% 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{mt25.tex}
\versiondate{31.5.03}

\def\chaptername{Product Measures}
\newchapter{25}
\def\chaptername{Product measures}

I come now to another chapter on `pure' measure theory, discussing a
fundamental construction -- or, as you may prefer to consider it, two
constructions, since the problems involved in forming the product of
two arbitrary measure spaces (\S251) are rather different
from those arising in the product of arbitrarily many probability spaces
($\S$254).   This work is going to stretch our technique to the utmost,
for while the fundamental theorems to which we are moving are natural
aims, the proofs are lengthy and there are many
pitfalls beside the true paths.

\allowmorestretch{500}{
The central idea is that of `repeated integration'.   You have
probably already seen formulae of the type\penalty-100 
`$\iint f(x,y)dxdy$' used
to calculate the integral of a function of two real variables over a
region in the plane.   One of the basic techniques of advanced calculus
is reversing the order of integration;  for instance, we expect
$\int_0^1(\int_y^1f(x,y)dx)dy$ to be equal to
$\int_0^1(\int_0^xf(x,y)dy)dx$.   As I have developed the subject, we
already have a third calculation to compare with these two:  $\int_Df$,
where $D=\{(x,y):0\le y\le x\le 1\}$ and the integral is taken with
respect to Lebesgue measure on the plane.   The first two sections of
this chapter are devoted to an analysis of the relationship between
one- and two-dimensional Lebesgue measure which makes these operations
valid -- some of the time;  part of the work has to be devoted to a
careful description of the exact conditions which must be imposed on $f$
and $D$ if we are to be safe.
}%end of allowmorestretch

Repeated integration, in one form or another, appears everywhere in
measure theory, and it is therefore necessary sooner or later to develop
the most general possible expression of the idea.   The standard method
is through the theory of products of general measure spaces.   Given
measure spaces $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$, the aim is to find a
measure $\lambda$ on $X\times Y$ which will, at least, give the right
measure $\mu E\cdot\nu F$ to a `rectangle' $E\times F$ where
$E\in\Sigma$
and $F\in\Tau$.   It turns out that there are already difficulties in
deciding what `the' product measure is, and to do the job properly I
find I need, even at this stage, to describe two related but
distinguishable constructions.   These constructions and their
elementary properties take up the whole of \S251.   In \S252 I turn to
integration over the product, with Fubini's and Tonelli's theorems
relating $\int fd\lambda$ with $\iint f(x,y)\mu(dx)\nu(dy)$.   Because
the construction of $\lambda$ is symmetric between the two factors, this
automatically provides theorems relating $\iint f(x,y)\mu(dx)\nu(dy)$
with $\iint f(x,y)\nu(dy)\mu(dx)$.   \S253 looks at the space
$L^1(\lambda)$ and its relationship with $L^1(\mu)$ and $L^1(\nu)$.

For general measure spaces, there are obstacles in the way of forming an
infinite product;  to start with, if $\sequencen{(X_n,\mu_n)}$ is a
sequence of measure spaces, then a product measure $\lambda$ on
$X=\prod_{n\in\Bbb N}X_n$ ought to set $\lambda
X=\prod_{n=0}^{\infty}\mu_nX_n$, and there is no guarantee that the
product will converge, or behave well when it does.   But for
probability spaces, when $\mu_nX_n=1$ for every $n$, this problem at
least evaporates.   It is possible to define the product of any family
of probability spaces;  this is the burden of \S254.

I end the chapter with three sections which are a preparation for
Chapters
27 and 28, but are also important in their own right as an investigation
of the way in which the group structure of $\BbbR^r$ interacts with
Lebesgue and other measures.   \S255 deals with the `convolution'
$f*g$ of two functions, where $(f*g)(x)=\int f(y)g(x-y)dy$ (the
integration being with respect to Lebesgue measure).   In \S257 I show
that some of the same ideas, suitably transformed, can be used to
describe a convolution $\nu_1*\nu_2$ of two measures on $\BbbR^r$;  in
preparation for this I include a section on Radon measures on $\BbbR^r$
(\S256).

\discrpage

\frfilename{mt251.tex}
\versiondate{10.11.06}
\copyrightdate{2000}

\def\chaptername{Product measures}
\def\sectionname{Finite products}

\newsection{251}

The first construction to set up is the product of a pair of measure
spaces.    It turns out that there are already substantial technical
difficulties in the way of finding a canonical universally applicable
method.   I find myself therefore describing two related, but distinct,
constructions, the `primitive' and `c.l.d.' product measures (251C,
251F).   After listing the fundamental properties of the c.l.d\ product
measure (251I-251J), I work through the identification of the product of
Lebesgue measure with itself (251N) and a fairly thorough discussion of
subspaces (251O-251S).

\vleader{48pt}{251A}{Definition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$
be two
measure spaces.   For $A\subseteq X\times Y$ set

\Centerline{$\theta A=\inf\{\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n:
  E_n\in\Sigma,\,F_n\in\Tau\,\Forall \,n\in\Bbb N,
  \,A\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n\}$.}

\cmmnt{\medskip

\noindent{\bf Remark} In the products $\mu E_n\cdot\nu F_n$,
$0\cdot\infty$ is to be taken as $0$, as in \S135.
}

\leader{251B}{Lemma} In the context of 251A, $\theta$ is an outer
measure on $X\times Y$.

\proof{{\bf (a)} Setting $E_n=F_n=\emptyset$ for every
$n\in\Bbb N$, we see that $\theta\emptyset = 0$.

\medskip

{\bf (b)} If $A\subseteq B\subseteq X\times Y$, then whenever
$B\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$ we shall have
$A\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$;  so $\theta A\le\theta B$.

\medskip

{\bf (c)} Let $\sequencen{A_n}$ be a sequence of subsets of $X\times Y$,
with union $A$.
For any $\epsilon>0$, we may choose, for each $n\in\Bbb N$, sequences
$\sequence{m}{E_{nm}}$ in $\Sigma$ and $\sequence{m}{F_{nm}}$ in $\Tau$
such that $A_n\subseteq\bigcup_{m\in\Bbb N}E_{nm}\times F_{nm}$ and
$\sum_{m=0}^{\infty}\mu E_{nm}\cdot\nu F_{nm}
\le\theta A_n+2^{-n}\epsilon$.
Because $\Bbb N\times\Bbb N$ is countable,  we have a bijection
$k\mapsto(n_k,m_k):\Bbb N\to\Bbb N\times\Bbb N$, and now

\Centerline{$A\subseteq\bigcup_{n,m\in\Bbb N}E_{nm}\times F_{nm}
=\bigcup_{k\in\Bbb N}E_{n_km_k}\times F_{n_km_k}$,}

\noindent so that

$$\eqalign{\theta A&\le\sum_{k=0}^{\infty}
  \mu E_{n_km_k}\cdot\nu F_{n_km_k}
=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\mu E_{nm}\cdot\nu F_{nm}\cr
&\le\sum_{n=0}^{\infty}\theta A_n+2^{-n}\epsilon
=2\epsilon+\sum_{n=0}^{\infty}\theta A_n.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\theta
A\le\sum_{n=0}^{\infty}\theta A_n$.

As $\sequencen{A_n}$ is arbitrary, $\theta$ is an outer measure.
}

\leader{251C}{Definition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces.   By the {\bf primitive product measure} on $X\times Y$
I shall mean the measure $\lambda_0$ derived by \Caratheodory's
method\cmmnt{ (113C)} from the outer measure $\theta$ defined in 251A.

\cmmnt{\medskip

\noindent{\bf Remark}
I ought to point out that there is no general agreement on what
`the' product measure on $X\times Y$ should be.   Indeed in 251F below I
will introduce an alternative one, and in the notes to this section I
will mention a third.
}

\leader{251D}{Definition}\cmmnt{ It is convenient to have a name for a
natural construction for $\sigma$-algebras.}   If $X$ and $Y$ are sets
with $\sigma$-algebras $\Sigma\subseteq\Cal PX$ and
$\Tau\subseteq\Cal PY$, I will write $\Sigma\tensorhat\Tau$ for the
$\sigma$-algebra of subsets of $X\times Y$ generated by
$\{E\times F:E\in\Sigma,\,F\in\Tau\}$.


\leader{251E}{Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces;  let $\lambda_0$  be the primitive product measure on
$X\times Y$, and $\Lambda$ its domain.
Then $\Sigma\tensorhat\Tau\subseteq\Lambda$  and
$\lambda_0(E\times F)=\mu E\cdot\nu F$ for all $E\in\Sigma$
and $F\in\Tau$.

\proof{{\bf (a)}   Suppose that $E\in\Sigma$ and $A\subseteq X\times Y$.
For any $\epsilon>0$, there are sequences $\sequencen{E_n}$ in $\Sigma$
and $\sequencen{F_n}$ in $\Tau$ such that
$A\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$ and
$\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n\le\theta A+\epsilon$.   Now

\Centerline{$A\cap(E\times Y)
\subseteq\bigcup_{n\in\Bbb N}(E_n\cap E)\times F_n$,
\quad$A\setminus(E\times Y)
\subseteq\bigcup_{n\in\Bbb N}(E_n\setminus E)\times F_n$,}

\noindent so

$$\eqalign{\theta(A\cap(E\times Y))+\theta(A\setminus(E\times Y))
&\le\sum_{n=0}^{\infty}\mu(E_n\cap E)\cdot\nu F_n
  +\sum_{n=0}^{\infty}\mu(E_n\setminus E)\cdot\nu F_n\cr
&=\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
\le\theta A+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta(A\cap(E\times Y))+\theta(A\setminus(E\times Y))\le\theta A$.
And this is enough to ensure that $E\times Y\in\Lambda$ (see 113D).

\medskip

{\bf (b)} Similarly, $X\times F\in\Lambda$ for every $F\in\Tau$,
so $E\times F=(E\times Y)\cap(X\times F)\in\Lambda$ for every
$E\in\Sigma$, $F\in\Tau$.

Because $\Lambda$ is a $\sigma$-algebra, it must include the smallest
$\sigma$-algebra containing all the products $E\times F$, that is,
$\Lambda\supseteq\Sigma\tensorhat\Tau$.

\medskip


{\bf (c)} Take $E\in\Sigma$, $F\in\Tau$.   We know that $E\times
F\in\Lambda$;  setting $E_0=E$, $F_0=F$, $E_n=F_n=\emptyset$ for $n\ge
1$ in the definition of $\theta$, we have

\Centerline{$\lambda_0(E\times F)
=\theta(E\times F)\le\mu E\cdot\nu F$.}

We have come to the central idea of the construction.   In fact
$\theta(E\times F)=\mu E\cdot\nu F$.   \Prf\ Suppose that
$E\times F\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$ where
$E_n\in\Sigma$ and $F_n\in\Tau$ for every $n$.   Set
$u=\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n$.   If $u=\infty$ or $\mu E=0$
or $\nu F=0$ then of course $\mu E\cdot\nu F\le u$.    Otherwise, set

\Centerline{$I=\{n:n\in\Bbb N,\,\mu E_n=0\}$,
\quad$J=\{n:n\in\Bbb N,\,\nu F_n=0\}$,
\quad$K=\Bbb N\setminus(I\cup J)$,}

\Centerline{$E'=E\setminus\bigcup_{n\in I}E_n$,
\quad$F'=F\setminus\bigcup_{n\in J}F_n$.}

\noindent Then $\mu E'=\mu E$ and $\nu F'=\nu F$;
$E'\times F'\subseteq\bigcup_{n\in K}E_n\times F_n$;  and for $n\in K$,
$\mu E_n<\infty$ and $\nu F_n<\infty$, since
$\mu E_n\cdot\nu F_n\le u<\infty$ and
neither $\mu E_n$ nor $\nu F_n$ is zero.   Set

\Centerline{$f_n=\nu F_n\chi E_n:X\to\Bbb R$}

\noindent if $n\in K$, and $f_n=\tbf{0}:X\to\Bbb R$ if $n\in I\cup J$.
Then $f_n$ is a simple function and
$\int f_n=\nu F_n\mu E_n$ for $n\in K$, $0$ otherwise, so

\Centerline{$\sum_{n=0}^{\infty}$
$\textfont3=\twelveex\int f_n(x)\mu(dx)$
$=\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
\le u$.}

\noindent By B.Levi's theorem (123A), applied to
$\sequencen{\sum_{k=0}^nf_k}$,
$g=\sum_{n=0}^{\infty}f_n$ is integrable and $\int g\,d\mu\le u$.
Write $E''$  for $\{x:x\in E',\,g(x)<\infty\}$, so that
$\mu E''=\mu E'=\mu E$.   Now take any $x\in E''$ and set
$K_x=\{n:n\in K,\,x\in E_n\}$.   Because
$E'\times F'\subseteq\bigcup_{n\in K}E_n\times F_n$,
$F'\subseteq\bigcup_{n\in K_x}F_n$ and

\Centerline{$\nu F=\nu F'\le\sum_{n\in K_x}\nu F_n
=\sum_{n=0}^{\infty}f_n(x)=g(x)$.}

\noindent Thus $g(x)\ge\nu F$ for every $x\in E''$.   We are
supposing that $0<\mu E=\mu E''$ and $0<\nu F$, so we must have
$\nu F<\infty$, $\mu E''<\infty$.   Now  $g\ge\nu F\chi E''$,  so

\Centerline{$\mu E\cdot\nu F=\mu E''\cdot\nu F
=\int\nu F\chi E''\le\int g\le u$
$=\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n$.}

\noindent As $\sequencen{E_n}$, $\sequencen{F_n}$ are arbitrary,
$\theta(E\times F)\ge\mu E\cdot\nu F$ and
$\theta(E\times F)=\mu E\cdot\nu F$.\ \Qed

Thus

\Centerline{$\lambda_0(E\times F)=\theta(E\times F)=\mu E\cdot\nu F$}

\noindent for all $E\in\Sigma$, $F\in\Tau$.
}%end of proof of 251E

\leaveitout{for (c), M.Burke offers:

$\chi E(x)\chi F(y)\le\sum_{n=0}^{\infty}\chi E_n(x)\chi F_n(y)$

by 135F-135G, integrating with respect to $y$, \query

$\chi E(x)\nu F\le\sum_{n=0}^{\infty}\chi E_n(x)\nu F_n$

and now

$\mu E\cdot\nu F\le\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n$.}


\leader{251F}{Definition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\lambda_0$ the primitive product
measure\cmmnt{ defined in 251C}.
By the {\bf c.l.d.\ product measure} on $X\times Y$ I shall mean the
function $\lambda:\dom\lambda_0\to[0,\infty]$ defined by setting

\Centerline{$\lambda W=\sup\{\lambda_0(W\cap(E\times F)):E\in\Sigma,\,
F\in\Tau,\,\mu E<\infty,\,\nu F<\infty\}$}

\noindent for $W\in\dom\lambda_0$.

\leader{251G}{Remark}\cmmnt{ I had better show at once
that} $\lambda$ is a
measure.   \prooflet{\Prf\ Of course its domain $\Lambda=\dom\lambda_0$
is a $\sigma$-algebra,
and $\lambda\emptyset=\lambda_0\emptyset=0$.   If $\sequencen{W_n}$ is a
disjoint sequence in $\Lambda$, then for any
$E\in\Sigma$, $F\in\Tau$ of finite measure

\Centerline{$\lambda_0(\bigcup_{n\in\Bbb N}W_n\cap(E\times F))
=\sum_{n=0}^{\infty}\lambda_0(W_n\cap(E\times F))
\le\sum_{n=0}^{\infty}\lambda W_n$,}

\noindent so
$\lambda(\bigcup_{n\in\Bbb N}W_n)\le\sum_{n=0}^{\infty}\lambda W_n$.
On the other hand, if
$a<\sum_{n=0}^{\infty}\lambda W_n$, then we can find $m\in\Bbb N$ and
$a_0,\ldots,a_m$ such that $a\le\sum_{n=0}^ma_n$ and $a_n<\lambda W_n$
for each $n\le m$;  now there are $E_0,\ldots,E_m\in\Sigma$ and
$F_0,\ldots,F_m\in\Tau$, all of finite measure, such that
$a_n\le\lambda_0(W_n\cap(E_n\times F_n))$ for each $n$.   Setting
$E=\bigcup_{n\le m}E_n$ and $F=\bigcup_{n\le m}F_n$, we have
$\mu E<\infty$ and $\nu F<\infty$, so
$$\eqalign{\lambda(\bigcup_{n\in\Bbb N}W_n)
&\ge\lambda_0(\bigcup_{n\in\Bbb N}W_n\cap(E\times F))
=\sum_{n=0}^{\infty}\lambda_0(W_n\cap(E\times F))\cr
&\ge\sum_{n=0}^m\lambda_0(W_n\cap(E_n\times F_n))
\ge\sum_{n=0}^ma_n
\ge a.\cr}$$

\noindent As $a$ is arbitrary, $\lambda(\bigcup_{n\in\Bbb
N}W_n)\ge\sum_{n=0}^{\infty}\lambda W_n$ and $\lambda(\bigcup_{n\in\Bbb
N}W_n)=\sum_{n=0}^{\infty}\lambda W_n$.   As $\sequencen{W_n}$ is
arbitrary, $\lambda$ is a measure.\ \Qed}

\leader{251H}{}\cmmnt{ We need a simple property of the measure
$\lambda_0$.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two
measure spaces;  let $\lambda_0$ be the primitive product measure on
$X\times Y$, and $\Lambda$ its domain.   If $H\subseteq X\times Y$ and
$H\cap(E\times F)\in\Lambda$ whenever
$\mu E<\infty$ and $\nu F<\infty$, then $H\in\Lambda$.

\proof{ Let $\theta$ be the outer measure described in 251A.   Suppose
that $A\subseteq X\times Y$ and $\theta A<\infty$.   Let
$\epsilon>0$.   Let $\sequencen{E_n}$, $\sequencen{F_n}$ be sequences in
$\Sigma$, $\Tau$ respectively such that $A\subseteq\bigcup_{n\in\Bbb
N}E_n\times F_n$ and $\sum_{n=0}^{\infty}\mu E_n\cdot\nu
F_n\le\theta A+\epsilon$.
Now, for each $n$, the product of the measures $\mu E_n$, $\nu E_n$ is
finite, so either one is zero or
both are finite.   If $\mu E_n=0$ or $\nu F_n=0$ then of course

\Centerline{$\mu E_n\cdot\nu F_n=0
=\theta((E_n\times F_n)\cap H)+\theta((E_n\times F_n)\setminus H)$.}

\noindent If $\mu E_n<\infty$ and $\nu F_n<\infty$ then

$$\eqalignno{\mu E_n\cdot\nu F_n
&=\lambda_0(E_n\times F_n)\cr
&=\lambda_0((E_n\times F_n)\cap H)+\lambda_0((E_n\times F_n)\setminus
H)\cr
&=\theta((E_n\times F_n)\cap H)+\theta((E_n\times F_n)\setminus H).\cr
}$$

\noindent Accordingly, because $\theta$ is an outer measure,

$$\eqalign{\theta(A\cap H)+\theta(A\setminus H)
&\le\sum_{n=0}^{\infty}\theta((E_n\times F_n)\cap H)
  +\sum_{n=0}^{\infty}\theta((E_n\times F_n)\setminus H)\cr
&=\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
\le\theta A+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\theta(A\cap H)+\theta
(A\setminus H)\le\theta A$.   As $A$ is arbitrary, $H\in\Lambda$.
}

\leader{251I}{}\cmmnt{ Now for the fundamental properties of the
c.l.d.\ product measure.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces;  let $\lambda$ be the c.l.d.\ product measure on
$X\times Y$, and $\Lambda$ its domain.   Then

(a) $\Sigma\tensorhat\Tau\subseteq\Lambda$ and
$\lambda(E\times F)=\mu E\cdot\nu F$ whenever $E\in\Sigma$, $F\in\Tau$
and $\mu E\cdot\nu F<\infty$;

(b) for every $W\in\Lambda$ there is a $V\in\Sigma\tensorhat\Tau$ such
that $V\subseteq W$ and $\lambda V=\lambda W$;

(c) $(X\times Y,\Lambda,\lambda)$ is complete and locally determined,
and in fact is the c.l.d.\ version of
$(X\times Y,\Lambda,\lambda_0)$\cmmnt{ as described in 213D-213E};  in
particular, $\lambda W=\lambda_0W$ whenever $\lambda_0W<\infty$;

(d) if $W\in\Lambda$ and $\lambda W>0$ then there are $E\in\Sigma$,
$F\in\Tau$ such that $\mu E<\infty$, $\nu F<\infty$ and
$\lambda(W\cap(E\times F))>0$;

(e) if $W\in\Lambda$ and $\lambda W<\infty$, then for every $\epsilon>0$
there are $E_0,\ldots,E_n\in\Sigma$, $F_0,\ldots,F_n\in\Tau$, all of
finite measure, such that
$\lambda(W\symmdiff\bigcup_{i\le n}(E_i\times F_i))\le\epsilon$.

\proof{ Take $\theta$ to be the outer measure of 251A and
$\lambda_0$ the primitive product measure of 251C.  Set
$\Sigma^f=\{E:E\in\Sigma,\,\mu E<\infty\}$ and
$\Tau^f=\{F:F\in\Tau,\,\nu F<\infty\}$.

\medskip

{\bf (a)} By 251E, $\Sigma\tensorhat\Tau\subseteq\Lambda$.   If
$E\in\Sigma$ and $F\in\Tau$ and $\mu E\cdot\nu F<\infty$, either
$\mu E\cdot\nu F=0$ and $\lambda(E\times F)=\lambda_0(E\times F)=0$ or
both $\mu E$ and $\nu F$ are finite and again
$\lambda(E\times F)=\lambda_0(E\times F)=\mu E\cdot\nu F$.

\medskip

{\bf (b)(i)} Take any $a<\lambda W$.   Then there are $E\in\Sigma^f$,
$F\in\Tau^f$ such that $\lambda_0(W\cap(E\times F))>a$ (251F);  now

$$\eqalign{\theta((E\times F)\setminus W)
&=\lambda_0((E\times F)\setminus W)\cr
&=\lambda_0(E\times F)-\lambda_0(W\cap(E\times F))
<\lambda_0(E\times F)-a.\cr}$$

\noindent Let $\sequencen{E_n}$, $\sequencen{F_n}$ be sequences in
$\Sigma$, $\Tau$ respectively such that $(E\times F)\setminus
W\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$ and $\sum_{n=0}^{\infty}\mu
E_n\cdot\nu F_n\le\lambda_0(E\times F)-a$.   Consider

\Centerline{$V=(E\times
F)\setminus\bigcup_{n\in\Bbb N}E_n\times F_n\in\Sigma\tensorhat\Tau$;}

\noindent then $V\subseteq W$, and

$$\eqalignno{\lambda V=\lambda_0V
&=\lambda_0(E\times F)-\lambda_0((E\times F)\setminus V)\cr
&\ge\lambda_0(E\times F)-\lambda_0(\bigcup_{n\in\Bbb N}E_n\times F_n)\cr
\noalign{\noindent (because $(E\times F)\setminus
V\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$)}
&\ge\lambda_0(E\times F)-\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
\ge a\cr}$$

\noindent (by the choice of the $E_n$, $F_n$).

\medskip

\quad{\bf (ii)} Thus for every $a<\lambda W$ there is a
$V\in\Sigma\tensorhat\Tau$ such that $V\subseteq W$ and $\lambda V\ge
a$.   Now choose a sequence $\sequencen{a_n}$ strictly increasing to
$\lambda W$, and for each $a_n$ a corresponding $V_n$;  then
$V=\bigcup_{n\in\Bbb N}V_n$ belongs to the $\sigma$-algebra
$\Sigma\tensorhat\Tau$, is included in $W$, and has measure at least
$\sup_{n\in\Bbb N}\lambda V_n$ and at most $\lambda W$;  so $\lambda
V=\lambda W$, as required.

\medskip

{\bf (c)(i)} If $H\subseteq X\times Y$ is $\lambda$-negligible, there is
a $W\in\Lambda$ such that $H\subseteq W$ and $\lambda W=0$.   If
$E\in\Sigma$, $F\in\Tau$ are of finite measure,
$\lambda_0(W\cap(E\times  F))=0$;  but $\lambda_0$, being derived from
the outer measure $\theta$
by \Caratheodory's method, is complete (212A), so
$H\cap(E\times F)\in\Lambda$ and $\lambda_0(H\cap(E\times F))=0$.
Because $E$ and $F$
are arbitrary, $H\in\Lambda$, by 251H.   As $H$ is arbitrary, $\lambda$
is complete.

\medskip

\quad{\bf (ii)} If $W\in\Lambda$ and $\lambda W=\infty$, then there must
be $E\in\Sigma$, $F\in\Tau$ such that $\mu E<\infty$, $\nu F<\infty$ and
$\lambda_0(W\cap(E\times F))>0$;  now

\Centerline{$0<\lambda(W\cap(E\times F))\le\mu E\cdot\nu F<\infty$.}

\noindent Thus $\lambda$ is semi-finite.

\medskip

\quad{\bf (iii)} If $H\subseteq X\times Y$ and $H\cap W\in\Lambda$
whenever $\lambda W<\infty$, then, in particular, $H\cap(E\times
F)\in\Lambda$ whenever $\mu E<\infty$ and $\nu F<\infty$;  by 251H
again, $H\in\Lambda$.   Thus $\lambda$ is locally determined.

\medskip

\quad{\bf (iv)} If $W\in\Lambda$ and $\lambda_0W<\infty$, then we have
sequences $\sequencen{E_n}$ in $\Sigma$, $\sequencen{F_n}$ in $\Tau$
such
that $W\subseteq\bigcup_{n\in\Bbb N}(E_n\times F_n)$ and
$\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n<\infty$.   Set

\Centerline{$I=\{n:\mu E_n=\infty\}$,
\quad$J=\{n:\nu F_n=\infty\}$,
\quad$K=\Bbb N\setminus (I\cup J)$;}

\noindent then $\nu(\bigcup_{n\in I}F_n)=\mu(\bigcup_{n\in J}E_n)=0$, so
$\lambda_0(W\setminus W')=0$, where

\Centerline{$W'=W\cap\bigcup_{n\in K}(E_n\times F_n)
\supseteq W\setminus((\bigcup_{n\in J}E_n\times Y)
  \cup(X\times\bigcup_{n\in I}F_n))$.}

\noindent Now set $E'_n=\bigcup_{i\in K,i\le n}E_i$, $F'_n=\bigcup_{i\in
K,i\le n}F_i$ for each $n$.   We have $W'=\bigcup_{n\in\Bbb
N}W'\cap(E'_n\times F'_n)$, so

\Centerline{$\lambda
W\le\lambda_0W=\lambda_0W'=\lim_{n\to\infty}\lambda_0(W'\cap(E'_n\times
F'_n))
\le \lambda W'
\le\lambda W$,}

\noindent and $\lambda W=\lambda_0W$.

\medskip

\quad{\bf (v)} Following the terminology of 213D, let us write

\Centerline{$\tilde\Lambda=\{W:W\subseteq X\times Y,\,W\cap V\in\Lambda$
whenever $V\in\Lambda$ and $\lambda_0V<\infty\}$,}

\Centerline{$\tilde\lambda W
=\sup\{\lambda_0(W\cap V):V\in\Lambda$, $\lambda_0V<\infty\}$.}

\noindent Because $\lambda_0(E\times F)<\infty$ whenever $\mu E<\infty$
and $\nu F<\infty$, $\tilde\Lambda\subseteq\Lambda$ and
$\tilde\Lambda=\Lambda$.


Now for any $W\in\Lambda$ we have

$$\eqalignno{\tilde\lambda W
&=\sup\{\lambda_0(W\cap V):V\in\Lambda,\,\lambda_0V<\infty\}\cr
&\ge\sup\{\lambda_0(W\cap (E\times F)):E\in\Sigma^f,\,F\in\Tau^f\}\cr
&=\lambda W\cr
&\ge\sup\{\lambda(W\cap V):V\in\Lambda,\,\lambda_0V<\infty\}\cr
&=\sup\{\lambda_0(W\cap V):V\in\Lambda,\,\lambda_0V<\infty\},\cr}$$

\noindent using (iv) just above, so that $\lambda=\tilde\lambda$ is the
c.l.d.\ version of $\lambda_0$.

\medskip

{\bf (d)} If $W\in\Lambda$ and $\lambda W>0$, there are
$E\in\Sigma^f$ and $F\in\Tau^f$ such that
$\lambda(W\cap(E\times F))=\lambda_0(W\cap(E\times F))>0$.

\medskip

{\bf (e)} There are $E\in\Sigma^f$, $F\in\Tau^f$ such that
$\lambda_0(W\cap(E\times F))\ge\lambda W-\bover13\epsilon$;   set
$V_1=W\cap (E\times F)$;  then

\Centerline{$\lambda(W\setminus V_1)=\lambda W-\lambda V_1
=\lambda W-\lambda_0V_1\le\Bover13\epsilon$.}

\noindent   There
are sequences $\sequencen{E'_n}$ in $\Sigma$, $\sequencen{F'_n}$ in
$\Tau$ such that $V_1\subseteq\bigcup_{n\in\Bbb
N}E'_n\times F'_n$ and $\sum_{n=0}^{\infty}\mu E'_n\cdot\nu
F'_n\le\lambda_0V_1+\bover13\epsilon$.   Replacing $E'_n$, $F'_n$ by
$E'_n\cap E$, $F'_n\cap F$ if necessary, we may suppose that
$E'_n\in\Sigma^f$ and $F'_n\in\Tau^f$ for every $n$.   Set
$V_2=\bigcup_{n\in\Bbb N}E'_n\times F'_n$;  then

\Centerline{$\lambda(V_2\setminus V_1)
\le\lambda_0(V_2\setminus V_1)
\le\sum_{n=0}^{\infty}\mu E'_n\cdot\nu F'_n-\lambda_0V_1
\le\Bover13\epsilon$.}

\noindent   Let $m\in\Bbb N$ be such that
$\sum_{n=m+1}^{\infty}\mu E'_n\cdot\nu F'_n\le\bover13\epsilon$, and set

\Centerline{$V=\bigcup_{n=0}^mE'_n\times F'_n$.}

\noindent Then

\Centerline{$\lambda(V_2\setminus V)
\le\sum_{n=m+1}^{\infty}\mu E'_n\cdot\nu F'_n\le\Bover13\epsilon$.}

Putting these together, we have
$W\symmdiff V\subseteq(W\setminus V_1)\cup(V_2\setminus
V_1)\cup(V_2\setminus V)$, so

\Centerline{$\lambda(W\symmdiff V)
\le\lambda(W\setminus V_1)+\lambda(V_2\setminus
V_1)+\lambda(V_2\setminus
V)\le\Bover13\epsilon+\Bover13\epsilon+\Bover13\epsilon=\epsilon$.}

\noindent And $V$ is of the required form.
}

\leader{251J}{Proposition} If $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are
semi-finite measure spaces and $\lambda$ is the c.l.d.\ product measure
on $X\times Y$, then $\lambda(E\times F)=\mu E\cdot\nu F$ for all
$E\in\Sigma$, $F\in\Tau$.

\proof{ Setting $\Sigma^f=\{E:E\in\Sigma,\,\mu E<\infty\}$,
$\Tau^f=\{F:F\in\Tau,\,\nu F<\infty\}$, we have

$$\eqalign{\lambda(E\times F)
&=\sup\{\lambda_0((E\cap E_0)\times(F\cap F_0)):
  E_0\in\Sigma^f,\,F_0\in\Tau^f\}\cr
&=\sup\{\mu(E\cap E_0)\cdot\nu(F\cap F_0)):
  E_0\in\Sigma^f,\,F_0\in\Tau^f\}\cr
&=\sup\{\mu(E\cap E_0):E_0\in\Sigma^f\}\cdot
  \sup\{\nu(F\cap F_0):F_0\in\Tau^f\}
=\mu E\cdot\nu F\cr}$$

\noindent (using 213A).
}%end of proof of 251J

\leader{251K}{$\sigma$-finite \dvrocolon{spaces}}\cmmnt{ Of course
most of the measure spaces we shall apply these results to are
$\sigma$-finite, and in this case there are some useful simplifications.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces.   Then the c.l.d.\ product measure on
$X\times Y$ is equal to the primitive product measure, and is the
completion of its restriction to $\Sigma\tensorhat\Tau$;  moreover, this
common product measure is $\sigma$-finite.

\proof{ Write $\lambda_0$, $\lambda$ for the primitive and c.l.d.\
product measures, as usual, and $\Lambda$ for their domain.   Let
$\sequencen{E_n}$, $\sequencen{F_n}$ be non-decreasing sequences of sets
of finite measure covering $X$, $Y$ respectively (see 211D).

\medskip

{\bf (a)} For each $n\in\Bbb N$, $\lambda(E_n\times F_n)=\mu E_n\cdot\nu
F_n$ is finite, by 251Ia.   Since $X\times Y=\bigcup_{n\in\Bbb
N}E_n\times F_n$, $\lambda$ is $\sigma$-finite.

\medskip

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

\Centerline{$\lambda_0W=\lim_{n\to\infty}\lambda_0(W\cap(E_n\times
F_n))=\lim_{n\to\infty}\lambda(W\cap(E_n\times F_n))=\lambda W$.}

\noindent So $\lambda=\lambda_0$.

\medskip

{\bf (c)} Write $\lambda_{\Cal B}$ for the restriction of
$\lambda=\lambda_0$ to $\Sigma\tensorhat\Tau$, and
$\hat\lambda_{\Cal B}$ for its completion.

\medskip

\quad{\bf (i)} Suppose that $W\in\dom\hat\lambda_{\Cal B}$.   Then there
are $W'$, $W''\in\Sigma\tensorhat\Tau$ such that
$W'\subseteq W\subseteq W''$ and $\lambda_{\Cal B}(W''\setminus W')=0$
(212C).   In this case,
$\lambda(W''\setminus W')=0$;  as $\lambda$ is complete, $W\in\Lambda$
and

\Centerline{$\lambda W=\lambda W'=\lambda_{\Cal B}W'
=\hat\lambda_{\Cal B}W$.}

\noindent Thus $\lambda$ extends $\hat\lambda_{\Cal B}$.

\medskip

\quad{\bf (ii)} If $W\in\Lambda$, then there is a
$V\in\Sigma\tensorhat\Tau$ such that $V\subseteq W$ and
$\lambda(W\setminus V)=0$.   \Prf\ For each $n\in\Bbb N$ there is a
$V_n\in\Sigma\tensorhat\Tau$ such that
$V_n\subseteq W\cap(E_n\times F_n)$ and
$\lambda V_n=\lambda(W\cap(E_n\times F_n))$ (251Ib).   But as
$\lambda(E_n\times F_n)=\mu E_n\cdot\nu F_n$ is finite, this means that
$\lambda(W\cap(E_n\times F_n)\setminus V_n)=0$.   So if we set
$V=\bigcup_{n\in\Bbb N}V_n$, we shall have $V\in\Sigma\tensorhat\Tau$,
$V\subseteq W$ and

\Centerline{$W\setminus V
=\bigcup_{n\in\Bbb N}W\cap(E_n\times F_n)\setminus V
\subseteq\bigcup_{n\in\Bbb N}W\cap(E_n\times F_n)\setminus V_n$}

\noindent is $\lambda$-negligible.\ \Qed

Similarly, there is a $V'\in\Sigma\tensorhat\Tau$ such that
$V'\subseteq(X\times Y)\setminus W$ and
$\lambda(((X\times Y)\setminus W)\setminus V')=0$.   Setting $V''=(X\times Y)\setminus V'$,
$V''\in\Sigma\tensorhat\Tau$, $W\subseteq V''$ and
$\lambda(V''\setminus W)=0$.   So

\Centerline{$\lambda_{\Cal B}(V''\setminus V)
=\lambda(V''\setminus V)
=\lambda(V''\setminus W)+\lambda(W\setminus V)=0$,}

\noindent and $W$ is measured by $\hat\lambda_{\Cal B}$, with
$\hat\lambda_{\Cal B}W=\lambda_{\Cal B}V=\lambda W$.   As $W$ is
arbitrary, $\hat\lambda_{\Cal B}=\lambda$.
}%end of proof of 251K

\leader{*251L}{}\cmmnt{ The following result fits in naturally here;
I star it because it will be used seldom (there is a more important version
of the same idea in 254G) and the proof
can be skipped until you come to need it.

\medskip

\noindent}{\bf Proposition}
Let $(X_1,\Sigma_1,\mu_1)$, $(X_2,\Sigma_2,\mu_2)$,
$(Y_1,\Tau_1,\nu_1)$ and $(Y_2,\Tau_2,\nu_2)$ be $\sigma$-finite
measure spaces;  let
$\lambda_1$, $\lambda_2$ be the product measures on $X_1\times Y_1$,
$X_2\times Y_2$ respectively.   Suppose that $f:X_1\to X_2$ and
$g:Y_1\to Y_2$ are \imp\ functions, and that
$h(x,y)=(f(x),g(y))$ for $x\in X_1$, $y\in Y_1$.   Then $h$ is \imp.
%251K

\proof{ Write $\Lambda_1$, $\Lambda_2$ for the domains of $\lambda_1$,
$\lambda_2$ respectively.

\medskip

{\bf (a)} Suppose that $E\in\Sigma_2$ and $F\in\Tau_2$ have finite
measure.   Then $\lambda_1h^{-1}[W\cap(E\times F)]$ is defined and equal to
$\lambda_2(W\cap(E\times F))$ for every $W\in\Lambda_2$.   \Prf\

$$\eqalign{\lambda_1h^{-1}[E\times F]
&=\lambda_1(f^{-1}[E]\times g^{-1}[F])
=\mu_1f^{-1}[E]\cdot\nu_1g^{-1}[F]\cr
&=\mu_2E\cdot\nu_2F
=\lambda_2(E\times F)\cr}$$

\noindent by 251E/251J.\ \Qed

\medskip

{\bf (b)} Take $E_0\in\Sigma_2$ and $F_0\in\Tau_2$ of finite measure.
Let $\tilde\lambda_1$, $\tilde\lambda_2$ be the subspace measures on
$f^{-1}[E_0]\times g^{-1}[F_0]$ and $E_0\times F_0$ respectively.
Set $\tilde h=h\restr f^{-1}[E_0]\times g^{-1}[F_0]$, and write
$\iota$ for the identity map from $E_0\times F_0$ to $X_2\times Y_2$;  let
$\lambda=\tilde\lambda_1\tilde h^{-1}$ and
$\lambda'=\tilde\lambda_2\iota^{-1}$ be the
image measures on $X_2\times Y_2$.   Then (a) tells us that

$$\eqalign{\lambda(E\times F)
&=\lambda_1(h^{-1}[(E\cap E_0)\times(F\cap F_0)])\cr
&=\lambda_2((E\cap E_0)\times(F\cap F_0))
=\lambda'(E\times F)\cr}$$

\noindent whenever $E\in\Sigma_2$ and $F\in\Tau_2$.   By the Monotone Class
Theorem (136C), $\lambda$ and $\lambda'$ agree on
$\Sigma_2\tensorhat\Tau_2$, that is,
$\lambda_1(h^{-1}[W\cap(E_0\times F_0)])=\lambda_2(W\cap(E_0\times F_0))$
for every $W\in\Sigma_2\tensorhat\Tau_2$.

If $W$ is any member of $\Lambda_2$,
there are $W'$, $W''\in\Sigma_2\tensorhat\Tau_2$ such
that $W'\subseteq W\subseteq W''$ and $\lambda_2(W''\setminus W')=0$
(251K).   Now we must have

\Centerline{$h^{-1}[W'\cap(E_0\times F_0)]
\subseteq h^{-1}[W\cap(E_0\times F_0)]
\subseteq h^{-1}[W''\cap(E_0\times F_0)]$,}

\Centerline{$\lambda_1(h^{-1}[W''\cap(E_0\times F_0)]\setminus
h^{-1}[W'\cap(E_0\times F_0)])
=\lambda_2((W''\setminus W')\cap(E_0\times F_0))=0$;}

\noindent because $\lambda_1$ is complete,
$\lambda_1h^{-1}[W\cap(E_0\times F_0)]$ is defined and equal to

\Centerline{$\lambda_1h^{-1}[W'\cap(E_0\times F_0)]
=\lambda_2(W'\cap(E_0\times F_0))
=\lambda_2(W\cap(E_0\times F_0))$.}

\medskip

{\bf (c)} Now suppose that $\sequencen{E_n}$, $\sequencen{F_n}$ are
non-decreasing sequences of sets of finite measure with union $X_2$, $Y_2$
respectively.   If $W\in\Lambda_2$,

\Centerline{$\lambda_1h^{-1}[W]
=\sup_{n\in\Bbb N}\lambda_1h^{-1}[W\cap(E_n\times F_n)]
=\sup_{n\in\Bbb N}\lambda_2(W\cap(E_n\times F_n))
=\lambda_2W$.}

\noindent So $h$ is \imp, as claimed.
}%end of proof of 251L

\leader{251M}{}\cmmnt{ It is time that I gave some examples.   Of
course the central
example is Lebesgue measure.   In this case we have the only reasonable
result.   I pause to describe the leading example of the product
$\Sigma\tensorhat\Tau$ introduced in 251D.

\medskip

\noindent}{\bf Proposition} Let $r$, $s\ge 1$ be integers.   Then we
have a natural bijection $\phi:\BbbR^r\times\BbbR^s\to\BbbR^{r+s}$,
defined by setting

\Centerline{$\phi((\xi_1,\ldots,\xi_r),(\eta_1,\ldots,\eta_s))
=(\xi_1,\ldots,\xi_r,\eta_1,\ldots,\eta_s)$}

\noindent for $\xi_1,\ldots,\xi_r,\eta_1,\ldots,\eta_s\in\Bbb R$.   If
we write $\Cal B_r$, $\Cal B_s$ and $\Cal B_{r+s}$ for the Borel
$\sigma$-algebras of $\BbbR^r$, $\BbbR^s$ and $\BbbR^{r+s}$
respectively, then $\phi$ identifies
$\Cal B_{r+s}$ with $\Cal B_r\tensorhat\Cal B_s$.

\proof{{\bf (a)} Write $\Cal B$ for the $\sigma$-algebra
$\{\phi^{-1}[W]:W\in\Cal B_{r+s}\}$ copied onto $\BbbR^r\times\BbbR^s$
by the bijection $\phi$;   we are seeking to prove that
$\Cal B=\Cal B_r\tensorhat\Cal B_s$.   We have maps
$\pi_1:\BbbR^{r+s}\to\BbbR^r$, $\pi_2:\BbbR^{r+s}\to\BbbR^s$ defined by
setting $\pi_1(\phi(x,y))=x$, $\pi_2(\phi(x,y))=y$.   Each co-ordinate
of $\pi_1$ is continuous, therefore Borel measurable (121Db), so
$\pi_1^{-1}[E]\in\Cal B_{r+s}$ for every $E\in\Cal B_r$, by 121K.
Similarly,
$\pi_2^{-1}[F]\in\Cal B_{r+s}$ for every $F\in\Cal B_s$.   So
$\phi[E\times F]=\pi_1^{-1}[E]\cap\pi_1^{-1}[F]$ belongs to
$\Cal B_{r+s}$, that is, $E\times F\in\Cal B$, whenever $E\in\Cal B_r$
and $F\in\Cal B_s$.   Because $\Cal B$ is a
$\sigma$-algebra, $\Cal B_r\tensorhat\Cal B_s\subseteq\Cal B$.

\medskip

{\bf (b)} Now examine sets of the form

\Centerline{$\{(x,y):\xi_i\le\alpha\}
=\{x:\xi_i\le\alpha\}\times\BbbR^s$,}

\Centerline{$\{(x,y):\eta_j\le\alpha\}
=\BbbR^r\times\{y:\eta_j\le\alpha\}$}

\noindent for $\alpha\in\Bbb R$, $i\le r$ and $j\le s$, taking
$x=(\xi_1,\ldots,\xi_r)$ and $y=(\eta_1,\ldots,\eta_s)$.   All of these
belong to $\Cal B_r\tensorhat\Cal B_s$.   But the $\sigma$-algebra they
generate is just $\Cal B$, by 121J.   So
$\Cal B\subseteq\Cal B_r\tensorhat\Cal B_s$ and
$\Cal B=\Cal B_r\tensorhat\Cal B_s$.
}%end of proof of 251M

\leader{251N}{Theorem}  Let $r$, $s\ge 1$ be integers.   Then the
bijection $\phi:\BbbR^r\times\BbbR^s\to\BbbR^{r+s}$ described in 251M
identifies Lebesgue measure on $\BbbR^{r+s}$ with the
c.l.d.\ product $\lambda$
of Lebesgue measure on $\BbbR^r$ and Lebesgue measure on $\BbbR^s$.

\proof{ Write $\mu_r$,
$\mu_s$, $\mu_{r+s}$ for the three versions of Lebesgue measure,
$\mu^*_r$, $\mu^*_s$ and $\mu^*_{r+s}$ for the corresponding outer
measures, and
$\theta$ for the outer measure on $\BbbR^r\times\BbbR^s$ derived from
$\mu_r$ and $\mu_s$ by the formula of 251A.

\medskip

{\bf (a)} If $I\subseteq\BbbR^r$ and $J\subseteq \BbbR^s$ are
half-open intervals, then
$\phi[I\times J]\subseteq\BbbR^{r+s}$ is also a half-open interval, and

\Centerline{$\mu_{r+s}(\phi[I\times J])=\mu_rI\cdot\mu_sJ$;}

\noindent this is immediate from the definition of the Lebesgue measure
of an interval.   (I speak of `half-open' intervals here, that is,
intervals of the form
$\prod_{1\le j\le r}\coint{\alpha_j,\beta_j}$, because I used them in the
definition of Lebesgue measure in \S115.   If you prefer
to work with open intervals or closed intervals it makes no difference.)
Note also that every half-open interval in $\BbbR^{r+s}$ is expressible
as $\phi[I\times J]$ for suitable $I$, $J$.

\medskip

{\bf (b)} For any $A\subseteq\BbbR^{r+s}$,
$\theta(\phi^{-1}[A])\le\mu^*_{r+s}(A)$.   \Prf\ For any $\epsilon>0$,
there is a sequence $\sequencen{K_n}$ of half-open intervals in $\Bbb
R^{r+s}$ such that $A\subseteq\bigcup_{n\in\Bbb N}K_n$ and
$\sum_{n=0}^{\infty}\mu_{r+s}(K_n)\le\mu^*_{r+s}(A)+\epsilon$.
Express each $K_n$ as $\phi[I_n\times J_n]$, where $I_n$ and $J_n$ are
half-open intervals in $\BbbR^r$ and $\BbbR^s$ respectively;  then
$\phi^{-1}[A]\subseteq\bigcup_{n\in\Bbb N}I_n\times J_n$, so that

\Centerline{$\theta(\phi^{-1}[A])
\le\sum_{n=0}^{\infty}\mu_rI_n\cdot\mu_sJ_n
=\sum_{n=0}^{\infty}\mu_{r+s}(K_n)\le\mu^*_{r+s}(A)+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (c)} If $E\subseteq\BbbR^r$ and $F\subseteq\BbbR^s$ are
measurable, then $\mu^*_{r+s}(\phi[E\times F])\le\mu_rE\cdot\mu_sF$.

\medskip

\Prf\ {\bf (i)} Consider first the case $\mu_rE<\infty$, $\mu_sF<\infty$.   In
this case, given $\epsilon>0$, there are sequences $\sequencen{I_n}$,
$\sequencen{J_n}$ of half-open intervals such that
$E\subseteq\bigcup_{n\in\Bbb N}I_n$,
$F\subseteq\bigcup_{n\in\Bbb N}F_n$,

\Centerline{$\sum_{n=0}^{\infty}\mu_rI_n\le\mu^*_rE+\epsilon
=\mu_rE+\epsilon$,}

\Centerline{$\sum_{n=0}^{\infty}\mu_sJ_n\le\mu^*_sF+\epsilon=\mu_sF
+\epsilon$.}

\noindent   Accordingly
$E\times F\subseteq\bigcup_{m,n\in\Bbb N}I_m\times J_n$ and
$\phi[E\times F]\subseteq\bigcup_{m,n\in\Bbb N}\phi[I_m\times J_n]$, so
that

$$\eqalign{\mu^*_{r+s}(\phi[E\times F])
&\le\sum_{m,n=0}^{\infty}\mu_{r+s}(\phi[I_m\times J_n])
=\sum_{m,n=0}^{\infty}\mu_rI_m\cdot\mu_sJ_n\cr
&=\sum_{m=0}^{\infty}\mu_rI_m\cdot\sum_{n=0}^{\infty}\mu_sJ_n
\le(\mu_rE+\epsilon)(\mu_sF+\epsilon).\cr}$$

\noindent As $\epsilon$ is arbitrary,
we have the result.

\medskip

\quad{\bf (ii)}
Next, if $\mu_rE=0$, there is a sequence $\sequencen{F_n}$ of sets
of finite measure covering $\BbbR^s\supseteq F$, so that

\Centerline{$\mu^*_{r+s}(\phi[E\times F])
\le\sum_{n=0}^{\infty}\mu^*_{r+s}(\phi[E\times F_n])
\le\sum_{n=0}^{\infty}\mu_rE\cdot\mu_sF_n
=0
=\mu_rE\cdot\mu_sF$.}

\noindent Similarly,
$\mu^*_{r+s}(\phi[E\times F])\le\mu_rE\cdot\mu_sF$ if $\mu_sF=0$.
The only remaining case is that in
which both of $\mu_rE$, $\mu_sF$ are strictly positive and one is
infinite;  but in this case $\mu_rE\cdot\mu_sF=\infty$, so surely
$\mu^*_{r+s}(\phi[E\times F])\le\mu_rE\cdot\mu_sF$.\ \Qed

\medskip

{\bf (d)} If $A\subseteq\BbbR^{r+s}$, then
$\mu^*_{r+s}(A)\le\theta(\phi^{-1}[A])$.   \Prf\ Given $\epsilon>0$,
there are sequences $\sequencen{E_n}$, $\sequencen{F_n}$ of measurable
sets in $\BbbR^r$, $\BbbR^s$ respectively such that
$\phi^{-1}[A]\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$ and
$\sum_{n=0}^{\infty}\mu_rE_n\cdot\mu_sF_n
\le\theta(\phi^{-1}[A])+\epsilon$.
Now $A\subseteq\bigcup_{n\in\Bbb N}\phi[E_n\times F_n]$, so

\Centerline{$\mu^*_{r+s}(A)
\le\sum_{n=0}^{\infty}\mu^*_{r+s}(\phi[E_n\times F_n])
\le\sum_{n=0}^{\infty}\mu_rE_n\cdot\mu_sF_n
\le\theta(\phi^{-1}[A])+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, we have the result.\ \Qed

\medskip

{\bf (e)} Putting (c) and (d) together,  we have
$\theta(\phi^{-1}[A])=\mu^*_{r+s}(A)$ for every $A\subseteq\Bbb
R^{r+s}$.    Thus $\theta$ on $\BbbR^r\times\BbbR^s$ corresponds
exactly to $\mu^*_{r+s}$ on $\BbbR^{r+s}$.   So the associated
measures $\lambda_0$, $\mu_{r+s}$ must correspond in the same way,
writing $\lambda_0$ for the primitive product measure.   But 251K tells
us that $\lambda_0=\lambda$, so we have the result.
}%end of proof of 251N

\vleader{48pt}{251O}{}\cmmnt{ In fact, a large proportion of the
applications
of the constructions here are to subspaces of Euclidean space, rather
than to the whole product $\BbbR^r\times\BbbR^s$.   It would not have
been especially difficult to write 251N out to deal with
arbitrary subspaces, but I prefer to give a more general description of
the product of subspace measures, as I feel that it illuminates the
method.   I start with a straightforward result on strictly localizable
spaces.

\medskip

\noindent}{\bf Proposition}  Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
strictly localizable measure spaces.   Then the c.l.d.\ product measure
on $X\times Y$ is strictly localizable;  moreover, if
$\langle X_i\rangle_{i\in I}$ and $\langle Y_j\rangle_{j\in J}$ are
decompositions of $X$ and $Y$ respectively,
$\langle X_i\times Y_j\rangle_{(i,j)\in I\times J}$ is a decomposition of
$X\times Y$.

\proof{ Let $\langle X_i\rangle_{i\in I}$ and
$\langle Y_j\rangle_{j\in J}$ be decompositions of $X$, $Y$ respectively.
Then $\langle X_i\times Y_j\rangle_{(i,j)\in I\times J}$ is a partition of
$X\times Y$ into measurable sets of finite measure.   If
$W\subseteq X\times Y$ and $\lambda W>0$, there are sets $E\in\Sigma$,
$F\in\Tau$ such that $\mu E<\infty$, $\nu F<\infty$ and
$\lambda(W\cap(E\times F))>0$.   We know that
$\mu E=\sum_{i\in I}\mu(E\cap X_i)$ and
$\mu F=\sum_{j\in J}\mu(F\cap Y_j)$, so there must be finite sets
$I_0\subseteq I$, $J_0\subseteq J$ such that

\Centerline{$\mu E\cdot\nu F
  -(\sum_{i\in I_0}\mu(E\cap X_i))(\sum_{j\in J_0}\nu(F\cap Y_j))
<\lambda(W\cap(E\times F))$.}

\noindent Setting $E'=\bigcup_{i\in I_0}X_i$ and $F'=\bigcup_{j\in J_0}Y_j$
we have

\Centerline{$\lambda((E\times F)\setminus(E'\times F'))
=\lambda(E\times F)-\lambda((E\cap E')\times(F\cap
F'))<\lambda(W\cap(E\times F))$,}

\noindent so that $\lambda(W\cap(E'\times F'))>0$.   There must
therefore be some $i\in I_0$, $j\in J_0$ such that
$\lambda(W\cap(X_i\times Y_j))>0$.

This shows that $\{X_i\times Y_j:i\in I,\,j\in J\}$ satisfies the
criterion of 213O, so that $\lambda$, being complete and locally
determined, must be strictly localizable.   Because
$\langle X_i\times Y_j\rangle_{(i,j)\in I\times J}$ covers $X\times Y$,
it is actually a decomposition of $X\times Y$ (213Ob).
}

\leader{251P}{Lemma} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\lambda$ the c.l.d.\ product measure on
$X\times Y$.   Let $\lambda^*$ be the corresponding
outer measure\cmmnt{ (132B)}.   Then

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

\noindent for every
$C\subseteq X\times Y$, where $\theta$ is the outer measure of 251A.

\proof{ Write $\Lambda$ for the domain of $\lambda$, $\Sigma^f$ for
$\{E:E\in\Sigma,\,\mu E<\infty\}$, $\Tau^f$ for $\{F:F\in\Tau,\,\nu
F<\infty\}$;  set $u=\sup\{\theta(C\cap(E\times F)):E\in\Sigma^f$,
$F\in\Tau^f\}$.

\medskip

{\bf (a)} If $C\subseteq W\in\Lambda$, $E\in\Sigma^f$ and $F\in\Tau^f$,
then

$$\eqalignno{\theta(C\cap(E\times F))
&\le\theta(W\cap(E\times F))
=\lambda_0(W\cap(E\times F))\cr
\noalign{\noindent (where $\lambda_0$ is the primitive product measure)}
&\le\lambda W.\cr}$$

\noindent As $E$ and $F$ are arbitrary, $u\le\lambda W$;  as $W$ is
arbitrary, $u\le\lambda^*C$.

\medskip

{\bf (b)} If $u=\infty$, then of course $\lambda^*C=u$.   Otherwise, let
$\sequencen{E_n}$, $\sequencen{F_n}$ be
sequences in $\Sigma^f$, $\Tau^f$ respectively such that

\Centerline{$u=\sup_{n\in\Bbb N}\theta(C\cap(E_n\times F_n))$.}

\noindent Consider $C'=C\setminus(\bigcup_{n\in\Bbb
N}E_n\times\bigcup_{n\in\Bbb
N}F_n)$.   If $E\in\Sigma^f$ and $F\in\Tau^f$, then for every $n\in\Bbb
N$ we have

$$\eqalignno{u
&\ge\theta(C\cap((E\cup E_n)\times(F\cup F_n)))\cr
&=\theta(C\cap((E\cup E_n)\times(F\cup F_n))\cap(E_n\times F_n))\cr
&{\hskip 10em}+\theta(C\cap((E\cup E_n)\times(F\cup F_n))
  \setminus(E_n\times F_n))\cr
\noalign{\noindent (because $E_n\times F_n\in\Lambda$, by 251E)}
&\ge\theta(C\cap(E_n\times F_n))
  +\theta(C'\cap(E\times F)).\cr}$$

\noindent  Taking the supremum of the right-hand expression as $n$
varies, we have $u\ge u+\theta(C'\cap(E\times F))$ so

\Centerline{$\lambda(C'\cap(E\times F))=\theta(C'\cap(E\times F))=0$.}

\noindent As $E$ and $F$ are arbitrary, $\lambda C'=0$.

But this means that

$$\eqalignno{\lambda^*C
&\le\lambda^*
  (C\cap(\bigcup_{n\in\Bbb N}E_n\times\bigcup_{n\in\Bbb N}F_n))
  +\lambda^*C'\cr
&=\lim_{n\to\infty}
  \lambda^*(C\cap(\bigcup_{i\le n}E_i\times\bigcup_{i\le n}F_i))\cr
\noalign{\noindent (using 132Ae)}
&\le u,\cr}$$

\noindent as required.
}%end of proof of 251P

\leader{251Q}{\bf Proposition}  Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$
be measure spaces, and $A\subseteq X$, $B\subseteq Y$ subsets;  write
$\mu_A$, $\nu_B$ for the subspace measures on $A$, $B$ respectively.
Let $\lambda$ be the c.l.d.\ product measure on $X\times Y$, and
$\lambda^{\#}$
the subspace measure it induces on $A\times B$.   Let $\tilde\lambda$ be
the c.l.d.\ product measure of $\mu_A$ and $\nu_B$ on $A\times B$.   Then

(i) $\tilde\lambda$ extends $\lambda^{\#}$.

(ii) If

\inset{{\it either} ($\alpha$) $A\in\Sigma$ and $B\in\Tau$}

\inset{{\it or} ($\beta$) $A$ and $B$ can both be covered by sequences
of sets of finite measure}

\inset{{\it or} ($\gamma$) $\mu$ and $\nu$ are both strictly
localizable,}

\noindent then $\tilde\lambda=\lambda^{\#}$.

\proof{ Let $\theta$ be the outer measure on $X\times Y$ defined from
$\mu$ and $\nu$ by the formula of 251A, and $\tilde\theta$ the outer
measure on $A\times B$ similarly defined from $\mu_A$ and $\nu_B$.
Write $\Lambda$ for the domain of $\lambda$,
$\tilde\Lambda$ for the domain of $\tilde\lambda$, and
$\Lambda^{\#}=\{W\cap(A\times B):W\in\Lambda\}$ for
the domain of $\lambda^{\#}$.
Set $\Sigma^f=\{E:\mu E<\infty\}$, $\Tau^f=\{F:\nu F<\infty\}$.

\medskip

{\bf (a)} The first point to observe is that $\tilde\theta C=\theta C$
for every $C\subseteq A\times B$.   \Prf\ (i) If $\sequencen{E_n}$
and $\sequencen{F_n}$ are sequences in $\Sigma$, $\Tau$ respectively
such that $C\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$, then

\Centerline{$C=C\cap(A\times B)\subseteq\bigcup_{n\in\Bbb N}(E_n\cap
A)\times(F_n\cap B)$,}

\noindent so

$$\eqalign{\tilde\theta C
&\le\sum_{n=0}^{\infty}\mu_A(E_n\cap A)\cdot\nu_B(F_n\cap B)\cr
&=\sum_{n=0}^{\infty}\mu^*(E_n\cap A)\cdot\nu^*(F_n\cap B)
\le\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n.\cr}$$

\noindent As $\sequencen{E_n}$ and $\sequencen{F_n}$ are arbitrary,
$\tilde\theta C\le\theta C$.   (ii) If $\sequencen{\tilde E_n}$,
$\sequencen{\tilde F_n}$ are sequences in $\Sigma_A=\dom\mu_A$,
$\Tau_B=\dom\nu_B$ respectively such that
$C\subseteq\bigcup_{n\in\Bbb N}\tilde E_n\times\tilde F_n$, then for
each $n\in\Bbb N$ we can choose
$E_n\in\Sigma$, $F_n\in\Tau$ such that

\Centerline{$\tilde E_n\subseteq E_n$,\quad $\mu E_n
=\mu^*\tilde E_n=\mu_A\tilde E_n$,}

\Centerline{$\tilde F_n\subseteq F_n$,
\quad $\nu F_n=\nu^*\tilde F_n=\nu_B\tilde F_n$,}

\noindent and now

\Centerline{$\theta C\le\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n
=\sum_{n=0}^{\infty}\mu_A\tilde E_n\cdot\nu_B\tilde F_n$.}

\noindent As $\sequencen{\tilde E_n}$, $\sequencen{\tilde F_n}$ are
arbitrary, $\theta C\le\tilde\theta C$.\ \Qed

\medskip

{\bf (b)} It follows that $\Lambda^{\#}\subseteq\tilde\Lambda$.   \Prf\
Suppose that $V\in\Lambda^{\#}$ and that $C\subseteq A\times B$.   In
this
case there is a $W\in\Lambda$ such that $V=W\cap(A\times B)$.   So

\Centerline{$\tilde\theta(C\cap V)+\tilde\theta(C\setminus V)
=\theta(C\cap W)+\theta(C\setminus W)
=\theta C
=\tilde\theta C$.}

\noindent As $C$ is arbitrary, $V\in\tilde\Lambda$.\ \Qed

Accordingly, for $V\in\Lambda^{\#}$,

$$\eqalignno{\lambda^{\#}V
&=\lambda^*V
=\sup\{\theta(V\cap(E\times F)):E\in\Sigma^f,\,F\in\Tau^f\}\cr
&=\sup\{\theta(V\cap(\tilde E\times\tilde F)):\tilde
E\in\Sigma_A,\,\tilde F\in\Tau_B,\,\mu_A\tilde E<\infty,\,\nu_B\tilde
F<\infty\}\cr
&=\sup\{\tilde\theta(V\cap(\tilde E\times\tilde F)):\tilde
E\in\Sigma_A,\,\tilde F\in\Tau_B,\,\mu_A\tilde E<\infty,\,\nu_B\tilde
F<\infty\}
=\tilde\lambda V,\cr}$$

\noindent using 251P twice.

This proves part (i) of the proposition.

\medskip

{\bf (c)} The next thing to check is that if $V\in\tilde\Lambda$ and
$V\subseteq E\times F$ where $E\in\Sigma^f$ and $F\in\Tau^f$, then
$V\in\Lambda^{\#}$.   \Prf\ Let $W\subseteq E\times F$ be a measurable
envelope of $V$ with respect to $\lambda$ (132Ee).    Then

$$\eqalignno{\theta(W\cap(A\times B)\setminus V)
&=\tilde\theta(W\cap(A\times B)\setminus V)
=\tilde\lambda(W\cap(A\times B)\setminus V)\cr
\noalign{\noindent (because $W\cap(A\times
B)\in\Lambda^{\#}\subseteq\tilde\Lambda$, $V\in\tilde\Lambda$)}
&=\tilde\lambda(W\cap(A\times B))-\tilde\lambda V
=\tilde\theta(W\cap(A\times B))-\tilde\theta V\cr
&=\theta(W\cap(A\times B))-\theta V
=\lambda^*(W\cap(A\times B))-\lambda^* V\cr
&\le\lambda W-\lambda^* V
=0.\cr}$$

\noindent But this means that $W'=W\cap(A\times B)\setminus V\in\Lambda$
and $V=(A\times B)\cap(W\setminus W')$ belongs to $\Lambda^{\#}$.\ \Qed

\medskip

{\bf (d)} Now fix any $V\in\tilde\Lambda$, and look at the conditions
($\alpha$)-($\gamma$) of part (ii) of the proposition.

\medskip

\quad\grheada\ If $A\in\Sigma$ and $B\in\Tau$, and $C\subseteq X\times Y$,
then $A\times B\in\Lambda$ (251E), so

$$\eqalignno{\theta(C\cap V)+\theta(C\setminus V)
&=\theta(C\cap V)+\theta((C\setminus V)\cap(A\times B))
  +\theta((C\setminus V)\setminus(A\times B))\cr
&=\tilde\theta(C\cap V)+\tilde\theta(C\cap(A\times B)\setminus V)
  +\theta(C\setminus(A\times B))\cr
&=\tilde\theta(C\cap(A\times B))
  +\theta(C\setminus(A\times B))\cr
&=\theta(C\cap(A\times B))
  +\theta(C\setminus(A\times B))
=\theta C.\cr}$$

\noindent As $C$ is arbitrary, $V\in\Lambda$, so
$V=V\cap(A\times B)$ belongs to $\Lambda^{\#}$.

\medskip

\quad\grheadb\ If $A\subseteq\bigcup_{n\in\Bbb N}E_n$ and
$B\subseteq\bigcup_{n\in\Bbb N}F_n$ where all the $E_n$, $F_n$ are of
finite measure, then
$V=\bigcup_{m,n\in\Bbb N}V\cap(E_m\times F_n)\in\Lambda^{\#}$, by (c).

\medskip

\quad\grheadc\ If $\familyiI{X_i}$, $\family{j}{J}{Y_j}$ are
decompositions
of $X$, $Y$ respectively, then for each $i\in I$, $j\in J$ we have
$V\cap(X_i\times Y_j)\in\Lambda^{\#}$, that is, there is a
$W_{ij}\in\Lambda$ such that
$V\cap(X_i\times Y_j)=W_{ij}\cap(A\times B)$.
Now $\family{(i,j)}{I\times J}{X_i\times Y_j}$ is a decomposition of
$X\times Y$ for $\lambda$ (251O), so that

\Centerline{$W=\bigcup_{i\in I,j\in J}W_{ij}\cap(X_i\times Y_j)
\in\Lambda$,}

\noindent and $V=W\cap(A\times B)\in\Lambda^{\#}$.

\medskip

{\bf (e)} Thus any of the three conditions is sufficient to ensure that
$\tilde\Lambda=\Lambda^{\#}$, in which case (a) tells us that
$\tilde\lambda=\lambda^{\#}$.
}%end of proof of 251Q

\leader{251R}{Corollary} Let $r$, $s\ge 1$ be integers, and
$\phi:\Bbb R^r\times\BbbR^s\to\BbbR^{r+s}$ the natural bijection.   If
$A\subseteq\BbbR^r$ and $B\subseteq\BbbR^s$, then the restriction of
$\phi$ to $A\times B$ identifies the product of Lebesgue measure on $A$
and Lebesgue measure on $B$ with Lebesgue measure on $\phi[A\times
B]\subseteq\BbbR^{r+s}$.

\cmmnt{\medskip

\noindent{\bf Remark} Note that by `Lebesgue measure on $A$' I mean
the subspace measure $\mu_{rA}$ on $A$ induced by $r$-dimensional
Lebesgue measure $\mu_r$ on $\BbbR^r$, whether or not $A$ is itself a
measurable set.
}%end of comment

\proof{ By 251Q, using either of the conditions
(ii-$\beta$) or (ii-$\gamma$), the product measure $\tilde\lambda$ on
$A\times B$ is just the subspace measure $\lambda^{\#}$ on $A\times B$
induced by the
product measure $\lambda$ on $\BbbR^r\times\BbbR^s$.   But by 251N we
know that $\phi$ is an isomorphism between $(\BbbR^r\times\Bbb
R^s,\lambda)$ and $(\BbbR^{r+s},\mu_{r+s})$;    so it must also
identify $\tilde\lambda$ with the subspace measure on $\phi[A\times B]$.
}

\leader{251S}{Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\lambda$ the c.l.d.\ product
measure on $X\times Y$.   If
$A\subseteq X$ and $B\subseteq Y$ can be covered by sequences of sets of
finite measure, then $\lambda^*(A\times
B)=\mu^*A\cdot\nu^*B$.

\proof{ In the language of 251Q,

$$\eqalignno{\lambda^*(A\times B)
&=\lambda^{\#}(A\times B)
=\tilde\lambda(A\times B)
=\mu_AA\cdot\nu_BB\cr
\noalign{\noindent (by 251K and 251E)}
&=\mu^*A\cdot\nu^*B.\cr}$$
}%end of proof of 251S

\leader{251T}{}\cmmnt{ The next proposition gives an idea of how the
technical definitions here fit together.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces.   Write $(X,\hat\Sigma,\hat\mu)$ and
$(X,\tilde\Sigma,\tilde\mu)$ for the completion and c.l.d.\ version of
$(X,\Sigma,\mu)$\cmmnt{ (212C, 213E)}.   Let $\lambda$, $\hat\lambda$
and $\tilde\lambda$ be the three c.l.d.\ product measures on $X\times Y$
obtained from the pairs $(\mu,\nu)$, $(\hat\mu,\nu)$ and
$(\tilde\mu,\nu)$
of factor measures.   Then $\lambda=\hat\lambda=\tilde\lambda$.

\proof{ Write $\Lambda$, $\hat\Lambda$ and $\tilde\Lambda$ for the
domains
of $\lambda$, $\hat\lambda$, $\tilde\lambda$ respectively;  and
$\theta$,
$\hat\theta$, $\tilde\theta$ for the outer measures on $X\times Y$
obtained
by the formula of 251A from the three pairs of factor measures.

\medskip

{\bf (a)} If $E\in\Sigma$ and $\mu E<\infty$, then $\theta$,
$\hat\theta$
and $\tilde\theta$ agree on subsets of $E\times Y$.   \Prf\ Take
$A\subseteq E\times Y$ and $\epsilon>0$.

\medskip

\quad{\bf (i)} There are sequences $\sequencen{E_n}$ in $\Sigma$,
$\sequencen{F_n}$ in $\Tau$ such that
$A\subseteq\bigcup_{n\in\Bbb N}E_n\times F_n$ and
$\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n\le\theta A+\epsilon$.   Now
$\tilde\mu E_n\le\mu E_n$ for every $n$ (213Fb), so

\Centerline{$\tilde\theta A
\le\sum_{n=0}^{\infty}\tilde\mu E_n\cdot\nu F_n
\le\sum_{n=0}^{\infty}\mu E_n\cdot\nu F_n\le\theta A+\epsilon$.}

\medskip

\quad{\bf (ii)} There are sequences $\sequencen{\hat E_n}$ in
$\hat\Sigma$, $\sequencen{\hat F_n}$ in $\Tau$ such that
$A\subseteq\bigcup_{n\in\Bbb N}\hat E_n\times\hat F_n$ and
$\sum_{n=0}^{\infty}\hat\mu\hat E_n\cdot\nu\hat F_n
\le\hat\theta A+\epsilon$.   Now for each $n$ there is an
$E'_n\in\Sigma$ such that $\hat E_n\subseteq E'_n$ and
$\mu E'_n=\hat\mu\hat E_n$, so that

\Centerline{$\theta A\le\sum_{n=0}^{\infty}\mu E'_n\cdot\nu\hat F_n
=\sum_{n=0}^{\infty}\hat\mu\hat E_n\cdot\nu\hat F_n
\le\hat\theta A+\epsilon$.}

\medskip

\quad{\bf (iii)} There are sequences $\sequencen{\tilde E_n}$ in
$\tilde\Sigma$, $\sequencen{\tilde F_n}$ in $\Tau$ such that
$A\subseteq\bigcup_{n\in\Bbb N}\tilde E_n\times\tilde F_n$ and
$\sum_{n=0}^{\infty}\tilde\mu\tilde E_n\cdot\nu\tilde F_n
\le\tilde\theta A+\epsilon$.   Now for each $n$,
$\tilde E_n\cap E\in\hat\Sigma$, so

\Centerline{$\hat\theta A
\le\sum_{n=0}^{\infty}\hat\mu(\tilde E_n\cap E)\cdot\nu\tilde F_n
\le\sum_{n=0}^{\infty}\tilde\mu\tilde E_n\cdot\nu\tilde F_n
\le\tilde\theta A+\epsilon$.}

\medskip

\quad{\bf (iv)} Since $A$ and $\epsilon$ are arbitrary,
$\theta=\hat\theta=\tilde\theta$ on $\Cal P(E\times Y)$.\ \Qed

\medskip

{\bf (b)} Consequently, the outer measures $\lambda^*$, $\hat\lambda^*$
and $\tilde\lambda^*$ are identical.   \Prf\ Use 251P.   Take
$A\subseteq X\times Y$, $E\in\Sigma$, $\hat E\in\hat\Sigma$,
$\tilde E\in\tilde\Sigma$,
$F\in\Tau$ such that $\mu E$, $\hat\mu\hat E$, $\tilde\mu\tilde E$ and
$\nu F$ are all finite.   Then

\medskip

\quad{\bf (i)}

\Centerline{$\theta(A\cap(E\times F))=\hat\theta(A\cap(E\times F))
\le\hat\lambda^*A$,
\quad$\theta(A\cap(E\times F))=\tilde\theta(A\cap(E\times F))
\le\tilde\lambda^*A$}

\noindent because $\hat\mu E$ and $\tilde\mu E$ are both finite.

\medskip

\quad{\bf (ii)} There is an $E'\in\Sigma$ such that $\hat E\subseteq E'$
and $\mu E'<\infty$, so that

\Centerline{$\hat\theta(A\cap(\hat E\times F))
\le\hat\theta(A\cap(E'\times F))
=\theta(A\cap(E'\times F))
\le\lambda^*A$.}

\medskip

\quad{\bf (iii)} There is an $E''\in\Sigma$ such that
$E''\subseteq\tilde E$ and $\tilde\mu(\tilde E\setminus E'')=0$ (213Fc),
so that
$\tilde\theta((\tilde E\setminus E'')\times Y)=0$ and $\mu E''<\infty$;
accordingly

\Centerline{$\tilde\theta(A\cap(\tilde E\times F))
=\tilde\theta(A\cap(E''\times F))
=\theta(A\cap(E''\times F))
\le\lambda^*A$.}

\medskip

\quad{\bf (iv)} Taking the suprema over $E$, $\hat E$, $\tilde E$ and
$F$,
we get

\Centerline{$\lambda^*A\le\hat\lambda^*A$,
\quad$\lambda^*A\le\tilde\lambda^*A$,
\quad$\hat\lambda^*A\le\lambda^*A$,
\quad$\tilde\lambda^*A\le\lambda^*A$.}

\noindent As $A$ is arbitrary,
$\lambda^*=\hat\lambda^*=\tilde\lambda^*$.\ \Qed

\medskip

{\bf (c)} Now $\lambda$, $\hat\lambda$ and $\tilde\lambda$ are all
complete and locally determined, so by 213C are the measures
defined by
\Caratheodory's method from their own outer measures, and are therefore
identical.
}%end of proof of 251T

\leader{251U}{}\cmmnt{ It is `obvious', and an easy consequence of
theorems so far proved, that the set $\{(x,x):x\in\Bbb R\}$ is
negligible for Lebesgue measure on $\BbbR^2$.   The corresponding result
is true in the square of any {\it atomless} measure space.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be an atomless measure
space, and let $\lambda$ be the c.l.d.\ measure on $X\times X$.   Then
$\Delta=\{(x,x):x\in X\}$ is $\lambda$-negligible.

\proof{ Let $E$, $F\in\Sigma$ be sets of finite measure, and
$n\in\Bbb N$.   Applying 215D repeatedly, we can find a disjoint family
$\ofamily{i}{n}{F_i}$ of measurable subsets of $F$ such that
$\mu F_i=\Bover{\mu F}{n+1}$ for each $i$;  setting
$F_n=F\setminus\bigcup_{i<n}F_i$, we also have
$\mu F_n=\Bover{\mu F}{n+1}$.   Now

\Centerline{$\Delta\cap(E\times F)
\subseteq\bigcup_{i\le n}(E\cap F_i)\times F_i$,}

\noindent so

\Centerline{$\lambda^*(\Delta\cap(E\times F))
\le\sum_{i=0}^n\mu(E\cap F_i)\cdot\mu F_i
=\Bover{\mu F}{n+1}\sum_{i=0}^n\mu(E\cap F_i)
\le\Bover1{n+1}\mu E\cdot\mu F$.}

\noindent As $n$ is arbitrary, $\lambda(\Delta\cap(E\times F))=0$;  as
$E$ and $F$ are arbitrary, $\lambda\Delta=0$.
}%end of proof of 251U

%\mu\times(\nu_1+\nu_2)=(\mu\times\nu_1)+(\mu\times\nu_2)
%may need restrictions?

\leader{*251W}{Products of more than two spaces}\cmmnt{ The whole of
this section can be repeated for arbitrary finite products.   The labour
is substantial but no new ideas are required.   By the time we need the
general construction in any formal way, it should come very naturally,
and I do not think it is necessary to work through the next
page before proceeding, especially as products of {\it probability}
spaces are dealt with in \S254.   However, for completeness, and to help
locate results when applications do appear, I list them here.   They do
of course constitute a very instructive set of exercises.   The most
important fragments are repeated in 251Xe-251Xf.

} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a finite family of measure
spaces, and set $X=\prod_{i\in I}X_i$.   Write
$\Sigma_i^f=\{E:E\in\Sigma_i,\,\mu_iE<\infty\}$ for each $i\in I$.

\spheader 251Wa For $A\subseteq X$ set

$$\theta A=\inf\{\sum_{n=0}^{\infty}\prod_{i\in I}\mu_iE_{ni}:
E_{ni}\in\Sigma_i\Forall i\in I,\,n\in\Bbb N,\,
A\subseteq\bigcup_{n\in\Bbb N}\prod_{i\in I}E_{ni}\}.$$

\noindent Then $\theta$ is an outer measure on $X$.   Let $\lambda_0$ be
the measure on $X$ derived by \Caratheodory's method from $\theta$, and
$\Lambda$ its domain.

\spheader 251Wb If $\familyiI{X_i}$ is a finite family of sets and
$\Sigma_i$ is a $\sigma$-algebra of subsets of $X_i$ for each $i\in I$,
then $\Tensorhat_{i\in I}\Sigma_i$ is the $\sigma$-algebra of subsets of
$X=\prod_{i\in I}X_i$ generated by $\{\prod_{i\in I}E_i:E_i\in\Sigma_i$
for every $i\in I\}$.   \cmmnt{(For the corresponding construction
when $I$ is infinite, see 254E.)}

\spheader 251Wc $\lambda_0(\prod_{i\in I}E_i)$ is defined and equal to
$\prod_{i\in I}\mu_iE_i$ whenever $E_i\in\Sigma_i$ for each $i\in I$.

\spheader 251Wd The {\bf c.l.d.\ product measure} on $X$ is the measure
$\lambda$ defined by setting

\Centerline{$\lambda W
=\sup\{\lambda_0(W\cap\prod_{i\in I}E_i):
  E_i\in\Sigma_i^f$ for each $i\in I\}$}

\noindent for $W\in\Lambda$.   If $I$ is empty, \cmmnt{so that
$X=\{\emptyset\}$, then the appropriate convention is to} set
$\lambda X=1$.

\spheader 251We If $H\subseteq X$, then $H\in\Lambda$ iff
$H\cap\prod_{i\in I}E_i\in\Lambda$ whenever $E_i\in\Sigma_i^f$ for each
$i\in I$.

\spheader 251Wf(i) $\Tensorhat_{i\in I}\Sigma_i\subseteq\Lambda$ and
$\lambda(\prod_{i\in I}E_i)=\prod_{i\in I}\mu_iE_i$ whenever
$E_i\in\Sigma_i^f$ for each $i$.

\quad(ii) For every $W\in\Lambda$ there is a $V\in\Tensorhat_{i\in
I}\Sigma_i$ such that $V\subseteq W$ and $\lambda V=\lambda W$.

\quad(iii) $\lambda$ is complete and locally determined, and is the
c.l.d.\ version of $\lambda_0$.

\quad(iv) If $W\in\Lambda$ and $\lambda W>0$ then there are
$E_i\in\Sigma_i^f$, for $i\in I$, such that
$\lambda(W\cap\prod_{i\in I}E_i)>0$.

\quad(v) If $W\in\Lambda$ and $\lambda W<\infty$, then for every
$\epsilon>0$ there are $n\in\Bbb N$ and
$E_{0i},\ldots,E_{ni}\in\Sigma_i^f$, for each $i\in I$, such that
$\lambda(W\symmdiff\bigcup_{k\le n}\prod_{i\in I}E_{ki})\le\epsilon$.

\spheader 251Wg If each $\mu_i$ is $\sigma$-finite, so is $\lambda$, and
$\lambda=\lambda_0$ is the completion of its restriction to
$\Tensorhat_{i\in I}\Sigma_i$.

\spheader 251Wh If $\family{j}{J}{I_j}$ is any partition of $I$, then
$\lambda$ can be identified with the c.l.d.\ product of
$\family{j}{J}{\lambda_j}$, where $\lambda_j$ is the c.l.d.\ product of
$\family{i}{I_j}{\mu_i}$.   \cmmnt{(See the arguments in 251N and also
in 254N below.)}

\spheader 251Wi If $I=\{1,\ldots,n\}$ and each $\mu_i$ is Lebesgue
measure on $\Bbb R$, then $\lambda$ can be identified with Lebesgue
measure on $\BbbR^n$.

\spheader 251Wj If, for each $i\in I$, we have a decomposition
$\family{j}{J_i}{X_{ij}}$ of $X_i$, then
$\langle\prod_{i\in I}X_{i,f(i)}\rangle_{f\in\prod_{i\in I}J_i}$ is a
decomposition of $X$.

\spheader 251Wk For any $C\subseteq X$,

\Centerline{$\lambda^*C
=\sup\{\theta(C\cap\prod_{i\in I}E_i):
  E_i\in\Sigma_i^f$ for every $i\in I\}$.}

\spheader 251Wl Suppose that $A_i\subseteq X_i$ for each $i\in I$.
Write $\lambda^{\#}$ for the subspace measure on $A=\prod_{i\in I}A_i$,
and $\tilde\lambda$ for the c.l.d.\ product of the subspace measures on
the $A_i$.   Then $\tilde\lambda$ extends $\lambda^{\#}$, and if

\quad{\it either} $A_i\in\Sigma_i$ for every $i$

\quad{\it or} every $A_i$ can be covered by a sequence of sets of finite
measure

\quad{\it or} every $\mu_i$ is strictly localizable,

\noindent then $\tilde\lambda=\lambda^{\#}$.

\spheader 251Wm If $A_i\subseteq X_i$ can be covered by a sequence of
sets of finite measure for each $i\in I$, then
$\lambda^*(\prod_{i\in I}A_i)=\prod_{i\in I}\mu_i^*A_i$.

\spheader 251Wn Writing $\hat\mu_i$, $\tilde\mu_i$ for the completion
and c.l.d.\ version of each $\mu_i$, $\lambda$ is the c.l.d.\ product of
$\familyiI{\hat\mu_i}$ and also of $\familyiI{\tilde\mu_i}$.

\spheader 251Wo If all the $(X_i,\Sigma_i,\mu_i)$ are the same atomless
measure space $(X,\Sigma,\mu)$, then
$\{x:x\in X$, $i\mapsto x(i)$ is injective$\}$ is
$\lambda$-conegligible.

\spheader 251Wp Now suppose that we have another family
$\familyiI{(Y_i,\Tau_i,\nu_i)}$ of measure spaces, with product
$(Y,\Lambda',\lambda')$, and \imp\ functions $f_i:X_i\to Y_i$ for each $i$;
define $f:X\to Y$ by saying that $f(x)(i)=f_i(x(i))$ for $x\in X$ and
$i\in I$.   If all the $\nu_i$ are $\sigma$-finite, then $f$ is \imp\ for
$\lambda$ and $\lambda'$.

\exercises{
\leader{251X}{Basic exercises (a)}
%\spheader 251Xa
Let $X$ and $Y$ be sets, $\Cal A\subseteq\Cal PX$ and
$\Cal B\subseteq\Cal PY$.   Let $\Sigma$ be the $\sigma$-algebra of subsets
of $X$ generated by $\Cal A$ and $\Tau$ the $\sigma$-algebra of subsets of
$Y$ generated by $\Cal B$.   Show that $\Sigma\tensorhat\Tau$ is the
$\sigma$-algebra of subsets of $X\times Y$ generated by
$\{A\times B:A\in\Cal A$, $B\in\Cal B\}$.
%251D query out of order

\spheader 251Xb Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be
measure spaces;  let $\lambda_0$ be the primitive product measure on
$X\times Y$, and $\lambda$ the c.l.d.\ product measure.   Show that
$\lambda_0W<\infty$ iff $\lambda W<\infty$ and $W$ is included in a set
of the form

\Centerline{$(E\times Y)\cup (X\times F)\cup\bigcup_{n\in\Bbb
N}E_n\times F_n$}

\noindent where $\mu E=\nu F=0$ and $\mu E_n<\infty$, $\nu F_n<\infty$
for every $n$.
%251F

\sqheader 251Xc Show that if $X$ and $Y$ are any sets, with
their respective counting measures, then the primitive and c.l.d.\
product measures on $X\times Y$ are both counting measure on $X\times
Y$.
%251F

\spheader 251Xd Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces;  let $\lambda_0$ be the primitive product measure on
$X\times Y$, and $\lambda$ the c.l.d.\ product measure.    Show that

$$\eqalign{\lambda_0&\text{ is locally determined}\cr
&\iff\,\lambda_0\text{ is semi-finite}\cr
&\iff\,\lambda_0=\lambda\cr
&\iff\,\lambda_0\text{ and }\lambda
  \text{ have the same negligible sets}.\cr}$$

\sqheader 251Xe (See 251W.)  Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in
I}$ be a family of measure spaces, where $I$ is a non-empty finite set.
Set $X=\prod_{i\in I}X_i$.   For $A\subseteq X$, set

\Centerline{$\theta A
=\inf\{\sum_{n=0}^{\infty}\prod_{i\in I}\mu_iE_{ni}:
E_{ni}\in\Sigma_i\Forall n\in\Bbb N,\,i\in
I,\,A\subseteq\bigcup_{n\in\Bbb N}\prod_{i\in I}E_{ni}\}$.}

\noindent Show that $\theta$ is an outer measure on $X$.   Let
$\lambda_0$ be the measure defined from $\theta$ by \Caratheodory's
method, and for $W\in\dom\lambda_0$ set

\Centerline{$\lambda W=\sup\{\lambda_0(W\cap\prod_{i\in I}E_i):
E_i\in\Sigma_i,\,\mu_iE_i<\infty$ for every $i\in I\}$.}

\noindent Show that $\lambda$ is a measure on $X$, and is the c.l.d.\
version of $\lambda_0$.
%251I

\sqheader 251Xf (See 251W.)  Let $I$ be a non-empty finite set and
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ a family of measure
spaces.   For non-empty $K\subseteq I$ set $X^{(K)}=\prod_{i\in K}X_i$
and let $\lambda_0^{(K)}$, $\lambda^{(K)}$ be the measures on $X^{(K)}$
constructed as in 251Xe.   Show that if $K$ is a non-empty proper
subset of $I$, then the natural bijection between $X^{(I)}$ and
$X^{(K)}\times X^{(I\setminus K)}$ identifies $\lambda_0^{(I)}$ with the
primitive product measure of $\lambda_0^{(K)}$ and
$\lambda_0^{(I\setminus K)}$, and $\lambda^{(I)}$ with the c.l.d.\
product measure of $\lambda^{(K)}$ and $\lambda^{(I\setminus K)}$.
%251Xe, 251I

\sqheader 251Xg Using 251Xe-251Xf above, or otherwise, show that if
$(X_1,\Sigma_1,\mu_1)$, $(X_2,\Sigma_2,\mu_2)$,
$(X_3,\Sigma_3,\mu_3)$ are measure spaces then the primitive and c.l.d.\
product measures
$\lambda_0$, $\lambda$ of $(X_1\times X_2)\times X_3$, constructed by
first taking the appropriate product measure on $X_1\times X_2$ and then
taking the product of this with the measure of $X_3$, are identified
with the corresponding product measures on $X_1\times(X_2\times X_3)$ by
the canonical bijection between the sets $(X_1\times X_2)\times X_3$ and
$X_1\times(X_2\times X_3)$.
%251Xf, 251Xe, 251I

\spheader 251Xh(i) What happens in 251Xe when $I$ is a
singleton?  (ii) Devise an appropriate convention to make 251Xe-251Xf
remain valid when one or more of the sets $I$, $K$, $I\setminus K$ there
is empty.
%251Xf, 251Xe, 251I

\sqheader 251Xi Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, and $I$ any non-empty set;  let $\nu$ be
counting measure on $I$.   Show that the c.l.d.\ product measure on
$X\times I$ is equal to (or at any rate identifiable with) the direct
sum measure of the family $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$,
if we set $(X_i,\Sigma_i,\mu_i)=(X,\Sigma,\mu)$ for every $i$.
%251I

\sqheader 251Xj Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of measure
spaces, with direct sum $(X,\Sigma,\mu)$
(214L).   Let $(Y,\Tau,\nu)$ be any measure space, and give $X\times Y$,
$X_i\times Y$ their c.l.d.\ product measures.   Show that the natural
bijection between $X\times Y$ and
$Z=\bigcup_{i\in I}(X_i\times Y)\times\{i\}$
is an isomorphism between the measure of $X\times Y$ and
the direct sum measure on $Z$.
%251Xi, 251I

\sqheader 251Xk Let $(X,\Sigma,\mu)$ be any measure space, and
$Y$ a singleton
set $\{y\}$;  let $\nu$ be the measure on $Y$ such that $\nu Y=1$.
Show that the natural bijection between $X\times\{y\}$ and $X$
identifies the primitive product measure on $X\times\{y\}$ with
$\check\mu$ as defined in 213Xa, and the c.l.d.\ product measure with
the c.l.d.\
version of $\mu$.   Explain how to put this together with 251Xg and
251Ic to prove 251T.
%251I

\sqheader 251Xl Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\lambda$ the c.l.d.\ product measure on $X\times Y$.   Show
that $\lambda$ is the c.l.d.\ version of its restriction to
$\Sigma\tensorhat\Tau$.
%251K

\spheader 251Xm Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, with primitive and c.l.d.\ product measures $\lambda_0$,
$\lambda$.   Let
$\lambda_1$ be any measure with domain $\Sigma\tensorhat\Tau$ such that
$\lambda_1(E\times F)=\mu E\cdot\nu F$ whenever $E\in\Sigma$ and
$F\in\Tau$.   Show that $\lambda W\le\lambda_1W\le\lambda_0W$ for every
$W\in\Sigma\tensorhat\Tau$.
%251K

\spheader 251Xn Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two measure
spaces, and $\lambda_0$ the primitive product measure on $X\times Y$.
Show that the corresponding outer measure $\lambda_0^*$ is just the
outer measure $\theta$ of 251A.
%251P

\spheader 251Xo Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$
be measure spaces, and $A\subseteq X$, $B\subseteq
Y$ subsets;  write $\mu_A$, $\nu_B$ for the subspace
measures.   Let $\lambda_0$ be the primitive
product measure on $X\times Y$, and $\lambda_0^{\#}$
the subspace measure it induces on $A\times B$.   Let $\tilde\lambda_0$
be the primitive product measure of
$\mu_A$ and $\nu_B$ on $A\times B$.   Show that $\tilde\lambda_0$
extends $\lambda_0^{\#}$.
Show that if
{\it either} ($\alpha$) $A\in\Sigma$ and $B\in\Tau$
{\it or} ($\beta$) $A$ and $B$ can both be covered by sequences of sets
of finite measure
{\it or} ($\gamma$) $\mu$ and $\nu$ are both strictly localizable,
then $\tilde\lambda_0=\lambda_0^{\#}$.
%251Q

\spheader 251Xp In 251Q, show that $\tilde\lambda$ and $\lambda^{\#}$
will have the same null ideals, even if none of the conditions of
251Q(ii) are satisfied.
%251Q query out of order

\spheader 251Xq Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be any measure
spaces, and $\lambda_0$ the primitive product measure on $X\times Y$.
Show that $\lambda_0^*(A\times B)=\mu^*A\cdot\nu^*B$ for any
$A\subseteq X$ and $B\subseteq Y$.
%251S

\spheader 251Xr Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\hat\mu$ the completion of $\mu$.
Show that $\mu$, $\nu$ and $\hat\mu$, $\nu$ have the same primitive
product measures.
%251T

\spheader 251Xs Let $(X,\Sigma,\mu)$ be a semi-finite measure space.
Show that $\mu$ is atomless iff the diagonal $\{(x,x):x\in X\}$ is
negligible for the c.l.d.\ product measure on $X\times X$.
%251U

\sqheader 251Xt Let $(X,\Sigma,\mu)$ be an atomless measure
space, and $(Y,\Tau,\nu)$ any measure space.   Show that the c.l.d.\
product measure on $X\times Y$ is atomless.
%251+

\sqheader 251Xu Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\lambda$ the c.l.d.\ product measure on $X\times Y$.
(i) Show that if $\mu$ and $\nu$ are purely atomic, so is $\lambda$.
(ii) Show that if $\mu$ and $\nu$ are point-supported, so is $\lambda$.
%251+  (ii) used in 491F

\leader{251Y}{Further exercises (a)}
%\spheader 251Ya
Let $X$, $Y$ be sets with $\sigma$-algebras of subsets $\Sigma$, $\Tau$.
Suppose that $h:X\times Y\to\Bbb R$ is $\Sigma\tensorhat\Tau$-measurable
and $\phi:X\to Y$ is $(\Sigma,\Tau)$-measurable (121Yc).   Show that
$x\mapsto h(x,\phi(x)):X\to\Bbb R$ is $\Sigma$-measurable.
%251D

\spheader 251Yb Show that there are
measure spaces $(X_1,\Sigma_1,\mu_1)$ and $(X_2,\Sigma_2,\mu_2)$,
a probability space $(Y,\Tau,\nu)$ and an \imp\ function $f:X_1\to X_2$
such that $h:X_1\times Y\to X_2\times Y$ is not \imp\ for the c.l.d.\
product measures on $X_1\times Y$ and $X_2\times Y$, where
$h(x,y)=(f(x),y)$ for $x\in X_1$ and $y\in Y$.
%251L

\spheader 251Yc Let $(X,\Sigma,\mu)$ be a complete
locally determined measure space with a subspace $A$ whose measure is
not locally determined (see 216Xb).      Set $Y=\{0\}$, $\nu Y=1$ and
consider the c.l.d.\ product measures on $X\times Y$ and $A\times Y$;
write $\Lambda$, $\tilde\Lambda$ for their domains.   Show that
$\tilde\Lambda$ properly includes $\{W\cap(A\times Y):W\in\Lambda\}$.
%251P

\spheader 251Yd Let $(X,\Sigma,\mu)$ be any measure space,
$(Y,\Tau,\nu)$ an atomless measure space, and $f:X\to Y$ a
$(\Sigma,\Tau)$-measurable function.   Show that $\{(x,f(x)):x\in X\}$
is negligible for the c.l.d.\ product measure on $X\times Y$.
%251U
}%end of exercises

\endnotes{
\Notesheader{251} There are real difficulties in deciding which
construction to declare as `the' product of two arbitrary measures.
My phrase `primitive product measure', and
notation $\lambda_0$, betray a bias;  my own preference is for the
c.l.d.\ product $\lambda$, for two principal reasons.   The first is
that $\lambda_0$ is likely to be `bad', in particular,
not semi-finite, even if $\mu$ and $\nu$ are `good' (251Xd, 252Yk),
while
$\lambda$ inherits some of the most important properties of $\mu$ and
$\nu$ (e.g., 251O);  the second is that in the case of topological
measure spaces $X$ and $Y$, there is often a canonical topological
measure on $X\times Y$, which is likely to be more closely related to
$\lambda$ than to $\lambda_0$.   But for elucidation of this point I
must ask you to wait until \S417 in Volume 4.

It would be possible to remove the `primitive' product measure entirely
from the exposition, or at least to relegate it to the exercises.   This
is indeed what I expect to do in the rest of this treatise, since (in my
view) all significant features of product measures on finitely many
factors can be expressed in terms of the c.l.d.\ product measure.   For
the first introduction to product measures, however, a direct approach
to the c.l.d.\ product measure (through the description of $\lambda^*$
in 251P, for instance) is an uncomfortably large bite, and I have some
sort of duty to present the most natural rival to the c.l.d.\ product
measure prominently enough for you to judge for yourself whether I am
right to dismiss it.   There certainly are results associated with the
primitive product measure (251Xn, 251Xq, 252Yc) which have an
agreeable simplicity.

The clash is avoided altogether, of course, if we specialize immediately
to $\sigma$-finite spaces, in which the two
constructions coincide (251K).   But even this does not solve all
problems.   There is a popular alternative measure often called `the'
product measure:  the restriction $\lambda_{0\Cal B}$ of $\lambda_0$ to
the $\sigma$-algebra $\Sigma\tensorhat\Tau$.   (See, for instance,
{\smc Halmos 50}.)   The advantage of this is that if a function $f$ on
$X\times Y$ is
$\Sigma\tensorhat\Tau$-measurable, then $x\mapsto f(x,y)$ is
$\Sigma$-measurable for every $y\in Y$.   (This is because

$$\{W:W\subseteq X\times Y,\,\{x:(x,y)\in
W\}\in\Sigma\,\Forall \,y\in Y\}$$

\noindent is a $\sigma$-algebra of subsets of $X\times Y$ containing
$E\times F$ whenever $E\in\Sigma$ and $F\in\Tau$, and therefore including
$\Sigma\tensorhat\Tau$.)   The primary objection, to my mind, is that
Lebesgue measure on $\BbbR^2$ is no longer `the' product of Lebesgue
measure on $\Bbb R$ with itself.   Generally, it is right to seek
measures which measure as many sets as possible, and I prefer to face up
to the technical problems (which I acknowledge are off-putting) by
seeking appropriate definitions on the approach to major theorems,
rather than rely on ad hoc fixes when the time comes to apply them.

I omit further examples of product measures for the
moment, because the investigation of particular examples will be much
easier with the aid of results from the next section.   Of course the
leading example, and the one which should come always to mind in
response to the words `product measure', is Lebesgue measure on
$\BbbR^2$, the case $r=s=1$ of 251N and 251R.   For an indication of
what can happen when
one of the factors is not $\sigma$-finite, you could look ahead to 252K.

I hope that you will see that the definition of the outer measure
$\theta$ in 251A corresponds to the standard definition of Lebesgue
outer measure, with `measurable rectangles' $E\times F$ taking the
place of intervals, and the functional $E\times F\mapsto\mu E\cdot\nu F$
taking the place of `length' or `volume' of an interval;
moreover, thinking of $E$ and $F$ as intervals, there is an obvious
relation between Lebesgue measure on $\BbbR^2$ and the product measure
on $\Bbb R\times\Bbb R$.   Of course an `obvious relationship' is not
the same thing as a proper theorem with exact hypotheses and
conclusions, but Theorem 251N is clearly central.   Long before that,
however, there is another parallel between the construction of 251A and
that of Lebesgue measure.   In both cases, the proof that we have an
outer measure comes directly from the defining formula (in 113Yd I gave
as an exercise a general result covering 251B), and consequently a very
general construction can lead us to a measure.   But the measure would
be of far less interest and value if it did not measure, and measure
correctly, the basic sets, in this case the measurable rectangles.
Thus 251E corresponds to the theorem that intervals are Lebesgue
measurable, with the right measure (114Db, 114G).   This is the real key
to the construction, and is one of the fundamental ideas of measure
theory.

Yet another parallel is in 251Xn;  the outer measure defining the
primitive product measure $\lambda_0$ is exactly equal to the outer
measure defined from $\lambda_0$.   I described the corresponding
phenomenon for Lebesgue measure in 132C.

Any construction which claims the title `canonical' must
satisfy a variety of natural requirements;  for instance, one expects
the canonical bijection between $X\times Y$ and $Y\times X$ to be an
isomorphism between the corresponding product measure spaces.
`Commutativity' of the product in this sense is I think obvious from
the definitions in 251A-251C.   It is obviously desirable -- not, I
think, obviously true -- that the product should be `associative' in
that the canonical bijection between $(X\times Y)\times Z$ and
$X\times(Y\times Z)$ should also be an isomorphism between the
corresponding products of product measures.   This is in fact valid for
both the primitive and c.l.d.\ product measures (251Wh, 251Xe-251Xg).

Working through the classification of measure spaces presented in
\S211, we find that the primitive product measure $\lambda_0$ of
arbitrary factor measures $\mu$, $\nu$ is complete, while the c.l.d.\
product measure $\lambda$ is always complete and locally determined.
$\lambda_0$ may not be
semi-finite, even if $\mu$ and $\nu$ are strictly localizable
(252Yk);  but $\lambda$ will be strictly localizable if $\mu$ and
$\nu$ are (251O).   Of course this is associated with the fact that the
c.l.d.\ product measure is distributive over direct sums (251Xj).   If
either $\mu$ or $\nu$ is atomless, so is $\lambda$ (251Xt).   Both
$\lambda$ and $\lambda_0$ are $\sigma$-finite if $\mu$
and $\nu$ are (251K).   It is possible for both $\mu$ and $\nu$ to be
localizable but $\lambda$ not (254U).

At least if you have worked through Chapter 21, you have now done enough
`pure' measure theory for this kind of investigation, however
straightforward, to raise a good many questions.   Apart from direct
sums, we also have the constructions of `completion', `subspace', `outer
measure' and (in particular) `c.l.d.\ version' to integrate into the new
ideas;  I offer some results in 251T and 251Xk.   Concerning
subspaces, some possibly surprising difficulties arise.   The problem is
that the
product measure on the product of two subspaces can have a larger domain
than one might expect.   I give a simple example in 251Yc and a more
elaborate one in 254Yg.   For strictly localizable spaces, there is
no problem (251Q);  but no other criterion drawn from the list of
properties considered in \S251 seems adequate to remove the
possibility of a disconcerting phenomenon.

}%end of notes

\discrpage

\frfilename{mt252.tex}
\versiondate{6.12.07}
\copyrightdate{2000}

\def\displaycause#1{\noalign{\noindent (#1)}}

\def\chaptername{Product measures}
\def\sectionname{Fubini's theorem}

\newsection{252}

Perhaps the most important feature of the concept of `product
measure' is the fact that we can use it to discuss repeated integrals.
In this section I give versions of Fubini's theorem and Tonelli's
theorem (252B, 252G) with a variety of corollaries, the most useful ones
being versions for $\sigma$-finite spaces (252C, 252H).   As
applications I describe the relationship between integration and
measuring ordinate sets (252N) and calculate
the $r$-dimensional volume of a ball in $\BbbR^r$
(252Q\cmmnt{, 252Xi}).
I mention counter-examples showing the difficulties which can arise with
non-$\sigma$-finite measures and non-integrable functions
(252K-252L\cmmnt{, 252Xf-252Xg}).

\leader{252A}{Repeated integrals}  Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be
measure spaces, and $f$ a real-valued function defined on a set
$\dom f\subseteq X\times Y$.   We can seek to form the {\bf repeated
integral}

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)
=\int\bigl(\int f(x,y)\nu(dy)\bigr)\mu(dx)$,}

\noindent which should be interpreted as follows:  set

\Centerline{$D=\{x:x\in X,\,\int f(x,y)\nu(dy)$ is defined in
$[-\infty,\infty]\}$,}

\Centerline{$g(x)=\int f(x,y)\nu(dy)$ for $x\in D$,}

\noindent and then write $\iint f(x,y)\nu(dy)\mu(dx)=\int g(x)\mu(dx)$
if this is defined.   \cmmnt{Of course the subset of $Y$ on which
$y\mapsto f(x,y)$ is defined may vary with $x$, but it must always be
conegligible, as must $D$.

Similarly, exchanging the roles of $X$ and $Y$, we can seek
a repeated integral

\Centerline{$\iint f(x,y)\mu(dx)\nu(dy)
=\int\bigl(\int f(x,y)\mu(dx)\bigr)\nu(dy)$.}

\noindent The point is that, under appropriate conditions on $\mu$ and
$\nu$, we can relate these repeated integrals to each other
by connecting them both with the integral of $f$ itself with respect to
the product measure on $X\times Y$.

As will become apparent shortly, it is essential here to allow oneself
to discuss the integral of a function which is not everywhere defined.   It
is of less importance whether one allows integrands and integrals to
take infinite values, but for definiteness let me say that I shall be
following the rules of 135F;  that is, $\int f=\int f^+-\int f^-$
provided that $f$ is defined almost everywhere, takes values in
$[-\infty,\infty]$ and is virtually measurable, and at most one of
$\int f^+$, $\int f^-$ is infinite.
}%end of comment

\leader{252B}{Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with c.l.d.\ product
$(X\times Y,\Lambda,\lambda)$\cmmnt{ (251F)}.   Suppose that $\nu$ is
$\sigma$-finite and that $\mu$ is {\it either} strictly localizable
{\it or} complete and locally determined.   Let $f$ be a
$[-\infty,\infty]$-valued function such that $\int fd\lambda$ is defined
in $[-\infty,\infty]$.   Then $\iint f(x,y)\nu(dy)\mu(dx)$ is defined
and is equal to $\int fd\lambda$.

\proof{ The proof of this result involves substantial technical
difficulties.   If you have not seen these ideas before, you should
almost certainly not go straight to the full generality of the version
announced above.   I will therefore start by writing out a proof in the
case in which both $\mu$ and $\nu$ are totally finite;  this is already
lengthy enough.   I will present it in such a way that only the central
section (part (b) below) needs to be amended in the general case, and
then, after completing the proof of the special case, I will give the
alternative version of (b) which is required for the full result.

\medskip

{\bf (a)} Write $\eusm L$ for the family of $[0,\infty]$-valued
functions $f$ such that $\int fd\lambda$ and
\discrversionA{\break}{}$\iint f(x,y)\nu(dy)\mu(dx)$ are defined and
equal.   My
aim is to show first that $f\in\eusm L$ whenever $f$ is non-negative and
$\int fd\lambda$ is defined, and then to look at differences of
functions in $\eusm L$.   To prove that enough functions belong to
$\eusm L$, my strategy will be to start with `elementary' functions and
work outwards through progressively larger classes.   It is most
efficient to begin by describing ways of building new members of $\eusm
L$ from old, as follows.

\medskip

\quad{\bf (i)} $f_1+f_2\in\eusm L$ for all $f_1$, $f_2\in\eusm L$, and
$cf\in\eusm L$ for all $f\in\eusm L$, $c\in\coint{0,\infty}$;  this is
because

\Centerline{$\int(f_1+f_2)(x,y)\nu(dy)
=\int f_1(x,y)\nu(dy)+\int f_2(x,y)\nu(dy)$,}

\Centerline{$\int (cf)(x,y)\nu(dy)=c\int f(x,y)\nu(dy)$}

\noindent whenever the right-hand sides are defined, which we are
supposing to be the case for almost every $x$, so that

$$\eqalign{\iint(f_1+f_2)(x,y)\nu(dy)\mu(dx)
&=\iint f_1(x,y)\nu(dy)\mu(dx)+\iint f_2(x,y)\nu(dy)\mu(dx)\cr
&=\int f_1d\lambda+\int f_2d\lambda
=\int(f_1+f_2)d\lambda,\cr}$$

$$\iint(cf)(x,y)\nu(dy)\mu(dx)
=c\int f(x,y)\nu(dy)\mu(dx)
=c\int fd\lambda
=\int(cf)d\lambda.$$

\medskip

\quad{\bf (ii)} If $\sequencen{f_n}$ is a sequence in $\eusm L$ such
that $f_n(x,y)\le f_{n+1}(x,y)$ whenever $n\in\Bbb N$ and
$(x,y)\in\dom f_n\cap\dom f_{n+1}$, then $\sup_{n\in\Bbb N}f_n\in\eusm
L$.   \Prf\
Set $f=\sup_{n\in\Bbb N}f_n$;  for $x\in X$, $n\in\Bbb N$ set
$g_n(x)=\int f_n(x,y)\nu(dy)$ when the integral is defined in
$[0,\infty]$.   Since here I am allowing $\infty$ as a value of a
function, it is natural to regard $f$ as defined on $\bigcap_{n\in\Bbb
N}\dom f_n$.   By
B.Levi's theorem, $\int fd\lambda=\sup_{n\in\Bbb N}\int f_nd\lambda$;
write $u$ for this common value in $[0,\infty]$.   Next,
because $f_n\le f_{n+1}$ wherever both are defined, $g_n\le g_{n+1}$
wherever both are defined, for each $n$;   we are supposing that
$f_n\in\eusm L$, so $g_n$ is defined $\mu$-almost everywhere for each
$n$, and

\Centerline{$\sup_{n\in\Bbb N}\int g_nd\mu
=\sup_{n\in\Bbb N}\int f_nd\lambda=u$.}

\noindent By B.Levi's theorem again, $\int g\,d\mu=u$, where
$g=\sup_{n\in\Bbb N}g_n$.   Now take any $x\in\bigcap_{n\in\Bbb N}\dom
g_n$, and consider the functions $f_{xn}$ on $Y$, setting
$f_{xn}(y)=f_n(x,y)$ whenever this is defined.   Each $f_{xn}$ has an
integral in $[0,\infty]$, and $f_{xn}(y)\le f_{x,n+1}(y)$ whenever both
are defined, and

\Centerline{$\sup_{n\in\Bbb N}\int f_{xn}d\nu=g(x)$;}

\noindent so, using B.Levi's theorem for a third time,
$\int(\sup_{n\in\Bbb
N}f_{xn})d\nu$ is defined and equal to $g(x)$, that is,

\Centerline{$\int f(x,y)\nu(dy)=g(x)$.}

\noindent This is true for almost every $x$, so

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)
=\int g\,d\mu=u=\int fd\lambda$.}

\noindent Thus $f\in\eusm L$, as claimed.\ \Qed

\medskip

\quad{\bf (iii)} The expression of the ideas in the next section of the
proof will go more smoothly if I introduce another term.   Write
$\Cal W$ for $\{W:W\subseteq X\times Y,\,\chi W\in\eusm L\}$.   Then

\inset{($\alpha$) if $W$, $W'\in\Cal W$ and $W\cap W'=\emptyset$,
$W\cup W'\in\Cal W$}

\noindent by (i), because $\chi(W\cup W')=\chi W+\chi W'$,

\inset{($\beta$) $\bigcup_{n\in\Bbb N}W_n\in\Cal W$ whenever
$\sequencen{W_n}$ is a non-decreasing sequence in $\Cal W$}

\noindent because $\sequencen{\chi W_n}\uparrow\chi W$, and we can use
(ii).

It is also helpful to note that, for any $W\subseteq X\times Y$ and any
$x\in X$, $\int\chi W(x,y)\nu(dy)=\nu W[\{x\}]$, at least whenever
$W[\{x\}]=\{y:(x,y)\in W\}$ is measured by $\nu$.   Moreover, because
$\lambda$ is complete, a set $W\subseteq X\times Y$ belongs to $\Lambda$
iff $\chi W$ is $\lambda$-virtually measurable iff
$\int\chi W\,d\lambda$ is defined in $[0,\infty]$, and in this case
$\lambda W=\int\chi W\,d\lambda$.

\medskip

\quad{\bf (iv)} Finally, we need to observe that, in appropriate
circumstances, the difference of two members of $\Cal W$ will belong to
$\Cal W$:  if $W$, $W'\in\Cal W$ and $W\subseteq W'$ and $\lambda
W'<\infty$, then $W'\setminus W\in\Cal W$.   \Prf\ We are supposing that
$g(x)=\int\chi W(x,y)\nu(dy)$ and $g'(x)=\int\chi W'(x,y)\nu(dy)$ are
defined for almost every $x$, and that $\int g\,d\mu=\lambda W$, $\int
g'd\mu=\lambda W'$.   Because $\lambda W'$ is finite, $g'$ must be
finite almost everywhere, and $D=\{x:x\in\dom g\cap\dom
g',\,g'(x)<\infty\}$ is conegligible.   Now, for any $x\in D$, both
$g(x)$ and $g'(x)$ are finite, so

\Centerline{$y\mapsto\chi(W'\setminus W)(x,y)=\chi W'(x,y)-\chi W(x,y)$}

\noindent is the difference of two integrable functions, and

$$\eqalign{\int\chi(W'\setminus W)(x,y)\nu(dy)
&=\int\chi W'(x,y)-\chi W(x,y)\nu(dy)\cr
&=\int\chi W'(x,y)\nu(dy)-\int\chi W(x,y)\nu(dy)
=g'(x)-g(x).\cr}$$

\noindent Accordingly

\Centerline{$\iint\chi(W'\setminus W)(x,y)\nu(dy)\mu(dx)
=\int g'(x)-g(x)\mu(dx)
=\lambda W'-\lambda W
=\lambda(W'\setminus W)$,}

\noindent and $W'\setminus W$ belongs to $\Cal W$.\ \Qed

(Of course the argument just above can be shortened by a few words if we
allow ourselves to assume that $\mu$ and $\nu$ are totally finite,
since then $g(x)$ and $g'(x)$ will be finite whenever they are defined;
but the key idea, that the difference of integrable functions is
integrable, is unchanged.)

\medskip

{\bf (b)} Now let us examine the class $\Cal W$, assuming that $\mu$ and
$\nu$ are totally finite.

\medskip

\quad{\bf (i)} $E\times F\in\Cal W$ for all $E\in\Sigma$, $F\in\Tau$.
\Prf\ $\lambda(E\times F)=\mu E\cdot\nu F$ (251J),  and

\Centerline{$\int\chi(E\times F)(x,y)\nu(dy)=\nu F\,\chi E(x)$}

\noindent for each $x$, so

$$\eqalign{\iint\chi(E\times F)(x,y)\nu(dy)\mu(dx)
&=\int(\nu F\,\chi E(x))\mu(dx)
=\mu E\cdot\nu F\cr
&=\lambda(E\times F)
=\int\chi(E\times F)d\lambda. \text{ \Qed}\cr}$$

\medskip

\quad{\bf (ii)} Let $\Cal C$ be $\{E\times F:E\in\Sigma,\,F\in\Tau\}$.
Then $\Cal C$ is closed under finite intersections (because
$(E\times F)\cap(E'\times F')=(E\cap E')\times(F\cap F')$)
and is included in
$\Cal W$.   In particular, $X\times Y\in\Cal W$.   But this, together
with (a-iv) and (a-iii-$\beta$) above, means that $\Cal W$ is a Dynkin
class (definition:  136A), so includes the $\sigma$-algebra of subsets
of $X\times Y$
generated by $\Cal C$, by the Monotone Class Theorem (136B);  that is,
$\Cal W\supseteq\Sigma\tensorhat\Tau$ (definition:  251D).

\medskip

\quad{\bf (iii)} Next, $W\in\Cal W$ whenever $W\subseteq X\times Y$ is
$\lambda$-negligible.   \Prf\ By 251Ib, there is a
$V\in\Sigma\tensorhat\Tau$ such that $V\subseteq(X\times Y)\setminus W$
and $\lambda V=\lambda((X\times Y)\setminus W)$.   Because
$\lambda(X\times Y)=\mu X\cdot\nu Y$ is finite, $V'=(X\times Y)\setminus
V$ is $\lambda$-negligible, and we have $W\subseteq
V'\in\Sigma\tensorhat\Tau$.   Consequently

\Centerline{$0=\lambda V'=\iint\chi V'(x,y)\nu(dy)\mu(dx)$.}

\noindent But this means that

\Centerline{$D=\{x:\int\chi V'(x,y)\nu(dy)$ is defined and equal
to $0\}$}

\noindent is conegligible.   If $x\in D$, then we must have
$\chi V'(x,y)=0$ for $\nu$-almost every $y$, that is, $V'[\{x\}]$ is
negligible;  in which case $W[\{x\}]\subseteq V'[\{x\}]$ also is
negligible, and $\int\chi W(x,y)\nu(dy)=0$.   And this is true for every
$x\in D$, so $\int\chi W(x,y)\nu(dy)$ is defined and equal to $0$ for
almost every $x$, and

\Centerline{$\iint\chi W(x,y)\nu(dy)\mu(dx)=0=\lambda W$,}

\noindent as required.\ \Qed

\medskip

\quad{\bf (iv)} It follows that $\Lambda\subseteq\Cal W$.   \Prf\ If
$W\in\Lambda$, then, by 251Ib again, there is a
$V\in\Sigma\tensorhat\Tau$ such that $V\subseteq W$ and
$\lambda V=\lambda W$, so that $\lambda(W\setminus V)=0$.   Now
$V\in\Cal W$ by
(ii) and $W\setminus V\in\Cal W$ by (iii), so $W\in\Cal W$ by
(a-iii-$\alpha$).\ \Qed

\medskip

{\bf (c)} I return to the class $\eusm L$.

\medskip

\quad{\bf (i)} If $f\in\eusm L$ and $g$ is a $[0,\infty]$-valued
function defined and equal to $f\,\,\lambda$-a.e., then $g\in\eusm L$.
\Prf\ Set

\Centerline{$W=(X\times Y)\setminus\{(x,y):(x,y)\in\dom f\cap\dom g$,
$f(x,y)=g(x,y)\}$,}

\noindent so that $\lambda W=0$.   (Remember that $\lambda$ is
complete.)   By (b), $\iint\chi W(x,y)\nu(dy)\mu(dx)=0$, that is,
$W[\{x\}]$ is $\nu$-negligible for $\mu$-almost every $x$.   Let $D$ be
$\{x:x\in X,\,W[\{x\}]$ is $\nu$-negligible$\}$.   Then $D$ is
$\mu$-conegligible.   If $x\in D$, then

\Centerline{$W[\{x\}]=Y\setminus\{y:(x,y)\in\dom f\cap\dom g$,
$f(x,y)=g(x,y)\}$}

\noindent is negligible, so that $\int f(x,y)\nu(dy)=\int g(x,y)\nu(dy)$
if either is defined.   Thus the functions
\discrcenter{468pt}{$x\mapsto\int f(x,y)\nu(dy)$,
\ifdim\pagewidth>467pt\quad\fi
$x\mapsto\int g(x,y)\nu(dy)$ }are equal almost everywhere, and

\Centerline{$\iint g(x,y)\nu(dy)\mu(dx)=\iint
f(x,y)\nu(dy)\mu(dx)
=\int fd\lambda=\int g\,d\lambda$,}

\noindent so that $g\in\eusm L$.\ \Qed

\medskip

\quad{\bf (ii)} Now let $f$ be any non-negative function such that
$\int fd\lambda$ is defined in $[0,\infty]$.  Then $f\in\eusm L$.
\Prf\ For $k$, $n\in\Bbb N$ set

\Centerline{$W_{nk}=\{(x,y):(x,y)\in\dom f,\,f(x,y)\ge 2^{-n}k\}$.}

\noindent Because $\lambda$ is complete and $f$ is $\lambda$-virtually
measurable and $\dom f$ is conegligible, every $W_{nk}$ belongs to
$\Lambda$, so $\chi W_{nk}\in\eusm L$, by (b).   Set
$f_n=\sum_{k=1}^{4^n}2^{-n}\chi W_{nk}$, so that

$$\eqalign{f_n(x,y)
&=2^{-n}k\text{ if }k\le 4^n\text{ and }2^{-n}k
                   \le f(x,y)<2^{-n}(k+1),\cr
&=2^n\text{ if }f(x,y)\ge 2^n,\cr
&=0\text{ if }(x,y)\in(X\times Y)\setminus\dom f.\cr}$$

\noindent By (a-i), $f_n\in\eusm L$ for every $n\in\Bbb N$, while
$\sequencen{f_n}$ is non-decreasing, so $f'=\sup_{n\in\Bbb N}f_n\in\eusm
L$, by (a-ii).   But $f\eae f'$, so $f\in\eusm L$, by (i) just above.\
\Qed

\medskip

\quad{\bf (iii)} Finally, let $f$ be any $[-\infty,\infty]$-valued
function such that $\int fd\lambda$ is defined in $[-\infty,\infty]$.
Then $\int f^+d\lambda$, $\int f^-d\lambda$ are both defined and at most
one is infinite.   By (ii), both $f^+$ and $f^-$ belong to $\eusm L$.
Set $g(x)=\int f^+(x,y)\nu(dy)$, $h(x)=\int f^-(x,y)\nu(dy)$ whenever
these are defined;  then $\int g\,d\mu=\int f^+d\lambda$ and $\int
h\,d\mu=\int f^-d\lambda$ are both defined in $[0,\infty]$.

Suppose first that $\int f^-d\lambda$ is finite.   Then $\int h\,d\mu$
is finite, so $h$ must be finite $\mu$-almost everywhere;  set

\Centerline{$D=\{x:x\in\dom g\cap\dom h,\,h(x)<\infty\}$.}

\noindent For any $x\in D$, $\int f^+(x,y)\nu(dy)$ and
$\int f^-(x,y)\nu(dy)$ are defined in $[0,\infty]$, and the latter is
finite;  so

\Centerline{$\int f(x,y)\nu(dy)
=\int f^+(x,y)\nu(dy)-\int f^-(x,y)\nu(dy)
=g(x)-h(x)$}

\noindent is defined in $\ocint{-\infty,\infty}$.   Because $D$ is
conegligible,

$$\eqalign{\iint f(x,y)\nu(dy)\mu(dx)
&=\int g(x)-h(x)\mu(dx)
=\int g\,d\mu-\int h\,d\mu\cr
&=\int f^+d\lambda-\int f^-d\lambda
=\int fd\lambda,\cr}$$

\noindent as required.

Thus we have the result when $\int f^-d\lambda$ is finite.   Similarly,
or by applying the argument above to $-f$, we see that
$\iint f(x,y)\nu(dy)\mu(dx)=\int fd\lambda$
if $\int f^+d\lambda$ is finite.

Thus the theorem is proved, at least when $\mu$ and $\nu$ are totally
finite.

\medskip

{\bf (b*)} The only point in the argument above where we needed to know
anything special about the measures $\mu$ and $\nu$ was in part (b),
when showing that $\Lambda\subseteq\Cal W$.   I now return to this point
under the hypotheses of the theorem as stated, that $\nu$ is
$\sigma$-finite and $\mu$ is either strictly localizable or complete and
locally determined.

\medskip

\quad{\bf (i)} It will be helpful to note that the completion $\hat\mu$
of $\mu$ (212C) is identical with its c.l.d.\ version $\tilde\mu$
(213E).   \Prf\ If $\mu$ is strictly localizable, then
$\hat\mu=\tilde\mu$ by 213Ha.   If $\mu$ is complete and locally
determined, then $\hat\mu=\mu=\tilde\mu$ (212D, 213Hf).\ \Qed

\medskip

\quad{\bf (ii)}  Write $\Sigma^f=\{G:G\in\Sigma,\,\mu G<\infty\}$,
$\Tau^f=\{H:H\in\Tau,\,\nu H<\infty\}$.   For $G\in\Sigma^f$,
$H\in\Tau^f$ let $\mu_G$, $\nu_H$ and $\lambda_{G\times H}$ be the
subspace measures on $G$, $H$ and $G\times H$ respectively;  then
$\lambda_{G\times H}$ is the c.l.d.\ product measure of $\mu_G$ and
$\nu_H$ (251Q(ii-$\alpha$)).   Now $W\cap(G\times H)\in\Cal W$ for every
$W\in\Lambda$.   \Prf\ $W\cap(G\times H)$ belongs to the domain of
$\lambda_{G\times H}$, so by (b) of this proof, applied to the totally
finite measures $\mu_G$ and $\nu_H$,

$$\eqalignno{\lambda(W\cap(G\times H))
&=\lambda_{G\times H}(W\cap(G\times H))\cr
&=\int_G\int_H\chi(W\cap(G\times H))(x,y)\nu_H(dy)\mu_G(dx)\cr
&=\int_G\int_Y\chi(W\cap(G\times H))(x,y)\nu(dy)\mu_G(dx)\cr
\displaycause{because $\chi(W\cap(G\times H))(x,y)=0$ if
$y\in Y\setminus H$, so we can use 131E}
&=\int_X\int_Y\chi(W\cap(G\times H))(x,y)\nu(dy)\mu(dx)\cr}$$

\noindent by 131E again, because
$\int_Y\chi(W\cap(G\times H))(x,y)\nu(dy)=0$ if
$x\in X\setminus G$.   So $W\cap(G\times H)\in\Cal W$.\ \Qed

\medskip

\quad{\bf (iii)} In fact, $W\in\Cal W$ for every $W\in\Lambda$.   \Prf\
Remember that we are supposing that $\nu$ is
$\sigma$-finite.   Let $\sequencen{Y_n}$ be a non-decreasing sequence in
$\Tau^f$ covering $Y$, and for each $n\in\Bbb N$ set
$W_n=W\cap(X\times Y_n)$, $g_n(x)=\int\chi W_n(x,y)\nu(dy)$ whenever
this is defined.   For any $G\in\Sigma^f$,

\Centerline{$\int_Gg_nd\mu
=\iint\chi(W\cap(G\times Y_n))(x,y)\nu(dy)\mu(dx)$}

\noindent is defined and equal to $\lambda(W\cap(G\times Y_n))$, by
(ii).   But this means, first, that $G\setminus\dom g_n$ is negligible,
that is, that $\hat\mu(G\setminus\dom g_n)=0$.   Since this is so
whenever $\mu G$ is finite, $\tilde\mu(X\setminus\dom g_n)=0$, and
$g_n$ is defined
$\tilde\mu$-a.e.;  but $\tilde\mu=\hat\mu$, so $g_n$ is defined
$\hat\mu$-a.e., that is, $\mu$-a.e.\ (212Eb).   Next, if we set
$E_{na}=\{x:x\in\dom g_n,\,g_n(x)\ge a\}$ for $a\in\Bbb R$, then
$E_{na}\cap G\in\hat\Sigma$ whenever $G\in\Sigma^f$, where $\hat\Sigma$
is the domain of $\hat\mu$;  by the definition in 213D, $E_{na}$ is
measured by $\tilde\mu=\hat\mu$.   As $a$ is arbitrary, $g_n$ is
$\mu$-virtually measurable (212Fa).

We can therefore speak of $\int g_nd\mu$.   Now

$$\eqalignno{\iint\chi W_n(x,y)\nu(dy)\mu(dx)
&=\int g_nd\mu
=\sup_{G\in\Sigma^f}\int_Gg_n\cr
\displaycause{213B, because $\mu$ is semi-finite}
&=\sup_{G\in\Sigma^f}\lambda(W\cap(G\times Y_n))
=\lambda(W\cap(X\times Y_n))\cr}$$

\noindent by the definition in 251F.   Thus
$W\cap(X\times Y_n)\in\Cal W$.

This is true for every $n\in\Bbb N$.   Because $\sequencen{Y_n}\uparrow
Y$, $W\in\Cal W$, by (a-iii-$\beta$).\ \Qed

\medskip

\quad{\bf (iv)} We can therefore return to part (c) of the argument
above and conclude as before.
}%end of proof of 252B

\leader{252C}{}\cmmnt{ The theorem above is of course asymmetric, in
that different hypotheses are imposed on the two factor measures $\mu$
and $\nu$.   If we want a `symmetric' theorem we have to suppose that
they are both $\sigma$-finite, as follows.

\medskip

\noindent}{\bf Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two
$\sigma$-finite measure spaces, and $\lambda$ the c.l.d.\ product
measure on $X\times Y$.   If $f$ is $\lambda$-integrable, then
$\iint f(x,y)\nu(dy)\mu(dx)$ and $\iint f(x,y)\mu(dx)\nu(dy)$ are defined, finite and equal.

\proof{ Since $\mu$ and $\nu$ are surely strictly localizable (211Lc),
we can apply 252B from either side to conclude that

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)
=\int f\,d\lambda
=\iint f(x,y)\mu(dx)\nu(dy)$.}
}%end of proof of 252C

\leader{252D}{}\cmmnt{ So many applications of Fubini's theorem are to
indicator functions that I take a few lines to spell out the form
which 252B takes in this case, as in parts (b)-(b*) of the proof there.

\medskip

\noindent}{\bf Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces and $\lambda$ the c.l.d.\ product measure on
$X\times Y$.   Suppose that $\nu$ is $\sigma$-finite and that $\mu$ is
either strictly localizable or complete and locally determined.

(i) If $W\in\dom\lambda$, then $\int\nu^*W[\{x\}]\mu(dx)$ is defined in
$[0,\infty]$ and equal to $\lambda W$.

(ii) If $\nu$ is complete, we can write $\int\nu W[\{x\}]\mu(dx)$ in
place of $\int\nu^*W[\{x\}]\mu(dx)$.

\proof{ The point is just that $\int\chi W(x,y)\nu(dy)=\hat\nu W[\{x\}]$
whenever either is defined, where $\hat\nu$ is the completion of $\nu$
(212F).   Now 252B tells us that

\Centerline{$\lambda W=\iint\chi W(x,y)\nu(dy)\mu(dx)
=\int\hat\nu W[\{x\}]\mu(dx)$.}

\noindent We always have $\hat\nu W[\{x\}]=\nu^*W[\{x\}]$, by the
definition of $\hat\nu$ (212C);  and if $\nu$ is complete, then
$\hat\nu=\nu$ so $\lambda W=\int\nu W[\{x\}]\mu(dx)$.
}%end of proof of 252D

\leader{252E}{Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, with c.l.d.\ product $(X\times Y,\Lambda,\lambda)$.
Suppose that $\nu$ is $\sigma$-finite and that $\mu$ has locally
determined negligible sets\cmmnt{ (213I)}.   Then if $f$ is a
$\Lambda$-measurable real-valued function defined on a subset of
$X\times Y$, $y\mapsto f(x,y)$ is
$\nu$-virtually measurable for $\mu$-almost every $x\in X$.

\proof{ Let $\tilde f$ be a $\Lambda$-measurable extension of $f$ to a
real-valued function defined everywhere in $X\times Y$ (121I), and set
$\tilde f_x(y)=\tilde f(x,y)$ for all $x\in X$, $y\in Y$,

\Centerline{$D=\{x:x\in X,\,\tilde f_x$ is $\nu$-virtually
measurable$\}$.}

If $G\in\Sigma$ and $\mu G<\infty$, then $G\setminus D$ is negligible.
\Prf\ Let $\sequencen{Y_n}$ be a non-decreasing sequence of sets
of finite measure covering $Y$ respectively, and set

$$\eqalign{\tilde f_n(x,y)
&=\tilde f(x,y)\text{ if }x\in G,\,y\in Y_n
  \text{ and }|\tilde f(x,y)|\le n,\cr
&=0\text{ for other }x\in X\times Y.\cr}$$

\noindent Then each $\tilde f_n$ is
$\lambda$-integrable, being bounded and $\Lambda$-measurable and zero
off $G\times Y_n$.
Consequently, setting $\tilde f_{nx}(y)=\tilde f_n(x,y)$,

\Centerline{$\int(\int\tilde f_{nx}d\nu)\mu(dx)$ exists
$=\int\tilde f_nd\lambda$.}

\noindent But this surely means that $\tilde f_{nx}$ is
$\nu$-integrable, therefore $\nu$-virtually measurable, for almost
every $x\in X$.   Set

\Centerline{$D_n=\{x:x\in X,\,\tilde f_{nx}$ is $\nu$-virtually
measurable$\}$;}

\noindent then every $D_n$ is $\mu$-conegligible, so
$\bigcap_{n\in\Bbb N}D_n$ is conegligible.   But for any
$x\in G\cap\bigcap_{n\in\Bbb N}D_n$,
$\tilde f_x=\lim_{n\to\infty}\tilde f_{nx}$ is $\nu$-virtually
measurable.   Thus
$G\setminus D\subseteq X\setminus\bigcap_{n\in\Bbb N}D_n$ is
negligible.\ \Qed

This is true whenever $\mu G<\infty$.   As $G$ is arbitrary and $\mu$ has
locally determined negligible sets, $D$ is conegligible.   But, for any
$x\in D$, $y\mapsto f(x,y)$ is a restriction of $\tilde f_x$ and must be
$\nu$-virtually measurable.
}%end of proof of 252E

\leader{252F}{}\cmmnt{ As a further corollary we can get some useful
information about the c.l.d.\ product measure for arbitrary measure
spaces.

\medskip

\noindent}{\bf Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be two
measure spaces, $\lambda$ the c.l.d.\ product measure on
$X\times Y$, and $\Lambda$ its domain.   Let $W\in\Lambda$ be such
that the vertical section $W[\{x\}]$ is $\nu$-negligible
for $\mu$-almost every $x\in X$.   Then $\lambda W=0$.

\proof{ Take $E\in\Sigma$, $F\in\Tau$ of finite measure.   Let
$\lambda_{E\times F}$ be the subspace measure on $E\times F$.   By
251Q(ii-$\alpha$) again,
this is just the product of the subspace measures $\mu_E$ and $\nu_F$.
We know that $W\cap(E\times F)$ is measured by $\lambda_{E\times F}$.
At the same time, the vertical section $(W\cap(E\times
F))[\{x\}]=W[\{x\}]\cap F$ is $\nu_F$-negligible for $\mu_E$-almost
every $x\in E$.   Applying 252B to $\mu_E$ and $\nu_F$ and
$\chi(W\cap(E\times F))$,

\Centerline{$\lambda(W\cap(E\times F))
=\lambda_{E\times F}(W\cap(E\times F))
=\int_E\nu_F(W[\{x\}]\cap F)\mu_E(dx)=0$.}

\noindent But looking at the definition in 251F, we see that this means
that $\lambda W=0$, as claimed.
}%end of proof of 252F

\leader{252G}{}\cmmnt{ Theorem 252B and its corollaries depend on the
factor measures $\mu$ and $\nu$ belonging to restricted classes.   There
is a partial result which applies to all c.l.d.\ product measures, as
follows.

\medskip

\noindent}{\bf Tonelli's theorem} Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be
measure spaces, and $(X\times Y,\Lambda,\lambda)$ their c.l.d.\ product.
Let $f$ be a $\Lambda$-measurable $[-\infty,\infty]$-valued function
defined on a member of $\Lambda$, and suppose that either
$\iint|f(x,y)|\mu(dx)\nu(dy)$ or $\iint|f(x,y)|\nu(dy)\mu(dx)$ exists in
$\Bbb R$.   Then $f$ is $\lambda$-integrable.

\proof{ Because the construction of the product measure is symmetric in
the two factors, it is enough to consider the case in which $\iint
|f(x,y)|\nu(dy)\mu(dx)$ is defined and finite, as the same ideas will
surely deal with the other case also.

\medskip

{\bf (a)} The first step is to check that $f$ is defined and finite
$\lambda$-a.e.   \Prf\ Set $W=\{(x,y):(x,y)\in\dom f,\,f(x,y)$ is
finite$\}$.   Then $W\in\Lambda$.   The hypothesis

\Centerline{$\iint|f(x,y)|\nu(dy)\mu(dx)$ is defined and finite}

\noindent includes the assertion

\Centerline{$\int|f(x,y)|\nu(dy)$ is defined and finite for
$\mu$-almost every $x$,}

\noindent which implies that

\Centerline{for $\mu$-almost every $x$, $f(x,y)$ is defined and finite
for $\nu$-almost every $y$;}

\noindent that is, that

\Centerline{for $\mu$-almost every $x$, $W[\{x\}]$ is
$\nu$-conegligible.}

\noindent But by 252F this implies
that $(X\times Y)\setminus W$ is $\lambda$-negligible, as required.\
\Qed

\medskip

{\bf (b)} Let $h$ be any non-negative $\lambda$-simple
function such that $h\le|f|\,\,\lambda$-a.e.   Then
\ifdim\pagewidth>467pt$\int h$ cannot be greater than
$\iint|f(x,y)|\nu(dy)\mu(dx)$.
\else $\int h$ is at most $\iint|f(x,y)|\nu(dy)\mu(dx)$.\fi
\Prf\ Set

\Centerline{$W=\{(x,y):(x,y)\in\dom f,\,h(x,y)\le|f(x,y)|\}$,
\quad$h'=h\times\chi W$;}

\noindent then $h'$ is a simple function and $h'\eae h$.   Express $h'$
as $\sum_{i=0}^na_i\chi W_i$ where $a_i\ge 0$ and $\lambda W_i<\infty$
for each $i$.
Let $\epsilon>0$.   For each $i\le n$ there are $E_i\in\Sigma$,
$F_i\in\Tau$ such that $\mu E_i<\infty$, $\nu F_i<\infty$ and
$\lambda(W_i\cap(E_i\times F_i))\ge\lambda W_i-\epsilon$.   Set
$E=\bigcup_{i\le n}E_i$ and $F=\bigcup_{i\le n}F_i$.   Consider the
subspace measures $\mu_E$ and $\nu_F$ and their product
$\lambda_{E\times F}$ on $E\times F$;  then $\lambda_{E\times F}$ is the
subspace measure on $E\times F$ defined from $\lambda$ (251Q(ii-$\alpha$)
once more).
Accordingly, applying 252B to the product $\mu_E\times\nu_F$,

\Centerline{$\int_{E\times F}h'\,d\lambda
=\int_{E\times F}h'\,d\lambda_{E\times F}
=\int_E\int_Fh'(x,y)\nu_F(dy)\mu_E(dx)$.}

For any $x$, we know that $h'(x,y)\le|f(x,y)|$ whenever $f(x,y)$ is
defined.   So we can be sure that

\Centerline{$\int_Fh'(x,y)\nu_F(dy)
=\int h'(x,y)\chi F(y)\nu(dy)
\le\int|f(x,y)|\nu(dy)$}

\noindent at least whenever $\int_Fh'(x,y)\nu_F(dy)$ and
$\int|f(x,y)|\nu(dy)$ are both defined, which is the case for almost
every $x\in E$.   Consequently

$$\eqalign{\int_{E\times F}h'\,d\lambda
&=\int_E\int_Fh'(x,y)\nu_F(dy)\mu_E(dx)\cr
&\le\int_E\int|f(x,y)|\nu(dy)\mu(dx)
\le\iint|f(x,y)|\nu(dy)\mu(dx).\cr}$$

On the other hand,

$$\eqalign{\int h'\,d\lambda-\int_{E\times F}h'\,d\lambda
&=\sum_{i=0}^na_i\lambda(W_i\setminus(E\times F))\cr
&\le\sum_{i=0}^na_i\lambda(W_i\setminus(E_i\times F_i))
\le\epsilon\sum_{i=0}^na_i.\cr}$$

\noindent So

\Centerline{$\int h\,d\lambda
=\int h'd\lambda
\le\iint|f(x,y)|\nu(dy)\mu(dx)$$+\epsilon\sum_{i=0}^na_i$.}

\noindent As $\epsilon$ is arbitrary,
$\int h\,d\lambda
\le\iint|f(x,y)|\nu(dy)\mu(dx)$, as claimed.\ \Qed

\medskip

{\bf (c)} This is true whenever $h$ is a $\lambda$-simple function
less than or equal to $|f|\,\,\lambda$-a.e.   But $|f|$ is
$\Lambda$-measurable and $\lambda$ is semi-finite (251Ic), so this is
enough to ensure that $|f|$ is $\lambda$-integrable (213B), which
(because $f$ is supposed to be $\Lambda$-measurable) in
turn implies that $f$ is $\lambda$-integrable.
}%end of proof of 252G

\leader{252H}{\bf Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces, $\lambda$ the c.l.d.\ product measure on
$X\times Y$, and $\Lambda$ its domain.

(a) Let $f$ be a
$\Lambda$-measurable $[-\infty,\infty]$-valued function defined on a member of
$\Lambda$.   Then if one of

\Centerline{$\int_{X\times Y}|f(x,y)|\lambda(d(x,y))$,
\quad$\int_Y\int_X|f(x,y)|\mu(dx)\nu(dy)$,
\quad$\int_X\int_Y|f(x,y)|\nu(dy)\mu(dx)$}

\noindent exists in $\Bbb R$, so do the other two, and in this case

\Centerline{$\int_{X\times Y}f(x,y)\lambda(d(x,y))
=\int_Y\int_Xf(x,y)\mu(dx)\nu(dy)
=\int_X\int_Yf(x,y)\nu(dy)\mu(dx)$.}

(b)\dvAnew{2013} Let $f$ be a
$\Lambda$-measurable $[0,\infty]$-valued function defined on a member of
$\Lambda$.   Then

\Centerline{$\int_{X\times Y}f(x,y)\lambda(d(x,y))
=\int_Y\int_Xf(x,y)\mu(dx)\nu(dy)
=\int_X\int_Yf(x,y)\nu(dy)\mu(dx)$}

\noindent in the sense that if one of the integrals is defined in
$[0,\infty]$ so are the other two, and all three are then equal.

\proof{{\bf (a)(i)} Suppose that $\int|f|d\lambda$ is finite.
Because both $\mu$ and $\nu$ are $\sigma$-finite, 252B tells us that

\Centerline{$\iint|f(x,y)|\mu(dx)\nu(dy)$,
\quad$\iint|f(x,y)|\nu(dy)\mu(dx)$}

\noindent both exist and are equal to
$\int|f|d\lambda$, while

\Centerline{$\iint f(x,y)\mu(dx)\nu(dy)$,
\quad$\iint f(x,y)\nu(dy)\mu(dx)$}

\noindent both exist and are equal to $\int fd\lambda$.

\medskip

\quad{\bf (ii)} Now suppose that $\iint|f(x,y)|\nu(dy)\mu(dx)$ exists in
$\Bbb R$.   Then 252G tells us that $|f|$ is
$\lambda$-integrable, so we can
use (i) to complete the argument.   Exchanging the coordinates, the same
argument applies if $\iint|f(x,y)|\mu(dx)\nu(dy)$ exists in $\Bbb R$.

\medskip

{\bf (b)(i)} If $\int fd\lambda$ is defined, the result is immediate from
252B.

\medskip

\quad{\bf (ii)} Suppose that $\iint_Xf(x,y)\nu(dy)\mu(dx)$ is defined.
As in part (a) of the proof of 252G, but this time setting
$W=\dom f$, we see that $W\in\Lambda$ and that $W[\{x\}]$ is
$\nu$-conegligible for $\mu$-almost every $x$, so that $W$ is
$\lambda$-conegligible.   Since $f$ is non-negative, $\Lambda$-measurable
and defined almost everywhere, $\int fd\lambda$ is defined in $[0,\infty]$
and we are in case (i).
}%end of proof of 252H

\leader{252I}{Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, $\lambda$ the c.l.d.\ product measure on $X\times
Y$, and $\Lambda$ its domain.   Take $W\in\Lambda$.   If
either of the integrals

\Centerline{$\int\mu^*W^{-1}[\{y\}]\nu(dy)$,
\quad$\int\nu^*W[\{x\}]\mu(dx)$}

\noindent exists and is finite, then $\lambda W<\infty$.

\proof{ Apply 252G with $f=\chi W$, remembering that

\Centerline{$\mu^*W^{-1}[\{y\}]=\int\chi W(x,y)\mu(dx)$,
\quad$\nu^*W[\{x\}]=\int\chi W(x,y)\nu(dy)$}

\noindent whenever the integrals are defined, as in the proof of 252D.
}%end of proof of 252I

\cmmnt{
\leader{252J}{Remarks} 252H is the basic form of Fubini's theorem;  it
is not a
coincidence that most authors avoid non-$\sigma$-finite spaces in this
context.    The next two examples exhibit some of the
difficulties which can arise if we leave the familiar territory of
more-or-less Borel measurable  functions on $\sigma$-finite spaces.
The first is a classic.
}%end of comment

\leader{252K}{Example} Let $(X,\Sigma,\mu)$ be $[0,1]$ with Lebesgue
measure, and let $(Y,\Tau,\nu)$ be $[0,1]$ with counting measure.
\cmmnt{

\spheader 252Ka} Consider the set

\Centerline{$W=\{(t,t):t\in[0,1]\}\subseteq X\times Y$.}

\cmmnt{
\noindent We observe that $W$ is expressible as

\Centerline{$\bigcap_{n\in\Bbb N}\bigcup_{k=0}^n
[\Bover{k}{n+1},\Bover{k+1}{n+1}]\times[\Bover{k}{n+1},
\Bover{k+1}{n+1}]\in\Sigma\tensorhat\Tau$.}

\noindent If we look at the sections

\Centerline{$W^{-1}[\{t\}]=W[\{t\}]=\{t\}$}

\noindent for $t\in [0,1]$, we have
}

\ifwithproofs
\Centerline{$\iint\chi W(x,y)\mu(dx)\nu(dy)
=\int\mu W^{-1}[\{y\}]\nu(dy)=\int 0\,\nu(dy)=0$,}

\Centerline{$\iint\chi W(x,y)\nu(dy)\mu(dx)
=\int\nu W[\{x\}]\mu(dx)=\int 1\,\mu(dx)=1$,}
\else
\Centerline{$\iint\chi W(x,y)\mu(dx)\nu(dy)
=0$,}

\Centerline{$\iint\chi W(x,y)\nu(dy)\mu(dx)
=1$,}
\fi

\noindent so the two repeated integrals differ.
\cmmnt{It is therefore not
generally possible to reverse the order of repeated integration, even
for a non-negative measurable function in which both repeated integrals
exist and are finite.
}%end of comment

\cmmnt{
\spheader 252Kb Because the set $W$ of part (a) actually belongs
to $\Sigma\tensorhat\Tau$, we know that it is measured by the c.l.d.\
product measure $\lambda$, and 252F (applied with the coordinates
reversed) tells us that $\lambda W=0$.
}%end of comment

\cmmnt{
\spheader 252Kc It is in fact easy to give a
full description of $\lambda$.

\medskip

\quad{\bf (i)} The point is that a set $W\subseteq[0,1]\times[0,1]$
belongs to the domain $\Lambda$ of $\lambda$ iff every horizontal
section $W^{-1}[\{y\}]$ is Lebesgue measurable.   \prooflet{\Prf\
($\alpha$) If $W\in\Lambda$, then, for every $b\in[0,1]$,
$\lambda([0,1]\times\{b\})$ is finite, so $W\cap([0,1]\times\{b\})$ is a
set of finite measure, and

\Centerline{$\lambda(W\cap([0,1]\times\{b\}))
=\int\mu(W\cap([0,1]\times\{b\}))^{-1}[\{y\}]\nu(dy)
=\mu W^{-1}[\{b\}]$}

\noindent by 252D, because $\mu$ is $\sigma$-finite, $\nu$ is both
strictly localizable and complete and locally determined, and

$$\eqalign{(W\cap([0,1]\times\{b\}))^{-1}[\{y\}]
&=W^{-1}[\{b\}]\text{ if }y=b,\cr
&=\emptyset\text{ otherwise.}\cr}$$

\noindent As $b$ is arbitrary, every horizontal section of $W$ is
measurable.   ($\beta$) If every horizontal section of $W$ is
measurable, let $F\subseteq[0,1]$ be any set of finite measure for
$\nu$;  then $F$ is finite, so

\Centerline{$W\cap([0,1]\times F)
=\bigcup_{y\in F}W^{-1}[\{y\}]\times\{y\}
\in\Sigma\tensorhat\Tau\subseteq\Lambda$.}

\noindent But it follows that $W$ itself belongs to $\Lambda$, by
251H.\ \Qed
}%end of prooflet

\medskip

\quad{\bf (ii)} Now some of the same calculations show that for every
$W\in\Lambda$,

\Centerline{$\lambda W=\sum_{y\in [0,1]}\mu W^{-1}[\{y\}]$.}

\prooflet{\noindent\Prf\ For any finite $F\subseteq[0,1]$,

$$\eqalign{\lambda(W\cap([0,1]\times F))
&=\int\mu(W\cap([0,1]\times F))^{-1}[\{y\}]\nu(dy)\cr
&=\int_F\mu W^{-1}[\{y\}]\nu(dy)
=\sum_{y\in F}\mu W^{-1}[\{y\}].\cr}$$

\noindent So

\Centerline{$\lambda W=\sup_{F\subseteq[0,1]\text{ is finite}}
\sum_{y\in F}\mu W^{-1}[\{y\}]=\sum_{y\in [0,1]}\mu W^{-1}[\{y\}]$.
\Qed}
}%end of prooflet
}%end of comment

\cmmnt{
\leader{252L}{Example} For the second example, I turn to a problem that
can arise if we neglect to check that a function is measurable as a
function of two variables.

Let $(X,\Sigma,\mu)=(Y,\Tau,\nu)$ be $\omega_1$, the first uncountable
ordinal (2A1Fc), with the countable-cocountable measure (211R).   Set

\Centerline{$W=\{(\xi,\eta):\xi\le\eta<\omega_1\}\subseteq X\times Y$.}

\noindent Then all the horizontal sections
$W^{-1}[\{\eta\}]=\{\xi:\xi\le\eta\}$ are countable, so

\Centerline{$\int\mu W^{-1}[\{\eta\}]\nu(d\eta)
=\int 0\,\nu(d\eta)
=0$,}

\noindent while all the vertical sections
$W[\{\xi\}]=\{\eta:\xi\le\eta<\omega_1\}$ are cocountable, so

\Centerline{$\int\nu W[\{\xi\}]\mu(d\xi)
=\int 1\,\mu(d\xi)
=1$.}

\noindent Because the two repeated integrals are different, they cannot
both be equal to the measure of $W$, and the sole resolution is to say
that $W$ is not measured by the product measure.
}%end of comment

\cmmnt{
\leader{252M}{Remark} A third kind of difficulty in the formula

\Centerline{$\iint f(x,y)dxdy=\iint f(x,y)dydx$}

\noindent can arise even on probability spaces with
$\Sigma\tensorhat\Tau$-measurable real-valued functions defined
everywhere if we neglect to check that $f$ is integrable with respect to
the product measure.   In 252H, we do need the hypothesis that one of

\Centerline{$\int_{X\times Y}|f(x,y)|\lambda(d(x,y))$,
\quad$\int_Y\int_X|f(x,y)|\mu(dx)\nu(dy)$,
\quad$\int_X\int_Y|f(x,y)|\nu(dy)\mu(dx)$}

\noindent is finite.   For examples to show this, see 252Xf and 252Xg.
}%end of comment

\leader{252N}{Integration through ordinate sets I:  Proposition} Let
$(X,\Sigma,\mu)$ be a complete locally determined measure space, and
$\lambda$ the c.l.d.\ product measure on $X\times\Bbb R$, where $\Bbb R$
is given Lebesgue measure;  write $\Lambda$ for the domain of $\lambda$.
For any
$[0,\infty]$-valued function $f$ defined on a conegligible subset of
$X$, write $\Omega_f$, $\Omega'_f$ for the {\bf ordinate sets}

\Centerline{$\Omega_f=\{(x,a):x\in\dom f,\,0\le a\le f(x)\}
\subseteq X\times\Bbb R$,}

\Centerline{$\Omega'_f=\{(x,a):x\in\dom f,\,0\le a<f(x)\}
\subseteq X\times\Bbb R$.}

\noindent Then

\Centerline{$\lambda\Omega_f=\lambda\Omega'_f=\int fd\mu$}

\noindent in the sense that if one of these is defined in $[0,\infty]$,
so are the other two, and they are equal.

\proof{{\bf (a)} If $\Omega_f\in\Lambda$, then

\Centerline{$\int f(x)\mu(dx)
=\int\nu\{y:(x,y)\in\Omega_f\}\mu(dx)
=\lambda\Omega_f$}

\noindent by 252D, writing $\mu$ for Lebesgue measure, because $f$ is
defined almost everywhere.   Similarly, if $\Omega'_f\in\Lambda$,

\Centerline{$\int f(x)\mu(dx)
=\int\nu\{y:(x,y)\in\Omega'_f\}\mu(dx)
=\lambda\Omega'_f$.}

\medskip

{\bf (b)} If $\int fd\mu$ is defined, then $f$ is $\mu$-virtually
measurable, therefore measurable (because $\mu$ is complete);
again because $\mu$ is complete, $\dom f\in\Sigma$.   So

\Centerline{$\Omega'_f
=\bigcup_{q\in\Bbb Q,q>0}\{x:x\in\dom f,\,f(x)>q\}\times[0,q]$,}

\Centerline{$\Omega_f
=\bigcap_{n\ge 1}\bigcup_{q\in\Bbb Q,q>0}
  \{x:x\in\dom f,\,f(x)\ge q-\Bover1n\}\times[0,q]$}

\noindent belong to $\Lambda$, so that $\lambda\Omega_f$ and
$\lambda\Omega'_f$ are defined.   Now both are equal to
$\int fd\mu$, by (a).
}%end of proof of 252N

\leader{252O}{Integration through ordinate sets II:  Proposition} Let
$(X,\Sigma,\mu)$ be a measure space, and $f$ a non-negative $\mu$-virtually
measurable function defined on a conegligible subset of $X$.   Then

\Centerline{$\int fd\mu
=\int_0^{\infty}\mu^*\{x:x\in\dom f,\,f(x)\ge t\}dt
=\int_0^{\infty}\mu^*\{x:x\in\dom f,\,f(x)>t\}dt$}

\noindent in $[0,\infty]$, where the integrals $\int\ldots dt$ are taken
with respect to Lebesgue measure.

\proof{ Completing $\mu$ does not change the integral of $f$ or the outer
measure $\mu^*$ (212Fb, 212Ea), so we may suppose that $\mu$ is complete,
in which case $\dom f$ and $f$ will be measurable.
For $n$, $k\in\Bbb N$ set $E_{nk}=\{x:x\in\dom f,\,f(x)>2^{-n}k\}$,
$g_n(x)=2^{-n}\sum_{k=1}^{4^n}\chi E_{nk}$.   Then $\sequencen{g_n}$ is
a non-decreasing sequence of measurable functions converging to $f$ at
every point of $\dom f$, so $\int fd\mu=\lim_{n\to\infty}\int g_nd\mu$
and $\mu\{x:f(x)>t\}=\lim_{n\to\infty}\mu\{x:g_n(x)>t\}$ for every
$t\ge 0$;  consequently

\Centerline{$\int_0^{\infty}\mu\{x:f(x)>t\}dt
=\lim_{n\to\infty}\int_0^{\infty}\mu\{x:g_n(x)>t\}dt$.}

\noindent On the other hand,
$\mu\{x:g_n(x)>t\}=\mu E_{nk}$ if $1\le k\le 4^n$ and
$2^{-n}(k-1)\le t<2^{-n}k$, $0$ if $t\ge 2^n$, so that

\Centerline{$\int_0^{\infty}\mu\{x:g_n(x)>t\}dt$
$=\sum_{k=1}^{4^n}2^{-n}\mu E_{nk}$
$=\int g_n\,d\mu$,}

\noindent for every $n\in\Bbb N$.   So
$\int_0^{\infty}\mu\{x:f(x)>t\}dt=\int fd\mu$.

Now $\mu\{x:f(x)\ge t\}=\mu\{x:f(x)>t\}$ for almost all $t$.
\Prf\ Set $C=\{t:\mu\{x:f(x)>t\}<\infty\}$, $h(t)=\mu\{x:f(x)>t\}$ for
$t\in C$.   If $C$ is not empty,
$h:C\to\coint{0,\infty}$ is monotonic, so is continuous
almost everywhere in $C$ (222A).   But at any point of
$C\setminus\{\inf C\}$ at which $h$ is continuous,

\Centerline{$\mu\{x:f(x)\ge t\}=\lim_{s\uparrow t}\mu\{x:f(x)>s\}
=\mu\{x:f(x)>t\}$.}

\noindent So we have the result, since
$\mu\{x:f(x)\ge t\}=\mu\{x:f(x)>t\}=\infty$ for any
$t\in\coint{0,\infty}\setminus C$.\ \Qed

Accordingly $\int_0^{\infty}\mu\{x:f(x)\ge t\}dt$ is also equal to
$\int fd\mu$.
}%end of proof of 252O

\leader{*252P}{}\cmmnt{ If we work through the ideas of 252B for
$\Sigma\tensorhat\Tau$-measurable functions, we get the following, which
is sometimes useful.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be a measure space, and
$(Y,\Tau,\nu)$ a $\sigma$-finite measure space.   Then for any
$\Sigma\tensorhat\Tau$-measurable function
$f:X\times Y\to[0,\infty]$,
$x\mapsto\int f(x,y)\nu(dy):X\to[0,\infty]$ is $\Sigma$-measurable.
If $\mu$ is semi-finite,
$\iint f(x,y)\nu(dy)\mu(dx)\discretionary{}{}{}=\int fd\lambda$,
where $\lambda$ is the c.l.d.\ product measure on $X\times Y$.

\proof{{\bf (a)} Let $\sequencen{Y_n}$ be a non-decreasing sequence of
subsets of $Y$ of finite measure with union $Y$.   Set

$$\eqalign{\Cal A
&=\{W:W\subseteq X\times Y,\,W[\{x\}]\in\Tau\text{ for every }x\in X,\cr
&\qquad\qquad\qquad\qquad
x\mapsto\nu(Y_n\cap W[\{x\}])\text{ is }\Sigma\text{-measurable for
every }n\in\Bbb N\}.\cr}$$

\noindent Then $\Cal A$ is a Dynkin class of subsets of $X\times Y$
including $\{E\times F:E\in\Sigma,\,F\in\Tau\}$, so includes
$\Sigma\tensorhat\Tau$, by the Monotone Class Theorem again (136B).

This means that if $W\in\Sigma\tensorhat\Tau$, then

\Centerline{$\mu W[\{x\}]=\sup_{n\in\Bbb N}\nu(Y_n\cap W[\{x\}])$}

\noindent is defined for every $x\in X$ and is a $\Sigma$-measurable
function of $x$.

\medskip

{\bf (b)} Now, for $n$, $k\in\Bbb N$, set

\Centerline{$W_{nk}=\{(x,y):f(x,y)\ge 2^{-n}k\}$,
\quad$g_n=\sum_{k=1}^{4^n}2^{-n}\chi W_{nk}$.}

\noindent Then if we set

\Centerline{$h_n(x)
=\int g_n(x,y)\nu(dy)$$=\sum_{k=1}^{4^n}2^{-n}\nu W_{nk}[\{x\}]$}

\noindent for $n\in\Bbb N$ and $x\in X$, $h_n:X\to[0,\infty]$ is
$\Sigma$-measurable, and

\Centerline{$\lim_{n\to\infty}h_n(x)
=\int(\lim_{n\to\infty}g_n(x,y))\nu(dy)=\int f(x,y)\nu(dy)$}

\noindent for every $x$, because $\sequencen{g_n(x,y)}$ is a
non-decreasing sequence with limit $f(x,y)$ for all $x\in X$, $y\in Y$.
So $x\mapsto\int f(x,y)\nu(dy)$ is defined everywhere in $X$ and is
$\Sigma$-measurable.

\medskip

{\bf (c)} If $E\subseteq X$ is measurable and has finite measure, then
$\int_E\int f(x,y)\nu(dy)\mu(dx)=\int_{E\times Y}fd\lambda$, applying
252B to the product of the subspace measure $\mu_E$ and $\nu$ (and using
251Q to check that the product of $\mu_E$ and $\nu$ is the subspace
measure on $E\times Y$).   Now if $\lambda W$ is defined and finite,
there must be a non-decreasing sequence $\sequencen{E_n}$ of subsets of
$X$ of finite measure such that
$\lambda W=\sup_{n\in\Bbb N}\lambda(W\cap(E_n\times Y))$, so that
$W\setminus\bigcup_{n\in\Bbb N}(E_n\times Y)$ is negligible, and

$$\eqalignno{\int_Wfd\lambda
&=\lim_{n\to\infty}\int_{W\cap(E_n\times Y)}fd\lambda\cr
\displaycause{by B.Levi's theorem applied to
$\sequencen{f\times\chi(W\cap(E_n\times Y))}$}
&\le\lim_{n\to\infty}\int_{E_n\times Y}fd\lambda
=\lim_{n\to\infty}\int_{E_n}\int f(x,y)\nu(dy)\mu(dx)\cr
&\le\iint f(x,y)\nu(dy)\mu(dx).\cr}$$

\noindent By 213B once more,

\Centerline{$\int fd\lambda
=\sup_{\lambda W<\infty}\int_Wfd\lambda
\le\iint f(x,y)\nu(dy)\mu(dx)$.}

\noindent But also, if $\mu$ is semi-finite,

\Centerline{$\iint f(x,y)\nu(dy)\mu(dx)
=\sup_{\mu E<\infty}\int_E\int f(x,y)\nu(dy)\mu(dx)
\le\int fd\lambda$,}

\noindent so $\int fd\lambda=\iint f(x,y)\nu(dy)\mu(dx)$, as claimed.
}%end of proof of 252P

\leader{252Q}{The volume of a ball}\dvro{ The Lebesgue measure of the
unit ball in $\BbbR^r$ is}
{ We now have all the essential
machinery to perform a little calculation which is, I suppose, desirable
simply as general knowledge:  the volume of the unit ball
$\{x:\|x\|\le 1\}=\{(\xi_1,\ldots,\xi_r):\sum_{i=1}^r\xi_i^2\le 1\}$ in
$\BbbR^r$.   In fact, from a theoretical point
of view, I think we could very nearly just call it $\beta_r$ and leave
it at that;  but since there is a general formula in terms of
$\beta_2=\pi$ and factorials, it seems shameful not to present it.   The
calculation has
nothing to do with Lebesgue integration, and I could dismiss it as mere
advanced calculus;  but since only a minority of mathematicians are now
taught calculus to this level with reasonable rigour before being
introduced to the Lebesgue integral, I do not doubt that many readers,
like myself, missed some of the subtleties involved.   I
therefore take the space to spell the details out in the style used
elsewhere in this volume, recognising that the machinery employed is a
good deal more elaborate than is really necessary for this result.

\header{252Qa}{\bf (a)} The first basic fact we need is that, for any
$n\ge 1$,

$$\eqalign{I_n=\int_{-\pi/2}^{\pi/2}\cos^nt\,dt
&=\Bover{(2k)!}{(2^kk!)^2}\pi\text{ if }n=2k\text{ is even},\cr
&=2\Bover{(2^kk!)^2}{(2k+1)!}\text{ if }n=2k+1\text{ is odd}.\cr}$$

\prooflet{\noindent \Prf\  For $n=0$, of course,

\Centerline{$I_0=\int_{-\pi/2}^{\pi/2}1\,dt=\pi
=\Bover{0!}{(2^00!)^2}\pi$,}

\noindent while for $n=1$ we have

\Centerline{$I_1=\sin\bover{\pi}2-\sin(-\bover{\pi}2)=2
=2\Bover{(2^00!)^2}{1!}$,}

\noindent using the Fundamental Theorem of Calculus (225L) and the fact
that $\sin'=\cos$ is bounded.   For the inductive step to $n+1\ge 2$, we
can use integration by parts (225F):

$$\eqalign{I_{n+1}
&=\int_{-\pi/2}^{\pi/2}\cos t\cos^nt\,dt\cr
&=\sin\Bover{\pi}2\cos^n\Bover{\pi}2
   -\sin(-\Bover{\pi}2)\cos^n(-\Bover{\pi}2)
   +\int_{-\pi/2}^{\pi/2}\sin t\cdot n\cos^{n-1}t\cdot\sin t\,dt\cr
&=n\int_{-\pi/2}^{\pi/2}(1-\cos^2t)\cos^{n-1}t\,dt
=n(I_{n-1}-I_{n+1}),\cr}$$

\noindent   so that $I_{n+1}=\Bover{n}{n+1}I_{n-1}$.
Now the given formulae follow by an easy induction.\ \Qed}

\spheader 252Qb The next result is that, for any $n\in\Bbb N$
and any $a\ge 0$,

\Centerline{$\int_{-a}^a(a^2-s^2)^{n/2}ds=I_{n+1}a^{n+1}$.}

\prooflet{\noindent\Prf\ Of course this is an integration by
substitution;  but the singularity of the integrand at $s=\pm a$
complicates the issue slightly.   I offer the following argument.   If
$a=0$ the
result is trivial;  take $a>0$.   For $-a\le b\le a$, set
$F(b)=\int_{-a}^b(a^2-s^2)^{n/2}ds$.   Because the integrand is
continuous, $F'(b)$ exists and is equal to $(a^2-b^2)^{n/2}$ for
$-a<b<a$ (222H).   Set $G(t)=F(a\sin t)$;  then $G$ is continuous and

\Centerline{$G'(t)=aF'(a\sin t)\cos t=a^{n+1}\cos^{n+1}t$}

\noindent for
$-\bover{\pi}2<t<\bover{\pi}2$.   Consequently

$$\eqalignno{\int_{-a}^a(a^2-s^2)^{n/2}ds
&=F(a)-F(-a)
=G(\Bover{\pi}2)-G(-\Bover{\pi}2)
=\int_{-\pi/2}^{\pi/2}G'(t)dt\cr
\noalign{\noindent (by 225L, as before)}
&=a^{n+1}I_{n+1},\cr}$$

\noindent as required.\ \Qed}

\spheader 252Qc Now at last we are ready for the balls
$B_r=\{x:x\in\BbbR^r,\,\|x\|\le 1\}$.   Let $\mu_r$ be Lebesgue
measure on $\BbbR^r$, and set $\beta_r=I_1I_2\ldots I_r$ for $r\ge 1$.
I claim that, writing

\Centerline{$B_r(a)=\{x:x\in\BbbR^r,\,\|x\|\le a\}$,}

\noindent we have $\mu_r(B_r(a))=\beta_ra^r$ for every $a\ge 0$.
\prooflet{\Prf\ Induce on $r$.   For $r=1$ we have $\beta_1=2$,
$B_1(a)=[-a,a]$, so the result is trivial.   For the inductive step to
$r+1$, we have

$$\eqalignno{\mu_{r+1}B_{r+1}(a)
&=\int\mu_r\{x:(x,t)\in B_{r+1}(a)\}dt \cr
\noalign{\noindent (putting 251N and 252D together, and using the fact
that $B_{r+1}(a)$ is closed, therefore measurable)}
&=\int_{-a}^a\mu_rB_r(\sqrt{a^2-t^2})dt\cr
\noalign{\noindent (because $(x,t)\in B_{r+1}(a)$ iff $|t|\le a$ and
$\|x\|\le\sqrt{a^2-t^2}$)}
&=\int_{-a}^a\beta_r(a^2-t^2)^{r/2}dt\cr
\noalign{\noindent (by the inductive hypothesis)}
&=\beta_ra^{r+1}I_{r+1}\cr
\noalign{\noindent (by (b) above)}
&=\beta_{r+1}a^{r+1}\cr}$$

\noindent (by the definition of $\beta_{r+1}$).   Thus the induction
continues.\ \Qed}

\spheader 252Qd In particular, the $r$-dimensional Lebesgue
measure of the $r$-dimensional ball $B_r=B_r(1)$ is just
$\beta_r=I_1\ldots I_r$.   Now an easy induction on $k$ shows that
}%end of \dvro

$$\eqalign{\beta_r
&=\Bover{1}{k!}\pi^k\text{ if }r=2k\text{ is even},\cr
&=\Bover{2^{2k+1}k!}{(2k+1)!}\pi^k\text{ if }r=2k+1\text{ is odd}.\cr}$$

\cmmnt{
\spheader 252Qe Note that in part (c) of the proof we saw that
$\{x:x\in\BbbR^r,\,\|x\|\le a\}$ has measure $\beta_ra^r$ for every
$a\ge 0$.

The formulae here are consistent with the assignation $\beta_0=1$;
which corresponds to saying that $\BbbR^0=\{\emptyset\}$, that
$\mu_0\BbbR^0=1$ and that $B_0=\{\emptyset\}$.   Taking $\mu_0\BbbR^0$ to
be $1$ is itself consistent with the idea that, following 251N, the product
measure $\mu_0\times\mu_r$ ought to match $\mu_{0+r}$ on $\BbbR^{0+r}$.
}%end of comment


\leader{252R}{Complex-valued functions}\cmmnt{ It is easy to apply the
results of 252B-252I
above to complex-valued functions, by considering their real and
imaginary parts.   Specifically:

\medskip

} {\bf (a)} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\lambda$ the c.l.d.\ product measure on $X\times
Y$.   Suppose that $\nu$ is $\sigma$-finite and that $\mu$ is either
strictly localizable or complete and locally determined.   Let $f$ be a
$\lambda$-integrable
complex-valued function.  Then $\iint f(x,y)\nu(dy)\mu(dx)$ is defined
and equal to $\int fd\lambda$.

\spheader 252Rb Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$
be measure spaces, $\lambda$ the c.l.d.\ product measure on $X\times
Y$, and $\Lambda$ its domain.
Let $f$ be a $\Lambda$-measurable complex-valued function defined on a
member of $\Lambda$, and suppose that either
$\iint|f(x,y)|\mu(dx)\nu(dy)$ or $\iint|f(x,y)|\nu(dy)\mu(dx)$ is
defined and finite.   Then $f$ is $\lambda$-integrable.

\spheader 252Rc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces, $\lambda$ the c.l.d.\ product
measure on $X\times Y$, and $\Lambda$ its domain.   Let $f$ be a
$\Lambda$-measurable complex-valued function defined on a member of
$\Lambda$.   Then if one of

\Centerline{$\int_{X\times Y}|f(x,y)|\lambda(d(x,y))$,
\quad$\int_Y\int_X|f(x,y)|\mu(dx)\nu(dy)$,
\quad$\int_X\int_Y|f(x,y)|\nu(dy)\mu(dx)$}

\noindent exists in $\Bbb R$, so do the other two, and in this case

\Centerline{$\int_{X\times Y}f(x,y)\lambda(d(x,y))
=\int_Y\int_Xf(x,y)\mu(dx)\nu(dy)
=\int_X\int_Yf(x,y)\nu(dy)\mu(dx)$.}

\exercises{
\leader{252X}{Basic exercises (a)}
%\spheader 252Xa
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and
$\lambda$ the c.l.d.\ product measure on $X\times Y$.   Let $f$ be a
$\lambda$-integrable real-valued function such that
$\int_{E\times F}f=0$ whenever $E\in\Sigma$, $F\in\Tau$, $\mu E<\infty$
and $\nu F<\infty$.   Show that $f=0\,\,\lambda$-a.e.   \Hint{use 251Ie
to show that $\int_Wf=0$ whenever $\lambda W<\infty$.}
%252B

\spheader 252Xb Let $f$, $g:\Bbb R\to\Bbb R$ be two
non-decreasing functions, and $\mu_f$, $\mu_g$ the associated
Lebesgue-Stieltjes measures (see 114Xa).   Set

\Centerline{$f(x^+)=\lim_{t\downarrow x}f(t)$,
\quad$f(x^-)=\lim_{t\uparrow x}f(t)$}

\noindent for each $x\in\Bbb R$, and define $g(x^+)$, $g(x^-)$
similarly.   Show that whenever $a\le b$ in $\Bbb R$,

$$\eqalign{\int_{[a,b]}f(x^-)\mu_g(dx)&+\int_{[a,b]}g(x^+)\mu_f(dx)
=g(b^+)f(b^+)-g(a^-)f(a^-)\cr
&=\int_{[a,b]}\Bover12(f(x^-)+f(x^+))\mu_g(dx)
  +\int_{[a,b]}\Bover12((g(x^-)+g(x^+))\mu_f(dx).\cr}$$

\noindent \Hint{find two expressions for
$(\mu_f\times\mu_g)\{(x,y):a\le x<y\le b\}$.}
%252C

\sqheader 252Xc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
complete locally determined measure spaces, $\lambda$ the c.l.d.\
product measure on $X\times Y$, and $\Lambda$ its domain.   Suppose that
$A\subseteq X$ and
$B\subseteq Y$.   Show that $A\times B\in\Lambda$ iff
either $\mu A=0$ or $\nu B=0$ or $A\in\Sigma$ and $B\in\Tau$.
\Hint{if $B$ is not negligible and $A\times B\in\Lambda$, take $H$ such
that $\nu H<\infty$ and $B\cap H$ is not negligible.   Then
$W=A\times(B\cap H)$ is measured by $\mu\times\nu_H$, where $\nu_H$ is
the subspace measure on $H$.   Now apply 252D to $\mu$, $\nu_H$ and $W$
to see that $A\in\Sigma$.}
%252D

\sqheader 252Xd Let $(X_1,\Sigma_1,\mu_1)$,
$(X_2,\Sigma_2,\mu_2)$, $(X_2,\Sigma_3,\mu_3)$ be three $\sigma$-finite
measure spaces, and $f$ a real-valued function defined almost everywhere
on $X_1\times X_2\times X_3$ and $\Lambda$-measurable, where $\Lambda$
is the domain of the product
measure described in 251W or 251Xg.   Show that if
$\iiint |f(x_1,x_2,x_3)|dx_1dx_2dx_3$ is defined in $\Bbb R$, then
$\iiint f(x_1,x_2,x_3)dx_2dx_3dx_1$ and
$\iiint f(x_1,x_2,x_3)dx_3dx_1dx_2$
exist and are equal.
%252H

\spheader 252Xe Give an example of strictly localizable measure spaces
$(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ and a $W\in\Sigma\tensorhat\Tau$ such
that $x\mapsto\nu W[\{x\}]$ is not $\Sigma$-measurable.   \Hint{in
252Kb, try $Y$ a proper subset of $[0,1]$.}
%252K

\sqheader 252Xf Set $f(x,y)=\sin(x-y)$ if $0\le y\le x\le y+2\pi$, $0$
for other $x$, $y\in\BbbR^2$.   Show that $\iint f(x,y)dx\,dy=0$ and
$\iint f(x,y)dy\,dx=2\pi$, taking all integrals with respect to Lebesgue
measure.
%252M

\sqheader 252Xg Set $f(x,y)=\Bover{x^2-y^2}{(x^2+y^2)^2}$ for $x$,
$y\in\ocint{0,1}$.   Show that
$\int_0^1\int_0^1f(x,y)dydx=\Bover{\pi}4$,
\discrversionA{\break}{}$\int_0^1\int_0^1f(x,y)dxdy=-\Bover{\pi}4$.
%252M

\sqheader 252Xh Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $f$ a $\Sigma\tensorhat\Tau$-measurable function
defined on a subset of $X\times Y$.   Show that $y\mapsto f(x,y)$ is
$\Tau$-measurable for every $x\in X$.
%252P

\spheader 252Xi Let $r\ge 1$ be an integer, and write $\beta_r$ for the
Lebesgue measure of the unit ball in $\BbbR^r$.   Set
$g_r(t)=r\beta_rt^{r-1}$ for $t\ge 0$, $\phi(x)=\|x\|$ for
$x\in\BbbR^r$.   (i) Writing $\mu_r$ for Lebesgue measure on
$\Bbb R^r$, show that $\mu_r\phi^{-1}[E]=\int_Er\beta_rt^{r-1}\mu_1(dt)$
for every Lebesgue measurable set $E\subseteq\coint{0,\infty}$.   ({\it
Hint\/}:  start with intervals $E$, noting from 115Xe that
$\mu_r\{x:\|x\|\le a\}=\beta_ra^r$ for $a\ge 0$, and progress to open
sets, negligible sets and general measurable sets.)  (ii) Using 235R,
show that

$$\eqalign{\int e^{-\|x\|^2/2}\mu_r(dx)
&=r\beta_r\int_0^{\infty}t^{r-1}e^{-t^2/2}\mu_1(dt)
=2^{(r-2)/2}r\beta_r\Gamma(\Bover{r}2)\cr
&=2^{r/2}\beta_r\Gamma(1+\Bover{r}2)
=(\sqrt{2}\Gamma(\Bover12))^r\cr}$$

\noindent where $\Gamma$ is the $\Gamma$-function (225Xh).   (iii) Show
that

\Centerline{$2\Gamma(\bover12)^2=2\beta_2\Gamma(2)
=2\beta_2\int_0^{\infty}te^{-t^2/2}dt=2\pi$,}

\noindent and hence that
$\beta_r=\Bover{\pi^{r/2}}{\Gamma(1+\bover{r}2)}$ and
$\int_{-\infty}^{\infty}e^{-t^2/2}dt=\sqrt{2\pi}$.
%252Q

\sqheader 252Xj\dvAnew{2016} Let $(X,\Sigma,\mu)$ be a measure space, and
$f:X\to\coint{0,\infty}$ a function.   Write $\Cal B$ for the Borel
$\sigma$-algebra of $\Bbb R$.   Show that the following are
equiveridical:  ($\alpha$) $f$ is $\Sigma$-measurable;  ($\beta$)
$\{(x,a):x\in X$, $0\le a\le f(x)\}\in\Sigma\tensorhat\Cal B$;
($\gamma$)
$\{(x,a):x\in X$, $0\le a<f(x)\}\in\Sigma\tensorhat\Cal B$.
%252N out of order query

\leader{252Y}{Further exercises (a)}
%\spheader 252Ya
Let $(X,\Sigma,\mu)$ be a measure space.   Show that the
following are equiveridical:  (i) the completion of $\mu$ is locally
determined;  (ii) the completion of $\mu$ coincides with the c.l.d.\
version of $\mu$;  (iii) whenever $(Y,\Tau,\nu)$ is a $\sigma$-finite
measure space and $\lambda$ the c.l.d.\ product measure on $X\times Y$
and $f$ is a function such that $\int fd\lambda$ is defined in
$[-\infty,\infty]$, then $\iint f(x,y)\nu(dy)\mu(dx)$ is defined and
equal to $\int fd\lambda$.
%252B

\spheader 252Yb Let $(X,\Sigma,\mu)$ be a measure space.   Show that the
following are equiveridical:  (i) $\mu$ has locally determined
negligible
sets;  (ii) whenever $(Y,\Tau,\nu)$ is a $\sigma$-finite measure
space and $\lambda$ the c.l.d.\ product measure on $X\times Y$, then
$\iint f(x,y)\nu(dy)\mu(dx)$ is defined and equal to $\int fd\lambda$
for any $\lambda$-integrable function $f$.
%252Ya, 252B

\spheader 252Yc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\lambda_0$
the primitive product measure on $X\times Y$ (251C).   Let $f$ be any
$\lambda_0$-integrable real-valued function.   Show that
$\iint f(x,y)\nu(dy)\mu(dx)=\int fd\lambda_0$.   \Hint{show that there
are sequences $\sequencen{G_n}$, $\sequencen{H_n}$ of sets of finite
measure such that $f(x,y)$ is defined and equal to $0$ for every
$(x,y)\in(X\times Y)\setminus\bigcup_{n\in\Bbb N}G_n\times H_n$.}
%252B

\spheader 252Yd Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be
measure spaces;  let $\lambda_0$ be the primitive product measure on
$X\times Y$, and $\lambda$ the c.l.d.\ product measure.   Show that  if
$f$ is a $\lambda_0$-integrable real-valued function, it is
$\lambda$-integrable, and $\int fd\lambda=\int fd\lambda_0$.
%252B

\spheader 252Ye\dvAformerly{252Yj}
Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $a<b$ in
$\Bbb R$, endowed with Lebesgue measure;  let $\Lambda$ be the domain of
the c.l.d.\ product measure $\lambda$ on $X\times[a,b]$.   Let
$f:X\times\ooint{a,b}\to\Bbb R$ be a $\Lambda$-measurable function such
that $t\mapsto f(x,t):[a,b]\to\Bbb R$ is continuous on $[a,b]$ and
differentiable on $\ooint{a,b}$ for every $x\in X$.   (i) Show that the
partial derivative $\Pd{f}{t}$ with respect to the second variable is
$\Lambda$-measurable.   (ii) Now suppose that $\Pd{f}{t}$ is
$\lambda$-integrable and that $\int f(x,t_0)\mu(dx)$ is defined and
finite for some $t_0\in\ooint{a,b}$.   Show that
$F(t)=\int f(x,t)\mu(dx)$ is defined in $\Bbb R$ for every $t\in[a,b]$,
that $F$ is absolutely continuous, and that
$F'(t)=\int\Pd{f}{t}(x,t)\mu(dx)$
for almost every $t\in\ooint{a,b}$.   ({\it Hint\/}:
$F(c)=F(a)+\int_{X\times[a,c]}\Pd{f}{t}d\lambda$ for every $c\in[a,b]$.)
%252C %225L

\spheader 252Yf Show that
$\Bover{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^1t^{a-1}(1-t)^{b-1}dt$
for all $a$, $b>0$.   ({\it Hint\/}:  show that

\Centerline{$\int_0^{\infty}t^{a-1}\int_t^{\infty}
  e^{-x}(x-t)^{b-1}dxdt
=\int_0^{\infty}e^{-x}\int_0^xt^{a-1}(x-t)^{b-1}dtdx$.)}
%252C

\spheader 252Yg Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces and $\lambda$ the c.l.d.\ product measure
on $X\times Y$.   Suppose that
$f\in\eusm L^0(\lambda)$ and that $1<p<\infty$.   Show that
$(\int|\int f(x,y)dx|^pdy)^{1/p}\le\int(\int|f(x,y)|^pdy)^{1/p}dx$.
({\it Hint\/}:
set $q=\bover{p}{p-1}$ and consider the integral
$\int|f(x,y)g(y)|\lambda(d(x,y))$ for $g\in\eusm L^q(\nu)$, using 244K.)
%252C

\spheader 252Yh Let $\nu$ be Lebesgue measure on $\coint{0,\infty}$;
suppose that $f\in\eusm L^p(\nu)$ where $1<p<\infty$.   Set
$F(y)=\Bover1y\int_0^yf$ for $y>0$.   Show that
$\|F\|_p\le\Bover{p}{p-1}\|f\|_p$.   ({\it Hint\/}:
$F(y)=\int_0^1f(xy)dx$;  use 252Yg with $X=[0,1]$,
$Y=\coint{0,\infty}$.)
%252Yg, 252C

\spheader 252Yi Show that if $p$ is any non-zero (real) polynomial in
$r$ variables, then $\{x:x\in\BbbR^r,\,p(x)=0\}$ is Lebesgue negligible.
%252D

\spheader 252Yj Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, with c.l.d.\ product $(X\times Y,\Lambda,\lambda)$.   Let $f$ be
a non-negative $\Lambda$-measurable real-valued
function defined on a $\lambda$-conegligible set, and suppose that

\Centerline{$
\overline{\intop}\bigl(\overline{\intop}f(x,y)\mu(dx)\bigr)
\nu(dy)$}

\noindent is finite.   Show that $f$ is $\lambda$-integrable.
%252G

\spheader 252Yk Let $(X,\Sigma,\mu)$ be the unit interval $[0,1]$ with
Lebesgue measure, and $(Y,\Tau,\nu)$ the interval with counting measure,
as in 252K;  let $\lambda_0$ be the primitive product measure on
$[0,1]^2$.   (i) Setting $\Delta=\{(t,t):t\in[0,1]\}$, show that
$\lambda_0\Delta=\infty$.   (ii) Show that
$\lambda_0$ is not semi-finite.   (iii) Show that if
$W\in\dom\lambda_0$, then $\lambda_0W=\sum_{y\in[0,1]}\mu W^{-1}[\{y\}]$
if there are a countable set $A\subseteq[0,1]$ and a Lebesgue negligible
set $E\subseteq[0,1]$ such that
$W\subseteq([0,1]\times A)\cup(E\times[0,1])$, $\infty$ otherwise.
%252K

\spheader 252Yl Let $(X,\Sigma,\mu)$ be a measure space, and
$\lambda_0$ the primitive product measure on
$X\times\Bbb R$, where $\Bbb R$ is given Lebesgue measure;  write
$\Lambda$ for its domain.   For any
$[0,\infty]$-valued function $f$ defined on a conegligible subset of
$X$, write $\Omega_f$, $\Omega'_f$ for the corresponding ordinate sets,
as in 252N.   Show that if any of $\lambda_0\Omega_f$,
$\lambda_0\Omega'_f$,
$\int fd\mu$ is defined and finite, so are the others, and all three are
equal.
%252B, 252N

\spheader 252Ym Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space, and $f$ a non-negative function defined on a conegligible
subset of $X$.   Write $\Omega_f$, $\Omega'_f$ for the corresponding
ordinate sets, as in 252N.   Let
$\lambda$ be the c.l.d.\ product measure on $X\times\Bbb R$, where
$\Bbb R$ is given Lebesgue measure.   Show that
$\overline{\int}f\,d\mu=\lambda^*\Omega_f=\lambda^*\Omega'_f$.
%252N, 252B

\spheader 252Yn Let $(X,\Sigma,\mu)$ be a measure space and
$f:X\to\coint{0,\infty}$ a function.   Show that
$\overline{\int}fd\mu=\int_0^{\infty}\mu^*\{x:f(x)\ge t\}dt$.
%252O

\spheader 252Yo Let $(X,\Sigma,\mu)$ be a complete measure space and
write $\eusm M^{0,\infty}$ for the set
$\{f:f\in\eusm L^0(\mu),\,\mu\{x:|f(x)|\ge a\}$ is finite for some
$a\in\coint{0,\infty}\}$.   (i) Show that for each
$f\in\eusm M^{0,\infty}$
there is a non-increasing $f^*:\ooint{0,\infty}\to\Bbb R$ such that
$\mu_L\{t:f^*(t)\ge\alpha\}=\mu\{x:|f(x)|\ge\alpha\}$ for
every $\alpha>0$, writing $\mu_L$ for Lebesgue measure.   (ii) Show that
$\int_E|f|d\mu\le\int_0^{\mu E}f^*d\mu_L$ for every $E\in\Sigma$
(allowing $\infty$).   ({\it Hint\/}:  $(f\times\chi E)^*\le f^*$.)
(iii) Show that $\|f^*\|_p=\|f\|_p$ for every $p\in[1,\infty]$,
$f\in\eusm M^{0,\infty}$.   ({\it
Hint\/}:  $(|f|^p)^*=(f^*)^p$.)   (iv) Show that if $f$,
$g\in\eusm M^{0,\infty}$ then
$\int|f\times g|d\mu\le\int f^*\times g^*d\mu_L$.   ({\it
Hint\/}:  look at simple functions first.)   (v) Show that if $\mu$ is
atomless then $\int_0^af^*d\mu_L=\sup_{E\in\Sigma,\mu E\le a}\int_E|f|$
for every $a\ge 0$.   ({\it Hint\/}:  215D.)   (vi) Show that
$A\subseteq\eusm L^1(\mu)$ is uniformly integrable iff $\{f^*:f\in A\}$
is uniformly integrable in $\eusm L^1(\mu_L)$.   ($f^*$ is called the
{\bf decreasing rearrangement} of $f$.)
%252O

\spheader 252Yp Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, and write $\nu$ for Lebesgue measure on
$[0,1]$.   Show
that the c.l.d.\ product measure $\lambda$ on $X\times[0,1]$ is
localizable iff $\mu$ is localizable.   ({\it Hints\/}: (i) if
$\Cal E\subseteq\Sigma$, show that $F\in\Sigma$ is an essential supremum
for $\Cal E$ in $\Sigma$ iff $F\times[0,1]$ is an essential supremum for
$\{E\times[0,1]:E\in\Cal E\}$ in $\Lambda=\dom\lambda$.   (ii) For
$W\in\Lambda$, $n\in\Bbb N$, $k<2^n$ set

\Centerline{$W_{nk}
=\{x:x\in X,\,\nu^*\{t:(x,t)\in W,\,2^{-n}k\le t\le 2^{-n}(k+1)\}
\ge 2^{-n-1}\}$.}

\noindent Show that if $\Cal W\subseteq\Lambda$ and $F_{nk}$ is an
essential supremum for $\{W_{nk}:W\in\Cal W\}$ in $\Sigma$ for all $n$,
$k$, then

\Centerline{$\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}
\bigcup_{k<2^{m}}F_{mk}\times[2^{-m}k,2^{-m}(k+1)]$}

\noindent is an essential supremum for $\Cal W$ in $\Lambda$.)
%252O

\spheader 252Yq Let $(X,\Sigma,\mu)$ be the space of Example
216D, and give Lebesgue measure to $[0,1]$.   Show that the c.l.d.\
product measure on $X\times [0,1]$ is complete, locally determined,
atomless and not localizable.
%252Yp 252O

\spheader 252Yr Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space and $(Y,\Tau,\nu)$ a semi-finite measure space
with $\nu Y>0$.   Show that if the c.l.d.\ product measure on
$X\times Y$ is strictly localizable, then $\mu$ is strictly localizable.
({\it Hint\/}:  take $F\in\Tau$, $0<\nu F<\infty$.   Let
$\langle W_i\rangle_{i\in I}$ be a decomposition of $X\times Y$.   For
$i\in I$, $n\in\Bbb N$
set $E_{in}=\{x:\nu^*\{y:y\in F,\,(x,y)\in W_i\}\ge 2^{-n}\}$.   Apply
213Yf to $\{E_{in}:i\in I,\,n\in\Bbb N\}$.)
%252Yq 252O

\spheader 252Ys Let $(X,\Sigma,\mu)$ be the space of Example
216E, and give Lebesgue measure to $[0,1]$.   Show that the c.l.d.\
product measure on $X\times[0,1]$ is complete, locally determined,
atomless and localizable, but not strictly localizable.
%252Yr, 252O

\spheader 252Yt Let $(X,\Sigma,\mu)$ be a measure space and $f$ a
$\mu$-integrable complex-valued function.   For
$\alpha\in\ocint{-\pi,\pi}$
set $H_{\alpha}=\{x:x\in\dom f,\,\Real(e^{-i\alpha}f(x))>0\}$.   Show
that $\int_{-\pi}^{\pi}\Real(e^{-i\alpha}\int_{H_{\alpha}}f)d\alpha
\discrversionA{\break}{}=2\int|f|$, and hence that there is some
$\alpha$ such that
$|\int_{H_{\alpha}}f|\ge\Bover1{\pi}\int|f|$.   (Compare 246F.)
%246F, 252R

\spheader 252Yu Set $f(t)=t-\ln(t+1)$ for $t>-1$.   (i) Show that
$\Gamma(a+1)=a^{a+1}e^{-a}\int_{-1}^{\infty}e^{-af(u)}du$ for every
$a>0$.   \Hint{substitute $u=\bover{t}{a}-1$ in 225Xh(iii).}   (ii) Show
that there is a $\delta>0$ such that $f(t)\ge\bover13t^2$ for
$-1\le t\le\delta$.   (iii)
Setting $\alpha=\bover12f(\delta)$, show that (for $a\ge 1$)

\Centerline{$\sqrt{a}\int_{\delta}^{\infty}e^{-af(t)}dt
\le\sqrt{a}e^{-a\alpha}\int_0^{\infty}e^{-f(t)/2}dt\to 0$}

\noindent as $a\to\infty$.   (iv) Set $g_a(t)=e^{-af(t/\sqrt{a})}$ if
$-\sqrt{a}<t\le\delta\sqrt{a}$, $0$ otherwise.   Show that
$g_a(t)\le e^{-t^2/3}$ for all $a$, $t$ and that
$\lim_{a\to\infty}g_a(t)=e^{-t^2/2}$ for all $t$, so that

$$\eqalign{\lim_{a\to\infty}\Bover{e^a\Gamma(a+1)}{a^{a+\bover12}}
&=\lim_{a\to\infty}\sqrt{a}\int_{-1}^{\infty}e^{-af(t)}dt
=\lim_{a\to\infty}\sqrt{a}\int_{-1}^{\delta}e^{-af(t)}dt\cr
&=\lim_{a\to\infty}\int_{-\infty}^{\infty}g_a(t)dt
=\int_{-\infty}^{\infty}e^{-t^2/2}dt=\sqrt{2\pi}.\cr}$$

\noindent (v) Show that $\lim_{n\to\infty}\Bover{n!}{e^{-n}n^n\sqrt{n}}
=\sqrt{2\pi}$.   (This is {\bf Stirling's formula}.)
%252+

\spheader 252Yv Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $f$, $g$ two real-valued, $\mu$-virtually measurable
functions defined almost everywhere in $X$.   (i) Let $\lambda$ be the
c.l.d.\ product of $\mu$ and Lebesgue measure on $\Bbb R$.
Setting $\Omega^*_f=\{(x,a):x\in\dom f$, $a\in\Bbb R$, $a\le f(x)\}$ and
$\Omega^*_g=\{(x,a):x\in\dom g$, $a\in\Bbb R$, $a\le g(x)\}$, show that
$\lambda(\Omega^*_f\setminus\Omega^*_g)=\int(f-g)^+d\mu$ and
$\lambda(\Omega^*\symmdiff\Omega^*_g)=\int|f-g|d\mu$.
(ii) Suppose that $\mu$ is $\sigma$-finite.   Show that

\Centerline{$\int|f-g|d\mu
=\int_{-\infty}^{\infty}
\mu(\{x:x\in\dom f\cap\dom g$, $(f(x)-a)(g(x)-a)<0\}da$.}

\noindent(iii) Suppose that $\mu$ is $\sigma$-finite, that $\Tau$ is a
$\sigma$-subalgebra of $\Sigma$, that $E\in\Sigma$ and that $g:X\to[0,1]$
is $\Tau$-measurable.   Show that there is an $F\in\Tau$ such that
$\mu(E\symmdiff F)\le\int|\chi E-g|d\mu$.
%252N
}%end of exercises

\endnotes{
\Notesheader{252} For a volume and a half now I have asked you to accept
the idea of integrating partially-defined functions, insisting that
sooner or later they would appear at the core of the subject.   The
moment has now come.   If we wish to apply Fubini's and Tonelli's
theorems in the most fundamental of all cases, with both factors equal
to Lebesgue measure on the unit interval, it is surely natural to
look at all functions which are integrable on the square for
two-dimensional Lebesgue measure.   Now two-dimensional Lebesgue measure
is a complete measure, so, in particular, assigns zero measure to any
set of the form $\{(x,b):x\in A\}$ or $\{(a,y):y\in A\}$, whether or not
the set $A$ is measured by one-dimensional measure.   Accordingly, if
$f$ is a function of two variables which is integrable for
two-dimensional Lebesgue measure, there is no reason why any particular
section $x\mapsto f(x,b)$ or $y\mapsto f(a,y)$ should be measurable, let
alone integrable.   Consequently, even if $f$ itself is defined
everywhere, the outer integral of $\iint f(x,y)dxdy$ is likely to be
applied to a function which is not defined for every $y$.
Let me remark that the problem does not concern `$\infty$';  the
awkward functions are those with sections so irregular that they cannot
be assigned an integral at all.

I have seen many approaches to this particular nettle, generally less
whole-hearted than the one I have determined on for this treatise.
Part of the
difficulty is that Fubini's theorem really is at the centre of measure
theory.   Over large parts of the subject, it is possible to assert that
a result is non-trivial if and only if it depends on Fubini's theorem.
I am therefore unwilling to insert any local fix, saying that `in
this chapter, we shall integrate functions which are not defined
everywhere';  before long, such a provision would have to be
interpolated into the preambles to half the best theorems, or an
explanation offered of why it wasn't necessary in their particular
contexts.   I suppose that one of the commonest responses is (like
{\smc Halmos 50}) to restrict attention to
$\Sigma\tensorhat\Tau$-measurable
functions, which eliminates measurability problems for the moment
(252Xh, 252P);   but
unhappily (or rather, to my mind, happily) there are crucial
applications in which the functions are not actually
$\Sigma\tensorhat\Tau$-measurable, but belong to some wider class, and
this restriction sooner or later leads to undignified contortions as we
are forced to adapt
limited results to unforeseen contexts.   Besides, it leaves unsaid the
really rather important information that if $f$ is a measurable function
of two variables then (under appropriate conditions) almost all its
sections are measurable (252E).

In 252B and its corollaries there is a clumsy restriction:  we assume
that one of the measures is $\sigma$-finite and the other is either
strictly localizable or complete and locally determined.   The obvious
question is, whether we need these hypotheses.   From 252K we see that
the hypothesis `$\sigma$-finite' on the second factor can certainly not
be abandoned, even when the first factor is a complete probability
measure.   The requirement `$\mu$ is either strictly localizable or
complete and locally determined' is in fact fractionally stronger than
what is needed, as well as disagreeably elaborate.   The `right'
hypothesis is that the completion of $\mu$ should be locally determined
(see 252Ya).   The point is that because the
product of two measures is the same as the product of their c.l.d.\
versions (251T), no theorem which leads from the product measure to the
factor measures can distinguish between a measure and its c.l.d.\
version;  so that, in 252B, we must expect to need $\mu$ and its c.l.d.\
version to give rise to the same integrals.   The proof of 252B would be
better focused if the hypothesis was simplified to `$\nu$ is
$\sigma$-finite and $\mu$ is complete and locally determined'.   But
this would just transfer part of the argument into the proof of 252C.

We also have to work a little harder in 252B in order to cover functions
and integrals taking the values $\pm\infty$.   Fubini's theorem is so
central to measure theory that I believe it is worth taking a bit of
extra trouble to state the results in maximal generality.   This is
especially important because we frequently apply it in multiply repeated
integrals, as in 252Xd, in which we have even less control than usual
over the intermediate functions to be integrated.

I have expressed all the main results of this section in terms of the
`c.l.d.'\ product measure.   In the case of $\sigma$-finite spaces, of
course, which is where the theory works best, we could just as well use
the `primitive' product measure.   Indeed, Fubini's theorem itself has a
version in terms of the primitive product measure which is rather more
elegant than 252B as stated (252Yc), and covers the great majority of
applications.   (Integrals with respect to the primitive and c.l.d.\
product measures are of course very closely related;  see 252Yd.)   But
we do sometimes need to look at
non-$\sigma$-finite spaces, and in these cases the asymmetric form in
252B is close to the best we can do.   Using the primitive product
measure does not help at all with the most substantial obstacle, the
phenomenon in 252K (see 252Yk).

The pre-calculus concept of an integral as `the area under a curve' is
given expression in 252N:  the integral of a non-negative function is
the measure of its ordinate set.   This is unsatisfactory as a
definition of the integral, not just because of the requirement that the
base space should be complete and locally determined (which can be dealt
with by using the primitive product measure, as in 252Yl), but because
the construction of the product measure involves integration (part (c)
of the proof of 251E).
The idea of 252N is to relate the measure of an ordinate set to the
integral of the measures of its vertical sections.   Curiously, if
instead we integrate the measures of its {\it horizontal} sections, as
in 252O, we get a more versatile result.   (Indeed this one does not
involve the concept of `product measure', and could have appeared at any
point after \S123.)   Note that the integral
$\int_0^{\infty}\ldots dt$ here is applied to a monotonic function, so
may be interpreted as an improper Riemann integral.   If you think you
know enough about the Riemann integral to
make this a tempting alternative to the construction in \S122, the
tricky bit now becomes the proof that the integral is additive.

A different line of argument is to use integration over sections to
define a product measure.   The difficulty with this approach is that
unless we take great care we may find ourselves with an asymmetric
construction.   My own view is that such an asymmetry is acceptable only
when there is no alternative.   But in Chapter 43 of Volume 4 I will
describe a couple of examples.

Of the two examples I give here, 252K is supposed to show that when I
call for $\sigma$-finite spaces they are really necessary, while 252L is
supposed to show that joint measurability is essential in Tonelli's
theorem and its corollaries.   The factor spaces in 252K, Lebesgue
measure and counting measure, are chosen to show that it is only the
lack of $\sigma$-finiteness that can be the problem;  they are otherwise
as regular as one can reasonably ask.   In 252L I have used the
countable-cocountable measure on $\omega_1$, which you may feel is fit
only for counter-examples;  and the question does arise, whether the
same phenomenon occurs with Lebesgue measure.   This leads into deep
water, and I will return to it in Chapter 53 of Volume 5.

I ought perhaps to note explicitly that in Fubini's theorem, we really
do need to have a function which is integrable for the product measure.
I include 252Xf and 252Xg to remind you that even in the best-regulated
circumstances,
the repeated integrals $\iint f\,dxdy$, $\iint f\,dydx$ may fail to be
equal if $f$ is not integrable as a function of two variables.

There are many ways to calculate the volume $\beta_r$ of an
$r$-dimensional ball;  the one I have used in 252Q follows a line that
would have been natural to me before I ever heard of measure theory.
In 252Xi I suggest another method.   The idea of
integration-by-substitution, used in part (b) of the argument for 252Q,
is there
supported by an ad hoc argument;  I will present a different, more
generally applicable, approach in Chapter 26.   Elsewhere (252Xi, 252Yf,
252Yh, 252Yu) I find myself taking for granted substitutions of the form
$t\mapsto at$, $t\mapsto a+t$, $t\mapsto t^2$;
for a systematic justification, see
\S263.   Of course an enormous number of other formulae of advanced
calculus are also based on repeated integration of one kind or another,
and I give a sample handful of such results (252Xb,
252Ye-252Yh, %252Ye 252Yf 252Yg 252Yh
252Yu).

}%end of notes

\discrpage

\leaveitout{%\spheader 252Y?
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces, and
$\lambda_0$
the primitive product measure on $X\times Y$.   Let $f$ be a real-valued
function defined $\lambda_0$-a.e.\ on $X\times Y$.   Show that

\Centerline{$
\overline{\intop}\bigl(\overline{\intop}f(x,y)\mu(dx)\bigr)\nu(dy)
\le\overline{\intop}f\,d\lambda_0$.}
%252B
}%end of leaveitout

\frfilename{mt253.tex}
\versiondate{18.4.08}

\def\chaptername{Product measures}
\def\sectionname{Tensor products}

\newsection{253}

The theorems of the last section show that the integrable functions on a
product of two measure spaces can be effectively studied in terms of
integration on each factor space separately.   In this section I present
a very striking relationship between the $L^1$ space of a product
measure and the $L^1$ spaces of its factors, which actually determines
the product $L^1$ up to isomorphism as Banach lattice.   I start with a
brief note on bilinear operators (253A) and a description of the canonical
bilinear operator from $L^1(\mu)\times L^1(\nu)$ to $L^1(\mu\times\nu)$
(253B-253E).   The main theorem of the section is 253F, showing that
this canonical map is universal for continuous bilinear operators from
$L^1(\mu)\times L^1(\nu)$ to Banach spaces;  it also determines the
ordering of $L^1(\mu\times\nu)$ (253G).   I end with a description of a
fundamental type of conditional expectation operator (253H) and notes
on products of indefinite-integral measures (253I) and upper integrals
of special kinds of function (253J, 253K).

\leader{253A}{Bilinear operators}\cmmnt{ Before looking at any of the
measure theory in this section, I introduce a concept from the theory of
linear spaces.

\medskip

} {\bf (a)} Let $U$, $V$ and $W$ be linear spaces over
$\Bbb  R$\cmmnt{ (or, indeed, any field)}.   A map
$\phi:U\times V\to W$ is {\bf bilinear} if it is linear in each variable
separately, that is,

\Centerline{$\phi(u_1+u_2,v)=\phi(u_1,v)+\phi(u_2,v)$,}

\Centerline{$\phi(u,v_1+v_2)=\phi(u,v_1)+\phi(u,v_2)$,}

\Centerline{$\phi(\alpha u,v)=\alpha\phi(u,v)=\phi(u,\alpha v)$}

\noindent for all $u$, $u_1$, $u_2\in U$, $v$, $v_1$, $v_2\in V$ and
scalars $\alpha$.   Observe that such a $\phi$ gives rise to, and in
turn can be defined by, a linear operator
$T:U\to\eurm L(V;W)$\cmmnt{, writing
$\eurm L(V;W)$ for the space of linear operators from $V$ to $W$}, where

\Centerline{$(Tu)(v)=\phi(u,v)$}

\noindent for all $u\in U$, $v\in V$.  \cmmnt{Hence, or otherwise, we
can see, for instance, that} $\phi(0,v)=\phi(u,0)=0$ whenever $u\in U$ and
$v\in V$.

If $W'$ is another linear space\cmmnt{ over the same field,} and
$S:W\to W'$ is a linear operator, then $S\phi:U\times V\to W'$ is
bilinear.

\spheader 253Ab Now suppose that $U$, $V$ and $W$ are
normed spaces, and $\phi:U\times V\to W$ a bilinear operator.
\cmmnt{Then we
say that} $\phi$ is {\bf bounded} if
$\sup\{\|\phi(u,v)\|:\|u\|\le 1,\,\|v\|\le 1\}$ is finite, and in this
case we call this supremum the
norm $\|\phi\|$ of $\phi$.    \cmmnt{Note that}
$\|\phi(u,v)\|\le\|\phi\|\|u\|\|v\|$ for all $u\in U$, $v\in
V$\prooflet{ (because

\Centerline{$\|\phi(u,v)\|=\alpha\beta\|\phi(\alpha^{-1}u,\beta^{-1}v)\|
\le\alpha\beta\|\phi\|$}

\noindent whenever $\alpha>\|u\|$, $\beta>\|v\|$)}.

If $W'$ is another normed space and $S:W\to W'$ is a bounded linear
operator, then $S\phi:U\times V\to W'$ is a bounded bilinear operator,
and $\|S\phi\|\le\|S\|\|\phi\|$.

\leader{253B}{Definition}\cmmnt{ The most important bilinear operators of
this section are based on the following idea.}   Let $f$ and $g$ be
real-valued functions.   I will write $f\otimes g$ for the function
$(x,y)\mapsto f(x)g(y):\dom f\times\dom g\to\Bbb R$.

\leader{253C}{Proposition}
(a) Let $X$ and $Y$ be sets, and $\Sigma$, $\Tau\,\,\sigma$-algebras of
subsets of $X$, $Y$ respectively.   If $f$ is a $\Sigma$-measurable
real-valued function defined on a subset of $X$, and $g$ is a
$\Tau$-measurable real-valued function defined on a subset of $Y$, then
$f\otimes g$\cmmnt{, as defined in 253B,} is
$\Sigma\tensorhat\Tau$-measurable.

(b) Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces, and
$\lambda$ the c.l.d.\ product measure on $X\times Y$.  If
$f\in\eusm L^0(\mu)$ and $g\in\eusm L^0(\nu)$, then
$f\otimes g\in\eusm L^0(\lambda)$.

\cmmnt{\medskip

\noindent{\bf Remark} Recall from 241A that $\eusm L^0(\mu)$ is the
space of $\mu$-virtually measurable real-valued functions defined
on $\mu$-conegligible subsets of $X$.}

\proof{{\bf (a)} The point is that $f\otimes\chi Y$ is
$\Sigma\tensorhat\Tau$-measurable, because for any $\alpha\in\Bbb R$
there is an $E\in\Sigma$ such that

\Centerline{$\{x:f(x)\ge\alpha\}=E\cap\dom f$,}

\noindent so that

\Centerline{$\{(x,y):(f\otimes\chi Y)(x,y)\ge\alpha\}
=(E\cap\dom f)\times Y=(E\times Y)\cap\dom(f\otimes\chi Y)$,}

\noindent and of course $E\times Y\in\Sigma\tensorhat\Tau$.   Similarly,
$\chi X\otimes g$ is $\Sigma\tensorhat\Tau$-measurable and
$f\otimes g=(f\otimes\chi Y)\times(\chi X\otimes g)$ is
$\Sigma\tensorhat\Tau$-measurable.

\medskip

{\bf (b)} Let $E\in\Sigma$, $F\in\Tau$ be conegligible subsets of $X$,
$Y$ respectively such that $E\subseteq\dom f$, $F\subseteq\dom g$,
$f\restr E$ is $\Sigma$-measurable and $g\restr F$ is $\Tau$-measurable.
Write $\Lambda$ for the domain of $\lambda$.   Then
$\Sigma\tensorhat\Tau\subseteq\Lambda$ (251Ia).   Also $E\times F$ is
$\lambda$-conegligible, because

$$\eqalign{\lambda((X\times Y)\setminus(E\times F))
&\le\lambda((X\setminus E)\times Y)+\lambda(X\times(Y\setminus F))\cr
&=\mu(X\setminus E)\cdot\nu Y+\mu X\cdot\nu(Y\setminus F)
=0\cr}$$

\noindent (also from 251Ia).   So $\dom(f\otimes g)\supseteq E\times F$
is conegligible.   Also, by (a),
$(f\otimes g)\restr(E\times F)
=(f\restr E)\otimes(g\restr F)$ is $\Sigma\tensorhat\Tau$-measurable,
therefore $\Lambda$-measurable, and $f\otimes g$ is virtually
measurable.   Thus $f\otimes g\in\eusm L^0(\lambda)$, as claimed.
}%end of proof of 253C

\leader{253D}{}\cmmnt{ Now we can apply the ideas of 253B-253C to
integrable functions.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be measure spaces, and write $\lambda$ for the c.l.d.\
product measure on $X\times Y$.   If $f\in\eusm L^1(\mu)$ and
$g\in\eusm L^1(\nu)$, then $f\otimes g\in\eusm L^1(\lambda)$ and
$\int f\otimes g\,d\lambda=\int f\,d\mu\int g\,d\nu$.

\cmmnt{\medskip

\noindent{\bf Remark} I follow \S242 in writing $\eusm L^1(\mu)$ for
the space of $\mu$-integrable real-valued functions.
}

\proof{{\bf (a)} Consider first the
case $f=\chi E$, $g=\chi F$ where $E\in\Sigma$, $F\in\Tau$ have finite
measure;  then $f\otimes g=\chi(E\times F)$ is $\lambda$-integrable with
integral

\Centerline{$\lambda(E\times F)=\mu E\cdot\nu F
=\int f\,d\mu\cdot\int g\,d\nu$,}

\noindent by 251Ia.

\medskip

{\bf (b)}  It follows at once that $f\otimes g$ is $\lambda$-simple,
with $\int f\otimes g\,d\lambda=\int f\,d\mu\int g\,d\nu$, whenever $f$
is a $\mu$-simple function and $g$ is a $\nu$-simple function.

\medskip

{\bf (c)} If $f$ and $g$ are non-negative integrable functions, there
are non-decreasing sequences $\sequencen{f_n}$, $\sequencen{g_n}$ of
non-negative simple
functions converging almost everywhere to $f$, $g$ respectively;   now
note that if $E\subseteq X$, $F\subseteq Y$ are conegligible, $E\times
F$ is conegligible in $X\times Y$, as remarked in the proof of 253C, so
the non-decreasing sequence $\sequencen{f_n\times g_n}$ of
$\lambda$-simple functions converges almost everywhere to $f\otimes g$,
and

\Centerline{$\int f\otimes g\,d\lambda
=\lim_{n\to\infty}\int f_n\otimes g_nd\lambda
=\lim_{n\to\infty}\int f_nd\mu\int g_nd\nu
=\int f\,d\mu\int g\,d\nu$}

\noindent by B.Levi's theorem.

\medskip

{\bf (d)} Finally, for general $f$ and $g$, we can express them as
the differences $f^+-f^-$, $g^+-g^-$ of non-negative integrable
functions, and see that

\Centerline{$\int f\otimes g\,d\lambda
=\int f^+\otimes g^+-f^+\otimes g^--f^-\otimes g^+
+f^-\otimes g^-d\lambda
=\int f\,d\mu\int g\,d\nu.$}
}%end of proof of 253D

\leader{253E}{The canonical map $L^1\times L^1\to L^1$}\cmmnt{ I
continue the argument from 253D.   Because $E\times F$ is conegligible
in $X\times Y$ whenever $E$ and $F$ are
conegligible subsets of $X$ and $Y$,
$f_1\otimes g_1=f\otimes g\,\,\lambda$-a.e. whenever $f=f_1\,\,\mu$-a.e.
and $g=g_1\,\,\nu$-a.e.}   We may\cmmnt{ therefore} define
$u\otimes v\in L^1(\lambda)$, for $u\in L^1(\mu)$ and $v\in L^1(\nu)$,
by saying that
$u\otimes v=(f\otimes g)^{\ssbullet}$ whenever $u=f^{\ssbullet}$ and
$v=g^{\ssbullet}$.   \dvro{The}{

Now if $f$, $f_1$, $f_2\in\eusm L^1(\mu)$, $g$, $g_1$,
$g_2\in\eusm L^1(\nu)$ and $a\in\Bbb R$,

\Centerline{$(f_1+f_2)\otimes g=(f_1\otimes g)+(f_2\otimes g)$,}

\Centerline{$f\otimes(g_1+g_2)=(f\otimes g_1)+(f\otimes g_2)$,}

\Centerline{$(af)\otimes g=a(f\otimes g)=f\otimes(ag)$.}

\noindent It follows at once that the} map $(u,v)\mapsto u\otimes v$ is
bilinear.

\cmmnt{Moreover, if $f\in\eusm L^1(\mu)$ and $g\in\eusm L^1(\nu)$,
$|f|\otimes|g|=|f\otimes g|$, so
$\int|f\otimes g|d\lambda=\int|f|d\mu\int|g|d\nu$.   Accordingly}

\Centerline{$\|u\otimes v\|_1=\|u\|_1\|v\|_1$}

\noindent for all $u\in L^1(\mu)$, $v\in L^1(\nu)$.   In particular, the
bilinear operator $\otimes$ is bounded, with norm $1$ (except in the trivial
case in which one of $L^1(\mu)$, $L^1(\nu)$ is $0$-dimensional).

\leader{253F}{}\cmmnt{ We are now ready for the main theorem of this
section.

\medskip

\noindent}{\bf Theorem}
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces, and let
$\lambda$ be the c.l.d.\ product measure on $X\times Y$.   Let $W$ be
any Banach space and $\phi:L^1(\mu)\times L^1(\nu)\to W$ a bounded bilinear
operator.   Then there is a unique bounded linear operator
$T:L^1(\lambda)\to W$ such that $T(u\otimes v)=\phi(u,v)$ for all
$u\in L^1(\mu)$ and $v\in L^1(\nu)$, and $\|T\|=\|\phi\|$.

\proof{{\bf (a)} The centre of the argument is the following
fact:   if $E_0,\ldots,E_n$ are measurable sets of finite measure in
$X$, $F_0,\ldots,F_n$ are measurable sets of finite measure in $Y$,
$a_0,\ldots,a_n\in\Bbb R$ and
$\sum_{i=0}^na_i\chi(E_i\times F_i)=0\,\,\lambda$-a.e., then
$\sum_{i=0}^na_i\phi(\chi E_i^{\ssbullet},\chi F_i^{\ssbullet})=0$ in $W$.
\Prf\ We
can find a disjoint family $\langle G_j\rangle_{j\le m}$ of measurable
sets of finite measure in $X$ such that each $E_i$ is expressible as a
union of some subfamily of the $G_j$;  so that $\chi E_i$ is
expressible in the form $\sum_{j=0}^mb_{ij}\chi G_j$ (see 122Ca).
Similarly, we
can find a disjoint family $\langle H_k\rangle_{k\le l}$ of measurable
sets of finite measure in $Y$ such that each $\chi F_i$ is expressible
as $\sum_{k=0}^lc_{ik}\chi H_k$.   Now

\Centerline{$\sum_{j=0}^m\sum_{k=0}^l
\bigl(\sum_{i=0}^na_ib_{ij}c_{ik}\bigr)
\chi(G_j\times H_k)
=\sum_{i=0}^na_i\chi(E_i\times F_i)=0\,\,\lambda\text{-a.e.}$}

\noindent Because the $G_j\times H_k$ are disjoint, and
$\lambda(G_j\times H_k)=\mu G_j\cdot\nu H_k$ for all $j$, $k$, it
follows that for
every $j\le m$, $k\le l$ we have either $\mu G_j=0$ or $\nu H_k=0$ or
$\sum_{i=0}^na_ib_{ij}c_{ik}=0$.   In any of these three cases,
$\sum_{i=0}^na_ib_{ij}c_{ik}\phi(\chi G_j^{\ssbullet},\chi
H_k^{\ssbullet})=0$ in $W$.   But this means that

\Centerline{$0=\sum_{j=0}^m\sum_{k=0}^l
\bigl(\sum_{i=0}^na_ib_{ij}c_{ik}\bigr)
\phi(\chi G_j^{\ssbullet},\chi H_k^{\ssbullet})
=\sum_{i=0}^na_i\phi(\chi E_i^{\ssbullet},\chi F_i^{\ssbullet}),$}

\noindent as claimed.\ \Qed

\medskip

{\bf (b)} It follows that if  $E_0,\ldots,E_n$, $E'_0,\ldots,E'_m$  are
measurable sets of finite measure in $X$, $F_0,\ldots,F_n$,
$F'_0,\ldots,F'_m$ are measurable sets of finite measure in $Y$,
$a_0,\ldots,a_n,a'_0,\ldots,a'_m\in\Bbb R$ and
$\sum_{i=0}^na_i\chi(E_i\times F_i)=\sum_{i=0}^ma'_i\chi(E'_i\times
F'_i)\,\,\lambda$-a.e., then

\Centerline{$\sum_{i=0}^na_i\phi(\chi
E_i^{\ssbullet},\chi F_i^{\ssbullet})=\sum_{i=0}^ma'_i\phi(\chi
{E'_i}^{\ssbullet},\chi {F'_i}^{\ssbullet})$}

\noindent in $W$.   Let $M$ be the linear
subspace of $L^1(\lambda)$ generated by

\Centerline{$\{\chi(E\times F)^{\ssbullet}:E\in\Sigma,\,\mu E<\infty,\,
  F\in\Tau,\,\nu F<\infty\};$}

\noindent then we have a unique map $T_0:M\to W$ such that

\Centerline{$T_0(\sum_{i=0}^na_i\chi(E_i\times F_i)^{\ssbullet})
=\sum_{i=0}^na_i\phi(\chi E_i^{\ssbullet},\chi F_i^{\ssbullet})$}

\noindent whenever $E_0,\ldots,E_n$  are measurable sets of finite
measure in $X$, $F_0,\ldots,F_n$ are measurable sets of finite measure
in $Y$ and $a_0,\ldots,a_n\in\Bbb R$.   Of course $T_0$ is linear.

\medskip

{\bf (c)} Some of the same calculations show that
$\|T_0u\|\le\|\phi\|\|u\|_1$ for every $u\in M$.   \Prf\ If $u\in M$,
then, by the arguments of (a), we can express $u$ as
$\sum_{j=0}^m\sum_{k=0}^la_{jk}\chi(G_j\times
H_k)^{\ssbullet}$, where $\langle G_j\rangle_{j\le m}$ and $\langle
H_k\rangle_{k\le l}$ are disjoint families of sets of finite measure.
Now

$$\eqalign{\|T_0u\|
&=\|\sum_{j=0}^m\sum_{k=0}^la_{jk}\phi(\chi G_j^{\ssbullet},\chi
  H_k^{\ssbullet})\|
\le\sum_{j=0}^m\sum_{k=0}^l|a_{jk}|\|\phi(\chi G_j^{\ssbullet},\chi
  H_k^{\ssbullet})\|\cr
&\le\sum_{j=0}^m\sum_{k=0}^l|a_{jk}|\|\phi\|\|\chi
  G_j^{\ssbullet}\|_1\|\chi H_k^{\ssbullet}\|_1
=\|\phi\|\sum_{j=0}^m\sum_{k=0}^l|a_{jk}|\mu G_j\cdot\nu H_k\cr
&=\|\phi\|\sum_{j=0}^m\sum_{k=0}^l|a_{jk}|\lambda(G_j\times H_k)
=\|\phi\|\|u\|_1,\cr}$$

\noindent as claimed.\ \Qed

\medskip

{\bf (d)} The next point is to observe that $M$ is dense in
$L^1(\lambda)$ for $\|\,\|_1$.   \Prf\ Repeating the ideas above once
again, we observe that if $E_0,\ldots,E_n$ are sets of finite measure in
$X$ and $F_0,\ldots,F_n$ are sets of finite measure in $Y$, then
$\chi(\bigcup_{i\le n}E_i\times F_i)^{\ssbullet}\in M$;  this is
because,
expressing each $E_i$ as a union of $G_j$, where the $G_j$ are disjoint,
we have

\Centerline{$\bigcup_{i\le n}E_i\times F_i=\bigcup_{j\le m}G_j\times
F'_j,$}

\noindent where $F'_j=\bigcup\{F_i:G_j\subseteq E_i\}$ for each $j$;
now $\langle G_j\times F'_j\rangle_{j\le m}$ is disjoint, so

\Centerline{$\chi(\bigcup_{j\le m}G_j\times F_j)^{\ssbullet}
=\sum_{j=0}^m\chi(G_j\times F'_j)^{\ssbullet}\in M.$}

\noindent So 251Ie tells us that whenever $\lambda H<\infty$ and
$\epsilon>0$ there is a $G$ such that $\lambda(H\symmdiff G)\le\epsilon$
and $\chi G^{\ssbullet}\in M$;  now

\Centerline{$\|\chi H^{\ssbullet}-\chi
G^{\ssbullet}\|_1=\lambda(G\symmdiff H)\le\epsilon$,}

\noindent so $\chi H^{\ssbullet}$
is approximated arbitrarily closely by members of $M$, and belongs to
the closure $\overline{M}$ of $M$ in $L^1(\lambda)$.   Because $M$ is a
linear subspace of $L^1(\lambda)$, so is $\overline{M}$ (2A4Cb);
accordingly $\overline{M}$ contains the equivalence classes of all
$\lambda$-simple functions;  but these are dense in $L^1(\lambda)$
(242Mb), so $\overline{M}=L^1(\lambda)$, as claimed.\ \Qed

\medskip

{\bf (e)} Because $W$ is a Banach space, it follows that there is a
bounded linear operator $T:L^1(\lambda)\to W$ extending $T_0$, with
$\|T\|=\|T_0\|\le\|\phi\|$ (2A4I).    Now $T(u\otimes v)=\phi(u,v)$ for
all $u\in L^1(\mu)$, $v\in L^1(\nu)$.   \Prf\ If $u=\chi E^{\ssbullet}$ and
$v=\chi F^{\ssbullet}$, where $E$, $F$ are measurable sets of finite
measure, then

\Centerline{$T(u\otimes v)=T(\chi(E\times F)^{\ssbullet})
=T_0(\chi(E\times F)^{\ssbullet})
=\phi(\chi E^{\ssbullet},\chi F^{\ssbullet})=\phi(u,v).$}

\noindent Because $\phi$ and $\otimes$ are bilinear and $T$ is linear,

\Centerline{$T(f^{\ssbullet}\otimes g^{\ssbullet})
=\phi(f^{\ssbullet},g^{\ssbullet})$}

\noindent whenever $f$ and $g$ are simple functions.   Now whenever
$u\in L^1(\mu)$, $v\in L^1(\nu)$ and $\epsilon>0$, there are simple
functions $f$, $g$ such that $\|u-f^{\ssbullet}\|_1\le\epsilon$,
$\|v-g^{\ssbullet}\|_1\le\epsilon$ (242Mb again);  so that

$$\eqalign{\|\phi(u,v)-\phi(f^{\ssbullet},g^{\ssbullet})\|
&\le\|\phi(u-f^{\ssbullet},v-g^{\ssbullet})\|
+\|\phi(u,g^{\ssbullet}-v)\|
+\|\phi(f^{\ssbullet}-u,v)\|\cr
&\le\|\phi\|(\epsilon^2+\epsilon\|u\|_1+\epsilon\|v\|_1).}$$

\noindent Similarly

\Centerline{$\|u\otimes v-f^{\ssbullet}\otimes g^{\ssbullet}\|_1
\le\epsilon(\epsilon+\|u\|_1+\|v\|_1),$}

\noindent so

\Centerline{$\|T(u\otimes v)-T(f^{\ssbullet}\otimes g^{\ssbullet})\|
\le\epsilon\|T\|(\epsilon+\|u\|_1+\|v\|_1);$}

\noindent because $T(f^{\ssbullet}\otimes g^{\ssbullet})
=\phi(f^{\ssbullet},g^{\ssbullet})$,

\Centerline{$\|T(u\otimes v)-\phi(u,v)\|
\le\epsilon(\|T\|+\|\phi\|)(\epsilon+\|u\|_1
+\|v\|_1).$}

\noindent As $\epsilon$ is arbitrary, $T(u\otimes v)=\phi(u,v)$, as
required.\ \Qed

\medskip

{\bf (f)} The argument of (e) ensured that $\|T\|\le\|\phi\|$.
Because $\|u\otimes v\|_1\le\|u\|_1\|v\|_1$ for all $u\in L^1(\mu)$ and
$v\in L^1(\nu)$, $\|\phi(u,v)\|\le\|T\|\|u\|_1\|v\|_1$ for all $u$, $v$,
and $\|\phi\|\le\|T\|$;  so $\|T\|=\|\phi\|$.

\medskip

{\bf (g)} Thus $T$ has the required properties.   To see that it is
unique, we have only to observe that any bounded linear operator
$S:L^1(\lambda)\to W$ such that $S(u\otimes v)=\phi(u,v)$ for all $u\in
L^1(\mu)$, $v\in L^1(\nu)$ must agree with $T$ on objects of the form
$\chi(E\times F)^{\ssbullet}$ where $E$ and $F$ are of finite measure,
and
therefore on every member of $M$;  because $M$ is dense and both $S$ and
$T$ are continuous, they agree everywhere in $L^1(\lambda)$.
}

\leader{253G}{The order structure of \dvrocolon{$L^1$}}\cmmnt{ In
253F I have treated the $L^1$ spaces exclusively as
normed linear spaces.   In general, however, the order structure of an
$L^1$ space  (see 242C) is as important as its norm.   The map
$\otimes:L^1(\mu)\times L^1(\nu)\to L^1(\lambda)$ respects the order
structures of the three spaces in the following strong sense.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\lambda$ the c.l.d.\ product measure on $X\times
Y$.   Then

(a)  $u\otimes v\ge 0$ in
$L^1(\lambda)$ whenever $u\ge 0$ in $L^1(\mu)$ and $v\ge 0$ in
$L^1(\nu)$.

(b) The positive cone $\{w:w\ge 0\}$ of
$L^1(\lambda)$ is precisely the closed
convex hull $C$ of $\{u\otimes v:u\ge 0,\,v\ge 0\}$ in $L^1(\lambda)$.

*(c) Let $W$ be any Banach lattice, and $T:L^1(\lambda)\to W$ a
bounded linear operator.   Then
the following are equiveridical:

\quad(i) $Tw\ge 0$ in $W$ whenever $w\ge 0$ in
$L^1(\lambda)$;

\quad(ii) $T(u\otimes v)\ge 0$ in $W$ whenever $u\ge 0$ in
$L^1(\mu)$ and $v\ge 0$ in $L^1(\nu)$.

\proof{{\bf (a)} If $u$, $v\ge 0$ then they are expressible as
$f^{\ssbullet}$, $g^{\ssbullet}$ where $f\in\eusm L^1(\mu)$,
$g\in\eusm L^1(\nu)$, $f\ge 0$ and $g\ge 0$.
Now $f\otimes g\ge 0$ so $u\otimes v=(f\otimes g)^{\ssbullet}\ge 0$.

\medskip

{\bf (b)(i)} Write $L^1(\lambda)^+$ for $\{w:w\in L^1(\lambda),\,w\ge
0\}$.   Then $L^1(\lambda)^+$ is a closed convex set in $L^1(\lambda)$
(242De);  by (a), it contains $u\otimes v$ whenever $u\in L^1(\mu)^+$ and
$v\in L^1(\nu)^+$, so it must include $C$.

\medskip

\quad{\bf (ii)}($\alpha$)
Of course $0=0\otimes 0\in C$.  ($\beta$) If $u\in M$, as defined in
the proof of 253F, and $u>0$, then $u$ is expressible as $\sum_{j\le
m,k\le l}a_{jk}\chi(G_j\times H_k)^{\ssbullet}$, where $G_0,\ldots,G_m$
and $H_0,\ldots,H_l$ are disjoint sequences of sets of finite measure,
as in (a) of the proof of 253F.   Now $a_{jk}$ can be negative only if
$\chi(G_j\times H_k)^{\ssbullet}=0$, so replacing every $a_{jk}$ by
$\max(0,a_{jk})$  if necessary, we can suppose that $a_{jk}\ge 0$ for
all $j$, $k$.     Not all the $a_{jk}$ can be zero, so $a=\sum_{j\le
m,k\le l}a_{jk}>0$, and

\Centerline{$u
=\sum_{j\le m,k\le l}
  \Bover{a_{jk}}a\cdot a\chi(G_j\times H_k)^{\ssbullet}
=\sum_{j\le m,k\le l}\Bover{a_{jk}}a
  \cdot(a\chi G_j^{\ssbullet})\otimes\chi H_k^{\ssbullet}
\in C$.}

\noindent ($\gamma$) If $w\in L^1(\lambda)^+$ and $\epsilon>0$, express
$w$ as $h^{\ssbullet}$ where $h\ge 0$ in $\eusm L^1(\lambda)$.   There
is a simple function $h_1\ge 0$ such that $h_1\leae h$ and
$\int h\le\int h_1+\epsilon$.   Express $h_1$ as
$\sum_{i=0}^na_i\chi H_i$
where $\lambda H_i<\infty$ and $a_i\ge 0$ for each $i$, and for each
$i\le n$ choose sets $G_{i0},\ldots,G_{im_i}\in\Sigma$,
$F_{i0},\ldots,F_{im_i}\in\Tau$, all of finite measure, such that
$G_{i0},\ldots,G_{im_i}$ are disjoint and
$\lambda(H_i\symmdiff\bigcup_{j\le m_i}G_{ij}\times
F_{ij})\le\epsilon/(n+1)(a_i+1)$, as in (d) of the proof of 253F.   Set

\Centerline{$w_0
=\sum_{i=0}^na_i\sum_{j=0}^{m_i}\chi(G_{ij}\times F_{ij})^{\ssbullet}$.}

\noindent Then $w_0\in C$ because $w_0\in M$ and $w_0\ge 0$.   Also

$$\eqalign{\|w-w_0\|_1
&\le\|w-h_1^{\ssbullet}\|_1+\|h_1^{\ssbullet}-w_0\|_1\cr
&\le\int(h-h_1)d\lambda
  +\sum_{i=0}^na_i\int|\chi H_i-\sum_{j=0}^{m_i}\chi(G_{ij}\times
  F_{ij})|d\lambda\cr
&\le\epsilon+\sum_{i=0}^na_i\lambda(H\symmdiff\bigcup_{j\le
m_i}G_{ij}\times F_{ij})
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary and $C$ is closed, $w\in C$.   As
$w$ is arbitrary, $L^1(\lambda)^+\subseteq C$ and $C=L^1(\lambda)^+$.

\medskip

{\bf (c)} Part (a) tells us that (i)$\Rightarrow$(ii).   For the reverse
implication, we need a fragment from the theory of Banach lattices:
$W^+=\{w:w\in W,\,w\ge 0\}$ is a closed set in $W$.
\Prf\ If $w$, $w'\in W$, then

\Centerline{$w=(w-w')+w'\le|w-w'|+w'\le|w-w'|+|w'|$,}

\Centerline{$-w=(w'-w)-w'\le|w-w'|-w'\le|w-w'|+|w'|$,}

\Centerline{$|w|\le|w-w'|+|w'|$,
\quad$|w|-|w'|\le|w-w'|$,}

\noindent because $|w|=w\vee(-w)$ and the order of $W$ is
translation-invariant (241Ec).   Similarly, $|w'|-|w|\le|w-w'|$ and
$||w|-|w'||\le|w-w'|$, so $\||w|-|w'|\|\le\|w-w'\|$, by the definition
of Banach lattice (242G).   Setting $\phi(w)=|w|-w$, we see that
$\|\phi(w)-\phi(w')\|\le 2\|w-w'\|$ for all $w$, $w'\in W$, so that
$\phi$ is continuous.

Now, because the order is invariant under multiplication by positive
scalars,

\Centerline{$w\ge 0\iff 2w\ge 0\iff w\ge-w\iff w=|w|\iff\phi(w)=0$,}

\noindent so $W^+=\{w:\phi(w)=0\}$ is closed.\ \Qed

Now suppose that (ii) is true, and set
$C_1=\{w:w\in L^1(\lambda),\,Tw\ge 0\}$.   Then $C_1$ contains
$u\otimes v$ whenever $u$, $v\ge 0$;  but also
it is convex, because $T$ is linear, and closed, because $T$ is
continuous and $C_1=T^{-1}[W^+]$.   By (b), $C_1$ includes
$\{w:w\in L^1(\lambda),\,w\ge 0\}$, as required by (i).
}%end of proof of 253G

\leader{253H}{Conditional \dvrocolon{expectations}}\cmmnt{ The ideas
of this section and the preceding one provide us with some of the most
important examples of conditional expectations.

\medskip

\noindent}{\bf Theorem} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
complete probability spaces, with c.l.d.\ product
$(X\times Y,\Lambda,\lambda)$.   Set $\Lambda_1=\{E\times Y:E\in\Sigma\}$.
Then $\Lambda_1$ is a $\sigma$-subalgebra of $\Lambda$.   Given a
$\lambda$-integrable real-valued function $f$, set

\Centerline{$g(x,y)=\int f(x,z)\nu(dz)$}

\noindent whenever $x\in X$, $y\in Y$ and the integral is defined in
$\Bbb R$.   Then $g$ is a conditional expectation of $f$ on $\Lambda_1$.

\proof{ We know that $\Lambda_1\subseteq\Lambda$, by 251Ia, and
$\Lambda_1$ is a $\sigma$-algebra of sets because $\Sigma$ is.
Fubini's theorem (252B, 252C)
tells us that $f_1(x)=\int f(x,z)\nu(dz)$ is defined for almost every
$x$, and therefore that $g=f_1\otimes\chi Y$ is defined almost everywhere
in $X\times Y$.   $f_1$ is $\mu$-virtually measurable;  because $\mu$ is
complete, $f_1$ is $\Sigma$-measurable, so $g$ is $\Lambda_1$-measurable
(since $\{(x,y):g(x,y)\le\alpha\}=\{x:f_1(x)\le\alpha\}\times Y$ for
every $\alpha\in\Bbb R$).   Finally, if $W\in\Lambda_1$, then
$W=E\times Y$ for some $E\in\Sigma$, so

$$\eqalignno{\int_Wg\,d\lambda
&=\int (f_1\otimes\chi Y)\times(\chi E\otimes\chi Y)d\lambda
=\int f_1\times\chi E\,d\mu\int\chi Y\,d\nu\cr
\noalign{\noindent (by 253D)}
&=\iint\chi E(x)f(x,y)\nu(dy)\mu(dx)
=\int f\times\chi(E\times Y)d\lambda\cr
\noalign{\noindent (by Fubini's theorem)}
&=\int_Wf\,d\lambda.\cr}$$

\noindent So $g$ is a conditional expectation of $f$.
}%end of proof of 253H

\leader{253I}{}\cmmnt{ This is a convenient moment to set out a
useful result on products of indefinite-integral measures.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $f\in\eusm L^0(\mu)$, $g\in\eusm L^0(\nu)$
non-negative functions.   Let $\mu'$, $\nuprime$ be the corresponding
indefinite-integral measures\cmmnt{ (see \S234)}.   Let $\lambda$ be
the c.l.d.\ product of $\mu$ and $\nu$, and $\lambda'$ the
indefinite-integral measure defined from $\lambda$ and
$f\otimes g\in\eusm L^0(\lambda)$\cmmnt{ (253Cb)}.   Then $\lambda'$ is
the c.l.d.\ product of $\mu'$ and $\nuprime$.

\proof{  Write $\theta$ for the c.l.d.\ product of $\mu'$ and $\nuprime$.

\medskip

{\bf (a)} If we replace $\mu$ by its completion, we do not change $\mu'$
(234Ke);  at the same time, we do not change $\lambda$, by 251T.
The same applies to $\nu$.   So it will be enough to prove the result on
the assumption that $\mu$ and $\nu$ are complete;  in which case $f$ and
$g$ are measurable and have measurable domains.

Set $F=\{x:x\in\dom f,\,f(x)>0\}$ and $G=\{y:y\in\dom g,\,g(y)>0\}$, so
that $F\times G=\{w:w\in\dom(f\otimes g),\,(f\otimes g)(w)>0\}$.   Then
$F$ is $\mu'$-conegligible and $G$ is $\nuprime$-conegligible, so
$F\times G$ is $\theta$-conegligible as well as
$\lambda'$-conegligible.   Because both $\theta$ and $\lambda'$ are
complete (251Ic, 234I), it will be enough to show that the subspace
measures $\theta_{F\times G}$, $\lambda'_{F\times G}$ on $F\times G$ are
equal.   But note that $\theta_{F\times G}$ can be identified with the
product of $\mu_F'$ and $\nu_G'$, where $\mu_F'$ and $\nu_G'$ are the
subspace measures on $F$, $G$ respectively (251Q(ii-$\alpha$)).   At the
same time, $\mu_F'$ is the indefinite-integral measure defined from the
subspace measure $\mu_F$ on $F$ and the function $f\restr F$,
$\nu_G'$ is the indefinite-integral measure defined from the subspace
measure $\nu_G$ on $G$ and $g\restr G$, and $\lambda'_{F\times G}$ is
defined from the subspace measure $\lambda_{F\times G}$ and
$(f\restr F)\otimes(g\restr G)$.   Finally, by 251Q again,
$\lambda_{F\times G}$ is the product of $\mu_F$ and $\nu_G$.

What all this means is that it will be enough to deal with the case in
which $F=X$ and $G=Y$, that is, $f$ and $g$ are everywhere defined and
strictly positive;  which is what I will suppose from now on.

\medskip

{\bf (b)} In this case $\dom\mu'=\Sigma$ and $\dom\nuprime=\Tau$ (234La).
Similarly, $\dom\lambda'=\Lambda$ is just the domain of $\lambda$.   Set

\Centerline{$F_n=\{x:x\in X,\,2^{-n}\le f(x)\le 2^n\}$,
\quad$G_n=\{y:y\in Y,\,2^{-n}\le g(y)\le 2^n\}$}

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

\medskip

{\bf (c)} Set

\Centerline{$\Cal A=\{W:W\in\dom\theta\cap\dom\lambda'$,
$\theta(W)=\lambda'(W)\}$.}

\noindent If $\mu'E$ and $\nuprime H$ are defined and finite, then
$f\times\chi E$ and $g\times\chi H$ are integrable, so

$$\eqalign{\lambda'(E\times H)
&=\int(f\otimes g)\times\chi(E\times H)d\lambda
=\int(f\times\chi E)\otimes(g\times\chi H)d\lambda\cr
&=\int f\times\chi E\,d\mu\cdot\int g\times\chi H\,d\nu
=\theta(E\times H)\cr}$$

\noindent by 253D and 251Ia, that is, $E\times H\in\Cal A$.   If we now
look at
$\Cal A_{EH}=\{W:W\subseteq X\times Y$, $W\cap(E\times H)\in\Cal A\}$,
then we see that

\inset{$\Cal A_{EH}$ contains $E'\times H'$ for every $E'\in\Sigma$,
$H'\in\Tau$,}

\inset{if $\sequencen{W_n}$ is a non-decreasing sequence in
$\Cal A_{EH}$ then $\bigcup_{n\in\Bbb N}W_n\in\Cal A_{EH}$,}

\inset{if $W$, $W'\in\Cal A_{EH}$ and $W\subseteq W'$ then
$W'\setminus W\in\Cal A_{EH}$.}

\noindent Thus $\Cal A_{EH}$ is a Dynkin class of subsets of
$X\times Y$, and by the Monotone Class Theorem (136B) includes
the $\sigma$-algebra generated by
$\{E'\times H':E'\in\Sigma,\,H'\in\Tau\}$, which is
$\Sigma\tensorhat\Tau$.

\medskip

{\bf (d)} Now suppose that $W\in\Lambda$.   In this case
$W\in\dom\theta$ and $\theta W\le\lambda'W$.   \Prf\ Take $n\in\Bbb N$,
and $E\in\Sigma$, $H\in\Tau$ such that $\mu'E$ and $\nuprime H$ are
both finite.   Set $E'=E\cap F_n$, $H'=H\cap G_n$ and $W'=W\cap(E'\times
H')$.   Then $W'\in\Lambda$, while $\mu E'\le 2^n\mu'E$ and $\nu H'\le
2^n\nuprime H$ are finite.   By 251Ib there is a $V\in\Sigma\tensorhat\Tau$
such that $V\subseteq W'$ and $\lambda V=\lambda W'$.  Similarly, there
is a $V'\in\Sigma\tensorhat\Tau$ such that $V'\subseteq(E'\times
H')\setminus W'$ and $\lambda V'=\lambda((E'\times H')\setminus W')$.
This means that $\lambda((E'\times H')\setminus(V\cup V'))=0$, so
$\lambda'((E'\times H')\setminus(V\cup V'))=0$.   But
$(E'\times H')\setminus(V\cup V')\in\Cal A$, by (c), so $\theta((E'\times H')\setminus(V\cup V'))=0$ and $W'\in\dom\theta$, while

\Centerline{$\theta W'=\theta V=\lambda'V
\le\lambda'W$.}

Since $E$ and $H$ are arbitrary, $W\cap(F_n\times G_n)\in\dom\theta$
(251H) and $\theta(W\cap(F_n\times G_n))\le\lambda'W$.   Since
$\sequencen{F_n}$, $\sequencen{G_n}$ are non-decreasing sequences with
unions $X$, $Y$ respectively,

\Centerline{$\theta W=\sup_{n\in\Bbb N}\theta(W\cap(F_n\times G_n))
\le\lambda'W$.  \Qed}

\medskip

{\bf (e)} In the same way, $\lambda'W$ is defined and less than or equal
to $\theta W$ for every $W\in\dom\theta$.   \Prf\ The arguments are very
similar, but a refinement seems to be necessary at the last stage.
Take $n\in\Bbb N$, and $E\in\Sigma$, $H\in\Tau$ such that $\mu E$
and $\nu H$ are both finite.   Set $E'=E\cap F_n$, $H'=H\cap G_n$ and
$W'=W\cap(E'\times H')$.   Then $W'\in\dom\theta$, while $\mu'E'\le
2^n\mu E$ and $\nuprime H'\le 2^n\nu H$ are finite.   This time, there are
$V$, $V'\in\Sigma\tensorhat\Tau$ such that $V\subseteq W'$,
$V'\subseteq(E'\times H')\setminus W'$, $\theta V=\theta W'$ and
$\theta V'=\theta((E'\times H')\setminus W')$.   Accordingly

\Centerline{$\lambda'V+\lambda'V'=\theta V+\theta V'
=\theta(E'\times H')=\lambda'(E'\times H')$,}

\noindent so that $\lambda'W'$ is defined and equal to $\theta W'$.

What this means is that $W\cap(F_n\times G_n)\cap(E\times H)\in\Cal A$
whenever $\mu E$ and $\nu H$ are finite.   So
$W\cap(F_n\times G_n)\in\Lambda$, by 251H;
as $n$ is arbitrary, $W\in\Lambda$ and $\lambda'W$ is defined.

\Quer\ Suppose, if possible, that $\lambda'W>\theta W$.   Then there is
some $n\in\Bbb N$ such that $\lambda'(W\cap(F_n\times G_n))>\theta W$.
Because $\lambda$ is semi-finite, 213B tells us that there is some
$\lambda$-simple function $h$ such that
$h\le(f\otimes g)\times\chi(W\cap(F_n\times G_n))$ and
$\int h\,d\lambda>\theta W$;  setting $V=\{(x,y):h(x,y)>0\}$, we see that $V\subseteq W\cap(F_n\times G_n)$,
$\lambda V$ is defined and finite and $\lambda'V>\theta W$.   Now there
must be sets $E\in\Sigma$, $H\in\Tau$ such that $\mu E$ and $\nu F$ are
both finite and
$\lambda(V\setminus(E\times H))<4^{-n}(\lambda'V-\theta W)$.   But in this case $V\in\Lambda\subseteq\dom\theta$ (by (d)), so we can apply the argument just above to $V$ and conclude that
$V\cap(E\times H)=V\cap(F_n\times G_n)\cap(E\times H)$ belongs to
$\Cal A$.   And now

$$\eqalign{\lambda'V
&=\lambda'(V\cap(E\times H))+\lambda'(V\setminus(E\times H))\cr
&\le\theta(V\cap(E\times H))+4^n\lambda(V\setminus(E\times H))
<\theta V+\lambda'V-\theta W
\le\lambda'V,\cr}$$

\noindent which is absurd.\ \Bang

So $\lambda'W$ is defined and not greater than $\theta W$.\
\Qed

\medskip

{\bf (f)} Putting this together with (d), we see that $\lambda'=\theta$,
as claimed.
}%end of proof of 253I

\cmmnt{\medskip

\noindent{\bf Remark} If $\mu'$ and $\nuprime$ are totally finite, so that
they are `truly continuous' with respect to $\mu$ and $\nu$ in the sense
of 232Ab, then $f$ and $g$ are integrable, so $f\otimes g$ is
$\lambda$-integrable, and $\theta=\lambda'$ is truly continuous with
respect to $\lambda$.

The proof above can be simplified using fragments of the general theory
of complete locally determined spaces, which will be given in \S412 %412M
in Volume 4.}%end of comment

\leader{*253J}{Upper \dvrocolon{integrals}}\cmmnt{ The idea of 253D
can be repeated in terms of upper integrals, as follows.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces, with c.l.d.\ product measure $\lambda$.
Then for any functions $f$ and $g$, defined on conegligible subsets of
$X$ and $Y$ respectively, and taking values in $[0,\infty]$,

\Centerline{$\overlineint f\otimes g\,d\lambda
=\overlineint fd\mu\cdot\overlineint g\,d\nu$.}

\cmmnt{\medskip

\noindent{\bf Remark} Here $(f\otimes g)(x,y)=f(x)g(y)$ for all
$x\in\dom f$ and $y\in\dom g$, taking $0\cdot\infty=0$, as in \S135.
}%end of comment

\proof{{\bf (a)} I show first that
$\overline{\int}f\otimes g\le\overline{\int}f\overline{\int}g$.   \Prf\
If $\overline{\int}f=0$, then $f=0$ a.e., so $f\otimes g=0$ a.e.\ and
the result is immediate.   The same argument applies if
$\overline{\int}g=0$.   If both $\overline{\int}f$ and
$\overline{\int}g$ are non-zero, and either is infinite, the result is
trivial.   So let us suppose that both are finite.   In this case there
are integrable $f_0$, $g_0$ such that $f\leae f_0$, $g\leae g_0$,
$\overline{\int}f=\int f_0$ and $\overline{\int}g=\int g_0$ (133Ja/135Ha).
So $f\otimes g\leae f_0\otimes g_0$, and

\Centerline{$
\overlineint f\otimes g\le\int f_0\otimes g_0
=\int f_0\int g_0=\overlineint f\overlineint g$,}

\noindent by 253D.\ \Qed

\medskip

{\bf (b)} For the reverse inequality, we need consider only the case in
which $\overline{\int}f\otimes g$ is finite, so that there is a
$\lambda$-integrable function $h$ such that $f\otimes g\leae h$ and
$\overline{\int}f\otimes g=\int h$.   Set

\Centerline{$f_0(x)=\int h(x,y)\nu(dy)$}

\noindent whenever this is defined in $\Bbb R$, which is almost
everywhere, by Fubini's theorem (252B-252C).   Then
$f_0(x)\ge f(x)\overline{\int}g\,d\nu$ for every
$x\in\dom f_0\cap\dom f$, which is a conegligible set in $X$;  so

\Centerline{$
\overlineint f\otimes g=\int h\,d\lambda
=\int f_0d\mu\ge\overlineint f\overlineint g$,}

\noindent as required.
}%end of proof of 253J

\leader{*253K}{}\cmmnt{ A similar argument applies to upper
integrals of sums, as follows.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces, with c.l.d.\ product measure $\lambda$.   Then for
any real-valued functions $f$, $g$ defined on conegligible subsets of
$X$, $Y$ respectively,

\Centerline{$
\overlineint f(x)+g(y)\,\lambda(d(x,y))
=\overlineint f(x)\mu(dx)+\overlineint g(y)\nu(dy)$,}

\noindent at least when the right-hand side is defined in
$[-\infty,\infty]$.

\proof{Set $h(x,y)=f(x)+g(y)$ for $x\in\dom f$ and $y\in\dom g$, so that
$\dom h$ is $\lambda$-conegligible.

\medskip

{\bf (a)} As in 253J, I start by showing that
$\overline{\int}h\le\overline{\int}f+\overline{\int}g$.   \Prf\ If
either $\overline{\int}f$ or $\overline{\int}g$ is $\infty$, this is
trivial.   Otherwise, take integrable functions $f_0$, $g_0$ such that
$f\leae f_0$ and $g\leae g_0$.   Set
$h_0=(f_0\otimes\chi Y)+(\chi X\otimes g_0)$;  then
$h\le h_0\,\,\lambda$-a.e., so

\Centerline{$
\overlineint h\,d\lambda\le\int h_0d\lambda
=\int f_0d\mu+\int g_0d\nu$.}

\noindent As $f_0$, $g_0$ are arbitrary,
$\overline{\int}h\le\overline{\int}f+\overline{\int}g$.\ \Qed

\medskip

{\bf (b)} For the reverse inequality, suppose that $h\le h_0$ for
$\lambda$-almost every $(x,y)$, where $h_0$ is $\lambda$-integrable.
Set $f_0(x)=\int h_0(x,y)\nu(dy)$ whenever this is defined in $\Bbb R$.
Then $f_0(x)\ge f(x)+\overline{\int}g\,d\nu$ whenever $x\in\dom
f\cap\dom f_0$, so

\Centerline{$\int h_0\,d\lambda
=\int f_0d\mu\ge\overlineint fd\mu+\overlineint g\,d\nu$.}

\noindent As $h_0$ is arbitrary,
$\overline{\int}h\ge\overline{\int}f+\overline{\int}g$, as required.
}%end of proof of 253K

\leader{253L}{Complex spaces} As usual, the ideas\cmmnt{ of 253F and
253H} apply essentially unchanged to complex $L^1$ spaces.   Writing
$L^1_{\Bbb C}(\mu)$, etc., for the complex $L^1$ spaces involved, we
have the following\cmmnt{ results}.
Throughout, let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$
be measure spaces, and $\lambda$ the c.l.d.\ product measure on
$X\times Y$.

\spheader 253La If $f\in\eusm L^0_{\Bbb C}(\mu)$ and
$g\in\eusm L^0_{\Bbb C}(\nu)$ then $f\otimes g$, defined by the formula
$(f\otimes g)(x,y)=f(x)g(y)$ for $x\in\dom f$ and $y\in\dom g$, belongs
to $\eusm L^0_{\Bbb C}(\lambda)$.

\spheader 253Lb If $f\in\eusm L^1_{\Bbb C}(\mu)$ and
$g\in\eusm L^1_{\Bbb C}(\nu)$ then
$f\otimes g\in\eusm L^1_{\Bbb C}(\lambda)$ and
$\int f\otimes g\,d\lambda=\int fd\mu\int g\,d\nu$.

\spheader 253Lc We have a bilinear operator $(u,v)\mapsto u\otimes v:
L^1_{\Bbb C}(\mu)\times L^1_{\Bbb C}(\nu)\to L^1_{\Bbb C}(\lambda)$
defined by writing $f^{\ssbullet}\otimes g^{\ssbullet}=(f\otimes
g)^{\ssbullet}$ for all $f\in\eusm L^1_{\Bbb C}(\mu)$, $g\in \eusm
L^1_{\Bbb C}(\nu)$.

\spheader 253Ld If $W$ is any complex Banach space and
$\phi:L^1_{\Bbb C}(\mu)\times L^1_{\Bbb C}(\nu)\to W$ is any bounded
bilinear operator, then there is a unique bounded linear operator
$T:L^1_{\Bbb C}(\lambda)\to W$ such that $T(u\otimes v)=\phi(u,v)$ for
every $u\in L^1_{\Bbb C}(\mu)$ and $v\in L^1_{\Bbb C}(\nu)$, and
$\|T\|=\|\phi\|$.

\spheader 253Le If $\mu$ and $\nu$ are complete probability measures,
and $\Lambda_1=\{E\times Y:E\in\Sigma\}$, then for any $f\in\eusm
L^1_{\Bbb C}(\lambda)$ we have a conditional expectation $g$ of $f$ on
$\Lambda_1$ given by setting $g(x,y)=\int f(x,z)\nu(dz)$ whenever this
is defined.

\exercises{
\leader{253X}{Basic exercises $\pmb{>}$(a)}
%\spheader 253Xa
Let $U$, $V$ and $W$ be linear
spaces.   Show that the set of
bilinear operators from $U\times V$ to $W$ has a natural linear structure
agreeing with those of $\eurm L(U;\eurm L(V;W))$ and $\eurm L(V;\eurm
L(U;W))$, writing $\eurm L(U;W)$
for the linear space of linear operators from $U$ to $W$.

\sqheader 253Xb Let $U$, $V$ and $W$ be normed spaces.   (i)
Show that for a
bilinear operator $\phi:U\times V\to W$ the following are equiveridical:
($\alpha$) $\phi$ is bounded in the sense of 253Ab;  ($\beta$) $\phi$ is
continuous;  ($\gamma$) $\phi$ is continuous at some point of $U\times
V$.   (ii) Show that the space of bounded bilinear operators from $U\times V$
to $W$ is a linear subspace of the space of all bilinear operators from
$U\times V$ to $W$, and that the functional $\|\,\|$ defined in 253Ab is
a norm, agreeing with the norms of $\eurm B(U;\eurm B(V;W))$ and $\eurm
B(V;\eurm B(U;W))$, writing $\eurm B(U;W)$ for the normed space of
bounded linear operators from $U$ to $W$.

\spheader 253Xc Let
$(X_1,\Sigma_1,\mu_1),\ldots,(X_n,\Sigma_n,\mu_n)$ be
measure spaces, and $\lambda$ the c.l.d.\ product measure on
$X_1\times\ldots\times X_n$, as described in 251W.   Let $W$ be a
Banach space, and suppose that
$\phi:L^1(\mu_1)\times\ldots\times L^1(\mu_n)\to W$ is {\bf multilinear}
(that is, linear in each variable
separately) and {\bf bounded} (that is,
$\|\phi\|=\sup\{\phi(u_1,\ldots,u_n):
\|u_i\|_1\le 1\Forall i\le n\}<\infty$).   Show that there is a unique
bounded linear operator
$T:L^1(\lambda)\to W$ such that $T\otimes=\phi$, where
$\otimes:L^1(\mu_1)\times\ldots\times L^1(\mu_n)\to L^1(\lambda)$ is a
canonical multilinear operator (to be defined).

\spheader 253Xd Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\lambda$ the c.l.d.\ product measure on $X\times Y$.   Show
that if $A\subseteq L^1(\mu)$ and $B\subseteq L^1(\nu)$ are both
uniformly integrable, then $\{u\otimes v:u\in A,\,v\in B\}$ is uniformly
integrable in $L^1(\lambda)$.

\sqheader 253Xe Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces and $\lambda$ the c.l.d.\ product measure on $X\times Y$.
Show that

\quad (i) we have a bilinear operator
$(u,v)\mapsto u\otimes v:L^0(\mu)\times L^0(\nu)\to L^0(\lambda)$
given by setting
$f^{\ssbullet}\otimes g^{\ssbullet}=(f\otimes g)^{\ssbullet}$ for all
$f\in\eusm L^0(\mu)$ and $g\in \eusm L^0(\nu)$;

\quad (ii) if $1\le p\le\infty$ then $u\otimes v\in L^p(\lambda)$ and
$\|u\otimes v\|_p=\|u\|_p\|v\|_p$ for all $u\in L^p(\mu)$ and
$v\in L^p(\nu)$;

\quad (iii) if $u$, $u'\in L^2(\mu)$ and $v$, $v'\in L^2(\nu)$ then the
inner product $\innerprod{u\otimes v}{u'\otimes v'}$, taken in
$L^2(\lambda)$, is just $\innerprod{u}{u'}\innerprod{v}{v'}$;

\quad (iv) the map
$(u,v)\mapsto u\otimes v:L^0(\mu)\times L^0(\nu)\to L^0(\lambda)$
is continuous if $L^0(\mu)$, $L^0(\nu)$ and $L^0(\lambda)$
are all given their topologies of convergence in measure.

\spheader 253Xf In 253Xe, assume that $\mu$ and $\nu$ are semi-finite.
Show that if $u_0,\ldots,u_n$ are linearly independent members of
$L^0(\mu)$ and $v_0,\ldots,v_n\in L^0(\nu)$ are not all $0$, then
$\sum_{i=0}^nu_i\otimes v_i\ne 0$ in $L^0(\lambda)$.   ({\it Hint\/}:
start by finding sets $E\in\Sigma$, $F\in\Tau$ of finite measure such
that $u_0\times\chi E^{\ssbullet},\ldots,u_n\times\chi E^{\ssbullet}$
are linearly independent and
$v_0\times\chi F^{\ssbullet},\ldots,v_n\times\chi F^{\ssbullet}$
are not all $0$.)

\spheader 253Xg In 253Xe, assume that $\mu$ and $\nu$ are semi-finite.
If $U$, $V$ are linear subspaces of $L^0(\mu)$ and $L^0(\nu)$
respectively, write $U\otimes V$ for the linear subspace of
$L^0(\lambda)$ generated by $\{u\otimes v:u\in U,\,v\in V\}$.
Show that if $W$ is any linear space and $\phi:U\times V\to W$ is a
bilinear operator, there is a unique linear operator $T:U\otimes V\to W$
such that $T(u\otimes v)=\phi(u,v)$ for all $u\in U$, $v\in V$.
({\it Hint\/}: start by showing that if $u_0,\ldots,u_n\in U$ and
$v_0,\ldots,v_n\in V$ are such that $\sum_{i=0}^nu_i\otimes v_i=0$, then
$\sum_{i=0}^n\phi(u_i,v_i)=0$ -- do this by expressing the $u_i$ as
linear combinations of some linearly independent family and applying
253Xf.)

\sqheader 253Xh Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be complete
probability spaces, with c.l.d.\ product measure $\lambda$.   Suppose
that $p\in[1,\infty]$ and that $f\in\eusm L^p(\lambda)$.   Set
$g(x)=\int f(x,y)\nu(dy)$ whenever this is defined.   Show that
$g\in\eusm L^p(\mu)$ and that $\|g\|_p\le\|f\|_p$.   ({\it Hint\/}:  253H,
244M.)

\spheader 253Xi Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, with c.l.d.\ product measure $\lambda$, and
$p\in\coint{1,\infty}$.   Show that $\{w:w\in L^p(\lambda),\,w\ge 0\}$
is the closed convex hull in $L^p(\lambda)$ of
$\{u\otimes v:u\in L^p(\mu),\,v\in L^p(\nu),\,u\ge 0,\,v\ge 0\}$
(see 253Xe(ii) above).

\leader{253Y}{Further exercises (a)}
%\spheader 253Ya
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure spaces, and
$\lambda_0$ the primitive product measure on $X\times Y$.  Show that if
$f\in\eusm L^0(\mu)$ and $g\in\eusm L^0(\nu)$, then
$f\otimes g\in\eusm L^0(\lambda_0)$.
%253C

\spheader 253Yb Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be measure spaces, and $\lambda_0$ the primitive
product measure on $X\times Y$.   Show that if $f\in\eusm L^1(\mu)$ and
$g\in\eusm L^1(\nu)$, then $f\otimes g\in\eusm L^1(\lambda_0)$ and
$\int f\otimes g\,d\lambda_0=\int f\,d\mu\int g\,d\nu$.
%253Ya, 253C

\spheader 253Yc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and $\lambda_0$, $\lambda$ the primitive and c.l.d.\
product measures on $X\times Y$.   Show that the embedding 
$\eusm L^1(\lambda_0)\embedsinto\eusm L^1(\lambda)$
induces a Banach lattice isomorphism between $L^1(\lambda_0)$ and
$L^1(\lambda)$.
%253D

\spheader 253Yd Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ be strictly
localizable measure spaces, with c.l.d.\ product measure $\lambda$.
Show that $L^{\infty}(\lambda)$ can be identified with $L^1(\lambda)^*$.
Show that under this identification
$\{w:w\in L^{\infty}(\lambda),\,w\ge 0\}$ is the weak*-closed convex hull of
$\{u\otimes v:u\in L^{\infty}(\mu),\,v\in L^{\infty}(\nu),
\,u\ge 0,\,v\ge 0\}$.
%253G, 253Xe

\spheader 253Ye Find a version of 253J valid when one of $\mu$, $\nu$
is not $\sigma$-finite.
%253J

\spheader 253Yf Let $(X,\Sigma,\mu)$ be any measure space and $V$ any
Banach space.   Write $\eusm L^1_V=\eusm L^1_V(\mu)$ for the set of
functions $f$
such that ($\alpha$) $\dom f$ is a conegligible subset of $X$ ($\beta$)
$f$ takes values in $V$ ($\gamma$) there is a conegligible set
$D\subseteq\dom
f$ such that $f[D]$ is separable and $D\cap f^{-1}[G]\in\Sigma$ for
every open set $G\subseteq V$ ($\delta$) the integral
$\int\|f(x)\|\mu(dx)$ is finite.   (These are the {\bf Bochner
integrable} functions from $X$ to
$V$.)   For $f$, $g\in\eusm L^1_V$ write $f\sim g$ if
$f=g\,\,\mu$-a.e.;  let $L^1_V$ be the set of equivalence classes
in $\eusm L^1_V$ under $\sim$.   Show that

\quad (i) $f+g$, $cf\in\eusm L^1_V$ for all $f$, 
$g\in\eusm L^1_V$, $c\in\Bbb R$;

\quad (ii) $L^1_V$ has a natural linear space structure, defined by
writing $f^{\ssbullet}+g^{\ssbullet}=(f+g)^{\ssbullet}$,
$cf^{\ssbullet}=(cf)^{\ssbullet}$ for $f$, $g\in\eusm L^1_V$ and
$c\in\Bbb R$;

\quad (iii) $L^1_V$ has a norm $\|\,\|$, defined by writing
$\|f^{\ssbullet}\|=\int\|f(x)\|\mu(dx)$ for $f\in\eusm L^1_V$;

\quad (iv) $L^1_V$ is a Banach space under this norm;

\quad (v) there is a natural map
$\otimes:\eusm L^1\times V\to\eusm L^1_V$ defined by writing
$(f\otimes v)(x)=f(x)v$ when $f\in\eusm L^1=\eusm L^1_{\Bbb R}(\mu)$,
$v\in V$ and $x\in\dom f$;

\quad (vi) there is a canonical bilinear operator
$\otimes:L^1\times V\to L^1_V$ defined by writing
$f^{\ssbullet}\otimes v=(f\otimes v)^{\ssbullet}$ for $f\in\eusm L^1$ and
$v\in V$;

\quad (vii) whenever $W$ is a Banach space and 
$\phi:L^1\times V\to W$ is a bounded bilinear operator, 
there is a unique bounded
linear operator $T:L^1_V\to W$ such that $T(u\otimes v)=\phi(u,v)$
for all $u\in L^1$ and $v\in V$, and $\|T\|=\|\phi\|$.   
%new 2008
(When $W=V$ and
$\phi(u,v)=(\int u)v$ for $u\in L^1$ and $v\in V$, $Tf^{\ssbullet}$ 
is called the {\bf Bochner integral} of $f$.)

%253+

\spheader 253Yg Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and
$\lambda_0$ the primitive product measure on $X\times Y$.   If $f$ is a
$\lambda_0$-integrable function, write $f_x(y)=f(x,y)$ whenever this is
defined.   Show that we have a map $x\mapsto f_x^{\ssbullet}$ from a
conegligible subset
$D_0$ of $X$ to $L^1(\nu)$.   Show that this map is a Bochner integrable
function, as defined in 253Yf, 
%new 2008
and that its Bochner integral is $\int fd\lambda_0$.
%253Yf, 253+

\spheader 253Yh Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
measure spaces, and
suppose that $\phi$ is a function from $X$ to a separable subset of
$L^1(\nu)$ which is measurable in the sense that $\phi^{-1}[G]\in\Sigma$
for every open $G\subseteq L^1(\nu)$.   Show that
there is a $\Lambda$-measurable function $f$ from $X\times Y$ to $\Bbb
R$, where $\Lambda$ is the domain of the c.l.d.\ product measure on
$X\times Y$, such that $\phi(x)=f_x^{\ssbullet}$ for every $x\in X$,
writing $f_x(y)=f(x,y)$ for $x\in X$, $y\in Y$.
%253+

\spheader 253Yi Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, and $\lambda$ the c.l.d.\ product measure on $X\times Y$.   Show
that 253Yg provides a canonical identification between $L^1(\lambda)$
and $L^1_{L^1(\nu)}(\mu)$.
%253Yg, 253Yi, 253+

\spheader 253Yj Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be complete
locally determined measure spaces, with c.l.d.\ product measure
$\lambda$.   (i) Suppose that $K\in\eusm L^2(\lambda)$,
$f\in\eusm L^2(\mu)$.   Show that $h(y)=\int K(x,y)f(x)dx$ is defined for almost
all $y\in Y$ and that $h\in\eusm L^2(\nu)$.   ({\it Hint\/}:  to see
that $h$ is defined a.e., consider $\int_{E\times F}K(x,y)f(x)d(x,y)$
for $\mu E$, $\nu F<\infty$;  to see that $h\in\eusm L^2$ consider
$\int h\times g$ where $g\in\eusm L^2(\nu)$.)   (ii) Show that the map
$f\mapsto h$ corresponds to a bounded linear operator
$T_K:L^2(\mu)\to L^2(\nu)$.   (iii) Show that the map $K\mapsto T_K$
corresponds to a bounded linear operator, of norm at most $1$, from
$L^2(\lambda)$ to $\eurm B(L^2(\mu);L^2(\nu))$.
%253+

\spheader 253Yk Suppose that $p$, $q\in[1,\infty]$ and that
$\bover1p+\bover1q=1$, interpreting $\bover1{\infty}$ as $0$ as usual.
Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ be complete locally determined
measure spaces with c.l.d.\ product measure $\lambda$.    Show that the
ideas of 253Yj can be used to define a bounded linear operator, of norm
at most $1$, from $L^p(\lambda)$ to $\eurm B(L^q(\mu);L^p(\nu))$.
%253Yj, 253+

\spheader 253Yl In 253Xc, suppose that $W$ is a Banach lattice.   Show
that the following are equiveridical:  (i) $Tu\ge 0$ whenever
$u\in L^1(\lambda)$;  (ii) $\phi(u_1,\ldots,u_n)\ge 0$ whenever
$u_i\ge 0$ in $L^1(\mu_i)$ for each $i\le n$.
%253G, 253Xc
}%end of exercises

\endnotes{
\Notesheader{253} Throughout the main arguments of this section, I have
written the results in terms of the c.l.d.\ product measure;  of course
the isomorphism noted in 253Yc means that they could just as well have
been expressed in terms of the primitive product measure.   The
more restricted notion of integrability with respect to the primitive
product measure is indeed the one appropriate for the ideas of 253Yg.

Theorem 253F is a `universal mapping theorem';
it asserts that every bounded bilinear operator on $L^1(\mu)\times
L^1(\nu)$ factors through $\otimes:L^1(\mu)\times L^1(\nu)\to
L^1(\lambda)$, at least if the range space is a Banach space.   It is
easy to see that this property defines the pair $(L^1(\lambda),\otimes)$
up to Banach space isomorphism, in the following sense:  if $V$ is a
Banach space, and $\psi:L^1(\mu)\times L^1(\nu)\to V$ is a bounded
bilinear operator such that for every bounded bilinear operator $\phi$ from
$L^1(\mu)\times L^1(\nu)$ to any Banach space $W$ there is a unique
bounded linear operator $T:V\to W$ such that $T\psi = \phi$ and
$\|T\|=\|\phi\|$, then there is an isometric Banach space isomorphism
$S:L^1(\lambda)\to V$ such that $S\otimes=\psi$.    There is of course a
general theory of bilinear operators between Banach spaces;  in the language
of this theory, $L^1(\lambda)$ is, or is isomorphic to, the
`projective tensor product' of $L^1(\mu)$ and $L^1(\nu)$.   For an
introduction to
this subject, see {\smc Defant \& Floret 93}, \S I.3, or 
{\smc Semadeni 71}, \S20.   
I should perhaps emphasise, for the sake of those who
have not encountered tensor products before, that this theorem is
special to $L^1$ spaces.   While some of the same ideas can be applied
to other function spaces (see 253Xe-253Xg), there is no other class to
which 253F applies.

There is also a theory of tensor products of Banach lattices, for which
I do not think we are quite ready (it needs general ideas about ordered
linear spaces for which I mean to wait until Chapter 35 in the next volume).   However
253G shows that the ordering, and therefore the Banach lattice
structure, of
$L^1(\lambda)$ is determined by the ordering of $L^1(\mu)$ and
$L^1(\nu)$ and the map
$\otimes:L^1(\mu)\times L^1(\nu)\to L^1(\lambda)$.

The conditional expectation operators described in 253H are of very
great importance, largely because in this special context we have a
realization of the conditional expectation operator as a function $P_0$
from $\eusm L^1(\lambda)$ to $\eusm L^1(\lambda\restr\Lambda_1)$, not
just as a function from
$L^1(\lambda)$ to $L^1(\lambda\restr\Lambda_1)$, as in 242J.
As described here, $P_0(f+f')$ need not be equal, in the strict sense,
to $P_0f+P_0f'$;  it can have a larger domain.   In applications,
however, one might be willing to restrict attention to the linear space
$\eusm U$  of bounded $\Sigma\tensorhat\Tau$-measurable functions
defined everywhere on $X\times Y$, so that $P_0$ becomes an operator
from $\eusm U$ to itself (see 252P).
}%end of notes

\discrpage

\frfilename{mt254.tex}
\versiondate{23.2.16}
\copyrightdate{2002}

\def\chaptername{Product measures}
\def\sectionname{Infinite products}

\newsection{254}

I come now to the second basic idea of this
chapter:  the description of a product measure on the product of a
(possibly large) family of probability spaces.   The section begins with
a construction on similar lines to that of \S251 (254A-254F) and its
defining property in terms of \imp\ functions (254G).   I discuss the
usual measure on $\{0,1\}^I$
(254J-254K), subspace measures (254L) and various properties of
subproducts (254M-254T), including a study of the associated conditional
expectation operators (254R-254T).

\leader{254A}{Definitions (a)} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be
a family of probability
spaces.   Set $X=\prod_{i\in I}X_i$, the family of functions $x$ with
domain $I$ such that $x(i)\in X_i$ for every $i\in I$.   In this
context, I will say that a {\bf measurable cylinder} is a subset of $X$
expressible in the form

\Centerline{$C=\prod_{i\in I}C_i,$}

\noindent where $C_i\in\Sigma_i$ for every $i\in I$ and
$\{i:C_i\ne X_i\}$ is finite.   Note that for a non-empty $C\subseteq X$
this expression is unique.   \prooflet{\Prf\ Suppose that
$C=\prod_{i\in I}C_i=\prod_{i\in I}C'_i$.   For each $i\in I$ set

\Centerline{$D_i=\{x(i):x\in C\}.$}

\noindent Of course $D_i\subseteq C_i$.   Because $C\ne\emptyset$, we
can fix on some $z\in C$.   If $i\in I$ and $t\in C_i$,
consider $x\in X$ defined by setting

\Centerline{$x(i)=t$, \quad$x(j)=z(j)\text{ for }j\ne i;$}

\noindent then $x\in C$ so $t=x(i)\in D_i$.   Thus $D_i=C_i$ for
$i\in I$.   Similarly, $D_i=C'_i$.\ \Qed}

\header{254Ab}{\bf (b)} We can therefore define a functional
$\theta_0:\Cal C\to[0,1]$, where $\Cal C$ is the set of measurable
cylinders, by setting

\Centerline{$\theta_0C=\prod_{i\in I}\mu_iC_i$}

\noindent whenever  $C_i\in\Sigma_i$ for every $i\in I$ and
$\{i:C_i\ne X_i\}$ is finite\cmmnt{,
noting that only finitely many terms in the product can differ from $1$,
so that it can safely be treated as a finite product.   If
$C=\emptyset$, one of the $C_i$ must be empty, so $\theta_0C$ is surely
$0$, even though the expression of $C$ as $\prod_{i\in I}C_i$
is no longer unique}.

\header{254Ac}{\bf (c)} Now define $\theta:\Cal PX\to[0,1]$ by setting

\Centerline{$\theta A
=\inf\{\sum_{n=0}^{\infty}\theta_0C_n:C_n\in\Cal C
  \text{ for every }n\in\Bbb N$, $A\subseteq\bigcup_{n\in\Bbb N}C_n\}$.}

\leader{254B}{Lemma} The functional $\theta$ defined in 254Ac is always
an outer measure on $X$.

\proof{ Use exactly the same arguments as those in 251B above.}

\leader{254C}{Definition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any indexed family of
probability spaces, and $X$ the Cartesian
product $\prod_{i\in I}X_i$.   The {\bf product measure} on $X$ is the
measure defined by \Caratheodory's method\cmmnt{ (113C)} from the
outer measure $\theta$ defined in 254A.

\cmmnt{
\leader{254D}{Remarks (a)} In 254Ab, I asserted that if $C\in\Cal C$ and
no $C_i$ is empty, then nor is $C=\prod_{i\in I}C_i$.   This is the
`Axiom of Choice':  the product of
any family $\langle C_i\rangle_{i\in I}$ of
non-empty sets is non-empty, that is, there is a `choice function'
$x$ with domain $I$ picking out a distinguished member $x(i)$ of each
$C_i$.   In this volume I have not attempted to be scrupulous in
indicating uses of the axiom of choice.   In fact the use here is not an
absolutely vital one;  I mean, the theory of infinite products, even
uncountable products, of probability spaces does not change character
completely in the absence of the full axiom of choice (provided, that
is, that we allow ourselves to use the countable axiom of choice).   The
point is that all we really need, in the present context, is that
$X=\prod_{i\in I}X_i$ should be non-empty;  and in many contexts we can
prove this, for the particular cases of interest, without using the
axiom of choice, by actually exhibiting a member of $X$.   The simplest
case in which this is difficult is when the $X_i$ are uncontrolled Borel
subsets of $[0,1]$;  and even then, if they are presented with coherent
descriptions, we may, with appropriate labour, be able to construct a
member of $X$.   But clearly such a process is liable to slow us down a
good deal, and for the moment I think there is no great virtue in taking
so much trouble.

\header{254Db}{\bf (b)}
I have given this section the title `infinite products', but it
is useful to be able to apply the ideas to finite $I$;  I should mention
in particular the cases $\#(I)\le 2$.

\medskip

\quad {\bf (i)}  If $I=\emptyset$, $X$ consists of the unique function
with domain $I$, the empty function.  If we identify a function with its
graph, then $X$ is actually $\{\emptyset\}$;  in any case, $X$ is to be
a singleton set, with $\lambda X=1$.

\medskip

\quad{\bf (ii)} If $I$ is a singleton $\{i\}$, then we can identify $X$
with $X_i$;  $\Cal C$ becomes identified with $\Sigma_i$ and $\theta_0$
with $\mu_i$, so that $\theta$ can be identified with $\mu_i^*$
and the `product measure' becomes the measure on $X_i$
defined from $\mu_i^*$, that is, the completion of $\mu_i$\cmmnt{
(see 213Xa(iv))}.

\medskip

\quad{\bf (iii)} If $I$ is a doubleton $\{i,j\}$, then we can identify
$X$ with $X_i\times X_j$;  in this case the definitions of 254A and 254C
match exactly with those of
251A and 251C, so that $\lambda$ here can be identified with the
primitive product measure as defined in 251C.   Because $\mu_i$ and
$\mu_j$ are both totally finite, this agrees with the c.l.d.\ product
measure of 251F.

\spheader 254Dc\dvAnew{2018}
In Volume 4, when considering products of probability spaces
endowed with certain kinds of topology, I will introduce some alternative
product measures.   In such contexts I may speak of the product measure here as
the `ordinary' product measure.
}%end of comment


\leader{254E}{Definition} Let $\langle X_i\rangle_{i\in I}$ be any
family of sets, and $X=\prod_{i\in I}X_i$.   If $\Sigma_i$ is a
$\sigma$-subalgebra of subsets of $X_i$ for each $i\in I$, I write
$\Tensorhat_{i\in I}\Sigma_i$ for the $\sigma$-algebra of subsets of $X$
generated by

\Centerline{$\{\{x:x\in X$, $x(i)\in E\}: i\in I$, $E\in\Sigma_i\}$.}

\cmmnt{\noindent (Compare 251D.)}

\leader{254F}{Theorem} Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of probability spaces, and let $\lambda$ be the product
measure on $X=\prod_{i\in I}X_i$\cmmnt{ defined as in 254C};  let
$\Lambda$ be its domain.

(a) $\lambda X=1$.

(b) If $E_i\in\Sigma_i$ for every $i\in I$, and $\{i:E_i\ne X_i\}$ is
countable, then $\prod_{i\in I}E_i\in\Lambda$, and
$\lambda(\prod_{i\in I}E_i)=\prod_{i\in I}\mu_iE_i$.   In particular,
$\lambda C=\theta_0C$
for every measurable cylinder $C$,\cmmnt{ as defined in 254A,} and if
$j\in I$ then $x\mapsto x(j):X\to X_j$ is inverse-measure-preserving.

(c) $\Tensorhat_{i\in I}\Sigma_i\subseteq\Lambda$.

(d) $\lambda$ is complete.

(e) For every $W\in\Lambda$ and $\epsilon>0$ there is a finite family
$C_0,\ldots,C_n$ of measurable cylinders such that
$\lambda(W\symmdiff\bigcup_{k\le n}C_k)\penalty-100\le\epsilon$.

(f) For every $W\in\Lambda$ there are $W_1$,
$W_2\in\Tensorhat_{i\in I}\Sigma_i$ such that
$W_1\subseteq W\subseteq W_2$ and $\lambda(W_2\setminus W_1)=0$.

\cmmnt{\medskip

\noindent{\bf Remark} Perhaps I should pause to interpret the
product $\prod_{i\in I}\mu_iE_i$.   Because all the $\mu_iE_i$ belong to
$[0,1]$, this is simply $\inf_{J\subseteq I,J\text{ is
finite}}\prod_{i\in J}\mu_iE_i$, taking the empty product to be $1$.
}%end of comment

\proof{ Throughout this proof, define $\Cal C$, $\theta_0$
and $\theta$ as in 254A.   I will write out an argument which applies to
finite $I$ as well as infinite $I$, but you may reasonably
prefer to assume that $I$ is infinite on first reading.

\medskip

{\bf (a)} Of course $\lambda X=\theta X$, so I have to show that
$\theta X=1$.   Because $X$, $\emptyset\in\Cal C$ and
$\theta_0X=\prod_{i\in I}\mu_iX_i=1$ and $\theta_0\emptyset=0$,

\Centerline{$\theta X\le\theta_0X+\theta_0\emptyset+\ldots=1$.}

\noindent   I therefore have to show
that $\theta X\ge 1$.   \Quer\  Suppose, if possible, otherwise.

\medskip

\quad{\bf (i)} There is a sequence $\sequencen{C_n}$ in $\Cal C$,
covering $X$, such that $\sum_{n=0}^{\infty}\theta_0C_n<1$.    For each
$n\in\Bbb N$, express $C_n$ as $\{x:x(i)\in E_{ni}\Forall i\in I\}$,
where every $E_{ni}\in\Sigma_i$ and $J_n=\{i:E_{ni}\ne X_i\}$ is finite.
No $J_n$ can be empty, because
$\theta_0C_n<1=\theta_0X$;  set $J=\bigcup_{n\in\Bbb N}J_n$.   Then $J$
is a countable non-empty subset of $I$.   Set $K=\Bbb N$ if $J$ is
infinite, $\{k:0\le k<\#(J)\}$ if $J$ is finite;   let
$k\mapsto i_k:K\to J$ be a bijection.

For each $k\in K$, set $L_k=\{i_j:j<k\}\subseteq J$, and set
$\alpha_{nk}=\prod_{i\in I\setminus L_k}\mu_iE_{ni}$ for $n\in\Bbb N$,
$k\in K$.   If $J$ is finite, then we can identify $L_{\#(J)}$ with $J$,
and set $\alpha_{n,\#(J)}=1$ for every
$n$.   We have $\alpha_{n0}=\theta_0C_n$ for each $n$, so
$\sum_{n=0}^{\infty}\alpha_{n0}<1$.   For $n\in\Bbb N$, $k\in K$ and
$t\in X_{i_k}$ set

$$\eqalign{f_{nk}(t)&=\alpha_{n,k+1}\text{ if }t\in E_{n,i_k},\cr
&=0\text{ otherwise}.\cr}$$

\noindent Then

\Centerline{$\int f_{nk}d\mu_{i_k}
=\alpha_{n,k+1}\mu_{i_k}E_{n,i_k}=\alpha_{nk}$.}

\medskip

\quad{\bf (ii)} Choose $t_k\in X_{i_k}$ inductively, for $k\in K$, as
follows.   The inductive hypothesis will be that
$\sum_{n\in M_k}\alpha_{nk}<1$, where
$M_k=\{n:n\in \Bbb N$, $t_j\in E_{n,i_j}\Forall j<k\}$;  of course
$M_0=\Bbb N$, so the induction starts.   Given that

\Centerline{$1>\sum_{n\in M_k}\alpha_{nk}
=\sum_{n\in M_k}\int f_{nk}d\mu_{i_k}
=\int(\sum_{n\in M_k}f_{nk})d\mu_{i_k}$}

\noindent (by B.Levi's theorem), there must be a $t_k\in X_{i_k}$ such
that $\sum_{n\in M_k}f_{nk}(t_{k})<1$.   Now for such a choice of
$t_k$, $\alpha_{n,k+1}=f_{nk}(t_k)$ for every $n\in M_{k+1}$, so that
$\sum_{n\in M_{k+1}}\alpha_{n,k+1}<1$, and the induction continues,
unless $J$ is finite and $k+1=\#(J)$.   In this last case we must just
have $M_{\#(J)}=\emptyset$, because $\alpha_{n,\#(J)}=1$ for every $n$.

\medskip

\quad{\bf (iii)} If $J$ is infinite, we obtain a full sequence
$\sequence{k}{t_k}$;  if $J$ is finite, we obtain just a finite sequence
$\langle t_k\rangle_{k<\#(J)}$.   In either case, there is an $x\in X$
such that $x(i_k)=t_k$ for each $k\in K$.   Now there must be some
$m\in\Bbb N$ such that $x\in C_m$.   Because $J_m=\{i:E_{mi}\ne X_i\}$
is finite, there is a $k\in\Bbb N$ such that $J_m\subseteq L_k$
(allowing $k=\#(J)$ if $J$ is finite).   Now $m\in M_k$, so in fact we
cannot have $k=\#(J)$, and
$\alpha_{mk}=1$, so $\sum_{n\in M_k}\alpha_{nk}\ge 1$, contrary to the
inductive hypothesis.\ \Bang

This contradiction shows that $\theta X=1$.

\medskip

{\bf (b)(i)} I take the particular case first.
Suppose that $j\in I$ and
$E\in\Sigma_j$, and let $C\in\Cal C$;  set $W=\{x:x\in X$, $x(j)\in
E\}$;
then $C\cap W$ and $C\setminus W$ both belong to $\Cal C$, and $\theta_0
C=\theta_0(C\cap W)+\theta_0(C\setminus W)$.   \Prf\ If
$C=\prod_{i\in I}C_i$, where $C_i\in\Sigma_i$ for each $i$, then
$C\cap W=\prod_{i\in I}C_i'$, where $C_i'=C_i$ if $i\ne j$, and
$C_j'=C_j\cap E$;  similarly,
$C\setminus W=\prod_{i\in I}C_i''$, where $C_i''=C_i$ if $i\ne j$, and
$C_j''=C_j\setminus E$.   So both belong to $\Cal C$, and

\Centerline{$\theta_0(C\cap W)+\theta_0(C\setminus W)
=(\mu_j(C_j\cap E)+\mu_j(C_j\setminus E))\prod_{i\ne j}\mu C_i
=\prod_{i\in I}\mu C_i
=\theta_0C$.  \Qed}

\medskip

\quad{\bf (ii)} Now suppose that $A\subseteq X$ is any set, and
$\epsilon>0$.   Then there is a sequence $\sequencen{C_n}$ in $\Cal C$
such that $A\subseteq\bigcup_{n\in\Bbb N}C_n$ and
$\sum_{n=0}^{\infty}\theta_0C_n\le\theta A+\epsilon$.   In this case

\Centerline{$A\cap W\subseteq\bigcup_{n\in\Bbb N}C_n\cap W$,\quad
$A\setminus W\subseteq\bigcup_{n\in\Bbb N}C_n\setminus W,$}

\noindent so

\Centerline{$\theta(A\cap W)\le\sum_{n=0}^{\infty}\theta_0(C_n\cap
W),\quad
\theta(A\setminus W)\le\sum_{n=0}^{\infty}\theta_0(C_n\setminus W),$}

\noindent and

\Centerline{$\theta(A\cap W)+\theta(A\setminus W)
\le\sum_{n=0}^{\infty}\theta_0(C_n\cap W)+\theta_0(C_n\setminus W)
=\sum_{n=0}^{\infty}\theta_0C_n
\le\theta A+\epsilon.$}
\noindent As $\epsilon$ is arbitrary, $\theta(A\cap W)+\theta(A\setminus
W)\le\theta A$;  as $A$ is arbitrary, $W\in\Lambda$.

\medskip

\quad{\bf (iii)}
I show next that if $J\subseteq I$ is finite and $C_i\in\Sigma_i$ for
each $i\in J$, and $C=\{x:x\in X$, $x(i)\in C_i\Forall i\in J\}$,
then $C\in\Lambda$ and  $\lambda C=\prod_{i\in J}\mu_iC_i$.   \Prf\
Induce on $\#(J)$.   If $\#(J)=0$, that is, $J=\emptyset$, then $C=X$
and this is part (a).   For the inductive step to $\#(J)=n+1$, take any
$j\in J$ and set $J'=J\setminus\{j\}$,

\Centerline{$C'=\{x:x\in X$, $x(i)\in C_i\Forall i\in J'\},$}

\Centerline{$C''=C'\setminus C=\{x:x\in C'$, $x(j)\in X_j\setminus
C_j\}.$}

\noindent Then $C$, $C'$, $C''$ all belong to $\Cal C$, and $\theta_0
C'=\prod_{i\in J'}\mu_iC_i=\alpha$ say, $\theta_0 C=\alpha\mu_jC_j$,
$\theta_0 C''=\alpha(1-\mu_jC_j)$.   Moreover, by the inductive
hypothesis, $C'\in\Lambda$ and $\alpha=\lambda C'=\theta C'$.   So
$C=C'\cap\{x:x(j)\in C_j\}\in\Lambda$ by (ii), and $C''=C'\setminus
C\in\Lambda$.

We surely have $\lambda C=\theta C\le\theta_0 C$, $\lambda
C''\le\theta_0 C''$;   but also

\Centerline{$\alpha=\lambda C'=\lambda C+\lambda C''\le\theta_0
C+\theta_0C''=\alpha,$}

\noindent so in fact

\Centerline{$\lambda C=\theta_0 C=\alpha\mu_jC_j
=\prod_{i\in J}\mu C_i$,}

\noindent and the induction proceeds.\ \Qed

\medskip

\quad{\bf (iv)} Now let us return to the general case of a set $W$ of
the form $\prod_{i\in I}E_i$ where $E_i\in\Sigma_i$ for each $i$, and
$K=\{i:E_i\ne X_i\}$ is countable.   If $K$ is finite then
$W=\{x:x(i)\in E_i\Forall i\in K\}$ so $W\in\Lambda$ and

\Centerline{$\lambda W
=\prod_{i\in K}\mu_iE_i=\prod_{i\in I}\mu_iE_i$.}

\noindent    Otherwise, let
$\sequencen{i_n}$ be an enumeration of $K$.   For each $n\in\Bbb N$ set
$W_n=\{x:x\in X$, $x(i_k)\in E_{i_k}\Forall k\le n\}$;  then  we know
that $W_n\in\Lambda$ and that
$\lambda W_n=\prod_{k=0}^n\mu_{i_k}E_{i_k}$.   But $\sequencen{W_n}$ is
a non-increasing sequence with intersection $W$, so $W\in\Lambda$ and

\Centerline{$\lambda W=\lim_{n\to\infty}\lambda W_n
=\prod_{i\in K}\mu_iE_i
=\prod_{i\in I}\mu_iE_i.$}

\medskip

{\bf (c)} is an immediate consequence of (b) and the definition of
$\Tensorhat_{i\in I}\Sigma_i$.

\medskip

{\bf (d)} Because $\lambda$ is constructed by \Caratheodory's method it
must be complete.

\medskip

{\bf (e)} Let $\sequencen{C_n}$ be a sequence in $\Cal C$ such that
$W\subseteq\bigcup_{n\in\Bbb N}C_n$ and
$\sum_{n=0}^{\infty}\theta_0C_n\le\theta W+\bover12\epsilon$.   Set
$V=\bigcup_{n\in\Bbb N}C_n$;  by (b), $V\in\Lambda$.   Let $n\in\Bbb N$
be
such that $\sum_{i=n+1}^{\infty}\theta_0C_i\le\bover12\epsilon$, and
consider $W'=\bigcup_{k\le n}C_k$.
Since $V\setminus W'\subseteq\bigcup_{i>n}C_i$,

$$\eqalign{\lambda(W\symmdiff W')
&\le\lambda(V\setminus W)+\lambda(V\setminus W')
=\lambda V-\lambda W+\lambda(V\setminus W')
=\theta V-\theta W+\theta(V\setminus W')\cr
&\le\sum_{i=0}^{\infty}\theta_0C_i-\theta W+\sum_{i=n+1}\theta_0C_i
\le\Bover12\epsilon+\Bover12\epsilon
=\epsilon.\cr}$$

\medskip

{\bf (f)(i)} If $W\in\Lambda$ and $\epsilon>0$ there is a
$V\in\Tensorhat_{i\in I}\Sigma_i$ such that $W\subseteq V$ and $\lambda
V\le\lambda W+\epsilon$.   \Prf\ Let $\sequencen{C_n}$ be a sequence in
$\Cal C$ such that $W\subseteq\bigcup_{n\in\Bbb N}C_n$ and
$\sum_{n=0}^{\infty}\theta_0C_n\le\theta W+\epsilon$.   Then
$C_n\in\Tensorhat_{i\in I}\Sigma_i$ for each $n$, so
$V=\bigcup_{n\in\Bbb N}C_n\in\Tensorhat_{i\in I}\Sigma_i$.   Now
$W\subseteq V$, and

\Centerline{$\lambda V=\theta V\le\sum_{n=0}^{\infty}\theta_0C_n
\le\theta W+\epsilon=\lambda W+\epsilon$.\ \Qed}

\medskip

\quad{\bf (ii)} Now, given $W\in\Lambda$, let $\sequencen{V_n}$ be a
sequence of sets in $\Tensorhat_{i\in I}\Sigma_i$ such that $W\subseteq
V_n$ and $\lambda V_n\le\lambda W+2^{-n}$ for each $n$;  then
$W_2=\bigcap_{n\in\Bbb N}V_n$ belongs to $\Tensorhat_{i\in I}\Sigma_i$
and
$W\subseteq W_2$ and $\lambda W_2=\lambda W$.   Similarly, there is a
$W'_2\in\Tensorhat_{i\in I}\Sigma_i$ such that $X\setminus W\subseteq
W'_2$ and $\lambda W'_2=\lambda(X\setminus W)$, so we may take
$W_1=X\setminus W'_2$ to complete the proof.
}

\leader{254G}{}\cmmnt{ The following is a fundamental, indeed
defining, property of product measures.   (Compare 251L.)

\medskip

\noindent}{\bf Theorem} Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a
family
of probability spaces with product $(X,\Lambda,\lambda)$.   Let
$(Y,\Tau,\nu)$ be a complete probability space and $\phi:Y\to X$ a
function.
Suppose that $\nu^*\phi^{-1}[C]\le\lambda C$ for every measurable
cylinder $C\subseteq X$.   Then $\phi$ is \imp.
In particular, $\phi$ is \imp\ iff $\phi^{-1}[C]\in\Tau$ and
$\nu\phi^{-1}[C]=\lambda C$ for every measurable cylinder
$C\subseteq X$.

\cmmnt{\medskip

\noindent{\bf Remark} By $\nu^*$ I mean the usual outer measure defined
from $\nu$ as in \S132.
}

\proof{{\bf (a)} First note that, writing $\theta$ for the outer measure
of 254A, $\nu^*\phi^{-1}[A]\le\theta A$ for every $A\subseteq X$.
\Prf\ Given $\epsilon>0$, there is a sequence $\sequencen{C_n}$ of
measurable cylinders such that $A\subseteq\bigcup_{n\in\Bbb N}C_n$ and
$\sum_{n=0}^{\infty}\theta_0C_n\le\theta A+\epsilon$, where $\theta_0$
is the functional of 254A.   But we know that $\theta_0C=\lambda C$ for
every measurable cylinder $C$ (254Fb), so

\Centerline{$\nu^*\phi^{-1}[A]
\le\nu^*(\bigcup_{n\in\Bbb N}\phi^{-1}[C_n])
\le\sum_{n=0}^{\infty}\nu^*\phi^{-1}[C_n]
\le\sum_{n=0}^{\infty}\lambda C_n
\le\theta A+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu^*\phi^{-1}[A]\le\theta A$.\
\Qed

\medskip

{\bf (b)} Now take any $W\in\Lambda$.   Then there are $F$, $F'\in\Tau$
such that

\Centerline{$\phi^{-1}[W]\subseteq F$, \quad$\phi^{-1}[X\setminus
W]\subseteq F'$,}

\Centerline{$\nu F=\nu^*\phi^{-1}[W]\le\theta W=\lambda W$,
\quad$\nu F'\le\lambda[X\setminus W]$.}

\noindent We have

\Centerline{$F\cup F'\supseteq\phi^{-1}[W]\cup\phi^{-1}[X\setminus
W]=Y$,}

\noindent so

\Centerline{$\nu(F\cap F')=\nu F+\nu F'-\nu(F\cup F')
\le\lambda W+\lambda(X\setminus W)-1=0$.}

\noindent Now

\Centerline{$F\setminus\phi^{-1}[W]\subseteq F\cap\phi^{-1}[X\setminus
W]\subseteq F\cap F'$}

\noindent is $\nu$-negligible.   Because $\nu$ is complete,
$F\setminus\phi^{-1}[W]\in\Tau$ and
$\phi^{-1}[W]=F\setminus(F\setminus\phi^{-1}[W])$ belongs to $\Tau$.
Moreover,

\Centerline{$1=\nu F+\nu F'\le\lambda W+\lambda(X\setminus W)=1$,}

\noindent so we must have $\nu F=\lambda W$;  but this means that
$\nu\phi^{-1}[W]=\nu W$.   As $W$ is arbitrary, $\phi$ is \imp.
}%end of proof of 254G

\leader{254H}{Corollary} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
and $\langle(Y_i,\Tau_i,\nu_i)\rangle_{i\in I}$ be two families of
probability spaces, with products $(X,\Lambda,\lambda)$ and
$(Y,\Lambda',\lambda')$.   Suppose that for each $i\in I$ we are given
an \imp\ function $\phi_i:X_i\to Y_i$.   Set
$\phi(x)=\familyiI{\phi_i(x(i))}$ for $x\in X$.   Then $\phi:X\to Y$ is
\imp.

\proof{ If $C=\prod_{i\in I}C_i$ is a measurable cylinder in $Y$, then
$\phi^{-1}[C]=\prod_{i\in I}\phi_i^{-1}[C_i]$ is a measurable cylinder
in
$X$, and

\Centerline{$\lambda\phi^{-1}[C]
=\prod_{i\in I}\mu_i\phi_i^{-1}[C_i]
=\prod_{i\in I}\nu_i C_i
=\lambda'C$.}

\noindent Since $\lambda$ is a complete probability measure, 254G tells
us that $\phi$ is \imp.
}%end of proof of 254H

\leader{254I}{}\cmmnt{ Corresponding to 251T we have the following.

\medskip

\noindent}{\bf Proposition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces, $\lambda$ the product measure
on $X=\prod_{i\in I}X_i$, and $\Lambda$ its domain.   Then $\lambda$ is
also the product of the
completions $\hat\mu_i$ of the $\mu_i$\cmmnt{ (212C)}.

\proof{ Write $\hat\lambda$ for the product of the $\hat\mu_i$, and
$\hat\Lambda$ for its domain.   (i) The identity map from $X_i$ to
itself is \imp\ if regarded as a map from $(X_i,\hat\mu_i)$ to
$(X_i,\mu_i)$, so
the identity map on $X$ is \imp\ if regarded as a map from
$(X,\hat\lambda)$ to $(X,\lambda)$, by 254H;  that is,
$\Lambda\subseteq\hat\Lambda$ and $\lambda=\hat\lambda\restr\Lambda$.
(ii) If $C$ is a measurable cylinder for $\langle\hat\mu_i\rangle_{i\in
I}$, that is, $C=\prod_{i\in I}C_i$ where $C_i\in\hat\Sigma_i$ for every
$i$ and $\{i:C_i\ne X_i\}$ is
finite, then for each $i\in I$ we can find a $C'_i\in\Sigma_i$ such that
$C_i\subseteq C'_i$ and $\mu_iC'_i=\hat\mu_iC_i$;  setting
$C'=\prod_{i\in I}C'_i$, we get

\Centerline{$\lambda^*C\le\lambda C'=\prod_{i\in I}\mu_iC'_i
=\prod_{i\in I}\hat\mu_iC_i=\hat\lambda C$.}

\noindent By 254G, $\lambda W$ must be defined and equal to
$\hat\lambda W$ whenever $W\in\hat\Lambda$.   Putting this
together with (i), we see that $\lambda=\hat\lambda$.
}%end of proof of 254I

\leader{254J}{}{\bf The product measure on $\{0,1\}^I$ (a)} Perhaps the
most important of all examples of infinite product measures is the case
in which each factor $X_i$ is just $\{0,1\}$ and each $\mu_i$ is the
`fair-coin' probability measure, setting

\Centerline{$\mu_i\{0\}=\mu_i\{1\}=\Bover12$.}

\noindent In this case, the product $X=\{0,1\}^I$ has a family
$\langle E_i\rangle_{i\in I}$ of measurable sets such that,
writing $\lambda$ for the product measure on $X$,

\Centerline{$\lambda(\bigcap_{i\in J}E_i)=2^{-\#(J)}$ if $J\subseteq I$
is finite.}

\noindent\prooflet{(Just take $E_i=\{x:x(i)=1\}$ for each $i$.)  }I
will call this
$\lambda$ the {\bf usual measure} on $\{0,1\}^I$.   Observe that if $I$
is finite then $\lambda\{x\}=2^{-\#(I)}$ for each
$x\in X$\cmmnt{ (using
254Fb)}.   On the other hand, if $I$ is infinite, then $\lambda\{x\}=0$
for every $x\in X$\prooflet{ (because, again using 254Fb,
$\lambda^*\{x\}\le 2^{-n}$ for every $n$)}.

\spheader 254Jb There is a natural bijection between $\{0,1\}^I$ and
$\Cal PI$, matching $x\in\{0,1\}^I$ with $\{i:i\in I$, $x(i)=1\}$.
So we get a standard measure $\tilde\lambda$ on $\Cal PI$, which I will
call the {\bf usual measure on $\Cal PI$}.   Note that for any finite
$b\subseteq I$ and any $c\subseteq b$ we have

\Centerline{$\tilde\lambda\{a:a\cap b=c\}=\lambda\{x:x(i)=1$ for $i\in
c$, $x(i)=0$ for $i\in b\setminus c\}=2^{-\#(b)}$.}

\spheader 254Jc Of course we can apply 254G to these measures; if
$(Y,\Tau,\nu)$ is a complete probability space, a function
$\phi:Y\to\{0,1\}^I$ is \imp\ iff

\Centerline{$\nu\{y:y\in Y$, $\phi(y)\restr J=z\}=2^{-\#(J)}$}

\noindent whenever $J\subseteq I$ is finite and
$z\in\{0,1\}^J$\prooflet{;  this is because the measurable cylinders in
$\{0,1\}^I$ are precisely the sets of the form $\{x:x\restr J=z\}$ where
$J\subseteq I$ is finite}.

\spheader 254Jd Define addition on $X$ by setting
$(x+y)(i)=x(i)+_2y(i)$ for every $i\in I$, $x$, $y\in X$, where
$0+_20=1+_21=0$, $0+_21=1+_20=1$.   If $y\in X$, the
map $x\mapsto x+y:X\to X$ is\cmmnt{ \imp.   \prooflet{\Prf\ If $J\subseteq I$
is finite and $z\in\{0,1\}^J$, set $z'=\family{j}{J}{z(j)+_2y(j)}$;  then

\Centerline{$\lambda\{x:(x+y)\restr J=z\}
=\lambda\{x:x\restr J=z'\}=2^{-\#(J)}$.}

\noindent As $J$ is arbitrary, (c) tells us that $x\mapsto x+y$ is \imp.\
\Qed}    Now since

\Centerline{$(x+y)+y=x+(y+y)=x+0=x$}

\noindent for every $x$, the map $x\mapsto x+y:X\to X$ is bijective and
equal to its inverse, so it is actually} a measure space automorphism
of $(X,\lambda)$.

%\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
%\def\header#1{
%     \wheader{#1}{10}{4}{4}{24pt}}

\header{254Je}{\bf *(e)}
Just because all the factors $(X_i,\mu_i)$ are the same, we
have another class of automorphisms of $(X,\lambda)$, corresponding to
permutations of $I$.   If $\pi:I\to I$ is any permutation, then
we have a corresponding function $x\mapsto x\pi:X\to X$.   \cmmnt{If
$J\subseteq I$ is finite and $z\in\{0,1\}^J$, set $J'=\pi[J]$ and
$z'=z\pi^{-1}\in\{0,1\}^{J'}$;  then

\Centerline{$\lambda\{x:(x\pi)\restr J=z\}=\lambda\{x:x\restr J'=z'\}
=2^{-\#(J')}=2^{-\#(J)}$.}

\noindent So $x\mapsto x\pi$ is \imp.   This time, its inverse is
$x\mapsto x\pi^{-1}$, which is again \imp;  so} $x\mapsto x\pi$ is a measure
space automorphism.

\leader{254K}{}\cmmnt{In the case of countably infinite $I$, we
have a very important
relationship between the usual product measure of $\{0,1\}^I$ and
Lebesgue measure on $[0,1]$.

\medskip

\noindent}{\bf Proposition} Let $\lambda$ be the usual measure on
$X=\{0,1\}^{\Bbb N}$, and let $\mu$ be Lebesgue measure on $[0,1]$;
write $\Lambda$ for the domain of $\lambda$ and $\Sigma$ for the domain
of $\mu$.

(i) For $x\in X$ set $\phi(x)=\sum_{i=0}^{\infty}2^{-i-1}x(i)$.   Then

\qquad $\phi^{-1}[E]\in\Lambda$ and $\lambda\phi^{-1}[E]=\mu E$
for every $E\in\Sigma$;

\qquad $\phi[F]\in\Sigma$ and $\mu\phi[F]=\lambda F$ for
every $F\in\Lambda$.

(ii) There is a bijection $\tilde\phi:X\to[0,1]$ which is equal to
$\phi$ at all but countably many points, and any such bijection is an
isomorphism between $(X,\Lambda,\lambda)$ and $([0,1],\Sigma,\mu)$.

\proof{{\bf (a)} The first point to observe is that $\phi$ itself
is nearly a bijection.   Setting

\Centerline{$H=\{x:x\in X$, $\exists\,m\in\Bbb
N$, $x(i)=x(m)\Forall i\ge m\}$,}

\Centerline{$H'=\{2^{-n}k:n\in\Bbb N$, $k\le 2^n\}$,}

\noindent then $H$ and $H'$ are countable and $\phi\restr X\setminus H$
is a bijection between $X\setminus H$ and $[0,1]\setminus H'$.   (For
$t\in[0,1]\setminus H'$, $\phi^{-1}(t)$ is the binary expansion of $t$.)
Because $H$ and $H'$ are countably infinite, there is a bijection
between them;  combining this with $\phi\restr X\setminus H$, we have a
bijection between $X$ and $[0,1]$ equal to $\phi$ except at countably
many points.   For the rest of this proof, let $\tilde\phi$ be any such
bijection.   Let $M$ be the countable set $\{x:x\in
X$, $\phi(x)\ne\tilde\phi(x)\}$, and $N$ the countable set
$\phi[M]\cup\tilde\phi[M]$;  then
$\phi[A]\symmdiff\tilde\phi[A]\subseteq N$ for every $A\subseteq X$.

\medskip

{\bf (b)} To see that $\lambda\tilde\phi^{-1}[E]$ exists and is equal to
$\mu E$ for every $E\in\Sigma$, I consider successively more complex
sets $E$.

\medskip

\quad\grheada\ If $E=\{t\}$ then
$\lambda\tilde\phi^{-1}[E]=\lambda\{\tilde\phi^{-1}(t)\}$ exists and is
zero.

\medskip

\quad\grheadb\ If $E$ is of the form
$\coint{2^{-n}k,2^{-n}(k+1)}$, where $n\in\Bbb N$ and $0\le k<2^n$, then
$\phi^{-1}[E]$ differs by at most two points from a set of the form
$\{x:x(i)=z(i)\Forall i<n\}$, so $\tilde\phi^{-1}[E]$ differs from
this by a countable set, and

\Centerline{$\lambda\tilde\phi^{-1}[E]=2^{-n}=\mu E$.}

\medskip

\quad\grheadc\ If $E$ is of the form
$\coint{2^{-n}k,2^{-n}l}$, where $n\in\Bbb N$ and $0\le k<l\le 2^n$,
then
\discrcenter{468pt}
{$E=\bigcup_{k\le i<l}\coint{2^{-n}i,2^{-n}(i+1)}$, }so

\Centerline{$\lambda\tilde\phi^{-1}[E]=2^{-n}(l-k)=\mu E$.}

\medskip

\quad\grheadd\ If $E$ is of the form $\coint{t,u}$, where
$0\le t<u\le 1$, then for each $n\in\Bbb N$ set
$k_n=\lfloor 2^nt\rfloor$, the integer
part of $2^nt$, $l_n=\lfloor 2^nu\rfloor$ and
$E_n=\coint{2^{-n}(k_n+1),2^{-n}l_n}$;  then $\sequencen{E_n}$ is a
non-decreasing sequence and $\bigcup_{n\in\Bbb N}E_n$ is $\ooint{t,u}$.
So (using ($\alpha$))

$$\eqalign{\lambda\tilde\phi^{-1}[E]
&=\lambda\tilde\phi^{-1}[\bigcup_{n\in\Bbb N}E_n]
=\lim_{n\to\infty}\lambda\tilde\phi^{-1}[E_n]\cr
&=\lim_{n\to\infty}\mu E_n
=\mu E.\cr}$$

\medskip

\quad\grheade\ If $E\in\Sigma$, then for any
$\epsilon>0$ there is a sequence $\sequencen{I_n}$ of
half-open subintervals of $\coint{0,1}$ such that
$E\setminus\{1\}\subseteq\bigcup_{n\in\Bbb N}I_n$ and
$\sum_{n=0}^{\infty}\mu I_n\le\mu E+\epsilon$;  now
$\tilde\phi^{-1}[E]
\subseteq \{\tilde\phi^{-1}(1)\}\cup\bigcup_{n\in\Bbb N}\phi^{-1}[I_n]$,
so

\Centerline{$\lambda^*\tilde\phi^{-1}[E]
\le\lambda(\bigcup_{n\in\Bbb N}\tilde\phi^{-1}[I_n])
\le \sum_{n=0}^{\infty}\lambda\tilde\phi^{-1}[I_n]
=\sum_{n=0}^{\infty}\mu I_n
\le\mu E+\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lambda^*\tilde\phi^{-1}[E]\le\mu E$,  and there
is a $V\in\Lambda$ such that $\tilde\phi^{-1}[E]\subseteq V$ and
$\lambda V\le\mu E$.

\medskip

\quad\grheadz\ Similarly, there is a $V'\in\Lambda$ such
that $V'\supseteq\tilde\phi^{-1}[[0,1]\setminus E]$ and
$\lambda V'\le\mu([0,1]\setminus E)$.   Now $V\cup V'=X$, so

\Centerline{$\lambda(V\cap V')=\lambda V+\lambda V'-\lambda(V\cup V')
\le\mu E+(1-\mu E)-1=0$}

\noindent and

\Centerline{$\tilde\phi^{-1}[E]=(X\setminus V')
\cup(V\cap V'\cap\tilde\phi^{-1}[E])$}

\noindent belongs to $\Lambda$, with

\Centerline{$\lambda\tilde\phi^{-1}[E]\le\lambda V\le\mu E$;}

\noindent at the same time,

\Centerline{$1-\lambda\tilde\phi^{-1}[E]\le\lambda V'\le 1-\mu E$}

\noindent so $\lambda\tilde\phi^{-1}[E]=\mu E$.

\medskip

{\bf (c)} Now suppose that $C\subseteq X$ is a measurable cylinder of
the special form $\{x:x(0)=\epsilon_0,\ldots,x(n)=\epsilon_n\}$ for some
$\epsilon_0,\ldots,\epsilon_n\in\{0,1\}$.   Then
$\phi[C]=[t,t+2^{-n-1}]$ where $t=\sum_{i=0}^n2^{-i-1}\epsilon_i$, so
that $\mu\phi[C]=\lambda C$.   Since
$\tilde\phi[C]\symmdiff\phi[C]\subseteq N$ is countable,
$\mu\tilde\phi[C]=\lambda C$.

If $C\subseteq X$ is any measurable cylinder, then it is of the form
$\{x:x\restr J=z\}$ for some finite $J\subseteq\Bbb N$;  taking $n$ so
large that $J\subseteq\{0,\ldots,n\}$, $C$ is expressible as a disjoint
union of $2^{n+1-\#(J)}$ sets of the form just considered, being just
those in which $\epsilon_i=z(i)$ for $i\in J$.   Summing their measures,
we again get $\mu\tilde\phi[C]=\lambda C$.   Now 254G tells us that
$\tilde\phi^{-1}:[0,1]\to X$ is \imp, that is, $\tilde\phi[W]$ is
Lebesgue measurable, with measure $\lambda W$, for every $W\in\Lambda$.

Putting this together with (b), $\tilde\phi$ must be an isomorphism
between $(X,\Lambda,\lambda)$ and $([0,1],\Sigma,\mu)$, as claimed in
(ii) of the proposition.

\medskip

{\bf (d)} As for (i), if $E\in\Sigma$ then
$\phi^{-1}[E]\symmdiff\tilde\phi^{-1}[E]\subseteq M$ is countable, so
$\lambda\phi^{-1}[E]=\lambda\tilde\phi^{-1}[E]=\mu E$.   While if
$W\in\Lambda$, $\phi[F]\symmdiff\tilde\phi[W]\subseteq N$ is countable,
so $\mu\phi[W]=\mu\tilde\phi[W]=\lambda W$.
}%end of proof of 254K

\leader{254L}{\dvrocolon{Subspaces}}\cmmnt{ Just as in 251Q, we can
consider the product of
subspace measures.   There is a simplification in the form of
the result because in the present context we are restricted to
probability measures.

\medskip

\noindent}{\bf Theorem} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of probability spaces, and $(X,\Lambda,\lambda)$ their
product.

(a) For each $i\in I$, let $A_i\subseteq X_i$ be a set of full outer
measure, and write $\tilde\mu_i$ for the subspace measure on
$A_i$\cmmnt{ (214B)}.
Let $\tilde\lambda$ be the product measure on $A=\prod_{i\in I}A_i$.
Then $\tilde\lambda$ is the subspace measure on $A$ induced by
$\lambda$.

(b) $\lambda^*(\prod_{i\in I}A_i)=\prod_{i\in I}\mu_i^*A_i$ whenever
$A_i\subseteq X_i$ for every $i$.

\proof{{\bf (a)} Write $\lambda_A$ for the subspace measure on $A$
defined from $\lambda$, and $\Lambda_A$ for its domain;  write
$\tilde\Lambda$ for the domain of $\tilde\lambda$.

\medskip

\quad{\bf (i)} Let $\phi:A\to X$ be the identity map.   If $C\subseteq
X$ is a measurable cylinder, say $C=\prod_{i\in I}C_i$ where
$C_i\in\Sigma_i$ for each $i$, then $\phi^{-1}[C]=\prod_{i\in I}(C_i\cap
A_i)$ is a measurable cylinder in $A$, and

\Centerline{$\tilde\lambda\phi^{-1}[C]=\prod_{i\in I}\tilde\mu_i(C_i\cap
A_i)\le\prod_{i\in I}\mu_iC_i=\mu C$.}

\noindent By 254G, $\phi$ is \imp, that is, $\tilde\lambda(A\cap
W)=\lambda W$ for every $W\in\Lambda$.   But this
means that $\tilde\lambda V$ is defined and equal to
$\lambda_AV=\lambda^*V$ for every $V\in\Lambda_A$, since
for any such $V$ there is a $W\in\Lambda$ such that $V=A\cap
W$ and $\lambda W=\lambda_AV$.   In particular, $\lambda_AA=1$.

\medskip

\quad{\bf (ii)} Now regard $\phi$ as a function from the measure space
$(A,\Lambda_A,\lambda_A)$ to $(A,\tilde\Lambda,\tilde\lambda)$.   If $D$
is a measurable
cylinder in $A$, we can express it as $\prod_{i\in I}D_i$ where every
$D_i$ belongs to the domain of $\tilde\mu_i$ and $D_i=A_i$ for all but
finitely many $i$.   Now for each $i$ we can find $C_i\in\Sigma_i$ such
that $D_i=C_i\cap A_i$ and $\mu C_i=\tilde\mu_iD_i$, and we can suppose
that $C_i=X_i$ whenever $D_i=A_i$.   In this case $C=\prod_{i\in
I}C_i\in\Lambda$ and

\Centerline{$\lambda C=\prod_{i\in I}\mu_iC_i=\prod_{i\in
I}\tilde\mu_iD_i=\tilde\lambda D$.}

\noindent Accordingly

\Centerline{$\lambda_A\phi^{-1}[D]=\lambda_A(A\cap C)\le\lambda C
=\tilde\lambda D$.}

\noindent By 254G again, $\phi$ is \imp\ in this manifestation, that is,
$\lambda_AV$ is defined and equal to $\tilde\lambda V$ for every
$V\in\tilde\Lambda$.   Putting this together with (i), we
have $\lambda_A=\tilde\lambda$, as claimed.

\medskip

{\bf (b)} For each $i\in I$, choose a set $E_i\in\Sigma_i$ such that
$A_i\subseteq E_i$ and $\mu_iE_i=\mu^*_iA_i$;  do this in such a way
that $E_i=X_i$ whenever $\mu^*_iA_i=1$.   Set
$B_i=A_i\cup(X_i\setminus E_i)$, so that $\mu^*_iB_i=1$ for each $i$ (if
$F\in\Sigma_i$ and
$F\supseteq B_i$ then $F\cap E_i\supseteq A_i$, so

\Centerline{$\mu_iF=\mu_i(F\cap E_i)+\mu_i(F\setminus E_i)
=\mu_i E_i+\mu_i(X_i\setminus E_i)=1$.)}

\noindent By (a), we can identify the subspace measure $\lambda_B$ on
$B=\prod_{i\in I}B_i$ with the product of the subspace measures
$\tilde\mu_i$ on $B_i$.   In particular, $\lambda^*B=\lambda_BB=1$.
Now $A_i=B_i\cap E_i$ so (writing $A=\prod_{i\in I}A_i$),
$A=B\cap\prod_{i\in I}E_i$.

If $\prod_{i\in I}\mu_i^*A_i=0$, then for every $\epsilon>0$ there is a
finite $J\subseteq I$ such that $\prod_{i\in J}\mu_i^*A_i\le\epsilon$;
consequently (using 254Fb)

\Centerline{$\lambda^*A\le\lambda\{x:x(i)\in E_i$ for every
$i\in J\}=\prod_{i\in J}\mu_iE_i\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lambda^*A=0$.  If
$\prod_{i\in I}\mu_i^*A_i>0$, then for every $n\in\Bbb N$ the set
$\{i:\mu^*A_i\le 1-2^{-n}\}$ must be finite, so

\Centerline{$J=\{i:\mu^*A_i<1\}=\{i:E_i\ne X_i\}$}

\noindent is countable.   By 254Fb again,
applied to $\langle E_i\cap B_i\rangle_{i\in I}$ in the product
$\prod_{i\in I}B_i$,

$$\eqalign{\lambda^*(\prod_{i\in I}A_i)
&=\lambda_B(\prod_{i\in I}A_i)
=\lambda_B\{x:x\in B,\,x(i)\in E_i\cap B_i\text{ for every }i\in J\}\cr
&=\prod_{i\in J}\tilde\mu_{i}(E_i\cap B_i)
=\prod_{i\in I}\mu_i^*A_i,\cr}$$

\noindent as required.
}%end of proof of 254L

\leader{254M}{}\cmmnt{ I now turn to the basic results which make it
possible to use these product measures effectively.   First, I offer a
vocabulary for
dealing with subproducts.}   Let $\langle X_i\rangle_{i\in I}$ be a
family of sets, with product $X$.

\spheader 254Ma For $J\subseteq I$, write $X_J$ for
$\prod_{i\in J}X_i$.   We have a canonical bijection
$x\mapsto(x\restr J,x\restr I\setminus J):
X\to X_I\times X_{I\setminus J}$.   Associated with this
we have the map  $x\mapsto\pi_J(x)=x\restr J:X\to X_J$.   Now I will say
that a set $W\subseteq X$ is {\bf determined by coordinates in $J$} if
there is a $V\subseteq X_J$ such that $W=\pi_J^{-1}[V]$;  that is, $W$
corresponds to
$V\times X_{I\setminus J}\subseteq X_J\times X_{I\setminus J}$.

\cmmnt{It is easy to see that}

$$\eqalign{W&\text{ is determined by coordinates in }J\cr
&\quad\iff x'\in W\text{ whenever }x\in W,\,x'\in X\text{ and }x'\restr
J=x\restr J\cr
&\quad\iff W=\pi_J^{-1}[\pi_J[W]].\cr}$$

\noindent It follows that if $W$ is determined by coordinates in $J$,
and $J\subseteq K\subseteq I$, $W$ is also determined by coordinates in
$K$.   The family $\Cal W_J$ of subsets of $X$ determined by coordinates
in $J$ is closed under complementation and arbitrary unions and
intersections.
\prooflet{\Prf\ If $W\in\Cal W_J$, then

\Centerline{$X\setminus W=X\setminus\pi_J^{-1}[\pi_J[W]]
=\pi_J^{-1}[X_J\setminus\pi_J[W]]\in\Cal W_J$.}

\noindent If $\Cal V\subseteq\Cal W_J$, then

\Centerline{$\bigcup\Cal V
=\bigcup_{V\in\Cal V}\pi_J^{-1}[\pi_J[V]]
=\pi_J^{-1}[\bigcup_{V\in\Cal V}\pi_J[V]]
\in\Cal W_J$.  \Qed}
}%end of prooflet

\spheader 254Mb \cmmnt{It follows that}

\Centerline{$\Cal W=\bigcup\{\Cal W_J:J\subseteq I$ is
countable$\}$\cmmnt{,}}

\noindent\cmmnt{the family of subsets of $X$ determined by coordinates
in some countable
set, }is a $\sigma$-algebra of subsets of $X$.
\prooflet{\Prf\ (i) $X$ and $\emptyset$ are determined by coordinates in
$\emptyset$ (recall that $X_{\emptyset}$ is a singleton, and that
$X=\pi_{\emptyset}^{-1}[X_{\emptyset}]$,
$\emptyset=\pi^{-1}_{\emptyset}[\emptyset]$).   (ii) If $W\in\Cal W$,
there is a countable $J\subseteq I$ such that $W\in\Cal W_J$;  now

\Centerline{$X\setminus W=\pi_J^{-1}[X_J\setminus\pi_J[W]]\in\Cal
W_J\subseteq\Cal W$.}

\noindent (iii) If $\sequencen{W_n}$
is a sequence in $\Cal W$, then for each $n\in\Bbb N$ there is a
countable $J_n\subseteq I$ such that $W\in\Cal W_{J_n}$.   Now
$J=\bigcup_{n\in\Bbb N}J_n$ is a countable subset of $I$,
and every $W_n$ belongs to $\Cal W_J$, so

\Centerline{$\bigcup_{n\in\Bbb N}W_n\in\Cal W_J\subseteq\Cal W$.  \Qed}
}%end of prooflet

\spheader 254Mc If $i\in I$ and $E\subseteq X_i$ then
$\{x:x\in X$, $x(i)\in E\}$ is determined by the single coordinate $i$,
so surely belongs to $\Cal W$;  accordingly $\Cal W$ must include
$\Tensorhat_{i\in I}\Cal PX_i$.
\cmmnt{{\it A fortiori}, if $\Sigma_i$ is a $\sigma$-algebra of
subsets of $X_i$ for each $i$,
$\Cal W\supseteq\Tensorhat_{i\in I}\Sigma_i$;  that
is, every member of $\Tensorhat_{i\in I}\Sigma_i$ is determined
by coordinates in some countable set.}

\vleader{72pt}{254N}{\bf Theorem} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be
a family of probability spaces and $\langle K_j\rangle_{j\in J}$ a
partition of $I$.   For each $j\in J$ let $\lambda_j$ be the product
measure on $Z_j=\prod_{i\in K_j}X_i$, and write $\lambda$ for the
product measure on $X=\prod_{i\in I}X_i$.   Then the natural bijection

\Centerline{$x\mapsto\phi(x)
=\langle x\restr K_j\rangle_{j\in J}:X\to\prod_{j\in J}Z_j$}

\noindent identifies $\lambda$ with the product of the family
$\langle\lambda_j\rangle_{j\in J}$.

In particular, if $K\subseteq I$ is any set, then $\lambda$ can be
identified with the c.l.d.\ product of the product measures on
$\prod_{i\in K}X_i$ and $\prod_{i\in I\setminus K}X_i$.

\proof{ (Compare 251N.) Write $Z=\prod_{j\in J}Z_j$ and
$\tilde\lambda$ for the product measure on $Z$;  let $\Lambda$,
$\tilde\Lambda$ be the domains of $\lambda$ and $\tilde\lambda$.

\medskip

{\bf (a)} Let $C\subseteq Z$ be a measurable cylinder.   Then
$\lambda^*\phi^{-1}[C]\le\tilde\lambda C$.   \Prf\ Express $C$ as
$\prod_{j\in J}C_j$ where $C_j\subseteq Z_j$ belongs to the domain
$\Lambda_j$ of $\lambda_j$ for each $j$.
Set $L=\{j:C_j\ne Z_j\}$, so that $L$ is finite.   Let $\epsilon>0$.
For each $j\in L$ let $\sequencen{C_{jn}}$ be a sequence of measurable
cylinders in $Z_j=\prod_{i\in K_j}X_i$ such that
$C_j\subseteq\bigcup_{n\in\Bbb N}C_{jn}$ and
$\sum_{n=0}^{\infty}\lambda_jC_{jn}\le\lambda C_j+\epsilon$.   Express
each $C_{jn}$ as $\prod_{i\in K_j}C_{jni}$ where $C_{jni}\in\Sigma_i$
for $i\in K_j$ (and $\{i:C_{jni}\ne X_i\}$ is finite).

For $f\in\BbbN^L$, set

\Centerline{$D_f
=\{x:x\in X$, $x(i)\in C_{j,f(j),i}$ whenever $j\in L$, $i\in K_j\}$.}

\noindent Because $\bigcup_{j\in L}\{i:C_{j,f(j),i}\ne X_i\}$ is finite,
$D_f$ is a measurable cylinder in $X$, and

\Centerline{$\lambda D_f=\prod_{j\in L}\prod_{i\in K_j}\mu_iC_{j,f(j),i}
=\prod_{j\in L}\lambda_jC_{j,f(j)}$.}

\noindent Also

\Centerline{$\bigcup\{D_f:f\in\BbbN^L\}\supseteq\phi^{-1}[C]$}

\noindent because if $\phi(x)\in C$ then $\phi(x)(j)\in C_j$ for each
$j\in L$, so there must be an $f\in\BbbN^L$ such that $\phi(x)(j)\in
C_{j,f(j)}$ for every $j\in L$.   But (because $\BbbN^L$ is countable)
this means that

$$\eqalign{\lambda^*\phi^{-1}[C]
&\le\sum_{f\in\BbbN^L}\lambda D_f
=\sum_{f\in \BbbN^L}\prod_{j\in L}\lambda_jC_{j,f(j)}\cr
&=\prod_{j\in L}\sum_{n=0}^{\infty}\lambda_jC_{jn}
\le\prod_{j\in L}(\lambda_jC_j+\epsilon).\cr}$$

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lambda^*\phi^{-1}[C]\le\prod_{j\in L}\lambda_jC_j
=\tilde\lambda C$.  \Qed}

By 254G, it follows that $\lambda\phi^{-1}[W]$ is defined, and equal to
$\tilde\lambda W$, whenever $W\in\tilde\Lambda$.
\medskip

{\bf (b)} Next, $\tilde\lambda\phi[D]=\lambda D$ for every measurable
cylinder $D\subseteq X$.   \Prf\ This is easy.   Express $D$ as
$\prod_{i\in I}D_i$ where $D_i\in\Sigma_i$ for every $i\in I$ and
$\{i:D_i\ne\Sigma_i\}$ is finite.   Then $\phi[D]=\prod_{j\in J}\tilde
D_j$, where $\tilde D_j=\prod_{i\in K_j}D_i$ is a measurable cylinder
for each $j\in J$.   Because $\{j:\tilde D_j\ne Z_j\}$ must also be
finite (in fact, it cannot have more members than the finite set
$\{i:D_i\ne X_i\}$), $\prod_{j\in J}\tilde D_j$ is itself a measurable
cylinder in $Z$, and

\Centerline{$\tilde\lambda\phi[D]
=\prod_{j\in J}\lambda_j\tilde D_j
=\prod_{j\in J}\prod_{i\in K_j}\mu D_i
=\lambda D$.   \Qed}

Applying 254G to $\phi^{-1}:Z\to X$, it follows that
$\tilde\lambda\phi[W]$ is defined, and equal to $\lambda W$, for every
$W\in\Lambda$.   But together with (a) this means that
for any $W\subseteq X$,

\inset{if $W\in\Lambda$ then $\phi[W]\in\tilde\Lambda$ and
$\tilde\lambda\phi[W]=\lambda W$,}

\inset{if $\phi[W]\in\tilde\Lambda$ then $W\in\Lambda$ and $\lambda
W=\tilde\lambda\phi[W]$.}

\noindent And of course this is just what is meant by saying that $\phi$
is an isomorphism between $(X,\Lambda,\lambda)$ and
$(Z,\tilde\Lambda,\tilde\lambda)$.
}%end of proof of 254N

\leader{254O}{Proposition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces.   For each $J\subseteq I$ let
$\lambda_J$ be the product probability measure on
$X_J=\prod_{i\in J}X_i$, and $\Lambda_J$ its domain;  write $X=X_I$,
$\lambda=\lambda_I$ and $\Lambda=\Lambda_I$.   For $x\in X$ and
$J\subseteq I$ set $\pi_J(x)=x\restr J\in X_J$.

(a) For every $J\subseteq I$, $\lambda_J$ is the image measure
$\lambda\pi_J^{-1}$\cmmnt{ (234D)};  in particular,
$\pi_J:X\to X_J$ is \imp\ for $\lambda$
and $\lambda_J$.

(b) If $J\subseteq I$ and $W\in\Lambda$ is
determined by coordinates in $J$\cmmnt{ (254M)}, then
$\lambda_J\pi_J[W]$ is defined and equal to $\lambda W$.   Consequently
there are $W_1$, $W_2$ belonging to the $\sigma$-algebra of subsets of
$X$ generated by

\Centerline{$\{\{x:x(i)\in E\}:i\in J$, $E\in\Sigma_i\}$}

\noindent such that $W_1\subseteq W\subseteq W_2$ and
$\lambda(W_2\setminus W_1)=0$.

(c) For every $W\in\Lambda$, we can find a
countable set $J$ and $W_1$, $W_2\in\Lambda$, both
determined by coordinates in $J$, such that
$W_1\subseteq W\subseteq W_2$ and $\lambda(W_2\setminus W_1)=0$.

(d) For every $W\in\Lambda$, there is a
countable set $J\subseteq I$ such that $\pi_J[W]\in\Lambda_J$ and
$\lambda_J\pi_J[W]=\lambda W$;  so that
$W'=\pi_J^{-1}[\pi_J[W]]$ belongs to $\Lambda$, and
$\lambda(W'\setminus W)=0$.

\proof{{\bf (a)(i)} By 254N, we can identify $\lambda$ with the
product of $\lambda_J$ and $\lambda_{I\setminus J}$ on
$X_J\times X_{I\setminus J}$.
Now $\pi_J^{-1}[E]\subseteq X$ corresponds to
$E\times X_{I\setminus J}\subseteq X_J\times X_{I\setminus J}$, so

\Centerline{$\lambda(\pi^{-1}[E])
=\lambda_JE\cdot\lambda_{I\setminus J}X_{I\setminus J}=\lambda_JE$,}

\noindent by 251E or 251Ia, whenever $E\in\Lambda_J$.   This shows that
$\pi_J$ is \imp.

\medskip

\quad{\bf (ii)} To see that $\lambda_J$ is actually the image measure,
suppose that $E\subseteq X_J$ is such that $\pi_J^{-1}[E]\in\Lambda$.
Identifying $\pi^{-1}_J[E]$ with $E\times X_{I\setminus J}$, as before,
we are supposing that $E\times X_{I\setminus J}$ is measured by the
product measure on $X_J\times X_{I\setminus J}$.   But this means that
for $\lambda_{I\setminus J}$-almost every
$z\in X_{I\setminus J}$, $E_z=\{y:(y,z)\in E\times X_{I\setminus J}\}$
belongs to $\Lambda_J$ (252D(ii), because $\lambda_J$ is complete).
Since $E_z=E$ for every $z$, $E$ itself belongs to $\Lambda_J$, as
claimed.

\medskip

{\bf (b)} If $W\in\Lambda$ is determined by coordinates
in $J$, set $H=\pi_J[W]$;  then $\pi_J^{-1}[H]=W$, so $H\in\Lambda_J$ by
(a) just above.   By 254Ff, there are $H_1$,
$H_2\in\Tensorhat_{i\in J}\Sigma_i$ such that
$H_1\subseteq H\subseteq H_2$ and $\lambda_J(H_2\setminus H_1)=0$.

Let $\Tau_J$ be the $\sigma$-algebra of subsets of $X$ generated by sets
of the form $\{x:x(i)\in E\}$ where $i\in J$ and $E\in\Sigma_J$.
Consider $\Tau'_J=\{G:G\subseteq X_J$, $\pi_J^{-1}[G]\in\Tau_J\}$.
This
is a $\sigma$-algebra of subsets of $X_J$, and it contains $\{y:y\in
X_J$, $y(i)\in E\}$ whenever $i\in J$, $E\in\Sigma_J$ (because

\Centerline{$\pi_J^{-1}[\{y:y\in X_J$, $y(i)\in E\}]=\{x:x\in X$,
$x(i)\in
E\}$}

\noindent whenever $i\in J$, $E\subseteq X_i$).   So $\Tau'_J$ must
include $\Tensorhat_{i\in J}\Sigma_i$.   In particular, $H_1$ and $H_2$
both belong to $\Tau'_J$, that is, $W_k=\pi_J^{-1}[H_k]$ belongs to
$\Tau_J$ for both $k$.   Of course $W_1\subseteq W\subseteq W_2$,
because $H_1\subseteq H\subseteq H_2$, and

\Centerline{$\lambda(W_2\setminus W_1)=\lambda_J(H_2\setminus H_1)=0$,}
\noindent as required.

\medskip

{\bf (c)} Now take any $W\in\Lambda$.   By
254Ff, there are $W_1$ and $W_2\in\Tensorhat_{i\in I}\Sigma_i$ such that
$W_1\subseteq W\subseteq W_2$ and $\lambda(W_2\setminus W_1)=0$.   By
254Mc, there are countable sets $J_1$, $J_2\subseteq I$ such that, for
each $k$, $W_k$ is determined by coordinates in $J_k$.   Setting
$J=J_1\cup J_2$, $J$ is a countable subset of $I$ and both $W_1$ and
$W_2$ are determined by coordinates in $J$.

\medskip

{\bf (d)} Continuing the argument from (c), $\pi_J[W_1]$,
$\pi_J[W_2]\in\Lambda_J$, by (b), and
$\lambda_J(\pi_J[W_2]\setminus\pi_J[W_1])=0$.   Since
$\pi_J[W_1]\subseteq\pi_J[W]\subseteq\pi_J[W_2]$, it follows that
$\pi_J[W]\in\Lambda_J$, with
$\lambda_J\pi_J[W]=\lambda_J\pi_J[W_2]$;  so that, setting
$W'=\pi_J^{-1}[\pi_J[W]]$, $W'\in\Lambda$, and

\Centerline{$\lambda W'=\lambda_J\pi_J[W]=\lambda_J\pi_J[W_2]
=\lambda\pi_J^{-1}[\pi_J[W_2]]=\lambda W_2=\lambda W$.}
}%end of proof of 254O

\leader{254P}{Proposition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of probability spaces, and for each $J\subseteq I$ let
$\lambda_J$ be the product probability measure on
$X_J=\prod_{i\in J}X_i$, and $\Lambda_J$ its domain;  write $X=X_I$,
$\Lambda=\Lambda_I$ and $\lambda=\lambda_I$.   For $x\in X$ and
$J\subseteq I$ set $\pi_J(x)=x\restr J\in X_J$.

(a) If $J\subseteq I$ and $g$ is a real-valued function defined on a
subset of $X_J$, then $g$ is $\Lambda_J$-measurable iff $g\pi_J$ is
$\Lambda$-measurable.

(b) Whenever $f$ is a $\Lambda$-measurable real-valued function defined
on a $\lambda$-conegligible subset of $X$, we can find a countable set
$J\subseteq I$ and a $\Lambda_J$-measurable function $g$ defined on a
$\lambda_J$-conegligible subset of $X_J$ such that $f$ extends $g\pi_J$.

\proof{{\bf (a)(i)} If $g$ is $\Lambda_J$-measurable and $a\in\Bbb R$,
there is an $H\in\Lambda_J$ such that $\{y:y\in\dom g$, $g(y)\ge
a\}=H\cap\dom g$.   Now $\pi_J^{-1}[H]\in\Lambda$, by 254Oa, and
$\{x:x\in\dom g\pi_J$, $g\pi_J(x)\ge a\}=\pi_J^{-1}[H]\cap\dom g\pi_J$.
So
$g\pi_J$ is $\Lambda$-measurable.

\medskip

\quad{\bf (ii)} If $g\pi_J$ is $\Lambda$-measurable and $a\in\Bbb R$,
then there is a $W\in\Lambda$ such that
$\{x:g\pi_J(x)\ge a\}=W\cap\dom g\pi_J$.   As in the proof of 254Oa, we
may identify $\lambda$ with the
product of $\lambda_J$ and $\lambda_{I\setminus J}$, and 252D(ii) tells
us that, if we identify $W$ with the corresponding subset of
$X_J\times X_{I\setminus J}$, there is at least one
$z\in X_{I\setminus J}$ such
that $W_z=\{y:y\in X_J$, $(y,z)\in W\}$ belongs to $\Lambda_J$.   But
since (on this convention) $g\pi_J(y,z)=g(y)$ for every $y\in X_J$, we
see that $\{y:y\in\dom g$, $g(y)\ge a\}=W_z\cap\dom g$.   As $a$ is
arbitrary, $g$ is $\Lambda_J$-measurable.

\medskip

{\bf (b)} For rational numbers $q$, set
$W_q=\{x:x\in\dom f$, $f(x)\ge q\}$.   By 254Oc we can find for each $q$
a countable set
$J_q\subseteq I$ and sets $W'_q$, $W_q''$, both determined by
coordinates
in $J_q$, such that $W'_q\subseteq W_q\subseteq W_q''$ and
$\lambda(W_q''\setminus W_q')=0$.   Set $J=\bigcup_{q\in\Bbb Q}J_q$,
$V=X\setminus\bigcup_{q\in\Bbb Q}(W_q''\setminus W_q')$;  then $J$ is a
countable subset of $I$ and $V$ is a conegligible subset of $X$;
moreover, $V$ is determined by coordinates in $J$ because all the
$W_q'$, $W_q''$ are.

For every $q\in\Bbb Q$, $W_q\cap V=W_q'\cap V$, because
$V\cap(W_q\setminus W_q')\subseteq V\cap(W_q''\setminus
W_q')=\emptyset$;  so $W_q\cap V$ is determined by coordinates in $J$.
Consequently $V\cap\dom f=\bigcup_{q\in\Bbb Q}V\cap W_q$ also is
determined by coordinates in $J$.   Also

\Centerline{$\{x:x\in V\cap\dom f$, $f(x)\ge a\}
=\bigcap_{q\le a}V\cap W_q$}

\noindent is determined by coordinates in $J$.   What this means is that
if $x$, $x'\in V$ and $\pi_Jx=\pi_Jx'$, then $x\in\dom f$ iff $x'\in\dom
f$ and in this case $f(x)=f(x')$.   Setting $H=\pi_J[V\cap\dom f]$, we
have $\pi_J^{-1}[H]=V\cap\dom f$ a conegligible subset of $X$, so
(because $\lambda_J=\lambda\pi_J^{-1}$) $H$ is conegligible in $X_J$.
Also, for
$y\in H$, $f(x)=f(x')$ whenever $\pi_Jx=\pi_Jx'=y$, so there is a
function $g:H\to\Bbb R$ defined by saying that $g\pi_J(x)=f(x)$ whenever
$x\in V\cap\dom f$.   Thus $g$ is defined almost everywhere in $X_J$ and
$f$ extends $g\pi_J$.   Finally, for any $a\in\Bbb R$,

\Centerline{$\pi_J^{-1}[\{y:g(y)\ge a\}]
=\{x:x\in V\cap\dom f$, $f(x)\ge a\}\in\Lambda$;}

\noindent by 254Oa, $\{y:g(y)\ge a\}\in\Lambda_J$;  as $a$ is arbitrary,
$g$ is measurable.
}%end of proof of 254P

\leader{254Q}{Proposition} Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in
I}$ be a family of probability spaces, and for each $J\subseteq I$ let
$\lambda_J$ be the product probability measure on $X_J=\prod_{i\in
J}X_i$;  write $X=X_I$, $\lambda=\lambda_I$.   For $x\in X$, $J\subseteq
I$ set $\pi_J(x)=x\restr J\in X_J$.

(a) Let $\eusm S$ be the linear subspace of $\Bbb R^X$ spanned by
$\{\chi C:C\subseteq X$ is a measurable cylinder$\}$.   Then for every
$\lambda$-integrable real-valued function $f$ and every $\epsilon>0$
there is a $g\in\eusm S$ such that $\int|f-g|d\lambda\le\epsilon$.

(b) Whenever $J\subseteq I$ and $g$ is a real-valued function defined on
a subset of $X_J$, then $\int g\,d\lambda_J=\int g\pi_Jd\lambda$ if
either integral is defined in $[-\infty,\infty]$.

(c) Whenever $f$ is a $\lambda$-integrable real-valued function, we can
find a countable set $J\subseteq X$ and a
$\lambda_J$-integrable function $g$ such that $f$ extends $g\pi_J$.

\proof{{\bf (a)(i)} Write $\overline{\eusm S}$ for the set of functions
$f$ satisfying the assertion, that is, such that for every $\epsilon>0$
there is a $g\in\eusm S$ such that $\int|f-g|\le\epsilon$.   Then
$f_1+f_2$ and $cf_1\in\overline{\eusm S}$ whenever $f_1$,
$f_2\in\overline{\eusm S}$.   \Prf\ Given $\epsilon>0$ there are $g_1$,
$g_2\in\eusm S$ such that $\int|f_1-g_1|\le\bover{\epsilon}{2+|c|}$,
$\int|f_2-g_2|\le\bover{\epsilon}2$;  now $g_1+g_2$, $cg_1\in\eusm S$
and $\int|(f_1+f_2)-(g_1+g_2)|\le\epsilon$,
$\int|cf_1-cg_1|\le\epsilon$.\ \QeD\  Also, of course,
$f\in\overline{\eusm S}$ whenever $f_0\in\overline{\eusm S}$ and
$f\eae f_0$.


\medskip

\quad{\bf (ii)} Write $\Cal W$ for
$\{W:W\subseteq X$, $\chi W\in\overline{\eusm S}\}$, and $\Cal C$ for
the
family of measurable
cylinders in $X$.   Then it is plain from the definition in 254A that
$C\cap C'\in\Cal C$ for all $C$, $C'\in\Cal C$, and of course
$C\in\Cal W$ for every $C\in\Cal C$, because $\chi C\in\eusm S$.   Next,
$W\setminus V\in\Cal W$ whenever $W$, $V\in\Cal W$ and $V\subseteq W$,
because then $\chi(W\setminus V)=\chi W-\chi V$.   Thirdly,
$\bigcup_{n\in\Bbb N}W_n\in\Cal W$ for every non-decreasing sequence
$\sequencen{W_n}$ in $\Cal W$.   \Prf\ Set $W=\bigcup_{n\in\Bbb N}W_n$.
Given $\epsilon>0$, there is an $n\in\Bbb N$ such that
$\lambda(W\setminus W_n)\le\bover{\epsilon}2$.   Now there is a
$g\in\eusm S$ such that $\int|\chi W_n-g|\le\bover{\epsilon}2$, so that
$\int|\chi W-g|\le\epsilon$.\ \QeD\  Thus $\Cal W$ is a Dynkin class of
subsets of $X$.

By the Monotone Class Theorem (136B), $\Cal W$ must include the
$\sigma$-algebra of subsets of $X$ generated by $\Cal C$, which is
$\Tensorhat_{i\in I}\Sigma_i$.   But this means that $\Cal W$ contains
every measurable subset of $X$, since by 254Ff any measurable set
differs by a negligible set from some member of $\Tensorhat_{i\in
I}\Sigma_i$.

\medskip

\quad{\bf (iii)} Thus $\overline{\eusm S}$ contains the indicator
function of any measurable subset of $X$.   Because it is closed under
addition and scalar multiplication, it contains all simple functions.
But this means that it must contain all integrable functions.   \Prf\ If
$f$ is a real-valued function which is integrable over $X$, and
$\epsilon>0$, there is a simple function $h:X\to\Bbb R$ such that
$\int|f-h|\le\bover{\epsilon}2$ (242M), and now there is a $g\in\eusm S$
such that $\int|h-g|\le\bover{\epsilon}2$, so that
$\int|f-g|\le\epsilon$.\ \Qed

This proves part (a) of the proposition.

\medskip

{\bf (b)} Put 254Oa and 235J together.

\medskip

{\bf (c)} By 254Pb, there are a countable $J\subseteq I$ and a
real-valued function $g$ defined on a conegligible subset of $X_J$ such
that $f$ extends $g\pi_J$.   Now $\dom(g\pi_J)=\pi_J^{-1}[\dom g]$ is
conegligible, so $f\eae g\pi_J$ and $g\pi_J$ is $\lambda$-integrable.
By (b), $g$ is $\lambda_J$-integrable.
}%end of proof of 254Q

\leader{254R}{Conditional
expectations \dvrocolon{again}}\cmmnt{ Putting the ideas of 253H
together with the work above, we obtain some results which are important
not only for their direct applications but for the light they throw on
the structures here.

\medskip

\noindent}{\bf Theorem} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces with product
$(X,\Lambda,\lambda)$.   For $J\subseteq I$ let
$\Lambda_J\subseteq\Lambda$ be the $\sigma$-subalgebra of sets
determined by coordinates in $J$\cmmnt{ (254Ma)}.   Then we may regard
$L^0(\lambda\restr\Lambda_J)$ as a subspace of
$L^0(\lambda)$\cmmnt{ (242Jh)}.    Let
$P_J:L^1(\lambda)\to L^1(\lambda\restr\Lambda_J)\subseteq L^1(\lambda)$
be the corresponding
conditional expectation operator\cmmnt{ (242Jd)}.   Then

(a) for any $J$, $K\subseteq I$, $P_{K\cap J}=P_KP_J$;

(b) for any $u\in L^1(\lambda)$, there is a countable set
$J^*\subseteq I$ such that $P_Ju=u$ iff $J\supseteq J^*$;

(c) for any $u\in L^0(\lambda)$, there is a unique smallest set
$J^*\subseteq I$ such that $u\in L^0(\lambda\restr\Lambda_{J^*})$, and
this $J^*$ is countable;

(d) for any $W\in\Lambda$ there is a unique smallest set
$J^*\subseteq I$
such that $W\symmdiff W'$ is negligible for some $W'\in\Lambda_{J^*}$,
and this $J^*$ is countable;

(e) for any $\Lambda$-measurable real-valued function $f:X\to\Bbb R$
there is a unique smallest set $J^*\subseteq I$ such that $f$ is equal
almost everywhere to a $\Lambda_J^*$-measurable function, and this $J^*$
is countable.

\proof{ For $J\subseteq I$, write $X_J=\prod_{i\in J}X_i$, let
$\lambda_J$ be the product measure on $X_J$, and set
$\pi_J(x)=x\restr J$ for $x\in X$.   Write $L^0_J$ for
$L^0(\lambda\restr\Lambda_J)$, regarded as a subset of $L^0=L^0_I$, and
$L^1_J$ for $L^1(\lambda\restr\Lambda_J)=L^1(\lambda)\cap L^0_J$, as in
242Jb;  thus $L^1_J$ is the set of values of the projection $P_J$.

\medskip

{\bf (a)(i)} Let $C\subseteq X$ be a measurable cylinder, expressed as
$\prod_{i\in I}C_i$ where $C_i\in\Sigma_i$ for every $i$ and
$L=\{i:C_i\ne X_i\}$ is finite.   Set

\Centerline{$C'_i=C_i$ for $i\in J$, $X_i$ for $i\in I\setminus J$,
\quad$C'=\prod_{i\in I}C'_i$,
\quad$\alpha=\prod_{i\in I\setminus J}\mu_iC_i$.}

\noindent Then $\alpha\chi C'$ is a conditional
expectation of $\chi C$ on $\Lambda_J$.   \Prf\ By
254N, we can identify $\lambda$ with the product of $\lambda_J$ and
$\lambda_{I\setminus J}$.   This identifies $\Lambda_J$ with
$\{E\times X_{I\setminus J}:E\in\dom\lambda_J\}$.   By 253H we have a
conditional expectation $g$ of $\chi C$ defined by setting

\Centerline{$g(y,z)=\int\chi C(y,t)\lambda_{I\setminus J}(dt)$}

\noindent for $y\in X_J$, $z\in X_{I\setminus J}$.   But $C$ is
identified with $C_J\times C_{I\setminus J}$, where
$C_J=\prod_{i\in J}C_i$, so that $g(y,z)=0$ if $y\notin C_J$ and
otherwise is
$\lambda_{I\setminus J}C_{I\setminus J}=\alpha$.   Thus
$g=\alpha\chi(C_J\times X_{I\setminus J})$.   But the identification
between $X_I\times X_{I\setminus J}$ and $X$ matches
$C_J\times X_{I\setminus J}$ with $C'$, as described above.   So $g$
becomes
identified with $\alpha\chi C'$ and $\alpha\chi C'$ is a conditional
expectation of $\chi C$.\ \Qed

\medskip

\quad{\bf (ii)} Next, setting

\Centerline{$C''_i=C'_i$ for $i\in K$, $X_i$ for $i\in I\setminus K$,
\quad$C''=\prod_{i\in I}C''_i$,}

\Centerline{$\beta=\prod_{i\in I\setminus K}\mu_iC'_i
=\prod_{i\in I\setminus(J\cup K)}\mu_iC_i$,}

\noindent the same arguments show that $\beta\chi C''$ is a conditional
expectation of $\chi C'$ on $\Lambda_K$.   So we have

\Centerline{$P_KP_J(\chi C)^{\ssbullet}
=\beta\alpha(\chi C'')^{\ssbullet}$.}

\noindent But if we look at $\beta\alpha$, this is just
$\prod_{i\in I\setminus(K\cap J)}\mu_iC_i$, while $C''_i=C_i$ if
$i\in K\cap J$,
$X_i$ for other $i$.   So $\beta\alpha\chi C''$ is a conditional
expectation of $\chi C$ on $\Lambda_{K\cap J}$, and

\Centerline{$P_KP_J(\chi C)^{\ssbullet}
=P_{K\cap J}(\chi C)^{\ssbullet}$.}

\medskip

\quad{\bf (iii)} Thus we see that the operators $P_KP_J$, $P_{K\cap J}$
agree on elements of the form $\chi C^{\ssbullet}$ where $C$ is a
measurable cylinder.   Because they are both linear, they agree on
linear combinations of these, that is, $P_KP_Jv=P_{K\cap J}v$ whenever
$v=g^{\ssbullet}$ for some $g$ in the space $\eusm S$ of 254Q.   But if
$u\in L^1(\lambda)$ and $\epsilon>0$, there is a $\lambda$-integrable
function  $f$ such that $f^{\ssbullet}=u$ and there is a $g\in\eusm S$
such that $\int|f-g|\le\epsilon$ (254Qa), so that
$\|u-v\|_1\le\epsilon$, where $v=g^{\ssbullet}$.   Since $P_J$, $P_K$
and $P_{K\cap J}$ are all linear operators of norm $1$,

\Centerline{$\|P_KP_Ju-P_{K\cap J}u\|_1
\le2\|u-v\|_1+\|P_KP_Jv-P_{K\cap J}v\|_1\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $P_KP_Ju=P_{K\cap J}u$;  as $u$
is arbitrary, $P_KP_J=P_{K\cap J}$.

\medskip

{\bf (b)} Take $u\in L^1(\lambda)$.   Let $\Cal J$ be the family of all
subsets $J$ of $I$ such that $P_Ju=u$.   By (a), $J\cap K\in\Cal J$ for
all $J$, $K\in\Cal J$.   Next, $\Cal J$ contains a countable set $J_0$.
\Prf\ Let $f$ be a $\lambda$-integrable function such that
$f^{\ssbullet}=u$.   By 254Qc, we can find a countable set
$J_0\subseteq I$ and a $\lambda_{J_0}$-integrable function $g$ such that
$f\eae g\pi_{J_0}$.   Now $g\pi_{J_0}$ is $\Lambda_{J_0}$-measurable
and $u=(g\pi_{J_0})^{\ssbullet}$ belongs to $L^1_{J_0}$, so
$J_0\in\Cal J$.\ \Qed

Write $J^*=\bigcap\Cal J$, so that $J^*\subseteq J_0$ is countable.
Then $J^*\in\Cal J$.   \Prf\ Let $\epsilon>0$.   As in the proof of (a)
above, there is a $g\in\eusm S$ such that $\|u-v\|_1\le\epsilon$, where
$v=g^{\ssbullet}$.   But because $g$ is a finite linear combination of
indicator functions of measurable cylinders, each determined by
coordinates in some
finite set, there is a finite $K\subseteq I$ such that
$g$ is $\Lambda_K$-measurable, so that $P_Kv=v$.
Because $K$ is finite, there must be $J_1,\ldots,J_n\in\Cal J$ such that
$J^*\cap K=\bigcap_{1\le i\le n}J_i\cap K$;  but as $\Cal J$ is closed
under finite intersections, $J=J_1\cap\ldots\cap J_n\in\Cal J$, and
$J^*\cap K=J\cap K$.

Now we have

\Centerline{$P_{J^*}v=P_{J^*}P_Kv=P_{J^*\cap K}v=P_{J\cap K}v=P_JP_Kv
=P_Jv$,}

\noindent using (a) twice.   Because both $P_J$ and $P_{J^*}$ have norm
$1$,

$$\eqalign{\|P_{J^*}u-u\|_1
&\le\|P_{J^*}u-P_{J^*}v\|_1
  +\|P_{J^*}v-P_Jv\|_1
  +\|P_Jv-P_Ju\|_1
  +\|P_Ju-u\|_1\cr
&\le\|u-v\|_1+0+\|u-v\|_1+0
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $P_{J^*}u=u$ and
$J^*\in\Cal J$.\ \Qed

Now, for any $J\subseteq I$,

$$\eqalign{P_Ju=u
&\Longrightarrow J\in\Cal J
\Longrightarrow J\supseteq J^*\cr
&\Longrightarrow P_Ju=P_JP_{J^*}u=P_{J\cap J^*}u=P_{J^*}u=u.\cr}$$

\noindent Thus $J^*$ has the required properties.

\medskip

{\bf (c)} Set $e=(\chi X)^{\ssbullet}$, $u_n=(-ne)\vee(u\wedge ne)$ for
each $n\in\Bbb N$.   Then, for any $J\subseteq I$, $u\in L^0_J$ iff
$u_n\in L^0_J$ for every $n$.   \Prf\ ($\alpha$) If $u\in L^0_J$, then
$u$ is expressible as $f^{\ssbullet}$ for some $\Lambda_J$-measurable
$f$;  now $f_n=(-n\chi X)\vee(f\wedge n\chi X)$ is
$\Lambda_J$-measurable, so $u_n=f_n^{\ssbullet}\in L^0_J$ for every
$n$.   ($\beta$) If $u_n\in L^0_J$ for each $n$, then for each $n$ we
can find a $\Lambda_J$-measurable function $f_n$ such that
$f_n^{\ssbullet}=u_n$.   But there is also a $\Lambda$-measurable
function $f$ such that $u=f^{\ssbullet}$, and we must have
$f_n\eae(-n\chi X)\vee(f\wedge n\chi X)$ for each $n$, so that
$f\eae\lim_{n\to\infty}f_n$ and
$u=(\lim_{n\to\infty}f_n)^{\ssbullet}$.   Since $\lim_{n\to\infty}f_n$
is $\Lambda_J$-measurable and defined on a
$\mu\restr\Lambda_J$-conegligible set, $u\in L^0_J$.\ \Qed

As every $u_n$ belongs to $L^1$, we know that

\Centerline{$u_n\in L^0_J\iff u_n\in L^1_J\iff P_Ju_n=u_n$.}

\noindent By (b), there is for each $n$ a countable $J_n^*$ such that
$P_{J}u_n=u_n$ iff $J\supseteq J_n^*$.   So we see that $u\in L^0_J$ iff
$J\supseteq J_n^*$ for every $n$, that is,
$J\supseteq\bigcup_{n\in\Bbb N}J_n^*$.   Thus
$J^*=\bigcup_{n\in\Bbb N}J_n^*$ has the property claimed.

\medskip

{\bf (d)} Applying (c) to $u=(\chi W)^{\ssbullet}$, we have a
(countable) unique smallest $J^*$ such that $u\in L^0_{J^*}$.   But if
$J\subseteq I$, then there is a $W'\in\Lambda_J$ such that
$W'\symmdiff W$ is negligible iff $u\in L^0_J$.   So this is the $J^*$
we are looking for.

\medskip

{\bf (e)} Again apply (c), this time to $f^{\ssbullet}$.
}%end of proof of 254R

\leader{254S}{Proposition} Let
$\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a family of probability
spaces, with product $(X,\Lambda,\lambda)$.

(a) If $A\subseteq X$ is
determined by coordinates in $I\setminus\{j\}$ for every $j\in I$,
then its outer measure $\lambda^*A$ must be either $0$ or $1$.

(b) If $W\in\Lambda$ and $\lambda W>0$, then for every $\epsilon>0$
there are a $W'\in\Lambda$ and a finite set $J\subseteq I$ such that
$\lambda W'\ge 1-\epsilon$ and for every $x\in W'$ there is a $y\in W$
such that $x\restr I\setminus J=y\restr I\setminus J$.

\proof{ For $J\subseteq I$ write $X_J$ for $\prod_{i\in J}X_i$ and
$\lambda_J$ for the product measure on $X_J$.

\medskip

{\bf (a)} Let $W$ be a measurable envelope of $A$.   By 254Rd, there is
a smallest $J\subseteq I$ for which
there is a $W'\in\Lambda$, determined by coordinates in $J$, with
$\lambda(W\symmdiff W')=0$.   Now $J=\emptyset$.   \Prf\ Take any
$j\in I$.   Then $A$ is determined by coordinates in $I\setminus\{j\}$,
that is, can be regarded as $X_j\times A'$ for some
$A'\subseteq X_{I\setminus\{j\}}$.   We can also think of
$\lambda$ as the product of $\lambda_{\{j\}}$ and
$\lambda_{I\setminus\{j\}}$ (254N).   Let $\Lambda_{I\setminus\{j\}}$ be
the domain of $\lambda_{I\setminus\{j\}}$.   By 251S,

\Centerline{$\lambda^*A
=\lambda_{\{j\}}^*X_j\cdot\lambda_{I\setminus\{j\}}^*A'
=\lambda_{I\setminus\{j\}}^*A'$.}

\noindent Let $V\in\Lambda_{I\setminus\{j\}}$ be measurable envelope of
$A'$. Then $W'=X_j\times V$ belongs to
$\Lambda$, includes $A$ and has measure $\lambda^*A$, so
$\lambda(W\cap W')=\lambda W=\lambda W'$ and $W\symmdiff W'$ is
negligible.   At the same time, $W'$ is determined by coordinates in
$I\setminus\{j\}$.   This means that $J$ must be included in
$I\setminus\{j\}$.   As $j$ is arbitrary, $J=\emptyset$.\ \Qed

But the only subsets of $X$ which are determined by coordinates in
$\emptyset$ are $X$ and $\emptyset$.   Since $W$ differs from one of
these by a negligible set, $\lambda^*A=\lambda W\in\{0,1\}$, as claimed.

\medskip

{\bf (b)} Set $\eta=\bover12\min(\epsilon,1)\lambda W$.   By 254Fe,
there is a measurable set $V$, determined by coordinates in a finite
subset $J$ of $I$, such that $\lambda(W\symmdiff V)\le\eta$.   Note that

\Centerline{$\lambda V\ge\lambda W-\eta\ge\Bover12\lambda W>0$,}

\noindent so

\Centerline{$\lambda(W\symmdiff V)\le\Bover12\epsilon\lambda W
\le\epsilon\lambda V$.}

\noindent We may identify
$\lambda$ with the c.l.d.\ product of $\lambda_J$ and
$\lambda_{I\setminus J}$ (254N).   Let $\tilde W$,
$\tilde V\subseteq X_I\times X_{I\setminus J}$ be the sets corresponding
to $W$, $V\subseteq X$.   Then $\tilde V$ can be expressed as
$U\times X_{I\setminus J}$ where $\lambda_JU=\lambda V>0$.  Set
$U'=\{z:z\in X_{I\setminus J}$, $\lambda_J\tilde W^{-1}[\{z\}]=0\}$.
Then $U'$ is measured by $\lambda_{I\setminus J}$ (252D(ii) again,
because both $\lambda_J$ and $\lambda_{I\setminus J}$ are complete), and

$$\eqalignno{\lambda_JU\cdot\lambda_{I\setminus J}U'
&\le\int\lambda_J(\tilde W^{-1}[\{z\}]\symmdiff U)
  \lambda_{I\setminus J}(dz)\cr
\displaycause{because if $z\in U'$ then
$\lambda_J(\tilde W^{-1}[\{z\}]\symmdiff U)=\lambda_JU$}
&=\int\lambda_J(\tilde W\symmdiff\tilde V)^{-1}[\{z\}]
  \lambda_{I\setminus J}(dz)\cr
&=(\lambda_J\times\lambda_{I\setminus J})(\tilde W\symmdiff\tilde V)\cr
\displaycause{252D once more}
&=\lambda(W\symmdiff V)
\le\epsilon\lambda V
=\epsilon\lambda_JU.\cr}$$

\noindent This means that $\lambda_{I\setminus J}U'\le\epsilon$.   Set
$W'=\{x:x\in X$, $x\restr I\setminus J\notin U'\}$;  then
$\lambda W'\ge 1-\epsilon$.   If $x\in W'$, then
$z=x\restr I\setminus J\notin U'$, so $\tilde W^{-1}[\{z\}]$ is not
empty, that is, there is a $y\in W$ such that $y\restr I\setminus J=z$.
So this $W'$ has the required properties.
}%end of proof of 254S

\leader{254T}{Remarks} \cmmnt{It is important to understand that the
results above apply to $L^0$ and $L^1$ and
measurable-sets-up-to-a-negligible-set, not to sets and functions
themselves.   One idea does apply to sets and functions, whether
measurable or not.

\header{254Ta}}{\bf (a)} Let $\langle X_i\rangle_{i\in I}$ be a family
of sets with Cartesian product $X$.   For each $J\subseteq I$ let
$\Cal W_J$
be the set of subsets of $X$ determined by coordinates in $J$.   Then
$\Cal W_J\cap\Cal W_K=\Cal W_{J\cap K}$ for all $J$, $K\subseteq I$.
\prooflet{\Prf\ Of course
$\Cal W_J\cap\Cal W_K\supseteq\Cal W_{J\cap K}$, because
$\Cal W_J\supseteq\Cal W_{J'}$ whenever $J'\subseteq J$.   On the
other hand, suppose $W\in\Cal W_J\cap\Cal W_K$, $x\in W$, $y\in X$
and $x\restr J\cap K=y\restr J\cap K$.   Set $z(i)=x(i)$ for $i\in J$,
$y(i)$ for $i\in I\setminus J$.   Then $z\restr J=x\restr J$ so $z\in
W$.   Also $y\restr K=z\restr K$ so $y\in W$.   As $x$, $y$ are
arbitrary, $W\in\Cal W_{J\cap K}$;  as $W$ is arbitrary,
$\Cal W_J\cap\Cal W_K\subseteq\Cal W_{J\cap K}$.\ \QeD}
Accordingly, for any $W\subseteq X$, $\Cal F=\{J:W\in\Cal W_J\}$ is a
filter on $I$ (unless $W=X$ or $W=\emptyset$, in which case
$\Cal F=\Cal PX$).   \cmmnt{But $\Cal F$ does not necessarily have a
least element, as the following example shows.}

\spheader 254Tb Set $X=\{0,1\}^{\Bbb N}$,

\Centerline{$W=\{x:x\in X$, $\lim_{i\to\infty}x(i)=0\}$.}

\noindent Then for every $n\in\Bbb N\,\,W$ is determined by coordinates
in
$J_n=\{i:i\ge n\}$.   But $W$ is not determined by coordinates in
$\bigcap_{n\in\Bbb N}J_n=\emptyset$.   \cmmnt{Note that

\Centerline{$W=\bigcup_{n\in\Bbb N}\bigcap_{i\ge n}\{x:x(i)=0\}$}

\noindent is measured by the usual measure on $X$.   But it is also
negligible (since it is countable);  in 254Rd we have $J^*=\emptyset$,
$W'=\emptyset$.}


\leader{*254U}{}\cmmnt{ I am now in a position to describe a
counter-example answering a natural question arising out of \S251.

\medskip

\noindent}{\bf Example} There are a localizable measure space
$(X,\Sigma,\mu)$ and a
probability space $(Y,\Tau,\nu)$ such that the c.l.d.\ product measure
$\lambda$ on $X\times Y$ is not localizable.

\proof{{\bf (a)} Take $(X,\Sigma,\mu)$ to be the space of 216E, so that
$X=\{0,1\}^I$, where $I=\Cal PC$ for some set $C$ with cardinal greater
than $\frak c$.   For each $\gamma\in C$ write $E_{\gamma}$ for
$\{x:x\in X$, $x(\{\gamma\})=1\}$ (that is, $G_{\{\gamma\}}$ in the
notation of 216Ec);  then $E_{\gamma}\in\Sigma$ and $\mu E_{\gamma}=1$;
also every measurable set of non-zero measure meets some $E_{\gamma}$ in
a set of non-zero measure, while $E_{\gamma}\cap E_{\delta}$ is
negligible for all distinct $\gamma$, $\delta$ (see 216Ee).

Let $(Y,\Tau,\nu)$ be $\{0,1\}^C$ with the usual measure (254J).   For
$\gamma\in C$, let $F_{\gamma}$ be $\{y:y\in Y$, $y(\gamma)=1\}$, so
that $\nu F_{\gamma}=\bover12$.
Let $\lambda$ be the c.l.d.\ product measure on $X\times Y$, and
$\Lambda$ its domain.

\medskip

{\bf (b)} Consider the family $\Cal W=\{E_{\gamma}\times
F_{\gamma}:\gamma\in C\}\subseteq\Lambda$.   \Quer\ Suppose, if
possible, that $V$ were an essential supremum of $\Cal W$ in $\Lambda$
in the sense of 211G.   For $\gamma\in C$ write
$H_{\gamma}=\{x:V[\{x\}]\symmdiff F_{\gamma}$ is negligible$\}$.
Because $F_{\gamma}\symmdiff F_{\delta}$ is non-negligible,
$H_{\gamma}\cap H_{\delta}=\emptyset$ for all $\gamma\ne\delta$.

Now $E_{\gamma}\setminus H_{\gamma}$ is $\mu$-negligible for every
$\gamma\in C$.   \Prf\
$\lambda((E_{\gamma}\times F_{\gamma})\setminus V)=0$, so
$F_{\gamma}\setminus V[\{x\}]$ is negligible for almost every
$x\in E_{\gamma}$, by 252D.   On the other hand, if we set
$F'_{\gamma}=Y\setminus F_{\gamma}$,
$W_{\gamma}=(X\times Y)\setminus(E_{\gamma}\times F'_{\gamma})$, then we
see that

\Centerline{$(E_{\gamma}\times F'_{\gamma})\cap(E_{\gamma}\times
F_{\gamma})=\emptyset$,
\quad $E_{\gamma}\times F_{\gamma}\subseteq W_{\gamma}$,}

\Centerline{$\lambda((E_{\delta}\times F_{\delta})\setminus W_{\gamma})
=\lambda((E_{\gamma}\times F'_{\gamma})
  \cap(E_{\delta}\times F_{\delta}))
\le\mu(E_{\gamma}\cap E_{\delta})
=0$}

\noindent for every $\delta\ne\gamma$, so $W_{\gamma}$ is an essential
upper bound for $\Cal W$ and
$V\cap(E_{\gamma}\times F'_{\gamma})=V\setminus W_{\gamma}$ must be
$\lambda$-negligible.    Accordingly
$V[\{x\}]\setminus F_{\gamma}=V[\{x\}]\cap F'_{\gamma}$
is $\nu$-negligible for $\mu$-almost every $x\in E_{\gamma}$.   But this
means that $V[\{x\}]\symmdiff F_{\gamma}$ is $\nu$-negligible for
$\mu$-almost every $x\in E_{\gamma}$, that is,
$\nu(E_{\gamma}\setminus H_{\gamma})=0$.\ \Qed

Now consider the family
$\langle E_{\gamma}\cap H_{\gamma}\rangle_{\gamma\in C}$.   This is a
disjoint family of sets of
finite measure in $X$.   If $E\in\Sigma$ has non-zero measure, there is
a $\gamma\in C$ such that
$\mu(E_{\gamma}\cap H_{\gamma}\cap E)=\nu(E_{\gamma}\cap E)>0$.   But
this means that $\Cal E=\{E_{\gamma}\cap H_{\gamma}:\gamma\in C\}$
satisfies the conditions of
213Oa, and $\mu$ must be strictly localizable;  which it isn't.\ \Bang

\medskip

{\bf (c)} Thus we have found a family $\Cal W\subseteq\Lambda$ with no
essential supremum in $\Lambda$, and $\lambda$ is not localizable.
}%end of proof of 254U

\cmmnt{\medskip

\noindent{\bf Remark} If $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ are any
localizable measure spaces with a non-localizable c.l.d.\ product
measure, then their c.l.d.\ versions are still localizable (213Hb) and
still have a
non-localizable product (251T), which cannot be strictly localizable;
so that at least one of the factors is not strictly localizable (251O).
Thus any
example of the type here must involve a complete locally determined
localizable space which is not strictly localizable, as in 216E.
}%end of comment

\leader{*254V}{}\cmmnt{ Corresponding to 251U and 251Wo, we have the
following result on countable powers of atomless probability spaces.

\medskip

\noindent}{\bf Proposition} Let $(X,\Sigma,\mu)$ be an atomless
probability space and $I$ a countable set.   Let $\lambda$ be the
product probability measure on $X^I$.   Then
$\{x:x\in X^I$, $x$ is injective$\}$ is $\lambda$-conegligible.
%used in 495

\proof{ For any pair $\{i,j\}$ of distinct elements of $X$, the set
$\{z:z\in X^{\{i,j\}}$, $z(i)=z(j)\}$ is negligible for the product
measure on $X^{\{i,j\}}$, by 251U.   By 254Oa,
$\{x:x\in X$, $x(i)=x(j)\}$ is $\lambda$-negligible.   Because $I$ is
countable, there are only countably many such pairs $\{i,j\}$, so
$\{x:x\in X$, $x(i)=x(j)$ for some distinct $i$, $j\in I\}$ is
negligible, and its complement is conegligible;  but this complement is
just the set of injective functions from $I$ to $X$.
}%end of proof of 254V

\exercises{
\leader{254X}{Basic exercises (a)}
%\spheader 254Xa
Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any family of
probability spaces, with product $(X,\Lambda,\mu)$.    Write $\Cal E$
for the family of subsets of $X$ expressible as the union of a finite
disjoint family of measurable cylinders.   (i) Show that if
$C\subseteq X$
is a measurable cylinder then $X\setminus C\in\Cal E$.   (ii) Show that
$W\cap V\in\Cal E$ for all $W$, $V\in\Cal E$.   (iii) Show that
$X\setminus W\in\Cal E$ for every $W\in\Cal E$.   (iv) Show that
$\Cal E$ is an algebra of subsets of $X$.   (v) Show that for any
$W\in\Lambda$, $\epsilon>0$ there is a $V\in\Cal E$ such that
$\lambda(W\symmdiff V)\le\epsilon^2$.   (vi) Show that for any
$W\in\Lambda$ and $\epsilon>0$ there are disjoint measurable cylinders
$C_0,\ldots,C_n$ such that
$\lambda(W\cap C_j)\ge(1-\epsilon)\lambda C_j$ for every $j$ and
$\lambda(W\setminus\bigcup_{j\le n}C_j)\le\epsilon$.
\Hint{select the $C_j$ from the
measurable cylinders composing a set $V$ as in (v).}   (vii) Show that
if $f$, $g$ are $\lambda$-integrable functions and $\int_Cf\le\int_Cg$
for every measurable cylinder $C\subseteq X$, then $f\leae g$.
\Hint{show that $\int_Wf\le\int_Wf$ for every $W\in\Lambda$.}
%254F

\sqheader 254Xb Let $\langle (X_i,\Sigma_i,\mu_i)$ be
a family of probability spaces, with product $(X,\Lambda,\lambda)$.
Show that the outer measure $\lambda^*$ defined by $\lambda$ is exactly
the outer measure $\theta$ described in 254A, that is, that $\theta$ is
a regular outer measure.
%254F

\spheader 254Xc Let $\langle (X_i,\Sigma_i,\mu_i)$ be
a family of probability
spaces, with product $(X,\Lambda,\lambda)$.   Write $\lambda_0$ for the
restriction of $\lambda$ to $\Tensorhat_{i\in I}\Sigma_i$, and $\Cal C$
for the family of measurable cylinders in $X$.   Suppose that
$(Y,\Tau,\nu)$ is a probability space and $\phi:Y\to X$ a function.
(i) Show that $\phi$ is \imp\ when regarded as a function from
$(Y,\Tau,\nu)$ to $(X,\Tensorhat_{i\in I}\Sigma_i,\lambda_0)$ iff
$\phi^{-1}[C]$ belongs to $\Tau$ and $\nu\phi^{-1}[C]=\lambda_0C$ for
every $C\in\Cal C$.   (ii) Show that $\lambda_0$ is the only measure on
$X$ with this property.   \Hint{136C.}
%254G

\sqheader 254Xd Let $I$ be a set and $(Y,\Tau,\nu)$ a complete
probability space.   Show that a function $\phi:Y\to\{0,1\}^I$ is \imp\
for $\nu$ and the usual measure on $\{0,1\}^I$ iff
$\nu\{y:\phi(y)(i)=1$ for every $i\in J\}=2^{-\#(J)}$ for every finite
$J\subseteq I$.
%254J

\sqheader 254Xe Let $I$ be any set and $\lambda$ the usual measure on
$X=\{0,1\}^I$.   Define addition on $X$ as in 254Jd.   Show that
the map $(x,y)\mapsto x+y:X\times X\to X$ is \imp, if $X\times X$ is
given its product measure.
%254J

\sqheader 254Xf Let $I$ be any set and $\lambda$ the usual measure on
$\Cal PI$.   (i) Show that the map
$a\mapsto a\symmdiff b:\Cal PI\to\Cal PI$ is \imp\ for any
$b\subseteq I$;  in particular, $a\mapsto I\setminus a$ is \imp.  (ii)
Show that the map
$(a,b)\mapsto a\symmdiff b:\Cal PI\times\Cal PI\to\Cal PI$ is \imp.
%254J

\sqheader 254Xg Show that for any $q\in[0,1]$ and any set $I$ there is a
measure $\lambda$ on $\Cal PI$ such that
$\lambda\{a:J\subseteq a\}=q^{\#(J)}$ for every finite $J\subseteq I$.
%254J

\sqheader 254Xh Let $(Y,\Tau,\nu)$ be a complete probability
space, and write $\mu$ for Lebesgue measure on $[0,1]$.   Suppose that
$\phi:Y\to[0,1]$ is a function such that $\nu\phi^{-1}[I]$ exists and is
equal to $\mu I$ for every interval $I$ of the form
$[2^{-n}k,2^{-n}(k+1)]$, where $n\in\Bbb N$ and $0\le k<2^n$.   Show
that $\phi$ is \imp\ for $\nu$ and $\mu$.
%254K

\spheader 254Xi Show that if $\tilde\phi:\{0,1\}^{\Bbb N}\to[0,1]$ is
any bijection constructed by the method of 254K, then
$\{\tilde\phi^{-1}[E]:E\subseteq[0,1]$ is a Borel set$\}$ is just the
$\sigma$-algebra of subsets of $\{0,1\}^{\Bbb N}$ generated by the sets
$\{x:x(i)=1\}$ for $i\in\Bbb N$.
%254K

\spheader 254Xj Let $\langle X_i\rangle_{i\in I}$ be a family of
sets, and for each $i\in I$ let $\Sigma_i$ be a $\sigma$-algebra of
subsets of $X_i$.   Show that for every
$E\in\Tensorhat_{i\in I}\Sigma_i$ there is a countable set
$J\subseteq I$ such that $E$ is
expressible as $\pi_J^{-1}[F]$ for some $F\in\Tensorhat_{i\in J}X_j$,
writing $\pi_J(x)=x\restr J\in\prod_{i\in J}X_i$ for
$x\in\prod_{i\in I}X_i$.
%254M

\spheader 254Xk(i) Let $\nu$ be the usual measure on
$X=\{0,1\}^{\Bbb N}$.   Show that for any $k\ge 1$, $(X,\nu)$ is
isomorphic to $(X^k,\nu_k)$, where $\nu_k$ is the measure on $X^k$ which
is the product measure obtained by giving each factor $X$ the measure
$\nu$.    (ii) Writing $\mu_{[0,1]}$ for Lebesgue measure on
$[0,1]$, etc., show that for any $k\ge 1$, $([0,1]^k,\mu_{[0,1]^k})$ is
isomorphic to $([0,1],\mu_{[0,1]})$.
%254K, 254N

\spheader 254Xl(i) Writing $\mu_{[0,1]}$ for Lebesgue measure on
$[0,1]$, etc., show that $([0,1],\mu_{[0,1]})$ is isomorphic to
$(\coint{0,1},\mu_{\coint{0,1}})$.    (ii) Show that for any $k\ge
1$, $(\coint{0,1}^k,\mu_{\coint{0,1}^k})$ is isomorphic to
$(\coint{0,1},\mu_{\coint{0,1}})$.   (iii) Show that for any $k\ge
1$, $(\Bbb R,\mu_{\Bbb R})$ is isomorphic to $(\BbbR^k,\mu_{\Bbb
R^k})$.
%254K, 254N

\spheader 254Xm Let $\mu$ be Lebesgue measure on $[0,1]$ and
$\lambda$ the product measure on $[0,1]^{\Bbb N}$.   Show that
$([0,1],\mu)$ and $([0,1]^{\Bbb N},\lambda)$ are isomorphic.
%254K, 254N

\spheader 254Xn Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$
be a family of complete probability spaces and $\lambda$ the product
measure on $\prod_{i\in I}X_i$, with domain $\Lambda$.   Suppose that
$A_i\subseteq X_i$ for
each $i\in I$.   Show that $\prod_{i\in I}A_i\in\Lambda$ iff either (i)
$\prod_{i\in I}\mu_i^*A_i=0$ or (ii) $A_i\in\Sigma_i$ for
every $i$ and $\{i:A_i\ne X_i\}$ is countable.   \Hint{assemble
ideas from 252Xc, 254F, 254L and 254N.}
%254F, 254L, 254N

\spheader 254Xo Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces with product $(X,\Lambda,\lambda)$.   (i) Show that,
for any $A\subseteq X$,

\Centerline{$\lambda^*A=\min\{\lambda_J^*\pi_J[A]:J\subseteq I$ is
countable$\}$,}

\noindent where for $J\subseteq I$ I write $\lambda_J$ for the product
probability measure on $X_J=\prod_{i\in J}X_i$ and $\pi_J:X\to X_J$ for
the canonical map.   (ii) Show that if $J$, $K\subseteq I$ are disjoint
and $A$, $B\subseteq X$ are determined by coordinates in $J$, $K$
respectively, then $\lambda^*(A\cap B)=\lambda^*A\cdot\lambda^*B$.
%254O

\spheader 254Xp Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces with product $(X,\Lambda,\lambda)$.   Let $\eusm S$
be the linear span of the set of indicator functions of measurable
cylinders in $X$, as in 254Q.   Show that
$\{f^{\ssbullet}:f\in\eusm S\}$ is dense in $L^p(\mu)$ for every
$p\in\coint{1,\infty}$.
%254Q

\spheader 254Xq Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be a
family of probability spaces, and $(X,\Lambda,\lambda)$ their product;
for $J\subseteq I$ let $\Lambda_J$ be the $\sigma$-algebra of members of
$\Lambda$ determined by coordinates in $J$ and
$P_J:L^1=L^1(\lambda)\to L^1_J=L^1(\lambda\restr\Lambda_J)$ the
corresponding conditional expectation.   (i) Show that if $u\in L^1_J$
and $v\in L^1_{I\setminus J}$ then $u\times v\in L^1$ and
$\int u\times v=\int u\cdot\int v$.
\Hint{253D.}   (ii) Show that if $\Cal J\subseteq\Cal PI$ is
non-empty, with $J^*=\bigcap\Cal J$, then
$L^1_{J^*}=\bigcap_{J\in\Cal J}L^1_J$.
%254R

\spheader 254Xr(i) Let $I$ be any set and $\lambda$ the usual measure on
$\Cal PI$.   Let $A\subseteq\Cal PI$ be such that $a\symmdiff b\in A$
whenever $a\in A$ and $b\subseteq I$ is finite.   Show that $\lambda^*A$
must be either $0$ or $1$.
(ii) Let $\lambda$ be the usual measure on $\{0,1\}^{\Bbb N}$, and
$\Lambda$ its domain.   Let $f:\{0,1\}^{\Bbb N}\to\Bbb R$ be a function
such that, for $x$, $y\in\{0,1\}^{\Bbb N}$,
$f(x)=f(y)\iff\{n:n\in\Bbb N$, $x(n)\ne y(n)\}$ is finite.
Show that $f$ is not $\Lambda$-measurable.
\Hint{for any $q\in\Bbb Q$, $\lambda^*\{x:f(x)\le q\}$ is either $0$ or
$1$.}
%254S

\spheader 254Xs Let $\familyiI{X_i}$ be any family of sets and
$A\subseteq B\subseteq\prod_{i\in I}X_i$.   Suppose that $A$ is
determined by coordinates in $J\subseteq I$ and that $B$ is determined
by coordinates in $K$.   Show that there is a set $C$ such that
$A\subseteq C\subseteq B$ and $C$ is determined by coordinates in
$J\cap K$.
%254T

\leader{254Y}{Further exercises (a)}
%\spheader 254Ya
Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of
probability spaces, and for $J\subseteq I$ let $\lambda_J$ be the
product
measure on $X_J=\prod_{i\in J}X_i$;  write $X=X_I$, $\lambda=\lambda_I$
and $\pi_J(x)=x\restr J$ for $x\in X$ and $J\subseteq I$.

\quad (i) Show that for $K\subseteq J\subseteq I$ we have a natural
linear, order-preserving and
norm-preserving map $T_{JK}:L^1(\lambda_K)\to L^1(\lambda_J)$ defined by
writing $T_{JK}(f^{\ssbullet})=(f\pi_{KJ})^{\ssbullet}$ for every
$\lambda_K$-integrable function $f$, where $\pi_{KJ}(y)=y\restr K$ for
$y\in X_J$.

\quad (ii) Write $\Cal K$ for the set of finite subsets of $I$.   Show
that if  $W$ is any Banach space and $\langle T_K\rangle_{K\in\Cal K}$
is a family such that ($\alpha$) $T_K$ is a bounded linear operator from
$L^1(\lambda_K)$ to $W$ for every $K\in\Cal K$ ($\beta$)
$T_K=T_JT_{JK}$ whenever $K\subseteq J\in \Cal K$ ($\gamma$)
$\sup_{K\in\Cal K}\|T_K\|<\infty$, then there is a unique bounded
linear operator $T:L^1(\lambda)\to W$ such that $T_K=TT_{IK}$ for every
$K\in\Cal K$.

\quad (iii) Write $\Cal J$ for the set of countable subsets of $I$.
Show that $L^1(\lambda)=\bigcup_{J\in\Cal J}T_{IJ}[L^1(\lambda_J)]$.
%254F

\spheader 254Yb Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$
be any family of measure spaces.
Set $X=\prod_{i\in I}X_i$ and let $\Cal F$ be a filter on the
set $[I]^{<\omega}$ of finite subsets of $I$ such that
$\{J:i\in J\in[I]^{<\omega}\}\in\Cal F$ for every $i\in I$.   Show that
there is a complete locally determined
measure $\lambda$ on $X$ such that $\lambda(\prod_{i\in I}E_i)$
is defined and equal to $\lim_{J\to\Cal F}\prod_{i\in J}\mu_iE_i$ whenever
$E_i\in\Sigma_i$ for every $i\in I$ and
$\lim_{J\to\Cal F}\prod_{i\in J}\mu_iE_i$ is defined in
$\coint{0,\infty}$.
\Hint{{\smc Baker 04}.}  %n10313.tex
%254F

\spheader 254Yc
Let $\familyiI{(X_i,\Sigma_i,\mu_i)}$ be a family of probability
spaces, and $\lambda$ a complete measure on $X=\prod_{i\in I}X_i$.
Suppose that for every complete probability space $(Y,\Tau,\nu)$ and
function
$\phi:Y\to X$, $\phi$ is \imp\ for $\nu$ and $\lambda$ iff
$\nu\phi^{-1}[C]$ is defined and
equal to $\theta_0 C$ for every measurable cylinder $C\subseteq X$,
writing $\theta_0$ for the functional of 254A.    Show that
$\lambda$ is the product measure on $X$.
%254G

\spheader 254Yd Let $I$ be a set, and $\lambda$ the usual
measure on $\{0,1\}^I$.   Show that $L^1(\lambda)$ is separable, in its
norm topology, iff $I$ is countable.
%254J

\spheader 254Ye Let $f:[0,1]\to[0,1]^2$ be a function
which is \imp\ for Lebesgue planar measure on $[0,1]^2$ and
Lebesgue linear measure on $[0,1]$, as in 134Yl;  let $f_1$, $f_2$ be the
coordinates of $f$.   Define $g:[0,1]\to[0,1]^{\Bbb N}$ by setting
$g(t)=\sequencen{f_1f_2^n(t)}$ for $0\le t\le 1$.   Show that $g$ is
\imp.   \Hint{show that $g_n:[0,1]\to[0,1]^{n+1}$ is \imp\ for every
$n\ge 1$, where
$g_n(t)=(f_1(t),f_1f_2(t),\ldots,\penalty-100f_1f_2^{n-1}(t),f_2^n(t))$
for $t\in[0,1]$.}
%254N 254G out of order query

\spheader 254Yf Let $I$ be a set, and $\lambda$ the usual measure on
$\Cal PI$.   Show that if $\Cal F$ is a non-principal ultrafilter on $I$
then $\lambda^*\Cal F=1$.   \Hint{254Xr, 254Xf.}
%254J, 254S

\spheader 254Yg Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ and $\lambda$ be as
in 254U.   Set $A=\{x_{\gamma}:\gamma\in C\}$ as defined in 216E.   Let
$\mu_A$ be the subspace measure on $A$, and $\tilde\lambda$ the c.l.d.\
product measure of $\mu_A$ and $\nu$ on $A\times Y$.   Show that
$\tilde\lambda$ is a proper extension of the subspace measure
$\lambda_{A\times Y}$.   \Hint{consider
$\tilde W=\{(x_{\gamma},y):\gamma\in C$, $y\in F_{\gamma}\}$, in the
notation of 254U.}
%254U

\spheader 254Yh Let $(X,\Sigma,\mu)$ be an atomless probability space,
$I$ a set with cardinal at most $\#(X)$, and $A$ the set of injective
functions from $I$ to $X$.   Show that $A$ has full outer measure for
the product measure on $X^I$.
%254V

}%end of exercises

\endnotes{
\Notesheader{254} While there are many reasons for studying infinite
products of probability spaces, one stands pre-eminent, from the point
of view of abstract measure theory:  they provide constructions of
essentially new kinds of measure space.   I cannot describe the nature
of this `newness' effectively without venturing into the territory of
Volume 3.   But the function spaces of Chapter 24 do give at least a
form of words we can use:  these are the first {\it probability} spaces
$(X,\Lambda,\lambda)$ we have seen for which $L^1(\lambda)$ need not be
separable for its norm topology (254Yd).

The formulae of 254A, like those of 251A, lead
very naturally to measures;  the point at
which they become more than a curiosity is when we find that the product
measure $\lambda$ is a
probability measure (254Fa), which must be regarded as the crucial
argument of this section, just as 251E is the essential basis of
\S251.   It is I think remarkable that it makes no difference to the
result here whether $I$ is finite, countably infinite or uncountable.
If you write out the proof for the case $I=\Bbb N$, it will seem natural
to expand the sets $J_n$ until they are initial segments of $I$ itself,
thereby avoiding altogether the auxiliary set $K$;  but this is a
misleading simplification, because it hides an essential feature of the
argument, which is that any sequence in $\Cal C$ involves only countably
many coordinates, so that as long as we are dealing with only one such
sequence the uncountability of the whole set $I$ is irrelevant.
This general principle naturally permeates the whole of the section;  in
254O I have tried to spell out the way in which many of the questions we
are interested in can be expressed in terms of countable subproducts of
the factor spaces $X_i$.   See also the exercises 254Xj, 254Xn and
254Ya(iii).

There is a slightly paradoxical side to this principle:  even the
best-behaved subsets $E_i$ of $X_i$ may fail to have measurable products
$\prod_{i\in I}E_i$ if $E_i\ne X_i$ for uncountably many $i$.   For
instance, $\ooint{0,1}^I$ is not a measurable subset of $[0,1]^I$ if $I$
is uncountable (254Xn).   It has full outer measure and its
own product measure is just the subspace measure (254L), but any
measurable subset must have measure zero.   The point is that the empty
set is the only member of $\Tensorhat_{i\in I}\Sigma_i$, where
$\Sigma_i$ is the algebra of Lebesgue measurable subsets of $[0,1]$ for
each $i$, which is included in $\ooint{0,1}^I$ (see 254Xj).

As in \S251, I use a construction which automatically produces a
complete measure on the product space.   I am sure that this is the best
choice for `the' product measure.   But there are occasions when its
restriction to the $\sigma$-algebra generated by the measurable
cylinders is worth looking at;  see 254Xc.

Lemma 254G is a result of a type which will be commoner in Volume 3
than in the present volume.   It describes the product measure in terms
not of what it {\it is} but of what it {\it does};
specifically, in terms of a property of the associated family of
\imp\ functions.  It
is therefore a `universal mapping theorem'.   (Compare 253F.)
Because this description is sufficient to determine the product measure
completely (254Yc), it is not surprising that I use it repeatedly.

The `usual measure' on $\{0,1\}^I$ (254J) is sometimes called
`coin-tossing measure' because it can be used to model the concept of
tossing a coin arbitrarily many times indexed by the set $I$, taking an
$x\in\{0,1\}^I$ to represent the outcome in which the coin is `heads'
for just those $i\in I$ for which $x(i)=1$.   The sets, or `events',
in the class $\Cal C$ are those which can be specified by declaring
the outcomes of
finitely many tosses, and the probability of any particular sequence of
$n$ results is $1/2^n$, regardless of which tosses we look at or in
which order.   In Chapter 27 I will return to the use of product
measures to represent probabilities involving independent events.

In 254K I come to the first case in this treatise of a non-trivial
isomorphism between two measure spaces.   If you have been brought up on
a conventional diet of modern abstract pure mathematics based on algebra
and topology, you may already have been struck by the absence of
emphasis on any concept of `homomorphism' or `isomorphism'.   Here
indeed I start to speak of `isomorphisms' between measure spaces
without even troubling to define them;  I hope it really is obvious that
an isomorphism between measure spaces $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ is a bijection $\phi:X\to Y$ such that
$\Tau=\{F:F\subseteq Y$, $\phi^{-1}[F]\in\Sigma\}$ and
$\nu F=\mu\phi^{-1}[F]$ for every $F\in\Tau$, so that $\Sigma$ is
necessarily $\{E:E\subseteq X$, $\phi[E]\in\Tau\}$ and
$\mu E=\nu\phi[E]$
for every $E\in\Sigma$.   Put like this, you may, if you worked through
the exercises of Volume 1, be reminded of some constructions of
$\sigma$-algebras in 111Xc-111Xd and of the `image measures' in
234C-234D.   The result in 254K (see also 134Yo) naturally leads to two
distinct notions of
`homomorphism' between two measure spaces $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$:

\inset{(i) a function $\phi:X\to Y$ such that $\phi^{-1}[F]\in\Sigma$
and $\mu\phi^{-1}[F]=\nu F$ for every $F\in\Tau$,}

\inset{(ii) a function $\phi:X\to Y$ such that $\phi[E]\in \Tau$ and
$\nu\phi[E]=\mu E$ for every $E\in\Sigma$.}

\noindent On either definition, we find that a bijection $\phi:X\to Y$
is an isomorphism iff $\phi$ and $\phi^{-1}$ are both homomorphisms.
(Also, of course, the composition of homomorphisms will be a
homomorphism.)   My own view is that (i) is the more important, and
in this treatise I study such functions at length, calling
them `\imp'.   But both have their uses.   The
function $\phi$ of 254K not only satisfies both definitions, but is also
`nearly' an isomorphism in several different ways, of which possibly
the most important is that there are conegligible sets
$X'\subseteq\{0,1\}^{\Bbb N}$,
$Y'\subseteq [0,1]$ such that $\phi\restr X'$ is an
isomorphism between $X'$ and $Y'$ when both are
given their subspace measures.

Having once established the isomorphism between $[0,1]$ and
$\{0,1\}^{\Bbb N}$, we are led immediately to many more;  see
254Xk-254Xm.   In fact Lebesgue measure on $[0,1]$ is isomorphic to a
large proportion of the probability spaces arising in applications.   In
Volumes 3 and 4 I will discuss these isomorphisms at length.

The general notion of `subproduct' is associated with some of the
deepest and most characteristic results in the theory of product
measures.   Because we are looking at products of arbitrary families of
probability spaces, the definition must ignore any possible structure in
the index set $I$ of 254A-254C.   But many applications, naturally
enough, deal with index sets with favoured subsets or partitions, and
the first essential step is the `associative law' (254N;  compare
251Xe-251Xf and 251Wh).   This is, for instance, the tool by which
we can apply Fubini's theorem within infinite products.   The natural
projection maps from $\prod_{i\in I}X_i$ to $\prod_{i\in J}X_i$, where
$J\subseteq I$, are related in a way which has already been used as the
basis of theorems in \S235;  the product measure on $\prod_{i\in J}X_i$
is precisely the image of the product measure on $\prod_{i\in I}X_i$
(254Oa).   In 254O-254Q I explore the consequences of this fact and the
fact already noted that all measurable sets in the product are
`essentially' determined by coordinates in some countable set.

In 254R I go more deeply into this notion of a set
$W\subseteq\prod_{i\in I}X_i$ `determined by coordinates in' a set
$J\subseteq I$.   In its primitive form this is a purely set-theoretic
notion (254M, 254Ta).   I think that even a three-element set $I$ can
give us surprises;  I invite you to try to visualize subsets of
$[0,1]^3$ which are determined by pairs of coordinates.   But the
interactions of this with measure-theoretic ideas, and in particular
with a willingness to add or discard negligible sets, lead to much
more, and in particular to the unique minimal sets of coordinates
associated with measurable sets and functions (254R).   Of course
these results can be elegantly and effectively described in terms of
$L^1$ and $L^0$ spaces, in which negligible sets are swept out of sight
as the spaces are constructed.   The basis of all this is the fact that
the conditional expectation operators associated with subproducts
multiply together in the simplest possible way (254Ra);  but some
further idea is needed to show that if $\Cal J$ is a non-empty family of
subsets of $I$, then $L^0_{\bigcap\Cal J}=\bigcap_{J\in\Cal J}L^0_J$
(see part (b) of the proof of 254R, and 254Xq(iii)).

254Sa is a version of the `zero-one law' (272O below).   254Sb is a
strong version of the principle that measurable sets in a
product must be approximable by sets determined by a {\it finite} set of
coordinates (254Fe, 254Qa, 254Xa).   Evidently it is not a coincidence
that the set $W$ of 254Tb is negligible.   In \S272 I will revisit many
of the ideas of 254R-254S and 254Xq, in particular, in the more
general context of `independent $\sigma$-algebras'.

Finally, 254U and 254Yg hardly belong to this section at all;  they are
unfinished business from \S251.   They are here because the construction
of 254A-254C is the simplest way to produce an adequately complex
probability space $(Y,\Tau,\nu)$.
}%end of notes

\discrpage
\frfilename{mt255.tex}
\versiondate{3.7.08}
\copyrightdate{2000}

\def\chaptername{Product measures}
\def\sectionname{Convolutions of functions}

\newsection{255}

I devote a section to a construction which is of great importance -- and
will in particular be very useful in Chapters 27 and 28 -- and may also
be regarded as a series of exercises on the work so far.

I find it difficult to know how much repetition to indulge in in this
section, because the natural unified expression of the ideas is in the
theory of topological
groups, and I do not think we are yet ready for the general theory (I
will come to it in Chapter 44 in Volume 4).   The groups we need for this volume are

\qquad $\Bbb R$;

\qquad $\BbbR^r$, for $r\ge 2$;

\qquad $S^1=\{z:z\in\Bbb C,\,|z|=1\}$, the `circle group';

\qquad $\Bbb Z$, the group of integers.

\noindent All the ideas already appear in the theory of convolutions on
$\Bbb R$, and I will therefore present this material in relatively
detailed form, before sketching the forms appropriate to the groups
$\BbbR^r$ and $S^1$ (or $\ocint{-\pi,\pi}$);  $\Bbb Z$ can I think be
safely left to the exercises.

\leader{255A}{}\cmmnt{ This being a book on measure theory, it is
perhaps appropriate for me to emphasize, as the basis of the theory of
convolutions, certain measure space isomorphisms.

\medskip

\noindent}{\bf Theorem} Let $\mu$ be Lebesgue measure on $\Bbb R$ and
$\mu_2$ Lebesgue measure on $\BbbR^2$;  write $\Sigma$, $\Sigma_2$ for
their domains.

(a) For any $a\in\Bbb R$, the map $x\mapsto a+x:\Bbb R\to\Bbb R$ is a
measure space automorphism of $(\Bbb R,\Sigma,\mu)$.

(b) The map $x\mapsto -x:\Bbb R\to\Bbb R$ is a measure space
automorphism of $(\Bbb R,\Sigma,\mu)$.

(c) For any $a\in\Bbb R$, the map $x\mapsto a-x:\Bbb R\to\Bbb R$ is a
measure space automorphism of $(\Bbb R,\Sigma,\mu)$.

(d) The map $(x,y)\mapsto(x+y,y):\BbbR^2\to\BbbR^2$ is a measure space
automorphism of $(\BbbR^2,\Sigma_2,\mu_2)$.

(e) The map $(x,y)\mapsto(x-y,y):\BbbR^2\to\BbbR^2$ is a measure space
automorphism of $(\BbbR^2,\Sigma_2,\mu_2)$.

\cmmnt{\medskip

\noindent{\bf Remark} I ought to remark that (b), (d) and (e) may be
regarded as simple
special cases of Theorem 263A in the next chapter.   I nevertheless feel
that it is worth writing out separate proofs here, partly because the
general case of linear operators dealt with in 263A requires some extra
machinery not needed here, but more because the result here has nothing
to do with the {\it linear} structure of $\Bbb R$ and $\BbbR^2$;  it is
exclusively dependent on the {\it group} structure of $\Bbb R$, together
with the links between its topology and measure, and the arguments I
give now are adaptable to the proper generalizations to abelian
topological groups.

}%end of comment

\proof{{\bf (a)} This is just the translation-invariance of Lebesgue
measure, dealt with in \S134.   There I showed that if $E\in\Sigma$
then $E+a\in\Sigma$ and $\mu(E+a)=\mu E$ (134Ab);  that is, writing
$\phi(x)=x+a$, $\mu(\phi[E])$ exists and is equal to $\mu E$ for every
$E\in\Sigma$.   But of course we also have

\Centerline{$\mu(\phi^{-1}[E])=\mu(E+(-a))=\mu E$}

\noindent for every $E\in\Sigma$, so $\phi$ is an automorphism.

\medskip

{\bf (b)} The point is that $\mu^*(A)=\mu^*(-A)$ for every
$A\subseteq\Bbb R$.   \Prf\ (I follow the definitions of Volume 1.)   If
$\epsilon>0$, there is a sequence $\sequencen{I_n}$ of half-open
intervals covering $A$ with $\sum_{n=0}^{\infty}\mu
I_n\le\mu^*A+\epsilon$.   Now $-A\subseteq\bigcup_{n\in\Bbb N}(-I_n)$.
But if $I_n=\coint{a_n,b_n}$ then $-I_n=\ocint{-b_n,a_n}$, so

\Centerline{$\mu^*(-A)\le\sum_{n=0}^{\infty}\mu(-I_n)
=\sum_{n=0}^{\infty}\max(0,-a_n-(-b_n))=\sum_{n=0}^{\infty}\mu I_n
\le\mu^*A+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\mu^*(-A)\le\mu^*A$.   Also of
course $\mu^*A\le\mu^*(-(-A))=\mu^*A$, so $\mu^*(-A)=\mu^*A$.   \Qed

This means that, setting $\phi(x)=-x$ this time, $\phi$ is an
automorphism of the structure $(\Bbb R,\mu^*)$.   But since $\mu$ is
defined from $\mu^*$ by the abstract procedure of \Caratheodory's
method, $\phi$ must also be an automorphism of the structure $(\Bbb
R,\Sigma,\mu)$.

\medskip

{\bf (c)} Put (a) and (b) together;  $x\mapsto a-x$ is the composition
of the automorphisms $x\mapsto -x$ and $x\mapsto a+x$, and the
composition of automorphisms is surely an automorphism.

\medskip

{\bf (d)(i)} Write $\Tau$ for the set
$\{E:E\in\Sigma_2,\,\phi[E]\in\Sigma_2\}$, where this time
$\phi(x,y)=(x+y,y)$ for $x$, $y\in\Bbb R$, so that 
$\phi:\BbbR^2\to\Bbb R^2$ is a permutation.   
Then $\Tau$ is a $\sigma$-algebra, being the
intersection of the $\sigma$-algebras $\Sigma_2$ and
$\{E:\phi[E]\in\Sigma_2\}=\{\phi^{-1}[F]:F\in\Sigma_2\}$.   Moreover,
$\mu_2E=\mu_2(\phi[E])$ for every $E\in\Tau$.   \Prf\  By 252D,
we have

\Centerline{$\mu_2E=\int\mu\{x:(x,y)\in E\}\mu(dy)$.}

\noindent But applying the same result to $\phi[E]$ we have

$$\eqalignno{\mu_2\phi[E]
&=\int\mu\{x:(x,y)\in\phi[E]\}\mu(dy)
=\int\mu\{x:(x-y,y)\in E\}\mu(dy)\cr
&=\int\mu(E^{-1}[\{y\}]+y)\mu(dy)
=\int\mu E^{-1}[\{y\}]\mu(dy)\cr
\noalign{\noindent (because Lebesgue measure is translation-invariant)}
&=\mu_2E. \text{  \Qed}\cr}$$

\medskip

\quad{\bf (ii)} Now $\phi$ and $\phi^{-1}$ are clearly continuous, so
that $\phi[G]$ is open, and therefore measurable, for every open $G$;
consequently all open sets must belong to $\Tau$.   Because $\Tau$ is a
$\sigma$-algebra, it contains all Borel sets.   Now let $E$ be any
measurable set.   Then there are Borel sets $H_1$, $H_2$ such that
$H_1\subseteq E\subseteq H_2$ and $\mu_2(H_2\setminus H_1)=0$ (134Fb).
We have $\phi[H_1]\subseteq\phi[E]\subseteq\phi[H_2]$ and

\Centerline{$\mu(\phi[H_2]\setminus\phi[H_1])
=\mu\phi[H_2\setminus H_1]=\mu(H_2\setminus H_1)=0$.}

\noindent Thus $\phi[E]\setminus\phi[H_1]$ must be negligible, therefore
measurable, and $\phi[E]=\phi[H_1]\cup(\phi[E]\setminus\phi[H_1])$ is
measurable.   This shows that $\phi[E]$ is measurable whenever $E$ is.

\medskip

\quad{\bf (iii)} Repeating the same arguments with $-y$ in the place of
$y$, we see that $\phi^{-1}[E]$ is measurable, and
$\mu_2\phi^{-1}[E]=\mu_2E$, for every $E\in\Sigma_2$.
So $\phi$ is an automorphism of the structure $(\BbbR^2,\Sigma_2,\mu_2)$.

\medskip

{\bf (e)} Of course this is an immediate corollary either of the proof
of (d) or of (d) itself as stated, since $(x,y)\mapsto(x-y,y)$ is just
the inverse of $(x,y)\mapsto(x+y,y)$.
}%end of proof of 255A

\leader{255B}{Corollary} (a) If $a\in\Bbb R$, then for any
complex-valued function $f$ defined on a subset of $\Bbb R$

\Centerline{$\int f(x)dx
=\int f(a+x)dx=\int f(-x)dx=\int f(a-x)dx$}

\noindent in the sense that if one of the integrals exists so do the
others, and they are then all equal.

(b) If $f$ is a complex-valued function defined on a subset of
$\BbbR^2$, then

\Centerline{$\int f(x+y,y)d(x,y)=\int f(x-y,y)d(x,y)
=\int f(x,y)d(x,y)$}

\noindent in the sense that if one of the integrals exists and is finite
so does the other, and they are then equal.

\cmmnt{
\leader{255C}{Remarks (a)}  I am not sure whether it ought to be
`obvious' that if $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ are measure spaces
and $\phi:X\to Y$ is an isomorphism, then for any function
$f$ defined on a subset of $Y$

\Centerline{$\int f(\phi(x))\mu(dx)=\int f(y)\nu(dy)$}

\noindent in the sense that if one is defined so is the other, and they
are then equal.   If it is obvious then the obviousness must be
contingent on the nature of the definition of integration:
integrability with respect to the measure $\mu$ is something which
depends on the structure $(X,\Sigma,\mu)$ and on no other properties of
$X$.   If it is not obvious then it is an easy deduction from Theorem
235A above, applied in turn to $\phi$ and $\phi^{-1}$ and to the real
and imaginary parts of $f$.   In any case the isomorphisms of 255A are
just those needed to prove 255B.

\header{255Cb}{\bf (b)} Note that in 255Bb I write $\int f(x,y)d(x,y)$
to emphasize that I am considering the integral of $f$ with respect to
two-dimensional Lebesgue measure.   The fact that

\Centerline{$\int\bigl(\int f(x,y)dx\bigr)dy
=\int\bigl(\int f(x+y,y)dx\bigr)dy=\int\bigl(\int f(x-y,y)dx\bigr)dy$}

\noindent is actually easier, being an immediate consequence of the
equality $\int f(a+x)dx=\int f(x)dx$.   But applications of this result
often depend essentially on the fact that the functions
$(x,y)\mapsto f(x+y,y)$, $(x,y)\mapsto f(x-y,y)$ are measurable as
functions of two variables.

\header{255Cc}{\bf (c)} I have moved directly to complex-valued
functions because these are necessary for the applications in Chapter
28.   If however
they give you any discomfort, either technically or aesthetically, all
the measure-theoretic ideas of this section are already to be found in
the real case, and you may wish at first to read it as if only real
numbers were involved.
}%end of comment

\leader{255D}{}\cmmnt{ A further corollary of 255A will be useful.

\medskip

\noindent}{\bf Corollary} Let $f$ be a complex-valued function defined
on a subset of $\Bbb R$.

(a) If $f$ is measurable, then the functions $(x,y)\mapsto f(x+y)$,
$(x,y)\mapsto f(x-y)$ are measurable.

(b) If $f$ is defined almost everywhere in $\Bbb R$, then the functions
$(x,y)\mapsto f(x+y)$, $(x,y)\mapsto f(x-y)$ are defined almost
everywhere in $\BbbR^2$.

\proof{ Writing $g_1(x,y)=f(x+y)$, $g_2(x,y)=f(x-y)$ whenever these are
defined, we have

\Centerline{$g_1(x,y)=(f\otimes\chi\Bbb R)(\phi(x,y))$,
\quad $g_2(x,y)=(f\otimes\chi\Bbb R)(\phi^{-1}(x,y))$,}

\noindent writing $\phi(x,y)=(x+y,y)$ as in 255B(d-e), and
$(f\otimes\chi\Bbb R)(x,y)=f(x)$, following the notation of 253B.   By
253C, $f\otimes\chi\Bbb R$ is measurable if $f$ is, and defined
almost everywhere if $f$ is.   Because $\phi$ is a measure space
automorphism, $(f\otimes\chi\Bbb R)\phi=g_1$ and
$(f\otimes\chi\Bbb R)\phi^{-1}=g_2$ are measurable, or defined almost
everywhere, if $f$ is.
}%end of proof of 255D

\vleader{48pt}{255E}{The basic formula} Let $f$ and $g$ be measurable
complex-valued
functions defined almost everywhere in $\Bbb R$.   Write $f*g$ for the
function defined by the formula

\Centerline{$(f*g)(x)=\int f(x-y)g(y)dy$}

\noindent whenever the integral exists\cmmnt{ (with respect to
Lebesgue measure, naturally)} as a complex number.   Then $f*g$ is the
{\bf convolution} of the functions $f$ and $g$.
\cmmnt{

Observe that} $\dom(|f|*|g|)=\dom(f*g)$, and\cmmnt{ that}
$|f*g|\le |f|*|g|$ everywhere on their common domain, for all $f$ and
$g$.

\cmmnt{\medskip

\noindent{\bf Remark} Note that I am here prepared to contemplate the
convolution of $f$ and $g$ for arbitrary members of
$\eusm L^0_{\Bbb C}$, the space of
almost-everywhere-defined measurable complex-valued functions, even
though the domain of $f*g$ may be empty.   }%end of comment

\leader{255F}{Elementary properties (a)} Because integration is linear,
we surely have


\Centerline{$((f_1+f_2)*g)(x)=(f_1*g)(x)+(f_2*g)(x)$,}

\Centerline{$(f*(g_1+g_2))(x)=(f*g_1)(x)+(f*g_2)(x)$,}

\Centerline{$(cf*g)(x)=(f*cg)(x)=c(f*g)(x)$}

\noindent whenever the right-hand sides of the formulae are defined.

\spheader 255Fb If $f$ and $g$ are measurable complex-valued
functions defined almost everywhere in $\Bbb R$, then
$f*g=g*f$\cmmnt{, in the
strict sense that they have the same domain and the same value at each
point of that common domain}.

\prooflet{\Prf\ Take $x\in\Bbb R$ and apply 255Ba to see that

$$\eqalign{(f*g)(x)
&=\int f(x-y)g(y)dy
=\int f(x-(x-y))g(x-y)dy\cr
&=\int f(y)g(x-y)dy
=(g*f)(x)\cr}$$

\noindent if either is defined.   \Qed}

\header{255Fc}{\bf (c)} If $f_1$, $f_2$, $g_1$, $g_2$ are measurable
complex-valued functions defined almost everywhere in $\Bbb R$,
$f_1\eae f_2$ and $g_1\eae g_2$, then $f_1*g_1=f_2*g_2$.
\prooflet{\Prf\ For every $x\in\Bbb R$ we
shall have $f_1(x-y)=f_2(x-y)$ for almost every $y\in\Bbb R$, by 255Ac.
Consequently $f_1(x-y)g_1(y)=f_2(x-y)g_2(y)$ for almost every $y$,
and $(f_1*g_1)(x)=(f_2*g_2)(x)$ in the sense that if one of these is
defined so is the other, and they are then equal.\ \Qed}

It follows that if $u$, $v\in L^0_{\Bbb C}$, then we have a function
$\theta(u,v)$
which is equal to $f*g$ whenever $f$, $g\in\eusm L^0_{\Bbb C}$
are such that $f^{\ssbullet}=u$ and $g^{\ssbullet}=u$.
\cmmnt{Observe that $\theta(u,v)=\theta(v,u)$, and
that $\theta(u_1+u_2,v)$ extends $\theta(u_1,v)+\theta(u_2,v)$,
$\theta(cu,v)$ extends $c\theta(u,v)$ for all $u$, $u_1$, $u_2$,
$v\in L^0_{\Bbb C}$ and $c\in\Bbb C$.}

\leader{255G}{}\cmmnt{ I have grouped 255Fa-255Fc together because
they depend
only on ideas up to and including 255Ac and 255Ba.   Using the second
halves of 255A and 255B we get much deeper.   I begin with what seems to
be the fundamental result.

\medskip

\noindent}{\bf Theorem} Let $f$, $g$ and $h$ be measurable
complex-valued functions defined almost everywhere in $\Bbb R$.

(a) Suppose that $\int h(x+y)f(x)g(y)d(x,y)$ exists in $\Bbb C$.
Then

$$\eqalign{\int h(x)(f*g)(x)dx
&=\int h(x+y)f(x)g(y)d(x,y)\cr
&=\iint h(x+y)f(x)g(y)dxdy
=\iint h(x+y)f(x)g(y)dydx\cr}$$

\noindent provided
that in the expression $h(x)(f*g)(x)$ we interpret the product as $0$ if
$h(x)=0$ and $(f*g)(x)$ is undefined.

(b) If, on a similar interpretation of $|h(x)|(|f|*|g|)(x)$, the integral
$\int |h(x)|(|f|*|g|)(x)dx$ is finite, then
$\int h(x+y)f(x)g(y)d(x,y)$ exists in $\Bbb C$.

\proof{ Consider the functions

\Centerline{$k_1(x,y)=h(x)f(x-y)g(y)$,\quad $k_2(x,y)=h(x+y)f(x)g(y)$}

\noindent wherever these are defined.   255D tells us that $k_1$ and
$k_2$ are measurable and defined almost everywhere.   Now setting
$\phi(x,y)=(x+y,y)$, we have $k_2=k_1\phi$, so that

\Centerline{$\int k_1(x,y)d(x,y)=\int k_2(x,y)d(x,y)$}

\noindent if either exists, by 255Bb.

If

\Centerline{$\int h(x+y)f(x)g(y)d(x,y)=\int k_2$}

\noindent exists, then by Fubini's theorem we have

\Centerline{$\int k_2=\int k_1(x,y)d(x,y)
=\int(\int h(x)f(x-y)g(y)dy)dx$}

\noindent so $\int h(x)f(x-y)g(y)dy$ exists almost everywhere, that is,
$(f*g)(x)$ exists for almost every $x$ such that $h(x)\ne 0$;  on the
interpretation I am using here, $h(x)(f*g)(x)$ exists almost everywhere,
and

$$\eqalign{\int h(x)(f*g)(x)dx
&=\int\bigl(\int h(x)f(x-y)g(y)dy\bigr)dx
=\int k_1\cr
&=\int k_2
=\int h(x+y)f(x)g(y)d(x,y)\cr
&=\iint h(x+y)f(x)g(y)dxdy
=\iint h(x+y)f(x)g(y)dydx\cr}$$

\noindent by Fubini's theorem again.

If (on the same interpretation) $|h|\times(|f|*|g|)$ is integrable,

\Centerline{$|k_1(x,y)|=|h(x)||f(x-y)||g(y)|$}

\noindent is measurable, and

\Centerline{$\iint|h(x)||f(x-y)||g(y)|dydx
=\int|h(x)|(|f|*|g|)(x)dx$}

\noindent is finite, so by Tonelli's theorem (252G, 252H) $k_1$ and
$k_2$ are integrable.
}%end of proof of 255G

\leader{255H}{}\cmmnt{ Certain standard results are now easy.

\wheader{255H}{4}{2}{2}{24pt}

\noindent}{\bf Corollary} If $f$, $g$ are complex-valued
functions which are integrable over $\Bbb R$, then $f*g$ is integrable,
with

\Centerline{$\int f*g =\int f\int g$,
\quad$\int|f*g|\le\int|f|\int|g|$.}

\proof{ In 255G, set $h(x)=1$ for every $x\in\Bbb R$;  then

\Centerline{$\int h(x+y)f(x)g(y)d(x,y)
=\int f(x)g(y)d(x,y)=\int f\int g$}

\noindent by 253D, so

\Centerline{$\int f*g
=\int h(x)(f*g)(x)dx=\int h(x+y)f(x)g(y)d(x,y)=\int f\int g$,}

\noindent as claimed.   Now

\Centerline{$\int|f*g|\le\int|f|*|g|=\int|f|\int|g|$.}
}%\end of proof of 255H

\leader{255I}{Corollary} For any measurable complex-valued functions
$f$, $g$ defined almost everywhere in $\Bbb R$, $f*g$ is measurable and
has measurable domain.

\proof{ Set $f_n(x)=f(x)$ if $x\in\dom f$, $|x|\le n$ and $|f(x)|\le n$,
and $0$ elsewhere in $\Bbb R$;  define $g_n$ similarly from $g$.   Then
$f_n$ and $g_n$ are integrable, $|f_n|\le|f|$ and $|g_n|\le|g|$ almost
everywhere, $f\eae\lim_{n\to\infty}f_n$ and $g\eae\lim_{n\to\infty}g_n$.
Consequently, by Lebesgue's Dominated Convergence Theorem,

$$\eqalign{(f*g)(x)
&=\int f(x-y)g(y)dy
=\int \lim_{n\to\infty}f_n(x-y)g_n(y)dy\cr
&=\lim_{n\to\infty}\int f_n(x-y)g_n(y)dy
=\lim_{n\to\infty}(f_n*g_n)(x)\cr}$$

\noindent for every $x\in\dom f*g$.   But $f_n*g_n$ is integrable,
therefore measurable, for every $n$, so that $f*g$ must be measurable.

As for the domain of $f*g$,

$$\eqalign{x\in\dom(f*g)
&\iff\int f(x-y)g(y)dy\text{ is defined in }\Bbb C\cr
&\iff\int|f(x-y)||g(y)|dy\text{ is defined in }\Bbb R\cr
&\iff\int|f_n(x-y)||g_n(y)|dy\text{ is defined in }\Bbb R
\text{ for every }n\cr
&\qquad\qquad\text{and }
\sup_{n\in\Bbb N}\int|f_n(x-y)||g_n(y)|dy<\infty.\cr
}$$

\noindent Because every $|f_n|*|g_n|$ is integrable, therefore
measurable and with measurable domain,

\Centerline{$\dom(f*g)=\{x:x\in\bigcap_{n\in\Bbb N}\dom(|f_n|*|g_n|),\,
\sup_{n\in\Bbb N}(|f_n|*|g_n|)(x)<\infty\}$}

\noindent is measurable.
}%end of proof of 255I

\leader{255J}{Theorem} Let $f$, $g$ and $h$ be complex-valued measurable
functions, defined almost everywhere in $\Bbb R$, such that $f*g$ and $g*h$
are defined a.e.   Suppose that
$x\in\Bbb R$ is such that one of $(|f|*(|g|*|h|))(x)$,
$((|f|*|g|)*|h|)(x)$ is defined in $\Bbb R$.   Then $f*(g*h)$ and
$(f*g)*h$ are defined and equal at $x$.

\proof{ Set $k(y)=f(x-y)$ when this is defined, so that $k$ is
measurable and defined almost everywhere (255D).

\medskip

{\bf (a)} If $(|f|*(|g|*|h|))(x)$ is defined, this is
$\int|k(y)|(|g|*|h|)(y)dy$, so by 255G we have

\Centerline{$\int k(y)(g*h)(y)dy=\int k(y+z)g(y)h(z)d(y,z)$,}

\noindent that is,

$$\eqalign{(f*(g*h))(x)
&=\int f(x-y)(g*h)(y)dy
=\int k(y)(g*h)(y)dy\cr
&=\int k(y+z)g(y)h(z)d(y,z)
=\iint k(y+z)g(y)h(z)dydz\cr
&=\iint f(x-y-z)g(y)h(z)dydz
=\int (f*g)(x-z)h(z)dz\cr
&=((f*g)*h)(x).\cr}$$

\medskip

{\bf (b)} If $((|f|*|g|)*|h|)(x)$ is defined, this is

$$\eqalign{\int(|f|*|g|)(x-z)|h(z)|dz
&=\iint|f(x-z-y)||g(y)||h(z)|dydz\cr
&=\iint|k(y+z)||g(y)||h(z)|dydz.\cr}$$

\noindent By 255D again, $(y,z)\mapsto k(y+z)$ is measurable, so we can
apply Tonelli's theorem to see that $\int k(y+z)g(y)h(z)d(y,z)$ is defined,
and is equal to $\int k(y)(g*h)(y)dy=(f*(g*h))(x)$ by 255Ga.   On the other
side, by the last two lines of the proof of (a),
$\int k(y+z)g(y)h(z)d(y,z)$ is also equal to $((f*g)*h)(x)$.
}%end of proof of  255J

\leader{255K}{}\cmmnt{ I do not think we shall need an exhaustive
discussion of
the question of just when $(f*g)(x)$ is defined;  this seems to be
complicated.   However there is a fundamental case in which we can be
sure that $(f*g)(x)$ is defined everywhere.

\medskip

\noindent}{\bf Proposition} Suppose that $f$, $g$ are measurable
complex-valued
functions defined almost everywhere in $\Bbb R$, and that
$f\in\eusm L^p_{\Bbb C}$,
$g\in\eusm L^q_{\Bbb C}$ where $p$, $q\in[1,\infty]$ and
$\bover1p+\bover1q=1$\cmmnt{ (writing $\bover{1}{\infty}=0$ as
usual)}.   Then $f*g$ is defined everywhere in $\Bbb R$, is uniformly
continuous, and

$$\eqalign{\sup_{x\in\Bbb R}|(f*g)(x)|
&\le\|f\|_p\|g\|_q\text{ if }1<p<\infty,\,1<q<\infty,\cr
&\le\|f\|_1\esssup|g|\text{ if }p=1,\,q=\infty,\cr
&\le\esssup|f|\cdot\|g\|_1\text{ if }p=\infty,\,q=1.\cr}$$

\proof{{\bf (a)} (For an introduction to $\eusm L^p$ spaces, see \S244.)
For any $x\in\Bbb R$, the function $f_x$, defined by setting
$f_x(y)=f(x-y)$ whenever $x-y\in\dom f$, must also belong to
$\eusm L^p$, because $f_x=f\phi$ for an automorphism $\phi$ of the
measure
space.   Consequently $(f*g)(x)=\int f_x\times g$ is defined, and of
modulus at most $\|f\|_p\|g\|_q$ or $\|f\|_1\esssup|g|$ or
$\esssup|f|\cdot\|g\|_1$, by 244Eb/244Pb and 243Fa/243K.

\medskip

{\bf (b)} To see that $f*g$ is uniformly continuous, argue as follows.
Suppose first that $p<\infty$.   Let $\epsilon>0$.   Let $\eta>0$ be
such that $(2+2^{1/p})\|g\|_q\eta\le\epsilon$.
Then there is a bounded continuous function $h:\Bbb R\to\Bbb C$ such
that $\{x:h(x)\ne 0\}$ is bounded and $\|f-h\|_p\le\eta$ (244Hb/244Pb);
let $M\ge 1$ be such that $h(x)=0$ whenever $|x|\ge M-1$.   Next,
$h$ is uniformly continuous, so there is a $\delta\in\ocint{0,1}$ such
that $|h(x)-h(x')|\le M^{-1/p}\eta$ whenever $|x-x'|\le\delta$.

Suppose that $|x-x'|\le\delta$.   Defining $h_x(y)=h(x-y)$, as before,
we have

$$\eqalignno{\int|h_x-h_{x'}|^p
&=\int|h(x-y)-h(x'-y)|^pdy
=\int|h(t)-h(x'-x+t)|^pdt\cr
\displaycause{substituting $t=x-y$}
&=\int_{-M}^M|h(t)-h(x'-x+t)|^pdt\cr
\displaycause{because $h(t)=h(x'-x+t)=0$ if $|t|\ge M$}
&\le 2M(M^{-1/p}\eta)^p\cr
\displaycause{because $|h(t)-h(x'-x+t)|\le M^{-1/p}\eta$ for every $t$}
&=2\eta^p.\cr}$$

\noindent So $\|h_x-h_{x'}\|_p\le 2^{1/p}\eta$.   On the other hand,

\Centerline{$\int|h_{x}-f_{x}|^p
=\int|h(x-y)-f(x-y)|^pdy
=\int|h(y)-f(y)|^pdy$,}

\noindent so $\|h_x-f_x\|_p=\|h-f\|_p\le\eta$, and similarly
$\|h_{x'}-f_{x'}\|_p\le\eta$.   So

\Centerline{$\|f_x-f_{x'}\|_p
\le\|f_x-h_x\|_p+|h_x-h_{x'}\|_p+\|h_{x'}-f_{x'}\|_p
\le(2+2^{1/p})\eta$.}

\noindent This means that

$$\eqalign{|(f*g)(x)-(f*g)(x')|
&=|\int f_x\times g-\int f_{x'}\times g|
=|\int(f_x-f_{x'})\times g|\cr
&\le\|f_x-f_{x'}|_p\|g\|_q
\le(2+2^{1/p})\|g\|_q\eta
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $f*g$ is uniformly continuous.

The argument here supposes that $p$ is finite.   But if $p=\infty$ then
$q=1$ is finite, so we can apply the method with $g$ in place of $f$ to
show that $g*f$ is uniformly continuous, and $f*g=g*f$  by 255Fb.
}%end of proof of 255K

\leader{255L}{The $r$-dimensional case} I have written 255A-255K out as
theorems about Lebesgue measure on $\Bbb R$.   However they all
apply\cmmnt{ equally well} to Lebesgue measure on $\BbbR^r$ for any
$r\ge 1$\cmmnt{, and the
modifications required are so small that I think I need do no more than
ask you to read through the arguments again, turning every $\Bbb R$ into
an $\BbbR^r$, and every $\BbbR^2$ into an $(\BbbR^r)^2$.   In 255A
and elsewhere, the measure $\mu_2$ should be read either as Lebesgue
measure on $\BbbR^{2r}$ or as the product measure on $(\BbbR^r)^2$;
by 251N the two may be identified.   There is a trivial modification
required in part (b) of the proof;  if $I_n=\coint{a_n,b_n}$ then

\Centerline{$\mu I_n=\mu(-I_n)
=\prod_{i=1}^r\max(0,\beta_{ni}-\alpha_{ni})$,}

\noindent writing $a_n=(\alpha_{n1},\ldots,\alpha_{nr})$.   In the proof
of 255I, the functions $f_n$ should be defined by saying that
$f_n(x)=f(x)$ if $|f(x)|\le n$ and $\|x\|\le n$, $0$ otherwise.

In quoting these results, therefore, I shall be uninhibited in referring
to the paragraphs 255A-255K as if they were actually written out for
general $r\ge 1$}.

\leader{255M}{The case of $\ocint{-\pi,\pi}$}\dvro{ {\bf (a)}}{ The same
ideas also apply
to the circle group $S^1$ and to the interval $\ocint{-\pi,\pi}$, but
here perhaps rather more explanation is in order.

\spheader 255Ma The first thing to establish is the appropriate
group operation.}   If we think of $S^1$ as the set 
$\{z:z\in\Bbb C,\,|z|=1\}$, then the group operation is complex 
multiplication, and in
the formulae above $x+y$ must be rendered as $xy$, while $x-y$ must be
rendered as $xy^{-1}$.   On the interval $\ocint{-\pi,\pi}$, the group
operation is $+_{2\pi}$, where for $x$, $y\in\ocint{-\pi,\pi}$ I write
$x+_{2\pi}y$ for whichever of $x+y$, $x+y+2\pi$, $x+y-2\pi$ belongs to
$\ocint{-\pi,\pi}$.   \cmmnt{To see that this is indeed a group
operation, one
method is to note that it corresponds to multiplication on $S^1$ if we
use the canonical bijection
$x\mapsto e^{ix}:\ocint{-\pi,\pi}\to S^1$;  another, to note that it
corresponds to the operation on the quotient group $\Bbb R/2\pi\Bbb Z$.
Thus in this interpretation of the ideas of 255A-255K, we shall wish to
replace $x+y$ by $x+_{2\pi}y$, $-x$ by $-_{2\pi}x$, and $x-y$ by
$x-_{2\pi}y$, where

\Centerline{$-_{2\pi}x=-x$ if $x\in\ooint{-\pi,\pi}$,
\quad$-_{2\pi}\pi=\pi$,}

\noindent and $x-_{2\pi}y$ is whichever of $x-y$, $x-y+2\pi$, $x-y-2\pi$
belongs to $\ocint{-\pi,\pi}$.
}%end of comment

\header{255Mb}{\bf (b)} As for the measure, the measure to use on
$\ocint{-\pi,\pi}$ is just Lebesgue measure.   \cmmnt{Note that
because
$\ocint{-\pi,\pi}$ is Lebesgue measurable, there will be no confusion
concerning the meaning of `measurable subset', as the relatively
measurable subsets of $\ocint{-\pi,\pi}$ are actually measured by
Lebesgue measure on $\Bbb R$.   Also we can identify the product measure
on $\ocint{-\pi,\pi}\times\ocint{-\pi,\pi}$ with the subspace measure
induced by Lebesgue measure on $\BbbR^2$ (251R).}

On $S^1$, we need the corresponding measure induced by the canonical
bijection between $S^1$ and $\ocint{-\pi,\pi}$\cmmnt{, which indeed is
often called `Lebesgue measure on $S^1$'}.   \cmmnt{(We shall see in
265E that it is
also equal to Hausdorff one-dimensional measure on $S^1$.)   We are very
close to the level at which it
would become reasonable to move to $S^1$ and this measure (or its
normalized version, in which it is reduced by a factor of $2\pi$, so as
to make $S^1$ a probability space).   However, the elementary theory of
Fourier series, which will be the principal application of this work in
the present volume, is generally done on intervals in $\Bbb R$, so that
formulae based on $\ocint{-\pi,\pi}$ are closer to the standard
expressions.   Henceforth, therefore, I will express the work in
terms of $\ocint{-\pi,\pi}$.}


\cmmnt{
\header{255Mc}{\bf (c)} The result corresponding to 255A now takes a
slightly different form, so I spell it out.
}

\leader{255N}{Theorem} Let $\mu$ be Lebesgue measure on
$\ocint{-\pi,\pi}$ and
$\mu_2$ Lebesgue measure on $\ocint{-\pi,\pi}\times\ocint{-\pi,\pi}$;
write $\Sigma$, $\Sigma_2$ for their domains.

(a) For any $a\in\ocint{-\pi,\pi}$, the map $x\mapsto
a+_{2\pi}x:\ocint{-\pi,\pi}\to\ocint{-\pi,\pi}$ is a
measure space automorphism of $(\ocint{-\pi,\pi},\Sigma,\mu)$.

(b) The map $x\mapsto -_{2\pi}x:\ocint{-\pi,\pi}\to\ocint{-\pi,\pi}$ is
a measure space automorphism of $(\ocint{-\pi,\pi},\Sigma,\mu)$.

(c) For any $a\in\ocint{-\pi,\pi}$, the map
$x\mapsto a-_{2\pi}x:\ocint{-\pi,\pi}\to\ocint{-\pi,\pi}$ is a
measure space automorphism of $(\ocint{-\pi,\pi},\Sigma,\mu)$.

(d) The map $(x,y)\mapsto(x+_{2\pi}y,y):\ocint{-\pi,\pi}^2\to
\ocint{-\pi,\pi}^2$ is a measure space automorphism of
$(\ocint{-\pi,\pi}^2,
\ifdim\pagewidth>467pt\penalty-50\fi
\Sigma_2,
\ifdim\pagewidth>467pt\penalty-10\fi
\mu_2)$.

(e) The map $(x,y)\mapsto(x-_{2\pi}y,y):\ocint{-\pi,\pi}^2\to
\ocint{-\pi,\pi}^2$ is a measure space automorphism of
$(\ocint{-\pi,\pi}^2,
\ifdim\pagewidth>467pt\penalty-50\fi
\Sigma_2,
\ifdim\pagewidth>467pt\penalty-10\fi
\mu_2)$.

\proof{{\bf (a)} Set $\phi(x)=a+_{2\pi}x$.   Then for any
$E\subseteq\ocint{-\pi,\pi}$,

\Centerline{$\phi[E]
=((E+a)\cap\ocint{-\pi,\pi})
\cup(((E+a)\cap\ocint{\pi,3\pi})-2\pi)
\cup(((E+a)\cap\ocint{-3\pi,-\pi})+2\pi)$,}

\noindent and these three sets are disjoint, so that

$$\eqalignno{\mu\phi[E]
&=\mu((E+a)\cap\ocint{-\pi,\pi})
+\mu(((E+a)\cap\ocint{\pi,3\pi})-2\pi)\cr
&\qquad\qquad\qquad\qquad\qquad\qquad
  +\mu(((E+a)\cap\ocint{-3\pi,-\pi})+2\pi)\cr
&=\mu_L((E+a)\cap\ocint{-\pi,\pi})
+\mu_L(((E+a)\cap\ocint{\pi,3\pi})-2\pi)\cr
&\qquad\qquad\qquad\qquad\qquad\qquad
  +\mu_L(((E+a)\cap\ocint{-3\pi,-\pi})+2\pi)\cr
\noalign{\noindent (writing $\mu_L$ for Lebesgue measure on $\Bbb R$)}
&=\mu_L((E+a)\cap\ocint{-\pi,\pi})
+\mu_L((E+a)\cap\ocint{\pi,3\pi})
+\mu_L((E+a)\cap\ocint{-3\pi,-\pi})\cr
&=\mu_L(E+a)
=\mu_LE
=\mu E.\cr}$$

\noindent Similarly, $\mu\phi^{-1}[E]$ is defined and equal to $\mu E$
for every $E\in\Sigma$, so that $\phi$ is an automorphism of
$(\ocint{-\pi,\pi},
\ifdim\pagewidth>467pt\penalty-50\fi
\Sigma,
\ifdim\pagewidth>467pt\penalty-10\fi
\mu)$.

\medskip

{\bf (b)} Of course this is quicker.   Setting $\phi(x)=-_{2\pi}x$ for
$x\in\ocint{-\pi,\pi}$,  we have

$$\eqalign{\mu(\phi[E])
&=\mu(\phi[E]\cap\ooint{-\pi,\pi})
=\mu(-(E\cap\ooint{-\pi,\pi})\cr
&=\mu_L(-(E\cap\ooint{-\pi,\pi}))
=\mu_L(E\cap\ooint{-\pi,\pi})\cr
&=\mu(E\cap\ooint{-\pi,\pi})
=\mu E\cr}$$

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

\medskip

{\bf (c)} This is just a matter of putting (a) and (b) together, as in
255A.

\medskip

{\bf (d)} We can argue as in (a), but with a little more elaboration.
If $E\in\Sigma_2$, and $\phi(x,y)=(x+_{2\pi}y,y)$ for $x$,
$y\in\ocint{-\pi,\pi}$, set $\psi(x,y)=(x+y,y)$ for $x$, $y\in\Bbb R$,
and write $c=(2\pi,0)\in\BbbR^2$, $H=\ocint{-\pi,\pi}^2$, $H'=H+c$,
$H''=H-c$.   Then for any $E\in\Sigma_2$,

\Centerline{$\phi[E]=(\psi[E]\cap H)\cup((\psi[E]\cap H')-c)
\cup((\psi[E]\cap H'')+c)$,}

\noindent so

$$\eqalignno{\mu_2\phi[E]
&=\mu_2(\psi[E]\cap H)
+\mu_2((\psi[E]\cap H')-c)
+\mu_2((\psi[E]\cap H'')+c)\cr
&=\mu_L(\psi[E]\cap H)
+\mu_L((\psi[E]\cap H')-c)
+\mu_L((\psi[E]\cap H'')+c)\cr
\noalign{\noindent (this time writing $\mu_L$ for Lebesgue measure on
$\BbbR^2$)}
&=\mu_L(\psi[E]\cap H)
+\mu_L(\psi[E]\cap H')
+\mu_L(\psi[E]\cap H'')\cr
&=\mu_L\psi[E]
=\mu_LE
=\mu_2 E.\cr}$$

\noindent In the same way, $\mu_2(\phi^{-1}[E])=\mu_2E$ for every
$E\in\Sigma_2$, so $\phi$ is an automorphism of
$(\ocint{-\pi,\pi}^2,\Sigma_2,\mu_2)$, as required.

\medskip

{\bf (e)} Finally, (e) is just a restatement of (d), as before.
}%end of proof of 255N

\leader{255O}{Convolutions on $\ocint{-\pi,\pi}$}\cmmnt{ With the
fundamental
result established, the same arguments as in 255B-255K now yield the
following.}   Write $\mu$ for Lebesgue measure on $\ocint{-\pi,\pi}$.


\header{255Oa}{\bf (a)} Let $f$ and $g$ be measurable
complex-valued
functions defined almost everywhere in $\ocint{-\pi,\pi}$.   Write $f*g$
for the function defined by the formula

\Centerline{$(f*g)(x)=\int_{-\pi}^{\pi}f(x-_{2\pi}y)g(y)dy$}

\noindent whenever $x\in\ocint{-\pi,\pi}$ and the integral exists as a
complex number.   Then $f*g$ is the {\bf
convolution} of the functions $f$ and $g$.

\header{255Ob}{\bf (b)} If $f$ and $g$ are measurable complex-valued
functions defined almost everywhere in $\ocint{-\pi,\pi}$, then
$f*g=g*f$.

\vspheader{60pt}255Oc Let $f$, $g$ and $h$ be measurable complex-valued
functions defined almost everywhere in $\ocint{-\pi,\pi}$.   Then

\quad{(i)}

\Centerline{$\int_{-\pi}^{\pi}h(x)(f*g)(x)dx
=\int_{\ocint{-\pi,\pi}^2}h(x+_{2\pi}y)f(x)g(y)d(x,y)$}

\noindent whenever the right-hand side exists and is finite, provided
that in the expression $h(x)(f*g)(x)$ we interpret the product as $0$ if
$h(x)=0$ and $(f*g)(x)$ is undefined.

\quad{(ii)} If, on the same interpretation of $|h(x)|(|f|*|g|)(x)$,
the integral  $\int_{-\pi}^{\pi}|h(x)|(|f|*|g|)(x)dx$ is
finite, then $\int_{\ocint{-\pi,\pi}^2}h(x+_{2\pi}y)f(x)g(y)d(x,y)$ exists in
$\Bbb C$, so again we shall have

\Centerline{$\int_{-\pi}^{\pi}h(x)(f*g)(x)dx
=\int_{\ocint{-\pi,\pi}^2}h(x+_{2\pi}y)f(x)g(y)d(x,y)$.}

\spheader 255Od If $f$, $g$ are complex-valued
functions which are integrable over $\ocint{-\pi,\pi}$, then $f*g$ is
integrable, with

\Centerline{$\int_{-\pi}^{\pi}f*g
=\int_{-\pi}^{\pi}f\int_{-\pi}^{\pi}g$,
\quad$\int_{-\pi}^{\pi}|f*g|
\le\int_{-\pi}^{\pi}|f|\int_{-\pi}^{\pi}|g|$.}

\spheader 255Oe Let $f$, $g$, $h$ be complex-valued measurable
functions defined almost everywhere in $\ocint{-\pi,\pi}$, such that $f*g$
and $g*h$ are also defined almost everywhere.   Suppose
that $x\in\ocint{-\pi,\pi}$ is such that one of $(|f|*(|g|*|h|))(x)$,
$((|f|*|g|)*|h|)(x)$ is defined in $\Bbb R$.   Then $f*(g*h)$ and
$(f*g)*h$ are defined and equal at $x$.

\spheader 255Of Suppose that $f\in\eusm L^p_{\Bbb C}(\mu)$,
$g\in\eusm L^{q}_{\Bbb C}(\mu)$ where $p$, $q\in[1,\infty]$ and
$\bover1p+\bover1{q}=1$.   Then $f*g$ is defined
everywhere in $\ocint{-\pi,\pi}$, and
$\sup_{x\in\ocint{-\pi,\pi}}|(f*g)(x)|\le\|f\|_p\|g\|_q$, interpreting
$\|\,\,\|_{\infty}$ as $\esssup |\,\,|$\cmmnt{, as in 255K}.

\exercises{
\leader{255X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 255Xa
Let $f$, $g$ be complex-valued
functions defined almost everywhere in $\Bbb R$.   Show that for any
$x\in\Bbb R$, $(f*g)(x)=\int f(x+y)g(-y)dy$ if either is defined.
%255E

\sqheader 255Xb Let $f$ and $g$ be complex-valued functions defined
almost everywhere in $\Bbb R$.   (i) Show that if $f$ and $g$ are even
functions, so is $f*g$.   (ii) Show that if $f$ is even and $g$ is odd
then $f*g$ is odd.   (iii) Show that if $f$ and $g$ are odd then $f*g$
is even.
%255E

\spheader 255Xc Suppose that $f$, $g$ are real-valued measurable
functions defined almost everywhere in $\BbbR^r$ and such that $f>0$
a.e., $g\ge 0$ a.e.\ and $\{x:g(x)>0\}$ is not negligible.   Show that
$f*g>0$ everywhere in $\dom(f*g)$.
%255F

\sqheader 255Xd Suppose that $f:\Bbb R\to\Bbb C$ is a bounded
differentiable function and that $f'$ is bounded.   Show that for any
integrable
complex-valued function $g$ on $\Bbb R$, $f*g$ is differentiable and
$(f*g)'=f'*g$ everywhere.   \Hint{123D.}
%255F

\spheader 255Xe A complex-valued function $g$ defined almost everywhere
in $\Bbb R$ is {\bf locally integrable} if $\int_a^bg$ is defined in
$\Bbb C$ whenever $a<b$ in $\Bbb R$.   Suppose that $g$ is such a
function and that $f:\Bbb R\to\Bbb C$ is a differentiable function, with
continuous derivative, such that $\{x:f(x)\ne 0\}$ is bounded.   Show
that $(f*g)'=f'*g$ everywhere.
%255F 255Xd

\sqheader 255Xf Set $\phi_{\delta}(x)=\exp(-\bover1{\delta^2-x^2})$ if
$|x|<\delta$, $0$ if $|x|\ge\delta$, as in 242Xi.   Set
$\alpha_{\delta}=\int\phi_{\delta}$,
$\psi_{\delta}=\alpha_{\delta}^{-1}\phi_{\delta}$.   Let $f$ be a
locally integrable
complex-valued function on $\Bbb R$.   (i) Show that $f*\psi_{\delta}$
is a smooth function defined everywhere on $\Bbb R$ for every
$\delta>0$.   (ii) Show that
$\lim_{\delta\downarrow 0}(f*\psi_{\delta})(x)=f(x)$ for almost every
$x\in\Bbb R$.   \Hint{223Yg.}   (iii) Show that if $f$ is integrable
then $\lim_{\delta\downarrow 0}\int|f-f*\psi_{\delta}|=0$.   \Hint{use
(ii) and 245H(a-ii) {\it or} look first at the case $f=\chi[a,b]$ and
use 242O, noting that $\int|f*\psi_{\delta}|\le\int|f|$.}   (iv) Show
that if $f$ is uniformly continuous and defined everywhere in $\Bbb R$
then
$\lim_{\delta\downarrow 0}\sup_{x\in\Bbb R}|f(x)-(f*\psi_{\delta})(x)|=0$.
%255F 255Xe

\sqheader 255Xg For $\alpha>0$, set
$g_{\alpha}(t)=\Bover1{\Gamma(\alpha)}t^{\alpha-1}$ for $t>0$, $0$ for
$t\le 0$.   Show that $g_{\alpha}*g_{\beta}=g_{\alpha+\beta}$ for all
$\alpha$, $\beta>0$.   \Hint{252Yf.}
%255F

\sqheader 255Xh Let $\mu$ be Lebesgue measure on $\Bbb R$.
For $u$, $v$,
$w\in L^0_{\Bbb C}=L^0_{\Bbb C}(\mu)$, say that
$u*v=w$ if $f*g$ is defined almost everywhere and $(f*g)^{\ssbullet}=w$
whenever $f$, $g\in\eusm L^0_{\Bbb C}(\mu)$, $f^{\ssbullet}=u$ and
$g^{\ssbullet}=w$.   (i) Show that $(u_1+u_2)*v=u_1*v+u_2*v$ whenever
$u_1$, $u_2$, $v\in L^0_{\Bbb C}$ and $u_1*v$ and $u_2*v$ are defined in
this sense.   (ii) Show that $u*v=v*u$ whenever $u$, $v\in L^0(\Bbb C)$ and
either $u*v$ or $v*u$ is defined.   (iii) Show that if $u$, $v$,
$w\in L^0_{\Bbb C}$, $u*v$ and $v*w$ are defined,
and either $|u|*(|v|*|w|)$ or $(|u|*|v|)*|w|$ is
defined, then then $u*(v*w)=(u*v)*w$ are defined and equal.
%255I

\sqheader 255Xi Let $\mu$ be Lebesgue measure on $\Bbb R$.
(i) Show that $u*v$, as defined in 255Xh, belongs to $L^1_{\Bbb C}(\mu)$
whenever $u$, $v\in L^1_{\Bbb C}(\mu)$.
(ii) Show that $L^1_{\Bbb C}$ is a
commutative Banach algebra under $*$ (definition: 2A4J).
%255Xh 255I

\spheader 255Xj(i) Show that if $h$ is an integrable
function on $\BbbR^2$, then   $(Th)(x)=\int h(x-y,y)dy$ exists for
almost every $x\in\Bbb R$, and that $\int (Th)(x)dx=\int h(x,y)d(x,y)$.
(ii) Write $\mu_2$ for Lebesgue measure on $\BbbR^2$, $\mu$ for
Lebesgue measure on $\Bbb R$.   Show that there is a linear operator
$\tilde T:L^1(\mu_2)\to L^1(\mu)$ defined by setting
$\tilde T(h^{\ssbullet})=(Th)^{\ssbullet}$ for every integrable function
$h$ on $\BbbR^2$.   (iii) Show that in the language of 253E and 255Xh,
$\tilde T(u\otimes v)=u*v$ for all $u$, $v\in L^1(\mu)$.
%255Xh 255I 255Xi

\sqheader 255Xk  For $\pmb{a}$, $\pmb{b}\in\Bbb C^{\Bbb Z}$ set
$(\pmb{a}*\pmb{b})(n)=\sum_{i\in\Bbb Z}\pmb{a}(n-i)\pmb{b}(i)$ whenever
$\sum_{i\in\Bbb Z}|\pmb{a}(n-i)\pmb{b}(i)|<\infty$.   Show that

\quad (i) $\pmb{a}*\pmb{b}=\pmb{b}*\pmb{a}$;

\quad (ii) $\sum_{i\in\Bbb Z}\pmb{c}(i)(\pmb{a}*\pmb{b})(i)
=\sum_{i,j\in\Bbb Z}\pmb{c}(i+j)\pmb{a}(i)\pmb{b}(j)$ if
$\sum_{i,j\in\Bbb Z}|\pmb{c}(i+j)\pmb{a}(i)\pmb{b}(j)|<\infty$;

\quad (iii) if $\pmb{a}$, $\pmb{b}\in\ell^1(\Bbb Z)$ then
$\pmb{a}*\pmb{b}\in\ell^1(\Bbb Z)$ and
$\|\pmb{a}*\pmb{b}\|_1\le\|\pmb{a}\|_1\|\pmb{b}\|_1$;

\quad (iv) If $\pmb{a}$, $\pmb{b}\in\ell^2(\Bbb Z)$ then
$\pmb{a}*\pmb{b}\in\ell^{\infty}(\Bbb Z)$ and
$\|\pmb{a}*\pmb{b}\|_{\infty}\le\|\pmb{a}\|_2\|\pmb{b}\|_2$;

\quad (v) if $\pmb{a}$, $\pmb{b}$, $\pmb{c}\in\Bbb C^{\Bbb Z}$ and
$(|\pmb{a}|*(|\pmb{b}|*|\pmb{c}|))(n)$ is well-defined, then
$(\pmb{a}*(\pmb{b}*\pmb{c}))(n)=((\pmb{a}*\pmb{b})*\pmb{c})(n)$.
%255K

\leader{255Y}{Further exercises (a)}
%\spheader 255Ya
Let $f$ be a complex-valued function which is integrable
over $\Bbb R$.   (i) Let $x$ be any point of the Lebesgue set of $f$.
Show that for any $\epsilon>0$ there is a $\delta>0$ such that
$|f(x)-(f*g)(x)|\le\epsilon$ whenever $g:\Bbb R\to\coint{0,\infty}$ is a
function which is non-decreasing on $\ocint{-\infty,0}$, non-decreasing
on $\coint{0,\infty}$, and has $\int g=1$ and
$\int_{-\delta}^{\delta}g\ge 1-\delta$.   (ii) Show that for any
$\epsilon>0$ there is a $\delta>0$ such that $\|f-f*g\|_1\le\epsilon$
whenever $g:\Bbb R\to\coint{0,\infty}$ is a function which is
non-decreasing on $\ocint{-\infty,0}$, non-decreasing on
$\coint{0,\infty}$, and has $\int g=1$ and
$\int_{-\delta}^{\delta}g\ge 1-\delta$.
%255Xf 255F

\spheader 255Yb Let $f$ be a complex-valued function which is integrable
over $\Bbb R$.   Show that, for almost every $x\in\Bbb R$,

\Centerline{$\lim_{a\to\infty}
  \Bover{a}{\pi}\int_{-\infty}^{\infty}
  \Bover{f(y)}{1+a^2(x-y)^2}dy$,
\quad$\lim_{a\to\infty}\Bover1a\int_x^{\infty}f(y)e^{-a(y-x)}dy$,}

\Centerline{$\lim_{\sigma\downarrow 0}\Bover1{\sigma\sqrt{2\pi}}
  \int_{-\infty}^{\infty}f(y)e^{-(y-x)^2/2\sigma^2}dy$}

\noindent all exist and are equal to $f(x)$.   \Hint{263G.}
%255Ya, 255Xf 255F

\spheader 255Yc
Set $f(x)=1$ for all $x\in\Bbb R$, $g(x)=\Bover{x}{|x|}$
for $0<|x|\le 1$ and $0$ otherwise, $h(x)=\tanh x$ for all $x\in\Bbb R$.
Show that $f*(g*h)$ and $(f*g)*h$ are both defined (and constant)
everywhere, and are different.
%255J

\spheader 255Yd Discuss what can happen if, in the context of 255J, we
know that $(|f|*(|g|*|h|))(x)$ is defined, but have no information on
the domain of $f*g$.
%255J

\spheader 255Ye Suppose that $p\in\coint{1,\infty}$
and that $f\in\eusm L^p_{\Bbb C}(\mu)$, where $\mu$ is Lebesgue measure
on $\BbbR^r$.   For $a\in\BbbR^r$ set $(S_af)(x)=f(a+x)$ whenever
$a+x\in\dom f$.   Show that $S_af\in\eusm L^p_{\Bbb C}(\mu)$, and that
for every $\epsilon>0$ there is a $\delta>0$ such that
$\|S_af-f\|_p\le\epsilon$ whenever $|a|\le\delta$.
%255K

\spheader 255Yf Suppose that $p$, $q\in\ooint{1,\infty}$ and
$\bover1p+\bover1q=1$.   Take $f\in\eusm L^p_{\Bbb C}(\mu)$ and
$g\in\eusm L^q_{\Bbb C}(\mu)$, where $\mu$ is Lebesgue measure on
$\BbbR^r$.   Show that $\lim_{\|x\|\to\infty}(f*g)(x)=0$.
\Hint{use 244Hb.}
%255K

\spheader 255Yg Repeat 255Ye and 255K, this time taking $\mu$ to
be Lebesgue measure on $\ocint{-\pi,\pi}$, and setting
$(S_af)(x)=f(a+_{2\pi}x)$ for $a\in\ocint{-\pi,\pi}$;  show that in the
new version of 255K,
$(f*g)(\pi)=\lim_{x\downarrow -\pi}(f*g)(x)$.
%255K, 255Ye

\spheader 255Yh Let $\mu$ be Lebesgue measure on $\Bbb R$.   For
$a\in\Bbb R$, $f\in\eusm L^0=\eusm L^0(\mu)$ set $(S_af)(x)=f(a+x)$
whenever $a+x\in\dom f$.

\quad (i) Show that $S_af\in\eusm L^0$ for every $f\in\eusm L^0$.

\quad  (ii) Show that we have a map $\tilde S_a:L^0\to L^0$ defined by
setting $\tilde S_a(f^{\ssbullet})=(S_af)^{\ssbullet}$ for every
$f\in\eusm L^0$.

\quad  (iii) Show that $\tilde S_a$ is a Riesz space isomorphism and is
a homeomorphism for the topology of convergence in measure;  moreover,
that $\tilde S_a(u\times v)=\tilde S_au\times\tilde S_av$ for all $u$,
$v\in L^0$.

\quad  (iv) Show that $\tilde S_{a+b}=\tilde S_a\tilde S_b$ for all $a$,
$b\in\Bbb R$.

\quad  (v) Show that $\lim_{a\to 0}\tilde S_au=u$ for the topology of
convergence in measure, for every $u\in L^0$.

\quad  (vi) Show that if $1\le p\le\infty$ then $\tilde S_a\restr L^p$
is an isometric isomorphism of the Banach lattice $L^p$.

\quad  (vii) Show that if $p\in\coint{1,\infty}$ then
$\lim_{a\to 0}\|\tilde S_au-u\|_p=0$ for every $u\in L^p$.

\quad  (viii) Show that if $A\subseteq L^1$ is uniformly integrable and
$M\ge 0$, then $\{\tilde S_au:u\in A,\,|a|\le M\}$ is uniformly
integrable.

\quad (ix) Suppose that $u$, $v\in L^0$ are such that $u*v$ is defined
in $L^0$ in the sense of 255Xh.   Show that
$\tilde S_a(u*v)=(\tilde S_au)*v=u*(\tilde S_av)$ for
every $a\in\Bbb R$.
%255Ye, 255K

\spheader 255Yi Prove 255Nd from 255Na by the method used to
prove 255Ad from 255Aa, rather than by quoting 255Ad.
%255N

\spheader 255Yj Let $\mu$ be Lebesgue measure on $\Bbb R$, and
$\phi:\Bbb R\to\Bbb R$ a convex function;  let
$\bar\phi:L^0\to L^0=L^0(\mu)$ be the associated operator (see 241I).
Show that if $u\in L^1=L^1(\mu)$, $v\in L^0$ are such that $u\ge 0$,
$\int u=1$ and $u*v$, $u*\bar\phi(v)$ are both defined in the sense of
255Xh, then $\bar\phi(u*v)\le u*\bar\phi(v)$.
\Hint{233I.}
%255Xh 255I

\spheader 255Yk Let $\mu$ be Lebesgue measure on $\Bbb R$, and
$p\in[1,\infty]$.   Let $f\in\eusm L^1_{\Bbb C}(\mu)$,
$g\in\eusm L^p_{\Bbb C}(\mu)$.   Show that
$f*g\in\eusm L^p_{\Bbb C}(\mu)$ and that
$\|f*g\|_p\le\|f\|_1\|g\|_p$.   \Hint{argue from 255Yj, as in 244M.}
%255Yj 255K 255I

\spheader 255Yl Suppose that $p$, $q$, $r\in\ooint{1,\infty}$ and that
$\bover1p+\bover1q=1+\bover1r$.   Let $\mu$ be Lebesgue measure on
$\Bbb R$.   (i) Show that

\Centerline{$\int f\times g
\le\|f\|_p^{1-p/r}\|g\|_q^{1-q/r}(\int f^p\times g^q)^{1/r}$}

\noindent whenever $f$, $g\ge 0$ and $f\in\eusm L^p(\mu)$,
$g\in\eusm L^q(\mu)$.   \Hint{set $p'=p/(p-1)$, etc.;  $f_1=f^{p/q'}$,
$g_1=g^{q/p'}$, $h=(f^p\times g^q)^{1/r}$.   Use 244Xc to see that
$\|f_1\times g_1\|_{r'}\le\|f_1\|_{q'}\|g_1\|_{p'}$, so that
$\int f_1\times g_1\times h\le\|f_1\|_{q'}\|g_1\|_{p'}\|h\|_r$.}   (ii)
Show that $f*g$ is defined a.e.\ and that
$\|f*g\|_r\le\|f\|_p\|g\|_q$ for all $f\in\eusm L^p(\mu)$,
$g\in\eusm L^q(\mu)$.   \Hint{take $f$, $g\ge 0$.   Use (i) to see that
$(f*g)(x)^r\le\|f\|_p^{r-p}\|g\|_q^{r-q}\int f(y)^pg(x-y)^qdy$, so that
$\|f*g\|_r^r\le\|f\|_p^{r-p}\|g\|_q^{r-q}\int f(y)^p\|g\|_q^qdy$.}
(This is {\bf Young's inequality}.)
%244Xc 255K

\spheader 255Ym Repeat the results of this section for the group
$(S^1)^r$, where $r\ge 2$, given its product measure.
%+

\spheader 255Yn Let $G$ be a group and $\mu$ a
$\sigma$-finite measure on $G$ such that ($\alpha$) for every $a\in G$,
the map $x\mapsto ax$ is an automorphism of $(G,\mu)$ ($\beta$) the map
$(x,y)\mapsto (x,xy)$ is an automorphism of $(G^2,\mu_2)$, where $\mu_2$
is the c.l.d.\ product measure on $G\times G$.   For $f$,
$g\in\eusm L^0_{\Bbb C}(\mu)$ write $(f*g)(x)=\int f(y)g(y^{-1}x)dy$
whenever this is defined.   Show that
%+

\quad (i) if $f$, $g$, $h\in\eusm L^0_{\Bbb C}(\mu)$ and $\int
h(xy)f(x)g(y)d(x,y)$ is defined in $\Bbb C$, then $\int h(x)(f*g)(x)dx$
exists and is equal to $\int h(xy)f(x)g(y)d(x,y)$, provided
that in the expression $h(x)(f*g)(x)$ we interpret the product as $0$ if
$h(x)=0$ and $(f*g)(x)$ is undefined;

\quad (ii) if $f$, $g\in\eusm L^1_{\Bbb C}(\mu)$ then $f*g\in\eusm
L^1_{\Bbb C}(\mu)$ and $\int f*g=\int f\int g$,
$\|f*g\|_1\le\|f\|_1\|g\|_1$;

\quad (iii) if $f$, $g$, $h\in\eusm L^1_{\Bbb C}(\mu)$ then
$f*(g*h)=(f*g)*h$.

\noindent (See {\smc Halmos 50}, \S59.)

\spheader 255Yo Repeat 255Yn for counting measure on any group $G$.
%+ 255Yn
}%end of exercises

\endnotes{
\Notesheader{255} I have tried to set this section out in such a way
that it will be clear that the basis of all the work here is 255A, and
the crucial application is 255G.   I hope that if and when you come to
look at general topological groups (for instance, in Chapter 44), you
will find it easy to trace
through the ideas in any abelian topological group for which
you can prove a version of 255A.   For non-abelian groups, of
course, rather more care is necessary, especially as in some important
examples we no longer have $\mu\{x^{-1}:x\in E\}=\mu E$ for every $E$;
see 255Yn-255Yo for a little of what can be done without using
topological ideas.

The critical point in 255A is the move from the one-dimensional results
in 255Aa-255Ac, which are just the translation- and
reflection-invariance of
Lebesgue measure, to the two-dimensional results in 255Ac-255Ad.   And
the living centre of the argument, as I present it, is the fact that the
shear transformation $\phi$ is an automorphism of the structure
$(\BbbR^2,\Sigma_2)$.   The actual calculation of $\mu_2\phi[E]$,
assuming that it is measurable, is an easy application of Fubini's and
Tonelli's theorems and the translation-invariance of $\mu$.   It is for
this step
that we absolutely need the topological properties of Lebesgue measure.
I should perhaps remind you that the fact that $\phi$ is a homeomorphism
is not sufficient;  in 134I I described a homeomorphism of the unit
interval which does not preserve measurability, and it is easy to adapt
this to produce a homeomorphism $\psi:\BbbR^2\to\BbbR^2$ such that
$\psi[E]$ is not always measurable for measurable $E$.   The argument of
255A is dependent on the special relationships between all three of the
measure, topology and group structure of $\Bbb R$.

I have already indulged in a few remarks on what ought, or ought not, to
be `obvious' (255C).   But perhaps I can add that such results as
255B and the later claim, in the proof of 255K, that a reflected version
of a function in $\eusm L^p$ is also in $\eusm L^p$, can only be trivial
consequences of results like 255A if every step in the construction of
the integral is done in the abstract context of general measure spaces.
Even though we are here working exclusively with the Lebesgue integral,
the argument will become untrustworthy if we have at any stage in the
definition of the integral even mentioned that we are thinking of
Lebesgue measure.   I advance this as a solid reason for defining
`integration' on abstract measure spaces from the beginning, as I did in
Volume 1.   Indeed, I suggest that generally in pure mathematics there
are good reasons for casting arguments into the forms appropriate to the
arguments themselves.

I am writing this book for readers who are interested in proofs, and as
elsewhere I have written the proofs of this section out in detail.   But
most of us find it useful to go through some
material in `advanced calculus' mode, by which I mean starting with a
formula such as

\Centerline{$(f*g)(x)=\int f(x-y)g(y)dy$,}

\noindent and then working out consequences by formal manipulations, for
instance

\Centerline{$\int h(x)(f*g)(x)dx=\iint h(x)f(x-y)g(y)dydx
=\iint h(x+y)f(x)g(y)dydx$,}

\noindent without troubling about the precise applicability of the
formulae to begin with.   In some ways this formula-driven approach can
be more truthful to the structure of the subject than the careful
analysis I habitually present.   The exact hypotheses necessary to make
the theorems strictly true are surely secondary, in such contexts as
this section, to the pattern formed by the ensemble of the theorems,
which can be adequately and elegantly expressed in straightforward
formulae.   Of course I do still insist that we cannot properly
appreciate the structure, nor safely use it, without mastering the ideas
of the proofs -- and as I have said elsewhere, I believe that mastery of
ideas necessarily includes mastery of the formal details, at least in
the sense of being able to reconstruct them fairly fluently
on demand.

Throughout the main exposition of this section, I have worked with
functions rather than equivalence classes of functions.    But all the
results here have interpretations of great importance for the theory of
the `function spaces' of Chapter 24.   In 255Xh and the succeeding
exercises, I have pointed to a definition of convolution as an operator
from a subset of $L^0\times L^0$ to $L^0$.
It is an interesting point
that if $u$, $v\in L^0$ then $u*v$ can be interpreted as a
{\it function}, not as a member of $L^0$ (255Fc).
Thus 255H can be regarded as saying that $u*v\in\eusm L^1$
for $u$, $v\in L^1$.   We cannot quite say that convolution is a
bilinear operator from $L^1\times L^1$ to $\eusm L^1$, because $\eusm L^1$,
as I define it,
is not strictly speaking a linear space.   If we want a bilinear
operator, then we have to regard convolution as a function from
$L^1\times L^1$ to $L^1$.   But when we look at convolution as a
function on $L^2\times L^2$, for instance, then our functions $u*v$ are
defined everywhere (255K), and indeed are continuous functions vanishing
at $\infty$ (255Ye-255Yf).   So in this case it seems more appropriate
to regard convolution as a bilinear operator from $L^2\times L^2$ to
some space of continuous functions, and not as an operator from
$L^2\times L^2$ to $L^{\infty}$.   For an example of an interesting
convolution which is not naturally representable in terms of an operator
on $L^p$ spaces, see 255Xg.

Because convolution acts as a continuous bilinear operator from
$L^1(\mu)\times L^1(\mu)$ to $L^1(\mu)$, where $\mu$ is Lebesgue measure
on $\Bbb R$, Theorem 253F tells us that it must correspond to a linear
operator from $L^1(\mu_2)$ to $L^1(\mu)$, where $\mu_2$ is Lebesgue
measure on $\BbbR^2$.  This is the operator $\tilde T$ of 255Xj.

So far in these notes I have written as though we were concerned only
with Lebesgue measure on $\Bbb R$.   However many applications of the
ideas involve $\BbbR^r$ or $\ocint{-\pi,\pi}$ or $S^1$.   The move to
$\BbbR^r$ should be elementary.   The move to $S^1$ does require a
re-formulation of the basic result 255A/255N.   It should also be clear
that there will be no new difficulties in moving to $\ocint{-\pi,\pi}^r$
or $(S^1)^r$.   Moreover, we can also go through the whole theory for
the groups $\Bbb Z$ and $\Bbb Z^r$, where the appropriate measure is now
counting measure, so that $L^0_{\Bbb C}$ becomes identified with
$\Bbb C^{\Bbb Z}$ or $\Bbb C^{\Bbb Z^r}$ (255Xk, 255Yo).


}%end of notes

\discrpage

\frfilename{mt256.tex}
\versiondate{6.8.15}
\copyrightdate{2000}

\def\chaptername{Product measures}
\def\sectionname{Radon measures on $\BbbR^r$}

\newsection{256}

\def\headlinesectionname{Radon measures on {$\eightBbb R^r$}}

In the next section, and again in Chapters 27 and 28,
we need to consider the principal class of measures on Euclidean spaces.
For a proper discussion of this class, and the interrelationships
between the measures and the topologies involved, we must wait until
Volume 4.   For the moment, therefore, I present definitions adapted to
the case in hand, warning you that the correct generalizations are not
quite obvious.   I give the definition (256A) and a characterization
(256C) of Radon measures on Euclidean spaces, and theorems on the
construction of
Radon measures as indefinite integrals (256E, 256J), as image measures
(256G) and as product measures (256K).   In passing I give a version of
Lusin's theorem concerning measurable functions on Radon measure spaces
(256F).

Throughout this section, $r$ and $s$ will be integers greater than or equal
to $1$.

\vleader{48pt}{256A}{Definitions} Let $\nu$ be a measure on $\BbbR^r$ 
and $\Sigma$ its domain.

\spheader 256Aa $\nu$ is a {\bf topological measure} if every open set
belongs to $\Sigma$.   \cmmnt{Note that in this case every Borel set,
and in particular every closed set, belongs to $\Sigma$.}

\spheader 256Ab $\nu$ is {\bf locally finite} if every bounded set has
finite outer measure.

\spheader 256Ac If $\nu$ is a topological measure, it is {\bf inner
regular with respect to the compact sets} if

\Centerline{$\nu E=\sup\{\nu K:K\subseteq E$ is compact$\}$}

\noindent for every $E\in\Sigma$.   \cmmnt{(Because $\nu$ is a
topological measure, and compact sets are closed (2A2Ec), $\nu K$ is
defined for every compact set $K$.)}

\spheader 256Ad $\nu$ is a {\bf Radon measure} if it is a complete
locally finite topological measure which is inner regular with respect
to the compact sets.


\leader{256B}{}\cmmnt{ It will be convenient to be able to call on the
following elementary facts.

\medskip

\noindent}{\bf Lemma} Let $\nu$ be a Radon measure on $\BbbR^r$, and
$\Sigma$ its domain.

(a) $\nu$ is $\sigma$-finite.

(b) For any $E\in\Sigma$ and any
$\epsilon>0$ there are a closed
set $F\subseteq E$ and an open set $G\supseteq E$ such that
$\nu(G\setminus F)\le\epsilon$.

(c) For every $E\in\Sigma$ there is a set
$H\subseteq E$, expressible as the union of a sequence of compact sets,
such that $\nu(E\setminus H)=0$.

(d) Every continuous real-valued function on $\BbbR^r$ is
$\Sigma$-measurable.

(e) If $h:\BbbR^r\to\Bbb R$ is continuous and has bounded support,
then $h$ is $\nu$-integrable.

\proof{{\bf (a)} For each $n\in\Bbb N$, $B(\tbf{0},n)=\{x:\|x\|\le n\}$
is a closed bounded set,
therefore Borel.   So if $\nu$ is a Radon measure on $\BbbR^r$,
$\sequencen{B(\tbf{0},n)}$ is a sequence of
sets of finite measurea covering $\BbbR^r$.

\medskip

{\bf (b)} Set $E_n=\{x:x\in E,\,n\le\|x\|<n+1\}$ for each $n$.   Then
$\nu E_n<\infty$, so there is a compact set $K_n\subseteq E_n$ such that
$\nu K_n\ge\nu E_n-2^{-n-2}\epsilon$.   Set $F=\bigcup_{n\in\Bbb N}K_n$;
then

\Centerline{$\nu(E\setminus F)
=\sum_{n=0}^{\infty}\nu(E_n\setminus K_n)\le\Bover12\epsilon$.}

\noindent Also $F\subseteq E$ and $F$ is closed because

\Centerline{$F\cap B(\tbf{0},n)=\bigcup_{i\le n}K_i\cap B(\tbf{0},n)$}

\noindent is closed for each $n$.

In the same way, there is a closed set $F'\subseteq\BbbR^r\setminus E$
such that $\nu((\BbbR^r\setminus E)\setminus F')\le\bover12\epsilon$.
Setting $G=\BbbR^r\setminus F'$, we see that $G$ is open, that
$G\supseteq E$ and that $\nu(G\setminus E)\le\bover12\epsilon$, so that
$\nu(G\setminus F)\le\epsilon$, as required.

\medskip

{\bf (c)} By (b), we can choose for each $n\in\Bbb N$ a closed set
$F_n\subseteq E$ such that $\nu(E\setminus F_n)\le 2^{-n}$.   Set
$H=\bigcup_{n\in\Bbb N}F_n$;  then $H\subseteq E$ and $\nu(E\setminus
H)=0$, and also $H=\bigcup_{m,n\in\Bbb N}B(\tbf{0},m)\cap F_n$ is a
countable union of compact sets.

\medskip

{\bf (d)} If $h:\BbbR^r\to\Bbb R$ is continuous, all the sets
$\{x:h(x)>a\}$ are open, so belong to $\Sigma$.

\medskip

{\bf (e)} By (d), $h$ is measurable.   Now we are supposing that there
is some $n\in\Bbb N$ such that $h(x)=0$ whenever $x\notin B(\tbf{0},n)$.
Since $B(\tbf{0},n)$ is compact (2A2F), $h$ is bounded on $B(\tbf{0},n)$
(2A2G), and we have $|h|\le \gamma\chi B(\tbf{0},n)$ for some $\gamma$;
since $\nu B(\tbf{0},n)$ is finite, $h$ is $\nu$-integrable.
}%end of proof of 256B

\leader{256C}{Theorem} A measure $\nu$ on $\BbbR^r$ is a Radon measure
iff it is the completion of a locally finite measure defined on the
$\sigma$-algebra\cmmnt{ $\Cal B$} of Borel subsets of $\BbbR^r$.

\proof{{\bf (a)} Suppose first that $\nu$ is a Radon measure.   Write
$\Sigma$ for its domain.

\medskip

\quad{\bf (i)} Set $\nu_0=\nu\restr\Cal B$.   Then $\nu_0$ is a measure
with domain $\Cal B$, and it is locally finite because
$\nu_0B(\tbf{0},n)=\nu B(\tbf{0},n)$ is finite for every $n$.   Let
$\hat\nu_0$ be the completion of $\nu_0$ (212C).

\medskip

\quad{\bf (ii)} If $\hat\nu_0$ measures $E$, there are $E_1$,
$E_2\in\Cal B$ such that $E_1\subseteq E\subseteq E_2$ and
$\nu_0(E_2\setminus E_1)=0$.   Now $E\setminus E_1\subseteq E_2\setminus
E_1$ must be
$\nu$-negligible;  as $\nu$ is complete, $E\in\Sigma$ and

\Centerline{$\nu E=\nu E_1=\nu_0 E_1=\hat\nu_0E$.}

\medskip

\quad{\bf (iii)} If $E\in\Sigma$, then by 256Bc there is a
Borel set $H\subseteq E$ such that $\nu(E\setminus H)=0$.   Equally,
there is a Borel set $H'\subseteq\BbbR^r\setminus E$ such that
$\nu((\BbbR^r\setminus E)\setminus H')=0$, so that we have $H\subseteq
E\subseteq\BbbR^r\setminus H'$ and

\Centerline{$\nu_0((\BbbR^r\setminus H')\setminus H)=\nu((\Bbb
R^r\setminus H')\setminus H)=0$.}

\noindent So $\hat\nu_0E$ is defined and equal to $\nu_0E_1=\nu E$.

This shows that $\nu=\hat\nu_0$ is the completion of the locally finite
Borel measure $\nu\restr\Cal B$.   And this is true for any Radon
measure $\nu$ on $\BbbR^r$.

\medskip

{\bf (b)} For the rest of the proof, I suppose that $\nu_0$ is a locally
finite measure with domain $\Cal B$ and $\nu$ is its completion.    Write
$\Sigma$ for the domain of $\nu$.   We say that a subset of $\BbbR^r$
is a {\bf K$_{\sigma}$ set} if it is expressible as the union of a
sequence of compact sets.   Note that every K$_{\sigma}$ set is a Borel
set, so belongs to $\Sigma$.   Set

\Centerline{$\Cal A=\{E:E\in\Sigma$, there is a K$_{\sigma}$ set
$H\subseteq E$ such that $\nu(E\setminus H)=0\}$,}

\Centerline{$\Sigma=\{E:E\in\Cal A,\,\BbbR^r\setminus E\in\Cal A\}$.}

\medskip

{\bf (c)(i)} Every open set is itself a K$_{\sigma}$ set, so belongs to
$\Cal A$.
\Prf\ Let $G\subseteq\BbbR^r$ be open.   If $G=\emptyset$ then $G$ is
compact and the result is trivial.   Otherwise, let $\Cal I$ be the set
of closed intervals of the form $[q,q']$, where $q$, $q'\in\Bbb Q^r$,
which are included in $G$.   Then all the members of $\Cal I$ are closed
and bounded, therefore compact.   If $x\in G$, there is a $\delta>0$
such that $B(x,\delta)=\{y:\|y-x\|\le\delta\}\subseteq G$;  now there is
an $I\in\Cal I$ such that $x\in I\subseteq B(x,\delta)$.   Thus
$G=\bigcup\Cal I$.   But $\Cal I$ is countable, so $G$ is K$_{\sigma}$.\
\Qed

\medskip

\quad{\bf (ii)} Every closed subset of $\Bbb R$ is K$_{\sigma}$, so
belongs to $\Cal A$.
\Prf\ If $F\subseteq\Bbb R$ is closed, then
$F=\bigcup_{n\in\Bbb N}F\cap B(\tbf{0},n)$;  but every $F\cap
B(\tbf{0},n)$ is closed and bounded, therefore compact.\ \Qed

\medskip

\quad{\bf (iii)} If $\sequencen{E_n}$ is any sequence in $\Cal A$, then
$E=\bigcup_{n\in\Bbb N}E_n$ belongs to
$\Cal A$.   \Prf\ For each $n\in\Bbb N$ we have a countable family
$\Cal K_n$ of compact subsets of $E_n$ such that $\nu(E_n\setminus\bigcup\Cal K_n)=0$;  now
$\Cal K=\bigcup_{n\in\Bbb N}\Cal K_n$ is a countable
family of compact subsets of $E$, and
$E\setminus\bigcup\Cal K
\subseteq\bigcup_{n\in\Bbb N}(E_n\setminus\bigcup\Cal K_n)$ is
$\nu$-negligible.\ \Qed

\medskip

\quad{\bf (iv)} If $\sequencen{E_n}$ is any sequence in $\Cal A$, then
$F=\bigcap_{n\in\Bbb N}E_n\in\Cal A$.   \Prf\ For each $n\in\Bbb N$, let
$\sequence{i}{K_{ni}}$ be a sequence of compact subsets of $E_n$ such
that $\nu(E_n\setminus\bigcup_{i\in\Bbb N}K_{ni})=0$.   Set
$K'_{nj}=\bigcup_{i\le j}K_{ni}$ for each $j$, so that

\Centerline{$\nu(E_n\cap H)=\lim_{j\to\infty}\nu(K'_{nj}\cap H)$}

\noindent for every $H\in\Sigma$.   Now, for each $m$, $n\in\Bbb N$,
choose $j(m,n)$ such that

\Centerline{$\nu(E_n\cap B(\tbf{0},m)\cap K'_{n,j(m,n)})
\ge\nu(E_n\cap B(\tbf{0},m))-2^{-(m+n)}$.}

\noindent Set $K_m=\bigcap_{n\in\Bbb N}K'_{n,j(m,n)}$;  then $K_m$ is
closed (being an intersection of closed sets) and bounded (being a
subset of $K'_{0,j(m,0)}$), therefore compact.   Also $K_m\subseteq F$,
because $K'_{n,j(m,n)}\subseteq E_n$ for each $n$, and

\Centerline{$\nu(F\cap B(\tbf{0},m)\setminus K_m)
\le\sum_{n=0}^{\infty}\nu(E_n\cap B(\tbf{0},m)\setminus K'_{n,j(m,n)})
\le\sum_{n=0}^{\infty}2^{-(m+n)}
=2^{-m+1}$.}

\noindent Consequently $H=\bigcup_{m\in\Bbb N}K_m$ is a K$_{\sigma}$
subset of $F$ and

\Centerline{$\nu(F\cap B(\tbf{0},m)\setminus H)
\le\inf_{k\ge m}\nu(F\cap B(\tbf{0},k)\setminus H_k)
=0$}

\noindent for every $m$, so $\nu(F\setminus H)=0$ and $F\in\Cal A$.\ \Qed

\medskip

{\bf (d)} $\Sigma$ is a $\sigma$-algebra of subsets of $\Bbb R$.
\Prf\ {(i)} $\emptyset$ and its complement are open, so belong to $\Cal
A$ and therefore to $\Sigma$.
{(ii)} If $E\in\Sigma$ then both $\BbbR^r\setminus E$ and $\Bbb
R^r\setminus(\BbbR^r\setminus E)=E$ belong to $\Cal A$, so $\Bbb
R^r\setminus E\in\Sigma$.   {(iii)} Let $\sequencen{E_n}$ be a
sequence in $\Sigma$ with union $E$.   By (a-iii) and (a-iv),

\Centerline{$E\in\Cal A$,
\quad$\BbbR^r\setminus E
=\bigcap_{n\in\Bbb N}(\BbbR^r\setminus E_n)\in\Cal A$,}

\noindent so $E\in\Sigma$.\  \Qed

\medskip

{\bf (e)} By (c-i) and (c-ii), every open set belongs to $\Sigma$;
consequently every Borel set
belongs to $\Sigma$ and therefore to $\Cal A$.   Now if $E$ is any
member
of $\Sigma$, there is a Borel set $E_1\subseteq E$ such that
$\nu(E\setminus E_1)=0$ and a K$_{\sigma}$ set $H\subseteq E_1$ such
that $\nu(E_1\setminus H)=0$.   Express $H$ as $\bigcup_{n\in\Bbb N}K_n$
where every $K_n$ is compact;  then

\Centerline{$\nu E
=\nu H
=\lim_{n\to\infty}\nu(\bigcup_{i\le n}K_i)
\le\sup_{K\subseteq E\text{ is compact}}\nu K
\le\nu E$}

\noindent because $\bigcup_{i\in n}K_i$ is a compact subset of $E$ for
every $n$.

\medskip

{\bf (f)} Thus $\nu$ is inner regular with respect to the compact sets.
But of course it is complete (being the completion of $\nu_0$) and  a
locally
finite topological measure (because $\nu_0$ is);  so it is a Radon
measure.   This completes the proof.
}%end of proof of 256C

\leader{256D}{Proposition} If $\nu$ and $\nuprime$ are two Radon
measures on $\BbbR^r$, the following are equiveridical:

(i) $\nu=\nuprime$;

(ii) $\nu K=\nuprime K$ for every compact set $K\subseteq\BbbR^r$;

(iii) $\nu G=\nuprime G$ for every open set $G\subseteq\BbbR^r$;

(iv) $\int h\,d\nu=\int h\,d\nuprime$ for every continuous function
$h:\BbbR^r\to\Bbb R$ with bounded support.

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

\medskip

{\bf (b)(iv)$\Rightarrow$(iii)} If (iv) is true, and $G\subseteq\BbbR^r$
is an open set, then for each $n\in\Bbb N$ set

\Centerline{$h_n(x)
=\min(1,2^n\inf_{y\in\BbbR^r\setminus(G\cap B(\tbf{0},n))}\|y-x\|)$}

\noindent for $x\in\BbbR^r$.   Then $h_n$ is continuous (in fact
$|h_n(x)-h_n(x')|\le 2^n\|x-x'\|$ for all $x$, $x'\in\BbbR^r$) and zero
outside $B(\tbf{0},n)$, so $\int h_nd\nu=\int h_nd\nuprime$.   Next,
$\sequencen{h_n(x)}$ is a non-decreasing sequence converging to
$\chi G(x)$ for every $x\in\BbbR^r$.   So

\Centerline{$\nu G=\lim_{n\to\infty}\int h_nd\nu
=\lim_{n\to\infty}\int h_nd\nuprime=\nuprime G$,}

\noindent by 135Ga.   As $G$ is arbitrary, (iii) is true.

\medskip

{\bf (c)(iii)$\Rightarrow$(ii)} If (iii) is true, and
$K\subseteq\BbbR^r$ is compact, let $n$ be so large that $\|x\|<n$ for
every $x\in K$.   Set $G=\{x:\|x\|<n\}$, $H=G\setminus K$.   Then $G$
and $H$ are open and $G$ is bounded, so $\nu G=\nuprime G$ is finite,
and

\Centerline{$\nu K=\nu G-\nu H=\nuprime G-\nuprime H=\nuprime K$.}

\noindent As $K$ is arbitrary, (ii) is true.

\medskip

{\bf (d)(ii)$\Rightarrow$(i)} If $\nu$, $\nuprime$ agree on the compact
sets, then

\Centerline{$\nu E=\sup_{K\subseteq E\text{ is compact}}\nu K
=\sup_{K\subseteq E\text{ is compact}}\nuprime K
=\nuprime E$}

\noindent for every Borel set $E$.   So
$\nu\restr\Cal B=\nuprime\restr\Cal B$, where $\Cal B$ is the algebra of Borel sets.
But since $\nu$ and $\nuprime$ are both the completions of their
restrictions to $\Cal B$, they are identical.
}%end of proof of 256D

\leader{256E}{}\cmmnt{ It is I suppose time I gave some examples of
Radon measures.   However it will save a few lines if I first establish
some basic constructions.   You may wish to glance ahead to 256H at this
point.

\medskip

\noindent}{\bf Theorem} Let $\nu$ be a Radon measure on $\BbbR^r$, with
domain $\Sigma$, and
$f$ a non-negative $\Sigma$-measurable function defined on a
$\nu$-conegligible subset of $\BbbR^r$.   Suppose that $f$ is
{\bf locally integrable} in the sense that $\int_Efd\nu<\infty$ for every
bounded set $E$.
Then the indefinite-integral measure $\nuprime$ on $\BbbR^r$ defined by
saying that

\Centerline{$\nuprime E=\int_Efd\nu$ whenever
$E\cap\{x:x\in\dom f,\,f(x)>0\}\in\Sigma$}

\noindent is a Radon measure on $\BbbR^r$.

\proof{ For the construction of $\nuprime$, see 234I-234L.
Indefinite-integral measures, as I define them, are always complete (234I).
$\nuprime$ is locally finite because $f$ is
locally integrable.   $\nuprime$ is a
topological measure because every open set belongs to $\Sigma$ and
therefore to the domain $\Sigma'$ of $\nuprime$.
To see that $\nuprime$ is inner regular with respect to the compact
sets, take any set $E\in\Sigma'$, and set
$E'=\{x:x\in E\cap\dom f,\,f(x)>0\}$.
Then $E'\in\Sigma$, so there is a set $H\subseteq E'$,
expressible as the union of a sequence of compact sets,
such that $\nu(E'\setminus H)=0$.   In this case

\Centerline{$\nuprime(E\setminus H)=\int_{E\setminus H}fd\nu=0$.}

\noindent Let $\sequencen{K_n}$ be a sequence of compact sets with union
$H$;  then

\Centerline{$\nuprime E=\nuprime H
=\lim_{n\to\infty}\nuprime(\bigcup_{i\le n}K_i)
\le\sup_{K\subseteq E\text{ is compact}}\nuprime K\le\nuprime E$.}

\noindent As $E$ is arbitrary, $\nuprime$ is inner regular with respect
to the compact sets.
}%end of proof of 256E

\leader{256F}{Theorem} Let $\nu$ be a Radon measure on $\BbbR^r$, and
$\Sigma$ its domain.   Let $f:D\to\Bbb R$ be a $\Sigma$-measurable
function, where $D\subseteq\BbbR^r$.   Then for every $\epsilon>0$
there is a closed set $F\subseteq\BbbR^r$ such that
$\nu(\BbbR^r\setminus F)\le\epsilon$ and $f\restr F$ is continuous.

\proof{ By 121I, there is a $\Sigma$-measurable function
$h:\BbbR^r\to\Bbb R$ extending $f$.
Enumerate $\Bbb Q$ as $\sequencen{q_n}$.
For each $n\in\Bbb N$ set $E_n=\{x:h(x)\le q_n\}$, $E'_n=\{x:h(x)>q_n\}$
and use 256Bb to choose closed sets  $F_n\subseteq E_n$,
$F'_n\subseteq E'_n$ such that
$\nu(E_n\setminus F_n)\le 2^{-n-2}\epsilon$ and
$\nu(E'_n\setminus F'_n)\le 2^{-n-2}\epsilon$.   Set
$F=\bigcap_{n\in\Bbb N}(F_n\cup F'_n)$;  then $F$ is closed and

\Centerline{$\nu(\BbbR^r\setminus F)
\le\sum_{n=0}^{\infty}\nu(\BbbR^r\setminus(F_n\cup F'_n))
\le\sum_{n=0}^{\infty}\nu(E_n\setminus F_n)+\nu(E'_n\setminus F'_n)
\le\epsilon$.}

I claim that $h\restr F$ is continuous.   \Prf\ Suppose that $x\in F$
and $\delta>0$.   Then there are $m$, $n\in\Bbb N$ such that

\Centerline{$h(x)-\delta\le q_m<h(x)\le q_n\le h(x)+\delta$.}

\noindent This means that $x\in E'_m\cap E_n$;  consequently
$x\notin F_m\cup F'_n$.
Because $F_m\cup F'_n$ is closed, there is an $\eta>0$
such that $y\notin F_m\cup F'_n$ whenever $\|y-x\|\le\eta$.
Now suppose that $y\in F$ and $\|y-x\|\le\eta$.   Then
$y\in(F_m\cup F'_m)\cap(F_n\cup F'_n)$ and $y\notin F_m\cup F'_n$,
so $y\in F'_m\cap F_n\subseteq E'_m\cap E_n$ and $q_m<h(y)\le q_n$.
Consequently
$|h(y)-h(x)|\le\delta$.   As $x$ and $\delta$ are arbitrary, $h\restr F$
is continuous.\ \QeD\
It follows that $f\restr F=(h\restr F)\restr D$ is continuous, as required.
}%end of proof of 256F

\leader{256G}{Theorem} Let $\nu$ be a Radon measure on $\BbbR^r$, with
domain $\Sigma$, and
suppose that $\phi:\BbbR^r\to\BbbR^s$ is measurable in the sense that
all its coordinates are $\Sigma$-measurable.   If the image measure
$\nuprime=\nu\phi^{-1}$\cmmnt{ (234D)} is locally finite, it is a
Radon measure.

\proof{ Write $\Sigma'$ for the
domain of $\nuprime$.   If $\phi=(\phi_1,\ldots,\phi_s)$, then

\Centerline{$\phi^{-1}[\{y:\eta_j\le\alpha\}]
=\{x:\phi_j(x)\le\alpha\}\in\Sigma$,}

\noindent so $\{y:\eta_j\le\alpha\}\in\Sigma'$
for every $j\le s$, $\alpha\in\Bbb R$, where I write
$y=(\eta_1,\ldots,\eta_s)$ for $y\in\BbbR^s$.   Consequently every
Borel subset of $\BbbR^s$ belongs to $\Sigma'$ (121J), and $\nuprime$
is a topological measure.   It is complete by 234Eb.

The point is of course that $\nuprime$ is inner regular with respect to
the compact
sets.   \Prf\ Suppose that $F\in\Sigma'$ and that $\gamma<\nuprime F$.
For each $j\le s$, there is a closed set $H_j\subseteq\BbbR^r$ such
that $\phi_j\restr H_j$ is continuous and
$\nu(\BbbR^r\setminus H_j)<\bover1{s}(\nuprime F-\gamma)$, by 256F.   Set $H=\bigcap_{j\le s}H_j$;  then $H$ is closed and $\phi\restr H$ is continuous and

\Centerline{$\nu(\BbbR^r\setminus H)<\nuprime F-\gamma
=\nu\phi^{-1}[F]-\gamma$,}

\noindent so that
$\nu(\phi^{-1}[F]\cap H)>\gamma$.   Let $K\subseteq\phi^{-1}[F]\cap H$
be a compact set such that $\nu K\ge\gamma$, and set $L=\phi[K]$.
Because $K\subseteq H$ and $\phi\restr H$ is continuous, $L$ is compact
(2A2Eb).   Of course $L\subseteq F$, and

\Centerline{$\nuprime L=\nu\phi^{-1}[L]\ge\nu K\ge\gamma$.}

\noindent As $F$ and $\gamma$ are arbitrary, $\nuprime$ is inner regular
with respect to the compact sets.\ \Qed

Since $\nuprime$ is locally finite by the hypothesis of the theorem, it
is a Radon measure.
}%end of proof of 256G

\leader{256H}{Examples}\cmmnt{ I come at last to the promised
examples.

\medskip

} {\bf (a)} Lebesgue measure on $\BbbR^r$ is a Radon measure.
\prooflet{(It is a topological measure by 115G, and inner regular with
respect to the compact sets by 134Fb.)}

\spheader 256Hb A point-supported measure on $\BbbR^r$ is a Radon measure
iff it is locally finite.   \prooflet{\Prf\ Let $\mu$ be a point-supported
measure on $\BbbR^r$.   If it is a Radon measure, then of course it is
locally finite.   If it is locally finite, then surely it is a 
complete topological
measure, since it measures every subset of $\BbbR^r$.   Let 
$h:\BbbR^r\to[0,\infty]$ be such that $\mu E=\sum_{x\in E}h(x)$ for every
$E\subseteq\BbbR^r$.   Take any $E\subseteq\BbbR^r$.   Then

$$\eqalign{\mu E
&=\sum_{x\in E}h(x)
=\sup_{I\subseteq E\text{ is finite}}\sum_{x\in I}h(x)\cr&
=\sup_{I\subseteq E\text{ is finite}}\mu I
\le\sup_{K\subseteq E\text{ is compact}}\mu K
\le\mu E
\cr}$$

\noindent so $\mu E=\sup_{K\subseteq E\text{ is compact}}\mu K$;
thus $\mu$ is inner regular with respect to the compact sets and is
a Radon measure.\ \Qed}%end of prooflet

\spheader 256Hc \cmmnt{Now we come to a new idea.}  Recall that the
Cantor set $C$\cmmnt{ (134G)} is a closed Lebesgue negligible subset of
$[0,1]$, and that the Cantor function\cmmnt{ (134H)} is a
non-decreasing
continuous function $f:[0,1]\to[0,1]$ such that $f(0)=0$, $f(1)=1$ and
$f$ is constant on each of the intervals composing $[0,1]\setminus C$.
It follows that if we set $g(x)=\bover12(x+f(x))$ for $x\in[0,1]$, then
$g:[0,1]\to[0,1]$ is a continuous permutation\cmmnt{ such that the
Lebesgue measure of $g[C]$ is $\bover12$ (134I);
consequently $g^{-1}:[0,1]\to[0,1]$ is
continuous}.   Now extend $g$ to a permutation $h:\Bbb R\to\Bbb R$ by
setting $h(x)=x$ for $x\in\Bbb R\setminus[0,1]$.
\cmmnt{Then $h$ and $h^{-1}$ are continuous.   Note that $h[C]=g[C]$
has Lebesgue measure $\bover12$.}

Let $\nu_1$ be the indefinite-integral measure defined from  Lebesgue
measure $\mu$ on $\Bbb R$ and
the function $2\chi(h[C])$\cmmnt{;  that is, $\nu_1E=2\mu(E\cap h[C])$ whenever
this is defined}.   \cmmnt{By 256E,} $\nu_1$ is a Radon measure, and
$\nu_1h[C]=\nu_1\Bbb R=1$.   Let $\nu$ be the measure $\nu_1(h^{-1})^{-1}$,
that is,
$\nu E=\nu_1h[E]$ for just those $E\subseteq\Bbb R$ such that
$h[E]\in\dom\nu_1$.   Then $\nu$ is a Radon probability measure on
$\Bbb R$\cmmnt{, by 256G}, and $\nu C=1$,
$\nu(\Bbb R\setminus C)=\mu C=0$.

\cmmnt{
\leader{256I}{Remarks (a)} The measure $\nu$ of 256Hc, sometimes called
{\bf Cantor measure}, is a classic example, and as such has many
constructions, some rather more natural than the one I use here (see
256Xk, and also 264Ym below).   But I choose the method above
because it
yields directly, without further investigation or any appeal to more
advanced general theory, the fact that $\nu$ is a Radon measure.

\spheader 256Ib The examples above are chosen to represent
the extremes under the `Lebesgue decomposition' described in
232I.   If $\nu$ is a (totally finite) Radon measure on $\BbbR^r$, we
can use 232Ib to express its restriction $\nu\restr\Cal B$ to the
Borel $\sigma$-algebra as $\nu_p+\nu_{ac}+\nu_{cs}$, where $\nu_p$ is
the `point-mass' or `atomic' part of $\nu\restr\Cal B$, $\nu_{ac}$ is
the `absolutely
continuous' part (with respect to Lebesgue measure), and $\nu_{cs}$ is
the `atomless singular part'.   In the example of 256Hb, we have
$\nu\restr\Cal B=\nu_p$;  in 256E, if we start from Lebesgue measure, we
have $\nu\restr\Cal B=\nu_{ac}$;
and in 256Hc we have $\nu\restr\Cal B=\nu_{cs}$.
}%end of comment

\leader{256J}{Absolutely continuous Radon \dvrocolon{measures}}\cmmnt{
It is worth pausing a moment over the indefinite-integral measures
described in 256E.

\medskip

\noindent}{\bf Proposition} Let $\nu$ be a Radon measure
on $\BbbR^r$, and write $\mu$ for Lebesgue measure on
$\BbbR^r$.   Then the following are equiveridical:

\quad(i) $\nu$ is an indefinite-integral measure over $\mu$;

\quad(ii) $\nu E=0$ whenever $E$ is a Borel subset of $\BbbR^r$ and
$\mu E=0$.

\noindent In this case, if $g\in\eusm L^0(\mu)$ and
$\int_Eg\,d\mu=\nu E$ for
every Borel set $E\subseteq\BbbR^r$, then $g$ is a Radon-Nikod\'ym
derivative of $\nu$ with respect to $\mu$\cmmnt{ in the sense of
232Hf}.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} If $f$ is a Radon-Nikod\'ym
derivative of $\nu$ with respect to $\mu$, then of course

\Centerline{$\nu E=\int_Efd\mu=0$}

\noindent whenever $\mu E=0$.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} If $\nu E=0$ for every $\mu$-negligible
Borel set $E$, then $\nu E$ is defined and equal to $0$ for
every $\mu$-negligible set $E$, because $\nu$ is complete and any
$\mu$-negligible set is included in a $\mu$-negligible Borel set.
Consequently $\dom\nu$ includes the
domain $\Sigma$ of $\mu$, since every Lebesgue measurable set is
expressible as the union of a Borel set and a negligible set.

For each $n\in\Bbb N$ set $E_n=\{x:n\le\|x\|<n+1\}$, so that
$\sequencen{E_n}$ is a partition of $\BbbR^r$ into bounded Borel sets.
Set $\nu_nE=\nu(E\cap E_n)$ for every Lebesgue measurable set $E$ and
every $n\in\Bbb N$.   Now $\nu_n$ is absolutely continuous with respect
to $\mu$ (232Ba), so by the Radon-Nikod\'ym theorem (in the form 232F)
there is a $\mu$-integrable function $f_n$ such that
$\int_Ef_nd\mu=\nu_n E$
for every Lebesgue measurable set $E$.   Because $\nu_n E\ge 0$ for
every $E\in\Sigma$,
$f_n\ge 0$ a.e.;  because $\nu_n(\BbbR^r\setminus E_n)=0$, $f_n=0$
a.e.\ on $\BbbR^r\setminus E_n$.   Now if we set

\Centerline{$f=\max(0,\sum_{n=0}^{\infty}f_n)$,}

\noindent $f$ will be defined $\mu$-a.e.\ and we shall have

\Centerline{$\int_Efd\mu
=\sum_{n=0}^{\infty}\int_Ef_nd\mu
=\sum_{n=0}^{\infty}\nu(E\cap E_n)=\nu E$}

\noindent for every Borel set $E$, so that the indefinite-integral
measure $\nuprime$ defined by $f$ and $\mu$ agrees with $\nu$ on the
Borel sets.   Since this ensures that $\nuprime$ is locally finite,
$\nuprime$ is a Radon measure, by 256E, and is equal to $\nu$, by 256D.
Accordingly $\nu$ is an indefinite-integral measure over $\mu$.

\medskip

{\bf (b)} As in (a-ii) above, $g$ must be locally integrable and the
indefinite-integral measure defined by $g$ agrees with $\nu$ on the
Borel sets, so is identical with $\nu$.
}%end of proof of 256J

\leader{256K}{\dvrocolon{Products}}\cmmnt{ The class of Radon measures
on Euclidean spaces is stable under a wide variety of operations, as we
have already seen;  in particular, we have the following.

\medskip

\noindent}{\bf Theorem} Let $\nu_1$, $\nu_2$ be Radon
measures on $\BbbR^r$ and $\BbbR^s$ respectively.
Let $\lambda$ be their c.l.d.\ product measure on
$\BbbR^r\times\BbbR^s$.   Then $\lambda$ is a Radon measure.

\cmmnt{\medskip

\noindent{\bf Remark} When I say that $\lambda$ is `Radon' according
to the definition in 256A, I am of course identifying
$\BbbR^r\times\BbbR^s$ with $\BbbR^{r+s}$, as in 251M-251N.
}

\proof{ I hope the following notation will seem natural.
Write $\Sigma_1$, $\Sigma_2$
for the domains of $\nu_1$, $\nu_2$;  $\Cal B_r$, $\Cal B_s$ for the
Borel $\sigma$-algebras of $\BbbR^r$, $\BbbR^s$;  $\Lambda$ for the
domain of $\lambda$;  and $\Cal B$ for the Borel $\sigma$-algebra of
$\BbbR^{r+s}$.

Because each $\nu_i$ is the completion of its restriction to the Borel
sets (256C), $\lambda$ is the product of $\nu_1\restr\Cal B_r$ and
$\nu_2\restr\Cal B_s$ (251T).   Because $\nu_1\restr\Cal B_r$ and
$\nu_2\restr\Cal B_s$ are $\sigma$-finite (256Ba, 212Ga), $\lambda$ must
be the completion of its restriction to $\Cal B_r\tensorhat\Cal B_s$,
which by 251M is identified with $\Cal B$.   Setting
$Q_n=\{(x,y):\|x\|\le n$, $\|y\|\le n\}$ we have

\Centerline{$\lambda Q_n=\nu_1\{x:\|x\|\le n\}\cdot\nu_2\{y:\|y\|\le n\}
<\infty$}

\noindent for every $n$, while every bounded subset of $\BbbR^{r+s}$ is
included in some $Q_n$.   So $\lambda\restr\Cal B$ is locally finite,
and its completion $\lambda$ is a Radon measure, by 256C.
}%end of proof of 256K

\cmmnt{
\leader{256L}{Remark} We see from 253I that if $\nu_1$ and
$\nu_2$ are Radon measures on $\BbbR^r$ and $\BbbR^s$ respectively, and
both are indefinite-integral measures over Lebesgue measure, then their
product measure on $\BbbR^{r+s}$ is also an indefinite-integral measure
over Lebesgue measure.
}%end of comment

\leader{*256M}{}\cmmnt{ For the sake of applications in \S286 below, I
include another result, which is in fact one of the fundamental
properties of Radon measures, as will appear in \S414.

\medskip

\noindent}{\bf Proposition} Let $\nu$ be a Radon measure on $\Bbb R^r$,
and $D$ any subset of $\Bbb R^r$.   Let $\Phi$ be a non-empty
upwards-directed family of non-negative continuous functions from $D$ to
$\Bbb R$.   For $x\in D$ set $g(x)=\sup_{f\in\Phi}f(x)$ in $[0,\infty]$.
Then

(a) $g:D\to[0,\infty]$ is lower semi-continuous,
therefore Borel measurable;

(b) $\int_Dg\,d\nu=\sup_{f\in\Phi}\int_Dfd\nu$.

\proof{{\bf (a)} For any $u\in[-\infty,\infty]$,

\Centerline{$\{x:x\in D,\,g(x)>u\}
=\bigcup_{f\in\Phi}\{x:x\in D,\,f(x)>u\}$}

\noindent is an open set for the subspace topology on $D$ (2A3C), so is
the intersection of $D$ with a Borel subset of $\Bbb R^r$.   This is
enough to show that $g$ is Borel measurable (121B-121C).

\medskip

{\bf (b)} Accordingly $\int_Dg\,d\nu$ will be defined in $[0,\infty]$,
and of course $\int_Dg\,d\nu\ge\sup_{f\in\Phi}\int_Dfd\nu$.

For the reverse inequality, observe that there is a countable set
$\Psi\subseteq\Phi$ such that $g(x)=\sup_{f\in\Psi}f(x)$ for every
$x\in D$.   \Prf\ For $a\in\Bbb Q$, $q$, $q'\in\BbbQ^r$ set

\Centerline{$\Phi_{aqq'}
=\{f:f\in\Phi,\,f(y)>a\text{ whenever }y\in D\cap[q,q']\}$,}

\noindent interpreting $[q,q']$ as in 115G.   Choose
$f_{aqq'}\in\Phi_{aqq'}$ if $\Phi_{aqq'}$ is not empty, and arbitrarily
in $\Phi$ otherwise;  and set
$\Psi=\{f_{aqq'}:a\in\Bbb Q,\,q,\,q'\in\BbbQ^r\}$, so that $\Psi$ is a
countable subset of $\Phi$.   If $x\in D$ and $b<g(x)$, there is an
$a\in\Bbb Q$ such that $b\le a<g(x)$;  there is an
$\hat f\in\Phi$ such that $\hat f(x)>a$;  because $\hat f$ is
continuous, there are $q$, $q'\in\Bbb Q^r$ such that $q\le x\le q'$ and
$\hat f(y)\ge a$ whenever $y\in D\cap[q,q']$;  so that
$\hat f\in\Phi_{aqq'}$, $\Phi_{aqq'}\ne\emptyset$,
$f_{aqq'}\in\Phi_{aqq'}$ and $\sup_{f\in\Psi}f(x)\ge f_{aqq'}(x)\ge b$.
As $b$ is
arbitrary, $g(x)=\sup_{f\in\Psi}f(x)$.\ \Qed

Let $\sequencen{f_n}$ be a sequence running over $\Psi$.   Because $\Phi$ is
upwards-directed, we can choose $\sequencen{f'_n}$ in $\Phi$ inductively
in such a way that $f'_{n+1}\ge\max(f'_n,f_n)$ for every $n\in\Bbb N$.
So $\sequencen{f'_n}$ is a non-decreasing sequence in $\Phi$ and
$\sup_{n\in\Bbb N}f'_n(x)\ge\sup_{f\in\Psi}f(x)=g(x)$ for every
$x\in D$.   By B.Levi's theorem,

\Centerline{$\int_Dg\,d\nu
\le\sup_{n\in\Bbb N}\int_Df'_nd\nu\le\sup_{f\in\Phi}\int_Dfd\nu$,}

\noindent and we have the required inequality.
}%end of proof of 256M

\exercises{
\leader{256X}{Basic exercises $\pmb{>}$(a)}
%\spheader 256Xa
Let $\nu$ be a measure on $\BbbR^r$.   (i) Show that it is locally
finite, in the sense of 256Ab, iff for every $x\in\BbbR^r$ there is a
$\delta>0$ such that $\nu^*B(x,\delta)<\infty$.   \Hint{the sets
$B(\tbf{0},n)$ are compact.}   (ii) Show that in this case $\nu$ is
$\sigma$-finite.
%256A

\sqheader 256Xb Let $\nu$ be a Radon measure on $\BbbR^r$ and $\Cal G$
a non-empty
upwards-directed family of open sets in $\BbbR^r$.   (i) Show that
$\nu(\bigcup\Cal G)=\sup_{G\in\Cal G}\nu G$.   \Hint{observe that if
$K\subseteq\bigcup\Cal G$ is compact, then $K\subseteq G$ for some
$G\in\Cal G$.}   (ii) Show that
$\nu(E\cap\bigcup\Cal G)=\sup_{G\in\Cal G}\nu(E\cap G)$ for every set $E$ which is measured by $\nu$.
%256A

\sqheader 256Xc Let $\nu$ be a Radon measure on $\BbbR^r$ and $\Cal F$
a non-empty downwards-directed family of closed sets in $\BbbR^r$ such
that $\inf_{F\in\Cal F}\nu F<\infty$.   (i) Show that
$\nu(\bigcap\Cal F)=\inf_{F\in\Cal F}\nu F$.   \Hint{apply 256Xb(ii) to $\Cal G=\{\BbbR^r\setminus F:F\in\Cal F\}$.}   (ii) Show that
$\nu(E\cap\bigcap\Cal F)=\inf_{F\in\Cal F}\nu(E\cap F)$ for every
$E$ in the domain of $\nu$.
%256A

\sqheader 256Xd Show that a Radon measure $\nu$ on $\BbbR^r$ is
atomless iff $\nu\{x\}=0$ for every $x\in\BbbR^r$.   \Hint{apply 256Xc
with $\Cal F=\{F:F\subseteq E$ is closed, not negligible$\}$.}
%256A

\spheader 256Xe Let $\nu_1$, $\nu_2$ be Radon measures on
$\BbbR^r$, and $\alpha_1$, $\alpha_2\in\ooint{0,\infty}$.   Set
$\Sigma=\dom\nu_1\cap\dom\nu_2$, and for $E\in\Sigma$ set
$\nu E=\alpha_1\nu_1E+\alpha_2\nu_2E$.   Show that $\nu$ is a Radon measure
on $\BbbR^r$.   Show that $\nu$ is an indefinite-integral measure over
Lebesgue measure iff $\nu_1$, $\nu_2$ are, and that in this case a linear
combination of
of Radon-Nikod\'ym derivatives of $\nu_1$ and $\nu_2$ is a
Radon-Nikod\'ym derivative of $\nu$.
%256A

\sqheader 256Xf Let $\nu$ be a Radon measure on $\BbbR^r$.   (i) Show
that there is a unique closed set $F\subseteq\BbbR^r$ such that, for
open sets $G\subseteq\BbbR^r$, $\nu G>0$ iff $G\cap F\ne\emptyset$.
($F$ is called the {\bf support} of $\nu$.)   (ii) Generally, a
set $A\subseteq\BbbR^r$ is called {\bf self-supporting} if $\nu^*(A\cap
G)>0$ whenever $G\subseteq\BbbR^r$ is an open set meeting $A$.   Show
that for every closed set $F\subseteq\BbbR^r$ there is a unique
self-supporting closed set $F'\subseteq F$ such that $\nu(F\setminus
F')=0$.
%256A

\sqheader 256Xg Show that a measure $\nu$ on $\Bbb R$ is a
Radon measure iff it is a Lebesgue-Stieltjes measure as
described in 114Xa.   Show that in this case $\nu$ is an
indefinite-integral measure over Lebesgue measure iff the function
$x\mapsto\nu[a,x]:[a,b]\to\Bbb R$ is
absolutely continuous whenever $a\le b$ in $\Bbb R$.
%256C

\spheader 256Xh Let $\nu$ be a Radon measure on $\BbbR^r$.   Let
$C_k$ be the space of continuous real-valued functions on
$\BbbR^r$ with bounded supports.   Show that for every
$\nu$-integrable function $f$ and every $\epsilon>0$ there is a
$g\in C_k$ such that $\int|f-g|d\nu\le\epsilon$.   \Hint{use
arguments from 242O, but in (a-i) of the proof there start with {\it
closed} intervals $I$.}
%256C

\spheader 256Xi Let $\nu$ be a Radon measure on $\BbbR^r$, and $\nu^*$ the
corresponding outer measure.   Show that
$\nu A=\inf\{\nu G:G\supseteq A$ is open$\}$ for every set
$A\subseteq\BbbR^r$.
%256D

\spheader 256Xj Let $\nu$, $\nuprime$ be two Radon measures on
$\BbbR^r$, and
suppose that $\nu I=\nuprime I$ for every half-open interval
$I\subseteq\BbbR^r$ (definition:  115Ab).   Show that $\nu=\nuprime$.
%256D

\spheader 256Xk Let $\nu$ be Cantor measure (256Hc).
(i) Show that if $C_n$ is the $n$th set used in the construction of the
Cantor set, so that $C_n$ consists of $2^n$ intervals of length
$3^{-n}$, then $\nu I=2^{-n}$ for each of the intervals $I$ composing
$C_n$.   (ii) Let $\lambda$ be the usual measure on $\{0,1\}^{\Bbb N}$
(254J).   Define $\phi:\{0,1\}^{\Bbb N}\to \Bbb R$ by setting
$\phi(x)=\bover23\sum_{n=0}^{\infty}3^{-n}x(n)$ for each
$x\in\{0,1\}^{\Bbb N}$.   Show that $\phi$ is a bijection between
$\{0,1\}^{\Bbb N}$ and $C$.   (iii) Show that if $\Cal B$ is the Borel
$\sigma$-algebra of $\Bbb R$, then $\{\phi^{-1}[E]:E\in\Cal B\}$ is
precisely the $\sigma$-algebra of subsets of $\{0,1\}^{\Bbb N}$
generated by the sets $\{x:x(n)=i\}$ for
$n\in\Bbb N$, $i\in\{0,1\}$.   (iv) Show that $\phi$ is an isomorphism
between $(\{0,1\}^{\Bbb N},\lambda)$ and $(C,\nu_C)$, where $\nu_C$ is
the subspace measure on $C$ induced by $\nu$.
%256H

\spheader 256Xl Let $\nu$ and $\nuprime$ be two Radon measures on
$\BbbR^r$.   Show that $\nuprime$ is an indefinite-integral measure over
$\nu$ iff $\nuprime E=0$ whenever $\nu E=0$, and in this case a function
$f$ is a Radon-Nikod\'ym derivative of $\nuprime$ with respect to $\nu$
iff $\int_Efd\nu=\nuprime E$ for every Borel set $E$.
%256J

\leader{256Y}{Further exercises (a)}
%\spheader 256Ya
Let $\nu$ be a Radon measure on $\BbbR^r$, and $X$ any subset of $\Bbb
R^r$;  let $\nu_X$ be the subspace measure on $X$ and $\Sigma_X$ its
domain, and give $X$ its
subspace topology.   Show that $\nu_X$ has the following
properties: (i) $\nu_X$ is complete and locally determined;   (ii) every
open subset of $X$ belongs to $\Sigma_X$;  (iii)
$\nu_XE=\sup\{\nu_XF:F\subseteq E$ is closed in $X\}$ for every
$E\in\Sigma_X$;   (iv) whenever $\Cal G$ is a non-empty
upwards-directed family of open subsets of $X$,
$\nu_X(\bigcup\Cal G)=\sup_{G\in\Cal G}\nu_XG$;
(v) every point of $X$ belongs to an open set of finite measure.
%256A

\spheader 256Yb Let $\nu$ be a Radon measure on $\BbbR^r$, with domain
$\Sigma$, and $f:\Bbb R^r\to\Bbb R$ a function.   Show that the following
are equiveridical:  (i) $f$ is $\Sigma$-measurable;
(ii) for every non-negligible set
$E\in\Sigma$ there is a non-negligible $F\in\Sigma$  such
that $F\subseteq E$ and $f\restr F$ is continuous;  (iii) for every set
$E\in\Sigma$, $\nu E=\sup_{K\in\Cal K_f,K\subseteq E}\nu K$, where
$\Cal K_f=\{K:K\subseteq\BbbR^r$ is compact, $f\restr K$ is continuous$\}$.
\Hint{for (ii)$\Rightarrow$(i), apply 215B(iv) to $\Cal K_f$.}
%256F

\spheader 256Yc Take $\nu$, $X$, $\nu_X$ and $\Sigma_X$ as in 256Ya.
Suppose that
$f:X\to\Bbb R$ is a function.   Show that $f$ is $\Sigma_X$-measurable
iff for every non-negligible measurable set $E\subseteq X$ there is a
non-negligible measurable $F\subseteq E$ such that $f\restr F$ is
continuous.
%256F

\spheader 256Yd\dvAnew{2011}(i) Let $\lambda$ be the usual measure on
$\{0,1\}^{\Bbb N}$.   Define $\psi:\{0,1\}^{\Bbb N}\to\{0,1\}^{\Bbb N}$ by
setting $\psi(x)(i)=x(i+1)$ for $x\in\{0,1\}^{\Bbb N}$ and $j\in\Bbb N$.
Show that $\psi$ is \imp.   (ii) Define $\theta:\Bbb R\to\Bbb R$ by setting
$\theta(t)=\fraction{3t}=3t-\lfloor 3t\rfloor$ for $t\in\Bbb R$.
Show that $\theta$ is \imp\ for Cantor measure as defined in 256Hc.
%256H 256Xk 

\spheader 256Ye Let $\sequencen{\nu_n}$ be a sequence of Radon measures
on $\BbbR^r$.   Show that there is a Radon measure $\nu$ on $\BbbR^r$
such that every $\nu_n$ is an indefinite-integral measure over $\nu$.
\Hint{find a sequence $\sequencen{\alpha_n}$ of
strictly positive numbers such that $\sum_{n=0}^{\infty}\alpha_n\nu_n
B(\tbf{0},k)<\infty$ for every $k$, and set
$\nu=\sum_{n=0}^{\infty}\alpha_n\nu_n$, using the idea of 256Xe.}
%256J, 256Xe, 256Xl

\spheader 256Yf A set
$G\subseteq\BbbR^{\Bbb N}$ is {\bf open} if for every $x\in G$ there are
$n\in\Bbb N$, $\delta>0$ such that

\Centerline{$\{y:y\in\BbbR^{\Bbb N},\,|y(i)-x(i)|<\delta$ for every
$i\le n\}\subseteq G$.}

\noindent The {\bf Borel $\sigma$-algebra} of $\BbbR^{\Bbb N}$ is the
$\sigma$-algebra $\Cal B$ of subsets of
$\BbbR^{\Bbb N}$ generated, in the sense of 111Gb, by the family
$\frak T$ of open sets.   (i) Show that $\frak T$ is a topology (2A3A).
(ii) Show that a filter $\Cal F$ on $\BbbR^{\Bbb N}$ converges to
$x\in\BbbR^{\Bbb N}$ iff $\pi_i[[\Cal F]]\to x(i)$ for every
$i\in\Bbb N$, where
$\pi_i(y)=y(i)$ for $i\in\Bbb N$, $y\in\BbbR^{\Bbb N}$.   (iii)
Show that $\Cal B$ is the $\sigma$-algebra generated by sets of the form
$\{x:x\in \BbbR^{\Bbb N},\,x(i)\le a\}$,
where $i$ runs over $\Bbb N$ and $a$ runs over $\Bbb R$.
(iv) Show that if $\alpha_i\ge 0$ for every $i\in\Bbb N$, then
$\{x:|x(i)|\le\alpha_i\Forall i\in\Bbb N\}$ is compact.
%2A3R
(v) Show that any open set in $\BbbR^{\Bbb N}$ is the union of a
sequence of closed sets.   \Hint{look at sets of the form
$\{x:q_i\le x(i)\le q'_i\Forall i\le n\}$, where $q_i$, $q'_i\in\Bbb Q$ for $i\le n$.}   (vi) Show that if $\nu_0$ is any probability  measure with domain
$\Cal B$, then its completion $\nu$ is inner regular with respect to the
compact sets, and therefore may be called a `Radon measure on
$\BbbR^{\Bbb N}$'.   \Hint{show that there are compact sets of measure
arbitrarily
close to $1$, and therefore that every open set, and every closed set,
includes a K$_{\sigma}$ set of the same measure.}
%256K  256Cprf
}%end of exercises


\endnotes{
\Notesheader{256} Radon measures on Euclidean spaces are very special,
and the results of this section do not give clear pointers to the
direction the theory takes when applied to other kinds of topological
space.   With the material here you could make
a stab at developing a theory of Radon measures on
complete separable metric spaces, provided you use 256Xa as the basis
for your definition of `locally finite'.   These are the spaces for
which a version of 256C is true.   (See 256Yf.)   But for
generalizations to other
types of topological space, and for the more
interesting parts of the theory on $\BbbR^r$, I must ask you to wait
for Volume 4.
My purpose in introducing Radon measures here is strictly
limited;  I wish only to give a basis for \S257 and \S271 sufficiently
solid not
to need later revision.   In fact I think that all we really need are
the Radon probability measures.

The chief technical difficulty in the definition of
`Radon measure' here lies
in the insistence on completeness.   It may well be that for
everything studied in this volume, it would be simpler to look at
locally finite measures with domain the algebra of Borel sets.
This would involve us in a number of circumlocutions when dealing with
Lebesgue measure itself and its derivates, since Lebesgue measure
is defined on a larger $\sigma$-algebra;  but the serious objection
arises in the more advanced theory, when non-Borel sets of
various kinds become central.   Since my aim in this book is to provide
secure foundations for the study of all aspects of measure theory, I ask
you to take a little extra trouble now in order to avoid the possibility
of having to re-work all your ideas later.
The extra trouble arises, for instance, in 256D, 256Xe and 256Xj;  since
different Radon measures are defined on different $\sigma$-algebras, we
have to check that two Radon measures which agree on the compact sets,
or on the open sets, have the same domains.   On the credit side, some
of the power of 256G arises from the fact that the Radon image measure
$\nu\phi^{-1}$ is defined on the whole $\sigma$-algebra
$\{F:\phi^{-1}[F]\in\dom(\nu)\}$, not just on the Borel sets.

The further technical point that Radon measures are expected to be
locally finite gives less difficulty;  its effect is that from most
points of view there is little difference between a general Radon
measure and a totally finite Radon measure.   The extra condition which
obviously has to be put into the hypotheses of such results as 256E and
256G is no burden on either intuition or memory.

In effect, we have two definitions of Radon measures on Euclidean
spaces:  they are the inner regular locally finite topological measures,
and they are also the completions of the locally finite Borel measures.
The equivalence of these definitions is Theorem 256C.   The latter
definition is the better adapted to 256K, and the former to 256G.    The
`inner regularity' of the basic definition refers to compact sets;
we also have forms of inner regularity with respect to closed sets
(256Bb) and K$_{\sigma}$ sets (256Bc), and a complementary notion of `outer
regularity' with respect to open sets (256Xi).

}%end of notes

\discrpage

\frfilename{mt257.tex} 
\versiondate{14.8.13} 
\copyrightdate{1995} 
 
\def\chaptername{Product measures} 
\def\sectionname{Convolutions of measures} 
 
\newsection{257} 
 
The ideas of this chapter can be brought together in a satisfying way in 
the theory of convolutions of Radon measures, 
which will be useful in \S272 and again in 
\S285.   I give just the definition (257A) and the central property 
(257B) of the convolution of totally finite Radon measures, with a few 
corollaries and a note on the relation between convolution of functions 
and convolution of measures (257F). 
 
\leader{257A}{Definition}  Let $r\ge 1$ be an 
integer and $\nu_1$, $\nu_2$ two totally finite Radon measures on 
$\Bbb R^r$.   Let $\lambda$ be the product measure on 
$\BbbR^r\times\BbbR^r$\cmmnt{;   then $\lambda$ also is a (totally 
finite) Radon measure, by 256K}.   Define 
$\phi:\BbbR^r\times\BbbR^r\to\BbbR^r$ by setting 
$\phi(x,y)=x+y$\cmmnt{;  then 
$\phi$ is continuous, therefore measurable in the sense of 
256G}.   The {\bf convolution} 
of $\nu_1$ and $\nu_2$, $\nu_1*\nu_2$, is the image measure 
$\lambda\phi^{-1}$;  \cmmnt{by 256G,} this is a Radon measure. 
 
Note that if $\nu_1$ and $\nu_2$ are Radon probability measures, then 
$\lambda$ and $\nu_1*\nu_2$ are also probability measures. 
 
\leader{257B}{Theorem} Let $r\ge 1$ be an integer, and $\nu_1$ and 
$\nu_2$ two totally finite Radon measures on $\BbbR^r$;  let 
$\nu=\nu_1*\nu_2$ be their convolution, and $\lambda$ their product on 
$\BbbR^r\times\BbbR^r$.   Then for any real-valued function $h$ defined 
on a subset of $\BbbR^r$, 
 
\Centerline{$\int h(x+y)\lambda(d(x,y))$ exists $=\int h(x)\nu(dx)$} 
 
\noindent if either integral is defined in $[-\infty,\infty]$. 
%will this work for $[-\infty,\infty]$-valued h?   
 
\proof{ Apply 235J with $J(x,y)=1$, $\phi(x,y)=x+y$ for all $x$, 
$y\in\BbbR^r$. 
}%end of proof of 257B 
 
\leader{257C}{Corollary} Let $r\ge 1$ be an integer, and $\nu_1$, 
$\nu_2$ two totally finite Radon measures on $\BbbR^r$;  let 
$\nu=\nu_1*\nu_2$ be their convolution, and $\lambda$ their product on 
$\BbbR^r\times\BbbR^r$;  write $\Lambda$ for the domain of $\lambda$. 
Let $h$ be a $\Lambda$-measurable function 
defined $\lambda$-almost everywhere in $\BbbR^r$.     Suppose that any 
one of the integrals 
 
\Centerline{$\iint|h(x+y)|\nu_1(dx)\nu_2(dy)$, 
\quad$\iint|h(x+y)|\nu_2(dy)\nu_1(dx)$, 
\quad$\int h(x+y)\lambda(d(x,y))$} 
 
\noindent exists and is finite.   Then $h$ is $\nu$-integrable and 
 
\Centerline{$\int h(x)\nu(dx)=\iint h(x+y)\nu_1(dx)\nu_2(dy) 
=\iint h(x+y)\nu_2(dy)\nu_1(dx)$.} 
 
\proof{ Put 257B together with Fubini's and Tonelli's theorems (252H). 
}%end of proof of 257C 
 
\vleader{48pt}{257D}{Corollary} If $\nu_1$ and $\nu_2$ are totally finite Radon 
measures on $\BbbR^r$, then $\nu_1*\nu_2=\nu_2*\nu_1$. 
 
\proof{ For any Borel set $E\subseteq\BbbR^r$, apply 257C to $h=\chi E$ 
to see that 
 
$$\eqalign{(\nu_1*\nu_2)(E) 
&=\iint\chi E(x+y)\nu_1(dx)\nu_2(dy) 
=\iint\chi E(x+y)\nu_2(dy)\nu_1(dx)\cr 
&=\iint\chi E(y+x)\nu_2(dy)\nu_1(dx) 
=(\nu_2*\nu_1)(E).\cr}$$ 
 
\noindent Thus $\nu_1*\nu_2$ and $\nu_2*\nu_1$ agree on the Borel sets 
of $\BbbR^r$;  because they are both Radon measures, they must be 
identical (256D). 
}%end of proof of 257D 
 
\leader{257E}{Corollary} If $\nu_1$, $\nu_2$ and $\nu_3$ are totally 
finite Radon measures on $\BbbR^r$, then 
$(\nu_1*\nu_2)*\nu_3=\nu_1*(\nu_2*\nu_3)$. 
 
\proof{ For any Borel set $E\subseteq\BbbR^r$, apply 257B to $h=\chi E$ 
to see that 
 
$$\eqalignno{((\nu_1*\nu_2)*\nu_3)(E) 
&=\iint\chi E(x+z)(\nu_1*\nu_2)(dx)\nu_3(dz)\cr 
&=\iiint\chi E(x+y+z)\nu_1(dx)\nu_2(dy)\nu_3(dz)\cr 
\noalign{\noindent (because $x\mapsto \chi E(x+z)$ is Borel measurable 
for every $z$)} 
&=\iint\chi E(x+y)\nu_1(dx)(\nu_2*\nu_3)(dy)\cr 
\noalign{\noindent (because $(x,y)\mapsto\chi E(x+y)$ is 
Borel measurable, so $y\mapsto\int\chi E(x+y)\nu_1(dx)$ is 
$(\nu_2*\nu_3)$-integrable)} 
&=(\nu_1*(\nu_2*\nu_3))(E).\cr}$$ 
 
\noindent Thus $(\nu_1*\nu_2)*\nu_3$ and $\nu_1*(\nu_2*\nu_3)$ agree on 
the Borel sets of $\BbbR^r$;  because they are both Radon measures, 
they must be identical. 
}%end of proof of 257E 
 
\leader{257F}{Theorem} Suppose that $\nu_1$ and $\nu_2$ are totally 
finite Radon measures on $\BbbR^r$ which are indefinite-integral 
measures over Lebesgue measure $\mu$.   Then $\nu_1*\nu_2$ also is an 
indefinite-integral measure over $\mu$;  if $f_1$ and $f_2$ are 
Radon-Nikod\'ym derivatives of $\nu_1$, $\nu_2$ respectively, then 
$f_1*f_2$ is a Radon-Nikod\'ym derivative of $\nu_1*\nu_2$. 
 
\proof{ By 255H/255L, $f_1*f_2$ is integrable with 
respect to $\mu$, 
with $\int f_1*f_2d\mu=1$, and of course $f_1*f_2$ is non-negative.   If 
$E\subseteq\BbbR^r$ is a Borel set, 
 
$$\eqalignno{\int_Ef_1*f_2d\mu 
&=\iint\chi E(x+y)f_1(x)f_2(y)\mu(dx)\mu(dy)\cr 
\noalign{\noindent (255G)} 
&=\iint\chi E(x+y)f_2(y)\nu_1(dx)\mu(dy)\cr 
\noalign{\noindent (because $x\mapsto\chi E(x+y)$ is Borel measurable)} 
&=\iint\chi E(x+y)\nu_1(dx)\nu_2(dy)\cr 
\noalign{\noindent (because $(x,y)\mapsto\chi E(x+y)$ is 
Borel measurable, so $y\mapsto\int\chi E(x+y)\nu_1(dx)$ 
is $\nu_2$-integrable)} 
&=(\nu_1*\nu_2)(E).\cr}$$ 
 
\noindent So $f_1*f_2$ is a Radon-Nikod\'ym derivative of $\nu$ with 
respect to $\mu$, by 256J. 
}%end of proof of 257F 
 
\exercises{\leader{257X}{Basic exercises $\pmb{>}$(a)} 
%\spheader 257Xa 
Let $r\ge 1$ be an integer.   Let $\delta_0$ be the 
Dirac measure on $\BbbR^r$ 
concentrated at $0$.   Show that $\delta_0$ is a Radon probability 
measure on $\BbbR^r$ and that 
$\delta_0*\nu=\nu$ for every totally finite Radon measure on $\BbbR^r$. 
%257A 
 
\spheader 257Xb Let $\mu$ and $\nu$ be totally finite Radon measures on 
$\BbbR^r$, and $E$ any set measured by their convolution $\mu*\nu$. 
Show that $\int\mu(E-y)\nu(dy)$ is defined 
in $[0,\infty]$ and equal to $(\mu*\nu)(E)$. 
%257B 
 
\spheader 257Xc Let $\nu_1,\ldots,\nu_n$ be totally finite Radon 
measures on $\BbbR^r$, and let $\nu$ be the convolution 
$\nu_1*\ldots*\nu_n$ (using 257E to see that such a bracketless 
expression is legitimate).   Show that 
 
\Centerline{$\int h(x)\nu(dx)=\int\ldots\int 
h(x_1+\ldots+x_n)\nu_1(dx_1)\ldots\nu_n(dx_n)$} 
 
\noindent for every $\nu$-integrable function $h$. 
%257E 
 
\spheader 257Xd Let $\nu_1$ and $\nu_2$ be totally finite Radon measures 
on $\BbbR^r$, with supports $F_1$, $F_2$ (256Xf).   Show that the 
support of $\nu_1*\nu_2$ is $\overline{\{x+y:x\in F_1,\,y\in F_2\}}$. 
%257E 
 
\sqheader 257Xe Let $\nu_1$ and $\nu_2$ be totally finite Radon measures 
on $\BbbR^r$, and suppose that $\nu_1$ has a Radon-Nikod\'ym derivative 
$f$ with respect 
to Lebesgue measure $\mu$.   Show that $\nu_1*\nu_2$ has a 
Radon-Nikod\'ym derivative $g$, 
where $g(x)=\int f(x-y)\nu_2(dy)$ for $\mu$-almost every $x\in\BbbR^r$. 
%257F 
 
\spheader 257Xf Suppose that $\nu_1$, $\nu_2$, $\nu_1'$ and $\nu_2'$ are 
totally finite Radon measures on $\BbbR^r$, and that $\nu_1'$, $\nu_2'$ 
are absolutely continuous with respect to $\nu_1$, $\nu_2$ respectively.   Show that 
$\nu_1'*\nu_2'$ is absolutely continuous with respect to $\nu_1*\nu_2$. 
%257F 
 
\leader{257Y}{Further exercises (a)} 
Let $M$ be the space of countably additive 
functionals defined on the algebra $\Cal B$ of Borel subsets of 
$\Bbb R$, with its norm $\|\nu\|=|\nu|(\Bbb R)$ (see 231Yh).   (i) Show 
that we have a unique bilinear operator $*:M\times M\to M$ such that 
$(\mu_1\restr\Cal B)*(\mu_2\restr\Cal B)=(\mu_1*\mu_2)\restr\Cal B$ for 
all totally finite Radon measures $\mu_1$, $\mu_2$ on $\Bbb R$.  (ii) 
Show that $*$ is commutative and associative.   (iii) Show that 
$\|\nu_1*\nu_2\|\le\|\nu_1\|\|\nu_2\|$ for all $\nu_1$, $\nu_2\in M$, so 
that $M$ is a Banach algebra under this multiplication.   (iv) Show that 
$M$ has a multiplicative identity.   (v) Show that $L^1(\mu)$ can be 
regarded as a closed subalgebra of $M$, where $\mu$ is Lebesgue measure 
on $\BbbR^r$ (cf.\ 255Xi). 
%257A 
 
\spheader 257Yb Let us say that a {\bf Radon 
measure on $\ocint{-\pi,\pi}$} is a 
complete measure $\nu$ on $\ocint{-\pi,\pi}$ 
such that (i) every Borel subset of 
$\ocint{-\pi,\pi}$ belongs to the domain $\Sigma$
of $\mu$ (ii) for every $E\in\Sigma$ there 
are Borel sets $E_1$, $E_2$ such that $E_1\subseteq E\subseteq E_2$ 
and $\nu(E_2\setminus E_1)=0$ (iii) every compact subset 
of $\ocint{-\pi,\pi}$ has finite measure.   Show that for any two 
totally finite Radon measures $\nu_1$, $\nu_2$ on $\ocint{-\pi,\pi}$ 
there is a unique totally finite Radon measure $\nu$ on 
$\ocint{-\pi,\pi}$ such that 
 
\Centerline{$\int h(x)\nu(dx)=\int h(x+_{2\pi}y)\nu_1(dx)\nu_2(dy)$} 
 
\noindent for every $\nu$-integrable function $h$, where $+_{2\pi}$ is 
defined as in 255Ma. 
}%end of exercises 
 
\endnotes{ 
\Notesheader{257} 
Of course convolution of functions and convolution of measures are very 
closely connected;  the obvious link being 257F, but the correspondence 
between 255G and 257B is also very marked.   In effect, they give us 
the same notion of convolution $u*v$ when $u$, $v$ are positive members 
of $L^1$ and $u*v$ is interpreted in $L^1$ rather than as a function 
(257Ya).   But 
we should have to go rather deeper than the arguments here to 
find ideas in the theory of convolution of measures to correspond to 
such results as 255K.   I will return to questions of this type in  
\S444 in Volume 4. 
 
All the theorems of this section can be extended to general abelian 
locally compact Hausdorff topological groups;  but for such generality 
we need much more advanced ideas (see \S444), and for the moment I leave 
only the 
suggestion in 257Yb that you should try to adapt the ideas here to 
$\ocint{-\pi,\pi}$ or $S^1$. 
}%end of comment 
 
\discrpage 
 
%\ifresultsonly\else
\frfilename{mt25conc.tex}
\versiondate{10.11.06}
\copyrightdate{2006}

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

\def\chaptername{Product measures}
\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

\leaveitout{2{}15 (product measures), referred to in {\smc Bogachev 07}, is
now \S251.}

\leaveitout{2{}51Id: {\smc Fremlin 00}, 251Id }%

\leader{2{}51N}{} Paragraph numbers in the second half of \S2{}51, referred
to in editions of Volumes 3 and 4 up to and including 2006, and in {\smc
Bogachev 07}, have been
changed, so that 2{}51M-2{}51S are now 251N-251T.
%2{}51M>251N 2{}51N>251O 2{}51P>251Q 2{}51R>251S 2{}51S>251T

\leaveitout{2{}52F: {\smc Fremlin 00}, 252F}%

\leader{2{}52Yf}{Exercise} This exercise, referred to in the first
edition of Volume 1, has been moved to 252Ym.
%upper integrals and outer measures of ordinate sets;
%133 notes

\leaveitout{\S2{}54 %254
usual measure on $\{0,1\}^I$ {\smc Fremlin \& Plebanek 03}}

\leaveitout{2{}54J: {\smc Fremlin 00}, {\smc Fremlin 03}
The usual measure on $\{0,1\}^A$ 254J}

\leaveitout{2{}54M:  {\smc Fremlin 00} Since every $F_n$, for $n\in L$, 
is determined by coordinates in $I\setminus\{i\}$, so is $F$.  254M}

\leaveitout{2{}54T:  {\smc Fremlin 00} Let $\Cal J$ be the family of those subsets $J$
of $I$ such that $F$ is determined by coordinates in $J$.   Then
$I\in\Cal J$ and $J\cap J'\in\Cal J$ for any $J$, $J'\in\Cal J$.   254T}

\leaveitout{2{}54U:  product of localizable not localizable;  {\smc
Bogachev 07};  254U.}

\leader{2{}54Yh}{Exercise} This exercise, referred to in the 2013
edition of Volume 4, has been moved to 254Ye.
%4{}16Yi

\bigskip

\noindent{\bf References}

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

\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

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\end



