% 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\tinyplus{{\vrule height 1.6pt depth -1.3pt width 2.5pt}
    \hskip-1.3pt{\vrule height 2.6pt depth -0.1pt width 0.3pt}}
\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{mt36.tex} 
\versiondate{13.8.98/27.7.17} 
\copyrightdate{1996} 
 
\def\chaptername{Function Spaces} 
\def\sectionname{Introduction} 
 
\newchapter{36} 
 
Chapter 24 of Volume 2 was devoted to the elementary theory of the \lq 
function spaces' $L^0$, $L^1$, $L^2$ and $L^{\infty}$ associated with a 
given measure space. 
In this chapter I return to these spaces to show how they can be related 
to the more abstract themes of the present volume.   In particular, I 
develop constructions to demonstrate, as clearly as I can, the way in 
which the function spaces associated with a measure space in fact 
depend only on its measure algebra;  and how many of their features can 
(in my view) best be understood in terms of constructions involving 
measure algebras. 
 
The chapter is very long, not because there are many essentially new ideas, but because the intuitions I seek to develop depend, 
for their logical foundations, on technically complex arguments.   This 
is perhaps best exemplified by \S364.   If two measure spaces 
$(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ have isomorphic measure algebras 
$(\frak A,\bar\mu)$, $(\frak B,\bar\nu)$ then the spaces $L^0(\mu)$, 
$L^0(\nu)$ are isomorphic as topological $f$-algebras;  and more:  for 
any isomorphism between $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ 
there is a unique corresponding isomorphism between the $L^0$ spaces. 
The intuition involved is in a way very simple.   If $f$, $g$ are 
measurable real-valued functions on $X$ and $Y$ respectively, then 
$f^{\ssbullet}\in L^0(\mu)$ will correspond to 
$g^{\ssbullet}\in L^0(\nu)$ if and only if 
$\Bvalue{f^{\ssbullet}>\alpha}=\{x:f(x)>\alpha\}^{\ssbullet}\in\frak A$ 
corresponds to 
$\Bvalue{g^{\ssbullet}>\alpha}=\{y:g(y)>\alpha\}^{\ssbullet}\in\frak B$ 
for every $\alpha$. 
But the check that this formula is consistent, and defines an 
isomorphism of the required kind, involves a good deal of detailed work. 
It turns out, in fact, that the {\it measures} $\mu$ and $\nu$ do not 
enter this part of the argument at all, except through their ideals of 
negligible sets (used in the construction of $\frak A$ and $\frak B$). 
This is already evident, if you look for it, in the theory of 
$L^0(\mu)$;  in \S241, as written out, you will find that the measure of an 
individual set is not once mentioned, except in the exercises. 
Consequently there is an invitation to develop the theory with algebras 
$\frak A$ which are not necessarily measure algebras.   Here is another 
reason for the length of the chapter;  substantial parts of the work are 
being done in greater generality than the corresponding sections of 
Chapter 24, necessitating a degree of repetition.   Of course this is 
not \lq measure theory' in the strict sense;  but for thirty years now 
measure theory has been coloured by the existence of these 
generalizations, and I think it is useful to understand which parts 
of the theory apply only to measure algebras, and which can be extended 
to other $\sigma$-complete Boolean algebras, like the algebraic theory 
of $L^0$, or even to all Boolean algebras, like the theory of 
$L^{\infty}$. 
 
Here, then, are two of the objectives of this chapter:  first, to 
express the ideas of Chapter 24 in ways making explicit their 
independence of particular measure spaces, by setting up constructions 
based exclusively on the measure algebras involved;  second, to set out 
some natural generalizations to other algebras.   But to justify the 
effort needed I ought to point to some mathematically significant idea 
which demands these constructions for its expression, and here I mention 
the categorical nature of the constructions.   Between Boolean algebras 
we have a variety of natural and important classes of \lq morphism'; 
for instance, the Boolean homomorphisms and the 
order-continuous Boolean homomorphisms;  while between measure algebras 
we have in addition the 
measure-preserving Boolean homomorphisms.   Now it turns out that if we 
construct the $L^p$ spaces in the natural ways then morphisms between 
the underlying algebras give rise to morphisms between their $L^p$ 
spaces.   For instance, any Boolean homomorphism from $\frak A$ to 
$\frak B$ produces a multiplicative norm-contractive Riesz homomorphism 
from $L^{\infty}(\frak A)$ to $L^{\infty}(\frak B)$;  if $\frak A$ and 
$\frak B$ are Dedekind $\sigma$-complete, then any sequentially 
order-continuous Boolean homomorphism from $\frak A$ to $\frak B$ 
produces a sequentially order-continuous multiplicative Riesz 
homomorphism from $L^0(\frak A)$ to $L^0(\frak B)$;  and if $(\frak 
A,\bar\mu)$ and $(\frak B,\bar\nu)$ are measure algebras, then any 
measure-preserving Boolean homomorphism from $\frak A$ to $\frak B$ 
produces norm-preserving Riesz homomorphisms from $L^p(\frak A,\bar\mu)$ 
to $L^p(\frak B,\bar\nu)$ for every $p\in[1,\infty]$.   All of 
these are \lq functors', that is, a composition of homomorphisms between 
algebras gives rise to a composition of the corresponding operators 
between their function spaces, and are \lq covariant', that is, a 
homomorphism from $\frak A$ to $\frak B$ leads to an operator from 
$L^p(\frak A)$ to $L^p(\frak B)$.   But the same constructions lead us 
to a functor which is \lq contravariant':  starting from an 
order-continuous Boolean homomorphism from a semi-finite measure algebra 
$(\frak A,\bar\mu)$ to a measure algebra $(\frak B,\bar\nu)$, we have an 
operator from $L^1(\frak B,\bar\nu)$ to $L^1(\frak A,\bar\mu)$.   This 
last is in fact a kind of conditional expectation operator.   In my view 
it is not possible to make sense of the theory of measure-preserving 
transformations without at least an intuitive grasp of these ideas. 
 
Another theme is the characterization of each construction in terms of 
universal mapping theorems:  for instance, each $L^p$ space, for $1\le 
p\le\infty$, can be characterized as Banach lattice in terms of 
factorizations of functions of an appropriate class from the underlying 
algebra to Banach lattices. 
 
Now let me try to sketch a route-map for the journey ahead.   I begin 
with two sections on the space $S(\frak A)$;  this construction applies 
to any Boolean algebra (indeed, any Boolean ring), and corresponds to 
the space of `simple functions' on a measure space.   Just because it 
is especially close to the algebra (or ring) $\frak A$, there is a 
particularly large number of universal mapping theorems corresponding to 
different aspects of its structure (\S361).   In \S362 I seek to relate 
ideas on additive functionals on Boolean algebras from Chapter 23 and 
\S\S326-327 to the theory of Riesz space duals in \S356.   I then 
turn to a systematic discussion of the function spaces of Chapter 24: 
$L^{\infty}$ (\S363), $L^0$ (\S364), $L^1$ (\S365) and other $L^p$ 
(\S366), followed by an account of convergence in measure (\S367). 
While all these sections are dominated by the objectives sketched in the 
paragraphs above, I do include a few major theorems not covered by the 
ideas of Volume 2, such as the Kelley-Nachbin characterization of the 
Banach spaces $L^{\infty}(\frak A)$ for Dedekind complete $\frak A$ 
(363R).   In the last two sections of the chapter I turn to 
the use of $L^0$ spaces in the representation of Archimedean Riesz 
spaces (\S368) and of Banach lattices which are separated by their 
order-continuous duals (\S369). 
 
\discrpage 
 
\frfilename{mt361.tex}
\versiondate{6.2.08}
\copyrightdate{1995}

\def\chaptername{Function spaces}
\def\sectionname{$S$}

\newsection{361}

This is the fundamental Riesz space associated with a Boolean ring
$\frak A$.   When $\frak A$ is a ring of sets, $S(\frak A)$ can be
regarded as the linear space of `simple functions' generated by the
indicator functions of members of $\frak A$ (361L).   Its most
important property is the universal mapping theorem 361F, which
establishes a one-to-one correspondence between (finitely) additive
functions on $\frak A$ (361B-361C) and linear operators on $S(\frak A)$.
Simple universal mapping theorems of this type can be interesting, but
do not by themselves lead to new insights;  what makes this one
important is the fact that $S(\frak A)$ has a canonical Riesz space
structure, norm and multiplication (361E).   From this we can deduce
universal mapping theorems for many other classes of function (361G,
361H, 361I, 361Xb).
\cmmnt{(Particularly important are countably additive and
completely additive real-valued functionals, which will be dealt with in
the next section.)}   While the exact construction of
$S(\frak A)$\cmmnt{ (and the associated map from $\frak A$ to
$S(\frak A)$)} can be
varied (361D, 361L, 361M, 361Ya), its structure is uniquely defined, so
homomorphisms between Boolean rings correspond to maps between their
$S(\,)$-spaces (361J), and (when $\frak A$ is a
Boolean algebra) $\frak A$ can be recovered from the
Riesz space $S(\frak A)$ as the algebra of its projection bands (361K).

\leader{361A}{Boolean rings} \cmmnt{In this section I speak of Boolean
{\it rings} rather than {\it algebras};   there are ideas in \S365 below
which are more naturally expressed in terms of the ring of elements of
finite measure in a measure algebra than in terms of the whole algebra.
I should perhaps therefore recall some of the ideas of \S311, which is
the last time when Boolean rings without identity were mentioned, and
set out some simple facts.

\spheader 361Aa
Any Boolean ring $\frak A$ can be represented as the ring of compact
open subsets of its Stone space $Z$, which is
a zero-dimensional locally compact Hausdorff space
(311I);  $Z$ is just the set of surjective ring homomorphisms from
$\frak A$ onto $\Bbb Z_2$ (311E).

\medskip

}{\bf (b)} If $\frak A$ and $\frak B$ are Boolean rings and
$\pi:\frak A\to\frak B$ is a function, then the following are
equiveridical:  (i) $\pi$ is a ring homomorphism;
(ii) $\pi(a\Bsetminus b)=\pi a\Bsetminus\pi b$ for all $a$,
$b\in\frak A$;   (iii) $\pi$ is a lattice homomorphism and $\pi 0=0$.
\prooflet{\Prf\ See 312H.   To prove
(ii)$\Rightarrow$(iii), observe that if $a$, $b\in\frak A$ then

\Centerline{$\pi(a\Bcap b)=\pi a\Bsetminus\pi(a\Bsetminus b)
=\pi a\Bsetminus(\pi a\Bsetminus\pi b) =\pi a\Bcap\pi b$,}

\Centerline{$\pi a=\pi((a\Bcup b)\cap a)
=\pi(a\Bcup b)\cap\pi a\Bsubseteq\pi(a\Bcup b)$,}

\Centerline{$\pi(b\Bsetminus a)=\pi((a\Bcup b)\Bsetminus a)
=\pi(a\Bcup b)\Bsetminus\pi a$,}

\Centerline{$\pi(a\Bcup b)=\pi a\Bcup\pi(b\Bsetminus a)
=\pi a\Bcup(\pi b\Bsetminus\pi a)=\pi a\Bcup\pi b$.  \Qed}
}


\spheader 361Ac If $\frak A$ and $\frak B$ are Boolean rings and
$\pi:\frak A\to\frak B$ is a ring homomorphism, then $\pi$ is
order-continuous iff $\inf\pi[A]=0$ whenever $A\subseteq\frak A$ is
non-empty and downwards-directed and $\inf A=0$ in $\frak A$;  while
$\pi$ is sequentially order-continuous iff $\inf_{n\in\Bbb N}\pi a_n=0$
whenever $\sequencen{a_n}$ is a non-increasing sequence in $\frak A$
with infimum $0$.   \prooflet{(See 313L.)}

\spheader 361Ad \cmmnt{ The following will be a particularly important
type of Boolean ring for us.}   If $(\frak A,\bar\mu)$ is a measure
algebra, then the ideal $\frak A^f=\{a:a\in\frak A,\,\bar\mu a<\infty\}$
is a Boolean ring in its own right.   Now suppose that
$(\frak B,\bar\nu)$ is another measure algebra and
$\frak B^f\subseteq\frak B$
the corresponding ring of elements of finite measure.   We can say that
a ring homomorphism $\pi:\frak A^f\to\frak B^f$ is
{\bf measure-preserving} if $\bar\nu\pi a=\bar\mu a$
for every $a\in\frak A^f$.   In this case $\pi$ is order-continuous.
\prooflet{\Prf\ If $A\subseteq\frak A^f$ is
non-empty, downwards-directed and has infimum $0$, then
$\inf_{a\in A}\bar\mu a=0$, by 321F;  but this means that
$\inf_{a\in A}\bar\nu\pi a=0$, and $\inf\pi[A]=0$ in $\frak B^f$.\ \Qed}

\leader{361B}{Definition} Let $\frak A$ be a Boolean ring and $U$
a linear space.   A function $\nu:\frak A\to U$ is
{\bf finitely additive}, or just {\bf additive}, if
$\nu(a\Bcup b)=\nu a+\nu b$ whenever $a$, $b\in\frak A$ and
$a\Bcap b=0$.

\leader{361C}{Elementary facts}\cmmnt{ We have the following immediate
consequences of this definition, corresponding to 326B and 313L.}   Let
$\frak A$ be a Boolean ring, $U$ a linear space and
$\nu:\frak A\to U$ an additive function.

\spheader 361Ca $\nu 0=0$\prooflet{ (because $\nu 0=\nu 0+\nu 0$)}.

\spheader 361Cb If $a_0,\ldots,a_m$ are disjoint in $\frak A$, then
$\nu(\sup_{j\le m}a_j)=\sum_{j=0}^m\nu a_j$.   \prooflet{(Induce on $m$.)}

\spheader 361Cc If $\frak B$ is another Boolean ring and
$\pi:\frak B\to\frak A$ is a ring homomorphism, then
$\nu\pi:\frak B\to U$ is additive.   In particular, if $\frak B$ is a
subring of $\frak A$, then
$\nu\restrp\frak{B}:\frak B\to U$ is  additive.

\spheader 361Cd If $V$ is another linear space and $T:U\to V$ is a
linear operator, then $T\nu:\frak A\to V$ is additive.

\spheader 361Ce If $U$ is a partially ordered linear space, then $\nu$
is order-preserving iff it is non-negative, that is, $\nu a\ge 0$ for
every $a\in\frak A$.   \prooflet{\Prf\ ($\alpha$) If $\nu$ is
order-preserving, then of course $0=\nu 0\le\nu a$ for every
$a\in\frak A$.   ($\beta$) If $\nu$ is non-negative, and $a\Bsubseteq b$ in $\frak A$, then

\Centerline{$\nu a\le\nu a+\nu(b\Bsetminus a)=\nu b$.  \Qed}
}%end of prooflet

\spheader 361Cf If $U$ is a partially ordered linear space and $\nu$ is
non-negative, then (i) $\nu$ is order-continuous iff $\inf\nu[A]=0$
whenever $A\subseteq\frak A$ is a non-empty downwards-directed set with
infimum $0$ (ii) $\nu$ is sequentially order-continuous iff
$\inf_{n\in\Bbb N}\nu a_n=0$ whenever $\sequencen{a_n}$ is a
non-increasing sequence in $\frak A$ with infimum $0$.

\medskip

\prooflet{\Prf{\bf (i)}
If $\nu$ is order-continuous, then of course $\inf\nu[A]=\nu 0=0$
whenever $A\subseteq\frak A$ is a non-empty downwards-directed set with
infimum $0$.   If $\nu$ satisfies the condition, and
$A\subseteq\frak A$ is a non-empty upwards-directed set with supremum $c$, then
$\{c\Bsetminus a:a\in\frak A\}$ is downwards-directed with infimum $0$
(313Aa), so that

$$\eqalignno{\sup_{a\in A}\nu a
&=\sup_{a\in A}\nu c-\nu(c\Bsetminus a)
=\nu c-\inf_{a\in A}\nu(c\Bsetminus a)\cr
\noalign{\noindent (by 351Db)}
&=\nu c.\cr}$$

\noindent Similarly, if $A\subseteq\frak A$ is a non-empty
downwards-directed set with infimum $c$, then

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

\noindent Putting these together, $\nu$ is order-continuous.

\medskip

\quad{\bf (ii)} If $\nu$ is sequentially order-continuous, then of course
$\inf_{n\in\Bbb N}\nu a_n=\nu 0=0$ whenever $\sequencen{a_n}$ is a
non-increasing
sequence in $\frak A$ with infimum $0$.   If $\nu$ satisfies the
condition, and $\sequencen{a_n}$ is a non-decreasing sequence in
$\frak A$ with supremum $c$, then $\sequencen{c\Bsetminus a_n}$ is
non-increasing and has infimum $0$, so that

\Centerline{$\sup_{n\in\Bbb N}\nu a_n
=\sup_{n\in\Bbb N}\nu c-\nu(c\Bsetminus a_n)
=\nu c-\inf_{n\in\Bbb N}\nu(c\Bsetminus a_n)
=\nu c$.}

\noindent Similarly, if $\sequencen{a_n}$ is a non-increasing sequence
in $\frak A$ with infimum $c$, then $\sequencen{a_n\Bsetminus c}$ is
non-increasing and has infimum $0$, so that

\Centerline{$\inf_{n\in\Bbb N}\nu a_n
=\inf_{n\in\Bbb N}\nu c+\nu(c\Bsetminus a_n)
=\nu c+\inf_{n\in\Bbb N}\nu(c\Bsetminus a_n)
=\nu c$.}

\noindent Thus $\nu$ is sequentially order-continuous.
\Qed}

\leader{361D}{Construction} Let $\frak A$ be a Boolean ring, and $Z$
its Stone space.   For $a\in\frak A$ write $\chi a$ for the
indicator function of the open-and-compact subset $\widehat{a}$ of
$Z$ corresponding to $a$.    Note that $\chi a=0$ iff $a=0$.
Let $S(\frak A)$ be the linear subspace of
$\Bbb R^Z$ generated by $\{\chi a:a\in\frak A\}$.   \cmmnt{Because
$\chi a$ is
a bounded function for every $a$,} $S(\frak A)$ is a subspace of the
$M$-space $\ell^{\infty}(Z)$ of all bounded real-valued functions on
$Z$\cmmnt{ (354Ha)},
and $\|\,\|_{\infty}$ is a norm on $S(\frak A)$.   \cmmnt{Because
$\chi a\times\chi b=\chi(a\Bcap b)$ for all $a$, $b\in \frak A$ (writing
$\times$ for pointwise multiplication of functions, as in 281B),}
$S(\frak A)$ is closed under $\times$.

\leader{361E}{}\cmmnt{ I give a portmanteau proposition running
through the elementary, mostly algebraic, properties of $S(\frak A)$.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean ring, with
Stone space $Z$.   Write $S$ for $S(\frak A)$.

(a) If $a_0,\ldots,a_n\in\frak A$,  there are disjoint $b_0,\ldots,b_m$
such that each $a_i$ is expressible as the supremum of some of the
$b_j$.

(b) If $u\in S$, it is expressible in the
form $\sum_{j=0}^m\beta_j\chi b_j$ where $b_0,\ldots,b_m$ are disjoint
members of $\frak A$ and $\beta_j\in\Bbb R$ for each $j$.   If all the
$b_j$ are non-zero then $\|u\|_{\infty}=\sup_{j\le m}|\beta_j|$.

(c) If $u\in S$ is non-negative, it is expressible in the
form $\sum_{j=0}^m\beta_j\chi b_j$ where $b_0,\ldots,b_m$ are disjoint
members of $\frak A$ and $\beta_j\ge 0$ for each $j$, and simultaneously
in the form $\sum_{j=0}^m\gamma_j\chi c_j$ where
$c_0\Bsupseteq c_1\Bsupseteq\ldots\Bsupseteq c_m$ and $\gamma_j\ge 0$ for every $j$.

(d) If $u=\sum_{j=0}^m\beta_j\chi b_j$ where $b_0,\ldots,b_m$ are
disjoint members of $\frak A$ and $\beta_j\in\Bbb R$ for each $j$, then
$|u|=\sum_{j=0}^m|\beta_j|\chi b_j\in S$.

(e) $S$ is a Riesz subspace of $\Bbb R^Z$;  in its own right, it is an
Archimedean Riesz space.   If $\frak A$ is a Boolean algebra, then $S$
has an order unit $\chi 1$ and
$\|u\|_{\infty}=\min\{\alpha:\alpha\ge 0,\,|u|\le\alpha\chi 1\}$ for every $u\in S$.

(f) The map $\chi:\frak A\to S$ is injective, additive, non-negative, a
lattice homomorphism and order-continuous.

(g) Suppose that $u\ge 0$ in $S$ and $\delta\ge 0$ in $\Bbb R$.   Then

\Centerline{$\Bvalue{u>\delta}
=\max\{a:a\in\frak A,\,(\delta+\eta)\chi a\le u$ for some $\eta>0\}$}

\noindent is defined in $\frak A$, and

\Centerline{$\delta\chi\Bvalue{u>\delta}\le u
\le\delta\chi\Bvalue{u>0}\vee\|u\|_{\infty}\Bvalue{u>\delta}$.}

\noindent In particular, $u\le\|u\|_{\infty}\chi\Bvalue{u>0}$ and there
is an $\eta>0$ such that $\eta\chi\Bvalue{u>0}\le u$.   If $u$, $v\ge 0$
in $S$ then $u\wedge v=0$ iff $\Bvalue{u>0}\Bcap\Bvalue{v>0}=0$.

(h) Under $\times$, $S$ is an $f$-algebra\cmmnt{ (352W)} and a
commutative normed algebra\cmmnt{ (2A4J)}.

(i) For any $u\in S$, $u\ge 0$ iff $u=v\times v$ for some $v\in S$.

\proof{ Write $\widehat{a}$ for the
open-and-compact subset of $Z$ corresponding to $a\in\frak A$.

\medskip

{\bf (a)} Induce on $n$.   If $n=0$ take $m=0$, $b_0=a_0$.    For the
inductive step to $n\ge 1$, take disjoint $b_0,\ldots,b_m$ such that
$a_i$ is the supremum of some of the $b_j$ for each $i<n$;  now replace
$b_0,\ldots,b_m$ with
$b_0\Bcap a_n,\ldots,b_m\Bcap a_n,b_0\Bsetminus a_n,\ldots,
b_m\Bsetminus a_n,\discretionary{}{}{}a_n\Bsetminus\sup_{j\le m}b_j$
to obtain a suitable string for $a_0,\ldots,a_n$.

\medskip

{\bf (b)} If $u=0$ set $m=0$, $b_0=0$, $\beta_0=0$.   Otherwise, express
$u$ as $\sum_{i=0}^n\alpha_i\chi a_i$ where $a_0,\ldots,a_n\in\frak A$
and $\alpha_0,\ldots,\alpha_n$ are real numbers.   Let $b_0,\ldots,b_m$
be disjoint and such that every $a_i$ is expressible
as the supremum of some of the $b_j$.   Set $\gamma_{ij}=1$ if
$b_j\subseteq a_i$, $0$
otherwise, so that, because the $b_j$ are disjoint,
$\chi a_i=\sum_{j=0}^m\gamma_{ij}\chi b_j$ for each $i$.   Then

\Centerline{$u=\sum_{i=0}^n\alpha_i\chi a_i
=\sum_{i=0}^n\sum_{j=0}^m\alpha_i\gamma_{ij}\chi b_j
=\sum_{j=0}^m\beta_j\chi b_j$,}

\noindent setting $\beta_j=\sum_{i=0}^n\alpha_i\gamma_{ij}$ for each
$j\le m$.

The expression for $\|u\|_{\infty}$ is now obvious.

\medskip

{\bf (c)(i)} If $u\ge 0$ in (b), we must have $\beta_j=u(z)\ge 0$
whenever $z\in\widehat{b}_j$, so that $\beta_j\ge 0$ whenever
$b_j\ne 0$;
consequently $u=\sum_{j=0}^m|\beta_j|\chi b_j$ is in the required form.

\medskip

\quad{\bf (ii)} If we suppose that every $\beta_j$ is non-negative, and
rearrange the terms of the
sum so that $\beta_0\le\ldots\le\beta_m$, then we may set
$\gamma_0=\beta_0$, $\gamma_j=\beta_j-\beta_{j-1}$ for $1\le j\le m$,
$c_j=\sup_{i\ge j}b_i$ to get

\Centerline{$\sum_{j=0}^m\gamma_j\chi
c_j=\sum_{j=0}^m\sum_{i=j}^m\gamma_j\chi b_i
=\sum_{i=0}^m\sum_{j=0}^i\gamma_j\chi b_i
=\sum_{i=0}^m\beta_i\chi b_i=u$.}

\medskip

{\bf (d)} is trivial, because $\widehat{b}_0,\ldots,\widehat{b}_n$ are
disjoint.

\medskip

{\bf (e)} By (d), $|u|\in S$ for every $u\in S$, so $S$ is a Riesz
subspace of $\Bbb R^Z$, and in itself is an Archimedean Riesz space.
If $\frak A$ is a Boolean algebra, then $\chi 1$, the constant function
with value $1$, belongs to $S$, and is an order unit of $S$;  while

\Centerline{$\|u\|_{\infty}
=\min\{\alpha:\alpha\ge 0,\,|u(z)|\le\alpha\Forall z\in Z\}
=\min\{\alpha:\alpha\ge 0,\,|u|\le\alpha\chi 1\}$}

\noindent for every $u\in S$.

\medskip

{\bf (f)} $\chi$ is injective because $\widehat{a}\ne\widehat{b}$
whenever $a\ne b$.   $\chi$ is additive because
$\widehat{a}\cap\widehat{b}=\emptyset$
whenever $a\Bcap b=0$.   Of course $\chi$ is non-negative.
It is a lattice
homomorphism because $a\mapsto\widehat{a}:\frak A\to\Cal PZ$ and
$E\mapsto\chi E:\Cal PZ\to\Bbb R^Z$ are.   To see that $\nu$
is order-continuous, take a
non-empty downwards-directed $A\subseteq\frak A$ with infimum $0$.
\Quer\  Suppose, if possible, that $\{\chi a:a\in A\}$ does not have
infimum $0$ in $S$.   Then there is a $u>0$ in $S$ such that
$u\le\chi a$ for every $a\in\frak A$.  Now $u$ can be expressed as
$\sum_{j=0}^m\beta_j\chi b_j$ where $b_0,\ldots,b_m$ are disjoint.   There
must be some $z_0\in Z$ such that $u(z_0)>0$;  take $j$ such that
$z_0\in\widehat{b}_j$, so that $b_j\ne 0$ and $\beta_j=u(z_0)>0$.   But
now, for any $z\in\widehat{b}_j$, $a\in A$,

\Centerline{$(\chi a)(z)\ge u(z)=\beta_j>0$}

\noindent and $z\in\widehat{a}$.   As $z$ is arbitrary,
$\widehat{b}_j\subseteq\widehat{a}$
and $b_j\Bsubseteq a$;  as $a$ is arbitrary,  $b_j$ is a non-zero lower
bound for $A$ in $\frak A$.\ \BanG\   So $\inf\chi[A]=0$ in $S$.   As
$A$ is arbitrary, $\chi$ is order-continuous, by the criterion of
361C(f-i).

\medskip

{\bf (g)} Express $u$ as $\sum_{j=0}^m\beta_j\chi b_j$ where
$b_0,\ldots,b_m$ are disjoint and every $\beta_j\ge 0$.   Then given
$\delta\ge 0$, $\eta>0$ and $a\in\frak A$ we have
$(\delta+\eta)\chi a\le u$ iff
$a\Bsubseteq\sup\{b_j:j\le m,\,\beta_j\ge\delta+\eta\}$.   So
$\Bvalue{u>\delta}=\sup\{b_j:j\le m,\,\beta_j>\delta\}$.   Writing
$c=\Bvalue{u>\delta}$, $d=\Bvalue{u>0}=\sup\{b_j:\beta_j>0\}$, we have

$$\eqalign{u(z)&\le\|u\|_{\infty}\text{ if }z\in\widehat{c},\cr
&\le\delta\text{ if }z\in\widehat{d}\setminus\widehat{c},\cr
&=0\text{ if }z\in Z\setminus\widehat{d}.\cr}$$

\noindent So

\Centerline{$\delta\chi c\le u\le\|u\|_{\infty}\chi c\vee\delta\chi d$,}

\noindent as claimed.   Taking $\delta=0$ we get
$u\le\|u\|_{\infty}\chi d$.   Set

\Centerline{$\eta=\min(\{1\}\cup\{\beta_j:j\le m$, $\beta_j>0\})$;}

\noindent then $\eta>0$ and $\eta\chi d\le u$.

If $u$, $v\in S^+$ take $\eta$, $\eta'>0$ such that

\Centerline{$\eta\chi\Bvalue{u>0}\le u$,
\quad$\eta'\chi\Bvalue{v>0}\le v$.}

\noindent Then

\Centerline{$\min(\eta,\eta')\chi(\Bvalue{u>0}\Bcap\Bvalue{v>0})
\le u\wedge v
\le\max(\|u\|_{\infty},\|v\|_{\infty})
\chi(\Bvalue{u>0}\Bcap\Bvalue{v>0})$.}

\noindent So

\Centerline{$u\wedge v=0
\Longrightarrow\Bvalue{u>0}\Bcap\Bvalue{v>0}=0
\Longrightarrow u\wedge v=0$.}

\medskip

{\bf (h)} $S$ is a commutative $f$-algebra and normed algebra just
because it is a Riesz subspace of the $f$-algebra and commutative
normed algebra $\ell^{\infty}(Z)$ and is closed under multiplication.

\medskip

{\bf (i)} If $u=\sum_{j=0}^m\beta_j\chi b_j$ where $b_0,\ldots,b_m$ are
disjoint and $\beta_j\ge 0$ for every $j$, then $u=v\times v$ where
$v=\sum_{j=0}^m\sqrt{\beta_j}\chi b_j$.
}% end of proof of 361E

\leader{361F}{}\cmmnt{ I now turn to the universal mapping
theorems which really define the construction.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Boolean ring, and $U$ any
linear space.   Then there is a one-to-one correspondence
between additive functions $\nu:\frak A\to U$ and linear operators
$T:S(\frak A)\to U$, given by the formula $\nu=T\chi$.

\proof{{\bf (a)} The core of the proof is the following observation.
Let $\nu:\frak A\to U$ be additive.   If $a_0,\ldots,a_n\in\frak A$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$ are such that
$\sum_{i=0}^n\alpha_i\chi a_i=0$ in $S=S(\frak A)$, then
$\sum_{i=0}^n\alpha_i\nu a_i=0$ in $U$.   \Prf\ By 361Ea, we can find
disjoint $b_0,\ldots,b_m$ such that each $a_i$ is the supremum of some
of the $b_j$;  set $\gamma_{ij}=1$ if $b_j\Bsubseteq a_i$, $0$
otherwise, so that $\chi a_i=\sum_{j=0}^m\gamma_{ij}\chi b_j$ and
$\nu a_i=\sum_{j=0}^m\gamma_{ij}\nu b_j$ for each $i$.   Set
$\beta_j=\sum_{i=0}^n\alpha_i\gamma_{ij}$ for each $j$, so that

\Centerline{$0=\sum_{i=0}^n\alpha_i\chi a_i
=\sum_{j=0}^m\beta_j\chi b_j$.}

\noindent Now $\beta_j\nu b_j=0$ in $U$ for each $j$, because either
$b_j=0$ and $\nu b_j=0$, or there is some $z\in\widehat{b}_j$ so that
$\beta_j$ must be $0$.   Accordingly

\Centerline{$0=\sum_{j=0}^m\beta_j\nu b_j
=\sum_{j=0}^m\sum_{i=0}^n\alpha_i\gamma_{ij}\nu b_j
=\sum_{i=0}^n\alpha_i\nu a_i$.  \Qed}

\medskip

{\bf (b)} It follows that if $u\in S$ is expressible simultaneously as
$\sum_{i=0}^n\alpha_i\chi a_i=\sum_{j=0}^m\beta_j\chi b_j$, then

\Centerline{$\sum_{i=0}^n\alpha_i\chi a_i+\sum_{j=0}^m(-\beta_j)\chi
b_j=0$ in $S$,}

\noindent so that

\Centerline{$\sum_{i=0}^n\alpha_i\nu a_i
+\sum_{j=0}^m(-\beta_j)\nu b_j=0$ in $U$,}

\noindent and

\Centerline{$\sum_{i=0}^n\alpha_i\nu a_i
=\sum_{j=0}^m\beta_j\nu b_j$.}

\noindent We can therefore define $T:S\to U$ by setting

\Centerline{$T(\sum_{i=0}^n\alpha_i\chi a_i)=\sum_{i=0}^n\alpha_i\nu
a_i$}

\noindent whenever $a_0,\ldots, a_n\in\frak A$ and
$\alpha_0,\ldots,\alpha_n\in\Bbb R$.

\medskip

{\bf (c)} It is now elementary to check that $T$ is linear, and that
$T\chi a=\nu a$ for every $a\in\frak A$.   Of course this last condition
uniquely defines $T$, because $\{\chi a:a\in\frak A\}$ spans the linear
space $S$.
}%end of proof of 361F

\leader{361G}{Theorem} Let $\frak A$ be a Boolean ring, and $U$ a
partially ordered linear space.   Let $\nu:\frak A\to U$ be an additive
function, and $T:S(\frak A)\to U$ the corresponding linear operator.

(a) $\nu$ is non-negative iff $T$ is positive.

(b) In this case,

\quad (i) if $T$ is order-continuous or sequentially order-continuous,
so is $\nu$;

\quad (ii) if $U$ is Archimedean and $\nu$ is order-continuous or
sequentially order-continuous, so is $T$.

(c) If $U$ is a Riesz space, then the following are equiveridical:

\quad (i) $T$ is a Riesz homomorphism;

\quad (ii) $\nu
a\wedge\nu b=0$ in $U$ whenever $a\Bcap b=0$ in $\frak A$;

\quad (iii) $\nu$ is a lattice homomorphism.

\proof{ Write $S$ for $S(\frak A)$.

\medskip

{\bf (a)} If $T$ is positive, then surely $\nu a=T\chi a\ge 0$ for every
$a\in\frak A$, so $\nu=T\chi$ is non-negative.
If $\nu$ is non-negative, and $u\ge 0$ in $S$, then $u$ is expressible
as $\sum_{j=0}^m\beta_j\chi b_j$ where $b_0,\ldots,b_m\in\frak A$ and
$\beta_j\ge 0$ for every $j$ (361Ec), so that

\Centerline{$Tu=\sum_{j=0}^m\beta_j\nu b_j\ge 0$.}

\noindent Thus $T$ is positive.

\medskip

{\bf (b)(i)} If $T$ is order-continuous (resp.\ sequentially
order-continuous) then $\nu=T\chi$ is the composition of two
order-continuous (resp.\ sequentially order-continuous) functions
(361Ef), so must be order-continuous (resp.\ sequentially
order-continuous).

\medskip

\quad{\bf (ii)} Assume now that $U$ is Archimedean.

\medskip

\qquad\grheada\ Suppose that $\nu$ is
order-continuous and that $A\subseteq S$ is non-empty,
downwards-directed and has infimum $0$.   Fix $u_0\in A$, set
$\alpha=\|u\|_{\infty}$ and $a_0=\Bvalue{u>0}$ (in the language of
361Eg).   If $\alpha=0$ then of course $\inf_{u\in A}Tu=Tu_0=0$.
Otherwise, take any $w\in U$ such that $w\not\le 0$.   Then there is some $\delta>0$ such
that $w\not\le\delta\nu a_0$, because $U$ is Archimedean.   Set
$A'=\{u:u\in A,\,u\le u_0\}$;
because $A$ is downwards-directed, $A'$ has the same lower bounds as
$A$, and $\inf A'=0$, while $A'$ is still downwards-directed.  For
$u\in A'$ set $c_u=\Bvalue{u>\delta}$, so that

\Centerline{$\delta\chi c_u\le u
\le\alpha\chi c_u+\delta\chi\Bvalue{u>0}
\le\alpha\chi c_u+\delta\chi a_0$}

\noindent (361Eg).   If $u$, $v\in A'$ and $u\le v$, then
$c_u\Bsubseteq c_v$, so $C=\{c_u:u\in A'\}$ is downwards-directed;  but if $c$ is any
lower bound for $C$ in $\frak A$, $\delta\chi c$ is a lower bound for
$A'$ in $S$, so is zero, and $c=0$ in $\frak A$.   Thus $\inf C=0$ in
$\frak A$, and $\inf_{u\in A'}\nu c_u=0$ in $U$.   But this means, in
particular, that $\bover1{\alpha}(w-\delta\nu a_0)$ is not a lower bound
for $\nu[C]$, and there is some $u\in A'$ such that
$\bover1{\alpha}(w-\delta\nu a_0)\not\le\nu c_u$, that is,
$w-\delta\nu a_0\not\le\alpha\nu c_u$, that is,
$w\not\le\delta\nu a_0+\alpha\nu c_u$.
As $u\le\alpha\chi c_u+\delta\chi a_0$,

\Centerline{$Tu\le T(\alpha\chi c_u+\delta\chi a_0)=\alpha\nu
c_u+\delta\nu a_0$,}

\noindent and $w\not\le Tu$.   Since $w$ is arbitrary, this means that
$0=\inf T[A]$;  as $A$ is arbitrary, $T$ is order-continuous.

\medskip

\qquad\grheadb\ The argument for sequential order-continuity is
essentially the same.   Suppose that $\nu$ is sequentially
order-continuous and that $\sequencen{u_n}$ is a non-increasing sequence
in $S$ with infimum $0$.   Again set $\alpha=\|u_0\|$,
$a_0=\Bvalue{u_0>0}$;  again we may suppose that $\alpha>0$;  again take
any $w\in U$ such that $w\not\le 0$.   As before, there is some
$\delta>0$ such that $w\not\le\delta\nu a_0$.   For $n\in\Bbb N$ set
$c_n=\Bvalue{u_n>\delta}$, so that

\Centerline{$\delta\chi c_n\le u_n\le\alpha\chi c_n+\delta\chi a_0$.}

\noindent   The sequence $\sequencen{c_n}$ is non-increasing because
$\sequencen{u_n}$ is,  and if $c\Bsubseteq c_n$ for every $n$, then
$\delta\chi c\le u_n$ for every $n$, so is zero, and $c=0$ in $\frak A$.
Thus $\inf_{n\in\Bbb N}c_n=0$ in $\frak A$, and
$\inf_{n\in\Bbb N}\nu c_n=0$ in $U$, because $\nu$ is sequentially order-continuous.
Replacing $A'$, $C$ in the argument above by $\{u_n:n\in\Bbb N\}$,
$\{c_n:n\in\Bbb N\}$ we find an $n$ such that $w\not\le Tu_n$.   Since
$w$ is arbitrary, this means that
$0=\inf_{n\in\Bbb N}Tu_n$;  as $\sequencen{u_n}$ is arbitrary, $T$ is
sequentially order-continuous.

\medskip

{\bf (c)(i)$\Rightarrow$(iii)} If $T:S(\frak A)\to U$ is a Riesz
homomorphism, and
$\nu=T\chi$, then surely $\nu$ is a lattice homomorphism because $T$ and
$\chi$ are.

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} is trivial.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} If $\nu a\wedge\nu b=0$ whenever $a\Bcap
b=0$, then for any $u\in S(\frak A)$ we
have an expression of $u$ as $\sum_{j=0}^m\beta_j\chi b_j$, where
$b_0,\ldots,b_m\in\frak A$ are disjoint.   Now

\Centerline{$|Tu|=|\sum_{j=0}^m\beta_j\nu b_j|
=\sum_{j=0}^m|\beta_j|\nu b_j=T(\sum_{j=0}^m|\beta_j|\chi b_j)
=T(|u|)$}

\noindent by 352Fb and 361Ed.   As $u$ is arbitrary, $T$ is a
Riesz homomorphism (352G).
}%end of proof of 361G

\leader{361H}{Theorem} Let $\frak A$ be a Boolean ring and $U$ a
Dedekind complete Riesz space.   Suppose that $\nu:\frak A\to U$ is an
additive function and $T:S=S(\frak A)\to U$ is the corresponding linear
operator.   Then $T\in\eurm L^{\sim}=\eurm L^{\sim}(S;U)$ iff $\{\nu
b:b\Bsubseteq a\}$ is order-bounded in $U$ for every $a\in\frak A$, and
in this case $|T|\in\eurm L^{\sim}$ corresponds to $|\nu|:\frak A\to U$,
defined by setting

$$\eqalign{|\nu|(a)
&=\sup\{\sum_{j=0}^n|\nu a_i|:a_0,\ldots,a_n\Bsubseteq a
\text{ are disjoint}\}\cr
&=\sup\{\nu b-\nu(a\Bsetminus b):b\Bsubseteq a\}\cr}$$

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

\proof{{\bf (a)} Suppose that $T\in\eurm L^{\sim}$ and $a\in\frak A$.
Then for any $b\Bsubseteq a$, we have $\chi b\le\chi a$ so

\Centerline{$|\nu b|=|T\chi b|\le|T|(\chi a)$.}

\noindent Accordingly $\{\nu b:b\Bsubseteq a\}$ is order-bounded in $U$.

\medskip

{\bf (b)} Now suppose that $\{\nu b:b\Bsubseteq a\}$ is order-bounded in
$U$ for every $a\in\frak A$.   Then for any $a\in\frak A$ we can define
$w_a=\sup\{|\nu b|:b\Bsubseteq a\}$;  in this case,
$\nu b-\nu(a\Bsetminus b)\le 2w_a$ whenever $b\Bsubseteq a$, so
$\theta a=\sup_{b\Bsubseteq a}\nu b-\nu(a\Bsetminus b)$ is defined
in $U$.   Considering $b=a$, $b=0$ we see that $\theta a\ge|\nu a|$.
Next, $\theta:\frak A\to U$ is additive.   \Prf\ Take $a_1$,
$a_2\in\frak A$ such that $a_1\Bcap a_2=0$;  set $a_0=a_1\Bcup a_2$.
For each $j\le 2$ set

\Centerline{$A_j=\{\nu(a_j\Bcap b)-\nu(a_j\Bsetminus b):b\in\frak
A\}\subseteq U$.}

\noindent   Then $A_0\subseteq A_1+A_2$, because

\Centerline{$\nu(a_0\Bcap b)-\nu(a_0\Bsetminus b)
=\nu(a_1\Bcap b)-\nu(a_1\Bsetminus b)
  +\nu(a_2\Bcap b)-\nu(a_2\Bsetminus b)$}

\noindent for every $b\in\frak A$.   But also $A_1+A_2\subseteq A_0$,
because if $b_1$, $b_2\in\frak A$ then

\Centerline{$\nu(a_1\Bcap b_1)-\nu(a_1\Bsetminus b_1)
+\nu(a_2\Bcap b_2)-\nu(a_2\Bsetminus b_2)
=\nu(a_0\Bcap b)-\nu(a_0\Bsetminus b)
$}

\noindent where $b=(a_1\Bcap b_1)\Bcup(a_2\Bcap b_2)$.   So
$A_0=A_1+A_2$, and

\Centerline{$\theta a_0=\sup A_0=\sup A_1+\sup A_2
=\theta a_1+\theta a_2$}

\noindent (351Dc).\ \Qed

We therefore have a corresponding positive operator $T_1:S\to U$ such
that $\theta=T_1\chi$.   But we also see that
$\theta a=\sup\{\sum_{i=0}^n|\nu a_i|:a_0,\ldots,a_n\Bsubseteq a$ are disjoint$\}$ for every $a\in\frak A$.   \Prf\ If $a_0,\ldots,a_n$ are
disjoint and included in $a$, then

\Centerline{$\sum_{i=0}^n|\nu a_i|\le\sum_{i=0}^n\theta a_i
=\theta(\sup_{i\le n}a_i)\le\theta a$.}

\noindent On the other hand,

\Centerline{$\theta a\le\sup_{b\Bsubseteq a}|\nu b|+|\nu(a\Bsetminus b)|
\le\sup\{\sum_{i=0}^n|\nu a_i|:a_0,\ldots,a_n\Bsubseteq a$
  are disjoint$\}$.  \Qed}

It follows that $T\in\eurm L^{\sim}$.   \Prf\ Take any $u\ge 0$ in $S$.
Set $a=\Bvalue{u>0}$ (361Eg) and $\alpha=\|u\|_{\infty}$.   If $0<|v|\le
u$, then $v$ is expressible as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n$ are disjoint and no $\alpha_i$ nor $a_i$ is zero.
Since $|v|\le\alpha\chi a$, we must have $|\alpha_i|\le\alpha$,
$a_i\Bsubseteq a$ for each $i$.   So

\Centerline{$|Tv|=|\sum_{i=0}^n\alpha_i\nu a_i|
\le\sum_{i=0}^n|\alpha_i||\nu a_i|
\le\alpha\sum_{i=0}^n|\nu a_i|
\le\alpha\theta a$.}

\noindent Thus $\{|Tv|:|v|\le u\}$ is bounded above by $\alpha\theta a$.
As $u$ is arbitrary, $T\in\eurm L^{\sim}$.
\Qed\

\medskip

{\bf (c)} Thus $T\in\eurm L^{\sim}$ iff $\nu$ is order-bounded on the
sets $\{b:b\Bsubseteq a\}$, and in this case the two formulae offered
for $|\nu|$ are consistent and make $|\nu|=\theta$.    Finally,
$\theta=|T|\chi$.   \Prf\ Take $a\in\frak A$.   If
$a_0,\ldots,a_n\Bsubseteq a$ are disjoint, then

\Centerline{$\sum_{i=0}^n|\nu a_i|=\sum_{i=0}^n|T\chi a_i|
\le\sum_{i=0}^n|T|(\chi a_i)\le|T|(\chi a)$;}

\noindent so $\theta a\le|T|(\chi a)$.   On the other hand, the argument
at the end of (b) above shows that $|T|(\chi a)\le\theta a$ for every
$a$.   Thus $|T|(\chi a)=\theta a$ for every
$a\in\frak A$, as required.\ \Qed
}%end of proof of 361H

\leader{361I}{Theorem} Let $\frak A$ be a Boolean ring, $U$ a normed
space and $\nu:\frak A\to U$
an additive function.   Give $S=S(\frak A)$ its norm $\|\,\|_{\infty}$,
and let $T:S\to U$ be the linear operator corresponding to $\nu$.   Then
$T$ is a bounded linear operator iff $\{\nu a:a\in\frak A\}$ is bounded,
and in this case $\|T\|=\sup_{a,b\in\frak A}\|\nu a-\nu b\|$.

\proof{{\bf (a)} If $T$ is bounded, then

\Centerline{$\|\nu a-\nu b\|=\|T(\chi a-\chi b)\|\le\|T\|\|\chi a-\chi
b\|_{\infty}\le
\|T\|$}

\noindent for every $a\in\frak A$, so $\nu$ is bounded and
$\sup_{a,b\in\frak A}\|\nu a-\nu b\|\le\|T\|$.

\medskip

{\bf (b)(i)} For the converse, we need a refinement of an idea in 361Ec.
If $u\in S$ and $u\ge 0$ and $\|u\|_{\infty}\le 1$, then $u$ is
expressible as $\sum_{i=0}^m\gamma_i\chi c_i$ where $\gamma_i\ge 0$ and
$\sum_{i=0}^m\gamma_i=1$.   \Prf\ If $u=0$, take $n=0$, $c_0=0$,
$\gamma_0=1$.   Otherwise, start from an expression
$u=\sum_{j=0}^n\gamma_j\chi c_j$ where
$c_0\Bsupseteq\ldots\Bsupseteq c_n$ and every $\gamma_j$ is
non-negative, as in 361Ec.   We may suppose
that $c_n\ne 0$, in which case

\Centerline{$\sum_{j=0}^n\gamma_j=u(z)\le 1$}

\noindent for every $z\in\widehat{c}_n\subseteq Z$, the Stone space of
$\frak A$.   Set $m=n+1$, $c_m=0$ and $\gamma_m=1-\sum_{j=0}^n\gamma_j$ to get the required form.\ \Qed

\medskip

\quad{\bf (ii)} The next fact we need is an elementary property of real
numbers:  if $\gamma_0,\ldots,\gamma_m$,
$\gamma'_0,\ldots,\gamma'_n\ge 0$
and $\sum_{i=0}^m\gamma_i=\sum_{j=0}^n\gamma'_j$, then there are
$\delta_{ij}\ge 0$ such that $\gamma_i=\sum_{j=0}^n\delta_{ij}$ for
every $i\le m$ and $\gamma'_j=\sum_{i=0}^m\delta_{ij}$ for every
$j\le n$.   \Prf\ This is just the case $U=\Bbb R$ of 352Fd.\ \Qed

\medskip

\quad{\bf (iii)} Now suppose that $\nu$ is bounded;  set
$\alpha_0=\sup_{a\in\frak A}\|\nu a\|<\infty$.   Then

\Centerline{$\alpha=\sup_{a,b\in\frak A}\|\nu a-\nu b\|\le 2\alpha_0$}

\noindent is also finite.
If $u\in S$ and $\|u\|_{\infty}\le 1$, then we can express $u$ as
$u^+-u^-$ where $u^+$, $u^-$ are
non-negative and also of norm at most $1$.   By (i), we can express
these as

\Centerline{$u^+=\sum_{i=0}^m\gamma_i\chi c_i$,
\quad $u^-=\sum_{j=0}^n\gamma_j'\chi c_j'$}

\noindent where all the $\gamma_i$, $\gamma_j'$ are non-negative and
$\sum_{i=0}^m\gamma_i=\sum_{j=0}^n\gamma'_j=1$.   Take
$\langle\delta_{ij}\rangle_{i\le m,j\le n}$ from (ii).   Set
$c_{ij}=c_i$, $c'_{ij}=c'_j$ for all $i$, $j$, so that

\Centerline{$u^+=\sum_{i=0}^m\sum_{j=0}^n\delta_{ij}\chi c_{ij}$,
\quad$u^-=\sum_{i=0}^m\sum_{j=0}^n\delta_{ij}\chi c'_{ij}$,}

\Centerline{$u=\sum_{i=0}^m\sum_{j=0}^n\delta_{ij}(\chi c_{ij}-\chi
c'_{ij})$,}

\Centerline{$Tu=\sum_{i=0}^m\sum_{j=0}^n\delta_{ij}(\nu c_{ij}-\nu
c'_{ij})$,}

\Centerline{$\|Tu\|
\le\sum_{i=0}^m\sum_{j=0}^n
\delta_{ij}\|\nu c_{ij}-\nu c'_{ij}\|
\le\sum_{i=0}^m\sum_{j=0}^n
\delta_{ij}\alpha=\alpha$.}

\noindent As $u$ is arbitrary, $T$ is a bounded linear operator and
$\|T\|\le\alpha$, as required.
}%end of proof of 361I

\leader{361J}{}\cmmnt{ The last few paragraphs describe the properties
of $S(\frak A)$ in terms of universal mapping theorems.   The next
theorem looks at the construction as a functor which converts Boolean
algebras into Riesz spaces and ring homomorphisms into Riesz
homomorphisms.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ and $\frak B$ be Boolean rings
and $\pi:\frak A\to\frak B$ a ring homomorphism.

(a) We have a Riesz homomorphism $T_{\pi}:S(\frak A)\to S(\frak B)$
given by the formula

\Centerline{$T_{\pi}(\chi a)=\chi(\pi a)$ for every $a\in\frak A$.}

\noindent For any $u\in S(\frak A)$, $\|T_{\pi}u\|_{\infty}
=\min\{\|u'\|_{\infty}:u'\in S(\frak A),\,T_{\pi}u'=T_{\pi}u\}$;
\cmmnt{in particular,} $\|T_{\pi}u\|_{\infty}\le\|u\|_{\infty}$.
\cmmnt{Moreover,} $T_{\pi}(u\times u')=T_{\pi}u\times T_{\pi}u'$
for all $u$, $u'\in S(\frak A)$.

(b) $T_{\pi}$ is surjective iff $\pi$ is surjective, and in this case
$\|v\|_{\infty}=\min\{\|u\|_{\infty}:u\in S(\frak A),\,T_{\pi}u=v\}$ for
every
$v\in S(\frak B)$.

(c) The kernel of $T_{\pi}$ is just the set of those $u\in S(\frak A)$
such that $\pi\Bvalue{|u|>0}=0$, defining $\Bvalue{\ldots>\ldots}$ as in
361Eg.

(d) $T_{\pi}$ is injective iff $\pi$ is injective, and in this case
$\|T_{\pi}u\|_{\infty}=\|u\|_{\infty}$ for every $u\in S(\frak A)$.

(e) $T_{\pi}$ is order-continuous iff $\pi$ is order-continuous.

(f) $T_{\pi}$ is sequentially order-continuous iff $\pi$ is sequentially
order-continuous.

(g) If $\frak C$ is another Boolean ring and $\phi:\frak B\to\frak C$
is another ring homomorphism, then
$T_{\phi\pi}=T_{\phi}T_{\pi}:S(\frak A)\to S(\frak C)$.

\proof{{\bf (a)} The map $\chi\pi:\frak A\to S(\frak B)$ is additive
(361Cc), so corresponds to a linear operator $T=T_{\pi}:S(\frak A)\to S(\frak
B)$, by 361F.   $\chi$ and $\pi$ are both lattice homomorphisms, so
$\chi\pi$ also is, and $T$ is a Riesz homomorphism (361Gc).   If
$u=\sum_{i=0}^n\alpha_i\chi a_i$,
where $a_0,\ldots,a_n$ are disjoint, then look at $I=\{i:i\le n,\,\pi
a_i\ne 0\}$.   We have

\Centerline{$Tu=\sum_{i=0}^n\alpha_i\chi(\pi a_i)=\sum_{i\in
I}\alpha_i\chi(\pi a_i)$}

\noindent and $\pi a_0,\ldots,\pi a_n$ are disjoint, so that

\Centerline{$\|Tu\|_{\infty}=\sup_{i\in I}|\alpha_i|=\|u'\|_{\infty}
\le\sup_{a_i\ne 0}|\alpha_i|\le\|u\|_{\infty}$,}

\noindent where $u'=\sum_{i\in I}\alpha_i\chi a_i$, so that $Tu'=Tu$.
If $a$, $a'\in\frak A$, then

\Centerline{$T(\chi a\times\chi a')=T\chi(a\Bcap a')
=\chi\pi(a\Bcap a')=\chi\pi a\times\chi\pi a'
=T\chi a\times T\chi a'$,}

\noindent so $T$ is multiplicative.


\medskip

{\bf (b)} If $\pi$ is surjective, then $T[S(\frak A)]$ must be the
linear span of

\Centerline{$\{T(\chi a):a\in\frak A\}=\{\chi(\pi a):a\in\frak
A\}=\{\chi b:b\in\frak B\}$,}

\noindent so is the whole of $S(\frak B)$.   If $T$ is surjective,
and $b\in\frak B$, then there must be a $u\in\frak A$ such that $Tu=\chi
b$.   We can express $u$ as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n$ are disjoint;  now

\Centerline{$\chi b=Tu=\sum_{i=0}^n\alpha_i\chi(\pi a_i)$,}

\noindent and $\pi a_0,\ldots,\pi a_n$ are disjoint in $\frak B$, so we
must have

\Centerline{$b=\sup_{i\in I}\pi a_i=\pi(\sup_{i\in I}a_i)\in\pi[\frak
A]$,}

\noindent where $I=\{i:\alpha_i=1\}$.   As $b$ is arbitrary, $\pi$ is
surjective.   Of course the formula for $\|v\|_{\infty}$ is a
consequence of the formula for $\|Tu\|_{\infty}$ in (a).

\medskip

{\bf (c)(i)} If $\pi\Bvalue{|u|>0}=0$ then
$|u|\le\alpha\chi a$, where $\alpha=\|u\|_{\infty}$,
and $a=\Bvalue{|u|>0}$, so

\Centerline{$|Tu|=T|u|\le\alpha T(\chi
a)=\alpha\chi(\pi a)=0$,}

\noindent and $Tu=0$.

\medskip

\quad{\bf (ii)} If $u\in S(\frak A)$ and
$Tu=0$, express $u$ as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n$ are disjoint and every $\alpha_i$ is non-zero (361Eb).
In this case

\Centerline{$0=|Tu|=T|u|=\sum_{i=0}^n|\alpha_i|\chi(\pi
a_i)$,}

\noindent so $\pi a_i=0$ for every $i$, and

\Centerline{$\pi\Bvalue{|u|>0}=\pi(\sup_{i\le n}a_i)=\sup_{i\le n}\pi
a_i=0$.}

\medskip

{\bf (d)} If $T$ is injective and $a\in\frak A\setminus\{0\}$,
then $\chi(\pi a)=T(\chi a)\ne 0$, so $\pi a\ne 0$;  as $a$ is
arbitrary, $\pi$ is injective.   If $\pi$ is injective then
$\pi\Bvalue{|u|>0}\ne 0$ for every non-zero $u\in S(\frak A)$, so $T$ is
injective, by (c).
In this case the formula in (a) shows that $T$ is norm-preserving.

\medskip

{\bf (e)(i)}  If $T$ is order-continuous and $A\subseteq\frak A$
is a non-empty downwards-directed set with infimum $0$ in $\frak A$, let
$b$ be any lower bound for $\pi[A]$ in $\frak B$.   Then

\Centerline{$\chi b\le\chi(\pi a)=T(\chi a)$}

\noindent for any $a\in A$.   But $T\chi$ is order-continuous, by
361Ef, so $\inf_{a\in A}T(\chi a)=0$, and $b$ must be $0$.   As
$b$ is arbitrary, $\inf_{a\in A}\pi a=0$;  as  $A$ is arbitrary, $\pi$
is order-continuous.

\medskip

\quad{\bf (ii)}  If $\pi$ is order-continuous, so is
$\chi\pi:\frak A\to S(\frak B)$, using 361Ef again;  but now by
361G(b-ii) $T$ must be order-continuous.

\medskip

{\bf (f)(i)}  If $T$ is sequentially order-continuous, and
$\sequencen{a_n}$ is a non-increasing sequence in $\frak A$ with infimum
$0$, let $b$ be any lower bound for $\{\pi a_n:n\in\Bbb N\}$ in
$\frak B$.   Then

\Centerline{$\chi b\le\chi(\pi a_n)=T(\chi a_n)$}

\noindent for any $a\in A$.   But $T\chi$ is sequentially
order-continuous so $\inf_{n\in\Bbb N}T(\chi a_n)=0$,
and $b$ must be $0$.   As $b$ is arbitrary, $\inf_{n\in\Bbb N}\pi
a_n=0$;  as  $A$ is arbitrary, $\pi$ is sequentially order-continuous.

\medskip

\quad{\bf (ii)}  If $\pi$ is sequentially order-continuous, so is
$\chi\pi:\frak A\to S(\frak B)$;  but now $T$ must be sequentially
order-continuous.

\medskip

{\bf (g)} We need only check that

\Centerline{$T_{\phi\pi}(\chi a)=\chi(\phi(\pi a))
=T_{\phi}(\chi(\pi a))=T_{\phi}T_{\pi}(\chi a)$}

\noindent for every $a\in\frak A$.
}%end of proof of 361J

\leader{361K}{Proposition} Let $\frak A$ be a Boolean algebra.   For
$a\in\frak A$ write $V_a$ for the solid linear subspace of $S(\frak A)$
generated by $\chi a$.   Then $a\mapsto V_a$ is a Boolean isomorphism
between $\frak A$ and the algebra of projection bands in $S(\frak A)$.

\proof{ Write $S$ for $S(\frak A)$.

\medskip

{\bf (a)} The point is that, for any $a\in\frak A$,

\quad(i) $|u|\wedge|v|=0$ whenever $u\in V_a$, $v\in V_{1\Bsetminus
a}$,

\quad(ii) $V_a+V_{1\Bsetminus a}=S$.

\noindent\Prf\ (i) is just because
$\chi a\wedge\chi(1\Bsetminus a)=0$.   As for (ii), if $w\in S$ then

\Centerline{$w=(w\times\chi a)+(w\times\chi(1\Bsetminus a))
\in V_a+V_{1\Bsetminus a}$.  \Qed}

\medskip

{\bf (b)} Accordingly $V_a+V_a^{\perp}\supseteq V_a+V_{1\Bsetminus a}=S$
and $V_a$ is a projection band (352R).   Next, any projection band
$U\subseteq S$ is of the
form $V_a$.   \Prf\ We know that $\chi 1=u+v$ where $u\in U$,
$v\in U^{\perp}$.   Since $|u|\wedge|v|=0$, $u$ and $v$ must be the
indicator functions of complementary subsets of $Z$, the Stone
space of $\frak A$.   But $\{z:u(z)\ne 0\}=\{z:u(z)\ge 1\}$ must be of
the form $\widehat{a}$, where $a=\Bvalue{u>0}$, in which case
$u=\chi a$ and $v=\chi(1\Bsetminus a)$.  Accordingly $U\supseteq V_a$
and $U^{\perp}\supseteq V_{1\Bsetminus a}$.   But this means that $U$
must be $V_a$ precisely.\ \Qed

\medskip

{\bf (c)} Thus $a\mapsto V_a$ is a surjective function from $\frak A$
onto the algebra of projection bands in $S$.   Now

\Centerline{$a\Bsubseteq b\iff \chi a\in V_b\iff V_a\subseteq V_b$,}

\noindent so $a\mapsto V_a$ is order-preserving and bijective.   By
312M it is a Boolean isomorphism.
}%end of proof of 361K


\leader{361L}{Proposition} Let $X$ be a set, and $\Sigma$ a ring of
subsets of $X$, that is, a subring of the Boolean ring $\Cal PX$.   Then
$S(\Sigma)$ can be identified, as ordered linear space,
with the linear subspace of $\ell^{\infty}(X)$ generated by the
indicator
functions of members of $\Sigma$, which is a Riesz subspace of
$\ell^{\infty}(X)$.   The norm of $S(\Sigma)$ corresponds to the uniform
norm on $\ell^{\infty}(X)$, and its multiplication to pointwise
multiplication of functions.

\proof{ Let $Z$ be the Stone space of $\Sigma$, and for $E\in\Sigma$
write $\chi E$ for the indicator function of $E$ as a subset of
$X$, $\hat\chi E$ for the indicator function of the
open-and-compact subset of $Z$ corresponding to $E$.   Of course
$\chi:\Sigma\to\ell^{\infty}(X)$
is additive, so by 361F there is a linear operator $T:S\to\ell^{\infty}(X)$, writing $S$ for $S(\Sigma)$,
such that $T(\hat\chi E)=\chi E$ for every $E\in\Sigma$.

If $u\in S$, $Tu\ge 0$ iff $u\ge 0$.   \Prf\ Express $u$ as
$\sum_{j=0}^m\beta_j\hat\chi E_j$ where $E_0,\ldots,E_m$ are disjoint.
Then $Tu=\sum_{j=0}^m\beta_j\chi E_j$, so


\Centerline{$u\ge 0
\iff \beta_j\ge 0\text{ whenever }E_j\ne\emptyset
\iff Tu\ge 0$.  \Qed}

\noindent But this means ($\alpha$) that

\Centerline{$Tu=0\iff Tu\ge 0 \,\,\&\,\, T(-u)\ge 0
\iff u\ge 0\,\,\&\,-u\ge 0\iff u=0$,}

\noindent so that $T$ is injective and is a linear space isomorphism
between $S$ and its image $\eusm S$, which must be the linear space
spanned by $\{\chi E:E\in\Sigma\}$ ($\beta$) that $T$ is an
order-isomorphism between $S$ and $\eusm S$.

Because $\chi E\wedge\chi F=0$ whenever $E$, $F\in\Sigma$ and
$E\cap F=\emptyset$, $T$ is a Riesz homomorphism and $\eusm S$ is a Riesz subspace of $\ell^{\infty}(X)$ (361Gc).   Now

\Centerline{$\|u\|_{\infty}
=\inf\{\alpha:|u|\le\alpha\hat\chi X\}
=\inf\{\alpha:|Tu|\le\alpha\chi X\}
=\|Tu\|_{\infty}$}

\noindent for every $u\in S$.   Finally,

\Centerline{$T(\hat\chi E\times\hat\chi F)
=T(\hat\chi(E\cap F))
=\chi(E\cap F)
=T(\hat\chi E)\times T(\hat\chi F)$}

\noindent for all $E$, $F\in\Sigma$, so $\eusm S$ is closed under
pointwise multiplication and the multiplications of $S$, $\eusm S$ are
identified by $T$.
}%end of proof of 361L

\leader{361M}{Proposition} Let $X$ be a set, $\Sigma$ a ring of
subsets of $X$, and $\Cal I$ an ideal of $\Sigma$;  write $\frak A$ for
the quotient ring $\Sigma/\Cal I$.   Let $\eusm S$ be the
linear span of $\{\chi E:E\in\Sigma\}$ in $\Bbb R^X$, and write

\Centerline{$V=\{f:f\in \eusm S,\,\{x:f(x)\ne 0\}\in\Cal I\}$.}

\noindent\cmmnt{ Then }$V$ is a solid linear subspace of $\eusm S$.
\cmmnt{Now} $S(\frak A)$ becomes identified with the quotient Riesz space
$\eusm S/V$, if for every $E\in\Sigma$ we identify
$\chi(E^{\ssbullet})\in
S(\frak A)$ with $(\chi E)^{\ssbullet}\in \eusm S/V$.   If we give
$\eusm S$ its
uniform norm inherited from $\ell^{\infty}(X)$, $V$ is a closed linear
subspace of $\eusm S$, and the quotient norm on $\eusm S/V$ corresponds
to the norm
of $S(\frak A)$:

\Centerline{$\|f^{\ssbullet}\|=\min\{\alpha:\{x:|f(x)|>\alpha\}\in\Cal
I\}$.}

\noindent If we write $\times$ for pointwise multiplication on $\eusm
S$, then
$V$ is an ideal of the ring $(\eusm S,+,\times)$, and the multiplication
induced on $\eusm S/V$ corresponds to the multiplication of $S(\frak
A)$.

\proof{ Use 361J and 361L.   We can identify $\eusm S$ with $S(\Sigma)$.
Now the canonical ring homomorphism $E\mapsto E^{\ssbullet}$ corresponds
to a surjective Riesz homomorphism $T$ from $S(\Sigma)$ to
$S(\frak A)$ which takes $\chi E$ to $\chi(E^{\ssbullet})$.   For
$f\in\eusm S$, $\Bvalue{|f|>0}$ is just $\{x:f(x)\ne 0\}$, so   the
kernel of $T$ is just the set of those $f\in \eusm S$ such that
$\{x:f(x)\ne 0\}\in\Cal I$, which is $V$.   So

\Centerline{$S(\frak A)=T[\eusm S]\cong \eusm S/V$.}

As noted in 361Ja, $T(f\times g)=Tf\times Tg$ for all $f$, $g\in \eusm
S$, so
the multiplications of $\eusm S/V$ and $S(\frak A)$ match.   As for the
norms, the norm of $S(\frak A)$ corresponds to the norm of $\eusm S/V$
by the formulae in 361Ja or 361Jb.   To see that $V$ is closed in
$\eusm S$, we need note only that if
$f\in\overline{V}$ then

\Centerline{$\|Tf\|_{\infty}=\inf_{g\in V}\|f+g\|_{\infty}=\inf_{g\in
V}\|f-g\|_{\infty}=0$,}

\noindent so that $Tf=0$ and $f\in V$.
To check the formula for $\|f^{\ssbullet}\|$,
take any $f\in\eusm S$.   Express it as $\sum_{i=0}^n\alpha_i\chi E_i$
where $E_0,\ldots,E_n\in\Sigma$ are disjoint.   Set
$I=\{i:E_i\notin\Cal I\}$;  then

\Centerline{$\|Tf\|_{\infty}=\max_{i\in
I}|\alpha_i|=\min\{\alpha:\{x:|f(x)|>\alpha\}\in\Cal I\}$.}
}%end of proof of 361M

\exercises{\leader{361X}{Basic exercises (a)}
%\spheader 361Xa
Let $\frak A$ be a Boolean ring and $U$ a linear space.   Show that a
function $\nu:\frak A\to U$ is additive iff $\nu 0=0$ and
$\nu(a\Bcup  b)+\nu(a\Bcap b)=\nu a+\nu b$ for all $a$, $b\in\frak A$.
%361C

\sqheader 361Xb
Let $U$ be an {\bf algebra over} $\Bbb R$, that is, a
real linear space endowed with a multiplication $\times$ such that
$(U,+,\times)$ is a ring and $\alpha(w\times z)=(\alpha w)\times
z=w\times(\alpha z)$ for all $w$, $z\in U$ and all $\alpha\in\Bbb R$.
Let $\frak A$ be a Boolean ring, $\nu:\frak A\to U$ an additive
function and $T:S(\frak A)\to U$ the corresponding linear operator.
Show that $T$ is multiplicative iff $\nu(a\Bcap b)=\nu a\times\nu b$ for
all $a$, $b\in\frak A$.
%361G

\sqheader 361Xc Let $\frak A$ be a Boolean ring, and $U$ a Dedekind
complete Riesz space.   Suppose that $\nu:\frak A\to U$ is an additive
function such that the corresponding linear operator $T:S(\frak A)\to U$
belongs to $\eurm L^{\sim}=\eurm L^{\sim}(S(\frak A);U)$.   Show that
$T^+\in\eurm L^{\sim}$ corresponds to $\nu^+:\frak A\to U$, where
$\nu^+a=\sup_{b\Bsubseteq a}\nu b$ for every $a\in\frak A$.
%361H

\spheader 361Xd Let $\frak A$ and $\frak
B$ be Boolean algebras.   Show that there is a natural one-to-one
correspondence between Boolean homomorphisms $\pi:\frak A\to\frak B$ and
Riesz homomorphisms $T:S(\frak A)\to S(\frak B)$ such that $T(\chi
1_{\frak A})=\chi 1_{\frak B}$, given by setting $T(\chi a)=\chi(\pi a)$
for every $a\in\frak A$.
%361J

\spheader 361Xe Let $\frak A$, $\frak B$ be Boolean rings and
$T:S(\frak A)\to S(\frak B)$ a linear operator such that $T(u\times
v)=Tu\times Tv$ for all $u$, $v\in S(\frak A)$.    Show that there is a
ring homomorphism $\pi:\frak A\to\frak B$ such that $T(\chi a)=\chi(\pi
a)$ for every $a\in\frak A$.
%361J

\spheader 361Xf Let $\frak A$ and $\frak B$ be Boolean rings.   Show
that any isomorphism of the algebras $S(\frak A)$ and $S(\frak B)$
(using the word `algebra' in the sense of 361Xb) must be a Riesz space
isomorphism, and therefore corresponds to an isomorphism between $\frak
A$ and $\frak B$.
%361G, 361Xb, 361Xe 361J

\spheader 361Xg Let $\frak A$, $\frak B$ be Boolean algebras and
$T:S(\frak A)\to S(\frak B)$ a Riesz homomorphism.    Show that there
are a ring homomorphism $\pi:\frak A\to\frak B$ and a non-negative
$v\in S(\frak B)$ such that $T(\chi a)=v\times\chi(\pi a)$ for every
$a\in\frak A$.
%361J

\spheader 361Xh Let $\frak A$ be a Boolean algebra,
$\pi:\frak A\to\frak A$ a Boolean homomorphism and
$T_{\pi}:S(\frak A)\to S(\frak A)$ the associated Riesz homomorphism.
Let $\frak C$ be the fixed-point subalgebra of $\pi$ (312K).
Show that $S(\frak C)$ may be identified with the linear subspace of
$S(\frak A)$ generated by $\{\chi c:c\in\frak C\}$, and that this is
$\{u:u\in S(\frak A)$, $T_{\pi}u=u\}$.
%361J

\spheader 361Xi Let $\frak A$ be a Boolean ring.   Show that for any
$u\in S(\frak A)$ the solid linear subspace of $S(\frak A)$ generated by
$u$ is a projection band in $S(\frak A)$.   Show that the set of such
bands is an ideal in the algebra of all projection bands, and is
isomorphic to $\frak A$.
%361K

\sqheader 361Xj Let $X$ be a set and $\Sigma$ a $\sigma$-algebra of subsets
of $X$.    Show that the linear span $\eusm S$ in $\Bbb R^X$ of
$\{\chi E:E\in\Sigma\}$ is just the set of $\Sigma$-measurable functions $f:X\to\Bbb R$ which take only finitely many values.
%361L

\spheader 361Xk For any Boolean ring $\frak A$, we may define its
`complex $S$-space' $S_{\Bbb C}(\frak A)$ as the linear span in
$\Bbb C^Z$ of the indicator functions of open-and-compact subsets of
the Stone space $Z$ of $\frak A$.   State and prove results corresponding
to 361Eb, 361Ed, 361Eh, 361F, 361L and 361M.
%361M

\spheader 361Xl\dvAnew{2013} Let $\frak A$ be a Boolean algebra, $U$ a
partially ordered linear space and $\nu:\frak A\to U$ a non-negative
additive function.   (i) Show that $\nu$ is order-continuous iff
$\nu 1=\sup_{J\subseteq I\text{ is finite}}\sum_{i\in J}\nu a_i$ whenever
$\familyiI{a_i}$ is a partition of unity in $U$.
(i) Show that $\nu$ is order-continuous iff
$\nu 1=\sup_{n\in\Bbb N}\sum_{i=0}^n\nu a_i$ whenever
$\sequence{i}{a_i}$ is a partition of unity in $U$.
%361C out of order query

\leader{361Y}{Further exercises (a)}
%\spheader 361Ya
Let $\frak A$ be a Boolean ring.   For $a\in\frak A$ let
$e_a\in\BbbR^{\frak A}$ be the function such that $e_a(a)=1$, $e_a(b)=0$
for $b\in\frak A\setminus\{a\}$;  let $V$ be the linear subspace of
$\BbbR^{\frak A}$ generated by $\{e_a:a\in\frak A\}$.
Let $W\subseteq V$ be the linear subspace
spanned by members of $V$ of the form $e_{a\Bcup b}-e_a-e_b$ where $a$,
$b\in\frak A$ are disjoint.   Define $\chi':\frak A\to V/W$ by taking
$\chi'a=e_a^{\ssbullet}$ to be the image in $V/W$ of $e_a\in V$.
Show, without using
the axiom of choice, that the pair $(V/W,\chi')$ has the universal
mapping property of $(S(\frak A),\chi)$ as described in 361F and that
$V/W$ has a Riesz space structure, a norm and a multiplicative structure
as described in 361D-361E.  Prove results corresponding to 361E-361M.
  %361E 361F 361G 361H 361I 361J 361K 361L 361M
%361D

\spheader 361Yb Let $\familyiI{\frak A_i}$ be a non-empty family
of Boolean algebras, with free product $\frak A$;  write
$\varepsilon_i:\frak A_i\to\frak A$ for the canonical maps, and

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

\noindent Suppose that $U$ is a linear space and $\theta:C\to U$ is such
that

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

\noindent whenever $c\in C$, $i\in I$ and $a\in\frak A_i$.
Show that there is a unique additive function $\nu:\frak A\to U$ extending
$\theta$.   \Hint{326E.}
%361F

\spheader 361Yc Let $\frak A$ be a Boolean ring and $U$ a Dedekind
complete Riesz space.   Let
$A\subseteq\eurm L^{\sim}=\eurm L^{\sim}(S(\frak A);U)$ be a non-empty set.   Suppose that
$\tilde T=\sup A$ is defined in $\eurm L^{\sim}$, and that $\tilde\nu=\tilde T\chi$.   Show that for any $a\in\frak A$,

\Centerline{$\tilde\nu a
=\sup\{\sum_{i=0}^nT_i(\chi a_i):T_0,\ldots,T_n\in A,
\,a_0,\ldots,a_n\Bsubseteq a$ are disjoint, $\sup_{i\le n}a_i=a\}$.}
%361H

\spheader 361Yd Let $\frak A$ be a Boolean algebra.   Show that the
algebra of all bands of $S(\frak A)$ can be identified with the Dedekind
completion of $\frak A$ (314U).
%361K

\spheader 361Ye Let $\frak A$ be a Boolean ring, and $U$ a complex
normed space.   Let $\nu:\frak A\to U$ be an additive function and
$T:S_{\Bbb C}(\frak A)\to U$ the corresponding linear operator (cf.\ 361Xk).   Show
that (giving $S_{\Bbb C}(\frak A)$ its usual norm $\|\,\|_{\infty}$)

\Centerline{$\|T\|
=\sup\{\|\sum_{j=0}^n\zeta_j\nu a_j\|:
 a_0,\ldots,a_n\in\frak A$ are disjoint, $|\zeta_j|=1$ for every $j\}$}

\noindent if either is finite.
%361M, 361Xk

\spheader 361Yf Let $U$ be a Riesz space.   Show that it is isomorphic
to $S(\frak A)$, for some Boolean algebra $\frak A$, iff it has an order
unit and every solid linear subspace of $U$ is a projection band.
%361M
}%end of exercises

\cmmnt{\Notesheader{361} The space
$S(\frak A)$ corresponds of course to the idea of `simple function'
which belongs to the very beginnings of the theory of integration
(122A).   All that 361D is trying to do is to set up a logically sound
description of this obvious concept which can be derived from the
Boolean ring $\frak A$ itself.   To my eye, there is a defect in the
construction there.   It relies on the axiom of choice, since it uses
the Stone space;  but none of the elementary properties of $S(\frak A)$
have anything to do with the axiom of choice.   In 361Ya I offer an
alternative construction which is in a formal sense more `elementary'.
If you work through the suggestion there you will find, however, that
the technical details become significantly more complicated, and would
be intolerable were it not for the intuition provided by the Stone space
construction.   Of course this intuition is chiefly valuable in the
finitistic arguments used in 361E, 361F and 361I;  and for these
arguments we really need the Stone representation only for finite
Boolean rings, which does not depend on the axiom of choice.

It is quite true that in most of this volume (and in most of this
chapter) I use the axiom of choice without scruple and without comment.
I mention it here only because I find myself using arguments dependent
on choice to prove theorems of a type to which the axiom cannot be
relevant.

The linear space structure of $S(\frak A)$, together with the map
$\chi$, are uniquely determined by the first universal mapping theorem
here, 361F.   This result says nothing about the order structure, which
needs the further refinement in 361Ga.   What is striking is that the
partial order defined by 361Ga is actually a lattice ordering, so that
we can have a universal mapping theorem for functions to Riesz spaces,
as in 361Gc and 361Ja.   Moreover, the same ordering provides a happy
abundance of results concerning order-continuous functions (361Gb,
361Je-361Jf).   When the codomain is a Dedekind complete Riesz space, so
that we have a Riesz space $\eurm L^{\sim}(S;U)$, and a corresponding
modulus function $T\mapsto|T|$ for linear operators, there are
reasonably natural formulae for $|T|\chi$ in terms of $T\chi$ (361H);
see also 361Xc and 361Yc.
The multiplicative structure of $S(\frak A)$ is defined
by 361Xb, and its norm by 361I.

The Boolean ring $\frak A$ cannot be recovered from the linear space
structure of $S(\frak A)$ alone (since this tells us only the
cardinality of $\frak A$), but if we add either the ordering or the
multiplication of $S(\frak A)$ then $\frak A$ is easy to identify (361K,
361Xf).

The most important Boolean algebras of measure theory arise either as
algebras of sets or as their quotients, so it is a welcome fact that in
such cases the spaces $S(\frak A)$ have straightforward representations
in terms of the construction of $\frak A$ (361L-361M).

In Chapter 24 I offered a paragraph in each section to sketch a version
of the theory based on the field of complex numbers rather than the
field of real numbers.   This was because so many of the most important
applications of these ideas involve complex numbers, even though (in my
view) the ideas themselves are most clearly and characteristically
expressed in terms of real numbers.   In the present chapter we are one
step farther away from these applications, and I therefore relegate
complex numbers to the exercises, as in 361Xk and 361Ye.

}%end of comment

\discrpage

\frfilename{mt362.tex}
\versiondate{31.12.10}
\copyrightdate{1996}

\def\chaptername{Function spaces}
\def\sectionname{{$S^{\sim}$}}

\newsection{362}

The next stage in our journey is the systematic investigation of linear
functionals on spaces $S=S(\frak A)$.   We already know that these
correspond to additive real-valued functionals on the algebra $\frak A$
(361F).   My purpose here is to show how the structure of the Riesz
space dual $S^{\sim}$ and its bands is related to the classes of
additive functionals introduced in \S\S326-327.   The first step is just
to check the identification of the linear and order structures of
$S^{\sim}$ and the space $M$ of bounded finitely additive functionals
(362A);   all the ideas needed for this have already been set out, and
the basic properties of $S^{\sim}$ are covered by the general results in
\S356.   Next, we need to be able to describe the operations on $M$
corresponding to the Riesz space operations $|\,\,|$, $\vee$, $\wedge$
on $S^{\sim}$, and the band projections from $S^{\sim}$
onto $S^{\sim}_c$ and $S^{\times}$;  these are dealt with in 362B, with
a supplementary remark in 362D.   In
the case of measure algebras, we have some further important bands which
present themselves in $M$, rather than in $S^{\sim}$, and which are
treated in 362C.   Since all these spaces are $L$-spaces, it is worth
taking a moment to identify their uniformly integrable subsets;  I do
this in 362E.

While some of the ideas here have interesting extensions to the case in
which $\frak A$ is a Boolean ring without identity, these can I think be
left to one side;  the work of this section will be done on the
assumption that every $\frak A$ is a Boolean algebra.

\leader{362A}{Theorem} Let $\frak A$ be a Boolean algebra.   Write $S$
for $S(\frak A)$.

(a) The partially ordered linear space of all finitely additive
real-valued functionals on $\frak A$ may be identified with the
partially ordered linear space of all real-valued linear functionals on
$S$.

(b) The linear space of bounded finitely additive real-valued
functionals on $\frak A$ may be identified with the $L$-space
$S^{\sim}$ of order-bounded linear functionals on $S$.   If
$f\in S^{\sim}$ corresponds to $\nu:\frak A\to\Bbb R$, then
$f^+\in S^{\sim}$ corresponds to $\nu^+$, where

\Centerline{$\nu^+a=\sup_{b\Bsubseteqshort a}\nu b$}

\noindent for every $a\in\frak A$, and

\Centerline{$\|f\|=\sup_{a\in\frak A}\nu a-\nu(1\Bsetminus a)$.}

(c) The linear space of bounded countably additive real-valued
functionals on $\frak A$ may be identified with the $L$-space
$S^{\sim}_c$.

(d) The linear space of completely additive real-valued functionals on
$\frak A$ may be identified with the $L$-space $S^{\times}$.

\proof{ By 361F, we have a canonical one-to-one correspondence between
linear functionals $f:S\to\Bbb R$ and additive functionals
$\nu_f:\frak A\to\Bbb R$, given by setting $\nu_f=f\chi$.

\medskip

{\bf (a)} Now
it is clear that $\nu_{f+g}=\nu_f+\nu_g$, $\nu_{\alpha f}=\alpha\nu_f$
for all $f$, $g$ and $\alpha$, so this one-to-one correspondence is a
linear space isomorphism.   To see that it is also an order-isomorphism,
we need note only that $\nu_f$ is non-negative iff $f$ is, by 361Ga.

\medskip

{\bf (b)} Recall from 356N that, because $S$ is a Riesz space with
order unit (361Ee), $S^{\sim}$ has a corresponding norm under which it
is an $L$-space.

\medskip

\quad {\bf (i)} If $f\in S^{\sim}$, then

\Centerline{$\sup_{b\in\frak A}|\nu_fb|
=\sup_{b\in\frak A}|f(\chi b)|\le\sup\{|f(u)|:u\in S,\,|u|\le\chi 1\}$}

\noindent is finite, and $\nu_f$ is bounded.

\medskip

\quad{\bf (ii)} Now suppose that $\nu_f$ is bounded and that $v\in S^+$.
Then there is an $\alpha\ge 0$ such that $v\le\alpha\chi 1$ (361Ee).
If $u\in S$ and $|u|\le v$, then we can express $u$ as
$\sum_{i=0}^n\alpha_i\chi a_i$ where $a_0,\ldots,a_n$ are disjoint
(361Eb);  now $|\alpha_i|\le\alpha$ whenever $a_i\ne 0$, so

\Centerline{$|f(u)|=|\sum_{i=0}^n\alpha_i\nu_fa_i|
\le\alpha\sum_{i=0}^n|\nu_fa_i|
=\alpha(\nu_fc_1-\nu_fc_2)
\le 2\alpha\sup_{b\in\frak A}|\nu_fb|$,}

\noindent setting $c_1=\sup\{a_i:i\le n,\,\nu_fa_i\ge 0\}$,
$c_2=\sup\{a_i:i\le n,\,\nu_fa_i<0\}$.   This shows that
$\{f(u):|u|\le v\}$ is bounded.   As $v$ is arbitrary, $f\in S^{\sim}$
(356Aa).

\medskip

\quad{\bf (iii)} To check the correspondence between $f^+$ and
$\nu_f^+$, refine the arguments of (i) and (ii)  as follows.   Take any
$f\in S^{\sim}$.    If $a\in\frak A$,

\Centerline{$\nu_f^+a=\sup_{b\Bsubseteqshort a}\nu_fb
=\sup_{b\Bsubseteqshort a}f(\chi b)\le\sup\{f(u):u\in S,\,0\le u\le\chi a\}
=f^+(\chi a)$.}

\noindent On the other hand, if $u\in S$ and $0\le u\le\chi a$, then we
can express $u$ as $\sum_{i=0}^n\alpha_i\chi a_i$ where $a_0,\ldots,a_n$
are disjoint;  now $0\le\alpha_i\le 1$ whenever $a_i\ne 0$, so

\Centerline{$f(u)=\sum_{i=0}^n\alpha_i\nu_fa_i\le\nu_fc
\le\nu_f^+a$,}

\noindent where $c=\sup\{a_i:i\le n,\,\nu_fa_i\ge 0\}$.
As $u$ is arbitrary, $f^+(\chi a)\le\nu_f^+a$.   This shows that
$\nu_f^+=f^+\chi$ is finitely additive, and that $\nu_f^+=\nu_{f^+}$, as
claimed.

\medskip

\quad{\bf (iv)} Now, for any $f\in S^{\sim}$,

$$\eqalignno{\|f\|
&=|f|(\chi 1)\cr
\noalign{\noindent (356N)}
&=(2f^+-f)(\chi 1)
=2\nu_f^+1-\nu_f1\cr
\noalign{\noindent (by (iii) just above)}
&=\sup_{a\in\frak A}2\nu_fa-\nu_f1
=\sup_{a\in\frak A}\nu_fa-\nu_f(1\Bsetminus a).\cr}$$

\medskip

{\bf (c)} If $f\ge 0$ in $S^{\sim}$, then $f$ is sequentially
order-continuous iff $\nu_f$ is sequentially order-continuous (361Gb),
that is, iff $\nu_f$ is countably additive (326Kc).   Generally, an
order-bounded linear functional belongs to $S^{\sim}_c$ iff it is
expressible as the difference of two sequentially
order-continuous positive linear functionals (356Ab), while a
bounded finitely additive functional is countably additive iff it is
expressible as the difference of two non-negative countably additive
functionals (326L);  so in the present context $f\in S^{\sim}_c$ iff
$\nu_f$ is bounded and countably additive.

\medskip

{\bf (d)} If $f\ge 0$ in $S^{\sim}$, then $f$ is
order-continuous iff $\nu_f$ is order-continuous (361Gb), that is, iff
$\nu_f$ is completely additive (326Oc).   Generally, an
order-bounded linear functional belongs to $S^{\times}$ iff it is
expressible as the difference of two
order-continuous positive linear functionals (356Ac), while a
finitely additive functional is completely additive iff it is
expressible as the difference of two non-negative completely additive
functionals (326Q);  so in the present context $f\in S^{\times}$ iff
$\nu_f$ is completely additive.
}%end of proof of 362A

\leader{362B}{Spaces of finitely additive
\dvrocolon{functionals}}\cmmnt{ The
identifications in the last theorem mean that we can relate the Riesz
space structure of $S(\frak A)^{\sim}$ to constructions involving
finitely additive functionals.   I have already set out the most useful
facts as exercises (326Yd, 326Ym, 326Yn, 326Yp, 326Yq);  it is now time
to repeat them more formally.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Boolean algebra.
Let $M$ be the Riesz space of bounded finitely
additive real-valued functionals on $\frak A$, $M_{\sigma}\subseteq M$
the space of bounded countably additive functionals, and
$M_{\tau}\subseteq M_{\sigma}$ the space of completely additive
functionals.

(a) For any $\mu$, $\nu\in M$, $\mu\vee\nu$, $\mu\wedge\nu$ and $|\nu|$
are defined by the formulae

\Centerline{$(\mu\vee\nu)(a)
=\sup_{b\Bsubseteqshort a}\mu b+\nu(a\Bsetminus b)$,}

\Centerline{$(\mu\wedge\nu)(a)=\inf_{b\Bsubseteqshort a}\mu b+\nu(a\Bsetminus
b)$,}

\Centerline{$|\nu|(a)=\sup_{b\Bsubseteqshort a}\nu b-\nu(a\Bsetminus b)
=\sup_{b,c\Bsubseteqshort a}\nu b-\nu c$}

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

\Centerline{$\|\nu\|=|\nu|(1)
=\sup_{a\in\frak A}\nu a-\nu(1\Bsetminus a)$,}

\noindent $M$ becomes an $L$-space.

(b) $M_{\sigma}$ and $M_{\tau}$ are projection bands in $M$, therefore
$L$-spaces in their own right.   In particular, $|\nu|\in M_{\sigma}$
for every $\nu\in M_{\sigma}$, and $|\nu|\in M_{\tau}$ for every
$\nu\in M_{\tau}$.

(c) The band projection $P_{\sigma}:M\to M_{\sigma}$ is defined by the
formula

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

\noindent whenever $c\in\frak A$ and $\nu\ge 0$ in $M$.

(d) The band projection $P_{\tau}:M\to M_{\tau}$ is defined by the
formula

\Centerline{$(P_{\tau}\nu)(c)=\inf\{\sup_{a\in A}\nu a:
 A$ is a non-empty upwards-directed set with supremum $c\}$}

\noindent whenever $c\in\frak A$ and $\nu\ge 0$ in $M$.

(e) If $A\subseteq M$ is upwards-directed, then $A$ is bounded above in
$M$ iff $\{\nu 1:\nu\in A\}$ is bounded above in $\Bbb R$, and in this
case (if $A\ne\emptyset$) $\sup A$ is defined by the formula

\Centerline{$(\sup A)(a)=\sup_{\nu\in A}\nu a$ for every $a\in\frak A$.}

\wheader{362B}{0}{0}{0}{60pt}

(f) Suppose that $\mu$, $\nu\in M$.

\quad(i) The following are equiveridical:

\qquad($\alpha$) $\nu$ belongs to the band in $M$ generated by $\mu$;

\qquad ($\beta$) for every $\epsilon>0$ there is a $\delta>0$ such that
$|\nu a|\le\epsilon$ whenever $|\mu|a\le\delta$;

\qquad ($\gamma$) $\lim_{n\to\infty}\nu a_n=0$ whenever
$\sequencen{a_n}$ is a non-increasing sequence in $\frak A$ such that
$\lim_{n\to\infty}|\mu|(a_n)=0$.

\quad(ii) Now suppose that $\mu$, $\nu\ge 0$, and let $\nu_1$, $\nu_2$
be the components of $\nu$ in the band generated by $\mu$ and its
complement.   Then

\Centerline{$\nu_1c
=\sup_{\delta>0}\inf_{\mu a\le\delta}\nu(c\Bsetminus a)$,
\quad$\nu_2c
=\inf_{\delta>0}\sup_{a\Bsubseteqshort c,\mu a\le\delta}\nu a$}

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

\proof{{\bf (a)} Of course $\mu\vee\nu=\nu+(\mu-\nu)^+$,
$\mu\wedge\nu=\nu-(\nu-\mu)^+$, $|\nu|=\nu\vee(-\nu)$ (352D), so the
formula of 362Ab gives

$$\eqalign{(\mu\vee\nu)(a)
&=\nu a+\sup_{b\Bsubseteqshort a}\mu b-\nu b
=\sup_{b\Bsubseteqshort a}\mu b+\nu(a\Bsetminus b),\cr
(\mu\wedge\nu)(a)
&=\nu a-\sup_{b\Bsubseteqshort a}\nu b-\mu b
=\inf_{b\Bsubseteqshort a}\mu b+\nu(a\Bsetminus b),\cr
|\nu|(a)
&=\sup_{b\Bsubseteqshort a}\nu b-\nu(a\Bsetminus b)
\le\sup_{b,c\Bsubseteqshort a}\nu b-\nu c
=\sup_{b,c\Bsubseteqshort a}\nu(b\Bsetminus c)-\nu(c\Bsetminus b)\cr
&\le\sup_{b,c\Bsubseteqshort a}|\nu|(b\Bsetminus c)+|\nu|(c\Bsetminus b)
=\sup_{b,c\Bsubseteqshort a}|\nu|(b\Bsymmdiff c)
\le|\nu|(a).\cr}$$

\noindent The formula offered for $\|\nu\|$ corresponds exactly to the
formula in 362Ab for the norm of the associated member of
$S(\frak A)^{\sim}$;  because $S(\frak A)^{\sim}$ is an $L$-space under
its norm, so is $M$.

\medskip

{\bf (b)} By 362Ac-362Ad, $M_{\sigma}$ and $M_{\tau}$ may be identified
with $S(\frak A)^{\sim}_c$ and $S(\frak A)^{\times}$, which are
bands in $S(\frak A)^{\sim}$ (356B), therefore projection bands (353J);
so that $M_{\sigma}$
and $M_{\tau}$ are projection bands in $M$, and are $L$-spaces in their
own right (354O).

\medskip

{\bf (c)} Take any $\nu\ge 0$ in $M$.   Set

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

\noindent for every $c\in\frak A$.   Then of course
$0\le\nu_{\sigma}c\le\nu c$ for every $c$.
The point is that $\nu_{\sigma}$ is countably additive.
\Prf\
Let $\sequence{i}{c_i}$ be a disjoint sequence in $\frak A$, with
supremum $c$.   Then for any $\epsilon>0$ we have non-decreasing
sequences $\sequencen{a_n}$, $\sequencen{a_{in}}$, for $i\in\Bbb N$,
such that

\Centerline{$\sup_{n\in\Bbb N}a_n=c$,
\quad$\sup_{n\in\Bbb N}a_{in}=c_i$ for $i\in\Bbb N$,}

\Centerline{$\sup_{n\in\Bbb N}\nu a_n\le \nu_{\sigma}c+\epsilon$,}

\Centerline{$\sup_{n\in\Bbb N}\nu a_{in}\le\nu_{\sigma}c_i
+2^{-i}\epsilon$ for every $i\in\Bbb N$.}

\noindent Set $b_n=\sup_{i\le n}a_{in}$ for each $n$;  then
$\sequencen{b_n}$ is non-decreasing, and

\Centerline{$\sup_{n\in\Bbb N}b_n=\sup_{i,n\in\Bbb N}a_{in}
=\sup_{i\in\Bbb N}c_i=c$,}

\noindent so

$$\eqalign{\nu_{\sigma}c
&\le\sup_{n\in\Bbb N}\nu b_n
=\sup_{n\in\Bbb N}\sum_{i=0}^n\nu a_{in}\cr
&=\sum_{i=0}^{\infty}\sup_{n\in\Bbb N}\nu a_{in}
\le\sum_{i=0}^{\infty}\nu_{\sigma}c_i+2^{-i}\epsilon
=\sum_{i=0}^{\infty}\nu_{\sigma}c_i+2\epsilon.\cr}$$

\noindent On the other hand, $\sequencen{a_n\Bcap c_i}$ is a
non-decreasing sequence with supremum $c\Bcap c_i=c_i$ for each $i$, so
$\nu_{\sigma}c_i\le\sup_{n\in\Bbb N}\nu(a_n\Bcap c_i)$, and

$$\eqalignno{\sum_{i=0}^{\infty}\nu_{\sigma}c_i
&\le\sum_{i=0}^{\infty}\sup_{n\in\Bbb N}\nu(a_n\Bcap c_i)
=\sup_{n\in\Bbb N}\sum_{i=0}^{\infty}\nu(a_n\Bcap c_i)\cr
\noalign{\noindent(because $\sequencen{a_n}$ is non-decreasing)}
&\le\sup_{n\in\Bbb N}\nu a_n\cr
\noalign{\noindent(because $\sequence{i}{c_i}$ is disjoint)}
&\le\nu_{\sigma}c+\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\nu_{\sigma}c=\sum_{i=0}^{\infty}\nu_{\sigma}c_i$;  as
$\sequence{i}{c_i}$ is arbitrary, $\nu_{\sigma}$ is countably
additive.\ \Qed

Thus $\nu_{\sigma}\in M_{\sigma}$.   On the other hand, if
$\nuprime\in M_{\sigma}$ and $0\le\nuprime\le \nu$, then whenever $c\in\frak A$
and $\sequencen{a_n}$ is a non-decreasing sequence with supremum $c$,

\Centerline{$\nuprime c=\sup_{n\in\Bbb N}\nuprime a_n\le\sup_{n\in\Bbb N}\nu
a_n$.}

\noindent So we must have $\nuprime c\le\nu_{\sigma}c$.   This means that

\Centerline{$\nu_{\sigma}=\sup\{\nuprime:\nuprime\in
M_{\sigma},\,\nuprime\le\nu\}=P_{\sigma}\nu$,}

\noindent as claimed.

\medskip

{\bf (d)} The same ideas, with essentially elementary modifications,
deal with the completely additive part.   Take any $\nu\ge 0$ in $M$.
Set

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

\noindent for every $c\in\frak A$.   Then of course
$0\le\nu_{\tau}c\le\nu c$ for every $c$.
The point is that $\nu_{\tau}$ is completely additive.
\Prf\ Note first that if $c\in\frak A$, $\epsilon>0$ there is a
non-empty upwards-directed $A$, with supremum $c$, such that $\sup_{a\in
A}\nu a\le\nu_{\tau}c+\epsilon\nu c$;  for if $\nu c=0$ we can take
$A=\{c\}$.   Now let $\langle c_i\rangle_{i\in I}$ be a partition of
unity in $\frak A$.
Then for any $\epsilon>0$ we have non-empty
upwards-directed sets $A$, $A_i$, for $i\in I$, such that

\Centerline{$\sup A=1$,
\quad$\sup A_i=c_i$ for $i\in I$,
\quad$\sup_{a\in A}\nu a\le \nu_{\tau}1+\epsilon\nu 1$,}

\Centerline{$\sup_{a\in A_i}\nu a\le\nu_{\tau}c_i+\epsilon\nu c_i$ for
every
$i\in I$.}

\noindent Set

\Centerline{$B=\{\sup_{i\in J}a_i:J\subseteq I$ is finite, $a_i\in A_i$
for every $i\in J\}$;}

\noindent then $B$ is non-empty and upwards-directed,
and

\Centerline{$\sup B=\sup(\bigcup_{i\in I}A_i)=1$,}

\noindent so

$$\eqalign{\nu_{\tau}1
&\le\sup_{b\in B}\nu b
=\sup\{\sum_{i\in J}\nu a_i:J\subseteq I\text{ is finite},\,
a_i\in A_i\Forall i\in J\}\cr
&\le\sum_{i\in I}\nu_{\tau}c_i+\epsilon\nu c_i
\le\epsilon\nu 1+\sum_{i\in I}\nu_{\tau}c_i.\cr}$$

\noindent On the other hand, $A'_i=\{a\Bcap c_i:a\in A\}$ is a
non-empty upwards-directed set with supremum $c_i$ for each $i$, so
$\nu_{\tau}c_i\le\sup_{a\in A'_i}\nu a$, and

$$\eqalignno{\sum_{i\in I}\nu_{\tau}c_i
&\le\sum_{i\in I}\sup_{a\in A}\nu(a\Bcap c_i)
=\sup_{a\in A}\sum_{i\in I}\nu(a\Bcap c_i)\cr
&\le\sup_{a\in A}\nu a
\le\nu_{\tau}1+\epsilon\nu 1.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\nu_{\tau}c=\sum_{i\in I}\nu_{\tau}c_i$;  as
$\langle c_i\rangle_{i\in I}$ is arbitrary, $\nu_{\tau}$ is completely
additive, by 326R.\ \Qed

Thus $\nu_{\tau}\in M_{\tau}$.   On the other hand, if
$\nuprime\in M_{\tau}$ and $0\le\nuprime\le \nu$, then whenever
$c\in\frak A$
and $A$ is a non-empty upwards-directed set with supremum $c$,

\Centerline{$\nuprime c=\sup_{a\in A}\nuprime a\le\sup_{a\in A}\nu a$}

\noindent (using 326Oc).   So we must have $\nuprime c\le\nu_{\tau}c$.   This
means that

\Centerline{$\nu_{\tau}=\sup\{\nuprime:\nuprime\in
M_{\tau},\,\nuprime\le\nu\}=P_{\tau}\nu$,}

\noindent as claimed.

\medskip

{\bf (e)} If $A$ is empty, of course it is bounded above in $M$, and
$\{\nu 1:\nu\in A\}=\emptyset$ is bounded above in $\Bbb R$;  so let us
suppose that $A$ is not empty.   In this case, if $\lambda_0\in M$ is an
upper bound for $A$, then $\lambda_01$ is an upper bound for $\{\nu
1:\nu\in A\}$.   On the other hand, if $\sup_{\nu\in A}\nu 1=\gamma$ is
finite, $\gamma^*=\sup\{\nu a:\nu\in A,\,a\in\frak A\}$ is finite.
\Prf\ Fix $\nu_0\in A$.   Set $\gamma_1=\sup_{a\in\frak
A}|\nu_0a|<\infty$.   Then for any $\nu\in A$ and $a\in\frak A$ there is a
$\nuprime\in A$ such that $\nu_0\vee\nu\le\nuprime$, so that

\Centerline{$\nu a\le\nuprime a=\nuprime1-\nuprime(1\Bsetminus a)
\le\gamma-\nu_0(1\Bsetminus a)\le\gamma+\gamma_1$.}

\noindent So

\Centerline{$\gamma^*\le\gamma+\gamma_1<\infty$.   \Qed}

Set $\lambda a=\sup_{\nu\in A}\nu a$ for every $a\in\frak A$.    Then
$\lambda:\frak A\to\Bbb R$ is additive.   \Prf\ If $a$, $b\in\frak A$
are disjoint, then

$$\eqalignno{\lambda(a\Bcup b)
&=\sup_{\nu\in A}\nu(a\Bcup b)
=\sup_{\nu\in A}\nu a+\nu b
=\sup_{\nu\in A}\nu a+\sup_{\nu\in A}\nu b\cr
\noalign{\noindent (because $A$ is upwards-directed)}
&=\lambda a+\lambda b.\text{ \Qed}\cr}$$

\noindent Also $\lambda a\le\gamma^*$ for every $a$, so

\Centerline{$|\lambda a|=\max(\lambda a,-\lambda a)
=\max(\lambda a,\lambda(1\Bsetminus a)-\lambda 1)
\le\gamma^*+|\lambda 1|$}

\noindent for every $a\in\frak A$, and $\lambda$ is bounded.

This shows that $\lambda\in M$, so that $A$ is bounded above in $M$.
Of course $\lambda$ must be actually the least upper bound of $A$ in
$M$.

\medskip

{\bf (f)(i)}\grheada$\Rightarrow$\grheadb\ Suppose that $\nu$ belongs to
the band in $M$ generated by $\mu$, that is,
$|\nu|=\sup_{n\in\Bbb N}|\nu|\wedge n|\mu|$ (352Vb).   Let $\epsilon>0$.
Then there is an
$n\in\Bbb N$ such that $|\nu|(1)\le\bover12\epsilon+(|\nu|\wedge
n|\mu|)(1)$ ((e) above).   Set $\delta=\bover1{2n+1}\epsilon>0$.   If
$|\mu|(a)\le\delta$, then

$$\eqalign{|\nu a|
&\le|\nu|(a)
=(|\nu|\wedge n|\mu|)(a)+(|\nu|-|\nu|\wedge n|\mu|)(a)\cr
&\le n|\mu|(a)+(|\nu|-|\nu|\wedge n|\mu|)(1)
\le n\delta+\Bover12\epsilon
\le\epsilon.\cr}$$

\noindent So ($\beta$) is satisfied.

\medskip

\quad{\bf not-($\pmb{\alpha}$)$\Rightarrow$not-($\pmb{\beta}$)}
Suppose that $\nu$ does not belong
to the band in $M$ generated by $|\mu|$.   Then there is a $\nu_1>0$
such that $\nu_1\le|\nu|$ and $\nu_1\wedge|\mu|=0$ (353C).   For any
$\delta>0$, there is an $a\in\frak A$ such that $\nu_1(1\Bsetminus
a)+|\mu|(a)\le\min(\delta,\bover12\nu_11)$ ((a) above);  now
$|\mu|(a)\le\delta$ but

\Centerline{$|\nu|(a)\ge\nu_1a=\nu_11-\nu_1(1\Bsetminus a)
\ge\nu_11-\bover12\nu_11=\bover12\nu_11$.}

\noindent Thus $\mu$, $\nu$ do not satisfy ($\beta$) (with
$\epsilon=\bover12\nu_11$).

\medskip

\quad\grheadb$\Rightarrow$\grheadc\  is trivial.

\medskip

\quad\grheadc$\Rightarrow$\grheada\  Observe first that if
$\sequence{k}{c_k}$ is a non-increasing sequence in $\frak A$ such that
$\lim_{k\to\infty}|\mu|c_k=0$, then $\lim_{k\to\infty}\nu^+c_k=0$.
\Prf\ Let $\epsilon>0$.   Because $\nu^+\wedge\nu^-=0$, there is a
$b\in\frak A$ such that $\nu^+b+\nu^-(1\Bsetminus b)\le\epsilon$, by
part (a).   Now $\sequence{k}{c_k\Bsetminus b}$ is non-increasing and
$\lim_{k\to\infty}|\mu|(c_k\Bsetminus b)=0$, so
$\lim_{k\to\infty}\nu(c_k\Bsetminus b)=0$ and

$$\eqalign{\limsup_{k\to\infty}\nu^+c_k
&=\limsup_{k\to\infty}\nu^+(c_k\Bcap b)+\nu(c_k\Bsetminus b)
   +\nu^-(c_k\Bsetminus b)\cr
&\le\nu^+b+\nu^-(1\Bsetminus b)\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\lim_{k\to\infty}\nu^+c_k=0$.\ \Qed

\Quer\ Now suppose, if possible, that $\nu^+$ does not belong to the
band generated by $\mu$.   Then there is a $\nu_1>0$ such that
$\nu_1\le\nu^+$ and $\nu_1\wedge|\mu|=0$.   Set
$\epsilon=\bover14\nu_11>0$.   For each $n\in\Bbb N$, we can choose
$a_n\in\frak A$ such that
$|\mu|a_n+\nu_1(1\Bsetminus a_n)\le 2^{-n}\epsilon$, by part (a) again.
For $n\ge k$, set $b_{kn}=\sup_{k\le i\le n}a_i$;  then

\Centerline{$|\mu|b_{kn}\le\sum_{i=k}^n|\mu|a_i\le 2^{-k+1}\epsilon$,}

\noindent and $\langle b_{kn}\rangle_{n\ge k}$ is non-decreasing.   Set
$\gamma_k=\sup_{n\ge k}\nu_1b_{kn}$ and choose $m(k)\ge k$
such that $\nu_1b_{k,m(k)}\ge\gamma_k-2^{-k}\epsilon$.   Setting
$b_k=b_{k,m(k)}$, we see that $b_k\Bcup b_{k+1}=b_{kn}$ where
$n=\max(m(k),m(k+1))$, so that

\Centerline{$\nu_1(b_k\Bcup b_{k+1})\le\gamma_k
\le\nu_1b_k+2^{-k}\epsilon$}

\noindent and $\nu_1(b_{k+1}\Bsetminus b_k)\le 2^{-k}\epsilon$.   Set
$c_k=\inf_{i\le k}b_i$ for each $k$;  then

\Centerline{$\nu_1(b_{k+1}\Bsetminus c_{k+1})
=\nu_1(b_{k+1}\Bsetminus c_k)
\le\nu_1(b_{k+1}\Bsetminus b_k)+\nu_1(b_k\Bsetminus c_k)
\le 2^{-k}\epsilon+\nu_1(b_k\Bsetminus c_k)$}

\noindent for each $k$;  inducing on $k$, we see that

\Centerline{$\nu_1(b_k\Bsetminus c_k)\le\sum_{i=0}^{k-1}2^{-i}\epsilon
\le 2\epsilon$}

\noindent for every $k$.   This means that

\Centerline{$\nu^+c_k\ge\nu_1c_k
\ge\nu_1b_k-2\epsilon\ge\nu_1a_k-2\epsilon
=\nu_11-\nu_1(1\Bsetminus a_k)-2\epsilon
\ge4\epsilon -\epsilon-2\epsilon=\epsilon$}

\noindent for every $k\in\Bbb N$.   On the other hand,
$\sequence{k}{c_k}$ is a non-increasing sequence and

\Centerline{$|\mu|c_k\le|\mu|b_k\le 2^{-k+1}\epsilon$}

\noindent for every $k$, which contradicts the paragraph just
above.\ \Bang

This means that $\nu^+$ must belong to the band generated by $\mu$.
Similarly $\nu^-=(-\nu)^+$ belongs to the band generated by $\mu$ and
$\nu=\nu^++\nu^-$ also does.

\medskip

\quad{\bf (ii)} Take $c\in\frak A$.   Set

\Centerline{$\beta_1=\sup_{\delta>0}\inf_{\mu a\le\delta}\nu(c\Bsetminus
a)$,
\quad$\beta_2=\inf_{\delta>0}\sup_{a\Bsubseteqshort c,\mu a\le\delta}\nu a$.}

\noindent Then

\Centerline{$\beta_1=\sup_{\delta>0}\inf_{a\Bsubseteqshort c,\mu
a\le\delta}\nu(c\Bsetminus a)=\nu c-\beta_2$.}

\noindent Take any $\epsilon>0$.   Because $\nu_1$ belongs to the band
generated by $\mu$, part (i) tells us that there is a $\delta>0$ such
that $\nu_1 a\le\epsilon$ whenever $\mu a\le\delta$.   In this case, if
$\mu a\le\delta$,

\Centerline{$\nu(c\Bsetminus a)=\nu c-\nu(c\Bcap a)\ge\nu c-\epsilon
\ge\nu_1c-\epsilon$;}

\noindent thus

\Centerline{$\beta_1\ge\inf_{\mu a\le\delta}\nu(c\Bsetminus a)
\ge\nu_1c-\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\beta_1\ge\nu_1c$.   On the other
hand, given $\epsilon$, $\delta>0$, there is an $a\Bsubseteq c$ such
that $\mu a+\nu_2(c\Bsetminus a)\le\min(\delta,\epsilon)$, because
$\mu\wedge\nu_2=0$ (using (a) again).   In this case, of course, $\mu
a\le\delta$, while

\Centerline{$\nu a\ge\nu_2a=\nu_2c-\nu_2(c\Bsetminus a)
\ge\nu_2c-\epsilon$.}

\noindent Thus $\sup_{a\Bsubseteqshort c,\mu a\le\delta}\nu a
\ge\nu_2c-\epsilon$.   As $\delta$ is arbitrary,
$\beta_2\ge\nu_2c-\epsilon$.   As $\epsilon$ is arbitrary,
$\beta_2\ge\nu_2c$;  but as

\Centerline{$\beta_1+\beta_2=\nu c=\nu_1c+\nu_2c$,}

\noindent $\beta_i=\nu_ic$ for both $i$, as claimed.
}%end of proof of 362B

\medskip

\noindent{\bf Remark} The $L$-space norm $\|\,\|$ on $M$, described in (a)
above, is the {\bf total variation norm}.

\leader{362C}{}\cmmnt{ The formula in 362B(f-i) has, I hope, already
reminded you of the concept of `absolutely continuous' additive
functional from the Radon-Nikod\'ym theorem (Chapter 23, \S327).   The
expressions in 362Bf are limited by the assumption that $\mu$, like
$\nu$, is finite-valued.   If we relax this we get an alternative
version of some of the same ideas.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra
and $M$ be the Riesz space of bounded finitely
additive real-valued functionals on $\frak A$.   Write

\Centerline{$M_{ac}=\{\nu:\nu\in M$ is absolutely continuous with
respect to $\bar\mu\}$\dvro{,}{}}

\cmmnt{\noindent (see 327A),}

\Centerline{$M_{tc}=\{\nu:\nu\in M$ is continuous with respect to the
measure-algebra topology on $\frak A\}$,}

\Centerline{$M_t=\{\nu:\nu\in M$,
$|\nu|1=\sup_{\bar\mu a<\infty}|\nu|a\}$.}

\noindent Then $M_{ac}$, $M_{tc}$ and $M_t$ are bands in $M$.

\proof{{\bf (a)(i)} It is easy to check that $M_{ac}$ is a linear
subspace of $M$.

\medskip

\quad{\bf (ii)} If $\nu\in M_{ac}$, $\nuprime\in M$ and $|\nuprime|\le|\nu|$
then $\nuprime\in M_{ac}$.   \Prf\ Given $\epsilon>0$ there is a $\delta>0$
such that $|\nu a|\le\bover12\epsilon$ whenever $\bar\mu a\le\delta$.
Now

\Centerline{$|\nuprime a|\le|\nuprime|(a)\le|\nu|(a)
\le 2\sup_{c\Bsubseteqshort a}|\nu c|\le\epsilon$}

\noindent (using the formula for $|\nu|$ in 362Ba) whenever $\bar\mu
a\le\delta$.   As $\epsilon$ is arbitrary, $\nuprime$ is absolutely
continuous.\ \Qed

\medskip

\quad{\bf (iii)} If $A\subseteq M_{ac}$ is non-empty and
upwards-directed
and $\nu=\sup A$ in $M$, then $\nu\in M_{ac}$.    \Prf\ Let
$\epsilon>0$.   Then there is a $\nuprime\in A$ such that $\nu
1\le\nuprime1+\bover12\epsilon$ (362Be).   Now there is a $\delta>0$ such
that $|\nu
a|\le\bover12\epsilon$ whenever $\bar\mu a\le\delta$.   If now $\bar\mu
a\le\delta$,

\Centerline{$|\nu a|\le|\nuprime a|+(\nu-\nuprime)(a)
\le\bover12\epsilon+(\nu-\nuprime)(1)
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu$ is absolutely continuous
with respect to $\bar\mu$.\ \Qed

Putting these together, we see that $M_{ac}$ is a band.

\medskip

{\bf (b)(i)} We know that $M_{tc}$ consists just of those $\nu\in M$
which are continuous at $0$ (327Bc).   Of course this is a linear
subspace of $M$.

\medskip

\quad{\bf (ii)} If $\nu\in M_{tc}$, $\nuprime\in M$ and $|\nuprime|\le|\nu|$
then $|\nu|\in M_{tc}$.   \Prf\ Write
$\frak A^f=\{d:d\in\frak A,\,\bar\mu d<\infty\}$.
Given $\epsilon>0$ there are $d\in\frak A^f$,
$\delta>0$ such that $|\nu a|\le\bover12\epsilon$
whenever $\bar\mu(a\Bcap d)\le\delta$.   Now

\Centerline{$|\nuprime a|\le|\nuprime|(a)\le|\nu|(a)
\le 2\sup_{c\Bsubseteqshort a}|\nu c|\le\epsilon$}

\noindent whenever $\bar\mu(a\Bcap d)\le\delta$.   As $\epsilon$ is
arbitrary, $\nuprime$ is continuous at $0$ and belongs to $M_{tc}$.\ \Qed

\medskip

\quad{\bf (iii)} If $A\subseteq M_{tc}$ is non-empty and
upwards-directed and
$\nu=\sup A$ in $M$, then $\nu\in M_{tc}$.    \Prf\ Let $\epsilon>0$.
Then there is a $\nuprime\in A$ such that
$\nu 1\le\nuprime1+\bover12\epsilon$.
There are $d\in\frak A^f$, $\delta>0$ such that
$|\nu a|\le\bover12\epsilon$ whenever $\bar\mu(a\Bcap d)\le\delta$.
If now $\bar\mu(a\Bcap d)\le\delta$,

\Centerline{$|\nu a|\le|\nuprime a|+(\nu-\nuprime)(a)
\le\bover12\epsilon+(\nu-\nuprime)(1)
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu$ is continuous at $0$,
therefore belongs to $M_{tc}$.\ \Qed

Putting these together, we see that $M_{tc}$ is a band.

\medskip

{\bf (c)(i)} $M_t$ is a linear subspace of $M$.   \Prf\ Suppose that
$\nu_1$, $\nu_2\in M_t$ and $\alpha\in\Bbb R$.   Given $\epsilon>0$,
there are $a_1$, $a_2\in\frak A^f$ such that
$|\nu_1|(1\Bsetminus a_1)\le\bover{\epsilon}{1+|\alpha|}$
and $|\nu_2|(1\Bsetminus a_2)\le\epsilon$.   Set $a=a_1\Bcup a_2$;
then $\bar\mu a<\infty$ and

\Centerline{$|\nu_1+\nu_2|(1\Bsetminus a)\le 2\epsilon$,
\quad$|\alpha\nu_1|(1\Bsetminus a)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu_1+\nu_2$ and $\alpha\nu_1$
belong to $M_t$;  as $\nu_1$, $\nu_2$ and $\alpha$ are arbitrary, $M_t$
is a linear subspace of $M$.\ \Qed

\medskip

\quad{\bf (ii)} If $\nu\in M_t$, $\nuprime\in M$ and $|\nuprime|\le|\nu|$ then

\Centerline{$\inf_{\bar\mu a<\infty}|\nuprime|(1\Bsetminus
a)\le\inf_{\bar\mu a<\infty}|\nu|(1\Bsetminus a)=0$,}

\noindent so $\nuprime\in M_t$.    Thus $M_t$ is a solid linear subspace of
$M$.


\medskip

\quad{\bf (iii)} If $A\subseteq M_t^+$ is non-empty and upwards-directed
and $\nu=\sup A$ is defined in $M$, then $\nu\in M_t$.   \Prf\

\Centerline{$|\nu|1
=\nu1
=\sup_{\nuprime\in A}\nuprime1
=\sup_{\nuprime\in A,\bar\mu a<\infty}\nuprime a
=\sup_{\bar\mu a<\infty}\nu a$.}

\noindent As $A$ is arbitrary, $\nu\in M_t$.\ \QeD\  Thus $M_t$ is a
band in $M$.
}%end of proof of 362C

\vleader{108pt}{362D}{}\cmmnt{ For semi-finite measure algebras, among
others, the formula of 362Bd takes a special form.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a \wsid\ Boolean algebra.
Let $M$ be the space of bounded finitely additive functionals on
$\frak A$, $M_{\tau}\subseteq M$ the space of completely additive
functionals,
and $P_{\tau}:M\to M_{\tau}$ the band projection\cmmnt{, as in 362B}.
Then for any $\nu\in M^+$ and $c\in\frak A$ there is a non-empty
upwards-directed set $A\subseteq\frak A$ with supremum $c$ such that
$(P_{\tau}\nu)(c)=\sup_{a\in A}\nu a$\cmmnt{;  that is, the `inf'
in 362Bd can be read as `min'}.

\proof{ By 362Bd, we can find for each $n$ a non-empty upwards-directed
$A_n$, with supremum $c$, such that
$\sup_{a\in A_n}\nu a\le(P_{\tau}\nu)(c)+2^{-n}$.   Set
$B_n=\{c\Bsetminus a:a\in A_n\}$ for
each $n$, so that $B_n$ is downwards-directed and has infimum $0$.
Because $\frak A$ is \wsid,

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

\noindent is also a downwards-directed set with infimum $0$.
Consequently $A=\{c\Bsetminus b:b\in B\}$ is upwards-directed and has
supremum $c$.   Moreover, for any $n\in\Bbb N$ and $a\in A$, there is an
$a'\in A_n$ such that $a\Bsubseteq a'$;  so, using 362Bd again and
referring to the choice of $A_n$,

\Centerline{$(P_{\tau}\nu)(c)
\le\sup_{a\in A}\nu a
\le\sup_{a'\in A_n}\nu a'
\le(P_{\tau}\nu)(c)+2^{-n}$.}

\noindent As $n$ is arbitrary, $A$ has the required property.
}%end of proof of 362D

\leader{362E}{Uniformly integrable\dvrocolon{ sets}}\cmmnt{ The spaces
$S^{\sim}$, $S^{\sim}_c$ and $S^{\times}$ of 362A, or, if you prefer,
the spaces $M$, $M_{\sigma}$, $M_{\tau}$, $M_{ac}$, $M_{tc}$, $M_t$ of
362B-362C, are all $L$-spaces, and any serious study of them must
involve a discussion of their uniformly integrable ( = relatively weakly
compact) subsets.   The basic work has been done in 356O;  I spell
out its application in this context.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Boolean algebra and $M$ the
$L$-space of bounded finitely additive functionals on $\frak A$.   Then
a norm-bounded set $C\subseteq M$ is uniformly integrable iff
$\lim_{n\to\infty}\sup_{\nu\in C}|\nu a_n|=0$ for every disjoint
sequence $\sequencen{a_n}$ in $\frak A$.

\proof{ Write $S$ for $S(\frak A)$ and $\tilde C$ for the set
$\{f:f\in S^{\sim},\,f\chi\in C\}$.   Because the map
$f\mapsto f\chi$ is a
normed Riesz space isomorphism between $S^{\sim}$ and $M$, $\tilde C$ is
uniformly integrable in $M$ iff $C$ is uniformly integrable in
$S^{\sim}$.

\medskip

{\bf (a)} Suppose that $C$ is uniformly integrable and that
$\sequencen{a_n}$ is a disjoint sequence in $\frak A$.   Then
$\sequencen{\chi a_n}$ is a disjoint order-bounded sequence in
$S^{\sim}$, while $\tilde C$ is uniformly integrable, so
$\lim_{n\to\infty}\sup_{f\in\tilde C}|f(\chi a_n)|=0$, by 356O;
but this means that $\lim_{n\to\infty}\sup_{\nu\in C}|\nu a_n|=0$.  Thus
the condition is satisfied.

\medskip

{\bf (b)} Now suppose that $C$ is not uniformly integrable.   By
356O, in the other direction, there is a disjoint sequence
$\sequencen{u_n}$ in $S$ such
that $0\le u_n\le\chi 1$ for each $n$ and
$\limsup_{n\to\infty}\sup_{f\in\tilde C}|f(u_n)|>0$.   For each $n$,
take $c_n=\Bvalue{u_n>0}$ (361Eg);  then $0\le u_n\le\chi c_n$ and
$\sequencen{c_n}$ is disjoint.   Now

$$\eqalign{\limsup_{n\to\infty}\sup_{\nu\in C}|\nu|(c_n)
&=\limsup_{n\to\infty}\sup_{f\in\tilde C}|f|(\chi c_n)\cr
&\ge\limsup_{n\to\infty}\sup_{f\in\tilde C}|f(u_n)|
>0.\cr}$$

\noindent So if we choose $\nu_n\in C$ such that
$|\nu_n|(c_n)\ge\bover12\sup_{\nu\in C}|\nu|(c_n)$, we shall have
$\limsup_{n\to\infty}|\nu_n|(c_n)>0$.   Next, for each $n$, we can find
$a_n\Bsubseteq c_n$ such that $|\nu_na_n|\ge\bover12|\nu_n|(c_n)$, so
that

\Centerline{$\limsup_{n\in\Bbb N}\sup_{\nu\in C}|\nu a_n|
\ge\limsup_{n\to\infty}|\nu_na_n|>0$.}

\noindent Since $\sequencen{a_n}$, like $\sequencen{c_n}$, is disjoint,
the condition is not satisfied.   This completes the proof.
}%end of proof of 362E

\exercises{\leader{362X}{Basic exercises $\pmb{>}$(a)} Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and $\nu_1$, $\nu_2$ two
countably additive functionals on $\frak A$.   Show that
$|\nu_1|\wedge|\nu_2|=0$ in the Riesz space of bounded finitely additive
functionals on $\frak A$ iff there is a $c\in\frak A$ such that
$\nu_1a=\nu_1(a\Bcap c)$ and $\nu_2a=\nu_2(a\Bsetminus c)$ for every
$a\in\frak A$.
%362B used in 531V

\spheader 362Xb Let $(\frak A,\bar\mu)$ be a measure algebra, and take
$M$, $M_{ac}$ as in 362C.
Show that for any non-negative $\nu\in M$, the component $\nu_{ac}$ of
$\nu$ in $M_{ac}$ is given by the formula

\Centerline{$\nu_{ac}c
=\sup_{\delta>0}\inf_{\bar\mu a\le\delta}\nu(c\Bsetminus a)$.}
%362C

\spheader 362Xc Let $(\frak A,\bar\mu)$ be a measure algebra, and take
$M$, $M_t$ as in 362C.   (i) Show that $M_t$ is just the set of those
$\nu\in M$ such that $\nu a=\lim_{b\to\Cal F}\nu(a\Bcap b)$ for every
$a\in\frak A$, where $\Cal F$ is the filter on $\frak A$ generated by
the sets $\{b:b\in\frak A^f,\,b\Bsupseteq b_0\}$ as
$b_0$ runs over the set $\frak A^f$ of elements of $\frak A$ of
finite measure.   (ii)
Show that the complementary band $M_t^{\perp}$ of $M_t$ in $M$ is just
the set of those $\nu\in M$ such that $\nu a=0$ for every $a\in\frak A^f$.
(iii) Show that for any $\nu\in M$, its component $\nu_t$ in $M_t$ is
given by the formula $\nu_ta=\lim_{b\to\Cal F}\nu(a\Bcap b)$ for every
$a\in \frak A$.
%362C

\spheader 362Xd Let $(\frak A,\bar\mu)$ be a measure algebra.   Write
$M$, $M_{\sigma}$, $M_{\tau}$, $M_{ac}$, $M_{tc}$ and $M_t$ as in
362B-362C.   Show that (i) $M_{\sigma}\subseteq M_{ac}$ (ii)
$M_{ac}\cap M_t=M_{tc}\subseteq M_{\tau}$ (iii) if $(\frak A,\bar\mu)$
is $\sigma$-finite, then $M_{\sigma}=M_{tc}$.
%362C

\spheader 362Xe Let $\frak A$ be a Boolean algebra, and $M$ the space of
bounded additive functionals on $\frak A$.   Let us say that a
non-zero finitely additive functional $\nu:\frak A\to\Bbb R$ is
{\bf atomic} if whenever $a$, $b\in\frak A$ and $a\Bcap b=0$ then at least
one of $\nu a$, $\nu b$ is zero.   (i) Show that for a non-zero finitely
additive functional $\nu$ on $\frak A$ the following are equiveridical:
($\alpha$) $\nu$ is
atomic; ($\beta$) $\nu\in M$ and $|\nu|$ is atomic;
($\gamma$) $\nu\in M$ and the corresponding linear functional
$f_{|\nu|}=|f_{\nu}|\in S(\frak A)^{\sim}$ is a Riesz homomorphism;
($\delta$) there are a
multiplicative linear functional $f:S(\frak A)\to\Bbb R$ and an
$\alpha\in\Bbb R$ such that $\nu a=\alpha f(\chi a)$ for every
$a\in\frak A$;  ($\epsilon$) $\nu\in M$
and the band in $M$ generated by $\nu$
is the set of multiples of $\nu$.
(ii) Show that a completely additive functional
$\nu:\frak A\to\Bbb R$ is atomic iff there are $a\in\frak A$ and
$\alpha\in\Bbb R\setminus\{0\}$ such that $a$ is an atom in $\frak A$
and $\nu b=\alpha$ when $a\Bsubseteq b$, $0$ when $a\Bcap b=0$.

\spheader 362Xf Let $\frak A$ be a Boolean algebra.   (i) Show that the
properly atomless functionals (definition:  326F) form a band $M_c$
in the Riesz space $M$ of all bounded finitely additive functionals on
$\frak A$.  (ii) Show that the complementary band $M_c^{\perp}$ consists
of just those $\nu\in M$ expressible as a sum $\sum_{i\in I}\nu_i$ of
countably many atomic functionals $\nu_i\in M$.
(iii) Show that if $\frak A$ is purely atomic then a
properly atomless completely additive functional on
$\frak A$ must be $0$.
%362Xe

\spheader 362Xg Let $X$ be a set and $\Sigma$ an algebra of subsets of
$X$.   Let $M$ be the Riesz space of bounded finitely additive
functionals on $\Sigma$, $M_{\tau}$ the space of completely additive
functionals and $M_p$ the space of functionals expressible in the form
$\nu E=\sum_{x\in E}\alpha_x$ for some absolutely summable family
$\langle\alpha_x\rangle_{x\in X}$ of real numbers.   (i) Show that $M_p$
is a band in $M$.    (ii) Show that if all singleton subsets of $X$
belong to $\Sigma$ then $M_p=M_{\tau}$.   (iii) Show that if $\Sigma$ is
a $\sigma$-algebra then every member of $M_p$ is countably additive.
(iv) Show that if $X$ is a compact
zero-dimensional Hausdorff space and $\Sigma$ is the algebra of
open-and-closed subsets of $X$ then the complementary band $M_p^{\perp}$
of $M_p$ in $M$ is the band $M_c$ of
properly atomless functionals described in 362Xf.
%362Xf

\spheader 362Xh Let $(X,\Sigma,\mu)$ be a measure space.   Let $M$ be
the Riesz space of bounded finitely additive functionals on $\Sigma$ and
$M_{\sigma}$ the space of bounded countably additive functionals.
Let $M_{tc}$, $M_{ac}$ be
the spaces of truly continuous and bounded absolutely continuous
additive functionals as defined in 232A.   Show that
$M_{tc}$ and $M_{ac}$ are bands in $M$ and that
$M_{tc}\subseteq M_{\sigma}\cap M_{ac}$.    Show that if $\mu$ is
$\sigma$-finite then $M_{tc}=M_{\sigma}\cap M_{ac}$.

\spheader 362Xi Let $\frak A$ be a Boolean algebra and
$M$ the Riesz space
of bounded finitely additive functionals on $\frak A$.  (i) For any
non-empty downwards-directed set $A\subseteq\frak A$ set
$N_A=\{\nu:\nu\in M,\,\inf_{a\in A}|\nu|a=0\}$.   Show that $N_A$ is a
band in $M$.   (ii) For any non-empty set $\Cal A$ of non-empty
downwards-directed sets in $\frak A$ set
$M_{\Cal A}=\{\nu:\nu\in M,\,\inf_{a\in A}|\nu|a=0\Forall A\in\Cal A\}$.
Show that $M_{\Cal A}$ is a band in $M$.
(iii)  Explain how to represent as such $M_{\Cal A}$ the bands
$M_{\sigma}$, $M_{\tau}$, $M_t$, $M_{ac}$, $M_{tc}$
described in 362B-362C, and also
any band generated by a single element of $M$.
(iv) Suppose, in (ii), that $\Cal A$ has the property
that for any $A$, $A'\in\Cal A$ there is a $B\in\Cal A$ such that for
every $b\in B$ there are $a\in A$, $a'\in A'$ such that $a\Bcup
a'\Bsubseteq b$.   Show that for any non-negative $\nu\in M$, the
component $\nu_1$ of $\nu$ in $M_{\Cal A}$ is given by the formula
$\nu_1c=\inf_{A\in\Cal A}\sup_{a\in A}\nu(c\Bsetminus a)$, so that the
component $\nu_2$ of $\nu$ in $M_{\Cal A}^{\perp}$ is given by the
formula $\nu_2c=\sup_{A\in\Cal A}\inf_{a\in A}\nu(c\Bcap a)$.
(Cf.\ 356Yb.)

\leader{362Y}{Further exercises (a)}
%\spheader 362Ya
Let $\frak A$ be a Boolean algebra.
Let $\frak C$ be the band algebra of the Riesz space $M$ of bounded
finitely additive functionals on $\frak A$ (353B).   Show that the
bands $M_{\sigma}$, $M_{\tau}$, $M_c$ (362B, 362Xf)  generate a
subalgebra $\frak C_0$ of $\frak C$ with at most six atoms.   Give an
example in which $\frak C_0$ has six atoms.   How many atoms can it have
if (i) $\frak A$ is atomless (ii) $\frak A$ is purely atomic (iii)
$\frak A$ is Dedekind $\sigma$-complete?
%362B

\spheader 362Yb Let $(\frak A,\bar\mu)$ be a measure algebra.   Let
$\frak C$ be the band algebra of the Riesz space $M$ of bounded
finitely additive functionals on $\frak A$.   Show that the bands
$M_{\sigma}$, $M_{\tau}$, $M_c$, $M_{ac}$, $M_{tc}$, $M_t$ (362B, 362C,
362Xf)  generate a subalgebra $\frak C_0$ of $\frak C$ with at
most twelve atoms.   Give an example in which $\frak C_0$ has twelve
atoms.   How many atoms can it have if (i) $\frak A$ is atomless (ii)
$\frak A$ is purely atomic (iii) $(\frak A,\bar\mu)$ is semi-finite (iv)
$(\frak A,\bar\mu)$ is localizable (v) $(\frak A,\bar\mu)$ is
$\sigma$-finite (vi) $(\frak A,\bar\mu)$ is totally finite?
%362C

\spheader 362Yc Give an example of a set $X$, a $\sigma$-algebra
$\Sigma$ of subsets of $X$, and a functional in $M_p$ (as defined in
362Xg) which is not completely additive.

\spheader 362Yd Let $U$ be a Riesz space and $f$, $g\in U^{\sim}$.
Show that the following are equiveridical:  ($\alpha$) $g$ is in the
band in $U^{\sim}$ generated by $f$;  ($\beta$) for every $u\in U^+$,
$\epsilon>0$ there is a $\delta>0$ such that $|g(v)|\le\epsilon$
whenever $0\le v\le u$ and $|f|(v)\le\delta$;   ($\gamma$)
$\lim_{n\to\infty}g(u_n)=0$ whenever $\sequencen{u_n}$ is a
non-increasing sequence in $U^+$ and $\lim_{n\to\infty}|f|(u_n)=0$.
%362B

\spheader 362Ye Let $\frak A$ be a weakly $\sigma$-distributive Boolean
algebra (316Ye).   Show that the `inf' in the formula for
$P_{\sigma}\nu$ in 362Bc can be replaced by `min'.
%362D

\spheader 362Yf Let $\frak A$ be any Boolean algebra and $M$ the space
of bounded finitely additive functionals on $\frak A$.   Let
$C\subseteq M$ be such that $\sup_{\nu\in C}|\nu a|<\infty$ for every
$a\in\frak A$.
(i) Suppose that $\sup_{n\in\Bbb N}\sup_{\nu\in C}|\nu a_n|$ is finite
for every disjoint sequence $\sequencen{a_n}$ in $\frak A$.   Show that
$C$ is norm-bounded.   (ii) Suppose that
$\lim_{n\to\infty}\sup_{\nu\in C}|\nu a_n|=0$ for every disjoint
sequence $\sequencen{a_n}$ in $\frak A$.   Show that $C$ is uniformly
integrable.
%362E

\spheader 362Yg Let $\frak A$ be a Boolean algebra and $M_{\tau}$ the
space of completely additive functionals on $\frak A$.   Let
$C\subseteq M_{\tau}$ be such that $\sup_{\nu\in C}|\nu a|<\infty$ for
every atom $a\in\frak A$.  (i) Suppose that
$\sup_{n\in\Bbb N}\sup_{\nu\in C}|\nu a_n|$ is finite for every disjoint
sequence $\sequencen{a_n}$ in $\frak A$.   Show that $C$ is
norm-bounded.   (ii) Suppose that
$\lim_{n\to\infty}\sup_{\nu\in C}|\nu a_n|=0$ for every disjoint
sequence $\sequencen{a_n}$ in $\frak A$.   Show that $C$ is uniformly
integrable.
%362E

\spheader 362Yh Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\sequencen{\nu_n}$ a sequence of countably additive
real-valued functionals on $\frak A$.   Suppose that
$\nu a=\lim_{n\to\infty}\nu_na$ is defined in $\Bbb R$ for every
$a\in\frak A$.   Show that $\nu$ is countably additive and that
$\{\nu_n:n\in\Bbb N\}$ is uniformly integrable.   \Hint{246Yg.}   Show
that if every $\nu_n$ is completely additive, so is $\nu$.
%362E

\spheader 362Yi Let $\frak A$ be a Boolean algebra, $M$ the Riesz space
of bounded finitely additive functionals on $\frak A$, and
$M_c\subseteq M$ the band of properly atomless functionals (362Xf).
Show that for a
non-negative $\nu\in M$ the component $\nu_c$ of $\nu$ in $M_c$ is given
by the formula

\Centerline{$\nu_ca=\inf_{\delta>0}\sup\{\sum_{i=0}^n\nu a_i:
a_0,\ldots,a_n\Bsubseteq a$ are disjoint, $\nu a_i\le\delta$ for every
$i\}$}

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

\spheader 362Yj Let $\frak A$ be a Boolean algebra and $M$ the
$L$-space of bounded additive real-valued functionals on $\frak A$.
Show that the complexification of $M$, as defined in 354Yl, can be
identified with the Banach space of bounded additive functionals
$\nu:\frak A\to\Bbb C$, writing

\Centerline{$\|\nu\|=\sup\{\sum_{i=0}^n|\nu a_i|:a_0,\ldots,a_n$ are
disjoint elements of $\frak A\}$}

\noindent for such $\nu$.

\spheader 362Yk Let $\frak A$ be a Boolean algebra and $M$ the $L$-space of
bounded additive real-valued functionals on $\frak A$.   Suppose that $M_0$
is a norm-closed linear subspace of $M$ and that
$a\mapsto\nu(a\Bcap c):\frak A\to\Bbb R$ belongs to $M_0$
whenever $\nu\in M_0$ and $c\in\frak A$.   Show that $M_0$ is a band in
$M$.   \Hint{436L.}
}%end of exercises

\cmmnt{
\Notesheader{362} The Boolean algebras most immediately important in
measure theory are of course $\sigma$-algebras of measurable sets and
their quotient measure algebras.   It is therefore natural to begin any
investigation by concentrating on Dedekind $\sigma$-complete algebras.
Nevertheless, in this section and the last (and in \S326), I have gone
to some trouble not to specialize to $\sigma$-complete algebras except
when necessary.   Partly this is just force of habit, but partly it is
because I wish to lay a foundation for a further step forward:  the
investigation of the ways in which additive functionals on general
Boolean algebras reflect the concepts of measure theory, and indeed can
generate them.   Some of the results in this direction can be
surprising.   I do not think it obvious that the condition ($\gamma$) in
362B(f-i), for instance, is sufficient in the absence of any hypothesis
of Dedekind $\sigma$-completeness or countable additivity.

Given a Boolean algebra $\frak A$ with the associated Riesz space
$M\cong S(\frak A)^{\sim}$ of bounded additive functionals on
$\frak A$, we now have a substantial list of bands in $M$:  $M_{\sigma}$,
$M_{\tau}$, $M_c$ (362Xf), and for a measure algebra the further bands
$M_{ac}$,
$M_{tc}$ and $M_t$;  for an algebra of sets we also have $M_p$ (362Xg).
These bands can be used to generate finite subalgebras of the band
algebra of $M$ (362Ya-362Yb), and for any such finite subalgebra we have
a corresponding decomposition of $M$ as a direct sum of the bands which
are the atoms of the subalgebra (352Tb).   This decomposition of $M$
can be regarded as a recipe for decomposing its members into finite sums
of functionals with special properties.   What I called the `Lebesgue
decomposition' in 232I is just such a recipe.   In that context I had
a measure space $(X,\Sigma,\mu)$ and was looking at the countably
additive functionals from $\Sigma$ to $\Bbb R$, that is, at $M_{\sigma}$
in the language of this section, and the bands involved in the
decomposition were $M_p$, $M_{ac}$ and $M_{tc}$.   But I hope that it
will be plain that these ideas can be refined indefinitely, as we refine
the classification of additive functionals.   At each stage, of course,
the exact enumeration of the subalgebra of bands generated by the
classification (as in 362Ya-362Yb) is a necessary check that we have
understood the relationships between the classes we have described.

These decompositions are of such importance that it is worth examining
the corresponding band projections.   I give formulae for the action of
band projections on (non-negative) functionals in 362Bc, 362Bd,
362B(f-ii),
362Xb, 362Xc(iii), 362Xi(iv) and 362Yi.   Of course these are readily
adapted to give formulae for the projections onto the complementary
bands, as in 362Bf and 362Xi.

If we have an algebra of sets, the completely additive functionals are
(usually) of relatively minor importance;  in the standard examples,
they correspond to functionals defined as weighted sums of point masses
(362Xg(ii)).   The point is that measure algebras $\frak A$ appear as
quotients of $\sigma$-algebras $\Sigma$ of sets by $\sigma$-ideals $\Cal
I$;  consequently the countably additive functionals on $\frak A$
correspond exactly to the countably additive functionals on $\Sigma$
which are zero on $\Cal I$;  but the canonical homomorphism from
$\Sigma$ to $\frak A$ is hardly ever order-continuous, so
completely additive functionals on $\frak A$ rarely correspond to
completely additive
functionals on $\Sigma$.
On the other hand, when we are looking at countably additive functionals
on $\Sigma$, we have to consider the possibility that they are singular
in the sense that they are carried on some member of $\Cal I$;  in the
measure algebra context this possibility disappears, and we can often be
sure that every countably additive functional is absolutely continuous,
as in 327Bb.

For any Boolean algebra $\frak A$, we can regard it as the algebra of
open-and-closed subsets of its Stone space $Z$;  the points of $Z$
correspond to Boolean homomorphisms from $\frak A$ to $\{0,1\}$, which
are the normalised `atomic elements'
in the space of additive functionals on
$\frak A$ (362Xe, 362Xg(iv)).   It is the case that all non-negative
additive functionals on a Boolean algebra $\frak A$ can be represented
by appropriate measures on its Stone space (see 416Q in Volume 4), but I
prefer to
hold this result back until it can take its place among other theorems
on representing functionals by measures and integrals.

It is one of the leitmotivs of this chapter, that Boolean algebras and
Riesz spaces are Siamese twins;  again and again, matching results are
proved by the application of identical ideas.   A typical example is the
pair 362B(f-i) and 362Yd.   Many of us have been tempted to try to
describe something which would provide a common generalization of
Boolean algebras and Riesz spaces (and lattice-ordered groups).   I have
not yet seen any such structure which was worth the trouble.   Most of
the time, in this chapter, I shall be using ideas from the general
theory of Riesz spaces to suggest and illuminate questions in measure
theory;  but if you pursue this subject you will surely find that
intuitions often come to you first in the context of Boolean algebras,
and the applications to Riesz spaces are secondary.

In 362E I give a condition for uniform integrability in terms of
disjoint sequences, following the pattern established in 246G and
repeated in 354R and 356O.   The condition of 362E assumes that
the set is norm-bounded;  but if you have 246G to hand, you will see
that it can be done with weaker assumptions involving atoms, as in
362Yf-362Yg.

I mention once again the Banach-Ulam problem:  if $\frak A$ is Dedekind
complete, can $S(\frak A)^{\sim}_c$ be different from
$S(\frak A)^{\times}$?   This is obviously equivalent to the form given
in the notes to \S326 above.   See 363S below.
}%end of comment

\discrpage

\frfilename{mt363.tex}
\versiondate{4.3.08}
\copyrightdate{2000}

\def\chaptername{Function spaces}
\def\sectionname{$L^{\infty}$}

\newsection{363}

In this section I set out to describe an abstract construction for
$L^{\infty}$ spaces on arbitrary Boolean algebras,
corresponding to the $L^{\infty}(\mu)$ spaces of \S243.   I begin with
the definition of $L^{\infty}(\frak A)$ (363A) and elementary facts
concerning its own structure and the embedding
$S(\frak A)\embedsinto L^{\infty}(\frak A)$ (363B-363D).   I give the basic universal mapping
theorems which define the Banach lattice structure of $L^{\infty}$
(363E) and a description of the action of Boolean homomorphisms on
$L^{\infty}$ spaces (363F-363G) before discussing the representation of
$L^{\infty}(\Sigma)$ and $L^{\infty}(\Sigma/\Cal I)$ for
$\sigma$-algebras $\Sigma$
and ideals $\Cal I$ of sets (363H).   This leads at once to the
identification of
$L^{\infty}(\mu)$, as defined in Volume 2, with $L^{\infty}(\frak A)$,
where $\frak A$ is the measure algebra of $\mu$ (363I).   Like
$S(\frak A)$, $L^{\infty}(\frak A)$ determines the algebra $\frak A$
(363J).   I
briefly discuss the dual spaces of $L^{\infty}$;  they correspond
exactly to the duals of $S$ described in \S362 (363K).   Linear
functionals on $L^{\infty}$ can for some purposes be treated as
`integrals' (363L).

In the second half of the section I present some of the theory of
Dedekind complete and
$\sigma$-complete algebras.   First, $L^{\infty}(\frak A)$ is Dedekind
($\sigma$-\nobreak)complete iff $\frak A$ is (363M).   The spaces
$L^{\infty}(\frak A)$, for Dedekind $\sigma$-complete $\frak A$, are
precisely the Dedekind $\sigma$-complete Riesz spaces with order unit
(363N-363P).   The spaces $L^{\infty}(\frak A)$, for Dedekind
complete $\frak A$, are precisely the normed spaces which may be put in
place of $\Bbb R$ in the Hahn-Banach theorem (363R).   Finally, I
mention some equivalent forms of the Banach-Ulam problem (363S).

\leader{363A}{Definition} Let $\frak A$ be a Boolean algebra, with Stone
space $Z$.   I will write $L^{\infty}(\frak A)$ for the space
$C(Z)=C_b(Z)$ of
continuous real-valued functions from $Z$ to $\Bbb R$, endowed with the
linear structure, order structure, norm and multiplication of
$C(Z)=C_b(Z)$.   \cmmnt{ (Recall that because $Z$ is compact (311I),
$\{u(z):z\in Z\}$
is bounded for every $u\in L^{\infty}(\frak A)=C(Z)$ (2A3N(b-iii)), that
is, $C(Z)=C_b(Z)$.   Of
course if $\frak A=\{0\}$, so that $Z=\emptyset$, then $C(Z)$ has just
one member, the empty function.)}

\leader{363B}{Theorem} Let $\frak A$ be any Boolean algebra;  write
$L^{\infty}$ for $L^{\infty}(\frak A)$.

(a) $L^{\infty}$ is an $M$-space;  its standard order unit is the
constant function taking the value $1$ at each point;  in particular,
$L^{\infty}$ is a Banach lattice with a Fatou norm and the Levi
property.

(b)  $L^{\infty}$ is a commutative Banach algebra and an $f$-algebra.

(c) If $u\in L^{\infty}$ then $u\ge 0$ iff there is a $v\in L^{\infty}$
such that $u=v\times v$.

\proof{{\bf (a)} See 354Hb and 354J.

\medskip

{\bf (b)-(c)} are obvious from the definitions of Banach algebra (2A4J)
and $f$-algebra (352W) and the ordering of $L^{\infty}=C(Z)$.
}%end of proof of 363B

\leader{363C}{Proposition} Let $\frak A$ be any Boolean algebra.   Then
$S(\frak A)$ is a norm-dense, order-dense Riesz subspace of
$L^{\infty}(\frak A)$, closed under multiplication.

\proof{ Let $Z$ be the Stone space of $\frak A$.   Using the definition
of $S=S(\frak A)$ set out in 361D, it is obvious that $S$ is a linear
subspace of $L^{\infty}=L^{\infty}(\frak A)=C(Z)$ closed under
multiplication.   Because $S$, like $L^{\infty}$, is a Riesz subspace of
$\Bbb R^Z$ (361Ee), $S$ is a Riesz subspace of $L^{\infty}$.   By the
Stone-Weierstrass theorem (in either of the forms given in 281A and
281E), $S$ is
norm-dense in $L^{\infty}$.   Consequently it is order-dense (354I).
}%end of proof of 363C

\leader{363D}{Proposition} Let $\frak A$ be a Boolean algebra.   If we
regard $\chi a\in S(\frak A)$\cmmnt{ (361D)} as a member of
$L^{\infty}(\frak A)$ for each
$a\in\frak A$, then $\chi:\frak A\to L^{\infty}(\frak A)$ is additive,
order-preserving, order-continuous and a lattice homomorphism.

\proof{ Because the embedding
$S=S(\frak A)\embedsinto L^{\infty}(\frak A)=L^{\infty}$ is a Riesz
homomorphism, $\chi:\frak A\to L^{\infty}$ is
additive and a lattice homomorphism (361F-361G).   Because $S$ is
order-dense in $L^{\infty}$ (363C), the embedding
$S\embedsinto L^{\infty}$ is order-continuous (352Nb), so
$\chi:\frak A\to L^{\infty}$ is order-continuous (361Gb).
}%end of proof of 363D

\leader{363E}{Theorem} Let $\frak A$ be a Boolean algebra, and $U$ a
Banach space.   Let $\nu:\frak A\to U$ be a bounded additive function.

(a) There is a unique bounded linear operator
$T:L^{\infty}(\frak A)\to U$ such that $T\chi=\nu$;  in this case
$\|T\|=\sup_{a,b\in\frak A}\|\nu a-\nu b\|$.

(b) If $U$ is a Banach lattice then $T$ is positive iff $\nu$ is
non-negative;  and in this case $T$ is order-continuous iff $\nu$ is
order-continuous, and sequentially order-continuous iff $\nu$ is
sequentially order-continuous.

(c) If $U$ is a Banach lattice then $T$ is a Riesz homomorphism iff
$\nu$ is a lattice homomorphism iff $\nu a\wedge\nu b=0$ whenever
$a\Bcap b=0$.

\proof{ Write $S=S(\frak A)$, $L^{\infty}=L^{\infty}(\frak A)$.

\medskip

{\bf (a)} By 361I there is a unique bounded linear operator $T_0:S\to U$
such that $T_0\chi=\nu$, and $\|T_0\|=\sup\{\|\nu a-\nu b\|:a$,
$b\in\frak A\}$.   But because $U$ is a Banach space and $S$ is dense in
$L^{\infty}$, $T_0$ has a unique extension to a bounded linear operator
$T:L^{\infty}\to U$ with the same norm (2A4I).

\medskip

{\bf (b)(i)} If $T$ is positive then $T_0$ is positive so $\nu$ is
non-negative, by 361Ga.

\medskip

\quad{\bf (ii)} If $\nu$ is non-negative then $T_0$ is positive, by
361Ga in the other direction.   But
if $u\in L^{\infty+}$ and $\epsilon>0$, then by 354I there is a
$v\in S^+$ such that
$\|u-v\|_{\infty}\le\epsilon$;  now $\|Tu-Tv\|\le\epsilon\|T\|$.   But
$Tv=T_0v$ belongs to the
positive cone $U^+$ of $U$.   As $\epsilon$ is arbitrary, $Tu$ belongs
to the closure of $U^+$, which is $U^+$ (354Bc).   As $u$ is
arbitrary, $T$ is positive.

\medskip

\quad{\bf (iii)} Now suppose that $\nu$ is order-continuous as well as
non-negative, and that $A\subseteq L^{\infty}$ is a non-empty
downwards-directed set with infimum $0$.   Set

\Centerline{$B=\{v:v\in S$, there is some $u\in A$ such that
$v\ge u\}$.}

\noindent Then $B$ is downwards-directed (indeed, $v_1\wedge v_2\in B$
for every $v_1$, $v_2\in B$), and $u=\inf\{v:v\in B,\,u\le v\}$ for
every $u\in A$ (354I again), so $B$ has the same lower bounds as $A$ and
$\inf B=0$ in $L^{\infty}$ and in $S$.   But we know from 361Gb that
$T_0$ is order-continuous, while any lower bound for $\{Tu:u\in A\}$ in
$U$ must also be a lower bound for $\{Tv:v\in B\}=\{T_0v:v\in B\}$, so
$\inf_{u\in A}Tu=\inf_{v\in B}T_0v=0$ in $U$.   As $A$ is arbitrary, $T$
is order-continuous (351Ga).

\medskip

\quad{\bf (iv)} Suppose next that $\nu$ is only sequentially
order-continuous, and that $\sequencen{u_n}$ is a non-increasing
sequence in $L^{\infty}$ with infimum $0$.   For each $n$, $k$ choose
$w_{nk}\in S$ such that $u_n\le w_{nk}$ and
$\|w_{nk}-u_n\|_{\infty}\le 2^{-k}$ (354I once more), and set
$w'_n=\inf_{j,k\le n}w_{jk}$ for each
$n$.   Then $\sequencen{w'_n}$ is a non-increasing sequence in $S$, and
any lower bound of $\{w'_n:n\in\Bbb N\}$ is also a lower bound of
$\{u_n:n\in\Bbb N\}$, so $0=\inf_{n\in\Bbb N}w'_n$ in $S$ and
$L^{\infty}$.   Since $T_0:S\to U$ is sequentially order-continuous
(361Gb),

\Centerline{$\inf_{n\in\Bbb N}Tu_n\le\inf_{n\in\Bbb N}Tw'_n
=\inf_{n\in\Bbb N}T_0w'_n=0$}

\noindent in $U$.   As $\sequencen{u_n}$ is arbitrary, $T$ is
sequentially order-continuous.

\medskip

\quad{\bf (v)} On the other hand, if $T$ is order-continuous or
sequentially order-continuous, so is $\nu=T\chi$, because $\chi$ is
order-continuous (363D).

\medskip

{\bf (c)} We know that $T_0:S\to U$ is a Riesz homomorphism iff $\nu$ is
a lattice homomorphism iff $\nu a\wedge\nu b=0$ whenever $a\Bcap b=0$,
by 361Gc.   But $T_0$ is a Riesz homomorphism iff $T$ is.   \Prf\ If $T$
is a Riesz homomorphism so is $T_0$, because the embedding
$S\embedsinto L^{\infty}$ is a Riesz homomorphism.   On the other hand, if
$T_0$ is a
Riesz homomorphism, then the functions $u\mapsto u^+\mapsto T(u^+)$,
$u\mapsto Tu\mapsto (Tu)^+$ are continuous (by 354Bb) and agree on $S$,
so agree on
$L^{\infty}$, and $T$ is a Riesz homomorphism, by 352G.\ \Qed
}%end of proof of 363E

\leader{363F}{Theorem} Let $\frak A$ and $\frak B$ be Boolean algebras,
and $\pi:\frak A\to\frak B$ a Boolean homomorphism.

(a) There is an associated multiplicative Riesz homomorphism
$T_{\pi}:L^{\infty}(\frak A)\to L^{\infty}(\frak B)$, of norm at most
$1$, defined by saying that $T_{\pi}(\chi a)=\chi(\pi a)$ for every
$a\in\frak A$.

(b) For any $u\in L^{\infty}(\frak A)$, there is a
$u'\in L^{\infty}(\frak A)$ such that $T_{\pi}u=T_{\pi}u'$ and
$\|u'\|_{\infty}=\|T_{\pi}u\|_{\infty}\le\|u\|_{\infty}$.

(c)(i) The kernel of $T_{\pi}$ is the norm-closed linear subspace of
$L^{\infty}(\frak A)$ generated by $\{\chi a:a\in\frak A,\,\pi a=0\}$.

\quad(ii) The set of values of $T_{\pi}$ is the norm-closed linear subspace
of $L^{\infty}(\frak B)$ generated by $\{\chi(\pi a):a\in\frak A\}$.

(d) $T_{\pi}$ is surjective iff $\pi$ is surjective, and in this case
$\|v\|_{\infty}=\min\{\|u\|_{\infty}:T_{\pi}u=v\}$ for every
$v\in L^{\infty}(\frak B)$.

(e) $T_{\pi}$ is injective iff $\pi$ is injective, and in this case
$\|T_{\pi}u\|_{\infty}=\|u\|_{\infty}$ for every
$u\in L^{\infty}(\frak A)$.

(f) $T_{\pi}$ is order-continuous, or sequentially order-continuous, iff
$\pi$ is.

(g) If $\frak C$ is another Boolean algebra and $\theta:\frak B\to\frak
C$ is another Boolean homomorphism, then $T_{\theta\pi}
=T_{\theta}T_{\pi}:L^{\infty}(\frak A)\to L^{\infty}(\frak C)$.

\proof{ Let $Z$ and $W$ be the Stone spaces of $\frak A$ and $\frak B$.
By 312Q there is a continuous function $\phi:W\to Z$ such that
$\widehat{\pi a}=\phi^{-1}[\widehat a]$ for every $a\in\frak A$, where
$\widehat{a}$ is the open-and-closed subset of $Z$ corresponding to
$a\in\frak A$.   Write $T$ for $T_{\pi}$.

\medskip

{\bf (a)} For $u\in L^{\infty}(\frak A)=C(Z)$, set
$Tu=u\phi:W\to\Bbb R$.   Then $Tu\in C(W)=L^{\infty}(\frak B)$.   It is
obvious, or at any rate very easy to check, that
$T:L^{\infty}(\frak A)\to L^{\infty}(\frak B)$ is linear,
multiplicative, a Riesz homomorphism and of norm $1$
unless $\frak B=\{0\}$, $W=\emptyset$.    If $a\in\frak A$, then

\Centerline{$T(\chi a)=(\chi a)\phi=(\chi\widehat a)\phi
=\chi(\phi^{-1}[\widehat a])=\chi(\pi a)$,}

\noindent identifying $\chi a\in L^{\infty}(\frak A)$ with the
indicator function $\chi\widehat a:Z\to\{0,1\}$ of the set
$\widehat a$.   Of course $T_{\pi}=T$ is the only continuous linear
operator with these properties, by 363Ea.

\medskip

{\bf (b)} Set $\alpha=\|Tu\|_{\infty}$,
$u'(z)=\med(-\alpha,u(z),\alpha)$ for $z\in Z$;  that is,
$u'=\med(-\alpha e,u,\alpha e)$ in $L^{\infty}(\frak A)$, where
$e$ is the standard order unit of $L^{\infty}(\frak A)$.   Then $Te$ is
the standard order unit of $L^{\infty}(\frak B)$, so

\Centerline{$Tu'=\med(-\alpha Te,Tu,\alpha Te)=Tu$}

\noindent (because $T$ is a lattice homomorphism, see 3A1Ic), while

\Centerline{$\|u'\|_{\infty}\le\alpha=\|Tu\|_{\infty}=\|Tu'\|_{\infty}
\le\|u'\|_{\infty}\le\|u\|_{\infty}$.}

\medskip

{\bf (c)(i)} Let $U$ be the closed linear subspace of
$L^{\infty}(\frak A)$ generated by $\{\chi a:\pi a=0\}$, and $U_0$ the
kernel of $T$.
Because $T$ is continuous and linear, $U_0$ is a closed linear subspace,
and $T(\chi a)=\chi 0=0$ whenever $\pi a=0$;  so $U\subseteq U_0$.   Now
take any $u\in U_0$ and $\epsilon>0$.   Then $T(u^+)=(Tu)^+=0$, so
$u^+\in U_0$.   By 354I there is a $u'\in S(\frak A)$ such that
$0\le u'\le u^+$ and $\|u^+-u'\|_{\infty}\le\epsilon$.   Now
$0\le Tu'\le Tu^+=0$, so $Tu'=0$.   Express $u'$ as
$\sum_{i=0}^n\alpha_i\chi a_i$ where
$\alpha_i\ge 0$ for each $i$.   For each $i$,
$\alpha_i\chi(\pi a_i)=T(\alpha_i\chi a_i)=0$, so $\pi a_i=0$ or
$\alpha_i=0$;  in either
case $\alpha_i\chi a_i\in U$.   Consequently $u'\in U$.   As $\epsilon$
is arbitrary and $U$ is closed, $u^+\in U$.   Similarly,
$u^-=(-u)^+\in U$ and $u=u^+-u^-\in U$.   As $u$ is arbitrary,
$U_0\subseteq U$ and $U_0=U$.

\medskip

\quad{\bf (ii)} Let $V$ be the closed linear subspace of
$L^{\infty}(\frak B)$ generated by $\{\chi(\pi a):a\in\frak A\}$, and
$V_0=T[L^{\infty}(\frak A)]$.   Then $T[S(\frak A)]\subseteq V$, so

\Centerline{$V_0
=T[\overline{S(\frak A)}]
\subseteq\overline{T[S(\frak A)]}
\subseteq\overline{V}
=V$.}

\noindent On the other hand, $V_0$ is a closed linear subspace in
$L^{\infty}(\frak B)$.   \Prf\ It is a linear
subspace because $T$ is a linear operator.   To see that it is closed,
take any $v\in\overline{V}_0$.   Then there is a sequence
$\sequencen{v_n}$ in $V_0$ such that
$\|v-v_n\|_{\infty}\le 2^{-n}$ for every $n\in\Bbb N$.
Choose $u_n\in L^{\infty}(\frak A)$ such that $Tu_0=v_0$, while
$Tu_n=v_n-v_{n-1}$ and $\|u_n\|_{\infty}=\|v_n-v_{n-1}\|_{\infty}$ for
$n\ge 1$ (using (b) above).   Then

\Centerline{$\sum_{n=1}^{\infty}\|u_n\|_{\infty}
\le\sum_{n=1}^{\infty}\|v-v_n\|_{\infty}+\|v-v_{n-1}\|_{\infty}$}

\noindent is finite, so $u=\lim_{n\to\infty}\sum_{i=0}^nu_i$ is defined
in the Banach space $L^{\infty}(\frak A)$, and

\Centerline{$Tu=\lim_{n\to\infty}\sum_{i=0}^nTu_i
=\lim_{n\to\infty}v_n=v$.}

\noindent As $v$ is arbitrary, $V_0$ is closed.\ \QeD\   Since
$\chi(\pi a)=T(\chi a)\in V_0$ for every $a\in\frak A$, $V\subseteq V_0$
and $V=V_0$, as required.

\medskip

{\bf (d)}  If $\pi$ is surjective, then $T$ is surjective, by (c-ii).
If $T$ is surjective and $b\in\frak B$, then there is a
$u\in L^{\infty}(\frak A)$ such that $Tu=\chi b$.   Now there is a
$u'\in S(\frak A)$ such that $\|u-u'\|_{\infty}\le\bover13$, so that
$\|Tu'-\chi b\|_{\infty}\le\bover13$.   Taking $a\in\frak A$ such that
$\{z:u'(z)\ge\bover12\}=\widehat a$, we must have $\pi a=b$, since

\Centerline{$\widehat b=\{w:(Tu')(w)\ge\bover12\}
=\phi^{-1}[\widehat a]=\widehat{\pi a}$.}

\noindent As $b$ is arbitrary, $\pi$ is surjective.

Now (b) tells us that in this case
$\|v\|_{\infty}=\min\{\|u\|_{\infty}:Tu=v\}$ for every
$v\in L^{\infty}(\frak B)$.

\medskip

{\bf (e)} By (c-i), $T$ is injective iff $\pi$ is injective.   In
this case, for any $u\in L^{\infty}(\frak A)$,

$$\eqalignno{\|Tu\|_{\infty}
&=\|T|u|\|_{\infty}\cr
\noalign{\noindent (because $T$ is a Riesz homomorphism)}
&\ge\sup\{\|Tu'\|_{\infty}:u'\in S(\frak A),\,u'\le|u|\}\cr
&=\sup\{\|u'\|_{\infty}:u'\in S(\frak A),\,u'\le|u|\}\cr
\noalign{\noindent (by 361Jd)}
&=\|u\|_{\infty}\cr
\noalign{\noindent (by 354I)}
&\ge\|Tu\|_{\infty},\cr}$$

\noindent and $\|Tu\|_{\infty}=\|u\|_{\infty}$.

\medskip

{\bf (f)} If $T$ is (sequentially) order-continuous then $\pi=T\chi$ is
(sequentially) order-continuous, by 363D.   If $\pi$ is (sequentially)
order-continuous then $\chi\pi:\frak A\to L^{\infty}(\frak B)$ is
(sequentially) order-continuous, so $T$ is (sequentially)
order-continuous, by 363Eb.

\medskip

{\bf (g)} This is elementary, in view of the uniqueness of
$T_{\theta\pi}$.
}%end of proof of 363F

\leader{363G}{Corollary}  Let $\frak A$ be a Boolean algebra.

(a) If $\frak C$ is a subalgebra of $\frak A$, then
$L^{\infty}(\frak C)$ can be identified, as Banach lattice and as Banach
algebra, with the
closed linear subspace of $L^{\infty}(\frak A)$ generated by
$\{\chi c:c\in\frak C\}$.

(b) If $\Cal I$ is an ideal of $\frak A$, then
$L^{\infty}(\frak A/\Cal I)$ can be identified, as Banach lattice and as
Banach algebra, with the
quotient space $L^{\infty}(\frak A)/V$, where $V$ is the closed linear
subspace of $L^{\infty}(\frak A)$ generated by $\{\chi a:a\in\Cal I\}$.

\proof{ Apply 363Fc-363Fd
to the identity map from $\frak C$ to $\frak A$ and
the canonical map from $\frak A$ onto $\frak A/\Cal I$.
}%end of proof of 363G

\leader{363H}{Representations of
\dvrocolon{$L^{\infty}(\frak A)$}}\cmmnt{ Much of
the importance of the concept of $L^{\infty}(\frak A)$ arises from the
way it is naturally represented in the contexts in which the most
familiar Boolean algebras appear.

\medskip

\noindent}{\bf Proposition} Let $X$ be a set and $\Sigma$ an
algebra of subsets of $X$.

(a) Write $S(\Sigma)$ for the linear subspace of $\ell^{\infty}(X)$
generated by the indicator functions of members of $\Sigma$,
and $\eusm L^{\infty}$ for its $\|\,\|_{\infty}$-closure
in $\ell^{\infty}(X)$.

\quad(i) $L^{\infty}(\Sigma)$ can be identified, as Banach lattice and
Banach algebra, with $\eusm L^{\infty}$;  if $E\in\Sigma$, then
$\chi E$, defined in $L^{\infty}(\Sigma)$ as in 361D, can be identified
with the indicator function of $E$ regarded as a subset of $X$.

\quad(ii) A bounded function $f:X\to\Bbb R$ belongs to $\eusm L^{\infty}$
iff whenever $\alpha<\beta$ in $\Bbb R$ there is an $E\in\Sigma$ such that
$\{x:f(x)>\beta\}\subseteq E\subseteq\{x:f(x)>\alpha\}$.

\quad(iii) In particular, $L^{\infty}(\Cal PX)$
can be identified with $\ell^{\infty}(X)$.

(b) Now suppose that $\Sigma$ is a $\sigma$-algebra of subsets of $X$.

\quad(i) $\eusm L^{\infty}$ is just the set of bounded $\Sigma$-measurable
real-valued functions on $X$.

\quad(ii) If $\frak A$ is a Dedekind $\sigma$-complete Boolean algebra and
$\pi:\Sigma\to\frak A$ is a surjective sequentially order-continuous
Boolean homomorphism with kernel $\Cal I$, then $L^{\infty}(\frak A)$
can be identified, as Banach lattice and Banach algebra, with
$\eusm L^{\infty}/\eusm W$, where
$\eusm W=\{f:f\in\eusm L^{\infty},\,\{x:f(x)\ne 0\}\in\Cal I\}$
is a solid linear subspace and closed ideal
of $\eusm L^{\infty}$.   For $f\in\eusm L^{\infty}$,

\Centerline{$\|f^{\ssbullet}\|_{\infty}
=\min\{\alpha:\alpha\ge 0,\,\{x:|f(x)|>\alpha\}\in\Cal I\}$.}

\quad(iii) In particular, if $\Cal I$ is any $\sigma$-ideal of $\Sigma$ and
$E\mapsto E^{\ssbullet}$ is the canonical homomorphism from $\Sigma$
onto $\frak A=\Sigma/\Cal I$, then we have an identification of
$L^{\infty}(\frak A)$ with a quotient of $\eusm L^{\infty}$, and for any
$E\in\Sigma$ we can identify $\chi(E^{\ssbullet})\in L^{\infty}(\frak
A)$ with the equivalence class $(\chi E)^{\ssbullet}\in\eusm
L^{\infty}/\eusm W$ of the indicator function $\chi E$.

\proof{{\bf (a)(i)} By 361L, $S(\Sigma)$, as described here, can be
identified with $S(\Sigma)$ as defined in 361D.   Because
the normed space $\ell^{\infty}(X)$ is complete,
$\eusm L^{\infty}$ can be identified with the
normed space completion of $S(\Sigma)$ for
$\|\,\|_{\infty}$;  but 363C shows that the same is true of
$L^{\infty}(\Sigma)$.   Thus we have a canonical Banach space isomorphism
between $\eusm L^{\infty}$ and $L^{\infty}(\Sigma)$.   Because
multiplication and the lattice operations are $\|\,\|_{\infty}$-continuous,
both in $\eusm L^{\infty}$ and in $L^{\infty}(\Sigma)$, this isomorphism is
multiplicative and order-preserving, that is, identifies
$\eusm L^{\infty}$ with $L^{\infty}(\Sigma)$ as Banach algebra and Banach
lattice.   In the language of 363E, $\eusm L^{\infty}$ is the image of
$L^{\infty}(\Sigma)$ in $\ell^{\infty}(X)$
under the operator associated with
the additive function $E\mapsto\chi E:\Sigma\to\ell^{\infty}(X)$.

\medskip

\quad{\bf (ii)}\grheada\ If $f\in\eusm L^{\infty}$ and $\alpha<\beta$ in
$\Bbb R$, let $g\in S(\Sigma)$ be such that
$\|f-g\|_{\infty}\le\bover12(\beta-\alpha)$.   Set
$E=\{x:g(x)>\bover12(\alpha+\beta)\}$;  by 361G or otherwise,
$E\in\Sigma$, and
$\{x:f(x)>\beta\}\subseteq E\subseteq\{x:f(x)>\alpha\}$.

\medskip

\qquad\grheadb\ If $f$ satisfies the condition, take any $\epsilon>0$.
Let $n\in\Bbb N$ be such that $\|f\|_{\infty}<n\epsilon$.   For
$-n\le i\le n$, let $E_i\in\Sigma$ be such that
$\{x:f(x)>(i+1)\epsilon\}\subseteq E_i\subseteq\{x:f(x)>i\epsilon\}$.
Set $g(x)=\epsilon\sum_{i=-n}^n\chi E_i-\epsilon n$ for $x\in X$;  then
$g\in S(\Sigma)$ and $\|f-g\|_{\infty}\le\epsilon$.   As $\epsilon$ is
arbitrary, $f\in\eusm L^{\infty}$.

\medskip

\quad{\bf (iii)} Now (ii) shows that if $\Sigma=\Cal PX$ we shall have
$\eusm L^{\infty}=\ell^{\infty}(X)$ and $L^{\infty}(\Cal PX)$ becomes
identified with $\ell^{\infty}(X)$.

\medskip

{\bf (b)(i)} If $\Sigma$ is a $\sigma$-algebra and $f:X\to\Bbb R$ is
bounded then

$$\eqalign{f\text{ is }\Sigma\text{-measurable}
&\iff\{x:f(x)>\alpha\}\in\Sigma\text{ for every }\alpha\in\Bbb R\cr
&\iff\text{ whenever }\alpha\in\Bbb R,\,n\in\Bbb N
  \text{ there is an }E\in\Sigma\cr
&\mskip50mu  \text{ such that }
  \{x:f(x)>\alpha+2^{-n}\}\subseteq E\subseteq\{x:f(x)>\alpha\}\cr
&\iff\text{ whenever }\beta>\alpha\text{ there is an }E\in\Sigma\cr
&\mskip50mu  \text{ such that }\{x:f(x)>\beta\}\subseteq
  E\subseteq\{x:f(x)>\alpha\}\cr
&\iff f\in\eusm L^{\infty}\cr}$$

\noindent by (a-ii) above.

\medskip

\quad{\bf (ii)}\grheada\
By 363F, we have a multiplicative Riesz homomorphism
$T=T_{\pi}$ from $L^{\infty}(\Sigma)$ to $L^{\infty}(\frak A)$ which is
surjective (363Fd) and has kernel the closed linear subspace $W$ of
$L^{\infty}(\Sigma)$ generated by $\{\chi E:E\in\Cal I\}$.   Now under
the identification described in (a), $W$ corresponds to $\eusm W$.   \Prf\
$\eusm W$ is a linear subspace of $\eusm L^{\infty}$ because

\Centerline{$\{x:(f+g)(x)\ne 0\}
\subseteq\{x:f(x)\ne 0\}\cup\{x:g(x)\ne 0\}\in\Cal I$,}

\Centerline{$\{x:(\alpha f)(x)\ne 0\}\subseteq\{x:f(x)\ne 0\}\in\Cal I$}

\noindent whenever $f$, $g\in\eusm W$ and $\alpha\in\Bbb R$.
If $\sequencen{f_n}$ is a sequence in $\eusm W$ converging to
$f\in\eusm L^{\infty}$, then

\Centerline{$\{x:f(x)\ne 0\}
\subseteq\bigcup_{n\in\Bbb N}\{x:f_n(x)\ne 0\}\in\Cal I$,}

\noindent so $f\in\eusm W$.   Thus $\eusm W$ is a closed linear subspace
of $\eusm L^{\infty}$.   If $E\in\Cal I$, then $\chi E$,
taken in $S(\Sigma)$ or $L^{\infty}(\Sigma)$, corresponds to the
function $\chi E:X\to\{0,1\}$, which belongs to $\eusm W$;  so that $W$
must correspond to the closed linear span in $\eusm L^{\infty}$ of such
indicator functions, which is a subspace of $\eusm W$.
On the other hand, if $f\in\eusm W$ and $\epsilon>0$, set

\Centerline{$E_n=\{x:n\epsilon<f(x)\le (n+1)\epsilon\}$,
\quad$E'_n=\{x:-(n+1)\epsilon\le f(x)<-n\epsilon\}$}

\noindent for $n\in\Bbb N$;  all these belong to $\Cal I$, so
$g=\epsilon\sum_{n=0}^{\infty}(\chi E_n-\chi E'_n)\in\eusm W$
corresponds to a
member of $W$, while $\|f-g\|_{\infty}\le\epsilon$.   As $W$ is closed,
$f$ also must correspond to some member of $W$.   As $f$ is arbitrary,
$W$ and $\eusm W$ match exactly.\ \Qed

\medskip

\qquad\grheadb\ Because $T$ is a multiplicative Riesz homomorphism,
$L^{\infty}(\frak A)\cong L^{\infty}(\Sigma)/W$
is matched canonically, in its linear, order
and multiplicative structures, with $\eusm L^{\infty}/\eusm W$.   We
know also that

\Centerline{$\|v\|_{\infty}
=\min\{\|u\|_{\infty}:u\in L^{\infty}(\Sigma),\,Tu=v\}$}

\noindent for every $v\in L^{\infty}(\frak A)$ (363Fd), that is, that
the norm of $L^{\infty}(\frak A)$ corresponds to the quotient norm on
$L^{\infty}(\Sigma)/W$.

As for
the given formula for the norm, take any $f\in\eusm L^{\infty}$.   There
is a $g\in \eusm L^{\infty}$ such that
$Tf=Tg$ and $\|Tf\|_{\infty}
=\|g\|_{\infty}$.   (Here I am treating $T$ as an operator from
$\eusm L^{\infty}$ onto $L^{\infty}(\frak A)$.)   In this case

\Centerline{$\{x:|f(x)|>\|Tf\|_{\infty}\}
\subseteq\{x:f(x)\ne g(x)\}\in\Cal I$.}

\noindent On the other hand, if $\alpha\ge 0$ and
$\{x:|f(x)|>\alpha\}\in\Cal I$, and we
set $h=\med(-\alpha\chi X,f,\alpha\chi X)$, then
$Th=Tf$, so $\|Tf\|_{\infty}\le\|h\|_{\infty}\le\alpha$.

\medskip

\quad{\bf (iii)} Put (a-i) and (ii) just above together.
}%end of proof of 363H

\leader{363I}{Corollary} Let $(X,\Sigma,\mu)$ be a measure space, with
measure algebra $\frak A$.   Then $L^{\infty}(\mu)$ can be identified,
as Banach lattice and Banach algebra, with $L^{\infty}(\frak A)$;  the
identification matches $(\chi E)^{\ssbullet}\in L^{\infty}(\mu)$ with
$\chi(E^{\ssbullet})\in L^{\infty}(\frak A)$, for every $E\in\Sigma$.

\cmmnt{\medskip

\noindent{\bf Remark} The space I called $\eusm L^{\infty}(\mu)$ in
Chapter 24 is not strictly speaking the space
$\eusm L^{\infty}\cong L^{\infty}(\Sigma)$ of 363H;  I took
$\eusm L^{\infty}(\mu)\subseteq\eusm L^0(\mu)$ to be the set of essentially
bounded, virtually measurable functions defined almost everywhere in
$X$, and in general this is larger.   But, as remarked in the notes to
\S243, $L^{\infty}(\mu)$ can equally well be regarded as a quotient of
what I there called $\eusm L^{\infty}_{\Sigma}$, which is the
$\eusm L^{\infty}$ above, because every function in
$\eusm L^{\infty}(\mu)$ is equal almost everywhere to some member of
$\eusm L^{\infty}_{\Sigma}$.
}%end of comment

\leader{363J}{Recovering the algebra $\frak A$:  Proposition}  Let
$\frak A$ be a Boolean algebra.   For $a\in\frak A$ write $V_a$ for the
solid linear subspace of $L^{\infty}(\frak A)$ generated by $\chi a$.
Then $a\mapsto V_a$ is a Boolean isomorphism between $\frak A$ and the
algebra of projection bands in $L^{\infty}(\frak A)$.

\proof{ The proof is nearly identical to that of 361K.   If
$a\in\frak A$, $u\in V_a$ and $v\in V_{1\Bsetminus a}$, then
$|u|\wedge|v|=0$ because $\chi a\wedge\chi(1\Bsetminus a)=0$;  and if
$w\in L^{\infty}(\frak A)$ then

\Centerline{$w=(w\times\chi a)+(w\times\chi(1\Bsetminus a))
\in V_a+V_{1\Bsetminus a}$}

\noindent because $|w\times\chi a|\le\|w\|_{\infty}\chi a$ and
$|w\times\chi(1\Bsetminus a)|\le\|w\|_{\infty}\chi(1\Bsetminus a)$.   So
$V_a$ and $V_{1\Bsetminus a}$ are complementary projection bands in
$L^{\infty}=L^{\infty}(\frak A)$.   Next, if $U\subseteq L^{\infty}$ is
a projection band, then $\chi 1$ is expressible as $u+v$ where $u\in U$,
$v\in U^{\perp}$;  thinking of $L^{\infty}$ as the space of continuous
real-valued functions on the Stone space $Z$ of $\frak A$, $u$ and $v$
must be the indicator functions of complementary subsets $E$, $F$
of $Z$, which must be open-and-closed, so that $E=\widehat a$,
$F=\widehat{1\Bsetminus a}$.   In this case $V_a\subseteq U$ and
$V_{1\Bsetminus a}\subseteq U^{\perp}$, so $U$ must be $V_a$ precisely.
Thus $a\mapsto V_a$ is surjective.   Finally, just as in 361K,
$a\Bsubseteq b\iff V_a\subseteq V_b$, so we have a Boolean isomorphism.
}%end of proof of 363J

\leader{363K}{Dual spaces of $L^{\infty}$\dvro{: }{}}\cmmnt{ The
questions treated in \S362 yield nothing new in the present context.
I spell out the details.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Boolean algebra.   Let
$M$, $M_{\sigma}$ and $M_{\tau}$ be the $L$-spaces of bounded finitely
additive functionals, bounded countably additive functionals and
completely additive functionals on $\frak A$.   Then the embedding
$S(\frak A)\embedsinto L^{\infty}(\frak A)$ induces Riesz space isomorphisms
between $S(\frak A)^{\sim}\cong M$ and
$L^{\infty}(\frak A)^{\sim}=L^{\infty}(\frak A)^*$,
$S(\frak A)^{\sim}_c\cong M_{\sigma}$
and $L^{\infty}(\frak A)^{\sim}_c$, and
$S(\frak A)^{\times}\cong M_{\tau}$ and $L^{\infty}(\frak A)^{\times}$.

\proof{ Write $S=S(\frak A)$, $L^{\infty}=L^{\infty}(\frak A)$.

\medskip

{\bf (a)}
For the identifications $S^{\sim}\cong M$, $S^{\sim}_c\cong M_{\sigma}$
and $S^{\times}\cong M_{\tau}$ see 362A.

\medskip

{\bf (b)} $L^{\infty*}=L^{\infty\sim}$ either because $L^{\infty}$ is a
Banach lattice (356Dc) or because $L^{\infty}$ has an order-unit
norm, so that a linear functional on $L^{\infty}$ is order-bounded iff
it is bounded on the unit ball.

\medskip

{\bf (c)}
If $f$ is a positive linear functional on $L^{\infty}$, then $f\restr S$
is a positive linear functional.   Because $S$ is order-dense in
$L^{\infty}$ (363C), the embedding is order-continuous (352Nb);
so if $f$ is (sequentially) order-continuous, so is $f\restr S$.
Accordingly the restriction operator $f\mapsto f\restr S$ gives maps
from $L^{\infty\sim}$ to $S^{\sim}$, $(L^{\infty})^{\sim}_c$ to
$S^{\sim}_c$ and $L^{\infty\times}$ to $S^{\times}$.   If
$f\in L^{\infty\sim}$ and $f\restr S\ge 0$, then $f(u^+)\ge 0$ for every
$u\in S$ and therefore for every $u\in L^{\infty}$, and $f\ge 0$;  so
all these restriction maps are injective positive linear operators.

\medskip

{\bf (d)} I need to show that they are surjective.

\medskip

\quad{\bf (i)} If $g\in S^{\sim}$, then $g$ is bounded on the unit ball
$\{u:u\in S,\,\|u\|_{\infty}\le 1\}$, so has an extension to a
continuous linear $f:L^{\infty}\to\Bbb R$ (2A4I);  thus
$S^{\sim}=\{f\restr S:f\in L^{\infty\sim}\}$.   This means that
$f\mapsto f\restr S$ is actually a Riesz space isomorphism between
$L^{\infty\sim}$ and $S^{\sim}$.   In particular,
$|f|\restr S=|f\restr S|$ for any $f\in L^{\infty\sim}$.

\medskip

\quad{\bf (ii)} If $f:L^{\infty}\to \Bbb R$ is a positive linear
operator and $f\restr S\in S^{\sim}_c$, let $\sequencen{u_n}$ be a
non-increasing sequence in $L^{\infty}$ with infimum $0$.   For each
$n$, $k\in\Bbb N$ there is a $v_{nk}\in S$ such that
$u_n\le v_{nk}\le u_n+2^{-k}e$, where $e$ is the standard order unit of
$L^{\infty}$ (354I, as usual);  set
$w_n=\inf_{i,k\le n}v_{ik}$;  then $\sequencen{w_n}$ is a non-increasing
sequence in $S$ with infimum $0$, so

\Centerline{$0\le\inf_{n\in\Bbb N}f(u_n)\le\inf_{n\in\Bbb N}f(w_n)=0$.}

\noindent As $\sequencen{u_n}$ is arbitrary,
$f\in(L^{\infty})^{\sim}_c$.   Consequently, for general
$f\in L^{\infty\sim}$,

\Centerline{$f\in (L^{\infty})^{\sim}_c \iff |f|\in(L^{\infty})^{\sim}_c
\iff |f\restr S|\in S^{\sim}_c\iff f\restr S\in S^{\sim}_c$,}

\noindent and the map $f\mapsto f\restr S:(L^{\infty})^{\sim}_c\to
S^{\sim}_c$ is a Riesz space isomorphism.

\medskip

\quad{\bf (iii)} Similarly, if $f\in L^{\infty\sim}$ is non-negative and
$f\restr S\in S^{\times}$, then whenever $A\subseteq L^{\infty}$ is
non-empty, downwards-directed and has infimum $0$,
$B=\{w:w\in S,\,\exists\,u\in A,\,w\ge u\}$ has infimum $0$, so
$\inf_{u\in A}f(u)\le\inf_{w\in B}f(w)\le 0$ and $f\in L^{\infty\times}$.
As in (ii), it follows that $f\mapsto f\restr S$ is a surjection from
$L^{\infty\times}$ onto $S^{\times}$.
}%end of proof of 363K

\leader{*363L}{Integration with respect to a finitely additive
functional (a)} If $\frak A$ is a Boolean algebra and
$\nu:\frak A\to\Bbb R$ is a bounded additive functional,
then\cmmnt{ by 363K} we
have a corresponding functional $f_{\nu}\in L^{\infty}(\frak A)^*$
defined by saying that $f_{\nu}(\chi a)=\nu a$ for every $a\in\frak A$.
There are contexts in which it is convenient\cmmnt{, and even
helpful,} to use
the formula $\dashint u\,d\nu$ in place of $f_{\nu}(u)$ for
$u\in L^{\infty}=L^{\infty}(\frak A)$.   \cmmnt{When doing so, we must
of course remember
that we may have lost some of the standard properties of `integration'.
But enough of our intuitions (including, for instance, the idea of
stochastic independence)
%exercise?
remain valid to make the formula a guide to interesting ideas.}
% have seen  \dashint  called the "Bartle integration operator"
% or "horizontal integral"

\spheader 363Lb Let $M$ be the $L$-space of bounded finitely additive
functionals on $\frak A$\cmmnt{ (362B)}.   Then we have a function
$(u,\nu)\mapsto\dashint u\,d\nu:L^{\infty}\times M\to\Bbb R$.
Now this map
is bilinear.   \prooflet{\Prf\ For $\mu$, $\nu\in M$, $u$,
$v\in L^{\infty}$ and $\alpha\in\Bbb R$,

\Centerline{$\dashint u+v\,d\nu=\dashint u\,d\nu+\dashint v\,d\nu$,
\quad$\dashint\alpha u\,d\nu=\alpha\dashint u\,d\nu$}

\noindent just because $f_{\nu}$ is linear.   On the other side, we have

\Centerline{$(f_{\mu}+f_{\nu})(\chi a)
=f_{\mu}(\chi a)+f_{\nu}(\chi a)=\mu a+\nu a
=(\mu+\nu)(a)=f_{\mu+\nu}(\chi a)$}

\noindent for every $a\in\frak A$, so that $f_{\mu}+f_{\nu}$ and
$f_{\mu+\nu}$ must agree on $S(\frak A)$ and therefore on $L^{\infty}$.
But this means that
$\dashint u\,d(\mu+\nu)=\dashint u\,d\mu+\dashint u\,d\nu$.   Similarly,
$\dashint u\,d(\alpha\mu)=\alpha\dashint u\,d\mu$.\ \Qed}

\spheader 363Lc If $\nu$ is non-negative, we have
$\dashint u\,d\nu\ge 0$ whenever $u\ge 0$\cmmnt{, as in part (c) of the
proof of 363K}.   \cmmnt{Consequently, for any $\nu\in M$ and
$u\in L^{\infty}$,

$$\eqalign{|\dashint u\,d\nu|
&=|\dashint u^+\,d\nu^+-\dashint u^-d\nu^+
  -\dashint u^+d\nu^-+\dashint u^-d\nu^-|\cr
&\le\dashint u^+\,d\nu^++\dashint u^-d\nu^++\dashint u^+d\nu^-+\dashint u^-d\nu^-\cr
&=\dashint|u|d|\nu|
\le\dashint\|u\|_{\infty}\chi 1\,d|\nu|
=\|u\|_{\infty}|\nu|(1)
=\|u\|_{\infty}\|\nu\|.\cr}$$

\noindent So} $(u,\nu)\mapsto\dashint u\,d\nu$ has norm\cmmnt{ (as defined
in 253Ab)} at most $1$.   If $\frak A\ne 0$, the norm is exactly $1$.
\cmmnt{(For this we need to know that there is a $\nu\in M^+$ such
that $\nu 1=1$.   Take any $z$ in the Stone space of $\frak A$ and set
$\nu a=1$ if $z\in\widehat{a}$, $0$ otherwise.)}

\cmmnt{\spheader 363Ld We do not have any result corresponding to
B.Levi's theorem in this language, because (even if
$\nu$ is non-negative and countably additive) there is no reason to
suppose that $\sup_{n\in\Bbb N}u_n$ is defined in $L^{\infty}$ just
because $\sup_{n\in\Bbb N}\dashint u_nd\nu$ is finite.   But if $\nu$ is
countably additive and $\frak A$ is Dedekind $\sigma$-complete, we have
something corresponding to
Lebesgue's Dominated Convergence Theorem (363Yg).}

\spheader 363Le \cmmnt{One formula which we can imitate in the
present context is that of 252O, where the ordinary integral is
represented in the form

\Centerline{$\dashint fd\mu=\int_0^{\infty}\mu\{x:f(x)\ge t\}dt$}

\noindent for non-negative $f$.
In the context of general Boolean algebras, we cannot directly
represent the set $\Bvalue{f\ge t}=\{x:f(x)\ge t\}$ (though in the next
section I will show that in Dedekind $\sigma$-complete Boolean algebras
there is an effective expression of this idea).
But what we can say is the following.}
If $\frak A$ is any Boolean algebra, and
$\nu:\frak A\to\coint{0,\infty}$ is a non-negative additive functional,
and $u\in L^{\infty}(\frak A)^+$, then

\Centerline{$\dashint u\,d\nu
=\int_0^{\infty}\sup\{\nu a:t\chi a\le u\}dt$\dvro{.}{,}}

\noindent\cmmnt{where the right-hand integral is taken with respect
to Lebesgue measure.}   \prooflet{\Prf\ (i) For $t\ge 0$ set
$h(t)=\sup\{\nu a:t\chi a\le u\}$.   Then $h$ is non-increasing and zero
for $t>\|u\|_{\infty}$, so $\int_0^{\infty}h(t)dt$ is defined in
$\Bbb R$.   If we set $h_n(t)=h(2^{-n}(k+1))$ whenever $k$, $n\in\Bbb N$
and $2^{-n}k\le t<2^{-n}(k+1)$, then $\sequencen{h_n(t)}$ is a
non-decreasing sequence which converges to $h(t)$ whenever $h$ is
continuous at $t$, which is almost everywhere (222A, or otherwise);  so
$\int_0^{\infty}h(t)dt=\lim_{n\to\infty}\int_0^{\infty}h_n(t)dt$.
Next, given $n\in\Bbb N$ and $\epsilon>0$, we can choose for each
$k\le k^*=\lfloor 2^n\|u\|_{\infty}\rfloor$ an $a_k$ such that
$2^{-n}(k+1)\chi a_k\le u$ and $\nu a_k\ge h(2^{-n}(k+1))-\epsilon$.
In this case $\sum_{k=0}^{k^*}2^{-n}\chi a_k\le u$, so

$$\eqalign{\int_0^{\infty}h_n(t)dt
&=2^{-n}\sum_{k=0}^{k^*}h(2^{-n}(k+1))
\le\|u\|_{\infty}\epsilon+2^{-n}\sum_{k=0}^{k^*}\nu a_k\cr
&=\|u\|_{\infty}\epsilon+\dashint\sum_{k=0}^{k^*}2^{-n}\chi a_kd\nu
\le\|u\|_{\infty}\epsilon+\dashint u\,d\nu.\cr}$$

\noindent As $n$ and $\epsilon$ are arbitrary,
$\int_0^{\infty}h(t)dt\le\dashint u\,d\nu$.   (ii) In the other direction,
there is for any $\epsilon>0$ a $v\in S(\frak A)$ such that
$v\le u\le v+\epsilon\chi 1$.   If we express $v$ as
$\sum_{j=0}^m\gamma_j\chi c_j$ where $c_0\Bsupseteq\ldots\Bsupseteq c_m$
and $\gamma_j\ge 0$ for every $j$ (361Ec), then we shall have
$h(t)\ge\nu c_k$ whenever $t\le\sum_{j=0}^k\gamma_j$, so

\Centerline{$\int_0^{\infty}h(t)dt\ge\sum_{k=0}^m\gamma_k\nu c_k
=\dashint v\,d\nu\ge\dashint u\,d\nu-\epsilon\nu 1$.}

\noindent As $\epsilon$ is arbitrary,
$\int_0^{\infty}h(t)dt\ge\dashint u\,d\nu$ and the two `integrals' are
equal.\ \Qed}

\cmmnt{\spheader 363Lf The formula $\dashint\,d\nu$ is especially natural
when $\frak A$ is an algebra of sets, so that $L^{\infty}$ can be
directly interpreted as a space of functions (363Ha);  better still,
when $\frak A$ is actually a $\sigma$-algebra of subsets of a set $X$,
$L^{\infty}$ can be identified with the space of bounded
$\frak A$-measurable functions on $X$, as in 363Hb.   So in such
contexts I may write $\dashint g\,d\nu$ or even $\dashint g(x)\nu(dx)$
when $g:X\to\Bbb R$ is bounded and
$\frak A$-measurable, and $\nu:\frak A\to\Bbb R$ is a bounded additive
functional.   But I will try to take care to signal any such deviation
from the normal principle that the symbol $\int$ refers to the
sequentially order-continuous integral defined in \S122 with the minor
modifications introduced in \S\S133 and 135.}

\vleader{108pt}{363M}{}\cmmnt{ Now I come to a fundamental fact underlying a
number of theorems in both this volume and the last.

\medskip

\noindent}{\bf Theorem} Let $\frak A$ be a Boolean algebra.

(a) $\frak A$ is Dedekind $\sigma$-complete iff $L^{\infty}(\frak A)$ is
Dedekind $\sigma$-complete.

(b) $\frak A$ is Dedekind complete iff $L^{\infty}(\frak A)$ is Dedekind
complete.

\proof{{\bf (a)(i)} Suppose that $\frak A$ is Dedekind
$\sigma$-complete.   By 314M, we may identify $\frak A$ with a quotient
$\Sigma/\Cal M$, where $\Cal M$ is the ideal of meager subsets of the
Stone space $Z$ of $\frak A$, and
$\Sigma=\{E\symmdiff A:E\in\Cal E,\,A\in\Cal M\}$, writing
$\Cal E=\{\widehat a:a\in\frak A\}$ for the
algebra of open-and-closed subsets of $Z$.   By 363Hb,
$L^{\infty}=L^{\infty}(\frak A)$ can be identified with
$\eusm L^{\infty}/\eusm V$, where $\eusm L^{\infty}$ is the space of
bounded $\Sigma$-measurable functions from $Z$ to $\Bbb R$, and
$\eusm V$ is the space of functions zero except on a member of $\Cal M$.

Now suppose that $\sequencen{u_n}$ is a sequence in $L^{\infty}$ with an
upper bound $u\in L^{\infty}$.   Express $u_n$, $u$ as
$f_n^{\ssbullet}$, $f^{\ssbullet}$ where $f_n$, $f\in\eusm L^{\infty}$.
Set $g(z)=\sup_{n\in\Bbb N}\min(f_n(z),f(z))$ for every $z\in Z$;  then
$g\in\eusm L^{\infty}$ (121F), so we have a corresponding member
$v=g^{\ssbullet}$ of $L^{\infty}$.   For each $n\in\Bbb N$, $u\ge u_n$
so $(f_n-f)^+\in\eusm V$,

\Centerline{$\{z:f_n(z)>g(z)\}\subseteq\{z:f_n(z)>f(z)\}\in\Cal M$}

\noindent and $v\ge u_n$.   If $w\in L^{\infty}$ and $w\ge u_n$ for
every $n$, then express $w$ as $h^{\ssbullet}$ where
$h\in\eusm L^{\infty}$;  we have $(f_n-h)^+\in\eusm V$ for every $n$, so

\Centerline{$\{z:g(z)>h(z)\}
\subseteq\bigcup_{n\in\Bbb N}\{z:f_n(z)>h(z)\}\in\Cal M$}

\noindent because $\Cal M$ is a $\sigma$-ideal, and $(g-h)^+\in\eusm V$,
so $w\ge v$.   Thus $v=\sup_{n\in\Bbb N}u_n$ in $L^{\infty}$.   As
$\sequencen{u_n}$ is arbitrary, $L^{\infty}$ is Dedekind
$\sigma$-complete (using 353H).

\medskip

\quad{\bf (ii)} Now suppose that $L^{\infty}$ is Dedekind
$\sigma$-complete, and that $A$ is a countable non-empty set in
$\frak A$.   In this case $\{\chi a:a\in A\}$ has a least upper bound
$u$ in $L^{\infty}$.   Take $v\in S(\frak A)$ such that $0\le v\le u$
and $\|u-v\|_{\infty}\le\bover13$;  set $b=\Bvalue{v>\bover13}$, as
defined in 361Eg.   If $a\in A$, then
$\|(\chi a-v)^+\|_{\infty}\le\|u-v\|_{\infty}\le\bover13$, so
$\bover23\chi a\le v$ and
$a\Bsubseteq b$.   If $c\in\frak A$ is any upper bound for $A$, then
$v\le u\le\chi c$ so $b\Bsubseteq c$.   Thus $b=\sup A$ in $\frak A$.
As $A$ is arbitrary, $\frak A$ is Dedekind $\sigma$-complete.

\medskip

{\bf (b)(i)} For the second half of this theorem I use an argument which
depends on joining the representation described in (a-i) above with the
original definition of $L^{\infty}$ in 363A.   The point is that
$C(Z)\subseteq\eusm L^{\infty}$, and for any
$f\in C(Z)=L^{\infty}(\frak A)$ its equivalence class $f^{\ssbullet}$ in
$\eusm L^{\infty}/\eusm V$
corresponds to $f$ itself.   \Prf\ Perhaps it will help to give a name
$T$ to the canonical isomorphism from $\eusm L^{\infty}/\eusm V$ to
$L^{\infty}$.   Then $V=\{f:Tf^{\ssbullet}=f\}$ is a
closed linear subspace of $C(Z)$, because $f\mapsto f^{\ssbullet}$ and
$T$ are continuous linear operators.   But if $a\in\frak A$, then
$(\widehat a)^{\ssbullet}$, the equivalence class of
$\widehat a\in\Sigma$ in $\Sigma/\Cal M$, corresponds to $a$ (see the
proof of 314M), so
$(\chi\widehat a)^{\ssbullet}\in\eusm L^{\infty}/\eusm V$
corresponds to $\chi a$;  that is,
$T(\chi\widehat a)^{\ssbullet}=\chi\widehat a$, if we identify
$\chi a\in L^{\infty}$ with $\chi\widehat a:Z\to\{0,1\}$.   So $V$
contains $\chi\widehat a$ for every $a\in\frak A$;  because $V$ is a
linear subspace, $S(\frak A)\subseteq V$;  because $V$ is closed,
$L^{\infty}\subseteq V$.\ \Qed

For a general $f\in\eusm L^{\infty}$, $g=Tf^{\ssbullet}$ must be the
unique member of $C(Z)$ such that $g^{\ssbullet}=f^{\ssbullet}$, that
is, such that $\{z:g(z)\ne f(z)\}$ is meager.

\medskip

\quad{\bf (ii)} Suppose now that $\frak A$ is actually Dedekind
complete.   In this case $Z$ is extremally disconnected (314S).
Consequently every open set belongs to $\Sigma$.   \Prf\ If $G$ is open,
then $\overline{G}$ is open-and-closed;  but $A=\overline{G}\setminus G$
is a closed set with empty interior, so is meager, and
$G=\overline{G}\symmdiff A\in\Sigma$.\ \Qed

Let $A\subseteq L^{\infty}=C(Z)$ be any
non-empty set with an upper bound in $C(Z)$.    For each $z\in Z$ set
$g(z)=\sup_{u\in A}u(z)$.   Then

\Centerline{$G_{\alpha}=\{z:g(z)>\alpha\}
=\bigcup_{u\in A}\{z:u(z)>\alpha\}$}

\noindent is open for every $\alpha\in\Bbb R$ (that is, $g$ is lower
semi-continuous).   Thus $G_{\alpha}\in \Sigma$ for every $\alpha$, so
$g\in\eusm L^{\infty}$, and $v=Tg^{\ssbullet}$ is defined in $C(Z)$.
For any $u\in A$, $g\ge u$ in $\eusm L^{\infty}$, so

\Centerline{$v=Tg^{\ssbullet}\ge Tu^{\ssbullet}=u$}

\noindent in $L^{\infty}$;  thus $v$ is an upper bound for $A$ in
$L^{\infty}$.   On the other hand, if $w$ is any upper bound for $A$ in
$L^{\infty}=C(Z)$, then surely $w(z)\ge u(z)$ for every $z\in Z$ and
$u\in A$, so $w\ge g$ and

\Centerline{$w=Tw^{\ssbullet}\ge Tg^{\ssbullet}=v$.}

\noindent This means that $v$ is the least upper bound of $A$.   As $A$
is arbitrary, $L^{\infty}$ is Dedekind complete.

\medskip

\quad{\bf (iii)} Finally, if $L^{\infty}$ is Dedekind complete, then the
argument of (a-ii), applied to arbitrary non-empty subsets $A$ of
$\frak A$, shows that $\frak A$ also is Dedekind complete.
}%end of proof of 363M

\leader{363N}{}\cmmnt{ Much of the importance of $L^{\infty}$ spaces
in the theory of Riesz spaces arises from the next result.

\medskip

\noindent}{\bf Proposition} Let $U$ be a Dedekind $\sigma$-complete
Riesz space with an order unit.   Then $U$ is isomorphic, as Riesz
space, to $L^{\infty}(\frak A)$, where $\frak A$ is the algebra of
projection bands in $U$.

\proof{{\bf (a)} By 353N, $U$ is isomorphic to a norm-dense Riesz
subspace of $C(X)$ for some compact Hausdorff space $X$;  for the rest
of this argument, therefore, we may suppose that $U$ actually is such a
subspace.

\medskip

{\bf (b)} $U=C(X)$.   \Prf\ If $g\in C(X)$ then by 354I there are
sequences $\sequencen{f_n}$, $\sequencen{f'_n}$ in $U$ such that
$f_n\le g\le g_n$ and $\|g_n-f_n\|_{\infty}\le 2^{-n}$ for every $n$.
Now $\{f_n:n\in\Bbb N\}$ has a least upper bound $f$ in $U$;  since we
must have $f_n\le f\le g_n$ for every $n$, $f=g$ and $g\in U$.\ \Qed

\medskip

{\bf (c)} Next, $X$ is zero-dimensional.   \Prf\ Suppose that
$G\subseteq X$ is open and $x\in G$.   Then there is an open set $G_1$
such that $x\in G_1\subseteq\overline{G}_1\subseteq G$ (3A3Bb).
There is an $f\in C(X)$ such that $0\le f\le\chi G_1$ and $f(x)>0$
(also by 3A3Bb);  write $H$ for $\{y:f(y)>0\}$.   Set
$g=\sup_{n\in\Bbb N}(nf\wedge\chi X)$, the supremum being taken in
$U=C(X)$.   For each
$y\in H$, we must have $g(y)\ge\min(1,nf(y))$ for every $n$, so that
$g(y)=1$.   On the other hand, if $y\in X\setminus\overline{H}$, there
is an $h\in C(X)$ such that $h(y)>0$ and $0\le
h\le\chi(X\setminus\overline{H})$;  now $h\wedge f=0$ so $h\wedge g=0$
and $g(y)=0$.   Thus $\chi H\le g\le\chi\overline{H}$.   The set
$\{y:g(y)\in\{0,1\}\}$ is closed and includes
$H\cup(X\setminus\overline{H})$ so must be the whole of $X$;  thus
$G_2=\{y:g(y)>\bover12\}=\{y:g(y)\ge\bover12\}$ is open-and-closed, and
we have

\Centerline{$x\in H\subseteq
G_2\subseteq\overline{H}\subseteq\overline{G}_1\subseteq G$.}

\noindent As $x$, $G$ are arbitrary, the set of open-and-closed subsets
of $X$ is a base for the topology of $X$, and $X$ is
zero-dimensional.\ \Qed

\medskip

{\bf (d)} We can therefore identify $X$ with the Stone space of its
algebra $\Cal E$ of open-and-closed sets (311J).   But in this case 363A
immediately identifies $U=C(X)$ with $L^{\infty}(\Cal E)$.   By 363J,
$\Cal E$ is isomorphic to $\frak A$, so $U\cong L^{\infty}(\frak A)$.
}%end of proof of 363N

\cmmnt{\medskip

\noindent{\bf Remark} Note that in part (c) of the argument above, we
have to take care over the interpretation of `sup'.   In the
space of all real-valued functions on $X$, the supremum of
$\{nf\wedge\chi X:n\in\Bbb N\}$ is just $\chi H$.   But $g$ is supposed
to be the least {\it continuous} function greater than or equal to
$nf\wedge\chi X$ for every $n$, and is therefore likely to be strictly
greater than $\chi H$, even though sandwiched between $\chi H$ and
$\chi\overline{H}$.
}%end of comment

\leader{363O}{Corollary} Let $U$ be a Dedekind $\sigma$-complete
$M$-space.   Then $U$ is isomorphic, as Banach lattice, to
$L^{\infty}(\frak A)$, where $\frak A$ is the algebra of projection
bands of $U$.

\proof{ This is merely the special case of 363N in which $U$ is known
from the start to be complete under an order-unit norm.}

\leader{363P}{Corollary} Let $U$ be any Dedekind $\sigma$-complete Riesz
space and $e\in U^+$.   Then the solid linear subspace $U_e$ of $U$
generated by $e$ is isomorphic, as Riesz space, to $L^{\infty}(\frak A)$
for some Dedekind $\sigma$-complete Boolean algebra $\frak A$;  and if
$U$ is Dedekind complete, so is $\frak A$.

\proof{ Because $U$ is Dedekind
$\sigma$-complete, so is $U_e$ (353K(a-i)).   Apply 363N to $U_e$ to see
that $U_e\cong L^{\infty}(\frak A)$
for some $\frak A$.    Because $U_e$ is Dedekind $\sigma$-complete, so
is $\frak A$, by
363Ma;  while if $U$ is Dedekind complete, so are $U_e$ and $\frak A$,
by 353K(b-i) and 363Mb.
}%end of proof of 363P

\leader{363Q}{}\cmmnt{ The next theorem will be a
striking characterization of
the Dedekind complete $L^{\infty}$ spaces as normed spaces.   As a
warming-up exercise I give a much simpler result concerning their nature
as Banach lattices.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Dedekind complete Boolean
algebra.   Then for any Banach lattice $U$, a linear operator $T:U\to
L^{\infty}=L^{\infty}(\frak A)$ is continuous iff it is order-bounded,
and in this case $\|T\|=\||T|\|$, where the modulus $|T|$ is taken in
$\eurm L^{\sim}(U;L^{\infty})$.

\proof{ It is generally true that order-bounded operators between Banach
lattices are continuous (355C).   If $T:U\to L^{\infty}$ is continuous,
then for any $w\in U^+$

\Centerline{$|u|\le w\Longrightarrow\|u\|\le\|w\|
\Longrightarrow \|Tu\|_{\infty}\le\|T\|\|w\|
\Longrightarrow |Tu|\le\|T\|\|w\|e e$,}

\noindent where $e$ is the standard order unit of $L^{\infty}$.
So $T$ is order-bounded.
As $L^{\infty}$ is Dedekind complete (363Mb), $|T|$ is defined in
$\eurm L^{\sim}(U;L^{\infty})$ (355Ea).   For any $w\in U$,

\Centerline{$|T||w|=\sup\{|Tu|:|u|\le|w|\}\le\|T\|\|w\|e$,}

\noindent so $\||T|(w)\|\le\|T\|\|w\|$;  accordingly $\||T|\|\le\|T\|$.
On the other hand, of course,

\Centerline{$|Tw|\le|T||w|\le\||T|\|\|w\|e$}

\noindent for every $w\in U$, so $\|T\|\le\||T|\|$ and the two norms are
equal.
}%end of proof of 363Q

\cmmnt{\medskip

\noindent{\bf Remark} Of course what is happening here is that the
spaces $L^{\infty}(\frak A)$, for Dedekind complete $\frak A$, are just
the Dedekind complete $M$-spaces;  this is an elementary consequence of
363N and 363M.}%end of comment

\leader{363R}{}\cmmnt{ Now for something much deeper.

\medskip

\noindent}{\bf Theorem} Let $U$ be a normed space over $\Bbb R$.   Then
the following are equiveridical:

(i)  there is a Dedekind complete Boolean algebra $\frak A$ such that
$U$ is isomorphic, as normed space, to $L^{\infty}(\frak A)$;

(ii) whenever $V$ is a normed space, $V_0$ a linear subspace of $V$, and
$T_0:V_0\to U$ is a bounded linear operator, there is an extension of
$T_0$ to a bounded linear operator $T:V\to U$ with $\|T\|=\|T_0\|$.

\proof{ For the purposes of the argument below, let us say that a normed
space $U$ satisfying the condition (ii) has the `Hahn-Banach
property'.

\medskip

{\bf Part A: (i)$\Rightarrow$(ii)} I have to show that $L^{\infty}(\frak
A)$ has the Hahn-Banach property for every Dedekind complete Boolean
algebra $\frak A$.   Let $V$ be a normed space, $V_0$ a linear subspace
of $V$, and $T_0:V_0\to L^{\infty}=L^{\infty}(\frak A)$ a bounded linear
operator.   Set $\gamma=\|T_0\|$.

Let $\frak P$ be the set of all functions $T$ such that $\dom T$ is a
linear subspace of $V$ including $V_0$ and $T:\dom T\to U$ is a bounded
linear operator extending $T_0$ and with norm at most $\gamma$.   Order
$\frak P$ by saying that $T_1\le T_2$ if $T_2$ extends $T_1$.   Then any
non-empty totally ordered subset $\frak Q$ of $\frak P$ has an upper
bound in $\frak P$.   \Prf\ Set $\dom T=\bigcup\{\dom T_1:T_1\in\frak
Q\}$, $Tv=T_1v$ whenever $T_1\in\frak Q$ and $v\in\dom T_1$;  it is
elementary to check that $T\in\frak P$, so that $T$ is an upper bound
for $\frak Q$ in $\frak P$.\ \Qed

By Zorn's Lemma, $\frak P$ has a maximal element $\tilde T$.   Now
$\dom\tilde T=V$.   \Prf\Quer\ Suppose, if possible, otherwise.   Write
$\tilde V=\dom \tilde T$ and take any $\tilde v\in V\setminus \tilde V$;
let $V_1$ be the linear span of $\tilde V\cup\{\tilde v\}$, that is,
$\{v+\alpha\tilde v:v\in \tilde V,\,\alpha\in\Bbb R\}$.

If $v_1$, $v_2\in \tilde V$ then, writing $e$ for the standard order
unit of $L^{\infty}$,

$$\eqalign{\tilde Tv_1+\tilde Tv_2
&=\tilde T(v_1+v_2)
\le\|\tilde T(v_1+v_2)\|_{\infty}e\cr
&\le\gamma\|v_1+v_2\|e
\le\gamma\|v_1-\tilde v\|e+\gamma\|v_2+\tilde v\|e,\cr}$$

\noindent so

\Centerline{$\tilde Tv_1-\gamma\|v_1-\tilde v\|e\le\gamma\|v_2+\tilde
v\|e-\tilde Tv_2$.}

\noindent Because $L^{\infty}$ is Dedekind complete (363Mb),

\Centerline{$\tilde u
=\sup_{v_1\in \tilde V}\tilde Tv_1-\gamma\|v_1-\tilde v\|e$}

\noindent is defined in $L^{\infty}$ and $\tilde u\le\gamma\|v_2+\tilde
v\|e-Tv_2$
for every $v_2\in \tilde V$.   Putting these together, we have

\Centerline{$\tilde Tv+\tilde u\le\gamma\|v+\tilde v\|e$,
\quad $\tilde Tv-\tilde u\le\gamma\|v-\tilde v\|e$}

\noindent for all $v\in \tilde V$.   Consequently, if $v\in \tilde V$,
then for $\alpha>0$

\Centerline{$\tilde Tv+\alpha \tilde u
=\alpha(\tilde T(\bover1{\alpha}v)+\tilde u)
\le\alpha\gamma\|\bover1{\alpha}v+\tilde v\|e
=\gamma\|v+\alpha \tilde v\|e$,}

\noindent while for $\alpha<0$

\Centerline{$\tilde Tv+\alpha \tilde u
=|\alpha|(\tilde T(-\bover1{\alpha}v)-\tilde u)
\le|\alpha|\gamma\|-\bover1{\alpha}v-\tilde v\|e
=\gamma\|v+\alpha \tilde v\|e$,}

\noindent and of course

\Centerline{$\tilde Tv\le\|\tilde Tv\|_{\infty}e\le\gamma\|v\|e$.}

\noindent So we have

\Centerline{$\tilde Tv+\alpha \tilde u\le\gamma\|v+\alpha \tilde v\|e$}

\noindent for every $v\in \tilde V$, $\alpha\in\Bbb R$.

Define $T_1:V_1\to L^{\infty}$ by setting $T_1(v+\alpha \tilde v)
=\tilde Tv+\alpha
\tilde u$ for every $v\in \tilde V$, $\alpha\in\Bbb R$.   (This is
well-defined because $\tilde v\notin\tilde V$, so any member of $V_1$ is
uniquely expressible as $v+\alpha\tilde v$ where $v\in\tilde V$ and
$\alpha\in\Bbb R$.)    Then $T_1$ is a linear
operator, extending $T_0$, from a linear subspace of $V$ to
$L^{\infty}$.
But from the calculations above we know that $T_1v\le\gamma\|v\|e$ for
every $v\in V_1$;  since we also have

\Centerline{$T_1v=-T_1(-v)\ge-\gamma\|-v\|e=-\gamma\|v\|e$,}

\noindent $\|T_1v\|_{\infty}\le\gamma\|v\|$ for every $v\in V_1$, and
$T_1\in\frak P$.   But now $T_1$ is a member of $\frak P$ properly
extending $\tilde T$, which is supposed to be impossible.\ \Bang\Qed

Accordingly $\tilde T:V\to L^{\infty}$ is an extension of $T_0$ to the
whole of $V$, with the same norm as $T_0$.   As $V$ and $T_0$ are
arbitrary, $L^{\infty}$ has the Hahn-Banach property.

\medskip

{\bf Part B: (ii)$\Rightarrow$(i)} Now let $U$ be a normed space with
the Hahn-Banach property.   If $U=\{0\}$ then of course it is isomorphic
to $L^{\infty}(\frak A)$, where $\frak A=\{0\}$, so henceforth I will
take it for granted that $U\ne\{0\}$.

\medskip

{\bf (a)} Let $Z$ be the unit ball of the dual $U^*$ of $U$, with the
weak* topology.   Then $Z$ is a compact Hausdorff space (3A5F).
For $u\in U$ set $Z_u=\{z:z\in Z,\,|z(u)|=\|u\|\}$;  then $Z_u$ is a
closed subset of $Z$ (because $f\mapsto f(u)$ is continuous), and is
non-empty, by the Hahn-Banach theorem (3A5Ab, or Part A above!)
Now let $\frak P$ be the set of those closed sets $X\subseteq Z$ such
that $X\cap Z_u\ne\emptyset$ for every $u\in U$.   If
$\frak Q\subseteq\frak P$ is non-empty and totally ordered, then
$\bigcap\frak Q\in\frak P$, because for any $u\in U$

\Centerline{$\{X\cap Z_u:X\in\frak Q\}$}

\noindent is a downwards-directed family of non-empty compact sets, so
must have non-empty intersection.   By Zorn's Lemma, upside down,
$\frak P$ has a minimal element $X$;  with its relative topology, $X$ is a
compact Hausdorff space.

\medskip

{\bf (b)} We have a linear operator $R:U\to C(X)$ given by setting
$(Ru)(x)=x(u)$ for every $u\in U$, $x\in X$;  because $X\subseteq Z$,
$\|Ru\|_{\infty}\le\|u\|$, and because $X\in\frak P$,
$\|Ru\|_{\infty}=\|u\|$, for every $u\in U$.   Moreover, if
$G\subseteq X$ is a non-empty open set (in the relative topology of $X$)
then
$X\setminus G$ cannot belong to $\frak P$, because $X$ is minimal, so
there is a (non-zero) $u\in U$ such that $|x(u)|<\|u\|$ for every
$x\in X\setminus G$.   Replacing $u$ by $\|u\|^{-1}u$ if need be, we may
suppose that $\|u\|=1$.

What this means is that $W=R[U]$ is a linear subspace of $C(X)$ which is
isomorphic, as normed space, to $U$, and has the property that whenever
$G\subseteq X$ is a non-empty relatively open set there is an $f\in W$
such that $\|f\|_{\infty}=1$ and $|f(x)|<1$ for every $x\in X\setminus G$.
Observe that, because $X\setminus G$ is
compact, there is now some $\alpha<1$ such that $|f(u)|\le\alpha$ for
every $f\in X\setminus G$.

Because $W$ is isomorphic to $U$, it has the Hahn-Banach property.

\medskip

{\bf (c)} Now consider $V=\ell^{\infty}(X)$, $V_0=W$,
$T_0:V_0\to W$ the identity map.   Because $W$ has the Hahn-Banach
property, there is a linear operator $T:\ell^{\infty}(X)\to W$,
extending $T_0$, and of norm $\|T_0\|=1$.

\medskip

{\bf (d)} If $h\in\ell^{\infty}(X)$ and $x_0\in
X\setminus\overline{\{x:h(x)\ne 0\}}$, then $(Th)(x_0)=0$.   \Prf\Quer\
Otherwise, set
$G=\{y:y\in X\setminus\overline{\{x:h(x)\ne 0\}},\,(Th)(y)\ne 0\}$.
This is a non-empty open set in $X$, so there
are $f\in W$, $\alpha<1$ such that $\|f\|_{\infty}=1$
and $|f(x)|\le\alpha$ for every $x\in X\setminus G$.

Because $\|f\|_{\infty}=1$, there must be some $x_1\in X$ such that
$|f(x_1)|=1$,
and of course $x_1\in G$, so that $(Th)(x_1)\ne 0$.   But let
$\delta>0$ be such that $\delta\|h\|_{\infty}\le 1-\alpha$.   Then,
because $h(x)=0$ for $x\in G$,
$|f(x)|+|\delta h(x)|\le 1$
for every $x\in X$, and $\|f+\delta h\|_{\infty}$,
$\|f-\delta h\|_{\infty}$ are both less than or equal to $1$.   As
$Tf=f$ and $\|T\|=1$,
this means that

\Centerline{$\|f+\delta Th\|_{\infty}\le 1$,
\quad $\|f-\delta Th\|_{\infty}\le 1$;}

\noindent consequently

\Centerline{$|f(x_1)|+\delta|(Th)(x_1)|
=\max(|(f+\delta Th)(x_1)|,|(f-\delta Th)(x_1)|)\le 1$.}

\noindent But $|f(x_1)|=1$ and $\delta(Th)(x_1)\ne 0$, so this is
impossible.\ \Bang\Qed

\medskip

{\bf (e)} It follows that $Th=h$ for every $h\in C(X)$.   \Prf\Quer\
Suppose, if possible, otherwise.   Then there is a $\delta>0$ such that
$G=\{x:|(Th)(x)-h(x)|>\delta\}$ is not empty.   Let $f\in W$ be such
that $\|f\|=1$ but $|f(x)|<1$ for every $x\in X\setminus G$.   Then
there is an $x_0\in X$ such that $|f(x_0)|=1$;  of course $x_0$ must
belong to $G$.   Set $f_1=\Bover{h(x_0)}{f(x_0)}f$, so that $f_1\in W$
and $f_1(x_0)=h(x_0)$.   Set

\Centerline{$h_1(x)=\med(h(x)-\delta,f_1(x),h(x)+\delta)$}

\noindent for $x\in X$.   Then $h_1\in C(X)$.   Setting

\Centerline{$H=\{x:|h(x)-h(x_0)|+|f_1(x)-f_1(x_0)|<\delta\}$,}

\noindent $H$ is an open set containing $x_0$ and

\Centerline{$|f_1(x)-h(x)|\le|f_1(x_0)-h(x_0)|+\delta=\delta$,
\quad $h_1(x)=f_1(x)$}

\noindent for
every $x\in H$.   Consequently
$x_0\notin\overline{\{x:(f_1-h_1)(x)\ne 0\}}$, and $T(f_1-h_1)(x_0)=0$,
by (d).   But this means that

\Centerline{$(Th_1)(x_0)=(Tf_1)(x_0)=f_1(x_0)=h(x_0)$,}

\noindent so that

\Centerline{$|h(x_0)-(Th)(x_0)|=|T(h_1-h)(x_0)|
\le\|T(h_1-h)\|_{\infty}\le\|h_1-h\|_{\infty}\le\delta$,}

\noindent which is impossible, because $x_0\in G$.\ \Bang\Qed

\medskip

{\bf (f)} This tells us at once that $W=C(X)$.   But (d) also tells us
that $X$ is extremally disconnected.   \Prf\  Let $G\subseteq X$ be any
open set.   Then $\chi X=\chi G+\chi(X\setminus G)$, so

\Centerline{$\chi X=T(\chi X)=h_1+h_2$,}

\noindent where $h_1=T(\chi G)$, $h_2=T(\chi(X\setminus G))$.   Now
from (d) we see that $h_1$
must be zero on $X\setminus\overline{G}$ while $h_2$ must be zero on
$G$.   Thus we have $h_1(x)=1$ for $x\in G$;  as $h_1$ is continuous,
$h_1(x)=1$ for $x\in\overline{G}$, and $h_1=\chi\overline G$.   Of
course it follows that $\overline{G}$ is open.   As $G$ is arbitrary,
$X$ is extremally disconnected.\ \Qed

\medskip

{\bf (g)} Being
also compact and Hausdorff, therefore regular (3A3Bb), $X$ is
zero-dimensional (3A3Bd).   We may therefore identify $X$
with the Stone space of its regular open algebra $\RO(X)$ (314S),
and $W=C(X)$ with $L^{\infty}(\RO(X))$.
Thus $R:U\to C(X)$ is a Banach space isomorphism between $U$ and
$C(X)\cong L^{\infty}(\RO(X))$;  so $U$ is of the type declared.
}%end of proof of 363R

\leader{363S}{The Banach-Ulam \dvrocolon{problem}}\cmmnt{ At a couple
of points already (232Hc, the notes to \S326) I have remarked on a
problem which was early recognised as a fundamental question in abstract
measure theory.   I now set out some formulations of the problem which
arise naturally from the work done so far.   I will do this by writing
down a list of equiveridical statements;  the `Banach-Ulam problem'
asks whether they are true.

I should remark that this is not generally counted as an `open' problem.
It is in fact believed by most of us that these statements are
independent of the usual axioms of Zermelo-Fraenkel set theory,
including the axiom of choice and even the continuum hypothesis.   As
such, this problem belongs to Volume 5 rather than anywhere earlier, but
its manifestations will become steadily more obtrusive as we continue
through this volume and the next, and I think it will be helpful to
begin collecting them now.
The ideas needed to show that the statements here imply each other are
already accessible;  in particular, they involve no set theory beyond
Zorn's Lemma.   These implications constitute the following theorem,
derived from {\smc Luxemburg 67a}.

\medskip

\noindent}{\bf Theorem} The following statements are equiveridical.

(i) There are a set $X$ and a probability measure $\nu$, with domain
$\Cal PX$, such that $\nu\{x\}=0$ for every $x\in X$.

(ii) There are a localizable measure space $(X,\Sigma,\mu)$ and an
absolutely continuous countably additive functional
$\nu:\Sigma\to\Bbb R$ which is not truly continuous, so has no
Radon-Nikod\'ym derivative\cmmnt{ (definitions:  232Ab, 232Hf)}.

(iii) There are a Dedekind complete Boolean algebra $\frak A$ and a
countably additive functional $\nu:\frak A\to\Bbb R$ which is not
completely additive.

(iv) There is a Dedekind complete Riesz space $U$ such that
$U^{\sim}_c\ne U^{\times}$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} Let $X$ be a set with a probability
measure $\nu$, defined on $\Cal PX$, such that $\nu\{x\}=0$ for every
$x\in X$.   Let $\mu$ be counting measure on $X$.   Then $(X,\Cal
PX,\mu)$ is strictly localizable, and $\nu:\Cal PX\to\Bbb R$ is
countably additive;  also $\nu E=0$ whenever $\mu E$ is finite, so $\nu$
is absolutely continuous with respect to $\mu$.   But if $\mu E<\infty$
then $E$ is finite and $\nu(X\setminus E)=1$, so $\nu$ is not truly
continuous, and has no Radon-Nikod\'ym derivative (232D).

\medskip

{\bf (b)(ii)$\Rightarrow$(iii)} Let $(X,\Sigma,\mu)$ be a localizable
measure space and $\nu:\Sigma\to\Bbb R$ an absolutely continuous
countably additive functional which is not truly continuous.   Let
$(\frak A,\bar\mu)$ be the measure algebra of $\mu$;  then we have an
absolutely continuous countably additive functional
$\bar\nu:\frak A\to\Bbb R$ defined by setting
$\bar\nu E^{\ssbullet}=\nu E$ for every
$E\in\Sigma$ (327C).   Since $\nu$ is not truly continuous, $\bar\nu$ is
not completely additive (327Ce).   Also $\frak A$ is Dedekind complete,
because $\mu$ is localizable, so $\frak A$ and $\bar\nu$ witness (iii).

\medskip

{\bf (c)(iii)$\Rightarrow$(i)} Let $\frak A$ be a Dedekind complete
Boolean algebra and $\nu:\frak A\to\Bbb R$ a countably additive
functional which is not completely additive.   Because $\nu$ is bounded
(326M), therefore expressible as the difference of non-negative
countably additive functionals (326L), there must be a non-negative
countably additive functional $\nuprime$ on $\frak A$ which is not
completely additive.

By 326R, there is a partition of unity $\familyiI{a_i}$ in $\frak A$
such that $\sum_{i\in I}\nuprime a_i<\nuprime1$.   Set
$K=\{i:i\in I,\,\nuprime a_i>0\}$;  then $K$ must be countable, so

\Centerline{$\nuprime(\sup_{i\in I\setminus K}a_i)
=\nuprime1-\nuprime(\sup_{i\in K}a_i)
=\nuprime1-\sum_{i\in K}\nuprime a_i
>0$.}

\noindent For $J\subseteq I$ set
$\mu J=\nuprime(\sup_{i\in J\setminus K}a_i)$;  the supremum is always
defined because $\frak A$ is Dedekind complete.   Because $\nuprime$ is
countably additive and non-negative, so is $\mu$;  because $\nuprime a_i=0$
for $i\in J\setminus K$, $\mu\{i\}=0$ for every $i\in I$.   Multiplying
$\mu$ by a suitable scalar, if need be, $(I,\Cal PI,\mu)$ witnesses that
(i) is true.

\medskip

{\bf (d)(iii)$\Rightarrow$(iv)} If $\frak A$ is a Dedekind complete
Boolean algebra with a countably additive functional which is not
completely additive, then $U=L^{\infty}(\frak A)$ is a Dedekind complete
Riesz space (363Mb) and $U^{\sim}_c\ne U^{\times}$, by 363K (recalling,
as in (c) above, that the functional must be bounded).

\medskip

{\bf (e)(iv)$\Rightarrow$(iii)} Let $U$ be a Dedekind complete Riesz
space such that $U^{\times}\ne U^{\sim}_c$.   Take
$f\in U^{\sim}_c\setminus U^{\times}$;  replacing $f$ by $|f|$ if need be,
we may suppose that $f\ge 0$ is sequentially order-continuous but not
order-continuous (355H, 355I).   Let $A$ be a non-empty
downwards-directed set in $U$, with infimum $0$, such that
$\inf_{u\in A}f(u)>0$ (351Ga).   Take $e\in A$, and consider the solid
linear subspace $U_e$ of $U$ generated by $e$;  write $g$ for the
restriction of $f$ to $U_e$.   Because the embedding of $U_e$ in $U$ is
order-continuous, $g\in (U_e)^{\sim}_c$;  because $A\cap U_e$ is
downwards-directed and has infimum $0$, and

\Centerline{$\inf_{u\in A\cap U_e}g(u)=\inf_{u\in A}f(u)>0$,}

\noindent $g\notin U_e^{\times}$.   But $U_e$ is a Riesz space with
order unit $e$, and is Dedekind complete because $U$ is;  so it
can be identified with $L^{\infty}(\frak A)$ for some Boolean algebra
$\frak A$ (363N), and $\frak A$ is Dedekind complete, by 363M.

Accordingly we have a Dedekind complete Boolean algebra $\frak A$ such
that $L^{\infty}(\frak A)^{\sim}_c\ne L^{\infty}(\frak A)^{\times}$.
By 363K, there is a (bounded) countably additive functional on $\frak A$
which is not completely additive, and (iii) is true.
}%end of proof of 363S

\exercises{\leader{363X}{Basic exercises (a)}
%\spheader 363Xa
Let $\frak A$ be a Boolean algebra and $U$ a Banach algebra.   Let
$\nu:\frak A\to U$ be a bounded additive function and
$T:L^{\infty}(\frak A)\to U$ the corresponding bounded linear operator.
Show that $T$ is multiplicative iff $\nu(a\Bcap b)=\nu a\times\nu b$ for
all $a$, $b\in\frak A$.
%363E

\sqheader 363Xb Let $\frak A$, $\frak B$ be Boolean algebras and
$T:L^{\infty}(\frak A)\to L^{\infty}(\frak A)$ a linear operator.
Show that the following are equiveridical:  (i) there is a Boolean
homomorphism $\pi:\frak A\to\frak A$ such that $T=T_{\pi}$ (ii)
$T(u\times v)=Tu\times Tv$ for all $u$, $v\in L^{\infty}(\frak A)$ (iii)
$T$ is a Riesz homomorphism and $Te_{\frak A}=e_{\frak B}$, where
$e_{\frak A}$ is the standard order unit of $L^{\infty}(\frak A)$.
%363F

\spheader 363Xc Let $\frak A$, $\frak B$ be Boolean algebras and
$T:L^{\infty}(\frak A)\to L^{\infty}(\frak B)$ a Riesz homomorphism.
Show that there are a Boolean homomorphism $\pi:\frak A\to\frak B$ and a
$v\ge 0$ in $L^{\infty}(\frak B)$ such that $Tu=v\times T_{\pi}u$ for
every $u\in L^{\infty}(\frak A)$, where $T_{\pi}$ is the operator
associated with $\pi$ (363F).
%363F, 363Xb


\spheader 363Xd Let $\frak A$ be a Boolean algebra and $\frak C$ a
subalgebra of $\frak A$.   Show that $L^{\infty}(\frak C)$, regarded as
a subspace of $L^{\infty}(\frak A)$ (363Ga), is order-dense in
$L^{\infty}(\frak A)$ iff $\frak C$ is order-dense in $\frak A$.
%363G

\sqheader 363Xe Let $(X,\Sigma,\mu)$ be a measure space with measure
algebra $\frak A$, and $\eusm L^{\infty}$ the space of bounded
$\Sigma$-measurable real-valued functions on $X$.  A {\bf linear
lifting} of $\mu$ is a positive linear operator
$T:L^{\infty}(\frak A)\to\eusm L^{\infty}$ such that
$T(\chi 1_{\frak A})=\chi X$ and
$(Tu)^{\ssbullet}=u$ for every $u\in L^{\infty}(\frak A)$, writing
$f\mapsto f^{\ssbullet}$ for the canonical map from $\eusm L^{\infty}$
to $L^{\infty}(\frak A)$ (363H-363I).   (i) Show that if
$\theta:\frak A\to\Sigma$ is a lifting in the sense of 341A then
$T_{\theta}$, as
defined in 363F, is a linear lifting.   (ii) Show that if
$T:L^{\infty}(\frak A)\to\eusm L^{\infty}$ is a linear lifting, then
there is a corresponding lower density
$\underline{\theta}:\frak A\to\Sigma$ defined by setting
$\underline{\theta}a=\{x:T(\chi a)(x)=1\}$ for each $a\in\frak A$.
(iii) Show that
$\underline{\theta}$, as defined in (ii), is a lifting iff $T$ is a
Riesz homomorphism iff $T$ is multiplicative.
%363I

\spheader 363Xf Let $U$ be any commutative ring with
multiplicative identity $1$.   Show that the set $A$ of
{\bf idempotents} in $U$ (that is, elements $a\in U$ such that $a^2=a$)
is a Boolean algebra with identity $1$, writing $a\Bcap b=ab$,
$1\Bsetminus a=1-a$ for $a$, $b\in A$.
%363J

\spheader 363Xg Let $\frak A$ be a Boolean algebra.   Show that
$\frak A$ is isomorphic to the Boolean algebras of multiplicative
idempotents of $S(\frak A)$ and $L^{\infty}(\frak A)$.
%363J, 363Xf

\spheader 363Xh Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.  (i) Show that for any $u\in L^{\infty}(\frak A)$,
$\alpha\in\Bbb R$ there are elements $\Bvalue{u\ge\alpha}$,
$\Bvalue{u>\alpha}\in\frak A$, where $\Bvalue{u\ge\alpha}$ is the
largest $a\in\frak A$ such that $u\times\chi a\ge\alpha\chi a$,
and $\Bvalue{u>\alpha}=\sup_{\beta>\alpha}\Bvalue{u\ge\beta}$.
(ii) Show that in the context of 363Hb, if $u$ corresponds to
$f^{\ssbullet}$ for $f\in\eusm L^{\infty}$, then
$\Bvalue{u\ge\alpha}=\{x:f(x)\ge\alpha\}^{\ssbullet}$,
$\Bvalue{u>\alpha}=\{x:f(x)>\alpha\}^{\ssbullet}$.   (iii) Show that if
$A\subseteq L^{\infty}$ is non-empty and $v\in L^{\infty}$, then $v=\sup
A$ iff $\Bvalue{v>\alpha}=\sup_{u\in A}\Bvalue{u>\alpha}$ for every
$\alpha\in\Bbb R$;  in particular, $v=u$ iff
$\Bvalue{v>\alpha}=\Bvalue{u>\alpha}$ for every $\alpha$. (iv) Show that
a function $\phi:\Bbb R\to\frak A$ is of the form
$\phi(\alpha)=\Bvalue{u>\alpha}$ iff ($\alpha$)
$\phi(\alpha)=\sup_{\beta>\alpha}\phi(\beta)$ for every $\alpha\in\Bbb
R$ ($\beta$) there is an $M$ such that $\phi(M)=0$, $\phi(-M)=1$.   (v)
Put (iii) and (iv) together to give a proof that $L^{\infty}$ is
Dedekind $\sigma$-complete if $\frak A$ is.
%363M

\spheader 363Xi Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $U\subseteq L^{\infty}(\frak A)$ a (sequentially)
order-closed Riesz subspace containing $\chi 1$.   Show that $U$ can be
identified with $L^{\infty}(\frak B)$ for some (sequentially)
order-closed subalgebra $\frak B\subseteq\frak A$.
\Hint{set $\frak B=\{b:\chi b\in U\}$ and use 363N.}
%363N

\leader{363Y}{Further exercises (a)}
%\spheader 363Ya
Let $\frak A$ be a Boolean algebra.
Given the linear structure, ordering, multiplication and norm of
$S(\frak A)$ as described in \S361, show that a norm completion of
$S(\frak A)$ will serve for $L^{\infty}(\frak A)$ in the sense that all
the results of 363B-363Q can be proved with no use of the axiom of
choice except an occasional appeal to countably many choices in
sequential forms of the theorems.
%363A

\spheader 363Yb Let $\frak A$ be a Boolean algebra.   Show that
$\frak A$ is ccc iff $L^{\infty}(\frak A)$ has the countable sup
property (241Ye, 353Yd).
%363C

\spheader 363Yc Let $X$ be an extremally disconnected topological space,
and $\RO(X)$ its regular open algebra.   Show that there is a natural
isomorphism between $L^{\infty}(\RO(X))$ and $C_b(X)$.
%363E

\spheader 363Yd Let $\frak A$ be a Boolean algebra.   (i) If
$u\in L^{\infty}=L^{\infty}(\frak A)$, show that $|u|=e$, the standard
order unit
of $L^{\infty}$, iff $\max(\|u+v\|_{\infty},\|u-v\|_{\infty})>1$
whenever $v\in L^{\infty}\setminus\{0\}$.   (ii) Show that if $u$,
$v\in L^{\infty}$ then $|u|\wedge|v|=0$ iff $\|\alpha u+v+w\|_{\infty}
\le\max(\|\alpha u+w\|_{\infty},\|v+w\|_{\infty})$ whenever
$\alpha=\pm 1$ and $w\in L^{\infty}$.
(iii) Show that if $T:L^{\infty}\to L^{\infty}$ is a normed space
automorphism then there are a Boolean automorphism
$\pi:\frak A\to\frak A$ and
a $w\in L^{\infty}$ such that $|w|=e$ and $Tu=w\times T_{\pi}u$ for
every $u\in L^{\infty}$.
%363Xb, 363F

\spheader 363Ye Let $X$ be a set, $\Sigma$ an algebra of subsets of $X$,
and $\Cal I$ an ideal in $\Sigma$, and
$\eusm L^{\infty}$ the set of bounded functions $f:X\to\Bbb R$ such that
whenever $\alpha<\beta$ in $\Bbb R$ there is an $E\in\Sigma$ such that
$\{x:f(x)\le\alpha\}\subseteq E\subseteq\{x:f(x)\le\beta\}$, as in 363H.
(i) Show
that $\eusm L^{\infty}=\{g\phi:g\in C(Z)\}$, where $Z$ is the Stone
space of $\Sigma$ and $\phi:X\to Z$ is a function (to be described).
(ii) Show
that $L^{\infty}(\Sigma/\Cal I)$ can be identified, as Banach lattice
and Banach algebra, with $\eusm L^{\infty}/\eusm V$, where $\eusm V$ is
the set of those functions $f\in\eusm L^{\infty}$ such that for every
$\epsilon>0$ there is a member of $\Cal I$ including
$\{x:|f(x)|\ge\epsilon\}$.
%363H

\spheader 363Yf\dvAformerly{3{}63Yg}
Let $(X,\Sigma,\mu)$ be a complete probability space
with measure algebra $\frak A$.   Let $\sequencen{\frak B_n}$ be a
non-decreasing sequence of closed subalgebras of $\frak A$ such that
$\frak A$ is the closed subalgebra of itself generated by
$\bigcup_{n\in\Bbb N}\frak B_n$, and set
$\Sigma_n=\{F:F^{\ssbullet}\in\frak B_n\}$ for each $n$.   Let
$P_n:L^1(\mu)\to L^1(\mu\restr\Sigma_n)$ be the conditional expectation
operator for each $n$, so that $P_n\restr L^{\infty}(\mu)$ is a positive
linear operator from $L^{\infty}(\mu)\cong L^{\infty}(\frak A)$ to
$L^{\infty}(\mu\restr\Sigma_n)\cong L^{\infty}(\frak B_n)$.   Suppose
that we are given for each $n$ a lifting $\theta_n:\frak B_n\to\Sigma_n$
and that $\theta_{n+1}b=\theta_nb$ whenever $n\in\Bbb N$ and
$b\in\frak B_n$.
Let $T_n:L^{\infty}(\frak B_n)\to\eusm L^{\infty}$ be the corresponding
linear liftings (363Xe), and $\Cal F$ any non-principal ultrafilter
on $\Bbb N$.   (i) Show that for any $u\in L^{\infty}(\frak A)$,
$\sequencen{T_nP_nu}$ converges almost everywhere.   (ii) For
$u\in L^{\infty}(\frak A)$ set $(Tu)(x)=\lim_{n\to\Cal F}(T_nP_nu)(x)$
for $x\in X$, $u\in L^{\infty}(\frak A)$.   Show that $T$ is a linear
lifting for $\mu$.   (iii) Use 363Xe(ii) and 341J to show that there is a
lifting $\theta$ of $\mu$ extending every $\theta_n$.   (iv) Use this as
the countable-cofinality inductive step in a proof of the Lifting
Theorem (using partial liftings rather than partial lower densities, as
suggested in 341Li).
%363I, 363Xe

\spheader 363Yg Let $\frak A$ be a Boolean algebra and
$\nu:\frak A\to\Bbb R$ a
bounded countably additive functional.   Suppose that
$\sequencen{u_n}$ is an order-bounded sequence in
$L^{\infty}(\frak A)$ such that $\inf_{n\in\Bbb N}\sup_{m\ge n}u_m$
and $\sup_{n\in\Bbb N}\inf_{m\ge n}u_m$ are defined in
$L^{\infty}(\frak A)$ and equal to $u$ say.   Show
that $\int u\,d\nu=\lim_{n\to\infty}\int u_nd\nu$.
%363L

\spheader 363Yh Let $\Sigma$ be the family of those sets
$E\subseteq[0,1]$ such that $\mu(\interior E)=\mu\overline{E}$, where
$\mu$ is Lebesgue measure.   (i) Show that $\Sigma$ is an algebra of
subsets of $[0,1]$ and that every member of $\Sigma$ is Lebesgue
measurable.   (ii) Show that if we identify $L^{\infty}(\Sigma)$ with a
set of real-valued functions on $[0,1]$, as in 363H, then we get just
the space of Riemann integrable functions.   (iii) Show that if we write
$\nu$ for $\mu\restr\Sigma$, then $\dashint d\nu$, as defined in 363L,
is just the Riemann integral.
%363L

\spheader 363Yi\dvAformerly{3{}63Yj}
Let $X$ be a compact Hausdorff
space.   Let us say that a linear subspace $U$ of $C(X)$ is
{\bf $\ell^{\infty}$-{\vthsp}complemented} in $C(X)$ if there is a linear
subspace $V$ such that $C(X)=U\oplus V$ and
$\|u+v\|_{\infty}=\max(\|u\|_{\infty},\|v\|_{\infty})$ for all $u\in U$,
$v\in\ V$.   Show that there is a one-to-one correspondence between such
subspaces $U$ and open-and-closed subsets $E$ of $X$, given by setting
$U=\{u:u\in C(X),\,u(x)=0\Forall x\in X\setminus E\}$.   Hence show
that if $\frak A$ is any Boolean algebra, there is a canonical
isomorphism between $\frak A$ and the partially ordered set of
$\ell^{\infty}$-complemented subspaces of $L^{\infty}(\frak A)$.
%363R mt36bits
}%end of exercises

\endnotes{
\Notesheader{363} As with $S(\frak A)$, I have chosen a definition of
$L^{\infty}(\frak A)$ in terms of the Stone space of $\frak A$;  but as
with $S(\frak A)$, this is optional (363Ya).   By and large the basic
properties of $L^{\infty}$ are derived very naturally from those of $S$.
The spaces $L^{\infty}(\frak A)$, for general Boolean algebras
$\frak A$, are not in fact particularly important;  they have too few
properties not shared by all the spaces $C(X)$ for compact Hausdorff
$X$.   The point at which it becomes helpful to interpret $C(X)$ as
$L^{\infty}(\frak A)$ is when $C(X)$ is Dedekind $\sigma$-complete.
The spaces $X$ for which this is true are difficult to picture,
and alternative representations of $L^{\infty}$ along the lines of
363H-363I can be easier on the imagination.

For Dedekind $\sigma$-complete $\frak A$, there is an alternative
description of members of $L^{\infty}(\frak A)$ in terms of objects
`$\Bvalue{u>\alpha}$' (363Xh);  I will return to this idea in the next
section.   For the moment I remark only that it gives an alternative
approach to 363M not necessarily depending on the representation of
$L^{\infty}$ as a quotient $\eusm L^{\infty}/\eusm V$ nor on an analysis
of a Stone space.   I used a version of such an argument in the proof of
363M which I gave in {\smc Fremlin 74a}, 43D.

I spend so much time on 363M not only because Dedekind completeness is
one of the basic properties of any lattice, but because it offers an
abstract expression of one of the central results of Chapter 24.   In
243H I showed that $L^{\infty}(\mu)$ is always Dedekind
$\sigma$-complete, and that it is Dedekind complete if $\mu$ is
localizable.   We can now relate this to the results of 321H and 322Be:
the measure algebra of any measure is Dedekind $\sigma$-complete,
and the measure algebra of a localizable measure is Dedekind complete.

The ideas of the proof of 363M can of course be rearranged in various
ways.   One uses 353Yb:  for completely regular spaces $X$, $C(X)$
is Dedekind complete iff $X$ is extremally disconnected;  while for
compact Hausdorff spaces, $X$ is extremally disconnected iff it is the
Stone space of a Dedekind complete algebra.   With the right
modification of the concept `extremally disconnected' (314Yf), the
same approach works for Dedekind $\sigma$-completeness.

363R is the `Nachbin-Kelley theorem';  it is
commonly phrased `a normed space $U$ has the
Hahn-Banach extension property iff it is isomorphic, as normed space, to
$C(X)$ for some compact extremally disconnected Hausdorff space $X$',
but the expression in terms of $L^{\infty}$ spaces seems natural in the
present context.   The implication in one direction (Part A of the
proof) calls for nothing but a check through one of the standard proofs
of the Hahn-Banach theorem to make sure that the argument applies in the
generalized form.   Part B of the proof has ideas in it;  I have tried
to set it out in a way suggesting that if you can remember the
construction of the set $X$ the rest is just a matter of a little
ingenuity.

One way of trying to understand the multiple structures of $L^{\infty}$
spaces is by looking at the corresponding automorphisms.   We observe,
for instance, that an operator $T$ from $L^{\infty}(\frak A)$ to itself
is a Banach algebra automorphism iff it is a Banach lattice automorphism
preserving the standard order unit iff it corresponds to an automorphism
of the algebra $\frak A$ (363Xb).   Of course there are Banach space
automorphisms of $L^{\infty}$ which do not respect the order or
multiplicative structure;  but they have to be closely related to
algebra isomorphisms (363Yd).

I devote a couple of exercises (363Xe, 363Yf) to indications of how the
ideas here are relevant to the Lifting Theorem.   If you found the
formulae of the proof of 341G obscure it may help to work through the
parallel argument.

A lecture by W.A.J.Luxemburg on the equivalence between (i) and (iv) in
363S was one of the turning points in my mathematical apprenticeship.
I introduce it here, even though the real importance of the Banach-Ulam
problem lies in the metamathematical ideas it has nourished, because
these formulations provide a focus for questions which arise naturally
in this volume and which otherwise might prove distracting.   The next
group of significant ideas in this context will appear in \S438.
}%end of comment

\discrpage

\frfilename{mt364.tex}
\versiondate{16.7.11}
\copyrightdate{1996}

\def\chaptername{Function spaces}
\def\sectionname{$L^0$}

\def\dottimes{\mskip5mu\dot\times\mskip5mu}
\def\frdotplus{\mskip5mu\dot+\mskip5mu}
\def\smalldashint{\vrule height2.6pt width4.3pt depth-2.4pt
  \mskip-8.5mu\smallint}

\newsection{364}

My next objective is to develop an abstract construction corresponding
to the $L^0(\mu)$ spaces of \S241.   These generalized $L^0$ spaces
will form the basis of the work of the rest of this chapter and also the
next;  partly because their own properties are remarkable, but even
more because they form a framework for the study of Archimedean Riesz
spaces in general (see \S368).   There seem to be significant new
difficulties, and I take the space to describe an approach which
can be made essentially independent of the route through Stone spaces
used in the last three sections\cmmnt{ (364Ya)}.
I embark directly on a definition in
the new language (364A), and relate it to the constructions of \S241
(364B-364D, 364I) and \S\S361-363 (364J).   The ideas of Chapter 27 can
also be expressed in this language;  I make a start on developing the
machinery for this in 364F-364G, with the formula `$\Bvalue{u\in E}$',
`the region in which $u$ belongs to $E$', and some exercises
(364Xe-364Xf).    Following through
the questions addressed in \S363, I discuss Dedekind completeness in
$L^0$ (364L-364M), properties of its multiplication (364N), the
expression of the original algebra in terms of $L^0$ (364O), the action
of Boolean homomorphisms on $L^0$ (364P) and product spaces (364R).   In
364S-364V I describe representations of the $L^0$ space of a regular
open algebra.

\leader{364A}{The set $L^0(\frak A)$ }{\bf (a) Definition}
Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra.   I will write $L^0(\frak A)$ for the
set of all functions $\alpha\mapsto\Bvalue{u>\alpha}:\Bbb R\to\frak A$
such that

\inset{($\alpha$) $\Bvalue{u>\alpha}=\sup_{\beta>\alpha}\Bvalue{u>\beta}$
in $\frak A$ for every $\alpha\in\Bbb R$,

($\beta$) $\inf_{\alpha\in\Bbb R}\Bvalue{u>\alpha}=0$,

($\gamma$) $\sup_{\alpha\in\Bbb R}\Bvalue{u>\alpha}=1$.}

\cmmnt{
\spheader 364Ab My reasons for using the notation
`$\Bvalue{u>\alpha}$' rather than `$u(\alpha)$' will I hope become
clear in the next few paragraphs.   For the moment, if you think of
$\frak A$ as a $\sigma$-algebra of sets and of $L^0(\frak A)$ as the
family of $\frak A$-measurable real-valued functions, then
$\Bvalue{u>\alpha}$ corresponds to the set $\{x:u(x)>\alpha\}$ (364Ia).

\spheader 364Ac Some readers will recognise the formula
`$\Bvalue{\ldots}$' as belonging to the language of forcing, so that
$\Bvalue{u>\alpha}$ could be read as `the Boolean value of the
proposition ``$u>\alpha${''}'.   But a vocalisation closer to my
intention might be `the region where $u>\alpha$'.

\spheader 364Ad Note that condition ($\alpha$) of (a) automatically
ensures that $\Bvalue{u>\alpha}\Bsubseteq\Bvalue{u>\alpha'}$ whenever
$\alpha'\le\alpha$ in $\Bbb R$.
}

\spheader 364Ae In fact it will sometimes be convenient to note that the
conditions of (a) can be replaced by

\inset{($\alpha'$) $\Bvalue{u>\alpha}
=\sup_{q\in\Bbb Q,q>\alpha}\Bvalue{u>q}$ for every
$\alpha\in\Bbb R$,

($\beta'$) $\inf_{n\in\Bbb N}\Bvalue{u>n}=0$,

($\gamma'$) $\sup_{n\in\Bbb N}\Bvalue{u>-n}=1$;}

\noindent the point being that we need look only at suprema and infima
of countable subsets of $\frak A$.

%\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\header{364Af}{\bf *(f)} Indeed, \cmmnt{because the function
$\alpha\mapsto\Bvalue{u>\alpha}$ is determined by its values on $\Bbb Q$,}
we have the option of declaring $L^0(\frak A)$ to be the set of functions
$\alpha\mapsto\Bvalue{u>\alpha}:\Bbb Q\to\frak A$ such that

\inset{($\alpha''$) $\Bvalue{u>q}
=\sup_{q'\in\Bbb Q,q'>q}\Bvalue{u>q'}$ for every
$q\in\Bbb Q$,

($\beta'$) $\inf_{n\in\Bbb N}\Bvalue{u>n}=0$,

($\gamma'$) $\sup_{n\in\Bbb N}\Bvalue{u>-n}=1$.}

\cmmnt{\noindent However I shall hold this in reserve until I come to
forcing constructions in Chapter 55 of Volume 5.}

\cmmnt{
\spheader 364Ag In order to integrate this construction into the framework
of the rest of this book, I
match it with an alternative route to the same object, based on
$\sigma$-algebras and $\sigma$-ideals of sets, as follows.
}%end of comment

\leader{364B}{Proposition} Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, and $\Cal I$ a $\sigma$-ideal of
$\Sigma$.

(a) Write $\eusm L^0=\eusm L^0_{\Sigma}$ for the space of all
$\Sigma$-measurable functions from $X$ to $\Bbb R$.   Then
$\eusm L^0$, with its linear structure, ordering and multiplication
inherited from $\Bbb R^X$, is a Dedekind
$\sigma$-complete $f$-algebra with multiplicative identity.

(b) Set

\Centerline{$\eusm W=\eusm W_{\Cal I}
=\{f:f\in\eusm L^0,\,\{x:f(x)\ne 0\}\in\Cal I\}$.}

\noindent Then

\quad (i) $\eusm W$ is a sequentially order-closed solid linear subspace
and ideal of $\eusm L^0$;

\quad (ii) the quotient space $\eusm L^0/\eusm W$, with its inherited
linear, order and multiplicative structures, is a Dedekind
$\sigma$-complete Riesz space and an $f$-algebra with a multiplicative
identity;

\quad (iii) for $f$, $g\in\eusm L^0$,
$f^{\ssbullet}\le g^{\ssbullet}$ in $\eusm L^0/\eusm W$ iff
$\{x:f(x)>g(x)\}\in\Cal I$,
and $f^{\ssbullet}=g^{\ssbullet}$ in $\eusm L^0/\eusm W$ iff
$\{x:f(x)\ne g(x)\}\in\Cal I$.

\proof{ (Compare 241A-241H.)  %241A 241B 241C 241D 241E 241H

\medskip

{\bf (a)} The point is just that $\eusm L^0$ is a sequentially
order-closed Riesz subspace and subalgebra of $\Bbb R^X$.   The facts we
need to know -- that constant functions belong to $\eusm L^0$, that
$f+g$, $\alpha f$, $f\times g$,
$\sup_{n\in\Bbb N}f_n$ belong to $\eusm L^0$ whenever $f$, $g$, $f_n$ do
and $\{f_n:n\in\Bbb N\}$ is bounded above -- are all covered by
121E-121F.
Its multiplicative identity is of course the constant function $\chi X$.

\medskip

{\bf (b)(i)} The necessary verifications are all elementary.

\medskip

\quad{\bf (ii)} Because $\eusm W$ is a solid linear subspace of the
Riesz space $\eusm L^0$, the quotient inherits a Riesz space structure
(351J, 352Jb);  because $\eusm W$ is an ideal of the ring
$(\eusm L^0,+,\times)$, $\eusm L^0/\eusm W$
inherits a multiplication;  it is a commutative algebra because
$\eusm L^0$ is;  and has a multiplicative identity
$e=\chi X^{\ssbullet}$ because $\chi X$ is the identity of $\eusm L^0$.

To check that $\eusm L^0/\eusm W$ is an $f$-algebra it is enough to
observe that, for any non-negative $f$, $g$ , $h\in\eusm L^0$,

\Centerline{$f^{\ssbullet}\times g^{\ssbullet}
=(f\times g)^{\ssbullet}\ge 0$,}

\noindent and if $f^{\ssbullet}\wedge g^{\ssbullet}=0$ then
$\{x:f(x)>0\}\cap\{x:g(x)>0\}\in\Cal I$, so that
$\{x:f(x)h(x)>0\}\cap\{x:g(x)>0\}\in\Cal I$ and

\Centerline{$(f^{\ssbullet}\times h^{\ssbullet})\wedge g^{\ssbullet}
=(h^{\ssbullet}\times f^{\ssbullet})\wedge g^{\ssbullet}=0$.}

\noindent Finally, $\eusm L^0/\eusm W$ is
Dedekind $\sigma$-complete, by 353K(a-iii).

\medskip

\quad{\bf (iii)}
For $f$, $g\in\eusm L^0$,

\Centerline{$f^{\ssbullet}\le g^{\ssbullet}
\iff (f-g)^+\in\eusm W
\iff\{x:f(x)>g(x)\}=\{x:(f-g)^+(x)\ne 0\}\in\Cal I$}

\noindent (using the fact that the canonical map from $\eusm L^0$ to
$\eusm L^0/\eusm W$ is a Riesz homomorphism, so that
$((f-g)^+)^{\ssbullet}=(f^{\ssbullet}-g^{\ssbullet})^+$).   Similarly

\Centerline{$f^{\ssbullet}=g^{\ssbullet}
\iff f-g\in\eusm W
\iff\{x:f(x)\ne g(x)\}=\{x:(f-g)(x)\ne 0\}\in\Cal I$.}
}%end of proof of 364B

\vleader{108pt}{364C}{\bf Theorem} Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\pi:\Sigma\to\frak A$ a
surjective Boolean homomorphism, with kernel a $\sigma$-ideal $\Cal I$;
define $\eusm L^0=\eusm L^0_{\Sigma}$ and $\eusm W=\eusm W_{\Cal I}$
as in 364B, so that
$U=\eusm L^0/\eusm W$ is a Dedekind $\sigma$-complete $f$-algebra with multiplicative
identity.

(a) We have a canonical bijection  $T:U\to L^0=L^0(\frak A)$ defined by
the formula

\Centerline{$\Bvalue{Tf^{\ssbullet}>\alpha}=\pi\{x:f(x)>\alpha\}$}

\noindent for every $f\in\eusm L^0$ and $\alpha\in\Bbb R$.

(b)(i) For any $u$, $v\in U$,

\Centerline{$\Bvalue{T(u+v)>\alpha}
=\sup_{q\in\Bbb Q}\Bvalue{Tu>q}\Bcap\Bvalue{Tv>\alpha-q}$}

\noindent for every $\alpha\in\Bbb R$.

\quad(ii) For any $u\in U$ and $\gamma>0$,

\Centerline{$\Bvalue{T(\gamma
u)>\alpha}=\Bvalue{Tu>\bover{\alpha}{\gamma}}$}

\noindent for every $\alpha\in\Bbb R$.

\quad(iii) For any $u$, $v\in U$,

\Centerline{$u\le v\iff\Bvalue{Tu>\alpha}\Bsubseteq\Bvalue{Tv>\alpha}$
for
every $\alpha\in\Bbb R$.}

\quad(iv) For any $u$, $v\in U^+$,

\Centerline{$\Bvalue{T(u\times v)>\alpha}
=\sup_{q\in\Bbb Q,q>0}\Bvalue{Tu>q}\Bcap\Bvalue{Tv>\bover{\alpha}q}$}

\noindent for every $\alpha\ge 0$.

\quad(v) Writing $e=(\chi X)^{\ssbullet}$ for the multiplicative
identity of $U$, we have

\Centerline{$\Bvalue{Te>\alpha}=1$ if $\alpha<1$, $0$ if $\alpha\ge 1$.}

\proof{{\bf (a)(i)} Given $f\in \eusm L^0$, set
$\zeta_f(\alpha)=\pi\{x:f(x)>\alpha\}$ for $\alpha\in\Bbb R$.   Then it
is easy to see that $\zeta_f$ satisfies the conditions
($\alpha$)$'$-($\gamma$)$'$ of 364Ae, because $\pi$ is sequentially
order-continuous (313Qb).    Moreover, if $f^{\ssbullet}=g^{\ssbullet}$
in $U$, then

\Centerline{$\zeta_f(\alpha)\Bsymmdiff \zeta_g(\alpha)
=\pi(\{x:f(x)>\alpha\}\symmdiff\{x:g(x)>\alpha\})
=0$}

\noindent for every $\alpha\in\Bbb R$, because

\Centerline{$\{x:f(x)>\alpha\}\symmdiff\{x:g(x)>\alpha\}
\subseteq\{x:f(x)\ne g(x)\}\in\Cal I$,}

\noindent and $\zeta_f=\zeta_g$.   So we have a well-defined member $Tu$
of $L^0$ defined by the given formula, for any $u\in U$.

\medskip

\quad{\bf (ii)} Next, given $w\in L^0$, there is a $u\in \eusm L^0/\eusm
W$ such that $Tu=w$.   \Prf\
For each $q\in\Bbb Q$, choose $F_q\in\Sigma$ such
that $\pi F_q=\Bvalue{w>q}$ in $\frak A$.    Note that if $q'\ge q$ then

\Centerline{$\pi(F_{q'}\setminus
F_q)=\Bvalue{u>q'}\Bsetminus\Bvalue{u>q}=0$,}


\noindent so $F_{q'}\setminus F_q\in\Cal I$.   Set

\Centerline{$H=\bigcup_{q\in\Bbb Q}F_q\setminus\bigcap_{n\in\Bbb
N}\bigcup_{q\in\Bbb Q,q\ge n}F_q\in\Sigma$,}

\noindent and for $x\in X$ set

$$\eqalign{f(x)&=\sup\{q:q\in\Bbb Q,\,x\in F_q\}\text{ if }x\in H,\cr
&=0\text{ otherwise.}\cr}$$

\noindent ($H$ is chosen just to make the formula here give a finite
value for every $x$.)   We have

$$\eqalignno{\pi H
&=\sup_{q\in\Bbb Q}\Bvalue{w>q}\Bsetminus\inf_{n\in\Bbb N}\sup_{q\in\Bbb
Q,q\ge n}\Bvalue{w>q}\cr
&=1_{\frak A}\Bsetminus\inf_{n\in\Bbb
N}\Bvalue{w>n}
=1_{\frak A}\Bsetminus 0_{\frak A}
=1_{\frak A},\cr}$$

\noindent so $X\setminus H\in\Cal I$.
Now, for any $\alpha\in\Bbb R$,

$$\eqalign{\{x:f(x)>\alpha\}
&=\bigcup_{q\in\Bbb Q,q>\alpha}F_q\cup(X\setminus H)
  \text{ if }\alpha<0,\cr
&=\bigcup_{q\in\Bbb Q,q>\alpha}F_q\setminus(X\setminus H)
\text{ if }\alpha\ge 0,\cr}$$

\noindent and in either case belongs to $\Sigma$;  so that $f\in\eusm
L^0$ and $f^{\ssbullet}$ is defined in $L^0$.
Next, for any $\alpha\in\Bbb R$,

$$\eqalign{\Bvalue{Tf^{\ssbullet}>\alpha}
&=\pi\{x:f(x)>\alpha\}
=\pi(\bigcup_{q\in\Bbb Q,q>\alpha}F_q)\cr
&=\sup_{q\in\Bbb Q,q>\alpha}\Bvalue{w>q}
=\Bvalue{w>\alpha},\cr}$$

\noindent and $Tf^{\ssbullet}=w$.\ \Qed

\medskip

\quad{\bf (iii)} Thus $T$ is surjective.   To see that it is injective,
observe that if $f$, $g\in\eusm L^0$, then

$$\eqalign{Tf^{\ssbullet}=Tg^{\ssbullet}
&\Longrightarrow\Bvalue{Tf^{\ssbullet}>\alpha}
  =\Bvalue{Tg^{\ssbullet}>\alpha}
   \text{ for every }\alpha\in\Bbb R\cr
&\Longrightarrow\pi\{x:f(x)>\alpha\}=\pi\{x:g(x)>\alpha\}
   \text{ for every }\alpha\in\Bbb R\cr
&\Longrightarrow\{x:f(x)>\alpha\}\symmdiff\{x:g(x)>\alpha\}\in\Cal I
   \text{ for every }\alpha\in\Bbb R\cr
&\Longrightarrow\{x:f(x)\ne g(x)\}
   =\bigcup_{q\in\Bbb Q}(\{x:f(x)>q\}\symmdiff\{x:g(x)>q\})\in\Cal I\cr
&\Longrightarrow f^{\ssbullet}=g^{\ssbullet}.\cr}$$

\noindent So we have the claimed bijection.

\medskip

{\bf (b)(i)} Let $f$, $g\in\eusm L^0$ be such that $u=f^{\ssbullet}$ and
$v=g^{\ssbullet}$, so that $u+v=(f+g)^{\ssbullet}$.
For any $\alpha\in\Bbb R$,

$$\eqalignno{\Bvalue{T(u+v)>\alpha}
&=\pi\{x:f(x)+g(x)>\alpha\}\cr
&=\pi(\bigcup_{q\in\Bbb Q}\{x:f(x)>q\}\cap\{x:g(x)>\alpha-q\})\cr
&=\sup_{q\in\Bbb Q}\pi\{x:f(x)>q\}\Bcap\pi\{x:g(x)>\alpha-q\}\cr
\noalign{\noindent (because $\pi$ is a sequentially order-continuous
Boolean homomorphism)}
&=\sup_{q\in\Bbb Q}\Bvalue{Tu>q}\Bcap\Bvalue{Tv>\alpha-q}.\cr}$$

\medskip

\quad{\bf (ii)} Let $f\in\eusm L^0$ be such that $f^{\ssbullet}=u$, so
that $(\gamma f)^{\ssbullet}=\gamma u$.   For any $\alpha\in\Bbb R$,

\Centerline{$\Bvalue{T(\gamma u)>\alpha}
=\pi\{x:\gamma f(x)>\alpha\}
=\pi\{x:f(x)>\Bover{\alpha}{\gamma}\}
=\Bvalue{Tu>\bover{\alpha}{\gamma}}$.}

\medskip

\quad{\bf (iii)} Let $f$, $g\in\eusm L^0$ be such that $f^{\ssbullet}=u$
and $g^{\ssbullet}=v$.   Then

$$\eqalignno{u\le v
&\iff\{x:f(x)>g(x)\}\in\Cal I\cr
\noalign{\noindent (see 364B(b-iii))}
&\iff\bigcup_{q\in\Bbb Q}\{x:f(x)>q\ge g(x)\}\in\Cal I\cr
&\iff \{x:f(x)>\alpha\}\setminus\{x:g(x)>\alpha\}\in\Cal I
   \text{ for every }\alpha\in\Bbb R\cr
&\iff \pi\{x:f(x)>\alpha\}\Bsetminus\pi\{x:g(x)>\alpha\}=0
   \text{ for every }\alpha\cr
&\iff \Bvalue{Tu>\alpha}\Bsubseteq\Bvalue{Tv>\alpha}
   \text{ for every }\alpha.\cr}$$

\medskip

\quad {\bf (iv)} Now suppose that $u$, $v\ge 0$, so that they can be
expressed as $f^{\ssbullet}$, $g^{\ssbullet}$ where $f$, $g\ge 0$ in
$\eusm L^0$ (351J), and $u\times v=(f\times g)^{\ssbullet}$.   If
$\alpha\ge0$, then

$$\eqalign{\Bvalue{T(u\times v)>\alpha}
&=\pi(\bigcup_{q\in\Bbb Q,q>0}
   \{x:f(x)>q\}\cap\{x:g(x)>\Bover{\alpha}q\})\cr
&=\sup_{q\in\Bbb Q,q>0}
   \pi\{x:f(x)>q\}\Bcap\pi\{x:g(x)>\Bover{\alpha}q\}\cr
&=\sup_{q\in\Bbb Q,q>0}
   \Bvalue{Tu>q}\Bcap\Bvalue{Tv>\bover{\alpha}q}.\cr}$$

\medskip

\quad{\bf (v)} This is trivial, because

$$\eqalign{\Bvalue{T(\chi X)^{\ssbullet}>\alpha}
&=\pi\{x:(\chi X)(x)>\alpha\}\cr
&=\pi X=1\text{ if }\alpha<1,\cr
&=\pi\emptyset=0\text{ if }\alpha\ge 1.\cr}$$
}%end of proof of 364C

\leader{364D}{Theorem} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra.   Then $L^0=L^0(\frak A)$ has the structure of a
Dedekind $\sigma$-complete $f$-algebra with multiplicative identity $e$,
defined by saying

\Centerline{$\Bvalue{u+v>\alpha}
=\sup_{q\in\Bbb Q}\Bvalue{u>q}\Bcap\Bvalue{v>\alpha-q}$,}

\noindent whenever $u$, $v\in L^0$ and $\alpha\in\Bbb R$,

\Centerline{$\Bvalue{\gamma
u>\alpha}=\Bvalue{u>\bover{\alpha}{\gamma}}$}

\noindent whenever $u\in L^0$, $\gamma\in\ooint{0,\infty}$ and
$\alpha\in\Bbb R$,

\Centerline{$u\le v\iff\Bvalue{u>\alpha}\Bsubseteq\Bvalue{v>\alpha}$ for
every $\alpha\in\Bbb R$,}

\Centerline{$\Bvalue{u\times v>\alpha}
=\sup_{q\in\Bbb Q,q>0}\Bvalue{u>q}\Bcap\Bvalue{v>\bover{\alpha}q}$}

\noindent whenever $u$, $v\ge 0$ in $L^0$ and $\alpha\ge 0$,

\Centerline{$\Bvalue{e>\alpha}=1$ if $\alpha<1$, $0$ if $\alpha\ge 1$.}

\proof{{\bf (a)} By the Loomis-Sikorski theorem (314M), we can find a
set $Z$ (the Stone space of $\frak A$), a
$\sigma$-algebra $\Sigma$ of subsets of $Z$ (the algebra generated by
the open-and-closed sets and the ideal $\Cal M$ of meager sets) and a
surjective sequentially order-continuous Boolean homomorphism
$\pi:\Sigma\to\frak A$ (corresponding to the identification between
$\frak A$ and the quotient $\Sigma/\Cal M$).   Consequently, defining
$\eusm L^0=\eusm L^0_{\Sigma}$ and $\eusm W=\eusm W_{\Cal M}$
as in 364B, we have a bijection between the
Dedekind $\sigma$-complete $f$-algebra $\eusm L^0/\eusm W$ and $L^0$
(364Ca).   Of course this endows $L^0$ itself with the structure of a
Dedekind $\sigma$-complete $f$-algebra;  and 364Cb tells us that the
description of the algebraic operations above is consistent with this
structure.

\medskip

{\bf (b)} In fact the $f$-algebra structure is completely defined by
the description offered.   For while scalar multiplication is not
described for $\gamma\le 0$, the assertion that $L^0$ is a Riesz space
implies that $0u=0$ and that $\gamma u=(-\gamma)(-u)$ for $\gamma<0$;
so if we have formulae to describe $u+v$ and $\gamma u$ for $\gamma>0$,
this suffices to define the linear structure of $L^0$.   Note that we
have an element $\underline{0}$ in $L^0$ defined by setting

\Centerline{$\Bvalue{\underline{0}>\alpha}=0$ if $\alpha\ge 0$, $1$ if
$\alpha<0$,}

\noindent and the formula for $u+v$ shows us that

\Centerline{$\Bvalue{\underline{0}+u>\alpha}
=\sup_{q\in\Bbb Q}\Bvalue{\underline{0}>q}\Bcap\Bvalue{u>\alpha-q}
=\sup_{q\in\Bbb Q,q<0}\Bvalue{u>\alpha-q}
=\Bvalue{u>\alpha}$}

\noindent for every $\alpha$, so that $\underline{0}$ is the zero of
$L^0$.   As for multiplication, if $L^0$ is to be an $f$-algebra we
must have

\Centerline{$\Bvalue{u\times v>\alpha}
\Bsupseteq\Bvalue{\underline{0}>\alpha}=1$}

\noindent whenever $u$, $v\in(L^0)^+$ and $\alpha<0$, because
$u\times v\ge\underline{0}$.   So the formula offered is sufficient to
determine $u\times v$ for non-negative $u$ and $v$;  and for others we
know that

\Centerline{$u\times v=(u^+\times v^+)-(u^+\times v^-)
-(u^-\times v^+)+(u^-\times v^-)$,}

\noindent so the whole of the multiplication of $L^0$ is defined.
}%end of proof of 364D

\leader{364E}{}\cmmnt{ The rest of this section will be devoted to
understanding the structure just established.   I start with a pair of
elementary facts.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra.

(a) If $u$, $v\in L^0=L^0(\frak A)$ and $\alpha$, $\beta\in\Bbb R$,

\Centerline{$\Bvalue{u+v>\alpha+\beta}
\Bsubseteq\Bvalue{u>\alpha}\Bcup\Bvalue{v>\beta}$.}

(b) If $u$, $v\ge 0$ in $L^0$ and $\alpha$, $\beta\ge 0$ in $\Bbb R$,

\Centerline{$\Bvalue{u\times v>\alpha\beta}
\Bsubseteq\Bvalue{u>\alpha}\Bcup\Bvalue{v>\beta}$.}

\proof{{\bf (a)} For any $q\in\Bbb Q$, either $q\ge\alpha$ and
$\Bvalue{u>q}\Bsubseteq\Bvalue{u>\alpha}$, or $q\le\alpha$ and
$\Bvalue{v>\alpha+\beta-q}$\vthsp$\Bsubseteq\Bvalue{v>\beta}$;  thus in
all cases

\Centerline{$\Bvalue{u>q}\Bcap\Bvalue{v>\alpha+\beta-q}
\Bsubseteq\Bvalue{u>\alpha}\Bcup\Bvalue{v>\beta}$;}

\noindent taking the supremum over $q$, we have the result.

\medskip

{\bf (b)} The same idea works, replacing $\alpha+\beta-q$ by
$\alpha\beta/q$ for $q>0$.
}%end of proof of 364E

\leader{364F}{}\cmmnt{ Yet another description of $L^0$ is sometimes
appropriate, and leads naturally to an important construction (364H).

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra.   Then there is a bijection between
$L^0=L^0(\frak A)$ and the set $\Phi$ of sequentially order-continuous
Boolean homomorphisms from the algebra $\Cal B$ of Borel subsets of
$\Bbb R$ to $\frak A$, defined by saying that $u\in L^0$ corresponds to
$\phi\in\Phi$ iff $\Bvalue{u>\alpha}=\phi(\ooint{\alpha,\infty})$ for
every $\alpha\in\Bbb R$.

\proof{{\bf (a)} If $\phi\in\Phi$, then the map
$\alpha\mapsto\phi(\ooint{\alpha,\infty})$ satisfies the conditions of
364Ae, so corresponds to an element $u_{\phi}$ of $L^0$.

\medskip

{\bf (b)} If $\phi$, $\psi\in \Phi$ and $u_{\phi}=u_{\psi}$, then
$\phi=\psi$.   \Prf\ Set $\Cal A=\{E:E\in\Cal B,\,\phi(E)=\psi(E)\}$.
Then $\Cal A$ is a
$\sigma$-subalgebra of $\Cal B$, because $\phi$ and $\psi$ are both
sequentially order-continuous Boolean homomorphisms, and contains
$\ooint{\alpha,\infty}$ for every $\alpha\in\Bbb R$.   Now $\Cal A$
contains $\ocint{-\infty,\alpha}$ for every $\alpha$, and therefore
includes $\Cal B$ (121J).   But this means that $\phi=\psi$.\ \Qed

\medskip

{\bf (c)} Thus $\phi\mapsto u_{\phi}$ is injective.   But it is also
surjective.   \Prf\ As in 364D, take a set $Z$, a $\sigma$-algebra
$\Sigma$ of subsets of $Z$ and a surjective sequentially
order-continuous Boolean homomorphism $\pi:\Sigma\to\frak A$;  let
$T:\eusm L^0_{\Sigma}/\eusm W_{\pi^{-1}[\{0\}]}\to L^0$
be the bijection described in 364C.   If
$u\in L^0$, there is an $f\in\eusm L^0_{\Sigma}$ such that
$Tf^{\ssbullet}=u$.
Now consider $\phi E=\pi f^{-1}[E]$ for $E\in\Cal B$.   $f^{-1}[E]$
always belongs to $\Sigma$ (121Ef), so $\phi E$ is always
well-defined;   $E\mapsto f^{-1}[E]$ and $\pi$ are sequentially
order-continuous, so $\phi$ also is;  and

\Centerline{$\phi(\ooint{\alpha,\infty})=\pi\{z:f(z)>\alpha\}
=\Bvalue{u>\alpha}$}

\noindent for every $\alpha$, so $u=u_{\phi}$.\ \Qed

Thus we have the declared bijection.
}%end of proof of 364F

\leader{364G}{Definitions (a)} In the context of 364F, I will write
$\Bvalue{u\in E}$, `the region where $u$ takes values in $E$', for
$\phi(E)$, where $\phi:\Cal B\to\frak A$ is the homomorphism
corresponding to $u\in L^0$.   Thus
$\Bvalue{u>\alpha}=\Bvalue{u\in\ooint{\alpha,\infty}\,}$.
\cmmnt{In the same spirit} I write $\Bvalue{u\ge\alpha}$ for
$\Bvalue{u\in\coint{\alpha,\infty}\,}
=\inf_{\beta<\alpha}\Bvalue{u>\beta}$,
$\Bvalue{u\ne 0}=\Bvalue{|u|>0}=\Bvalue{u>0}\Bcup\Bvalue{u<0}$
and so on, so that\cmmnt{ (for instance)}
$\Bvalue{u=\alpha}=\Bvalue{u\in\{\alpha\}}
=\Bvalue{u\ge\alpha}\Bsetminus\Bvalue{u>\alpha}$ for $u\in L^0$ and
$\alpha\in\Bbb R$.

\spheader 364Gb\dvAformerly{364Xd} If $(\frak A,\bar\mu)$ is a probability
algebra, $\bar\mu\phi:\Cal B\to[0,1]$ is a probability
measure, so that its completion $\nu$ is a Radon probability measure on
$\Bbb R$\cmmnt{ (256C)};   I will call $\nu$ the {\bf distribution} of
$u$\cmmnt{ (cf.\ 271C)}.

\leader{364H}{Proposition} Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra, $E\subseteq\Bbb R$ a Borel set, and
$h:E\to\Bbb R$ a Borel measurable function.   Then whenever
$u\in L^0=L^0(\frak A)$ is such that $\Bvalue{u\in E}=1$, there is an
element $\bar h(u)$ of $L^0$ defined by saying that
$\Bvalue{\bar h(u)\in F}=\Bvalue{u\in h^{-1}[F]}$ for every Borel set
$F\subseteq\Bbb R$.

\proof{ All we have to observe is that
$F\mapsto \Bvalue{u\in h^{-1}[F]}$ is a sequentially order-continuous
Boolean homomorphism.   (The condition `$\Bvalue{u\in E}=1$' ensures
that $\Bvalue{u\in h^{-1}[\Bbb R]}=1$.)
}%end of proof of 364H

\leader{364I}{Examples} \cmmnt{Perhaps I should spell out the most
important contexts in which we apply these ideas, even though they have
in effect already been mentioned.

\medskip

}{\bf (a)} Let $X$ be a set and $\Sigma$ a $\sigma$-algebra of
subsets of $X$.   Then we may identify $L^0(\Sigma)$ with the space
$\eusm L^0=\eusm L^0_{\Sigma}$ of
$\Sigma$-measurable real-valued functions on $X$.
\cmmnt{(This is the case $\frak A=\Sigma$ of 364C.)}   For
$f\in\eusm L^0$,
$\Bvalue{f\in E}$\cmmnt{ (364G)} is just $f^{-1}[E]$, for any
Borel set $E\subseteq\Bbb R$;  and if $h$ is a Borel measurable
function, $\bar h(f)$\cmmnt{ (364H)} is just the composition $hf$, for
any $f\in\eusm L^0$.

\spheader 364Ib Now suppose that $\Cal I$ is a $\sigma$-ideal of
$\Sigma$ and that $\frak A=\Sigma/\Cal I$.   Then, as in 364C, we
identify $L^0(\frak A)$ with a quotient $\eusm L^0/\eusm W_{\Cal I}$.
For
$f\in\eusm L^0$, $\Bvalue{f^{\ssbullet}\in E}=f^{-1}[E]^{\ssbullet}$,
and $\bar h(f^{\ssbullet})=(hf)^{\ssbullet}$, for any Borel set $E$ and
any Borel measurable function $h:\Bbb R\to\Bbb R$.

\spheader 364Ic In particular, if $(X,\Sigma,\mu)$ is a measure space
with measure algebra $\frak A$, then $L^0(\frak A)$ becomes identified
with $L^0(\mu)$ as defined in \S241, and the distribution of
$f\in\eusm L^0(\mu)$, as defined in 271C, is the same as the distribution of
$f^{\ssbullet}\in L^0(\mu)\cong L^0(\frak A)$, as defined in 364Gb.

\cmmnt{The same remarks as in 363I apply here;  the space
$\eusm L^0(\mu)$ of 241A is larger than the space
$\eusm L^0=\eusm L^0_{\Sigma}$ considered here.
But for every $f\in\eusm L^0(\mu)$ there is a
$g\in\eusm L^0_{\Sigma}$ such that $g\eae f$ (241Bk), so that
$L^0(\mu)\mskip2mu=\mskip2mu\eusm L^0(\mu)/\mskip-4mu\eae$
can be identified with $\eusm L^0_{\Sigma}/\eusm N$, where $\eusm N$ is
the set of functions in $\eusm L^0$ which are zero almost
everywhere (241Yc).
}

\leader{364J}{Embedding $S$ and $L^{\infty}$ in $L^0$:  Proposition} Let
$\frak A$ be a Dedekind $\sigma$-complete Boolean algebra.

(a) We have a canonical embedding of $L^{\infty}=L^{\infty}(\frak A)$ as
an order-dense solid linear subspace of $L^0=L^0(\frak A)$;  it is the
solid linear subspace generated by the multiplicative identity $e$ of
$L^0$.   Consequently $S=S(\frak A)$ also is embedded as an order-dense
Riesz subspace and subalgebra of $L^0$.

(b) This embedding respects the linear, lattice and multiplicative
structures of $L^{\infty}$ and $S$, and the definition of
$\Bvalue{u>\delta}$, for $u\in S^+$ and $\delta\ge 0$, given in 361Eg.

(c) For $a\in\frak A$, $\chi a$, when regarded as a member of $L^0$, can
be described by the formula

$$\eqalign{\Bvalue{\chi a>\alpha}
&=1\text{ if }\alpha<0,\cr
&=a\text{ if }0\le\alpha<1,\cr
&=0\text{ if }1\le\alpha.\cr}$$

\noindent The function $\chi:\frak A\to L^0$ is additive, injective,
order-continuous and a lattice homomorphism.

(d) For every $u\in (L^0)^+$ there is a non-decreasing sequence
$\sequencen{u_n}$ in $S$ such that $u_0\ge 0$ and
$\sup_{n\in\Bbb N}u_n=u$.

\proof{ Let $Z$, $\Sigma$, $\Cal M$,
$\eusm L^0=\eusm L^0_{\Sigma}$, $\eusm W=\eusm W_{\Cal M}$
and $\pi$ be as in the
proof of 364D.   I defined $L^{\infty}$
to be the space $C(Z)$ of continuous real-valued functions on $Z$
(363A);  but because $\frak A$ is Dedekind
$\sigma$-complete, there is an alternative representation as
$\eusm L^{\infty}/\eusm W\cap\eusm L^{\infty}$, where $\eusm L^{\infty}$
is the space of bounded $\Sigma$-measurable functions from $Z$ to
$\Bbb R$ (363Hb).   Put like this, we clearly have an embedding of
$L^{\infty}\cong\eusm L^{\infty}/\eusm W\cap\eusm L^{\infty}$ in
$L^0\cong\eusm L^0/\eusm W$;  and this embedding represents
$L^{\infty}$ as a Riesz subspace and subalgebra of $L^0$, because
$\eusm L^{\infty}$ is a Riesz subspace and subalgebra of $\eusm L^0$.
$L^{\infty}$ becomes the solid linear subspace of $L^0$ generated by
$(\chi Z)^{\ssbullet}=e$, because $\eusm L^{\infty}$ is the solid linear
subspace of $\eusm L^0$ generated by $\chi Z$.   To see that
$L^{\infty}$ is order-dense in $L^0$, we have only to note that
$f=\sup_{n\in\Bbb N}f\wedge n\chi Z$ in $\eusm L^0$ for every
$f\in\eusm L^0$, and therefore (because the
map $f\mapsto f^{\ssbullet}$ is sequentially order-continuous)
$u=\sup_{n\in\Bbb N}u\wedge ne$ in $L^0$ for every $u\in L^0$.

To identify $\chi a$, we have the formula
$\chi(\pi E)=(\chi E)^{\ssbullet}$, as in 363H(b-iii);
but this means that, if $a=\pi E$,

$$\eqalign{\Bvalue{\chi a>\alpha}=\pi\{z:\chi E(z)>\alpha\}
&=\pi Z=1\text{ if }\alpha<0,\cr
&=\pi E=a\text{ if }0\le\alpha<1,\cr
&=\pi\emptyset=0\text{ if }\alpha\ge 1,\cr}$$

\noindent using the formula in 364Ca.   Evidently $\chi$ is injective.

Because $S$ is an order-dense Riesz subspace and subalgebra of
$L^{\infty}$ (363C), the same embedding represents it as an
order-dense
Riesz subspace and subalgebra of $L^0$.   (For `order-dense', use
352N(c-iii).)   Concerning the formula $\Bvalue{u>\delta}$, suppose that
$u\in S^+$ and $\delta\ge 0$;  express $u$ as
$\sum_{j=0}^m\beta_j\chi b_j$, where
$b_0,\ldots,b_m\in\frak A$ are disjoint and $\beta_j\ge 0$ for every $j$.
Then we have disjoint sets $F_0,\ldots,F_m\in\Sigma$ such that
$\pi F_j=b_j$ for every $j$, and $u$ is identified with
$(\sum_{j=0}^m\beta_j\chi F_j)^{\ssbullet}$.   Using 364Ca, we have

\Centerline{$\Bvalue{u>\delta}
=\pi\{z:\sum_{j=0}^m\beta_j\chi F_j(z)>\delta\}
=\pi(\bigcup\{F_j:\beta_j>\delta\})
=\sup\{b_j:\beta_j>\delta\}$,}

\noindent matching the expression in the proof of 361Eg.   So the new
interpretation of $\Bvalue{\ldots}$ matches the former definition in the
special case envisaged in 361E.

Because $\chi:\frak A\to L^{\infty}$ is additive, order-continuous and a
lattice homomorphism (363D), and the embedding map
$L^{\infty}\embedsinto L^0$ also is, $\chi:\frak A\to L^0$ has the same
properties.

Finally, if $u\ge 0$ in $L^0$, we can represent it as $f^{\ssbullet}$
where $f\ge 0$ in $\eusm L^0$.   For $n\in\Bbb N$ set

$$\eqalign{f_n(z)
&=2^{-n}k\text{ if }2^{-n}k\le f(z)<2^{-n}(k+1)
   \text{ where }0\le k<4^n,\cr
&=0\text{ if }f(z)\ge 2^n;\cr}$$

\noindent then $\sequencen{f_n^{\ssbullet}}$ is a non-decreasing
sequence in $S^+$ with supremum $u$.
}%end of proof of 364J

\leader{364K}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra.
Then $S(\frak A^f)$ can be embedded as a Riesz subspace of
$L^0(\frak A)$, which is order-dense iff $(\frak A,\bar\mu)$ is
semi-finite.

\proof{ (Recall that $\frak A^f$ is the ring $\{a:\bar\mu a<\infty\}$.)
The embedding $\frak A^f\embedsinto\frak A$ is an injective ring
homomorphism, so induces an embedding of $S(\frak A^f)$ as a Riesz
subspace of $S(\frak A)$, by 361J.   Now $S(\frak A^f)$ is order-dense
in $S(\frak A)$ iff $(\frak A,\bar\mu)$ is semi-finite.   \Prf\ (i) If
$(\frak A,\bar\mu)$ is semi-finite and $v>0$ in $S(\frak A)$, then $v$
is expressible as $\sum_{j=0}^n\beta_j\chi b_j$ where $\beta_j\ge 0$ for
each $j$ and some $\beta_j\chi b_j$ is non-zero;  now there is a
non-zero $a\in\frak A^f$ such that $a\Bsubseteq b_j$, so that
$0<\beta_j\chi a\in S(\frak A^f)$ and $\beta_j\chi a\le v$.   As $v$ is
arbitrary, $S(\frak A^f)$ is quasi-order-dense, therefor order-dense
(353A).  (ii) If $S(\frak A^f)$ is order-dense in $S(\frak A)$ and
$b\in\frak A\setminus\{0\}$, there is a $u>0$ in $S(\frak A^f)$ such
that $u\le\chi b$;  now there are $\alpha>0$,
$a\in\frak A^f\setminus\{0\}$ such that
$\alpha\chi a\le u$, in which case $a\Bsubseteq b$.\ \Qed

Now because $S(\frak A^f)\subseteq S(\frak A)$ and $S(\frak A)$ is
order-dense in $L^0(\frak A)$, we must have

$$\eqalign{S(\frak A^f)\text{ is order-dense in }L^0(\frak A)
&\iff S(\frak A^f)\text{ is order-dense in }S(\frak A)\cr
&\iff (\frak A,\bar\mu)\text{ is semi-finite}.\cr}$$
}%end of proof of 364K

\leader{364L}{Suprema and infima in \dvrocolon{$L^0$}}\cmmnt{ We know
that any $L^0(\frak A)$ is a Dedekind $\sigma$-complete partially
ordered set.   There is a useful description of
suprema for this ordering in (a) of the next result.
We do not have such a simple formula for
general infima (though see 364Xm), but facts in (b) are useful.

\medskip

\noindent}{\bf Proposition}
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra, and
$L^0=L^0(\frak A)$.

(a) Let $A$ be a subset of $L^0$.

\quad(i) $A$ is bounded above in $L^0$ iff there is a sequence
$\sequencen{c_n}$ in $\frak A$, with infimum $0$, such that
$\Bvalue{u>n}\Bsubseteq c_n$ for every $u\in A$.

\quad(ii) If $A$ is non-empty, then $A$ has a supremum in $L^0$ iff
$c_{\alpha}=\sup_{u\in A}\Bvalue{u>\alpha}$ is defined in $\frak A$ for
every $\alpha\in\Bbb R$ and $\inf_{n\in\Bbb N}c_n=0$;  and in this case
$c_{\alpha}=\Bvalue{\sup A>\alpha}$ for every $\alpha$.

\quad(iii) If $A$ is non-empty and bounded above, then $A$ has a supremum in
$L^0$ iff $\sup_{u\in A}\Bvalue{u>\alpha}$ is defined in $\frak A$ for
every $\alpha\in\Bbb R$.

(b)(i) If $u$, $v\in L^0$, then
$\Bvalue{u\wedge v>\alpha}=\Bvalue{u>\alpha}\Bcap\Bvalue{v>\alpha}$ for
every $\alpha\in\Bbb R$.

\quad(ii) If $A$ is a non-empty subset of $(L^0)^+$,
then $\inf A=0$ in $L^0$ iff $\inf_{u\in A}\Bvalue{u>\alpha}=0$ in
$\frak A$ for every $\alpha>0$.

\proof{{\bf (a)(i)}\grheada\ If $A$ has an upper bound $u_0$, set
$c_n=\Bvalue{u_0>n}$ for each $n$;  then $\sequencen{c_n}$ satisfies the
conditions.

\medskip

\quad\grheadb\ If $\sequencen{c_n}$ satisfies the conditions, set

$$\eqalign{\phi(\alpha)&=1\text{ if }\alpha<0,\cr
&=\inf_{i\le n}c_i\text{ if }n\in\Bbb N,\,\alpha\in\coint{n,n+1}.\cr}$$

\noindent    Then it is easy to check that $\phi$ satisfies the
conditions of 364Aa, since $\inf_{n\in\Bbb N}c_n=0$.   So there is a
$u_0\in L^0$ such that $\phi(\alpha)=\Bvalue{u_0>\alpha}$ for each
$\alpha$.   Now, given $u\in
A$ and $\alpha\in\Bbb R$,

$$\eqalign{\Bvalue{u>\alpha}&\Bsubseteq 1=\Bvalue{u_0>\alpha}
\text{ if }\alpha<0,\cr
&\Bsubseteq\inf_{i\le n}\Bvalue{u>i}\Bsubseteq\inf_{i\le n}c_i
=\Bvalue{u_0>\alpha}\text{ if }n\in\Bbb N,\,
\alpha\in\coint{n,n+1}.\cr}$$

\noindent Thus $u_0$ is an upper bound for $A$ in $L^0$.

\medskip

\quad{\bf (ii)}\grheada\ Suppose that
$c_{\alpha}=\sup_{u\in A}\Bvalue{u>\alpha}$ is
defined in $\frak A$ for every $\alpha$, and
that $\inf_{n\in\Bbb N}c_n=0$.   Then, for any $\alpha$,

\Centerline{$\sup_{q\in\Bbb Q,q>\alpha}c_q
=\sup_{u\in A,q\in\Bbb Q,q>\alpha}\Bvalue{u>q}
=\sup_{u\in A}\Bvalue{u>\alpha}
=c_{\alpha}$.}

\noindent Also, we are supposing that $A$ contains some $u_0$, so that

\Centerline{$\sup_{n\in\Bbb N}c_{-n}
\Bsupseteq\sup_{n\in\Bbb N}\Bvalue{u_0>-n}
=1$.}

\noindent Accordingly there is a $u^*\in L^0$ such that
$\Bvalue{u^*>\alpha}=c_{\alpha}$ for every $\alpha\in\Bbb R$.
But now, for $v\in L^0$,

$$\eqalign{v\text{ is an upper bound for }A
&\iff\Bvalue{u>\alpha}\Bsubseteq\Bvalue{v>\alpha}
\text{ for every }u\in A,\,\alpha\in\Bbb R\cr
&\iff\Bvalue{u^*>\alpha}\Bsubseteq\Bvalue{v>\alpha}
\text{ for every }\alpha\cr
&\iff u^*\le v,\cr}$$

\noindent so that $u^*=\sup A$ in $L^0$.

\medskip

\qquad\grheadb Now suppose that $u^*=\sup A$ is defined in $L^0$.   Of
course $\Bvalue{u^*>\alpha}$ must be an upper bound for
$\{\Bvalue{u>\alpha}:u\in A\}$ for every $\alpha$.   \Quer\ Suppose we
have an $\alpha$ for which it is not the least upper bound, that is,
there is a $c\Bsubset\Bvalue{u^*>\alpha}$ which is an upper bound for
$\{\Bvalue{u>\alpha}:u\in A\}$.   Define $\phi:\Bbb R\to\frak A$ by
setting

$$\eqalign{\phi(\beta)
&=c\Bcap\Bvalue{u^*>\beta}\text{ if }\beta\ge\alpha,\cr
&=\Bvalue{u^*>\beta}\text{ if }\beta<\alpha.\cr}$$

\noindent It is easy to see that $\phi$ satisfies the conditions of 364Aa
(we need the distributive law 313Ba to check that
$\phi(\beta)=\sup_{\gamma>\beta}\phi(\gamma)$ if $\beta\ge\alpha$), so
corresponds to a member $v$ of $L^0$.    But we now find that $v$ is an
upper bound for $A$ (because if $u\in A$ and $\beta\ge\alpha$ then

\Centerline{$\Bvalue{u>\beta}
\Bsubseteq\Bvalue{u>\alpha}\Bcap\Bvalue{u^*>\beta}
\Bsubseteq c\Bcap\Bvalue{u^*>\beta}=\Bvalue{v>\beta}$,)}

\noindent that $v\le u^*$ and that $v\ne u^*$ (because
$\Bvalue{v>\alpha}=c\ne\Bvalue{u^*>\alpha}$);  but this is impossible,
because $u^*$ is supposed to be the supremum of $A$.\ \BanG\  Thus if
$u^*=\sup A$ is defined in $L^0$, then
$\sup_{u\in A}\Bvalue{u>\alpha}=\Bvalue{u^*>\alpha}$
is defined in $\frak A$ for every $\alpha\in\Bbb R$.   Also, of course,

\Centerline{$\inf_{n\in\Bbb N}\sup_{u\in A}\Bvalue{u>n}
=\inf_{n\in\Bbb N}\Bvalue{u^*>n}=0$.}

\medskip

\quad{\bf (iii)} This is now easy.   If $A$ has a supremum, then surely it
satisfies the condition, by (b).   If $A$ satisfies the condition, then
we have a family $\langle c_{\alpha}\rangle_{\alpha\in\Bbb R}$ as
required in (b);  but also, by (a) or otherwise, there is a sequence
$\sequencen{c'_n}$ such that $c_n\Bsubseteq c'_n$ for every $n$ and
$\inf_{n\in\Bbb N}c'_n=0$, so $\inf_{n\in\Bbb N}c_n$ also is $0$, and
both conditions in (b) are satisfied, so $A$ has a supremum.

\medskip

{\bf (b)(i)} Take $Z$, $\eusm L^0$ and $\pi$ as in the proof of
364D.   Express $u$ as $f^{\ssbullet}$, $v$ as $g^{\ssbullet}$ where
$f$, $g\in\eusm L^0$, so that $u\wedge v=(f\wedge g)^{\ssbullet}$,
because the canonical map from $\eusm L^0$ to $L^0$ is a Riesz
homomorphism (351J).   Then

$$\eqalign{\Bvalue{u\wedge v>\alpha}
&=\pi\{z:\min(f(z),g(z))>\alpha\}
=\pi(\{z:f(z)>\alpha\}\cap\{z:g(z)>\alpha\})\cr
&=\pi\{z:f(z)>\alpha\}\Bcap\pi\{z:g(z)>\alpha\}
=\Bvalue{u>\alpha}\Bcap\Bvalue{v>\alpha}\cr}$$

\noindent for every $\alpha$.

\medskip

\quad{\bf (ii)}\grheada\
If $\inf_{u\in A}\Bvalue{u>\alpha}=0$ for every $\alpha>0$,
and $v$ is any lower bound
for $A$, then $\Bvalue{v>\alpha}$ must be $0$ for every $\alpha>0$, so
that $\Bvalue{v>0}=0$;  now since $\Bvalue{0>\alpha}=0$ for
$\alpha\ge 0$, $1$ for $\alpha<0$, $v\le 0$.   As $v$ is arbitrary,
$\inf A=0$.

\medskip

\qquad\grheadb If $\alpha>0$ is such that $\inf_{u\in A}\Bvalue{u>\alpha}$
is undefined, or not equal to $0$, let $c\in\frak A$ be such that
$0\ne c\Bsubseteq\Bvalue{u>\alpha}$ for every $u\in A$, and consider
$v=\alpha\chi c$.   Then $\Bvalue{v>\beta}=\Bvalue{\chi
c>\bover{\beta}{\alpha}}$ is $1$ if $\beta<0$, $c$ if $0\le\beta<\alpha$
and $0$ if $\beta\ge\alpha$.   If $u\in A$ then $\Bvalue{u>\beta}$ is
$1$ if $\beta<0$ (since $u\ge 0$), at least $\Bvalue{u>\alpha}\Bsupseteq
c$ if $0\le\beta<\alpha$, and always includes $0$;   so that
$v\le u$.   As $u$ is arbitrary, $\inf A$ is either undefined in $L^0$
or not $0$.
}%end of proof of 364L

\leader{364M}{}\cmmnt{ Now we have a reward for our labour, in that
the following basic theorem is easy.

\medskip

\noindent}{\bf Theorem} For a Dedekind $\sigma$-complete Boolean algebra
$\frak A$, $L^0=L^0(\frak A)$ is Dedekind complete iff $\frak A$ is.

\proof{ The description of suprema in 364L(a-iii) makes it obvious that if
$\frak A$ is Dedekind complete, so that $\sup_{u\in A}\Bvalue{u>\alpha}$
is always defined, then $L^0$ must be Dedekind complete.
On the other hand, if $L^0$ is Dedekind complete, then so is
$L^{\infty}(\frak A)$ (by 364J and 353K(b-i)), so that $\frak A$ also is
Dedekind complete, by 363Mb.
}%end of proof of 364M

\leader{364N}{The multiplication of \dvrocolon{$L^0$}}\cmmnt{ I have
already observed that $L^0$ is always an $f$-algebra with identity;  in
particular
(because $L^0$ is surely Archimedean) the map $u\mapsto u\times v$ is
order-continuous for every $v\ge 0$ (353Pa), and multiplication is
commutative (353Pb, or otherwise).   The multiplicative identity is
$\chi 1$ (364D, 364Jc).  By 353Qb, or otherwise, $u\times v=0$ iff
$|u|\wedge|v|=0$.   There is one special feature of multiplication in
$L^0$ which I can mention here.

\medskip

\noindent}{\bf Proposition}\dvAformerly{3{}64P} Let $\frak A$
be a Dedekind $\sigma$-complete Boolean algebra.
Then an element $u$ of $L^0=L^0(\frak A)$ has a multiplicative inverse in
$L^0$ iff $|u|$ is a weak order unit in $L^0$ iff $\Bvalue{|u|>0}=1$.

\proof{ If $u$ is invertible, then $|u|$ is a weak order unit, by
353Qc or otherwise.   In this case, setting
$c=1\Bsetminus\Bvalue{|u|>0}$, we see that

\Centerline{$\Bvalue{|u|\wedge\chi c>0}=\Bvalue{|u|>0}\Bcap c=0$}

\noindent (364L(b-i)), so that $|u|\wedge\chi c\le 0$ and $\chi c=0$, that
is, $c=0$;  so $\Bvalue{|u|>0}$ must be $1$.   To complete the circuit,
suppose that $\Bvalue{|u|>0}=1$.   Let $Z$, $\Sigma$,
$\eusm L^0=\eusm L^0_{\Sigma}$,
$\pi$, $\Cal M$ be as in the proof of
364D, and $S:\eusm L^0\to L^0$ the canonical map, so that
$\Bvalue{Sh>\alpha}=\pi\{z:h(z)>\alpha\}$ for every $h\in\eusm L^0$,
$\alpha\in\Bbb R$.   Express $u$ as $Sf$ where $f\in\eusm L^0$.   Then
$\pi\{z:|f(z)|>0\}=\Bvalue{S|f|>0}=1$, so $\{z:f(z)=0\}\in\Cal M$.   Set

\Centerline{$g(z)=\Bover1{f(z)}$ if $f(z)\ne 0$,
\quad$g(z)=0$ if $f(z)=0$.}

\noindent Then $\{z:f(z)g(z)\ne 1\}\in\Cal M$ so

\Centerline{$u\times Sg=S(f\times g)=S(\chi Z)=\chi 1$}

\noindent and $u$ is invertible.
}%end of proof of 364N

\cmmnt{\medskip

\noindent{\bf Remark} The repeated phrase `by 353x or otherwise'
reflects the fact that the abstract methods there can all be replaced in
this case by simple direct arguments based on the construction in
364B-364D.
}%end of comment

\leader{364O}{Recovering the algebra:  Proposition} Let
$\frak A$ be a Dedekind $\sigma$-complete Boolean algebra.   For
$a\in\frak A$
write $V_a$ for the band in $L^0=L^0(\frak A)$ generated by $\chi a$.
Then $a\mapsto V_a$ is a Boolean isomorphism between $\frak A$ and the
algebra of projection bands in $L^0$.

\proof{ I copy from the argument for 363J, itself based on 361K.   If
$a\in\frak A$ and $w\in L^0$ then $w\times\chi a\in V_a$.
\Prf\ If $v\in V_a^{\perp}$ then $|\chi a|\wedge|v|=0$, so
$\chi a\times v=0$, so $(w\times\chi a)\times v=0$, so
$|w\times\chi a|\wedge|v|=0$;
thus $w\times\chi a\in V_a^{\perp\perp}$, which is equal to $V_a$
because $L^0$ is Archimedean (353Ba).\ \QeD\  Now, if
$a\in\frak A$, $u\in V_a$ and $v\in V_{1\Bsetminus a}$, then
$|u|\wedge|v|=0$
because $\chi a\wedge\chi(1\Bsetminus a)=0$;  and if $w\in
L^{0}(\frak A)$ then

\Centerline{$w=(w\times\chi a)+(w\times\chi(1\Bsetminus a))\in
V_a+V_{1\Bsetminus a}$.}

\noindent So
$V_a$ and $V_{1\Bsetminus a}$ are complementary projection bands in
$L^0$.   Next, if $U\subseteq L^0$ is
a projection band, then $\chi 1$ is expressible as $u+v=u\vee v$ where
$u\in U$, $v\in U^{\perp}$.   Setting $a=\Bvalue{u>\bover12}$,
$a'=\Bvalue{v>\bover12}$ we must have  $a\Bcup a'=1$ and $a\Bcap a'=0$
(using 364L), so that $a'=1\Bsetminus a$;  also
$\bover12\chi a\le u$, so that $\chi a\in U$, and
similarly $\chi(1\Bsetminus a)\in U^{\perp}$.
In this case $V_a\subseteq U$ and
$V_{1\Bsetminus a}\subseteq U^{\perp}$, so $U$ must be $V_a$ precisely.
Thus $a\mapsto V_a$ is surjective.   Finally, just as in 361K,
$a\Bsubseteq b\iff V_a\subseteq V_b$, so we have a Boolean isomorphism.
}%end of proof of 364O

\leader{364P}{}\cmmnt{ I come at last to the result corresponding to
361J and 363F.

\medskip

\noindent}{\bf Theorem}
Let $\frak A$ and $\frak B$ be Dedekind $\sigma$-complete Boolean algebras,
and $\pi:\frak A\to\frak B$ a sequentially order-continuous Boolean
homomorphism.

(a) We have a multiplicative sequentially order-continuous Riesz
homomorphism $T_{\pi}:L^0(\frak A)\to L^0(\frak B)$ defined by the
formula

\Centerline{$\Bvalue{T_{\pi}u>\alpha}=\pi\Bvalue{u>\alpha}$}

\noindent whenever $\alpha\in\Bbb R$ and $u\in L^0(\frak A)$.

(b) Defining $\chi a\in L^0(\frak A)$ as in 364J,
$T_{\pi}(\chi a)=\chi(\pi a)$ in $L^0(\frak B)$ for every $a\in\frak A$.
If we regard $L^{\infty}(\frak A)$ and $L^{\infty}(\frak B)$ as embedded
in $L^0(\frak A)$ and $L^0(\frak B)$ respectively, then $T_{\pi}$, as
defined here, agrees on $L^{\infty}(\frak A)$ with $T_{\pi}$ as defined
in 363F.

(c) $T_{\pi}$ is order-continuous iff $\pi$ is order-continuous,
injective iff $\pi$ is injective, surjective iff $\pi$ is surjective.

(d) $\Bvalue{T_{\pi}u\in E}=\pi\Bvalue{u\in E}$ for every $u\in
L^0(\frak A)$ and every Borel set $E\Bsubseteq\Bbb R$;  consequently
$\bar hT_{\pi}=T_{\pi}\bar h$ for every Borel measurable
$h:\Bbb R\to\Bbb R$, writing $\bar h$ indifferently for the associated
maps from $L^0(\frak A)$ to itself and from $L^0(\frak B)$ to
itself\cmmnt{ (364H)}.

(e) If $\frak C$ is another Dedekind $\sigma$-complete Boolean algebra
and $\theta:\frak B\to\frak C$ another sequentially order-continuous
Boolean homomorphism then $T_{\theta\pi}=T_{\theta}T_{\pi}:L^0(\frak
A)\to L^0(\frak C)$.

\proof{ I write $T$ for $T_{\pi}$.

\medskip

{\bf (a)(i)} To see that $Tu$ is well-defined in $L^0(\frak B)$ for
every $u\in L^0(\frak A)$, all we need to do is to check that the map
$\alpha\mapsto\pi\Bvalue{u>\alpha}:\Bbb R\to\frak B$ satisfies the
conditions of 364Ae, and this is easy, because $\pi$ preserves all
countable suprema and infima.

\medskip

\quad{\bf (ii)} To see that $T$ is linear and order-preserving and
multiplicative, we can use the formulae of 364D.  For instance, if $u$,
$v\in L^0(\frak A)$, then

$$\eqalign{\Bvalue{Tu+Tv>\alpha}
&=\sup_{q\in\Bbb Q}\Bvalue{Tu>q}\Bcap\Bvalue{Tv>\alpha-q}
=\sup_{q\in\Bbb Q}\pi\Bvalue{u>q}\Bcap\pi\Bvalue{v>\alpha-q}\cr
&=\pi(\sup_{q\in\Bbb Q}\Bvalue{u>q}\Bcap\Bvalue{v>\alpha-q})
=\pi\Bvalue{u+v>\alpha}
=\Bvalue{T(u+v)>\alpha}\cr}$$

\noindent for every $\alpha\in\Bbb R$, so that $Tu+Tv=T(u+v)$.   In the
same way,

\Centerline{$T(\gamma u)=\gamma Tu$ whenever $\gamma>0$,}

\Centerline{$Tu\le Tv$ whenever $u\le v$,}

\Centerline{$Tu\times Tv=T(u\times v)$ whenever $u$, $v\ge 0$,}

\noindent so that, using the distributive laws, $T$ is linear and
multiplicative.

To see that $T$ is a sequentially order-continuous Riesz homomorphism,
suppose that $A\subseteq L^0(\frak A)$ is a countable non-empty set with
a supremum $u^*\in L^0(\frak A)$;  then
$T[A]$ is a non-empty subset of $L^0(\frak B)$ with an upper bound
$Tu^*$, and

$$\eqalignno{\sup_{u\in A}\Bvalue{Tu>\alpha}
&=\sup_{u\in A}\pi\Bvalue{u>\alpha}
=\pi(\sup_{u\in A}\Bvalue{u>\alpha})
=\pi\Bvalue{u^*>\alpha}\cr
\noalign{\noindent (using 364La)}
&=\Bvalue{Tu^*>\alpha}\cr}$$

\noindent for every $\alpha\in\Bbb R$.   So using 364La again,
$Tu^*=\sup_{u\in A}Tu$.   Now this is true, in particular, for doubleton
sets $A$, so that $T$ is a Riesz homomorphism;  and also for
non-decreasing sequences, so that $T$ is sequentially
order-continuous.

\medskip

{\bf (b)} The identification of $T(\chi a)$ with $\chi(\pi a)$ is
another almost trivial verification.   It follows that $T$ agrees with
the map of 363F on $S(\frak A)$, if we think of $S(\frak A)$ as a
subspace of $L^0(\frak A)$.   Next, if
$u\in L^{\infty}(\frak A)\subseteq L^0(\frak A)$, and
$\gamma=\|u\|_{\infty}$, then $|u|\le \gamma\chi 1_{\frak A}$, so that
$|Tu|\le \gamma\chi 1_{\frak B}$, and
$Tu\in L^{\infty}(\frak B)$, with $\|Tu\|_{\infty}\le\gamma$.   Thus
$T\restr L^{\infty}(\frak A)$ has norm at most $1$.   As it agrees with
the map of 363F on $S(\frak A)$, which is $\|\,\|_{\infty}$-dense in
$L^{\infty}(\frak A)$ (363C), and both are continuous, they must agree
on the whole of $L^{\infty}(\frak A)$.

\medskip

{\bf (c)(i)}\grheada\ Suppose that $\pi$ is order-continuous, and that
$A\subseteq L^0(\frak A)$ is a non-empty set with a supremum
$u^*\in L^0(\frak A)$.   Then for any $\alpha\in\Bbb R$,

$$\eqalignno{\Bvalue{Tu^*>\alpha}
&=\pi\Bvalue{u^*>\alpha}
=\pi(\sup_{u\in A}\Bvalue{u>\alpha})\cr
\noalign{\noindent (by 364La)}
&=\sup_{u\in A}\pi\Bvalue{u>\alpha}\cr
\noalign{\noindent (because $\pi$ is order-continuous)}
&=\sup_{u\in A}\Bvalue{Tu>\alpha}.\cr}$$

\noindent As $\alpha$ is arbitrary, $Tu^*=\sup T[A]$, by 364La again.
As $A$ is arbitrary, $T$ is order-continuous (351Ga).

\medskip

\qquad\grheadb\ Now suppose that $T$ is order-continuous and that
$A\subseteq\frak A$ is a non-empty set with supremum $c$ in $\frak A$.
Then $\chi c=\sup_{a\in A}\chi a$ (364Jc) so

\Centerline{$\chi(\pi c)=T(\chi c)=\sup_{a\in A}T(\chi a)
=\sup_{a\in A}\chi(\pi a)$.}

\noindent But now

\Centerline{$\pi c
=\Bvalue{\chi(\pi c)>0}
=\sup_{a\in A}\Bvalue{\chi(\pi a)>0}
=\sup_{a\in A}\pi a$.}

\noindent As $A$ is arbitrary, $\pi$ is order-continuous.

\medskip

\quad{\bf (ii)}\grheada\ If $\pi$ is injective and $u$, $v$ are distinct
elements of $L^0(\frak A)$, then there must be some $\alpha$ such that
$\Bvalue{u>\alpha}\ne\Bvalue{v>\alpha}$, in which case
$\Bvalue{Tu>\alpha}\ne\Bvalue{Tv>\alpha}$ and $Tu\ne Tv$.

\medskip

\qquad\grheadb\ Now suppose that $T$ is injective.   It is easy to see
that $\chi:\frak A\to L^0(\frak A)$ is injective, so that $T\chi:\frak
A\to L^0(\frak B)$ is injective;  but this is the same as $\chi\pi$ (by
(b)), so $\pi$ must also be injective.

\medskip

\quad{\bf (iii)}\grheada\ Suppose that $\pi$ is surjective.   Let
$\Sigma$ be a $\sigma$-algebra of sets such that there is a sequentially
order-continuous Boolean surjection $\phi:\Sigma\to\frak A$.   Then
$\pi\phi:\Sigma\to\frak B$ is surjective.   So given
$w\in L^0(\frak B)$, there is an $f\in\eusm L^0_{\Sigma}$ such that
$\Bvalue{w>\alpha}=\pi\phi\{x:f(x)>\alpha\}$ for every $\alpha\in\Bbb R$
(364C).   But, also by 364C, there is a $u\in L^0(\frak A)$ such that
$\Bvalue{u>\alpha}=\phi\{x:f(x)>\alpha\}$ for every $\alpha$.   And now
of course $Tu=w$.   As $w$ is arbitrary, $T$ is surjective.

\medskip

\qquad\grheadb\ If $T$ is surjective, and $b\in\frak B$, there must be
some $u\in L^0(\frak A)$ such that $Tu=\chi b$.   Now set
$a=\Bvalue{u>0}$ and see that $\pi a=\Bvalue{\chi b>0}=b$.   As $b$ is
arbitrary, $\pi$ is surjective.

\medskip

{\bf (d)} The map $E\mapsto\pi\Bvalue{u\in E}$ is a sequentially
order-continuous Boolean homomorphism, equal to $\Bvalue{Tu\in E}$ when
$E$ is of the form $\ooint{\alpha,\infty}$;  so by 364F the two are
equal for all Borel sets $E$.

If $h:\Bbb R\to\Bbb R$ is a Borel measurable function, $u\in L^0(\frak
A)$ and $E\subseteq\Bbb R$ is a Borel set, then

$$\eqalign{\Bvalue{\bar h(Tu)\in E}
&=\Bvalue{Tu\in h^{-1}[E]}
=\pi\Bvalue{u\in h^{-1}[E]}\cr
&=\pi\Bvalue{\bar h(u)\in E}
=\Bvalue{T(\bar h(u))\in E}.\cr}$$

\noindent As $E$ and $u$ are arbitrary, $T\bar h=\bar hT$.

\medskip

{\bf (e)} This is immediate from (a).
}%end of proof of 364P

\leader{364Q}{Proposition} Let $X$ and $Y$ be sets, $\Sigma$, $\Tau$
$\sigma$-algebras of subsets of $X$, $Y$ respectively, and $\Cal I$,
$\Cal J\,\,\sigma$-ideals of $\Sigma$, $\Tau$.   Set
$\frak A=\Sigma/\Cal I$ and $\frak B=\Tau/\Cal J$.   Suppose that
$\phi:X\to Y$ is
a function such that $\phi^{-1}[F]\in\Sigma$ for every $F\in\Tau$ and
$\phi^{-1}[F]\in\Cal I$ for every $F\in\Cal J$.

(a) There is a
sequentially order-continuous Boolean homomorphism
$\pi:\frak B\to\frak A$ defined by saying that
$\pi F^{\ssbullet}=\phi^{-1}[F]^{\ssbullet}$
for every $F\in\Tau$.

(b) Let $T_{\pi}:L^0(\frak B)\to L^0(\frak A)$ be the
Riesz homomorphism corresponding to $\pi$, as defined in 364P.
If we identify
$L^0(\frak B)$ with $\eusm L^0_{\Tau}/\eusm W_{\Cal J}$ and
$L^0(\frak A)$ with $\eusm L^0_{\Sigma}/\eusm W_{\Cal I}$ in the manner
of 364B-364C, then $T_{\pi}(g^{\ssbullet})=(g\phi)^{\ssbullet}$ for every
$g\in\eusm L^0_{\Tau}$.

(c) Let $Z$ be a third set, $\Upsilon$ a $\sigma$-algebra of subsets of
$Z$, $\Cal K$ a $\sigma$-ideal of $\Upsilon$, and $\psi:Y\to Z$ a function
such that $\psi^{-1}[G]\in\Tau$ for every $G\in\Upsilon$ and
$\psi^{-1}[G]\in\Cal J$ for every $G\in\Cal K$.   Let
$\theta:\frak C\to\frak B$ and $T_{\theta}:L^0(\frak C)\to L^0(\frak B)$ be
the homomorphisms corresponding to $\psi$ as in (a)-(b).   Then
$\pi\theta:\frak C\to\frak A$ and
$T_{\pi}T_{\theta}:L^0(\frak C)\to L^0(\frak A)$ correspond to
$\psi\phi:X\to Y$ in the same way.

(d) Now suppose that $\mu$ and $\nu$ are measures with domains $\Sigma$,
$\Tau$ and null ideals $\Cal N(\mu)$, $\Cal N(\nu)$ respectively, and that
$\Cal I=\Sigma\cap\Cal N(\mu)$ and $\Cal J=\Tau\cap\Cal N(\nu)$.   In this
case, identifying $L^0(\frak A)$, $L^0(\frak B)$ with $L^0(\mu)$ and
$L^0(\nu)$ as in 364Ic, we have $g\phi\in\eusm L^0(\mu)$ and
$T_{\pi}(g^{\ssbullet})=(g\phi)^{\ssbullet}$ for every
$g\in\eusm L^0(\nu)$.

\proof{{\bf (a)} The argument is essentially that of 324A-324B, somewhat
simplified.   Explicitly:  if $F_1$, $F_2\in\Tau$ and
$F_1^{\ssbullet}=F_2^{\ssbullet}$, then $F_1\symmdiff F_2\in\Cal J$ so
$\phi^{-1}[F_1]\symmdiff\phi^{-1}[F_2]=\phi^{-1}[F_1\symmdiff F_2]$ belongs
to $\Cal I$ and $\phi^{-1}[F_1]^{\ssbullet}=\phi^{-1}[F_2]^{\ssbullet}$.
So the formula offered defines a map $\pi:\frak B\to\frak A$.
It is a Boolean homomorphism, because if $F_1$, $F_2\in\Tau$ then

$$\eqalign{\pi F_1^{\ssbullet}\Bsymmdiff\pi F_2^{\ssbullet}
&=\phi^{-1}[F_1]^{\ssbullet}\Bsymmdiff\phi^{-1}[F_2]^{\ssbullet}
=(\phi^{-1}[F_1]\symmdiff\phi^{-1}[F_2])^{\ssbullet}\cr
&=\phi^{-1}[F_1\symmdiff F_2]^{\ssbullet}
=\pi(F_1\symmdiff F_2)^{\ssbullet}
=\pi(F_1^{\ssbullet}\Bsymmdiff F_2^{\ssbullet}),\cr}$$

\noindent so $\pi(b_1\Bsymmdiff b_2)=\pi b_1\Bsymmdiff b_2$ for all $b_1$,
$b_2\in\frak B$.
Similarly $\pi(b_1\Bcap b_2)=\pi b_1\Bcap b_2$ for all $b_1$,
$b_2\in\frak B$, and of course

\Centerline{$\pi 1_{\frak B}=\pi Y^{\ssbullet}
=\phi^{-1}[Y]^{\ssbullet}=X^{\ssbullet}=1_{\frak A}$.}

To see that $\pi$ is sequentially order-continuous, let
$\sequencen{b_n}$ be a sequence in $\frak B$.  For each $n$ we may
choose an $F_n\in\Tau$ such that $F_n^{\ssbullet}=b_n$, and set
$F=\bigcup_{n\in\Bbb N}F_n$.   As the map
$H\mapsto H^{\ssbullet}:\Tau\to\frak B$ is sequentially order-continuous
(313Qb), $F^{\ssbullet}=\sup_{n\in\Bbb N}b_n$ in $\frak B$.   Now

$$\eqalign{\pi(\sup_{n\in\Bbb N}b_n)
&=\pi F^{\ssbullet}
=\phi^{-1}[F]^{\ssbullet}
=(\bigcup_{n\in\Bbb N}\phi^{-1}[F_n])^{\ssbullet}\cr
&=\sup_{n\in\Bbb N}\phi^{-1}[F_n]^{\ssbullet}
=\sup_{n\in\Bbb N}\pi F_n^{\ssbullet}
=\sup_{n\in\Bbb N}\pi b_n.\cr}$$

\noindent So $\pi$ is sequentially order-continuous, by 313Lc.

\medskip

{\bf (b)} Now suppose that $g:Y\to\Bbb R$ is $\Tau$-measurable;  write
$v$ for $g^{\ssbullet}$ in
$\eusm L^0_{\Tau}/\eusm W_{\Cal J}\cong L^0(\frak B)$.
Set $f=g\phi$;  then

\Centerline{$\{x:f(x)>\alpha\}=\phi^{-1}[\{y:g(y)>\alpha\}]$}

\noindent belongs to $\Sigma$ for every $\alpha\in\Bbb R$, so
$f$ is $\Sigma$-measurable and we can speak of
$u=f^{\ssbullet}$ in
$\eusm L^0_{\Sigma}/\eusm W_{\Cal I}\cong L^0(\frak A)$.   Now, by 364Ca,

$$\eqalign{\Bvalue{u>\alpha}
&=\{x:f(x)>\alpha\}^{\ssbullet}
=\phi^{-1}[\{y:g(y)>\alpha\}]^{\ssbullet}\cr
&=\pi\{y:g(y)>\alpha\}^{\ssbullet}
=\pi\Bvalue{v>\alpha}
=\Bvalue{T_{\pi}v>\alpha}\cr}$$

\noindent for every $\alpha\in\Bbb R$, and

\Centerline{$(g\phi)^{\ssbullet}=f^{\ssbullet}=u=T_{\pi}v=T_{\pi}g^{\ssbullet}$,}

\noindent as claimed.

\medskip

{\bf (c)} Starting from the
facts that $(\psi\phi)^{-1}[G]=\phi^{-1}[\psi^{-1}[G]]$ for every
$G\in\Upsilon$ and $h(\psi\phi)=(h\psi)\phi$ for every
$h\in\eusm L^0_{\Upsilon}$, we just have to run through the formulae.

\medskip

{\bf (d)} If $g\in\eusm L^0(\nu)$, there are a
$g_0\in\eusm L^0_{\Tau}$ and an $F\in\Cal J$ such that $g(y)$ is defined
and equal to $g_0(y)$ for every $y\in Y\setminus F$.   In this case,
$\phi^{-1}[F]\in\Cal I$ and
$g\phi(x)$ is defined and equal to $g_0\phi(x)$ for every
$x\in X\setminus\phi^{-1}[F]$, so $g\phi\in\eusm L^0(\mu)$ and

\Centerline{$(g\phi)^{\ssbullet}=(g_0\phi)^{\ssbullet}
=T_{\pi}(g_0^{\ssbullet})=T_{\pi}(g^{\ssbullet})$}

\noindent by (b).
}%end of proof of 364Q

\leader{364R}{Products:  Proposition} Let
$\langle\frak A_i\rangle_{i\in I}$ be a family of Dedekind
$\sigma$-complete Boolean algebras, with simple product $\frak A$.   If
$\pi_i:\frak A\to\frak A_i$ is the coordinate map for each $i$, and
$T_i:L^0(\frak A)\to L^0(\frak A_i)$ the corresponding homomorphism,
then $u\mapsto Tu=\langle T_iu\rangle_{i\in I}:
L^0(\frak A)\to\prod_{i\in I}L^0(\frak A_i)$ is a multiplicative
Riesz space isomorphism,
so $L^0(\frak A)$ may be identified with the $f$-algebra product
$\prod_{i\in I}L^0(\frak A_i)$\cmmnt{ (352Wc)}.

\proof{ Because each $\pi_i$ is a surjective order-continuous Boolean
homomorphism, 364P assures us that there are corresponding surjective
multiplicative Riesz homomorphisms $T_i$.   So all we need to check is
that the multiplicative Riesz homomorphism
$T:L^0(\frak A)\to\prod_{i\in I}L^0(\frak A_i)$ is a bijection.

If $u$, $v\in L^0(\frak A)$ are distinct, there must be some
$\alpha\in\Bbb R$ such that $\Bvalue{u>\alpha}\ne\Bvalue{v>\alpha}$.
In this case there must be an $i\in I$ such that
$\pi_i\Bvalue{u>\alpha}\ne\pi_i\Bvalue{v>\alpha}$, that is,
$\Bvalue{T_iu>\alpha}\ne\Bvalue{T_iv>\alpha}$.   So $T_iu\ne T_iv$ and
$Tu\ne Tv$.   As $u$, $v$ are arbitrary, $T$ is injective.

If $w=\langle w_i\rangle_{i\in I}$ is any member of
$\prod_{i\in I}L^0(\frak A_i)$, then for $\alpha\in\Bbb R$ set

\Centerline{$\phi(\alpha)=\langle\Bvalue{w_i>\alpha}\rangle_{i\in
I}\in\frak A$.}

\noindent It is easy to check that $\phi$ satisfies the conditions of
364Aa, because, for instance,

\Centerline{$\sup_{\beta>\alpha}\pi_i\phi(\beta)
=\sup_{\beta>\alpha}\Bvalue{w_i>\beta}=\Bvalue{w_i>\alpha}
=\pi_i\phi(\alpha)$}

\noindent for every $i$, so that
$\sup_{\beta>\alpha}\phi(\beta)=\phi(\alpha)$, for every $\alpha\in\Bbb
R$;  and the other two conditions are also satisfied because they are
satisfied coordinate-by-coordinate.   So there is a $u\in L^0(\frak A)$
such that $\phi(\alpha)=\Bvalue{u>\alpha}$ for every $\alpha$, that is,
$\pi_i\Bvalue{u>\alpha}=\Bvalue{w_i>\alpha}$ for all $\alpha$, $i$, that
is, $T_iu=w_i$ for every $i$, that is, $Tu=w$.   As $w$ is arbitrary,
$T$ is surjective and we are done.
}%end of proof of 364R

\leader{*364S}{Regular open \dvrocolon{algebras}}\cmmnt{ I noted in
314P that for every topological space $X$ there is a corresponding
Dedekind complete Boolean algebra $\RO(X)$ of regular open sets.   We
have an identification of $L^0(\RO(X))$ as a space of equivalence
classes of
functions, different in kind from the representations above, as follows.
This is hard work (especially if we do it in full generality), but
instructive.   I start with a temporary definition.

\wheader{*364S}{4}{2}{2}{36pt}

\noindent}{\bf Definition} Let $(X,\frak T)$ be a topological space and
$f:X\to\Bbb R$ a function.   For $x\in X$ write

\Centerline{$\omega(f,x)
=\inf_{G\in\frak T,x\in G}\sup_{y,z\in G}|f(y)-f(z)|$}

\noindent (allowing $\infty$).

\leader{*364T}{Theorem} Let $X$ be any topological space, and
$\RO(X)$ its regular open algebra.   Let $U$ be the set of functions
$f:X\to\Bbb R$ such that $\{x:\omega(f,x)<\epsilon\}$ is dense in $X$
for every $\epsilon>0$.  Then $U$ is a Riesz subspace of $\Bbb R^X$,
closed under multiplication, and we have a surjective multiplicative
Riesz homomorphism $T:U\to L^0(\RO(X))$ defined by writing

\Centerline{$\Bvalue{Tf>\alpha}
=\sup_{\beta>\alpha}\interior\overline{\{x:f(x)>\beta\}}$,}

\noindent the supremum being taken in $\RO(X)$, for every
$\alpha\in\Bbb R$ and $f\in U$.   The kernel of $T$ is
the set $W$ of functions $f:X\to\Bbb R$ such that
$\interior\{x:|f(x)|\le\epsilon\}$ is dense for every $\epsilon>0$, so
$L^0(\RO(X))$ can be identified, as $f$-algebra, with the
quotient space $U/W$.

\proof{{\bf (a)(i)}\grheada\ The first thing to observe is that for any
$f\in\Bbb R^X$ and $\epsilon>0$ the set

$$\eqalign{\{x:\omega(f,x)<\epsilon\}
&=\bigcup\{G:G\subseteq X\text{ is open and non-empty}\cr
&\qquad\qquad\qquad\qquad
  \text{ and }\sup_{y,z\in G}|f(y)-f(z)|<\epsilon\}\cr}$$

\noindent is open.

\medskip

\qquad\grheadb\ Next, it is easy to see that

\Centerline{$\omega(f+g,x)\le\omega(f,x)+\omega(g,x)$,}

\Centerline{$\omega(\gamma f,x)=|\gamma|\omega(f,x)$,}

\Centerline{$\omega(|f|,x)\le\omega(f,x)$,}

\noindent for all $f$, $g\in\Bbb R^X$ and $\gamma\in\Bbb R$.

\medskip

\qquad\grheadc\ Thirdly, it is useful to know that if $f\in U$ and
$G\subseteq X$ is a non-empty open set, then there is a non-empty open
set $G'\subseteq G$ on which $f$ is bounded.   \Prf\ Take any $x_0\in G$
such that $\omega(f,x_0)<1$;  then there is a non-empty open set $G'$
containing $x_0$ such that $|f(y)-f(z)|<1$ for all $y$, $z\in G'$, and
we may suppose that $G'\subseteq G$.   But now $|f(x)|\le 1+|f(x_0)|$
for every $x\in G'$.\ \Qed

\medskip

\quad{\bf (ii)} So if $f$, $g\in U$ and $\gamma\in\Bbb R$ then

\Centerline{$\{x:\omega(f+g,x)<\epsilon\}
\supseteq\{x:\omega(f,x)<\Bover12\epsilon\}
   \cap\{x:\omega(g,x)<\Bover12\epsilon\}$}

\noindent is the intersection of two dense open sets and is therefore
dense, while

\Centerline{$\{x:\omega(\gamma f,x)<\epsilon\}
\supseteq\{x:\omega(f,x)<\Bover{\epsilon}{1+|\gamma|}\}$,}

\Centerline{$\{x:\omega(|f|,x)<\epsilon\}
\supseteq\{x:\omega(f,x)<\epsilon\}$}

\noindent are also dense.   As $\epsilon$ is arbitrary, $f+g$,
$\gamma f$ and $|f|$ all belong to $U$;  as $f$, $g$ and $\gamma$ are arbitrary, $U$ is a Riesz subspace of $\Bbb R^X$.

\medskip

\quad{\bf (iii)} If $f$, $g\in U$ then $f\times g\in U$.   \Prf\ Take
$\epsilon>0$ and let $G_0$ be a non-empty open subset of $X$.   By the
last remark in (i) above, there is a non-empty open set
$G_1\subseteq G_0$ such that $|f|\vee|g|$ is bounded on $G_1$;  say
$\max(|f(x)|,|g(x)|)\le\gamma$ for every $x\in G_1$.

Set $\delta=\Bover{\epsilon}{2\gamma+1}>0$.   Then there is an
$x\in G_1$ such that $\omega(f,x)<\delta$ and $\omega(g,x)<\delta$.
Let $H$, $H'$ be open sets containing $x$ such that $|f(y)-f(z)|\le\delta$
for all $y$, $z\in H$ and $|g(y)-g(z)|\le\delta$ for all $y$, $z\in H'$.
Consider $G=G_1\cap H\cap H'$.   This is an open set containing $x$, and
if $y$, $z\in G$ then

$$\eqalign{|f(y)g(y)-f(z)g(z)|
&\le|f(y)-f(z)||g(z)|+|f(z)||g(y)-g(z)|\cr
&\le\delta\gamma+\gamma\delta.\cr}$$

\noindent Accordingly

\Centerline{$\omega(f\times g,x)\le 2\delta\gamma<\epsilon$,}

\noindent while $x\in G_0$.   As $G_0$ is arbitrary,
$\{x:\omega(f\times g,x)<\epsilon\}$ is dense;  as $\epsilon$ is
arbitrary, $f\times g\in U$.\ \Qed

Thus $U$ is a subalgebra of $\Bbb R^X$.

\woddheader{*364T}{6}{2}{2}{30pt}

{\bf (b)} Now, for $f\in U$, consider the map $\phi_f:\Bbb R\to\RO(X)$
defined by setting

\Centerline{$\phi_f(\alpha)
=\sup_{\beta>\alpha}\interior\overline{\{x:f(x)>\beta\}}$}

\noindent for every $\alpha\in\Bbb R$.   Then $\phi_f$ satisfies the
conditions of 364Aa.   \Prf\ (See 314P for the calculation of suprema and
infima in $\RO(X)$.)  (i) If $\alpha\in\Bbb R$ then

$$\eqalign{\phi_f(\alpha)
&=\sup_{\beta>\alpha}\interior\overline{\{x:f(x)>\beta\}}
=\sup_{\gamma>\beta>\alpha}\interior\overline{\{x:f(x)>\gamma\}}\cr
&=\sup_{\beta>\alpha}\sup_{\gamma>\beta}
   \interior\overline{\{x:f(x)>\gamma\}}
=\sup_{\beta>\alpha}\phi_f(\beta).\cr}$$

\noindent (ii) If $G_0\subseteq X$ is a non-empty open set, then there
is a non-empty open set $G_1\subseteq G_0$ such that $f$ is bounded on
$G_1$;  say $|f(x)|<\gamma$ for every $x\in G_1$.   If $\beta>\gamma$
then $G_1$ does not meet $\{x:f(x)>\beta\}$, so
$G_1\cap\interior\overline{\{x:f(x)>\gamma\}}=\emptyset$;  as $\beta$ is
arbitrary, $G_1\cap\phi_f(\gamma)=\emptyset$ and
$G_0\not\subseteq\inf_{\alpha\in\Bbb R}\phi_f(\alpha)$.   On the other
hand, $G_1\subseteq\{x:f(x)>-\gamma\}$, so

\Centerline{$G_1\subseteq\interior\overline{\{x:f(x)>-\gamma\}}
\subseteq\phi_f(-\gamma)$}

\noindent and $G_0\cap\sup_{\alpha\in\Bbb R}\phi_f(\alpha)\ne\emptyset$.
As $G_0$ is arbitrary, $\inf_{\alpha\in\Bbb R}\phi_f(\alpha)=\emptyset$
and $\sup_{\alpha\in\Bbb R}\phi_f(\alpha)=X$.\ \Qed

\medskip

{\bf (c)} Thus we have a map $T:U\to L^0=L^0(\RO(X))$ defined by
setting $\Bvalue{Tf>\alpha}=\phi_f(\alpha)$ whenever $\alpha\in\Bbb R$ and
$f\in U$.

It is worth noting that

\Centerline{$\{x:f(x)>\alpha+\omega(f,x)\}
\subseteq\Bvalue{Tf>\alpha}
\subseteq\{x:f(x)+\omega(f,x)\ge\alpha\}$}

\noindent for every $f\in U$ and $\alpha\in\Bbb R$.   \Prf\ (i) If
$f(x)>\alpha+\omega(f,x)$, set
$\delta=\bover12(f(x)-\alpha-\omega(f,x))>0$.   Then there is an open
set $G$ containing $x$ such that $|f(y)-f(z)|<\omega(f,x)+\delta$ for
every $y$, $z\in G$, so that $f(y)>\alpha+\delta$ for every $y\in G$,
and

\Centerline{$x\in\interior\{y:f(y)>\alpha+\delta\}
\subseteq\Bvalue{Tf>\alpha}$.}

\noindent (ii) If $f(x)+\omega(f,x)<\alpha$, set
$\delta=\bover12(\alpha-f(x)-\omega(f,x))>0$;  then there is an open
neighbourhood $G$ of $x$ such that $|f(y)-f(z)|<\omega(f,x)+\delta$ for
every $y$, $z\in G$, so that $f(y)<\alpha$ for every $y\in G$.
Accordingly $G$ does not meet $\{y:f(y)>\beta\}$ nor
$\overline{\{y:f(y)>\beta\}}$ for any $\beta>\alpha$,
$G\cap\Bvalue{Tf>\alpha}=\emptyset$ and
$x\notin\Bvalue{Tf>\alpha}$.\ \Qed

\medskip

{\bf (d)} $T$ is additive.   \Prf\ Let $f$, $g\in U$ and
$\alpha<\beta\in\Bbb R$.
Set $\delta=\bover15(\beta-\alpha)>0$,
$H=\{x:\omega(f,x)<\delta,\,\omega(g,x)<\delta\}$;
then $H$ is the intersection of two dense open
sets, so is itself dense and open.

\medskip

\quad{\bf (i)} If $x\in H\cap\Bvalue{T(f+g)>\beta}$, then
$(f+g)(x)+\omega(f+g,x)\ge\beta$;  but $\omega(f+g,x)\le 2\delta$ (see
(a-i-$\beta$) above), so $f(x)+g(x)\ge\beta-2\delta>\alpha+2\delta$ and
there is a $q\in\Bbb Q$ such that

\Centerline{$f(x)>q+\delta\ge q+\omega(f,x)$,
\quad$g(x)>\alpha-q+\delta\ge\alpha-q+\omega(g,x)$.}

\noindent Accordingly

\Centerline{$x\in\Bvalue{Tf>q}\cap\Bvalue{Tg>\alpha-q}
\subseteq\Bvalue{Tf+Tg>\alpha}$.}

\noindent Thus
$H\cap\Bvalue{T(f+g)>\beta}\subseteq\Bvalue{Tf+Tg>\alpha}$.   Because
$H$ is dense, $\Bvalue{T(f+g)>\beta}\subseteq\Bvalue{Tf+Tg>\alpha}$.

\medskip

\quad{\bf (ii)} If $x\in H$, then

$$\eqalignno{x\in\bigcup_{q\in\Bbb Q}
&(\Bvalue{Tf>q}\cap\Bvalue{Tg>\beta-q})\cr
&\Longrightarrow\exists\,q\in\Bbb Q,\,f(x)+\omega(f,x)\ge q,\,
   g(x)+\omega(g,x)\ge\beta-q\cr
&\Longrightarrow f(x)+g(x)+2\delta\ge\beta\cr
&\Longrightarrow (f+g)(x)\ge\alpha+3\delta>\alpha+\omega(f+g,x)\cr
&\Longrightarrow x\in\Bvalue{T(f+g)>\alpha}.\cr}$$

\noindent Thus

\Centerline{$H\cap\bigcup_{q\in\Bbb
Q}(\Bvalue{Tf>q}\cap\Bvalue{Tg>\beta-q})
\subseteq\Bvalue{T(f+g)>\alpha}$.}

\noindent Because $H$ is dense and
$\bigcup_{q\in\Bbb Q}(\Bvalue{Tf>q}\cap\Bvalue{Tg>\beta-q})$ is open,

$$\eqalign{\Bvalue{Tf+Tg>\beta}
&=\interior\overline{\bigcup_{q\in\Bbb Q}\Bvalue{Tf>q}
   \cap\Bvalue{Tg>\beta-q}}\cr
&\subseteq\interior\overline{\Bvalue{T(f+g)>\alpha}}
=\Bvalue{T(f+g)>\alpha}.\cr}$$

\medskip

\quad{\bf (iii)} Now let $\beta\downarrow\alpha$;  we have

$$\eqalign{\Bvalue{T(f+g)>\alpha}
&=\sup_{\beta>\alpha}\Bvalue{T(f+g)>\beta}
\subseteq\Bvalue{Tf+Tg>\alpha}\cr
&=\sup_{\beta>\alpha}\Bvalue{Tf+Tg>\beta}
\subseteq\Bvalue{T(f+g)>\alpha},\cr}$$

\noindent so $\Bvalue{T(f+g)>\alpha}=\Bvalue{Tf+Tg>\alpha}$.   As
$\alpha$ is arbitrary, $T(f+g)=Tf+Tg$;  as $f$ and $g$ are arbitrary,
$T$ is additive.\ \Qed

\medskip

{\bf (e)} It is now easy to see that $T$ is linear.   \Prf\ If
$\gamma>0$, $f\in U$ and $\alpha\in\Bbb R$ then

$$\eqalign{\Bvalue{T(\gamma f)>\alpha}
&=\sup_{\beta>\alpha}\interior\overline{\{x:\gamma f(x)>\beta\}}
=\sup_{\beta>\alpha}
  \interior\overline{\{x:f(x)>\Bover{\beta}{\gamma}\}}\cr
&=\sup_{\beta>\alpha/\gamma}\interior\overline{\{x:f(x)>\beta\}}
=\Bvalue{Tf>\Bover{\alpha}{\gamma}}
=\Bvalue{\gamma Tf>\alpha}.\cr}$$

\noindent As $\alpha$ is arbitrary, $T(\gamma f)=\gamma Tf$;  because we
already know that $T$ is additive, this is enough to show that $T$ is
linear.\ \Qed

\medskip

{\bf (f)} In fact $T$ is a Riesz homomorphism.   \Prf\ If $f\in U$ and
$\alpha\ge 0$ then

$$\eqalign{\Bvalue{T(f^+)>\alpha}
&=\sup_{\beta>\alpha}\interior\overline{\{x:f^+(x)>\beta\}}
=\sup_{\beta>\alpha}\interior\overline{\{x:f(x)>\beta\}}\cr
&=\Bvalue{Tf>\alpha}
=\Bvalue{(Tf)^+>\alpha}.\cr}$$

\noindent If $\alpha<0$ then

\Centerline{$\Bvalue{T(f^+)>\alpha}
=\sup_{\beta>\alpha}\interior\overline{\{x:f^+(x)>\beta\}}
=X=\Bvalue{(Tf)^+>\alpha}$.   \Qed}

\medskip

{\bf (g)} Of course the constant function $\chi X$ belongs to $U$, and
is its multiplicative identity;  and $T(\chi X)$ is the multiplicative
identity of $L^0$, because

$$\eqalign{\Bvalue{T(\chi X)>\alpha}
&=\sup_{\beta>\alpha}\interior\overline{\{x:(\chi X)(x)>\beta\}}\cr
&=X\text{ if }\alpha<1,\,\emptyset\text{ if }\alpha\ge 1.\cr}$$

\noindent By 353Qd, or otherwise, $T$ is multiplicative.

\medskip

{\bf (h)}  The kernel of $T$ is $W$.   \Prf\ (i) For $f\in U$,

$$\eqalignno{Tf=0
&\Longrightarrow\Bvalue{T|f|>0}=\Bvalue{|Tf|>0}=\emptyset\cr
&\Longrightarrow\{x:|f(x)|>\omega(|f|,x)\}=\emptyset\cr
&\Longrightarrow\interior\{x:|f(x)|\le\epsilon\}
  \supseteq\{x:\omega(|f|,x)<\epsilon\}
  \text{ is dense for every }\epsilon>0\cr
&\Longrightarrow f\in W.\cr}$$

\noindent (ii) If $f\in W$, then, first,

\Centerline{$\{x:\omega(f,x)<\epsilon\}
\supseteq\interior\{x:|f(x)|\le\Bover13\epsilon\}$}

\noindent is dense for every $\epsilon>0$, so $f\in U$;  and next, for
any $\beta>0$, $\overline{\{x:|f(x)|>\beta\}}$ does not meet the dense
open set $\interior\{x:|f(x)|\le\beta\}$, so

\Centerline{$\Bvalue{|Tf|>0}=\Bvalue{T|f|>0}
=\sup_{\beta>0}\interior\overline{\{x:|f(x)|>\beta\}}=\emptyset$}

\noindent and $Tf=0$.\ \Qed

\medskip

{\bf (i)} Finally, $T$ is surjective.   \Prf\ Take any $u\in L^0$.
Define $\tilde f:X\to[-\infty,\infty]$ by setting
$\tilde f(x)=\sup\{\alpha:x\in\Bvalue{u>\alpha}\}$ for each $x$,
counting $\inf\emptyset$ as $-\infty$.   Then

\Centerline{$\{x:\tilde
f(x)>\alpha\}=\bigcup_{\beta>\alpha}\Bvalue{u>\beta}$}

\noindent is open, for every $\alpha\in\Bbb R$.   The set

\Centerline{$\{x:\tilde f(x)=\infty\}
=\bigcap_{\alpha\in\Bbb R}\Bvalue{u>\alpha}$}

\noindent is nowhere dense, because
$\inf_{\alpha\in\Bbb R}\Bvalue{u>\alpha}=\emptyset$ in $\RO(X)$;  while

\Centerline{$\{x:\tilde f(x)=-\infty\}
=X\setminus\bigcup_{\alpha\in\Bbb R}\Bvalue{u>\alpha}$}

\noindent also is nowhere dense, because
$\sup_{\alpha\in\Bbb R}\Bvalue{u>\alpha}=X$ in $\RO(X)$.   Accordingly
$E=\interior\{x:\tilde f(x)\in\Bbb R\}$ is dense.    Set
$f(x)=\tilde f(x)$ for $x\in E$, $0$ for $x\in X\setminus E$.

Let $\epsilon>0$.   If $G\subseteq X$ is a non-empty open set, there is
an $\alpha\in\Bbb R$ such that $G\not\subseteq\Bvalue{u>\alpha}$, so
$G_1=G\setminus\overline{\Bvalue{u>\alpha}}\ne\emptyset$, and
$\tilde f(x)\le\alpha$ for every $x\in G_1$.   Set

\Centerline{$\alpha'=\sup_{x\in G_1}\tilde f(x)\le\alpha<\infty$.}

\noindent Because $E$ meets $G_1$, $\alpha'>-\infty$.   Then
$G_2=G_1\cap\Bvalue{u>\alpha'-\bover12\epsilon}$ is a
non-empty open subset of $G$ and
$\alpha'-\bover12\epsilon\le\tilde f(x)\le\alpha'$ for
every $x\in G_2$.   Accordingly $|f(y)-f(z)|\le\bover12\epsilon$ for all
$y$, $z\in G_2$, and $\omega(f,x)<\epsilon$ for all $x\in G_2$.
As $G$ is arbitrary, $\{x:\omega(f,x)<\epsilon\}$ is dense;  as
$\epsilon$ is arbitrary, $f\in U$.

Take $\alpha<\beta$ in $\Bbb R$, and set
$\delta=\bover12(\beta-\alpha)$.   Then
$H=E\cap\{x:\omega(f,x)<\delta\}$ is a dense open set, and

$$\eqalign{H\cap\Bvalue{Tf>\beta}
&\subseteq H\cap\{x:f(x)+\omega(f,x)\ge\beta\}
\subseteq E\cap\{x:f(x)>\alpha\}\cr
&\subseteq\{x:\tilde f(x)>\alpha\}
\subseteq\Bvalue{u>\alpha}.\cr}$$

\noindent As $H$ is dense,
$\Bvalue{Tf>\beta}\subseteq\Bvalue{u>\alpha}$.   In the other direction

$$\eqalign{H\cap\Bvalue{u>\beta}
&\subseteq H\cap\{x:\tilde f(x)\ge\beta\}
=H\cap\{x:f(x)\ge\beta\}\cr
&\subseteq\{x:f(x)>\alpha+\omega(f,x)\}
\subseteq\Bvalue{Tf>\alpha},\cr}$$

\noindent so $\Bvalue{u>\beta}\subseteq\Bvalue{Tf>\alpha}$.   Just as in
(d) above, this is enough to show that $Tf=u$.   As $u$ is arbitrary,
$T$ is surjective.\ \Qed

This completes the proof.
}%end of proof of 364T

\leader{*364U}{Compact spaces} Suppose now that $X$ is a compact
Hausdorff topological space.
In this case the space $U$ of 364T is just the space of functions
$f:X\to\Bbb R$ such that $\{x:f$ is continuous at $x\}$ is dense in $X$.
\prooflet{\Prf\ It is easy to see that

\Centerline{$\{x:f$ is continuous at
$x\}=\{x:\omega(f,x)=0\}=\bigcap_{n\in\Bbb N}H_n$}

\noindent where $H_n=\{x:\omega(f,x)<2^{-n}\}$ for
each $n$.   Each $H_n$ is an open set (see part (a-i-$\alpha$) of the
proof of 364T), so by Baire's theorem (3A3G) $\bigcap_{n\in\Bbb N}H_n$
is dense iff every $H_n$ is dense, that is, iff $f\in U$.\ \Qed}

Now $W$, as defined in 364T,
becomes $\{f:f\in U,\,\{x:f(x)=0\}$ is dense$\}$.   \prooflet{\Prf\ (i) If
$f\in W$, then $T|f|=0$, so (by the formula in (c) of the proof of 364T)
$|f(x)|\le\omega(|f|,x)$ for every $x$.   But $\{x:\omega(f,x)=0\}$ is
dense, because $f\in U$, so $\{x:f(x)=0\}$ also is dense.   (ii) If
$f\in U$ and $\{x:f(x)=0\}$ is dense, then

\Centerline{$\omega(f,x)
\ge\inf_{x\in G\text{ is open}}\sup_{y\in G}|f(y)-f(x)|
\ge|f(x)|$}

\noindent for every $x\in X$.   So for any $\epsilon>0$,
$\interior\{x:|f(x)|\le\epsilon\}\supseteq\{x:\omega(f,x)<\epsilon\}$ is
dense, and $f\in W$.\ \Qed}


\cmmnt{In the case of extremally disconnected spaces, we can go
farther.}

\leader{*364V}{Theorem} Let $X$ be a compact Hausdorff extremally
disconnected space, and $\RO(X)$ its regular open algebra.   Write
$C^{\infty}=C^{\infty}(X)$ for the space of continuous functions
$g:X\to[-\infty,\infty]$ such that $\{x:g(x)=\pm\infty\}$ is nowhere
dense.   Then we have a bijection $S:C^{\infty}\to L^0=L^0(\RO(X))$
defined by saying that

\Centerline{$\Bvalue{Sg>\alpha}=\overline{\{x:g(x)>\alpha\}}$}

\noindent for every $\alpha\in\Bbb R$.   Addition and multiplication in
$L^0$ correspond to the operations $\frdotplus$, $\dottimes$ on
$C^{\infty}$ defined by saying that $g\frdotplus h$, $g\dottimes h$ are
the unique elements of $C^{\infty}$ agreeing with $g+h$, $g\times h$ on
$\{x:g(x),\,h(x)$ are both finite$\}$.   Scalar multiplication in $L^0$
corresponds to the operation

\Centerline{$(\gamma g)(x)=\gamma g(x)$ for $x\in X$, $g\in C^{\infty}$,
$\gamma\in\Bbb R$}

\noindent on $C^{\infty}$ (counting $0\cdot\infty$ as $0$), while the
ordering of $L^0$ corresponds to the relation

\Centerline{$g\le h\iff g(x)\le h(x)$ for every $x\in X$.}

\proof{{\bf (a)} For $g\in C^{\infty}$, set $H_g=\{x:g(x)\in\Bbb R\}$,
so that $H_g$ is a dense open set, and define $Rg:X\to\Bbb R$ by
setting $(Rg)(x)=g(x)$ if $x\in H_g$, $0$ if $x\in X\setminus H_g$.
Then $Rg$ is continuous at every point of $H_g$, so belongs to the space
$U$ of 364T-364U.   Set $Sg=T(Rg)$, where $T:U\to L^0$ is the map
of 364T.   Then

\Centerline{$\Bvalue{Sg>\alpha}=\overline{\{x:g(x)>\alpha\}}$}

\noindent for every $\alpha\in\Bbb R$.   \Prf\ (i) $\omega(g,x)=0$ for
every $x\in H_g$, so, if $\beta>\alpha$,

\Centerline{$H_g\cap\Bvalue{Sg>\beta}
\subseteq\{x:x\in H_g,\,(Rg)(x)\ge\beta\}
\subseteq\{x:g(x)\ge\beta\}$}

\noindent by the formula in part (c) of the proof of 364T.   As
$\Bvalue{Sg>\beta}$ is open and $H_g$ is dense,

\Centerline{$\Bvalue{Sg>\beta}
\subseteq\overline{H_g\cap\Bvalue{Sg>\beta}}\subseteq\{x:g(x)\ge\beta\}
\subseteq\{x:g(x)>\alpha\}$.}

\noindent Now

\Centerline{$\Bvalue{Sg>\alpha}=\sup_{\beta>\alpha}\Bvalue{Sg>\beta}
=\interior\overline{\bigcupop_{\beta>\alpha}\Bvalue{Sg>\beta}}
\subseteq\overline{\{x:g(x)>\alpha\}}$.}

\noindent (ii) In the other direction,
$H_g\cap\{x:g(x)>\alpha\}\subseteq\Bvalue{Sg>\alpha}$, by the other half
of the formula in the proof of 364T.   Again because $\{x:g(x)>\alpha\}$
is open and $H_g$ is dense,

\Centerline{$\overline{\{x:g(x)>\alpha\}}
\subseteq\overline{\Bvalue{Sg>\alpha}}=\Bvalue{Sg>\alpha}$}

\noindent because $X$ is extremally disconnected (see 314S).\ \Qed

\medskip

{\bf (b)} Thus $S=TR$ defined by the formula offered.
Now if $g$, $h\in C^{\infty}$ and $g\le h$, we surely have
$\{x:g(x)>\alpha\}
\subseteq\{x:h(x)>\alpha\}$ for every $\alpha$, so
$\Bvalue{Sg>\alpha}\subseteq\Bvalue{Sh>\alpha}$ for every $\alpha$ and
$Sg\le Sh$.   On the other hand, if $g\not\le h$ then $Sg\not\le Sh$.
\Prf\ Take $x_0$ such that $g(x_0)>h(x_0)$, and $\alpha\in\Bbb R$ such
that $g(x_0)>\alpha>h(x_0)$;  set $H=\{x:g(x)>\alpha>h(x)\}$;  this is a
non-empty open set and $H\subseteq\Bvalue{Sg>\alpha}$.   On the other
hand, $H\cap\{x:h(x)>\alpha\}=\emptyset$ so
$H\cap\Bvalue{Sh>\alpha}=\emptyset$.   Thus
$\Bvalue{Sg>\alpha}\not\subseteq\Bvalue{Sh>\alpha}$ and
$Sg\not\le Sh$.\ \QeD\   In particular, $S$ is injective.

\medskip

{\bf (c)} $S$ is surjective.   \Prf\ If $u\in L^0$, set

\Centerline{$g(x)
=\sup\{\alpha:x\in\Bvalue{u>\alpha}\}\in[-\infty,\infty]$}

\noindent for every $x\in X$, taking $\sup\emptyset=-\infty$.   Then,
for any $\alpha\in\Bbb R$,
$\{x:g(x)>\alpha\}=\bigcup_{\beta>\alpha}\Bvalue{u>\alpha}$ is open.
On the other hand,

\Centerline{$\{x:g(x)<\alpha\}
=\bigcup_{\beta<\alpha}\{x:x\notin\Bvalue{u>\beta}\}$}

\noindent also is open, because all the sets $\Bvalue{u>\beta}$ are
open-and-closed.   So $g:X\to[-\infty,\infty]$ is continuous.
Also

\Centerline{$\{x:g(x)>-\infty\}
=\bigcup_{\alpha\in\Bbb R}\Bvalue{u>\alpha}$,}

\Centerline{$\{x:g(x)<\infty\}
=\bigcup_{\alpha\in\Bbb R}X\setminus\Bvalue{u>\alpha}$}

\noindent are dense, so $g\in C^{\infty}$.    Now, for
any $\alpha\in\Bbb R$,

$$\eqalign{\Bvalue{Sg>\alpha}
&=\overline{\{x:g(x)>\alpha\}}
=\overline{\bigcup_{\beta>\alpha}\Bvalue{u>\beta}}\cr
&=\interior\overline{\bigcup_{\beta>\alpha}\Bvalue{u>\beta}}
=\sup_{\beta>\alpha}\Bvalue{u>\beta}
=\Bvalue{u>\alpha}.\cr}$$

\noindent So $Sg=u$.   As $u$ is arbitrary, $S$ is surjective.\ \Qed

\medskip

{\bf (d)} Accordingly $S$ is a bijection.   I have already checked (in
part (b)) that it is an isomorphism of the order structures.   For the
algebraic operations, observe that if $g$, $h\in C^{\infty}$ then there
are $f_1$, $f_2\in C^{\infty}$ such that $Sg+Sh=Sf_1$ and $Sg\times
Sh=Sf_2$, that is,

\Centerline{$T(Rg+Rh)=TRg+TRh=TRf_1$,
\quad$T(Rg\times Rh)=TRg\times TRh=TRf_2$.}

\noindent   But this means that

\Centerline{$T(Rg+Rh-Rf_1)=T((Rg\times Rh)-Rf_2)=0$,}

\noindent so that $Rg+Rh-Rf_1$, $(Rg\times Rh)-Rf_2$ belong to $W$, as
defined in 364T-364U, and
are zero on dense sets (364U).   Since we know also that the set
$G=\{x:g(x)$, $h(x)$ are both finite$\}$ is a dense open set, while $g$,
$h$, $f_1$ and $f_2$ are all continuous, we must have
$f_1(x)=g(x)+h(x)$, $f_2(x)=g(x)h(x)$ for every $x\in G$.   And of
course this uniquely specifies $f_1$ and $f_2$ as members of
$C^{\infty}$.

Thus we do have operations $\frdotplus$, $\dottimes$ as
described, rendering $S$ additive and multiplicative.   As for scalar
multiplication, it is easy to check that $R(\gamma g)=\gamma Rg$ (at
least, unless $\gamma=0$, which is trivial), so that
$S(\gamma g)=\gamma Sg$ for every $g\in C^{\infty}$.
}%end of proof of 364V

\exercises{\leader{364X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 364Xa
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   For $u$, $v\in L^0=L^0(\frak A)$ set
$\Bvalue{u<v}=\Bvalue{v>u}=\Bvalue{v-u>0}$,
$\Bvalue{u\le v}=\Bvalue{v\ge u}=1\Bsetminus\Bvalue{v<u}$,
$\Bvalue{u=v}=\Bvalue{u\le v}\Bcap\Bvalue{v\le u}$.   (i) Show that
$(\Bvalue{u<v},\Bvalue{u=v},\Bvalue{u>v})$ is always a partition of
unity in $\frak A$.   (ii) Show that for any $u$, $u'$, $v$,
$v'\in L^0$, $\Bvalue{u\le u'}\Bcap\Bvalue{v\le v'}
\discretionary{}{}{}\Bsubseteq\Bvalue{u+v\le u'+v'}$ and
$\Bvalue{u=u'}\Bcap\Bvalue{v=v'}\Bsubseteq\Bvalue{u\times v=u'\times v'}$.
%364F

\spheader 364Xb Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   (i) Show that if $u$, $v\in L^0=L^0(\frak A)$ and $\alpha$,
$\beta\in\Bbb R$ then $\Bvalue{u+v\ge\alpha+\beta}
\Bsubseteq\Bvalue{u\ge\alpha}\Bcup\Bvalue{v\ge\beta}$.   (ii) Show that
if $u$, $v\in(L^0)^+$ and $\alpha$, $\beta\ge 0$ then
$\Bvalue{u\times v\ge\alpha\beta}
\Bsubseteq\Bvalue{u\ge\alpha}\Bcup\Bvalue{v\ge\beta}$.
%364F, 364E

\spheader 364Xc Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $u\in L^0(\frak A)$.   Show that
$\{\Bvalue{u\in E}:E\subseteq\Bbb R$ is Borel$\}$ is
the $\sigma$-subalgebra of $\frak A$ generated by
$\{\Bvalue{u>\alpha}:\alpha\in\Bbb R\}$.
%364F

\sqheader 364Xe Let $(\frak A,\bar\mu)$ be a probability algebra, and
$\langle u_i\rangle_{i\in I}$ any family in $L^0(\frak A)$;  for each
$i\in I$ let $\frak B_i$ be the closed subalgebra of $\frak A$ generated
by $\{\Bvalue{u_i>\alpha}:\alpha\in\Bbb R\}$.   Show that the following
are equiveridical:  (i) $\bar\mu(\inf_{i\in J}\Bvalue{u_i>\alpha_i})
=\prod_{i\in J}\bar\mu\Bvalue{u_i>\alpha_i}$
whenever $J\subseteq I$ is finite and $\alpha_i\in\Bbb R$ for each
$i\in J$ (ii) $\langle\frak B_i\rangle_{i\in I}$ is
stochastically independent in the
sense of 325L.   (In this case we may call $\langle u_i\rangle_{i\in I}$
{\bf independent}.)
%364F used in vol 6

\sqheader 364Xf Let $(\frak A,\bar\mu)$ be a probability algebra and
$u$, $v$ two $\bar\mu$-independent members of $L^0(\frak A)$.
Show that the distribution of their sum is the convolution of their
distributions.    \Hint{272T}.
%364F used in chap 63

\sqheader 364Xg Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $g$, $h:\Bbb R\to\Bbb R$  Borel measurable functions.   (i)
Show that $\bar g\bar h=\bar{gh}$, where $\bar g$, $\bar h:L^0\to L^0$
are defined as in 364H.   (ii) Show that
$\overline{g+h}(u)=\bar g(u)+\bar h(u)$,
$\overline{g\times h}(u)=\bar g(u)\times\bar h(u)$ for
every $u\in L^0=L^0(\frak A)$.   (iii) Show that if $\sequencen{h_n}$ is
a sequence of Borel measurable functions on $\Bbb R$ and
$\sup_{n\in\Bbb N}h_n=h$, then
$\sup_{n\in\Bbb N}\bar{h}_n(u)=\bar h(u)$ for
every $u\in L^0$.   (iv) Show that if $h$ is non-decreasing and
continuous on the left, then $\bar h(\sup A)=\sup\bar h[A]$ whenever
$A\subseteq L^0$ is a non-empty set with a supremum in $L^0$.
%364H

\spheader 364Xh Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   (i) Show that $S(\frak A)$ can be identified ($\alpha$) with
the set of
those $u\in L^0=L^0(\frak A)$ such that
$\{\Bvalue{u>\alpha}:\alpha\in\Bbb R\}$ is finite ($\beta$) with the set
of those $u\in L^0$ such that $\Bvalue{u\in I}=1$ for some finite
$I\subseteq\Bbb R$.   (ii) Show that
$L^{\infty}(\frak A)$ can be identified with the set of those $u\in L^0$
such that $\Bvalue{u\in[-\alpha,\alpha]}=1$ for some $\alpha\ge 0$, and
that $\|u\|_{\infty}$ is the smallest such $\alpha$.
%364J

\spheader 364Xi Show that if $\frak A$ is a Dedekind $\sigma$-complete
Boolean algebra, and $u\in L^0(\frak A)$, then for any $\alpha\in\Bbb R$

\Centerline{$\Bvalue{u>\alpha}
=\inf_{\beta>\alpha}
  \sup\{a:a\in\frak A,\,u\times\chi a\ge\beta\chi a\}$}

\noindent (compare 363Xh).
%364J

\sqheader 364Xj\dvAnew{2008;  originally 3{}64Xw}
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\nu:\frak A\to\Bbb R$ a non-negative
finitely additive functional.
Let $\smalldashint:L^{\infty}(\frak A)\to\Bbb R$
be the corresponding linear
functional, as in 363L.   Write $U$ for the set of those
$u\in L^0(\frak A)$ such that
$\sup\{\smalldashint v:v\in L^{\infty}(\frak A)$, $v\le|u|\}$ is finite.
Show that $\smalldashint$ has an
extension to a non-negative linear functional on $U$.
%364J

\spheader 364Xk Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $u\ge 0$ in $L^0=L^0(\frak A)$.   Show that
$u=\sup_{q\in\Bbb Q}q\chi\Bvalue{u>q}$ in $L^0$.
%364L

\spheader 364Xl(i) Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $A\subseteq L^0(\frak A)$ a non-empty countable set
with supremum $w$.   Show that
$\Bvalue{w\in G}\Bsubseteq\sup_{u\in A}\Bvalue{u\in G}$ for every open
set $G\subseteq\Bbb R$.
(ii) Let $(\frak A,\bar\mu)$ be a localizable measure algebra and
$A\subseteq L^0(\frak A)$ a non-empty set with supremum $w$.   Show that
$\Bvalue{w\in G}\Bsubseteq\sup_{u\in A}\Bvalue{u\in G}$ for every open
set $G\subseteq\Bbb R$.
%364L

\spheader 364Xm
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra and
$A\subseteq L^0=L^0(\frak A)$ a non-empty set which is bounded below in
$L^0$.   Suppose that $\phi_0(\alpha)=\inf_{u\in A}\Bvalue{u>\alpha}$ is
defined in $\frak A$ for every $\alpha\in\Bbb R$.   Show that $v=\inf A$
is defined in $L^0$, and that
$\Bvalue{v>\alpha}=\sup_{\beta>\alpha}\phi_0(\beta)$ for every
$\alpha\in\Bbb R$.
%364L

\sqheader 364Xn Let $(X,\Sigma,\mu)$ be a measure space and
$f:X\to\coint{0,\infty}$ a function;  set
$A=\{g^{\ssbullet}:g\in\eusm L^0(\mu),\,g\leae f\}$.   (i) Show that if
$(X,\Sigma,\mu)$
either is localizable or has the measurable envelope property
(213Xl), then $\sup A$ is defined in $L^0(\mu)$.   (ii) Show that if
$(X,\Sigma,\mu)$ is complete and locally determined and $w=\sup A$ is
defined in $L^0(\mu)$, then $w\in A$.
%used in 614R

\spheader 364Xo Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   Show that if $u$, $v\in L^0=L^0(\frak A)$ then the
following are equiveridical:  ($\alpha$)
$\Bvalue{|v|>0}\Bsubseteq\Bvalue{|u|>0}$ ($\beta$) $v$ belongs to the
band in $L^0$ generated by $u$ ($\gamma$) there is a $w\in L^0$ such
that $u\times w=v$.
%364N

\sqheader 364Xp Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $a\in\frak A$;  let $\frak A_a$ be the principal ideal of
$\frak A$ generated by $a$.   Show that $L^0(\frak A_a)$ can be
identified, as $f$-algebra, with the band in $L^0(\frak A)$ generated
by $\chi a$.
%364O

\spheader 364Xq Let $\frak A$ and $\frak B$ be Dedekind
$\sigma$-complete Boolean algebras, and $\pi:\frak A\to\frak B$ a
sequentially order-continuous Boolean homomorphism.   Let
$T:L^0(\frak A)\to L^0(\frak B)$ be the corresponding Riesz homomorphism
(364P).   Show that (i) the kernel of $T$ is the sequentially
order-closed solid linear subspace of $L^0(\frak A)$ generated by
$\{\chi a:a\in\frak A,\,\pi a=0\}$   (ii) the set of values of $T$
is the sequentially
order-closed linear subspace of $L^0(\frak B)$ generated by
$\{\chi(\pi a):a\in\frak A\}$.
%364P

\spheader 364Xr Let $\frak A$ and $\frak B$ be Dedekind
$\sigma$-complete Boolean algebras, and $\pi:\frak A\to\frak B$ a
sequentially order-continuous Boolean homomorphism, with
$T:L^0(\frak A)\to L^0(\frak B)$ the associated operator.   Suppose that
$h$ is a Borel measurable real-valued function defined on a Borel subset
of $\Bbb R$.   Show that $\bar h(Tu)=T\bar h(u)$ whenever
$u\in L^0(\frak A)$ and $\bar h(u)$ is defined in the sense of 364H.
%364P

\spheader 364Xs Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
probability algebras, and $\pi:\frak A\to\frak B$ a measure-preserving
Boolean homomorphism;  let $T:L^0(\frak A)\to L^0(\frak B)$ be the
corresponding Riesz homomorphism.   Show that if $\familyiI{u_i}$ is a
family in $L^0(\frak A)$, it is $\bar\mu$-independent iff
$\familyiI{Tu_i}$ is $\bar\nu$-independent.
%364P 364Xe

\sqheader 364Xt Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\frak B$ a $\sigma$-subalgebra of $\frak A$.   Show that
$L^0(\frak B)$ can be identified with the sequentially order-closed
Riesz subspace of $L^0(\frak A)$ generated by $\{\chi b:b\in\frak B\}$.
%364P

\spheader 364Xu Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\pi:\frak A\to\frak A$ a sequentially order-continuous
Boolean homomorphism;  let $T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ be the
corresponding Riesz homomorphism.   Let $\frak C$ be the fixed-point
subalgebra of $\pi$.   Show that $\{u:u\in L^0(\frak A)$, $T_{\pi}u=u\}$
can be identified with $L^0(\frak C)$.
%364P 364Xt

\spheader 364Xv Use the ideas of part (d) of the proof of 364T to show
that the operator $T$ there is multiplicative, without appealing to
353Q.
%364T

\spheader 364Xw\dvAnew{2012} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra and $\frak B$ an order-closed subalgebra of $\frak A$.
Show that $L^0(\frak B)$, regarded as a subset of $L^0(\frak A)$,
is order-closed in $L^0(\frak A)$.
%364Xt 364P out of order query

\spheader 364Xx\dvAnew{2018} (W.Ricker) Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, and $\Cal I$ a $\sigma$-ideal of $\Sigma$
such that $\Sigma/\Cal I$ is Dedekind complete.   Suppose that $\Phi$ is a
family of $\Sigma$-measurable real-valued
functions, all with domains belonging to $\Sigma$, such that
$\{x:x\in\dom f\cap\dom g$, $f(x)\ne g(x)\}\in\Cal I$ whenever
$f$, $g\in\Phi$.
Show that there is a $\Sigma$-measurable function $h:X\to\Bbb R$ such that
$\{x:x\in\dom f$, $f(x)\ne h(x)\}\in\Cal I$ for every $f\in\Phi$.
\Hint{213N.}
%364M

\leader{364Y}{Further exercises $\pmb{>}$(a)}%
%\spheader 364Ya
(i) Show directly, without using the Loomis-Sikorski theorem or the
Stone representation,
that if $\frak A$ is any Dedekind $\sigma$-complete Boolean algebra then
the formulae of 364D define a group operation $+$ on $L^0(\frak A)$, and
generally an $f$-algebra structure.   (ii) Defining
$\chi:\frak A\to L^0(\frak A)$ by the formula in 364Jc, show that
$S(\frak A)$ and $L^{\infty}(\frak A)$ can be identified with the linear
span of $\{\chi a:a\in\frak A\}$ and the solid linear subspace of
$L^0(\frak A)$ generated by $e=\chi 1$.   (iii) Still without using the
Loomis-Sikorski theorem, explain how to define
$\bar h:L^0(\frak A)\to L^0(\frak A)$ for continuous functions
$h:\Bbb R\to\Bbb R$.   (iv) Check that these ideas are sufficient to yield
364L-364R, %364L 364M 364N 364O 364P 364R
except that in 364Pd we may have difficulty with arbitrary
Borel functions $h$.
%364D, 364J

\spheader 364Yb Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\pmb{u}=(u_1,\ldots,u_n)$ a member of $L^0(\frak A)^n$.
Write $\Cal B_n$ for the algebra of Borel sets in $\BbbR^n$.
(i) Show that there
is a unique sequentially order-continuous Boolean homomorphism
$E\mapsto\Bvalue{\pmb{u}\in E}:\Cal B_n\to\frak A$ such that
$\Bvalue{\pmb{u}\in E}=\inf_{i\le n}\Bvalue{u_i>\alpha_i}$ when
$E=\prod_{i\le n}\ooint{\alpha_i,\infty}$.
(ii) Show that for every sequentially order-continuous Boolean
homomorphism $\phi:\Cal B_n\to\frak A$ there is a unique
$\pmb{u}\in L^0(\frak A)^n$ such that
$\phi E=\Bvalue{\pmb{u}\in E}$ for every $E\in\Cal B_n$.
%364F

\spheader 364Yc Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra, $n\ge 1$ and $h:\BbbR^n\to\Bbb R$ a Borel measurable function.
Show that we have a corresponding function
$\bar h:L^0(\frak A)^n\to L^0(\frak A)$ defined by saying that
$\Bvalue{\bar h(\pmb{u})\in E}
=\Bvalue{\pmb{u}\in h^{-1}[E]}$ for every Borel set
$E\subseteq\Bbb R$ and $\pmb{u}\in L^0(\frak A)^n$.
%364H

\spheader 364Yd Suppose that $h_1(x,y)=x+y$, $h_2(x,y)=xy$,
$h_3(x,y)=\max(x,y)$ for all $x$, $y\in\Bbb R$.   Show that, in the
language of 364Yc, $\bar h_1(u,v)=u+v$, $\bar h_2(u,v)=u\times v$, $\bar
h_3(u,v)=u\vee v$ for all $u$, $v\in L^0$.
%364Yc, 364H

\spheader 364Ye Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   Show that $\frak A$ is ccc iff $L^0(\frak A)$ has the
countable sup property.
%364J 363Yb

\spheader 364Yf 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 such that $Te=e'$, where $e$, $e'$
are the multiplicative identities of $L^0(\frak A)$, $L^0(\frak B)$
respectively.   Show that there is a unique sequentially
order-continuous Boolean homomorphism $\pi:\frak A\to\frak B$ such that
$T=T_{\pi}$ in the sense of 364P.   %\Hint{use 353Qd.}
(Compare 375A below.)
%364P

\spheader 364Yg Let $\frak A$ and $\frak B$ be
Dedekind $\sigma$-complete Boolean algebras
and $\pi:\frak A\to\frak B$ a sequentially order-continuous ring
homomorphism.
(i) Show that we have a multiplicative sequentially order-continuous Riesz
homomorphism $T_{\pi}:L^0(\frak A)\to L^0(\frak B)$ defined by the
formula

\Centerline{$\Bvalue{T_{\pi}u>\alpha}=\pi\Bvalue{u>\alpha}$}

\noindent whenever $u\in L^0(\frak A)$ and $\alpha>0$.
(ii) Show that $T_{\pi}$ is order-continuous iff $\pi$ is order-continuous,
injective iff $\pi$ is injective, and surjective iff $\pi$ is surjective.
(iii) Show that if
$\frak C$ is another Dedekind $\sigma$-complete Boolean algebra
and $\theta:\frak B\to\frak C$ another sequentially order-continuous
ring homomorphism then
$T_{\theta\pi}=T_{\theta}T_{\pi}:L^0(\frak A)\to L^0(\frak C)$.
%364P out of order

\spheader 364Yh Suppose, in 364T, that $X=\Bbb Q$.   (i) Show that there
is an $f\in W$ such that $f(q)>0$ for every $q\in\Bbb Q$.   (ii) Show
that there is a $u\in L^0$ such that no $f\in U$ representing $u$ can be
continuous at any point of $\Bbb Q$.
%364T

\spheader 364Yi Let $X$ and $Y$ be topological spaces and $\phi:X\to Y$
a continuous function such that $\phi^{-1}[M]$ is nowhere dense in $X$
for every nowhere dense subset $M$ of $Y$.   (Cf.\ 313R.)   (i) Show
that we have an order-continuous Boolean homomorphism $\pi$ from the
regular open algebra $\RO(Y)$ of $Y$ to the regular open algebra
$\RO(X)$ of $X$ defined by the formula
$\pi G=\interior\overline{\phi^{-1}[G]}$ for every $G\in\RO(Y)$.
(ii) Show that if $U_X$, $U_Y$
are the function spaces of 364T  then $g\phi\in U_X$ for every
$g\in U_Y$.   (iii) Show that if $T_X:U_X\to L^0(\RO(X))$,
$T_Y:U_Y\to L^0(\RO(Y))$ are the canonical surjections, and $T:L^0(\RO(Y))\to L^0(\RO(X))$ is the homomorphism corresponding to $\pi$, then
$T(T_Yg)=T_X(g\phi)$ for every $g\in U_Y$.   (iv) Rewrite these ideas
for the special case in which $X$ is a dense subset of $Y$ and $\phi$ is
the identity map, showing that in this case $\pi$ and $T$ are
isomorphisms.
%364T

\spheader 364Yj Let $X$ be a Baire space, $\RO(X)$ its algebra of
regular open sets, $\Cal M$ its ideal of meager sets, and
$\widehat{\Cal B}$ the Baire-property
$\sigma$-algebra $\{G\symmdiff A:G\subseteq X$ is open, $A\in\Cal M\}$,
so that $\RO(X)$ can be identified with $\widehat{\Cal B}/\Cal M$
(314Yd).   (i)
Repeat the arguments of 364U in this context.  (ii) Show that the space
$U$ of 364T-364U is a subspace of
$\eusm L^0=\eusm L^0_{\widehat{\Cal B}}$, and
that $W=U\cap\eusm W$ where
$\eusm W=\{f:f\in\Bbb R^X,\,\{x:f(x)\ne 0\}\in\Cal M\}$, so that the
representations of $L^0(\RO(X))$ as $U/W$,
$\eusm L^0/\eusm W$ are consistent.
%364U

\spheader 364Yk Work through the arguments of 364T and 364Yj for the
case of compact Hausdorff $X$, seeking simplifications based on 364U.
%364U, 364Yj

\spheader 364Yl Let $X$ be an extremally disconnected compact Hausdorff
space with regular open algebra $\RO(X)$.   Let $U_0$ be the space of
real-valued functions $f:X\to\Bbb R$ such that $\interior\{x:f$ is
continuous at $x\}$ is dense.   Show that $U_0$ is a Riesz subspace of
the space $U$ of 364T, and that every member of $L^0(\RO(X))$ is
represented by a member of $U_0$.
%364V

\spheader 364Ym Let $X$ be a Baire space.   Let $Q$ be the set of all
continuous real-valued functions defined on subsets of $X$, and $Q^*$
the set of all members of $Q$ which are maximal in the sense that there
is no member of $Q$ properly extending them.   (i) Show that the domain
of any member of $Q^*$ is a dense G$_{\delta}$ set.   (ii) Show that we
can define addition and multiplication and scalar multiplication on
$Q^*$ by saying that $f\frdotplus g$, $f\dottimes g$, $\gamma.f$ are to
be the unique members of $Q^*$ extending the partially-defined functions
$f+g$, $f\times g$, $\gamma f$, and that these definitions render $Q^*$
an $f$-algebra if we say that $f\le g$ iff $f(x)\le g(x)$ for every
$x\in\dom f\cap\dom g$.   (iii) Show that every member of $Q^*$ has an
extension to a member of $U$, as defined in 364T, and that these
extensions define an isomorphism between $Q^*$ and $L^0(\RO(X))$, where
$\RO(X)$ is the regular open algebra of $X$.   (iv) Show that if $X$ is
compact, Hausdorff and extremally disconnected, then every member of
$Q^*$ has a unique extension to a member of $C^{\infty}(X)$, as defined
in 364V.
%364V

\spheader 364Yn Let $X$ be an extremally disconnected Hausdorff space,
and $Z$ any compact Hausdorff space.   Show that if $D\subseteq X$ is
dense and $f:D\to Z$ is continuous, there is a continuous $g:X\to Z$
extending $f$.
%364V

\spheader 364Yo Let $(\frak A,\bar\mu)$ be a probability algebra.
(i) Show that for any $\pmb{u}=(u_1,\ldots,u_n)\in L^0(\frak A)^n$
there is a unique Radon probability measure $\nu$ on
$\Bbb R^n$ such that
$\nu(\prod_{1\le i\le n}\ooint{\alpha,\infty})
=\bar\mu(\inf_{1\le i\le n}\Bvalue{u_i>\alpha_i})$ for all
$\alpha_1,\ldots,\alpha_n\in\Bbb R$, and that now
$\nu E=\bar\mu\Bvalue{\pmb{u}\in E}$ for every Borel set
$E\subseteq\BbbR^n$.
I will call $\nu$ the {\bf distribution} of $\pmb{u}$.
(ii) Show that $(u_1,\ldots,u_n)$ is
independent iff $\nu$ is expressible as $\prod_{1\le i\le n}\nu_i$ where
$\nu_i$ is a Radon probability measure on $\Bbb R$ for each $i$.
(iii) Write $\frak A_{\pmb{u}}$ for the closed subalgebra
$\{\Bvalue{\pmb{u}\in E}:E\subseteq\BbbR^n$ is a Borel set$\}$;  check that
$u_i\in L^0(\frak A_{\pmb{u}})$ for every $i$.
Suppose that $(\frak B,\bar\nu)$ is another probability algebra and
that $\pmb{v}=(v_1,\ldots,v_n)\in(L^0(\frak B))^n$.
Show that the following are equiveridical:  ($\alpha$) there
is a measure-preserving
isomorphism $\pi:\frak A_{\pmb{u}}\to\frak B_{\pmb{v}}$ such that
$T_{\pi}u_i=v_i$ for every $i$ ($\beta$) $\pmb{u}$ and $\pmb{v}$ have the
same distribution.
%364Xe 364Yb 364F 364P out of order query

\spheader 364Yp\dvAnew{2018} $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$ and $\Cal I$ a $\sigma$-ideal of $\Sigma$;  let $\frak A$ be the
quotient algebra $\Sigma/\Cal I$.   Set $\Cal L^0=\Cal L^0_{\Sigma}$ as
as in 364B-364C;  for $f\in\Cal L^0$ write $Tf^{\ssbullet}$
for the corresponding member of $L^0=L^0(\frak A)$ (364C).
Suppose that $\frak A$ is ccc.
Let $g:X\to\coint{0,\infty}$ be any function.   Show that
$\{Tf^{\ssbullet}:f\in\Cal L^0$, $f\le g\}$ is bounded above in $L^0$.
%364C out of order query
}%end of exercises

\endnotes{
\Notesheader{364} This has been a long section, and so far all we have
is a supposedly thorough grasp of the construction of $L^0$ spaces;
discussion of their properties still lies ahead.   The difficulties seem
to stem from a variety of causes.   First, $L^0$ spaces have a rich
structure, being linear ordered spaces with multiplications;
consequently all the main theorems have to check rather a lot of
different aspects.   Second, unlike $L^{\infty}$ spaces, they are not
accessible by means of the theory of normed spaces, so I must expect to
do more of the work here rather than in an appendix.   But this is in
fact a crucial difference, because it affects the proof of the central
theorem 364D.   The point is that a given algebra $\frak A$ will be
expressible in the form $\Sigma/\Cal I$ for a variety of algebras
$\Sigma$ of sets.   Consequently any definition of $L^0(\frak A)$ as a
quotient $\eusm L^0_{\Sigma}/\eusm W_{\Cal I}$
must include a check that the
structure produced is independent of the particular pair $\Sigma$,
$\Cal I$ chosen.

The same question arises with $S(\frak A)$ and $L^{\infty}(\frak A)$.
But in the case of $S$, I was able to use a general theory of additive
functions on $\frak A$ (see the proof of 361L), while in the case of
$L^{\infty}$ I could quote the result for $S$ and a little theory of
normed spaces (see the proof of 363H).
The theorems of \S368 will show, among other things, that a
similar approach (describing $L^0$ as a special kind of extension of $S$
or $L^{\infty}$) can be made to work in the present situation.
I have chosen, however, an alternative route using a novel technique.
The price is the time required to develop skill in the technique, and to
relate it to the earlier approach (364C, 364D, 364J).   The reward is a
construction which is based directly on the algebra $\frak A$,
independent of any representation (364A), and methods of dealing with it
which are complementary to those of the previous three sections.
In particular, they can be used in the absence of the full axiom of
choice (364Ya).

I have deliberately chosen the notation $\Bvalue{u>\alpha}$ from the
theory of forcing.   I do not propose to try to explain myself here, but
I remark that much of the labour of this section is a necessary basis
for understanding real analysis in Boolean-valued models of set theory.
The idea is that just as a function $f:X\to\Bbb R$ can be described in
terms of the sets $\{x:f(x)>\alpha\}$, so can an element $u$ of
$L^0(\frak A)$ be described in terms of the regions $\Bvalue{u>\alpha}$
of $\frak A$ where in some sense $u$ is greater than $\alpha$.   This
description is well adapted to discussion of the order struction of
$L^0(\frak A)$ (see 364L-364M), but rather ill-adapted to discussion of
its
linear and multiplicative structures, which leads to a large part of the
length of the work above.   Once we have succeeded in describing the
algebraic operations on $L^0$ in terms of the values of
$\Bvalue{u>\alpha}$, however, as in 364D, the fundamental result on the
action of Boolean homomorphisms (364P) is elegant and reasonably
straightforward.

The concept `$\Bvalue{u>\alpha}$' can be dramatically generalized to
the concept `$\Bvalue{(u_1,\ldots,u_n)\in E}$', where $E$ is a Borel
subset of $\BbbR^n$ and $u_1,\ldots,u_n\in L^0(\frak A)$ (364G,
364Yb).   This is supposed to recall the notation $\Pr(X\in E)$,
already used in Chapter 27.   If, as sometimes seems reasonable, we wish
to regard a random variable as a member of $L^0(\mu)$ rather than of
$\eusm L^0(\mu)$, then `$\Bvalue{u\in E}$' is the appropriate
translation of `$X^{-1}[E]$'.   The reasons why we can reach all Borel
sets $E$ here,
but then have to stop, seem to lie fairly deep;  I will return to this
question in 566O in Volume 5.   We see that we have
here another potential definition of $L^0(\frak A)$, as the set of
sequentially order-continuous Boolean homomorphisms from the Borel
$\sigma$-algebra of $\Bbb R$ to $\frak A$.   This is suitably
independent of realizations of $\frak A$, but makes the $f$-algebra
structure of $L^0$ difficult to elucidate, unless we move to a further
level of abstraction in the definitions, as in 364Yd.

I take the space to describe the $L^0$ spaces of general regular open
algebras in detail (364T) partly to offer a demonstration of an
appropriate technique, and partly to show that we are not limited to
$\sigma$-algebras of sets and their quotients.   This really is a new
representation;  for instance, it does not meld in any straightforward
way with the constructions of 364F-364H.   Of course the most
important examples are compact Hausdorff spaces, for which alternative
methods are available (364U-364V, 364Yj, 364Yl, 364Ym);  from the point
of view of applications, indeed, it is worth working through the details of
the theory for compact
Hausdorff spaces (364Yk).   The version in 364V is derived
from {\smc Vulikh 67}.   But I have starred everything from 364S on,
because I shall not rely on this work later for anything essential.
%368G uses 364V
}%end of comment

\discrpage

\frfilename{mt365.tex}
\versiondate{20.1.15}
\copyrightdate{1996}

\def\chaptername{Function spaces}
\def\sectionname{$L^1$}

\newsection{365}

Continuing my programme of developing the ideas of Chapter 24 at a
deeper level of abstraction, I arrive at last at $L^1$.   As usual, the
first step is to establish a definition which can be matched both with
the constructions of the previous sections and with the definition of
$L^1(\mu)$ in \S242 (365A-365C, 365F).   Next, I give what I regard as
the most characteristic internal properties of $L^1$ spaces, including
versions of the Radon-Nikod\'ym theorem (365E),
before turning to abstract versions of theorems in \S235 (365H, 365S)
and the duality between $L^1$ and $L^{\infty}$
(365K-365M).   As in \S\S361 and 363, the construction is associated
with universal mapping theorems (365I-365J) which define the Banach
lattice structure of $L^1$.   As in \S\S361, 363 and 364, homomorphisms
between measure algebras correspond to operators between their $L^1$
spaces;  but now the duality theory gives us two types of operators
(365N-365P), of which one class can be thought of as abstract
conditional expectations (365Q).   For localizable measure algebras, the
underlying algebra can be recovered from its $L^1$ space (365R), but the
measure cannot.

\leader{365A}{Definition} Let $(\frak A,\bar\mu)$ be a measure algebra.
For $u\in L^0(\frak A)$, write

\Centerline{$\|u\|_1
=\int_0^{\infty}\bar\mu\Bvalue{|u|>\alpha}\,d\alpha$,}

\noindent the integral being with respect to Lebesgue measure on
$\Bbb R$, and allowing $\infty$ as a value of the integral.
\cmmnt{(Because
the integrand is monotonic, it is certainly measurable.)}   Set
$L^1_{\bar\mu}=L^1(\frak A,\bar\mu)
=\{u:u\in L^0(\frak A),\,\|u\|_1<\infty\}$.

\cmmnt{It is convenient to note at once that if
$u\in L^1_{\bar\mu}$, then $\mu\Bvalue{|u|>\alpha}$ must be
finite for almost
every $\alpha>0$, and therefore for every $\alpha>0$, since it is a
non-increasing function of $\alpha$;  so that $\Bvalue{u>\alpha}$ also
belongs to the Boolean ring $\frak A^f=\{a:\bar\mu a<\infty\}$ for every
$\alpha>0$.
}

\leader{365B}{Theorem} Let $(X,\Sigma,\mu)$ be a measure space with
measure algebra $(\frak A,\bar\mu)$.   Then the canonical isomorphism
between $L^0(\mu)$ and $L^0(\frak A)$\cmmnt{ (364Ic)}
matches $L^1(\mu)\subseteq L^0(\mu)$\cmmnt{, defined in \S242,} with
$L^1(\frak A,\bar\mu)\subseteq L^0(\frak A)$, and the standard norm of
$L^1(\mu)$ with
$\|\,\|_1:L^1(\frak A,\bar\mu)\to\coint{0,\infty}$\cmmnt{, as
defined in 365A}.

\proof{ Take any $\Sigma$-measurable function $f:X\to\Bbb R$ (364B);
write $f^{\ssbullet}$
for its equivalence class in $L^0(\mu)$, and $u$ for the corresponding
element of $L^0(\frak A)$.   Then
$\Bvalue{|u|>\alpha}=\{x:|f(x)|>\alpha\}^{\ssbullet}$ in $\frak A$ for
every $\alpha\in\Bbb R$, and

\Centerline{$\|u\|_1
=\int_0^{\infty}\mu\{x:|f(x)|>\alpha\}\,d\alpha=\int|f|d\mu$}

\noindent by 252O.   In particular, $u\in L^1(\frak A,\bar\mu)$ iff
$f\in\eusm L^1(\mu)$ iff $f^{\ssbullet}\in L^1(\mu)$, and in this case
$\|u\|_1=\|f^{\ssbullet}\|_1$.
}%end of proof of 365B

\leader{365C}{}\cmmnt{ Accordingly we can apply everything we know
about $L^1(\mu)$ spaces to $L^1_{\bar\mu}$ spaces.   For instance:

\medskip

\noindent}{\bf Theorem} For any measure algebra $(\frak A,\bar\mu)$,
$L^1(\frak A,\bar\mu)$ is a solid linear subspace of
$L^0(\frak A)$, and
$\|\,\|_1$ is a norm on $L^1(\frak A,\bar\mu)$ under which
$L^1(\frak A,\bar\mu)$ is
an $L$-space.   Consequently $L^1(\frak A,\bar\mu)$ is a perfect Riesz
space with an order-continuous norm
which has the Levi property, and if $\sequencen{u_n}$ is a
non-decreasing norm-bounded sequence in $L^1(\frak A,\bar\mu)$ then it
converges for $\|\,\|_1$ to $\sup_{n\in\Bbb N}u_n$.

\proof{ $(\frak A,\bar\mu)$ is isomorphic to the measure algebra of some
measure space $(X,\Sigma,\mu)$ (321J).   $L^1(\mu)$ is a solid linear
subspace of $L^0(\mu)$ (242Cb), so $L^1_{\bar\mu}$ is a solid
linear subspace of $L^0(\frak A)$.   $L^1(\mu)$ is an $L$-space
(354M), so $L^1_{\bar\mu}$ also is.   The rest of the properties
claimed are general features of $L$-spaces (354N, 354E, 356P).
}%end of proof of 365C

\leader{365D}{Integration} Let $(\frak A,\bar\mu)$ be any measure
algebra.

\spheader 365Da If $u\in L^1=L^1(\frak A,\bar\mu)$, then $u^+$ and $u^-$,
calculated in $L^0=L^0(\frak A)$, belong to $L^1$, and we may set

\Centerline{$\int u
=\|u^+\|_1-\|u^-\|_1
=\int_0^{\infty}\bar\mu\Bvalue{u>\alpha}\,d\alpha
   -\int_0^{\infty}\bar\mu\Bvalue{-u>\alpha}\,d\alpha$.}

\noindent Now $\int:L^1\to\Bbb R$ is an order-continuous positive linear
functional\cmmnt{ (356Pc), and under the translation of 365B matches
the integral on $L^1(\mu)$ as defined in 242Ab}.   \cmmnt{Note that if
$a\in\frak A$ then

\Centerline{$\int\chi a
=\int_0^{\infty}\bar\mu\Bvalue{\chi a>\alpha}d\alpha
=\int_0^1\bar\mu a\,d\alpha=\bar\mu a$,}

\noindent so that if $\bar\mu$ is totally finite then the integral here
agrees with that of 363L on $L^{\infty}(\frak A)$.
I will sometimes write $\int u\,d\bar\mu$ if it seems helpful to
indicate the measure.}

\spheader 365Db\cmmnt{ Of course} $\|u\|_1=\int|u|\ge|\int u|$ for
every $u\in L^1$.

\spheader 365Dc If $u\in L^1$ and $a\in\frak A$ we may set
$\int_au=\int u\times\chi a$.   \cmmnt{(Compare 242Ac.)}   If
$\gamma>0$ and $0\ne a\Bsubseteq\Bvalue{u>\gamma}$ then\cmmnt{ there
is a $\delta>\gamma$ such that $a'=a\Bcap\Bvalue{u>\delta}\ne 0$, so
that}

\Centerline{$\int_au
\cmmnt{\mskip5mu
=\intop_0^{\infty}\bar\mu(a\Bcap\Bvalue{u>\alpha})d\alpha
\ge\intop_0^{\gamma}\bar\mu a\,d\alpha+\intop_{\gamma}^{\delta}\bar\mu a'}
>\gamma\bar\mu a$.}

\noindent In particular,\cmmnt{ setting $a=\Bvalue{u>\gamma}$,}
$\bar\mu\Bvalue{u>\gamma}$ must be finite.

\spheader 365Dd{\bf (i)} If $u\in L^1$ then $u\ge 0$
iff $\int_au\ge 0$ for every $a\in\frak A^f$\cmmnt{, writing
$\frak A^f=\{a:\bar\mu a<\infty\}$, as usual}.   \prooflet{\Prf\ If
$u\ge 0$
then $u\times\chi a\ge 0$ and $\int_au\ge 0$ for every $a\in\frak A$.   If
$u\not\ge 0$, then $\Bvalue{u^->0}\ne 0$ and there is an $\alpha>0$ such
that $a=\Bvalue{u^->\alpha}\ne 0$.   But now
$\bar\mu a$ is finite ((c) above) and

\Centerline{$\int u\times\chi a=-\int u^-\times\chi a
=-\int\bar\mu(a\Bcap\Bvalue{u^-\ge\beta})d\beta
\le-\alpha\bar\mu a<0$,}

\noindent so $\int_au<0$.\ \QeD}

\medskip

\quad{\bf (ii)} If $u$, $v\in L^1$ and $\int_au=\int_av$ for every
$a\in\frak A^f$ then $u=v$\cmmnt{ (cf.\ 242Ce)}.

\medskip

\quad{\bf (iii)} If $u\ge 0$ in $L^1$ then
$\int u=\sup\{\int_au:a\in\frak A^f\}$.   \prooflet{\Prf\ Of course
$u\times\chi a\le u$ so $\int_au\le u$ for every $a\in\frak A$.   On the
other hand,
setting $a_n=\Bvalue{u>2^{-n}}$, $\sequencen{u\times\chi a_n}$ is a
non-decreasing sequence with supremum $u$, so
$\int u=\lim_{n\to\infty}\int_{a_n}u$, while $\bar\mu a_n$ is finite for
every $n$.\ \Qed}

\spheader 365De If $u\in L^1$, $u\ge 0$ and $\int u=0$ then
$u=0$\prooflet{ (put 365B and 122Rc together)}.   If
$u\in L^1$, $u\ge 0$ and $\int_au=0$ then $u\times\chi a=0$, that is,
$a\Bcap\Bvalue{u>0}=0$.

\spheader 365Df If $C\subseteq L^1$ is non-empty and
upwards-directed and $\sup_{v\in C}\int v$ is finite, then $\sup C$ is
defined in $L^1$ and
$\int\sup C=\sup_{v\in C}\int v$\prooflet{ (356Pc)}.

\spheader 365Dg\cmmnt{ It will occasionally be convenient to adapt the
conventions of \S133 to the new context;  so that} I may write
$\int u=\infty$ if $u\in L^0$, $u^-\in L^1$ and
$u^+\notin L^1$, while $\int u=-\infty$ if $u^+\in L^1$ and
$u^-\notin L^1$.

\spheader 365Dh On this convention,\cmmnt{ we can restate (f) as
follows:}  if
$C\subseteq(L^0)^+$ is non-empty and upwards-directed and has a supremum
$u$ in $L^0$, then $\int u=\sup_{v\in C}\int v$ in $[0,\infty]$.
\prooflet{\Prf\ For if $\sup_{v\in C}\int v$ is infinite, then surely
$\int u=\infty$;  while otherwise we can apply (f).\ \Qed}

\leader{365E}{The Radon-Nikod\'ym theorem again} (a) Let
$(\frak A,\bar\mu)$ be a semi-finite measure algebra and
$\nu:\frak A\to\Bbb R$
an additive functional.   Then the following are equiveridical:

\inset{(i) there is a $u\in L^1=L^1(\frak A,\bar\mu)$ such that
$\nu a=\int_au$ for every $a\in\frak A$;

(ii) $\nu$ is additive and continuous for the measure-algebra
topology on $\frak A$;

(iii) $\nu$ is completely additive.}

(b) Let $(\frak A,\bar\mu)$ be any measure algebra, and
$\nu:\frak A^f\to\Bbb R$ a function.   Then the following are
equiveridical:

\inset{(i) $\nu$ is additive and bounded and $\inf_{a\in A}|\nu a|=0$
whenever $A\subseteq\frak A^f$ is downwards-directed and has infimum
$0$;

(ii) there is a $u\in L^1$ such that
$\nu a=\int_au$ for every $a\in\frak A^f$.}

\proof{{\bf (a)} The equivalence of (ii) and (iii) is 327Bd.
The equivalence of (i) and (iii) is just a translation of 327D into
the new context, since $(\frak A,\bar\mu)$ is isomorphic to the measure
algebra of a measure space which by 322Bd will be semi-finite.

\medskip

{\bf (b)(i)$\Rightarrow$(ii)}\grheada\ Set
$M=\sup_{a\in\frak A^f}|\nu a|<\infty$.

Let $D\subseteq\frak A^f$ be a maximal disjoint set.   For each $d\in
D$, write $\frak A_d$ for the principal ideal of $\frak A$ generated by
$d$, and $\bar\mu_d$ for the restriction of $\bar\mu$ to $\frak A_d$, so
that $(\frak A_d,\bar\mu_d)$ is a totally finite measure algebra.   Set
$\nu_d=\nu\restrp\frak A_d$;  then $\nu_d:\frak A_d\to\Bbb R$ is
completely additive.   By (a), there is a
$u_d\in L^1(\frak A_d,\bar\mu_d)$
such that $\int_au_d=\nu_da=\nu a$ for every $a\Bsubseteq d$.

Now $u_d^+\in L^0(\frak A_d)$ corresponds to a member $\tilde u_d^+$ of
$L^0(\frak A)^+$ defined by saying

$$\eqalign{\Bvalue{\tilde u_d^+>\alpha}&=\Bvalue{u_d^+>\alpha}
=\Bvalue{u_d>\alpha}\text{ if }\alpha\ge 0,\cr
&=1\text{ if }\alpha<0.\cr}$$

\noindent If $a\in\frak A$, then

\Centerline{$\int_a\tilde u_d^+d\bar\mu
=\int_0^{\infty}\bar\mu(a\Bcap\Bvalue{\tilde u_d^+>\alpha})d\alpha
=\int_0^{\infty}\bar\mu_d(a\Bcap\Bvalue{u_d^+>\alpha})d\alpha
=\int_{a\Bcap d}u_d^+d\bar\mu_d$;}

\noindent taking $a=1$, we see that
$\|\tilde u_d^+\|_1=\|u_d^+\|_1=\nu\Bvalue{u_d>0}$ is
finite, so that $\tilde u_d^+\in L^1$.

\medskip

\qquad\grheadb\ For any finite $I\subseteq D$, set
$v_I=\sum_{d\in I}\tilde u_d^+$.   Then

\Centerline{$\int v_I=\nu(\sup_{d\in I}\Bvalue{u_d>0})\le M$;}

\noindent consequently the upwards-directed set $A=\{v_I:I\subseteq D$
is finite$\}$ is bounded above in $L^1$, and we can set
$v=\sup A$ in $L^1$.
If $a\in\frak A$, then
$\int_av_I=\sum_{d\in I}\int_{a\Bcap d}u_d^+$ for each finite
$I\subseteq D$, so $\int_av=\sum_{d\in D}\int_{a\Bcap d}u_d^+$.

Applying the same arguments to $-\nu$, there is a $w\in L^1$
such that

\Centerline{$\int_aw=\sum_{d\in D}\int_{a\Bcap d}u_d^-$}

\noindent for every $a\in\frak A$.   Try $u=v-w$;  then

\Centerline{$\int_au
=\sum_{d\in D}\int_{a\Bcap d}u_d^+-\int_{a\Bcap d}u_d^- %
=\sum_{d\in D}\int_{a\Bcap d}u_d
=\sum_{d\in D}\nu(a\Bcap d)$}

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

\medskip

\qquad\grheadc\ Now take any $a\in\frak A^f$.   For
$J\subseteq D$ set $a_J=\sup_{d\in J}a\Bcap d$.   Let $\epsilon>0$.
Then there is a finite $I\subseteq D$ such that

\Centerline{$|\int_au-\nu a_J|
=|\sum_{d\in D}\nu(a\Bcap d)-\sum_{d\in J}\nu(a\Bcap d)|
\le\epsilon$}

\noindent whenever $I\subseteq J\subseteq D$ and $J$ is finite.   But
now consider

\Centerline{$A=\{a\Bsetminus a_J:I\subseteq J\subseteq D,\,J$ is
finite$\}$.}

\noindent Then $\inf A=0$, so there is a finite $J$ such that
$I\subseteq J\subseteq D$ and

\Centerline{$|\nu a-\nu a_J|=|\nu(a\Bsetminus a_J)|\le\epsilon$.}

\noindent Consequently

\Centerline{$|\nu a-\int_au|
\le|\nu a-\nu a_J|+|\int_au-\nu a_J|\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\nu a=\int_au$.   As $a$ is
arbitrary, (ii) is proved.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} From where we now are, this is nearly
trivial.   Thinking of $\nu a$ as $\int u\times\chi a$, $\nu$ is surely
additive and bounded.   Also $|\nu a|\le\int|u|\times\chi a$.   If
$A\subseteq\frak A^f$ is non-empty, downwards-directed and has infimum
$0$, the same is true of $\{|u|\times\chi a:a\in A\}$, because $a\mapsto
|u|\times\chi a$ is order-continuous, so

\Centerline{$\inf_{a\in A}|\nu a|\le\inf_{a\in A}\int|u|\times\chi
a=\inf_{a\in A}\||u|\times\chi a\|_1=0$.}

}%end of proof of 365E

\leader{365F}{}\cmmnt{ It will be useful later to have spelt out the
following elementary facts.

\medskip

\noindent}{\bf Lemma} Let $(\frak A,\bar\mu)$ be a measure algebra.
Write $S^f$ for the intersection $S(\frak A)\cap L^1(\frak A,\bar\mu)$.
Then $S^f$ is a norm-dense and order-dense Riesz subspace of
$L^1(\frak A,\bar\mu)$, and can be identified with $S(\frak A^f)$.
The function $\chi:\frak A^f\to S^f\subseteq L^1(\frak A,\bar\mu)$ is an
injective
order-continuous additive lattice homomorphism.   If $u\ge 0$ in
$L^1(\frak A,\bar\mu)$,
there is a non-decreasing sequence $\sequencen{u_n}$ in $(S^f)^+$ such
that $u=\sup_{n\in\Bbb N}u_n=\lim_{n\to\infty}u_n$.

\proof{ As in 364K, we can think of $S(\frak A^f)$ as a Riesz
subspace of $S=S(\frak A)$, embedded in $L^0(\frak A)$.
If $u\in S$, it is
expressible as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n\in\frak A$ are disjoint and no $\alpha_i$ is zero.
Now $|u|=\sum_{i=0}^n|\alpha_i|\chi a_i$, so $u\in L^1$ iff
$\bar\mu a_i<\infty$ for every $i$, that is, iff $u\in S(\frak A^f)$;  thus
$S^f\cong S(\frak A^f)$.

Now suppose that $u\ge 0$ in $L^1$.   By 364Jd, there is a
non-decreasing sequence $\sequencen{u_n}$ in $S(\frak A)^+$ such that
$u_0\ge 0$ and $u=\sup_{n\in\Bbb N}u_n$ in $L^0$.   Because $L^1$ is a
solid linear subspace of $L^0$, every $u_n$ belongs to $L^1$ and
therefore to $S^f$.   By 365C, $\sequencen{u_n}$ is norm-convergent to
$u$.   This shows also that $S^f$ is order-dense in $L^1$.

The map $\chi:\frak A^f\to S^f$ is an injective order-continuous
additive lattice homomorphism;  because $S^f$ is regularly embedded in
$L^1$ (352Ne), $\chi$ has the same properties when regarded as a map
into $L^1$.

For general $u\in L^1$, there are sequences in $S^f$ converging to $u^+$
and to $u^-$, so that their difference is a sequence in $S^f$ converging
to $u$, and $u$ belongs to the closure of $S^f$;  thus $S^f$ is
norm-dense in $L^1$.
}%end of proof of 365F

\cmmnt{\medskip

\noindent{\bf Remark} Of course $S^f$ here corresponds to the space of
(equivalence classes of) simple functions, as in 242Mb.
}

\leader{365G}{Semi-finite algebras:  Lemma} Let $(\frak A,\bar\mu)$ be a
measure algebra.

(a) $(\frak A,\bar\mu)$ is semi-finite iff $L^1=L^1(\frak A,\bar\mu)$ is
order-dense in $L^0=L^0(\frak A)$.

(b) In this case, writing
$S^f=S(\frak A)\cap L^1$\cmmnt{ (as in 365F)},
$\int u=\sup\{\int v:v\in S^f,\,0\le v\le u\}$ in $[0,\infty]$ for every
$u\in(L^0)^+$.


\proof{{\bf (a)} If $(\frak A,\bar\mu)$ is semi-finite then $S^f$ is
order-dense in $L^0$ (364K), so $L^1$ also must be.   If $L^1$ is
order-dense in $L^0$, then so is $S^f$, by 365F and 352Nc, so
($\frak A,\bar\mu)$ is semi-finite, by 364K in the other direction.

\medskip

{\bf (b)} Set $C=\{v:v\in S^f,\,0\le v\le u\}$.   Then $C$ is an
upwards-directed set with supremum $u$, because $S^f$ is order-dense in
$L^0$.   So $\int u=\sup_{v\in C}\int v$ by 365Dh.
}%end of proof of 365G

\leader{365H}{Measurable \dvrocolon{transformations}}\cmmnt{ We have a
generalization of the ideas of \S235 in this abstract context.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be measure algebras, and $\pi:\frak A\to\frak B$ a sequentially
order-continuous Boolean homomorphism.   Let
$T:L^0(\frak A)\to L^0(\frak B)$ be the sequentially order-continuous
Riesz homomorphism associated with $\pi$\cmmnt{ (364P)}.

(a) Suppose that $w\ge 0$ in $L^0(\frak B)$ is such that
$\int_{\pi a}w\,d\bar\nu=\bar\mu a$ whenever $a\in\frak A$ and
$\bar\mu a<\infty$.
Then for any $u\in L^1(\frak A,\bar\mu)$ and $a\in\frak A$,
$\int_{\pi a}Tu\times w\,d\bar\nu$ is defined and equal to
$\int_au\,d\bar\mu$.

(b) Suppose that $w'\ge 0$ in $L^0(\frak A)$ is such that
$\int_aw'd\bar\mu=\bar\nu(\pi a)$ for every $a\in\frak A$.   Then
$\int Tu\,d\bar\nu=\int u\times w'\,d\bar\mu$ whenever
$u\in L^0(\frak A)$ and
either integral is defined in $[-\infty,\infty]$.

\cmmnt{\medskip

\noindent{\bf Remark} I am using the convention of 365Dg concerning
`$\infty$' as the value of an integral.}

\proof{{\bf (a)} If $u\in S^f=L^1_{\bar\mu}\cap S(\frak A)$ then
$u$ is expressible as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n$ have finite measure, so that
$Tu=\sum_{i=0}^n\alpha_i\chi(\pi a_i)$ and

\Centerline{$\int Tu\times w\,d\bar\nu
=\sum_{i=0}^n\alpha_i\int w\times\chi\pi a_i\,d\bar\nu
=\sum_{i=0}^n\alpha_i\bar\mu a_i=\int u\,d\bar\mu$.}

\noindent If $u\ge 0$ in $L^1_{\bar\mu}$ there is a
non-decreasing sequence $\sequencen{u_n}$ in $S^f$ with supremum $u$, so
that $Tu=\sup_{n\in\Bbb N}Tu_n$ and
$w\times Tu=\sup_{n\in\Bbb N}w\times Tu_n$ in $L^0(\frak B)$, and

\Centerline{$\int Tu\times w=\sup_{n\in\Bbb N}\int Tu_n\times w
=\sup_{n\in\Bbb N}\int u_n=\int u$.}

\noindent (365Df tells us that in this context
$Tu\times w\in L^1_{\bar\nu}$.)   Finally, for general
$u\in L^1_{\bar\mu}$,

\Centerline{$\int Tu\times w=\int Tu^+\times w-\int Tu^-\times w
=\int u^+-\int u^-=\int u$.}

\medskip

{\bf (b)} The argument follows the same lines:  start with $u=\chi a$
for $a\in\frak A$, then with $u\in S(\frak A)$, then with $u\in
L^0(\frak A)^+$ and conclude with general $u\in L^0(\frak A)$.   The
point is that $T$ is a Riesz homomorphism, so that at the last step

$$\eqalign{\int Tu
&=\int(Tu)^+-\int(Tu)^- %
=\int T(u^+)-\int T(u^-)\cr
&=\int u^+\times w'-\int u^-\times w'
=\int(u\times w')^+-\int(u\times w')^- %
=\int u\times w'\cr}$$

\noindent whenever either side is defined in $[-\infty,\infty]$.
}%end of proof of 365H

\leader{365I}{Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra and
$U$ a Banach space.   Let $\nu:\frak A^f\to U$ be a function.   Then the
following are equiveridical:

\inset{(i) there is a continuous linear operator $T$ from
$\cmmnt{L^1=\mskip5mu}L^1(\frak A,\bar\mu)$ to $U$ such that
$\nu a=T(\chi a)$ for every $a\in\frak A^f$;

(ii)($\alpha$) $\nu$ is additive

\quad($\beta$) there is an $M\ge 0$ such
that $\|\nu a\|\le M\bar\mu a$ for every $a\in\frak A^f$.}

\noindent Moreover, in this case, $T$ is unique and $\|T\|$ is the
smallest number $M$ satisfying the condition in (ii-$\beta$).

\proof{{\bf (a)(i)$\Rightarrow$(ii)} If $T:L^1\to U$ is a continuous
linear operator, then $\chi a\in L^1$ for every
$a\in\frak A^f$, so
$\nu=T\chi$ is a function from $\frak A^f$ to $U$.   If $a$,
$b\in\frak A^f$ and $a\Bcap b=0$, then $\chi(a\Bcup b)=\chi a+\chi b$ in
$L^0=L^0(\frak A)$ and therefore in $L^1$, so

\Centerline{$\nu(a\Bcup b)=T\chi(a\Bcup b)=T(\chi a+\chi b)=T(\chi
a)+T(\chi b)=\nu a+\nu b$.}

\noindent If $a\in\frak A^f$ then $\|\chi a\|_1=\bar\mu a$ (using the
formula in 365A, or otherwise), so

\Centerline{$\|\nu a\|=\|T(\chi a)\|\le\|T\|\|\chi a\|_1=\|T\|\bar\mu
a$.}

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Now suppose that $\nu:\frak A^f\to U$ is
additive and that $\|\nu a\|\le M\bar\mu a$ for every $a\in\frak A^f$.
Write $S^f$ for $L^1\cap S(\frak A)$, as in 365F.   Then there is a linear
operator $T_0:S^f\to U$ such that $T_0(\chi a)=\nu a$ for every
$a\in\frak A^f$ (361F).
Next, $\|T_0u\|\le M\|u\|_1$ for every $u\in S^f$.   \Prf\ If
$u\in S^f\cong S(\frak A^f)$, then $u$ is expressible as
$\sum_{j=0}^m\beta_j\chi b_j$ where
$b_0,\ldots,b_m\in\frak A^f$ are disjoint (361Eb).   So

\Centerline{$\|T_0u\|=\|\sum_{j=0}^m\beta_j\nu b_j\|
\le M\sum_{j=0}^m|\beta_j|\bar\mu b_j=M\|u\|_1$.   \Qed}

\noindent There is therefore a continuous linear operator $T:L^1\to U$,
extending
$T_0$, and with $\|T\|\le\|T_0\|\le M$ (2A4I).   Of course we
still have $\nu=T\chi$.

\medskip

{\bf (c)} The argument in (b) shows that $T_0=T\restr S^f$ and $T$ are
uniquely defined from $\nu$.   We have also seen that if $T$, $\nu$
correspond to each other then

\Centerline{$\|\nu a\|\le\|T\|\bar\mu a$ for every $a\in\frak A^f$,}

\Centerline{$\|T\|\le M$ whenever $\|\nu a\|\le M\bar\mu a$ for every
$a\in\frak A^f$,}

\noindent so that

\Centerline{$\|T\|=\min\{M:M\ge 0,\,\|\nu a\|\le M\bar\mu a$ for every
$a\in\frak A^f\}$.}
}%end of proof of 365I

\leader{365J}{Theorem}\dvAformerly{3{}56K}
Let $(\frak A,\bar\mu)$ be a measure algebra, $U$
a Banach lattice, and $T$ a bounded linear operator from
$L^1=L^1(\frak A,\bar\mu)$ to $U$.   Let $\nu:\frak A^f\to U$ be the
corresponding additive function, as in 365I.

(a) $T$ is a positive linear operator iff $\nu a\ge 0$ in $U$ for every
$a\in\frak A^f$;  in this case, $T$ is order-continuous.

(b) If $U$ is Dedekind complete and
$T\in\eurm L^{\sim}(L^1;U)$, then
$|T|:L^1\to U$ corresponds to $|\nu|:\frak A^f\to U$, where

\Centerline{$|\nu|(a)=\sup\{\sum_{i=0}^n|\nu
a_i|:a_0,\ldots,a_n\Bsubseteq a$ are disjoint$\}$}

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

(c) $T$ is a Riesz homomorphism iff $\nu$ is a lattice homomorphism.

\proof{ As in 365F, let $S^f$ be $L^1\cap S(\frak A)$, identified with
$S(\frak A^f)$.

\medskip

{\bf (a)(i)} If $T$ is a positive linear operator and $a\in\frak A^f$,
then $\chi a\ge 0$ in $L^1$, so $\nu a=T(\chi a)\ge 0$ in $U$.

\medskip

\quad{\bf (ii)} Now suppose that $\nu a\ge 0$ in $U$ for every
$a\in\frak A^f$, and take $u\ge 0$ in $L^1$, $\epsilon>0$ in $\Bbb R$.
Then there is a $v\in S^f$ such that $0\le v\le u$ and
$\|u-v\|_1\le\epsilon$ (365F).   Express $v$ as
$\sum_{i=0}^n\alpha_i\chi a_i$ where $a_i\in\frak A^f$, $\alpha_i\ge 0$
for each $i$.   Now

\Centerline{$\|Tu-Tv\|\le\|T\|\|u-v\|_1
\le\epsilon\|T\|$.}

\noindent On the other hand,

\Centerline{$Tv=\sum_{i=0}^n\alpha_i\nu a_i\in U^+$.}

\noindent As $U^+$ is norm-closed in $U$ (354Bc), and $\epsilon$ is
arbitrary,
$Tu\in U^+$.   As $u$ is arbitrary, $T$ is a positive linear operator.

\medskip

\quad{\bf (iii)} By 355Ka, $T$ is order-continuous.

\medskip

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

\Centerline{$|\nu b|=|T(\chi b)|\le|T|(\chi b)\le|T|(\chi a)$}

\noindent for every $b\Bsubseteq a$, so $\{\nu b:b\Bsubseteq a\}$ is
order-bounded in $U$.   As $a$ is arbitrary, we have an additive function
$|\nu|:\frak A^f\to U$ given by the proposed formula, by 361H.
Next, $|T|:L^1\to U$ is a bounded linear operator (355C), so
we can speak of $\||T|\|$;  and we also have an additive function
$\nu_1:\frak A^f\to U$ corresponding to $|T|$.

If $b\Bsubseteq a\in\frak A^f$, then

\Centerline{$\nu b-\nu(a\Bsetminus b)
=T(\chi b)-T(\chi(a\Bsetminus b))
\le |T|(\chi b)+|T|(\chi(a\Bsetminus b))
=|T|(\chi a)=\nu_1a$;}

\noindent taking the supremum over $b$, the other formula in 361H tells us
that $|\nu|a\le\nu_1a$, so

\Centerline{$\||\nu|a\|\le\|\nu_1a\|=\||T|(\chi a)\|
\le\||T|\|\|\chi a\|_1=\||T|\|\bar\mu a$.}

\noindent By 365I, there is a bounded linear operator
$S:L^1\to U$ such that $S(\chi a)=|\nu|a$ for every $a\in\frak A^f$.

We now have $(S-T)(\chi a)=|\nu|a-\nu a\ge 0$ for every $a\in\frak A^f$, so
$S-T\ge 0$ in $\eurm L^{\sim}(L^1;U)$, by (a) above, and $T\le S$;
similarly, $-T\le S$ and
$|T|\le S$.   On the other hand, $|\nu|a\le\nu_1a$ for every $a$, so
the same argument shows that $S\le|T|$.   Thus $S=|T|$ and $|\nu|$
corresponds to $|T|$, as claimed.

\medskip

{\bf (c)(i)}  If $T$ is a lattice homomorphism, then so is $\nu=T\chi$,
because $\chi:\frak A^f\to S^f$ is a lattice homomorphism.

\medskip

\quad{\bf (ii)} Now suppose that $\chi$ is a lattice homomorphism.   In
this case $T\restr S^f$ is a Riesz homomorphism (361Gc), that is,
$|Tv|=T|v|$ for every $v\in S^f$.   Because $S^f$
is norm-dense in $L^1$ and the map $u\mapsto|u|$ is continuous both in
$L^1$
and in $U$ (354Bb), $|Tu|=T|u|$ for every $u\in L^1$, and $T$ is a Riesz
homomorphism.
}%end of proof of 365J

\leader{365K}{The duality between $L^1$ and
$L^{\infty}$}\dvAformerly{3{}65L}  Let
$(\frak A,\bar\mu)$ be a measure algebra, and set
$L^1=L^1(\frak A,\bar\mu)$, $L^{\infty}=L^{\infty}(\frak A)$.   If we
identify $L^{\infty}$ with the solid linear subspace of
$L^0=L^0(\frak A)$ generated by $e=\chi 1_{\frak A}$\cmmnt{ (364J)},
then we have a bilinear operator $(u,v)\mapsto u\times v:
L^1\times L^{\infty}\to L^1$\prooflet{, because
$|u\times v|\le\|v\|_{\infty}|u|$ and
$L^1$ is a solid linear subspace of $L^0$}.
\cmmnt{Note that} $\|u\times v\|_1\le\|u\|_1\|v\|_{\infty}$,
so\cmmnt{ that}
the bilinear operator $(u,v)\mapsto u\times v$ has norm at most
$1$\cmmnt{ (253Ab, 253E)}.   Consequently we have a bilinear functional
$(u,v)\mapsto\int u\times v:L^1\times L^{\infty}\to\Bbb R$, which also
has norm at most $1$, corresponding to
linear operators $S:L^1\to (L^{\infty})^*$ and $T:L^{\infty}\to(L^1)^*$,
both of norm at most $1$, defined by the formula

\Centerline{$(Su)(v)=(Tv)(u)=\int u\times v$ for $u\in L^1$,
$v\in L^{\infty}$.}

\noindent\cmmnt{ Because $L^1$ and $L^{\infty}$ are
both Banach lattices, we have }$(L^1)^*=(L^1)^{\sim}$ and
$(L^{\infty})^*=(L^{\infty})^{\sim}$\cmmnt{ (356Dc)}.
\cmmnt{Because the norm of
$L^1$ is order-continuous,} $(L^1)^*=(L^1)^{\times}$\cmmnt{ (356Dd)}.

\leader{365L}{Theorem}\dvAformerly{3{}65M} Let $(\frak A,\bar\mu)$ be a measure algebra, and
set $L^1=L^1(\frak A,\bar\mu)$, $L^{\infty}=L^{\infty}(\frak A)$.   Let
$S:L^1\to(L^{\infty})^*=(L^{\infty})^{\sim}$,
$T:L^{\infty}\to (L^1)^*=(L^1)^{\sim}=(L^1)^{\times}$ be the canonical
maps defined by
the duality between $L^1$ and $L^{\infty}$\cmmnt{, as in 365K}.   Then

(a) $S$ and $T$ are order-continuous Riesz homomorphisms,
$S[L^1]\subseteq(L^{\infty})^{\times}$, $S$ is
norm-preserving and $T[L^{\infty}]$ is order-dense in $(L^1)^{\sim}$.

(b) $(\frak A,\bar\mu)$ is semi-finite iff $T$ is injective, and in this
case $T$ is norm-preserving, while $S$ is a normed Riesz space
isomorphism between $L^1$ and $(L^{\infty})^{\times}$.

(c) $(\frak A,\bar\mu)$ is localizable iff $T$ is bijective, and in this
case $T$ is a normed Riesz space isomorphism between $L^{\infty}$ and
$(L^1)^*=(L^1)^{\sim}=(L^1)^{\times}$.

\proof{{\bf (a)(i)} If $u\ge 0$ in $L^1$ and $v\ge 0$ in $L^{\infty}$
then $u\times v\ge 0$ and

\Centerline{$(Tv)(u)=\int u\times v\ge 0$.}

\noindent As $u$ is arbitrary, $Tv\ge 0$ in $(L^1)^{\times}$;  as $v$ is
arbitrary, $T$ is a positive linear operator.

If $v\in L^{\infty}$, set $a=\Bvalue{v>0}\in\frak A$.   (Remember that
we are identifying $L^0(\mu)$, as defined in \S241, with $L^0(\frak A)$,
as defined in \S364.)   Then $v^+=v\times\chi a$, so for any $u\ge 0$ in
$L^1$

\Centerline{$(Tv^+)(u)=\int u\times v\times\chi a
=(Tv)(u\times\chi a)\le(Tv)^+(u)$.}

\noindent As $u$ is arbitrary, $Tv^+\le(Tv)^+$.   On the other hand,
because $T$ is a positive linear operator, $Tv^+\ge Tv$ and $Tv^+\ge 0$,
so $Tv^+\ge(Tv)^+$.   Thus $Tv^+=(Tv)^+$.   As $v$ is arbitrary, $T$ is
a Riesz homomorphism (352G).

\medskip

\quad{\bf (ii)} Exactly the same arguments show that $S$ is a Riesz
homomorphism.

\medskip

\quad{\bf (iii)} Given $u\in L^1$, set $a=\Bvalue{u>0}$;  then

\Centerline{$\|Su\|\ge (Su)(\chi a-\chi(1\Bsetminus a))
=\int_au-\int_{1\Bsetminus a}u=\int|u|=\|u\|_1\ge\|Su\|$.}

\noindent So $S$ is norm-preserving.

\medskip

\quad{\bf (iv)} By 355Ka, $S$ is order-continuous.

\medskip

\quad{\bf (v)} If $A\subseteq L^{\infty}$ is a non-empty
downwards-directed set with infimum $0$, and $u\in(L^1)^+$, then
$\inf_{v\in A}u\times v=0$ for every $u\in(L^1)^+$, because $v\mapsto
u\times v:L^0\to L^0$ is
order-continuous.   So

\Centerline{$\inf_{v\in A}(Tv)(u)
=\inf_{v\in A}\int u\times v=\inf_{v\in A}\|u\times v\|_1=0$.}

\noindent As $a$ is arbitrary,
the only possible non-negative lower bound for $T[A]$ in
$(L^1)^{\times}$ is $0$.   As $A$ is arbitrary, $T$ is order-continuous.

\medskip

\quad{\bf (vi)} The ideas of (v) show also that
$S[L^1]\subseteq(L^{\infty})^{\times}$.   \Prf\ If $u\in(L^1)^+$ and
$A\subseteq L^{\infty}$ is non-empty, downwards-directed and has infimum
$0$, then

\Centerline{$\inf_{v\in A}(Su)(v)=\inf_{v\in A}\int u\times v=0$.}

\noindent As $A$ is arbitrary, $Su$ is order-continuous.   For general
$u\in L^1$, $Su=Su^+-Su^-$ belongs to $(L^{\infty})^{\times}$.\ \Qed

\medskip

\quad{\bf (vii)} Now suppose that $h>0$ in
$(L^1)^{\sim}=(L^1)^*=(L^1)^{\times}$.
By 365Ja, applied to $-h$, there must be an $a\in\frak A^f$ such that
$h(\chi a)>0$.   Set $\nu b=h(\chi(a\Bcap b))$ for
$b\in\frak A^f$.   Then $\nu$ is additive and non-negative and bounded
by $\|h\|\bar\mu a$.   If $A\subseteq\frak A^f$ is a non-empty
downwards-directed set with infimum $0$, then $C=\{\chi b:b\in A\}$ is
downwards-directed and has infimum $0$ in $L^0(\frak A)$ (364Jc), so
$\inf_{b\in A}\nu b=\inf_{u\in C}h(u)=0$.   By 365Eb, there is a
$v\in L^1$ such that
$\nu b=\int_bv$ for every $b\in\frak A^f$.   As $\int_bv\ge 0$ for every
$b\in\frak A^f$, $v\ge 0$ (365C(d-i)).   Setting $b=\Bvalue{v>\|h\|}$,
we have

\Centerline{$\int_bv\le h(\chi b)\le\|h\|\|\chi b\|_1=\|h\|\bar\mu b$;}

\noindent so $b=0$ (365Cc).   Accordingly $0\le v\le\|h\|\chi 1$ and
$v\in L^{\infty}$.   Consider $Tv\in(L^1)^{\times}$.   We have
$Tv\ge 0$ because $T$ is positive;  also

\Centerline{$(Tv)(\chi a)=\int_av=\nu a=h(\chi a)>0$,}

\noindent so $Tv>0$.   Next, for every $b\in\frak A^f$,

\Centerline{$(Tv)(\chi b)=\int_bv=h(\chi(a\Bcap b))\le h(\chi b)$.}

\noindent By 365Ja again, $h-Tv\ge 0$, that is, $Tv\le h$.   As $h$ is
arbitrary, $T[L^{\infty}]$ is quasi-order-dense in $(L^1)^*$, therefore
order-dense (353A).

\medskip

{\bf (b)(i)}  If $(\frak A,\bar\mu)$ is not semi-finite, let
$a\in\frak A$ be such that $\bar\mu a=\infty$ and $\bar\mu b=\infty$
whenever $0\ne b\Bsubseteq a$.   If $u\in L^1$, then
$\Bvalue{|u|>\bover1n}$ has finite
measure for every $n\ge 1$, so must be disjoint from $a$;  accordingly

\Centerline{$a\Bcap\Bvalue{|u|>0}=\sup_{n\ge
1}a\Bcap\Bvalue{|u|>\bover1n}=0$.}

\noindent This means that $\int u\times\chi a=0$ for every $u\in L^1$.
Accordingly $T(\chi a)=0$ and $T$ is not injective.

\medskip

\quad{\bf (ii)} If $(\frak A,\bar\mu)$ is semi-finite, take any $v\in
L^{\infty}$.   Then if $0\le\delta<\|v\|_{\infty}$,
$a=\Bvalue{|v|>\delta}\ne 0$.   Let $b\subseteq a$ be such that
$0<\bar\mu b<\infty$.   Then $\chi b\in L^1$, and

\Centerline{$\|Tv\|=\||Tv|\|=\|T|v|\|
\ge\Bover{(T|v|)(\chi b)}{\|\chi b\|_1}\ge\delta$}

\noindent because $|v|\times\chi b\ge\delta\chi b$, so

\Centerline{$(T|v|)(\chi b)\ge\delta\bar\mu b=\delta\|\chi b\|_1$.}

\noindent   As $\delta$ is arbitrary, $\|Tv\|\ge\|v\|_{\infty}$.   But
we already know that $\|Tv\|\le\|v\|_{\infty}$, so the two are equal.
As $v$ is arbitrary, $T$ is norm-preserving (and, in particular, is
injective).

\medskip

\quad{\bf (iii)} Still supposing that $(\frak A,\bar\mu)$ is
semi-finite, $S[L^1]=(L^{\infty})^{\times}$.
\Prf\ Take any $h\in(L^{\infty})^{\times}$.    For $a\in\frak A$, set
$\nu a=h(\chi a)$.   By 363K, $\nu:\frak A\to\Bbb R$
is completely additive.   By 365Ea, there is a $u\in L^1$ such that

\Centerline{$(Su)(\chi a)=\int u\times\chi a=\int_au=\nu a
=h(\chi a)$}

\noindent for every $a\in\frak A$.   Because $Su$ and $h$ are both
linear functionals on $L^{\infty}$, they must agree on $S(\frak A)$;
because they are continuous and $S(\frak A)$ is dense in $L^{\infty}$
(363C), $Su=h$.   As $h$ is arbitrary, $S$ is surjective.\ \Qed

\medskip

{\bf (c)} Using (b), we know that if either $T$ is bijective or
$(\frak A,\bar\mu)$ is localizable, then $(\frak A,\bar\mu)$ is
semi-finite.   Given this, if $T$
is bijective, then it is a Riesz space isomorphism between $L^{\infty}$
and $(L^1)^{\sim}$, which is Dedekind complete (356B);  so 363Mb tells
us that $\frak A$ is Dedekind complete and $(\frak A,\bar\mu)$ is
localizable.   In the other direction, if $(\frak A,\bar\mu)$ is
localizable, then $L^{\infty}$ is Dedekind complete.   As $T$ is
injective, $T[L^{\infty}]$ is, in itself, Dedekind complete;  being an
order-dense Riesz subspace of $(L^1)^{\sim}$ (by (a) here) it must be
solid (353L);  as it contains $T(\chi 1)$, which is the standard order
unit of the $M$-space $(L^1)^{\sim}$, it is the whole of $(L^1)^{\sim}$,
and $T$ is bijective.
}%end of proof of 365L

\leader{365M}{Corollary}\dvAformerly{3{}65N}
If $(\frak A,\bar\mu)$ is a localizable measure
algebra, $L^{\infty}(\frak A)$ is a perfect Riesz space.

\proof{ By 365L(b)-(c), we can identify $L^{\infty}$ with
$(L^1_{\bar\mu})^{\times}\cong(L^{\infty})^{\times\times}$.
}%end of proof of 365M

\leader{365N}{Theorem}\dvAformerly{3{}65O}
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be measure algebras.   Let $\pi:\frak A^f\to\frak B^f$ be a
measure-preserving ring homomorphism.

(a) There is a unique order-continuous norm-preserving Riesz
homomorphism $T_{\pi}:L^1(\frak A,\bar\mu)\to L^1(\frak B,\bar\nu)$ such
that
$T_{\pi}(\chi a)=\chi(\pi a)$ whenever $a\in\frak A^f$.   We have
$T_{\pi}(u\times\chi a)=T_{\pi}u\times\chi(\pi a)$ whenever
$a\in\frak A^f$ and $u\in L^1(\frak A,\bar\mu)$.

(b) $\int T_{\pi}u=\int u$ and $\int_{\pi a}T_{\pi}u=\int_au$ for every
$u\in L^1(\frak A,\bar\mu)$ and $a\in\frak A^f$.

(c) $\Bvalue{T_{\pi}u>\alpha}=\pi\Bvalue{u>\alpha}$ for every
$u\in L^1(\frak A,\bar\mu)$ and $\alpha>0$.

(d) $T_{\pi}$ is surjective iff $\pi$ is.

(e) If $(\frak C,\bar\lambda)$ is another measure algebra and
$\theta:\frak B^f\to\frak C^f$ another measure-preserving ring
homomorphism, then $T_{\theta\pi}=T_{\theta}T_{\pi}:
L^1(\frak A,\bar\mu)\to L^1(\frak C,\bar\lambda)$.

\proof{ Throughout the
proof I will write $T$ for $T_{\pi}$ and $S^f$ for
$S(\frak A)\cap L^1_{\bar\mu}\cong S(\frak A^f)$ (see 365F).

\medskip

{\bf (a)(i)} We have a map $\psi:\frak A^f\to L^1_{\bar\nu}$ defined by
writing
$\psi a=\chi(\pi a)$ for $a\in\frak A^f$.   Because

\Centerline{$\chi\pi(a\Bcup b)=\chi(\pi a\Bcup\pi b)
=\chi\pi a+\chi\pi b,\quad\|\chi(\pi a)\|_1=\bar\nu(\pi a)=\bar\mu a$}

\noindent whenever $a$, $b\in\frak A^f$ and $a\Bcap b=0$, we get a
(unique) corresponding bounded linear operator
$T:L^1_{\bar\mu}\to L^1_{\bar\nu}$ such that $T\chi=\chi\pi$ on
$\frak A^f$ (365I).   Because
$\pi:\frak A^f\to\frak B^f$ and $\chi:\frak B^f\to L^1_{\bar\nu}$ are
lattice homomorphisms, so is $\psi$, and $T$ is a Riesz homomorphism
(365Jc).

\medskip

\quad{\bf (ii)}
If $u\in S^f$, express it as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n$ are disjoint in $\frak A^f$.   Then
$Tu=\sum_{i=0}^n\alpha_i\chi(\pi a_i)$ and $\pi a_0,\ldots,\pi a_n$ are
disjoint in $\frak B^f$, so

\Centerline{$\|Tu\|_1
=\sum_{i=0}^n|\alpha_i|\bar\nu(\pi a_i)
=\sum_{i=0}^n|\alpha_i|\bar\mu a_i
=\|u\|_1$.}

\noindent Because $S^f$ is norm-dense in $L^1_{\bar\mu}$ and
$u\mapsto\|u\|_1$ is continuous (in both $L^1_{\bar\mu}$ and
$L^1_{\bar\nu}$), $\|Tu\|_1=\|u\|_1$ for every $u\in L^1_{\bar\mu}$,
that is, $T$ is norm-preserving.   As noted in 365Ja, $T$ is
order-continuous.

\medskip

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

\Centerline{$T(\chi a\times\chi b)=T(\chi(a\Bcap b))=\chi\pi(a\Bcap b)
=\chi(\pi a\Bcap\pi b)=\chi\pi a\times\chi\pi b
=\chi\pi a\times T(\chi b)$.}

\noindent Because $T$ is linear and $\times$ is bilinear,
$T(\chi a\times u)=\chi\pi a\times Tu$ for every $u\in S^f$.
Because the maps
$u\mapsto u\times\chi a:L^1_{\bar\mu}\to L^1_{\bar\mu}$,
$T:L^1_{\bar\mu}\to L^1_{\bar\nu}$ and
$v\mapsto v\times\chi\pi a:L^1_{\bar\nu}\to L^1_{\bar\nu}$ are all
continuous,
$Tu\times\chi\pi a=T(u\times\chi a)$ for every $u\in L^1_{\bar\mu}$.

\medskip

\quad{\bf (iv)} $T$ is unique because the formula $T(\chi a)=\chi\pi a$
defines $T$ on the norm-dense and order-dense subspace $S^f$.

\medskip

{\bf (b)} Because $T$ is positive,

\Centerline{$\int Tu=\|Tu^+\|_1-\|Tu^-\|_1
=\|u^+\|_1-\|u^-\|_1=\int u$.}

\noindent For $a\in\frak A^f$,

\Centerline{$\int_{\pi a}Tu=\int Tu\times\chi\pi a
=\int T(u\times\chi a)=\int u\times\chi a=\int_au$.}

\medskip

{\bf (c)}  If $u\in S^f$, express it as $\sum_{i=0}^n\alpha_i\chi a_i$
where $a_0,\ldots,a_n$ are disjoint;  then

\Centerline{$\pi\Bvalue{u>\alpha}
=\pi(\sup_{i\in I}a_i)
=\sup_{i\in I}\pi a_i
=\Bvalue{Tu>\alpha}$}

\noindent where $I=\{i:i\le n,\,\alpha_i>\alpha\}$.   For
$u\in(L^1_{\bar\mu})^+$,
take a sequence $\sequencen{u_n}$ in $S^f$ with supremum $u$;  then
$\sup_{n\in\Bbb N}Tu_n=Tu$, so

$$\eqalignno{\pi\Bvalue{u>\alpha}
&=\pi(\sup_{n\in\Bbb N}\Bvalue{u_n>\alpha})\cr
\noalign{\noindent (364L(a-ii);  $\Bvalue{u>\alpha}\in\frak A^f$ by 365A)}
&=\sup_{n\in\Bbb N}\pi\Bvalue{u_n>\alpha}\cr
\noalign{\noindent (because $\pi$ is order-continuous, see 361Ad)}
&=\sup_{n\in\Bbb N}\Bvalue{Tu_n>\alpha}
=\Bvalue{Tu>\alpha}\cr}$$

\noindent because $T$ is order-continuous.
For general $u\in L^1_{\bar\mu}$,

\Centerline{$\pi\Bvalue{u>\alpha}=\pi\Bvalue{u^+>\alpha}
=\Bvalue{T(u^+)>\alpha}
=\Bvalue{(Tu)^+>\alpha}
=\Bvalue{Tu>\alpha}$}

\noindent because $T$ is a Riesz homomorphism.

\medskip

{\bf (d)(i)} Suppose that $T$ is surjective and that $b\in\frak B^f$.
Then there is a $u\in L^1_{\bar\mu}$ such that $Tu=\chi b$.   Now

\Centerline{$b=\Bvalue{Tu>\bover12}
=\pi\Bvalue{u>\bover12}\in\pi[\frak A^f]$;}

\noindent as $b$ is arbitrary, $\pi$ is surjective.

\medskip

\quad{\bf (ii)} Suppose now that $\pi$ is surjective.   Then
$T[L^1_{\bar\mu}]$ is a linear subspace of $L^1_{\bar\nu}$ containing
$\chi b$ for every $b\in\frak B^f$, so includes $S(\frak B^f)$.   If
$v\in(L^1_{\bar\nu})^+$ there is a sequence $\sequencen{v_n}$ in
$S(\frak B^f)^+$ with supremum $v$.   For each $n$, choose $u_n$ such
that
$Tu_n=v_n$.   Setting $u'_n=\sup_{i\le n}u_i$, we get a
non-decreasing sequence $\sequencen{u'_n}$ such that $v_n\le Tu'_n\le v$
for every $n\in\Bbb N$.   So

\Centerline{$\sup_{n\in\Bbb N}\|u'_n\|_1=\sup_{n\in\Bbb
N}\|Tu'_n\|_1\le\|v\|_1<\infty$}

\noindent and $u=\sup_{n\in\Bbb N}u'_n$ is defined in $L^1_{\bar\mu}$,
with

\Centerline{$Tu=\sup_{n\in\Bbb N}Tu'_n=v$.}

\noindent Thus $(L^1_{\bar\nu})^+\subseteq T[L^1_{\bar\mu}]$;
consequently $L^1_{\bar\nu}\subseteq T[L^1_{\bar\mu}]$ and $T$ is
surjective.

\medskip

{\bf (e)} This is an immediate consequence of the `uniqueness'
assertion in (i), because for any $a\in\frak A^f$

\Centerline{$T_{\theta}T_{\pi}(\chi a)
=T_{\theta}\chi(\pi a)
=\chi(\theta\pi a)$,}

\noindent so that $T_{\theta}T_{\pi}:L^1_{\bar\mu}\to L^1_{\bar\lambda}$
is a bounded linear operator taking the right values at elements
$\chi a$, and must therefore be equal to $T_{\theta\pi}$.
}%end of proof of 365N

\leader{365O}{Theorem}\dvAformerly{3{}65P}
Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be measure algebras, and $\pi:\frak A^f\to\frak B$ an order-continuous
ring homomorphism.

(a) There is a unique positive linear operator
$P_{\pi}:L^1(\frak B,\bar\nu)\to L^1(\frak A,\bar\mu)$ such that
$\int_aP_{\pi}v=\int_{\pi a}v$ for every $v\in L^1(\frak B,\bar\nu)$ and
$a\in\frak A^f$.

(b) $P_{\pi}$ is order-continuous and norm-continuous, and
$\|P_{\pi}\|\le 1$.

(c) If $a\in\frak A^f$ and $v\in L^1(\frak B,\bar\nu)$ then
$P_{\pi}(v\times\chi\pi a)=P_{\pi}v\times\chi a$.

(d) If $\pi[\frak A^f]$ is order-dense in $\frak B$ then $P_{\pi}$ is a
norm-preserving
Riesz homomorphism;  in particular, $P_{\pi}$ is injective.

(e) If $(\frak B,\bar\nu)$ is semi-finite and $\pi$ is
injective, then $P_{\pi}$ is surjective, and there is for every
$u\in L^1(\frak A,\bar\mu)$ a $v\in L^1(\frak B,\bar\nu)$ such that
$P_{\pi}v=u$ and $\|v\|_1=\|u\|_1$.

(f) Suppose again that $(\frak B,\bar\nu)$ is semi-finite.   If
$(\frak C,\bar\lambda)$
is another measure algebra and $\theta:\frak B\to\frak C$ an
order-continuous Boolean homomorphism, then
$P_{\theta\pi}=P_{\pi}P_{\theta'}:
L^1(\frak C,\bar\lambda)\to L^1(\frak A,\bar\mu)$, where I write
$\theta'$ for the restriction of $\theta$ to $\frak B^f$.

\proof{ I write $P$ for $P_{\pi}$.

\medskip

{\bf (a)-(b)} For $v\in L^1_{\bar\nu}$ and $a\in\frak A^f$ set
$\nu_v(a)=\int_{\pi a}v$.   Then $\nu_v:\frak A^f\to\Bbb R$ is additive,
bounded (by $\|v\|_1$) and if $A\subseteq\frak A^f$ is non-empty,
downwards-directed and has infimum $0$, then

\Centerline{$\inf_{a\in A}|\nu_v(a)|
\le\inf_{a\in A}\int|v|\times\chi\pi a
=0$}

\noindent because $a\mapsto\int|v|\times\chi\pi a$ is a composition of
order-continuous functions, therefore
order-continuous.   So 365Eb tells us that there is a $Pv\in
L^1_{\bar\mu}$ such that $\int_aPv=\nu_v(a)=\int_{\pi a}v$ for every
$a\in\frak A^f$.   By 365D(d-ii), this formula defines $Pv$ uniquely.
Consequently $P$ must be linear (since $Pv_1+Pv_2$, $\alpha Pv$ will
always have the properties defining $P(v_1+v_2)$, $P(\alpha v)$).

If $v\ge 0$ in $L^1_{\bar\nu}$, then $\int_aPv=\int_{\pi a}v\ge 0$ for
every $a\in\frak A^f$, so $Pv\ge 0$ (365D(d-i));  thus $P$ is positive.
It must therefore be norm-continuous and order-continuous (355C, 355Ka).

Again supposing that $v\ge 0$, we have

\Centerline{$\|Pv\|_1=\int Pv=\sup_{a\in\frak A^f}\int_aPv
=\sup_{a\in\frak A^f}\int_{\pi a}v\le\|v\|_1$}

\noindent (using 365D(d-iii)).   For general $v\in L^1_{\bar\nu}$,

\Centerline{$\|Pv\|_1=\||Pv|\|_1\le\|P|v|\|_1\le\|v\|_1$.}

\medskip

{\bf (c)} For any $c\in\frak A^f$,

\Centerline{$\int_cPv\times\chi a
=\int_{c\Bcap a}Pv
=\int_{\pi(c\Bcap a)}v
=\int_{\pi c}v\times\chi\pi a
=\int_cP(v\times\chi\pi a)$.}

\medskip

{\bf (d)} Now suppose that $\pi[\frak A^f]$ is order-dense.   Take any
$v$, $v'\in L^1_{\bar\nu}$ such that $v\wedge v'=0$.   \Quer\ Suppose,
if possible, that $u=Pv\wedge Pv'>0$.   Take $\alpha>0$ such that
$a=\Bvalue{u>\alpha}$ is non-zero.   Since

\Centerline{$\int_{\pi a}v=\int_aPv\ge\int_au>0$,}

\noindent $b=\pi a\Bcap\Bvalue{v>0}$ is non-zero.
Let $c\in\frak A^f$ be such
that $0\ne \pi c\Bsubseteq b$;  then $\pi(a\Bcap c)=\pi c\ne 0$, so
$a\Bcap c\ne 0$, and

\Centerline{$0<\int_{a\Bcap c}u\le\int_{a\Bcap c}Pv'
\le\int_{\pi c}v'$.}

\noindent But $\pi c\Bsubseteq\Bvalue{v>0}$ and $v\wedge v'=0$ so
$\int_{\pi c}v'=0$.\ \Bang

So $Pv\wedge Pv'=0$.   As $v$, $v'$ are arbitrary, $P$ is a Riesz
homomorphism (352G).

Next, if $v\ge 0$ in $L^1_{\bar\nu}$,

\Centerline{$\int Pv=\sup_{a\in\frak A^f}\int_aPv=\sup_{a\in\frak
A^f}\int_{\pi a}v=\int v$}

\noindent because $\pi[\frak A^f]$ is upwards-directed and has supremum
$1$ in $\frak B$.   So, for general $v\in L^1_{\bar\nu}$,

\Centerline{$\|Pv\|_1=\int|Pv|=\int P|v|=\int|v|=\|v\|_1$,}

\noindent and $P$ is norm-preserving.

\medskip

{\bf (e)} Next suppose that $(\frak B,\bar\nu)$ is semi-finite and that
$\pi$ is injective.

\medskip

\quad{\bf (i)} If $u>0$ in $L^1_{\bar\mu}$, there is a $v>0$ in
$L^1_{\bar\nu}$ such that $Pv\le u$ and $\int Pv\ge\int v$.   \Prf\ Let
$\delta>0$ be such that
$a=\Bvalue{u>\delta}\ne 0$.   Then $\pi a\ne 0$.   Because $(\frak
B,\bar\nu)$ is semi-finite, there is a non-zero $b\in\frak B^f$ such
that $b\Bsubseteq \pi a$.   Set $u_1=P(\chi b)$.   Then $u_1\ge 0$,
$\int_au_1=\bar\nu b>0$ and

\Centerline{$\int_{1\Bsetminus a}u_1
=\sup_{c\in\frak A^f}\int_{c\Bsetminus a}u_1
=\sup_{c\in\frak A^f}\int_{\pi c\Bsetminus\pi a}\chi b
=0$.}

\noindent So $u_1\times\chi(1\Bsetminus a)=0$ and
$0\ne\Bvalue{u_1>0}\Bsubseteq a$.
Let $\gamma>0$ be such that $\Bvalue{u_1>\gamma}\ne\Bvalue{u_1>0}$, and
set $a_1=a\Bsetminus\Bvalue{u_1>\gamma}$,
$v=\Bover{\delta}{\gamma}\chi(b\Bcap\pi a_1)$.   Then

\Centerline{$Pv
=\Bover{\delta}{\gamma} P(\chi b\times\chi(\pi a_1))
=\Bover{\delta}{\gamma} P(\chi b)\times\chi a_1
=\Bover{\delta}{\gamma}u_1\times\chi a_1
\le\delta\chi a\le u$,}

\noindent because

\Centerline{$\Bvalue{u_1\times\chi a_1
>\gamma}\Bsubseteq\Bvalue{u_1>\gamma}\Bcap a_1=0$}

\noindent so

\Centerline{$u_1\times\chi a_1\le\gamma\chi\Bvalue{u_1>0}
\le\gamma\chi a$.}

\noindent Also $a_1\Bcap\Bvalue{u_1>0}\ne 0$, so $Pv$ and $v$ are
non-zero;  and

\Centerline{$\int Pv\ge\int_{a_1}Pv=\int_{\pi a_1}v=\int v$.
\Qed}

\medskip

\quad{\bf (ii)} Now take any $u\ge 0$ in $L^1_{\bar\mu}$, and set
$B=\{v:v\in L^1_{\bar\nu},\,v\ge 0,\,Pv\le u,\,\int v\le\int Pv\}$.
$B$ is not empty because it contains $0$.
If $C\subseteq B$ is non-empty and upwards-directed, then
$\sup_{v\in C}\int v\le\int u$ is finite, so $C$ has a supremum in
$L^1_{\bar\nu}$ (365Df).   Because $P$ is order-continuous,
$P(\sup C)=\sup P[C]\le u$;  also

\Centerline{$\int\sup C=\sup_{v\in C}\int v
\le\sup_{v\in C}\int Pv\le\int P(\sup C)$.}

\noindent Thus $\sup C\in B$.   As $C$ is arbitrary, $B$ satisfies the
conditions of Zorn's Lemma, and has a maximal element $v_0$ say.

\Quer\ Suppose, if possible, that $Pv_0\ne u$.   By ($\alpha$),
there is a $v_1>0$ such that $Pv_1\le u-Pv_0$ and $\int v_1\le\int Pv_1$.
In this case,
$v_0<v_0+v_1\in B$, which is impossible.\ \Bang\ Thus
$Pv_0=u$;  also

\Centerline{$\|v_0\|_1=\int v_0\le\int Pv_0=\|Pv_0\|_1$.}

\medskip

\quad{\bf (iii)} Finally, take any $u\in L^1_{\bar\mu}$.   By (ii),
there are non-negative $v_1$, $v_2\in L^1_{\bar\nu}$ such that
$Pv_1=u^+$,
$Pv_2=u^-$, $\|v_1\|_1\le\|u^+\|_1$ and $\|v_2\|_1\le\|u^-\|_1$.
Setting $v=v_1-v_2$, we have $Pv=u$.   Also we must have

\Centerline{$\|v\|_1\le\|v_1\|_1+\|v_2\|_1\le\|u^+\|_1+\|u^-\|_1
=\|u\|_1\le\|P\|\|v\|_1=\|v\|_1$,}

\noindent so $\|v\|_1=\|u\|_1$, as required.

\medskip

{\bf (f)} As usual, this is a consequence of the
uniqueness of $P$.   However (because I do not assume that
$\pi[\frak A^f]\subseteq\frak B^f$) there is an extra refinement:  we
need to know
that $\int_bP_{\theta'}w=\int_{\theta b}w$ for every $b\in\frak B$ and
$w\in L^1_{\bar\lambda}$.   \Prf\ Because $\theta$ is order-continuous
and $(\frak B,\bar\nu)$ is semi-finite,
$\theta b=\sup\{\theta b':b'\in\frak B^f,\,b'\Bsubseteq b\}$, so if
$w\ge 0$ then

\Centerline{$\int_{\theta b}w
=\sup_{b'\in\frak B^f,b'\Bsubseteq b}\int_{\theta b'}w
=\sup_{b'\in\frak B^f,b'\Bsubseteq b}\int_{b'}P_{\theta'}w
=\int_bP_{\theta'}w$.}

\noindent Expressing $w$ as $w^+-w^-$, we see that the same is true for
every $w\in L^1_{\bar\nu}$.\ \Qed

Now we can say that $PP_{\theta'}$ is a positive linear operator
from $L^1_{\bar\lambda}$ to $L^1_{\bar\mu}$ such that

\Centerline{$\int_aPP_{\theta'}w
=\int_{\pi a}P_{\theta'}w
=\int_{\theta\pi a}w
=\int_aP_{\theta\pi}w$}

\noindent whenever $a\in\frak A^f$ and $w\in L^1_{\bar\lambda}$, and
must be equal to $P_{\theta\pi}$.
}%end of proof of 365O

\leader{365P}{Proposition}\dvAformerly{3{}65Q}
Let $(\frak A,\bar\mu)$ and
$(\frak B,\bar\mu)$ be measure algebras and $\pi:\frak A^f\to\frak B^f$
a measure-preserving ring homomorphism.

(a) In the language of 365N-365O above,
$P_{\pi}T_{\pi}$ is the identity operator on $L^1(\frak A,\bar\mu)$.

(b) If $\pi$ is surjective\cmmnt{ (so that it is an isomorphism
between $\frak A^f$ and $\frak B^f$)} then
$P_{\pi}=T_{\pi}^{-1}=T_{\pi^{-1}}$ and
$T_{\pi}=P_{\pi}^{-1}=P_{\pi^{-1}}$.

\proof{{\bf (a)} If $u\in L^1_{\bar\mu}$ and $a\in\frak A^f$ then

\Centerline{$\int_aP_{\pi}T_{\pi}u=\int_{\pi a}T_{\pi}u=\int_au$.}

\noindent So $u=P_{\pi}T_{\pi}u$, by 365D(d-ii).

\medskip

{\bf (b)} From 365Nd, we know that $T_{\pi}$ is
surjective, while $P_{\pi}T_{\pi}$ is the identity, so that
$P_{\pi}=T_{\pi}^{-1}$ and $T_{\pi}=P_{\pi}^{-1}$.
As for $T_{\pi^{-1}}$, 365Ne tells us that
$T_{\pi^{-1}}=T_{\pi}^{-1}$;  so

\Centerline{$P_{\pi^{-1}}=T_{\pi^{-1}}^{-1}=T_{\pi}$.}

}%end of proof of 365P

\leader{365Q}{Conditional expectations}\dvAformerly{3{}65R} \cmmnt{It is
a nearly
universal rule that any investigation of $L^1$ spaces must include a
look at conditional expectations.   In the present context, they take
the following form.

\medskip

}{\bf (a)} Let $(\frak A,\bar\mu)$ be a probability algebra
and $\frak B$ a closed subalgebra;  write $\bar\nu$ for the restriction
$\bar\mu\restrp\frak B$.   The identity map from $\frak B$ to $\frak A$
induces operators $T:L^1(\frak B,\bar\nu)\to L^1(\frak A,\bar\mu)$ and
$P:L^1(\frak A,\bar\mu)\to L^1(\frak B,\bar\nu)$.   If we take
$L^0(\frak A)$ to be
defined as the set of functions from $\Bbb R$ to $\frak A$ described in
364Aa, then $L^0(\frak B)$
becomes a subset of $L^0(\frak A)$ in the literal sense, and $T$ is
actually the identity operator associated with the subset
$L^1(\frak B,\bar\nu)\subseteq L^1(\frak A,\bar\mu)$;
$L^1(\frak B,\bar\nu)$ is a
norm-closed and order-closed Riesz subspace of $L^1(\frak A,\bar\mu)$.
$P$ is a positive linear operator, while $PT$ is the identity, so $P$ is
a projection from $L^1(\frak A,\bar\mu)$ onto $L^1(\frak B,\bar\nu)$.
$P$ is defined by the\cmmnt{ familiar} formula

\Centerline{$\int_bPu=\int_bu$ for every
$u\in L^1(\frak A,\bar\mu)$, $b\in\frak B$,}

\noindent so is the conditional expectation operator in the sense of
242J.   \cmmnt{Observe that the formula in 365A tells us that}
$L^1(\frak B,\bar\nu)$ is just $L^1(\frak A,\bar\mu)\cap L^0(\frak B)$.
\cmmnt{Translating 233K into this language, we see that}
$P(u\times v)=Pu\times v$ whenever $u\in L^1(\frak A,\bar\mu)$,
$v\in L^0(\frak B)$ and $u\times v\in L^1(\frak A,\bar\mu)$.


\spheader 365Qb\cmmnt{ Just as in 233I-233J and 242K, we have a version of
Jensen's inequality.}  Let $h:\Bbb R\to\Bbb R$ be a convex
function and $\bar h:L^0(\frak A)\to L^0(\frak A)$ the corresponding
map\cmmnt{ (364H)}.   If
$u\in L^1(\frak A,\bar\mu)$, then $h(\int u)\le\int\bar h(u)$;  and if
$\bar h(u)\in L^1(\frak A,\bar\mu)$, then $\bar h(Pu)\le P(\bar h(u))$.
\prooflet{\Prf\ I repeat the proof of 233I-233J. For each $q\in\Bbb Q$,
take $\beta_q\in\Bbb R$ such that $h(t)\ge h_q(t)=h(q)+\beta_q(t-q)$ for
every $t\in\Bbb R$, so that
$h(t)=\sup_{q\in\Bbb Q}h_q(t)$ for every $t\in\Bbb R$, and
$\bar h(u)=\sup_{q\in\Bbb Q}\bar h_q(u)$ for every
$u\in L^0(\frak A)$.   (This is because

$$\eqalign{\Bvalue{\bar h(u)>\alpha}
&=\Bvalue{u\in h^{-1}[\,\ooint{\alpha,\infty}\,]}
=\Bvalue{u\in\bigcup_{q\in\Bbb Q}h_q^{-1}[\,\ooint{\alpha,\infty}\,]}\cr
&=\sup_{q\in\Bbb Q}\Bvalue{u\in h_q^{-1}[\,\ooint{\alpha,\infty}\,]}
=\sup_{q\in\Bbb Q}\Bvalue{\bar h_q(u)>\alpha}\cr}$$

\noindent for every $\alpha\in\Bbb R$.)   But setting $e=\chi 1$, we see
that $\bar h_q(u)=h(q)e+\beta_q(u-qe)$ for every $u\in L^0(\frak A)$, so
that

\Centerline{$\int\bar h_q(u)=h(q)+\beta_q(\int u-q)=h_q(\int u)$,}

\Centerline{$P(\bar h_q(u))=h(q)e+\beta_q(Pu-qe)=\bar h_q(Pu)$}

\noindent because $\int e=1$ and $Pe=e$.   Taking the
supremum over $q$, we get

\Centerline{$h(\int u)
=\sup_{q\in\Bbb Q}h_q(\int u)
=\sup_{q\in\Bbb Q}\int\bar h_q(u)
\le\int\bar h(u)$,}

\noindent and if $\bar h(u)\in L^1_{\bar\mu}$ then

\Centerline{$\bar h(Pu)
=\sup_{q\in\Bbb Q}\bar h_q(Pu)
=\sup_{q\in\Bbb Q}P(\bar h_q(u))
\le P(\bar h(u))$.   \Qed}}%end of prooflet

\cmmnt{Of course the result in this form can also be deduced from
233I-233J if we represent $(\frak A,\bar\mu)$ as the measure algebra of
a probability space $(X,\Sigma,\mu)$ and set
$\Tau=\{E:E\in\Sigma,\,E^{\ssbullet}\in\frak B\}$.}

\spheader 365Qc\dvAnew{2016?}\cmmnt{ I note here a fact which is occasionally
useful.}   If
$u\in L^1(\frak A,\bar\mu)$ is non-negative, then
$\Bvalue{Pu>0}=\upr(\Bvalue{u>0},\frak B)$, the upper envelope of
$\Bvalue{u>0}$ in $\frak B$\cmmnt{ as defined in 313S}.
\prooflet{\Prf\ We have only to observe that, for $b\in\frak B$,

$$\eqalign{b\Bcap\Bvalue{Pu>0}=0
&\iff\chi b\times Pu=0
\iff\int_bPu=0\cr
&\iff\int_bu=0
\iff b\Bcap\Bvalue{u>0}=0.\cr}$$

\noindent Taking complements, $b\Bsupseteq\Bvalue{Pu>0}$ iff
$b\Bsupseteq\Bvalue{u>0}$.\ \Qed}

\spheader 365Qd\dvAnew{2016?} Suppose now that $(\frak C,\bar\lambda)$ is
another probability algebra and $\pi:\frak A\to\frak C$ is a
measure-preserving Boolean homomorphism.   Then $\frak D=\pi[\frak B]$
is a closed subalgebra of $\frak C$\cmmnt{ (314F(a-i))}.   Let
$Q:L^1(\frak C,\bar\lambda)\to L^1(\frak D,\bar\lambda\restrp\frak D)
\subseteq L^1(\frak C,\bar\lambda)$
be the conditional expectation associated with $\frak D$, and
$T_{\pi}:L^1(\frak A,\bar\mu)\to L^1(\frak C,\bar\lambda)$
the norm-preserving Riesz homomorphism defined by $\pi$.
Then $T_{\pi}P=QT_{\pi}$.\prooflet{ \Prf\ Take $u\in L^1(\frak A,\bar\mu)$.
Then

\Centerline{$\Bvalue{T_{\pi}Pu>\alpha}
=\pi\Bvalue{Pu>\alpha}\in\pi[\frak B]=\frak D$}

\noindent for every $\alpha\in\Bbb R$, so $T_{\pi}Pu\in L^0(\frak D)$.
If $d\in\frak D$, set $b=\pi^{-1}d\in\frak B$;  then

$$\eqalign{\int_dT_{\pi}Pu
&=\int T_{\pi}Pu\times\chi d
=\int T_{\pi}Pu\times T_{\pi}\chi b
=\int T_{\pi}(Pu\times\chi b)\cr
&=\int Pu\times\chi b
=\int_bPu
=\int_bu
=\int u\times\chi b\cr
&=\int T_{\pi}(u\times\chi b)
=\int T_{\pi}u\times T_{\pi}\chi b
=\int T_{\pi}u\times\chi d
=\int_d T_{\pi}u.\cr}$$

\noindent As $d$ is arbitrary, $T_{\pi}Pu$ satisfies the defining formula
for $QT_{\pi}u$ and $T_{\pi}Pu=QT_{\pi}u$;  as $u$ is arbitrary,
$T_{\pi}P=QT_{\pi}$.\ \Qed}

\leader{365R}{Recovering the algebra:  Proposition}\dvAformerly{3{}65S}
(a) Let
$(\frak A,\bar\mu)$ be a localizable measure algebra.   Then $\frak A$
is isomorphic to the band algebra of $L^1(\frak A,\bar\mu)$.

(b) Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra, and
$\bar\mu$, $\bar\nu$ two measures on $\frak A$ such that
$(\frak A,\bar\mu)$ and $(\frak A,\bar\nu)$ are both semi-finite measure
algebras.   Then $L^1(\frak A,\bar\mu)$ is isomorphic, as Banach
lattice, to $L^1(\frak A,\bar\nu)$.

\proof{{\bf (a)} Because $(\frak A,\bar\mu)$ is semi-finite,
$L^1_{\bar\mu}$ is order-dense in $L^0=L^0(\frak A)$ (365G).
Consequently, $L^1_{\bar\mu}$ and
$L^0$ have isomorphic band algebras (353D).   But the band algebra of
$L^0$ is just its algebra of projection bands (because
$\frak A$ and therefore $L^0$ are Dedekind complete, see 364M and 353J),
which is isomorphic to $\frak A$ (364O).

\medskip

{\bf (b)} Let $\pi:\frak A\to\frak A$ be the identity map.   Regarding
$\pi$ as an order-continuous Boolean homomorphism from
$\frak A^f_{\bar\mu}=\{a:\bar\mu a<\infty\}$ to $(\frak A,\bar\nu)$, we
have an associated positive linear
operator $P=P_{\pi}:L^1_{\bar\nu}\to L^1_{\bar\mu}$;
similarly, we have
$Q=P_{\pi^{-1}}:L^1_{\bar\mu}\to L^1_{\bar\nu}$, and both
$P$ and $Q$ have
norm at most $1$ (365Ob).   Now 365Of assures us that $QP$ is the
identity operator on $L^1_{\bar\nu}$ and $PQ$ is the identity
operator on $L^1_{\bar\mu}$.   So $P$ and $Q$ are the two halves
of a Banach lattice isomorphism between $L^1_{\bar\mu}$ and
$L^1_{\bar\nu}$.
}%end of proof of 365R

\leader{365S}{}\dvAformerly{3{}65T}\cmmnt{ Having opened the question of
varying measures
on a single Boolean algebra, this seems an appropriate moment
for a general description of how they interact.

\medskip

\noindent}{\bf Proposition} Let $\frak A$ be a Dedekind complete Boolean
algebra, and $\bar\mu:\frak A\to[0,\infty]$,
$\bar\nu:\frak A\to[0,\infty]$ two functions such that
$(\frak A,\bar\mu)$ and $(\frak A,\bar\nu)$ are both
semi-finite\cmmnt{ (therefore localizable)} measure algebras.

(a) There is a
unique $u\in L^0=L^0(\frak A)$ such that\cmmnt{ (if we allow $\infty$
as a value of the integral)} $\int_au\,d\bar\mu=\bar\nu a$
for every $a\in\frak A$.

(b) For $v\in L^0(\frak A)$,
$\int v\,d\bar\nu=\int u\times v\,d\bar\mu$ if either is defined in
$[-\infty,\infty]$.

(c) $u$ is strictly positive\cmmnt{ (i.e.,
$\Bvalue{u>0}=1$)} and, writing $\bover{1}{u}$ for the multiplicative
inverse of $u$, $\int_a\bover{1}{u}d\bar\nu=\bar\mu a$ for every
$a\in\frak A$.

\proof{{\bf (a)} Because $(\frak A,\bar\nu)$ is semi-finite, there is a
partition
of unity $D\subseteq \frak A$ such that $\bar\nu d<\infty$ for every
$d\in D$.   For each $d\in D$, the functional
$a\mapsto\bar\nu(a\Bcap d):\frak A\to\Bbb R$ is completely additive, so
there is a $u_d\in L^1_{\bar\mu}$ such that
$\int_au_dd\bar\mu=\bar\nu(a\Bcap d)$ for every $a\in\frak A$.   Because
$\int_au_dd\bar\mu\ge 0$ for every $a$, $u_d\ge 0$.   Because
$\int_{1\Bsetminus d}u_d=0$, $\Bvalue{u_d>0}\Bsubseteq d$.   Now
$u=\sup_{d\in D}u_d$ is defined in $L^0$.   \Prf\ (This is a special
case of 368K below.)
For $n\in\Bbb N$, set $c_n=\sup_{d\in D}\Bvalue{u_d>n}$.   If $d$,
$d'\in D$ are distinct, then $d\Bcap\Bvalue{u_{d'}>n}=0$, so
$d\Bcap c_n=\Bvalue{u_d>n}$.   Set $c=\inf_{n\in\Bbb N}c_n$.   If
$d\in D$, then

\Centerline{$d\Bcap c
=\inf_{n\in\Bbb N}d\Bcap c_n
=\inf_{n\in\Bbb N}\Bvalue{u_d>n}
=0$.}

\noindent But $c\Bsubseteq c_0\Bsubseteq\sup D$, so $c=0$.   By 364L(a-i),
$\{u_d:d\in D\}$ is bounded above in $L^0$, so has a supremum, because
$L^0$ is Dedekind complete, by 364M.\ \Qed

For finite
$I\subseteq D$ set $\tilde u_I=\sum_{d\in I}u_d=\sup_{d\in I}u_d$
(because $u_d\wedge u_c=0$ for distinct $c$, $d\in D$).   Then
$u=\sup\{\tilde u_I:I\subseteq D,\,I$ is
finite$\}$.
So, for any $a\in\frak A$,

$$\eqalignno{\int_au\,d\bar\mu
&=\sup_{I\subseteq D\text{ is finite}}\int_a\tilde u_Id\bar\mu\cr
\noalign{\noindent (365Dh)}
&=\sup_{I\subseteq D\text{ is finite}}\sum_{d\in
I}\int_au_dd\bar\mu
=\sup_{I\subseteq D\text{ is finite}}\sum_{d\in I}\bar\nu(a\Bcap
d)
=\bar\nu a.\cr}$$

Note that if $a\in\frak A$ is non-zero, then $\bar\nu a>0$, so
$a\Bcap\Bvalue{u>0}\ne 0$;  consequently $\Bvalue{u>0}=1$.

To see that $u$ is unique, observe that if $u'$ has the same property
then for any $d\in D$

\Centerline{$\int_au\times\chi d\,d\bar\mu
=\bar\nu(a\Bcap d)=\int_au'\times\chi d\,d\bar\mu$}

\noindent for every $a\in\frak A$, so that
$u\times\chi d=u'\times\chi d$;  because $\sup D=1$ in $\frak A$, $u$
must be equal to $u'$.

\medskip

{\bf (b)} Use 365Hb, with $\pi$ and $T$ the identity maps.

\medskip

{\bf (c)} In the same way, there is a $w\in L^0$ such that
$\int_aw\,d\bar\nu=\bar\mu a$ for every $a\in\frak A$.   To relate $u$
and $w$, observe that applying (b) above we get

\Centerline{$\int w\times\chi a\times u\,d\bar\mu
=\int w\times\chi a\,d\bar\nu$}

\noindent for every $a\in\frak A$, that is, $\int_aw\times
u\,d\bar\mu=\bar\mu a$ for every $a$.   But from this we see that
$w\times u\times\chi b=\chi b$ at least when $\bar\mu b<\infty$, so that
$w\times u=\chi 1$ is the multiplicative identity of $L^0$, and
$w=\bover1{u}$.
}%end of proof of 365S

\leader{365T}{Uniform \dvrocolon{integrability}}\dvAformerly{3{}65U}\cmmnt{
Continuing the
programme in 365C, we can transcribe the ideas of  \S\S246, 247, 354 and
356 into the new context.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra.
Set $L^1=L^1(\frak A,\bar\mu)$.

(a) For a non-empty subset $A$ of $L^1$, the following are equiveridical:

\quad(i) $A$ is uniformly integrable in the sense of 354P;

\quad(ii) for every $\epsilon>0$ there are an $a\in\frak A^f$ and an
$M\ge 0$ such that $\int(|u|-M\chi a)^+\le\epsilon$
for every $u\in\frak A$;

\quad(iii)($\alpha$)
$\sup_{u\in A}|\int_au|$ is finite for every atom $a\in\frak A$,

\qquad($\beta$) for every $\epsilon>0$ there are $c\in\frak A^f$ and
$\delta>0$ such that $|\int_au|\le\epsilon$ whenever $u\in A$,
$a\in\frak A$ and $\bar\mu(a\Bcap c)\le\delta$;

\quad(iv)($\alpha$)
$\sup_{u\in A}|\int_au|$ is finite for every atom $a\in\frak A$,

\qquad($\beta$) $\lim_{n\to\infty}\sup_{u\in A}|\int_{a_n}u|=0$ for
every disjoint sequence $\sequencen{a_n}$ in $\frak A$;

\quad(v) $A$ is relatively weakly compact in $L^1$.

(b) If $(\frak A,\bar\mu)$ is a probability algebra and $A\subseteq L^1$
is uniformly integrable, then there is a solid convex norm-closed uniformly
integrable set $C\supseteq A$ such that $P[C]\subseteq C$ whenever
$P:L^1\to L^1$ is the conditional expectation operator associated with a
closed subalgebra of $\frak A$.

\proof{ 354Q, 354R, 356Q and 246D, with a little help from 246C and 246G.
}%end of proof of 365T

\exercises{\leader{365X}{Basic exercises (a)}
%\spheader 365Xa
Let $(\frak A,\bar\mu)$ be a measure algebra, and $u\in L^1_{\bar\mu}$.
Show that

\Centerline{$\int u=\int_0^{\infty}\bar\mu\Bvalue{u>\alpha}\,d\alpha
  -\int_{-\infty}^0\bar\mu(1\Bsetminus\Bvalue{u>\alpha})\,d\alpha$.}
%365D

\sqheader 365Xb Let $(\frak A,\bar\mu)$ be any measure algebra, and
$u\in L^1_{\bar\mu}$.   (i) Show that
$\|u\|_1\le 2\sup_{a\in\frak A^f}|\int_au|$.   \Hint{246F.}   (ii) Show
that for any $\epsilon>0$ there is an $a\in\frak A^f$ such that
$|\int u-\int_bu|\le\epsilon$ whenever $a\Bsubseteq b\in\frak A$.
%365D

\sqheader 365Xc Let $U$ be an $L$-space.   If $\sequencen{u_n}$ is any
norm-bounded sequence in $U^+$, show that

\Centerline{$\liminf_{n\to\infty}u_n
=\sup_{n\in\Bbb N}\inf_{m\ge n}u_m$}
%\Centerline needed for Lulu version

\noindent is defined in $U$, and that
$\int\liminf_{n\to\infty}u_n\le\liminf_{n\to\infty}\int u_n$.
%365D

\spheader 365Xd Let $U$ be an $L$-space.   Let $\Cal F$ be a filter on
$U$ such that $\{u:u\ge 0,\,\|u\|\le k\}$ belongs to $\Cal F$ for some
$k\in\Bbb N$.   Show that $u_0=\sup_{F\in\Cal F,F\subseteq U^+}\inf F$
is defined in
$U$, and that $\int u_0\le\sup_{F\in\Cal F}\inf_{u\in F}\int u$.
%365D

\spheader 365Xe Let $(\frak A,\bar\mu)$ be a measure algebra and
$A\subseteq L^1_{\bar\mu}$ a non-empty set.   Show that $A$ is bounded
above in $L^1_{\bar\mu}$ iff

\Centerline{$\sup\{\sum_{i=0}^n\int_{a_i}u_i:a_0,\ldots,a_n$ is a
partition of unity in $\frak A$, $u_0,\ldots,u_n\in A\}$}

\noindent is finite, and that in this case the supremum is $\int\sup A$.
\Hint{given $u_0,\ldots,u_n\in A$, set $b_i=\inf_j\Bvalue{u_i\ge u_j}$,
$a_i=b_i\Bsetminus\sup_{j<i}b_j$, and show
that $\int\sup_{i\le n}u_i=\sum_{i=0}^n\int_{a_i}u_i$.}
%365D

\spheader 365Xf\dvAnew{2014}
Let $(\frak A,\bar\mu)$ be a measure algebra and
$u$, $v\in L^0(\frak A)^+$.   Show that
$\int u\times v\,d\bar\mu
=\int_0^{\infty}(\int_{\Bvalue{u>\alpha}}v\,d\bar\mu)d\alpha$.
\Hint{start with $u\in S(\frak A)^+$.}
%365D out of order query

\spheader 365Xg Let $(\frak A,\bar\mu)$ be any measure algebra and
$\nu:\frak A^f\to\Bbb R$ a bounded additive functional.   Show that the
following are equiveridical:  (i) there is a $u\in L^1_{\bar\mu}$ such
that
$\nu a=\int_au$ for every $a\in\frak A^f$;  (ii) for every $\epsilon>0$
there is a $\delta>0$ such that $|\nu a|\le\epsilon$ whenever
$\bar\mu a\le\delta$;  (iii) for every $\epsilon>0$, $c\in\frak A^f$
there is a
$\delta>0$ such that $\nu a\le\epsilon$ whenever $a\Bsubseteq c$ and
$\bar\mu a\le\delta$;  (iv) for every $\epsilon>0$ there are
$c\in\frak A^f$, $\delta>0$ such that $|\nu a|\le\epsilon$ whenever
$a\in\frak A^f$
and $\bar\mu(a\Bcap c)\le\delta$;  (v) $\lim_{n\to\infty}\nu a_n=0$
whenever $\sequencen{a_n}$ is a
non-increasing sequence in $\frak A^f$ with infimum $0$.
%365E

\spheader 365Xh Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
measure algebras, and $\pi:\frak A\to\frak B$ a sequentially
order-continuous Boolean homomorphism.   Let
$T:L^0(\frak A)\to L^0(\frak B)$ be the Riesz homomorphism associated
with $\pi$ (364P).
Suppose that $w\ge 0$ in $L^0(\frak B)$ is such that
$\int_{\pi a}w\,d\bar\nu=\bar\mu a$ whenever $a\in\frak A$.   Show that
for any $u\in L^0(\frak A,\bar\mu)$,
$\int Tu\times w\,d\bar\nu=\int u\,d\bar\mu$ whenever either is defined
in $[-\infty,\infty]$.
%365H

\sqheader 365Xi Let $(\frak A,\bar\mu)$ be a measure algebra and
$a\in\frak A$;  write $\frak A_a$ for the principal ideal it generates.
Show that if $\pi$ is the identity embedding of $\frak A^f\cap\frak A_a$
into $\frak A^f$, then $T_{\pi}$, as defined in 365N, identifies
$L^1(\frak A_a,\bar\mu\restrp\frak A_a)$ with a band in
$L^1_{\bar\mu}$.
%365N

\sqheader 365Xj Let $(X,\Sigma,\mu)$ and $(Y,\Tau,\nu)$ be measure
spaces, with measure algebras $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$.   Let $\phi:X\to Y$ be an \imp\ function and
$\pi:\frak B\to\frak A$ the corresponding
measure-preserving homomorphism (324M).   Show that
$T_{\pi}:L^1_{\bar\nu}\to L^1_{\bar\mu}$ (365N) corresponds to the map
$g^{\ssbullet}\mapsto(g\phi)^{\ssbullet}:L^1(\nu)\to L^1(\mu)$ of
242Xd.
%365N

\spheader 365Xk Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be measure algebras.   Let $\pi:\frak A^f\to\frak B^f$ be a
ring homomorphism such that, for some $\gamma>0$, $\bar\nu(\pi
a)\le\gamma\bar\mu a$ for every $a\in\frak A^f$.  (i) Show that there is
a unique order-continuous Riesz
homomorphism $T:L^1_{\bar\mu}\to L^1_{\bar\nu}$ such that
$T(\chi a)=\chi(\pi a)$ whenever $a\in\frak A^f$, and that
$\|T\|\le\gamma$.   (ii) Show that
$\Bvalue{Tu>\alpha}=\pi\Bvalue{u>\alpha}$ whenever
$u\in L^1_{\bar\mu}$ and $\alpha>0$.
(iii) Show that $T$ is surjective iff $\pi$ is, injective iff $\pi$ is.
(iv) Show that $T$ is norm-preserving iff $\bar\nu(\pi a)=\bar\mu a$ for
every $a\in\frak A^f$.
%365N

\spheader 365Xl 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 $T:L^1_{\bar\mu}\to L^1_{\bar\nu}$ and
$P:L^1_{\bar\nu}\to L^1_{\bar\mu}$ be the operators corresponding to
$\pi\restrp\frak A^f$, as described in 365N-365O, and
$\tilde T:L^{\infty}(\frak A)\to L^{\infty}(\frak B)$ the operator
corresponding to $\pi$, as described in 363F.   (i) Show that
$T(u\times v)=Tu\times \tilde Tv$ for every $u\in L^1_{\bar\mu}$,
$v\in L^{\infty}(\frak A)$.
(ii) Show that if $\pi$ is order-continuous, then
$\int Pv\times u=\int v\times\tilde Tu$ for every
$u\in L^{\infty}(\frak A)$, $v\in L^1_{\bar\nu}$.
%365N

\sqheader 365Xm Let $(X,\Sigma,\mu)$ be a probability space, with
measure algebra $(\frak A,\bar\mu)$, and let $\Tau$ be a
$\sigma$-subalgebra of $\Sigma$.   Set $\nu=\mu\restrp\Tau$, $\frak
B=\{F^{\ssbullet}:F\in\Tau\}\subseteq\frak A$,
$\bar\nu=\bar\mu\restrp\frak B$, so that
$(\frak B,\bar\nu)$ is a measure algebra.   Let $\pi:\frak B\to\frak A$
be the identity homomorphism.    Show that
$T_{\pi}:L^1_{\bar\nu}\to L^1_{\bar\mu}$ (365N) corresponds to the
canonical
embedding of $L^1(\nu)$ in $L^1(\mu)$ described in 242Jb, while
$P_{\pi}:L^1_{\bar\mu}\to L^1_{\bar\nu}$ (365O)
corresponds to the conditional expectation operator described in
242Jd.
%365O

\spheader 365Xn Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
probability algebras, $\pi:\frak A\to\frak B$ a measure-preserving Boolean
homomorphism, and $T:L^0(\frak A)\to L^0(\frak B)$ the corresponding Riesz
homomorphism.   Let $\frak C$ be a closed subalgebra of $\frak A$ and
$P:L^1(\frak A,\bar\mu)\to L^1(\frak C,\bar\mu\restrp\frak C)
\subseteq L^1(\frak A,\bar\mu)$,
$Q:L^1(\frak B,\bar\nu)\to L^1(\frak B,\bar\nu)$ the conditional
expectation operators defined from $\frak C\embedsinto\frak A$
and $\pi[\frak C]\embedsinto\frak B$.   Show that $TP=QT$.
%365Q

\spheader 365Xo Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $(\widehat{\frak A},\hat\mu)$ its localization (322Q).
Show that the natural embedding of $\frak A$ in $\widehat{\frak A}$
induces a Banach lattice isomorphism between $L^1_{\bar\mu}$ and
$L^1_{\hat\mu}$, so that the band algebra of
$L^1_{\bar\mu}$ can be identified with the Dedekind completion
$\widehat{\frak A}$ of $\frak A$.
%365N, 365R

\spheader 365Xp Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\bar\mu$, $\bar\nu$ two functions such that $(\frak
A,\bar\mu)$, $(\frak A,\bar\nu)$ are measure algebras.   Show that
$L^1_{\bar\mu}\subseteq L^1_{\bar\nu}$ (as subsets of $L^0(\frak A)$)
iff there is a $\gamma>0$ such that $\bar\nu a\le\gamma\bar\mu a$ for
every $a\in\frak A$.   \Hint{show that the identity operator from
$L^1_{\bar\mu}$ to $L^1_{\bar\nu}$ is bounded.}
%365R

\spheader 365Xq Let $(\frak A,\bar\mu)$ be a measure algebra,
$I_{\infty}$ the ideal of `purely infinite' elements of $\frak A$
together with $0$ and
$\bar\mu_{\text{sf}}$ the measure on $\frak B=\frak A/I_{\infty}$
as defined in 322Xa.
Let $\pi:\frak A\to\frak B$ be the canonical map.   Show that $T_{\pi}$,
as defined in 365N, is a Banach lattice isomorphism between
$L^1_{\bar\mu}$ and $L^1(\frak B,\bar\mu_{\text{sf}})$.
%365N, 365Xo, 365R

\spheader 365Xr Let $(X,\Sigma,\mu)$ be a a semi-finite measure space.
Show that $L^1(\mu)$ is separable iff $\mu$ is $\sigma$-finite and has
countable Maharam type.
%365R

\leader{365Y}{Further exercises (a)}
%\spheader 365Ya
Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, not $\{0\}$.   Show that the topological density of
$L^1_{\bar\mu}$ is $\max(\omega,\tau(\frak A),c(\frak A))$, where
$\tau(\frak A)$, $c(\frak A)$ are the Maharam type and cellularity of
$\frak A$.
%365Xr, 365R

\discrversionA{\spheader 365Yb ({\smc de Pauw p19})\footnote{under embargo}
%Prop.\ 4.6
Let $(\frak A,\bar\mu)$ be a measure algebra and
$T:L^{\infty}(\frak A)\to L^1(\frak A,\bar\mu)^*$ the canonical map.
As in 322Xa, let $I_{\infty}$ be the $\sigma$-ideal of those $a\in\frak A$ 
which are either $0$ or purely infinite.   Show that $T$ is surjective iff
the quotient algebra $\frak A/I_{\infty}$ is Dedekind complete.}{}   
%365L 322Xa 363Fd 
}%end of exercises

\cmmnt{
\Notesheader{365} You should not suppose that $L^1$ spaces appear in the
second half of this chapter because they are of secondary importance.
Indeed I regard them as the most important of all function spaces.   I
have delayed the discussion of them for so long because it is here that
for the first time we need measure algebras in an essential way.

The actual definition of $L^1_{\bar\mu}$ which I give is designed
for speed rather than illumination;  I seek only a formula, visibly
independent of any particular representation of $(\frak A,\bar\mu)$ as
the measure algebra of a measure space, from which I can prove 365B.
365C-365D and 365Ea are now elementary.   In 365Eb I take a
page to describe a form of the Radon-Nikod\'ym theorem which is
applicable to arbitrary measure algebras, at the cost of dealing with
functionals on the ring $\frak A^f$ rather than on the whole algebra
$\frak A$.    This is less for the sake of applications than to
emphasize one of the central properties of $L^1$:  it depends only on
$\frak A^f$ and $\bar\mu\restrp\frak A^f$.   For alternative versions of
the condition 365Eb(i) see 365Xg.

The convergence theorems (B.Levi's theorem, Fatou's lemma and Lebesgue's
dominated convergence theorem) are so central to the theory of
integrable functions that it is natural to look for versions in the
language here.   Corresponding to B.Levi's theorem is the Levi property
of a norm in an $L$-space;  note how the abstract formulation makes it
natural to speak of general upwards-directed families rather than of
non-decreasing sequences, though the sequential form is so often used
that I have spelt it out (365C).   In the same way, the integral becomes
order-continuous rather than just sequentially order-continuous (365Da).
Corresponding to Fatou's lemma we have 365Xc-365Xd.   For abstract
versions of Lebesgue's theorem I will wait until \S367.
%367I

In 365H I have deliberately followed the hypotheses of 235A and 235R.
Of course 365H can be deduced from these if we use the Stone
representations of $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$, so that
$\pi$ can be represented by a function between the Stone spaces
(312Q).   But 365H is essentially simpler, because the technical
problems concerning measurability which took up so much of \S235 have
been swept under the carpet.   In the same way, 365Xh corresponds to
235E.   Here we have a fair example of the way in which the abstract
expression in terms of measure algebras can be tidier than the
expression in terms of measure spaces.   But in my view this is because
here, at least, some of the mathematics has been left out.

365I-365J %365I 365J
correspond closely to 361F-361H %361F 361G 361H
and 363E.
365L is a re-run of 243G, but with the additional refinement that I
examine the action of $L^1$ on $L^{\infty}$ (the operator $S$) as well
as the action of $L^{\infty}$ on $L^1$ (the operator $T$).   Of course
365Lc is just the abstract version of 243Hb, and can easily be proved
from it.   Note that while the proof of 365L does not itself involve any
representation of $(\frak A,\bar\mu)$ as the measure algebra of a
measure space, (a-vii) and (b-iii) of the proof of 365L depend on the
Radon-Nikod\'ym theorem through 327D and 365E.   For a development of
the theory of $L^1(\frak A,\bar\mu)$ which does not (in a formal sense)
depend on measure spaces, see {\smc Fremlin 74a}, 63J.

Theorems 365N-365P lie at the centre of my picture of $L^1$ spaces, and
are supposed to show their dual nature.   Starting from a semi-finite
measure algebra $(\frak A,\bar\mu)$ we have two essentially different
routes to the $L^1$-space:  we can either build it up from
indicator functions of elements of finite measure, so that it is
naturally embedded in $L^0(\frak A)$, or we can think of it as the
order-continuous dual of $L^{\infty}(\frak A)$.   The first is a
`covariant' construction (signalled by the formula
$T_{\theta\pi}=T_{\theta}T_{\pi}$ in 365Ne) and the second is
`contravariant' (so that $P_{\theta\pi}=P_{\pi}P_{\theta'}$ in
365Of).   The first construction is the natural one if we are
seeking to copy the ideas of \S242, but the second arises inevitably if
we follow the ordinary paths of functional analysis and study dual
spaces whenever they appear.   The link between them is the
Radon-Nikod\'ym theorem.

I have deliberately written out 365N and 365O with different
hypotheses on the homomorphism $\pi$ in the hope of showing that the two
routes to $L^1$ really are different, and can be expected to tell us
different things about it.
I use the letter $P$ in 365O in order to echo the language of 242J;
in the most important context, in which $\frak A$ is actually a
subalgebra of $\frak B$ and $\pi$ is the identity map, $P$ is a kind of
conditional expectation operator (365Q).   I note that in the proof of
365Oe I have returned to first principles, using some of the ideas of
the Radon-Nikod\'ym theorem (232E), but a different approach to the
exhaustion step (converting `for every $u>0$ there is a $v>0$ such
that $Pv\le u$' into `$P$ is surjective').   I chose the somewhat
cruder method in 232E (part (c) of the proof) in order to use the
weakest possible form of the axiom of choice.   In the present context
such scruples seem absurd.

I used the words `covariant' and `contravariant' above;  of course
this distinction depends on the side of the mirror on which we are
standing;  if our measure-preserving homomorphism is derived
(contravariantly) from an inverse-measure-preserving transformation,
then the $T$'s become contravariant (365Xj).   An important
component of this work, for me, is the fact that not all
measure-preserving homomorphisms between measure algebras can be
represented by inverse-measure-preserving functions (343Jb, 343M).

I have noted at various points (e.g., 242Yd) that the properties of
$L^1(\mu)$ are not much affected by peculiarities in a measure space
$(X,\Sigma,\mu)$.   In this section I offer an explanation:  unlike
$L^0$ or $L^{\infty}$, $L^1$ really
depends only on $\frak A^f$, the ring of elements of finite measure in
the measure algebra.   (See 365N-365P, %365N 365O 365P
365Xo and 365Xq.) %365Xo 365Xp 365Xq
Note that while the
algebra $\frak A$ is uniquely determined (given that $(\frak A,\bar\mu)$
is localizable, 365Ra), the measure $\bar\mu$ is not;  if $\frak A$ is
any algebra carrying two non-isomorphic semi-finite measures, the
corresponding $L^1$ spaces are still isomorphic (365Rb).   For instance,
the $L^1$-spaces of Lebesgue measure $\mu$ on $\Bbb R$, and the subspace
measure $\mu_{[0,1]}$ on $[0,1]$, are isomorphic, though their measure
algebras are not.

In 365T I have recapitulated the results in \S\S246, 247, 354 and
356 concerning
uniform integrability and weak compactness, but I make no attempt to add to
them.   Once we have left measure
spaces behind, these ideas belong to the theory of Banach lattices, and
there is little to relate them to the questions dealt with in this
section.   But see 373Xj and 373Xn below.
}%end of notes

\discrpage

\frfilename{mt366.tex}
\versiondate{10.11.08}
\copyrightdate{1995}

\def\chaptername{Function spaces}
\def\sectionname{$L^p$}

\newsection{366}

In this section I apply the methods of this chapter to $L^p$ spaces,
where $1<p<\infty$.   The constructions proceed without surprises up to
366E, translating the ideas of \S244 by the methods used in \S365.
Turning to the action of Boolean homomorphisms on $L^p$ spaces, I
introduce a
space $M^0$, which can be regarded as the part of $L^0$ that can be
determined from the ring $\frak A^f$ of elements of $\frak A$ of finite
measure (366F), and which includes $L^p$ whenever $1\le p<\infty$.   Now
a measure-preserving ring homomorphism from $\frak A^f$ to $\frak B^f$
acts on the $M^0$ spaces in a way which includes injective Riesz
homomorphisms from $L^p(\frak A,\bar\mu)$ to $L^p(\frak B,\bar\nu)$ and
surjective positive linear operators from $L^p(\frak B,\bar\nu)$ to
$L^p(\frak A,\bar\mu)$ (366H).   The latter may be regarded as
conditional expectation operators (366J).   The case $p=2$ (366K-366L)
is of course by far the most important.   As with the familiar spaces
$L^p(\mu)$ of Chapter 24, we have complex versions
$L^p_{\Bbb C}(\frak A,\bar\mu)$ with the expected properties (366M).

\leader{366A}{Definition}  Let $(\frak A,\bar\mu)$ be a measure algebra
and suppose that $1<p<\infty$.   For $u\in L^0(\frak A)$, define
$|u|^p\in L^0(\frak A)$ by setting

$$\eqalign{\Bvalue{|u|^p>\alpha}
&=\Bvalue{|u|>\alpha^{1/p}}\text{ if }\alpha\ge 0,\cr
&=1\text{ if }\alpha<0.\cr}$$

\noindent\cmmnt{(In the language of 364H, $|u|^p=\bar h(u)$,
where $h(t)=|t|^p$ for $t\in\Bbb R$.)   }Set

\Centerline{$L^p_{\bar\mu}=L^p(\frak A,\bar\mu)
=\{u:u\in L^0(\frak A),\,|u|^p\in L^1(\frak A,\bar\mu)\}$,}

\noindent and for $u\in L^0(\frak A)$ set

\Centerline{$\|u\|_p=(\int|u|^p)^{1/p}=\||u|^p\|_1^{1/p}$,}

\noindent counting $\infty^{1/p}$ as $\infty$\cmmnt{, so that
$L^p_{\bar\mu}=\{u:u\in L^0(\frak A)$, $\|u\|_p<\infty\}$.}

\leader{366B}{Theorem} Let $(X,\Sigma,\mu)$ be a measure space, and
$(\frak A,\bar\mu)$ its measure algebra.   Then the canonical
isomorphism between $L^0(\mu)$ and $L^0(\frak A)$\cmmnt{ (364Ic)}
makes $L^p(\mu)$\cmmnt{, as defined in \S244,} correspond to
$L^p(\frak A,\bar\mu)$.

\proof{ What we really have to check is that if $w\in L^0(\mu)$
corresponds to $u\in L^0(\frak A)$, then $|w|^p$, as defined in 244A,
corresponds to $|u|^p$ as defined in 366A.   But this was noted in
364Ib.

Now, because the isomorphism between $L^0(\mu)$ and $L^0(\frak A)$
matches $L^1(\mu)$ with $L^1_{\bar\mu}$ (365B), we can be sure
that $|w|^p\in L^1(\mu)$ iff $|u|^p\in L^1_{\bar\mu}$, and that
in this case

\Centerline{$\|w\|_p=\bigl(\int|w|^p\bigr)^{1/p}
=\bigl(\int|u|^p\bigr)^{1/p}=\|u\|_p$,}

\noindent as required.
}%end of proof of 366B

\leader{366C}{Corollary} For any measure algebra $(\frak A,\bar\mu)$ and
$p\in\ooint{1,\infty}$, $L^p=L^p(\frak A,\bar\mu)$ is a solid linear
subspace of $L^0(\frak A)$.   It is a Dedekind complete Banach lattice
under its uniformly convex
norm $\|\,\|_p$.   Setting $q=p/(p-1)$, $(L^p)^*$ is
identified with $L^q(\frak A,\bar\mu)$ by the duality
$(u,v)\mapsto\int u\times v$.   Writing $\frak A^f$ for the ring
$\{a:a\in\frak A,\,\bar\mu a<\infty\}$, $S(\frak A^f)$ is norm-dense in
$L^p$.

\proof{ Because we can find a measure space $(X,\Sigma,\mu)$ such that
$(\frak A,\bar\mu)$ is isomorphic to the measure algebra of $\mu$
(321J), this is just a digest of the results in 244B,
244E-244H, %244E 244F 244G 244H
244K, 244L and 244O\Latereditions.
(Of course $S(\frak A^f)$ corresponds to
the space $S$ of equivalence classes of simple functions in 244Ha,
just as in 365F.)
}%end of proof of 366C

\leader{366D}{}\cmmnt{ I can add a little more, corresponding to
365C and 365L.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra,
and $p\in\ooint{1,\infty}$.

(a) The norm $\|\,\|_p$ on $L^p=L^p(\frak A,\bar\mu)$ is
order-continuous.

(b) $L^p$ has the Levi property.

(c) Setting $q=p/(p-1)$, the canonical identification of
$L^q=L^q(\frak A,\bar\mu)$ with $(L^p)^*$ is a Riesz space isomorphism
between $L^q$ and $(L^p)^{\sim}=(L^p)^{\times}$.

(d) $L^p$ is a perfect Riesz space.

\proof{{\bf (a)} Suppose that $A\subseteq L^p$ is non-empty,
downwards-directed and has infimum $0$.   For $u$, $v\ge 0$ in $L^p$,
$u\le v\Rightarrow u^p\le v^p$ (by the definition in 366A, or
otherwise), so $B=\{u^p:u\in A\}$ is downwards-directed.   If
$v_0=\inf B$ in $L^1=L^1(\frak A,\bar\mu)$, then $v_0^{1/p}$ (defined by
the formula in 366A, or otherwise) is less than or equal to every member
of $A$, so must be $0$, and $v_0=0$.   Accordingly $\inf B=0$ in $L^1$.
Because $\|\,\|_1$ is order-continuous (365C),

\Centerline{$\inf_{u\in A}\|u\|_p=\inf_{u\in A}\|u^p\|_1^{1/p}
=(\inf_{v\in B}\|v\|_1)^{1/p}=0$.}

\noindent As $A$ is arbitrary, $\|\,\|_p$ is order-continuous.

\medskip

{\bf (b)} Now suppose that $A\subseteq (L^p)^+$ is non-empty,
upwards-directed and norm-bounded.   Then $B=\{u^p:u\in A\}$ is
non-empty, upwards-directed and norm-bounded in $L^1$.   So $v_0=\sup B$
is defined in $L^1$, and $v_0^{1/p}$ is an upper bound for $A$ in $L^p$.

\medskip

{\bf (c)} By 356Dd, $(L^p)^*=(L^p)^{\sim}=(L^p)^{\times}$.   The
extra information we need is that the identification of $L^q$ with
$(L^p)^*$ is an order-isomorphism.   \Prf\ ($\alpha$) If $w\in (L^q)^+$
and $u\in (L^p)^+$ then $u\times w\ge 0$ in $L^1$, so
$(Tw)(u)=\int u\times w\ge 0$, writing $T:L^q\to(L^p)^*$ for the
canonical bijection.   As $u$
is arbitrary, $Tw\ge 0$.  As $w$ is arbitrary, $T$ is a positive linear
operator.   ($\beta$) If $w\in L^q$ and $Tw\ge 0$, consider
$u=(w^-)^{q/p}$.   Then $u\ge 0$ in $L^p$ and $w^+\times u=0$
(because $\Bvalue{w^+>0}\Bcap\Bvalue{u>0}
=\Bvalue{w^+>0}\Bcap\Bvalue{w^->0}=0$), so

\Centerline{$0\le (Tw)(u)=\int w\times u=-\int w^-\times u
=-\int(w^-)^q\le 0$,}

\noindent and $\int(w^-)^q=0$.   But as $(w^-)^q\ge 0$ in $L^1$, this
means that $(w^-)^q$ and $w^-$ must be $0$, that is, $w\ge 0$.   As $w$
is arbitrary, $T^{-1}$ is positive and $T$ is an order-isomorphism.\
\Qed

\medskip

{\bf (d)} This is an immediate consequence of (c), since $p=q/(q-1)$, so
that $L^p$ can be identified with $(L^q)^*=(L^q)^{\times}$.   From
356M we see that it is also a consequence of (a) and (b).
}%end of proof of 366D

\leader{366E}{Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra, and $p\in[1,\infty]$.   Set $q=p/(p-1)$ if
$1<p<\infty$, $q=\infty$ if $p=1$ and $q=1$ if $p=\infty$.   Then

\Centerline{$L^q(\frak A,\bar\mu)
=\{u:u\in L^0(\frak A),\,u\times v\in L^1(\frak A,\bar\mu)$ for every
$v\in L^p(\frak A,\bar\mu)\}$.}

\proof{{\bf (a)} We already know that if $u\in L^p=L^p(\frak A,\bar\mu)$
and $v\in L^q=L^q(\frak A,\bar\mu)$ then
$u\times v\in L^1=L^1(\frak A,\bar\mu)$;  this is elementary if
$p\in\{1,\infty\}$ and otherwise is covered by 366C.

\medskip

{\bf (b)} So suppose that $u\in L^0\setminus L^p$.   If $p=1$ then of
course $\chi 1\in L^{\infty}=L^q$ and $u\times\chi 1\notin L^1$.   If
$p>1$ set

\Centerline{$A=\{w:w\in S(\frak A^f),\,0\le w\le|u|\}$.}

\noindent Because $\bar\mu$ is semi-finite, $S(\frak A^f)$ is
order-dense in $L^0$ (364K), and $|u|=\sup A$.   Because the
norm on $L^p$ has the Levi property (365C, 366Db, 363Ba) and $A$ is not
bounded above in $L^p$, $\sup_{w\in A}\|w\|_p=\infty$.

For each $n\in\Bbb N$ choose $w_n\in A$ with $\|w_n\|_p>4^n$.   Then
there is a $v_n\in L^q$ such that $\|v_n\|_q=1$ and $\int w_n\times
v_n\ge 4^n$.   \Prf\ ($\alpha$) If $p<\infty$ this is covered by 366C,
since $\|w_n\|_p=\sup\{\int w_n\times v:\|v\|_q\le 1\}$.   ($\beta$) If
$p=\infty$ then $\Bvalue{w_n>4^n}\ne 0$;  because $\bar\mu$ is
semi-finite, there is a $b\Bsubseteq\Bvalue{w_n>4^n}$ such that
$0<\bar\mu b<\infty$, and $\|\bover1{\bar\mu b}\chi b\|_1=1$, while
$\int w_n\times\bover1{\bar\mu b}\chi b\ge 4^n$.\ \Qed

Because $L^q$ is complete (363Ba, 366C),
$v=\sum_{n=0}^{\infty}2^{-n}|v_n|$ is defined in $L^q$.   But now

\Centerline{$\int|u|\times v\ge2^{-n}\int w_n\times v_n\ge 2^n$}

\noindent for every $n$, so $u\times v\notin L^1$.
}%end of proof of 366E

\cmmnt{\medskip

\noindent{\bf Remark} This result is characteristic of perfect subspaces
of $L^0$;  see 369C and 369J.}

\leader{366F}{}\cmmnt{ The next step is to look at the action of
Boolean homomorphisms, as in 365N.   It will be convenient to be able
to deal with all $L^p$ spaces at once by introducing names for a pair of
spaces which include all of them.

\medskip

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

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

\Centerline{$M^{1,0}_{\bar\mu}
=M^{1,0}(\frak A,\bar\mu)
=\{u:u\in M^0_{\bar\mu},\,u\times\chi a\in L^1(\frak A,\bar\mu)$
  whenever $\bar\mu a<\infty\}$.}

\leader{366G}{Lemma} Let $(\frak A,\bar\mu)$ be any measure algebra.
Write $M^0=M^0(\frak A,\bar\mu)$, etc.

(a) $M^0$ and $M^{1,0}$ are Dedekind complete solid linear subspaces of
$L^0$ which include $L^p$ for every $p\in\coint{1,\infty}$;  moreover,
$M^0$ is closed under multiplication.

(b) If $u\in M^0$ and $u\ge 0$, there is a
non-decreasing sequence $\sequencen{u_n}$ in $S(\frak A^f)$ such that
$u=\sup_{n\in\Bbb N}u_n$.

(c) $M^{1,0}
=\{u:u\in L^0,\,(|u|-\epsilon\chi 1)^+\in L^1$ for every $\epsilon>0\}
=L^1+(L^{\infty}\cap M^0)$.

(d) If $u$, $v\in M^{1,0}$ and $\int_au\le\int_av$ whenever
$\bar\mu a<\infty$, then $u\le v$;  so if $\int_au=\int_av$ whenever
$\bar\mu a<\infty$, $u=v$.

\proof{{\bf (a)} If $u$, $v\in M^0$ and $\gamma\in\Bbb R$, then for any
$\alpha>0$

\Centerline{$\Bvalue{|u+v|>\alpha}
\Bsubseteq\Bvalue{|u|>\bover12\alpha}
\Bcup\Bvalue{|v|>\bover12\alpha}$,}

\Centerline{$\Bvalue{|\gamma u|>\alpha}
\Bsubseteq\Bvalue{|u|>\bover{\alpha}{1+|\gamma|}}$,}

\Centerline{$\Bvalue{|u\times v|>\alpha}
\Bsubseteq\Bvalue{|u|>\sqrt{\alpha}\thinspace}
\Bcup\Bvalue{|v|>\sqrt{\alpha}\thinspace}$}

\noindent (364E) are of finite measure.   So $u+v$, $\gamma u$ and
$u\times v$ belong to $M^0$.   Thus $M^0$ is a linear subspace of $L^0$
closed under
multiplication.   If $u\in M^0$, $|v|\le |u|$ and $\alpha>0$, then
$\Bvalue{|v|>\alpha}\Bsubseteq\Bvalue{|u|>\alpha}$ has finite measure;
thus $v\in M^0$ and $M^0$ is a solid linear subspace of $L^0$.   It
follows that $M^{1,0}$ also is.   If $u\in L^p=L^p(\frak A,\bar\mu)$,
where $p<\infty$, and $\alpha>0$, then
$\Bvalue{|u|>\alpha}=\Bvalue{|u|^p>\alpha^p}$ has finite measure, so
$u\in M^0$;  moreover, if $\bar\mu a<\infty$, then $\chi a\in L^q$,
where $q=p/(p-1)$, so $u\times\chi a\in L^1$;  thus $u\in M^{1,0}$.

To see that $M^0$ is Dedekind complete, observe that if
$A\subseteq(M^0)^+$ is non-empty and bounded above by $u_0\in M^0$,
and $\alpha>0$, then
$\{\Bvalue{u>\alpha}:u\in A\}$ is bounded above by
$\Bvalue{u_0>\alpha}\in\frak A^f$, so has a
supremum in $\frak A$ (321C).   Accordingly $\sup A$ is defined in $L^0$
(364L(a-iii)) and belongs to $M^0$.
Finally, $M^{1,0}$, being a solid linear
subspace of $M^0$, must also be Dedekind complete.

\medskip

{\bf (b)} If $u\ge 0$ in $M^0$, then there is a non-decreasing sequence
$\sequencen{u_n}$ in $S=S(\frak A)$ such that $u=\sup_{n\in\Bbb N}u_n$
and $u_0\ge 0$ (364Jd).   But now every $u_n$ belongs to
$S\cap M^0=S(\frak A^f)$, just as in 365F.

\medskip

{\bf (c)(i)} If $u\in M^{1,0}$ and $\epsilon>0$, then
$a=\Bvalue{|u|>\epsilon}\in\frak A^f$, so $u\times\chi a\in
L^1=L^1(\frak A,\bar\mu)$;  but $(|u|-\epsilon\chi 1)^+\le |u|\times\chi
a$, so $(|u|-\epsilon\chi 1)^+\in L^1$.

\medskip

\quad{\bf (ii)} Suppose that $u\in L^0$ and $(|u|-\epsilon\chi 1)^+\in
L^1$ for every $\epsilon>0$.   Then, given $\epsilon>0$,
$v=(|u|-\bover12\epsilon\chi 1)^+\in L^1$, and
$\bar\mu\Bvalue{v>\bover12\epsilon}<\infty$;  but
$\Bvalue{|u|>\epsilon}\Bsubseteq\Bvalue{v>\bover12\epsilon}$, so also
has finite measure.   Thus $u\in M^0$.   Next, if $a\in\frak A^f$, then
$|u\times\chi a|\le\chi a+(|u|-\chi 1)^+\in L^1$, so $u\in M^{1,0}$.

\medskip

\quad{\bf (iii)} Of course $L^1$ and $L^{\infty}\cap M^0$ are included
in $M^{1,0}$, so their linear sum also is.   On the other hand, if $u\in
M^{1,0}$, then

\Centerline{$u=(u^+-\chi 1)^+-(u^--\chi 1)^++(u^+\wedge\chi 1)
-(u^-\wedge\chi 1)\in L^1+(L^{\infty}\cap M^0)$.}

\medskip

{\bf (d)} Take $\alpha>0$ and set $a=\Bvalue{u-v>\alpha}$.   Because
both $u$ and
$v$ belong to $M^{1,0}_{\bar\mu}$, $\bar\mu a<\infty$ and
$\int_au\le\int_av$, that is, $\int_au-v\le 0$;  so $a$ must be $0$
(365Dc).   As $\alpha$ is arbitrary, $u-v\le 0$ and $u\le v$.
If $\int_au=\int_av$ for every $a\in\frak A^f$, then $v\le u$ so
$u=v$.
}%end of proof of 366G

\leader{366H}{Theorem} Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$
be measure algebras.   Let $\pi:\frak A^f\to\frak B^f$ be a
measure-preserving ring homomorphism.

(a)(i) We have a unique order-continuous Riesz homomorphism
$T=T_{\pi}:M^0(\frak A,\bar\mu)\to M^0(\frak B,\bar\nu)$ such that
$T(\chi a)=\chi(\pi a)$ for every $a\in\frak A^f$.

\quad(ii) $\Bvalue{Tu>\alpha}=\pi\Bvalue{u>\alpha}$ for every
$u\in M^0(\frak A,\bar\mu)$ and $\alpha>0$.

\quad(iii) $T$ is injective and multiplicative.

\quad(iv) For $p\in[1,\infty]$ and $u\in M^0(\frak A,\bar\mu)$,
$\|Tu\|_p=\|u\|_p$;  in particular, $Tu\in L^p(\frak B,\bar\nu)$ iff
$u\in L^p(\frak A,\bar\mu)$.   \cmmnt{Consequently }$\int Tu=\int u$
whenever $u\in L^1(\frak A,\bar\mu)$.

\quad(v) For $u\in M^0(\frak A,\bar\mu)$,
$Tu\in M^{1,0}(\frak B,\bar\nu)$ iff $u\in M^{1,0}(\frak A,\bar\mu)$.

(b)(i) We have a unique order-continuous positive linear operator
$P=P_{\pi}:M^{1,0}(\frak B,\bar\nu)\to M^{1,0}(\frak A,\bar\mu)$ such
that
$\int_aPv=\int_{\pi a}v$ whenever $v\in M^{1,0}(\frak B,\bar\nu)$ and
$a\in\frak A^f$.

\quad(ii) If $u\in M^0(\frak A,\bar\mu)$,
$v\in M^{1,0}(\frak B,\bar\nu)$ and
$v\times Tu\in M^{1,0}(\frak B,\bar\nu)$, then
$P(v\times Tu)=u\times Pv$.

\quad(iii) If $q\in\coint{1,\infty}$ and $v\in L^q(\frak B,\bar\nu)$,
then
$Pv\in L^q(\frak A,\bar\mu)$ and $\|Pv\|_q\le\|v\|_q$;  if
$v\in L^{\infty}(\frak B)\cap M^0(\frak B,\bar\nu)$, then
$Pv\in L^{\infty}(\frak A)$ and $\|Pv\|_{\infty}\le\|v\|_{\infty}$.

\quad(iv) $PTu=u$ for every $u\in M^{1,0}(\frak A,\bar\mu)$;  in
particular, $P[L^p(\frak B,\bar\nu)]=L^p(\frak A,\bar\mu)$ for every
$p\in\coint{1,\infty}$.

(c) If $(\frak C,\lambda)$ is another measure algebra and
$\theta:\frak B^f\to\frak C^f$ another measure-preserving ring
homomorphism, then $T_{\theta\pi}
=T_{\theta}T_{\pi}:M^0(\frak A,\bar\mu)\to M^0(\frak C,\bar\lambda)$ and
$P_{\theta\pi}=P_{\pi}P_{\theta}:M^{1,0}(\frak C,\bar\lambda)\to
M^{1,0}(\frak A,\bar\mu)$.

(d) Now suppose that $\pi[\frak A^f]=\frak B^f$, so that $\pi$ is a
measure-preserving isomorphism between the rings $\frak A^f$ and
$\frak B^f$.

\quad(i) $T$ is a Riesz space isomorphism between $M^0(\frak A,\bar\mu)$
and $M^0(\frak B,\bar\nu)$, and its inverse is $T_{\pi^{-1}}$.

\quad(ii) $P$ is a Riesz space isomorphism between
$M^{1,0}(\frak B,\bar\nu)$ and $M^{1,0}(\frak A,\bar\mu)$, and its
inverse is $P_{\pi^{-1}}$.

\quad(iii) The restriction of $T$ to $M^{1,0}(\frak A,\bar\mu)$ is
$P^{-1}=P_{\pi^{-1}}$;  the restriction of
$T^{-1}=T_{\pi^{-1}}$ to $M^{1,0}(\frak B,\bar\nu)$ is $P$.

\quad(iv) For any $p\in\coint{1,\infty}$,
$T\restr L^p(\frak A,\bar\mu)=P_{\pi^{-1}}\restr L^p(\frak A,\bar\mu)$
and $P\restr L^p(\frak B,\bar\nu)
=T_{\pi^{-1}}\restr L^p(\frak B,\bar\nu)$ are
the two halves of a Banach lattice isomorphism between
$L^p(\frak A,\bar\mu)$ and $L^p(\frak B,\bar\nu)$.

\proof{{\bf (a)(i)} By 361J, $\pi$ induces a multiplicative Riesz
homomorphism $T_0:S(\frak A^f)\to S(\frak B^f)$ which is
order-continuous because $\pi$ is (361Ad, 361Je).   If
$u\in S(\frak A^f)$ and $\alpha>0$, then
$\Bvalue{T_0u>\alpha}=\pi\Bvalue{u>\alpha}$.
\Prf\ Express $u$ as $\sum_{i=0}^n\alpha_i\chi a_i$ where
$a_0,\ldots,a_n$ are disjoint in $\frak A^f$;  then
$T_0u=\sum_{i=0}^n\alpha_i\chi(\pi a_i)$, so

\Centerline{$\Bvalue{T_0u>\alpha}
=\sup\{\pi a_i:i\le n,\,\alpha_i>\alpha\}
=\pi(\sup\{a_i:i\le n,\,\alpha_i>\alpha\})
=\pi\Bvalue{u>\alpha}$.   \Qed}

Now if $u_0\ge 0$ in $M^0_{\bar\mu}$,
$\sup\{T_0u:u\in S(\frak A^f),\,0\le u\le u_0\}$ is defined in
$M^0_{\bar\nu}$.   \Prf\ Set
$A=\{u:u\in S(\frak A^f),\,0\le u\le u_0\}$.   Because $u_0=\sup A$
(366Gb),

\Centerline{$\sup_{u\in A}\Bvalue{Tu>\alpha}
=\sup_{u\in A}\pi\Bvalue{u>\alpha}
=\pi(\sup_{u\in A}\Bvalue{u>\alpha})
=\pi\Bvalue{u_0>\alpha}$}

\noindent is defined and belongs to $\frak B^f$ for any $\alpha>0$.
Also

\Centerline{$\inf_{n\ge 1}\sup_{u\in A}\Bvalue{Tu>n}
=\pi(\inf_{n\ge 1}\Bvalue{u_0>n})=0$.}

\noindent By 364L(a-ii), $v_0=\sup T_0[A]$ is defined in $L^0(\frak B)$,
and $\Bvalue{v_0>\alpha}=\pi\Bvalue{u_0>\alpha}\in\frak B^f$ for every
$\alpha>0$, so $v_0\in M^0_{\bar\nu}$, as required.\ \Qed

Consequently $T_0$ has a unique extension to an order-continuous Riesz
homomorphism $T:M^0_{\bar\mu}\to M^0_{\bar\nu}$ (355F).

\medskip

\quad{\bf (ii)} If $u_0\in M^0_{\bar\mu}$ and $\alpha>0$, then

$$\eqalignno{\Bvalue{Tu_0>\alpha}
&=\Bvalue{Tu_0^+>\alpha}\cr
\noalign{\noindent (because $T$ is a Riesz homomorphism)}
&=\sup_{u\in S(\frak A^f),0\le u\le u_0^+}\Bvalue{Tu>\alpha}\cr
\noalign{\noindent (because $T$ is order-continuous and $S(\frak A^f)$
is order-dense in $M^0_{\bar\mu}$)}
&=\pi\Bvalue{u_0>\alpha}\cr}$$

\noindent by the argument used in (i).

\medskip

\quad{\bf (iii)} I have already remarked, at the beginning of the proof
of (i), that $T(u\times u')=Tu\times Tu'$ for $u$, $u'\in S(\frak A^f)$.
Because both $T$ and $\times$ are order-continuous and $S(\frak A^f)$ is
order-dense in $M^0_{\bar\mu}$,

$$\eqalign{T(u_0\times u_1)
&=\sup\{T(u\times u'):u,\,u'\in S(\frak A^f),\,0\le u\le u_0,\,
   0\le u'\le u_1\}\cr
&=\sup_{u,u'}Tu\times Tu'
=Tu_0\times Tu_1\cr}$$

\noindent whenever $u_0$, $u_1\ge 0$ in $M^0_{\bar\mu}$.   Because $T$
is linear and $\times$ is bilinear, it follows that $T$ is
multiplicative on $M^0_{\bar\mu}$.

To see that it is injective, observe that if $u\ne 0$ in $M^0_{\bar\mu}$
then there is some $\alpha>0$ such that $a=\Bvalue{|u|>\alpha}\ne 0$, so
that $0<\alpha\chi\pi a\le T|u|=|Tu|$ and $Tu\ne 0$.

\medskip

\quad{\bf (iv)}\grheada\ For any $\alpha>0$,

\Centerline{$\Bvalue{|Tu|^p>\alpha}
=\Bvalue{T|u|>\alpha^{1/p}}
=\pi\Bvalue{|u|>\alpha^{1/p}}
=\pi\Bvalue{|u|^p>\alpha}$.}

\noindent So

\Centerline{$\||Tu|^p\|_1
=\int_0^{\infty}\bar\nu\Bvalue{|Tu|^p>\alpha}\,d\alpha
=\int_0^{\infty}\bar\mu\Bvalue{|u|^p>\alpha}\,d\alpha
=\||u|^p\|_1$.}

\noindent If $p<\infty$ then, taking $p$th roots, $\|Tu\|_p=\|u\|_p$.

\medskip

\qquad\grheadb\ As for the case $p=\infty$, if
$u\in L^{\infty}(\frak A)$ and $\gamma=\|u\|_{\infty}>0$ then
$\Bvalue{|u|>\gamma}=0$, so
$\Bvalue{|Tu|>\gamma}=\pi\Bvalue{|u|>\gamma}=0$.   This shows that
$\|Tu\|_{\infty}\le\gamma$.   On the other hand, if $0<\alpha<\gamma$
then $a=\Bvalue{|u|>\alpha}\ne 0$, and
$\alpha\chi a\le|u|$ so $\alpha\chi(\pi a)\le|Tu|$;  as
$\pi a\ne 0$ (because $\bar\nu(\pi a)=\bar\mu a>0$),
$\|Tu\|_{\infty}>\alpha$.   This shows that
$\|Tu\|_{\infty}=\|u\|_{\infty}$, at least when $u\ne 0$;  but the case
$u=0$ is trivial.

\medskip

\qquad\grheadc\ If $u\in L^1_{\bar\mu}$, then

\Centerline{$\int Tu=\|(Tu)^+\|_1-\|(Tu)^-\|_1=\|Tu^+\|_1
-\|Tu^-\|_1=\|u^+\|_1-\|u^-\|_1=\int u$.}

\medskip

\quad{\bf (v)} If $u\in M^{1,0}_{\bar\mu}$ and $\epsilon>0$, then
$T(|u|\wedge\epsilon\chi 1_{\frak A})=|Tu|\wedge\epsilon\chi 1_{\frak
B}$.   \Prf\ Set $a=\Bvalue{|u|>\epsilon}\in\frak A^f$.   Then
$|u|\wedge\epsilon\chi 1_{\frak A}=\epsilon\chi a+|u|-|u|\times\chi a$
and
$\Bvalue{|Tu|>\epsilon}=\pi a$.   So

$$\eqalign{T(|u|\wedge\epsilon\chi 1_{\frak A})
&=T(\epsilon\chi a)+T|u|-T(|u|\times\chi a)\cr
&=\epsilon\chi(\pi a)+|Tu|-|Tu|\times\chi(\pi a)
=|Tu|\wedge\epsilon\chi 1_{\frak B}.  \text{ \Qed}}$$

Consequently

\Centerline{$T(|u|-\epsilon\chi 1_{\frak A})^+
=T(|u|-|u|\wedge\epsilon\chi 1_{\frak A})
=(|Tu|-\epsilon\chi 1_{\frak B})^+$.}

\noindent But this means that $(|u|-\epsilon\chi 1_{\frak A})^+\in
L^1_{\bar\mu}$ iff $(|Tu|-\epsilon\chi 1_{\frak B})^+\in L^1_{\bar\nu}$.
Since this is true for every $\epsilon>0$, 366Gc tells us that $u\in
M^{1,0}_{\bar\mu}$ iff $Tu\in M^{1,0}_{\bar\nu}$.

\medskip


{\bf (b)(i)}\grheada\ By 365Oa, we have an order-continuous positive
linear operator $P_0:L^1_{\bar\nu}\to L^1_{\bar\mu}$ such that
$\int_aP_0v=\int_{\pi a}v$ for every $v\in L^1_{\bar\nu}$ and
$a\in\frak A^f$.

\medskip

\qquad\grheadb\ We now find that if $v_0\ge 0$ in $M^{1,0}_{\bar\nu}$
and $B=\{v:v\in L^1_{\bar\nu},\,0\le v\le v_0\}$, then $P_0[B]$ has a
supremum in $L^0(\frak A)$ which belongs to $M^{1,0}_{\bar\mu}$.   \Prf\
Because $B$ is upwards-directed and $P_0$ is order-preserving, $P_0[B]$
is upwards-directed.   If $\alpha>0$ and $v\in B$ and
$a=\Bvalue{P_0v>\alpha}$, then

\Centerline{$v
\le(v_0-\bover{\alpha}2\chi 1_{\frak B})^+
  +\Bover{\alpha}2\chi 1_{\frak B}$,}

\noindent so

$$\eqalign{\alpha\bar\mu a
&\le\int_aP_0v
=\int_{\pi a}v
\le\int(v_0-\Bover{\alpha}2\chi 1_{\frak B})^+
 +\Bover{\alpha}2\bar\nu(\pi a)\cr
&=\int(v_0-\Bover{\alpha}2\chi 1_{\frak B})^+
 +\Bover{\alpha}2\bar\mu a\cr}$$

\noindent and

\Centerline{$\bar\mu\Bvalue{P_0v>\alpha}
\le\Bover2{\alpha}\int(v_0-\bover{\alpha}2\chi 1_{\frak B})^+$.}

\noindent Thus $\{\Bvalue{P_0v>\alpha}:v\in B\}$ is an upwards-directed
set in $\frak A^f$ with measures bounded above in $\Bbb R$, and

\Centerline{$c_{\alpha}=\sup_{v\in B}\Bvalue{P_0v>\alpha}$}
%\Centerline needed for Lulu version

\noindent is defined in $\frak A^f$.  Also

\Centerline{$\inf_{n\ge 1}\bar\mu c_n
\le\inf_{n\ge 1}\Bover2n\int(v_0-\bover{n}2\chi 1_{\frak B})^+
=0$.}

\noindent So $\inf_{n\in\Bbb N}c_n=0$ and $P_0[B]$ has a supremum
$u_0\in L^0(\frak A)$ (364L(a-ii)).   As
$\Bvalue{u_0>\alpha}=c_{\alpha}\in\frak A^f$ for every $\alpha>0$,
$u_0\in M^0_{\bar\mu}$.   If $c\in\frak A^f$, then

\Centerline{$\int_cu_0=\sup_{v\in B}\int_cP_0v
=\sup_{v\in B}\int_{\pi c}v\le\int_{\pi c}v_0<\infty$,}

\noindent so $u_0\in M^{1,0}_{\bar\mu}$.\ \Qed

\medskip

\qquad\grheadc\ Now 355F tells us that $P_0$ has a unique extension
to an order-continuous positive linear operator
$P:M^{1,0}_{\bar\nu}\to M^{1,0}_{\bar\mu}$.   If $v_0\ge 0$ in
$M^{1,0}_{\bar\nu}$ and $a\in\frak A^f$, then, as remarked above,

$$\eqalign{\int_aPv_0
&=\sup\{\int_aP_0v:v\in L^1_{\bar\nu},\,0\le v\le v_0\}\cr
&=\sup\{\int_{\pi a}v:v\in L^1_{\bar\nu},\,0\le v\le v_0\}
=\int_{\pi a}v_0;\cr}$$

\noindent because $P$ is linear, $\int_aPv=\int_{\pi a}v$ for every
$v\in M^{1,0}_{\bar\nu}$, $a\in\frak A^f$.

\medskip

\qquad\grheadd\ By 366Gd, $P$ is uniquely defined by the formula

\Centerline{$\int_aPv=\int_{\pi a}v$ whenever
$v\in M^{1,0}_{\bar\nu}$ and $a\in\frak A^f$.}

\medskip

\quad{\bf (ii)} Because $M^0_{\bar\mu}$ is closed under multiplication,
$u\times Pv$ certainly belongs to $M^0_{\bar\mu}$.

\medskip

\qquad\grheada\ Suppose that $u$, $v\ge 0$.   Fix $c\in\frak A^f$ for
the moment.
Suppose that $u'\in S(\frak A^f)$.    Then
we can express $u'$ as $\sum_{i=0}^n\alpha_i\chi a_i$ where $a_i\in\frak
A^f$ for every $i\le n$.   Accordingly

\Centerline{$\int_c u'\times Pv
=\sum_{i=0}^n\alpha_i\int_{c\Bcap a_i}Pv
=\sum_{i=0}^n\alpha_i\int v\times\chi(\pi a_i)\times\chi(\pi c)
=\int_{\pi c} v\times Tu'$.}

\noindent Next, we
can find a non-decreasing sequence $\sequencen{u_n}$ in $S(\frak A^f)^+$
with supremum $u$, and

$$\eqalign{\sup_{n\in\Bbb N}\int_c u_n\times Pv
&=\sup_{n\in\Bbb N}\int_{\pi c} v\times Tu_n
=\int_{\pi c}\sup_{n\in\Bbb N}v\times Tu_n\cr
&=\int_{\pi c} v\times\sup_{n\in\Bbb N}Tu_n
=\int_{\pi c} v\times Tu,\cr}$$

\noindent using the order-continuity of $T$, $\int$ and $\times$.
But this means that $u\times Pv=\sup_{n\in\Bbb N}u_n\times Pv$ is
integrable over $c$ and that $\int_c u\times Pv=\int_{\pi c} v\times
Tu$.   As $c$ is arbitrary, $u\times Pv=P(v\times Tu)\in
M^{1,0}_{\bar\mu}$.

\medskip

\qquad\grheadb\ For general $u$, $v$,

\Centerline{$v^+\times Tu^+\,+\,v^+\times Tu^-\,+\,v^-\times
Tu^+\,+\,v^-\times Tu^-=|v|\times T|u|=|v\times Tu|\in
M^{1,0}_{\bar\nu}$}

\noindent (because $T$ is a Riesz
homomorphism), so we may apply ($\alpha$) to each of the four products;
combining them, we get $P(v\times Tu)=u\times Pv$, as required.

\medskip

\quad{\bf (iii)} Because $P$ is a positive operator, we surely have
$|Pv|\le P|v|$, so it will be enough to show that $\|Pv\|_q\le\|v\|_q$
for $v\ge 0$ in $L^q_{\bar\nu}$.

\medskip

\qquad \grheada\ I take the case $q=1$ first.   In this case, for any
$a\in\frak A^f$, we have $\int_aPv=\int_{\pi a}v\le\|v\|_1$.
In particular, setting $a_n=\Bvalue{Pv>2^{-n}}$,
$\int_{a_n}Pv\le\|v\|_1$.   But $Pv=\sup_{n\in\Bbb N}Pv\times\chi a_n$,
so

\Centerline{$\|Pv\|_1=\sup_{n\in\Bbb N}\int_{a_n}Pv\le\|v\|_1$.}

\medskip

\qquad\grheadb\ Next, suppose that $q=\infty$, so that
$v\in L^{\infty}(\frak B)^+$;  say $\|v\|_{\infty}=\gamma$.   \Quer\ If
$\gamma>0$ and $a=\Bvalue{Pv>\gamma}\ne 0$, then

\Centerline{$\gamma\bar\mu a<\int_aPv
=\int_{\pi a}v\le\gamma\bar\nu(\pi a)=\gamma\bar\mu a$.   \Bang}

\noindent So $\Bvalue{Pv>\gamma}=0$ and $Pv\in L^{\infty}(\frak A)$,
with $\|Pv\|_{\infty}\le\|v\|_{\infty}$, at least when
$\|v\|_{\infty}>0$;
but the case $\|v\|_{\infty}=0$ is trivial.

\medskip

\qquad\grheadc\ I come at last to the `general' case
$q\in\ooint{1,\infty}$, $v\in L^q_{\bar\nu}$.   In this case set
$p=q/(q-1)$.   If $u\in L^p_{\bar\mu}$ then $Tu\in L^p_{\bar\nu}$ so
$Tu\times v\in L^1_{\bar\nu}$ and

$$\eqalignno{|\int u\times Pv|
&\le\|u\times Pv\|_1
=\|P(Tu\times v)\|_1\cr
\noalign{\noindent (by (ii))}
&\le\|Tu\times v\|_1\cr
\noalign{\noindent (by ($\alpha$) just above)}
&=\int|Tu|\times|v|
\le\|Tu\|_p\|v\|_q
=\|u\|_p\|v\|_q\cr
}$$

\noindent by (a-iii) of this theorem.   But this means that $u\mapsto
\int u\times Pv$ is a bounded linear functional on $L^p_{\bar\mu}$, and
is therefore represented by some $w\in L^q_{\bar\mu}$ with
$\|w\|_q\le\|v\|_q$.   If $a\in\frak A^f$ then
$\chi a\in L^p_{\bar\mu}$, so
$\int_aw=\int_aPv$;  accordingly $Pv$ is actually equal to $w$ (by
366Gd) and
$\|Pv\|_q=\|w\|_q\le\|v\|_q$, as claimed.

\medskip

\quad{\bf (iv)} If $u\in M^{1,0}_{\bar\mu}$ and $a\in\frak A^f$, we must
have

\Centerline{$\int_aPTu=\int_{\pi a}Tu=\int T(\chi a)\times Tu=\int
T(\chi a\times u)=\int \chi a\times u=\int_au$,}

\noindent using (a-iv) to see that $\int\chi a\times u$ is defined and
equal to $\int T(\chi a\times u)$.   As $a$ is arbitrary, $u\in
M^{1,0}_{\bar\mu}$ and $PTu=u$.

\medskip

{\bf (c)} As usual, in view of the uniqueness of $T_{\theta\pi}$ and
$P_{\theta\pi}$, all we have to check is that

\Centerline{$T_{\theta}T(\chi a)=T_{\theta}\chi(\pi a)
=\chi(\theta\pi a)=T_{\theta\pi}(\chi a)$,}

\Centerline{$\int_aPP_{\theta}w=\int_{\pi a}P_{\theta}w
=\int_{\theta\pi a}w=\int_aP_{\theta\pi}w$}

\noindent whenever $a\in\frak A^f$ and $w\in M^{1,0}_{\bar\lambda}$.

\medskip

{\bf (d)(i)} By (c), $T_{\pi^{-1}}T=T_{\pi^{-1}\pi}$ must be the
identity operator on $M^0_{\bar\mu}$;  similarly, $TT_{\pi^{-1}}$ is the
identity operator on $M^0_{\bar\nu}$.   Because $T$ and $T_{\pi^{-1}}$
are Riesz homomorphisms, they must be the two halves of a Riesz space
isomorphism.

\medskip

\quad{\bf (ii)} In the same way, $P$ and $P_{\pi^{-1}}$ must be the two
halves of an ordered linear space isomorphism between
$M^{1,0}_{\bar\mu}$ and $M^{1,0}_{\bar\nu}$, and are therefore both
Riesz homomorphisms.

\medskip

\quad{\bf (iii)} By (b-iv), $PTu=u$ for every $u\in M^{1,0}_{\bar\mu}$,
so $T\restr M^{1,0}_{\bar\mu}$ must be $P^{-1}$.
Similarly $P=P_{\pi^{-1}}^{-1}$ is the restriction of
$T^{-1}=T_{\pi^{-1}}$ to $M^{1,0}_{\bar\nu}$.

\medskip

\quad{\bf (iv)} Because $T^{-1}[L^p_{\bar\nu}]=L^p_{\bar\mu}$
(by (a-iv)), and $T$ is a bijection between $M^0_{\bar\mu}$ and
$M^0_{\bar\nu}$, $T\restr L^p_{\bar\mu}$ must be a Riesz space
isomorphism between $L^p_{\bar\mu}$ and $L^p_{\bar\nu}$;  (a-iv) also
tells us that it is norm-preserving.   Now its inverse is
$P\restr L^p_{\bar\nu}$, by (iii) here.
}%end of proof of 366H

\leader{366I}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra,
and $\frak B$ a $\sigma$-subalgebra of $\frak A$.
Then, for any $p\in\coint{1,\infty}$,
$L^p(\frak B,\bar\mu\restrp\frak B)$
can be identified, as Banach lattice, with the closed linear subspace of
$L^p(\frak A,\bar\mu)$ generated by
$\{\chi b:b\in\frak B,\,\bar\mu b<\infty\}$.

\proof{ The identity map $b\mapsto b:\frak B\to\frak A$ induces an
injective Riesz homomorphism $T:L^0(\frak B)\to L^0(\frak A)$ (364P)
such that $Tu\in L^p_{\frak A}=L^p(\frak A,\bar\mu)$ and
$\|Tu\|_p=\|u\|_p$ whenever $p\in\coint{1,\infty}$ and
$u\in L^p_{\frak B}=L^p(\frak B,\bar\mu\restrp\frak B)$ (366H(a-iv)).
Because $S(\frak B^f)$, the linear span of
$\{\chi b:b\in\frak B,\,\bar\mu b<\infty\}$,
is dense in $L^p_{\frak B}$ (366C), the image of $L^p_{\frak B}$ in
$L^p_{\frak A}$ must be the closure of the image of
$S(\frak B^f)$ in $L^p_{\frak A}$, that is, the closed linear span of
$\{\chi b:b\in\frak B^f\}$ interpreted as a subset of $L^p_{\frak A}$.
}%end of proof of 366I

\leader{366J}{Corollary} If $(\frak A,\bar\mu)$ is a probability
algebra, $\frak B$ is a closed subalgebra of $\frak A$, and
$P:L^1(\frak A,\bar\mu)\to L^1(\frak B,\bar\mu\restrp\frak B)$ is the
conditional
expectation operator\cmmnt{ (365Q)}, then $\|Pu\|_p\le\|u\|_p$
whenever $p\in[1,\infty]$ and $u\in L^p(\frak A,\bar\mu)$.

\proof{ Because $(\frak A,\bar\mu)$ is totally finite,
$M^{1,0}(\frak A,\bar\mu)=L^1_{\bar\mu}$, so that the operator $P$ of
366Hb can be identified with the conditional expectation operator of
365Q.   Now 366H(b-iii) gives the result.
}%end of proof of 366J

\cmmnt{\medskip

\noindent{\bf Remark} Of course this is also covered by 244M.
}

\vleader{72pt}{366K}{Corollary} Let $(\frak A,\bar\mu)$ and
$(\frak B,\bar\nu)$ be measure algebras, and $\pi:\frak A^f\to\frak B^f$ a
measure-preserving ring homomorphism.   Let
$T:L^2(\frak A,\bar\mu)\to L^2(\frak B,\bar\nu)$ and
$P:L^2(\frak B,\bar\nu)\to L^2(\frak A,\bar\mu)$ be the
corresponding operators\cmmnt{, as in 366H}.   Then
$TP:L^2(\frak B,\bar\nu)\to L^2(\frak B,\bar\nu)$ is an orthogonal
projection, its range $TP[L^2(\frak B,\bar\nu)]$
being isomorphic, as Banach lattice, to $L^2(\frak A,\bar\mu)$.   The
kernel of $TP$ is just

\Centerline{$\{v:v\in L^2(\frak B,\bar\nu),\,\int_{\pi a}v=0$ for
every $a\in\frak A^f\}$.}

\proof{ Most of this is simply because $T$ is a
norm-preserving Riesz homomorphism (so that $T[L^2_{\bar\mu}]$ is
isomorphic to $L^2_{\bar\mu}$), $PT$ is the identity on $L^2_{\bar\mu}$
(so that $(TP)^2=TP$) and $\|P\|\le 1$ (so that $\|TP\|\le 1$).   These
are enough to ensure that $TP$ is a projection of norm at most $1$, that
is, an orthogonal projection.   Also

$$\eqalign{TPv=0
&\iff Pv=0\iff\int_aPv=0\text{ for every }a\in\frak A^f\cr
&\iff \int_{\pi a}v=0\text{ for every }a\in\frak A^f.\cr}$$
}%end of proof of 366K

\leader{366L}{Corollary} Let $(\frak A,\bar\mu)$ be a measure algebra,
and $\pi:\frak A^f\to\frak A^f$ a measure-preserving ring automorphism.
Then there is a corresponding Banach lattice isomorphism $T$ of
$L^2=L^2(\frak A,\bar\mu)$ defined by writing $T(\chi a)=\chi(\pi a)$
for every $a\in\frak A^f$.   Its inverse is defined by the formula

\Centerline{$\int_aT^{-1}u=\int_{\pi a}u$ for every $u\in L^2$,
$a\in\frak A^f$.}

\proof{ In the language of 366H, $T=T_{\pi}$ and $T^{-1}=P_{\pi}$.}

%could add corollaries corresponding to 242L, 244Nb

\leader{*366M}{Complex $L^p$ spaces}\dvAnew{2011}
{\bf (a)} Just as in \S\S241-244,
%241J, 242P, 243K and 244P\formerly{2{}44O},
we have `complex' versions of all the spaces
considered in this chapter.\cmmnt{   Using the representation theorems for
Boolean algebras, we can get effective descriptions of
these matching the ones in Chapter 24.}   Thus for any Boolean algebra
$\frak A$ with Stone space $Z$, we can identify
$L^{\infty}_{\Bbb C}(\frak A)$ with the space $C(Z;\Bbb C)$ of continuous
functions from $Z$ to $\Bbb C$;  inside this, we have a
$\|\,\|_{\infty}$-dense subspace $S_{\frak C}(\frak A)$ consisting of
complex
linear combinations of indicator functions of open-and-closed sets.
If $\frak A$ is a Dedekind
$\sigma$-complete Boolean algebra, identified with a quotient
$\Sigma/\Cal M$ where $\Sigma$ is a $\sigma$-algebra of subsets of a set $Z$
and $\Cal M$ is a $\sigma$-ideal of $\Sigma$, then we can write
$\eusm L^0_{\Bbb C}$ for the set of functions from $Z$ to $\Bbb C$ such
that their real and imaginary parts are both $\Sigma$-measurable,
$\Cal W_{\Bbb C}$ for the set of those $f\in\eusm L^0_{\Bbb C}$ such that
$\{z:f(z)\ne 0\}$ belongs to $\Cal M$, and
$L^0_{\Bbb C}=L^0_{\Bbb C}(\frak A)$ for
the linear space quotient $\eusm L^0_{\Bbb C}/\Cal W_{\Bbb C}$.
As in 241J, we\cmmnt{ find that we} have
a natural embedding of $L^0=L^0(\frak A)$ in
$L^0_{\Bbb C}$ and functions

\Centerline{$\Real:L^0_{\Bbb C}\to L^0$,
\quad$\Imag:L^0_{\Bbb C}\to L^0$,
\quad$|\mskip10mu|:L^0_{\Bbb C}\to L^0$,
\quad$\bar{\phantom{u}}:L^0_{\Bbb C}\to L^0_{\Bbb C}$}

\noindent such that

\Centerline{$u=\Real(u)+i\Imag(u)$,
\quad$\Real(u+v)=\Real(u)+\Real(v)$,
\quad$\Imag(u+v)=\Imag(u)+\Imag(v)$,}

\Centerline{$\Real(\alpha u)=\Real(\alpha)\Real(u)-\Imag(\alpha)\Imag(u)$,
\quad$\Imag(\alpha u)=\Real(\alpha)\Imag(u)+\Imag(\alpha)\Real(u)$,}

\Centerline{$|\alpha u|=|\alpha||u|$,
\quad$|u+v|\le|u|+|v|$,
\quad$|u|=\sup_{|\gamma|=1}\Real(\gamma u)$,}

\Centerline{$\bar u=\Real(u)-i\Imag(u)$,
\quad$\overline{u+v}=\bar u+\bar v$,
\quad$\overline{\alpha u}=\bar\alpha\bar u$}

\noindent for all $u$, $v\in L^0_{\Bbb C}$ and $\alpha\in\Bbb C$.

I seem to have omitted to mention it in 241J, but of course we also have a
multiplication

\Centerline{$u\times v=(\Real(u)\times\Real(v)-\Imag(u)\times\Imag(v))
  +i(\Real(u)\times\Imag(v)+\Imag(u)\times\Real(v))$,}

\noindent for which we have\cmmnt{ the expected formulae}

\Centerline{$u\times v=v\times u$,
\quad$u\times(v\times w)=(u\times v)\times w$,
\quad$u\times(v+w)=(u\times v)+(u\times w)$,}

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

\Centerline{$\overline{u\times v}=\bar u\times\bar v$,
\quad$|u\times v|=|u|\times|v|$,
\quad$u\times\bar u=|u|^2=(\Real(u))^2+(\Imag(u))^2$}

\noindent for $u$, $v\in L^0_{\Bbb C}$ and $\alpha\in\Bbb C$.

\spheader 366Mb
If $(\frak A,\bar\mu)$ is a measure algebra and $1\le p<\infty$, we can
think of $L^p_{\Bbb C}(\frak A,\bar\mu)$ as the set of those
$u\in L^0_{\Bbb C}$ such that $|u|\in L^p(\frak A,\bar\mu)$, with
its norm defined by the formula $\|u\|_p=\||u|\|_p$;  this will make
$L^p_{\Bbb C}(\frak A,\bar\mu)$ a Banach space\cmmnt{ (cf.\ 242Pb,
244Pb\formerly{2{}44O})}, with dual $L^q(\frak A,\bar\mu)$ where
$\bover1p+\bover1q=1$ if $p>1$\cmmnt{ (244Pb again)}.   (Similarly, if
$(\frak A,\bar\mu)$ is localizable, the dual of
$L^1_{\Bbb C}(\frak A,\bar\mu)$ can be identified with
$L^{\infty}_{\Bbb C}$\cmmnt{, as in 365Lc}.)

Writing $S_{\Bbb C}(\frak A^f)$ for the space of linear combinations of
indicator functions of elements of $\frak A$ of finite measure,
$S_{\Bbb C}(\frak A^f)$ is dense in $L^p_{\Bbb C}(\frak A,\bar\mu)$
whenever $1\le p<\infty$\cmmnt{, as in 366C}.

\spheader 366Mc
\cmmnt{Of course }$L^1$- and $L^2$-spaces
have\cmmnt{ special additional features,
their} integrals and inner products.   Here we\cmmnt{ can} set

\Centerline{$\int u=\int\Real(u)+i\int\Imag(u)$}

\noindent for $u\in L^1_{\Bbb C}(\frak A,\bar\mu)$, and
$\int:L^1_{\Bbb C}(\frak A,\bar\mu)\to\Bbb C$ becomes a
$\Bbb C$-linear functional.
As for $L^2$,\cmmnt{ we see at once from the formulae above that}

\Centerline{$|u\times v|=|u|\times|v|\in L^1(\frak A,\bar\mu)$,
\quad$u\times v\in L^1_{\Bbb C}(\frak A,\bar\mu)$,
\quad$\int u\times\bar u=\|u\|_2^2$}

\noindent for $u$, $v\in L^2_{\Bbb C}(\frak A,\bar\mu)$.   So if we set

\Centerline{$\innerprod{u}{v}=\int u\times\bar v$}

\noindent for $u$, $v\in L^2_{\Bbb C}(\frak A,\bar\mu)$,
$L^2_{\Bbb C}(\frak A,\bar\mu)$ becomes a complex Hilbert space.

\spheader 366Md\cmmnt{ In
the language of the present chapter we have something else to look at.}
If $\frak A$, $\frak B$ are Dedekind $\sigma$-complete Boolean algebras
and $\pi:\frak A\to\frak B$ is
a sequentially order-continuous Boolean homomorphism, then we have a
linear operator $T_{\pi}:L^0_{\Bbb C}(\frak A)\to L^0_{\Bbb C}(\frak B)$
defined by setting
$T_{\pi}u
=T_{\pi}^{\text{real}}(\Real(u))+iT^{\text{real}}_{\pi}(\Imag(u))$,
where $T^{\text{real}}_{\pi}:L^0(\frak A)\to L^0(\frak B)$
is the Riesz homomorphism
described in 364P.   \cmmnt{Of course }$T_{\pi}$,
like $T^{\text{real}}_{\pi}$, will be
multiplicative;  \cmmnt{hence, or otherwise, }$T_{\pi}|u|=|T_{\pi}u|$
for every
$u\in L^0_{\Bbb C}(\frak A)$.   \cmmnt{Observe that}
$T_{\pi}\bar u=\overline{T_{\pi}u}$ for every $u\in L^0_{\Bbb C}(\frak A)$.
Also,\cmmnt{ as in 364Pe,} if $\frak C$ is another
Dedekind $\sigma$-complete
Boolean algebra and $\pi:\frak A\to\frak B$ and $\phi:\frak B\to\frak C$
are sequentially order-continuous Boolean homomorphisms,
$T_{\phi\pi}=T_{\phi}T_{\pi}$.   So if $\pi:\frak A\to\frak A$ is a
Boolean automorphism, $T_{\pi}$ will be a bijection with inverse
$T_{\pi^{-1}}$.

\spheader 366Me
Similarly, if $(\frak A,\bar\mu)$ is a measure algebra and
$\pi:\frak A\to\frak A$ is a measure-preserving Boolean homomorphism,
$\int T_{\pi}u=\int u$ for every $u\in L^1_{\Bbb C}(\frak A,\bar\mu)$.
If $u$, $v\in L^2_{\Bbb C}(\frak A,\bar\mu)$, then

\Centerline{$\innerprod{T_{\pi}u}{T_{\pi}v}
=\int T_{\pi}u\times\overline{T_{\pi}v}
=\int T_{\pi}u\times T_{\pi}\bar v
=\int T_{\pi}(u\times\bar v)
=\int u\times\bar v
=\innerprod{u}{v}$.}

\noindent If $\pi$ is\cmmnt{ actually} a measure-preserving Boolean
automorphism, we shall have

\Centerline{$\innerprod{T_{\pi}u}{v}
=\innerprod{T_{\pi^{-1}}T_{\pi}u}{T_{\pi^{-1}}v}
=\innerprod{u}{T_{\pi}^{-1}v}$}

\noindent for all $u$, $v\in L^2_{\Bbb C}(\frak A,\bar\mu)$.

\exercises{\leader{366X}{Basic exercises (a)}
%\spheader 366Xa
Let $(\frak A,\bar\mu)$ be a measure algebra and $p\in\ooint{1,\infty}$.
Show that $\|u\|_p^p
=p\int_0^{\infty}\alpha^{p-1}\bar\mu\Bvalue{|u|>\alpha}\,d\alpha$
for every $u\in L^0(\frak A)$.  (Cf.\ 263Xa.)
%366A

\sqheader 366Xb Let $(\frak A,\bar\mu)$ be a localizable measure algebra
and $p\in[1,\infty]$.   Show that the band algebra of
$L^p_{\bar\mu}$ is isomorphic to $\frak A$.   (Cf.\ 365R.)

\spheader 366Xc Let $(\frak A,\bar\mu)$ be a measure algebra and
$p\in\ooint{1,\infty}$.   Show that $L^p_{\bar\mu}$ is separable
iff $L^1_{\bar\mu}$ is.
%366C

\spheader 366Xd
Let $(\frak A,\bar\mu)$ be a measure algebra.   (i) Show that
$L^{\infty}(\frak A)\cap M^0_{\bar\mu}$ and
$L^{\infty}(\frak A)\cap M^{1,0}_{\bar\mu}$, as defined in 366F,
are equal.   (ii)
Call this intersection $M^{\infty,0}_{\bar\mu}$.   Show that it
is a norm-closed solid linear subspace of $L^{\infty}(\frak A)$,
therefore a Banach lattice in its own right.
%366G

\spheader 366Xe Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
and $(\widehat{\frak A},\hat\mu)$ its localization (322Q).   Show that
the natural embedding of $\frak A$ in $\widehat{\frak A}$ induces a
Banach lattice isomorphism between $L^p_{\bar\mu}$ and
$L^p_{\hat\mu}$ for every $p\in\coint{1,\infty}$, so
that the band algebra of $L^p_{\bar\mu}$ can be identified
with $\widehat{\frak A}$.
%366H

\spheader 366Xf Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
which is not localizable (cf.\ 211Ye, 216D), and
$(\widehat{\frak A},\hat\mu)$ its localization.   Let
$\pi:\frak A\to\widehat{\frak A}$ be the identity embedding, so that
$\pi$ is an order-continuous
measure-preserving Boolean homomorphism.   Show that if we set
$v=\chi b$ where $b\in\widehat{\frak A}\setminus\frak A$, then there is
no $u\in L^{\infty}(\frak A)$ such that $\int_au=\int_{\pi a}v$ whenever
$\bar\mu a<\infty$.
%366H

\spheader 366Xg In 366H, show that $\Bvalue{Tu\in E}=\pi\Bvalue{u\in E}$
(notation:  364G)
whenever $u\in M^0_{\bar\mu}$ and $E\subseteq\Bbb R$ is a Borel set such
that $0\notin\overline{E}$.
%366H

\sqheader 366Xh Let $(\frak A,\bar\mu)$ be a measure algebra and let $G$
be the group of all measure-preserving ring automorphisms of
$\frak A^f$.   Let $H$ be the group of all Banach lattice automorphisms
of $L^2_{\bar\mu}$.   Show that the map $\pi\mapsto T$ of 366L is
an injective group homomorphism from $G$ to $H$, so that $G$ is
represented as a subgroup of $H$.
%366L

\spheader 366Xi Let $\langle(\frak A_i,\bar\mu_i)\rangle_{i\in I}$ be
any family of measure algebras, with simple product $(\frak A,\bar\mu)$
(322L).   Show that for any $p\in\coint{1,\infty}$,
$L^p_{\bar\mu}$ can be
identified, as normed Riesz space, with the solid linear subspace

\Centerline{$\{u:\|u\|
=\bigl(\sum_{i\in I}\|u(i)\|_p^p\bigr)^{1/p}<\infty\}$}

\noindent of $\prod_{i\in I}L^p_{\bar\mu_i}$.

\spheader 366Xj Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\bar\mu$, $\bar\nu$ two functionals rendering $\frak A$ a
semi-finite measure algebra.   Show that for any $p\in\coint{1,\infty}$,
$L^p_{\bar\mu}$ and $L^p_{\bar\nu}$ are isomorphic as
normed Riesz spaces.   \Hint{use 366Xe to reduce to the case in which
$\frak A$ is Dedekind complete.   Take $w\in L^0(\frak A)$ such that
$\int_aw\,d\bar\mu=\bar\nu a$ for every $a\in\frak A$ (365S).   Set
$Tu=w^{1/p}\times u$ for $u\in L^p_{\bar\mu}$.}

\spheader 366Xk Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure
algebras, and $p\in\coint{1,\infty}$.   Show that the following are
equiveridical:  (i) $L^p_{\bar\mu}$ and $L^p_{\bar\nu}$ are isomorphic
as
Banach lattices;  (ii) $L^p_{\bar\mu}$ and $L^p_{\bar\nu}$ are
isomorphic as
Riesz spaces;  (iii) $\frak A$ and $\frak B$ have isomorphic Dedekind
completions.
%366Xj

\spheader 366Xl\dvAformerly{3{}63Xj}
For a Boolean algebra $\frak A$, state and prove
results corresponding to 363C, 363Ea and 363F-363I %363F 363G 363H 363I
for $L^{\infty}_{\Bbb C}(\frak A)$ as defined in 366Ma.
%366M

\leader{366Y}{Further exercises (a)}
Let $(\frak A,\bar\mu)$ be a measure algebra and suppose that $0<p<1$.
Write $L^p=L^p_{\bar\mu}=L^p(\frak A,\bar\mu)$ for
$\{u:u\in L^0(\frak A),\,|u|^p\in L^1_{\bar\mu}\}$, and for
$u\in L^p$ set $\tau(u)=\int|u|^p$.
(i) Show that $\tau$ is an F-seminorm (2A5B\Latereditions) and
defines a Hausdorff linear space topology on $L^p$.
(ii) Show that if $A\subseteq L^p$ is non-empty, downwards-directed and
has infimum $0$ then $\inf_{u\in A}\tau(u)=0$.   (iii) Show that if
$A\subseteq L^p$ is non-empty, upwards-directed and bounded in the
linear topological space sense then $A$ is bounded above.
(iv) Show that $(L^p)^{\sim}=(L^p)^{\times}$ is just the set of
continuous linear functionals from $L^p$ to $\Bbb R$, and is $\{0\}$ iff
$\frak A$ has no atom of finite measure.
%366A

\spheader 366Yb Let $(\frak A,\bar\mu)$ be a measure algebra.   Show
that $M^0(\frak A,\bar\mu)$ has the countable sup property.
%366F

\spheader 366Yc Let $(\frak A,\bar\mu)$ be a measure algebra and define
$M^{\infty,0}_{\bar\mu}$ as in 366Xd.   Show that
$(M^{\infty,0}_{\bar\mu})^{\times}$ can be identified with
$L^1_{\bar\mu}$.
%366G

\spheader 366Yd In 366H, show that if $\tilde T:M^0(\frak A,\bar\mu)\to
M^0(\frak B,\bar\nu)$ is any positive linear operator such that
$\tilde T(\chi a)=\chi(\pi a)$ for every $a\in\frak A^f$, then
$\tilde T$ is order-continuous, so is equal to $T_{\pi}$.
%366H

\spheader 366Ye Let $(\frak A,\bar\mu)$ be a measure algebra.   (i)
Show that there is a natural one-to-one correspondence between
$M^{1,0}(\frak A,\bar\mu)$ and the set of additive functionals
$\nu:\frak A^f\to\Bbb R$ such that $\nu<\backstep{4}<\mu$
in the double sense that
for every $\epsilon>0$ there are $\delta$, $M>0$ such that $|\nu
a|\le\epsilon$ whenever $\mu a\le\delta$ and $|\nu a|\le\epsilon\mu a$
whenever $\mu a\ge M$.   (ii) Use this description of $M^{1,0}$ to prove
366H(b-i).
%366H

\spheader 366Yf In 366H, show that the following are equiveridical:
($\alpha$) $\pi[\frak A^f]=\frak B^f$;  ($\beta$) $T=T_{\pi}$ is
surjective;  ($\gamma$) $P=P_{\pi}$ is injective;  ($\delta$) $P$ is
a Riesz homomorphism;  ($\epsilon$) there is some $q\in[1,\infty]$ such
that $\|Pv\|_q=\|v\|_q$ for every $v\in L^q_{\bar\nu}$;  ($\zeta$)
$TPv=v$ for every $v\in M^{1,0}_{\bar\nu}$.
%366H

\spheader 366Yg Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
measure algebras, and suppose that $\pi:\frak A^f\to\frak B^f$ is a
measure-preserving ring homomorphism, as in 366H;  let
$T:M^0_{\bar\mu}\to M^0_{\bar\nu}$ be the associated
linear operator.   Show that if $0<p<1$ (as in 366Ya) then
$L^p_{\bar\mu}\subseteq M^0_{\bar\mu}$ and
$T^{-1}[L^p_{\bar\nu}]=L^p_{\bar\mu}$.
%366H, 366Ya

\spheader 366Yh Let $(\frak A,\bar\mu)$ be a totally finite measure
algebra.   (i) For each Boolean automorphism $\pi:\frak A\to\frak A$,
let $T_{\pi}:L^0(\frak A)\to L^0(\frak A)$ be the associated Riesz space
isomorphism, and let $w_{\pi}\in(L^1_{\bar\mu})^+$ be such that
$\int_aw_{\pi}=\mu(\pi^{-1}a)$ for every $a\in\frak A$ (365Ea).   Set
$Q_{\pi}u=T_{\pi}u\times\sqrt{w_{\pi}}$ for $u\in L^0(\frak A)$.   Show
that $\|Q_{\pi}u\|_2=\|u\|_2$ for every $u\in L^2_{\bar\mu}$.   (ii)
Show that if $\pi$, $\phi:\frak A\to\frak A$ are Boolean automorphisms
then $Q_{\pi\phi}=Q_{\pi}Q_{\phi}$.

\spheader 366Yi Let $(\frak A,\bar\mu)$ be a measure algebra, and
$\pi:\frak A^f\to\frak A^f$ a measure-preserving Boolean homomorphism,
with associated linear operator
$T_{\pi}:M^0_{\bar\mu}\to M^0_{\bar\mu}$.   Show that the following are
equiveridical:  (i) there is some $p\in\coint{1,\infty}$ such that
$\{T_{\pi}^n\restr L^p_{\bar\mu}:n\in\Bbb N\}$ is relatively compact in
$\eurm B(L^p_{\bar\mu};L^p_{\bar\mu})$ for the strong operator topology;
(ii) for every $p\in\coint{1,\infty}$,
$\{T_{\pi}^n\restr L^p_{\bar\mu}:n\in\Bbb N\}$ is relatively compact in
$\eurm B(L^p_{\bar\mu};L^p_{\bar\mu})$ for the strong operator topology;
(iii) $\{\pi^na:n\in\Bbb N\}$ is relatively compact in $\frak A^f$, for
the strong measure-algebra topology, for every $a\in\frak A^f$.

\spheader 366Yj\dvAformerly{3{}64Yn}
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra.   Show that $L^0_{\Bbb C}(\frak A)$ can be identified with the
complexification of $L^0(\frak A)$ as defined in 354Yl.
%366M

\spheader 366Yk\dvAformerly{3{}64Yn}
Write $\Cal B(\Bbb C)$ for the Borel $\sigma$-algebra of
$\Bbb C\cong\BbbR^2$ as defined in 111Gd.   Show that if $\frak A$ is a
Dedekind $\sigma$-complete Boolean algebra, we have a unique function
$(u,E)\mapsto\Bvalue{u\in E}:
L^0_{\Bbb C}(\frak A)\times\Cal B(\Bbb C)\to\frak A$ such that
(i) for any $u\in L^0_{\Bbb C}(\frak A)$, the function
$E\mapsto\Bvalue{u\in E}$ is a sequentially order-continuous Boolean
homomorphism from $\Cal B(\Bbb C)$ to $\frak A$ (ii) if $E_0$,
$E_1\subseteq\Bbb R$ are Borel sets, then
$\Bvalue{u\in E_0\times E_1}
=\Bvalue{\Real(u)\in E_0}\Bcap\Bvalue{\Imag(u)\in E_1}$ for every
$u\in L^0_{\Bbb C}(\frak A)$ (iii) if $\phi:\Cal B(\Bbb C)\to\frak A$ is a
sequentially order-continuous Boolean homomorphism, then there is a unique
$u\in L^0_{\Bbb C}(\frak A)$ such that $\phi(E)=\Bvalue{u\in E}$ for every
$E\in\Cal B(\Bbb C)$.
%366M 364G

\spheader 366Yl\dvAformerly{3{}64Yn}
A function $h:\Bbb C\to\Bbb C$ is called {\bf Borel
measurable} if its real and imaginary parts are
$\Cal B(\Bbb C)$-measurable, where
$\Cal B(\Bbb C)$ is the Borel $\sigma$-algebra of $\Bbb C$.
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra.
(i) Show that for every Borel measurable $h:\Bbb C\to\Bbb C$ and
$u\in L^0_{\Bbb C}(\frak A)$ we have an element
$\bar h(u)\in L^0_{\Bbb C}(\frak A)$ such that
$\Bvalue{\bar h(u)\in E}=\Bvalue{u\in h^{-1}[E]}$ for every
$E\in\Cal B(\Bbb C)$.   (ii) Show that if $\pi:\frak A\to\frak A$ is a
sequentially order-continuous Boolean homomorphism and
$T:L^0_{\Bbb C}(\frak A)\to L^0_{\Bbb C}(\frak A)$ the corresponding linear
operator (366Mc), then $T\bar h=\bar hT$ for every Borel measurable
$h:\Bbb C\to\Bbb C$.
%366M 366Yk 364G 364Pd

\spheader 366Ym\dvAformerly{3{}63Yi}
Show that a normed space over $\Bbb C$ has the
Hahn-Banach property of 363R for complex spaces iff it is isomorphic to
$L^{\infty}_{\Bbb C}(\frak A)$ for some Dedekind
complete Boolean algebra $\frak A$.
%363R
}%end of exercises


\cmmnt{
\Notesheader{366}
The $L^p$ spaces, for $1\le p\le\infty$, constitute the most important
family of leading examples for the theory of Banach lattices, and it is
not to be wondered at that their properties reflect a wide variety of
general results.   Thus 366Dd and 366E can both be regarded as special
cases of theorems about perfect Riesz spaces (356M and 369D).   In
a different direction, the concept of `Orlicz space' (369Xd below)
generalizes the $L^p$ spaces if they are regarded as normed subspaces of
$L^0$ invariant
under measure-preserving automorphisms of the underlying algebra.   Yet
another generalization looks at the (non-locally-convex) spaces $L^p$
for $0<p<1$ (366Ya).

In 366H and its associated results I try to emphasize the way in which
measure-preserving homomorphisms of the underlying algebras induce both
`direct' and `dual' operators on $L^p$ spaces.   We have already
seen the phenomenon in 365N-365O.   I express this in a slightly different
form in 366H, noting that we really do need the
homomorphisms to be measure-preserving, for the dual operators as well
as the direct operators, so we no longer have the shift in the
hypotheses which
appears between 365N and 365O.    Of course all these refinements in the
hypotheses are irrelevant to the principal applications of the results,
and they make substantial demands on the reader;  but I believe that the
demands are actually demands to expand one's imagination, to encompass
the different
ways in which the spaces depend on the underlying measure algebras.

In the context of 366H, $L^{\infty}$ is set apart from the other $L^p$
spaces, because $L^{\infty}(\frak A)$ is not in general determined by
the ideal $\frak A^f$, and the hypotheses of 366H do not look outside
$\frak A^f$.   366H(a-iv) and 366H(b-iii) reach only the space
$M^{\infty,0}$ as defined in 366Xd.    To deal with $L^{\infty}$ we need
slightly stronger hypotheses.   If we are given a measure-preserving
Boolean homomorphism from $\frak A$ to $\frak B$, rather than from
$\frak A^f$ to
$\frak B^f$, then of course the direct operator $T$ has a natural
version acting on $L^{\infty}(\frak A)$ and indeed on
$M^{1,\infty}_{\bar\mu}$, as in 363F and 369Xi.    If we know
that $(\frak A,\bar\mu)$ is
localizable, then $\frak A$ can be recovered from $\frak A^f$, and the
dual operator $P$ acts on $L^{\infty}(\frak B)$, as in 369Xi.   But in
general we can't expect this to work (366Xf).

Of course 366H can be applied to many other spaces;  for reasons which
will appear in \S\S371 and 374, the archetypes are not really $L^p$
spaces at all, but the spaces $M^{1,0}$ (366F) and  $M^{1,\infty}$.

I include 366L and 366Yh as pointers to one of the important
applications of these
ideas:  the investigation of properties of a measure-preserving
homomorphism in terms of its action on $L^p$ spaces.   The case $p=2$ is
the most useful because the group of unitary operators (that is, the
normed space automorphisms) of $L^2$ has been studied intensively.

}%end of comment

\discrpage

\frfilename{mt367.tex}
\versiondate{2.5.16/7.9.18}
\copyrightdate{1998}
\long\def\varinset#1{{\narrower{#1}}}
\def\starto{\to^*}

\def\chaptername{Function spaces}
\def\sectionname{Convergence in measure}

\newsection{367}

Continuing through the ideas of Chapter 24, I come to `convergence in
measure'.   The basic results of \S245
all translate easily into the new language (367L-367M, 367P).   The
associated concept of (sequential) order-convergence can also be
expressed in abstract terms (367A), and I take the trouble to do this in
the context of general lattices (367A-367B), since the
concept can be applied in many ways (367C-367E,
367K\cmmnt{, 367Xa-367Xp}).   In
the particular case of $L^0$ spaces, which are the first aim of this
section, the idea is most naturally expressed by 367F.   It enables us
to express some of the basic theorems in Volumes 1 and 2 in the
language of this chapter (367I-367J).

In 367N and 367O I give two of the most characteristic properties of the
topology of convergence in measure on $L^0$;  it is one of the
fundamental types of topological Riesz space.   Another striking fact is
the way it is determined by the Riesz space structure (367T).   In 367U
I set out a theorem which is the basis of many remarkable applications
of the concept;  for the sake of a result in \S369 I give one such
application (367V).

\leader{367A}{\dvrocolon{Order*-convergence}}\cmmnt{ As I have
remarked before, the function spaces of measure theory have three
interdependent structures:  they are linear spaces, they have a variety
of interesting topologies, and they are ordered spaces.   Ordinary
elementary functional analysis studies interactions between topologies
and linear structures, in the theory of normed spaces and, more
generally, of linear topological spaces.   Chapter 35 in this volume
looked at interactions between linear and order structures.   It is
natural to seek to complete the triangle with a theory of topological
ordered spaces.   The relative obscurity of any such theory is in part
due to the difficulty of finding convincing definitions;  that is,
isolating concepts which lead to elegant and useful general
theorems.   Among the many rival ideas, however, I believe it is
possible to identify one which is particularly important in the context
of measure theory.

In its natural home in the theory of $L^0$ spaces, this notion of
`order*-convergence' has a particularly straightforward expression
(367F).   But, suitably interpreted, the same idea can be applied
in other contexts, some of which will be very useful to us, and I
therefore begin with a definition which is applicable in any lattice.

\medskip

\noindent}{\bf Definition} Let $P$ be a lattice, $p$ an element of $P$ and $\sequencen{p_n}$ a sequence in $P$.   I will say that
$\sequencen{p_n}$ {\bf order*-converges} to $p$ if

$$\eqalign{p
&=\inf\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee p_i)\wedge p''
\Forall i\ge n\}\cr
&=\sup\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(p_i\wedge p'')
\Forall i\ge n\}\cr}$$

\noindent whenever $p'\le p\le p''$ in $P$.

\cmmnt{\medskip

\noindent{\bf Remark} In the formulae above, we always have
$p'\vee(p_i\wedge p'')\le(p'\vee p_i)\wedge p''$, because $p'\le p''$.
If $P$ is a distributive lattice, both are equal to $\med(p',p_i,p'')$.
}

\vleader{108pt}{367B}{Lemma} Let $P$ be a lattice.

(a) A sequence in $P$ can order*-converge to at most one point.

(b) A constant sequence order*-converges to its constant value.

(c) Any subsequence of an order*-convergent sequence is
order*-convergent, with the same limit.

(d) If $\sequencen{p_n}$ and $\sequencen{p'_n}$ both order*-converge
to $p$, and $p_n\le q_n\le p'_n$ for every $n$, then $\sequencen{q_n}$
order*-converges to $p$.

(e) If $\sequencen{p_n}$ is an order-bounded sequence in $P$, then it
order*-converges to $p\in P$ iff

$$\eqalign{p
&=\inf\{q:\,\exists\,n\in\Bbb N,\,q\ge p_i\Forall i\ge n\}\cr
&=\sup\{q:\,\exists\,n\in\Bbb N,\,q\le p_i\Forall i\ge n\}.\cr}$$

(f) If $P$ is a Dedekind $\sigma$-complete lattice\cmmnt{ (314Ab)} and
$\sequencen{p_n}$ is an order-bounded sequence in $P$, then it
order*-converges to $p\in P$ iff

\Centerline{$p=\sup_{n\in\Bbb N}\inf_{i\ge n}p_i
=\inf_{n\in\Bbb N}\sup_{i\ge n}p_i$.}

\proof{{\bf (a)} Suppose that $\sequencen{p_n}$ is order*-convergent
to both $p$ and $\tilde p$.   Set $p'=p\wedge\tilde p$, $p''=p\vee\tilde
p$;  then

\Centerline{$p
=\inf\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee p_i)\wedge
p''\Forall i\ge n\}
=\tilde p$.}

\medskip

{\bf (b)} is trivial.

\medskip

{\bf (c)} Suppose that $\sequencen{p_n}$ is order*-convergent to $p$,
and that $\sequencen{p'_n}$ is a subsequence of $\sequencen{p_n}$.
Take $p'$, $p''$ such that $p'\le p\le p''$, and set

\Centerline{$B=\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(p_i\wedge
p'')\Forall i\ge n\}$,}

\Centerline{$B'=\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(p'_i\wedge
p'')\Forall i\ge n\}$,}

\Centerline{$C=\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee p_i)\wedge
p''\Forall i\ge n\}$,}

\Centerline{$C'=\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee p'_i)\wedge
p''\Forall i\ge n\}$.}

\noindent If $q\in B'$ and $q'\in C$, then for all sufficiently large
$i$

\Centerline{$q\le p'\vee(p'_i\wedge p'')\le(p'\vee p'_i)\wedge p''\le
q'$.}

\noindent As $p=\inf C$, we must have $q\le p$;  thus $p$ is an upper
bound for $B'$.   On the other hand,
$\{p'_i:i\ge n\}\subseteq\{p_i:i\ge n\}$ for every $n$, so $B\subseteq
B'$ and $p$ must be the least upper bound of $B'$, since $p=\sup B$.

Similarly, $p=\inf C'$.   As $p'$ and $p''$ are arbitrary,
$\sequencen{p'_n}$ order*-converges to $p$.

\medskip

{\bf (d)} Take $p'$, $p''$ such that $p'\le p\le p''$, and set

\Centerline{$B=\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(p_i\wedge
p'')\Forall i\ge n\}$,}

\Centerline{$B'=\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(q_i\wedge
p'')\Forall i\ge n\}$,}

\Centerline{$C=\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee p'_i)\wedge
p''\Forall i\ge n\}$,}

\Centerline{$C'=\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee q_i)\wedge
p''\Forall i\ge n\}$.}

\noindent If $q\in B'$ and $q'\in C$, then for all sufficiently large
$i$

\Centerline{$q\le p'\vee(q_i\wedge p'')\le(p'\vee p'_i)\wedge p''\le
q'$.}

\noindent As $p=\inf C$, we must have $q\le p$;  thus $p$ is an upper
bound for $B'$.   On the other hand,
$p'\vee(p_i\wedge p'')\le p'\vee(q_i\wedge p'')$ for every $i$, so
$B\subseteq B'$ and $p=\sup B'$.
Similarly, $p=\inf C'$.   As $p'$ and $p''$ are arbitrary,
$\sequencen{q_n}$ order*-converges to $p$.

\medskip

{\bf (e)} Set

\Centerline{$B=\{q:\,\exists\,n\in\Bbb N,\,q\le p_i\Forall i\ge
n\}$,}

\Centerline{$C=\{q:\,\exists\,n\in\Bbb N,\,q\ge p_i\Forall i\ge
n\}$.}

\medskip

\quad{\bf (i)} Suppose that $\sequencen{p_n}$ order*-converges to
$p$.   Let $p'$, $p''$ be such that $p'\le p_n\le p''$ for every
$n\in\Bbb N$ and $p'\le p\le p''$.   Then

\Centerline{$B
=\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(p_i\wedge p'')\Forall i\ge
n\}$,}

\noindent so $\sup B=p$.   Similarly, $\inf C=p$, so the condition is
satisfied.

\medskip

\quad{\bf (ii)} Suppose that $\sup B=\inf C=p$.   Take any $p'$, $p''$
such that $p'\le p\le p''$ and set

\Centerline{$B'=\{q:\,\exists\,n\in\Bbb N,\,q\le p'\vee(p_i\wedge
p'')\Forall i\ge n\}$,}

\Centerline{$C'=\{q:\,\exists\,n\in\Bbb N,\,q\ge(p'\vee p_i)\wedge
p''\Forall i\ge n\}$.}

\noindent If $q\in B'$ and $q'\in C$, then for all large enough $i$

\Centerline{$q\le p'\vee(p_i\wedge p'')\le p'\vee q'=q'$}

\noindent because $p\le q'$.   As $\inf C=p$, $p$ is an upper bound for
$B'$.   On the other hand,
if $q\in B$, then $q\le p$, so $q\le p'\vee(p_i\wedge p'')$ whenever
$q\le p_i$, which is so for all sufficiently large $i$, and $q\in B'$.
Thus $B'\supseteq B$ and $p$ must be the supremum of $B'$.   Similarly,
$p=\inf C'$;  as $p'$ and $p''$ are arbitrary, $\sequencen{p_n}$
order*-converges to $p$.

\medskip

{\bf (f)} This follows at once from (e).   Setting

\Centerline{$B
=\{q:\,\exists\,n\in\Bbb N,\,q\le p_i\Forall i\ge n\}$,
\quad$B'=\{\inf_{i\ge n}p_i:i\in\Bbb N\}$,}

\noindent then $B'\subseteq B$ and for every $q\in B$ there is a $q'\in
B'$ such that $q\le q'$;  so $\sup B=\sup B'$ if either is defined.
Similarly,

\Centerline{$\inf\{q:\,\exists\,n\in\Bbb N,\,q\ge p_i\Forall i\ge n\}
=\inf_{n\in\Bbb N}\sup_{i\ge n}p_i$}

\noindent if either is defined.
}%end of proof of 367B

\cmmnt{\medskip

\noindent{\bf Remark} Part (b) above tells us that we may speak of `the'
order*-limit of an order*-convergent sequence.
}

\leader{367C}{Proposition}\dvArevised{2016} Let $U$ be a Riesz space.

(a) Suppose that
$\sequencen{u_n}$, $\sequencen{v_n}$ are two sequences in $U$
order*-converging to $u$, $v$ respectively.

\quad(i) $\sequencen{u_n+w}$
order*-converges to $u+w$ for every $w\in U$,
and $\alpha u_n$ order*-converges to
$\alpha u$ for every $\alpha\in\Bbb R$.

\quad(ii) $\sequencen{u_n\vee v_n}$ order*-converges to $u\vee v$
and $\sequencen{u_n\wedge v_n}$ order*-converges to $u\wedge v$.

\quad(iii) If $\sequencen{w_n}$ is any sequence in $U$, then it
order*-converges to $w\in U$ iff $\sequencen{|w_n-w|}$
order*-converges to $0$.

\quad(iv) $\sequencen{u_n+v_n}$ order*-converges to $u+v$.

\quad(v)\dvAnew{2020} If $\sequencen{w_n}$ and
$\sequencen{z_n}$ are sequences in $U$, $\sequencen{w_n}$ order*-converges 
to $0$ and $|z_n|\le|w_n|$ for every $n$, then $\sequencen{z_n}$
order*-converges to $0$.

(b) Now suppose that $U$ is Archimedean.

\quad(i) If $\sequencen{\alpha_n}$ is a sequence in
$\Bbb R$ converging to
$\alpha\in\Bbb R$, and $\sequencen{u_n}$ is a sequence in $U$
order*-converging to $u\in U$, then $\sequencen{\alpha_nu_n}$
order*-converges to $\alpha u$.

\quad(ii) A sequence
$\sequencen{w_n}$ in $U^+$ is {\it not} order*-convergent to $0$ iff
there is a $\tilde w>0$ such that
$\tilde w=\sup_{i\ge n}\tilde w\wedge w_i$ for every $n\in\Bbb N$.

\quad(iii) If $\sequencen{u_n}$ is a sequence in $U^+$ such that
$\{\sum_{i=0}^nu_i:n\in\Bbb N\}$ is bounded above, then
$\sequencen{u_n}$ order*-converges to $0$.

\proof{{\bf (a)(i)}\grheada\
$\sequencen{u_n+w}$ order*-converges to $u+w$
because the ordering of $U$ is translation-invariant;  the map
$w'\mapsto w'+w$ is an order-isomorphism.

\medskip

\qquad\grheadb\ If $\alpha>0$, then the map
$w'\mapsto\alpha w'$ is an order-isomorphism, so
$\sequencen{\alpha u_n}$
order*-converges to $\alpha u$.

\medskip

\qquad\grheadc\ If $\alpha=0$ then $\sequencen{\alpha u_n}$
order*-converges to $\alpha u=0$ by 367Bb.

\medskip

\qquad\grheadd\ If $w'\le -u\le w''$ then $-w''\le u\le -w'$ so

$$\eqalign{u
&=\inf\{w:\,\exists\,n\in\Bbb N,\,
w\ge((-w'')\vee u_i)\wedge (-w')\Forall i\ge n\}\cr
&=\sup\{w:\,\exists\,n\in\Bbb N,\,
w\le (-w'')\vee(u_i\wedge (-w'))\Forall i\ge n\}.\cr}$$

\noindent Turning these formulae upside down,

$$\eqalign{-u
&=\sup\{w:\,\exists\,n\in\Bbb N,\,
w\le(w''\wedge(-u_i))\vee w'\Forall i\ge n\}\cr
&=\inf\{w:\,\exists\,n\in\Bbb N,\,
w\ge w''\wedge((-u_i)\vee w')\Forall i\ge n\}.\cr}$$

\noindent As $w'$ and $w''$ are arbitrary, $\sequencen{-u_n}$
order*-converges to $-u$.

\medskip

\qquad\grheade\ Putting ($\beta$) and ($\delta$) together,
$\sequencen{\alpha u_n}$ order*-converges to $\alpha u$ for every
$\alpha<0$.

\medskip

\quad{\bf (ii)} Suppose that $w'\le u\vee v\le w''$.   Set

\Centerline{$B=\{w:\,\exists\,n\in\Bbb N,\,w\le w'\vee((u_i\vee
v_i)\wedge w'')\Forall i\ge n\}$,}

\Centerline{$C=\{w:\,\exists\,n\in\Bbb N,\,w\ge(w'\vee(u_i\vee
v_i))\wedge w''\Forall i\ge n\}$,}

\Centerline{$B_1=\{w:\,\exists\,n\in\Bbb N,\,w\le (w'\wedge
u)\vee(u_i\wedge w'')\Forall i\ge n\}$,}

\Centerline{$B_2=\{w:\,\exists\,n\in\Bbb N,\,w\le (w'\wedge
v)\vee(v_i\wedge w'')\Forall i\ge n\}$,}

\Centerline{$C_1=\{w:\,\exists\,n\in\Bbb N,\,w\ge((w'\wedge u)\vee
u_i)\wedge w''\Forall i\ge n\}$,}

\Centerline{$C_2=\{w:\,\exists\,n\in\Bbb N,\,w\ge((w'\wedge v)\vee
v_i)\wedge w''\Forall i\ge n\}$,}

If $w_1\in B_1$ and $w_2\in B_2$ then $w_1\vee w_2\in B$.   \Prf\ There
is an $n\in\Bbb N$ such that $w_1\le (w'\wedge u)\vee(u_i\wedge w'')$
for every $i\ge n$, while $w_2\le (w'\wedge v)\vee(v_i\wedge w'')$ for
every $i\ge n$.   So

$$\eqalignno{w_1\vee w_2
&\le(w'\wedge u)\vee(w'\wedge v)\vee(u_i\wedge w'')
  \vee(v_i\wedge w'')\cr
&=(w'\wedge(u\vee v))\vee((u_i\vee v_i)\wedge w'')\cr
\noalign{\noindent (352Ec)}
&=w'\vee((u_i\vee v_i)\wedge w'')\cr}$$

\noindent for every $i\ge n$, and $w_1\vee w_2\in B$.\ \Qed

Similarly, if $w_1\in C_1$ and $w_2\in C_2$ then $w_1\vee w_2\in C$.
\Prf\ There is an $n\in\Bbb N$ such that
$w_1\ge((w'\wedge u)\vee u_i)\wedge w''$ and
$w_2\ge((w'\wedge v)\vee v_i)\wedge w''$ for every $i\ge n$.   So

$$\eqalign{w_1\vee w_2
&\ge(((w'\wedge u)\vee u_i)\wedge w'')
  \vee(((w'\wedge v)\vee v_i)\wedge w'')\cr
&=((w'\wedge u)\vee u_i\vee(w'\wedge v)\vee v_i)\wedge w''\cr
&=((w'\wedge(u\vee v))\vee(u_i\vee v_i))\wedge w''\cr
&=(w'\vee(u_i\vee v_i))\wedge w''\cr}$$

\noindent for every $i\ge n$, so $w_1\vee w_2\in C$.\ \Qed

At the same time, of course, $w\le\tilde w$ whenever $w\in B$ and
$\tilde w\in C$, since there is some $i\in\Bbb N$ such that

\Centerline{$w\le w'\vee((u_i\vee v_i)\wedge w'')
\le(w'\vee(u_i\vee v_i))\wedge w''
\le\tilde w$.}

\noindent Since

\Centerline{$\sup\{w_1\vee w_2:w_1\in B_1,\,w_2\in B_2\}
=(\sup B_1)\vee(\sup B_2)
=u\vee v$,}

\Centerline{$\inf\{w_1\vee w_2:w_1\in C_1,\,w_2\in C_2\}
=(\inf C_1)\vee(\inf C_2)
=u\vee v$}

\noindent (using the generalized distributive laws in 352E), we must
have $\sup B=\inf C=u\vee v$.   As $w'$ and $w''$ are arbitrary,
$\sequencen{u_n\vee v_n}$ is order*-convergent to $u\vee v$.

Putting this together with (i), we see that
$\sequencen{u_n\wedge v_n}=\sequencen{-((-u_n)\vee(-v_n))}$
order*-converges to $-((-u)\vee(-v))=u\wedge v$.

\medskip

\quad{\bf (iii)} The hard parts are over.   ($\alpha$) If
$\sequencen{w_n}$
order*-converges to $w$, then $\sequencen{w_n-w}$,
$\sequencen{w-w_n}$ and
$\sequencen{|w_n-w|}
=\sequencen{(w_n-w)\vee(w-w_n)}$ all order*-converge to $0$, putting
(i) and (ii) together.   ($\beta$) If $\sequencen{|w_n-w|}$
order*-converges to $0$, then so do $\sequencen{-|w_n-w|}$ and
$\sequencen{w_n-w}$, by (i) and 367Bd;  so $\sequencen{w_n}$
order*-converges to $0$, by (i) again.

\medskip

\quad{\bf (iv)} $\sequencen{|u_n-u|}$ and $\sequencen{|v_n-v|}$
order*-converge to $0$, by (iii), so
$\sequencen{2(|u_n-u|\vee|v_n-v|)}$ also order*-converges to $0$, by
(ii) and (i).   But

\Centerline{$0\le|(u_n+v_n)-(u+v)|\le|u_n-u|+|v_n-v|
\le 2(|u_n-u|\vee|v_n-v|)$}

\noindent for every $n$, so $\sequencen{|(u_n+v_n)-(u+v)|}$
order*-converges to $0$, by 367Bb and 367Bd, and
$\sequencen{u_n+v_n}$ order*-converges to $u+v$.

\medskip

\quad{\bf (v)} By (iii), $\sequencen{|w_n|}$ order*-converges to $0$.
So $\inf_{n\in\Bbb N}\sup_{m\ge n}|w_m|$ is defined and equal to $0$;
consequently $\inf_{n\in\Bbb N}\sup_{m\ge n}|z_m|=0$,
$\sequencen{|z_n|}$ order*-converges to $0$ and
$\sequencen{z_n}$ order*-converges to $0$, by (iii) again.

\medskip

{\bf (b)(i)} Set $\beta_n=\sup_{i\ge n}|\alpha_i-\alpha|$ for each $n$.
Then $\sequencen{\beta_n}\to 0$, so $\inf_{n\in\Bbb N}\beta_n|u|=0$,
because $U$ is Archimedean.   Consequently $\sequencen{\beta_n|u|}$
order*-converges to $0$, by 367Be.   But we also have
$\beta_0|u_n-u|$ order*-converging to $0$, by (a-iii) and (a-i), so
$\sequencen{\beta_0|u_n-u|+\beta_n|u|}$ order*-converges to $0$, by
(a-iv).   As $|\alpha_nu_n-\alpha u|\le\beta_0|u_n-u|+\beta_n|u|$
for every
$n$, $\sequencen{\alpha_nu_n}$ order*-converges to $\alpha u$, as
required.

\medskip

\quad{\bf (ii)}\grheada\
Suppose that $\sequencen{w_n}$ is not order*-convergent
to $0$.   Then there are $w'$, $w''$ such that $w'\le 0\le w''$ and
either

\Centerline{$B=\{w:\,\exists\,n\in\Bbb N,\,w\le w'\vee(w_i\wedge
w'')\Forall i\ge n\}$}

\noindent does not have supremum $0$, or

\Centerline{$C=\{w:\,\exists\,n\in\Bbb N,\,w\ge(w'\vee w_i)\wedge
w''\Forall i\ge n\}$}

\noindent does not have infimum $0$.   Now $0\in B$, because every
$w_i\ge 0$, and every member of $B$ is a lower bound for $C$;  so $0$
cannot be the greatest lower bound of $C$.   Let $\tilde w>0$ be a lower
bound for $C$.

Let $n\in\Bbb N$, and set

\Centerline{$C_n=\{w:w\ge(w'\vee w_i)\wedge w''\Forall i\ge n\}
=\{w:w\ge w_i\wedge w''\Forall i\ge n\}$.}

\noindent (Recall that $U$ is a distributive lattice.)
Because $U$ is Archimedean, we know that $\inf(C_n-A_n)=0$,
where $A_n=\{w_i\wedge w'':i\ge n\}$ (353F).   Now $\tilde w$ is a lower
bound for $C_n$, so


$$\eqalign{\inf_{i\ge n}(\tilde w-w_i)^+
&\le\inf\{(w-w_i)^+:w\in C,\,i\ge n\}\cr
&\le\inf\{(w-(w_i\wedge w''))^+:w\in C,\,i\ge n\}\cr
&=\inf\{w-(w_i\wedge w''):w\in C,\,i\ge n\}
=\inf(C_n-A_n)
=0.\cr}$$

\noindent As this is true for every $n\in\Bbb N$, $\tilde w$ has the
property declared.

\medskip

\qquad\grheadb\ If $\tilde w>0$ is such that
$\tilde w=\sup_{i\ge n}\tilde w\wedge w_i$ for every $n\in\Bbb N$, then

\Centerline{$\{w:\,\exists\,n\in\Bbb N,\,w\ge(0\vee w_i)\wedge\tilde
w\Forall i\ge n\}$}

\noindent cannot have infimum $0$, and $\sequencen{w_n}$ is not
order*-convergent to $0$.

\medskip

\quad{\bf (iii)} Set $v_n=\sum_{i=0}^nu_i$ for $n\in\Bbb N$.
Let $C$ be the set of upper bounds of
$\{v_n:n\in\Bbb N\}$, and write $B$ for
$\{w-v_n:w\in C$, $n\in\Bbb N\}$.   Then $\inf B=0$
(353F).   But if $n\in\Bbb N$ and $w\in C$ then
$u_i=v_{i+1}-v_i\le w-v_n$ for every $i\ge n$.   So

\Centerline{$\{u:\,\exists\,n\in\Bbb N,\,u\ge u_i\Forall i\ge n\}$}

\noindent includes $B$ and must have infimum $0$.   On the other side,

\Centerline{$\{u:\,\exists\,n\in\Bbb N,\,u\le u_i\Forall i\ge n\}$}

\noindent contains $0$ and must have supremum $0$.   By 367Be,
$\sequencen{u_n}$ order*-converges to $0$.
}%end of proof of 367C

\leader{367D}{}\cmmnt{ As an example of the use of this concept in a
moderately general setting, I offer the following.

\medskip

\noindent}{\bf Proposition}\dvArevised{2016}
Let $U$ be a Riesz space with a Riesz norm $\|\,\|$.

(a) If a sequence in $U$ is both order*-convergent and
norm-convergent, the two limits are the same.

(b) $\|\,\|$ is order-continuous iff every order-bounded
order*-convergent sequence in $U$ is norm-convergent.

\leaveitout{(b) Let $U$ be a Banach lattice and
$\sequencen{u_n}$ a sequence in $U$ which is norm-convergent to
$u\in U$.   Then $\sequencen{u_n}$ has a subsequence which is
order-bounded
and order*-convergent to $u$.   So if $\sequencen{u_n}$ itself is
order*-convergent, its order*-limit is $u$.
}

\proof{{\bf (a)} Let $\sequencen{u_n}$ be a sequence in $U$ which is
order*-convergent to $v$ and norm-convergent to $w$.   Then
$\sequencen{\med(u_n,v,w)}
=\sequencen{((v\wedge w)\vee u_n)\wedge(v\vee w)}$ is
order*-convergent to $\med(v,v,w)=v$ (367C(a-ii)) and norm-convergent to
$\med(w,v,w)=w$ (354Xc, or 354B with 352D).   So

$$\eqalign{v
&=\inf\{u:\,\exists\,n\in\Bbb N,\,u\ge\med(u_i,v,w)
\Forall i\ge n\}\cr
&=\sup\{u:\,\exists\,n\in\Bbb N,\,u\le\med(u_i,v,w)
\Forall i\ge n\}.\cr}$$

\noindent But if $n\in\Bbb N$ and $u\ge\med(u_i,v,w)$ for every
$i\ge n$, then $u\ge\lim_{i\to\infty}\med(u_i,v,w)=w$, because
$\{u':u'\le u\}$ is norm-closed (354Bc).   As $u$ is arbitrary,
$w\le v$.   Similarly, because $\{u':u'\ge u\}$ is norm-closed for
every $u$, $w\ge v$.   So $w=v$, as claimed.

\medskip

{\bf (b)(i)} Suppose that $\|\,\|$ is order-continuous, and that
an order-bounded sequence $\sequencen{u_n}$ order*-converges to $u$.
Then $\sequencen{|u_n-u|}$ is order-bounded and
order*-convergent to $0$ (367C(a-iii)), so

\Centerline{$C
=\{v:\,\exists\,n\in\Bbb N,\,v\ge|u_i-u|\Forall i\ge n\}$}

\noindent has infimum $0$ (367Be).   Because $U$ is a lattice, $C$ is
downwards-directed, so $\inf_{v\in C}\|v\|=0$.   But

\Centerline{$\inf_{v\in C}\|v\|
\ge\inf_{n\in\Bbb N}\sup_{i\ge n}\|u_i-u\|$,}

\noindent so $\lim_{n\to\infty}\|u_n-u\|=0$, that is, $\sequencen{u_n}$
is norm-convergent to $u$.

\medskip

\quad{\bf (ii)} Suppose that all order-bounded order*-convergent
sequences in $U$ are norm-convergent,

\medskip

\qquad\grheada\ Let $\sequencen{u_n}$ be a non-increasing
sequence in $U^+$, and set $v_n=u_n-u_{n+1}$ for each $n$.   Then
$\sequencen{v_n}$ is norm-convergent to $0$.   \Prf\ By 367C(b-iii),
applied to $\sequencen{-v_n}$, $\sequencen{v_n}$ is order*-convergent
to $0$, so must be norm-convergent, and (a) here tells us that the
norm limit is $0$.\ \Qed

\medskip

\qquad\grheadb\ Now suppose that $A\subseteq U^+$ is a non-empty
downwards-directed set with infimum $0$.   Choose $\sequencen{u_n}$,
$\sequencen{\gamma_n}$ inductively, as follows.   Start with any
$u_0\in A$.   Given $u_n\in A$, set
$\gamma_n=\sup_{u\in A\cap[0,u_n]}\|u_n-u\|$ and choose
$u_{n+1}\in A\cap[0,u_n]$ such that
$\|u_n-u_{n+1}\|\ge\bover12\gamma_n$;  continue.

By ($\alpha$), $\lim_{n\to\infty}\|u_n-u_{n+1}\|=0$, so
$\lim_{n\to\infty}\gamma_n=0$.   Now suppose that $v\in U$ is any
lower bound of $\{u_n:n\in\Bbb N\}$.   Then $v\le 0$.
\Prf\ Take $u\in A$ and $n\in\Bbb N$.   Then there is a
$u'\in A$ such that $u'\le u\wedge u_n$, because $A$ is
downwards-directed.   So

$$\eqalignno{\|v-u\wedge v\|
&\le\|v-u'\wedge v\|
=\|u'\vee v-u'\|
\Displaycause{because $u'\vee v+u'\wedge v=u'+v$, as noted in 352D}
\le\|u_n-u'\|
\le\gamma_n.\cr}$$

\noindent As $n$ is arbitrary, $v-u\wedge v=0$ and $v\le u$.   As $u$
is arbitrary, $v$ is a lower bound of $A$ and must be less than or
equal to $\inf A=0$.\ \Qed

Thus $\sequencen{u_n}$ is a non-increasing sequence with infimum $0$,
and order*-converges to $0$, by 367Be.   Accordingly it
norm-converges to $0$, and
$\inf_{u\in A}\|u\|=\inf_{n\in\Bbb N}\|u_n\|=0$.   As $A$ is
arbitrary, $\|\,\|$ is order-continuous,

\leaveitout{(c) Let $\sequencen{u'_n}$ be a subsequence of
$\sequencen{u_n}$
such that $\|u'_n-u\|\le 2^{-n}$ for each $n\in\Bbb N$.   Then
$v_n=\sup_{i\ge n}|u'_i-u|$ is defined in $U$, and $\|v_n\|\le 2^{-n+1}$,
for each $n$ (354C).   Because $\inf_{n\in\Bbb N}\|v_n\|=0$,
$\inf_{n\in\Bbb N}v_n$ must be $0$, while $u-v_n\le u'_i\le u+v_n$
whenever $i\ge n$;  so $\sequencen{u'_n}$ order*-converges to $u$, by
367Be.

Now if $\sequencen{u_n}$ has an order*-limit, this must be $u$, by
367Ba and 367Bc.}

As $A$ is arbitrary, the norm of $U$ is order-continuous.
}%end of proof of 367D

\vleader{48pt}{367E}{}\cmmnt{ One of the fundamental obstacles to the
development of any satisfying general theory of ordered topological
spaces is the erratic nature of the relations between subspace
topologies of order topologies and order topologies on subspaces.   The
particular virtue of
order*-convergence in the context of function spaces is that it is
relatively robust when transferred to the subspaces we are interested
in.

\medskip

\noindent}{\bf Proposition} Let $U$ be an Archimedean Riesz space and
$V$ a regularly embedded Riesz subspace.   \cmmnt{(For instance, $V$
might be either solid or order-dense.)}   If $\sequencen{v_n}$ is a
sequence in $V$ and $v\in V$, then $\sequencen{v_n}$ order*-converges
to $v$ when regarded as a sequence in $V$, iff it order*-converges to
$v$ when regarded as a sequence in $U$.

\proof{{\bf (a)} Since, in either $V$ or $U$, $\sequencen{v_n}$
order*-converges to $v$ iff $\sequencen{|v_n-v|}$ order*-converges
to $0$ (367C(a-iii)), it is enough to consider the case $v_n\ge 0$,
$v=0$.

\medskip

{\bf (b)} If $\sequencen{v_n}$ is {\it not} order*-convergent to $0$
in $U$, then, by 367C(b-ii), there is a $u>0$ in $U$ such that
$u=\sup_{i\ge n}u\wedge v_i$ for every $n\in\Bbb N$
(the supremum being taken in $U$, of course).
In particular, there is a $k\in\Bbb N$ such that $u\wedge v_k>0$.
Now consider the set

\Centerline{$C=\{w:w\in V,\,\exists\,n\in\Bbb N,\,
w\ge v_i\wedge v_k\Forall i\ge n\}$.}

\noindent Then for any $w\in C$,

\Centerline{$u\wedge v_k=\sup_{i\ge n}u\wedge v_i\wedge v_k\le w$,}

\noindent using the generalized distributive law in $U$, so $0$ is not
the greatest lower bound of $C$ in $U$.   But as the embedding of $V$ in
$U$ is order-continuous, $0$ is not the greatest lower bound of $C$ in
$V$, and $\sequencen{v_n}$ cannot be
order*-convergent to $0$ in $V$.

\medskip

{\bf (c)} Now suppose that $\sequencen{v_n}$ is not order*-convergent
to $0$ in $V$.   Because $V$, like $U$, is Archimedean (351Rc), there is a
$w>0$ in $V$ such that $w=\sup_{i\ge n}w\wedge v_i$ for every
$n\in\Bbb N$, the suprema being taken in $V$.   Again because $V$ is regularly
embedded in $U$, we have the same suprema in $U$, so, by 367C(b-ii)
in the
other direction, $\sequencen{v_n}$ is not order*-convergent to $0$ in
$U$.
}%end of proof of 367E

\leader{367F}{}\cmmnt{ I now spell out the connexion between the
definition above and the concepts introduced in 245C.

\medskip

\noindent}{\bf Proposition} Let $X$ be a set, $\Sigma$ a
$\sigma$-algebra of subsets of $X$, $\frak A$ a Boolean algebra and
$\pi:\Sigma\to\frak A$ a sequentially order-continuous surjective Boolean
homomorphism;  let
$\Cal I$ be its kernel.   Write $\eusm L^0$ for the space of
$\Sigma$-measurable functions from $X$ to $\Bbb R$, and let
$T=T_{\pi}:\eusm L^0\to L^0=L^0(\frak A)$ be
the canonical Riesz homomorphism\cmmnt{ (364C, 364P)}.   Then for any
$\sequencen{f_n}$ and $f$ in $\eusm L^0$, $\sequencen{Tf_n}$
order*-converges to $Tf$ in $L^0$ iff
$X\setminus\{x:f(x)=\lim_{n\to\infty}f_n(x)\}\in\Cal I$.

\proof{ Set $H=\{x:\lim_{n\to\infty}f_n(x)$ exists $=f(x)\}$;  of course
$H\in\Sigma$.
Set $g_n(x)=|f_n(x)-f(x)|$ for $n\in\Bbb N$ and $x\in X$.

\medskip

{\bf (a)} If $X\setminus H\in\Cal I$, set
$h_n(x)=\sup_{i\ge n}g_i(x)$ for $x\in H$ and $h_n(x)=0$ for $x\in
X\setminus H$.   Then $\sequencen{h_n}$ is a non-increasing sequence
with infimum $0$ in $\eusm L^0$, so $\inf_{n\in\Bbb N}Th_n=0$ in $L^0$,
because $T$ is sequentially order-continuous (364Pa).   But as
$X\setminus H\in\Cal I$, $Th_n\ge Tg_i=|Tf_i-Tf|$ whenever $i\ge n$, so
$\sequencen{|Tf_n-Tf|}$ order*-converges to $0$, by 367Be or 367Bf,
and $\sequencen{Tf_n}$ order*-converges to $Tf$, by 367C(a-iii).

\medskip

{\bf (b)} Now suppose that $\sequencen{Tf_n}$ order*-converges to
$Tf$.   Set $g'_n(x)=\min(1,g_n(x))$ for $n\in\Bbb N$,
$x\in X$;  then $\sequencen{Tg'_n}=\sequencen{e\wedge|Tf_n-Tf|}$
order*-converges to $0$, where $e=T(\chi X)$.
By 367Bf, $\inf_{n\in\Bbb N}\sup_{i\ge n}Tg'_i=0$ in $L^0$.   But $T$ is
a sequentially order-continuous Riesz homomorphism, so
$T(\inf_{n\in\Bbb N}\sup_{i\ge n}g'_i)=0$, that is,

\Centerline{$X\setminus H=\{x:\inf_{n\in\Bbb N}\sup_{i\ge n}g'_i>0\}$}

\noindent belongs to $\Cal I$.
}%end of proof of 367F

\leader{367G}{Corollary}
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra.

(a) Any order*-convergent sequence in $L^0=L^0(\frak A)$ is
order-bounded.

(b) If $\sequencen{u_n}$ is a sequence in $L^0$, then it is
order*-convergent to $u\in L^0$ iff

\Centerline{$u=\inf_{n\in\Bbb N}\sup_{i\ge n}u_i
=\sup_{n\in\Bbb N}\inf_{i\ge n}u_i$.}

\proof{{\bf (a)} We can express $\frak A$ as a quotient $\Sigma/\Cal I$
of a $\sigma$-algebra of sets, in which case $L^0$ can be identified
with the
canonical image of $\eusm L^0=\eusm L^0(\Sigma)$ (364C).   If
$\sequencen{u_n}$ is an order*-convergent sequence in $L^0$, then it
is expressible as $\sequencen{Tf_n}$, where $T:\eusm L^0\to L^0$ is the
canonical map, and 367F tells us that $\sequencen{f_n(x)}$ converges for
every $x\in H$, where $X\setminus H\in\Cal I$.    If we set
$h(x)=\sup_{n\in\Bbb N}|f_n(x)|$ for $x\in H$, $0$ for
$x\in X\setminus H$, then we see that $|u_n|\le Th$ for every
$n\in\Bbb N$, so that $\sequencen{u_n}$ is order-bounded in $L^0$.

\medskip

{\bf (b)} This now follows from 367Bf, because $L^0$ is Dedekind
$\sigma$-complete.
}%end of proof of 367G

\leader{367H}{Proposition}
Suppose that $E\subseteq\Bbb R$ is a Borel
set and $h:E\to\Bbb R$ is a continuous function.   Let $\frak A$ be a
Dedekind $\sigma$-complete Boolean algebra and set
$Q_E=\{u:u\in L^0,\,\Bvalue{u\in E}=1\}$, where $L^0=L^0(\frak A)$.
Let $\bar h:Q_E\to L^0$ be the function defined by $h$\cmmnt{ (364H)}.
Then $\sequencen{\bar h(u_n)}$ order*-converges to $\bar h(u)$
whenever $\sequencen{u_n}$ is a sequence in $Q_E$ order*-converging
to $u\in Q_E$.

\proof{ This is an easy consequence of 367F.   We can represent
$\frak A$ as $\Sigma/\Cal I$ where $\Sigma$ is a $\sigma$-algebra of
subsets of
some set $X$ and $\Cal I$ is a $\sigma$-ideal of $\Sigma$ (314M);  let
$T:\eusm L^0\to L^0(\frak A)$ be the corresponding homomorphism
(364C, 367F).   Now we can find $\Sigma$-measurable functions
$\sequencen{f_n}$, $f$ such that $Tf_n=u_n$, $Tf=u$, as in 367F;   and
the hypothesis $\Bvalue{u_n\in E}=1$, $\Bvalue{u\in E}=1$ means just
that, adjusting $f_n$ and $f$ on a member of $\Cal I$ if necessary, we
can suppose that $f_n(x)$, $f(x)\in E$ for every $x\in X$.   (I am
passing over the trivial case $E=\emptyset$, $X\in\Cal I$,
$\frak A=\{0\}$.)   Accordingly
$\bar h(u_n)=T(hf_n)$ and $\bar h(u)=T(hf)$, and (because $h$ is
continuous)

\Centerline{$\{x:h(f(x))\ne\lim_{n\to\infty}h(f_n(x))\}
\subseteq\{x:f(x)\ne\lim_{n\to\infty}f_n(x)\}\in\Cal I$,}

\noindent so $\sequencen{\bar h(u_n)}$ order*-converges to
$\bar h(u)$.
}%end of proof of 367H

\leader{367I}{Dominated \dvrocolon{convergence}}\cmmnt{ We now have a
suitable language in which to express an abstract version of Lebesgue's
Dominated Convergence Theorem.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a measure algebra.
If $\sequencen{u_n}$ is a sequence in
$L^1=L^1_{\bar\mu}\cmmnt{\mskip5mu=L^1(\frak A,\bar\mu)}$ which
is order-bounded and order*-convergent in $L^1$, then
$\sequencen{u_n}$ is norm-convergent to some $u\in L^1$;  in particular,
$\int u=\lim_{n\to\infty}\int u_n$.

\proof{ The norm of $L^1$ is order-continuous (365C), so
$\sequencen{u_n}$ is norm-convergent to $u$, by 367Da.   As $\int$ is
norm-continuous, $\int u=\lim_{n\to\infty}\int u_n$.
}%end of proof of 367I

\leader{367J}{The Martingale Theorem}\cmmnt{ In the same way, we can
re-write theorems from \S275 in this language.

\medskip

\noindent{\bf Theorem}} Let $(\frak A,\bar\mu)$ be a probability
algebra, and $\sequencen{\frak B_n}$ a non-decreasing sequence of closed
subalgebras of $\frak A$.   For each $n\in\Bbb N$ let
$P_n:L^1=L^1_{\bar\mu}\to L^1\cap L^0(\frak B_n)$ be the
conditional expectation
operator\cmmnt{ (365Q)};  let $\frak B$ be the closed subalgebra of
$\frak A$ generated by $\bigcup_{n\in\Bbb N}\frak B_n$, and $P$ the
conditional expectation operator onto $L^1\cap L^0(\frak B)$.

(a) If $\sequencen{u_n}$ is a norm-bounded sequence in $L^1$ such that
$P_n(u_{n+1})=u_n$ for every $n\in\Bbb N$, then $\sequencen{u_n}$ is
order*-convergent in $L^1$.

(b) If $u\in L^1$ then $\sequencen{P_nu}$ is order*-convergent and
$\|\,\|_1$-convergent to $Pu$.

\proof{ If we represent $(\frak A,\bar\mu)$ as the measure algebra of a
probability space, these become mere translations of 275G and 275I.
(Note that this argument relies on the description of
order*-convergence in $L^0$ in terms of a.e.\ convergence of
functions, as in 367F;  so that we need to know that
order*-convergence in $L^1$ matches order*-convergence in $L^0$,
which is what 367E is for.)
}%end of proof of 367J

\cmmnt{\medskip

\noindent{\bf Remark} See also 367Q below.
}

\leader{367K}{}\cmmnt{ Some of the most important applications of
these ideas concern spaces of continuous functions.   I do not think
that this is
the time to go very far along this road, but one particular fact will be
useful in \S376.

\medskip

\noindent}{\bf Proposition} Let $X$ be a locally compact Hausdorff
space, and $\sequencen{u_n}$ a sequence in $C(X)$\cmmnt{, the space of
continuous real-valued functions on $X$}.   Then $\sequencen{u_n}$
order*-converges to $0$ in $C(X)$ iff
$\{x:x\in X,\,\limsup_{n\to\infty}|u_n(x)|>0\}$ is
meager.   In particular, $\sequencen{u_n}$ order*-converges to $0$ if
$\lim_{n\to\infty}u_n(x)=0$ for every $x$.

\proof{{\bf (a)} The following elementary fact is worth noting:  if
$A\subseteq C(X)^+$ is non-empty and $\inf A=0$ in $C(X)$, then
$G=\bigcup_{u\in A}\{x:u(x)<\epsilon\}$ is dense for every $\epsilon>0$.
\Prf\Quer\ If not, take $x_0\in X\setminus\overline{G}$.   Because $X$
is completely regular (3A3Bb), there is a continuous function
$w:X\to[0,1]$
such that $w(x_0)=1$ and $w(x)=0$ for every $x\in\overline{G}$.   But in
this case $0<\epsilon w\le u$ for every $u\in A$, which is
impossible.\ \Bang\Qed\

\medskip

{\bf (b)} Suppose that $\sequencen{u_n}$ order*-converges to $0$.
Set $v_n=|u_n|\wedge\chi X$, so that $\sequencen{v_n}$
order*-converges to $0$ (using 367Ca, as usual).   Set

\Centerline{$B=\{v:v\in C(X),\,\exists\,n\in\Bbb N,\,
v_i\le v\Forall i\ge n\}$,}

\noindent so that $\inf B=0$ in
$C(X)$ (367Be).   For each $k\in\Bbb N$, set
$G_k=\bigcup_{v\in B}\{x:v(x)<2^{-k}\}$;  then $G_k$ is dense, by (a),
and of course is open.   So $H=\bigcup_{k\in\Bbb N}X\setminus G_k$ is a
countable union of
nowhere dense sets and is meager.   But this means that

$$\eqalign{\{x:\limsup_{n\to\infty}|u_n(x)|>0\}
&=\{x:\limsup_{n\to\infty}v_n(x)>0\}\cr
&\subseteq\{x:\inf_{v\in B}v(x)>0\}
\subseteq H\cr}$$

\noindent is meager.

\medskip

{\bf (c)} Now suppose that $\sequencen{u_n}$ does not order*-converge
to $0$.   By 367C(b-ii), there is a $w>0$ in $C(X)$ such that
$w=\sup_{i\ge n}w\wedge|u_i|$ for every $n\in\Bbb N$;  that is,
$\inf_{i\ge n}(w-|u_i|)^+=0$ for every $n$.   Set

\Centerline{$G_n
=\{x:\inf_{i\ge n}(w-|u_i|)^+(x)<2^{-n}\}
=\{x:\sup_{i\ge n}|u_i(x)|>w(x)-2^{-n}\}$}

\noindent for each $n$.   Then

\Centerline{$H=\bigcap_{n\in\Bbb N}G_n
=\{x:\limsup_{n\to\infty}u_n(x)\ge w(x)\}$}

\noindent is the intersection of a sequence of dense open sets, and its
complement is meager.

Let $G$ be the non-empty open set $\{x:w(x)>0\}$.   Then $G$ is not
meager, by Baire's theorem (3A3Ha);  so $G\cap H$ cannot be meager.
But $\{x:\limsup_{n\to\infty}|u_n(x)|>0\}$ includes $G\cap H$, so is
also not meager.
}%end of proof of 367K

\cmmnt{\medskip

\noindent{\bf Remark} Unless the topology of $X$ is discrete, $C(X)$ is
not regularly embedded in $\Bbb R^X$, and we expect to find sequences in
$C(X)$ which order*-converge to $0$ in $C(X)$ but not in $\Bbb R^X$.
But the proposition tells us that if we have a sequence in $C(X)$ which
order*-converges in $\Bbb R^X$ to a member of $C(X)$, then it
order*-converges in $C(X)$.
}%end of comment

\leader{367L}{}\cmmnt{ Everything above concerns a particular notion
of sequential convergence.   There is inevitably a suggestion that there
ought to be a topological interpretation of this convergence (see
367Yb, 367Yk, 3A3P), but I have taken care to avoid spelling one out at
this stage;  I will return to the point in \S393.
(For a general discussion in the context of Boolean algebras,
see {\smc Vladimirov 02}, chap.\ 4.)   I come
now to something which really is a topology, and is as closely involved
with order-convergence as any.

\medskip

\noindent}{\bf Convergence in measure} Let $(\frak A,\bar\mu)$ be a
measure algebra.   For $a\in\frak A^f=\{a:\bar\mu a<\infty\}$,
$u\in L^0=L^0(\frak A)$ and $\epsilon>0$ set
$\tau_a(u)=\int|u|\wedge\chi a$
and $\tau_{a\epsilon}(u)=\bar\mu(a\Bcap\Bvalue{|u|>\epsilon})$.
Then the {\bf topology of convergence in measure} on $L^0$ is defined
{\it either} as the topology generated by the F-seminorms
$\tau_a$ {\it or} by saying that $G\subseteq L^0$ is
open iff for every $u\in G$ there are $a\in\frak A^f$ and $\epsilon>0$
such that $v\in G$ whenever $\tau_{a\epsilon}(u-v)\le\epsilon$.

\cmmnt{\medskip

\noindent{\bf Remark} The sentences above include a number of assertions
which need proving.   But at this point, rather than write out any of
the relevant arguments, I refer you to \S245.   Since we
know that $L^0(\frak A)$ can be identified with $L^0(\mu)$ for a
suitable measure space $(X,\Sigma,\mu)$ (321J, 364Ic), everything we
know about general spaces $L^0(\mu)$ can be applied directly to
$L^0(\frak A)$ for measure algebras $(\frak A,\bar\mu)$;  and that is
what I will
do for the next few paragraphs.   So far, all I have done is to write
$\tau_a$ in place of the $\bar\tau_F$ of 245Ac, and call on the
remarks in 245Bb and 245F.
}%end of comment

\leader{367M}{Theorem}
(a) For any measure algebra $(\frak A,\bar\mu)$,
the topology $\frak T$ of convergence in measure on $L^0=L^0(\frak A)$
is a linear space topology, and any order*-convergent sequence in
$L^0$ is $\frak T$-convergent to the same limit.

(b) $u\mapsto |u|:L^0\to L^0$ and $(u,v)\mapsto u\vee v$,
$(u,v)\mapsto u\times v:L^0\times L^0\to L^0$ are continuous.

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

(d) $(\frak A,\bar\mu)$ is localizable iff $\frak T$ is Hausdorff and
$L^0$ is complete under the uniformity corresponding to $\frak T$.

(e) $(\frak A,\bar\mu)$ is $\sigma$-finite iff $\frak T$ is metrizable.

\proof{ 245D, 245Cb, 245E.   Of course we need 322B to assure us that
the phrases `semi-finite', `localizable', `$\sigma$-finite'
here correspond to the same phrases used in \S245, and 367F to identify
order*-convergence in $L^0$ with the order-convergence studied in
\S245.
}%end of proof of 367M

\leader{367N}{Proposition}
Let $(\frak A,\bar\mu)$ be a measure algebra
and give $L^0=L^0(\frak A)$ its topology of convergence in measure.

(a) If $A\subseteq L^0$ is a non-empty, downwards-directed set with
infimum
$0$, then for every neighbourhood $G$ of $0$ in $L^0$ there is a
$u\in A$ such that $v\in G$ whenever $|v|\le u$.

(b) If $U\subseteq L^0$ is an order-dense Riesz subspace, it is
topologically dense.

(c) In particular, $S(\frak A)$ and $L^{\infty}(\frak A)$ are
topologically dense.

\proof{{\bf (a)} Let $a\in\frak A^f$, $\epsilon>0$ be such that $u\in G$
whenever $\int|u|\wedge\chi a\le\epsilon$ (see 245Bb).   Since
$\{u\wedge\chi a:u\in A\}$ is a downwards-directed set in
$L^1=L^1_{\bar\mu}$ with infimum $0$
in $L^1$, there must be a $u\in A$ such that
$\int u\wedge\chi a\le\epsilon$ (365Da).   But now $[-u,u]\subseteq G$,
as required.

\medskip

{\bf (b)} Write $\overline{U}$ for the closure of $U$.   Then
$(L^0)^+\subseteq\overline{U}$.   \Prf\ If $v\in(L^0)^+$, then
$\{u:u\in U$, $u\le v\}$ is an upwards-directed set with supremum $u$,
so $A=\{v-u:u\in U$, $u\le v\}$ is a downwards-directed set with infimum
$0$ (351Db).   By (a), every neighbourhood of $0$ meets $A$, and
(because subtraction is continuous) every neighbourhood of $v$ meets
$U$, that is, $v\in\overline{U}$.\ \Qed

Since $\overline{U}$ is a linear subspace of $L^0$ (2A5Ec), it includes
$(L^0)^+-(L^0)^+=L^0$ (352D).

\medskip

{\bf (c)} By 364Ja, $S(\frak A)$ and $L^{\infty}(\frak A)$ are
order-dense Riesz subspaces of $L^0$.
}%end of proof of 367N

\leader{367O}{Theorem} Let $U$ be a Banach lattice and
$(\frak A,\bar\mu)$ a measure algebra.   Give $L^0=L^0(\frak A)$ its
topology of convergence in measure.   If $T:U\to L^0$ is a
positive linear operator, then it is continuous.

\proof{ Take any open set $G\subseteq L^0$.   \Quer\ Suppose, if
possible, that $T^{-1}[G]$ is not open.   Then we can find $u$,
$\sequencen{u_n}\in U$ such that $Tu\in G$ and $\|u_n-u\|\le 2^{-n}$,
$Tu_n\notin G$ for every $n$.   Set $H=G-Tu$;  then $H$ is an open set
containing $0$ but not $T(u_n-u)$, for any $n\in\Bbb N$.   Since
$\sum_{n=0}^{\infty}n\|u_n-u\|<\infty$, $v=\sum_{n=0}^{\infty}n|u_n-u|$
is defined in $U$, and $|T(u_n-u)|\le\bover1nTv$ for every $n\ge 1$.
But by 367Na (or otherwise) we know that there is some $n$ such that
$w\in H$
whenever $|w|\le\bover1nTv$, so that $T(u_n-u)\in H$ for some $n$, which
is impossible.\ \Bang
}%end of proof of 367O

\vleader{60pt}{367P}{Proposition}
Let $(\frak A,\bar\mu)$ be a $\sigma$-finite measure algebra.

(a) A sequence $\sequencen{u_n}$ in $L^0=L^0(\frak A)$ converges in
measure to $u\in L^0$ iff every subsequence of $\sequencen{u_n}$ has a
sub-subsequence which order*-converges to $u$.

(b) A set $F\subseteq L^0$ is closed for the topology of convergence in
measure iff $u\in F$ whenever there is a sequence $\sequencen{u_n}$ in
$F$ order*-converging to $u\in L^0$.

\proof{ 245K, 245L.
}%end of proof of 367P

\leader{367Q}{}\cmmnt{ As an example of the power of the language we now
have available, I give abstract versions of some martingale
convergence theorems.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a probability algebra;
for each closed subalgebra $\frak B$ of $\frak A$, let
$P_{\frak B}$ be the corresponding conditional expectation operator
from $L^1=L^1_{\bar\mu}$ to $L^1\cap L^0(\frak B)=L^1_{\bar\mu\restrp\frak B}$.

(a) If $\Bbb B$ is a non-empty downwards-directed family of closed
subalgebras of $\frak A$ with intersection $\frak C$, and
$u\in L^1=L^1_{\bar\mu}$, then
$P_{\frak C}u$ is the $\|\,\|_1$-limit 
$\lim_{\frak B\to\Cal F(\Bbb B\closedownarrow)}P_{\frak B}u$, where
$\Cal F(\Bbb B\closedownarrow)$ is the filter on $\Bbb B$ generated by
$\{\{\frak B:\frak B_0\supseteq\frak B\in\Bbb B\}:\frak B_0\in\Bbb B\}$. 

(b) If $\Bbb B$ is a non-empty upwards-directed family of closed
subalgebras of $\frak A$ and $\frak C$ is the closed subalgebra generated
by $\bigcup\Bbb B$, then for every
$u\in L^1$, $P_{\frak C}u$ is the $\|\,\|_1$-limit 
$\lim_{\frak B\to\Cal F(\Bbb B\closeuparrow)}P_{\frak B}u$, where
$\Cal F(\Bbb B\closeuparrow)$ is the filter on $\Bbb B$ generated by
$\{\{\frak B:\frak B_0\subseteq\frak B\in\Bbb B\}:\frak B_0\in\Bbb B\}$. as
$\frak B$ decreases through $\Bbb B$.

(c) Suppose that $\Bbb B$ is a non-empty upwards-directed family of closed
subalgebras of $\frak A$, and $\family{\frak B}{\Bbb B}{u_{\frak B}}$ is a
$\|\,\|_1$-bounded family in $L^1$ such that
$u_{\frak B}=P_{\frak B}u_{\frak C}$ whenever $\frak B$,
$\frak C\in\Bbb B$ and $\frak B\subseteq\frak C$.   Then
$\lim_{\frak B\to\Cal F(\Bbb B\closeuparrow)}u_{\frak B}$
is defined for the topology of convergence in measure and belongs to $L^1$.

%do we have full pointwise limits?\query don't believe it

\proof{{\bf (a)(i)} Note first that $\{P_{\frak B}u:\frak B\in\Bbb B\}$ is
uniformly integrable (246D, or directly), therefore relatively weakly
compact in $L^1$ (247C/356Q).   Consequently there must be a
$v\in L^1$ which is a weak cluster point of $P_{\frak B}u$ as $\frak B$
decreases through $\Bbb B$, in the sense that
$v$ belongs to the weak closure
$\overline{\{P_{\frak B}u:\frak B\in\Bbb B,\,\frak B\subseteq\frak B_0\}}$
for every $\frak B_0\in\Bbb B$.

It follows that $v=P_{\frak C}u$.   \Prf\ For every $\frak B_0\in\Bbb B$,
$L^1\cap L^0(\frak B_0)=L^1_{\bar\mu\restrp\frak B_0}$ is a norm-closed linear
subspace of $L^1$ containing $P_{\frak B}u$ whenever
$\frak B\subseteq\frak B_0$.   It is therefore weakly closed (3A5Ee)
and contains
$v$.   Consequently $\Bvalue{v>\alpha}\in\frak B_0$ for every
$\alpha\in\Bbb R$.   As $\frak B_0$ is arbitrary,
$\Bvalue{v>\alpha}\in\frak C$ for every $\alpha\in\Bbb R$, and
$v\in L^1_{\bar\mu\restrp\frak C}$.   Next, if $c\in\frak C$,
then

\Centerline{$\int_cv\in\overline{\{\intop_cP_{\frak B}u:\frak B\in\Bbb B\}}
=\{\int_cu\}$;}

\noindent so $v=P_{\frak C}u$.\ \Qed

\medskip

\quad{\bf (ii)} Now take $\epsilon>0$.   Then
there is a $\frak B_0\in\Bbb B$ such
that $\|P_{\frak B}u-P_{\frak B_0}u\|_1\le\bover12\epsilon$ whenever
$\frak B\in\Bbb B$ and $\frak B\subseteq\frak B_0$.
\Prf\Quer\ Otherwise, we can find a non-increasing sequence
$\sequencen{\frak B_n}$ in $\Bbb B$ such that
$\|P_{\frak B_{n+1}}u-P_{\frak B_n}u\|_1>\bover12\epsilon$
for every $n\in\Bbb N$.
By the reverse martingale theorem (275K), $\sequencen{P_{\frak B_n}u}$ is
order*-convergent to $w$ say.   But as $\{P_{\frak B_n}u:n\in\Bbb N\}$ is
uniformly integrable, $\sequencen{P_{\frak B_n}u}$ is $\|\,\|_1$-convergent
to $w$ (246Ja), and
$\lim_{n\to\infty}\|P_{\frak B_{n+1}}u-P_{\frak B_n}u\|_1=0$.\ \Bang\Qed

At this point, however, observe that
$C=\{w:\|w-P_{\frak B_0}u\|_1\le\bover12\epsilon\}$ is convex and
$\|\,\|_1$-closed, therefore weakly closed, in $L^1$.   Since it
contains $P_{\frak B}u$ whenever $\frak B\in\Bbb B$ and
$\frak B\subseteq\frak B_0$, it contains $v=P_{\frak C}u$.   Consequently

\Centerline{$\|P_{\frak B}u-P_{\frak C}u\|_1
\le\|P_{\frak B}u-P_{\frak B_0}u\|_1+\|P_{\frak B_0}u-v\|_1
\le\epsilon$}

\noindent whenever $\frak B\in\Bbb B$ and $\frak B\subseteq\frak B_0$.
As $\epsilon$ and $u$ are arbitrary, (a) is true.

\medskip

{\bf (b)} We can use the same method.   Again take any $u\in L^1$.

\medskip

\quad{\bf (i)} This time, observe that $P_{\frak B}u$ must have a weak
cluster point $v$ as $\frak B$ increases through $\Bbb B$.   Since
$P_{\frak B}u$ belongs to $L^1\cap L^0(\frak C)$ for every
$\frak B\in\Bbb B$, so does $v$.   Next, if $b\in\frak B_0\in\Bbb B$, then
$\int_bP_{\frak B}u=\int_bu$ whenever $\frak B\supseteq\frak B_0$, so
$\int_bv=\int_bu$.   Thus $\frak D=\{b:b\in\frak A$, $\int_bv=\int_bu\}$
includes $\bigcup\Bbb B$.   But $\frak D$ is closed for the measure algebra
topology of $\frak A$, so $\frak D\supseteq\frak C$ and $\int_cv=\int_cu$
for every $c\in\frak C$.   Thus once again we have $v=P_{\frak C}u$.

\medskip

\quad{\bf (ii)} Now repeat the argument of (a-ii) almost word for word, but
taking `$\frak B\supseteq\frak B'$' in place of every
`$\frak B\subseteq\frak B'$', and quoting the ordinary martingale theorem
instead of the reverse martingale theorem.

\medskip

{\bf (c)(i)} If $\sequencen{\frak B_n}$ is a non-decreasing sequence in
$\Bbb B$, then $\sequencen{u_{\frak B_n}}$ is order*-convergent, by
Doob's martingale theorem (367Ja).

\medskip

\quad{\bf (ii)}
It follows that the image $\Cal G$ of $\Cal F(\Bbb B\closeuparrow)$
under the map $\frak B\mapsto u_{\frak B}:\Bbb B\to L^0$ is
Cauchy for the linear space topology $\frak T$ of convergence in measure.
\Prf\Quer\ Otherwise, set $\tau(v)=\int|v|\wedge\chi 1$ for $v\in L^0$.
There is an $\epsilon>0$ such that $\sup_{v,v'\in C}\tau(v-v')>2\epsilon$
for every $C\in\Cal G$;  in which case, for any $\frak B\in\Bbb B$, there
must be a $\frak C\in\Bbb B$ such that $\frak C\supseteq\frak B$
$\tau(u_{\frak C}-u_{\frak B})\ge\epsilon$.   But now there will be a
non-decreasing sequence $\sequencen{\frak B_n}$ in $\Bbb B$ such that
$\tau(u_{\frak B_{n+1}}-u_{\frak B_n})\ge\epsilon$ for every $n\in\Bbb N$
and $\sequencen{u_{\frak B_n}}$ cannot be order*-convergent.\ \Bang\Qed

\medskip

\quad{\bf (iii)} By 367Mc,
$u=\lim\Cal G=\lim_{\frak B\to\Cal F(\Bbb B\closeuparrow)}u_{\frak B}$
is defined in $L^0$ for $\frak T$.   But as $u$ belongs to the
$\frak T$-closure of the $\|\,\|_1$-bounded set
$\{u_{\frak B}:\frak B\in\Bbb B\}$, $u\in L^1$, by 245J(b-i).
}%end of proof of 367Q

\leader{367R}{}\cmmnt{ It will be useful later to be able to quote
the following straightforward facts.

\medskip

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ be a measure
algebra.   Give $\frak A$ its measure-algebra topology\cmmnt{ (323A)}
and $L^0=L^0(\frak A)$ the topology of convergence in measure.

(a) The map $\chi:\frak A\to L^0$ is a homeomorphism between $\frak A$ and
its image in $L^0$.

(b) If $\frak A$ has countable Maharam type, then $L^0$ is separable.

(c) Suppose that $\frak B$ is a subalgebra of $\frak A$ which
is closed for the measure-algebra topology.   Then
$L^0(\frak B)$ is closed in $L^0(\frak A)$.

(d)\dvAnew{2014} A non-empty set $A\subseteq L^0$ is bounded in the linear
topological space sense\cmmnt{ (3A5N)} iff
$\inf_{k\in\Bbb N}\sup_{u\in A}\bar\mu(a\Bcap\Bvalue{|u|>k})=0$ for every
$a\in\frak A^f$.

\proof{{\bf (a)} Of course $\chi$ is injective (364Jc).
The measure-algebra
topology of $\frak A$ is defined by the pseudometrics
$\rho_a(b,c)=\bar\mu(a\Bcap(b\symmdiff c))$, while the topology of $L^0$
is defined by the pseudometrics $\sigma_a(u,v)=\int|u-v|\wedge\chi a$,
in both cases taking $a$ to run over elements of $\frak A$ of finite
measure;  as $\sigma_a(\chi b,\chi c)$ is always equal to $\rho_a(b,c)$,
we have the result.

\medskip

{\bf (b)} By 331O, $\frak A$ is separable in its measure-algebra topology;
let $B\subseteq\frak A$ be a countable dense set.   Set

\Centerline{$B^*=\{\sum_{i=0}^n\alpha_i\chi b_i:n\in\Bbb N$,
$\alpha_0,\ldots,\alpha_n\in\Bbb Q$, $b_0,\ldots,b_n\in B\}$.}

\noindent $B^*$ is a countable subset of $L^0$;   let $V$ be its
closure.   Then $V$ includes $S(\frak A)$.   \Prf\ For any $n\in\Bbb N$,
the function
$(\alpha_0,\ldots,\alpha_n,a_0,\ldots,a_n)\mapsto
\sum_{i=0}^n\alpha_i\chi a_i:\BbbR^{n+1}\times\frak A^{n+1}\to L^0$ is
continuous, just because $\chi:\frak A\to L^0$ and
addition and scalar multiplication in $L^0$ are continuous ((a) above,
367M).   So

\Centerline{$D_n=\{(\alpha_0,\ldots,\alpha_n,a_0,\ldots,a_n):
\sum_{i=0}^n\alpha_i\chi a_i\in V\}$}

\noindent is a closed subset of $\BbbR^{n+1}\times\frak A^{n+1}$ including
$\BbbQ^{n+1}\times B^{n+1}$.  But $\BbbQ^{n+1}\times B^{n+1}$ is dense in
$\BbbR^{n+1}\times\frak A^{n+1}$ (3A3Ie), so
$D_n=\BbbR^{n+1}\times\frak A^{n+1}$, that is,
$\sum_{i=0}^n\alpha_i\chi a_i\in V$ whenever
$\alpha_0,\ldots,\alpha_n\in\Bbb R$ and $a_0,\ldots,a_n\in V$.   As $n$ is
arbitrary, $S(\frak A)\subseteq V$.\ \Qed

Since $S(\frak A)$ is dense in $L^0$ (367Nc), $V=L^0$, $B^*$ is dense in
$L^0$ and $L^0$ is separable.

\medskip

{\bf (c)} Note first that $\frak B$ is order-closed in $\frak A$
(323D(c-i)), so that $L^0(\frak B)$, defined as in 364A, is a subset of
$L^0(\frak A)$ (cf.\ 364Xt).   Applying 364P to the identity map
$\frak B\embedsinto\frak A$, we see that the map
$L^0(\frak B)\embedsinto L^0(\frak A)$ identifies the operations
of addition, scalar multiplication and supremum in $L^0(\frak B)$ with the
restrictions of the corresponding operations on $L^0(\frak A)$.

Suppose that $u\in L^0(\frak A)$ is in the closure of $L^0(\frak B)$,
and $\alpha\in\Bbb R$;  let $n\in\Bbb N$ be such that $|\alpha|<n$, and
fix $a\in\frak A^f$ for the moment.
For each $k\in\Bbb N$, choose
$v_k\in L^0(\frak B)$ such that
$\int|u-v_k|\wedge\chi a\le 2^{-k}$ (367L).
Consider $v'_k=\med(-n\chi 1,v_k,n\chi 1)$ for $k\in\Bbb N$, and
$v=\inf_{k\in\Bbb N}\sup_{j\ge k}v'_k$.   We do not
need to ask whether the operations here are being performed in
$L^0(\frak A)$ or in $L^0(\frak B)$, and $v$ will belong to $L^0(\frak B)$.
Accordingly, now necessarily working in $L^0(\frak A)$,
we shall have

\Centerline{$v\times\chi a
=\inf_{k\in\Bbb N}\sup_{j\ge k}v'_k\times\chi a$.}

\noindent Now observe that, for each $k$,
$w_k=2n\sup_{j\ge k}|u-v_j|\wedge\chi a$ is defined in
$L^1_{\bar\mu}$ and $\int w_k\le 2^{-k+2}n$.   Set
$u'=\med(-n\chi 1,u,n\chi 1)$.   For $j\ge k$,

$$\eqalign{|u'\times\chi a-v'_j\times\chi a|
&=|\med(-n\chi 1,u\times\chi a,n\chi 1)
   -\med(-n\chi 1,v_j\times\chi a,n\chi 1)|\cr
&\le|u-v_j|\wedge 2n\chi a
\le w_k.\cr}$$

\noindent So, for any $m\in\Bbb N$,

$$\eqalign{u'\times\chi a-v\times\chi a
&=\sup_{k\in\Bbb N}\inf_{j\ge k}u'\times\chi a-v'_k\times\chi a\cr
&=\sup_{k\ge m}\inf_{j\ge k}u'\times\chi a-v'_k\times\chi a
\le\sup_{k\ge m}w_k
=w_m,\cr
v\times\chi a-u'\times\chi a
&=\inf_{k\in\Bbb N}\sup_{j\ge k}v'_k\times\chi a-u'\times\chi a\cr
&\le\sup_{j\ge m}v'_k\times\chi a-u'\times\chi a
\le w_m.\cr}$$

\noindent Putting these together,

\Centerline{$|u'\times\chi a-v\times\chi a|\le w_m$}

\noindent for every $m\in\Bbb N$, and $u'\times\chi a=v\times\chi a$.
But this means that
$a\Bcap\Bvalue{u'>\alpha}=a\Bcap\Bvalue{v>\alpha}$;  at the same time,
because $-n<\alpha<n$, $\Bvalue{u'>\alpha}=\Bvalue{u>\alpha}$.

Thus we see that for every $a\in\frak A^f$ there is a $b\in\frak B$ such
that $a\Bcap(b\Bsymmdiff\Bvalue{u>\alpha})=0$.   It follows at once that
$\Bvalue{u>\alpha}$ belongs to the closure of $\frak B$, which is
$\frak B$ itself.   As $\alpha$ is arbitrary, $u\in L^0(\frak B)$;  as
$u$ is arbitrary, $L^0(\frak B)$ is closed.

\medskip

{\bf (d)(i)} Suppose that $A$ is topologically bounded, $a\in\frak A^f$
and $\epsilon>0$.   Then $G=\{v:v\in L^0$,
$\bar\mu(a\Bcap\Bvalue{|v|>1})\le\epsilon$ is a neighbourhood of $0$, so
there is a $k\in\Bbb N$ such that $A\subseteq kG$.   If $u\in A$, there is
a $v\in U$ such that $u=kv$, so that

\Centerline{$\bar\mu(a\Bcap\Bvalue{|u|>k})
=\bar\mu(a\Bcap\Bvalue{|v|>1})\le\epsilon$.}

\noindent Thus
$\inf_{k\in\Bbb N}\sup_{u\in A}\bar\mu(a\Bcap\Bvalue{|u|>k})\le\epsilon$;
as $a$ and $\epsilon$ are arbitrary, the condition is satisfied.

\medskip

\quad{\bf (ii)} Suppose that the condition is satisfied.   Let
$G$ be a neighbourhood of of $0$ in $L^0$.   Then there are an
$a\in\frak A^f$ and an $\epsilon>0$
such that $v\in G$ whenever
$\bar\mu(a\Bcap\Bvalue{|v|>\epsilon})\le\epsilon$.   Now there is a
$k\in\Bbb N$ such that $\bar\mu(a\Bcap\Bvalue{|u|>k})\le\epsilon$ for every
$u\in A$.   Let $n\ge 1$ be such that $n\epsilon\ge k$;  then

\Centerline{$\bar\mu(a\Bcap\Bvalue{\bover1n|u|>\epsilon})
=\bar\mu(a\Bcap\Bvalue{|u|>n\epsilon})
\le\bar\mu(a\Bcap\Bvalue{|u|>k})
\le\epsilon$}

\noindent for every $u\in A$, so $\Bover1nA\subseteq G$ and
$A\subseteq nG$.   As $G$ is arbitrary, $A$ is topologically bounded.
}%end of proof of 367R

\leader{367S}{Proposition} Let $E\subseteq\Bbb R$ be a Borel set, and
$h:E\to\Bbb R$ a continuous function.   Let $(\frak A,\bar\mu)$ be a
measure algebra, and $\bar h:Q_E\to L^0=L^0(\frak A)$ the associated
function, where
$Q_E=\{u:u\in L^0$, $\Bvalue{u\in E}=1\}$\cmmnt{ (364H)}.   Then
$\bar h$ is continuous for the topology of convergence in measure.

\proof{ (Compare 245Dd.)   Express $(\frak A,\bar\mu)$ as the measure
algebra of a measure space $(X,\Sigma,\mu)$.   Take
any $u\in Q_E$, any $a\in\frak A$ such that $\bar\mu a<\infty$, and any
$\epsilon>0$.
Express $u$ as $f^{\ssbullet}$ where $f:X\to E$ is a measurable
function, and $a$ as $F^{\ssbullet}$ where $F\in\Sigma$.
Set $\eta=\epsilon/(2+\mu F)$.
For each $n\in\Bbb N$, write $E_n$ for

\Centerline{$\{t:t\in E$, $|h(s)-h(t)|\le\eta$ whenever
$s\in E$ and $|s-t|\le 2^{-n}\}$.}

\noindent Then $\sequencen{E_n}$ is a non-decreasing
sequence of Borel sets with union $E$, so there is an $n$ such that
$\mu\{x:x\in F$, $f(x)\notin E_n\}\le\eta$.

Now suppose that $v\in Q_E$ is such that
$\int|v-u|\wedge\chi a\le 2^{-n}\eta$.   Express $v$ as
$g^{\ssbullet}$ where
$g:X\to E$ is a measurable function.   Then

\Centerline{$\int_F\min(1,|g(x)-f(x)|)\mu(dx)\le 2^{-n}\eta$,}

\noindent so $\mu\{x:x\in F,\,|f(x)-g(x)|>2^{-n}\}\le\eta$, and

$$\eqalign{\{x:x\in F,\,|&h(g(x))-h(f(x))|>\eta\}\cr
&\subseteq\{x:x\in F,\,f(x)\notin E_n\}
\cup\{x:x\in F,\,|f(x)-g(x)|>2^{-n}\}\cr}$$

\noindent has measure at most $2\eta$.   But this means that

\Centerline{$\int|\bar h(v)-\bar h(u)|\wedge\chi a
=\int_F\min(1,|hg(x)-hf(x)|)\mu(dx)\le 2\eta+\eta\mu F=\epsilon$.}

\noindent As $u$, $a$ and $\epsilon$ are arbitrary, $\bar h$ is
continuous.
}%end of proof of 367S

\leader{367T}{Intrinsic description of convergence in
\dvrocolon{measure}}\cmmnt{ It is a remarkable fact that the
topology of convergence in measure, not only on $L^0$ but on its
order-dense Riesz subspaces, can be described in terms of the Riesz
space structure alone, without referring at all to the underlying
measure algebra or to integration.   (Compare 324H.)   There is more
than one way of doing this.   As far as
I know, none is outstandingly convincing;  I present a formulation which
seems to me to exhibit some, at least, of the essence of the phenomenon.

\wheader{367T}{4}{2}{2}{72pt}

\noindent}{\bf Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra, and $U$ an order-dense Riesz subspace of
$L^0=L^0(\frak A)$.   Suppose that $A\subseteq U$ and $u^*\in U$.   Then
$u^*$ belongs
to the closure of $A$ for the topology of convergence in measure iff

\inset{\noindent there is an order-dense Riesz subspace $V$ of $U$ such
that

\varinset{\noindent for every $v\in V^+$ there is a non-empty
downwards-directed $B\subseteq U$, with infimum $0$, such that

\varinset{\noindent for every $w\in B$ there is a $u\in A$ such that

\Centerline{$|u-u^*|\wedge v\le w$.}}}}

\proof{{\bf (a)} Suppose first that $u^*\in\overline{A}$.   Take $V$ to
be $U\cap L^1_{\bar\mu}$;  then $V$ is an order-dense Riesz
subspace of $L^0$, by 352Nc and 353A,
% 352N(c-i)+353A o-d iff q-o-d 352N(c-ii)
and is therefore order-dense in $U$.
(This is where I use the hypothesis that $(\frak A,\bar\mu)$ is
semi-finite, so that $L^1_{\bar\mu}$ is order-dense in $L^0$, by 365Ga.)

Take any $v\in V^+$.   For each $n\in\Bbb N$, set
$a_n=\Bvalue{v>2^{-n}}\in\frak A^f$.   Because $u^*\in\overline{A}$,
there is a $u_n\in A$ such that $\bar\mu b_n\le 2^{-n}$, where

\Centerline{$b_n=a_n\Bcap\Bvalue{|u_n-u^*|>2^{-n}}
=\Bvalue{|u_n-u^*|\wedge v>2^{-n}}$.}

\noindent Set $c_n=\sup_{i\ge n}b_i$;  then $\bar\mu c_n\le 2^{-n+1}$
for each $n$, so $\inf_{n\in\Bbb N}c_n=0$ and $\inf_{n\in\Bbb N}w_n=0$
in $L^0$, where $w_n=v\times\chi c_n+2^{-n}\chi 1$.   Also
$|u_n-u^*|\wedge v\le w_n$ for each $n$.

The $w_n$ need not belong to $U$, so we cannot set $B=\{w_n:n\in\Bbb
N\}$.   But if instead we write
\Centerline{$B=\{w:w\in U,\,w\ge v\wedge w_n$ for some $n\in\Bbb N\}$,}

\noindent then $B$ is non-empty and downwards-directed (because
$\sequencen{w_n}$ is non-increasing);  and
$$\eqalignno{\inf B
&=v-\sup\{v-w:w\in B\}\cr
&=v-\sup\{w:w\in U,\,w\le (v-w_n)^+\text{ for some }n\in\Bbb N\}\cr
&=v-\sup_{n\in\Bbb N}(v-w_n)^+\cr
\noalign{\noindent (because $U$ is order-dense in $L^0$)}
&=0.\cr}$$

\noindent Since for every $w\in B$ there is an $n$ such that
$w\ge v\wedge w_n\ge v\wedge|u_n-u^*|$, $B$ witnesses that the condition
is satisfied.

\medskip

{\bf (b)} Now suppose that the condition is satisfied.   Fix
$a\in\frak A^f$ and $\epsilon>0$.   Because $V$ is order-dense in $U$ and therefore in
$L^0$, there is a $v\in V$ such that $0\le v\le\chi a$ and
$\int v\ge\bar\mu a-\epsilon$.   Let $B$ be a downwards-directed set,
with infimum $0$, such that for every $w\in B$ there is a $u\in A$ with
$v\wedge|u-u^*|\le w$.   Then there is a $w\in B$ such that
$\int w\wedge v\le\epsilon$.   Now there is a $u\in A$ such that
$|u-u^*|\wedge v\le w$, so that

\Centerline{$\int|u-u^*|\wedge\chi a\le\epsilon+\int|u-u^*|\wedge
v
\le\epsilon+\int w\wedge v\le 2\epsilon$.}

\noindent As $a$ and $\epsilon$ are arbitrary, $u^*\in\overline{A}$.
}%end of proof of 367T

\leader{*367U}{Theorem} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra;  write $L^1$ for $L^1_{\bar\mu}$.   Let
$P:(L^1)^{**}\to L^1$ be the linear operator corresponding to the band
projection from
$(L^1)^{**}=(L^1)^{\times\sim}$ onto $(L^1)^{\times\times}$ and the
canonical isomorphism between $L^1$ and $(L^1)^{\times\times}$.   For
$A\subseteq L^1$ write $A^*$ for the weak* closure of the image of
$A$ in $(L^1)^{**}$.   Then for every $A\subseteq L^1$

\Centerline{$P[A^*]\subseteq\overline{\Gamma(A)}$,}

\noindent where $\Gamma(A)$ is the convex hull of $A$ and
$\overline{\Gamma(A)}$ is the closure of $\Gamma(A)$ in
$L^0=L^0(\frak A)$ for the topology of convergence in measure.

\proof{{\bf (a)} The statement of the theorem includes a number of
assertions:  that $(L^1)^*=(L^1)^{\times}$;  that
$(L^1)^{**}=((L^1)^*)^{\sim}$;  that the natural embedding of $L^1$ into
$(L^1)^{**}=(L^1)^{\times\sim}$ identifies $L^1$ with
$(L^1)^{\times\times}$;  and that $(L^1)^{\times\times}$ is a
projection band in $(L^1)^{\times\sim}$.   For proofs of these see 365C,
% L^1 an L-space, octs norm, Levi prop
356P,
% (L^1)^*=(L^1)^{\times}=(L^1)^{\sim}
356B
% U^{\times} projection band in U^{\sim}
and 356D.
% U^*=U^{\sim} for B lattice U

Now for the new argument.   First, observe that the statement of the
theorem involves the measure algebra $(\frak A,\bar\mu)$ and the space
$L^0$ only in the definition of `convergence in measure';  everything
else depends only on the Banach lattice structure of $L^1$.   And since
we are concerned only with the question of whether members of $P[A^*]$,
which is surely a subset of $L^1$, belong to $\overline{\Gamma(A)}$,
367T shows that this also can be answered in terms of the Riesz space
structure of $L^1$.   What this means is that we can suppose that
$(\frak A,\bar\mu)$ is localizable.   \Prf\ Let
$(\widehat{\frak A},\tilde\mu)$ be the localization of
$(\frak A,\bar\mu)$ (322Q).   The
natural expression of $\frak A$ as an order-dense subalgebra of
$\widehat{\frak A}$ identifies
$\frak A^f=\{a:a\in\frak A,\,\bar\mu a<\infty\}$ with
$\widehat{\frak A}^f$ (322P), so that $L^1_{\bar\mu}$
becomes identified with $L^1_{\tilde\mu}$, by 365Nd.
Thus we can think of $L^1$ as $L^1_{\tilde\mu}$,
and $(\widehat{\frak A},\tilde\mu)$ is localizable.\ \Qed

\medskip

{\bf (b)} We need a version of a result in \S362.   As we are supposing that
$(\frak A,\bar\mu)$ is localizable, we can identify
$(L^1)^*$, as Banach lattice, with $L^{\infty}=L^{\infty}(\frak A)$ (365Lc).
Take any $\phi\in(L^1)^{**}\cong(L^{\infty})^*$ such that
$\phi\ge 0$ and $P\phi=0$.   Then
$C=\{c:c\in\frak A$, $\phi(\chi c)=0\}$ is an order-dense ideal of $\frak A$.
\Prf\ Just because $\phi$ is a positive linear operator,
$C$ is an ideal of $\frak A$.   We have an $L$-space isomorphism between
$(L^{\infty})^*=(L^{\infty})^{\sim}$ and the space $M$ of bounded additive
functionals on $\frak A$, and this isomorphism matches $(L^{\infty})^{\times}$
with the projection band $M_{\tau}$ of completely additive functionals (363K,
362B).   So if we write $P_{\tau}:M\to M_{\tau}$ for the band projection onto
$M_{\tau}$, $P_{\tau}$ must correspond to the band projection
$P:(L^{\infty})^{\sim}\to(L^{\infty})^{\times}$.   Let $\nu$ be the member of
$M$ corresponding to $\phi$, so that $\nu a=\phi(\chi a)$ for every
$a\in\frak A$.   Then $(P_{\tau}\nu)(1)=(P\phi)(\chi 1)=0$, that is,
$P_{\tau}\nu=0$.   Now $\frak A$ is \wsid\ (322F), so there is an
upwards-directed set $D\subseteq\frak A$, with supremum $1$, such that
$\sup_{d\in D}\nu d=(P_{\tau}\nu)(1)=0$, that is,
$0=\nu d=\phi(\chi d)$ for every
$d\in D$, and $D\subseteq C$.   So $\sup C=1$, that is, $C$ is order-dense, as
claimed.\ \Qed

\medskip

{\bf (c)} Now take $\phi\in A^*$ and set $u_0=P\phi$;  I have to show that
$u_0\in\overline{\Gamma(A)}$.   Write $R$ for the canonical map from
$L^1$ to $(L^1)^{**}$, so that $\phi$ belongs to the weak* closure of
$R[A]$.

\medskip

\quad{\bf (i)} Consider first the case $u_0=0$, that is, $P\phi =0$.   Since
$P$ is a band projection, $P|\phi|=0$.   By (b),
$C=\{c:(P|\phi)|(\chi c)=0\}$ is an order-dense ideal in $\frak A$.
Take any $a\in\frak A^f$ and $\epsilon>0$.
Then $\{c:c\in C$, $c\Bsubseteq a\}$ is upwards-directed and has supremum $a$,
so there is a $c\in C$ such that $\bar\mu(a\Bsetminus c)\le\epsilon$.

Consider the map $Q:L^1\to L^1$ defined by setting $Qw=w\times\chi c$
for every $w\in L^1$.   Then its adjoint $Q':L^{\infty}\to L^{\infty}$
(3A5Ed) can be defined by the same formula:  $Q'v=v\times\chi c$ for
every $v\in L^{\infty}$.   Now

\Centerline{$|\phi(Q'v)|\le\|v\|_{\infty}|\phi|(\chi c)=0$}

\noindent for every $v\in L^{\infty}$, and $Q''\phi=0$, where
$Q'':(L^{\infty})^*\to(L^{\infty})^*$ is the adjoint of $Q'$.   Since
$Q''$ is continuous for the weak* topology on $(L^{\infty})^*$,
$0\in \overline{Q''[R[A]]}$, where $\overline{Q''R[A]}$ is the closure for the
weak* topology of $(L^{\infty})^*$.   But of course $Q''R=RQ$, while
the weak* topology of $(L^{\infty})^*$ corresponds, on the image
$R[L^1]$ of $L^1$, to the weak topology of $L^1$;  so that $0$ belongs
to the closure of $Q[A]$ for the weak topology of $L^1$.

Because $Q$ is linear, $Q[\Gamma(A)]$ is convex.   Since $0$ belongs to
the closure of $Q[\Gamma(A)]$ for the weak topology of $L^1$, it belongs
to the closure of $Q[\Gamma(A)]$ for the norm topology (3A5Ee).   So
there is a $w\in\Gamma(A)$ such that $\|w\times\chi c\|_1\le\epsilon^2$.
But this means that $\bar\mu(c\Bcap\Bvalue{|w|\ge\epsilon})\le\epsilon$
and $\bar\mu(a\Bcap\Bvalue{|w|\ge\epsilon})\le 2\epsilon$.   Since $a$
and $\epsilon$ are arbitrary, $0\in\overline{\Gamma(A)}$.

\medskip

\quad{\bf (ii)} This deals with the case $u_0=0$.   Now the general case
follows at once if we set $B=A-u_0$ and observe that $\phi-Ru_0\in B^*$,
so

\Centerline{$0=P(\phi-Ru_0)\in\overline{\Gamma(B)}
=\overline{\Gamma(A)-u_0}=\overline{\Gamma(A)}-u_0$}

\noindent because the topology of convergence in measure is a linear space
topology.
}%end of proof of 367U

\cmmnt{\medskip

\noindent{\bf Remark} This is a version of a theorem from
{\smc Bukhvalov 95}.
}

\leader{*367V}{Corollary} Let $(\frak A,\bar\mu)$ be a localizable
measure algebra.  Let $\Cal C$ be a family of convex subsets of
$L^0=L^0(\frak A)$, all closed for the topology of convergence in
measure, with the finite intersection property, and suppose that for
every non-zero $a\in\frak A$ there are a
non-zero $b\subseteq a$ and a $C\in\Cal C$ such that
$\sup_{u\in C}\int_b|u|<\infty$.   Then $\bigcap\Cal C\ne\emptyset$.

\proof{ Because $\Cal C$ has the finite intersection property, there is
an ultrafilter $\Cal F$ on $L^0$ including $\Cal C$.   Set

\Centerline{$I
=\{a:a\in\frak A,\,\inf_{F\in\Cal F}\sup_{u\in F}\int_a|u|<\infty\}$;}

\noindent because $\Cal F$ is a filter, $I$ is an ideal in $\frak A$,
and the condition on $\Cal C$ tells us that $I$ is order-dense.   For
each $a\in I$, define $Q_a:L^0\to L^0$ by setting $Q_au=u\times\chi a$.
Then there is an $F\in\Cal F$ such that $Q_a[F]$ is a norm-bounded set
in $L^1=L^1_{\bar\mu}$, so $\phi_a=\lim_{u\to\Cal F}RQ_au$ is defined in
$(L^{\infty})^*=L^{\infty}(\frak A)^*$ for the weak* topology on
$(L^{\infty})^*$, writing
$R$ for the canonical map from $L^1$ to $(L^{\infty})^*\cong(L^1)^{**}$.
(Once again, we can identify $(L^1)^*$ with $L^{\infty}$ because
$(\frak A,\bar\mu)$ is localizable.)
If $P:(L^{\infty})^*\to L^1$ is the map corresponding to the
band projection $\tilde P$ from $(L^{\infty})^{\sim}$ onto
$(L^{\infty})^{\times}$, as in 367U, and $C\in\Cal C$, then 367U tells
us that $P\phi_a$
must belong to the closure of the convex set $Q_a[C]$ for the topology
of convergence in measure.   Moreover, if $a\Bsubseteq b\in I$, so that
$Q_a=Q_aQ_b$, then $P\phi_a=Q_aP\phi_b$.   \Prf\ Observe that

\Centerline{$\tilde P=RP:(L^{\infty})^*\to(L^{\infty})^{\times}$,
\quad$Q''_aR=RQ_a\restr L^1$,
\quad$Q''_a\restr(L^{\infty})^{\times}=RQ_aR^{-1}$.}

\noindent $Q_a\restr L^1$ is
a band projection on $L^1$, so its adjoint $Q'_a$ is a band projection
on $L^{\infty}\cong(L^1)^{\sim}$ (356C) and $Q''_a$ is a band
projection on $(L^{\infty})^*\cong(L^{\infty})^{\sim}$.   This means
that $Q''_a$ will commute with $\tilde P$ (352Sb).   But also
$Q''_a$ is continuous for the weak* topology of $(L^{\infty})^*$, so

\Centerline{$Q''_a\phi_b=\lim_{u\to\Cal F}Q''_aRQ_bu
=\lim_{u\to\Cal F}RQ_aQ_bu=\phi_a$}

\noindent and

\Centerline{$P\phi_a=R^{-1}\tilde P\phi_a
=R^{-1}\tilde PQ''_a(\phi_b)=R^{-1}Q''_a\tilde P\phi_b
=Q_aR^{-1}\tilde P\phi_b=Q_aP\phi_b$.   \Qed}

Generally, if $a$, $b\in I$, then

\Centerline{$Q_aP\phi_b=Q_aQ_bP\phi_b=Q_{a\Bcap b}P\phi_b
=P\phi_{a\Bcap b}=Q_bP\phi_a$.}

\noindent
What this means is that if we take a partition $D$ of unity included in
$I$ (313K), so that $L^0\cong\prod_{d\in D}L^0(\frak A_d)$ (315F(iii),
364R), and define $w\in L^0$ by saying that $w\times\chi d=P\phi_d$
for every $d\in D$, then we shall have

\Centerline{$w\times\chi a\times\chi d=Q_aP\phi_d=Q_dP\phi_a
=P\phi_a\times\chi d$}

\noindent whenever $a\in I$ and $d\in D$.   Consequently

\Centerline{$w\times\chi a=P\phi_a\in\overline{Q_a[C]}$}

\noindent for every $a\in I$ and $C\in\Cal C$.
But now, given $a\in\frak A^f$ and $\epsilon>0$ and
$C\in\Cal C$, there is a $b\in I$ such that
$\bar\mu(a\Bsetminus b)\le\epsilon$;
$w\times\chi b\in\overline{Q_b[C]}$, so there is a
$u\in C$ such that
$\bar\mu(b\Bcap\Bvalue{|w-u|\ge\epsilon})\le\epsilon$;  and
$\bar\mu(a\Bcap\Bvalue{|w-u|\ge\epsilon})\le 2\epsilon$.   As $a$ and
$\epsilon$ are arbitrary and $C$ is closed, $w\in C$;  as $C$ is
arbitrary, $w\in\bigcap\Cal C$ and $\bigcap\Cal C\ne\emptyset$.
}%end of proof of 367V

\leader{*367W}{\dvrocolon{Independence}}\cmmnt{ I have given myself
very little room in this chapter to discuss stochastic independence.
There are direct translations of results from \S272 in 364Xe-364Xf.
However the language here is adapted to a significant result not
presented in \S272.   I had better begin by repeating a definition from
364Xe.}   Let $(\frak A,\bar\mu)$ be a probability algebra.   Then a
family $\familyiI{u_i}$ in $L^0(\frak A)$ is {\bf stochastically
independent}
if $\bar\mu(\inf_{i\in J}\Bvalue{u_i>\alpha_i})
=\prod_{i\in J}\bar\mu\Bvalue{u_i>\alpha_i}$ whenever $J\subseteq I$ is
a non-empty finite set and $\alpha_i\in\Bbb R$ for every $i\in I$.
\cmmnt{(The direct translation of the definition in 272Ac would rather
be `$\bar\mu(\inf_{i\in J}\Bvalue{u_i\le\alpha_i})
=\prod_{i\in J}\bar\mu\Bvalue{u_i\le\alpha_i}$ whenever $J\subseteq I$
is a non-empty finite set and $\alpha_i\in\Bbb R$ for every $i\in I$',
intepreting $\Bvalue{u_i\le\alpha_i}$ as in 364Xa.   Of course 272F
tells us that this comes to the same thing.)}   \cmmnt{Now the new
fact is the following.}

\medskip

\noindent{\bf Proposition} Let $(\frak A,\bar\mu)$ be a probability
algebra, and $I$ any set.   Give $L^0=L^0(\frak A)$ its topology of
convergence in measure.   Then the collection of independent families
$\familyiI{u_i}$ is a closed set in $(L^0)^I$.

\proof{ Suppose that $\familyiI{u_i}\in(L^0)^I$ is not independent.
Then there are a finite set $J\subseteq I$ and a family
$\family{i}{J}{\alpha_i}$ of real numbers such that
$\bar\mu(\inf_{i\in J}\Bvalue{u_i>\alpha_i})
\ne\prod_{i\in J}\bar\mu\Bvalue{u_i>\alpha_i}$.   Set
$a_i=\Bvalue{u_i>\alpha_i}$ for each $i$.   Let $\delta>0$ be such that
$\gamma\ne\prod_{i\in J}\gamma_i$ whenever
$|\gamma-\bar\mu(\inf_{i\in J}a_i)|\le 2\delta\#(J)$ and
$|\gamma_i-\bar\mu a_i|\le 2\delta$ for every
$i\in J$.   Let $\eta\in\ocint{0,1}$ be such that
$\bar\mu\Bvalue{u_i>\alpha_i+2\eta}
\ge\bar\mu a_i-\delta$ for every $i\in J$.

Now if $\familyiI{v_i}\in(L^0)^I$ and
$\bar\mu\Bvalue{|v_i-u_i|>\eta}\le\delta$ for each $i\in J$,
$\familyiI{v_i}$ is not independent.   \Prf\ For each $i\in J$, consider
$b_i=\Bvalue{v_i>\alpha_i+\eta}$, $a'_i=\Bvalue{u_i>\alpha_i+2\eta}$.
We have

\Centerline{$a'_i=\Bvalue{u_i>\alpha_i+2\eta}
\Bsubseteq\Bvalue{v_i>\alpha_i+\eta}\Bcup\Bvalue{u_i-v_i>\eta}
\Bsubseteq b_i\Bcup\Bvalue{|u_i-v_i|>\eta}$}

\noindent (364Ea), and

\Centerline{$b_i=\Bvalue{v_i>\alpha_i+\eta}
\Bsubseteq\Bvalue{u_i>\alpha_i}\Bcup\Bvalue{v_i-u_i>\eta}
\Bsubseteq a_i\Bcup\Bvalue{|v_i-u_i|>\eta}$,}

\noindent so

\Centerline{$b_i\Bsymmdiff a_i=(b_i\Bsetminus a_i)\Bcup(a_i\Bsetminus
b_i)
\Bsubseteq\Bvalue{|v_i-u_i|>\eta}\Bcup(a_i\Bsetminus a'_i)$}

\noindent has measure at most $2\delta$.   It follows that
$(\inf_{i\in J}b_i)\Bsymmdiff(\inf_{i\in J}a_i)$ has measure at most
$2\delta\#(J)$, and
$|\bar\mu(\inf_{i\in J}b_i)\penalty-100-\bar\mu(\inf_{i\in J}a_i)|\le 2\delta\#(J)$.
At the same time, for each $i\in J$,
$|\bar\mu b_i-\bar\mu a_i|\le 2\delta$.   By the choice of $\delta$,
$\bar\mu(\inf_{i\in J}b_i)\ne\prod_{i\in J}\bar\mu b_i$, and
$\familyiI{v_i}$ is not independent.\ \Qed

This shows that the set of non-independent families is open in
$(L^0)^I$, so that the set of independent families is closed, as
claimed.
}%end of proof of 367W

\exercises{\leader{367X}{Basic exercises $\pmb{>}$(a)}
%\spheader 367Xa
Let $P$ be a lattice.   (i) Show that if $p\in P$ and $\sequencen{p_n}$
is a non-decreasing sequence in $P$, then $\sequencen{p_n}$ is
order*-convergent to $p$ iff $p=\sup_{n\in\Bbb N}p_n$.
(ii) Suppose that $\sequencen{p_n}$ is a sequence in
$P$ order*-converging to $p\in P$.   Show that
$p=\sup_{n\in\Bbb N}p\wedge p_n=\inf_{n\in\Bbb N}p\vee p_n$.
(iii) Let $\sequencen{p_n}$, $\sequencen{q_n}$ be two sequences in $P$
which are order*-convergent to $p$, $q$ respectively.   Show that if
$p_n\le q_n$ for every $n$ then $p\le q$.
(iv) Let $\sequencen{p_n}$ be a sequence in
$P$.   Show that $\sequencen{p_n}$ order*-converges to
$p\in P$ iff $\sequencen{p_n\vee p}$ and $\sequencen{p_n\wedge p}$
both order*-converge to $p$.
%367B

\spheader 367Xb Let $P$ and $Q$ be lattices, and $f:P\to Q$ an
order-preserving function.   Suppose that $\sequencen{p_n}$ is an
order-bounded sequence which
order*-converges to $p$ in $P$.   Show that $\sequencen{f(p_n)}$
order*-converges to $f(p)$ in $Q$ if {\it either} $f$ is
order-continuous {\it or} $P$ is Dedekind $\sigma$-complete and $f$ is
sequentially order-continuous.
%367B

\spheader 367Xc Let $P$ be {\it either} a Boolean algebra {\it or} a
Riesz space.   Suppose that $\sequencen{p_n}$ is a sequence in $P$ such
that $\sequencen{p_{2n}}$ and $\sequencen{p_{2n+1}}$ are both
order*-convergent to $p\in P$.   Show that $\sequencen{p_n}$ is
order*-convergent to $p$.   \Hint{313B, 352E.}
%367B

\spheader 367Xd\dvAnew{2016}
Let $\family{i}{I}{P_i}$ be a family of lattices with
product $P$ (315Xd).   Show that a sequence $\sequencen{p_n}$ in $P$
order*-converges to $p\in P$ iff $\sequencen{p_n(i)}$ order*-converges
to $p(i)$ in $P_i$ for every $i\in I$.
%367B 315Xd

\sqheader 367Xe Let $\frak A$ be a Boolean algebra and
$\sequencen{a_n}$, $\sequencen{b_n}$ two sequences in $\frak A$
order*-converging to $a$, $b$ respectively.   Show that
$\sequencen{a_n\Bcup b_n}$, $\sequencen{a_n\Bcap b_n}$,
$\sequencen{a_n\Bsetminus b_n}$, $\sequencen{a_n\Bsymmdiff b_n}$
order*-converge to $a\Bcup b$, $a\Bcap b$, $a\Bsetminus b$ and
$a\Bsymmdiff b$ respectively.
%367C

\spheader 367Xf Let $\frak A$ be a Boolean algebra and $\sequencen{a_n}$
a sequence in $\frak A$.   Show that $\sequencen{a_n}$ does not
order*-converge to $0$ iff there is a non-zero $a\in\frak A$ such
that $a=\sup_{i\ge n}a\wedge a_i$ for every $n\in\Bbb N$.
%367C

\sqheader 367Xg(i) Let $U$ be a Riesz space and $\sequencen{u_n}$ an
order*-convergent sequence in $U^+$ with limit $u$.
Show that $h(u)\le\liminf_{n\to\infty}h(u_n)$ for every
$h\in (U^{\times})^+$.
(ii) Let $U$ be a Riesz space and $\sequencen{u_n}$ an
order-bounded order*-convergent sequence in $U$ with limit $u$.
Show that $h(u)=\lim_{n\to\infty}h(u_n)$ for every $h\in U^{\times}$.
(Compare 356Xd.)
%367C

\spheader 367Xh\dvAformerly{3{}67Yq}(i) Show that if $U$ is a
Banach lattice, every
norm-convergent sequence has a subsequence which is order-bounded and
order*-convergent.   \Hint{consider the case in which
$\sum_{n=0}^{\infty}\|u_n-u\|$ is finite.}
(ii) Find a Riesz norm on C([0,1]) for which there is
an order-bounded norm-convergent sequence which has no
order*-convergent subsequence.
%\|\,\|_1
%367D

\sqheader 367Xi Let $U$ be a Riesz space with a Fatou norm $\|\,\,\|$.
(i) Show that if $\sequencen{u_n}$ is an order*-convergent sequence
in $U$ with limit $u$, then $\|u\|\le\liminf_{n\to\infty}\|u_n\|$.
\Hint{$\sequencen{|u_n|\wedge|u|}$ is order*-convergent to $|u|$.}
(ii) Show that if $\sequencen{u_n}$ is a norm-convergent sequence in $U$
it has an order*-convergent subsequence.   \Hint{if
$\sum_{n=0}^{\infty}\|u_n\|<\infty$ then $\sequencen{u_n}$
order*-converges to $0$.}
%367D 367Xh

\spheader 367Xj Let $U$ and $V$ be Archimedean Riesz spaces and
$T:U\to V$ an order-continuous Riesz homomorphism.   Show that if
$\sequencen{u_n}$ is a sequence in $U$ which order*-converges to
$u\in U$, then $\sequencen{Tu_n}$ order*-converges to $Tu$ in $V$.
%367E 367Ye(i)

\spheader 367Xk Let $\frak A$ be a Boolean algebra and $\frak B$
a regularly embedded subalgebra.
Show that if $\sequencen{b_n}$ is a sequence
in $\frak B$ and $b\in\frak B$, then $\sequencen{b_n}$
order*-converges to $b$ in $\frak B$ iff it order*-converges to
$b$ in $\frak A$.
%367E

\spheader 367Xl Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\sequencen{u_n}$, $\sequencen{v_n}$ two sequences in
$L^0(\frak A)$ which are order*-convergent to $u$, $v$
respectively.   Show that $\sequencen{u_n\times v_n}$
order*-converges
to $u\times v$.   Show that if $u$, $u_n$ all have
multiplicative inverses $u^{-1}$, $u_n^{-1}$ then
$\sequencen{u_n^{-1}}$ order*-converges to $u^{-1}$.
%367F

\spheader 367Xm Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $\Cal I$ a $\sigma$-ideal of $\frak A$.   Show that for any
$\sequencen{a_n}\in\frak A^{\Bbb N}$ and $a\in\frak A$,
$\sequencen{a_n^{\ssbullet}}$
order*-converges to $a^{\ssbullet}$ in $\frak A/\Cal I$ iff
$\inf_{n\in\Bbb N}\sup_{m\ge n}a_m\symmdiff a\in\Cal I$.
%367F 367Xf

\sqheader 367Xn Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra, and $\sequencen{h_n}$ a sequence of Borel measurable functions
from $\Bbb R$ to itself such that $h(t)=\lim_{n\to\infty}h_n(t)$ is
defined for every $t\in\Bbb R$.   Show that $\sequencen{\bar h_n(u)}$
order*-converges to $\bar h(u)$ for every $u\in L^0=L^0(\frak A)$,
where $\bar h_n$, $\bar h:L^0\to L^0$ are defined as in 364H.
%367F

\spheader 367Xo Let $U$ be an $L$-space and
$\sequencen{u_n}$ a sequence in $U$ which is
order*-convergent to $u\in U$.   Show that
$\sequencen{u_n}$ is norm-convergent to $u$ iff $\{u_n:n\in\Bbb N\}$ is
uniformly integrable iff $\|u\|_1=\lim_{n\to\infty}\|u_n\|_1$.
\Hint{245H, 246J.}
%367H

\spheader 367Xp Let $U$ be an $L$-space and $\sequencen{u_n}$ a
norm-bounded sequence in $U$.   Show that there are a $v\in U$ and a
subsequence $\sequencen{v_n}$ of $\sequencen{u_n}$ such that
$\sequencen{\bover1{n+1}\sum_{i=0}^nw_i}$ order*-converges to $v$ for
every subsequence $\sequencen{w_n}$ of $\sequencen{v_n}$.   \Hint{276H.}
%367J

\spheader 367Xq Let $(\frak A,\bar\mu)$ be a measure algebra and
$p\in\coint{1,\infty}$.   For $v\in (L^p)^+=(L^p_{\bar\mu})^+$
define $\tau_v:L^0\to\coint{0,\infty}$ by setting
$\tau_v(u)=\||u|\wedge v\|_p$ for $u\in U$.
Show that each $\tau_v$ is an F-seminorm and that the topology on
$L^0(\frak A)$ defined by $\{\tau_v:v\in(L^p)^+\}$ is the topology
of convergence in measure.
%367L

\spheader 367Xr Let $(\frak A,\bar\mu)$ be a $\sigma$-finite measure
algebra.   Suppose we have a double sequence
$\family{(i,j)}{\Bbb N\times\Bbb N}{u_{ij}}$ in $L^0=L^0(\frak A)$
such that $\sequence{j}{u_{ij}}$ order*-converges to $u_i$ in $L^0$
for each
$i$, while $\sequence{i}{u_i}$ order*-converges to $u$.   Show that
there is a strictly
increasing sequence $\sequence{i}{n(i)}$ such that
$\sequence{i}{u_{i,n(i)}}$ order*-converges to $u$.
%367M

\spheader 367Xs Let $(X,\Sigma,\mu)$ be a semi-finite measure space.
Show that $L^0(\mu)$ is separable for the topology of convergence in
measure iff $\mu$ is $\sigma$-finite and has countable Maharam type.
(Cf.\ 365Xr.)
%367M

\spheader 367Xt Let $(\frak A,\bar\mu)$ be a measure algebra, and
give $L^0=L^0(\frak A)$ its topology of convergence in measure.
Show that if $(\frak A,\bar\mu)$ is semi-finite, then a set
$A\subseteq L^0$ is bounded in the linear-topological-space sense
iff $\{\alpha_nx_n:n\in\Bbb N\}$
is order-bounded for every sequence $\sequencen{x_n}$ in $A$ and every
sequence $\sequencen{\alpha_n}$ in $\Bbb R$ converging to $0$.
%367N

\spheader 367Xu Let $(\frak A,\bar\mu)$ be a measure algebra, and
write $\frak T$ for its measure-algebra topology.   (i) Show
that if $\sequencen{a_n}$ is order*-convergent to $a\in\frak A$, then
$\sequencen{a_n}\to a$ for $\frak T$.
(ii) Show that if $(\frak A,\bar\mu)$ is $\sigma$-finite, then
($\alpha$) a sequence converges to $a$ for $\frak T$ iff
every subsequence has a sub-subsequence which is order*-convergent to
to $a$ ($\beta$) a set $F\subseteq\frak A$ is $\frak T$-closed
iff $a\in F$ whenever there is a sequence $\sequencen{a_n}$ in
$F$ which is order*-convergent to $a\in\frak A$.   (iii) Show that if
$(\frak A,\bar\mu)$ is semi-finite but not $\sigma$-finite, there is
an $A\subseteq L^0$ such that the limit of any order*-convergent
sequence in $A$ belongs to $A$, but $A$ is not $\frak T$-closed.
%367P

\spheader 367Xv Let $U$ be a Banach lattice with an order-continuous
norm.   (i) Show that a sequence $\sequencen{u_n}$ is norm-convergent to
$u\in U$ iff every subsequence has a sub-subsequence which is
order-bounded and order*-convergent to $u$.   (ii) Show that a set
$F\subseteq U$ is closed for
the norm topology iff $u\in F$ whenever there is an order-bounded
sequence $\sequencen{u_n}$ in $F$ order*-converging to $u\in U$.
%367P

\sqheader 367Xw Let $(\frak A,\bar\mu)$ be a probability algebra.   For
$u\in L^0=L^0(\frak A)$ let $\nu_u$ be the distribution of $u$ (364Gb).
Show that $u\mapsto\nu_u$ is continuous when $L^0$ is given the topology
of convergence in measure and the space of probability distributions on
$\Bbb R$ is given the vague topology (274Ld).
%367P sort of;  would be useful in Vol 6 (L\'evy processes)

\spheader 367Xx Let $(\frak A,\bar\mu)$ be a probability algebra and
$\sequencen{u_n}$ a stochastically independent sequence in
$L^0(\frak A)$, all
with the Cauchy distribution $\nu_{C,1}$ with centre $0$ and
scale parameter
$1$ (285Xp).   For each $n$ let $C_n$ be the convex hull of
$\{u_i:i\ge n\}$, and $\overline{C_n}$ its closure for the topology of
convergence in measure.   Show that every $u\in\overline{C_0}$ has
distribution $\nu_{C,1}$.   \Hint{consider first $u\in C_0$.}   Show
that $\overline{C_0}$ is bounded for the topology of convergence in
measure.   Show that $\bigcap_{n\in\Bbb N}\overline{C_n}=\emptyset$.
%367V

\spheader 367Xy If $U$ is a linear space and $C\subseteq U$ is a convex
set, a function $f:C\to\Bbb R$ is {\bf convex} if
$f({\alpha}x+(1-{\alpha})y)\le {\alpha}f(x)+(1-{\alpha})f(y)$
whenever $x$, $y\in C$ and $\alpha\in[0,1]$.
Let $(\frak A,\bar\mu)$ be a localizable measure algebra and
$C\subseteq L^1_{\bar\mu}$ a non-empty convex
norm-bounded set which is closed in $L^0(\frak A)$ for the topology of
convergence in measure.   Show that any convex function $f:C\to\Bbb R$
which is lower semi-continuous for the topology of convergence in
measure is bounded below and attains its infimum.
%367V

\spheader 367Xz Let $(\frak A,\bar\mu)$ be the measure algebra of
Lebesgue measure on $[0,1]$.   Show that there are a sequence
$\sequencen{u_n}$ in $L^1=L^1_{\bar\mu}$ and $u$, $v\in L^1$ such
that $u_n$ and $v$ are independent for every $n$, $\sequencen{u_n}$
converges weakly to $u$, but $u$ and $v$ are not independent.
%367W

\leaveitout{\spheader 367X? Let $(\frak A,\bar\mu)$ be a totally finite
measure algebra.   Show that $(\alpha,u)\mapsto\bar\mu\Bvalue{u>\alpha}:
\Bbb R\times L^0\to\Bbb R$ is lower semi-continuous if we give 
$L^0=L^0(\frak A)$ its topology of convergence in measure.

can we find room for this sometime?  \query}

\leader{367Y}{Further exercises (a)}
%\spheader 367Ya
Give an example of an Archimedean Riesz space $U$ and an
order-bounded sequence $\sequencen{u_n}$ in $U$ which is
order*-convergent to $0$, but such that there is no non-increasing
sequence $\sequencen{v_n}$, with infimum $0$, such that $u_n\le v_n$ for
every $n\in\Bbb N$.
%367A mt36bits

\spheader 367Yb Let $P$ be any lattice.  (i) Show that there is a
topology on $P$ for which a set $A\subseteq P$ is closed iff $p\in A$
whenever there is a sequence in $A$ which is order*-convergent to
$p$.   Show that any closed set for this topology is sequentially
order-closed.   (ii) Now let $Q$ be another lattice, with the
topology defined in the same way, and $f:P\to Q$ an order-preserving
function.  Show that if $f$ is topologically
continuous it is sequentially order-continuous.
%367A

\spheader 367Yc Give an example of a distributive lattice $P$ with
$p$, $q\in P$ and a sequence $\sequencen{p_n}$, order*-convergent to
$p$, such that $\sequencen{p_n\wedge q}$ is not order*-convergent to
$p\wedge q$.
%367B mt36bits

\spheader 367Yd Let us say that a lattice $P$ is {\bf
$(2,\infty)$-distributive} if ($\alpha$) whenever $A$, $B\subseteq P$
are non-empty sets with infima $p$, $q$ respectively, then $\inf\{a\vee
b:a\in A,\,b\in B\}=p\vee q$
($\beta$) whenever $A$, $B\subseteq P$ are non-empty sets with suprema
$p$, $q$ respectively, then $\sup\{a\wedge b:a\in A,\,b\in B\}=p\wedge
q$.   Show that, in this case, if $\sequencen{p_n}$ order*-converges
to $p$ and $\sequencen{q_n}$ order*-converges to $q$,
$\sequencen{p_n\vee q_n}$ order*-converges to $p\vee q$.
%367C

\spheader 367Ye(i) Give an example of a Riesz space $U$ with an
order-dense Riesz subspace $V$ of $U$ and a sequence $\sequencen{v_n}$
in $V$ such that
$\sequencen{v_n}$ order*-converges to $0$ in $V$ but does not
order*-converge in $U$.
(ii) Give an example of a Riesz space $U$ with an order-dense Riesz
subspace $V$ of $U$ and a sequence $\sequencen{v_n}$ in $V$,
order-bounded in $V$, such that
$\sequencen{v_n}$ order*-converges to $0$ in $U$ but does not
order*-converge in $V$.
%367E mt36bits

\spheader 367Yf Let $U$ be an Archimedean $f$-algebra.   Show that if
$\sequencen{u_n}$, $\sequencen{v_n}$ are sequences in $U$
order*-converging to $u$, $v$ respectively, then
$\sequencen{u_n\times v_n}$ order*-converges to $u\times v$.
%367F, 367Xl

\spheader 367Yg Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $r\ge 1$.   Let $E\subseteq\BbbR^r$ be a Borel set and
write $Q_E=\{(u_1,\ldots,u_r):\Bvalue{(u_1,\ldots,u_r)\in E}=1\}
\subseteq L^0(\frak A)^r$ (364Yb).   Let $h:E\to\Bbb R$ be a
continuous function and $\bar h:Q_E\to L^0=L^0(\frak A)$ the
corresponding map
(364Yc).   Show that if $\sequencen{\tbf{u}_n}$ is a sequence in $Q_E$
which is order*-convergent to $\tbf{u}\in Q_E$ (in the lattice
$(L^0)^r$), then $\sequencen{\bar h(\tbf{u}_n)}$ is
order*-convergent to $\bar h(\tbf{u})$.
%367G

\spheader 367Yh Let $X$ be a completely regular Baire space (definition:
314Yd), and $\sequencen{u_n}$ a sequence in $C(X)$.   Show
that $\sequencen{u_n}$ order*-converges to $0$ in $C(X)$ iff
$\{x:\limsup_{n\to\infty}|u_n(x)|>0\}$ is meager in $X$.
%367J

\spheader 367Yi(i) Give an example of a sequence $\sequencen{u_n}$ in
$C([0,1])$ such that $\lim_{n\to\infty}u_n(x)=0$ for every $x\in[0,1]$,
but $\{u_n:n\in\Bbb N\}$ is not order-bounded in $C([0,1])$.   (ii) Give
an example of an order-bounded sequence $\sequencen{u_n}$ in $C(\Bbb Q)$
such that $\lim_{n\to\infty}u_n(q)=0$ for every $q\in\Bbb Q$, but
$\sup_{i\ge
n}u_i=\chi\Bbb Q$ in $C(\Bbb Q)$ for every $n\in\Bbb N$.   (iii) Give an
example of a sequence $\sequencen{u_n}$ in $C([0,1])$ such that
$\sequencen{u_n}$ order*-converges to $0$ in $C([0,1])$, but
$\lim_{n\to\infty}u_n(q)>0$ for every $q\in\Bbb Q\cap[0,1]$.
%367Yh, 367J

\spheader 367Yj Write out an alternative proof of 367J/367Yh based on
the fact that, for a Baire space $X$, $C(X)$ can be identified with an
order-dense Riesz subspace of a quotient of the space of
$\widehat{\Cal B}$-measurable functions, where $\widehat{\Cal B}$ is the
Baire-property algebra of $X$.
%367J, 367Yh

\spheader 367Yk Let $\frak A$ be a ccc \wsid\
Boolean algebra.   Show that there is a topology on $\frak A$ such that
the closure of any $A\subseteq\frak A$ is precisely the set of limits of
order*-convergent sequences in $A$.
%367K see \S393

\spheader 367Yl Give an example of a set $X$ and a double sequence
$\langle u_{mn}\rangle_{m,n\in\Bbb N}$ in $\Bbb R^X$ such that
$\lim_{n\to\infty}u_{mn}(x)=u_m(x)$ exists for every $m\in\Bbb N$ and
$x\in X$, $\lim_{m\to\infty}u_m(x)=0$ for every $x\in X$, but there is
no sequence $\sequence{k}{v_k}$ in $\{u_{mn}:m,\,n\in\Bbb N\}$ such that
$\lim_{k\to\infty}v_k(x)=0$ for every $x$.
%367Yk, 367K

\spheader 367Ym Let $U$ be a Riesz space with a Riesz norm $\|\,\|$.
For $v\in U^+$ define $\tau_v:U\to\coint{0,\infty}$ by
setting $\tau_v(u)=\||u|\wedge v\|$ for every $u\in U$.
Show that every $\tau_v$ is an F-seminorm on $U$,
and that $\{\tau_v:v\in U^+\}$ defines a Hausdorff linear space
topology on $U$.
%367K

\spheader 367Yn Let $U$ be any Riesz space.   For $h\in U^{\sim+}$
(356Ab) and $v\in U^+$ define $\tau_{vh}:U\to\coint{0,\infty}$ by
setting $\tau_{vh}(u)=h(|u|\wedge v)$ for every $u\in U$.
Show that $\tau_{vh}$ is an F-seminorm on $U$.
%367K

\spheader 367Yo Let $(\frak A,\bar\mu)$ be a $\sigma$-finite measure
algebra.   Show that the function
$(\alpha,u)\mapsto\Bvalue{u>\alpha}:\Bbb R\times L^0\to\frak A$ is Borel
measurable when $L^0=L^0(\frak A)$ is given the topology of convergence
in measure and $\frak A$ is given its measure-algebra topology.
%367Me 36bits

\spheader 367Yp Let $\frak G$ be the regular open algebra of $\Bbb R$.
Show that there is no Hausdorff topology $\frak T$ on $L^0(\frak G)$
such that $\sequencen{u_n}$ is $\frak T$-convergent to $u$ whenever
$\sequencen{u_n}$ is order*-convergent to $u$.
\leaveitout{\Hint{Let $H$ be any
$\frak T$-open set containing $0$.   Enumerate $\Bbb Q$ as
$\sequencen{q_n}$.   Find inductively a non-decreasing sequence
$\sequencen{G_n}$ in $\frak G$ such that $\chi G_n\in H$, $q_n\in G_n$
for every $n$.   Conclude that $\chi\Bbb R\in\overline{H}$.}}
%367P

\spheader 367Yq In 367Qc, show that
$u=\lim_{\frak B\to\Cal F(\Bbb B\closeuparrow)}u_{\frak B}$ for the norm
topology of $L^1$ iff $\{u_{\frak B}:\frak B\in\Bbb B\}$
is uniformly integrable, and that in this case
$u_{\frak B}=P_{\frak B}u$ for every $\frak B\in\Bbb B$.
%367Q

\spheader 367Yr Let $\frak A$ be a Dedekind $\sigma$-complete Boolean
algebra and $r\ge 1$.   Let $E\subseteq\BbbR^r$ be a Borel set and
write $Q_E=\{(u_1,\ldots,u_r):\Bvalue{(u_1,\ldots,u_r)\in E}=1\}
\subseteq L^0(\frak A)^r$, as in 367Yg.   Let $h:E\to\Bbb R$ be a
continuous function and $\bar h:Q_E\to L^0=L^0(\frak A)$ the
corresponding map.   Show that if $\bar h$ is continuous if $L^0$ is
given its
topology of convergence in measure and $(L^0)^r$ the product topology.
%367S

\spheader 367Ys Show that 367U is true for all measure algebras,
whether semi-finite or not.
%367U

\spheader 367Yt\dvAnew{2018}
Let $(\frak A,\bar\mu)$ be a measure algebra.   For
$a\in\frak A^f$, $n\in\Bbb N$ and $u$, $v\in L^0(\frak A)$, set
$\rho_{an}(u,v)=\int_{-n}^n
  \bar\mu(a\Bcap(\Bvalue{u>\alpha}\Bsymmdiff\Bvalue{v>\alpha}))d\alpha$, 
the integral being with respect to Lebesgue measure.   (i) Show that the
integral is always defined.   (ii) Show that each 
$\rho_{an}:L^0(\frak A)\times L^0(\frak A)\to\coint{0,\infty}$ is a
pseudometric.   (iii) Show that $\{\rho_{an}:a\in\frak A^f$, $n\in\Bbb N\}$
defines the topology of convergence in measure on $L^0(\frak A)$.
}%end of exercises

\cmmnt{
\Notesheader{367} I have given a very general definition of
`order*-convergence'.   The general theory of convergence structures
on ordered spaces is complex and has many traps for the unwary.   I have
tried to lay out a safe path to
the results which are important in the context of this book.   But the
propositions here are necessarily full of little conditions (e.g.,
the requirement that $U$ should be Archimedean in 367E) whose
significance may not be immediately obvious.   In particular, the
definition is very much better adapted to distributive lattices than to
others (367Yc, 367Yd).   It is useful in the study of Riesz
spaces and Boolean algebras largely because these satisfy strong
distributive laws (313B, 352E).   The special feature which
distinguishes the definition here from other definitions of
order-convergence is the fact that it can be applied to sequences which
are not order-bounded.   For order-bounded sequences there are useful
simplifications (367Be-f), but the Martingale Theorem (367J), for
instance, if we want to express it in terms of its natural home in the
Riesz space
$L^1$, refers to sequences which are hardly ever order-bounded.

The * in the phrase `order*-convergent' is supposed to be a
warning that it may not represent exactly the concept you expect.   I
think nearly any author using the phrase `order-convergent' would accept
sequences fulfilling the conditions of 367Bf;  but beyond this no
standard definitions have taken root.

The fact that order*-convergent sequences in an $L^0$ space are
order-bounded (367G) is actually one of the characteristic properties
of $L^0$.   Related ideas will be important in the next section (368A,
368M).

It is one of the outstanding characteristics of measure algebras in this
context that they provide non-trivial linear space topologies on their
$L^0$ spaces, related in striking ways to the order structure.   Not all
$L^0$ spaces have such topologies (367Yp).   A
topology corresponding to `convergence in measure' can be defined
on $L^0(\frak A)$ for any Maharam algebra $\frak A$;  see 393K below.

367T shows that the topology of convergence in measure on $L^0(\frak A)$
is (at least for semi-finite measure algebras) determined by the Riesz
space structure of $L^0$;  and that indeed the same is true of its
order-dense Riesz subspaces.   This fact is important for a full
understanding of the representation theorems in \S369 below.   If a
Riesz space $U$ can be embedded as an order-dense subspace of any such
$L^0$, then there is already a `topology of convergence in measure'
on $U$, independent of the embedding.   It is therefore not surprising
that there should be alternative descriptions of the topology of
convergence in measure on the important subspaces of $L^0$
(367Xq, 367Ym).

For $\sigma$-finite measure algebras, the topology of convergence in
measure is easily described in terms of order-convergence (367P).
For other measure algebras, the formula fails (367Xu(iii)).
367Yp shows that trying to apply the same ideas to Riesz spaces in
general gives rise to some very curious phenomena.

367V enables us to prove results which would ordinarily be associated
with some form of compactness.   Of course compactness is indeed
involved, as the proof through 367U makes clear;  but it is weak*
compactness in $(L^1)^{**}$, rather than in the space immediately to
hand.

I hardly mention `uniform integrability' in this section, not because
it is uninteresting, but because I have nothing to add at this point to
246J and the exercises in \S246.   But I do include translations of
Lebesgue's Dominated Convergence Theorem (367I) and the Martingale
Theorem (367J) to show how these can be expressed in the language of
this chapter.
}%end of comment

\discrpage

\frfilename{mt368.tex}
\versiondate{16.9.09}
\copyrightdate{1997}

\def\chaptername{Function spaces}
\def\sectionname{Embedding Riesz spaces in $L^0$}

\newsection{368}

In this section I turn to the representation of Archimedean Riesz spaces
as function spaces.   Any Archimedean Riesz space $U$ can be represented
as an order-dense subspace of $L^0(\frak A)$, where $\frak A$ is its
band algebra (368E).   Consequently we get representations of
Archimedean Riesz
spaces as quotients of subspaces of $\Bbb R^X$ (368F) and as subspaces
of $C^{\infty}(X)$ (368G), and a notion of `Dedekind completion'
(368I-368J).   Closely associated with these is the fact that we have a
very general extension theorem for order-continuous Riesz homomorphisms
into $L^0$ spaces (368B).   I give a characterization of $L^0$
spaces in terms of lateral completeness (368M\cmmnt{, 368Yd}), and I
discuss \wsid\ Riesz spaces (368N-368S).


\leader{368A}{Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra, and $A\subseteq(L^0)^+$ a set with no upper bound in
$L^0$, where
$L^0=L^0(\frak A)$.   If {\it either} $A$ is countable {\it or}
$\frak A$ is Dedekind complete, there is a $v>0$ in $L^0$ such that
$nv=\sup_{u\in A}u\wedge nv$ for every $n\in\Bbb N$.

\proof{ The hypothesis `$A$ is countable or $\frak A$ is Dedekind
complete' ensures that $c_{\alpha}=\sup_{u\in A}\Bvalue{u>\alpha}$ is
defined for each $\alpha$.   By 364L(a-i),
$c=\inf_{n\in\Bbb N}c_n=\inf_{\alpha\in\Bbb R}c_{\alpha}$ is non-zero.
Now for any $n\ge 1$, $\alpha\in\Bbb R$

\Centerline{$\Bvalue{\sup_{u\in A}(u\wedge n\chi c)>\alpha}
=\sup_{u\in A}\Bvalue{u>\alpha}\Bcap\Bvalue{\chi c>\bover{\alpha}n}
=\Bvalue{\chi c>\bover{\alpha}n},$}

\noindent because if $\alpha\ge 0$ then

\Centerline{$\sup_{u\in A}\Bvalue{u>\alpha}
=c_{\alpha}\Bsupseteq c\Bsupseteq\Bvalue{\chi c>\bover{\alpha}n}$,}

\noindent while if $\alpha<0$ then (because $A$ is a non-empty subset of
$(L^0)^+$)

\Centerline{$\sup_{u\in A}\Bvalue{u>\alpha}=1
=\Bvalue{\chi c>\bover{\alpha}n}$.}

\noindent So $\sup_{u\in A}u\wedge n\chi c=n\chi c$ for every $n\ge 1$,
and
we can take $v=\chi c$.   (The case $n=0$ is of course trivial.)
}%end of proof of 368A

\leader{368B}{Theorem} Let $\frak A$ be a Dedekind complete Boolean
algebra, $U$ an Archimedean Riesz space, $V$ an order-dense Riesz
subspace of $U$ and $T:V\to L^0=L^0(\frak A)$ an order-continuous Riesz
homomorphism.   Then $T$ has a unique extension to an
order-continuous Riesz homomorphism $\tilde T:U\to L^0$.
\proof{{\bf (a)}  The key to the proof is the following:  if $u\ge 0$ in
$U$, then $\{Tv:v\in V,\,0\le v\le u\}$ is bounded above in $L^0$.
\Prf\Quer\ Suppose, if possible, otherwise.   Then by 368A there is a
$w>0$ in $L^0$ such that $nw=\sup_{v\in A}nw\wedge Tv$ for every
$n\in\Bbb N$,
where $A=\{v:v\in V,\,0\le v\le u\}$.   In particular, there is a
$v_0\in A$ such that $w_0=w\wedge Tv_0>0$.   Because $U$ is Archimedean,
$\inf_{k\ge 1}\bover1ku=0$, so
$v_0=\sup_{k\ge 1}(v_0-\bover1ku)^+$.
Because $V$ is order-dense in $U$, $v_0=\sup B$ where

\Centerline{$B
=\{v:v\in V,\,0\le v\le (v_0-\Bover1ku)^+$ for some $k\ge 1\}$.}

\noindent Because $T$ is order-continuous, $Tv_0=\sup T[B]$ in $L^0$,
and there is a $v_1\in B$ such that $w_1=w_0\wedge Tv_1>0$.   Let
$k\ge 1$
be such that $v_1\le (v_0-\bover1ku)^+$.   Then for any $m\in\Bbb N$,

$$\eqalignno{mv_1\wedge u
&\le(mv_1\wedge kv_0)+(mv_1\wedge(u-kv_0)^+)\cr
\displaycause{352F(a-i)}
&\le kv_0+(m+k)(v_1\wedge(\Bover1ku-v_0)^+)
=kv_0.\cr}$$

\noindent So for any $v\in A$, $m\in\Bbb N$,

\Centerline{$mw_1\wedge Tv=mw_1\wedge mTv_1\wedge Tv
\le T(mv_1\wedge v)\le T(mv_1\wedge u)\le T(kv_0)=kTv_0$.}

\noindent But this means that, for $m\in\Bbb N$,

\Centerline{$mw_1=mw_1\wedge mw=\sup_{v\in A}mw_1\wedge(mw\wedge Tv)
=\sup_{v\in A}mw_1\wedge Tv\le kTv_0$,}

\noindent which is impossible because $L^0$ is Archimedean
and $w_1>0$.\ \Bang\Qed

\medskip

{\bf (b)}  Because $L^0$ is Dedekind complete,
$\sup\{Tv:v\in V,\,0\le v\le u\}$ is defined in $L^0$ for every
$u\in U$.   By 355F, $T$ has a
unique extension to an order-continuous Riesz homomorphism from $U$ to
$L^0$.
}%end of proof of 368B

\leader{368C}{Corollary} Let $\frak A$ and $\frak B$ be Dedekind
complete Boolean algebras and $U$, $V$ order-dense Riesz subspaces of
$L^0(\frak A)$, $L^0(\frak B)$ respectively.   Then any Riesz space
isomorphism
between $U$ and $V$ extends uniquely to a Riesz space isomorphism
between $L^0(\frak A)$ and $L^0(\frak B)$;  and in this case $\frak A$
and $\frak B$ must be isomorphic as Boolean algebras.

\proof{ If $T:U\to V$ is a Riesz space isomorphism, then 368B tells us
that we have (unique)
order-continuous Riesz homomorphisms $\tilde T:L^0(\frak A)\to L^0(\frak B)$
and $\tilde T':L^0(\frak B)\to L^0(\frak A)$ extending $T$, $T^{-1}$
respectively.   Now $\tilde T'\tilde T:L^0(\frak A)\to L^0(\frak A)$ is an
order-continuous Riesz homomorphism agreeing with the identity on $U$,
so must be the identity on $L^0(\frak A)$;  similarly $\tilde T\tilde T'$
is the
identity on $L^0(\frak B)$, and $\tilde T$ is a Riesz space isomorphism.
To see that $\frak A$ and $\frak B$ are isomorphic, recall that by 364O
they can be identified with the algebras of projection bands of
$L^0(\frak A)$ and $L^0(\frak B)$, which must be isomorphic.
}%end of proof of 368C

\leader{368D}{Corollary} Suppose that $\frak A$ is a Dedekind
$\sigma$-complete Boolean algebra, and that $U$ is an order-dense Riesz
subspace of $L^0(\frak A)$ which is isomorphic, as Riesz space, to
$L^0(\frak B)$ for some Dedekind complete Boolean algebra $\frak B$.
Then $U=L^0(\frak A)$ and $\frak A$ is isomorphic to $\frak B$ (so, in
particular, is Dedekind complete).

\proof{ The identity mapping $U\to U$ is surely an order-continuous
Riesz homomorphism, so by 368B extends to an order-continuous Riesz
homomorphism
$\tilde T:L^0(\frak A)\to U$.   Now $\tilde T$ must be injective,
because if $u\ne 0$ in $L^0(\frak A)$ there is a $u'\in U$ such that
$0<u'\le|u|$,
so that $0<u'\le|\tilde Tu|$.   So we must have $U=L^0(\frak A)$ and
$\tilde T$ the identity map.   By 364O again, $\frak A\cong\frak B$.
}%end of proof of 368D

\leader{368E}{Theorem} Let $U$ be any Archimedean Riesz space, and
$\frak A$ its band algebra\cmmnt{ (353B)}.   Then $U$ can be
embedded as an order-dense Riesz subspace of $L^0(\frak A)$.

\proof{{\bf (a)} If $U=\{0\}$ then $\frak A=\{0\}$,
$L^0=L^0(\frak A)=\{0\}$ and the result is trivial;  I shall therefore
suppose henceforth
that $U$ is non-trivial.   Note that by 352Q $\frak A$ is Dedekind
complete.

Let $C\subseteq U^+\setminus\{0\}$ be a maximal disjoint set (in the
sense of 352C);  to obtain such a set apply Zorn's lemma to the family
of all disjoint subsets of $U^+\setminus\{0\}$.   Now I can write down
the formula for the
embedding $T:U\to L^0$ immediately, though there will be a good deal of
work to do in justification:  for $u\in U$ and $\alpha\in\Bbb R$,
$\Bvalue{Tu>\alpha}$ will be the band in $U$ generated by

\Centerline{$\{e\wedge(u-\alpha e)^+: e\in C\}$.}

\noindent (For once, I allow myself to use the formula $\Bvalue{\ldots}$
without checking immediately that it represents a member of $L^0$;  all
I
claim for the moment is that $\Bvalue{Tu>\alpha}$ is a member of $\frak
A$
determined by $u$ and $\alpha$.)

\medskip

{\bf (b)} Before getting down to the main argument, I make some remarks
which will be useful later.

\medskip

\quad{\bf (i)} If $u>0$ in $U$, then there is some $e\in C$ such that
$u\wedge e>0$, since otherwise we ought to have added $u$ to $C$.   Thus
$C^{\perp}=\{0\}$.
\medskip

\quad{\bf (ii)} If $u\in U$ and $e\in C$ and $\alpha\in\Bbb R$, then
$v=e\wedge(\alpha e-u)^+$ belongs to $\Bvalue{Tu>\alpha}^{\perp}$.
\Prf\
If $e'\in C$, then either $e'\ne e$ so

\Centerline{$v\wedge e'\wedge(u-\alpha e')^+\le e\wedge e'=0$,}

\noindent or $e'=e$ and

\Centerline{$v\wedge e'\wedge(u-\alpha e')^+
\le (\alpha e-u)^+\wedge(u-\alpha e)^+=0$.}

\noindent   Accordingly $\Bvalue{Tu>\alpha}$ is included in the band
$\{v\}^{\perp}$ and $v\in\Bvalue{Tu>\alpha}^{\perp}$.\ \Qed

\medskip

{\bf (c)} Now I must confirm that the formula given for
$\Bvalue{Tu>\alpha}$ is consistent with the conditions laid down in
364Aa.   \Prf\ Take $u\in U$.

\medskip

\quad{\bf (i)} If $\alpha\le\beta$ then

\Centerline{$0\le e\wedge(u-\beta e)^+
\le e\wedge(u-\alpha e)^+\in\Bvalue{Tu>\alpha}$}

\noindent so $e\wedge(u-\beta e)^+\in\Bvalue{Tu>\alpha}$, for every
$e\in C$, and $\Bvalue{Tu>\beta}\subseteq\Bvalue{Tu>\alpha}$.

\medskip

\quad{\bf (ii)} Given
$\alpha\in\Bbb R$, set $W=\sup_{\beta>\alpha}\Bvalue{Tu>\beta}$ in
$\frak A$, that is, the band in $U$ generated by
$\{e\wedge(u-\beta e)^+:e\in C,\,\beta>\alpha\}$.   Then for each
$e\in C$,

\Centerline{$\sup_{\beta>\alpha}e\wedge(u-\beta e)^+
=e\wedge(u-\inf_{\beta>\alpha}\beta e)^+
=e\wedge(u-\alpha e)^+$}

\noindent using the general distributive laws in $U$ (352E), the
translation-invariance of the order (351D) and the fact that $U$ is
Archimedean (to see that $\alpha e=\inf_{\beta>\alpha}\beta e$).   So
$e\wedge(u-\alpha e)^+\in W$;  as $e$ is arbitrary,
$\Bvalue{Tu>\alpha}\subseteq W$ and $\Bvalue{Tu>\alpha}=W$.

\medskip

\quad{\bf (iii)} Now set $W=\inf_{n\in\Bbb N}\Bvalue{Tu>n}$.   For any
$e\in C$, $n\in\Bbb N$ we have

\Centerline{$e\wedge(ne-u)^+\in\Bvalue{Tu>n}^{\perp}\subseteq
W^{\perp}$,}

\noindent so that

\Centerline{$e\wedge(e-\bover1nu^+)^+\le e\wedge(e-\bover1nu)^+\in
W^{\perp}$}

\noindent for every $n\ge 1$ and

\Centerline{$e=\sup_{n\ge 1}e\wedge(e-\bover1nu^+)^+\in W^{\perp}$. }

\noindent Thus $C\subseteq W^{\perp}$ and $W\subseteq C^{\perp}=\{0\}$.
So we have $\inf_{n\in\Bbb N}\Bvalue{Tu>n}=0$.

\medskip

\quad{\bf (iv)} Finally, set
$W=\sup_{n\in\Bbb N}\Bvalue{Tu>-n}$.   Then

\Centerline{$e\wedge(e-\bover1nu^-)^+\le e\wedge(e+\bover1nu)^+\le
e\wedge(u+ne)^+\in W$}

\noindent for every $n\ge 1$ and $e\in C$, so

\Centerline{$e=\sup_{n\ge 1}e\wedge(e-\bover1nu^-)^+\in W$}

\noindent for every $e\in C$ and $W^{\perp}=\{0\}$, $W=U$.   Thus all
three conditions of 364Aa are satisfied.\ \Qed

\medskip

{\bf (d)} Thus we have a well-defined map $T:U\to L^0$.
I show next that $T(u+v)=Tu+Tv$ for all $u$, $v\in U$.   \Prf\ I rely on
the formulae in 364D and 364Ea, and on partitions of unity in $\frak A$,
constructed as follows.   Fix $n\ge 1$ for the moment.   Then we know
that

\Centerline{$\sup_{i\in\Bbb Z}\Bvalue{Tu>\bover{i}{n}}=1$,
\quad$\inf_{i\in\Bbb Z}\Bvalue{Tu>\bover{i}{n}}=0$.}

\noindent So setting

\Centerline{$V_i
=\Bvalue{Tu>\bover{i}n}\Bsetminus\Bvalue{Tu>\bover{i+1}n}
=\Bvalue{Tu>\bover{i}n}\cap\Bvalue{Tu>\bover{i+1}n}^{\perp}$,}

\noindent $\langle V_i\rangle_{i\in\Bbb Z}$ is a partition of unity in
$\frak A$.   Similarly, $\langle W_i\rangle_{i\in\Bbb Z}$ is a partition
of unity, where

\Centerline{$W_i
=\Bvalue{Tv>\bover{i}n}\cap\Bvalue{Tv>\bover{i+1}n}^{\perp}$.}

\noindent Now, for any $i$, $j$, $k\in\Bbb Z$ such that $i+j\ge k$,

\Centerline{$V_i\cap W_j
\subseteq\Bvalue{Tu>\bover{i}n}\cap\Bvalue{Tv>\bover{j}n}
\subseteq\Bvalue{Tu+Tv>\bover{i+j}n}
\subseteq\Bvalue{Tu+Tv>\bover{k}n}$;}

\noindent thus

\Centerline{$\Bvalue{Tu+Tv>\bover{k}n}
\supseteq\sup_{i+j\ge k}V_i\cap W_j$.}

\noindent On the other hand, if $q\in\Bbb Q$ and $k\in\Bbb Z$, there is
an $i\in\Bbb Z$ such that $\bover{i}n\le q<\bover{i+1}n$, so that

\Centerline{$\Bvalue{Tu>q}\cap\Bvalue{Tv>\bover{k+1}n-q}
\subseteq\Bvalue{Tu>\bover{i}n}\cap\Bvalue{Tv>\bover{k-i}n}
\subseteq\sup_{i+j\ge k}V_i\cap W_j$;}

\noindent thus for any $k\in\Bbb Z$
\Centerline{$\Bvalue{Tu+Tv>\bover{k+1}n}\subseteq\sup_{i+j\ge k}V_i\cap
W_j\subseteq\Bvalue{Tu+Tv>\bover{k}n}$.}


Also, if $0<w\in V_i\cap W_j$ and $e\in C$ then

\Centerline{$w\wedge e\wedge(u-\Bover{i+1}ne)^+
=w\wedge e\wedge(v-\Bover{j+1}ne)^+=0$,}

\noindent so that

\Centerline{$w\wedge e\wedge(u+v-\Bover{i+j+2}ne)^+=0$}

\noindent because

\Centerline{$(u+v-\Bover{i+j+2}ne)^+
\le(u-\Bover{i+1}ne)^++(v-\Bover{j+1}ne)^+$}

\noindent by 352Fc.   But this means that
$V_i\cap W_j\cap\Bvalue{T(u+v)>\bover{i+j+2}n}=\{0\}$.   Turning this
round,

\Centerline{$\Bvalue{T(u+v)>\bover{k+1}n}\cap\sup_{i+j\le k-1}V_i\cap
W_j=0$,}

\noindent and because $\sup_{i,j\in\Bbb Z}V_i\cap W_j=U$ in $\frak A$,

\Centerline{$\Bvalue{T(u+v)>\bover{k+1}n}\subseteq\sup_{i+j\ge k}V_i\cap
W_j$.}

\noindent Finally, if $i+j\ge k$ and $0<w\in V_i\cap V_j$, then there is
an $e\in C$ such that $w_1=w\wedge e\wedge(u-\bover{i}ne)^+>0$;  there
is an $e'\in C$ such that $w_2=w_1\wedge e'\wedge(v-\bover{j}ne')^+>0$;
of course $e=e'$, and

$$\eqalign{0
<w_2
\le e\wedge(u-\Bover{i}ne)^+\wedge(v-\Bover{j}n)^+
&\le e\wedge(u+v-\Bover{i+j}ne)^+\cr
&\in\Bvalue{T(u+v)>\Bover{i+j}n}
\subseteq\Bvalue{T(u+v)>\Bover{k}n}\cr}$$

\noindent using 352Fc again.   This shows that
$w\not\in\Bvalue{T(u+v)>\bover{k}n}^{\perp}$;  as $w$ is arbitrary,
$V_i\cap W_j\subseteq\Bvalue{T(u+v)>\bover{k}n}$;  so we get

\Centerline{$\sup_{i+j\ge k}V_i\cap W_j\subseteq
\Bvalue{T(u+v)>\bover{k}{n}}$.}

Putting these four facts together, we see that

\Centerline{$\Bvalue{T(u+v)>\bover{k+1}n}\subseteq\sup_{i+j\ge k}V_i\cap
W_j\subseteq\Bvalue{Tu+Tv>\bover{k}n}$,}
\Centerline{$\Bvalue{Tu+Tv>\bover{k+1}n}\subseteq\sup_{i+j\ge k}V_i\cap
W_j\subseteq\Bvalue{T(u+v)>\bover{k}n}$}

\noindent for all $n\ge 1$ and $k\in\Bbb Z$.   But this means that we must
have

\Centerline{$\Bvalue{T(u+v)>\beta}\subseteq\Bvalue{Tu+Tv>\alpha}$,
\quad$\Bvalue{Tu+Tv>\beta}\subseteq\Bvalue{T(u+v)>\alpha}$}

\noindent whenever $\alpha<\beta$.   Consequently

$$\eqalign{\Bvalue{Tu+Tv>\alpha}
&=\sup_{\beta>\alpha}\Bvalue{Tu+Tv>\beta}
\subseteq\Bvalue{T(u+v)>\alpha}\cr
&=\sup_{\beta>\alpha}\Bvalue{T(u+v)>\beta}
\subseteq\Bvalue{Tu+Tv>\alpha}\cr}$$

\noindent and $\Bvalue{Tu+Tv>\alpha}=\Bvalue{T(u+v)>\alpha}$ for every
$\alpha$, that is, $T(u+v)=Tu+Tv$.\ \Qed

\medskip

{\bf (e)} The hardest part is over.   If $u\in U$, $\gamma>0$ and
$\alpha\in\Bbb R$, then for any $e\in C$

\Centerline{$\min(1,\Bover1{\gamma})(e\wedge(\gamma u-\alpha e)^+)
\le e\wedge(u-\Bover{\alpha}{\gamma}e)^+
\le\max(1,\Bover1{\gamma})(e\wedge(\gamma u-\alpha e)^+)$,}

\noindent so

\Centerline{$\Bvalue{T(\gamma
u)>\alpha}=\Bvalue{Tu>\bover{\alpha}{\gamma}}
=\Bvalue{\gamma Tu>\alpha}$;}

\noindent as $\alpha$ is arbitrary, $\gamma Tu=T(\gamma u)$;  as
$\gamma$ and $u$ are arbitrary, $T$ is linear.   (We need only check
linearity for
$\gamma>0$ because we know from the additivity of $T$ that $T(-u)=-Tu$
for every $u$.)

\medskip

{\bf (f)} To see that $T$ is a Riesz homomorphism, take any $u\in U$ and
$\alpha\in\Bbb R$ and consider the band
$\Bvalue{Tu>\alpha}\Bcup\Bvalue{-Tu>\alpha}\penalty-100=\Bvalue{|Tu|>\alpha}$ (by
364L(a-ii)).    This is the band generated by
$\{e\wedge(u-\alpha e)^+:e\in C\}\cup\{e\wedge(-u-\alpha e)^+:e\in C\}$.
But this must also be the band generated by

\Centerline{$\{(e\wedge(u-\alpha e)^+)\vee(e\wedge(-u-\alpha e)^+):
e\in C\}=\{e\wedge(|u|-\alpha e)^+:e\in C\}$,}

\noindent which is $\Bvalue{T|u|>\alpha}$.   Thus
$\Bvalue{|Tu|>\alpha}=\Bvalue{T|u|>\alpha}$ for every $\alpha$ and
$|Tu|=T|u|$.   As $u$ is arbitrary, $T$ is a Riesz homomorphism.

\medskip

{\bf (g)} To see that $T$ is injective, take any non-zero $u\in U$.
Then
there must be some $e\in C$ such that $|u|\wedge e\ne 0$, and some
$\alpha>0$ such that $|u|\wedge e\not\le\alpha e$, so that
$e\wedge(|u|-\alpha e)^+\ne 0$ and $\Bvalue{T|u|>\alpha}\ne\{0\}$ and
$T|u|\ne 0$ and $Tu\ne 0$.

Thus $T$ embeds $U$ as a Riesz subspace of $L^0$.

\medskip

{\bf (h)} Finally, I must check that $T[U]$ is order-dense in $L^0$.
\Prf\ Let $p>0$ in $L^0$.   Then there is some $\alpha>0$ such that
$V=\Bvalue{p>\alpha}\ne 0$.   Take $u>0$ in $V$.   Let $e\in C$ be such
that $u\wedge e>0$.   Then $v=u\wedge\alpha e>0$.   Now
$e\wedge(v-\alpha e)^+=0$;  but also $e'\wedge v=0$ for every $e'\in C$
distinct from $e$, so that $\Bvalue{Tv>\alpha}=\{0\}$.   Next, $v\in V$,
so $e'\wedge(v-\beta e')^+\in V$ whenever $e'\in C$ and $\beta\ge 0$,
and $\Bvalue{Tv>\beta}\subseteq V$ for every $\beta\ge 0$.   Accordingly
we have

$$\eqalign{\Bvalue{Tv>\beta}&=\{0\}\subseteq\Bvalue{p>\beta}
  \text{ if }\beta\ge\alpha,\cr
&\subseteq V\subseteq\Bvalue{p>\beta}
\text{ if }0\le\beta<\alpha,\cr
&=U=\Bvalue{p>\beta}
\text{ if }\beta<0,\cr}$$

\noindent and $Tv\le p$.   Also $Tv>0$, by (g).   As $p$ is arbitrary,
$T[U]$ is order-dense in $L^0$.\ \Qed
}%end of proof of 368E

\leader{368F}{Corollary} A Riesz space $U$ is Archimedean iff it is
isomorphic to a Riesz subspace of some reduced power $\Bbb R^X|\Cal F$,
where $X$ is a set and $\Cal F$ is a filter on $X$ such that
$\bigcap_{n\in\Bbb N}F_n\in\Cal F$ whenever $\sequencen{F_n}$ is a
sequence in $\Cal F$.

\proof{{\bf (a)} If $U$ is an Archimedean Riesz space, then by 368E
there is a space of the form $L^0=L^0(\frak A)$ such that $U$ can be
embedded into $L^0$.   As in the proof of 364D, $L^0$ is isomorphic to
some space of the form $\eusm L^0(\Sigma)/\eusm W$, where $\Sigma$ is a
$\sigma$-algebra of subsets of some set $X$ and
$\eusm W=\{f:f\in\eusm L^0,\,\{x:f(x)\ne 0\}\in\Cal I\}$, $\Cal I$ being
a $\sigma$-ideal of
$\Sigma$.   But now $\Cal F=\{A:A\cup E=X$ for some $E\in\Cal I\}$ is a
filter on $X$ such that $\bigcap_{n\in\Bbb N}F_n\in\Cal F$ for every
sequence $\sequencen{F_n}$ in $\Cal F$.   (I am passing over the trivial
case $X\in\Cal I$, since then $U$ must be $\{0\}$.)   And
$\eusm L^0/\eusm W$ is
(isomorphic to) the image of $\eusm L^0$ in $\Bbb R^X|\Cal F$, since
$\eusm W=\{f:f\in\eusm L^0,\,\{x:f(x)=0\}\in\Cal F\}$.   Thus $U$ is
isomorphic to a Riesz subspace of $\Bbb R^X|\Cal F$.

\medskip

{\bf (b)} On the other hand, if $\Cal F$ is a filter on $X$ closed under
countable intersections, then
$\eusm W=\{f:f\in\Bbb R^X,\,\{x:f(x)=0\}\in\Cal F\}$ is a sequentially
order-closed solid linear
subspace of the Dedekind $\sigma$-complete Riesz space $\Bbb R^X$, so
that $\Bbb R^X|\Cal F=\Bbb R^X/\eusm W$ is Dedekind
$\sigma$-complete (353K(a-iii)) and all its Riesz subspaces must be
Archimedean (353Ia, 351Rc).
}%end of proof of 368F

\leader{368G}{Corollary} Every Archimedean Riesz space $U$ is isomorphic
to an order-dense Riesz subspace of some space
$C^{\infty}(X)$\cmmnt{ (definition:  364V)}, where
$X$ is an extremally disconnected compact Hausdorff space.

\proof{ Let $Z$ be the Stone space of the band algebra $\frak A$ of $U$.
Because $\frak A$ is Dedekind complete (352Q again), $Z$ is extremally
disconnected and $\frak A$ can be identified
with the regular open algebra $\RO(Z)$ of $Z$ (314S).   By 364V,
$L^0(\RO(Z))$ can be identified with $C^{\infty}(Z)$.    So an
embedding
of $U$ as an order-dense Riesz subspace of $L^0(\frak A)$ (368E) can be
regarded as an embedding of $U$ as an order-dense Riesz subspace of
$C^{\infty}(Z)$.
}%end of proof of 368G

\leader{368H}{Corollary} Any Dedekind complete Riesz space $U$ is
isomorphic to an order-dense solid linear subspace of $L^0(\frak A)$ for
some Dedekind complete Boolean algebra $\frak A$.

\proof{ Embed $U$ in $L^0=L^0(\frak A)$ as in 368E;  because $U$ is
order-dense in $L^0$ and (in itself) Dedekind complete, it is solid
(353L).
}%end of proof of 368H

\leader{368I}{Corollary} Let $U$ be an Archimedean Riesz space.   Then
$U$ can be embedded as an order-dense Riesz subspace of a Dedekind
complete
Riesz space $V$ in such a way that the solid linear subspace of $V$
generated by $U$ is $V$ itself,
and this can be done in essentially only one way.
If $W$ is any other Dedekind complete Riesz space and $T:U\to W$ is an
order-continuous positive linear operator, there is a unique positive
linear operator $\tilde T:V\to W$ extending $T$.

\proof{ By 368E, we may suppose that $U$ is actually an
order-dense Riesz subspace of $L^0(\frak A)$, where $\frak A$ is a
Dedekind complete Boolean algebra.   In this case, we can take $V$ to be
the solid linear subspace generated by $U$, that is, $\{v:|v|\le u$ for
some $u\in U\}$;  being a solid linear subspace of the Dedekind complete
Riesz space $L^0(\frak A)$, $V$ is Dedekind complete, and of course $U$
is order-dense in $V$.

If $W$ is any other Dedekind complete Riesz space and $T:U\to W$ is an
order-continuous positive linear operator, then for any $v\in V^+$ there
is a $u_0\in U$ such that $v\le u_0$, so that $Tu_0$ is an upper bound
for $\{Tu:u\in U,\,0\le u\le v\}$;  as $W$ is Dedekind complete,
$\sup_{u\in U,0\le u\le v}Tu$ is defined in $W$.   By 355Fa, $T$ has a
unique extension to an order-continuous positive linear operator from
$V$ to $W$.

In particular, if $V_1$ is another Dedekind complete Riesz space in
which $U$ can be embedded as an order-dense Riesz subspace, this
embedding of $U$
extends to an embedding of $V$;  since $V$ is Dedekind complete, its
copy in $V_1$ must be a solid linear subspace, so if $V_1$ is the solid
linear subspace of itself generated by $U$, we get an identification
between $V$
and $V_1$, uniquely determined by the embeddings of $U$ in $V$ and
$V_1$.
}%end of proof of 368I

\leader{368J}{Definition} If $U$ is an Archimedean Riesz space, a {\bf
Dedekind completion} of $U$ is a Dedekind complete Riesz space $V$
together
with an embedding of $U$ in $V$ as an order-dense Riesz subspace of $V$
such that the solid linear subspace of $V$ generated by $U$ is $V$
itself.
\cmmnt{368I tells us that every Archimedean Riesz space $U$ has an
essentially unique Dedekind completion, so that we may speak of `the'
Dedekind completion of $U$.}

\leader{368K}{}\cmmnt{ This is a convenient point at which to give a
characterization of the Riesz spaces $L^0(\frak A)$.

\medskip

\noindent}{\bf Lemma} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra.
Suppose that $A\subseteq L^0(\frak A)^+$ is disjoint.   If {\it either}
$A$ is countable {\it or} $\frak A$ is Dedekind complete,
$A$ is bounded above in $L^0(\frak A)$.

\proof{ If $A=\emptyset$, this is trivial;  suppose that $A$ is not
empty.   For $n\in\Bbb N$, set $a_n=\sup_{u\in A}\Bvalue{u>n}$;  this is
always defined;  set $a=\inf_{n\in\Bbb N}a_n$.   Now $a=0$.   \Prf\Quer\
Otherwise, there must be a $u\in A$ such that $a'=a\Bcap\Bvalue{u>0}\ne
0$, since $a\Bsubseteq a_0$.   But now, for any $n$, and any $v\in
A\setminus\{u\}$,

\Centerline{$a'\Bcap\Bvalue{v>n}\Bsubseteq\Bvalue{u>0}\Bcap\Bvalue{v>0}
=0$,}

\noindent so that $a'\Bsubseteq\Bvalue{u>n}$.   As $n$ is arbitrary,
$\inf_{n\in\Bbb N}\Bvalue{u>n}\ne 0$, which is impossible.\ \Bang\Qed\

By 364L(a-i), $A$ is bounded above.
}%end of proof of 368K

\leader{368L}{Definition} A Riesz space $U$ is called {\bf laterally
complete} or {\bf universally complete} if $A$ is bounded above whenever
$A\subseteq U^+$ is disjoint.

\vleader{108pt}{368M}{Theorem} Let $U$ be an Archimedean Riesz space.   Then the
following are equiveridical:

(i) there is a Dedekind complete Boolean algebra $\frak A$ such that $U$
is isomorphic to $L^0(\frak A)$;

(ii) $U$ is Dedekind $\sigma$-complete and laterally complete;

(iii) whenever $V$ is an Archimedean Riesz space, $V_0$ is an
order-dense Riesz subspace of $V$ and $T:V_0\to U$ is an
order-continuous Riesz homomorphism, there is a positive linear operator
$\tilde T:V\to U$ extending $T$.

\proof{{\bf (a)(i)$\Rightarrow$(ii)} and {\bf (i)$\Rightarrow$(iii)} are
covered by 368K and 368B.

\medskip

{\bf (b)(ii)$\Rightarrow$(i)} Assume (ii).   By 368E, we may suppose
that $U$ is actually an order-dense Riesz subspace of $L^0=L^0(\frak A)$
for a Dedekind complete Boolean algebra $\frak A$.

\medskip

\quad\grheada\ If $u\in U^+$ and $a\in\frak A$ then
$u\times\chi a\in U$.
\Prf\ Set $A=\{v:v\in U,\,0\le v\le\chi a\}$, and let $C\subseteq A$ be
a maximal disjoint set;  then $w=\sup C$ is defined in $U$, and is also
the supremum in $L^0$.   Set $b=\Bvalue{w>0}$.   As $w\le\chi a$,
$b\Bsubseteq a$.   \Quer\ If $b\ne a$, then $\chi(a\Bsetminus b)>0$, and
there is a
$v'\in U$ such that $0<v'\le\chi(a\Bsetminus b)$;  but now $v'\in A$ and
$v'\wedge w=0$, so $v'\wedge v=0$ for every $v\in C$, and we ought to
have added $v'$ to $C$.\ \BanG\  Thus $\Bvalue{w>0}=a$.

Now consider $u'=\sup_{n\in\Bbb N}u\wedge nw$;  as $U$ is Dedekind
$\sigma$-complete, $u'\in U$.   Since $\Bvalue{u'>0}\Bsubseteq a$,
$u'\le u\times\chi a$.   On the other hand,

\Centerline{$u\times\chi\Bvalue{w>\bover1n}\times\chi\Bvalue{u\le n}
\le u\wedge n^2w\le u'$}

\noindent for every $n\ge 1$, so, taking the supremum over $n$,
$u\times\chi a\le u'$.   Accordingly

\Centerline{$u\times\chi a=u'\in U$,}

\noindent as required.\ \Qed

\medskip

\quad\grheadb\ If $w\ge 0$ in $L^0$, there is a $u\in U$ such that
$\bover12w\le u\le w$.   \Prf\ Set

\Centerline{$A=\{u:u\in U,\,0\le u\le w\}$,}

\Centerline{$C=\{a:a\in\frak A,\,a\Bsubseteq\Bvalue{u-\bover12w\ge 0}$
for
some $u\in A\}$.}

\noindent Then $\sup A=w$, so $C$ is order-dense in $\frak A$.   (If
$a\in\frak A\setminus\{0\}$, either $a\Bcap\Bvalue{w>0}=0$ and
$a\subseteq\Bvalue{0-\bover12w\ge 0}$, so $a\in C$, or there is a $u\in
U$
such that $0<u\le w\times\chi a$.   In the latter case there is some $n$
such that $2^nu\le w$ and $2^{n+1}u\not\le w$, and now
$c=a\Bcap\Bvalue{2^nu-\bover12w\ge 0}$ is a non-zero member of $C$
included in $a$.)   Let $D\subseteq C$ be a partition of unity and for
each $d\in D$
choose $u_d\in A$ such that $d\Bsubseteq\Bvalue{u_d-\bover12w\ge 0}$.
By ($\alpha$), $u_d\times\chi d\in U$ for every $d\in D$, so
$u=\sup_{d\in D}u_d\times\chi d\in U$.   Now $u\le w$, but also
$\Bvalue{u-\bover12w\ge 0}\Bsupseteq d$ for every $d\in D$, so is equal
to $1$, and $u\ge\bover12w$, as required.\ \Qed

\medskip

\quad\grheadc\ Given $w\ge 0$ in $L^0$, we can therefore choose
$\sequencen{u_n}$, $\sequencen{v_n}$ inductively such that $v_0=0$ and

\Centerline{$u_n\in U$,
\quad$\Bover12(w-v_n)\le u_n\le w-v_n$,
\quad$v_{n+1}=v_n+u_n$}

\noindent for every $n\in\Bbb N$.   Now $\sequencen{v_n}$ is a
non-decreasing sequence in $U$ and $w-v_n\le 2^{-n}w$ for every $n$, so
$w=\sup_{n\in\Bbb N}v_n\in U$.

As $w$ is arbitrary, $(L^0)^+\subseteq U$ and $U=L^0$ is of the right
form.

\medskip

{\bf (c)(iii)$\Rightarrow$(i)} As in (b), we may suppose that $U$ is an
order-dense Riesz subspace of $L^0$.   But now apply condition (iii)
with $V=L^0$, $V_0=U$ and $T$ the identity operator.   There is an
extension $\tilde T:L^0\to U$.   If $v\ge 0$ in $L^0$, $\tilde Tv\ge Tu=u$
whenever
$u\in U$ and $u\le v$, so $\tilde Tv\ge v$,
since $v=\sup\{u:u\in U,\,0\le u\le v\}$ in $L^0$.   Similarly,
$\tilde T(\tilde Tv-v)\ge\tilde Tv-v$.   But as $\tilde Tv\in U$,
$\tilde T(\tilde Tv)=T(\tilde Tv)=\tilde Tv$ and
$\tilde T(\tilde Tv-v)=0$, so
$v=\tilde Tv\in U$.   As $v$ is arbitrary, $U=L^0$.
}%end of proof of 368M

\leader{368N}{Weakly $(\sigma,\infty)$-distributive Riesz
\dvrocolon{spaces}}\cmmnt{ We are now ready to look at the class of
Riesz spaces corresponding to the \wsid\ Boolean algebras of \S316.

\medskip

\noindent}{\bf Definition} Let $U$ be a Riesz space.   Then $U$ is {\bf
\wsid} if whenever $\sequencen{A_n}$ is a sequence of non-empty
downwards-directed subsets of $U^+$, each with infimum $0$, and
$\bigcup_{n\in\Bbb N}A_n$ has an upper bound in $U$, then

\Centerline{$\{u:u\in U$, for every $n\in\Bbb N$ there is a $v\in A_n$
such that $v\le u\}$}

\noindent has infimum $0$ in $U$.

\cmmnt{\medskip

\noindent{\bf Remark} Because the definition looks only at sequences
$\sequencen{A_n}$ such that $\bigcup_{n\in\Bbb N}A_n$ is order-bounded,
we can invert it, as follows:  a Riesz space $U$ is \wsid\ iff whenever
$\sequencen{A_n}$ is a sequence of
non-empty upwards-directed subsets of $U^+$, all with supremum $u_0$,
then

\Centerline{$\{u:u\in U^+$, for every $n\in\Bbb N$ there is a $v\in A_n$
such that $u\le v\}$}

\noindent also has supremum $u_0$.
}%end of comment

\leader{368O}{Lemma} Let $U$ be an Archimedean Riesz space.   Then the
following are equiveridical:

(i) $U$ is not \wsid;

(ii) there are a $u>0$ in $U$ and a sequence $\sequencen{A_n}$ of
non-empty downwards-directed sets, all with infimum $0$, such that
$\sup_{n\in\Bbb N}u_n=u$ whenever $u_n\in A_n$ for every $n\in\Bbb N$.

\proof{(ii)$\Rightarrow$(i) is immediate from the definition of `\wsid'.
For (i)$\Rightarrow$(ii), suppose that $U$ is not \wsid.   Then there is
a sequence $\sequencen{A_n}$ of non-empty downwards-directed sets, all
with infimum $0$, such that $\bigcup_{n\in\Bbb N}A_n$ is bounded above,
but

\Centerline{$A=\{w:w\in U$, for every $n\in\Bbb N$ there is a $v\in A_n$
such that $v\le w\}$}

\noindent does not have infimum $0$.   Let $u>0$ be a lower bound for
$A$, and set $A'_n=\{u\wedge v:v\in A_n\}$ for each $n\in\Bbb N$.   Then
each $A'_n$ is a non-empty downwards-directed set with infimum $0$.
Let $\sequencen{u_n}$ be a sequence such that $u_n\in A'_n$ for every
$n$.   Express each $u_n$ as $u\wedge v_n$ where $v_n\in A_n$.   Let $B$
be the set of upper bounds of $\{v_n:n\in\Bbb N\}$.   Then
$\inf_{w\in B,n\in\Bbb N}w-v_n=0$, because $U$ is Archimedean (353F),
while $B\subseteq A$, so
$u\le w$ for every $w\in B$.   If $u'$ is any upper bound for
$\{u_n:n\in\Bbb N\}$, then

\Centerline{$u-u'\le u-u\wedge v_n=(u-v_n)^+\le(w-v_n)^+=w-v_n$}

\noindent whenever $n\in\Bbb N$ and $w\in B$.   So $u'\ge u$.   Thus
$u=\sup_{n\in\Bbb N}u_n$.   As $\sequencen{u_n}$ is arbitrary, $u$ and
$\sequencen{A'_n}$ witness that (ii) is true.
}%end of proof of 368O

\leader{368P}{Proposition} (a) A regularly embedded Riesz subspace of an
Archimedean \wsid\ Riesz space is \wsid.

%query:  do we need "Archimedean" here?  for order-dense ssps, no pb
%(368Ye)

(b) An Archimedean Riesz space with a \wsid\ order-dense Riesz subspace
is \wsid.

%query:  do we need "Archimedean" here?

(c) If $U$ is a Riesz space such that $U^{\times}$ separates the points
of $U$, then $U$ is \wsid;  in particular, $U^{\sim}$ and $U^{\times}$
are \wsid\ for every Riesz space $U$.

\proof{{\bf (a)} Suppose that $U$ is an Archimedean Riesz space and that
$V\subseteq U$ is a regularly embedded Riesz subspace which is not
\wsid.   Then 368O tells us that there are a $v>0$ in $V$ and a sequence
$\sequencen{A_n}$ of non-empty downwards-directed subsets of $V$, all
with infimum $0$ in $V$, such that $\sup_{n\in\Bbb N}v_n=v$ in $V$ whenever $v_n\in A_n$ for every $n\in\Bbb N$.   Because $V$ is regularly embedded in
$U$, $\inf A_n=0$ in $U$ for every $n$ and $\sup_{n\in\Bbb N}v_n=v$ in
$U$ for every sequence $\sequencen{v_n}\in\prod_{n\in\Bbb N}A_n$, so $U$ is not \wsid.   Turning this round, we have (a).

\medskip

{\bf (b)} Let $U$ be an Archimedean Riesz space which is not \wsid, and
$V$ an order-dense Riesz subspace of $U$.   By 368O again, there are a
$u^*>0$ in $U$ and a sequence $\sequencen{A_n}$ of non-empty
downwards-directed sets in $U$, all with infimum $0$, such that
$\sup_{n\in\Bbb N}u_n=u^*$ whenever
$u_n\in A_n$ for every $n$.   Let $v\in V$ be such that $0<v\le u^*$.
Set

\Centerline{$B_n=\{w:w\in V$, there is some $u\in A_n$ such that
$v\wedge u\le w\le v\}$}

\noindent for each $n\in\Bbb N$.   Because $A_n$ is downwards-directed,
$w\wedge w'\in B_n$ for all $w$, $w'\in B_n$;  $v\in B_n$, so
$B_n\ne\emptyset$;  and $\inf B_n=0$ in $V$.   \Prf\ Setting

\Centerline{$C=\{w:w\in V^+$, there is some $u\in A_n$ such that
$w\le(v-u)^+\}$,}

\noindent then (because $V$ is order-dense) any upper bound for $C$ in
$U$ is also an upper bound of $\{(v-u)^+:u\in A_n\}$.   But

\Centerline{$\sup_{u\in A_n}(v-u)^+=(v-\inf A_n)^+=v$,}

\noindent so $v=\sup C$ in $U$ and $\inf B_n=\inf\{v-w:w\in C\}=0$ in
$U$ and in $V$.\ \Qed

Now if $v_n\in B_n$ for every $n\in\Bbb N$, we can choose $u_n\in A_n$
such that $v\wedge u_n\le v_n\le v$ for every $n$, so that

\Centerline{$v=v\wedge u^*=v\wedge\sup_{n\in\Bbb N}u_n
=\sup_{n\in\Bbb N}v\wedge u_n\le\sup_{n\in\Bbb N}v_n\le v$,}

\noindent and $v=\sup_{n\in\Bbb N}v_n$.   Thus $\sequencen{B_n}$
witnesses that $V$ is not \wsid.

\medskip

{\bf (c)} Now suppose that $U^{\times}$ separates the points of $U$.
In this case $U$ is surely Archimedean (356G).   \Quer\ If $U$ is not
\wsid, there are a $u>0$ in $U$ and a sequence $\sequencen{A_n}$ of
non-empty downwards-directed sets, all with infimum $0$, such that
$\sup_{n\in\Bbb N}u_n=u$ whenever $u_n\in A_n$ for each $n$.   Take
$f\in U^{\times}$ such
that $f(u)\ne 0$;  replacing $f$ by $|f|$ if necessary, we may suppose
that $f>0$.   Set $\delta=f(u)>0$.   For each $n\in\Bbb N$, there is a
$u_n\in A_n$ such that $f(u_n)\le 2^{-n-2}\delta$.   But in this case
$\sequencen{\sup_{i<n}u_i}$ is a
non-decreasing sequence with supremum $u$, so

\Centerline{$f(u)=\lim_{n\to\infty}f(\sup_{i\le
n}u_i)\le\sum_{i=0}^{\infty}f(u_i)\le\bover12\delta<f(u)$,}

\noindent which is absurd.\ \BanG\  Thus $U$ is \wsid.

For any Riesz space $U$, $U$ acts on $U^{\sim}$ as a subspace of
$U^{\sim\times}$ (356F);  as $U$ surely separates the points of
$U^{\sim}$, so does $U^{\sim\times}$.   So $U^{\sim}$ is \wsid.   Now
$U^{\times}$ is a
band in $U^{\sim}$ (356B), so is regularly embedded, and must also be
\wsid, by (a) above.
}%end of proof of 368P

\leader{368Q}{Theorem} (a) For any Boolean algebra $\frak A$, $\frak A$
is \wsid\ iff $S(\frak A)$ is \wsid\ iff $L^{\infty}(\frak A)$ is \wsid.

(b) For a Dedekind $\sigma$-complete Boolean algebra $\frak A$,
$L^0(\frak A)$ is \wsid\ iff $\frak A$ is \wsid.

\proof{{\bf (a)(i)} \Quer\ Suppose, if possible, that $\frak A$ is
\wsid\ but $S=S(\frak A)$ is not.   By 368O, as usual, we have a $u>0$
in $S$ and
a sequence $\sequencen{A_n}$ of non-empty downwards-directed sets in
$S$, all with
infimum $0$, such that $u=\sup_{n\in\Bbb N}u_n$ whenever $u_n\in A_n$
for every $n$.   Let $\alpha>0$ be such that $c=\Bvalue{u>\alpha}\ne 0$
(361Eg), and consider

\Centerline{$B_n=\{\Bvalue{v>\alpha}:v\in A_n\}\subseteq\frak A$}

\noindent for each $n\in\Bbb N$.   Then each $B_n$ is
downwards-directed (because $A_n$ is), and $\inf B_n=0$ in $\frak A$
(because if $b$ is a lower bound of $B_n$, $\alpha\chi b\le v$ for every
$v\in A_n$).   Because $\frak A$ is \wsid, there must be some $a\in\frak
A$
such that $a\notBsupseteq c$ but there is, for every $n\in\Bbb N$, a
$b_n\in B_n$ such that $a\Bsupseteq b_n$.   Take $v_n\in A_n$ such that
$b_n=\Bvalue{v_n>\alpha}$, so that

\Centerline{$v_n\le\alpha\chi 1\vee\|v_n\|_{\infty}\chi b_n
\le\alpha\chi 1\vee\|u\|_{\infty}\chi a$.}

\noindent Since $u=\sup_{n\in\Bbb N}v_n$, $u\le\alpha\chi
1\vee\|u\|_{\infty}\chi a$.   But in this case

\Centerline{$c=\Bvalue{u>\alpha}\Bsubseteq a$,}

\noindent contradicting the choice of $a$.\ \Bang

Thus $S$ must be \wsid\ if $\frak A$ is.

\medskip

\quad{\bf (ii)} Now suppose that $S$ is \wsid, and let $\sequencen{B_n}$
be
a sequence of non-empty downwards-directed subsets of $\frak A$, all
with
infimum $0$.   Set $A_n=\{\chi b:b\in B_n\}$ for each $n$;  then
$A_n\subseteq S$ is non-empty, downwards-directed and has infimum $0$ in
$S$, because $\chi:\frak A\to S$ is order-continuous (361Ef).   Set

\Centerline{$A=\{v:v\in S$, for every $n\in\Bbb N$ there is a $u\in A_n$
such that $u\le v\}$,}

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

\noindent \Quer\ If $0$ is not the greatest lower bound of $B$, take a
non-zero lower bound $c$.   Because $S$ is \wsid, $\inf A=0$, and there
is a $v\in A$ such that $\chi c\not\le v$.   Express $v$ as
$\sum_{i=0}^n\alpha_i\chi a_i$, where $\langle a_i\rangle_{i\le n}$ is
disjoint, and set $a=\sup\{a_i:i\le n,\,\alpha_i\ge 1\}$;  then
$\chi a\le v$, so $c\notBsubseteq a$.   For each $n$ there is a
$b_n\in B_n$ such
that $\chi b_n\le v$.   But in this case $b_n\Bsubseteq a$ for each
$n\in\Bbb N$, so that $a\in B$;  which means that $c$ is not a lower
bound for $B$.\ \Bang

Thus $\inf B=0$ in $\frak A$.   As $\sequencen{B_n}$ is arbitrary,
$\frak A$ is \wsid.

\medskip

\quad{\bf (iii)} Thus $S$ is \wsid\ iff $\frak A$ is.   But $S$ is
order-dense in $L^{\infty}=L^{\infty}(\frak A)$ (363C), therefore
regularly embedded (352Ne), so 368Pa-b tell us that $S$ is \wsid\ iff
$L^{\infty}$ is.

\medskip

{\bf (b)} In the same way, because $S$ can be regarded as an order-dense
Riesz subspace of $L^0=L^0(\frak A)$ (364Ja), $L^0$ is \wsid\ iff $S$ is,
that is, iff $\frak A$ is.
}%end of proof of 368Q

\leader{368R}{Corollary} An Archimedean Riesz space is \wsid\ iff its
band algebra is \wsid.

\proof{ Let $U$ be an Archimedean Riesz space and $\frak A$ its band
algebra.   By 368E, $U$ is isomorphic to an order-dense Riesz subspace
of $L^0=L^0(\frak A)$.   By 368P, $U$ is \wsid\ iff $L^0$ is;  and by
368Qb $L^0$ is \wsid\ iff $\frak A$ is.
}%end of proof of 368R

\leader{368S}{Corollary} If $(\frak A,\bar\mu)$ is a semi-finite measure
algebra, any regularly embedded Riesz subspace (in particular, any solid
linear subspace and any order-dense Riesz subspace) of $L^0(\frak A)$ is
\wsid.

\proof{ By 322F, $\frak A$ is \wsid;  by 368Qb, $L^0(\frak A)$ is \wsid;
by 368Pa, any regularly embedded Riesz subspace is \wsid.
}%end of proof of 368S

\exercises{\leader{368X}{Basic exercises (a)}
%\spheader 368Xa
Let $X$ be an uncountable set and $\Sigma$ the countable-cocountable
$\sigma$-algebra of subsets of $X$.   Show that there is a family
$A\subseteq L^0=L^0(\Sigma)$ such that $u\wedge v=0$ for all distinct
$u$, $v\in A$ but $A$ has no upper bound in $L^0$.
Show moreover that if $w>0$ in $L^0$ then there is an $n\in\Bbb N$ such
that $nw\ne\sup_{u\in A}u\wedge nw$.
%368A

\spheader 368Xb\dvAnew{2009}
Let $U$ be a linear space, $\frak A$ a Dedekind complete
Boolean algebra, and $p:U\to L^0=L^0(\frak A)$ a function such that
$p(u+v)\le p(u)+p(v)$ and $p(\alpha u)=\alpha p(u)$ whenever $u$, $v\in U$
and $\alpha\ge 0$.   Suppose that $V\subseteq U$ is a linear subspace and
$T:V\to L^0$ is a linear operator such that $Tv\le p(v)$ for every
$v\in V$.   Show that there is a linear operator
$\tilde T:U\to L^0$, extending
$T$, such that $\tilde Tu\le p(u)$ for every $u\in U$.
\Hint{part A of the proof of 363R.}
%368B

\spheader 368Xc\dvAformerly{3{}68Xb}
Let $\frak A$ be any Boolean algebra, and
$\widehat{\frak A}$ its Dedekind completion (314U).   Show that
$L^{\infty}(\widehat{\frak A})$ can be identified with the Dedekind
completions of $S(\frak A)$ and $L^{\infty}(\frak A)$.
%368J

\spheader 368Xd\dvAformerly{3{}68Xc} Explain how to prove 368K from 368A.
%368K

\spheader 368Xe\dvAformerly{3{}68Xd} Show that any product of \wsid\ Riesz spaces is \wsid.
%368N

\spheader 368Xf\dvAformerly{3{}68Xe}
Let $\frak A$ be a Dedekind complete \wsid\ Boolean
algebra.   Show that a
set $A\subseteq L^0=L^0(\frak A)$ is order-bounded iff
$\sequencen{2^{-n}u_n}$ order*-converges to $0$ in $L^0$ whenever
$\sequencen{u_n}$ is a sequence in $A$.   \Hint{use 368A.   If $v>0$ and
$v=\sup_{u\in A}v\wedge 2^{-n}u$ for every $n$, we can find a $w>0$ and
a sequence $\sequencen{u_n}$ in $A$ such that $w\le 2^{-n}u_n$ for every
$n$.}
%368R

\spheader 368Xg\dvAformerly{3{}68Xf}
Give a direct proof of 368S, using the ideas of 322F,
but not relying on it or on 368Q.
%368S

\leader{368Y}{Further exercises (a)}
%\spheader 368Ya
(i) Use 364T-364U to show that if $X$ is any compact Hausdorff space
then $C(X)$ can be regarded as an order-dense Riesz subspace of
$L^0(\RO(X))$, where $\RO(X)$ is the regular open algebra of $X$.
(ii) Use 353N
to show that any Archimedean Riesz space with order unit can be embedded
as an order-dense Riesz subspace of some $L^0(\RO(X))$.   (iii) Let $U$
be an Archimedean Riesz space and $C\subseteq U^+$ a maximal disjoint
set, as in part (a) of the proof of 368E.   For $e\in C$ let $U_e$ be
the solid linear
subspace of $U$ generated by $e$, and let $V$ be the solid linear
subspace of $U$ generated by $C$.   Show that $V$ can be embedded as an
order-dense
Riesz subspace of $\prod_{e\in C}U_e$ and therefore in $\prod_{e\in
C}L^0(\RO(X_e))\cong L^0(\prod_{e\in C}\RO(X_e))$ for a suitable
family of regular open algebras $\RO(X_e)$.   (iv) Now use 368B to
complete a proof of 368E.
%368E

\spheader 368Yb Let $U$ be any Archimedean Riesz space.   Let $\Cal V$
be the family of pairs $(A,B)$ of non-empty subsets of $U$ such that $B$
is the set of upper bounds of $A$ and $A$ is the set of lower bounds of
$B$.
Show that $\Cal V$ can be given the structure of a Dedekind complete
Riesz
space defined by the formulae

\Centerline{$(A_1,B_1)+(A_2,B_2)=(A,B)$ iff $A_1+A_2\subseteq A$,
$B_1+B_2\subseteq B$,}

\Centerline{$\alpha(A,B)=(\alpha A,\alpha B)$ if $\alpha>0$,}

\Centerline{$(A_1,B_1)\le(A_2,B_2)$ iff $A_1\subseteq A_2$.}

\noindent Show that $u\mapsto(\ocint{-\infty,u},\coint{u,\infty})$
defines
an embedding of $U$ as an order-dense Riesz subspace of $\Cal V$, so
that
$\Cal V$ may be identified with the Dedekind completion of $U$.
%368J

\spheader 368Yc Work through the proof of 364T when $X$ is compact,
Hausdorff and extremally disconnected, and show that it is easier than
the general case.   Hence show that 368Yb can be used to shorten the
proof of 368E sketched in 368Ya.
%368J, 368Yb

\spheader 368Yd Let $U$ be a Riesz space.   Show that the following are
equiveridical:  (i) $U$ is isomorphic, as Riesz space, to $L^0(\frak A)$
for some Dedekind $\sigma$-complete Boolean algebra $\frak A$ (ii) $U$
is Dedekind $\sigma$-complete and has a weak order unit and whenever
$A\subseteq U^+$ is countable and disjoint then $A$ is
bounded above in $U$.
%368M

\spheader 368Ye Let $U$ be a \wsid\ Riesz space and $V$ a Riesz subspace
of $U$ which is {\it either} solid {\it or} order-dense.   Show that $V$
is \wsid.
%368P

\spheader 368Yf Show that $C([0,1])$ is not \wsid.   (Compare 316J.)
%368R

\spheader 368Yg Let $\frak A$ be a ccc \wsid\ Boolean algebra.
Suppose we have a double sequence
$\family{(i,j)}{\Bbb N\times\Bbb N}{a_{ij}}$ in
$\frak A$ such that $\sequence{j}{a_{ij}}$ order*-converges to $a_i$
in $\frak A$
for each $i$, while $\sequence{i}{a_i}$ order*-converges to $a$.
Show that there
is a strictly increasing sequence $\sequence{i}{n(i)}$ such that
$\sequence{i}{a_{i,n(i)}}$ order*-converges to $a$.
%368R

\spheader 368Yh Let $U$ be a \wsid\ Riesz space with the countable sup
property.   Suppose we have an order-bounded double sequence
$\family{(i,j)}{\Bbb N\times\Bbb N}{u_{ij}}$ in
$U$ such that $\sequence{j}{u_{ij}}$ order*-converges to $u_i$ in $U$
for each $i$, while $\sequence{i}{u_i}$ order*-converges to $u$.
Show that there
is a strictly increasing sequence $\sequence{i}{n(i)}$ such that
$\sequence{i}{u_{i,n(i)}}$ order*-converges to $u$.
%368R

\spheader 368Yi Let $\frak A$ be a ccc \wsid\ Dedekind complete Boolean
algebra.   Show that there is a topology on $L^0=L^0(\frak A)$ such that
the closure of any $A\subseteq L^0$ is precisely the set of
order*-limits of sequences in $A$.   (Cf.\ 367Yk.)
%368Yh 368R

\spheader 368Yj Let $U$ be a \wsid\ Riesz space and $f:U\to\Bbb R$ a
positive linear functional;  write $f_{\tau}$ for the component of $f$
in $U^{\times}$.   (i) Show that for any $u\in U^+$ there is an
upwards-directed $A\subseteq[0,u]$, with supremum $u$, such that
$f_{\tau}(u)=\sup_{v\in A}f(v)$.   (See 356Xe, 362D.)  (ii) Show that if
$f$ is strictly positive, so is $f_{\tau}$.   (Compare 391D.)
%368R
}%end of exercises

\cmmnt{
\Notesheader{368} 368A-368B are manifestations of a principle which will
reappear in \S375:   Dedekind complete $L^0$ spaces are in some
sense `maximal'.   If we have an order-dense subspace $U$ of such an
$L^0$, then any Archimedean Riesz space including $U$ as an order-dense
subspace can itself be embedded in $L^0$ (368B).   In fact this property
characterizes Dedekind complete $L^0$ spaces (368M).   Moreover, any
Archimedean Riesz space $U$ can be embedded in this way (368E);
by 368C, the $L^0$ space (though not the embedding) is unique up to
isomorphism.   If $U$ and $V$ are Archimedean Riesz spaces, each
embedded as an order-dense Riesz subspace of a Dedekind complete $L^0$
space, then
any order-continuous Riesz homomorphism from $U$ to $V$ extends uniquely
to the $L^0$ spaces (368B).   If one Dedekind complete $L^0$ space is
embedded as an
order-dense Riesz subspace of another, they must in fact be the same
(368D).   Thus we can say that every Archimedean Riesz space $U$ can be
extended to a Dedekind complete $L^0$ space, in a way which respects
order-continuous Riesz homomorphisms, and that this extension is
maximal, in that $U$ cannot be order-dense in any larger space.

The proof of 368E which I give is long because I am using a bare-hands
approach.   Alternative methods shift the burdens.   For instance, if we
take the trouble to develop a direct construction of the `Dedekind
completion' of a Riesz space (368Yb), then we need prove the theorem
only for Dedekind complete Riesz spaces.   A more substantial aid is the
representation theorem for Archimedean Riesz spaces with order unit
(353N);  I sketch an argument in 368Ya.   The drawback to this
approach is the proof of Theorem 364T, which seems to be quite as long
as the direct proof of 368E which I give here.   Of course we need 364T
only for compact Hausdorff spaces, which are usefully easier than the
general case (364U, 368Yc).

368G is a version of Ogasawara's representation theorem for Archimedean
Riesz spaces.   Both this and 368F can be regarded as expressions of the
principle that an Archimedean Riesz space is `nearly' a space of
functions.

I have remarked before on the parallels between the theories of Boolean
algebras and Archimedean Riesz spaces.   The notion of
`weak $(\sigma,\infty)$-distributivity' is one of the more striking
correspondences.   (Compare, for instance, 316Xi(i) with 368Pa.)   What is
really important to us, of course, is the fact that the function spaces
of measure theory are mostly \wsid, by 368S.   Of course this is easy to
prove directly (368Xg), but I think that the argument through 368Q gives
a better idea of what is really happening here.   Some of the features
of `order*-convergence', as defined in \S367, are related to weak
$(\sigma,\infty)$-distributivity (compare 367Yi, 367Yp);
in 368Yi I describe a topology which
can be thought of as an abstract version of the topology of convergence
in measure on the $L^0$ space of a $\sigma$-finite measure algebra
(367M).
}%end of comment

\discrpage

\frfilename{mt369.tex}
\versiondate{23.11.16}
\copyrightdate{1996}

\def\chaptername{Function spaces}
\def\sectionname{Banach function spaces}

\newsection{369}

In this section I continue the work of \S368 with results which involve
measure algebras.   The first step is a modification of the basic
representation theorem for Archimedean Riesz spaces.   If $U$ is any
Archimedean Riesz space, it can be represented as a subspace of
$L^0=L^0(\frak A)$, where $\frak A$ is its band algebra (368E);
now if $U^{\times}$ separates the points of $U$, there is a measure
rendering $\frak A$ a localizable measure algebra
(369A\cmmnt{, 369Xa}).
Moreover, we get a simultaneous representation of $U^{\times}$ as a
subspace of $L^0$ (369C-369D), the duality between $U$ and $U^{\times}$ corresponding exactly to the familiar duality between $L^p$ and $L^q$.
In particular, every $L$-space can be represented as an $L^1$-space (369E).

Still drawing inspiration from the classical $L^p$ spaces, we have a
general theory of `associated Fatou norms' (369F-369M, 369R).   I
include notes on the spaces $M^{1,\infty}$, $M^{\infty,1}$ and $M^{1,0}$
(369N-369Q), which will be particularly useful in the next chapter.

\leader{369A}{Theorem} Let $U$ be a Riesz space such that $U^{\times}$
separates the points of $U$.   Then $U$ can be embedded as an
order-dense Riesz subspace of $\cmmnt{L^0=}L^0(\frak A)$ for some
localizable measure algebra $(\frak A,\bar\mu)$.

\proof{{\bf (a)} Consider the canonical map $S:U\to U^{\times\times}$.
We know that this is a Riesz homomorphism onto an order-dense Riesz
subspace of $U^{\times\times}$ (356I).   Because $U^{\times}$
separates the points of $U$, $S$ is injective.   Let $\frak A$ be the
band algebra of $U^{\times\times}$ and
$T:U^{\times\times}\to L^0$ an injective Riesz homomorphism onto an
order-dense Riesz subspace $V$ of $L^0$, as in 368E.   The composition
$TS:U\to L^0$ is now an injective Riesz homomorphism,
so embeds $U$ as a Riesz subspace of $L^0$,
which is order-dense because $V$ is order-dense
in $L^0$ and $TS[U]$ is order-dense in $V$ (352N(c-iii)).
Thus all that we need to find is a measure $\bar\mu$ on $\frak A$
rendering it a localizable measure algebra.

\medskip

{\bf (b)} Note that $V$ is isomorphic, as Riesz space, to $U^{\times\times}$,
which is Dedekind complete (356B), so $V$ must be solid in $L^0$
(353L).   Also $V^{\times}$ must separate the points of $V$
(356Lb).

Let $D$ be the set of those $d\in\frak A$ such that the principal ideal
$\frak A_d$ is measurable in the sense that there is some $\bar\nu$ for
which $(\frak A_d,\bar\nu)$ is a totally finite measure algebra.  Then $D$
is order-dense in $\frak A$.   \Prf\ Take any non-zero $a\in\frak A$.
Because $V$ is order-dense, there is a non-zero $v\in V$ such that
$v\le\chi a$.   Take $h\ge 0$ in $V^{\times}$ such that $h(v)>0$.   Then
there is a $v'$ such that $0<v'\le v$ and $h(w)>0$ whenever $0<w\le v'$
in $V$ (356H).
Let $\alpha>0$ be such that $d=\Bvalue{v'>\alpha}\ne 0$.   Then $\chi
b\le\bover1{\alpha}v'\in V$ whenever $b\in\frak A_d$.   Set $\bar\nu
b=h(\chi b)\in\coint{0,\infty}$ for $b\in\frak A_d$.   Because the map
$b\mapsto\chi b:\frak A\to L^0$ is additive and order-continuous, the
map $b\mapsto\chi b:\frak A_d\to V$ also is, and $\bar\nu=h\chi$ must be
additive and order-continuous;  in particular,
$\bar\nu(\sup_{n\in\Bbb N}b_n)=\sum_{n=0}^{\infty}\bar\nu b_n$ whenever
$\sequencen{b_n}$ is a
disjoint sequence in $\frak A_d$.   Moreover, if $b\in\frak A_d$ is
non-zero, then $0<\alpha\chi b\le v'$, so $\bar\nu b=h(\chi b)>0$.
Thus $(\frak A_d,\bar\nu)$ is a totally finite measure algebra, and
$d\in D$, while $0\ne d\Bsubseteq a$.   As $a$ is arbitrary, $D$ is
order-dense.\ \Qed

\medskip

{\bf (c)} By 313K, there is a partition of unity $C\subseteq D$.   For
each $c\in C$, let $\bar\nu_c:\frak A_c\to\coint{0,\infty}$ be a
functional such that $(\frak A_c,\bar\nu_c)$ is a totally finite measure
algebra.   Define $\bar\mu:\frak A\to[0,\infty]$ by setting
$\bar\mu a=\sum_{c\in C}\bar\nu_c(a\Bcap c)$ for every $a\in\frak A$.   Then $(\frak A,\bar\mu)$ is a localizable measure algebra.   \Prf\ (i)
$\bar\mu 0=\sum_{c\in C}\bar\nu 0=0$.   (ii) If $\sequencen{a_n}$ is a
disjoint sequence in $\frak A$ with supremum $a$, then

\Centerline{$\bar\mu a
=\sum_{c\in C}\bar\nu_c(a\Bcap c)
=\sum_{c\in C,n\in\Bbb N}\bar\nu_c(a_n\Bcap c)
=\sum_{n=0}^{\infty}\bar\mu a_n$.}

\noindent (iii) If $a\in\frak A\setminus\{0\}$, then there is a $c\in C$
such that $a\Bcap c\ne 0$, so that $\bar\mu a\ge\bar\nu_c(a\Bcap c)>0$.
Thus $(\frak A,\bar\mu)$ is a measure algebra.   (iv) Moreover, in
(iii), $\bar\mu(a\Bcap c)=\bar\nu_c(a\Bcap c)$ is finite.
So $(\frak A,\bar\mu)$ is semi-finite.
(v) $\frak A$ is Dedekind complete, being
a band algebra (352Q), so $(\frak A,\bar\mu)$ is localizable.\ \Qed
}%end of proof of 369A

\leader{369B}{Corollary} Let $U$ be a Banach lattice with
order-continuous norm.   Then $U$ can be embedded as an order-dense
solid linear subspace of $L^0(\frak A)$ for some localizable measure
algebra $(\frak A,\bar\mu)$.

\proof{ By 356Dd, $U^{\times}=U^*$, which separates the points of
$U$, by the Hahn-Banach theorem (3A5Ae).   So 369A tells us that $U$ can
be embedded as an order-dense Riesz subspace of an appropriate
$L^0(\frak A)$.    But also $U$ is Dedekind complete (354Ee), so its
copy in $L^0(\frak A)$ must be solid, as in 368H.
}%end of proof of 369B

\leader{369C}{}\cmmnt{ The representation in 369A is complemented by
the following result, which is a kind of generalization of 365L and
366Dc.

\medskip

\noindent}{\bf Theorem} 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.   Set

\Centerline{$V=\{v:v\in L^0,\,v\times u\in L^1$ for every $u\in U\}$,}

\noindent writing $L^1$ for $L^1(\frak A,\bar\mu)\subseteq L^0$.   Then
$V$ is a solid linear subspace of $L^0$, and we have an order-continuous
injective Riesz homomorphism $T:V\to U^{\times}$ defined by setting

\Centerline{$(Tv)(u)=\int u\times v$ for all
$u\in U$, $v\in V$.}

\noindent The image of $V$ is order-dense in $U^{\times}$.   If
$(\frak A,\bar\mu)$ is localizable, then $T$ is surjective, so is a Riesz space isomorphism between $V$ and $U^{\times}$.

\proof{{\bf (a)(i)} Because $\times:L^0\times L^0\to L^0$ is bilinear
and $L^1$ is a linear subspace of $L^0$, $V$ is a linear subspace of
$L^0$.   If $u\in U$, $v\in V$, $w\in L^0$ and $|w|\le|v|$, then

\Centerline{$|w\times u|=|w|\times|u|\le|v|\times|u|=|v\times u|\in
L^1$;}

\noindent  as $L^1$ is solid, $w\times u\in L^1$;  as $u$ is arbitrary,
$w\in V$;  this shows that $V$ is solid.

\medskip

\quad{\bf (ii)} By the definition of $V$, $(Tv)(u)$ is defined in
$\Bbb R$ whenever $u\in U$ and $v\in V$.
Because $\times$ is bilinear and $\int$
is linear, $Tv:U\to\Bbb R$ is linear for every $v\in V$, and $T$ is a
linear functional from $V$ to the space of linear operators from $U$ to
$\Bbb R$.

\medskip

\quad{\bf (iii)}
If $u\ge 0$ in $U$ and $v\ge 0$ in $V$, then $u\times v\ge 0$ in $L^1$
and $(Tv)(u)=\int u\times v\ge 0$.   This shows that $T$ is a positive
linear operator from $V$ to $U^{\sim}$.

\medskip

\quad{\bf (iv)}
If $v\ge 0$ in $V$ and $A\subseteq U$ is a non-empty downwards-directed
set with infimum $0$ in $U$, then $\inf A=0$ in $L^0$, because $U$ is
order-dense (352Nb).   Consequently $\inf_{u\in A}u\times v=0$ in
$L^0$ and in $L^1$ (364B(b-ii), 353Pa), and

\Centerline{$\inf_{u\in A}(Tv)(u)=\inf_{u\in A}\int u\times v=0$}

\noindent (because $\int$ is order-continuous).   As $A$ is arbitrary,
$Tv$ is order-continuous.   As $v$ is arbitrary,
$T[V]\subseteq U^{\times}$.

\medskip

\quad{\bf (v)}
If $v\in V$ and $u_0\ge 0$ in $U$, set $a=\Bvalue{v>0}$.   Then
$v^+=v\times\chi a$.   Set $A=\{u:u\in U,\,0\le u\le u_0\times\chi a\}$.
Because $U$ is order-dense in $L^0$, $u_0\times\chi a=\sup A$ in $L^0$.
Because $\times$ and $\int$ are order-continuous,

$$\eqalign{(Tv)^+(u_0)
&\ge\sup_{u\in A}(Tv)(u)
=\sup_{u\in A}\int v\times u\cr
&=\int v\times u_0\times\chi a
=\int v^+\times u_0
=T(v^+)(u_0).\cr}$$

\noindent As $u_0$ is arbitrary, $(Tv)^+\ge Tv^+$.   But because $T$ is
a positive linear operator, we must have $Tv^+\ge(Tv)^+$, so that
$Tv^+=(Tv)^+$.   As $v$ is arbitrary, $T$ is a Riesz homomorphism.

\medskip

\quad{\bf (vi)} Now $T$ is injective.   \Prf\ If $v\ne 0$ in $V$, there
is a $u>0$ in $U$ such that $u\le|v|$, because $U$ is order-dense.   In
this case $u\times|v|>0$ so $\int u\times|v|>0$.   Accordingly
$|Tv|=T|v|\ne 0$ and $Tv\ne 0$.\ \Qed

\medskip

{\bf (b)} Putting (a-i) to (a-vi) together, we see that $T$ is an
injective Riesz homomorphism from $V$ to $U^{\times}$.   All this is
easy.    The point of the theorem is the fact that $T[V]$ is order-dense
in $U^{\times}$.

\Prf\  Take $h>0$ in $U^{\times}$.
Let $U_1$ be the solid linear subspace of $L^0$ generated by $U$.   Then
$U$ is an order-dense Riesz subspace of $U_1$, $h:U\to\Bbb R$ is an
order-continuous positive linear functional, and
$\sup\{h(u):u\in U,\,0\le u\le v\}$ is defined in $\Bbb R$ for every
$v\ge 0$ in $U_1$;  so we have an
extension $\tilde h$ of $h$ to $U_1$ such that $\tilde h\in U_1^{\times}$
(355F).

Set $S_1=S(\frak A)\cap U_1$;  then $S_1$ is an order-dense Riesz
subspace of $U_1$, because $S(\frak A)$ is order-dense in $L^0$ and
$U_1$ is solid in $L^0$.   Note
that $S_1$ is the linear span of $\{\chi c:c\in I\}$, where
$I=\{c:c\in\frak A,\,\chi c\in U_1\}$, and that $I$ is an ideal in
$\frak A$.

Because $h\ne 0$, $\tilde h\ne 0$;  there must therefore be a
$u_0\in S_1$ such that $\tilde h(u_0)>0$, and a $d\in I$ such that $\tilde h(\chi d)>0$.   For $a\in\frak A$, set
$\nu a=\tilde h\chi(d\Bcap a)$.
Because $\Bcap$, $\chi$ and $\tilde h$ are all order-continuous, so is
$\nu$, and $\nu:\frak A\to\Bbb R$ is a non-negative completely additive
functional.

By 365Ea, there is a $v\in L^1$ such that

\Centerline{$\int_av=\nu a=\tilde h\chi(d\Bcap a)$}

\noindent for every $a\in\frak A$;  of course $v\ge 0$.   We have
$\int u\times v\le \tilde h(u)$ whenever $u=\chi a$ for $a\in I$, and
therefore for every $u\in S_1^+$.   If $u\in U^+$, then
$A=\{u':u'\in S_1,\,0\le u'\le u\}$ is upwards-directed, $\sup A=u$ and

\Centerline{$\sup_{u'\in A}\int v\times u'\le\sup_{u'\in A}\tilde h(u')
=\tilde h(u)=h(u)$}

\noindent is finite, so $v\times u=\sup_{u\in A'}v\times u'$ belongs to
$L^1$ (365Df) and $\int v\times u\le h(u)$.   As $u$
is arbitrary, $v\in V$ and $Tv\le h$.   At the same time, because
$\chi d\in U_1$, there is a $w\in U$ such that $\chi d\le w$ and

\Centerline{$(Tv)(w)=\int v\times w\ge\int_dv=\tilde h(\chi d)>0$}

\noindent and $Tv>0$.   As $h$ is arbitrary, $T[V]$
is order-dense.\ \Qed

It follows that $T$ is order-continuous (352Nb again), as can also be easily
proved by the argument of (a-iv) above.

\medskip

{\bf (c)} Now suppose that $(\frak A,\bar\mu)$ is localizable, that is,
that $\frak A$ is Dedekind complete.   $T^{-1}:T[V]\to V$ is a Riesz
space isomorphism, so
certainly an order-continuous Riesz homomorphism;  because $V$ is a
solid linear subspace of $L^0$, $T^{-1}$ is still an injective
order-continuous Riesz homomorphism when regarded as a map from $T[V]$
to $L^0$.   Since $T[V]$ is order-dense in $U^{\times}$, $T^{-1}$ has an
extension to an order-continuous Riesz homomorphism
$Q:U^{\times}\to L^0$ (368B).
But $Q[U^{\times}]\subseteq V$.   \Prf\ Take $h\ge 0$ in $U^{\times}$
and $u\ge 0$ in $U$.   Then $B=\{g:g\in T[V],\,0\le g\le h\}$ is
upwards-directed and has supremum $h$.   For $g\in B$, we know that
$u\times T^{-1}g\in L^1$ and $\int u\times T^{-1}g=g(u)$, by the
definition of $T$.   But this means that

\Centerline{$\sup_{g\in B}\int u\times T^{-1}g
=\sup_{g\in B}g(u)=h(u)<\infty$.}

\noindent Since $\{u\times T^{-1}g:g\in B\}$ is upwards-directed, it
follows that

\Centerline{$u\times Qh=\sup_{g\in B}u\times Qg
=\sup_{g\in B}u\times T^{-1}g\in L^1$}

\noindent by 365Df again.  As $u$ is arbitrary, $Qh\in V$.   As $h$
is arbitrary (and $Q$ is linear), $Q[U^{\times}]\subseteq V$.\ \Qed

Also $Q$ is injective.   \Prf\ If $h\in U^{\times}$ is non-zero, there
is a
$v\in V$ such that $0<Tv\le|h|$, so that

\Centerline{$|Qh|=Q|h|\ge QTv=v>0$}

\noindent and $Qh\ne 0$.\ \QeD\   Since $QT$ is the identity on $V$, $Q$
and $T$ must be the two halves of a Riesz space isomorphism between $V$
and $U^{\times}$.
}%end of proof of 369C

\leader{369D}{Corollary} Let $U$ be any Riesz space such that
$U^{\times}$ separates the points of $U$.   Then there is a localizable
measure algebra $(\frak A,\bar\mu)$ such that the pair $(U,U^{\times})$
can be represented by a pair $(V,W)$ of order-dense Riesz subspaces of
$L^0=L^0(\frak A)$ such that
$W=\{w:w\in L^0,\,v\times w\in L^1$ for every $v\in V\}$, writing $L^1$ for $L^1(\frak A,\bar\mu)$.
In this case, $U^{\times\times}$ becomes represented by $\tilde
V=\{v:v\in L^0,\,v\times w\in L^1$ for every $w\in W\}\supseteq V$.

\proof{ Put 369A and 369C together.   The construction of 369A finds
$(\frak A,\bar\mu)$ and an order-dense $V$ which is isomorphic to $U$,
and 369C identifies $W$ with $V^{\times}$ and $W^{\times}$ with
$\tilde V$.   To check that $W$ is order-dense, take any $u>0$ in $L^0$.    There is a $v\in V$ such that
$0<v\le u$.   There is an $h\in(V^{\times})^+$ such that $h(v)>0$, so
there is a $w\in W^+$ such that $w\times v\ne 0$, that is,
$w\wedge v\ne 0$.
But now $w\wedge v\in W$, because $W$ is solid, and $0<w\wedge v\le u$.
}%end of proof of 369D

\cmmnt{\medskip

\noindent{\bf Remark} Thus the canonical embedding of $U$ in
$U^{\times\times}$ (356I) is represented by the embedding
$V\embedsinto\tilde V$;  $U$, or $V$, is `perfect' iff $V=\tilde V$.
}%end of comment

\leader{369E}{Kakutani's theorem}\cmmnt{ ({\smc Kakutani 1941})} If $U$
is any $L$-space, there is a
localizable measure algebra $(\frak A,\bar\mu)$ such that $U$ is
isomorphic, as Banach lattice, to $\cmmnt{L^1=}L^1(\frak A,\bar\mu)$.

\proof{ $U$ is a perfect Riesz space, and $U^{\times}=U^*$ has an order
unit $\int$ defined by saying that $\int u=\|u\|$ for $u\ge 0$ (356P).
By 369D, we can find a localizable measure algebra $(\frak A,\bar\mu)$
and an
identification of the pair $(U,U^{\times})$, as dual Riesz spaces, with
a pair $(V,W)$ of subspaces of $L^0=L^0(\frak A)$;  and $V$ will be
$\{v:v\times w\in L^1$ for every $w\in W\}$.   But $W$, like
$U^{\times}$, must have an order unit;  call it $e$.   Because $W$ is
order-dense, $\Bvalue{e>0}$ must be $1$ and $e$ must have a
multiplicative inverse
$\bover1e$ in $L^0$ (364N).   This means that $V$ must be
$\{v:v\times e\in L^1\}$, so that $v\mapsto v\times e$ is a Riesz space isomorphism
between $V$ and $L^1$, which gives a Riesz space isomorphism between $U$
and $L^1$.
Moreover, if we write $\|\,\|'$ for the norm on $V$ corresponding to the
norm of $U$, we have

\Centerline{$\|v\|'=\int|v|\times e=\int|v\times e|
=\|v\times e\|_1$ for $v\in V$.}

\noindent Thus the Riesz space isomorphism between $U$ and $L^1$ is
norm-preserving, and $U$ and $L^1$ are isomorphic as Banach lattices.
}%end of proof of 369E

\leader{369F}{}\cmmnt{ The $L^p$ spaces are leading examples for a
general theory of normed subspaces of $L^0$, which I proceed to sketch
in the rest of the section.

\medskip

\noindent}{\bf Definition} Let $\frak A$ be a Dedekind $\sigma$-complete
Boolean algebra.   An {\bf extended Fatou norm} on $L^0=L^0(\frak A)$ is
a function $\tau:L^0\to[0,\infty]$ such that

(i) $\tau(u+v)\le\tau(u)+\tau(v)$ for all $u$, $v\in L^0$;

(ii) $\tau(\alpha u)=|\alpha|\tau(u)$ whenever $u\in L^0$ and
$\alpha\in\Bbb R$\cmmnt{ (counting $0\cdot\infty$ as $0$, as usual)};

(iii) $\tau(u)\le\tau(v)$ whenever $|u|\le|v|$ in $L^0$;

(iv) $\sup_{u\in A}\tau(u)=\tau(v)$ whenever $A\subseteq (L^0)^+$ is a
non-empty upwards-directed set with supremum $v$ in $L^0$;

(v) $\tau(u)>0$ for every non-zero $u\in L^0$;

(vi) whenever $u>0$ in $L^0$ there is a $v\in L^0$ such that $0<v\le u$
and $\tau(v)<\infty$.

\leader{369G}{Proposition} Let $\frak A$ be a Dedekind
$\sigma$-complete Boolean algebra and $\tau$ an extended Fatou norm on
$L^0=L^0(\frak A)$.
Then $L^{\tau}=\{u:u\in L^0,\,\tau(u)<\infty\}$ is an order-dense solid
linear subspace of $L^0$, and $\tau$, restricted to $L^{\tau}$, is a
Fatou norm under which $L^{\tau}$ is a Banach lattice.   If
$\sequencen{u_n}$ is a non-decreasing norm-bounded sequence in
$(L^{\tau})^+$, then it has a supremum in $L^{\tau}$;  if
$\frak A$ is Dedekind complete, then $L^{\tau}$ has the Levi property.

\proof{{\bf (a)} By (i), (ii) and (iii) of 369F, $L^{\tau}$ is a solid
linear subspace of $L^0$;  by (vi), it is order-dense.   Hypotheses (i),
(ii), (iii) and (v) show that $\tau$ is a Riesz norm on $L^{\tau}$,
while (iv) shows that it is a Fatou norm.

\medskip

{\bf (b)(i)} Suppose that $\sequencen{u_n}$ is a non-decreasing
norm-bounded sequence in $(L^{\tau})^+$.   Then $u=\sup_{n\in\Bbb N}u_n$
is defined in $L^0$.   \Prf\Quer\ Otherwise, there is a $v>0$ in $L^0$
such that $kv=\sup_{n\in\Bbb N}kv\wedge u_n$ for every $k\in\Bbb N$
(368A).   By (v)-(vi) of 369F, there is a $v'$ such that $0<v'\le v$
and $0<\tau(v')<\infty$.   Now $kv'=\sup_{n\in\Bbb N}kv'\wedge u_n$ for
every $k$, so

\Centerline{$k\tau(v')=\tau(kv')=\sup_{n\in\Bbb N}\tau(kv'\wedge
u_n)\le\sup_{n\in\Bbb N}\tau(u_n)$}

\noindent for every $k$, using 369F(iv), and $\sup_{n\in\Bbb
N}\tau(u_n)=\infty$, contrary to hypothesis.\ \Bang\QeD\   By 369F(iv)
again, $\tau(u)=\sup_{n\in\Bbb N}\tau(u_n)<\infty$, so that
$u\in L^{\tau}$ and $u=\sup_{n\in\Bbb N}u_n$ in $L^{\tau}$.

\medskip

\quad{\bf (ii)} It follows that $L^{\tau}$ is complete under $\tau$.
\Prf\ Let $\sequencen{u_n}$ be a sequence in $L^{\tau}$ such that
$\tau(u_{n+1}-u_n)\le 2^{-n}$ for every $n\in\Bbb N$.   Set
$v_{mn}=\sum_{i=m}^n|u_{i+1}-u_i|$ for $m\le n$;  then
$\tau(v_{mn})\le 2^{-m+1}$ for every $n$, so by (i) just above $v_m=\sup_{n\in\Bbb N}v_{mn}$ is defined in $L^{\tau}$, and $\tau(v_m)\le 2^{-m+1}$.   Now
$v_m=|u_{m+1}-u_m|+v_{m+1}$ for each $m$, so $\sequence{m}{u_m-v_m}$ is
non-decreasing and $\sequence{m}{u_m+v_m}$ is non-increasing, while
$u_m-v_m\le u_m\le u_m+v_m$ for every $m$.   Accordingly
$u=\sup_{m\in\Bbb N}u_m-v_m$ is defined in $L^{\tau}$ and
$|u-u_m|\le v_m$ for every $m$.   But this means that
$\lim_{m\to\infty}\tau(u-u_m)\le\lim_{m\to\infty}\tau(v_m)=0$ and
$u=\lim_{m\to\infty}u_m$ in $L^{\tau}$.   As $\sequencen{u_n}$ is
arbitrary, $L^{\tau}$ is complete.\ \Qed

\medskip

{\bf (c)} Now suppose that $\frak A$ is Dedekind complete and
$A\subseteq (L^{\tau})^+$ is a non-empty upwards-directed norm-bounded
set in $L^{\tau}$.   By the argument of (b-i) above, using the other
half of 368A, $\sup A$ is defined in $L^0$ and belongs to $L^{\tau}$.
As $A$ is arbitrary, $L^{\tau}$ has the Levi property.
}%end of proof of 369G

\leader{369H}{Associate norms:  Definition} Let $(\frak A,\bar\mu)$ be a
semi-finite measure algebra, and $\tau$ an extended Fatou norm on
$L^0=L^0(\frak A)$.   Define $\tau':L^0\to[0,\infty]$ by setting

\Centerline{$\tau'(u)=\sup\{\|u\times v\|_1:v\in L^0,\,\tau(v)\le 1\}$}

\noindent for every $u\in L^0$;  then $\tau'$ is the {\bf associate} of
$\tau$.   \cmmnt{(The word suggests a symmetric relationship;  it is
justified by the next theorem.)}

\leader{369I}{Theorem} Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $\tau$ an extended Fatou norm on $L^0=L^0(\frak A)$.   Then

(i) its associate $\tau'$ is also an extended Fatou norm on $L^0$;

(ii) $\tau$ is the associate of $\tau'$;

(iii) $\|u\times v\|_1\le\tau(u)\tau'(v)$ for all $u$, $v\in L^0$.

\proof{{\bf (a)} Before embarking on the proof that $\tau'$ is an
extended Fatou seminorm on $L^0$, I give the greater part of the
argument needed to show that $\tau=\tau''$, where

\Centerline{$\tau''(u)=\sup\{\|u\times w\|_1:w\in L^0,\,
  \tau'(w)\le 1\}$}

\noindent for every $u\in L^0$.

\medskip

\quad\grheada\ Set

\Centerline{$B=\{u:u\in L^1,\,\tau(u)\le 1\}$,}

\noindent writing $L^1$ for $L^1(\frak A,\bar\mu)$.   Then $B$ is a convex set in $L^1$ and is
closed for the norm topology of $L^1$.   \Prf\ Suppose that $u$ belongs
to the closure of $B$ in $L^1$.   Then for each $n\in\Bbb N$ we can
choose $u_n\in B$ such that $\|u-u_n\|_1\le 2^{-n}$.   Set
$v_{mn}=\inf_{m\le i\le n}|u_i|$ for $m\le n$, and

\Centerline{$v_m=\inf_{n\ge m}v_{mn}=\inf_{n\ge m}|u_n|\le|u|$}

\noindent for $m\in\Bbb N$.  The sequence $\sequence{m}{v_m}$ is
non-decreasing, $\tau(v_m)\le\tau(u_m)\le 1$ for every $m$, and

\Centerline{$\||u|-v_m\|_1\le\sup_{n\ge m}\||u|-v_{mn}\|_1
\le\sum_{i=m}^{\infty}\||u|-|u_i|\|_1
\le\sum_{i=m}^{\infty}\|u-u_i\|_1\to 0$}

\noindent as $m\to\infty$.   So $|u|=\sup_{m\in\Bbb N}v_m$ in $L^0$,

\Centerline{$\tau(u)=\tau(|u|)=\sup_{m\in\Bbb N}\tau(v_m)\le 1$}

\noindent and $u\in B$.\ \Qed

\medskip

\quad\grheadb\ Now take any $u_0\in L^0$ such that $\tau(u_0)>1$.
Then, writing $\frak A^f$ for $\{a:\bar\mu a<\infty\}$,

\Centerline{$A=\{u:u\in S(\frak A^f),\,0\le u\le u_0\}$}
\noindent is an upwards-directed set with supremum $u_0$ (this is where
I use the hypothesis that $(\frak A,\bar\mu)$ is semi-finite, so that
$S(\frak A^f)$ is order-dense in $L^0$), and $\sup_{u\in
A}\tau(u)=\tau(u_0)>1$.   Take $u_1\in A$ such that $\tau(u_1)>1$, that
is, $u_1\notin B$.   By the Hahn-Banach theorem (3A5Cc), there is a
continuous linear functional $f:L^1\to\Bbb R$ such that $f(u_1)>1$ but
$f(u)\le 1$ for every $u\in B$.   Because $(L^1)^*=(L^1)^{\sim}$
(356Dc), $|f|$ is defined in $(L^1)^*$, and of course

\Centerline{$|f|(u_1)\ge f(u_1)>1$,
\quad$|f|(u)=\sup\{f(v):|v|\le u\}\le 1$}

\noindent whenever $u\in B$ and $u\ge 0$.   Set $c=\Bvalue{u_1>0}$, so
that $\bar\mu c<\infty$, and define

\Centerline{$\nu a=|f|(\chi(a\Bcap c))$}

\noindent for every $a\in\frak A$.   Then $\nu$ is a completely additive
real-valued functional on $\frak A$, so there is a $w\in L^1$ such that
$\nu a=\int_aw$ for every $a\in\frak A$ (365Ea).  Because
$\nu a\ge 0$ for every $a$, $w\ge 0$.   Now

\Centerline{$\int_aw=|f|(\chi a\times\chi c)$}

\noindent for every $a\in\frak A$, so

\Centerline{$\int w\times u=|f|(u\times\chi c)\le|f|(u)\le 1$}

\noindent for every $u\in S(\frak A)^+\cap B$.   But if $\tau(v)\le 1$,
then

\Centerline{$A_v=\{u:u\in S(\frak A)^+\cap B,\,u\le|v|\}$}

\noindent is an upwards-directed set with supremum $|v|$, so that

\Centerline{$\|w\times v\|_1=\sup_{u\in A_v}\int w\times u\le 1$.}

\noindent Thus $\tau'(w)\le 1$.   On the other hand,

\Centerline{$\|w\times u_0\|_1\ge\int w\times u_0
\ge\int w\times u_1=|f|(u_1)>1$,}

\noindent so $\tau''(u_0)>1$.

\medskip

\quad\grheadc\ This shows that, for $u\in L^0$,

\Centerline{$\tau''(u)\le 1\Longrightarrow\tau(u)\le 1$.}

\medskip

{\bf (c)} Now I return to the proof that $\tau'$ is an extended Fatou
norm.   It is easy to check that it satisfies conditions (i)-(iv) of
369F;  in effect, these depend only on the fact that $\|\,\|_1$ is an
extended Fatou norm.   For (v)-(vi), take $v>0$ in $L^0$.   Then there
is a $u$ such that $0\le u\le v$ and $0<\tau(u)<\infty$;   set
$\alpha=1/\tau(u)$.   Then $\tau(2\alpha u)>1$, so that
$\tau''(2\alpha u)>1$ and there is a $w\in L^0$ such that
$\tau'(w)\le 1$ and $\|2\alpha u\times w\|_1>1$.
But now set $v_1=v\wedge|w|$;  then

\Centerline{$v\ge v_1\ge u\wedge|w|>0$,}

\noindent while $\tau'(v_1)<\infty$.   Also $v\wedge\alpha u\ne 0$ so

\Centerline{$\tau'(v)\ge\|v\times\alpha u\|_1>0$.}

\noindent As $v$ is arbitrary, $\tau'$ satisfies 369F(v)-(vi).

\medskip

{\bf (d)} Accordingly $\tau''$ also is an extended Fatou norm.   Now in
(a) I showed that

\Centerline{$\tau''(u)\le 1\Longrightarrow\tau(u)\le 1$.}

\noindent It follows easily that $\tau(u)\le\tau''(u)$ for every $u$
(since otherwise there would be some $\alpha$ such that

\Centerline{$\tau''(\alpha
u)=\alpha\tau''(u)<1<\alpha\tau(u)=\tau(\alpha u)$).}

\noindent On the other hand, we surely have

\Centerline{$\tau(u)\le 1
\Longrightarrow\|u\times v\|_1\le 1\text{ whenever }\tau'(v)\le 1
\Longrightarrow\tau''(u)\le 1$,}

\noindent so we must also have $\tau''(u)\le\tau(u)$ for every $u$.
Thus $\tau''=\tau$, as claimed.

\medskip

{\bf (e)} Of course we have $\|u\times v\|_1\le 1$ whenever
$\tau(u)\le 1$ and $\tau'(v)\le 1$.   It follows easily that
$\|u\times v\|_1\le\tau(u)\tau'(v)$ whenever $u$, $v\in L^0$ and both $\tau(u)$, $\tau'(v)$ are non-zero.   But if one of them is zero, then $u\times v=0$,
because both $\tau$ and $\tau'$ satisfy (v) of 369F, so the result is
trivial.
}%end of proof of 369I

\leader{369J}{Theorem} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra, and $\tau$ an extended Fatou norm on
$L^0=L^0(\frak A)$, with associate $\theta$.   Then

\Centerline{$L^{\theta}=\{v:v\in L^0,\,
  u\times v\in L^1(\frak A,\bar\mu)$ for every $u\in L^{\tau}\}$.}

\proof{{\bf (a)} If $v\in L^{\theta}$ and $u\in L^{\tau}$, then
$\|u\times v\|_1$ is finite, by 369I(iii), so
$u\times v\in L^1=L^1(\frak A,\bar\mu)$.

\medskip

{\bf (b)} If $v\notin L^{\theta}$ then for every $n\in\Bbb N$ there is a
$u_n$ such that $\tau(u_n)\le 1$ and $\|u_n\times v\|_1\ge 2^n$.   Set
$w_n=\sum_{i=0}^n2^{-i}|u_i|$ for each $n$.   Then $\sequencen{w_n}$ is
a non-decreasing sequence and $\tau(w_n)\le 2$ for each $n$, so
$w=\sup_{n\in\Bbb N}w_n$ is defined in $L^{\tau}$, by 369G;  now
$\int w\times|v|\ge n+1$ for every $n$, so $w\times v\notin L^1$.
}%end of proof of 369J


\leader{369K}{Corollary} Let $(\frak A,\bar\mu)$ be a localizable
measure algebra, and $\tau$ an extended Fatou norm on
$\cmmnt{L^0=}L^0(\frak A)$, with associate $\theta$.   Then $L^{\theta}$ may be identified, as normed Riesz space, with
$(L^{\tau})^{\times}\subseteq(L^{\tau})^*$, and $L^{\tau}$ is a perfect
Riesz space.

\proof{ Putting 369J and 369C together, we have an identification
between $L^{\theta}$ and $(L^{\tau})^{\times}$.   Now 369I tells us that
$\tau$ is the associate of $\theta$, so that we can identify $L^{\tau}$
with $(L^{\theta})^{\times}$, and $L^{\tau}$ is perfect, as in 369D.

By the definition of $\theta$, we have, for any $v\in L^{\theta}$,

$$\eqalign{\theta(v)
&=\sup_{\tau(u)\le 1}\|u\times v\|_1\cr
&=\sup_{\tau(u)\le 1,\|w\|_{\infty}\le 1}\int u\times v\times w
=\sup_{\tau(u)\le 1}\int u\times v,\cr}$$

\noindent which is the norm of the linear functional on $L^{\tau}$
corresponding to $v$.
}%end of proof of 369K

\leader{369L}{$\pmb{L^p}$}\cmmnt{ I remarked above that the $L^p$ spaces are
leading examples for this theory;  perhaps I should spell out the
details.}   Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra and
$p\in[1,\infty]$.   Then $\|\,\|_p$ is an extended Fatou norm.
\prooflet{\Prf\ Conditions (i)-(iii) and (v) of 369F are satisfied just because
$L^p=L^p_{\bar\mu}$ is a solid linear subspace of $L^0(\frak A)$
on which $\|\,\|_p$ is a Riesz norm, (iv)
because $\|\,\|_p$ is a Fatou norm with the Levi property (363Ba,
365C, 366D), and (vi) because $S(\frak A^f)$ is included in
$L^p$ and order-dense in $L^0=L^0(\frak A)$ (364K).\ \Qed}

As usual, set $q=p/(p-1)$ if $1<p<\infty$, $\infty$ if $p=1$, and $1$ if
$p=\infty$.   Then $\|\,\|_q$ is the associate extended Fatou norm of
$\|\,\|_p$.   \prooflet{\Prf\ By 365Lb and 366C,
$\|v\|_q=\sup\{\|u\times v\|_1:\|u\|_p\le 1\}$ for every
$v\in L^q=L^q_{\bar\mu}$.   But as $L^q$ is order-dense in $L^0$,

$$\eqalign{\|v\|_q
=\sup_{w\in L^q,|w|\le v}\|w\|_q
&=\sup\{\int|u|\times|w|:w\in L^q,\,w\le|v|,\,\|u\|_p\le 1\}\cr
&=\sup\{\int|u|\times|v|:\|u\|_p\le 1\}\cr}$$

\noindent for every $v\in L^0$.\ \Qed}

\vleader{108pt}{369M}{Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra and $\tau$ an extended Fatou norm on
$L^0=L^0(\frak A)$.   Then

(a) the embedding $L^{\tau}\embedsinto L^0$ is continuous for the norm
topology of $L^{\tau}$ and the topology of convergence in measure on
$L^0$;

(b) $\tau:L^0\to[0,\infty]$ is lower semi-continuous\cmmnt{, that is, all the balls $\{u:\tau(u)\le\gamma\}$ are closed for the topology
of convergence in measure};

(c) if $\sequencen{u_n}$ is a sequence in $L^0$ which is
order*-convergent to $u\in L^0$\cmmnt{ (definition:  367A)}, then
$\tau(u)$ is at most $\liminf_{n\to\infty}\tau(u_n)$.

\proof{{\bf (a)} This is a special case of 367O.

\medskip

{\bf (b)} Set $B_{\gamma}=\{u:\tau(u)\le\gamma\}$.   If
$u\in L^0\setminus B_{\gamma}$, then

\Centerline{$A=\{|u|\times\chi a:a\in\frak A^f\}$}

\noindent is an upwards-directed set with supremum $|u|$, so there is an
$a\in\frak A^f$ such that $\tau(u\times\chi a)>\gamma$.   \Quer\ If $u$
is in the closure of $B_{\gamma}$ for the topology of convergence in
measure, then for every $k\in\Bbb N$ there is a $v_k\in B_{\gamma}$ such
that $\bar\mu(a\Bcap\Bvalue{|u-v_k|>2^{-k}})\le 2^{-k}$ (see the
formulae in 367L).   Set

\Centerline{$v'_k=|u|\wedge\inf_{i\ge k}|v_i|$}

\noindent for each $k$, and $v^*=\sup_{k\in\Bbb N}v'_k$.   Then
$\tau(v'_k)\le\tau(v_k)\le\gamma$ for each $k$, and $\sequence{k}{v_k}$
is non-decreasing, so $\tau(v^*)\le\gamma$.   But

\Centerline{$a\Bcap\Bvalue{|u|-v^*>2^{-k}}
\Bsubseteq a\Bcap\sup_{i\ge k}\Bvalue{|u-v_i|>2^{-k}}$}

\noindent has measure at most $\sum_{i=k}^{\infty}2^{-i}$ for each $k$,
so $a\Bcap\Bvalue{|u|-v^*>0}$ must be $0$, that is, $|u|\times\chi a\le
v^*$ and $\tau(|u|\times\chi a)\le\gamma$;  contrary to the choice of
$a$.\ \BanG\   Thus $u$ cannot belong to the closure of $B_{\gamma}$.
As $u$ is arbitrary, $B_{\gamma}$ is closed.

\medskip

{\bf (c)} If $\sequencen{u_n}$ order*-converges to $u$, it converges
in measure (367Ma).   If $\gamma>\liminf_{n\to\infty}\tau(u_n)$, there
is a subsequence of $\sequencen{u_n}$ in $B_{\gamma}$, and
$\tau(u)\le\gamma$,
by (b).   As $\gamma$ is arbitary,
$\tau(u)\le\liminf_{n\to\infty}\tau(u_n)$.
}%end of proof of 369M

\leader{369N}{}\cmmnt{ I now turn to another special case which we
have already had occasion to consider in other contexts.

\medskip

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

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

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

\noindent and

\Centerline{$\|u\|_{\infty,1}=\max(\|u\|_1,\|u\|_{\infty})$}

\noindent for $u\in L^0(\frak A)$.

\cmmnt{\medskip

\noindent{\bf Remark} I hope that the notation I have chosen here will
not completely overload your short-term memory.   The idea is that in
$M^{p,q}$ the symbol $p$ is supposed to indicate the `local' nature
of the space, that is, the nature of $u\times\chi a$ where
$u\in M^{p,q}$ and $\bar\mu a<\infty$, while $q$ indicates the nature of
$|u|\wedge\chi 1$ for $u\in M^{p,q}$.   Thus $M^{1,\infty}$ is the space
of $u$ such that $u\times\chi a\in L^1$ for every $a\in\frak A^f$ and
$|u|\wedge\chi 1\in L^{\infty}$;  in $M^{1,0}$ we demand further that
$|u|\wedge\chi 1\in M^0$ (366F);  while in $M^{\infty,1}$ we ask that
$|u|\wedge\chi 1\in L^1$ and that $u\times\chi a\in L^{\infty}$ for every $a\in\frak A^f$.
}%end of comment

\leader{369O}{Proposition} Let $(\frak A,\bar\mu)$ be a semi-finite
measure algebra.

(a) $\|\,\|_{\infty,1}$ is an extended Fatou norm on $L^0=L^0(\frak A)$,
and the corresponding Banach lattice is $M^{\infty,1}(\frak A,\bar\mu)$.

(b) The associate of $\|\,\|_{\infty,1}$ is $\|\,\|_{1,\infty}$, which
may be defined by any of the formulae

$$\eqalign{\|u\|_{1,\infty}
&=\sup\{\|u\times v\|_1:v\in L^0,\,\|v\|_{\infty,1}\le 1\}\cr
&=\min\{\|v\|_1+\|w\|_{\infty}:v\in L^1,\,w\in L^{\infty},\,v+w=u\}\cr
&=\min\{\alpha+\int(|u|-\alpha\chi 1)^+:\alpha\ge 0\}\cr
&=\int_0^{\infty}\min(1,\bar\mu\Bvalue{|u|>\alpha})d\alpha\cr}$$

\noindent for every $u\in L^0$, writing $L^1=L^1(\frak A,\bar\mu)$,
$L^{\infty}=L^{\infty}(\frak A)$.

(c)

\Centerline{$\{u:u\in L^0,\,\|u\|_{1,\infty}<\infty\}
=M^{1,\infty}=M^{1,\infty}(\frak A,\bar\mu)$,}

\Centerline{$\{u:u\in L^0,\,\|u\|_{\infty,1}<\infty\}=M^{\infty,1}
=M^{\infty,1}(\frak A,\bar\mu)$.}

(d) Writing $\frak A^f=\{a:\bar\mu a<\infty\}$, $S(\frak A^f)$ is
norm-dense in $M^{\infty,1}$ and $S(\frak A)$ is norm-dense in
$M^{1,\infty}$.

(e) For any $p\in[1,\infty]$,

\Centerline{$\|u\|_{1,\infty}\le\|u\|_p\le\|u\|_{\infty,1}$}

\noindent for every $u\in L^0$.

\cmmnt{\medskip

\noindent{\bf Remark} By writing `min' rather than `inf' in the
formulae of part (b) I mean to assert that the infima are attained.
}

\proof{{\bf (a)} This is easy;  to see that $\|\,\|_{\infty,1}$ is an
extended Fatou norm all we need to know is that $\|\,\|_1$
and $\|\,\|_{\infty}$ are extended Fatou norms, and work through the
clauses of 369F.   And obviously

\Centerline{$M^{\infty,1}=\{u:\|u\|_1<\infty$, $\|u\|_{\infty}<\infty\}
=\{u:\|u\|_{\infty,1}<\infty\}$.}

\medskip

{\bf (b)} We have four functionals on $L^0$ to look at;  let me give
them names:

\Centerline{$\tau_1(u)=\sup\{\|u\times v\|_1:\|v\|_{\infty,1}\le 1\}$,}

\Centerline{$\tau_2(u)=\inf\{\|u'\|_1+\|u''\|_{\infty}:u=u'+u''\}$,}

\Centerline{$\tau_3(u)
=\inf_{\alpha\ge 0}(\alpha+\int(|u|-\alpha\chi 1)^+)$,}

\Centerline{$\tau_4(u)
=\int_0^{\infty}\min(1,\bar\mu\Bvalue{|u|>\alpha})d\alpha$.}

\noindent (I write `$\inf$' here to avoid the question of attainment
for the moment.)   Now we have the following.

\medskip

\quad{\bf (i)} $\tau_1(u)\le\tau_2(u)$.   \Prf\ If
$\|v\|_{\infty,1}\le 1$ and $u=u'+u''$, then

\Centerline{$\|u\times v\|_1
\le\|u'\times v\|_1+\|u''\times v\|_1
\le\|u'\|_1\|v\|_{\infty}+\|u''\|_{\infty}\|v\|_1
\le\|u'\|_1+\|u''\|_{\infty}$.}

\noindent Taking the supremum over $v$ and the infinum over $u'$ and
$u''$, $\tau_1(u)\le\tau_2(u)$.\ \Qed

\medskip

\quad{\bf (ii)} $\tau_2(u)\le\tau_4(u)$.   \Prf\ If $\tau_4(u)=\infty$
this is trivial.   Otherwise, take $w$ such that $\|w\|_{\infty}\le 1$
and $u=|u|\times w$.   Set
$\alpha_0=\inf\{\alpha:\bar\mu\Bvalue{|u|>\alpha}\le 1\}$, and try

\Centerline{$u'=w\times(|u|-\alpha_0\chi 1)^+$,
\quad$u''=w\times(|u|\wedge\alpha_0\chi 1)$.}

\noindent Then $u=u'+u''$, $|u'|\le(|u|-\alpha_0\chi 1)^+$,

$$\eqalign{\|u'\|_1
&=\int_0^{\infty}\bar\mu\Bvalue{|u'|>\alpha}d\alpha\
=\int_0^{\infty}\bar\mu\Bvalue{|u|>\alpha+\alpha_0}d\alpha\cr
&=\int_{\alpha_0}^{\infty}\bar\mu\Bvalue{|u|>\alpha}d\alpha
=\int_{\alpha_0}^{\infty}
   \min(1,\bar\mu\Bvalue{|u|>\alpha})d\alpha\cr}$$

\noindent and

\Centerline{$\|u''\|_{\infty}
\le\alpha_0
=\int_0^{\alpha_0}\min(1,\Bvalue{|u|>\alpha})d\alpha$,}

\noindent so

\Centerline{$\tau_2(u)\le\|u'\|_1+\|u''\|_{\infty}\le\tau_4(u)$.
\Qed}

\medskip

\quad{\bf (iii)} $\tau_4(u)\le\tau_3(u)$.   \Prf\ For any $\alpha\ge 0$,

$$\eqalign{\tau_4(u)
&=\int_0^{\alpha}\min(1,\bar\mu\Bvalue{|u|>\beta})d\beta
   +\int_{\alpha}^{\infty}\min(1,\bar\mu\Bvalue{|u|>\beta})d\beta\cr
&\le\alpha+\int_0^{\infty}\bar\mu\Bvalue{|u|>\alpha+\beta}d\beta\cr
&=\alpha
   +\int_0^{\infty}\bar\mu\Bvalue{(|u|-\alpha\chi 1)^+>\beta}d\beta
=\alpha+\int(|u|-\alpha\chi 1)^+.\cr}$$

\noindent Taking the infimum over $\alpha$,
$\tau_4(u)\le\tau_3(u)$.\ \Qed

\medskip

\quad{\bf (iv)} $\tau_3(u)\le\tau_1(u)$.

\medskip

\quad\Prf\grheada\  It is enough to consider the case
$0<\tau_1(u)<\infty$,
because if $\tau_1(u)=0$ then $u=0$ and evidently $\tau_3(0)=0$, while
if $\tau_1(u)=\infty$ the required inequality is trivial.   Furthermore,
since $\tau_3(u)=\tau_3(|u|)$ and $\tau_1(u)=\tau_1(|u|)$, it is enough
to consider the case $u\ge 0$.

\medskip

\qquad\grheadb\ Note next that if $\bar\mu a<\infty$, then
$\|\bover1{\max(1,\bar\mu a)}\chi a\|_{\infty,1}\le 1$, so that
$\int_au\le\max(1,\bar\mu a)\tau_1(u)$.

\medskip

\qquad\grheadc\ Set $c=\Bvalue{u>2\tau_1(u)}$.   If $a\Bsubseteq c$ and
$\bar\mu a<\infty$, then

\Centerline{$2\tau_1(u)\bar\mu a\le\int_au
\le\max(1,\bar\mu a)\tau_1(u)$,}

\noindent so $\bar\mu a\le\bover12$.   As $(\frak A,\bar\mu)$ is
semi-finite, it follows that $\bar\mu c\le\bover12$ (322Eb).

\medskip

\qquad\grheadd\ I may therefore write

\Centerline{$\alpha_0=\inf\{\alpha:\alpha\ge
0,\,\bar\mu\Bvalue{u>\alpha}\le 1\}$.}

\noindent Now
$\Bvalue{u>\alpha_0}=\sup_{\alpha>\alpha_0}\Bvalue{u>\alpha}$, so


\Centerline{$\bar\mu\Bvalue{u>\alpha_0}
=\sup_{\alpha>\alpha_0}\bar\mu\Bvalue{u>\alpha}\le 1$.}

\medskip

\qquad\grheade\ If $\alpha\ge\alpha_0$ then

\Centerline{$(u-\alpha_0\chi 1)^+
\le(\alpha-\alpha_0)\chi\Bvalue{u>\alpha_0}+(u-\alpha\chi 1)^+$,}
\noindent so

$$\eqalign{\alpha_0+\int(u-\alpha_0\chi 1)^+
&\le\alpha_0+(\alpha-\alpha_0)\bar\mu\Bvalue{u>\alpha_0}
   +\int(u-\alpha\chi 1)^+\cr
&\le\alpha+\int(u-\alpha\chi 1)^+.\cr}$$

\noindent If $0\le\alpha<\alpha_0$ then, for every
$\beta\in\coint{0,\alpha_0-\alpha}$,

\Centerline{$(u-\alpha_0\chi 1)^++\beta\Bvalue{u>\alpha+\beta}
\le(u-\alpha\chi 1)^+$,}

\noindent while $\bar\mu\Bvalue{u>\alpha+\beta}>1$, so

\Centerline{$\int(u-\alpha_0\chi 1)^++\beta+\alpha
\le\alpha+\int(u-\alpha\chi 1)^+$;}

\noindent taking the supremum over $\beta$,

\Centerline{$\alpha_0+\int(u-\alpha_0\chi 1)^+
\le\alpha+\int(u-\alpha\chi 1)^+$.}

\noindent Thus $\alpha_0+\int(u-\alpha_0\chi 1)^+=\tau_3(u)$.

\medskip

\qquad{\bf ($\pmb{\zeta}$)} If $\alpha_0=0$, take $v=\chi\Bvalue{u>0}$;
then $\|v\|_{\infty,1}=\bar\mu\Bvalue{u>0}\le 1$ and

\Centerline{$\tau_3(u)=\int u=\|u\times v\|_1\le\tau_1(u)$.}

\qquad{\bf ($\pmb{\eta}$)} If $\alpha_0>0$, set
$\gamma=\bar\mu\Bvalue{u>\alpha_0}$.
Take any $\beta\in\coint{0,\alpha_0}$.   Then
$\bar\mu(\Bvalue{u>\beta}\Bsetminus\Bvalue{u>\alpha_0})>1-\gamma$, so
there is a $b\Bsubseteq\Bvalue{u>\beta}\Bsetminus\Bvalue{u>\alpha_0}$
such that $1-\gamma<\bar\mu b<\infty$.   Set
$v=\chi\Bvalue{u>\alpha_0}+\bover{1-\gamma}{\bar\mu b}\chi b$.   Then
$\|v\|_{\infty,1}=1$ so

\Centerline{$\tau_1(u)\ge\int u\times v
\ge\int(u-\alpha_0\chi 1)^+
+\alpha_0\gamma+\beta\Bover{1-\gamma}{\bar\mu b}\bar\mu b
=\tau_3(u)-(1-\gamma)(\alpha_0-\beta)$.}

\noindent As $\beta$ is arbitrary, $\tau_1(u)\ge\tau_3(u)$ in this case
also.\ \Qed

\medskip

\quad{\bf (v)} Thus $\tau_1(u)=\tau_2(u)=\tau_3(u)=\tau_4(u)$ for every
$u\in L^0$, and I may write $\|u\|_{1,\infty}$ for their common value;
being the associate of $\|\,\|_{\infty,1}$, $\|\,\|_{1,\infty}$ is an
extended Fatou norm.   As for the attainment of the infima, the argument
of (iv-$\epsilon$) above shows that, at least when
$0<\|u\|_{1,\infty}<\infty$, there is an $\alpha_0$ such that
$\alpha_0+\int(|u|-\alpha_0)^+=\|u\|_{1,\infty}$.   This omits the cases
$\|u\|_{1,\infty}\in\{0,\infty\}$;  but in either of these cases we can
set $\alpha_0=0$ to see that the infimum is attained for trivial
reasons.
For the other infimum, observe that the argument of (ii) produces $u'$,
$u''$ such that $u=u'+u''$ and $\|u'\|_1+\|u''\|_{\infty}\le\tau_4(u)$.

\medskip

{\bf (c)} This is now obvious from the definition of $\|\,\|_{\infty,1}$
and the characterization of $\|\,\|_{1,\infty}$ in terms of $\|\,\|_1$
and $\|\,\|_{\infty}$.

\medskip

{\bf (d)} To see that $S=S(\frak A)$ is norm-dense in $M^{1,\infty}$, we
need only note that $S$ is dense in $L^{\infty}$ and $S\cap L^1$ is
dense in $L^1$;  so that given $v\in L^1$, $w\in L^{\infty}$ and
$\epsilon>0$ there are $v'$, $w'\in S$ such that

\Centerline{$\|(v+w)-(v'+w')\|_{1,\infty}\le
\|v-v'\|_1+\|w-w'\|_{\infty}\le\epsilon$.}

\noindent As for $M^{\infty,1}$, if $u\ge 0$ in $M^{\infty,1}$ and
$r\in\Bbb N$, set $v_r=\sup_{k\in\Bbb N}2^{-r}k\chi\Bvalue{u>2^{-r}k}$;
then each $v_r$ belongs to $S^f=S(\frak A^f)$ and
$\|u-v_r\|_{\infty}\le 2^{-r}$, while $\sequence{r}{v_r}$ is a
non-decreasing sequence with supremum $u$, so that
$\lim_{r\to\infty}\int v_r=\int u$ and
$\lim_{r\to\infty}\|u-v_r\|_{\infty,1}=0$.   Thus $(S^f)^+$ is dense in
$(M^{\infty,1})^+$.   As usual, it follows that $S^f=(S^f)^+-(S^f)^+$ is
dense in $M^{\infty,1}=(M^{\infty,1})^+-(M^{\infty,1})^+$.

\medskip

{\bf (e)(i)} If $p=1$ or $p=\infty$ this is immediate from the
definition
of $\|\,\|_{\infty,1}$ and the characterization of $\|\,\|_{1,\infty}$
in
(b).   So suppose henceforth that $1<p<\infty$.

\medskip

\quad{\bf (ii)} If $\|u\|_{\infty,1}\le 1$ then $\|u\|_p\le 1$.   \Prf\
Because $\|u\|_{\infty}\le 1$, $|u|^p\le|u|$, so that
$\int|u|^p\le\|u\|_1\le 1$ and $\|u\|_p\le 1$.\ \Qed

On considering scalar multiples of $u$, we see at once that
$\|u\|_p\le\|u\|_{\infty,1}$ for every $u\in L^0$.

\medskip

\quad{\bf (ii)} Now set $q=p/(p-1)$.   Then

$$\eqalignno{\|u\|_p&=\sup\{\|u\times v\|_1:\|v\|_q\le 1\}\cr
\noalign{\noindent (369L)}
&\ge\sup\{\|u\times v\|_1:\|v\|_{\infty,1}\le 1\}
=\|u\|_{1,\infty}\cr}$$

\noindent because $\|\,\|_{1,\infty}$ is the associate of
$\|\,\|_{\infty,1}$.   This completes the proof.
}%end of proof of 369O

\leader{369P}{}\cmmnt{ In preparation for some ideas in \S372, I go a
little farther with $M^{1,0}$, as defined in 366F.

\medskip

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

(a) $M^{1,0}=M^{1,0}(\frak A,\bar\mu)$ is a norm-closed solid linear
subspace of $\cmmnt{M^{1,\infty}=}M^{1,\infty}(\frak A,\bar\mu)$.

(b) The norm $\|\,\|_{1,\infty}$ is order-continuous on
$M^{1,0}$.

(c) $S(\frak A^f)$ and $L^1(\frak A,\bar\mu)$ are norm-dense and
order-dense in $M^{1,0}$.

\proof{{\bf (a)} Of course $M^{1,0}$, being a solid linear subspace
of $L^0=L^0(\frak A)$ included in $M^{1,\infty}$, is a solid linear subspace of
$M^{1,\infty}$.   To see that it is norm-closed, take any point $u$ of
its closure.   Then for any $\epsilon>0$ there is a $v\in M^{1,0}$ such
that $\|u-v\|_{1,\infty}\le\epsilon$;  now
$(|u-v|-\epsilon\chi 1)^+\in L^1=L^1_{\bar\mu}$, so
$\Bvalue{|u-v|>2\epsilon}$ has finite measure;  also
$\Bvalue{|v|>\epsilon}$ has finite measure, so

\Centerline{$\Bvalue{|u|>3\epsilon}
\Bsubseteq\Bvalue{|u-v|>2\epsilon}\Bcup\Bvalue{|v|>\epsilon}$}

\noindent (364Ea) has finite measure.   As $\epsilon$ is
arbitrary, $u\in M^{1,0}$;  as $u$ is arbitrary, $M^{1,0}$ is closed.

\medskip

{\bf (b)} Suppose that $A\subseteq M^{1,0}$ is non-empty and
downwards-directed and has infimum $0$.   Let $\epsilon>0$.   Set
$B=\{(u-\epsilon\chi 1)^+:u\in A\}$.   Then $B\subseteq L^1$ (by 366Gc);
$B$ is
non-empty and downwards-directed and has infimum $0$.   Because
$\|\,\|_1$ is order-continuous (365C), $\inf_{v\in B}\|v\|_1=0$ and
there is a $u\in A$ such that $\|(u-\epsilon\chi 1)^+\|_1\le\epsilon$,
so that $\|u\|_{1,\infty}\le 2\epsilon$.   As $\epsilon$ is arbitrary,
$\inf_{u\in A}\|u\|_{1,\infty}=0$;  as $A$ is arbitrary,
$\|\,\|_{1,\infty}$ is order-continuous on $M^{1,0}$.

\medskip

{\bf (c)} By 366Gb, $S(\frak A^f)$ is order-dense in $M^{1,0}$.
Because the norm of $M^{1,0}$ is order-continuous, $S(\frak A^f)$ is
also norm-dense (354Ef).   Now
$S(\frak A^f)\subseteq L^1\subseteq M^{1,0}$, so $L^1$ must also be norm-dense and order-dense.
}%end of proof of 369P

\leader{369Q}{Corollary} Let $(\frak A,\bar\mu)$ be a localizable
measure algebra.   Set $M^{1,\infty}=M^{1,\infty}(\frak A,\bar\mu)$,
etc.

(a) $(M^{1,\infty})^{\times}$ and $(M^{1,0})^{\times}$ can both be
identified with $M^{\infty,1}$.

(b) $(M^{\infty,1})^{\times}$ can be identified with $M^{1,\infty}$;
$M^{1,\infty}$ and $M^{\infty,1}$ are perfect Riesz spaces.

\proof{ Everything is covered by 369O and 369K except the identification
of $(M^{1,0})^{\times}$ with $M^{\infty,1}$.   For this I return to
369C.   Of course $M^{1,0}$ is order-dense in $L^0$, because it includes
$L^1$, or otherwise.   Setting

\Centerline{$V=\{v:v\in L^0,\,u\times v\in L^1$ for every $u\in
M^{1,0}\}$,}

\noindent 369C identifies $V$ with $(M^{1,0})^{\times}$.   Of course
$M^{\infty,1}\subseteq V$ just because $M^{1,0}\subseteq M^{1,\infty}$.

Also $V\subseteq M^{\infty,1}$.   \Prf\ Because $L^1\subseteq M^{1,0}$ and $\|\,\|_{\infty}$ is the associate of $\|\,\|_1$,
$V\subseteq L^{\infty}$.   \Quer\ If there is a $v\in V\setminus L^1$,
then (because $(\frak A,\bar\mu)$ is semi-finite,
so that $|v|=\sup_{a\in\frak A^f}|v|\times\chi a$)
$\sup_{a\in\frak A^f}\int_a|v|=\infty$.   For each $n\in\Bbb N$ choose $a_n\in\frak A^f$
such that $\int_{a_n}|v|\ge 4^n$, and set
$u=\sup_{n\in\Bbb N}2^{-n}\chi a_n\in M^{1,0}$;  then
$\int u\times|v|\ge 2^n$ for each $n$, so
$v\notin V$.\ \BanG\  Thus $V\subseteq L^1$ and
$V\subseteq M^{\infty,1}$.\ \Qed

So $M^{\infty,1}=V$ can be identified with $(M^{1,0})^{\times}$.
}%end of proof of 369Q

\leader{369R}{}\cmmnt{ The detailed formulae of 369O are of course
special to the norms $\|\,\|_1$, $\|\,\|_{\infty}$, but the general
phenomenon is not.

\medskip

\noindent}{\bf Theorem} Let $(\frak A,\bar\mu)$ be a localizable measure
algebra, and $\tau_1$, $\tau_2$ two extended Fatou norms on
$L^0=L^0(\frak A)$ with associates $\tau'_1$, $\tau'_2$.   Then we have
an extended Fatou norm $\tau$ defined by the formula

\Centerline{$\tau(u)=\min\{\tau_1(v)+\tau_2(w):v,\,w\in L^0,\,v+w=u\}$}

\noindent for every $u\in L^0$, and its associate $\tau'$ is given by
the formula

\Centerline{$\tau'(u)=\max(\tau'_1(u),\tau'_2(u))$}

\noindent for every $u\in L^0$.   Moreover, the corresponding function
spaces are

\Centerline{$L^{\tau}=L^{\tau_1}+L^{\tau_2}$,
\quad$L^{\tau'}=L^{\tau'_1}\cap L^{\tau'_2}$.}

\proof{{\bf (a)} For the moment, define $\tau$ by setting

\Centerline{$\tau(u)=\inf\{\tau_1(v)+\tau_2(w):v+w=u\}$}

\noindent for $u\in L^0$.   It is easy to check that, for $u$,
$u'\in L^0$ and $\alpha\in\Bbb R$,

\Centerline{$\tau(u+u')\le\tau(u)+\tau(u')$,
\quad$\tau(\alpha u)=|\alpha|\tau(u)$,
\quad$\tau(u)\le\tau(u')$ if $|u|\le|u'|$.}

\noindent (For the last, remember that in this case $u=u'\times z$ where
$\|z\|_{\infty}\le 1$.)    Note also that if $u\ge 0$ then
$\tau(u)=\inf\{\tau_1(v)+\tau_2(u-v):0\le v\le u\}$.

\medskip

{\bf (b)} Take any non-empty, upwards-directed set $A\subseteq(L^0)^+$,
with supremum $u_0$.   Suppose that
$\gamma=\sup_{u\in A}\tau(u)<\infty$.    For $u\in A$ and $n\in\Bbb N$ set

\Centerline{$C_{un}=\{v:v\in L^0,\,0\le v\le u_0,\,
\tau_1(v)+\tau_2(u-v)^+\le\gamma+2^{-n}\}$.}

\medskip

\quad{\bf (i)} Every $C_{un}$ is non-empty (because $\tau(u)\le\gamma$).

\medskip

\quad{\bf (ii)} Every $C_{un}$ is convex (because if $v_1$, $v_2\in C_{un}$ and
$\alpha\in[0,1]$ and $v=\alpha v_1+(1-\alpha)v_2$, then

\Centerline{$(u-v)^+=(\alpha(u-v_1)+(1-\alpha)(u-v_2))^+
\le\alpha(u-v_1)^++(1-\alpha)(u-v_2)^+$,}

\noindent so

$$\eqalign{\tau_1(v)+\tau_2(u-v)^+
&\le\alpha\tau_1(v_1)+(1-\alpha)\tau_1(v_2)+\alpha\tau_2(u-v_1)^+
+(1-\alpha)\tau_2(u-v_2)^+\cr
&\le\gamma+2^{-n}).\cr}$$

\medskip

\quad{\bf (iii)} if $u$, $u'\in A$, $m$, $n\in\Bbb N$, $u\le u'$ and $m\le n$
then  $C_{u'n}\subseteq C_{um}$.

\medskip

\quad{\bf (iv)} Every $C_{un}$ is closed for the topology of convergence in
measure.   \Prf\Quer\ Suppose otherwise.   Then we can find a $v$ in the
closure of $C_{un}$ for the topology of convergence in measure, but such
that $\tau_1(v)+\tau_2(u-v)^+>\gamma+2^{-n}$.   In this case

\Centerline{$\tau_1(v)=\sup\{\tau_1(v\times\chi a):a\in\frak A^f\}$,
\quad$\tau_2(u-v)^+=\sup\{\tau_2((u-v)^+\times\chi a):a\in\frak A^f\}$,}

\noindent so there is an $a\in\frak A^f$ such that

\Centerline{$\tau_1(v\times\chi a)+\tau_2((u-v)^+\times\chi a)
>\gamma+2^{-n}$.}

\noindent Now there is a sequence $\sequence{k}{v_k}$ in $C_{un}$ such
that $\bar\mu(a\Bcap\Bvalue{|v-v_k|\ge2^{-k}})\le 2^{-k}$ for every $k$.
Setting

\Centerline{$v'_k=\inf_{i\ge k}v_i$,
\quad$w_k=\inf_{i\ge k}(u-v_i)^+$}

\noindent we have

\Centerline{$\tau_1(v'_k)+\tau_2(w_k)\le\tau_1(v_k)+\tau_2(u-v_k)^+
\le\gamma+2^{-n}$}

\noindent for each $k$, and $\sequence{k}{v'_k}$, $\sequence{k}{w_k}$
are non-decreasing.   So setting $v^*=\sup_{k\in\Bbb N}v\wedge v'_k$,
$w^*=\sup_{k\in\Bbb N}(u-v)^+\wedge w_k$, we get

\Centerline{$\tau_1(v^*)+\tau_2(w^*)\le\gamma+2^{-n}$.}

\noindent But $v^*\ge v\times\chi a$ and $w^*\ge(u-v)^+\times\chi a$,
so

\Centerline{$\tau_1(v\times\chi a)+\tau_2((u-v)^+\times\chi a)\le
\gamma+2^{-n}$,}

\noindent contrary to the choice of $a$.\ \Bang\Qed

\medskip

\quad{\bf (v)} If $a\in\frak A\setminus\{0\}$, there is a non-zero
$b\Bsubseteq a$ such that $\bar\mu b<\infty$ and
$b\Bsubseteq\Bvalue{u_0\le\alpha}$ for some $\alpha>0$.   Take any $u\in A$;
then $\sup_{v\in C_{u0}}\int_b|v|$ is finite.

\medskip

{\bf (c)} Thus $\{C_{un}:u\in A$, $n\in\Bbb N\}$ satisfies
all the conditions of 367V, and $\bigcap_{u\in A,n\in\Bbb N}C_{un}$ is
non-empty.   If $v$ belongs to the intersection, then

\Centerline{$\tau_1(v)+\tau_2(u-v)^+\le\gamma$}

\noindent for every $u\in A$;  since $\{(u-v)^+:u\in A\}$ is an
upwards-directed set with supremum $(u_0-v)^+$, and $\tau_2$ is an
extended Fatou norm,

\Centerline{$\tau_1(v)+\tau_2(u_0-v)^+\le\gamma$.}

\medskip

{\bf (d)} This shows both that the infimum in the definition of
$\tau(u)$ is always attained (since this is trivial if $\tau(u)=\infty$,
and otherwise we consider $A=\{|u|\}$), and also that $\tau(\sup
A)=\sup_{u\in A}\tau(u)$ whenever $A\subseteq (L^0)^+$ is a non-empty
upwards-directed set with a supremum.   Thus $\tau$ satisfies conditions
(i)-(iv) of 369F.   Condition (vi) there is trivial, since (for
instance) $\tau(v)\le\tau_1(v)$ for every $v$.
As for 369F(v), suppose that $u>0$ in $L^0$.   Take $u_1$ such that
$0<u_1\le u$ and $\tau'_1(u_1)\le 1$, and $u_2$ such that $0<u_2\le u_1$ and
$\tau'_2(u_2)\le 1$.   In this case, if $u_2=v+w$, we must have

\Centerline{$\tau_1(v)+\tau_2(w)
\ge\|v\times u_1\|_1+\|w\times u_2\|_1
\ge\|u_2\times u_2\|_1$;}

\noindent so that

\Centerline{$\tau(u)\ge\|u_2\times u_2\|_1>0$.}

Thus all the conditions of 369F are satisfied, and $\tau$ is an extended
Fatou norm on $L^0$.

\medskip

{\bf (e)} The calculation of $\tau'$ is now very easy.   Since surely we
have $\tau\le\tau_i$ for both $i$, we must have $\tau'\ge\tau'_i$ for
both $i$.   On the other hand, if $u$, $z\in L^0$, then there are $v$,
$w$ such that $u=v+w$ and $\tau(u)=\tau_1(v)+\tau_2(w)$, so that

\Centerline{$\|u\times z\|_1\le\|v\times z\|_1+\|w\times z\|_1
\le\tau_1(v)\tau'_1(z)+\tau_2(w)\tau'_2(z)
\le\tau(u)\max(\tau'_1(z),\tau'_2(z))$;}

\noindent as $u$ is arbitrary, $\tau'(z)\le\max(\tau'_1(z),\tau'_2(z))$.
So $\tau'=\max(\tau'_1,\tau'_2)$, as claimed.

\medskip

{\bf (f)} Finally, it is obvious that

\Centerline{$L^{\tau'}=\{z:\tau'(z)<\infty\}=\{z:\tau'_1(z)<\infty,\,
\tau'_2(z)<\infty\}=L^{\tau'_1}\cap L^{\tau'_2}$,}

\noindent while the fact that the infimum in the definition of $\tau$ is
always attained means that $L^{\tau}\subseteq L^{\tau_1}+L^{\tau_2}$, so
that we have equality here also.
}%end of proof of 369R

\exercises{\leader{369X}{Basic exercises $\pmb{>}$(a)}
%\spheader 369Xa
Let $\frak A$ be a Dedekind $\sigma$-complete Boolean algebra.  Show
that the following are equiveridical:  (i) there is a function $\bar\mu$
such that $(\frak A,\bar\mu)$ is a semi-finite measure algebra;  (ii)
$(L^{\infty})^{\times}$ separates the points of
$L^{\infty}=L^{\infty}(\frak A)$;   (iii) for every non-zero
$a\in\frak A$ there is a completely additive functional
$\nu:\frak A\to\Bbb R$ such
that $\nu a\ne 0$;   (iv) there is some order-dense Riesz subspace $U$
of $L^0=L^0(\frak A)$ such that $U^{\times}$ separates the points of
$U$;  (v) for every order-dense Riesz subspace $U$ of $L^0$ there is an
order-dense Riesz subspace $V$ of $U$ such that $V^{\times}$ separates
the points of $V$.
%369A

\spheader 369Xb Let us say that a function
$\phi:\Bbb R\to\ocint{-\infty,\infty}$ is {\bf convex} if
$\phi(\alpha s+(1-\alpha)t)\le\alpha\phi(s)+(1-\alpha)\phi(t)$
for all $s$, $t\in\Bbb R$
and $\alpha\in[0,1]$, interpreting $0\cdot\infty$ as $0$, as usual.
For any convex function $\phi:\Bbb R\to\ocint{-\infty,\infty}$ which is
not always infinite, set $\phi^*(t)=\sup_{s\in\Bbb R}st-\phi(s)$ for
every $t\in\Bbb R$.   (i) Show that
$\phi^*:\Bbb R\to\ocint{-\infty,\infty}$ is convex and lower
semi-continuous and not always infinite.   \Hint{233Xh.}
(ii) Show that if $\phi$ is
lower semi-continuous then $\phi=\phi^{**}$.   \Hint{It is easy to check
that $\phi^{**}\le\phi$.   For the reverse inequality, consider first the
case $\phi(t)=\alpha t+\beta$, and use 233Ha).}
%369Xb

\sqheader 369Xc For the purposes of this exercise and the next, say that
a {\bf Young's function} is a
non-negative non-constant lower semi-continuous convex function
$\phi:\coint{0,\infty}\to[0,\infty]$ such that $\phi(0)=0$ and $\phi(t)$
is finite for some $t>0$.   ({\bf Warning!} the phrase `Young's
function' has other meanings.)   (i) Show that in this case $\phi$ is
non-decreasing and continuous on the left and $\phi^*$, defined by
saying that $\phi^*(t)=\sup_{s\ge 0}st-\phi(s)$ for every $t\ge 0$, is
again a Young's function.   (ii) Show that $\phi^{**}=\phi$.   Say that
$\phi$ and $\phi^*$ are {\bf complementary}.
(iii) Compute $\phi^*$ in the cases ($\alpha$) $\phi(t)=t$ ($\beta$)
$\phi(t)=\max(0,t-1)$ ($\gamma$) $\phi(t)=t^2$ ($\delta$) $\phi(t)=t^p$
where $1<p<\infty$.
%369Xc

\sqheader 369Xd  Let $\phi$, $\psi=\phi^*$ be complementary Young's
functions in the sense of 369Xc, and
$(\frak A,\bar\mu)$ a semi-finite measure algebra.   Set

\Centerline{$B=\{u:u\in L^0,\,\int\bar\phi(|u|)\le 1\}$,
\quad$C=\{v:v\in L^0,\,\int\bar\psi(|v|)\le 1\}$.}

\noindent (For finite-valued $\phi$, $\bar\phi:(L^0)^+\to L^0$ is given
by 364H.   Devise an appropriate convention for the case in which
$\phi$ takes the value $\infty$.)   (i) Show that $B$ and $C$ are
order-closed solid convex sets, and that $\int|u\times v|\le 2$ for all
$u\in B$, $v\in C$.   \Hint{for `order-closed', use 364Xg(iv).}
(ii) Show that there is a unique extended Fatou norm $\tau_{\phi}$ on
$L^0$ for which $B$ is the unit ball.  (iii) Show that if
$u\in L^0\setminus B$ there is a $v\in C$ such that $\int|u\times v|>1$.
\Hint{start with the case in which
$u\in S(\frak A)^+$.}   (iv) Show that
$\tau_{\psi}\le\tau'_{\phi}\le 2\tau_{\psi}$, where $\tau_{\psi}$ is the
extended Fatou norm corresponding to $\psi$ and $\tau'_{\phi}$ is the
associate of $\tau_{\phi}$, so that $\tau_{\psi}$ and $\tau'_{\phi}$ can
be interpreted as equivalent norms on the same Banach space.

($U$ and $V$ are complementary {\bf Orlicz spaces};  I will call
$\tau_{\phi}$, $\tau_{\psi}$ {\bf Orlicz norms}.)
%369F, 369H

\spheader 369Xe Let $U$ be a Riesz space such that $U^{\times}$
separates the points of $U$, and suppose that $\|\,\|$ is a Fatou norm
on $U$.  (i) Show that there is a localizable measure algebra
$(\frak A,\bar\mu)$ with an extended Fatou norm $\tau$ on $L^0(\frak A)$
such that $U$ can be identified, as normed Riesz space, with an
order-dense Riesz subspace of $L^{\tau}$.    (ii) Hence, or otherwise,
show that $\|u\|=\sup_{f\in U^{\times},\,\|f\|\le 1}|f(u)|$ for every
$u\in U$.
(iii) Show that if $U$ is Dedekind complete and has the Levi property,
then $U$ becomes identified with $L^{\tau}$ itself, and in particular is
a Banach lattice (cf.\ 354Xn).
%369I

\spheader 369Xf Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $\tau$ an extended Fatou norm on $L^0(\frak A)$.   Show
that the norm of $L^{\tau}$ is order-continuous iff the norm topology of
$L^{\tau}$ agrees with the topology of convergence in measure on any
order-bounded subset of $L^{\tau}$.
%369M

\spheader 369Xg Let $(\frak A,\bar\mu)$ be a $\sigma$-finite measure
algebra of countable Maharam type, and $\tau$ an extended Fatou norm on
$L^0(\frak A)$ such that the norm of $L^{\tau}$ is order-continuous.
Show that $L^{\tau}$ is separable in its norm topology.
%369Xf  369M

\spheader 369Xh Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure algebras, and $\pi:\frak A^f\to\frak B^f$ a
measure-preserving ring homomorphism, as in 366H, with associated maps
$T:M^0_{\bar\mu}\to M^0_{\bar\nu}$ and
$P:M^{1,0}_{\bar\nu}\to M^{1,0}_{\bar\mu}$.   Show that
$\|Tu\|_{\infty,1}=\|u\|_{\infty,1}$ for every
$u\in M^{\infty,1}_{\bar\mu}$ and $\|Pv\|_{\infty,1}\le\|v\|_{\infty,1}$
for every $v\in M^{\infty,1}_{\bar\nu}$.
%369N

\spheader 369Xi 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.   (i) Show that there is a unique Riesz
homomorphism $T:M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\nu}$ such
that $T(\chi a)=\chi(\pi a)$ for every $a\in\frak A$
and $\|Tu\|_{1,\infty}=\|u\|_{1,\infty}$ for every
$u\in M^{1,\infty}_{\bar\mu}$.   (ii) Now suppose that
$(\frak A,\bar\mu)$ is localizable and $\pi$ is order-continuous.
Show that there is a unique positive linear operator
$P:M^{1,\infty}_{\bar\nu}\to M^{1,\infty}_{\bar\mu}$ such that
$\int_aPv=\int_{\pi a}v$ for every $a\in\frak A^f$ and
$v\in M^{1,\infty}_{\bar\nu}$, and that
$\|Pv\|_{\infty}\le\|v\|_{\infty}$ for every
$v\in L^{\infty}(\frak B)$, %referred in 366 notes
$\|Pv\|_{\infty,1}\le\|v\|_{\infty,1}$ for every
$v\in M^{\infty,1}_{\bar\nu}$,
$\|Pv\|_{1,\infty}\le\|v\|_{1,\infty}$ for
every $v\in M^{1,\infty}_{\bar\nu}$.   (Compare 365O.)
%369N

\spheader 369Xj Let $(\frak A,\bar\mu)$ be an atomless semi-finite
measure algebra.   Show that
$\|u\|_{1,\infty}=\max\{\int_a|u|:a\in\frak A,\,\bar\mu a\le 1\}$ for
every $u\in L^0(\frak A)$.   \Hint{take
$a\Bsupseteq\Bvalue{|u|>\alpha_0}$ in part (b-iv) of the proof of 369O.}
%369O

\spheader 369Xk Let $(\frak A,\bar\mu)$ be any semi-finite measure
algebra.   Show that if $\tau_{\phi}$ is any Orlicz
norm on $L^0=L^0(\frak A)$, then there is a $\gamma>0$ such that
$\|u\|_{1,\infty}\le\gamma\tau_{\phi}(u)\le\gamma^2\|u\|_{\infty,1}$ for
every $u\in L^0$, so that
$M^{\infty,1}_{\bar\mu}\subseteq L^{\tau_{\phi}}
\subseteq M^{1,\infty}_{\bar\mu}$.
%369O

\spheader 369Xl Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra.   Show that the subspaces $M^{1,\infty}_{\bar\mu}$,
$M^{\infty,1}_{\bar\mu}$ of $L^0(\frak A)$ can be expressed as a
complementary pair
of Orlicz spaces, and that the norm $\|\,\|_{\infty,1}$ can be
represented as an Orlicz norm, but that if $\frak A$ is atomless and
$\bar\mu$ is not totally finite. $\|\,\|_{1,\infty}$ cannot be represented as an
Orlicz norm.
%369O

\sqheader 369Xm Let $(\frak A,\bar\mu)$ be a measure algebra and $U$ a
Banach space.   (i) Suppose that $\nu:\frak A\to U$ is an additive
function such that $\|\nu a\|\le\min(1,\bar\mu a)$ for every $a\in\frak
A$.   Show that there is a unique bounded linear operator
$T:M^{1,\infty}_{\bar\mu}\to U$ such that $T(\chi a)=\nu a$ for
every $a\in\frak A$.   (ii) Suppose that $\nu:\frak A^f\to U$ is an
additive function such that $\|\nu a\|\le\max(1,\bar\mu a)$ for every
$a\in\frak A^f$.   Show that there is a unique bounded linear operator
$T:M^{\infty,1}_{\bar\mu}\to U$ such that $T(\chi a)=\nu a$ for
every $a\in\frak A^f$.
%369O

\spheader 369Xn Let $(\frak A,\bar\mu)$ and $(\frak B,\bar\nu)$ be
semi-finite measure algebras, and $\phi:\coint{0,\infty}\to[0,\infty]$ a
Young's function;  write $\tau_{\phi}$ for the
corresponding
Orlicz norm on either $L^0(\frak A)$ or $L^0(\frak B)$.
Let $\pi:\frak A\to\frak B$ be a
measure-preserving Boolean homomorphism, with associated map
$T:M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\nu}$, as
in 369Xi.   (i) Show that $\tau_{\phi}(Tu)=\tau_{\phi}(u)$ for every
$u\in M^{1,\infty}_{\bar\mu}$.
(ii) Show that if $(\frak A,\bar\mu)$ is localizable, $\pi$ is
order-continuous and
$P:M^{1,\infty}_{\bar\nu}\to M^{1,\infty}_{\bar\mu}$ is the map of 369Xi(ii),
then $\tau_{\phi}(Pv)\le\tau_{\phi}(v)$ for every
$v\in M^{1,\infty}_{\bar\nu}$.
\Hint{365Q.}
%369N

\sqheader 369Xo Let $(\frak A,\bar\mu)$ be any semi-finite measure
algebra and $\tau_1$, $\tau_2$ two extended Fatou norms on
$L^0(\frak A)$.   Show that $u\mapsto\max(\tau_1(u),\tau_2(u))$ is an extended Fatou norm.
%369O

\spheader 369Xp Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $(\widehat{\frak A},\tilde\mu)$ its localization (322Q).
Show that the Dedekind completion of $M^{1,\infty}(\frak A,\bar\mu)$ can
be identified with $M^{1,\infty}(\widehat{\frak A},\tilde\mu)$.
%369Q

\spheader 369Xq Let $(\frak A,\bar\mu)$ be a localizable measure
algebra.   (i) Show that if $\frak B$ is any closed subalgebra of
$\frak A$ such that $\sup\{b:b\in\frak B,\,\bar\mu b<\infty\}=1$ in
$\frak A$, we have an order-continuous positive linear operator
$P_{\frak B}:
M^{1,\infty}_{\bar\mu}\to M^{1,\infty}_{\bar\mu\restrp\frak B}$
such that $\int_bP_{\frak B}u=\int_bu$ whenever
$u\in M^{1,\infty}_{\bar\mu}$, $b\in\frak B$ and $\bar\mu b<\infty$.
(ii) Show that if $\sequencen{\frak B_n}$ is a
non-decreasing sequence of closed subalgebras of $\frak A$ such that
$\sup\{b:b\in\frak B_0,\,\bar\mu b<\infty\}=1$ in $\frak A$, and
$\frak B$ is the closure of $\bigcup_{n\in\Bbb N}\frak B_n$, then
$\sequencen{P_{\frak B_n}u}$ is order*-convergent to $P_{\frak B}u$
for every $u\in M^{1,\infty}_{\bar\mu}$.   (Cf.\ 367J.)
%369Q

\spheader 369Xr Let $\phi_1$ and $\phi_2$ be Young's functions
and $(\frak A,\bar\mu)$ a semi-finite measure algebra.   Set
$\phi(t)=\max(\phi_1(t),\penalty-100\phi_2(t))$ for $t\in\coint{0,\infty}$.
(i) Show that $\phi$ is a Young's function.   (ii) Writing $\tau_{\phi_1}$
$\tau_{\phi_2}$, $\tau_{\phi}$ for the corresponding extended Fatou
norms on $L^0(\frak A)$ (369Xd), show that
$\tau_{\phi}\ge\max(\tau_{\phi_1},\tau_{\phi_2})\ge\bover12\tau_{\phi}$,
so that $L^{\tau_{\phi}}=L^{\tau_{\phi_1}}\cap L^{\tau_{\phi_2}}$ and
$L^{\tau_{\phi^*}}=L^{\tau_{\phi_1^*}}+L^{\tau_{\phi_2^*}}$, writing
$\phi^*$ for the Young's function complementary to $\phi$.   (iii)
Repeat with $\psi=\phi_1+\phi_2$ in place of $\phi$.
%369R

\leader{369Y}{Further exercises (a)}
%\spheader 369Ya
Let $(\frak A,\bar\mu)$ be a localizable measure algebra
and $A\subseteq L^0=L^0(\frak A)$ a countable set.   Show that the solid
linear subspace $U$ of $L^0$ generated by $A$ is a perfect Riesz space.
\Hint{reduce to the case in which $U$ is order-dense.   If
$A=\{u_n:n\in\Bbb N\}$, $w\in(L^0)^+\setminus U$ find $v_n\in(L^0)^+$
such that $\int v_n\times w\ge 2^n\ge 4^n\int v_n\times|u_i|$ for every
$i\le n$.   Show that $v=\sup_{n\in\Bbb N}v_n$ is defined in $L^0$ and
corresponds to a member of $U^{\times}$.}
%369C

\spheader 369Yb Let $U$ be a Banach lattice and suppose that
$p\in\coint{1,\infty}$ is such that $\|u+v\|^p=\|u\|^p+\|v\|^p$ whenever
$u$, $v\in U$ and $|u|\wedge|v|=0$.   Show that $U$ is isomorphic, as
Banach lattice, to $L^p_{\bar\mu}$ for some localizable measure
algebra $(\frak A,\bar\mu)$.   \Hint{start by using 354Yb to show that
the norm of $U$ is order-continuous, as in 354Yk.}
%369E

\spheader 369Yc Let $(\frak A,\bar\mu)$ be a semi-finite measure
algebra, and $\tau$ an Orlicz norm on $L^0(\frak A)$.   Show that
$L^{\tau}$ has the Levi property, whether or not $\frak A$ is Dedekind
complete.
%369Xd 369H

\spheader 369Yd
Let $\phi:\coint{0,\infty}\to\coint{0,\infty}$ be a strictly increasing
Young's function such that $\sup_{t>0}\phi(2t)/\phi(t)$
is finite.   Show that the associated Orlicz norms $\tau_{\phi}$
are always order-continuous on their function spaces.
%369Xd, 369H

\spheader 369Ye
Let $\phi:\coint{0,\infty}\to[0,\infty]$ be a Young's function, and
suppose that the corresponding Orlicz norm on
$L^0(\frak A)$, where $(\frak A,\bar\mu)$ is an atomless measure algebra which
is not totally finite, is order-continuous on its function
space $L^{\tau_{\phi}}$.   Show that there is an $M\ge 0$ such that
$\phi(2t)\le M\phi(t)$ for every $t\ge 0$.
%369Xd, 369H

\spheader 369Yf Let $(\frak A,\bar\mu)$ be a semi-finite measure algebra
and $\phi:\coint{0,\infty}\to\coint{0,\infty}$ be a strictly increasing
Young's function such that $\sup_{t>0}\phi(2t)/\phi(t)$ is finite.
Show that if $\Cal F$ is a filter on $L^{\tau_{\phi}}$, then
$\Cal F\to u\in L^{\tau_{\phi}}$ for the norm $\tau_{\phi}$ iff (i)
$\Cal F\to u$ for the topology of convergence in measure (ii)
$\limsup_{v\to\Cal F}\tau_{\phi}(v)\le\tau_{\phi}(u)$.   (Compare
245Xl.)
%369M mt36bits

\spheader 369Yg Give examples of extended Fatou norms $\tau$ on
measure spaces $L^0(\frak A)$, where $(\frak A,\bar\mu)$ is a semi-finite
measure algebra, such that ($\alpha$) $\tau\restr L^{\tau}$ is
order-continuous ($\beta$) there is a
sequence $\sequencen{u_n}$ in $L^{\tau}$, converging in measure to
$u\in L^{\tau}$, such that $\lim_{n\to\infty}\tau(u_n)=\tau(u)$ but
$\sequencen{u_n}$ does not converge to $u$ for the norm on $L^{\tau}$.
Do this (i) with $\tau$ an Orlicz norm (ii) with $(\frak A,\bar\mu)$ the measure
algebra of Lebesgue measure on $[0,1]$.
%369M %mt36bits

\spheader 369Yh Let $(\frak A,\bar\mu)$ be any measure algebra.   Show
that $(M^{1,0}_{\bar\mu})^{\times}$ can be identified with
$M^{\infty,1}_{\bar\mu}$.   \Hint{show that neither $M^{1,0}$ nor
$M^{\infty,1}$ is changed by moving first to the semi-finite version of
$(\frak A,\bar\mu)$, as described in 322Xa, and then to its
localization.}
%369Q

\spheader 369Yi Give an example to show that the result of 369R may fail
if $(\frak A,\bar\mu)$ is only semi-finite, not localizable.
%369R
}%end of exercises

\endnotes{
\Notesheader{369} The representation theorems 369A-369D give a
concrete form to the notion of `perfect' Riesz space:  it is just one
which can be expressed as a subspace of $L^0(\frak A)$, for some
localizable measure algebra $(\frak A,\bar\mu)$, in such a way that it
is its own second dual, where the duality here is between subspaces of
$L^0$, taking $V=\{v:u\times v\in L^1$ for every $u\in U\}$.   (I see
that in this expression I ought somewhere to mention that both $U$ and
$V$ are assumed to be order-dense in $L^0$.)   Indeed I believe that
the original perfect spaces were the `vollkommene R\"aume' of
G.K\"othe, which were subspaces of $\BbbR^{\Bbb N}$, corresponding to
the measure algebra $\Cal P\Bbb N$ with counting measure, so that $V$
or $U^{\times}$ was $\{v:u\times v\in\ell^1$ for every $u\in U\}$.

I have presented Kakutani's theorem on the representation of $L$-spaces
as a corollary of 369A and 369C.   As usual in such things, this is a
reversal of the historical relationship;  Kakutani's theorem was one of
the results which led to the general theory.   The complete
list of localizable measure algebras provided by Maharam's theorem
(332B, 332J) now gives us a complete list of $L$-spaces.

Just as perfect Riesz spaces come in dual pairs, so do some of the most
important Banach lattices:  those with Fatou norms and the Levi property
for which the order-continuous dual separates the points.   (Note that
the dual of any space with a Riesz norm has these properties;  see
356Da.)   I leave the details of representing such spaces to you
(369Xe).   The machinery of 369F-369K gives a solid basis for studying
such pairs.

Among the extended Fatou norms of 369F the Orlicz norms (369Xd,
369Yd-369Yf) form a
significant subfamily.   Because they are defined in a way which is to
some extent independent of the measure algebra involved, these spaces
have some of the same properties as $L^p$ spaces in relation to
measure-preserving homomorphisms (369Xi-369Xn).   In \S\S373-374 I will
elaborate on these ideas.   Among the Orlicz spaces, we have a
largest and a smallest;  these are just $M^{1,\infty}=L^1+L^{\infty}$
and $M^{\infty,1}=L^1\cap L^{\infty}$ (369N-369O, 369Xk, 369Xl).   Of
course these two are particularly important.

There is an interesting phenomenon here.   It is easy to see that
$\|\,\|_{\infty,1}=\max(\|\,\|_1,\|\,\|_{\infty})$ is an extended Fatou
norm and that the corresponding Banach lattice is $L^1\cap L^{\infty}$;
and that the same ideas work for any pair of extended Fatou norms
(369Xo).   To check that the dual of $L^1\cap L^{\infty}$ is precisely
the linear sum $L^{\infty}+L^1$ a little more is needed, and the
generalization of this fact to other extended Fatou norms (369R) seems
to go quite deep.   In view of our ordinary expectation that properties
of these normed function spaces should be reflected in perfect Riesz
spaces in general, I mention that I believe I have found an example,
dependent on the continuum hypothesis, of
two perfect Riesz subspaces $U$, $V$ of $\BbbR^{\Bbb N}$ such that
their linear sum $U+V$ is not perfect.
% mt36bits
}%end of comment

\frnewpage

%\ifresultsonly\else
\frfilename{mt36conc.tex}
\versiondate{20.7.11}
\copyrightdate{2011}
     
\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
     
\def\chaptername{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{}64Be}{$L^0(\frak A)$} This re-phrasing of the definition
of $L^0(\frak A)$, 
referred to in the 2008 edition of Volume 5, is now 364Af.

\leader{3{}64D}{$L^0$ as $f$-algebra} This paragraph, 
referred to in the 2008 edition of Volume 5, is now 364C.

\leader{3{}64E}{Algebraic operations on $L^0$} This paragraph, 
referred to in the 2008 edition of Volume 5, is now 364D.

\leader{3{}64G}{} The identification of $L^0(\frak A)$ with the set of
sequentially order-continuous Boolean homomorphisms from $\Cal B(\Bbb R)$
to $\frak A$, referred to in the 2008 edition of Volume 5, is now
364F.

\leader{3{}64I}{Action of Borel functions on $L^0$}
This paragraph, 
referred to in the 2003 and 2006 editions of Volume 4, is now 364H.

\leader{3{}64J}{$L^0(\Sigma/\Cal I)$} The identification of
$L^0(\Sigma/\Cal I)$ as a space of equivalence classes of functions,
referred to in the 2003 and 2206 editions of Volume 4 and the
2008 edition of Volume 5, is now 364I.

\leader{3{}64K}{Embedding $S$ and $L^{\infty}$ in $L^0$}
This paragraph, 
referred to in the 2003 and 2006 editions of Volume 4, is now 364J.

\leader{3{}64M-3{}64N}{Suprema and infima in $L^0(\frak A)$} These 
paragraphs, referred to in the 2003 and 2006 editions of Volume 4 and
the 2008 edition of Volume 5, have now been amalgamated as 364L.

\leader{3{}64O}{Dedekind completeness of $L^0$} This paragraph, 
referred to in the 2008 edition of Volume 5, is now 364M.

\leader{3{}64P}{Multiplicative inverses in $L^0$}
This paragraph, 
referred to in the 2003 and 2006 editions of Volume 4, is now 364J.

\leader{3{}64R}{Action of Boolean homomorphisms on $L^0$} This paragraph,
referred to  the 2003 and 2006 editions of Volume 4 and in the 
2008 edition of Volume 5, is now 364P.

\leader{3{}64Xw}{Extension of $\dashint$} This exercise, 
referred to in the 2008 edition of Volume 5, is now 364Xj.

\leader{3{}64Yn}{$L^0_{\Bbb C}(\frak A)$} This exercise on complex $L^0$
spaces, referred to in the 2003 and 2006 editions of Volume 4,
has been moved to 366M.

\leader{3{}65K}{Additive functions on $\frak A^f$ and linear operators 
on $L^1$} This theorem, referred to in the 2003, 2006 and 2013 printings of
Volume 4, is now 365J.

\leader{3{}65M}{$L^1$ and $L^{\infty}$} This theorem, referred to in the 
2008 printing of Volume 5, is now 365L.

\leader{3{}65O}{Ring homomorphisms on $\frak A^f$ and Riesz homomorphisms
on $L^1$} This theorem, referred to in the 2013 printing of Volume 4 and
the 2008 printing of Volume 5, is now 365N.

\leader{3{}65P}{Order-continuous ring homomorphisms on $\frak A^f$ and
conditional expectations} This theorem, referred to in the 2008 printing of
Volume 5, is now 365O.

\leader{3{}65R}{Conditional expectations} These notes, referred to in the
2006 and 2013 printings of Volume 4 and the 2008 printing of Volume 5, is
now 365Q.
    
\leader{3{}65T}{Change of measure} This proposition, 
referred to in the 2008 printing of
Volume 5, is now 365S.

\frnewpage
%\fi
     
\references{
\frfilename{mt3r.tex}
\versiondate{25.8.17}
\copyrightdate{1999}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 3}

\medskip

%A

Anderson I.\ [87] {\it Combinatorics of Finite Sets.}  Oxford U.P., 1987.
\cmmnt{[332Xk, 3A1K.]}

% Anzai H.\ [51] `Ergodic skew product transformations of the torus',
% % Osaka Math.\ J.\ 3 (1951) 83-99.

Anzai H.\ [51] `On an example of a measure-preserving transformation
which is not conjugate to its inverse',
Proc.\ Japanese Acad.\ Sci.\ 27 (1951) 517-522.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip
%B

Balcar B., G\l\'owczy\'nski W. \& Jech T.\ [98] `The sequential
topology on complete Boolean algebras', Fundamenta Math.\ 155 (1998)
59-78.
\cmmnt{[393L, 393Q.]}

Balcar B., Jech T.\ \& Paz\'ak T.\ [05]
`Complete ccc Boolean algebras, the order sequential topology,
and a problem of von Neumann',
Bull.\ London Math.\ Soc.\ 37 (2005) 885-898.
\cmmnt{[393L, 393Q.]}
% arXiv.math.LO/0312473

Becker H.\ \& Kechris A.S.\ [96] {\it The descriptive set theory of
Polish group actions.}  Cambridge U.P., 1996
(London Math.\ Soc.\ Lecture Note Series 232).
\cmmnt{[\S395 {\it notes\/}.]}

Bekkali M.\ \& Bonnet R.\ [89] `Rigid Boolean algebras', pp.\ 637-678 in
{\smc Monk 89}.
\cmmnt{[384L.]}

Bellow A.\ \& K\"olzow D.\ [76] {\it Measure Theory, Oberwolfach 1975.}
Springer, 1976 (Lecture Notes in Mathematics 541).

Bhaskara Rao, K.P.S.\ \& Bhaskara Rao, M.\ {\it Theory of Charges.}
Academic, 1983.
%326H

Billingsley P.\ [65] {\it Ergodic Theory and Information.}  Wiley, 1965.
\cmmnt{[\S372 {\it notes\/}.]}

Bollob\'as B.\ [79] {\it Graph Theory.}   Springer, 1979.
\cmmnt{[332Xk, 3A1K.]}

Bose R.C.\ \& Manvel B.\ [84] {\it Introduction to Combinatorial Theory.}   Wiley, 1984.
\cmmnt{[3A1K.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[\S3A3, \S3A4.]}

Bourbaki N. [68] {\it Theory of Sets.}   Hermann/Addison-Wesley, 1968.
\cmmnt{[\S315 {\it notes\/}.]}

Bourbaki N. [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[\S3A5.]}

Bukhvalov A.V.\ [95] `Optimization without compactness, and its
applications', pp.\ 95-112 in
{\smc Huijsmans Kaashoek Luxemburg \& Pagter 95}.
\cmmnt{[367U.]}

Burke M.R.\ [93] `Liftings for Lebesgue measure', pp.\ 119-150 in
{\smc Judah 93}.   \cmmnt{[341L, 345F.]}

Burke M.R.\ [n95] `Consistent liftings', privately circulated, 1995.
\cmmnt{[346Ya.]}

Burnside W.\ [1911] {\it Theory of Groups of Finite Order.}   Cambridge
U.P., 1911 (reprinted by Dover, 1955).
\cmmnt{[\S384 {\it notes\/}.]}

\medskip
%C

Chacon R.V. [69]
`Weakly mixing transformations which are not strongly mixing',
Proc.\ Amer.\ Math.\ Soc.\ 22 (1969) 559-562.
\cmmnt{[372R.]}

Chacon R.V.\ \& Krengel U.\ [64] `Linear modulus of a linear
operator', Proc.\ Amer.\ Math.\ Soc.\ 15 (1964) 553-559.
\cmmnt{[\S371 {\it notes\/}.]}

Choksi J.R.\ \& Prasad V.S.\ [82] `Ergodic theory of homogeneous measure
algebras', pp.\ 367-408 of {\smc K\"olzow \& Maharam-Stone 82}.
\cmmnt{[383I.]}

%Christenson C.O.\ \& Voxmann L.\ [77] {\it Aspects of Topology.}
%Dekker, 1977.

Cohn H.\ [06] `A short proof of the simple continued fraction expansion of
$e$', Amer.\ Math.\ Monthly 113 (2006) 57-62;
{\tt arXiv:math.NT/0601660}.
\cmmnt{[372L.]}

Coleman A.J.\ \& Ribenboim P.\ [67] (eds.) {\it Proceedings of the Symposium in Analysis, Queen's University, June 1967.}  Queen's University, Kingston, Ontario, 1967.

Comfort W.W.\ \& Negrepontis S.\ [82] {\it Chain Conditions in
Topology.}   Cambridge U.P., 1982.   \cmmnt{[\S391 {\it notes\/}.]}

\leaveitout{
Cornfeld I.P., Fomin S.V.\ \& Sina\v\i Ya.G.\ [82] {\it Ergodic Theory.}
Springer, 1982 (Grundlehren der mathematischen Wissenschaften 245)
}

Czisz\'ar I.\ [67] `Information-type measures of difference of
probability distributions and indirect observations',
Studia Scientiarum Math.\ Hungarica 2 (1967) 299-318.   \cmmnt{[386G.]}

\medskip
%D

Davey B.A. \& Priestley H.A. [90] `Introduction to Lattices and Order',
Cambridge U.P., 1990.  \cmmnt{[3A6C.]}

%Dow A.\ \& Stepr\=ans J.\ [93] `The $\sigma$-linkedness of the measure algebra', Canad.\ Math.\ Bulletin 37 (1993) 42-45.

Dugundji J.\ [66] {\it Topology.}  Allyn \& Bacon, 1966.
\cmmnt{[\S3A3, 3A4Bb.]}

Dunford N.\ [1936] `Integration and linear operators',
Trans.\ Amer.\ Math.\ Soc.\ 40 (1936) 474-494.
\cmmnt{[\S376 {\it notes\/}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).
\cmmnt{[\S356 {\it notes\/}, \S371 {\it notes\/}, \S376 {\it notes\/},
\S3A5.]}

Dye H.A.\ [59] `On groups of measure preserving transformations I',
Amer.\ J.\ Math.\ 81 (1959) 119-159.   \cmmnt{[\S388 {\it notes\/}.]}

\medskip
%E

Eigen S.J.\ [82] `The group of measure-preserving transformations of
$[0,1]$ has no outer automorphisms', Math.\ Ann.\ 259 (1982) 259-270.
\cmmnt{[384D.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).  \cmmnt{[\S3A3, \S3A4.]}

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S3A1.]}
%QA248.E6  7850 2996 6

Erd\H{o}s P.\ \& Oxtoby J.C.\ [55] `Partitions of the plane into sets
having positive measure in every non-null product set', Trans.\ Amer.\
Math.\ Soc.\ 79 (1955) 91-102.   \cmmnt{[\S325 {\it notes\/}.]}

\medskip
%F

%Farah I.\ [04] `Examples of $\epsilon$-exhaustive pathological
%submeasures', Fundamenta Math.\ 181 (2004) 257-272.

Fathi A.\ [78] `Le groupe des transformations de $[0,1]$ qui
pr\'eservent la mesure de Lebesgue est un groupe simple', Israel J.\
Math.\ 29 (1978) 302-308.  \cmmnt{[\S382 {\it notes\/}, 383I.]}

Fremlin D.H.\ [74a] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[Chap.\ 35 {\it intro\/}., 354Yb, \S355 {\it notes\/},
\S356 {\it notes\/}, \S363 {\it notes\/}, \S365 {\it notes\/},
\S371 {\it notes\/}, \S376 {\it notes\/}.]}

Fremlin D.H.\ [74b] `A characterization of $L$-spaces', Indag.\ Math.\
36 (1974) 270-275.  \cmmnt{[\S371 {\it notes\/}.]}

%Fremlin D.H.\ [84] {\it Consequences of Martin's Axiom.}
%Cambridge U.P., 1984.

\leaveitout{Fremlin D.H.\ [89] `Measure algebras', pp.\ 876-980 in
{\smc Monk 89}.}

Fremlin D.H., de Pagter B.\ \&  Ricker W.J.\ [05]
`Sequential closedness of Boolean algebras of projections in Banach
spaces', Studia Math.\ 167 (2005) 45-62.
\cmmnt{[\S323 {\it notes\/}.]}

Frol\'\i k Z.\ [68] `Fixed points of maps of extremally disconnected
spaces and complete Boolean algebras', Bull.\ Acad.\ Polon.\ Sci.\ 16
(1968) 269-275.  \cmmnt{[382E.]}

\medskip
%G

Gaal S.A.\ [64] {\it Point Set Topology.}  Academic, 1964.
\cmmnt{[\S3A3, \S3A4.]}

Gaifman H.\ [64] `Concerning measures on Boolean algebras', Pacific J.\
Math.\ 14 (1964) 61-73.  \cmmnt{[\S391 {\it notes\/}.]}

Gale D. [60] {\it The theory of linear economic models.}  McGraw-Hill,
1960.  \cmmnt{[3A5D.]}

Gnedenko B.V.\ \& Kolmogorov A.N.\ [54] {\it Limit Distributions for
Sums of Independent Random Variables.}  Addison-Wesley, 1954.
\cmmnt{[\S342 {\it notes\/}.]}

Graf S.\ \& Weizs\"acker H.von [76] `On the existence of lower densities in non-complete measure spaces',
pp.\ 155-158 in {\smc Bellow \& K\"olzow 76}.  \cmmnt{[341L.]}

\medskip%H

Hajian A.\ \& Ito Y.\ [69]
`Weakly wandering sets and invariant measures for a group of
transformations', J.\ of Math.\ and Mech.\ 18 (1969) 1203-1216.
\cmmnt{[396B.]}

Hajian A., Ito Y.\ \& Kakutani S.\ [75] `Full groups and a theorem of
Dye', Advances in Math.\ 17 (1975) 48-59.
\cmmnt{[\S388 {\it notes\/}.]}

Halmos P.R.\ [1948] `The range of a vector measure', Bull.\ Amer.\ Math.\
Soc.\ 54 (1948) 416-421.
\cmmnt{[326Yk.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[3A1D.]}

\leaveitout{
Hopf E.\ [1932] `Theory of measures and invariant integrals', Trans.\
Amer.\ Math.\ Soc.\ 34 (1932) 372-393.
}

Huijsmans C.B., Kaashoek M.A., Luxemburg W.A.J.\ \& de Pagter B.\ [95]
(eds.) {\it Operator Theory in Function Spaces and Banach Lattices.}
Birkh\"auser, 1995.

\medskip %I

Ionescu Tulcea C.\ \& Ionescu Tulcea A.\ [69] {\it Topics in the Theory
of Lifting.}  Springer, 1969.  \cmmnt{[\S341 {\it notes\/}.]}

\medskip %J

James I.M.\ [87] {\it Topological and Uniform Spaces.}  Springer, 1987.
\cmmnt{[\S3A3, \S3A4.]}

Jech T.\ [03] {\it Set Theory, Millennium Edition.}  Springer, 2002.
\cmmnt{[\S3A1.]}

Jech T.\ [08] `Algebraic characterizations of measure algebras',
Proc.\ Amer.\ Math.\ Soc.\ 136 (2008) 1285-1294.
%{\tt http://www.math.psu.edu/jech/preprints/papers.html}
\cmmnt{[393Xj.]}

Johnson R.A.\ [80] `Strong liftings which are not Borel liftings',
Proc.\ Amer.\ Math.\ Soc.\ 80 (1980) 234-236.   \cmmnt{[345F.]}
%Russell A. Johnson

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

\medskip %K

Kakutani S.\ [1941] `Concrete representation of abstract $L$-spaces
and the mean ergodic theorem', Annals of Math. 42 (1941) 523-537.
\cmmnt{[369E.]}

Kalton N.J., Peck N.T.\ \& Roberts J.W.\ [84] `An F-space sampler',
Cambridge U.P., 1984.  \cmmnt{[\S375 {\it notes\/}.]}

Kalton N.J.\ \& Roberts J.W.\ [83] `Uniformly exhaustive submeasures and
nearly additive set functions', Trans.\ Amer.\ Math.\ Soc.\ 278 (1983)
803-816.  \cmmnt{[392D, \S392 {\it notes\/}.]}

Kantorovich L.V., Vulikh B.Z. \& Pinsker A.G.\ [50]
{\it Functional Analysis in
Partially Ordered Spaces}, Gostekhizdat, 1950.
\cmmnt{[391D.]}

Kawada Y.\ [1944] `\"Uber die Existenz der invarianten Integrale', Jap.\
J.\ Math.\ 19 (1944) 81-95.  \cmmnt{[\S395 {\it notes\/}.]}

Kelley J.L.\ [59] `Measures on Boolean algebras', Pacific J.\ Math.\ 9
(1959) 1165-1177.  \cmmnt{[\S391 {\it notes\/}.]}

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

\leaveitout{
King J.L.F.\ [00] `The generic transformation has roots of all orders',
Colloquium Mathematicum 84/85  (2000) 521-547.
  % if  \frak A  is the Lebesgue measure algebra
  % a comeager set in the Polish group
  % \Aut_{\bar\mu}\frak A   has roots of all orders
}

Kolmogorov A.N. [58] `New metric invariants of transitive dynamical
systems and automorphisms of Lebesgue spaces', Dokl.\ Akad.\ Nauk SSSR
119 (1958) 861-864.  \cmmnt{[385P.]}

K\"olzow D.\ \& Maharam-Stone D.\ [82] (eds.) {\it Measure Theory
Oberwolfach 1981.}   Springer, 1982 (Lecture Notes in Math.\ 945).

Koppelberg S.\ [89] {\it General Theory of Boolean
Algebras\/}, vol.\ 1 of {\smc Monk 89}.
\cmmnt{[Chap.\ 31 {\it intro\/}., \S332 {\it notes\/}.]}

K\"othe G.\ [69] {\it Topological Vector Spaces I.}   Springer, 1969.
\cmmnt{[\S356 {\it notes\/}, 3A5N.]}

Kranz P.\ \& Labuda I.\ [93] (eds.) {\it Proceedings of the Orlicz
Memorial Conference, 1991\/}, unpublished manuscript, available from
first editor ({\tt mmkranz{\@}olemiss.edu}).

Krengel, U.\ [63] `"Uber den Absolutbetrag stetiger linearer
Operatoren und seine Anwendung auf ergodische Zerlegungen', Math.\
Scand.\ 13 (1963) 151-187.   \cmmnt{[371Xb.]}

Krieger W.\ [76] `On ergodic flows and the isomorphism of factors',
Math.\ Ann.\ 223 (1976) 19-70.   [\S388 {\it notes}.]
   %Dye's theorem in general form, von-Neumann-algebra version

Krivine J.-L.\ [71] {\it Introduction to Axiomatic Set Theory.}
D.Reidel, 1971.  \cmmnt{[\S3A1.]}

Kullback S.\ [67]
`A lower bound for discrimination information in terms of variation',
IEEE Trans.\ on Information Theory 13 (1967) 126-127.
\cmmnt{[386G.]}
%Wendt S IN5 EL24 T IT

Kunen K.\ [80] {\it Set Theory.}  North-Holland, 1980.
\cmmnt{[\S3A1.]}

Kwapien S.\ [73] `On the form of a linear operator on the space of
all measurable functions', Bull.\ Acad.\ Polon.\ Sci.\ 21 (1973)
951-954.  \cmmnt{[\S375 {\it notes\/}.]}

\medskip %L

Lang S.\ [93] {\it Real and Functional Analysis.}  Springer, 1993.
\cmmnt{[\S3A5.]}

Liapounoff A.A.\ [1940] `Sur les fonctions-vecteurs compl\`etement
additives', Bull.\ Acad.\ Sci.\ URSS (Izvestia Akad.\ Nauk SSSR) 4 (1940)
465-478.
\cmmnt{[326H.]}

Lindenstrauss J.\ \& Tzafriri L.\ [79] {\it Classical Banach Spaces II.}
Springer, 1979, reprinted in {\smc Lindenstrauss \& Tzafriri 96}.
\cmmnt{[\S354 {\it notes\/}, 374Xj.]}

Lindenstrauss J.\ \& Tzafriri L.\ [96] {\it Classical Banach Spaces I \& II.}  Springer, 1996.

Lipschutz S.\ [64] {\it Set Theory and Related Topics.}  McGraw-Hill,
1964 (Schaum's Outline Series).  \cmmnt{[3A1D.]}

Luxemburg W.A.J.\ [67a] `Is every integral normal?',
Bull.\ Amer.\ Math.\ Soc.\ 73 (1967) 685-688.   \cmmnt{[363S.]}

Luxemburg W.A.J.\ [67b] `Rearrangement-invariant Banach function spaces',
pp.\ 83-144 in {\smc Coleman \& Ribenboim 67}.
\cmmnt{[\S374 {\it notes}.]}

Luxemburg W.A.J.\ \& Zaanen A.C.\ [71] {\it Riesz Spaces
I.} North-Holland, 1971.  \cmmnt{[Chap.\ 35 {\it intro\/}.]}

\medskip %M

Macheras N.D., Musia\l\ K.\ \& Strauss W.\ [99] `On products of
admissible liftings and densities',
J.\ for Analysis and its Applications 18 (1999) 651-668. \cmmnt{[346G.]}

Macheras N.D.\ \& Strauss W.\ [95] `Products of lower densities', J.\
for Analysis and its Applications 14 (1995) 25-32.  \cmmnt{[346Xf.]}

Macheras N.D.\  \& Strauss W.\ [96a] `On products of almost strong
liftings', J.\ Australian Math.\ Soc.\ (A) 60 (1996) 1-23.
\cmmnt{[346Yc.]}

Macheras N.D.\ \& Strauss W.\ [96b] `The product lifting for arbitrary products of complete
probability spaces', Atti Sem.\ Math.\ Fis.\ Univ.\ Modena 44 (1996) 485-496.
\cmmnt{[346H, 346Yd.]}

Maharam D.\ [1942] `On homogeneous measure algebras', Proc.\ Nat.\ Acad.\
Sci.\ U.S.A.\ 28 (1942) 108-111.  \cmmnt{[331F, 332B.]}

Maharam D.\ [1947] `An algebraic characterization of measure algebras',
Ann.\ Math.\ 48 (1947) 154-167.
\cmmnt{[393J.]}

Maharam D.\ [58] `On a theorem of von Neumann', Proc.\ Amer.\ Math.\
Soc.\ 9 (1958) 987-994.
\cmmnt{[\S341 {\it notes\/}, \S346 {\it notes\/}.]}

Marczewski E.\ [53] `On compact measures',
Fund.\ Math.\ 40 (1953) 113-124.  \cmmnt{[342A.]}

McCune W.\ [97] `Solution of the Robbins problem',
J.\ Automated Reasoning 19 (1997) 263-276.
\cmmnt{[311Yc.]}

Miller B.D.\ [04] PhD Thesis, University of California, Berkeley, 2004.
%\query
\cmmnt{[382Xc, \S382 {\it notes\/}.]}

Monk J.D.\ [89] (ed.) {\it Handbook of Boolean Algebra.}
North-Holland, 1989.

\medskip %N

Nadkarni M.G.\ [90] `On the existence of a finite invariant measure',
Proc.\ Indian Acad.\ Sci., Math.\ Sci.\ 100 (1990) 203-220.
\cmmnt{[\S395 {\it notes\/}.]}

\medskip %O

Ornstein D.S.\ [74] {\it Ergodic Theory, Randomness and Dynamical Systems.}  Yale U.P., 1974.
\cmmnt{[\S387 {\it notes\/}.]}

Ornstein D.S.\ \& Shields P.C.\ [73] `An uncountable family of
$K$-automorphisms', Advances in Math.\ 10 (1973) 63-88.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip%P

%Pedersen G.K.\ [89] {\it Analysis Now.}  Springer, 1989.

Perovi\'c \v{Z}.\ \& Veli\v{c}kovi\'c B.\ [18] `Ranks of Maharam algebras',
Advances in Math.\ 330 (2018) 253-279.
\cmmnt{[394A.]}

Petersen K.\ [83] {\it Ergodic Theory.}   Cambridge U.P., 1983.
\cmmnt{[328Xa, 385C, \S385 {\it notes\/}, 386E.]}

\medskip %Q%R

Roberts J.W.\ [93] `Maharam's problem', in {\smc Kranz \& Labuda 93}.
\cmmnt{[\S394 {\it notes\/}.]}

Rotman J.J.\ [84] `An Introduction to the Theory of Groups', Allyn \&
Bacon, 1984.
\cmmnt{[\S384 {\it notes\/}, 3A6B.]}

Rudin W. [91] {\it Functional Analysis.}  McGraw-Hill, 1991.
\cmmnt{[\S3A5.]}

Ryzhikov V.V.\ [93] `Factorization of an automorphism of a complete Boolean algebra into a product of three involutions', Mat.\ Zametki (=Math.\ Notes of Russian Acad.\ Sci.) 54 (1993) 79-84.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip %S

Sazonov V.V.\ [66] `On perfect measures', A.M.S.\ Translations (2) 48
(1966) 229-254.   \cmmnt{[\S342 {\it notes\/}.]}

Schaefer H.H.\ [66] {\it Topological Vector Spaces.}  MacMillan, 1966;  reprinted with corrections Springer, 1971.
\cmmnt{[3A4A, 3A5J, 3A5N.]}
%Store A	QA 611.S3    6640 0159 9

Schaefer H.H.\ [74] {\it Banach Lattices and Positive Operators.}
Springer, 1974.   \cmmnt{[Chap.\ 35 {\it intro\/}., \S354 {\it notes\/}.]}

Schubert H.\ [68] {\it Topology.}  Allyn \& Bacon, 1968.
\cmmnt{[\S3A3, \S3A4.]}

Shelah S.\ [98] `The lifting problem with the full ideal',
J.\ Applied Analysis 4 (1998) 1-17.   \cmmnt{[341L.]}

Sikorski R.\ [64] {\it Boolean Algebras.}  Springer, 1964.
\cmmnt{[Chap.\ 31 {\it intro\/}.]}

Sina\v\i\ Ya.G.\ [59] `The notion of entropy of a dynamical system',
Dokl.\ Akad.\ Nauk SSSR 125 (1959) 768-771.  \cmmnt{[385P.]}

Sina\v\i\ Ya.G.\ [62] `Weak isomorphism of transformations with an invariant measure', Soviet Math.\ 3 (1962) 1725-1729.
\cmmnt{[387E.]}
% ({\cyr Dokl.\ Akad.\ Nauk SSSR 147 (1962) 797-800}).

Smorodinsky M.\ [71] {\it Ergodic Theory, Entropy.}
Springer, 1971 (Lecture Notes in Math., 214).
\cmmnt{[\S387 {\it notes\/}.]}

\v St\v ep\'anek P.\ \& Rubin M.\ [89] `Homogeneous Boolean algebras',
pp.\ 679-715 in {\smc Monk 89}.   \cmmnt{[382S, \S382 {\it notes\/}.]}

\medskip %T

Talagrand M.\ [82a] `Closed convex hull of set of measurable functions,
Riemann-\vthsp measurable functions and measurability of translations',
Ann.\ Institut Fourier (Grenoble) 32 (1982) 39-69.
\cmmnt{[\S346 {\it notes\/}.]}

Talagrand M.\ [82b] `La pathologie des rel\grv{e}vements invariants',
Proc.\ Amer.\ Math.\ Soc.\ 84 (1982) 379-382.
\cmmnt{[345F.]}

Talagrand M.\ [84] {\it Pettis integral and measure theory.}
Mem.\ Amer.\ Math.\ Soc.\  307 (1984).
\cmmnt{[\S346 {\it notes\/}.]}

%Talagrand M.\ [06] `Maharam's problem', C.R.Acad.Sci.\ Paris 342 (2006)
%501-503.

Talagrand M.\ [08] `Maharam's problem', Annals of Math.\ 168 (2008)
981-1009.
\cmmnt{[\S394.]}

Taylor A.E.\ [64] {\it Introduction to Functional Analysis.}
Wiley, 1964.
\cmmnt{[\S3A5.]}

Todor\v{c}evi\'c S.\ [04]
`A problem of von Neumann and Maharam about algebras supporting
continuous submeasures',
Fund.\ Math.\ 183 (2004) 169-183.
\cmmnt{[393S.]}

Truss J.K.\ [89] `Infinite permutation groups I:  products of conjugacy
classes',
J.\ Algebra 120 (1989) 454-493.
\cmmnt{[\S382 {\it notes\/}.]}

\medskip %U%V

Vladimirov D.A.\ [02] {\it Boolean algebras in analysis.}
Kluwer, 2002 (Math.\ and its Appl.\ 540).
\cmmnt{[367L.]}

Vulikh B.C.\ [67] {\it Introduction to the Theory of Partially Ordered
Vector Spaces.}   Wolters-Noordhoff, 1967.
\cmmnt{[\S364 {\it notes\/}.]}

\medskip %W

Wagon S. [85] {\it The Banach-Tarski Paradox.}   Cambridge U.P., 1985.
\cmmnt{[\S395 {\it notes\/}.]}

Wilansky A. [64] {\it Functional Analysis.}  Blaisdell, 1964.
\cmmnt{[\S3A5.]}

\medskip %X%Y%Z

Zaanen A.C.\ [83] {\it Riesz Spaces II.}  North-Holland, 1983.
\cmmnt{[Chap.\ 35 {\it intro.\/}, 376K, \S376 {\it notes\/}.]}


\frnewpage

}%end of references


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

\end



