% 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{mt37.tex} 
\versiondate{27.6.16} 
\copyrightdate{1996} 
 
\def\chaptername{Linear operators between function spaces} 
\def\sectionname{Introduction} 
 
\newchapter{37} 
 
As everywhere in functional analysis, the function spaces of measure 
theory cannot be properly understood without investigating linear 
operators between them.   In this chapter I have collected a number of 
results which rely on, or illuminate, the measure-theoretic aspects of 
the theory.   \S371 is devoted to a fundamental property of linear 
operators on $L$-spaces, if considered abstractly, that is, of 
$L^1$-spaces, if considered in the languages of Chapters 24 and 36, and to an 
introduction to the class $\Cal T$ of operators which are 
norm-decreasing for both $\|\,\|_1$ and $\|\,\|_{\infty}$.   This makes 
it possible to prove a version of Birkhoff's Ergodic Theorem for 
operators which need not be positive (372D).   In \S372 I give various 
forms of this theorem, for linear operators between function spaces, for 
measure-preserving Boolean homomorphisms between measure algebras, and 
for \imp\ functions between measure spaces, with an excursion into the 
theory of continued fractions.   In \S373 I make a fuller analysis of 
the class $\Cal T$, with a complete characterization of those $u$, $v$ 
such that $v=Tu$ for some $T\in\Cal T$.   Using this we can describe  
`rearrangement-invariant' function spaces and extended Fatou norms 
(\S374).   Returning to ideas left on one side in \S\S364 and 368, I 
investigate positive linear operators defined on $L^0$ spaces (\S375). 
In the penultimate section of the chapter (\S376),  
I look at operators which 
can be defined in terms of kernels on product spaces.   Finally, in \S377, 
I examine the function spaces of reduced products, projective limits 
and inductive limits of probability algebras. 
 
\discrpage 
 
\frfilename{mt371.tex}
\versiondate{13.12.06}
\copyrightdate{1995}

\def\chaptername{Linear operators between function spaces}
\def\sectionname{The Chacon-Krengel theorem}

\newsection{371}

The first topic I wish to treat is a remarkable property of $L$-spaces:
if $U$ and $V$ are $L$-spaces, then every continuous linear operator
$T:U\to V$ is order-bounded, and $\||T|\|=\|T\|$ (371D).   This
generalizes in
various ways to other $V$ (371B, 371C).   I apply the result to a
special type of operator between $M^{1,0}$ spaces which will be
conspicuous in the next section (371F-371H).

\leader{371A}{Lemma} Let $U$ be an $L$-space, $V$ a Banach lattice and
$T:U\to V$ a bounded linear operator.   Take $u\ge 0$ in $U$ and set

\Centerline{$B=\{\sum_{i=0}^n|Tu_i|:u_0,\ldots,u_n\in U^+,\,
\sum_{i=0}^nu_i=u\}\subseteq V^+$.}

\noindent Then $B$ is upwards-directed and
$\sup_{v\in B}\|v\|\le\|T\|\|u\|$.

\proof{{\bf (a)} Suppose that $v$, $v'\in B$.   Then we have
$u_0,\ldots,u_m,u'_0,\ldots,u'_n\in U^+$ such that
$\sum_{i=0}^mu_i=\sum_{j=0}^nu'_j=u$, $v=\sum_{i=0}^m|Tu_i|$ and
$v'=\sum_{j=0}^n|Tu'_j|$.   Now there are $v_{ij}\ge 0$ in $U$, for
$i\le m$ and $j\le n$, such that $u_i=\sum_{j=0}^nv_{ij}$ for $i\le m$
and $u'_j=\sum_{i=0}^mv_{ij}$ for $j\le n$ (352Fd).   We have
$u=\sum_{i=0}^m\sum_{j=0}^nv_{ij}$, so that
$v''=\sum_{i=0}^m\sum_{j=0}^n|Tv_{ij}|\in B$.   But

\Centerline{$v=\sum_{i=0}^m|Tu_i|=\sum_{i=0}^m|T(\sum_{j=0}^nv_{ij})|
\le\sum_{i=0}^m\sum_{j=0}^m|Tv_{ij}|=v''$,}

\noindent and similarly $v'\le v''$.   As $v$ and $v'$ are arbitrary,
$B$ is upwards-directed.

\medskip

{\bf (b)} The other part is easy.   If $v\in B$ is expressed as
$\sum_{i=0}^n|Tu_i|$ where $u_i\ge 0$ for every $i$
and $\sum_{i=0}^nu_i=u$, then

\Centerline{$\|v\|\le\sum_{i=0}^n\|Tu_i\|\le\|T\|\sum_{i=0}^n\|u_i\|
=\|T\|\|u\|$}

\noindent because $U$ is an $L$-space.
}%end of proof of 371A

\leader{371B}{Theorem} Let $U$ be an
$L$-space and $V$ a Dedekind complete Banach
lattice $U$ with a Fatou norm.   Then the Riesz space
$\eurm L^{\sim}(U;V)=\eurm L^{\times}(U;V)$ is a closed linear
subspace of the Banach space $\eurm B(U;V)$ and is in itself a Banach
lattice with a Fatou norm.

\proof{{\bf (a)} I start by noting that
$\eurm L^{\sim}(U;V)=\eurm L^{\times}(U;V)\subseteq\eurm B(U;V)$ just
because $V$ has a Riesz norm
and $U$ is a Banach lattice with an order-continuous norm (355Kb, 355C).

\medskip

{\bf (b)} The first new step is to check that $\||T|\|\le\|T\|$ for
any $T\in\eurm L^{\sim}(U;V)$.   \Prf\ Start with any $u\in U^+$.
Set

\Centerline{$B=\{\sum_{i=0}^n|Tu_i|:u_0,\ldots,u_n\in U^+,\,
\sum_{i=0}^nu_i=u\}\subseteq V^+$,}

\noindent as in 371A.   If $u_0,\ldots,u_n\ge 0$ are such that
$\sum_{i=0}^nu_i=u$, then $|Tu_i|\le|T|u_i$ for each $i$, so that
$\sum_{i=0}^n|Tu_i|\le\sum_{i=0}^n|T|u_i=|T|u$;  thus $B$ is bounded
above by $|T|u$ and $\sup B\le|T|u$.   On the other hand, if $|v|\le u$
in $U$, then $v^++v^-+(u-|v|)=u$, so $|Tv^+|+|Tv^-|+|T(u-|v|)|\in B$
and

\Centerline{$|Tv|=|Tv^++Tv^-|\le|Tv^+|+|Tv^-|\le\sup B$.}

\noindent As $v$ is arbitrary, $|T|u\le\sup B$ and $|T|u=\sup B$.
Consequently

\Centerline{$\||T|u\|\le\|\sup B\|=\sup_{w\in B}\|w\|\le\|T\|\|u\|$}

\noindent because $V$ has a Fatou norm and $B$ is upwards-directed.

For general $u\in U$,

\Centerline{$\||T|u\|\le\||T||u|\|\le\|T\|\||u|\|=\|T\|\|u\|$.}

\noindent This shows that $\||T|\|\le\|T\|$.  \Qed

\medskip

{\bf (c)} Now if $|S|\le|T|$ in $\eurm L^{\sim}(U;V)$, and $u\in U$,
we must have

\Centerline{$\|Su\|\le\||S||u|\|\le\||T||u|\|\le\||T|\|\||u|\|
\le\|T\|\|u\|$;}

\noindent as $u$ is arbitrary, $\|S\|\le\|T\|$.   This shows that the
norm of $\eurm L^{\sim}(U;V)$, inherited from $\eurm B(U;V)$, is a
Riesz norm.

\medskip

{\bf (d)} Suppose next that $T\in\eurm B(U;V)$ belongs to the
norm-closure of $\eurm L^{\sim}(U;V)$.   For each $n\in\Bbb N$ choose
$T_n\in\eurm L^{\sim}(U;V)$ such that $\|T-T_n\|\le 2^{-n}$.   Set
$S_n=|T_{n+1}-T_n|\in\eurm L^{\sim}(U;V)$ for each $n$.   Then

\Centerline{$\|S_n\|=\|T_{n+1}-T_n\|\le 3\cdot 2^{-n-1}$}

\noindent for each $n$, so $S=\sum_{n=0}^{\infty}S_n$ is defined in the
Banach space $\eurm B(U;V)$.   But if $u\in U^+$, we surely have

\Centerline{$Su=\sum_{n=0}^{\infty}S_nu\ge 0$}

\noindent in $V$.   Moreover, if $u\in U^+$ and $|v|\le u$, then for
any $n\in\Bbb N$

\Centerline{$|T_{n+1}v-T_0v|
=|\sum_{i=0}^n(T_{i+1}-T_i)v|
\le\sum_{i=0}^nS_iu
\le Su$,}

\noindent and $T_0v-Su\le T_{n+1}v\le T_0v+Su$;  letting $n\to\infty$,
we see that

\Centerline{$-|T_0|u-Su\le T_0v-Su\le Tv\le T_0v+Su\le|T_0|u+Su$.}

\noindent So $|Tv|\le|T_0|u+Su$ whenever $|v|\le u$.   As
$u$ is arbitrary, $T\in\eurm L^{\sim}(U;V)$.

This shows that $\eurm L^{\sim}(U;V)$ is closed in $\eurm B(U;V)$
and is therefore a Banach space in its own right;  putting this together
with (b), we see that it is a Banach lattice.

\medskip

{\bf (e)} Finally, the norm of $\eurm L^{\sim}(U;V)$ is a Fatou norm.
\Prf\ Let $A\subseteq\eurm L^{\sim}(U;V)^+$ be a non-empty,
upwards-directed set with supremum $T_0\in\eurm L^{\sim}(U;V)$.   For
any $u\in U$,

\Centerline{$\|T_0u\|=\||T_0u|\|\le\|T_0|u|\|=\|\sup_{T\in A}T|u|\|$}

\noindent by 355Ed.   But $\{T|u|:T\in A\}$ is upwards-directed and
the norm of $V$ is a Fatou norm, so

\Centerline{$\|T_0u\|\le\sup_{T\in A}\|T|u|\|\le\sup_{T\in
A}\|T\|\|u\|$.}

\noindent As $u$ is arbitrary, $\|T_0\|\le\sup_{T\in A}\|T\|$.   As $A$
is arbitrary, the norm of $\eurm L^{\sim}(U;V)$ is Fatou.  \Qed
}%end of proof of 371B

\leader{371C}{Theorem} Let $U$ be an $L$-space and $V$ a Dedekind
complete Banach
lattice with a Fatou norm and the Levi property.   Then $\eurm
B(U;V)=\eurm L^{\sim}(U;V)=\eurm L^{\times}(U;V)$ is a Dedekind complete
Banach lattice with a Fatou norm
and the Levi property.   In particular, $|T|$ is defined and
$\||T|\|=\|T\|$ for every $T\in\eurm B(U;V)$.

\proof{{\bf (a)} Let $T:U\to V$ be any bounded linear operator.   Then
$T\in\eurm L^{\sim}(U;V)$.   \Prf\ Take any $u\ge 0$ in $U$.   Set

\Centerline{$B=\{\sum_{i=0}^n|Tu_i|:u_0,\ldots,u_n\in U^+,\,
\sum_{i=0}^nu_i=u\}\subseteq V^+$}

\noindent as in 371A.   Then 371A tells us that $B$ is
upwards-directed and norm-bounded.   Because $V$ has the Levi property,
$B$ is bounded above.   But just as in part (b) of the proof of 371B,
any upper bound of $B$ is also an upper bound of $\{Tv:|v|\le u\}$.   As
$u$ is arbitrary, $T\in\eurm L^{\sim}(U;V)$.
\Qed

\medskip

{\bf (b)} Accordingly $\eurm L^{\sim}(U;V)=\eurm B(U;V)$.   By 371B,
this is a Banach lattice with a Fatou norm, and equal to $\eurm
L^{\times}(U;V)$.   To see that it also has
the Levi property, let $A\subseteq\eurm L^{\sim}(U;V)$ be any
non-empty norm-bounded upwards-directed set.   For $u\in U^+$,
$\{Tu:T\in A\}$ is non-empty, norm-bounded and upwards-directed in $V$,
so is bounded above in $V$.  By 355Ed, $A$ is bounded above in
$\eurm L^{\sim}(U;V)$.
}%end of proof of 371C

\leader{371D}{Corollary} Let $U$ and $V$ be $L$-spaces.   Then
$\eurm L^{\sim}(U;V)=\eurm L^{\times}(U;V)=\eurm B(U;V)$ is a Dedekind
complete Banach lattice with a Fatou norm and the Levi property.

\cmmnt{
\leader{371E}{Remarks} Note that both these theorems show that
$\eurm L^{\sim}(U;V)$ is a Banach lattice with properties similar to
those of $V$ whenever $U$ is an $L$-space.   They can therefore be
applied repeatedly, to give facts about
$\eurm L^{\sim}(U_1;\eurm L^{\sim}(U_2;V))$ where $U_1$, $U_2$ are
$L$-spaces and $V$ is a Banach
lattice, for instance.   I hope that this formula will recall some of
those in the theory of bilinear operators and tensor products (see
253Xa-253Xb).
}%end of comment

\leader{371F}{The class \dvrocolon{$\Cal T^{(0)}$}}\cmmnt{ For the
sake of applications in the next section, I introduce now a class of
operators of great intrinsic interest.

\medskip

\noindent}{\bf Definition} Let $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$
be measure algebras.   \cmmnt{Recall that $M^{1,0}(\frak A,\bar\mu)$
is the space of those $u\in L^1(\frak A,\bar\mu)+L^{\infty}(\frak A)$
such that $\bar\mu\Bvalue{|u|>\alpha}<\infty$ for every $\alpha>0$
(366F-366G, 369P).}
Write $\Cal T^{(0)}=\Cal T^{(0)}_{\bar\mu,\bar\nu}$ for the set of all
linear operators
$T:M^{1,0}(\frak A,\bar\mu)\to M^{1,0}(\frak B,\bar\nu)$ such that
$Tu\in L^1(\frak B,\bar\nu)$ and
$\|Tu\|_1\le\|u\|_1$ for every
$u\in L^1(\frak A,\bar\mu)$, $Tu\in L^{\infty}(\frak B)$ and
$\|Tu\|_{\infty}\le\|u\|_{\infty}$ for
every $u\in L^{\infty}(\frak A)\cap M^{1,0}(\frak A,\bar\mu)$.

\leader{371G}{Proposition} Let $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ be measure algebras.

(a) $\Cal T^{(0)}=\Cal T^{(0)}_{\bar\mu,\bar\nu}$
is a convex set in the unit ball of
$\eurm B(M^{1,0}(\frak A,\bar\mu);M^{1,0}(\frak B,\bar\nu))$.   If
$T_0:L^1(\frak A,\bar\mu)\to L^1(\frak B,\bar\nu)$ is a linear operator
of norm at most $1$, and $T_0u\in L^{\infty}(\frak B)$ and
$\|T_0u\|_{\infty}\le\|u\|_{\infty}$ for
every $u\in L^1(\frak A,\bar\mu)\cap L^{\infty}(\frak A)$, then $T_0$
has a unique extension to a member of $\Cal T^{(0)}$.

(b) If $T\in\Cal T^{(0)}$ then $T$ is order-bounded and $|T|$, taken in

\Centerline{$\eurm L^{\sim}(M^{1,0}(\frak A,\bar\mu);
  M^{1,0}(\frak B,\bar\nu))
=\eurm L^{\times}(M^{1,0}(\frak A,\bar\mu);M^{1,0}(\frak B,\bar\nu))$,}

\noindent also belongs to $\Cal T^{(0)}$.

(c) If $T\in\Cal T^{(0)}$ then $\|Tu\|_{1,\infty}\le\|u\|_{1,\infty}$
for every $u\in M^{1,0}(\frak A,\bar\mu)$.

(d) If $T\in\Cal T^{(0)}$, $p\in\coint{1,\infty}$ and
$w\in L^p(\frak A,\bar\mu)$ then
$Tw\in L^p(\frak B,\bar\nu)$ and $\|Tw\|_p\le\|w\|_p$.

(e) If $(\frak C,\bar\lambda)$ is another measure algebra then
$ST\in\Cal T^{(0)}_{\bar\mu,\bar\lambda}$ whenever
$T\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$ and
$S\in\Cal T^{(0)}_{\bar\nu,\bar\lambda}$.

\proof{ I write $M^{1,0}_{\bar\mu}$, $L^p_{\bar\nu}$ for
$M^{1,0}_{\bar\mu}$, $L^p(\frak B,\bar\nu)$, etc.

\medskip

{\bf (a)(i)} If $T\in\Cal T^{(0)}$ and $u\in M^{1,0}_{\bar\mu}$ then
there are $v\in L^1_{\bar\mu}$,
$w\in L^{\infty}_{\bar\mu}$ such that $u=v+w$ and
$\|v\|_1+\|w\|_{\infty}=\|u\|_{1,\infty}$ (369Ob);  so that

\Centerline{$\|Tu\|_{1,\infty}\le\|Tv\|_1+\|Tw\|_{\infty}
\le\|v\|_1+\|w\|_{\infty}\le\|u\|_{1,\infty}$.}

\noindent As $u$ is arbitrary, $T$ is in the unit ball of
$\eurm B(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\nu})$.

\medskip

\quad{\bf (ii)} Because the unit balls of
$\eurm B(L^1_{\bar\mu};L^1_{\bar\nu})$ and
$\eurm B(L^{\infty}_{\bar\mu};L^{\infty}_{\bar\nu})$ are convex, so is
$\Cal T^{(0)}$.

\medskip

\quad{\bf (iii)} Now suppose that $T_0:L^1_{\bar\mu}\to L^1_{\bar\nu}$
is a linear operator of norm at most $1$ such that
$\|T_0u\|_{\infty}\le\|u\|_{\infty}$ for every
$u\in L^1_{\bar\mu}\cap L^{\infty}_{\bar\mu}$.   By the argument of (i),
$T_0$ is a bounded operator for
the $\|\,\|_{1,\infty}$ norms;  since $L^1_{\bar\mu}$ is dense in
$M^{1,0}_{\bar\mu}$ (369Pc), $T_0$ has a unique extension to a
bounded linear operator $T:M^{1,0}_{\bar\mu}\to M^{1,0}_{\bar\nu}$.   Of
course $\|Tu\|_1=\|T_0u\|_1\le\|u\|_1$ for every $u\in L^1_{\bar\mu}$.

Now suppose that $u\in L^{\infty}_{\bar\mu}\cap M^{1,0}_{\bar\mu}$;  set
$\gamma=\|u\|_{\infty}$.   Let $\epsilon>0$, and set

\Centerline{$v=(u^+-\epsilon\chi 1)^+-(u^--\epsilon\chi 1)^+$;}

\noindent then $|v|\le|u|$ and $\|u-v\|_{\infty}\le\epsilon$ and
$v\in L^1_{\bar\mu}\cap L^{\infty}_{\bar\mu}$.   Accordingly

\Centerline{$\|Tu-Tv\|_{1,\infty}\le\|u-v\|_{1,\infty}\le\epsilon$,
\quad$\|Tv\|_{\infty}=\|T_0v\|_{\infty}
\le\|v\|_{\infty}\le\gamma$.}
\noindent So if we set $w=(|Tu-Tv|-\epsilon\chi 1)^+\in L^1_{\bar\nu}$,
$\|w\|_1\le\epsilon$;  while

\Centerline{$|Tu|\le|Tv|+w+\epsilon\chi 1\le(\gamma+\epsilon)\chi 1+w$,}

\noindent so

\Centerline{$\|(|Tu|-(\gamma+\epsilon)\chi 1)^+\|_1
\le\|w\|_1\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $|Tu|\le\gamma\chi 1$, that
is, $\|Tu\|_{\infty}\le\|u\|_{\infty}$.   As $u$ is arbitrary,
$T\in\Cal T^{(0)}$.

\medskip

{\bf (b)} Because $M^{1,0}_{\bar\mu}$ has an order-continuous norm
(369Pb), $\eurm L^{\sim}(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\nu})
=\eurm L^{\times}(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\nu})$ (355Kb).   Take
any $T\in\Cal T^{(0)}$ and consider
$T_0=T\restr L^1_{\bar\mu}:L^1_{\bar\mu}\to L^1_{\bar\nu}$.   This is an
operator of norm at most $1$.   By
371D, $T_0$ is order-bounded, and $\||T_0|\|\le 1$, where $|T_0|$
is taken in $\eurm L^{\sim}(L^1_{\bar\mu};L^1_{\bar\mu})
=\eurm B(L^1_{\bar\mu};L^1_{\bar\nu})$.   Now if
$u\in L^1_{\bar\mu}\cap L^{\infty}_{\bar\nu}$,

\Centerline{$||T_0|u|\le|T_0||u|
=\sup_{|u'|\le|u|}|T_0u'|\le\|u\|_{\infty}\chi 1$,}

\noindent so $\||T_0|u\|_{\infty}\le\|u\|_{\infty}$.   By (a), there is
a unique $S\in\Cal T^{(0)}$ extending $|T_0|$.   Now $Su^+\ge 0$ for
every $u\in L^1_{\bar\mu}$, so $Su^+\ge 0$ for every
$u\in M^{1,0}_{\bar\mu}$ (since the function
$u\mapsto (Su^+)^+-Su^+:M^{1,0}_{\bar\mu}\to M^{1,0}_{\bar\nu}$ is
continuous and zero on the dense set $L^1_{\bar\mu}$), that is, $S$ is a
positive operator;  also $S|u|\ge|Tu|$ for every $u\in L^1_{\bar\mu}$,
so $Sv\ge S|u|\ge|Tu|$ whenever $u$, $v\in M^{1,0}_{\bar\mu}$ and
$|u|\le v$.   This means
that $T:M^{1,0}_{\bar\mu}\to M^{1,0}_{\bar\nu}$ is order-bounded.
Because $M^{1,0}_{\bar\nu}$ is
Dedekind complete (366Ga), $|T|$ is defined in
$\eurm L^{\sim}(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\mu})$.

If $v\ge 0$ in $L^1_{\bar\mu}$, then

\Centerline{$|T|v=\sup_{|u|\le v}Tu=\sup_{|u|\le v}T_0u=|T_0|v=Sv$.}

\noindent Thus $|T|$ agrees with $S$ on $L^1_{\bar\mu}$.   Because
$M^{1,0}_{\bar\mu}$ is a
Banach lattice (or otherwise), $|T|$ is a bounded operator, therefore
continuous (2A4Fc), so $|T|=S\in\Cal T^{(0)}$, which is what we needed
to know.

\medskip

{\bf (c)} We can express $u$ as $v+w$ where
$\|v\|_1+\|w\|_{\infty}=\|u\|_{1,\infty}$;  now
$w=u-v\in M^{1,0}_{\bar\mu}$, so we can speak of $Tw$, and

\Centerline{$\|Tu\|_{1,\infty}=\|Tv+Tw\|_{1,\infty}
\le\|Tv\|_1+\|Tw\|_{\infty}\le\|v\|_1+\|w\|_{\infty}=\|u\|_{1,\infty}$,}

\noindent as required.

\medskip

{\bf (d)} This can be thought of as a generalization of 244M. 
We need to revisit the proof of Jensen's inequality in 
233H-233J.  %233H 233I 233J

\medskip

\quad{\bf (i)} Suppose that $T$, $p$, $w$ are as described, and that in
addition $T$ is positive.   As in the proof of 244M, 
the function $t\mapsto|t|^p$ is convex, 
so we can find families $\langle\beta_q\rangle_{q\in\Bbb Q}$,
$\langle\gamma_q\rangle_{q\in\Bbb Q}$ of real numbers such that
$|t|^p=\sup_{q\in\Bbb Q}\beta_q+\gamma_q(t-q)$ for every $t\in\Bbb R$
(233Hb).   Then
$|u|^p=\sup_{q\in\Bbb Q}\beta_q\chi 1+\gamma_q(u-q\chi 1)$ for every
$u\in L^0$.   (The easiest way to check this is perhaps to think of
$L^0$ as a quotient of a space of functions, as in 364C;  it is also a
consequence of 364Xg(iii).)   We know that
$|w|^p\in L^1_{\bar\mu}$, so we may speak of $T(|w|^p)$;  while
$w\in M^{1,0}_{\bar\mu}$ (366Ga), so we may speak of $Tw$.

For any $q\in\Bbb Q$, $0^p\ge\beta_q-q\gamma_q$, that is,
$q\gamma_q-\beta_q\ge 0$, while
$\gamma_qw-|w|^p\le(q\gamma_q-\beta_q)\chi 1$ and
$\|(\gamma_qw-|w|^p)^+\|_{\infty}\le q\gamma_q-\beta_q$.   Now this
means that

$$\eqalign{T(\gamma_qw-|w|^p)
&\le T(\gamma_qw-|w|^p)^+
\le\|T(\gamma_qw-|w|^p)^+\|_{\infty}\chi 1\cr
&\le\|(\gamma_qw-|w|^p)^+\|_{\infty}\chi 1
\le (q\gamma_q-\beta_q)\chi 1.\cr}$$

\noindent Turning this round again,

\Centerline{$\beta_q\chi 1+\gamma_q(Tw-q\chi 1)\le T(|w|^p)$.}

\noindent Taking the supremum over $q$, $|Tw|^p\le T(|w|^p)$, so that
$\int|Tw|^p\le\int|w|^p$ (because $\|Tv\|_1\le\|v\|_1$ for every
$v\in L^1$).   Thus $Tw\in L^p$ and $\|Tw\|_p\le\|w\|_p$.

\medskip

\quad{\bf (ii)} For a general $T\in\Cal T^{(0)}$, we have
$|T|\in\Cal T^{(0)}$, by
(b), and $|Tw|\le|T||w|$, so that $\|Tw\|_p\le\||T||w|\|_p\le\|w\|_p$,
as required.

\medskip

{\bf (e)} This is elementary, because

\Centerline{$\|STu\|_1\le\|Tu\|_1\le\|u\|_1$,
\quad$\|STv\|_{\infty}\le\|Tu\|_{\infty}\le\|u\|_{\infty}$}

\noindent whenever $u\in L^1_{\bar\mu}$ and
$v\in L^{\infty}_{\bar\mu}\cap M^{1,0}_{\bar\mu}$.
}%end of proof of 371G

\cmmnt{
\leader{371H}{Remark} In the context of 366H,
$T_{\pi}\restr M^{1,0}_{\bar\mu}\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$,
while $P_{\pi}\in\Cal T^{(0)}_{\bar\nu,\bar\mu}$.   Thus 366H(a-iv) and
366H(b-iii) are special cases of 371Gd.
}

\exercises{\leader{371X}{Basic exercises $\pmb{>}$(a)}
%\spheader 371Xa
Let $U$ be an $L$-space, $V$ a Banach lattice with an order-continuous
norm and $T:U\to V$ a bounded linear operator.   Let $B$ be the unit
ball of $U$.   Show that $|T|[B]\subseteq\overline{T[B]}$.
%371A

\spheader 371Xb Let $U$ and $V$ be Banach spaces.  (i) Show that the
space $\eurm K(U;V)$ of compact linear operators from $U$ to
$V$ (definition:  3A5La) is a closed
linear subspace of $\eurm B(U;V)$.   (ii) Show that if $U$ is an
$L$-space and $V$ is a Banach lattice with an order-continuous norm,
then $\eurm K(U;V)$ is a norm-closed Riesz subspace of
$\eurm L^{\sim}(U;V)$.  (See {\smc Krengel 63}.)
%371A, 371Xa

\spheader 371Xc Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
and set $U=L^1(\frak A,\bar\mu)$.   Show that
$\eurm L^{\sim}(U;U)=\eurm B(U;U)$ is a Banach lattice with a Fatou norm
and the Levi property.
Show that its norm is order-continuous iff $\frak A$ is finite.
\Hint{consider operators $u\mapsto u\times\chi a$, where $a\in\frak A$.}
%371C

\sqheader 371Xd Let $U$ be a Banach lattice, and $V$ a Dedekind complete
$M$-space.   Show that $\eurm L^{\sim}(U;V)=\eurm B(U;V)$ is a Banach
lattice with a Fatou norm and the Levi property.

\spheader 371Xe Let $U$ and $V$ be Riesz spaces, of which $V$ is
Dedekind complete, and let $T\in\eurm L^{\sim}(U;V)$.   Define
$T'\in\eurm L^{\sim}(V^{\sim};U^{\sim})$ by writing $T'(h)=hT$ for
$h\in V^{\sim}$.   (i) Show that $|T|'\ge|T'|$ in
$\eurm L^{\sim}(V^{\sim};U^{\sim})$.   (ii) Show that $|T|'h=|T'|h$ for
every $h\in V^{\times}$.   ({\it Hint\/}:  show that if $u\in U^+$ and
$h\in(V^{\times})^+$ then $(|T'|h)(u)$ and $h(|T|u)$ are both equal to
$\sup\{\sum_{i=0}^ng_i(Tu_i):|g_i|\le h,\,u_i\ge 0,
\,\sum_{i=0}^nu_i=u\}$.)
%371C

\sqheader 371Xf Using 371D, but nothing about uniformly integrable sets
beyond the definition (354P), show that if $U$ and $V$ are $L$-spaces,
$A\subseteq U$ is uniformly integrable in $U$, and $T:U\to V$ is a
bounded linear operator, then $T[A]$ is uniformly integrable in $V$.
%371D

\leader{371Y}{Further exercises (a)}
%\spheader 371Ya
Let $U$ and $V$ be Banach spaces.   (i) Show that the
space $\eurm K_w(U;V)$ of weakly compact linear operators from
$U$ to $V$ (definition:  3A5Lb) is a closed linear subspace of
$\eurm B(U;V)$.   (ii) Show
that if $U$ is an $L$-space and $V$ is a Banach lattice with an
order-continuous norm, then $\eurm K_w(U;V)$ is a norm-closed Riesz
subspace of $\eurm L^{\sim}(U;V)$.
%371A, 371Xa

\spheader 371Yb Let $(\frak A,\bar\mu)$ be a measure algebra, $U$ a
Banach space, and $T:L^1(\frak A,\bar\mu)\to U$ a bounded linear
operator.   Show that $T$ is a compact linear operator iff
$\{\Bover1{\bar\mu a}T(\chi a):a\in\frak A,\,0<\bar\mu a<\infty\}$ is
relatively compact in $U$.
%could be anywhere;  needs only that unit ball is
%closed convex hull of these

\spheader 371Yc Let $(\frak A,\bar\mu)$ be a probability algebra, and
set $L^1=L^1(\frak A,\bar\mu)$.   Let $\sequencen{a_n}$ be a
stochastically
independent sequence of elements of $\frak A$ of measure $\bover12$, and
define $T:L^1\to\BbbR^{\Bbb N}$ by setting
$Tu(n)=\int u-2\int_{a_n}u$ for each $n$.   Show that
$T\in\eurm B(L^1;\pmb{c}_0)\setminus\eurm L^{\sim}(L^1;\pmb{c}_0)$,
where $\pmb{c}_0$
is the Banach lattice of sequences converging to $0$.   (See
272Ye\formerly{2{}72Yd}.)
%371B

\spheader 371Yd Regarding $T$ of 371Yc as a map from $L^1$ to
$\ell^{\infty}$, show that $|T'|\ne|T|'$ in 
$\eurm L^{\sim}((\ell^{\infty})^*,L^{\infty}(\frak A))$.
%371Yc, 371B

\spheader 371Ye(i) In $\ell^2$ define $e_i$ by setting $e_i(i)=1$,
$e_i(j)=0$ if $j\ne i$.   Show that if
$T\in\eurm L^{\sim}(\ell^2;\ell^2)$ then
$\innerprod{|T|e_i}{e_j}=|\innerprod{Te_i}{e_j}|$ for all $i$,
$j\in\Bbb N$.   (ii) Show that for each $n\in\Bbb N$ there is an
orthogonal
$(2^n\times 2^n)$-matrix $\tbf{A}_n$ such that every coefficient of
$\tbf{A}_n$ has modulus $2^{-n/2}$.   ({\it Hint\/}:  $\tbf{A}_{n+1}
=\Bover1{\sqrt 2}\Matrix{\tbf{A}_n&\tbf{A}_n\\-\tbf{A}_n&\tbf{A}_n}$.)
(iii) Show that there is a linear isometry $S:\ell^2\to\ell^2$ such that
$|\innerprod{Se_i}{e_j}|=2^{-n/2}$ if $2^n\le i,\,j<2^{n+1}$.   (iv)
Show that $S\notin\eurm L^{\sim}(\ell^2;\ell^2)$.
%371C

}%end of exercises

\endnotes{
\Notesheader{371}
The `Chacon-Krengel theorem', properly speaking 
({\smc Chacon \& Krengel 64}), is 371D in the case in which 
$U=L^1(\mu)$, $V=L^1(\nu)$;
of course no new ideas are required in the generalizations here, which I
have copied from {\smc Fremlin 74a}.

Anyone with a training in functional analysis will automatically seek to
investigate properties of operators $T:U\to V$ in terms of properties of
their adjoints $T':V^*\to U^*$, as in 371Xe and 371Yd.   When $U$ is an
$L$-space, then $U^*$ is a Dedekind complete $M$-space, and it is easy
to see that this forces $T'$ to be order-bounded, for any Banach lattice
$V$ (371Xd).   But since no important $L$-space is reflexive, this
approach cannot reach 371B-371D without a new idea of some kind.   It
can however be adapted to the special case in 371Gb
({\smc Dunford \& Schwartz 57}, VIII.6.4).

In fact the results of 371B-371C are characteristic of
$L$-spaces ({\smc Fremlin 74b}).   To see that they fail in the simplest
cases in which $U$ is not an $L$-space and $V$ is not an $M$-space, see
371Yc-371Ye.

}%end of comment

\discrpage

\frfilename{mt372.tex}
\versiondate{7.12.08/17.7.11}
\copyrightdate{1995}

\def\chaptername{Linear operators between function spaces}
\def\sectionname{The ergodic theorem}

\newsection{372}

I come now to one of the most remarkable topics in measure theory.   I
cannot do it justice in the space I have allowed for it here, but I can
give the basic theorem (372D, 372F) and a variety of the corollaries
through which it is regularly used (372E,
372G-372J), %372G 372H 372J
together with brief
notes on one of its most famous and characteristic applications (to
continued fractions, 372L-372N) %372L 372M 372N
and on `ergodic' and `mixing'
transformations (372O-372S). %372O 372Q 372S
In the first half of the section (down to
372G) I express the arguments in the abstract language of measure
algebras and their associated function spaces, as developed in Chapter
36;  the second half, from 372H onwards,
contains translations of the
results into the language of measure spaces and measurable functions,
the more traditional, and more readily applicable, forms.

\leader{372A}{Lemma} Let $U$ be a reflexive Banach space and $T:U\to U$
a bounded linear operator of norm at most $1$.   Then

\Centerline{$V=\{u+v-Tu:u,\,v\in U,\,Tv=v\}$}

\noindent is dense in $U$.

\proof{ Of course $V$ is a linear subspace of $U$.   \Quer\ Suppose, if
possible, that it is not dense.   Then there is a non-zero $h\in U^*$
such that $h(v)=0$ for every $v\in V$ (3A5Ad).
Take $u\in U$ such that $h(u)\ne 0$.   Set

\Centerline{$u_n=\Bover1{n+1}\sum_{i=0}^nT^iu$}

\noindent for each $n\in\Bbb N$, taking $T^0$ to be the identity
operator;  because

\Centerline{$\|T^iu\|\le\|T^i\|\|u\|\le\|T\|^i\|u\|\le\|u\|$}

\noindent for each $i$, $\|u_n\|\le\|u\|$ for every $n$.   Note also
that $T^{i+1}u-T^iu\in V$ for every $i$, so that
$h(T^{i+1}u-T^iu)=0$;  accordingly $h(T^iu)=h(u)$ for every $i$, and
$h(u_n)=h(u)$ for every $n$.

Let $\Cal F$ be any non-principal ultrafilter on $\Bbb N$.   Because $U$
is reflexive, $v=\lim_{n\to\Cal F}u_n$ is defined in $U$ for the weak
topology on $U$ (3A5Gc).   Now $Tv=v$.   \Prf\ For each $n\in\Bbb
N$,

\Centerline{$Tu_n-u_n
=\Bover1{n+1}\sum_{i=0}^n(T^{i+1}u-T^iu)
=\Bover1{n+1}(T^{n+1}u-u)$}

\noindent has norm at most $\bover2{n+1}\|u\|$.   So
$\sequencen{Tu_n-u_n}\to 0$ for the norm topology $U$ and therefore for
the weak topology, and surely $\lim_{n\to\Cal F}Tu_n-u_n=0$.   On the
other hand (because $T$ is continuous for the weak topology, 2A5If)

\Centerline{$Tv=\lim_{n\to\Cal F}Tu_n=\lim_{n\to\Cal F}(Tu_n-u_n)
+\lim_{n\to\Cal F}u_n=0+v=v$,}

\noindent where all the limits are taken for the weak topology.\ \Qed

But this means that $v\in V$, while

\Centerline{$h(v)=\lim_{n\to\Cal F}h(u_n)=h(u)\ne 0$,}

\noindent contradicting the assumption that $h\in V^{\smallcirc}$.\
\Bang
}%end of proof of 372A

\leader{372B}{Lemma} Let $(\frak A,\bar\mu)$ be a measure algebra, and
$T:L^1\to L^1$ a positive linear operator of norm at most $1$, where
$L^1=L^1(\frak A,\bar\mu)$.   Take any $u\in L^1$ and $m\in\Bbb N$, and
set

\Centerline{$a=\Bvalue{u>0}\Bcup\Bvalue{u+Tu>0}\Bcup\Bvalue{u+Tu+T^2u>0}
\Bcup\ldots\Bcup\Bvalue{u+Tu+\ldots+T^mu>0}$.}

\noindent Then $\int_au\ge 0$.

\proof{ Set $u_0=u$, $u_1=u+Tu,\ldots,u_m=u+Tu+\ldots+T^mu$,
$v=\sup_{i\le m}u_i$, so that $a=\Bvalue{v>0}$.   Consider $u+T(v^+)$.
We have $T(v^+)\ge Tv\ge Tu_i$ for every $i\le m$
(because $T$ is positive), so that $u+T(v^+)\ge u+Tu_i=u_{i+1}$ for
$i<m$,
and $u+T(v^+)\ge\sup_{1\le i\le m}u_i$.   Also $u+T(v^+)\ge u$ because
$T(v^+)\ge 0$, so $u+T(v^+)\ge v$.   Accordingly

\Centerline{$\int_au\ge\int_av-\int_aT(v^+)=\int v^+-\int_aT(v^+)
\ge\|v^+\|_1-\|Tv^+\|_1\ge 0$}

\noindent because $\|T\|\le 1$.
}%end of proof of 372B

\leader{372C}{Maximal Ergodic Theorem}
Let $(\frak A,\bar\mu)$ be a measure algebra, and $T:L^1\to L^1$ a
linear operator, where $L^1=L^1(\frak A,\bar\mu)$, such that
$\|Tu\|_1\le\|u\|_1$ for every $u\in L^1$ and
$\|Tu\|_{\infty}\le\|u\|_{\infty}$ for every
$u\in L^1\cap L^{\infty}(\frak A)$.
Set $A_n=\bover1{n+1}\sum_{i=0}^nT^i$ for each
$n\in\Bbb N$.   Then for any $u\in L^1$, $u^*=\sup_{n\in\Bbb N}A_nu$ is
defined in $L^0(\frak A)$, and
$\alpha\bar\mu\Bvalue{u^*>\alpha}\le\|u\|_1$ for every $\alpha>0$.

\proof{{\bf (a)} To begin with, suppose that $T$ is positive and that
$u\ge 0$ in $L^1$.   Note that if $v\in L^1\cap L^{\infty}$, then
$\|T^iv\|_{\infty}\le\|v\|_{\infty}$ for every $i\in\Bbb N$, so
$\|A_nv\|_{\infty}\le\|v\|_{\infty}$ for every $n$;  in particular,
$A_n(\chi a)\le\chi 1$ for every $n$ and every $a$ of finite measure.

For $m\in\Bbb N$ and $\alpha>0$, set

\Centerline{$a_{m\alpha}=\Bvalue{\sup_{i\le m}A_iu>\alpha}$.}

\noindent Then $\alpha\bar\mu a_{m\alpha}\le\|u\|_1$.   \Prf\ Set
$a=a_{m\alpha}$,
$w=u-\alpha\chi a$.   Of course $\sup_{i\le m}A_iu$ belongs to
$L^1$, so $\bar\mu a$ is finite and $w\in L^1$.   For any $i\le
m$,

\Centerline{$A_iw=A_iu-\alpha A_i(\chi a)\ge A_iu-\alpha\chi 1$,}

\noindent so $\Bvalue{A_iw>0}\Bsupseteq\Bvalue{A_iu>\alpha}$.
Accordingly $a\Bsubseteq b$, where

\Centerline{$b=\sup_{i\le m}\Bvalue{A_iw>0}
=\sup_{i\le m}\Bvalue{w+Tw+\ldots+T^iw>0}$.}

\noindent By 372B, $\int_bw\ge 0$.   But this means that

\Centerline{$\alpha\bar\mu a=\alpha\int_b\chi a
=\int_bu-\int_bw\le\int_bu\le\|u\|_1$,}

\noindent as claimed.\   \Qed

It follows that if we set $c_{\alpha}=\sup_{n\in\Bbb N}a_{n\alpha}$,
$\bar\mu c_{\alpha}\le\alpha^{-1}\|u\|_1$ for every $\alpha>0$ and
$\inf_{\alpha>0}c_{\alpha}=0$.   But this is exactly the criterion in
364L(a-ii) for $u^*=\sup_{n\in\Bbb N}A_nu$ to be defined in $L^0$.   And
$\Bvalue{u^*>\alpha}=c_{\alpha}$, so
$\alpha\bar\mu\Bvalue{u^*>\alpha}\le\|u\|_1$ for every $\alpha>0$, as
required.

\medskip

{\bf (b)} Now consider the case of general $T$, $u$.   In this case $T$
is order-bounded and $\||T|\|\le 1$, where $|T|$ is the modulus of $T$
in $\eurm L^{\sim}(L^1;L^1)=\eurm B(L^1;L^1)$ (371D).   If
$w\in L^1\cap L^{\infty}$, then

\Centerline{$||T|w|\le |T||w|
=\sup_{|w'|\le|w|}|Tw'|\le\|w\|_{\infty}\chi 1$,}

\noindent so $\||T|w\|_{\infty}\le\|w\|_{\infty}$.   Thus $|T|$ also
satisfies the conditions of the theorem.
Setting $B_n=\bover1{n+1}\sum_{i=0}^n|T|^i$, $B_n\ge A_n$ in
$\eurm L^{\sim}(L^1;L^1)$ and $B_n|u|\ge A_nu$ for every $n$.   But by
(a), $v=\sup_{n\in\Bbb N}B_n|u|$ is defined in $L^0$ and
$\alpha\bar\mu\Bvalue{v>\alpha}\le\||u|\|_1=\|u\|_1$ for every
$\alpha>0$.   Consequently $u^*=\sup_{n\in\Bbb N}A_nu$ is defined in
$L^0$ and $u^*\le v$, so that
$\alpha\bar\mu\Bvalue{u^*>\alpha}\le\|u\|_1$ for every $\alpha>0$.
}%end of proof of 372C


\leader{372D}{}\cmmnt{ We are now ready for a very general form of the
Ergodic Theorem.   I express it in terms of the space $M^{1,0}$ from
366F and the class $\Cal T^{(0)}$ of operators from 371F.   If these
formulae are unfamiliar, you may like to glance at the statement of 372F
before looking them up.

\medskip

\noindent}{\bf The Ergodic Theorem:  first form} Let $(\frak A,\bar\mu)$
be a measure algebra, and set $M^{1,0}=M^{1,0}(\frak A,\bar\mu)$,
$\Cal T^{(0)}=\Cal T^{(0)}_{\bar\mu,\bar\mu}
\subseteq\eurm B(M^{1,0};M^{1,0})$\cmmnt{ as in
371F-371G}.   Take any $T\in\Cal T^{(0)}$, and set
$A_n=\bover1{n+1}\sum_{i=0}^nT^i:M^{1,0}\to M^{1,0}$ for every $n$.
Then for any $u\in M^{1,0}$, $\sequencen{A_nu}$ is
order*-convergent\cmmnt{ (definition:  367A)} and
$\|\,\|_{1,\infty}$-convergent to a member $Pu$ of $M^{1,0}$.
The operator $P:M^{1,0}\to M^{1,0}$ is a projection onto the linear
subspace $\{u:u\in M^{1,0},\,Tu=u\}$, and $P\in\Cal T^{(0)}$.

\proof{{\bf (a)}  It will be convenient to start with some elementary
remarks.   First, every $A_n$ belongs to $\Cal T^{(0)}$, by 371Ge and
371Ga.    Next, $\sequencen{A_nu}$ is order-bounded in
$L^0=L^0(\frak A)$ for any $u\in M^{1,0}$;  this is because if $u=v+w$,
where $v\in L^1=L^1(\frak A,\bar\mu)$ and
$w\in L^{\infty}=L^{\infty}(\frak A)$,
then $\sequencen{A_nv}$ and $\sequencen{A_n(-v)}$ are bounded above, by
372C, while $\sequencen{A_nw}$ is norm- and order-bounded in
$L^{\infty}$.   Accordingly I can uninhibitedly speak of
$P^*(u)=\inf_{n\in\Bbb N}\sup_{i\ge n}A_iu$ and
$P_*(u)=\sup_{n\in\Bbb N}\inf_{i\ge n}A_iu$
for any $u\in M^{1,0}$, these both being defined in $L^0$.

\medskip

{\bf (b)} Write $V_1$ for the set of those $u\in M^{1,0}$ such that
$\sequencen{A_nu}$ is order*-convergent in $L^0$;
that is, $P^*(u)=P_*(u)$ (367Be).   It is easy to see that $V_1$ is a
linear subspace of $M^{1,0}$ (use 367Ca).   Also it is closed
for $\|\,\|_{1,\infty}$.

\Prf\ We know that $|T|$, taken in $\eurm L^{\sim}(M^{1,0};M^{1,0})$,
belongs to $\Cal T^{(0)}$ (371Gb);  set
$B_n=\bover1{n+1}\sum_{i=0}^n|T|^i$ for each $i$.

Suppose that $u_0\in\overline{V}_1$.   Then for any $\epsilon>0$ there
is a $u\in V_1$ such that $\|u_0-u\|_{1,\infty}\le\epsilon^2$.   Write
$Pu=P^*(u)=P_*(u)$ for the order*-limit of
$\sequencen{A_nu}$.   Express $u_0-u$ as $v+w$ where $v\in L^1$,
$w\in L^{\infty}$ and $\|v\|_1+\|w\|_{\infty}\le 2\epsilon^2$.

Set $v^*=\sup_{n\in\Bbb N}B_n|v|$.   Then
$\bar\mu\Bvalue{v^*>\epsilon}\le2\epsilon$, by 372C.   Next, if
$w^*=\sup_{n\in\Bbb N}B_n|w|$, we surely have
$w^*\le 2\epsilon^2\chi 1$.   Now

\Centerline{$|A_nu_0-A_nu|=|A_nv+A_nw|\le B_n|v|+B_n|w|\le v^*+w^*$}

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

\Centerline{$A_nu-v^*-w^*\le A_nu_0\le A_nu+v^*+w^*$}

\noindent for every $n$.   Because $\sequencen{A_nu}$
order*-converges to $Pu$,

\Centerline{$Pu-v^*-w^*\le P_*(u_0)\le P^*(u_0)\le Pu+v^*+w^*$,}

\noindent and $P^*(u_0)-P_*(u_0)\le 2(v^*+w^*)$.   On the other hand,

\Centerline{$\bar\mu\Bvalue{2(v^*+w^*)>2\epsilon+4\epsilon^2}
\le\bar\mu\Bvalue{v^*>\epsilon}+\bar\mu\Bvalue{w^*>2\epsilon^2}
=\bar\mu\Bvalue{v^*>\epsilon}\le 2\epsilon$}

\noindent (using 364Ea for the first inequality).   So

\Centerline{$\bar\mu\Bvalue{P^*(u_0)-P_*(u_0)>2\epsilon(1+2\epsilon)}\le
2\epsilon$.}

\noindent Since $\epsilon$ is arbitrary, $\sequencen{A_nu_0}$
order*-converges to $P^*(u_0)=P_*(u_0)$, and $u_0\in V_1$.   As $u_0$ is
arbitrary, $V_1$ is closed.\   \Qed

\medskip


{\bf (c)} Similarly, the set $V_2$ of those $u\in M^{1,0}$ for which
$\sequencen{A_nu}$ is norm-convergent is a linear subspace of $M^{1,0}$,
and it also is closed.   \Prf\ This is a standard argument.   If
$u_0\in\overline{V}_2$ and $\epsilon>0$, there is a $u\in V_2$ such that
$\|u_0-u\|_{1,\infty}\le\epsilon$.   There is an $n\in\Bbb N$ such that
$\|A_iu-A_ju\|_{1,\infty}\le\epsilon$ for all $i$, $j\ge n$, and now
$\|A_iu_0-A_ju_0\|_{1,\infty}\le 3\epsilon$ for all $i$, $j\ge n$,
because every $A_i$ has norm at most $1$ in $\eurm B(M^{1,0};M^{1,0})$
(371Gc).   As $\epsilon$ is arbitrary,
$\sequencen{A_iu_0}$ is Cauchy;  because $M^{1,0}$ is complete,
it is convergent, and $u_0\in V_2$.   As $u_0$ is arbitrary, $V_2$ is
closed.\   \Qed

\medskip

{\bf (d)} Now let $V$ be
$\{u+v-Tu:u\in M^{1,0}\cap L^{\infty},\,v\in M^{1,0},\,Tv=v\}$.   Then
$V\subseteq V_1\cap V_2$.   \Prf\ If $u\in M^{1,0}\cap L^{\infty}$, then
for any $n\in\Bbb N$

\Centerline{$A_n(u-Tu)=\bover1{n+1}(u-T^{n+1}u)\to 0$}

\noindent for $\|\,\|_{\infty}$, and therefore is both
order*-convergent and convergent for
$\|\,\|_{1,\infty}$;  so $u-Tu\in V_1\cap V_2$.   On the other hand, if
$Tv=v$, then of course $A_nv=v$ for every $n$, so again
$v\in V_1\cap V_2$.\   \Qed

\medskip

{\bf (e)} Consequently $L^2=L^2(\frak A,\bar\mu)\subseteq V_1\cap V_2$.
\Prf\ $L^2\cap V_1\cap V_2$ is a linear subspace;  but also it is closed
for the norm topology of $L^2$, because the identity map from $L^2$ to
$M^{1,0}$ is continuous (369Oe).   We know also that $T\restr L^2$
is an operator of norm at
most $1$ from $L^2$ to itself (371Gd).   Consequently
$W=\{u+v-Tu:u,\,v\in L^2,\,Tv=v\}$ is dense in $L^2$ (372A).   On the
other hand, given $u\in L^2$ and $\epsilon>0$, there is a $u'\in L^2\cap
L^{\infty}$ such that $\|u-u'\|_2\le\epsilon$ (take
$u'=(u\wedge\gamma\chi 1)\vee(-\gamma\chi 1)$ for any $\gamma$ large
enough), and now $\|(u-Tu)-(u'-Tu')\|_2\le 2\epsilon$.   Thus
$W'=\{u'+v-Tu':u'\in L^2\cap L^{\infty},\,v\in L^2,\,Tv=v\}$ is dense in
$L^2$.   But $W'\subseteq V_1\cap V_2$, by (d) above.   Thus
$L^2\cap V_1\cap V_2$ is dense in $L^2$, and is therefore the whole of
$L^2$.\
\Qed

\medskip

{\bf (f)} $L^2\supseteq S(\frak A^f)$ is dense in $M^{1,0}$, by 369Pc,
so $V_1=V_2=M^{1,0}$.   This shows that $\sequencen{A_nu}$ is
norm-convergent and order*-convergent for every $u\in M^{1,0}$.   By
367Da, the limits are the same.   Write $Pu$
for the common value of the limits.

\medskip

{\bf (g)} Of course we now have

\Centerline{$\|Pu\|_{\infty}\le\sup_{n\in\Bbb N}\|A_nu\|_{\infty}
\le\|u\|_{\infty}$}

\noindent for every $u\in L^{\infty}\cap M^{1,0}$, while

\Centerline{$\|Pu\|_1\le\liminf_{n\to\infty}\|A_nu\|_1\le \|u\|_1$}

\noindent for every $u\in L^1$, by Fatou's Lemma.   So $P\in\Cal
T^{(0)}$.
If $u\in M^{1,0}$ and $Tu=u$, then surely $Pu=u$, because $A_nu=u$
for every $u$.   On the other hand, for any $u\in M^{1,0}$, $TPu=Pu$.
\Prf\   Because $\sequencen{A_nu}$ is norm-convergent to $Pu$,

$$\eqalign{\|TPu-Pu\|_{1,\infty}
&=\lim_{n\to\infty}\|TA_nu-A_nu\|_{1,\infty}\cr
&=\lim_{n\to\infty}\Bover1{n+1}\|T^{n+1}u-u\|_{1,\infty}
=0.  \text{ \Qed}\cr}$$

\noindent Thus, writing $U=\{u:Tu=u\}$, $P[M^{1,0}]=U$ and $Pu=u$ for
every $u\in U$.
}%end of proof of 372D

\leader{372E}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra,
and $\pi:\frak A^f\to\frak A^f$ a measure-preserving ring homomorphism,
where $\frak A^f=\{a:\bar\mu a<\infty\}$.   Let $T:M^{1,0}\to M^{1,0}$
be the corresponding Riesz homomorphism, where
$M^{1,0}=M^{1,0}(\frak A,\bar\mu)$\cmmnt{ (366H, in particular part
(a-v))}.   Set
$A_n=\bover1{n+1}\sum_{i=0}^nT^i$ for $n\in\Bbb N$.   Then for every
$u\in M^{1,0}$, $\sequencen{A_nu}$ is order*-convergent and
$\|\,\|_{1,\infty}$-convergent to some $v$ such that $Tv=v$.

\proof{ By 366H(a-iv), $T\in\Cal T^{(0)}$, as defined in 371F.   So the
result follows at once from 372D.
}%end of proof of 372E

\leader{372F}{The Ergodic Theorem:  second form} Let $(\frak A,\bar\mu)$
be a measure algebra, and let $T:L^1\to L^1$, where
$L^1=L^1(\frak A,\bar\mu)$, be a linear operator of norm at most $1$
such that $Tu\in L^{\infty}=L^{\infty}(\frak A)$ and
$\|Tu\|_{\infty}\le\|u\|_{\infty}$
whenever $u\in L^1\cap L^{\infty}$.   Set
$A_n=\bover1{n+1}\sum_{i=0}^nT^i:L^1\to L^1$ for every $n$.   Then for
any $u\in L^1$, $\sequencen{A_nu}$ is order*-convergent to an element
$Pu$ of $L^1$.   The operator $P:L^1\to L^1$ is a projection of norm at
most $1$ onto the linear subspace $\{u:u\in L^1,\,Tu=u\}$.

\proof{ By 371Ga, there is an extension of $T$ to a member $\tilde T$ of
$\Cal T^{(0)}$.   So 372D tells us that $\sequencen{A_nu}$ is
order*-convergent to some $Pu\in L^1$ for every $u\in L^1$, and
$P:L^1\to L^1$ is a projection of norm at most $1$, because $P$ is the
restriction of a projection $\tilde P\in\Cal T^{(0)}$.   Also we still
have $TPu=Pu$ for every $u\in L^1$, and $Pu=u$ whenever $Tu=u$, so the
set of values $P[L^1]$ of $P$ must be exactly $\{u:u\in L^1,\,Tu=u\}$.
}%end of proof of 372F

\cmmnt{\medskip

\noindent{\bf Remark} In 372D and 372F I have used the phrase
`order*-convergent' from \S367 without always being specific about the
partially ordered set in which it is to be interpreted.   But, as
remarked in 367E, the notion is robust enough for the omission to be
immaterial here.   Since both $M^{1,0}$ and $L^1$ are solid linear
subspaces of $L^0$, a sequence in $M^{1,0}$ is order*-convergent to a
member of $M^{1,0}$ (when order*-convergence is interpreted in the
partially ordered set $M^{1,0}$) iff it is order*-convergent to the same
point (when convergence is interpreted in the set $L^0$);  and the same
applies to $L^1$ in place of $M^{1,0}$.
}%end of comment

\leader{372G}{Corollary} Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\pi:\frak A\to\frak A$ a measure-preserving Boolean
homomorphism.   Let $T:L^1\to L^1$ be the corresponding Riesz
homomorphism, where $L^1=L^1(\frak A,\bar\mu)$.   Set
$A_n=\bover1{n+1}\sum_{i=0}^nT^i$ for $n\in\Bbb N$.   Then for every
$u\in L^1$, $\sequencen{A_nu}$ is order*-convergent and
$\|\,\,\|_1$-convergent.   If we set $Pu=\lim_{n\to\infty}A_nu$ for each
$u$, $P$ is the conditional expectation operator corresponding to the
fixed-point subalgebra $\frak C=\{a:\pi a=a\}$ of $\frak A$.

\proof{{\bf (a)} The first part is just a special case of 372E;  the
point is that because $(\frak A,\bar\mu)$ is totally finite,
$L^{\infty}(\frak A)\subseteq L^1$, so $M^{1,0}(\frak A,\bar\mu)=L^1$.
Also (because $\bar\mu 1=1$) $\|u\|_{\infty}\le\|u\|_1$ for every $u\in
L^{\infty}$, so the norm $\|\,\|_{1,\infty}$ is actually equal to
$\|\,\|_1$.

\medskip

{\bf (b)} For the last sentence, recall that $\frak C$ is a closed
subalgebra of $\frak A$ (cf.\ 333R).   By 372D or 372F, $P$ is a
projection operator onto the subspace $\{u:Tu=u\}$.   Now
$\Bvalue{Tu>\alpha}=\pi\Bvalue{u>\alpha}$ (365Nc), so $Tu=u$ iff
$\Bvalue{u>\alpha}\in\frak C$ for every $\alpha\in\Bbb R$, that is, iff
$u$ belongs to the canonical image of
$L^1(\frak C,\bar\mu\restrp\frak C)$ in $L^1$ (365Q).   To identify $Pu$
further, observe that if $u\in L^1$ and $a\in\frak C$ then

\Centerline{$\int_aTu=\int_{\pi a}Tu=\int_au$}

\noindent (365Nb).   Consequently $\int_aT^iu=\int_au$ for every
$i\in\Bbb N$, $\int_aA_nu=\int_au$ for every $n\in\Bbb N$, and
$\int_aPu=\int_au$ (because $Pu$ is the limit of $\sequencen{A_nu}$ for
$\|\,\|_1$).   But this is enough to define $Pu$ as the conditional
expectation of $u$ on $\frak C$ (365Q).
}%end of proof of 372G

\leader{372H}{}\cmmnt{ The Ergodic Theorem is most often expressed in
terms of transformations of measure spaces.   In the next few
corollaries I will formulate such expressions.   The translation is
straightforward.

\medskip

\noindent}{\bf Corollary\dvAformerly{3{}72I}}
Let $(X,\Sigma,\mu)$ be a measure space and
$\phi:X\to X$ an \imp\ function.   Let $f$ be a real-valued function
which is integrable over $X$.   Then

\Centerline{$g(x)
=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))$}

\noindent is defined for almost every $x\in X$, and $g\phi(x)=g(x)$ for
almost every $x$.

\proof{ Let $(\frak A,\bar\mu)$ be the measure algebra of
$(X,\Sigma,\mu)$, and $\pi:\frak A\to\frak A$,
$T:L^0(\frak A)\to L^0(\frak A)$ the homomorphisms corresponding to
$\phi$, as in 364Qd.
Set $u=f^{\ssbullet}$ in $L^1(\frak A,\bar\mu)$.   Then for any
$i\in\Bbb N$, $T^iu=(f\phi^i)^{\ssbullet}$ (364Q(c)-(d)), so setting
$A_n=\bover1{n+1}\sum_{i=0}^nT^i$, $A_nu=g_n^{\ssbullet}$, where
$g_n(x)=\bover1{n+1}\sum_{i=0}^nf(\phi^i(x))$ whenever this is
defined.   Now we know from 372F or 372E that $\sequencen{A_nu}$ is
order*-convergent to some $v$ such that $Tv=v$, so $\sequencen{g_n}$
must be convergent almost everywhere (367F), and taking
$g=\lim_{n\to\infty}g_n$ where this is defined, $g^{\ssbullet}=v$.
Accordingly $(g\phi)^{\ssbullet}=Tv=v=g^{\ssbullet}$ and $g\phi\eae g$,
as claimed.
}%end of proof of 372H

\leader{372I}{}\cmmnt{ The following facts will be
useful in the next version of the theorem, and elsewhere.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a measure space with
measure algebra $(\frak A,\bar\mu)$.   Let $\phi:X\to X$ be an \imp\
function and $\pi:\frak A\to\frak A$ the associated
homomorphism\cmmnt{, as in 343A and 364Qd}.
Set $\frak C=\{c:c\in\frak A,\,\pi c=c\}$,
$\Tau=\{E:E\in\Sigma,\,\phi^{-1}[E]\symmdiff E$ is negligible$\}$ and
$\Tau_0=\{E:E\in\Sigma,\,\phi^{-1}[E]=E\}$.   Then $\Tau$ and $\Tau_0$
are $\sigma$-subalgebras of $\Sigma$;  $\Tau_0\subseteq\Tau$,
$\Tau=\{E:E\in\Sigma,\,E^{\ssbullet}\in \frak C\}$, and
$\frak C=\{E^{\ssbullet}:E\in \Tau_0\}$.

\proof{ It is easy to see that $\Tau$ and $\Tau_0$ are
$\sigma$-subalgebras of $\Sigma$ and that
$\Tau_0\subseteq\Tau=\{E:E^{\ssbullet}\in\frak C\}$.   So we have only
to check that if $c\in\frak C$ there is an
$E\in\Tau_0$ such that $E^{\ssbullet}=c$.   \Prf\ Start with any
$F\in\Sigma$ such that $F^{\ssbullet}=c$.
Now $F\symmdiff\phi^{-i}[F]$ is negligible for every $i\in\Bbb N$,
because $(\phi^{-i}[F])^{\ssbullet}=\pi^ic=c$.   So if we set

$$\eqalign{E
&=\bigcup_{n\in\Bbb N}\bigcap_{i\ge n}\phi^{-i}[F]\cr
&=\{x:\text{ there is an }n\in\Bbb N\text{ such that }\phi^i(x)\in F
  \text{ for every }i\ge n\},\cr}$$

\noindent $E^{\ssbullet}=c$.
On the other hand, it is easy to check that $E\in\Tau_0$.\   \Qed
}%end of proof of 372I

\leader{372J}{The Ergodic Theorem:  third form}\dvAformerly{3{}72K}
Let $(X,\Sigma,\mu)$ be a probability space and
$\phi:X\to X$ an \imp\ function.
Let $f$ be a real-valued function which is integrable over $X$.   Then

\Centerline{$g(x)
=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))$}

\noindent is defined for almost every $x\in X$;  $g\phi\eae g$, and $g$
is a conditional expectation of $f$ on the $\sigma$-algebra
$\Tau=\{E:E\in\Sigma,\,\phi^{-1}[E]\symmdiff E$ is negligible$\}$.
If {\it either} $f$ is $\Sigma$-measurable and defined everywhere in $X$
{\it or} $\phi[E]$ is negligible for every negligible set $E$, then $g$
is a conditional expectation of $f$ on the $\sigma$-algebra
$\Tau_0=\{E:E\in\Sigma,\,\phi^{-1}[E]=E\}$.

\proof{{\bf (a)} We know by 372H that $g$ is defined almost everywhere
and that $g\phi\eae g$.   In the language of the proof of 372H,
$g^{\ssbullet}=v$ is the conditional expectation of $u=f^{\ssbullet}$ on
the closed subalgebra

\Centerline{$\frak C=\{a:a\in\frak A,\,\pi a=a\}
=\{F^{\ssbullet}:F\in\Tau\}
=\{F^{\ssbullet}:F\in\Tau_0\}$,}

\noindent by 372G and 372I.   So $v$ must be expressible as
$h^{\ssbullet}$ where $h:X\to\Bbb R$ is
$\Tau_0$-measurable and is a conditional expectation of $f$ on $\Tau_0$
(and also on $\Tau$).
Since every set of measure zero belongs to $\Tau$,
$g=h\,\,\mu\restrp\Tau$-a.e., and $g$ also is a conditional expectation
of $f$ on $\Tau$.

\medskip

{\bf (b)} Suppose now that $f$ is defined everywhere and
$\Sigma$-measurable.   Here I come to a technical
obstruction.   The definition of `conditional expectation' in 233D
asks for $g$ to be $\mu\restrp\Tau_0$-integrable, and since
$\mu$-negligible sets do not
need to be $\mu\restrp\Tau_0$-negligible we have some more checking to
do, to confirm that $\{x:x\in\dom g,\,g(x)=h(x)\}$ is
$\mu\restrp\Tau_0$-conegligible as well as $\mu$-conegligible.

\medskip

\quad{\bf (i)} For $n\in\Bbb N$, set
$\Sigma_n=\{\phi^{-n}[E]:E\in\Sigma\}$;  then
$\Sigma_n$ is a $\sigma$-subalgebra of $\Sigma$, including $\Tau_0$.
Set $\Sigma_{\infty}=\bigcap_{n\in\Bbb N}\Sigma_n$, still a
$\sigma$-algebra including $\Tau_0$.   Now any negligible set
$E\in\Sigma_{\infty}$ is $\mu\restrp\Tau_0$-negligible.   \Prf\ For each
$n\in\Bbb N$ choose $F_n\in\Sigma$ such that $E=\phi^{-n}[F_n]$.
Because $\phi$ is \imp, every $F_n$ is negligible, so that

\Centerline{$E^*=\bigcap_{m\in\Bbb N}\bigcup_{n\in\Bbb N,j\ge m}
\phi^{-j}[F_n]$}

\noindent is negligible.   Of course
$E=\bigcap_{m\in\Bbb N}\phi^{-m}[F_m]$ is included in $E^*$.   Now

\Centerline{$\phi^{-1}[E^*]
=\bigcap_{m\in\Bbb N}\bigcup_{n\in\Bbb N,j\ge m}\phi^{-j-1}[F_n]
=\bigcap_{m\ge 1}\bigcup_{n\in\Bbb N,j\ge m}\phi^{-j}[F_n]
=E^*$}

\noindent because

\Centerline{$\bigcup_{n\in\Bbb N,j\ge 1}\phi^{-j}[F_n]
\subseteq\bigcup_{n\in\Bbb N,j\ge 0}\phi^{-j}[F_n]$.}

\noindent So $E^*\in\Tau_0$ and $E$ is included in a negligible member
of $\Tau_0$, which is what we needed to know.\   \Qed

\medskip

\quad{\bf (ii)} We are assuming that $f$ is $\Sigma$-measurable and
defined everywhere, so that
$g_n=\bover1{n+1}\sum_{i=0}^nf\frsmallcirc\phi^i$ is
$\Sigma$-measurable and defined everywhere.   If we set
$g^*=\limsup_{n\to\infty}g_n$, then $g^*:X\to[-\infty,\infty]$ is
$\Sigma_{\infty}$-measurable.   \Prf\ For any $m\in\Bbb N$,
$f\frsmallcirc\phi^i$ is $\Sigma_m$-measurable for every $i\ge m$, since
$\{x:f(\phi^i(x))>\alpha\}=\phi^{-m}[\{x:f(\phi^{i-m}(x))>\alpha\}]$ for
every $\alpha$.   Accordingly

\Centerline{$g^*
=\limsup_{n\to\infty}\Bover1{n+1}\sum_{i=m}^nf\frsmallcirc\phi^i$}

\noindent is $\Sigma_m$-measurable.   As $m$ is arbitrary, $g^*$ is
$\Sigma_{\infty}$-measurable.\   \Qed

Since $h$ is surely $\Sigma_{\infty}$-measurable, and
$h=g^*\,\,\mu$-a.e., (i) tells us that $h=g^*\,\,\mu\restrp\Tau_0$-a.e.
But similarly $h=\liminf_{n\to\infty}g_n\,\,\mu\restrp\Tau_0$-a.e., so
we
must have $h=g\,\,\mu\restrp\Tau_0$-a.e.; and $g$, like $h$, is a
conditional expectation of $f$ on $\Tau_0$.

\medskip

{\bf (c)} Finally, suppose that $\phi[E]$ is negligible for every
negligible set $E$.
Then every $\mu$-negligible set is $\mu\restrp\Tau_0$-negligible.
\Prf\
If $E$ is $\mu$-negligible, then $\phi[E]$,
$\phi^2[E]=\phi[\phi[E]],\ldots$ are all negligible, so
$E^*=\bigcup_{n\in\Bbb N}\phi^n[E]$ is negligible, and there is a
measurable negligible set $F\supseteq E^*$.   Now $F_*=\bigcup_{m\in\Bbb
N}\bigcap_{n\ge m}\phi^{-n}[F]$ is a negligible set in $\Tau_0$
including $E$, so $E$ is $\mu\restrp\Tau_0$-negligible.\ \Qed\
Consequently $g=h\,\,\mu\restrp\Tau_0$-a.e., and in this case also $g$
is a conditional expectation of $f$ on $\Tau_0$.
}%end of proof of 372J

\cmmnt{
\leader{372K}{Remark} Parts (b)-(c) of the proof above are
dominated by the technical question of the exact definition of
`conditional expectation of $f$ on $\Tau_0$', and it is natural to be
impatient with such details.   The kind of example I am concerned about
is the following.   Let $C\subseteq[0,1]$ be the Cantor set (134G), and
$\phi:[0,1]\to[0,1]$ a Borel measurable function such that
$\phi[C]=[0,1]$ and $\phi(x)=x$ for $x\in[0,1]\setminus C$.   (For
instance, we could take $\phi$ agreeing with the Cantor function on $C$
(134H).)   Because $C$ is negligible, $\phi$ is \imp\ for Lebesgue
measure $\mu$, and if $f$ is any Lebesgue integrable function then
$g(x)=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))$ is defined
and equal to $f(x)$ for every $x\in\dom f\setminus C$.   But for
$x\in C$ we can, by manipulating $\phi$, arrange for $g(x)$ to be almost
anything;  and if $f$ is undefined on $C$ then $g$ will also be
undefined on $C$.   On the other hand, $C$ is not
$\mu\restrp\Tau_0$-negligible, because the only member of $\Tau_0$
including $C$ is $[0,1]$.   So we cannot be sure of being able to form
$\int g\,d(\mu\restrp\Tau_0)$.

If instead of Lebesgue measure itself we took its restriction
$\mu_{\Cal B}$ to the algebra of Borel subsets of $[0,1]$, then $\phi$
would still
be \imp\ for $\mu_{\Cal B}$, but we should now have to worry about the
possibility that $f\restr C$ was non-measurable, so that $g\restr C$
came out to be non-measurable, even if everywhere defined, and $g$ was
not $\mu_{\Cal B}\restrp\Tau_0$-virtually measurable.

In the statement of 372J I have offered two ways of being sure that the
problem does not arise:  check that $\phi[E]$ is negligible whenever $E$
is negligible (so that all negligible sets are
$\mu\restrp\Tau_0$-negligible), or check that $f$ is defined everywhere
and $\Sigma$-measurable.   Even if these conditions are not immediately
satisfied in a particular application, it may be possible to modify the
problem so that they are.   For instance, completing the measure will
leave $\phi$ \imp\ (234Ba\formerly{2{}35Hc}),
will not change the integrable functions but
will make them all measurable (212F, 212Bc), and may enlarge $\Tau_0$
enough to
make a difference.   If our function $f$ is measurable (because the
measure is complete, or otherwise) we can extend it to a measurable
function defined everywhere (121I) and the corresponding extension of
$g$ will be $\mu\restrp\Tau_0$-integrable.   Alternatively, if the
difficulty seems to lie in the behaviour of $\phi$ rather than in the
behaviour of $f$ (as in the example above), it may help to modify $\phi$
on a negligible set.
}%end of comment

\leader{372L}{Continued fractions }\cmmnt{A particularly delightful
application of the results above is to a question which belongs as much
to number theory as to analysis.   It takes a bit of space to describe,
but I hope you will agree with me that it is well worth knowing in
itself, and that it also illuminates some of the ideas above.

\medskip

}{\bf (a)} Set $X=[0,1]\setminus\Bbb Q$.   For $x\in X$, set
$\phi(x)=\fraction{\bover1x}$, the fractional part of $\bover1x$, and
$k_1(x)=\bover1x-\phi(x)=\lfloor\bover1x\rfloor$, the integer part of
$\bover1x$;  then $\phi(x)\in X$ for each $x\in X$, so we may define
$k_n(x)=k_1(\phi^{n-1}(x))$ for every $n\ge 1$.   The strictly positive
integers $k_1(x)$, $k_2(x)$,
$k_3(x),\ldots$ are the {\bf continued fraction coefficients} of $x$.
\cmmnt{Of course }$k_{n+1}(x)=k_n(\phi(x))$ for every $n\ge 1$.
Now define
$\sequencen{p_n(x)}$, $\sequencen{q_n(x)}$ inductively by setting

\Centerline{$p_0(x)=0$,
\quad $p_1(x)=1$,
\quad$p_{n}(x)=p_{n-2}(x)+k_n(x)p_{n-1}(x)$ for $n\ge 1$,}

\Centerline{$q_0(x)=1$,
\quad$q_1(x)=k_1(x)$,
\quad$q_{n}(x)=q_{n-2}(x)+k_n(x)q_{n-1}(x)$ for $n\ge 1$.}

\noindent The {\bf continued fraction approximations}\cmmnt{ or
{\bf convergents}} to $x$ are the quotients $p_n(x)/q_n(x)$.

\cmmnt{(I do not discuss rational $x$, because for my purposes here
these are merely distracting.   But if we set $k_1(0)=\infty$,
$\phi(0)=0$ then the formulae above produce the conventional values for
$k_n(x)$ for rational $x\in\coint{0,1}$.   As for the $p_n$ and $q_n$,
use the formulae above until you get to $x=p_n(x)/q_n(x)$,
$\phi^n(x)=0$, $k_{n+1}(x)=\infty$, and then  set $p_{m}(x)=p_{n}(x)$,
$q_{m}(x)=q_{n}(x)$ for $m\ge n$.)
}

\cmmnt{\spheader 372Lb The point is that the quotients
$r_n(x)=p_n(x)/q_n(x)$ are, in a strong sense, good rational
approximations to $x$.   (See 372Xl(v).)   We have
$r_n(x)<x<r_{n+1}(x)$
for every even $n$ (372Xl).   If $x=\pi - 3$, then the first few
coefficients are

\Centerline{$k_1=7$,\quad $k_2=15$, \quad$k_3=1$,}
%then  292  1  1  1  2  1  3  1  14  2  1  1  2
%http://functions.wolfram.com/Constant/Pi/10

\Centerline{$r_1=\Bover17$,  \quad$r_2=\Bover{15}{106}$,
\quad$r_3=\Bover{16}{113}$;}

\noindent the first and third of these corresponding to the classical
approximations $\pi\bumpeq\Bover{22}7$, $\pi\bumpeq\Bover{355}{113}$.
Or if we take $x=e-2$, we get

\Centerline{$k_1=1$, \quad$k_2=2$, \quad$k_3=1$, \quad$k_4=1$,
\quad$k_5=4$, \quad$k_6=1$, \quad$k_7=1$,}
%then  6  1  1  8  1  1  10  1  1  12  1  1  ...

\Centerline{$r_1=1$,  \quad$r_2=\Bover23$,
\quad$r_3=\Bover34$, \quad$r_4=\Bover57$, \quad$r_5=\Bover{23}{32}$,
\quad$r_6=\Bover{28}{39}$, \quad$r_7=\Bover{51}{71}$;}

\noindent note that the obvious approximations $\bover{17}{24}$,
$\bover{86}{120}$ derived from the series for $e$ are not in fact as
close as the even terms $\bover57$, $\bover{28}{39}$ above, and involve
larger numbers\footnote{There is a remarkable expression for the continued
fraction expansion of $e$, due essentially to Euler;
$k_{3m-1}=2m$, $k_{3m}=k_{3m+1}=1$ for $m\ge 2$.  See {\smc Cohn 06}.}.
}

\spheader 372Lc\cmmnt{ Now we need a variety of miscellaneous facts
about these coefficients, which I list here.

\medskip

\quad}{\bf (i)} For any $x\in X$, $n\ge 1$ we have

\Centerline{$p_{n-1}(x)q_n(x)-p_n(x)q_{n-1}(x)=(-1)^n$,
\quad$\phi^n(x)
=\Bover{p_n(x)-xq_n(x)}{xq_{n-1}(x)-p_{n-1}(x)}$\dvro{,}{}}

\cmmnt{\noindent (induce on $n$), so}

\Centerline{$x=\Bover{p_n(x)+p_{n-1}(x)\phi^n(x)}
{q_n(x)+q_{n-1}(x)\phi^n(x)}$.}

\medskip

\quad{\bf (ii)}\dvro{ For}{ Another easy induction on $n$ shows that
for} any finite string $\tbf{m}=(m_1,\ldots,m_n)$ of strictly positive
integers the set $D_{\tbf{m}}=\{x:x\in X,\,k_i(x)=m_i$ for $1\le i\le
n\}$ is an interval in $X$ on which $\phi^n$ is monotonic, being
strictly increasing if $n$ is even and strictly decreasing if $n$ is
odd.   \prooflet{(For the inductive step, note just that

\Centerline{$D_{(m_1,\ldots,m_n)}
=[\bover1{m_1+1},\bover1{m_1}]
\cap\phi^{-1}[D_{(m_2,\ldots,m_n)}]$.)}
}

\medskip

\quad{\bf (iii)} We also need to know that\cmmnt{ the intervals
$D_{\tbf{m}}$
of (ii) are small;  specifically, that} if $\tbf{m}=(m_1,\ldots,m_n)$,
the length of $D_{\tbf{m}}$ is at most $2^{-n+1}$.   \prooflet{\Prf\ All
the coefficients $p_i$, $q_i$, for $i\le n$, take  constant values
$p_i^*$, $q_i^*$ on $D_{\tbf{m}}$, since they are determined from the
coefficients $k_i$ which are constant on $D_{\tbf{m}}$ by definition.
Now every $x\in D_{\tbf{m}}$ is of the form
$(p^*_n+tp^*_{n-1})/(q^*_n+tq^*_{n-1})$ for some $t\in X$ (see (i)
above) and therefore lies between
$p^*_{n-1}/q^*_{n-1}$ and $p^*_n/q^*_n$.   But the distance between
these is

\Centerline{$\bigl|\Bover{p^*_nq^*_{n-1}-p^*_{n-1}q^*_n}
{q^*_nq^*_{n-1}}\bigr|
=\Bover1{q^*_nq^*_{n-1}}$,}

\noindent by the first formula in (i).   Next, noting that $q^*_{i}\ge
q^*_{i-1}+q^*_{i-2}$ for each $i\ge 2$, we see that $q^*_iq^*_{i-1}\ge
2q^*_{i-1}q^*_{i-2}$ for $i\ge 2$, and therefore that $q_n^*q_{n-1}^*\ge
2^{n-1}$, so that the length of $D_{\tbf{m}}$ is at most
$2^{-n+1}$.\ \Qed
}%end of prooflet

\leader{372M}{Theorem} Set $X=[0,1]\setminus\Bbb Q$, and define
$\phi:X\to X$ as in 372L.   Then for every Lebesgue integrable function
$f$ on $X$,

\Centerline{$\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))
=\Bover1{\ln 2}\int_{0}^{1}\Bover{f(t)}{1+t}dt$}

\noindent for almost every $x\in X$.

\proof{{\bf (a)} The integral just written, and the phrase `almost
every', refer of course to Lebesgue measure;  but the first step is to
introduce another measure, so I had better give a name $\mu_L$ to
Lebesgue measure on $X$.   Let $\nu$ be the indefinite-integral measure
on $X$ defined by saying that $\nu E=\bover1{\ln
2}\int_E\bover1{1+x}\mu_L(dx)$ whenever this is defined.   The
coefficient $\bover1{\ln 2}$ is of course chosen to make $\nu X=1$.
Because $\bover1{1+x}>0$ for every $x\in X$, $\dom\nu=\dom\mu_L$ and
$\nu$ has just the same negligible sets as $\mu_L$
(234Lc\formerly{2{}34D});   I can
therefore safely use the terms `measurable set', `almost
everywhere' and `negligible' without declaring which measure I have
in mind each time.

\medskip

{\bf (b)} Now $\phi$ is \imp\ when regarded as a function from $(X,\nu)$
to itself.   \Prf\ For each $k\ge 1$, set
$I_k=\coint{\bover1{k+1},\bover1{k}}$.   On $X\cap I_k$,
$\phi(x)=\bover1x-k$.   Observe that $\phi\restr I_k:X\cap I_k\to X$ is
bijective and differentiable relative to its domain in the sense of
262Fb.   Consider, for any measurable $E\subseteq X$,

$$\eqalign{\int_E\Bover1{(y+k)(y+k+1)}\mu_L(dy)
&=\int_{I_k\cap\phi^{-1}[E]}
\Bover1{(\phi(x)+k)(\phi(x)+k+1)}|\phi'(x)|\mu_L(dx)\cr
&=\int_{I_k\cap\phi^{-1}[E]}
\Bover{x^2}{x+1}\Bover1{x^2}\mu_L(dx)
=\ln 2\cdot\nu(I_k\cap\phi^{-1}[E]),\cr}$$

\noindent using 263D (or more primitive results, of course).   But

\Centerline{$\sum_{k=1}^{\infty}\Bover1{(y+k)(y+k+1)}
=\sum_{k=1}^{\infty}\Bover1{y+k}-\Bover1{y+k+1}=\Bover1{y+1}$}

\noindent for every $y\in[0,1]$, so

$$\nu E=\Bover1{\ln
2}\sum_{k=1}^{\infty}\int_E\Bover1{(y+k)(y+k+1)}\mu_L(dy)
=\sum_{k=1}^{\infty}\nu(I_k\cap\phi^{-1}[E])=\nu\phi^{-1}[E].$$

\noindent As $E$ is arbitrary, $\phi$ is \imp.\   \Qed

\medskip

{\bf (c)} The next thing we need to know is that if $E\subseteq X$ and
$\phi^{-1}[E]=E$ then $E$ is either negligible or conegligible.   \Prf\
I use the sets $D_{\tbf{m}}$ of 372L(c-ii).

\medskip

\quad{\bf (i)} For any string $\tbf{m}=(m_1,\ldots,m_n)$ of strictly
positive integers, we have

\Centerline{$x=\Bover{p^*_n+p^*_{n-1}\phi^n(x)}
{q_n^*+q_{n-1}^*\phi^n(x)}$}

\noindent for every $x\in D_{\tbf{m}}$, where $p_n^*$, etc., are defined
from $\tbf{m}$ as in 372L(c-iii).   Recall also that $\phi^n$ is
strictly monotonic on $D_{\tbf{m}}$.   So for any interval
$I\subseteq[0,1]$ (open, closed or half-open) with endpoints
$\alpha<\beta$, $\phi^{-n}[I]\cap D_{\tbf{m}}$ will be of the form
$X\cap J$, where $J$ is an interval with endpoints
$(p^*_n+p^*_{n-1}\alpha)/(q^*_n+q^*_{n-1}\alpha)$,
$(p^*_n+p^*_{n-1}\beta)/(q^*_n+q^*_{n-1}\beta)$ in some order.   This
means that we can estimate $\mu_L(\phi^{-n}[I]\cap
D_{\tbf{m}})/\mu_LD_{\tbf{m}}$, because it is

$$\bover{
\bigl|\bover{p^*_n+p^*_{n-1}\alpha}{q^*_n+q^*_{n-1}\alpha}
-\bover{p^*_n+p^*_{n-1}\beta}{q^*_n+q^*_{n-1}\beta}\bigr|}
{\bigl|\bover{p^*_n}{q^*_n}
-\bover{p^*_n+p^*_{n-1}}{q_n^*+q^*_{n-1}}\bigr|}
=\Bover{(\beta-\alpha)q^*_n(q_n^*+q^*_{n-1})}
{(q_n^*+q_{n-1}^*\alpha)(q_n^*+q_{n-1}^*\beta)}
\ge\Bover{(\beta-\alpha)q_n^*}{q_n^*+q_{n-1}^*}
\ge\Bover12(\beta-\alpha).$$

\noindent Now look at

\Centerline{$\Cal A=\{E:E\subseteq [0,1]$ is Lebesgue measurable,
 $\mu_L(\phi^{-n}[E]\cap D_{\tbf{m}})
  \ge\Bover12\mu_LE\cdot\mu_LD_{\tbf{m}}\}$.}

\noindent Clearly the union of two disjoint members of $\Cal A$ belongs
to $\Cal A$.   Because $\Cal A$ contains every subinterval of $[0,1]$
it includes the algebra $\Cal E$ of subsets of $[0,1]$ consisting of
finite unions of intervals.   Next, the union of any
non-decreasing sequence in $\Cal A$ belongs to $\Cal A$, and the
intersection of a non-increasing sequence likewise.   But this means
that $\Cal A$ must include the
$\sigma$-algebra generated by $\Cal E$ (136G), that is, the Borel
$\sigma$-algebra.   But also, if $E\in\Cal A$ and $H\subseteq[0,1]$ is
negligible, then

\Centerline{$\mu_L(\phi^{-n}[E\symmdiff H]\cap D_{\tbf{m}})
=\mu_L(\phi^{-n}[E]\cap D_{\tbf{m}})
\ge\Bover12\mu_LE\cdot\mu_LD_{\tbf{m}}
=\Bover12\mu_L(E\symmdiff H)\cdot\mu_LD_{\tbf{m}}$}

\noindent and $E\symmdiff H\in\Cal A$.   And this means that every
Lebesgue measurable subset of $[0,1]$ belongs to $\Cal A$ (134Fb).

\medskip

\quad{\bf (ii)} \Quer\ Now suppose, if possible, that $E$ is a
measurable subset of $X$ and that $\phi^{-1}[E]=E$ and $E$ is neither
negligible nor conegligible in $X$.    Set $\gamma=\bover12\mu_LE>0$.
By Lebesgue's density theorem (223B) there is some $x\in X\setminus E$
such that $\lim_{\delta\downarrow 0}\psi(\delta)=0$, where
$\psi(\delta)=\Bover1{2\delta}\mu_L(E\cap[x-\delta,x+\delta])$ for
$\delta>0$.
Take $n$ so large that $\psi(\delta)<\bover12\gamma$ whenever
$0<\delta\le 2^{-n+1}$, and set $m_i=k_i(x)$ for $i\le n$, so that $x\in
D_{\tbf{m}}$.   Taking the least $\delta$ such that
$D_{\tbf{m}}\subseteq[x-\delta,x+\delta]$, we must have
$\delta\le 2^{-n+1}$, because the length of $D_{\tbf{m}}$ is at most
$2^{-n+1}$ (372L(c-iii)), while $\mu_LD_{\tbf{m}}\ge\delta$, because
$D_{\tbf{m}}$ is an interval.   Accordingly

\Centerline{$\mu_L(E\cap D_{\tbf{m}})\le\mu_L(E\cap[x-\delta,x+\delta])
=2\delta\psi(\delta)<\gamma\delta\le\gamma\mu_LD_{\tbf{m}}$.}

\noindent But we also have

\Centerline{$\mu_L(E\cap D_{\tbf{m}})=\mu_L(\phi^{-n}[E]\cap
D_{\tbf{m}})\ge\gamma\mu_L D_{\tbf{m}}$,}

\noindent by (i).\   \Bang

This proves the result.\   \Qed

\medskip

{\bf (d)} The final fact we need in preparation is that $\phi[E]$ is
negligible for every negligible $E\subseteq X$.   This is because $\phi$
is differentiable relative to its domain (see 263D(ii)).

\medskip

{\bf (e)} Now let $f$ be any $\mu_L$-integrable function.   Because
$\bover1{1+x}\le 1$ for every $x$, $f$ is also $\nu$-integrable
(235K\formerly{2{}35M});   consequently, using (b) above and 372J,

\Centerline{$g(x)
=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))$}

\noindent is defined for almost every $x\in X$, and is a conditional
expectation of $f$ (with respect to the measure $\nu$) on the
$\sigma$-algebra $\Tau_0=\{E:E$ is measurable, $\phi^{-1}[E]=E\}$.
But we have just seen that $\Tau_0$ consists only of negligible and
conegligible sets, so $g$ must be essentially constant;  since
$\int g\,d\nu=\int fd\nu$, we must have

$$\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))
=\int fd\nu
=\Bover1{\ln 2}\int_0^1\Bover{f(t)}{1+t}\mu_L(dt)$$

\noindent for almost every $x$ (using 235K to calculate $\int fd\nu$).
}%end of proof of 372M

\leader{372N}{Corollary} For almost every $x\in [0,1]\setminus\Bbb Q$,

\Centerline{$\lim_{n\to\infty}\Bover1n\#(\{i:1\le i\le n,\,k_i(x)=k\})
=\Bover1{\ln 2}(2\ln(k+1)-\ln k-\ln(k+2))$}

\noindent for every $k\ge 1$, where $k_1(x),\ldots$ are the continued
fraction coefficients of $x$.

\proof{ In 372M, set $f=\chi(X\cap[\bover1{k+1},\bover1k])$.   Then (for
$i\ge 1$) $f(\phi^i(x))=1$ if $k_i(x)=k$ and zero otherwise.   So

$$\eqalign{\lim_{n\to\infty}&\Bover1n\#(\{i:1\le i\le n,\,k_i(x)=k\})\cr
&=\lim_{n\to\infty}\Bover1{n}\sum_{i=1}^nf(\phi^i(x))
=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf(\phi^i(x))\cr
&=\Bover1{\ln 2}\int_0^1\Bover{f(t)}{1+t}dt
=\Bover1{\ln 2}\int_{1/k+1}^{1/k}\Bover1{1+t}dt\cr
&=\Bover1{\ln 2}(\ln(1+\Bover1k)-\ln(1+\Bover1{k+1}))
=\Bover1{\ln 2}(2\ln(k+1)-\ln k-\ln(k+2)),\cr}$$

\noindent for almost every $x\in X$.
}%end of proof of 372N

\leader{372O}{Mixing and
ergodic \dvrocolon{transformations}}\cmmnt{ This seems an
appropriate moment for some brief notes on three special types of
measure-preserving homomorphism or \imp\ function.

\medskip

\noindent}{\bf Definitions (a)(i)}\discrversionA{\footnote{Amended
2010.}}{} Let $\frak A$ be a Boolean algebra.
Then a Boolean homomorphism $\pi:\frak A\to\frak A$ is {\bf ergodic} if
whenever $a$, $b\in\frak A\setminus\{0\}$ there are $m$, $n\in\Bbb N$ such
that $\pi^ma\Bcap\pi^nb\ne 0$.

\medskip

\quad{\bf (ii)} Let $(\frak A,\bar\mu)$ be a probability
algebra and $\pi:\frak A\to\frak A$ a measure-preserving Boolean
homomorphism.   Then $\pi$ is {\bf mixing}\cmmnt{ (sometimes called {\bf
strongly mixing})}
if $\lim_{n\to\infty}\bar\mu(\pi^na\Bcap b)=\bar\mu a\cdot\bar\mu b$ for
all $a$, $b\in\frak A$.

\medskip

\quad{\bf (iii)}\dvAnew{2011} Let $(\frak A,\bar\mu)$ be a probability
algebra and $\pi:\frak A\to\frak A$ a measure-preserving Boolean
homomorphism.   Then $\pi$ is {\bf weakly mixing}
if $\lim_{n\to\infty}\Bover1n\sum_{i=0}^{n-1}
|\bar\mu(\pi^na\Bcap b)-\bar\mu a\cdot\bar\mu b|=0$ for
all $a$, $b\in\frak A$.

\medskip

\spheader 372Ob Let $(X,\Sigma,\mu)$ be a probability space and
$\phi:X\to X$ an \imp\ function.

\medskip

\quad{\bf (i)} $\phi$ is {\bf ergodic}\cmmnt{ (also called {\bf
metrically
transitive}, {\bf indecomposable})} if every measurable set $E$ such
that $\phi^{-1}[E]=E$ is either negligible or conegligible.

\medskip

\quad{\bf (ii)} $\phi$ is {\bf mixing} if
$\lim_{n\to\infty}\mu(F\cap\phi^{-n}[E])=\mu E\cdot\mu F$ for all $E$,
$F\in\Sigma$.

\medskip

\quad{\bf (iii)}\dvAnew{2011} $\phi$ is {\bf weakly mixing} if
if $\lim_{n\to\infty}\Bover1n\sum_{i=0}^{n-1}
|\mu(F\cap\phi^{-n}[E])-\mu E\cdot\mu F|=0$ for
all $E$, $F\in\Sigma$.


\leader{372P}{}\dvAnew{2010}\cmmnt{ For the principal applications of the
idea in 372O(a-i),
we have an alternative definition in terms of fixed-point subalgebras.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra and
$\pi:\frak A\to\frak A$ a Boolean homomorphism, with fixed-point subalgebra
$\frak C$.

(a) If $\pi$ is ergodic, then $\frak C=\{0,1\}$.

(b) If $\pi$ is an automorphism, then
$\pi$ is ergodic iff
$\sup_{n\in\Bbb Z}\pi^na=1$ for every $a\in\frak A\setminus\{0\}$.

(c) If $\pi$ is an automorphism and $\frak A$ is Dedekind
$\sigma$-complete, then $\pi$ is ergodic iff $\frak C=\{0,1\}$.

\proof{{\bf (a)} If $c\in\frak C$, then $\pi^mc=c$ is disjoint from
$\pi^n(1\Bsetminus c)=1\Bsetminus c$ for all $m$, $n\in\Bbb N$, so one of
$c$, $1\Bsetminus c$ must be zero.

\medskip

{\bf (b)(i)} If $\pi$ is ergodic and $a\ne 0$ and $b\Bcap\pi^na=0$ for
every $n\in\Bbb Z$, then $\pi^mb\Bcap\pi^na=\pi^m(b\Bcap\pi^{n-m}a)=0$ for
all $m$, $n\in\Bbb N$, so $b=0$.   As $b$ is arbitrary,
$\sup_{n\in\Bbb Z}\pi^na=1$;  as $a$ is arbitrary, $\pi$ satisfies the
condition.

\medskip

\quad{\bf (ii)} If $\pi$ satisfies the condition, and $a$,
$b\in\frak A\setminus\{0\}$, then there is an $m\in\Bbb Z$ such that
$\pi^ma\Bcap b\ne 0$;  setting $n=\max(-m,0)$,
$\pi^{m+n}a\Bcap\pi^nb\ne 0$, while $m+n$ and $n$ both belong to $\Bbb N$.
As $a$ and $b$ are arbitrary, $\pi$ is ergodic.

\medskip

{\bf (c)} If $\pi$ is ergodic then $\frak C=\{0,1\}$, by (a).
If $\frak C=\{0,1\}$ and $a\in\frak A\setminus\{0\}$, consider
$c=\sup_{n\in\Bbb Z}\pi^na$, which is defined because $\frak A$ is Dedekind
$\sigma$-complete.   Being an automorphism, $\pi$ is
order-continuous (313Ld), so
$\pi c=\sup_{n\in\Bbb Z}\pi^{n+1}a=c$ and $c\in\frak C$.
Since $c\Bsupseteq a$ is non-zero, $c=1$.   As $a$ is arbitrary,
$\pi$ is ergodic, by (b).
}%end of proof of 372P

\leader{372Q}{}\cmmnt{ The following facts are equally straightforward.

\medskip

\noindent}{\bf Proposition}\dvArevised{ and expanded 2011}
(a) Let $(\frak A,\bar\mu)$ be a probability
algebra, $\pi:\frak A\to\frak A$ a measure-preserving Boolean
homomorphism, and $T:L^0=L^0(\frak A)\to L^0$ the Riesz homomorphism such
that $T(\chi a)=\chi\pi a$ for every $a\in\frak A$.

\quad(i) If $\pi$ is mixing, it is weakly mixing.

\quad(ii) If $\pi$ is weakly mixing, it is ergodic.

\quad(iii) The following
are equiveridical:  ($\alpha$) $\pi$ is
ergodic;  ($\beta$) the only $u\in L^0$ such that $Tu=u$ are the
multiples of
$\chi 1$;   ($\gamma$) for every $u\in L^1=L^1(\frak A,\bar\mu)$,
$\sequencen{\bover1{n+1}\sum_{i=0}^nT^iu}$ order*-converges to
$(\int u)\chi 1$.

\quad(iv)
The following are equiveridical:  ($\alpha$) $\pi$ is mixing;
($\beta$) $\lim_{n\to\infty}\innerprod{T^nu}{v}=\int u\int v$ for all $u$,
$v\in L^2(\frak A,\bar\mu)$.

\quad(v)
The following are equiveridical:  ($\alpha$) $\pi$ is weakly mixing;
($\beta$)
$\lim_{n\to\infty}\Bover1n\sum_{k=0}^{n-1}
  |\innerprod{T^ku}{v}-\int u\int v|=0$ for all $u$,
$v\in L^2(\frak A,\bar\mu)$.

(b) Let $(X,\Sigma,\mu)$ be a probability space, with measure algebra
$(\frak A,\bar\mu)$.   Let $\phi:X\to X$ be an \imp\ function and
$\pi:\frak A\to\frak A$ the associated homomorphism such that
$\pi E^{\ssbullet}=(\phi^{-1}[E])^{\ssbullet}$ for every $E\in\Sigma$.

\quad (i) The following are equiveridical:  ($\alpha$) $\phi$ is
ergodic;
($\beta$) $\pi$ is ergodic;  ($\gamma$) for every $\mu$-integrable
real-valued function $f$,
$\sequencen{\bover1{n+1}\sum_{i=0}^nf(\phi^i(x))}$ converges to $\int f$
for almost every $x\in X$.

\quad (ii) $\phi$ is mixing iff $\pi$ is, and in this case $\phi$ is
weakly mixing.

\quad (iii) $\phi$ is weakly mixing iff $\pi$ is, and in this case $\phi$
is ergodic.

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

\medskip

\quad{\bf (iii)\grheada$\Rightarrow$\grheadb} $Tu=u$ iff
$\pi\Bvalue{u>\alpha}=\Bvalue{u>\alpha}$ for
every $\alpha$;  if $\pi$ is ergodic, this means that
$\Bvalue{u>\alpha}\in\{0,1\}$ for every $\alpha$, by 372Pa,
and $u$ must be of the
form $\gamma\chi 1$, where $\gamma=\inf\{\alpha:\Bvalue{u>\alpha}=0\}$.

\medskip

\qquad{\bf \grheadb$\Rightarrow$\grheadc} If ($\beta$) is true and
$u\in L^1$, then we know from 372G that
$\sequencen{\bover1{n+1}\sum_{i=0}^nT^iu}$ is order*-convergent and
$\|\,\|_1$-convergent to some $v$ such that $Tv=v$;  by ($\beta$), $v$
is of the form $\gamma\chi 1$;  and

\Centerline{$\gamma
=\int v=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^n\int T^iu=\int
u$.}

\qquad\grheadc$\Rightarrow$\grheada\  Assuming ($\gamma$), take any
$a\in\frak A$ such that $\pi a=a$, and consider $u=\chi a$.   Then
$T^iu=\chi a$ for every $i$, so

\Centerline{$\chi a=\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nT^iu
=(\int u)\chi 1=\bar\mu a\cdot\chi 1$,}

\noindent and $a$ must be either $0$ or $1$.   By 372Pc, $\pi$ is ergodic.

\medskip

\quad{\bf (iv)}\grheada$\Rightarrow$\grheadb\ Since $\pi$ is mixing,

$$\eqalign{\lim_{n\to\infty}\innerprod{T^n\chi a}{\chi b}
&=\lim_{n\to\infty}\innerprod{\chi\pi^n a}{\chi b}
=\lim_{n\to\infty}\bar\mu(\pi^na\Bcap b)\cr
&=\bar\mu a\cdot\bar\mu b
=\int\chi a\int\chi b\cr}$$

\noindent for all $a$, $b\in\frak A$.   Because
$(u,v)\mapsto\innerprod{T^nu}{v}$ and $(u,v)\mapsto\int u\int v$ are
both bilinear,

\Centerline{$\lim_{n\to\infty}\innerprod{T^nu}{v}=\int u\int v$}

\noindent for all $u$, $v\in S(\frak A)$.   For general $u$,
$v\in L^2(\frak A,\bar\mu)$, take any $\epsilon>0$.   Then there are $u'$,
$v'\in S(\frak A)$ such that

\Centerline{$(\|u-u'\|_2+\|v-v'\|_2)\max(\|u\|_2,\|v\|_2+\|v-v'\|_2)
\le\epsilon$}

\noindent (366C), so that

$$\eqalignno{|\innerprod{T^nu}{v}-\innerprod{T^nu'}{v'}|
&\le|\innerprod{T^nu}{v-v'}|+|\innerprod{T^nu-T^nu'}{v'}|\cr
&\le\|T^nu\|_2\|v-v'\|_2+\|T^nu-T^nu'\|_2\|v'\|_2\cr
&\le\|u\|_2\|v-v'\|_2+\|u-u'\|_2(\|v\|_2+\|v-v'\|_2)\cr
\displaycause{366H(a-iv)}
&\le\epsilon,\cr
|\int u\int v-\int u'\int v'|
&\le|\int u||\int v-v'|+|\int u-u'||\int v'|\cr
&\le\|u\|_2\|v-v'\|_2+\|u-u'\|_2\|v'\|_2
\le\epsilon\cr}$$

\noindent for every $n$, and

\Centerline{$\limsup_{n\to\infty}|\innerprod{T^nu}{v}-\int u\int v|
\le 2\epsilon+\lim_{n\to\infty}|\innerprod{T^nu'}{v'}-\int u'\int v'|
=2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lim_{n\to\infty}\innerprod{T^nu}{v}=\int u\int v$, as required.

\medskip

\qquad\grheadb$\Rightarrow$\grheada\ This is elementary, as ($\alpha$) is
just the case $u=\chi a$, $v=\chi b$ of ($\beta$).

\medskip

\quad{\bf (v)} The argument is essentially the same as in (iv);
($\alpha$) is a special case of ($\beta$);  if ($\alpha$) is true, then by
linearity ($\beta$) is true when $u$, $v\in S(\frak A)$, and the functional
$(u,v)\mapsto\limsup_{n\to\infty}\Bover1n\sum_{k=0}^{n-1}
   |\innerprod{T^ku}{v}-\int u\int v|$ is continuous.

\medskip

{\bf (b)(i)}\grheada$\Rightarrow$\grheadb\  If $\pi a=a$
there is an $E$ such that $\phi^{-1}[E]=E$ and $E^{\ssbullet}=a$, by
372I;  now $\bar\mu a=\mu E\in\{0,1\}$, so $a\in\{0,1\}$.   Thus
the fixed-point subalgebra of $\pi$ is $\{0,1\}$;  by
372Pc again, $\pi$ is ergodic.

\medskip

\qquad\grheadb$\Rightarrow$\grheadc\ Set $u=f^{\ssbullet}\in L^1$.   In
the language of (a), $T^iu=(f\phi^i)^{\ssbullet}$ for each $i$, as in the
proof of 372H, so that

\Centerline{$(\Bover1{n+1}\sum_{i=0}^nf\phi^i)^{\ssbullet}
=\Bover1{n+1}\sum_{i=0}^nT^iu$}

\noindent is order*-convergent to $(\int u)\chi 1=(\int f)\chi 1$, and
$\bover1{n+1}\sum_{i=0}^nf\phi^i\to\int f$ a.e.

\medskip

\qquad\grheadc$\Rightarrow$\grheada\ If $\phi^{-1}[E]=E$ then, applying
($\gamma$) to $f=\chi E$, we see that $\chi E\eae\mu E\cdot\chi X$, so
that $E$ is either negligible or conegligible.

\medskip

\quad{\bf (ii)-(iii)} Simply translating the definitions, we see that
$\pi$ is mixing, or weakly mixing, iff $\phi$ is.
So the results here are reformulations of (a-i) and (a-ii).
}%end of proof of 372Q

\cmmnt{\leader{372R}{Remarks}\dvAformerly{3{}72Pc}
{\bf (a)} The reason for introducing
`ergodic' homomorphisms in this section is of course 372G/372J;  if
$\pi$ in 372G, or $\phi$ in 372J, is ergodic, then the limit $Pu$ or $g$
must be (essentially) constant, being a conditional expectation on a
trivial subalgebra.

\spheader 372Rb In the definition 372O(b-i) I should perhaps emphasize that
we look only at {\it measurable} sets $E$.   We certainly expect that
there will generally be many sets $E$ for which $\phi^{-1}[E]=E$, since
any union of orbits of $\phi$ will have this property.

\spheader 372Rc Part (c) of the proof of 372M was devoted to showing
that the function $\phi$ there was ergodic;  see also 372Xm.   For
another ergodic transformation see 372Xr.   For examples of mixing
transformations see 333P, 372Xp, 372Xq, 372Xt, 372Xw and 372Xx.

\spheader 372Rd It seems to be difficult to display explicitly
a weakly mixing transformation which is not mixing.   There is an
example in {\smc Chacon 69}, and I give another in 494F in Volume 4.
In a certain sense, however, `most' measure-preserving automorphisms of the
Lebesgue probability algebra are weakly mixing but not mixing;  I will
return to this in 494E.
}%end of comment

\leader{372S}{}\cmmnt{ There is a useful sufficient condition for a
homomorphism or function to be mixing.

\medskip

\noindent}{\bf Proposition} (a) Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\pi:\frak A\to\frak A$ a measure-preserving Boolean
homomorphism.   If $\bigcap_{n\in\Bbb N}\pi^n[\frak A]=\{0,1\}$, then
$\pi$ is mixing.

(b) Let $(X,\Sigma,\mu)$ be a probability space, and $\phi:X\to X$ an
\imp\ function.   Set

\Centerline{$\Tau=\{E:$ for every $n\in\Bbb N$ there is an $F\in\Sigma$
such that $E=\phi^{-n}[F]\}$.}

\noindent If every member of $\Tau$ is either negligible or
conegligible, $\phi$ is mixing.

\proof{{\bf (a)} Let $T:L^0=L^0(\frak A)\to L^0$ be the Riesz
homomorphism associated with $\pi$.   Take any $a$, $b\in\frak A$ and
any non-principal ultrafilter $\Cal F$ on $\Bbb N$.   Then
$\sequencen{T^n(\chi a)}$ is a bounded sequence in the reflexive space
$L^2_{\bar\mu}=L^2(\frak A,\bar\mu)$, so
$v=\lim_{n\to\Cal F}T^n(\chi a)$ is defined for the weak topology of
$L^2_{\bar\mu}$.   Now for each
$n\in\Bbb N$ set $\frak B_n=\pi^n[\frak A]$.   This is a closed
subalgebra of $\frak A$ (314F(a-i)), and contains $\pi^ia$ for every
$i\ge n$.   So if we identify $L^2(\frak B_n,\bar\mu\restrp\frak B_n)$
with the corresponding subspace of $L^2_{\bar\mu}$ (366I), it
contains $T^i(\chi a)$ for every $i\ge n$;  but also it is norm-closed,
therefore weakly closed (3A5Ee), so contains $v$.   This means that
$\Bvalue{v>\alpha}$ must belong to $\frak B_n$ for every $\alpha$ and
every $n$.   But in this case
$\Bvalue{v>\alpha}\in\bigcap_{n\in\Bbb N}\frak B_n=\{0,1\}$ for every
$\alpha$, and $v$ is of the form $\gamma\chi 1$.   Also

\Centerline{$\gamma=\int v=\lim_{n\to\Cal F}\int T^n(\chi a)
=\bar\mu a$.}

\noindent So

\Centerline{$\lim_{n\to\Cal F}\bar\mu(\pi^n a\Bcap b)
=\lim_{n\to\Cal F}\int T^n(\chi a)\times\chi b
=\int v\times\chi b
=\gamma\bar\mu b
=\bar\mu a\cdot\bar\mu b$.}

\noindent But this is true of every non-principal ultrafilter $\Cal F$
on $\Bbb N$, so we must have
$\lim_{n\to\infty}\bar\mu(\pi^na\Bcap b)=\bar\mu a\cdot\bar\mu b$
(3A3Lc).   As $a$ and $b$ are arbitrary, $\pi$ is mixing.

\medskip

{\bf (b)} Let $(\frak A,\bar\mu)$ be the measure algebra of
$(X,\Sigma,\mu)$, and $\pi:\frak A\to\frak A$ the measure-preserving
homomorphism corresponding to $\phi$.
The point is that if $a\in\bigcap_{n\in\Bbb N}\pi^n[\frak A]$,
there is an $E\in\Tau$ such that $E^{\ssbullet}=a$.   \Prf\ For each
$n\in\Bbb N$ there is an $a_n\in\frak A$ such that $\pi^na_n=a$;  say
$a_n=F_n^{\ssbullet}$ where $F_n\in\Sigma$.   Then
$\phi^{-n}[F_n]^{\ssbullet}=a$.   Set

\Centerline{$E=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}\phi^{-n}[F_n]$,
\quad $E_k=\bigcup_{m\ge k}\bigcap_{n\ge m}\phi^{-(n-k)}[F_n]$}

\noindent for each $k$;  then $E^{\ssbullet}=a$ and

\Centerline{$\phi^{-k}[E_k]
=\bigcup_{m\ge k}\bigcap_{n\ge m}\phi^{-n}[F_n]
=\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}\phi^{-n}[F_n]=E$}

\noindent for every $k$, so $E\in\Tau$.\  \Qed

So $\bigcap_{n\in\Bbb N}\frak A_n=\{0,1\}$ and $\pi$ and $\phi$ are
mixing.
}%end of proof of 372S

\exercises{\leader{372X}{Basic exercises (a)}
%\spheader 372Xa
Let $U$ be any reflexive Banach space, and $T:U\to U$ an operator of
norm at most $1$.   Set $A_n=\bover1{n+1}\sum_{i=0}^nT^i$ for each
$n\in\Bbb N$.   Show that $Pu=\lim_{n\to\infty}A_nu$ is defined (as a
limit for the norm topology) for every $u\in U$, and that $P:U\to U$ is
a projection onto $\{u:Tu=u\}$.  \Hint{show that $\{u:Pu$ is defined$\}$
is a closed linear subspace of $U$ containing $Tu-u$ for every $u\in
U$.}

(This is a version of the {\bf mean ergodic theorem}.)
%372A

\sqheader 372Xb Let $(\frak A,\bar\mu)$ be a measure algebra, and
$T\in\Cal T^{(0)}_{\bar\mu,\bar\mu}$;  set
$A_n=\bover1{n+1}\sum_{i=0}^nT^i$ for $n\in\Bbb N$.   Take any
$p\in\coint{1,\infty}$ and $u\in L^p=L^p(\frak A,\bar\mu)$.   Show that
$\sequencen{A_nu}$ is order*-convergent and
$\|\,\|_p$-convergent to some
$v\in L^p$.   \Hint{put 372Xa together with 372D.}
%372D

\spheader 372Xc Let $(\frak A,\bar\mu)$ be a probability algebra, and
$\pi:\frak A\to\frak A$ a measure-preserving Boolean homomorphism.   Let
$P:L^1\to L^1$ be the operator defined as in 365O/366Hb, where
$L^1=L^1_{\bar\mu}$, so that $\int_aPu=\int_{\pi a}u$ for
$u\in L^1$ and $a\in\frak A$.   Set
$A_n=\bover1{n+1}\sum_{i=0}^nP^i:L^1\to L^1$
for each $i$.   Show that for any $u\in L^1$, $\sequencen{A_nu}$
is order*-convergent and $\|\,\,\|_1$-convergent to the conditional
expectation of $u$ on the subalgebra $\{a:\pi a=a\}$.
%372G

\spheader 372Xd Show that if $f$ is any Lebesgue integrable function on
$\Bbb R$, and $y\in\Bbb R\setminus\{0\}$, then

\Centerline{$\lim_{n\to\infty}\bover1{n+1}\sum_{k=0}^nf(x+ky)=0$}

\noindent for almost every $x\in\Bbb R$.
%372H

\spheader 372Xe Let $(X,\Sigma,\mu)$ be a measure space and
$\phi:X\to X$ an \imp\ function.   Set
$\Tau=\{E:E\in\Sigma,\,\mu(\phi^{-1}[E]\symmdiff E)=0\}$,
$\Tau_0=\{E:E\in\Sigma,\,\phi^{-1}[E]=E\}$.   (i) Show that
$\Tau=\{E\symmdiff F:E\in\Tau_0,\,F\in\Sigma,\,\mu F=0\}$.   (ii) Show
that a set $A\subseteq X$ is $\mu\restrp\Tau_0$-negligible iff
$\phi^n[A]$ is $\mu$-negligible for every $n\in\Bbb N$.
%372J

\sqheader 372Xf Let $\nu$ be a Radon probability measure on $\Bbb R$
such that $\int|t|\nu(dt)$ is finite (cf.\ 271F).   On
$X=\Bbb R^{\Bbb N}$ let $\lambda$ be the product measure obtained when
each factor is given the measure $\nu$.  Define $\phi:X\to X$ by setting
$\phi(x)(n)=x(n+1)$ for $x\in X$, $n\in\Bbb N$.   (i) Show that $\phi$
is \imp.   \Hint{254G.   See also 372Xw below.}    (iii) Set
$\gamma=\int t\nu(dt)$, the
expectation of the distribution $\nu$.   By considering
$\bover1{n+1}\sum_{i=0}^nf\frsmallcirc\phi^i$, where $f(x)=x(0)$ for
$x\in X$, show that
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^nx(i)=\gamma$ for
$\lambda$-almost every $x\in X$.
%372J

\sqheader 372Xg Use the Ergodic Theorem to prove Kolmogorov's Strong Law
of Large Numbers (273I), as follows.   Given a complete probability
space $(\Omega,\Sigma,\mu)$ and an independent identically distributed
sequence $\sequencen{f_n}$ of measurable functions from $\Omega$ to
$\Bbb R$, set $X=\BbbR^{\Bbb N}$ and
$f(\omega)=\sequencen{f_n(\omega)}$ for $\omega\in\Omega$.   Show that
if we give each copy of $\Bbb R$ the distribution of $f_0$ then $f$ is
\imp\ for $\mu$ and the product measure $\lambda$ on $X$.   Now use
372Xf.
%372Xf, 372J

\sqheader 372Xh Let
$\sequencen{f_n}$ be a sequence of real-valued random variables with finite
expectation such that
$(f_0,f_1,\ldots,f_n)$ has the same joint distribution as
$(f_1,f_2,\ldots,f_{n+1})$ for every $n\in\Bbb N$.   Show that
$\sequencen{\Bover1{n+1}\sum_{i=0}^nf_i}$ converges a.e.   \Hint{Let
$(X,\Sigma,\mu)$ be the underlying probability space.   Reduce to the
case in which every $f_i$ is measurable and defined everywhere in $X$.
Define $\theta:X\to\BbbR^{\Bbb N}$ by setting $\theta(x)(n)=f_n(x)$ for
$x\in X$,
$n\in\Bbb N$.   Let $\lambda$ be the image measure $\mu\theta^{-1}$.
Set $\phi(z)(n)=z(n+1)$ for $z\in\BbbR^{\Bbb N}$ and $n\in\Bbb N$.
Show that $\phi$ is \imp\ for $\lambda$, and apply 372J.}
%372J 372Xg out of order
% strong law for stationary processes

\spheader 372Xi Show that the continued fraction coefficients of
$\Bover1{\sqrt 2}$ are $1$, $2$, $2$, $2,\ldots$.
%372L

\sqheader 372Xj For $x\in X=[0,1]\setminus\Bbb Q$ let
$k_1(x),k_2(x),\ldots$ be its continued-fraction coefficients.
Show that $x\mapsto\sequencen{k_{n+1}(x)-1}$ is a bijection between $X$
and $\BbbN^{\Bbb N}$ which is a homeomorphism if $X$ is given its usual
topology (as a subset of $\Bbb R$) and $\BbbN^{\Bbb N}$ is given its
usual product topology (each copy of $\Bbb N$ being given its discrete
topology).
%372L

\spheader 372Xk Set $x=\bover12(\sqrt5-1)$.
Show that, in the notation of 372L, $k_n(x)=1$ and
$q_n(x)=p_{n-1}(x)$ for every $n\ge 1$ and that $\sequencen{p_n(x)}$ is the
Fibonacci sequence.
%372L

\spheader 372Xl For any irrational $x\in[0,1]$ let
$k_1(x),k_2(x),\ldots$ be its continued-fraction coefficients and
$p_n(x)$, $q_n(x)$ the numerators and denominators of its
continued-fraction approximations, as described in 372L.   Write
$r_n(x)=p_n(x)/q_n(x)$.   (i) Show that $x$ lies between $r_n(x)$ and
$r_{n+1}(x)$ for every $n\in\Bbb N$.   (ii) Show that
$r_{n+1}(x)-r_n(x)=(-1)^n/q_n(x)q_{n+1}(x)$ for every $n\in\Bbb N$.
(iii) Show that $|x-r_n(x)|\le 1/q_n(x)^2k_{n+1}(x)$ for every $n\ge 1$.
(iv) Hence show that for almost every $\gamma\in\Bbb R$, the set
$\{(p,q):p\in\Bbb Z,\,q\ge 1,\,|\gamma-\bover{p}{q}|\le\epsilon/q^2\}$
is infinite for every $\epsilon>0$.  (v) Show that if $n\ge 1$,
$p$, $q\in\Bbb N$ and $0<q\le q_n(x)$, then
$|x-\bover{p}{q}|\ge|x-r_n(x)|$, with equality only when $p=p_n(x)$ and
$q=q_n(x)$.
%372L, 372M

\spheader 372Xm In 372M, let $\Tau_1$ be the family $\{E:$ for every
$n\in\Bbb N$ there is a measurable set $F\subseteq X$ such that
$\phi^{-n}[F]=E\}$.   Show that every member of $\Tau_1$ is either
negligible or conegligible.   \Hint{the argument of part (c) of the
proof of 372M still works.}   Hence show that $\phi$ is mixing for the
measure $\nu$.
%372M, 372S

\spheader 372Xn Let $(\frak A,\bar\mu)$ be an atomless probability
algebra.   Show that the following are equiveridical:  (i) $\frak A$ is
homogeneous; (ii) there is an ergodic measure-preserving Boolean
homomorphism $\pi:\frak A\to\frak A$;  (iii) there is a mixing
measure-preserving automorphism $\pi:\frak A\to\frak A$.   \Hint{333P.}
%372O

\spheader 372Xo\dvArevised{2011}
Let $(\frak A,\bar\mu)$ be a probability algebra, and
$\pi:\frak A\to\frak A$ a measure-preserving Boolean homomorphism.   (i)
Show that if $n\ge 1$ then $\pi$ is mixing iff $\pi^n$ is mixing.   (ii)
Show that if $n\ge 1$ then $\pi$ is weakly mixing iff $\pi^n$ is weakly
mixing.   (iii)
Show that if $n\ge 1$ and $\pi^n$ is ergodic then $\pi$ is ergodic.
(iv) Show that if $\pi$ is an automorphism then it is ergodic, or
mixing, or weakly mixing, iff $\pi^{-1}$ is.
%372O

\sqheader 372Xp Consider the {\bf tent map}
$\phi_{\alpha}(x)=\alpha\min(x,1-x)$ for $x\in[0,1]$,
$\alpha\in[0,2]$.   Show that $\phi_2$ is \imp\ and mixing for Lebesgue
measure on $[0,1]$.   \Hint{show that
$\phi_2^{n+1}(x)=\phi_2(\fraction{2^nx})$ for
$n\ge 1$, and hence that $\mu(I\cap\phi_2^{-n}[J])=\mu I\cdot\mu J$
whenever $I$ is of the form
$[2^{-n}k,2^{-n}(k+1)]$ and $J$ is an interval.}
%372O

\spheader 372Xq Consider the {\bf logistic map}
$\psi_{\beta}(x)=\beta x(1-x)$ for $x\in [0,1]$, $\beta\in[0,4]$.   Show
that $\psi_4$ is \imp\ and mixing for the Radon measure on $[0,1]$ with
density function
$t\mapsto\Bover1{\pi\sqrt{t(1-t)}}$.   \Hint{show that the transformation
$t\mapsto\sin^2\bover{\pi t}2$ matches it with the tent map.}   Show that
for almost every $x$,

\Centerline{$\lim_{n\to\infty}\Bover1{n+1}\#(\{i:i\le
n,\,\psi_4^i(x)\le\alpha\})=\Bover2{\pi}\arcsin\sqrt{\alpha}$}

\noindent for every $\alpha\in[0,1]$.
%372O

\spheader 372Xr Let $\mu$ be Lebesgue measure on
$\coint{0,1}$, and fix an irrational number $\alpha\in\coint{0,1}$.
(i) Set $\phi(x)=x+_1\alpha$ for every
$x\in\coint{0,1}$, where $x+_1\alpha$ is whichever of $x+\alpha$,
$x+\alpha-1$ belongs to $\coint{0,1}$.    Show that $\phi$ is \imp.
(ii) Show that if $I\subseteq\coint{0,1}$ is an interval then
$\lim_{n\to\infty}\bover1{n+1}\sum_{i=0}^n\chi I(\phi^i(x))=\mu I$ for
almost every $x\in\coint{0,1}$.   \Hint{this is Weyl's Equidistribution
Theorem (281N).}   (iii) Show that $\phi$ is ergodic.   \Hint{take the
conditional expectation operator $P$ of 372G, and look at
$P(\chi I^{\ssbullet})$ for intervals $I$.}   (iv) Show that $\phi^n$ is
ergodic for any $n\in\Bbb Z\setminus\{0\}$.   (v) Show that $\phi$ is
not weakly mixing.
%372G, 372Q

\spheader 372Xs Let $p$, $q\in [1,\infty]$ be such that
$\bover1p+\bover1q=1$.
(i) Let $(\frak A,\bar\mu)$ be a probability algebra,
$\pi:\frak A\to\frak A$ a mixing measure-preserving homomorphism, and
$T:L^0(\frak A)\to L^0(\frak A)$ the corresponding homomorphism.
Show that $\lim_{n\to\infty}\int T^nu\times v=\int u\int v$ whenever
$u\in L^p(\frak A,\bar\mu)$ and $v\in L^q(\frak A,\bar\mu)$.
\Hint{start with $u$, $v\in S(\frak A)$.}
(ii) Let $(X,\Sigma,\mu)$ be a probability space and
$\phi:X\to X$ a mixing \imp\ function.   Show that
$\lim_{n\to\infty}\int f(\phi^n(x))g(x)dx=\int f\int g$ whenever
$f\in\eusm L^p(\mu)$ and $g\in\eusm L^q(\mu)$.
%372Q

\spheader 372Xt Give $\coint{0,1}$ Lebesgue measure $\mu$, and let
$k\ge 2$ be an integer.   Define $\phi:\coint{0,1}\to\coint{0,1}$ by
setting
$\phi(x)=\fraction{kx}$, the fractional part of $kx$.   Show that $\phi$
is \imp.    Show that $\phi$ is mixing.   \Hint{if
$I=\coint{k^{-n}i,k^{-n}(i+1)}$, $J=\coint{k^{-n}j,k^{-n}(j+1)}$ then
$\mu(I\cap\phi^{-m}[J])=\mu I\cdot\mu J$ for all $m\ge n$.}
%372Q

\spheader 372Xu Let $(X,\Sigma,\mu)$ be a probability space and
$\phi:X\to X$ an ergodic \imp\ function.   Let $f$ be a
$\mu$-virtually measurable function defined almost everywhere in $X$
such that $\int fd\mu=\infty$.   Show that
$\lim_{n\to\infty}\Bover1{n+1}\sum_{i=0}^nf\phi^i$ is infinite a.e.
\Hint{look at the corresponding limits for $f_m=f\wedge m\chi X$.}
%372Q

\spheader 372Xv For irrational $x\in[0,1]$, write $k_1(x),k_2(x),\ldots$
for the continued-fraction coefficients of $x$.   Show that the limit
$\lim_{n\to\infty}\bover1n\sum_{i=1}^nk_i(x)$ is infinite
for almost every
$x$.   \Hint{take $\phi$, $\nu$ as in 372M, and show that
$\int k_1d\nu=\infty$.}
%372Q, 372Xu

\spheader 372Xw Let $(X,\Sigma,\mu)$ be any probability space, and let
$\lambda$ be the product measure on $X^{\Bbb N}$.   Define
$\phi:X^{\Bbb N}\to X^{\Bbb N}$ by setting $\phi(x)(n)=x(n+1)$.  Show
that $\phi$ is
\imp.   Show that $\phi$ satisfies the conditions of 372S, so is mixing.
%372S

\spheader 372Xx Let $(X,\Sigma,\mu)$ be any probability space, and
$\lambda$ the product measure on $X^{\Bbb Z}$.   Define
$\phi:X^{\Bbb Z}\to X^{\Bbb Z}$ by setting $\phi(x)(n)=x(n+1)$.  Show
that $\phi$ is
\imp.   Show that $\phi$ is mixing.   \Hint{show that if $C$, $C'$ are
basic cylinder sets then $\mu(C\cap\phi^{-n}[C'])=\mu C\cdot\mu C'$ for
all $n$ large enough.}   Show that $\phi$ does not
ordinarily satisfy the conditions of 372S.   (Compare 333P.)
%372S

\spheader 372Xy\dvAnew{2011}(i) Let $\frak A$ be a Boolean algebra,
$\pi:\frak A\to\frak A$ a Boolean homomorphism, and
$\phi:\frak A\to\frak A$ an automorphism.   Show that if
$\pi$ is ergodic then $\phi\pi\phi^{-1}$ is ergodic.
(ii) Let $(\frak A,\bar\mu)$ be a probability algebra,
$\pi:\frak A\to\frak A$ a measure-preserving Boolean homomorphism, and
$\phi:\frak A\to\frak A$ a measure-preserving
Boolean automorphism.   Show that if
$\pi$ is mixing, or weakly mixing, then so is $\phi\pi\phi^{-1}$.
%372O out of order query

\leader{372Y}{Further exercises (a)}
%\spheader 372Ya
In 372D, show that the null space of the limit operator $P$ is precisely
the closure in $M^{1,0}$ of the subspace $\{Tu-u:u\in M^{1,0}\}$.

\spheader 372Yb Let $(\frak A,\bar\mu)$ be a measure algebra,
$T\in\Cal T^{(0)}_{\bar\mu,\bar\mu}$, $p\in\ooint{1,\infty}$ and
$u\in L^p(\frak A,\bar\mu)$.  Set
$u^*=\sup_{n\in\Bbb N}\bover1{n+1}\sum_{i=0}^n|T^iu|$.   (i) Show that
for any $\gamma>0$,

\Centerline{$\bar\mu\Bvalue{u^*>\gamma}
\le\Bover2{\gamma}\int_{\Bvalue{|u|>\gamma/2}}|u|$.}

\noindent\Hint{apply 372C to $(|u|-\bover12\gamma\chi 1)^+$.}   (ii)
Show that $\|u^*\|_p\le 2(\Bover{p}{p-1})^{1/p}\|u\|_p$.   ({\it
Hint\/}: show that
$\int_{\Bvalue{|u|>\alpha}}|u|=\alpha\bar\mu\Bvalue{|u|>\alpha}
+\int_{\alpha}^{\infty}\bar\mu\Bvalue{|u|>\beta}d\beta$;  see 365A.
Use 366Xa to show that

\Centerline{$\|u^*\|_p^p\le 2p\int_0^{\infty}\gamma^{p-2}
\int_{\gamma/2}^{\infty}\bar\mu\Bvalue{|u|>\beta}d\beta d\gamma
+2^p\|u\|_p^p$,}

\noindent and reverse the order of integration.   Compare
275Yd.)   (This is {\bf Wiener's Dominated Ergodic Theorem}.)
% 372D

\spheader 372Yc Let $(\frak A,\bar\mu)$ be a probability algebra and $T$
an operator in $\Cal T^{(0)}_{\bar\mu,\bar\mu}$.   Take
$u\in L^1=L^1(\frak A,\bar\mu)$ such that $h(|u|)\in L^1$, where
$h(t)=t\ln t$ for
$t\ge 1$, $0$ for $t\le 1$, and $\bar h$ is the corresponding function
from $L^0(\frak A)$ to itself.   Set
$u^*=\sup_{n\in\Bbb N}\bover1{n+1}\sum_{i=0}^n|T^iu|$.   Show that
$u^*\in L^1$.   \Hint{use
the method of 372Yb to show that
$\int_2^{\infty}\bar\mu\Bvalue{u^*>\gamma}d\gamma\le 2\int\bar h(u)$.}
%372D, 372Yb

\spheader 372Yd Let $U$ be a Banach space, $(\frak A,\bar\mu)$ a
semi-finite measure algebra and $\sequencen{T_n}$ a sequence of
continuous linear operators from $U$ to $L^0=L^0(\frak A)$ with its
topology of convergence in measure.   Suppose that
$\sup_{n\in\Bbb N}T_nu$ is defined in $L^0$ for every $u\in U$.
Show that $\{u:u\in U$, $\sequencen{T_nu}$ is order*-convergent in $L^0\}$
is a norm-closed linear subspace of $U$.
%372D

\spheader 372Ye In 372G, suppose that $\frak A$ is atomless.   Show that
there is always an $a\in\frak A$ such that $\bar\mu a\le\bover12$ and
$\inf_{i\le n}\pi^ia\ne 0$ for every $n$, so that (except in trivial
cases) $\sequencen{A_n(\chi a)}$ will not be
$\|\,\|_{\infty}$-convergent.
%372G

\spheader 372Yf Let $(X,\Sigma,\mu)$ be a measure space with
measure algebra $(\frak A,\bar\mu)$.   Let $\Phi$ be a family of \imp\
functions from $X$ to itself, and for $\phi\in\Phi$ let
$\pi_{\phi}:\frak A\to\frak A$ be the associated
homomorphism.   Set
$\frak C=\{c:c\in\frak A,\,\pi_{\phi}c=c$ for every $\phi\in\Phi\}$,
$\Tau=\{E:E\in\Sigma,\,\phi^{-1}[E]\symmdiff E$ is negligible for every
$\phi\in\Phi\}$ and
$\Tau_0=\{E:E\in\Sigma,\,\phi^{-1}[E]=E$ for every $\phi\in\Phi\}$.
Show that (i) $\Tau$ and $\Tau_0$
are $\sigma$-subalgebras of $\Sigma$ (ii) $\Tau_0\subseteq\Tau$ (iii)
$\Tau=\{E:E\in\Sigma,\,E^{\ssbullet}\in\frak C\}$ (iv) if $\Phi$ is
countable and $\phi\psi=\psi\phi$ for all $\phi$, $\psi\in\Phi$, then
$\frak C=\{E^{\ssbullet}:E\in\Tau_0\}$.
%372I

\spheader 372Yg  Show that an irrational $x\in\ooint{0,1}$ has an
eventually periodic sequence of continued fraction coefficients iff it
is a solution of a quadratic equation with integral coefficients.
%372L  mt37bits

\spheader 372Yh In the language of 372L-372N and 372Xl, show the
following.   (i) For any $x\in X$ and $n\ge 2$,
$q_n(x)q_{n-1}(x)\ge 2^{n-1}$ and $p_{n}(x)p_{n+1}(x)\ge 2^{n-1}$, so that
$q_{n+1}(x)p_{n}(x)\ge 2^{n-1}$, $|1-x/r_n(x)|\le 2^{-n+1}$ and
$|\ln x-\ln r_n(x)|\le 2^{-n+2}$.   Also
$|x-r_n(x)|\ge 1/q_n(x)q_{n+2}(x)$.   (ii) For any $x\in X$, $n\ge 1$,
$p_{n+1}(x)=q_n(\phi(x))$ and
$q_n(x)\prod_{i=0}^{n-1}r_{n-i}(\phi^{i}(x))=1$.   (iii) For any
$x\in X$, $n\ge 1$, $|\ln q_n(x)+\sum_{i=0}^{n-1}\ln\phi^i(x)|\le 4$.
(iv) For almost every $x\in X$,

\Centerline{$\lim_{n\to\infty}\Bover1{n}\ln q_n(x)
=-\Bover1{\ln 2}\int_0^1\Bover{\ln t}{1+t}dt
=\Bover{\pi^2}{12\ln 2}$.}

\noindent \Hint{225Xg, 282Xo.}  (v) For almost every $x\in X$,
$\lim_{n\to\infty}\bover1n\ln|x-r_n(x)|=-\Bover{\pi^2}{6\ln 2}$.
(vi) For almost every $x\in X$, $11^{-n}\le|x-r_n(x)|\le 10^{-n}$ and
$3^n\le q_n(x)\le 4^n$ for all but finitely many $n$.
%372N, 372Xl

\spheader 372Yi\dvArevised{2011}(i)
Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
probability spaces, with c.l.d.\ product $(X\times Y,\Lambda,\lambda)$.
Suppose that
$\phi:X\to X$ is a weakly mixing \imp\ function and
$\psi:Y\to Y$ is an ergodic
\imp\ function.   Define $\theta:X\times Y\to X\times Y$ by setting
$\theta(x,y)=(\phi(x),\psi(y))$ for all $x$, $y$.   Show that $\theta$
is an ergodic \imp\ function.
(ii) Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be probability
algebras, with probability algebra free product $(\frak C,\bar\lambda)$.
Suppose that $\phi:\frak A\to\frak A$ is a
weakly mixing measure-preserving
Boolean homomorphism and $\psi:\frak B\to\frak B$ is an ergodic
measure-preserving Boolean
homomorphism.   Let $\theta:\frak C\to\frak C$ be the measure-preserving
Boolean homomorphism such that $\theta(a\otimes b)=\phi a\otimes\psi b$
for all $a\in\frak A$ and $b\in\frak B$ (325Xe).   ($\alpha$)
Show that $\theta$ is ergodic.
($\beta$)\dvAnew{2016} Show that if $\psi$ is weakly mixing
then $\theta$ is weakly mixing.
($\gamma$)\dvAnew{2016} Show that if $\phi$ and $\psi$ are mixing then
$\theta$ is mixing.
%372O

\spheader 372Yj Let $\langle(X_i,\Sigma_i,\mu_i)\rangle_{i\in I}$ be any
family of probability spaces, with product $(X,\Lambda,\lambda)$.
Suppose that for each $i\in I$ we are given an \imp\ function
$\phi_i:X_i\to X_i$.   (i) Show that there is a corresponding \imp\
function
$\phi:X\to X$ given by setting $\phi(x)(i)=\phi_i(x(i))$ for $x\in X$,
$i\in I$.   (ii) Show that if every $\phi_i$ is mixing so is $\phi$.
(iii)\dvAnew{2011}
Show that if every $\phi_i$ is weakly mixing so is $\phi$.
%372O 372Yi

\spheader 372Yk Give an example of an ergodic measure-preserving
automorphism $\phi:\coint{0,1}\to\coint{0,1}$ such that $\phi^2$ is not
ergodic.   \Hint{set $\phi(x)=\bover12(1+\phi_0(2x))$ for $x<\bover12$,
$x-\bover12$ for $x\ge\bover12$.   See also 388Xg.}
%von Neumann transf
%can we have ergodic $\phi$ such that $\phi^n$ not ergodic for any
% n\ge 2 ?  I suppose so
%372O, 372Xo

\spheader 372Yl Show that there is an ergodic $\phi:[0,1]\to[0,1]$ such
that $(\xi_1,\xi_2)\mapsto(\phi(\xi_1),\phi(\xi_2)):[0,1]^2\to[0,1]^2$
is not ergodic.   \Hint{372Xr.}
%372Yj, 372Xr, 372O, 372Q

\spheader 372Ym Let $M$ be an $r\times r$ matrix with integer
coefficients and non-zero determinant, where $r\ge 1$.   Let
$\phi:\coint{0,1}^r\to\coint{0,1}^r$ be the function such that
$\phi(x)-Mx\in\Bbb Z^r$ for every $x\in\coint{0,1}^r$.   Show that
$\phi$ is \imp\ for Lebesgue measure on $\coint{0,1}^r$.
%372Q  372Xt %mt37bits

\spheader 372Yn(i)\dvArevised{2011}
Let $(\frak A,\bar\mu)$ be a probability algebra,
$\pi:\frak A\to\frak A$ a weakly mixing measure-preserving Boolean
homomorphism, and $T=T_{\pi}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ the
corresponding linear operator (365N).   Show that if
$u\in L^1_{\bar\mu}$ is such that $\{T^nu:n\in\Bbb N\}$ is relatively
compact for the norm topology, then $u=\alpha\chi 1$ for some $\alpha$.
(ii) Let $\mu$ be Lebesgue measure on $\coint{0,1}$, $(\frak A,\bar\mu)$
its measure algebra, $\alpha\in\coint{0,1}$ an irrational number,
$\phi(x)=x+_1\alpha$ for $x\in\coint{0,1}$ (as in 372Xr), and
$T:L^1(\mu)\to L^1(\mu)$ the linear operator defined by setting
$Tg^{\ssbullet}=(g\phi)^{\ssbullet}$ for $g\in\eusm L^1(\mu)$.   Show
that $\{T^n:n\in\Bbb Z\}$ is relatively compact for the strong operator
topology on $\eurm B(L^1(\mu);L^1(\mu))$.
%372Xr 372Q

\spheader 372Yo In 372M, show that for any measurable set
$E\subseteq X$, $\lim_{n\to\infty}\mu_L\phi^{-n}[E]=\nu E$.
\Hint{recall that
$\phi$ is mixing for $\nu$ (372Xm).   Hence show that
$\lim_{n\to\infty}\int_{\phi^{-n}[E]}g\,d\nu=\nu E\cdot\int g\,d\nu$ for
any integrable $g$.   Apply this to a Radon-Nikod\'ym derivative of
$\mu_L$ with respect to $\nu$.}   (I understand that this result is due
to Gauss.)
%372L, 372S, 372Xm

\spheader 372Yp\dvAnew{2010} (i) Show that there are a Boolean algebra
$\frak A$ and an automorphism $\pi:\frak A\to\frak A$ which is not
ergodic, but has fixed-point algebra $\{0,1\}$.
%finite-cofinite algebra of  \Bbb Z\times\{0,1\}
(ii) Show that there are a $\sigma$-finite measure algebra
$(\frak A,\bar\mu)$ and a measure-preserving Boolean homomorphism
$\pi:\frak A\to\frak A$ which is not ergodic, but has fixed-point algebra
$\{0,1\}$.
%372P mt37bits out of order query

\spheader 372Yq\dvAnew{2010}
For a Boolean algebra $\frak A$ and a Boolean homomorphism
$\pi:\frak A\to\frak A$, write $T_{\pi}$ for the corresponding operator
from $L^{\infty}(\frak A)$ to itself, as defined in 363F.
(i) Suppose that $\frak A$ is a Boolean algebra, $\pi:\frak A\to\frak A$ is
a Boolean homomorphism, $u\in L^{\infty}(\frak A)$ and $T_{\pi}u=u$.
Show that if either $\pi$ is ergodic or $\frak A$ is
Dedekind $\sigma$-complete and the fixed-point subalgebra of $\pi$ is
$\{0,1\}$, then $u$ must be a multiple of $\chi 1$.
(ii)\dvAformerly{3{}63Yk}
Find a Boolean algebra $\frak A$, an automorphism
$\pi:\frak A\to\frak A$ with fixed-point algebra $\{0,1\}$,
and a $u\in L^{\infty}(\frak A)$, not a
multiple of $\chi 1$, such that $T_{\pi}u=u$.
%372P mt37bits out of order query

\spheader 372Yr\dvAnew{2011} Set $\Cal F_d
=\{I:I\subseteq\Bbb N$, $\lim_{n\to\infty}\Bover1n\#(I\cap n)=1\}$.
(i) Show that $\Cal F_d$ is a filter on $\Bbb N$.   (ii) Show that for a
bounded sequence $\sequencen{\alpha_n}$ in $\Bbb R$, the following are
equiveridical:  ($\alpha$) $\lim_{n\to\Cal F_d}\alpha_n=0$;   ($\beta$)
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n|\alpha_k|=0$;
($\gamma$)
$\lim_{n\to\infty}\Bover1{n+1}\sum_{k=0}^n\alpha^2_k=0$.
($\Cal F_d$ is called the {\bf (asymptotic) density filter}.)
%372O out of order query

\spheader 372Ys\dvAnew{2011} Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\phi:\frak A\to\frak A$ a measure-preserving Boolean
homomorphism.   (i) Show that there are a probability algebra
$(\frak C,\bar\lambda)$, a measure-preserving Boolean homomorphism
$\pi:\frak A\to\frak C$, and a measure-preserving automorphism
$\tilde\phi:\frak C\to\frak C$ such that $\tilde\phi\pi=\pi\phi$ and
$\frak C$ is the closure of
$\bigcup_{n\in\Bbb N}\tilde\phi^{-n}[\pi[\frak A]]$ for the
measure-algebra topology.   \Hint{328J.}   (ii) Show that
$\tilde\phi$ is ergodic, or mixing, or weakly mixing iff $\phi$ is.
%372Q

\leaveitout{\spheader 372Y* Let $\Cal F$ be a filter on $\Bbb N$.   If
$(\frak A,\bar\mu)$ is a probability algebra and $\pi:\frak A\to\frak A$
is a measure-preserving Boolean homomorphism, say that $\pi$ is
{\bf $\Cal F$-mixing} if
$\lim_{n\to\Cal F}\bar\mu(\pi^na\Bcap b)=\bar\mu a\cdot\bar\mu b$
for all $a$, $b\in\frak A$;  if
$(X,\Sigma,\mu)$
is a probability space and $\phi:X\to X$ is an \imp\ function, say that
$\phi$ is
{\bf $\Cal F$-mixing} if $\lim_{n\to\Cal F}\mu(E\cap\phi^{-n}[F])
=\mu E\cdot\mu F$ for all $a$, $b\in\frak A$.    (i) Show that in this
case
no finite set can belong to $\Cal F$.   (ii) Show that $\pi$ is mixing
iff it is $\Cal F$-mixing for every non-principal ultrafilter $\Cal F$
on $\Bbb N$.   (iii) Show that if $\pi$ is $\Cal F$-mixing for any
$\Cal F$, it is ergodic.   (iv) In 372Yj, show that if every $\phi_i$ is
$\Cal F$-mixing, so is $\phi$.   (v) In 372Xr, show that $\phi$ is not
$\Cal F$-mixing for any $\Cal F$.   (vi) Show that if
$\Cal F=\{I:\lim_{n\to\infty}\Bover1n\#(I\cap n)=1\}$, then
$\pi$ is $\Cal F$-mixing iff it is weakly mixing.
%?example of $\Cal F$-mixing, not mixing?
%372Q
}%end of leaveitout
}%end of exercises

\cmmnt{\Notesheader{372} I have chosen an entirely conventional route
to the Ergodic Theorem here, through the Mean Ergodic Theorem (372Xa)
or, rather, the fundamental lemma underlying it (372A), and the Maximal
Ergodic Theorem (372B-372C).   What is not to be found in every
presentation is the generality here.   I speak of arbitrary $T\in\Cal
T^{(0)}$, the operators which are contractions both for $\|\,\|_1$ and
for $\|\,\|_{\infty}$, not requiring $T$ to be positive, let alone
correspond to a measure-preserving homomorphism.   (I do not mention
$\Cal T^{(0)}$ in the statement of 372C, but of course it is present in
spirit.)   The work we
have done up to this point puts this extra generality within easy reach,
but as the rest of the section shows, it is not needed for the principal
examples.   Only in 372Xc do I offer an application not associated in the
usual way with a measure-preserving homomorphism or an \imp\ function.

The Ergodic Theorem is an `almost-everywhere pointwise convergence
theorem', like the strong law(s) of large numbers and the martingale
theorem(s) (\S273, \S275).   Indeed Kolmogorov's form of the strong law
can be derived from the Ergodic Theorem (372Xg).   There are some very
strong family resemblances.   For instance, the Maximal Ergodic Theorem
corresponds to the most basic of all the martingale inequalities (275D).
Consequently we have similar results, obtained by similar methods,
concerning the domination of sequences starting from members of $L^p$
(372Yb, 275Yd), though the inequalities are not identical.   (Compare
also 372Yc with 275Ye.)   There are some tantalising reflections of
these traits in results surrounding Carleson's theorem on the pointwise
convergence of square-integrable Fourier series (see \S286 {\it
notes\/}), but Carleson's theorem seems to be much harder than the
others.   Other forms of the strong law (273D, 273H) do not appear to
fit into quite the same pattern, but I note that here, as with the
Ergodic Theorem, we begin with a study of square-integrable functions
(see part (e) of the proof of 372D).

After 372D, there is a contraction and concentration in the scope of the
results, starting with a simple replacement of $M^{1,0}$ with $L^1$
(372F).   Of course it is almost as easy to prove 372D from 372F as the
other way about;  I give precedence to 372D only because $M^{1,0}$ is
the space naturally associated with the class $\Cal T^{(0)}$ of
operators to which these methods apply.   Following this I turn to the
special family of operators to which the rest of the section is devoted,
those associated with measure-preserving homomorphisms (372E), generally
on probability spaces (372G).   This is the point at which we can begin
to identify the limit as a conditional expectation as well as an
invariant element.

Next comes the translation into the language of measure spaces and \imp\
functions, all perfectly straightforward in view of
372I.   These turn 372E into 372H and 372G into the main part of
372J.

In 372J-372K I find myself writing at some length about a technical
problem.   The root of the difficulty is in the definition of
`conditional expectation'.   Now it is generally accepted that any pure
mathematician has `Humpty Dumpty's privilege':  `When {\it I} use
a word, it means just what I choose it to
mean -- neither more nor less'.   With any privilege come duties and
responsibilities;  here, the duty to be self-consistent, and
the responsibility to try to use terms in ways which will not mystify or
mislead the unprepared reader.   Having written down a definition of
`conditional expectation' in Volume 2, I must either stick to it, or go
back and change it, or very carefully explain exactly what modification
I wish to make here.   I don't wish to suggest that absolute consistency
-- in terminology or anything else -- is supreme among mathematical
virtues.   Surely it is better to give local meanings to words, or
tolerate ambiguities, than to suppress ideas which cannot be formulated
effectively otherwise, and among `ideas' I wish to include the
analogies and resonances which a suitable language can suggest.   But I
do say that it is always best to be conscious of what one is doing -- I
go farther:  one of the things which mathematics is for, is to raise our
consciousness of what our thoughts really are.   So I believe it is
right to pause occasionally over such questions.

In 372L-372N (see also 372Xl, 372Xv, 372Xm, 372Xk, 372Yh, 372Yo) I make an
excursion into number theory.   This is a remarkable example of the
power of advanced measure theory to give striking results in other
branches of mathematics.   Everything here is derived from 
{\smc Billingsley 65}, who goes farther than I have space for, and gives
references to more.   Here let me point to 372Xj;  almost accidentally,
the construction offers a useful formula for a homeomorphism
between two of the most important spaces of descriptive set theory,
which will be important to us in Volume 4.

I end the section by introducing three terms, `ergodic', `mixing'
and `weakly mixing'
transformations, not because I wish to use them for any new ideas (apart
from the elementary 372P-372S, %372P 372Q 372R 372S
these must wait for \S\S385-387 %385S 385T 386D 387C 387E 387K
below and \S494 in Volume 4), %494D 494E 494F
but because
it may help if I immediately classify some of the \imp\ functions we
have seen (372Xp-372Xr, %372Xp 372Xq 372Xr
372Xt, 372Xw, 372Xx).
Of course in any application of any ergodic theorem it
is of great importance to be able to identify the limits promised by the
theorem, and the point about an ergodic transformation is just that our
averages converge to constant limits (372Q).   Actually proving that a
given \imp\ function is ergodic is rarely quite trivial (see 372M,
372Xq, 372Xr), though a handful of standard techniques cover a large
number of cases, and it is usually obvious when a map is {\it not}
ergodic, so that if an invariant region does not leap to the eye one has
a good hope of ergodicity.   The extra concept of `weakly mixing'
transformation is hardly relevant to anything in this volume (though see
372Yi-372Yj), but
is associated with a remarkable topological fact about automorphism groups
of probability algebras, to come in 494E.

I ought to remark on the odd shift between the definitions of `ergodic
Boolean homomorphism' and `ergodic \imp\ function' in 372O.   The point is
that the version in 372O(b-i) is the standard formulation in this context,
but that its natural translation into the version `a Boolean homomorphism
from a probability algebra to itself is ergodic if its fixed-point
subalgebra is trivial', although perfectly satisfactory in that context,
allows unwelcome phenomena if applied to general Boolean algebras
(372Yp, 372Yq).   The definition in 372O(a-i) is
rather closer to the essential idea
of ergodicity of a dynamical system, which asks that the system should
always evolve along a path which approximates all possible states.
In practice, however,
we shall nearly always be dealing with automorphisms of
Dedekind $\sigma$-complete algebras, for which we can use the
fixed-point criterion of 372Pc.

I take the opportunity to mention two famous functions from $[0,1]$ to
itself, the `tent' and `logistic' maps (372Xp, 372Xq).   In the
formulae $\phi_{\alpha}$, $\psi_{\beta}$ I include redundant parameters;
this is because the real importance of these functions lies in the way
their behaviour depends, in bewildering complexity, on these parameters.
It is only at the extreme values $\alpha=2$, $\beta=4$ that the methods
of this volume can tell us anything interesting.
}%end of comment

\discrpage

\frfilename{mt373.tex} 
\versiondate{25.5.16}
\copyrightdate{1996} 
      
\def\chaptername{Linear operators between function spaces} 
\def\sectionname{Decreasing rearrangements} 
\hyphenation{re-arrange-ments}
      
\newsection{373} 
      
I take a section to discuss operators in the class $\Cal T^{(0)}$ of 
371F-371H %371F 371G 371H
and two associated classes $\Cal T$, $\Cal T^{\times}$ (373A).   These turn 
out to be intimately related to the idea of `decreasing 
rearrangement' (373C).   In 373D-373F I give elementary properties of 
decreasing rearrangements;  then in 373G-373O I show how they may be 
used to characterize the set $\{Tu:T\in\Cal T\}$ for a given $u$. 
The argument uses a natural topology on the set $\Cal T$ (373K). 
I conclude with remarks on the possible values of $\int Tu\times v$ for 
$T\in\Cal T$ (373P-373Q) and identifications between $\Cal 
T^{(0)}_{\bar\mu,\bar\nu}$, $\Cal T^{(0)}_{\bar\nu,\bar\mu}$ and $\Cal 
T^{\times}_{\bar\mu,\bar\nu}$ (373R-373T). 
      
\vleader{60pt}{373A}{Definition} Let $(\frak A,\bar\mu)$ and  
$(\frak B,\bar\nu)$ be measure algebras.   \cmmnt{Recall that 
$M^{1,\infty}(\frak A,\bar\mu)=L^1(\frak A,\bar\mu)+L^{\infty}(\frak A)$ 
is the set of those $u\in L^0(\frak A)$ such that $(|u|-\alpha\chi 1)^+$ 
is integrable for some $\alpha$, its norm $\|\,\|_{1,\infty}$ being 
defined by the formulae 
      
$$\eqalign{\|u\|_{1,\infty} 
&=\min\{\|v\|_1+\|w\|_{\infty}:v\in L^1,\,w\in L^{\infty},\,v+w=u\}\cr 
&=\min\{\alpha+\int(|u|-\alpha\chi 1)^+:\alpha\ge 0\}\cr}$$ 
      
\noindent (369Ob).}%end of comment 
      
\spheader 373Aa $\Cal T_{\bar\mu,\bar\nu}$ will be 
the space of linear operators  
$T:M^{1,\infty}(\frak A,\bar\mu)\to M^{1,\infty}(\frak B,\bar\nu)$ such that 
$\|Tu\|_1\le\|u\|_1$ for every 
$u\in L^1(\frak A,\bar\mu)$ and $\|Tu\|_{\infty}\le\|u\|_{\infty}$ for 
every $u\in L^{\infty}(\frak A)$.   \cmmnt{(Compare the definition of 
$\Cal T^{(0)}$ in 371F.)} 
      
\spheader 373Ab If $\frak B$ is Dedekind complete\cmmnt{, so that 
$M^{1,\infty}(\frak A,\bar\mu)$, 
being a solid linear subspace of the Dedekind complete space $L^0(\frak B)$, 
is Dedekind complete},  
$\Cal T^{\times}_{\bar\mu,\bar\nu}$ will be  
$\Cal T_{\bar\mu,\bar\nu} 
\cap\eurm L^{\times}(M^{1,\infty}(\frak A,\bar\mu); 
\discretionary{}{}{}M^{1,\infty}(\frak A,\bar\mu))$. 
      
\vleader{108pt}{373B}{Proposition} Let $(\frak A,\bar\mu)$ and  
$(\frak B,\bar\nu)$ be measure algebras. 
      
(a) $\Cal T=\Cal T_{\bar\mu,\bar\nu}$ is a convex subset of the unit 
ball of  
$\eurm B(M^{1,\infty}(\frak A,\bar\mu);M^{1,\infty}(\frak B,\bar\nu))$. 
      
(b) If $T\in\Cal T$ then $T\restr M^{1,0}(\frak A,\bar\mu)$ belongs to 
$\Cal T^{(0)}_{\bar\mu,\bar\nu}$. 
So if $T\in\Cal T$, $p\in\coint{1,\infty}$ and  
$u\in L^p(\frak A,\bar\mu)$ then 
$Tu\in L^p(\frak B,\bar\nu)$ and $\|Tu\|_p\le\|u\|_p$. 
      
(c) If $\frak B$ is Dedekind complete, then $\Cal T$ is a solid subset of
$\eurm L^{\sim}(M^{1,\infty}(\frak A,\bar\mu);M^{1,\infty}(\frak B,\bar\nu))$.
      
(d) If $\pi:\frak A\to\frak B$ is a measure-preserving Boolean 
homomorphism, 
then we have a corresponding operator $T\in\Cal T$ defined by saying 
that $T(\chi a)=\chi(\pi a)$ for every $a\in\frak A$.   If $\pi$ is 
order-continuous, then so is $T$. 
      
(e) If $(\frak C,\bar\lambda)$ is another measure algebra,
$T\in\Cal T$ and $S\in\Cal T_{\bar\nu,\bar\lambda}$ then  
$ST\in\Cal T_{\bar\mu,\bar\lambda}$. 
      
\proof{{\bf (a)} As 371G, parts (a-i) and (a-ii) of the proof. 
      
\medskip 
      
{\bf (b)} If $u\in M^{1,0}_{\bar\mu}$ and $\epsilon>0$, then $u$ is 
expressible as $u'+u''$ where $\|u''\|_{\infty}\le\epsilon$ and $u'\in 
L^1_{\bar\mu}$.   (Set 
      
\Centerline{$u''=(u^+\wedge\epsilon\chi 1)-(u^-\wedge\epsilon\chi 1)$.)} 
      
\noindent So 
      
\Centerline{$(|Tu|-\epsilon\chi 1)^+
\le(|Tu|-|Tu''|)^+
\le||Tu|-|Tu''||
\le|Tu-Tu''|
=|Tu'|\in L^1_{\bar\nu}$.} 
      
\noindent As $\epsilon$ is arbitrary, $Tu\in M^{1,0}_{\bar\nu}$;  as $u$ 
is arbitrary, $T\restr M^{1,0}_{\bar\mu}\in \Cal T^{(0)}$.   Now the 
rest is a consequence of 371Gd. 
      
\medskip 
      
{\bf (c)(i)} Because $M^{1,\infty}_{\bar\nu}$ is a solid 
linear subspace of $L^0(\frak B)$, which is Dedekind complete because 
$\frak B$ is,  
$\eurm L^{\sim}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$  
is a Riesz space (355Ea). 
      
\medskip

\quad{\bf (ii)} 
$\Cal T\subseteq\eurm L^{\sim}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$.
\Prf\ Suppose that $T\in\Cal T$.
Take any $u\ge 0$ in $M^{1,\infty}_{\bar\mu}$.   Let 
$\alpha\ge 0$ be such that $(u-\alpha\chi 1)^+\in L^1_{\bar\mu}$. 
Because $T\restr L^1_{\bar\mu}$ belongs to 
$\eurm B(L^1_{\bar\mu};L^1_{\bar\nu}) 
=\eurm L^{\sim}(L^1_{\bar\mu};L^1_{\bar\nu})$ (371D), 
$w_0=\sup\{Tv:v\in L^1_{\bar\mu},\,0\le v\le(u-\alpha\chi 1)^+\}$ is 
defined in $L^1_{\bar\nu}$.   Now if $v\in M^{1,\infty}_{\bar\mu}$ and 
$0\le v\le u$, we must have 
      
\Centerline{$Tv=T(v-\alpha\chi 1)^++T(v\wedge\alpha\chi 1) 
\le w_0+\alpha\chi 1\in M^{1,\infty}_{\bar\nu}$.} 
      
\noindent Thus $\{Tv:0\le v\le u\}$ is bounded above in 
$M^{1,\infty}_{\bar\nu}$.   As $u$ is arbitrary, 
$T\in\eurm L^{\sim}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$ 
(355Ba).\ \Qed 

\medskip
      
\quad{\bf (iii)} $\Cal T$ is solid in
$\eurm L^{\sim}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$.
\Prf\ Suppose that $T\in\Cal T$,
$T_1\eurm L^{\sim}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$ and
$|T_1|\le|T|$.   Then
      
$$\eqalignno{\|T_1u\|_1 
&\le\||T_1||u|\|_1 
\Displaycause{355Eb}
\le\||T||u|\|_1 
\le\||T|\restr L^1_{\bar\mu}\|\||u|\|_1 
=\|T\restr L^1_{\bar\mu}\|\|u\|_1
\Displaycause{371D} 
\le\|u\|_1\cr}$$ 
      
\noindent for every $u\in L^1_{\bar\mu}$.   At the same time, if 
$u\in L^{\infty}(\frak A)$, then 
      
$$\eqalign{|T_1u| 
&\le|T_1||u| 
\le|T||u| 
=\sup_{|v|\le|u|}Tv\cr 
&\le\sup_{|v|\le|u|}\|Tv\|_{\infty}\chi 1 
\le\sup_{|v|\le|u|}\|v\|_{\infty}\chi 1 
=\|u\|_{\infty}\chi 1,\cr}$$ 
      
\noindent so $\|T_1u\|_{\infty}\le\|u\|_{\infty}$.   Thus $T_1\in\Cal T$.   
By 352Ja, this is enough to show that $\Cal T$ is solid.\ \Qed 
      
\medskip 
      
{\bf (d)} By 365N and 363F, we have norm-preserving positive linear 
operators $T_1:L^1_{\bar\mu}\to L^1_{\bar\nu}$ and 
$T_{\infty}:L^{\infty}(\frak A)\to L^{\infty}(\frak B)$ defined by 
saying that $T_1(\chi a)=\chi(\pi a)$ whenever $\bar\mu a<\infty$ and 
$T_{\infty}(\chi a)=\chi(\pi a)$ for every $a\in\frak A$.   If  
$u\in S(\frak A^f)=L^1_{\bar\mu}\cap S(\frak A)$ (365F), then 
$T_1u=T_{\infty}u$, because 
both $T_1$ and $T_{\infty}$ are linear and they agree on  
$\{\chi a:\bar\mu a<\infty\}$.   If $u\ge 0$ in 
$M^{\infty,1}_{\bar\mu}=L^1_{\bar\mu}\cap L^{\infty}(\frak A)$, 
there is a non-decreasing sequence $\sequencen{u_n}$ 
in $S(\frak A^f)$ such that $u=\sup_{n\in\Bbb N}u_n$ and 
      
\Centerline{$\lim_{n\to\infty}\|u-u_n\|_1 
=\lim_{n\to\infty}\|u-u_n\|_{\infty}=0$} 
      
\noindent (see the proof of 369Od), so that 
      
\Centerline{$T_1u=\sup_{n\in\Bbb N}T_1u_n 
=\sup_{n\in\Bbb N}T_{\infty}u_n=T_{\infty}u$.} 
      
\noindent Accordingly $T_1$ and $T_{\infty}$ agree on  
$L^1_{\bar\mu}\cap L^{\infty}(\frak A)$.   But this means that if  
$u\in M^{1,\infty}_{\bar\mu}$ 
is expressed as $v+w=v'+w'$, where $v$, $v'\in L^1_{\bar\mu}$ and $w$, 
$w'\in L^{\infty}(\frak A)$, we shall have 
      
\Centerline{$T_1v'+T_{\infty}w' 
=T_1v+T_{\infty}w+T_1(v'-v)-T_{\infty}(w-w') 
=T_1v+T_{\infty}w$,} 
      
\noindent because $v'-v=w-w'\in M^{\infty,1}_{\bar\mu}$.   Accordingly 
we have an operator $T:M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\nu}$ 
defined by setting 
      
\Centerline{$T(v+w)=T_1v+T_{\infty}w$ whenever $v\in L^1_{\bar\mu}$, 
$w\in L^{\infty}(\frak A)$.} 
      
\noindent This formula makes it easy to check that $T$ is linear 
and positive, and it clearly belongs to $\Cal T$. 
      
To see that $T$ is uniquely defined, observe that $T\restr 
L^1_{\bar\mu}$ 
and $T\restr L^{\infty}(\frak A)$ are uniquely defined by the values $T$ 
takes on $S(\frak A^f)$, $S(\frak A)$ respectively, because these spaces 
are dense for the appropriate norms. 
      
Now suppose that $\pi$ is order-continuous.   Then $T_1$ and 
$T_{\infty}$ are also order-continuous (365Na, 363Ff).   If  
$A\subseteq M^{1,\infty}_{\bar\mu}$ is non-empty and downwards-directed  
and has infimum $0$, take $u_0\in A$ 
and $\gamma>0$ such that $(u_0-\gamma\chi 1)^+\in L^1_{\bar\mu}$.   Set 
      
\Centerline{$A_1=\{(u-\gamma\chi 1)^+:u\in A,\,u\le u_0\}$, 
\quad $A_{\infty}=\{u\wedge\gamma\chi 1:u\in A\}$.} 
      
\noindent Then $A_1\subseteq L^1_{\bar\mu}$ and $A_{\infty}\subseteq 
L^{\infty}(\frak A)$ are both downwards-directed and have infimum $0$, 
so $\inf T_1[A_1]=\inf T_{\infty}[A_{\infty}]=0$ in $L^0(\frak B)$. 
But this means that $\inf(T_1[A_1]+T_{\infty}[A_{\infty}])=0$ (351Dc). 
Now any $w\in 
T_1[A_1]+T_{\infty}[A_{\infty}]$ is expressible as $T(u-\gamma\chi 
1)^++T(u'\wedge\gamma\chi 1)$ where $u$, $u'\in A$;  because $A$ is 
downwards-directed, there is a $v\in A$ such that $v\le u\wedge u'$, in 
which case $Tv\le w$.   Accordingly $T[A]$ must also have infimum $0$. 
As $A$ is arbitrary, $T$ is order-continuous. 
      
\medskip 
      
{\bf (e)} is obvious, as usual. 
}%end of proof of 373B 
      
\leader{373C}{Decreasing rearrangements}\cmmnt{ The following concept 
is fundamental to any understanding of the class $\Cal T$.}  Let 
$(\frak A,\bar\mu)$ be a measure algebra.   Write 
$\cmmnt{M^{0,\infty}_{\bar\mu}=\mskip5mu}M^{0,\infty}(\frak A,\bar\mu)$ 
for the set of those $u\in L^0(\frak A)$ such that 
$\bar\mu\Bvalue{|u|>\alpha}$ is finite for some $\alpha\in\Bbb R$. 
\cmmnt{(See 369N for the ideology of this notation.)}   \cmmnt{It is 
easy to see that} $M^{0,\infty}(\frak A,\bar\mu)$ is a solid linear 
subspace of $L^0(\frak A)$.   Let $(\frak A_L,\bar\mu_L)$ be the measure 
algebra of Lebesgue measure on $\coint{0,\infty}$.   For 
$u\in M^{0,\infty}(\frak A,\bar\mu)$ its {\bf decreasing rearrangement} is 
$u^*\in \cmmnt{M^{0,\infty}_{\bar\mu_L}=\mskip5mu}M^{0,\infty}(\frak A_L,\bar\mu_L)$, 
defined by setting 
$u^*=g^{\ssbullet}$, where 
      
\Centerline{$g(t) 
=\min\{\alpha:\alpha\ge 0,\,\bar\mu\Bvalue{|u|>\alpha}\le t\}$} 
      
\noindent for every $t>0$.   \cmmnt{(The infimum is always finite
because $\inf_{\alpha\in\Bbb R}\bar\mu\Bvalue{|u|>\alpha}=0$, by 
364Aa($\beta$) and 321F, and by 364Aa($\alpha$) the infimum is attained.)} 
      
\cmmnt{I will 
maintain this usage of the symbols $\frak A_L$, $\bar\mu_L$, $u^*$ for 
the rest of this section.} 
      
\leader{373D}{Lemma} Let $(\frak A,\bar\mu)$ be a measure algebra. 
      
(a) For any $u\in M^{0,\infty}(\frak A,\bar\mu)$, its decreasing 
rearrangement $u^*$ may be defined by the formula 
      
\Centerline{$\Bvalue{u^*>\alpha} 
=\coint{0,\bar\mu\Bvalue{|u|>\alpha}}^{\ssbullet}$ for every  
$\alpha\ge 0$,} 
      
\noindent that is, 
      
\Centerline{$\bar\mu_L\Bvalue{u^*>\alpha}=\bar\mu\Bvalue{|u|>\alpha}$ 
for every $\alpha\ge 0$.} 
      
(b) If $|u|\le|v|$ in $M^{0,\infty}(\frak A,\bar\mu)$, then  
$u^*\le v^*$;  \cmmnt{in particular,} $|u|^*=u^*$. 
      
(c)(i) If $u=\sum_{i=0}^n\alpha_i\chi a_i$, where  
$a_0\Bsupseteq a_1\Bsupseteq\ldots\Bsupseteq a_n$ and $\alpha_i\ge 0$ for each $i$, 
then $u^*=\sum_{i=0}^n\alpha_i\chi\coint{0,\bar\mu a_i}^{\ssbullet}$. 
      
\quad(ii) If $u=\sum_{i=0}^n\alpha_i\chi a_i$ where $a_0,\ldots,a_n$ are 
disjoint and $|\alpha_0|\ge|\alpha_1|\ge\ldots\ge|\alpha_n|$, then 
$u^*=\sum_{i=0}^n|\alpha_i|\chi\coint{\beta_i,\beta_{i+1}}^{\ssbullet}$, 
where $\beta_i=\sum_{j<i}\bar\mu a_i$ for $i\le n+1$. 
      
(d) If $E\subseteq\ooint{0,\infty}$ is any Borel set, and  
$u\in M^0(\frak A,\bar\mu)$, then   
$\bar\mu_L\Bvalue{u^*\in E}=\bar\mu\Bvalue{|u|\in E}$. 
      
(e) Let $h:\coint{0,\infty}\to\coint{0,\infty}$ be a non-decreasing 
function such that $h(0)=0$, and write $\bar h$ for the corresponding 
functions on $L^0(\frak A)^+$ and $L^0(\frak A_L)^+$\cmmnt{ (364H)}.   
Then $(\bar h(u))^*=\bar h(u^*)$ whenever $u\ge 0$ in  
$M^0(\frak A,\bar\mu)$.   If $h$ is 
continuous on the left, $(\bar h(u))^*=\bar h(u^*)$ whenever $u\ge 0$ in 
$M^{0,\infty}(\frak A,\bar\mu)$. 
      
(f) If $u\in M^{0,\infty}(\frak A,\bar\mu)$ and $\alpha\ge 0$, then 
      
\Centerline{$(u^*-\alpha\chi 1)^+=((|u|-\alpha\chi 1)^+)^*$.} 
 
\woddheader{373D}{0}{0}{0}{24pt} 
      
(g) If $u\in M^{0,\infty}(\frak A,\bar\mu)$, then for any $t>0$ 
      
\Centerline{$\int_0^tu^* 
=\inf_{\alpha\ge 0}\alpha t+\int(|u|-\alpha\chi 1)^+$.} 
      
(h) If $A\subseteq (M^{0,\infty}(\frak A,\bar\mu))^+$ is non-empty and 
upwards-directed and has supremum  
$u_0\in M^{0,\infty}(\frak A,\bar\mu)$, then 
$u_0^*=\sup_{u\in A}u^*$. 
      
\wheader{373D}{0}{0}{0}{24pt} 
      
\proof{{\bf (a)} Set 
      
\Centerline{$g(t)=\inf\{\alpha:\bar\mu\Bvalue{|u|>\alpha}\le t\}$} 
      
\noindent as in 373C.  If $\alpha\ge 0$, 
      
\Centerline{$g(t)>\alpha 
\iff\bar\mu\Bvalue{|u|>\beta}>t\text{ for some }\beta>\alpha 
\iff\bar\mu\Bvalue{|u|>\alpha}>t$} 
      
\noindent (because 
$\Bvalue{|u|>\alpha}=\sup_{\beta>\alpha}\Bvalue{|u|>\beta}$), so 
      
\Centerline{$\Bvalue{u^*>\alpha}=\{t:g(t)>\alpha\}^{\ssbullet} 
=\coint{0,\bar\mu\Bvalue{|u|>\alpha}}^{\ssbullet}$.} 
      
\noindent Of course this formula defines $u^*$. 
      
\medskip 
      
{\bf (b)} This is obvious, either from the definition in 373C or from 
(a) just above. 
      
\medskip 
      
{\bf (c)(i)} Setting $v=\sum_{i=0}^n\alpha_i\chi\coint{0,\bar\mu 
a_i}^{\ssbullet}$,  we have 
      
$$\eqalign{\Bvalue{v>\alpha} 
&=0\text{ if }\sum_{i=0}^n\alpha_i\le\alpha,\cr 
&=\coint{0,\bar\mu a_j}^{\ssbullet} 
\text{ if }\sum_{i=0}^{j-1}\alpha_i\le\alpha<\sum_{i=0}^j\alpha_i,\cr 
&=\coint{0,\bar\mu a_0}^{\ssbullet} 
\text{ if }0\le\alpha<\alpha_0,\cr}$$ 
      
\noindent and in all cases is equal to 
$\coint{0,\bar\mu\Bvalue{|u|>\alpha}}^{\ssbullet}$. 
      
\medskip 
      
\quad{\bf (ii)} A similar argument applies.   (If any $a_j$ has infinite 
measure, then $a_i$ is irrelevant for $i>j$.) 
      
\medskip 
      
{\bf (d)} Fix $\gamma>0$ for the moment, and consider 
      
\Centerline{$\Cal A=\{E:E\subseteq\ooint{\gamma,\infty}$ is a Borel set, 
$\bar\mu_L\Bvalue{u^*\in E}=\bar\mu\Bvalue{|u|\in E}\}$,} 
      
\Centerline{$\Cal I=\{\ooint{\alpha,\infty}:\alpha\ge\gamma\}$.} 
      
\noindent Then $\Cal I\subseteq\Cal A$ (by (a)), $I\cap J\in\Cal I$ for 
all $I$, $J\in\Cal I$, $E\setminus F\in\Cal A$ whenever $E$, 
$F\in\Cal A$ and $F\subseteq E$ (because $u\in M^0_{\bar\mu}$, so 
$\bar\mu\Bvalue{|u|\in E}<\infty$), and 
$\bigcup_{n\in\Bbb N}E_n\in\Cal A$ whenever $\sequencen{E_n}$ is a 
non-decreasing sequence in $\Cal A$.   So, by the Monotone Class Theorem 
(136B), $\Cal A$ includes the $\sigma$-algebra of subsets of 
$\ooint{\gamma,\infty}$ generated by $\Cal I$;  but this must contain 
$E\cap\ooint{\gamma,\infty}$ for every Borel set $E\subseteq\Bbb R$. 
      
Accordingly, for any Borel set $E\subseteq\ooint{0,\infty}$, 
      
\Centerline{$\bar\mu_L\Bvalue{u^*\in E} 
=\sup_{n\in\Bbb N}\bar\mu_L 
  \Bvalue{u^*\in E\cap\ooint{2^{-n},\infty}\thinspace} 
=\bar\mu\Bvalue{|u|\in E}$.} 
      
\medskip 
      
{\bf (e)} For any $\alpha>0$, $E_{\alpha}=\{t:h(t)>\alpha\}$ is a Borel 
subset of 
$\ooint{0,\infty}$.   If $u\in M^0_{\bar\mu}$ then, using (d) above, 
      
\Centerline{$\bar\mu_L\Bvalue{\bar h(u^*)>\alpha} 
=\bar\mu_L\Bvalue{u^*\in E_{\alpha}} 
=\bar\mu\Bvalue{u\in E_{\alpha}} 
=\bar\mu\Bvalue{\bar h(u)>\alpha} 
=\bar\mu_L\Bvalue{(\bar h(u))^*>\alpha}$.} 
      
\noindent As both $(\bar h(u))^*$ and $\bar h(u^*)$ are equivalence 
classes of non-increasing functions, they must be equal. 
      
If $h$ is continuous on the left, then 
$E_{\alpha}=\ooint{\gamma,\infty}$ for some $\gamma$, so we no longer 
need to use (d), and the argument 
works for any $u\in (M^{0,\infty}_{\bar\mu})^+$. 
      
\medskip 
      
{\bf (f)} Apply (e) with $h(\beta)=\max(0,\beta-\alpha)$. 
      
\medskip 
      
{\bf (g)} Express $u^*$ as $g^{\ssbullet}$, where 
      
\Centerline{$g(s)=\inf\{\alpha:\bar\mu\Bvalue{|u|>\alpha}\le s\}$} 
      
\noindent for every $s>0$.   Because $g$ is non-increasing, it is easy 
to check that, for $t>0$, 
      
\Centerline{$\int_0^tg=tg(t)+\int_0^{\infty}\max(0,g(s)-g(t))ds 
\le\alpha t+\int_0^{\infty}\max(0,g(s)-\alpha)ds$} 
      
\noindent for every $\alpha\ge 0$;  so that 
      
\Centerline{$\int_0^tu^*=\min_{\alpha\ge 0}\alpha t+\int(u^*-\alpha\chi 
1)^+$.} 
      
\noindent Now 
      
$$\eqalign{\int(u^*-\alpha\chi 1)^+ 
&=\int_0^{\infty}\bar\mu_L\Bvalue{(u^*-\alpha\chi 1)^+>\beta}d\beta\cr 
&=\int_0^{\infty}\bar\mu\Bvalue{(|u|-\alpha\chi 1)^+>\beta}d\beta 
=\int(|u|-\alpha\chi 1)^+\cr}$$ 
      
\noindent for every $\alpha\ge 0$, using (f) and 365A, and 
      
\Centerline{$\int_0^tu^*=\min_{\alpha\ge 0}\alpha t+\int(|u|-\alpha\chi 
1)^+$.} 
      
\medskip 
      
{\bf (h)} 
      
\Centerline{$\bar\mu\Bvalue{u_0>\alpha} 
=\bar\mu(\sup_{u\in A}\Bvalue{u>\alpha}) 
=\sup_{u\in A}\bar\mu\Bvalue{u>\alpha}$} 
      
\noindent for any $\alpha>0$, using 364L(a-ii) and 321D.   So 
      
\Centerline{$\Bvalue{u_0^*>\alpha} 
=\coint{0,\bar\mu\Bvalue{u_0>\alpha}}^{\ssbullet} 
=\sup_{u\in A}\coint{0,\bar\mu\Bvalue{u>\alpha}}^{\ssbullet} 
=\sup_{u\in A}\Bvalue{u^*>\alpha}$} 
      
\noindent for every $\alpha$, and $u_0^*=\sup_{u\in A}u^*$. 
}%end of proof of 373D 
      
\leader{373E}{Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra. 
Then $\int|u\times v|\le\int u^*\times v^*$ for all $u$,  
$v\in M^{0,\infty}(\frak A,\bar\mu)$. 
      
\proof{{\bf (a)} Consider first the case $u$, $v\ge 0$ in $S(\frak A)$. 
Then we may express $u$, $v$ as $\sum_{i=0}^m\alpha_i\chi a_i$, 
$\sum_{j=0}^n\beta_j\chi b_j$ where  
$a_0\Bsupseteq a_1\Bsupseteq\ldots\Bsupseteq a_m$,  
$b_0\Bsupseteq\ldots\Bsupseteq b_n$ 
in $\frak A$ and $\alpha_i$, $\beta_j\ge 0$ for all $i$, $j$ (361Ec). 
Now $u^*$, $v^*$ are given by 
      
\Centerline{$u^*=\sum_{i=0}^m\alpha_i\chi\coint{0,\bar\mu 
a_i}^{\ssbullet}$, 
\quad$v^*=\sum_{j=0}^n\beta_j\chi\coint{0,\bar\mu b_j}^{\ssbullet}$} 
      
\noindent (373Dc).   So 
      
$$\eqalign{\int u\times v 
&=\sum_{i=0}^m\sum_{j=0}^n\alpha_i\beta_j\bar\mu(a_i\Bcap b_j) 
\le\sum_{i=0}^m\sum_{j=0}^n\alpha_i\beta_j 
  \min(\bar\mu a_i,\bar\mu b_j)\cr 
&=\sum_{i=0}^m\sum_{j=0}^n\alpha_i\beta_j\mu_L(\coint{0,\bar\mu 
  a_i}\cap\coint{0,\bar\mu b_j}) 
=\int u^*\times v^*.\cr}$$ 
      
\medskip 
      
{\bf (b)} For the general case, we have non-decreasing sequences 
$\sequencen{u_n}$, $\sequencen{v_n}$ in $S(\frak A)^+$ with suprema 
$|u|$, $|v|$ respectively (364Jd), so that 
      
\Centerline{$|u\times v|=|u|\times|v|=\sup_{n\in\Bbb N}|u|\times v_n 
=\sup_{m,n\in\Bbb N}u_m\times v_n=\sup_{n\in\Bbb N}u_n\times v_n$} 
      
\noindent and 
      
\Centerline{$\int|u\times v| 
=\int\sup_{n\in\Bbb N}u_n\times v_n 
=\sup_{n\in\Bbb N}\int u_n\times v_n 
\le\sup_{n\in\Bbb N}\int u^*_n\times v^*_n 
\le\int u^*\times v^*$,} 
      
\noindent using 373Db. 
}%end of proof of 373E 
      
\leader{373F}{Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra, and 
$u$ any member of $M^{0,\infty}(\frak A,\bar\mu)$. 
      
(a) For any $p\in[1,\infty]$, $u\in L^p(\frak A,\bar\mu)$ iff  
$u^*\in L^p(\frak A_L,\bar\mu_L)$, and in this case $\|u\|_p=\|u^*\|_p$. 
      
(b)(i) $u\in M^0(\frak A,\bar\mu)$ iff  
$u^*\in M^0(\frak A_L,\bar\mu_L)$; 
      
\quad(ii) $u\in M^{1,\infty}(\frak A,\bar\mu)$ iff  
$u^*\in M^{1,\infty}(\frak A_L,\bar\mu_L)$, and in this case 
$\|u\|_{1,\infty}=\|u^*\|_{1,\infty}$; 
      
\quad(iii) $u\in M^{1,0}(\frak A,\bar\mu)$ iff  
$u^*\in M^{1,0}(\frak A_L,\bar\mu_L)$; 
      
\quad(iv) $u\in M^{\infty,1}(\frak A,\bar\mu)$ iff  
$u^*\in M^{\infty,1}(\frak A_L,\bar\mu_L)$, and in this case 
$\|u\|_{\infty,1}=\|u^*\|_{\infty,1}$. 
      
\proof{{\bf (a)}(i) Consider first the case $p=1$.   In this case 
      
\Centerline{$\int|u|=\int_0^{\infty}\bar\mu\Bvalue{|u|>\alpha}d\alpha 
=\int_0^{\infty}\bar\mu_L\Bvalue{u^*>\alpha}d\alpha 
=\int u^*$.} 
\noindent (ii) If $1<p<\infty$, then by 373De we have 
$(|u|^p)^*=(u^*)^p$, so that 
      
\Centerline{$\|u\|^p_p=\int|u|^p=\int(|u|^p)^*=\int(u^*)^p=\|u^*\|_p^p$} 
      
\noindent if either $\|u\|_p$ or $\|u^*\|_p$ is finite.   (iii) As for 
$p=\infty$, 
      
\Centerline{$\|u\|_{\infty}\le\gamma 
\iff\Bvalue{|u|>\gamma}=0 
\iff\Bvalue{u^*>\gamma}=0 
\iff\|u^*\|_{\infty}\le\gamma$.} 
      
\wheader{373F}{6}{2}{2}{36pt} %p206 
      
{\bf (b)(i)} 
      
$$\eqalign{u\in M^0_{\bar\mu} 
&\iff\bar\mu\Bvalue{|u|>\alpha}<\infty\text{ for every }\alpha>0\cr 
&\iff\bar\mu_L\Bvalue{u^*>\alpha}<\infty\text{ for every }\alpha>0 
\iff u^*\in M^0_{\bar\mu_L}.\cr}$$ 
      
\medskip 
      
\quad{\bf (ii)} For any $\alpha\ge 0$, 
      
\Centerline{$\int(|u|-\alpha\chi 1)^+=\int(u^*-\alpha\chi 1)^+$} 
      
\noindent as in the proof of 373Dg.   So 
$\|u\|_{1,\infty}=\|u^*\|_{1,\infty}$ if either is finite, 
by the formula in 369Ob. 
      
\medskip 
      
\quad{\bf (iii)} This follows from (i) and (ii), because 
$M^{1,0}=M^0\cap M^{1,\infty}$. 
      
\medskip 
      
\quad{\bf (iv)} Allowing $\infty$ as a value of an integral, we have 
      
$$\eqalign{\|u\|_{1,\infty} 
&=\min\{\alpha+\int(|u|-\alpha\chi 1)^+:\alpha\ge 0\}\cr 
&=\min\{\alpha+\int(u^*-\alpha\chi 1)^+:\alpha\ge 0\} 
=\|u^*\|_{1,\infty}\cr}$$ 
      
\noindent by 369Ob;  in particular, $u\in M^{1,\infty}_{\bar\mu}$ iff 
$u^*\in M^{1,\infty}_{\bar\mu_L}$. 
}%end of proof of 373F 
      
\leader{373G}{Lemma} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras.   If 
      
\inset{{\it either} $u\in M^{1,\infty}(\frak A,\bar\mu)$ and  
$T\in\Cal T_{\bar\mu,\bar\nu}$} 
      
\inset{{\it or} $u\in M^{1,0}(\frak A,\bar\mu)$ and  
$T\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$,} 
      
\noindent  then  $\int_0^t(Tu)^*\le\int_0^tu^*$ for every $t\ge 0$. 
      
\proof{ Set $T_1=T\restr L^1_{\bar\mu}$, so that $\|T_1\|\le 1$ in 
$\eurm B(L^1_{\bar\mu};L^1_{\bar\nu})$, and $|T_1|$ is defined in  
$\eurm B(L^1_{\bar\mu};L^1_{\bar\nu})$, also with norm at most $1$.   If 
$\alpha\ge 0$, then we can express $u$ as $u_1+u_2$ where 
$|u_1|\le(|u|-\alpha\chi 1)^+$ and $|u_2|\le\alpha\chi 1$.   (Let  
$w\in L^{\infty}(\frak A)$ be such that $\|w\|_{\infty}\le 1$, $u=|u|\times w$;  set $u_2=w\times(|u|\wedge\alpha\chi 1)$.)   So if 
$\int(|u|-\alpha\chi 1)^+<\infty$, 
      
\Centerline{$|Tu|\le|Tu_1|+|Tu_2|\le|T_1||u_1|+\alpha\chi 1$} 
      
\noindent and 
      
\Centerline{$\int(|Tu|-\alpha\chi 
1)^+\le\int|T_1||u_1|\le\int|u_1|\le\int(|u|-\alpha\chi 1)^+$.} 
      
\noindent The formula of 373Dg now tells us that 
$\int_0^t(Tu)^*\le\int_0^tu^*$ for every $t$. 
}%end of proof of 373G 
      
\vleader{60pt}{373H}{Lemma} Let $(\frak A,\bar\mu)$ be a measure algebra, and 
$\theta:\frak A^f\to\Bbb R$ an additive functional, where  
$\frak A^f=\{a:\bar\mu a<\infty\}$. 
      
(a) The following are equiveridical: 
      
\inset{($\alpha$)  
$\lim_{t\downarrow 0}\sup_{\bar\mu a\le t}|\theta a| 
=\lim_{t\to\infty}\Bover1t\sup_{\bar\mu a\le t}|\theta a|=0$,} 
      
\inset{($\beta$) there is some $u\in M^{1,0}(\frak A,\bar\mu)$ 
such that $\theta a =\int_au$ for every $a\in\frak A^f$,} 
      
\noindent and in this case $u$ is uniquely defined. 
      
(b) Now suppose that $(\frak A,\bar\mu)$ is localizable.   Then the 
following are equiveridical: 
      
\inset{($\alpha$)  
$\lim_{t\downarrow 0}\sup_{\bar\mu a\le t}|\theta a|=0$,\quad 
$\limsup_{t\to\infty}\Bover1t\sup_{\bar\mu a\le t}|\theta a|<\infty$,} 
      
\inset{($\beta$) there is some  
$u\in M^{1,\infty}(\frak A,\bar\mu)$ 
such that $\theta a =\int_au$ for every $a\in\frak A^f$,} 
      
\noindent and again this $u$ is uniquely defined. 
      
\proof{{\bf (a)(i)} Assume ($\alpha$).   For $a$, $c\in\frak A^f$, set 
$\theta_c(a)=\theta(a\Bcap c)$.   Then for each $c\in\frak A^f$, there 
is a unique $u_c\in L^1_{\bar\mu}$ such that 
$\theta_ca=\int_au_c$ for every $a\in\frak A^f$ (365Eb). 
Because $u_c$ is unique we must have $u_c=u_d\times\chi c$ whenever 
$c\Bsubseteq d\in\frak A^f$.   Next, given $\alpha>0$, there is a 
$t_0\ge 0$ such that $|\theta a|\le\alpha\bar\mu a$ whenever $a\in\frak 
A^f$ and $\bar\mu a\ge t_0$;  so that $\bar\mu\Bvalue{u_c>\alpha}\le 
t_0$ for every $c\in\frak A^f$, and $e(\alpha)=\sup_{c\in\frak 
A^f}\Bvalue{u_c^+>\alpha}$ is defined in $\frak A^f$.   Of course 
$e(\alpha)=\Bvalue{u_{e(1)}^+>\alpha}$ for every $\alpha\ge 1$, so 
$\inf_{\alpha\in\Bbb R}e(\alpha)=0$, and 
$v_1=\sup_{c\in\frak A^f}u_c^+$ is defined in $L^0=L^0(\frak A)$ 
(364L(a-ii) again). 
Because $\Bvalue{v_1>\alpha}=e(\alpha)\in\frak A^f$ for each $\alpha>0$, 
$v_1\in M^0_{\bar\mu}$.   For any $a\in\frak A^f$, 
      
\Centerline{$v_1\times\chi a=\sup_{c\in\frak A^f}u_c^+\times\chi a 
=u_a^+$,} 
      
\noindent so $v_1\in M^{1,0}_{\bar\mu}$ and $\int_av_1=\int_au_a^+$ for every $a\in\frak A^f$. 
      
Similarly, $v_2=\sup_{c\in\frak A^f}u_c^-$ is defined in $M^{1,0}_{\bar\mu}$ and 
$\int_av_2=\int_au_a^-$ for every $a\in\frak A^f$.   So we can set 
$u=v_1-v_2\in M^{1,0}_{\bar\mu}$ and get 
      
\Centerline{$\int_au=\int_au_a=\theta a$} 
      
\noindent for every $a\in\frak A^f$.   Thus ($\beta$) is true. 
      
\medskip 
      
\quad{\bf (ii)} Assume ($\beta$). 
If $\epsilon>0$, there is a $\delta>0$ such that 
$\int_a(|u|-\epsilon\chi 1)^+\le\epsilon$ whenever $\bar\mu a\le\delta$ 
(365Ea), so 
that $|\int_au|\le\epsilon(1+\bar\mu a)$ whenever $\bar\mu a\le\delta$. 
As $\epsilon$ is arbitrary, $\lim_{t\downarrow 0}\sup_{\bar\mu a\le 
t}|\int_au|=0$.   Moreover, whenever $t>0$ and $\bar\mu a\le t$, 
$\bover1t|\int_au|\le\epsilon+\bover1t\int(|u|-\epsilon\chi 1)^+$. 
Thus 
      
\Centerline{$\limsup_{t\to\infty}\bover1t\sup_{\bar\mu a\le t}|\int_a 
u|\le\epsilon$.} 
      
\noindent As $\epsilon$ is arbitrary, $\theta$ satisfies the conditions 
in ($\alpha$). 
      
\medskip 
      
\quad{\bf (iii)} The uniqueness of $u$ is a consequence of 366Gd. 
      
\medskip 
      
{\bf (b)} The argument for (b) uses the same ideas. 
      
\medskip 
      
\quad{\bf (i)} Assume ($\alpha$).   Again, for each $c\in\frak A^f$, we 
have a unique $u_c\in L^1_{\bar\mu}$ such that $\theta_ca=\int_au_c$ for every 
$a\in\frak A^f$;  again, set  
$e(\alpha)=\sup_{c\in\frak A^f}\Bvalue{u_c^+>\alpha}$, 
which is still defined because $\frak A$ is supposed to be Dedekind complete. 
This time, there are $t_0$, $\gamma\ge 0$ such that  
$|\theta a|\le\gamma\bar\mu a$ whenever $a\in\frak A^f$ and  
$\bar\mu a\ge t_0$; 
so that $\bar\mu\Bvalue{u_c>\gamma}\le t_0$ for every $c\in\frak A^f$, 
and $\bar\mu e(\gamma)<\infty$.   Accordingly 
      
\Centerline{$\inf_{\alpha\ge\gamma}e(\alpha) 
=\inf_{\alpha\ge\gamma}\Bvalue{u_{e(\gamma)}^+>\alpha}=0$,} 
      
\noindent and once more $v_1=\sup_{c\in\frak A^f}u_c^+$ is defined in 
$L^0=L^0(\frak A)$.   As before, 
$v_1\times\chi a=u_a^+\in L^1_{\bar\mu}$ for any $a\in\frak A^f$, 
Because $\Bvalue{v_1>\gamma}=e(\gamma)\in\frak A^f$,  
$v_1\in M^{1,\infty}_{\bar\mu}$.   Similarly,  
$v_2=\sup_{c\in\frak A^f}u_c^-$ is defined 
in $M^{1,\infty}_{\bar\mu}$, with $v_2\times\chi a=u_a^-$ for every $a\in\frak A^f$.   So $u=v_1-v_2\in M^{1,\infty}_{\bar\mu}$, and 
      
\Centerline{$\int_au=\int_au_a=\theta a$} 
      
\noindent for every $a\in\frak A^f$. 
      
\medskip 
      
\quad{\bf (ii)} Assume ($\beta$).   Take $\gamma\ge 0$ such that 
$\beta=\int(|u|-\gamma\chi 1)^+$ is finite.   If $\epsilon>0$, there is 
a $\delta>0$ such that $\int_a(|u|-\gamma\chi 1)^+\le\epsilon$ whenever 
$\bar\mu a\le\delta$, so that $|\int_au|\le\epsilon+\gamma\bar\mu a$ 
whenever $\bar\mu a\le\delta$.   As $\epsilon$ is arbitrary, 
$\lim_{t\downarrow 0}\sup_{\bar\mu a\le t}|\int_au|=0$.   Moreover, 
whenever $t>0$ and $\bar\mu a\le t$, then 
$\bover1t|\int_au|\le\gamma+\bover1t\int(|u|-\epsilon\chi 1)^+$.   Thus 
      
\Centerline{$\limsup_{t\to\infty}\bover1t\sup_{\bar\mu a\le t}|\int_a 
u|\le\gamma<\infty$,} 
      
\noindent and the function $a\mapsto\int_au$ satisfies the conditions in 
($\beta$). 
      
\medskip 
      
\quad{\bf (iii)} $u$ is uniquely defined because $u\times\chi a$ must be 
$u_a$, as defined in (i), for every $a\in\frak A^f$, and  
$(\frak A,\bar\mu)$ is semi-finite. 
}%end of proof of 373H 
      
\leader{373I}{Lemma} Suppose that $u$, $v$,  
$w\in M^{0,\infty}(\frak A_L,\bar\mu_L)$ 
are all equivalence classes of non-negative 
non-increasing functions.   If 
$\int_0^tu\le\int_0^tv$ for every $t\ge 0$, then  
$\int u\times w\le\int v\times w$. 
      
\proof{ For $n\in\Bbb N$, $i\le 4^n$ set 
$a_{ni}=\Bvalue{w>2^{-n}i}$;  set $w_n=\sum_{i=1}^{4^n}2^{-n}\chi 
a_{ni}$.   Then each $a_{ni}$ is of the form $[0,t]^{\ssbullet}$, so 
      
\Centerline{$\int u\times w_n=\sum_{i=1}^{4^n}2^{-n}\int_{a_{ni}}u 
\le\sum_{i=1}^{4^n}2^{-n}\int_{a_{ni}}v=\int v\times w_n$.} 
      
\noindent Also $\sequencen{w_n}$ is a non-decreasing sequence with 
supremum $w$, so 
      
\Centerline{$\int u\times w=\sup_{n\in\Bbb N}\int u\times w_n 
\le\sup_{n\in\Bbb N}\int v\times w_n=\int v\times w$.} 
}%end of proof of 373I 
      
\leader{373J}{Corollary} Suppose that $(\frak A,\bar\mu)$ and  
$(\frak B,\bar\nu)$ are measure algebras and  
$v\in M^{0,\infty}(\frak B,\bar\nu)$.   If 
      
\inset{{\it either} $u\in M^{1,0}(\frak A,\bar\mu)$ and  
$T\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$} 
      
\inset{{\it or} $u\in M^{1,\infty}(\frak A,\bar\mu)$ and  
$T\in\Cal T_{\bar\mu,\bar\nu}$} 
      
\noindent then $\int|Tu\times v|\le\int u^*\times v^*$. 
      
\proof{ Put 373E, 373G and 373I together. 
}%end of proof of 373J 
      
\vleader{48pt}{373K}{The very weak operator topology} Let  
$(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be two measure algebras.      
For $u\in M^{1,\infty}(\frak A,\bar\mu)$ and
$w\in M^{\infty,1}(\frak B,\bar\nu)$ set 
      
\Centerline{$\tau_{uw}(T) 
=|\int Tu\times w|$ for $T\in\eurm B
=\eurm B(M^{1,\infty}(\frak A,\bar\mu);M^{1,\infty}(\frak B,\bar\nu))$.} 
      
\noindent Then $\tau_{uw}$ is a seminorm on $\eurm B$.   I will call 
the topology generated by  
$\{\tau_{uw}:u\in M^{1,\infty}(\frak A,\bar\mu),\, 
w\in M^{\infty,1}(\frak B,\bar\nu)\}$\cmmnt{ (2A5B)} the 
{\bf very weak operator topology} on $\eurm B$. 
      
\leader{373L}{Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra and 
$(\frak B,\bar\nu)$ a localizable measure algebra.   Then  
$\cmmnt{\Cal T=\mskip5mu}\Cal T_{\bar\mu,\bar\nu}$ 
is compact in the very weak operator topology. 
      
\proof{ Let $\Cal F$ be an ultrafilter on $\Cal T$.   If  
$u\in M^{1,\infty}_{\bar\mu}$ and $w\in M^{\infty,1}_{\bar\nu}$ then 
      
\Centerline{$|\int Tu\times w|\le\int u^*\times w^*<\infty$} 
      
\noindent for every $T\in\Cal T$ (373J);  $\int u^*\times w^*$ is finite 
because $u^*\in M_{\bar\mu_L}^{1,\infty}$ and 
$w^*\in M_{\bar\mu_L}^{\infty,1}$ (373F). 
      
In particular, $\{\int Tu\times w:T\in\Cal T\}$ is bounded. 
Consequently $h_u(w)=\lim_{T\to\Cal F}\int Tu\times w$ is defined in 
$\Bbb R$ (2A3Se). 
Because $w\mapsto\int Tu\times w$ is additive for every $T\in\Cal T$, so 
is $h_u$.   Also 
      
\Centerline{$|h_u(w)|\le\int u^*\times 
w^*\le\|u^*\|_{1,\infty}\|w^*\|_{\infty,1} 
=\|u\|_{1,\infty}\|w\|_{\infty,1}$} 
      
\noindent for every $w\in M^{\infty,1}_{\bar\nu}$. 
      
$|h_u(\chi b)|\le\int_0^tu^*$ whenever $b\in\frak B^f$ and $\bar\nu b\le 
t$.   So 
      
\Centerline{$\lim_{t\downarrow 0}\sup_{\bar\nu b\le t}|h_u(\chi b)| 
\le\lim_{t\downarrow 0}\int_0^tu^*=0$,} 
      
\Centerline{$\limsup_{t\to\infty}\bover1t\sup_{\bar\nu b\le t}|h_u(\chi 
b)|\le\limsup_{t\to\infty}\bover1t\int_0^tu^*<\infty$.} 
      
\noindent Of course $b\mapsto h_u(\chi b)$ is additive, so by 373Hb there 
is a unique $Su\in M^{1,\infty}_{\bar\nu}$ such that $h_u(\chi 
b)=\int_bSu$ for every $b\in\frak B^f$.   Since both $h_u$ and $w\mapsto 
\int Su\times w$ are linear and continuous on $M^{\infty,1}_{\bar\nu}$, 
and $S(\frak B^f)$ is dense in $M^{\infty,1}_{\bar\nu}$ (369Od), 
      
\Centerline{$\int Su\times w=h_u(w)=\lim_{T\to\Cal F}\int Tu\times w$} 
      
\noindent for every $w\in 
M^{\infty,1}_{\bar\nu}$. 
And this is true for every $u\in M^{1,\infty}_{\bar\mu}$. 
      
For any particular $w\in M^{\infty,1}_{\bar\nu}$, all the maps 
$u\mapsto \int Tu\times w$ are linear, so $u\mapsto\int Su\times w$ also 
is;  that is, $S:M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\nu}$ is 
linear. 
      
Now $S\in\Cal T$.   \Prf\ ($\alpha$) If $u\in L^1_{\bar\mu}$ and 
$b$, $c\in\frak B^f$, then 
      
$$\eqalign{\int_bSu-\int_cSu 
&=\lim_{T\to\Cal F}\int Tu\times(\chi b-\chi c) 
\le\sup_{T\in\Cal T}\int Tu\times(\chi b-\chi c)\cr 
&\le\sup_{T\in\Cal T}\|Tu\|_1\|\chi b-\chi c\|_{\infty} 
\le\|u\|_1.\cr}$$ 
      
\noindent But, setting $e=\Bvalue{Su>0}$, we have 
      
$$\eqalign{\int|Su| 
&=\int_eSu-\int_{1\Bsetminus e}Su\cr 
&=\sup_{b\in\frak B^f,b\Bsubseteq e}\int_bSu
 +\sup_{c\in\frak B^f,c\Bsubseteq 1\Bsetminus e}\int_c(-Su)
\le\|u\|_1.\cr}$$ 
      
\noindent ($\beta$) If $u\in L^{\infty}(\frak A)$, then 
      
\Centerline{$|\int_bSu| 
\le\sup_{T\in\Cal T}|\int Tu\times\chi b| 
\le\sup_{T\in\Cal T}\|Tu\|_{\infty}\bar\nu b 
\le\|u\|_{\infty}\bar\nu b$} 
      
\noindent for every $b\in\frak B^f$.   So 
$\Bvalue{Su>\|u\|_{\infty}}=\Bvalue{-Su>\|u\|_{\infty}}=0$ and 
$\|Su\|_{\infty}\le\|u\|_{\infty}$.   (Note that both parts of this 
argument depend on knowing that $(\frak B,\bar\nu)$ is semi-finite, so 
that we cannot be troubled by purely infinite elements of 
$\frak B$.)\ \Qed 
      
Of course we now have $\lim_{T\to\Cal F}\tau_{uw}(T-S)=0$ for all 
$u\in M^{1,\infty}_{\bar\mu}$, $w\in M^{\infty,1}_{\bar\nu}$, so that 
$S=\lim\Cal F$ in $\Cal T$.   As $\Cal F$ is arbitrary, $\Cal T$ is 
compact (2A3R). 
}%end of proof of 373L 
      
\leader{373M}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra 
and $(\frak B,\bar\nu)$ a localizable measure algebra, and $u$ any 
member of $M^{1,\infty}(\frak A,\bar\mu)$.   Then  
$B=\{Tu:T\in\Cal T_{\bar\mu,\bar\nu}\}$ is compact in $M^{1,\infty}(\frak B,\bar\nu)$ for the topology 
$\frak T_s(M^{1,\infty}(\frak B,\bar\nu), 
M^{\infty,1}(\frak B,\bar\nu))$. 
      
\proof{ The point is just that the map  
$T\mapsto Tu:\Cal T_{\bar\mu,\bar\nu}\to M^{1,\infty}_{\bar\nu}$ 
is continuous for 
the very weak operator topology on $\Cal T_{\bar\mu,\bar\nu}$ and 
$\frak T_s(M^{1,\infty}_{\bar\nu},M^{\infty,1}_{\bar\nu})$.   So $B$ 
is a continuous image of a compact set, therefore compact (2A3N(b-ii)). 
}%end of proof of 373M 
      
\leader{373N}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra, 
$(\frak B,\bar\nu)$ a localizable measure algebra and $u$ any 
member of $M^{1,\infty}(\frak A,\bar\mu)$;  set  
$B=\{Tu:T\in\Cal T_{\bar\mu,\bar\nu}\}$.   If $\sequencen{v_n}$ is any non-decreasing 
sequence in $B$, then $\sup_{n\in\Bbb N}v_n$ is defined in 
$M^{1,\infty}(\frak B,\bar\nu)$ and belongs to $B$. 
      
\proof{{\bf (a)} The point is that $(M^{1,\infty}_{\bar\nu})^+$ is a closed set
for $\frak T_s(M^{1,\infty}_{\bar\nu},M^{\infty,1}_{\bar\nu})$.   \Prf\ If
$w\in M^{1,\infty}_{\bar\nu}$ and $w\not\ge 0$, then $b=\Bvalue{-w>0}\ne 0$.
As $(\frak B,\bar\nu)$ is semi-finite, there is a non-zero $c\in\frak B^f$ with
$c\Bsubseteq b$, and $\int_c(-w)>0$, that is, $\int w\times\chi c<0$.
Now $\chi c\in M^{\infty,1}_{\bar\nu}$ so
$\{w':\int_cw'<0\}$ is a neighbourhood of $w$ disjoint from
$(M^{1,\infty}_{\bar\nu})^+$.   Thus 
$M^{1,\infty}_{\bar\nu}\setminus(M^{1,\infty}_{\bar\nu})^+$ is open and
$(M^{1,\infty}_{\bar\nu})^+$ is closed.\ \Qed

\medskip

{\bf (b)} Because $\frak T_s(M^{1,\infty}_{\bar\nu},M^{\infty,1}_{\bar\nu})$ is
a linear space topology, the sets 
$\{w:v\le w\}=\{w:w-v\in(M^{1,\infty}_{\bar\nu})^+\}$
and $\{w:w\le v\}=\{w:v-w\in(M^{1,\infty}_{\bar\nu})^+\}$ are closed for every 
$w\in M^{1,\infty}_{\bar\nu}$.   Now consider the given sequence
$\sequencen{v_n}$ in $B$.   By 373M, it has a cluster point $v\in B$.
Since $\{w:v_n\le w\}$ is a closed set containing $v_i$ whenever $i\ge n$,
$v_n\le v$, for every $n\in\Bbb N$.   On the other hand, if $v'$ is any
upper bound of $\{v_n:n\in\Bbb N\}$ in $M^{1,\infty}_{\bar\nu}$ then
$v\le v'$ because $\{w:w\le v'\}$.   Accordingly
$\sup_{n\in\Bbb N}v_n=v$ is defined and belongs to $B$.
}%end of proof of 373N 
      
\leader{373O}{Theorem} Suppose that $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ are 
measure algebras, $u\in M^{1,\infty}(\frak A,\bar\mu)$ and
$v\in M^{1,\infty}(\frak B,\bar\nu)$.   Then the following are equiveridical: 
      
(i) there is a $T\in\Cal T_{\bar\mu,\bar\nu}$ such that $Tu=v$, 
      
(ii) $\int_0^tv^*\le\int_0^tu^*$ for every $t\ge 0$. 
      
\noindent In particular, given $u\in M^{1,\infty}(\frak A,\bar\mu)$, 
there are $S\in\Cal T_{\bar\mu,\bar\mu_L}$, 
$T\in\Cal T_{\bar\mu_L,\bar\mu}$ such that $Su=u^*$ and $Tu^*=u$. 
      
\proof{ (i)$\Rightarrow$(ii) is covered by Lemma 373G.   Accordingly I shall devote 
the rest of the proof to showing that (ii)$\Rightarrow$(i). 
      
\medskip 
      
{\bf (a)} If $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ are measure 
algebras, $u\in M^{1,\infty}_{\bar\mu}$ and $v\in M^{1,\infty}_{\bar\nu}$, 
I will say that $v\preccurlyeq u$ if 
there is a $T\in\Cal T_{\bar\mu,\bar\nu}$ such that $Tu=v$, and that 
$v\sim u$ if $v\preccurlyeq u$ and $u\preccurlyeq v$.   (Properly 
speaking, I ought to write $(u,\bar\mu)\preccurlyeq (v,\bar\nu)$, 
because we could in principle have two different measures on the same 
algebra.   But I do not think any confusion is likely to arise in the 
argument which follows.)   By 373Be, $\preccurlyeq$ is transitive and 
$\sim$ is an equivalence relation.   Now we have the following facts. 
      
\medskip 
      
{\bf (b)} If $(\frak A,\bar\mu)$ is a measure algebra and $u_1$, $u_2\in 
M^{1,\infty}_{\bar\mu}$ are such that $|u_1|\le|u_2|$, then 
$u_1\preccurlyeq u_2$.   \Prf\ There is a $w\in L^{\infty}(\frak A)$ 
such that $u_1=w\times u_2$ and $\|w\|_{\infty}\le 1$.   Set $Tv=w\times 
v$ for  for $v\in M^{1,\infty}_{\bar\mu}$;  then $T\in\Cal 
T_{\bar\mu,\bar\mu}$ and $Tu_2=u_1$.\ \QeD\  So $u\sim|u|$ for every 
$u\in M^{1,\infty}_{\bar\mu}$. 
      
\medskip 
      
{\bf (c)(i)} If $(\frak A,\bar\mu)$ is a measure algebra and $u\ge 0$ in 
$S(\frak A)$, then $u^*\preccurlyeq u$.   \Prf\ If $u=0$ this is 
trivial.   Otherwise, express $u$ as $\sum_{i=0}^n\alpha_i\chi a_i$ 
where $a_0,\ldots,a_n$ are disjoint and non-zero and 
$\alpha_0>\alpha_1\ldots>\alpha_n>0\in\Bbb R$.   If $\bar\mu a_i=\infty$ 
for any $i$, take $m$ to be minimal subject to $\bar\mu a_m=\infty$; 
otherwise, set $m=n$.   Then 
$u^*=\sum_{i=0}^m\alpha_i\chi\coint{\beta_i,\beta_{i+1}}^{\ssbullet}$, 
where $\beta_j=\sum_{i=0}^{j-1}\bar\mu a_i$ for $j\le m+1$. 
      
For $i<m$, and for $i=m$ if $\bar\mu a_m<\infty$, define 
$h_i:M^{1,\infty}_{\bar\mu}\to\Bbb R$ by setting 
      
\Centerline{$h_i(v)=\Bover1{\bar\mu a_i}\int_{a_i}v$} 
      
\noindent for every $v\in M^{1,\infty}_{\bar\mu}$.   If $\bar\mu 
a_m=\infty$, then we need a different idea to define $h_m$, as follows. 
Let $I$ be $\{a:a\in\frak A,\,\bar\mu(a\cap a_m)<\infty\}$.   Then $I$ 
is an ideal of $\frak A$ not containing $a_m$, so there is a Boolean 
homomorphism 
$\pi:\frak A\to\{0,1\}$ such that $\pi a=0$ for $a\in I$ and $\pi a_m=1$ 
(311D).   This induces a corresponding $\|\,\|_{\infty}$-continuous 
linear operator 
$h:L^{\infty}(\frak A)\to L^{\infty}(\{0,1\})\cong\Bbb R$, as in 363F. 
Now $h(\chi a)=0$ whenever $\bar\mu a<\infty$, and accordingly $h(v)=0$ 
whenever $v\in M^{\infty,1}_{\bar\mu}$, since $S(\frak A^f)$ is dense in 
$M^{\infty,1}_{\bar\mu}$ for $\|\,\|_{\infty,1}$ and therefore also for 
$\|\,\|_{\infty}$.   But this means 
that $h$ has a unique extension to a linear functional 
$h_m:M^{1,\infty}_{\bar\mu}\to\Bbb R$ such that $h_m(v)=0$ for every 
$v\in L^1_{\bar\mu}$, while $h_m(\chi a_m)=1$ and 
$|h_m(v)|\le\|v\|_{\infty}$ for every $v\in L^{\infty}(\frak A)$. 
      
Having defined $h_i$ for every $i\le m$, define 
$T:M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\mu_L}$ by setting 
      
\Centerline{$Tv=\sum_{i=0}^mh_i(v) 
\chi\coint{\beta_i,\beta_{i+1}}^{\ssbullet}$} 
      
\noindent for every $v\in M^{1,\infty}_{\bar\mu}$. 
      
For any $i\le m$ and $v\in L^1_{\bar\mu}$, 
      
\Centerline{$\int_{\beta_i}^{\beta_{i+1}}|Tv| 
=|h_i(v)|\bar\mu a_i\le\int_{a_i}|v|$;} 
      
\noindent summing over $i$, $\|Tv\|_1\le\|v\|_1$.   Similarly, for any 
$i\le m$ and $v\in L^{\infty}(\frak B)$, $|h_i(v)|\le\|v\|_{\infty}$, so 
$\|Tv\|_{\infty}\le\|v\|_{\infty}$. 
      
Thus $T\in\Cal T_{\bar\mu,\bar\mu_L}$.   Since $u^*=Tu$, we conclude 
that $u^*\preccurlyeq u$, as claimed.\ \Qed 
      
\medskip 
      
\quad{\bf (ii)} If $(\frak A,\bar\mu)$ is a measure algebra and $u\ge 0$ in 
$M^{1,\infty}_{\bar\mu}$, then $u^*\preccurlyeq u$.   \Prf\ Let 
$\sequencen{u_n}$ be a non-decreasing sequence in $S(\frak A)$ with 
$u_0\ge 0$ and $\sup_{n\in\Bbb N}u_n=u$.   Then $\sequencen{u_n^*}$ is a 
non-decreasing sequence in $M^{1,\infty}_{\bar\mu_L}$ with supremum 
$u^*$, by 373Db and 373Dh.   Also $u_n^*\preccurlyeq u_n\preccurlyeq u$ 
for every $n$, by (b) above and (i) here.   By 373N, 
$u^*\preccurlyeq u$.\ \Qed 
      
\medskip 
      
{\bf (d)(i)} If $(\frak A,\bar\mu)$ is a measure algebra and $u\ge 0$ in 
$S(\frak A)$, then $u\preccurlyeq u^*$.   \Prf\ The argument is very 
similar to that of (c-i).   Again, the result is trivial if $u=0$; 
suppose that $u>0$ and define $\alpha_i$, $a_i$, $m$, $\beta_i$ as 
before.   This time, set $a'_i=a_i$ for $i<m$, $a'_m=\sup_{m\le j\le 
n}a_j$, $\tilde u=\sum_{i=0}^m\alpha_i\chi a'_i$;  then $u\le\tilde u$ 
and $\tilde u^*=u^*$.   Set 
      
\Centerline{$h_i(v) 
=\Bover1{\beta_{i+1}-\beta_i}\int_{\beta_i}^{\beta_{i+1}}v$} 
      
\noindent if $i\le m$, $\beta_{i+1}<\infty$ (that is, $\bar\mu 
a_i<\infty$) and $v\in M^{1,\infty}_{\bar\mu_L}$;  and if 
$\bar\mu a_m=\infty$, set 
      
\Centerline{$h_m(v)=\lim_{k\to\Cal F}\Bover1k\int_{0}^{k}v$} 
      
\noindent for some non-principal ultrafilter $\Cal F$ on $\Bbb N$.   As 
before, we have 
      
\Centerline{$|h_i(v)|\bar\mu a'_i\le\int_{\beta_i}^{\beta_{i+1}}|v|$,} 
      
\noindent whenever $v\in L^1_{\bar\mu_L}$ and $i\le m$, while 
$|h_i(v)|\le\|v\|_{\infty}$ whenever $v\in L^{\infty}(\frak A_L)$ and 
$i\le m$.   So we can define $T\in\Cal T_{\bar\mu_L,\bar\mu}$ by setting 
$Tv=\sum_{i=0}^mh_i(v)\chi a'_i$ for every $v\in M^{1,\infty}_{\bar\mu_L}$, and get 
      
\Centerline{$u\preccurlyeq \tilde u=Tu^*\preccurlyeq u^*$.   \Qed} 
      
\medskip 
      
\quad{\bf (ii)} If $(\frak A,\bar\mu)$ is a measure algebra and $u\ge 0$ in 
$M^{1,\infty}_{\bar\mu}$, then $u\preccurlyeq u^*$.   \Prf\ This time I 
seek to copy the ideas of (c-ii);  there is a new obstacle to circumvent, 
since $(\frak A,\bar\mu)$ might not be localizable.   Set 
      
\Centerline{$\alpha_0=\inf\{\alpha:\alpha\ge 
0,\,\bar\mu\Bvalue{u>\alpha}<\infty\}$, \quad$e=\Bvalue{u>\alpha_0}$.} 
      
\noindent Then $e=\sup_{n\in\Bbb N}\Bvalue{u>\alpha_0+2^{-n}}$ is a 
countable supremum of elements of finite measure, so the principal ideal 
$\frak A_e$, with its induced measure $\bar\mu_e$, is $\sigma$-finite. 
Now let $\sequencen{u_n}$ be a non-decreasing sequence in $S(\frak A)$ 
with $u_0\ge 0$ and $\sup_{n\in\Bbb N}u_n=u$;  set $\tilde u=u\times\chi 
e$ and $\tilde u_n=u_n\times\chi e$, regarded as members of $S(\frak 
A_e)$, for each $n$.   In this case 
      
\Centerline{$\tilde u_n\preccurlyeq\tilde u_n^*\preccurlyeq u^*$} 
\noindent for every $n$.   Because $(\frak A_e,\bar\mu_e)$ is 
$\sigma$-finite, therefore localizable, 373N tells us that $\tilde 
u\preccurlyeq u^*$. 
      
Let $S\in\Cal T_{\bar\mu_L,\bar\mu_e}$ be such that $Su^*=\tilde u$. 
As in (i), choose a non-principal ultrafilter $\Cal F$ on $\Bbb N$ 
and set 
      
\Centerline{$h(v)=\lim_{k\to\Cal F}\Bover1k\int_0^kv$} 
      
\noindent for $v\in M^{1,\infty}_{\bar\mu_L}$.   Now define 
$T:M^{1,\infty}_{\bar\mu_L}\to M^{1,\infty}_{\bar\mu}$ by setting 
      
\Centerline{$Tv=Sv+h(v)\chi(1\Bsetminus e)$,} 
      
\noindent here regarding $Sv$ as a member of $M^{1,\infty}_{\bar\mu}$. 
(I am taking it to be obvious that $M^{1,\infty}_{\bar\mu_e}$ can be 
identified with $\{w\times\chi e:w\in M^{1,\infty}_{\bar\mu}\}$.)   Then 
it is easy to see that $T\in\Cal T_{\bar\mu_L,\bar\mu}$.   Also $u\le 
Tu^*$, because 
      
\Centerline{$h(u^*)=\inf\{\alpha:\bar\mu_L\Bvalue{u^*>\alpha}<\infty\} 
=\alpha_0$,} 
      
\noindent while $u\times\chi(1\Bsetminus e)\le\alpha_0\chi(1\Bsetminus 
e)$.    So we get $u\preccurlyeq Tu^*\preccurlyeq u^*$.\ \Qed 
      
\medskip 
      
{\bf (e)(i)} Now suppose that $u$, $v\ge 0$ in $M^{1,\infty}_{\bar\mu_L}$, 
that $\int_0^tu^*\ge\int_0^tv^*$ for every $t\ge 0$, and that $v$ is of 
the form $\sum_{i=1}^n\alpha_i\chi a_i$ where 
$\alpha_1>\ldots>\alpha_n>0$, $a_1,\ldots,a_n\in\frak A_L$ are disjoint 
and $0<\bar\mu_La_i<\infty$ for each $i$. 
Then $v\preccurlyeq u$.   \Prf\ Induce on $n$.   If $n=0$ then $v=0$ and 
the result is trivial.   For the inductive step to $n\ge 1$, if 
$v^*\le u^*$ we have 
      
\Centerline{$v\sim v^*\preccurlyeq u^*\sim u$,} 
      
\noindent using (b)-(d) above.   Otherwise, look at 
$\phi(t)=\bover1t\int_0^tu^*$ for $t>0$.   We have 
      
\Centerline{$\phi(t)\ge\Bover1t\int_0^tv^*=\alpha_1$} 
      
\noindent for $t\le \beta=\bar\mu a_1$, while 
$\lim_{t\to\infty}\phi(t)<\alpha_1$, because 
$(\lim_{t\to\infty}\phi(t))\chi 1\le u^*$ and $v^*\le\alpha_1\chi 1$ and 
$v^*\not\le u^*$.   Because $\phi$ is continuous, there is a 
$\gamma\ge\beta$ such that 
$\phi(\gamma)=\alpha_1$.   Define $T_0\in\Cal T_{\bar\mu_L,\bar\mu_L}$ 
by setting 
      
\Centerline{$ T_0w=(\Bover1{\gamma}\int_0^{\gamma}w) 
\chi\coint{0,\gamma}^{\ssbullet} 
+(w\times\chi\coint{\gamma,\infty}^{\ssbullet})$} 
      
\noindent for every $w\in M^{1,\infty}_{\bar\mu_L}$.   Then 
$T_0u^*\preccurlyeq u^*\sim u$, and 
      
\Centerline{$ T_0u^*\times\chi\coint{0,\gamma}^{\ssbullet} 
=(\Bover1{\gamma}\int_0^{\gamma}u^*)\chi\coint{0,\gamma}^{\ssbullet} 
=\alpha_1\chi\coint{0,\gamma}^{\ssbullet}$.} 
      
We need to know that $\int_0^tT_0u^*\ge\int_0^tv^*$ for every $t$;  this 
is because 
      
$$\eqalign{\int_0^tT_0u^* 
&=\alpha_1t\ge\int_0^tv^*\text{ whenever }t\le\gamma,\cr 
&=\int_0^{\gamma}T_0u^*+\int_{\gamma}^tT_0u^* 
=\int_0^tu^*\ge\int_0^tv^*\text{ whenever }t\ge\gamma.\cr}$$ 
      
Set 
      
\Centerline{$u_1=T_0u^*\times\chi\coint{\beta,\infty}^{\ssbullet}$, 
\quad$v_1=v^*\times\chi\coint{\beta,\infty}^{\ssbullet}$.} 
      
\noindent Then $u_1^*$, $v_1^*$ are just translations of $T_0u^*$, $v^*$ 
to the left, so that 
      
\Centerline{$\int_0^tu_1^*=\int_{\beta}^{\beta+t}T_0u^* 
=\int_0^{\beta+t}T_0u^*-\alpha_1\beta 
\ge\int_0^{\beta+t}v^*-\alpha_1\beta 
=\int_{\beta}^{\beta+t}v^*=\int_0^tv_1^*$} 
      
\noindent for every $t\ge 0$.   Also 
$v_1=\sum_{i=2}^n\alpha_i\chi\coint{\beta_{i-1},\beta_i}^{\ssbullet}$ 
where $\beta_i=\sum_{j=1}^i\bar\mu a_j$ for each $j$.   So by the 
inductive hypothesis, $v_1\preccurlyeq u_1$. 
      
Let $S\in\Cal T_{\bar\mu_L,\bar\mu_L}$ be such that $Su_1=v_1$, and 
define $T\in\Cal T_{\bar\mu_L,\bar\mu_L}$ by setting 
      
\Centerline{$Tw=w\times\chi\coint{0,\beta}^{\ssbullet} 
+S(w\times\chi\coint{\beta,\infty}^{\ssbullet}) 
\times\chi\coint{\beta,\infty}^{\ssbullet}$} 
      
\noindent for every $w\in M^{1,\infty}_{\bar\mu_L,\bar\mu_L}$.   Then 
$TT_0u^*=v^*$, so $v\sim v^*\preccurlyeq u^*\sim u$, as required.\ \Qed 
      
\medskip 
      
\quad{\bf (ii)} We are nearly home.   If $u$, $v\ge 0$ in 
$M^{1,\infty}_{\bar\mu_L}$ and $\int_0^tv^*\le\int_0^tu^*$ for every 
$t\ge 0$, then $v\preccurlyeq u$.   \Prf\ Let $\sequencen{v_n}$ be a 
non-decreasing sequence in $S(\frak A^f_L)^+$ with supremum $v$.   Then 
$v_n^*\le v^*$ for each $n$, so (i) tells us that $v_n\preccurlyeq u$ 
for every $n$.   By 373N, for the last time, $v\preccurlyeq u$.\ \Qed 
      
\medskip 
      
{\bf (f)} Finally, suppose that $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ 
are arbitrary measure algebras and that $u\in M^{1,\infty}_{\bar\mu}$, 
$v\in M^{1,\infty}_{\bar\nu}$ are such that 
$\int_0^tv^*\le\int_0^tu^*$ for every $t\ge 0$.   Then

$$\eqalignno{v
&\sim|v|\preccurlyeq|v|^*=v^*
\Displaycause{by (b), (d) and 373Db}
\preccurlyeq u^*
\Displaycause{by (e)}
=|u|^*\preccurlyeq|u|
\Displaycause{by (c)}
\sim u\cr}$$

\noindent and $v\preccurlyeq u$, as claimed
}%end of proof of 373O 
      
\leader{373P}{Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra 
and $(\frak B,\bar\nu)$ a semi-finite measure algebra.   Then for any 
$u\in M^{1,\infty}(\frak A,\bar\mu)$ and $v\in M^0(\frak B,\bar\nu)$, 
there is a $T\in\cmmnt{\Cal T=\mskip5mu}\Cal T_{\bar\mu,\bar\nu}$ such that 
$\int Tu\times v=\int u^*\times v^*$. 
 
\proof{{\bf (a)} It is convenient to dispose immediately of some 
elementary questions. 
      
\medskip 
      
\quad{\bf (i)} We need only find a $T\in\Cal T$ such that $\int|Tu\times 
v|\ge\int u^*\times v^*$.   \Prf\ Take $v_0\in L^{\infty}(\frak B)$ such 
that $|Tu\times v|=v_0\times Tu\times v$ and $\|v_0\|_{\infty}\le 1$, 
and set $T_1w=v_0\times Tw$ for $w\in M^{1,\infty}_{\bar\mu}$;  then 
$T_1\in\Cal T$ and 
      
\Centerline{$\int T_1u\times v=\int|Tu\times v| 
\ge\int u^*\times v^*\ge\int T_1u\times v$} 
      
\noindent by 373J.\ \Qed 
      
\medskip 
      
\quad{\bf (ii)} Consequently it will be enough to consider $v\ge 0$, 
since of course $\int|Tu\times v|=\int|Tu\times|v||$, while $|v|^*=v^*$. 
      
\medskip 
      
\quad{\bf (iii)} It will be enough to consider $u=u^*$.   \Prf\ If we 
can find $T\in\Cal T_{\bar\mu_L,\bar\nu}$ such that $\int Tu^*\times 
v=\int(u^*)^*\times v^*$, then we know from 373O that there is an 
$S\in\Cal T_{\bar\mu,\bar\mu_L}$ such that $Su=u^*$, so that $TS\in\Cal 
T$ and 
      
\Centerline{$\int TSu\times v=\int(u^*)^*\times v^*=\int u^*\times v^*$. 
\Qed} 
      
\medskip 
      
\quad{\bf (iv)} It will be enough to consider localizable $(\frak 
B,\bar\nu)$.   \Prf\ Assuming that $v\ge 0$, following (ii) above, set 
$e=\Bvalue{v>0}=\sup_{n\in\Bbb N}\Bvalue{v>2^{-n}}$, and let $\bar\nu_e$ 
be the restriction of $\bar\nu$ to the principal ideal $\frak B_e$ 
generated by $e$.   Then if we write $\tilde v$ for the member of 
$L^0(\frak B_e)$ corresponding to $v$ (so that $\Bvalue{\tilde 
v>\alpha}=\Bvalue{v>\alpha}$ for every $\alpha>0$), $\tilde v^*=v^*$. 
Also $(\frak B_e,\bar\nu_e)$ is $\sigma$-finite, therefore localizable. 
Now if we can find $T\in \Cal T_{\bar\mu,\bar\nu_e}$ such that $\int 
Tu\times\tilde v=\int u^*\times\tilde v^*$, then $ST$ will belong to 
$\Cal T_{\bar\mu,\bar\nu}$, where $S:L^0(\frak B_e)\to L^0(\frak B)$ is 
the canonical embedding defined by the formula 
      
$$\eqalign{\Bvalue{Sw>\alpha}&=\Bvalue{w>\alpha} 
  \text{ if }\alpha\ge 0,\cr 
&=\Bvalue{w>\alpha}\Bcup(1\Bsetminus e)\text{ if }\alpha<0,}$$ 
      
\noindent and 
      
\Centerline{$\int STu\times v=\int Tu\times\tilde v=\int u^*\times 
\tilde v^*=\int u^*\times v^*$.  \Qed} 
      
      
\medskip 
      
{\bf (b)} So let us suppose henceforth that $\bar\mu=\bar\mu_L$, $u=u^*$ 
is the equivalence class of a non-increasing non-negative function, 
$v\ge 0$ and $(\frak B,\bar\nu)$ is localizable. 
      
For $n$, $i\in\Bbb N$ set 
      
\Centerline{$b_{ni}=\Bvalue{v>2^{-n}i}$, 
\quad$\beta_{ni}=\bar\nu b_{ni}$, 
\quad$c_{ni}=b_{ni}\Bsetminus b_{n,i+1}$, 
\quad$\gamma_{ni}=\bar\nu c_{ni}=\beta_{ni}-\beta_{n,i+1}$} 
      
\noindent (because $\beta_{ni}<\infty$ if $i>0$;  this is really where I 
use the hypothesis that $v\in M^0$).   For $n\in\Bbb N$ set 
      
\Centerline{$K_n=\{i:i\ge 1,\,\gamma_{ni}>0\}$,} 
      
$$T_nw=\sum_{i\in 
K_n}\bigl(\bover1{\gamma_{ni}}\int_{\beta_{n,i+1}}^{\beta_{ni}}w\bigr) 
\chi c_{ni}$$ 
      
\noindent for $w\in M^{1,\infty}_{\bar\mu_L}$;  this is defined 
in $L^0(\frak B)$ because $K_n$ is countable and $\sequence{i}{c_{ni}}$ 
is disjoint.   Of course $T_n:M^{1,\infty}_{\bar\mu_L}\to 
L^0(\frak B)$ is linear.   If $w\in L^{\infty}(\frak A_L)$ then 
      
\Centerline{$\|T_nw\|_{\infty}=\sup_{i\in 
K_n}\bigl|\Bover1{\gamma_{ni}}\int_{\beta_{n,i+1}}^{\beta_{ni}}w\bigr| 
\le\|w\|_{\infty}$,} 
      
\noindent and if $w\in L^1_{\bar\mu_L}$ then 
      
\Centerline{$\|T_nw\|_1 
=\sum_{i\in K_n}\bigl|\Bover1{\gamma_{ni}} 
$$\textfont3=\twelveex\int_{\beta_{n,i+1}}^{\beta_{ni}}w\bigr|\bar\nu c_{ni} 
=$$\sum_{i\in K_n}\bigl|$$\textfont3=\twelveex\int_{\beta_{n,i+1}}^{\beta_{ni}}w\bigr| 
\le\|w\|_1$;} 
      
\noindent so $T_nw\in M^{1,\infty}_{\bar\nu}$ for every $w\in 
M^{1,\infty}_{\bar\mu_L}$, and $T_n\in\Cal T$.    It will be helpful to 
observe that 
      
\Centerline{$\int_{c_{ni}}T_nw=\int_{\beta_{n,i+1}}^{\beta_{ni}}w$} 
      
\noindent whenever $i\ge 1$, since if $i\notin K_n$ then both sides are 
$0$. 
      
Note next that for every $n$, $i\in\Bbb N$, 
      
\Centerline{$b_{ni}=b_{n+1,2i}$, 
\quad$\beta_{ni}=\beta_{n+1,2i}$, 
\quad$c_{ni}=c_{n+1,2i}\Bcup c_{n+1,2i+1}$, 
\quad $\gamma_{ni}=\gamma_{n+1,2i}+\gamma_{n+1,2i+1}$,} 
      
\noindent so that, for $i\ge 1$, 
      
\Centerline{$\int_{c_{ni}}T_nu=\int_{\beta_{n,i+1}}^{\beta_{ni}}u 
=\int_{c_{ni}}T_{n+1}u$.} 
      
\noindent This means that if $T$ is any cluster point of 
$\sequencen{T_n}$ in $\Cal T$ for the very weak operator topology (and 
such a cluster point exists, by 
373L), $\int_{c_{mi}}Tu$ must be a cluster point of 
$\sequencen{\int_{c_{mi}}T_nu}$, and therefore equal to 
$\int_{c_{mi}}T_mu$, whenever $m\in\Bbb N$ and $i\ge 1$. 
      
Consequently, if $m\in\Bbb N$, 
      
$$\eqalignno{\int|Tu\times v| 
&\ge\sum_{i=0}^{\infty}\int_{c_{mi}}|Tu|\times v 
\ge\sum_{i=0}^{\infty}2^{-m}i\int_{c_{mi}}|Tu|\cr 
\noalign{\noindent (because $c_{mi}\Bsubseteq\Bvalue{v>2^{-m}i}$)} 
&\ge\sum_{i=1}^{\infty}2^{-m}i|\int_{c_{mi}}Tu| 
=\sum_{i=1}^{\infty}2^{-m}i\int_{c_{mi}}T_mu\cr 
&=\sum_{i=0}^{\infty}2^{-m}i\int_{\beta_{m,i+1}}^{\beta_{mi}}u 
\ge\int u\times(v^*-2^{-m}\chi 1)^+\cr}$$ 
      
\noindent because 
      
\Centerline{$[\beta_{m,i+1},\beta_{mi}]^{\ssbullet} 
\Bsubseteq\Bvalue{v^*\le 2^{-m}(i+1)}=\Bvalue{(v^*-2^{-m}\chi 1)^+\le 
2^{-m}i}$} 
      
\noindent for each $i\in\Bbb N$.    But letting $m\to\infty$, we have 
      
\Centerline{$\int|Tu\times v|\ge\lim_{m\to\infty} 
\int u\times(v^*-2^{-m}\chi 1)^+ 
=\int u\times v^*$} 
      
\noindent because $\sequence{m}{u\times(v^*-2^{-m}\chi 1)^+}$ is a 
non-decreasing sequence with supremum $u\times v^*$.   In view of the 
reductions in (a) above, this is enough to complete the proof. 
}%end of proof of 373P 
      
\leader{373Q}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra, 
$(\frak B,\bar\nu)$ a semi-finite measure algebra,  
$u\in M^{1,\infty}(\frak A,\bar\mu)$ and  
$v\in M^{0,\infty}(\frak B,\bar\nu)$.   Then 
      
\Centerline{$\int u^*\times v^* 
=\sup\{\int|Tu\times v|:T\in\Cal T_{\bar\mu,\bar\nu}\} 
=\sup\{\int Tu\times v:T\in\Cal T_{\bar\mu,\bar\nu}\} 
$.} 
      
\proof{ There is a non-decreasing sequence $\sequencen{c_n}$ in $\frak B^f$ 
such that $v^*=\sup_{n\in\Bbb N}v_n^*$, where $v_n=v\times\chi c_n$ for each
$n$.   \Prf\ For each rational $q>0$, we can find a 
countable non-empty set $B_q\subseteq\frak B$ such that 
      
\Centerline{$b\Bsubseteq\Bvalue{|v|>q}$, 
$\bar\nu b<\infty$ for every $b\in B_q$,} 
      
\Centerline{$\sup_{b\in B_q}\bar\nu b=\bar\nu\Bvalue{|v|>q}$} 
      
\noindent (because $(\frak B,\bar\nu)$ is semi-finite).   Let 
$\sequencen{b_n}$ be a sequence running over $\bigcup_{q\in\Bbb 
Q,q>0}B_q$ and set $c_n=\sup_{i\le n}b_i$, $v_n=v\times\chi c_n$ for 
each $n$. 
Then $\sequencen{|v_n|}$ and $\sequencen{v_n^*}$ are non-decreasing and 
$\sup_{n\in\Bbb N}v_n^*\le v^*$ in $L^0(\frak A_L)$.   But in fact 
$\sup_{n\in\Bbb N}v_n^*=v^*$, because 
      
\Centerline{$\bar\mu_L\Bvalue{v^*>q} 
=\bar\mu\Bvalue{|v|>q} 
=\sup_{n\in\Bbb N}\bar\mu\Bvalue{v_n>q} 
=\sup_{n\in\Bbb N}\bar\mu_L\Bvalue{v_n^*>q} 
=\bar\mu_L\Bvalue{\sup_{n\in\Bbb N}v_n^*>q}$} 
      
\noindent for every rational $q>0$, by 373Da.\ \Qed 
      
For each $n\in\Bbb N$ we have a $T_n\in\Cal T_{\bar\mu,\bar\nu}$ such 
that $\int T_nu\times v_n=\int u^*\times v_n^*$ (373P).   Set 
$S_nw=T_nw\times\chi c_n$ for $n\in\Bbb N$, $w\in 
M^{1,\infty}_{\bar\mu}$;  then every $S_n$ belongs to $\Cal 
T_{\bar\mu,\bar\nu}$, so 
      
$$\eqalign{\sup\{\int Tu\times v:T\in\Cal T_{\bar\mu,\bar\nu}\} 
&\ge\sup_{n\in\Bbb N}\int S_nu\times v 
=\sup_{n\in\Bbb N}\int T_nu\times v_n\cr 
&=\sup_{n\in\Bbb N}\int u^*\times v_n^* 
=\int u^*\times v^*\cr 
&\ge\sup\{\int|Tu\times v|:T\in\Cal T_{\bar\mu,\bar\nu}\} 
\ge\sup\{\int Tu\times v:T\in\Cal T_{\bar\mu,\bar\nu}\} 
\cr}$$ 
      
\noindent by 373J, as usual. 
}%end of proof of 373Q 
      
\leader{373R}{Order-continuous operators:  Proposition} Let  
$(\frak A,\bar\mu)$ be a measure algebra, $(\frak B,\bar\nu)$ 
a localizable measure algebra, and 
$T_0\in\cmmnt{\Cal T^{(0)}=\mskip5mu}\Cal T^{(0)}_{\bar\mu,\bar\nu}$.   Then 
there is a $T\in\Cal T^{\times}=\Cal T^{\times}_{\bar\mu,\bar\nu}$ 
extending $T_0$.   If $(\frak A,\bar\mu)$ is semi-finite, $T$ is 
uniquely defined. 
      
\proof{{\bf (a)} Suppose first that $T_0\in\Cal T^{(0)}$ is 
non-negative, regarded as a member of $\eurm 
L^{\sim}(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\nu})$.   In this 
case $T_0$ has an extension to an 
order-continuous positive linear operator 
$T:M^{1,\infty}_{\bar\mu}\to L^0(\frak B)$ defined by saying that 
$Tw=\sup\{T_0u:u\in M^{1,0}_{\bar\mu},\,0\le u\le w\}$ for every 
$w\ge 0$ in 
$M^{1,\infty}_{\bar\mu}$.   \Prf\ I use 355F.   $M^{1,0}_{\bar\mu}$ is a 
solid linear subspace of $M^{1,\infty}_{\bar\mu}$. 
$T_0$ is order-continuous when its codomain is taken to be 
$M^{1,0}_{\bar\nu}$, as noted in 371Gb, and therefore if its codomain is 
taken to 
be $L^0(\frak B)$, because $M^{1,0}$ is a solid linear subspace in 
$L^0$, so the embedding is order-continuous.    If 
$w\ge 0$ in $M^{1,\infty}_{\bar\mu}$, let 
$\gamma\ge 0$ be such that $u_1=(w-\gamma\chi 1)^+$ is integrable.   If 
$u\in M^{1,0}_{\bar\mu}$ and $0\le u\le w$, then 
$(u-\gamma\chi 1)^+\le u_1$, so 
      
\Centerline{$T_0u=T_0(u-\gamma\chi 1)^++T_0(u\wedge\gamma\chi 1) 
\le T_0u_1+\gamma\chi 1\in L^0(\frak B)$.} 
      
\noindent Thus $\{T_0u:u\in M^{1,0}_{\bar\nu},\,0\le u\le w\}$ is 
bounded above in $L^0(\frak B)$, for any $w\ge 0$ in 
$M^{1,\infty}_{\bar\mu}$.     $L^0(\frak B)$ is Dedekind 
complete, because $(\frak B,\bar\nu)$ is localizable, so 
$\sup\{T_0u:0\le 
u\le w\}$ is defined in $L^0(\frak B)$;  and this is true for every 
$w\in(M^{1,\infty}_{\bar\mu})^+$.   Thus the conditions of 355F are 
satisfied and we have the result.\ \Qed 
      
\medskip 
      
{\bf (b)} Now suppose that $T_0$ is any member of $\Cal T^{(0)}$.   Then 
$T_0$ has an extension to a member of $\Cal T^{\times}$.   \Prf\ 
$|T_0|$, $T_0^+=\bover12(|T_0|+T_0)$ and $T_0^-=\bover12(|T_0|-T_0)$, 
taken in $\eurm L^{\sim}(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\nu})$, all 
belong to $\Cal T^{(0)}$ (371G), so have  extensions $S$, $S_1$ and 
$S_2$ to order-continuous positive linear operators from 
$M^{1,\infty}_{\bar\mu}$ to $L^0(\frak B)$ as defined in (a).   Now for 
any $w\in L^1_{\bar\mu}$, 
      
\Centerline{$\|Sw\|_1=\||T_0|w\|_1\le\|w\|_1$,} 
      
\noindent and for any $w\in L^{\infty}(\frak A)$, 
      
\Centerline{$|Sw|\le S|w|=\sup\{|T_0|u:u\in M^{1,0}_{\bar\mu},\,0\le 
u\le w\}\le\|w\|_{\infty}\chi 1$,} 
      
\noindent so $\|Sw\|_{\infty}\le\|w\|_{\infty}$.   Thus $S\in\Cal T$; 
similarly, $S_1$ and $S_2$ can be regarded as operators from 
$M^{1,\infty}_{\bar\mu}$ to $M^{1,\infty}_{\bar\nu}$, and as such belong 
to $\Cal T$.   Next, for $w\ge 0$ in $M^{1,\infty}_{\bar\mu}$, 
      
$$\eqalign{S_1w+S_2w 
&=\sup\{T_0^+u:u\in M^{1,0}_{\bar\mu},\,0\le u\le w\} 
  +\sup\{T_0^-u:u\in M^{1,0}_{\bar\mu},\,0\le u\le w\}\cr 
&=\sup\{T_0^+u+T_0^-u:u\in M^{1,0}_{\bar\mu},\,0\le u\le w\} 
=Sw.\cr}$$ 
      
\noindent But this means that 
      
\Centerline{$S=S_1+S_2\ge|S_1-S_2|$} 
      
\noindent and $T=S_1-S_2\in\Cal T$, by 373Bc;  while of course $T$ 
extends $T_0^+-T_0^-=T_0$.   Finally, because $S_1$ and $S_2$ are 
order-continuous, 
$T\in\eurm L^{\times}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$, 
so $T\in\Cal T^{\times}$.\ \Qed 
      
\medskip 
      
{\bf (c)} If $(\frak A,\bar\mu)$ is semi-finite, then 
$M^{1,0}_{\bar\mu}$ is order-dense in $M^{1,\infty}_{\bar\mu}$ (because 
it includes $L^1_{\bar\mu}$, which is order-dense in $L^0(\frak A)$); 
so that the extension $T$ is unique, by 355Fe. 
}%end of proof of 373R 
      
\leader{373S}{Adjoints in $\Cal T^{(0)}$:  Theorem}  Let  
$(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be measure algebras, and $T$ any member of $\Cal T^{(0)}_{\bar\mu,\bar\nu}$.   Then there is a unique 
operator $T'\in\Cal T^{(0)}_{\bar\nu,\bar\mu}$ such that  
$\int_aT'(\chi b)=\int_bT(\chi a)$ whenever $a\in\frak A^f$ and
$b\in\frak B^f$, and now 
$\int u\times T'v=\int Tu\times v$ whenever  
$u\in M^{1,0}(\frak A,\bar\mu)$, 
$v\in M^{1,0}(\frak B,\bar\nu)$ are such that  
$\int u^*\times v^*<\infty$. 
      
\proof{{\bf (a)} For each $v\in M^{1,0}_{\bar\nu}$ we can 
define $T'v\in M^{1,0}_{\bar\mu}$ by the formula 
      
\Centerline{$\int_aT'v=\int T(\chi a)\times v$} 
      
\noindent for every $a\in\frak A^f$.   \Prf\ Set $\theta a=\int T(\chi 
a)\times v$ for each $a\in\frak A^f$;  because $\int(\chi a)^*\times 
v^*<\infty$, the integral is defined and finite (373J).   Of course 
$\theta:\frak A^f\to\Bbb R$ is additive because $\chi$ is additive and 
$T$, $\times$ and $\int$ are linear.   Also 
      
\Centerline{$\lim_{t\downarrow 0}\sup_{\bar\mu a\le t}|\theta a| 
\le\lim_{t\downarrow 0}\int_0^tv^*=0$,} 
      
\Centerline{$\lim_{t\to\infty}\Bover1t\sup_{\bar\mu a\le t}|\theta a| 
\le\lim_{t\to\infty}\bover1t\int_0^tv^*=0$} 
      
\noindent because $v\in M^{1,0}_{\bar\nu}$, so $v^*\in 
M^{1,0}_{\bar\mu_L}$.   By 373Ha, there is a unique $T'v\in 
M^{1,0}_{\bar\mu}$ such 
that $\int_aT'v=\theta a$ for every $a\in\frak A^f$.\ \Qed 
      
\medskip 
      
{\bf (b)} Because the formula uniquely determines $T'v$, we see that 
$T':M^{1,0}_{\bar\nu}\to M^{1,0}_{\bar\mu}$ is linear.   Now $T'\in\Cal 
T^{(0)}_{\bar\nu,\bar\mu}$.   \Prf\ (i) If $v\in L^1_{\bar\nu}$, then 
(because $T'v\in M^{1,0}_{\bar\mu}$) $|T'v|=\sup_{a\in\frak 
A^f}|T'v|\times\chi a$, and 
      
$$\eqalign{\|T'v\|_1 
&=\int|T'v| 
=\sup_{a\in\frak A^f}\int_a|T'v| 
=\sup_{b,c\in\frak A^f}(\int_bT'v-\int_cT'v)\cr 
&=\sup_{b,c\in\frak A^f}\int T(\chi b-\chi c)\times v 
\le\sup_{b,c\in\frak A^f}\int(\chi b-\chi c)^*\times v^*\cr 
&=\int v^* 
=\|v\|_1.\cr}$$ 
      
\noindent (ii) Now suppose that $v\in L^{\infty}(\frak B)\cap 
M^{1,0}_{\bar\nu}$, and set $\gamma=\|v\|_{\infty}$.   \Quer\ If 
$a=\Bvalue{|T'v|>\gamma}\ne 0$, then $T'v\ne 0$ so $v\ne 0$ and 
$\gamma>0$ and $\bar\mu a<\infty$, because $T'v\in M^{1,0}_{\bar\mu}$. 
Set $b=\Bvalue{(T'v)^+>\gamma}$, $c=\Bvalue{(T'v)^->\gamma}$;  then 
      
$$\eqalign{\gamma\bar\mu a 
&<\int_a|T'v| 
=\int_bT'v-\int_cT'v 
=\int T(\chi b-\chi c)\times v\cr 
&\le\gamma\|T(\chi b-\chi c)\|_1 
\le\gamma\|\chi b-\chi c\|_1 
=\gamma\bar\mu a,\cr}$$ 
      
\noindent which is impossible.\ \BanG\ 
Thus $\Bvalue{|T'v|>\gamma}=0$ and 
$\|T'v\|_{\infty}\le\gamma=\|v\|_{\infty}$. 
      
Putting this together with (i), we see that $T'\in \Cal 
T^{(0)}_{\bar\nu,\bar\mu}$.\ \Qed 
      
\medskip 
      
{\bf (c)} Let $|T|$ be the modulus of $T$ in 
$\eurm L^{\sim}(M^{1,0}_{\bar\mu};M^{1,0}_{\bar\nu})$, so that 
$|T|\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$, by 371Gb.   If $u\ge 0$ in 
$M^{1,0}_{\bar\mu}$, $v\ge 0$ in $M^{1,0}_{\bar\nu}$ are such that $\int 
u^*\times v^*<\infty$, let $\sequencen{u_n}$ be a 
non-decreasing sequence in $S(\frak A^f)^+$ with supremum $u$.   In this 
case $|T|u=\sup_{n\in\Bbb N}|T|u_n$, so $\int|T|u\times v=\sup_{n\in\Bbb 
N}\int|T|u_n\times v$ and 
      
\Centerline{$|\int Tu\times v-\int Tu_n\times v| 
\le\int|T|(u-u_n)\times v\to 0$} 
      
\noindent as $n\to\infty$, because 
      
\Centerline{$\int|T|u\times v\le\int u^*\times v^*<\infty$.} 
      
\noindent   At the same time, 
      
\Centerline{$|\int u\times T'v-\int u_n\times T'v| 
\le\int(u-u_n)\times|T'v|\to 0$} 
      
\noindent because $\int u\times|T'v|\le\int u^*\times v^*<\infty$.   So 
      
\Centerline{$\int Tu\times v=\lim_{n\to\infty}\int Tu_n\times v 
=\lim_{n\to\infty}\int u_n\times T'v=\int u\times T'v$,} 
      
\noindent the middle equality being valid because each $u_n$ is a linear 
combination of indicator functions. 
      
Because $T$ and $T'$ are linear, it follows at once that 
$\int u\times T'v=\int Tu\times v$ whenever $u\in M^{1,0}_{\bar\mu}$, 
$v\in M^{1,0}_{\bar\nu}$ are such that $\int u^*\times v^*<\infty$. 
      
\medskip 
      
{\bf (d)} Finally, to see that $T'$ is uniquely defined by the formula 
in the statement of the theorem, observe that this surely defines 
$T'(\chi b)$ for every $b\in\frak B^f$, by the remarks in (a). 
Consequently it defines $T'$ on $S(\frak B^f)$.   Since $S(\frak B^f)$ 
is order-dense in $M^{1,0}_{\bar\nu}$, and any member of 
$\Cal T^{(0)}_{\bar\nu,\bar\mu}$ must belong to 
$\eurm L^{\times}(M^{1,0}_{\bar\nu};M^{1,0}_{\bar\mu})$ (371Gb), the 
restriction of $T'$ to $S(\frak B^f)$ determines $T'$ (355J). 
}%end of proof of 373S 
      
\leader{373T}{Corollary} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ 
be localizable measure algebras.   Then for any  
$T\in\Cal T^{\times}_{\bar\mu,\bar\nu}$ there is a unique 
$T'\in\Cal T^{\times}_{\bar\nu,\bar\mu}$ such that  
$\int u\times T'v=\int Tu\times v$ whenever  
$u\in M^{1,\infty}(\frak A,\bar\mu)$, 
$v\in M^{1,\infty}(\frak B,\bar\nu)$ are such that  
$\int u^*\times v^*<\infty$. 
      
\proof{ The restriction $T\restr M^{1,0}_{\bar\mu}$ belongs to 
$\Cal T^{(0)}_{\bar\mu,\bar\nu}$ (373Bb), so there is a unique $S\in\Cal 
T^{(0)}_{\bar\nu,\bar\mu}$ such that $\int u\times Sv=\int Tu\times v$ 
whenever 
$u\in M^{1,0}_{\bar\mu}$, $v\in M^{1,0}_{\bar\nu}$ are 
such that $\int u^*\times v^*<\infty$ (373S).   Now there is a unique 
$T'\in\Cal T^{\times}_{\bar\nu,\bar\mu}$ extending $S$ (373R).   If 
$u\ge 0$ in $M^{1,\infty}_{\bar\mu}$, $v\ge 0$ in 
$M^{1,\infty}_{\bar\nu}$ are such that $\int u^*\times v^*<\infty$, then 
$\int u\times T'v=\int Tu\times v$.  \Prf\ If $T\ge 0$, then both are 
      
\Centerline{$\sup\{\int u_0\times T'v_0:u_0\in M^{1,0}_{\bar\mu},\,v\in 
M^{1,0}_{\bar\nu},\,0\le u_0\le u,\,0\le v_0\le v\}$} 
      
\noindent because both $T$ and $T'$ are (order-\nobreak)continuous.   In 
general, we can apply the same argument to $T^+$ and $T^-$, taken in 
$\eurm L^{\sim}(M^{1,\infty}_{\bar\mu};M^{1,\infty}_{\bar\nu})$, since these 
belong to $\Cal T^{\times}_{\bar\mu,\bar\nu}$, by 373B and 355H, 
and we shall surely have $T'=(T^+)'-(T^-)'$.\ \QeD\ 
As in 373S, it follows that $\int u\times T'v=\int Tu\times v$ whenever 
$u\in M^{1,\infty}_{\bar\mu}$, $v \in M^{1,\infty}_{\bar\nu}$ are such 
that $\int u^*\times v^*<\infty$. 
}%end of proof of 373T 
      
\leader{373U}{Corollary} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ 
be localizable measure algebras, and $\pi:\frak A\to\frak B$ an 
order-continuous measure-preserving Boolean homomorphism.   Then the 
associated map $T\in\Cal T^{\times}_{\bar\mu,\bar\nu}$\cmmnt{ (373Bd)} 
has an adjoint $P\in\Cal T^{\times}_{\bar\nu,\bar\mu}$ defined by the 
formula $\int_aP(\chi b)=\bar\nu(b\Bcap\pi a)$ for $a\in\frak A^f$, 
$b\in\frak B^f$. 
      
\proof{ By 373T, $T$ has an adjoint $P=T'$ such that
      
\Centerline{$\int_aP(\chi b)=\int \chi a\times P(\chi b) 
=\int T(\chi a)\times\chi b=\int\chi(\pi a)\times\chi b 
=\bar\nu(\pi a\Bcap b)$} 
      
\noindent whenever $a\in\frak A^f$ and $b\in\frak B^f$.   To see that this 
defines $P$ uniquely, let $S\in\Cal T^{\times}_{\bar\nu,\bar\mu}$ be any 
other operator with the same property.   By 373Hb, $S(\chi b)=P(\chi b)$ 
for every $b\in\frak B^f$, so $S$ and $P$ agree on $S(\frak B^f)$. 
Because both $P$ and $S$ are supposed to belong to 
$\eurm L^{\times}(M^{1,\infty}_{\bar\nu};M^{1,\infty}_{\bar\mu})$, and  
$S(\frak B^f)$ is order-dense in $M^{1,\infty}_{\bar\nu}$, $S=P$, by 355J. 
}%end of proof of 373U 
      
\exercises{\leader{373X}{Basic exercises (a)} 
%\spheader 373Xa 
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be measure algebras, and 
$\pi:\frak A\to\frak B$ a {\it ring} homomorphism such that  
$\bar\nu\pi a\le\bar\mu a$ for every $a\in\frak A$.   (i) Show that there is a 
unique $T\in\Cal T_{\bar\mu,\bar\nu}$ such that $T(\chi a)=\chi(\pi a)$ 
for every $a\in\frak A$, and that $T$ is a Riesz homomorphism.   (ii) 
Show that $T$ is (sequentially) order-continuous iff $\pi$ is. 
%373B 
      
\sqheader 373Xb Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\mu)$ be 
measure algebras, and $\phi:\Bbb R\to\Bbb R$ a convex function such that 
$\phi(0)\le 0$.   Show that if $T\in\Cal T_{\bar\mu,\bar\nu}$ and $T\ge 0$, 
then $\bar\phi(Tu)\le T(\bar\phi(u))$ whenever $u\in M^{1,\infty}_{\bar\mu}$ 
is such that $\bar\phi(u)\in M^{1,\infty}_{\bar\mu}$.
\Hint{371Gd.}
%373B 
      
\spheader 373Xc  Let $(\frak A,\bar\mu)$ be a measure 
algebra.    Show that if $w\in L^{\infty}(\frak A)$ and 
$\|w\|_{\infty}\le 1$ then  
$u\mapsto u\times w:M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\mu}$ belongs to $\Cal T^{\times}_{\bar\mu,\bar\mu}$. 
%373B 
      
\spheader 373Xd Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras.    Show 
that if $\langle a_i\rangle_{i\in I}$, $\langle b_i\rangle_{i\in I}$ are 
disjoint families in $\frak A$, $\frak B$ respectively, and $\langle 
T_i\rangle_{i\in I}$ is any family in $\Cal T_{\bar\mu,\bar\nu}$, and 
{\it either} $I$ is countable {\it or} $\frak B$ is Dedekind complete, 
then we have an operator $T\in\Cal T_{\bar\mu,\bar\nu}$ such that 
$Tu\times\chi 
b_i=T_i(u\times\chi a_i)\times\chi b_i$ for every $u\in 
M^{1,\infty}_{\bar\mu,\bar\nu}$, $i\in I$. 
%373B 
      
\sqheader 373Xe Let $I$, $J$ be sets and write $\mu=\bar\mu$, 
$\nu=\bar\nu$ 
for counting measure on $I$, $J$ respectively.   Show that there is a 
natural one-to-one correspondence between $\Cal 
T^{\times}_{\bar\mu,\bar\nu}$ and the set of matrices $\langle 
a_{ij}\rangle_{i\in I,j\in J}$ such that $\sum_{i\in I}|a_{ij}|\le 1$ 
for every $j\in J$, $\sum_{j\in J}|a_{ij}|\le 1$ for every $i\in I$. 
%373B 
      
\sqheader 373Xf Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be 
$\sigma$-finite measure spaces, with measure algebras $(\frak 
A,\bar\mu)$ and $(\frak B,\bar\nu)$, and product measure $\lambda$ on 
$X\times Y$.   Let 
$h:X\times Y\to\Bbb R$ be a measurable function such that 
$\int|h(x,y)|dx\le 
1$ for $\nu$-almost every $y\in Y$ and $\int|h(x,y)|dy\le 1$ for 
$\mu$-almost every $x\in X$.   Show that there is a corresponding 
$T\in\Cal T^{\times}_{\bar\mu,\bar\nu}$ defined by writing 
$T(f^{\ssbullet})=g^{\ssbullet}$ whenever $f\in\eusm L^1(\mu)+\eusm 
L^{\infty}(\mu)$ and $g(y)=\int h(x,y)f(x)dx$ for almost every $y$. 
%373B 
      
\sqheader 373Xg Let $\mu$ be Lebesgue measure on $\Bbb R$, and  
$(\frak A,\bar\mu)$ its measure algebra.   Show that for any  
$\mu$-integrable function $h$ with $\int|h|d\mu\le 1$ we have a  
corresponding $T\in\Cal T^{\times}_{\bar\mu,\bar\mu}$ defined by setting 
$T(f^{\ssbullet})=(h*f)^{\ssbullet}$ whenever  
$g\in\eusm L^1(\mu)+\eusm L^{\infty}(\mu)$, writing $h*f$ for the  
convolution of $h$ and $f$ 
(255E).   Explain how this may be regarded as a special case of 373Xf. 
%373B 
      
\sqheader 373Xh Let $(\frak A,\bar\mu)$ be a probability algebra and 
$u\in L^0(\frak A)^+$;  let $\nu_u$ be its distribution (364GB).   
Show that each of $u^*$, $\nu_u$ is uniquely determined by the other. 
%373C 
      
\spheader 373Xi Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras, and $\pi:\frak A\to\frak B$ a measure-preserving 
Boolean homomorphism;  let $T:M^{1,\infty}_{\bar\mu}\to 
M^{1,\infty}_{\bar\nu}$ be the corresponding operator (373Bd).   Show 
that $(Tu)^*=u^*$ for every $u\in M^{1,\infty}_{\bar\mu}$. 
%373C 
      
\spheader 373Xj Let $(\frak A,\bar\mu)$ be a totally finite measure 
algebra, and $A$ a subset of $L^1_{\bar\mu}$.   Show that the following 
are equiveridical:  (i) $A$ is uniformly integrable; 
(ii) $\{u^*:u\in A\}$ is uniformly integrable in $L^1_{\bar\mu_L}$; 
(iii) $\lim_{t\downarrow 0}\sup_{u\in A}\int_0^tu^*=0$. 
%373D 
      
\spheader 373Xk Let $(\frak A,\bar\mu)$ be a measure algebra, and 
$A\subseteq (M^0_{\bar\mu})^+$ a non-empty downwards-directed set. 
Show that $(\inf A)^*=\inf_{u\in A}u^*$ in $L^0(\frak A_L)$. 
%373D 
      
\spheader 373Xl Let $(\frak A,\bar\mu)$ be a measure algebra.   Show 
that $\|u\|_{1,\infty}=\int_0^1u^*$ for every 
$u\in M^{1,\infty}(\frak A,\bar\mu)$. 
%373F 
      
\spheader 373Xm Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras, and $\phi$ a Young's function 
(369Xc).   Write $U_{\phi,\bar\mu}\subseteq L^0(\frak A)$, 
$U_{\phi,\bar\nu}\subseteq L^0(\frak B)$ for the corresponding Orlicz 
spaces.   (i) Show that if $T\in\Cal T_{\bar\mu,\bar\nu}$ and $u\in 
U_{\phi,\bar\mu}$, then $Tu\in U_{\phi,\bar\nu}$ and 
$\|Tu\|_{\phi}\le\|u\|_{\phi}$.  (ii) Show that $u\in U_{\phi,\bar\mu}$ 
iff $u^*\in U_{\phi,\bar\mu_L}$, and in this case 
$\|u\|_{\phi}=\|u^*\|_{\phi}$. 
%373F 
      
\sqheader 373Xn Let $(\frak A,\bar\mu)$ be a measure algebra and  
$(\frak B,\bar\nu)$ a totally finite measure algebra.   Show that if  
$A\subseteq L^1_{\bar\mu}$ is uniformly integrable, then  
$\{Tu:u\in A,\,T\in\Cal T_{\bar\mu,\bar\nu}\}$ is uniformly integrable in 
$L^1_{\bar\nu}$. 
%373G 
      
\spheader 373Xo(i) Give examples of $u$, $v\in L^1(\frak A_L)$ such 
that $(u+v)^*\not\le u^*+v^*$.   (ii) Show that if $(\frak A,\bar\mu)$ 
is any measure algebra and $u$, $v\in M^{0,\infty}_{\bar\mu}$, 
then $\int_0^t(u+v)^*\le\int_0^tu^*+v^*$ for every $t\ge 0$. 
%373G 
      
\spheader 373Xp Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be two 
measure algebras.   For $u\in M^{1,0}_{\bar\mu}$, 
$w\in M^{\infty,1}_{\bar\nu}$ set 
      
\Centerline{$\rho_{uw}(S,T)=|\int Su\times w-\int Tu\times w|$ for 
$S$, $T\in\Cal T^{(0)}=\Cal T^{(0)}_{\bar\mu,\bar\nu}$.} 
      
\noindent The topology generated 
by the pseudometrics $\rho_{uw}$ is the {\bf very weak operator 
topology} on $\Cal T^{(0)}$.   Show that $\Cal T^{(0)}$ is compact in 
this topology. 
%373L 
      
\spheader 373Xq Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras and let $u\in M^{1,0}_{\bar\mu}$.  (i) Show that 
$B=\{Tu:T\in\Cal T^{(0)}_{\bar\mu,\bar\nu}\}$ is compact for the 
topology $\frak T_s(M^{1,0}_{\bar\nu},M^{\infty,1}_{\bar\nu})$. 
(ii) Show that any non-decreasing sequence in $B$ has a 
supremum in $L^0(\frak B)$ which belongs to $B$. 
%373M 
      
\spheader 373Xr Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras, and $u\in M^{1,0}_{\bar\mu}$, $v\in 
M^{1,0}_{\bar\nu}$. 
Show that the following are equiveridical:  (i) there is a 
$T\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$ such that $Tu=v$;  (ii) 
$\int_0^tu^*\le\int_0^tv^*$ for every $t\ge 0$. 
%373O 
      
\spheader 373Xs Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras.   Suppose that $u_1$, $u_2\in M^{1,\infty}_{\bar\mu}$ 
and $v\in M^{1,\infty}_{\bar\nu}$ are such that 
$\int_0^tv^*\le\int_0^t(u_1+u_2)^*$ 
for every $t\ge 0$.   Show that there are $v_1$, $v_2\in 
M^{1,\infty}_{\bar\nu}$ such that $v_1+v_2=v$ and 
$\int_0^tv_i^*\le\int_0^tu_i^*$ for both $i$, every $t\ge 0$. 
%373O 
      
\sqheader 373Xt Set $g(t)=t/(t+1)$ for $t\ge 0$, and set 
$v=g^{\ssbullet}$, 
$u=\chi[0,1]^{\ssbullet}\in L^{\infty}(\frak A_L)$.   Show that $\int 
u^*\times v^*=1>\int Tu\times v$ for every $T\in\Cal 
T_{\bar\mu_L,\bar\mu_L}$. 
%373P, 373Q 
      
\spheader 373Xu Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras, and for $T\in\Cal T^{(0)}_{\bar\mu,\bar\nu}$ define 
$T'\in\Cal T^{(0)}_{\bar\nu,\bar\mu}$ as in 373S.   Show that $T''=T$. 
%373S 
      
\spheader 373Xv Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras, and give $\Cal T^{(0)}_{\bar\mu,\bar\nu}$, $\Cal 
T^{(0)}_{\bar\nu,\bar\mu}$ their very weak operator topologies (373Xp). 
Show that the map $T\mapsto T':\Cal T^{(0)}_{\bar\mu,\bar\nu}\to\Cal 
T^{(0)}_{\bar\nu,\bar\mu}$ is an isomorphism for the convex, order and 
topological structures of the two spaces.   (By the `convex 
structure' of a convex set $C$ in a linear space I mean the operation 
$(x,y,t)\mapsto tx+(1-t)y:C\times C\times[0,1]\to C$.) 
%373S 
      
      
\leader{373Y}{Further exercises (a)} 
%\spheader 373Ya 
Let $(\frak A,\bar\mu)$ be the measure algebra of Lebesgue measure on 
$[0,1]$.   Set $u=f^{\ssbullet}$ and $v=g^{\ssbullet}$ in $L^0(\frak 
A)$, where $f(t)=t$, 
$g(t)=1-2|t-\bover12|$ for $t\in[0,1]$.   Show that $u^*=v^*$, but that 
there is no measure-preserving Boolean homomorphism $\pi:\frak A\to\frak 
A$ such that $T_{\pi}v=u$, writing  
$T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ for the operator induced by $\pi$,  
as in 364P.   \Hint{show that 
$\{\Bvalue{v>\alpha}:\alpha\in\Bbb R\}$ does not $\tau$-generate $\frak 
A$.} 
%373O 
      
\spheader 373Yb Let $(\frak A,\bar\mu)$ be a totally finite homogeneous 
measure algebra of uncountable Maharam type.   Let $u$, 
$v\in(M^{1,\infty}_{\bar\mu})^+$ be such that $u^*=v^*$.   Show 
that there is a 
measure-preserving automorphism $\pi:\frak A\to\frak A$ such that 
$T_{\pi}u=v$. 
%373O 
      
\spheader 373Yc Let $u$, $v\in M^{1,\infty}_{\bar\mu_L}$ be such that 
$u=u^*$, $v=v^*$ and 
$\int_0^tv\le\int_0^tu$ for every $t\ge 0$.   (i) Show that there is a 
non-negative $T\in\Cal T_{\bar\mu_L,\bar\mu_L}$ such that $Tu=v$ and 
$\int_0^tTw\le\int_0^tw$ for every $w\in(M^{1,\infty}_{\bar\mu_L})^+$.   
(ii) Show 
that any such $T$ must belong to $\Cal T^{\times}_{\bar\mu_L,\bar\mu_L}$. 
%373O mt37bits
      
\spheader 373Yd Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be 
measure algebras, and $u\in M^{1,\infty}_{\bar\mu}$.  (i) Suppose that 
$w\in S(\frak B^f)$.   Show directly (without quoting the result of 
373O, but possibly using some of the ideas of the proof) that for every 
$\gamma<\int u^*\times w^*$ there is a $T\in\Cal T_{\bar\mu,\bar\nu}$ 
such that $\int Tu\times w\ge\gamma$. 
(ii) Suppose that $(\frak B,\bar\nu)$ is localizable and that $v\in 
M^{1,\infty}_{\bar\nu}\setminus\{Tu:T\in\Cal 
T_{\bar\mu,\bar\nu}\}$.   Show that there is a $w\in S(\frak B^f)$ such 
that $\int v\times w>\sup\{\int Tu\times w:T\in\Cal 
T_{\bar\mu,\bar\nu}\}$.   \Hint{use 373M and the Hahn-Banach theorem in 
the following form:  if $U$ is a linear space with the topology  
$\frak T_s(U,V)$ defined by 
a linear subspace $V$ of $\eurm L(U;\Bbb R)$, $C\subseteq U$ is a 
non-empty closed convex set, and $v\in U\setminus C$, then there is an 
$f\in V$ such that $f(v)>\sup_{u\in C}f(u)$.}   (iii) Hence prove 373O 
for localizable $(\frak B,\bar\nu)$.   (iv) Now prove 373O for general 
$(\frak B,\bar\nu)$. 
      
\spheader 373Ye(i) Define $v\in L^{\infty}(\frak A_L)$ as in 373Xt. 
Show that there is no $T\in\Cal T^{\times}_{\bar\mu_L,\bar\mu_L}$ such 
that $Tv=v^*$.   (ii) Set $h(t)=1+\max(0,\bover{\sin t}t)$ for $t>0$, 
$w=h^{\ssbullet}\in L^{\infty}(\frak A_L)$. 
Show that there is no $T\in\Cal T^{\times}_{\bar\mu_L,\bar\mu_L}$ such 
that $Tw^*=w$. 
%373Xt, 373P 
      
\spheader 373Yf Let $(\frak A,\bar\mu)$ be the measure algebra of 
Lebesgue 
measure on $[0,1]$.   Show that $\Cal T_{\bar\mu,\bar\mu_L}=\Cal 
T^{\times}_{\bar\mu,\bar\mu_L}$ can be identified, as convex ordered 
space, 
with $\Cal T^{\times}_{\bar\mu_L,\bar\mu}$, and that this is a proper 
subset 
of $\Cal T_{\bar\mu_L,\bar\mu}$. 
%373T 
      
\spheader 373Yg Show that the adjoint operation of 373T is not as a rule 
continuous for the very weak operator topologies of 
$\Cal T^{\times}_{\bar\mu,\bar\nu}$, $\Cal T^{\times}_{\bar\nu,\bar\mu}$. 
%373T   mt37bits
}%end of exercises 
      
\cmmnt{ 
\Notesheader{373} 373A-373B are just alternative expressions of concepts 
already treated in 371F-371H.   My use of the simpler formula 
$\Cal T_{\bar\mu,\bar\nu}$ symbolizes my view that $\Cal T$, rather than 
$\Cal T^{(0)}$ or $\Cal T^{\times}$, is the most natural vehicle for 
these ideas;  I used $\Cal T^{(0)}$ in \S\S371 only because that 
made it 
possible to give theorems which applied to all measure algebras, without 
demanding localizability (compare 371Gb with 373Bc). 
      
The obvious examples of operators in $\Cal T$ are those derived from 
measure-preserving Boolean homomorphisms, as in 373Bd, and their 
adjoints (373U).   Note that the latter include conditional expectation 
operators.   In return, we find that operators in $\Cal T$ share some of 
the characteristic properties of the operators derived from Boolean 
homomorphisms (373Bb, 373Xb, 373Xm). 
Other examples are multiplication operators (373Xc), operators obtained 
by piecing others together (373Xd) and kernel operators of the type 
described in 373Xe-373Xf, including convolution operators (373Xg). 
(For a general theory of kernel operators, see \S376 below.) 
      
Most of the section is devoted to the relationships between the classes 
$\Cal T$ of operators and the `decreasing rearrangements' of 373C. 
If you like, the decreasing rearrangement $u^*$ of $u$ describes the 
`distribution' of $|u|$ (373Xh);  but for $u\notin M^0$ it loses some 
information (373Xt, 373Ye).   It is important to be conscious that even 
when $u\in L^0(\frak A_L)$, $u^*$ is not necessarily obtained by 
`rearranging' the elements of the algebra $\frak A_L$ by a 
measure-preserving automorphism (which would, of course, correspond to 
an automorphism of the measure space $(\coint{0,\infty},\mu_L)$, by 
344C).   I will treat `rearrangements' of this narrower type in the 
next section;  for the moment, see 373Ya.   Apart from this, the basic 
properties of decreasing rearrangements are straightforward enough 
(373D-373F).   The only obscure area concerns the relationship between 
$(u+v)^*$ and $u^*$, $v^*$ (see 373Xo). 
      
In 373G I embark on results involving both decreasing rearrangements and 
operators in $\Cal T$, leading to the characterization of the sets 
$\{Tu:T\in\Cal T\}$ in 373O.   In one direction this is easy, and is the 
content of 373G.   In the other direction it depends on a deeper 
analysis, and the easiest method seems to be through studying the 
`very weak operator topology' on $\Cal T$ (373K-373L), even though this 
is an effective tool only when one of the algebras involved is 
localizable (373L).   A functional analyst is likely to feel that the 
method is both natural and illuminating;  but from the point of view of 
a measure theorist it is not perfectly satisfactory, because it is 
essentially non-constructive.   While it tells us that there are 
operators $T\in\Cal T$ acting in the required ways, it gives only the 
vaguest of hints concerning what they actually look like. 
      
Of course the very weak operator topology is interesting in its own 
right;  and see also 373Xp-373Xq. 
      
The proof of 373O can be thought of as consisting of three steps. 
Given that $\int_0^tv^*\le\int_0^tu^*$ for every $t$, then I set out to 
show that $v$ is expressible as $T_1v^*$ (parts (c)-(d) of the proof), 
that $v^*$ is expressible as $T_2u^*$ (part (g)) and that $u^*$ is 
expressible as $T_3u$ 
(parts (e)-(f)), each $T_i$ belonging to an appropriate $\Cal T$.   In 
all three steps the general case follows easily from the case in which 
$u\in S(\frak A)$ and $v\in S(\frak B)$.   If we are willing to use a more 
sophisticated version of the 
Hahn-Banach theorem than those given in 3A5A and 363R, there is 
an alternative route (373Yd).   I note that the central step above, from 
$u^*$ to $v^*$, can be performed with an order-continuous $T_2$ (373Yc), 
but that in general neither of the other steps can (373Ye), so that we 
cannot use $\Cal T^{\times}$ in place of $\Cal T$ here. 
      
A companion result to 373O, in that it also shows that 
$\{Tu:T\in\Cal T\}$ is large enough to reach natural bounds, is 373P; 
given $u$ and 
$v$, we can find $T$ such that $\int Tu\times v$ is as large as 
possible.   In this form the result is valid only for $v\in M^{(0)}$ 
(373Xt).  But if we do not demand that the supremum should be attained, 
we can deal with other $v$ (373Q). 
      
We already know that every operator in $\Cal T^{(0)}$ is a difference of 
order-continuous operators, just because $M^{1,0}$ has an 
order-continuous norm (371Gb).   It is therefore not surprising that 
members of $\Cal T^{(0)}$ can be extended to members of $\Cal T^{\times}$,  
at least when the codomain $M^{1,\infty}_{\bar\nu}$ is 
Dedekind complete (373R).   It is also very natural to look for a 
correspondence between $\Cal T_{\bar\mu,\bar\nu}$ and  
$\Cal T_{\bar\nu,\bar\mu}$, because if $T\in\Cal T_{\bar\mu,\bar\nu}$  
we shall surely have an adjoint operator $(T\restr L^1_{\bar\mu})'$ from 
$(L^1_{\bar\nu})^*$ to $(L^1_{\bar\mu})^*$, and we can hope that this 
will correspond to some member of $\Cal T_{\bar\nu,\bar\mu}$.   But when 
we come to the details, the normed-space properties of a general member 
of $\Cal T$ are not enough (373Yf), and we need some kind of 
order-continuity.   For members of $\Cal T^{(0)}$ this is automatically 
present (373S), and now the canonical isomorphism between $\Cal T^{(0)}$ 
and $\Cal T^{\times}$ gives us an isomorphism between  
$\Cal T^{\times}_{\bar\mu,\bar\nu}$ and  
$\Cal T^{\times}_{\bar\nu,\bar\mu}$ 
when $\bar\mu$ and $\bar\nu$ are localizable (373T). 
}%end of comment 
      
\discrpage 
      
\frfilename{mt374.tex}
\versiondate{15.6.09}
\copyrightdate{1996}

\def\chaptername{Linear operators between function spaces}
\def\sectionname{Rearrangement-invariant spaces}

\def\ri{rearrangement{\vthsp}-{\vthsp}invariant}

\newsection{374}

As is to be expected, many of the most important function spaces of
analysis are symmetric in various ways;  in particular, they share the
symmetries of the underlying measure algebras.   The natural expression
of this is to say that they are `\ri' (374E).
In fact it turns out that in many cases they have the stronger property
of `$\Cal T$-invariance' (374A).   In this section I give a brief
account of the most important properties of these two kinds of
invariance.   In particular, $\Cal T$-invariance is related to a kind of
transfer mechanism, enabling us to associate function spaces on
different measure algebras (374C-374D).   As for
rearrangement-{\vthsp}invariance, the salient fact is that on the most
important measure algebras many \ri\ spaces are
$\Cal T$-invariant (374K, 374M).

\leader{374A}{$\Cal T$-invariance:  Definitions} Let
$(\frak A,\bar\mu)$ be a measure algebra.   \cmmnt{Recall that I write

\Centerline{$M^{1,\infty}_{\bar\mu}
=L^1_{\bar\mu}+L^{\infty}(\frak A)\subseteq L^0(\frak A)$,}

\Centerline{$M^{\infty,1}_{\bar\mu}
=L^1_{\bar\mu}\cap L^{\infty}(\frak A)$,}

\Centerline{$M^{0,\infty}_{\bar\mu}=\{u:u\in L^0(\frak A),
  \,\inf_{\alpha>0}\bar\mu\Bvalue{|u|>\alpha}<\infty\}$,}

\noindent (369N, 373C).}%end of comment

\spheader 374Aa I will say that a subset $A$ of
$M^{1,\infty}_{\bar\mu}$ is
{\bf $\Cal T$-invariant} if $Tu\in A$ whenever $u\in A$ and
$T\in\Cal T=\Cal T_{\bar\mu,\bar\mu}$\cmmnt{ (definition:  373Aa)}.

\spheader 374Ab An extended Fatou norm $\tau$ on $L^0$
is {\bf $\Cal T$-invariant}\cmmnt{ or {\bf fully symmetric}} if
$\tau(Tu)\le\tau(u)$ whenever $u\in M^{1,\infty}_{\bar\mu}$ and
$T\in\Cal T$.

%`fully symmetric' in Dodds Schl\"uchtermann & Sukochev 01

\spheader 374Ac \cmmnt{As in \S373,} I will write
$(\frak A_L,\bar\mu_L)$ for the measure algebra of Lebesgue measure on
$\coint{0,\infty}$, and $u^*\in M^{0,\infty}_{\bar\mu_L}$ for
the decreasing rearrangement of any $u$ belonging to any
$M^{0,\infty}_{\bar\mu}$\cmmnt{ (373C)}.

\leader{374B}{}\cmmnt{ The first step is to show that the associate of
a $\Cal T$-invariant norm is $\Cal T$-invariant.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra and $\tau$ a $\Cal T$-invariant extended Fatou norm on
$L^0(\frak A)$.   Let $L^{\tau}$ be the Banach lattice
defined from $\tau$\cmmnt{ (369G)}, and $\tau'$ the associate extended
Fatou norm\cmmnt{ (369H-369I)}.   Then

(i) $M^{\infty,1}_{\bar\mu}\subseteq L^{\tau}
\subseteq M^{1,\infty}_{\bar\mu}$;

(ii)  $\tau'$ is
also $\Cal T$-invariant, and $\int u^*\times v^*\le\tau(u)\tau'(v)$ for
all $u$, $v\in M^{0,\infty}_{\bar\mu}$.

\proof{{\bf (a)} I check first that
$L^{\tau}\subseteq M^{0,\infty}_{\bar\mu}$.
\Prf\ Take any $u\in L^0(\frak A)\setminus M^{0,\infty}_{\bar\mu}$.
There is surely some $w>0$ in $L^{\tau}$, and we can suppose that
$w=\chi a$ for some $a$ of finite measure.   Now, for any $n\in\Bbb N$,

\Centerline{$(|u|\wedge n\chi 1)^*=n\chi 1\ge nw^*$}

\noindent in $L^0(\frak A_L)$, because
$\bar\mu\Bvalue{|u|>n}=\infty$.   So there is a
$T\in\Cal T_{\bar\mu,\bar\mu}$ such that $T(|u|\wedge n\chi 1)=nw$, by
373O, and

\Centerline{$\tau(u)\ge\tau(|u|\wedge n\chi 1)
\ge\tau(T(|u|\wedge n\chi 1))=\tau(nw)=n\tau(w)$.}

\noindent As $n$ is arbitrary, $\tau(u)=\infty$.   As $u$ is arbitrary,
$L^{\tau}\subseteq M^{0,\infty}_{\bar\mu}$.\ \Qed

\medskip

{\bf (b)} Next, $\int u^*\times v^*\le\tau(u)\tau'(v)$ for all $u$,
$v\in M^{0,\infty}_{\bar\mu}$.   \Prf\ If $u\in M^{1,\infty}_{\bar\mu}$,
then

$$\eqalignno{\int u^*\times v^*
&=\sup\{\int|Tu\times v|:T\in\Cal T_{\bar\mu,\bar\mu}\}\cr
\noalign{\noindent (373Q)}
&\le\sup\{\tau(Tu)\tau'(v):T\in\Cal T_{\bar\mu,\bar\mu}\}
=\tau(u)\tau'(v).\cr}$$

\noindent Generally, setting $u_n=|u|\wedge n\chi 1$,
$\sequencen{u_n^*}$ is a non-decreasing sequence with
supremum $u^*$ (373Db, 373Dh), so

\Centerline{$\int u^*\times v^*
=\sup_{n\in\Bbb N}\int u_n^*\times v^*
\le\sup_{n\in\Bbb N}\tau(u_n)\tau'(v)
=\tau(u)\tau'(v)$.  \Qed}

\medskip

{\bf (c)} Consequently, $L^{\tau}\subseteq M^{1,\infty}_{\bar\mu}$.
\Prf\ If $\frak A=\{0\}$, this is trivial.   Otherwise, take
$u\in L^{\tau}$.   There is surely some non-zero $a$ such that
$\tau'(\chi a)<\infty$;  now, setting $v=\chi a$,

\Centerline{$\int_0^{\bar\mu a}u^*=\int u^*\times v^*
\le\tau(u)\tau'(v)<\infty$}

\noindent by (b) above.   But this means that
$u^*\in M^{1,\infty}_{\bar\mu}$, so that $u\in M^{1,\infty}_{\bar\mu}$
(373F(b-ii)).\ \Qed

\medskip

{\bf (d)} Next, $\tau'$ is $\Cal T$-invariant.   \Prf\  Suppose
that $v\in M^{1,\infty}_{\bar\mu}$, $T\in\Cal T_{\bar\mu,\bar\mu}$,
$u\in L^0(\frak A)$ and $\tau(u)\le 1$.   Then
$u\in M^{1,\infty}_{\bar\mu}$, by (c), so

\Centerline{$\int|u\times Tv|\le\int u^*\times v^*
\le\tau(u)\tau'(v)\le\tau'(v)$,}

\noindent using 373J for the first inequality.   Taking the supremum
over $u$, we see that $\tau'(Tv)\le\tau'(v)$;  as $T$ and $v$ are
arbitrary, $\tau'$ is $\Cal T$-invariant.\ \Qed

\medskip

{\bf (e)} Finally, putting (d) and (c) together,
$L^{\tau'}\subseteq M^{1,\infty}_{\bar\mu}$, so that
$L^{\tau}\supseteq M^{\infty,1}_{\bar\mu}$, using 369J and 369O.
}%end of proof of 374B

\leader{374C}{}\cmmnt{ For any $\Cal T$-invariant extended Fatou norm
on $L^0(\frak A_L)$ there are corresponding norms on $L^0(\frak A)$ for
any semi-finite measure algebra, as follows.

\medskip

\noindent}{\bf Theorem} Let $\theta$ be a $\Cal T$-invariant extended
Fatou norm on $L^0(\frak A_L)$, and $(\frak A,\bar\mu)$ a
semi-finite measure algebra.

(a) There is a $\Cal T$-invariant extended Fatou norm $\tau$ on
$L^0(\frak A)$ defined by setting

$$\eqalign{\tau(u)
&=\theta(u^*)\text{ if }u\in M^{0,\infty}_{\bar\mu},\cr
&=\infty\text{ if }u\in L^0(\frak A)\setminus M^{0,\infty}_{\bar\mu}.\cr}$$

(b) Writing $\theta'$, $\tau'$ for the associates of $\theta$ and
$\tau$, we now have

$$\eqalign{\tau'(v)&=\theta'(v^*)
  \text{ if }v\in M^{0,\infty}_{\bar\mu},\cr
&=\infty\text{ if }v\in L^0(\frak A)\setminus M^{0,\infty}_{\bar\mu}.\cr}$$

(c) If $\theta$ is an order-continuous norm on the Banach lattice
$L^{\theta}$, then $\tau$ is an order-continuous norm on $L^{\tau}$.

\wheader{374C}{0}{0}{0}{24pt}

\proof{{\bf (a)(i)} The argument seems to run better if I use a
different formula to define $\tau$:  set

\Centerline{$\tau(u)
=\sup\{\int|u\times Tw|:T\in\Cal T_{\bar\mu_L,\bar\mu},
   \,w\in L^0(\frak A_L),\,\theta'(w)\le 1\}$}

\noindent for $u\in L^0(\frak A)$.   (By 374B(i),
$w\in M^{1,\infty}_{\bar\mu_L}$ whenever $\theta'(w)\le 1$, so
there is no difficulty in defining $Tw$.)   Now $\tau(u)=\theta(u^*)$
for every $u\in M^{0,\infty}_{\bar\mu}$.   \Prf\
($\alpha$) If $w\in L^0(\frak A_L)$ and $\theta'(w)\le 1$, then
$w\in M^{1,\infty}_{\bar\mu_L}$, so there is
an $S\in\Cal T_{\bar\mu_L,\bar\mu_L}$ such that $Sw=w^*$ (373O).
Accordingly $\theta'(w^*)\le\theta'(w)$ (because $\theta'$ is
$\Cal T$-invariant, by 374B);  now

\Centerline{$\int|u\times Tw|\le\int u^*\times w^*
\le\theta(u^*)\theta'(w^*)\le\theta(u^*)\theta'(w)\le\theta(u^*)$;}

\noindent as $w$ is arbitrary, $\tau(u)\le\theta(u^*)$.  ($\beta$) If
$w\in L^0(\frak A_L)$ and $\theta'(w)\le 1$, then

$$\eqalignno{\int|u^*\times w|
&\le\int(u^*)^*\times w^*\cr
\noalign{\noindent (373E)}
&=\int u^*\times w^*
=\sup\{\int|u\times Tw|:T\in\Cal T_{\bar\mu_L,\bar\mu}\}\cr
\noalign{\noindent (373Q)}
&\le\tau(u).\cr}$$

\noindent But because $\theta$
is the associate of $\theta'$ (369I(ii)), this means that
$\theta(u^*)\le\tau(u)$.\   \Qed

\medskip

\quad{\bf (ii)} Now $\tau$ is an extended Fatou norm on $L^0(\frak A)$.
\Prf\ Of the conditions in 369F, (i)-(iv) are satisfied just because
$\tau(u)=\sup_{v\in B}\int|u\times v|$ for some set $B\subseteq L^0$.
As for (v) and (vi), observe that if $u\in M^{\infty,1}_{\bar\mu}$ then
$u^*\in M^{\infty,1}_{\bar\mu_L}$ (373F(b-iv)), so
that $\tau(u)=\theta(u^*)<\infty$, by 374B(i), while also

\Centerline{$u\ne 0\Longrightarrow u^*\ne 0\Longrightarrow
\tau(u)=\theta(u^*)>0$.}

\noindent As $M^{\infty,1}_{\bar\mu}$ is order-dense in
$L^0(\frak A)$ (this is where I use the hypothesis that
$(\frak A,\bar\mu)$ is semi-finite), 369F(v)-(vi) are satisfied, and
$\tau$ is an extended Fatou norm.\   \Qed

\medskip

\quad{\bf (iii)} $\tau$ is $\Cal T$-invariant.   \Prf\
Take $u\in M^{1,\infty}_{\bar\mu}$ and
$T\in\Cal T_{\bar\mu,\bar\mu}$.    There are
$S_0\in\Cal T_{\bar\mu_L,\bar\mu}$ and
$S_1\in\Cal T_{\bar\mu,\bar\mu_L}$ such that $S_0u^*=u$, $S_1Tu=(Tu)^*$
(373O);  now $S_1TS_0\in\Cal T_{\bar\mu_L,\bar\mu_L}$ (373Be), so

\Centerline{$\tau(Tu)=\theta((Tu)^*)=\theta(S_1TS_0u^*)\le\theta(u^*)
=\tau(u)$}

\noindent because $\theta$ is $\Cal T$-invariant.\ \Qed

\medskip

\quad{\bf (iv)} We can now return to the definition of $\tau$.   I have
already remarked that $\tau(u)=\theta(u^*)$ if $u\in
M^{0,\infty}_{\bar\mu}$.   For other $u$, we must have $\tau(u)=\infty$
just because $\tau$ is a $\Cal T$-invariant extended Fatou norm
(374B(i)).   So the definitions in the statement of the theorem and (i)
above coincide.

\medskip

{\bf (b)} We surely have $\tau'(v)=\infty$ if $v\in L^0(\frak
A)\setminus M^{0,\infty}_{\bar\mu}$, by 374B, because $\tau'$,
like $\tau$, is a $\Cal T$-invariant extended Fatou norm.   So take
$v\in M^{0,\infty}_{\bar\mu}$.

\medskip

\quad{\bf (i)} If $u\in L^0(\frak A)$ and $\tau(u)\le 1$, then

\Centerline{$\int|v\times u|
\le\int v^*\times u^*
\le\theta'(v^*)\theta(u^*)
=\theta'(v^*)\tau(u)
\le\theta'(v^*)$;}

\noindent as $u$ is arbitrary, $\tau'(v)\le\theta'(v^*)$.

\medskip

\quad{\bf (ii)} If $w\in L^0(\frak A_L)$ and $\theta(w)\le 1$, then

$$\eqalignno{\int|v^*\times w|
&\le\int v^*\times w^*
=\sup\{\int|v\times Tw|:T\in\Cal T_{\bar\mu_L,\bar\mu}\}\cr
\noalign{\noindent (373Q)}
&\le\sup\{\tau'(v)\tau(Tw):T\in\Cal T_{\bar\mu_L,\bar\mu}\}
=\sup\{\tau'(v)\theta((Tw)^*):T\in\Cal T_{\bar\mu_L,\bar\mu}\}\cr
&\le\sup\{\tau'(v)\theta(STw):T\in\Cal T_{\bar\mu_L,\bar\mu},
  \,S\in\Cal T_{\bar\mu,\bar\mu_L}\}\cr
\noalign{\noindent (because, given $T$, we can find an $S$ such that
$STw=(Tw)^*$, by 373O)}
&\le\sup\{\tau'(v)\theta(Tw):T\in\Cal T_{\bar\mu_L,\bar\mu_L}\}
\le\tau'(v).\cr}$$

\noindent As $w$ is arbitrary, $\theta'(v^*)\le\tau'(v)$ and the two are
equal.   This completes the proof of (b).

\medskip

{\bf (c)(i)} The first step is to note that
$L^{\tau}\subseteq M^0_{\bar\mu}$.   \Prf\Quer\ Suppose that
$u\in L^{\tau}\setminus M^0_{\bar\mu}$, that is, that
$\bar\mu\Bvalue{|u|>\alpha}=\infty$ for
some $\alpha>0$.   Then $u^*\ge\alpha\chi 1$ in $L^0(\frak A_L)$, so
$L^{\infty}(\frak A_L)\subseteq L^{\theta}$.   For each $n\in\Bbb N$,
set $v_n=\chi\coint{n,\infty}^{\ssbullet}$.   Then $v_n^*=v_0$, so we
can find a $T_n\in\Cal T_{\bar\mu_L,\bar\mu_L}$ such that $T_nv_n=v_0$
(373O), and $\theta(v_n)\ge\theta(v_0)$ for every $n$.   But as
$\sequencen{v_n}$ is a decreasing sequence with infimum $0$, this means
that $\theta$ is not an order-continuous norm.\ \Bang\Qed

\medskip

\quad{\bf (ii)} Now suppose that $A\subseteq L^{\tau}$ is non-empty and
downwards-directed and has infimum $0$.   Then
$\inf_{u\in A}\bar\mu\Bvalue{u>\alpha}=0$ for every $\alpha>0$ (put
364L(b-ii) and 321F
together).   But this means that $B=\{u^*:u\in A\}$ must have infimum
$0$;  since $B$ is surely downwards-directed,
$\inf_{v\in B}\theta(v)=0$, that is, $\inf_{u\in A}\tau(u)=0$.   As $A$
is arbitrary, $\tau$ is an order-continuous norm.
}%end of proof of 374C

\leader{374D}{}\cmmnt{ What is more, every $\Cal T$-invariant extended
Fatou norm can be represented in this way.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $\tau$ a $\Cal T$-invariant extended Fatou norm on
$L^0(\frak A)$.   Then there is a $\Cal T$-invariant extended Fatou norm
$\theta$ on $L^0(\frak A_L)$ such that $\tau(u)=\theta(u^*)$
for every $u\in M^{0,\infty}_{\bar\mu}$.

\proof{ I use the method of 374C.   If $\frak A=\{0\}$ the result is
trivial;  assume that $\frak A\ne\{0\}$.

\woddheader{374D}{6}{2}{2}{24pt}

{\bf (a)}   Set

\Centerline{$\theta(w)=\sup\{\int|w\times Tv|:
  T\in\Cal T_{\bar\mu,\bar\mu_L},\,v\in L^0(\frak A),\,\tau'(v)\le 1\}$}

\noindent for $w\in L^0(\frak A_L)$. Note that

\Centerline{$\theta(w)
=\sup\{\int w^*\times v^*:v\in L^0(\frak A),\,\tau'(v)\le 1\}$}

\noindent for every $w\in M^{0,\infty}_{\bar\mu_L}$, by 373Q again.

$\theta$ is an extended Fatou norm on $L^0(\frak A_L)$.   \Prf\ As in
374C, the conditions 369F(i)-(iv) are elementary.   If $w>0$ in
$L^0(\frak A_L)$, take any $v\in L^0(\frak A)$ such that
$0<\tau'(v)\le 1$;  then $w^*\times v^*\ne 0$ so
$\theta(w)\ge\int w^*\times v^*>0$.
So 369F(v) is satisfied.   As for 369F(vi), if $w>0$ in
$L^0(\frak A_L)$, take a non-zero $a\in\frak A$ of finite measure such
that $\alpha=\tau(\chi a)<\infty$.   Let $\beta>0$, $b\in\frak A_L$ be
such that $0<\bar\mu_Lb\le\bar\mu a$ and $\beta\chi b\le w$;  then

\Centerline{$\theta(\chi b)
=\sup_{\tau'(v)\le 1}\int(\chi b)^*\times v^*
\le\sup_{\tau'(v)\le 1}\int(\chi a)^*\times v^*
\le\tau(\chi a)<\infty$}

\noindent by 374B(ii).   So $\theta(\beta\chi b)<\infty$ and 369F(vi) is
satisfied.   Thus $\theta$ is an extended Fatou norm.\   \Qed

\medskip

{\bf (b)} $\theta$ is $\Cal T$-invariant.   \Prf\ If
$T\in\Cal T_{\bar\mu_L,\bar\mu_L}$ and $w\in M^{1,\infty}_{\bar\mu_L}$,
then

\Centerline{$\theta(Tw)=\sup_{\tau'(v)\le 1}\int(Tw)^*\times v^*
\le\sup_{\tau'(v)\le 1}\int w^*\times v^*=\theta(w)$}

\noindent by 373G and 373I.\ \Qed

\medskip

{\bf (c)} $\theta(u^*)=\tau(u)$ for every $u\in M^{0,\infty}_{\bar\mu}$.
\Prf\ We have

\Centerline{$\tau(u)=\sup_{\tau'(v)\le 1}\int|u\times v|
\le\sup_{\tau'(v)\le 1}\int u^*\times v^*\le\tau(u)$,}

\noindent using 369I, 373E and 374B.   So

\Centerline{$\theta(u^*)=\sup_{\tau'(v)\le 1}\int u^*\times v^*
=\tau(u)$}

\noindent by the remark in (a) above.\ \Qed
}%end of proof of 374D

\leader{374E}{}\cmmnt{ I turn now to
rearrangement{\vthsp}-invariance.}
Let $(\frak A,\bar\mu)$ be a measure algebra.

\spheader 374Ea I will say that a subset $A$ of $L^0=L^0(\frak A)$ is
{\bf \ri} if $T_{\pi}u\in A$ whenever $u\in A$ and
$\pi:\frak A\to\frak A$ is a measure-preserving Boolean automorphism,
writing $T_{\pi}:L^0\to L^0$ for the isomorphism corresponding to
$\pi$\cmmnt{ (364P)}.

\spheader 374Eb I will say that an extended Fatou norm $\tau$ on $L^0$
is {\bf \ri} if $\tau(T_{\pi}u)=\tau(u)$ whenever
$u\in L^0$ and $\pi:\frak A\to\frak A$ is a measure-preserving
automorphism.

\leader{374F}{Remarks\cmmnt{ (a)}} If $(\frak A,\bar\mu)$ is a
semi-finite measure algebra
and $\pi:\frak A\to\frak A$ is a sequentially order-continuous
measure-preserving Boolean homomorphism, then
$T_{\pi}\restr M^{1,\infty}_{\bar\mu}$ belongs to
$\Cal T_{\bar\mu,\bar\mu}$\cmmnt{;  this is obvious from the
definition of
$M^{1,\infty}=L^1+L^{\infty}$ and the basic properties of $T_{\pi}$
(364P)}.   Accordingly, any $\Cal T$-invariant extended Fatou norm
$\tau$ on $L^0(\frak A)$ must be \ri\cmmnt{, since
(by 374B) we shall have $\tau(u)=\tau(T_{\pi}(u))=\infty$ when
$u\notin M^{1,\infty}_{\bar\mu}$}.   Similarly, any $\Cal T$-invariant
subset of $M^{1,\infty}_{\bar\mu}$ will be \ri.

\cmmnt{\spheader 374Fb I seek to describe cases in which
rearrangement-{\vthsp}invariance implies
$\Cal T$-invariance.   This happens only for certain measure
algebras;  in order to shorten the statements of the main theorems I
introduce a special phrase.}

\leader{374G}{Definition} I say that a measure algebra
$(\frak A,\bar\mu)$ is {\bf quasi-homogeneous} if for any non-zero $a$,
$b\in\frak A$ there is a measure-preserving Boolean automorphism
$\pi:\frak A\to\frak A$ such that $\pi a\Bcap b\ne 0$.

\leader{374H}{Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra.   Then the following are equiveridical:

(i) $(\frak A,\bar\mu)$ is quasi-homogeneous;

(ii) {\it either} $\frak A$ is purely atomic and every atom of $\frak A$
has the same measure {\it or} there is a $\kappa\ge\omega$ such that the
principal ideal $\frak A_a$ is homogeneous, with Maharam type $\kappa$,
for every $a\in\frak A$ of non-zero finite measure.

\proof{{\bf (i)$\Rightarrow$(ii)} Suppose that $(\frak A,\bar\mu)$ is
quasi-homogeneous.

\medskip

\quad\grheada\ Suppose that $\frak A$ has an atom $a$.   In this case,
for any $b\in\frak A\setminus\{0\}$ there is an automorphism $\pi$ of
$(\frak A,\bar\mu)$ such that $\pi a\Bcap b\ne 0$;  now $\pi a$ must be
an atom, so $\pi a=\pi a\Bcap b$ and $\pi a$ is an atom included in $b$.
As $b$ is arbitrary, $\frak A$ is purely atomic;  moreover, if $b$ is an
atom, then it must be equal to $\pi a$ and therefore of the same measure
as $a$, so all atoms of $\frak A$ have the same measure.

\medskip

\quad\grheadb\ Now suppose that $\frak A$ is atomless.   In this case,
if $a\in\frak A$ has finite non-zero measure, $\frak A_a$ is
homogeneous.   \Prf\Quer\ Otherwise, there are non-zero $b$,
$c\Bsubseteq a$ such that the principal ideals $\frak A_b$, $\frak A_c$
are homogeneous and of different Maharam types, by Maharam's theorem
(332B, 332H).   But now there is supposed to be an automorphism $\pi$
such that $\pi b\Bcap c\ne 0$, in which case $\frak A_b$,
$\frak A_{\pi b}$, $\frak A_{\pi b\Bcap c}$ and $\frak A_c$ must all
have the same Maharam type.\   \Bang\Qed

Consequently, if $a$, $b\in\frak A$ are both of non-zero finite measure,
the Maharam types of $\frak A_a$, $\frak A_{a\Bcup b}$ and $\frak A_b$
must all be the same infinite cardinal $\kappa$.

\medskip

{\bf (ii)$\Rightarrow$(i)} Assume (ii), and take $a$,
$b\in\frak A\setminus\{0\}$.   If $a\Bcap b\ne 0$ we can take $\pi$ to
be the identity automorphism and stop.   So let us suppose that
$a\Bcap b=0$.

\medskip

\quad\grheada\ If $\frak A$ is purely atomic and every atom has the same
measure, then there are atoms $a_0\Bsubseteq a$, $b_0\Bsubseteq b$.
Set

$$\eqalign{\pi c
&=c\text{ if }c\Bsupseteq a_0\Bcup b_0
  \text{ or }c\Bcap(a_0\Bcup b_0)=0,\cr
&=c\Bsymmdiff(a_0\Bcup b_0)\text{ otherwise.}\cr}$$

\noindent Then it is easy to check that $\pi$ is a measure-preserving
automorphism of $\frak A$ such that $\pi a_0=b_0$, so that
$\pi a\Bcap b\ne 0$.

\medskip

\quad\grheadb\ If $\frak A_c$ is \Mth\ with the same
infinite Maharam type $\kappa$ for every non-zero $c$ of finite measure,
set $\gamma=\min(1,\bar\mu a,\bar\mu b)>0$.   Because $\frak A$ is
atomless, there are $a_0\Bsubseteq a$, $b_0\Bsubseteq b$ with
$\bar\mu a_0=\bar\mu b_0=\gamma$ (331C).   Now
$\frak A_{a_0}$ and $\frak A_{b_0}$
are homogeneous with the same Maharam type and the same magnitude, so by
Maharam's theorem (331I) there is a measure-preserving isomorphism
$\pi_0:\frak A_{a_0}\to\frak A_{b_0}$.   Define $\pi:\frak A\to\frak A$
by setting

\Centerline{$\pi c
=(c\Bsetminus(a_0\Bcup b_0))\Bcup\pi_0(c\Bcap a_0)
  \Bcup\pi_0^{-1}(c\Bcap b_0)$}

\noindent for $c\in\frak A$;
then it is easy to see that $\pi$ is a measure-preserving
automorphism of $\frak A$ and that $\pi a\Bcap b\ne 0$.
}%end of proof of 374H

\cmmnt{\medskip

\noindent{\bf Remark} We shall return to these ideas in Chapter 38.   In
particular, the construction of $\pi$ from $\pi_0$ in the last part of
the proof will be of great importance;  in the language of 381R,
$\pi=\cycle{a_0\,_{\pi_0}\,b_0}$.
}%end of comment

\leader{374I}{Corollary} Let $(\frak A,\bar\mu)$ be a quasi-homogeneous
semi-finite measure algebra.   Then

(a) whenever $a$, $b\in\frak A$ have the same finite measure, the
principal ideals $\frak A_a$, $\frak A_b$ are isomorphic as measure
algebras;

(b) there is a subgroup $\Gamma$ of the additive group $\Bbb R$ such
that ($\alpha$) $\bar\mu a\in\Gamma$ whenever $a\in\frak A$ and
$\bar\mu a<\infty$ ($\beta$) whenever $a\in\frak A$, $\gamma\in\Gamma$
and $0\le\gamma\le\bar\mu a$ then there is a $c\Bsubseteq a$ such that
$\bar\mu c=\gamma$.

\proof{ If $\frak A$ is purely atomic, with all its atoms of measure
$\gamma_0$, set $\Gamma=\gamma_0\Bbb Z$, and the results are elementary.
If $\frak A$ is atomless, set $\Gamma=\Bbb R$;  then (a) is a
consequence of Maharam's theorem, and (b) is a consequence of 331C,
already used in the proof of 374H.
}%end of proof of 374I

\leader{374J}{Lemma} Let $(\frak A,\bar\mu)$ be a quasi-homogeneous
semi-finite measure algebra and $u$,
$v\in M^{0,\infty}_{\bar\mu}$.   Let $\Aut_{\bar\mu}$ be the group
of measure-preserving automorphisms of $\frak A$.   Then

\Centerline{$\int u^*\times v^*
=\sup_{\pi\in\Aut_{\bar\mu}}\int|u\times T_{\pi}v|$,}

\noindent where $T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ is the
isomorphism corresponding to $\pi$.

\proof{{\bf (a)} Suppose first that $u$, $v$ are non-negative and belong
to $S(\frak A^f)$, where $\frak A^f$ is the ring $\{a:\bar\mu
a<\infty\}$, as usual.   Then they can be expressed as
$u=\sum_{i=0}^m\alpha_i\chi a_i$, $v=\sum_{j=0}^n\beta_j\chi b_j$ where
$\alpha_0\ge\ldots\alpha_m\ge 0$, $\beta_0\ge\ldots\ge\beta_n\ge 0$,
$a_0,\ldots,a_m$ are disjoint and of finite measure, and
$b_0,\ldots,b_n$ are disjoint and of finite measure.   Extending each
list by a final term having a coefficient of $0$, if need be, we may
suppose that $\sup_{i\le m}a_i=\sup_{j\le n}b_j$.

Let $(t_0,\ldots,t_s)$ enumerate in ascending order the set

\Centerline{$\{0\}\cup\{\sum_{i=0}^k\bar\mu a_i:k\le m\}
\cup\{\sum_{j=0}^k\bar\mu b_j:k\le n\}$.}

\noindent Then every $t_r$ belongs to the subgroup $\Gamma$ of 374Ib,
and $t_s=\sum_{i=0}^m\bar\mu a_i=\sum_{j=0}^n\bar\mu b_j$.
For $1\le r\le s$ let $k(r)$, $l(r)$ be minimal subject to the
requirements $t_r\le\sum_{i=0}^{k(r)}\bar\mu a_i$,
$t_r\le\sum_{j=0}^{l(r)}\bar\mu b_j$.   Then
$\bar\mu a_i=\sum_{k(r)=i}t_r-t_{r-1}$, so (using 374Ib) we can find a
disjoint family $\langle c_r\rangle_{1\le r\le s}$ such that
$c_r\Bsubseteq a_{k(r)}$ and $\bar\mu c_r=t_r-t_{r-1}$ for each $r$.
Similarly, there is
a disjoint family $\langle d_r\rangle_{1\le r\le s}$ such that
$d_r\Bsubseteq b_{l(r)}$ and $\bar\mu d_r=t_r-t_{r-1}$ for each $r$.
Now the principal ideals $\frak A_{c_r}$, $\frak A_{d_r}$ are isomorphic
for every $r$, by 374Ia;  let $\pi_r:\frak A_{d_r}\to\frak A_{c_r}$ be
measure-preserving isomorphisms.   Define $\pi:\frak A\to\frak A$ by
setting

\Centerline{$\pi a=(a\Bsetminus\sup_{1\le r\le s}d_r)\Bcup\sup_{1\le
r\le s}\pi_r(a\Bcap d_r)$;}

\noindent because

\Centerline{$\sup_{r\le s}c_r=\sup_{i\le m}a_i=\sup_{j\le n}b_j
=\sup_{r\le s}d_r$,}

\noindent $\pi:\frak A\to\frak A$ is a measure-preserving automorphism.

Now

\Centerline{$u=\sum_{r=1}^s\alpha_{k(r)}\chi c_r$,
\quad$v=\sum_{r=1}^s\beta_{l(r)}\chi d_r$,}

\Centerline{$u^*=\sum_{r=1}^s\alpha_{k(r)}
   \chi\coint{t_{r-1},t_r}^{\ssbullet}$,
\quad$v^*=\sum_{r=1}^s\beta_{l(r)}
   \chi\coint{t_{r-1},t_r}^{\ssbullet}$,}

\noindent so

\Centerline{$\int u\times T_{\pi}v
=\sum_{r=1}^s\alpha_{k(r)}\beta_{l(r)}\bar\mu c_r
=\sum_{r=1}^s\alpha_{k(r)}\beta_{l(r)}(t_r-t_{r-1})
=\int u^*\times v^*$.}

\medskip

{\bf (b)} Now take any $u_0$, $v_0\in M^{0,\infty}_{\bar\mu}$.   Set

\Centerline{$A=\{u:u\in S(\frak A^f),\,0\le u\le|u_0|\}$,
\quad$B=\{v:v\in S(\frak A^f),\,0\le v\le|v_0|\}$.}

\noindent Then $A$ is an upwards-directed set with supremum $|u_0|$,
because $(\frak A,\bar\mu)$ is semi-finite, so $\{u^*:u\in A\}$ is an
upwards-directed set with supremum $|u_0|^*=u_0^*$ (373Db, 373Dh).
Similarly $\{v^*:v\in B\}$ is upwards-directed and has supremum $v_0^*$,
so $\{u^*\times v^*:u\in A,\,v\in B\}$ is upwards-directed and has
supremum $u_0^*\times v_0^*$.

Consequently, if $\gamma<\int u_0^*\times v_0^*$, there are $u\in A$,
$v\in B$ such that $\gamma\le\int u^*\times v^*$.   Now, by (a), there
is a $\pi\in\Aut_{\bar\mu}$ such that

\Centerline{$\gamma\le\int u\times T_{\pi}v
\le\int|u_0|\times T_{\pi}|v_0|=\int|u_0\times T_{\pi}v_0|$}

\noindent because $T_{\pi}$ is a Riesz homomorphism.   As $\gamma$ is
arbitrary,

\Centerline{$\int u_0^*\times v_0^*
\le\sup_{\pi\in\Aut_{\bar\mu}}\int|u_0\times T_{\pi}v_0|$.}

\noindent But the reverse inequality is immediate from 373J.
}%end of proof of 374J

\leader{374K}{Theorem} Let $(\frak A,\bar\mu)$ be a quasi-homogeneous
semi-finite measure algebra, and $\tau$ a \ri\
extended Fatou norm on $L^0=L^0(\frak A)$.   Then $\tau$ is
$\Cal T$-invariant.

\proof{ Write $\tau'$ for the associate of $\tau$.   Then 374J tells us
that for any $u$, $v\in M^{0,\infty}_{\bar\mu}$,

\Centerline{$\int u^*\times v^*
=\sup_{\pi\in\Aut_{\bar\mu}}\int|T_{\pi}u\times v|
\le\sup_{\pi\in\Aut_{\bar\mu}}\tau(T_{\pi}u)\tau'(v)
=\tau(u)\tau'(v)$,}

\noindent writing $u^*$, $v^*$ for the decreasing rearrangements of $u$
and $v$, and $\Aut_{\bar\mu}$ for the group of measure-preserving automorphisms of
$(\frak A,\bar\mu)$.   But now, if $u\in M^{1,\infty}_{\bar\mu}$
and $T\in\Cal T_{\bar\mu,\bar\mu}$,

$$\eqalignno{\tau(Tu)
&=\sup\{\int|Tu\times v|:\tau'(v)\le 1\}\cr
\noalign{\noindent (369I)}
&\le\sup\{\int u^*\times v^*:\tau'(v)\le 1\}\cr
\noalign{\noindent (373J)}
&\le\tau(u).\cr}$$

\noindent As $T$, $u$ are arbitrary, $\tau$ is $\Cal T$-invariant.
}%end of proof of 374K

\leader{374L}{Lemma} Let $(\frak A,\bar\mu)$ be a quasi-homogeneous
semi-finite measure algebra.   Suppose that $u$,
$v\in (M^{0,\infty}_{\bar\mu})^+$
are such that $\int u^*\times v^*=\infty$.   Then there is
a measure-preserving automorphism $\pi:\frak A\to\frak A$ such that
$\int u\times T_{\pi}v=\infty$.

\proof{ I take three cases separately.

\medskip

{\bf (a)} Suppose that $\frak A$ is purely atomic;  then $u$,
$v\in L^{\infty}(\frak A)$ and $u^*$, $v^*\in L^{\infty}(\frak A_L)$, so
neither $u^*$ nor $v^*$ can belong to $L^1_{\bar\mu_L}$ and neither $u$ nor
$v$ can belong to $L^1_{\bar\mu}$.
Let $\gamma$ be the common measure of the atoms of $\frak A$.
For each $n\in\Bbb N$, set

\Centerline{$\alpha_n=\inf\{\alpha:\alpha\ge 0,\,
\bar\mu\Bvalue{u>\alpha}\le 3^n\gamma\}$,
\quad$\tilde a_n=\Bvalue{u>\bover12\alpha_n}$.}

\noindent Then
$\bar\mu\Bvalue{u>\alpha_n}\le 3^n\gamma$;  also
$\alpha_n>0$, since otherwise $u$ would belong to $L^1_{\bar\mu}$, so
$\bar\mu\tilde a_n\ge 3^n\gamma$.  We can therefore choose
$\sequencen{a'_n}$ inductively such that $a'_n\Bsubseteq\tilde a_n$ and
$\bar\mu a'_n=3^n\gamma$ for each $n$  (using 374Ib).
For each $n\ge 1$, set $a''_n=a'_n\Bsetminus\sup_{i<n}a'_i$;  then
$\bar\mu a''_n\ge\Bover12\cdot 3^{-n}\gamma$, so we can choose an
$a_n\Bsubseteq a''_n$ such that $\bar\mu a_n=3^{n-1}\gamma$.

Also, of course,
$\sequencen{\alpha_n}$ is non-increasing.   We now see that

\Centerline{$\langle a_n\rangle_{n\ge 1}$ is disjoint,
\quad$u\ge\Bover12\alpha_n\chi a_n$ for every $n\ge 1$,}

\Centerline{$u^*\le\|u\|_{\infty}\chi\coint{0,\gamma}^{\ssbullet}
   \vee\sup_{n\in\Bbb N}
   \alpha_n\chi\coint{3^n\gamma,3^{n+1}\gamma}^{\ssbullet}$.}

Similarly, there are a non-increasing sequence $\sequencen{\beta_n}$
in $\coint{0,\infty}$ and a disjoint sequence $\langle b_n\rangle_{n\ge 1}$
in $\frak A$ such that

\Centerline{$\bar\mu b_n=3^{n-1}\gamma$,
\quad$v\ge\Bover12\beta_n\chi b_n$ for every $n\ge 1$,}

\Centerline{$v^*\le\|v\|_{\infty}\chi\coint{0,\gamma}^{\ssbullet}
   \vee\sup_{n\in\Bbb N}
   \beta_n\chi\coint{3^n\gamma,3^{n+1}\gamma}^{\ssbullet}$.}

\noindent We are supposing that

$$\eqalign{\infty
&=\int u^*\times v^*
=\gamma\|u\|_{\infty}\|v\|_{\infty}
  +\sum_{n=0}^{\infty}2\cdot 3^n\gamma\alpha_n\beta_n\cr
&=\gamma\|u\|_{\infty}\|v\|_{\infty}+2\gamma\alpha_0\beta_0
  +2\gamma\sum_{n=0}^{\infty}3^{2n+1}(\alpha_{2n+1}\beta_{2n+1}
       +3\alpha_{2n+2}\beta_{2n+2})\cr
&\le\gamma\|u\|_{\infty}\|v\|_{\infty}+2\gamma\alpha_0\beta_0
  +24\sum_{n=0}^{\infty}3^{2n}\gamma\alpha_{2n+1}\beta_{2n+1},\cr}$$

\noindent so $\sum_{n=0}^{\infty}3^{2n}\alpha_{2n+1}\beta_{2n+1}=\infty$.

At this point, recall that we are dealing with a purely atomic algebra
in which every atom has measure $\gamma$.
Let $A_n$, $B_n$ be the sets of
atoms included in $a_n$, $b_n$ for each $n\ge 1$, and
$A=\bigcup_{n\ge 1}A_n\cup B_n$.   Then $\#(A_n)=\#(B_n)=3^{n-1}$ for
each $n\ge 1$.   We
therefore have a permutation $\phi:A\to A$ such that
$\phi[B_{2n+1}]=A_{2n+1}$
for every $n$.   (The point is that
$A\setminus\bigcup_{n\in\Bbb N}A_{2n+1}$ and
$A\setminus\bigcup_{n\in\Bbb N}B_{2n+1}$ are both countably
infinite.)   Define $\pi:\frak A\to\frak A$ by setting

\Centerline{$\pi c
=(c\Bsetminus\sup A)\Bcup\sup_{a\in A,a\Bsubseteq c}\phi a$}

\noindent for $c\in\frak A$.
Then $\pi$ is well-defined (because $A$ is countable), and it
is easy to check that it is a measure-preserving Boolean automorphism
(because it is just a permutation of the atoms);  and
$\pi b_{2n+1}=a_{2n+1}$ for every $n$.   Consequently

\Centerline{$\int u\times T_{\pi}v
\ge\sum_{n=0}^{\infty}\Bover14\alpha_{2n+1}\beta_{2n+1}
   \bar\mu a_{2n+1}
=\Bover14\gamma\sum_{n=0}^{\infty}3^{2n}\alpha_{2n+1}\beta_{2n+1}
=\infty$.}

\noindent So we have found a suitable automorphism.

\medskip

{\bf (b)} Next, consider the case in which $(\frak A,\bar\mu)$ is
atomless and of finite magnitude $\gamma$.   Of course $\gamma>0$.   For
each $n\in\Bbb N$ set

\Centerline{$\alpha_n=\inf\{\alpha:\alpha\ge 0,\,
  \bar\mu\Bvalue{u>\alpha}\le 3^{-n}\gamma\}$,
\quad$\tilde a_n=\Bvalue{u>\Bover12\alpha_n}$.}

\noindent Then $\sequencen{\alpha_n}$ is non-decreasing and

\Centerline{$u^*\le\sup_{n\in\Bbb N}\alpha_{n+1}
\chi\coint{3^{-n-1}\gamma,3^{-n}\gamma}^{\ssbullet}$.}

\noindent This time, $\bar\mu\tilde a_n\ge 3^{-n}\gamma$,
and we are in an atomless measure algebra, so we can choose
$a'_n\Bsubseteq\tilde a_n$ such that $\bar\mu a'_n=3^{-n}\gamma$;
taking $a''_n=a'_n\Bsetminus\sup_{i>n}a'_i$,
$\bar\mu a''_n\ge\Bover12\cdot 3^{-n}\gamma$, and we can choose
$a_n\Bsubseteq a''_n$ such that $\bar\mu a_n=3^{-n-1}\gamma$ for every $n$.
As before, $u\ge\Bover12\alpha_n\chi a_n$ for every $n$, and
$\sequencen{a_n}$ is disjoint.

In the same way, we can find $\sequencen{\beta_n}$, $\sequencen{b_n}$
such that $\sequencen{b_n}$ is disjoint,

\Centerline{$v^*\le\sup_{n\in\Bbb N}\beta_{n+1}
   \chi\coint{3^{-n-1}\gamma,3^{-n}\gamma}^{\ssbullet}$,
\quad$v\ge\sup_{n\in\Bbb N}\Bover12\beta_n\chi b_n$}

\noindent and $\bar\mu b_n=3^{-n-1}\gamma$ for each $n$.   In this case, we
have

\Centerline{$\infty
=\int u^*\times v^*
\le\sum_{n=0}^{\infty}2\cdot 3^{-n-1}\gamma\alpha_{n+1}\beta_{n+1}$,}

\noindent and $\sum_{n=0}^{\infty}3^{-n}\alpha_n\beta_n$ is infinite.

Now all the principal ideals $\frak A_{a_n}$, $\frak A_{b_n}$ are
homogeneous and of the same Maharam type, so there are
measure-preserving isomorphisms $\pi_n:\frak A_{b_n}\to\frak A_{a_n}$;
similarly, setting $\tilde a=1\Bsetminus\sup_{n\in\Bbb N}a_n$ and
$\tilde b=1\Bsetminus\sup_{n\in\Bbb N}b_n$, there is a measure-preserving
isomorphism $\tilde\pi:\frak A_{\tilde b}\to\frak A_{\tilde a}$.
Define $\pi:\frak A\to\frak A$ by setting

\Centerline{$\pi c
=\tilde\pi(c\Bcap\tilde b)\Bcup\sup_{n\in\Bbb N}\pi_n(c\Bcap a_n)$}

\noindent for every $c\in\frak A$;
then $\pi$ is a measure-preserving automorphism
of $\frak A$, and $\pi b_n=a_n$ for each $n$.   In this case,

\Centerline{$\int u\times T_{\pi}v
\ge\Bover14\sum_{n=0}^{\infty}3^{-n-1}\gamma\alpha_n\beta_n=\infty$,}

\noindent and again we have a suitable automorphism.

\medskip

{\bf (c)} Thirdly, consider the case in which $\frak A$ is atomless and
not totally finite;  take $\kappa$ to be the common Maharam type of
all the principal ideals $\frak A_a$ where $0<\bar\mu a<\infty$.   In
this case, set

\Centerline{$\alpha_n
=\inf\{\alpha:\bar\mu\Bvalue{u>\alpha}\le 3^n\}$,
\quad$\beta_n
=\inf\{\alpha:\bar\mu\Bvalue{v>\alpha}\le 3^n\}$
}

\noindent for each $n\in\Bbb Z$.   This time

\Centerline{$u^*
\le\sup_{n\in\Bbb Z}\alpha_n
\chi\coint{3^n,3^{n+1}}^{\ssbullet}$,
\quad$v^*
\le\sup_{n\in\Bbb Z}\beta_n
\chi\coint{3^n,3^{n+1}}^{\ssbullet}$,}

\noindent so

\Centerline{$\infty=\int u^*\times v^*
=2\sum_{n=-\infty}^{\infty}3^n\alpha_n\beta_n
\le 8\sum_{n=-\infty}^{\infty}3^{2n}\alpha_{2n}\beta_{2n}$.}

For each $n\in\Bbb Z$, $3^n\le\bar\mu\Bvalue{u>\bover12\alpha_n}$,
so there is an $a''_n$ such that

\Centerline{$a''_n\Bsubseteq\Bvalue{u>\bover12\alpha_n}$,
\quad$\bar\mu a''_n=3^n$.}

\noindent Set
$a'_n=a''_n\Bsetminus\sup_{-\infty<i<n}a''_i$;  then
$\bar\mu a'_n\ge\Bover12\cdot 3^n$ for each $n$; choose
$a_n\Bsubseteq a'_n$ such that $\bar\mu a_n=3^{n-1}$.   Then
$\sequencen{a_n}$ is disjoint and $u\ge\Bover12\alpha_n\chi a_n$ for each
$n$.

Similarly, there is a disjoint sequence $\sequencen{b_n}$ such that

\Centerline{$\bar\mu b_n=3^{n-1}$,
\quad$v\ge\Bover12\beta_n\chi b_n$}

\noindent for each $n\in\Bbb N$.

Set $d^*=\sup_{n\in\Bbb Z}a_n\Bcup\sup_{n\in\Bbb Z}b_n$.   Then

\Centerline{$\tilde a=d^*\Bsetminus\sup_{n\in\Bbb Z}a_{2n}$,
\quad$\tilde b=d^*\Bsetminus\sup_{n\in\Bbb Z}b_{2n}$}

\noindent both have magnitude $\omega$ and Maharam type $\kappa$.   So
there is a measure-preserving isomorphism
$\tilde\pi:\frak A_{\tilde b}\to\frak A_{\tilde a}$ (332J).
At the same time,
for each $n\in\Bbb Z$ there is a measure-preserving isomorphism
$\pi_n:\frak A_{b_{2n}}\to\frak A_{a_{2n}}$.   So once again we can
assemble these to
form a measure-preserving automorphism $\pi:\frak A\to\frak A$, defined
by the formula

\Centerline{$\pi c
=(c\Bsetminus d^*)\Bcup\tilde\pi(c\Bcap\tilde b)
   \Bcup\sup_{n\in\Bbb Z}\pi_n(c\Bcap b_{2n})$.}

Just as in (a) and (b) above,

\Centerline{$\int u\times T_{\pi}v
\ge\sum_{n=-\infty}^{\infty}\Bover1{4}\cdot 3^{2n-1}\alpha_{2n}\beta_{2n}
=\infty$.}

\noindent
Thus we have a suitable $\pi$ in any of the cases allowed by 374H.
}%end of proof of 374L

\leader{374M}{Proposition} Let $(\frak A,\bar\mu)$ be a
quasi-homogeneous localizable measure algebra, and
$U\subseteq L^0=L^0(\frak A)$ a solid linear subspace which, regarded as
a Riesz space, is perfect.   If $U$ is \ri\ and
$M^{\infty,1}_{\bar\mu}\subseteq U\subseteq M^{1,\infty}_{\bar\mu}$,
then $U$ is $\Cal T$-invariant.

\proof{ Set $V=\{v:u\times v\in L^1$ for every $u\in U\}$, so that $V$
is a solid linear subspace of $L^0$ which can be identified with
$U^{\times}$ (369C),
and $U$ becomes $\{u:u\times v\in L^1$ for every $v\in V\}$;  note that
$M^{\infty,1}_{\bar\mu}\subseteq V\subseteq M^{1,\infty}_{\bar\mu}$
(using 369Q).

If $u\in U^+$, $v\in V^+$ and $\pi:\frak A\to\frak A$ is a
measure-preserving automorphism, then $T_{\pi}u\in U$, so
$\int v\times T_{\pi}u<\infty$;  by 374L, $\int u^*\times v^*$ is
finite.   But this
means that if $u\in U$, $v\in V$ and $T\in\Cal T_{\bar\mu,\bar\mu}$,

\Centerline{$\int|Tu\times v|\le\int u^*\times v^*<\infty$.}

\noindent As $v$ is arbitrary, $Tu\in U$;  as $T$ and $u$ are arbitrary,
$U$ is $\Cal T$-invariant.
}%end of proof of 374M

\exercises{\leader{374X}{Basic exercises $\pmb{>}$(a)}
%\spheader 374Xa
Let $(\frak A,\bar\mu)$ be a measure algebra and
$A\subseteq M^{1,\infty}_{\bar\mu}$ a $\Cal T$-invariant set.   (i) Show
that $A$ is solid.   (ii) Show that if $A$ is a linear subspace and not
$\{0\}$, then it includes $M^{\infty,1}_{\bar\mu}$.   (iii) Show
that if $u\in A$, $v\in M^{0,\infty}_{\bar\mu}$ and
$\int_0^tv^*\le\int_0^tu^*$ for every $t>0$, then $v\in A$.   (iv) Show
that if $(\frak B,\bar\nu)$ is any other
measure algebra, then $B=\{Tu:u\in A,\,T\in\Cal T_{\bar\mu,\bar\nu}\}$
and $C=\{v:v\in M^{1,\infty}_{\bar\nu},\,Tv\in A$ for every
$T\in\Cal T_{\bar\nu,\bar\mu}\}$ are $\Cal T$-invariant subsets of
$M^{1,\infty}_{\bar\nu}$, and that $B\subseteq C$.   Give two
examples in which $B\subset C$.    Show that if
$(\frak A,\bar\mu)=(\frak A_L,\bar\mu_L)$ then $B=C$.
%374A

\sqheader 374Xb Let $(\frak A,\bar\mu)$ be a measure algebra.   Show
that the extended Fatou norm $\|\,\|_p$ on $L^0(\frak A)$ is
$\Cal T$-invariant for every $p\in[1,\infty]$.   \Hint{371Gd.}
%374A

\spheader 374Xc  Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure algebras, and $\phi$ a Young's function (369Xc).
Let $\tau_{\phi}$, $\tilde\tau_{\phi}$ be the
corresponding Orlicz norms on $L^0(\frak A)$, $L^0(\frak B)$.   Show
that $\tilde\tau_{\phi}(Tu)\le\tau_{\phi}(u)$ for every
$u\in L^0(\frak A)$, $T\in\Cal T_{\bar\mu,\bar\nu}$.   \Hint{369Xn,
373Xm.}   In particular, $\tau_{\phi}$ is $\Cal T$-invariant.
%374A

\spheader 374Xd Show that if $(\frak A,\bar\mu)$ is a semi-finite
measure algebra and $\tau$ is a $\Cal T$-invariant extended Fatou norm
on $L^0(\frak A)$, then the Banach lattice $L^{\tau}$ defined from
$\tau$ is $\Cal T$-invariant.
%374B

\spheader 374Xe Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
and $\tau$ a $\Cal T$-invariant extended Fatou norm on $L^0(\frak A)$
which is an order-continuous norm on $L^{\tau}$.   Show that
$L^{\tau}\subseteq M^{1,0}_{\bar\mu}$.
%374B

\spheader 374Xf Let $\theta$ be a $\Cal T$-invariant extended Fatou
norm on $L^0(\frak A_L)$ and $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$
two semi-finite measure algebras.   Let $\tau_1$, $\tau_2$ be the
extended Fatou norms on $L^0(\frak A)$, $L^0(\frak B)$ defined from
$\theta$ by the method of 374C.   Show that  $\tau_2(Tu)\le\tau_1(u)$
whenever $u\in M^{1,\infty}_{\bar\mu}$ and
$T\in\Cal T_{\bar\mu,\bar\nu}$.
%374C

\sqheader 374Xg Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, not $\{0\}$, and set
$\tau(u)=\sup_{0<\bar\mu a<\infty}\Bover1{\sqrt{\bar\mu a}}\int_a|u|$ for
$u\in L^0(\frak A)$.
Show that $\tau$ is a $\Cal T$-invariant extended Fatou norm.   Find
examples of
$(\frak A,\bar\mu)$ for which $\tau$ is, and is not, order-continuous on
$L^{\tau}$.
%374C

\spheader 374Xh Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure algebras and $\tau$ a $\Cal T$-invariant extended
Fatou norm on $L^0(\frak A)$.   (i) Show that there is a
$\Cal T$-invariant extended Fatou norm $\theta$ on $L^0(\frak B)$
defined by setting
$\theta(v)=\sup\{\tau(Tv):T\in\Cal T_{\bar\nu,\bar\mu}\}$ for
$v\in M^{1,\infty}_{\bar\nu}$.    (ii) Show that when
$(\frak A,\bar\mu)=(\frak A_L,\bar\mu_L)$ then $\theta(v)=\tau(v^*)$ for
every $v\in M^{0,\infty}_{\bar\nu}$.   (iii) Show that when
$(\frak B,\bar\nu)=(\frak A_L,\bar\mu_L)$ then $\tau(u)=\theta(u^*)$ for
every $u\in M^{0,\infty}_{\bar\mu}$.
%374D

\spheader 374Xi Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
and $\tau$ an extended Fatou norm on $L^0=L^0(\frak A)$.   Suppose that
$L^{\tau}$ is a $\Cal T$-invariant subset of $L^0$.
Show that there is a $\Cal T$-invariant extended Fatou norm $\tilde\tau$
which is equivalent to $\tau$ in the sense that, for some $M>0$,
$\tilde\tau(u)\le M\tau(u)\le M^2\tilde\tau(u)$ for every $u\in L^0$.
\Hint{show first that $\int u^*\times v^*<\infty$ for every
$u\in L^{\tau}$ and $v\in L^{\tau'}$, then that
$\sup_{\tau(u)\le 1,\tau'(v)\le 1}\int u^*\times v^*<\infty$.}
%374D

\spheader 374Xj Suppose that $\tau$ is a $\Cal T$-invariant extended
Fatou norm on $L^0(\frak A_L)$, and that $0<w=w^*\in
M^{1,\infty}_{\bar\mu_L}$.   Let $(\frak A,\bar\mu)$ be any semi-finite
measure algebra.   Show that the function $u\mapsto\tau(w\times u^*)$
extends to a $\Cal T$-invariant extended Fatou norm $\theta$ on
$L^0(\frak A)$.   \Hint{$\tau(w\times u^*)
=\sup\{\tau(w\times Tu):T\in\Cal T_{\bar\mu,\bar\mu_L}\}$ for
$u\in M^{1,\infty}_{\bar\mu_L}$.}   (When
$\tau=\|\,\|_p$ these norms are called {\bf Lorentz norms};  see 
{\smc Lindenstrauss \& Tzafriri 79}, p.\ 121.)
%374D

\spheader 374Xk  Let $(\frak A,\bar\mu)$ be $\Cal P\Bbb N$ with counting
measure.   Identify $L^0(\frak A)$ with $\BbbR^{\Bbb N}$.
Let $U$ be $\{u:u\in\BbbR^{\Bbb N},\,\{n:u(n)\ne 0\}$ is finite$\}$.
Show that $U$ is a perfect Riesz space, and is \ri\ but not
$\Cal T$-invariant.
%374M

\spheader 374Xl Let $(\frak A,\bar\mu)$ be an atomless quasi-homogeneous
localizable measure algebra, and $U\subseteq L^0(\frak A)$ a
\ri\ solid linear subspace which is a perfect Riesz
space.   Show that $U\subseteq M^{1,\infty}_{\bar\mu}$ and that $U$ is
$\Cal T$-invariant.   \Hint{assume $U\ne\{0\}$.   Show that (i)
$\chi a\in U$ whenever $\bar\mu a<\infty$
(ii) $V=\{v:v\times u\in L^1\Forall u\in U\}$ is \ri\ (iii) $U$,
$V\subseteq M^{1,\infty}$.}
%374M

\leader{374Y}{Further exercises (a)}
%\spheader 374Ya
Let $(\frak A,\bar\mu)$ be a localizable measure algebra and
$U\subseteq M^{1,\infty}_{\bar\mu}$ a non-zero
$\Cal T$-invariant Riesz subspace which, regarded as a Riesz space, is
perfect.   (i) Show that $U$ includes $M^{\infty,1}_{\bar\mu}$.   (ii)
Show that its dual
$\{v:v\in L^0(\frak A),\,v\times u\in L^1_{\bar\mu}\Forall u\in U\}$
(which in this exercise I will denote by $U^{\times}$) is also
$\Cal T$-invariant, and is $\{v:v\in M^{0,\infty}_{\bar\mu},\,\int
u^*\times v^*<\infty\Forall u\in U\}$.   (iii) Show that for any
localizable measure algebra $(\frak B,\bar\nu)$ the set
$V=\{v:v\in M^{1,\infty}_{\bar\nu},\,Tv\in U
\Forall T\in\Cal T_{\bar\nu,\bar\mu}\}$ is a perfect Riesz subspace of
$L^0(\frak B)$,
and that $V^{\times}=\{v:v\in M^{1,\infty}_{\bar\nu},\,Tv\in
U^{\times}\Forall T\in\Cal T_{\bar\nu,\bar\mu}\}$.   (iv) Show that
if, in (i)-(iii), $(\frak A,\bar\mu)=(\frak A_L,\bar\mu_L)$, then
$V=\{v:v\in M^{0,\infty},\,v^*\in U\}$.   (v) Show that if, in (iii),
$(\frak B,\bar\nu)=(\frak A_L,\bar\mu_L)$, then
$U=\{u:u\in M^{0,\infty}_{\bar\mu},\,u^*\in V\}$.
%374Xh, 374C, 374D

\spheader 374Yb Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and
suppose that $1\le q\le p<\infty$.   Let $w_{pq}\in L^0(\frak A_L)$ be
the equivalence class of the function $t\mapsto t^{(q-p)/p}$.   (i) Show
that for any $u\in L^0(\frak A)$,

\Centerline{$\int w_{pq}\times(u^*)^q
=p\int_0^{\infty}t^{q-1}(\bar\mu\Bvalue{|u|>t})^{q/p}dt$.}

\noindent (ii) Show that we have an extended Fatou norm $\|\,\|_{p,q}$
on $L^0(\frak A)$ defined by setting

\Centerline{$\|u\|_{p,q}
=\bigl(p\int_0^{\infty}t^{q-1}
(\bar\mu\Bvalue{|u|>t})^{q/p}dt\bigr)^{1/q}$}

\noindent for every $u\in L^0(\frak A)$.   \Hint{use 374Xj with
$w=w_{pq}^{1/q}$, $\|\,\|=\|\,\|_q$.}   (iii) Show that if
$(\frak B,\bar\nu)$ is another semi-finite measure algebra and
$T\in\Cal T_{\bar\mu,\bar\nu}$, then $\|Tu\|_{p,q}\le\|u\|_{p,q}$ for
every $u\in M^{1,\infty}_{\bar\mu}$.   (iv) Show that $\|\,\|_{p,q}$ is
an order-continuous norm on $L^{\|\,\|_{p,q}}$.
%374Xj, 374D

\spheader 374Yc Let $(\frak A,\bar\mu)$ be a homogeneous measure
algebra of uncountable Maharam type, and $u$, $v\ge 0$ in
$M^0_{\bar\mu}$ such that $u^*=v^*$.   Show that there
is a measure-preserving automorphism $\pi$ of $\frak A$ such that
$T_{\pi}u=v$, where $T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ is the
isomorphism corresponding to $\pi$.
%374E

\spheader 374Yd In $L^0(\frak A_L)$ let $u$ be the equivalence class of
the function $f(t)=te^{-t}$.   Show that there is no Boolean
automorphism $\pi$ of $\frak A_L$ such that $T_{\pi}u=u^*$.   \Hint{show
that $\frak A_L$ is $\tau$-generated by
$\{\Bvalue{u^*>\alpha}:\alpha>0\}$.}
%374E

\spheader 374Ye Let $(\frak A,\bar\mu)$ be a quasi-homogeneous
semi-finite measure algebra and $C\subseteq L^0(\frak A)$ a solid convex
order-closed \ri\ set.   Show that $C\cap M^{1,\infty}_{\bar\mu}$ is
$\Cal T$-invariant.
%374K

}%end of exercises

\leaveitout{\leader{374Z}{Problem} In 374D, if we are told that $\tau$
is an order-continuous norm on $L^{\tau}$, can we arrange that $\theta$
should be an order-continuous norm on $L^{\theta}$?
}%end of leaveitout

\endnotes{
\Notesheader{374} I gave this section the title
`\ri\ spaces' because it looks good on the
Contents page, and it follows what has been common practice since 
{\smc Luxemburg 67b};  but actually I think that it's $\Cal T$-invariance
which matters, and that \ri\ spaces are significant largely
because the important ones are $\Cal T$-invariant.   The particular
quality of $\Cal T$-invariance which I have tried to bring out here is
its transferability from one measure algebra (or measure space, of
course) to another.   This is what I take at a relatively leisurely pace
in 374B-374D and 374Xf, and then encapsulate in 374Xh and 374Ya.   The
special place of the Lebesgue algebra $(\frak A_L,\bar\mu_L)$ arises
from its being more or less the simplest algebra over which every
$\Cal T$-invariant set can be described;  see 374Xa.

I don't think this work is particularly easy, and (as in \S373) there
are rather a lot of unattractive names in it;  but once one has achieved
a reasonable familiarity with the concepts, the techniques used can be
seen to amount to half a dozen ideas -- non-trivial ideas,
to be sure -- from \S\S369 and 373.
From \S369 I take concepts of duality:  the
symmetric relationship between a perfect Riesz space $U\subseteq L^0$
and the representation of its dual (369C-369D), and the notion of
associate extended Fatou norms (369H-369K). %369H 369I 369J 369K
From \S373 I take the idea
of `decreasing rearrangement' and theorems guaranteeing the existence
of useful members of $\Cal T_{\bar\mu,\bar\nu}$
(373O-373Q). %373O 373P 373Q
The results of the present section all depend on repeated use of these
facts, assembled in a variety of patterns.

There is one new method here, but an easy one:  the construction of
measure-preserving automorphisms by joining isomorphisms together, as in
the proofs of 374H and 374J.   I shall return to this idea, in greater
generality and more systematically investigated, in \S381.   I hope that
the special cases here will give no difficulty.

While $\Cal T$-invariance is a similar phenomenon for both extended
Fatou norms and perfect Riesz spaces (see 374Xh, 374Ya), the former seem
easier to deal with.  The essential difference is I think in 374B(i);
with a $\Cal T$-invariant extended Fatou norm, we are necessarily
confined to $M^{1,\infty}$, the natural domain of the methods used here.
For perfect Riesz spaces we have examples like $\BbbR^{\Bbb N}\cong
L^0(\Cal P\Bbb N)$ and its dual, the space of eventually-zero sequences
(374Xk);  these are \ri\ but not $\Cal T$-invariant,
as I have defined it.   This problem does not
arise over atomless algebras (374Xl).

I think it is obvious that for algebras which are not
quasi-homogeneous (374G)
rearrange\discretionary{-}{}{}ment-{\vthsp}invariance is going to be
of limited interest;  there will be regions between which there is no
communication by means of
measure-preserving automorphisms, and the best we can hope for is a
discussion of quasi-homogeneous components, if they exist, corresponding
to the partition of unity used in the proof of 332J.
There is a special difficulty concerning
rearrangement-{\vthsp}invariance
in $L^0(\frak A_L)$:  two elements can have the same decreasing
rearrangement without being rearrangements of each other in the strict
sense (373Ya, 374Yd).   The phenomenon of 373Ya is specific to algebras
of countable Maharam type (374Yc).   You will see that some of the
labour of 374L is because we have to make room for the pieces to move
in.   374J is easier just because in that context we can settle for a
supremum, rather than
an actual infinity, so the rearrangement needed (part (a) of the proof)
can be based on a region of finite measure.
}%end of comment

\discrpage


\frfilename{mt375.tex}
\versiondate{30.1.10}
\copyrightdate{1996}

\def\ssbhm{$\sigma$-\vthsp{}subhomomorphism}

\def\chaptername{Linear operators between function spaces}
\def\sectionname{Kwapien's theorem}

\newsection{375}

In \S368 and the first part of \S369 I examined maps from various types
of Riesz space into $L^0$
spaces.   There are equally striking results about maps out of $L^0$
spaces.   I start with some relatively elementary facts about positive
linear operators from $L^0$ spaces to Archimedean Riesz spaces in
general (375A-375D), and then turn to a remarkable analysis, due
essentially to S.Kwapien, of the positive linear operators from a
general $L^0$ space
to the $L^0$ space of a semi-finite measure algebra (375J), with a
couple of simple corollaries.

\leader{375A}{Theorem} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $W$ an Archimedean Riesz space.   If
$T:L^0(\frak A)\to W$ is a positive linear operator, it is sequentially
order-continuous.

\proof{{\bf (a)} The first step is to observe that if $\sequencen{u_n}$
is any non-increasing sequence in $L^0=L^0(\frak A)$ with infimum $0$,
and $\epsilon>0$, then $\{n(u_n-\epsilon u_0):n\in\Bbb N\}$ is bounded
above in $L^0$.   \Prf\ For $k\in\Bbb N$ set
$a_k=\sup_{n\in\Bbb N}\Bvalue{n(u_n-\epsilon u_0)>k}$;  set
$a=\inf_{k\in\Bbb N}a_k$.   \Quer\ Suppose, if possible, that $a\ne 0$.
Because $u_n\le u_0$, $n(u_n-\epsilon u_0)\le nu_0$ for every $n$ and

\Centerline{$a\Bsubseteq a_0\Bsubseteq\Bvalue{u_0>0}
=\Bvalue{\epsilon u_0>0}
=\sup_{n\in\Bbb N}\Bvalue{\epsilon u_0-u_n>0}$.}

\noindent So there is some $m\in\Bbb N$ such that
$a'=a\Bcap\Bvalue{\epsilon u_0-u_m>0}\ne 0$.   Now, for any $n\ge m$,
any $k\in\Bbb N$,

\Centerline{$a'\Bcap\Bvalue{n(u_n-\epsilon u_0)>k}
\Bsubseteq\Bvalue{\epsilon u_0-u_m>0}\Bcap\Bvalue{u_m-\epsilon u_0>0}
=0$.}

\noindent But $a'\Bsubseteq\sup_{n\in\Bbb N}\Bvalue{n(u_n-\epsilon
u_0)>k}$, so in fact

\Centerline{$a'\Bsubseteq\sup_{n\le m}\Bvalue{n(u_n-\epsilon
u_0)>k}=\Bvalue{v>k}$,}

\noindent where $v=\sup_{n\le m}n(u_n-\epsilon u_0)$.   And
this means that $\inf_{k\in\Bbb N}\Bvalue{v>k}\Bsupseteq a'\ne 0$, which
is impossible.\   \Bang\   Accordingly $a=0$;  by 364L(a-i),
$\{n(u_n-\epsilon u_0):n\in\Bbb N\}$ is bounded above.\   \Qed

\medskip

{\bf (b)} Now suppose that $\sequencen{u_n}$ is a
non-increasing sequence in $L^0$ with infimum $0$, and that $w\in W$ is
a lower bound for $\{Tu_n:n\in\Bbb N\}$.
Take any $\epsilon>0$.   By (a), $\{n(u_n-\epsilon u_0):n\in\Bbb N\}$
has an upper bound $v$ in $L^0$.   Because $T$ is positive,

\Centerline{$w\le Tu_n=T(u_n-\epsilon u_0)+T(\epsilon u_0)\le
T(\Bover1nv)+T(\epsilon u_0)=\Bover1nTv+\epsilon Tu_0$}

\noindent for every $n\ge 1$.   Because $W$ is Archimedean,
$w\le\epsilon Tu_0$.   But this is true for every $\epsilon>0$, so
(again because $W$ is Archimedean) $w\le 0$.   As $w$ is arbitrary,
$\inf_{n\in\Bbb N}Tu_n=0$.   As $\sequencen{u_n}$ is arbitrary, $T$ is
sequentially order-continuous (351Gb).
}%end of proof of 375A

\leader{375B}{Proposition} Let $\frak A$ be an atomless Dedekind
$\sigma$-complete Boolean algebra.   Then $L^0(\frak A)^{\times}=\{0\}$.

\proof{ \Quer\ Suppose, if possible, that $h:L^0(\frak A)\to\Bbb R$ is a
non-zero order-continuous positive linear functional.   Then there is a
$u>0$ in $L^0$ such that $h(v)>0$ whenever $0<v\le u$ (356H).
Because $\frak A$ is atomless, there is a disjoint sequence
$\sequencen{a_n}$ such that $a_n\Bsubseteq\Bvalue{u>0}$ for each $n$, so
that $u_n=u\times\chi a_n>0$, while $u_m\wedge u_n=0$ if $m\ne n$.   Now
however

\Centerline{$v=\sup_{n\in\Bbb N}\Bover{n}{h(u_n)}u_n$}

\noindent is defined in $L^0$, by 368K, and $h(v)\ge n$ for every $n$,
which is impossible.\ \Bang
}%end of proof of 375B

\leader{375C}{Theorem} Let $\frak A$ be a Dedekind complete Boolean
algebra, $W$ an Archimedean Riesz space, and $T:L^0(\frak A)\to W$ an
order-continuous Riesz homomorphism.   Then $V=T[L^0(\frak A)]$ is an
order-closed Riesz subspace of $W$.

\proof{ The kernel $U$ of $T$ is a band in $L^0=L^0(\frak A)$
(352Oe), and must be a projection band (353J), because $L^0$ is Dedekind
complete (364M).   Since $U+U^{\perp}=L^0$, $T[U]+T[U^{\perp}]=V$, that
is, $T[U^{\perp}]=V$;  since $U\cap U^{\perp}=\{0\}$, $T$ is an
isomorphism between $U^{\perp}$ and $V$.   Now
suppose that $A\subseteq V$ is upwards-directed and has a least upper
bound $w\in W$.   Then $B=\{u:u\in U^{\perp},\,Tu\in A\}$ is
upwards-directed and $T[B]=A$.   The point is that $B$ is bounded above
in $L^0$.   \Prf\Quer\ If not, then $\{u^+:u\in B\}$ cannot be bounded
above, so there is a $u_0>0$ in $L^0$ such that
$nu_0=\sup_{u\in B}nu_0\wedge u^+$
for every $n\in\Bbb N$ (368A).   Since $B\subseteq U^{\perp}$,
$u_0\in U^{\perp}$ and $Tu_0>0$.   But now, because $T$ is an
order-continuous Riesz homomorphism,

\Centerline{$nTu_0
=\sup_{u\in B}T(nu_0\wedge u^+)
=\sup_{v\in A}nTu_0\wedge v^+\le w^+$}

\noindent for every $n\in\Bbb N$, which is impossible.\   \Bang\Qed\

Set $u^*=\sup B$;  then $Tu^*=\sup A=w$ and $w\in V$.   As $A$ is
arbitrary, $V$ is order-closed.
}%end of proof of 375C

\leader{375D}{Corollary} Let $W$ be a Riesz space and $V$
an order-dense Riesz subspace which is isomorphic to $L^0(\frak A)$ for
some Dedekind complete Boolean algebra $\frak A$.   Then $V=W$.

\proof{ By 353G, $W$ is Archimedean.
So we can apply 375C to an isomorphism $T:L^0(\frak A)\to V$ to see that
$V$ is order-closed in $W$.
}%end of proof of 375D

\leader{375E}{Theorem}\dvAnew{2010}
Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, $(\frak B,\bar\nu)$ any measure algebra, and
$T:L^0(\frak A)\to L^0(\frak B)$ an order-continuous positive linear
operator.   Then $T$ is continuous for the topologies of convergence in
measure.

\proof{ \Quer\ Otherwise, we can find $w\in L^0(\frak A)$,
$b\in\frak B^f$ and $\epsilon>0$
such that whenever $a\in\frak A^f$ and $\delta>0$ there is a
$u\in L^0(\frak A)$ such that
$\bar\mu(a\Bcap\Bvalue{|u-w|>\delta})\le\delta$
and $\bar\nu(b\Bcap\Bvalue{|Tu-Tw|>\epsilon})\ge\epsilon$ (367L, 2A3H).
%4A5Fa would help here
Of course it follows that whenever $a\in\frak A^f$ and $\delta>0$ there
is a $u\in L^0(\frak A)$ such that
$\bar\mu(a\Bcap\Bvalue{|u|>\delta})\le\delta$
and $\bar\nu(b\Bcap\Bvalue{|Tu|>\epsilon})\ge\epsilon$.   Choose
$\sequencen{a_n}$ and $\sequencen{u_n}$ inductively, as follows.   $a_0=0$.
Given that $a_n\in\frak A^f$, let $u_n\in L^0(\frak A)$ be such that
$\bar\mu(a_n\Bcap\Bvalue{|u_n|>2^{-n}})\le 2^{-n}$ and
$\bar\nu(b\Bcap\Bvalue{|Tu_n|>\epsilon})\ge\epsilon$.
Of course it follows that
$\bar\nu(b\Bcap\Bvalue{T|u_n|>\epsilon})\ge\epsilon$.  Because
$(\frak A,\bar\mu)$ is semi-finite,
$|u_n|=\sup_{a\in\frak A^f}|u_n|\times\chi a)$;  because $T$ is
order-continuous,
$T|u_n|=\sup_{a\in\frak A^f}T(|u_n|\times\chi a)$, and we can find
$a_{n+1}\in\frak A^f$ such that
$\bar\nu(b\Bcap\Bvalue{T(|u_n|\times\chi a_{n+1})>\epsilon})
\ge\bover12\epsilon$.   Enlarging $a_{n+1}$ if necessary, arrange that
$a_{n+1}\Bsupseteq a_n$.   Continue.

At the end of the induction, set $v_n=2^n|u_n|\times\chi a_{n+1}$;
then $\bar\mu(a_n\Bcap\Bvalue{v_n>1})\le 2^{-n}$, for each $n\in\Bbb N$.
It follows that $\{v_n:n\in\Bbb N\}$ is bounded above.   \Prf\ For
$k\in\Bbb N$,  set $c_k=\sup_{n\in\Bbb N}\Bvalue{v_n>k}$.   Then
$c_k\Bsubseteq\sup_{n\in\Bbb N}a_n$.   If $n\in\Bbb N$ and $\delta>0$, let
$m\ge n$ be such that $2^{-m+1}\le\delta$, and $k\ge 1$ such that
$\bar\mu(a_n\Bcap\Bvalue{\sup_{m<n}v_m>k})\le\delta$.   Then

\Centerline{$\bar\mu(a_n\Bcap c_k)
\le\bar\mu(a_n\Bcap\Bvalue{\sup_{m<n}v_m>k})
  +\sum_{i=m}^{\infty}\bar\mu(a_i\Bcap\Bvalue{v_i>1})
\le 2\delta$.}

\noindent As $\delta$ is arbitrary, $a_n\Bcap\inf_{k\in\Bbb N}c_k=0$;
as $n$ is arbitrary, $\inf_{k\in\Bbb N}c_k=0$;  by 364L(a-i) again,
$\{v_n:n\in\Bbb N\}$ is bounded above.\ \Qed

Set $v=\sup_{n\in\Bbb N}v_n$.   Then $2^{-n}v\ge|u_n|\times\chi a_{n+1}$,
so $2^{-n}Tv\ge T(|u_n|\times\chi a_{n+1})$ and
$\bar\nu(b\Bcap\Bvalue{2^{-n}Tv>\epsilon})\ge\bover12\epsilon$, for each
$n\in\Bbb N$.   But $\inf_{n\in\Bbb N}2^{-n}Tv=0$, so
$\inf_{n\in\Bbb N}\Bvalue{2^{-n}Tv>\epsilon}=0$ (364L(b-ii)) and
$\inf_{n\in\Bbb N}\bar\nu(b\Bcap\Bvalue{2^{-n}Tv>\epsilon})=0$.\ \Bang

So we have the result.
}

\leader{375F}{}\cmmnt{ I come now to the deepest result of this
section, concerning positive linear operators from $L^0(\frak A)$ to
$L^0(\frak B)$ where $\frak B$ is a measure algebra.   I approach
through a couple of lemmas which are striking enough in their own right.

The following temporary definition will be useful.

\medskip

\noindent}{\bf Definition} Let $\frak A$ and $\frak B$ be Boolean
algebras.   I will say that a function $\phi:\frak A\to\frak B$ is a
{\bf \ssbhm} if

\inset{$\phi(a\Bcup a')=\phi(a)\Bcup\phi(a')$ for all $a$,
$a'\in\frak A$,}

\inset{$\inf_{n\in\Bbb N}\phi(a_n)=0$ whenever $\sequencen{a_n}$ is a
non-increasing sequence in $\frak A$ with infimum $0$.}

\cmmnt{\noindent Now we have the following easy facts.}

\leader{375G}{Lemma} Let $\frak A$ and $\frak B$ be Boolean algebras and
$\phi:\frak A\to\frak B$ a \ssbhm.

(a) $\phi(0)=0$, $\phi(a)\Bsubseteq\phi(a')$ whenever $a\Bsubseteq a'$,
and $\phi(a)\Bsetminus\phi(a')\Bsubseteq\phi(a\Bsetminus a')$ for every
$a$, $a'\in\frak A$.

(b) If $\bar\mu$, $\bar\nu$ are measures such that $(\frak A,\bar\mu)$
and $(\frak B,\bar\nu)$ are totally finite measure algebras, then for
every $\epsilon>0$ there is a $\delta>0$ such that
$\bar\nu\phi(a)\le\epsilon$ whenever $\bar\mu a\le\delta$.

\proof{{\bf (a)} This is elementary.   Set every $a_n=0$ in the second
clause of the definition 375F to see that $\phi(0)=0$.   The other two
parts are immediate consequences of the first clause.

\medskip

{\bf (b)} (Compare 232Ba, 327Bb.)   \Quer\ Suppose, if possible,
otherwise.   Then for every $n\in\Bbb N$ there is an $a_n\in\frak A$
such that $\bar\mu a_n\le 2^{-n}$ and $\bar\nu\phi(a_n)\ge\epsilon$.
Set $c_n=\sup_{i\ge n}a_i$ for each $n$;  then $\sequencen{c_n}$ is
non-increasing and has infimum $0$ (since $\bar\mu c_n\le 2^{-n+1}$ for
each $n$), but $\bar\nu\phi(c_n)\ge\epsilon$ for every $n$, so
$\inf_{n\in\Bbb N}\phi c_n$ cannot be $0$.\ \Bang
}%end of proof of 375G

\leader{375H}{Lemma} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
totally finite measure algebras and $\phi:\frak A\to\frak B$ a
\ssbhm.   Then for every non-zero $b_0\in\frak B$
there are a non-zero $b\Bsubseteq b_0$ and an $m\in\Bbb N$ such that
$b\Bcap\inf_{j\le m}\phi(a_j)=0$ whenever $a_0,\ldots,a_m\in\frak A$ are
disjoint.

\proof{{\bf (a)}  Suppose first that $\frak A$ is atomless and that
$\bar\mu 1=1$.

Set $\epsilon=\bover15\bar\nu b_0$ and let $m\ge 1$ be such that
$\bar\nu\phi(a)\le\epsilon$ whenever $\bar\mu a\le\bover1m$ (375Gb).
We need to know that
$(1-\bover1m)^m\le\bover12$;   this is because (if $m\ge 2$)
$\ln m-\ln(m-1)\ge\bover1m$, so $m\ln(1-\bover1m)\le-1\le-\ln 2$.

Set

\Centerline{$C=\{\inf_{j\le m}\phi(a_j):a_0,\ldots,a_m\in\frak A$ are
disjoint$\}$.}

\noindent\Quer\ Suppose, if possible, that $b_0\Bsubseteq\sup C$.   Then
there are $c_0,\ldots,c_k\in C$ such that
$\bar\nu(b_0\Bcap\sup_{i\le k}c_i)\discretionary{}{}{}\ge 4\epsilon$.
For each $i\le k$
choose disjoint $a_{i0},\ldots,a_{im}\in\frak A$ such that
$c_i=\inf_{j\le m}\phi(a_{ij})$.   Let $D$ be the set of atoms of the
finite subalgebra
of $\frak A$ generated by $\{a_{ij}:i\le k,\,j\le m\}$, so that $D$ is a
finite partition of unity in $\frak A$, and every $a_{ij}$ is the join
of the members of $D$ it includes.   Set $p=\#(D)$, and for each
$d\in D$ take a maximal disjoint set
$E_d\subseteq\{e:e\Bsubseteq d,\,\bar\mu e=\bover1{pm}\}$, so that
$\bar\mu(d\Bsetminus\sup E_d)<\bover1{pm}$;  set

\Centerline{$d^*=1\Bsetminus\sup(\bigcup_{d\in D}E_d)
=\sup_{d\in D}(d\Bsetminus\sup E_d)$,}

\noindent so that $\bar\mu d^*$ is a multiple of $\bover1{pm}$ and is
less than $\bover1m$.   Let $E^*$ be a disjoint set of elements of
measure $\bover1{pm}$ with union $d^*$, and take
$E=E^*\cup\bigcup_{d\in D}E_d$, so that $E$ is a partition of unity in
$\frak A$, $\bar\mu
e=\bover1{pm}$ for every $e\in E$, and $a_{ij}\Bsetminus d^*$ is the
join of the members of $E$ it includes for every $i\le k$ and $j\le m$.

Set

\Centerline{$\Cal K=\{K:K\subseteq E,\,\#(K)=p\}$,
\quad $M=\#(\Cal K)=\Bover{(mp)!}{p!(mp-p)!}$.}

\noindent For every $K\in\Cal K$, $\bar\mu(\sup K)=\bover1m$ so
$\bar\nu\phi(\sup K)\le\epsilon$.   So if we set

\Centerline{$v=\sum_{K\in\Cal K}\chi\phi(\sup K)$,}

\noindent $\int v\le\epsilon M$.   On the other hand,

\Centerline{$\bar\nu(b_0\Bcap\sup_{i\le k}c_i)\ge 4\epsilon$,
\quad$\bar\nu\phi(d^*)\le\epsilon$,}

\noindent so $\bar\nu b_1\ge 3\epsilon$, where

\Centerline{$b_1=b_0\Bcap\sup_{i\le k}c_i\Bsetminus\phi(d^*)$.}

\noindent Accordingly $\int v\le\bover13M\bar\nu b_1$ and

\Centerline{$b_2=b_1\Bcap\Bvalue{v<\bover12M}$}

\noindent is non-zero.

Because $b_2\Bsubseteq b_1$, there is an $i\le k$ such that $b_2\Bcap
c_i\ne 0$.   Now

\Centerline{$b_2\Bcap c_i
\Bsubseteq c_i\Bsetminus\phi(d^*)
=\inf_{j\le m}\phi(a_{ij})\Bsetminus\phi(d^*)
\Bsubseteq\inf_{j\le m}\phi(a_{ij}\Bsetminus d^*)$.}

\noindent But every $a_{ij}\Bsetminus d^*$ is the join of the members of
$E$ it includes, so

$$\eqalign{b_2\Bcap c_i
&\Bsubseteq\inf_{j\le m}\phi(a_{ij}\Bsetminus d^*)
\Bsubseteq\inf_{j\le m}\phi(\sup\{e:e\in E,\,e\Bsubseteq a_{ij}\})\cr
&=\inf_{j\le m}\sup\{\phi(e):e\in E,\,e\Bsubseteq a_{ij}\}\cr
&=\sup\{\inf_{j\le m}\phi(e_j):e_0,\ldots,e_m\in E
   \text{ and }e_j\Bsubseteq a_{ij}\text{ for every }j\}.\cr}$$

\noindent So there are $e_0,\ldots,e_m\in E$ such that $e_j\Bsubseteq
a_{ij}$ for each $j$ and $b_3=b_2\Bcap\inf_{j\le m}\phi(e_j)\ne 0$.
Because $a_{i0},\ldots,a_{im}$ are disjoint, $e_0,\ldots,e_m$ are
distinct;  set $J=\{e_0,\ldots,e_m\}$.   Then whenever $K\in\Cal K$ and
$K\cap J\ne\emptyset$, $b_3\Bsubseteq\phi(\sup K)$.

So let us calculate the size of $\Cal K_1=\{K:K\in\Cal K,\,K\cap
J\ne\emptyset\}$.   This is

$$\eqalign{M-\Bover{(mp-m-1)!}{p!(mp-p-m-1)!}
&=M\bigl(1-\Bover{(mp-p)(mp-p-1)\ldots(mp-p-m)}
  {mp(mp-1)\ldots(mp-m)}\bigr)\cr
&\ge M\bigl(1-(\Bover{mp-p}{mp})^{m+1}\bigr)
\ge\Bover12M.\cr}$$

\noindent But this means that $b_3\Bsubseteq\Bvalue{v\ge\bover12M}$,
while also $b_3\Bsubseteq\Bvalue{v<\bover12M}$;  which is surely
impossible.\  \Bang

Accordingly $b_0\notBsubseteq\sup C$, and we can take
$b=b_0\Bsetminus\sup C$.

\medskip

{\bf (b)} Now for the general case.   Let $A$ be the set of atoms of
$\frak A$, and set $d=1\Bsetminus\sup A$.   Then the principal
ideal $\frak A_d$ is
atomless, so there are a non-zero $b_1\Bsubseteq b_0$ and an
$n\in\Bbb N$ such that $b_1\Bcap\inf_{j\le n}\phi(a_j)=0$ whenever
$a_0,\ldots,a_n\in\frak A_d$ are disjoint.   \Prf\ If $\bar\mu d>0$ this
follows from (a), if we apply it to $\phi\restrp\frak A_d$ and
$(\bar\mu d)^{-1}\bar\mu\restrp\frak A_d$.   If $\bar\mu d=0$ then we can
just take $b_1=b_0$ and $n=0$.\  \Qed

Let $\delta>0$ be such that $\bar\nu\phi(a)<\bar\nu b_1$ whenever
$\bar\mu a\le\delta$.   Let $A_1\subseteq A$ be a finite set such that
$\bar\mu(\sup A_1)\ge\bar\mu(\sup A)-\delta$, and set $r=\#(A)$,
$d^*=\sup(A\Bsetminus A_1)$.   Then $\bar\mu d^*\le\delta$ so
$b=b_1\Bsetminus\phi(d^*)\ne 0$.   Try $m=n+r$.   If $a_0,\ldots,a_m$
are disjoint, then at most $r$ of them can meet $\sup A_1$, so
(re-ordering if necessary) we can suppose that $a_0,\ldots,a_n$ are
disjoint from $\sup A_1$, in which case $a_j\Bsetminus d^*\Bsubseteq d$
for each $j\le m$.   But in this case (because $b\Bcap\phi(d^*)=0$)

\Centerline{$b\Bcap\inf_{j\le m}\phi(a_j)
\Bsubseteq b\Bcap\inf_{j\le n}\phi(a_j)
=b\Bcap\inf_{j\le n}\phi(a_j\Bcap d)
=0$}

\noindent by the choice of $n$ and $b_1$.

Thus in the general case also we can find appropriate $b$ and $m$.
}%end of proof of 375H

\leader{375I}{Lemma} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
totally finite measure algebras and $\phi:\frak A\to\frak B$ a
\ssbhm.   Then for every non-zero $b_0\in\frak B$
there are a non-zero $b\Bsubseteq b_0$ and a finite partition of unity
$C\subseteq\frak A$ such that $a\mapsto b\Bcap\phi(a\Bcap c)$ is a ring
homomorphism for every $c\in C$.

\proof{ By 375H, we can find $b_1$, $m$ such that
$0\ne b_1\Bsubseteq b_0$ and $b_1\Bcap\inf_{j\le m}\phi(a_j)=0$ whenever
$a_0,\ldots,a_m\in\frak A$ are disjoint.   Do this with the smallest
possible $m$.   If $m=0$ then $b_1\Bcap\phi(1)=0$, so we can take
$b=b_1$, $C=\{1\}$.   Otherwise, because $m$ is minimal, there must be
disjoint $c_1,\ldots,c_m\in\frak A$ such that
$b=b_1\Bcap\inf_{1\le j\le m}\phi(c_j)\ne 0$.   Set
$c_0=1\Bsetminus\sup_{1\le j\le m}c_j$,
$C=\{c_0,c_1,\ldots,c_m\}$;  then $C$ is a partition of unity in
$\frak A$.   Set $\pi_j(a)=b\Bcap\phi(a\Bcap c_j)$ for each
$a\in\frak A$ and $j\le m$.   Then we always have
$\pi_j(a\Bcup a')=\pi_j(a)\Bcup\pi_j(a')$ for all $a$, $a'\in\frak A$,
because $\phi$ is a subhomomorphism.

To see that every $\pi_j$ is a ring homomorphism, we need only check
that $\pi_j(a\Bcap a')=0$ whenever $a\Bcap a'=0$.   (Compare
312H(iv).)   In the case $j=0$, we actually have $\pi_0(a)=0$ for
every $a$, because
$b\Bcap\phi(c_0)=b_1\Bcap\inf_{0\le j\le m}\phi(c_j)=0$ by the choice of
$b_1$ and $m$.   When $1\le j\le m$, if $a\Bcap a'=0$, then

\Centerline{$\pi_j(a)\Bcap\pi_j(a')
=b_1\Bcap\inf_{1\le i\le m,i\ne j}\phi(c_j)\Bcap\phi(a)\Bcap\phi(a')$}

\noindent is again $0$, because $a$, $a'$,
$c_1,\ldots,c_{j-1},c_{j+1},\ldots,c_m$ are disjoint.   So we have a
suitable pair $b$, $C$.
}%end of proof of 375I

\leader{375J}{Theorem} Let $\frak A$ be any Dedekind $\sigma$-complete
Boolean algebra and $(\frak B,\bar\nu)$ a semi-finite measure algebra.
Let $T:L^0(\frak A)\to L^0(\frak B)$ be a positive linear operator.
Then we can find $B$, $\langle A_b\rangle_{b\in B}$ such that $B$ is a
partition of unity in $\frak B$, each $A_b$ is a finite partition of
unity in $\frak A$, and $u\mapsto T(u\times\chi a)\times\chi b$ is a
Riesz homomorphism whenever $b\in B$ and $a\in A_b$.

\proof{{\bf (a)} Write $B^*$ for the set of potential members of $B$;
that is, the set of those $b\in \frak B$ such that there is a finite
partition of unity $A\subseteq\frak A$ such that $T_{ab}$ is a Riesz
homomorphism for every $a\in A$, writing $T_{ab}(u)=T(u\times\chi
a)\times\chi b$.   If I can show that $B^*$ is order-dense in $\frak B$,
this will suffice, since there will then be a partition of unity
$B\subseteq B^*$.

\medskip

{\bf (b)} So let $b_0$ be any non-zero member of $\frak B$;  I seek a
non-zero member of $B^*$ included in $b_0$.   Of course there is a
non-zero $b_1\Bsubseteq b_0$ with $\bar\nu b_1<\infty$.   Let $\gamma>0$
be such that $b_2=b_1\Bcap\Bvalue{T(\chi 1)\le\gamma}$ is non-zero.
Define $\mu:\frak A\to\coint{0,\infty}$ by setting
$\mu a=\int_{b_2}T(\chi a)$
for every $a\in\frak A$.   Then $\mu$ is countably additive, because
$\chi$, $T$ and $\int$ are all additive and sequentially
order-continuous (using 375A).   Set $\Cal N=\{a:\mu a=0\}$;  then
$\Cal N$ is a $\sigma$-ideal of $\frak A$, and $(\frak C,\bar\mu)$ is a
totally finite measure algebra, where $\frak C=\frak A/\Cal N$ and
$\bar\mu a^{\ssbullet}=\mu a$ for every $a\in\frak A$ (just as in 321H).

\medskip

{\bf (c)} We have a function $\phi$ from $\frak C$ to the principal
ideal $\frak B_{b_2}$ defined by saying that
$\phi a^{\ssbullet}=b_2\Bcap\Bvalue{T(\chi a)>0}$ for every
$a\in\frak A$.   \Prf\ If $a_1$, $a_2\in\frak A$ are such that
$a_1^{\ssbullet}=a_2^{\ssbullet}$ in $\frak C$, this means that
$a_1\Bsymmdiff a_2\in\Cal N$;  now

$$\eqalign{\Bvalue{T(\chi a_1)>0}\Bsymmdiff\Bvalue{T(\chi a_2)>0}
&\Bsubseteq\Bvalue{|T(\chi a_1)-T(\chi a_2)|>0}\cr
&\Bsubseteq\Bvalue{T(|\chi a_1-\chi a_2|)>0}
=\Bvalue{T\chi(a_1\Bsymmdiff a_2)>0}\cr}$$

\noindent is disjoint from $b_2$ because $\int_{b_2}T\chi(a_1\Bsymmdiff
a_2)=0$.   Accordingly $b_2\Bcap\Bvalue{T(\chi a_1)>0}
=b_2\Bcap\Bvalue{T(\chi a_2)>0}$ and we can take this common value for
$\phi(a_1^{\ssbullet})=\phi(a_2^{\ssbullet})$.\  \Qed

\medskip

{\bf (d)} Now $\phi$ is a \ssbhm.   \Prf\ (i) For any
$a_1$, $a_2\in\frak A$ we have

\Centerline{$\Bvalue{T\chi(a_1\Bcup a_2)>0}
=\Bvalue{T(\chi a_1)>0}\Bcup\Bvalue{T(\chi a_2)>0}$}

\noindent because

\Centerline{$T(\chi a_1)\vee T(\chi a_2)\le T\chi(a_1\Bcup a_2)
\le T(\chi a_1)+T(\chi a_2)$.}

\noindent So $\phi(c_1\Bcup c_2)=\phi(c_1)\Bcup\phi(c_2)$ for all $c_1$,
$c_2\in\frak C$.   (ii) If $\sequencen{c_n}$ is a non-increasing
sequence in $\frak C$ with infimum $0$, choose $a_n\in\frak A$ such that
$a_n^{\ssbullet}=c_n$ for each $n$, and set
$\tilde a_n=\inf_{i\le n}a_i\Bsetminus\inf_{i\in\Bbb N}a_i$ for each
$n$;  then $\tilde a_n^{\ssbullet}=c_n$ so
$\phi(c_n)=\Bvalue{T(\chi\tilde a_n)>0}$ for
each $n$, while $\sequencen{\tilde a_n}$ is non-increasing and
$\inf_{n\in\Bbb N}\tilde a_n=0$.   \Quer\ Suppose, if possible, that
$b'=\inf_{n\in\Bbb N}\phi(c_n)\ne 0$;  set
$\epsilon=\bover12\bar\nu b'$.   Then
$\bar\nu(b_2\Bcap\Bvalue{T(\chi\tilde a_n)>0})\ge 2\epsilon$ for every
$n\in\Bbb N$.   For each $n$, take $\alpha_n>0$ such that
$\bar\nu(b_2\Bcap\Bvalue{T(\chi\tilde a_n)>\alpha_n})\ge\epsilon$.
Then $u=\sup_{n\in\Bbb N}n\alpha_n^{-1}\chi\tilde a_n$ is defined in
$L^0(\frak A)$ (because
$\sup_{n\in\Bbb N}\Bvalue{n\alpha_n^{-1}\chi\tilde a_n>k}
\subseteq\tilde a_m$ if $k\ge\max_{i\le m}i\alpha_i^{-1}$, so
$\inf_{k\in\Bbb N}\sup_{n\in\Bbb N}
\Bvalue{n\alpha_n^{-1}\chi\tilde a_n>k}=0$).   But now

\Centerline{$\bar\nu(b_2\Bcap\Bvalue{Tu>n})
\ge\bar\nu(b_2\Bcap\Bvalue{T(\chi\tilde a_n)>\alpha_n})\ge\epsilon$}

\noindent for every $n$, so $\inf_{n\in\Bbb N}\Bvalue{Tu>n}\ne 0$, which
is impossible.\ \BanG\   Thus $\inf_{n\in\Bbb N}\phi(c_n)=0$;  as
$\sequencen{c_n}$ is arbitrary, $\phi$ is a \ssbhm.\
\Qed

\medskip

{\bf (e)} By 375I, there are a non-zero $b\in\frak B_{b_2}$ and a finite
partition of unity $C\subseteq\frak C$ such that $d\mapsto
b\Bcap\phi(d\Bcap c)$ is a ring homomorphism for every $c\in C$.   There
is a partition of unity $A\subseteq\frak A$, of the same size as $C$,
such that $C=\{a^{\ssbullet}:a\in A\}$.   Now $T_{ab}$ is a Riesz
homomorphism for every $a\in A$.   \Prf\ It is surely a positive linear
operator.   If $u_1$, $u_2\in L^0(\frak A)$ and $u_1\wedge u_2=0$, set
$e_i=\Bvalue{u_i>0}$ for each $i$, so that $e_1\Bcap e_2=0$.   Observe
that $u_i=\sup_{n\in\Bbb N}u_i\wedge n\chi e_i$, so that

\Centerline{$\Bvalue{T_{ab}u_i>0}=\sup_{n\in\Bbb
N}\Bvalue{T_{ab}(u_i\wedge n\chi e_i)>0}\Bsubseteq\Bvalue{T_{ab}(\chi
e_i)>0}=b\Bcap\Bvalue{T\chi(e_i\Bcap a)>0}$}

\noindent for both $i$ (of course $T_{ab}$, like $T$, is sequentially
order-continuous).   But this means that

$$\eqalign{\Bvalue{T_{ab}u_1>0}\Bcap\Bvalue{T_{ab}u_2>0}
&\Bsubseteq b\Bcap\Bvalue{T\chi(e_1\Bcap
a)>0}\Bcap\Bvalue{T\chi(e_2\Bcap a)>0}\cr
&=b\Bcap\phi(e_1^{\ssbullet}\Bcap
   a^{\ssbullet})\Bcap\phi(e_2^{\ssbullet}\Bcap a^{\ssbullet})
=0\cr}$$

\noindent because $a^{\ssbullet}\in C$, so
$d\mapsto b\Bcap\phi(d\Bcap a^{\ssbullet})$ is a ring homomorphism, while
$e_1^{\ssbullet}\Bcap e_2^{\ssbullet}=0$.   So
$T_{ab}u_1\wedge T_{ab}u_2=0$.   As $u_1$ and $u_2$ are arbitrary,
$T_{ab}$ is a Riesz homomorphism (352G(iv)).\ \Qed

\medskip

{\bf (f)} Thus $b\in B^*$.   As $b_0$ is arbitrary, $B^*$ is
order-dense, and we're home.
}%end of proof of 375J

\leader{375K}{Corollary} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $U$ a Dedekind complete Riesz space such that
$U^{\times}$ separates the points of $U$.   If $T:L^0(\frak A)\to U$ is
a positive linear operator, there is a sequence $\sequencen{T_n}$ of
Riesz homomorphisms from $L^0(\frak A)$ to $U$ such that
$T=\sum_{n=0}^{\infty}T_n$, in the sense that $Tu=\sup_{n\in\Bbb
N}\sum_{i=0}^nT_iu$ for every $u\ge 0$ in $L^0(\frak A)$.

\proof{ By 369A, $U$ can be embedded as an order-dense Riesz subspace of
$L^0(\frak B)$ for some localizable measure algebra $(\frak B,\bar\nu)$;
being Dedekind complete, it is solid in $L^0(\frak B)$ (353L).   Regard
$T$ as an operator from $L^0(\frak A)$ to $L^0(\frak B)$, and take $B$,
$\langle A_b\rangle_{b\in B}$ as in 375J.   Note that $L^0(\frak B)$ can
be identified with $\prod_{b\in B}L^0(\frak B_b)$ (364R, 322L).   For
each $b\in B$ let $f_b:A_b\to\Bbb N$ be an injection.   If $b\in B$ and
$n\in f_b[A_b]$, set $T_{nb}(u)=\chi b\times T(u\times\chi f_b^{-1}(n))$;
otherwise set $T_{nb}=0$.   Then $T_{nb}:L^0(\frak A)\to L^0(\frak B_b)$
is a Riesz homomorphism;   because $A_b$ is a finite partition of unity,
$\sum_{n=0}^{\infty}T_{nb}u=\chi b\times Tu$
for every $u\in L^0(\frak A)$.   But this means that if we set
$T_nu=\langle T_{nb}u\rangle_{b\in B}$,

\Centerline{$T_n:L^0(\frak A)\to\prod_{b\in
B}L^0(\frak B_b)\cong L^0(\frak B)$}

\noindent is a Riesz homomorphism for each
$n$;  and $T=\sum_{n=0}^{\infty}T_n$.   Of course every $T_n$ is an
operator from $L^0(\frak A)$ to $U$ because $|T_nu|\le T|u|\in U$ for
every $u\in L^0(\frak A)$.
}%end of proof of 375K

\leader{375L}{Corollary} (a) If $\frak A$ is a Dedekind
$\sigma$-complete Boolean algebra, $(\frak B,\bar\nu)$ is a semi-finite
measure algebra, and there is any non-zero positive linear operator from
$L^0(\frak A)$ to $L^0(\frak B)$, then there is a non-trivial
sequentially order-continuous ring homomorphism from $\frak A$ to
$\frak B$.

(b) If $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ are homogeneous
probability algebras and $\tau(\frak A)>\tau(\frak B)$, then
$\eurm L^{\sim}(L^0(\frak A);L^0(\frak B))=\{0\}$.

\proof{{\bf (a)} It is probably quickest to look at the proof of 375J:
starting from a non-zero positive linear operator $T:L^0(\frak A)\to
L^0(\frak B)$, we move to a non-zero \ssbhm\
$\phi:\frak A/\Cal N\to\frak B$ and thence to a non-zero ring
homomorphism from $\frak A/\Cal N$ to $\frak B$, corresponding to a
non-zero ring homomorphism from $\frak A$ to $\frak B$, which is
sequentially order-continuous because it is dominated by $\phi$.
Alternatively, quoting 375J, we have a non-zero Riesz
homomorphism $T_1:L^0(\frak A)\to L^0(\frak B)$, and it is easy to check
that $a\mapsto\Bvalue{T(\chi a)>0}$ is a non-zero sequentially
order-continuous ring homomorphism.

\medskip

{\bf (b)} Use (a) and 331J.
}%end of proof of 375L

\exercises{
\leader{375X}{Basic exercises (a)}
%\spheader 375Xa
Let $\frak A$ be a Dedekind complete Boolean algebra and $W$ an
Archimedean Riesz space.   Let $T:L^0(\frak A)\to W$ be a positive
linear operator.   Show that $T$ is order-continuous iff
$T\chi:\frak A\to W$ is order-continuous.
%375A

\spheader 375Xb Let $\frak A$ be an atomless Dedekind $\sigma$-complete
Boolean algebra and $W$ a Banach lattice.   Show that the only
order-continuous positive linear operator from $L^0(\frak A)$ to $W$ is
the zero operator.
%375B

\spheader 375Xc Let $\frak A$ be a Dedekind complete Boolean algebra and
$W$ a Riesz space.   Let $T:L^0(\frak A)\to W$ be an
order-continuous Riesz homomorphism such that $T[L^0(\frak A)]$ is
order-dense in $W$.   Show that $T$ is surjective.
%375D

\sqheader 375Xd Let $\frak A$ and $\frak B$ be Boolean algebras and
$\phi:\frak A\to\frak B$ a \ssbhm\ as defined in 375F.
Show that $\phi$ is sequentially order-continuous.
%375F

\sqheader 375Xe Let $\frak A$ be the measure algebra of Lebesgue measure
on $[0,1]$ and $\frak G$ the regular open algebra of $\Bbb R$.  (i) Show
that there is no non-zero positive linear operator from $L^0(\frak G)$
to $L^0(\frak A)$.   \Hint{suppose $T:L^0(\frak G)\to L^0(\frak A)$ were
such an operator.   Reduce to the case $T(\chi 1)\le\chi 1$.   Let
$\sequencen{b_n}$ enumerate
an order-dense subset of $\frak G$ (316Yo).   For each $n\in\Bbb N$ take
non-zero $b'_n\Bsubseteq b_n$ such that
$\int T(\chi b'_n)\le 2^{-n-2}\int T(\chi 1)$ and consider
$T\chi(\sup_{n\in\Bbb N}b'_n)$.  See also 375Yf-375Ye.}
(ii) Show that there is no non-zero
positive linear operator from $L^0(\frak A)$ to $L^0(\frak G)$.
\Hint{suppose $T:L^0(\frak A)\to L^0(\frak G)$ were such an operator.
For each $n\in \Bbb N$ choose $a_n\in\frak A$,
$\alpha_n>0$ such that $\bar\mu a_n\le 2^{-n}$  and if
$b_n\Bsubseteq\Bvalue{T(\chi 1)>0}$ then
$b_n\Bcap\Bvalue{T(\chi a_n)>\alpha_n}\ne 0$.   Consider $Tu$ where
$u=\sum_{n=0}^{\infty}n\alpha_n^{-1}\chi a_n$.}
%375J

\spheader 375Xf In 375K, show that for any $u\in L^0(\frak A)$

\Centerline{$\inf_{n\in\Bbb N}\sup_{m\ge n}
\Bvalue{|Tu-\sum_{i=0}^mT_iu|>0}=0$.}
%375K

\sqheader 375Xg Prove directly, without quoting
375F-375L, that if $\frak A$ is a Dedekind $\sigma$-complete Boolean
algebra then every positive linear functional from $L^0(\frak A)$ to
$\Bbb R$ is a finite sum of Riesz homomorphisms.
%375L

\spheader 375Xh Let $\frak A$ and $\frak B$ be Dedekind $\sigma$-complete
Boolean algebras, and $T:L^0(\frak A)\to L^0(\frak B)$ a Riesz
homomorphism.   Show that there are a
sequentially order-continuous ring homomorphism $\pi:\frak A\to\frak B$ and
a $w\in L^0(\frak B)^+$ such that $Tu=w\times T_{\pi}u$ for every
$u\in L^0(\frak A)$, where $T_{\pi}:L^0(\frak A)\to L^0(\frak B)$ is
defined as in 364Yg.
%375L

\leader{375Y}{Further exercises (a)}\dvAnew{2010}
%\spheader 375Ya
Let $\frak A$ and $\frak B$ be Dedekind $\sigma$-complete
Boolean algebras, and $T:L^0(\frak A)\to L^0(\frak B)$
a linear operator.   (i) Show that if $T$ is order-bounded, then
$\sequencen{Tu_n}$ order*-converges to $0$ in
$L^0(\frak B)$ (definition: 367A)
whenever $\sequencen{u_n}$ order*-converges to $0$ in $L^0(\frak A)$.
(ii) Show that if $\frak B$ is ccc and \wsid\ and
$\sequencen{Tu_n}$ order*-converges to $0$ in $L^0(\frak B)$
whenever $\sequencen{u_n}$ order*-converges to $0$ in $L^0(\frak A)$, then
$T$ is order-bounded.
%375A mt37bits

\spheader 375Yb
Show that the following are equiveridical:  (i) there is a purely atomic
probability space $(X,\Sigma,\mu)$
such that $\Sigma=\Cal PX$ and $\mu\{x\}=0$ for every $x\in X$;  (ii)
there are a set $X$ and a Riesz homomorphism $f:\Bbb R^X\to\Bbb R$ which
is not order-continuous;   (iii) there are a Dedekind complete Boolean
algebra $\frak A$ and a positive linear operator
$f:L^0(\frak A)\to\Bbb R$ which is not order-continuous;  (iv) there are
a Dedekind complete Boolean algebra $\frak A$ and a sequentially
order-continuous Boolean homomorphism $\pi:\frak A\to\{0,1\}$ which is
not order-continuous;  (v) there are a Dedekind complete Riesz space $U$
and a sequentially order-continuous Riesz homomorphism $f:U\to\Bbb R$
which is not order-continuous;
*(vi) there are an atomless Dedekind complete Boolean algebra $\frak A$
and a sequentially
order-continuous Boolean homomorphism $\pi:\frak A\to\{0,1\}$ which is
not order-continuous.   (Compare 363S.)
%375B mt37bits

\spheader 375Yc Give an example of an atomless Dedekind
$\sigma$-complete Boolean algebra $\frak A$ such that
$L^0(\frak A)^{\sim}\ne\{0\}$.
%375B

\spheader 375Yd\discrversionA{\footnote{Reversed 2010.}}{}
Let $\frak A$ be the measure algebra of Lebesgue measure on $[0,1]$, and
set $L^0=L^0(\frak A)$.   Show that there is a positive linear operator
$T:L^0\to L^0$ such that $T[L^0]$ is not order-closed in $L^0$.
%375C

\spheader 375Ye\dvAnew{2010}
Show that the following are equiveridical:
(i) there is a probability space $(X,\Sigma,\mu)$
such that $\Sigma=\Cal PX$ and $\mu\{x\}=0$ for every $x\in X$;
(ii) there are localizable measure algebras $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ and a positive linear operator
$T:L^0(\frak A)\to L^0(\frak B)$
which is not order-continuous.
%375Yb 375E

\spheader 375Yf
Let $\frak A$, $\frak B$ be Dedekind $\sigma$-complete Boolean algebras
of which $\frak B$ is weakly $\sigma$-distributive.   Let
$T:L^0(\frak A)\to L^0(\frak B)$ be a positive linear operator.   Show
that $a\mapsto\Bvalue{T(\chi a)>0}:\frak A\to\frak B$ is a
\ssbhm.
%375J

\spheader 375Yg\dvAnew{2010}
Let $\frak A$, $\frak B$ be Dedekind $\sigma$-complete Boolean algebras
of which $\frak B$ is weakly $\sigma$-distributive.   Let
$\phi:\frak A\to\frak B$ be a \ssbhm\ such that $\pi a\ne 0$ whenever
$a\in\frak A\setminus\{0\}$.   Show that $\frak A$ is weakly
$\sigma$-distributive.
%375Yf 375J

\spheader 375Yh Let $\frak A$ and $\frak B$ be Dedekind complete Boolean
algebras, and $\phi:\frak A\to\frak B$ a \ssbhm\ such that
$\phi 1_{\frak A}=1_{\frak B}$.   Show that there
is a sequentially order-continuous Boolean homomorphism
$\pi:\frak A\to\frak B$ such that $\pi a\Bsubseteq\phi a$ for every
$a\in\frak A$.
%375L

\spheader 375Yi Let $\frak G$ be the regular open algebra of $\Bbb R$,
and $L^0=L^0(\frak G)$.   Give an example of a non-zero positive linear
operator $T:L^0\to L^0$ such that there is no non-zero Riesz homomorphism
$S:L^0\to L^0$ with $S\le T$.
%375L mt37bits
}%end of exercises

\leader{375Z}{Problem} Let $\frak G$ be the regular open algebra of
$\Bbb R$, and $L^0=L^0(\frak G)$.   If $T:L^0\to L^0$ is a positive
linear operator, must $T[L^0]$ be order-closed?

\endnotes{
\Notesheader{375} Both this section, and the earlier work on linear
operators into $L^0$ spaces, can be regarded as describing different
aspects of a single fact:  $L^0$ spaces are very large.   The most
explicit statements of this principle are 368E and 375D:  every
Archimedean Riesz space can be embedded into a Dedekind complete $L^0$
space, but no such $L^0$ space can be properly embedded as an
order-dense Riesz subspace of any other Archimedean Riesz space.
Consequently there are many maps into $L^0$ spaces (368B).   But by the
same token there are few
maps out of them (375B, 375Lb), and those which do exist have a variety
of special properties (375A, 375J).

The original version of Kwapien's theorem ({\smc Kwapien 73}) was the
special case of 375J in which $\frak A$ is the Lebesgue measure algebra.
The ideas of the proof here are mostly taken
from {\smc Kalton Peck \& Roberts 84}.   I have based my account on
the concept of `subhomomorphism' (375F);  this seems to be an
effective tool when $\frak B$ is \wsid\ (375Yf),
but less useful in other cases.   The case $\frak B=\{0,1\}$,
$L^0(\frak B)\cong\Bbb R$ is not entirely trivial and is worth working
through on its own (375Xg).
}%end of notes

\discrpage


%\showhyphens{\ssbhm}
\frfilename{mt376.tex}
\versiondate{8.4.10}
\copyrightdate{1996}
\def\esssup{\mathop{\text{ess sup}}}

\def\chaptername{Linear operators between function spaces}
\def\sectionname{Kernel operators}

\newsection{376}

The theory of linear integral equations is in large part the theory of
operators $T$ defined from formulae of the type

\Centerline{$(Tf)(y)=\int k(x,y)f(x)dx$}

\noindent for some function $k$ of two variables.   I make no attempt to
study the general theory here.   However, the concepts developed in this
book make it easy to discuss certain aspects of such operators defined
between the `function spaces' of measure theory, meaning spaces of
equivalence classes of functions, and indeed allow us to do some of the
work in the abstract theory of Riesz spaces, omitting all formal mention
of measures (376D, 376H, 376P).   I give a very brief account of two
theorems characterizing kernel operators in the abstract (376E, 376H),
with corollaries to show the form these theorems can take in the
ordinary language of integral kernels (376J, 376N).   To give an idea of
the kind of results we can hope for in this area, I go a bit farther
with operators with domain $L^1$ (376Mb, 376P, 376S).

I take the opportunity to spell out versions of results from \S253 in
the language of this volume (376B-376C).

\cmmnt{
\leader{376A}{Kernel operators} To give an idea of where this
section is going, I will try to describe the central idea in a
relatively concrete special case.   Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be $\sigma$-finite measure spaces;  you can take them
both to be $[0,1]$ with Lebesgue measure if you like.   Let $\lambda$ be
the product measure on $X\times Y$.   If $k\in\eusm L^1(\lambda)$, then
$\int k(x,y)dx$ is defined for almost every $y$, by Fubini's theorem;
so if $f\in\eusm L^{\infty}(\mu)$ then $g(y)=\int k(x,y)f(x)dx$
is defined for almost every $y$.   Also

\Centerline{$\int g(y)dy=\int k(x,y)f(x)dxdy$}

\noindent is defined, because $(x,y)\mapsto k(x,y)f(x)$ is
$\lambda$-virtually measurable, defined $\lambda$-a.e.\ and is
dominated by a multiple of the integrable function $k$.
Thus $k$ defines a function from $\eusm L^{\infty}(\mu)$ to $\eusm
L^1(\nu)$.   Changing $f$ on a set of measure $0$ will not change $g$,
so we can think of this as an operator from $L^{\infty}(\mu)$ to $\eusm
L^1(\nu)$;  and of course we can move immediately to the equivalence
class of $g$ in $L^1(\nu)$, so getting an operator $T_k$ from
$L^{\infty}(\mu)$ to $L^1(\nu)$.   This operator is plainly linear;
also it is easy to check that $\pm T_k\le T_{|k|}$, so that $T_k\in\eurm
L^{\sim}(L^{\infty}(\mu);L^1(\nu))$, and that $\|T_k\|\le\int|k|$.
Moreover, changing $k$ on a $\lambda$-negligible set does not change
$T_k$, so that in fact we can speak of $T_w$ for any $w\in
L^1(\lambda)$.

I think it is obvious, even before investigating them, that operators
representable in this way will be important.   We can immediately ask
what their properties will be and whether there is any straightforward
way of recognising them.   We can look at the properties of the map
$w\mapsto T_w:L^1(\lambda)\to\eurm L^{\sim}(L^{\infty}(\mu);L^1(\nu))$.
And we can ask what happens when $L^{\infty}(\mu)$ and $L^1(\nu)$ are
replaced by other function spaces, defined by extended Fatou norms or
otherwise.   Theorems 376E and 376H are answers to questions of this kind.

It turns out that the formula $g(y)=\int k(x,y)f(x)dx$ gives rise to a
variety of technical problems, and it is much easier to characterize
$Tu$ in terms of its action on the dual.   In the language of the
special case above, if $h\in\eusm L^{\infty}(\nu)$, then we shall have

\Centerline{$\int k(x,y)f(x)h(y)d(x,y)=\int g(y)h(y)dy$;}

\noindent since $g^{\ssbullet}\in L^1(\nu)$ is entirely determined by
the integrals $\int g(y)h(y)dy$ as $h$ runs over $\eusm
L^{\infty}(\nu)$, we can define the operator $T$ in terms of the
functional $(f,h)\mapsto\int k(x,y)f(x)h(y)d(x,y)$.   This enables us to
extend the results from the case of $\sigma$-finite spaces to general
strictly localizable spaces;  perhaps more to the point in the present
context, it gives them natural expressions in terms of function spaces
defined from measure algebras rather than measure spaces, as in 376E.

Before going farther along this road, however, I give a couple of
results relating the theorems of \S253 to the methods of this volume.
}%end of comment

\leader{376B}{The canonical map $L^0\times L^0\to L^0$:  Proposition}
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be semi-finite measure
algebras, and $(\frak C,\bar\lambda)$ their localizable measure algebra
free product\cmmnt{ (325E)}.   Then we have a bilinear operator
$(u,v)\mapsto u\otimes v:
L^0(\frak A)\times L^0(\frak B)\to L^0(\frak C)$ with the following
properties.

(a) For any $u\in L^0(\frak A)$, $v\in L^0(\frak B)$ and $\alpha\in\Bbb R$,

\Centerline{$\Bvalue{u\otimes\chi 1_{\frak B}>\alpha}
=\Bvalue{u>\alpha}\otimes 1_{\frak B}$,
\quad $\Bvalue{\chi 1_{\frak A}\otimes v>\alpha}
=1_{\frak A}\otimes\Bvalue{v>\alpha}$}

\noindent where for $a\in\frak A$, $b\in\frak B$ I write $a\otimes b$
for the corresponding member of
$\frak A\otimes\frak B$\cmmnt{ (315N)}, identified with a subalgebra
of $\frak C$\cmmnt{ (325Dc)}.

(b)(i) For any $u\in L^0(\frak A)^+$, the map
$v\mapsto u\otimes v:L^0(\frak B)\to L^0(\frak C)$ is an
order-continuous multiplicative Riesz homomorphism.

\quad (ii) For any $v\in L^0(\frak B)^+$, the map
$u\mapsto u\otimes v:L^0(\frak A)\to L^0(\frak C)$ is an
order-continuous multiplicative Riesz homomorphism.

(c) In particular, $|u\otimes v|=|u|\otimes|v|$ for all
$u\in L^0(\frak A)$ and $v\in L^0(\frak B)$.

(d) For any $u\in L^0(\frak A)^+$ and $v\in L^0(\frak B)^+$,
$\Bvalue{u\otimes v>0}=\Bvalue{u>0}\otimes\Bvalue{v>0}$.

\proof{ The canonical maps $a\mapsto a\otimes 1_{\frak B}$,
$b\mapsto 1_{\frak A}\otimes b$ from $\frak A$, $\frak B$ to $\frak C$
are order-continuous Boolean homomorphisms (325Da), so induce
order-continuous multiplicative Riesz homomorphisms from $L^0(\frak A)$
and $L^0(\frak B)$ to $L^0(\frak C)$ (364P);  write $\tilde u$,
$\tilde v$ for the images of
$u\in L^0(\frak A)$, $v\in L^0(\frak B)$.   Observe that
$|\tilde u|=|u|^{\sim}$, $|\tilde v|=|v|^{\sim}$ and
$(\chi 1_{\frak A})^{\sim}=(\chi 1_{\frak B})^{\sim}=\chi 1_{\frak C}$.
Now set $u\otimes v=\tilde u\times\tilde v$.   The properties
listed in (a)-(c) are just a matter of putting the definition in
364Pa together with the fact that $L^0(\frak C)$ is an $f$-algebra
(364D).   As for
$\Bvalue{u\otimes v>0}=\Bvalue{\tilde u\times\tilde v>0}$, this is (for
non-negative $u$, $v$) just

\Centerline{$\Bvalue{\tilde u>0}\Bcap\Bvalue{\tilde v>0}
=(\Bvalue{u>0}\otimes 1_{\frak B})
  \Bcap(1_{\frak A}\otimes\Bvalue{v>0})
=\Bvalue{u>0}\otimes\Bvalue{v>0}$.}
}%end of proof of 376B

\leader{376C}{}\cmmnt{ For $L^1$ spaces we have a similar result, with
additions corresponding to the Banach lattice structures of the three
spaces.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be semi-finite measure algebras with localizable measure algebra free
product $(\frak C,\bar\lambda)$.

(a) If $u\in L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$ and
$v\in L^1_{\bar\nu}=L^1(\frak B,\bar\nu)$ then
$u\otimes v\in L^1_{\bar\lambda}=L^1(\frak C,\bar\lambda)$ and

\Centerline{$\int u\otimes v=\int u\int v$,
\quad $\|u\otimes v\|_1=\|u\|_1\|v\|_1$.}

(b) Let $W$ be a Banach space and
$\phi:L^1_{\bar\mu}\times L^1_{\bar\nu}\to W$ a bounded bilinear operator.
Then there is a unique
bounded linear operator $T:L^1_{\bar\lambda}\to W$ such that
$T(u\otimes v)=\phi(u,v)$ for all $u\in L^1_{\bar\mu}$ and
$v\in L^1_{\bar\nu}$, and $\|T\|=\|\phi\|$.

(c) Suppose, in (b), that $W$ is a Banach lattice.   Then

\quad(i) $T$ is positive iff $\phi(u,v)\ge 0$ for all $u$, $v\ge 0$;

\quad(ii) $T$ is a Riesz homomorphism iff
$u\mapsto\phi(u,v_0):L^1_{\bar\mu}\to W$ and
$v\mapsto\phi(u_0,v):L^1_{\bar\nu}\to W$ are Riesz homomorphisms for all
$v_0\ge 0$ in $L^1_{\bar\nu}$ and $u_0\ge 0$ in $L^1_{\bar\mu}$.

\proof{{\bf (a)} I refer to the proof of 325D.   Let $(X,\Sigma,\mu)$
and $(Y,\Tau,\nu)$ be the Stone spaces of $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ (321K), so that $(\frak C,\bar\lambda)$ can be
identified with the measure algebra of the c.l.d.\ product measure
$\lambda$ on $X\times Y$ (part (a) of the proof of 325D), and
$L^1_{\bar\mu}$, $L^1_{\bar\nu}$, $L^1_{\bar\lambda}$ can be identified
with $L^1(\mu)$, $L^1(\nu)$ and $L^1(\lambda)$ (365B).    Now if
$f\in\eusm L^0(\mu)$ and $g\in\eusm L^0(\nu)$ then
$f\otimes g\in\eusm L^0(\lambda)$ (253Cb), and it is easy to check that
$(f\otimes g)^{\ssbullet}\in L^0(\bar\lambda)$ corresponds to
$f^{\ssbullet}\otimes g^{\ssbullet}$ as defined in 376B.   (Look first
at the cases in which
one of $f$, $g$ is a constant function with value $1$.)     By 253E, we
have a canonical map
$(f^{\ssbullet},g^{\ssbullet})\mapsto(f\otimes g)^{\ssbullet}$ from
$L^1(\mu)\times L^1(\nu)$ to $L^1(\lambda)$, with
$\int f\otimes g=\int f\int g$ (253D);   so that if
$u\in L^1_{\bar\mu}$ and $v\in L^1_{\bar\nu}$ we must have
$u\otimes v\in L^1_{\bar\lambda}$, with $\int u\otimes v=\int u\int v$.
As in 253E, it follows that $\|u\otimes v\|_1=\|u\|_1\|v\|_1$.

\medskip

{\bf (b)} In view of the situation described in (a) above, this is now
just a translation of the same result about $L^1(\mu)$, $L^1(\nu)$ and
$L^1(\lambda)$, which is Theorem 253F.

\medskip

{\bf (c)} Identifying the algebraic free product $\frak A\otimes\frak B$
with its canonical image in $\frak C$ (325Dc), I write
$(\frak A\otimes\frak B)^f$ for
$\{c:c\in\frak A\otimes\frak B,\,\bar\lambda c<\infty\}$,
so that $(\frak A\otimes\frak B)^f$ is a subring of $\frak C$.
Recall that any member of $\frak A\otimes\frak B$ is expressible
as $\sup_{i\le n}a_i\otimes b_i$ where $a_0,\ldots,a_n$ are disjoint
(315Oa);  evidently this will belong to $(\frak A\otimes\frak B)^f$ iff
$\bar\mu a_i\cdot\bar\nu b_i$ is finite for every $i$.

The next fact to lift from previous theorems is in part (e) of the proof
of 253F:  the linear span $M$ of
$\{\chi(a\otimes b):a\in\frak A^f,\,b\in\frak B^f\}$ is norm-dense in
$L^1_{\bar\lambda}$.    Of
course $M$ can also be regarded as the linear span of
$\{\chi c:c\in(\frak A\otimes\frak B)^f\}$, or
$S(\frak A\otimes\frak B)^f$.
(Strictly speaking, this last remark relies on 361J;  the identity map
from $(\frak A\otimes\frak B)^f$ to $\frak C$ induces an injective Riesz
homomorphism from $S(\frak A\otimes\frak B)^f$ into
$S(\frak C)\subseteq L^0(\frak C)$.   To see that $\chi c\in M$ for
every $c\in(\frak A\otimes\frak B)^f$, we need to know that $c$ can be
expressed as a
disjoint union of members of $\frak A\otimes\frak B$, as noted above.)

\medskip

\quad{\bf (i)} If $T$ is positive then of course
$\phi(u,v)=T(u\otimes v)\ge 0$ whenever $u$, $v\ge 0$, since
$u\otimes v\ge 0$.   On the other
hand, if $\phi$ is non-negative on $U^+\times V^+$, then, in particular,
$T\chi(a\otimes b)=\phi(\chi a,\chi b)\ge 0$ whenever
$\bar\mu a\cdot\bar\nu b<\infty$.   Consequently $T(\chi c)\ge 0$ for
every $c\in(\frak A\otimes\frak B)^f$ and $Tw\ge 0$ whenever $w\ge 0$ in
$M\cong S(\frak A\otimes\frak B)^f$, as in 361Ga.

Now this means that $T|w|\ge 0$ whenever $w\in M$.   But as $M$ is
norm-dense in $L^1_{\bar\lambda}$, $w\mapsto T|w|$ is continuous and
$W^+$ is closed, it follows that $T|w|\ge 0$ for every
$w\in L^1_{\bar\lambda}$, that is, that $T$ is positive.

\medskip

\quad{\bf (ii)} If $T$ is a Riesz homomorphism then of course
$u\mapsto\phi(u,v_0)=T(u\otimes v_0)$ and
$v\mapsto\phi(u_0,v)=T(u_0\otimes
v)$ are Riesz homomorphisms for $v_0$, $u_0\ge 0$.   On the other hand,
if all these maps are Riesz homomorphisms, then, in particular,

$$\eqalign{T\chi(a\otimes b)\wedge T\chi(a'\otimes b')
&=\phi(\chi a,\chi b)\wedge\phi(\chi a',\chi b')\cr
&\le \phi(\chi a,\chi b+\chi b')\wedge\phi(\chi a',\chi b+\chi b')\cr
&=\phi(\chi a\wedge\chi a',\chi b+\chi b')
=0\cr}$$

\noindent whenever $a$, $a'\in\frak A^f$, $b$, $b'\in\frak B^f$ and
$a\Bcap a'=0$.   Similarly, $T\chi(a\otimes b)\wedge T\chi(a'\otimes
b')=0$ if $b\Bcap b'=0$.   But this means that $T\chi c\wedge T\chi
c'=0$ whenever $c$, $c'\in(\frak A\otimes\frak B)^f$ and $c\Bcap c'=0$.
\Prf\ Express $c$, $c'$ as $\sup_{i\le m}a_i\otimes b_i$, $\sup_{j\le
n}a'_j\otimes b'_j$ where $a_i$, $a'_j$, $b_i$, $b'_j$ all have finite
measure.   Now if $i\le m$, $j\le n$, $(a_i\Bcap a'_j)\otimes(b_i\Bcap
b'_j)=(a_i\otimes b_i)\Bcap(a'_j\otimes b'_j)=0$, so one of $a_i\Bcap
a'_j$, $b_i\Bcap b'_j$ must be zero, and in either case
$T\chi(a_i\otimes b_i)\wedge T\chi(a'_j\otimes b'_j)=0$.   Accordingly

$$\eqalign{T\chi c\wedge T\chi c'
&\le(\sum_{i=0}^mT\chi(a_i\otimes b_i))
   \wedge(\sum_{j=0}^nT\chi(a'_j\times b'_j))\cr
&\le\sum_{i=0}^m\sum_{j=0}^nT\chi(a_i\otimes b_i)
   \wedge T\chi(a'_j\otimes b'_j)
=0,\cr}$$

\noindent using 352F(a-ii) for the second inequality.\ \Qed

This implies that $T\restr M$ must be a Riesz homomorphism (361Gc), that
is, $T|w|=|Tw|$ for all $w\in M$.   Again because $M$ is dense in
$L^1_{\bar\lambda}$, $T|w|=|Tw|$ for every $w\in L^1_{\bar\lambda}$, and
$T$ is a Riesz homomorphism.
}%end of proof of 376C

\leader{376D}{Abstract integral operators:  Definition}\cmmnt{ The
following concept will be used repeatedly in the theorems below;  it is
perhaps worth giving it a name.}   Let $U$ be a Riesz space and $V$ a
Dedekind complete Riesz space\cmmnt{, so that $\eurm L^{\times}(U;V)$
is a Dedekind complete Riesz space (355H)}.   If $f\in U^{\times}$ and
$v\in V$ write $P_{fv}u=f(u)v$ for each $u\in U$;  then
$P_{fv}\in\eurm L^{\times}(U;V)$.   \prooflet{\Prf\ If $f\ge 0$ in
$U^{\times}$ and $v\ge 0$ in $V^{\times}$ then $P_{fv}$ is a positive
linear operator from $U$ to $V$ which is
order-continuous because if $A\subseteq U$ is non-empty,
downwards-directed and has infimum $0$, then (as $V$ is Archimedean)

\Centerline{$\inf_{u\in A}P_{fv}(u)=\inf_{u\in A}f(u)v=0$.}

\noindent Of course $(f,g)\mapsto P_{fg}$ is bilinear, so
$P_{fv}\in \eurm L^{\times}(U;V)$ for every $f\in U^{\times}$,
$v\in V$.\ \QeD\ }%end of prooflet
\cmmnt{Now} I call a linear operator from $U$ to $V$ an {\bf abstract integral
operator} if it is in the band in $\eurm L^{\times}(U;V)$ generated by
$\{P_{fv}:f\in U^{\times},\,v\in V\}$.

\cmmnt{The first result describes these operators when $U$, $V$ are
expressed as subspaces of $L^0(\frak A)$, $L^0(\frak B)$ for measure
algebras $\frak A$, $\frak B$ and $V$ is perfect.}

\leader{376E}{Theorem} Let $(\frak
A,\bar\mu)$ and $(\frak B,\bar\nu)$ be semi-finite measure algebras,
with localizable measure algebra free product $(\frak C,\bar\lambda)$,
and $U\subseteq L^0(\frak A)$, $V\subseteq L^0(\frak B)$ order-dense
Riesz subspaces.   Write $W$ for the set of those $w\in L^0(\frak C)$
such that $w\times(u\otimes v)$ is integrable for every $u\in U$ and
$v\in V$.   Then we have an operator $w\mapsto T_w:W\to \eurm
L^{\times}(U;V^{\times})$ defined by setting

\Centerline{$T_w(u)(v)=\int w\times(u\otimes v)$}

\noindent for every $w\in W$, $u\in U$ and $v\in V$.   The map
$w\mapsto T_w$ is a Riesz space isomorphism between $W$ and the band of
abstract integral operators in $\eurm L^{\times}(U;V^{\times})$.

\proof{{\bf (a)} The first thing to check is that the formula offered
does define a member $T_w(u)$ of $V^{\times}$ for any $w\in W$ and
$u\in U$.   \Prf\    Of course $T_w(u)$ is a linear operator because
$\int$ is linear and $\otimes$ and $\times$ are bilinear.   It belongs to
$V^{\sim}$ because, writing $g(v)=\int |w|\times(|u|\otimes v)$, $g$ is
a positive linear operator and $|T_w(u)(v)|\le g(|v|)$ for every $v$.
(I am here using 376Bc to see that $|w\times(u\otimes
Fv)|=|w|\times(|u|\otimes|v|)$.)   Also $g\in V^{\times}$ because
$v\mapsto|u|\otimes v$, $w'\mapsto |w|\times w'$ and $\int$ are all
order-continuous;  so $T_w(u)$ also belongs to $V^{\times}$.\ \Qed

\medskip

{\bf (b)} Next, for any given $w\in W$, the map $T_w:U\to V^{\times}$ is
linear (again because $\otimes$ and $\times$ are bilinear).   It is
helpful to note that $W$ is a solid linear subspace of $L^0(\frak C)$.
Now if $w\ge 0$ in $W$, then $T_w\in\eurm L^{\times}(U;V^{\times})$.
\Prf\ If $u$, $v\ge 0$ then $u\otimes v\ge 0$,
$w\times(u\otimes v)\ge 0$ and $T_w(u)(v)\ge 0$;   as $v$ is arbitrary,
$T_w(u)\ge 0$ whenever $u\ge 0$;  as
$u$ is arbitrary, $T_w$ is positive.   If $A\subseteq U$ is non-empty,
downwards-directed and has infimum $0$, then $T_w[A]$ is
downwards-directed, and for any $v\in V^+$

\Centerline{$(\inf T_w[A])(v)=\inf_{u\in A}T_w(u)(v)
=\inf_{u\in A}\int w\times(u\otimes v)=0$}

\noindent because $u\mapsto u\otimes v$ is order-continuous.   So $\inf
T_w[A]=0$;  as $A$ is arbitrary, $T_w$ is order-continuous.\ \Qed\

For general $w\in W$, we now have $T_w=T_{w^+}-T_{w^-}\in
L^{\times}(U;V^{\times})$.

\medskip

{\bf (c)} Ths shows that $w\mapsto T_w$ is a map from $W$ to
$L^{\times}(U;V^{\times})$.   Running through the formulae once again,
it is linear, positive and order-continuous;  this last because, given a
non-empty downwards-directed $C\subseteq W$ with infimum $0$, then for
any $u\in U^+$, $v\in V^+$

\Centerline{$(\inf_{w\in C}T_w)(u)(v)
\le\inf_{w\in C}\int w\times(u\otimes v)=0$}

\noindent (because $\int$ and $\times$ are order-continuous);  as $v$ is
arbitrary, $(\inf_{w\in C}T_w)(u)=0$;  as $u$ is arbitrary, $\inf_{w\in
C}T_w=0$.

\medskip

{\bf (d)} All this is easy, being nothing but a string of applications
of the elementary properties of $\otimes$, $\times$ and $\int$.   But I
think a new idea is needed for the next fact:  the map $w\mapsto
T_w:W\to\eurm L^{\times}(U;V^{\times})$ is a Riesz homomorphism.   \Prf\
Write $\frak D$ for the set of those $d\in\frak C$ such that $T_w\wedge
T_{w'}=0$ whenever $w$, $w'\in W^+$, $\Bvalue{w>0}\Bsubseteq d$ and
$\Bvalue{w'>0}\Bsubseteq 1_{\frak C}\Bsetminus d$.
(i) If $d_1$, $d_2\in \frak D$, $w$, $w'\in W^+$,
$\Bvalue{w>0}\Bsubseteq d_1\Bcup d_2$ and $\Bvalue{w'>0}\Bcap(d_1\Bcup
d_2)=0$, then set $w_1=w\times\chi d_1$, $w_2=w-w_1$.   In this case

\Centerline{$\Bvalue{w_1>0}\Bsubseteq d_1$,
\quad$\Bvalue{w_2>0}\Bsubseteq d_2$,}

\noindent so

\Centerline{$T_{w_1}\wedge T_{w'}=T_{w_2}\wedge T_{w'}=0$,
\quad$T_w\wedge T_{w'}
\le(T_{w_1}\wedge T_{w'})+(T_{w_2}\wedge T_{w'})=0$.}

\noindent As $w$, $w'$ are arbitrary, $d_1\Bcup d_2\in \frak D$.   Thus
$\frak D$ is closed under $\Bcup$.   (ii) The symmetry of the definition
of $\frak D$ means that $1_{\frak C}\Bsetminus d\in \frak D$ whenever
$d\in \frak D$.   (iii) Of course $0\in \frak D$, just because $T_w=0$
if $w\in W^+$ and $\Bvalue{w>0}=0$;  so $\frak D$ is a subalgebra of
$\frak C$.   (iv) If $D\subseteq \frak D$ is non-empty and
upwards-directed, with supremum $c$ in $\frak C$, and if $w$, $w'\in
W^+$ are such that $\Bvalue{w>0}\Bsubseteq c$, $\Bvalue{w'>0}\Bcap c=0$,
then consider $\{w\times\chi d:d\in D\}$.   This is upwards-directed,
with supremum $w$;  so $T_w=\sup_{d\in D}T_{w\times\chi d}$, because the
map $q\mapsto T_q$ is order-continuous.   Also $T_{w\times\chi d}\wedge
T_{w'}=0$ for every $d\in D$, so $T_w\wedge T_{w'}=0$.   As $w$, $w'$
are arbitrary, $c\in\frak D$;  as $D$ is arbitrary, $\frak D$ is an
order-closed subalgebra of $\frak C$.   (v) If $a\in\frak A$ and $w$,
$w'\in W^+$ are such that $\Bvalue{w>0}\Bsubseteq a\otimes 1_{\frak B}$ and
$\Bvalue{w'>0}\Bcap(a\otimes 1_{\frak B})=0$, then any $u\in U^+$ is
expressible as $u_1+u_2$ where $u_1=u\times\chi a$,
$u_2=u\times\chi(1_{\frak A}\Bsetminus a)$.   Now

\Centerline{$T_w(u_2)(v)=\int w\times(u_2\otimes v)
=\int w\times\chi(a\otimes 1_{\frak B})\times(u\otimes v)
  \times\chi((1_{\frak A}\Bsetminus a)\otimes 1_{\frak B})
=0$}

\noindent for every $v\in V$, so $T_w(u_2)=0$.   Similarly,
$T_{w'}(u_1)=0$.   But this means that

\Centerline{$(T_w\wedge T_{w'})(u)\le T_w(u_2)+T_{w'}(u_1)=0$.}

\noindent As $u$ is arbitrary, $T_w\wedge T_{w'}=0$;  as $w$ and $w'$
are arbitrary, $a\otimes 1_{\frak B}\in \frak D$.   (vi)  Now suppose
that $b\in\frak B$ and that $w$, $w'\in W^+$ are such that
$\Bvalue{w>0}\Bsubseteq 1_{\frak A}\otimes b$ and
$\Bvalue{w'>0}\Bcap(1_{\frak A}\otimes b)=0$.   If $u\in U^+$ and
$v\in V^+$ then

\Centerline{$(T_w\wedge T_{w'})(u)(v)
\le\int w\times(u\otimes(v\times\chi(1_{\frak B}\Bsetminus b)))
  +\int w'\times(u\otimes(v\times\chi b))
=0$.}

\noindent As $u$, $v$ are arbitrary, $T_w\wedge T_{w'}=0$;  as $w$ and
$w'$ are arbitrary, $1_{\frak A}\otimes b\in\frak D$.   (vii) This means
that $\frak D$ is an order-closed subalgebra of $\frak C$ including
$\frak A\otimes\frak B$, and is therefore the whole of $\frak C$
(325D(c-ii)).   (viii) Now take any $w$, $w'\in W$ such that
$w\wedge w'=0$, and consider $c=\Bvalue{w>0}$.   Then
$\Bvalue{w'>0}\Bsubseteq 1_{\frak C}\Bsetminus c$ and $c\in\frak D$,
so $T_w\wedge T_{w'}=0$.   This is
what we need to be sure that $w\mapsto T_w$ is a Riesz homomorphism
(352G).\ \Qed

\medskip

{\bf (e)} The map $w\mapsto T_w$ is injective.   \Prf\ (i) If $w>0$ in
$W$, then consider

\Centerline{$A=\{a:a\in\frak A,\,\exists\,u\in U,\,\chi a\le u\}$,
\quad$B=\{b:b\in\frak B,\,\exists\,v\in V,\,\chi b\le v\}$.}

\noindent Because $U$ and $V$ are order-dense in $L^0(\frak A)$ and
$L^0(\frak B)$ respectively, $A$ and $B$ are order-dense in $\frak A$
and $\frak B$.   Also both are upwards-directed.   So
$\sup_{a\in A,b\in B}a\otimes b=1_{\frak C}$ and
$0<\int w=\sup_{a\in A,b\in B}\int_{a\otimes b}w$.   Take $a\in A$,
$b\in B$ such that
$\int_{a\otimes b}w>0$;  then there are $u\in U$, $v\in V$ such that
$\chi a\le u$ and $\chi b\le v$, so that

\Centerline{$T_w(u)(v)\ge\int_{a\otimes b}w>0$}

\noindent and $T_w>0$.   (ii) For general non-zero $w\in W$, we now have
$|T_w|=T_{|w|}>0$ so $T_w\ne 0$.\ \Qed

Thus $w\mapsto T_w$ is an order-continuous injective Riesz homomorphism.

\medskip

{\bf (f)} Write $\tilde W$ for $\{T_w:w\in W\}$, so that $\tilde W$ is a
Riesz subspace of $\eurm L^{\times}(U;V^{\times})$ isomorphic to $W$,
and $\widehat{W}$ for the band it generates in
$\eurm L^{\times}(U;V^{\times})$.   Then $\tilde W$ is order-dense in
$\widehat{W}$.   \Prf\ Suppose that $S>0$ in
$\widehat{W}=\tilde W^{\perp\perp}$ (353Ba).   Then
$S\notin\tilde W^{\perp}$, so
there is a $w\in W$ such that $S\wedge T_w>0$.
Set $w_1=w\wedge\chi 1_{\frak C}$.   Then
$w=\sup_{n\in\Bbb N}w\wedge nw_1$, so
$T_w=\sup_{n\in\Bbb N}T_w\wedge nT_{w_1}$ and $R=S\wedge T_{w_1}>0$.

Set $U_1=U\cap L^1(\frak A,\bar\mu)$.   Because $U$ is an order-dense
Riesz subspace of $L^0(\frak A)$, $U_1$ is an order-dense Riesz subspace
of $L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$, therefore also norm-dense.
Similarly $V_1=V\cap L^1(\frak B,\bar\nu)$ is a norm-dense Riesz
subspace of $L^1_{\bar\nu}=L^1(\frak B,\bar\nu)$.   Define
$\phi_0:U_1\times V_1\to\Bbb R$ by setting $\phi_0(u,v)=R(u)(v)$ for
$u\in U_1$ and $v\in V_1$.   Then $\phi_0$ is bilinear, and

$$\eqalign{|\phi_0(u,v)|
&=|R(u)(v)|\le|R(u)|(|v|)
\le R(|u|)(|v|)
\le T_{w_1}(|u|)(|v|)\cr
&=\biggerint w_1\times(|u|\otimes|v|)
\le\biggerint|u|\otimes|v|
=\|u\|_1\|v\|_1\cr}$$

\noindent for all $u\in U_1$, $v\in V_1$, because $0\le R\le T_{w_1}$ in
$\eurm L^{\times}(U;V^{\times})$.   Because $U_1$, $V_1$ are
norm-dense in $L^1_{\bar\mu}$, $L^1_{\bar\nu}$ respectively, $\phi_0$
has a unique extension to a continuous bilinear operator
$\phi:L^1_{\bar\mu}\times L^1_{\bar\nu}\to\Bbb R$.    (To reduce this to
standard results on linear operators, think of $R$ as a function from
$U_1$ to $V_1^*$;  since every member of $V_1^*$ has a unique extension
to a member of $(L^1_{\bar\nu})^*$, we get a corresponding function
$R_1:U_1\to(L^1_{\bar\nu})^*$ which is continuous and linear, so has a
unique extension to a continuous linear operator
$R_2:L^1_{\bar\mu}\to(L^1_{\bar\nu})^*$, and we set
$\phi(u,v)=R_2(u)(v)$.)

By 376C, there is a unique
$h\in(L^1_{\bar\lambda})^*=L^1(\frak C,\bar\lambda)^*$
such that $h(u\otimes v)=\phi(u,v)$ for every $u\in L^1_{\bar\mu}$ and
$v\in L^1_{\bar\nu}$.   Because $(\frak C,\bar\lambda)$
is localizable, this $h$ corresponds to a $w'\in L^{\infty}(\frak C)$
(365Lc), and

\Centerline{$\int w'\times(u\otimes v)=h(u\otimes v)=\phi_0(u,v)
=R(u)(v)$}

\noindent for every $u\in U_1$, $v\in V_1$.

Because $U_1$ is norm-dense in $L^1_{\bar\mu}$, $U_1^+$ is dense in
$(L^1_{\bar\mu})^+$, and similarly $V_1^+$ is dense in
$(L^1_{\bar\nu})^+$, so $U_1^+\times V_1^+$ is dense in
$(L^1_{\bar\mu})^+\times(L^1_{\bar\nu})^+$;  now $\phi_0$ is
non-negative on $U_1^+\times V_1^+$, so $\phi$ (being continuous) is
non-negative on $(L^1_{\bar\mu})^+\times(L^1_{\bar\nu})^+$.   By 376Cc,
$h\ge 0$ in $(L^1_{\bar\lambda})^*$ and $w'\ge 0$ in
$L^{\infty}(\frak C)$.   In the same way, because
$\phi_0(u,v)\le T_w(u)(v)$ for
$u\in U_1^+$ and $v\in V_1^+$, $w'\le w_1\le w$ in $L^0(\frak C)$, so
$w'\in W$.   We have

\Centerline{$T_{w'}(u)(v)=\int w'\times(u\otimes v)=R(u)(v)$}

\noindent for all $u\in U_1$, $v\in V_1$.   If $u\in U_1^+$, then
$T_{w'}(u)$ and $R(u)$ are both order-continuous, so must be identical,
since $V_1$ is order-dense in $V$.   This means that $T_{w'}$ and $R$
agree on $U_1$.   But as both are themselves
order-continuous linear operators, and $U_1$ is order-dense in $U$, they
must be equal.

Thus $0<T_{w'}\le S$ in $\eurm L^{\times}(U;V^{\times})$.   As $S$ is
arbitrary, $\tilde W$ is quasi-order-dense in $\widehat{W}$, therefore
order-dense (353A).\ \Qed

\medskip

{\bf (g)} Because $w\mapsto T_w:W\mapsto\tilde W$ is an injective Riesz
homomorphism, we have an inverse map $Q:\tilde W\to L^0(\frak C)$,
setting $Q(T_w)=w$;  this is a Riesz homomorphism, and it is
order-continuous because $W$ is solid in $L^0(\frak C)$, so that the
embedding $W\embedsinto L^0(\frak C)$ is order-continuous.   By 368B, $Q$
has an extension to an order-continuous Riesz homomorphism
$\tilde Q:\widehat{W}\to L^0(\frak C)$.
Because $Q(S)>0$ whenever $S>0$ in
$\tilde W$, $\tilde Q(S)>0$ whenever $S>0$ in $\widehat{W}$, so
$\tilde Q$ is injective.
Now $\tilde Q(S)\in W$ for every $S\in\widehat{W}$.
\Prf\ It is enough to look at non-negative $S$.   In this case,
$\tilde Q(S)$ must be $\sup\{\tilde Q(T_w):w\in W,\,T_w\le S\}=\sup C$,
where $C=\{w:T_w\le S\}\subseteq W$.   Take $u\in U^+$ and $v\in V^+$.
Then $\{w\times(u\otimes v):w\in C\}$ is upwards-directed, because $C$ is,
and

\Centerline{$\sup_{w\in C}\int w\times(u\otimes v)
=\sup_{w\in C}T_w(u)(v)
\le S(u)(v)<\infty$.}

\noindent So
$\tilde Q(S)\times(u\otimes v)=\sup_{w\in C}w\times(u\otimes v)$ belongs
to $L^1_{\bar\lambda}$ (365Df).   As $u$ and $v$
are arbitrary, $\tilde Q(S)\in W$.\ \Qed

\medskip

{\bf (h)} Of course this means that $\tilde W=\widehat{W}$ and
$\tilde Q=Q$,
that is, that $w\mapsto T_w:W\mapsto\widehat{W}$ is a Riesz space
isomorphism.

\medskip

{\bf (i)} I have still to check on the identification of $\widehat{W}$
as the band $Z$ of abstract integral operators in
$\eurm L^{\times}(U;V^{\times})$.   Write $P_{fg}(u)=f(u)g$ for
$f\in U^{\times}$, $g\in V^{\times}$ and $u\in U$.

Set

\Centerline{$U^{\#}=\{u:u\in L^0(\frak A),\,u\times u'\in L^1_{\bar\mu}$
for every $u'\in U\}$,}

\Centerline{$V^{\#}=\{v:v\in L^0(\frak B),\,v\times v'\in L^1_{\bar\nu}$
for every $v'\in V\}$.}

\noindent From 369C we know that if we set $f_u(u')=\int u\times u'$ for
$u\in U^{\#}$ and $u'\in U$, then $f_u\in U^{\times}$ for every
$u\in U^{\#}$, and $u\mapsto f_u$ is an isomorphism between $U^{\#}$ and
an order-dense Riesz subspace of $U^{\times}$.   Similarly, setting
$g_v(v')=\int v\times v'$ for $v\in V^{\#}$ and $v'\in V$,
$v\mapsto g_v$
is an isomorphism between $V^{\#}$ and an order-dense Riesz subspace of
$V^{\times}$.

If $u\in U^{\#}$ and $v\in V^{\#}$ then

\Centerline{$\int(u\otimes v)\times(u'\otimes v')
=\int(u\times u')\otimes(v\times v')
=(\int u\times u')(\int v\times v')
=f_u(u')g_v(v')$}

\noindent for every $u'\in U$, $v'\in V$, so $u\otimes v\in W$ and
$T_{u\otimes v}=P_{f_ug_v}$.

Now take $f\in(U^{\times})^+$ and $g\in (V^{\times})^+$.
Set $A=\{u:u\in U^{\#},\,u\ge 0,\,f_u\le f\}$ and
$B=\{v:v\in V^{\#},\,v\ge 0,\,g_v\le g\}$.   These are upwards-directed,
so $C=\{u\otimes v:u\in A,\,v\in B\}$ is upwards-directed in
$L^0(\frak C)$.   Because $\{f_u:u\in U^{\#}\}$ is order-dense in
$U^{\times}$, $f=\sup_{u\in A}f_u$;  by 355Ed,
$f(u')=\sup_{u\in A}f_u(u')$ for every $u'\in U^+$.  Similarly,
$g(v')=\sup_{v\in B}f_v(v')$ for every $v'\in V^+$.

\Quer\ Suppose, if possible, that $C$ is not bounded above in $L^0(\frak
C)$.   Because $\frak C$ and $L^0(\frak C)$ are Dedekind complete,

\Centerline{$c
=\inf_{n\in\Bbb N}\sup_{u\in A,v\in B}\Bvalue{u\otimes v\ge n}$}

\noindent must be non-zero (364L(a-i)).   Because $U$ and $V$ are
order-dense
in $L^0(\frak A)$, $L^0(\frak B)$ respectively,

\Centerline{$1_{\frak A}=\sup\{\Bvalue{u'>0}:u'\in U\}$,
\quad$1_{\frak B}=\sup\{\Bvalue{v'>0}:v'\in V\}$,}

\noindent and there are $u'\in U^+$, $v'\in V^+$ such that
$c\Bcap\Bvalue{u'>0}\otimes\Bvalue{v'>0}\ne 0$, so that
$\int_cu'\otimes v'>0$.   But now, for any $n\in\Bbb N$,

$$\eqalignno{f(u')g(v')
&\ge\sup_{u\in A,v\in B}f_u(u')g_v(v')\cr
&=\sup_{u\in A,v\in B}\int(u\otimes v)\times(u'\otimes v')\cr
&\ge\sup_{u\in A,v\in B}\int((u\otimes v)\wedge n\chi c)\times(u'\otimes
v')\cr
&=\int\sup_{u\in A,v\in B}((u\otimes v)\wedge n\chi c)\times(u'\otimes
v')\cr
\noalign{\noindent (because $w\mapsto\int w\times(u'\otimes v')$ is
order-continuous)}
&=\int(n\chi c)\times(u'\otimes v')
=n\int_cu'\otimes v',\cr}$$

\noindent which is impossible.\ \Bang

Thus $C$ is bounded above in $L^0(\frak C)$, and has a supremum $w\in
L^0(\frak C)$.   If $u'\in U^+$, $v'\in V^+$ then

$$\eqalign{\int w\times(u'\otimes v')
&=\sup_{u\in A,v\in B}\int(u\otimes v)\times(u'\otimes v')\cr
&=\sup_{u\in A,v\in B}f_u(u')g_v(v')
=f(u')g(v')
=P_{fg}(u')(v').\cr}$$

\noindent Thus $w\in W$ and

\Centerline{$P_{fg}=T_w\in\tilde W\subseteq\widehat{W}$.}

\noindent And this is true for any non-negative $f\in U^{\times}$ and
$g\in V^{\times}$.   Of course it follows that $P_{fg}\in\widehat{W}$ for
every $f\in U^{\times}$, $g\in V^{\times}$;  as $\widehat{W}$ is a band,
it must include $Z$.

\medskip

{\bf (j)} Finally, $\widehat{W}\subseteq Z$.   \Prf\ Since
$Z=Z^{\perp\perp}$, it is enough to show that $\widehat{W}\cap
Z^{\perp}=\{0\}$.   Take any $T>0$ in $\widehat{W}$.   There are
$u'_0\in U^+$, $v'_0\in V^+$ such that $T(u'_0)(v'_0)>0$.   So there is
a $v\in V^{\#}$ such that $0\le g_v\le T(u'_0)$ and $g_v(v'_0)>0$, that
is, $\int v\times v'_0>0$.   Because $V$ is order-dense in
$L^0(\frak B)$, there is a $v'_1\in V$ such that
$0<v'_1\le v'_0\times\chi\Bvalue{v>0}$, so that

\Centerline{$0<\int v\times v'_1=g_v(v'_1)\le T(u'_0)(v'_1)$}

\noindent and $\Bvalue{v'_1>0}\Bsubseteq\Bvalue{v>0}$.

Now consider the functional $u'\mapsto h(u')=T(u')(v'_1):U\to\Bbb R$.
This belongs to $(U^{\times})^+$ and $h(u'_0)>0$, so there is a
$u\in U^{\#}$ such that $0\le f_u\le h$ and $f_u(u'_0)>0$.   This time,
$\int u\times u'_0>0$ so (because $U$ is order-dense in $L^0(\frak A)$)
there is a $u'_1\in U$ such that $h(u'_1)>0$ and
$\Bvalue{u'_1>0}\Bsubseteq\Bvalue{u>0}$.

We can express $T$ as $T_w$ where $w\in W^+$.   In this case, we have

\Centerline{$\int w\times(u'_1\otimes v'_1)=T(u'_1)(v'_1)=h(u'_1)>0$,}

\noindent so

$$\eqalign{0
&\ne\Bvalue{w>0}\cap\Bvalue{u'_1\otimes v'_1>0}
=\Bvalue{w>0}\cap(\Bvalue{u'_1>0}\otimes\Bvalue{v'_1>0})\cr
&\Bsubseteq\Bvalue{w>0}\cap(\Bvalue{u>0}\otimes\Bvalue{v>0})
=\Bvalue{w>0}\cap\Bvalue{u\otimes v>0},\cr}$$

\noindent and $w\wedge(u\otimes v)>0$, so

\Centerline{$T_w\wedge P_{f_ug_v}
=T_w\wedge T_{u\otimes v}
=T_{w\wedge(u\otimes v)}>0$.}

\noindent Thus $T\notin Z^{\perp}$.   Accordingly
$\widehat{W}\cap Z^{\perp}=\{0\}$ and
$\widehat{W}\subseteq Z^{\perp\perp}=Z$.\ \Qed

Since we already know that $Z\subseteq\widehat{W}$, this completes the
proof.
}%end of proof of 376E

\vleader{72pt}{376F}{Corollary} Let $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$
be localizable measure algebras, with localizable measure algebra free
product $(\frak C,\bar\lambda)$.   Let $U\subseteq L^0(\frak A)$,
$V\subseteq L^0(\frak B)$ be perfect order-dense solid linear subspaces,
and $T:U\to V$ a linear operator.   Then the following are
equiveridical:

(i) $T$ is an abstract integral operator;

(ii) there is a $w\in L^0(\frak C)$ such that $\int w\times(u\otimes
v')$ is defined and equal to $\int Tu\times v'$ whenever $u\in U$ and
$v'\in L^0(\frak B)$ is such that $v'\times v$ is integrable for every
$v\in V$.

\proof{ Setting $V^{\#}=\{v':v'\in L^0(\frak B),\,v\times v'\in L^1$ for
every $v\in V\}$, we know that we can identify $V^{\#}$ with
$V^{\times}$ and $V$ with $(V^{\#})^{\times}$ (369C).   So the
equivalence of (i) and (ii) is just 376E applied to $V^{\#}$ in place of
$V$.
}%end of proof of 376F

\leader{376G}{Lemma} Let $U$ be a Riesz space, $V$ an Archimedean Riesz
space, $T:U\to V$ a linear operator, $f\in (U^{\sim})^+$ and $e\in V^+$.
Suppose that $0\le Tu\le f(u)e$ for every $u\in U^+$.   Then if
$\sequencen{u_n}$ is a sequence in $U$ such that
$\lim_{n\to\infty}g(u_n)=0$ whenever $g\in U^{\sim}$ and $|g|\le f$,
$\sequencen{Tu_n}$ order*-converges to $0$ in
$V$\cmmnt{ (definition:  367A)}.

\proof{ Let $V_e$ be the solid linear subspace of $V$ generated by $e$;
then $Tu\in V_e$ for every $u\in U$.   We can identify $V_e$ with an
order-dense and norm-dense Riesz subspace of $C(X)$, where $X$ is a
compact Hausdorff space, with $e$ corresponding to $\chi X$ (353N).
For $x\in X$, set $g_x(u)=(Tu)(x)$ for every $u\in U$;  then
$0\le g_x(u)\le f(u)$ for $u\ge 0$, so $|g_x|\le f$ and
$\lim_{n\to\infty}(Tu_n)(x)=0$.   As $x$ is arbitrary,
$\sequencen{Tu_n}$ order*-converges to $0$ in $C(X)$, by 367K, and
therefore in $V_e$, because $V_e$ is
order-dense in $C(X)$ (367E).   But $V_e$, regarded as a subspace of
$V$, is solid, so 367E tells us also that $\sequencen{Tu_n}$
order*-converges to $0$ in $V$.
}%end of proof of 376G

\leader{376H}{Theorem} Let $U$ be a Riesz space and $V$ a
\wsid\ Dedekind complete Riesz space\cmmnt{ (definition: 368N)}.
Suppose that $T\in\eurm L^{\times}(U;V)$.
Then the following are equiveridical:

(i) $T$ is an abstract integral operator;

(ii) whenever $\sequencen{u_n}$ is an order-bounded sequence in $U^+$
and $\lim_{n\to\infty}f(u_n)=0$ for every $f\in U^{\times}$, then
$\sequencen{Tu_n}$ order*-converges to $0$ in $V$;

(iii) whenever $\sequencen{u_n}$ is an order-bounded sequence in $U$
and $\lim_{n\to\infty}f(u_n)=0$ for every $f\in U^{\times}$, then
$\sequencen{Tu_n}$ order*-converges to $0$ in $V$.

\proof{ For $f\in U^{\times}$, $v\in V$ and $u\in U$ set
$P_{fv}(u)=f(u)v$.   Write $Z\subseteq\eurm L^{\times}(U;V)$ for the
band of abstract integral operators.

\medskip

{\bf (a)(i)$\Rightarrow$(iii)} Suppose that $T\in Z^+$, and that
$\sequencen{u_n}$ is an order-bounded sequence in $U$ such that
$\lim_{n\to\infty}f(u_n)=0$ for every $f\in U^{\times}$.   Note that
$\{P_{fv}:f\in U^{\times+},\,v\in V^+\}$ is upwards-directed, so that
$T=\sup\{T\wedge P_{fv}:f\in U^{\times+},\,v\in V^+\}$ (352Va).

Take $u^*\in U^+$ such that $|u_n|\le u^*$ for every $n$, and set
$w=\inf_{n\in\Bbb N}\sup_{m\ge n}Tu_m$, which is defined because
$|Tu_n|\le Tu^*$ for every $n$.   Now $w\le(T-P_{fv})^+(u^*)$ for
every $f\in U^{\times+}$ and $v\in V^+$.   \Prf\ Setting
$T_1=T\wedge P_{fv}$, $w_0=(T-P_{fv})^+(u^*)$ we have

\Centerline{$Tu_n-T_1u_n
\le|T-T_1|(u^*)
=(T-P_{fv})^+(u^*)=w_0$}

\noindent for every $n\in\Bbb N$, so $Tu_n\le w_0+T_1u_n$.   On the
other hand, $0\le T_1u\le f(u)v$ for every $u\in U^+$, so by 376G we
must have $\inf_{n\in\Bbb N}\sup_{m\ge n}T_1u_m=0$.   Accordingly

\Centerline{$w\le w_0+\inf_{n\in\Bbb N}\sup_{m\ge n}T_1u_m=w_0$.   \Qed}

\noindent But as $\inf\{(T-P_{fv})^+:f\in U^{\times+},\,v\in V^+\}=0$,
$w\le 0$.    Similarly (or applying the same argument to
$\sequencen{-u_n}$), $\sup_{n\in\Bbb N}\inf_{n\in\Bbb N}Tu_n\ge 0$ and
$\sequencen{Tu_n}$ order*-converges to zero.

For general $T\in Z$, this shows that $\sequencen{T^+u_n}$ and
$\sequencen{T^-u_n}$ both order*-converge to $0$, so
$\sequencen{Tu_n}$ order*-converges to $0$, by 367C(a-iv).
As $\sequencen{u_n}$ is arbitrary, (iii) is satisfied.

\medskip

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

\medskip

{\bf (c)(ii)$\Rightarrow$(i)} \Quer\ Now suppose, if possible, that (ii)
is satisfied, but that $T\notin Z$.   Because $\eurm L^{\times}(U;V)$ is
Dedekind complete (355H), $Z$ is a projection band (353J), so $T$ is
expressible as
$T_1+T_2$ where $T_1\in Z$, $T_2\in Z^{\perp}$ and $T_2\ne 0$.   At
least one of $T_2^+$, $T_2^-$ is non-zero;  replacing $T$ by $-T$ if
need be, we may suppose that $T_2^+>0$.

Because $T_2^+$, like $T$, belongs to $\eurm L^{\times}(U;V)$, its
kernel $U_0$ is a band in $U$, which cannot be the whole of $U$, and
there is a $u_0>0$ in $U_0^{\perp}$.   In this case $T_2^+u_0>0$;
because $T_2^+\wedge(T_2^-+|T_1|)=0$, there is a $u_1\in[0,u_0]$ such
that $T_2^+(u_0-u_1)+(T_2^-+|T_1|)(u_1)\not\ge T_2^+u_0$, so that

\Centerline{$Tu_1\ge T_2u_1-|T_1|(u_1)\not\le 0$}

\noindent and $Tu_1\ne 0$.   Now this means that the sequence
$(Tu_1,Tu_1,\ldots)$ is not order*-convergent to zero, so there must
be some $f\in U^{\times}$ such that $(f(u_1),f(u_1),\ldots)$ does not
converge to $0$, that is, $f(u_1)\ne 0$;  replacing $f$ by $|f|$ if
necessary, we may suppose that $f\ge 0$ and that $f(u_1)>0$.

By 356H, there is a $u_2$ such that $0<u_2\le u_1$ and $g(u_2)=0$
whenever $g\in U^{\times}$ and $g\wedge f=0$.
Because $0<u_2\le u_0$, $u_2\in U_0^{\perp}$ and $v_0=T_2^+u_2>0$.
Consider $P_{fv_0}\in Z$.   Because $T_2\in Z^{\perp}$, $T_2^+\wedge
P_{fv_0}=0$;  set $S=P_{fv_0}+T_2^-$, so that $T_2^+\wedge S=0$.   Then

\Centerline{$\inf_{u\in[0,u_2]}T_2^+(u_2-u)+Su=0$,
\quad$\sup_{u\in[0,u_2]}T_2^+u-Su=v_0$}

\noindent (use 355Ec for the first equality, and then subtract both
sides from $v_0$).   Now $Su\ge f(u)v_0$ for every $u\ge 0$, so that for
any $\epsilon>0$

\Centerline{$\sup_{u\in[0,u_2],f(u)\ge\epsilon}T_2^+u-Su
\le(1-\epsilon)v_0$}

\noindent and accordingly

\Centerline{$\sup_{u\in[0,u_2],f(u)\le\epsilon}T_2^+u=v_0$,}

\noindent since the join of these two suprema is surely at least $v_0$,
while the second is at most $v_0$.
Note also that

\Centerline{$v_0=\sup_{u\in[0,u_2],f(u)\le\epsilon}T_2^+u
=\sup_{0\le u'\le u\le u_2,f(u)\le\epsilon}T_2u'
=\sup_{0\le u'\le u_2,f(u')\le\epsilon}T_2u'$.}

For $k\in\Bbb N$ set $A_k=\{u:0\le u\le u_2,\,f(u)\le 2^{-k}\}$.   We
know that

\Centerline{$B_k=\{\sup_{u\in I}T_2u:I\subseteq A_k$ is finite$\}$}

\noindent is an upwards-directed set with supremum $v_0$ for each $k$.
Because $V$ is \wsid, we can find a sequence $\sequence{k}{v'_k}$ such
that $v'_k\in B_k$ for every $k$ and $v_1=\inf_{k\in\Bbb N}v'_k>0$.
For each $k$ let $I_k\subseteq A_k$ be a finite set such that
$v'_k=\sup_{u\in I_k}T_2u$.

Because each $I_k$ is finite, we can build a sequence $\sequencen{u'_n}$
in $[0,u_2]$ enumerating each in turn, so that
$\lim_{n\to\infty}f(u'_n)=0$ (since $f(u)\le 2^{-k}$ if $u\in I_k$)
while $\sup_{m\ge n}T_2u'_m\ge v_1$ for every $n$ (since $\{u'_m:m\ge
n\}$ always includes some $I_k$).   Now $\sequencen{T_2u'_n}$ does not
order*-converge to $0$.

However, $\lim_{n\to\infty}g(u'_n)=0$ for every $g\in U^{\times}$.
\Prf\ Express $|g|$ as $g_1+g_2$ where $g_1$ belongs to the band of
$U^{\times}$ generated by $f$ and $g_2\wedge f=0$ (353Ic).   Then
$g_2(u'_n)=g_2(u_2)=0$ for every $n$, by the choice of $u_2$.   Also
$g_1=\sup_{n\in\Bbb N}g_1\wedge nf$ (352Vb);  so, given
$\epsilon>0$, there is an $m\in\Bbb N$ such that
$(g_1-mf)^+(u_2)\le\epsilon$ and $(g_1-mf)^+(u'_n)\le\epsilon$ for every
$n\in\Bbb N$.   But this means that

\Centerline{$|g(u'_n)|\le|g|(u'_n)\le\epsilon+mf(u'_n)$}

\noindent for every $n$, and $\limsup_{n\to\infty}|g(u'_n)|\le\epsilon$;
as $\epsilon$ is arbitrary, $\lim_{n\to\infty}g(u'_n)=0$.\ \Qed

Now, however, part (a) of this proof tells us that $\sequencen{T_1u'_n}$
is order*-convergent to $0$, because $T_1\in Z$, while
$\sequencen{Tu'_n}$ is order*-convergent to $0$, by hypothesis;  so
$\sequencen{T_2u'_n}=\sequencen{Tu'_n-T_1u'_n}$ order*-converges to
$0$.\ \Bang

This contradiction shows that every operator satisfying the condition
(ii) must be in $Z$.
}%end of proof of 376H

\leader{376I}{}\cmmnt{ The following elementary remark will be useful
for the next corollary and also for Theorem 376S.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure
space and $U$ an order-dense solid linear subspace of $L^0(\mu)$.   Then
there is a non-decreasing sequence $\sequencen{X_n}$ of measurable
subsets of $X$, with union $X$, such that $\chi X_n^{\ssbullet}\in U$
for every $n\in\Bbb N$.

\proof{ Write $\frak A$ for the measure
algebra of $\mu$, so that $L^0(\mu)$ can be identified with
$L^0(\frak A)$ (364Ic).   $A=\{a:a\in\frak A\setminus\{0\},\,\chi a\in U\}$
is order-dense in $\frak A$, so includes a partition of unity
$\langle a_i\rangle_{i\in I}$.
Because $\mu$ is $\sigma$-finite, $\frak A$ is
ccc (322G) and $I$ is countable, so we can take $I$ to be a subset of
$\Bbb N$.   Choose $E_i\in\Sigma$ such that $E_i^{\ssbullet}=a_i$ for
$i\in I$;  set $E=X\setminus\bigcup_{i\in I}E_i$,
$X_n=E\cup\bigcup_{i\in I,i\le n}E_i$ for $n\in\Bbb N$.
}%end of proof of 376I

\leader{376J}{Corollary} Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces, with product measure $\lambda$ on
$X\times Y$.   Let $U\subseteq L^0(\mu)$,
$V\subseteq L^0(\nu)$ be perfect order-dense solid linear subspaces, and
$T:U\to V$ a linear operator.   Write
$\eusm U=\{f:f\in\eusm L^0(\mu),\,f^{\ssbullet}\in U\}$,
$\eusm V^{\#}=\{h:h\in\eusm L^0(\nu),\,h^{\ssbullet}\times v\in L^1$ for
every $v\in V\}$.   Then the following are equiveridical:

(i) $T$ is an abstract integral operator;

(ii) there is a $k\in\eusm L^0(\lambda)$ such that

\quad ($\alpha$) $\int|k(x,y)f(x)h(y)|d(x,y)<\infty$ for every
$f\in\eusm U$, $h\in\eusm V^{\#}$,

\quad ($\beta$) if $f\in\eusm U$ and we set $g(y)=\int k(x,y)f(x)dx$
wherever this is defined, then $g\in\eusm L^0(\nu)$ and
$Tf^{\ssbullet}=g^{\ssbullet}$;

(iii) $T\in\eurm L^{\sim}(U;V)$ and whenever $\sequencen{u_n}$ is an
order-bounded sequence in $U^+$ and $\lim_{n\to\infty}h(u_n)=0$ for
every $h\in U^{\times}$, then $\sequencen{Tu_n}$ order*-converges to
$0$ in $V$.

\cmmnt{\medskip

\noindent{\bf Remark} I write `$d(x,y)$' above to indicate integration
with respect to the product measure $\lambda$.   Recall that in the
terminology of \S251, $\lambda$ can be taken to be either the
`primitive' or `c.l.d.' product measure (251K).}

\proof{ The idea is of course to identify $L^0(\mu)$ and $L^0(\nu)$ with
$L^0(\frak A)$ and $L^0(\frak B)$, where $(\frak A,\bar\mu)$ and $(\frak
B,\bar\nu)$ are the measure algebras of $\mu$ and $\nu$, so that their
localizable measure algebra free product can be identified with the
measure algebra of $\lambda$ (325E), while
$V^{\#}=\{h^{\ssbullet}:h\in\eusm V^{\#}\}$ can be identified with
$V^{\times}$, because $(T,\Tau,\nu)$ is localizable (see the last sentence
in 369C).

\medskip

{\bf (a)(i)$\Rightarrow$(ii)}
By 376F, there is a $w\in L^0(\lambda)$ such
that $\int w\times(u\otimes v')$ is defined and equal to
$\int Tu\times v'$ whenever $u\in U$ and $v'\in V^{\#}$.   Express $w$ as
$k^{\ssbullet}$ where $k\in\eusm L^0(\lambda)$.   If $f\in\eusm U$ and
$h\in\eusm V^{\#}$ then $\int|k(x,y)f(x)h(y)|d(x,y)
=\int|w\times(f^{\ssbullet}\otimes h^{\ssbullet}|$ is finite, so
(ii-$\alpha$) is satisfied.

Now take any $f\in\eusm U$, and set $g(y)=\int k(x,y)f(x)dx$ whenever
this is defined in $\Bbb R$.
Write $\Cal F$ for the set of those $F\in\Tau$ such that
$\chi F\in \eusm V^{\#}$.   Then for any $F\in\Cal F$, $g$ is defined
almost everywhere in $F$ and $g\restr F$ is $\nu$-virtually measurable.
\Prf\ $\int k(x,y)f(x)\chi F(y)d(x,y)$ is defined in $\Bbb R$, so by
Fubini's theorem (252B, 252C) $g_F(y)=\int k(x,y)f(x)\chi F(y)dx$ is
defined for almost every $y$, and is
$\nu$-virtually measurable;  now $g\restr F=g_F\restr F$.\ \Qed\
Next, there is a sequence $\sequencen{F_n}$ in $\Cal F$ with union $Y$,
by 376I, because $V$ is perfect and order-dense, so $V^{\#}$ must also
be order-dense in $L^0(\nu)$.

For each $n\in\Bbb N$, there is a measurable set $F'_n\subseteq
F_n\cap\dom g$ such that $g\restr F_n$ is measurable and $F_n\setminus
F'_n$ is negligible.   Setting $G=\bigcup_{n\in\Bbb N}F'_n$, $G$ is
conegligible and $g\restr G$ is measurable, so $g\in\eusm L^0(\nu)$.

If $\tilde g\in L^0(\nu)$ represents $Tu\in L^0(\nu)$, then for any
$F\in\Cal F$

\Centerline{$\int_F\tilde g=\int Tu\times(\chi F)^{\ssbullet}
=\int_Fg$.}

\noindent In particular, this is true whenever $F\in\Tau$ and
$F\subseteq F_n$.   So $g$ and $\tilde g$ agree almost everywhere in
$F_n$, for each $n$, and $g\eae\tilde g$.   Thus $g$ also represents
$Tu$, as required in (ii-$\beta$).

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Set $w=k^{\ssbullet}$ in $L^0(\lambda)$.
If $f\in\eusm U$ and $h\in\eusm V^{\#}$ the hypothesis ($\alpha$)
tells us that $(x,y)\mapsto k(x,y)f(x)h(y)$ is integrable (because it
surely belongs to $\eusm L^0(\lambda)$).   By Fubini's theorem,

\Centerline{$\int k(x,y)f(x)h(y)d(x,y)
=\int g(y)h(y)dy$}

\noindent where $g(y)=\int k(x,y)f(x)dx$ for almost every $y$, so that
$Tf^{\ssbullet}=g^{\ssbullet}$, by ($\beta$).   But this means that,
setting $u=f^{\ssbullet}$ and $v'=h^{\ssbullet}$,

\Centerline{$\int w\times(u\otimes v')=\int Tu\times v'$;}

\noindent and this is true for every $u\in U$, $v'\in V^{\#}$.

Thus $T$ satisfies the condition 376F(ii), and is an abstract integral
operator.

\medskip

{\bf (b)(i)$\Rightarrow$(iii)} Because $V$ is \wsid\ (368S), this is
covered by 376H(i)$\Rightarrow$(iii).

\medskip

{\bf (c)(iii)$\Rightarrow$(i)} Suppose that $T$ satisfies (iii).   The
point is that $T^+$ is order-continuous.   \Prf\Quer\
Otherwise, let $A\subseteq U$ be a
non-empty downwards-directed set, with infimum $0$, such that
$v_0=\inf_{u\in A}T^+(u)>0$.   Let $\sequencen{X_n}$ be a non-decreasing
sequence of sets of finite measure covering $X$, and set
$a_n=X_n^{\ssbullet}$ for each $n$.   For each $n$,
$\inf_{u\in A}\Bvalue{u>2^{-n}}=0$, so we can find $\tilde u_n\in A$
such that $\bar\mu(a_n\Bcap\Bvalue{\tilde u_n>2^{-n}})\le 2^{-n}$.   Set
$u_n=\inf_{i\le n}\tilde u_i$ for each $n$;  then $\sequencen{u_n}$ is
non-increasing and has infimum $0$;  also, $[0,u_n]$ meets $A$ for each
$n$, so that $v_0\le\sup\{Tu:0\le u\le u_n\}$ for each $n$.   Because
$V$ is \wsid, we can find a sequence $\sequencen{I_n}$
of finite sets such that $I_n\subseteq[0,u_n]$ for each $n$ and
$v_1=\inf_{n\in\Bbb N}\sup_{u\in I_n}(Tu)^+>0$.   Enumerating
$\bigcup_{n\in\Bbb N}I_n$ as $\sequencen{u'_n}$, as in part (c) of the
proof of 376H, we see that $\sequencen{u'_n}$ is order-bounded and
$\lim_{n\to\infty}f(u'_n)=0$ for every $f\in U^{\times}$ (indeed,
$\sequencen{u'_n}$ order*-converges to $0$ in $U$), while
$\sequencen{Tu'_n}\not\to^* 0$ in $V$.\ \Bang\Qed

Similarly, $T^-$ is order-continuous, so $T\in\eurm L^{\times}(U;V)$.
Accordingly $T$ is an abstract integral operator by condition (ii) of
376H.
}%end of proof of 376J

\leader{376K}{}\cmmnt{ As an application of the ideas above, I give
a result due to N.Dunford (376N) which was one of the inspirations
underlying the theory.   Following the method of {\smc Zaanen 83},  I
begin with a couple of elementary lemmas.

\wheader{376K}{4}{2}{2}{60pt}

\noindent}{\bf Lemma} Let $U$ and $V$ be Riesz spaces.   Then there is a
Riesz space isomorphism $T\mapsto T':\eurm
L^{\times}(U;V^{\times})\to\eurm L^{\times}(V;U^{\times})$ defined by
the formula

\Centerline{$(T'v)(u)=(Tu)(v)$ for every $u\in U$, $v\in V$.}

\noindent If we write
$P_{fg}(u)=f(u)g$ for $f\in U^{\times}$, $g\in V^{\times}$ and $u\in U$,
then $P_{fg}\in\eurm L^{\times}(U;V^{\times})$ and $P'_{fg}=P_{gf}$ in
$\eurm L^{\times}(V;U^{\times})$.
Consequently $T$ is an abstract integral operator iff $T'$ is.

\proof{ All the ideas involved have already appeared.   For positive
$T\in \eurm L^{\times}(U;V^{\times})$ the functional $(u,v)\mapsto
(Tu)(v)$ is bilinear and order-continuous in each variable separately;
so (just as in the first part of the proof of 376E) corresponds to a
$T'\in\eurm L^{\times}(V;U^{\times})$.   The map $T\mapsto T':
\eurm L^{\times}(U;V^{\times})^+\to\eurm L^{\times}(V;U^{\times})^+$ is
evidently an additive, order-preserving bijection, so extends to an
isomorphism between $\eurm L^{\times}(U;V^{\times})$ and $\eurm
L^{\times}(V;U^{\times})$ given by the same formula.   I remarked in
part (i) of the proof of 376E that every $P_{fg}$ belongs to $\eurm
L^{\times}(U;V^{\times})$, and the identification $P'_{fg}=P_{gf}$ is
just a matter of checking the formulae.   Of course it follows at once
that the bands of abstract integral operators must also be matched by
the map $T\mapsto T'$.
}%end of proof of 376K

\leader{376L}{Lemma} Let $U$ be a Banach lattice with an order-continuous
norm.   If $w\in U^+$ there is a $g\in(U^{\times})^+$ such that for every
$\epsilon>0$ there is a $\delta>0$ such that $\|u\|\le\epsilon$ whenever
$0\le u\le w$ and $g(u)\le\delta$.

\woddheader{376L}{0}{0}{0}{24pt} %p243

\proof{{\bf (a)}  As remarked in 356D, $U^*=U^{\sim}=U^{\times}$.
Set

\Centerline{$A=\{v:v\in U$ and there is an $f\in(U^{\times})^+$ such
that $f(u)>0$ whenever $0<u\le|v|\}$.}

\noindent Then $v'\in A$ whenever $|v'|\le|v|\in A$ and $v+v'\in A$ for
all $v$, $v'\in A$ (if $f(u)>0$ whenever $0<u\le|v|$ and $f'(u)>0$
whenever $0<u\le|v'|$, then $(f+f')(u)>0$ whenever $0<u\le|v+v'|$);
moreover, if $v_0>0$ in $U$, there is a $v\in A$ such that $0<v\le v_0$.
\Prf\ Because $U^{\times}=U^*$ separates the points of $U$, there is a
$g>0$ in $U^{\times}$ such that $g(v_0)>0$;  now by 356H there is a
$v\in\ocint{0,v_0}$ such that $g$ is strictly positive on $\ocint{0,v}$,
so that $v\in A$.\ \QeD\   But this means that $A$ is an order-dense
solid linear subspace of $U$.

\medskip

{\bf (b)} In fact $w\in A$.   \Prf\ $w=\sup B$, where $B=A\cap[0,w]$.
Because $B$ is upwards-directed, $w\in\overline{B}$ (354Ea), and there
is a sequence $\sequencen{u'_n}$
in $B$ converging to $w$ for the norm.   For each $n$, choose
$f_n\in(U^{\times})^+$
such that $f_n(u)>0$ whenever $0<u\le u'_n$.   Set

\Centerline{$f=\sum_{n=0}^{\infty}\Bover1{2^{n}(1+\|f_n\|)}f_n$}

\noindent in $U^*=U^{\times}$.   Then
whenever $0<u\le w$ there is some $n\in\Bbb N$ such that $u\wedge
u'_n>0$, so that $f_n(u)>0$ and $f(u)>0$.   So $f$ witnesses that $w\in
A$.\ \Qed

\medskip

{\bf (c)} Take $g\in(U^{\times})^+$ such that $g(u)>0$ whenever
$0<u\le w$.   This $g$ serves.   \Prf\Quer\  Otherwise, there is some
$\epsilon>0$ such that for every $n\in\Bbb N$ we can find a
$u_n\in[0,w]$ with $g(u_n)\le 2^{-n}$ and $\|u_n\|\ge\epsilon$.   Set
$v_n=\sup_{i\ge n}u_i$;  then $0\le v_n\le w$, $g(v_n)\le 2^{-n+1}$ and
$\|v_n\|\ge\epsilon$ for every $n\in\Bbb N$.   But $\sequencen{v_n}$ is
non-decreasing, so $v=\inf_{n\in\Bbb N}v_n$ must be non-zero, while
$0\le v\le w$ and $g(v)=0$;  which is impossible.\ \Bang\Qed

Thus we have found an appropriate $g$.
}%end of proof of 376L

\leader{376M}{Theorem} (a) Let $U$ be a Banach lattice with an
order-continuous norm and $V$ a Dedekind complete $M$-space.
Then every bounded linear operator from $U$ to $V$ is an abstract
integral operator.

(b) Let $U$ be an $L$-space and $V$ a Banach lattice with
order-continuous
norm.   Then every bounded linear operator from $U$ to $V^{\times}$ is
an abstract integral operator.

\proof{{\bf (a)} By 355Kb and 355C,
$\eurm L^{\times}(U;V)=\eurm L^{\sim}(U;V)\subseteq\eurm B(U;V)$;  but
since norm-bounded sets in $V$ are also order-bounded,
$\{Tu:|u|\le u_0\}$ is bounded above in $V$ for every
$T\in\eurm B(U;V)$ and $u_0\in U^+$, and
$\eurm B(U;V)=\eurm L^{\times}(U;V)$.

I repeat ideas from the proof of 376H.   (I cannot quote 376H directly
as I am not assuming that $V$ is \wsid.)   \Quer\ Suppose, if possible,
that $\eurm B(U;V)$ is not the band $Z$ of abstract integral operators.
In this case there is a $T>0$ in
$Z^{\perp}$.   Take $u_1\ge 0$ such that $v_0=Tu_1$ is non-zero.
Let $f\ge 0$ in
$U^{\times}$ be such that for every $\epsilon>0$ there is a $\delta>0$
such that $\|u\|\le\epsilon$ whenever $0\le u\le u_1$ and
$f(u)\le\delta$ (376L).   Then, just as in part (c) of
the proof of 376H,

\Centerline{$\sup_{u\in[0,u_1],f(u)\le\delta}Tu=v_0$}

\noindent for every $\delta>0$.   But there is a $\delta>0$ such that
$\|T\|\|u\|\le\bover12\|v_0\|$ whenever $0\le u\le u_1$ and
$f(u)\le\delta$;  in which case
$\|\sup_{u\in[0,u_1],f(u)\le\delta}Tu\|\le\bover12\|v_0\|$, which is
impossible.\ \Bang

Thus $Z=\eurm B(U;V)$, as required.

\medskip

{\bf (b)} Because $V$ has an order-continuous norm,
$V^*=V^{\times}=V^{\sim}$;  and the norm of $V^*$ is a Fatou norm with
the Levi property (356Da).   So
$\eurm B(U;V^*)=\eurm L^{\times}(U;V^{\times})$, by 371C.
By 376K, this is canonically
isomorphic to $\eurm L^{\times}(V;U^{\times})$.   Now $U^{\times}=U^*$
is an $M$-space (356Pb).   By (a), every member of
$\eurm L^{\times}(V;U^{\times})$ is an abstract integral operator;
but the isomorphism between $\eurm L^{\times}(V;U^{\times})$ and
$\eurm L^{\times}(U;V^{\times})$ matches the abstract integral operators in
each space (376K), so
every member of $\eurm B(U;V^*)$ is also an abstract integral operator,
as claimed.
}%end of proof of 376M

\leader{376N}{Corollary:  Dunford's theorem} Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be $\sigma$-finite measure spaces and
$T:L^1(\mu)\to L^p(\nu)$ a bounded linear operator, where $1<p\le\infty$.
Then there is a measurable function $k:X\times Y\to\Bbb R$ such that
$Tf^{\ssbullet}=g_f^{\ssbullet}$, where
$g_f(y)=\int k(x,y)f(x)dx$ almost everywhere, for every
$f\in\eusm L^1(\mu)$.

\proof{ Set $q=\bover{p}{p-1}$ if $p$ is finite, $1$ if $p=\infty$.   We
can identify $L^p(\nu)$ with $V^{\times}$, where
$V=L^q(\nu)\cong L^p(\nu)^{\times}$ (366Dc, 365Lc) has an
order-continuous norm because $1\le q<\infty$.   By 376Mb, $T$ is an
abstract integral operator.   By 376F/376J, $T$ is
represented by a kernel, as claimed.
}%end of proof of 376N

\leader{376O}{}\cmmnt{ Under the right conditions, weakly compact
operators are abstract integral operators.

\medskip

\noindent}{\bf Lemma} Let $U$ be a Riesz space, and $W$ a solid linear
subspace of $U^{\sim}$.   If $C\subseteq U$ is relatively compact for
the weak topology $\frak T_s(U,W)$\cmmnt{ (3A5E)}, then for
every $g\in W^+$ and $\epsilon>0$ there is a $u^*\in U^+$ such that
$g(|u|-u^*)^+\le\epsilon$ for every $u\in C$.

\proof{ Let $W_g$ be the solid linear subspace of $W$ generated by $g$.
Then $W_g$ is an Archimedean Riesz space with order unit, so
$W_g^{\times}$ is a band in the $L$-space $W_g^*=W_g^{\sim}$ (356Na),
and is therefore an $L$-space in its own right (354O).   For $u\in U$,
$h\in W^{\times}_g$ set $(Tu)(h)=h(u)$;  then $T$ is an order-continuous
Riesz homomomorphism from $U$ to $W^{\times}_g$ (356F).

Now $W_g$ is perfect.
\Prf\ I use 356K.   $W_g$ is Dedekind complete because it is a solid
linear subspace of the Dedekind complete space $U^{\sim}$.
$W_g^{\times}$ separates the points of $W$ because $T[U]$ does.
If $A\subseteq W_g$ is
upwards-directed and $\sup_{h\in A}\phi(h)$ is finite for every
$\phi\in W_g^{\times}$, then $A$ acts on $W_g^{\times}$ as a set of
bounded linear functionals which, by the Uniform Boundedness Theorem
(3A5Ha), is uniformly bounded;  that is, there is some $M\ge 0$ such
that
$\sup_{h\in A}|\phi(h)|\le M\|\phi\|$ for every $\phi\in W_g^{\times}$.
Because $g$ is the standard order unit of $W_g$, we have
$\|\phi\|=|\phi|(g)$ and $|\phi(h)|\le M|\phi|(g)$ for every
$\phi\in W_g^{\times}$ and $h\in A$.   In particular,

\Centerline{$h(u)\le|h(u)|=|(Tu)(h)|\le M|Tu|(g)=M(Tu)(g)=Mg(u)$}

\noindent for every $h\in A$ and $u\in U^+$.   But this means that
$h\le Mg$ for every $h\in A$ and $A$ is bounded above in $W_g$.   Thus
all the conditions of 356K are satisfied and $W_g$ is perfect.\ \Qed

Accordingly $T$ is continuous for the topologies $\frak T_s(U,W)$ and
$\frak T_s(W_g^{\times},W_g^{\times\times})$, because every element
$\phi$ of $W_g^{\times\times}$ corresponds to a member of
$W_g\subseteq W$, so 3A5Ec applies.

Now we are supposing that $C$ is relatively compact for
$\frak T_s(U,W)$, that is, is included in some compact set $C'$;
accordingly $T[C']$ is compact and $T[C]$ is relatively compact for
$\frak T_s(W_g^{\times},W_g^{\times\times})$.   Since $W_g^{\times}$ is
an $L$-space, $T[C]$ is uniformly integrable (356Q);  consequently
(ignoring the trivial case $C=\emptyset$) there are
$\phi_0,\ldots,\phi_n\in T[C]$ such that
$\|(|\phi|-\sup_{i\le n}|\phi_i|)^+\|\le\epsilon$ for every
$\phi\in T[C]$ (354Rb), so that
$(|\phi|-\sup_{i\le n}|\phi_i|)^+(g)\le\epsilon$ for every
$\phi\in T[C]$.

Translating this back into terms of $C$ itself, and recalling that $T$
is a Riesz homomorphism, we see that there are $u_0,\ldots,u_n\in C$
such that $g(|u|-\sup_{i\le n}|u_i|)^+\le\epsilon$ for every $u\in C$.
Setting $u^*=\sup_{i\le n}|u_i|$ we have the result.
}%end of proof of 376O

\leader{376P}{Theorem} Let $U$ be an $L$-space and $V$ a perfect Riesz
space.   If $T:U\to V$ is a linear operator such that
$\{Tu:u\in U,\,\|u\|\le 1\}$ is relatively compact for the weak topology
$\frak T_s(V,V^{\times})$, then $T$ is an abstract integral operator.

\proof{{\bf (a)} For any $g\ge 0$ in $V^{\times}$,
$M_g=\sup_{\|u\|\le 1}g(|Tu|)$ is finite.   \Prf\ By 376O, there is a
$v^*\in V^+$ such that
$g(|Tu|-v^*)^+\le 1$ whenever $\|u\|\le 1$;  now $M_g\le g(v^*)+1$.\
\Qed\   Considering $\|u\|^{-1}u$, we see that $g(|Tu|)\le M_g\|u\|$ for
every $u\in U$.

Next, we find that $T\in\eurm L^{\sim}(U;V)$.   \Prf\ Take $u\in U^+$.
Set

\Centerline{$B=\{\sum_{i=0}^n|Tu_i|:u_0,\ldots,u_n\in U^+,\,
\sum_{i=0}^nu_i=u\}\subseteq V^+$.}

\noindent Then $B$ is upwards-directed.   (Cf.\ 371A.)   If $g\ge 0$ in
$V^{\times}$,

$$\eqalign{\sup_{v\in B}g(v)
&=\sup\{\sum_{i=0}^ng(|Tu_i|):\sum_{i=0}^nu_i=u\}\cr
&\le\sup\{\sum_{i=0}^nM_g\|u_i\|:\sum_{i=0}^nu_i=u\}
=M_g\|u\|\cr}$$

\noindent is finite.   By 356K, $B$ is bounded above in $V$;  and of
course any upper bound for $B$ is also an upper bound for
$\{Tu':0\le u'\le u\}$.   As $u$ is arbitrary, $T$ is order-bounded.\
\Qed

Because $U$ is a Banach lattice with an order-continuous norm,
$T\in\eurm L^{\times}(U;V)$ (355Kb).

\medskip

{\bf (b)} Since we can identify $\eurm L^{\times}(U;V)$ with
$\eurm L^{\times}(U;V^{\times\times})$, we have an adjoint operator
$T'\in\eurm L^{\times}(V^{\times};U^{\times})$, as in 376K.   Now if
$g\ge 0$ in
$V^{\times}$ and $\sequencen{g_n}$ is a sequence in $[0,g]$ such that
$\lim_{n\to\infty}g_n(v)=0$ for every $v\in V$, $\sequencen{T'g_n}$
order*-converges to $0$ in $U^{\times}$.   \Prf\ For any
$\epsilon>0$, there is a
$v^*\in V^+$ such that $g(|Tu|-v^*)^+\le\epsilon$ whenever $\|u\|\le 1$;
consequently

$$\eqalign{\|T'g_n\|
&=\sup_{\|u\|\le 1}(T'g_n)(u)
=\sup_{\|u\|\le 1}g_n(Tu)\cr
&\le g_n(v^*)+\sup_{\|u\|\le 1}g_n(|Tu|-v^*)^+\cr
&\le g_n(v^*)+\sup_{\|u\|\le 1}g(|Tu|-v^*)^+
\le g_n(v^*)+\epsilon\cr}$$

\noindent for every $n\in\Bbb N$.   As $\lim_{n\to\infty}g_n(v^*)=0$,
$\limsup_{n\to\infty}\|T'g_n\|\le\epsilon$;  as $\epsilon$ is arbitrary,
$\sequencen{\|T'g_n\|}\to 0$.   But as $U^{\times}$ is an $M$-space
(356Pb), it follows that $\sequencen{T'g_n}$ order*-converges to
$0$.\ \Qed

By 368Pc, $U^{\times}$ is \wsid.   By 376H, $T'$ is an abstract integral
operator, so $T$ also is, by 376K.
}%end of proof of 376P

\leader{376Q}{Corollary} Let $(X,\Sigma,\mu)$ and
$(Y,\Tau,\nu)$ be $\sigma$-finite measure spaces and
$T:L^1(\mu)\to L^1(\nu)$ a weakly compact linear operator.
Then there is a function $k:X\times Y\to\Bbb R$ such that
$Tf^{\ssbullet}=g_f^{\ssbullet}$, where
$g_f(y)=\int k(x,y)f(x)dx$ almost everywhere, for every
$f\in\eusm L^1(\mu)$.

\proof{ This follows from 376P and 376J, just as in 376N.
}%end of proof of 376Q

\leader{376R}{}\cmmnt{ So far I have mentioned actual kernel functions
$k(x,y)$ only as a way of giving slightly more concrete form to the
abstract kernels of 376E.   But of course they can provide new
structures and insights.   I give one result as an example.   The
following lemma is useful.

\medskip

\noindent}{\bf Lemma} Let $(X,\Sigma,\mu)$ be a measure space,
$(Y,\Tau,\nu)$ a $\sigma$-finite measure space, and $\lambda$ the
c.l.d.\ product measure on $X\times Y$.   Suppose that $k$ is a
$\lambda$-integrable real-valued function.   Then for any $\epsilon>0$
there is a finite partition $E_0,\ldots,E_n$ of $X$ into measurable sets
such that
$\|k-k_1\|_1\le\epsilon$, where

$$\eqalign{k_1(x,y)&=\Bover1{\mu E_i}\int_{E_i}k(t,y)dt
\text{ whenever }x\in E_i,\,0<\mu E_i<\infty\cr
&\hskip12em\text{ and the integral is defined in }\Bbb R,\cr
&=0\text{ in all other cases}.\cr}$$

\proof{ Once again I refer to the proof of 253F:  there are sets
$H_0,\ldots,H_r$ of finite measure in $X$, sets $F_0,\ldots,F_r$ of
finite measure in $Y$, and $\alpha_0,\ldots,\alpha_r$ such that
$\|k-k_2\|_1\le\bover12\epsilon$, where
$k_2=\sum_{j=0}^r\alpha_i\chi(H_j\times F_j)$.   Let $E_0,\ldots,E_n$ be
the partition of $X$ generated by $\{H_i:i\le r\}$.    Then for any
$i\le n$, $\int_{E_i\times Y}|k-k_1|$ is defined and is at most
$2\int_{E_i\times Y}|k-k_2|$.   \Prf\ If $\mu E_i=0$, this is trivial,
as both are zero.   If $\mu E_i=\infty$, then again the result is
elementary, since both $k_1$ and $k_2$ are zero on $E_i\times Y$.   So
let us suppose that $0<\mu E_i<\infty$.   In this case
$\int_{E_i}k(t,y)dt$ must be defined for almost every $y$, by Fubini's
theorem.   So $k_1$ is defined
almost everywhere in $E_i\times Y$, and

\Centerline{$\int_{E_i\times Y}|k-k_1|
=\int_Y\int_{E_i}|k(x,y)-k_1(x,y)|dxdy$.}

\noindent Now take some fixed $y\in Y$ such that

\Centerline{$\beta=\Bover1{\mu E_i}\int_{E_i}k(t,y)dt$}

\noindent is defined.   Then $\beta=k_1(x,y)$ for every $x\in E_i$.
For every $x\in E_i$, we must have
$k_2(x,y)=\alpha$ where $\alpha=\sum\{\alpha_j:E_i\subseteq H_j,\,y\in
F_j\}$.   But in this case, because
$\int_{E_i}k(x,y)-\beta\,dx=0$, we have

\Centerline{$\int_{E_i}\max(0,k(x,y)-\beta)dx
=\int_{E_i}\max(0,\beta-k(x,y))dx
=\Bover12\int_{E_i}|k(x,y)-k_1(x,y)|dx$.}

\noindent If $\beta\ge\alpha$,

\Centerline{$\int_{E_i}\max(0,k(x,y)-\beta)dx
\le\int_{E_i}\max(0,k(x,y)-\alpha)dx\le\int_{E_i}|k(x,y)-k_2(x,y)|dx$;}

\noindent if $\beta\le\alpha$,

\Centerline{$\int_{E_i}\max(0,\beta-k(x,y))dx
\le\int_{E_i}\max(0,\alpha-k(x,y))dx\le\int_{E_i}|k(x,y)-k_2(x,y)|dx$;}

\noindent in either case,

\Centerline{$\Bover12\int_{E_i}|k(x,y)-k_1(x,y)|dx
\le\int_{E_i}|k(x,y)-k_2(x,y)|dx$.}

\noindent This is true for almost every $y$, so integrating with respect
to $y$ we get the result.\ \Qed

Now, summing over $i$, we get

\Centerline{$\int|k-k_1|\le 2\int|k-k_2|\le\epsilon$,}

\noindent as required.
}%end of proof of 376R

\leader{376S}{Theorem} Let $(X,\Sigma,\mu)$ be a complete locally
determined measure space, $(Y,\Tau,\nu)$ a $\sigma$-finite measure
space, and $\lambda$ the c.l.d.\ product measure on $X\times Y$.   Let
$\tau$ be an extended Fatou norm on $L^0(\nu)$ and write
$\eusm L^{\tau'}$ for
$\{g:g\in\eusm L^0(\nu),\,\tau'(g^{\ssbullet})<\infty\}$,
where $\tau'$ is the associate extended Fatou norm of
$\tau$\cmmnt{ (369H-369I)}.   Suppose that $k\in\eusm L^0(\lambda)$ is
such that $k\times(f\otimes g)$ is integrable whenever
$f\in\eusm L^1(\mu)$ and $g\in\eusm L^{\tau'}$.
Then we have a corresponding linear operator $T:L^1(\mu)\to L^{\tau}$
defined by saying that
$\int(Tf^{\ssbullet})\times g^{\ssbullet}=\int k\times(f\otimes g)$
whenever $f\in\eusm L^1(\mu)$ and $g\in\eusm L^{\tau'}$.

For $x\in X$ set $k_x(y)=k(x,y)$ whenever this is defined.    Then
$k_x\in\eusm L^0(\nu)$ for almost every $x$;  set
$v_x=k_x^{\ssbullet}\in L^0(\nu)$ for such $x$.   In this case
$x\mapsto\tau(v_x)$ is measurable and defined and finite almost
everywhere, and $\|T\|=\esssup_x\tau(v_x)$.

\cmmnt{\medskip

\noindent{\bf Remarks} The discussion of extended Fatou norms in \S369
regarded them as functionals on spaces of the form $L^0(\frak A)$.   I
trust that no-one will be offended if I now speak of an extended Fatou
norm on $L^0(\nu)$, with the associated function spaces $L^{\tau}$,
$L^{\tau'}\subseteq L^0$, taking for granted the identification in
364Ic.

Recall that $(f\otimes g)(x,y)=f(x)g(y)$ for $x\in\dom f$ and $y\in\dom g$
(253B).

\wheader{376S}{0}{0}{0}{24pt}

By `$\esssup_x\tau(v_x)$' I mean

\Centerline{$\inf\{M:M\ge 0,\,\{x:v_x$ is defined and $\tau(v_x)\le M\}$
is conegligible$\}$}

\noindent (see 243D).
}%end of comment

\proof{{\bf (a)} To see that the formula
$(f,g)\mapsto\int k\times(f\otimes g)$ gives rise to an operator in
$\eurm L^{\times}(U;(L^{\tau'})^{\times})$, it is perhaps quickest to
repeat the
argument of parts (a) and (b) of the proof of 376E.   (We are not quite
in a position to quote 376E, as stated, because the localizable measure
algebra free product there might be strictly larger than the measure
algebra of $\lambda$;  see 325B.)   The first step, of course, is to
note that changing $f$ or $g$ on a negligible set does not affect the
integral $\int k\times(f\otimes g)$, so that we have a bilinear
functional on $L^1\times L^{\tau'}$;  and the other essential element is
the fact that the maps
$f^{\ssbullet}\mapsto(f\otimes\chi Y)^{\ssbullet}$,
$g^{\ssbullet}\mapsto (\chi X\otimes g)^{\ssbullet}$
are order-continuous (put 325A and 364Pc together).

By 369K, we can identify $(L^{\tau'})^{\times}$ with $L^{\tau}$, so that
$T$ becomes an operator in $\eurm L^{\times}(U;L^{\tau})$.   Note that
it must be norm-bounded (355C).

\medskip

{\bf (b)} By 376I, there is a non-decreasing sequence $\sequencen{Y_n}$
of measurable sets in $Y$, covering $Y$, such that
$\chi Y_n\in\eusm L^{\tau'}$ for every $n$.
Set $X_0=\{x:x\in X,\,k_x\in\eusm L^0(\nu)\}$.   Then $X_0$ is
conegligible in $X$.   \Prf\ Let $E\in\Sigma$ be any set of finite
measure.   Then for any $n\in\Bbb N$, $k\times(\chi E\otimes\chi Y_n)$
is integrable, that is, $\int_{E\times Y_n}k$ is defined and finite;  so
by Fubini's theorem $\int_{Y_n}k_x$ is defined and finite for almost
every $x\in E$.   Consequently, for almost every $x\in E$,
$k_x\times\chi Y_n\in\eusm L^0(\nu)$ for every $n\in\Bbb N$, that is,
$k_x\in\eusm L^0(\nu)$, that is, $x\in X_0$.

Thus $E\setminus X_0$ is negligible for every set $E$ of finite measure.
Because $\mu$ is complete and locally determined, $X_0$ is
conegligible.\ \Qed

This means that $v_x$ and $\tau(v_x)$ are defined for almost every $x$.

\medskip

{\bf (c)} $\tau(v_x)\le\|T\|$ for almost every $x$.   \Prf\ Take any
$E\in\Sigma$ of finite measure, and $n\in\Bbb N$.   Then
$k\times\chi(E\times Y_n)$ is integrable.   For each $r\in\Bbb N$, there
is a finite partition $E_{r0},\ldots,E_{r,m(r)}$ of $E$ into measurable
sets such that $\int_{E\times Y_n}|k-k^{(r)}|\le 2^{-r}$, where

$$\eqalign{k^{(r)}(x,y)&=\Bover1{\mu E_{ri}}\int_{E_{ri}}k(t,y)dt
\text{ whenever }y\in Y_n,\,x\in E_{ri},\,\mu E_{ri}>0\cr
&\hskip12em\text{ and the integral is defined in }\Bbb R\cr
&=0\text{ otherwise}\cr}$$

\noindent (376R).   Now $k^{(r)}$ also is integrable over $E\times Y_n$,
so $k^{(r)}_x\in\eusm L^0(\nu)$ for almost every $x\in E$, writing
$k^{(r)}_x(y)=k^{(r)}(x,y)$, and we can speak of
$v^{(r)}_x=(k^{(r)}_x)^{\ssbullet}$ for almost every $x$.   Note that
$k^{(r)}_x=k^{(r)}_{x'}$ whenever $x$, $x'$ belong to the same $E_{ri}$.

If $\mu E_{ri}>0$, then $v^{(r)}_x$ must be defined for every $x\in
E_{ri}$.   If $v'\in L^{\tau'}$ is represented by $g\in\eusm L^{\tau'}$
then

$$\eqalign{\int k\times(\chi E_{ri}\otimes(g\times\chi Y_n))
&=\int_{E_{ri}\times Y_n}k(t,y)g(y)d(t,y)\cr
&=\mu E_{ri}\int k^{(r)}(x,y)g(y)dy
=\mu E_{ri}\int v^{(r)}_x\times v'\cr}$$

\noindent for any $x\in E_{ri}$.   But this means that

\Centerline{$\mu E_{ri}\int v^{(r)}_x\times v'=\int T(\chi
E_{ri}^{\ssbullet})\times v'\times\chi Y_n^{\ssbullet}$}

\noindent for every $v'\in L^{\tau'}$, so

\Centerline{$v^{(r)}_x=\Bover1{\mu E_{ri}}T(\chi
E_{ri}^{\ssbullet})\times\chi Y_n^{\ssbullet}$,
\quad$\tau(v^{(r)}_x)\le\Bover1{\mu E_{ri}}\|T\|\|\chi
E_{ri}^{\ssbullet}\|_1=\|T\|$}

\noindent for every $x\in E_{ri}$.   This is true whenever
$\mu E_{ri}>0$, so in fact $\tau(v^{(r)}_x)\le\|T\|$ for almost every
$x\in E$.

Because $\sum_{r\in\Bbb N}\int_{E\times Y_n}|k-k^{(r)}|<\infty$, we must
have $k(x,y)=\lim_{r\to\infty}k^{(r)}(x,y)$ for almost every
$(x,y)\in E\times Y_n$.   Consequently, for almost every $x\in E$,
$k(x,y)=\lim_{r\to\infty}k^{(r)}(x,y)$ for almost every $y\in Y_n$, that
is, $\sequence{r}{v^{(r)}_x}$ order*-converges to
$v_x\times\chi Y_n^{\ssbullet}$
(in $L^0(\nu)$) for almost every $x\in E$.   But this means that, for
almost every $x\in E$,

\Centerline{$\tau(v_x\times\chi Y_n^{\ssbullet})
\le\liminf_{r\to\infty}\tau(v^{(r)}_x)\le\|T\|$}

\noindent (369Mc).   Now

\Centerline{$\tau(v_x)=\lim_{n\to\infty}\tau(v_x\times\chi
Y_n^{\ssbullet})\le\|T\|$}

\noindent for almost every $x\in E$.

As in (b), this implies (since $E$ is arbitrary) that
$\tau(v_x)\le\|T\|$ for almost every $x\in X$.\ \Qed

\medskip

{\bf (d)} I now show that $x\mapsto\tau(v_x)$ is measurable.   \Prf\
Take $\gamma\in\coint{0,\infty}$ and set
$A=\{x:x\in X_0,\,\tau(v_x)\le\gamma\}$.   Suppose that $\mu E<\infty$.
Let $G$ be a measurable envelope of $A\cap E$ (132Ee).   Set
$\tilde k(x,y)=k(x,y)$ when $x\in G$ and $(x,y)\in\dom k$, $0$
otherwise.   If $f\in\eusm L^1(\mu)$ and $g\in\eusm L^{\tau'}$, then

\Centerline{$\int\tilde k(x,y)f(x)g(y)d(x,y)
=\int_{G\times Y}k(x,y)f(x)g(y)d(x,y)=\int_Gf(x)\int_Yk(x,y)g(y)dydx$}

\noindent is defined.

Take any $g\in\eusm L^{\tau'}$.
For $x\in X_0$, set $h(x)=\int|\tilde k(x,y)g(y)|dy$.   Then $h$ is
finite almost everywhere and measurable.   For $x\in A\cap E$,

\Centerline{$\int|\tilde k(x,y)g(y)|dy
=\int|v_x\times g^{\ssbullet}|\le\gamma\tau'(g^{\ssbullet})$.}

\noindent So the measurable set
$G'=\{x:h(x)\le\gamma\tau'(g^{\ssbullet})\}$ includes $A\cap E$, and
$\mu(G\setminus G')=0$.   Consequently

\Centerline{$|\int\tilde k(x,y)f(x)g(y)d(x,y)|\le\int_G|f(x)|h(x)dx
\le\gamma\|f\|_1\tau'(g^{\ssbullet})$,}

\noindent and this is true whenever $f\in\eusm L^1(\mu)$.

Now we have an operator $\tilde T:L^1(\mu)\to L^{\tau}$ defined by the
formula

\Centerline{$\int(\tilde Tf^{\ssbullet})\times g^{\ssbullet}
=\int\tilde k\times(f\otimes g)$ when $f\in\eusm L^1(\nu)$ and
$g\in\eusm L^{\tau'}$,}

\noindent and the formula just above tells us that
$|\int\tilde Tu\times v'|\le\gamma\|u\|_1\tau'(v')$ for every
$u\in L^1(\nu)$ and $v'\in L^{\tau'}$;  that is,
$\tau(\tilde Tu)\le\gamma\|u\|_1$ for every $u\in L^1(\mu)$;  that is,
$\|\tilde T\|\le\gamma$.   But now (c) tells us
that $\tau(\tilde v_x)\le\gamma$ for almost every $x\in X$, where
$\tilde v_x$ is the equivalence class of $y\mapsto \tilde k(x,y)$, that
is, $\tilde v_x=v_x$ for $x\in G\cap X_0$, $0$ for $x\in X\setminus G$.
So $\tau(v_x)\le\gamma$ for almost every $x\in G$, and $G\setminus A$ is
negligible.   But this means that $A\cap E$ is measurable.   As $E$ is
arbitrary, $A$ is measurable;  as $\gamma$ is arbitrary,
$x\mapsto\tau(v_x)$ is measurable.\ \Qed

\medskip

{\bf (e)} Finally, the ideas in (d) show that
$\|T\|\le\esssup_x\tau(v_x)$.   \Prf\ Set $M=\esssup_x\tau(v_x)$.   If
$f\in\eusm L^1(\mu)$ and $g\in\eusm L^{\tau'}$, then

\Centerline{$\int|k(x,y)f(x)g(y)|d(x,y)
\le\int|f(x)|\tau(v_x)\tau'(g^{\ssbullet})dx
\le M\|f\|_1\tau'(g^{\ssbullet})$;}

\noindent as $g$ is arbitrary, $\tau(Tf^{\ssbullet})\le M\|f\|_1$;  as
$f$ is arbitrary, $\|T\|\le M$.\ \Qed
}%end of proof of 376S

\exercises{\leader{376X}{Basic exercises $\pmb{>}$(a)}
%\spheader 376Xa
Let $\mu$ be Lebesgue measure on $\Bbb R$.   Let $h$ be a
$\mu$-integrable
real-valued function with $\|h\|_1\le 1$, and set $k(x,y)=h(y-x)$
whenever this is defined.   Show that if $f$ is in either
$\eusm L^1(\mu)$ or $\eusm L^{\infty}(\mu)$ then
$g(y)=\int k(x,y)f(x)dx$ is
defined for almost every $y\in\Bbb R$, and that this formula gives rise
to an operator $T\in\Cal T^{\times}_{\bar\mu,\bar\mu}$ as defined in
373Ab.   \Hint{255H.}
%376A

\spheader 376Xb Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure algebras with localizable measure algebra free
product $(\frak C,\bar\lambda)$, and take $p\in[1,\infty]$.   Show that
if $u\in L^p(\frak A,\bar\mu)$ and $v\in L^p(\frak B,\bar\nu)$ then
$u\otimes v\in L^p(\frak C,\bar\lambda)$ and
$\|u\otimes v\|_p=\|u\|_p\|v\|_p$.
%376C

\sqheader 376Xc Let $U$, $V$, $W$ be Riesz spaces, of which $V$ and $W$
are Dedekind complete, and suppose that $T\in\eurm L^{\times}(U;V)$ and
$S\in\eurm L^{\times}(V;W)$.   Show that if either $S$ or $T$ is an
abstract integral operator, so is $ST$.
%376D

\spheader 376Xd Let $h$ be a Lebesgue integrable function on $\Bbb R$,
and $f$ a square-integrable function.   Suppose that $\sequencen{f_n}$
is a sequence of measurable functions such that ($\alpha$) $|f_n|\le f$
for every $n$ ($\beta$) $\lim_{n\to\infty}\int_Ef_n=0$ for every
measurable set $E$ of finite measure.   Show that
$\lim_{n\to\infty}(h*f_n)(y)=0$ for almost every $y\in\Bbb R$, where
$h*f_n$ is the convolution of $h$ and $f_n$.   \Hint{376Xa, 376H.}
%376H, 376Xa

\spheader 376Xe Let $U$ and $V$ be Riesz spaces, of which $V$ is
Dedekind complete.   Suppose that $W\subseteq U^{\sim}$ is a solid
linear subspace, and that $T$ belongs to the band in $\eurm L^{\sim}(U;V)$
generated by operators of the form $u\mapsto f(u)v$,
where $f\in W$ and $v\in V$.   Show that whenever $\sequencen{u_n}$ is
an order-bounded sequence in $U$ such that $\lim_{n\to\infty}f(u_n)=0$
for every $f\in W$, then $\sequencen{Tu_n}$ order*-converges to $0$
in $V$.
%376H

\spheader 376Xf Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
and $U\subseteq L^0=L^0(\frak A)$ an order-dense Riesz subspace such
that $U^{\times}$ separates the points of $U$.   Let $\sequencen{u_n}$
be an order-bounded sequence in $U$.   Show that the following are
equiveridical:  (i) $\lim_{n\to\infty}f(|u_n|)=0$ for every $f\in U^{\times}$;
(ii) $\sequencen{u_n}\to 0$ for the topology of
convergence in measure on $L^0$.   \Hint{by 367T, condition (ii) is
intrinsic to $U$, so we can replace $(\frak A,\bar\mu)$ by a localizable
algebra and use the representation in 369D.}
%376H

\spheader 376Xg Let $U$ be a Banach lattice with an order-continuous
norm, and $V$ a \wsid\ Riesz space.  Show that for $T\in\eurm
L^{\sim}(U;V)$ the following are equiveridical:  (i) $T$ belongs to the
band in $\eurm L^{\sim}(U;V)$ generated by operators of the form
$u\mapsto f(u)v$ where $f\in U^{\sim}$, $v\in V$;  (ii)
$\sequencen{Tu_n}$ order*-converges to $0$ in $V$ whenever
$\sequencen{u_n}$ is
an order-bounded sequence in $U^+$ which is norm-convergent to $0$;
(iii) $\sequencen{Tu_n}$ order*-converges to $0$ in $V$ whenever
$\sequencen{u_n}$ is an order-bounded sequence in $U$ which is weakly
convergent to $0$.
%376H

\spheader 376Xh Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be
$\sigma$-finite measure spaces, with product measure $\lambda$ on
$X\times Y$, and measure algebras $(\frak A,\bar\mu)$,
$(\frak B,\bar\nu)$.   Suppose that $k\in\eusm L^0(\lambda)$.   Show
that the
following are equiveridical:  (i)($\alpha$) if $f\in\eusm L^1(\mu)$ then
$g_f(y)=\int k(x,y)f(x)dx$ is defined for almost every $y$ and
$g_f\in\eusm L^1(\nu)$ ($\beta$) there is an operator
$T\in\Cal T^{\times}_{\bar\mu,\bar\nu}$ defined by setting
$Tf^{\ssbullet}=g_f^{\ssbullet}$
for every $f\in \eusm L^1(\mu)$;  (ii) $\int|k(x,y)|dy\le 1$ for almost
every $x\in X$, $\int|k(x,y)|dx\le 1$ for almost every $y\in Y$.
%376J

\sqheader 376Xi(i) Show that there is a compact linear operator from
$\ell^2$ to itself which is not in $\eurm L^{\sim}(\ell^2;\ell^2)$.
\Hint{start from the operator $S$ of 371Ye.}   (ii) Show that the
identity operator on $\ell^2$ is an abstract integral operator.
%376J

\sqheader 376Xj Let $\mu$ be Lebesgue measure on $[0,1]$.   (i) Give an
example of a measurable function $k:[0,1]^2\to\Bbb R$ such that, for any
$f\in\eusm L^2(\mu)$, $g_f(y)=\int k(x,y)f(x)dx$ is defined
for every $y$ and $\|g_f\|_2=\|f\|_2$, but $k$ is not integrable, so the
linear isometry on $L^2=L^2(\mu)$ defined by $k$ does not belong to
$\eurm L^{\sim}(L^2;L^2)$.   (ii) Show that the identity operator on $L^2$
is not an abstract integral operator.
%376Xi %376J

\spheader 376Xk Let $(X,\Sigma,\mu)$ be a $\sigma$-finite measure space
and $(Y,\Tau,\nu)$ a complete locally determined measure space.   Let
$U\subseteq L^0(\mu)$, $V\subseteq L^0(\nu)$ be solid linear subspaces,
of which $V$ is order-dense;  write
$V^{\#}=\{v:v\in L^0(\nu),\,v\times v'$ is integrable for every
$v'\in V\}$,
$\eusm U=\{f:f\in\eusm L^0(\nu),\,f^{\ssbullet}\in U\}$,
$\eusm V=\{g:g\in\eusm L^0(\nu),\,g^{\ssbullet}\in V\}$,
$\eusm V^{\#}=\{h:h\in\eusm L^0(\nu),\,h^{\ssbullet}\in V^{\#}\}$.
Let $\lambda$ be the c.l.d.\ product measure on $X\times Y$, and
$k\in\eusm L^0(\lambda)$ a function such that $k\times(f\otimes g)$ is
integrable for whenever $f\in\eusm U$ and $g\in\eusm V$.
(i) Show that for
any $f\in\eusm U$, $h_f(y)=\int k(x,y)f(x)dx$ is defined for almost
every $y\in Y$, and that $h_f\in\eusm V^{\#}$.   (ii) Show that we have
a map $T\in\eurm L^{\times}(U;V^{\#})$ defined {\it either} by writing
$Tf^{\ssbullet}=h_f^{\ssbullet}$ for every $f\in\eusm U$ {\it or} by
writing
$\int(Tf^{\ssbullet})\times g^{\ssbullet}=\int k\times(f\otimes g)$ for
every $f\in\eusm U$ and $g\in\eusm V$.
%376J

\spheader 376Xl Let $(X,\Sigma,\mu)$, $(Y,\Tau,\nu)$ and
$(Z,\Lambda,\lambda)$ be $\sigma$-finite measure spaces, and $U$, $V$,
$W$ perfect order-dense solid linear subspaces of $L^0(\mu)$, $L^0(\nu)$
and $L^0(\lambda)$ respectively.   Suppose that $T:U\to V$ and $S:V\to
W$ are abstract integral operators corresponding to kernels
$k_1\in\eusm L^0(\mu\times\nu)$, $k_2\in\eusm L^0(\nu\times\lambda)$,
writing
$\mu\times\nu$ for the (c.l.d.\ or primitive) product measure on
$X\times Y$.   Show that $ST:U\to W$ is represented by the kernel
$k\in\eusm L^0(\mu\times\lambda)$ defined by setting
$k(x,z)=\int k_1(x,y)k_2(y,z)dy$ whenever this integral is defined.
%376J

\spheader 376Xm Let $U$ be a perfect Riesz space.   Show that a set
$C\subseteq U$ is relatively compact for $\frak T_s(U,U^{\times})$ iff
for every $g\in(U^{\times})^+$, $\epsilon>0$ there is a $u^*\in U$ such
that $g(|u|-u^*)^+\le\epsilon$ for every $u\in C$.   \Hint{376O and the
proof of 356Q.}

\sqheader 376Xn Let $\mu$ be Lebesgue measure on $[0,1]$, and $\nu$
counting measure on $[0,1]$.   Set $k(x,y)=1$ if $x=y$, $0$ otherwise.
Show that 376S fails in this context (with, e.g.,
$\tau=\|\,\|_{\infty}$).
%376S

\spheader 376Xo Suppose, in 376Xk, that $U=L^{\tau}$ for some extended
Fatou norm on $L^0(\mu)$ and that $V=L^1(\nu)$, so that
$V^{\#}=L^{\infty}(\nu)$.   Set $k_y(x)=k(x,y)$ whenever this is
defined, $w_y=k_y^{\ssbullet}$ whenever $k_y\in\eusm L^0(\mu)$.   Show
that $w_y\in L^{\tau'}$ for almost every $y\in Y$, and that the norm of
$T$ in $\eurm B(L^{\tau};L^{\infty})$ is $\esssup_y\tau'(w_y)$.
\Hint{do the case of totally finite $Y$ first.}
%376S, 376Xk

\leader{376Y}{Further exercises (a)}
%\spheader 376Ya
Let $U$, $V$ and $W$ be linear spaces (over any field $F$) and
$\phi:U\times V\to W$ a bilinear operator.   Let $W_0$ be the linear subspace
of $W$ generated by $\phi[U\times V]$.   Show that the following are
equiveridical:  (i) for every linear space $Z$ over $F$ and every
bilinear
$\psi:U\times V\to Z$, there is a (unique) linear operator $T:W_0\to Z$
such
that $T\phi=\psi$ (ii) whenever $u_0,\ldots,u_n\in U$ are linearly
independent and $v_0,\ldots,v_n\in V$ are non-zero,
$\sum_{i=0}^n\phi(u_i,v_i)\ne 0$
(iii) whenever $u_0,\ldots,u_n\in U$ are non-zero and
$v_0,\ldots,v_n\in V$ are linearly independent,
$\sum_{i=0}^n\phi(u_i,v_i)\ne 0$ (iv) for
any Hamel bases $\langle u_i\rangle_{i\in I}$,
$\langle v_j\rangle_{j\in J}$ of $U$ and $V$,
$\langle\phi(u_i,v_j)\rangle_{i\in I,j\in J}$ is a
Hamel basis of $W_0$
(v) for some pair $\langle u_i\rangle_{i\in I}$,
$\langle v_j\rangle_{j\in J}$ of Hamel bases of $U$ and $V$,
$\langle\phi(u_i,v_j)\rangle_{i\in I,j\in J}$ is a Hamel basis of $W_0$.
%`Hamel basis' not defined until v4

\spheader 376Yb
Let $(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ be semi-finite measure
algebras, and $(\frak C,\bar\lambda)$ their localizable measure algebra
free product.   Show that $\otimes:L^0(\frak A)\times L^0(\frak B)\to
L^0(\frak C)$ satisfies the equivalent conditions of 376Ya.
%376B, 376Ya

\spheader 376Yc Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be semi-finite
measure spaces and $\lambda$ the c.l.d.\ product measure on $X\times Y$.
Show that the map $(f,g)\mapsto f\otimes g:
\eusm L^0(\mu)\times\eusm L^0(\nu)\to\eusm L^0(\lambda)$ induces a map
$(u,v)\mapsto u\otimes v:L^0(\mu)\times L^0(\nu)\to L^0(\lambda)$
possessing all the properties described in 376B and 376Ya, subject to
a suitable interpretation of the formula
$\otimes:\frak A\times\frak B\to\frak C$.
%376B, 376Ya

\spheader 376Yd Let $(\frak B_{\omega_1},\bar\nu_{\omega_1})$
be the measure algebra of
$\{0,1\}^{\omega_1}$ with its usual measure, and
$\langle a_{\xi}\rangle_{\xi<\omega_1}$ a stochastically independent
(definition:  325Xf) family
of elements of measure $\bover12$ in $\frak B_{\omega_1}$.   Set
$U=L^2(\frak B_{\omega_1},\bar\nu_{\omega_1})$ and
$V=\{v:v\in\BbbR^{\omega_1},\,\{\xi:v(\xi)\ne 0\}$ is
countable$\}$.   Define $T:U\to\BbbR^{\omega_1}$ by setting
$Tu(\xi)=2\int_{a_{\xi}}u-\int u$ for $\xi<\omega_1$, $u\in U$.     Show
that (i) $Tu\in V$ for every $u\in U$ (ii)
$\sequencen{Tu_n}$ order*-converges to $0$ in $V$ whenever
$\sequencen{u_n}$ is a sequence in $U$ such that
$\lim_{n\to\infty}f(u_n)=0$ for every $f\in U^{\times}$ (iii)
$T\notin\eurm L^{\sim}(U;V)$.
%376H

\spheader 376Ye Let $U$ be a Riesz space with the countable sup property
(definition:  241Ye)
such that $U^{\times}$ separates the points of $U$, and
$\sequencen{u_n}$ a sequence in $U$.   Show that the
following are equiveridical:  (i) $\lim_{n\to\infty}f(v\wedge|u_n|)=0$
for every $f\in U^{\times}$, $v\in U^+$;  (ii) every subsequence of
$\sequencen{u_n}$ has a
sub-subsequence which is order*-convergent to $0$.
%376H

\spheader 376Yf Let $U$ be an Archimedean Riesz space and $\frak A$ a
\wsid\ Dedekind complete Boolean algebra.   Suppose that $T:U\to
L^0=L^0(\frak A)$ is a linear operator such that
$\sequencen{|Tu_n|}$ order*-converges to $0$ in $L^0$ whenever
$\sequencen{u_n}$ is order-bounded and order*-convergent to $0$ in
$U$.   Show that $T\in\eurm L^{\sim}_c(U;L^0)$ (definition:  355G),
so that if $U$ has the
countable sup property then $T\in\eurm L^{\times}(U;L^0)$.
%376H

\spheader 376Yg Suppose that $(Y,\Tau,\nu)$ is a probability space in
which $\Tau=\Cal PY$ and $\nu\{y\}=0$ for every $y\in Y$.    (See
363S.)   Take $X=Y$ and let $\mu$ be counting measure on $X$;  let
$\lambda$ be the c.l.d.\ product measure on $X\times Y$, and set
$k(x,y)=1$ if $x=y$, $0$ otherwise.   Show that we have an operator
$T:L^{\infty}(\mu)\to L^{\infty}(\nu)$ defined by setting
$Tf=g^{\ssbullet}$ whenever $f\in L^{\infty}(\mu)\cong\ell^{\infty}(X)$
and $g(y)=\int k(x,y)f(x)dx=f(y)$ for every $y\in Y$.   Show that $T$
satisfies the conditions (ii) and (iii) of 376J but does not belong to
$\eurm L^{\times}(L^{\infty}(\mu);L^{\infty}(\nu))$.
%376J

\spheader 376Yh Give an example of an abstract integral operator
$T:\ell^2\to L^1(\mu)$, where $\mu$ is Lebesgue measure on $[0,1]$, such
that $\sequencen{Te_n}$ is not order*-convergent in $L^1(\mu)$, where
$\sequencen{e_n}$ is the standard orthonormal sequence in $\ell^2$.
%376J

\spheader 376Yi Set $k(m,n)=1/\pi(n-m+\bover12)$ for $m$, $n\in\Bbb Z$.
(i) Show that
$\sum_{n=-\infty}^{\infty}k(m,n)^2=1$ and
$\sum_{n=-\infty}^{\infty}k(m,n)k(m',n)=0$ for all distinct $m$,
$m'\in\Bbb Z$.   \Hint{find the Fourier series of
$x\mapsto e^{i(m+\bover12)x}$ and use 282K.}
(ii) Show that there is a norm-preserving linear operator $T$ from
$\ell^2=\ell^2(\Bbb Z)$ to itself given by the formula
$(Tu)(n)=\sum_{m=-\infty}^{\infty}k(m,n)u(m)$.   (iii) Show
that $T^2$ is the identity operator on $\ell^2$.   (iv) Show that
$T\notin\eurm L^{\sim}(\ell^2;\ell^2)$.   \Hint{consider
$\sum_{m,n=-\infty}^{\infty}|k(m,n)|x(m)x(n)$ where
$x(n)=1/\sqrt{|n|}\ln|n|$ for $|n|\ge 2$.}   ($T$ is a form of the
{\bf Hilbert transform.})
%376J

\spheader 376Yj Let $U$ be an $L$-space and $V$ a Banach lattice with an
order-continuous norm.   Let $T\in\eurm L^{\sim}(U;V)$.   Show that the
following are equiveridical:  (i) $T$ is an abstract integral operator;
(ii) $T[C]$ is norm-compact in $V$ whenever $C$ is weakly compact in
$U$.   \Hint{start with the case in which $C$ is order-bounded, and
remember that it is weakly sequentially compact.}
%376M

\spheader 376Yk Let $(X,\Sigma,\mu)$ be a complete locally determined
measure space and $(Y,\Tau,\nu)$, $(Z,\Lambda,\lambda)$ two
$\sigma$-finite measure spaces.   Suppose that $\tau$, $\theta$ are
extended Fatou norms on $L^0(\nu)$, $L^0(\lambda)$ respectively, and
that $T:L^1(\mu)\to L^{\tau}$ is an abstract integral operator, with
corresponding kernel $k\in\eusm L^0(\mu\times\nu)$, while $S\in\eurm
L^{\times}(L^{\tau};L^{\theta})$, so that $ST:L^1(\mu)\to L^{\theta}$ is
an abstract integral operator (376Xc);  let
$\tilde k\in\eusm L^0(\mu\times\lambda)$ be the corresponding kernel.
For $x\in X$ set
$v_x=k_x^{\ssbullet}$ when this is defined in $L^{\tau}$, as in 376S,
and similarly take $w_x=\tilde k_x^{\ssbullet}\in L^{\theta}$.   Show
that $Sv_x=w_x$ for almost every $x\in X$.
%376S
}%end of exercises

\endnotes{
\Notesheader{376}
I leave 376Yb to the exercises because I do not rely on it for any of
the work here, but of course it is an essential aspect of the map
$\otimes:L^0(\frak A)\times L^0(\frak B)\to L^0(\frak C)$ I discuss in
this section.   The conditions in 376Ya are characterizations of the
`tensor product' of two linear spaces, a construction of great
importance
in abstract linear algebra (and, indeed, in modern applied linear
algebra;  it is by no means trivial even in the finite-dimensional
case).   In particular, note that conditions (ii), (iii) of 376Ya apply
to arbitrary subspaces of $U$ and $V$ if they apply to $U$ and $V$
themselves.

The principal ideas used in 376B-376C have already been set out in
\S\S253 and 325.   Here I do little more than list the references.   I
remark however that it is quite striking that $L^1(\frak C,\bar\lambda)$
should have no fewer than three universal mapping theorems attached to
it (376Cb, 376C(c-i) and 376C(c-ii)).

The real work of this section begins in 376E.
As usual, much of the proof is taken up with relatively straightforward
verifications, as in parts (a) and (b), while part (i) is just a
manoeuvre to show that it doesn't matter if $\frak A$ and $\frak B$
aren't Dedekind complete, because $\frak C$ is.   But I think that parts
(d), (f) and (j) have ideas in them.   In particular, part (f) is a kind
of application of the Radon-Nikod\'ym theorem (through the
identification of $L^1(\frak C,\bar\lambda)^*$ with
$L^{\infty}(\frak C)$).

I have split 376E from 376H because the former demands the language of
measure algebras, while the latter can be put into the language of pure
Riesz space theory.   Asking for a \wsid\ space $V$ in 376H is a way of
applying the ideas to $V=L^0$ as well as to Banach function spaces.
(When $V=L^0$, indeed, variations on the hypotheses are possible, using
376Yf.)   But it is a reminder of one of the directions in which it is
often possible to find generalizations of ideas beginning in measure
theory.

The condition `$\lim_{n\to\infty}f(u_n)=0$ for every
$f\in U^{\times}$' (376H(ii)) seems natural in this context, and
gives marginally greater generality than some alternatives (because it
does the right thing when $U^{\times}$ does not separate the points of
$U$), but it is not the only way of expressing the idea;  see 376Xf and
376Ye.   Note that the conditions (ii) and (iii) of 376H are
significantly
different.   In 376H(iii) we could easily have $|u_n|=u^*$ for every
$n$;  for instance, if $u_n=2\chi a_n-\chi 1$ for some stochastically
independent sequence $\sequencen{a_n}$ of elements of measure $\bover12$
in a probability algebra (272Ye).

If you have studied compact linear operators between Banach spaces
(definition:  3A5La), you will have encountered the condition
`$Tu_n\to 0$ strongly whenever $u_n\to 0$ weakly'.   The conditions in
376H and
376J are of this type.   If a sequence $\sequencen{u_n}$ in a Riesz
space $U$ is order-bounded and order*-convergent to $0$, then
$\lim_{n\to\infty}f(u_n)=0$ for every $f\in U^{\times}$ (367Xg).
Visibly this latter condition is associated
with weak convergence, and order*-convergence is (in Banach lattices)
closely related to norm convergence (367D).   In the context of
376H, an abstract integral operator is one which transforms convergent
sequences of a weak type into convergent sequences of a stronger type.
The relationship between the classes of (weakly) compact operators and
abstract integral operators is interesting, but outside the scope of
this book;  I leave you with 376P-376Q and 376Y, and a pair of
elementary examples to guard against extravagant conjecture (376Xi).

376O belongs to an extensive general theory of weak compactness in
perfect Riesz spaces, based on adaptations of the concept of `uniform
integrability'.  I give the next step in 376Xm.   For more information
see {\smc Fremlin 74a}, chap.\ 8.

Note that 376Mb and 376P overlap when $V^{\times}$ in 376Mb is reflexive
-- for instance, when $V$ is an $L^p$ space for some
$p\in\ooint{1,\infty}$ -- since then every bounded linear operator from
$L^1$ to $V^{\times}$ must be weakly compact.   For more information on the
representation of operators see {\smc Dunford \& Schwartz 57},
particularly Table VI in the notes to Chapter VI.

As soon as we leave formulations in terms of the spaces $L^0(\frak A)$
and their subspaces, and return to the original conception of a
kernel operator in terms of integrating functions against sections of
a kernel, we are necessarily involved in the pathology of Fubini's
theorem for general measure spaces.   In general, the repeated integrals
$\iint k(x,y)dxdy$, $\iint k(x,y)dydx$ need not be equal, and something
has to give (376Xn).   Of course this particular worry disappears if the
spaces are $\sigma$-finite, as in 376J.  In 376S I take the trouble to
offer a more general condition,
mostly as a reminder that the techniques developed in Volume 2
do enable us sometimes to go beyond the $\sigma$-finite case.   Note
that this is one of the many contexts in which anything we can prove
about probability spaces will be true of all $\sigma$-finite spaces;
but that we cannot make the next step, to all strictly localizable
spaces.

376S verges on the theory of integration of vector-valued functions,
which I don't wish to enter here;  but it also seems to have a natural
place in the context of this chapter.   It is of course a special
property of $L^1$ spaces.   The formula
$\|T_k\|=\esssup_x\tau(k_x^{\ssbullet})$ shows that
$\|T_{|k|}\|=\|T_k\|$;
now we know fron 376E that $T_{|k|}=|T_k|$, so we get a special case of
the Chacon-Krengel theorem (371D).   Reversing the roles of $X$ and
$Y$, we find ourselves with an operator from $L^{\tau}$ to $L^{\infty}$
(376Xo), which is the other standard context in which $\|T\|=\||T|\|$
(371Xd).   I include two exercises on $L^2$ spaces (376Xj, 376Yi)
designed to emphasize the fact that $\eurm B(U;V)$ is included in
$\eurm L^{\sim}(U;V)$ only in very special cases.

The history of the theory here is even more confusing than that of
mathematics in general, because so many of the ideas were developed in
national schools in very imperfect contact with each other.   My own
account gives no hint of how this material arose;  I ought in
particular to note that 376N is one of the oldest results, coming
(essentially) from {\smc Dunford 1936}.   For further references,
see {\smc Zaanen 83}, chap.\ 13.
}%end of comment

\discrpage

\frfilename{mt377.tex}
\versiondate{30.12.09}
\copyrightdate{2008}

\def\chaptername{Linear operators between function spaces}
\def\sectionname{Function spaces of reduced products}

\newsection{*377\dvAnew{2008}}

In \S328 I introduced `reduced products' of probability algebras.   In this
section I seek to describe the function spaces of reduced products as
images of subspaces of products of function spaces of the original
algebras.   I add a group of universal mapping theorems associated
with the constructions of projective and inductive limits of directed
families of probability algebras (377G-377H).

\leader{377A}{Proposition} If $\familyiI{\frak A_i}$ is a
non-empty family of Boolean
algebras with simple product $\frak A$, then $L^{\infty}(\frak A)$ can be
identified, as normed space and $f$-algebra, with the subspace
$W_{\infty}$ of $\prod_{i\in I}L^{\infty}(\frak A_i)$
consisting of families $u=\familyiI{u_i}$ such that
$\|u\|_{\infty}=\sup_{i\in I}\|u_i\|_{\infty}$ is finite.

\proof{{\bf (a)} I begin by noting that $W_{\infty}$ is, in itself, an
Archimedean
$f$-algebra and $\|\,\|_{\infty}$ is a Riesz norm on $W_{\infty}$.
\Prf\ $W_{\infty}$ is a solid linear subspace of
$\prod_{i\in I}L^{\infty}(\frak A_i)$, so inherits a Riesz space structure
(352K, 352Ia).   Now it is easy to check that
$e=\familyiI{\chi 1_{\frak A_i}}$ is an order unit in $W_{\infty}$ and
that $\|\,\|_{\infty}$ is the corresponding order-unit norm (354F-354G).
Finally, because $W_{\infty}$ is the solid linear subspace
of $\prod_{i\in I}L^{\infty}(\frak A_i)$ generated by
$e$, and $e$ is the multiplicative identity of
$\prod_{i\in I}L^{\infty}(\frak A_i)$, $W_{\infty}$ is closed under
multiplication, and is an $f$-algebra.\ \Qed

\medskip

{\bf (b)} We have a natural function
$\theta:\frak A\to W_{\infty}$ defined by saying that
$\theta a=\familyiI{\chi a_i}$ whenever
$a=\familyiI{a_i}\in\frak A$.   Clearly $\theta$ is additive and
$\|\theta a\|_{\infty}\le 1$ for every $a\in\frak A$;  moreover,
$\theta a\wedge\theta b=0$ when $a$, $b\in\frak A$ are disjoint.
By 363E, we have a corresponding Riesz homomorphism
$T:L^{\infty}(\frak A)\to W_{\infty}$ of norm at most $1$.

\medskip

{\bf (c)} In fact $\|Tw\|_{\infty}=\|w\|_{\infty}$ for every
$w\in L^{\infty}(\frak A)$.   \Prf\ If $w=0$, this is trivial.
If $w\in S(\frak A)\setminus\{0\}$, express it as
$\sum_{k=0}^n\alpha_k\chi a^{(k)}$ where $\langle a^{(k)}\rangle_{k\le n}$
is a disjoint family of non-zero elements.
Expressing each $a^{(k)}$ as $\familyiI{a_{ki}}$,

\Centerline{$Tw=\familyiI{\sum_{k=0}^n\alpha_k\chi a_{ki}}$.}

\noindent There must be a $j$ such that $|\alpha_j|=\|w\|_{\infty}$;  now
there is an $i$ such that $a_{ji}\ne 0$;  as
$\langle a_{ki}\rangle_{k\le n}$ is disjoint,

\Centerline{$\|Tw\|_{\infty}
\ge\|\sum_{k=0}^n\alpha_k\chi a_{ki}\|_{\infty}
\ge|\alpha_j|=\|w\|_{\infty}$.}

If now $w$ is any member of $L^{\infty}(\frak A)$,

$$\eqalign{\|w\|_{\infty}
&=\sup\{\|w'\|_{\infty}:w'\in S(\frak A),\,|w'|\le|w|\}\cr
&=\sup\{\|Tw'\|_{\infty}:w'\in S(\frak A),\,|w'|\le|w|\}
\le\|Tw\|_{\infty}\cr}$$

\noindent because $T$ is a Riesz homomorphism.\ \Qed

Thus $T$ is norm-preserving, therefore injective.

\medskip

{\bf (d)} Next, $T$ is surjective.   \Prf\
Suppose that $\familyiI{u_i}\in W_{\infty}^+$ is non-negative, and that
$\epsilon>0$.   Let $n\in\Bbb N$ be such that
$n\epsilon\ge\sup_{i\in I}\|u_i\|_{\infty}$, and for $k\le n$, $i\in I$ set
$a_{ki}=\Bvalue{u_i>k\epsilon}$.   Set
$w=\epsilon\sum_{k=1}^n\chi(\familyiI{a_{ki}})$.   Then
$w\in L^{\infty}(\frak A)$ and $Tw=\familyiI{v_i}$, where
$v_i=\epsilon\sum_{k=1}^n\chi a_{ki}$, so that $v_i\le u_i$ and
$\|u_i-v_i\|_{\infty}\le\epsilon$, for every $i\in I$.
Thus $\|Tw-\familyiI{u_i}\|_{\infty}\le\epsilon$.

As $\familyiI{u_i}$ and $\epsilon$ are arbitrary,
$T[L^{\infty}(\frak A)]\cap W_{\infty}^+$ is norm-dense in $W_{\infty}^+$.
But $T$ is an isometry and $L^{\infty}(\frak A)$ is norm-complete, so
$T[L^{\infty}(\frak A)]$ is closed in $W_{\infty}$ and includes
$W_{\infty}^+$ and therefore $W_{\infty}$;  that is, $T$ is surjective.\
\Qed

So $T$ is a norm-preserving bijective Riesz homomorphism, that is, a normed
Riesz space isomorphism.   Finally, by 353Qd or otherwise, $T$ is
multiplicative, so is an $f$-algebra isomorphism.
}%end of proof of 377A

\leader{377B}{Theorem} Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a
non-empty
family of probability algebras, and $(\frak B,\bar\nu)$ a probability
algebra.   Let $\frak A$ be the simple product of $\familyiI{\frak A_i}$,
and $\pi:\frak A\to\frak B$ a Boolean homomorphism such that
$\bar\nu\pi(\familyiI{a_i})\le\sup_{i\in I}\bar\mu_ia_i$ whenever
$\familyiI{a_i}\in\frak A$.
Let $W_0$ be the subspace of $\prod_{i\in I}L^0(\frak A_i)$ consisting
of families $\familyiI{u_i}$ such that
$\inf_{k\in\Bbb N}\sup_{i\in I}\bar\mu_i\Bvalue{|u_i|>k}=0$.

(a) $W_0$ is
a solid linear subspace and a subalgebra of $\prod_{i\in I}L^0(\frak A_i)$,
and there is a unique Riesz homomorphism
$T:W_0\to L^0(\frak B)$ such that
$T(\familyiI{\chi a_i})=\chi\pi(\familyiI{a_i})$ whenever
$\familyiI{a_i}\in\frak A$.   Moreover, $T$ is multiplicative,
and $\Bvalue{Tu>0}\Bsubseteq\pi(\familyiI{\Bvalue{u_i>0}})$ whenever
$u=\familyiI{u_i}$ belongs to $W_0$.

(b) If $h:\Bbb R\to\Bbb R$ is a continuous function, and we write
$\bar h$ for the corresponding maps from $L^0$ to itself for any of the
spaces $L^0=L^0(\frak A_i)$, $L^0=L^0(\frak B)$\cmmnt{ (364H)}, then
$\familyiI{\bar h(u_i)}\in W_0$ and
$T(\familyiI{\bar h(u_i)})=\bar h(Tu)$ whenever
$u=\familyiI{u_i}$ belongs to $W_0$.

\proof{{\bf (a)} For
$u=\familyiI{u_i}\in\prod_{i\in I}L^0(\frak A_i)$ and $k\in\Bbb N$, set
$\gamma_k(u)=\sup_{i\in I}\bar\mu_i\Bvalue{|u_i|>k}$.

\medskip

\quad{\bf (i)} $W_0$ is a solid linear
subspace and subalgebra of the $f$-algebra
$\prod_{i\in I}L^0(\frak A_i)$.   \Prf\ For $k\in\Bbb N$ and $u$,
$v\in\prod_{i\in I}L^0(\frak A_i)$,

\Centerline{$\gamma_k(u)\le\gamma_k(v)$ whenever $|u|\le|v|$,}

\Centerline{$\gamma_{2k}(u+v)\le\gamma_k(u)+\gamma_k(v)$,}

\Centerline{$\gamma_{k^2}(u\times v)\le\gamma_k(u)+\gamma_k(v)$}

\noindent for all $u$, $v\in\prod_{i\in I}L^0(\frak A_i)$ and $k\in\Bbb N$.
So $W_0$ is solid, is closed under addition, and is closed under
multiplication.\ \Qed

\medskip

\quad{\bf (ii)}
Let $W_{\infty}\subseteq W_0$ be the set of families
$\familyiI{u_i}\in\prod_{i\in I}L^{\infty}(\frak A_i)$ such that
$\sup_{i\in I}\|u_i\|_{\infty}$ is finite;  by 377A, we can identify
$W_{\infty}$ with $L^{\infty}(\frak A)$.   We therefore have a
corresponding multiplicative
Riesz homomorphism $S:W_{\infty}\to L^{\infty}(\frak B)$ such
that $S(\familyiI{\chi a_i})=\chi\pi(\familyiI{a_i})$ whenever
$\familyiI{a_i}\in\frak A$ (363F);  note that
$S(\familyiI{\chi 1_{\frak A_i}})=\chi 1_{\frak B}$.

\medskip

\quad{\bf (iii)} If $u=\familyiI{u_i}\in W_{\infty}$ and $k\in\Bbb N$, then
$\Bvalue{Su>k}\Bsubseteq\pi(\familyiI{\Bvalue{u_i>k}})$.   \Prf\ Setting
$a_i=\Bvalue{u_i>k}$, we have
$u_i\times\chi(1_{\frak A_i}\Bsetminus a_i)\le k\chi 1_{\frak A_i}$ for
every $i$.   Set $a=\familyiI{a_i}$.
Since $S$ is a multiplicative Riesz homomorphism,

$$\eqalignno{Su\times\chi(1_{\frak B}\Bsetminus\pi a)
&=Su\times\chi\pi(\familyiI{1_{\frak A_i}\Bsetminus a_i})
=S(\familyiI{u_i})\times S(\familyiI{\chi(1_{\frak A_i}\Bsetminus a_i)})\cr
&=S(\familyiI{u_i}\times\familyiI{\chi(1_{\frak A_i}\Bsetminus a_i}))
=S(\familyiI{u_i\times\chi(1_{\frak A_i}\Bsetminus a_i)})\cr
&\le S(\familyiI{k\chi 1_{\frak A_i}})
=k\chi 1_{\frak B}\cr}$$

\noindent and $\Bvalue{Su>k}\Bsubseteq\pi a$, as claimed.\ \Qed


\medskip

\quad{\bf (iv)} If $u=\familyiI{u_i}\in W_0^+$, then
$\sup\{Sv:v\in W_{\infty}$, $0\le v\le u\}$ is defined in
$L^0(\frak B)$.   \Prf\ Set $A_u=S[W_{\infty}\cap[0,u]\,]$.
Because $W_{\infty}\cap[0,u]$ is upwards-directed, so is $A$.
If $v=\familyiI{v_i}\in W_{\infty}\cap[0,u]$, then
$\Bvalue{Sv>k}\Bsubseteq\pi(\familyiI{\Bvalue{v_i>k}})$, by (iii), so

\Centerline{$\bar\nu\Bvalue{Sv>k}\le\sup_{i\in I}\bar\mu_i\Bvalue{v_i>k}
\le\gamma_k(u)$.}

\noindent Thus $\bar\nu\Bvalue{w>k}\le\gamma_k(u)$ for every $w\in A$.
Since $u\in W_0$, $\lim_{k\to\infty}\gamma_k(u)=0$;  so 364L(a-ii)
tells us that $\sup A_u$ is defined in $L^0(\frak B)$.\ \QeD\

By 355F, there is a Riesz homomorphism $T:W_0\to L^0(\frak B)$
extending $S$ and such that $Tu=A_u$ for every
$u\in W_0^+$.   By 353Qd, $T$ is multiplicative.

\medskip

\quad{\bf (v)} Because $T$ is multiplicative, we can repeat the
calculations of (iii), with $T$ in place of $S$, to see that

\Centerline{$\Bvalue{Tu>k}\Bsubseteq\pi(\familyiI{\Bvalue{u_i>k}})$}

\noindent whenever $u=\familyiI{u_i}\in W_0$;  in particular,
$\Bvalue{Tu>0}\Bsubseteq\pi(\familyiI{\Bvalue{u_i>0}})$.

\medskip

\quad{\bf (vi)} To see that $T$ is uniquely defined, let
$T':W_0\to L^0(\frak B)$ be another Riesz homomorphism agreeing with $T$ on
families of the form $\familyiI{\chi a_i}$.
Then $T$ and $T'$ agree on $W_{\infty}\cong L^{\infty}(\frak A)$,
by the uniqueness guaranteed in
363Fa, and $T'$ also is multiplicative, by 353Qd once more.   As in (v), we
therefore have

\Centerline{$\Bvalue{Tu>k}\Bcup\Bvalue{T'u>k}
\Bsubseteq\pi(\familyiI{\Bvalue{u_i>k}})$,
\quad$\bar\nu(\Bvalue{Tu>k}\Bcup\Bvalue{T'u>k})\le\gamma_k(u)$}

\noindent whenever $u\in W_0$ and $k\in\Bbb N$.

Suppose that $u\in W_0^+$ and $\epsilon>0$.   Then there is a
$k\in\Bbb N$ such that $\gamma_k(u)\le\epsilon$.   Set
$v_i=u_i\wedge k\chi 1_{\frak A_i}$ for each $i$, and
$v=\familyiI{v_i}$.   Then $Tv=T'v$, so

\Centerline{$\bar\nu\Bvalue{|Tu-T'u|>0}
\le\bar\nu(\Bvalue{Tu-Tv>0}\Bcup\Bvalue{T'u-T'v>0})
\le\gamma_0(u-v)=\gamma_k(u)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$Tu=T'u$;  as $u$ is arbitrary, $T=T'$.

\medskip

{\bf (b)(i)} If $\epsilon>0$, there is a $k\in\Bbb N$ such that
$\bar\mu_i\Bvalue{|u_i|>k}\le\epsilon$ for every $i\in I$.   Now there is
an $l\in\Bbb N$ such that $|h(t)|\le l$ whenever $|t|\le k$.   So
$\Bvalue{|\bar h(u_i)|>l}\Bsubseteq\Bvalue{|u_i|>k}$ and
$\bar\mu_i\Bvalue{|\bar h(u_i)|>l}\le\epsilon$ for every $i\in I$.
As $\epsilon$ is arbitrary, $\familyiI{\bar h(u_i)}\in W_0$.

\medskip

\quad{\bf (ii)} Again take any $\epsilon>0$.   Let $k\in\Bbb N$ be such
that $\bar\mu_ia_i\le\epsilon$ for every $i\in I$, where
$a_i=\Bvalue{|u_i|>k}$.   By the Stone-Weierstrass theorem in the form
281E, there is a polynomial $g:\Bbb R\to\Bbb R$ such that
$|g(t)-h(t)|\le\epsilon$ whenever $|t|\le k$.   Setting
$v_i=\bar h(u_i)$, $v'_i=\bar g(u_i)$, $v=\familyiI{u_i}$ and
$v'=\familyiI{v'_i}$, we have
$\Bvalue{|v_i-v'_i|>\epsilon}\Bsubseteq a_i$ for every $i$
(use 364Ib for a quick check of the calculation).   Because $T$ is
multiplicative (and $T(\familyiI{\chi 1_{\frak A_i}})=\chi 1_{\frak B}$),
$Tv'=\bar g(Tu)$.   So

$$\eqalignno{\Bvalue{|Tv-\bar h(Tu)|>2\epsilon}
&\Bsubseteq\Bvalue{T|v-v'|>\epsilon}
   \Bcup\Bvalue{|\bar g(Tu)-\bar h(Tu)|>\epsilon}\cr
&\Bsubseteq\pi(\familyiI{\Bvalue{|v_i-v'_i|>\epsilon}})
   \Bcup\Bvalue{|Tu|>k}\cr
\displaycause{using (b)}
&\Bsubseteq\pi(\familyiI{a_i})\cr}$$

\noindent (see (a-v) above),
which has measure at most $\epsilon$.   As $\epsilon$ is
arbitrary, $Tv=\bar h(Tu)$, as claimed.
}%end of proof of 377B

\leader{377C}{Theorem} Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a
non-empty
family of probability algebras, $(\frak B,\bar\nu)$ a probability algebra,
and $\pi:\prod_{i\in I}\frak A_i\to\frak B$ a Boolean homomorphism such
that $\bar\nu\pi(\familyiI{a_i})\le\sup_{i\in I}\bar\mu_ia_i$ whenever
$\familyiI{a_i}\in\prod_{i\in I}\frak A_i$.
Let $W_0\subseteq\prod_{i\in I}L^0(\frak A_i)$ and $T:W_0\to L^0(\frak B)$
be as in 377B.
Suppose {\it either} that every $\frak A_i$ is atomless {\it or} that
there is an ultrafilter $\Cal F$ on $I$ such that
$\bar\nu\pi(\familyiI{a_i})=\lim_{i\to\Cal F}\bar\mu_ia_i$ whenever
$\familyiI{a_i}$ in $\prod_{i\in I}\frak A_i$.
For $1\le p\le\infty$
let $W_p$ be the subspace of $\prod_{i\in I}L^0(\frak A_i)$
consisting of families $\familyiI{u_i}$ such that
$\sup_{i\in I}\|u_i\|_p$ is finite.   Then
$T[W_p]\subseteq L^p(\frak B,\bar\nu)$, and
$\|Tu\|_p\le\sup_{i\in I}\|u_i\|_p$ whenever $u=\familyiI{u_i}$ belongs to
$W_p$.

\proof{{\bf (a)} I should begin by explaining why $W_1\subseteq W_0$.
All we need to observe is that if $u=\familyiI{u_i}$ belongs to $W_1$,
so that $\gamma=\sup_{i\in I}\|u_i\|_1$ is finite, then

\Centerline{$\inf_{k\ge 1}\sup_{i\in I}\bar\mu_i\Bvalue{u_i>k}
\le\inf_{k\ge 1}\Bover{\gamma}k=0$,}

\noindent so $u\in W_0$.   Of course we now have $W_p\subseteq W_1$ for
$p\ge 1$, because every $(\frak A_i,\bar\mu_i)$ is a probability algebra.

\medskip

{\bf (b)} I start real work on the proof
with a note on the case in which every $\frak A_i$
is atomless.   Suppose that this is so, and that we are given a family
$\familyiI{a_i}\in\prod_{i\in I}\frak A_i$ and $\gamma\in\Bbb Q\cap[0,1]$.
Then there is a family $\familyiI{a'_i}$ such that
$a'_i\Bsubseteq a_i$ and $\bar\mu_ia'_i=\gamma\mu_ia_i$ for every $i\in I$,
and

\Centerline{$\gamma\bar\nu\pi(\familyiI{a_i})
\le\bar\nu\pi(\familyiI{a'_i})$.}

\Prf\ For each $i\in I$, we can find a non-decreasing
family $\langle a_{it}\rangle_{t\in[0,1]}$ in $\frak A_i$ such that
$a_{i1}=a_i$ and $\bar\mu_ia_{it}=t\bar\mu a_i$ for every $t\in[0,1]$.
Set $b(t)=\pi(\familyiI{a_{it}})$ and
$\beta(t)=\bar\nu b(t)$ for $t\in[0,1]$;  then
$\beta(s)\le\beta(t)\le\beta(s)+t-s$ for $0\le s\le t\le 1$, because

\Centerline{$\beta(t)-\beta(s)
=\bar\nu\pi(\familyiI{a_{it}\Bsetminus a_{is}})
\le\sup_{i\in I}\bar\mu_i(a_{it}\Bsetminus a_{is})
=(t-s)\sup_{i\in I}\bar\mu_ia_i
\le t-s$.}

\noindent Let $n\ge 1$ be such that $\Bover1n\le\epsilon$ and $m=n\gamma$
is an integer, and set
$\alpha_i=\beta(\bover{i+1}n)-\beta(\bover{i}n)$ for $i<n$;  then

\Centerline{$\sum_{i=0}^{n-1}\alpha_i=\beta(1)=\bar\nu b(1)$.}

\noindent Consider the possible values of
$\gamma_K=\sum_{k\in K}\alpha_k$ for
sets $K\in[n]^m$.
(I am thinking of $n$ as the set $\{0,1,\ldots,n-1\}$.)
The average value of $\gamma_K$ over all $m$-element subsets of $n$ is just
$\bover{m}n\beta(1)=\gamma\beta(1)$, so there is some $K$ such that
$\gamma_K\ge\gamma\beta(1)$.

Set

\Centerline{$a'_i=\sup_{k\in K}a_{i,(k+1)/n}\Bsetminus a_{i,k/n}$}

\noindent for $i\in I$.   Then $\bar\mu_ia'_i=\gamma\bar\mu_ia_i$ for
every $i$, while

\Centerline{$\bar\nu\pi(\familyiI{a'_i})
=\sup_{k\in K}\bar\nu(b(\bover{k+1}n)\Bsetminus b(\bover{k}n))
=\sum_{k\in K}\alpha_k$}

\noindent is at least $\gamma\beta(1)$, as required.\ \Qed

\medskip

{\bf (c)} We find now that under either of the hypotheses proposed,

\Centerline{$\sum_{k=0}^n\gamma_k\bar\nu\pi(\familyiI{a_{ki}})
\le\sup_{i\in I}\sum_{k=0}^n\gamma_k\mu_ia_{ki}$}

\noindent whenever $\gamma_0,\ldots,\gamma_n\ge 0$ are rational and
$\langle a_{ki}\rangle_{k\le n}$ is a disjoint family in $\frak A_i$ for
each $i\in I$.

\medskip

\Prf{\bf (i)} Consider first the case in which every $\frak A_i$ is
atomless and every $\gamma_k$ is between $0$ and $1$.   In this case, given
$\epsilon>0$, (b) above tells us that
we can find $a'_{ki}\Bsubseteq a_{ki}$, for $i\in I$ and
$k\le n$, such that $\bar\mu_ia'_{ki}=\gamma_k\bar\mu_ia_{ki}$ and

\Centerline{$\gamma_k\bar\nu\pi(\familyiI{a_{ki}})
\le\bar\nu\pi(\familyiI{a'_{ki}})$.}

\noindent Set $c_i=\sup_{k\le n}a'_{ki}$ for $i\in I$;  then

$$\eqalign{\sum_{k=0}^n\gamma_k\bar\nu\pi(\familyiI{a_{ki}})
&\le\sum_{k=0}^n\bar\nu\pi(\familyiI{a'_{ki}})
=\bar\nu\pi(\sup_{k\le n}\familyiI{a'_{ki}})
=\bar\nu\pi(\familyiI{c_i})\cr
&\le\sup_{i\in I}\bar\mu_ic_i
=\sup_{i\in I}\sum_{k=0}^n\bar\mu_ia'_{ki}
=\sup_{i\in I}\sum_{k=0}^n\gamma_k\bar\mu_ia_{ki},\cr}$$

\noindent as required.

\medskip

\quad{\bf (ii)} Because $T$ is linear, it follows at once that the result
is true for any rational $\gamma_0,\ldots,\gamma_n\ge 0$, if every
$\frak A_i$ is atomless.

\medskip

\quad{\bf (iii)} Now consider the case in which there is an ultrafilter
$\Cal F$ on $I$ such that
$\bar\nu\pi(\familyiI{a_i})=\lim_{i\to\Cal F}\bar\mu_ia_i$ for every
$\familyiI{a_i}$.   In this case, given $\epsilon>0$, the set

\Centerline{$J=\{j:j\in I$,
$\bar\nu(\familyiI{a_{ki}})\le\bar\mu_ja_{kj}+\epsilon$
for every $k\le n\}$}

\noindent belongs to $\Cal F$ and is not empty.   Take any $j\in J$;  then

\Centerline{$\sum_{k=0}^n\gamma_k\bar\nu\pi(\familyiI{a_{ki}})
\le\sum_{k=0}^n\gamma_k(\bar\mu_ja_{kj}+\epsilon)
\le\epsilon\sum_{k=0}^n\gamma_k
  +\sup_{i\in I}\sum_{k=0}^n\gamma_k\bar\mu_ia_{ki}$.}

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

\medskip

{\bf (d)} Next,
$\int Tu\le\sup_{i\in I}\int u_i$ whenever $u=\familyiI{u_i}$ belongs to
$W_{\infty}^+$.   \Prf\ Let $\epsilon>0$ and let $n\in\Bbb N$ be such that
$\|u_i\|_{\infty}\le n\epsilon$ for every $i\in I$.   For $i\in I$ and
$k\le n$, set
$a_{ki}=\Bvalue{u_i>k\epsilon}\Bsetminus\Bvalue{u_i>(k+1)\epsilon}$;
for $i\in I$, set $u'_i=\sum_{k=0}^nk\epsilon\chi a_{ki}$;
then $u'_i\le u_i\le u'_i+\epsilon\chi 1_{\frak A_i}$.   Setting
$u'=\familyiI{u'_i}$, $Tu\le Tu'+\epsilon\chi 1_{\frak B}$, so

$$\eqalignno{\int Tu-\epsilon
&\le\int Tu'
=\int\sum_{k=0}^nk\epsilon\chi\pi(\familyiI{a_{ki}})\cr
&=\sum_{k=0}^nk\epsilon\bar\nu\pi(\familyiI{a_{ki}})
\le\sup_{i\in I}\sum_{k=0}^nk\epsilon\bar\mu_ia_{ki}\cr
\displaycause{by (c)}
&=\sup_{i\in I}\int u'_i
\le\sup_{i\in I}\int u_i.\cr}$$

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

\medskip

{\bf (d)} It follows that $Tu\in L^1(\frak B,\bar\nu)$ and
$\int Tu\le\sup_{i\in I}\int u_i$ whenever $u=\familyiI{u_i}$ belongs to
$W_1^+$ and $u\ge 0$.
\Prf\ Set $\gamma=\sup_{i\in I}\int u_i$.   Let
$\epsilon>0$.   Set $\gamma'=\gamma/\epsilon$.
For $i\in I$ set
$v_i=u_i\wedge\gamma'\chi 1_{\frak A_i}$;  set $v=\familyiI{v_i}$.   Then
$v\in W_{\infty}$ and

\Centerline{$\int Tv\le\sup_{i\in I}\int v_i
\le\sup_{i\in I}\int u_i=\gamma$}

\noindent by (c) above.   Also
$\Bvalue{Tu-Tv>0}\Bsubseteq\pi(\familyiI{\Bvalue{u_i>\gamma'}})$,
by 377Ba, so

\Centerline{$\bar\nu\Bvalue{Tu-Tv>0}
\le\sup_{i\in I}\bar\mu_i\Bvalue{u_i>\gamma'}\le\epsilon$.}

Thus for each $n\in\Bbb N$ we can find a $w_n\in L^{\infty}(\frak B)$ such
that $0\le w_n\le Tu$, $\int w_n\le\gamma$ and
$\bar\nu\Bvalue{Tu-w_n>0}\le 2^{-n}$.
Set $w'_n=\inf_{i\ge n}w_i$ for each
$n$;  then $\sequencen{w'_n}$ is a non-decreasing sequence with
supremum $Tu$ in $L^0(\frak B)$,
while $\int w'_n\le\gamma$ for every $n$.   Consequently
$Tu\in L^1(\frak B,\bar\nu)$ and $\int Tu\le\gamma$, as claimed.\ \Qed

\medskip

{\bf (e)} Because $T$ is a Riesz homomorphism, $Tu\in L^1(\frak B,\bar\nu)$
and $\|Tu\|_1=\int T|u|$ is at most
$\sup_{i\in I}\int|u_i|=\sup_{i\in I}\|u_i\|_1$ for every $u\in W_1$.

\medskip

{\bf (f)} Now suppose that $p\in\ooint{1,\infty}$ and that
$u=\familyiI{u_i}$ belongs to $W_p$.   In this case, $\familyiI{|u_i|^p}$
belongs to $W_1$, so $T(\familyiI{|u_i|^p})\in L^1(\frak B,\bar\nu)$ and
$\int T(\familyiI{|u_i|^p})\le\sup_{i\in I}\int|u_i|^p$.   By 377Bb,
with $h(t)=|t|^p$, $T(\familyiI{|u_i|^p})=|Tu|^p$.   So
$Tu\in L^p(\frak B,\bar\nu)$ and

\Centerline{$\|Tu\|_p
=(\int|Tu|^p)^{1/p}
\le\sup_{i\in I}(\int|u_i|^p)^{1/p}
=\sup_{i\in I}\|u_i\|_p$}

\noindent as claimed.
}%end of proof of 377C

\leader{377D}{}\cmmnt{ The original motivation for the work of this
section was to understand the function spaces associated with the reduced
products of \S328.   For these we have various simplifications in addition
to that observed in 377C.

\medskip

\noindent}{\bf Theorem} Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a
family of probability algebras, $\Cal F$ an ultrafilter on $I$,
and $(\frak B,\bar\nu)$ a probability
algebra.   Let $\frak A$ be the simple product $\prod_{i\in I}\frak A_i$
and $\pi:\frak A\to\frak B$ a Boolean homomorphism such that
$\bar\nu\pi(\familyiI{a_i})=\lim_{i\to\Cal F}\bar\mu_ia_i$
whenever $\familyiI{a_i}\in\frak A$.   Let
$W_0\subseteq\prod_{i\in I}L^0(\frak A_i)$ and $T:W_0\to L^0(\frak B)$ be
as in 377B-377C.

(a) If $u=\familyiI{u_i}$ belongs to $W_0$ and $\{i:i\in I$, $u_i=0\}\in\Cal F$, then
$Tu=0$.

(b) For $1\le p\le\infty$, write $W_p$ for the set of those families
$\familyiI{u_i}\in\prod_{i\in I}L^p(\frak A_i,\bar\mu_i)$ such that
$\sup_{i\in I}\|u_i\|_p$ is finite.   Then $Tu\in L^p(\frak B,\bar\nu)$
and $\|Tu\|_p\le\lim_{i\to\Cal F}\|u_i\|_p$ whenever
$u=\familyiI{u_i}$ belongs to $W_p$.

(c) Let $W_{ui}$ be the subspace of
$\prod_{i\in I}L^1(\frak A_i,\bar\mu_i)$
consisting of families $\familyiI{u_i}$ such
that $\inf_{k\in\Bbb N}\sup_{i\in I}\int(|u_i|-k\chi 1_{\frak A_i})^+=0$.
Then $\int Tu=\lim_{i\to\Cal F}\int u_i$ and
$\|Tu\|_1=\lim_{i\to\Cal F}\|u_i\|_1$
whenever $u=\familyiI{u_i}$ belongs to $W_{ui}$.

(d) Suppose now that $\pi[\frak A]=\frak B$.

\quad(i) $T[W_0]=L^0(\frak B)$.

\quad(ii) $T[W_{ui}]=L^1(\frak B,\bar\nu)$.

\quad(iii) If $p\in[1,\infty]$, then $T[W_p]=L^p(\frak B,\bar\nu)$ and for
every $w\in L^p(\frak B,\bar\nu)$ there is a
$u=\familyiI{u_i}$ in $W_p$ such that $Tu=w$ and
$\sup_{i\in I}\|u_i\|_p=\|w\|_p$.

\proof{{\bf (a)} Setting

$$\eqalign{a_i&=1_{\frak A_i}\text{ if }u_i\ne 0,\cr
&=0\text{ if }u_i=0,\cr}$$

\noindent $\familyiI{a_i}\in\frak A$ and
$\bar\nu\pi(\familyiI{a_i})=\lim_{i\to\Cal F}\bar\mu_ia_i=0$, so
$\pi(\familyiI{a_i})=0$.   Accordingly

\Centerline{$Tu=T(\familyiI{u_i\times\chi a_i})
=Tu\times T(\familyiI{\chi a_i})=Tu\times\chi\pi(\familyiI{a_i})=0$.}

\medskip

{\bf (b)} Suppose that $u=\familyiI{u_i}\in W_p$ and that $J\in\Cal F$.
Set

$$\eqalign{v_i&=u_i\text{ if }i\in J,\cr
&=0\text{ if }i\in I\setminus J;\cr}$$

\noindent then, putting (a) and 377C together,

\Centerline{$\|Tu\|_p=\|Tv\|_p\le\sup_{i\in I}\|v_i\|_p
=\sup_{i\in J}\|u_i\|_p$.}

\noindent As $J$ is arbitrary,
$\|Tu\|_p\le\lim_{i\to\Cal F}\|u_i\|_p$.

\medskip

{\bf (c)(i)} Clearly $W_{ui}$ is a
solid linear subspace of $W_1$.
Suppose that $u=\familyiI{u_i}\in W_{ui}^+$ and $\epsilon>0$.
Let $n\ge 1$ be such that
$\int(u_i-n\epsilon\chi 1_{\frak A_i})^+\le\epsilon$ for every $i\in I$.
For $i\in I$ and $k\le n$, set $a_{ki}=\Bvalue{u_i>k\epsilon}$;
set $v_i=\sum_{k=1}^nk\epsilon\chi a_{ki}$, so that

\Centerline{$v_i\le u_i
\le v_i+\epsilon\chi 1_{\frak A_i}+(u_i-n\epsilon\chi 1_{\frak A_i})^+$,
\quad$\int u_i\le\int v_i+2\epsilon$.}

\noindent If $v=\familyiI{v_i}$, then

$$\eqalign{\int Tu
&=\|Tu\|_1
\le\lim_{i\to\Cal F}\|u_i\|_1
=\lim_{i\to\Cal F}\int u_i\cr
&\le 2\epsilon+\lim_{i\to\Cal F}\int v_i
=2\epsilon+\sum_{k=1}^nk\epsilon\lim_{i\to\Cal F}\bar\mu_ia_{ki}\cr
&=2\epsilon+\sum_{k=1}^nk\epsilon\bar\nu\pi(\familyiI{a_{ki}})
=2\epsilon+\int\sum_{k=1}^nk\epsilon\chi\pi(\familyiI{a_{ki}})\cr
&=2\epsilon+\int Tv
\le 2\epsilon+\int Tu.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\int Tu=\lim_{i\to\Cal F}\int u_i$.

\medskip

\quad{\bf (ii)} It follows at once that
$\int Tu=\lim_{i\to\Cal F}\int u_i$
and that

\Centerline{$\|Tu\|_1=\int|Tu|=\int T|u|=\lim_{i\to\Cal F}\int|u_i|
=\lim_{i\to\Cal F}\|u_i\|_1$}

\noindent whenever $u=\familyiI{u_i}\in W_{ui}$.

\medskip

{\bf (d)(i)}\grheada\
Let $T_{\pi}:L^{\infty}(\frak A)\to L^{\infty}(\frak B)$
be the Riesz homomorphism associated with the Boolean homomorphism
$\pi:\frak A\to\frak B$.   Since $\pi$ is surjective,
363Fd tells us that $T_{\pi}$ is surjective.
Identifying $W_{\infty}$ with
$L^{\infty}(\frak A)$, and $T\restr W_{\infty}$ with
$T_{\pi}$, as in part (a) of the
the proof of 377B, we see that $T[W_{\infty}]=L^{\infty}(\frak B)$.
Moreover, 363Fd tells us also that if $w\in L^{\infty}$ there is a
$v\in L^{\infty}(\frak A)$ such that $T_{\pi}v=w$ and
$\|v\|_{\infty}=\|w\|_{\infty}$;  translating this into terms of
$W_{\infty}$, we have a $u=\familyiI{u_i}\in W_{\infty}$ such that
$Tu=w$ and $\sup_{i\in I}\|u_i\|_{\infty}=\|w\|_{\infty}$.

It will be useful to know that if $b\in\frak B$ and $\epsilon>0$ there is a
family $\familyiI{a_i}\in\frak A$ such that $\pi(\familyiI{a_i})=b$ and
$\sup_{i\in I}\bar\mu_ia_i\le\bar\nu b+\epsilon$.   \Prf\ By hypothesis,
there is a family $\familyiI{a'_i}\in\frak A$ such that
$\pi(\familyiI{a'_i})=b$, and $\bar\nu b=\lim_{i\to\Cal F}\bar\mu_ia_i$.
Set

$$\eqalign{a_i
&=a'_i\text{ if }\bar\mu_ia_i\le\bar\nu b+\epsilon,\cr
&=0\text{ for other }i\in I.\cr}$$

\noindent Then $\lim_{i\to\Cal F}\bar\mu_i(a'_i\Bsymmdiff a_i)=0$ so
$\pi(\familyiI{a'_i\Bsymmdiff a_i})=0$ and $\pi(\familyiI{a_i})=b$, while
$\bar\mu_ia_i\le\bar\nu b+\epsilon$ for every $i\in I$.\ \Qed

\medskip

\qquad\grheadb\ Now suppose that $w\in L^0(\frak B)^+$.   For each
$n\in\Bbb N$, set $w_n=w\wedge n\chi 1_{\frak B}$ and
let $u^{(n)}=\familyiI{u_{ni}}\in W_{\infty}$ be such that
$Tu^{(n)}=w_{n+1}-w_n$ and $\|u_{ni}\|_{\infty}\le 1$ for
every $i\in I$.   Next, for each $n$, set $b_n=\Bvalue{w_{n+1}-w_n>0}$,
and let
$\familyiI{a_{ni}}\in\frak A$ be such that $\pi(\familyiI{a_{ni}})=b_n$ and
$\sup_{i\in I}\bar\mu_ia_{ni}\le\bar\nu b_n+2^{-n}$.   If we set
$a'_{ni}=\inf_{m\le n}a_{mi}$ and
$u'_{ni}=u_{ni}\times\chi a'_{ni}$, we shall have

$$\eqalign{T(\familyiI{u'_{ni}})
&=T(\familyiI{u_{ni}})\times\chi\pi(\familyiI{a'_{ni}})\cr
&=(w_{n+1}-w_n)\times\inf_{m\le n}\chi b_m
=w_{n+1}-w_n\cr}$$

\noindent for every $n$.   Also, for each $i\in I$, $\sequencen{a'_{ni}}$
is non-increasing and

\Centerline{$\lim_{n\to\infty}\bar\mu_ia'_{ni}
\le\lim_{n\to\infty}\bar\nu b_n+2^{-n}=0$.}

\noindent So $v_i=\sup_{n\in\Bbb N}\sum_{m=0}^nu'_{ni}$ is defined in
$L^0(\frak A_i)$, and

\Centerline{$
\inf_{k\in\Bbb N}\sup_{i\in I}\bar\mu_i\Bvalue{v_i>k}
\le\inf_{k\in\Bbb N}\sup_{i\in I}\bar\mu_ia'_{ki}=0$.}

\noindent Thus $v=\familyiI{v_i}$ belongs to $W_0$ and we can speak of
$Tv$.   Of course

\Centerline{$Tv\ge\sum_{m=0}^nT(\familyiI{u'_{ni}})=w_{n+1}$}

\noindent for every $n$, so $Tv\ge w$.   On the other hand, for any
$n\in\Bbb N$,

\Centerline{$\Bvalue{v_i-\sum_{m=0}^nu'_{ni}>0}\Bsubseteq a'_{ni}$}

\noindent for every $i$, so $\Bvalue{Tv-w_{n+1}>0}\Bsubseteq b_n$, by 377B;  as
$\inf_{n\in\Bbb N}b_n=0$, $Tv=\sup_{n\in\Bbb N}w_n=w$.

\medskip

\qquad\grheadc\ Thus $T[W_0]\supseteq L^0(\frak B)^+$;  as $T$ is linear,
$T[W_0]=L^0(\frak B)$.

\medskip

\quad{\bf (ii)} Now suppose that $w\in L^1(\frak B,\bar\nu)^+$.
In this case, repeat the process of (i-$\beta$) above.
This time, observe that as $\chi b_{n+1}\le w_{n+1}-w_n$ for every $n$,
$\sum_{n=0}^{\infty}\bar\nu b_n\le 1+\int w$ is finite.   Consequently, in
the first place,

\Centerline{$\sum_{n=0}^{\infty}\int u'_{ni}
\le\sum_{n=0}^{\infty}\bar\mu_ia_{ni}
\le\sum_{n=0}^{\infty}\bar\nu b_n+2^{-n}$}

\noindent is finite, and $v_i\in L^1(\frak A_i,\bar\mu_i)$, for every
$i\in I$.   But also, for any $k\in\Bbb N$ and $i\in I$,

\Centerline{$\int(v_i-k\chi 1_{\frak A_i})^+
\le\sum_{n=k}^{\infty}\int u'_{ni}
\le\sum_{n=k}^{\infty}\bar\nu b_n+2^{-n}
\to 0$}

\noindent $k\to\infty$.   So $v\in W_{ui}$ and $w\in T[W_{ui}]$.   Because
$W_{ui}$ is a linear subspace of $W_0$, $T[W_{ui}]=L^1(\frak B,\bar\nu)$.

\medskip

\quad{\bf (iii)}\grheada\ If $p=\infty$ the result has already been dealt
with in (i-$\alpha$) above.

\medskip

\qquad\grheadb\ For the case $p=1$, take $w\in L^1(\frak B,\bar\nu)$.
Let $v=\familyiI{v_i}\in W_{ui}$ be such that $Tv=w$.   For $i\in I$ set

$$\eqalign{u_i&=\Bover{\|w\|_1}{\|v_i\|_1}v_i
  \text{ if }\|v_i\|_1>\|w\|_1,\cr
&=v_i\text{ otherwise}.\cr}$$

\noindent Then

\Centerline{$\bar\mu\Bvalue{(|u_i|-k>0}
\le\bar\mu\Bvalue{|v_i|-k>0}$}

\noindent for all $k\in\Bbb N$ and $i\in I$, so
$u=\familyiI{u_i}\in W_{ui}$.   Since $\lim_{i\to\Cal F}\|v_i\|_1=\|w\|_1$,
by (c) above, $\lim_{i\to\Cal F}\|u_i-v_i\|_1=0$ and $Tu=Tv=w$, by (b).
And of course $\|u_i\|_1\le\|w\|_1$ for every $i$.

\medskip

\qquad\grheadc\ Now suppose that $1<p<\infty$ and that
$w\in L^p(\frak B,\bar\nu)$.   By ($\beta$), there is a
$v=\familyiI{v_i}\in W_1$ such that $Tv=|w|^p$ and
$\sup_{i\in I}\|v_i\|_1=\|w\|_p^p$.   Set $v'_i=|v_i|^{1/p}$ for each
$i$;  then $v'=\familyiI{v'_i}\in W_p$ and $Tv'=|w|$, by 377Bb.   Next, $w$
is expressible as $|w|\times\tilde w$, where
$\tilde w\in L^{\infty}(\frak B)$ and $\|\tilde w\|_{\infty}\le 1$.
There is a $\tilde v=\familyiI{\tilde v_i}\in W_{\infty}$ such that
$T\tilde v=\tilde w$ and $\sup_{i\in I}\|\tilde v_i\|_{\infty}=1$.
Set $u_i=v'_i\times\tilde v_i$ for each $i$;  then $u=\familyiI{u_i}$
belongs to $W_p$, $\|u_i\|_p\le\|w\|_p$ for every $i$, and $Tu=w$.
}%end of proof of 377D

\leader{377E}{}\cmmnt{ In the case of a reduced power of a probability
algebra we can express these ideas in a slightly different way.

\medskip

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be probability algebras, $I$ a set and $\Cal F$ an ultrafilter on $I$.
Let $\pi:\frak A^I\to\frak B$ be a Boolean homomorphism such that
$\bar\nu\pi(\familyiI{a_i})=\lim_{i\to\Cal F}\bar\mu a_i$ whenever
$\familyiI{a_i}\in\frak A^I$.   Let $W_0$ be the set of families in
$L^0(\frak A)^I$ which are bounded for the topology of convergence in
measure on $L^0(\frak A)$.

(a)(i) $W_0$ is a solid linear subspace and a subalgebra of
$L^0(\frak A)^I$,
and there is a unique multiplicative Riesz homomorphism
$T:W_0\to L^0(\frak B)$ such that
$T(\familyiI{\chi a_i})=\chi\pi(\familyiI{a_i})$ whenever
$\familyiI{a_i}\in\frak A^I$.

\quad(ii) $\Bvalue{Tu>0}\Bsubseteq\pi(\familyiI{\Bvalue{u_i>0}})$ whenever
$u=\familyiI{u_i}$ belongs to $W_0$.

\quad(iii) If $h:\Bbb R\to\Bbb R$ is a continuous function, and we write
$\bar h$ for the corresponding maps from $L^0$ to itself for either of the
spaces $L^0=L^0(\frak A)$, $L^0=L^0(\frak B)$, then
$\familyiI{\bar h(u_i)}\in W_0$ and
$T(\familyiI{\bar h(u_i)})=\bar h(Tu)$ whenever
$u=\familyiI{u_i}$ belongs to $W_0$.

(b)(i) For $1\le p\le\infty$
let $W_p$ be the subspace of $L^p(\frak A,\bar\mu)^I$
consisting of $\|\,\|_p$-bounded families.   Then
$T[W_p]\subseteq L^p(\frak B,\bar\nu)$, and
$\|Tu\|_p\le\lim_{i\to\Cal F}\|u_i\|_p$
whenever $u=\familyiI{u_i}$ belongs to $W_p$.

\quad(ii) Let $W_{ui}$ be the subspace of
$L^1(\frak A_i,\bar\mu_i)^I$
consisting of uniformly integrable families.
Then $\int Tu=\lim_{i\to\Cal F}\int u_i$ and
$\|Tu\|_1=\lim_{i\to\Cal F}\|u_i\|_1$
whenever $u=\familyiI{u_i}$ belongs to $W_{ui}$.

(c)(i) We have a measure-preserving Boolean homomorphism
$\tilde\pi:\frak A\to\frak B$ defined by setting
$\tilde\pi a=\pi(\familyiI{a})$ for each $a\in\frak A$.

\quad(ii) Let $P_{\tilde\pi}:L^1(\frak B,\bar\nu)\to L^1(\frak A,\bar\mu)$
be the conditional-expectation operator corresponding to
$\tilde\pi:\frak A\to\frak B$\cmmnt{ (365O)}.
If $\familyiI{u_i}$ is a uniformly
integrable family in $L^1(\frak A)$, then $P_{\tilde\pi}T(\familyiI{u_i})$
is the limit $\lim_{i\to\Cal F}u_i$ for the weak topology of
$L^1(\frak A,\bar\mu)$.

\quad(iii) Suppose that $1<p<\infty$ and that $\familyiI{u_i}$ is a
bounded family
in $L^p(\frak A,\bar\mu)$.   Then $P_{\tilde\pi}T(\familyiI{u_i})$ is the
limit $\lim_{i\to\Cal F}u_i$ for the weak topology of
$L^p(\frak A,\bar\mu)$.

\proof{{\bf (a)} By 367Rd, a family $\familyiI{u_i}$ in
$L^0(\frak A)$ is bounded for the topology of convergence in measure iff
$\inf_{k\in\Bbb N}\sup_{i\in I}\bar\mu\Bvalue{|u_i|>k}=0$.
So we just have a special case of 377B.

\medskip

{\bf (b)} Similarly, the condition
`$\inf_{k\in\Bbb N}\sup_{i\in I}\int(|u_i|-k\chi 1_{\frak A_i})^+=0$'
translates into
`$\{u_i:i\in I\}$ is uniformly integrable' (cf.\ 246Bd), so we are
looking at a special case of 377Db-377Dc.

\medskip

{\bf (c)(i)} $\tilde\pi$ is a Boolean homomorphism just because the
function taking $a\in\frak A$ into the constant family with value $a$ is a
Boolean homomorphism from $\frak A$ to $\frak A^I$.   The formula
`$\bar\nu\pi(\familyiI{a_i})=\lim_{i\to\Cal F}\bar\mu a_i$' now ensures
that $\tilde\pi$ is measure-preserving.

\medskip

\quad{\bf (ii)} By the defining formula for $P_{\tilde\pi}$ (365Oa),

$$\eqalignno{\int_aP_{\tilde\pi}T(\familyiI{u_i})
&=\int T(\familyiI{u_i})\times\chi\tilde\pi(a)
=\int T(\familyiI{u_i})\times\chi\pi(\familyiI{a})\cr
&=\int T(\familyiI{u_i})\times T(\familyiI{\chi a})\cr
&=\int T(\familyiI{u_i\times\chi a})
=\lim_{i\to\Cal F}\int u_i\times\chi a\cr
\displaycause{because $\{u_i\times\chi a:i\in I\}$ is uniformly integrable}
&=\lim_{i\to\Cal F}\int_au_i\cr}$$

\noindent for every $a\in\frak A$.   It follows that
$P_{\tilde\pi}T(\familyiI{u_i})=\lim_{i\to\Cal F}u_i$.   \Prf\ We have

\Centerline{$\int P_{\tilde\pi}T(\familyiI{u_i})\times v
=\lim_{i\to\Cal F}\int u_i\times v$}

\noindent whenever $v=\chi a$, for any $a\in\frak A$;  by linearity,
whenever $v\in S(\frak A)$, the space of $\frak A$-simple functions;
and by continuity, whenever $v\in L^{\infty}(\frak A)$ (because
$\{u_i:i\in I\}$ is $\|\,\|_1$-bounded, and $S(\frak A)$ is
$\|\,\|_{\infty}$-dense in $L^{\infty}(\frak A)$).
Since $L^{\infty}(\frak A)$ can be
identified with the dual of $L^1(\frak A,\bar\mu)$ (365Lc), we have the
required weak convergence.\ \Qed

\medskip

\quad{\bf (iii)} If $\{u_i:i\in I\}$ is $\|\,\|_p$-bounded,
where $1<p<\infty$, then it is uniformly integrable.   \Prf\
Set $q=\Bover{p}{p-1}$.   If $k\ge 1$,

\Centerline{$\inf_{k\ge 1}\sup_{i\in I}\int(|u_i|-k\chi 1_{\frak A})^+
\le\inf_{k\ge 1}\Bover1{k^{p-1}}\sup_{i\in I}\|u_i\|_p^p=0$.  \Qed}

\noindent So

\Centerline{$\int P_{\tilde\pi}T(\familyiI{u_i})\times v
=\lim_{i\to\Cal F}\int u_i\times v$}

\noindent for every $v\in S(\frak A)$, and therefore for every
$v\in L^q(\frak A,\bar\mu)$, since $v$ can be $\|\,\|_q$-approximated by
members of
$S(\frak A)$ (366C).   Since $L^q(\frak A,\bar\mu)$ can be identified with
$L^p(\frak A,\bar\mu)^*$, we again have weak convergence.
}%end of proof of 377E

\leader{377F}{Proposition} Let $(\frak A,\bar\mu)$ and
$(\frak A',\bar\mu')$ be probability algebras, $I$ a set and $\Cal F$ an
ultrafilter on $I$;  let $(\frak B,\bar\nu)$ and $(\frak B',\bar\nu')$ be
the reduced powers $(\frak A,\bar\mu)^I|\Cal F$,
$(\frak A',\bar\mu')^I|\Cal F$\cmmnt{ as described in
328A-328C}, %328A 328B 328C
with corresponding homomorphisms $\pi:\frak A^I\to\frak B$ and
$\pi':{\frak A'}^I\to\frak B'$.

(a) Writing $W_0$, $W_0'$ for the spaces of topologically
bounded families in $L^0(\frak A)^I$, $L^0(\frak A')^I$ respectively,
we have unique Riesz homomorphisms $T:W_0\to L^0(\frak B)$ and
$T':W_0'\to L^0(\frak B')$ such that
$T(\familyiI{\chi a_i})=\chi\pi(\familyiI{a_i})$,
$T'(\familyiI{\chi a'_i})=\chi\pi'(\familyiI{a'_i})$ whenever
$\familyiI{a_i}\in\frak A^I$ and $\familyiI{a'_i}\in(\frak A')^I$.

(b) Suppose that $S:L^1(\frak A,\bar\mu)\to L^1(\frak A',\bar\mu')$ is a
bounded linear operator.   Then we have a unique bounded linear operator
$\hat S:L^1(\frak B,\bar\nu)\to L^1(\frak B',\bar\nu')$ such that
$\hat ST(\familyiI{u_i})=T'(\familyiI{Su_i})$ whenever $\familyiI{u_i}$
is a uniformly integrable family in $L^1(\frak A,\bar\mu)$.

(c) The map
$S\mapsto\hat S$ is a norm-preserving Riesz homomorphism from
$\eurm B(L^1(\frak A,\bar\mu);L^1(\frak A',\bar\mu'))$ to
$\eurm B(L^1(\frak B,\bar\nu);\penalty-100L^1(\frak B',\bar\nu'))$.


\proof{{\bf (a)} Once again, this is nothing but a specialization of the
corresponding fragments of 377Ba and 377Ea.

\medskip

{\bf (b)} Write $W_{ui}$ for the space of uniformly integrable families
in $L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$.   If $\familyiI{u_i}\in W_{ui}$,
then $\familyiI{Su_i}$ is uniformly integrable in
$L^1_{\bar\mu'}=L^1(\frak A',\bar\mu')$ (because $\{u_i:i\in I\}$ and
$\{Su_i:i\in I\}$ are relatively weakly compact, as in 247D),
so belongs to $W'_0$, and we can speak of
$T'(\familyiI{Su_i})$.   If moreover $T(\familyiI{u_i})=0$, then
$\lim_{i\to\Cal F}\|u_i\|_1=0$ (377E(b-ii)), so
$\lim_{i\to\Cal F}\|Su_i\|_1=0$ and $T'(\familyiI{Su_i})=0$.
Finally, $T[W_{ui}]=L^1_{\bar\nu}=L^1(\frak B,\bar\nu)$ by 377D(d-ii).
So the given formula defines a linear operator
$\hat S:L^1_{\bar\nu}\to L^1_{\bar\nu'}=L^1(\frak B',\bar\nu')$.
Next, if $w\in L^1_{\bar\nu}$, we can take any family
$\familyiI{u_i}\in W_{ui}$ such that $T(\familyiI{u_i})=w$, and

$$\eqalignno{\|\hat Sw\|_1
&=\|T'(\familyiI{Su_i})\|_1
=\lim_{i\to\Cal F}\|Su_i\|_1\cr
\displaycause{377E(b-ii)}
&\le\|S\|\lim_{i\to\Cal F}\|u_i\|_1
=\|S\|\|w\|_1.\cr}$$

\noindent As $w$ is arbitrary, $\hat S$ is a bounded linear operator, and
$\|\hat S\|\le\|S\|$.   On the other hand, if
$u\in L^1_{\bar\mu}$ and $\|u\|_1\le 1$,
$\|T(\familyiI{u})\|_1\le 1$ so

\Centerline{$\|\hat S\|\ge\|\hat ST(\familyiI{u})\|
=\|T'(\familyiI{Su})\|_1=\|Su\|_1$;}

\noindent as $u$ is arbitrary, $\|\hat S\|\ge\|S\|$.

\medskip

{\bf (c)(i)} Recall from 371D that the Banach space
$\eurm B(L^1_{\bar\mu};L^1_{\bar\mu'})$ of continuous
linear operators is also the Dedekind complete Riesz space
$L^{\sim}(L^1_{\bar\mu};L^1_{\bar\mu'})$ of order-bounded linear operators,
and its norm is a Riesz norm;  similarly,
$\eurm B(L^1_{\bar\nu};L^1_{\bar\nu'})
=L^{\sim}(L^1_{\bar\nu};L^1_{\bar\nu'})$.
We have already seen that $S\mapsto\hat S$ is norm-preserving, and it is
clearly linear.   If $w\in(L^1_{\bar\nu})^+$, then, by
377D(d-ii), $w=T(\familyiI{u_i})$ for a family
$\familyiI{u_i}\in W_{ui}$;  since $T$ is a Riesz homomorphism,
$w=w^+=T(\familyiI{u_i^+}$;  so that
if $S\ge 0$ we shall have $\hat Sw=T'(\familyiI{Su_i^+})\ge 0$.
This shows
that $\hat S\ge 0$ whenever $S\ge 0$, so that $S\mapsto\hat S$ is a
positive linear operator.

\medskip

\quad{\bf (ii)} To show that $S\mapsto\hat S$ is a Riesz homomorphism, I
argue as follows.   Take any bounded linear operator
$S:L^1_{\bar\mu}\to L^1_{\bar\mu'}$ and $\epsilon>0$.   Then

\Centerline{$B=\{\sum_{k=0}^n|Sv_k|:
  v_0,\ldots,v_k\in(L^1_{\bar\mu})^+$,
$\sum_{k=0}^nv_k=\chi 1_{\frak A}\}$}

\noindent is an upwards-directed set in $L^1_{\bar\mu'}$ with
supremum $|S|(\chi 1_{\frak A})$ (371A, part (b) of the proof of 371B).
So we can find $v_0,\ldots,v_n\in(L^1_{\bar\mu})^+$ such that
$\sum_{k=0}^nv_k=\chi 1_{\frak A}$ and $\|v'\|_1\le\epsilon$, where
$v'=|S|(\chi 1_{\frak A})-\sum_{k=0}^n|Sv_k|\ge 0$.

Next, if $0\le u\le\chi 1_{\frak A}$ in $L^1_{\bar\mu}$, set
$u'=\chi 1_{\frak A}-u$;  we have

$$\eqalign{|S|(\chi 1_{\frak A})-v'
&=\sum_{k=0}^n|Sv_k|
\le\sum_{k=0}^n|S(u\times v_k)|+\sum_{k=0}^n|S(u'\times v_k)|\cr
&\le|S|(u)+|S|(u')
=|S|(\chi 1_{\frak A}).\cr}$$

\noindent So $|S|(u)-\sum_{k=0}^n|S(u\times v_k)|\le v'$ and
$\||S|(u)-\sum_{k=0}^n|S(u\times v_k)|\|_1\le\epsilon$.

Now take any $w\in L^1_{\bar\nu}$ such that $0\le w\le\chi 1_{\frak B}$.
Again because $T$ is a Riesz homomorphism and
$T(\familyiI{\chi 1_{\frak A}})=\chi 1_{\frak B}$, we can express $w$ as
$T(\familyiI{u_i})$ where
$0\le u_i\le\chi 1_{\frak A}$ for every $i$.   Consequently, setting
$v'_i=|S|u_i-\sum_{k=0}^n|S(u_i\times v_k)|$ for each $i$,
and $w'=T'(\familyiI{v'_i})$,

$$\eqalign{|S|\sphat\mskip4mu(w)
&=T'(\familyiI{|S|u_i})
=T'(\familyiI{\sum_{k=0}^n|S(u_i\times v_k)|+v'_i})\cr
&=\sum_{k=0}^n|T'(\familyiI{S(u_i\times v_k)})|+T'(\familyiI{v'_i})\cr
&=\sum_{k=0}^n|\hat ST(\familyiI{u_i\times v_k})|+w'
=\sum_{k=0}^n|\hat S\bigl(T(\familyiI{u_i})\times T(\familyiI{v_k})\bigr)|
   +w'\cr
&\le\sum_{k=0}^n
  |\hat S|\bigl(T(\familyiI{u_i})\times T(\familyiI{v_k})\bigr)
   +w'
=|\hat S|(w)+w'\cr}$$

\noindent because

\Centerline{$\sum_{k=0}^nT(\familyiI{v_k})=T(\familyiI{\chi 1_{\frak A}})
=\chi 1_{\frak B}$.}

\noindent But we also have
$\|w'\|_1=\lim_{i\to\Cal F}\|v'_i\|_1\le\epsilon$, while
$|\hat S|\le|S|\sphat$.   So we conclude that
$\||S|\sphat\mskip4mu(w)-|\hat S|(w)\|_1\le\epsilon$;  as $\epsilon$ is arbitrary,
$|S|\sphat\mskip4mu(w)=|\hat S|(w)$.

This is true whenever $0\le w\le\chi 1_{\frak B}$.   But as both
$|S|\sphat\mskip4mu$ and $|\hat S|$ are continuous linear operators, and
$L^{\infty}(\frak B)$ is dense in $L^1_{\bar\nu}$, $|S|\sphat=|\hat S|$.
As $S$ is arbitrary, we have a Riesz homomorphism (352G).
}%end of proof of 377F

\leader{377G}{Projective limits:  Proposition}
Let $(I,\le)$, $\familyiI{(\frak A_i,\bar\mu_i)}$ and
$\langle\pi_{ij}\rangle_{i\le j}$ be such that $(I,\le)$ is a non-empty
upwards-directed partially ordered
set, every $(\frak A_i,\bar\mu_i)$ is a probability
algebra, $\pi_{ij}:\frak A_j\to\frak A_i$ is a measure-preserving
Boolean homomorphism whenever $i\le j$ in $I$, and
$\pi_{ik}=\pi_{ij}\pi_{jk}$ whenever $i\le j\le k$.   Let
$(\frak C,\bar\lambda,\familyiI{\pi_i})$ be the corresponding projective
limit\cmmnt{ (328I)}.   Write $L^1_{\bar\mu_i}$ for
$L^1(\frak A_i,\bar\mu_i)$ and $L^1_{\bar\lambda}$ for
$L^1(\frak C,\bar\lambda)$.   For $i\le j$ in $I$, let
$T_{ij}:L^1_{\bar\mu_j}\to L^1_{\bar\mu_i}$ and
$P_{ij}:L^1_{\bar\mu_i}\to L^1_{\bar\mu_j}$ be the
norm-preserving Riesz homomorphism and the positive linear operator
corresponding to $\pi_{ij}:\frak A_j\to\frak A_i$\cmmnt{ (365N, 365O)},
and $T_i:L^1_{\bar\lambda}\to L^1_{\bar\mu_i}$,
$P_i:L^1_{\bar\mu_i}\to L^1_{\bar\lambda}$ the operators corresponding to
$\pi_i:\frak C\to\frak A_i$.   Let $X$ be any set.

(a) Suppose that for each $i\in I$ we are given a function
$S_i:L^1_{\bar\mu_i}\to X$ such that $S_iT_{ij}=S_j$ whenever $i\le j$
in $I$.   Then there is a unique function $S:L^1_{\bar\lambda}\to X$ such
that $S=S_iT_i$ for every $i\in I$.

(b) Suppose that for each $i\in I$ we are given a function
$S_i:X\to L^1_{\bar\mu_i}$ such that $T_{ij}S_j=S_i$ whenever $i\le j$
in $I$.   Then there is a unique function $S:X\to L^1_{\bar\lambda}$ such
that $T_iS=S_i$ for every $i\in I$.

(c) Suppose that $X$ is a topological space, and
for each $i\in I$ we are given a norm-continuous function
$S_i:L^1_{\bar\mu_i}\to X$ such that $S_jP_{ij}=S_i$ whenever $i\le j$
in $I$.   Then there is a unique function $S:L^1_{\bar\lambda}\to X$ such
that $SP_i=S_i$ for every $i\in I$.

(d) Suppose that for each $i\in I$ we are given a function
$S_i:X\to L^1_{\bar\mu_i}$ such that $P_{ij}S_i=S_j$ whenever $i\le j$ in
$I$.   Then there is a unique function $S:X\to L^1_{\bar\lambda}$ such that
$S=P_iS_i$ for every $i\in I$.

\ifwithproofs\medskip

\noindent{\bf proof: preliminary remarks (i)} It will be helpful to
recall some basic facts from \S\S328 and 365.
If $i\le j$ in $I$, then by the definition of `projective limit' we have
$\pi_{ij}\pi_j=\pi_i$ so
$T_{ij}T_j=T_i$ and $P_jP_{ij}=P_i$.
Also $P_{ij}T_{ij}$ is the identity operator on $L^1_{\bar\mu_j}$, and
$P_iT_i$ is the identity operator on $L^1_{\bar\lambda}$.

\medskip

\quad{\bf (ii)}
At a deeper level, we have useful concretizations of
$(\frak C,\bar\lambda)$, as follows.   Fix $i\in I$ for the moment.
For $j\ge i$, set $\frak B_j=\pi_{ij}[\frak A_j]$,
$\bar\nu_j=\bar\mu_i\restrp\frak B_j$;  then
$\frak B_j$ is a closed subalgebra of $\frak A_i$, isomorphic (as
probability algebra) to $\frak A_j$.   If $u\in L^1_{\bar\mu_i}$ and
$b\in\frak B_j$, set $b'=\pi_{ij}^{-1}b\in\frak A_j$;  then

\Centerline{$\int_bu=\int_{\pi_{ij}b'}u=\int_{b'}P_{ij}u
=\int_bT_{ij}P_{ij}u$;}

\noindent thus $T_{ij}P_{ij}$ is the conditional expectation
$P_{\frak B_j}:L^1_{\bar\mu_i}\to L^1(\frak B_j,\bar\nu_j)$,
identifying $L^1(\frak B_j,\bar\nu_j)$ with
$L^1_{\bar\mu_i}\cap L^0(\frak B_j)$ as in 365Qa.

If $k\ge j$, then $\pi_{ik}=\pi_{ij}\pi_{jk}$ so
$\frak B_k\subseteq\frak B_j$;  thus $\Bbb B=\{\frak B_j:j\ge i\}$ is
downwards-directed.   Set $\frak D=\bigcap\Bbb B$,
$\bar\nu=\bar\mu\restrp\frak D$.

For $k\ge i$, set $\phi_k=\pi_{ik}^{-1}\restrp\frak D:\frak D\to\frak A_k$;
then $\phi_k$ is a measure-preserving Boolean homomorphism, and
$\phi_j=\pi_{jk}\phi_k$ whenever $i\le j\le k$.   We can therefore define
$\phi_j:\frak D\to\frak A_j$, for any $j\in I$, by saying that
$\phi_j=\pi_{jk}\phi_k$ whenever $k\in I$ is greater than or equal to both
$i$ and $j$, and we shall have $\phi_j=\pi_{jk}\phi_k$ whenever
$j\le k$ in $I$.   \Prf\ If $j\in I$ and
$k_0$, $k_1$ are two upper bounds of $\{i,j\}$
in $I$, take an upper bound $k$ of $\{k_0,k_1\}$;  then

\Centerline{$\pi_{jk_0}\phi_{k_0}
=\pi_{jk_0}\pi_{k_0k}\phi_k
=\pi_{jk}\phi_k
=\pi_{jk_1}\pi_{k_1k}\phi_k
=\pi_{jk_1}\phi_{k_1}$,}

\noindent so $\phi_j$ is well-defined.   If $j$, $k\in I$ and
$j\le k$, let $k'$ be an upper bound of $\{i,k\}$;  then

\Centerline{$\pi_{jk}\phi_k
=\pi_{jk}\pi_{kk'}\phi_{k'}
=\pi_{jk'}\phi_{k'}
=\phi_j$.  \Qed}

\noindent Of course every $\phi_j$ is a measure-preserving Boolean
homomorphism.

By the definition of $(\frak C,\bar\lambda)$, there is a
measure-preserving Boolean homomorphism $\phi:\frak D\to\frak C$ such that
$\pi_j\phi=\phi_j$ for every $j\in I$.   In this case, $\pi_i\phi=\phi_i$
is the identity embedding of $\frak D$ in $\frak A_i$, and
$\pi_i[\frak C]=\frak D$.   Accordingly $P_{\frak D}=T_iP_i$.   By
the generalized reverse martingale theorem 367Qa,
$T_iP_i$ is the limit of $P_{\frak B}$ as $\frak B$ decreases in
$\Bbb B$, in the sense that for every $u\in L^1(\frak A_i)$ and
$\epsilon>0$ there is a $j\ge i$ in $I$ such that

\Centerline{$\|T_iP_iu-T_{ik}P_{ik}u\|_1
=\|P_{\frak D}u-P_{\frak B_k}u\|_1\le\epsilon$}

\noindent whenever $k\ge j$ in $I$.   If we write $\Cal F(I\closeuparrow)$ for
the filter on $I$ generated by $\{\{k:k\ge j\}:j\in I\}$, we have

\Centerline{$T_iP_iu=\lim_{j\to\Cal F(I\uparrow)}T_{ij}P_{ij}u$,}

\noindent for the norm in $L^1_{\bar\mu_i}$, for every
$u\in L^1_{\bar\mu_i}$.

Now let us turn to (a)-(d) as listed above.

\medskip

{\bf (a)} All we have to know is that

\Centerline{$S_iT_i=S_iT_{ij}T_j=S_jT_j$}

\noindent whenever $i\le j$ in $I$;  because $I$ is upwards-directed,
$S_iT_i=S_jT_j$ for all $i$, $j\in I$, and we have a sound definition for
$S$.

\medskip

{\bf (b)} The point is that $T_iP_iS_i=S_i$ for every
$i\in I$.   \Prf\ For $j\ge i$,

\Centerline{$T_{ij}P_{ij}S_i=T_{ij}P_{ij}T_{ij}S_j=T_{ij}S_j=S_i$.}

\noindent If $x\in X$,

\Centerline{$T_iP_iS_ix=\lim_{j\to\Cal F(I\uparrow)}T_{ij}P_{ij}S_ix
=S_ix$.  \Qed}

\noindent If now $i\le j$ in $I$,

\Centerline{$P_iS_i
=P_jP_{ij}T_{ij}S_j=P_jS_j$.}

\noindent As $I$ is upwards-directed, $P_iS_i=P_jS_j$ for all
$i$, $j\in I$;  write $S$ for this common value.   Then

\Centerline{$T_iS=T_iP_iS_i=S_i$}

\noindent for every $i\in I$.
As $T_i$ is injective for every $i\in I$,
the formula uniquely defines the function $S$.

\medskip

{\bf (c)} This time, we have $S_iT_iP_i=S_i$ for every $i\in I$.
\Prf\ For any $u\in L^1_{\bar\lambda}$,

$$\eqalignno{S_iT_iP_iu
&=\lim_{j\to\Cal F(I\uparrow)}S_iT_{ij}P_{ij}u\cr
\displaycause{because $S_i$ is continuous}
&=\lim_{j\to\Cal F(I\uparrow)}S_jP_{ij}T_{ij}P_{ij}u
=\lim_{j\to\Cal F(I\uparrow)}S_jP_{ij}u
=S_iu.  \text{ \Qed}\cr}$$

\noindent If $i\le j$ in $I$,

\Centerline{$S_iT_i=S_jP_{ij}T_{ij}T_j=S_jT_j$;}

\noindent consequently $S_iT_i=S_jT_j$ for all $i$, $j\in I$, and we can
call this common function $S$.   In this case, $SP_i=S_iT_iP_i=S_i$ for
every $i\in I$.   Since $P_i[L^1_{\bar\mu_i}]=L^1_{\bar\lambda}$,
this defines $S$ uniquely.

\medskip

{\bf (d)} As in (a), all we have to check is that if $i\le j$ in $I$ then

\Centerline{$P_jS_j=P_jP_{ij}S_i=P_iS_i$.}
\else\fi%end of proof of 377G

\leader{377H}{Inductive limits:  Proposition}
Let $(I,\le)$, $\familyiI{(\frak A_i,\bar\mu_i)}$ and
$\langle\pi_{ji}\rangle_{i\le j}$ be such that $(I,\le)$ is a non-empty
upwards-directed partially ordered
set, every $(\frak A_i,\bar\mu_i)$ is a probability
algebra, $\pi_{ji}:\frak A_i\to\frak A_j$ is a measure-preserving
Boolean homomorphism whenever $i\le j$ in $I$, and
$\pi_{ki}=\pi_{kj}\pi_{ji}$ whenever $i\le j\le k$.   Let
$(\frak C,\bar\lambda,\familyiI{\pi_i})$ be the corresponding inductive
limit\cmmnt{ (328H)}.   Write $L^1_{\bar\mu_i}$ for
$L^1(\frak A_i,\bar\mu_i)$ and $L^1_{\bar\lambda}$ for
$L^1(\frak C,\bar\lambda)$.   For $i\le j$ in $I$, let
$T_{ji}:L^1_{\bar\mu_i}\to L^1_{\bar\mu_j}$ and
$P_{ji}:L^1_{\bar\mu_j}\to L^1_{\bar\mu_i}$ be the Riesz homomorphism
and the positive linear operator corresponding to
$\pi_{ji}:\frak A_i\to\frak A_j$, and
$T_i:L^1_{\bar\mu_i}\to L^1_{\bar\lambda}$,
$P_i:L^1_{\bar\lambda}\to L^1_{\bar\mu_i}$ the operators corresponding to
$\pi_i:\frak A_i\to\frak C$.   Let $X$ be a set.

(a) Suppose that for each $i\in I$ we are given a function
$S_i:L^1_{\bar\mu_i}\to X$ such that
$S_jT_{ji}=S_i$ whenever $i\le j$
in $I$.   Then there is a
function $S:L^1_{\bar\lambda}\to X$ such
that $S_i=ST_i$ for every $i\in I$.

(b) Suppose that for each $i\in I$ we are given a function
$S_i:X\to L^1_{\bar\mu_i}$ such that $T_{ji}S_i=S_j$ whenever $i\le j$
in $I$.   Then there is a unique function $S:X\to L^1_{\bar\lambda}$ such
that $T_iS_i=S$ for every $i\in I$.

(c) Suppose that for each $i\in I$ we are given a function
$S_i:L^1_{\bar\mu_i}\to X$ such that
$S_iP_{ji}=S_j$ whenever $i\le j$ in $I$.   Then there is a unique
function $S:L^1_{\bar\lambda}\to X$ such
that $S=S_iP_i$ for every $i\in I$.

(d) Suppose that for each $i\in I$ we are given a function
$S_i:X\to L^1_{\bar\mu_i}$ such that $P_{ji}S_j=S_i$ whenever $i\le j$ in
$I$, and that

\Centerline{$\inf_{k\in\Bbb N}\sup_{i\in I}
  \int(|S_ix|-k\chi 1_{\frak A_i})^+=0$}

\noindent for every $x\in X$.
Then there is a unique function $S:X\to L^1_{\bar\lambda}$ such that
$S_i=P_iS$ for every $i\in I$.

\proof{ We can follow the same programme as in
the proof of 377G, but with a couple of new twists.

\medskip

{\bf preliminary remarks (i)}
If $i\le j$ in $I$, then by the definition of `inductive limit' we have
$\pi_j\pi_{ji}=\pi_i$ so
$T_jT_{ji}=T_i$ and $P_{ji}P_j=P_i$.
$P_{ji}T_{ji}$ and $P_iT_i$ are the identity
operator on $L^1_{\bar\mu_i}$.

\medskip

\quad{\bf (ii)} Let $\Cal F(I\closeuparrow)$ be the filter on $I$ generated by
$\{\{k:k\ge j\}:j\in I\}$.   Then
$\lim_{i\to\Cal F(I\uparrow)}T_iP_iu=u$ for every $u\in L^1_{\bar\lambda}$.
\Prf\ Setting $\frak B_i=T_i[\frak A_i]$ for each $i\in I$,
$\Bbb B=\{\frak B_i:i\in I\}$ is an upwards-directed family of closed
subalgebras of $\frak C$;  set $\frak D=\overline{\bigcup\Bbb B}$ and
$\bar\nu=\bar\lambda\restrp\frak D$, so that $(\frak D,\bar\nu)$ is a
probability algebra.   Since $\pi_i:\frak A_i\to\frak D$ is a
measure-preserving Boolean homomorphism and $\pi_i=\pi_j\pi_{ji}$ whenever
$i\le j$ in $I$, there is a measure-preserving Boolean homomorphism
$\phi:\frak C\to\frak D$ such that $\phi\pi_i=\pi_i$ for every $i$.
But this means that $\frak C=\frak D$.

As in 377G, we can identify
each $T_iP_i:L^1_{\bar\lambda}\to L^1_{\bar\lambda}$
with the conditional expectation $P_{\frak B_i}$.
This time, 367Qb tells us that $P_{\frak B}u\to P_{\frak D}u=u$ as
$\frak B$ increases through $\Bbb B$, that is,
$u=\lim_{i\to\Cal F(I\uparrow)}T_iP_iu$, for every
$u\in L^1_{\bar\lambda}$.\ \Qed

\medskip

{\bf (a)} The point is that if $i$, $j\in I$, $u\in L^1_{\bar\mu_i}$,
$v\in L^1_{\bar\mu_j}$ and $T_iu=T_jv$, then $S_iu=S_jv$.   \Prf\
Let $k\in I$ be such that $i\le k$ and $j\le k$.   Then

\Centerline{$T_kT_{ki}u=T_iu=T_jv=T_kT_{kj}v$;}

\noindent since $T_k$ is injective, $T_{ki}u=T_{kj}v$.   Accordingly

\Centerline{$S_iu=S_kT_{ki}u=S_kT_{kj}v=S_jv$.  \Qed}

\noindent There is therefore a function
$S':\bigcup_{i\in I}S_i[L^1_{\bar\mu_i}]\to X$ defined by saying that
$S(T_iu)=S_iu$ whenever $i\in I$ and $u\in L^1_{\bar\mu_i}$;  extending
$S'$ arbitrarily to a function $S:L^1_{\bar\lambda}\to X$, we get the
result.

\medskip

{\bf (b)} All we have to do is to check that if $i\le j$ in $I$, then

\Centerline{$T_iS_i=T_jT_{ji}S_i=T_jS_j$.}

\medskip

{\bf (c)} In this case, we have

\Centerline{$S_jP_j=S_iP_{ji}P_j=S_iP_i$}

\noindent whenever $i\le j$ in $I$.

\medskip

{\bf (d)(i)} For each $x\in X$,
$\{T_iS_ix:i\in I\}\subseteq L^1_{\bar\lambda}$ is uniformly integrable.
\Prf\ If $k\in\Bbb N$ and $i\in I$,

\Centerline{$|T_iS_ix|
\le T_i(|S_ix|-k\chi 1_{\frak A_i})^++T_i(k\chi 1_{\frak A_i})
\le T_i(|S_ix|-k\chi 1_{\frak A_i})^++k\chi 1_{\frak C}$,}

\noindent so

\Centerline{$\int(|T_iS_ix|-k\chi 1_{\frak C})^+
\le\int T_i(|S_ix|-k\chi 1_{\frak A_i})^+
=\int(|S_ix|-k\chi 1_{\frak A_i})^+$.}

\noindent Accordingly

\Centerline{$\inf_{k\in\Bbb N}\sup_{i\in I}
  \int(|T_iS_ix|-k\chi 1_{\frak C})^+
\le\inf_{k\in\Bbb N}\sup_{i\in I}\int(|S_ix|-k\chi 1_{\frak A_i})^+
=0$.  \Qed}

\medskip

\quad{\bf (ii)} Fix an ultrafilter $\Cal G$ on $I$ including
$\Cal F(I\closeuparrow)$.   For each $x\in X$, $\{T_iS_ix:i\in I\}$ is
relatively weakly compact in $L^1_{\bar\lambda}$, so
$Sx=\lim_{i\to\Cal G}T_iS_ix$ is defined for the weak topology on
$L^1_{\bar\lambda}$.   Now for any $i\in I$,

$$\eqalignno{P_iSx
&=\lim_{j\to\Cal G}P_iT_jS_jx\cr
\displaycause{for the weak topology on $L^1_{\bar\mu_i}$}
&=\lim_{j\to\Cal G}P_{ji}P_jT_jS_jx\cr
\displaycause{because $\{j:j\ge i\}\in\Cal G$}
&=\lim_{j\to\Cal G}P_{ji}S_jx
=S_ix.\cr}$$

\medskip

\quad{\bf (iii)} To see that $S$ is uniquely defined, it is enough to
recall that

\Centerline{$Sx
=\lim_{i\to\Cal F(I\uparrow)}T_iP_iSx
=\lim_{i\to\Cal F(I\uparrow)}T_iS_ix$}

\noindent is uniquely defined by the family $\familyiI{S_ix}$, for every
$x\in X$.
}%end of proof of 377H

\exercises{\leader{377X}{Basic exercises (a)}
%\spheader 377Xa
In 377B, show that $\familyiI{u_i}\in\prod_{i\in I}L^0(\frak A_i)$ belongs
to $W_0$ iff $\{u_i^*:i\in I\}$ is bounded above in $L^0(\frak A_L)$, where
$\frak A_L$ is the measure algebra of Lebesgue measure on
$\coint{0,\infty}$, and
$u_i^*$ is the decreasing rearrangement of $u_i$ for each $i$ (373C).
%377B

\spheader 377Xb In 377D, suppose that $u=\familyiI{u_i}$ and
$v=\familyiI{v_i}$ belong to $W_2$, and that at least one of
$|u|^2$, $|v|^2$ belongs to $W_{ui}$.
Show that $\innerprod{Tu}{Tv}=\lim_{i\to\Cal F}\innerprod{u_i}{v_i}$.
%377D

\spheader 377Xc Let $\familyiI{(\frak A_i,\bar\mu_i)}$ be a family of
probability algebras, and suppose that we have
$u_i\in L^1(\frak A_i,\bar\mu_i)$ for each $i$.   Show that the following
are equiveridical:  (i)
$\inf_{k\in\Bbb N}\sup_{i\in I}\int(|u_i|-k\chi 1_{\frak A_i})^+=0$;
(ii) $\{u_i^*:i\in I\}$ is uniformly integrable in $L^1(\mu_L)$, where
$\mu_L$ is Lebesgue measure on $\coint{0,\infty}$, and $u_i^*$ is the
decreasing rearrangement of $u_i$ for each $i\in I$.
%377D

\spheader 377Xd Take any $p\in\ooint{1,\infty}$.   Show that 377G remains
true if we replace every `$L^1$' by `$L^p$'.
%377G

\spheader 377Xe Take any $p\in\ooint{1,\infty}$.   Show that 377H remains
true if we replace every `$L^1$' by `$L^p$' and in
part (d) we replace
`$\inf_{k\in\Bbb N}\sup_{i\in I}\int(|S_ix|-k\chi 1_{\frak A_i})^+=0$' by
`$\sup_{i\in I}\|S_ix\|_p<\infty$'.
%377G

\spheader 377Xf In 377Ha, suppose that $X$ has a metric $\rho$ under which
it is complete, and that $\familyiI{S_i}$ is uniformly equicontinuous in
the sense that for every $\epsilon>0$ there is a $\delta>0$ such that
$\rho(S_iu,S_iv)\le\epsilon$ whenever $i\in I$, $u$, $v\in L^1_{\bar\mu_i}$
and $\|u-v\|_1\le\delta$.   Show that there is a unique continuous function
$S:L^1_{\bar\lambda}\to X$ such that $S_i=ST_i$ for every $i\in I$.
%377H

\leader{377Y}{Further exercises (a)}
%\spheader 377Ya
Find a non-empty family $\familyiI{(\frak A_i,\bar\mu_i)}$
of probability algebras, a probability algebra $(\frak B,\bar\nu)$, a
Boolean homomorphism $\pi:\prod_{i\in I}\frak A_i\to\frak B$ such
that $\bar\nu\pi(\familyiI{a_i})\le\sup_{i\in I}\bar\mu_ia_i$ whenever
$\familyiI{a_i}\in\prod_{i\in I}\frak A_i$, and an element
$u=\familyiI{u_i}$ of $W_0^+$, as described in 377B, such that
$\|Tu\|_1>\sup_{i\in I}\|u_i\|_1$, where $T:W_0\to L^0(\frak B)$ is
the Riesz homomorphism of 377B-377C.   \Hint{$\#(I)=2$.}
%377C

\spheader 377Yb Show that if, in 377Gc, we omit the hypothesis that the
$S_i$ are to be continuous, then the result can fail.
%377G mt37bits

\spheader 377Yc Let $\familyiI{U_i}$ be a non-empty family of $L$-spaces
and $\Cal F$ an ultrafilter on $I$.   (i) Show that $\prod_{i\in I}U_i$ is a
Dedekind complete Riesz space (see 352K) in which
$W_{\infty}=\{\familyiI{u_i}:\sup_{i\in I}\|u_i\|<\infty\}$ is a solid
linear subspace.   (ii) Let $W_0\subseteq W_{\infty}$ be
$\{\{\familyiI{u_i}:\sup_{i\in I}\|u_i\|<\infty$,
$\lim_{i\to\Cal F}\|u_i\|=0\}$;  show that $W_0$ is a solid linear subspace
of $W_{\infty}$.   (iii) Let $U$ be the quotient Riesz space
$W_{\infty}/W_0$ (352U).   Show that $U$ is an $L$-space under the norm
$\|\familyiI{u_i}^{\ssbullet}\|=\lim_{i\to\Cal F}\|u_i\|$ for
$\familyiI{u_i}\in W_{\infty}$.
%377D out of order query

\spheader 377Yd Let $V$ be a normed space, and suppose that for every
finite-dimensional subspace $V_0$ of $V$ there are an $L$-space $U$ and a
norm-preserving linear map $T:V_0\to U$.   Show that there are an $L$-space
$U$ and a norm-preserving linear map $T:V\to U$.
%377Yc
}%end of exercises


\endnotes{
\Notesheader{377} Although my main target in this section has been to
understand the function spaces of reduced products of probability algebras,
I have as usual felt that the ideas are clearer if each is developed in
a context closer to the most general case in which it is applicable.
Only in part (b) of the proof of 377C, I think, does this involve us in
extra work.

The new techniques of this section are forced on us by the fact that we are
looking at Boolean homomorphisms $\pi:\prod_{i\in I}\frak A_i\to\frak B$
which are not normally sequentially order-continuous.    While we have
a natural Riesz homomorphism from $L^{\infty}(\prod_{i\in I}\frak A_i)$ to
$L^{\infty}(\frak B)$, as in 363F, we cannot expect
a similar operator from the whole of
$L^0(\prod_{i\in I}\frak A_i)\cong\prod_{i\in I}L^0(\frak A_i)$ to
$L^0(\frak B)$.   However the condition
`$\bar\nu\pi(\familyiI{a_i})\le\sup_{i\in I}\bar\mu_ia_i$' ensures that
there is a space $W_0\subseteq\prod_{i\in I}L^0(\frak A_i)$ on which an
operator to $L^0(\frak B)$ can be defined, and which is large enough to
give us a method of investigating the spaces $L^p(\frak B,\bar\nu)$ as
images of subspaces $W_p$ of products
$\prod_{i\in I}L^p(\frak A_i,\bar\mu_i)$.

In 377E, the case $p=1$ is
special because we can identify $W_{ui}$ as the space of relatively weakly
compact families in $L^1(\frak A,\bar\mu)$, and for such a family
$u=\familyiI{u_i}$ we have $\|Tu\|_1=\lim_{i\to\Cal F}\|u_i\|_1$.
So the Banach space $L^1(\frak B,\bar\nu)$ is a kind of reduced power,
describable in terms of the normed space $L^1(\frak A,\bar\nu)$.   For
other $L^p$ spaces we need to know something more, e.g., the lattice
structure, if we are to identify those $u\in W_p$ such that $Tu=0$.
The difference becomes significant when we come to look at morphisms of
$L^p(\frak B,\bar\nu)$ corresponding to morphisms of
$L^p(\frak A,\bar\mu)$, as in 377F.

In 377G-377H I give a string of results which are visibly mass-produced.
What is striking is that in eight cases out of eight we have a
straightforward formula corresponding to the idea that
$(\frak C,\bar\lambda)$ is a limit of $\familyiI{(\frak A_i,\bar\mu_i)}$.
What is curious is that in two of the eight cases (377Gc, 377Hd)
we have to impose different special conditions on the functions $S_i$ which
the target $S$ is supposed to approximate, and in just one case (377Ha) the
target $S$ is not uniquely defined in the absence of further constraints
(377Xf).   I think the ideas take up
enough room when given only in their application to $L^1$ spaces, but of
course there are versions, only slightly modified, which apply to other
$L^p$ spaces (377Xd-377Xe).

The repeated conditions of the form

\Centerline{$\inf_{k\in\Bbb N}\sup_{i\in I}\bar\mu_i\Bvalue{|u_i|>k}=0$,}

\Centerline{$\inf_{k\in\Bbb N}\sup_{i\in I}
  \int(|u_i|-k\chi 1_{\frak A_i})^+=0$,}

\noindent (377B, 377Dc, 377Hd) both have expressions in terms of
decreasing rearrangements (377Xa, 377Xc).   The latter is clearly
associated with uniform integrability and weak compactness, and
unsurprisingly we use it to show that a weak limit will be defined.
The former is there to ensure that a set appearing in an $L^0$ space will
be bounded above, so that we can apply 355F to extend a Riesz homomorphism.
}%end of notes

\discrpage

%\ifresultsonly\else
\frfilename{mt37conc.tex}
\versiondate{7.12.08}
\copyrightdate{2008}

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

\def\chaptername{Linear operators between function spaces}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my
knowledge) appeared in print in references to this chapter, and which
have since been changed.

\medskip

\leader{3{}72I}{} The version of the Ergodic Theorem in 3{}72I, referred to
in the 2003 and 2006 editions of Volume 4, is now 372H.

\leader{3{}72K}{} The version of the Ergodic Theorem in 3{}72K, referred to
in the 2003 and 2006 editions of Volume 4, is now 372J.

\leader{3{}72P}{Mixing and ergodic transformations} The definitions in
3{}72P are now in 372O.

\leader{}{3{}72Xm} The tent map, 
referred to in the 2003 and 2006 editions of Volume 4, is now in
372Xp.

\frnewpage
%\fi

\references{
\frfilename{mt3r.tex}
\versiondate{25.8.17}
\copyrightdate{1999}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 3}

\medskip

%A

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P., 1987.
\cmmnt{[332Xk, 3A1K.]}

% Anzai H.\ [51] `Ergodic skew product transformations of the torus',
% % Osaka Math.\ J.\ 3 (1951) 83-99.

Anzai H.\ [51] `On an example of a measure-preserving transformation
which is not conjugate to its inverse',
Proc.\ Japanese Acad.\ Sci.\ 27 (1951) 517-522.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip
%B

Balcar B., G\l\'owczy\'nski W. \& Jech T.\ [98] `The sequential
topology on complete Boolean algebras', Fundamenta Math.\ 155 (1998)
59-78.
\cmmnt{[393L, 393Q.]}

Balcar B., Jech T.\ \& Paz\'ak T.\ [05]
`Complete ccc Boolean algebras, the order sequential topology,
and a problem of von Neumann',
Bull.\ London Math.\ Soc.\ 37 (2005) 885-898.
\cmmnt{[393L, 393Q.]}
% arXiv.math.LO/0312473

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996
(London Math.\ Soc.\ Lecture Note Series 232).
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\cmmnt{[384L.]}

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%326H

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\medskip
%C

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\medskip
%D

Davey B.A. \& Priestley H.A. [90] `Introduction to Lattices and Order',
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Dugundji J.\ [66] {\it Topology.}  Allyn \& Bacon, 1966.
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\medskip
%E

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%QA248.E6  7850 2996 6

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\medskip
%F

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%Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
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\leaveitout{Fremlin D.H.\ [89] `Measure algebras', pp.\ 876-980 in
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Fremlin D.H., de Pagter B.\ \&  Ricker W.J.\ [05]
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\medskip
%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
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\medskip%H

Hajian A.\ \& Ito Y.\ [69]
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\leaveitout{
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}

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\medskip %I

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\medskip %J

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%{\tt http://www.math.psu.edu/jech/preprints/papers.html}
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%Russell A. Johnson

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\medskip %K

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\leaveitout{
Kelley J.L. \& Namioka I.\ [76] {\it Linear Topological Spaces.}
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\leaveitout{
King J.L.F.\ [00] `The generic transformation has roots of all orders',
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  % if  \frak A  is the Lebesgue measure algebra
  % a comeager set in the Polish group
  % \Aut_{\bar\mu}\frak A   has roots of all orders
}

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\cmmnt{[386G.]}
%Wendt S IN5 EL24 T IT

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\medskip %L

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\medskip %M

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\medskip %N

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\medskip %O

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\medskip%P

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\medskip %Q%R

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\medskip %S

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%Store A	QA 611.S3    6640 0159 9

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\medskip %T

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\medskip %U%V

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\medskip %W

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\medskip %X%Y%Z

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\frnewpage

}%end of references


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

\end



