% The following TeXfile comprises part of

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

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

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

%        %   \amstexsupportedfalse

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

%        \input amstex
%        \documentstyle{amsppt}

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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


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

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

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

   \Centerline{#1}

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

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

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

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

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

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

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

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

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

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

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

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

%End of lines for missing AmSTeX

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

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

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

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

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


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

%hyphenations
\hyphenation{Archi-me-dean}

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

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

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

%Headline assignments

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

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

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

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

\newdimen\pagetotalplus
\newdimen\pagetotalminus

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt35.tex} 
\versiondate{4.9.09} 
\copyrightdate{1995} 
 
\def\chaptername{Riesz spaces} 
\def\sectionname{Introduction} 
 
\newchapter{35} 
 
The next three chapters are devoted to an abstract description of the 
`function spaces' described in Chapter 24, this time concentrating on 
their internal structure and relationships with their associated 
measure algebras.   I find that any convincing account of these must 
involve a substantial amount of general theory concerning partially 
ordered linear spaces, and in particular various types of Riesz space or 
vector lattice.   I therefore provide an introduction to this theory, a 
kind of appendix built into the middle of the volume.   The relation of 
this chapter to the next two is very like the relation of Chapter 31 to 
Chapter 32.   As with Chapter 31, it is not really meant to be read for 
its own sake;  those with a particular interest in Riesz spaces might be 
better served by {\smc Luxemburg \& Zaanen 71}, {\smc Schaefer 74}, 
{\smc Zaanen 83} or my own book {\smc Fremlin 74a}. 
 
I begin with three sections in an easy gradation towards the particular 
class of spaces which we need to understand:  partially ordered linear 
spaces (\S351), general Riesz spaces (\S352) and Archimedean Riesz 
spaces (\S353);  the last includes notes on Dedekind 
($\sigma$-\nobreak)\penalty-100complete 
spaces.   These sections cover the fragments of the algebraic theory of 
Riesz spaces which I will use.   In the second half of the chapter, I deal 
with normed Riesz spaces (in particular, $L$- and $M$-spaces)(\S354), 
spaces of linear operators (\S355) and dual Riesz spaces (\S356). 
 
\discrpage 
 
 
\frfilename{mt351.tex}
\versiondate{16.10.07}
\copyrightdate{1996}

\def\chaptername{Riesz spaces}
\def\sectionname{Partially ordered linear spaces}
\def\pols{partially ordered linear space}
\def\polss{partially ordered linear spaces}

\newsection{351}

I begin with an account of the most basic structures which involve an
order relation on a linear space, \polss.   As often in this volume, I
find myself impelled to do some of the work in very much greater
generality than is strictly required, in order to show more clearly the
nature of the arguments being used.   I give the definition (351A) and
most elementary properties (351B-351L) of \polss;  then I describe a
general representation theorem for arbitrary \polss\ as subspaces of
reduced powers of $\Bbb R$ (351M-351Q).   I end with a brief note on
Archimedean \polss\ (351R).


\leader{351A}{Definition}\cmmnt{ I repeat a definition mentioned in
241E.}   A {\bf \pols} is a linear space $(U,+,\cdot)$ over
$\Bbb R$ together with a partial order $\le$ on $U$ such that

\Centerline{$u\le v\Longrightarrow u+w\le v+w$,}

\Centerline{$u\ge 0$, $\alpha\ge 0\Longrightarrow \alpha u\ge 0$}

\noindent for $u$, $v$, $w\in U$ and $\alpha\in\Bbb R$.

\leader{351B}{Elementary facts} Let $U$ be a \pols.  \cmmnt{ We have
the following elementary consequences of the definition above,
corresponding to the familiar rules for manipulating inequalities among
real numbers.}

\spheader 351Ba For $u$, $v\in U$,

\dvro{\Centerline{$u\le v\iff 0\le v-u\iff -v\le-u$.}}
{\Centerline{$u\le v\Longrightarrow 0=u+(-u)\le v+(-u)=v-u
\Longrightarrow u=0+u\le v-u+u=v$,}

\Centerline{$u\le v\Longrightarrow -v=u+(-v-u)\le v+(-v-u)=-u$.}}

\spheader 351Bb Suppose that $u$, $v\in U$ and $u\le v$.   Then
$\alpha u\le\alpha v$ for every $\alpha\ge 0$ and $\alpha v\le\alpha u$ for
every $\alpha\le 0$.   \prooflet{\Prf\ (i) If $\alpha\ge 0$, then
$\alpha(v-u)\ge 0$ so $\alpha v\ge\alpha u$.   (ii) If $\alpha\le 0$
then $(-\alpha)u\le(-\alpha)v$ so

\Centerline{$\alpha v=-(-\alpha)v\le -(-\alpha u)=u$.   \Qed}
}

\spheader 351Bc If $u\ge 0$ and $\alpha\le\beta$ in $\Bbb R$,
then\cmmnt{ $(\beta-\alpha)u\ge 0$, so} $\alpha u\le\beta u$.
If $0\le u\le v$ in $U$ and $0\le\alpha\le\beta$ in $\Bbb R$, then
$\alpha u\cmmnt{\mskip5mu\le\beta u}\le\beta v$.

\leader{351C}{Positive cones} Let $U$ be a \pols.

\spheader 351Ca I will write $U^+$ for the {\bf positive cone} of $U$,
the set $\{u:u\in U,\,u\ge 0\}$.

\spheader 351Cb \cmmnt{By 351Ba, the ordering is determined by the
positive cone $U^+$, in the sense that}
$u\le v\iff v-u\in U^+$.

\spheader 351Cc \cmmnt{It is easy to characterize positive cones.}
If $U$ is a real linear space,  a set $C\subseteq U$ is the positive
cone for some ordering rendering $U$ a \pols\ iff

\Centerline{$u+v\in C$,
\quad$\alpha u\in C$ whenever $u$, $v\in C$ and $\alpha\ge 0$,}

\Centerline{$0\in C$,\quad $u\in C\,\&\,-u\in C\Longrightarrow u=0$.}

\prooflet{\noindent \Prf\ (i) If $C=U^+$ for some \pols\ ordering $\le$
of $U$, then

\Centerline{$u$, $v\in C\Longrightarrow 0\le u\le u+v\Longrightarrow u+v\in C$,}

\Centerline{$u\in C$, $\alpha\ge 0\Longrightarrow \alpha u\ge 0$, i.e.,
$\alpha u\in C$,}

\Centerline{$0\le 0$ so $0\in C$,}

\Centerline{$u$, $-u\in C\Longrightarrow u=0+u\le(-u)+u=0\le u\Longrightarrow
u=0$.}

\noindent (ii) On the other hand, if $C$ satisfies the conditions,
define the relation $\le$ by writing $u\le v\iff v-u\in C$;  then

\Centerline{$u-u=0\in C$ so $u\le u$ for every $u\in U$,}

\Centerline{if $u\le v$ and $v\le w$ then $w-u=(w-v)+(v-u)\in C$ so
$u\le w$,}

\Centerline{if $u\le v$ and $v\le u$ then $u-v$, $v-u\in C$ so $u-v=0$
and $u=v$}

\noindent and $\le$ is a partial order;  moreover,

\Centerline{if $u\le v$ and $w\in U$ then $(v+w)-(u+w)=v-u\in C$ and
$u+w\le v+w$,}

\Centerline{if $u$, $\alpha\ge 0$ then $\alpha u\in C$ and $\alpha u\ge
0$,}

\Centerline{$u\ge 0\iff u\in C$.}

\noindent So $\le$ makes $U$ a \pols\ in which $C$ is the positive
cone.\ \Qed}

\spheader 351Cd \cmmnt{An incidental useful fact.}  Let $U$ be a
\pols, and $u\in U$.   Then $u\ge 0$ iff $u\ge -u$.   \prooflet{\Prf\ If
$u\ge 0$ then $0\ge -u$ so $u\ge -u$.   If $u\ge -u$ then $2u\ge 0$ so
$u=\bover12\cdot 2u\ge 0$.\ \Qed}

\spheader 351Ce \cmmnt{I have called $U^+$ a `positive cone'
without defining the term `cone'.  I think this is something we can
pass by for the moment;  but it will be useful to recognise that} $U^+$
is always
convex\cmmnt{, for if $u$, $v\in U^+$ and $\alpha\in[0,1]$ then
$\alpha u$, $(1-\alpha)v\ge 0$ and $\alpha u+(1-\alpha)v\in U^+$, so is a
`convex cone' as defined in 3A5Ba}.

\leader{351D}{Suprema and infima}  Let $U$ be a \pols.

\spheader 351Da \cmmnt{The
definition of \lq\pols' implies that} $u\mapsto u+w$ is always an
order-isomorphism;  \cmmnt{on the other hand,} $u\mapsto -u$ is
order-reversing\cmmnt{, by 351Ba}.

\spheader 351Db \dvro{If}{It follows that if} $A\subseteq U$ and $v\in U$
then

\Centerline{$\sup_{u\in A}(v+u)=v+\sup A$ if either side is defined,}

\Centerline{$\inf_{u\in A}(v+u)=v+\inf A$ if either side is defined,}

\Centerline{$\sup_{u\in A}(v-u)=v-\inf A$ if either side is defined,}

\Centerline{$\inf_{u\in A}(v-u)=v-\sup A$ if either side is defined.}

\spheader 351Dc \dvro{If}{Moreover, we find that if} $A$, $B\subseteq U$
and $\sup A$ and $\sup B$ are defined, then $\sup(A+B)$ is defined and
equal to $\sup A+\sup B$\cmmnt{, writing $A+B=\{u+v:u\in A,\,v\in B\}$
as usual}.   \prooflet{\Prf\ Set $u_0=\sup A$, $v_0=\sup B$.
Using (b), we have

$$\eqalign{u_0+v_0
&=\sup_{u\in A}(u+v_0)\cr
&=\sup_{u\in A}(\sup_{v\in B}(u+v))
=\sup(A+B).\text{ \Qed}\cr}$$
}

\noindent Similarly, if $A$, $B\subseteq U$ and $\inf A$, $\inf B$ are
defined then $\inf(A+B)=\inf A+\inf B$.
%end of 351Dc

\spheader 351Dd If $\alpha>0$ then\cmmnt{ $u\mapsto\alpha u$ is an
order-isomorphism, so we have} $\sup(\alpha A)=\alpha\sup A$ if either
side is defined;  similarly, $\inf(\alpha A)=\alpha\inf A$.

\leader{351E}{Linear subspaces} If $U$ is a \pols, and $V$ is any linear
subspace of $U$, then $V$, with the induced linear and order structures,
is a \pols\cmmnt{;  this is obvious from the definition}.

\leader{351F}{Positive linear operators} Let $U$ and $V$ be \polss, and
write $\eurm L(U;V)$ for the linear space of all linear operators from
$U$ to $V$.
For $S$, $T\in\eurm L(U;V)$ say that $S\le T$ iff $Su\le Tu$ for every
$u\in U^+$.   Under this ordering, $\eurm L(U;V)$ is a \pols;  its
positive cone is
$\{T:Tu\ge 0$ for every $u\in U^+\}$.   \prooflet{\Prf\ This is an
elementary verification.\ \Qed}   \dvro{For}{Note that, for}
$T\in\eurm L(U;V)$,\cmmnt{

$$\eqalign{T\ge 0
&\Longrightarrow Tu\le Tu+T(v-u)=Tv\text{ whenever }u\le v\text{ in }U\cr
&\Longrightarrow 0=T0\le Tu\text{ for every }u\in U^+\cr
&\Longrightarrow T\ge 0,\cr}$$

\noindent so that} $T\ge 0$ iff $T$ is order-preserving.   In this case
we say that $T$ is a {\bf positive} linear operator.

\cmmnt{Clearly }$ST$ is a positive linear operator whenever $U$, $V$
and $W$ are \polss\ and $T:U\to V$, $S:V\to W$ are positive linear
operators\cmmnt{ (cf.\ 313Ia)}.
%end of 351F

\leader{351G}{Order-continuous positive linear operators:  Proposition}
Let $U$ and $V$ be \polss\ and $T:U\to V$ a positive linear operator.

(a) The following are equiveridical:

\quad(i) $T$ is order-continuous;

\quad(ii) $\inf T[A]=0$ in $V$ whenever $A\subseteq U$ is a non-empty
downwards-directed set with infimum $0$ in $U$;

\quad(iii) $\sup T[A]=Tw$ in
$V$ whenever $A\subseteq U^+$ is a non-empty upwards-directed set with
supremum $w$ in $U$.

(b) The following are equiveridical:

\quad(i) $T$ is sequentially
order-continuous;

\quad(ii) $\inf_{n\in\Bbb N}Tu_n=0$ in $V$ whenever
$\sequencen{u_n}$ is a non-increasing sequence in $U$ with infimum $0$
in $U$;

\quad(iii) $\sup_{n\in\Bbb N}Tu_n=Tw$ in $V$ whenever
$\sequencen{u_n}$ is a non-decreasing sequence in $U^+$ with supremum
$w$ in $U$.

\proof{{\bf (a)(i)$\Rightarrow$(iii)} is trivial.

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} Assuming (iii), and given that $A$ is
non-empty, downwards-directed and has infimum $0$, take any $u_0\in A$
and consider $A'=\{u:u\in A,\,u\le u_0\}$,
$B=u_0-A'$.   Then $A'$ is non-empty, downwards-directed and has infimum
$0$, so $B$ is non-empty, upwards-directed and has supremum $u_0$ (using
351Db);  by (iii), $\sup T[B]=Tu_0$ and (inverting again)

\Centerline{$\inf T[A']=\inf T[u_0-B]=\inf(Tu_0-T[B])=Tu_0-\sup T[B]=0$.}

\noindent But (because $T$ is positive) $0$ is surely a lower bound for
$T[A]$, so it is also the infimum of $T[A]$.   As $A$ is arbitrary, (ii)
is true.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} Suppose now that (ii) is true.
\grheada\ If $A\subseteq U$ is non-empty, downwards-directed and has
infimum $w$, then $A-w$ is non-empty, downwards-directed and has infimum
$0$, so

\Centerline{$\inf T[A-w]=0$,\quad$\inf T[A]=\inf(T[A-w]+Tw)
=Tw+\inf T[A-w]=Tw$.}

\noindent\grheadb\ If $A\subseteq U$ is non-empty, upwards-directed and
has supremum $w$, then $-A$ is non-empty, downwards-directed and has
infimum $-w$, so

\Centerline{$\sup T[A]=-\inf(-T[A])=-\inf T[-A]=-T(-w)=Tw$.}

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

\medskip

{\bf (b)} The arguments are identical, replacing each directed set by an
appropriate sequence.
}%end of proof of 351G

\leader{351H}{Riesz homomorphisms (a)}\cmmnt{ For the sake of a
representation theorem below (351Q), I introduce the
following definition.}  Let $U$, $V$ be \polss.   A {\bf Riesz
homomorphism} from $U$ to $V$ is a linear operator $T:U\to V$ such that
whenever $A\subseteq U$ is a finite non-empty set and $\inf A=0$ in $U$,
then $\inf T[A]=0$ in $V$.   \cmmnt{The following facts are now nearly
obvious.}

\spheader 351Hb Any Riesz homomorphism is a positive linear operator.
\prooflet{(For if $T$ is a Riesz homomorphism and $u\ge 0$, then
$\inf\{0,u\}=0$ so $\inf\{0,Tu\}=0$ and $Tu\ge 0$.)}

\spheader 351Hc Let $U$ and $V$ be \polss\ and $T:U\to V$ a Riesz
homomorphism.   Then

\Centerline{$\inf T[A]$ exists $=T(\inf A)$, \quad $\sup T[A]$ exists
$=T(\sup A)$}

\noindent whenever $A\subseteq U$ is a finite non-empty set and
$\inf A$, $\sup A$ exist.   \prooflet{(Apply the definition in (a) to

\Centerline{$A'=\{u-\inf A:u\in A\}$,\quad $A''=\{\sup A-u:u\in A\}$.)}
}

\spheader 351Hd If $U$, $V$ and $W$ are \polss\ and $T:U\to V$,
$S:V\to W$
are Riesz homomorphisms then $ST:U\to W$ is a Riesz homomorphism.

\leader{351I}{Solid sets} Let $U$ be a \pols.   I will say
that a subset $A$ of $U$ is {\bf solid} if

\dvro{\Centerline{$A=\bigcup_{u\in A}[-u,u]$.}}
{\Centerline{$A=\{v:v\in U,\,-u\le v\le u$ for some $u\in A\}
=\bigcup_{u\in A}[-u,u]$}}

\cmmnt{\noindent in the notation of 2A1Ab.   (I should perhaps remark
that while this definition
is well established in the case of Riesz spaces (352J), the extension
to general \polss\ is not standard.   See 351Yb for a warning.)}

\leader{351J}{Proposition} Let $U$ be a \pols\ and $V$ a solid linear
subspace of $U$.   Then the quotient linear space $U/V$ has a \pols\
structure defined by either of the rules

\inset{$u^{\ssbullet}\le w^{\ssbullet}$ iff there is some $v\in V$ such
that $u\le v+w$,}

\inset{$(U/V)^+=\{u^{\ssbullet}:u\in U^+\}$,}

\noindent and for this partial order on $U/V$ the map
$u\mapsto u^{\ssbullet}:U\to U/V$ is a Riesz homomorphism.

\proof{{\bf (a)} I had better start by giving priority to one of the
descriptions of the relation $\le$ on $U/V$;  I choose the first.   To
see that this makes $U/V$ a \pols, we have to check the following.

(i) $0\in V$ and $u\le u+0$, so $u^{\ssbullet}\le u^{\ssbullet}$ for
every $u\in U$.

(ii) If $u_1$, $u_2$, $u_3\in U$ and
$u_1^{\ssbullet}\le u_2^{\ssbullet}$,
$u_2^{\ssbullet}\le u_3^{\ssbullet}$ then there are
$v_1$, $v_2\in V$ such that $u_1\le u_2+v_1$, $u_2\le u_3+v_2$;  in
which case $v_1+v_2\in V$ and $u_1\le u_3+v_1+v_2$, so
$u_1^{\ssbullet}\le u_3^{\ssbullet}$.

(iii) If $u$, $w\in U$ and $u^{\ssbullet}\le w^{\ssbullet}$,
$w^{\ssbullet}\le u^{\ssbullet}$ then there are $v$, $v'\in V$ such that
$u\le w+v$, $w\le u+v'$.   Now there are $v_0$, $v'_0\in V$ such that
$-v_0\le v\le v_0$, $-v'_0\le v'\le v'_0$, and in this case $v_0$,
$v'_0\ge 0$ (351Cd), so

\Centerline{$-v_0-v'_0\le -v'\le u-w\le v\le v_0+v'_0\in V$,}.

\noindent Accordingly  $u-w\in V$ and $u^{\ssbullet}=w^{\ssbullet}$.
Thus $U/V$ is a partially ordered set.

(iv) If $u_1$, $u_2$, $w\in U$ and $u_1^{\ssbullet}\le u_2^{\ssbullet}$,
then there is a $v\in V$ such that $u_1\le u_2+v$, in which case
$u_1+w\le u_2+w+v$ and $u_1^{\ssbullet}+w^{\ssbullet}\le
u_2^{\ssbullet}+w^{\ssbullet}$.

(v) If $u\in U$, $\alpha\in \Bbb R$, $u^{\ssbullet}\ge 0$ and
$\alpha\ge 0$ then there is a $v\in V$ such that $u+v\ge 0$;
now $\alpha v\in V$ and $\alpha u+\alpha v\ge 0$, so
$\alpha u^{\ssbullet}=(\alpha u)^{\ssbullet}\ge 0$.

Thus $U/V$ is a \pols.

\medskip

{\bf (b)} Now $(U/V)^+=\{u^{\ssbullet}:u\ge 0\}$.   \Prf\ If $u\ge 0$
then of course $u^{\ssbullet}\ge 0$ because $0\in V$ and $u+0\ge 0$.
On the other hand, if we have any element $p$ of $(U/V)^+$, there are
$u\in U$, $v\in V$ such that $u^{\ssbullet}=p$ and $u+v\ge 0$;  but now
$p=(u+v)^{\ssbullet}$ is of the required form.\ \Qed

\medskip

{\bf (c)} Finally, $u\mapsto u^{\ssbullet}$ is a Riesz homomorphism.
\Prf\ Suppose that $A\subseteq U$ is a non-empty finite set and that
$\inf A=0$ in $U$.   Then $u^{\ssbullet}\ge 0$ for every $u\in A$, that
is, $0$ is a lower bound for $\{u^{\ssbullet}:u\in A\}$.   Let $p\in
U/V$ be any other lower bound for
$\{u^{\ssbullet}:u\in A\}$.   Express $p$ as $w^{\ssbullet}$ where $w\in
U$.   For each $u\in A$, $w^{\ssbullet}\le u^{\ssbullet}$ so there is a
$v_u\in V$ such that $w\le u+v_u$.   Next, there is a $v'_u\in V$ such
that $-v'_u\le v_u\le v'_u$.   Set $v^*=\sum_{u\in A}v'_u\in V$.   Then
$v_u\le v'_u\le v^*$ so $w\le u+v^*$ for every $u\in A$, and $w-v^*$ is
a lower bound for $A$ in $U$.   Accordingly $w-v^*\le 0$, $w\le0 + v^*$
and $p=w^{\ssbullet}\le 0$.   As $p$ is arbitrary,
$\inf\{u^{\ssbullet}:u\in A\}=0$;  as $A$ is arbitrary, $u\mapsto
u^{\ssbullet}$ is a Riesz homomorphism.\ \Qed
}%end of proof of 351J

\leader{351K}{Lemma} Suppose that $U$ is a \pols, and that $W$, $V$ are
solid linear subspaces of $U$ such that $W\subseteq V$.   Then
$V_1=\{v^{\ssbullet}:v\in V\}$ is a solid linear subspace of $U/W$.

\proof{ (i) Because the map $u\mapsto u^{\ssbullet}$ is linear, $V_1$ is
a linear subspace of $U/W$.   (ii) If $p\in V_1$, there is a $v\in V$
such that $p=v^{\ssbullet}$;  because $V$ is solid in $U$, there is a
$v_0\in V$ such that $-v_0\le v\le v_0$;  now $v_0^{\ssbullet}\in V_1$
and $-v_0^{\ssbullet}\le p\le v_0^{\ssbullet}$.   (iii) If $p\in V_1$,
$q\in U/W$ and $-p\le q\le p$, take $v_0\in V$, $u\in U$ such that
$v_0^{\ssbullet}=p$ and $u^{\ssbullet}=q$.   Because
$-v_0^{\ssbullet}\le u^{\ssbullet}\le v_0^{\ssbullet}$, there are $w$,
$w'\in W$ such that
$-v_0-w\le u\le v_0+w'$.   Now $-v_0-w$, $v_0+w'$ both belong to $V$,
which is solid, so $u\in V$
and $q=u^{\ssbullet}\in V_1$.   (iv) Putting (ii) and (iii) together,
$V_1$ is solid.
}%end of proof of 351K

\leader{351L}{Products} If $\langle U_i\rangle_{i\in I}$ is any family
of \polss, we have a product linear space $U=\prod_{i\in I}U_i$;  if we
set $u\le v$ in $U$ iff $u(i)\le v(i)$ for every $i\in I$, $U$ becomes a
\pols, with positive cone $\{u:u(i)\ge 0$ for every $i\in I\}$.   For
each $i\in I$ the map $u\mapsto u(i):U\to U_i$ is an order-continuous
Riesz
homomorphism\cmmnt{ (in fact, it preserves arbitrary suprema and
infima)}.

\leader{351M}{Reduced powers of $\Bbb R$ (a)} Let $X$ be any set.   Then
$\Bbb R^X$ is a \pols\ if we say that $f\le g$ means that $f(x)\le g(x)$
for every $x\in X$\cmmnt{, as in 351L}.   If now $\Cal F$ is a filter
on $X$, we have a corresponding set

\Centerline{$V=\{f:f\in\Bbb R^X,\,\{x:f(x)=0\}\in\Cal F\}$;}

\noindent\cmmnt{it is easy to see that }$V$ is a linear subspace of
$\Bbb R^X$, and is solid\cmmnt{ because $f\in V$ iff $|f|\in V$}.
By the {\bf reduced power} $\Bbb R^X|\Cal F$ I shall mean the quotient
\pols\ $\Bbb R^X/V$.

\spheader 351Mb \dvro{For}{Note that for} $f\in\Bbb R^X$,

\Centerline{$f^{\ssbullet}\ge 0$ in
$\Bbb R^X|\Cal F\iff\{x:f(x)\ge 0\}\in \Cal F$.}

\prooflet{\noindent \Prf\ (i) If $f^{\ssbullet}\ge 0$, there is a
$g\in V$ such that $f+g\ge 0$;  now

\Centerline{$\{x:f(x)\ge 0\}\supseteq\{x:g(x)=0\}\in\Cal F$.}

\noindent (ii) If $\{x:f(x)\ge 0\}\in\Cal F$, then
$\{x:(|f|-f)(x)=0\}\in\Cal F$, so
$f^{\ssbullet}=|f|^{\ssbullet}\ge 0$.\ \Qed}

\leader{351N}{}\cmmnt{ On the way to the next theorem, the main result
(in terms of mathematical content)
of this section, we need a string of lemmas.

\medskip

\noindent}{\bf Lemma} Let $U$ be a \pols.   If $u$,
$v_0,\ldots,v_n\in U$ are such that $u\ne 0$ and $v_0,\ldots,v_n\ge 0$
then there is a
linear functional $f:U\to\Bbb R$
such that $f(u)\ne 0$ and $f(v_i)\ge 0$ for every $i$.

\proof{ The point is that at most one of $u$, $-u$ can belong to the
convex cone $C$ generated by $\{v_0,\ldots,v_n\}$, because this is
included in the convex cone set $U^+$, and since $u\ne 0$ at most one of
$u$, $-u$ can belong to $U^+$.

Now however the Hahn-Banach theorem, in the form 3A5D, tells us
that if $u\notin C$ there is a linear functional $f:U\to\Bbb R$ such
that $f(u)<0$ and $f(v_i)\ge 0$ for every $i$;  while if $-u\notin C$ we
can get $f(-u)<0$ and $f(v_i)\ge 0$ for every $i$.   Thus in either case
we have a functional of the required type.
}%end of proof of 351N

\leader{351O}{Lemma} Let $U$ be a \pols, and $u_0$ a non-zero member of
$U$.   Then there is a solid linear subspace $V$ of $U$ such that
$u_0\notin V$ and whenever $A\subseteq U$ is finite, non-empty and has
infimum $0$ then $A\cap V\ne\emptyset$.

\proof{{\bf (a)} Let $\Cal W$ be the family of all solid linear
subspaces of $U$ not containing $u_0$.   Then any non-empty totally
ordered $\Cal V\subseteq\Cal W$ has an upper bound $\bigcup\Cal V$ in
$\Cal W$.    By Zorn's Lemma, $\Cal W$ has a maximal element $V$ say.
This is surely a solid linear subspace of $U$ not containing $u_0$.

\medskip

{\bf (b)} Now
for any $w\in U^+\setminus V$ there are $\alpha\ge 0$, $v\in V^+$ such
that $-\alpha w-v\le u_0\le \alpha w+v$.   \Prf\ Let $V_1$ be

\Centerline{$\{u:u\in U$, there are $\alpha\ge 0$, $v\in V^+$ such that
$-\alpha w-v\le u\le\alpha w+v\}$.}

\noindent Then it is easy to check that $V_1$ is a solid linear subspace
of $U$, including $V$, and containing $w$;  because $w\notin V$,
$V_1\ne V$, so $V_1\notin\Cal W$ and $u\in V_1$, as claimed.\ \Qed

\medskip

{\bf (c)} It follows that if $A\subseteq U$ is finite and non-empty and
$\inf A=0$ in $U$ then $A\cap V\ne\emptyset$.   \Prf\Quer\ Otherwise, for every
$w\in A$ there must be $\alpha_w\ge 0$, $v_w\in V^+$ such that
$-\alpha_ww-v_w\le u_0\le\alpha_ww+v_w$.   Set
$\alpha=1+\sum_{w\in A}\alpha_w$, $v=\sum_{w\in A}v_w\in V$;  then
$-\alpha w-v\le u_0\le\alpha w+v$ for every $w\in A$.   Accordingly
$\bover1{\alpha}(u_0-v)\le w$ for
every $w\in A$ and $\bover1{\alpha}(u_0-v)\le 0$, so $u_0\le v$.
Similarly,
$-\bover1{\alpha}(v+u_0)\le w$ for every $w\in A$ and $-v\le u_0$.   But
(because $V$ is solid) this means that $u_0\in V$, which is not so.\
\Bang\Qed

Accordingly $V$ has the required properties.
}%end of proof of 351O

\leader{351P}{Lemma} Let $U$ be a \pols\ and $u$ a non-zero element of $U$, and suppose that $A_0,\ldots,A_n$ are finite non-empty subsets of $U$ such that $\inf A_j=0$ for every $j\le n$.   Then there is a linear
functional $f:U\to\Bbb R$ such that $f(u)\ne 0$ and $\min f[A_j]=0$ for
every $j\le n$.

\proof{ By 351O, there is a solid linear subspace $V$ of $U$ such that
$u\notin V$ and $A_j\cap V\ne 0$ for every $j\le n$.   Give the quotient
space $U/V$ its standard partial ordering (351J), and in $U/V$ set
$C=\{v^{\ssbullet}:v\in\bigcup_{j\le n}A_j\}$.   Then $C$ is a finite
subset of $(U/V)^+$, while $u^{\ssbullet}\ne 0$, so by 351N there is a
linear functional $g:U/V\to\Bbb R$ such that $g(u^{\ssbullet})\ne 0$ but
$g(p)\ge 0$ for every $p\in C$.   Set $f(v)=g(v^{\ssbullet})$ for $v\in
U$;  then $f:U\to\Bbb R$ is linear, $f(u)\ne 0$ and $f(v)\ge 0$ for
every $v\in\bigcup_{j\le n}A_j$.   But also, for each $j\le n$, there is
a $v_j\in A_j\cap V$, so that $f(v_j)=0$;  and this means that
$\min f[A_j]$ must be $0$, as required.
}%end of proof of 351P

\leader{351Q}{}\cmmnt{ Now we are ready for the theorem.

\medskip

\noindent}{\bf Theorem} Let $U$ be any \pols.   Then we can find a set
$X$, a filter $\Cal F$ on $X$ and an injective Riesz homomorphism from
$U$ to the reduced power
$\Bbb R^X|\Cal F$\cmmnt{ described in 351M}.

\proof{ Let $X=U'$ be the set of all linear functionals $f:U\to\Bbb R$;
for $u\in U$ define $\hat u\in\Bbb R^X$ by setting
$\hat u(f)=f(u)$ whenever $f\in X$ and $u\in U$.   Let $\Cal A$ be the
family of non-empty finite sets $A\subseteq U$ such that $\inf A=0$.
For $A\in\Cal A$ let $F_A$ be the set of those $f\in X$ such that
$\min f[A]=0$.   Since $0\in F_A$ for every $A\in\Cal A$, the set

\Centerline{$\Cal F=\{F:F\subseteq X$, there are
$A_0,\ldots,A_n\in\Cal A$ such that
$F\supseteq\bigcap_{j\le n}F_{A_j}\}$}

\noindent is a filter on $X$.   Set
$\psi(u)=\hat u^{\ssbullet}\in \Bbb R^X|\Cal F$ for $u\in U$.
Then $\psi:U\to\Bbb R^X|\Cal F$ is an injective Riesz homomorphism.

\Prf\ (i) $\psi$ is linear because $u\mapsto\hat u:U\to\BbbR^X$ and
$h\mapsto h^{\ssbullet}:\Bbb R^X\to\BbbR^X|\Cal F$ are linear.
(ii) If $A\in\Cal A$,
then $F_A\in\Cal F$.   So, first, if $v\in A$, then
$\{f:\hat v(f)\ge 0\}\in\Cal F$, so that
$\psi(v)=\hat v^{\ssbullet}\ge 0$ in
$\Bbb R^X|\Cal F$
(351Mb).   Next, if $w\in\Bbb R^X|\Cal F$ and $w\le\psi(v)$ for every
$v\in A$, we can express $w$ as $h^{\ssbullet}$ where
$h^{\ssbullet}\le\hat v^{\ssbullet}$ for every $v\in A$, that is,
$H_v=\{f:h(f)\le\hat v(f)\}\in\Cal F$ for every $v\in A$.   But now
$H=F_A\cap\bigcap_{v\in A}H_v\in\Cal F$, and for $f\in H$ we have
$h(f)\le\min_{v\in A}f(v)=0$.   This means that $w=h^{\ssbullet}\le 0$.
As $w$ is arbitrary, $\inf\psi[A]=0$.   As $A$ is arbitrary, $\psi$ is a
Riesz homomorphism.   (iii) Finally, \Quer\ suppose, if possible, that
there is a non-zero $u\in U$ such that $\psi(u)=0$.   Then
$F=\{f:f(u)=0\}\in\Cal F$, and there are $A_0,\ldots,A_n\in\Cal A$
such that $F\supseteq\bigcap_{j\le n}F_{A_j}$.   By 351P, there is an
$f\in\bigcap_{j\le n}F_{A_j}$ such that $f(u)\ne 0$;  which is
impossible.\ \Bang\
Accordingly $\psi$ is injective, as claimed.\ \Qed
}%end of proof of 351Q

\vleader{72pt}{351R}{Archimedean spaces (a)} For a \pols\ $U$, the following
are equiveridical:  (i) if $u$, $v\in U$ are such that $nu\le v$ for every
$n\in\Bbb N$ then $u\le 0$ (ii) if $u\ge 0$ in $U$ then
$\inf_{\delta>0}\delta u=0$.   \prooflet{\Prf{\bf (i)$\Rightarrow$(ii)} If
(i) is true and $u\ge 0$, then of course $\delta u\ge 0$ for every
$\delta>0$;  on the other hand, if $v\le\delta u$ for every $\delta>0$,
then $nv\le n\cdot\bover1nu=u$ for every $n\ge 1$, while of course
$0v=0\le u$, so $v\le 0$.   Thus $0$ is the greatest lower bound of
$\{\delta u:\delta>0\}$.   {\bf (ii)$\Rightarrow$(i)} If (ii) is true
and $nu\le v$ for every $n\in\Bbb N$, then $0\le v$ and $u\le\bover1nv$
for every $n\ge 1$.   If now $\delta>0$, then there is an $n\ge 1$ such
that $\bover1n\le\delta$, so that $u\le\bover1nv\le\delta v$ (351Bc).
Accordingly $u$ is a lower bound for $\{\delta v:\delta>0\}$ and
$u\le 0$.\ \Qed}

\spheader 351Rb I will say that \polss\ satisfying the equiveridical
conditions of (a) above are {\bf Archimedean}.

\spheader 351Rc Any linear subspace of an Archimedean \pols, with the
induced \pols\ structure, is Archimedean.

\spheader 351Rd Any product of Archimedean \polss\ is Archimedean.
\prooflet{\Prf\ If $U=\prod_{i\in I}U_i$ is a product of Archimedean
spaces, and $nu\le v$ in $U$ for every $n\in\Bbb N$, then for each $i\in
I$ we must have $nu(i)\le v(i)$ for every $n$, so that $u(i)\le 0$;
accordingly $u\le 0$.\ \Qed}
In particular, $\Bbb R^X$ is Archimedean for any set $X$.

\exercises{\leader{351X}{Basic exercises $\pmb{>}$(a)}
Let $\zeta$ be any ordinal.   The {\bf lexicographic ordering} of
$\Bbb R^{\zeta}$ is defined by saying that $f\le g$ iff either $f=g$
or there
is a $\xi<\zeta$ such that $f(\eta)=g(\eta)$ for $\eta<\xi$ and
$f(\xi)<g(\xi)$.   Show that this is a total order on $\BbbR^{\zeta}$
which renders $\BbbR^{\zeta}$ a \pols.
%351A

\spheader 351Xb Let $U$ be a \pols\ and $V$ a linear subspace of $U$.
Show that the formulae of 351J define a \pols\ structure on the quotient
$U/V$ iff $V$ is {\bf order-convex}, that is, $u\in V$ whenever $v_1$,
$v_2\in V$ and $v_1\le u\le v_2$.
%351J

\spheader 351Xc Let $\langle U_i\rangle_{i\in I}$ be a family of \polss\
with product $U$.   For $i\in I$, define $T_i:U_i\to U$ by setting
$T_ix=u$ where
$u(i)=x$, $u(j)=0$ for $j\ne i$.   Show that $T_i$ is an injective
order-continuous Riesz homomorphism.
%351L

\sqheader 351Xd Let $U$ be a \pols\ and $\langle V_i\rangle_{i\in I}$ a
family of \polss\ with product $V$.   Show that $\eurm L(U;V)$ can be
identified, as \pols, with $\prod_{i\in I}\eurm L(U;V_i)$.
%351L

\sqheader 351Xe Show that if $U$, $V$ are \polss\ and $V$ is
Archimedean, then $\eurm L(U;V)$ is Archimedean.
%351R

\leader{351Y}{Further exercises (a)}
%\spheader 351Ya
Give an example of two \polss\ $U$ and $V$ and a bijective
Riesz homomorphism $T:U\to V$ such that $T^{-1}:V\to U$ is not a Riesz
homomorphism.
%351H mt35bits

\spheader 351Yb(i) Let $U$ be a \pols.   Show that $U$ is a solid
subset of itself (on the definition 351I) iff $U=U^+-U^+$.   (ii) Give
an example of a \pols\ $U$ satisfying this condition with an element
$u\in U$ such that the intersection of the solid sets containing $u$ is
not solid.
%351I mt35bits

\spheader 351Yc Show that a reduced power $\Bbb R^X|\Cal F$, as
described in 351M, is totally ordered iff $\Cal F$ is an ultrafilter,
and in this case has a natural structure as a totally ordered field.
%351M

\spheader 351Yd Let $U$ be a \pols, and suppose that $A$, $B\subseteq U$
are two non-empty finite sets such that ($\alpha$) $u\vee v=\sup\{u,v\}$
is defined for every $u\in A$, $v\in B$ ($\beta$) $\inf A$ and $\inf B$
and $(\inf A)\vee(\inf B)$ are defined.   Show that
$\inf\{u\vee v:u\in A,\,v\in B\}=(\inf A)\vee(\inf B)$.   ({\it Hint\/}:
show that this is true if $U=\Bbb R$, if
$U=\Bbb R^X$ and if $U=\Bbb R^X|\Cal F$, and use 351Q.)
%351Q

\spheader 351Ye Show that a reduced power $\Bbb R^X|\Cal F$, as
described in 351M, is Archimedean iff $\bigcap_{n\in\Bbb N}F_n\in\Cal F$
whenever $\sequencen{F_n}$ is a sequence in $\Cal F$.
%351R
}%end of exercises

\cmmnt{\Notesheader{351} The idea of `\pols' is a very natural
abstraction from the elementary examples of $\Bbb R^X$ and its
subspaces, and the only possible difficulty lies in guessing the exact
boundary at which one's standard manipulations with such familiar spaces
cease to be valid in the general case.   (For instance, most people's
favourite examples are Archimedean, in the sense of 351R, so it is
prudent to check your intuitions against a non-Archimedean space like
that of 351Xa.)   There is really no room for any deep idea to appear in
351B-351F.   When I come to what I call `Riesz homomorphisms',
however (351H), there are some more interesting possibilities in the
background.

I shall not discuss the applications of Theorem 351Q to general \polss;
it is here for the sake of its application to Riesz spaces in the next
section.   But I think it is a very striking fact that not only does any
\pols\ $U$ appear as a linear subspace of some reduced power of
$\Bbb R$, but the embedding can be taken to preserve any suprema and infima of
finite sets which exist in $U$.   This is in a sense a result of the
same kind as the Stone representation theorem for Boolean algebras;  it
gives us a chance to confirm that an intuition valid for $\Bbb R$ or
$\BbbR^X$ may in fact apply to arbitrary \polss.   If you like, this
provides a metamathematical foundation for such results as those in
351B.   I have to say that for \polss\ it is generally quicker to find a
proof directly from the definition than to trace through an argument
relying on 351Q;  but this is not always the case for Riesz spaces.   I
offer 351Yd as an example of a result where a direct proof does at least
call for a moment's thought, while the argument through 351Q is
straightforward.

`Reduced powers' are of course of great importance for other reasons;
I mention 351Yc as a hint of what can be done.

}%end of notes
\discrpage

\frfilename{mt352.tex}
\versiondate{9.6.16/16.8.22}
\copyrightdate{1998}

\def\chaptername{Riesz spaces}
\def\sectionname{Riesz spaces}
\def\pols{partially ordered linear space}
\def\med{\mathop{\text{med}}}

\newsection{352}

In this section I sketch those fragments of the theory we need which can
be expressed as theorems about general Riesz spaces or vector lattices.
I begin with the definition (352A) and most elementary properties
(352C-352F).
In 352G-352J I discuss Riesz homomorphisms and the associated subspaces
(Riesz subspaces, solid linear subspaces);  I mention product spaces
(352K, 352T) and quotient spaces (352Jb, 352U) and the form the
representation theorem 351Q takes in the present context
(352L-352M).   Most of the second half of the section
concerns the theory of `bands' in Riesz spaces, with the algebras of
complemented bands (352Q) and projection bands (352S) and a description
of bands generated by upwards-directed sets (352V).   I conclude with a
description of `$f$-algebras' (352W).

\leader{352A}{}\cmmnt{ I repeat a definition from 241E.

\medskip

\noindent}{\bf Definition} A {\bf Riesz space} or {\bf vector lattice}
is a partially ordered linear space which is a lattice.

\leader{352B}{Lemma} If $U$ is a \pols, then it is a Riesz space iff
$\sup\{0,u\}$ is defined for every $u\in U$.

\proof{ If $U$ is a lattice, then of course $\sup\{u,0\}$ is defined for
every $u$.   If $\sup\{u,0\}$ is defined for every $u$, and $v_1$, $v_2$
are any two members of $U$, consider $w=v_1+\sup\{0,v_2-v_1\}$;  by
351Db, $w=\sup\{v_1,v_2\}$.   Next,

\Centerline{$\inf\{v_1,v_2\}=-\sup\{-v_1,-v_2\}$}

\noindent must also be defined in $U$, because $v\mapsto -v$ is
order-reversing;  as $v_1$ and $v_2$ are arbitrary, $U$ is a lattice.
}%end of proof of 352B

\leader{352C}{Notation} In any Riesz space $U$ I will write

\Centerline{$u^+=u\vee 0$,\quad $u^-=(-u)\vee 0=(-u)^+$,\quad
$|u|=u\vee(-u)$}

\noindent where\cmmnt{ (as in any lattice)} $u\vee v=\sup\{u,v\}$
(and $u\wedge v=\inf\{u,v\}$).

\dvro{A}{I mention immediately a term which will be useful:  a} family
$\familyiI{u_i}$ in $U$ is {\bf disjoint} if
$|u_i|\wedge|u_j|=0$ for all distinct $i$, $j\in I$.   Similarly, a set
$C\subseteq U$ is {\bf disjoint} if $|u|\wedge|v|=0$ for all distinct
$u$, $v\in C$.

\leader{352D}{Elementary identities} Let $U$ be a Riesz space.
\cmmnt{ The translation-invariance of the order, and its invariance
under positive scalar multiplication, reversal under negative
multiplication, lead directly to the following, which are in effect
special cases of 351D:}

\Centerline{$u+(v\vee w)=(u+v)\vee(u+w)$,
\quad$u+(v\wedge w)=(u+v)\wedge(u+w)$,}

\Centerline{$\alpha(u\vee v)=\alpha u\vee\alpha v$ and
$\alpha(u\wedge v)=\alpha u\wedge\alpha v$ if $\alpha\ge 0$,}

\Centerline{$-(u\vee v)=(-u)\wedge(-v)$.}

\noindent Combining and elaborating on these facts, we get

\Centerline{$u^+-u^-
\prooflet{\mskip5mu =(u\vee 0)-((-u)\vee 0)
=u+(0\vee(-u))-((-u)\vee 0)}
=u$,}

\Centerline{$u^++u^-
\prooflet{\mskip5mu =2u^+-u
=(2u\vee 0)-u
=u\vee(-u)}
=|u|$,}

\Centerline{$u\ge 0
\iff -u\le 0
\iff u^-=0
\iff u=u^+
\iff u=|u|$,}

\Centerline{$|-u|=|u|$,\quad$|\,|u|\,|=|u|$,\quad$|\alpha u|
=|\alpha||u|$\dvro{,}{}}

\prooflet{\hfill (looking at the cases $\alpha\ge 0$, $\alpha\le 0$
separately),}

\ifwithproofs{
$$\eqalign{u\vee v+u\wedge v
&=u+(0\vee(v-u))+v+((u-v)\wedge 0)\cr
&=u+(0\vee(v-u))+v-((v-u)\vee 0)
=u+v,\cr}$$
}\else{
$$\eqalign{u\vee v+u\wedge v
=u+v,\cr}$$
}\fi

\Centerline{$u\vee v
\prooflet{\mskip5mu =u+(0\vee(v-u))}
=u+(v-u)^+$,}

\Centerline{$u\wedge v
\prooflet{\mskip5mu =u+(0\wedge(v-u))
=u-(-0\vee(u-v))}
=u-(u-v)^+$,}

\Centerline{$u\vee v
\prooflet{\mskip5mu =\Bover12(2u\vee 2v)
=\Bover12(u+v+(u-v)\vee(v-u))}
=\Bover12(u+v+|u-v|)$,}

\Centerline{$u\wedge v
\prooflet{\mskip5mu =u+v-u\vee v}
=\Bover12(u+v-|u-v|)$,}

\Centerline{$u^+\vee u^-
\prooflet{\mskip5mu =u\vee(-u)\vee 0}
=|u|$,
\quad$u^+\wedge u^-
\prooflet{\mskip5mu =u^++u^--(u^+\vee u^-)}
=0$,}

\ifwithproofs
$$\eqalignno{|u+v|=(u+v)\vee((-u)+(-v))&
\le(|u|+|v|)\vee(|u|+|v|)=|u|+|v|,\cr
\cr
||u|-|v||
=(|u|-|v|)\vee(|v|-|u|)&
\le(|u-v|+|v|-|v|)\vee(|v-u|+|u|-|u|)\cr&
=|u-v|\vee|v-u|=|u-v|,\cr
\cr
|u\vee v|&\le|u|+|v|\cr&&
(\text{because }
-|u|\le u\vee v\le|u|\vee|v|\le|u|+|v|)
}$$
\else
\Centerline{$|u+v|\le|u|+|v|$,
\quad$||u|-|v||\le|u-v|$,
\quad$|u\vee v|\le|u|+|v|$}
\fi

\noindent for $u$, $v\in U$ and $\alpha\in\Bbb R$.

\leader{352E}{Distributive laws} Let $U$ be a Riesz space.

\spheader 352Ea If $A$, $B\subseteq U$ have suprema $a$, $b$ in $U$,
then $C=\{u\wedge v:u\in A,\,v\in B\}$ has supremum $a\wedge b$.
\prooflet{\Prf\ Of course $u\wedge v\le a\wedge b$ for all $u\in A$,
$v\in B$, so $a\wedge b$ is an upper bound for $C$.   Now suppose that
$c$ is any upper bound for $C$.   If $u\in A$ and $v\in B$ then

\Centerline{$u-(u-v)^+=u\wedge v\le c$,
\quad$u\le c+(u-v)^+\le c+(a-v)^+$}

\noindent (because $(a-v)^+=\sup\{a-v,0\}\ge\sup\{u-v,0\}=(u-v)^+$).
As $u$ is arbitrary, $a\le c+(a-v)^+$ and $a\wedge v\le c$.   Now turn
the argument round:

\Centerline{$v=(a\wedge v)+(v-a)^+\le c+(v-a)^+\le c+(b-a)^+$,}

\noindent and this is true for every $v\in B$, so $b\le c+(b-a)^+$ and
$a\wedge b\le c$.   As $c$ is arbitrary, $a\wedge b=\sup C$, as
claimed.\ \Qed\ }

\spheader 352Eb\cmmnt{ Similarly, or applying (a) to $-A$ and $-B$,}
$\inf\{u\vee v:u\in A,\,v\in B\}=\inf A\vee\inf B$ whenever $A$,
$B\subseteq U$ and the right-hand-side is defined.

\spheader 352Ec\cmmnt{ In particular,} $U$ is a distributive
lattice\cmmnt{ (definition:  3A1Ic)}.

\leader{352F}{Further identities
and \dvrocolon{inequalities}}\cmmnt{ At a
slightly deeper level we have the following facts.

\medskip

\noindent}{\bf Proposition} Let $U$ be a Riesz space.

(a)(i) If $u$, $v$, $w\ge 0$ in $U$ then
$u\wedge(v+w)\le (u\wedge v)+(u\wedge w)$.

\quad(ii) If $v_0,\ldots,v_m,w_0,\ldots,w_n\in U^+$ then

\Centerline{$\sum_{i=0}^mv_i\wedge\sum_{j=0}^nw_j
\le\sum_{i=0}^m\sum_{j=0}^nv_i\wedge w_j$.}

(b) If $u_0,\ldots,u_n\in U$ are disjoint, then
$|\sum_{i=0}^n\alpha_i u_i|=\sum_{i=0}^n|\alpha_i||u_i|$ for any
$\alpha_0,\ldots,\alpha_n\in\Bbb R$.

(c) If $u$, $v\in U$ then

\Centerline{$u^+\wedge v^+\le(u+v)^+\le u^++v^+$.}

(d) If $u_0,\ldots,u_m,v_0,\ldots,v_n\in U^+$ and
$\sum_{i=0}^mu_i=\sum_{j=0}^nv_j$, then there is a family $\langle
w_{ij}\rangle_{i\le m,j\le n}$ in $U^+$ such that
$\sum_{i=0}^mw_{ij}=v_j$ for every $j\le n$ and $\sum_{j=0}^nw_{ij}=u_i$
for every $i\le m$.

\proof{{\bf (a)(i)}

$$\eqalign{u\wedge(v+w)
&=\bigl((u+w)\wedge(v+w)\bigr)\wedge u\cr
&\le\bigl((u\wedge v)+w\bigr)\wedge\bigl((u\wedge v)+u\bigr)
=(u\wedge v)+(u\wedge w).\cr}$$

\medskip

\quad{\bf (ii)} Inducing on $n$, we see that

\Centerline{$u\wedge\sum_{j=0}^nw_i\le\sum_{j=0}^nu\wedge w_i$}

\noindent for every $u\ge 0$;  so that

\Centerline{$\sum_{i=0}^mv_i\wedge\sum_{j=0}^nw_j
\le\sum_{j=0}^n(\sum_{i=0}^mv_i)\wedge w_j
\le\sum_{j=0}^n\sum_{i=0}^mv_i\wedge w_j$.}

\medskip

{\bf (b)(i)}\grheada\ If $u\wedge v=0$ then

\Centerline{$(u-v)^+=u-(u\wedge v)=u$,
\quad$(u-v)^-=(v-u)^+=v-(v\wedge u)=v$,}

\Centerline{$|u-v|=(u-v)^++(u-v)^-=u+v=|u+v|$,}

\noindent so if $|u|\wedge|v|=0$ then

$$\eqalign{(u^++v^+)\wedge(u^-+v^-)
&\le (u^+\wedge u^-)+(u^+\wedge v^-)+(v^+\wedge u^-)+(v^+\wedge v^-)\cr
&\le 0+(|u|\wedge|v|)+(|v|\wedge|u|)+0=0\cr}$$

\noindent and

\Centerline{$|u+v|=|(u^++v^+)-(u^-+v^-)|=u^++v^++u^-+v^-=|u|+|v|$.}

\medskip

\qquad\grheadb\ Now if $|u|\wedge|v|=0$ and $\alpha$,
$\beta\in\Bbb R$,

\Centerline{$|\alpha u|\wedge|\beta v|
=|\alpha||u|\wedge|\beta||v|
\le(|\alpha|+|\beta|)|u|\wedge(|\alpha|+|\beta|)|v|
=(|\alpha|+|\beta|)(|u|\wedge|v|)
=0$.}

\medskip

\quad{\bf (ii)} We may therefore proceed by induction.   The case $n=0$
is trivial.   For the inductive step to $n+1$, setting
$u'_i=\alpha_iu_i$ we have $|u'_i|\wedge|u'_j|=0$ for all $i\ne j$, by
(i-$\gamma$).   By (i-$\alpha$),

\Centerline{$|u'_{n+1}|\wedge|\sum_{i=0}^nu'_i|
\le\sum_{i=0}^n|u'_{n+1}|\wedge|u'_i|=0$,}

\noindent so by (i-$\beta$) and the inductive hypothesis

\Centerline{$|\sum_{i=0}^{n+1}u'_i|
=|u'_{n+1}|+|\sum_{i=0}^nu'_i|
=\sum_{i=0}^{n+1}|u'_i|$}

\noindent as required.

\medskip

{\bf (c)} By 352E,

\Centerline{$u^+\wedge v^+=(u\vee 0)\wedge(v\vee 0)=(u\wedge v)\vee 0$.}

\noindent Now

\Centerline{$u\wedge v=\Bover12(u+v-|u-v|)\le\Bover12(u+v+|u+v|)
=(u+v)^+$,}

\noindent and of course $0\le(u+v)^+$, so $u^+\wedge v^+\le(u+v)^+$.

For the other inequality we need only note that $u+v\le u^++v^+$
(because $u\le u^+$, $v\le v^+$) and $0\le u^++v^+$.

\medskip

{\bf (d)} Write $w$ for the common value of $\sum_{i=0}^mu_i$ and
$\sum_{j=0}^nv_j$.

Induce on
$k=\#(\{(i,j):i\le m,\,j\le n,\,u_i\wedge v_j>0\})$.   If $k=0$, that
is, $u_i\wedge v_j=0$ for all $i$, $j$, then, by (a-ii),
we must have $w\wedge w=0$, that is, $w=0$,
and we can take $w_{ij}=0$ for all $i$, $j$.   For
the inductive step to $k\ge 1$, take $i^*$, $j^*$ such that
$w^*=u_{i^*}\wedge v_{j^*}>0$.   Set

\Centerline{$\tilde u_{i^*}=u_{i^*}-w^*$,\quad $\tilde u_i=u_i$ for
$i\ne i^*$,}

\Centerline{$\tilde v_{j^*}=v_{j^*}-w^*$,\quad $\tilde v_j=v_j$ for
$j\ne j^*$.}

\noindent Then
$\sum_{i=0}^m\tilde u_i=\sum_{j=0}^n\tilde v_j=w-w^*$ and
$\tilde u_i\wedge\tilde v_j\le u_i\wedge v_j$ for all $i$, $j$,
while $\tilde u_{i^*}\wedge\tilde v_{j^*}=0$;  so that

\Centerline{$\#(\{(i,j):\tilde u_i\wedge\tilde v_j>0\})<k$.}

\noindent By the inductive hypothesis, there are $\tilde w_{ij}\ge 0$,
for $i\le m$ and $j\le n$, such that
$\tilde u_i=\sum_{j=0}^n\tilde w_{ij}$ for each $i$ and
$\tilde v_j=\sum_{i=0}^m\tilde w_{ij}$ for each
$j$.   Set $w_{i^*j^*}=\tilde w_{i^*j^*}+w^*$,
$w_{ij}=\tilde w_{ij}$ for
$(i,j)\ne(i^*,j^*)$;  then $u_i=\sum_{j=0}^nw_{ij}$ and
$v_j=\sum_{i=0}^mw_{ij}$, so the induction proceeds.
}%end of proof of 352F

\leader{352G}{Riesz homomorphisms:  Proposition} Let $U$ be a Riesz
space, $V$ a \pols\ and $T:U\to V$ a linear operator.   Then the
following
are equiveridical:

(i) $T$ is a Riesz homomorphism in the sense of 351H;

(ii) $(Tu)^+=\sup\{Tu,0\}$ is defined and equal to $T(u^+)$ for every
$u\in U$;

(iii) $\sup\{Tu,-Tu\}$ is defined and equal to $T|u|$ for every
$u\in U$;

(iv) $\inf\{Tu,Tv\}=0$ in $V$ whenever $u\wedge v=0$ in $U$.

\proof{ (i)$\Rightarrow$(iii) and (i)$\Rightarrow$(iv) are special cases
of 351Hc.   For
(iii)$\Rightarrow$(ii) we have

\Centerline{$\sup\{Tu,0\}=\Bover12Tu+\sup\{\Bover12Tu,-\Bover12Tu\}
=\Bover12Tu+\Bover12T|u|=T(u^+)$.}


\noindent For (ii)$\Rightarrow$(i), argue as follows.   If (ii) is true
and $u$, $v\in U$, then

\Centerline{$Tu\wedge Tv
=\inf\{Tu,Tv\}
=Tu+\inf\{0,Tv-Tu\}
=Tu-\sup\{0,T(u-v)\}$}

\noindent is defined and equal to


\Centerline{$Tu-T((u-v)^+)
=T(u-(u-v)^+)
=T(u\wedge v)$.}

\noindent Inducing on $n$,

\Centerline{$\inf_{i\le n}Tu_i=T(\inf_{i\le n}u_i)$}

\noindent for all $u_0,\ldots,u_n\in U$;  in particular, if $\inf_{i\le
n}u_i=0$ then $\inf_{i\le n}Tu_i=0$;  which is the definition I gave of
Riesz homomorphism.

Finally, for (iv)$\Rightarrow$(ii), we know from (iv) that
$0=\inf\{T(u^+),T(u^-)\}$, so $-T(u^+)=\inf\{0,-Tu\}$ and
$T(u^+)=\sup\{0,Tu\}$.
}%end of proof of 352G

\leader{352H}{Proposition} If $U$ and $V$ are Riesz spaces and
$T:U\to V$ is a bijective Riesz homomorphism, then $T$ is a
partially-ordered-linear-space isomorphism, and $T^{-1}:V\to U$ is a
Riesz homomorphism.

\proof{ Use 352G(ii).  If $v\in V$, set $u=T^{-1}v$;  then $T(u^+)=v^+$
so $T^{-1}(v^+)=u^+=(T^{-1}v)^+$.   Thus $T^{-1}$ is a Riesz
homomorphism;  in particular, it is order-preserving, so $T$ is an
isomorphism for the order structures as well as for the linear
structures.
}%end of proof of 352H

\leader{352I}{Riesz subspaces (a)} If $U$ is a \pols, a {\bf Riesz
subspace} of $U$ is a linear subspace $V$ such that $\sup\{u,v\}$ and
$\inf\{u,v\}$ are defined in $U$ and belong to $V$ for every $u$,
$v\in V$.   In this case\cmmnt{ they are the supremum and infimum of
$\{u,v\}$
in $V$, so} $V$, with the induced order and linear structure, is a Riesz
space in its own right, and the embedding map $u\mapsto u:V\to U$ is a
Riesz homomorphism.

\spheader 352Ib Generally, if $U$ is a Riesz space, $V$ is a \pols\ and
$T:U\to V$ is a Riesz homomorphism, then $T[U]$ is a Riesz subspace of
$V$\prooflet{ (because, by 351Hc, $Tu\vee Tu'=T(u\vee u')$,
$Tu\wedge Tu'=T(u\wedge u')$ are defined in $V$ and
belong to $T[U]$ for all $u$, $u'\in U$)}.

\spheader 352Ic If $U$ is a Riesz space and $V$ is a linear subspace of
$U$, then $V$ is a Riesz subspace of $U$ iff $|u|\in V$ for every
$u\in V$.   \prooflet{\Prf\ In this case,

\Centerline{$u\vee v=\Bover12(u+v+|u-v|)$,
\quad$u\wedge v=\Bover12(u+v-|u-v|)$}

\noindent belong to $V$ for all $u$, $v\in V$.\ \Qed}

\leader{352J}{Solid subsets (a)} If $U$ is a Riesz space, a
subset $A$ of $U$ is solid\cmmnt{ (in the sense of 351I)} iff
$v\in A$ whenever $u\in A$ and $|v|\le|u|$.   \prooflet{\Prf\ ($\alpha$)
If $A$
is solid, $u\in A$ and $|v|\le|u|$, then there is some $w\in A$ such
that $-w\le u\le w$;  in this case $|v|\le|u|\le w$ and $-w\le v\le w$
and $v\in A$.   ($\beta$) Suppose that $A$ satisfies the condition.   If
$u\in A$, then $|u|\in A$ and $-|u|\le u\le|u|$.   If $w\in A$ and
$-w\le u\le w$ then $-u\le w$, $|u|\le w=|w|$ and $u\in A$.   Thus $A$
is solid.\ \Qed\ }   In particular, if $A$ is
solid, then $v\in A$ iff $|v|\in A$.

For any set $A\subseteq U$, the set

\Centerline{$\{u:$ there is some $v\in A$ such that $|u|\le|v|\}$}

\noindent is a solid subset of $U$, the smallest solid set including
$A$;  we call it the {\bf solid hull} of $A$ in $U$.

Any solid linear subspace of $U$ is a Riesz subspace\prooflet{ (use
352Ic)}.   If $V\subseteq U$ is a Riesz subspace,
then the solid hull of $V$ in $U$ is

\Centerline{$\{u:$ there is some $v\in V$ such that $|u|\le v\}$}

\noindent and is a solid linear subspace of $U$.

\spheader 352Jb If $T$ is a Riesz homomorphism from a Riesz space $U$ to
a \pols\ $V$, then its kernel $W$ is a solid linear subspace of $U$.
\prooflet{\Prf\ If $u\in W$ and $|v|\le|u|$, then
$T|u|=\sup\{Tu,T(-u)\}=0$, while $-|u|\le v\le |u|$, so that $-0\le
Tv\le 0$ and $v\in W$.\ \Qed\ }

Now the quotient space $U/W$\cmmnt{,
as defined in 351J,} isomorphic, as \pols, to
$T[U]$, and\cmmnt{ in particular} is a Riesz space.
\prooflet{\Prf\ Because $U/W$ is the linear
space quotient of $V$ by the kernel of
the linear operator $T$, we have an induced linear space isomorphism
$S:U/W\to T[U]$ given by setting $Su^{\ssbullet}=Tu$ for every $u\in U$.
If $p\ge 0$ in $U/W$ there is a $u\in U^+$ such that $u^{\ssbullet}=p$
(351J), so that $Sp=Tu\ge 0$.   On the other hand, if $p\in U/W$ and
$Sp\ge 0$, take $u\in U$ such that $u^{\ssbullet}=p$.   By 352G, we have

\Centerline{$T(u^+)=(Tu)^+=\sup\{Sp,0\}=Sp=Tu$,}

\noindent so that $T(u^-)=Tu^+-Tu=0$, $u^-\in W$ and
$p=(u^+)^{\ssbullet}\ge 0$.   Thus $Sp\ge 0$ iff $p\ge 0$, and $S$ is a
partially-ordered-linear-space isomorphism.   We know from 352Ib that
$T[U]$ is a Riesz space, so $U/W$ also is.\ \Qed}

\cmmnt{
\spheader 352Jc Because a subset of a Riesz space is a solid linear
subspace iff it is the kernel of a Riesz homomorphism (see 352U below), such subspaces are sometimes called {\bf ideals}.
}%end of comment

\leader{352K}{Products} If $\langle U_i\rangle_{i\in I}$ is any family
of Riesz spaces, then the product \pols\
$U=\prod_{i\in I}U_i$\cmmnt{ (351L)} is a Riesz space, with

\Centerline{$u\vee v=\langle u(i)\vee v(i)\rangle_{i\in I}$,
\quad$u\wedge v=\langle u(i)\wedge v(i)\rangle_{i\in I}$,
\quad$|u|=\langle|u(i)|\rangle_{i\in I}$}

\noindent for all $u$, $v\in U$.

\leader{352L}{Theorem} Let $U$ be any Riesz space.   Then there are a
set $X$, a filter $\Cal F$ on $X$ and a Riesz subspace of the Riesz
space $\Bbb R^X|\Cal F$\cmmnt{ (definition:  351M)} which is
isomorphic, as Riesz space, to $U$.

\proof{ By 351Q, we can find such $X$ and $\Cal F$ and an injective
Riesz homomorphism $T:U\to\Bbb R^X|\Cal F$.   By 352K, or otherwise,
$\Bbb R^X$ is a Riesz space;  by 352Jb, $\Bbb R^X|\Cal F$ is a Riesz
space (recall that it is a quotient of $\Bbb R^X$ by a solid linear
subspace, as explained in 351M);  by 352Ib, $T[U]$ is a Riesz subspace
of $\Bbb R^X|\Cal F$;  and by 352H it is isomorphic to $U$.
}%end of proof of 352L

\leader{352M}{Corollary} Any identity involving the operations $+$, $-$,
$\vee$, $\wedge$, $^+$, $^-$, $|\,\,|$ and scalar multiplication, and
the relation $\le$, which is valid in $\Bbb R$, is valid in all Riesz
spaces.

\cmmnt{\medskip

\noindent{\bf Remark} I suppose some would say that a strict proof of
this must begin with a formal description of what the phrase \lq any
identity involving the operations$\ldots$' means.   However I think it
is clear in practice what is involved.   Given a proposed identity like

\Centerline{$0\le\sum_{i=0}^n|\alpha_i||u_i|-|\sum_{i=0}^n\alpha_iu_i|
\le\sum_{i\ne j}(|\alpha_i|+|\alpha_j|)(|u_i|\wedge|u_j|)$,}

\noindent (compare 352Fb), then to check that it is valid in all Riesz
spaces you need only check (i) that it is true in $\Bbb R$ (ii) that it
is true in $\Bbb R^X$ (iii) that it is true in any $\Bbb R^X|\Cal F$
(iv) that it is true in any Riesz subspace of $\Bbb R^X|\Cal F$;  and
you can hope that
the arguments for (ii)-(iv) will be nearly trivial, since
(ii) is generally nothing but a coordinate-by-coordinate repetition of
(i), and (iii) and (iv) involve only transformations of the formula by
Riesz homomorphisms which preserve its structure.
%mt35bits
}%end of comment

\leader{352N}{Order-density and order-continuity} Let $U$
be a Riesz space.

\spheader 352Na {\bf Definition}
A Riesz subspace $V$ of $U$ is
{\bf quasi-order-dense} if for every $u>0$ in $U$ there is a $v\in V$
such that $0<v\le u$;  it is {\bf order-dense} if $u=\sup\{v:v\in
V,\,0\le v\le u\}$ for every $u\in U^+$.

\spheader 352Nb If $U$ is a Riesz space and $V$ is a quasi-order-dense
Riesz subspace of $U$, then the embedding $V\embedsinto U$ is
order-continuous.   \prooflet{\Prf\ Let $A\subseteq V$ be a non-empty
set such that $\inf A=0$ in $V$.   \Quer\ If $0$ is not the infimum of
$A$ in $U$, then there is a $u>0$ such that $u$ is a lower bound for $A$
in $U$;  now there is a
$v\in V$ such that $0<v\le u$, and $v$ is a lower bound for $A$ in
$V$ which is strictly greater than $0$.\ \Bang\   Thus $0=\inf A$ in
$U$.   As $A$ is arbitrary, the embedding is order-continuous,
by 351Ga.\ \Qed}

\spheader 352Nc(i) If $V\subseteq U$ is an order-dense Riesz subspace,
it is quasi-order-dense.   (ii) If $V$ is a quasi-order-dense Riesz
subspace of $U$ and $W$ is a quasi-order-dense Riesz subspace of $V$,
then $W$ is a
quasi-order-dense Riesz subspace of $U$.   (iii) If $V$ is an
order-dense Riesz subspace of $U$ and $W$ is an order-dense Riesz
subspace of $V$, then $W$ is an order-dense Riesz subspace of $U$.
\cmmnt{(Use (b).)}
(iv) If $V$ is a quasi-order-dense solid linear subspace of $U$ and $W$
is a quasi-order-dense Riesz subspace of $U$ then $V\cap W$ is
quasi-order-dense\cmmnt{ in $V$, therefore} in $U$.

\spheader 352Nd \dvro{A}{I ought somewhere to remark that a} Riesz
homomorphism\cmmnt{, being a lattice homomorphism,} is
order-continuous iff it preserves arbitrary suprema and infima\cmmnt{;
compare 313L(b-iv) and (b-v)}.

\spheader 352Ne If $V$ is a Riesz subspace of $U$, we say that it is
{\bf regularly embedded} in $U$ if the identity map from $V$ to $U$ is
order-continuous\cmmnt{, that is, whenever $A\subseteq V$ is non-empty
and has infimum $0$ in $V$, then $0$ is still its
greatest lower bound in $U$.   Thus quasi-order-dense Riesz subspaces
and solid linear subspaces are regularly embedded}.

\leader{352O}{Bands} Let $U$ be a Riesz space.

\spheader 352Oa {\bf Definition}    A {\bf
band} or {\bf normal subspace} of $U$ is an order-closed solid linear
subspace.

\spheader 352Ob If $V\subseteq U$ is a solid
linear subspace then it is a band iff $\sup A\in V$ whenever
$A\subseteq V^+$ is a non-empty, upwards-directed subset of $V$ with a
supremum in
$U$.   \prooflet{\Prf\ Of course the condition is necessary;  I have to
show that it is sufficient.   (i) Let $A\subseteq V$ be any non-empty
upwards-directed set with a supremum in $V$.   Take any $u_0\in A$ and
set
$A_1=\{u-u_0:u\in A,\,u\ge u_0\}$.   Then $A_1$ is a non-empty
upwards-directed subset of $V^+$, and $u_0+A_1=\{u:u\in A,\,u\ge u_0\}$
has the same upper bounds as $A$, so $\sup A_1=\sup A-u_0$ is defined in
$U$ and belongs to $V$.   Now $\sup A=u_0+\sup A_1$ also belongs to $V$.
(ii) If $A\subseteq V$ is non-empty, downwards-directed and has an
infimum in $U$, then $-A\subseteq V$ is upwards-directed, so $\inf
A=\sup(-A)$ belongs to $V$.   Thus $V$ is order-closed.\ \Qed\ }

\spheader 352Oc For any set $A\subseteq U$
set $A^{\perp}=\{v:v\in U,\,|u|\wedge|v|=0$ for every $u\in A\}$.   Then
$A^{\perp}$ is a band.  \prooflet{\Prf\ (i) Of course $0\in A^{\perp}$.
(ii) If $v$, $w\in A^{\perp}$ and $u\in A$, then

\Centerline{$0\le|u|\wedge|v+w|\le(|u|\wedge|v|)+(|u|\wedge|w|)=0$,}

\noindent so $v+w\in A^{\perp}$.   (iii) If $v\in A^{\perp}$ and
$|w|\le|v|$ then

\Centerline{$0\le|u|\wedge|w|\le|u|\wedge|v|=0$}

\noindent for every $u\in A$, so $w\in A^{\perp}$.   (iv) If $v\in
A^{\perp}$ then $nv\in A^{\perp}$ for every $n$, by (ii).   So if
$\alpha\in\Bbb R$, take $n\in\Bbb N$ such that $|\alpha|\le n$;  then

\Centerline{$|\alpha v|=|\alpha||v|\le n|v|\in A^{\perp}$}

\noindent and $\alpha v\in A^{\perp}$.   Thus $A^{\perp}$ is a solid
linear subspace of $U$.   (v) If $B\subseteq (A^{\perp})^+$ is non-empty
and upwards-directed and has a supremum $w$ in $U$, then

\Centerline{$|u|\wedge|w|=|u|\wedge w=\sup_{v\in B}|u|\wedge v=0$}

\noindent by 352Ea, so $w\in A^{\perp}$.   Thus $A^{\perp}$ is a
band.\ \Qed\ }

\spheader 352Od For any $A\subseteq U$, $A\subseteq
(A^{\perp})^{\perp}$.   Also $B^{\perp}\subseteq A^{\perp}$ whenever
$A\subseteq B$.   So\prooflet{

\Centerline{$A^{\perp\perp\perp}\subseteq A^{\perp}\subseteq
A^{\perp\perp\perp}$}

\noindent and} $A^{\perp}=A^{\perp\perp\perp}$.

\spheader 352Oe If $W$ is another Riesz space and $T:U\to W$ is an
order-continuous Riesz homomorphism then its kernel is a band.
\prooflet{(For $\{0\}$ is order-closed in $W$ and the inverse image of
an order-closed set under an order-continuous order-preserving function is
order-closed (313Id).)}

\leader{352P}{Complemented bands} Let $U$ be a Riesz space.
A band $V\subseteq U$ is {\bf complemented} if $V^{\perp\perp}=V$, that
is, if $V$ is of the form $A^{\perp}$ for some
$A\subseteq U$\cmmnt{ (352Od)}.
In this case its {\bf complement} is the complemented band $V^{\perp}$.

\vleader{36pt}{352Q}{Theorem} In any Riesz space $U$, the set $\frak C$
of complemented bands forms a Dedekind complete Boolean algebra, with

\Centerline{$V\hskip.2em\Bcapshort_{\frak C}\hskip.2em W=V\cap W$,
\quad$V\hskip.2em\Bcupshort_{\frak C}\hskip.2em W=(V+W)^{\perp\perp}$,}

\Centerline{$1_{\frak C}=U$,
\quad $0_{\frak C}=\{0\}$,
\quad$1_{\frak C}\hskip.2em\Bsetminusshort_{\frak C}\hskip.2em
V=V^{\perp}$,}

\Centerline{$V\hskip.2em\Bsubseteqshort_{\frak C}\hskip.2em W\iff
V\subseteq W$}

\noindent for $V$, $W\in\frak C$.

\proof{ To show that $\frak C$ is a Boolean algebra, I use the
identification of Boolean algebras with complemented distributive
lattices (311L).

\medskip

{\bf (a)} Of course $\frak C$ is partially ordered by $\subseteq$.   If
$V$, $W\in\frak C$ then

\Centerline{$V\cap W=V^{\perp\perp}\cap W^{\perp\perp}
=(V^{\perp}\cup W^{\perp})^{\perp}\in\frak C$,}

\noindent and $V\cap W$ must be $\inf\{V,W\}$ in $\frak C$.   The map
$V\mapsto V^{\perp}:\frak C\to\frak C$ is an order-reversing permutation,
so that $V\subseteq W$ iff $W^{\perp}\subseteq V^{\perp}$ and
$V\vee W=\sup\{V,W\}$ will be $(V^{\perp}\cap W^{\perp})^{\perp}$;  thus
$\frak C$ is a lattice.   Note also that $V\vee W$ must be the smallest
complemented band including $V+W$, that is, it is $(V+W)^{\perp\perp}$.

\medskip

{\bf (b)} If $V_1$, $V_2$, $W\in\frak C$ then
$(V_1\vee V_2)\wedge W=(V_1\wedge W)\vee(V_2\wedge W)$.   \Prf\ Of course
$(V_1\vee V_2)\wedge W\supseteq(V_1\wedge W)\vee(V_2\wedge W)$.   \Quer\
Suppose, if possible, that there is a $u\in(V_1\vee V_2)\cap
W\setminus((V_1\cap W)\vee(V_2\cap W))$.   Then $u\notin((V_1\cap
W)^{\perp}\cap(V_2\cap W)^{\perp})^{\perp}$, so there is a $v\in(V_1\cap
W)^{\perp}\cap(V_2\cap W)^{\perp}$ such that $u_1=|u|\wedge|v|>0$.   Now
$u_1\in V_1\vee V_2=(V_1^{\perp}\cap V_2^{\perp})^{\perp}$ so $u_1\notin
V_1^{\perp}\cap V_2^{\perp}$;  say $u_1\notin V_j^{\perp}$, and there is
a $v_j\in V_j$ such that $u_2=u_1\wedge|v_j|>0$.   In this case we still
have $u_2\in (V_j\cap W)^{\perp}$, because $u_2\le|v|$, but also $u_2\in
V_j$ and $u_2\in W$ because $u_2\le|u|$;  but this means that
$u_2=u_2\wedge u_2=0$, which is absurd.\ \Bang\  Thus
$(V_1\vee V_2)\wedge W\subseteq(V_1\wedge W)\vee(V_2\wedge W)$ and the
two are equal.\ \Qed

\medskip

{\bf (c)} Now if $V\in\frak C$,

\Centerline{$V\wedge V^{\perp}=\{0\}$}

\noindent is the least member of $\frak C$, because if
$v\in V\cap V^{\perp}$ then $|v|=|v|\wedge|v|=0$.   By 311L, $\frak C$
has a Boolean algebra structure, with the Boolean relations described;
by 312M, this structure is uniquely defined.

\medskip

{\bf (d)} Finally, if $\Cal V\subseteq\frak C$ is non-empty, then

\Centerline{$\bigcap\Cal V=(\bigcup_{V\in\Cal
V}V^{\perp})^{\perp}\in\frak C$}

\noindent and is $\inf\Cal V$ in $\frak C$.   So $\frak C$ is Dedekind
complete.
}%end of proof of 352Q

\leader{352R}{Projection bands} Let $U$ be a Riesz space.

\spheader 352Ra A {\bf projection band} in $U$ is a set
$V\subseteq U$ such that $V+V^{\perp}=U$.   In this case $V$ is a
complemented band.   \prooflet{\Prf\ If $v\in V^{\perp\perp}$ then $v$
is expressible as $v_1+v_2$ where $v_1\in V$ and $v_2\in V^{\perp}$.
Now $|v|=|v_1|+|v_2|\ge|v_2|$ (352Fb), so

\Centerline{$|v_2|=|v_2|\wedge|v_2|\le|v_2|\wedge|v|=0$}

\noindent and $v=v_1\in V$.  Thus $V=V^{\perp\perp}$ is a complemented
band.\ \QeD}   Observe that $U=V^{\perp}+V^{\perp\perp}$ so $V^{\perp}$
also is a projection band.

\spheader 352Rb\cmmnt{ Because $V\cap V^{\perp}$ is always $\{0\}$, we
must have} $U=V\oplus V^{\perp}$ for any projection band $V\subseteq U$;
\cmmnt{accordingly} there is a corresponding {\bf band projection}
$P_V:U\to U$ defined by setting $P(v+w)=v$ whenever $v\in V$, $w\in
V^{\perp}$.   In this context I will say that $v$ is the {\bf component}
of $v+w$ in $V$.   The kernel of $P$ is $V^{\perp}$, the set of values
is $V$, and $P^2=P$.   \cmmnt{Moreover,}
$P$ is an order-continuous Riesz homomorphism.   \prooflet{\Prf\ (i) $P$
is a linear operator because $V$ and $V^{\perp}$ are linear subspaces.
(ii) If $v\in V$ and $w\in V^{\perp}$ then $|v+w|=|v|+|w|$, by 352Fb, so
$P|v+w|=|v|=|P(v+w)|$;  consequently $P$ is a Riesz homomorphism (352G).
(iii) If $A\subseteq U$ is downwards-directed and has infimum $0$, then
$Pu\le u$ for every $u\in A$, so $\inf P[A]=0$;  thus $P$ is
order-continuous.\ \Qed\ }

\spheader 352Rc Note that for any band projection $P$, and any $u\in U$,
we have\cmmnt{ $|Pu|\wedge|u-Pu|=0$, so that $|u|=|Pu|+|u-Pu|$ and (in
particular)} $|Pu|\le|u|$;  \cmmnt{consequently} $P[W]\subseteq W$ for
any solid linear subspace $W$ of $U$.

\spheader 352Rd A linear operator $P:U\to U$ is a band projection iff
$Pu\wedge(u-Pu)=0$ for every $u\in U^+$.   \prooflet{\Prf\ I remarked in
(c) that the condition is satisfied for any band projection.
Now suppose that $P$ has the property.   (i) For any $u\in U^+$,
$Pu\ge 0$ and $u-Pu\ge 0$;  in particular, $P$ is a positive linear
operator.   (ii) If $u$, $v\in U^+$ then
$u-Pu\le(u+v)-P(u+v)$, so

\Centerline{$Pv\wedge(u-Pu)\le P(u+v)\wedge((u+v)-P(u+v))=0$}

\noindent and $Pv\wedge(u-Pu)=0$.   (iii) If $u$, $v\in U$ then $|Pv|\le
P|v|$, $|u-Pu|\le|u|-P|u|$ (because $w\mapsto w-Pw$ is a positive linear
operator), so

\Centerline{$|Pv|\wedge|u-Pu|\le P|v|\wedge(|u|-P|u|)=0$.}

\noindent (iv) Setting $V=P[U]$, we see that $u-Pu\in V^{\perp}$ for
every $u\in U$, so that

\Centerline{$u=u+(u-Pu)\in V+V^{\perp}$}

\noindent for every $u$, and $U=V+V^{\perp}$;  thus $V$ is a projection
band.   (v) Since $Pu\in V$ and $u-Pu\in V^{\perp}$ for every $u\in U$,
$P$ is the band projection onto $V$.\ \Qed}

\leader{352S}{Proposition} Let $U$ be any Riesz space.

(a) The family $\frak B$ of projection bands in $U$ is a subalgebra of
the Boolean algebra $\frak C$ of complemented bands in $U$.

(b) For $V\in\frak B$ let $P_V:U\to V$ be the corresponding projection.
Then for any $e\in U^+$,

\Centerline{$P_{V\cap W}e=P_Ve\wedge P_We=P_VP_We$,
\quad$P_{V\vee W}e=P_Ve\vee P_We$}

\noindent for all $V$, $W\in\frak B$.   In particular, band projections
commute.

(c) If $V\in\frak B$ then the algebra of projection bands in $V$ is just
the principal ideal of $\frak B$ generated by $V$.

\proof{{\bf (a)} Of course $0_{\frak C}=\{0\}\in\frak B$.   If
$V\in \frak B$ then $V^{\perp}=1_{\frak C}\Bsetminus V$ belongs to
$\frak B$.   If now $W$ is another member of $\frak B$, then

\Centerline{$(V\cap W)+(V\cap W)^{\perp}
\supseteq (V\cap W)+V^{\perp}+W^{\perp}$.}

\noindent But if $u\in U$ then we can express $u$ as $v+v'$, where $v\in
V$ and $v'\in V^{\perp}$, and $v$ as $w+w'$, where $w\in W$ and $w'\in
W^{\perp}$;  and as $|w|\le|v|$, we also have $w\in V$, so that

\Centerline{$u=w+v'+w'\in (V\cap W)+V^{\perp}+W^{\perp}$.}

\noindent This shows that $V\cap W\in\frak B$.   Thus $\frak B$ is
closed under intersection and complements
and is a subalgebra of $\frak C$.

\medskip

(b) If $V$, $W\in\frak B$ and $e\in U^+$, we have $e=e_1+e_2+e_3+e_4$
where

\Centerline{$e_1=P_WP_Ve\in V\cap W$,
\quad$e_2=P_{W^{\perp}}P_Ve\in V\cap W^{\perp}$,}

\Centerline{$e_3=P_WP_{V^{\perp}}e\in V^{\perp}\cap W$,
\quad$e_4=P_{W^{\perp}}P_{V^{\perp}}e\in V^{\perp}\cap W^{\perp}$,}

\Centerline{$e_1+e_2=P_Ve$,
\quad$e_1+e_3=P_We$.}

\noindent Now $e_2+e_3+e_4$ belongs to $(V\cap W)^{\perp}$, so $e_1$
must be the component of $e$ in $V\cap W$;  similarly $e_4$ is the
component of $e$ in $V^{\perp}\cap W^{\perp}$, and $e_1+e_2+e_3$ is the
component of $e$ in $V\vee W$.    But as $e_2\wedge e_3=0$, we have

\Centerline{$P_{V\cap W}e=e_1=(e_1+e_2)\wedge(e_1+e_3)
=P_Ve\wedge P_We$,}

\Centerline{$P_{V\vee W}e=e_1+e_2+e_3=(e_1+e_2)\vee(e_1+e_3)=P_Ve\vee
P_We$,}

\noindent as required.

It follows that

\Centerline{$P_VP_W=P_{V\cap W}=P_{W\cap V}=P_WP_V$.}


\medskip

{\bf (c)} If $V$, $W\in\frak B$ and $W\subseteq V$, then of course $W$
is a band in the Riesz space $V$ (because $V$ is order-closed in $U$, so
that for any set $A\subseteq W$ its supremum in $U$ will be its supremum
in $V$ if either is defined).
For any $v\in V$, we have an expression of it as $w+w'$,
where $w\in W$ and $w'\in W^{\perp}$, taken in $U$;  but as
$|w|+|w'|=|w+w'|=|v|\in V$, $w'$ belongs to $V$, and is in
$W^{\perp}_V$, the band in $V$ orthogonal to $W$.   Thus
$W+W^{\perp}_V=V$ and $W$ is a projection band in $V$.   Conversely, if
$W$ is a projection band in $V$, then $W^{\perp}$ (taken in $U$)
includes $W^{\perp}_V+V^{\perp}$, so that

\Centerline{$W+W^{\perp}\supseteq
W+W^{\perp}_V+V^{\perp}=V+V^{\perp}=U$}

\noindent and $W\in\frak
B$.

Thus the algebra of projection bands in $V$ is, as a set, equal to the
principal ideal $\frak B_V$;  because their orderings agree, or
otherwise, their Boolean algebra structures coincide.
}%end of proof of 352S

\vleader{60pt}{352T}{Products again (a)} If $U=\prod_{i\in I}U_i$ is a product
of Riesz spaces, then for any $J\subseteq I$ we have a subspace

\Centerline{$V_J=\{u:u\in U,\,u(i)=0$ for all $i\in I\setminus J\}$}

\noindent of $U$, canonically isomorphic to $\prod_{i\in J}U_i$.   Each
$V_J$ is a projection band, its complement being $V_{I\setminus J}$;
the map $J\mapsto V_J$ is a Boolean homomorphism from $\Cal PI$ to the
algebra $\frak B$ of projection bands in $U$, and
$\langle V_{\{i\}}\rangle_{i\in I}$ is a partition of unity in $\frak B$.

\spheader 352Tb Conversely, if $U$ is a Riesz space and
$(V_0,\ldots,V_n)$ is a {\it finite} partition of unity in the algebra
$\frak B$ of projection bands in $U$, then every element of $U$ is
uniquely expressible as $\sum_{i=0}^nu_i$ where $u_i\in V_i$ for each
$i$.   \prooflet{(Induce on $n$, using 352Rb for the case $n=2$,
and 352Sc in the inductive step.)}   This decomposition corresponds to a
Riesz space isomorphism between $U$ and $\prod_{i\le n}V_i$.

\leader{352U}{Quotient spaces (a)} If $U$ is a Riesz space and $V$ is a
solid linear subspace, then the canonical map from $U$ onto the
quotient \pols\ $U/V$ is a Riesz homomorphism\cmmnt{ (351J)}, so
$U/V$ is a Riesz space\cmmnt{ (352Ib)}.   I have already noted that
if $U$ and $W$ are Riesz spaces and $T:U\to W$ a
Riesz homomorphism, then the kernel $V$ of $T$ is a solid linear
subspace of
$U$ and the Riesz subspace $T[U]$ of $W$ is isomorphic to $U/V$\cmmnt{
(352Jb)}.

\spheader 352Ub Suppose that $U$ is a Riesz space and $V$ a solid linear
subspace.   Then the canonical map from $U$ to $U/V$ is
order-continuous iff $V$ is a band.   \prooflet{\Prf\ (i) If $u\mapsto
u^{\ssbullet}$ is order-continuous, its kernel $V$ is a band, by 352Oe.
(ii) If $V$ is a band, and $A\subseteq U$ is non-empty and
downwards-directed and has infimum $0$, let $p\in U/V$ be any lower
bound for $\{u^{\ssbullet}:u\in A\}$.   Express $p$ as $w^{\ssbullet}$.
Then $((w-u)^+)^{\ssbullet}=(w^{\ssbullet}-u^{\ssbullet})^+=0$, that is,
$(w-u)^+\in V$ for every $u\in A$.   But this means that

\Centerline{$w^+=\sup_{u\in A}(w-u)^+\in V$,
\quad$p^+=(w^+)^{\ssbullet}=0$,}

\noindent that is, $p\le 0$.   As $p$ is arbitrary, $\inf_{u\in
A}u^{\ssbullet}=0$;  as $A$ is arbitrary, $u\mapsto u^{\ssbullet}$ is
order-continuous.\ \Qed
}%end of prooflet

\leader{352V}{Principal bands} Let $U$ be a Riesz space.   Evidently the
intersection of any family of Riesz subspaces of $U$ is a Riesz
subspace, the intersection of any family of solid linear subspaces is a
solid linear subspace and the intersection of any family of bands is a
band;  we may\cmmnt{ therefore} speak of the band generated by a subset $A$ of
$U$, the intersection of all the bands including $A$.  \cmmnt{Now we
have the following description of the band generated by a single
element.}

\medskip

\noindent{\bf Lemma} Let $U$ be a Riesz space.

(a) If $A\subseteq U^+$ is upwards-directed and $2w\in A$ for every
$w\in A$, then an element $u$ of $U$ belongs to the band generated by
$A$ iff $|u|=\sup_{w\in A}|u|\wedge w$.

(b) If $u\in U$ and $w\in U^+$, then $u$ belongs to the band in $U$
generated by $w$ iff $|u|=\sup_{n\in\Bbb N}|u|\wedge nw$.

\proof{{\bf (a)} Let $W$ be the band generated by $A$ and $W'$ the set
of elements of $U$ satisfying the condition.

\medskip

\quad{\bf (i)} If $u\in W'$ then $|u|\wedge w\in W$ for every $w\in A$,
because $W$ is a solid linear subspace;  because $W$ is also
order-closed, $|u|$ and $u$ belong to $W$.   Thus $W'\subseteq W$.

\medskip

\quad{\bf (ii)} Now $W'$ is a band.

\medskip

\Prf\grheada\ If $u\in W'$ and $|v|\le|u|$ then

\Centerline{$\sup_{w\in A}|v|\wedge w
=\sup_{w\in A}|v|\wedge|u|\wedge w
=|v|\wedge\sup_{w\in A}|u|\wedge w
=|v|\wedge|u|=|v|$}

\noindent by 352Ea, so $v\in W'$.

\medskip

\qquad\grheadb\ If $u$, $v\in W'$ then, for any $w_1$, $w_2\in A$ there
is a $w\in A$ such that $w\ge w_1\vee w_2$.   Now $w_1+w_2\le 2w\in
A$, and

\Centerline{$(|u|+|v|)\wedge 2w
\ge(|u|\wedge w_1)+(|v|\wedge w_2)$.}

\noindent So any upper bound for $\{(|u|+|v|)\wedge w:w\in A\}$ must
also be an upper bound for
$\{|u|\wedge w:w\in A\}+\{|v|\wedge w:w\in A\}$ and therefore greater
than or equal to

$$\eqalign{\sup(\{|u|\wedge w:w\in A\}+\{|v|\wedge w:w\in A\})
&=\sup_{w\in A}|u|\wedge w+\sup_{w\in A}|v|\wedge w\cr
&=|u|+|v|\cr}$$

\noindent (351Dc).   But this means that
$\sup_{w\in A}(|u|+|v|)\wedge w$ must be $|u|+|v|$, and $|u|+|v|$
belongs to $W'$;  it follows from ($\alpha$) that $u+v$ belongs to $W'$.

\medskip

\qquad\grheadc\ Just as in 352Oc, we now have

\Centerline{$nu\in W'$ for every $n\in\Bbb N$, $u\in W'$,}

\noindent and therefore $\alpha u\in W'$ for every $\alpha\in\Bbb R$,
$u\in W'$, since $|\alpha u|\le|nu|$ if $|\alpha|\le n$.   Thus $W'$ is
a solid linear subspace of $U$.

\medskip

\qquad\grheadd\  Now suppose that $C\subseteq (W')^+$ has a supremum $v$
in $U$.   Then any upper bound of $\{v\wedge w:w\in A\}$ must also be an
upper bound of $\{u\wedge w:u\in C,\,w\in A\}$ and greater than or equal
to $u=\sup_{w\in A}u\wedge w$ for every $u\in C$, therefore greater than
or equal to $v=\sup C$.   Thus $v=\sup_{w\in A}v\wedge w$ and $v\in W'$.
As $C$ is arbitrary, $W'$ is a band (352Ob).\ \Qed

\medskip

\quad{\bf (iii)} Since $A$ is obviously included in $W'$, $W'$ must
include $W$;  putting this together with (i), $W=W'$, as claimed.

\medskip

{\bf (b)} Apply (a) with $A=\{nw:n\in\Bbb N\}$.
}%end of proof of 352V

\leader{352W}{$f$-algebras}\cmmnt{ Some of the most important Riesz
spaces have multiplicative structures as well as their order and linear
structures.   A particular class of these structures appears
sufficiently often for it to be useful to develop a little of its
theory.   The following definition is a common approach.

\medskip

} {\bf (a) Definition} An {\bf $f$-algebra} is a Riesz space
$U$ with a multiplication $\times:U\times U\to U$ such that

\Centerline{$u\times(v\times w)=(u\times v)\times w$,}

\Centerline{$(u+v)\times w=(u\times w)+(v\times w)$,
\quad$u\times(v+w)=(u\times v)+(u\times w)$,}

\Centerline{$\alpha(u\times v)=(\alpha u)\times v=u\times(\alpha v)$}

\noindent for all $u$, $v$, $w\in U$ and $\alpha\in\Bbb R$, and

\Centerline{$u\times v\ge 0$ whenever $u$, $v\ge 0$,}

\Centerline{if $u\wedge v=0$ then
$(u\times w)\wedge v=(w\times u)\wedge v=0$ for every $w\ge 0$.}

\noindent An $f$-algebra is {\bf commutative} if $u\times v=v\times u$
for all $u$, $v$.

\spheader 352Wb Let $U$ be an $f$-algebra.

\medskip

\quad{\bf (i)} If $u\wedge v=0$ in $U$, then
$u\times v=0$.   \prooflet{\Prf\ $v\ge 0$ so
$v\wedge(u\times v)=(u\times v)\wedge v=0$ and
$u\times v=(u\times v)\wedge(u\times v)=0$.\ \Qed}

\medskip

\quad{\bf (ii)} $u\times u\ge 0$ for every $u\in U$.   \prooflet{\Prf\

$$\eqalign{(u^+-u^-)\times(u^+-u^-)
&=u^+\times u^+-u^+\times u^--u^-\times u^++u^-\times u^-\cr
&=u^+\times u^++u^-\times u^-
\ge 0.\text{ \Qed}\cr}$$
}%end of prooflet

\medskip

\quad{\bf (iii)} If $u$, $v\in U$ then $|u\times v|=|u|\times|v|$.
\prooflet{\Prf\ $u^+\times v^+$, $u^+\times v^-$, $u^-\times v^+$ and
$u^+\times v^-$ are disjoint, so

$$\eqalign{|u\times v|
&=|u^+\times v^+-u^+\times v^--u^-\times v^++u^-\times v^-|\cr
&=u^+\times v^++u^+\times v^-+u^-\times v^++u^-\times v^-\cr
&=|u|\times|v|\cr}$$

\noindent by 352Fb.\ \Qed}

\medskip

\quad{\bf (iv)} If $v\in U^+$ the
maps $u\mapsto u\times v$, $u\mapsto v\times u:U\to U$ are Riesz
homomorphisms.
\prooflet{\Prf\ The first four clauses of the definition in (a) ensure
that they are linear operators.   If $u\in U$, then

\Centerline{$|u|\times v=|u\times v|$,
\quad$v\times|u|=|v\times u|$}

\noindent by (iii), so we have Riesz homomorphisms, by 352G(iii).\ \Qed}

\medskip

{\bf (c)} Let $\familyiI{U_i}$ be a family of $f$-algebras, with Riesz
space product $U$ (352K).   If we set
$u\times v=\familyiI{u(i)\times v(i)}$ for all $u$, $v\in U$, then $U$
becomes an $f$-algebra.

\exercises{\leader{352X}{Basic exercises $\pmb{>}$(a)}
%\spheader 352Xa
Let $U$ be any Riesz space.   Show that
$|u^+-v^+|\le|u-v|$ for all $u$, $v\in U$.
%352F

\sqheader 352Xb
Let $U$, $V$ be Riesz spaces and $T:U\to V$ a linear operator.   Show that
the following are equiveridical:  (i) $T$ is a Riesz homomorphism;
(ii) $T(u\vee v)=Tu\vee Tv$ for all $u$, $v\in U$;
(iii) $T(u\wedge v)=Tu\wedge Tv$ for all $u$, $v\in U$;
(iv) $|Tu|=T|u|$ for every $u\in U$.
%352G

\spheader 352Xc Let $U$ be a Riesz space and $V$ a solid linear
subspace;  for $u\in U$ write $u^{\ssbullet}$ for the corresponding
element of $U/V$.   Show that if $A\subseteq U$ is solid then
$\{u^{\ssbullet}:u\in A\}$ is solid in $U/W$.
%352J

\spheader 352Xd Let $U$ be a Riesz space.   Show that
$\med(\alpha u,\alpha v,\alpha w)=\alpha\med(u,v,w)$ for all $u$, $v$,
$w\in U$ and all $\alpha\in\Bbb R$.   \Hint{3A1Ic, 352M.}
%352L

\spheader 352Xe Let $U$ and $V$ be Riesz spaces and $T:U\to V$ a Riesz
homomorphism with kernel $W$.   Show that if $W$ is a band in $U$ and
$T[U]$ is regularly embedded in $V$ then $T$ is order-continuous.
%352O

\spheader 352Xf Give $U=\BbbR^2$ its lexicographic ordering (351Xa).
Show that it has a band $V$ which is not complemented.
%352P  [Riesz space because totally ordered]

\spheader 352Xg Let $U$ be a Riesz space and $\frak C$ the algebra of
complemented bands in $U$.   Show that for any $V\in\frak C$ the algebra
of complemented bands in $V$ is just the principal ideal of $\frak C$
generated by $V$.
%352Q

\sqheader 352Xh Let $U=C([0,1])$ be the space of continuous functions
from $[0,1]$ to $\Bbb R$, with its usual linear and order structures, so
that it is a Riesz subspace of $\BbbR^{[0,1]}$.   Set $V=\{u:u\in
U,\,u(t)=0$ if $t\le\bover12\}$.   Show that $V$ is a band in $U$ and
that $V^{\perp}=\{u:u(t)=0$ if $t\ge\bover12\}$, so that $V$ is
complemented but is not a projection band.
%352R

\spheader 352Xi Show that the Boolean homomorphism $J\mapsto V_J:\Cal
PI\to\frak B$ of 352Ta is order-continuous.
%352T

\spheader 352Xj Let $U$ be a Riesz space and $A\subseteq U^+$ an
upwards-directed set.   Show that the band generated by $A$ is
$\{u:|u|=\sup_{n\in\Bbb N,w\in A}|u|\wedge nw\}$.
%352V

\sqheader 352Xk(i) Let $X$ be any set.   Setting
$(u\times v)(x)=u(x)v(x)$ for $u$, $v\in\Bbb R^X$, $x\in X$, show that
$\Bbb R^X$
is a commutative $f$-algebra.   (ii) With the same definition of
$\times$, show
that $\ell^{\infty}(X)$ is an $f$-algebra.   (iii) If $X$ is a
topological space, show that $C(X)$, $C_b(X)$ (definition:  281A, 354Hb)
are $f$-algebras.   (iv)
If $(X,\Sigma,\mu)$ is a measure space, show that $L^0(\mu)$ and
$L^{\infty}(\mu)$ (\S241, \S243) are $f$-algebras.
%352W

\spheader 352Xl  Let $U\subseteq\BbbR^{\Bbb Z}$ be the set of sequences
$u$ such that $\{n:u(n)\ne 0\}$ is bounded above in $\Bbb Z$.   For $u$,
$v\in U$ (i) say that $u\le v$ if either $u=v$ or there is an $n\in\Bbb
Z$ such that $u(n)<v(n)$, $u(i)=v(i)$ for every $i>n$ (ii) say that
$(u*v)(n)=\sum_{i=-\infty}^{\infty}u(i)v(n-i)$ for every $n\in\Bbb Z$.
Show that $U$ is a commutative
$f$-algebra under this ordering and multiplication, and that $(U,+,*)$ is a
field.
%352W

\spheader 352Xm Let $U$ be an $f$-algebra.  (i) Show that any
complemented band in $U$
is an ideal in the ring $(U,+,\times)$.   (ii) Show that if $P:U\to U$
is a band projection, then $P(u\times v)=Pu\times Pv$ for every $u$,
$v\in U$.
%352W

\spheader 352Xn Let $U$ be an $f$-algebra with multiplicative identity
$e$.   Show that $u-\gamma e\le\Bover1{2\gamma}u^2$ for every $u\in U$,
$\gamma>0$.   \Hint{$(u^+-\gamma e)^2\ge 0$.}
%352W

\spheader 352Xo\dvAnew{2022} Let $U$ be a Riesz space.   
(i) Show that if $u$, $v$, $u'$,
$v'\in U$ then $|u\vee v-u'\vee v'|\le|u-u'|\vee|v-v'|$.   (ii) Show that if
$u$, $v$, $w$, $u'$, $v'$, $w'\in U$ then
$|\med(u,v,w)-\med(u',v',w')|\le|u-u'|\vee|v-v'|\vee|w-w'|$.

\leader{352Y}{Further exercises (a)}\dvAnew{2016}
%\spheader 352Ya
Find an $f$-algebra with a non-commutative multiplication.
%352W U=\BbbR^2, lexicographic, u\times v=u(0)v.   See 353Pb

\spheader 352Yb\dvAnew{2016} Let $U$ be an $f$-algebra.
Show that the multiplication of $U$
is commutative iff $u\times v=(u\wedge v)\times(u\vee v)$ for all $u$, $v\in U$.
%352W 352Ya 35bits

\spheader 352Yc\dvAnew{2016} Let $U$ be an $f$-algebra.   Show that
$u\times\med(v_1,v_2,v_3)=\med(u\times v_1,u\times v_2,u\times v_3)$
whenever $u$, $v_1$, $v_2$, $v_3\in U$.
%352W easy for u\ge 0;  now if |v_1|+|v_2|  and  |w_1|+|w_2|  are disjoint
% then  $(v_1\vee v_2)+(w_1\vee w_2)=(v_1+v_2)\vee(w_1\vee w_2)
}%end of exercises

\cmmnt{
\Notesheader{352} In this section we begin to see a striking
characteristic of the theory of Riesz spaces:  repeated reflections of
results in Boolean algebra.   Without spelling out a complete list, I
mention the distributive laws (313Bc, 352Ea) and the behaviour of
order-continuous homomorphisms (313Pa, 313Qa, 352N, 352Oe, 352Ub,
352Xe).
Riesz subspaces correspond to subalgebras, solid linear subspaces to
ideals and Riesz homomorphisms to Boolean homomorphisms.
We even have a correspondence, though a weaker one, between the
representation theorems available;  every Boolean algebra is isomorphic
to a subalgebra of a power of $\Bbb Z_2$ (311D-311E), while every
Riesz space is isomorphic to a Riesz subspace of a quotient of a power
of $\Bbb R$ (352L).   It would be a closer parallel if every Riesz space
were embeddable in some $\Bbb R^X$;  I must emphasize that the
differences are as important as the agreements.   Subspaces of
$\BbbR^X$ are of great importance, but are by no means adequate for our
needs.   And of course the details -- for instance, the identities in
352D-352F, or 352V -- frequently involve new techniques in the case of
Riesz spaces.   Elsewhere, as in 352G, I find myself arguing rather from
the opposite side, when applying results from the theory of general
partially ordered linear spaces, which has little to do with Boolean
algebra.

In the theory of bands in Riesz spaces -- corresponding to order-closed
ideals in Boolean algebras -- we have a new complication in the form of
bands which are not complemented, which does not arise in the Boolean
algebra context;  but it will disappear again when we come to specialize to
Archimedean Riesz spaces (353B).   (Similarly, order-density and
quasi-order-density coincide in both Boolean algebras (313K) and
Archimedean Riesz spaces (353A).)   Otherwise the algebra of
complemented bands in a Riesz space looks very like the algebra of
order-closed ideals in a Boolean algebra (314Yh, 352Q).   The algebra
of projection bands in a Riesz space (352S) would correspond, in a
Boolean algebra, to the algebra itself.

I draw your attention to 352H.   The result is nearly trivial, but it
amounts to saying that the theory of Riesz spaces will be `algebraic',
like the theories of groups or linear spaces, rather than `analytic',
like the theories of \pols s or topological spaces, in which we can have
bijective morphisms which are not isomorphisms.

}%end of comment

\discrpage

\frfilename{mt353.tex}
\versiondate{16.2.17}
\copyrightdate{1998}

\def\chaptername{Riesz spaces}
\def\sectionname{Archimedean and Dedekind complete Riesz spaces}

\newsection{353}

I take a few pages over elementary properties of Archimedean
and Dedekind ($\sigma$)-complete Riesz spaces.

\leader{353A}{Proposition} Let $U$ be an Archimedean Riesz space.   Then
every quasi-order-dense Riesz subspace of $U$ is order-dense.

\proof{ Let $V\subseteq U$ be a quasi-order-dense Riesz subspace, and
$u\ge 0$ in $U$.   Set $A=\{v:v\in V,\,v\le u\}$.   \Quer\ Suppose, if
possible, that $u$ is not the least upper bound of $A$.   Then there is
a $u_1<u$ such that $v\le u_1$ for every $v\in A$.   Because $0\in A$,
$u_1\ge 0$.   Because $V$ is quasi-order-dense, there is a $v\in V$
such that $0<v\le u-u_1$.   Now $nv\le u_1$ for every $n\in\Bbb N$.
\Prf\ Induce on $n$.   For $n=0$ this is trivial.   For the inductive
step, given $nv\le u_1$, then $(n+1)v\le u_1+v\le u$, so $(n+1)v\in A$
and $(n+1)v\le u_1$.   Thus the induction proceeds.\ \QeD\ But this is
impossible, because $v>0$ and $U$ is supposed to be Archimedean.\ \Bang

So $u=\sup A$.   As $u$ is arbitrary, $V$ is order-dense.
}%end of proof of 353A

\vleader{108pt}{353B}{Proposition} Let $U$ be an Archimedean Riesz space.   Then

(a) for every $A\subseteq U$, the band generated by $A$ is
$A^{\perp\perp}$,

(b) every band in $U$ is complemented.

\proof{{\bf (a)} Let $V$ be the band generated by $A$.   Then $V$ is
surely included in $A^{\perp\perp}$, because this is a band including
$A$ (352O).   \Quer\ Suppose, if possible,
that $V\ne A^{\perp\perp}$.   Then there is a
$w\in A^{\perp\perp}\setminus V$, so that $|w|\notin V$.   Set
$B=\{v:v\in V,\,v\le |w|\}$;  then $B$ is upwards-directed and
non-empty.  Because $V$ is order-closed, $|w|$ cannot be the supremum of
$A$, and there is a $u_0>0$ such that $|w|-u_0\ge v$ for every $v\in B$.
Now $u_0\wedge|w|\ne 0$, so $u_0\notin A^{\perp}$, and there is a
$u_1\in A$
such that $v=u_0\wedge|u_1|>0$.   In this case $nv\in B$ for every
$n\in\Bbb N$.   \Prf\ Induce on $n$.   For $n=0$ this is trivial.   For
the inductive step, given that $nv\in B$, then $nv\le |w|-u_0$ so $(n+1)v\le nv+u_0\le|w|$;  but also $(n+1)v\le nv+|u_1|\in V$, so $(n+1)v\in B$.\
\QeD\  But this means that $nv\le|w|$ for every $n$, which is
impossible, because $U$ is Archimedean.\ \BanG

\medskip

{\bf (b)} Now if $V\subseteq U$ is any band, it is surely the band
generated by itself, so is equal to $V^{\perp\perp}$, and is
complemented (352P).
}%end of proof of 353B

\cmmnt{\medskip

\noindent{\bf Remark} We may therefore speak of the {\bf band algebra}
of an Archimedean Riesz space, rather than the `complemented band
algebra' (352Q).}

\leader{353C}{Corollary} Let $U$ be an Archimedean Riesz space and $v\in
U$.   Let $V$ be the band in $U$ generated by $v$.   If $u\in U$, then
$u\in V$ iff there is no $w$ such that $0<w\le|u|$ and $w\wedge|v|=0$.

\proof{ By 353B, $V=\{v\}^{\perp\perp}$.
Now, for $u\in U$,

\Centerline{$u\notin V
\iff \,\exists\,w\in\{v\}^{\perp},\,|u|\wedge|w|>0
\iff \,\exists\,w\in\{v\}^{\perp},\,0<w\le|u|$.}

\noindent Turning this round, we have the condition announced.
}%end of proof of 353C

\leader{353D}{Proposition} Let $U$ be an Archimedean Riesz space and
$V$ an order-dense Riesz subspace of $U$.   Then the map
$W\mapsto W\cap V$ is an isomorphism between the band algebras of $U$ and $V$.

\proof{ If $W\subseteq U$ is a band, then $W\cap V$ is surely a band in
$V$ (it is order-closed in $V$ because it is the inverse image of the
order-closed set $W$ under the embedding $V\embedsinto U$, which is
order-continuous by 352Nc and 352Nb).   If $W$, $W'$ are distinct bands
in $U$, say
$W'\not\subseteq W$, then $W'\not\subseteq W^{\perp\perp}$, by 353B, so
$W'\cap W^{\perp}\ne\{0\}$;  because $V$ is order-dense, $V\cap W'\cap
W^{\perp}\ne\{0\}$, and $V\cap W'\ne V\cap W$.   Thus $W\mapsto W\cap V$
is injective.

If $Q\subseteq V$ is a band in $V$, then its complementary band in $V$
is just $Q^{\perp}\cap V$, where $Q^{\perp}$ is taken in $U$.   So
(because $V$, like $U$, is Archimedean, by 351Rc)
$Q=(Q^{\perp}\cap V)^{\perp}\cap V=W\cap V$, where
$W=(Q^{\perp}\cap V)^{\perp}$ is a band in $U$.   Thus
the map $W\mapsto W\cap V$ is an order-preserving bijection between the
two band algebras.   By 312M, it is a Boolean isomorphism, as claimed.
}%end of proof of 353D

\leader{353E}{Lemma} Let $U$ be an Archimedean Riesz space and
$V\subseteq U$ a band such that $\sup\{v:v\in V,\,0\le v\le u\}$ is
defined for every $u\in U^+$.   Then $V$ is a projection band.

\proof{ Take any $u\in U^+$ and set $v=\sup\{v':v'\in V^+,\,v'\le u\}$,
$w=u-v$.   $v\in V$ because $V$ is a band.   Also $w\in V^{\perp}$.
\Prf\Quer\ If not, there is some $v_0\in V^+$ such that $w\wedge v_0>0$.
Now for any $n\in \Bbb N$ we see that

\Centerline{$n(w\wedge v_0)\le u\Longrightarrow n(w\wedge v_0)\le v
\Longrightarrow (n+1)(w\wedge v_0)\le v+w=u$,}

\noindent so an induction on $n$ shows that $n(w\wedge v_0)\le u$ for every $n$;
which is impossible, because $U$ is supposed to be
Archimedean.\ \Bang\QeD\  Accordingly $u=v+w\in V+V^{\perp}$.   As $u$
is arbitrary,
$U^+\subseteq V+V^{\perp}$, and $V$ is a projection band (352R).
}%end of proof of 353E

\leader{353F}{Lemma} Let $U$ be an Archimedean Riesz space.   If
$A\subseteq U$ is non-empty and bounded above and $B$ is the set of its upper bounds, then $\inf(B-A)=0$.
\proof{ \Quer\ If not, let $w>0$ be a lower bound for $B-A$.   If
$u\in A$ and $v\in B$, then $v-u\ge w$, that is, $u\le v-w$;  as $u$ is
arbitrary,
$v-w\in B$.   Take any $u_0\in A$ and $v_0\in B$.   Inducing on $n$, we see
that $v_0-nw\in B$ for every $n\in\Bbb N$, so that $v_0-nw\ge u_0$,
$nw\le v_0-u_0$ for every $n$;  but this is impossible, because $U$ is supposed to be Archimedean.\ \Bang
}%end of proof of 353F

\leader{353G}{Proposition}\dvAformerly{3{}53Q}
Let $U$ be a Riesz space and $V$ an
order-dense Riesz subspace of $U$.   If $V$ is Archimedean, so is $U$.

\proof{ \Quer\ Otherwise, let $u'$, $u\in U$ be such that $u'>0$ and
$nu'\le u$ for every $n\in\Bbb N$.   Let $v'\in V$ be such that
$0<v'\le u'$;  set $\tilde u=u-v'$;  let $v\in V$ be such that $v\le u$
but $v\not\le\tilde u$.   (This
is where we need $V$ to be order-dense rather than just quasi-order-dense.)
Let $w\in V$ be such that $w>0$ and $w\le(v-\tilde u)^+$;  note that
$w\le u-\tilde u=v'$.

Because $V$ is Archimedean, there is an $n\ge 1$ such that $nw\not\le v$.
In this case,

\Centerline{$0<(nw-v)^+\le((n+1)v'-(v+v'))^+\le (u-(v+v'))^+
=(\tilde u-v)^+$}

\noindent but

\Centerline{$(nw-v)^+\wedge(\tilde u-v)^+\le nw\wedge n(\tilde u-v)^+
\le n((v-\tilde u)^+\wedge(\tilde u-v)^+)=0$,}

\noindent which is impossible.\ \Bang
}%end of proof of 353G

\leader{353H}{Dedekind completeness}\dvAformerly{3{}53G}\cmmnt{ Recall that a
partially ordered set $P$ is Dedekind ($\sigma$)-complete if (countable)
non-empty sets with upper and lower bounds have suprema and infima in
$P$ (314A).}  For a Riesz space $U$, $U$ is Dedekind complete iff every
non-empty upwards-directed subset of $U^+$ with an upper bound has a
least upper bound, and is Dedekind $\sigma$-complete iff every
non-decreasing sequence in $U^+$ with an upper bound has a least upper
bound.   \prooflet{\Prf\ (Compare 314Bc.)  (i) Suppose that any
non-empty
upwards-directed order-bounded subset of $U^+$ has an upper bound, and
that $A\subseteq U$ is any
non-empty set with an upper bound.   Take $u_0\in A$ and set

\Centerline{$B=\{u_0\vee u_1\vee\ldots\vee u_n-u_0:u_1,\ldots,u_n\in
A\}$.}

\noindent Then $B$ is an upwards-directed subset of $U^+$, and if $w$ is
an upper bound of $A$ then $w-u_0$ is an upper bound of $B$.   So $\sup
B$ is defined in $U$, and in this case $u_0+\sup B=\sup A$.   As $A$ is
arbitrary, $U$ is Dedekind complete.   (ii) Suppose that
order-bounded non-decreasing sequences in $U^+$ have suprema, and that
$A\subseteq U$ is any countable non-empty set with an upper bound.   Let
$\sequencen{u_n}$ be a sequence running over $A$, and set
$v_n=\sup_{i\le n}u_i-u_0$ for each $n$.   Then $\sequencen{v_n}$ is a
non-decreasing order-bounded sequence in $U^+$, and $u_0+\sup_{n\in\Bbb
N}v_n=\sup A$.   (iii) Finally, still supposing that order-bounded
non-decreasing sequences in $U^+$ have suprema, if $A\subseteq U$ is
non-empty, countable and bounded below, $\inf A$ will be defined and
equal to $-\sup(-A)$.\ \Qed}%end of prooflet

\leader{353I}{Proposition} Let $U$ be a 
Dedekind $\sigma$-complete Riesz space.

(a) $U$ is Archimedean.

(b) For any $v\in U$ the band generated by $v$ is a projection band.

(c) If $u$, $v\in U$, then $u$ is uniquely expressible as $u_1+u_2$,
where $u_1$ belongs to the band generated by $v$ and $|u_2|\wedge|v|=0$.

\proof{{\bf (a)} Suppose that $u$, $v\in U$ are such that $nu\le v$ for
every $n\in\Bbb
N$.   Then $nu^+\le v^+$ for every $n$, and $A=\{nu^+:n\in\Bbb N\}$ is a
countable non-empty upwards-directed set with an upper bound;  say
$w=\sup A$.   Since $A+u^+\subseteq A$, $w+u^+=\sup(A+u^+)\le w$, and
$u\le u^+\le 0$.   As $u$, $v$ are arbitrary, $U$ is Archimedean.

\medskip

{\bf (b)} Let $V$ be the band generated by $v$.   Take any $u\in U^+$
and set $A=\{v':v'\in V,\,0\le v'\le u\}$.   Then
$\{u\wedge n|v|:n\in\Bbb N\}$ is a countable set with an upper bound,
so has a supremum $u_1$ say
in $U$.   Now $u_1$ is an upper bound for $A$.   \Prf\ If $v'\in A$,
then

\Centerline{$v'=\sup_{n\in\Bbb N}v'\wedge n|v|\le u_1$}

\noindent by 352Vb.\ \QeD\   Since $u\wedge n|v|\in A$ for
every $n$, $u_1=\sup A$.

As $u$ is arbitrary, 353E tells us that $V$ is a projection band.

\medskip

{\bf (c)} Again let $V$ be the band generated by $v$.   Then
$\{v\}^{\perp\perp}$ is a band containing $v$, so

\Centerline{$\{v\}\subseteq V\subseteq\{v\}^{\perp\perp}$,
\quad$\{v\}^{\perp}\supseteq V^{\perp}
  \supseteq\{v\}^{\perp\perp\perp}=\{v\}^{\perp}$}

\noindent (352Od), and $V^{\perp}=\{v\}^{\perp}$.

Now, if $u\in U$, $u$ is uniquely expressible in the form $u_1+u_2$
where $u_1\in V$ and $u_2\in V^{\perp}$, by (b).   But

\Centerline{$u_2\in V^{\perp}\iff u_2\in\{v\}^{\perp}\iff
|u_2|\wedge|v|=0$.}

\noindent So we have the result.
}%end of proof of 353I

\leader{353J}{Proposition}\dvAformerly{3{}53I} In a Dedekind complete 
Riesz space, all bands are projection bands.

\proof{ Use 353E, noting that the sets $\{v:v\in V,\,0\le v\le u\}$
there are always non-empty, upwards-directed and bounded above, so
always have suprema.
}%end of proof of 353J

\vleader{72pt}{353K}{Proposition} (a) Let $U$ be a Dedekind
$\sigma$-complete Riesz space.

\quad(i) If $V$ is a solid linear subspace of $U$,   then $V$
is (in itself) Dedekind $\sigma$-complete.

\quad(ii) If $V$ is a
sequentially order-closed Riesz subspace of $U$ then $V$ is Dedekind
$\sigma$-complete.

\quad(iii) If $V$ is a sequentially
order-closed solid linear subspace of $U$, the canonical map from $U$ to
the quotient space $U/V$ is sequentially order-continuous, and $U/V$
also is Dedekind $\sigma$-complete.

(b) Let $U$ be a Dedekind complete Riesz space.

\quad(i) If $V$ is a solid
linear subspace of $U$, then $V$ is Dedekind complete.

\quad(ii) If $V\subseteq U$ is an order-closed Riesz subspace then $V$
is Dedekind complete.

\proof{{\bf (a)(i)} If $\sequencen{u_n}$ is a non-decreasing sequence in
$V^+$ with an upper bound $v\in V$, then $w=\sup_{n\in\Bbb N}u_n$ is
defined in $U$;  but as $0\le w\le v$, $w\in V$ and
$w=\sup_{n\in\Bbb N}u_n$ in $V$.   Thus $V$ is Dedekind $\sigma$-complete.

\medskip

{\bf (ii)} If $\sequencen{u_n}$ is a non-decreasing order-bounded
sequence in $W$, then $u=\sup_{n\in\Bbb N}u_n$ is defined in $U$;  but
because $V$ is sequentially order-closed, $u\in V$ and
$u=\sup_{n\in\Bbb N}u_n$ in $V$.

\medskip

{\bf (iii)} Let $\sequencen{u_n}$ be a non-decreasing sequence in $U$
with supremum $u$.   Then of course $u^{\ssbullet}$ is an upper bound
for $A=\{u_n^{\ssbullet}:n\in\Bbb N\}$ in $U/V$.   Now let $p$ be any
other upper bound for $A$.   Express $p$ as $v^{\ssbullet}$.   Then for
each $n\in\Bbb N$ we have $u_n^{\ssbullet}\le p$, so that
$(u_n-v)^+\in V$.   Because $V$ is sequentially order-closed,
$(u-v)^+=\sup_{n\in\Bbb N}(u_n-v)^+\in V$ and $u^{\ssbullet}\le p$.
Thus $u^{\ssbullet}$ is the least upper bound of $A$.   By 351Gb,
$u\mapsto u^{\ssbullet}$ is sequentially order-continuous.

Now suppose that $\sequencen{p_n}$ is a non-decreasing sequence in
$(U/V)^+$ with an upper bound $p\in (U/V)^+$.   Let $u\in U^+$ be such
that $u^{\ssbullet}=p$, and for each $n\in\Bbb N$ let $u_n\in U^+$ be
such that $u_n^{\ssbullet}=p_n$.   Set $v_n=u\wedge\sup_{i\le n}u_i$ for
each $n$;  then $v_n^{\ssbullet}=p_n$ for each $n$, and
$\sequencen{v_n}$ is a non-decreasing order-bounded sequence in $U$.
Set $v=\sup_{n\in\Bbb N}v_n$;  by the last paragraph,
$v^{\ssbullet}=\sup_{n\in\Bbb N}p_n$ in $U/V$.   As $\sequencen{p_n}$ is
arbitrary, $U/V$ is Dedekind $\sigma$-complete, as claimed.

\medskip

{\bf (b)} The argument is the same as parts (i) and (ii) of the proof of
(a).
}%end of proof of 353K

\leader{353L}{Proposition}\dvAformerly{3{}53K} Let $U$ be a Riesz space and 
$V$ a quasi-order-dense Riesz subspace of $U$ which is (in itself) Dedekind
complete.   Then $V$ is a solid linear subspace of $U$.

\proof{ Suppose that $v\in V$, $u\in U$ and $|u|\le|v|$.   Consider
$A=\{w:w\in V,\,0\le w\le u^+\}$.   Then $A$ is a non-empty subset of
$V$ with an upper bound in $V$ (viz., $|v|$).   So $A$ has a supremum
$v_0$ in $V$.   Because the embedding $V\embedsinto U$ is order-continuous
(352Nb), $v_0$ is the supremum of $A$ in $U$.   But as $V$ is
order-dense (353A),
$v_0=u^+$ and $u^+\in V$.   Similarly, $u^-\in V$ and $u\in V$.   As
$u$ and $v$ are arbitrary, $V$ is solid.
}%end of proof of 353L

\leader{353M}{Order units} Let $U$ be a Riesz space.

\spheader 353Ma  An element $e$ of $U^+$ is an {\bf order unit} in $U$
if $U$ is the solid linear subspace of itself generated by $e$\cmmnt{;
that is, if for every $u\in U$ there is an $n\in\Bbb N$ such that
$|u|\le ne$}.
\prooflet{(For  the solid linear subspace generated by $v\in U^+$ is
$\bigcup_{n\in\Bbb N}[-nv,nv]$.)}

\spheader 353Mb An element $e$ of $U^+$ is a {\bf weak order unit} in
$U$ if $U$ is the principal band generated by $e$\cmmnt{;  that is, if
$u=\sup_{n\in\Bbb N}u\wedge ne$ for every $u\in U^+$ (352Vb)}.

Of course an order unit is a weak order unit.

\spheader 353Mc If $U$ is Archimedean, then an element $e$ of $U^+$ is a
weak order unit iff\prooflet{ $\{e\}^{\perp\perp}=U$ (353B), that is,
iff
$\{e\}^{\perp}=\{0\}$ (because

\Centerline{$\{e\}^{\perp}=\{0\}
\Longrightarrow\{e\}^{\perp\perp}=\{0\}^{\perp}=U
\Longrightarrow\{e\}^{\perp}=\{e\}^{\perp\perp\perp}=U^{\perp}=\{0\}$,)}

\noindent that is, iff} $u\wedge e>0$ whenever $u>0$.

\leader{353N}{Theorem}\dvAformerly{3{}53M} Let $U$ be an Archimedean Riesz 
space with order
unit $e$.   Then it can be embedded as an order-dense and norm-dense
Riesz subspace of $C(X)$, where $X$ is a
compact Hausdorff space, in such a way that $e$ corresponds to $\chi X$;
moreover, this embedding is essentially unique.

\cmmnt{\medskip

\noindent{\bf Remark} Here $C(X)$ is the space of all continuous
functions from $X$ to $\Bbb R$;  because $X$ is compact, they are all
bounded, so that $\chi X$ is an order unit in $C(X)$.
}%end of comment

\proof{{\bf (a)} Let $X$ be the set of Riesz homomorphisms $x$ from $U$
to $\Bbb R$ such that $x(e)=1$.   Define $T:U\to\Bbb R^X$ by setting
$(Tu)(x)=x(u)$ for $x\in X$, $u\in U$;  then it is easy to check that
$T$ is a Riesz homomorphism, just because every member of $X$ is a Riesz
homomorphism, and of course $Te=\chi X$.

\medskip

{\bf (b)} The key to the proof is the fact that $X$ separates the points
of $U$, that is, that $T$ is injective.   I choose the following method
to show this.
Suppose that $w\in U$ and $w>0$.   Because $U$ is Archimedean, there is
a $\delta>0$ such that $(w-\delta e)^+\ne 0$.   Now there is an $x\in X$ such that $x(w)\ge\delta$.
\Prf\ (i) By 351O, there is a solid linear subspace $V$ of $U$ such that
$(w-\delta e)^+\notin V$ and whenever $u\wedge v=0$ in $U$ then one of
$u$, $v$ belongs to $V$.   (ii) Because $V\ne U$, $e\notin V$, so no
non-zero multiple of $e$ can belong to $V$.   Also observe that if $u$,
$v\in U\setminus V$, then one of $(u-v)^+$, $(v-u)^+$ must belong to
$V$, while neither $u=u\wedge v+(u-v)^+$ nor $v=u\wedge v+(v-u)^+$ does;
so $u\wedge v\notin V$.   (iii) For each $u\in U$ set
$A_u=\{\alpha:\alpha\in\Bbb R,\,(u-\alpha e)^+\in V\}$.   Then

\Centerline{$\alpha\ge\beta\in A_u
\Longrightarrow 0\le(u-\alpha e)^+\le(u-\beta e)^+\in V
\Longrightarrow \alpha\in A_u$.}

\noindent Also $A_u$ is
non-empty and bounded below, because if $\alpha\ge 0$ is such that
$-\alpha e\le u\le\alpha e$ then $\alpha\in A_u$ and $-\alpha-1\notin
A_u$ (since $(u-(-\alpha-1)e)^+\ge e\notin V$).
(iv) Set $x(u)=\inf A_u$ for every $u\in U$;  then $\alpha\in
A_u$ for every $\alpha>x(u)$, $\alpha\notin A_u$ for every
$\alpha<x(u)$.   (v) If $u$, $v\in U$, $\alpha>x(u)$ and $\beta>x(v)$
then

\Centerline{$((u+v)-(\alpha+\beta)e)^+\le(u-\alpha e)^++(v-\beta e)^+
\in V$}

\noindent (352Fc), so $\alpha+\beta\in A_{u+v}$;  as $\alpha$ and
$\beta$ are arbitrary, $x(u+v)\le x(u)+x(v)$.   (vi) If $u$, $v\in U$,
$\alpha<x(u)$ and $\beta<x(v)$ then

\Centerline{$((u+v)-(\alpha+\beta)e)^+
\ge(u-\alpha e)^+\wedge(v-\beta e)^+\notin V$,}

\noindent using (ii) of this argument and the other part of 352Fc, so
$\alpha+\beta\notin A_{u+v}$.   As $\alpha$ and $\beta$ are arbitrary,
$x(u+v)\ge x(u)+x(v)$.   (vii) Thus $x:U\to\Bbb R$ is additive.   (viii)
If $u\in U$ and $\gamma>0$ then

\Centerline{$\alpha\in A_u
\Longrightarrow (\gamma u-\alpha\gamma e)^+=\gamma(u-\alpha e)^+\in V
\Longrightarrow \gamma\alpha\in A_{\gamma u}$;}

\noindent thus $A_{\gamma u}\supseteq\gamma A_u$;  similarly,
$A_u\supseteq\gamma^{-1}A_{\gamma u}$ so $A_{\gamma u}=\gamma A_u$ and
$x(\gamma u)=\gamma x(u)$.   (ix) Consequently $x$ is linear, since we
know already from (vii) that $x(0u)=0.x(u)$, $x(-u)=-x(u)$.   (x) If
$u\ge 0$ then $u+\alpha e\ge\alpha e\notin V$ for every $\alpha>0$, that
is, $-\alpha\notin A_u$ for every $\alpha>0$, and $x(u)\ge 0$;  thus $x$
is a positive linear functional.   (xi) If $u\wedge v=0$, then one of
$u$, $v$ belongs to $V$, so $\min(x(u),x(v))\le 0$ and (using (x))
$\min(x(u),x(v))=0$;  thus $x$ is a Riesz homomorphism (352G(iv)).
(xii) $A_e=\coint{1,\infty}$ so $x(e)=1$.   Thus $x\in X$.   (xiii)
$\delta\notin A_w$ so $x(w)\ge\delta$.\ \Qed

\medskip

{\bf (c)} Thus $Tw\ne 0$ whenever $w>0$;  consequently $|Tw|=T|w|\ne 0$
whenever $w\ne 0$, and $T$ is injective.   I now
have to define the topology of $X$.   This is just the subspace topology
on $X$ if we regard $X$ as a subset of $\BbbR^U$ with its product
topology.
To see that $X$ is compact, observe that if for each $u\in U$ we choose
an $\alpha_u$ such that $|u|\le\alpha_ue$, then $X$ is a subspace of
$Q=\prod_{u\in U}[-\alpha_u,\alpha_u]$.
Because $Q$ is a product of compact spaces, it is compact, by
Tychonoff's theorem (3A3J).   Now $X$ is a closed subset of $Q$.
\Prf\ $X$ is just the intersection of the sets

\Centerline{$\{x:x(u+v)=x(u)+x(v)\}$,
\quad$\{x:x(\alpha u)=\alpha x(u)\}$,}

\Centerline{$\{x:x(u^+)=\max(x(u),0)\}$,
\quad$\{x:x(e)=1\}$}

\noindent as $u$, $v$ run over $U$ and $\alpha$ over $\Bbb R$;  and each
of these is closed, so $X$ is an intersection of closed sets and
therefore itself closed.\ \QeD\  Consequently $X$ also is compact.
Moreover, the coordinate functionals $x\mapsto x(u)$ are continuous on
$Q$, therefore on $X$ also, that is, $Tu:X\to\Bbb R$ is a continuous
function for every $u\in U$.

Note also that because $Q$ is a product of Hausdorff spaces, $Q$ and $X$
are Hausdorff (3A3Id, 3A3Bh).

\medskip

{\bf (d)} So $T$ is a Riesz homomorphism from $U$ to $C(X)$.   Now
$T[U]$ is a Riesz subspace of $C(X)$, containing $\chi X$, and such that
if $x$, $y\in X$ are distinct there is an $f\in T[U]$ such that
$f(x)\ne f(y)$ (because there is surely a $u\in U$ such that
$x(u)\ne y(u)$).   By the Stone-Weierstrass theorem (281A), $T[U]$ is
$\|\,\|_{\infty}$-dense in $C(X)$.

Consequently it is also order-dense.   \Prf\ If $f>0$ in $C(X)$, set
$\epsilon=\bover13\|f\|_{\infty}$, and let $u\in U$ be such that
$\|f-Tu\|_{\infty}\le\epsilon$;  set $v=(u-\epsilon e)^+$.   Since

\Centerline{$0<(f-2\epsilon\chi X)^+\le(Tu-\epsilon\chi X)^+\le f^+=f$,}

\noindent $0<Tv\le f$.   As $f$ is arbitrary, $T[U]$ is
quasi-order-dense, therefore order-dense (353A).\ \Qed

\medskip

{\bf (e)} I have still to show that the representation is (essentially)
unique.   Suppose, then, that we have another representation of $U$ as a
norm-dense Riesz subspace of $C(Z)$, with $e$ this time corresponding to
$\chi Z$;  to simplify the notation, let us suppose that $U$ is actually
a subspace of $C(Z)$.   Then for each $z\in Z$, we have a functional
$\hat z:U\to\Bbb R$ defined by setting $\hat z(u)=u(z)$ for every $u\in
U$;  of course $\hat z$ is a Riesz homomorphism such that $\hat z(e)=1$,
that is, $\hat z\in X$.   Thus we have a function $z\mapsto\hat z:Z\to
X$.   For any $u\in U$, the function
$z\mapsto\hat z(u)=u(z)$ is continuous, so the function $z\mapsto\hat z$
is continuous (3A3Ib).   If $z_1$, $z_2$ are distinct members of $Z$,
there is an $f\in C(Z)$ such that $f(z_1)\ne f(z_2)$ (3A3Bf);  now
there is a $u\in U$ such that
$\|f-u\|_{\infty}\le\bover13|f(z_1)-f(z_2)|$, so that $u(z_1)\ne u(z_2)$
and $\hat z_1\ne\hat z_2$.   Thus $z\mapsto\hat z$ is injective.
Finally, it is also surjective.   \Prf\ Suppose that $x\in X$.   Set
$V=\{u:u\in U,\,x(u)=0\}$;  then $V$ is a solid linear subspace of $U$
(352Jb), not containing $e$.   For $z\in V^+$ set $G_v=\{z:v(z)>1\}$.
Because $e\notin V$, $G_v\ne Z$.    $\Cal G=\{G_v:v\in V^+\}$ is an
upwards-directed family of open sets in $Z$, not containing $Z$;
consequently, because $Z$ is compact, $\Cal G$ cannot be an open cover
of $Z$.   Take $z\in Z\setminus\bigcup\Cal G$.   Then $v(z)\le 1$ for
every $v\in V^+$;  because $\alpha|v|\in V^+$ whenever $v\in V$ and
$\alpha\ge 0$, we must have $v(z)=0$ for every $v\in V$.   Now, given
any $u\in U$, consider $v=u-x(u)e$.   Then $x(v)=0$ so $v\in V$ and
$v(z)=0$, that is,

\Centerline{$u(z)=(v+x(u)e)(z)=v(z)+x(u)e(z)=x(u)$.}

\noindent  As $u$ is arbitrary, $\hat z=x$;  as $x$ is arbitrary, we
have the result.\ \Qed

Thus $z\mapsto\hat z$ is a continuous bijection from the compact
Hausdorff space $Z$ to the compact Hausdorff space $X$;  it must
therefore be a homeomorphism (3A3Dd).

This argument shows that if $U$ is embedded as a norm-dense Riesz
subspace of $C(Z)$, where $Z$ is compact and Hausdorff, then $Z$ must be
homeomorphic to $X$.   But it shows also that a homeomorphism is
canonically defined by the embedding;  $z\in Z$ corresponds to the Riesz
homomorphism $u\mapsto u(z)$ in $X$.
}%end of proof of 353N

\leader{353O}{Lemma} Let $U$ be a Riesz space, $V$ an Archimedean Riesz
space and $S$, $T:U\to V$ Riesz homomorphisms such that $Su\wedge Tu'=0$
in $V$ whenever $u\wedge u'=0$ in $U$.   Set $W=\{u:Su=Tu\}$.   Then $W$
is a solid linear subspace of $U$;  if $S$ and $T$ are order-continuous,
$W$ is a band.

\proof{{\bf (a)} It is easy to check that, because $S$ and $T$ are Riesz
homomorphisms, $W$ is a Riesz subspace of $U$.

\medskip

{\bf (b)} If $w\in W$ and $0\le u\le w$ in $U$, then $Su\le Tu$.
\Prf\Quer\ Otherwise, set $e=Sw=Tw$, and let $V_e$ be the solid linear
subspace of $V$ generated by $e$, so that $V_e$ is an Archimedean Riesz
space with order unit, containing both $Su$ and $Tu$.   By 353N (or its
proof), there is a Riesz homomorphism $x:V_e\to\Bbb R$ such that
$x(e)=1$ and $x(Su)>x(Tu)$.   Take $\alpha$ such that
$x(Su)>\alpha>x(Tu)$, and consider $u'=(u-\alpha w)^+$,
$u''=(\alpha w-u)^+$.    Then

\Centerline{$x(Su')=\max(0,x(Su)-\alpha x(Sw))=\max(0,x(Su)-\alpha)>0$,}

\Centerline{$x(Tu'')=\max(0,\alpha x(Tw)-x(Tu))
=\max(0,\alpha-x(Tu))>0$,}

\noindent so

\Centerline{$x(Su'\wedge Tu'')=\min(x(Su'),x(Tu''))>0$}

\noindent and $Su'\wedge Tu''>0$, while $u'\wedge u''=0$.\ \Bang\Qed

Similarly, $Tu\le Su$ and $u\in W$.   As $u$ and $w$ are arbitrary, $W$
is a solid linear subspace.

\medskip

{\bf (c)} Finally, suppose that $S$ and $T$ are order-continuous, and
that $A\subseteq W$ is a non-empty upwards-directed set with supremum
$u$ in $U$.   Then

\Centerline{$Su=\sup S[A]=\sup T[A]=Tu$}

\noindent and $u\in W$.   As $u$ and $A$ are arbitrary, $W$ is a band
(352Ob).
}%end of proof of 353O

\leader{353P}{\dvrocolon{$f$-algebras}}\cmmnt{ I give two results on
$f$-algebras, intended to clarify the connexions between the
multiplicative and lattice structures of the Riesz spaces in Chapter 36.

\wheader{353P}{4}{2}{2}{72pt}

\noindent}{\bf Proposition} Let $U$ be an Archimedean
$f$-algebra\cmmnt{ (352W)}.   Then

(a) the multiplication is separately
order-continuous in the sense that the maps $u\mapsto u\times w$,
$u\mapsto w\times u$ are
order-continuous for every $w\in U^+$;

(b) the multiplication is commutative.

\proof{{\bf (a)} Let $A\subseteq U$ be a non-empty set with infimum $0$,
and $v_0\in U^+$ a lower bound for $\{u\times w:u\in A\}$.   Fix
$u_0\in A$.   If $u\in A$ and $\delta>0$, then
$v_0\wedge(u_0-\bover1{\delta}u)^+\le\delta
u_0\times w$.   \Prf\ Set $v=v_0\wedge(u_0-\bover1{\delta}u)^+$.   Then

\Centerline{$\delta v\wedge(u-\delta u_0)^+
\le(\delta u_0-u)^+\wedge(u-\delta u_0)^+=0$,}

\noindent so $v\wedge(u-\delta u_0)^+=0$ and
$v\wedge((u-\delta u_0)^+\times w)=0$.   But

\Centerline{$v\le v_0\le u\times w
\le(u-\delta u_0)^+\times w+\delta u_0\times w$,}

\noindent so

\Centerline{$v
\le((u-\delta u_0)^+\times w)\wedge v+(\delta u_0\times w)\wedge v
\le\delta u_0\times w$,}

\noindent by 352Fa.\ \Qed

\woddheader{353P}{0}{0}{0}{24pt}

Taking the infimum over $u$, and using the distributive laws (352E), we
get

\Centerline{$v_0\wedge u_0\le\delta u_0\times w$.}

\noindent Taking the infimum over $\delta$, and using the hypothesis
that $U$ is Archimedean,

\Centerline{$v_0\wedge u_0=0$.}

\noindent But this means that $v_0\wedge(u_0\times w)=0$, while
$v_0\le u_0\times w$, so $v_0=0$.   As $v_0$ is arbitrary, $\inf_{u\in
A}u\times w=0$;  as $A$ is arbitrary, $u\mapsto u\times w$ is
order-continuous.   Similarly, $u\mapsto w\times u$ is order-continuous.

\medskip

{\bf (b)(i)} Fix $v\in U^+$, and for $u\in U$ set

\Centerline{$Su=u\times v$,\quad $Tu=v\times u$.}

\noindent Then $S$ and $T$ are both order-continuous Riesz homomorphisms
from $U$ to itself (352W(b-iv) and (a) above).   Also, $Su\wedge Tu'=0$
whenever $u\wedge u'=0$.   \Prf\

\Centerline{$0=(u\times v)\wedge u'=(u\times v)\wedge(v\times u')$.
\Qed}

\noindent So $W=\{u:u\times v=v\times u\}$ is a band in $U$ (353O).
Of course $v\in W$ (because $Sv=Tv=v^2$).   If $u\in W^{\perp}$, then
$v\wedge|u|=0$ so $Su=Tu=0$ (352W(b-i)), and $u\in W$;  but this means
that $W^{\perp}=\{0\}$ and $W=W^{\perp\perp}=U$ (353Bb).   Thus $v\times
u=u\times v$ for every $u\in U$.

\medskip

\quad{\bf (ii)} This is true for every $v\in U^+$.   Of course it
follows that $v\times u=u\times v$ for every $u$, $v\in U$, so that
multiplication is commutative.
}%end of proof of 353P

\leader{353Q}{Proposition}\dvAformerly{3{}53P} Let $U$ be an Archimedean 
$f$-algebra with multiplicative identity $e$.

(a) $e$ is a weak order unit in $U$.

(b) If $u$, $v\in U$ then $u\times v=0$ iff $|u|\wedge|v|=0$.

(c) If $u\in U$ has a multiplicative inverse $u^{-1}$ then $|u|$ also
has a multiplicative inverse;  if $u\ge 0$ then $u^{-1}\ge 0$ and $u$ is
a weak order unit.

(d) If $V$ is another Archimedean $f$-algebra with multiplicative
identity $e'$, and $T:U\to V$ is a positive linear operator such that
$Te=e'$, then $T$ is a Riesz homomorphism iff $T(u\times v)=Tu\times Tv$
for all $u$, $v\in U$.

\proof{{\bf (a)} $e=e^2\ge 0$ by 352W(b-ii).   If $u\in U$ and
$e\wedge|u|=0$ then $|u|=(e\times|u|)\wedge|u|=0$;  by 353Mc, $e$ is a
weak order unit.

\medskip

{\bf (b)} If $|u|\wedge|v|=0$ then $u\times v=0$, by 352Wb.   If
$w=|u|\wedge|v|>0$, then $w^2\le|u|\times|v|$.   Let $n\in\Bbb N$ be
such that $nw\not\le e$, and set $w_1=(nw-e)^+$, $w_2=(e-nw)^+$.   Then

$$\eqalign{0
&\ne w_1
=w_1\times e
=w_1\times w_2+w_1\times(e\wedge nw)\cr
&=w_1\times(e\wedge nw)
\le (nw)^2
\le n^2|u|\times|v|
=n^2|u\times v|,\cr}$$

\noindent so $u\times v\ne 0$.

\medskip

{\bf (c)} $u\times u^{-1}=e$ so $|u|\times|u^{-1}|=|e|=e$ (352W(b-iii)),
and $|u^{-1}|=|u|^{-1}$.   (Recall that inverses in any semigroup with
identity are unique, so that we need have no inhibitions in using the
formulae $u^{-1}$, $|u|^{-1}$.)

Now suppose that $u\ge 0$.
Then $u^{-1}=|u^{-1}|\ge 0$.   If $u\wedge|v|=0$ then

\Centerline{$e\wedge|v|=(u\times u^{-1})\wedge|v|=0$,}

\noindent so $v=0$;  accordingly $u$ is a weak order unit.

\medskip

{\bf (d)(i)} If $T$ is multiplicative, and $u\wedge v=0$ in $U$, then
$Tu\times Tv=T(u\times v)=0$ and $Tu\wedge Tv=0$, by (b).  So $T$ is a
Riesz homomorphism, by 352G.

\medskip

\quad{\bf (ii)} Accordingly I shall henceforth assume that $T$ is a
Riesz homomorphism and seek to show that it is multiplicative.

If $u$, $v\in U^+$, then $T(u\times v)$ and $Tu\times Tv$ both belong to
the band generated by $Tu$.
\Prf\ Write $W$ for this band.   ($\alpha$) For any $n\ge 1$ we have
$(v-ne)^2\ge 0$, that is, $2nv\le v^2+n^2e$, so

\Centerline{$n(v-ne)\le 2nv-n^2e\le v^2$.}

\noindent Consequently

\Centerline{$T(u\times v)-nTu
=T(u\times v)-nT(u\times e)
=T(u\times(v-ne))
\le\Bover1nT(u\times v^2)$}

\noindent because $v'\mapsto T(u\times v')$ is a positive linear
operator;  as $V$ is Archimedean,
$\inf_{n\in\Bbb N}(T(u\times v)-nTu)^+=0$ and $T(u\times
v)=\sup_{n\in\Bbb N}T(u\times v)\wedge nTu$ belongs to $W$.
($\beta$) If $w\wedge|Tu|=0$ then

\Centerline{$w\wedge|Tu\times Tv|=w\wedge(|Tu|\times|Tv|)=0$;}

\noindent so $Tu\times Tv\in W^{\perp\perp}=W$.\ \Qed

\medskip

\quad{\bf (iii)} Fix
$v\in U^+$.   For $u\in U$, set $S_1u=Tu\times Tv$ and
$S_2u=T(u\times v)$.   (Cf.\ (b-i) of the proof of 353P.)
Then $S_1$ and $S_2$ are both Riesz homomorphisms from $U$ to
$V$.   If $u\wedge u'=0$ in $U$, then $S_1u\wedge S_2u'=0$ in $V$,
because (by (ii) just above) $S_1u$ belongs to the band generated by
$Tu$, while $S_2u'$ belongs to the band generated by $Tu'$, and
$Tu\wedge Tu'=T(u\wedge u')=0$.   By 353O, $W=\{u:S_1u=S_2u\}$ is a
solid linear subspace of $U$.   Of course it contains $e$, since

\Centerline{$S_1e=Te\times Tv=e'\times Tv=Tv=T(e\times v)=S_2e$.}

\noindent In fact $u\in W$ for every $u\in U^+$.   \Prf\ As noted in
(ii) just above,
$u-ne\le\bover1nu^2$ for every $n\ge 1$.   So

$$\eqalign{|S_1u-S_2u|
&=|S_1(u-ne)^++S_1(u\wedge ne)-S_2(u-ne)^+-S_2(u\wedge ne)|\cr
&\le S_1(u-ne)^+ + S_2(u-ne)^+
\le\Bover1n(S_1u^2+S_2u^2)\cr}$$

\noindent for every $n\ge 1$, and $|S_1u-S_2u|=0$, that is,
$S_1u=S_2u$.\ \Qed

So $W=U$, that is, $Tu\times Tv=T(u\times v)$ for every $u\in U$.  And
this is true for every $v\in U^+$.   It follows at once that it is true
for every $v\in U$, so that $T$ is multiplicative, as claimed.
}%end of proof of 353Q

\exercises{
\leader{353X}{Basic exercises $\pmb{>}$(a)}
%\spheader 353Xa
Let $U$ be a Riesz space in which every band is complemented.   Show
that $U$ is Archimedean.
%353B

\spheader 353Xb A Riesz space $U$ has the {\bf principal projection
property} iff the
band generated by any single member of $U$ is a projection band.   Show
that any Dedekind $\sigma$-complete Riesz space has the principal
projection property, and that any Riesz space with the principal
projection property is Archimedean.
%353I

\sqheader 353Xc Fill in the missing part (b-iii) of 353K.
%353K

\spheader 353Xd Let $U$ be an Archimedean $f$-algebra with an
order-unit which is a multiplicative identity.   Show that $U$ can be
identified, as $f$-algebra, with a subspace of $C(X)$ for some compact
Hausdorff space
$X$.
%353Q

\leader{353Y}{Further exercises (a)}
%\spheader 353Ya
Let $U$ be a Riesz space in which every quasi-order-dense solid linear
subspace is order-dense.   Show that $U$ is Archimedean.
%353A

\spheader 353Yb Let $X$ be a completely regular Hausdorff space.   Show
that $C(X)$ is Dedekind complete iff $C_b(X)$
is Dedekind complete iff $X$ is extremally disconnected.
%353H

\spheader 353Yc Let $X$ be a compact Hausdorff space.   Show that $C(X)$
is Dedekind $\sigma$-complete iff $\overline{G}$ is open for every
cozero set $G\subseteq X$.   (Cf.\ 314Yf.)   Show that in this case $X$
is zero-dimensional.
%353Yb 353H
%works for normal spaces I think, but these not yet defined

\spheader 353Yd Let $U$ be an Archimedean Riesz space.   Show that the
following are equiveridical:  (i) $U$ has the countable sup property
(241Ye) (ii) for every
$A\subseteq U$ there is a countable $B\subseteq A$ such that $A$ and $B$
have the same upper bounds
(iii) every order-bounded disjoint subset of $U^+$ is countable.
%353L+

\spheader 353Ye Let $U$ be an Archimedean Riesz space such that
every order-bounded disjoint sequence in $U^+$ has a supremum in $U$.
Show that $U$ has the
principal projection property, but need not be Dedekind
$\sigma$-complete.
%353N, 35bits

\spheader 353Yf Let $U$ be an Archimedean $f$-algebra.   Show that an
element $e$ of $U$ is a multiplicative identity iff $e^2=e$ and $e$ is a
weak order unit.
%353P 35bits

\spheader 353Yg Let $U$ be an Archimedean $f$-algebra with a
multiplicative identity.   Show that if $u\in U$ then $u$ is invertible
iff $|u|$ is invertible.
%353Q
% u^{-1} = u\times|u|^{-2}
}%end of exercises

\cmmnt{
\Notesheader{353} As in the last section, many of the results above have
parallels in the theory of Boolean algebras;  thus 353A corresponds to
313K, 353H corresponds in part to remarks in 314Bc and 314Xa, and
353K corresponds to 314C-314E.   Riesz spaces are more complicated;
for instance, principal ideals in Boolean algebras are straightforward,
while in Riesz spaces we have to distinguish between the solid linear
subspace generated by an element and the band generated by the same
element.   Thus an `order unit' in a Boolean ring would just be an
identity, while in a Riesz space we must distinguish between `order
unit' and `weak
order unit'.   As this remark may suggest to you, (Archimedean) Riesz
spaces are actually closer in spirit to arbitrary Boolean rings than to
the Boolean
algebras we have been concentrating on so far;  to the point that in
\S361 below I will return briefly to general Boolean rings.

Note that the standard definition of `order-dense' in Boolean
algebras, as given in 313J, corresponds to the definition of
`quasi-order-dense' in Riesz spaces (352Na);  the point here being that
Boolean algebras behave like Archimedean Riesz spaces, in which there is
no need to make a distinction.

I give the representation theorem 353N more for completeness than
because we need it in any formal sense.   In 351Q and 352L I have given
representation theorems for general partially ordered linear spaces, and
general Riesz spaces, as quotients of spaces of functions;  in 368F
below I give a theorem for Archimedean Riesz spaces corresponding rather
more closely to the expressions of the $L^p$ spaces as quotients of
spaces of measurable functions.   In 353N, by contrast, we have a
theorem expressing Archimedean Riesz spaces with order units as true
spaces of functions, rather than as spaces of equivalence classes of
functions.   All these theorems are important in forming an appropriate
mental picture of ordered linear spaces, as in 352M.

I give a bare-handed proof of 353N, using only the Riesz space structure
of $C(X)$;  if you know a little about extreme points of dual unit balls
you can approach from that direction instead, using 354Yj.   The point
is that (as part (d) of the proof of 353N
makes clear) the space $X$ can be
regarded as a subset of the normed space dual $U^*$ of $U$ with its
weak* topology.   In this treatise generally, and in the present
chapter in particular, I allow myself to be slightly prejudiced against
normed-space methods;  you can find them in any book on functional
analysis, and I prefer here to develop techniques like those in part (b)
of the proof of 353N, which will be a useful preparation for such
theorems as 368E.

There is a very close analogy between 353N and the Stone representation
of Boolean algebras (311E, 311I-311K). %311I 311J 311K
Just as the proof of 311E
looked at the set of ring homomorphisms from $\frak A$ to the elementary
Boolean algebra $\Bbb Z_2$, so the proof of 353N looks at Riesz
homomorphisms from $U$ to the elementary $M$-space $\Bbb R$.   Later on,
the most important $M$-spaces, from the point of view of this treatise,
will be the $L^{\infty}$ spaces of \S363, explicitly defined in terms of
Stone representations (363A).

Of the two parts of 353P, it is (a) which is most important for the
purposes of this book.   The $f$-algebras we shall encounter in Chapter
36 can be seen to be commutative for different, and more elementary,
reasons.  The (separate) order-continuity of multiplication, however, is
not always immediately obvious.   Similarly, the uniferent Riesz
homomorphisms we shall encounter can generally be seen to be
multiplicative without relying on the arguments of 353Qd.
}%end of notes


\discrpage

\frfilename{mt354.tex}
\versiondate{18.8.08}
\copyrightdate{2000}

\def\chaptername{Riesz Spaces}
\def\sectionname{Banach lattices}

\newsection{354}

The next step is a brief discussion of norms on Riesz spaces.   I start
with the essential definitions (354A, 354D) with the principal
properties of
general Riesz norms (354B-354C) and order-continuous norms (354E).   I
then describe two of the most important classes of Banach lattice:
$M$-spaces (354F-354L) %354F 354G 354H 354I 354J 354K 354L
and $L$-spaces (354M-354R), %354M 354N 354O 354P 354Q 354R
with their elementary
properties.   For $M$-spaces I give the basic representation theorem
(354K-354L), and for $L$-spaces I give a note on uniform integrability
(354P-354R).  %354P 354Q 354R

\leader{354A}{Definitions (a)} If $U$ is a Riesz space, a {\bf Riesz
norm} or {\bf lattice norm}
%Deville Godefroy & Zizler
on $U$ is a norm $\|\,\|$ such that $\|u\|\le\|v\|$ whenever
$|u|\le|v|$\cmmnt{;  that is, a norm such that $\||u|\|=\|u\|$ for
every $u$ and $\|u\|\le\|v\|$ whenever $0\le u\le v$}.

\spheader 354Ab A {\bf Banach lattice} is a Riesz space with a Riesz
norm under which it is complete.

\cmmnt{\medskip

\noindent{\bf Remark} We have already seen many examples of Banach
lattices;  I list some in 354Xa below.
}

\leader{354B}{Lemma} Let $U$ be a Riesz space with a Riesz norm
$\|\,\|$.

(a) $U$ is Archimedean.

(b) The maps $u\mapsto|u|$ and $u\mapsto u^+$ are uniformly continuous.

(c) For any $u\in U$, the sets $\{v:v\le u\}$ and $\{v:v\ge u\}$ are
closed;  in particular, the positive cone of $U$ is closed.

(d) Any band in $U$ is closed.

(e) If $V$ is a norm-dense Riesz subspace of $U$, then $V^+=\{v:v\in
V,\,v\ge 0\}$ is norm-dense in the positive cone $U^+$ of $U$.

\proof{{\bf (a)} If $u$, $v\in U$ are such that $nu\le v$ for every
$n\in\Bbb N$, then $nu^+\le v^+$ so $n\|u^+\|\le\|v^+\|$ for every $n$,
and $\|u^+\|=0$, that is, $u^+=0$ and $u\le 0$.   As $u$, $v$ are
arbitrary, $U$ is Archimedean.

\medskip

{\bf (b)} For any $u$, $v\in U$, $||u|-|v||\le|u-v|$ (352D), so
$\||u|-|v|\|\le\|u-v\|$;  thus $u\mapsto|u|$ is uniformly continuous.
Consequently $u\mapsto \bover12(u+|u|)=u^+$ is uniformly continuous.

\medskip

{\bf (c)} Now $\{v:v\le u\}=\{v:(v-u)^+=0\}$ is closed because the
function $v\mapsto (v-u)^+$ is continuous and $\{0\}$ is closed.
Similarly $\{v:v\ge u\}=\{v:(u-v)^+=0\}$ is closed.

\medskip

{\bf (d)} If $V\subseteq U$ is a band, then $V=V^{\perp\perp}$ (353B),
that is, $V=\{v:|v|\wedge|w|=0$ for every $w\in V^{\perp}\}$.   Because
the function $v\mapsto |v|\wedge|w|=\bover12(|v|+|w|-||v|-|w||)$ is
continuous, all the sets $\{v:|v|\wedge|w|=0\}$ are closed, and so is
their intersection $V$.

\medskip

{\bf (e)} Observe that $V^+=\{v^+:v\in V\}$ and $U^+=\{u^+:u\in U\}$;
recall that $u\mapsto u^+$ is continuous, and apply 3A3Eb.
}%end of proof of 354B

\leader{354C}{Lemma} If $U$ is a Banach lattice and $\sequencen{u_n}$ is
a sequence in $U$ such that $\sum_{n=0}^{\infty}\|u_n\|<\infty$, then
$\sup_{n\in\Bbb N}u_n$ is defined in $U$, with
$\|\sup_{n\in\Bbb N}u_n\|\le\sum_{n=0}^{\infty}\|u_n\|$.

\proof{ Set $v_n=\sup_{i\le n}u_i$ for each $n$.   Then

\Centerline{$0\le v_{n+1}-v_n\le(u_{n+1}-u_n)^+\le|u_{n+1}-u_n|$}

\noindent for each $n\in\Bbb N$, so

\Centerline{$\sum_{n=0}^{\infty}\|v_{n+1}-v_n\|
\le\sum_{n=0}^{\infty}\|u_{n+1}-u_n\|
\le\sum_{n=0}^{\infty}\|u_{n+1}\|+\|u_n\|$}

\noindent is finite, and $\sequencen{v_n}$ is Cauchy.   Let $u$ be its
limit;  because $\sequencen{v_n}$ is non-decreasing, and the sets
$\{v:v\ge v_n\}$ are all closed, $u\ge v_n$ for each $n\in\Bbb N$.   On
the other hand, if $v\ge v_n$ for every $n$, then

\Centerline{$(u-v)^+=\lim_{n\to\infty}(v_n-v)^+=0$,}

\noindent and $u\le v$.   So

\Centerline{$u=\sup_{n\in\Bbb N}v_n=\sup_{n\in\Bbb N}u_n$}

\noindent is the required supremum.

To estimate its norm, observe that $|v_n|\le\sum_{i=0}^n|u_i|$ for each
$n$ (induce on $n$, using the last item in 352D for the inductive step),
so that

\Centerline{$\|u\|=\lim_{n\to\infty}\|v_n\|
\le\sum_{i=0}^{\infty}\||u_i|\|=\sum_{i=0}^{\infty}\|u_i\|$.}
}%end of proof of 354C

\leader{354D}{}\cmmnt{ I come now to the basic properties according to
which we classify Riesz norms.

\medskip

\noindent}{\bf Definitions (a)} A {\bf Fatou norm} on a Riesz space $U$
is a Riesz norm on $U$ such that whenever $A\subseteq U^+$ is non-empty
and upwards-directed and has a least upper bound in $U$, then
$\|\sup A\|=\sup_{u\in A}\|u\|$.   \cmmnt{(Observe that, once we know
that $\|\,\|$ is a Riesz norm, we can be sure that $\|u\|\le\|\sup A\|$
for every $u\in A$, so that all we shall need to check is that
$\|\sup A\|\le\sup_{u\in A}\|u\|$.)}

\spheader 354Db A Riesz norm on a Riesz space $U$ has the {\bf Levi
property} if every upwards-directed norm-bounded set is bounded above.

\spheader 354Dc A Riesz norm on a Riesz space $U$ is
{\bf order-continuous} if $\inf_{u\in A}\|u\|=0$ whenever $A\subseteq U$
is a non-empty downwards-directed set with infimum $0$.

\leader{354E}{Proposition} Let $U$ be a Riesz space with an
order-continuous Riesz norm $\|\,\|$.

(a) If $A\subseteq U$ is non-empty and upwards-directed and has a
supremum, then $\sup A\in\overline{A}$.

(b) $\|\,\|$ is Fatou.

(c) If $A\subseteq U$ is non-empty and upwards-directed and bounded
above, then for every $\epsilon>0$ there is a $u\in A$ such that
$\|(v-u)^+\|\le\epsilon$ for every $v\in A$;
%new 2008
that is, the filter $\Cal F(A\closeuparrow)$ on $U$ generated by
$\{\{v:v\in A$, $u\le v\}:u\in A\}$ is a Cauchy filter.

(d) Any non-decreasing order-bounded sequence in $U$ is Cauchy.

(e) If $U$ is a Banach lattice it is Dedekind complete.

(f) Every order-dense Riesz subspace of $U$ is norm-dense.

%new 2008
(g) Every norm-closed solid linear subspace of $U$ is a band.

\proof{{\bf (a)} Suppose that $A\subseteq U$ is non-empty and
upwards-directed and has a least upper bound $u_0$.    Then
$B=\{u_0-u:u\in A\}$ is downwards-directed and has infimum $0$.   So
$\inf_{u\in A}\|u_0-u\|=0$, and $u_0\in\overline{A}$.

\medskip

{\bf (b)} If, in (a), $A\subseteq U^+$, then we must have

\Centerline{$\|u_0\|\le\inf_{u\in A}\|u\|+\|u-u_0\|
\le\sup_{u\in A}\|u\|$.}

\noindent As $A$ is arbitary, $\|\,\|$ is a Fatou norm.

\medskip

{\bf (c)} Let $B$ be the set of upper bounds for $A$.   Then $B$ is
downwards-directed;  because $A$ is upwards-directed,
$B-A=\{v-u:v\in B,\,u\in A\}$ is downwards-directed.  By 353F,
$\inf(B-A)=0$.   So
there are $w\in B$, $u\in A$ such that $\|w-u\|\le\epsilon$.   Now if
$v\in A$,

\Centerline{$(v-u)^+=(v\vee u)-u\le w-u$,}

\noindent so $\|(v-u)^+\|\le\epsilon$.

In terms of the filter $\Cal F(A\closeuparrow)$, this tells us
that if $v_0$, $v_1$ belong to $F_u=\{v:v\in A$, $v\ge u\}$ then
$|v_0-v_1|\le w-u$ so $\|v_0-v_1\|\le\epsilon$ and the diameter of $F_u$ is
at most $\epsilon$.   As $\epsilon$ is arbitrary,
$\Cal F(A\closeuparrow)$ is a Cauchy filter.

\medskip


{\bf (d)} If $\sequencen{u_n}$ is a non-decreasing order-bounded
sequence, and $\epsilon>0$, then, applying (c)
to $\{u_n:n\in\Bbb N\}$, we find
that there is an $m\in\Bbb N$ such that $\|u_m-u_n\|\le\epsilon$
whenever $m\ge n$.

\medskip

{\bf (e)} Now suppose that $U$ is a Banach lattice.   Let $A\subseteq U$
be any non-empty set with an upper bound.   Set
$B=\{u_0\vee\ldots\vee u_n:u_0,\ldots,u_n\in A\}$, so that $B$ is
upwards-directed and has the same upper bounds as $A$.
Let $\Cal F(B\closeuparrow)$ be the filter on $U$ generated by
$\{B\cap\coint{v,\infty}:v\in B\}$.   By (c), this is a Cauchy filter with
a limit $u^*$ say.   For every $u\in A$, $\coint{u,\infty}$ is a closed set
belonging to $\Cal F(B\closeuparrow)$, so contains $u^*$;  thus $u^*$ is an
upper bound for $A$.   If $w$ is any upper bound for $A$,
$\ocint{-\infty,w}$ is a closed set belonging to $\Cal F(B\closeuparrow)$,
so contains $u^*$;  thus $u^*=\sup A$ and $A$ has a supremum.

\medskip

{\bf (f)} If $V$ is an order-dense Riesz subspace of $U$ and $u\in U^+$,
set $A=\{v:v\in V,\,v\le u\}$.   Then $A$ is upwards-directed and
has supremum $u$, so $u\in\overline{A}\subseteq\overline{V}$, by (a).
Thus $U^+\subseteq\overline{V}$;  it follows at once that
$U=U^+-U^+\subseteq\overline{V}$.

\medskip

{\bf (g)} If $V$ is a norm-closed solid linear subspace of $U$,
and $A\subseteq V^+$ is a non-empty, upwards-directed subset of $V$ with a
supremum in $U$, then $\sup A\in\overline{A}\subseteq V$, by (a);  by
352Ob, $V$ is a band.
}%end of proof of 354E

\vleader{72pt}{354F}{Lemma} If $U$ is an Archimedean Riesz space with an
order unit $e$\cmmnt{ (definition:  353M)}, there is a Riesz norm
$\|\,\|_e$ defined on $U$ by the formula

\Centerline{$\|u\|_e=\min\{\alpha:\alpha\ge 0$, $|u|\le\alpha e\}$}

\noindent for every $u\in U$.

\proof{ This is a routine verification.   Because $e$ is an order-unit,
$\{\alpha:\alpha\ge 0$, $|u|\le\alpha e\}$ is always non-empty,
so always has an
infimum $\alpha_0$ say;  now $|u|-\alpha_0e\le\delta e$ for every
$\delta>0$, so (because $U$ is Archimedean) $|u|-\alpha_0e\le 0$ and
$|u|\le\alpha_0e$, so that the minimum is attained.   In particular,
$\|u\|_e=0$ iff $u=0$.   The subadditivity and homogeneity of $\|\,\|_e$
are immediate from the facts that $|u+v|\le|u|+|v|$, $|\alpha
u|=|\alpha||u|$.
}%end of proof

\leader{354G}{Definitions (a)} If $U$ is an Archimedean Riesz space and
$e$ an order unit in $U$, the norm $\|\,\|_e$ as defined in 354F is the
{\bf order-unit norm} on $U$ associated with $e$.

\spheader 354Gb An {\bf $M$-space} is a Banach lattice in which the norm
is an order-unit norm.

\spheader 354Gc If $U$ is an $M$-space, its {\bf standard order unit} is
the order unit $e$ such that $\|\,\|_e$ is the norm of $U$.
\cmmnt{(To see that $e$ is uniquely defined, observe that it is
$\sup\{u:u\in U,\,\|u\|\le 1\}$.)}

\leader{354H}{Examples (a)} For any set $X$, $\ell^{\infty}(X)$ is an
$M$-space with standard order unit $\chi X$.   \cmmnt{(As remarked in
243Xl, the completeness of $\ell^{\infty}(X)$ can be regarded as the
special case of 243E in which $X$ is given counting measure.)}

\spheader 354Hb For any topological space $X$, the space $C_b(X)$ of
bounded continuous real-valued functions on $X$ is an $M$-space with
standard order
unit $\chi X$.   \cmmnt{(It is a Riesz subspace of $\ell^{\infty}(X)$
containing the order unit of $\ell^{\infty}(X)$, therefore in its own
right an Archimedean Riesz space with order unit.   To see that it is
complete, it is enough to observe that it is closed in
$\ell^{\infty}(X)$ because a uniform limit of continuous functions is
continuous (3A3Nb).)}

\spheader 354Hc For any measure space $(X,\Sigma,\mu)$, the space
$L^{\infty}(\mu)$ is an $M$-space with standard order unit $\chi
X^{\ssbullet}$.

\leader{354I}{Lemma} Let $U$ be an Archimedean Riesz space with order
unit $e$, and $V$ a subset of $U$ which is dense for the
order-unit norm $\|\,\|_e$.   Then for any $u\in U$ there are sequences
$\sequencen{v_n}$, $\sequencen{w_n}$ in $V$ such that
$v_n\le v_{n+1}\le u\le w_{n+1}\le w_n$ and $\|w_n-v_n\|_e\le 2^{-n}$
for every $n$;  so that $u=\sup_{n\in\Bbb N}v_n=\inf_{n\in\Bbb N}w_n$ in
$U$.

If $V$ is a Riesz subspace of $U$, and $u\ge 0$, we may suppose that
$v_n\ge 0$ for every $n$.   Consequently $V$ is order-dense in $U$.

\proof{ For each $n\in\Bbb N$,
take $v_n$, $w_n\in V$ such that

\Centerline{$\|u-\Bover3{2^{n+3}}e-v_n\|_e\le\Bover1{2^{n+3}}$,
\quad$\|u+\Bover3{2^{n+3}}e-w_n\|_e\le\Bover1{2^{n+3}}$.}

\noindent Then

\Centerline{$u-\Bover1{2^{n+1}}e\le v_n\le u-\Bover1{2^{n+2}}e\le u
\le u+\Bover1{2^{n+2}}e\le w_n\le u+\Bover1{2^{n+1}}e$.}

\noindent Accordingly $\sequencen{v_n}$ is non-decreasing,
$\sequencen{w_n}$ is non-increasing and $\|w_n-v_n\|_e\le 2^{-n}$ for
every $n$.   Because $U$ is Archimedean,
$\sup_{n\in\Bbb N}v_n=\inf_{n\in\Bbb N}w_n=u$.

If $V$ is a Riesz subspace of $U$, then replacing $v_n$ by $v_n^+$ if
necessary we may suppose that every $v_n$ is non-negative;  and $V$ is
order-dense by the definition in 352Na.
}%end of proof of 354I

\leader{354J}{Proposition} Let $U$ be an Archimedean Riesz space with an
order unit $e$.   Then $\|\,\|_e$ is Fatou and has the Levi property.

\proof{ This is elementary.   If $A\subseteq U^+$ is non-empty,
upwards-directed and norm-bounded, then it is bounded above by
$\alpha e$, where $\alpha=\sup_{u\in A}\|u\|_e$.   This is all that is
called for in the Levi property.   If
moreover $\sup A$ is defined, then $\sup A\le\alpha e$ so
$\|\sup A\|\le\alpha$, as required in the Fatou property.
}%end of proof of 354J

\leader{354K}{Theorem} Let $U$ be an Archimedean Riesz space with order
unit $e$.   Then it can be embedded as an
order-dense and norm-dense Riesz subspace of $C(X)$, where $X$ is a
compact Hausdorff space, in such a way that $e$ corresponds to $\chi X$
and $\|\,\|_e$ corresponds to $\|\,\|_{\infty}$;  moreover, this
embedding is essentially unique.

\proof{ This is nearly word-for-word a repetition of 353N.   The only
addition is the mention of the norms.   Let $X$ and $T:U\to C(X)$ be
as in 353N.   Then, for any $u\in U$, $|u|\le\|u\|_ee$, so that

\Centerline{$|Tu|=T|u|\le\|u\|_eTe=\|u\|_e\chi X$,}

\noindent and $\|Tu\|_{\infty}\le\|u\|_e$.   On the other hand, if
$0<\delta<\|u\|_e$ then $u_1=(|u|-\delta e)^+>0$, so that
$Tu_1=(|Tu|-\delta\chi X)^+>0$ and $\|Tu\|_{\infty}\ge\delta$;  as
$\delta$ is arbitrary, $\|Tu\|_{\infty}\ge\|u\|_e$.
}%end of proof of 354K

\leader{354L}{Corollary} Any $M$-space $U$ is isomorphic,
as Banach lattice,
to $C(X)$ for some compact Hausdorff $X$, and the isomorphism is
essentially unique.
$X$ can be identified with the set of Riesz homomorphisms $x:U\to\Bbb R$
such that $x(e)=1$, where $e$ is the standard order unit of $U$,
with the topology induced by the product topology on $\BbbR^U$.

\proof{ By 354K,
there are a compact Hausdorff space $X$ and an embedding of $U$ as a
norm-dense Riesz subspace of $C(X)$ matching $\|\,\|_e$ to
$\|\,\|_{\infty}$.   Since $U$ is complete under $\|\,\|_e$, its image
is closed in $C(X)$
(3A4Ff), and must be the whole of $C(X)$.   The expression is
unique just in so far as the expression of 353N/354K is unique.
In particular, we may, if we wish, take $X$ to be the set of normalized
Riesz homomorphisms from $U$ to $\Bbb R$, as in the proof of 353N.
}%end of proof of 354L

\cmmnt{\medskip

\noindent{\bf Remark} The set of uniferent Riesz
homomorphisms from $U$ to $\Bbb R$ is sometimes called the {\bf
spectrum} of $U$.
}%end of comment

\leader{354M}{}\cmmnt{ I come now to a second fundamental class of
Banach lattices, in a strong sense `dual' to the class of $M$-spaces,
as will appear in \S356.

\medskip

\noindent}{\bf Definition} An {\bf $L$-space} is a Banach lattice $U$
such that $\|u+v\|=\|u\|+\|v\|$ whenever $u$, $v\in U^+$.

\medskip

\noindent{\bf Example} If $(X,\Sigma,\mu)$ is any measure space, then
$L^1(\mu)$, with its norm $\|\,\|_1$, is an
$L$-space\cmmnt{ (242D, 242F)}.   \cmmnt{In
particular, taking $\mu$ to be counting measure on $\Bbb N$,} $\ell^1$
is an $L$-space\cmmnt{ (242Xa)}.

\leader{354N}{Theorem} If $U$ is an $L$-space, then its norm is
order-continuous and has the Levi property.

\proof{{\bf (a)} Both of these are consequences of the following fact:
if $A\subseteq U$ is norm-bounded and non-empty and upwards-directed,
then $\sup A$ is defined in $U$ and belongs to the norm-closure of $A$
in $U$.   \Prf\ Fix $u_0\in A$;  set $B=\{u-u_0:u\in A$, $u\ge u_0\}$.
Then $B\subseteq U^+$ is norm-bounded, non-empty and upwards-directed.
Set $\gamma=\sup_{u\in B}\|u\|$.
Consider the filter $\Cal F(B\closeuparrow)$
on $U$ generated by sets of the form $\{v:v\in B$, $v\ge u\}$ for $u\in B$.
If $\epsilon>0$ there is a $u\in B$ such that $\|u\|\ge\gamma-\epsilon$;
now if $v$, $v'\in B\cap\coint{u,\infty}$, there is a $w\in B$ such that
$v\vee v'\le w$, so that

\Centerline{$\|v-v'\|\le\|w-u\|=\|w\|-\|u\|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\Cal F(B\closeuparrow)$ is
Cauchy and has
a limit $u^*$ say.   If $u\in B$, $\coint{u,\infty}$ is a closed set
belonging to $\Cal F(B\closeuparrow)$, so contains $u^*$;
thus $u^*$ is an
upper bound for $B$.   If $w$ is an upper bound for $B$, then
$\ocint{-\infty,w}$ is a closed set belonging to $\Cal F(B\closeuparrow)$, so
contains $u^*$;  thus $u^*$ is the least upper bound of $B$.   And
$B\in\Cal F(B\closeuparrow)$, so $u^*\in\overline{B}$.

Because $u\mapsto u_0$ is an order-preserving homeomorphism,

\Centerline{$u^*+u_0=\sup\{u:u_0\le u\in A\}=\sup A$}

\noindent and $u^*+u_0\in\overline{A}$, as required.\ \Qed

\medskip

{\bf (b)} This shows immediately that the norm has the Levi property.
But also it must be order-continuous.   \Prf\ If $A\subseteq U$ is
non-empty and downwards-directed and has infimum $0$, take any
$u_0\in A$ and consider $B=\{u_0-u:u\in A,\,u\le u_0\}$.   Then $B$ is
upwards-directed and has supremum $u_0$, so $u_0\in\overline{B}$ and

\Centerline{$\inf_{u\in A}\|u\|\le\inf_{v\in B}\|u_0-v\|=0$.  \Qed}
}%end of proof of 354N

\leader{354O}{Proposition} If $U$ is an $L$-space and $V$ is a
norm-closed Riesz subspace of $U$, then $V$ is an $L$-space in its own
right.   In particular, any band in $U$ is an $L$-space.

\proof{ For any Riesz subspace $V$ of $U$, we surely have
$\|u+v\|=\|u|+\|v\|$ whenever $u$, $v\in V^+$;  so if $V$ is
norm-closed, therefore a Banach lattice, it must be an $L$-space.   But
in any Banach lattice, a band is norm-closed (354Bd), so a band in an
$L$-space is again an $L$-space.
}%end of proof of 354O

\leader{354P}{Uniform integrability
in \dvrocolon{$L$-spaces}}\cmmnt{ Some of the ideas of \S246 can be
readily expressed in this abstract context.

\medskip

\noindent}{\bf Definition} Let $U$ be an $L$-space.   A set
$A\subseteq U$ is {\bf uniformly integrable} if for every $\epsilon>0$
there is a
$w\in U^+$ such that $\|(|u|-w)^+\|\le\epsilon$ for every $u\in A$.

\leader{354Q}{}\cmmnt{ Since I have already used the phrase
`uniformly integrable' based on a different formula, I had better check
instantly that the two definitions are consistent.

\medskip

\noindent}{\bf Proposition} If $(X,\Sigma,\mu)$ is any measure space,
then a subset of $L^1=L^1(\mu)$ is uniformly integrable in the sense of
354P iff it is uniformly integrable in the sense of 246A.

\proof{{\bf (a)} If $A\subseteq L^1$ is uniformly integrable in the
sense of 246A, then for any $\epsilon>0$ there are $M\ge 0$,
$E\in\Sigma$ such that $\mu E<\infty$ and
$\int(|u|-M\chi E^{\ssbullet})^+\le\epsilon$ for every $u\in A$;  now $w=M\chi E^{\ssbullet}$ belongs to $(L^1)^+$ and
$\|(|u|-w)^+\|\le\epsilon$ for
every $u\in A$.   As $\epsilon$ is arbitrary, $A$ is uniformly
integrable in the sense of 354P.

\medskip

{\bf (b)} Now suppose that $A$ is uniformly integrable in the sense of
354P.   Let $\epsilon>0$.   Then there is a $w\in (L^1)^+$ such that
$\|(|u|-w)^+\|\le\bover12\epsilon$ for every $u\in A$.   There is a
simple function $h:X\to\Bbb R$ such that
$\|w-h^{\ssbullet}\|\le\bover12\epsilon$ (242Mb);  now take
$E=\{x:h(x)\ne 0\}$, $M=\sup_{x\in X}|h(x)|$ (I pass over the trivial
case $X=\emptyset$), so that $h\le M\chi E$ and

\Centerline{$(|u|-M\chi E^{\ssbullet})^+
\le(|u|-w)^++(w-M\chi E^{\ssbullet})^+
\le(|u|-w)^++(w-h^{\ssbullet})^+$,}

\Centerline{$\int(|u|-M\chi E^{\ssbullet})^+
\le\|(|u|-w)^+\|+\|w-h^{\ssbullet}\|\le\epsilon$}

\noindent for every $u\in A$.   As $\epsilon$ is arbitrary, $A$ is
uniformly integrable in the sense of 354P.
}%end of proof of 354Q

\leader{354R}{}\cmmnt{ I give abstract versions of the easiest results
from \S246.

\medskip

\noindent}{\bf Theorem} Let $U$ be an $L$-space.

(a) If $A\subseteq U$ is uniformly integrable, then

\quad (i) $A$ is norm-bounded;

\quad (ii) every subset of $A$ is uniformly integrable;

\quad (iii) for any $\alpha\in\Bbb R$, $\alpha A$ is uniformly
integrable;

\quad (iv) there is a uniformly integrable, solid, convex, norm-closed
set $C\supseteq A$;

\quad (v) for any other uniformly integrable set $B\subseteq U$, $A\cup
B$ and $A+B$ are uniformly integrable.

(b) For any set $A\subseteq U$, the following are equiveridical:

\quad (i) $A$ is uniformly integrable;

\quad (ii) $\lim_{n\to\infty}(|u_n|-\sup_{i<n}|u_i|)^+=0$ for every
sequence $\sequencen{u_n}$ in $A$;

\quad (iii) {\it either} $A$ is empty {\it or} for every $\epsilon>0$
there are $u_0,\ldots,u_n\in A$ such that
\ifdim\pagewidth=390pt\penalty-1000\fi
$\|(|u|-\sup_{i\le n}|u_i|)^+\|\le\epsilon$ for every $u\in A$;

\quad (iv) $A$ is norm-bounded and any disjoint sequence in the solid
hull of $A$ is norm-convergent to $0$.

(c) If $V\subseteq U$ is a closed Riesz subspace then a subset of $V$ is
uniformly integrable when regarded as a subset of $V$ iff it is
uniformly integrable when regarded as a subset of $U$.

\proof{{\bf (a)(i)} There must be a $w\in U^+$ such that
$\int(|u|-w)^+\le 1$ for every $u\in A$;  now

\Centerline{$|u|\le|u|-w+|w|\le(|u|-w)^++|w|$,
\quad$\|u\|\le\|(|u|-w)^+\|
+\|w\|\le 1+\|w\|$}

\noindent for every $u\in A$, so $A$ is norm-bounded.

\medskip

\quad{\bf (ii)} This is immediate from the definition.

\medskip

\quad{\bf (iii)} Given $\epsilon>0$, we can find $w\in U^+$ such
that $|\alpha|\|(|u|-w)^+\|\le\epsilon$ for every
$u\in A$;  now $\|(|v|-|\alpha|w)^+\|\le\epsilon$ for every
$v\in\alpha A$.

\medskip

\quad{\bf (iv)} If $A$ is empty, take $C=A$.   Otherwise, try

\Centerline{$C=\{v:v\in U,\,\|(|v|-w)^+\|
  \le\sup_{u\in A}\|(|u|-w)^+\|$ for every $w\in U^+\}$.}

\noindent Evidently $A\subseteq C$, and $C$ satisfies the definition
354M because $A$ does.   The functionals

\Centerline{$v\mapsto\|(|v|-w)^+\|:U\to\Bbb R$}

\noindent are all continuous for $\|\,\|$ (because the operators
$v\mapsto|v|$, $v\mapsto v-w$, $v\mapsto v^+$, $v\mapsto\|v\|$ are
continuous), so $C$ is closed.   If $|v'|\le|v|$ and $v\in C$, then

\Centerline{$\|(|v'|-w)^+\|\le\|(|v|-w)^+\|
\le\sup_{u\in A}\|(|u|-w)^+\|$}

\noindent for every $w$, and $v'\in C$.   If $v=\alpha v_1+\beta v_2$
where  $v_1$, $v_2\in C$, $\alpha\in[0,1]$ and $\beta=1-\alpha$, then
$|v|\le\alpha|v_1|+\beta|v_2|$, so

\Centerline{$|v|-w\le(\alpha|v_1|-\alpha w)+(\beta|v_2|-\beta w)
\le(\alpha|v_1|-\alpha w)^++(\beta|v_2|-\beta w)^+$}

\noindent and

\Centerline{$(|v|-w)^+\le\alpha(|v_1|-w)^++\beta(|v_2|-w)^+$}

\noindent for every $w$;  accordingly

$$\eqalign{\|(|v|-w)^+\|
&\le\alpha\|(|v_1|-w)^+\|+\beta\|(|v_2|-w)^+\|\cr
&\le(\alpha+\beta)\sup_{u\in A}\|(|u|-w)^+\|
=\sup_{u\in A}\|(|u|-w)^+\|\cr}$$

\noindent for every $w$, and $v\in C$.

Thus $C$ has all the required properties.

\medskip

\quad{\bf (v)} I show first that $A\cup B$ is uniformly integrable.
\Prf\ Given $\epsilon>0$, let $w_1$, $w_2\in U^+$ be such that

\Centerline{$\|(|u|-w_1)^+\|\le\epsilon$ for every $u\in A$,
\quad$\|(|u|-w_2)^+\|\le\epsilon$ for every $u\in B$.}

\noindent Set $w=w_1\vee w_2$;  then
$\|(|u|-w)^+\|\le\epsilon$ for every $u\in A\cup B$.   As $\epsilon$ is
arbitrary, $A\cup B$ is uniformly integrable.\ \Qed

Now (iv) tells us that there is a convex uniformly integrable set $C$
including $A\cup B$, and in this case $A+B\subseteq 2C$, so $A+B$ is
also uniformly integrable, using (ii) and (iii).

\medskip

{\bf (b)(i)$\Rightarrow$(ii)\&(iv)} Suppose that $A$ is uniformly
integrable and that $\sequencen{u_n}$ is any sequence in the solid hull
of $A$.   Set $v_n=\sup_{i\le n}|u_i|$ for $n\in\Bbb N$ and

\Centerline{$v'_0=v_0=|u_0|$,
\quad$v'_n=v_n-v_{n-1}=(|u_n|-\sup_{i<n}|u_i|)^+$}

\noindent for $n\ge 1$.   Given $\epsilon>0$, there is a $w\in U^+$ such
that $\|(|u|-w)^+\|\le\epsilon$ for every $u\in A$, and therefore for
every $u$ in the solid hull of $A$.   Of course
$\sup_{n\in\Bbb N}\|v_n\wedge w\|\le\|w\|$ is finite, so there is an
$n\in\Bbb N$ such
that $\|v_i\wedge w\|\le\epsilon+\|v_n\wedge w\|$ for every
$i\in\Bbb N$.   But now, for $m>n$,

$$\eqalign{v'_m\le(|u_m|-v_n)^+
&\le(|u_m|-|u_m|\wedge w)^++((|u_m|\wedge w)-v_n)^+\cr
&\le(|u_m|-w)^++(v_m\wedge w)-(v_n\wedge w),\cr}$$

\noindent so that

$$\eqalign{\|v'_m\|
&\le\|(|u_m|-w)^+\|+\|(v_m\wedge w)-(v_n\wedge w)\|\cr
&=\|(|u_m|-w)^+\|+\|v_m\wedge w\|-\|v_n\wedge w\|
\le 2\epsilon,\cr}$$

\noindent using the $L$-space property of the norm for the equality in
the middle.   As $\epsilon$ is arbitrary, $\lim_{n\to\infty}v'_n=0$.
As $\sequencen{u_n}$ is arbitrary, condition (ii) is satisfied;  but so
is condition (iv), because we know from (a-i) that $A$ is norm-bounded,
and if $\sequencen{u_n}$ is disjoint then $v'_n=|u_n|$ for every $n$, so
that in this case $\lim_{n\to\infty}u_n=0$.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)$\Rightarrow$(i)} are elementary.

\medskip

\quad{\bf not-(i)$\Rightarrow$not-(iv)} Now suppose that $A$
is not uniformly integrable.   If it is not norm-bounded, we can stop.
Otherwise, there is some $\epsilon>0$ such that
$\sup_{u\in A}\|(|u|-w)^+\|>\epsilon$ for every $w\in U^+$.
Consequently we shall be able to choose inductively a sequence
$\sequencen{u_n}$ in $A$ such that
$\|(|u_n|-2^n\sup_{i<n}|u_i|)^+\|>\epsilon$ for every $n\ge 1$.
Because $A$ is norm-bounded,
$\sum_{i=0}^{\infty}2^{-i}\|u_i\|$ is finite, and we can set

\Centerline{$v_n
=(|u_n|-2^n\sup_{i<n}|u_i|-\sum_{i=n+1}^{\infty}2^{-i}|u_i|)^+$}

\noindent for each $n$.   (The sum $\sum_{i=n+1}^{\infty}2^{-i}|u_i|$ is
defined because $\langle\sum_{i=n+1}^m2^{-i}|u_i|\rangle_{m\ge n+1}$ is
a Cauchy sequence.)   We have $v_m\le|u_m|$,

$$\eqalign{v_m\wedge v_n
&\le(|u_m|-2^{-n}|u_n|)^+\wedge(|u_n|-2^n|u_m|)^+\cr
&\le(2^n|u_m|-|u_n|)^+\wedge(|u_n|-2^n|u_m|)^+=0\cr}$$

\noindent whenever $m<n$, so $\sequencen{v_n}$ is a disjoint sequence in
the solid hull of $A$;  while

\Centerline{$\|v_n\|
\ge\|(|u_n|-2^n\sup_{i<n}|u_i|)^+\|-\sum_{i=n+1}^{\infty}2^{-i}\|u_i\|
\ge\epsilon-2^{-n}\sup_{u\in A}\|u\|\to\epsilon$}

\noindent as $n\to\infty$, so condition (iv) is not satisfied.

\medskip

{\bf (c)} Now this follows at once, because conditions (b-ii) and (b-iv)
are satisfied in $V$ iff they are satsified in $U$.
}%end of proof of 354R

\exercises{\leader{354X}{Basic exercises $\pmb{>}$(a)}
%\spheader 354Xa
Work through the proofs that the following are all Banach lattices.  (i)
$\BbbR^r$ with ($\alpha$) $\|x\|_1=\sum_{i=1}^r|\xi_i|$ ($\beta$)
$\|x\|_2=\sqrt{\sum_{i=1}^r|\xi_i|^2}$ ($\gamma$)
$\|x\|_{\infty}=\max_{i\le r}|\xi_i|$, where $x=(\xi_1,\ldots,\xi_r)$.
(ii) $\ell^p(X)$, for any set $X$ and any $p\in[1,\infty]$ (242Xa,
243Xl, 244Xn).   (iii) $L^p(\mu)$, for any measure space
$(X,\Sigma,\mu)$ and any $p\in[1,\infty]$ (242F, 243E, 244G).   (iv)
$\pmb{c}_0$, the space of sequences convergent to $0$, with the norm
$\|\,\|_{\infty}$ inherited from $\ell^{\infty}$.
%354A

\spheader 354Xb Let $\langle U_i\rangle_{i\in I}$ be any family of
Banach lattices.   Write $U$ for their Riesz space product (352K), and
in $U$ set

\Centerline{$\|u\|_1=\sum_{i\in I}\|u(i)\|$,
\quad$V_1=\{u:\|u\|_1<\infty\}$,}

\Centerline{$\|u\|_{\infty}=\sup_{i\in I}\|u(i)\|$ (counting
$\sup\emptyset$ as $0$),
\quad$V_{\infty}=\{u:\|u\|_{\infty}<\infty\}$.}

\noindent Show that $V_1$, $V_{\infty}$ are solid linear subspaces of
$U$ and are Banach lattices under their norms $\|\,\|_1$,
$\|\,\|_{\infty}$.
%354A

\sqheader 354Xc Let $U$ be a Riesz space with a Riesz norm.   Show that
the maps $\wedge:U^2\to U$, $\vee:U^2\to U$
and $\med:U^3\to U$ are all uniformly continuous.
%354B used in 367D

\sqheader 354Xd Let $U$ be a Riesz space with a Riesz norm.   (i) Show
that any order-bounded set in $U$ is norm-bounded.   (ii) Show that in
$\BbbR^r$, with any of the standard Riesz norms (354Xa(i)),
norm-bounded sets are order-bounded.   (iii) Show that in
$\ell^1(\Bbb N)$ there is a
sequence converging to $0$ (for the norm) which is not order-bounded.
(iv) Show that in $\pmb{c}_0$ any sequence converging to $0$ is
order-bounded, but there is a norm-bounded set which is not
order-bounded.
%354B

\spheader 354Xe Let $U$ be a Riesz space with a Riesz norm.   Show that
it is a Banach lattice iff non-decreasing Cauchy sequences are
convergent.   \Hint{if $\|u_{n+1}-u_n\|\le 2^{-n}$ for every $n$, show
that $\sequencen{\sup_{i\le n}u_i}$ is Cauchy, and that
$\sequencen{u_n}$ converges to $\inf_{n\in\Bbb N}\sup_{m\ge n}u_m$.}
%354C

\spheader 354Xf Let $U$ be a Riesz space with a Riesz norm.   Show that
$U$ is a Banach lattice iff every non-decreasing Cauchy sequence
$\sequencen{u_n}$ in $U^+$ has a least upper bound $u$ with
$\|u\|=\lim_{n\to\infty}\|u_n\|$.
%354C, 354Xe

\spheader 354Xg Let $U$ be a Banach lattice.   Suppose that
$B\subseteq U$ is solid and $\sup_{n\in\Bbb N}u_n\in B$ whenever $\sequencen{u_n}$
is a non-decreasing sequence in $B$ with a supremum in $U$.   Show that
$B$ is closed.   \Hint{show first that $u\in B$ whenever there is a
sequence $\sequencen{u_n}$ in $B\cap U^+$ such that
$\|u-u_n\|\le 2^{-n}$ for every $n$;  do this by considering
$v_m=\inf_{n\ge m}u_n$.}
%354C

\spheader 354Xh Let $U$ be any Riesz space with a Riesz norm.   Show
that the Banach space completion of $U$ (3A5Jb) has a unique
partial ordering under which it is a Banach lattice.

\sqheader 354Xi Show that $\pmb{c}_0$ is a Banach lattice with an
order-continuous norm which does not have the Levi property.
%354D

\sqheader 354Xj Show that $\ell^{\infty}$, with $\|\,\|_{\infty}$, is a
Banach lattice with a Fatou norm which has the Levi property but is not
order-continuous.
%354D

\spheader 354Xk Let $U$ be a Riesz space with a Fatou norm.   Show that
if $V\subseteq U$ is a regularly embedded Riesz subspace then the induced norm on $V$ is a Fatou norm.
%354D

\spheader 354Xl Let $U$ be a Riesz space and $\|\,\|$ a Riesz norm on
$U$ which is order-continuous in the sense of 354Dc.   Show that its
restriction to $U^+$ is order-continuous in the sense of 313H.
%354D

\spheader 354Xm Let $U$ be a Riesz space with an order-continuous norm.
Show that if $V\subseteq U$ is a regularly embedded Riesz subspace then
the induced norm on $V$ is order-continuous.
%354D

\spheader 354Xn Let $U$ be a Dedekind $\sigma$-complete Riesz space with
a Fatou norm which has the Levi property.   Show that it is a Banach
lattice.   \Hint{354Xf.}
%354D

\spheader 354Xo Let $\langle U_i\rangle_{i\in I}$ be any family of
Banach lattices and let $V_1$, $V_{\infty}$ be the subspaces of
$U=\prod_{i\in I}U_i$ as described in 354Xb.   (i) Show that $V_1$,
$V_{\infty}$ have norms which are Fatou, or have the Levi property,
iff every $U_i$ has.   (ii) Show that the norm of $V_1$ is order-continuous
iff the norm of every $U_i$ is.    (iii) Show that $V_{\infty}$ is an
$M$-space iff every $U_i$ is.   (iv) Show that $V_1$ is an $L$-space iff
every $U_i$ is.
%354M

\spheader 354Xp Let $U$ be a Banach lattice with an order-continuous
norm.   (i) Show that a sublattice of
$U$ is norm-closed iff it is order-closed in the sense of 313Da.   (ii) Show
that a norm-closed Riesz subspace of $U$
is itself a Banach lattice with an order-continuous norm.
%354E

\sqheader 354Xq Let $U$ be an $M$-space and $V$ a norm-closed Riesz
subspace of $U$ containing the standard order unit of $U$.   (i) Show
that $V$, with the induced norm, is an $M$-space.   (ii) Deduce that the
space $\pmb{c}$ of convergent sequences is an $M$-space if given the
norm $\|\,\|_{\infty}$ inherited from $\ell^{\infty}$.
%354G

\spheader 354Xr Show that a Banach lattice $U$ is an $M$-space iff
($\alpha$) its norm is a Fatou norm with the Levi property ($\beta$)
$\|u\vee v\|=\max(\|u\|,\|v\|)$ for all $u$, $v\in U^+$.
%354J

\sqheader 354Xs Describe a topological space $X$ such that the space
$\pmb{c}$ of convergent sequences (354Xq) can be identified with $C(X)$.
%354K, 354Xq

\spheader 354Xt Let $D\subseteq\Bbb R$ be any non-empty set, and $V$ the
space of functions $f:D\to\Bbb R$ of bounded variation (\S224).   For
$f\in V$ set $\|f\|=\sup\{|f(t_0)|+\sum_{i=1}^n|f(t_i)-f(t_{i-1})|:
t_0\le t_1\le\ldots\le t_n$ in $D\}$ (224Yb).   Let $C$ be the set of
bounded non-decreasing functions from $D$ to $\coint{0,\infty}$.
Show that $C$ is the
positive cone of $V$ for a Riesz space ordering under which $V$ is an
$L$-space.
%354M

\leader{354Y}{Further exercises (a)}
%\spheader 354Ya
Let $U$ be a Riesz space with a Riesz norm, and $V$ a norm-dense Riesz
subspace of $U$.   Suppose that the induced norm on $V$ is Fatou, when
regarded as a norm on the Riesz space $V$.   Show (i) that $V$ is
order-dense in $U$ (ii) that the norm of $U$ is Fatou.   \Hint{for (i),
show
that if $u\in U^+$, $v_n\in V^+$ and $\|u-v_n\|\le 2^{-n-2}\|u\|$ for
every $n$, then $\|v_0-\inf_{i\le n}v_i\|\le\bover12\|u\|$ for every
$n$, so that $0$ cannot be $\inf_{n\in\Bbb N}v_n$ in $V$.}
%354D

\spheader 354Yb
Let $U$ be a Riesz space with a Riesz norm.   Show that the following
are equiveridical:  (i) $\lim_{n\to\infty}u_n=0$ whenever $\sequencen{u_n}$
is a disjoint order-bounded sequence in $U^+$ (ii)
$\lim_{n\to\infty}u_{n+1}-u_n=0$ for every order-bounded non-decreasing
sequence $\sequencen{u_n}$ in $U$ (iii) whenever $A\subseteq U^+$ is a
non-empty downwards-directed set in $U^+$ with infimum $0$,
$\inf_{u\in A}\sup_{v\in A,v\le u}\|u-v\|=0$.   \Hint{for
(i)$\Rightarrow$(ii), show by
induction that $\lim_{n\to\infty}u_n=0$ whenever $\sequencen{u_n}$ is an
order-bounded sequence such that, for some fixed $k\ge 1$,
$\inf_{i\in K}u_i=0$ for every $K\subseteq \Bbb N$ with $k$ members;
now show that if $\sequencen{u_n}$ is non-decreasing and $0\le u_n\le u$
for every $n$, then $\inf_{i\in K}(u_{i+1}-u_i-\bover1ku)^+=0$ whenever
$K\subseteq\Bbb N$ and $\#(K)=k\ge 1$.   For (iii)$\Rightarrow$(i), set
$A=\{u:\exists\,n,\,u\ge u_i\Forall i\ge n\}$.   See 
{\smc Fremlin 74a}, 24H.}
%354E

\spheader 354Yc Show that any Riesz space with an order-continuous norm
has the countable sup property (definition:  241Ye).
%354E, 354Yb

\spheader 354Yd Let $U$ be a Banach lattice.   Show that the following
are equiveridical:  (i) the norm on $U$ is order-continuous;  (ii) $U$
satisfies the conditions of 354Yb;   (iii) every order-bounded monotonic
sequence in $U$ is Cauchy.
%354E, 354Yb

\spheader 354Ye Let $U$ be a Riesz space with a Fatou norm.   Show that
the norm on $U$ is order-continuous iff it satisfies the conditions of
354Yb.
%354E, 354Yb

\spheader 354Yf For $f\in C([0,1])$, set $\|f\|_1=\int|f(x)|dx$.   Show
that $\|\,\|_1$ is a Riesz norm on $C([0,1])$ satisfying the conditions
of 354Yb, but is not order-continuous.
%354E, 354Yb

\spheader 354Yg Let $U$ be a Riesz space with a Riesz norm $\|\,\|$.
Show that $(U,\|\,\|)$ satisfies the conditions of 354Yb iff the norm of
its completion is order-continuous.
%354E, 354Yb

\spheader 354Yh Let $U$ be a Riesz space with a Riesz norm, and
$V\subseteq U$ a norm-dense Riesz subspace such that the induced norm on
$V$ is order-continuous.   Show that the norm of $U$ is
order-continuous.   \Hint{use 354Ya.}
%354E, 354Yb

\spheader 354Yi Let $U$ be an Archimedean Riesz space.   For any
$e\in U^+$, let $U_e$ be the solid linear subspace of $U$ generated by
$e$, so that $e$ is an order unit in $U_e$, and let $\|\,\|_e$ be the
corresponding order-unit norm on $U_e$.   We say that $U$ is {\bf
uniformly complete} if $U_e$ is complete under $\|\,\|_e$ for every
$e\in U^+$.   (i) Show that any Banach lattice is uniformly complete.
(ii) Show that any Dedekind $\sigma$-complete Riesz space is uniformly
complete (cf.\ 354Xn).   (iii) Show that if $U$ is a uniformly complete
Riesz space with a Riesz norm which has the Levi property, then $U$ is a
Banach lattice.   (iv) Show that if $U$ is a Banach lattice then a set
$A\subseteq U$ is closed, for the norm topology, iff $A\cap U_e$ is
$\|\,\|_e$-closed for every $e\in U^+$.   (v) Let $V$ be a solid linear
subspace of $U$.   Show that the quotient Riesz space $U/V$ is
Archimedean iff $V\cap U_e$ is $\|\,\|_e$-closed for every $e\in U^+$.
(vi) Show that if $U$ is uniformly complete and $V\subseteq U$ is a
solid linear subspace such that $U/V$ is Archimedean, then $U/V$ is
uniformly complete.   (vii) Show that $U$ is Dedekind $\sigma$-complete
iff it is uniformly complete and has the principal projection property
(353Xb).   %\Hint{for (vii), use 353Yc.}
%354H, 353Yc

\spheader 354Yj Let $U$ be an Archimedean Riesz space with an order unit,
endowed with its order-unit norm.   Let $Z$ be the unit ball
of $U^*$.   Show that for a linear functional $f:U\to\Bbb R$ the
following are equiveridical:  (i) $f$ is an {\bf extreme point} of $Z$,
that is, $f\in Z$ and $Z\setminus\{f\}$ is convex (ii) $|f(e)|=1$ and
one of $f$, $-f$ is a Riesz homomorphism.
%353N

\spheader 354Yk Let $U$ be a Banach lattice such that
$\|u+v\|=\|u\|+\|v\|$ whenever $u\wedge v=0$.   Show that $U$ is an
$L$-space.   \Hint{by 354Yd, the norm is order-continuous, so $U$ is
Dedekind complete.   If $u$, $v\ge 0$, set $e=u+v$, and represent $U_e$
as $C(X)$ where $X$ is extremally disconnected (353Yb);  now approximate
$u$ and $v$ by functions taking only finitely many values to show that
$\|u+v\|=\|u\|+\|v\|$.}
%354M

\spheader 354Yl Let $U$ be a uniformly complete Archimedean Riesz space
(354Yi).   Set $U_{\Bbb C}=U\times U$ with the complex linear structure
defined
by identifying $(u,v)\in U\times U$ as $u+iv\in U_{\Bbb C}$, so that
$u=\Real(u+iv)$, $v=\Imag(u+iv)$ and
$(\alpha+i\beta)(u+iv)=(\alpha u-\beta v)+i(\alpha v+\beta u)$.   (i)
Show that for $w\in U_{\Bbb C}$ we can define $|w|\in U$ by setting
$|w|=\sup_{|\zeta|=1}\Real(\zeta w)$.   (ii) Show that if $U$ is a
uniformly complete Riesz subspace of $\BbbR^X$ for some set $X$, then we
can identify $U_{\Bbb C}$ with the linear subspace of $\Bbb C^X$
generated by $U$.   (iii) Show that
$|w+w'|\le|w|+|w'|$, $|\gamma w|=|\gamma||w|$ for all $w\in U_{\Bbb C}$,
$\gamma\in\Bbb C$.   (iv) Show that if $w\in U_{\Bbb C}$ and
$|w|\le u_1+u_2$, where $u_1$, $u_2\in U^+$, then $w$ is expressible as
$w_1+w_2$ where $|w_j|\le u_j$ for both $j$.   \Hint{set $e=u_1+u_2$ and
represent $U_e$ as $C(X)$.}   (v) Show that if $U_0$ is a solid linear
subspace of $U$, then, for $w\in U_{\Bbb C}$, $|w|\in U_0$ iff $\Real
w$, $\Imag w$ both belong to $U_0$.   (vi) Show that if $U$ has a Riesz
norm then we have a norm on $U_{\Bbb C}$ defined by setting
$\|w\|=\||w|\|$, and that if $U$ is a Banach lattice then $U_{\Bbb C}$
is a (complex) Banach space.   (vii) Show
that if $U=L^p(\mu)$, where $(X,\Sigma,\mu)$ is a measure space and
$p\in[1,\infty]$, then $U_{\Bbb C}$ can be identified with
$L^p_{\Bbb C}(\mu)$ as
defined in 242P, 243K and 244P.   (We may call $U_{\Bbb C}$ the {\bf
complexification} of the Riesz space $U$.)

\spheader 354Ym Let $(X,\Sigma,\mu)$ be a measure space and $V$ a Banach
lattice.   Write $\eusm L^1_V$ for the space of Bochner integrable
functions from conegligible subsets of $X$ to $V$, and $L^1_V$ for the
corresponding set of equivalence classes (253Yf).   (i) Show that
$L^1_V$ is a Banach lattice under the ordering defined by saying that
$f^{\ssbullet}\le g^{\ssbullet}$ iff $f(x)\le g(x)$ in $V$ for
$\mu$-almost every $x\in X$.  (ii) Show that when $V=L^1(\nu)$, for some
other measure space $(Y,\Tau,\nu)$, then this ordering of $L^1_V$ agrees
with the ordering of $L^1(\lambda)$ where $\lambda$ is the (c.l.d.)
product measure on $X\times Y$ and we identify $L^1_V$ with
$L^1(\lambda)$, as in 253Yi.   (iii) Show that if $V$ has an
order-continuous norm, so has $L^1_V$.   \Hint{354Yd.}   (iv) Show
that if $\mu$ is Lebesgue measure on $[0,1]$ and $V=\ell^{\infty}$, then
$L^1_V$ is not Dedekind $\sigma$-complete.
}%end of exercises

\endnotes{
\Notesheader{354} Apart from some of the exercises, the material of this
section is pretty strictly confined to ideas which will be useful later
in this volume.   The basic Banach lattices of measure theory are the
$L^p$ spaces of Chapter 24;  these all have Fatou norms with the Levi
property (244Yf-244Yg), and for $p<\infty$ their norms are
order-continuous (244Ye).   In Chapter 36 I will return to these spaces
in a more abstract context.   Here I am mostly concerned to establish a
vocabulary in which their various properties, and the relationships
between these properties, can be expressed.

In normed Riesz spaces we have a very rich mixture of structures, and
must take particular care over the concepts of `boundedness',
`convergence' and `density', which have more than one possible
interpretation.
In particular, we must scrupulously distinguish between
`order-bounded' and `norm-bounded' sets.   I have not yet formally
introduced any of the various concepts of order-convergence (see \S367),
but I think that even so it is best to get into the habit of reminding
oneself, when a convergent sequence appears, that it is convergent for
the norm topology, rather than in any sense related directly to the
order structure.

I should perhaps warn you that for the study of
$M$-spaces 354L is not as helpful as it may look.   The trouble is that
apart from a few special cases (as in 354Xs) the topological space used
in the representation is actually more complicated and mysterious than
the $M$-space it is representing.

After the introduction of $M$-spaces, this section becomes a natural
place for `uniformly complete' spaces (354Yi).   For the moment I
leave these in the exercises.   But I mention them now because they
offer a straightforward route towards a theory of `complex Riesz
spaces' (354Yl).   In large parts of functional analysis it is natural,
and in some parts it is necessary, to work with normed spaces over
$\Bbb C$ rather than over $\Bbb R$, and for $L^2$ spaces in particular
it is useful to have a proper grasp of the complex case.   And while the
insights offered by the theory of Riesz spaces are not especially
important in such areas, I think we should always seek connexions
between different topics.   So it is worth remembering that uniformly
complete Riesz spaces have complexifications.

I shall have a great deal more to say about $L$-spaces when I come to
spaces of additive functionals (\S362) and to $L^1$ spaces again (\S365)
and to linear operators on them (\S371);  and before that, there will be
something in the next section on their duals, and on $L$-spaces which
are themselves dual spaces.   For the moment I just give some easy
results, direct translations of the corresponding facts in \S246, which
have natural expressions in the language of this section, holding deeper
ideas over.   In particular, the characterization of uniformly
integrable sets as relatively weakly compact sets (247C) is valid in
general $L$-spaces (356Q).

For an extensive treatment of Banach lattices, going very much deeper
than I have space for in this volume, see 
{\smc Lindenstrauss \& Tzafriri 79}.   
For a careful exposition of a great deal of useful
information, see {\smc Schaefer 74}.
}%end of comment

\discrpage

\frfilename{mt355.tex}
\versiondate{1.12.07}
\copyrightdate{1996}

\def\chaptername{Riesz spaces}
\def\sectionname{Spaces of linear operators}

\newsection{355}

We come now to a discussion of linear operators between Riesz spaces.
Linear operators are central to any kind of functional
analysis, and a feature of the theory of Riesz spaces is the way the
order structure picks out certain classes of operators for special
consideration.   Here I concentrate on positive and order-continuous
operators, with a brief mention of sequential order-continuity.   It
turns out, in fact, that we need to work with operators which are
differences of positive operators or of order-continuous positive
operators.   I define the basic spaces $\eurm L^{\sim}$, $\eurm
L^{\times}$ and $\eurm L^{\sim}_c$ (355A, 355G), with their most
important properties (355B, 355E, 355H-355I)
and some remarks on the special case of Banach lattices (355C, 355K).
At the same time I give an important theorem on extension of operators
(355F) and a corollary (355J).

The most important case is of course that in which the codomain is
$\Bbb R$, so that our operators become real-valued functionals;
I shall come to these in the next section.

\leader{355A}{Definition}  Let $U$ and $V$ be Riesz spaces.   A linear
operator $T:U\to V$ is
{\bf order-bounded} if $T[A]$ is order-bounded in $V$ for every
order-bounded $A\subseteq U$.

I will write $\eurm L^{\sim}(U;V)$ for the set of order-bounded linear
operators from $U$ to $V$.

\vleader{108pt}{355B}{Lemma} If $U$ and $V$ are Riesz spaces,

(a) a linear operator $T:U\to V$ is order-bounded iff
$\{Tu:0\le u\le w\}$ is bounded above in $V$ for every $w\in U^+$;

(b) in particular, any positive linear operator from $U$ to $V$ belongs
to $\eurm L^{\sim}=\eurm L^{\sim}(U;V)$;

(c) $\eurm L^{\sim}$ is a linear space;

(d) if $W$ is another Riesz space and $T:U\to V$ and $S:V\to W$ are
order-bounded linear operators, then $ST:U\to W$ is order-bounded.

\proof{{\bf (a)} This is elementary.   If $T\in\eurm L^{\sim}$ and
$w\in U^+$, $[0,w]$ is
order-bounded, so its image must be order-bounded in $V$, and in
particular bounded above.   On the other hand, if $T$ satisfies the
condition, and $A$ is order-bounded, then $A\subseteq[u_1,u_2]$ for some
$u_1\le u_2$, and

\Centerline{$T[A]\subseteq T[u_1+[0,u_2-u_1]]=Tu_1+T[[0,u_2-u_1]]$}

\noindent is bounded above;  similarly, $T[-A]$ is bounded above, so
$T[A]$ is bounded below;  as $A$ is arbitrary, $T$ is order-bounded.

\medskip

{\bf (b)} If $T$ is positive then $\{Tu:0\le u\le w\}$ is bounded above
by $Tw$ for every $w\ge 0$, so $T\in\eurm L^{\sim}$.

\medskip

{\bf (c)} If $T_1$, $T_2\in\eurm L^{\sim}$, $\alpha\in\Bbb R$ and
$A\subseteq U$ is order-bounded, then there are $v_1$, $v_2\in V$ such
that $T_i[A]\subseteq[-v_i,v_i]$ for both $i$.   Setting
$v=(1+|\alpha|)v_1+v_2$,
$(\alpha T_1+T_2)[A]\subseteq[-v,v]$;  as $A$ is arbitrary, $\alpha
T_1+T_2$ belongs to $\eurm L^{\sim}$;  as $\alpha$, $T_1$, $T_2$ are
arbitrary, and since the zero operator surely belongs to $\eurm
L^{\sim}$, $\eurm L^{\sim}$ is a linear subspace of the space of all
linear operators from $U$ to $V$.

\medskip

{\bf (d)} This is immediate from the definition;  if $A\subseteq U$ is
order-bounded, then $T[A]\subseteq V$ and $(ST)[A]=S[T[A]]\subseteq W$
are order-bounded.
}%end of proof of 355B

\leader{355C}{Theorem} If $U$ and $V$ are Banach lattices then every
order-bounded linear operator\cmmnt{ (in particular, every positive
linear operator)} from $U$ to $V$ is continuous.

\proof{ \Quer\ Suppose, if possible, that $T:U\to V$ is an order-bounded
linear operator which is not continuous.   Then for each $n\in\Bbb N$ we
can find a $u_n\in U$ such that $\|u_n\|\le 2^{-n}$ but $\|Tu_n\|\ge n$.
Now $u=\sup_{n\in\Bbb N}|u_n|$ is defined in $U$ (354C), and there is a
$v\in V$ such that $-v\le Tw\le v$ whenever $-u\le w\le u$;  but this
means that $\|v\|\ge\|Tu_n\|\ge n$ for every $n$, which is impossible.\
\Bang
}%end of proof of 355C

\leader{355D}{Lemma} Let $U$ be a Riesz space and $V$ any linear space
over $\Bbb R$.   Then a function $T:U^+\to V$ extends to a linear
operator from $U$ to $V$ iff

\Centerline{$T(u+u')=Tu+Tu'$,\quad $T(\alpha u)=\alpha Tu$}

\noindent for all $u$, $u'\in U^+$ and every $\alpha>0$, and in this
case the extension is unique.

\proof{ For in this case we can, and must, set

\Centerline{$T_1u=Tu_1-Tu_2$ whenever $u_1$, $u_2\in U^+$ and
$u=u_1-u_2$;}

\noindent it is elementary to check that this defines $T_1u$ uniquely
for every $u\in U$, and that $T_1$ is a linear operator extending $T$.
}%end of proof of 355D

\leader{355E}{Theorem}  Let $U$ be a Riesz space and $V$ a Dedekind
complete Riesz space.

(a) The space $\eurm L^{\sim}$ of
order-bounded linear operators from $U$ to $V$ is a Dedekind complete
Riesz space;  its positive cone is the set of positive linear operators
from $U$ to $V$.   In particular, every order-bounded linear operator
from $U$ to $V$ is expressible as the difference of positive linear
operators.


(b) For $T\in\eurm L^{\sim}$, $T^+$ and $|T|$ are defined in the Riesz
space $\eurm L^{\sim}$ by the formulae

\Centerline{$T^+(w)=\sup\{Tu:0\le u\le w\}$,}

\Centerline{$|T|(w)=\sup\{Tu:|u|\le
w\}=\sup\{\sum_{i=0}^n|Tu_i|:\sum_{i=0}^n|u_i|=w\}$}

\noindent for every $w\in U^+$.

(c) If $S$, $T\in\eurm L^{\sim}$ then

\Centerline{$(S\vee T)(w)=\sup_{0\le u\le w}Su+T(w-u)$,
\quad$(S\wedge T)(w)=\inf_{0\le u\le w}Su+T(w-u)$}

\noindent for every $w\in U^+$.

(d) Suppose that $A\subseteq \eurm L^{\sim}$ is non-empty and
upwards-directed.   Then $A$ is bounded above in $\eurm L^{\sim}$ iff
$\{Tu:T\in A\}$ is bounded above in $V$ for every $u\in U^+$, and in
this case $(\sup A)(u)=\sup_{T\in A}Tu$ for every $u\ge 0$.

(e) Suppose that $A\subseteq(\eurm L^{\sim})^+$ is non-empty and
downwards-directed.   Then $\inf A=0$ in $\eurm L^{\sim}$ iff
$\inf_{T\in A}Tu=0$ in $V$ for every $u\in U^+$.

\proof{{\bf (a)(i)} Suppose that $T\in\eurm L^{\sim}$.   For $w\in U^+$
set $R_T(w)=\sup\{Tu:0\le u\le w\}$;  this is defined because $V$ is
Dedekind complete and $\{Tu:0\le u\le w\}$ is bounded above in $V$.
Then $R_T(w_1+w_2)=R_Tw_1+R_Tw_2$ for all $w_1$, $w_2\in U^+$.   \Prf\
Setting $A_i=[0,w_i]$ for each $i$, $w=w_1+w_2$
and $A=[0,w]$, then of course
$A_1+A_2\subseteq A$;  but also $A\subseteq A_1+A_2$, because if
$u\in A$ then $u=(u\wedge w_1)+(u-w_1)^+$, and
$0\le(u-w_1)^+\le(w-w_1)^+=w_2$, so $u\in A_1+A_2$.   Consequently

$$\eqalign{R_Tw
&=\sup T[A]
=\sup T[A_1+A_2]=\sup(T[A_1]+T[A_2])\cr
&=\sup T[A_1]+\sup T[A_2]
=R_Tw_1+R_Tw_2\cr}$$

\noindent by 351Dc.\ \QeD\   Next, it is easy to see that
$R_T(\alpha w)=\alpha R_Tw$ for $w\in U^+$ and $\alpha>0$,
just because $u\mapsto\alpha u$, $v\mapsto\alpha v$ are isomorphisms of the partially ordered linear spaces $U$ and $V$.   It follows from 355D that we can
extend $R_T$ to a linear operator from $U$ to $V$.

Because $R_Tu\ge T0=0$ for every $u\in U^+$, $R_T$ is a positive linear
operator.   But also $R_Tu\ge Tu$ for every $u\in U^+$, so $R_T-T$ is
also positive, and $T=R_T-(R_T-T)$ is the difference of two positive
linear operators.

\medskip

\quad{\bf (ii)} This shows that every order-bounded operator is a
difference of positive operators.   But of course if $T_1$ and $T_2$ are
positive, then $(T_1-T_2)u\le T_1w$ whenever $0\le u\le w$ in $U$, so
that $T_1-T_2$ is order-bounded, by the criterion in 355Ba.   Thus
$\eurm
L^{\sim}$ is precisely the set of differences of positive operators.

\medskip

\quad{\bf (iii)} Just as in 351F, $\eurm L^{\sim}$ is a partially
ordered linear space if we say that $S\le T$ iff $Su\le Tu$ for every
$u\in U^+$.   Now it is a Riesz space.   \Prf\ Take any
$T\in\eurm L^{\sim}$.   Then $R_T$, as defined in (i), is an upper bound for $\{0,T\}$ in $\eurm L^{\sim}$.   If $S\in\eurm L^{\sim}$ is any other upper bound for
$\{0,T\}$, then for any $w\in U^+$ we must have
$Sw\ge Su\ge Tu$ whenever $u\in[0,w]$, so that $Sw\ge R_Tw$;  as $w$ is
arbitrary, $S\ge R_T$;  as $S$ is arbitrary, $R_T=\sup\{0,T\}$ in
$\eurm L^{\sim}$.   Thus $\sup\{0,T\}$ is defined in $\eurm L^{\sim}$ for every
$T\in\eurm L^{\sim}$;  by 352B, $\eurm L^{\sim}$ is a Riesz space.\ \Qed

(I defer the proof that it is Dedekind complete to (d-ii) below.)

\medskip

{\bf (b)} As remarked in (a-iii), $R_T=T^+$ for each
$T\in\eurm L^{\sim}$;  but this is just the formula given for $T^+$.   Now, if $T\in\eurm L^{\sim}$ and $w\in U^+$,

$$\eqalign{|T|(w)
&=2T^+w-Tw
=2\sup_{u\in[0,w]}Tu-Tw\cr
&=\sup_{u\in[0,w]}T(2u-w)
=\sup_{u\in[-w,w]}Tu,\cr}$$

\noindent which is the first formula offered for $|T|$.   In particular,
if $|u|\le w$ then $Tu$, $-Tu=T(-u)$ are both less than or equal to
$|T|(w)$, so that $|Tu|\le|T|(w)$.   So if $u_0,\ldots,u_n$ are such
that $\sum_{i=0}^n|u_i|=w$, then

\Centerline{$\sum_{i=0}^n|Tu_i|
\le\sum_{i=0}^n|T|(|u_i|)
=|T|(w)$.}

\noindent Thus $B=\{\sum_{i=0}^n|Tu_i|:\sum_{i=0}^n|u_i|=w\}$ is bounded
above by $|T|(w)$.   On the other hand, if $v$ is an upper bound for $B$
and $|u|\le w$, then

\Centerline{$Tu\le|Tu|+|T(w-|u|)|\le v$;}

\noindent as $u$ is arbitrary, $|T|(w)\le v$;  thus $|T|(w)$ is the
least upper bound for $B$.   This completes the proof of part (b) of
the theorem.

\medskip

{\bf (c)} We know that $S\vee T=T+(S-T)^+$ (352D), so that

$$\eqalign{(S\vee T)(w)
&=Tw+(S-T)^+(w)
=Tw+\sup_{0\le u\le w}(S-T)(u)\cr
&=\sup_{0\le u\le w}Tw+(S-T)(u)
=\sup_{0\le u\le w}Su+T(w-u)\cr}$$

\noindent for every $w\in U^+$, by the formula in (b).   Also from 352D
we have $S\wedge T=S+T-T\vee S$, so that

$$\eqalignno{(S\wedge T)(w)
&=Sw+Tw-\sup_{0\le u\le w}Tu+S(w-u)\cr
&=\inf_{0\le u\le w}Sw+Tw-Tu-S(w-u)\cr
\noalign{\noindent (351Db)}
&=\inf_{0\le u\le w}Su+T(w-u)\cr}$$

\noindent for $w\in U^+$.

\medskip

{\bf (d)(i)} Now suppose that $A\subseteq\eurm L^{\sim}$ is non-empty
and upwards-directed and that $\{Tu:T\in A\}$ is bounded above in $V$
for every $u\in U^+$.   In this case, because $V$ is Dedekind complete,
we may set $Ru=\sup_{T\in A}Tu$ for every $u\in U^+$.   Now
$R(u_1+u_2)=Ru_1+Ru_2$ for all $u_1$, $u_2\in U^+$.   \Prf\  Set
$B_i=\{Tu_i:T\in A\}$ for each $i$, $B=\{T(u_1+u_2):T\in A\}$.   Then
$B\subseteq B_1+B_2$, so


\Centerline{$R(u_1+u_2)
=\sup B\le\sup(B_1+B_2)=\sup B_1+\sup B_2=Ru_1+Ru_2$.}

\noindent On the other hand, if $v_i\in B_i$ for both $i$, there are
$T_i\in A$ such that $v_i=T_iu_i$ for each $i$;  because $A$ is
upwards-directed, there is a $T\in A$ such that $T\ge T_i$ for both $i$,
and now

\Centerline{$R(u_1+u_2)\ge T(u_1+u_2)=Tu_1+Tu_2\ge
T_1u_1+T_2u_2=v_1+v_2$.}

\noindent As $v_1$, $v_2$ are arbitrary,

\Centerline{$R(u_1+u_2)\ge\sup(B_1+B_2)=\sup B_1+\sup B_2=Ru_1+Ru_2$.
\Qed}

\noindent It is also easy to see that $R(\alpha u)=\alpha Ru$ for every
$u\in U^+$ and $\alpha>0$.   So, using 355D again, $R$ has an extension
to a linear operator from $U$ to $V$.

If we fix any $T_0\in A$, we have $T_0u\le Ru$ for every $u\in U^+$,
so $R-T_0$ is a positive linear operator, and $R=(R-T_0)+T_0$ belongs to
$\eurm L^{\sim}$.   Again, $Tu\le Ru$ for every $T\in A$ and $u\in U^+$, so
$R$ is an upper bound for $A$ in $\eurm L^{\sim}$;  and, finally, if $S$
is any upper bound for $A$ in $\eurm L^{\sim}$, then
$Su$ is an upper bound for $\{Tu:T\in A\}$, and must be greater than or
equal to $Ru$, for every $u\in U^+$;  so that $R\le S$ and $R=\sup A$ in
$\eurm L^{\sim}$.

\medskip

\quad{\bf (ii)} Consequently
$\eurm L^{\sim}$ is Dedekind complete.   \Prf\ If
$A\subseteq\eurm L^{\sim}$ is
non-empty and bounded above by $S$ say, then $A'=\{T_0\vee
T_1\vee\ldots\vee T_n:T_0,\ldots,T_n\in A\}$ is upwards-directed and
bounded above by $S$, so $\{Tu:T\in A'\}$ is bounded above by $Su$ for
every $u\in U^+$;  by (i) just above, $A'$ has a supremum in $\eurm
L^{\sim}$, which will also be the supremum of $A$.\ \Qed

\medskip

{\bf (e)} Suppose that $A\subseteq(\eurm L^{\sim})^+$ is non-empty and
downwards-directed.   Then $-A=\{-T:T\in A\}$ is non-empty and
upwards-directed, so

$$\eqalign{\inf A=0
&\iff\sup(-A)=0\cr
&\iff\sup_{T\in A}(-Tu)=0\text{ for every }u\in U^+\cr
&\iff\inf_{T\in A}Tu=0\text{ for every }u\in U^+.\cr}$$
}%end of proof of 355E

\leader{355F}{Theorem} Let $U$ and $V$ be Riesz spaces,
$U_0\subseteq U$ a Riesz subspace and $T_0:U_0\to V$ a
positive linear operator such that
$Su=\sup\{T_0w:w\in U_0,\,0\le w\le u\}$ is defined in $V$ for every
$u\in U^+$.   Suppose {\it either} that
$U_0$ is order-dense and that $T_0$ is order-continuous {\it or}
that $U_0$ is solid.

(a) There is a unique positive linear operator $T:U\to V$, extending $T_0$,
which agrees with $S$ on $U^+$.

(b) If $T_0$ is a Riesz homomorphism so is $T$.

(c) If $T_0$ is order-continuous so is $T$.

(d) If $U_0$ is order-dense and $T_0$ is an injective Riesz
homomorphism, then $T$ is injective.

(e) If $U_0$ is order-dense and $T_0$ is order-continuous then $T$ is
the only order-continuous positive linear operator from $U$ to $V$
extending $T_0$.

\proof{{\bf (a)(i)} (The key.) If $u$, $u'\in U^+$ then $S(u+u')=Su+Su'$.
\Prf\ If $w$, $w'\in U_0^+$, $w\le u$ and $w'\le u'$, then
$w+w'\le u+u'$, so

\Centerline{$T_0w+T_0w'=T_0(w+w')\le S(u+u')$;}

\noindent as $w$ and $w'$ are arbitrary, $Su+Su'\le S(u+u')$ (351Dc).   In
the other direction, suppose that $w\in U_0^+$ and $w\le u+u'$.

\medskip

\qquad{\bf case 1} Suppose that $U_0$ is solid.   Then $w\wedge u$ and
$(w-u)^+$ belong to $U_0$, while $w\wedge u\le u$ and
$(w-u)^+\le(u+u'-u)^+=u'$;  so

\Centerline{$T_0w=T_0(w\wedge u+(w-u)^+)=T_0(w\wedge u)+T_0(w-u)^+
\le Su+Su'$;}

\noindent as $w$ is arbitrary, $S(u+u')\le Su+Su'$ and we must have
equality.

\medskip

\qquad{\bf case 2} Suppose that $U_0$ is order-dense and $T_0$ is
order-continuous.   Set $A=\{v:v\in U_0^+$, $v\le w\wedge u\}$ and
$B=\{v:v\in U_0^+$, $v\le(w-u)^+\}$.   Then (taking the suprema in $U$)
$w\wedge u=\sup A$ and $(w-u)^+=\sup B$, because $U_0$ is order-dense;
by 351Dc again, $w=\sup(A+B)$ in $U$ and therefore $w=\sup(A+B)$ in $U_0$.
Also both $A$ and $B$ are upwards-directed, so $A+B$ also is.
Because $T_0$ is order-continuous,

\Centerline{$T_0w=\sup T_0[A+B]=\sup(T_0[A]+T_0[B])\le Su+Su'$.}

\noindent So once again we must have $S(u+u')\le Su+Su'$ and therefore
$S(u+u')=Su+Su'$.\ \Qed

\medskip

\quad{\bf (ii)} Of course $S(\alpha u)=\alpha Su$ whenever $u\in U^+$ and
$\alpha\ge 0$.   By 355D, $S$ has a unique extension to a linear operator
$T:U\to V$.   As $Tu=Su\ge 0$ whenever $u\ge 0$, $T$ is positive.   If
$u\in U_0^+$ then $Su=T_0u$, so $T$ extends $T_0$.

\medskip

{\bf (b)} Suppose that $T_0$ is a Riesz homomorphism.   If $u\wedge u'=0$
in $U$, then $w\wedge w'=0$ and $T_0w\wedge T_0w'=0$ whenever
$w\in U_0\cap[0,u]$ and $w'\in U_0\cap[0,u']$.   By 352Ea,
$Tu\wedge Tu'=Su\wedge Su'=0$ in $V$.   By 352G(iv), $T$ is a Riesz
homomorphism.

\medskip

{\bf (c)} Now suppose that $T_0$ is order-continuous.
Suppose that $B\subseteq U^+$ is non-empty and upwards-directed and has a
supremum $u_0\in U$.   Of course $Tu\le Tu_0$ for every $u\in B$, so $Tu_0$
is an upper bound for $T[B]$.   On the other hand, suppose that
$v$ is an upper bound for $T[B]$.   If $w\in U^+_0$ and
$u\in U^+$, $w\wedge u=\sup\{w':w'\in U_0$, $0\le w'\le w\wedge u\}$.
\Prf\ If $U_0$ is solid, $w\wedge u\in U_0$;  and otherwise $U_0$ is
order-dense.\ \QeD\  So if $w\in U_0$ and $0\le w\le u_0$,

\Centerline{$w=w\wedge\sup B=\sup_{u\in B}w\wedge u
=\sup_{u\in B}\sup(U_0\cap[0,w\wedge u])=\sup C$,}

\noindent where

\Centerline{$C=\{w':w'\in U_0^+$, $w'\le w\wedge u$ for some $u\in B\}$.}

\noindent Since $C$ is upwards-directed,

\Centerline{$T_0w=\sup T_0[C]\le v$.}

\noindent As $w$ is arbitrary, $Tu_0\le v$;  as $v$ is arbitrary,
$Tu_0=\sup T[B]$;  as $B$ is arbitrary, $T$ is order-continuous (351Ga).

\medskip

{\bf (d)} If $U_0$ is order-dense and $T_0$ is an injective
Riesz homomorphism, then for any non-zero $u\in U$ there is a non-zero
$w\in U_0$ such that $|w|\le|u|$;  so that

\Centerline{$|Tu|=T|u|\ge T_0|w|>0$}

\noindent because $T$ is a Riesz homomorphism, by (b).   As $u$ is
arbitrary, $T$ is injective.

\medskip

{\bf (e)} Finally, if $U_0$ is order-dense then any order-continuous
positive linear operator extending $T_0$ must agree with $S$ on $U^+$ and
is therefore equal to $T$.
}%end of proof of 355F

\vleader{48pt}{355G}{Definition} Let $U$ be a Riesz space and $V$ a Dedekind
complete Riesz space.   Then $\eurm L^{\times}(U;V)$ will be the set of
those $T\in\eurm L^{\sim}(U;V)$ expressible as the difference of
order-continuous positive linear operators, and $\eurm L^{\sim}_c(U;V)$
will be the set of those $T\in\eurm L^{\sim}(U;V)$ expressible as the
difference of sequentially order-continuous positive linear operators.

\cmmnt{Because a composition of (sequentially) order-continuous
functions is (sequentially) order-continuous, we shall have}

\Centerline{$ST\in\eurm L^{\times}(U;W)$ whenever
$S\in\eurm L^{\times}(V;W)$, $T\in\eurm L^{\times}(U;V)$,}

\Centerline{$ST\in\eurm L^{\sim}_c(U;W)$ whenever
$S\in\eurm L^{\sim}_c(V;W)$, $T\in\eurm L^{\sim}_c(U;V)$,}

\noindent for all Riesz spaces $U$ and all Dedekind complete Riesz
spaces $V$, $W$.

\leader{355H}{Theorem} Let $U$ be a Riesz space and $V$ a Dedekind
complete Riesz space.   Then

(i) $\eurm L^{\times}=\eurm L^{\times}(U;V)$ is a projection band in
$\eurm L^{\sim}=\eurm L^{\sim}(U;V)$,
therefore a Dedekind complete Riesz space in its own right;

(ii) a member $T$ of $\eurm L^{\sim}$ belongs to $\eurm L^{\times}$ iff
$|T|$ is order-continuous.

\proof{ There is a fair bit to check, but each individual step is easy
enough.

\medskip

{\bf (a)} Suppose that $S$, $T$ are order-continuous positive linear
operators from $U$ to $V$.   Then $S+T$ is order-continuous.   \Prf\ If
$A\subseteq U$ is non-empty, downwards-directed and has infimum $0$,
then for any $u_1$, $u_2\in A$ there is a $u\in A$ such that $u\le u_1$,
$u\le u_2$, and now
$(S+T)(u)\le Su_1+Tu_2$.   Consequently any lower bound for $(S+T)[A]$
must also be a lower bound for $S[A]+T[A]$.   But since

\Centerline{$\inf(S[A]+T[A])=\inf S[A]+\inf T[A]=0$}

\noindent (351Dc), $\inf (S+T)[A]$ must also be $0$;  as $A$ is
arbitrary, $S+T$ is order-continuous, by 351Ga.\ \Qed

\medskip

{\bf (b)} Consequently $S+T\in\eurm L^{\times}$ for all $S$, $T\in\eurm
L^{\times}$.   Since $-T$ and $\alpha T$ belong to $\eurm L^{\times}$
for every $T\in\eurm L^{\times}$ and $\alpha\ge 0$, we see that
$\eurm L^{\times}$ is a linear subspace of $\eurm L^{\sim}$.

\medskip

{\bf (c)} If $T:U\to V$ is an order-continuous linear operator,
$S:U\to V$ is linear and $0\le S\le T$, then $S$ is order-continuous.
\Prf\ If
$A\subseteq U$ is non-empty, downwards-directed and has infimum $0$,
then any lower bound of $S[A]$ must also be a lower bound of $T[A]$, so
$\inf S[A]=0$;  as $A$ is arbitrary, $S$ is order-continuous.\ \Qed

It follows that $\eurm L^{\times}$ is a solid linear subspace of
$\eurm L^{\sim}$.   \Prf\ If $T\in\eurm L^{\times}$ and $|S|\le|T|$ in
$\eurm L^{\sim}$, express $T$ as $T_1-T_2$ where $T_1$, $T_2$ are
order-continuous positive linear operators.   Then

\Centerline{$S^+$, $S^-\le|S|\le|T|\le T_1+T_2$,}

\noindent so $S^+$ and $S^-$ are order-continuous and
$S=S^+-S^-\in\eurm L^{\times}$.\ \Qed

Accordingly $\eurm L^{\times}$ is a Dedekind complete Riesz space in its
own right (353K(b-i)).

\wheader{355H}{4}{2}{2}{36pt}

{\bf (d)} The argument of (c) also shows that if $T\in\eurm L^{\times}$
then $|T|$ is order-continuous;  so that for $T\in\eurm L^{\sim}$,

\Centerline{$T\in\eurm L^{\times}\iff|T|\in\eurm L^{\times}\iff |T|$ is
order-continuous.}

\medskip

{\bf (e)} If $C\subseteq(\eurm L^{\times})^+$ is non-empty,
upwards-directed and has a supremum $T\in L^{\sim}$, then $T$ is
order-continuous, so belongs to $\eurm L^{\times}$.   \Prf\ Suppose that
$A\subseteq U^+$ is non-empty, upwards-directed and has supremum $w$.
Then

\Centerline{$Tw=\sup_{S\in C}Sw=\sup_{S\in C}\sup_{u\in A}Su
=\sup_{u\in A}Tu$,}

\noindent putting 355Ed and 351G(a-iii) together.   So (using 351Ga
again) $T$ is order-continuous.\ \QeD\  Consequently $\eurm L^{\times}$
is a band in $\eurm L^{\sim}$ (352Ob), and it is a projection band
because $\eurm L^{\sim}$ is Dedekind complete (353J).

This completes the proof.
}%end of proof of 355H

\leader{355I}{Theorem} Let $U$ be a Riesz space and $V$ a Dedekind
complete Riesz space.   Then $\eurm L^{\sim}_c(U;V)$ is a projection band in
$\eurm L^{\sim}(U;V)$, and a member $T$ of $\eurm L^{\sim}(U;V)$ belongs to
$\eurm L^{\sim}_c(U;V)$ iff $|T|$ is sequentially order-continuous.

\proof{ Copy the arguments of 355H.}

\leader{355J}{Proposition} Let $U$ be a Riesz space and $V$ a Dedekind
complete Riesz space.   Let $U_0\subseteq U$ be an order-dense Riesz
subspace;  then $T\mapsto T\restr U_0$ is an embedding of
$\eurm L^{\times}(U;V)$ as a solid linear subspace of
$\eurm L^{\times}(U_0;V)$.   In particular, any operator in
$\eurm L^{\times}(U_0;V)$
has at most one extension in $\eurm L^{\times}(U;V)$.

\proof{{\bf (a)} Because the embedding $U_0\embedsinto U$ is
positive and order-continuous (352Nb), $T\restr U_0$ is positive and
order-continuous whenever $T$ is;  so
$T\restr U_0\in\eurm L^{\times}(U_0;V)$ whenever
$T\in\eurm L^{\times}(U;V)$.   Because the map $T\mapsto T\restr U_0$ is
linear, the image $W$ of $\eurm L^{\times}(U;V)$ is a linear subspace of
$\eurm L^{\times}(U_0;V)$.

\medskip

{\bf (b)} If $T\in\eurm L^{\times}(U;V)$ and $T\restr U_0\ge 0$, then
$T\ge 0$.   \Prf\Quer\ Suppose, if possible, that there is a $u\in U^+$
such that $Tu\not\ge 0$.   Because $|T|\in\eurm L^{\times}(U;V)$ is
order-continuous and $A=\{v:v\in U_0,\,v\le u\}$ is an upwards-directed
set with supremum $u$, $\inf\{|T|(u-v):v\in A\}=0$ and there is a $v\in
A$ such that $Tu+|T|(u-v)\not\ge 0$.   But $Tv=Tu+T(v-u)\le Tu+|T|(u-v)$
so $Tv\not\ge 0$ and $T\restr U_0\not\ge 0$.\ \Bang\Qed

This shows that the map $T\mapsto T\restr U_0$ is an order-isomorphism
between $\eurm L^{\times}(U;V)$ and $W$, and in particular is injective.

\medskip

{\bf (c)} Now suppose that $S_0\in W$ and that $|S|\le|S_0|$ in
$\eurm L^{\times}(U_0;V)$.   Then $S\in W$.   \Prf\ Take
$T_0\in\eurm L^{\times}(U;V)$ such that $T_0\restr U_0=S_0$.
Then $S_1=|T_0|\restr U_0$ is a positive member of $W$ such that
$S_0\le S_1$ and $-S_0\le S_1$,
so $S^+\le S_1$.   Consequently, for any $u\in U^+$,

\Centerline{$\sup\{S^+v:v\in U_0,\,0\le v\le u\}
\le\sup\{S_1v:v\in U_0,\,0\le v\le u\}\le|T_0|(u)$}

\noindent is defined in $V$ (recall that we are assuming that $V$ is
Dedekind complete).   But this means that $S^+$ has an extension to an
order-continuous positive linear operator from $U$ to $V$ (355F), and
belongs to $W$.   Similarly, $S^-\in W$, so $S\in W$.\ \Qed

This shows that $W$ is a solid linear subspace of $\eurm
L^{\times}(U_0;V)$, as claimed.
}%end of proof of 355J

\leader{355K}{Proposition} Let $U$ be a Banach lattice with an
order-continuous norm.

(a) If $V$ is any Archimedean Riesz space and $T:U\to V$ is a positive
linear operator, then $T$ is order-continuous.

(b) If $V$ is a Dedekind complete Riesz space then $\eurm
L^{\times}(U;V)=\eurm L^{\sim}(U;V)$.

\proof{{\bf (a)} Suppose that $A\subseteq U^+$ is non-empty and
downwards-directed and has infimum $0$.   Then for each $n\in\Bbb N$
there is a $u_n\in A$ such that $\|u_n\|\le 4^{-n}$.   By 354C,
$u=\sup_{n\in\Bbb N}2^nu_n$ is defined in $U$.   Now $Tu_n\le 2^{-n}Tu$
for every $n$, so any lower bound for $T[A]$ must also be a lower bound
for $\{2^{-n}Tu:n\in\Bbb N\}$ and therefore (because $V$ is Archimedean)
less than or equal to $0$.   Thus $\inf T[A]=0$;  as $A$ is arbitrary,
$T$ is order-continuous.

\medskip

{\bf (b)} This is now immediate from 355Ea and the definition of
$\eurm L^{\times}$.
}%end of proof of 355K

\exercises{\leader{355X}{Basic exercises $\pmb{>}$(a)}
%\spheader 355Xa
Let $U$ and $V$ be arbitrary Riesz spaces.  (i) Show that the set $\eurm
L(U;V)$ of all linear operators from $U$ to $V$ is a partially ordered
linear space if we say that $S\le T$ whenever $Su\le Tu$ for every $u\in
U^+$.   (ii) Show that if $U$ and $V$ are Banach lattices then the set
of positive operators is closed in the normed space $\eurm B(U;V)$ of
bounded linear operators from $U$ to $V$.
%355C

\sqheader 355Xb  If $U$ is a Riesz space and $\|\,\|$, $\|\,\|'$ are two
norms on $U$ both rendering it a Banach lattice, show that they are
equivalent, that is, give rise to the same topology.
%355C

\spheader 355Xc Let $U$ be a Riesz space with a Riesz norm, $V$ an
Archimedean Riesz space with an order unit, and $T:U\to V$ a linear
operator which is continuous for the given norm on $U$ and the
order-unit norm on $V$.   Show that $T$ is order-bounded.
%355C

\spheader 355Xd Let $U$ be a Riesz space, $V$ an Archimedean Riesz
space, and $T:U^+\to V^+$ a map such that $T(u_1+u_2)=Tu_1+Tu_2$ for all
$u_1$, $u_2\in U^+$.   Show that $T$ has an extension to a linear
operator from $U$ to $V$.
%355D

\sqheader 355Xe Show that if $r$, $s\ge 1$ are integers then the Riesz
space $\eurm L^{\sim}(\BbbR^r;\BbbR^s)$ can be identified with the space
of real $s\times r$ matrices, saying that a matrix is positive iff every
coefficient is positive, so that if
$T=\langle\tau_{ij}\rangle_{1\le i\le s,1\le j\le r}$ then $|T|$, taken
in $\eurm L^{\sim}(\BbbR^r;\BbbR^s)$, is
$\langle|\tau_{ij}|\rangle_{1\le i\le s,1\le j\le r}$.   Show that a
positive matrix represents a Riesz
homomorphism iff each row has at most one non-zero coefficient.
%355E

\sqheader 355Xf Let $U$ be a Riesz space and $V$ a Dedekind complete
Riesz space.   Show that if $T_0,\ldots,T_n\in\eurm L^{\sim}(U;V)$ then

\Centerline{$(T_0\vee\ldots\vee T_n)(w)
=\sup\{\sum_{i=0}^nT_iu_i:u_i\ge 0\Forall i\le
n,\,\sum_{i=0}^nu_i=w\}$}

\noindent for every $w\in U^+$.
%355E

\sqheader 355Xg Let $U$ be a Riesz space, $V$ a Dedekind complete Riesz
space, and $A\subseteq\eurm L^{\sim}(U;V)$ a non-empty set.   Show that
$A$ is bounded above in $\eurm L^{\sim}(U;V)$ iff
$C_w=\{\sum_{i=0}^nT_iu_i:T_0,\ldots,T_n\in A,\,u_0,\ldots,u_n\in
U^+,\,\sum_{i=0}^nu_i=w\}$ is bounded above in $V$ for every $w\in U^+$,
and in this case $(\sup A)(w)=\sup C_w$ for every $w\in U^+$.
%355E

\leader{355Y}{Further exercises (a)}
%\spheader 355Ya
Let $U$ and $V$ be Banach lattices.   For
$T\in\eurm L^{\sim}=\eurm L^{\sim}(U;V)$, set

\Centerline{$\|T\|_{\sim}
=\sup_{w\in U^+,\|w\|\le 1}\inf\{\|v\|:|Tu|\le v$ whenever $|u|\le w\}$.}

\noindent Show that
$\|\,\|_{\sim}$ is a norm on $\eurm L^{\sim}$ under which $\eurm
L^{\sim}$ is a Banach space, and that the set of positive linear
operators is closed in $\eurm L^{\sim}$.
%355C

\spheader 355Yb Give an example of a continuous linear operator from
$\ell^2$ to itself which is not order-bounded.
%355C

\spheader 355Yc Let $U$ and $V$ be Riesz spaces and $T:U\to V$ a linear
operator.   (i) Show that for any $w\in U^+$,
$C_w=\{\sum_{i=0}^n|Tu_i|:u_0,\ldots,u_n\in U^+,\,\sum_{i=0}^nu_i=w\}$
is upwards-directed, and has the same upper bounds as $\{Tu:|u|\le w\}$.
\Hint{352Fd.}
(ii) Show that if $\sup C_w$ is defined for every
$w\in U^+$, then $S=T\vee(-T)$ is defined in the partially ordered
linear space $\eurm L^{\sim}(U;V)$ and $Sw=\sup C_w$ for every
$w\in U^+$.
%355E

\spheader 355Yd Let $U$, $V$ and $W$ be Riesz spaces, of which $V$ and
$W$ are Dedekind complete.   (i) Show that for any $S\in\eurm
L^{\times}(V;W)$, the map $T\mapsto ST:L^{\sim}(U;V)\to\eurm
L^{\sim}(U;W)$ belongs to $\eurm L^{\times}(\eurm L^{\sim}(U;V);\eurm
L^{\sim}(U;W))$, and is a Riesz homomorphism if $S$ is.   \Hint{355Yc.}
(ii) Show that for any
$T\in\eurm L^{\sim}(U;V)$, the map $S\mapsto ST:L^{\sim}(V;W)\to\eurm
L^{\sim}(U;W)$ belongs to $\eurm L^{\times}(L^{\sim}(V;W);\eurm
L^{\sim}(U;W))$.
%355E, 355Yc

\spheader 355Ye Let $\nu_{\Bbb N}$ be the usual measure on
$\{0,1\}^{\Bbb N}$ and
$\pmb{c}$ the Banach lattice of convergent sequences.   Find a
linear operator $T:L^2(\nu_{\Bbb N})\to\pmb{c}$ which is norm-continuous,
therefore
order-bounded, such that $0$ and $T$ have no common upper bound in the
partially ordered linear space
of all linear operators from $L^2(\nu_{\Bbb N})$ to $\pmb{c}$.
%355E (solution in mt35bits)

\spheader 355Yf Let $U$ and $V$ be Banach lattices.   Let
$\eurm L^{\text{reg}}$ be the linear space of operators from $U$ to $V$
expressible as the difference of positive operators.   For
$T\in\eurm L^{\text{reg}}$ let $\|T\|_{\text{reg}}$ be

\Centerline{$\inf\{\|T_1+T_2\|:T_1$, $T_2:U\to V$ are positive,
$T=T_1-T_2\}$.}

\noindent Show that $\|\,\|_{\text{reg}}$ is a norm under which
$\eurm L^{\text{reg}}$ is complete.
%355E

\spheader 355Yg Let $U$ and $V$ be Riesz spaces.   For this exercise
only, say that $\eurm L^{\times}(U;V)$ is to be the set
of linear operators $T:U\to V$ such that whenever $A\subseteq U$ is
non-empty,
downwards-directed and has infimum $0$ then $\{v:v\in
V^+,\,\exists\,w\in
A,\,|Tu|\le v$ whenever $|u|\le w\}$ has infimum $0$ in $V$.   (i) Show
that $\eurm L^{\times}(U;V)$ is a linear space.   (ii) Show that if $U$
is Archimedean then $\eurm L^{\times}(U;V)\subseteq\eurm L^{\sim}(U;V)$.
(iii) Show that if $U$ is Archimedean and $V$ is Dedekind complete then
this definition agrees with that of 355G.   (iv) Show that for any Riesz
spaces $U$, $V$ and $W$, $ST\in\eurm L^{\times}(U;W)$ for every
$S\in\eurm L^{\times}(V;W)$ and $T\in\eurm L^{\times}(U;V)$.   (v) Show
that if $U$ and $V$ are Banach lattices, then $\eurm L^{\times}(U;V)$ is
closed in $\eurm L^{\sim}(U;V)$ for the norm $\|\,\|_{\sim}$
of 355Ya.   (vi) Show that if $V$ is Archimedean and $U$ is a Banach
lattice with an order-continuous norm, then $\eurm L^{\times}(U;V)=\eurm
L^{\sim}(U;V)$.
%355G

\spheader 355Yh Let $U$ be a Riesz space and $V$ a Dedekind complete
Riesz space.   Show that the band projection
$P:\eurm L^{\sim}(U;V)\to\eurm L^{\times}(U;V)$ is given by the formula

$$\eqalign{(PT)(w)=\inf\{\sup_{u\in A}Tu:A\subseteq U^+
\text{ is non-empty},&\text{ upwards-directed}\cr
&\text{ and has supremum }w\}\cr}$$

\noindent for every $w\in U^+$, $T\in(\eurm L^{\sim}(U;V))^+$.   (Cf.\
362Bd.)
%355H mt35bits


\spheader 355Yi Show that if $U$ is a Riesz space with the countable sup
property
(241Ye), then $\eurm L^{\sim}_c(U;V)=\eurm L^{\times}(U;V)$ for every
Dedekind complete Riesz space $V$.

\spheader 355Yj Let $U$ and $V$ be Riesz spaces, of which $V$ is
Dedekind complete, and $U_0$ a solid linear subspace of $U$.   Show that
the map $T\mapsto T\restr U_0$ is an order-continuous Riesz homomorphism
from $\eurm L^{\times}(U;V)$ onto a solid linear subspace of
$\eurm L^{\times}(U_0;V)$.
%355J

\spheader 355Yk Let $U$ be a uniformly complete Riesz space (354Yi) and
$V$ a Dedekind complete Riesz space.   Let $U_{\Bbb C}$, $V_{\Bbb C}$ be
their complexifications (354Yl).   Show that the complexification of
$\eurm L^{\sim}(U;V)$ can be identified with the complex linear space of
linear operators $T:U_{\Bbb C}\to V_{\Bbb C}$ such that
$B_T(w)=\{|Tu|:|u|\le w\}$ is bounded above in $V$ for every $w\in U^+$,
and that now $|T|(w)=\sup B_T(w)$ for every
$T\in\eurm L^{\sim}(U;V)_{\Bbb C}$ and $w\in U^+$.
\Hint{if $u$, $v\in U$ and
$|u+iv|=w$, then $u$ and $v$ can be simultaneously approximated for the
order-unit norm $\|\,\|_w$ on the solid linear subspace generated by $w$
by finite sums $\sum_{j=0}^n(\cos\theta_j)w_j$,
$\sum_{j=0}^n(\sin\theta_j)w_j$ where $w_j\in U^+$, $\sum_{j=0}^nw_j=w$.
Consequently $|T(u+iv)|\le|T|(w)$ for every $T\in\eurm L^{\sim}_{\Bbb
C}$.}
%+
}%end of exercises

\cmmnt{
\Notesheader{355} I have had to make some choices in the basic
definitions of this chapter (355A, 355G).   For Dedekind complete
codomains $V$, there is no doubt what $\eurm L^{\sim}(U;V)$ should be,
since the order-bounded operators (in the sense of 355A) are just the
differences of positive operators (355Ea).   (These are sometimes called
`{\bf regular}' operators.)
When $V$ is not Dedekind complete, we have to choose between the two
notions, as not every order-bounded operator need be regular (355Ye).
In my previous book ({\smc Fremlin 74a}) I chose the regular operators;
I have still not encountered any really persuasive reason to settle
definitively on
either class.   In 355G the technical complications in dealing with any
natural equivalent of the larger space (see 355Yg) are such that I have
settled for the narrower class, but explicitly restricting the
definition to the case in which $V$ is Dedekind complete.   In the
applications in this book, the codomains are nearly always Dedekind
complete, so we can pass these questions by.

The elementary extension technique in 355D may recall the definition of
the Lebesgue integral (122L-122M).   In the same way, 351G may remind
you of the theorem that a linear operator between normed spaces is
continuous everywhere if it is continuous anywhere, or of the
corresponding results
about Boolean homomorphisms and additive functionals on Boolean algebras
(313L, 326Ka, 326R).

Of course 355Ea is the central fact about the space
$\eurm L^{\sim}(U;V)$
for Dedekind complete $V$;  because we get a new Riesz space from old
ones, the prospect of indefinite recursion immediately presents itself.
For Banach lattices, $\eurm L^{\sim}(U;V)$ is a linear subspace of the
space $\eurm B(U;V)$ of bounded linear operators (355C);  the question
of when the two are equal will be of great importance to us.   I give
only the vaguest hints on how to show that they can be different (355Yb,
355Ye), but these should be enough to make it plain that equality is the
exception rather than the rule.   It is also very useful that we have
effective formulae to describe the Riesz space operations on
$\eurm L^{\sim}(U;V)$ (355E, 355Xf-355Xg, 355Yc).   You may wish to compare
these with the corresponding formulae for additive functionals on
Boolean algebras in 326Yd and 362B.

If we think of $\eurm L^{\sim}$ as somehow corresponding to the space of
bounded additive functionals on a Boolean algebra, the bands
$\eurm L^{\sim}_c$ and $\eurm L^{\times}$ correspond to the spaces of
countably
additive and completely additive functionals.   In fact (as will appear
in \S362) this correspondence is very close indeed.   For the moment,
all I have sought to establish is that $\eurm L^{\sim}_c$ and
$\eurm L^{\times}$ are indeed bands.   Of course any case in which
$\eurm L^{\sim}(U;V)=\eurm L^{\sim}_c(U;V)$ or
$\eurm L^{\sim}_c(U;V)=\eurm L^{\times}(U;V)$ is of interest
(355Kb, 355Yi).

Between Banach lattices, positive linear operators are continuous
(355C);  it follows at once that the Riesz space structure determines
the topology (355Xb), so that it is not to be wondered at that there are
further connexions between the norm and the spaces $\eurm L^{\sim}$ and
$\eurm L^{\times}$, as in 355K.

355F will be a basic tool in the theory of representations of Riesz
spaces;  if we can represent an order-dense Riesz subspace of $U$ as a
subspace of a Dedekind complete space $V$, we have at least some chance
of expressing $U$ also as a subspace of $V$.   Of course it has other
applications, starting with analysis of the dual spaces.
}%end of comment

\discrpage

\frfilename{mt356.tex}
\versiondate{5.10.04}
\copyrightdate{1996}

\def\chaptername{Riesz spaces}
\def\sectionname{Dual spaces}

\newsection{356}

As always in functional analysis, large parts of the theory of Riesz
spaces are based on the study of linear functionals.   Following the
scheme of the last section, I define spaces $U^{\sim}$, $U^{\sim}_c$ and
$U^{\times}$, the `order-bounded', `sequentially order-continuous'
and `order-continuous' duals of a Riesz space $U$ (356A).   These are
Dedekind complete Riesz spaces (356B).   If $U$ carries a Riesz norm
they are closely connected with the normed space dual $U^*$, which is
itself a Banach lattice (356D).   For each of them, we have a canonical
Riesz homomorphism from $U$ to the corresponding bidual.   The map from
$U$ to $U^{\times\times}$ is particularly
important (356I);   when this map is an isomorphism we call $U$
`perfect' (356J).   The last third of the section deals with $L$- and
$M$-spaces and the duality between them (356N, 356P), with two important
theorems on uniform integrability (356O, 356Q).

\leader{356A}{Definition} Let $U$ be a Riesz space.

\spheader 356Aa I write $U^{\sim}$ for the space
$\eusm L^{\sim}(U;\Bbb R)$ of order-bounded real-valued linear functionals on
$U$, the {\bf order-bounded dual} of $U$.

\spheader 356Ab $U^{\sim}_c$ will be the space $\eusm L^{\sim}_c(U;\Bbb
R)$ of differences of sequentially order-continuous positive real-valued
linear functionals on $U$, the {\bf sequentially order-continuous dual}
of $U$.

\spheader 356Ac $U^{\times}$ will be the space $\eusm L^{\times}(U;\Bbb
R)$ of differences of order-continuous positive real-valued linear
functionals on $U$, the {\bf order-continuous dual} of $U$.

\cmmnt{\medskip

\noindent{\bf Remark} It is easy to check that the three spaces
$U^{\sim}$, $U^{\sim}_c$ and $U^{\times}$ are in general different
(356Xa-356Xc).   But the examples there leave open the question:  can we
find a Riesz space $U$, for which $U^{\sim}_c\ne U^{\times}$, and which
is actually Dedekind complete, rather than just Dedekind
$\sigma$-complete, as in 356Xc?   This leads to unexpectedly deep water;
it is yet another form of the Banach-Ulam problem.   Really this is a
question for Volume 5, but in 363S below I collect the relevant ideas
which are within the scope of the present volume.
}%end of comment

\leader{356B}{Theorem} For any Riesz space $U$,\cmmnt{ its
order-bounded dual} $U^{\sim}$ is a Dedekind
complete Riesz space in which $U^{\sim}_c$ and $U^{\times}$ are
projection bands,
therefore Dedekind complete Riesz spaces in their own right.   For
$f\in U^{\sim}$, $f^+$ and $|f|\in U^{\sim}$ are defined by the formulae

\Centerline{$f^+(w)=\sup\{f(u):0\le u\le w\}$,
\quad$|f|(w)=\sup\{f(u):|u|\le w\}$}

\noindent for every $w\in U^+$.   A non-empty upwards-directed set
$A\subseteq U^{\sim}$ is bounded above iff $\sup_{f\in A}f(u)$ is finite
for every $u\in U$, and in this case $(\sup A)(u)=\sup_{f\in A}f(u)$ for
every $u\in U^+$.

\proof{ 355E, 355H, 355I.
}%end of proof of 356B

\leader{356C}{Proposition} Let $U$ be any Riesz space and $P$ a band
projection on $U$.   Then its adjoint $P':U^{\sim}\to U^{\sim}$, defined
by setting $P'(f)=fP$ for every $f\in U^{\sim}$, is a band projection on
$U^{\sim}$.

\proof{ Because $P:U\to U$ is a positive linear operator,
$P'f\in U^{\sim}$ for every $f\in U^{\sim}$ (355Bd), and $P'$ is a
positive linear operator from $U^{\sim}$ to itself.   Set $Q=I-P$, the
complementary band projection on $U$;  then $Q'$ is another positive
linear operator on $U^{\sim}$, and $P'f+Q'f=f$ for every $f$.   Now
$P'f\wedge Q'f=0$ for every $f\ge 0$.   \Prf\ For any $w\in U^+$,

$$\eqalignno{(P'f-Q'f)^+(w)
&=\sup_{0\le u\le w}(P'f-Q'f)(u)\cr
&=\sup_{0\le u\le w}f(Pu-Qu)
=f(Pw)\cr
\displaycause{because $Pu-Qu\le Pu\le Pw=P(Pw)-Q(Pw)$ whenever
$0\le u\le w$}
&=(P'f)(w),\cr}$$

\noindent so $(P'f-Q'f)^+=P'f$, that is, $P'f\wedge Q'f=0$.\ \QeD\  By
352Rd, $P'$ is a band projection.
}%end of proof of 356C

\leader{356D}{Proposition} Let $U$ be a Riesz space with a Riesz norm.

(a) The normed space dual $U^*$ of $U$ is a solid linear subspace of
$U^{\sim}$, and in itself is a Banach lattice with a Fatou norm and has
the Levi property.

(b) The norm of $U$ is order-continuous iff $U^*\subseteq U^{\times}$.

(c) If $U$ is a Banach lattice, then $U^*=U^{\sim}$, so that $U^{\sim}$,
$U^{\times}$ and $U^{\sim}_c$ are all Banach lattices.

(d) If $U$ is a Banach lattice with order-continuous norm then
$U^*=U^{\times}=U^{\sim}$.

\proof{{\bf (a)(i)} If $f\in U^*$ then

\Centerline{$\sup_{|u|\le w}f(u)\le\sup_{|u|\le w}\|f\|\|u\|
=\|f\|\|w\|<\infty$}

\noindent for every $w\in U^+$, so $f\in U^{\sim}$ (355Ba).   Thus
$U^*\subseteq U^{\sim}$.

\medskip

\quad{\bf (ii)} If $f\in U^{\sim}$, $g\in U^*$ and $|f|\le|g|$, then
for any $w\in U$

\Centerline{$|f(w)|\le|f|(|w|)\le|g|(|w|)=\sup_{|u|\le|w|}g(u)
\le\sup_{|u|\le|w|}\|g\|\|u\|\le\|g\|\|w\|$.}

\noindent As $w$ is arbitrary, $f\in U^*$ and $\|f\|\le\|g\|$;  as $f$
and $g$ are arbitrary, $U^*$ is a solid linear subspace of $U^{\sim}$
and the norm of $U^*$ is a Riesz norm.   Because $U^*$ is a Banach space
it is also a Banach lattice.

\medskip

\quad{\bf (iii)} If $A\subseteq(U^*)^+$ is non-empty and upwards-directed
and $M=\sup_{f\in A}\|f\|$ is finite, then $\sup_{f\in A}f(u)\le M\|u\|$
is finite for every $u\in U^+$, so $g=\sup A$ is defined in $U^{\sim}$
(355Ed).   Now $g(u)=\sup_{f\in A}f(u)$ for every $u\in U^+$, as also
noted in 355Ed, so

\Centerline{$|g(u)|\le g(|u|)\le M\||u|\|=M\|u\|$}

\noindent for every $u\in U$, and $\|g\|\le M$.   But as $A$ is
arbitrary, this proves simultaneously that the norm of $U^{\sim}$ is
Fatou and has the Levi property.

\medskip

{\bf (b)(i)} Suppose that the norm of $U$ is order-continuous.
If $f\in U^*$
and $A\subseteq U$ is a non-empty downwards-directed set with infimum
$0$, then

\Centerline{$\inf_{u\in A}|f|(u)\le\inf_{u\in A}\|f\|\|u\|=0$,}

\noindent so $|f|\in U^{\times}$ and $f\in U^{\times}$.   Thus
$U^*\subseteq U^{\times}$.

\medskip

\quad{\bf (ii)} Now suppose that the norm is not order-continuous.
Then there is a non-empty downwards-directed set $A\subseteq U$, with
infimum $0$, such that $\inf_{u\in A}\|u\|=\delta>0$.   Set

\Centerline{$B=\{v:v\ge u$ for some $u\in A\}$.}

\noindent Then $B$ is convex.   \Prf\ If $v_1$, $v_2\in B$ and
$\alpha\in[0,1]$, there are $u_1$, $u_2\in A$ such that $v_i\ge u_i$ for
both $i$;  now there is a $u\in A$ such that $u\le u_1\wedge u_2$, so
that

\Centerline{$u=\alpha u+(1-\alpha)u\le\alpha v_1+(1-\alpha)v_2$,}

\noindent and $\alpha v_1+(1-\alpha)v_2\in B$.\ \QeD\  Also
$\inf_{v\in B}\|v\|=\delta>0$.   By the Hahn-Banach theorem (3A5Cb), there is an
$f\in U^*$ such that $\inf_{v\in B}f(v)>0$.   But now

\Centerline{$\inf_{u\in A}|f|(u)\ge\inf_{u\in A}f(u)>0$}

\noindent and $|f|$ is not order-continuous;  so $U^*\not\subseteq
U^{\times}$.

\medskip

{\bf (c)}  By 355C, $U^{\sim}\subseteq U^*$, so $U^{\sim}=U^*$.   Now
$U^{\times}$ and $U^{\sim}_c$, being bands, are closed linear subspaces
(354Bd), so are Banach lattices in their own right.

\medskip

{\bf (d)} Put (b) and (c) together.
}%end of proof of 356D


\leader{356E}{\dvrocolon{Biduals}}\cmmnt{ If you have studied any
functional analysis at all, it will come as no surprise that
duals-of-duals are important in the theory of Riesz spaces.   I start
with a simple lemma.

\medskip

\noindent}{\bf Lemma} Let $U$ be a Riesz space and $f:U\to\Bbb R$ a
positive
linear functional.   Then for any $u\in U^+$ there is a positive linear
functional $g:U\to\Bbb R$ such that $0\le g\le f$, $g(u)=f(u)$ and
$g(v)=0$ whenever $u\wedge v=0$.

\proof{ Set $g(v)=\sup_{\alpha\ge 0}f(v\wedge\alpha u)$ for every $v\in
U^+$.   Then it is easy to see that $g(\beta v)=\beta g(v)$ for every
$v\in U^+$, $\beta\in\coint{0,\infty}$.   If $v$, $w\in U^+$ then

\Centerline{$(v\wedge\alpha u)+(w\wedge\alpha u)
\le(v+w)\wedge 2\alpha u
\le (v\wedge 2\alpha u)+(w\wedge 2\alpha u)$}

\noindent for every $\alpha\ge 0$ (352Fa), so $g(v+w)=g(v)+g(w)$.
Accordingly $g$ has an extension to a linear functional from $U$ to
$\Bbb R$ (355D).   Of course $0\le g(v)\le f(v)$ for $v\ge 0$, so
$0\le g\le f$ in $U^{\sim}$.   We have $g(u)=f(u)$, while if $u\wedge v=0$
then $\alpha u\wedge v=0$ for every $\alpha\ge 0$, so $g(v)=0$.
}%end of proof of 356E

\leader{356F}{Theorem} Let $U$ be a Riesz space and $V$ a solid linear
subspace of $U^{\sim}$.   For $u\in U$ define $\hat u:V\to\Bbb R$ by
setting $\hat u(f)=f(u)$ for every $f\in V$.   Then $u\mapsto\hat u$ is
a Riesz homomorphism from $U$ to $V^{\times}$.

\proof{{\bf (a)} By the definition of addition and scalar multiplication
in $V$, $\hat u$ is linear for every $u$;  also
$\widehat{\alpha u}=\alpha\hat u$ and $(u_1+u_2)\sphat=\hat u_1+\hat u_2$
for all $u$, $u_1$,
$u_2\in U$ and $\alpha\in\Bbb R$.   If $u\ge 0$ then
$\hat u(f)=f(u)\ge  0$ for every $f\in V^+$, so $\hat u\ge 0$;
accordingly every $\hat u$
is the difference of two positive functionals, and $u\mapsto\hat u$ is a
linear operator from $U$ to $V^{\sim}$.

\medskip

{\bf (b)} If $B\subseteq V$ is a non-empty downwards-directed set with
infimum $0$, then $\inf_{f\in B}f(u)=0$ for every $u\in U^+$, by
355Ee.   But this means that $\hat u$ is order-continuous for every
$u\in U^+$, so that $\hat u\in V^{\times}$ for every $u\in U$.

\medskip

{\bf (c)} If $u\wedge v=0$ in $U$, then for any $f\in V^+$ there is a
$g\in[0,f]$ such that $g(u)=f(u)$ and $g(v)=0$ (356E).   So

\Centerline{$(\hat u\wedge\hat v)(f)\le\hat u(f-g)+\hat v(g)
=f(u)-g(u)+g(v)=0$.}

\noindent As $f$ is arbitrary, $\hat u\wedge\hat v=0$.   As $u$ and $v$
are arbitrary, $u\mapsto\hat u$ is a Riesz homomorphism (352G).
}%end of proof of 356F

\leader{356G}{Lemma} Suppose that $U$ is a Riesz space such that
$U^{\sim}$ separates the points of $U$.   Then $U$ is Archimedean.

\proof{ \Quer\ Otherwise, there are $u$, $v\in U$ such that $v>0$ and
$nv\le u$ for every $n\in\Bbb N$.   Now there is an $f\in U^{\sim}$ such
that $f(v)\ne 0$;  but $|f(v)|\le|f|(v)\le\bover1n|f|(u)$ for every $n$,
so this is impossible.\ \Bang
}%end of proof of 356G

\leader{356H}{Lemma} Let $U$ be an Archimedean Riesz space and $f>0$ in
$U^{\times}$.   Then there is a $u\in U$ such that (i) $u>0$ (ii)
$f(v)>0$ whenever $0<v\le u$ (iii) $g(u)=0$ whenever $g\wedge f=0$ in
$U^{\times}$.   Moreover, if $u_0\in U^+$ is such that $f(u_0)>0$, we
can arrange that $u\le u_0$.

\proof{{\bf (a)} Because $f>0$ there is certainly some $u_0\in U$ such that
$f(u_0)>0$.   Set $A=\{v:0\le v\le u_0,\,f(v)=0\}$.   Then
$(v_1+v_2)\wedge u_0\in A$ for all $v_1$, $v_2\in A$, so $A$ is
upwards-directed.   Because $f(u_0)>0=\sup f[A]$ and $f$ is
order-continuous,
$u_0$ cannot be the least upper bound of $A$, and there is another upper
bound $u_1$ of $A$ strictly less than $u_0$.

Set $u=u_0-u_1>0$.   If $0\le v\le u$ and $f(v)=0$, then

\Centerline{$w\in A\Longrightarrow w\le u_1\Longrightarrow w+v\le u_0
\Longrightarrow w+v\in A$;}

\noindent consequently $nv\in A$ and $nv\le u_0$ for every $n\in\Bbb N$,
so $v=0$.   Thus $u$ has properties (i) and (ii).

\medskip

{\bf (b)} Now suppose that $g\wedge f=0$ in $U^{\times}$.   Let
$\epsilon>0$.   Then for each $n\in\Bbb N$ there is a $v_n\in[0,u]$ such
that $f(v_n)+g(u-v_n)\le 2^{-n}\epsilon$ (355Ec).   If $v\le v_n$ for
every $n\in\Bbb N$ then $f(v)=0$ so $v=0$;  thus $\inf_{n\in\Bbb
N}v_n=0$.
Set $w_n=\inf_{i\le n}v_i$ for each $n\in\Bbb N$;  then
$\sequencen{w_n}$ is non-increasing and has infimum $0$ so (because $g$
is order-continuous) $\inf_{n\in\Bbb N}g(w_n)=0$.   But

\Centerline{$u-w_n=\sup_{i\le n}u-v_i\le\sum_{i=0}^nu-v_i$,}

\noindent so

\Centerline{$g(u-w_n)\le\sum_{i=0}^ng(u-v_i)\le 2\epsilon$}

\noindent for every $n$, and

\Centerline{$g(u)\le 2\epsilon+\inf_{n\in\Bbb N}g(w_n)=2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $g(u)=0$;  as $g$ is arbitrary,
$u$ has the third required property.
}%end of proof of 356H

\leader{356I}{Theorem} Let $U$ be any Archimedean Riesz space.   Then
the canonical map from $U$ to $U^{\times\times}$\cmmnt{ (356F)} is an
order-continuous Riesz homomorphism from $U$ onto an order-dense Riesz
subspace of $U^{\times\times}$.   If $U$ is Dedekind complete, its image
in $U^{\times\times}$ is solid.

\proof{{\bf (a)} By 356F, $u\mapsto\hat u:U\to U^{\times\times}$ is a
Riesz homomorphism.

To see that it is order-continuous, take any non-empty
downwards-directed set $A\subseteq U$ with infimum $0$.   Then
$C=\{\hat u:u\in A\}$ is downwards-directed, and for any
$f\in(U^{\times})^+$

\Centerline{$\inf_{\phi\in C}\phi(f)=\inf_{u\in A}f(u)=0$}

\noindent because $f$ is order-continuous.   As $f$ is arbitrary,
$\inf C=0$ (355Ee);  as $A$ is arbitrary, $u\mapsto\hat u$ is
order-continuous (351Ga).

\medskip

{\bf (b)} Now suppose that $\phi>0$ in $U^{\times\times}$.   By 356H,
there is an $f>0$ in $U^{\times}$ such that $\phi(f)>0$ and $\phi(g)=0$
whenever $g\wedge f=0$.   Next, there is a $u>0$ in $U$ such that
$f(u)>0$.   Since $u\ge 0$, $\hat u\ge 0$;  since $\hat u(f)>0$,
$\hat u\wedge\phi>0$.

Because $U^{\times\times}$ (being Dedekind complete) is Archimedean,
$\inf_{\alpha>0}\alpha\hat u=0$, and there is an $\alpha>0$ such that

\Centerline{$\psi=(\hat u\wedge\phi-\alpha\hat u)^+>0$.}

\noindent Let $g\in (U^{\times})^+$ be such that $\psi(g)>0$
%and $\psi(h)>0$ whenever $0<h\le g$
and $\theta(g)=0$ whenever $\theta\wedge\psi =0$ in $U^{\times\times}$.
Let $v\in U^+$ be such that $g(v)>0$
%and $g(w)>0$ whenever $0<w\le v$
and $h(v)=0$ whenever $h\wedge g=0$ in $U^{\times}$.

Because $\hat v(g)=g(v)>0$, $\hat v\wedge\psi>0$.   As $\psi\le\hat u$,
$\hat v\wedge\hat u>0$ and $\hat v\wedge\alpha\hat u>0$.   Set
$w=v\wedge\alpha u$;  then $\hat w=\hat v\wedge\alpha\hat u$, by
356F, so $\hat w>0$.

\Quer\ Suppose, if possible, that $\hat w\not\le\phi$.   Then
$\theta=(\hat w-\phi)^+>0$, so there is an $h\in(U^{\times})^+$ such
that $\theta(h)>0$ and $\theta(h')>0$ whenever $0<h'\le h$ (356H, for
the fourth and last time).   Now examine

$$\eqalignno{\theta(h\wedge g)
&\le(\alpha\hat u-\phi\wedge\hat u)^+(g)\cr
\noalign{\noindent (because $\hat w\le\alpha\hat u$, $\phi\wedge\hat
u\le\phi$, $h\wedge g\le g$)}
&=0\cr}$$

\noindent because
$(\alpha\hat u-\phi\wedge\hat u)^+\wedge\psi=0$.
So $h\wedge g=0$ and $h(v)=0$.   But this means that

\Centerline{$\theta(h)\le\hat w(h)\le\hat v(h)=0$,}

\noindent which is impossible.\ \Bang

Thus $0<\hat w\le\phi$.  As $\phi$ is arbitrary, the image $\hat U$ of
$U$ is quasi-order-dense in $U^{\times\times}$, therefore order-dense
(353A).

\medskip

{\bf (c)} Now suppose that $U$ is Dedekind complete and that
$0\le\phi\le\psi\in\hat U$.   Express $\psi$ as $\hat u$ where $u\in U$,
and set $A=\{v:v\in U,\,v\le u^+,\,\hat v\le\phi\}$.   If $v\in U$ and
$0\le\hat v\le\phi$, then $w=v^+\wedge u^+\in A$ and $\hat w=\hat v$;
thus $\phi=\sup\{\hat v:v\in A\}=\hat v_0$, where $v_0=\sup A$.   So
$\phi\in\hat U$.   As $\phi$ and $\psi$ are arbitrary, $\hat U$ is solid
in $U^{\times\times}$.
}%end of proof of 356I

\leader{356J}{Definition} A Riesz space $U$ is {\bf perfect} if the
canonical map from $U$ to $U^{\times\times}$ is an isomorphism.

\leader{356K}{Proposition} A Riesz space $U$ is perfect iff (i) it is
Dedekind complete (ii) $U^{\times}$ separates the points of $U$ (iii)
whenever $A\subseteq U$ is non-empty and upwards-directed and
$\{f(u):u\in A\}$ is bounded for every $f\in U^{\times}$, then $A$ is
bounded above in $U$.

\proof{{\bf (a)} Suppose that $U$ is perfect.   Because it is isomorphic
to $U^{\times\times}$, which is surely Dedekind complete, $U$ also is
Dedekind complete.   Because the map $u\mapsto\hat u:U\to
U^{\times\times}$ is injective, $U^{\times}$ separates the points of
$U$.   If $A\subseteq U$ is non-empty and upwards-directed ad
$\{f(u):u\in A\}$ is bounded above for every $f\in U^{\times}$, then
$B=\{\hat u:u\in A\}$ is non-empty and upwards-directed and
$\sup_{\phi\in B}\phi(f)<\infty$ for every $f\in U^{\times}$, so
$\sup B$ is defined in $U^{\times\sim}$ (355Ed);  but $U^{\times\times}$ is
a band in $U^{\times\sim}$, so $\sup B$ belongs to
$U^{\times\times}$ and is of
the form $\hat w$ for some $w\in U$.   Because $u\mapsto\hat u$ is a
Riesz space isomorphism, $w=\sup A$ in $U$.   Thus $U$ satisfies the
three conditions.

\medskip

{\bf (b)} Suppose that $U$ satisfies the three conditions.   We know
that $u\mapsto\hat u$ is an order-continuous Riesz homomorphism
onto an order-dense Riesz
subspace of $U^{\times\times}$ (356I).   It is injective because
$U^{\times}$ separates the points of $U$.   If $\phi\ge 0$ in
$U^{\times\times}$, set $A=\{u:u\in U^+,\,\hat u\le\phi\}$.   Then $A$
is non-empty and upwards-directed and for any $f\in U^{\times}$

\Centerline{$\sup_{u\in A}f(u)\le\sup_{u\in A}|f|(u)
\le\sup_{u\in A}\hat u(|f|)\le\phi(|f|)<\infty$,}

\noindent so by condition (iii) $A$ has an upper bound in $U$.   Since
$U$ is Dedekind complete, $w=\sup A$ is defined in $U$.   Now

\Centerline{$\hat w=\sup_{u\in A}\hat u=\phi$.}

\noindent As $\phi$ is arbitrary, the image of $U$ includes
$(U^{\times\times})^+$, therefore is the whole of $U^{\times\times}$,
and $u\mapsto\hat u$ is a bijective Riesz homomorphism, that is, a Riesz
space isomorphism.
}%end of proof of 356K

\leader{356L}{Proposition} (a) Any band in a perfect Riesz space is a
perfect Riesz space in its own right.

(b) For any Riesz space $U$, $U^{\sim}$ is perfect;  consequently $U^{\sim}_c$ and $U^{\times}$ are perfect.

\proof{{\bf (a)} I use the criterion of 356K.   Let $U$ be a perfect
Riesz space and $V$ a band in $U$.   Then $V$ is Dedekind complete
because $U$ is (353Kb).   If $v\in V\setminus\{0\}$ there is an
$f\in U^{\times}$ such that $f(v)\ne 0$;  but the embedding
$V\embedsinto U$ is
order-continuous (352N), so $g=f\restrp V$ belongs to $V^{\times}$,
and $g(v)\ne 0$.   Thus $V^{\times}$ separates the points of $V$.   If
$A\subseteq V$ is
non-empty and upwards-directed and $\sup_{v\in A}g(v)$ is finite for
every $g\in V^{\times}$, then $\sup_{v\in A}f(v)<\infty$ for every $f\in
U^{\times}$ (again because $f\restrp V\in V^{\times}$), so $A$ has an
upper bound in $U$;  because $U$ is Dedekind complete, $\sup A$ is
defined in $U$;  because $V$ is a band, $\sup A\in V$ and is an upper
bound for $A$ in $V$.   Thus $V$ satisfies the conditions of 356K and is
perfect.

\medskip

{\bf (b)} $U^{\sim}$ is Dedekind complete, by 355Ea.   If
$f\in U^{\sim}\setminus\{0\}$, there is a $u\in U$ such that $f(u)\ne 0$;  now
$\hat u(f)\ne 0$, where $\hat u\in U^{\sim\times}$ (356F).   Thus
$U^{\sim\times}$ separates the points of $U^{\sim}$.   If $A\subseteq
U^{\sim}$ is non-empty and upwards-directed and $\sup_{f\in A}\phi(f)$
is finite for every $\phi\in U^{\sim\times}$, then, in particular,

\Centerline{$\sup_{f\in A}f(u)=\sup_{f\in A}\hat u(f)<\infty$}

\noindent for every $u\in U$, so $A$ is bounded above in $U^{\sim}$, by
355Ed.   Thus $U^{\sim}$ satisfies the conditions of 356K and is
perfect.

By (a), it follows at once that $U^{\times}$ and $U^{\sim}_c$ are
perfect.
}%end of proof of 356L

\leader{356M}{Proposition} If $U$ is a Banach lattice in which the norm
is order-continuous and has the Levi property, then $U$ is perfect.

\proof{ By 356Db, $U^*=U^{\times}$;  since $U^*$ surely separates the
points of $U$, so does $U^{\times}$.   By 354Ee, $U$ is Dedekind
complete.   If $A\subseteq U$ is non-empty and upwards-directed and
$f[A]$ is bounded for every $f\in U^{\times}$, then $A$ is norm-bounded,
by the Uniform Boundedness Theorem (3A5Hb).   Because the norm is
supposed to have the Levi property, $A$ is bounded above in $U$.
Thus $U$ satisfies all the conditions of 356K and is perfect.
}%end of proof of 356M

\leader{356N}{$L$- and \dvrocolon{$M$-spaces}}\cmmnt{ I come now to
the duality between $L$-spaces and $M$-spaces which I hinted at in
\S354.

\medskip

\noindent}{\bf Proposition} Let $U$ be an Archimedean Riesz space with
an order-unit norm.

(a) $U^*=U^{\sim}$ is an $L$-space.

(b) If $e$ is the
standard order unit of $U$, then $\|f\|=|f|(e)$ for every $f\in U^*$.

(c) A linear functional $f:U\to\Bbb R$ is positive iff it belongs to
$U^*$ and $\|f\|=f(e)$.

(d) If $e\ne 0$ there is a positive linear functional $f$ on $U$ such
that $f(e)=1$.

\proof{{\bf (a)-(b)} We know already that $U^*\subseteq U^{\sim}$ is a
Banach lattice (356Da).   If $f\in U^{\sim}$ then

\Centerline{$\sup\{|f(u)|:\|u\|\le 1\}=\sup\{|f(u)|:|u|\le e\}=|f|(e)$,}

\noindent so $f\in U^*$ and $\|f\|=|f|(e)$;  thus $U^{\sim}=U^*$.   If
$f$, $g\ge 0$ in $U^*$, then

\Centerline{$\|f+g\|=(f+g)(e)=f(e)+g(e)=\|f\|+\|g\|$;}

\noindent thus $U^*$ is an $L$-space.

\medskip

{\bf (c)} As already remarked, if $f$ is positive then $f\in U^*$ and
$\|f\|=f(e)$.   On the other hand, if $f\in U^*$ and $\|f\|=f(e)$, take
any $u\ge 0$.   Set $v=(1+\|u\|)^{-1}u$.   Then $0\le v\le e$ and
$\|e-v\|\le 1$ and

\Centerline{$f(e-v)\le|f(e-v)|\le \|f\|=f(e)$.}

\noindent But this means that $f(v)\ge 0$ so $f(u)\ge 0$.   As $u$ is
arbitrary, $f\ge 0$.

\medskip

{\bf (d)} By the Hahn-Banach theorem (3A5Ac), there is an $f\in U^*$
such that $f(e)=\|f\|=1$;  by (c), $f$ is positive.
}%end of proof of 356N

\leader{356O}{Theorem} Let $U$ be an Archimedean Riesz space with
order-unit norm.   Then a set $A\subseteq U^*=U^{\sim}$ is uniformly
integrable iff it is norm-bounded and
$\lim_{n\to\infty}\sup_{f\in A}|f(u_n)|=0$ for every order-bounded disjoint sequence
$\sequencen{u_n}$ in $U^+$.

\proof{{\bf (a)} Suppose that $A$ is uniformly integrable.   Then it is
surely norm-bounded (354Ra).   If $\sequencen{u_n}$ is a disjoint
sequence in $U^+$ bounded above by $w$, then for any $\epsilon>0$ we can
find an $h\ge 0$ in $U^*$ such that $\|(|f|-h)^+\|\le\epsilon$ for every
$f\in A$.   Now $\sum_{i=0}^nh(u_i)\le h(w)$ for every $n$, and
$\lim_{n\to 0}h(u_n)=0$;  since at the same time

\Centerline{$|f(u_n)|\le|f|(u_n)\le h(u_n)+(|f|-h)^+(u_n)
\le h(u_n)+\epsilon\|u_n\|\le h(u_n)+\epsilon\|w\|$}

\noindent for every $f\in A$ and $n\in\Bbb N$,
$\limsup_{n\to\infty}\sup_{f\in A}|f|(u_n)\le\epsilon\|w\|$.   As
$\epsilon$ is arbitrary,

\Centerline{$\lim_{n\to\infty}\sup_{f\in A}|f|(u_n)=0$,}

\noindent and the conditions are satisfied.

\medskip

{\bf (b)(i)} Now suppose that $A$ is norm-bounded but not uniformly
integrable.   Write $B$ for the solid hull of $A$, $M$ for
$\sup_{f\in A}\|f\|=\sup_{f\in B}\|f\|$;  then there is a disjoint sequence
$\sequencen{g_n}$ in $B\cap(U^*)^+$ which is not
norm-convergent to $0$ (354R(b-iv)), that is,

\Centerline{$\delta=\bover12\limsup_{n\to\infty}g_n(e)
=\bover12\limsup_{n\to\infty}\|g_n\|>0$,}

\noindent where $e$ is the standard order unit of $U$.

\medskip

\quad{\bf (ii)} Set

\Centerline{$C=\{v:0\le v\le e,\,\sup_{g\in B}g(v)\ge\delta\}$,}

\Centerline{$D=\{w:0\le w\le e,\,\limsup_{n\to\infty}g_n(w)>\delta\}$.}

\noindent Then for any $u\in D$ we can find $v\in C$ and $w\in D$ such that
$v\wedge w=0$.   \Prf\ Set $\delta'=\limsup_{n\to\infty}g_n(u)$,
$\eta=(\delta'-\delta)/(3+M)>0$;  take $k\in\Bbb N$ so large that
$k\eta\ge M$.

Because $g_n(u)\ge\delta'-\eta$ for infinitely many $n$, we can find a
set $K\subseteq\Bbb N$, with $k$ members, such that $g_i(u)\ge\delta'-\eta$
for every $i\in K$.   Now we know that, for each $i\in K$,
$g_i\wedge k\sum_{j\in K,j\ne i}g_j=0$, so there is a $v_i\le u$ such that
$g_i(u-v_i)+k\sum_{j\in K,j\ne i}g_j(v_i)\le\eta$ (355Ec).   Now

\Centerline{$g_i(v_i)\ge g_i(u)-\eta\ge\delta'-2\eta$,
\quad$g_i(v_j)\le\Bover{\eta}k$ for $i$, $j\in K$, $i\ne j$.}

Set $v_i'=(v_i-\sum_{j\in K,j\ne i}v_j)^+$ for each $i\in K$;  then

\Centerline{$g_i(v'_i)\ge g_i(v_i)-\sum_{j\in K,j\ne
i}g_i(v_j)\ge\delta'-3\eta$}

\noindent for every $i\in K$, while $v'_j\wedge v'_i=0$ for distinct
$i$, $j\in K$.

For each $n\in\Bbb N$,

\Centerline{$\sum_{i\in K}g_n(u\wedge\Bover1{\eta}v'_i)
\le g_n(u)\le\|g_n\|\le\eta k$,}

\noindent so there is some $i(n)\in K$ such that

\Centerline{$g_n(u\wedge\Bover1{\eta}v'_{i(n)})\le\eta$,
\quad $g_n(u-\Bover1{\eta}v'_{i(n)})^+\ge g_n(u)-\eta$.}

\noindent Since $\{n:g_n(u)\ge\delta+2\eta\}$ is infinite, there is some
$m\in K$ such that $J=\{n:g_n(u)\ge\delta+2\eta,\,i(n)=m\}$ is infinite.
Try

\Centerline{$v=(v'_m-\eta u)^+$,
\quad$w=(u-\Bover1{\eta}v'_m)^+$.}

\noindent Then $v$, $w\in[0,u]$ and $v\wedge w=0$.   Next,

\Centerline{$g_m(v)\ge g_m(v'_m)-\eta M\ge\delta'-3\eta-\eta M=\delta$,}

\noindent so $v\in C$, while for any $n\in J$

\Centerline{$g_n(w)=g_n(u-\Bover1{\eta}v'_{i(n)})^+
\ge g_n(u)-\eta\ge\delta+\eta$;}

\noindent since $J$ is infinite,

\Centerline{$\limsup_{n\to\infty}g_n(w)\ge\delta+\eta>\delta$}

\noindent and $w\in D$.\ \Qed

\medskip

\quad{\bf (iii)} Since $e\in D$, we can choose inductively sequences
$\sequencen{w_n}$ in $D$, $\langle v_n\rangle_{n\in\Bbb N}$ in $C$ such
that $w_0=e$, $v_n\wedge w_{n+1}=0$, $v_n\vee w_{n+1}\le w_n$ for every
$n\in\Bbb N$.   But in this case $\sequencen{v_n}$ is a disjoint
order-bounded sequence in $[0,u]$, while
for each $n\in\Bbb N$, we can find $f_n\in A$ such that
$|f_n|(v_n)>\bover23\delta$.   Now there is a $u_n\in[0,v_n]$ such that
$|f_n(u_n)|\ge\bover13\delta$.   \Prf\ Set $\gamma=\sup_{0\le v\le
v_n}|f_n(v)|$.   Then $f_n^+(v_n)$, $f^-_n(v_n)$ are both less than or
equal to $\gamma$, so $|f_n|(v_n)\le 2\gamma$ and
$\gamma>\bover13\delta$;  so there is a $u_n\in[0,v_n]$ such that
$|f_n(u_n)|\ge\bover13\delta$.\ \Qed

Accordingly we have a disjoint sequence $\sequencen{u_n}$ in $[0,e]$
such that $\sup_{f\in A}|f(u_n)|\ge\bover13\delta$ for every
$n\in\Bbb N$.

\medskip

\quad{\bf (iv)} All this is on the assumption that $A$ is norm-bounded and
not uniformly integrable.   So, turning it round, we see that if $A$ is
norm-bounded and $\lim_{n\to\infty}\sup_{f\in A}|f(u_n)|=0$ for every
order-bounded disjoint sequence $\sequencen{u_n}$, $A$ must be uniformly
integrable.

This completes the proof.
}%end of proof of 356O

\leader{356P}{Proposition} Let $U$ be an $L$-space.

(a) $U$ is perfect.

(b) $U^*=U^{\sim}=U^{\times}$ is an $M$-space;  its standard
order unit is the functional $\int$ defined by setting
$\int u=\|u^+\|-\|u^-\|$ for every $u\in U$.

(c) If $A\subseteq U$ is non-empty and upwards-directed and
$\sup_{u\in A}\int u$ is finite, then $\sup A$ is defined in $U$ and $\int\sup A=\sup_{u\in A}\int u$.

\proof{{\bf (a)} By 354N we know that the norm on $U$ is
order-continuous and has
the Levi property, so 356M tells us that $U$ is perfect.

\medskip

{\bf (b)} 356Dd tells us that $U^*=U^{\sim}=U^{\times}$.

The $L$-space property tells us that the functional
$u\mapsto\|u\|:U^+\to\Bbb R$ is additive;  of course it is also
homogeneous, so by 355D it has an extension to a linear functional
$\int:U\to\Bbb R$ satisfying the given formula.   Because
$\int u=\|u\|\ge 0$ for $u\ge 0$, $\int\in(U^{\sim})^+$.   For
$f\in U^{\sim}$,

$$\eqalign{|f|\le\int
&\iff|f|(u)\le\int u\text{ for every }u\in U^+\cr
&\iff|f(v)|\le\|u\|\text{ whenever }|v|\le u\in U\cr
&\iff|f(v)|\le\|v\|\text{ for every }v\in U\cr
&\iff\|f\|\le 1,\cr}$$

\noindent so the norm on $U^*=U^{\sim}$ is the order-unit norm defined
from $\int$, and $U^{\sim}$ is an $M$-space, as claimed.

\wheader{356P}{4}{2}{2}{60pt}

{\bf (c)} Fix $u_0\in A$, and set $B=\{u^+:u\in A,\,u\ge u_0\}$.   Then
$B\subseteq U^+$ is upwards-directed, and

$$\eqalign{\sup_{v\in B}\|v\|
&=\sup_{u\in A,u\ge u_0}\int u^+
=\sup_{u\in A,u\ge u_0}\int u+\int u^-\cr
&\le\sup_{u\in A,u\ge u_0}\int u+\int u_0^-  %
<\infty.\cr}$$

\noindent Because $\|\,\|$ has the Levi property, $B$ is bounded above.
But (because $A$ is upwards-directed) every member of $A$ is dominated
by some member of $B$, so $A$ also is bounded above.   Because $U$ is
Dedekind complete, $\sup A$ is defined in $U$.
Finally, $\int\sup A=\sup_{u\in A}\int u$ because $\int$, being a
positive member of $U^{\times}$, is order-continuous.
}%end of proof of 356P

\leader{356Q}{Theorem} Let $U$ be any $L$-space.   Then a subset of $U$
is uniformly integrable iff it is relatively weakly compact.

\proof{{\bf (a)} Let $A\subseteq U$ be a uniformly integrable set.

\medskip

\quad{\bf (i)} Suppose that $\Cal F$ is an ultrafilter on $X$ containing
$A$.   Then $A\ne\emptyset$.      Because $A$ is norm-bounded,
$\sup_{u\in A}|f(u)|<\infty$ and $\phi(f)=\lim_{u\to\Cal F}f(u)$ is
defined in $\Bbb R$ for every $f\in U^*$ (2A3Se).

If $f$, $g\in U^*$ then

\Centerline{$\phi(f+g)=\lim_{u\to\Cal F}f(u)+g(u)
=\lim_{u\to\Cal F}f(u)+\lim_{u\to\Cal F}g(u)
=\phi(f)+\phi(g)$}

\noindent (2A3Sf).   Similarly,

\Centerline{$\phi(\alpha f)=\lim_{u\to\Cal F}\alpha f(u)
=\alpha\phi(f)$}

\noindent whenever $f\in U^*$ and $\alpha\in\Bbb R$.   Thus
$\phi:U^*\to\Bbb R$ is linear.   Also

\Centerline{$|\phi(f)|\le\sup_{u\in A}|f(u)|\le\|f\|\sup_{u\in
A}\|u\|$,}

\noindent so $\phi\in U^{**}=U^{*\sim}$.

\medskip

\quad{\bf (ii)}
Now the point of this argument is that $\phi\in U^{*\times}$.   \Prf\
Suppose that $B\subseteq U^*$ is non-empty and downwards-directed and
has infimum $0$.   Fix $f_0\in B$.   Let $\epsilon>0$.   Then there is a
$w\in U^+$ such that $\|(|u|-w)^+\|\le\epsilon$ for every $u\in A$,
which means that

\Centerline{$|f(u)|\le |f|(|u|)\le |f|(w)+|f|(|u|-w)^+\le
|f|(w)+\epsilon\|f\|$}

\noindent for every $f\in U^*$ and every $u\in A$.   Accordingly
$|\phi(f)|\le |f|(w)+\epsilon\|f\|$ for every $f\in U^*$.   Now
$\inf_{f\in B}f(w)=0$ (using 355Ee, as usual), so there is an
$f_1\in B$ such that $f_1\le f_0$ and $f_1(w)\le\epsilon$.   In this case

\Centerline{$|\phi|(f_1)
=\sup_{|f|\le f_1}|\phi(f)|
\le\sup_{|f|\le f_1}|f|(w)+\epsilon\|f\|
\le f_1(w)+\epsilon\|f_1\|
\le \epsilon(1+\|f_0\|)$.}

\noindent As $\epsilon$ is arbitrary, $\inf_{f\in B}|\phi|(f)=0$;  as
$B$ is arbitrary, $|\phi|$ is order-continuous and $\phi\in
U^{*\times}$.\ \Qed

\medskip

\quad{\bf (iii)} At this point, we recall that $U^*=U^{\times}$ and that
the canonical map from $U$ to $U^{\times\times}$ is surjective (356P).
So there is a $u_0\in U$ such that $\hat u_0=\phi$.   But now we see
that

\Centerline{$f(u_0)=\phi(f)=\lim_{u\to\Cal F}f(u)$}

\noindent for every $f\in U^*$;  which is just what is meant by saying
that $\Cal F\to u_0$ for the weak topology on $U$ (2A3Sd).

Accordingly every ultrafilter on $U$ containing $A$ has a limit in $U$.
But because the weak topology on $U$ is regular (3A3Be), it follows
that the closure of $A$ for the weak topology is compact (3A3De),
so that $A$ is relatively weakly compact.

\medskip

{\bf (b)} For the converse I use the criterion of 354R(b-iv).  Suppose
that $A\subseteq U$ is relatively weakly compact.   Then $A$ is
norm-bounded, by the Uniform Boundedness Theorem.   Now let
$\sequencen{u_n}$ be any disjoint sequence in the solid hull of $A$.
For each $n$, let $U_n$ be the band in $U$ generated by $u_n$.   Let
$P_n$ be the band projection from $U$ onto $U_n$ (353Ib).   Let
$v_n\in A$ be such that $|u_n|\le|v_n|$;  then

\Centerline{$|u_n|=P_n|u_n|\le P_n|v_n|=|P_nv_n|$,}

\noindent so $\|u_n\|\le\|P_nv_n\|$ for each $n$.   Let $g_n\in U^*$ be
such that $\|g_n\|=1$ and $g_n(P_nv_n)=\|P_nv_n\|$.

Define $T:U\to\BbbR^{\Bbb N}$ by setting $Tu=\sequencen{g_n(P_nu)}$ for
each $u\in U$.   Then $T$ is a continuous linear operator from $U$ to
$\ell^1$.   \Prf\ For $m\ne n$, $U_m\cap U_n=\{0\}$, because
$|u_m|\wedge|u_n|=0$.   So, for any $u\in U$, $\sequencen{P_nu}$ is a
disjoint sequence in $U$, and

\Centerline{$\sum_{i=0}^n\|P_iu\|=\|\sum_{i=0}^n|P_iu|\|
=\|\sup_{i\le n}|P_iu|\|\le\|u\|$}

\noindent for every $n$;  accordingly

\Centerline{$\|Tu\|_1=\sum_{i=0}^{\infty}|g_iP_iu|
\le\sum_{i=0}^{\infty}\|P_iu\|\le\|u\|$.}

\noindent Since $T$ is certainly a linear operator (because every
coordinate functional $g_iP_i$ is linear), we have the result.\ \Qed

Consequently $T[A]$ is relatively weakly compact in $\ell^1$, because
$T$ is continuous for the weak topologies (2A5If).   But $\ell^1$ can be
identified with $L^1(\mu)$, where $\mu$ is counting measure on $\Bbb N$.
So $T[A]$ is uniformly integrable in $\ell^1$, by 247C, and in
particular $\lim_{n\to\infty}\sup_{w\in T[A]}|w(n)|=0$.   But this means
that

\Centerline{$\lim_{n\to\infty}\|u_n\|
\le\lim_{n\to\infty}|g_n(P_nv_n)|
=\lim_{n\to\infty}|(Tv_n)(n)|
=0$.}

\noindent As $\sequencen{u_n}$ is arbitrary, $A$ satisfies the
conditions of 354R(b-iv) and is uniformly integrable.
}%end of proof of 356Q

\exercises{\leader{356X}{Basic exercises (a)}
%\spheader 356Xa
Show that if $U=\ell^{\infty}$ then
$U^{\times}=U^{\sim}_c$ can be identified with $\ell^1$, and is properly
included in $U^{\sim}$.   \Hint{show that if $f\in U^{\sim}_c$ then
$f(u)=\sum_{n=0}^{\infty}u(n)f(e_n)$, where $e_n(n)=1$, $e_n(i)=0$ for
$i\ne n$.}
%356A

\spheader 356Xb Show that if $U=C([0,1])$ then
$U^{\times}=U^{\sim}_c=\{0\}$.   \Hint{show that if $f\in
(U^{\sim}_c)^+$ and $\sequencen{q_n}$ enumerates $\Bbb Q\cap[0,1]$, then
for each $n\in\Bbb N$ there is a $u_n\in U^+$ such that $u_n(q_n)=1$ and
$f(u_n)\le 2^{-n}$.}
%356A

\spheader 356Xc Let $X$ be an uncountable set, $\mu$ the
countable-cocountable measure on $X$ and $\Sigma$ its domain (211R).   Let $U$ be the space of bounded $\Sigma$-measurable real-valued functions on
$X$.   Show that $U$ is a Dedekind $\sigma$-complete Banach lattice if
given the supremum norm $\|\,\|_{\infty}$.   Show that $U^{\times}$ can
be identified with $\ell^1(X)$ (cf.\ 356Xa), and that
$u\mapsto\int u\,d\mu$ belongs to $U^{\sim}_c\setminus U^{\times}$.
%356A

\spheader 356Xd Let $U$ be a Dedekind $\sigma$-complete Riesz space and
$f\in U^{\sim}_c$.   Let $\sequencen{u_n}$ be an order-bounded sequence
in $U$ which is order-convergent to $u\in U$ in the sense that
$u=\inf_{n\in\Bbb N}\sup_{m\ge n}u_m=\sup_{n\in\Bbb N}\inf_{m\ge n}u_m$.
Show that $\lim_{n\to\infty}f(u_n)$ exists and is equal to $f(u)$.
%356A

\spheader 356Xe Let $U$ be any Riesz space.   Show that the band
projection $P:U^{\sim}\to U^{\times}$ is defined by the formula

$$\eqalign{(Pf)(u)=\inf\{\sup_{v\in A}f(v):A\subseteq U
&\text{ is non-empty, upwards-directed}\cr
&\qquad\qquad\qquad\qquad\text{and has supremum }u\}\cr}$$

\noindent for every $f\in(U^{\sim})^+$, $u\in U^+$.   \Hint{show that
the formula for $Pf$ always defines an order-continuous linear
functional.   Compare 355Yh, 356Yb and 362Bd.}
%356B

\spheader 356Xf Let $U$ be any Riesz space.   Show that the band
projection $P:U^{\sim}\to U^{\sim}_c$ is defined by the formula

\Centerline{$(Pf)(u)=\inf\{\sup_{n\in\Bbb N}f(v_n):\sequencen{v_n}$ is a
non-decreasing sequence with supremum $u\}$}

\noindent for every $f\in(U^{\sim})^+$, $u\in U^+$.
%356B

\spheader 356Xg Let $U$ be a Riesz space with a Riesz norm.   Show that
$U^*$ is perfect.
%356D

\spheader 356Xh  Let $U$ be a Riesz space with a Riesz norm.   Show that
the canonical map from $U$ to $U^{**}$ is a Riesz homomorphism.
%356F

\spheader 356Xi Let $V$ be a perfect Riesz space and $U$ any Riesz
space.   Show that $\eusm L^{\sim}(U;V)$ is perfect.   \Hint{show that
if $u\in U$ and $g\in V^{\times}$ then $T\mapsto g(Tu)$ belongs to
$\eusm L^{\sim}(U;V)^{\times}$.}
%356K

\spheader 356Xj Let $U$ be an $M$-space.   Show that it is perfect iff
it is Dedekind complete and $U^{\times}$ separates the points of $U$.
%356K

\spheader 356Xk Let $U$ be a Banach lattice which, as a Riesz space, is
perfect.   Show that its norm has the Levi property.
%356M

\spheader 356Xl Write out a proof from first principles that if
$\sequencen{u_n}$ is a sequence in $\ell^1$ such that
$|u_n(n)|\ge\delta>0$ for every $n\in\Bbb N$, then $\{u_n:n\in\Bbb N\}$
is not relatively weakly compact.
%356Q

\spheader 356Xm Let $U$ be an $L$-space and $A\subseteq U$ a non-empty
set.   Show that the following are equiveridical:  (i) $A$ is uniformly
integrable (ii) $\inf_{f\in B}\sup_{u\in A}|f(u)|$ for every non-empty
downwards-directed set $B\subseteq U^{\times}$ with infimum $0$ (iii)
$\inf_{n\in\Bbb N}\sup_{u\in A}|f_n(u)|=0$ for every non-increasing
sequence $\sequencen{f_n}$ in $U^{\times}$ with infimum $0$ (iv) $A$ is
norm-bounded and $\lim_{n\to\infty}\sup_{u\in A}|f_n(u)|=0$ for every
disjoint order-bounded sequence $\sequencen{f_n}$ in $U^{\times}$.
%356Q

\leader{356Y}{Further exercises (a)}
%\spheader 356Ya
Let $U$ be a Riesz space with the countable sup
property.   Show that $U^{\times}=U^{\sim}_c$.
%356A

\spheader 356Yb Let $U$ be a Riesz space, and $\Cal A$ a family of
non-empty downwards-directed subsets of $U^+$ all with infimum $0$.
(i) Show that $U^{\sim}_{\Cal A}=\{f:f\in U^{\sim},\,\inf_{u\in
A}|f|(u)=0$ for every $A\in\Cal A\}$ is a band in $U^{\sim}$.   (ii) Set
$\Cal A^*=\{A_0+\ldots+A_n:A_0,\ldots,A_n\in\Cal A\}$.   Show that
$U^{\sim}_{\Cal A}=U^{\sim}_{\Cal A^*}$.    (iii) Take any
$f\in(U^{\sim})^+$, and let $g$, $h$ be the components of $f$ in
$U^{\sim}_{\Cal A}$, $(U^{\sim}_{\Cal A})^{\perp}$ respectively.   Show
that

\Centerline{$g(u)=\inf_{A\in\Cal A^*}\sup_{v\in A}f(u-v)^+$,
\quad$h(u)=\sup_{A\in\Cal A^*}\inf_{v\in A}f(u\wedge v)$}

\noindent for every $u\in U^+$.   (Cf.\ 362Xi.)
%356Xe, 356B

\spheader 356Yc Let $U$ be a Riesz space.   For any band $V\subseteq U$
write $V^{\smallcirc}$ for
$\{f:f\in U^{\times}$, $f(v)\le 1$ for every $v\in V\}
=\{f:f\in U^{\times},\,f(v)=0$ for every $v\in V\}$.
Show that $V\mapsto (V^{\perp})^{\smallcirc}$ is a surjective
order-continuous
Boolean homomorphism from the algebra of complemented bands of $U$ onto
the band algebra of $U^{\times}$, and that it is injective iff
$U^{\times}$ separates the points of $U$.
%356C

\spheader 356Yd Let $U$ be a Dedekind complete Riesz space such that
$U^{\times}$ separates the points of $U$ and $U$ is the solid linear
subspace of itself generated by a countable set.   Show that $U$ is
perfect.
%356K, 356Xj

\spheader 356Ye  Let $U$ be an $L$-space and
$\sequencen{u_n}$ a sequence in $U$ such that $\sequencen{f(u_n)}$ is
Cauchy for every $f\in U^*$.   Show that $\sequencen{u_n}$ is convergent
for the weak topology of $U$.   \Hint{use 356Xm(iv) to show that
$\{u_n:n\in\Bbb N\}$ is relatively weakly compact.}
%356Xm, 356Q

\spheader 356Yf  Let $U$ be a perfect Banach lattice with
order-continuous norm and
$\sequencen{u_n}$ a sequence in $U$ such that $\sequencen{f(u_n)}$ is
Cauchy for every $f\in U^*$.   Show that $\sequencen{u_n}$ is convergent
for the weak topology of $U$.   \Hint{set
$\phi(f)=\lim_{n\to\infty}f_n(u)$.   For any $g\in(U^*)^+$ let $V_g$ be
the solid linear subspace of $U^*$ generated by $g$,
$W_g=\{u:g(|u|)=0\}^{\perp}$, $\|u\|_g=g(|u|)$ for $u\in W_g$.   Show
that the completion of $W_g$ under $\|\,\|_g$ is an $L$-space with dual
isomorphic to $V_g$, and hence (using 356Ye) that $\phi\restrp V_g$
belongs to $V_g^{\times}$;  as $g$ is arbitrary, $\phi\in V^{\times}$
and may be identified with an element of $U$.}
%356Ye, 356Xm, 356Q

\spheader 356Yg Let $U$ be a uniformly complete Archimedean Riesz space
with complexification $V$ (354Yl).   (i) Show that the complexification
of $U^{\sim}$ can be identified with the space of linear functionals
$f:V\to\Bbb C$ such that $\sup_{|v|\le u}|f(v)|$ is finite for every
$u\in U^+$.   (ii) Show that if $U$ is a Banach lattice, then the
complexification of $U^{\sim}=U^*$ can be identified (as normed space)
with $V^*$.   (See 355Yk.)
%+

\spheader 356Yh Let $U$ be a perfect Banach lattice.
Show that the family of closed balls in $U$ is a compact class.
\Hint{342Ya.}
}%end of exercises

\endnotes{
\Notesheader{356} The section starts easily enough, with special cases
of results in \S355 (356B).   When $U$ has a Riesz norm,  the
identification of $U^*$ as a subspace of $U^{\sim}$, and the
characterization of order-continuous norms (356D) are pleasingly
comprehensive and straightforward.   Coming to biduals, we need to think
a little (356F), but there is still no real difficulty at first.   In
356H-356I, however, something more substantial is happening.   I have
written these arguments out in what seems to be the shortest route to
the main theorem, at the cost perhaps of neglecting any intuitive
foundation.   What I think we are really doing is matching bands in $U$,
$U^{\times}$ and $U^{\times\times}$, as in 356Yc.

From now on, almost the first thing we shall ask of any new Riesz space
will be whether it is perfect, and if not, which of the three conditions
of 356K it fails to satisfy.   For reasons which will I hope appear in
the next chapter, perfect Riesz spaces are especially important in
measure theory;  in particular, all $L^p$ spaces for
$p\in\coint{1,\infty}$  are perfect (366Dd), as are the $L^{\infty}$
spaces of localizable measure spaces (365M).   Further examples
will be discussed in \S369 and \S374.   Of course we have to remember
that there are also important Riesz spaces which are not perfect, of
which $C([0,1])$ and $\pmb{c}_0$ are two of the simplest examples.

The duality between $L$- and $M$-spaces (356N, 356P) is natural and
satisfying.   We are now in a position to make a determined attempt to
tidy up the notion of `uniform integrability'.   I give two major
theorems.   The first is yet another `disjoint-sequence'
characterization of uniformly integrable sets, to go with 246G and 354R.
The essential difference here is that we are looking at disjoint
sequences in a predual;  in a sense, this means that the result is a
sharper one, because the $M$-space $U$ need not be Dedekind complete
(for instance, it could be $C([0,1])$ -- this indeed is the archetype
for applications of the theorem) and therefore need not have as many
disjoint sequences as its dual.   (For instance, in the dual of
$C([0,1])$ we have all the point masses $\delta_t$, where
$\delta_t(u)=u(t)$;  these form a disjoint family in $C([0,1])^{\sim}$
not corresponding to any disjoint family in $C([0,1])$.)   The essence
of the proof is a device to extract a disjoint sequence in $U$ to match
approximately a subsequence of a given disjoint sequence in $U^{\sim}$.
In the example just suggested, this would correspond, given a sequence
$\sequencen{t_n}$ of distinct points in $[0,1]$, to finding a
subsequence $\sequence{i}{t_{n(i)}}$ which is discrete, so that we can
find disjoint $u_i\in C([0,1])$ with $u_i(t_{n(i)})=1$ for each $i$.

The second theorem, 356Q, is a new version of a result already given in
\S247:  in any $L$-space, uniform integrability is the same as relative
weak compactness.   I hope you are not exasperated by having been asked,
in Volume 2, to master a complex argument (one of the more difficult
sections of that volume) which was going to be superseded.   Actually it
is worse than that.   A theorem of Kakutani (369E) tells us that
every $L$-space is isomorphic to an $L^1$ space.   So 356Q is itself a
consequence of 247C.   I do at least owe you an explanation for writing
out two proofs.   The first point is that the result is sufficiently
important for it to be well worth while spending time in its
neighbourhood, and the contrasts and similarities between the two
arguments are instructive.   The second is that the proof I have just
given was not really accessible at the level of Volume 2.   It does not
rely on every single page of this chapter, but the key idea (that $U$ is
isomorphic to $U^{\times\times}$, so it will be enough if we can show
that $A$ is relatively compact in $U^{\times\times}$) depends
essentially on 356I, which lies pretty deep in the abstract theory of
Riesz spaces.   The third is an aesthetic one:  a theorem about
$L$-spaces ought to be proved in the category of normed Riesz spaces,
without calling on a large body of theory outside.   Of course this is a
book on measure theory, so I did the measure theory first, but if you
look at everything that went into it, the proof in \S247 is I believe
longer, in the formal sense, than the one here, even setting aside the
labour of proving Kakutani's theorem.

Let us examine the ideas in the two proofs.   First, concerning the
proof that uniformly integrable sets are relatively compact, the method
here is very smooth and natural;  the definition I chose of `uniform
integrability' is exactly adapted to showing that uniformly integrable
sets are relatively compact in the order-continuous bidual;  all the
effort goes into the proof that $L$-spaces are perfect.   The previous
argument depended on identifying the dual of $L^1$ as $L^{\infty}$ --
and was disagreeably complicated by the fact that the identification is
not always valid, so that I needed to reduce the problem to the
$\sigma$-finite case (part (b-ii) of the proof of 247C).   After that,
the Radon-Nikod\'ym theorem did the trick.   Actually Kakutani's theorem
shows that the side-step to $\sigma$-finite spaces is irrelevant.   It
directly represents an abstract $L$-space as $L^1(\mu)$ for a
localizable measure $\mu$, in which case $(L^1)^*\cong L^{\infty}$
exactly.

In the other direction, both arguments depend on a disjoint-sequence
criterion for uniform integrability (246G(iii) or 354R(b-iv)).   These
criteria belong to the `easy' side of the topic;  straightforward
Riesz space arguments do the job, whether written out in that language
or not.   (Of course the new one in this section, 356O, lies a little
deeper.)   I go a bit faster this time because I feel that you ought by
now to be happy with the Hahn-Banach theorem and the Uniform Boundedness
Theorem, which I was avoiding in Volume 2.   And then of course I quote
the result for $\ell^1$.   This looks like cheating.   But $\ell^1$
really is easier, as you will find if you just write out part (a) of the
proof of 247C for this case.   It is not exactly that you can dispense
with any particular element of the argument;  rather it is that the
formulae become much more direct when you can write $u(i)$ in place of
$\int_{F_i}u$, and `cluster points for the weak topology' become
pointwise limits of subsequences, so that the key step (the `sliding
hump', in which $u_{k(j)}(n(k(j)))$ is the only significant coordinate
of $u_{k(j)}$), is easier to find.

We now have a wide enough variety of conditions equivalent to uniform
integrability for it to be easy to find others;  I give a couple in
356Xm, corresponding in a way to those in 246G.   You may have noticed,
in the proof of 247C, that in fact the full strength of the hypothesis
`relatively weakly compact' is never used;  all that is demanded is
that a couple of sequences should have cluster points for the weak
topology.   So we see that a set $A$ is uniformly integrable iff every
sequence in $A$ has a weak cluster point.   But this extra refinement is
nothing to do with $L$-spaces;  it is generally true, in any normed
space $U$, that a set $A\subseteq U$ is relatively weakly compact iff
every sequence in $A$ has a cluster point in $U$ for the weak topology
(`Eberlein's theorem';  see 462D in Volume 4, {\smc K\"othe 69}, 24.2.1,
or {\smc Dunford \& Schwartz 57}, V.6.1).

There is a very rich theory concerning weak compactness in perfect Riesz
spaces, based on the ideas here;  some of it is explored in 
{\smc Fremlin 74a}.   As a sample, I give one of the basic properties of
perfect Banach lattices with order-continuous norms:  they are `weakly
sequentially complete' (356Yf).
}%end of comment

\frnewpage

%\ifresultsonly\else
\frfilename{mt35conc.tex}
\versiondate{16.2.17}
\copyrightdate{2004}
     
\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
     
\def\chaptername{Riesz 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

\leaveitout{\leader{3{}52D}{Identities} %352D
$u\vee v=u+(v-u)^+$ {\smc Pachl 13}, P.19
}

\leader{3{}53H}{Principal bands} This proposition, referred to in the 2003,
2006 and 2013 printings of Volume 4, is now 353I.

\leader{3{}53I}{Projection bands} This proposition, referred to in the
2003, 2006 and 2013 printings of Volume 4 and the
2008 and 2015 printings of Volume 5, is now 353J.

\leader{3{}53K}{Solid linear subspaces} This proposition, referred to in the
2008 and 2015 printings of Volume 5, is now 353L.

\leader{3{}53M}{Riesz spaces with order units} This theorem, 
referred to in the 2003, 2006 and 2013 printings of Volume 4, is now 353N.

\leader{3{}53P}{$f$-algebras with identity} This proposition, 
referred to in the 2003,
2006 and 2013 printings of Volume 4, is now 353Q.

\leaveitout{\leader{3{}54E}{Order-continuous norms} %354E
{\smc Pachl 13}, P.21
}

\leaveitout{\leader{3{}54N}{$L$-spaces} %354N
{\smc Pachl 13}, P.21
}

     
\leader{3{}54Yk}{Complexifications of normed Riesz spaces} 
This exercise, referred to in the 2003 edition of Volume 4, is now 354Yl.

\leaveitout{\leader{3{}56B}{$U^{\sim}$} %356B
Formulae for $f^+$, $|f|$ referred to in {\smc Pachl 13}, P.19.
} 

\leaveitout{\leader{3{}56N}{$L$- and $M$-spaces} %356N
(a)-(b) in {\smc Pachl 13}, P.21
}

%Pachl J.\ [13] {\it Uniform Spaces and Measures.}   Springer, 2013.
 

\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



