Let $\Sigma$ be the
countable-cocountable $\sigma$-algebra of subsets of $X$ and $\nu$ the
countable-cocountable measure on $X$ (211R).   Let $\mu$ be the
restriction to $\Sigma$ of counting measure on $X$.   If $\mu E=0$ then
$E=\emptyset$ and $\nu E=0$, so
$\nu$ is absolutely continuous.   But for any $E$ of finite measure we
have $\nu(X\setminus E)=1$, so $\nu$ is not truly continuous.   See also
232Xf(i).

\header{232Hc}{\bf *(c)} The space $(X,\Sigma,\mu)$ of this example
is, in terms of
the classification developed in Chapter 21, somewhat irregular;  for
instance, it is neither locally determined nor localizable, and
therefore not strictly localizable, though it is complete and
semi-finite.   Can this phenomenon occur in
a strictly localizable measure space?   We are led here into a
fascinating question.   Suppose, in (b), I used the same idea, but with
$\Sigma=\Cal PX$.   No difficulty arises in constructing $\mu$;  but can
there now be a $\nu$ with the required properties, that is, a non-zero
countably additive functional from $\Cal PX$ to $\Bbb R$ which is zero
on all finite sets?   This is the `Banach-Ulam problem', on which I
have written extensively elsewhere ({\smc Fremlin 93}), and to which I
will return in Chapter 54 in Volume 5.
The present question is touched on again in 363S in Volume 3.

\spheader 232Hd Following the Radon-Nikod\'ym theorem, the
question immediately arises:  for a given $\nu$,
how much possible variation is there in the corresponding $f$?   The
answer is straightforward enough:  two integrable functions $f$ and $g$
give rise to the same indefinite integral iff they are equal almost
everywhere (131Hb).

\spheader 232He I have stated the Radon-Nikod\'ym theorem in
terms of arbitrary integrable functions, meaning to interpret
`integrability'
in a wide sense, according to the conventions of Volume 1.   However,
given a truly continuous additive functional $\nu$, we can ask
whether there is in any sense a canonical integrable function
representing it.   The answer is no.   But we certainly do not need to
take arbitrary integrable functions of the type considered in Chapter
12.   If $f$ is any integrable function, there is a conegligible set $E$
such that $f\restr E$ is measurable, and now we can find a
conegligible measurable set $G\subseteq E\cap\dom f$;  if we set
$g(x)=f(x)$ for $x\in G$, $0$ for $x\in X\setminus G$, then $f\eae g$,
so $g$ has the same indefinite integral as $f$ (as noted in (d) just
above), while $g$ is measurable and has domain $X$.   Thus we can make a
trivial, but sometimes convenient, refinement to the theorem:  if
$(X,\Sigma,\mu)$ is a measure space, and $\nu:\Sigma\to\Bbb R$ is
finitely additive and truly continuous with respect to $\mu$, then there
is a $\Sigma$-measurable $\mu$-integrable function $g:X\to\Bbb R$ such
that $\int_Eg=\nu E$ for every $E\in\Sigma$.

\spheader 232Hf
It is convenient to introduce now a general definition.} %end of comment
If $(X,\Sigma,\mu)$ is a measure space and $\nu$ is a
$[-\infty,\infty]$-valued
functional defined on a family of subsets of $X$, I will say that
a $[-\infty,\infty]$-valued function $f$ defined on a subset of $X$ is a
{\bf Radon-Nikod\'ym derivative} of $\nu$ with respect to $\mu$ if
$\int_Efd\mu$ is defined\cmmnt{ (in the sense of 214D)} and
equal to $\nu E$ for
every $E\in\dom\nu$.   \cmmnt{Thus the integrable functions called $f$ in
232E-232G %232E 232F 232G
are all `Radon-Nikod\'ym derivatives';  later on we shall have less
well-regulated examples.}

\cmmnt{When $\nu$ is a measure and $f$ is non-negative,
$f$ may be called a {\bf density function}.}

\cmmnt{\spheader 232Hg Throughout the work above I have taken
it that $\nu$ is defined on the whole domain $\Sigma$ of $\mu$.   In
some of the
most important applications, however, $\nu$ is defined only on some
smaller $\sigma$-algebra $\Tau$.   In this case we commonly seek to
apply the same results with $\mu\restrp\Tau$ in place of $\mu$.
}%end of comment

\leader{232I}{}{\bf The Lebesgue decomposition of a countably additive
functional:  Proposition} (a) Let $(X,\Sigma,\mu)$ be a measure
space and $\nu:\Sigma\to\Bbb R$ a countably additive functional.
Then $\nu$ has unique expressions as

\Centerline{$\nu=\nu_s+\nu_{ac}=\nu_s+\nu_{tc}+\nu_e$,}

\noindent where $\nu_s$ is singular with respect to $\mu$,
$\nu_{ac}$ is absolutely continuous with respect to $\mu$, $\nu_{tc}$
is truly continuous with respect to $\mu$,
and $\nu_e$ is absolutely continuous
with respect to $\mu$ and zero on every set of finite measure.

(b) If $X=\BbbR^r$, $\Sigma$ is the algebra of Borel sets in $\BbbR^r$
and $\mu$ is the restriction of Lebesgue measure to $\Sigma$, then $\nu$
is uniquely expressible as $\nu_p+\nu_{cs}+\nu_{ac}$ where $\nu_{ac}$ is
absolutely continuous with respect to $\mu$, $\nu_{cs}$ is singular with
respect to
$\mu$ and zero on singletons, and $\nu_pE=\sum_{x\in E}\nu_p\{x\}$ for
every $E\in\Sigma$.

\proof{{\bf (a)(i)} Suppose first that $\nu$ is non-negative.   In
this case, set

\Centerline{$\nu_sE=\sup\{\nu(E\cap F):F\in\Sigma,\,\mu F=0\}$,}

\Centerline{$\nu_tE=\sup\{\nu(E\cap F):F\in\Sigma,\,\mu F<\infty\}$.}

\noindent Then both $\nu_s$ and $\nu_t$ are countably additive.
\prooflet{\Prf\
Surely $\nu_s\emptyset=\nu_t\emptyset=0$.   Let $\sequencen{E_n}$ be a
disjoint sequence in $\Sigma$ with union $E$.   \grheada\   If
$F\in\Sigma$
and $\mu F=0$, then

\Centerline{$\nu(E\cap F)=\sum_{n=0}^{\infty}\nu(E_n\cap F)
\le\sum_{n=0}^{\infty}\nu_s(E_n)$;}

\noindent  as $F$ is arbitrary,

\Centerline{$\nu_sE\le\sum_{n=0}^{\infty}\nu_sE_n$.}


\noindent \grheadb\ If $F\in\Sigma$ and $\mu F<\infty$, then

\Centerline{$\nu(E\cap F)=\sum_{n=0}^{\infty}\nu(E_n\cap F)
\le\sum_{n=0}^{\infty}\nu_t(E_n)$;}

\noindent as $F$ is arbitrary,

\Centerline{$\nu_tE\le\sum_{n=0}^{\infty}\nu_tE_n$.}

\noindent  \grheadc\ If
$\epsilon>0$, then (because $\sum_{n=0}^{\infty}\nu E_n=\nu E<\infty$)
there is an $n\in\Bbb N$ such that $\sum_{k=n+1}^{\infty}\nu
E_k\le\epsilon$.   Now, for each $k\le n$, there is an $F_k\in\Sigma$
such that $\mu F_k=0$ and
$\nu(E_k\cap F_k)\ge\nu_sE_k-{{\epsilon}\over{n+1}}$.   In this case,
$F=\bigcup_{k\le n}F_k\in\Sigma$, $\mu F=0$ and

\Centerline{$\nu_sE\ge\nu(E\cap F)
\ge\sum_{k=0}^n\nu(E_k\cap F_k)\ge\sum_{k=0}^n\nu_sE_k-\epsilon
\ge\sum_{k=0}^{\infty}\nu_sE_k-2\epsilon$,}

\noindent because

\Centerline{$\sum_{k=n+1}^{\infty}\nu_sE_k
\le\sum_{k=n+1}^{\infty}\nu E_k\le\epsilon$.}

\noindent   As $\epsilon$ is arbitrary,

\Centerline{$\nu_sE\ge\sum_{k=0}^{\infty}\nu_sE_k$.}

\noindent
\grheadd\ Similarly, for each $k\le n$, there is an $F'_k\in\Sigma$
such that
$\mu F'_k<\infty$ and
$\nu(E_k\cap F'_k)\ge\nu_tE_k-{{\epsilon}\over{n+1}}$.   In this case,
$F'=\bigcup_{k\le n}F'_k\in\Sigma$, $\mu F'<\infty$ and

\Centerline{$\nu_tE\ge\nu(E\cap F')\ge\sum_{k=0}^n\nu(E_k\cap F'_k)
\ge\sum_{k=0}^n\nu_tE_k-\epsilon
\ge\sum_{k=0}^{\infty}\nu_tE_k-2\epsilon$,}

\noindent because

\Centerline{$\sum_{k=n+1}^{\infty}\nu_tE_k
\le\sum_{k=n+1}^{\infty}\nu E_k\le\epsilon$.}

\noindent   As $\epsilon$ is arbitrary,

\Centerline{$\nu_tE\ge\sum_{k=0}^{\infty}\nu_tE_k$.}

\noindent \grheade\ Putting these together,
$\nu_sE=\sum_{n=0}^{\infty}\nu_sE_n$ and
$\nu_tE=\sum_{n=0}^{\infty}\nu_tE_n$.   As $\sequencen{E_n}$ is
arbitrary, $\nu_s$ and $\nu_t$ are countably additive.\ \Qed}


\medskip

\quad{\bf (ii)}  Still supposing that $\nu$ is
non-negative, if we choose a sequence $\sequencen{F_n}$ in
$\Sigma$ such
that $\mu F_n=0$ for each $n$ and $\lim_{n\to\infty}\nu F_n=\nu_sX$,
then $F^*=\bigcup_{n\in\Bbb N}F_n$ has $\mu F^*=0$ and
$\nu_sF^*=\nu F^*=\nu_sX$;
so that $\nu_s(X\setminus F^*)=0$, and $\nu_s$ is singular with respect
to $\mu$ in the sense of 232Ac.

Note that $\nu_sF=\nu F$ whenever $\mu F=0$.   So if we write
$\nu_{ac}=\nu-\nu_s$, then $\nu_{ac}$ is a countably additive functional
and $\nu_{ac}F=0$ whenever $\mu F=0$;  that is, $\nu_{ac}$ is absolutely
continuous with respect to $\mu$.

If we write $\nu_{tc}=\nu_t-\nu_s$, then $\nu_{tc}$ is a non-negative
countably additive functional;  $\nu_{tc}F=0$ whenever $\mu F=0$, and if
$\nu_{tc}E>0$ there is a set $F$ with $\mu F<\infty$ and
$\nu_{tc}(E\cap F)>0$.   So $\nu_{tc}$ is truly continuous with respect
to $\mu$, by 232Bb.   Set $\nu_e=\nu-\nu_t=\nu_{ac}-\nu_{tc}$.

Thus for any non-negative countably additive functional $\nu$, we have
expressions

\Centerline{$\nu=\nu_s+\nu_{ac}$,\quad $\nu_{ac}=\nu_{tc}+\nu_e$}

\noindent where $\nu_s$, $\nu_{ac}$, $\nu_{tc}$ and $\nu_e$ are all
non-negative
countably additive functionals, $\nu_s$ is singular with respect to
$\mu$, $\nu_{ac}$ and $\nu_e$ are absolutely continuous with respect to
$\mu$, $\nu_{tc}$ is truly continuous with respect to $\mu$, and
$\nu_eF=0$ whenever $\mu F<\infty$.


\medskip

\quad{\bf (iii)} For general countably additive functionals
$\nu:\Sigma\to\Bbb R$, we can express $\nu$ as
$\nuprime-\nu\mskip0.5mu''$,
where $\nuprime$ and $\nu\mskip0.5mu''$ are
non-negative countably additive functionals.   If we define
$\nuprime_s$,
$\nu_s'',\ldots,\nu_e''$ as in (i)-(ii), we get countably additive
functionals

\Centerline{$\nu_s=\nu_s'-\nu_s''$,
\quad $\nu_{ac}=\nu_{ac}'-\nu_{ac}''$,
\quad $\nu_{tc}=\nu_{tc}'-\nu_{tc}''$,
\quad $\nu_e=\nu_e'-\nu_e''$}

\noindent such that $\nu_s$ is singular with respect to $\mu$ (if $F'$,
$F''$ are such that

\Centerline{$\mu F=\mu F'=\nu_s'(X\setminus F)=\nu_s''(X\setminus
F)=0$,}

\noindent then $\mu(F'\cup F'')=0$ and $\nu_sE=0$ whenever $E\subseteq
X\setminus(F'\cup F'')$), $\nu_{ac}$ is absolutely continuous with
respect to $\mu$, $\nu_{tc}$ is truly continuous with respect to $\mu$,
and $\nu_eF=0$ whenever $\mu F<\infty$, while

\Centerline{$\nu=\nu_s+\nu_{ac}=\nu_s+\nu_{tc}+\nu_e$.}

\medskip

\quad{\bf (iv)} Moreover, these decompositions are unique.
\prooflet{\Prf\grheada\
If, for instance, $\nu=\tilde\nu_s+\tilde\nu_{ac}$, where $\tilde\nu_s$
is singular and $\tilde\nu_{ac}$ is absolutely continuous with respect
to $\mu$, let $F$, $\tilde F$ be such that $\mu F=\mu\tilde F=0$ and
$\tilde\nu_sE=0$ whenever $E\cap \tilde F=\emptyset$, $\nu_sE=0$
whenever $E\cap F=\emptyset$.   Set $F^*=F\cup\tilde F$.   If $E\in\Sigma$,
$\nu_{ac}(E\cap F^*)=0$ because $\mu F^*=0$ while $\nu_s(E\setminus F^*)=0$
because $E\setminus F^*$ is disjoint from $F$, so

\Centerline{$\nu_sE=\nu_s(E\cap F^*)+\nu_s(E\setminus F^*)
=\nu(E\cap F^*)-\nu_{ac}(E\cap F^*)+0=\nu(E\cap F^*)$.}

\noindent Similarly, $\tilde\nu_sE=\nu(E\cap F^*)$ and $\tilde\nu_sE=\nu_sE$;
as $E$ is arbitrary, $\tilde\nu_s=\nu_s$ and $\tilde\nu_{ac}=\nu_{ac}$.

\medskip

\qquad\grheadb\  Similarly, if $\nu_{ac}=\tilde\nu_{tc}+\tilde\nu_e$
where $\tilde\nu_{tc}$ is truly continuous with respect to $\mu$ and
$\tilde\nu_eF=0$ whenever $\mu F<\infty$, then there are sequences
$\sequencen{F_n}$, $\sequencen{\tilde F_n}$ of sets of finite measure
such that $\nu_{tc}F=0$ whenever
$F\cap\bigcup_{n\in\Bbb N}F_n=\emptyset$ and
$\tilde\nu_{tc}F=0$ whenever
$F\cap\bigcup_{n\in\Bbb N}\tilde F_n=\emptyset$.   Write
$F^*=\bigcup_{n\in\Bbb N}(F_n\cup\tilde F_n)$;
then $\tilde\nu_eE=\nu_eE=0$ whenever $E\subseteq F^*$ and
$\tilde\nu_{tc}E=\nu_{tc}E=0$ whenever $E\cap F^*=\emptyset$, so
$\nu_{tc}E=\nu_{ac}(E\cap F^*)=\tilde\nu_{tc}E$ for every $E\in\Sigma$,
$\nu_{tc}=\tilde\nu_{tc}$ and $\nu_e=\tilde\nu_e$.\ \Qed}

\medskip

{\bf (b)} In this case, $\mu$ is $\sigma$-finite (cf.\ 211P), so every
absolutely continuous countably additive functional is truly continuous
(232Bc), and we
shall always have $\nu_e=0$, $\nu_{ac}=\nu_{tc}$.   But in the other
direction we know that singleton sets, and therefore countable sets, are
all measurable.   We therefore have a
further decomposition $\nu_s=\nu_p+\nu_{cs}$, where there is a countable
set $K\subseteq\BbbR^r$ with $\nu_pE=0$ whenever $E\in\Sigma$ and
$E\cap K=\emptyset$, and $\nu_{cs}$ is singular with respect to $\mu$
and zero on countable sets.   \prooflet{\Prf\ (i) If $\nu\ge 0$, set

\Centerline{$\nu_pE=\sup\{\nu(E\cap K):K\subseteq\BbbR^r$ is
countable$\}$;}

\noindent   just
as with $\nu_s$, dealt with in (a) above, $\nu_p$ is countably
additive and there is a countable $K\subseteq\BbbR^r$ such that
$\nu_pE=\nu(E\cap K)$ for every $E\in\Sigma$.    (ii) For general $\nu$,
we can express $\nu$ as $\nuprime-\nu\mskip0.5mu''$ where $\nuprime$ and
$\nu\mskip0.5mu''$ are
non-negative, and write $\nu_p=\nu_p'-\nu_p''$.   (iii) $\nu_p$ is
characterized by saying that there is a countable set $K$ such that
$\nu_pE=\nu(E\cap K)$ for every $E\in\Sigma$ and $\nu\{x\}=0$ for every
$x\in \BbbR^r\setminus K$.   (iv) So if we set $\nu_{cs}=\nu_s-\nu_p$,
$\nu_{cs}$ will be singular with respect to $\mu$ and zero on countable
sets.\ \Qed}

Now, for any $E\in\Sigma$,

\Centerline{$\nu_pE=\nu(E\cap K)=\sum_{x\in K\cap E}\nu\{x\}
=\sum_{x\in E}\nu\{x\}$.}

}%end of proof of 232I

\medskip

\noindent{\bf Remark} The expression $\nu=\nu_p+\nu_{cs}+\nu_{ac}$ of
(b) is the {\bf Lebesgue decomposition} of $\nu$.

\exercises{
\leader{232X}{Basic exercises (a)}
%\spheader 232Xa
Let $(X,\Sigma,\mu)$ be a measure space and $\nu:\Sigma\to\Bbb R$
a countably additive functional which is absolutely continuous with
respect to $\mu$.   Show that the following are equiveridical:  (i)
$\nu$ is truly continuous with respect to $\mu$;  (ii) there is a sequence
$\sequencen{E_n}$ in $\Sigma$ such that $\mu E_n<\infty$ for every
$n\in\Bbb N$ and $\nu F=0$ whenever $F\in\Sigma$ and
$F\cap\bigcup_{n\in\Bbb N}E_n=\emptyset$.
%232B

\sqheader 232Xb Let $g:\Bbb R\to\Bbb R$ be a bounded
non-decreasing function and
$\mu_g$ the associated Lebesgue-Stieltjes measure (114Xa).   Show
that $\mu_g$ is absolutely continuous (equivalently, truly continuous)
with respect to Lebesgue measure iff the restriction of $g$
to any closed bounded interval is absolutely continuous in the sense of
225B.
%232B

\spheader 232Xc Let $(X,\Sigma,\mu)$ be a measure space, and $\nu$ and
$\lambda$ additive functionals on $\Sigma$ of which $\nu$ is positive
and countably additive, so that $(X,\Sigma,\nu)$ also is a measure
space.   (i) Show that if $\nu$ is absolutely continuous with respect to
$\mu$ and $\lambda$ is absolutely continuous with respect to $\nu$, then
$\lambda$ is absolutely continuous with respect to $\mu$.   (ii) Show
that if $\nu$ is truly continuous with respect to $\mu$ and $\lambda$ is
absolutely continuous with respect to $\nu$ then $\lambda$ is truly
continuous with respect to $\mu$.
%232B

\sqheader 232Xd Let $X$ be a non-empty set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Show that for any sequence
$\sequencen{\nu_n}$ of countably additive functionals on $\Sigma$ there
is a probability measure $\mu$ on $X$, with domain $\Sigma$, such that
every $\nu_n$ is absolutely continuous with respect to $\mu$.
\Hint{start with the case $\nu_n\ge 0$.}
%232B

\spheader 232Xe Let $(X,\Sigma,\mu)$ be a measure space and
$(X,\hat\Sigma,\hat\mu)$ its completion (212C).   Let
$\nu:\Sigma\to\Bbb R$ be an additive functional such that $\nu E=0$
whenever $\mu E=0$.
Show that $\nu$ has a unique extension to an additive functional
$\hat\nu:\hat\Sigma\to\Bbb R$ such that $\hat\nu E=0$ whenever
$\hat\mu E=0$.
%232B

\spheader 232Xf Let $\Cal F$ be an ultrafilter on $\Bbb N$
including the filter $\{\Bbb N\setminus I:I\subseteq\Bbb N$ is
finite$\}$ (2A1O).    Define $\nu:\Cal P\Bbb N\to\{0,1\}$ by setting
$\nu E=1$ if $E\in\Cal F$, $0$ for $E\in\Cal P\Bbb N\setminus\Cal F$.
(i) Let $\mu_1$ be counting measure on $\Cal P\Bbb N$.   Show that $\nu$
is additive and absolutely continuous with respect to $\mu_1$, but is
not truly continuous.   (ii) Define $\mu_2:\Cal P\Bbb N\to[0,1]$ by
setting $\mu_2E=\sum_{n\in E}2^{-n-1}$.   Show that $\nu$ is zero on
$\mu_2$-negligible sets, but is not absolutely continuous with respect
to $\mu_2$.
%232B

\spheader 232Xg Let $X$ be a set and $\Sigma$ a $\sigma$-algebra
of subsets of
$X$;  let $\nu:\Sigma\to\Bbb R$ be a countably additive functional.
Let $\Cal I$ be an {\bf ideal} of $\Sigma$, that is, a subset of
$\Sigma$ such that ($\alpha$) $\emptyset\in\Cal I$ ($\beta$)
$E\cup F\in\Cal I$ for all $E$, $F\in\Cal I$ ($\gamma$) if $E\in\Sigma$,
$F\in\Cal I$ and $E\subseteq F$ then $E\in\Cal I$.   Show that $\nu$ has
a unique decomposition as $\nu=\nu_{\Cal I}+\nuprime_{\Cal I}$, where
$\nu_{\Cal I}$ and $\nuprime_{\Cal I}$ are countably additive
functionals,
$\nuprime_{\Cal I}E=0$ for every $E\in\Cal I$, and whenever
$E\in\Sigma$ and $\nu_{\Cal I}E\ne 0$ there is an $F\in\Cal I$ such that
$\nu_{\Cal I}(E\cap F)\ne 0$.
%232I

\spheader 232Xh Rewrite this section in terms of complex-valued
additive functionals.
%232+

\leader{232Y}{Further exercises (a)}
%\spheader 232Ya
Let $(X,\Sigma,\mu)$ be a measure
space and $\nu:\Sigma\to\Bbb R$ a finitely additive functional.   If
$E$, $F$, $H\in\Sigma$ and $\mu H<\infty$ set
$\rho_H(E,F)=\mu(H\cap(E\symmdiff F))$.   (i) Show that $\rho_H$ is a
pseudometric on $\Sigma$ (2A3Fa).   (ii)
Let $\frak T$ be the topology on $\Sigma$ generated by
$\{\rho_H:H\in\Sigma,\,\mu H<\infty\}$ (2A3Fc).   Show that $\nu$ is
continuous for $\frak T$
iff it is truly continuous in the sense of 232Ab.
($\frak T$ is the topology of {\bf convergence in
measure} on $\Sigma$.)
%232A

\spheader 232Yb Let $(X,\Sigma,\mu)$ be a measure space and
$(X,\tilde\Sigma,\tilde\mu)$ its c.l.d.\ version (213E).   Let
$\nu:\Sigma\to\Bbb R$ be an additive functional which is truly
continuous with respect to $\mu$.   Show that $\nu$ has a unique
extension to a functional $\tilde\nu:\tilde\Sigma\to\Bbb R$ which is
truly continuous with respect to $\tilde\mu$.
%232B 232Xe

\spheader 232Yc (H.K\"onig)
Let $X$ be a set and $\mu$, $\nu$ two measures on $X$ with
the same domain $\Sigma$.   For $\alpha\ge 0$, $E\in\Sigma$ set
$(\alpha\mu\wedge\nu)(E)
=\inf\{\alpha\mu(E\cap F)+\nu(E\setminus F):F\in\Sigma\}$ (cf.\
112Ya).
Show that the following are equiveridical:  (i) $\nu E=0$ whenever
$\mu E=0$;  (ii) $\sup_{\alpha\ge 0}(\alpha\mu\wedge\nu)(E)=\nu E$ for
every $E\in\Sigma$.
%232B

\spheader 232Yd Let $(X,\Sigma,\mu)$ be a measure space and $f$
a $\mu$-integrable real-valued function.   Show that the indefinite
integral of $f$ is the unique additive functional
$\nu:\Sigma\to\Bbb R$ such that whenever $E\in\Sigma$ and $f(x)\in[a,b]$
for almost every $x\in E$, then $a\mu E\le\nu E\le b\mu E$.
%232D

\spheader 232Ye Let $(X,\Sigma,\mu)$ be a measure space and $f$ a
non-negative real-valued function which is integrable over $X$;  let
$\nu$ be its indefinite integral.   Show that for any function
$g:X\to\Bbb R$, $\int g\,d\nu=\int f\times g\,d\mu$ in the sense that if
one of these is defined in $[-\infty,\infty]$ so is the other, and they
are then equal.   \Hint{start with simple functions $g$.}
%232D

\spheader 232Yf Let $(X,\Sigma,\mu)$ be a measure space, $f$ an
integrable function, and $\nu:\Sigma\to\Bbb R$ the indefinite integral
of $f$.   Show that $|\nu|$, as defined in 231Ya, is the
indefinite integral of $|f|$.
%232D

\spheader 232Yg Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of subsets
of $X$, and $\nu:\Sigma\to\Bbb R$ a countably additive functional.
Show that $\nu$ has a Radon-Nikod\'ym
derivative with respect to $|\nu|$ as defined in 231Ya, and that any
such derivative has modulus equal to $1\,\,|\nu|$-a.e.
%232D

\spheader 232Yh For a non-decreasing function
$F:[a,b]\to\Bbb R$, where $a<b$,
let $\nu_F$ be the corresponding Lebesgue-Stieltjes measure.   Show that
if we define
$(\nu_F)_{ac}$, etc., with regard to Lebesgue measure on $[a,b]$, as in
232I, then

\Centerline{$(\nu_F)_p=\nu_{F_p}$, \quad$(\nu_F)_{ac}=\nu_{F_{ac}}$,
\quad$(\nu_F)_{cs}=\nu_{F_{cs}}$,}

\noindent where $F_p$, $F_{cs}$ and $F_{ac}$ are defined as in 226C.
%232I

\spheader 232Yi Extend the idea of (h) %232Yh
to general functions $F$ of bounded variation.
%232Yh 232I

\spheader 232Yj Extend the ideas of (h) and (i) %232Yh 232Yi
to open, half-open and unbounded intervals (cf.\ 226Yb).
%232Yh 232Yi 232I

\spheader 232Yk Say that two bounded additive functionals
$\nu_1$, $\nu_2$ on an algebra $\Sigma$ of sets are {\bf mutually
singular} if for any $\epsilon>0$ there is an $H\in\Sigma$ such that

\Centerline{$\sup\{|\nu_1F|:F\in\Sigma,\,F\subseteq H\}\le\epsilon$,}

\Centerline{$\sup\{|\nu_2F|:F\in\Sigma,\,F\cap H=\emptyset\}
\le\epsilon$.}

\quad {(i)} Show that $\nu_1$ and $\nu_2$ are mutually singular iff, in
the language of 231Ya-231Yb, $|\nu_1|\wedge|\nu_2|=0$.

\quad{(ii)} Show that if $\Sigma$ is a $\sigma$-algebra and $\nu_1$ and
$\nu_2$ are countably additive, then they are mutually singular iff
there is an $H\in\Sigma$ such that $\nu_1F=0$ whenever $F\in\Sigma$ and
$F\subseteq H$, while $\nu_2F=0$ whenever $F\in\Sigma$ and
$F\cap H=\emptyset$.

\quad{(iii)} Show that if $\nu_s$, $\nu_{tc}$ and $\nu_e$ are defined
from $\nu$ and $\mu$ as in 232I, then each pair of the three are
mutually singular.
%232I
}%end of exercises

\endnotes{
\Notesheader{232} The Radon-Nikod\'ym theorem must be on any list of the
half-dozen most important theorems of measure theory, and not only the
theorem itself, but the techniques necessary to prove it, are at the
heart of the subject.   In my book {\smc Fremlin 74} I discussed a
variety of more or less abstract versions of the theorem and of the
method, to some of which I will return in \S\S327 and 365 of the
next volume.

As I have presented it here, the essence of the proof is split between
231E and 232E.   I think we can distinguish the following elements.
Let $\nu$ be a countably additive functional.

\inset{(i) $\nu$ is bounded (231Ea).}

\inset{(ii) $\nu$ is expressible as the difference of non-negative
functionals (231F).}

\noindent   (I gave this as a corollary of 231Eb, but it can also be
proved by simpler methods, as in 231Ya.)

\inset{(iii) If $\nu>0$, there is an integrable $f$ such that
$0<\nu_f\le\nu$,}

\noindent writing $\nu_f$ for the indefinite integral of $f$.   (This is
the point at which we really do need the Hahn decomposition 231Eb.)

\inset{(iv) The set $\Psi=\{f:\nu_f\le\nu\}$ is closed under countable
suprema, so there is an $f\in\Psi$ maximising $\int f$.}

\noindent (In part (b) of the proof of 232E, I spoke of simple
functions;  but this was solely to simplify the technical details, and
the same argument works if we apply it to $\Psi$ instead of $\Phi$.
Note the use here of B.Levi's theorem.)

\inset{(v) Take $f$ from (iv) and use (iii) to show that $\nu-\nu_f=0$.}

\noindent Each of the steps (i)-(iv) requires a non-trivial idea, and
the importance of the theorem lies not only in its remarkable direct
consequences in the rest of this chapter and elsewhere, but in the
versatility and power of these ideas.

I introduce the idea of `truly continuous' functional in order to
give a reasonably straightforward account of the status of the
Radon-Nikod\'ym theorem in non-$\sigma$-finite measure spaces.   Of
course the whole point is that a truly continuous functional, like an
indefinite integral, must be concentrated on a $\sigma$-finite part of
the space (232Xa), so that 232E, as stated, can be deduced easily from
the standard form 232F.   I dare to use the word `truly' in this
context because this kind of continuity does indeed correspond to a
topological notion (232Ya).

There is a possible trap in the definition I give of `absolutely
continuous' functional.   Many authors use the condition of 232Ba as a
definition, saying that $\nu$ is absolutely continuous with respect to
$\mu$ if $\nu E=0$ whenever $\mu E=0$.   For countably additive
functionals this coincides with the $\epsilon$-$\delta$ formulation in
232Aa;  but for other additive functionals this
need not be so (232Xf(ii)).   Mostly the distinction is insignificant,
but I
note that in 232Bd it is critical, since $\nu$ there is not assumed to
be countably additive.

In 232I I describe one of the many ways of decomposing a countably
additive functional into mutually singular parts with special
properties.   In 231Yf-231Yg I have already suggested a method of
decomposing an additive functional into the sum of a countably additive
part and a `purely finitely additive' part.   All these results have
natural expressions in terms of the ordered linear space of bounded
additive functionals on an algebra (231Yc).
}%end of notes

\discrpage


\frfilename{mt233.tex} 
\versiondate{16.6.02} 
\copyrightdate{1994} 
 
\def\chaptername{The Radon-Nikod\'ym theorem} 
\def\sectionname{Conditional expectations} 
 
\newsection{233} 
 
I devote a section to a first look at one of the principal applications 
of the Radon-Nikod\'ym theorem.   It is one of the most 
vital ideas of measure theory, and will appear repeatedly in one form or 
another.   Here I give the definition and most basic properties of 
conditional expectations as they arise in abstract probability theory, 
with notes on convex functions and a version of Jensen's inequality 
(233I-233J). 
 
\leader{233A}{$\sigma$-subalgebras} Let $X$ be a set and $\Sigma$ a 
$\sigma$-algebra of subsets of $X$.   A {\bf $\sigma$-subalgebra} of 
$\Sigma$ is a $\sigma$-algebra $\Tau$ of subsets of $X$ such that 
$\Tau\subseteq\Sigma$.   If $(X,\Sigma,\mu)$ is a measure space and 
$\Tau$ is a $\sigma$-subalgebra of $\Sigma$, then 
$(X,\Tau,\mu\restrp\Tau)$ is again a measure space\cmmnt{;  this is 
immediate from the definition (112A).   Now we have the 
following straightforward lemma.   It is a special case of 235G below, 
but I give a separate proof in case you do not wish as yet 
to embark on the general investigation pursued in \S235}. 
 
\leader{233B}{Lemma} Let $(X,\Sigma,\mu)$ be a measure space and $\Tau$ 
a $\sigma$-subalgebra of $\Sigma$.   A real-valued function $f$ defined 
on a subset of $X$ 
is $\mu\restrp\Tau$-integrable iff (i) it is $\mu$-integrable (ii) 
$\dom f$ is $\mu\restrp\Tau$-conegligible (iii) $f$ is 
$\mu\restrp\Tau$-virtually measurable;  and in this case 
$\int fd(\mu\restrp\Tau)=\int fd\mu$. 
 
\proof{{\bf (a)}  Note first that if $f$ is a $\mu\restrp\Tau$-simple 
function, that is, is expressible as $\sum_{i=0}^na_i\chi E_i$ where 
$a_i\in\Bbb R$, $E_i\in\Tau$ and $(\mu\restrp\Tau)E_i<\infty$ for each 
$i$, then $f$ is $\mu$-simple and 
 
\Centerline{$\int f d\mu=\sum_{i=0}^na_i\mu E_i 
=\int f d(\mu\restrp\Tau)$.} 
 
\medskip 
 
{\bf (b)} Let $U_{\mu}$ be the set of non-negative 
$\mu$-integrable functions and $U_{\mu\restrp\Tau}$ the set of 
non-negative $\mu\restrp\Tau$-integrable functions. 
 
Suppose $f\in U_{\mu\restrp\Tau}$.   Then there is a 
non-decreasing sequence $\sequencen{f_n}$ of 
$\mu\restrp\Tau$-simple functions such that 
$f(x)=\lim_{n\to\infty}f_n\,\mu\restrp\Tau$-a.e.\ and 
 
\Centerline{$\int fd(\mu\restrp\Tau) 
=\lim_{n\to\infty}\int f_nd(\mu\restrp\Tau)$.} 
 
\noindent  But now 
every $f_n$ is also $\mu$-simple, and 
$\int f_nd\mu=\int f_nd(\mu\restrp\Tau)$ for every $n$,  and 
$f=\lim_{n\to\infty}f_n\,\mu$-a.e.   So $f\in U_{\mu}$ and 
$\int fd\mu=\int fd(\mu\restrp\Tau)$. 
 
\medskip 
 
{\bf (c)} Now suppose that $f$ is 
$\mu\restrp\Tau$-integrable.   Then it is the difference of two members 
of $U_{\mu\restrp\Tau}$, so is $\mu$-integrable, and 
$\int fd\mu=\int fd(\mu\restrp\Tau)$.   Also conditions (ii) and (iii) 
are satisfied, 
according to the conventions established in Volume 1 (122Nc, 122P-122Q). 
 
\medskip 
 
{\bf (d)} Suppose that $f$ satisfies conditions 
(i)-(iii).   Then $|f|\in U_{\mu}$, and  there is a conegligible set 
$E\subseteq\dom f$ such that $E\in\Tau$ and $f\restr E$ is 
$\Tau$-measurable.   Accordingly $|f|\restr E$ is $\Tau$-measurable. 
Now, if $\epsilon>0$, then 
 
\Centerline{$(\mu\restrp\Tau)\{x:x\in E,\,|f|(x)\ge\epsilon\} 
=\mu\{x:x\in E,\,|f|(x)\ge\epsilon\} 
\le{1\over\epsilon}\int|f|d\mu<\infty$;} 
 
\noindent moreover, 
 
$$\eqalign{\sup\{\int g\,d(\mu\restrp\Tau):g&\text{ is a } 
  \mu\restrp\Tau\text{-simple function}, 
  \,g\le |f|\,\mu\restrp\Tau\text{-a.e.}\}\cr 
&=\sup\{\int g\,d\mu:g\text{ is a } 
  \mu\restrp\Tau\text{-simple function}, 
  \,g\le |f|\,\mu\restrp\Tau\text{-a.e.}\}\cr 
&\le\sup\{\int g\,d\mu:g\text{ is a } 
  \mu\text{-simple function},\,g\le |f|\,\mu\text{-a.e.}\}\cr 
&\le\int |f|d\mu<\infty.\cr}$$ 
 
\noindent By the criterion of 122Ja, $|f|\in U_{\mu\restrp\Tau}$. 
Consequently $f$, being $\mu\restrp\Tau$-virtually $\Tau$-measurable, 
is $\mu\restrp\Tau$-integrable, by 122P.   This completes the proof. 
}%end of proof of 233B 
 
\cmmnt{ 
\leader{233C}{Remarks (a)} My argument just above is detailed 
to the point of 
pedantry.   I think, however, that while I can be accused of wasting 
paper by writing everything down, every element of the argument is 
necessary to the result.   To be sure, some of the details are needed 
only because I use such a wide notion of `integrable function';  if 
you restrict the notion of `integrability' to measurable functions 
defined on the whole measure space, there are simplifications at this 
stage, to be paid for later when you discover that many of the principal 
applications are to functions defined by formulae which do not apply on 
the whole underlying space. 
 
The essential point which does have to be grasped is that while a 
$\mu\restrp\Tau$-negligible set is always $\mu$-negligible, a 
$\mu$-negligible set need not be $\mu\restrp\Tau$-negligible. 
 
\header{233Cb}{\bf (b)} As the simplest possible example of the problems 
which can arise, I offer the following.   Let $(X,\Sigma,\mu)$ be 
$[0,1]^2$ with Lebesgue measure.   Let $\Tau$ be the set of those 
members of $\Sigma$ expressible as $F\times[0,1]$ for some 
$F\subseteq[0,1]$;  it is easy to see that $\Tau$ is a 
$\sigma$-subalgebra of $\Sigma$.   Consider $f$, $g:X\to[0,1]$ defined 
by saying that 
 
\Centerline{$f(t,u)=1$ if $u>0$, $0$ otherwise,} 
 
\Centerline{$g(t,u)=1$ if $t>0$, $0$ otherwise.} 
 
\noindent Then both $f$ and $g$ are $\mu$-integrable, being constant 
$\mu$-a.e.   But only $g$ is $\mu\restrp\Tau$-integrable, because any 
non-negligible $E\in\Tau$ includes a complete vertical section 
$\{t\}\times[0,1]$, so that $f$ takes both values $0$ and $1$ on $E$. 
If we set 
 
\Centerline{$h(t,u)=1$ if $u>0$, undefined otherwise,} 
 
\noindent then again (on the conventions I use) $h$ is 
$\mu$-integrable but not $\mu\restrp\Tau$-integrable, as there is no 
conegligible member of $\Tau$ included in the domain of $h$. 
 
\header{233Cc}{\bf (c)} If $f$ is defined everywhere in $X$, and 
$\mu\restrp\Tau$ is complete, then of course $f$ is 
$\mu\restrp\Tau$-integrable iff it is $\mu$-integrable and 
$\Tau$-measurable.   But note that in the example just above, which is 
one of the archetypes for this topic, $\mu\restrp\Tau$ is not complete, 
as singleton sets are negligible but not measurable. 
}%end of comment 
 
\leader{233D}{Conditional expectations} Let $(X,\Sigma,\mu)$ be a 
probability 
space\cmmnt{, that is, a measure space with $\mu X=1$}. 
\cmmnt{(Nearly all the ideas here work 
perfectly well for any totally finite measure space, but there seems 
nothing to be gained from the extension, and the traditional phrase 
`conditional expectation' demands a probability space.)} 
Let $\Tau\subseteq\Sigma$ be a $\sigma$-subalgebra. 
\dvro{ If $f$ is a $\mu$-integrable real-valued function, there is a 
$\mu\restrp\Tau$-integrable function $g$ such that $\int_Fg=\int_Ff$ for 
every $F\in\Tau$;  such  a function is {\bf a conditional expectation} 
of $f$ on $\Tau$.}{} 
 
\cmmnt{\header{233Da}{\bf (a)} For any 
$\mu$-integrable real-valued function $f$ defined on a conegligible 
subset of $X$, we have a corresponding indefinite integral 
$\nu_f:\Sigma\to\Bbb R$ given by the formula $\nu_fE=\int_Ef$ for every 
$E\in\Sigma$.   We know that $\nu_f$ is countably additive and truly 
continuous with respect to $\mu$, which in the present context is the 
same as saying that it is absolutely continuous (232Bc-232Bd).   Now 
consider the restrictions $\mu\restrp\Tau$, $\nu_f\restrp\Tau$ of $\mu$ 
and $\nu_f$ to the $\sigma$-algebra $\Tau$.   It follows 
directly from the definitions of `countably additive' and 
`absolutely continuous' that 
$\nu_f\restrp\Tau$ is countably additive and absolutely continuous with 
respect to $\mu\restrp\Tau$, therefore truly continuous with respect to 
$\mu\restrp\Tau$.   Consequently, the 
Radon-Nikod\'ym theorem (232E) tells us that there is a 
$\mu\restrp\Tau$-integrable function $g$ such that 
$(\nu_f\restrp\Tau)F=\int_Fg\,d(\mu\restrp\Tau)$ for every $F\in\Tau$. 
}%end of comment 
 
\cmmnt{\header{233Db}{\bf (b)} Let us define {\bf a conditional 
expectation of $f$ on} $\Tau$ to be such a function; that is, a 
$\mu\restrp\Tau$-integrable 
function $g$ such that $\int_Fg\,d(\mu\restrp\Tau)=\int_Ffd\mu$ for 
every $F\in\Tau$.   Looking back at 233B, we see that for such a $g$ we 
have 
 
\Centerline{$\int_Fg\,d(\mu\restrp\Tau) 
=\int g\times\chi F\,d(\mu\restrp\Tau) 
=\int g\times\chi F\, d\mu=\int_Fg\,d\mu$} 
 
\noindent for every $F\in\Tau$;  also, that $g$ is almost everywhere 
equal to a $\Tau$-measurable function defined everywhere in $X$ which is 
also a conditional expectation of $f$ on $\Tau$ (232He). 
}%end of comment 
 
\cmmnt{\header{233Dc}{\bf (c)} I set the word `a' of the phrase 
`a conditional 
expectation' in bold type to emphasize that there is nothing unique 
about the function $g$.   In 242J I will return to this 
point, and describe an object which could properly be called `the' 
conditional expectation of $f$ on $\Tau$.   $g$ is `essentially 
unique' only in the sense that if $g_1$, $g_2$ are both conditional 
expectations of $f$ on $\Tau$ then $g_1=g_2\,\,\mu\restrp\Tau$-a.e.\ 
(131Hb).   This does of course 
mean that a very large number of its properties -- for instance, the 
distribution function $G(a)=\hat\mu\{x:g(x)\le a\}$, where $\hat\mu$ is 
the completion of $\mu$ (212C) -- are independent of which $g$ we 
take. 
}%end of comment 
 
\cmmnt{\header{233Dd}{\bf (d)} A word of explanation of the phrase 
`conditional expectation' is in order.   This derives from the standard 
identification of probability with measure, due to Kolmogorov, which I 
will discuss more fully in Chapter 27.   A real-valued random 
variable may be regarded as a measurable, or virtually measurable, 
function $f$ on a probability space $(X,\Sigma,\mu)$;  its 
`expectation' becomes identified with $\int f d\mu$, supposing that this 
exists.   If $F\in\Sigma$ and $\mu F>0$ then the `conditional 
expectation of $f$ given $F$' is ${1\over{\mu F}}\int_Ff$.   If 
$F_0,\ldots,F_n$ is a partition of $X$ into measurable sets of 
non-zero measure, then the function $g$ given by 
 
\Centerline{$g(x)=\Bover1{\mu F_i}\int_{F_i}f$ if $x\in F_i$} 
 
\noindent is a kind of anticipated conditional expectation;  if we are 
one day told that $x\in F_i$, then $g(x)$ will be our subsequent 
estimate of the expectation of $f$.   In the terms of the definition 
above, $g$ is a conditional expectation of $f$ on the finite algebra 
$\Tau$ generated by $\{F_0,\ldots,F_n\}$.   An appropriate intuition for 
general $\sigma$-algebras $\Tau$ is that they consist of the events 
which we shall be able to observe at some stated future time $t_0$, 
while the whole algebra $\Sigma$ consists of all events, including those 
not observable until times later than $t_0$, if ever. 
}%end of comment 
 
\leader{233E}{}\cmmnt{ I list some of the elementary facts concerning 
conditional expectations. 
 
\medskip 
 
\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a probability space 
and $\Tau$ a 
$\sigma$-subalgebra of $\Sigma$.   Let $\sequencen{f_n}$ be a sequence 
of $\mu$-integrable real-valued functions, and for each $n$ let $g_n$ be 
a conditional expectation of $f_n$ on 
$\Tau$.   Then 
 
(a) $g_1+g_2$ is a conditional expectation of $f_1+f_2$ on $\Tau$; 
 
(b) for any $c\in\Bbb R$, $cg_0$ is a conditional expectation of 
$cf_0$ on $\Tau$; 
 
(c) if $f_1\leae f_2$ then $g_1\leae g_2$; 
 
(d) if $\sequencen{f_n}$ is non-decreasing a.e.\ and 
$f=\lim_{n\to\infty}f_n$ is $\mu$-integrable, then 
$\lim_{n\to\infty}g_n$ is a conditional expectation of $f$ on $\Tau$; 
 
(e) if $f=\lim_{n\to\infty}f_n$ is defined a.e.\ and there is a 
$\mu$-integrable function $h$ such that $|f_n|\leae h$ for every 
$n$, 
then $\lim_{n\to\infty}g_n$ is a conditional expectation of $f$ on 
$\Tau$; 
 
(f) if $F\in\Tau$ then $g_0\times \chi F$ is a conditional 
expectation of $f_0\times\chi F$ on $\Tau$; 
 
(g) if $h$ is a bounded, $\mu\restrp\Tau$-virtually measurable 
real-valued function 
defined $\mu\restrp\Tau$-almost everywhere in $X$, then $g_0\times h$ is 
a conditional expectation of $f_0\times h$ on $\Tau$; 
 
(h) if $\Upsilon$ is a $\sigma$-subalgebra of $\Tau$, then a function 
$h_0$ is a conditional expectation of $f_0$ on $\Upsilon$ iff it is a 
conditional expectation of $g_0$ on $\Upsilon$. 
 
\proof{{\bf (a)-(b)} We have only to observe that 
 
\Centerline{$\int_Fg_1+g_2d(\mu\restrp\Tau) 
=\int_Fg_1d(\mu\restrp\Tau)+\int_Fg_2d(\mu\restrp\Tau) 
=\int_Ff_1d\mu+\int_Ff_2d\mu=\int_Ff_1+f_2d\mu$,} 
 
\Centerline{$\int_Fcg_0d(\mu\restrp\Tau) 
=c\int_Fg_0d(\mu\restrp\Tau)=c\int_Ff_0d\mu=\int_Fcf_0d\mu$} 
 
\noindent for every $F\in\Tau$. 
 
\medskip 
 
{\bf (c)} If $F\in\Tau$ then 
 
\Centerline{$\int_Fg_1d(\mu\restrp\Tau)=\int_Ff_1d\mu\le\int_Ff_2d 
\mu 
=\int_Fg_2d(\mu\restrp\Tau)$} 
 
\noindent for every $F\in\Tau$;  consequently 
$g_1\le g_2\,\mu\restrp\Tau$-a.e.\ (131Ha). 
 
\medskip 
 
{\bf (d)} By (c), $\sequencen{g_n}$ is non-decreasing 
$\mu\restrp\Tau$-a.e.;   moreover, 
 
\Centerline{$\sup_{n\in\Bbb N}\int g_nd(\mu\restrp\Tau) 
=\sup_{n\in\Bbb N}\int f_nd\mu 
=\int fd\mu<\infty$.} 
 
\noindent By B.Levi's theorem, $g=\lim_{n\to\infty}g_n$ is defined 
$\mu\restrp\Tau$-almost everywhere, and 
 
\Centerline{$\int_Fg\,d(\mu\restrp\Tau) 
=\lim_{n\to\infty}\int_Fg_nd(\mu\restrp\Tau) 
=\lim_{n\to\infty}\int_Ff_nd\mu 
=\int_Ffd\mu$} 
 
\noindent for every $F\in\Tau$, so $g$ is a conditional expectation of 
$f$ on $\Tau$. 
 
\medskip 
 
{\bf (e)} Set $f'_n=\inf_{m\ge n}f_m$, $f''_n=\sup_{m\ge n}f_m$ for each 
$n\in\Bbb N$.   Then we have 
 
\Centerline{$-h\leae f'_n\le f_n\le f''_n\leae h$,} 
 
\noindent and $\sequencen{f'_n}$, $\sequencen{f''_n}$ are 
almost-everywhere-monotonic sequences of functions both converging 
almost everywhere to $f$.   For each $n$, let $g'_n$, $g''_n$ be 
conditional expectations of $f'_n$, $f''_n$ on $\Tau$.   By (iii) and 
(iv), $\sequencen{g'_n}$ and $\sequencen{g''_n}$ are 
almost-everywhere-monotonic sequences converging almost everywhere to 
conditional expectations $g'$, $g''$ of $f$.   Of course 
$g'=g''\,\,\mu\restrp\Tau$-a.e.\ (233Dc).   Also, for each $n$, 
$g'_n\leae g_n\leae g''_n$, so $\sequencen{g_n}$ converges to 
$g'\,\,\mu\restrp\Tau$-a.e., and $g=\lim_{n\to\infty}g_n$ is defined 
almost everywhere and is a conditional expectation of $f$ on $\Tau$. 
 
\medskip 
 
{\bf (f)} For any $H\in\Tau$, 
 
\Centerline{$\int_Hg_0\times\chi F\,d(\mu\restrp\Tau) 
=\int_{H\cap F}g_0d(\mu\restrp\Tau) 
=\int_{H\cap F}f_0d\mu 
=\int_Hf_0\times\chi F\,d\mu$.} 
 
\medskip 
 
{\bf (g)}(i) If $h$ is actually $(\mu\restrp\Tau)$-simple, say 
$h=\sum_{i=0}^na_i\chi F_i$ where $F_i\in\Tau$ for each $i$, then 
 
\Centerline{$\int_Fg_0\times h\,d(\mu\restrp\Tau) 
=\sum_{i=0}^na_i\int_Fg_0\times\chi F_id(\mu\restrp\Tau) 
=\sum_{i=0}^na_i\int_Ff\times\chi F_i\,d\mu 
=\int_Ff\times h\,d\mu$} 
 
\noindent for every $F\in\Tau$.   (ii) For the general case, if 
$h$ is $\mu\restrp\Tau$-virtually measurable and 
$|h(x)|\le M\,\,\mu\restrp\Tau$-almost 
everywhere, then there is a sequence $\sequencen{h_n}$ of 
$\mu\restrp\Tau$-simple functions converging to $h$ almost everywhere, 
and with $|h_n(x)|\le M$ for every $x$, $n$.   Now 
$f_0\times h_n\to f_0\times h$ a.e.\ and $|f_0\times h_n|\leae M|f_0|$ 
for each $n$, 
while $g_0\times h_n$ is a conditional expectation of $f_0\times h_n$ 
for every $n$, so by (e) we see that $\lim_{n\to\infty}g_0\times h_n$ 
will be a conditional expectation of $f_0\times h$;  but this is equal 
almost everywhere to $g_0\times h$. 
 
\medskip 
 
{\bf (h)} We need note only that 
$\int_Hg_0d(\mu\restrp\Tau)=\int_Hf_0d\mu$ for every $H\in\Upsilon$, so 
 
$$\eqalign{\int_Hh_0&d(\mu\restrp\Upsilon)=\int_Hg_0d(\mu\restrp\Tau) 
\text{ for every }H\in\Upsilon\cr 
&\iff\int_Hh_0d(\mu\restrp\Upsilon)=\int_Hf_0d\mu 
\text{ for every }H\in\Upsilon.\cr}$$ 
}%end of proof of 233E 
 
\cmmnt{ 
\leader{233F}{Remarks} Of course the results above are 
individually nearly trivial (though I think (e) and (g) might give you 
pause for 
thought if they were offered without previous preparation of the 
ground).   Cumulatively they amount to some quite strong properties. 
In \S242 I will restate them in language which is syntactically 
more direct, but relies on a deeper level of abstraction. 
 
As an illustration of the power of conditional 
expectations to surprise us, I offer the next proposition, which depends 
on the concept of `convex' function. 
}%end of comment 
 
\leader{233G}{Convex functions} Recall that a real-valued function 
$\phi$ defined on an interval $I\subseteq\Bbb R$ is {\bf convex} if 
 
\Centerline{$\phi(tb+(1-t)c)\le t\phi(b)+(1-t)\phi(c)$} 
 
\noindent whenever $b$, $c\in I$ and $t\in[0,1]$. 
 
\cmmnt{\medskip 
 
\noindent{\bf Examples} The formulae $|x|$, $x^2$, $e^{\pm x}\pm x$ 
define convex functions on $\Bbb R$;  on $\ooint{-1,1}$ we have 
$1/(1-x^2)$; on $\ooint{0,\infty}$ we have $1/x$ and $x\ln x$;  on 
$[0,1]$ we have the function which is zero on $\ooint{0,1}$ and $1$ on 
$\{0,1\}$. 
}%end of comment 
 
\leader{233H}{}\cmmnt{ The general theory of convex functions is both 
extensive and important;  I list a few of their more salient properties 
in 233Xe. 
For the moment the following lemma covers what we need. 
 
\medskip 
 
\noindent}{\bf Lemma} Let $I\subseteq\Bbb R$ be a non-empty open 
interval\cmmnt{ (bounded or unbounded)} and $\phi:I\to\Bbb R$ a convex 
function. 
 
(a) For every $a\in I$ there is a $b\in\Bbb R$ such that 
$\phi(x)\ge\phi(a)+b(x-a)$ for every $x\in I$. 
 
(b) If we take, for each $q\in I\cap\Bbb Q$, a $b_q\in\Bbb R$ such that 
$\phi(x)\ge\phi(q)+b_q(x-q)$ for every $x\in I$, then 
 
\Centerline{$\phi(x)=\sup_{q\in I\cap\Bbb Q}\phi(q)+b_q(x-q)$} 
 
\noindent for every $x\in I$. 
 
(c) $\phi$ is Borel measurable. 
 
\proof{{\bf (a)} If $c$, $c'\in I$ and $c<a<c'$, 
then $a$ is expressible as $dc+(1-d)c'$ for some $d\in\ooint{0,1}$, so 
that $\phi(a)\le d\phi(c)+(1-d)\phi(c')$ and 
 
$$\eqalign{{\Bover{\phi(a)-\phi(c)}{a-c}} 
&\le{\Bover{d\phi(c)+(1-d)\phi(c')-\phi(c)}{dc+(1-d)c'-c}} 
={\Bover{(1-d)(\phi(c')-\phi(c))}{(1-d)(c'-c)}}\cr 
&={\Bover{d(\phi(c')-\phi(c))}{d(c'-c)}} 
={\Bover{\phi(c')-d\phi(c)-(1-d)\phi(c')}{c'-dc-(1-d)c'}} 
\le{\Bover{\phi(c')-\phi(a)}{c'-a}}.\cr}$$ 
 
\noindent This means that 
 
\Centerline{$b=\sup_{c<a,c\in I}\Bover{\phi(a)-\phi(c)}{a-c}$} 
 
\noindent is finite, and $b\le\Bover{\phi(c')-\phi(a)}{c'-a}$ whenever 
$a<c'\in I$;  accordingly $\phi(x)\ge\phi(a)+b(x-a)$ for every $x\in I$. 
 
\medskip 
 
{\bf (b)}  By the choice of the $b_q$, $\phi(x)\ge\sup_{q\in\Bbb 
Q}\phi_q(x)$. 
On the other hand, given $x\in I$, fix $y\in I$ such that $x<y$ and let 
$b\in\Bbb R$ be such that $\phi(z)\ge\phi(x)+b(z-x)$ for every $z\in I$. 
If $q\in\Bbb Q$ and $x<q<y$, we have $\phi(y)\ge\phi(q)+b_q(y-q)$, so 
that 
$b_q\le\bover{\phi(y)-\phi(q)}{y-q}$ and 
 
$$\eqalign{\phi(q)+b_q(x-q) 
&=\phi(q)-b_q(q-x) 
\ge\phi(q)-\Bover{\phi(y)-\phi(q)}{y-q}(q-x)\cr 
&=\Bover{y-x}{y-q}\phi(q)-\Bover{q-x}{y-q}\phi(y) 
\ge\Bover{y-x}{y-q}(\phi(x)+b(q-x))-\Bover{q-x}{y-q}\phi(y).\cr}$$ 
 
\noindent Now 
 
 
$$\eqalign{\phi(x) 
&=\lim_{q\downarrow x} 
   \Bover{y-x}{y-q}(\phi(x)+b(q-x))-\Bover{q-x}{y-q}\phi(y)\cr 
&\le\sup_{q\in\Bbb Q\cap\ooint{x,y}} 
   \Bover{y-x}{y-q}(\phi(x)+b(q-x))-\Bover{q-x}{y-q}\phi(y)\cr 
&\le\sup_{q\in\Bbb Q\cap\ooint{x,y}}\phi(q)+b_q(x-q) 
\le\sup_{q\in\Bbb Q\cap I}\phi(q)+b_q(x-q).\cr}$$ 
 
\medskip 
 
{\bf (c)} Writing $\phi_q(x)=\phi(q)+b_q(x-q)$ for every  
$q\in\Bbb Q\cap I$, every $\phi_q$ is a Borel measurable function, and  
$\phi=\sup_{q\in I\cap\Bbb Q}\phi_q$ is the supremum of a countable  
family of Borel measurable functions, so is Borel measurable. 
}%end of proof of 233H 
 
\leader{233I}{Jensen's inequality} Let $(X,\Sigma,\mu)$ be a measure 
space and $\phi:\Bbb R\to\Bbb R$ a convex function. 
 
(a) Suppose that $f$ and $g$ are real-valued $\mu$-virtually measurable 
functions defined almost everywhere in $X$ and that  
$g\ge 0$ almost everywhere, 
$\int g=1$ and $g\times f$ is integrable.   Then 
$\phi(\int g\times f)\le\int g\times\phi f$, 
where we may need to interpret the right-hand integral as 
$\infty$. 
 
(b) In particular, if $\mu X=1$ and $f$ is a real-valued function which 
is integrable over $X$, then $\phi(\int f)\le\int\phi f$. 
 
\proof{{\bf (a)} For each $q\in\Bbb Q$ take $b_q$ such that 
$\phi(t)\ge\phi_q(t)=\phi(q)+b_q(t-q)$ for every $t\in\Bbb R$ (233Ha). 
Because $\phi$ is Borel measurable (233Hc), $\phi f$ is $\mu$-virtually 
measurable (121H), so $g\times\phi f$ also is;   since $g\times\phi f$ 
is defined almost everywhere and almost everywhere greater than or equal 
to the integrable function $g\times\phi_0f$, $\int g\times\phi f$ is 
defined in $\ocint{-\infty,\infty}$.   Now 
 
$$\eqalign{\phi_q\bigl(\intop g\times f\bigr) 
&=\phi(q)+b_q\int g\times f-b_qq\cr 
&=\int g\times(b_qf+(\phi(q)-b_qq)\chi X) 
=\int g\times\phi_qf 
\le\int g\times\phi f,\cr}$$ 
 
\noindent because $\int g=1$ and $g\ge 0$ a.e.   By 233Hb, 
 
\Centerline{$\phi(\intop g\times f) 
=\sup_{q\in\Bbb Q}\phi_q(\intop g\times f) 
\le\int g\times\phi f$.} 
 
\medskip 
 
{\bf (b)} Take $g$ to be the constant function with value $1$. 
}%end of proof of 233I 
 
\leader{233J}{}\cmmnt{ Even the special case 233Ib of Jensen's 
inequality is already very useful.   It can be extended as follows. 
 
\medskip 
 
\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a 
probability space and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Let 
$\phi:\Bbb R\to\Bbb R$ be a convex function and $f$ a $\mu$-integrable 
real-valued function defined almost everywhere in $X$ such that the 
composition $\phi f$ is also 
integrable.   If $g$ and $h$ are conditional expectations on $\Tau$ of 
$f$, $\phi f$ respectively, then $\phi g\leae h$.   Consequently 
$\int\phi g\le\int\phi f$. 
 
\proof{ We use the same ideas as in 233I.   For each $q\in\Bbb Q$ take a 
$b_q\in\Bbb R$ such 
that $\phi(t)\ge\phi_q(t)=\phi(q)+b_q(t-q)$ for every $t\in\Bbb R$, so 
that $\phi(t)=\sup_{q\in\Bbb Q}\phi_q(t)$ for every $t\in\Bbb R$.   Now 
setting 
 
\Centerline{$\psi_q(x)=\phi(q)+b_q(g(x)-q)$} 
 
\noindent for $x\in\dom g$, we see that $\psi_q=\phi_q g$ is a 
conditional expectation of $\phi_q f$, and as 
$\phi_q f\leae\phi f$ we must have $\psi_q\leae h$.   But also  
$\phi g=\sup_{q\in\Bbb Q}\psi_q$ wherever $g$ is 
defined, so $\phi g\leae h$, as claimed. 
 
It follows at once that $\int\phi g\le\int h=\int\phi f$. 
}%end of proof of 233J 
 
\leader{233K}{}\cmmnt{ I give the following proposition, an 
elaboration of 
233Eg, in a very general form, as its applications can turn up anywhere. 
 
\medskip 
 
\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a probability space, 
and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Suppose that $f$ is a 
$\mu$-integrable function and $h$ is a $(\mu\restrp\Tau)$-virtually 
measurable real-valued function defined $(\mu\restrp\Tau)$-almost 
everywhere in $X$.   Let $g$, $g_0$ be conditional expectations of $f$ 
and $|f|$ on $\Tau$.   Then $f\times h$ is integrable iff $g_0\times h$ 
is integrable, and in this case $g\times h$ is a conditional expectation 
of $f\times h$ on $\Tau$. 
 
\proof{{\bf (a)} Suppose that $h$ is a $\mu\restrp\Tau$-simple function. 
Then surely $f\times h$ and $g_0\times h$ are integrable, and $g\times 
h$ is a conditional expectation of $f\times h$ as in 233Eg. 
 
\medskip 
 
{\bf (b)} Now suppose that $f$, $h\ge 0$.   Then $g=g_0\ge 0$ a.e. 
(233Ec).  Let $\tilde h$ be a non-negative 
$\Tau$-measurable function defined everywhere in $X$ such that 
$h\eae\tilde h$.    For each $n\in\Bbb N$ set 
 
$$\eqalign{h_n(x) 
&=2^{-n}k\text{ if }0\le k<4^n 
  \text{ and }2^{-n}k\le\tilde h(x)<2^{-n}(k+1),\cr 
&=2^n\text{ if }\tilde h(x)\ge 2^{-n}.\cr}$$ 
 
\noindent Then $h_n$ is a $(\mu\restrp\Tau)$-simple function, so 
$g\times h_n$ is a conditional expectation of $f\times h_n$. 
Both $\sequencen{f\times h_n}$ and $\sequencen{g\times h_n}$ are almost 
everywhere non-decreasing sequences of integrable functions, with limits 
$f\times h$ and $g\times h$ respectively.   By B.Levi's theorem, 
 
$$\eqalignno{f\times h\text{ is integrable} 
&\iff\,f\times\tilde h\text{ is integrable}\cr 
&\iff\,\sup_{n\in\Bbb N}\int f\times h_n<\infty 
\iff\,\sup_{n\in\Bbb N}\int g\times h_n<\infty\cr 
\noalign{\noindent (because $\int g\times h_n=\int f\times h_n$ for each 
$n$)} 
&\iff\,g\times h\text{ is integrable} 
\iff\,g_0\times h\text{ is integrable}.\cr}$$ 
 
\noindent Moreover, in this case 
 
$$\eqalign{\int_Ef\times h 
&=\int_Ef\times\tilde h 
=\lim_{n\to\infty}\int_Ef\times h_n\cr 
&=\lim_{n\to\infty}\int_Eg\times h_n 
=\int_Eg\times\tilde h 
=\int_Eg\times h\cr}$$ 
 
\noindent for every $E\in\Tau$, while $g\times h$ is 
$(\mu\restrp\Tau)$-virtually measurable, so $g\times h$ is a 
conditional expectation of $f\times h$. 
 
\medskip 
 
{\bf (c)} Finally, consider the general case of integrable $f$ and 
virtually measurable $h$.   Set $f^+=f\vee 0$, $f^-=(-f)\vee 0$, 
so that $f=f^+-f^-$ and $0\le f^+,\,f^-\le|f|$;  similarly, set 
$h^+=h\vee 0$, $h^-=(-h)\vee 0$.   Let $g_1$, $g_2$ be conditional 
expectations of $f^+$, $f^-$ on $\Tau$.   Because $0\le f^+$, 
$f^-\le|f|$, $0\le g_1$, $g_2\leae g_0$,  while $g\eae g_1-g_2$. 
 
We see that 
 
$$\eqalign{f\times h\text{ is integrable} 
&\iff\,|f|\times|h|=|f\times h|\text{ is integrable}\cr 
&\iff\,g_0\times |h|\text{ is integrable}\cr 
&\iff\,g_0\times h\text{ is integrable}.\cr}$$ 
 
\noindent And in this case all four of 
$f^+\times h^+,\ldots,f^-\times h^-$ are integrable, so 
 
\Centerline{$(g_1-g_2)\times h 
=g_1\times h^+-g_2\times h^+-g_1\times h^-+g_2\times h^-$} 
 
\noindent is a conditional expectation of 
 
\Centerline{$f^+\times h^+-f^-\times h^+-f^+\times h^-+f^-\times h^- % 
=f\times h$.} 
 
\noindent Since $g\times h\eae(g_1-g_2)\times h$, this also is a 
conditional expectation of $f\times h$, and we're done. 
}%end of proof of 233K 
 
\exercises{ 
\leader{233X}{Basic exercises (a)} 
%\spheader 233Xa 
Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$ a 
$\sigma$-subalgebra of $\Sigma$.   Let $\sequencen{f_n}$ be a sequence 
of non-negative $\mu$-integrable functions and suppose that $g_n$ is a 
conditional expectation of $f_n$ on $\Tau$ for each $n$.   Suppose that 
$f=\liminf_{n\to\infty}f_n$ is integrable and has a conditional 
expectation $g$.   Show that $g\leae\liminf_{n\to\infty}g_n$. 
%233E 
 
\spheader 233Xb Let $I\subseteq\Bbb R$ be an interval, and 
$\phi:I\to\Bbb R$ a function.   Show that $\phi$ is convex iff 
$\{x:x\in I,\,\phi(x)+bx\le c\}$ is an interval for every $b$, 
$c\in\Bbb R$. 
%233G 
 
\sqheader 233Xc Let $I\subseteq\Bbb R$ be an open interval and 
$\phi:I\to\Bbb R$ a function.   (i) Show that if $\phi$ is 
differentiable then it is convex iff $\phi'$ is non-decreasing.   (ii) 
Show that if $\phi$ is absolutely continuous on every bounded closed 
subinterval of $I$ then $\phi$ is convex iff 
$\phi'$ is non-decreasing on its domain. 
 
\spheader 233Xd For any $r\ge 1$, a subset $C$ of $\BbbR^r$ is 
{\bf convex} if $tx + (1-t)y\in C$ for all $x$, $y\in C$ and 
$t\in[0,1]$.   If $C\subseteq\BbbR^r$ is convex, then a function 
$\phi:C\to\Bbb R$ is {\bf convex} if 
$\phi(tx+(1-t)y)\le t\phi(x)+(1-t)\phi(y)$ for all $x$, $y\in C$ and 
$t\in[0,1]$. 
 
Let $C\subseteq\BbbR^r$ be a convex set and $\phi:C\to\Bbb R$ a 
function.   Show that the following are equiveridical:   (i) the 
function $\phi$ is convex; (ii) the set 
$\{(x,t):x\in C,\,t\in\Bbb R,\,t\ge\phi(x)\}$ is convex in 
$\BbbR^{r+1}$;  (iii) the set 
$\{x:x\in C,\,\phi(x)+b\dotproduct x\le c\}$ is convex in $\BbbR^r$ for 
every $b\in\BbbR^r$ and $c\in\Bbb R$, writing 
$b\dotproduct x=\sum_{i=1}^r\beta_i\xi_i$ if 
$b=(\beta_1,\ldots,\beta_r)$ and $x=(\xi_1,\ldots,\xi_r)$. 
%233G 
 
\spheader 233Xe Let $I\subseteq\Bbb R$ be an interval 
and $\phi:I\to\Bbb R$ a convex function. 
 
\quad (i) Show that if $a$, $d\in I$ and $a<b\le c<d$ then 
 
\Centerline{$\Bover{\phi(b)-\phi(a)}{b-a} 
\le\Bover{\phi(d)-\phi(c)}{d-c}$.} 
 
\quad (ii) Show that $\phi$ is continuous at every interior point of 
$I$. 
 
\quad (iii) Show that {\it either} $\phi$ is monotonic on $I$ {\it or} 
there is a $c\in I$ such that $\phi(c)=\min_{x\in I}\phi(x)$ and $\phi$ 
is non-increasing on $I\cap\ocint{-\infty,c}$, monotonic 
non-decreasing on $I\cap\coint{c,\infty}$. 
 
\quad (iv) Show that $\phi$ is differentiable at all but countably many 
points of $I$, and that its derivative is non-decreasing in 
the sense that $\phi'(x)\le\phi'(y)$ whenever $x$, $y\in\dom\phi'$ and 
$x\le y$. 
 
\quad (v) Show that if $I$ is closed and bounded and $\phi$ is 
continuous then $\phi$ is absolutely continuous. 
 
\quad (vi) Show that if $I$ is closed and bounded and $\psi:I\to\Bbb R$ 
is absolutely continuous with a non-decreasing derivative then 
$\psi$ is convex. 
%233H 
 
\spheader 233Xf Show that if $I\subseteq\Bbb R$ is an interval and 
$\phi$, $\psi:I\to\Bbb R$ are convex functions so is $a\phi+b\psi$ for 
any $a$, $b\ge 0$. 
%233H 
 
\spheader 233Xg In the context of 233K, give an example in which 
$g\times h$ is integrable but $f\times h$ is not.  \Hint{take 
$X$, $\mu$, $\Tau$ as in 233Cb, and arrange for $g$ to be $0$.} 
%233K 
 
\spheader 233Xh Let $I\subseteq\Bbb R$ be an interval and $\Phi$ a 
non-empty family of convex real-valued functions on $I$ such that 
$\psi(x)=\sup_{\phi\in\Phi}\phi(x)$ is finite for every $x\in I$.   Show 
that $\psi$ is convex. 
%233H 
 
\leader{233Y}{Further exercises (a)} If $I\subseteq\Bbb R$ is an 
interval, a function $\phi:I\to\Bbb R$ is {\bf mid-convex} if 
$\phi(\bover{x+y}2)\le\bover12(\phi(x)+\phi(y))$ for all $x$, $y\in I$. 
Show that a mid-convex function which is bounded on any non-trivial 
subinterval of $I$ is convex. 
%233G 
 
\spheader 233Yb Generalize 233Xd to arbitrary normed spaces in 
place of $\BbbR^r$. 
%233G 
 
\spheader 233Yc Let $(X,\Sigma,\mu)$ be a probability space and 
$\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Let $\phi$ be a convex 
real-valued function with domain an interval $I\subseteq\Bbb R$, and $f$ 
an integrable real-valued function on $X$ such that $f(x)\in I$ for 
almost every $x\in X$ and $\phi f$ is integrable.   Let $g$, 
$h$ be conditional expectations on $\Tau$ of $f$, $\phi f$ 
respectively.   Show that $g(x)\in I$ for almost every $x$ and that 
$\phi g\leae h$. 
%233J 
 
\spheader 233Yd(i) Show that if $I\subseteq\Bbb R$ is a bounded 
interval, $E\subseteq I$ is Lebesgue measurable, and 
$\mu E>\bover23\mu I$ where $\mu$ is Lebesgue measure, then for every 
$x\in I$ there are $y$, $z\in E$ such that $z=\bover{x+y}2$.   \Hint{by 
134Ya/263A, $\mu(x+E)+\mu(2E)>\mu(2I)$.}   (ii) Show that if 
$f:[0,1]\to\Bbb R$ is a mid-convex Lebesgue measurable function 
(definition:  233Ya), $a>0$, and $E=\{x:x\in[0,1]$, $a\le f(x)<2a\}$ is 
not negligible, then there is a non-trivial interval $I\subseteq[0,1]$ 
such that $f(x)>0$ for every $x\in I$.   \Hint{223B.}   (iii) Suppose 
that $f:[0,1]\to\Bbb R$ is a mid-convex function such that $f\le 0$ 
almost everywhere in $[0,1]$.   Show that $f\le 0$ everywhere in 
$\ooint{0,1}$.   \Hint{for every $x\in\ooint{0,1}$, 
$\max(f(x-t),f(x+t))\le 0$ for almost every $t\in[0,\min(x,1-x)]$.} 
(iv) Suppose that $f:[0,1]\to\Bbb R$ is a mid-convex Lebesgue measurable 
function such that $f(0)=f(1)=0$.   Show that $f(x)\le 0$ for every 
$x\in[0,1]$.   \Hint{show that $\{x:f(x)\le 0\}$ is dense in $[0,1]$, 
use (ii) to show that it is conegligible in $[0,1]$ and apply (iii).} 
(v) Show that if $I\subseteq\Bbb R$ is an interval and $f:I\to\Bbb R$ is 
a mid-convex Lebesgue measurable function then it is convex. 
%233Ya 233G not in 2001 edition 
 
\spheader 233Ye Let $(X,\Sigma,\mu)$ be a probability space, $\Tau$ a 
$\sigma$-subalgebra of subsets of $X$, and $f:X\to[0,\infty]$ a 
$\Sigma$-measurable function.   Show that (i) there is a $\Tau$-measurable 
$g:X\to[0,\infty]$ such that $\int_Fg=\int_Ff$ for every $F\in\Tau$ 
(ii) any two such functions are equal a.e. 
 
\spheader 233Yf Suppose that $r\ge 1$ and $C\subseteq\BbbR^r\setminus\{0\}$ 
is a convex set.   Show that there is a non-zero $b\in\BbbR^r$  
such that $\varinnerprod{b}{z}\ge 0$ for every $z\in C$. 
\Hint{if $r=2$, identify $\BbbR^2$ with $\Bbb C$;  reduce to the case in 
which $C$ contains no points which are real and negative;  set 
$\theta=\sup\{\arg z:z\in C\}$ and $b=-ie^{i\theta}$.   Now induce on $r$.} 
%233Xd 
 
\spheader 233Yg Suppose that $r\ge 1$, $C\subseteq\BbbR^r$ is a convex set 
and $\phi:C\to\Bbb R$ is a convex function.   Show that there is a function 
$h:\BbbR^r\to\coint{-\infty,\infty}$ such that 
$\phi(z)=\sup\{h(y)+\varinnerprod{z}{y}:y\in\BbbR^r\}$ for every $z\in C$. 
\Hint{try $h(y)=\inf\{\phi(z)-\varinnerprod{z}{y}:z\in C\}$, and apply 
233Yf to a translate of $\{(z,t):\phi(z)\le t\}$.} 
%233Yf 233Xd 233H 
 
\spheader 233Yh Let $(X,\Sigma,\mu)$ be a probability space, $r\ge 1$ an 
integer and $C\subseteq\BbbR^r$ a convex set.    
Let $f_1,\ldots,f_r$ be $\mu$-integrable real-valued functions 
and suppose that $\{x:x\in\bigcap_{j\le r}\dom f_j$, 
$(f_1(x),\ldots,f_r(x))\in C\}$ is a conegligible subset of $X$. 
Show that $(\int f_1,\ldots,\int f_r)\in C$. 
\Hint{induce on $r$.} 
%233Yg 233H 
 
\spheader 233Yi Let $(X,\Sigma,\mu)$ be a probability space, $r\ge 1$ an 
integer, $C\subseteq\BbbR^r$ a convex set and $\phi:C\to\BbbR^r$ a convex 
function.   Let $f_1,\ldots,f_r$ be $\mu$-integrable real-valued functions 
and suppose that $\{x:x\in\bigcap_{j\le r}\dom f_j$, 
$(f_1(x),\ldots,f_r(x))\in C\}$ is a conegligible subset of $X$. 
Show that $\phi(\int f_1,\ldots,\int f_r)\le\int\phi(f_1,\ldots,f_r)$. 
%233Yg 233Yh 233H 
 
\spheader 233Yj Let $(X,\Sigma,\mu)$ be a measure space with $\mu X>0$,  
$r\ge 1$ an 
integer, $C\subseteq\BbbR^r$ a convex set such that $tz\in C$ whenever 
$z\in C$ and $t>0$, and $\phi:C\to\Bbb R$ a convex function. 
Let $f_1,\ldots,f_r$ be $\mu$-integrable real-valued functions 
and suppose that $\{x:x\in\bigcap_{j\le r}\dom f_j$, 
$(f_1(x),\ldots,f_r(x))\in C\}$ is a conegligible subset of $X$. 
Show that $(\int f_1,\ldots,\int f_r)\in C$ and that 
$\phi(\int f_1,\ldots,\int f_r)\le\int\phi(f_1,\ldots,f_r)$. 
\Hint{putting 215B(viii) and 235K below together, show that there are a 
probability measure $\nu$ on $X$ and a function $h:X\to\coint{0,\infty}$ 
such that $\int f_jd\mu=\int f_j\times h\,d\nu$ for every $j$.} 
%233Yh 233Yi 233H 
 
}%end of exercises 
 
\endnotes{ 
\Notesheader{233} The concept of `conditional expectation' is 
fundamental in probability theory, and will reappear in Chapter 27 in 
its natural context.   I hope that even as an exercise in technique, 
however, it will strike you as worth taking note of. 
 
I introduced 233E as a `list of elementary facts', and they are indeed 
straightforward.   But below the surface there are some remarkable ideas 
waiting for expression.   If you take $\Tau$ to be the trivial algebra 
$\{\emptyset,X\}$, so that the (unique) conditional expectation of an 
integrable function $f$ is the constant function $(\int f)\chi X$, then 
233Ed and 233Ee become versions of B.Levi's theorem and Lebesgue's 
Dominated Convergence Theorem.   (Fatou's Lemma is in 233Xa.)   Even 
233Eg can be thought of as a generalization of the result that 
$\int cf=c\int f$, where the constant $c$ has been replaced by a bounded 
$\Tau$-measurable function.   A recurrent theme in the later parts of 
this treatise will be the search for `conditioned' versions of theorems. 
The proof of 233Ee is a typical example of an argument which has been 
translated from a proof of the original `unconditioned' result. 
 
I suggested that 233I-233J are surprising, and I think 
that most of us find them so, even applied to the list of convex 
functions given in 233G.   But I should remark that in a way 233J has 
very little to 
do with conditional expectations.   The only properties of conditional 
expectations used in the proof are (i) that if $g$ is a conditional 
expectation of $f$, then $a\chi X+bg$ is a conditional expectation of 
$a\chi X+bf$ for all real $a$, $b$ (ii) if $g_1$, $g_2$ are conditional 
expectations of $f_1$, $f_2$ and $f_1\leae f_2$, then $g_1\leae g_2$.    
See 244Xm below.   Jensen's inequality has an interesting extension to the 
multidimensional case, explored in  
233Yf-233Yj.   %233Yf 233Yg 233Yh 233Yi 233Yj 
If you have encountered `geometric' forms of the Hahn-Banach theorem (see 
3A5C in Volume 3) you will find 233Yf and 233Yg very natural, and you may 
notice that the finite-dimensional case is slightly different from the 
infinite-dimensional case you have probably been taught.    
I think that in fact the most 
delicate step is in 233Yh. 
 
Note that 233Ib can be regarded as the special case of 233J in which 
$\Tau=\{\emptyset,X\}$.   In fact 233Ia can be derived from 233Ib 
applied to the measure $\nu$ where $\nu E=\int_Eg$ for every 
$E\in\Sigma$. 
 
Like 233B, 233K seems to have rather a lot of technical detail in the 
argument.   The point of this result is that we can deduce the 
integrability of $f\times h$ from that of $g_0\times h$ (but not from 
the integrability of $g\times h$;  see 233Xg). 
Otherwise it should be routine. 
}%end of notes 
 
\discrpage 
 
\frfilename{mt234.tex}
\versiondate{11.4.09}
\copyrightdate{2008}

\def\chaptername{The Radon-Nikod\'ym theorem}
\def\sectionname{Operations on measures}

\newsection{234}

I take a few pages to describe some standard constructions.   The ideas are
straightforward, but a number of details need to be worked
out if they are to be securely integrated into the general framework I
employ.   The first step is to formally introduce \imp\ functions
(234A-234B), the most important class of transformations between measure
spaces.   For construction of new measures, we have the notions of
image measure (234C-234E), %234C 234D 234E
sum of measures (234G-234H) and indefinite-integral measure
(234I-234O).  %234I 234J 234K 234L 234M 234N 234O
Finally I mention a way of ordering the measures on a given set
(234P-234Q).


\leader{234A}{Inverse-measure-preserving functions}\cmmnt{ It is high
time that I introduced the nearest thing in measure theory to a
`morphism'.}   If $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are measure
spaces, a function $\phi:X\to Y$ is {\bf \imp} if
$\phi^{-1}[F]\in\Sigma$ and $\mu(\phi^{-1}[F])=\nu F$ for every
$F\in\Tau$.

\leader{234B}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\phi:X\to Y$ an \imp\ function.


(a) If $\hat\mu$, $\hat\nu$ are the completions of $\mu$, $\nu$
respectively, $\phi$ is also \imp\ for $\hat\mu$ and $\hat\nu$.

(b) $\mu$ is a probability measure iff $\nu$ is a probability measure.

(c) $\mu$ is totally finite iff $\nu$ is totally finite.

(d)(i) If $\nu$ is $\sigma$-finite, then $\mu$ is $\sigma$-finite.

\quad(ii) If $\nu$ is semi-finite and $\mu$ is $\sigma$-finite, then $\nu$
is $\sigma$-finite.

(e)(i)
If $\nu$ is $\sigma$-finite and atomless, then $\mu$ is atomless.

\quad(ii) If $\nu$ is semi-finite and $\mu$ is purely atomic, then
$\nu$ is purely atomic.

(f)(i) $\mu^*\phi^{-1}[B]\le\nu^*B$ for
every $B\subseteq Y$.

\quad(ii) $\mu^*A\le\nu^*\phi[A]$ for every $A\subseteq X$.

(g) If $(Z,\Lambda,\lambda)$ is another measure space, and
$\psi:Y\to Z$ is \imp,  then $\psi\phi:X\to Z$ is \imp.

\proof{{\bf (a)} If $\hat\nu$ measures $F$, there are $F'$, $F''\in\Tau$
such that $F'\subseteq F\subseteq F''$ and $\nu(F''\setminus F')=0$.
Now

\Centerline{$\phi^{-1}[F']\subseteq\phi^{-1}[F]\subseteq\phi^{-1}[F'']$,
\quad$\mu(\phi^{-1}[F'']\setminus\phi^{-1}[F'])=\nu(F''\setminus F')=0$,}

\noindent so $\hat\mu$ measures $\phi^{-1}[F]$ and

\Centerline{$\hat\mu(\phi^{-1}[F])=\mu\phi^{-1}[F']=\nu F'=\hat\nu F$.}

\noindent As $F$ is arbitrary, $\phi$ is \imp\ for $\hat\mu$ and $\hat\nu$.

\medskip

{\bf (b)-(c)} are surely obvious.

\medskip

{\bf (d)(i)} If $\sequencen{F_n}$ is a cover of $Y$ by sets of finite
measure for $\nu$, then $\sequencen{\phi^{-1}[F_n]}$ is a cover of $X$ by
sets of finite measure for $\mu$.

\medskip

\quad{\bf (ii)} Let $\Cal F\subseteq\Tau$ be a disjoint family of
non-$\nu$-negligible sets.   Then $\family{F}{\Cal F}{\phi^{-1}[F]}$ is a
disjoint family of non-$\mu$-negligible sets.   By 215B(iii), $\Cal F$ is
countable.   By 215B(iii) in the opposite direction, $\nu$ is
$\sigma$-finite.

\medskip

{\bf (e)(i)} Suppose that $E\in\Sigma$ and $\mu E>0$.   Let
$\sequencen{F_n}$ be a cover of $Y$ by sets of finite measure for $\nu$.
Because $\nu$ is atomless, we can find, for each $n$, a
finite partition $\family{i}{I_n}{F_{ni}}$
of $F_n$ such that $\nu F_{ni}<\mu E$ for every $i\in I_n$ (use 215D
repeatedly).   Now
$X=\bigcup_{n\in\Bbb N,i\in I_n}\phi^{-1}[F_{ni}]$, so there are
$n\in\Bbb N$ and $i\in I_n$ with

\Centerline{$0<\mu(E\cap\phi^{-1}[F_{ni})
\le\mu\phi^{-1}[F_{ni}]=\nu F_{ni}<\mu E$,}

\noindent and $E$ is not a $\mu$-atom.   As $E$ is arbitrary, $\mu$ is
atomless.

\medskip

\quad{\bf (ii)} Suppose that $F\in\Tau$ and $\nu F>0$.   Because $\nu$ is
semi-finite, there is an $F_1\subseteq F$ such that $0<\nu F_1<\infty$.
Now $\mu\phi^{-1}[F_1]>0$;  because $\mu$ is purely atomic, there is a
$\mu$-atom $E\subseteq\phi^{-1}[F_1]$.

Let $\Cal G$ be the set of those $G\in\Tau$ such that $G\subseteq F_1$ and
$\mu(E\cap\phi^{-1}[G])=0$.   Then the union of any sequence in $\Cal G$
belongs to $\Cal G$, so by 215Ac there is an $H\in\Cal G$ such that
$\nu(G\setminus H)=0$ whenever $G\in\Cal G$.   Consider $F_1\setminus H$.
We have

\Centerline{$\nu(F_1\setminus H)
=\mu(\phi^{-1}[F_1]\setminus\phi^{-1}[H])
\ge\mu(E\setminus\phi^{-1}[H])
=\mu E>0$.}

\noindent If $G\in\Tau$ and $G\subseteq F_1\setminus H$,
then one of $E\cap\phi^{-1}[G]$, $E\setminus\phi^{-1}[G]$ is
$\mu$-negligible.
In the former case, $G\in\Cal G$ and $G=G\setminus H$ is $\nu$-negligible.
In the
latter case, $F_1\setminus G\in\Cal G$ and $(F_1\setminus H)\setminus G$ is
$\nu$-negligible.   As $G$ is arbitrary, $F_1\setminus H$ is a $\nu$-atom
included in $F$;  as $F$ is arbitrary, $\nu$ is purely atomic.

\medskip

{\bf (f)(i)} Let $F\in\Tau$ be such that $B\subseteq F$ and $\nu^*B=\nu F$
(132Aa);  then $\phi^{-1}[B]\subseteq\phi^{-1}[F]$ so

\Centerline{$\mu^*\phi^{-1}[B]\le\mu\phi^{-1}[F]=\nu F=\nu^*B$.}

\medskip

\quad{\bf (ii)} $\mu^*A\le\mu^*(\phi^{-1}[\phi[A]])\le\nu^*\phi[A]$ by (i).

\medskip

{\bf (g)} For any $W\in\Lambda$,

\Centerline{$\mu(\psi\phi)^{-1}[W]=\mu\phi^{-1}[\psi^{-1}[W]]
=\nu\psi^{-1}[W]=\lambda W$.}
}%end of proof of 234B


\leader{234C}{Image \dvrocolon{measures}}\cmmnt{ The following
construction is one of the commonest ways in which new measure spaces
appear.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, $Y$
any set, and $\phi:X\to Y$ a function.   Set

\Centerline{$\Tau=\{F:F\subseteq Y,\,\phi^{-1}[F]\in\Sigma\}$,
\quad$\nu F=\mu(\phi^{-1}[F])$ for every $F\in\Tau$.}

\noindent Then $(Y,\Tau,\nu)$ is a measure space.

\proof{{\bf (a)} $\emptyset=\phi^{-1}[\emptyset]\in\Sigma$ so
$\emptyset\in\Tau$.

\medskip

{\bf (b)} If $F\in\Tau$, then $\phi^{-1}[F]\in\Sigma$, so
$X\setminus\phi^{-1}[F]\in\Sigma$;  but
$X\setminus\phi^{-1}[F]=\phi^{-1}[Y\setminus F]$, so $Y\setminus
F\in\Tau$.

\medskip

{\bf (c)} If $\sequencen{F_n}$ is a sequence in $\Tau$,
then $\phi^{-1}[F_n]\in\Sigma$ for every $n$, so
$\bigcup_{n\in\Bbb N}\phi^{-1}[F_n]\in\Sigma$;  but
$\phi^{-1}[\bigcup_{n\in\Bbb N}F_n]=\bigcup_{n\in\Bbb N}\phi^{-1}[F_n]$,
so $\bigcup_{n\in\Bbb N}F_n\in\Tau$.

Thus $\Tau$ is a $\sigma$-algebra.

\medskip

{\bf (d)} $\nu\emptyset=\mu\phi^{-1}[\emptyset]=\mu\emptyset=0$.

\medskip

{\bf (e)} If $\sequencen{F_n}$ is a disjoint sequence in $\Tau$, then
$\sequencen{\phi^{-1}[F_n]}$ is a disjoint sequence in $\Sigma$, so

\Centerline{$\nu(\bigcup_{n\in\Bbb N}F_n)
=\mu\phi^{-1}[\bigcup_{n\in\Bbb N}F_n]
=\mu(\bigcup_{n\in\Bbb N}\phi^{-1}[F_n])
=\sum_{n=0}^{\infty}\mu\phi^{-1}[F_n]
=\sum_{n=0}^{\infty}\nu F_n$.}

\noindent So $\nu$ is a measure.

}%end of proof of 234C

\leader{234D}{Definition} In the context of 234C, $\nu$ is called
the {\bf image measure}\cmmnt{ or {\bf push-forward measure}};  
I will denote it $\mu\phi^{-1}$.

\cmmnt{\medskip

\noindent{\bf Remark} I ought perhaps to say that this construction does
not always produce exactly the `right' measure on $Y$;  there are
circumstances in which some modification of the measure $\mu\phi^{-1}$
described here is more useful.   But I will note these explicitly when
they occur;  when I use the unadorned phrase `image measure' I shall
mean the measure constructed above.
}%end of comment

\leader{234E}{Proposition} Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a
set and $\phi:X\to Y$ a function;  let $\mu\phi^{-1}$ be the image measure
on $Y$.

(a) $\phi$ is \imp\ for $\mu$ and $\mu\phi^{-1}$.

(b) If $\mu$ is complete, so is $\mu\phi^{-1}$.

(c) If $Z$ is another set, and $\psi:Y\to Z$ a function,
then the image measures $\mu(\psi\phi)^{-1}$ and $(\mu\phi^{-1})\psi^{-1}$
on $Z$ are the same.

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

\medskip

{\bf (b)} Write $\nu$ for $\mu\phi^{-1}$ and $\Tau$ for its domain.   If
$\nu^*B=0$, then $\mu^*\phi^{-1}[B]=0$, by 234B(f-i);  as $\mu$ is
complete, $\phi^{-1}[B]\in\Sigma$, so $B\in\Tau$.   As $B$ is arbitrary,
$\nu$ is complete.

\medskip

{\bf (c)} For $G\subseteq Z$ and $u\in[0,\infty]$,

$$\eqalign{(\mu(\psi\phi)^{-1})(G)&\text{ is defined and equal to }u\cr
&\iff\mu((\psi\phi)^{-1}[G])\text{ is defined and equal to }u\cr
&\iff\mu(\phi^{-1}[\psi^{-1}[G]])\text{ is defined and equal to }u\cr
&\iff(\mu\phi^{-1})(\psi^{-1}[G]))\text{ is defined and equal to }u\cr
&\iff((\mu\phi^{-1})\psi^{-1})(G)\text{ is defined and equal to }u.\cr}$$

}%end of proof of 234E


\leader{*234F}{}\cmmnt{ In the opposite direction, the
following construction of a pull-back measure is sometimes useful.

\medskip

\noindent}{\bf Proposition} Let $X$ be a set, $(Y,\Tau,\nu)$ a measure
space, and
$\phi:X\to Y$ a function such that $\phi [X]$ has full outer measure in $Y$.
Then there is a measure $\mu$ on $X$, with domain
$\Sigma=\{\phi^{-1}[F]:F\in\Tau\}$, such that $\phi$ is \imp\ for $\mu$ and
$\nu$.

\proof{ The check that $\Sigma$ is a $\sigma$-algebra of subsets of
$X$ is straightforward;  all we need to know is that
$\phi^{-1}[\emptyset]=\emptyset$,
$X\setminus\phi^{-1}[F]=\phi^{-1}[Y\setminus F]$ for every $F\subseteq Y$,
and that
$\phi^{-1}[\bigcup_{n\in\Bbb N}F_n]=\bigcup_{n\in\Bbb N}\phi^{-1}[F_n]$ for
every sequence $\sequencen{F_n}$ of subsets of $Y$.   The key fact is
that if $F_1$, $F_2\in\Tau$ and $\phi^{-1}[F_1]=\phi^{-1}[F_2]$,
then $\phi[X]$ does not meet $F_1\symmdiff F_2$;  because $\phi[X]$ has full
outer measure, $F_1\symmdiff F_2$ is $\nu$-negligible and
$\nu F_1=\nu F_2$.   Accordingly the formula $\mu\phi^{-1}[F]=\nu F$
does define a function $\mu:\Sigma\to[0,\infty]$.   Now

\Centerline{$\mu\emptyset=\mu\phi^{-1}[\emptyset]=\nu\emptyset=0$.}

\noindent Next, if $\sequencen{E_n}$ is a disjoint sequence in $\Sigma$,
choose $F_n\in\Tau$ such that $E_n=\phi^{-1}[F_n]$ for each $n\in\Bbb N$.
The sequence $\sequencen{F_n}$ need not be disjoint, but if we set
$F'_n=F_n\setminus\bigcup_{i<n}F_i$ for each $n\in\Bbb N$, then
$\sequencen{F'_n}$ is disjoint and

\Centerline{$E_n=E_n\setminus\bigcup_{i<n}E_i=\phi^{-1}[F'_n]$}

\noindent for each $n$;  so

\Centerline{$\mu(\bigcup_{n\in\Bbb N}E_n)
=\nu(\bigcup_{n\in\Bbb N}F'_n)
=\sum_{n=0}^{\infty}\nu F'_n
=\sum_{n=0}^{\infty}\mu E_n$.}

\noindent As $\sequencen{E_n}$ is arbitrary, $\mu$ is a measure on
$X$, as required.
}%end of proof of 234F


\leader{234G}{Sums of \dvrocolon{measures}}\cmmnt{ I come now to a quite
different way of building measures.   The idea is an obvious one, but the
technical details, in the general case I wish to examine, need watching.

\medskip

\noindent}{\bf Proposition}
Let $X$ be a set, and $\familyiI{\mu_i}$ a family of measures on $X$.
For each $i\in I$, let $\Sigma_i$ be the
domain of $\mu_i$.   Set $\Sigma=\Cal PX\cap\bigcap_{i\in I}\Sigma_i$ and
define $\mu:\Sigma\to[0,\infty]$ by setting $\mu E=\sum_{i\in I}\mu_iE$ for
every $E\in\Sigma$.   Then $\mu$ is a measure on $X$.

\proof{ $\Sigma$ is a $\sigma$-algebra of subsets of $X$ because every
$\Sigma_i$ is.   (Apply 111Ga with
$\frak S=\{\Sigma_i:i\in I\}\cup\{\Cal PX\}$.)
Of course $\mu$ takes values in $[0,\infty]$ (226A).
$\mu\emptyset=0$ because $\mu_i\emptyset=0$ for every $i$.   If
$\sequencen{E_n}$ is a disjoint sequence in $\Sigma$ with union $E$, then

$$\eqalignno{\mu E
&=\sum_{i\in I}\mu_iE
=\sum_{i\in I}\sum_{n=0}^{\infty}\mu_iE_n
=\sum_{n=0}^{\infty}\sum_{i\in I}\mu_iE_n\cr
\displaycause{226Af}
&=\sum_{n=0}^{\infty}\mu E_n.\cr}$$

\noindent So $\mu$ is a measure.
}%end of proof of 234G

\medskip

\noindent{\bf Remark} In this context, I will call $\mu$ the {\bf sum} of
the family $\familyiI{\mu_i}$.

\leader{234H}{Proposition} Let $X$ be a set and $\familyiI{\mu_i}$ a family
of complete measures on $X$ with sum $\mu$.

(a) $\mu$ is complete.

(b)(i) A subset of $X$ is $\mu$-negligible iff it is $\mu_i$-negligible for
every $i\in I$.

\quad(ii) A subset of $X$ is $\mu$-conegligible iff it is
$\mu_i$-conegligible for every $i\in I$.

(c) Let $f$ be a function from a subset of $X$ to $[-\infty,\infty]$.
Then $\int fd\mu$ is defined in $[-\infty,\infty]$ iff $\int fd\mu_i$ is
defined in $[-\infty,\infty]$ for every $i$ and one of
$\sum_{i\in I}f^+d\mu_i$, $\sum_{i\in I}f^-d\mu_i$ is finite, and in this
case $\int fd\mu=\sum_{i\in I}\int fd\mu_i$.

\proof{ Write $\Sigma_i=\dom\mu_i$ for $i\in I$,
$\Sigma=\Cal PX\cap\bigcap_{i\in I}\Sigma_i=\dom\mu$.

\medskip

{\bf (a)} If $E\subseteq F\in\Sigma$ and $\mu F=0$, then $\mu_iF=0$ for
every $i\in I$;  because $\mu_i$ is complete, $E_i\in\Sigma_i$ for
every $i\in I$, and $E\in\Sigma$.

\medskip

{\bf (b)} This now follows at once, since a set $A\subseteq X$ is
$\mu$-negligible iff $\mu A=0$.

\medskip

{\bf (c)(i)} Note first that (b-ii) tells us that, under either hypothesis,
$\dom f$ is conegligible both for $\mu$ and for every $\mu_i$, so that
if we extend $f$ to $X$ by giving it the value $0$ on $X\setminus\dom f$
then neither $\int fd\mu$ nor $\sum_{i\in I}\int fd\mu_i$ is affected.   So
let us assume from now on that $f$ is defined everywhere on $X$.   Now it
is plain that
either hypothesis ensures that $f$ is $\Sigma$-measurable, that is, is
$\Sigma_i$-measurable for every $i\in I$.

\medskip

\quad{\bf (ii)} Suppose that $f$ is non-negative.
For $n\in\Bbb N$ set
$f_n(x)=\sum_{k=1}^{4^n}2^{-n}\chi\{x:f(x)\ge 2^{-n}k\}$, so that
$\sequencen{f_n}$ is a non-decreasing sequence with supremum $f$.  We have

$$\eqalign{\int f_nd\mu
&=\sum_{k=1}^{4^n}2^{-n}\mu\{x:f(x)\ge 2^{-n}k\}
=\sum_{k=1}^{4^n}\sum_{i\in I}2^{-n}\mu_i\{x:f(x)\ge 2^{-n}k\}\cr
&=\sum_{i\in I}\sum_{k=1}^{4^n}2^{-n}\mu_i\{x:f(x)\ge 2^{-n}k\}
=\sum_{i\in I}\int f_nd\mu_i\cr}$$

\noindent for every $n$, so

$$\eqalign{\int fd\mu
&=\sup_{n\in\Bbb N}\int f_nd\mu
=\sup_{n\in\Bbb N}\sup_{J\subseteq I\text{ is finite}}
   \sum_{i\in J}\int f_nd\mu_i
=\sup_{J\subseteq I\text{ is finite}}\sup_{n\in\Bbb N}
   \sum_{i\in J}\int f_nd\mu_i\cr
&=\sup_{J\subseteq I\text{ is finite}}
   \lim_{n\to\infty}\sum_{i\in J}\int f_nd\mu_i
=\sup_{J\subseteq I\text{ is finite}}
   \sum_{i\in J}\lim_{n\to\infty}\int f_nd\mu_i
=\sum_{i\in I}\int fd\mu_i.\cr}$$

\medskip

\quad{\bf (iii)} Generally,

$$\eqalign{\int fd\mu&\text{ is defined in }[\infty,\infty]\cr
&\iff\int f^+d\mu\text{ and }\int f^-d\mu
  \text{ are defined and at most one is infinite}\cr
&\iff\sum_{i\in I}\int f^+d\mu_i\text{ and }\sum_{i\in I}\int f^-d\mu_i
  \text{ are defined and at most one is infinite}\cr
&\iff\int fd\mu_i\text{ is defined for every }i
  \text{ and at most one of }\sum_{i\in I}\int f^+d\mu_i,\cr
&\mskip400mu
  \sum_{i\in I}\int f^-d\mu_i\text{ is infinite},\cr}$$

\noindent and in this case

\Centerline{$\int fd\mu
=\int f^+d\mu-\int f^-d\mu
=\sum_{i\in I}\int f^+d\mu_i-\sum_{i\in I}\int f^-d\mu_i
=\sum_{i\in I}\int fd\mu_i$.}
}%end of proof of 234H

\leader{234I}{Indefinite-integral
\dvrocolon{measures}}\cmmnt{ Extending an
idea already used in 232D, we are led to the following construction;  once
again, we need to take care over the formal details if we want
to get full value from it.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a measure
space, and $f$ a non-negative $\mu$-virtually measurable real-valued
function defined on a conegligible subset of $X$.   Write
$\nu F=\int f\times\chi F\,d\mu$ whenever $F\subseteq X$ is such that
the integral is
defined in $[0,\infty]$\cmmnt{ according to the conventions of 133A}.
Then $\nu$ is a complete measure on $X$, and its domain includes
$\Sigma$.
%the trouble with taking [0,\infty]-valued functions is that if
%I allow  f=\infty  on a non-negligible set then  \nu  becomes
%non-semi-finite for a silly reason

\proof{{\bf (a)} Write $\Tau$ for the domain of $\nu$, that is, the
family of sets $F\subseteq X$ such that $\int f\times\chi F\,d\mu$ is
defined in
$[0,\infty]$, that is, $f\times\chi F$ is $\mu$-virtually measurable
(133A).   Then $\Tau$ is a $\sigma$-algebra of subsets of $X$.   \Prf\
For each $F\in\Tau$ let $H_F\subseteq X$ be a $\mu$-conegligible set
such that $f\times\chi F\restr H_F$ is $\Sigma$-measurable.   Because
$f$ itself is $\mu$-virtually measurable, $X\in\Tau$.   If $F\in\Tau$,
then

\Centerline{$f\times\chi(X\setminus F)\restr(H_X\cap H_F)
=f\restr(H_X\cap H_F)-(f\times\chi F)\restr(H_X\cap H_F)$}

\noindent is $\Sigma$-measurable, while $H_X\cap H_F$ is
$\mu$-conegligible, so $X\setminus F\in\Tau$.   If $\sequencen{F_n}$ is
a sequence in $\Tau$ with union $F$, set
$H=\bigcap_{n\in\Bbb N}H_{F_n}$;
then $H$ is conegligible, $f\times\chi F_n\restr H$ is $\Sigma$-measurable
for every $n\in\Bbb N$, and
$f\times\chi F=\sup_{n\in\Bbb N}f\times\chi F_n$, so
$f\times\chi F\restr H$ is $\Sigma$-measurable, and $F\in\Tau$.   Thus
$\Tau$ is a $\sigma$-algebra.   If $F\in\Sigma$, then
$f\times\chi F\restr H_X$ is $\Sigma$-measurable, so $F\in\Tau$.\ \Qed

\medskip

{\bf (b)} Next, $\nu$ is a measure.   \Prf\ Of course
$\nu F\in[0,\infty]$
for every $F\in\Tau$.   $f\times\chi\emptyset=0$ wherever it is defined,
so $\nu\emptyset=0$.   If $\sequencen{F_n}$ is a disjoint sequence in
$\Tau$ with union $F$, then
$f\times\chi F=\sum_{n=0}^{\infty}f\times\chi F_n$.
If $\nu F_m=\infty$ for some $m$, then we surely have
$\nu F=\infty=\sum_{n=0}^{\infty}\nu F_n$.   If $\nu F_m<\infty$ for
each $m$ but $\sum_{n=0}^{\infty}\nu F_n=\infty$, then

\Centerline{$\int f\times\chi(\bigcup_{n\le m}F_n)
=\sum_{n=0}^m\int f\times\chi F_n\to\infty$}

\noindent as $m\to\infty$, so again
$\nu F=\infty=\sum_{n=0}^{\infty}\nu F_n$.   If
$\sum_{n=0}^{\infty}\nu F_n<\infty$ then by B.Levi's theorem

\Centerline{$\nu F=\int\sum_{n=0}^{\infty}f\times\chi F_n
=\sum_{n=0}^{\infty}\int f\times\chi F_n
=\sum_{n=0}^{\infty}\nu F_n$.
\Qed}

\medskip

{\bf (c)} Finally, $\nu$ is complete.   \Prf\ If $A\subseteq F\in\Tau$
and $\nu F=0$, then $f\times\chi F=0$ a.e., so $f\times\chi A=0$ a.e.\
and $\nu A$ is defined and equal to zero.\ \Qed
}%end of proof of 234I

\leader{234J}{Definition} Let $(X,\Sigma,\mu)$ be a measure space, and
$\nu$ another measure on $X$ with domain $\Tau$.   I will call $\nu$ an
{\bf indefinite-integral measure} over $\mu$, or sometimes a {\bf
completed indefinite-integral measure}, if it can be obtained by the
method of 234I
from some non-negative virtually measurable function $f$ defined almost
everywhere on $X$.   In this case, $f$ is a Radon-Nikod\'ym derivative 
of $\nu$ with respect to $\mu$ in the sense of 232Hf.
\cmmnt{As in 232Hf, the phrase {\bf density function} is also used in
this context.}

\leader{234K}{Remarks} Let $(X,\Sigma,\mu)$ be a measure space, and $f$
a $\mu$-virtually measurable non-negative real-valued function defined
almost everywhere
on $X$;  let $\nu$ be the associated indefinite-integral measure.

\spheader 234Ka\cmmnt{ There is a $\Sigma$-measurable function
$g:X\to\coint{0,\infty}$ such that $f=g\,\,\mu$-a.e.   \Prf\ Let
$H\subseteq\dom f$ be a measurable conegligible set such that
$f\restr H$ is measurable, and set $g(x)=f(x)$ for $x\in H$, $g(x)=0$
for $x\in X\setminus H$.\ \QeD\    In
this case, $\int f\times\chi E\,d\mu=\int g\times\chi E\,d\mu$ if either
is defined.   So $g$ is a Radon-Nikod\'ym derivative of $\nu$, and}
$\nu$ has a
Radon-Nikod\'ym derivative which is $\Sigma$-measurable and defined
everywhere.

%includes old 2{}34Ce
\spheader 234Kb If $E$ is $\mu$-negligible, 
then\cmmnt{ $f\times\chi E=0\,\,\mu$-a.e., so} $\nu E=0$.   
\cmmnt{Many authors are prepared to say `$\nu$ is
absolutely continuous with respect to $\mu$' in this context.   But if
$\nu$ is not totally finite, it need not be
absolutely continuous in the
$\epsilon$-$\delta$ sense of 232Aa (234Xh), and
further difficulties can arise if $\mu$ or $\nu$ is not $\sigma$-finite
(see 234Yk, 234Ym).
}%end of comment

\cmmnt{\spheader 234Kc I have defined `indefinite-integral measure' in
such a way as to produce a complete measure.   In my view this is what
makes best sense in most applications.   There are occasions on which it
seems more appropriate to use the measure $\nu_0:\Sigma\to[0,\infty]$
defined by
setting $\nu_0 E=\int_Efd\mu=\int f\times\chi E\,d\mu$ for $E\in\Sigma$.
I suppose I would call this the {\bf uncompleted indefinite-integral
measure} over $\mu$ defined by $f$.   ($\nu$ is always the
completion of $\nu_0$;  see 234Lb.)
}%end of comment

\spheader 234Kd\cmmnt{ Note the way in which I formulated the
definition of $\nu$:  `$\nu E=\int f\times\chi E\,d\mu$ if the integral
is defined',
rather than `$\nu E=\int_Efd\mu$'.   The point is that the longer
formula gives a rule for deciding what the domain of $\nu$ must be.   Of
course it is the case that} $\nu E=\int_Efd\mu$ for every
$E\in\dom\nu$\cmmnt{ (apply 214F to $f\times\chi E$)}.

\spheader 234Ke\cmmnt{ Because} $\mu$ and its completion define the 
same\cmmnt{ virtually measurable functions, the same null ideals and the same
integrals\cmmnt{ (212Eb, 212F)},
they define the same} indefinite-integral measures.

\leader{234L}{The domain of an indefinite-integral
\dvrocolon{measure}}\cmmnt{ It is sometimes useful
to have an explicit description of the domain of a measure constructed
in this way.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, $f$
a non-negative $\mu$-virtually measurable function defined almost
everywhere
in $X$, and $\nu$ the associated indefinite-integral measure.   Set
$G=\{x:x\in\dom f$, $f(x)>0\}$, and let $\hat\Sigma$ be the domain of the
completion $\hat\mu$ of $\mu$.

(a) The domain $\Tau$ of $\nu$ is
$\{E:E\subseteq X,\,E\cap G\in\hat\Sigma\}$;  \cmmnt{in particular,}
$\Tau\supseteq\hat\Sigma\supseteq\Sigma$.

(b) $\nu$ is the completion of its restriction to $\Sigma$.

(c) A set $A\subseteq X$ is $\nu$-negligible iff $A\cap G$ is
$\mu$-negligible.

(d) In particular, if $\mu$\cmmnt{ itself} is complete, then
$\Tau=\{E:E\subseteq X,\,E\cap G\in\Sigma\}$ and $\nu A=0$ iff
$\mu(A\cap G)=0$.

\proof{{\bf (a)(i)} If $E\in\Tau$, then $f\times\chi E$ is virtually
measurable, so there is a conegligible measurable set $H\subseteq\dom f$
such that $f\times\chi E\restr H$ is measurable.   Now $E\cap G\cap
H=\{x:x\in H,\,(f\times\chi E)(x)>0\}$ must belong to $\Sigma$, while
$E\cap G\setminus H$ is negligible, so belongs to $\hat\Sigma$, and
$E\cap G\in\hat\Sigma$.

\medskip

\quad{\bf (ii)} If $E\cap G\in\hat\Sigma$, let $F_1$, $F_2\in\Sigma$ be
such that $F_1\subseteq E\cap G\subseteq F_2$ and $F_2\setminus F_1$ is
negligible.   Let $H\subseteq\dom f$ be a conegligible set such that
$f\restr H$ is measurable.   Then $H'=H\setminus(F_2\setminus F_1)$ is
conegligible and $f\times\chi E\restr H'=f\times\chi F_1\restr H'$ is
measurable, so $f\times\chi E$ is virtually measurable and $E\in\Tau$.

\medskip

{\bf (b)} Thus the given formula does indeed describe $\Tau$.   If
$E\in\Tau$, let $F_1$, $F_2\in\Sigma$ be such that
$F_1\subseteq E\cap G\subseteq F_2$ and $\mu(F_2\setminus F_1)=0$.
Because $G$ itself also
belongs to $\hat\Sigma$, there are $G_1$, $G_2\in\Sigma$ such that
$G_1\subseteq G\subseteq G_2$ and $\mu(G_2\setminus G_1)=0$.   Set
$F_2'=F_2\cup(X\setminus G_1)$.   Then $F_2'\in\Sigma$ and
$F_1\subseteq E\subseteq F'_2$.   But
$(F'_2\setminus F_1)\cap G
\subseteq(G_2\setminus G_1)\cup(F_2\setminus F_1)$ is $\mu$-negligible, so
$\nu(F'_2\setminus F_1)=0$.

This shows that if $\nuprime$ is the completion of $\nu\restr\Sigma$
and $\Tau'$ is its domain, then $\Tau\subseteq\Tau'$.   But as
$\nu$ is complete, it surely extends $\nuprime$, so $\nu=\nuprime$, as
claimed.

\medskip

{\bf (c)} Now take any $A\subseteq X$.   Because $\nu$ is complete,

$$\eqalign{A\text{ is }\nu\text{-negligible}
&\iff\nu A=0\cr
&\iff\int f\times\chi A\,d\mu=0\cr
&\iff f\times\chi A=0\,\,\mu\text{-a.e.}\cr
&\iff A\cap G\text{ is }\mu\text{-negligible.}\cr}$$

\medskip

{\bf (d)} This is just a restatement of (a) and (c) when $\mu=\hat\mu$.
}%end of proof of 234L

\leader{234M}{Corollary} If $(X,\Sigma,\mu)$ is a complete measure
space and $G\in\Sigma$, then the indefinite-integral measure over $\mu$
defined by $\chi G$ is just the measure $\mu\LLcorner G$ defined by
setting

\Centerline{$(\mu\LLcorner G)(F)=\mu(F\cap G)$ whenever $F\subseteq X$
and $F\cap G\in\Sigma$.}

\proof{ 234Ld.
}%end of proof of 234M

\leader{*234N}{}\cmmnt{ The next two results will not be relied on in
this
volume, but I include them for future reference, and to give an idea of
the scope of these ideas.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$\nu$ an indefinite-integral measure over $\mu$.

(a) If $\mu$ is semi-finite, so is $\nu$.

(b) If $\mu$ is complete and locally determined, so is $\nu$.

(c) If $\mu$ is localizable, so is $\nu$.

(d) If $\mu$ is strictly localizable, so is $\nu$.

(e) If $\mu$ is $\sigma$-finite, so is $\nu$.

(f) If $\mu$ is atomless, so is $\nu$.

\proof{ By 234Ka, we may express $\nu$ as the indefinite integral of a
$\Sigma$-measurable function $f:X\to\coint{0,\infty}$.   Let $\Tau$ be
the domain of $\nu$, and $\hat\Sigma$ the domain of the completion
$\hat\mu$ of
$\mu$;  set $G=\{x:x\in X,\,f(x)>0\}\in\Sigma$.

\medskip

{\bf (a)} Suppose that $E\in\Tau$ and that $\nu E=\infty$.   Then
$E\cap G$ cannot be $\mu$-negligible.
Because $\mu$ is semi-finite, there is a non-negligible $F\in\Sigma$
such that $F\subseteq E\cap G$ and $\mu F<\infty$.   Now
$F=\bigcup_{n\in\Bbb N}\{x:x\in F$, $2^{-n}\le f(x)\le n\}$, so there
is an $n\in\Bbb N$ such
that $F'=\{x:x\in F$, $2^{-n}\le f(x)\le n\}$ is non-negligible.
Because $f$ is measurable, $F'\in\Sigma\subseteq\Tau$ and
$2^{-n}\mu F'\le\nu F'\le n\mu F'$.   Thus we have found an
$F'\subseteq E$ such
that $0<\nu F'<\infty$.   As $E$ is arbitrary, $\nu$ is semi-finite.

\medskip

{\bf (b)} We already know that $\nu$ is complete (234Lb) and semi-finite.   Now
suppose that $E\subseteq X$ is such that $E\cap F\in\Tau$, that is,
$E\cap F\cap G\in\Sigma$ (234Ld), whenever $F\in\Tau$ and $\nu F<\infty$.   Then
$E\cap G\cap F\in\Sigma$ whenever
$F\in\Sigma$ and $\mu F<\infty$.   \Prf\ Set
$F_n=\{x:x\in F\cap G,\,f(x)\le n\}$.   Then
$\nu F_n\le n\mu F<\infty$, so $E\cap G\cap F_n\in\Sigma$ for every $n$.
But this means that
$E\cap G\cap F=\bigcup_{n\in\Bbb N}E\cap G\cap F_n\in\Sigma$.\ \QeD\   
Because $\mu$ is locally determined, $E\cap G\in\Sigma$ and $E\in\Tau$.   
As $E$ is arbitrary, $\nu$ is locally determined.

\medskip

{\bf (c)} Let $\Cal F\subseteq\Tau$ be any set.   Set
$\Cal E=\{F\cap G:F\in\Cal F\}$, so that $\Cal E\subseteq\hat\Sigma$.   By
212Ga, $\hat\mu$
is localizable, so $\Cal E$ has an essential supremum $H\in\hat\Sigma$.
But now, for any $H'\in\Tau$,
$H'\cup(X\setminus G)=(H'\cap G)\cup(X\setminus G)$ belongs to
$\hat\Sigma$, so

$$\eqalign{\nu(F\setminus H')&=0\text{ for every }F\in\Cal F\cr
&\iff\hat\mu(F\cap G\setminus H')=0\text{ for every }F\in\Cal F\cr
&\iff\hat\mu(E\setminus H')=0\text{ for every }E\in\Cal E\cr
&\iff\hat\mu(E\setminus(H'\cup(X\setminus G)))=0
  \text{ for every }E\in\Cal E\cr
&\iff\hat\mu(H\setminus((H'\cup(X\setminus G)))=0\cr
&\iff\hat\mu(H\cap G\setminus H')=0\cr
&\iff\nu(H\setminus H')=0.\cr}$$

\noindent Thus $H$ is also an essential supremum of $\Cal F$ in $\Tau$.
As $\Cal F$ is arbitrary, $\nu$ is localizable.

\medskip

{\bf (d)} Let
$\langle X_i\rangle_{i\in I}$ be a decomposition of $X$ for $\mu$;
then it is also a decomposition for $\hat\mu$ (212Gb).
Set $F_0=X\setminus G$, $F_n=\{x:x\in G,\,n-1<f(x)\le n\}$ for $n\ge 1$.
Then $\langle X_i\cap F_n\rangle_{i\in I,n\in\Bbb N}$ is a decomposition
for $\nu$.   \Prf\ (i) $\langle X_i\rangle_{i\in I}$ and
$\langle F_n\rangle_{n\in\Bbb N}$ are partitions of $X$ into members of $\Sigma\subseteq\Tau$, so
$\langle X_i\cap F_n\rangle_{i\in I,n\in\Bbb N}$ also is.
(ii)  $\nu(X_i\cap F_0)=0$,
$\nu(X_i\cap F_n)\le n\mu X_i<\infty$ for $i\in I$, $n\ge 1$.
(iii) If $E\subseteq X$ and $E\cap X_i\cap F_n\in\Tau$ for every
$i\in I$ and $n\in\Bbb N$ then
$E\cap X_i\cap G=\bigcup_{n\in\Bbb N}E\cap X_i\cap F_n\cap G$ belongs to
$\hat\Sigma$ for every $i$, so $E\cap G\in\hat\Sigma$ and
$E\in\Tau$.
(iv) If $E\in\Tau$, then of course

\Centerline{$\sum_{i\in I,n\in\Bbb N}\nu(E\cap X_i\cap F_n)
=\sup_{J\subseteq I\times\Bbb N\text{ is finite}}
  \sum_{(i,n)\in J}\nu(E\cap X_i\cap F_n)\le\nu E$.}

\noindent So if $\sum_{i\in I,n\in\Bbb N}\nu(E\cap X_i\cap F_n)=\infty$
it is surely equal to $\nu E$.   If the sum is finite, then
$K=\{i:i\in I,\,\nu(E\cap X_i)>0\}$ must be countable.   But for
$i\in I\setminus K$,
$\int_{E\cap X_i}fd\mu=0$, so $f=0\,\,\mu$-a.e.\ on $E\cap X_i$, that
is, $\hat\mu(E\cap G\cap X_i)=0$.   Because
$\langle X_i\rangle_{i\in I}$ is a decomposition for $\hat\mu$,
$\hat\mu(E\cap G\cap\bigcup_{i\in I\setminus K}X_i)=0$ and
$\nu(E\cap\bigcup_{i\in I\setminus K}X_i)=0$.   But this
means that

\Centerline{$\nu E=\sum_{i\in K}\nu(E\cap X_i)
=\sum_{i\in K,n\in\Bbb N}\nu(E\cap X_i\cap F_n)
=\sum_{i\in I,n\in\Bbb N}\nu(E\cap X_i\cap F_n)$.}

\noindent As $E$ is arbitrary,
$\langle X_i\cap F_n\rangle_{i\in I,n\in\Bbb N}$ is a decomposition for
$\nu$.\ \QeD\   So $\nu$ is strictly localizable.

\medskip

{\bf (e)} If $\mu$ is $\sigma$-finite, then in (d) we can take $I$ to be
countable, so that $I\times\Bbb N$ also is countable, and $\nu$ will be
$\sigma$-finite.

\medskip

{\bf (f)} If $\mu$ is atomless, so is $\hat\mu$ (212Gd).   If $E\in\Tau$
and $\nu E>0$, then $\hat\mu(E\cap G)>0$, so there is an $F\in\hat\Sigma$
such that $F\subseteq E\cap G$ and
neither $F$ nor $E\cap G\setminus F$ is $\hat\mu$-negligible.   But in this case both $\nu F=\int_Ffd\mu$ and
$\nu(E\setminus F)=\int_{E\setminus F}fd\mu$ must be greater than $0$
(122Rc).   As $E$ is arbitrary, $\nu$ is atomless.
}%end of proof of 234N

\leader{*234O}{}\cmmnt{ For localizable measures, there is a
straightforward description of the associated indefinite-integral
measures.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ be a localizable measure
space.   Then a measure $\nu$, with domain $\Tau\supseteq\Sigma$, is an
indefinite-integral measure over $\mu$ iff ($\alpha$) $\nu$ is
semi-finite
and zero on $\mu$-negligible sets ($\beta$) $\nu$ is the completion of
its restriction to $\Sigma$ ($\gamma$) whenever $\nu E>0$ there is an
$F\subseteq E$ such that $F\in\Sigma$, $\mu F<\infty$ and $\nu F>0$.

\proof{{\bf (a)} If $\nu$ is an indefinite-integral measure over $\nu$,
then by 234Na, 234Kb and 234Lb it is semi-finite, zero on
$\mu$-negligible sets and the completion of its restriction to $\Sigma$.
Now suppose that $E\in\Tau$ and $\nu E>0$.   Then there is an
$E_0\in\Sigma$
such that $E_0\subseteq E$ and $\nu E_0=\nu E$, by 234Lb.   If
$f:X\to\Bbb R$ is a $\Sigma$-measurable
Radon-Nikod\'ym derivative of $\nu$ (234Ka), and $G=\{x:f(x)>0\}$, then
$\mu(G\cap E_0)>0$;  because $\mu$ is semi-finite, there is an
$F\in\Sigma$ such that
$F\subseteq G\cap E_0$ and $0<\mu F<\infty$, in which case $\nu F>0$.

\medskip

{\bf (b)} So now suppose that $\nu$ satisfies the conditions.

\medskip

\quad{\bf (i)} Set $\Cal E=\{E:E\in\Sigma,\,\nu E<\infty\}$.   For
each $E\in\Cal E$, consider $\nu_E:\Sigma\to\Bbb R$, setting
$\nu_EG=\nu(G\cap E)$ for every $G\in\Sigma$.   Then $\nu_E$ is
countably additive and truly continuous with respect to $\mu$.   \Prf\
$\nu_E$ is countably additive, just as in 231De.   Because $\nu$ is zero
on $\mu$-negligible sets, $\nu_E$ must
be absolutely continuous with respect to $\mu$, by 232Ba.   Since
$\nu_E$ clearly satisfies condition ($\gamma$) of 232Bb,
it must be truly continuous.\ \Qed

By 232E, there is a $\mu$-integrable function $f_E$ such that
$\nu_EG=\int_Gf_Ed\mu$ for every $G\in\Sigma$, and we may suppose that
$f_E$ is $\Sigma$-measurable (232He).   
Because $\nu_E$ is non-negative, $f_E\ge 0\,\,\mu$-almost everywhere.

\medskip

\quad{\bf (ii)} If $E$, $F\in\Cal E$ then $f_E=f_F\,\,\mu$-a.e. on
$E\cap F$, because

\Centerline{$\int_Gf_Ed\mu=\nu G=\int_Gf_Fd\mu$}

\noindent whenever $G\in\Sigma$ and $G\subseteq E\cap F$.   Because
$(X,\Sigma,\mu)$ is localizable, there is a measurable $f:X\to\Bbb R$
such that $f_E=f\,\,\mu$-a.e.\ on $E$ for every $E\in\Cal E$ (213N).
Because every $f_E$ is non-negative almost everywhere, we may suppose
that
$f$ is non-negative, since surely $f_E=f\vee\tbf{0}\,\,\mu$-a.e. on $E$
for
every $E\in\Cal E$.

\medskip

\quad{\bf (iii)} Let $\nuprime$ be the indefinite-integral measure
defined by $f$.   If $E\in\Cal E$ then

\Centerline{$\nu E=\int_Ef_Ed\mu=\int_Efd\mu=\nuprime E$.}
\noindent For $E\in\Sigma\setminus\Cal E$, we have

\Centerline{$\nuprime E
\ge\sup\{\nuprime F:F\in\Cal E,\,F\subseteq E\}
=\sup\{\nu F:F\in\Cal E,\,F\subseteq E\}=\nu E=\infty$}

\noindent because $\nu$ is semi-finite.   Thus $\nuprime$ and $\nu$
agree on $\Sigma$.  But since each is the completion of its restriction
to $\Sigma$, they must be equal.
}%end of proof of 234O

\leader{234P}{Ordering \dvrocolon{measures}}\cmmnt{ There are many ways in
which one measure can dominate another.   Here I will describe one of the
simplest.

\medskip

\noindent}{\bf Definition}
Let $\mu$, $\nu$ be two measures on a set $X$.
I will say that $\mu\le\nu$ if $\mu E$ is defined, and $\mu E\le\nu E$,
whenever $\nu$ measures $E$.

\leader{234Q}{Proposition} Let $X$ be a set, and write $\Mu$ for the set of
all measures on $X$.

(a) Defining $\le$ as in 234P, $(\Mu,\le)$ is a partially ordered set.

(b) If $\mu$, $\nu\in\Mu$, then $\mu\le\nu$ iff there is a $\lambda\in\Mu$
such that $\mu+\lambda=\nu$.

(c) If $\mu\le\nu$ in $\Mu$ and $f$ is a $[-\infty,\infty]$-valued
function,
defined on a subset of $X$, such that $\int fd\nu$ is defined in
$[-\infty,\infty]$, then $\int fd\mu$ is defined;  if $f$ is non-negative,
$\int fd\mu\le\int fd\nu$.

\proof{{\bf (a)} Of course $\mu\le\mu$ for every $\mu\in\Mu$.
If $\mu\le\nu$ and $\nu\le\lambda$ in $\Mu$, then
$\dom\lambda\subseteq\dom\nu\subseteq\dom\mu$, and
$\mu E\le\nu E\le\lambda E$ whenever $\lambda$ measures $E$.
If $\mu\le\nu$ and $\nu\le\mu$ then
$\dom\mu\subseteq\dom\nu\subseteq\dom\mu$ and $\mu E\le\nu E\le\mu E$ for
every $E$ in their common domain, so $\mu=\nu$.

\medskip

{\bf (b)(i)}
If $\mu+\lambda=\nu$, then the definitions in 234G and 234P make
it plain that $\mu\le\nu$.

\medskip

\quad{\bf (ii)}\grheada\ In the reverse direction, if $\mu\le\nu$,
write $\Tau$ for the domain of $\nu$.
Define $\lambda:\Tau\to[0,\infty]$ by setting

\Centerline{$\lambda G
=\sup\{\nu F-\mu F:F\in\Tau,\,F\subseteq G,\,\mu F<\infty\}$}

\noindent for $G\in\Tau$.
Then $\lambda\in\Mu$.   \Prf\ Of course $\dom\lambda=\Tau$ is a
$\sigma$-algebra, and $\lambda\emptyset=0$.   Suppose that
$\sequencen{G_n}$ is a
disjoint sequence in $\Tau$ with union $G$.
If $F\in\Tau$, $F\subseteq G$ and $\mu F<\infty$, then

$$\eqalign{\nu F-\mu F
&=\sum_{n=0}^{\infty}\nu(F\cap G_n)-\sum_{n=0}^{\infty}\mu(F\cap G_n)\cr
&=\sum_{n=0}^{\infty}\nu(F\cap G_n)-\mu(F\cap G_n)
\le\sum_{n=0}^{\infty}\lambda G_n;\cr}$$

\noindent as $F$ is arbitrary,
$\lambda G\le\sum_{n=0}^{\infty}\lambda G_n$.   If
$\gamma<\sum_{n=0}^{\infty}\lambda G_n$, there are an $m\in\Bbb N$ such
that $\gamma<\sum_{n=0}^m\lambda G_n$, and $F_0,\ldots,F_m$ such that
$F_n\in\Tau$, $F_n\subseteq G_n$ and $\mu F_n<\infty$ for every $n\le m$,
while $\sum_{n=0}^m\nu F_n-\mu F_n\ge\gamma$.   Set
$F=\bigcup_{n\le m}F_n$;  then $F\in\Tau$, $F\subseteq G$ and
$\mu F<\infty$, so

\Centerline{$\lambda G\ge\nu F-\mu F
=\sum_{n=0}^m\nu F_n-\mu F_n\ge\gamma$.}

\noindent As $\gamma$ is arbitrary,
$\lambda G\ge\sum_{n=0}^{\infty}\lambda G_n$;  as $\sequencen{G_n}$ is
arbitrary, $\lambda$ is countably additive.\ \Qed

\medskip

\qquad\grheadb\ Now $\mu+\lambda=\nu$.   \Prf\ The domain of $\mu+\lambda$
is $\dom\mu\cap\dom\lambda=\Tau=\dom\nu$.    Take $G\in\Tau$.   If
$\mu G=\infty$, then $\nu G=\infty=(\mu+\lambda)G$.   Otherwise,

\Centerline{$(\mu+\lambda)G\ge\mu G+\nu G-\mu G=\nu G$.}

\noindent So if $\nu G=\infty$ we shall certainly have
$\nu G=(\mu+\lambda)G$.   Finally, if $\nu G<\infty$ then

$$\eqalign{(\mu+\lambda)G
&=\mu G+\sup\{\nu F-\mu F:F\in\Tau,\,F\subseteq G\}\cr
&=\sup\{\nu F+\mu(G\setminus F):F\in\Tau,\,F\subseteq G\}\cr
&\le\sup\{\nu F+\nu(G\setminus F):F\in\Tau,\,F\subseteq G\}
=\nu G,\cr}$$

\noindent so again we have equality.\ \Qed

Thus we have an appropriate expression of $\nu$ as a sum of measures.

\medskip

{\bf (c)(i)} If $f$ is non-negative, put (b) and 234Hc together.

\medskip

\quad{\bf (ii)} In general, if $\int fd\nu$ is defined, so are both
$\int f^+d\nu$ and $\int f^-d\nu$, and at most one is infinite;
so $\int f^+d\mu$ and $\int f^-d\mu$ are defined
and at most one is infinite.
}%end of proof of 234Q

\exercises{
\leader{234X}{Basic exercises (a)}
%\spheader 234Xa
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces, and
$\phi:X\to Y$ an \imp\ function.   Let $A\subseteq X$ be a set of full
outer measure in $X$.   Show that
$\phi[A]$ has full outer measure in $Y$, and
that $\phi\restr A$ is \imp\ for the subspace measures on $A$ and
$\phi[A]$.
%234B

\spheader 234Xb Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a set and
$\phi:X\to Y$ a function.   Show that if $\mu$ is point-supported, so is
the image measure $\mu\phi^{-1}$.
%234D

\spheader 234Xc Give an example of a probability space $(X,\Sigma,\mu)$, a
set $Y$, and a function $\phi:X\to Y$ such that the completion of the image
measure $\mu\phi^{-1}$ is not the image of the completion of $\mu$.
\Hint{$\#(X)=3$.}
%234E

\spheader 234Xd Let $X$, $Y$ be sets, $\phi:X\to Y$ a function
and $\familyiI{\mu_i}$ a family of
measures on $X$ with sum $\mu$.   Writing $\mu_i\phi^{-1}$,
$\mu\phi^{-1}$ for the image measures on $Y$, show that
$\mu\phi^{-1}=\sum_{i\in I}\mu_i\phi^{-1}$.
%234G

\spheader 234Xe Let $X$ be a set.
(i) Show that if $\familyiI{\mu_i}$ is a countable family of
$\sigma$-finite measures on $X$, and $\mu=\sum_{i\in I}\mu_i$ is
semi-finite, then $\mu$ is $\sigma$-finite.
(ii) Show that if $\familyiI{\mu_i}$ is a family of purely atomic measures
on $X$, and $\mu=\sum_{i\in I}\mu_i$ is semi-finite, then $\mu$ is purely
atomic.
(iii) Show that if $\familyiI{\mu_i}$ is any family of point-supported
measures on $X$, then $\sum_{i\in I}\mu_i$ is point-supported.
%234G

\sqheader 234Xf
Let $X$ be a set, and write $\Mu$ for the set of all measures on $X$.   For
$\mu\in\Mu$ and $\alpha\in\coint{0,\infty}$, define $\alpha\mu$ by saying
that if $\alpha>0$ then
$(\alpha\mu)(E)=\alpha\mu E$ for $E\in\dom\mu$, while if $\alpha=0$ then
$(\alpha\mu)(E)=0$ for every $E\subseteq X$.   (i) Show that
$\alpha\mu\in\Mu$ for all $\alpha\in\coint{0,\infty}$ and $\mu\in\Mu$.
(ii) Show that $(\alpha+\beta)\mu=\alpha\mu+\beta\mu$,
$\alpha(\beta\mu)=(\alpha\beta)\mu$, $\alpha(\mu+\nu)=\alpha\mu+\alpha\nu$
for all $\alpha$, $\beta\in\coint{0,\infty}$ and $\mu$, $\nu\in\Mu$.
%234G

\spheader 234Xg Let $X$ be a set, and $\familyiI{\mu_i}$ a family of
complete measures on $X$ with sum $\mu$.
Show that a $[-\infty,\infty]$-valued
function $f$ defined on a subset of $X$ is $\mu$-integrable iff it is
$\mu_i$-integrable for every $i\in I$ and $\sum_{i\in I}\int|f|d\mu_i$ is
finite.
%234H

\spheader 234Xh
Let $\mu$ be Lebesgue measure on $[0,1]$, and set $f(x)=\bover1x$ for
$x>0$. Let $\nu$ be the associated indefinite-integral measure.   Show
that the
domain of $\nu$ is equal to the domain of $\mu$.   Show that for every
$\delta\in\ocint{0,\bover12}$ there is a measurable set $E$ such that
$\mu E=\delta$ but $\nu E=\bover1{\delta}$.
%234K

\spheader 234Xi Let $(X,\Sigma,\mu)$ be a measure space.   (i) Show that if
$\nu_1$ and $\nu_2$ are indefinite-integral measures over $\mu$, so is
$\nu_1+\nu_2$.   (ii) Show that if $\familyiI{\nu_i}$ is a countable family
of indefinite-integral measures over $\mu$, and $\nu=\sum_{i\in I}\nu_i$ is
semi-finite, then $\nu$ is an indefinite-integral measure over $\mu$.
%234L

\spheader 234Xj Let $(X,\Sigma,\mu)$ be a measure space, and $\nu$ an
indefinite-integral measure over $\mu$.   Show that if $\mu$ is purely
atomic, so is $\nu$.
%234N

\spheader 234Xk Let $\mu$ be a point-supported measure.   Show that any
indefinite-integral measure over $\mu$ is point-supported.
%234N

\spheader 234Xl Let $X$ be a set, and $\Mu$ the set of measures on $X$,
with the partial ordering defined in 234P.   Show that (i) $\Mu$ has
greatest and least members (to be described);
(ii) if $\familyiI{\mu_i}$
and $\familyiI{\nu_i}$ are families in $\Mu$ such that $\mu_i\le\nu_i$ for
every $i$, then $\sum_{i\in I}\mu_i\le\sum_{i\in I}\nu_i$;   (iii) if we
define scalar multiplication as in 234Xf, then
$\alpha\mu\le\mu$ whenever $\mu\in\Mu$ and $\alpha\in[0,1]$;   (iv) writing
$\hat\mu$ for the completion of $\mu$, $\hat\mu\le\mu$ and
$\hat\mu\le\hat\nu$ whenever $\mu$, $\nu\in\Mu$ and $\mu\le\nu$;  (v)
writing $\tilde\mu$ for the c.l.d.\ version of $\mu$, $\tilde\mu\le\mu$ for
every $\mu\in\Mu$;  (vi)\dvArevised{2016} every subset of $\Mu$ has 
a least upper bound in $\Mu$ (cf.\ 112Yd);  
(vii)\dvAnew{2016} every subset of $\Mu$ has a greatest lower bound in $\Mu$.
%234Q

\spheader 234Xm Write out an elementary direct proof of 234Qc not depending
on 234Qb.
%234Q

\spheader 234Xn Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be measure spaces and $\phi:X\to Y$ an \imp\ function.
Show that if $\mu$ is $\sigma$-finite and purely atomic then
$\nu$ is purely atomic.

\leader{234Y}{Further exercises (a)}
%\spheader 234Ya
Write $\nu$ for Lebesgue measure on
$Y=[0,1]$, and $\Tau$ for its domain.    Let $A\subseteq[0,1]$ be a set
such that $\nu^*A=\nu^*([0,1]\setminus A)=1$, and set
$X=[0,1]\cup\{x+1:x\in A\}\cup\{x+2:x\in[0,1]\setminus A\}$.   Let
$\mu_{LX}$ be the subspace measure induced on $X$ by Lebesgue measure,
and set $\mu E=\bover13\mu_{LX}E$ for $E\in\Sigma=\dom\mu_{LX}$.
Define $\phi:X\to Y$ by writing $\phi(x)=x$ if $x\in[0,1]$,
$\phi(x)=x-1$ if $x\in X\cap\ocint{1,2}$ and $\phi(x)=x-2$ if
$x\in X\cap\ocint{2,3}$.   Show that $\nu$ is the image measure
$\mu\phi^{-1}$, but that $\nu^*A>\mu^*\phi^{-1}[A]$.
%234B

\spheader 234Yb Look for interesting examples of probability spaces
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ for which there are functions
$\phi:X\to Y$ such that $\phi[E]\in\Tau$ and $\nu\phi[E]=\mu E$ for every
$E\in\Sigma$.   \Hint{254K, 343J.}
%234B

\spheader 234Yc Let $\mu$ be
two-dimensional Lebesgue measure on the unit square $[0,1]^2$, and let
$\phi:[0,1]^2\to[0,1]$ be the projection onto the first coordinate, so that
$\phi(\xi_1,\xi_2)=\xi_1$ for $\xi_1$, $\xi_2\in[0,1]$.   Show that the
image measure $\mu\phi^{-1}$ is Lebesgue
one-dimensional measure on $[0,1]$.
%234E

\spheader 234Yd In 234F, show that the image measure $\mu\phi^{-1}$
extends $\nu$, and is equal to $\nu$ if and only if $F\in\Tau$ for
every $F\subseteq Y\setminus\phi[X]$.
%234F

\spheader 234Ye Let $(Y,\Tau,\nu)$ be a complete measure space, $X$ a set
and $\phi:X\to Y$ a surjection.   Set

\Centerline{$\Sigma=\{E:E\subseteq X$, $\phi[E]\in\Tau$,
$\nu(\phi[E]\cap\phi[X\setminus E])=0\}$,
\quad$\mu E=\nu\phi[E]$ for $E\in\Tau$.}

\noindent Show that $\mu$ is the
completion of the measure constructed by the process of 234F.
%234F

\spheader 234Yf Let $X$ be a set, and $\Mu$ the set of measures on $X$.
Show that $\Mu$, with addition as defined for two measures by the formulae
of 234G, is a commutative semigroup with identity;  describe the identity.
%234G

\spheader 234Yg Give an example of a set $X$, probability measures
$\mu_1$, $\mu_2$ on $X$ and a set $A\subseteq X$ such that $A$ is
both $\mu_1$-negligible and $\mu_2$-negligible, but is not
$\mu$-negligible, where $\mu=\mu_1+\mu_2$.
%234G 

\spheader 234Yh In 214O, show that if we set
$\nu E=\sup_{I\in\Cal I}\mu^*(E\cap I)$ for every $E\in\Sigma$, then
$\nu$ is a measure, while $\mu=\nu+\lambda$.
%234H

\spheader 234Yi Let $(X,\Sigma,\mu)$ be an atomless semi-finite measure
space and
$\nu$ an indefinite-integral measure over $\mu$.   Show that the
following are equiveridical:  (i) for every $\epsilon>0$ there is a
$\delta>0$ such that $\nu E\le\epsilon$ whenever $\mu E\le\delta$ (ii)
$\nu$ has a Radon-Nikod\'ym derivative expressible as the sum of a
bounded function and an integrable function.
%234K

\spheader 234Yj Let $(X,\Sigma,\mu)$ be a measure space and $\nu$ an
indefinite-integral measure over $\mu$, with Radon-Nikod\'ym derivative
$f$.   Show that the c.l.d.\ version of $\nu$ is the indefinite-integral
measure defined by $f$ over the c.l.d.\ version of $\mu$.
%234N

\spheader 234Yk Let $(X,\Sigma,\mu)$ be a semi-finite measure space which is not
localizable.   Show that there is a measure $\nu:\Sigma\to[0,\infty]$
such that $\nu E\le\mu E$ for every $E\in\Sigma$ but there is no
measurable
function $f$ such that $\nu E=\int_Efd\mu$ for every $E\in\Sigma$.
%234O

\spheader 234Yl Let $(X,\Sigma,\mu)$ be a localizable measure space with
locally determined negligible sets.   Show that a measure $\nu$,
with domain $\Tau\supseteq\Sigma$, is an indefinite-integral measure
over $\mu$ iff ($\alpha$) $\nu$ is complete and semi-finite and zero on
$\mu$-negligible sets
($\beta$) whenever $\nu E>0$ there is an $F\subseteq E$ such that
$F\in\Sigma$ and $\mu F<\infty$ and $\nu F>0$.
%234O

\spheader 234Ym Give an example of a localizable measure space
$(X,\Sigma,\mu)$ and a complete semi-finite measure $\nu$ on $X$,
defined on
a $\sigma$-algebra $\Tau\supseteq\Sigma$, zero on $\mu$-negligible sets,
and such that whenever $\nu E>0$ there
is an $F\subseteq E$ such that $F\in\Sigma$ and $\mu F<\infty$ and
$\nu F>0$, but $\nu$ is not an indefinite-integral measure over $\mu$.
\Hint{216Yb.}
%234O 23bits

\spheader 234Yn Let $(X,\Sigma,\mu)$ be a localizable measure space, and
$\nu$ a complete localizable measure on $X$, with domain
$\Tau\supseteq\Sigma$, which is the completion of its restriction to
$\Sigma$.   Show that if we set
$\nu_1F=\sup\{\nu(F\cap E):E\in\Sigma$, $\mu E<\infty\}$ for every
$F\in\Tau$, then $\nu_1$ is an indefinite-integral measure over $\mu$,
and there is an $H\in\Sigma$ such that $\nu_1F=\nu(F\cap H)$ for every
$F\in\Tau$.
%234O

\spheader 234Yo Let $X$ be a set, and $\Mu_{\text{sf}}$ the set of
semi-finite measures
on $X$.   For $\mu$, $\nu\in\Mu_{\text{sf}}$ say that $\mu\preccurlyeq\nu$
if $\dom\nu\subseteq\dom\mu$, $\mu F\le\nu F$ for every $F\in\dom\nu$, and
whenever $E\in\dom\mu$ and $\mu E>0$ there is an $F\in\dom\nu$ such that
$F\subseteq E$ and $0<\mu F<\infty$.   (i) Show that
$(\Mu_{\text{sf}},\preccurlyeq)$ is a
partially ordered set.   (ii) Show that if $A\subseteq\Mu_{\text{sf}}$
is a non-empty
set with an upper bound in $\Mu_{\text{sf}}$, then it has a least upper
bound $\lambda$
defined by saying that $\dom\lambda=\bigcap_{\mu\in A}\dom\mu$ and, for
$E\in\dom\lambda$,

$$\eqalign{\lambda E
&=\sup\{\sum_{i=0}^n\mu_iF_i:\mu_0,\ldots,\mu_n\in A,\,
   \langle F_i\rangle_{i\le n}\text{ is a partition of }E,\cr
&\mskip200mu
   F_i\in\dom\lambda\text{ for every }i\le n\}\cr
&=\sup\{\sum_{i=0}^n\mu_iF_i:\mu_0,\ldots,\mu_n\in A,\,
   F_0,\ldots,F_n\text{ are disjoint},\cr
&\mskip200mu
   F_i\in\dom\mu_i\text{ and }F_i\subseteq E\text{ for every }i\le n\}.
   \cr}$$

\noindent(iii) Suppose that $\mu$, $\nu\in\Mu_{\text{sf}}$ have completions $\hat\mu$,
$\hat\nu$ and c.l.d.\ versions $\tilde\mu$, $\tilde\nu$.
Show that $\tilde\mu\preccurlyeq\hat\mu\preccurlyeq\mu$.   Show that if
$\mu\preccurlyeq\nu$ then
$\hat\mu\preccurlyeq\hat\nu$ and $\tilde\mu\preccurlyeq\tilde\nu$.
%234P
}%end of exercises

\endnotes{\Notesheader{234}
One of the striking features of measure theory, compared with other
comparably abstract branches of pure mathematics, is the relative
unimportance of any notion of `morphism'.   The theory of groups, for
instance, is dominated by the concept of `homomorphism', and general
topology gives a similar place to `continuous function'.   In my view,
the nearest equivalent in measure theory is the idea of
`\imp\ function' (234A).   I mean in Volumes 3 and
4 to explore this concept more thoroughly.   In this volume I will
content myself with signalling such functions when they arise, and with
the basic facts listed in 234B.

Naturally linked with the idea of \imp\ function is the construction of
`image measures' (234C).   These appear everywhere in the subject, starting
with the not-quite-elementary 234Yc.   They are of such importance that it
is natural to explore variations, as in 234F and 234Yb, but in my view none
are of comparable significance.

Nearly half the section is taken up with `indefinite-integral measures'.
I have taken this part very carefully
because the ideas I wish to express here, in so far as they extend the
work of \S232, rely critically on the details of the formulation in 234I,
and it is easy to make a false step once we have left the relatively
sheltered context of complete $\sigma$-finite measures.   I believe that
if we take a little trouble at this point we can develop a theory
(234K-234N) %234K 234L 234M 234N
which will
offer a smooth path to later applications;  to see what I have in mind,
you can refer to the entries under
`indefinite-integral measure' in the index.   For the moment I mention
only a kind of Radon-Nikod\'ym theorem for localizable measures (234O).

The partial ordering described in 234P-234Q is only one of many which can
be considered, and for some purposes it seems unsatisfactory.   The most
important examples will appear in Chapter 41 of Volume 4, and have a
variety of special features for which it might be worth setting out further
abstractions.   However the version here has the merit of simplicity and
supports at least some of the relevant ideas (234Xl).   For an alternative
notion, see 234Yo.

}%end of notes

%look for ideas to add to \S412

\discrpage

\frfilename{mt235.tex}
\versiondate{30.3.03/20.8.08}
\copyrightdate{1994}

\def\chaptername{The Radon-Nikod\'ym theorem}
\def\sectionname{Measurable transformations}
\def\tildeTau{\tilde{\text{T}}}

\newsection{235}

I turn now to a topic which is separate from the
Radon-Nikod\'ym theorem, but which seems to fit better here than in
either of the next two chapters.   I seek to give results which will
generalize the basic formula of calculus

\Centerline{$\int g(y)dy=\int g(\phi(x))\phi'(x)dx$}

\noindent in the context of a general transformation $\phi$ between
measure spaces.   The principal results are I suppose 235A/235E, which
are very similar expressions of the basic idea, and 235J, which gives a
general criterion for a stronger result.   A formulation from a
different direction is in 235R.

\leader{235A}{}\cmmnt{ I start with the basic result, which is already
sufficient for a large proportion of the applications I have in mind.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\phi:D_{\phi}\to Y$, $J:D_J\to\coint{0,\infty}$
functions defined on conegligible subsets $D_{\phi}$, $D_J$ of $X$
such that

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu$ exists $=\nu F$}

\noindent whenever $F\in\Tau$ and $\nu F<\infty$.   Then

\Centerline{$\int_{\phi^{-1}[H]}J\times g\phi\,d\mu$ exists =
$\int_Hg\,d\nu$}

\noindent for every $\nu$-integrable function $g$ taking values in
$[-\infty,\infty]$ and every $H\in\Tau$, provided that
we interpret $(J\times g\phi)(x)$ as $0$ when $J(x)=0$ and $g(\phi(x))$
is undefined.   Consequently, interpreting $J\times f\phi$ in the same way,

\Centerline{$\underline{\intop}fd\nu
\le\underline{\intop}J\times f\phi\,d\mu
\le\overline{\intop}J\times f\phi\,d\mu\le\overline{\intop}fd\nu$}

\noindent for every $[-\infty,\infty]$-valued function $f$ defined almost
everywhere in $Y$.

\proof{{\bf (a)} If $g$ is a simple function, say
$g=\sum_{i=0}^na_i\chi F_i$ where $\nu F_i<\infty$ for each $i$, then

\Centerline{$\int J\times g\phi\,d\mu
=\sum_{i=0}^na_i\int J\times\chi(\phi^{-1}[F_i])\,d\mu
=\sum_{i=0}^na_i\nu F_i=\int g\,d\nu$.}

\medskip

{\bf (b)} If $\nu F=0$ then $\int J\times\chi(\phi^{-1}[F])=0$ so $J=0$
a.e.\ on
$\phi^{-1}[F]$.   So if $g$ is defined $\nu$-a.e., $J=0\,\,\mu$-a.e.\ on
$X\setminus\dom(g\phi)=(X\setminus D_{\phi})\cup\phi^{-1}[Y\setminus\dom
g]$, and, on the convention proposed, $J\times g\phi$ is defined
$\mu$-a.e.   Moreover, if $\lim_{n\to\infty}g_n
=g\,\,\nu$-a.e., then $\lim_{n\to\infty}J\times g_n\phi=J\times
g\phi\,\,\mu$-a.e.   So if $\sequencen{g_n}$ is a non-decreasing
sequence of simple functions converging almost everywhere to $g$,
$\sequencen{J\times g_n\phi}$ will be a non-decreasing sequence of
integrable functions converging almost everywhere to
$J\times g\phi$;  by B.Levi's theorem,

\Centerline{$\int J\times g\phi\,d\mu$ exists $=\lim_{n\to\infty}
\int J\times g_n\phi\,d\mu=\lim_{n\to\infty}\int g_nd\nu
=\int g\,d\nu$.}

\medskip

{\bf (c)} If $g=g^+-g^-$, where $g^+$ and $g^-$ are $\nu$-integrable
functions, then

\Centerline{$\int J\times g\phi\,d\mu
=\int J\times g^+\phi\,d\mu-\int J\times g^-\phi\,d\mu
=\int g^+d\nu-\int g^-d\nu
=\int g\,d\nu$.}

\medskip

{\bf (d)} This deals with the case $H=Y$.   For the general case, we
have

$$\eqalignno{\int_Hg\,d\nu
&=\int(g\times\chi H)d\nu\cr
\displaycause{131Fa}
&=\int J\times(g\times\chi H)\phi\,d\mu
=\int J\times g\phi\times\chi(\phi^{-1}[H])d\mu
=\int_{\phi^{-1}[H]}J\times g\phi\,d\mu\cr}$$

\noindent by 214F.

\medskip

{\bf (e)} For the upper and lower integrals, I note first that if $F$ is
$\nu$-negligible then $\int J\times\chi(\phi^{-1}[F])d\mu=0$, so that
$J=0\,\,\mu$-a.e.\ on $\phi^{-1}[F]$.   It follows that if $f$ and $g$ are
$[-\infty,\infty]$-valued functions on subsets of $Y$ and $f\leae g$, then
$J\times f\phi\leae J\times g\phi$.   Now if $\overline{\int}fd\nu=\infty$,
we surely have 
$\overline{\int}J\times f\phi\,d\mu\le\overline{\int}fd\nu$.   Otherwise, 

$$\eqalign{\overline{\int}fd\nu
&=\inf\{\int g\,d\nu:g\text{ is }\nu-\text{integrable and }f\leae g\}\cr
&=\inf\{\int J\times g\phi\,d\mu:
  g\text{ is }\nu\text{-integrable and }f\leae g\}\cr
&\le\inf\{\int h\,d\mu:
  h\text{ is }\mu\text{-integrable and }J\times f\phi\leae h\}
=\overline{\intop}J\times f\phi\,d\mu.\cr}$$

\noindent Similarly, or applying this argument to $-f$, we have
$\underline{\int}J\times f\phi\,d\mu\le\underline{\int}f\,d\nu$.
}%end of proof of 235A

\cmmnt{
\leader{235B}{Remarks (a)} Note the particular convention

\Centerline{$0\times\text{undefined}=0$}

\noindent which I am applying to the interpretation of $J\times g\phi$.
This is the first of a number
of technical points which will concern us in this section.   The point
is that if $g$ is defined $\nu$-almost everywhere, then for any
extension of $g$ to a function $g_1:Y\to\Bbb R$ we shall, on this
convention, have $J\times g\phi=J\times g_1\phi$ except on
$\{x:J(x)>0,\,\phi(x)\in Y\setminus\dom g\}$, which is negligible;  so
that

\Centerline{$\int J\times g\phi\,d\mu
=\int J\times g_1\phi\,d\mu
=\int g_1d\nu
=\int g\,d\nu$}

\noindent if $g$ and $g_1$ are integrable.   Thus the convention is
appropriate here, and while it adds a phrase to the statements of many
of the results of this section, it makes their application smoother.
(But I ought to insist that I am using this as a local convention only,
and the ordinary rule $0\times\text{undefined}=\text{undefined}$ will
stand elsewhere in this treatise unless explicitly overruled.)

\header{235Bb}{\bf (b)} I have had
to take care in the formulation of this theorem to distinguish between
the hypothesis

\Centerline{$\int J(x)\chi(\phi^{-1}[F])(x)\mu(dx)$ exists $=\nu F$
whenever $\nu F<\infty$}

\noindent and the perhaps more elegant alternative

\Centerline{$\int_{\phi^{-1}[F]}J(x)\mu(dx)$ exists $=\nu F$
whenever
$\nu F<\infty$,}

\noindent which is not quite adequate for the theorem.   (See 235Q
below.)   Recall that by $\int_Af$ I mean $\int(f\restr A)d\mu_A$, where
$\mu_A$ is the subspace measure on $A$ (214D).   It is possible for
$\int_A(f\restr A)d\mu_A$ to be defined even
when $\int f\times\chi A\,d\mu$ is not;  for instance, take $\mu$ to be
Lebesgue measure on $[0,1]$, $A$ any non-measurable subset of $[0,1]$,
and $f$ the constant function with value $1$;  then $\int_Af=\mu^*A$,
but $f\times\chi A=\chi A$ is not $\mu$-integrable.   It is however the
case that if $\int f\times\chi A\,d\mu$ is defined, then so is
$\int_Af$, and the two are equal;  this is a consequence of 214F.
While 235P shows that in most of the cases relevant to the present
volume the distinction can be passed over, it is important to avoid
assuming that $\phi^{-1}[F]$ is measurable for every $F\in\Tau$.   A
simple example
is the following.   Set $X=Y=[0,1]$.   Let $\mu$ be Lebesgue measure on
$[0,1]$, and define $\nu$ by setting

\Centerline{$\Tau=\{F:F\subseteq[0,1],\,F\cap[0,\bover12]$ is Lebesgue
measurable$\}$,}

\Centerline{$\nu F=2\mu(F\cap[0,\bover12])$ for every $F\in\Tau$.}

\noindent Set $\phi(x)=x$ for every $x\in[0,1]$.   Then we have

\Centerline{$\nu F=\int_FJ\,d\mu
=\int J\times\chi(\phi^{-1}[F])d\mu$}

\noindent for every $F\in\Tau$, where $J(x)=2$ for $x\in[0,\bover12]$
and $J(x)=0$ for $x\in\ocint{\bover12,1}$.   But of course there are
subsets $F$ of $[\bover12,1]$ which are not  Lebesgue measurable (see
134D), and such an $F$ necessarily belongs to $\Tau$, even though
$\phi^{-1}[F]$ does not belong to the domain $\Sigma$ of $\mu$.

The point here is that if $\nu F_0=0$  then we expect to have $J=0$ on
$\phi^{-1}[F_0]$, and it is of no importance whether $\phi^{-1}[F]$ is
measurable for $F\subseteq F_0$.
}%end of comment

\leader{235C}{}\cmmnt{ Theorem 235A is concerned with integration, and
accordingly the hypothesis $\int J\times\chi(\phi^{-1}[F])d\mu=\nu F$
looks only at sets $F$ of finite measure.   If we wish to consider
measurability of non-integrable functions, we need a slightly stronger
hypothesis.   I approach this version more gently, with a couple of
lemmas.

\medskip

\noindent}{\bf Lemma} Let $\Sigma$, $\Tau$ be
$\sigma$-algebras of subsets of $X$ and $Y$ respectively.   Suppose that
$D\subseteq X$ and that $\phi:D\to Y$ is a function such that
$\phi^{-1}[F]\in\Sigma_D$, the subspace $\sigma$-algebra, for every
$F\in\Tau$.   Then $g\phi$ is $\Sigma$-measurable for every
$[-\infty,\infty]$-valued
$\Tau$-measurable function $g$ defined on a subset of $Y$.

\proof{ Set $C=\dom g$ and $B=\dom g\phi=\phi^{-1}[C]$.   If
$a\in\Bbb R$,
then there is an $F\in\Tau$ such that $\{y:g(y)\le a\}=F\cap C$.   Now
there is an $E\in\Sigma$ such that $\phi^{-1}[F]=E\cap D$.   So

\Centerline{$\{x:g\phi(x)\le a\}=B\cap E\in\Sigma_B$.}

\noindent As $a$ is arbitrary, $g\phi$ is $\Sigma$-measurable.
}%end of proof of 235C

\leader{235D}{}\cmmnt{ Some of the results below are easier when we
can move freely between measure spaces and their completions (212C).
The next lemma is what we need.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with completions $(X,\hat\Sigma,\hat\mu)$ and
$(Y,\hat\Tau,\hat\nu)$.   Let $\phi:D_{\phi}\to Y$, $J:D_J\to
\coint{0,\infty}$ be functions defined on conegligible subsets of $X$.

(a) If $\int J\times\chi(\phi^{-1}[F])d\mu=\nu F$ whenever $F\in\Tau$
and $\nu F<\infty$, then $\int J\times\chi(\phi^{-1}[F])d\hat\mu=\hat\nu
F$ whenever $F\in\hat\Tau$ and $\hat\nu F<\infty$.

(b) If $\int J\times\chi(\phi^{-1}[F])d\mu=\nu F$ whenever $F\in\Tau$,
then $\int J\times\chi(\phi^{-1}[F])d\hat\mu=\hat\nu F$ whenever
$F\in\hat\Tau$.

\proof{ Both rely on the fact that either hypothesis is enough to ensure
that $\int J\times\chi(\phi^{-1}[F])d\mu=0$ whenever $\nu F=0$.
Accordingly, if $F$ is $\nu$-negligible, so that there is an $F'\in\Tau$
such that $F\subseteq F'$ and $\nu F'=0$, we shall have

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu
=\int J\times\chi(\phi^{-1}[F'])d\mu=0$.}

\noindent But now, given any $F\in\hat\Tau$, there is an $F_0\in\Tau$
such that $F_0\subseteq F$ and $\hat\nu(F\setminus F_0)=0$, so that

$$\eqalign{\int J\times\chi(\phi^{-1}[F])d\hat\mu
&=\int J\times\chi(\phi^{-1}[F])d\mu\cr
&=\int J\times\chi(\phi^{-1}[F_0])d\mu
   +\int J\times\chi(\phi^{-1}[F\setminus F_0])d\mu\cr
&=\nu F_0=\hat\nu F,\cr}$$

\noindent provided (for part (a)) that $\hat\nu F<\infty$.
}%end of proof of 235D

\cmmnt{\medskip

\noindent{\bf Remark} Thus if we have the hypotheses of any of the
principal results of this section valid for a pair of non-complete
measure spaces, we can expect to be able to draw some conclusion by
applying the result to the completions of the given spaces.
}%end of comment

\leader{235E}{}\cmmnt{ Now I come to the alternative version of 235A.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\phi:D_{\phi}\to Y$,
$J:D_J\to\coint{0,\infty}$ two functions defined on conegligible subsets
of $X$ such that

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu=\nu F$}

\noindent for every $F\in\Tau$, allowing $\infty$ as a value of the
integral.

(a) $J\times g\phi$ is $\mu$-virtually measurable for every
$\nu$-virtually measurable function $g$ defined on a subset of $Y$.

(b) Let $g$ be a $\nu$-virtually measurable $[-\infty,\infty]$-valued
function defined on a conegligible subset of $Y$.   Then
$\int J\times g\phi\,d\mu=\int g\,d\nu$ whenever either integral is
defined in
$[-\infty,\infty]$, if we interpret $(J\times g\phi)(x)$ as $0$ when
$J(x)=0$ and $g(\phi(x))$ is undefined.

\proof{ Let $(X,\hat\Sigma,\hat\mu)$ and $(Y,\hat\Tau,\hat\nu)$ be the
completions of $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$.   By 235D,

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\hat\mu=\hat\nu F$}

\noindent for every $F\in\hat\Tau$.   Recalling that a real-valued
function is $\mu$-virtually measurable iff it is
$\hat\Sigma$-measurable (212Fa), and that $\int fd\mu=\int fd\hat\mu$ if
either is defined in $[-\infty,\infty]$ (212Fb), the conclusions we are
seeking are

\inset{(a)$'\,\,J\times g\phi$ is $\hat\Sigma$-measurable for every
$\hat\Tau$-measurable function $g$ defined on a subset of $Y$;}

\inset{(b)$'\,\,\int J\times g\phi\,d\hat\mu=\int g\,d\hat\nu$ whenever
$g$ is a $\hat\Tau$-measurable function defined
almost everywhere in $Y$ and either integral is defined in
$[-\infty,\infty]$.}

\medskip

{\bf (a)} When I write

\Centerline{$\int J\times\chi D_{\phi}d\mu
=\int J\times\chi(\phi^{-1}[Y])d\mu=\nu Y$,}

\noindent which is part of the hypothesis of this theorem, I mean to
imply that $J\times\chi D_{\phi}$ is $\mu$-virtually measurable, that
is, is $\hat\Sigma$-measurable.   Because $D_{\phi}$ is conegligible, it
follows that $J$ is
$\hat\Sigma$-measurable, and its domain $D_J$, being conegligible, also
belongs to $\hat\Sigma$.   Set $G=\{x:x\in D_J,\,J(x)>0\}\in\hat\Sigma$.
Then for any set $A\subseteq X$, $J\times\chi A$ is
$\hat\Sigma$-measurable iff
$A\cap G\in\hat\Sigma$.   So the hypothesis is just that
$G\cap\phi^{-1}[F]\in\hat\Sigma$ for every $F\in\hat\Tau$.

Now let $g$ be a $[-\infty,\infty]$-valued function, defined on a subset
$C$ of $Y$,
which is $\hat\Tau$-measurable.   Applying 235C to $\phi\restr G$, we
see that $g\phi\restr G$ is $\hat\Sigma$-measurable, so
$(J\times g\phi)\restr G$ is $\hat\Sigma$-measurable.   On the other
hand, $J\times g\phi$ is zero almost everywhere in $X\setminus G$, so
(because $G\in\hat\Sigma$) $J\times g\phi$ is $\hat\Sigma$-measurable,
as required.

\medskip

{\bf (b)(i)} Suppose first that $g\ge 0$.   Then $J\times g\phi\ge 0$,
so (a) tells us that $\int J\times g\phi$ is defined in $[0,\infty]$.

\medskip

\qquad\grheada\ If $\int g\,d\hat\nu<\infty$ then
$\int J\times g\phi\,d\hat\mu=\int g \,d\hat\nu$ by 235A.

\medskip

\qquad\grheadb\ If there is some $\epsilon>0$ such that
$\hat\nu H=\infty$, where $H=\{y:g(y)\ge\epsilon\}$, then

\Centerline{$\int J\times g\phi\,d\hat\mu
\ge\epsilon\int J\times\chi(\phi^{-1}[H])d\hat\mu
=\epsilon\hat\nu H=\infty$,}

\noindent so

\Centerline{$\int J\times g\phi\,d\hat\mu=\infty=\int
g\,d\hat\nu$.}

\medskip

\qquad\grheadc\ Otherwise,

$$\eqalign{\int J\times g\phi\,d\hat\mu
&\ge\sup\{\int J\times h\phi\,d\hat\mu:
   h\text{ is }\hat\nu\text{-integrable},
   \,0\le h\le g\}\cr
&=\sup\{\int h\,d\hat\nu:h\text{ is }\hat\nu\text{-integrable},
   \,0\le h\le g\}
=\int g\,d\hat\nu
=\infty,\cr}$$

\noindent so once again $\int J\times \phi\,d\hat\mu=\int g\,d\hat\nu$.

\medskip

\quad{\bf (ii)} For general real-valued $g$, apply (i) to $g^+$ and
$g^-$ where $g^+=\bover12(|g|+g)$, $g^-=\bover12(|g|-g)$;   the point is
that $(J\times g\phi)^+=J\times g^+\phi$ and
$(J\times g\phi)^-=J\times g^-\phi$, so that

\Centerline{$\int J\times g\phi
=\int J\times g^+\phi-\int J\times g^-\phi
=\int g^+-\int g^- %
=\int g$}

\noindent if either side is defined in $[-\infty,\infty]$.
}%end of proof of 235E

\cmmnt{
\leader{235F}{Remarks (a)} Of course there are two special
cases of this
theorem which between them carry all its content:  the case $J=1$ a.e.\
and the case in which $X=Y$ and $\phi$ is the identity function.   If
$J=\chi X$ we are very close to 235G below, and if $\phi$ is the
identity function we are close to the indefinite-integral measures of
\S234.

\header{235Fb}{\bf (b)} As in 235A, we can strengthen the
conclusion of (b) in 235E to

\Centerline{$\int_{\phi^{-1}[F]}J\times
g\phi\,d\mu=\int_Fg\,d\nu$}

\noindent whenever $F\in\Tau$ and $\int_Fg\,d\nu$ is defined in
$[-\infty,\infty]$.
}%end of comment

\leader{235G}{Theorem} Let $(X,\Sigma,\mu)$  and $(Y,\Tau,\nu)$ be
measure spaces and $\phi:X\to Y$ an \imp\ function.   Then

(a) if $g$ is a $\nu$-virtually measurable $[-\infty,\infty]$-valued
function
defined on a subset of $Y$, $g\phi$ is $\mu$-virtually measurable;

(b) if $g$ is a $\nu$-virtually measurable $[-\infty,\infty]$-valued
function
defined on a conegligible subset of $Y$, $\int g\phi\,d\mu=\int g\,d\nu$
if either integral is defined in $[-\infty,\infty]$;

(c) if $g$ is a $\nu$-virtually measurable $[-\infty,\infty]$-valued
function defined on a conegligible subset of $Y$, and $F\in\Tau$, then
$\int_{\phi^{-1}[F]}g\phi\,d\mu=\int_Fg\,d\nu$ if either integral is
defined in $[-\infty,\infty]$.

\proof{{\bf (a)} This follows immediately from 234Ba and 235C;   taking
$\hat\Sigma$, $\hat\Tau$ to be the domains of the completions of $\mu$,
$\nu$ respectively, $\phi^{-1}[F]\in\hat\Sigma$ for every $F\in\hat\Tau$,
so if $g$ is $\hat\Tau$-measurable then $g\phi$ will be
$\hat\Sigma$-measurable.

\medskip

{\bf (b)} Apply 235E with $J=\chi X$;  we have

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu
=\mu\phi^{-1}[F]=\nu F$}

\noindent for every $F\in\Tau$, so

\Centerline{$\int g\phi=\int J\times g\phi=\int g$}

\noindent if either integral is defined in $[-\infty,\infty]$.

\medskip

{\bf (c)} Apply (b) to $g\times\chi F$.
}%end of proof of 235G

\cmmnt{
\leader{235H}{The image measure catastrophe}
Applications of 235A would run much more smoothly if we could say

\inset{`$\int g\,d\nu$ exists and is equal to 
$\int J\times g\phi\,d\mu$
for every $g:Y\to\Bbb R$ such that $J\times g\phi$ is
$\mu$-integrable'.}

\noindent Unhappily there is no hope of a universally applicable result
in this direction.   Suppose, for instance, that $\nu$ is Lebesgue
measure on $Y=[0,1]$, that $X\subseteq[0,1]$ is a
non-Lebesgue-measurable set of outer measure $1$ (134D), that $\mu$ is
the subspace measure $\nu_X$ on $X$, and that $\phi(x)=x$ for $x\in X$.
Then

\Centerline{$\mu\phi^{-1}F=\nu^*(X\cap F)=\nu F$}

\noindent for every  Lebesgue measurable set $F\subseteq Y$, so we can
take $J=\chi X$ and the hypotheses of 235A and 235E will be satisfied.
But if we write $g=\chi X:[0,1]\to\{0,1\}$, then $\int g\phi\,d\mu$ is
defined even though $\int g\,d\nu$ is not.

The point here is that there is a set $A\subseteq Y$ such that (in the
language of 235A/235E) $\phi^{-1}[A]\in\Sigma$ but $A\notin\hat\Tau$.
This is the {\bf image measure catastrophe}.   
The search for contexts in which we can be sure that it does not occur 
will be one of the motive
themes of Volume 4.   For the moment, I will offer some general remarks
(235I-235J), and describe one of the important cases in which the
problem does not arise (235K).
}%end of comment


\leader{235I}{Lemma}
Let $\Sigma$, $\Tau$ be $\sigma$-algebras of
subsets of $X$, $Y$ respectively, and $\phi$ a function from a subset
$D$ of $X$ to $Y$.   Suppose that $G\subseteq X$ and that

\Centerline{$\Tau=\{F:F\subseteq Y,\,G\cap\phi^{-1}[F]\in\Sigma\}$.}

\noindent Then a real-valued function $g$, defined on a member of
$\Tau$, is $\Tau$-measurable iff $\chi G\times g\phi$ is
$\Sigma$-measurable.

\proof{ Because surely $Y\in\Tau$, the hypothesis implies that $G\cap
D=G\cap\phi^{-1}[Y]$ belongs to $\Sigma$.

Let $g:C\to\Bbb R$ be a function, where $C\in\Tau$.   Set
$B=\dom(g\phi)=\phi^{-1}[C]$, and for $a\in\Bbb R$ set
$F_a=\{y:g(y)\ge a\}$,

\Centerline{$E_a=G\cap\phi^{-1}[F_a]
=\{x:x\in G\cap B,\,g\phi(x)\ge a\}$.}

\noindent Note that $G\cap B\in\Sigma$ because $C\in\Tau$.

\medskip

\quad{\bf (i)} If $g$ is $\Tau$-measurable, then
$F_a\in\Tau$ and $E_a\in\Sigma$ for every $a$.   Now

\Centerline{$G\cap\{x:x\in B,\,g\phi(x)\ge a\}=G\cap\phi^{-1}[F_a]
=E_a$,}

\noindent so $\{x:x\in B,\,(\chi G\times g\phi)(x)\ge a\}$ is either
$E_a$ or
$E_a\cup(B\setminus G)$, and in either case is relatively
$\Sigma$-measurable in $B$.   As $a$ is arbitrary, $\chi G\times g\phi$
is $\Sigma$-measurable.

\medskip

\quad{\bf (ii)} If $\chi G\times g\phi$ is $\Sigma$-measurable, then,
for any $a\in\Bbb R$,

\Centerline{$E_a=\{x:x\in G\cap B,\,(\chi G\times g\phi)(x)\ge
a\}\in\Sigma$}

\noindent because $G\cap B\in\Sigma$ and $\chi G\times g\phi$ is
$\Sigma$-measurable.
So $F_a\in\Tau$.   As $a$ is arbitrary, $g$ is
$\Tau$-measurable.
}%end of proof of 235I

\leader{235J}{Theorem}
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
complete measure spaces.   Let $\phi:D_{\phi}\to Y$,
$J:D_J\to\coint{0,\infty}$ be functions defined on conegligible subsets
of $X$, and set $G=\{x:x\in D_J,\,J(x)>0\}$.   Suppose that

\Centerline{$\Tau
=\{F:F\subseteq Y,\,G\cap\phi^{-1}[F]\in\Sigma\}$,}

\Centerline{$\nu F=\int J\times\chi(\phi^{-1}[F])d\mu$ for every
$F\in\Tau$.}

\noindent Then, for any real-valued function $g$ defined on a subset
of $Y$,
$\int J\times g\phi\,d\mu=\int g\,d\nu$ whenever either integral is
defined in $[-\infty,\infty]$, provided that we interpret
$(J\times g\phi)(x)$ as $0$ when $J(x)=0$ and $g(\phi(x))$ is undefined.

\proof{ If $g$ is $\Tau$-measurable and defined almost everywhere, this
is a consequence of 235E.   So I have to show that if $J\times g\phi$ is
measurable and defined almost everywhere, so is $g$.
Set $W=Y\setminus\dom g$.   Then $J\times g\phi$ is undefined on
$G\cap\phi^{-1}[W]$, because $g\phi$ is undefined there and we cannot
take advantage of the escape clause available when $J=0$;  so
$G\cap\phi^{-1}[W]$ must be negligible, therefore measurable, and
$W\in\Tau$.   Next,

\Centerline{$\nu W=\int J\times\chi(\phi^{-1}[W])=0$}

\noindent because $J\times\chi(\phi^{-1}[W])$ can be non-zero only on
the negligible set $G\cap\phi^{-1}[W]$.   So $g$ is defined almost
everywhere.

Note that the hypothesis surely implies that $J\times\chi
D_{\phi}=J\times\chi(\phi^{-1}[Y])$ is measurable, so that $J$ is
measurable (because $D_{\phi}$ is conegligible) and $G\in\Sigma$.
Writing $K(x)=1/J(x)$ for $x\in G$, $0$ for $x\in X\setminus G$, the
function $K:X\to\Bbb R$ is measurable, and

\Centerline{$\chi G\times g\phi=K\times J\times g\phi$}

\noindent is measurable.   So 235I tells
us that $g$ must be measurable, and we're done.
}%end of proof of 235J

\cmmnt{\medskip

\noindent{\bf Remark} When $J=\chi X$, the hypothesis of this theorem
becomes

\Centerline{$\Tau=\{F:F\subseteq Y,\,\phi^{-1}[F]\in\Sigma\}$,
\quad $\nu F=\mu\phi^{-1}[F]$ for every $F\in\Tau$;}

\noindent that is, $\nu$ is the image measure $\mu\phi^{-1}$ as defined
in 234D.
}%end of comment

\leader{235K}{Corollary}
Let $(X,\Sigma,\mu)$ be a complete measure
space, and $J$ a non-negative measurable function defined on a
conegligible subset of $X$.   Let $\nu$ be the associated
indefinite-integral measure, and $\Tau$ its domain.   Then for any
real-valued function $g$ defined on a subset of $X$,
$g$ is $\Tau$-measurable iff $J\times g$ is $\Sigma$-measurable, and
$\int g\,d\nu=\int J\times g\,d\mu$ if either integral is defined in
$[-\infty,\infty]$, provided that we interpret $(J\times g)(x)$ as $0$
when $J(x)=0$ and $g(x)$ is undefined.

\proof{ Put 235J, taking $Y=X$ and $\phi$ the identity function,
together with 234Ld.
}%end of proof of 235K

\cmmnt{
\leader{235L}{Applying the Radon-Nikod\'ym theorem}
In order to
use 235A-235J effectively, we need to be able to find suitable functions
$J$.   This can be difficult -- some very special examples will take up
most of Chapter 26 below.   But there are many circumstances in
which we can be sure that such $J$ exist, even if we do not know what
they are.    A minimal requirement is that if $\nu F<\infty$ and
$\mu^*\phi^{-1}[F]=0$ then $\nu F=0$, because
$\int J\times\chi(\phi^{-1}[F])d\mu$ will be zero for any $J$.   A
sufficient condition, in the special case of indefinite-integral
measures, is in 234O.   Another is the following.
}%end of comment

\leader{235M}{Theorem}
Let $(X,\Sigma,\mu)$ be a
$\sigma$-finite measure space, $(Y,\Tau,\nu)$ a semi-finite measure
space, and $\phi:D\to Y$ a function such that

(i) $D$ is a conegligible subset of $X$,

(ii) $\phi^{-1}[F]\in\Sigma$ for every $F\in\Tau$;

(iii) $\mu\phi^{-1}[F]>0$ whenever $F\in\Tau$ and $\nu F>0$.

\noindent Then there is a $\Sigma$-measurable function
$J:X\to\coint{0,\infty}$
such that $\int J\times\chi\phi^{-1}[F]\,d\mu=\nu F$ for every
$F\in\Tau$.

\proof{{\bf (a)} To begin with (down to the end of (c) below) let us
suppose that $D=X$ and that $\nu$ is totally finite.

Set $\tildeTau=\{\phi^{-1}[F]:F\in\Tau\}\subseteq\Sigma$.   Then
$\tildeTau$ is a $\sigma$-algebra of subsets of $X$.   \Prf\ (i)

\Centerline{$\emptyset=\phi^{-1}[\emptyset]\in\tildeTau$.}

\noindent (ii) If $E\in\tildeTau$, take $F\in\Tau$ such that
$E=\phi^{-1}[F]$, so that

\Centerline{$X\setminus E=\phi^{-1}[Y\setminus F]\in\tildeTau$.}

\noindent (iii) If $\sequencen{E_n}$ is any sequence in $\tildeTau$,
then for each $n\in\Bbb N$ choose $F_n\in\Tau$ such that
$E_n=\phi^{-1}[F_n]$;  then

\Centerline{$\bigcup_{n\in\Bbb N}E_n=\phi^{-1}[\bigcup_{n\in\Bbb
N}F_n]\in\tildeTau$.  \Qed}

\noindent Next, we have a totally finite measure
$\tilde\nu:\tildeTau\to[0,\nu Y]$ given by setting

\Centerline{$\tilde\nu(\phi^{-1}[F])=\nu F$ for every $F\in\Tau$.}

\noindent\Prf\ (i)  If $F$, $F'\in\Tau$ and
$\phi^{-1}[F]=\phi^{-1}[F']$, then $\phi^{-1}[F\symmdiff F']=\emptyset$,
so $\mu(\phi^{-1}[F\symmdiff F'])=0$ and $\nu(F\symmdiff F')=0$;
consequently $\nu F=\nu F'$.   This shows that $\tilde\nu$ is
well-defined.   (ii) Now

\Centerline{$\tilde\nu\emptyset
=\tilde\nu(\phi^{-1}[\emptyset])
=\nu\emptyset=0$.}

\noindent (iii) If $\sequencen{E_n}$ is a disjoint sequence in
$\tildeTau$, let $\sequencen{F_n}$ be a sequence in $\Tau$ such that
$E_n=\phi^{-1}[F_n]$ for each $n$;  set
$F'_n=F_n\setminus\bigcup_{m<n}F_m$ for each $n$;  then
$E_n=\phi^{-1}[F'_n]$ for each $n$, so

\Centerline{$\tilde\nu(\bigcup_{n\in\Bbb N}E_n)
=\tilde\nu(\phi^{-1}[\bigcup_{n\in\Bbb N}F'_n])
=\nu(\bigcup_{n\in\Bbb N}F'_n)
=\sum_{n=0}^{\infty}\nu F'_n
=\sum_{n=0}^{\infty}\tilde\nu E_n$.   \Qed}

Finally, observe that if $\tilde\nu E>0$ then $\mu E>0$, because
$E=\phi^{-1}[F]$ where $\nu F>0$.

\medskip

{\bf (b)} By 215B(ix) there is a $\Sigma$-measurable function
$h:X\to\ooint{0,\infty}$ such that $\int h\,d\mu$ is finite.   Define
$\tilde\mu:\tildeTau\to\coint{0,\infty}$ by setting
$\tilde\mu E=\int_Eh\,d\mu$ for every $E\in\tildeTau$;  then $\tilde\mu$
is a totally
finite measure.   If $E\in\tildeTau$ and $\tilde\mu E=0$, then (because
$h$ is strictly positive) $\mu E=0$ and $\tilde\nu E=0$.
Accordingly we may apply the Radon-Nikod\'ym theorem to $\tilde\mu$ and
$\tilde\nu$ to see that there is a $\tildeTau$-measurable function
$g:X\to\Bbb R$
such that $\int_Eg\,d\tilde\mu=\tilde\nu E$ for every $E\in\tildeTau$.
Because $\tilde\nu$ is non-negative, we may suppose that $g\ge 0$.

\medskip

{\bf (c)} Applying 235A to $\mu$, $\tilde\mu$, $h$ and the identity
function from $X$ to itself, we see that

\Centerline{$\int_Eg\times h\,d\mu=\int_Eg\,d\tilde\mu
=\tilde\nu E$}

\noindent for every $E\in\tildeTau$, that is, that

\Centerline{$\int J\times\chi(\phi^{-1}[F])d\mu = \nu F$}

\noindent for every $F\in\Tau$, writing $J=g\times h$.

\medskip

{\bf (d)} This completes the proof when $\nu$ is totally finite and
$D=X$.   For the general case, if $Y=\emptyset$ then $\mu X=0$ and the
result is trivial.   Otherwise, let $\hat\phi$ be any extension of
$\phi$ to a function from $X$ to $Y$ which is constant on
$X\setminus D$;  then
$\hat\phi^{-1}[F]\in\Sigma$ for every $F\in\Tau$, because
$D=\phi^{-1}[Y]\in\Sigma$ and $\hat\phi^{-1}[F]$ is always either
$\phi^{-1}[F]$ or $(X\setminus D)\cup\phi^{-1}[F]$.   Now $\nu$ must be
$\sigma$-finite.
\Prf\ Use the criterion of 215B(ii).   If $\Cal F$ is a disjoint family
in $\{F:F\in\Tau,\,0<\nu F<\infty\}$, then
$\Cal E=\{\hat\phi^{-1}[F]:F\in\Cal F\}$ is a disjoint family in
$\{E:\mu E>0\}$, so $\Cal E$ and $\Cal F$ are countable.   \Qed

Let $\sequencen{Y_n}$ be a partition of $Y$ into sets of finite
$\nu$-measure, and for each $n\in\Bbb N$ set $\nu_nF=\nu(F\cap Y_n)$ for
every $F\in\Tau$.   Then $\nu_n$ is a totally finite measure on $Y$, and
if $\nu_nF>0$ then $\nu F>0$ so

\Centerline{$\mu\hat\phi^{-1}[F]=\mu\phi^{-1}[F]>0$.}

\noindent Accordingly $\mu$, $\hat\phi$ and $\nu_n$ satisfy the
assumptions of the theorem together with those of (a) above, and there
is a $\Sigma$-measurable function $J_n:X\to\coint{0,\infty}$ such that

\Centerline{$\nu_nF=\int J_n\times\chi(\phi^{-1}[F])d\mu$}

\noindent for every $F\in\Tau$.   Now set
$J=\sum_{n=0}^{\infty}J_n\times\chi(\phi^{-1}[Y_n])$, so that
$J:X\to\coint{0,\infty}$ is $\Sigma$-measurable.   If $F\in\Tau$, then

$$\eqalign{\int J\times\chi(\phi^{-1}[F])d\mu
&=\sum_{n=0}^{\infty}
  \int J_n\times\chi(\phi^{-1}[Y_n])\times\chi(\phi^{-1}[F])d\mu\cr
&=\sum_{n=0}^{\infty}\int J_n\times\chi(\phi^{-1}[F\cap Y_n])d\mu
=\sum_{n=0}^{\infty}\nu(F\cap Y_n)
=\nu F,\cr}$$

\noindent as required.
}%end of proof of 235M

\leader{235N}{Remark}
Theorem 235M can fail if $\mu$ is only strictly
localizable rather than $\sigma$-finite.   \prooflet{\Prf\ Let $X=Y$ be
an uncountable set, $\Sigma=\Cal PX$,  $\mu$ counting measure on $X$
(112Bd), $\Tau$ the countable-cocountable $\sigma$-algebra of $Y$, $\nu$
the countable-cocountable measure on $Y$ (211R), $\phi:X\to Y$ the
identity map.   Then $\phi^{-1}[F]\in\Sigma$ and $\mu\phi^{-1}[F]>0$
whenever $\nu F>0$.   But if $J$ is any
$\mu$-integrable function on $X$, then $F=\{x:J(x)\ne 0\}$ is countable
and

\Centerline{$\nu(Y\setminus F)=1\ne 0
=\int_{\phi^{-1}[Y\setminus F]}J\,d\mu$.  \Qed}
}

\leader{*235O}{}\cmmnt{ There are some simplifications 
in the
case of $\sigma$-finite spaces;  in particular, 235A and 235E become
conflated.   I will give an adaptation of the hypotheses of 235A which
may be used in the $\sigma$-finite case.   First a lemma.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a measure space and $f$ a
non-negative integrable function on $X$.   If $A\subseteq X$ is such
that $\int_Af+\int_{X\setminus A}f=\int f$, then $f\times\chi A$ is
integrable.

\proof{ By 214Eb, there are $\mu$-integrable functions
$f_1$, $f_2$ such that $f_1$ extends $f\restr A$, $f_2$ extends
$f\restr X\setminus A$, and

\Centerline{$\int_Ef_1=\int_{E\cap A}f$,
\quad $\int_Ef_2=\int_{E\setminus A}f$}

\noindent for every $E\in\Sigma$.   Because $f$ is non-negative,
$\int_Ef_1$ and $\int_Ef_2$ are non-negative for every $E\in\Sigma$, and
$f_1$, $f_2$ are non-negative a.e.   Accordingly we have
$f\times\chi A\leae f_1$ and $f\times\chi(X\setminus A)\leae f_2$, so
that $f\leae f_1+f_2$.   But also

\Centerline{$\int f_1+f_2
=\int_Xf_1+\int_Xf_2=\int_Af+\int_{X\setminus A}f=\int f$,}

\noindent so $f\eae f_1+f_2$.   Accordingly

\Centerline{$f_1\eae f-f_2\leae f-f\times\chi(X\setminus A)
=f\times\chi A\leae f_1$}

\noindent and $f\times\chi A\eae f_1$ is integrable.
}%end of proof of 235O

\leader{*235P}{Proposition}
Let $(X,\Sigma,\mu)$ be a complete
measure space and $(Y,\Tau,\nu)$ a complete $\sigma$-finite measure
space.   Suppose that $\phi:D_{\phi}\to Y$,
$J:D_J\to\coint{0,\infty}$ are functions defined on conegligible subsets
$D_{\phi}$, $D_J$ of $X$ such that
$\int_{\phi^{-1}[F]}J\,d\mu$ exists and is equal to $\nu F$ whenever
$F\in\Tau$ and $\nu F<\infty$.

(a) $J\times g\phi$ is $\Sigma$-measurable for every
$\Tau$-measurable real-valued function $g$ defined on a subset of $Y$.

(b) If $g$ is a $\Tau$-measurable real-valued function defined almost
everywhere in $Y$, then $\int J\times g\phi\,d\mu=\int g\,d\nu$ whenever
either integral is defined in $[-\infty,\infty]$,  interpreting
$(J\times g\phi)(x)$ as $0$ when $J(x)=0$, $g(\phi(x))$ is undefined.

\proof{ The point is that the hypotheses of 235E are
satisfied.   To see this, let us write $\Sigma_C=\{E\cap C:E\in\Sigma\}$
and $\mu_C=\mu^*\restr \Sigma_C$ for the subspace measure on $C$, for
each $C\subseteq X$.   Let $\sequencen{Y_n}$ be a non-decreasing
sequence of sets with union $Y$ and with $\nu Y_n<\infty$ for every
$n\in\Bbb N$, starting from $Y_0=\emptyset$.

\medskip

{\bf (i)} Take any $F\in\Tau$ with $\nu F<\infty$, and set
$F_n=F\cup Y_n$ for each $n\in\Bbb N$;  write $C_n=\phi^{-1}[F_n]$.

Fix $n$ for the moment.   Then our hypothesis implies that

\Centerline{$\int_{C_0} J\,d\mu+\int_{C_n\setminus C_0} J\,d\mu
=\nu F+\nu(F_n\setminus F)=\nu F_n=\int_{C_n}J\,d\mu$.}

\noindent If we regard the subspace measures on $C_0$ and
$C_n\setminus C_0$ as derived from the measure $\mu_{C_n}$ of $C_n$
(214Ce), then 235O
tells us that $J\times\chi C_0$ is $\mu_{C_n}$-integrable, and there is
a $\mu$-integrable function $h_n$ such that  $h_n$ extends
$(J\times\chi C_0)\restr C_n$.

Let $E$ be a $\mu$-conegligible set, included in the domain $D_{\phi}$
of $\phi$, such that $h_n\restr E$ is $\Sigma$-measurable for every $n$.
Because $\sequencen{C_n}$ is a non-decreasing sequence with union
$\phi^{-1}[\bigcup_{n\in\Bbb N}F_n]
\ifdim\pagewidth=390pt\penalty-100\fi
=D_{\phi}$,

\Centerline{$(J\times\chi C_0)(x)=\lim_{n\to\infty}h_n(x)$}

\noindent for every $x\in E$, and $(J\times\chi C_0)\restr E$ is
measurable.   At the same time, we know that there is a $\mu$-integrable
$h$ extending $J\restr C_0$, and $0\leae J\times\chi C_0\leae|h|$.
Accordingly $J\times \chi C_0$ is integrable, and (using 214F)

\Centerline{$\int J\times\chi\phi^{-1}[F]\,d\mu
=\int J\times\chi C_0\,d\mu
=\int_{C_0}J\restr C_0\,d\mu_{C_0}
=\nu F$.}

\medskip

{\bf (ii)} This deals with $F$ of finite measure.   For general
$F\in\Tau$,

\Centerline{$\int J\times\chi(\phi^{-1}[F])\,d\mu
=\lim_{n\to\infty}\int J\times\chi(\phi^{-1}[F\cap Y_n])\,d\mu
=\lim_{n\to\infty}\nu(F\cap Y_n)
=\nu F$.}

\noindent So the hypotheses of 235E are satisfied, and the result
follows at once.
}%end of proof of 235P

\leader{*235Q}{}\cmmnt{ I remarked in 235Bb that a 
difficulty can
arise in 235A, for general measure spaces, if we speak of
$\int_{\phi^{-1}[F]}J\,d\mu$ in the hypothesis, in
place of $\int J\times\chi(\phi^{-1}[F])d\mu$.   Here is an example.

\medskip

\noindent}{\bf Example} Set $X=Y=[0,2]$.   Write $\Sigma_L$ for the
algebra of Lebesgue measurable subsets of $\Bbb R$, and $\mu_L$ for
Lebesgue measure;  write $\mu_c$ for counting measure on $\Bbb R$.   Set

\Centerline{$\Sigma=\Tau
=\{E:E\subseteq[0,2],\,E\cap\coint{0,1}\in\Sigma_L\}$\dvro{.}{;}}

\noindent\cmmnt{of course this is a $\sigma$-algebra of subsets of
$[0,2]$.  }For $E\in\Sigma=\Tau$, set

\Centerline{$\mu E=\nu
E=\mu_L(E\cap\coint{0,1})+\mu_c(E\cap[1,2])$\dvro{.}{;}}

\noindent\cmmnt{then $\mu$ is a complete measure -- in effect, it is
the direct sum of Lebesgue measure on $\coint{0,1}$ and counting measure
on $[1,2]$ (see 214L).   It is easy to see that

\Centerline{$\mu^*B=\mu_L^*(B\cap\coint{0,1})+\mu_c(B\cap[1,2])$}

\noindent for every $B\subseteq[0,2]$.   }Let $A\subseteq\coint{0,1}$ be
a non-Lebesgue-measurable set such that $\mu_L^*(E\setminus A)=\mu_LE$
for every Lebesgue measurable
$E\subseteq\coint{0,1}$\cmmnt{ (see 134D)}.   Define
$\phi:[0,2]\to[0,2]$ by
setting $\phi(x)=x+1$ if $x\in A$, $\phi(x)=x$ if $x\in[0,2]\setminus
A$.

If $F\in\Sigma$, then $\mu^*(\phi^{-1}[F])=\mu F$.   \prooflet{\Prf\
{\bf (i)} If
$F\cap[1,2]$ is finite, then $\mu F=\mu_L(F\cap[0,1])+\#(F\cap[1,2])$.
Now

\Centerline{$\phi^{-1}[F]=(F\cap\coint{0,1}\setminus
A)\cup(F\cap[1,2])\cup\{x:x\in A,\,x+1\in F\}$;}

\noindent as the last set is finite, therefore $\mu$-negligible,

\Centerline{$\mu^*(\phi^{-1}[F])
=\mu_L^*(F\cap\coint{0,1}\setminus A)+\#(F\cap[1,2])
=\mu_L(F\cap\coint{0,1})+\#(F\cap[1,2])
=\mu F$.}

\noindent {\bf (ii)} If $F\cap[1,2]$ is infinite, so is
$\phi^{-1}[F]\cap[1,2]$, so

\Centerline{$\mu^*(\phi^{-1}[F])=\infty=\mu F$.   \Qed}
}

This means that if we set $J(x)=1$ for every $x\in[0,2]$,
\cmmnt{

\Centerline{$\int_{\phi^{-1}[F]}J\,d\mu
=\mu_{\phi^{-1}[F]}(\phi^{-1}[F])
=\mu^*(\phi^{-1}[F])
=\mu F$}

\noindent for every $F\in\Sigma$, and} $\phi$, $J$ satisfy the amended
hypotheses for 235A.   But if we set $g=\chi\coint{0,1}$, then $g$ is
$\mu$-integrable, with $\int g\,d\mu=1$, while
\cmmnt{

\Centerline{$J(x)g(\phi(x))=1$ if $x\in[0,1]\setminus A$, $0$
otherwise,}

\noindent so, because $A\notin\Sigma$,} $J\times g\phi$ is
not\cmmnt{ measurable, and therefore (since $\mu$ is complete)
not} $\mu$-integrable.

\leader{235R}{Reversing the \dvrocolon{burden}}\cmmnt{ 
Throughout the work above, I have been using the formula

\Centerline{$\int J\times g\phi=\int g$,}

\noindent as being the natural extension of the formula

\Centerline{$\int g=\int g\phi\times \phi'$}

\noindent of ordinary advanced calculus.   But we can also move the
`derivative' $J$ to the other side of the equation, as follows.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ be measure
spaces and $\phi:X\to Y$, $J:Y\to\coint{0,\infty}$ functions such that
$\int_FJ\,d\nu$ and $\mu\phi^{-1}[F]$ are defined in $[0,\infty]$ and
equal for every $F\in\Tau$.   Then
$\int g\phi\,d\mu=\int J\times g\,d\nu$
whenever $g$ is $\nu$-virtually measurable and defined $\nu$-almost
everywhere and either integral is defined in $[-\infty,\infty]$.

\proof{ Let $\nu_1$ be the indefinite-integral measure over $\nu$
defined by $J$, and $\hat\mu$ the completion of $\mu$.   Then $\phi$ is
\imp\ for $\hat\mu$ and $\nu_1$.   \Prf\ If $F\in\Tau$, then
$\nu_1F=\int_FJ\,d\nu=\mu\phi^{-1}[F]$;  that is, $\phi$ is \imp\ for
$\mu$ and $\nu_1\restr\Tau$.   Since $\nu_1$ is the completion of
$\nu_1\restr\Tau$ (234Lb), $\phi$ is \imp\ for $\mu$ and $\nu_1$
(234Ba).\ \Qed

Of course we can also regard $\nu_1$ as being an indefinite-integral
measure over the completion $\hat\nu$ of $\nu$ (212Fb).   So if $g$ is
$\nu$-virtually measurable and defined $\nu$-almost everywhere,

\Centerline{$\int J\times g\,d\nu
=\int J\times g\,d\hat\nu
=\int g\,d\nu_1
=\int g\phi\,d\hat\mu
=\int g\phi\,d\mu$}

\noindent if any of the five integrals is defined in $[-\infty,\infty]$,
by 235K, 235Gb and 212Fb again.
}%end of proof of 235R

\exercises{
\leader{235X}{Basic exercises (a)}
%\spheader 235Xa
Explain what 235A tells us when $X=Y$,
$\Tau=\Sigma$, $\phi$ is the identity function and $\nu E=\alpha\mu E$
for every $E\in\Sigma$.
%235A

\spheader 235Xb Let $(X,\Sigma,\mu)$ be a measure space, $J$ an
integrable non-negative real-valued function on $X$, and
$\phi:D_{\phi}\to\Bbb R$ a measurable function, where $D_{\phi}$ is a
conegligible subset of $X$.   Set

\Centerline{$g(a)=\int_{\{x:\phi(x)\le a\}}J$}

\noindent for $a\in\Bbb R$, and let $\mu_g$ be the Lebesgue-Stieltjes
measure associated with $g$.   Show that
$\int J\times f\phi\,d\mu=\int fd\mu_g$ for every $\mu_g$-integrable
real function $f$.
%235A

\spheader 235Xc Let $\Sigma$, $\Tau$ and $\Lambda$ be
$\sigma$-algebras of subsets of $X$, $Y$ and $Z$ respectively.   Let us
say that a function $\phi:A\to Y$, where $A\subseteq X$, is
$(\Sigma,\Tau)$-measurable if $\phi^{-1}[F]\in\Sigma_A$, the subspace
$\sigma$-algebra of $A$, for every $F\in\Tau$.   Suppose that
$A\subseteq X$, $B\subseteq Y$, $\phi:A\to Y$ is
$(\Sigma,\Tau)$-measurable and $\psi:B\to Z$ is
$(\Tau,\Lambda)$-measurable.   Show that $\psi\phi$ is
$(\Sigma,\Lambda)$-measurable.   Deduce 235C.
%235C

\spheader 235Xd Let $(X,\Sigma,\mu)$ be a measure
space and $(Y,\Tau,\nu)$ a semi-finite measure space.   Let
$\phi:D_{\phi}\to Y$ and $J:D_J\to\coint{0,\infty}$ be
functions defined on conegligible subsets $D_{\phi}$, $D_J$ of $X$
such that $\int J\times\chi(\phi^{-1}[F])d\mu$ exists $=\nu F$ whenever
$F\in\Tau$ and $\nu F<\infty$.   Let $g$ be a
$\Tau$-measurable
real-valued function, defined on a conegligible subset of $Y$.   Show
that $J\times g\phi$ is $\mu$-integrable iff $g$ is $\nu$-integrable,
and the integrals are then equal, provided we interpret
$(J\times g\phi)(x)$ as
$0$ when $J(x)=0$ and $g(\phi(x))$ is undefined.
%235E

\spheader 235Xe Let $(X,\Sigma,\mu)$ be a measure space and
$E\in\Sigma$.   Define a measure $\mu\LLcorner E$ on $X$ by setting
$(\mu\LLcorner E)(F)=\mu(E\cap F)$ whenever $F\subseteq X$ is such that
$F\cap E\in\Sigma$.   Show that, for any function $f$ from a subset of
$X$ to $[-\infty,\infty]$, $\int fd(\mu\LLcorner E)=\int_Efd\mu$ if
either is defined in $[-\infty,\infty]$.
%235G

\sqheader 235Xf Let $g:\Bbb R\to\Bbb R$ be a
non-decreasing function which is absolutely continuous on
every closed bounded interval, and $\mu_g$ the associated
Lebesgue-Stieltjes measure (114Xa, 225Xd).   Write $\mu$ for Lebesgue
measure on $\Bbb R$, and let $f:\Bbb R\to\Bbb R$ be a function.   Show
that $\int f\times g'\,d\mu=\int f\,d\mu_g$ in the sense that if one of
the integrals exists, finite or infinite, so does the other, and they
are then equal.
%235J

\spheader 235Xg Let $g:\Bbb R\to\Bbb R$ be a
non-decreasing function and $J$ a non-negative real-valued
$\mu_g$-integrable function, where $\mu_g$ is the Lebesgue-Stieltjes
measure defined from $g$.   Set
$h(x)=\int_{\ocint{-\infty,x}}J\,d\mu_g$ for each $x\in\Bbb R$, and let
$\mu_h$ be the Lebesgue-Stieltjes measure associated with $h$.   Show
that, for any $f:\Bbb R\to\Bbb R$,
$\int f\times J\,d\mu_g=\int fd\mu_h$, in the sense that if one of the
integrals is defined in $[-\infty,\infty]$ so is the other, and they are
then equal.
%235J, 235Xf

\sqheader 235Xh Let $X$ be a set and $\lambda$, $\mu$, $\nu$ three
measures on $X$ such that $\mu$ is an indefinite-integral measure over
$\lambda$, with Radon-Nikod\'ym derivative $f$, and $\nu$ is an
indefinite-integral measure over $\mu$, with Radon-Nikod\'ym derivative
$g$.   Show that $\nu$ is an indefinite-integral measure over $\lambda$,
and that $f\times g$ is a Radon-Nikod\'ym derivative of $\nu$ with
respect to $\lambda$, provided we interpret $(f\times g)(x)$ as $0$ when
$f(x)=0$ and $g(x)$ is undefined.
%235K

\spheader 235Xi In 235M, if $\nu$ is not semi-finite, show that
we can still find a $J$ such that $\int_{\phi^{-1}[F]}J\,d\mu=\nu F$ for
every set $F$ of finite measure.   \Hint{use the `semi-finite version'
of $\nu$, as described in 213Xc.}
%235M

\spheader 235Xj Let $(X,\Sigma,\mu)$ be a $\sigma$-finite
measure space, and $\Tau$ a $\sigma$-subalgebra of $\Sigma$.   Let
$\nu:\Tau\to\Bbb R$ be a countably additive functional such that
$\nu F=0$ whenever $F\in\Tau$ and $\mu F=0$.   Show that there is a
$\mu$-integrable function $f$ such that $\int_Ffd\mu=\nu F$ for every
$F\in\Tau$.   \Hint{use the method of 235M, applied to the positive and
negative parts of $\nu$.}
%235M

\spheader 235Xk Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with completions $(X,\hat\Sigma,\hat\mu)$ and
$(Y,\hat\Tau,\hat\nu)$.   Let $\phi:D_{\phi}\to Y$,
$J:D_J\to \coint{0,\infty}$ be functions defined on conegligible subsets
of $X$. Show that if $\int_{\phi^{-1}[F]}J\,d\mu=\nu F$ whenever
$F\in\Tau$ and
$\nu F<\infty$, then $\int_{\phi^{-1}[F]}J\,d\mu=\nu F$ whenever
$F\in\hat\Tau$ and $\hat\nu F<\infty$.   Hence, or otherwise, show that
235Pb is valid for non-complete spaces $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$.
%235P

\spheader 235Xl Let $(X,\Sigma,\mu)$ be a complete measure space, 
$Y$ a set, $\phi:X\to Y$ a function and $\nu=\mu\phi^{-1}$ the 
corresponding image measure on $Y$.   Let $\nu_1$ be an
indefinite-integral measure over $\nu$.   Show that there is an
indefinite-integral measure $\mu_1$ over $\mu$ such that $\nu_1$ is the 
image measure $\mu_1\phi^{-1}$.
%??

\spheader 235Xm
 Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces,  
and $\phi:X\to Y$ an \imp\ function.   Let $\nu_1$ be an 
indefinite-integral measure over $\nu$.   Show that there is an 
indefinite-integral measure $\mu_1$ over $\mu$ such that $\phi$ is \imp\  
for $\mu_1$ and $\nu_1$. 
%235G out of order query 

\spheader 235Xn\dvAnew{2013}  
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be 
measure spaces, and $\phi:X\to Y$ an \imp\ function.   Show that
$\overline{\int}h\phi\,d\mu\le\overline{\int}h\,d\nu$ for every
real-valued function $h$ defined almost everywhere in $Y$.
(Compare 234Bf.)
%235G 

\leader{235Y}{Further exercises (a)} 
%\spheader 235Ya
Write $\Tau$ for the algebra of Borel subsets of
$Y=[0,1]$, and $\nu$ for the restriction of Lebesgue measure to $\Tau$.
Let $A\subseteq[0,1]$ be a set such that both $A$ and $[0,1]\setminus A$
have Lebesgue outer measure $1$, and set $X=A\cup[1,2]$.   Let $\Sigma$
be the algebra of relatively Borel subsets of $X$, and set 
$\mu E=\mu_A(A\cap E)$ for $E\in\Sigma$, where $\mu_A$ is the subspace
measure induced on $A$ by Lebesgue measure.   Define $\phi:X\to Y$ by
setting $\phi(x)=x$ if $x\in A$, $x-1$ if $x\in X\setminus A$.   Show
that $\nu$ is the image measure $\mu\phi^{-1}$, but that,
setting $g=\chi([0,1]\setminus A)$, $g\phi$ is $\mu$-integrable while
$g$ is not $\nu$-integrable.

\spheader 235Yb Let $(X,\Sigma,\mu)$ be a probability space and $\Tau$ a
$\sigma$-subalgebra of $\Sigma$.   Let $f$ be a non-negative
$\mu$-integrable function with $\int fd\mu=1$, so that its
indefinite-integral measure $\nu$ is a probability measure.   Let $g$ be
a $\nu$-integrable real-valued function and set $h=f\times g$,
intepreting
$h(x)$ as $0$ if $f(x)=0$ and $g(x)$ is undefined.   Let $f_1$, $h_1$ be
conditional expectations of $f$, $h$ on $\Tau$ with respect to the
measure $\mu$, and set $g_1=h_1/f_1$, interpreting $g_1(x)$ as $0$ if
$h_1(x)=0$ and $f_1(x)$ is either $0$ or undefined.   Show that $g_1$ is
a conditional expectation of $g$ on $\Tau$ with respect to the measure
$\nu$.
}%end of exercises

\endnotes{
\Notesheader{235} I see that I have taken up a great deal of space in
this section with technicalities;  the hypotheses of the theorems vary
erratically, with completeness, in particular, being invoked at
apparently arbitrary intervals, and ideas repeat themselves in a
haphazard pattern.   There
is nothing deep, and most of the work consists in laboriously verifying
details.   The trouble with this topic is that it is useful.   The
results here are abstract expressions of integration-by-substitution;
they have applications all over measure theory.
I cannot therefore content myself with theorems which will elegantly
express the underlying ideas, but must seek formulations which I can
quote in later arguments.

I hope that the examples in 235Bb, 235H, 235N, 235Q, 234Ya and
235Ya will go some way to persuade you that there are real traps for the
unwary, and that the careful verifications written out at such length
are necessary.   On the other hand, it is happily the case that in
simple contexts, in which the measures $\mu$, $\nu$ are $\sigma$-finite
and the transformations $\phi$ are Borel isomorphisms, no insuperable
difficulties arise, and in particular the image measure catastrophe does
not trouble us.   But for further work in this direction I refer you to
the applications in \S263, \S265 and \S271, and to Volume 4.
}%end of notes

\frnewpage

\ifresultsonly\else
\frfilename{mt23conc.tex}
\versiondate{20.8.08}
\copyrightdate{2008}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{The Radon-Nikod\'ym theorem}
\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{\S2{}34}{} Section \S2{}34 has been rewritten, with a good deal of
new material.   The former paragraphs 2{}34A-2{}34G, referred to in the
2002 and 2004 editions of Volume 3 and the 2003 and 2006 editions of
Volume 4, are now 234I-234O. %234I 234J 234K 234L 234M 234N 234O
% only 2{}34D in Vol 3
% all but 2{}34E in Vol4

\leader{\S2{}35}{} Section \S2{}35 has been re-organized, with
some material moved to \S2{}34.   Specifically, 
2{}35H, 2{}35I, 2{}35J, 2{}35L, 2{}35M, 2{}35T and 2{}35Xe, referred to in
the 2002 and 2004 editions of Volume 3 and the 2003 and 2006 editions of
Volume 4, are now dealt with in
234B, 235G, 235H, 235J, 235K, 235R and 234A.

\frnewpage

\fi
\references{
\frfilename{mt2r.tex}
\versiondate{17.12.12}
\copyrightdate{2000}

% 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 2}

\medskip

%A

Alexits G.\ [78] (ed.) {\it Fourier Analysis and Approximation Theory.}
North-Holland, 1978 (Colloq.\ Math.\ Soc.\ Janos Bolyai 19).

Antonov N.Yu.\ [96] `Convergence of Fourier series', East J.\ Approx.\ 7 (1996) 187-196.
\cmmnt{[\S286 {\it notes\/}.]}

Arias de Reyna J.\ [02] {\it Pointwise Convergence of Fourier Series.} Springer, 2002 (Lecture Notes in Mathematics 1785).
\cmmnt{[\S286 {\it notes\/}.]}

\medskip%B

Baker R.\ [04] `{}``Lebesgue measure'' on $\Bbb R^{\infty}$, II',
Proc.\ Amer.\ Math.\ Soc.\ 132 (2004) 2577-2591.
\cmmnt{[254Yb.]}

Bergelson V., March P.\ \& Rosenblatt J.\ [96] (eds.) {\it Convergence
in Ergodic Theory and Probability.}  de Gruyter, 1996.

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

du Bois-Reymond P.\ [1876] `Untersuchungen \"uber die Convergenz und
Divergenz der Fouriersche Darstellungformeln', Abh.\ Akad.\ M\"unchen 12
(1876) 1-103.   \cmmnt{[\S282 {\it notes}.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[2A5F.]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[2A5E.]}

\medskip%C

Carleson L.\ [66] `On convergence and growth of partial sums of
Fourier series', Acta Math.\ 116 (1966) 135-157.  \cmmnt{[\S282 {\it
notes}, \S286 {\it intro.}, \S286 {\it notes}.]}

Clarkson J.A.\ [1936] `Uniformly convex spaces', Trans.\ Amer.\ Math.\ Soc.\
40 (1936) 396-414.
\cmmnt{[244O.]}

\medskip%D

Defant A.\ \& Floret K.\ [93] {\it Tensor Norms and Operator Ideals},
North-Holland, 1993.   \cmmnt{[\S253 {\it notes}.]}

Doob J.L.\ [53] {\it Stochastic Processes.}  Wiley, 1953.
%used for 275Ye

Dudley R.M.\ [89] {\it Real Analysis and Probability.} Wadsworth \&
Brooks/Cole, 1989.   \cmmnt{[\S282 {\it notes}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).  \cmmnt{[\S244 {\it notes}, 2A5J.]}

\medskip%E

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S2A1.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).   \cmmnt{[2A5F.]}

Etemadi N.\ [96] `On convergence of partial sums of independent random
variables', pp.\ 137-144 in {\smc Bergelson March \& Rosenblatt 96}.
\cmmnt{[272V.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[263Ec, \S265 {\it notes}.]}
%Kirszbraun's theorem mentioned but not proved

\medskip%F

Federer H.\ [69] {\it Geometric Measure Theory.}  Springer, 1969
(reprinted 1996).
\cmmnt{[262C, 263Ec, \S264 {\it notes}, \S265 {\it notes},
\S266 {\it notes}.]}

Feller W.\ [66] {\it An Introduction to Probability Theory and its
Applications,} vol.\ II.   Wiley, 1966.  \cmmnt{[Chap.\ 27 {\it
intro.}, 274H, 275Xc, 285N.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S232 {\it notes}, 241F, \S244 {\it notes}, \S245 {\it notes},
\S247 {\it notes}.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304 in
{\smc Judah 93}.  \cmmnt{[232H.]}

\medskip%G\medskip%H

Haimo D.T.\ [67] (ed.) {\it Orthogonal Expansions and their Continuous
Analogues.}   Southern Illinois University Press, 1967.

Hall P.\ [82] {\it Rates of Convergence in the Central Limit Theorem.}
Pitman, 1982.  \cmmnt{[274H.]}

Halmos P.R.\ [50] {\it Measure Theory.}  Van Nostrand, 1950.
\cmmnt{[\S251 {\it notes}, \S252 {\it notes}, 255Yn.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[\S2A1.]}

Hanner O.\ [56] `On the uniform convexity of $L^p$ and $l^p$',
Arkiv f\"or Matematik 3 (1956) 239-244.
\cmmnt{[244O.]}

Henle J.M.\ [86] {\it An Outline of Set Theory.}  Springer, 1986.
\cmmnt{[\S2A1.]}

Hoeffding W.\ [63]
`Probability inequalities for sums of bounded random variables',
J.\ Amer.\ Statistical Association 58 (1963) 13-30.
\cmmnt{[272W.]}

Hunt R.A.\ [67] `On the convergence of Fourier series', pp.\ 235-255
in {\smc Haimo 67}.  \cmmnt{[\S286 {\it notes}.]}

\medskip%I\medskip%J

Jorsb{\o}e O.G.\ \& Mejlbro L.\ [82] {\it The Carleson-Hunt Theorem on
Fourier Series.}  Springer, 1982 (Lecture Notes in Mathematics 911).
\cmmnt{[\S286 {\it notes}.]}

Judah H.\ [93] (ed.) {Proceedings of the Bar-Ilan Conference on Set
Theory and the Reals, 1991.} Amer.\ Math.\ Soc.\ (Israel Mathematical
Conference Proceedings 6), 1993.

\medskip%K

\indexheader{Kelley}
Kelley J.L.\ [55] {\it General Topology.} Van Nostrand, 1955.
\cmmnt{[2A5F.]}

Kelley J.L.\ \& Namioka I.\ [76] {\it Linear Topological Spaces.}
Springer, 1976.   \cmmnt{[2A5C.]}

\indexheader{Kirszbraun}
Kirszbraun M.D.\ [1934] `\"Uber die zusammenziehenden und Lipschitzian
Transformationen', Fund.\ Math.\ 22 (1934) 77-108.  \cmmnt{[262C.]}

Kolmogorov A.N.\ [1926] `Une s\'erie de Fourier-Lebesgue divergente
partout', C.\ R.\ Acad.\ Sci.\ Paris 183 (1926) 1327-1328.
\cmmnt{[\S282 {\it notes}.]}

Koml\'os J.\ [67] `A generalization of a problem of Steinhaus', Acta
Math.\ Acad.\ Sci.\ Hung.\ 18 (1967) 217-229.  \cmmnt{[276H.]}

K\"orner T.W.\ [88] {\it Fourier Analysis.}  Cambridge U.P., 1988.
\cmmnt{[\S282 {\it notes}.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[2A5C, 2A5E, 2A5J.]}

Krivine J.-L.\ [71] {\it Introduction to Axiomatic Set Theory.}
D.~Reidel, 1971.  \cmmnt{[\S2A1.]}

\medskip%L

Lacey M.T.\ [05] {\it Carleson's Theorem: Proof, Complements, Variations.}
{\tt http://arxiv.org/pdf/math{\bsp}0307008v4{\penalty-100}.pdf}.
%expanded from 
% Publ.\ Mat.\ 48 (2004) 251-307
\cmmnt{[\S286 {\it notes}.]}

Lacey M.T.\ \& Thiele C.M.\ [00] `A proof of boundedness of the Carleson
operator', Math.\ Research Letters 7 (2000) 1-10.
\cmmnt{[\S286 {\it intro.}, 286H.]}

Lebesgue H.\ [72] {\it Oeuvres Scientifiques.}   L'Enseignement
Math\'ematique, Institut de Math\'ematiques, Univ.\ de Gen\`eve, 1972.
\cmmnt{[Chap.\ 27 {\it intro.}]}

Liapounoff A.\ [1901] `Nouvelle forme du th\'eor\`eme sur la limite
de probabilit\'e', M\'em.\ Acad.\ Imp.\ Sci.\ St-P\'etersbourg 12(5)
(1901) 1-24. \cmmnt{[274Xh.]}

Lighthill M.J.\ [59] {\it Introduction to Fourier Analysis and
Generalised Functions.}  Cambridge U.P., 1959.
\cmmnt{[\S284 {\it notes}.]}

Lindeberg J.W.\ [1922] `Eine neue Herleitung des Exponentialgesetzes in
der Wahrscheinlichkeitsrechnung', Math.\ Zeitschrift 15 (1922) 211-225.
\cmmnt{[274H, \S274 {\it notes}.]}

Lipschutz S.\ [64] {\it Set Theory and Related Topics.}  McGraw-Hill,
1964 (Schaum's Outline Series).  \cmmnt{[\S2A1.]}

Lo\`eve M.\ [77] {\it Probability Theory I.} Springer, 1977.
\cmmnt{[Chap.\ 27 {\it intro.}, 274H.]}

Luxemburg W.A.J.\ \& Zaanen A.C.\ [71] {\it Riesz Spaces
I.} North-Holland, 1971.
\cmmnt{[241F.]}


\medskip%M

Mozzochi C.J.\ [71] {\it On the Pointwise Convergence of Fourier
Series.}   Springer, 1971 (Lecture Notes in Mathematics 199).
\cmmnt{[\S286 {\it notes}.]}

\medskip%N

Naor A.\ [04] `Proof of the uniform convexity lemma',
{\tt http://www.cims.nyu.edu/\penalty-100$^{\sim}$naor/homepage files/\ifnum\stylenumber=12\break\fi{\vthsp}inequality\discretionary{}{}{}.pdf},
26.2.04.
\cmmnt{[244O.]}
%anon04.PDF

\medskip%O\medskip%P\medskip%Q\medskip%R

R\'enyi A.\ [70] {\it Probability Theory.}  North-Holland, 1970.
\cmmnt{[274H.]}

Roitman J.\ [90] {\it  An Introduction to Set Theory.}  Wiley, 1990.
\cmmnt{[\S2A1.]}

Roselli P.\ \& Willem M.\ [02] `A convexity inequality',
Amer.\ Math.\ Monthly 109 (2002) 64-70.
\cmmnt{[244Ym.]}

\medskip%S

%Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', 
%  Fundamenta Math.\ 5 (1924) 98-104.
% D-Y-S theorem
% books/saks.24.pdf   http://matwbn.icm.edu.pl/ksiazki/fm/fm5/fm5112.pdf

Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', Fund

Schipp F.\ [78] `On Carleson's method', pp.\ 679-695 in
{\smc Alexits 78}.   \cmmnt{[\S286 {\it notes}.]}

Semadeni Z.\ [71] {\it Banach spaces of continuous functions I.}
Polish Scientific Publishers, 1971.  \cmmnt{[\S253 {\it notes}.]}

Shiryayev A.N.\ [84] {\it Probability.} Springer, 1984. \cmmnt{[285N.]}

Steele J.M.\ [86] `An Efron-Stein inequality of nonsymmetric statistics',
Annals of Statistics 14 (1986) 753-758.
\cmmnt{[274Ya.]}

\medskip%T\medskip%U\medskip%V\medskip%W\medskip%Y\medskip%Z

Zaanen A.C.\ [83] {\it Riesz Spaces II.} North-Holland, 1983.
\cmmnt{[241F.]}

Zygmund A. [59] {\it Trigonometric Series.}   Cambridge U.P, 1959.
\cmmnt{[\S244 {\it notes}, \S282 {\it notes}, 284Xk.]}

\frnewpage

}%end of references

%---------------------------------------------------------------%

\end



