% 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{mt64.tex}
\versiondate{3.2.14/4.5.20}
\copyrightdate{2014}

\def\chaptername{The fundamental theorem of martingales and the
S-integral}

\newchapter{64}

To my mind, the `Riemann-sum' integral of \S613 is the natural starting
point for a theory of stochastic integration, and it has a rich assortment
of properties.   But if you are acquainted with the
Lebesgue-Stieltjes integral, you will have noticed that I have not
given results corresponding to the standard convergence theorems of
\S123, and if you have taken the trouble to check, you will have noticed
that they aren't true of the integral as presented so far.   If we make the
right modifications, however, we do have a kind of sequential smoothness
(644C) which can, with some difficulty, be used as the basis for
what I will call the `S-integral' (645P).   In fact the S-integral is much
closer than the Riemann-sum integral to the standard stochastic integral
developed in {\smc Protter 05}.

To do this we need to know quite a lot more about stochastic processes.
In \S641 I describe the `previsible version' of a near-simple process, which
corresponds to the c{\`a}gl{\`a}d function equal except at jump points to a
\cadlag\ function of a real variable.   Looking at pointwise limits of
sequences of previsible versions, we are led to the previsible processes of
\S642, which have the kind of measurability demanded of an integrand in the
S-integral (645I).   But the really important fact, if we are going to
have the S-integral for martingale integrators which are not jump-free, is
the fundamental theorem of martingales:  under certain
conditions, an integrator can be expressed not just as the sum of a
virtually local martingale and a process of bounded variation, as in the
Bichteler-Dellacherie theorem (627J), but as such a sum in which the
virtually local martingale has small residual oscillation (643M).

With the S-integral defined, we can look at its properties, which by
and large correspond to those of the Riemann-sum integral as established in
chapters 61-63.   Many of the details are not trivial, and I work through them
in \S646-648, with an S-integral version of It\^o's formula (646T).

I end the chapter with a brief note (\S649) left over from Chapter 63,
on Riemann-sum integrals, in the classical context of
progressively measurable stochastic processes defined on a probability
space, which can be calculated from sample paths, one path at a time;
for non-decreasing integrators, we can use a Stieltjes integral on each
path to calculate the S-integral.

\discrpage

\frfilename{mt641.tex}
\versiondate{20.1.21/29.9.23}
\copyrightdate{2018}

%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
%   \phantom{\tbf 1^{(#1)}}}
\def\varll{<\mskip-5mu<}  %tested in subscript only

\def\chaptername{The S-integral}
\def\sectionname{Previsible versions}

\newsection{641}
In \S618 I introduced `jump-free' processes without going into the question
of what the `jumps' were which they were free of.   We now need to look
at the structure of processes which are not jump-free.   In the standard
model of locally near-simple processes as those representable by
processes with \cadlag\ sample paths (631D), we have direct descriptions of
$\sigma$-algebras $\Sigma_{h^-}$ and random variables $X_{h^-}$ defined in
terms of observations taken {\it before} a stopping time $h$, rather than
at the stopping time, as in 612H.   I present these descriptions in 642E,
following corresponding definitions in the more abstract language I favour
in this volume (641B, 641F).   Once we have got hold of the previsible
version $\pmb u_<$ of a near-simple process, we have an expression for the
residual oscillation of $\pmb u$ in terms of $\pmb u-\pmb u_<$ (641Nb,
642Ga).   For moderately oscillatory processes which are not near-simple,
we do not have such a direct
description of their jumps, but the construction of the previsible version
still works (641L), and we have effective results on indefinite integrals
(641Q) and quadratic variations (641R).

\cmmnt{
\leader{641A}{Notation} \Asissix\ will be a stochastic integration structure,
as described in the notes to \S613.
If $\familyiI{\frak B_i}$ is a family of closed subalgebras of $\frak A$,
$\bigvee_{i\in I}\frak B_i$ will be the smallest closed subalgebra
including every $\frak B_i$.   $\theta$ will be the standard functional
defining the topology of convergence in measure on $L^0(\frak A)$
(613B).
If $t\in T$ then $\check t$ will be the constant stopping time at $t$.
If $\Cal S$ is a sublattice of $\Cal T$,
$\CalIS$ will be the upwards-directed set of finite sublattices of
$\Cal S$;   $\Mfa(\Cal S)$, $\Mob(\Cal S)$, $\Mmo(\Cal S)$, $\Mlmo(\Cal S)$
and $\Mns(\Cal S)$ will be the spaces of fully adapted,
order-bounded, moderately oscillatory, locally moderately oscillatory
and near-simple processes with domain $\Cal S$.
If $\pmb u=\sfamilysS{u}$ belongs to $\Mob(\Cal S)$,
$\sup|\pmb u|$ will be the supremum
$\sup_{\sigma\in\Cal S}|u_{\sigma}|$ in $L^0(\frak A)$, and
$\Osclln(\pmb u)$ will be the residual oscillation of $\pmb u$ (618B).
For sublattices $I$, $J$ of $\Cal T$, $I\sqcup J$ will be the sublattice of
$\Cal T$ generated by $I\cup J$.

}%end of comment

\leader{641B}{The algebras $\frak A_{\Cal S<\tau}$ (a) Definition}
If $\Cal S$ is a sublattice of $\Cal T$ and $\tau\in\Cal T$,
let $\frak A_{\Cal S<\tau}$ be the
closed subalgebra of $\frak A$ generated by
$\{a\Bcap\Bvalue{\sigma<\tau}:\sigma\in\Cal S$, $a\in\frak A_{\sigma}\}$.
\cmmnt{Note that}
$\frak A_{\Cal S<\tau}\subseteq\frak A_{\tau}$\prooflet{, by
611H(c-i)}.   I will write $\frak A_{<\tau}$ for $\frak A_{\Cal T<\tau}$.

\spheader 641Bb Let $\Cal S$ be a sublattice of $\Cal T$.

\medskip

\quad{\bf (i)} If $\sigma\le\tau$ in $\Cal T$ then
$\frak A_{\Cal S<\sigma}\subseteq\frak A_{\Cal S<\tau}$.   \prooflet{\Prf\
If $\rho\in\Cal S$ then $\Bvalue{\rho<\sigma}\in\frak A_{\rho}$ and
$\Bvalue{\rho<\sigma}\Bsubseteq\Bvalue{\rho<\tau}$ so

$$\eqalign{\{a\Bcap\Bvalue{\rho<\sigma}:\rho\in\Cal S,\,a\in\frak A_{\rho}\}&
=\{a\Bcap\Bvalue{\rho<\sigma}\Bcap\Bvalue{\rho<\tau}:
\rho\in\Cal S,\,a\in\frak A_{\rho}\}\cr&
\subseteq\{a\Bcap\Bvalue{\rho<\tau}:\rho\in\Cal S,\,a\in\frak A_{\rho}\}.
\text{ \Qed}\cr}$$
}

\medskip

\quad{\bf (ii)} If $\sigma\in\Cal S$, $\tau\in\Cal T$ and
$u\in L^0(\frak A_{\sigma})$, then
$u\times\chi\Bvalue{\sigma<\tau}\in L^0(\frak A_{\Cal S<\tau})$.
\prooflet{\Prf\ Write $u'$ for $u\times\chi\Bvalue{\sigma<\tau}$.
For $\alpha\ge 0$,
$\Bvalue{u'>\alpha}=\Bvalue{u>\alpha}\Bcap\Bvalue{\sigma<\tau}$ and
$\Bvalue{u'<-\alpha}\Bcap\Bvalue{\sigma<\tau}$ belong to
$\frak A_{\Cal S<\tau}$.   Now if $\alpha<0$

\Centerline{$\Bvalue{u'>\alpha}
=\sup_{\beta\in\ocint{\alpha,0}}\Bvalue{u'\ge\beta}
=1\Bsetminus\inf_{\beta\in\ocint{\alpha,0}}\Bvalue{u'<\beta}$}

\noindent again belongs to $\frak A_{\Cal S<\tau}$, so
$u'\in L^0(\frak A_{\Cal S<\tau})$.\ \Qed}

\medskip

\quad{\bf (iii)} If $\Cal S'$ is a sublattice of $\Cal T$ covering
$\Cal S$, then $\frak A_{\Cal S<\tau}\subseteq\frak A_{\Cal S'<\tau}$ for every
$\tau\in\Cal T$.
\prooflet{\Prf\ If $\sigma\in\Cal S$ and $a\in\frak A_{\sigma}$, then
$a\Bcap\Bvalue{\rho=\sigma}\in\frak A_{\rho}$ for every $\rho\in\Cal S'$
(611Hc), so

$$\eqalign{a\Bcap\Bvalue{\sigma<\tau}&
=a\Bcap\Bvalue{\sigma<\tau}\Bcap\sup_{\rho\in\Cal S}\Bvalue{\rho=\sigma}
=\sup_{\rho\in\Cal S}a\Bcap\Bvalue{\sigma<\tau}\Bcap\Bvalue{\rho=\sigma}\cr&
=\sup_{\rho\in\Cal S}a\Bcap\Bvalue{\rho=\sigma}\Bcap\Bvalue{\rho<\tau}
\in\frak A_{\Cal S'<\tau}.  \text{ \Qed}\cr}$$
}

In particular, $\frak A_{\Cal S<\tau}\subseteq\frak A_{\Cal S'<\tau}$ whenever
$\Cal S\subseteq\Cal S'$ and $\tau\in\Cal T$.

\medskip

\quad{\bf (iv)} Now suppose that $\Cal S$ is finitely
full and that $\tau\in\Cal S$.   Then
$\frak A_{\Cal S<\tau}$ is the closed subalgebra $\frak C$ generated by
$\{\Bvalue{\sigma<\tau}:\sigma\in\Cal S\}$.   \prooflet{\Prf\ Of course
$1\Bcap\Bvalue{\sigma<\tau}\in\frak A_{\Cal S<\tau}$ for every
$\sigma\in\Cal S$, so
$\frak C\subseteq\frak A_{\Cal S<\sigma}$.   On the other hand, if
$\sigma\in\Cal S$ and $a\in\frak A_{\sigma}$
then $a'=a\Bcap\Bvalue{\sigma<\tau}$ belongs to
$\frak A_{\sigma}\Bcap\frak A_{\tau}$
by 611H(c-iii).   By 611I, there is a $\rho\in\Cal T$ such that
$a'\Bsubseteq\Bvalue{\rho=\sigma}$
and $1\Bsetminus a'\Bsubseteq\Bvalue{\rho=\tau}$;  now $\rho\in\Cal S$ and
$a'=\Bvalue{\rho<\tau}\in\frak C$.   So
$\frak A_{\Cal S<\sigma}\subseteq\frak C$ and we have equality.\ \Qed}

\medskip

\spheader 641Bc Suppose that $\tau\in\Cal T$,
$I$ is a non-empty finite sublattice of $\Cal T$
and $(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells (611L).
Let $\frak B$ be the set of those $b\in\frak A$ such that

\inset{$b\Bcap\Bvalue{\tau\le\tau_0}$ is either $\Bvalue{\tau\le\tau_0}$ or
$0$,
\quad$b\Bcap\Bvalue{\tau_n<\tau}\in\frak A_{\tau_n}$,

for every $i<n$ there is an $a\in\frak A_{\tau_i}$ such that
$b\Bcap\Bvalue{\tau_i<\tau}\Bsetminus\Bvalue{\tau_{i+1}<\tau}
=a\Bcap\Bvalue{\tau_i<\tau}\Bsetminus\Bvalue{\tau_{i+1}<\tau}$.}

\noindent Then $\frak A_{I<\tau}=\frak B$.
\prooflet{\Prf\ Writing $J$ for $\{\tau_0,\ldots,\tau_n\}$,
$J$ covers $I$ (611Ke) so $\frak A_{I<\tau}=\frak A_{J<\tau}$ ((b-iii) above).
Set

\Centerline{$d_{-1}=\Bvalue{\tau\le\tau_0}=1\Bsetminus{\tau_0<\tau}$,
\quad$d_n=\Bvalue{\tau_n<\tau}$,}

\Centerline{$d_i=\Bvalue{\tau_i<\tau}\Bsetminus\Bvalue{\tau_{i+1}<\tau}$ for
$i<n$.}

\noindent Then $\{d_{-1},d_0,\ldots,d_n\}$ is a partition of unity in $\frak A$,
and $d_i\in\frak A_{J<\tau}$ whenever $-1\le i\le n$.

Set $\frak B_{-1}=\{0,d_{-1}\}$ and for $0\le i\le n$ set
$\frak B_i=\{d_i\Bcap a:a\in\frak A_{\tau_i}\}$.   Then $\frak B_i$ is
always an order-closed subalgebra of the principal ideal of $\frak A$
generated by $d_i$,
while $\frak B=\{b:b\Bcap d_i\in\frak B_i$ whenever $-1\le i\le n\}$.   So
$\frak B$ is an order-closed subalgebra of $\frak A$.
Since $\frak B_i\subseteq\frak A_{J<\tau}$ whenever $-1\le i\le n$,
$\frak B\subseteq\frak A_{J<\tau}$.

If $-1\le i\le n$, $j\le n$ and $a\in\frak A_{\tau_j}$, then
$d_i\Bcap a\Bcap\Bvalue{\tau_j<\tau}$ belongs to $\frak B_i$, because
if $i<j$ then $d_i\Bcap\Bvalue{\tau_j<\tau}=0$, while if $j\le i$ then
$a\Bcap\Bvalue{\tau_j<\tau}\in\frak A_{\tau_i}$.   So if we write
$A=\{a\Bcap\Bvalue{\tau_j<\tau}:j\le n$, $a\in\frak A_{\tau_j}\}$ for the
generating subset of $\frak A_{J<\tau}$, we see that
$a\Bcap d_i\in\frak B_i$ whenever $a\in A$ and $-1\le i\le n$, that is,
$A\subseteq\frak B$.   Accordingly $\frak A_{J<\tau}\subseteq\frak B$ and
$\frak B=\frak A_{J<\tau}=\frak A_{I<\tau}$, as claimed.\ \Qed}

\spheader 641Bd If $t\in T$ then
$\frak A_{<\check{t}}=\bigvee_{s<t}\frak A_s$.
\prooflet{\Prf\ If $s<t$, then $\Bvalue{\check s<\check t}=1$ so
$\frak A_s=\frak A_{\check s}\subseteq\frak A_{<\check t}$.
On the other hand, if $\sigma\in\Cal T$,

\Centerline{$\Bvalue{\sigma<\check{t}}
=\sup_{s\in T}\Bvalue{\check{t}>s}\Bsetminus\Bvalue{\sigma>s}
=\sup_{s<t}1\Bsetminus\Bvalue{\sigma>s}$}

\noindent belongs to $\bigvee_{s<t}\frak A_s$.\ \Qed}

\spheader 641Be If $\tau\in\Cal T$ then $\frak A_{<\tau}$ is the closed
subalgebra of $\frak A$ generated by

\Centerline{$\cmmnt{A=\mskip5mu}\{a:$ there is a $t\in T$ such that
$a\in\frak A_t$ and $a\Bsubseteq\Bvalue{\tau>t}\}$.}

\prooflet{\noindent\Prf\
Let $\frak B$ be the closed subalgebra generated by
$A$.   If $t\in T$, $a\in\frak A_t$ and $a\Bsubseteq\Bvalue{\tau>t}$, then
$a\in\frak A_{\check{t}}$ and $a\Bsubseteq\Bvalue{\check{t}<\tau}$
(611E(a-i-$\delta$)), so $a\in\frak A_{<\tau}$;  consequently
$\frak B\subseteq\frak A_{<\tau}$.   If $\sigma\in\Cal T$ and $t\in T$,
then $\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}$ belongs to
$\frak A_t$ and is included in $\Bvalue{\tau>t}$, so belongs to
$A\Bsubseteq\frak B$;
taking the supremum over $t$, $\Bvalue{\sigma<\tau}\in\frak B$.
As $\sigma$ is arbitrary and $\Cal T$ is full, (b-iv) tells us that
$\frak A_{<\tau}\subseteq\frak B$ and we have
equality.\ \Qed}

\leader{641C}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$C\subseteq\Cal T$ a non-empty set with supremum $\tau$.

(a) $\frak A_{\Cal S<\tau}=\bigvee_{\sigma\in C}\frak A_{\Cal S<\sigma}$.

(b) Now suppose that $C\subseteq\Cal S$.
Set $a=\inf_{\sigma\in C}\Bvalue{\sigma<\tau}$.   Then

\Centerline{$\bigvee_{\sigma\in C}\frak A_{\sigma}
=\{(b\Bsetminus a)\Bcup(c\Bcap a):b\in\frak A_{\tau}$,
$c\in\frak A_{\Cal S<\tau}\}$.}

\proof{{\bf (a)} By 641B(b-i),
$\bigvee_{\sigma\in C}\frak A_{\Cal S<\sigma}\subseteq\frak A_{\Cal S<\tau}$.
In the other direction, if $\rho\in\Cal S$ and $a\in\frak A_{\rho}$ then
$a\Bcap\Bvalue{\rho<\tau}=\sup_{\sigma\in C}a\Bcap\Bvalue{\rho<\sigma}$ (611Eb),
so belongs to $\bigvee_{\sigma\in C}\frak A_{\Cal S<\sigma}$.   As $\rho$ is
arbitrary,
$\frak A_{\Cal S<\tau}\subseteq\bigvee_{\sigma\in C}\frak A_{\Cal S<\sigma}$.

\medskip

{\bf (b)} Write $\frak B$ for $\bigvee_{\sigma\in C}\frak A_{\sigma}$ and
$\frak B'$ for
$\{(b\Bsetminus a)\Bcup(c\Bcap a):b\in\frak A_{\tau}$,
$c\in\frak A_{\Cal S<\tau}\}$.

\medskip

\quad{\bf (i)} Note first that

\Centerline{$\Bvalue{\sigma<\tau}\in\frak A_{\sigma}\Bcap\frak A_{\Cal S<\tau}
\Bsubseteq\frak B\cap\frak A_{\Cal S<\tau}$}

\noindent for every $\sigma\in C$, so $a\in\frak B\cap\frak A_{\Cal S<\tau}$.

\medskip

\quad{\bf (ii)} If $b\in\frak A_{\tau}$, then

\Centerline{$b\Bsetminus\Bvalue{\sigma<\tau}\in\frak A_{\sigma}
\subseteq\frak B$}

\noindent for every $\sigma\in C$ (using 611H(c-iii) again),
so $b\Bsetminus a\in\frak B$.

If $\rho\in\Cal S$, $d\in\frak A_{\rho}$ and $\rho\le\tau$, then
$d\Bcap\Bvalue{\rho<\sigma}\in\frak A_{\sigma}\subseteq\frak B$ for every
$\sigma\in C$, so $\frak B$ contains
$\sup_{\sigma\in C}d\Bcap\Bvalue{\rho<\sigma}=d\Bcap\Bvalue{\rho<\tau}$
(611Eb again).   Accordingly $\frak A_{\Cal S<\tau}\subseteq\frak B$.
By (i), it follows that $c\Bcap a\in\frak B$ whenever
$c\in\frak A_{\Cal S<\tau}$.

Thus $\frak B'\subseteq\frak B$.

\medskip

\quad{\bf (iii)} $\frak B'$ is a closed subalgebra of $\frak A$.   \Prf\
(Cf.\ 314Ja.)   The map $(b,c)\mapsto(b\Bsetminus a)\Bcup(c\Bcap a):
\frak A\times\frak A\to\frak A$ is an order-continuous Boolean
homomorphism, while $\frak A_{\tau}\times\frak A_{\Cal S<\tau}$ is a closed
subalgebra of $\frak A\times\frak A$, so the image $\frak B'$ is a closed
subalgebra of $\frak A$ (314F(a-i)).\ \Qed

If $\sigma\in C$ and $d\in\frak A_{\sigma}$ then $d\in\frak A_{\tau}$ and
$d\Bcap\Bvalue{\sigma<\tau}\in\frak A_{\Cal S<\tau}$.
So

\Centerline{$d=(d\Bsetminus a)\Bcup(d\Bcap a)
=(d\Bsetminus a)\Bcup(d\Bcap\Bvalue{\sigma<\tau}\Bcap a)$}

\noindent belongs to $\frak B'$.   As $\sigma$ and $d$ are arbitrary,
$\bigcup_{\sigma\in C}\frak A_{\sigma}\subseteq\frak B'$;  as $\frak B'$ is
a closed subalgebra, $\frak B\subseteq\frak B'$ and we have equality.
}%end of proof of 641C

\leader{641D}{Proposition} Let $\Cal S$ be a
sublattice of $\Cal T$ and $\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ an
$L^1$-process with a previsible variation
$\pmb v^{\#}=\sfamilysS{v^{\#}}$\cmmnt{ (626L)}.
Then $v^{\#}_{\tau}\in L^0(\frak A_{\Cal S<\tau})$ for every $\tau\in\Cal S$.

\proof{ Consider the formulae

\Centerline{$\Delta_{c(\sigma,\sigma')}(\tbf{1},Pd\pmb v)
=P_{\sigma}v_{\sigma'}-v_{\sigma}$,}

\Centerline{$v^{\#}_{\tau}
=\w1lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\tbf{1},Pd\pmb v)$.}

\noindent If $\sigma\le\sigma'$ in $\Cal S\wedge\tau$,
$\Delta_{c(\sigma,\sigma')}(\tbf{1},Pd\pmb v)\in L^0(\frak A_{\sigma})$ and

\Centerline{$\Bvalue{\Delta_{c(\sigma,\sigma')}(\tbf{1},Pd\pmb v)\ne 0}
\Bsubseteq\Bvalue{\sigma<\sigma'}\Bsubseteq\Bvalue{\sigma<\tau}$,}

\noindent so
$\Delta_{c(\sigma,\sigma')}(\tbf{1},Pd\pmb v)\in L^0(\frak A_{\Cal S<\tau})$,
by 641B(b-ii).   Accordingly
$S_{I\wedge\tau}(\tbf{1},Pd\pmb v)\in L^0(\frak A_{\Cal S<\tau})$ for every
$I\in\Cal I(\Cal S\wedge\tau)$.
Since $L^1_{\bar\mu}\cap L^0(\frak A_{\Cal S<\tau})$
is closed in $L^1_{\bar\mu}$ for the weak topology of $L^1_{\bar\mu}$
(613B(i-ii)), $v^{\#}_{\tau}\in L^0(\frak A_{\Cal S<\tau})$.
}%end of proof of 641D

\leader{641E}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a fully adapted process and $I$ a non-empty finite
sublattice of $\Cal S$.

(a) For any $\tau\in\Cal T$ there is an
element $u_{I<\tau}$ of $L^0(\frak A_{I<\tau})$ defined by saying that
$\Bvalue{\tau\le\min I}\penalty-100\Bsubseteq\Bvalue{u_{I<\tau}=0}$ and

\Centerline{$\Bvalue{\sigma<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau})
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma}}$}

\noindent for every $\sigma\in I$.

(b) If
$(\sigma_0,\ldots,\sigma_n)$ linearly generates the $I$-cells, then

\Centerline{$\Bvalue{\tau\le\sigma_0}\Bsubseteq\Bvalue{u_{I<\tau}=0}$,
\quad$\Bvalue{\sigma_n<\tau}\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_n}}$,}

\Centerline{$\Bvalue{\sigma_i<\tau}\Bcap\Bvalue{\tau\le\sigma_{i+1}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_i}}$ for every $i<n$.}

(c) The process $\family{\tau}{\Cal T}{u_{I<\tau}}$ is fully adapted.

(d) If $J$ is a maximal totally ordered subset of $I$, then
$u_{J<\tau}=u_{I<\tau}$ for every $\tau\in\Cal T$.

(e) If $\tau\in\Cal S$ then $u_{I<\tau}=u_{(I\wedge\tau)<\tau}$.

\proof{{\bf (a)} For $\sigma\in I$ set

\Centerline{$a_{\sigma}
=\Bvalue{\sigma<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau})
\in\frak A_{I<\tau}$.}

\noindent Let $B\subseteq\frak A_{I<\tau}$ 
be the set of atoms of the subalgebra generated by
$\{a_{\sigma}:\sigma\in I\}$.   If $b\in B$, either $b\Bcap a_{\sigma}=0$ for
every $\sigma\in I$ and

\Centerline{$b=1\Bsetminus\sup_{\sigma\in I}a_{\sigma}
\Bsupseteq\Bvalue{\tau\le\min I}$}

\noindent or $J_b=\{\sigma:\sigma\in I$, $b\Bsubseteq a_{\sigma}\}$ is
non-empty.   In the former case, set $v_b=0$.   In the latter case, 
if $\sigma$, $\sigma'\in J_b$,
$b$ is included in $\Bvalue{\sigma<\tau}\Bcap\Bvalue{\sigma'<\tau}$ so
cannot meet
either $\Bvalue{\sigma<\sigma'}$ or $\Bvalue{\sigma'<\sigma}$ for any
$\sigma$, $\sigma'\in J_b$, and

\Centerline{$b\Bsubseteq\Bvalue{\sigma=\sigma'}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma'}}$}

\noindent for any $\sigma$, $\sigma'\in J_b$.   Accordingly
we can write $v_b$ for the
common value of $u_{\sigma}\times\chi b$ for any $\sigma\in J_b$.   Now
if $\sigma\in J_b$ and $a\in\frak A_{\sigma}$, then
$a\Bcap b=a\Bcap\Bvalue{\sigma<\tau}\Bcap b$ belongs to $\frak A_{I<\tau}$ for
every $a\in\frak A_{\sigma}$, so
$v_b=u_{\sigma}\times\chi b$ belongs to $L^0(\frak A_{I<\tau})$.   We therefore
have an element $v=\sum_{b\in B}v_b$ of $L^0(\frak A_{I<\tau})$ such that
$v\times\chi b=v_b\times\chi b$ for every $b\in B$ and consequently
$v\times\chi a_{\sigma}=u_{\sigma}\times\chi a_{\sigma}$, that is,
$a_{\sigma}\Bsubseteq\Bvalue{v=u_{\sigma}}$, for every
$\sigma\in I$.   Since also
$\Bvalue{\tau\le\min I}\Bsubseteq 1\Bsetminus\sup_{\sigma\in I}a_{\sigma}
\Bsubseteq\Bvalue{v=0}$, this $v$ has the property
required of $u_{I<\tau}$, and the formula given does indeed define a member of
$L^0(\frak A_{I<\tau})$.

\medskip

{\bf (b)} Of course

\Centerline{$\Bvalue{\tau\le\sigma_0}=\Bvalue{\tau\le\min I}
\Bsubseteq\Bvalue{u_{I<\tau}=0}$.}

\noindent If $i<n$ and $\sigma\in I$ then

\Centerline{$\Bvalue{\tau\le\sigma_{i+1}}
   \Bcap\Bvalue{\sigma_i<\sigma}\Bcap\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\sigma_i<\sigma}\Bcap\Bvalue{\sigma<\sigma_{i+1}}
=0$}

\noindent (611Kd).   So

\Centerline{$\Bvalue{\sigma_i<\tau}\Bcap\Bvalue{\tau\le\sigma_{i+1}}
\Bsubseteq\Bvalue{\sigma_i<\tau}\Bsetminus\sup_{\sigma\in I}
   (\Bvalue{\sigma_i<\sigma}\Bcap\Bvalue{\sigma<\tau})
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_i}}$.}

\noindent At the top end, since $\sigma_n=\max I$,
$\Bvalue{\sigma_n<\sigma}\Bcap\Bvalue{\sigma<\tau}=0$ for every
$\sigma\in I$, so $\Bvalue{\sigma_n<\tau}\penalty-100
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_n}}$.

\medskip

{\bf (c)} As $\frak A_{I<\tau}\subseteq\frak A_{\tau}$,
$u_{I<\tau}\in L^0(\frak A_{\tau})$ for every $\tau\in\Cal T$.
If $\tau$, $\tau'\in\Cal T$, $c=\Bvalue{\tau=\tau'}$ and
$\sigma\in I$,
then $c\Bcap\Bvalue{\sigma<\tau}=c\Bcap\Bvalue{\sigma<\tau'}$ for every
$\sigma\in I$ (611E(c-iii-$\gamma$)), so

$$\eqalign{&c\Bcap(\Bvalue{\sigma<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau}))\cr&
\mskip150mu
=c\Bcap(\Bvalue{\sigma<\tau'}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau'}))\cr&
\mskip150mu
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma}}\Bcap\Bvalue{u_{I<\tau'}=u_{\sigma}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{I<\tau'}}.\cr}$$

\noindent At the same time

$$\eqalign{c\Bcap\Bvalue{\tau\le\min I}&
=c\Bcap\Bvalue{\tau'\le\min I}\cr&
\Bsubseteq\Bvalue{u_{I<\tau}=0}\Bcap\Bvalue{u_{I<\tau'}=0}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{I<\tau'}}.\cr}$$

\noindent So $c\Bsubseteq\Bvalue{u_{I<\tau}=u_{I<\tau'}}$;
as $\tau$ and $\tau'$ are arbitrary, $\family{\tau}{\Cal T}{u_{I<\tau}}$
is fully adapted.

\medskip

{\bf (d)} If $\langle\sigma_i\rangle_{i\le n}$ is the increasing enumeration of
$J$, then it linearly generates both the $I$-cells and the $J$-cells, so
(b) gives the result.

\medskip

{\bf (e)} For any $\sigma\in I$,

$$\eqalign{&\Bvalue{\sigma<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau})\cr&
\mskip80mu
=\Bvalue{\sigma\wedge\tau<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma\wedge\tau<\sigma'\wedge\tau}
      \Bcap\Bvalue{\sigma'\wedge\tau<\tau})\cr&
\mskip80mu
=\Bvalue{\sigma\wedge\tau<\tau}\Bsetminus\sup_{\sigma'\in I\wedge\tau}
   (\Bvalue{\sigma\wedge\tau<\sigma'}\Bcap\Bvalue{\sigma'<\tau})\cr&
\mskip80mu
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma\wedge\tau}}
   \Bcap\Bvalue{u_{(I\wedge\tau)<\tau}=u_{\sigma\wedge\tau}}\cr}$$

\noindent while

\Centerline{$\Bvalue{\tau\le\min I}=\Bvalue{\tau\le\min(I\wedge\tau)}
\Bsubseteq\Bvalue{u_{I<\tau}=0}\Bcap\Bvalue{u_{(I\wedge\tau)<\tau}=0}$.}
}%end of proof of 641E

\leader{641F}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u=\sfamilysS{u}$ a fully adapted process.   For $\tau\in\Cal S$,
set

\Centerline{$u_{<\tau}=\lim_{I\uparrow\CalIS}u_{I<\tau}$}

\noindent when the limit exists in $L^0(\frak A)$\cmmnt{, defining
$u_{I<\tau}$ as in 641E}.   If $u_{<\tau}$ is defined for every
$\tau\in\Cal S$, I will call $\pmb u_<=\family{\tau}{\Cal S}{u_{<\tau}}$ the
{\bf previsible version} of $\pmb u$.

\leader{641G}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.

(a) Let
$\pmb u=\sfamilysS{u}$ be a fully adapted process with a previsible version
$\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$.

\quad(i) $u_{<\sigma}\in L^0(\frak A_{\Cal S<\sigma})$ for every
$\sigma\in\Cal S$.

\quad(ii) $\pmb u_<$ is fully adapted to $\family{t}{T}{\frak A_t}$.

\quad(iii) $\Bvalue{\pmb u_<\ne\tbf 0}\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}$.
%654L

\quad(iv) If
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$, then
$z\pmb u$\cmmnt{ (definition: 612De)} has a previsible version, which is
$z\pmb u_<$.
%645D 645K

\quad(v) If $\Cal S$ has a least element, then $u_{<\min\Cal S}=0$.
%641P

\quad(vi) If $\Cal S'$ is a sublattice of $\Cal S$ which covers $\Cal S$, then
$\pmb u\restr\Cal S'$ has a previsible version, which is
$\pmb u_<\restr\Cal S'$.

\quad(vii) Suppose that $\pmb u$ is order-bounded.

\qquad($\alpha$) For any $\tau\in\Cal S$, $|u_{<\tau}|
\le\sup_{\sigma\in\Cal S}(|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau})$.

\qquad($\beta$) For any $\tau$, $\tau'\in\Cal S$,

\Centerline{$|u_{<\tau'}|\times\chi\Bvalue{\tau<\tau'}
\le\sup_{\sigma\in\Cal S\vee\tau}(|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau'})$.}

\qquad($\gamma$) $\pmb u_<$ is order-bounded
and $\sup|\pmb u_<|\le\sup|\pmb u|$.

\quad(viii) If $\pmb u$ is locally order-bounded then $\pmb u_<$ is locally
order-bounded.

(b) Writing $\tbf 1^{(\Cal S)}$ for the constant process with value
$\chi 1$ and domain $\Cal S$, its previsible version
$\tbfoneminus{\Cal S}$ is defined and equal to $\sfamilysS{\chi e}$, where
$e_{\sigma}=\sup_{\sigma'\in\Cal S}\Bvalue{\sigma'<\sigma}$ for
$\sigma\in\Cal S$.
%646I 647I

(c) Suppose that $\pmb u=\sfamilysS{u}$ is a fully adapted process.

\quad(i) $\pmb u$ has a previsible version iff
$\pmb u\restr\Cal S\wedge\tau$ has a previsible
version for every $\tau\in\Cal S$.

\quad(ii) In this case,
$(\pmb u\restr\Cal S\wedge\tau)_<=\pmb u_<\restr\Cal S\wedge\tau$ and

\Centerline{$(\pmb u\restr\Cal S\vee\tau)_<
=(\pmb u_<\restr\Cal S\vee\tau)
  \times\family{\sigma}{\Cal S\vee\tau}{\chi\Bvalue{\tau<\sigma}}
=(\pmb u_<\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}$}

\noindent for every $\tau\in\Cal S$.
%646I 654M

(d) Suppose that $k\ge 1$ is an integer, and
$h:\BbbR^k\to\Bbb R$ is a continuous function.   Take
$\pmb U=\ofamily{i}{k}{\pmb u_i}$ where each $\pmb u_i$ is a fully adapted
process with domain $\Cal S$ with a previsible version $\pmb u_{i<}$, and set
$\pmb U_<=\ofamily{i}{k}{\pmb u_{i<}}$ where $\pmb u_{i<}$ is the
previsible version of $\pmb u_i$ for each $i$.
Define $\bar h:(L^0)^k\to L^0$ and $\bar h\pmb U=\bar h\frsmallcirc\pmb U$
as in 619E-619F.   Then $\bar h\pmb U$ has a previsible version
$(\bar h\pmb U)_<=\bar h\frsmallcirc(\pmb U_<)\times\tbfoneminus{\Cal S}$.
%645D 645K
If $h(0,\ldots,0)=0$, then $(\bar h\pmb U)_<=\bar h\frsmallcirc\pmb U_<$.
%642D 654L

(e) Let $M$ be the set of those order-bounded processes $\pmb u$ with domain
$\Cal S$ such that $\pmb u$ has a previsible version $\pmb u_<$.

\quad(i) $M$ is an $f$-subalgebra of $\Mob(\Cal S)$, and
$\pmb u\mapsto\pmb u_<:M\to\Mob(\Cal S)$ is an $f$-algebra homomorphism.

\quad(ii) $M$ is closed for the ucp topology on
$\Mob(\Cal S)$, and $\pmb u\mapsto\pmb u_<:M\to\Mob(\Cal S)$
is continuous.

\proof{{\bf (a)(i)} We know that $u_{I<\sigma}$, as defined in
641E, belongs to $L^0(\frak A_{I<\sigma})\subseteq L^0(\frak A_{\Cal S<\sigma}$
for every non-empty $I\in\CalIS$ (641Ea, 641B(b-iii)).
Since $L^0(\frak A_{\Cal S<\sigma})$ is closed in $L^0(\frak A)$
(613B(i-i)), $u_{<\sigma}\in L^0(\frak A_{\Cal S<\sigma})$.

\medskip

\quad{\bf (ii)} By (i),
$u_{<\sigma}\in L^0(\frak A_{\Cal S<\sigma}\subseteq L^0(\frak A_{\sigma}$
for every
$\sigma\in\Cal S$,   Now suppose that $\sigma$, $\sigma'\in\Cal S$ and write
$c$ for $\Bvalue{\sigma=\sigma'}$.   For each
$I\in\CalIS\setminus\{\emptyset\}$,
$u_{I<\sigma}\times\chi c=u_{I<\sigma'}\times\chi c$ (641Ec), so
$u_{<\sigma}\times\chi c=u_{<\sigma'}\times\chi c$, that is,
$c\Bsubseteq\Bvalue{u_{<\sigma}=u_{<\sigma'}}$.   As $\sigma$ and $\sigma'$ are
arbitrary, $\pmb u_<$ is fully adapted.

\medskip

\quad{\bf (iii)} Set

\Centerline{$b=1\Bsetminus\Bvalue{\pmb u\ne\tbf 0}
=\inf_{\sigma\in\Cal S}\Bvalue{u_{\sigma}=0}$.}

\noindent Looking at the formula in 641Ea, we see that
$b\Bsubseteq\Bvalue{u_{I<\tau}=0}$ whenever $\tau\in\Cal S$ and
$I\in\CalIS$ is non-empty.   So $b\Bsubseteq\Bvalue{u_{<\tau}=0}$
whenever $\tau\in\Cal S$ and
$\Bvalue{\pmb u_<\ne\tbf 0}\Bsubseteq 1\Bsetminus b=\Bvalue{\pmb u\ne\tbf 0}$.

\medskip

\quad{\bf (iv)} Set $v_{\sigma}=z\times u_{\sigma}$ for $\sigma\in\Cal S$, so
that $\sfamilysS{v}=z\pmb u$ is fully adapted (612D(e-i)), and define
$v_{I<\sigma}$, for $\sigma\in\Cal S$ and non-empty $I\in\CalIS$,
as in 641E.   Again referring to the formula in 641Ea, we see that
$v_{I<\sigma}=z\times u_{I<\sigma}$ for all $\sigma$ and $I$.   But now we have

\Centerline{$\lim_{I\uparrow\CalIS}v_{I<\sigma}
=\lim_{I\uparrow\CalIS}z\times u_{I<\sigma}
=z\times\lim_{I\uparrow\CalIS}u_{I<\sigma}
=z\times u_{<\sigma}$}

\noindent for every $\sigma\in\Cal S$, so $(z\pmb u)_<$ is defined and equal to
$z\pmb u_{<}$.

\medskip

\quad{\bf (v)} If $I\in\CalIS$ is non-empty, then
$\min\Cal S\le\min I$ so $u_{I<\min\Cal S}=0$;  taking the limit as $I$
increases, $u_{<(\min\Cal S)}=0$.

\medskip

\quad{\bf (vi)} Take $\tau\in\Cal S'$ and $\epsilon>0$.   Then there is a
non-empty $I_0\in\CalIS$ such that
$\theta(u_{<\tau}-u_{I<\tau})\le\epsilon$ whenever $I\in\CalIS$
includes $I_0$.   Let $J_0$ be a finite sublattice of $\Cal S'$, containing
$\tau$, such that $\bar\mu a\le\epsilon$ where

\Centerline{$a=\sup_{\sigma\in I_0}
  (1\Bsetminus\sup_{\rho\in J_0}\Bvalue{\sigma=\rho})$.}

\noindent Take any $J\in\Cal I(\Cal S')$ including $J_0$,
and set $I=J\sqcup I_0$.   Then
$\Bvalue{u_{I<\tau}\ne u_{J<\tau}}\Bsubseteq a$.   \Prf\ Let $\frak B$ be
the subalgebra of $\frak A$ generated by

\Centerline{$\{a\}\cup\{\Bvalue{\sigma=\rho}:\sigma\in I_0$, $\rho\in J\}
\Bcup\{\Bvalue{\sigma\le\rho}:\sigma$, $\rho\in I\}$}

\noindent
and let $b$ be any atom of $\frak B$ disjoint from $a$.   Then for every
$\sigma\in I_0$ there is a $\rho\in J$ such that $b$ meets
$\Bvalue{\sigma=\rho}$, in which case $b\Bsubseteq\Bvalue{\sigma=\rho}$.
The set

\Centerline{$\bigcup_{\rho\in J}\{\sigma:\sigma\in\Cal S$,
$b\Bsubseteq\Bvalue{\sigma=\rho}\}$}

\noindent is a sublattice of $\Cal S$ including $I_0\cup J$ and therefore
including $I$.   For each $\sigma\in I$, $b$ is included in one of
$\Bvalue{\tau\le\sigma}$, $\Bvalue{\sigma<\tau}$ because
$\tau\in J_0\subseteq I$.   If $b\Bsubseteq\Bvalue{\tau\le\sigma}$ for every
$\sigma\in I$, then

\Centerline{$b\Bsubseteq\Bvalue{\tau\le\min I}\Bsubseteq\Bvalue{\tau\le\min J}
\Bsubseteq\Bvalue{u_{I<\tau}=0}\Bcap\Bvalue{u_{J<\tau}=0}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{J<\tau}}$.}

\noindent Otherwise, set
$\rho=\sup\{\sigma:\sigma\in I$, $b\Bsubseteq\Bvalue{\sigma<\tau}\}$;
then $b\Bsubseteq\Bvalue{\rho<\tau}$, and if $\sigma\in I$ and
$\sigma\not\le\rho$, $b\Bsubseteq\Bvalue{\tau\le\sigma}$.   So
$b\Bsubseteq\Bvalue{u_{I<\tau}=u_{\rho}}$.   Next, let $\rho'\in J$ be such
that $b\Bsubseteq\Bvalue{\rho=\rho'}$.   Then $b\Bsubseteq\Bvalue{\rho'<\tau}$,
and if $\sigma'\in J$ and $\sigma'\not\le\rho'$,
$b\Bsubseteq\Bvalue{\tau\le\sigma'}$.   So
$b\Bsubseteq\Bvalue{u_{J<\tau}=u_{\rho'}}$.   Putting these together,
$b\Bsubseteq\Bvalue{u_{I<\tau}=u_{J<\tau}}$ in this case also.   As
$b$ is arbitrary, $\Bvalue{u_{I<\tau}=u_{J<\tau}}$ includes $1\Bsetminus a$ and
$\Bvalue{u_{I<\tau}\ne u_{J<\tau}}\Bsubseteq a$.\ \Qed

It follows that

\Centerline{$\theta(u_{<\tau}-u_{J<\tau})
\le\theta(u_{<\tau}-u_{I<\tau})+\bar\mu\epsilon\le 2\epsilon$,}

\noindent and this is true whenever $J_0\subseteq J\in\Cal I(\Cal S')$.
As $\epsilon$ is arbitrary,
$u_{<\tau}=\lim_{J\uparrow\Cal I(\Cal S')}u_{J<\tau}$.   As $\tau$ is
arbitrary, $(\pmb u\restr\Cal S')_<$ is defined and equal to
$\pmb u_<\restr\Cal S'$.

\medskip

\quad{\bf (vii)}\grheada\ In the formulae of 641Ea, we see that if
$\sigma\in I\in\Cal I(\Cal S)$ then

\Centerline{$\Bvalue{\sigma<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau})
\Bsubseteq\Bvalue{|u_{I<\tau}|=|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau}}$,}

\noindent while $\Bvalue{\tau\le\min I}\Bsubseteq\Bvalue{u_{I<\tau}=0}$.   So
$|u_{I<\tau}|
\le\sup_{\sigma\in I}(|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau})$;  in the
limit as $I\uparrow\Cal I(\Cal S)$,
$|u_{<\tau}|
\le\sup_{\sigma\in\Cal S}(|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau})$.

\medskip

\qquad\grheadb\ By the same argument, we see that if
$\tau\in I\in\Cal I(\Cal S)$ then
$|u_{I<\tau'}|\times\chi\Bvalue{\tau<\tau'}
\le\sup_{\sigma\in\Cal S\vee\tau}|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau'}$,
so $|u_{<\tau'}|\times\chi\Bvalue{\tau<\tau'}
\le\sup_{\sigma\in\Cal S\vee\tau}|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau'}$.

\medskip

\qquad\grheadc\ And ($\alpha$) tells us that
$|\pmb u_<|$ is bounded above by $\sup|\pmb u|$.

\medskip

\quad{\bf (viii)} Similarly, if $\pmb u$ is locally order-bounded and
$\tau\in\Cal S$, then $|u_{I<\sigma}|\le\sup|\pmb u\restr\Cal S\wedge\tau|$
whenever $\sigma\in\Cal S\wedge\tau$ and
$I\in\CalIS$ is non-empty, so
$\sup|\pmb u_<\restr\Cal S\wedge\tau|\le\sup|\pmb u\restr\Cal S\wedge\tau|$.
(Or see (c) below.)

\medskip

{\bf (b)} Set $u_{\sigma}=\chi 1$ for $\sigma\in\Cal S$, so that
$\tbf 1^{(\Cal S)}=\sfamilysS{u}$.
This time we see that for any $\sigma\in\Cal S$ and non-empty
$I\in\CalIS$,

\Centerline{$\Bvalue{\min I<\sigma}\Bsubseteq\Bvalue{u_{I<\sigma}=\chi 1}$,
\quad$\Bvalue{\sigma\le\min I}\Bsubseteq\Bvalue{u_{I<\sigma}=0}$,}

\noindent so in fact $u_{I<\sigma}=\chi\Bvalue{\min I<\sigma}$.   Now

$$\eqalign{u_{<\sigma}&
=\lim_{I\uparrow\CalIS}\chi\Bvalue{\min I<\sigma}
=\sup_{I\in\CalIS}\chi\Bvalue{\min I<\sigma}\cr&
=\sup_{\sigma'\in\Cal S}\chi\Bvalue{\sigma'<\sigma}
=\chi(\sup_{\sigma'\in\Cal S}\Bvalue{\sigma'<\sigma})
=\chi e_{\sigma},\cr}$$

\noindent as claimed.

\medskip

{\bf (c)(i)}\grheada\ Suppose that $\pmb u$ has a previsible version
$\pmb u_<$, and $\tau\in\Cal S$.
If $\sigma\in\Cal S\wedge\tau$ and $I\in\CalIS$
contains $\tau$, then $u_{I<\sigma}=u_{(I\wedge\tau)<\sigma}$.   So

\Centerline{$\lim_{J\uparrow\Cal I(\Cal S\wedge\tau)}u_{J<\sigma}
=\lim_{I\uparrow\CalIS}u_{(I\wedge\tau)<\sigma}
=\lim_{I\uparrow\CalIS}u_{I<\sigma}
=u_{<\sigma}$.}

\noindent As $\sigma$ is arbitrary, $\pmb u\restr\Cal S\wedge\tau$ has a
previsible version, which is in fact $\pmb u_<\restr\Cal S\wedge\tau$.

\medskip

\qquad\grheadb\ Now suppose that $\tau\in\Cal S$ and
$\pmb w=\pmb u\restr\Cal S\wedge\tau$ has a previsible version
$\pmb w_<=\family{\sigma}{\Cal S\wedge\tau}{w_{<\sigma}}$.   Then
$u_{I<\tau}=w_{(I\wedge\tau)<\tau}$ for every $I\in\CalIS$ containing
$\tau$, and $\lim_{I\uparrow\CalIS}u_{I<\tau}=
\lim_{J\uparrow\Cal I(\Cal S\wedge\tau)}w_{J<\tau}=w_{<\tau}$ is defined.

So if $\pmb u\restr\Cal S\wedge\tau$ has a previsible version for every
$\tau\in\Cal S$, $\lim_{I\uparrow\CalIS}u_{I<\tau}$ is defined for every
$\tau\in\Cal S$ and $\pmb u$ has a previsible version.

\medskip

\quad{\bf (ii)}\grheada\ We saw in (i-$\alpha$) above that
$(\pmb v\restr\Cal S\wedge\tau)_<=\pmb v_<\restr\Cal S\wedge\tau$.

\medskip

\qquad\grheadb\ Take $\sigma\in\Cal S\vee\tau$, and set
$c=\Bvalue{\tau<\sigma}$.   If $I$ is a
finite sublattice of $\Cal S$ containing $\sigma$ and $\tau$,
and $J=I\vee\tau$, then for $\sigma'\in I$ set

\Centerline{$a_{\sigma'}=\Bvalue{\sigma'<\sigma}
  \Bsetminus\sup_{\sigma''\in I}
    (\Bvalue{\sigma'<\sigma''}\Bcap\Bvalue{\sigma''<\sigma})$}

\noindent and for $\sigma'\in J$ set

\Centerline{$b_{\sigma'}=\Bvalue{\sigma'<\sigma}
  \Bsetminus\sup_{\sigma''\in J}
    (\Bvalue{\sigma'<\sigma''}\Bcap\Bvalue{\sigma''<\sigma})$.}

\noindent Observe that $b_{\sigma'}=a_{\sigma'}\Bsubseteq c$ for every
$\sigma'\in J$.   So, for $\sigma'\in J$, we have

\Centerline{$b_{\sigma'}
\Bsubseteq\Bvalue{u_{I<\sigma}=u_{\sigma'}}
  \Bcap\Bvalue{u_{J<\sigma}=u_{\sigma'}}
\Bsubseteq\Bvalue{u_{I<\sigma}=u_{J<\sigma}}$.}

\noindent Since

\Centerline{$c=\Bvalue{\tau<\sigma}=\Bvalue{\min J<\sigma}
=\sup_{\sigma'\in J}b_{\sigma'}$}

\noindent (see the construction in 641Ea),
$c\Bsubseteq\Bvalue{u_{I<\sigma}=u_{J<\sigma}}$ and
$c\Bsubseteq\Bvalue{u_{J<\sigma}=\chi c\times u_{I<\sigma}}$.

On the other hand,

\Centerline{$1\Bsetminus c=\Bvalue{\tau=\sigma}
\Bsubseteq\Bvalue{u_{J<\sigma}=0}$}

\noindent and
$1\Bsetminus c\Bsubseteq\Bvalue{u_{J<\sigma}=\chi c\times u_{I<\sigma}}$.
So $u_{J<\sigma}=\chi c\times u_{I<\sigma}$.
Taking the limit as $I$ increases through $\CalIS$,

\Centerline{$\lim_{J\uparrow\Cal I(\Cal S\vee\tau)}u_{J<\sigma}
=\lim_{I\uparrow\CalIS}u_{(I\vee\tau)<\sigma}
=\chi c\times u_{<\sigma}$.}

As $\sigma$ is arbitrary,

\Centerline{$(\pmb u\restr\Cal S\vee\tau)_<
=(\pmb u_<\restr\Cal S\vee\tau)
  \times\family{\sigma}{\Cal S\vee\tau}{\chi\Bvalue{\tau<\sigma}}
=(\pmb u_<\restr\Cal S\vee\tau)
  \times\tbfoneminus{\Cal S\vee\tau}$}

\noindent by (b) applied in $\Cal S\vee\tau$.

\medskip

{\bf (d)} For each $i$, express $\pmb u_i$ as
$\family{\sigma}{\Cal S}{u_{i\sigma}}$
and $\pmb u_{i<}$ as $\family{\sigma}{\Cal S}{u_{i<\sigma}}$
where $u_{i<\sigma}=\lim_{I\uparrow\CalIS}u_{i,I<\sigma}$.
Write $U_{I<\sigma}=\ofamily{i}{k}{u_{i,I<\sigma}}$ for
$\sigma\in\Cal S$ and non-empty $I\in\CalIS$.   Then

\Centerline{$\lim_{I\uparrow\CalIS}U_{I<\sigma}
=\ofamily{i}{k}{\lim_{I\uparrow\CalIS}u_{i,I<\sigma}}
=\ofamily{i}{k}{u_{i<\sigma}}$}

\noindent is defined for any $\sigma\in\Cal S$;  I will call it $U_{<\sigma}$.

Set $v_{\sigma}=\bar h(\ofamily{i}{k}{u_{i\sigma}})$
and $z_{\sigma}=\chi 1$ for $\sigma\in\Cal S$, so
that $\bar h\pmb U=\sfamilysS{v}$ and $\tbf 1^{(\Cal S)}=\sfamilysS{z}$.
If $\tau\in I\in\CalIS$ then

\Centerline{$\Bvalue{\min I=\tau}
\Bsubseteq\Bvalue{v_{I<\tau}=0}\Bcap\Bvalue{z_{I<\tau}=0}
\Bsubseteq\Bvalue{v_{I<\tau}=\bar h(U_{I<\tau})\times z_{I<\tau}}$,}

\Centerline{$\Bvalue{\min I<\tau}
\Bsubseteq\Bvalue{v_{I<\tau}=\bar h(U_{I<\tau})}
   \Bcap\Bvalue{z_{I<\tau}=\chi 1}
\Bsubseteq\Bvalue{v_{I<\tau}=\bar h(U_{I<\tau})\times z_{I<\tau}}$}

\noindent while if $h(0,\ldots,0)=0$ we have

$$\eqalign{\Bvalue{\min I=\tau}
&\Bsubseteq\Bvalue{v_{I<\tau}=0}\Bcap\Bvalue{U_{I<\tau}=0}
\cr&\Bsubseteq\Bvalue{v_{I<\tau}=0}\Bcap\Bvalue{\bar h(U_{I<\tau})=0}
\Bsubseteq\Bvalue{v_{I<\tau}=\bar h(U_{I<\tau})}.
\cr}$$

\noindent In the limit, we see that

\Centerline{$v_{<\tau}
=\lim_{I\uparrow\CalIS}\bar h(U_{I<\tau})\times z_{I<\tau}
=\bar h(U_{<\tau})\times z_{<\tau}$}

\noindent because $\bar h:(L^0)^k\to L^0$ and $\times:L^0\times L^0$ are
continuous (619Ed, 613Ba), while if $h(0,\ldots,0)=0$ then

\Centerline{$v_{<\tau}
=\lim_{I\uparrow\CalIS}\bar h(U_{I<\tau})
=\bar h(U_{<\tau})$.}

\noindent As $\tau$ is arbitrary,

\Centerline{$(\bar h\pmb U)_<=\pmb v_<
=(\bar h\pmb U_<)\times\tbfoneminus{\Cal S}$}

\noindent and $(\bar h\pmb U)_<=\bar h\pmb U_<$ if $h(0,\ldots,0)=0$.

\medskip

{\bf (e)(i)} This is immediate from (d), applied to addition and multiplication
and the lattice operations.

\medskip

\quad{\bf (ii)}\grheada\
Suppose that $\pmb u=\sfamilysS{u}$ belongs to the closure of
$M\cap\Mob(\Cal S)$ in $\Mob(\Cal S)$, $\sigma\in\Cal S$
and $\epsilon>0$.   Then there are $\pmb w\in M\cap\Mob(\Cal S)$ and
non-empty $I\in\CalIS$ such that

\Centerline{$\theta(\sup|\pmb u-\pmb w|)\le\epsilon$,
\quad$\theta|w_{J<\sigma}-w_{I<\sigma}|\le\epsilon$ whenever
$I\subseteq J\in\CalIS$.}

\noindent Now the same calculation as in (a-vii) shows that

\Centerline{$|u_{J<\sigma}-w_{J<\sigma}|\le\sup|\pmb u-\pmb w|$}

\noindent for every non-empty $J\in\CalIS$, so
$\theta(u_{J<\sigma}-u_{I<\sigma})\le 3\epsilon$ whenever
$I\subseteq J\in\CalIS$.   As $\epsilon$ is arbitrary,
$\lim_{I\uparrow\CalIS}u_{I<\sigma}$ is defined;
as $\sigma$ is arbitrary, $\pmb u\in M$.

\medskip

\qquad\grheadb\ And (a-vii) and (d) tell us again that
$\sup|\pmb v_<-\pmb w_<|\le\sup|\pmb v-\pmb w|$ for every $\pmb v$,
$\pmb w\in M$, so $\pmb v\mapsto\pmb v_<$ is continuous.
}%end of proof of 641G

\leader{641H}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\Cal S'$ a sublattice of $\Cal S$ which sfeparates $\Cal S$.   If 
$\pmb u=\sfamilysS{u}$ is fully adapted and $\tau\in\Cal S$ is such that
$u_{<\tau}$ is defined, then 
$u_{<\tau}=\lim_{I\uparrow\Cal I(\Cal S')}u_{I<\tau}$.

\proof{ Let $\epsilon>0$.   Then there is a non-empty
$I\in\Cal I(\Cal S)$ such that
$\theta(u_{<\tau}-u_{J<\tau})\le\epsilon$ whenever $J\in\Cal I(\Cal S)$ includes
$I$.   Next, there is a finite set $C\subseteq\Cal S'$ such that

\Centerline{$\sum_{\sigma\in I}\bar\mu(\Bvalue{\sigma<\tau}
  \Bsetminus\sup_{\rho\in C}(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}))
\le\epsilon$;}

\noindent  let $K_0$ be the sublattice of $\Cal S'$ generated by $C$.
Set 

\Centerline{$a=\sup_{\sigma\in I}(\Bvalue{\sigma<\tau}
  \Bsetminus\sup_{\rho\in C}(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}))$,}

\noindent so that $\bar\mu a\le\epsilon$.   Take any $K\in\Cal I(\Cal S')$
including $K_0$ and set $J=I\sqcup K$.   Then
$1\Bsetminus a\Bsubseteq\Bvalue{u_{J<\tau}=u_{K<\tau}}$.   \Prf\ Setting

\Centerline{$a_{\sigma}=\Bvalue{\sigma<\tau}
\Bsetminus\sup_{\rho\in J}(\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})$}

\noindent for $\sigma\in J$, as in part(a) of the proof of 641E, we see that
if $\sigma$, $\sigma'\in J$ then 
$a_{\sigma}\Bcap\Bvalue{\sigma=\sigma'}\Bsubseteq a_{\sigma'}$.   Now
if $\sigma\in K$ then

$$\eqalign{a_{\sigma}&
=a_{\sigma}\Bcap\Bvalue{\sigma<\tau}
   \Bsetminus\sup_{\rho\in K}(\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})\cr&
\Bsubseteq\Bvalue{u_{J<\tau}=u_{\sigma}}\Bcap\Bvalue{u_{K<\tau}=u_{\sigma}}
\Bsubseteq\Bvalue{u_{J<\tau}=u_{K<\tau}}.\cr}$$

\noindent And if $\sigma\in I$ then

$$\eqalign{a_{\sigma}\Bsetminus a&
\Bsubseteq a_{\sigma}
   \Bcap\sup_{\rho\in C}(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau})
   \Bsetminus\sup_{\rho\in C}(\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})\cr&
\Bsubseteq\sup_{\rho\in C}(a_{\sigma}\Bcap\Bvalue{\sigma=\rho})
\Bsubseteq\sup_{\rho\in C}a_{\rho}
\Bsubseteq\Bvalue{u_{J<\tau}=u_{K<\tau}}.\cr}$$

But we know that the covered envelope of $I\cup K$
is a sublattice (611?\query), so includes $J$.   So for any $\sigma\in J$ we
have

\Centerline{$a_{\sigma}\Bsetminus a
=\sup_{\rho\in I\cup K}a_{\sigma}\Bcap\Bvalue{\sigma=\rho}\Bsetminus a
\Bsubseteq\sup_{\rho\in I\cup K}a_{\rho}\Bsetminus a
\Bsubseteq\Bvalue{u_{J<\tau}=u_{K<\tau}}$.}

\noindent And we certainly have

$$\eqalign{\Bvalue{\tau\le\min J}&
=\Bvalue{\tau\le\min J}\Bcap\Bvalue{\tau\le\min K}\cr&
\Bsubseteq\Bvalue{u_{J<\tau}=0}\Bcap\Bvalue{u_{K<\tau}=0}
\Bsubseteq\Bvalue{u_{J<\tau}=u_{K<\tau}}.\cr}$$

\noindent Since $\Bvalue{\tau\le\min J}\Bcup\sup_{\sigma\in J}a_{\sigma}=1$,
we see that $1\Bsetminus a\Bsubseteq\Bvalue{u_{J<\tau}=u_{K<\tau}}$, as
claimed.\ \Qed

It follows that

$$\eqalignno{\theta(u_{K<\tau}-u_{<\tau})&
\le\theta(u_{K<\tau}-u_{J<\tau})+\theta(u_{J<\tau}-u_{<\tau})
\le\bar\mu\Bvalue{u_{K<\tau}\ne u_{J<\tau}}+\epsilon
\Displaycause{as $J\supseteq I$}
\le\bar\mu a+\epsilon
\le 2\epsilon.\cr}$$

\noindent And this is true whenever $K\in\Cal I(\Cal S')$ includes $K_0$.
As $\epsilon$ is arbitrary, 
$u_{<\tau}=\lim_{K\uparrow\Cal I(\Cal S')}u_{K<\tau}$.
}%end of proof of 641H

\leader{641I}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a simple process with breakpoint string
$(\tau_0,\ldots,\tau_n)$ and starting value $u_{\downarrow}$\cmmnt{ (614Ba)}.

(a) $u_{<\tau}$ is defined and

\Centerline{$\inf_{\sigma\in\Cal S}\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{u_{<\tau}=0}$,
\quad$\Bvalue{\sigma<\tau}\Bcap\Bvalue{\tau\le\tau_0}
\Bsubseteq\Bvalue{u_{<\tau}=u_{\downarrow}}$ for every $\sigma\in\Cal S$,}

\Centerline{$\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau\le\tau_{i+1}}
\Bsubseteq\Bvalue{u_{<\tau}=u_{\tau_i}}$ for every $i<n$,
\quad$\Bvalue{\tau_n<\tau}
  \Bsubseteq\Bvalue{u_{<\tau}=u_{\tau_n}}$}

\noindent for every $\tau\in\Cal S$.

(b) Writing $\pmb u_<$ for the previsible version of $\pmb u$,

\Centerline{$\sup|\pmb u|=|u_{\tau_n}|\vee\sup|\pmb u_<|$.}

\proof{{\bf (a)} Take any $I\in\CalIS$
such that $\tau_i\in I$ for every $i\le n$.
We have $\sigma_0\le\ldots\le\sigma_m$ and
$0\le k_0\le\ldots\le k_n\le m$ such that

\Centerline{$\sigma_{k_i}=\tau_i$ for $i\le n$,}

\Centerline{$(\sigma_0,\ldots,\sigma_{k_0})$ linearly generates the
$I\wedge\tau_0$-cells,}

\Centerline{$(\sigma_{k_i},\ldots,\sigma_{k_{i+1}})$
linearly generates the $I\cap[\tau_i,\tau_{i+1}]$-cells for each $i\le n$,}

\Centerline{$(\sigma_{k_n},\ldots,\sigma_m)$ linearly generates the
$I\vee\tau_n$-cells,}

\noindent and therefore $(\sigma_0,\ldots,\sigma_m)$ linearly generates the
$I$-cells.   Then

$$\eqalign{\Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau\le\sigma_{j+1}}
&\Bsubseteq\Bvalue{\sigma_j<\tau_0}
\Bsubseteq\Bvalue{u_{\sigma_j}=u_{\downarrow}}\text{ if }j<k_0,\cr
&\Bsubseteq\Bvalue{\tau_i\le\sigma_j}\Bcap\Bvalue{\sigma_j<\tau_{i+1}}
\Bsubseteq\Bvalue{u_{\sigma_j}=u_{\tau_i}}\cr
&\mskip100mu\text{ if }i\le n\text{ and }k_i\le j<k_{i+1},\cr
&\Bsubseteq\Bvalue{\tau_n\le\sigma_j}
\Bsubseteq\Bvalue{u_{\sigma_j}=u_{\tau_n}}\text{ if }k_n\le i<m.\cr}$$

By 641Eb, we have

\Centerline{$\Bvalue{\tau\le\sigma_0}\Bsubseteq\Bvalue{u_{I<\tau}=0}$,}

$$\eqalign{\Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau\le\sigma_{j+1}}
&\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_j}}
\Bcap\Bvalue{u_{\sigma_j}=u_{\downarrow}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\downarrow}}\text{ if }j<k_0,\cr
&\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_j}}
  \Bcap\Bvalue{u_{\sigma_j}=u_{\tau_i}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\tau_i}}\cr&
\mskip100mu\text{ if }i\le n\text{ and }k_i\le j<k_{i+1},\cr
&\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_j}}
  \Bcap\Bvalue{u_{\sigma_j}=u_{\tau_n}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\tau_n}}
  \text{ if }k_n\le i<m\cr}$$

\noindent and

\Centerline{$\Bvalue{\sigma_m<\tau}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma_m}}\Bcap\Bvalue{\tau_n\le\sigma_m}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\tau_n}}$.}

\noindent Since

\Centerline{$\Bvalue{\sigma_0\le\tau}\Bcap\Bvalue{\tau<\tau_0}
=\sup_{j<k_0}\Bvalue{\sigma_j\le\tau}\Bcap\Bvalue{\tau<\sigma_{j+1}}$,}

\Centerline{$\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau\le\tau_{i+1}}
=\sup_{k_i\le j<k_{i+1}}
   \Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau\le\sigma_{j+1}}$,}

\Centerline{$\Bvalue{\tau_n<\tau}
=\sup_{k_n\le j}
   \Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau\le\sigma_{j+1}}
   \Bcup\Bvalue{\sigma_m<\tau}$,}

\noindent we get

\Centerline{$\Bvalue{\tau\le\min I}\Bsubseteq\Bvalue{u_{I<\tau}=0}$,
\quad$\Bvalue{\min I<\tau}\Bcap\Bvalue{\tau<\tau_0}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\downarrow}}$,}

\Centerline{$\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau\le\tau_{i+1}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\tau_i}}$,
\quad$\Bvalue{\tau_n<\tau}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\tau_n}}$.}

If therefore we define $v$ by the given formula, so that

\Centerline{$\inf_{\sigma\in\Cal S}\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{v=0}$,
\quad$\Bvalue{\sigma<\tau}\Bcap\Bvalue{\tau\le\tau_0}
\Bsubseteq\Bvalue{v=u_{\downarrow}}$ for every $\sigma\in\Cal S$,}

\Centerline{$\Bvalue{\sigma_i<\tau}\Bcap\Bvalue{\tau\le\tau_{i+1}}
\Bsubseteq\Bvalue{v=u_{\tau_i}}$ for every $i<n$,
\quad$\Bvalue{\tau_n<\tau}
  \Bsubseteq\Bvalue{v=u_{\tau_n}}$,}

\noindent and set $c=\inf_{\sigma\in\Cal S}\Bvalue{\tau\le\sigma}$,
then

\Centerline{$c\Bcup\Bvalue{\min I<\tau}
\Bsubseteq\Bvalue{u_{I<\tau}=v}$}

\noindent whenever $\{\tau_0,\ldots,\tau_n\}\subseteq I\in\CalIS$.
It follows that, for any $\sigma\in\Cal S$,

\Centerline{$c\Bcup\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{u_{I<\tau}=v}$}

\noindent whenever
$\{\sigma,\tau_0,\ldots,\tau_n\}\subseteq I\in\CalIS$.
Consequently we shall have

\Centerline{$\lim_{I\uparrow\CalIS}
   u_{I<\tau}\times\chi(c\Bcup\Bvalue{\sigma<\tau})
=v\times\chi(c\Bcup\Bvalue{\sigma<\tau})$}

\noindent for every $\sigma\in\Cal S$, and therefore

\Centerline{$\lim_{I\uparrow\CalIS}
   u_{I<\tau}\times\chi(c\Bcup\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau})
=v\times\chi(c\Bcup\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau})$,}

\noindent that is,

\Centerline{$\lim_{I\uparrow\CalIS}u_{I<\tau}=v$.}

\medskip

{\bf (b)} Of course $|u_{\tau_n}|\le\sup|\pmb u|$, and
$\sup|\pmb u_<|\le\sup|\pmb u|$ by 641G(a-vii), so
$\sup|\pmb u|\ge|u_{\tau_n}|\vee\sup|\pmb u_<|$.

In the other direction, if $\sigma\in\Cal S$ then

\Centerline{$(\Bvalue{\sigma<\tau_0},
\Bvalue{\tau_0\le\sigma}\Bcap\Bvalue{\sigma<\tau_1}.
\ldots,\Bvalue{\sigma_{n-1}\le\sigma}\Bcap\Bvalue{\sigma<\tau_n},
\Bcap{\tau_n\le\sigma}$}

\noindent is a partition of unity in $\frak A$, so

$$\eqalign{|u_{\sigma}|&
=(|u_{\sigma}|\times\chi\Bvalue{\sigma<\tau_0})
   \vee\sup_{i<n}(|u_{\sigma}|
     \times\chi(\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}))\cr&
\mskip200mu
   \vee(|u_{\sigma}|\times\chi\Bvalue{\tau_n\le\sigma})\cr&
=(|u_{\downarrow}|\times\chi\Bvalue{\sigma<\tau_0})
   \vee\sup_{i<n}(|u_{\tau_i}|
     \times\chi(\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}))\cr&
\mskip200mu
   \vee(|u_{\tau_n}|\times\chi\Bvalue{\tau_n\le\sigma})\cr&
=(|u_{<\tau_0}|\times\chi\Bvalue{\sigma<\tau_0})
   \vee\sup_{i<n}(|u_{<\tau_{i+1}}|
     \times\chi(\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}))\cr&
\mskip200mu
   \vee(|u_{\tau_n}|\times\chi\Bvalue{\tau_n\le\sigma})\cr&
\le\sup|\pmb u_<|\vee|u_{\tau_n}|.\cr}$$

\noindent As $\sigma$ is arbitrary,
$\sup|\pmb u|\le|u_{\tau_n}|\vee\sup|\pmb u_<|$.
}%end of proof of 641I

\leader{641J}{}\cmmnt{ We can now complement 614C with a formula for
an integral
$\int_{\Cal S}\pmb u\,d\pmb v$ when it is the {\it integrator}
$\pmb v$ which is simple.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ a simple process with starting value
$v_{\downarrow}$ and breakpoint string $(\tau_0,\ldots,\tau_n)$, and
$\pmb u=\sfamilysS{u}$ a fully adapted process which has a
previsible version $\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$.
Then

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v
=u_{<\tau_0}\times(v_{\tau_0}-v_{\downarrow})
   +\sum_{i=1}^nu_{<\tau_i}\times(v_{\tau_i}-v_{\tau_{i-1}})$.}

\proof{{\bf (a)} Suppose that $i\le n$ and that

\inset{{\it either} $i=0$, $\Cal S'=\Cal S\wedge\tau_0$,
$\tau_0\in I\in\Cal I(\Cal S')$ and
$v=v_{\downarrow}$

{\it or} $1\le i\le n$, $\Cal S'=\Cal S\cap[\tau_{i-1},\tau_i])$,
$\tau_i\in I\in\Cal I(\Cal S')$ and $v=v_{\tau_{i-1}}$}

\noindent then

\Centerline{$S_I(\pmb u,d\pmb v)
=u_{I<\tau_i}\times(v_{\tau_i}-v)$.}

\noindent\Prf\ Let $(\sigma_0,\ldots,\sigma_k)$ linearly generate the
$I$-cells.   Then

$$\eqalignno{S_I(\pmb u,d\pmb v)
&=\sum_{j=0}^{k-1}u_{\sigma_j}\times(v_{\sigma_{j+1}}-v_{\sigma_j})\cr&
=\sum_{j=0}^{k-1}u_{\sigma_j}\times(v_{\tau_i}-v)
   \times\chi\Bvalue{\sigma_{j+1}=\tau_i}
   \times\chi\Bvalue{\sigma_j<\tau_i}
\Displaycause{because $\Bvalue{v_{\sigma_{j+1}}\ne v_{\sigma_j}}
\Bsubseteq\Bvalue{\sigma_j<\tau_i}\Bcap\Bvalue{\sigma_{j+1}=\tau_i}
\Bsubseteq\Bvalue{v=v_{\sigma_j}}\Bcap\Bvalue{v_{\tau_i}=v_{\sigma_{j+1}}}$}
=\sum_{j=0}^{k-1}u_{I<\tau_i}\times(v_{\tau_i}-v)
   \times\chi\Bvalue{\sigma_{j+1}=\tau_i}
   \times\chi\Bvalue{\sigma_j<\tau_i}
\Displaycause{641Eb}
=u_{I<\tau_i}\times(v_{\tau_i}-v)\cr}$$

\noindent because

\Centerline{$\Bvalue{u_{I<\tau_i}\ne 0}
\Bsubseteq\Bvalue{\sigma_0<\tau_i}
=\sup_{j<k}(\Bvalue{\sigma_j<\tau_i}\Bcap\Bvalue{\sigma_{j+1}=\tau_i})$.
\Qed}

\noindent Taking the limit as
$I\uparrow\Cal I(\Cal S')$, $\int_{\Cal S'}\pmb u\,d\pmb v$ is defined and
equal to $u_{<\tau_i}\times(v_{\tau_i}-v)$.

\medskip

{\bf (b)} At the top end, $\int_{\Cal S\vee\tau_n}\pmb u\,d\pmb v=0$
because $\pmb v$ is constant on $\Cal S\vee\tau_n$.   So

$$\eqalign{\int_{\Cal S}\pmb u\,d\pmb v
&=\int_{\Cal S\wedge\tau_0}\pmb u\,d\pmb v
    +\sum_{i=1}^n\int_{\Cal S\cap[\tau_{i-1},\tau_i]}\pmb u\,d\pmb v
    +\int_{\Cal S\vee\tau_n}\pmb u\,d\pmb v\cr
&=u_{<\tau_0}\times(v_{\tau_0}-v_{\downarrow})
   +\sum_{i=1}^nu_{<\tau_i}\times(v_{\tau_i}-v_{\tau_{i-1}}).\cr}$$

}%end of proof of 641J

\leader{641K}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, 
$\pmb u=\sfamilysS{u}$ a non-decreasing non-negative
fully adapted process and $\tau$ a member of $\Cal S$.

(a) If $I$ is a non-empty finite sublattice of $\Cal S$ then
$u_{I<\tau}=\sup_{\rho\in I}(u_{\rho}\times\chi\Bvalue{\rho<\tau})$.

(b) $u_{<\tau}$ is defined and equal to
$\sup_{\rho\in\Cal S}(u_{\rho}\times\chi\Bvalue{\rho<\tau})$.

\proof{{\bf (a)} I return to the formula of 641Ea.   Set 
$x=\sup_{\rho\in I}(u_{\rho}\times\chi\Bvalue{\rho<\tau})$ and

\Centerline{$a_{\sigma}
=\Bvalue{\sigma<\tau}\Bsetminus\sup_{\sigma'\in I}
   (\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau})$}

\noindent for $\sigma\in I$.   Let $b$ be an atom of the subalgebra generated by
$\{\Bvalue{\sigma<\rho}:\sigma$, $\rho\in I\cup\{\tau\}\}$.
If $b\Bcap a_{\sigma}=0$ for every $\sigma\in I$ then 

\Centerline{$b\Bsubseteq\Bvalue{u_{I<\tau}=0}\Bcap\Bvalue{x=0}
\Bsubseteq\Bvalue{u_{I<\tau}=x}$.}

\noindent If $\sigma\in I$ is such that $b\Bsubseteq a_{\sigma}$ then
$b\Bsubseteq\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{u_{\sigma}\le x}$.   At the
same time, if $\rho\in I$ and $b\Bsubseteq\Bvalue{\rho<\tau}$, then
$b\Bcap\Bvalue{\sigma<\rho}=0$ so

$$\eqalignno{b\Bsubseteq\Bvalue{\rho\le\sigma}&
\Bsubseteq\Bvalue{u_{\rho}\le u_{\sigma}}
\Displaycause{614Kb}
\Bsubseteq\Bvalue{u_{\rho}\times\chi\Bvalue{\rho<\tau}\le u_{\sigma}}\cr}$$

\noindent because $u_{\sigma}\ge 0$;  accordingly 
$b\Bsubseteq\Bvalue{x\le u_{\sigma}}$ and

\Centerline{$b\Bsubseteq a_{\sigma}\Bcap\Bvalue{x=u_{\sigma}}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma}}\Bcap\Bvalue{x=u_{\sigma}}
\Bsubseteq\Bvalue{u_{I<\tau}=x}$}

\noindent by 641Ea.   As $b$ is arbitrary, $u_{I<\tau}=x$, as claimed.

\medskip

{\bf (b)} It follows at once that if $I$, $J\in\Cal I(\Cal S)$ and
$I\subseteq J$, then $u_{I<\tau}\le u_{J<\tau}$.   Now

\Centerline{$u_{I<\tau}
=\sup_{\sigma\in I}(u_{\sigma}\times\chi\Bvalue{\sigma<\tau})
\le\sup_{\sigma\in I}u_{\sigma}
\le u_{\tau}$}

\noindent for every $I\in\CalIS$, so
$\sup_{I\in\Cal I(\Cal S)}u_{I<\tau}$ is defined and equal to
$\lim_{I\uparrow\Cal I(\Cal S)}u_{I<\tau}$.   And of course
$\sup_{I\in\Cal I(\Cal S)}u_{I<\tau}
=\sup_{\rho\in\Cal S}(u_{\rho}\times\chi\Bvalue{\rho<\tau})$.
}%end of proof of 641K

\leader{641L}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u$ a fully adapted process with domain $\Cal S$.

(a) If $\pmb u$ is non-decreasing and non-negative, 
it has a previsible version $\pmb u_<$;  $\pmb u_<$ is non-decreasing and
$\pmb u_<\le\pmb u$.

(b) If $\pmb u$ is (locally) of bounded variation, it has a previsible version
which is (locally) of bounded variation.

(c) If $\pmb u$ is (locally) moderately oscillatory, it has a previsible
version which is (locally) moderately oscillatory.

\proof{{\bf (a)} Express $\pmb u$ as $\sfamilysS{u}$.   By 641K, 
$u_{<\tau}=\sup_{\rho\in\Cal S}(u_{\rho}\times\chi\Bvalue{\rho<\tau})$ 
is defined for every $\tau\in\Cal S$;  the formula shows at once that
$u_{<\tau}\le u_{<\tau'}\le u_{\tau'}$ whenever $\tau\le\tau'$ in $\Cal S$.

\medskip

{\bf (b)} If $\pmb u$ is of bounded variation, it is expressible as the
difference $\pmb v'-\pmb v''$ of two order-bounded non-negative
non-decreasing processes (614I-614J), and now $\pmb u_<=\pmb v'_<-\pmb v''_<$
(641G(e-i)) is again the difference of two order-bounded non-negative
non-decreasing processes, by (a) here and 641G(a-vii), so is of bounded
variation.   If $\pmb u$ is locally of bounded variation then the same argument,
applied to $\pmb u\restr\Cal S\wedge\tau$ for $\tau\in\Cal S$, 
shows that $\pmb u_<$ is defined
and locally of bounded variation (using 641Gc).

\medskip

{\bf (c)} Now we know that $\Mmo(\Cal S)$ is the closure in $\Mob(\Cal S)$
of the space $\Mbv(\Cal S)$ of processes of bounded variation (615Ea), so
641G(e-i) tells us that every moderately oscillatory process on $\Cal S$ has a
previsible version.   Moreover, because $\pmb u_<\in\Mbv(\Cal S)$ for every
$\pmb u\in\Mbv(\Cal S)$, by (b), and the map $\pmb u\mapsto\pmb u_<$ is
continuous for the ucp topology (641G(e-ii)), $\pmb u_<$ will be moderately
oscillatory whenever $\pmb u$ is.

As in (b), it follows that $\pmb u_<$ is defined, and is locally moderately
oscillatory, whenever $\pmb u$ is locally moderately oscillatory.
}%end of proof of 641L

\leader{641M}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a locally
moderately oscillatory process with previsible version
$\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$.   Suppose that
$\sequencen{\tau_n}$ is a non-decreasing sequence in $\Cal S$ with supremum
$\tau$ (taken in $\Cal T$) which belongs to $\Cal S$.   Set
$a=\inf_{n\in\Bbb N}\Bvalue{\tau_n<\tau}$.   Then
$a\Bsubseteq\Bvalue{u_{<\tau}=\lim_{n\to\infty}u_{\tau_n}}$.

\proof{ Because $\pmb u$ is moderately oscillatory, $\pmb u_<$ and
$w=\lim_{n\to\infty}u_{\tau_n}$ are both defined (641L).   Take $\epsilon>0$.
Then there is a finite sublattice $I$ of $\Cal S$ such that
$\theta(u_{<\tau}-u_{J<\tau})\le\epsilon$ whenever $J\in\Cal I(\Cal S)$
includes $I$.   For each $\sigma\in I$,
$\Bvalue{\sigma<\tau}=\sup_{n\in\Bbb N}\Bvalue{\sigma<\tau_n}$
(611Eb once more), while of
course $\sequencen{\Bvalue{\sigma<\tau_n}}$ is non-decreasing.   There is
therefore an $n\in\Bbb N$ such that $\bar\mu b\le\epsilon$, where

\Centerline{$b
=\sup_{\sigma\in I}(\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma<\tau_n})$.}

\noindent Now take any $m\ge n$ and consider the finite sublattice $J$ of
$\Cal S$ generated by $I\cup\{\tau_m,\tau\}$.   It is easy to see that

\Centerline{$\{\sigma:\sigma\in\Cal T$,
$\Bvalue{\tau_m<\sigma}\Bcap\Bvalue{\sigma<\tau}\Bsubseteq b\}$}

\noindent is full, therefore a sublattice of $\Cal T$ (611M(b-i) again).
As it includes $I\cup\{\tau_m,\tau\}$, it includes $J$.   So

\Centerline{$\Bvalue{u_{J<\tau}=u_{\tau_m}}
\Bsupseteq\Bvalue{\tau_m<\tau}
   \Bsetminus\sup_{\sigma\in J}(\Bvalue{\tau_m<\sigma}\Bcap\Bvalue{\sigma<\tau})
\Bsupseteq a\Bsetminus b$.}

\noindent Accordingly

\Centerline{$\theta(\chi a\times(u_{J<\tau}-u_{\tau_m}))
\le\bar\mu(a\Bcap\Bvalue{u_{J<\tau}\ne u_{\tau_m}})
\le\bar\mu b
\le\epsilon$.}

\noindent But now

$$\eqalign{\theta(\chi a\times(u_{<\tau}-u_{\tau_m}))&
\le\theta(\chi a\times(u_{<\tau}-u_{J<\tau}))
  +\theta(\chi a\times(u_{J<\tau}-u_{\tau_m}))\cr&
\le\theta(u_{<\tau}-u_{J<\tau})+\epsilon
\le 2\epsilon,\cr}$$

\noindent and this is true for every $m\ge n$.   Consequently
$\theta(\chi a\times(u_{<\tau}-w))\le 2\epsilon$.   As $\epsilon$ is arbitrary.
$\chi a\times(u_{<\tau}-w)=0$ and $a\Bsubseteq\Bvalue{u_{<\tau}=w}$, as
claimed.
}%end of proof of 641M

\leader{641N}{}\cmmnt{ The natural idea of a `jump' in a real function
$f$ is a point $t$ at which the left and right limits $f(t^-)$, $f(t^+)$
of the function are defined
and different.   (See 226B.)   For a \cadlag\ function $f$, the limits
exist (at least, on the interior of $\dom f$)
with $f(t^+)=f(t)$, so we are looking at $f(t)-f(t^-)$.
In a near-simple process $\pmb u$, this corresponds to
a non-zero value
of $\pmb u-\pmb u_<$.   Subject to a necessary close look at
the bottom end of the domain of $\pmb u$,
we find that a jump-free process, in the sense of
\S618, is one for which $\pmb u$ is essentially equal to $\pmb u_<$;
and in fact we can get an expression for the residual oscillation
$\Osclln(\pmb u)$.

\medskip

\noindent}{\bf Proposition} Let $\Cal S$ be a non-empty sublattice of
$\Cal T$, and $\pmb u=\sfamilysS{u}$ an order-bounded
process with a previsible version
$\pmb u_<=\family{\tau}{\Cal S}{u_{<\tau}}$.

(a) For $\tau\in\Cal S$ set
$e_{\tau}=\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau}$.
Then $\chi e_{\tau}\times|u_{\tau}-u_{<\tau}|
\le\Osclln(\pmb u\restr\Cal S\wedge\tau)$.

(b) If $\pmb u$ is near-simple, then
$\Osclln(\pmb u)=\sup_{\tau\in\Cal S}\chi e_{\tau}\times|u_{\tau}-u_{<\tau}|$.

\proof{{\bf (a)}
Suppose that $\tau\in I\subseteq J$ in $\Cal I(\Cal S\wedge\tau)$.   Take
$(\sigma_0,\ldots,\sigma_n)$ linearly generating the $J$-cells, and set
$a_i=\Bvalue{\sigma_i<\sigma_{i+1}}\Bcap\Bvalue{\sigma_{i+1}=\tau}$ for
$i<n$.   Then

\Centerline{$a_i
\Bsubseteq\Bvalue{|u_{\tau}-u_{J<\tau}|=|u_{\sigma_{i+1}}-u_{\sigma_i}|}
\Bsubseteq\Bvalue{|u_{\tau}-u_{J<\tau}|\le\Osclln_J(\pmb u)}$}

\noindent for $i<n$, while

\Centerline{$\Bvalue{\min I<\tau}
\Bsubseteq\Bvalue{\sigma_0<\tau}\Bsubseteq\sup_{i<n}a_i$}

\noindent because $\sigma_n=\tau$.   So

\Centerline{$|u_{\tau}-u_{J<\tau}|\times\chi\Bvalue{\min I<\tau}
\le\Osclln_J(\pmb u)\le\Osclln^*_I(\pmb u\restr\Cal S\wedge\tau)$.}

\noindent Letting $J\uparrow\Cal I(\Cal S\wedge\tau)$, we get

\Centerline{$|u_{\tau}-u_{<\tau}|\times\chi\Bvalue{\min I<\tau}
\le\Osclln^*_I(\pmb u\restr\Cal S\wedge\tau)$.}

\noindent Now

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}\Bvalue{\min I<\tau}
=\lim_{\sigma\downarrow\Cal S}\Bvalue{\sigma<\tau}
=\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau}
=e_{\tau}$,}

\noindent so
$\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}\chi\Bvalue{\min I<\tau}
=\chi e_{\tau}$, and

$$\eqalign{|u_{\tau}-u_{<\tau}|\times\chi e_{\tau}
&=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}
   (|u_{\tau}-u_{<\tau}|\times\chi\Bvalue{\min I<\tau})\cr
&\le\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}
   \Osclln^*_I(\pmb u\restr\Cal S\wedge\tau)
=\inf_{I\in\Cal I(\Cal S\wedge\tau)}
   \Osclln^*_I(\pmb u\restr\Cal S\wedge\tau)\cr
&=\Osclln(\pmb u\restr\Cal S\wedge\tau),\cr}$$

\noindent as required.

\medskip

{\bf (b)} Note that as
$\pmb u$ is moderately oscillatory (631Ca), it has a previsible version
(641L).   Write $\omega(\pmb u)$ for
$\sup_{\tau\in\Cal S}\chi e_{\tau}\times|u_{\tau}-u_{<\tau}|$.

\medskip

\quad{\bf (i)} Suppose that $\pmb u$ is simple, with breakpoint string
$(\tau_0,\ldots,\tau_n)$ in $\Cal S$ and starting value $u_{\downarrow}$.
Take $\sigma\le\sigma'$ in $\Cal S$.   Then, using the formulae in 641Ia,

\inset{----- if $\sigma'\le\tau_0$,

$$\eqalign{|u_{\sigma'}-u_{\sigma}|
&=|u_{\sigma'}-u_{\sigma}|\times\chi\Bvalue{\sigma<\sigma'}
=|u_{\sigma'}-u_{\downarrow}|\times\chi\Bvalue{\sigma<\sigma'}\cr
&\le|u_{\sigma'}-u_{\downarrow}|\times\chi e_{\sigma'}
=|u_{\sigma'}-u_{<\sigma'}|\times\chi e_{\sigma'}
\le\omega(\pmb u);\cr}$$

----- if $i<n$ and $\tau_i\le\sigma\le\sigma'\le\tau_{i+1}$,

$$\eqalignno{|u_{\sigma'}-u_{\sigma}|
&\le|u_{\tau_{i+1}}-u_{\tau_i}|
=|u_{\tau_{i+1}}-u_{\tau_i}|\times\chi\Bvalue{\tau_i<\tau_{i+1}}\cr&
=|u_{\tau_{i+1}}-u_{<\tau_{i+1}}|\times\chi\Bvalue{\tau_i<\tau_{i+1}}\cr&
\le|u_{\tau_{i+1}}-u_{<\tau_{i+1}}|\times\chi e_{\tau_{i+1}}
\le\omega(\pmb u);\cr}$$

----- if $\tau_n\le\sigma$, $|u_{\sigma'}-u_{\sigma}|=0$.}

\noindent By the first formula in 618Ca,

\Centerline{$\Osclln(\pmb u)\le\Osclln^*_I(\pmb u)\le\omega(\pmb u)$,}

\noindent where $I=\{\tau_0,\ldots,\tau_n\}$.

\medskip

\quad{\bf (ii)} Generally, if $\pmb u$ is near-simple,
take any $\epsilon>0$.   Then there is a simple process
$\pmb u'=\sfamilysS{u'}$ such that $\theta(\bar u)\le\epsilon$, where
$\bar u=\sup|\pmb u-\pmb u'|$.   Now

$$\eqalignno{\Osclln(\pmb u)
&\le\Osclln(\pmb u')+\Osclln(\pmb u-\pmb u')\cr
\displaycause{618B(c-ii)}
&\le\omega(\pmb u')+2\bar u
\Displaycause{(i) above and 618B(b-ii)}
\le\omega(\pmb u)+\omega(\pmb u'-\pmb u)+2\bar u
\Displaycause{because $\pmb v\mapsto\pmb v-\pmb v_<$ is linear, by 641Ge}
\le\omega(\pmb u)+4\bar u.\cr}$$

\noindent So

\Centerline{$\theta((\Osclln(\pmb u)-\omega(\pmb u))^+)
\le\theta(4\bar u)\le 4\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\theta((\Osclln(\pmb u)-\omega(\pmb u))^+)=0$ and
$\Osclln(\pmb u)\le\omega(\pmb u)$ in this case also.

\medskip

\quad{\bf (iii)} Putting this together with (a), we see that
$\omega(\pmb u)=\Osclln(\pmb u)$, as claimed.
}%end of proof of 641N

\leader{641O}{Corollary} Let $\Cal S$ be a non-empty sublattice of
$\Cal T$, and $\pmb u$ a locally jump-free process with domain $\Cal S$.
Then $\pmb u_<=\pmb u\times\tbfoneminus{\Cal S}$.

\proof{{\bf (a)} To begin with, suppose that $\pmb u$ is jump-free, therefore
near-simple (631Cb).
Then $\Osclln(\pmb u)=0$ so, in the language of
641N, $\chi e_{\tau}\times(u_{\tau}\times u_{<\tau})=0$ for every
$\tau\in\Cal S$, that is,
$\tbfoneminus{\Cal S}\times(\pmb u-\pmb u_<)=0$ (641Gb) and

$$\eqalignno{\pmb u\times\tbfoneminus{\Cal S}
&=\pmb u_<\times\tbfoneminus{\Cal S}
=(\pmb u\times\tbf{1}^{(\Cal S)})_<
\Displaycause{641G(e-i)}
=\pmb u_<.\cr}$$

\medskip

{\bf (b)} In general, take any $\tau\in\Cal S$;  then
$\pmb u\restr\Cal S\wedge\tau$ is jump-free, so

$$\eqalignno{\pmb u_<\restr\Cal S\wedge\tau
&=(\pmb u\restr\Cal S\wedge\tau)_<
\Displaycause{641G(c-ii)}
=(\pmb u\restr\Cal S\wedge\tau)\times(\tbf 1^{(\Cal S)}\restr\Cal S\wedge\tau)_<
  \cr&
=(\pmb u\restr\Cal S\wedge\tau)
  \times(\tbfoneminus{\Cal S}\restr\Cal S\wedge\tau)
=(\pmb u\times\tbfoneminus{\Cal S})\restr\Cal S\wedge\tau).
\cr}$$

\noindent As $\tau$ is arbitrary,
$\pmb u_<=\pmb u\times\tbfoneminus{\Cal S}$.
}%end of proof of 641O

\leader{641P}{Corollary} Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
and $\pmb u=\sfamilysS{u}$ a near-simple process
with starting value $u_{\downarrow}=0$.   Then
$\Osclln(\pmb u)=\sup_{\tau\in\Cal S}|u_{\tau}-u_{<\tau}|$.

\proof{ Of course $\pmb u$ is moderately oscillatory (631Ca again)
so has a starting value (615H).   In the language of
641Nb, $\Osclln(\pmb u)
=\sup_{\tau\in\Cal S}\chi e_{\tau}\times|u_{\tau}-u_{<\tau}|$.   But if
$\tau\in\Cal S$,

$$\eqalign{1\Bsetminus e_{\tau}&
=\inf_{\sigma\in\Cal S}\Bvalue{\tau\le\sigma}
=\inf_{\sigma\in\Cal S}\Bvalue{\tau=\sigma\wedge\tau}\cr&
\Bsubseteq\Bvalue{u_{\tau}=\lim_{\sigma\downarrow\Cal S}u_{\sigma\wedge\tau}}
=\Bvalue{u_{\tau}=u_{\downarrow}}
=\Bvalue{u_{\tau}=0}\cr&}$$

\noindent while of course we also have
$1\Bsetminus e_{\tau}\Bsubseteq\Bvalue{u_{I<\tau}=0}$ for every non-empty
$I\in\CalIS$, so $1\Bsetminus e_{\tau}\Bsubseteq\Bvalue{u_{<\tau}=0}$.
Thus $1\Bsetminus e_{\tau}\Bsubseteq\Bvalue{u_{\tau}=u_{<\tau}}$
and $\chi e_{\tau}\times|u_{\tau}-u_{<\tau}|=|u_{\tau}-u_{<\tau}|$
for every $\tau\in\Cal S$, so
$\Osclln(\pmb u)=\sup_{\tau\in\Cal S}|u_{\tau}-u_{<\tau}|$.
}%end of proof of 641P

\leader{641Q}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$ a locally moderately oscillatory process with previsible version
$\pmb u_<$, and $\pmb v$ a local integrator with previsible version
$\pmb v_<$.   Set $\pmb w=\iivu$.   Then
$\pmb w-\pmb w_<=\pmb u_<\times(\pmb v-\pmb v_<)$.

\proof{ Because $\pmb u$, $\pmb v$ and $\pmb w$ are all
locally moderately oscillatory (616Ib, 616J),
all the previsible versions are defined
(641L).   Express $\pmb u$, $\pmb v$, $\pmb u_<$, $\pmb v_<$ and
$\pmb w_<$ as $\sfamilysS{u}$, $\sfamilysS{v}$,
$\family{\sigma}{\Cal S}{u_{<\sigma}}$, $\family{\sigma}{\Cal S}{v_{<\sigma}}$
and $\family{\sigma}{\Cal S}{w_{<\sigma}}$.

\medskip

{\bf (a)} Suppose to begin with that $\pmb u$ and $\pmb v$ are both
non-negative and non-decreasing.   Then $\pmb w$ is non-negative and
non-decreasing (616Rb).   Take any $\tau\in\Cal S$.

\medskip

\quad{\bf (i)} If $\sigma\in\Cal S\wedge\tau$ then

\Centerline{$u_{\sigma}\times(v_{\tau}-v_{\sigma})
\le\int_{\Cal S\cap[\sigma,\tau]}\pmb u\,d\pmb v
\le u_{<\tau}\times(v_{\tau}-v_{\sigma})$.}

\noindent\Prf\ If $J$ is a finite sublattice of $\Cal S\cap[\sigma,\tau]$
containing both $\sigma$ and $\tau$, take a sequence
$(\sigma_0,\ldots,\sigma_n)$ linearly generating the $J$-cells.   Then

$$\eqalignno{u_{\sigma}\times(v_{\tau}-v_{\sigma})&
=u_{\sigma}\times\sum_{i=0}^{n-1}v_{\sigma_{i+1}}-v_{\sigma_i}
\le\sum_{i=0}^{n-1}u_{\sigma_i}\times(v_{\sigma_{i+1}}-v_{\sigma_i})\cr&
=\sum_{i=0}^{n-1}u_{\sigma_i}\times\chi\Bvalue{\sigma_i<\tau}
    \times(v_{\sigma_{i+1}}-v_{\sigma_i})
\Displaycause{because $\Bvalue{\sigma_i=\tau}
\Bsubseteq\Bvalue{\sigma_i=\sigma_{i+1}}
\Bsubseteq\Bvalue{v_{\sigma_i}=v_{\sigma_{i+1}}}$ for each $i$}
=\sum_{i=0}^{n-1}u_{\{\sigma_i\}<\tau}\times\chi\Bvalue{\sigma_i<\tau}
    \times(v_{\sigma_{i+1}}-v_{\sigma_i})
\Displaycause{by the formula in 641Ea}
\le\sum_{i=0}^{n-1}u_{<\tau}\times(v_{\sigma_{i+1}}-v_{\sigma_i})
\Displaycause{by the argument in part (a-ii) of the proof of 641L}
=u_{<\tau}\times(v_{\tau}-v_{\sigma}),\cr}$$

\noindent and

\Centerline{$u_{\sigma}\times(v_{\tau}-v_{\sigma})
\le S_J(\pmb u,d\pmb v)
\le u_{<\tau}\times(v_{\tau}-v_{\sigma})$.}

\noindent In the limit as $J\uparrow\Cal I(\Cal S\cap[\sigma,\tau])$,

\Centerline{$u_{\sigma}\times(v_{\tau}-v_{\sigma})
\le\int_{\Cal S\cap[\sigma,\tau]}\pmb u\,d\pmb v
\le u_{<\tau}\times(v_{\tau}-v_{\sigma})$.  \Qed}

\medskip

\quad{\bf (ii)} Now let $I$ be a non-empty finite sublattice of
$\Cal S$.   Then

\Centerline{$u_{I<\tau}\times(v_{\tau}-v_{I<\tau})
\le(w_{\tau}-w_{I<\tau})\times\chi\Bvalue{\min I<\tau}
\le u_{<\tau}\times(v_{\tau}-v_{I<\tau})$.}

\noindent\Prf\ Take a sequence $(\tau_0,\ldots,\tau_n)$ linearly generating the
$I$-cells.   As $\Bvalue{\tau\le\tau_0}
\Bsubseteq\Bvalue{u_{I<\tau}=0}\Bcap\Bvalue{\chi\Bvalue{\min I<\tau}=0}$,
we certainly have

$$\eqalign{\Bvalue{\tau\le\tau_0}&
\Bsubseteq\Bvalue{u_{I<\tau}\times(v_{\tau}-v_{I<\tau})
   \le(w_{\tau}-w_{I<\tau})\times\chi\Bvalue{\min I<\tau}}\cr&
\mskip50mu\Bcap\Bvalue{(w_{\tau}-w_{I<\tau})\times\chi\Bvalue{\min I<\tau}
   \le u_{<\tau}\times(v_{\tau}-v_{I<\tau})}.\cr}$$

\noindent Set
$d_i=\Bvalue{\tau_i<\tau}\Bsetminus\Bvalue{\tau_{i+1}<\tau}$ for $i<n$ and
$d_n=\Bvalue{\tau_n<\tau}$.   Then for any $i\le n$ we have
$d_i\Bsubseteq\Bvalue{w_{I<\tau}=w_{\tau_i}}
=\Bvalue{w_{\tau}-w_{I<\tau}=w_{\tau}-w_{\tau_i}}$
so (i) tells us that

$$\eqalign{d_i&
\Bsubseteq\Bvalue{u_{I<\tau}\times(v_{\tau}-v_{I<\tau})
   \le(w_{\tau}-w_{I<\tau})\times\chi\Bvalue{\min I<\tau}}\cr&
\mskip50mu\Bcap\Bvalue{(w_{\tau}-w_{I<\tau})\times\chi\Bvalue{\min I<\tau}
   \le u_{<\tau}\times(v_{\tau}-v_{I<\tau})}.\cr}$$

\noindent Since $\Bvalue{\tau\le\tau_0}\Bcup\sup_{i\le n}d_i=1$,
we have the result.\ \Qed

\medskip

\quad{\bf (iii)} Taking the limit as $I\uparrow\Cal I(\Cal S)$ in (ii),

\Centerline{$u_{<\tau}\times(v_{\tau}-v_{<\tau})
\le(w_{\tau}-w_{<\tau})\times\chi e
\le u_{<\tau}\times(v_{\tau}-v_{<\tau})$}

\noindent where $e=\lim_{I\uparrow\Cal I(\Cal S)}\Bvalue{\min I<\tau}
=\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau}$.   But now
$u_{<\tau}\times(v_{\tau}-v_{<\tau})\times\chi e
=(w_{\tau}-w_{<\tau})\times\chi e$ while

\Centerline{$1\Bsetminus e\Bsubseteq\Bvalue{w_{\tau}=0}
   \Bcap\Bvalue{w_{<\tau}=0}\Bcap\Bvalue{u_{<\tau}=0}$,}

\noindent so
$u_{<\tau}\times(v_{\tau}-v_{<\tau})\times\chi(1\Bsetminus e)
=(w_{\tau}-w_{<\tau})\times\chi(1\Bsetminus e)$;  adding, we get
$u_{<\tau}\times(v_{\tau}-v_{<\tau})=(w_{\tau}-w_{<\tau})$.

\medskip

\quad{\bf (iv)} As $\tau$ is arbitrary, we have
$\pmb w-\pmb w_<=\pmb u_<\times(\pmb v-\pmb v_<)$ whenever $\pmb u$
and $\pmb v$ are non-negative and non-decreasing.

\medskip

{\bf (b)} Since the operator $(\pmb u,\pmb v)\mapsto\iivu$ is bilinear and the
operator $\pmb u\mapsto\pmb u_<$ is linear, it follows that
$\pmb w-\pmb w_<=\pmb u_<\times(\pmb v-\pmb v_<)$ whenever $\pmb u$
and $\pmb v$ are of bounded variation (614I-614J).   Now if $\pmb u$ is
of bounded variation,
the map $\pmb v\mapsto\iivu:\Mmo(\Cal S)\to\Mmo(\Cal S)$
is continuous for the ucp topology on $\Mmo(\Cal S)$ (615Rb).   Since the
map $\pmb v\mapsto\pmb v_<$ is also continuous (641G(e-ii)) and the ucp topology
on $\Mmo$ is Hausdorff, $\pmb w-\pmb w_<=\pmb u_<\times(\pmb v-\pmb v_<)$
whenever $\pmb u$ is of bounded variation and $\pmb v$ is moderately
oscillatory.   And if $\pmb v$ is an integrator, then
$\pmb u\mapsto\iivu:\Mmo\to\Mob$ is continuous (616J) and takes values in
$\Mmo$ (616J, 616Ib), so now we see that
$\pmb w-\pmb w_<=\pmb u_<\times(\pmb v-\pmb v_<)$ whenever $\pmb u$ is
moderately oscillatory and $\pmb v$ is an integrator.

\medskip

{\bf (c)} For the general case in which $\pmb u$ is locally moderately
oscillatory and $\pmb v$ is a local integrator, apply (b) to
$\pmb u\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\wedge\tau$ to see
that $u_{<\tau}\times(v_{\tau}-v_{<\tau})=(w_{\tau}-w_{<\tau})$ for every
$\tau\in\Cal S$;  of course this depends on the fact that
$\pmb u_<\restr\Cal S\wedge\tau=(\pmb u\restr\Cal S\wedge\tau)$ (641Gc).
}%end of proof of 641Q

\leader{641R}{Corollary} Let $\Cal S$ be a non-empty sublattice of $\Cal T$ and
$\pmb v$ a local integrator with domain $\Cal S$, starting value $0$ and
quadratic variation $\pmb v^*$\cmmnt{ (617H)}.
Write $\pmb v_<$, $\pmb v^*_<$ for the previsible
versions of $\pmb v$ and $\pmb v^*$.
Then $\pmb v^*-\pmb v^*_<=(\pmb v-\pmb v_<)^2$.

\proof{ We know that $\pmb v^*=\pmb v^2-2\iiv(\pmb v)$ (617Ka).   So

\Centerline{$\pmb v^*-\pmb v^*_<
=\pmb v^2-\pmb v^2_<-2\pmb v_<\times(\pmb v-\pmb v_<)
=(\pmb v-\pmb v_<)^2$,}

\noindent using 641Gd and 641Q.
}%end of proof of 641R

\leader{641T}{}\cmmnt{ Since the previsible version of a moderately oscillatory
process is moderately oscillatory (641L), an integral
$\int_{\Cal S}\pmb u_<d\pmb v$ will often be defined.   There are
a couple of useful conditions which will ensure it is actually equal to
$\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

\noindent}{\bf Proposition} Let $\Cal S$ be a sublattice of $\Cal S$,
$\pmb u$ a moderately oscillatory process with previsible
version $\pmb u_<$ and
$\pmb v$ an integrator.   If either $\pmb v$ is jump-free or
$\pmb u$ is jump-free and has starting value $0$, then
$\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.


\proof{{\bf (a)} If $\pmb u$ is jump-free and has starting value $0$, then
in fact $\pmb u_<=\pmb u$, by 641O or 641P, and the result is trivial.   So
in the rest of this proof I will assume that $\pmb v$ is jump-free.
Express $\pmb u$, $\pmb u_<$ and $\pmb v$ as $\sfamilysS{u}$,
$\family{\sigma}{\Cal S}{u_{<\sigma}}$ and $\sfamilysS{v}$.

\medskip

{\bf (b)} Suppose that $\pmb u$ is non-negative and non-decreasing.
Recall that if $\sigma$, $\tau$ in $\Cal S$ then

\Centerline{$0\le u_{\sigma}\times\chi\Bvalue{\sigma<\tau}=u_{\{\sigma\}<\tau}
\le u_{<\tau}\le u_{\tau}$,}

\noindent as
in the proof of 641La.   So if $I\in\CalIS$ is non-empty and
$(\tau_0,\ldots,\tau_n)$
linearly generates the $I$-cells,

$$\eqalign{|S_I(\pmb u-\pmb u_<,d\pmb v)|&
=|\sum_{i=0}^{n-1}(u_{\tau_i}-u_{<\tau_i})\times(v_{\tau_{i+1}}-v_{\tau_i})|\cr&
\le\sum_{i=0}^{n-1}(u_{\tau_i}-u_{<\tau_i})
   \times\chi\Bvalue{\tau_i<\tau_{i+1}}\times|v_{\tau_{i+1}}-v_{\tau_i}|\cr&
\le\Osclln_I(\pmb v)\times\sum_{i=0}^{n-1}(u_{\tau_i}-u_{<\tau_i})
   \times\chi\Bvalue{\tau_i<\tau_{i+1}}\cr&
\le\Osclln_I(\pmb v)\times(u_{\tau_0}
    +\sum_{i=1}^{n-1}u_{\tau_i}-u_{\tau_{i-1}})\cr&
=\Osclln_I(\pmb v)\times u_{\tau_{n-1}}
\le\Osclln_I(\pmb v)\times\sup|\pmb u|\cr&
\to 0\text{ as }I\uparrow\CalIS\cr}$$

\noindent because $\pmb v$ is jump-free.  So in this case we have
$\int_{\Cal S}(\pmb u-\pmb u_<)d\pmb v=0$ and
$\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

{\bf (c)} Now, as in part (b) of the proof of 641Q, the linearity of
integrations and of the operator $\pmb u\mapsto\pmb u_<$ ensures that
$\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$ whenever
$\pmb u$ is of bounded variation and $\pmb v$ is jump-free, and the continuity
of the operators $\pmb u\mapsto\pmb u<$ and
$\pmb u\mapsto\int_{\Cal S}\pmb u\,d\pmb v$ implies that
$\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$ whenever
$\pmb u$ is moderately oscillatory and $\pmb v$ is jump-free.
}%end of proof of 641T

\leader{641U}{}\cmmnt{ For the next result of this kind,
I begin with a lemma which casts a little light on
the difference between discrete-time processes and continuous-time
processes.

\medskip

\noindent}{\bf Lemma} Suppose that $T$ has no points isolated on the right.
Then whenever $\tau\le\tau'$ in $\Cal T$ and $\epsilon>0$ there is a
$\sigma\in[\tau,\tau']$ such that
$\Bvalue{\tau<\sigma}=\Bvalue{\tau<\tau'}$ and
$\bar\mu(\Bvalue{\tau<\tau'}\Bsetminus\Bvalue{\sigma<\tau'})\le\epsilon$.

\proof{ Set $A=\{\sigma:\sigma\in[\tau,\tau']$,
$\Bvalue{\tau<\sigma}=\Bvalue{\tau<\tau'}\}$.   $A$ is non-empty
(because $\tau'\in A$) and downwards-directed (by 611E(c-i-$\beta$));
moreover,
$\sup_{\sigma\in A}\Bvalue{\sigma<\tau'}=\Bvalue{\tau<\tau'}$.
\Prf\Quer\ Otherwise,
$b=\Bvalue{\tau<\tau'}\Bsetminus\sup_{\sigma\in A}\Bvalue{\sigma<\tau'}$ is
non-zero.   Now
$b\Bsubseteq\sup_{t\in T}\Bvalue{\tau'>t}\Bsetminus\Bvalue{\tau>t}$
(611D);  let $t\in T$ be such that
$b\Bcap\Bvalue{\tau'>t}\Bsetminus\Bvalue{\tau>t}$ is non-zero.
Because $t$ is not isolated on the right,
$\Bvalue{\tau'>t}=\sup_{s>t}\Bvalue{\tau'>s}$ (611A(b-i)), and there is
an $s>t$ such that $b\Bcap\Bvalue{\tau'>s}\Bsetminus\Bvalue{\tau>t}$
is non-zero.

Set $a=\Bvalue{\tau'>s}\Bsetminus\Bvalue{\tau>t}$.   Then
$a\in\frak A_s=\frak A_{\check s}$ (611Hb) and
$a\Bsubseteq\Bvalue{\tau'>s}$, so
$a\in\frak A_{\check s}\cap\frak A_{\tau'}$ (611H(a-i)).
There is a therefore a $\sigma\in\Cal T$ such that

\Centerline{$a\Bsubseteq\Bvalue{\sigma=\check s}$,
\quad$1\Bsetminus a\Bsubseteq\Bvalue{\sigma=\tau'}$,
\quad$\check s\wedge\tau'\le\sigma\le\check s\vee\tau'$}

\noindent (611I again).   We see that

$$\eqalignno{\Bvalue{\sigma<\tau}
&=\Bvalue{\sigma<\tau}\Bcap(\Bvalue{\check s<\tau}\Bcup\Bvalue{\tau'<\tau})
=\Bvalue{\sigma<\tau}\Bcap\Bvalue{\check s<\tau}
\cr&\Bsubseteq\Bvalue{\sigma<\tau'}\Bcap\Bvalue{\check s<\tau}
\Bsubseteq a\Bcap\Bvalue{\check s<\tau}
\Bsubseteq\Bvalue{\tau>s}\Bsetminus\Bvalue{\tau>t}
\cr\displaycause{611E(a-i-$\delta$) once more}&
=0,
\cr}$$

\noindent that is, $\tau\le\sigma$, while also

$$\eqalign{\Bvalue{\tau'<\sigma}
&=\Bvalue{\tau'<\sigma}
  \Bcap(\Bvalue{\tau'<\check s}\Bcup\Bvalue{\tau'<\tau'})\cr&
=\Bvalue{\tau'<\sigma}\Bcap\Bvalue{\tau'<\check s}
\Bsubseteq a\Bcap\Bvalue{\tau'<\check s}
=0\cr}$$

\noindent and $\sigma\le\tau'$.   Moreover,

\Centerline{$a\Bsubseteq\Bvalue{\sigma=\check s}\Bsetminus\Bvalue{\tau>t}
=\Bvalue{\sigma=\check s}\Bcap\Bvalue{\tau\le\check t}
\Bsubseteq\Bvalue{\tau<\sigma}$}

\noindent (use 611E(a-i-$\zeta$)), while

\Centerline{$\Bvalue{\tau<\tau'}\Bsetminus a
\Bsubseteq\Bvalue{\tau<\tau'}\Bcap\Bvalue{\sigma=\tau'}
\Bsubseteq\Bvalue{\tau<\sigma}$.}

\noindent So
$\Bvalue{\tau<\tau'}\Bsubseteq\Bvalue{\tau<\sigma}=\Bvalue{\tau<\tau'}$
and $\sigma\in A$.

On the other hand,

\Centerline{$a
\Bsubseteq\Bvalue{\sigma=\check s}\Bcap\Bvalue{\check s<\tau'}
\Bsubseteq\Bvalue{\sigma<\tau'}$,}

\noindent so $a\Bcap b=0$.   But we chose $t$ and $s$ so that
$a$ should meet $b$.\ \Bang\Qed

There is therefore a $\sigma\in A$ such that
$\bar\mu(\Bvalue{\tau<\tau'}\Bsetminus\Bvalue{\sigma<\tau'})\le\epsilon$,
which is what we are looking for.
}%end of proof of 641U

\leader{641V}{Proposition} Suppose that $T$ has no points isolated on the right.
Let $\Cal S$ be an order-convex sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a moderately oscillatory process.   Then
$\pmb u_{\varll}=\pmb u_<$.

\cmmnt{\medskip

\noindent{\bf Remark} I hope it is clear that
$\pmb u_{\varll}=\family{\sigma}{\Cal S}{u_{\varll\sigma}}$ is the previsible version
$(\pmb u_<)_<$ of the previsible version of $\pmb u$, so that
$\pmb u_{\varll\tau}=\lim_{I\uparrow\CalIS}u_{<,I<\tau}$ for every
$\tau\in\Cal S$.}

\proof{{\bf (a)} Suppose to begin with (down to the end of (g) below)
that $\Cal S$ has a greatest element
$\tau$ and that $\epsilon>0$.   Let $J\in\CalIS$ be such that $\tau\in J$ and

\Centerline{$\theta(u_{<\tau}-u_{I<\tau})\le\epsilon$,
\quad$\theta(u_{\varll\tau}-u_{<,I<\tau})\le\epsilon$}

\noindent whenever $J\subseteq I\in\CalIS$.   Let $J_0$ be a maximal totally
ordered subset of $J$.

\medskip

{\bf (b)} We need to know that if $I\in\CalIS$ includes $J_0$ then
$\theta(u_{<\tau}-u_{I<\tau})\le\epsilon$ and
$\theta(u_{\varll\tau}-u_{<,I<\tau})\le\epsilon$.   \Prf\ Let $K$ be a maximal
totally ordered subset of $I$ including $J_0$.   The set

$$\eqalign{L&=\{\sigma:\sigma\in\Cal S,\,\min K\le\sigma\le\max K,\,
\med(\rho,\sigma,\rho')\in\{\rho,\rho'\}\cr&
\mskip200mu
\text{whenever }\rho,\,\rho'\text{ are successive members of }K\}\cr}$$

\noindent is a sublattice of $\Cal S$ including $J$ and $K$ is a maximal
totally ordered subset of $L$, so
$u_{I<\tau}=u_{K<\tau}=u_{L<\tau}$ (641Ed) and
$\theta(u_{<\tau}-u_{I<\tau})=\theta(u_{<\tau}-u_{L<\tau})\le\epsilon$.
Similarly, $\theta(u_{\varll\tau}-u_{<,I<\tau})\le\epsilon$.\ \Qed

\medskip

{\bf (c)} Enumerate $J_0$ in increasing order as
$\langle\tau_i\rangle_{i\le n}$, so that $\tau_n=\tau$.
For each $i<n$ set $c_i=\Bvalue{\tau_i<\tau_{i+1}}$ and
choose $\tau'_i\in\Cal T$ such that
$\tau_i\le\tau'_i\le\tau_{i+1}$, $\Bvalue{\tau_i<\tau'_i}=c_i$ and
$\bar\mu(c_i\Bsetminus\Bvalue{\tau'_i<\tau_{i+1}})
\le\Bover{\epsilon}{n+1}$, as in 641U.   Because $\Cal S$ is order-convex,
$\tau'_i\in\Cal S$.   Set $\tau'_n=\tau$ and
$a=\sup_{i<n}(c_i\Bsetminus\Bvalue{\tau'_i<\tau_{i+1}})$, so that
$\bar\mu a\le\epsilon$.

\medskip

{\bf (d)} Let $i<n$.   Then

\Centerline{$\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau'_i=\tau}
\Bsubseteq a$,}

\Centerline{$\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}\Bsetminus a
=\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau}\Bsetminus a
=\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}\Bsetminus a$.}

\noindent\Prf\

$$\eqalign{\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau'_i=\tau}&
=\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau'_i=\tau_{i+1}}
     \Bcap\Bvalue{\tau_{i+1}=\tau}\cr&
\Bsubseteq\Bvalue{\tau_i<\tau_{i+1}}\Bcap\Bvalue{\tau'_i=\tau_{i+1}}
\Bsubseteq a.\cr}$$

\noindent Next,

$$\eqalign{&(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau})
   \Bsymmdiff(\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau})\cr&
\mskip100mu
=(\Bvalue{\tau'_i<\tau_{i+1}}\Bcap\Bvalue{\tau_{i+1}=\tau})
   \Bsymmdiff(\Bvalue{\tau_i<\tau_{i+1}}\Bcap\Bvalue{\tau_{i+1}=\tau})\cr&
\mskip100mu
\Bsubseteq\Bvalue{\tau'_i<\tau_{i+1}}
   \Bsymmdiff\Bvalue{\tau_i<\tau_{i+1}}
=\Bvalue{\tau_i<\tau_{i+1}}
   \Bsetminus\Bvalue{\tau'_i<\tau_{i+1}}
\Bsubseteq a\cr}$$

\noindent so $\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}\Bsetminus a
=\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}\Bsetminus a$.
Finally, if $i=n-1$ then $\tau_{i+1}=\tau'_{i+1}=\tau$ so surely
$\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}\Bsetminus a
=\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau}\Bsetminus a$.
If $i<n-1$ then

\Centerline{$\Bvalue{\tau'_{i+1}=\tau}
=\Bvalue{\tau'_{i+1}=\tau_{i+2}}\Bcap\Bvalue{\tau_{i+2}=\tau}$,
\quad$\Bvalue{\tau_{i+1}=\tau}
=\Bvalue{\tau_{i+1}=\tau_{i+2}}\Bcap\Bvalue{\tau_{i+2}=\tau}$,}

\noindent so

$$\eqalign{\Bvalue{\tau'_{i+1}=\tau}\Bsymmdiff\Bvalue{\tau_{i+1}=\tau}&
\Bsubseteq\Bvalue{\tau'_{i+1}=\tau_{i+2}}
   \Bsymmdiff\Bvalue{\tau_{i+1}=\tau_{i+2}}
=\Bvalue{\tau'_{i+1}<\tau_{i+2}}
   \Bsymmdiff\Bvalue{\tau_{i+1}<\tau_{i+2}}\cr&
=\Bvalue{\tau_{i+1}<\tau_{i+2}}
   \Bsetminus\Bvalue{\tau'_{i+1}<\tau_{i+2}}
\Bsubseteq a\cr}$$

\noindent and
$\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}\Bsetminus a
=\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau}\Bsetminus a$.\ \Qed

\medskip

{\bf (e)} Set $L=\{\tau'_i:i\le n\}$.   Then
$\theta(u_{\varll\tau}-u_{<,L<\tau})\le 2\epsilon$.   \Prf\ Set
$L'=\{\tau_0,\tau'_0,\tau_1,\tau'_1,\ldots,\tau_n\}$.   Then

\Centerline{$u_{<,L'<\tau}
=\sum_{i=0}^{n-1}
  u_{<\tau_i}\times\chi(\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau'_i=\tau})
   +u_{<\tau'_i}\times\chi(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau})$}

\noindent (641Eb).   Similarly,

\Centerline{$u_{<,L<\tau}
=\sum_{i=0}^{n-1}
  u_{<\tau'_i}\times\chi(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau})$.}

\noindent Since $\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau'_i=\tau}$ and
$(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau})
\Bsymmdiff(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau})$ are both
included in $a$ for every $i<n$ ((d) above),
$\Bvalue{u_{<,L'<\tau}\ne u_{<,L<\tau}}\Bsubseteq a$ and

\Centerline{$\theta(u_{\varll\tau}-u_{<,L<\tau})
\le\theta(u_{\varll\tau}-u_{<,L'<\tau})+\epsilon
\le 2\epsilon$}

\noindent because $J_0\subseteq L'$.\ \Qed

\medskip

{\bf (f)} Let $I\in\CalIS$ be such that $J\cup\{\tau'_i:i<n\}\subseteq I$ and
$\theta(u_{<\tau'_i}-u_{I<\tau'_i})\le\Bover{\epsilon}{n+1}$ for every $i<n$.
Take a maximal totally subset $I_0$ of $I$ including
$\{\tau_i:i\le n\}\cup\{\tau'_i:i<n\}$.
For each $i<n$, set $K_i=I_0\cap[\tau_i,\tau'_i]$.   If $\sigma\in K_i$,
we have

\Centerline{$\Bvalue{\sigma<\tau'_i}
\in\frak A_{\sigma}\Bcap\frak A_{\tau'_i}\Bsubseteq\frak A_{\tau_{i+1}}$}

\noindent so we can define $\tilde\sigma\in\Cal T$ by saying that

\Centerline{$\Bvalue{\sigma<\tau'_i}\Bsubseteq\Bvalue{\tilde\sigma=\sigma}$,
\quad$\Bvalue{\tau'_i\le\sigma}\Bsubseteq\Bvalue{\tilde\sigma=\tau_{i+1}}$}

\noindent (611I once more).   Now $\tau_i\le\tilde\sigma\le\tau_{i+1}$ so
$\tilde\sigma\in\Cal S$.   If $\rho\le\sigma$ in $K_i$ then

\Centerline{$\Bvalue{\sigma<\tau'_i}
=\Bvalue{\sigma<\tau'_i}\Bcap\Bvalue{\rho<\tau'_i}
\Bsubseteq\Bvalue{\tilde\rho\le\tilde\sigma}$,
\quad
$\Bvalue{\tau'_i\le\sigma}\Bsubseteq\Bvalue{\tilde\sigma=\tau_{i+1}}
\Bsubseteq\Bvalue{\tilde\rho\le\tilde\sigma}$,}

\noindent so $\tilde\rho\le\tilde\sigma$;  it follows that
$\tilde K_i=\{\tilde\sigma:\sigma\in K_i\}$ is totally ordered.   Because
$\Bvalue{\tau_i<\tau'_i}=\Bvalue{\tau_i<\tau_{i+1}}$, $\tilde\tau_i=\tau_i$,
so $\min K_i=\tau_i=\min\tilde K_i$.   Also
$\max\tilde K_i=\tilde\tau'_i=\tau_{i+1}$.

The point of this is that
$\Bvalue{\sigma<\tau'_i}=\Bvalue{\tilde\sigma<\tau_{i+1}}$ for every
$\sigma\in K_i$.   It follows that $u_{K_i<\tau'_i}=u_{\tilde K_i<\tau_{i+1}}$,
and this is true for every $i<n$.

Now set $K=\bigcup_{i<n}K_i$ and $\tilde K=\bigcup_{i<n}\tilde K_i$.   Both of these
are totally ordered sets including $J_0$ with $\min K=\min\tilde K=\tau_0$.
Since $K\cap[\tau_i,\tau'_i]=K_i=I_0\cap[\tau_i,\tau'_i]$,

\Centerline{$c_i=\Bvalue{\tau_i<\tau'_i}
\Bsubseteq\Bvalue{u_{I_0<\tau'_i}=u_{K_i<\tau'_i}}
=\Bvalue{u_{I<\tau'_i}=u_{K_i<\tau'_i}}$;}

\noindent
similarly, $\tilde K\cap[\tau_i,\tau_{i+1}]=\tilde K_i$, so
$c_i=\Bvalue{\tau_i<\tau_{i+1}}
\Bsubseteq\Bvalue{u_{\tilde K<\tau_{i+1}}=u_{\tilde K_i<\tau_{i+1}}}$;
accordingly $c_i\Bsubseteq\Bvalue{u_{I<\tau'_i}=u_{\tilde K<\tau_{i+1}}}$.
Next, $\Bvalue{\tau_{i+1}=\tau}
\Bsubseteq\Bvalue{u_{\tilde K<\tau}=u_{\tilde K<\tau_{i+1}}}$ (641Ec), so if we set
$b_i=c_i\Bcap\Bvalue{\tau_{i+1}=\tau}$, we have
$b_i\Bsubseteq\Bvalue{u_{I<\tau'_i}=u_{\tilde K<\tau}}$.   By the choice of $I$,
$\theta((u_{<\tau'_i}-u_{\tilde K<\tau})\times\chi b_i)\le\Bover{\epsilon}{n+1}$.

\medskip

{\bf (g)} Now we find that

$$\eqalignno{\theta(u_{\varll\tau}-u_{<\tau})&
\le 2\epsilon+\theta(u_{<,L<\tau}-u_{<\tau})
\Displaycause{by (e)}
\le 3\epsilon+\theta(u_{<,L<\tau}-u_{\tilde K<\tau})
\Displaycause{by (b), because $J_0\subseteq\tilde K$}
=3\epsilon+\theta(u_{<,L<\tau}-u_{\tilde K<\tau}\times\chi\Bvalue{\tau_0<\tau})
\Displaycause{because $\tau_0=\min\tilde K$, so
$\Bvalue{\tau_0=\tau}\Bsubseteq\Bvalue{u_{\tilde K<\tau}=0}$}
=3\epsilon+\theta(u_{<,L<\tau}-\sum_{i=0}^{n-1}u_{\tilde K<\tau}\times
   \chi(\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}))\cr&
\le 4\epsilon+\theta(u_{<,L<\tau}-\sum_{i=0}^{n-1}u_{<\tau'_i}\times
   \chi(\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau}))\cr&
\Displaycause{by (f)}
\le 5\epsilon+\theta(u_{<,L<\tau}-\sum_{i=0}^{n-1}u_{\tau'_i}\times
   \chi(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau}))
\Displaycause{because
$\sup_{i<n}(\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_{i+1}=\tau})
   \Bsymmdiff(\Bvalue{\tau'_i<\tau}\Bcap\Bvalue{\tau'_{i+1}=\tau})
\Bsubseteq a$ has measure at most $\epsilon$, by (d)}
=5\epsilon\cr}$$

\noindent as noted in (e).

\medskip

{\bf (h)} At this point, we recall that $\epsilon>0$ was arbitrary, so we see
that $u_{\varll\tau}=u_{<\tau}$.   This was all on the assumption that
$\tau=\max\Cal S$.   But generally, given that $\Cal S$ is an order-convex
subset of $\Cal T$ and that $\tau\in\Cal S$, we can apply the argument so far to
$\pmb u\restr\Cal S\wedge\tau$,
$(\pmb u\restr\Cal S\wedge\tau)_<=\pmb u_<\restr\Cal S\wedge\tau$
and $(\pmb u\restr\Cal S\wedge\tau)_{\varll}
=\pmb u_{\varll}\restr\Cal S\wedge\tau$ (641G(c-ii)) to see that we still have
$u_{\varll\tau}=u_{\tau}$.   Thus $\pmb u_{\varll}=\pmb u_<$.
}%end of proof of 641V

\leader{641W}{Proposition} Suppose that $T$ has no points isolated on the right.
Let $\Cal S$ be an order-convex sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a moderately oscillatory process and
$\pmb v=\sfamilysS{v}$ a near-simple integrator.   Then
$\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.

\proof{{\bf (a)} If $\pmb v$ is simple, we can use the formula of 641J to see
that $\int_{\Cal S}\pmb u\,d\pmb v$ is determined by the previsible version
$\pmb u_<$.   Since in this context we have $\pmb u_{\varll}=\pmb u_<$, by 641V,
we shall have $\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

{\bf (b)} If $\pmb u$ is of bounded variation, then
$\pmb v\mapsto\int_{\Cal S}\pmb u\,d\pmb v$ is continuous for the ucp topology
(631H(a-ii)).   At the same time, $\pmb u_<$ is of bounded variation (641Lb),
so $\{\pmb v:\pmb v\in\Mns(\Cal S)$,
$\int_{\Cal S}\pmb u\,d\pmb v=\int_{\Cal S}\pmb u_<d\pmb v\}$ is closed for the
ucp topology.   By (a), it contains all simple functions, so it is the whole of
$\Mns(\Cal S)$, which is more than we need.

\medskip

{\bf (c)} In general, take any $\epsilon>0$.   Let $\delta>0$ be such that
$\theta(\int_{\Cal S}\pmb w\,d\pmb v)\le\epsilon$ whenever $I\in\CalIS$ and
$\pmb w\in\Mmo(\Cal S)$ is such that $\theta(\sup|\pmb w|)\le\delta$ (616J,
applied to $\Delta\pmb v$).
Then we have a process $\tildepmb u$ of bounded variation such that
$\theta(\sup|\pmb u-\tildepmb u|)\le\delta$.   As
$\tildepmb u$ is moderately oscillatory it has a previsible
version $\tildepmb u_<$.   Now
$\int_{\Cal S}\tildepmb u\,d\pmb v=\int_{\Cal S}\tildepmb u_<d\pmb v$, by
(b), while
$\theta(\int_{\Cal S}\pmb u\,d\pmb v-\int_{\Cal S}\tildepmb u\,d\pmb v)
\le\epsilon$.   Next, $\sup|\pmb u_<-\tildepmb u_<|\le\sup|\pmb u-\tildepmb u|)$
and $\theta(\int_{\Cal S}\pmb u_<d\pmb v-\int_{\Cal S}\tildepmb u_<d\pmb v)
\le\epsilon$.   Putting these together, we see that
$\theta(\int_{\Cal S}\pmb u\,d\pmb v-\int_{\Cal S}\pmb u_<d\pmb v)
\le 2\epsilon$.   As $\epsilon$ is arbitrary,
$\int_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.
}%end of proof of 641W

%\leader{641S}{}

\exercises{\leader{641X}{Basic exercises (a)}
%\spheader 641Xa
In 641B, show that $\frak A_{<\min\Cal T}=\{0,1\}$ and
$\frak A_{<\max\Cal T}=\bigvee_{t\in T}\frak A_t$.
%641C

\spheader 641Xi Give an example of a structure  \Asissix, a martingale
$\family{\sigma}{\Cal T}{u_{\sigma}}$, a $\tau\in\Cal T$ and a finite sublattice
$I$ of $\Cal T$ such that $u_{I<\tau}$ is not the conditional expectation of
$u_{\tau}$ on the algebra $\frak A_{I<\tau}$.
%641Bc     u_{\tau_1}=u_{\tau}=1 on $d_1$, $-1$ on $d_0$;
%   u_{\tau_0}=0

\spheader 641Xb Let $\Cal S$ be a non-empty sublattice of $\Cal T$, and
$\pmb v$ a near-simple process with domain $\Cal S$ and previsible version
$\pmb v_<$.   Show that $\sup|\pmb v|=\sup|\pmb v_<|\vee|v_{\uparrow}|$, where
$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$.
%641Ib

\spheader 641Xh Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a locally
moderately oscillatory process with previsible version
$\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$.   Suppose that
$A$ is a non-empty upwards-directed subset of $\Cal S$ such that
$\tau=\sup A$ belongs to $\Cal S$.   Set
$a=\inf_{\sigma\in A}\Bvalue{\sigma<\tau}$.   Show that
$a\Bsubseteq\Bvalue{u_{<\tau}=\lim_{\sigma\uparrow A}u_{\sigma}}$.
%641M out of order query

\spheader 641Xc Let $\Cal S$ be a sublattice of $\Cal T$ and $\pmb v$,
$\pmb w$ near-simple processes with domain $\Cal S$ such that
$(\pmb v-\pmb v_<)\times(\pmb w-\pmb w_<)=\tbf 0$.   Show that
$\Osclln(\pmb v+\pmb w)=\Osclln(\pmb v)\vee\Osclln(\pmb w)$.
%641Nb

\spheader 641Xd Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a process of
locally bounded variation with domain $\Cal S$ and starting value $0$.   Let
$\pmb v^{\uparrow}=\family{\tau}{\Cal S}
{\int_{\Cal S\wedge\tau}|d\pmb v|}$ be the cumulative variation of $\pmb v$
(614O).   Show that
$\pmb v^{\uparrow}-\pmb v^{\uparrow}_<=|\pmb v-\pmb v_<|$.
\Hint{614P(c-ii).}
%641Q 614P

\spheader 641Xe Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$, $\pmb w$ two local integrators with domain
$\Cal S$, at least one of which has starting value $0$.   Write $\pmb z$ for
their covariation.   Show that
$\pmb z-\pmb z_<=(\pmb v-\pmb v_<)\times(\pmb w-\pmb w_<)$.
%641R

\spheader 641Xf Show that if $T$ has an element which is isolated on the
right, then there are simple processes $\pmb u$, $\pmb v$ defined on the
whole of $\Cal T$ such that
$\int_{\Cal T}\pmb u_<\,d\pmb v\ne\int_{\Cal T}\pmb u\,d\pmb v$.
%641W

\spheader 641Xg Suppose that $T$ has no points isolated on the right.
Let $\Cal S$ be an order-convex sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a moderately oscillatory process and
$\pmb v=\sfamilysS{v}$ a near-simple integrator.   Show that
$\iivu=\iiv(\pmb u_<)$.
%641W

\spheader 641Xh Let $\Cal S$ be a sublattice of $\Cal T$ and $\pmb u$ a
non-decreasing fully adapted process with domain $\Cal S$.   Show that
$\pmb u_<$ is defined everywhere in $\Cal S$.
%641L

\leader{641Y}{Further exercises (a)}
%\spheader 641Ya
Give an example, in the context of 641B, of a structure
$(\frak A,T,\family{t}{T}{\frak A_t},\Cal T)$ with stopping times
$\sigma$, $\tau\in\Cal T$ such that
$\frak A_{<(\sigma\wedge\tau)}\ne\frak A_{<\sigma}\cap\frak A_{<\tau}$.
%mt64bits 641B 634N 642Yb

\spheader 641Yb Suppose that $\frak B$ is a closed subalgebra of $\frak A$
which is coordinated with $\family{t}{T}{\frak A_t}$ in the sense of
634Fb.   Set $\frak B_t=\frak B\cap\frak A_t$ for $t\in T$, and let
$\Cal T_{\frak B}\subseteq\Cal T$ be the sublattice of stopping times for
$(\frak B,\family{t}{T}{\frak B_t})$, as in 634C.   Show that if
$\tau\in\Cal T_{\frak B}$ then $\frak B_{<\tau}$, defined from
$\tau$ and $\Cal T_{\frak B}$ by the formula of 641B, is equal to
$\frak B\cap\frak A_{<\tau}$.
%641B mt64bits

\spheader 641Yd(i) Let $\pmb u$ be a simple process defined on a non-empty
sublattice of $\Cal T$.
Show that $\pmb u_<$ is previsibly simple in the sense
of 612Ye.
%641Ia
(ii) Let $\pmb u$ be a previsibly simple process on a sublattice
$\Cal S$ of $\Cal T$
defined from sequences $(\tau_0,\ldots,\tau_n)$ and $u_*,u_0,\ldots,,u_n$
by the formulae of 612Ye.   Suppose that $\pmb v=\sfamilysS{v}$
is an integrator with
domain $\Cal S$ and that {\it either} $\pmb v$ is jump-free {\it or} $\Cal S$ is
order-convex and no member of $T$ is isolated on the right.   Show that

$$\eqalignno{&\int_{\Cal S}\pmb u\,d\pmb v
=u_*\times(v_{\tau_0}-v_{\downarrow})
  +\sum_{i=0}^{n-1}u_i\times(v_{\tau_{i+1}}-v_{\tau_i})
  +u_{\tau_n}\times(v_{\uparrow}-v_{\tau_n})\cr}$$

\noindent where $v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ and
$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$.
%641T  641I  612Ye
}%end of exercises

\endnotes{
\Notesheader{641} While $\family{t}{T}{\frak A_{<\check t}}$ is a
filtration of closed subalgebras of $\frak A$,
you must not think of $<\mskip-4mu\tau$, in
$\frak A_{<\tau}$ and $u_{<\tau}$, as a stopping time.   We have to re-examine
all our standard formulae, as in 641Ca and 641Ya.   The familiar appearance of
641G(a-ii) is not exactly an accident, but it certainly demands a proper check.

The concept of `previsibility' is awkward at the bottom end.
In the definitions in 641E-641F, I have chosen to set $u_{<\tau}=0$ if
$\tau=\min\dom\pmb u$.   This will cascade through the clauses
`$h(0,\ldots,0)=0$' and `$g(0)=0$' in the statements of 641Gd and 642Da,
`$=0$ otherwise' in the
statement of 642E, `$=0$ if $h(\omega)=0$' in the statements of
642Fb and 642La, and `$=0$ if $t=0$' in the statement of 642K, all of which
seem a little arbitrary.   (There is also a subformula
`$\chi e_{\tau}\times$' in the statement of 641N which is there for the
same reason.)   But it seems to me that there are similar difficulties,
sometimes acknowledged\footnote{Cf.\ {\smc Rogers \& Williams 00},
intro.\ to Chap.\ IV.}
% we consign it to hell
and sometimes not, in the alternative presentations
I have seen.   On the ideological ground that a `previsible' process
$\pmb u=\sfamilysS{u}$ is one in which
each value $u_{\tau}$ can be determined from
observations of $u_{\sigma}$ in regions $\Bvalue{\sigma<\tau}$,
I have taken the view that in the region
$\inf_{\sigma\in\Cal S}\Bvalue{\tau\le\sigma}$, where we have no
previous information, we must go to a default value, and that the only
sensible general default value is $0$.   If you have any reason to choose
another, you should model this with a process starting at a time less than
$\tau$.   In the meantime, I ask you to put up with the fact that the
previsible version $\tbfoneminus{\Cal S}$ of a constant process
$\tbf 1^{(\Cal S)}$ might not
be $\tbf 1^{(\Cal S)}$.   After all, at the time of the first observation, its
constancy is {\it not} previsible.

Up to this point it has seemed safe to write $\tbf 1$ for a constant
process with value $\chi 1$ without declaring a domain other than the
default $\Cal T$ of 612D(e-i).   But if (as will be necessary when we come
to define the S-integral in \S645) we want to talk about $\tbf 1_<$, it will be
essential to understand exactly what its domain is.   The notation
$\tbf 1^{(\Cal S)}$ is inelegant but, I hope, easily interpreted.

Note that in 641W we have a fundamental difference between
$T=\Bbb N$ and $T=\coint{0,\infty}$ (see 641Xf).   The point here is that when
we come to the S-integral we shall have a basic formula

\Centerline{$\Sint_{\Cal S}\pmb u_<\,d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$}

\noindent (645R(a-i)),
so it is interesting to know when this happens to be equal to
the Riemann-sum integral $\int_{\Cal S}\pmb u_<\,d\pmb v$.

In 641Q-641R I return to a theme from \S617.   The formulae

\Centerline{$d(\iivu)\sim\pmb u\,d\pmb v$,
\quad$d\pmb v^*\sim(d\pmb v)^2$,
\quad$d\covar{\pmb v}{\pmb w}\sim d\pmb v\,d\pmb w$}

\noindent reappear, transformed, in formulae for the jumps of $\iivu$
(641Q), $\pmb v^*$ (641R) and $\covar{\pmb v}{\pmb w}$ (641Xe).
Similarly, $d\pmb v^{\uparrow}\sim|d\pmb v|$ (616T) turns into the formula
of 641Xd, a refinement of 618U.

In 641U-641W %641U 641V 641W
I look at some results which depend on particular assumptions concerning the
totally ordered set $T$ of the stochastic integration structure and the domain
of the processes under consideration.   While in a sense these are restrictive,
they are in fact satisfied by all the leading examples of the theory once we
have left the case $T=\Bbb N$.   It is fair to consider them as the normal case.
Elsewhere (641Nb, 641P, 641W) I call on processes to be near-simple.
This is indeed a restriction (for instance, previsible versions are often not
near-simple).   Its necessity is clear in 641P;  the previsible version can tell
us only about jumps as we approach a limit from below, not as we approach from
above.


}%end of notes

\discrpage

\frfilename{mt642.tex}
\versiondate{14.12.16/29.9.23}
\copyrightdate{2013}

\def\chaptername{The S-integral}
\def\sectionname{Previsible processes}

\newsection{642}
I continue the work of \S641 with a description of the previsible version
of a process defined in the standard way from a probability space and a
filtration of $\sigma$-algebras (642E-642G).
The other objective of the section is to make a step towards
a general theory of
`previsible' processes.   The point is that among such processes, starting
with those of the form $\pmb u_<$, a form of sequential convergence
(the order*-convergence of 642B) has striking connections with stochastic
integration.   I will come to this in \S644.   For the moment, I present
a definition of the space $\Mpv$ of previsible processes, with some of its
elementary properties (642D) and a description in terms of suitably
measurable processes in the case in which $T=\coint{0,\infty}$ (642L).

\cmmnt{
\leader{642A}{Notation} \Asissix\ will be a stochastic integration structure.
$\theta$ will be the standard functional
defining the topology of convergence in measure on $L^0(\frak A)$
(613B).   If $t\in T$ then $\check t$ will be the constant stopping time at $t$.
If $\Cal S$ is a sublattice of $\Cal T$,
$\Cal I(\Cal S)$ will be the upwards-directed set of finite sublattices of
$\Cal S$, and $\Cal S\wedge\tau$ will be the lattice
$\{\sigma\wedge\tau:\sigma\in\Cal S\}$ for every $\tau\in\Cal S$.
$\Mfa(\Cal S)$, $\Msimp(\Cal S)$, $\Mob(\Cal S)$, $\Mns(\Cal S)$ and
$\Mmo(\Cal S)$
will be the spaces of fully adapted
processes, simple processes, order-bounded processes, near-simple
processes and moderately oscillatory with domain $\Cal S$.
If $\pmb u=\sfamilysS{u}$ is an order-bounded process
(614E), $\sup|\pmb u|$ will be the supremum
$\sup_{\sigma\in\Cal S}|u_{\sigma}|$ in $L^0(\frak A)$, and
$\Osclln(\pmb u)$ will be the residual oscillation of $\pmb u$
(618B).


}%end of comment

\leader{642B}{Order*-convergence in $L^0$ and $(L^0)^{\Cal S}$ (a)}
In 367A, I gave a definition of order*-convergent sequence in arbitrary
lattices\cmmnt{ which is more elaborate than we need here}.   For our present
purposes, it will be enough to know that\cmmnt{, because
$\frak A$ is a Dedekind complete Boolean algebra,}
a sequence $\sequencen{u_n}$ in $L^0(\frak A)$ is order*-convergent to
$u\in L^0(\frak A)$ iff it is order-bounded and

\Centerline{$u=\inf_{n\in\Bbb N}\sup_{i\ge n}u_i
=\sup_{n\in\Bbb N}\inf_{i\ge n}u_i$\dvro{.}{}}

\noindent\cmmnt{(367Gb). }   Another way of expressing this is to say that
$\sequencen{u_n}$ is order*-convergent to $u$ iff there are a
non-decreasing sequence $\sequencen{v_n}$ and a non-increasing sequence
$\sequencen{w_n}$ such that $\sup_{n\in\Bbb N}v_n=\inf_{n\in\Bbb N}w_n=u$
and $v_n\le u_n\le w_n$ for every $n$.
In this case, $u$ is the limit of $\sequencen{u_n}$
for the topology of convergence in measure\cmmnt{ (367Pa)}.

Note that $\sequencen{u_n}$ is order*-convergent to $u$ iff $\sequencen{u_n-u}$
is order*-convergent to $0$, and that if $\sequencen{u_n}$ is order-convergent
to $0$ and $|v_n|\le|u_n|$ for every $n$, then $\sequencen{v_n}$ is
order-convergent to $0$ (367Ca).

If $\frak A$ is expressed as the measure algebra of a
probability space $(\Omega,\Sigma,\mu)$,
and each $u_n$, $u$ is represented as $f_n^{\ssbullet}$, $f^{\ssbullet}$
where the $f_n$, $f$ are measurable functions defined on $\Omega$, then
$\sequencen{u_n}$ is order*-convergent to $u$ iff
$f(\omega)=\lim_{n\to\infty}f_n(\omega)$ for almost every $\omega$\cmmnt{
(367F)}.   Hence\cmmnt{, or otherwise,}
we see that if $\sequencen{u_n}$ is order*-convergent to $u$ and
$\sequencen{v_n}$ is order*-convergent to $v$, then
$\sequencen{u_n+v_n}$, $\sequencen{\bar g(u_n)}$
and $\sequencen{u_n\times v_n}$ are order*-convergent to $\alpha u$,
$u+v$, $\bar g(u)$ and $u\times v$ respectively, for any
continuous $g:\Bbb R\to\Bbb R$.   More generally, if $\sequencen{u_{in}}$
is order*-convergent to $u_i$ for $1\le i\le k$, and
$g:\BbbR^k\to\Bbb R$ is continuous, then
$\sequencen{\bar g(u_{1n},\ldots,u_{kn})}$\cmmnt{, defined as in 619Eb}, is
order*-convergent to $\bar g(u_1,\ldots,u_k)$.

\spheader 642Bb Now suppose that we have
a sublattice $\Cal S$ of $\Cal T$, a sequence
$\sequencen{\pmb u_n}=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$
of processes with domain $\Cal S$, and another process
$\pmb u=\sfamilysS{u}$ with domain $\Cal S$.
Then $L^0(\frak A)^{\Cal S}$ is isomorphic, as
$f$-algebra, to $L^0(\frak A^{\Cal S})$, where $\frak A^{\Cal S}$ is the
simple product\cmmnt{ (315A, 364R)
and is itself a Dedekind complete Boolean algebra
(315D(e-i))}.   So\cmmnt{ we shall be able to use the same formula for
sequences of processes: } if $\Cal S$ is a sublattice of $\Cal T$,
$\sequencen{\pmb u_n}=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$ is a
sequence
of processes with domain $\Cal S$, and $\pmb u=\sfamilysS{u}$ is another
process with domain $\Cal S$, then

$$\eqalign{&\sequencen{\pmb u_n}\text{ is order*-convergent to }
\pmb u\cr
&\mskip50mu
\iff\sequencen{\pmb u_n}\text{ is order-bounded and }
\pmb u=\inf_{n\in\Bbb N}\sup_{i\ge n}\pmb u_i
=\sup_{n\in\Bbb N}\inf_{i\ge n}\pmb u_i\cr
&\mskip50mu
\iff\sequencen{u_{n\sigma}}\text{ is order-bounded and }
u_{\sigma}=\inf_{n\in\Bbb N}\sup_{i\ge n}u_{i\sigma}
=\sup_{n\in\Bbb N}\inf_{i\ge n}u_{i\sigma}\cr
&\mskip500mu
\text{ for every } \sigma\in\Cal S\cr&
\mskip50mu
\iff\sequencen{u_{n\sigma}}\text{ is order*-convergent to }u_{\sigma}
\text{ for every }\sigma\in\Cal S.\cr}$$

\noindent Note that if we have an order*-convergent sequence of
fully adapted processes, the limit will also be fully adapted.
\prooflet{\Prf\ Suppose that
$\sequencen{\pmb u_n}=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$
is a sequence of fully adapted processes order*-convergent to
$\pmb u=\sfamilysS{u}$.   For each $\sigma\in\Cal S$,
$\sequencen{u_{n\sigma}}$ is a sequence in $L^0(\frak A_{\sigma})$
which is order*-convergent, therefore topologically convergent, to
$u_{\sigma}$, and $u_{\sigma}\in L^0(\frak A_{\sigma})$ because
$L^0(\frak A_{\sigma})$ is topologically closed.   If $\sigma$,
$\tau\in\Cal S$ and $\Bvalue{\sigma=\tau}=a$, then

\Centerline{$u_{\sigma}\times\chi a
=\lim_{n\to\infty}u_{n\sigma}\times\chi a
=\lim_{n\to\infty}u_{n\tau}\times\chi a
=u_{\tau}\times\chi a$,}

\noindent so $a\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$.   Thus
$\pmb u$ is fully adapted.\ \Qed}

\spheader 642Bc A topologically
convergent sequence need not be order*-convergent\cmmnt{ (245Cc)}, but if
$\sequencen{u_n}$ is a sequence in $L^0$ such that
$\sum_{n=0}^{\infty}\theta(u_n)$ is finite, then $\sequencen{u_n}$ is
order*-convergent to $0$.   \prooflet{\Prf\ I check first that
$\{|u_n|:n\in\Bbb N\}$ is bounded above.   If $\epsilon>0$, there is an
$n\in\Bbb N$ such that

\Centerline{$\sum_{m=n}^{\infty}\bar\mu\Bvalue{|u_m|\ge 1}
\le\sum_{m=n}^{\infty}\theta(u_m)\le\epsilon$;}

\noindent now there is a $k\ge 1$ such that
$\sum_{m=0}^{n-1}\bar\mu\Bvalue{|u_m|>k}\le\epsilon$, so that

\Centerline{$\bar\mu(\sup_{m\in\Bbb N}\Bvalue{|u_m|>k})
\le\sum_{m=0}^{n-1}\bar\mu\Bvalue{|u_m|>k}
  +\sum_{m=n}^{\infty}\bar\mu\Bvalue{|u_m|\ge 1}
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$
\bar\mu(\inf_{\alpha\in\Bbb R}\sup_{m\in\Bbb N}\Bvalue{|u_m|>\alpha})
=\lim_{\alpha\to\infty}\bar\mu(\sup_{m\in\Bbb N}\Bvalue{|u_m|>\alpha})
=0$}

\noindent and $\sequencen{|u_n|}$ is bounded above, by the criterion in
364L(a-ii).

Accordingly we may speak of $\sup_{m\ge n}|u_m|$ for each $n$.   Given
$\epsilon>0$, let $n\in\Bbb N$ be such that
$\sum_{m=n}^{\infty}\theta(u_m)\le\epsilon^2$;  then

\Centerline{$\bar\mu\Bvalue{\sup_{m\ge n}|u_m|>\epsilon}
\le\sum_{m=n}^{\infty}\bar\mu\Bvalue{|u_m|>\epsilon}
\le\Bover1{\epsilon}\sum_{m=n}^{\infty}\theta(u_m)
\le\epsilon$,}

\noindent so $\theta(\sup_{m\ge n}|u_m|)\le 2\epsilon$.   As $\epsilon$ is
arbitrary, $\inf_{n\in\Bbb N}\sup_{m\ge n}|u_m|=0$ and
$\sequencen{u_n}$ is order*-convergent to $0$.\ \Qed}

\medskip

{\bf (d)} The description of order*-convergence in $L^0$ in terms of
pointwise convergence in $\eusm L^0$\cmmnt{, given in (a) above,} makes
it easy to see that if $\sequencen{h_n}$ is a sequence of
Borel measurable functions from $\BbbR^k$ to $\Bbb R$, and
$h(x)=\lim_{n\to\infty}h_n(x)$ is defined in $\Bbb R$ for every
$x\in\BbbR^k$, then $\sequencen{\bar h_n(u_1,\ldots,u_k)}$ is
order*-convergent to $\bar h(u_1,\ldots,u_k)$ whenever
$u_1,\ldots,u_k\in L^0$.   \cmmnt{(See 619Ee-619Ef.)}

\cmmnt{We are now ready for the next definition.}

\leader{642C}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$.
I will say that a process $\pmb x$ with domain $\Cal S$ is {\bf previsible} if
it belongs to the smallest subspace of $(L^0)^{\Cal S}$ which contains
$\pmb u_<$ for every simple process $\pmb u$ and which is closed under
order*-convergence of sequences in $(L^0)^{\Cal S}$.

\leader{642D}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\Mpv(\Cal S)$
the space of previsible processes with domain $\Cal S$.

(a) $\Mpv(\Cal S)$ is an $f$-subalgebra of $\Mfa(\Cal S)$,
and $\bar g\pmb u\in\Mpv(\Cal S)$ whenever
$\pmb u\in\Mpv(\Cal S)$ and $g:\Bbb R\to\Bbb R$ is a continuous function
such that $g(0)=0$.

(b) $\Mpv(\Cal S)\cap\Mob(\Cal S)$ is closed in $\Mob(\Cal S)$ for the
ucp topology.   \cmmnt{Consequently }$\pmb u_<\in\Mpv(\Cal S)$
for every $\pmb u\in\Mns(\Cal S)$.

%\discrversionA{what about $\pmb u\in\Mmo(\Cal S)$?  see 642P \query}

(c) If $\tau\in\Cal S$, then $\Mpv(\Cal S\wedge\tau)
=\{\pmb x\restr\Cal S\wedge\tau:\pmb x\in\Mpv(\Cal S)\}$.

\proof{{\bf (a)} Because $\pmb u_<$ is defined and fully adapted
for every simple process
$\pmb u$ with domain $\Cal S$ (641I, 641G(a-ii)), and the limit of an
order*-convergent sequence in $\Mfa(\Cal S)$ belongs to $\Mfa(\Cal S)$ (642Bb),
$\Mpv(\Cal S)\subseteq\Mfa(\Cal S)$.
If $g:\Bbb R\to\Bbb R$ is continuous and $g(0)=0$, then
$\{\pmb x:\pmb x\in\Mpv(\Cal S)$, $\bar g\pmb x\in\Mpv(\Cal S)\}$
is closed under
order*-convergence, by the last remark in 642Ba, and contains
$\pmb u_<$ for every $\pmb u\in\Msimp(\Cal S)$, by 612La and 641Gd,
so must be the
whole of $\Mpv(\Cal S)$.   Similarly, if $\pmb u\in\Msimp(\Cal S)$,
$\{\pmb x:\pmb x\in\Mpv(\Cal S)$,
$\pmb u_<+\pmb x\in\Mpv(\Cal S)\}=\Mpv(\Cal S)$;  so
if $\pmb x\in\Mpv(\Cal S)$,
$\{\pmb u:\pmb u\in\Mpv(\Cal S)$, $\pmb u+\pmb x\in\Mpv(\Cal S)\}
=\Mpv(\Cal S)$, and
$\Mpv(\Cal S)$ is closed under addition.   Putting these together,
$\Mpv(\Cal S)$ is an $f$-subalgebra of $\Mfa(\Cal S)$.

\medskip

{\bf (b)(i)} Suppose that $\pmb x=\sfamilysS{x}$ belongs to the closure of
$\Mpv(\Cal S)\cap\Mob(\Cal S)$ in
$\Mob(\Cal S)$ for the ucp topology.    Then for each $n\in\Bbb N$
there is a
$\pmb x_n\in\Mpv(\Cal S)$ such that $\theta(\bar x_n)\le 2^{-n}$, where
$\bar x_n=\sup|\pmb x_n-\pmb x|$.   Now $\sequencen{\pmb x_n}$ is
order*-convergent to $\pmb x$.   \Prf\ Express each $\pmb x_n$ as
$\family{\sigma}{\Cal S}{x_{n\sigma}}$.   Setting $w_n=\sum_{i=0}^n\bar x_i$
for $n\in\Bbb N$, $\sequencen{w_n}$ is Cauchy for the linear space topology
of convergence in measure, so is convergent to $w$ say;
$w\ge w_n$ for every $n$ and $\sequencen{w-w_n}$ is a non-increasing
sequence with infimum $0$.   For any $\sigma\in\Cal S$, we have
$|x_{n\sigma}-x_{\sigma}|\le w_n$ for every $n$, so

\Centerline{$\inf_{n\in\Bbb N}\sup_{i\ge n}x_{i\sigma}
\le\inf_{n\in\Bbb N}\sup_{i\ge n}w_i+x_{\sigma}
=x_{\sigma}$}

\noindent and similarly

\Centerline{$\sup_{n\in\Bbb N}\inf_{i\ge n}x_{i\sigma}
\ge x_{\sigma}$;}

\noindent of course

\Centerline{$\sup_{n\in\Bbb N}\inf_{i\ge n}x_{i\sigma}
\le\inf_{n\in\Bbb N}\sup_{i\ge n}x_{i\sigma}$,}

\noindent so $\sequencen{x_{n\sigma}}$ is order*-convergent to
$x_{\sigma}$.\ \Qed

Thus $\pmb x\in\Mpv(\Cal S)$.   As $\pmb x$ is arbitrary,
$\Mpv(\Cal S)\cap\Mob(\Cal S)$ is ucp-closed.

\medskip

\quad{\bf (ii)} Since $\pmb u\mapsto\pmb u_<:\Mmo(\Cal S)\to\Mob(\Cal S)$ is
continuous (641G(e-ii)),
$\{\pmb u:\pmb u\in\Mmo(\Cal S)$, $\pmb u_<\in\Mpv(\Cal S)\}$ is relatively
ucp-closed in $\Mmo(\Cal S)$;  as it certainly includes $\Msimp(\Cal S)$, it
must include the relative closure
$\overline{\Msimp(\Cal S)}\cap\Mmo(\Cal S)=\Mns(\Cal S)$ (631Ba, 631Ca).

\medskip

{\bf (c)(i)} If $\pmb v\in\Msimp(\Cal S)$ then
$\pmb v\restr\Cal S\wedge\tau\in\Msimp(\Cal S\wedge\tau)$ (612K(d-ii)) and
$\pmb v_<\restr\Cal S\wedge\tau=(\pmb v\restr\Cal S\wedge\tau)_<$
(641G(c-ii)) belongs to $\Mpv(\Cal S\wedge\tau)$.  Now

\Centerline{$\{\pmb x:\pmb x\in\Mpv(\Cal S)$,
$\pmb x\restr\Cal S\in\Mpv(\Cal S\wedge\tau)\}$}

\noindent is closed under order*-convergence;  since it
contains $\pmb v_<$ for every $\pmb v\in\Msimp(\Cal S)$, it is the whole
of $\Mpv(\Cal S)$, that is,
$\pmb x\restr\Cal S\wedge\tau\in\Mpv(\Cal S\wedge\tau)$ for every
$\pmb x\in\Mpv(\Cal S)$.

\medskip

\quad{\bf (ii)}\grheada\ For $\pmb u\in \Mfa(\Cal S\wedge\tau)$, define
$\tilde{\pmb u}=\sfamilysS{\tilde u}$ by saying that
$\tilde u_{\sigma}=u_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma\le\tau}$
for every $\sigma\in\Cal S$.   Then $\tilde{\pmb u}$ is fully adapted.
\Prf\ As $\Bvalue{\sigma\le\tau}\in\frak A_{\sigma}$
(611H(c-i)) and $u_{\sigma\wedge\tau}\in L^0(\frak A_{\sigma\wedge\tau})
\subseteq L^0(\frak A_{\sigma})$,
$\tilde u_{\sigma}\in L^0(\frak A_{\sigma})$ for every $\sigma\in\Cal S$.
If $\sigma$, $\sigma'\in\Cal S$ and $a=\Bvalue{\sigma=\sigma'}$,
then $a\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma'\wedge\tau}$
and $a\Bcap\Bvalue{\sigma\le\tau}=a\Bcap\Bvalue{\sigma'\le\tau}$
(611E(c-v-$\alpha$, 611E(c-iv-$\alpha$))) so

\Centerline{$\tilde u_{\sigma}\times\chi a
=u_{\sigma\wedge\tau}
  \times\chi(\Bvalue{\sigma\le\tau}\Bcap\Bvalue{\sigma=\sigma'})
=u_{\sigma'\wedge\tau}
  \times\chi(\Bvalue{\sigma'\le\tau}\Bcap\Bvalue{\sigma=\sigma'})
=\tilde u_{\sigma'}\times\chi a$}

\noindent and $a\Bsubseteq\Bvalue{\tilde u_{\sigma}=\tilde u_{\sigma'}}$.\
\Qed

Of course $\tilde{\pmb u}\restr\Cal S\wedge\tau=\pmb u$.

\medskip

\qquad\grheadb\ If $\pmb v=\family{\sigma}{\Cal S\wedge\tau}{v_{\sigma}}$
belongs to $\Msimp(\Cal S\wedge\tau)$, there is a
$\pmb w\in\Msimp(\Cal S)$ such that $\pmb w_<=(\pmb v_<)\sptilde$.
\Prf\ There is certainly a $\pmb w'\in\Msimp(\Cal S)$ extending
$\pmb v$ (631Ma, or otherwise).   Now define
$\pmb w''=\sfamilysS{w''}\in\Msimp(\Cal S)$ by saying that

\Centerline{$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{w''_{\sigma}=\chi 1}$,
\quad$\Bvalue{\tau\le\sigma}\Bsubseteq\Bvalue{w''_{\sigma}=0}$}

\noindent for every $\sigma\in\Cal S$ (612Ka), and set
$\pmb w=\pmb w'\times\pmb w''$.   Expressing $\pmb w$ as
$\sfamilysS{w}$,

\Centerline{$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{w_{\sigma}=v_{\sigma}}$,
\quad$\Bvalue{\tau\le\sigma}\Bsubseteq\Bvalue{w_{\sigma}=0}$}

\noindent for every $\sigma\in\Cal S$.

If $\sigma\le\tau$, then
$\Bvalue{\sigma'<\sigma}\Bsubseteq\Bvalue{w_{\sigma'}=v_{\sigma'}}$ for
every $\sigma'$, so, in the language of 641E,
$w_{I<\sigma}=v_{I<\sigma}$ for every $I\in\Cal I(\Cal S\wedge\sigma)$
and $w_{<\sigma}=v_{<\sigma}$.   Thus
$\pmb w_<\restr\Cal S\wedge\tau=\pmb v_<$.   For general
$\sigma\in\Cal S$, if $\sigma\wedge\tau\in I\in\Cal I(\Cal S\wedge\sigma)$,

\Centerline{$\Bvalue{\sigma'<\sigma}
  \Bsetminus(\Bvalue{\sigma'<\sigma\wedge\tau}
    \Bcap\Bvalue{\sigma\wedge\tau<\sigma})
\Bsubseteq\Bvalue{\tau\le\sigma'}\Bcup\Bvalue{\sigma\le\tau}
\Bsubseteq\Bvalue{w_{\sigma'}=0}\Bcup\Bvalue{\sigma\le\tau}$}

\noindent for every $\sigma'\in I$, so
$w_{I<\sigma}\times\chi\Bvalue{\tau<\sigma}=0$.   Taking the limit as
$I\uparrow\Cal I(\Cal S\wedge\sigma)$,
$w_{<\sigma}\times\chi\Bvalue{\tau<\sigma}=0$, so

$$\eqalignno{w_{<\sigma}
&=w_{<\sigma}\times\chi\Bvalue{\sigma\le\tau}
=w_{<(\sigma\wedge\tau)}\times\chi\Bvalue{\sigma\le\tau}\cr
\displaycause{because $\pmb w_<$ is fully adapted}
&=v_{<(\sigma\wedge\tau)}\times\chi\Bvalue{\sigma\le\tau}.\cr}$$

\noindent Thus $\pmb w_<=(\pmb v_<)\sptilde$, as required.\ \Qed

\medskip

\qquad\grheadc\ If $\sequencen{\pmb u_n}$ is a sequence in
$\Mfa(\Cal S\wedge\tau)$ which is order*-convergent to
$\pmb u\in \Mfa(\Cal S\wedge\tau)$, then
$\sequencen{\tilde{\pmb u}_n}$ is order*-convergent to
$\tilde{\pmb u}$.   \Prf\ Expressing $\pmb u_n$ as
$\family{\sigma}{\Cal S\wedge\tau}{u_{n\sigma}}$, etc.,
we have

\Centerline{$\inf_{n\in\Bbb N}\sup_{i\ge n}\tilde u_{i\sigma}
=\inf_{n\in\Bbb N}\sup_{i\ge n}
  u_{i,\sigma\wedge\tau}\times\chi\Bvalue{\sigma\le\tau}
=u_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma\le\tau}
=\tilde u_{\sigma}$}

\noindent and similarly

\Centerline{$\sup_{n\in\Bbb N}\inf_{i\ge n}\tilde u_{i\sigma}
=\tilde u_{\sigma}$}

\noindent for every $\sigma\in\Cal S$.\ \Qed

\medskip

\qquad\grheadd\ So

\Centerline{$\{\pmb x:\pmb x\in\Mpv(\Cal S\wedge\tau)$,
$\tilde{\pmb x}\in\Mpv(\Cal S)\}$}

\noindent is closed under order*-convergence and contains
$\pmb v_<$ for every $\pmb v\in\Msimp(\Cal S\wedge\tau)$, and is the
whole of $\Mpv(\Cal S\wedge\tau)$.   So

\Centerline{$\Mpv(\Cal S\wedge\tau)
=\{\tilde{\pmb x}\restr\Cal S\wedge\tau:
   \pmb x\in\Mpv(\Cal S\wedge\tau)\}
\subseteq\{\pmb x\restr\Cal S\wedge\tau:\pmb x\in\Mpv(\Cal S)\}$.}

\noindent With (i), this shows that we have equality.
}%end of proof of 642D

\leader{642E}{Previsible versions in the standard model of
near-simple \dvrocolon{processes}}\cmmnt{ I had better work through the
connections between the constructions of 641B and 641E and the
ideas on processes with \cadlag\ sample paths from which they were derived.

\medskip

\noindent}{\bf Proposition}
Let $(\Omega,\Sigma,\mu)$ be a complete probability space and
$\family{t}{\coint{0,\infty}}{\Sigma_t}$ a
right-continuous filtration of $\sigma$-subalgebras of $\Sigma$
all containing every negligible subset of $\Omega$.
Suppose that we are given a family $\langle U_t\rangle_{t\ge 0}$
of real-valued functions on $\Omega$ such that
$U_t$ is $\Sigma_t$-measurable for every $t$ and
$t\mapsto U_t(\omega):\coint{0,\infty}\to\Bbb R$ is \cadlag\ for
every $\omega\in\Omega$.   Let
$h:\Omega\to\coint{0,\infty}$ be a stopping time, and $\Sigma_{h^-}$
the $\sigma$-subalgebra of $\Sigma$ generated by
$\{E:$ there is a $t\in\coint{0,\infty}$
such that $E\in\Sigma_t$ and $h(\omega)>t$ for
every $\omega\in E\}$;  define $U_{h^-}:\Omega\to\Bbb R$ by setting

$$\eqalign{U_{h^-}(\omega)
&=\lim_{t\uparrow h(\omega)}U_t(\omega)\text{ if }h(\omega)>0,\cr
&=0\text{ otherwise.}\cr}$$

Suppose that $(\frak A,\bar\mu,\family{t}{\coint{0,\infty}}{\frak A_t})$
and $\pmb u=\family{\sigma}{\Cal T_f}{u_{\sigma}}$
are defined from
$(\Omega,\Sigma,\mu,\family{t}{\coint{0,\infty}}{\Sigma_t})$ and
$\family{t}{\coint{0,\infty}}{U_t}$ as in 612H.
Let $\tau$ be the stopping time represented by
$h$\cmmnt{ (612Ha)}, and
$\pmb u_<=\family{\sigma}{\Cal T_f}{u_{<\sigma}}$ the previsible version
of $\pmb u$.   Then

(a) $\frak A_{<\tau}=\{E^{\ssbullet}:E\in\Sigma_{h^-}\}$,

(b) $u_{<\tau}=U_{h^-}^{\ssbullet}$ in $L^0(\frak A)$.

\proof{ Note that $\pmb u$ is locally near-simple, by 631D, so we can speak of
its previsible variation.

\medskip

{\bf (a)} This is a direct translation of the description of
$\frak A_{<\tau}$ in 641Bd.

\medskip

{\bf (b)(i)} For $n\in\Bbb N$, define $f_n:\Omega\to\Bbb R$ by setting

$$\eqalign{f_n(\omega)
&=0\text{ if }h(\omega)=0,\cr
&=U_{2^{-n}k}(\omega)\text{ if }k<4^n
   \text{ and }2^{-n}k<h(\omega)\le 2^{-n}(k+1),\cr
&=U_{2^n}(\omega)\text{ if }2^n<h(\omega).\cr}$$

\noindent Then $U_{h^-}(\omega)=\lim_{n\to\infty}f_n(\omega)$ for every
$\omega$.

\medskip

\quad{\bf (ii)} In fact we know more than just that
$U_{h^-}=\lim_{n\to\infty}f_n$.   Given $\epsilon>0$, there is
for each $\omega\in\Omega$ an $n\in\Bbb N$ such that
$|U_{h^-}(\omega)-U_t(\omega)|\le\epsilon$ whenever
$\max(0,h(\omega)-2^{-n})\le t<h(\omega)$.   So there is an $n$ such that
$\mu F\ge 1-\epsilon$, where

$$\eqalign{F
&=\{\omega:h(\omega)\le 2^n\text{ and }
   |U_{h^-}(\omega)-U_t(\omega)|\le\epsilon\cr
&\mskip100mu\text{ whenever }
  t\in\Bbb Q\cap\coint{\max(0,h(\omega)-2^{-n},h(\omega))}\}\cr
&=\{\omega:h(\omega)\le 2^n\text{ and }
   |U_{h^-}(\omega)-U_t(\omega)|\le\epsilon\cr
&\mskip100mu\text{ whenever }
  \max(0,h(\omega)-2^{-n})\le t<h(\omega)\}.\cr}$$

\noindent Now suppose that $I\in\Cal I(\Cal T_f)$ includes the finite
sublattice $I_n=\{(2^{-n}k)\spcheck:k\le 4^n\}$ of constant stopping
times.
Then there is a sequence $(\sigma_0,\ldots,\sigma_m)$, linearly generating
the $I$-cells, such that the totally ordered set $I_n$ is included in
$\{\sigma_j:j\le m\}$.
Take stopping times $g_0,\ldots,g_m:\Omega\to\coint{0,\infty}$ representing
$\sigma_0,\ldots,\sigma_m$ and adjusted so that $g_j\le g_{j+1}$ for every
$j$ and whenever $\sigma_j=\check t$ then
$g_j(\omega)=t$ for every $\omega$.
Observe that $\{g_j(\omega):j\le m\}\supseteq\{2^{-n}k:k\le 4^n\}$ for
every $\omega$.
We can now calculate $u_{I<\tau}$ from 641Eb, and find that
$u_{I<\tau}=\tilde f^{\ssbullet}$, where

$$\eqalign{\tilde f(\omega)
&=0\text{ if }h(\omega)=0,\cr
&=U_{g_j}(\omega)\text{ if }j<m
   \text{ and }g_j(\omega)<h(\omega)\le g_{j+1}(\omega),\cr
&=U_{g_m}(\omega)\text{ if }g_m(\omega)<h(\omega).\cr}$$

If $\omega\in F$, then either $h(\omega)=0$ and
$\tilde f(\omega)=0$, or there is a $k<4^n$ such that
$2^{-n}k<h(\omega)$ and $|U_{h^-}(\omega)-U_t(\omega)|\le\epsilon$ whenever
$2^{-n}k\le t<h(\omega)$.   In the latter case, let $j\le m$ be maximal
such that $g_j(\omega)<h(\omega)$.  Then $2^{-n}k\le g_j(\omega)$
and

\Centerline{$|\tilde f(\omega)-U_{h^-}(\omega)|
=|U_{g_j}(\omega)-U_{h^-}(\omega)|
=|U_{g_j(\omega)}(\omega)-U_{h^-}(\omega)|
\le\epsilon$.}

\noindent So we see that
$F\subseteq\{\omega:|\tilde f(\omega)-U_{h^-}(\omega)|\le\epsilon\}$ and

\Centerline{$\theta(u_{I<\tau}-U_{h^-}^{\ssbullet})
=\theta((\tilde f-U_{h^-})^{\ssbullet})
\le\epsilon+\mu(\Omega\setminus F)
\le 2\epsilon$.}

\noindent And this is true whenever $I_n\subseteq I\in\Cal I(\Cal T_f)$.

\medskip

\quad{\bf (iii)} As $\epsilon$ is arbitrary,

\Centerline{$U_{h^-}^{\ssbullet}
=\lim_{I\uparrow\Cal I(\Cal T_f)}u_{I<\tau}
=u_{<\tau}$,}

\noindent as claimed.
}%end of proof of 642E

\leader{642F}{}\cmmnt{ I spell out what amounts to a special case (though
here we do
not need a right-continuous filtration) so as to be able to quote it later.

\medskip

\noindent}{\bf Corollary} Suppose that $(\Omega,\Sigma,\mu)$ is a
complete probability
space, $\langle\Sigma_t\rangle_{t\ge 0}$ a filtration of $\sigma$-subalgebras
of $\Sigma$ such that every $\mu$-negligible set belongs to $\Sigma_0$,
and \Asissixt\ the associated real-time stochastic integration structure.

(a) Suppose that $\Cal S$ is a sublattice of $\Cal T_f$ containing
$\check 0$ and that
$\pmb u=\sfamilysS{u}$ is a simple process with breakpoint string
$(\tau_0,\ldots,\tau_n)$ in $\Cal S$ starting from $\tau_0=\check 0$.
Suppose that $h_0,\ldots,h_n:\Omega\to\coint{0,\infty}$ are stopping times
representing $\tau_0,\ldots,\tau_n$ respectively, starting from
$h_0(\omega)=0$ for every $\omega$, and such that
$h_0\le\ldots\le h_n$.   For $i\le n$, let
$f_i:\Omega\to\Bbb R$ be a measurable function representing
$u_{\tau_i}\in L^0(\frak A)$.   If $h:\Omega\to\coint{0,\infty}$ is any
stopping time representing a member $\sigma$ of $\Cal S$, and we set

$$\eqalign{f(\omega)
&=f_i(\omega)\text{ if }i<n\text{ and }
   h_i(\omega)\le h(\omega)<h_{i+1}(\omega),\cr
&=f_n(\omega)\text{ if }h_n(\omega)\le h(\omega),\cr}$$

\noindent then $f^{\ssbullet}=u_{\sigma}$ in $L^0(\frak A)$.

(b) Now suppose that $\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$
is the previsible version of $\pmb u$.
If $h:\Omega\to\coint{0,\infty}$ is any
stopping time representing a member $\sigma$ of $\Cal S$, and we set

$$\eqalign{f_-(\omega)
&=0\text{ if }h(\omega)=0,\cr
&=f_i(\omega)\text{ if }i<n\text{ and }
   h_i(\omega)<h(\omega)\le h_{i+1}(\omega),\cr
&=f_n(\omega)\text{ if }h_n(\omega)<h(\omega),\cr}$$

\noindent then $f_-^{\ssbullet}=u_{<\sigma}$ in $L^0(\frak A)$.

\proof{{\bf (a)} I had better check that $f$ is measurable;
this is because all the sets
$\{\omega:h_i(\omega)<h(\omega)\}$ are measurable, as is
$\{\omega:h(\omega)=0\}$.   Now we have

$$\eqalignno{\Bvalue{f^{\ssbullet}=u_{\sigma}}
&\Bsupseteq\sup_{i\le n}\Bvalue{f^{\ssbullet}=f_i^{\ssbullet}}
   \Bcap\Bvalue{u_{\tau_i}=u_{\sigma}}\cr
\displaycause{because $f_i^{\ssbullet}=u_{\tau_i}$ for each $i$}
&\Bsupseteq\sup_{i\le n}\{\omega:f(\omega)=f_i(\omega)\}^{\ssbullet}
   \Bcap\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}\cr
\displaycause{612H(a-iv), taking $\tau_{n+1}=\max\Cal T$, so that
$\Bvalue{\sigma<\tau_{n+1}}=1$, because $\sigma\in\Cal T_f$}
&=\sup_{i\le n}\{\omega:f(\omega)=f_i(\omega)\}^{\ssbullet}
   \Bcap\{\omega:h_i(\omega)\le h(\omega)\}^{\ssbullet}
   \Bcap\{\omega:h(\omega)<h_{i+1}(\omega)\}^{\ssbullet}\cr
\displaycause{counting $h_{n+1}(\omega)$ as $\infty$}
&=\sup_{i\le n}\{\omega:h_i(\omega)\le h(\omega)
      <h_{i+1}(\omega)\}^{\ssbullet}
=\{\omega:h_0(\omega)\le h(\omega)\}^{\ssbullet}
=\Omega^{\ssbullet}
=1,\cr}$$

\noindent so $f^{\ssbullet}=u_{\sigma}$.

\medskip

{\bf (b)} Similarly,

$$\eqalignno{\Bvalue{f_-^{\ssbullet}=u_{<\sigma}}
&\Bsupseteq(\Bvalue{f_-^{\ssbullet}=0}\Bcap\Bvalue{u_{<\sigma}=0})
  \Bcup\sup_{i\le n}(\Bvalue{f_-^{\ssbullet}=f_i^{\ssbullet}}
    \Bcap\Bvalue{u_{\tau_i}=u_{<\sigma}})\cr
&\Bsupseteq(\{\omega:f_-(\omega)=0\}^{\ssbullet}
    \Bcap\Bvalue{\sigma=\check 0})\cr
&\mskip50mu
  \Bcup\sup_{i\le n}(\{\omega:f_-(\omega)=f_i(\omega)\}^{\ssbullet}
    \Bcap\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}})\cr
\displaycause{641I), again taking $\tau_{n+1}=\max\Cal T$}
&\Bsupseteq(\{\omega:h(\omega)=0\}^{\ssbullet}
    \Bcap\Bvalue{\sigma=\check 0})\cr
&\mskip50mu
 \Bcup\sup_{i\le n}(\{\omega:f_-(\omega)=f_i(\omega)\}^{\ssbullet}
   \Bcap\{\omega:h_i(\omega)<h(\omega)\le h_{i+1}(\omega)\}^{\ssbullet})\cr
\displaycause{again taking $h_{n+1}(\omega)=\infty$ for every $\omega$}
&=\{\omega:h(\omega)=0\}^{\ssbullet}
 \Bcup\sup_{i\le n}\{\omega:h_i(\omega)<h(\omega)
      \le h_{i+1}(\omega)\}^{\ssbullet}\cr
&=\{\omega:h_0(\omega)\le h(\omega)\}^{\ssbullet}
=1,\cr}$$

\noindent so $f_-^{\ssbullet}=u_{<\sigma}$.
}%end of proof of 642F

\leader{642G}{Corollary} Suppose that
$(\Omega,\Sigma,\mu,\family{t}{\coint{0,\infty}}{\Sigma_t})$,
$(\frak A,\bar\mu,\family{t}{\coint{0,\infty}}{\frak A_t})$,
$\langle U_t\rangle_{t\ge 0}$ and
$\pmb u=\family{\tau}{\Cal T_f}{u_{\tau}}$ are as
in\cmmnt{ 631D and} 642E\cmmnt{, that is, with right-continuous filtrations
and \cadlag\ sample paths}.

(a) If $h:\Omega\to\coint{0,\infty}$ is a stopping time
representing $\tau\in\Cal T_f$, and

\Centerline{$f(\omega)
=\sup_{0<t\le h(\omega)}|U_t(\omega)-\lim_{s\uparrow t}U_s(\omega)|$}

\noindent for $\omega\in\Omega$, then
$f^{\ssbullet}=\Osclln(\pmb u\restr[\check 0,\tau])$
in $L^0(\frak A)$.

(b) $\pmb u$ is locally jump-free iff
$\omega\mapsto U_t(\omega):\coint{0,\infty}\to\Bbb R$ is continuous for
almost every $\omega$.

\proof{{\bf (a)(i)} I had better check that $f$ is measurable.
\Prf\ Given $\alpha\in\Bbb R$, consider the set

$$\eqalign{H
&=\{(t,\omega):\omega\in\Omega,\,0<t\le h(\omega),\,
    |U_t(\omega)-\lim_{s\uparrow t}U_s(\omega)|>\alpha\}\cr
&=\bigcup_{\Atop{n\in\Bbb N}{q\in\Bbb Q\cap\coint{0,\infty}}}
    \bigcap_{q'\in\Bbb Q,q'\ge q}\{(t,\omega):q<t\le h(\omega)\}\cr
&\mskip180mu
       \cap(\{(t,\omega):t\le q'\}
          \cup\{(t,\omega):|U_t(\omega)-U_{q'}(\omega)|
	     \ge\alpha+2^{-n}\})\cr
&\in\Cal B\tensorhat\Sigma\cr}$$

\noindent where $\Cal B$ is the Borel $\sigma$-algebra of
$\coint{0,\infty}$, because the ordinate set
$\{(t,\omega):0\le t\le h(\omega)\}$ belongs to $\Cal B\tensorhat\Sigma$
(252Xj\Latereditions) and $(t,\omega)\mapsto U_t(\omega)$
is $\Cal B\tensorhat\Sigma$-measurable (631D).
Set $\pi_2(t,\omega)=\omega$ for
$\omega\in\Omega$ and $t\ge 0$.   Because $\mu$ is complete, $\Sigma$ is closed
under Souslin's operation (431A);  because
$\coint{0,\infty}$ is Polish,

\Centerline{$E=\{\omega:f(\omega)>\alpha\}=\pi_2[H]$}

\noindent belongs to $\Sigma$ (423O\Latereditions).
As $\alpha$ is arbitrary, $f$ is measurable.\ \Qed

\medskip

\quad{\bf (ii)} Refining these ideas, we can learn more.

\medskip

\qquad\grheada\ Fix $\alpha>0$ for a moment.
Define $H$ as in (i) above, so that
$E=\{\omega:f(\omega)>\alpha\}$ is the projection of $H$ on $\Omega$.
Now, for any $\omega\in E$, the \cadlag\ function
$t\mapsto U_t(\omega):[0,h(\omega)]\to\Bbb R$ can have at most finitely
many jumps of size $\alpha$ or more, so there is a least
$g(\omega)\in[0,h(\omega)]$ such that $(\omega,g(\omega))\in H$.
For $\omega\in X\setminus E$, set $g(\omega)=h(\omega)$.

\medskip

\qquad\grheadb\ $g$ is a stopping time.   \Prf\
For any $s\in\coint{0,\infty}$,

\Centerline{$\{\omega:\omega\in\Omega$, $g(\omega)\le s\}
=\{\omega:h(\omega)\le s\}\cup\pi_2[H_s]$}

\noindent where

$$\eqalign{H_s
&=\{(t,\omega):\omega\in\Omega,\,0<t\le s,\,
    |U_t(\omega)-\lim_{t'\uparrow t}U_{t'}(\omega)|>\alpha\}\cr
&=\bigcup_{\Atop{n\in\Bbb N}{q\in\Bbb Q\cap\coint{0,\infty}}}
    \bigcap_{q'\in\Bbb Q,q\le q'\le s}\{(t,\omega):q<t\le s\}\cr
&\mskip150mu
       \cap(\{(t,\omega):t\le q'\}
          \cup\{(t,\omega):|U_t(\omega)-U_{q'}(\omega)|
	     \ge\alpha+2^{-n}\})\cr
&\in\Cal B\tensorhat\Sigma_s\cr}$$

\noindent because
$(t,\omega)\mapsto U_t(\omega):[0,s]\times\Omega\to\Bbb R$ is
$(\Cal B\tensorhat\Sigma_s)$-measurable, as observed in part (a) of the
proof of 631D.   Now one of the assumptions transferred from 631D is that
$\Sigma_s$ contains all $\mu$-negligible sets, so that
$(\Omega,\Sigma_s,\mu\restr\Sigma_s)$ is complete, $\Sigma_s$ is closed
under Souslin's operation and $\pi_2[H_s]\in\Sigma_s$.   Since
$\{\omega:h(\omega)\le s\}$ certainly belongs to $\Sigma_s$,
$\{\omega:g(\omega)\le s\}\in\Sigma_s$.   As $s$ arbitrary, $g$ is a
stopping time.\ \Qed

\medskip

\qquad\grheadc\
Let $\sigma\in\Cal T$ be the stopping time represented by $g$.   Then
$\sigma\le\tau$ and
$|u_{\sigma}-u_{<\sigma}|\le\Osclln(\pmb u\restr[\check{0},\tau])$.
But

\Centerline{$E=\{\omega:
|U_{g(\omega)}(\omega)
  -\lim_{t'\uparrow g(\omega)}U_{t'}(\omega)|>\alpha\}$,}

\noindent so

\Centerline{$\Bvalue{|u_{\sigma}-u_{<\sigma}|>\alpha}
=E^{\ssbullet}=\Bvalue{f^{\ssbullet}>\alpha}$}

\noindent by 642E.   It follows that $\Bvalue{f^{\ssbullet}>\alpha}
\Bsubseteq\Bvalue{\Osclln(\pmb u\restr[\check{0},\tau])>\alpha}$.
And this is true for every $\alpha>0$.

\medskip

\qquad\grheadd\ Taking the supremum over $\alpha>0$, we get
$\Bvalue{f^{\ssbullet}>0}
\Bsubseteq\Bvalue{\Osclln(\pmb u\restr[\check{0},\tau])>0}$.   And of
course
$\Bvalue{f^{\ssbullet}>\alpha}=1
=\Bvalue{\Osclln(\pmb u\restr[\check{0},\tau])>\alpha}$ if $\alpha<0$.
So in fact we have
$f^{\ssbullet}\le\Osclln(\pmb u\restr[\check{0},\tau])$.

\medskip

\quad{\bf (iii)} In the other direction,
if $\sigma\le\tau$ in $\Cal T$, we have a stopping time
$g:\Omega\to\coint{0,\infty}$ representing $\sigma$, and we can suppose
that $g\le h$.   Now, for any $\omega\in\Omega$ such that $g(\omega)>0$,

\Centerline{$|U_g(\omega)-U_{g^-}(\omega)|
=\lim_{s\uparrow g(\omega)}|U_{g(\omega)}(\omega)-U_s(\omega)|
\le f(\omega)$}

\noindent and

\Centerline{$f^{\ssbullet}
\ge|U_g^{\ssbullet}-U_{g^-}^{\ssbullet}|
=|u_{\sigma}-u_{<\sigma}|$}

\noindent by 642Eb.   By 641N,
$\Osclln(\pmb u\restr[\check 0,\tau])\le f^{\ssbullet}$ and we have equality.

\medskip

{\bf (b)} Setting

\Centerline{$f_n(\omega)
=\sup_{0<t\le n}|U_t(\omega)-\lim_{s\uparrow t}U_s(\omega)|$}

\noindent for $n\ge 1$ and $\omega\in\Omega$, we see that

$$\eqalign{&\omega\mapsto U_t(\omega):\coint{0,\infty}\to\Bbb R
\text{ is continuous for almost every }\omega\cr
&\mskip100mu\iff f_n=0\text{ a.e.\ for every }n\ge 1\cr
&\mskip100mu\iff\Osclln(\pmb u\restr[\check 0,\check n])=0
   \text{ for every }n\ge 1\cr}$$

\noindent by (a).   Setting $v_{\tau}=\Osclln(\pmb u\restr[\check 0,\tau])$
for $\tau\in\Cal T_f$, 618Da tells us that $\family{\tau}{\Cal T_f}{v_{\tau}}$
is a non-decreasing fully adapted process;  conseqently

$$\eqalignno{\pmb u\text{ is locally jump-free}&
\iff v_{\tau}=0\text{ for every }\tau\in\Cal T_f\cr&
\iff v_{\tau\wedge\check n}=0\text{ for every }\tau\in\Cal T_f
   \text{ and }n\in\Bbb N
\Displaycause{because if $\tau\in\Cal T_f$ then
$\sup_{n\in\Bbb N}\Bvalue{\tau=\tau\wedge\check n}=1$}
\iff v_{\check n}=0\text{ for every }n\ge 1\cr&
\iff\Osclln(\pmb u\restr[\check 0,\check n])=0
   \text{ for every }n\ge 1\cr&
\iff\omega\mapsto U_t(\omega):\coint{0,\infty}\to\Bbb R
\text{ is continuous for almost every }\omega.\cr}$$

}%end of proof of 642G

\leader{642H}{Previsible \dvrocolon{$\sigma$-algebras}}\cmmnt{ In the
context of 642E-642G, we have an important $\sigma$-algebra of subsets of
$\coint{0,\infty}\times\Omega$.

\medskip

\noindent}{\bf Definitions (a)}
Given a probability space $(\Omega,\Sigma,\mu)$ and a filtration
$\langle\Sigma_t\rangle_{t\ge 0}$ of $\sigma$-subalgebras of $\Sigma$, the
{\bf previsible $\sigma$-algebra} is the $\sigma$-algebra
$\Lambdapv$ of subsets of
$\coint{0,\infty}\times\Omega$ generated by
sets $\ooint{s,\infty}\times E$ where $s\ge 0$ and $E\in\Sigma_s$.

\spheader 642Hb I will say that a family $\langle X_t\rangle_{t\ge 0}$ of
real-valued functions on $\Omega$ is {\bf previsibly measurable} if
$(t,\omega)\mapsto X_t(\omega):\coint{0,\infty}\times\Omega\to\Bbb R$ is
$\Lambdapv$-measurable.

\leader{642I}{Proposition} Previsibly measurable processes are
progressively measurable.

\proof{ Given a probability space $(\Omega,\Sigma,\mu)$ and a filtration
$\langle\Sigma_t\rangle_{t\ge 0}$ of $\sigma$-subalgebras of $\Sigma$,
consider the set

\Centerline{$\Lambda
=\{W:W\subseteq\coint{0,\infty}\times\Omega$,
$W\cap([0,t]\times\Omega)\in\Cal B([0,t])\tensorhat\Sigma_t$ for every
$t\ge 0\}$}

\noindent where $\Cal B([0,t])$ is the Borel $\sigma$-algebra of
$[0,t]$.   Then $\Lambda$ is a $\sigma$-algebra of subsets of
$\coint{0,\infty}\times\Omega$ containing $\ooint{s,\infty}\times E$
whenever $s\ge 0$ and $E\in\Sigma_s$, so $\Lambdapv\subseteq\Lambda$.
Now a family $\langle X_t\rangle_{t\ge 0}$ of real-valued functions on
$\Omega$ is a previsibly measurable process iff
$(s,\omega)\mapsto X_s(\omega):\coint{0,\infty}\times\Omega\to\Bbb R$
is $\Lambdapv$-measurable;  but in this
case it is $\Lambda$-measurable and
$(s,\omega)\mapsto X_s(\omega):[0,t]\times\Omega\to\Bbb R$ is
$\Cal B([0,t])\tensorhat\Sigma_t$-measurable for every $t$, that is,
$\langle X_t\rangle_{t\ge 0}$ is progressively measurable.
}%end of proof of 642I

\leader{642J}{Lemma} Let $(\Omega,\Sigma,\mu)$ be a
complete probability space,
$\langle\Sigma_t\rangle_{t\ge 0}$ a filtration
of $\sigma$-subalgebras of $\Sigma$, $\Lambdapv$ the
associated previsible $\sigma$-algebra and $W$ a member of
$\Lambdapv$.

(a) If $h:\Omega\to[0,\infty]$ is a stopping time, then
$\{(t,\omega):h(\omega)<t\}\in\Lambdapv$.

(b) $F=\pi_2[W]$ belongs to $\Sigma$, where $\pi_2(t,\omega)=\omega$ for
$\omega\in\Omega$ and $t\ge 0$.

(c) Now suppose that every $\Sigma_t$ contains every negligible set.
If $F$ is not negligible there is a
stopping time $h:\Omega\to[0,\infty]$
such that $\{\omega:(t,h(\omega))\in W\}$ is not negligible.

\proof{{\bf (a)}

\Centerline{$\{(t,\omega):h(\omega)<t\}=\bigcup_{q\in\Bbb Q,q\ge 0}
  \ooint{q,\infty}\times\{\omega:h(\omega)\le q\}$.}

\medskip

{\bf (b)} Take $\Cal A_0\subseteq\Lambdapv$ to be the family

\Centerline{$\{\{0\}\times\Omega\}
\cup\{\ocint{s,t}\times E:s<t,\,E\in\Sigma_s\}
\cup\{\ooint{s,\infty}\times E:E\in\Sigma_s\}$.}

\noindent The complement of any member of $\Cal A_0$ is expressible as a
finite union of members of $\Cal A_0$, so
$\Cal S(\Cal A_0)\supseteq\Lambdapv$ (421F).
By 423O again, $F=\pi_1[W]\in\Cal S(\Sigma)=\Sigma$.

\medskip

{\bf (c)(i)} Set

\Centerline{$\Cal A=\{[s,t]\times E:0<s\le t$,
  $E\in\bigcup_{s'<s}\Sigma_{s'}\}\cup\{\{0\}\times\Omega\}$}

\noindent then

\Centerline{$\Cal S(\Cal A)
=\Cal S(\Cal S(\Cal A))\supseteq\Cal S(\Cal A_0)
\supseteq\Lambdapv$.}

\noindent Note that the
intersection of two members of $\Cal A$ belongs to $\Cal A$.
(If $s'_0<s_0\le t_0$, $s'_1<s_1\le t_1$, $E_0\in\Sigma_{s'_0}$ and
$E_1\in\Sigma_{s'_1}$ and $([s_0,t_0]\times E_0)\cap([s_1,t_1]\times E_1)$
is not empty, it is equal to
$[\max(s_0,s_1),\min(t_0,t_1)]\times(E_0\cap E_1)$
where $\max(s'_0,s'_1)<\max(s_0,s_1)\le\min(t_0,t_1)$ and
$E_0\cap E_1\in\Sigma_{\max(s'_0,s'_1)}$.)

\medskip

\quad{\bf (ii)} Express $W$ as the kernel of a Souslin scheme
$\family{\rho}{S^*}{A_{\rho}}$
where $A_{\rho}\in\Cal A$ for $\rho\in S^*=\bigcup_{n\ge 1}\BbbN^n$.
Because $\Cal A$ is closed under finite intersections, we
can suppose that $A_{\rho'}\subseteq A_{\rho}$ if $\rho\subseteq\rho'$.
For each $\rho$, we can express $A_{\rho}$ as $I_{\rho}\times E_{\rho}$
where $E_{\rho}\in\Sigma$ and $I_{\rho}$ is a non-empty closed subinterval
of $\coint{0,\infty}$ such that $E_{\rho}\in\Sigma_{\min I_{\rho}}$.
It follows that for any $K\subseteq\NN$,

\Centerline{$\pi_1[\bigcup_{\phi\in K}\bigcap_{n\ge 1}A_{\phi\restr n}]
=\bigcup_{\phi\in K}\bigcap_{n\ge 1}E_{\phi\restr n}$.}

\noindent
As $F$ is not negligible, there is a compact $K\subseteq\NN$ such that
$F'$ is not negligible, where
$F'=\bigcup_{\phi\in K}\bigcap_{n\ge 1}E_{\phi\restr n}$
(431D).   Set $W'=\bigcup_{\phi\in K}\bigcap_{n\ge 1}A_{\phi\restr n}$, so
that $F'=\pi_1[W']$.   Note that if $\omega\in F'$, then

\Centerline{$K_{\omega}
=\{\phi:\phi\in K,\,\omega\in E_{\phi\restr n}$ for every
  $n\ge 1\}$}

\noindent is compact in $\NN$, so

\Centerline{$W'[\{\omega\}]
=\bigcup_{\phi\in K}\{t:\omega\in\bigcap_{n\ge 1}E_{\phi\restr n},
  \,t\in\bigcap_{n\ge 1}I_{\phi\restr n}\}
=\bigcup_{\phi\in K_{\omega}}\bigcap_{n\ge 1}I_{\phi\restr n}$}

\noindent is compact (421M).

\medskip

\quad{\bf (iii)} Define $h:\Omega\to[0,\infty]$ by setting

\Centerline{$h(\omega)=\inf W'[\{\omega\}]$ for every $\omega\in\Omega$}

\noindent(interpreting $\inf\emptyset$ as $\infty$).   Then $h$ is a
stopping time.   \Prf\ Take $t\ge 0$.   Set

\Centerline{$L=\{\phi:\phi\in K$, $E_{\phi\restr n}\in\Sigma_t$ for every
$n\ge 1\}$,}

$$\eqalign{W'_t
&=W'\cap([0,t]\times\Omega)
=\bigcup_{\phi\in K}\bigcap_{n\ge 1}
  (I_{\rho\restr n}\cap[0,t])\times E_{\phi\restr n}\cr
&=\bigcup_{\phi\in L}\bigcap_{n\ge 1}
  (I_{\rho\restr n}\cap[0,t])\times E_{\phi\restr n}\cr}$$

\noindent because if $\phi\in K\setminus L$, there is an
$n\ge 1$ such that $E_{\phi\restr n}\notin\Sigma_t$, in which case
$\min I_{\phi\restr n}>t$ and
$I_{\phi\restr n}\cap[0,t]$ is empty.   Now

$$\eqalignno{\{\omega:h(\omega)\le t\}
&=\{\omega:W'[\{\omega\}]\cap[0,t]\ne\emptyset\}\cr
\displaycause{because $W'[\{\omega\}]$ is compact}
&=\pi_1[W'_t]
=\bigcup_{\phi\in L}\bigcap_{n\ge 1}E_{\phi\restr n}
\in\Sigma_t\cr}$$

\noindent because $\mu\restr\Sigma_t$ is complete so $\Sigma_t$ is closed under
Souslin's operation (431A again).
Thus $h$ is a stopping time in the sense of 455La and 612H.\ \QeD\   And

\Centerline{$\{\omega:(\omega,h(\omega))\in W\}
=\{\omega:h(\omega)<\infty\}
=\pi_1[W']=F'$}

\noindent is non-negligible.
}%end of proof of 642J

\leader{642K}{Proposition} Let
$(\Omega,\Sigma,\mu,\langle\Sigma_t\rangle_{t\ge 0},
\langle U_t\rangle_{t\ge 0})$ be as in 642E.
For $\omega\in\Omega$ set

$$\eqalign{U_{t^-}(\omega)
&=\lim_{s\uparrow t}U_s(\omega)\text{ if }t>0,\cr
&=0\text{ if }t=0.\cr}$$

(a) If we take a stopping time $h:\Omega\to\coint{0,\infty}$ and define
$U_{h^-}$ as in 642E, we have $U_{h^-}(\omega)=U_{h(\omega)^-}(\omega)$
for every $\omega$;

(b) $\langle U_{t^-}\rangle_{t\ge 0}$ is previsibly measurable.

\proof{{\bf (a)} We just have to read the definitions.

\medskip

{\bf (b)} For each $n\in\Bbb N$, $\omega\in\Omega$ and $t\ge 0$ set

$$\eqalign{f_n(t,\omega)
&=0\text{ if }t=0,\cr
&=U_{2^{-n}k}(\omega)\text{ if }k<4^n\text{ and }2^{-n}k<t\le
2^{-n}(k+1),\cr
&=U_{2^n}(\omega)\text{ if }2^n<t.\cr}$$

\noindent Then

$$\eqalign{\{(t,\omega):f_n(t,\omega)>\alpha\}
=W
&\cup\bigcup_{k<2^n}\bigl(
  (\ooint{2^{-k}n,\infty}\times\{\omega:U_{2^{-n}k}(\omega)>\alpha\})\cr
&\mskip100mu
  \setminus(\ooint{2^{-n}(k+1),\infty}
     \times\{\omega:U_{2^{-n}k}(\omega)>\alpha\})\bigr)\cr
&  \cup(\ooint{2^n,\infty}\times\{\omega:U_{2^n}(\omega)>\alpha\})\cr}$$

\noindent where

$$\eqalign{W
&=(\coint{0,\infty}\times\Omega)\setminus(\ooint{0,\infty}\times\Omega)
    \text{ if }\alpha<0,\cr
&=\emptyset\text{ if }\alpha\ge 0.\cr}$$

\noindent These sets all belong to $\Lambdapv$, so $f_n$ is
$\Lambdapv$-measurable, for every $n\in\Bbb N$.   Now
$U_{t^-}(\omega)=\lim_{n\to\infty}f_n(t,\omega)$ for all $\omega$ and
$t$, so $(t,\omega)\mapsto U_{t^-}(\omega)$ is $\Lambdapv$-measurable.
}%end of proof of 642K

\leader{642L}{Theorem}
Let $(\Omega,\Sigma,\mu)$ be a complete probability space and
$\family{t}{\coint{0,\infty}}{\Sigma_t}$ a
filtration of $\sigma$-subalgebras of $\Sigma$
all containing every negligible subset of $\Omega$.
Let $\Lambdapv$ be the corresponding previsible $\sigma$-algebra.
Suppose that $(\frak A,\bar\mu,\family{t}{\coint{0,\infty}}{\frak A_t})$
is defined
from $(\Omega,\Sigma,\mu,\family{t}{\coint{0,\infty}}{\Sigma_t})$
as in 612H.

(a) Write $\eusm L^0=\eusm L^0(\Lambdapv)$ for the $f$-algebra of
$\Lambdapv$-measurable functions from $\coint{0,\infty}\times\Omega$
to $\Bbb R$.
For every $\phi\in\eusm L^0$, there is a fully adapted process
$\pmb x_{\phi}=\family{\sigma}{\Cal T_f}{x_{\phi\sigma}}$ defined by
saying that $x_{\phi\sigma}=\phi_h^{\ssbullet}$ whenever
$h:\Omega\to\coint{0,\infty}$ is a
stopping time representing $\sigma\in\Cal T_f$, where

$$\eqalign{\phi_h(\omega)
&=0\text{ if }h(\omega)=0,\cr
&=\phi(h(\omega),\omega)\text { for other }\omega\in\Omega,\cr}$$

\noindent and now $\pmb x_{\phi}\in\Mpv=\Mpv(\Cal T_f)$
as defined in 642D.

(b) The map
$\phi\mapsto\pmb x_{\phi}:\eusm L^0\to\Mpv$ is
a surjective $f$-algebra homomorphism with kernel

$$\eqalign{&\{\phi:\phi\in\eusm L^0\text{ and there is a }
\mu\text{-conegligible set }E\cr
&\mskip100mu
\text{such that }\phi(t,\omega)=0\text{ whenever }\omega\in E
\text{ and }t>0\},\cr}$$

\noindent and $\pmb x_{g\phi}=\bar g\pmb x_{\phi}$ for every
$\phi\in\eusm L^0$ and every Borel measurable $g:\Bbb R\to\Bbb R$ such that
$g(0)=0$.

(c)(i) If $\sequencen{\phi_n}$ is a sequence in $\eusm L^0$
converging pointwise to $\phi\in\eusm L^0$, then
$\sequencen{\pmb x_{\phi_n}}$ is order*-convergent to $\pmb x_{\phi}$.

\quad(ii) If $\sequencen{\pmb x_n}$ is a sequence in
$\Mpv$ which is order*-convergent to $\pmb x\in\Mpv$, there is a
pointwise convergent sequence $\sequencen{\phi_n}$ in
$\eusm L^0$ such that $\pmb x_n=\phi_n$ for every
$n\in\Bbb N$.

\cmmnt{\medskip

\noindent{\bf Remark} Of course we have an immediate identification
between $\eusm L^0$ here and the set of previsibly measurable processes as
defined in 642Hb.}

\proof{{\bf (a)(i)} Consider the set $\Phi$ of functions
$\phi:\coint{0,\infty}\times\Omega\to\Bbb R$ such that
$\phi_h$ is $\Sigma_h$-measurable for every stopping time
$h:\Omega\to\coint{0,\infty}$.   Then $\Phi$ is a linear space closed under
multiplication
and pointwise limits of sequences.   If $s\ge 0$, $E\in\Sigma_s$ and
$\phi=\chi(\ooint{s,\infty}\times E)$, then for any
finite stopping time $h$ and any $t\ge 0$,

\Centerline{$\{\omega:\phi_h(\omega)=1\}\cap\{\omega:h(\omega)\le t\}
=\{\omega:\omega\in E,\,h(\omega)>s\}\cap\{\omega:h(\omega)\le t\}$}

\noindent certainly belongs to $\Sigma_t$ if $t\ge s$, and otherwise is
empty, so still belongs to $\Sigma_t$.   As $t$ is arbitrary,
$\{\omega:\phi_h(\omega)=1\}\in\Sigma_h$ and $\phi_h$ is
$\Sigma_h$-measurable.
As $h$ is arbitrary, $\chi(\ooint{s,\infty}\times E)\in\Phi$.

Of course $\chi(\coint{0,\infty}\times\Omega)\in\Phi$.   Since
$\{W:\chi W\in\Phi\}$ is a Dynkin class (136A) closed under finite
intersection, it includes the $\sigma$-algebra generated by
sets of the form $\ooint{s,\infty}\times E$ where $E\in\Sigma_s$,
that is, $\chi W\in\Phi$ for every $W\in\Lambdapv$.   Consequently every
$\Lambdapv$-measurable real-valued function belongs to $\Phi$, that is,
$\eusm L^0\subseteq\Phi$.

\medskip

\quad{\bf (ii)} Now take $\phi\in\eusm L^0$.
If $h_0$, $h_1:\Omega\to\coint{0,\infty}$ are stopping times
representing the
same member of $\Cal T_f$, they are equal almost everywhere
(612H(a-iv)), in which case $\phi_{h_0}\eae\phi_{h_1}$ and
$\phi_{h_0}^{\ssbullet}=\phi_{h_1}^{\ssbullet}$;  we therefore have, for
each $\sigma\in\Cal T_f$, a unique member $x_{\phi\sigma}$ of
$L^0(\frak A)$ such that $x_{\phi\sigma}=\phi_h^{\ssbullet}$ whenever
$h^{\ssbullet}=\sigma$.   Just as in 612H(b-ii),
$\pmb x=\family{\sigma}{\Cal T_f}{x_{\phi\sigma}}$ is fully adapted.

\medskip

\quad{\bf (iii)} The map
$\phi\mapsto\pmb x_{\phi}:\eusm L^0\to L^0(\frak A)^{\Cal T_f}$ is a
linear operator;  moreover, if $g:\Bbb R\to\Bbb R$ is Borel measurable
and $g(0)=0$, then $\pmb x_{g\phi}=\bar g\pmb x_{\phi}$ for every
$\phi\in\eusm L^0$.   (In the language of (i) above,

\Centerline{$x_{g\phi,\sigma}=((g\phi)_h)^{\ssbullet}
=(g\phi_h)^{\ssbullet}=\bar g(x_{\phi\sigma})$}

\noindent whenever a finite
stopping time $h$ represents $\sigma\in\Cal T_f$.)
So $\phi\mapsto\pmb x_{\phi}$ is a multiplicative Riesz homomorphism.
Next, if $\sequencen{\phi_n}$ is a sequence in $\eusm L^0$
converging pointwise to $\phi\in\eusm L^0$,
$\sequencen{\pmb x_{\phi_n}}$ is order*-convergent to $\pmb x_{\phi}$.
(This time, if $h$ is a finite stopping time representing
$\sigma\in\Cal T_f$,
$\lim_{n\to\infty}(\phi_n)_h(\omega)=\phi_h(\omega)$ for every $\omega$,
so $\sequencen{x_{\phi_n\sigma}}=\sequencen{(\phi_n)_h^{\ssbullet}}$ is
order*-convergent to $\phi_h^{\ssbullet}=x_{\phi\sigma}$.)

\medskip

\quad{\bf (iv)} If $s\ge 0$, $E\in\Sigma_s$ and
$\phi=\chi(\ooint{s,\infty}\times E)$,
then $\pmb x_{\phi}\in\Mpv$.   \Prf\ Let
$\pmb v=\family{\sigma}{\Cal T_f}{v_{\sigma}}$ be the simple process
defined by the formulae of 612Ka with $n=0$, $\tau_0=\check s$,
$u_*=0$ and $u_0=\chi E^{\ssbullet}$.   Then
$\pmb v_<=\family{\sigma}{\Cal T_f}{v_{<\sigma}}$ is defined by saying
that

\Centerline{$\Bvalue{\sigma\le\check s}\Bsubseteq\Bvalue{v_{<\sigma}=0}$,
\quad$\Bvalue{\check s<\sigma}
\Bsubseteq\Bvalue{v_{<\sigma}=\chi E^{\ssbullet}}$}

\noindent for every $\sigma\in\Cal T_f$ (641I) again).
On the other hand, if a finite stopping time $h$ represents $\sigma$,
then

$$\eqalign{\phi_h(\omega)
&=0\text{ if }h(\omega)\le s,\cr
&=\chi E(\omega)\text{ if }h(\omega)>s.\cr}$$

\noindent Since

\Centerline{$\Bvalue{\check s<\sigma}
=\Bvalue{\sigma>s}=\{\omega:h(\omega)>s\}^{\ssbullet}$}

\noindent (611E(a-i-$\delta$), 612H(a-i)), $\phi_h^{\ssbullet}=v_{<\sigma}$.
As $\sigma$ is arbitrary, $\pmb x_{\phi}=\pmb v_<$ belongs to
$\Mpv$.\ \Qed

\medskip

\quad{\bf (v)} If $\phi=\chi(\{0\}\times\Omega)$, then
$\pmb x_{\phi}=0$.   So

\Centerline{$\pmb x_{\chi(\coint{0,\infty}\times\Omega)}
=\pmb x_{\chi(\ooint{0,\infty}\times\Omega)}
\in\Mpv$.}

\medskip

\quad{\bf (vi)} By (iii) and (v),
$\{W:W\in\Lambdapv$, $\pmb x_{\chi W}\in\Mpv\}$ is a Dynkin
class closed under finite intersections;
by (iv), it includes
$\{\ooint{s,\infty}\times E:s\ge 0$, $E\in\Sigma_s\}$, so is the whole of
$\Lambdapv$.   Now
$\{\phi:\phi\in\eusm L^0$, $\pmb x_{\phi}\in\Mpv\}$ is
a linear subspace of $\eusm L^0$ closed under
pointwise convergence of sequences, and includes
$\{\chi W:W\in\Lambdapv\}$, so is the whole of $\eusm L^0$.

This concludes the proof of (a).

\medskip

{\bf (b)(i)} In the course of (a) above, I showed that
$\phi\mapsto\pmb x_{\phi}$ is a multiplicative Riesz homomorphism and
that $\pmb x_{g\phi}=\bar g\pmb x_{\phi}$ for Borel
measurable functions $g$ such that $g(0)=0$.

\medskip

\quad{\bf (ii)}\grheada\ If $\phi\in\eusm L^0$
and there is a negligible set $F$ such that
$\phi$ is zero except on $\ooint{0,\infty}\times F$, then
$\phi(h(\omega),\omega)=0$ whenever $\omega\in\Omega\setminus F$
and $h$ is a
finite stopping time, so $x_{\phi\sigma}=0$ for every $\sigma\in\Cal T_f$
and $\pmb x_{\phi}=0$.

\medskip

\qquad\grheadb\ If $\phi\in\eusm L^0$ and $\pmb x_{\phi}=0$,
set $W=\{(t,\omega):\omega\in\Omega$, $t>0$, $\phi(t,\omega)\ne 0\}$ and
$F=\pi_1[W]$, as in 642J.   If $h:\Omega\to\coint{0,\infty}$ is a stopping
time representing $\sigma\in\Cal T_f$, then

\Centerline{$\{\omega:(\omega,h(\omega))\in W\}^{\ssbullet}
=\Bvalue{x_{\phi\sigma}\ne 0}=0$,}

\noindent that is, $\{\omega:(\omega,h(\omega))\in W\}$ is negligible.
By 642Jc, $F$ is negligible, while
$\phi$ is zero on $\ooint{0,\infty}\times(\Omega\setminus F)$.
Thus the declared set is the kernel of the operator
$\phi\mapsto\pmb x_{\phi}$.

\medskip

\quad{\bf (iii)} Let $\sequencen{\phi_n}$ be a sequence in
$\eusm L^0$ such that $\sequencen{\pmb x_{\phi_n}}$ is
order*-convergent in $\Mpv$.   Then there is a
pointwise convergent sequence
$\sequencen{\phi'_n}$ in $\eusm L^0$ such that
$\pmb x_{\phi'_n}=\pmb x_{\phi_n}$ for every $n$.   \Prf\ Set

\Centerline{$W=\{(t,\omega):\omega\in\Omega$, $t>0$,
$\sequencen{\phi_n(t,\omega)}$ is not convergent$\}$.}

\noindent
Then $W\in\Lambdapv$.   If $h:\Omega\to\coint{0,\infty}$ is a stopping
time representing $\sigma\in\Cal T_f$, then

\Centerline{$\{\omega:(\omega,h(\omega))\in W\}
=\{\omega:\sequencen{\phi_n(\omega,h(\omega))}$ is not convergent$\}$}

\noindent is negligible because $\sequencen{x_{\phi_n\sigma}}$ is
order*-convergent in $L^0(\frak A)$.   By 642Jc again,
$F=\pi_1[W]$ is negligible.   For $n\in\Bbb N$, $\omega\in\Omega$ and
$t\ge 0$, set

$$\eqalign{\phi'_n(t,\omega)
&=0\text{ if }t=0\text{ or }\omega\in F,\cr
&=\phi_n(t,\omega)\text{ otherwise}.\cr}$$

\noindent Since $\Sigma_0$ contains every negligible set,
$\ocint{0,\infty}\times(\Omega\setminus F)\in\Lambdapv$ and
$\phi'_n\in\eusm L^0$ for every $n$.   Also, by (ii-$\alpha$) above,
$\pmb x_{\phi'_n}=\pmb x_{\phi_n}$ for every $n$, while
$\sequencen{\phi'_n}$ is pointwise convergent everywhere in
$\coint{0,\infty}\times\Omega$.\ \Qed

\medskip

\quad{\bf (iv)} Suppose that
$\pmb v=\family{\sigma}{\Cal T_f}{v_{\sigma}}$ is a simple process.

\medskip

\qquad\grheada\
Let $(\tau_0,\ldots,\tau_k)$ be a breakpoint string for $\pmb v$, starting
from $\tau_0=\check 0$ (use 612Kb), and for $i\le k$
choose measurable $f_i:\Omega\to\Bbb R$ such that
$f_i^{\ssbullet}=v_{\tau_i}$.   For each $i\le k$, there is a
stopping time $h_i:\Omega\to\coint{0,\infty}$ representing $\tau_i$.
We can take $h_0=0$, and
since $\sup_{i\le j}h_i$ is always a stopping time (455L(c-iv)), we can
suppose that $h_0\le\ldots\le h_k$.   Now set

$$\eqalign{\phi(t,\omega)
&=0\text{ if }t=0,\cr
&=f_i(\omega)\text{ if }i<k
   \text{ and }h_i(\omega)<t\le h_{i+1}(\omega),\cr
&=f_k(\omega)\text{ if }h_k(\omega)\le t.\cr}$$

\medskip

\qquad\grheadb\ $\phi\in\eusm L^0$.
\Prf\ For $\alpha\in\Bbb R$,
$i\le k$ and $q\in\Bbb Q$ set $E_{\alpha qi}
=\{\omega:h_i(\omega)\le q$, $f_i(\omega)>\alpha\}$.
Then

\Centerline{$E_{\alpha qi}^{\ssbullet}
=\Bvalue{v_{\tau_i}>\alpha}\Bsetminus\Bvalue{\tau_i>q}
\in\frak A_q$}

\noindent because
$\Bvalue{v_{\tau_i}>\alpha}\in\frak A_{\tau_i}$.   (See the
definition in 611G.)   So $E_{\alpha qi}\in\Sigma_q$ and

\Centerline{$\ooint{q,\infty}\times E_{\alpha qi}
=\{(t,\omega):h_i(\omega)\le q<t$, $f_i(\omega)>\alpha\}$}

\noindent belongs to $\Lambdapv$.   Taking the union over
$q\in\Bbb Q\cap\coint{0,\infty}$,

\Centerline{$\{(t,\omega):
  h_i(\omega)<t$, $f_i(\omega)>\alpha\}
\in\Lambdapv$.}

\noindent As already noted in 642Ja,
$\{(t,\omega):h_{i+1}(\omega)<t\}\in\Lambdapv$ for every $i<k$.
Accordingly

$$\eqalign{&\{(t,\omega):t>0,\,\phi(t,\omega)>\alpha\}\cr
&\mskip50mu
=\bigcup_{i<k}(\{(t,\omega):h_i(\omega)<t,\,f_i(\omega)>\alpha\}
  \setminus\{(t,\omega):h_{i+1}(\omega)<t\})\cr
&\mskip120mu
  \cup\{(t,\omega):h_k(\omega)<t,\,f_k(\omega)>\alpha\}\cr
&\mskip50mu
\in\Lambdapv.\cr}$$

\noindent Of course $\{(t,\omega):t=0$, $\phi(t,\omega)>\alpha\}$ is either
$\{0\}\times\Omega$ or $\emptyset$, and in either case belongs to
$\Lambdapv$.   So $\phi$ is $\Lambdapv$-measurable.\ \Qed

\medskip

\qquad\grheadc\ If $h:\Omega\to\coint{0,\infty}$
is a stopping time representing $\sigma\in\Cal T_f$, then

$$\eqalign{\phi_h(\omega)
&=0\text{ if }h(\omega)=0,\cr
&=f_i(\omega)\text{ if }i<k
   \text{ and }h_i(\omega)<h(\omega)\le h_{i+1}(\omega),\cr
&=f_k(\omega)\text{ if }h_k(\omega)\le h(\omega).\cr}$$

\noindent So

\Centerline{$\Bvalue{\sigma=\check 0}
=\{\omega:h(\omega)=0\}^{\ssbullet}
\Bsubseteq\{\omega:\phi_h(\omega)=0\}^{\ssbullet}
=\Bvalue{\phi_h^{\ssbullet}=0}$,}

$$\eqalign{\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}&
=\{\omega:h_i(\omega)<h(\omega)\le h_{i+1}(\omega)\}^{\ssbullet}\cr&
\Bsubseteq\{\omega:\phi_h(\omega)=f_i(\omega)\}^{\ssbullet}
=\Bvalue{\phi_h^{\ssbullet}=v_{\tau_i}}\cr}$$

\noindent for $i<k$,

\Centerline{$\Bvalue{\tau_k<\sigma}
=\{\omega:h_k(\omega)<h(\omega)\}^{\ssbullet}
\Bsubseteq\{\omega:\phi_h(\omega)=f_k(\omega)\}^{\ssbullet}
=\Bvalue{\phi_h^{\ssbullet}=v_{\tau_k}}$.}

\noindent Matching this with 641I, we see that
$\phi_h^{\ssbullet}=v_{<\sigma}$.   As $h$ is arbitrary,
$\pmb x_{\phi}=\pmb v_<$.

\medskip

\quad{\bf (v)} Consider now the set

\Centerline{$\{\pmb x_{\phi}:\phi\in\eusm L^0\}\subseteq\Mpv$.}

\noindent From (iv), we see that this contains $\pmb v_<$ for every
$\pmb v\in\Msimp(\Cal T_f)$;   from (iii) here and (a-iii) above, we see
that it is closed under sequential order*-convergence;  by the definition
in 642C, it must be the whole of $\Mpv$.   Thus
$\phi\mapsto\pmb x_{\phi}:\eusm L^0\to\Mpv$ is surjective.

This completes the proof of (b).

\medskip

{\bf (c)} These facts have been dealt with in (a-iii) and (b-iii) above.
}%end of proof of 642L.

\leader{642N}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full
sublattice of $\Cal T$ such that
$\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and bounded below
in $\Cal S$.
If $\pmb u=\sfamilysS{u}$ is moderately oscillatory,
there is a $\pmb u'\in\Mns(\Cal S)$ such that $\pmb u_<=\pmb u'_<$.

\proof{ If $\Cal S$ is empty, this is trivial, so let us suppose otherwise.

\medskip

{\bf (a)} If $\Cal S$ has greatest and least members and $\delta>0$, then there
is a simple process $\pmb u'=\sfamilysS{u'}$ such that
$\theta(\sup|\pmb u_<-\pmb u'_<|)\le 2\delta$ and
$\theta(\sup|\pmb u'|)\le\theta(\sup|\pmb u_<|)+2\delta$.
\Prf\ Construct
$\sequence{i}{D_i}$, $\sequence{i}{y_i}$, $\sequence{i}{d_i}$,
$\langle c_{i\sigma}\rangle_{i\in\Bbb N,\sigma\in\Cal S}$
and $\tildepmb u=\sfamilysS{\tilde u}$ from
$\pmb u$ and $\delta$ as in 615M.   For $i\in\Bbb N$, set $\tau_i=\inf D_i$, so
that $\tau_i\in\Cal S$ and
$y_i\in L^0(\bigcap_{\sigma\in D_i}\frak A_{\sigma})=L^0(\frak A_{\tau_i})$
(615Mb, 632C(a-iii)).   As $D_0=\Cal S$ and
$\tau_i\le\sigma$ whenever $i\in\Bbb N$ and
$\sigma\in D_{i+1}$ (615Ma), $\tau_0=\min\Cal S$ and
$\tau_i\le\tau_{i+1}$ for every $i$.
Let $n\ge 1$ be such that $\bar\mu d_n\le\delta$ (615M(c-ii)),
and take $\pmb u'=\sfamilysS{u'}$
to be the simple process with domain $\Cal S$ and breakpoint
string $(\tau_0,\ldots,\tau_n)$ such that
$\Bvalue{\tau_i<\tau_{i+1}}\Bsubseteq\Bvalue{u'_{\tau_i}=y_i}$ for $i\le n$ and
$u'_{\tau_n}=0$ (612Ka).

If $\sigma\in\Cal S$, $i<n$ then

\Centerline{$\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\inf_{\tau\in D_{i+1}}\Bvalue{\sigma<\tau}
\Bsubseteq 1\Bsetminus c_{i+1,\sigma}$}

\noindent (615M(d-i)).   If $\sigma\in\Cal S$, $i<n$, $\tau\in D_i$
and $I\in\Cal I(\Cal S)$ contains $\tau$,
then for any $\sigma'\in I$

\Centerline{$\Bvalue{\tau\le\sigma'}\Bcap\Bvalue{\sigma'<\sigma}
   \Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Bsubseteq c_{i\sigma'}\Bsetminus c_{i+1,\sigma'}
\Bsubseteq\Bvalue{\tilde u_{\sigma'}=y_i}$}

\noindent (615Me).   As $\sigma'$ is arbitrary,

\Centerline{$\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Bsubseteq\Bvalue{\tilde u_{I<\sigma}=y_i}$,}

\noindent defining $\tilde u_{I<\sigma}$ as in 641Ea.   Taking the limit
as $I\uparrow\Cal I(\Cal S)$,

\Centerline{$\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Bsubseteq\Bvalue{\tilde u_{<\sigma}=y_i}$.}

\noindent Now

$$\eqalignno{\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}&
=\sup_{\tau\in D_i}\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Displaycause{632C(a-ii)}
\Bsubseteq\Bvalue{\tilde u_{<\sigma}=y_i};\cr}$$

\noindent and at the same time we have

\Centerline{$\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Bsubseteq\Bvalue{u'_{<\sigma}=y_i}$}

\noindent by 641I.   So
$\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Bsubseteq\Bvalue{u'_{<\sigma}=\tilde u_{<\sigma}}$, and this is true
whenever $i<n$.   At the bottom,

\Centerline{$\Bvalue{\tau_0=\sigma}
\Bsubseteq\Bvalue{u'_{<\sigma}=0}\Bcap\Bvalue{\tilde u_{<\sigma}=0}
\Bsubseteq\Bvalue{u'_{<\sigma}=\tilde u_{<\sigma}}$,}

\noindent so $\Bvalue{\sigma\le\tau_n}
\Bsubseteq\Bvalue{u'_{<\sigma}=\tilde u_{<\sigma}}$.   But

\Centerline{$\Bvalue{\tau_n<\sigma}
=\sup_{\tau\in D_n}\Bvalue{\tau<\sigma}\Bsubseteq d_n$.}

\noindent As $\sigma$ is arbitrary,
$\Bvalue{\pmb u'_<\ne\tildepmb u_<}\Bsubseteq d_n$ and
$\theta(\sup|\pmb u'_<-\tildepmb u_<|)\le\bar\mu d_n\le\delta$.

We know also that

$$\eqalignno{\sup|\tildepmb u_<-\pmb u_<|&
=\sup|(\tildepmb u-\pmb u)_<|
\Displaycause{641G(e-i)}
\le\sup|\tildepmb u-\pmb u|
\Displaycause{641G(a-vii)}
\le\delta\chi 1\cr}$$

\noindent (615Mf), so
$\theta(\sup|\pmb u'_<-\pmb u_<|)\le 2\delta$.

As for $\sup|\pmb u'|$, this is $\sup|\pmb u'_<|$ by 641Ib, because
$u'_{\tau_n}=0$.   So

\Centerline{$\theta(\sup|\pmb u'|)
=\theta(\sup|\pmb u'_<|)
\le\theta(\sup|\pmb u'_<-\pmb u_<|)+\theta(\sup|\pmb u_<|)
\le\theta(\sup|\pmb u_<|)+2\delta$,}

\noindent as required.\ \Qed

\medskip

{\bf (b)} With a bit more effort we have a similar result in the general case.
Again, take any $\delta>0$.   By 615G,
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ and
$u_{\uparrow}=\lim_{\uparrow\Cal S}u_{\sigma}$ are defined, and there are
$\tau^*$, $\tau_*\in\Cal S$ such that

\Centerline{$\theta(\sup_{\sigma\in\Cal S\vee\tau^*}|u_{\sigma}-u_{\uparrow}|)
\le\delta$,\quad
$\theta(\sup_{\sigma\in\Cal S\wedge\tau_*}|u_{\sigma}-u_{\downarrow}|)
\le\delta$;}

\noindent we can suppose that $\tau_*\le\tau^*$.   We now have

\Centerline{$\theta(\sup_{\sigma\in\Cal S\vee\tau^*}|u_{\sigma}-u_{\tau^*}|)
\le 2\delta$.}

\noindent $\Cal S_1=\Cal S\cap[\tau_*,\tau^*]$, like $\Cal S$,
is closed under
arbitrary infima, and $\pmb u\restr\Cal S_1$ is moderately oscillatory
(615F(a-i)), so (a) tells us that there is a simple process
$\pmb u'_1=\family{\sigma}{\Cal S_1}{u'_{1\sigma}}$ such that
$\theta(\sup|(\pmb u\restr\Cal S_1)_<-\pmb u'_{1<}|)\le 2\delta$.
Let $(\tau_0,\ldots,\tau_n)$ be a breakpoint string for $\pmb u'_1$ starting
from $\tau_0=\tau_*$ and ending with $\tau_n=\tau^*$.   Let
$\pmb u'=\sfamilysS{u'}$ be the simple process defined by the formula
of 612Ja from $\tau_0,\ldots,\tau_n$ and

\Centerline{$u_*=u_{\downarrow}$,
$u'_{1\tau_0},\ldots,u'_{1\tau_{n-1}}$,
$u_{\tau_n}\times\chi(\sup_{\sigma\in\Cal S}\Bvalue{\tau_n<\sigma})$.}

\noindent (Because $\Bvalue{\tau_n<\sigma}\in\frak A_{\tau_n}$ for every
$\sigma\in\Cal S$,
$\sup_{\sigma\in\Cal S}\Bvalue{\tau_n<\sigma}\in\frak A_{\tau_n}$ and
$u'_{1\tau_n}\times\chi(\sup_{\sigma\in\Cal S}\Bvalue{\tau_n<\sigma})
\in L^0(\frak A_{\tau_n})$.)   Set

\Centerline{$w=\sup_{\sigma\in\Cal S\wedge\tau_0}|u_{\sigma}-u_{\downarrow}|
      \vee\sup_{\sigma\in\Cal S_1}|u_{<\sigma}-u'_{1<\sigma}|
      \vee\sup_{\sigma\in\Cal S\vee\tau_n}|u_{\sigma}-u_{\tau_n}|$;}

\noindent then $\theta(w)\le 5\delta$.   Fix $\tau\in\Cal S$.

\medskip

\quad{\bf (i)} For any $\sigma\in\Cal S$,

$$\eqalign{\Bvalue{\tau\le\tau_0}\Bcap\Bvalue{\sigma<\tau}&
\Bsubseteq\Bvalue{\sigma<\tau_0}\cr&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau_0}}
    \Bcap\Bvalue{u'_{\sigma}=u_{\downarrow}}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-u'_{\sigma}|
    =|u_{\sigma\wedge\tau_0}-u_{\downarrow}}
\Bsubseteq\Bvalue{|u_{\sigma}-u'_{\sigma}|\le w}\cr}$$

\noindent By 641G(a-vii-$\alpha$), applied to $\pmb u-\pmb u'$,
$\Bvalue{\tau\le\tau_0}\Bsubseteq\Bvalue{|u_{<\tau}-u'_{<\tau}|\le w}$.

\medskip

\quad{\bf (ii)} Take $i<n$ and $\sigma\in\Cal S$;  set
$\sigma'=\med(\tau_i,\sigma,\tau_{i+1})$.
Applying 641I to the simple processes $\pmb u'$, $\pmb u'_0$, we have

$$\eqalign{\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}&
\Bsubseteq\Bvalue{u'_{<\sigma}=u'_{\tau_i}}
   \Bcap\Bvalue{u'_{0<\sigma'}=u'_{0\tau_i}}
   \Bcap\Bvalue{u'_{\tau_i}=u'_{0\tau_i}}
   \Bcap\Bvalue{u_{<\sigma}=u_{<\sigma'}}\cr&
\Bsubseteq\Bvalue{|u_{<\sigma}-u'_{<\sigma}|=|u_{<\sigma'}-u'_{0<\sigma'}|}\cr&
\Bsubseteq\Bvalue{|u_{<\sigma}-u'_{<\sigma}|\le w}.\cr}$$

\medskip

\quad{\bf (iii)} At the top end, setting $\sigma'=\sigma\vee\tau_n$, we have
$\Bvalue{\tau_n<\sigma'}\Bsubseteq\Bvalue{u'_{\rho}=u_{\tau_n}}$
whenever $\rho\in\Cal S\vee\tau_n$.   Applying 641G(a-vii-$\beta$) to
$\pmb u-\pmb u'$,

\Centerline{$\Bvalue{\tau_n<\sigma'}
\Bsubseteq\Bvalue{|u_{<\sigma'}-u'_{<\sigma'}|
     \le\sup_{\rho\in\Cal S\vee\tau_n}
            |u_{\rho}-u'_{\rho}|\times\chi\Bvalue{\rho<\sigma'}}$,}

\noindent so

$$\eqalign{\Bvalue{\tau_n<\sigma'}&
\Bsubseteq\Bvalue{|u_{<\sigma'}-u'_{<\sigma'}|
     \le\sup_{\rho\in\Cal S\vee\tau_n}
            |u_{\rho}-u_{\tau_n}|}\cr&
\Bsubseteq\Bvalue{|u_{<\sigma'}-u'_{<\sigma'}|\le w}.\cr}$$

\noindent Since

\Centerline{$\Bvalue{\tau_n<\sigma}
   =\Bvalue{\tau_n<\sigma'}
      \Bcap\Bvalue{u_{<\sigma}-u'_{<\sigma'}=u_{<\sigma}-u'_{<\sigma}}$,}

\noindent we see that $\Bvalue{\tau_n<\sigma}
\Bsubseteq\Bvalue{|u_{<\sigma}-u'_{<\sigma}|\le w}$.

\medskip

\quad{\bf (iv)} Since

\Centerline{$(\Bvalue{\sigma\le\tau_0}$,
$\Bvalue{\tau_0<\sigma}\Bcap\Bvalue{\sigma\le\tau_1},\ldots,
\Bvalue{\tau_{n-1}<\sigma}\Bcap\Bvalue{\sigma\le\tau_n}$,
$\Bvalue{\tau_n<\sigma})$}

\noindent is a partition of unity in $\frak A$, we see that
$|u_{<\sigma}-u'_{<\sigma}|\le w$;  as $\sigma$ is arbitrary,
$\sup|\pmb u_<-\pmb u'_<|\le w$ and
$\theta(\sup|\pmb u_<-\pmb u'_<|)\le 5\delta$.

\medskip

\quad{\bf (v)} A further feature of the construction here is that
$\sup|\pmb u'|\le\sup|\pmb u'_<|$.   \Prf\ By 641Ib,
$\sup|\pmb u'|=\sup|\pmb u'_<|\vee|u'_{\tau_n}|$.   But

\Centerline{$u'_{\tau_n}
=u_{\tau_n}\times\chi(\sup_{\sigma\in\Cal S}\Bvalue{\tau_n<\sigma})
=u'_{\tau_n}\times\chi(\sup_{\sigma\in\Cal S}\Bvalue{\tau_n<\sigma})$.}

\noindent And if $\sigma\in\Cal S$, then
$\Bvalue{\tau_n<\sigma}\Bsubseteq\Bvalue{u'_{<\sigma}=u'_{\tau_n}}$,
so

\Centerline{$|u'_{\tau_n}|\times\chi\Bvalue{\tau_n<\sigma}\le|u'_{<\sigma}|
\le\sup|\pmb u'_<|$.}

\noindent Taking the supremum over $\sigma$, $|u'_{\tau_n}|\le\sup|\pmb u'_<|$
and $\sup|\pmb u'|=\sup|\pmb u'_<|$.

\medskip

{\bf (c)} Once more supposing that $\Cal S$ is a non-empty
finitely full sublattice of
$\Cal T$ such that $\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty
and bounded below in $\Cal S$, take any $\pmb u\in\Mmo(\Cal S)$.
Choose simple processes $\pmb u_n$, $\pmb v_n$ as follows.   Start with
$\pmb u_0=\tbf 0$, the constant process on $\Cal S$ with value $0$.
Given that $\pmb u_n$ is a simple process with domain $\Cal S$, (b) tells us
that there is a simple process $\pmb v_n$ such that

\Centerline{$\theta(\sup|\pmb u_<-\pmb u_{n<}-\pmb v_{n<}|)\le 2^{-n}$,
\quad$\sup|\pmb v_n|=\sup|\pmb v_{n<}|$.\hskip 2truecm(*)}

\noindent Set $\pmb u_{n+1}=\pmb u_n+\pmb v_n$ and continue.

The construction ensures that

\Centerline{$\sup|\pmb u_{n+1}-\pmb u_n|
=\sup|\pmb v_n|=\sup|\pmb v_{<n}|$,
\quad$\theta(\sup|\pmb u_<-\pmb u_{(n+1)<}|)\le 2^{-n}$}

\noindent for every $n$.   Consequently (in the notation of 615B)

$$\eqalign{\widehat{\theta}(\pmb u_{n+1}-\pmb u_n)&
=\widehat{\theta}(\pmb v_{n+1})
=\widehat{\theta}(\pmb v_{(n+1)<})\cr&
=\widehat{\theta}(\pmb u_<-\pmb u_{(n+2)<}-\pmb u_<+\pmb u_{(n+1)<})
\le 2^{-n+1}\cr}$$

\noindent for every $n$, and
$\sum_{n=0}^{\infty}\widehat{\theta}(\pmb u_{n+1}-\pmb u_n)<\infty$.
Because $\Mob(\Cal S)$ is complete under the ucp uniformity (615Cc),
$\sequencen{\pmb u_n}$ has a limit $\pmb u'$ say for the ucp topology,
and $\pmb u'$ is near-simple.

Since

\Centerline{$\widehat{\theta}(\pmb u'_<-\pmb u_{n<})
\le\widehat{\theta}(\pmb u'-\pmb u_n)\to 0$}

\noindent as $n\to\infty$,

\Centerline{$\lim_{n\to\infty}\pmb v_{n<}
=\lim_{n\to\infty}\pmb u_{(n+1)<}-\pmb u_{n<}=0$.}

\noindent By (*) above,

\Centerline{$\pmb u_<=\lim_{n\to\infty}\pmb u_{n<}+\pmb v_{n<}=\pmb u'_<$.}

\noindent So $\pmb u'$ witnesses that the theorem is true.
}%end of proof of 642N

\leader{642O}{Corollary} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full
sublattice of $\Cal T$ such that
$\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and bounded below
in $\Cal S$.
If $\pmb u=\sfamilysS{u}$ is moderately oscillatory, $\pmb u_<$ is a previsible
process.

\proof{ By 642N, there is a $\pmb v\in\Mns(\Cal S)$ such that
$\pmb u_<=\pmb v_<$.   Now
there is a sequence $\sequencen{\pmb v_n}$ in $\Msimp(\Cal S)$ such that
$\lim_{n\to\infty}\theta(\sup|\pmb v_n-\pmb v|)=0$;  taking a subsequence if
necessary, we can arrange that
$\sum_{n\in\Bbb N}\theta(\sup|\pmb v_n-\pmb v|)$ is finite.
Since $\sup|\pmb v_{n<}-\pmb v_<|\le\sup|\pmb v_n-\pmb v|$ for each $n$
(641G(a-vii-$\gamma$)),
$\sum_{n\in\Bbb N}\theta(\sup|\pmb v_{n<}-\pmb v_<|)$ is finite and
$\sequencen{\sup|\pmb v_{n<}-\pmb v_<|}$ is order*-convergent to $0$ (642Bc).
It follows at once that if we express $\pmb v_<$ as
$\family{\sigma}{\Cal S}{v_{<\sigma}}$ and each $\pmb v_{n<}$ as
$\family{\sigma}{\Cal S}{v_{n<\sigma}}$,
$\sequencen{v_{n<\sigma}}$ is order*-convergent to $v_{<\sigma}$ for every
$\sigma\in\Cal S$ and $\sequencen{\pmb v_{n<}}$ is order*-convergent to
$\pmb v_<$ (642Bb).   So $\pmb v_<\in\Mpv(\Cal S)$ and $\pmb u_<$ is previsible.
}%end of proof of 642O

\leader{642P}{Proposition} (a)
Let $\Cal S$ be a sublattice of $\Cal T$ such that
there is a countable set $A\subseteq\Cal S$ which separates $\Cal S$.
Then $\pmb u_<$ is previsible whenever $\pmb u\in\Mfa(\Cal S)$ and
$\pmb u_<$ is defined.

(b) If $T\subseteq\Bbb R$, then for every sublattice
$\Cal S$ of $\Cal T$ there is a countable subset of $\Cal S$ which separates
$\Cal S$.

\proof{{\bf (a)} If $\Cal S$ is empty or a singleton, this is trivial;  so
suppose otherwise.   Then $A\ne\emptyset$.  Write $\Cal S'$ for the sublattice
of $\Cal S$ generated by $A$, so that $\Cal S'$ is countable and not empty and
separates $\Cal S$;  let $\sequencen{I_n}$ be a non-decreasing sequence of
finite sublattices of $\Cal S'$ with union $\Cal S'$.   For each $n$, enumerate
a maximal totally ordered subset of $I_n$ in increaasing order as
$(\tau_{n0},\ldots,\tau_{n,k(n)})$, and let $\pmb u^{(n)}=\sfamilysS{u^{(n)}}$
be the simple process with domain $\Cal S$ defined by saying that

\Centerline{$\Bvalue{\sigma<\tau_{n0}}\Bsubseteq\Bvalue{u^{(n)}_{\sigma}=0}$,
\quad$\Bvalue{\tau_{n,k(n)}\le\sigma}
\Bsubseteq\Bvalue{u^{(n)}_{\sigma}=u_{\tau_{n,k(n)}}}$,}

\Centerline{$\Bvalue{\tau_{ni}\le\sigma}\Bcap\Bvalue{\sigma<\tau_{n,i+1}}
\Bsubseteq\Bvalue{u^{(n)}_{\sigma}}=u_{\tau_{ni}}$ for $0\le i<k(n)$.}

If $\tau\in\Cal S$ and $n\in\Bbb N$, then

\Centerline{$\Bvalue{\tau\le\tau_{n0}}\Bsubseteq\Bvalue{u_{I_n<\tau}=0}$,
\quad$\Bvalue{\tau_{n,k(n)}<\tau}
   \Bsubseteq\Bvalue{u_{I_n<\tau}=u_{\tau_{n,k(n)}}}$,}

\Centerline{$\Bvalue{\tau_{ni}<\tau}\Bcap\Bvalue{\tau\le\tau_{n,i+1}}
\Bsubseteq\Bvalue{u_{I_n<\tau}=u_{\tau_{ni}}}$ for every $i<k(n)$}

\noindent by 641Eb.   On the other hand, 641I tells us that

$$\eqalign{\inf_{\sigma\in\Cal S}\Bvalue{\tau\le\sigma}&
  \Bsubseteq\Bvalue{u^{(n)}_{<\tau}=0},\cr
\Bvalue{\sigma<\tau}\Bcap\Bvalue{\tau\le\tau_{n0}}&
  \Bsubseteq\Bvalue{u^{(n)}_{<\tau}=0}\text{ for every }\sigma\in\Cal S,\cr}$$

\noindent so $\Bvalue{\tau\le\tau_{n0}}\Bsubseteq\Bvalue{u^{(n)}_{<\tau}=0}$,
while

$$\eqalign{\Bvalue{\tau_k(n)<\tau}&
  \Bsubseteq\Bvalue{u^{(n)}_{<\tau}=u_{\tau_{kn}}},\cr
\Bvalue{\tau_{ni}<\tau}\Bcap\Bvalue{\tau\le\tau_{n,i+1}}&
  \Bsubseteq\Bvalue{u^{(n)}_{<\tau}=u_{\tau_{ni}}}\text{ for every }i<n.\cr}$$

\noindent So we see that $u_{I_n<\tau}=u^{(n)}_{<\tau}$.

Finally, given $\tau\in\Cal S$,
we know from 641I that $u_{<\tau}=\lim_{I\uparrow\Cal I(\Cal S')}u_{I<\tau}$.
But any finite subset of $\Cal S'$ is included in $I_n$ for all $n$ large
enough, so

\Centerline{$u_{<\tau}=\lim_{n\to\infty}u_{I_n<\tau}
=\lim_{n\to\infty}u^{(n)}_{<\tau}$.}

\noindent As $\tau$ is arbitrary, $\sequencen{\pmb u^{(n)}_<}$
order*-converges to $\pmb u_<$.   Since every $\pmb u^{(n)}$ is simple,
every $\pmb u^{(n)}_<$ is previsible, and $\pmb u_<$ is previsible.

\medskip

{\bf (b)} The set $T_0$ of points of $T$ which are isolated on the right in
$T$ is countable, so there is a countable dense subset $Q$ of $T$ including
$T_0$.

\medskip

\quad{\bf (i)} For $q\in Q$ and $\tau\in\Cal S$, set

\Centerline{$c_{\tau q}=\Bvalue{\tau\le\check q}\Bsetminus
  \sup_{\sigma\in\Cal S}(\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\check q})$.}

\noindent Note that if $\sigma$, $\tau\in\Cal S$ then
$c_{\sigma q}\Bcap c_{\tau q}\Bsubseteq\Bvalue{\sigma=\tau}$.   \Prf\

$$\eqalign{c_{\sigma q}\Bcap c_{\tau q}&
\Bsubseteq\Bvalue{\sigma\le\check q}\Bcap\Bvalue{\tau\le\check q}\Bsetminus
   (\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\check q})\cr&
=\Bvalue{\sigma\le\check q}\Bcap\Bvalue{\tau\le\check q}\Bcap
   (\Bvalue{\sigma\le\tau}\Bcup\Bvalue{\check q<\sigma})
\Bsubseteq\Bvalue{\sigma\le\tau},\cr}$$

\noindent and similarly
$c_{\sigma q}\Bcap c_{\tau q}\Bsubseteq\Bvalue{\tau\le\sigma}$.\ \QeD\
Choose a countable $C_q\subseteq\Cal S$  such that
$\sup_{\tau\in\Cal C_q}c_{\tau q}=\sup_{\tau\in\Cal S}c_{\tau q}$.

Next, for $q$, $r\in Q$, set
$b_{qr}=\sup_{\tau\in\Cal S}(\Bvalue{\tau>q}\Bsetminus\Bvalue{\tau>r})$, and
choose a countable $B_{qr}\subseteq\Cal S$ such that
$b_{qr}=\sup_{\tau\in B_{qr}}(\Bvalue{\tau>q}\Bsetminus\Bvalue{\tau>r})$.   Set
$A=\bigcup_{q,r\in Q}B_{qr}\cup\bigcup_{q\in Q}C_q$, so that $A$ is a countable
subset of $\Cal S$.

\medskip

\quad{\bf (ii)} \Quer\
Suppose, if possible, that $A$ does not separate $\Cal S$.
Then there are $\tau$, $\tau'\in\Cal S$ such that

\Centerline{$a=\Bvalue{\tau<\tau'}\Bsetminus
  \sup_{\rho\in A}(\Bvalue{\tau\le\rho}\Bcap\Bvalue{\rho<\tau'})$}

\noindent is non-zero.   Note that if $\rho\in A$, then
$\Bvalue{\tau=\sigma}\Bsubseteq\Bvalue{\tau\le\rho}\Bcap\Bvalue{\rho<\tau'}$
does not meet $a$.

As $a\subseteq\Bvalue{\tau<\tau'}$, and $\{\check q:q\in Q\}$ separates
$\Cal T$ (633Da), there is a $q\in Q$ such that
$a_1=a\Bcap\Bvalue{\tau\le\check q}\Bcap\Bvalue{\check q<\tau'}$ is non-zero.
Now we know that

$$\eqalign{0&
=a_1\Bcap\sup_{\rho\in C_q}\Bvalue{\rho=\tau}
\Bsupseteq\sup_{\rho\in C_q}(c_{\tau q}\Bcap c_{\rho q})
=\sup_{\sigma\in S}(c_{\tau q}\Bcap c_{\sigma q})\cr&
=c_{\tau q}
=\Bvalue{\tau\le\check q}\Bsetminus
  \sup_{\sigma\in\Cal S}(\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\check q})
\cr}$$

\noindent so there is a $\sigma\in S$ such that
$a_2=a_1\Bcap\Bvalue{\tau<\sigma}\Bcap\Bvalue{\sigma\le\check q}$ is non-zero.
Again because $\{\check q:q\in Q\}$ separates $\Cal T$, there is an
$r\in Q$ such that
$a_3=a_2\Bcap\Bvalue{\tau\le\check r}\Bcap\Bvalue{\check r<\sigma}$ is non-zero.
Now

\Centerline{$a_3\Bsubseteq\Bvalue{\check r<\sigma}\Bcap\Bvalue{\sigma\le\check q}
\Bsubseteq b_{rq}
=\sup_{\rho\in A}(\Bvalue{\check r<\rho}\Bcap\Bvalue{\rho\le\check q})$}

\noindent and there is a $\rho\in A$ such that
$a_4=a_3\Bcap\Bvalue{\check r<\rho}\Bcap\Bvalue{\rho\le\check q}$ is non-zero.
But in this case

$$\eqalign{0&
\ne a_4
\Bsubseteq a\Bcap\Bvalue{\tau\le\check r}\Bcap\Bvalue{\check r<\rho}
  \Bcap\Bvalue{\rho\le\check q}\Bvalue{\check q<\tau'}\cr&
\Bsubseteq a\Bcap\Bvalue{\tau\le\rho}\Bcap\Bvalue{\rho<\tau'}\cr}$$

\noindent which is impossible, by the definition of $a$.\ \Bang

\medskip

\quad{\bf (iii)} So the countable set $A$ separates $\Cal S$, as required.
}%end of proof of 642P

%\leader{642Q}
%\leader{642R}
%\leader{642S}
%\leader{642T}
%\leader{642U}
%\leader{642V}
%\leader{642W}

\exercises{\leader{642X}{Basic exercises (a)}
%\spheader 642Xa
Let $\Cal S$ be a sublattice of $\Cal T$, and
$\sequencen{\pmb u_n}=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$
a sequence in $(L^0)^{\Cal S}$ which is
order*-convergent to $\pmb u\in(L^0)^{\Cal S}$.
(i) Show that if every $\pmb u_n$ is non-decreasing, then
$\pmb u$ is non-decreasing.   (ii)
Show that if every $\pmb u_n$ is fully adapted and of
bounded variation and $\sup_{n\in\Bbb N}\int_{\Cal S}|d\pmb u_n|$ is
defined in $L^0$, then $\pmb u$ is of bounded variation.   (iii) Give an
example in which every $\pmb u_n$ is simple, but $\pmb u$ is neither
near-simple nor an integrator.   (iv) Show that if every $\pmb u_n$ is
a martingale, and $\{u_{n\sigma}:n\in\Bbb N\}$ is uniformly integrable
for every $\sigma\in\Cal S$, then $\pmb u$ is a martingale.
\Hint{367Xo.}
%642B

\spheader 642Xb Let $\Cal S$ be a sublattice of $\Cal T$, and
$\sequencen{\pmb u_n}$ a sequence of order-bounded processes with
domain $\Cal S$ such that
$\sum_{n=0}^{\infty}\widehat{\theta}_{\Cal S}(\pmb u_n)$ is finite.
Show that $\sequencen{\pmb u_n}$ is order*-convergent to $0$.
%642B

\spheader 642Xc Suppose that $T=\Bbb N$.   Show that $\Mpv(\Cal T_f)$ can
be identified with the space of sequences $\sequencen{x_n}$ in
$L^0(\frak A)$ such that $x_0=0$ and $x_n\in L^0(\frak A_{n-1})$ for
$n\ge 1$.
%642C

\spheader 642Xd Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa, 617Xb, 618Xa, 622Xd, 626Xa and 627Xa.
Let $f:\coint{0,\infty}\to\Bbb R$ be a function
and $\pmb u$ the corresponding process on $\Cal T_f$.
(i) Show that
if $f$ is of bounded variation and is continuous on the right,
$\pmb u_<$ corresponds to the function $g$ where $g(0)=0$ and
$g(t)=\lim_{s\uparrow t}f(s)$ for $t>0$.
(ii) Show that
order*-convergence in $L^0(\frak A)$ corresponds to ordinary sequential
convergence in $\Bbb R$.
(iii) Show that $\pmb u$ is previsible iff $f(0)=0$ and $f$ is Borel
measurable.
%642C

\spheader 642Xe Let $\Cal S$ be a sublattice of $\Cal T$
and $z$ a member of
$L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Show that if
$\pmb x\in\Mpv(\Cal S)$ then $z\pmb x$ (definition:  612De) belongs to
$\Mpv(\Cal S)$.
%642C

\spheader 642Xf Show that if $\langle X_t\rangle_{t\ge 0}$ is a previsibly
measurable process then $X_0$ is constant.
%642H

\spheader 642Xg
Suppose that $T=\coint{0,\infty}$ and that
$\pmb v=\family{\sigma}{\Cal T_f}{v_{\sigma}}$
is the standard Poisson process (612U).   (i) Let
$\sequencen{\tau_n}$ be the sequence of jump times of $\pmb v$ (612Ue-612Uf).
Show that $\Bvalue{v_{<\sigma}=0}=\Bvalue{\sigma\le\tau_1}$
and $\Bvalue{v_{<\sigma}=n\chi 1}
=\Bvalue{\tau_n<\sigma}\Bcap\Bvalue{\sigma\le\tau_{n+1}}$ for every
$n\ge 1$.
(ii) Show that $v_{<\check t}=v_{\check t}$ for every $t\ge 0$.
(iii) Show that $t\mapsto v_{\check t}:T\to L^0(\frak A)$ is continuous.
%642E

\leader{642Y}{Further exercises (a)}
%\spheader 642Ya
Suppose that $T=\omega_1$, the first uncountable ordinal.
(i) Show that for any previsible process
$\family{\sigma}{\Cal T}{x_{\sigma}}$ there is a $\xi<\omega_1$ such that
$x_{\check\eta}=x_{\max\Cal T}$ whenever $\xi\le\eta<\omega_1$.
(ii) Define $\pmb v=\family{\sigma}{\Cal T}{v_{\sigma}}$
by saying that $v_{\sigma}=\chi\Bvalue{\sigma=\max\Cal T}$
for $\sigma\in\Cal T$.   Show that $\pmb v$ is a simple process and
a submartingale, but is not previsible.
%any jump process can do that, eg Poisson process
(iii) Show that if $\frak A=\overline{\bigcup_{\xi<\omega_1}\frak A_{\xi}}$ then
$\pmb v$ is equal to its previsible variation.
%642C 626B 626L mt64bits

\spheader 642Yb Let $\pmb w=\family{\sigma}{\Cal T_f}{w_{\sigma}}$
be the local martingale of 634N, a difference
of independent Poisson processes.   Show that $\pmb w$ is not a previsible
process.   \Hint{show that if $\tau_1$ is the stopping time at which
$\pmb w$ makes its first jump, then
$\Expn(w'_{\tau_1}\times w_{\tau_1})=0$ for every
$\|\,\|_{\infty}$-bounded previsible process
$\pmb w'=\family{\sigma}{\Cal T_f}{w'_{\sigma}}$.}
% 642+
% Distribution of $w_{\tau_1}$ is $\pm\bover12$ with equal probability and
% independent of $\frak A_{<\tau_1}$

\spheader 642Yc Give an example of a structure \Asissix\ and a moderately
oscillatory process $\pmb u$ with domain $\Cal T$ such that $\pmb u_<$ is not
previsible.
%mt64bits 642P
}%end of exercises

\endnotes{
\Notesheader{642} The idea of `jump' arises most naturally from
the representation of a
locally near-simple process by a classical stochastic process with
\cadlag\ sample
paths, as in 642E.   There is a paradoxical element in 641N and
642G which is
is of the greatest importance.   When we have a classical stochastic
process with \cadlag\ sample paths, we can detect jumps directly from the
basic process $\langle U_t\rangle_{t\ge 0}$.   But if we turn to
the corresponding process $\family{t}{\coint{0,\infty}}{U_t^{\ssbullet}}
=\family{t}{\coint{0,\infty}}{u_{\check t}}$, we may well find that
$u_{\check t}=u_{\check t^-}=\lim_{s\uparrow t}u_{\check s}$
for every $t>0$, as in 642Xg.   To discuss jumps in the language I have
chosen for the development of the general theory, we have to look for
jumps at arbitrary stopping times, which is the idea of part (a-ii) of the
proof of 642G.

You will see that I have not mentioned martingales in this section
or the last, and the
`previsible variations' of \S626 appear only in the example 642Ya.
There is a good deal more to be said about both, of course, but
most of it will
have to wait until we have some further tools, starting with 643B.
For the moment, however, I call your attention to 642Yb;  martingales which
are not jump-free may well not be previsible.
%actually:  is every previsible martingale jump-free?

The idea of 642L is that `previsible processes' in the sense of
642C correspond to `previsibly measurable processes' in the sense of 642H.
I will try to use these phrases consistently to distinguish between the
two concepts.   Note that the correspondence is not exact because I find
myself requiring previsible processes to start with $0$ (see the notes to
\S641) but previsibly
measurable processes can start with any constant function.   We shall find
that for all the important things we call on previsible processes to do the
starting value is irrelevant.

}%end of notes

\discrpage
\frfilename{mt643.tex}
\versiondate{24.7.17/9.6.21}
\copyrightdate{2013}

\def\acc{\mathop{\text{acc}}}
\def\app{\mathop{\text{app}}}

\def\chaptername{The S-integral}
\def\sectionname{The fundamental theorem of martingales}

\def\Jump{\mathop{\text{Jump}}\nolimits}

\newsection{643}

I come at last to one of the most remarkable properties of martingales:
under moderately restrictive conditions, a martingale can be expressed as
the sum of a local martingale with small jumps and a
process of locally bounded variation (643M).   In fact I express the
result in terms of the `residual oscillations' introduced in \S618,
but these are intimately connected with `jumps' in sample paths, if we use
the standard representation of locally near-simple
processes\cmmnt{ (631D, 642E-642G)}.
The proof depends on the notion of `accessibility' of a stopping time
(643C).

\cmmnt{\leader{643A}{Notation} We have the usual foundations;
\Asissix\ is a stochastic integration structure, and
$L^0(\frak A)$ is given its
topology of convergence in measure, defined by the functional $\theta$
where $\theta(u)=\Expn(|u|\wedge\chi 1)$ for $u\in L^0(\frak A)$.
For $t\in T$, $\check t$ is the constant stopping time at $t$.
For $\tau\in\Cal T$,
$P_{\tau}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ is the conditional expectation
associated with $\frak A_{\tau}$.   In addition, we shall need the
closed subalgebras $\frak A_{\Cal S<\tau}$ defined in 641B.   For a locally
moderately oscillatory process $\pmb u$,
$\pmb u_<$ will be its previsible version (641L).   For a sublattice $\Cal S$ of
$\Cal T$, $\Cal I(\Cal S)$ is the set of finite sublattices of $\Cal S$,
and if $\tau\in\Cal S$, then
$\Cal S\wedge\tau=\{\sigma\wedge\tau:\sigma\in\Cal S\}$ and
$\Cal S\vee\tau=\{\sigma\vee\tau:\sigma\in\Cal S\}$.
}

\leader{643B}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a locally moderately oscillatory process and
$C$ a non-empty upwards-directed
subset of $\Cal S$ with supremum $\tau\in\Cal S$.   Write
$\family{\sigma}{\Cal S}{u_{<\sigma}}$ for the previsible version of $\pmb u$
and $a$ for $\inf_{\sigma\in C}\Bvalue{\sigma<\tau}$.

(a) Set $w=\lim_{\sigma\uparrow C}u_{\sigma}$.   Then

\Centerline{$\Bvalue{w=u_{<\tau}}\Bsupseteq a$,
\quad$\Bvalue{w=u_{\tau}}\Bsupseteq 1\Bsetminus a$.}

(b) Now suppose that $\pmb u$ is a
martingale.   Write $P_{\Cal S<\tau}:L^1_{\bar\mu}\to L^1_{\bar\mu}$
for the conditional expectation associated with
$\frak A_{\Cal S<\tau}$.   Then
$a\Bsubseteq\Bvalue{u_{<\tau}=P_{\Cal S<\tau}u_{\tau}}$.

\proof{ Of course we know from 641L that $\pmb u$ has a previsible version.

\medskip

{\bf (a)(i)} 615Ga, applied in $\Cal S\wedge\tau$,
tells us that $w$ is well-defined.
Let $\epsilon>0$.   Then there is a finite sublattice $J$ of
$\Cal S$, containing $\min\Cal S$,
such that $\theta(u_{<\tau}-u_{I<\tau})\le\epsilon$
whenever $J\subseteq I\in\Cal I(\Cal S)$.   Let
$\sigma_0\in C$ be such that $\bar\mu d\le\epsilon$, where
$d=\sup_{\sigma'\in J}
  \Bvalue{\sigma'<\tau}\Bsetminus\Bvalue{\sigma'<\sigma_0}$ (611Eb).
Take $\sigma\in C$ such that $\sigma_0\le\sigma$ and
$\theta(w-u_{\sigma})\le\epsilon$, and let
$I\in\Cal I(\Cal S)$ be the sublattice generated by
$J\cup\{\sigma\}$.   Using 611E(c-i) and 611E(c-ii), it is easy to check
that

\Centerline{$\{\sigma':\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau}
\Bsubseteq d\}$}

\noindent is a sublattice of $\Cal T$;  as it includes $J\cup\{\sigma\}$,
it includes $I$.   But this means that

\Centerline{$a\Bsetminus d
\Bsubseteq\Bvalue{\sigma<\tau}\Bsetminus d
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma}}$,}

\noindent and

$$\eqalign{\theta((w-u_{<\tau})\times\chi a)
&\le\theta(w-u_{\sigma})+\theta((u_{\sigma}-u_{I<\tau})\times\chi a)
   +\theta(u_{I<\tau}-u_{<\tau})\cr
&\le\epsilon+\bar\mu d+\epsilon
\le 3\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta((w-u_{<\tau})\times\chi a)=0$ and
$a\Bsubseteq\Bvalue{w=u_{<\tau}}$.

\medskip

\quad{\bf (ii)} On the other side, if $\sigma\in C$ then
$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{u_{\sigma'}=u_{\tau}}$ whenever
$\sigma'\in C$ and $\sigma\le\sigma'$, so
$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{w=u_{\tau}}$.   As $\sigma$ is
arbitrary, $1\Bsetminus a\Bsubseteq\Bvalue{w=u_{\tau}}$.

\medskip

{\bf (b)} Write $\frak B$ for the closed subalgebra
$\bigvee_{\sigma\in C}\frak A_{\sigma}$ generated by
$\bigcup_{\sigma\in C}\frak A_{\sigma}$.
By 621C(g-ii), the conditional expectation of $u_{\tau}$ on
$\frak B$ is the $\|\,\|_1$-limit
$\1lim_{\sigma\uparrow C}P_{\sigma}u_{\tau}=\1lim_{\sigma\uparrow C}u_{\sigma}$,
which is equal to $w$ (613B(d-i)).   If $c\in\frak A_{\Cal S<\tau}$ and
$c\Bsubseteq a$, then $c\in\frak B$, by 641Cb, so

\Centerline{$\Expn(\chi c\times P_{\Cal S<\tau}u_{\tau})
=\Expn(\chi c\times u_{\tau})
=\Expn(\chi c\times w)
=\Expn(\chi c\times u_{<\tau})$}

\noindent because $a\Bsubseteq\Bvalue{w=u_{<\tau}}$.
As both $P_{\Cal S<\tau}u_{\tau}$ and $u_{<\tau}$ belong to
$L^0(\frak A_{\Cal S<\tau})$ (641G(a-i)), and $a\in\frak A_{\Cal S<\tau}$
(by the definition in 641Ba), this shows that
$\chi a\times P_{\Cal S<\tau}u_{\tau}=\chi a\times u_{<\tau}$, that is,
$a\Bsubseteq\Bvalue{P_{\Cal S<\tau}u_{\tau}=u_{<\tau}}$.
}%end of proof of 643B

\leader{643C}{Approachability and accessibility} Suppose that $\tau\in\Cal T$.

\medskip

{\bf (a)} The {\bf region of accessibility} of $\tau$ is

\Centerline{$\acc(\tau)=\sup_{\emptyset\ne C\subseteq\Cal T\wedge\tau}
  (\Bvalue{\sup C=\tau}\Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau})$.}

\medskip

{\bf (b)} For $\sigma\in\Cal T$, write $\sigma\ll\tau$ for
$\sup_{\rho\in\Cal T}(\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})$.
The {\bf region of approachability} of $\tau$ is

\Centerline{$\app(\tau)=\inf_{\sigma\le\tau}
   (\Bvalue{\sigma=\tau}\Bcup\Bvalue{\sigma\ll\tau})$}

\noindent so that

\Centerline{$1\Bsetminus\app(\tau)=\sup_{\sigma\le\tau}
   (\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma\ll\tau})$.}

\medskip

{\bf (c)} $\acc(\tau)\Bsubseteq\app(\tau)$.   \prooflet{\Prf\ Suppose that
$C\subseteq\Cal T\wedge\tau$ is non-empty and that $\sigma\le\tau$.   Then

$$\eqalignno{\Bvalue{\sup C=\tau}
   \Bcap\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma\ll\tau}&
\Bsubseteq\Bvalue{\sigma<\sup C}\Bsetminus\Bvalue{\sigma\ll\tau}
=\sup_{\rho\in C}\Bvalue{\sigma<\rho}\Bsetminus\Bvalue{\sigma\ll\tau}
\Displaycause{611Eb}
\Bsubseteq\sup_{\rho\in C}\Bvalue{\sigma<\rho}
   \Bsetminus(\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})
\Bsubseteq\sup_{\rho\in C}\Bvalue{\rho=\tau}\cr}$$

\noindent and

\Centerline{$(\Bvalue{\sup C=\tau}\Bsetminus\sup_{\rho\in C}\Bvalue{\rho=\tau})
\Bcap(\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma\ll\tau})=0$.}

\noindent As $C$ and
$\sigma$ are arbitrary, $\acc(\tau)\Bcap(1\Bsetminus\app(\tau))=0$, that is,
$\acc(\tau)\Bsubseteq\app(\tau)$.\ \Qed}

\spheader 643Cd{\bf (i)} If $\tau\in\Cal T$ then

\Centerline{$\Bvalue{\sup C=\tau}
   \Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau}
=(1\Bsetminus\Bvalue{\sup C<\tau})\Bcap\inf_{\sigma\in C}\Bvalue{\sigma<\tau}
\in\frak A_{<\tau}$}

\noindent whenever $\emptyset\ne C\subseteq\Cal T\wedge\tau$,
so $\acc(\tau)\in\frak A_{<\tau}$.

\medskip

\quad{\bf (ii)} If $\sigma$, $\rho$, $\tau\in\Cal T$ then

\Centerline{$\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau}
=\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau}\Bcap\Bvalue{\sigma<\tau}
=\Bvalue{\sigma<\rho\wedge\tau}\Bcap\Bvalue{\rho<\tau}
\in\frak A_{<\tau}$,}

\noindent so $\Bvalue{\sigma\ll\tau}\in\frak A_{<\tau}$ and
$\app(\tau)\in\frak A_{<\tau}$.

\medskip

\quad{\bf (iii)} Note that $\acc(\min\Cal T)=0$\cmmnt{, because the only
non-empty subset of $\Cal T\wedge\min\Cal T$ is
$\{\min\Cal T\}$}, while $\app(\min\Cal T)=1$.

\leader{643D}{Proposition} For $t\in T$, let
$\check t$ be the constant stopping time at $t$.
Let $T_{\text{r-i}}$ be the set of those
$t\in T$ which are isolated on the right, and for $t\in T_{\text{r-i}}$ define
$\check t^+\in\Cal T$ by saying that

$$\eqalign{\check t^+
&=\max\Cal T\text{ if }t=\max T\text{ is the greatest element of }T,\cr
&=\check s\text{ if }t\text{ is not the greatest element of }T\cr
&\mskip150mu
  \text{ and }s\text{ is the least element of }T\text{ greater than }t.\cr}$$

\noindent Then
$\app(\tau)=1\Bsetminus\sup_{t\in T_{\text{r-i}}}\Bvalue{\tau=\check t^+}$ for
every $\tau\in\Cal T$.

\proof{{\bf (a)(i)} Note that if $\sigma$, $\tau'$, $\tau''\in\Cal T$ then

$$\eqalign{\Bvalue{\tau'\le\tau''}\Bcap\Bvalue{\sigma\ll\tau'}&
=\sup_{\rho\in\Cal T}
   \Bvalue{\tau'\le\tau''}\Bcap\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau'}\cr&
\Bsubseteq\sup_{\rho\in\Cal T}\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau''}
=\Bvalue{\sigma\ll\tau''},\cr}$$

\noindent just as in 611E(c-iv-$\alpha$).

\medskip

\quad{\bf (ii)} If $t\in T_{\text{r-i}}$ and $\rho\in\Cal T$,

$$\eqalignno{\Bvalue{\check t<\rho}\Bcap\Bvalue{\rho<\check t^+}&
=\sup_{s,s'\in T}(\Bvalue{\rho>s}\Bsetminus\Bvalue{\check t>s})
    \Bcap(\Bvalue{\check t^+>s'}\Bsetminus\Bvalue{\rho>s'})\cr&
\Bsubseteq\sup_{s\ge t,s'\le t}\Bvalue{\rho>s}\Bsetminus\Bvalue{\rho>s'}
\Displaycause{because if $s<t$ then $\Bvalue{\check t>s}=1$ and if
$s'>t$ then $\Bvalue{\check t^+>s'}=0$}
\Bsubseteq\sup_{s'\le s}\Bvalue{\rho>s}\Bsetminus\Bvalue{\rho>s'}
=0.\cr}$$

\noindent So $\Bvalue{\check t\ll\check t^+}=0$.   And of course

\Centerline{$\Bvalue{\check t<\check t^+}
\Bsupseteq\Bvalue{\check t^+>t}\Bsetminus\Bvalue{\check t^+>t}=1$.}

\noindent Now

$$\eqalignno{\Bvalue{\tau=\check t^+}&
=\Bvalue{\tau\le\check t^+}\Bcap\Bvalue{\check t<\check t^+}
   \Bsetminus(\Bvalue{\check t^+\le\tau}\Bcap\Bvalue{\check t\ll\check t^+})
\Displaycause{611E(c-iv-$\alpha$) and (i) just above}
\Bsubseteq\Bvalue{\check t<\tau}\Bsetminus\Bvalue{\check t\ll\tau}
\Bsubseteq 1\Bsetminus\app(\tau).\cr}$$

\noindent As $t$ is arbitrary,

\Centerline{$\app(\tau)
\Bsubseteq 1\Bsetminus\sup_{t\in T_{\text{r-i}}}\Bvalue{\tau=\check t^+}$.}

\medskip

{\bf (b)} If $\sigma\le\tau$ and
$b\Bsubseteq\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma\ll\tau}$ is non-zero,
there is a $t\in T$ such that

\Centerline{$b'=b\Bcap\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}$}

\noindent is non-zero.   \Quer\ If $t\notin T_{\text{r-i}}$,
$\Bvalue{\tau>t}=\sup_{s>t}\Bvalue{\tau>s}$ (611A(b-i)), and there is
an $s>t$ such that $b'\Bcap\Bvalue{\tau>s}\ne 0$.   But now

$$\eqalign{b'\Bcap\Bvalue{\tau>s}&
=\Bsubseteq b\Bcap\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>t}\cr&
\Bsubseteq b\Bcap\Bvalue{\check s<\tau}\Bcap\Bvalue{\check t<\check s}
     \Bcap\Bvalue{\sigma\le\check t}
\Bsubseteq b\Bcap\Bvalue{\sigma\ll\tau},\cr}$$

\noindent contrary to the hypothesis on $b$.\ \Bang

Thus $t\in T_{\text{r-i}}$.   Now we know from (a-ii) that
$\Bvalue{\check t<\tau}\Bcap\Bvalue{\tau<\check t^+}=0$, so
$\Bvalue{\check t^+\le\tau}=\Bvalue{\check t<\tau}\Bsupseteq b'$.   At the same
time, $b'$ is disjoint from

\Centerline{$\Bvalue{\sigma\ll\tau}
\Bsupseteq\Bvalue{\sigma<\check t^+}\Bcap\Bvalue{\check t^+<\tau}
\Bsupseteq\Bvalue{\sigma\le\check t}\Bcap\Bvalue{\check t^+<\tau}
\Bsupseteq b'\Bcap\Bvalue{\check t^+<\tau}$}

\noindent so $b'\Bsubseteq\Bvalue{\tau\le\check t^+}$ and
$b'\Bsubseteq b\Bcap\Bvalue{\tau=\check t^+}$.

As $b$ is arbitrary,
$\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma\ll\tau}
\Bsubseteq\sup_{t\in T_{\text{r-i}}}\Bvalue{\tau=\check t^+}$.   As $\sigma$ is
arbitrary,

\Centerline{$1\Bsetminus\app(\tau)
\Bsubseteq\sup_{t\in T_{\text{r-i}}}\Bvalue{\tau=\check t^+}$,
\quad$\app(\tau)\Bsupseteq
1\Bsetminus\sup_{t\in T_{\text{r-i}}}\Bvalue{\tau=\check t^+}$.}

\noindent Putting this together with (a), we have equality, as claimed.
}%end of proof of 643D

\leader{643E}{Proposition} Suppose that $\tau\in\Cal T$.
For non-empty $C\subseteq\Cal T\wedge\tau$, set

\Centerline{$a_C=\Bvalue{\sup C=\tau}
   \Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau}$.}

(a) $a_C=\inf_{\sigma\in C}\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sup C<\tau}$
belongs to $\frak A_{<\tau}$ whenever
$\emptyset\ne C\subseteq\Cal T\wedge\tau$;  so
$\acc(\tau)\in\frak A_{<\tau}$.

(b) For every non-empty $C\subseteq\Cal T\wedge\tau$ there is a non-empty
upwards-directed $D\subseteq\Cal T\wedge\tau$ such that $a_D=a_C$.
\cmmnt{Consequently}

$$\eqalign{\acc(\tau)&
=\sup\{\Bvalue{\sup C=\tau}\Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau}:\cr&
\mskip180mu
C\subseteq\Cal T\wedge\tau\text{ is non-empty and upwards-directed}\}.\cr}$$

(c) If $\upsilon\in\Cal T$,
$\acc(\tau)\Bcap\Bvalue{\upsilon=\tau}
=\acc(\upsilon)\Bcap\Bvalue{\upsilon=\tau}$.

\proof{{\bf (a)} Immediate from the definitions and the fact that
$b\Bsetminus c=(1\Bsetminus c)\Bsetminus(1\Bsetminus b)$ for all
$b$, $c\in\frak A$.

\medskip

{\bf (b)} Set $D=\{\sigma':\sigma'\le\sup C$,
$\Bvalue{\sigma'<\tau}\Bsupseteq\inf_{\sigma\in C}\Bvalue{\sigma<\tau}$.
By 611E(c-ii), $D$ is closed under $\vee$.   Because $C\subseteq D$,
$\sup D=\sup C$ and
$\inf_{\sigma\in D}\Bvalue{\sigma<\tau}=\inf_{\sigma\in C}\Bvalue{\sigma<\tau}$,
so $a_D=a_C$.

\medskip

{\bf (c)} Write $c$ for $\Bvalue{\upsilon=\tau}$.
If $\emptyset\ne C\subseteq\Cal T\wedge\tau$ set
$D=\{\sigma\wedge\upsilon:\sigma\in C\}$, so that
$\emptyset\ne D\subseteq\Cal T\wedge\upsilon$, and

$$\eqalignno{c\Bcap a_C&
=c\Bcap\Bvalue{\sup C=\tau}
   \Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau}
=c\Bcap\Bvalue{\tau\wedge\sup C=\tau}
   \Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma\wedge\tau=\tau}\cr&
\Bsubseteq\Bvalue{\upsilon\wedge\sup C=\upsilon}
   \Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma\wedge\upsilon=\upsilon}
=\Bvalue{\sup D=\upsilon}
   \Bsetminus\sup_{\sigma\in D}\Bvalue{\sigma=\upsilon}
\Displaycause{611Cd}
\Bsubseteq b.\cr}$$

\noindent Taking the supremum over $C$,
$c\Bcap\acc(\tau)\Bsubseteq\acc(\upsilon)$.   Similarly,
$c\Bcap\acc(\upsilon)\Bsubseteq\acc(\tau)$ and
$c\Bcap\acc(\tau)=c\Bcap\acc(\upsilon)$, as claimed.
}%end of proof of 643E

\leader{643F}{Lemma} Suppose that
$\family{t}{T}{\frak A_t}$ is right-continuous.   Take
$\tau\in\Cal T$ and $\epsilon>0$.
For $I\in\Cal I(\Cal T\wedge\tau)$ and $\sigma\le\tau$, set

\Centerline{$d_{\sigma I}
=(\app(\tau)\Bsetminus\acc(\tau))\Bcap\Bvalue{\sigma<\tau}
    \Bsetminus\sup_{\rho\in I}
         (\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})$,
\quad$w_{\sigma I}=P_{\sigma}\chi d_{\sigma I}$.}

\noindent For $I\in\Cal I(\Cal T\wedge\tau)$ set
$w_I=\sup_{\sigma\le\tau}w_{\sigma I}$.

(a)(i) $d_{\sigma_0I}\Bcap\Bvalue{\sigma_0=\sigma_1}
=d_{\sigma_1I}\Bcap\Bvalue{\sigma_0=\sigma_1}$
whenever $I\in\Cal I(\Cal T\wedge\tau)$ and $\sigma_0$, $\sigma_1\le\tau$.

\quad(ii) For any $\sigma\le\tau$,
$\lim_{I\uparrow\Cal I(\Cal T\wedge\tau)}\bar\mu d_{\sigma I}=0$.

(b) For any $I\in\Cal I(\Cal T\wedge\tau)$,
$\langle w_{\sigma I}\rangle_{\sigma\le\tau}$ is fully adapted.

(c) If $I\subseteq J$ in $\Cal I(\Cal T\wedge\tau)$, then
$d_{\sigma I}\Bsupseteq d_{\sigma J}$ and
$w_{\sigma I}\ge w_{\sigma J}$ for every $\sigma\le\tau$, and
$w_I\ge w_J$.

(d) For $I\in\Cal I(\Cal T\wedge\tau)$, set

\Centerline{$A_I=\{\sigma:\sigma\le\tau$,
$\Bvalue{w_{\sigma I}>\epsilon}
\Bsupseteq\Bvalue{\sigma<\tau}\}$.}

\quad(i) $\Bvalue{w_I>\epsilon}
=\sup_{\sigma\in A_I}\Bvalue{\sigma<\tau}$.

\quad(ii) $A_I$ is closed under $\wedge$.

\quad(iii) Set $\bar\sigma_I=\inf A_I$.   Then

\qquad($\alpha$) $\bar\sigma_I\le\tau$;

\qquad($\beta$) $d_{\bar\sigma_II}$ is the limit
$\lim_{\sigma\downarrow A_I}d_{\sigma I}$ for the measure-algebra topology
of $\frak A$;
%needs right-continuity

\qquad($\gamma$) $w_{\bar\sigma_II}$ is the limit
$\1lim_{\sigma\downarrow A_I}w_{\sigma I}$ for the norm topology of
$L^1_{\bar\mu}$;
%needs right-continuity

\qquad($\delta$)
$\Bvalue{w_I>\epsilon}\Bsubseteq\Bvalue{w_{\bar\sigma_II}\ge\epsilon}$.
%needs right-continuity

\quad(iv) If $I\subseteq J$ in $\Cal I(\Cal T\wedge\tau)$
then $A_I\supseteq A_J$ and $\bar\sigma_I\le\bar\sigma_J$.

(e) There is an $I\in\Cal I(\Cal T\wedge\tau)$ such that
$\Expn(w_I)\le 3\epsilon$.
%Protter p 108 Lemma 2   %proof needs right-continuity

\proof{{\bf (a)(i)} Setting $b=\Bvalue{\sigma_0=\sigma_1}$,
$b\Bcap\Bvalue{\sigma_0<\sigma}=b\Bcap\Bvalue{\sigma_1<\sigma}$ for every
$\sigma\in\Cal T$ (611E(c-v), or otherwise), so

$$\eqalign{b\Bcap d_{\sigma_0I}&
=(\app(\tau)\Bsetminus\acc(\tau))\Bcap b\Bcap\Bvalue{\sigma_0<\tau}
   \Bsetminus\sup_{\rho\in I}
      (b\Bcap\Bvalue{\sigma_0<\rho}\Bcap\Bvalue{\rho<\tau})\cr&
=(\app(\tau)\Bsetminus\acc(\tau))\Bcap b\Bcap\Bvalue{\sigma_1<\tau}
   \Bsetminus\sup_{\rho\in I}
      (b\Bcap\Bvalue{\sigma_1<\rho}\Bcap\Bvalue{\rho<\tau})\cr&
=b\Bcap d_{\sigma_1I}.\cr}$$

\medskip

\quad{\bf (ii)} Take any $\eta>0$.   Since
$\app(\tau)\Bcap\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\sigma\ll\tau}$,
there is a finite set $J\subseteq\Cal T$ such that

$$\eqalign{\eta&
\ge\bar\mu(\app(\tau)\Bcap\Bvalue{\sigma<\tau}
   \Bsetminus\sup_{\rho\in J}\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau})\cr&
=\bar\mu(\app(\tau)\Bcap\Bvalue{\sigma<\tau}
   \Bsetminus\sup_{\rho\in J}\Bvalue{\sigma<\rho\wedge\tau}
      \Bcap\Bvalue{\rho\wedge\tau<\tau}).\cr}$$

\noindent Now if $I\in\Cal I(\Cal T\wedge\tau)$ includes
$\{\rho\wedge\tau:\rho\in J\}$,

\Centerline{$d_{\sigma I}\Bsubseteq\app(\tau)\Bcap\Bvalue{\sigma<\tau}
   \Bsetminus\sup_{\rho\in J}\Bvalue{\sigma<\rho\wedge\tau}
      \Bcap\Bvalue{\rho\wedge\tau<\tau}$}

\noindent has measure at most $\eta$.
As $\eta$ is arbitrary,
$\lim_{I\uparrow\Cal I(\Cal T\wedge\tau)}\bar\mu d_{\sigma I}=0$.

\medskip

{\bf (b)} $w_{\sigma I}=P_{\sigma}\chi d_{\sigma I}$ certainly belongs to
$L^0(\frak A_{\sigma})$ for every $\sigma\le\tau$.
If $\sigma_0$, $\sigma_1\le\tau$ and
$b=\Bvalue{\sigma_0=\sigma_1}$, then we saw in (a-i) that
$b\Bcap d_{\sigma_0I}=b\Bcap d_{\sigma_1I}$, so

$$\eqalignno{\chi b\times w_{\sigma_0I}
&=\chi b\times P_{\sigma_0}\chi d_{\sigma_0I}
=\chi b\times P_{\sigma_1}\chi d_{\sigma_0I}\cr
\displaycause{622Bb}
&=P_{\sigma_1}\chi(b\Bcap d_{\sigma_0I})\cr
\displaycause{because $b\in\frak A_{\sigma_1}$, by 611H(c-i)}
&=P_{\sigma_1}\chi(b\Bcap d_{\sigma_1I})
=\chi b\times w_{\sigma_1I},\cr}$$

\noindent and $b\Bsubseteq\Bvalue{w_{\sigma_0I}=w_{\sigma_1I}}$.
As $\sigma_0$ and $\sigma_1$ are arbitrary,
$\langle w_{\sigma I}\rangle_{\sigma\le\tau}$ is fully adapted.

\medskip

{\bf (c)} Immediate from the definitions.   Perhaps I should note
here that $w_{\sigma I}\le\chi 1$ for every $\sigma$ and $I$, so that
$w_I$ is always defined and less than or equal to $\chi 1$.

\medskip

{\bf (d)(i)} If $\sigma\le\tau$, then there is a $\rho\in A_I$ such that
$\Bvalue{\rho<\tau}
=\Bvalue{w_{\sigma I}>\epsilon}\Bcap\Bvalue{\sigma<\tau}$.
\Prf\ $b=\Bvalue{w_{\sigma I}>\epsilon}\Bcap\Bvalue{\sigma<\tau}$
belongs to $\frak A_{\sigma}$, so there is a $\rho\le\tau$
such that $b\Bsubseteq\Bvalue{\rho=\sigma}$ and
$1\Bsetminus b\Bsubseteq\Bvalue{\rho=\tau}$ (611I).    As
$b\Bsubseteq\Bvalue{\sigma<\tau}$,
$b=\Bvalue{\rho=\sigma}=\Bvalue{\rho<\tau}$.   Now

\Centerline{$\Bvalue{w_{\rho I}>\epsilon}
\Bsupseteq\Bvalue{\rho=\sigma}\Bcap\Bvalue{w_{\sigma I}>\epsilon}
=b=\Bvalue{\rho<\tau}$}

\noindent and $\rho\in A_I$.\ \Qed

Accordingly

$$\sup_{\sigma\le\tau}(\Bvalue{w_{\sigma I}>\epsilon}
   \Bcap\Bvalue{\sigma<\tau})
\Bsubseteq\sup_{\sigma\in A_I}\Bvalue{\sigma<\tau}
\Bsubseteq\sup_{\sigma\in A_I}\Bvalue{w_{\sigma I}>\epsilon}
\Bsubseteq\Bvalue{w_I>\epsilon}.$$

\noindent Now, by 364L(a-ii),

$$\eqalignno{\Bvalue{w_I>\epsilon}
&=\sup_{\sigma\le\tau}
  \Bvalue{w_{\sigma I}>\epsilon}
=\sup_{\sigma\le\tau}
  (\Bvalue{w_{\sigma I}>\epsilon}\Bcap\Bvalue{\sigma<\tau})\cr
\displaycause{because $d_{\tau I}\Bsubseteq\Bvalue{\tau<\tau}=0$, so
$w_{\tau I}=0$ and
$\Bvalue{w_{\sigma I}>\epsilon}\Bcap\Bvalue{\sigma=\tau}
\Bsubseteq\Bvalue{w_{\tau I}>\epsilon}=0$ for every $\sigma\le\tau$}
&\Bsubseteq\sup_{\sigma\in A_I}\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{w_I>\epsilon}\cr}$$

\noindent and we have equality.

\medskip

\quad{\bf (ii)} Suppose that $\sigma_0$,
$\sigma_1\in A_I$, and set $\sigma=\sigma_0\wedge\sigma_1$,
$b=\Bvalue{\sigma=\sigma_0}\Bsubseteq\Bvalue{w_{\sigma I}=w_{\sigma_0I}}$.
Then

\Centerline{$\Bvalue{w_{\sigma I}>\epsilon}
\Bsupseteq b\Bcap\Bvalue{w_{\sigma_0I}>\epsilon}
\Bsupseteq b\Bcap\Bvalue{\sigma_0<\tau}
=b\Bcap\Bvalue{\sigma<\tau}$.}

\noindent Similarly,

\Centerline{$\Bvalue{w_{\sigma I}>\epsilon}
\Bsupseteq\Bvalue{\sigma=\sigma_1}\Bcap\Bvalue{\sigma<\tau}$;}

\noindent as $\Bvalue{\sigma=\sigma_0}\Bcup\Bvalue{\sigma=\sigma_1}=1$
(611E(a-ii-$\gamma$)),
$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{w_{\sigma I}>\epsilon}$ and
$\sigma\in A_I$.

\medskip

\quad{\bf (iii)}\grheada\ Of course $\tau\in A_I$.
So $\bar\sigma_I\le\tau$.

\medskip

\qquad\grheadb\
Because $\family{t}{T}{\frak A_t}$ is right-continuous, we have
$\Bvalue{\bar\sigma_I<\rho}=\sup_{\sigma\in A_I}\Bvalue{\sigma<\rho}$
for any $\rho\in\Cal T$ (632C(a-ii));  because
$A_I$ is downwards-directed,
$\family{\sigma}{A_I}{\Bvalue{\sigma<\rho}}$
is upwards-directed and
$\Bvalue{\bar\sigma_I<\rho}
=\lim_{\sigma\downarrow A_I}\Bvalue{\sigma<\rho}$ for the measure-algebra
topology
(323D(a-ii)).   Because $\Bcap$ and $\Bsetminus$ are continuous for the
measure-algebra topology (323Ba),

$$\eqalign{d_{\bar\sigma_II}
&=(\app(\tau)\Bsetminus\acc(\tau))\Bcap\Bvalue{\bar\sigma_I<\tau}
  \Bsetminus\sup_{\rho\in I}
     (\Bvalue{\bar\sigma_I<\rho}\Bcap\Bvalue{\rho<\tau})\cr
&=\lim_{\sigma\downarrow A_I}(\app(\tau)\Bsetminus\acc(\tau))
   \Bcap\Bvalue{\sigma<\tau}\Bsetminus\sup_{\rho\in I}
     (\Bvalue{\sigma<\rho}\Bcap\Bvalue{\rho<\tau}))
=\lim_{\sigma\downarrow A_I}d_{\sigma I}.\cr}$$

\medskip

\qquad\grheadc\

$$\eqalignno{&\limsup_{\sigma\downarrow A_I}
\|w_{\bar\sigma_II}-w_{\sigma I}\|_1\cr
&\mskip50mu
\le\limsup_{\sigma\downarrow A_I}\|P_{\bar\sigma_I}\chi d_{\bar\sigma_II}
        -P_{\sigma}\chi d_{\bar\sigma_II}\|_1
  +\limsup_{\sigma\downarrow A_I}\|P_{\sigma}\chi d_{\bar\sigma_II}
        -P_{\sigma}\chi d_{\sigma I}\|_1\cr
&\mskip50mu
\le 0
  +\limsup_{\sigma\downarrow A_I}
    \|\chi d_{\bar\sigma_II}-\chi d_{\sigma I}\|_1\cr
\displaycause{621B(g-i)}
&\mskip50mu
=\limsup_{\sigma\downarrow A_I}
  \bar\mu(d_{\bar\sigma_II}\Bsymmdiff d_{\sigma I})
=0.\cr}$$

\noindent Thus
$w_{\bar\sigma_II}=\1lim_{\sigma\downarrow A_I}w_{\sigma I}$.

\medskip

\qquad\grheadd\ The point is that
$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{w_{\bar\sigma_II}>\gamma}$ whenever
$\sigma\in A_I$ and $\gamma<\epsilon$.   \Prf\Quer\ Otherwise, set
$\eta
=\bar\mu(\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{w_{\bar\sigma_II}>\gamma})
>0$.   By ($\gamma$) just above, there is a $\rho\in A_I$ such that
$\rho\le\sigma$ and
$\|w_{\rho I}-w_{\bar\sigma_II}\|_1<\eta(\epsilon-\gamma)$.   In this
case,

$$\eqalign{\eta
&\le\bar\mu(\Bvalue{\rho<\tau}
   \Bsetminus\Bvalue{w_{\bar\sigma_II}>\gamma})
\le\bar\mu(\Bvalue{w_{\rho I}>\epsilon}
   \Bsetminus\Bvalue{w_{\bar\sigma_II}>\gamma})\cr
&\le\Bover1{\epsilon-\gamma}\|w_{\rho I}-w_{\bar\sigma_II}\|_1
<\eta\cr}$$

\noindent which is absurd.\ \Bang\Qed

As $\gamma$ is arbitrary,
$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{w_{\bar\sigma_II}\ge\epsilon}$
for every $\sigma\in A_I$.   So

\Centerline{$\Bvalue{w_I>\epsilon}
=\sup_{\sigma\in A_I}\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{w_{\bar\sigma_II}\ge\epsilon}$,}

\noindent as claimed.

\medskip

\quad{\bf (iv)} If $\sigma\in A_J$ then

\Centerline{$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{w_{\sigma J}>\epsilon}
\Bsubseteq\Bvalue{w_{\sigma I}>\epsilon}$}

\noindent because $w_{\sigma J}\le w_{\sigma I}$ (by (c) above).
So $\sigma\in A_I$.   Thus $A_J\subseteq A_I$ and
$\bar\sigma_I=\inf A_I\le\inf A_J=\bar\sigma_J$.

\medskip

{\bf (e)} Set
$\sigma^*=\sup_{I\in\Cal I(\Cal T\wedge\tau)}\bar\sigma_I\le\tau$.
By the definition of $\acc(\tau)$, $\Bvalue{\sigma^*=\tau}\Bsetminus\acc(\tau)$
is included in $\sup_{I\in\Cal I(\Cal T\wedge\tau)}\Bvalue{\bar\sigma_I=\tau}$;
because
$\family{I}{\Cal I(\Cal T\wedge\tau)}{\bar\sigma_I}$ and
$\family{I}{\Cal I(\Cal T\wedge\tau)}{\Bvalue{\bar\sigma_I=\tau}}$
are non-decreasing, this is the limit
$\lim_{I\uparrow\Cal I(\Cal T\wedge\tau)}\Bvalue{\bar\sigma_I=\tau}$ for the
measure-algebra topology.

Putting this together with (a-ii), we see that there is an
$I\in\Cal I(\Cal T\wedge\tau)$ such that $\sigma^*\in I$,
$\bar\mu d_{\sigma^*I}\le\epsilon^2$ and
$\bar\mu((\Bvalue{\sigma^*=\tau}\Bsetminus\acc(\tau))
   \Bsetminus\Bvalue{\bar\sigma_I=\tau})\le\epsilon^2$.

Consider

$$\eqalignno{\epsilon\bar\mu\Bvalue{w_I>\epsilon}&
\le\epsilon\bar\mu\Bvalue{w_{\bar\sigma_II}\ge\epsilon}
\Displaycause{(d-iii-$\delta$) above}
\le\epsilon\bar\mu\Bvalue{w_{\bar\sigma_II}
     \times\chi\Bvalue{\bar\sigma_I<\sigma^*}\ge\epsilon}
   +\epsilon\bar\mu\Bvalue{w_{\sigma^*I}\ge\epsilon}
\Displaycause{because $\Bvalue{\bar\sigma_I\ge\sigma^*}
  =\Bvalue{\bar\sigma_I=\sigma^*}
  \Bsubseteq\Bvalue{w_{\bar\sigma_II}=w_{\sigma^*I}}$}
\le\Expn(w_{\bar\sigma_II}
     \times\chi\Bvalue{\bar\sigma_I<\sigma^*})
   +\Expn(w_{\sigma^*I})\cr&
=\bar\mu(d_{\bar\sigma_II}\Bcap\Bvalue{\bar\sigma_I<\sigma^*})
   +\bar\mu d_{\sigma^*I}
\Displaycause{because
  $w_{\bar\sigma_II}=P_{\bar\sigma_I}\chi d_{\bar\sigma_II}$ and
  $\Bvalue{\bar\sigma_I<\sigma^*}\in\frak A_{\bar\sigma_I}$}
\le\bar\mu(d_{\bar\sigma_II}\Bcap\Bvalue{\bar\sigma_I<\sigma^*}
      \Bcap\Bvalue{\sigma^*=\tau})
   +\epsilon^2
\Displaycause{because $\sigma^*\in I$, so
  $d_{\bar\sigma_II}\Bcap\Bvalue{\bar\sigma_I<\sigma^*}
  \Bcap\Bvalue{\sigma^*<\tau}=0$}
\le\bar\mu((\Bvalue{\sigma^*=\tau}\Bsetminus a)
  \Bsetminus\Bvalue{\bar\sigma_I=\tau})
   +\epsilon^2
\le 2\epsilon^2,\cr}$$

\noindent so $\bar\mu\Bvalue{w_I>\epsilon}\le 2\epsilon$ and
$\Expn(w_I)\le 3\epsilon$.
}%end of proof of 643F

\leader{643G}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a non-negative non-decreasing
$\|\,\|_{\infty}$-bounded process.
Suppose that for every $\epsilon>0$ there are an $I\in\Cal I(\Cal S)$
and a $w\in L^0(\frak A)$ such that $\|w\|_1\le\epsilon$ and
$P_{\sigma}v_{\tau}-v_{\sigma}\le w$ whenever $\sigma\le\tau$ in $\Cal S$ and
$\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau}=0$ for every
$\sigma'\in I$.   Let $\pmb v^{\#}$ be the previsible variation of
$\pmb v$\cmmnt{ (626O)}.   Then $\pmb v^{\#}$ is jump-free.

\proof{ The case $\Cal S=\emptyset$ is trivial;  suppose that $\Cal S$ is
non-empty.

\medskip

{\bf (a)} Something is worth noting straight away.   Of course we shall have
$P_{\sigma}v_{\tau}-v_{\sigma}=P_{\sigma}(v_{\tau}-v_{\sigma})\ge 0$
whenever $\sigma\le\tau$ in $\Cal S$;  that is, $\pmb v$ is a submartingale
and we can speak of its previsible version.   Since $\pmb v$ is
$\|\,\|_{\infty}$-bounded, $P_{\sigma}v_{\tau}-v_{\sigma}$ is always
square-integrable.   Now suppose that
$\sigma\le\tau$ in $\Cal S$, and that

\Centerline{$\gamma
=\sup_{\sigma=\sigma_0\le\ldots\le\sigma_n=\tau}
   \|\sum_{j=0}^{n-1}P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j}\|_2$}

\noindent is finite.   Then
$\|v^{\#}_{\tau}-v^{\#}_{\sigma}\|_2\le\gamma$.   \Prf\
By 626M(f-ii), $v^{\#}_{\tau}-v^{\#}_{\sigma}$ is the value of the previsible
version of $\pmb v\restr\Cal S\vee\sigma$ at $\tau$, so belongs to the
weak closure of $A=\{S_I(\tbf 1,Pd\pmb v):
\{\sigma,\tau\}\subseteq I\in\Cal I(\Cal S\cap[\sigma,\tau])\}$ in
$L^1_{\bar\mu}$.
If $\{\sigma,\tau\}\subseteq I\in\Cal I(\Cal S\cap[\sigma,\tau])$,
let $(\sigma_0,\ldots,\sigma_n)$ linearly generate the $I$-cells, so that
$S_I(\tbf 1,Pd\pmb v)
=\sum_{j=0}^{n-1}P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j}$ and we have
$\|S_I(\tbf 1,Pd\pmb v)\|_2\le\gamma$.   Thus $\|y\|_2\le\gamma$ for
every $y\in A$.
Now if $z\in L^{\infty}(\frak A)$, $\|z\|_2\le 1$ and $\eta>0$,
there is a $y\in A$ such that

\Centerline{$\Expn(z\times(v^{\#}_{\tau}-v^{\#}_{\sigma}))
\le\eta+\Expn(z\times y)
\le\eta+\|y\|_2
\le\eta+\gamma$;}

\noindent as $\eta$ and $z$ are arbitrary,
$\|v^{\#}_{\tau}-v^{\#}_{\sigma}\|_2\le\gamma$.\ \Qed

Squaring, we have

\Centerline{$\Expn((v^{\#}_{\tau}-v^{\#}_{\sigma})^2)
\le\sup_{\sigma=\sigma_0\le\ldots\le\sigma_n=\tau}
   \Expn((\sum_{j=0}^{n-1}P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j})^2)$.}

\noindent Moreover, whenever $\sigma=\sigma_0\le\ldots\le\sigma_n\le\tau$,
we have

$$\eqalign{
\Expn((\sum_{j=0}^{n-1}P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j})^2)
&\le\Expn(2\sum_{j=0}^{n-1}
  (\sum_{k=j}^{n-1}P_{\sigma_k}v_{\sigma_{k+1}}-v_{\sigma_k})
     \times(P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j}))\cr
&=2\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}
  \Expn((P_{\sigma_k}v_{\sigma_{k+1}}-v_{\sigma_k})
     \times(P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j}))\cr
&=2\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}
  \Expn(P_{\sigma_j}(P_{\sigma_k}v_{\sigma_{k+1}}-v_{\sigma_k})
     \times(P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j}))\cr
&=2\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}
  \Expn(P_{\sigma_j}(v_{\sigma_{k+1}}-v_{\sigma_k})
     \times(P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j}))\cr
&=2\sum_{j=0}^{n-1}
  \Expn(P_{\sigma_j}(v_{\tau}-v_{\sigma_j})
     \times(P_{\sigma_j}v_{\sigma_{j+1}}-v_{\sigma_j})).\cr}$$

\medskip

{\bf (b)} Now for the main line of this proof.
Let $M>0$ be such that $\|v_{\sigma}\|_{\infty}\le M$ for
every $\sigma\in\Cal S$.   Let $\epsilon>0$.
Let $I\in\Cal I(\Cal S)$ and $w\in L^0(\frak A)$ be such that
$I\ne\emptyset$, $\|w\|_1\le\epsilon$ and
$P_{\sigma}v_{\tau}-v_{\sigma}\le w$ whenever
$\sigma\le\tau$ in $\Cal S$ and
$\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau}=0$ for every
$\sigma'\in I$;  we can suppose that
$w\le M\chi 1$.   Let $(\tau_0,\ldots,\tau_n)$ linearly generate the
$I$-cells.   Then we shall have $0\le P_{\sigma}v_{\tau}-v_{\sigma}\le w$
whenever

\inset{{\it either} $\sigma\le\tau\le\tau_0$

{\it or} $\tau_{i-1}\le\sigma\le\tau\le\tau_i$ where $1\le i\le n$,

{\it or} $\tau_n\le\sigma\le\tau$.}

\noindent Set $w'=\sup_{\sigma\in\Cal S}P_{\sigma}w$.  Then

$$\eqalignno{\Expn(w')&
\le\|w'\|_2
\le 2\sup_{\sigma\in\Cal S}\|P_{\sigma}w\|_2
\Displaycause{623M, since $\pmb Pw$ is a martingale}
\le 2\|w\|_2
=2\sqrt{\Expn(w^2)}
\le 2\sqrt{M\Expn(w)}
\le 2\sqrt{\epsilon M}.\cr}$$

\medskip

{\bf (c)} Suppose that $\langle k_i\rangle_{i\le n+1}$,
$\langle\sigma_{ij}\rangle_{i\le n+1,j\le k_i}$ are such that
$\sigma_{ij}\in\Cal S$ whenever $i\le n+1$ and $j\le k_i$ and

\Centerline{$\sigma_{00}\le\sigma_{01}\le\ldots\le\sigma_{0k_0}=\tau_0,$}

\Centerline{$\tau_{i-1}=\sigma_{i0}\le\ldots\le\sigma_{ik_i}=\tau_i$
for $1\le i\le n$,}

\Centerline{$\tau_n=\sigma_{n+1,0}\le\ldots\le\sigma_{n+1,k_{n+1}}$.}

\noindent Then, using the last formula in (a),

$$\eqalignno{
&\Expn(\sum_{i=0}^{n+1}(\sum_{j=0}^{k_i-1}P_{\sigma_{ij}}v_{\sigma_{i,j+1}}
  -v_{\sigma_{ij}})^2)\cr
&\mskip100mu\le 2\sum_{i=0}^{n+1}\sum_{j=0}^{k_i-1}
   \Expn((P_{\sigma_{ij}}v_{\sigma_{ik_i}}-v_{\sigma_{ij}})
   \times(P_{\sigma_{ij}}v_{\sigma_{i,j+1}}-v_{\sigma_{ij}}))\cr
&\mskip100mu\le 2\sum_{i=0}^{n+1}\sum_{j=0}^{k_i-1}
   \Expn(w\times(P_{\sigma_{ij}}v_{\sigma_{i,j+1}}-v_{\sigma_{ij}}))\cr
&\mskip100mu=2\sum_{i=0}^{n+1}\sum_{j=0}^{k_i-1}
   \Expn(P_{\sigma_{ij}}w\times(v_{\sigma_{i,j+1}}-v_{\sigma_{ij}}))\cr
&\mskip100mu\le 2\sum_{i=0}^{n+1}\sum_{j=0}^{k_i-1}
   \Expn(w'\times(v_{\sigma_{i,j+1}}-v_{\sigma_{ij}}))\cr
&\mskip100mu=2\Expn(w'\times(v_{\sigma_{n+1,k_{n+1}}}-v_{\sigma_{00}}))
\le 4M\Expn(w')
\le 8M\sqrt{\epsilon M}.\cr}$$

\medskip

{\bf (d)} Putting this together with (a), we see that if
$\sigma\le\tau_0$ and $\tau_n\le\tau$ then

\Centerline{$\Expn\bigl((v^{\#}_{\tau_0}-v^{\#}_{\sigma})^2
   +\sum_{i=1}^n(v^{\#}_{\tau_i}-v^{\#}_{\tau_{i-1}})^2
   +(v^{\#}_{\tau}-v^{\#}_{\tau_n})^2\bigr)
\le 8M\sqrt{\epsilon M}$.}

\noindent Now consider $\Osclln_I^*(\pmb v^{\#})$.   By 618Ca,
this is

$$\eqalignno{&\sup\{|v^{\#}_{\sigma'}-v^{\#}_{\sigma}|:
   \sigma,\,\sigma'\in\Cal S
\text{ and either }\sigma\le\sigma'\le\tau_0\cr
&\mskip200mu\text{ or there is an }i
\text{ such that }\tau_{i-1}\le\sigma\le\sigma'\le\tau_i\cr
&\mskip200mu\text{ or }\tau_n\le\sigma\le\sigma'\}\cr
&\mskip100mu
=\sup_{\sigma\in\Cal S\wedge\tau_0}(v^{\#}_{\tau_0}-v^{\#}_{\sigma})
   \vee\sup_{1\le i\le n}(v^{\#}_{\tau_i}-v^{\#}_{\tau_{i-1}})
   \vee\sup_{\tau\in\Cal S\vee\tau_n}(v^{\#}_{\tau}-v^{\#}_{\tau_n}).\cr}$$

\noindent So

$$\eqalign{\Osclln_I^*(\pmb v^{\#})^2
&=\sup_{\sigma\in\Cal S\wedge\tau_0}(v^{\#}_{\tau_0}-v^{\#}_{\sigma})^2
   \vee\sup_{1\le i\le n}(v^{\#}_{\tau_i}-v^{\#}_{\tau_{i-1}})^2
   \vee\sup_{\tau\in\Cal S\vee\tau_n}(v^{\#}_{\tau}-v^{\#}_{\tau_n})^2\cr
&\le\sup_{\Atop{\sigma\in\Cal S\wedge\tau_0}{\tau\in\Cal S\vee\tau_n}}
  \bigl((v^{\#}_{\tau_0}-v^{\#}_{\sigma})^2
   +\sum_{1\le i\le n}(v^{\#}_{\tau_i}-v^{\#}_{\tau_{i-1}})^2
   +(v^{\#}_{\tau}-v^{\#}_{\tau_n})^2\bigr).\cr}$$


\noindent As the sum here increases as $\sigma$ decreases and $\tau$
increases,

$$\eqalign{\Expn((\Osclln_I^*(\pmb v^{\#}))^2)
&\le\sup_{\Atop{\sigma\in\Cal S\wedge\tau_0}{\tau\in\Cal S\vee\tau_n}}
  \Expn\bigl((v^{\#}_{\tau_0}-v^{\#}_{\sigma})^2
   +\sum_{1\le i\le n}(v^{\#}_{\tau_i}-v^{\#}_{\tau_{i-1}})^2
   +(v^{\#}_{\tau}-v^{\#}_{\tau_n})^2\bigr)\cr
&\le 8M\sqrt{\epsilon M}.\cr}$$

\noindent Accordingly

\Centerline{$\theta(\Osclln^*_I(\pmb v^{\#}))
\le\Expn(\Osclln^*_I(\pmb v^{\#}))
\le\|\Osclln^*_I(\pmb v^{\#})\|_2
\le\sqrt{8M\sqrt{\epsilon M}}$.}

\noindent As $\epsilon$ is arbitrary,
$\inf_{I\in\Cal I(\Cal S)}\theta(\Osclln^*_I(\pmb v^{\#}))=0$ and
$\pmb v^{\#}$ is jump-free.
}%end of proof of 643G

\leader{643H}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Take $\tau_1$ in $\Cal T$ and a non-negative
$v\in L^0(\frak A_{\tau_1})\cap L^1_{\bar\mu}$.
Set $v_{\sigma}=v\times\chi\Bvalue{\sigma=\tau_1}$ for
$\sigma\in\Cal T\wedge\tau_1$.   Then
$\pmb v=\family{\sigma}{\Cal T\wedge\tau_1}{v_{\sigma}}$ is a non-negative
non-decreasing submartingale.
Let $\pmb v^{\#}=\family{\sigma}{\Cal T\wedge\tau_1}{v^{\#}_{\sigma}}$
be its previsible variation,
and $\pmb v^{\#}_<=\family{\sigma}{\Cal T\wedge\tau_1}{v^{\#}_{<\sigma}}$
the previsible version of $\pmb v^{\#}$.
If $\tau\le\tau_1$ then
$\app(\tau)\Bsetminus\acc(\tau)\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}$.

\proof{ Write $a^*$ for $\app(\tau)\Bsetminus\acc(\tau)$.

\medskip

{\bf (a)} By 612CF, $v_{\sigma}\in L^0(\frak A_{\sigma})$ for every
$\sigma\le\tau_1$.   If $\sigma$, $\sigma'\in\Cal T\wedge\tau_1$ then

$$\eqalign{v_{\sigma}\times\chi\Bvalue{\sigma=\sigma'}
&=v\times\chi(\Bvalue{\tau_1\le\sigma}\Bcap\Bvalue{\sigma=\sigma'})\cr
&=v\times\chi(\Bvalue{\tau_1\le\sigma'}\Bcap\Bvalue{\sigma=\sigma'})
=v_{\sigma'}\times\chi\Bvalue{\sigma=\sigma'}\cr}$$

\noindent so
$\Bvalue{\sigma=\sigma'}\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma'}}$;  thus
$\pmb v$ is fully adapted.   Because $v\in L^1_{\bar\mu}$,
$\pmb v$ is an $L^1$-process.   If $\sigma\le\sigma'$, then
$\Bvalue{\tau_1\le\sigma}\Bsubseteq\Bvalue{\tau_1\le\sigma'}$ so
(because $v\ge 0$) $v_{\sigma}\le v_{\sigma'}$;  thus $\pmb v$ is
non-decreasing, therefore a submartingale (626B), and has a previsible
version (626O), which is itself non-negative and non-decreasing,
therefore locally moderately
oscillatory (616Ra, or otherwise), and has a previsible version (641L).

\medskip

{\bf (b)} Suppose first that $v\le\chi a^*$.

\medskip

\quad{\bf (i)} Take $I\in\Cal I(\Cal T\wedge\tau)$, and for $\sigma\le\tau$
define $d_{\sigma I}$, $w_{\sigma I}$ and $w_I$ from $\tau$ as in 643F.
Suppose that $\sigma_0\le\sigma_1\le\tau$ and
$\Bvalue{\sigma_0<\sigma}\Bcap\Bvalue{\sigma<\sigma_1}=0$ for every
$\sigma\in I$.   Then $P_{\sigma_0}v_{\sigma_1}-v_{\sigma_0}\le w_I$.
\Prf\

\Centerline{$\Bvalue{\sigma_0<\sigma}\Bcap\Bvalue{\sigma<\tau}
   \Bcap\Bvalue{\sigma_1=\tau}
\Bsubseteq\Bvalue{\sigma_0<\sigma}\Bcap\Bvalue{\sigma<\sigma_1}=0$,}

\noindent that is, $\Bvalue{\sigma_0<\sigma}\Bcap\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\sigma_1<\tau}$ for every $\sigma\in I$, and

$$\eqalign{&a^*
    \Bcap\Bvalue{\sigma_0<\tau}\Bsetminus\Bvalue{\sigma_1<\tau}\cr&
\mskip50mu
\Bsubseteq a^*
    \Bcap\Bvalue{\sigma_0<\tau}\Bsetminus\sup_{\sigma\in I}
        (\Bvalue{\sigma_0<\sigma}\Bcap\Bvalue{\sigma<\tau})\cr&
\mskip50mu
=d_{\sigma_0I}.\cr}$$

\noindent Consequently

$$\eqalign{P_{\sigma_0}v_{\sigma_1}-v_{\sigma_0}&
=P_{\sigma_0}(v\times\chi\Bvalue{\sigma_1=\tau_1}
       -v\times\chi\Bvalue{\sigma_0=\tau_1})\cr&
=P_{\sigma_0}(v\times\chi(\Bvalue{\tau=\tau_1}\Bcap\Bvalue{\sigma_1=\tau}
       \Bsetminus\Bvalue{\sigma_0=\tau}))\cr&
\le P_{\sigma_0}
   (v\times\chi(\Bvalue{\sigma_0<\tau}\Bsetminus\Bvalue{\sigma_1<\tau}))\cr&
\le P_{\sigma_0}\chi d_{\sigma_0I}
=w_{\sigma_0 I}
\le w_I.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (ii)} Now 643Fe tells us that for every $\epsilon>0$ there is an
$I\in\Cal I(\Cal T\wedge\tau)$ such that $\Expn(w_I)\le\epsilon$.
So we can apply 643G to see that
$(\pmb v\restr\Cal T\wedge\tau)^{\#}$ is jump-free, that is, that
$\pmb v^{\#}\restr\Cal T\wedge\tau$ is jump-free (626M(f-i)).   It follows
that $v^{\#}_{<\tau}=v^{\#}_{\tau}$ (641O), so surely we have
$a^*\Bsubseteq\Bvalue{v^{\#}_{<\tau}=v^{\#}_{\tau}}$.

\medskip

{\bf (c)} Next suppose that $\emptyset\ne C\subseteq\Cal T\wedge\tau$ and
$v\le\chi a_C$, where

\Centerline{$a_C=\Bvalue{\sup C=\tau}
   \Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau}\Bsubseteq\acc(\tau)$}

\noindent as in 643E.

\medskip

\quad{\bf (i)} If $\rho\le\sigma\in C$,

\Centerline{$v_{\rho}=v\times\chi\Bvalue{\rho=\tau_1}
\le\chi(a_C\Bcap\Bvalue{\rho=\tau_1})
\le\chi(a_C\Bcap\Bvalue{\sigma=\tau})=0$}

\noindent so $\pmb v\restr\Cal T\wedge\sigma=0$,
$\pmb v^{\#}\restr\Cal T\wedge\sigma=(\pmb v\restr\Cal T\wedge\sigma)^{\#}=0$
(626M(f-i)), $v^{\#}_{\sigma}=v^{\#}_{<\sigma}=0$
and $\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}$
(because $\pmb v^{\#}_<$ is fully adapted, by 641G(a-ii)).

\medskip

\quad{\bf (ii)} If $\sup C\le\sigma\le\sigma'\le\tau$ then

\Centerline{$\Bvalue{\sigma<\sigma'}\Bsubseteq\Bvalue{\sup C<\tau}
\Bsubseteq\Bvalue{v_{\sigma}=0}\Bcap\Bvalue{v_{\sigma'}=0}$}

\noindent so

\Centerline{$\Expn(P_{\sigma}v_{\sigma'}-v_{\sigma})
=\Expn(v_{\sigma'}-v_{\sigma})
=\Expn((v_{\sigma'}-v_{\sigma})\times\chi\Bvalue{\sigma<\sigma'})
=0$;}

\noindent as $v_{\sigma}\le P_{\sigma}v_{\sigma'}$,
$P_{\sigma}v_{\sigma'}-v_{\sigma}=0$.   It follows that
$S_I(\tbf 1,Pd\pmb v)=0$ for every finite sublattice $I$ of $[\sup C,\tau]$
and $\pmb v^{\#}$ is constant on $[\sup C,\tau]$.   Accordingly

\Centerline{$\Bvalue{\sup C<\tau}
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{\sup C}}
   \Bcap\Bvalue{v^{\#}_{<\tau}=v^{\#}_{\sup C}}
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}$.}

\medskip

\quad{\bf (iii)} Putting these together,

\Centerline{$a^*\Bsubseteq 1\Bsetminus a_C
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}$.}

\medskip

{\bf (d)} Thirdly, suppose that $t\in T_{\text{r-i}}$ and that
$v\le\chi\Bvalue{\tau=\check t^+}$ where $\check t^+$ is defined as in 643D.
If $\tau\wedge\check t^+\le\rho\le\sigma\le\tau$ then

\Centerline{$\Bvalue{\rho<\sigma}
\Bsubseteq\Bvalue{\rho<\tau_1}\Bcap\Bvalue{\check t^+<\sigma}
\Bsubseteq\Bvalue{v_{\rho}=0}\Bcap\Bvalue{v_{\sigma}=0}$}

\noindent because

\Centerline{$v_{\sigma}
\le\chi(\Bvalue{\tau=\check t^+}\Bcap\Bvalue{\sigma=\tau_1})
\le\chi(\Bvalue{\tau=\check t^+}\Bcap\Bvalue{\sigma=\tau})$.}

\noindent So

\Centerline{$P_{\rho}v_{\sigma}-v_{\rho}
=(P_{\rho}v_{\sigma}-v_{\rho})\times\chi\Bvalue{\rho<\sigma}
=P_{\rho}((v_{\sigma}-v_{\rho}))\times\chi\Bvalue{\rho<\sigma})
=0$.}

\noindent Thus $\pmb v^{\#}$ is constant on $[\tau\wedge\check t^+,\tau]$
and $\Bvalue{\tau\wedge\check t^+<\tau}
\Bsubseteq\Bvalue{v^{\#}_{<\tau}=v^{\#}_{\tau}}$.

If $\sigma\le\tau\wedge\check t$ then
$\Bvalue{\tau=\check t^+}
\Bsubseteq\Bvalue{\sigma<\tau_1}\Bsubseteq\Bvalue{v_{\sigma}=0}$,
so $v^{\#}_{\tau\wedge\check t}=0=v^{\#}_{<\tau\wedge\check t}$.   Now

$$\eqalign{\Bvalue{\tau<\check t^+}&
\Bsubseteq\Bvalue{\tau\le\check t}
   \Bcup(\Bvalue{\check t<\tau}\Bcap\Bvalue{\tau<\check t^+})
=\Bvalue{\tau\le\check t}\cr&
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{\tau\wedge\check t}}
    \Bcap\Bvalue{v^{\#}_{<\tau}=v^{\#}_{<\tau\wedge\check t}}
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}\cr}$$

\noindent and

\Centerline{$\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}
\Bsupseteq 1\Bsetminus\Bvalue{\tau=\check t^+}
\Bsupseteq a^*$.}

\medskip

{\bf (e)(i)} If we think of
$v\mapsto\pmb v$, $\pmb v\mapsto\pmb v^{\#}$, $\pmb v^{\#}\mapsto v^{\#}_{\tau}$
and $\pmb v^{\#}\mapsto v^{\#}_{<\tau}$ as functions, then
$v\mapsto v^{\#}_{\tau}$ and $v\mapsto v^{\#}_{<\tau}$ are additive, and the
set

\Centerline{$A=\{v:v\in L^0(\frak A_{\tau_1})\cap L^1_{\bar\mu}$, $v\ge 0$,
$a^*
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}\}$}

\noindent is
closed under addition and multiplication by non-negative scalars.   So

\Centerline{$D=\{d:d\in\frak A_{\tau_1}$, $v\in A$ whenever
$v\in L^0(\frak A_{\tau_1})\cap L^1_{\bar\mu}$ and $0\le v\le\chi d\}$}

\noindent is an ideal of $\frak A_{\tau_1}$.   (If $d_0$, $d_1\in D$,
$d\in\frak A_{\tau_1}$,
$d\Bsubseteq d_0\Bcup d_1$, $v\in L^0(\frak A_{\tau_1})\cap L^1_{\bar\mu}$
and $0\le v\le\chi d$, then $v=v\times\chi d_0+v\times\chi(d_1\Bsetminus d_0)$
is the sum of two members of $A$.)   We know also that
$a^*\in D$ ((b) above, with 643Cd),
$a_C\in D$ whenever
$\emptyset\ne C\subseteq\Cal T\wedge\tau$ (by (c)) and
$\Bvalue{\tau=\check t^+}\in D$ whenever $t\in T_{\text{r-i}}$ (by (d)).
So

\Centerline{$\sup D
\Bsupseteq a^*\Bcup\acc(\tau)
    \Bcup(1\Bsetminus\app(\tau))
=1$}

\noindent by the definition of $\acc(\tau)$ and 643D,
and $\sup_{d\in D}\bar\mu d=1$.

\medskip

\quad{\bf (ii)} Now suppose just that $v\in L^0(\frak A_{\tau_1})$
is non-negative and integrable.
Let $\delta>0$ be such that $\Expn(v\times\chi a)\le\epsilon$
whenever $\bar\mu a\le\delta$.   Let $d\in D$ be such that
$\bar\mu(1\Bsetminus d)\le\delta$, and $M>0$ such that
$\Expn((v-M\chi 1)^+)\le\epsilon$.   Set $w=\Bover1M(v\wedge M\chi d)$.
Then $w\in L^0(\frak A_{\tau_1})$, $0\le w\le v$ and
$v-w\le(v-M\chi 1)^++v\times\chi(\acc(\tau)\Bsetminus d)$ so
$\Expn(v-w)\le 2\epsilon$.   Also $w\in A$ (using (d) above), and
$a^*\Bsubseteq\Bvalue{w^{\#}_{\tau}=w^{\#}_{<\tau}}$.
We know too that $\pmb v^{\#}-\pmb w^{\#}=(\pmb v-\pmb w)^{\#}$.   Since
$\pmb v-\pmb w$ is non-negative, 626O tells us that

\Centerline{$\Expn(v^{\#}_{\tau}-w^{\#}_{\tau})
\le\Expn(v_{\tau}-w_{\tau})\le 2\epsilon$.}

\noindent Now $w^{\#}_{<\tau}\le v^{\#}_{<\tau}\le v^{\#}_{\tau}$
and $w^{\#}_{<\tau}\le w^{\#}_{\tau}\le v^{\#}_{\tau}$ so

\Centerline{$\Expn((v^{\#}_{\tau}-v^{\#}_{<\tau})\times\chi a^*)
\le\Expn(v^{\#}_{\tau}-w^{\#}_{\tau})
   +\Expn((w^{\#}_{\tau}-w^{\#}_{<\tau})\times\chi a^*)
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$a^*\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}$ and the proof is
complete.
}%end of proof of 643H

\leader{643I}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Take $\tau_1\in\Cal T$ and a martingale
$\pmb u=\langle u_{\sigma}\rangle_{\sigma\le\tau_1}$.
% such that $u_{\min\Cal T}=0$.   not needed?
Suppose that $\epsilon>0$ is such that
$\Bvalue{\sigma<\tau_1}\Bsubseteq\Bvalue{|u_{\sigma}|\le\epsilon}$ for every
$\sigma\le\tau_1$.   Then there is a martingale
$\tildepmb u=\langle\tilde u_{\sigma}\rangle_{\sigma\le\tau_1}$
such that $\Osclln(\tildepmb u)\le 2\epsilon\chi 1$ and
$\pmb u-\tildepmb u$ is of bounded variation.

\proof{ Write $v$ for $u_{\tau_1}$.

\medskip

{\bf (a)(i)} Set $\acute v=v^+$ and $\grave v=v^-$, so
that $\acute v$ and $\grave v$ are non-negative integrable members of
$L^0(\frak A_{\tau_1})$.   As in 643H, set

\Centerline{$\acute v_{\sigma}=\acute v\times\Bvalue{\sigma=\tau_1}$,
\quad$\grave v_{\sigma}=\grave v\times\Bvalue{\sigma=\tau_1}$,
\quad$v_{\sigma}=v\times\Bvalue{\sigma=\tau_1}$}

\noindent for $\sigma\le\tau_1$,

\Centerline{$\acute{\pmb v}=\langle\acute v_{\sigma}\rangle_{\sigma\le\tau_1}$,
\quad$\grave{\pmb v}=\langle\grave v_{\sigma}\rangle_{\sigma\le\tau_1}$,
\quad$\pmb v=\langle v_{\sigma}\rangle_{\sigma\le\tau_1}$,}

\noindent and let $\acute{\pmb v}^{\#}$, $\grave{\pmb v}^{\#}$
be the previsible variations of $\acute{\pmb v}$ and $\grave{\pmb v}$
respectively;  write $\pmb v^{\#}$ for
$\acute{\pmb v}^{\#}-\grave{\pmb v}^{\#}$, the previsible version of $\pmb v$.
Then $\pmb v-\pmb v^{\#}$ is a martingale (626O).   Set
$\tildepmb u=\pmb u-\pmb v+\pmb v^{\#}$, so that $\tildepmb u$ is a martingale.
Note that $\acute{\pmb v}$, $\grave{\pmb v}$, $\acute{\pmb v}^{\#}$
and $\grave{\pmb v}^{\#}$ are non-negative monotonic
processes so are of bounded variation;  accordingly
$\pmb v$, $\pmb v^{\#}$
and $\pmb u-\tildepmb u=\pmb v-\pmb v^{\#}$ are of bounded variation.

\medskip

\quad{\bf (ii)} $u_{\tau_1}=v_{\tau_1}$, so if $\sigma\le\tau_1$ then

\Centerline{$\Bvalue{\sigma=\tau_1}\Bsubseteq\Bvalue{u_{\sigma}-v_{\sigma}=0}$}

\noindent while

\Centerline{$\Bvalue{\sigma<\tau_1}
  \Bsubseteq\Bvalue{|u_{\sigma}|\le\epsilon}\Bcap\Bvalue{v_{\sigma}=0}
  \Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le\epsilon}$;}

\noindent thus $|u_{\sigma}-v_{\sigma}|\le\epsilon\chi 1$.   It follows at
once that $|u_{<\sigma}-v_{<\sigma}|\le\epsilon\chi 1$ and that
$|u_{\sigma}-v_{\sigma}-u_{<\sigma}+v_{<\sigma}|\le 2\epsilon\chi 1$
for every $\sigma\le\tau_1$.

\medskip

\quad{\bf (iii)} If $\sigma\le\tau_1$ and
$P_{<\sigma}$ is the conditional expectation
associated with the subalgebra $\frak A_{<\sigma}$,
$|\tilde u_{\sigma}-P_{<\sigma}\tilde u_{\sigma}|\le 2\epsilon\chi 1$.
\Prf\ As $v^{\#}_{\sigma}\in L^0(\frak A_{<\sigma})$ (641D),
$P_{<\sigma}v^{\#}_{\sigma}=v^{\#}_{\sigma}$ and

$$\eqalign{|\tilde u_{\sigma}-P_{<\sigma}\tilde u_{\sigma}|&
=|u_{\sigma}-v_{\sigma}-P_{<\sigma}u_{\sigma}+P_{<\sigma}v_{\sigma}
    +v^{\#}_{\sigma}-P_{<\sigma}v^{\#}_{\sigma}|\cr&
\le|u_{\sigma}-v_{\sigma}|+|P_{<\sigma}(u_{\sigma}-v_{\sigma})|
\le|u_{\sigma}-v_{\sigma}|+P_{<\sigma}|u_{\sigma}-v_{\sigma}|\cr&
\le\epsilon\chi 1+P_{<\sigma}(\epsilon\chi 1)
=2\epsilon\chi 1.  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} Take $\tau\le\tau_1$.

\medskip

\quad{\bf (i)} Set $a^*=\app(\tau)\Bsetminus\acc(\tau)$.
Then 643H and the last remark in (i) above tell us that

$$\eqalign{a^*&
\Bsubseteq\Bvalue{\acute v^{\#}_{\tau}=\acute v^{\#}_{<\tau}}
  \Bcap\Bvalue{\acute v^{\#}_{\tau}=\acute v^{\#}_{<\tau}}
\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{<\tau}}\cr&
\Bsubseteq\Bvalue{\tilde u_{\tau}-\tilde u_{<\tau}
  =u_{\tau}-u_{<\tau}-v_{\tau}+v_{<\tau}}
\Bsubseteq\Bvalue{|\tilde u_{\tau}-\tilde u_{<\tau}|\le 2\epsilon\chi 1}.\cr}$$

\medskip

\quad{\bf (ii)} Now suppose that $C$ is a non-empty upwards-directed subset of
$\Cal T\wedge\tau$ with supremum $\rho$ and
that $a_C=\Bvalue{\rho=\tau}\Bsetminus\sup_{\sigma\in C}\Bvalue{\sigma=\tau}$,
as in 643F and 643H.   Of course $a_C$ is also expressible as
$\Bvalue{\rho=\tau}\Bcap\inf_{\sigma\in C}\Bvalue{\sigma<\rho}$.   Next,
$a_C\Bsubseteq\Bvalue{\tilde u_{<\rho}=P_{<\rho}\tilde u_{\rho}}$
by 643Bb, because $\tildepmb u$ is a martingale.   Now

$$\eqalignno{a_C&
\Bsubseteq\Bvalue{\tilde u_{<\rho}=P_{<\rho}\tilde u_{\rho}}
   \Bcap\Bvalue{\tilde u_{\tau}=\tilde u_{\rho}}
   \Bcap\Bvalue{\tilde u_{<\tau}=\tilde u_{<\rho}}
\Displaycause{using 641L) again}
\Bsubseteq\Bvalue{|\tilde u_{\tau}-\tilde u_{<\tau}|
     =|\tilde u_{\rho}-P_{<\rho}\tilde u_{\rho}|}
\Bsubseteq\Bvalue{|\tilde u_{\tau}-\tilde u_{<\tau}|\le 2\epsilon\chi 1}\cr}$$

\noindent by (a-iii) above.

\medskip

\quad{\bf (iii)} Next, suppose that $t\in T$ is isolated on the right, and
define $\check t^+$ as in 643D.   Set $\rho=\tau\wedge\check t^+$ and
$c=\Bvalue{\check t<\rho}=\Bvalue{\rho>t}$ (611E(a-i-$\delta$).

\medskip

\qquad\grheada\ $c\Bsubseteq\Bvalue{\tilde u_{<\rho}=\tilde u_{\check t}}$.
\Prf\ For any $\sigma\in\Cal T$,

\Centerline{$\Bvalue{\check t<\sigma}\Bcap\Bvalue{\sigma<\rho}
\Bsubseteq\Bvalue{\check t<\sigma}\Bcap\Bvalue{\sigma<\check t^+}=0$.}

\noindent So if $I\in\Cal I(\Cal T\wedge\tau_1)$ contains $\check t$, and
$\tilde u_{I<\tau}$ is defined from $\tildepmb u$ as in 641Ea, then
$c=\Bvalue{\check t<\rho}\Bsubseteq\Bvalue{\tilde u_{I<\tau}=\tilde u_{\check t}}$.
Taking the limit as $I\uparrow\Cal I(\Cal T\wedge\tau_1)$,
$c\Bsubseteq\Bvalue{\tilde u_{<\rho}=\tilde u_{\check t}}$.\ \Qed

\medskip

\qquad\grheadb\ $c\in\frak A_{\check t}\Bcap\frak A_{\rho}$.   Now if
$a\in\frak A$ and $a\Bsubseteq c$, then $a\in\frak A_{\check t}=\frak A_t$
(611Hb) iff
$a\in\frak A_{\rho}$.   \Prf\ If $a\in\frak A_{\check t}$ then
$a\in\frak A_{\rho}$ by 611H(a-i).   In the other direction, set
$\frak B=\{b:b\in\frak A$, $c\Bcap b\in\frak A_{\check t}\}$.   Then
$\frak B$ is a closed subalgebra of $\frak A$.   If $\sigma\in\Cal T$, then

$$\eqalignno{c\Bcap\Bvalue{\sigma<\rho}&
=c\Bcap\Bvalue{\sigma<\check t^+}
\Displaycause{because
$c=\Bvalue{\check t^+\le\rho}=\Bvalue{\check t^+=\rho}$}
=c\Bcap\Bvalue{\sigma\le\check t}
\in\frak A_{\check t},\cr}$$

\noindent so $\Bvalue{\sigma<\rho}\in\frak B$.   By the definition of
$\frak A_{<\rho}$ (641Ba), $\frak A_{<\rho}\subseteq\frak B$
and $c\Bcap a\in\frak A_{\check t}$.\ \Qed

\medskip

\qquad\grheadc\ $c\Bsubseteq\Bvalue{P_{<\rho}\tilde u_{\rho}=\tilde u_{\check t}}$.
\Prf\ For $\alpha\in\Bbb R$,

$$\eqalign{\Bvalue{P_{<\rho}\tilde u_{\rho}\times\chi c>\alpha}
&=\Bvalue{P_{<\rho}\tilde u_{\rho}>\alpha}\Bcap c\text{ if }\alpha\ge 0,\cr
&=\Bvalue{P_{<\rho}\tilde u_{\rho}>\alpha}\Bcup(1\Bsetminus c)
   \text{ if }\alpha\le 0,\cr}$$

\noindent and in either case belongs to $\frak A_{\check t}$.   So
$P_{<\rho}\tilde u_{\rho}\times\chi c\in L^0(\frak A_{\check t})$.
Now if $a\in\frak A_{\check t}$,

$$\eqalignno{\Expn(((\tilde u_{\check t}\times\chi c)\times\chi a)&
=\Expn(\tilde u_{\check t}\times\chi(c\Bcap a))
=\Expn(\tilde u_{\check t^+}\times\chi(c\Bcap a))
\Displaycause{because $\tilde u$ is a martingale}
=\Expn(\tilde u_{\rho}\times\chi(c\Bcap a))
\Displaycause{because $c\Bsubseteq\Bvalue{\rho=\check t^+}$}
=\Expn(P_{<\rho}\tilde u_{\rho}\times\chi(c\Bcap a))
\Displaycause{because $c\Bcap a\in\frak A_{<\rho}$}
=\Expn((P_{<\rho}\tilde u_{\rho}\times\chi c)\times\chi a)).\cr}$$

\noindent So
$\tilde u_{\check t}\times\chi c=P_{<\rho}\tilde u_{\rho}\times\chi c$, that is,
$c\Bsubseteq\Bvalue{P_{<\rho}\tilde u_{\rho}=\tilde u_{\check t}}$.\ \Qed

\medskip

\qquad\grheadd\ Putting ($\alpha$) and ($\gamma$) together,

$$\eqalign{\Bvalue{\tau=\check t^+}&
=\Bvalue{\rho=\check t^+}\Bcap\Bvalue{\rho=\tau}
\Bsubseteq c\Bcap\Bvalue{\tilde u_{\tau}=\tilde u_{\rho}}
    \Bcap\Bvalue{\tilde u_{<\tau}=\tilde u_{<\rho}}\cr&
\Bsubseteq\Bvalue{\tilde u_{<\rho}=P_{<\rho}\tilde u_{\rho}}
    \Bcap\Bvalue{\tilde u_{\tau}=\tilde u_{\rho}}
    \Bcap\Bvalue{\tilde u_{<\tau}=\tilde u_{<\rho}}\cr&
\Bsubseteq\Bvalue{|\tilde u_{\tau}-\tilde u_{<\tau}|
        =|\tilde u_{\rho}-P_{\rho}\tilde u_{\rho}|}
\Bsubseteq\Bvalue{|\tilde u_{\tau}-\tilde u_{<\tau}|\le 2\epsilon\chi 1}\cr}$$

\noindent by (a-iii) again.

\medskip

\quad{\bf (iv)} Assembling (i)-(iii), we see that

\Centerline{$1
=a^*\Bcup\sup_{\emptyset\ne C\subseteq\Cal T\wedge\tau}a_C
     \Bcup\sup_{t\in T_{\text{r-i}}}\Bvalue{\tau=\check t^+}
\Bsubseteq\Bvalue{|\tilde u_{\tau}-\tilde u_{<\tau}|\le 2\epsilon\chi 1}$}

\noindent and $|\tilde u_{\tau}-\tilde u_{<\tau}|\le 2\epsilon\chi 1$.
As $\tau$ is arbitrary and $\tildepmb u$ is near-simple (632I),
$\Osclln(\tildepmb u)\le 2\epsilon\chi 1$ (641Nb), and we have found a suitable
auxiliary martingale.
}%end of proof of 643I

\leader{643J}{Corollary} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Take an interval
$\Cal S=[\tau,\tau']$ where $\tau\le\tau'$ in $\Cal T$,
and a martingale $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$.
Suppose that $\epsilon>0$ is such that
$\Bvalue{\sigma<\tau'}\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau}|\le\epsilon}$
for every $\sigma\in\Cal S$.   Then there is a martingale
$\tildepmb u=\family{\sigma}{\Cal S}{\tilde u_{\sigma}}$
such that $\Osclln(\tildepmb u)\le\epsilon\chi 1$, $\tilde u_{\tau}=0$ and
$\pmb u-\tildepmb u$ is of bounded variation.

\proof{ Set $u'_{\sigma}=P_{\sigma}(u_{\tau'}-u_{\tau})$ for
$\sigma\le\tau'$, so that
$\pmb u'=\langle u'_{\sigma}\rangle_{\sigma\le\tau'}$ is a martingale.
Then $\Bvalue{\sigma<\tau'}\Bsubseteq\Bvalue{|u'_{\sigma}|\le\epsilon}$
for every $\sigma\le\tau'$.   \Prf\ We have

$$\eqalignno{\Bvalue{\sigma\le\tau}&
=\Bvalue{\sigma=\sigma\wedge\tau}
\Bsubseteq\Bvalue{P_{\sigma}(u_{\tau'}-u_{\tau})
   =P_{\sigma\wedge\tau}(u_{\tau'}-u_{\tau})}
\Displaycause{622Bb once more}
=\Bvalue{u'_{\sigma}
   =P_{\sigma}P_{\tau}(u_{\tau'}-u_{\tau})}
\Displaycause{622Ba}
\Bsubseteq\Bvalue{u'_{\sigma}=0}
\Bsubseteq\Bvalue{|u'_{\sigma}|\le\epsilon},\cr}$$

\noindent while also

$$\eqalign{\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'}&
=\Bvalue{\sigma=\sigma\vee\tau}
   \Bcap\Bvalue{\sigma\vee\tau<\tau'}\cr&
\Bsubseteq\Bvalue{u'_{\sigma}
         =P_{\sigma\vee\tau}(u_{\tau'}-u_{\tau})}
   \Bcap\Bvalue{|u_{\sigma\vee\tau}-u_{\tau}|\le\epsilon}\cr&
\Bsubseteq\Bvalue{u'_{\sigma}=u_{\sigma\vee\tau}-u_{\tau}}
   \Bcap\Bvalue{|u_{\sigma\vee\tau}-u_{\tau}|\le\epsilon}
\Bsubseteq\Bvalue{|u'_{\sigma}|\le\epsilon}.\cr}$$

\noindent Now

\Centerline{$\Bvalue{\sigma<\tau'}
\Bsubseteq\Bvalue{\sigma\le\tau}
  \Bcup(\Bvalue{\tau\le\sigma}\Bcup\Bvalue{\sigma<\tau'})
\Bsubseteq\Bvalue{|u'_{\sigma}|\le\epsilon}$,}

\noindent as claimed.\ \Qed

By 643I, there is a martingale
$\tildepmb u'=\langle\tilde u'_{\sigma}\rangle_{\sigma\le\tau'}$ such that
$\Osclln(\tildepmb u')\le 2\epsilon\chi 1$
and $\pmb u'-\tildepmb u'$ is of bounded
variation.   Set
$\tildepmb u=\family{\sigma}{\Cal S}{\tilde u_{\sigma}}$ where
$\tilde u_{\sigma}=\tilde u'_{\sigma}-\tilde u'_{\tau}$ for
$\sigma\in\Cal S$.   Then $\tildepmb u$ is a martingale and
$\tilde u_{\tau}=0$.   In the language of
613Cc, $\Delta\tildepmb u=\Delta(\tildepmb u'\restr\Cal S)$, so

\Centerline{$\Osclln(\tildepmb u)
=\Osclln(\tildepmb u'\restr\Cal S)
\le\Osclln(\tildepmb u')\le 2\epsilon$}

\noindent (using 618D(b-i)).   We see also that
$u'_{\sigma}=u_{\sigma}-u_{\tau}$ for every $\sigma\in\Cal S$,
so $u_{\sigma}-\tilde u_{\sigma}
=u'_{\sigma}-\tilde u'_{\sigma}+u_{\tau}+\tilde u'_{\tau}$ for $\sigma\in\Cal S$,
$\Delta(\pmb u-\tildepmb u)=\Delta((\pmb u'-\tildepmb u')\restr\Cal S)$ and
$\int_{\Cal S}|d(\pmb u-\tildepmb u)|$ is defined in $L^0(\frak A)$ and equal to
$\int_{\Cal S}|d(\pmb u'-\tildepmb u')|$;  thus
$\pmb u-\tildepmb u$ is of bounded variation, as required.
}%end of proof of 643J

\leader{643K}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\sequencen{\tau_n}$ a non-decreasing sequence in $\Cal S$ such that
$\Cal S\subseteq\bigcup_{n\in\Bbb N}[\tau_0,\tau_n]$.
Suppose that for each $n\in\Bbb N$ we are given a
fully adapted process
$\pmb u_n=\family{\sigma}{\Cal S\cap[\tau_n,\tau_{n+1}]}{u_{n\sigma}}$
starting from $u_{n\tau_n}=0$.

(a) There is a unique fully adapted process $\pmb u=\sfamilysS{u}$ such
that

\Centerline{$u_{\sigma}
=u_{n\sigma}+\sum_{i=0}^{n-1}u_{i\tau_{i+1}}$
whenever $\sigma\in\Cal S$, $n\in\Bbb N$ and $\tau_n\le\sigma\le\tau_{n+1}$.
\hfil(*)}

(b) If every $\pmb u_n$ is a martingale, then $\pmb u$ is a
martingale.

(c) If every $\pmb u_n$ is order-bounded, then
$\pmb u$ is locally order-bounded and
$\Osclln(\pmb u\restr\Cal S\wedge\tau_n)=\sup_{i<n}\Osclln(\pmb u_i)$
for every $n\in\Bbb N$.

\proof{ Set $\Cal S_0=\bigcup_{n\in\Bbb N}\Cal S\cap[\tau_n,\tau_{n+1}]$.

\medskip

{\bf (a)} $\Cal S_0$ is a sublattice of $\Cal T$.   \Prf\ If $\sigma$,
$\sigma'\in\Cal S_0$, let $m$, $n\in\Bbb N$ be such that
$\tau_m\le\sigma\le\tau_{m+1}$ and $\tau_n\le\sigma'\le\tau_{n+1}$.
We can suppose that $m\le n$.
In this case $\sigma\wedge\sigma'\in[\tau_m,\tau_{m+1}]$ and
$\sigma\vee\sigma'\in[\tau_n,\tau_{n+1}]$ belong to $\Cal S_0$.\ \Qed

The formula (*) defines a fully adapted process $\pmb u^*$ on $\Cal S_0$.
\Prf\ If $n\in\Bbb N$ and $\tau_n\le\sigma\le\tau_{n+1}$ then
$u_{n\sigma}+\sum_{i=0}^{n-1}u_{i\tau_{i+1}}
\in L^0(\frak A_{\sigma})$ because
$L^0(\frak A_{\tau_{i+1}})\subseteq L^0(\frak A_{\sigma})$ for every
$i<n$.   If $m\le n$ in $\Bbb N$, $\sigma\in[\tau_m,\tau_{m+1}]$,
$\sigma'\in[\tau_n,\tau_{n+1}]$ and $c=\Bvalue{\sigma=\sigma'}$, then

\inset{----- if $m=n$ then
$c\Bsubseteq\Bvalue{u_{n\sigma}=u_{n\sigma'}}
=\Bvalue{u_{n\sigma}+\sum_{i=0}^{n-1}u_{i\tau_{i+1}}
=u_{n\sigma'}+\sum_{i=0}^{n-1}u_{i\tau_{i+1}}}$;

----- if $m<n$ then

$$\eqalign{c
&\Bsubseteq\Bvalue{\sigma=\tau_{m+1}}
   \Bcap\inf_{m<i<n}\Bvalue{\tau_i=\tau_{i+1}}
   \Bcap\Bvalue{\sigma'=\tau_n}\cr
&\Bsubseteq\Bvalue{u_{m\sigma}=u_{m\tau_{m+1}}}
   \Bcap\inf_{m<i<n}\Bvalue{u_{i\tau_{i+1}}=0}
   \Bcap\Bvalue{u_{n\sigma'}=0}\cr
&\Bsubseteq\Bvalue{u_{m\sigma}+\sum_{i=0}^{m-1}u_{i\tau_{i+1}}
=u_{n\sigma'}+\sum_{i=0}^{n-1}u_{i\tau_{i+1}}},\cr}$$
}

\noindent In particular, this shows that if $\sigma=\sigma'$ belongs to
$[\tau_n,\tau_{n+1}]$ for more than one $n$, then (*) gives a well-assigned
value for $u_{\sigma}$;  and in general we have
$\Bvalue{\sigma=\sigma'}\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma'}}$, so that
$\pmb u^*$ is fully adapted.\ \Qed

Next, $\Cal S_0$ covers $\Cal S\wedge\tau_n$
for every $n$.   \Prf\ Induce on $n$.
For $n=0$ this is trivial.   For the inductive step to $n+1\ge 1$, if
$\sigma\in\Cal S\wedge\tau_{n+1}$ then
$\{\sigma\}$ is covered by
$\{\sigma\wedge\tau_n,\sigma\vee\tau_n\}
\subseteq\Cal S\wedge\tau_n\cup\Cal S\cap[\tau_n,\tau_{n+1]}$
and therefore by $\Cal S_0$.
As $\sigma$ is arbitrary, $\Cal S\wedge\tau_{n+1}$ is covered by
$\Cal S_0$.\ \Qed

Consequently $\Cal S_0$ covers $\Cal S$.   Writing $\tildepmb u^*$ for the fully
adapted extension of $\pmb u^*$ to the covered envelope $\tilde\Cal S_0$ of
$\Cal S_0$, we have $\Cal S\subseteq\tilde\Cal S_0$ and
$\pmb u=\tildepmb u^*\restr\Cal S$ is a fully adapted process satisfying the
condition (*).   And because (*) determines the process
$\pmb u^*=\pmb u\restr\Cal S_0$, it determines $\pmb u$ (612R).

\medskip

{\bf (b)} $\pmb u\restr\Cal S_0$ is a martingale.   \Prf\ Suppose that
$m$, $n\in\Bbb N$,
$\sigma\in[\tau_m,\tau_{m+1}]$, $\sigma'\in[\tau_n,\tau_{n+1}]$ and
$\sigma\le\sigma'$.   If $m>n$ then $\sigma=\sigma'$ and of course
$P_{\sigma}u_{\sigma'}=u_{\sigma}$.   If $m=n$ then

\Centerline{$P_{\sigma}u_{\sigma'}
=\sum_{i=0}^{m-1}P_{\sigma}u_{i\tau_{i+1}}
   +P_{\sigma}u_{m\sigma'}
=\sum_{i=0}^{m-1}u_{i\tau_{i+1}}
   +u_{m\sigma}
=u_{\sigma}$.}

\noindent If $m<n$ then

$$\eqalign{P_{\sigma}u_{\sigma'}
&=\sum_{i=0}^{m-1}P_{\sigma}u_{i\tau_{i+1}}
  +\sum_{i=m}^{n-1}P_{\sigma}u_{i\tau_{i+1}}
  +P_{\sigma}u_{n\sigma'}\cr
&=\sum_{i=0}^{m-1}u_{i\tau_{i+1}}
  +P_{\sigma}u_{m\tau_{m+1}}
  +\sum_{i=m+1}^{n-1}P_{\sigma}u_{i\tau_{i+1}}
  +P_{\sigma}u_{n\sigma'}\cr
&=\sum_{i=0}^{m-1}u_{i\tau_{i+1}}
  +u_{m\sigma}
  +\sum_{i=m+1}^{n-1}P_{\sigma}P_{\tau_i}u_{i\tau_{i+1}}
  +P_{\sigma}P_{\tau_n}u_{n\sigma'}\cr
&=u_{\sigma}
  +\sum_{i=m+1}^{n-1}P_{\sigma}u_{i\tau_i}
  +P_{\sigma}u_{n\tau_n}
=u_{\sigma}.\cr}$$

\noindent As $\sigma$ and $\sigma'$ are arbitrary, $\pmb u\restr\Cal S_0$
is a martingale.\ \Qed

Because $\Cal S_0$ covers $\Cal S$ and is cofinal with $\Cal S$,
$\pmb u$ is a martingale.   \Prf\ By 622Oa, there is a martingale
$\pmb u'$ on
the ideal $\Cal S'$ of $\Cal T$ generated by $\Cal S_0$ which extends
$\pmb u\restr\Cal S_0$.   Now $\Cal S\subseteq\Cal S'$ so $\pmb u'\restr\Cal S$
is an extension of $\pmb u\restr\Cal S_0$, and $\pmb u=\pmb u'\restr\Cal S$ is a
martingale.\ \Qed

\medskip

{\bf (c)} Since
$|u_{n\sigma'}-u_{n\sigma}|=|u_{\sigma'}-u_{\sigma}|$ whenever
$\tau_n\le\sigma\le\sigma'\le\tau_{n+1}$,
$\Osclln(\pmb u\restr[\tau_n,\tau_{n+1}])
=\Osclln(\pmb u_n)$ for every $n$.
Now an easy induction on $n$, using 618Db for the inductive step, shows
that

\Centerline{$\Osclln(\pmb u\restr[\tau_0,\tau_n])
=\sup_{i<n}\Osclln(\pmb u\restr[\tau_i,\tau_{i+1}])
=\sup_{i<n}\Osclln(\pmb u_i)$}

\noindent for every $n$.
}%end of proof of 643K

\leader{643L}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$
with a least element and a non-decreasing sequence $\sequencen{\tau_n}$
such that $\{\tau_n:n\in\Bbb N\}$ is cofinal with $\Cal S$,
and $\pmb u=\sfamilysS{u}$ a martingale.
Then for any $\epsilon>0$ there is a local martingale
$\tildepmb u=\sfamilysS{\tilde u}$ such that
$\Osclln(\tildepmb u\restr\Cal S\wedge\tau)\le\epsilon\chi 1$ for every
$\tau\in\Cal S$ and $\pmb u-\tildepmb u$ is locally of bounded variation.

\proof{{\bf (a)} Choose $\sequencen{k_n}$ and $\sequence{i}{\sigma_i}$
as follows.
Start with $k_0=0$ and $\sigma_0=\min\Cal S$.   Given $k_n\in\Bbb N$ and
$\sigma_{k_n}\in\Cal S\wedge\tau_n$, then,
because $\pmb u$ is locally near-simple (632Ia),
there is a non-decreasing $\sequence{j}{\sigma_{nj}}$ in
$[\sigma_{k_n},\tau_n]$ such that $\sigma_{n0}=\sigma_{k_n}$,
$\inf_{j\in\Bbb N}\Bvalue{\sigma_{kj}<\tau_n}=0$ and
$\Bvalue{\sigma<\sigma_{n,j+1}}\penalty-100
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\sigma_{nj}}|\le\bover14\epsilon}$
whenever $j\in\Bbb N$ and $\sigma_{nj}\le\sigma\le\sigma_{n,j+1}$
(631Ra).   Let $k\in\Bbb N$ be such that
$\bar\mu\Bvalue{\sigma_{nk}<\tau_n}\le 2^{-n}$;  set $k_{n+1}=k_n+k$ and
$\sigma_{k_n+i}=\sigma_{ni}$ for $i\le k$.   Continue.

\medskip

{\bf (b)} We see that $\sequence{i}{\sigma_i}$ is a non-decreasing sequence
in $\Cal S$, that

\Centerline{$\lim_{i\to\infty}\bar\mu\Bvalue{\sigma_i<\tau_n}
=\lim_{m\to\infty}\bar\mu\Bvalue{\sigma_{k_{m+1}}<\tau_n}
\le\lim_{m\to\infty}\bar\mu\Bvalue{\sigma_{k_{m+1}}<\tau_m}
=0$}

\noindent for every $n\in\Bbb N$, and that
$\Bvalue{\sigma<\sigma_{i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\sigma_i}|\le\bover14\epsilon}$
whenever $i\in\Bbb N$ and $\sigma_i\le\sigma\le\sigma_{i+1}$.
Set $\Cal S_0=\bigcup_{i\in\Bbb N}\Cal S\wedge\sigma_i$.   Then
$\Cal S_0$ is an ideal of $\Cal S$ covering every $\tau_n$ and therefore
covering $\Cal S$.

\medskip

{\bf (c)} For each $i\in\Bbb N$, we can apply 643J to
$\pmb u\restr[\sigma_i,\sigma_{i+1}]$ to see that there is a martingale
$\tildepmb u_i=\langle\tilde u_{i\sigma}\rangle_{\sigma_i\le\sigma\le\sigma_{i+1}}$
such that $\Osclln(\tildepmb u_i)\le 2^{-i-1}\epsilon$, $\tilde u_{i\sigma_i}=0$ and
$\tildepmb u_i-\pmb u\restr[\sigma_i,\sigma_{i+1}]$ is of bounded variation.
By 643K, we have a martingale
$\tildepmb u'=\family{\sigma}{\Cal S_0}{\tilde u'_{\sigma}}$
such that $\Osclln(\tildepmb u'\restr\Cal S\wedge\sigma_i)\le\epsilon$
for every $i\in\Bbb N$;  it follows that
$\Osclln(\tildepmb u'\restr\Cal S\wedge\sigma)\le\epsilon$ for every
$\sigma\in\Cal S_0$.

Consider the difference $(\pmb u\restr\Cal S_0)-\tildepmb u'$.
If $i\in\Bbb N$ and $\sigma\in[\sigma_i,\sigma_{i+1}]$, this is given by

\Centerline{$u_{\sigma}-\tilde u'_{\sigma}
=u_{\sigma}-\tilde u_{i\sigma}-\tilde u'_{\sigma_i}$.}

\noindent So, for any $i$,

\Centerline{$\int_{\Cal S\wedge\sigma_i}|d(\pmb u-\tildepmb u')|
=\sum_{j=0}^{i-1}\int_{[\sigma_j,\sigma_{j+1}]}|d(\pmb u-\tildepmb u')|
=\sum_{j=0}^{i-1}\int_{[\sigma_j,\sigma_{j+1}]}
   |d(\pmb u_i-\tildepmb u_i)|$}

\noindent is well-defined, that is,
$(\pmb u-\tildepmb u')\restr\Cal S\wedge\sigma_i$ is of bounded
variation.   Thus $\pmb u\restr\Cal S_0-\tildepmb u'$ is locally of
bounded variation.

\medskip

{\bf (e)} For any $i\in\Bbb N$,

$$\eqalignno{\Osclln(\tildepmb u'\restr\Cal S\wedge\sigma_i)
&=\sup_{j<i}\Osclln(\tildepmb u'\restr[\sigma_j,\sigma_{j+1}])\cr
\displaycause{using 618Db repeatedly}
&=\sup_{j<i}\Osclln(\tildepmb u_j)
\le\epsilon\chi 1.\cr}$$

\noindent So $\Osclln(\pmb u'\restr\Cal S\wedge\sigma)\le\epsilon\chi 1$
for any $\sigma\in\Cal S_0$ (618Da again).

\medskip

{\bf (f)} As $\Cal S_0\subseteq\Cal S$ covers $\Cal S$, they have the same
covered envelope, and we have a
unique fully adapted process $\tildepmb u$ with domain $\Cal S$ extending
$\tildepmb u'$.
Since $\Cal S_0$ is a covering ideal of $\Cal S$,
$\tildepmb u$ is a local martingale.
$\pmb u-\tildepmb u$ is locally of bounded variation
because $(\pmb u-\tildepmb u)\restr\Cal S_0$ is locally of bounded
variation and we can apply 614Q(b-v).   Finally,

\Centerline{$
\sup_{\tau\in\Cal S}\Osclln(\tildepmb u\restr\Cal S\wedge\tau)
=\sup_{\sigma\in\Cal S'}\Osclln(\tildepmb u'\restr\Cal S_0\wedge\sigma)
\le\epsilon\chi 1$}

\noindent by 618Lc.
}%end of proof of 643L

\leader{643M}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$
with a least element, and $\pmb v=\sfamilysS{v}$ a \sm.
Then for any $\epsilon>0$ there is a local martingale
$\tildepmb v=\sfamilysS{\tilde v}$ such that
$\sup_{\tau\in\Cal S}\Osclln(\tildepmb v\restr\Cal S\wedge\tau)
\le\epsilon\chi 1$ and
$\pmb v-\tildepmb v$ is locally of bounded variation.

\proof{{\bf (a)} By 611Ob, $\Cal S$ is finitely full.
Because the sum of two processes which are locally of
bounded variation is again locally of bounded variation (614Q(b-iii)),
it is enough to consider the case in which $\pmb v$ is a virtually
local martingale, in which case it will be a local martingale and
locally near-simple (632I).   
Let $\Cal S_0$ be a covering ideal of $\Cal S$ such that
$\pmb v\restr\Cal S_0$ is a martingale.   By 627N, there are a
non-decreasing sequence $\sequencen{\tau_n}$ in $\Cal S_0$ and a
non-decreasing sequence $\sequencen{d_n}$ in $\frak A$ such that

\Centerline{$d_n\in\frak A_{\tau_n}$,
\quad$d_n\Bsubseteq\Bvalue{\tau_{n+1}=\tau_n}$}

\noindent for every $n\in\Bbb N$, and

\Centerline{$\sup_{n\in\Bbb N}d_n\Bcup\Bvalue{\tau\le\tau_n}=1$,
\quad$v_{\tau}=\lim_{n\to\infty}v_{\tau\wedge\tau_n}$}

\noindent for every $\tau\in\Cal S$.

\medskip

{\bf (b)} Set $\Cal S'=\bigcup_{n\in\Bbb N}\Cal S\wedge\tau_n$.
Then $\pmb v\restr\Cal S'$ is a martingale.   By 643L, there is a local
martingale $\tildepmb u=\family{\sigma}{\Cal S'}{\tilde u_{\sigma}}$
such that
$\Osclln(\tildepmb u\restr\Cal S'\wedge\sigma)\le\epsilon\chi 1$
for every $\sigma\in\Cal S'$ and
$\pmb v-\tildepmb u$ is locally of bounded variation.

\medskip

{\bf (c)} By 627O there are processes
$\tildepmb v=\family{\tau}{\Cal S}{\tilde v_{\tau}}$ and
$\pmb v^{\#}=\family{\tau}{\Cal S}{v^{\#}_{\tau}}$ such that

\Centerline{$\tilde v_{\tau}
=\lim_{n\to\infty}\tilde u_{\tau\wedge\tau_n}$,
\quad$v^{\#}_{\tau}
=\lim_{n\to\infty}v_{\tau\wedge\tau_n}-\tilde u_{\tau\wedge\tau_n}$}

\noindent for every $\tau\in\Cal S$.   Observe that

\Centerline{$\tilde v_{\tau}+v^{\#}_{\tau}
=\lim_{n\to\infty}v_{\tau\wedge\tau_n}
=v_{\tau}$}

\noindent for every $\tau\in\Cal S$.   Now 627Od tells us that
$\tildepmb v$ is a local martingale;  627Oe tells us that
$\pmb v-\tildepmb v=\pmb v^{\#}$ is locally of bounded variation;  and
627Of tells us that
$\Osclln(\tildepmb v\restr\Cal S\wedge\tau)\le\epsilon\chi 1$ for every
$\tau\in\Cal S$.   So $\tildepmb v$ has the required properties.
}%end of proof of 643M

\cmmnt{\medskip

\noindent{\bf Remark} Recall that a \sm\ is
the same thing as a local integrator (627Q).
}

\leader{643N}{}\cmmnt{ I have more than once noted (622P, 624H, 626V)
that $L^2$-martingales have
a special place in the theory, and later in this chapter they will become of
prime importance.   We are now ready for a general result showing why.

\medskip

\noindent}{\bf Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a sublattice of $\Cal T$ with a greatest
element, and $\pmb v=\sfamilysS{v}$ an integrator.   Then for every
$\epsilon>0$ there are an $L^{\infty}$-martingale
$\tildepmb v$ and a process
$\pmb v'$ of bounded variation, both with domain $\Cal S$, such that
$\bar\mu\Bvalue{\pmb v\ne\tildepmb v+\pmb v'}\le\epsilon$.

\proof{{\bf (a)} By 627J/627Q, $\pmb v$ is a \sm;  let
$\pmb v_1=\family{\sigma}{\Cal S}{v_{1\sigma}}$ be a
virtually local martingale such that $\pmb v-\pmb v_1$ is locally
of bounded variation.   As $\Cal S$ has a greatest element,
$\pmb v-\pmb v_1$ is actually of bounded variation.   Let
$\hatpmb v_1=\family{\sigma}{\hat\Cal S}{\hat v_{1\sigma}}$
be the fully adapted extension of $\pmb v_1$ to the
covered envelope $\hat\Cal S$ of $\Cal S$.
Then there is a non-empty
downwards-directed set $A\subseteq\hat\Cal S$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\max\Cal S}\le\bover13\epsilon$ and
$\pmb v_2=R_A(\hatpmb v_1)$, as defined in 623B,is a martingale.
Set $b=\sup_{\rho\in A}\Bvalue{\rho<\max\Cal S}$, so that
$\bar\mu b\le\bover13\epsilon$.   Note that
$\max\Cal S$ is also the greatest element of $\hat\Cal S$ (611M(b-ii)).
If $\sigma\in\hat\Cal S$ and $\rho\in A$, then

\Centerline{$1\Bsetminus b\Bsubseteq\Bvalue{\rho=\max\Cal S}
\Bsubseteq\Bvalue{\sigma\le\rho}
\Bsubseteq\Bvalue{\hat v_{1,\sigma\wedge\rho}=\hat v_{1\sigma}}$;}

\noindent taking the limit as $\rho\downarrow A$,

\Centerline{$1\Bsetminus b\Bsubseteq
\Bvalue{\hat v_{1,\sigma}
         =\lim_{\rho\downarrow A}\hat v_{1,\sigma\wedge\rho}}$,
\quad$\Bvalue{\hat v_{1,\sigma}
        \ne\lim_{\rho\downarrow A}\hat v_{1,\sigma\wedge\rho}}
            \Bsubseteq b$,}

\noindent As $\sigma$ is arbitrary,
$\Bvalue{\pmb v_2\ne\hatpmb v_1}\Bsubseteq b$.

Express $\pmb v_2$ as $\family{\sigma}{\hat\Cal S}{v_{2\sigma}}$.
Consider the martingale $\pmb v_3=\pmb Pv_{2,\max\Cal S}$ (622F).
This extends
$\pmb v_2$ to $\Cal T$.   By 643L there is a local martingale
$\pmb w=\family{\sigma}{\Cal T}{w_{\sigma}}$
such that $\Osclln(\pmb w)\le\chi 1$ and
$\pmb v_3-\pmb w$ is of bounded variation;  replacing $\pmb w$
by $\pmb w-w_{\min\Cal T}\tbf 1$ if necessary. we can arrange that
$w_{\min\Cal T}=0$.
Now there is a $\tau'\in\Cal T$ such that $\pmb w\restr\Cal T\wedge\tau'$
is a martingale and $b'=\Bvalue{\tau'<\max\Cal T}$ has measure at most
$\bover13\epsilon$.   Also $\pmb w$ is near-simple (632Ia again).   By
631Ra again, $\SL_1(\pmb w)$ is true and there is a non-decreasing sequence
$\sequence{i}{\tau_i}$ in $\Cal T$ such that $\tau_0=\min\Cal T$,
$\inf_{i\in\Bbb N}\Bvalue{\tau_i<\max\Cal T}=0$ and
$\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{|w_{\sigma}-w_{\tau_i}|<1}$ whenever
$i\in\Bbb N$ and $\sigma\in[\tau_i,\tau_{i+1}]$.   Accordingly
$|w_{\tau_{n+1}}-w_{\tau_n}|\le 2\chi 1$ whenever $n\in\Bbb N$.
\Prf\ Looking at the definition in 641Ea, we see that
$|w_{I<\tau_{n+1}}-w_{\tau_n}|\times\chi\Bvalue{\tau_n<\tau_{n+1}}\le\chi 1$
for every
finite sublattice $I$ of $\Cal S$ containing $\tau_n$ and $\tau_{n+1}$,
so that
$|w_{<\tau_{n+1}}-w_{\tau_n}|\times\chi\Bvalue{\tau_n<\tau_{n+1}}\le\chi 1$;
while
$|w_{\tau_{n+1}}-w_{<\tau_{n+1}}|\times\chi\Bvalue{\tau_n<\tau_{n+1}}
\le\Osclln(\pmb v\restr\Cal S\wedge\tau_{n+1})$,
by 641Na.   So

$$\eqalign{|w_{\tau_{n+1}}-w_{\tau_n}|\times\chi\Bvalue{\tau_n<\tau_{n+1}}&
\le\chi 1+\Osclln(\pmb w\restr\Cal S\wedge\tau_{n+1})\cr&
\le\chi 1+\Osclln(\pmb w)
\le 2\chi 1\cr}$$

\noindent (using 618D(b-i)).   It follows at once that
$\|w_{\tau_{n+1}}-w_{\tau_n}\|_{\infty}\le 2$.\ \Qed

As $w_{\tau_0}=0$,
$\|w_{\tau_n}\|_{\infty}\le 2n$ for every $n\in\Bbb N$.
Next, there is an $n\in\Bbb N$ such that $b''=\Bvalue{\tau_n<\max\Cal T}$ has
measure at most $\bover13\epsilon$.   Try

\Centerline{$\tildepmb v=\pmb Pw_{\tau_n\wedge\tau'}\restr\Cal S$,
\quad$\pmb v'=\pmb v_1-\pmb v+(\pmb w-\pmb v_3)\restr\Cal S$,
\quad$c=b\Bcup b'\Bcup b''$.}

\noindent Then $\tildepmb v$ is an $L^{\infty}$-martingale, $\pmb v'$ is of
bounded variation and $\bar\mu c\le\epsilon$.   We have

$$\eqalignno{\Bvalue{\tildepmb v\ne\pmb w\restr\Cal S}&
\Bsubseteq\Bvalue{\tau_n\wedge\tau'<\max\Cal S}
\Displaycause{because $\pmb w\restr\Cal T\wedge\tau_n\wedge\tau'$ is a
martingale}
\Bsubseteq\Bvalue{\tau_n<\max\Cal T}\Bcap\Bvalue{\tau'<\max\Cal T}
\Bsubseteq b'\Bcup b''.\cr}$$

\noindent Consequently

$$\eqalignno{\Bvalue{\tildepmb v\ne\pmb v+\pmb v'}&
\Bsubseteq\Bvalue{\tildepmb v\ne\pmb w\restr\Cal S}
    \Bcup\Bvalue{\pmb w\restr\Cal S\ne\pmb v_1+(\pmb w-\pmb v_3)\restr\Cal S}
  \cr&
\Bsubseteq b'\Bcup b''
    \Bcup\Bvalue{\pmb v_3\restr\Cal S\ne\pmb v_1}
\Bsubseteq b'\Bcup b''\Bcup\Bvalue{\pmb v_3\restr\hat\Cal S\ne\hatpmb v_1}
\Bsubseteq b'\Bcup b''\Bcup\Bvalue{\pmb v_2\ne\hatpmb v_1}
\Displaycause{because $\pmb v_3$ extends $\pmb v_2$}
\Bsubseteq b'\Bcup b''\Bcup b
= c\cr}$$

\noindent has measure at most $\epsilon$, as required.
}%end of proof of 643N

\leader{643O}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, and that $\Cal S$ is a non-empty finitely full sublattice of
$\Cal T$ with a greatest member
such that $\inf A\in\Cal S$ for every non-empty $A\subseteq\Cal S$.   If
$\pmb v$ is a near-simple integrator with domain $\Cal S$, there are an
$L^{\infty}$-martingale $\tildepmb v$ and a near-simple process $\pmb v'$ of
bounded variation, both with domain $\Cal S$, such that
$\Bvalue{\pmb v\ne\tildepmb v+\pmb v'}$ has measure at most $\epsilon$.

\proof{ By 643N, we have an $L^{\infty}$-martingale $\tildepmb v$ and a
process $\pmb v_0'$ of
bounded variation, both with domain $\Cal S$, such that
$a=\Bvalue{\pmb v\ne\tildepmb v+\pmb v_0'}$ has measure at most
$\epsilon$.   Express $\pmb v$, $\tildepmb v$ and $\pmb v_0'$ as
$\sfamilysS{v}$, $\sfamilysS{\tilde v}$ and
$\family{\sigma}{\Cal S}{v'_{0\sigma}}$.
By 632Ia once more, $\tildepmb v$ is locally near-simple;
as $\Cal S$ has a greatest member, $\tildepmb v$ is actually near-simple.   Set
$\pmb u=\sfamilysS{u}$ where
$u_{\sigma}=\chi(\upr(1\Bsetminus a,\frak A_{\sigma}))$ for
$\sigma\in\Cal S$, and
$\pmb v'=\pmb u\times(\pmb v-\tildepmb v)$.   Then $\pmb u$ is near-simple, by
632G, so $\pmb v'$ is near-simple, by 631F(a-ii).
Now for any $\sigma\in\Cal S$,

\Centerline{$1\Bsetminus a
   \Bsubseteq\Bvalue{v_{\sigma}-\tilde v_{\sigma}-v'_{0\sigma}=0}
   \in\frak A_{\sigma}$,}

\noindent so

\Centerline{$\Bvalue{u_{\sigma}\ne 0}=\upr(1\Bsetminus a,\frak A_{\sigma})
   \Bsubseteq\Bvalue{v_{\sigma}-\tilde v_{\sigma}-v'_{0\sigma}=0}$}

\noindent and $\pmb u\times(\pmb v-\tildepmb v-\pmb v'_0)=0$, that is,
$\pmb v'=\pmb u\times\pmb v'_0$.   By 614Q(a-ii), $\pmb v'$ is of bounded
variation.   Finally,

$$\eqalign{\Bvalue{\pmb v\ne\tildepmb v+\pmb v'}&
\Bsubseteq\Bvalue{\pmb v\ne\tildepmb v+\pmb v_0'}
    \Bcup\Bvalue{\pmb v_0'\ne\pmb v'}
\Bsubseteq a\Bcup\Bvalue{\pmb u\ne\tbf 1}\cr&
=a\Bcup\sup_{\sigma\in\Cal S}(1\Bsetminus\upr(1\Bsetminus a,\frak A_{\sigma})
\Bsubseteq a\Bcup\sup_{\sigma\in\Cal S}(1\Bsetminus(1\Bsetminus a))
=a\cr}$$

\noindent has measure at most $\epsilon$.
}%end of proof of 643O

%\leader{643Q}{}
%\leader{643R}{}
%\leader{643S}{}
%\leader{643T}{}
%\leader{643U}{}
%\leader{643V}{}
%\leader{643W}{}

\exercises{\leader{643X}{Basic exercises (a)}
%\spheader 643Xa
Show that if $t\in T$ then the region of accessibility
of $\check t$ is $1$ if $\{s:s<t\}$ is non-empty and has supremum $t$,
$0$ otherwise.
%643E

\spheader 643Xb
Let $\sequencen{\tau_n}$ be the sequence of jump times associated with
the standard Poisson process, as described in 612U.   Show that the
region of accessibility of $\tau_n$ is $0$ for every $n\ge 1$.
%643E

\spheader 643Xc Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process with a previsible version
$\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$.   Show that whenever
$\sigma$, $\tau\in\Cal S$,
$\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma\ll\tau}
\Bsubseteq\Bvalue{u_{<\tau}=u_{\sigma}}$.
%643I

\leader{643Y}{Further exercises (a)}
%\spheader 643Ya
Suppose that $\tau\in\Cal T$ and that
$C_0,\ldots,C_n$ are non-empty subsets of $\Cal T\wedge\tau$.   Show that there
are $D_0,\ldots,D_n$ such that

\Centerline{$\emptyset\ne D_0\subseteq\ldots\subseteq D_n
\subseteq\Cal T\wedge\tau$,
\quad$\sup_{i\le n}a_{D_i}\supseteq\sup_{i\le n}a_{C_i}$.}
%643E

\spheader 643Yb Show that 643L can be used in place of
621Hf or 628D to show that if the filtration is right-continuous
a martingale is a local integrator;  now use the construction of 633Ya to
see that a martingale is still a local integrator even when the filtration is
not right-continuous.
%643L
}%end of exercises

\endnotes{
\Notesheader{643} This section seems to be hard work, and I do not know of
any route to Theorem 643L which doesn't use most of the ideas here, though
(as usual) my own exposition is designed for line-by-line checking rather
than transparency.   Of course the complications involving regions of
approachability are quite unnecessary if we have a
real-time stochastic integration structure, in which case $\app(\tau)=1$ for
ecery $\tau$, by 643D.

The proof here is derived, at some remove, from
{\smc Protter 05}, Chap.\ III.   I note that while it depends on some deep
ideas about stochastic processes, it uses little
of the theory of integration developed in
\S\S613-617, %\S613 \S614 \S616 \S614 \S617
so could in
principle be used in a proof of Theorem 622H, as suggested in 643Yb.

\leaveitout{In 643F-643I %643F 643H 643I
I make an excursion through the special case in which there are no
times isolated on the right, that is, there are no gaps in the totally ordered
set $T$.   Of course this is the case for which the theory here was devised.
If $T=\Bbb N$ then the results of this section are not elementary, but are very
much easier.   It would be astonishing if there were a
major theorem which was true in both the cases $T=\coint{0,\infty}$ and
$T=\Bbb N$ and which failed in the general context of my presentation.
Nevertheless, I have not found an expression of the ideas in 643F and
643H which does not depend on an absence of gaps in $T$, and I have therefore
devised a method (633Q) for showing that this hypothesis will
be immaterial.}
%this seems to refer to something now deleted \query
}%end of notes

\discrpage

\frfilename{mt644.tex}
\versiondate{25.8.20}
\copyrightdate{2013}

\def\chaptername{The S-integral}
\def\sectionname{Pointwise convergence}
\def\rank{\mathop{\text{rank}}\nolimits}
\def\sm{semi-martingale}

\newsection{644}

It is a remarkable fact that while the Riemann-sum
integral, as defined in \S613, is
not `sequentially smooth' in the most natural adaptation of the definition
in 436A\cmmnt{ (644Xb)}, a variation on this concept (Theorem 644H)
gives us a route to a
Daniell-type integral, which I will develop in \S645.

\cmmnt{\leader{644A}{Notation} I repeat the familiar list.
\Asissix\ is a stochastic integration structure.   $\Cal T_f\subseteq\Cal T$
is the ideal of finite stopping
times.   If $\Cal S$ is a sublattice of $\Cal T$,
$\Cal I(\Cal S)$ is the set of finite sublattices of $\Cal S$ and
$\Cal S\wedge\tau=\{\sigma\wedge\tau:\sigma\in\Cal S\}$,
$\Cal S\vee\tau=\{\sigma\vee\tau:\sigma\in\Cal S\}$ for
$\tau\in\Cal T$.   $\Mmo(\Cal S)$ will be the space of moderately oscillatory
processes with domain $\Cal S$, and $\Mns(\Cal S)$ the space of near-simple
processes.   $L^0$ will be $L^0(\frak A)$, with the topology of
convergence in measure defined by the functional $\theta$ where
$\theta(w)=\Expn(|w|\wedge\chi 1)$ for $w\in L^0$.
If $\Cal S$ is a
sublattice of $\Cal T$ and $\pmb u=\sfamilysS{u}$ is an order-bounded
fully adapted process, I write $\sup\pmb u$ for
$\sup_{\sigma\in\Cal S}u_{\sigma}$;   $\|\pmb u\|_{\infty}$ will be
$\sup_{\sigma\in\Cal S}\|u_{\sigma}\|_{\infty}$;  if $\pmb u$ is
locally moderately oscillatory, $\pmb u_<$ will be its previsible version.
}%end of comment

\leader{644B}{Definitions (a)}\cmmnt{ It will be useful to have a short phrase
for the following.}  Let $\Cal S$ be a sublattice of $\Cal T$.
A family $A$ of processes with domain $\Cal S$ is
{\bf uniformly order-bounded} if $\sup_{\pmb u\in A}\sup|\pmb u|$ is
defined in $L^0$\cmmnt{;  that is, there is a $\bar u\in L^0$ such that
$|u_{\sigma}|\le\bar u$ for every $\sigma\in\Cal S$ whenever
$\sfamilysS{u}\in A$}.

\spheader 644Bb\cmmnt{ Similarly, a special class of
integrators will be prominent in the rest of the chapter.}  If $\Cal S$ is a
sublattice of $\Cal T$, $\Mnsu(\Cal S)$ will be the family of non-negative
non-decreasing near-simple processes with domain $\Cal S$.   Any member of
$\Mnsu(\Cal S)$\cmmnt{, being near-simple,} will be
order-bounded\cmmnt{ (631Ba)};  being
non-decreasing, it will be an integrator\cmmnt{ (616Ra)}.

\leader{644C}{Lemma}\cmmnt{ (The key.)}
Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\sup D\in\Cal S$ whenever $D\subseteq\Cal S$ is countable, non-empty
and bounded above in $\Cal S$.   Let $\sequencen{\pmb u_n}
=\sequencen{\langle u_{n\sigma}\rangle_{\sigma\in\Cal S}}$ be
a non-increasing sequence of non-negative moderately oscillatory
processes such that
$\inf_{n\in\Bbb N}\pmb u_{n<}$, taken in $(L^0)^{\Cal S}$, is zero.   Then
$\inf_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v=0$ for every
$\pmb v\in\Mnsu(\Cal S)$.

\proof{{\bf (a)} To begin with, suppose that $\Cal S$ has greatest and least
elements and that $\pmb v=\sfamilysS{v}$ is jump-free.

\medskip

\quad{\bf (i)} \Quer\ If $w=\inf_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v$ is
non-zero, there is an $\epsilon\in\ocint{0,1}$ such that

\Centerline{$\bar\mu\Bvalue{2\epsilon(v_{\max\Cal S}-v_{\min\Cal S})<w}
  \ge 2\epsilon$.}

\noindent \Prf\ Take $\delta>0$ such that $b=\Bvalue{w>\delta}$ is non-zero,
$M\ge 1$ such that
$\bar\mu\Bvalue{v_{\max\Cal S}-v_{\min\Cal S}\ge M}\le\bover13\bar\mu b$ and set
$\epsilon=\min(\Bover{\bar\mu b}3,\Bover{\delta}{2M})$.\ \Qed

\medskip

\quad{\bf (ii)} Now choose
a non-decreasing sequence $\sequencen{\tau_n}$ in $\Cal S$ and
sequences $\sequencen{w_n}$ in $L^0$, $\sequencen{a_n}$ in $\frak A$
inductively, as follows.   The inductive hypothesis is that
$w_n=\inf_{i\in\Bbb N}\int_{\Cal S\vee\tau_n}\pmb u_id\pmb v$,
$a_n\Bsubseteq\Bvalue{(1+2^{-n})\epsilon(v_{\max\Cal S}-v_{\tau_n})<w_n}$
and $\bar\mu a_n\ge(1+3^{-n})\epsilon$.

\medskip

\qquad\grheada\ Start with $\tau_0=\min\Cal S$, $w_0=w$ and
$a_0=\Bvalue{2\epsilon(v_{\max\Cal S}-v_{\min\Cal S})<w}$.

\medskip

\qquad\grheadb\ Given $\tau_n$, $w_n$ and $a_n$, let
$\delta\in\ocint{0,1}$ be such that

\Centerline{$\Bvalue{v_{\max\Cal S}-v_{\tau_n}>0}
  \Bsetminus\Bvalue{2^{-n-1}\epsilon(v_{\max\Cal S}-v_{\tau_n})>\delta}$}

\noindent has measure at most $3^{-n-1}\epsilon$, and set
$\eta=3^{-n-2}\epsilon\delta$.
Now $\iiv(\pmb u_n)$ is jump-free (618Q).
By 618E, there is a $\tau'_n\in\Cal S\vee\tau_n$ such that
$\Bvalue{\tau_n<\tau'_n}=\Bvalue{\tau_n<\max\Cal S}$ and, setting
$y=\int_{\Cal S\wedge\tau'_n}\pmb u_nd\pmb v
-\int_{\Cal S\wedge\tau_n}\pmb u_nd\pmb v$,
$\theta(y)\le\eta$.

\medskip

\qquad\grheadc\ Now there is an $I\in\Cal I(\Cal S\vee\tau'_n)$, containing
$\tau'_n$ and $\max\Cal S$, such that
$\theta(S_K(\pmb u_n,d\pmb v)-S_{K'}(\pmb u_n,d\pmb v))\le\eta$ whenever
$K$, $K'\in\Cal I(\Cal S\vee\tau'_n)$ include $I$, in which case
$\theta(S_{I\wedge\tau}(\pmb u_n,d\pmb v)
   -\int_{\Cal S\cap[\tau'_n,\tau]}\pmb u_nd\pmb v)\le 2\eta$
for every $\tau\in\Cal S\vee\tau'_n$ (613V(ii-$\beta$)).

Take $(\rho_0,\ldots,\rho_k)$ linearly generating the $I$-cells (611L).
By 611I, as usual, there is a $\tau_{n+1}\in\Cal T$ such that

\Centerline{$\Bvalue{\tau_{n+1}=\rho_i}
\Bsupseteq\Bvalue{u_{n\rho_i}\ge\epsilon}
    \Bsetminus\sup_{j<i}\Bvalue{u_{n\rho_j}\ge\epsilon}$}

\noindent for $i<k$,

\Centerline{$\Bvalue{\tau_{n+1}=\rho_k}
\Bsupseteq 1\Bsetminus\sup_{j<i}\Bvalue{u_{n\rho_j}\ge\epsilon}$,}

\Centerline{$\tau'_n=\rho_0\le\tau_{n+1}\le\rho_k=\max\Cal S$}

\noindent and $\tau_{n+1}\in\Cal S$ because $\Cal S$ is finitely full.
Now

\Centerline{$\Bvalue{\tau_{n+1}<\max\Cal S}
\Bsubseteq\Bvalue{u_{n\tau_{n+1}}\ge\epsilon}$}

\noindent and

\Centerline{$\Bvalue{\rho_i\wedge\tau_{n+1}<\rho_{i+1}\wedge\tau_{n+1}}
\Bsubseteq\Bvalue{\rho_i<\tau_{n+1}}
\Bsubseteq\sup_{i<j\le k}\Bvalue{\tau_{n+1}=\rho_j}
\Bsubseteq\Bvalue{u_{n,\rho_i\wedge\tau_n}<\epsilon}$}

\noindent for every $i<k$.   Accordingly

$$\eqalignno{S_{I\wedge\tau_{n+1}}(\pmb u_n,d\pmb v)&
=\sum_{i=0}^{k-1}u_{n,\rho_i\wedge\tau_{n+1}}
   \times(v_{\rho_{i+1}\wedge\tau_{n+1}}-v_{\rho_i\wedge\tau_{n+1}})
\Displaycause{because $(\rho_0\wedge\tau_{n+1},\ldots,\rho_k\wedge\tau_{n+1})$
linearly generates the $I\wedge\tau_{n+1}$-cells, by 611Kg}
\le\sum_{i=0}^{k-1}
   \epsilon(v_{\rho_{i+1}\wedge\tau_{n+1}}-v_{\rho_i\wedge\tau_{n+1}})
=\epsilon(v_{\tau_{n+1}}-v_{\tau'_n}).\cr}$$

\noindent Setting

\Centerline{$z=(\int_{\Cal S\cap[\tau'_n,\tau_{n+1}]}\pmb u_nd\pmb v
                       -\epsilon(v_{\tau_{n+1}}-v_{\tau'_n}))^+
\le(\int_{\Cal S\cap[\tau'_n,\tau_{n+1}]}\pmb u_nd\pmb v
                       -S_{I\wedge\tau_{n+1}}(\pmb u_n,d\pmb v))^+$,}

\noindent $\theta(z)\le 2\eta$, so $\theta(y+z)\le 3\eta$, while

\Centerline{$\int_{\Cal S\cap[\tau_n.\tau_{n+1}]}\pmb u_nd\pmb v
\le y+z+\epsilon(v_{\tau_{n+1}}-v_{\tau'_n})$.}

\medskip

\qquad\grheadd\ Set
$a_{n+1}=a_n\Bcap\Bvalue{y+z\le 2^{-n-1}\epsilon(v_{\max\Cal S}-v_{\tau_n})}$.
Then

$$\eqalignno{\bar\mu(a_n\Bsetminus a_{n+1})&
\le\bar\mu(\Bvalue{v_{\max\Cal S}-v_{\tau_n}>0}
   \Bcap\Bvalue{y+z\ge 2^{-n-1}\epsilon(v_{\max\Cal S}-v_{\tau_n})}
\Displaycause{because $a_n\Bsubseteq\Bvalue{w_n>0}
\Bsubseteq\Bvalue{v_{\max\Cal S}-v_{\tau_n}>0}$}
\le\bar\mu(\Bvalue{v_{\max\Cal S}-v_{\tau_n}>0}
        \Bcap\Bvalue{2^{-n-1}\epsilon(v_{\max\Cal S}-v_{\tau_n})\le\delta})
   \cr&
\mskip100mu +\bar\mu\Bvalue{y+z\ge\delta}\cr&
\le 3^{-n-1}\epsilon+\bar\mu\Bvalue{y+z\ge\delta}
\le 2\cdot 3^{-n-1}\epsilon\cr}$$

\noindent because $\theta(y+z)\le 3^{-n-1}\epsilon\delta$.   Accordingly
$\bar\mu a_{n+1}\ge(1+3^{-n-1})\epsilon$.

\medskip

\qquad\grheade\ For any $i\ge n$,

$$\eqalign{\int_{\Cal S\vee\tau_{n+1}}\pmb u_id\pmb v&
=\int_{\Cal S\vee\tau_n}\pmb u_id\pmb v
   -\int_{\Cal S\cap[\tau_n,\tau_{n+1}]}\pmb u_id\pmb v\cr&
\ge w_n-\int_{\Cal S\cap[\tau_n,\tau_{n+1}]}\pmb u_nd\pmb v
\ge w_n-y-z-\epsilon(v_{\tau_{n+1}}-v_{\tau'_n}).\cr}$$

\noindent So, setting
$w_{n+1}=\inf_{i\in\Bbb N}\int_{\Cal S\vee\tau_{n+1}}\pmb u_id\pmb v$,

\Centerline{$w_{n+1}\ge w_n-y-z-\epsilon(v_{\tau_{n+1}}-v_{\tau'_n})
\ge w_n-y-z-\epsilon(v_{\tau_{n+1}}-v_{\tau_n})$.}

\noindent Now

$$\eqalign{a_{n+1}&
\Bsubseteq\Bvalue{(1+2^{-n})\epsilon(v_{\max\Cal S}-v_{\tau_n})<w_n}
              \Bcap\Bvalue{y+z\le 2^{-n-1}\epsilon(v_{\max\Cal S}-v_{\tau_n}})
    \cr&
\Bsubseteq\Bvalue{(1+2^{-n-1})\epsilon(v_{\max\Cal S}-v_{\tau_n})
                       <w_n-y-z}\cr&
\Bsubseteq\Bvalue{(1+2^{-n-1})\epsilon(v_{\max\Cal S}-v_{\tau_{n+1}})
                       <w_n-y-z-\epsilon(v_{\tau_{n+1}}-v_{\tau_n})}\cr&
\Bsubseteq\Bvalue{(1+2^{-n-1})\epsilon(v_{\max\Cal S}-v_{\tau_{n+1}})<w_{n+1}},
\cr}$$

\noindent and the induction proceeds.

\medskip

\quad{\bf (iii)} At the end of the induction,
set $\tau=\sup_{n\in\Bbb N}\tau_n$;  by hypothesis, $\tau\in\Cal S$.
Since $\sequencen{\tau_n}$ is non-decreasing, $\sequencen{w_n}$ and
$\sequencen{v_{\max\Cal S}-v_{\tau_n}}$ are non-increasing.   The construction
in (ii) above arranged that $a_{n+1}\Bsubseteq a_n$ and
$\bar\mu a_n\ge\epsilon$ for every $n$.   At the same time we have

\Centerline{$a_n\Bsubseteq\Bvalue{w_n>0}\Bsubseteq\Bvalue{\tau_n<\max\Cal S}
=\Bvalue{\tau_n<\tau'_n}\Bsubseteq\Bvalue{\tau_n<\tau}$}

\noindent and

\Centerline{$a_{n+1}\Bsubseteq\Bvalue{u_{n\tau_{n+1}}\ge\epsilon}
\Bsubseteq\Bvalue{u_{i\tau_{n+1}}\ge\epsilon}$}

\noindent whenever $n\ge i$.   So for any $i\in\Bbb N$,
$\langle\tau_n\rangle_{n>i}$ is a non-decreasing sequence with supremum $\tau$
and

\Centerline{$a\Bsubseteq\Bvalue{\tau_n<\tau}
\Bcap\Bvalue{u_{i\tau_n}\ge\epsilon}$}

\noindent for every $n>i$.   By 641M or otherwise,
$a\Bsubseteq\Bvalue{u_{i<\tau}\ge\epsilon}$;  as $i$ is arbitrary,
$a\Bsubseteq\Bvalue{\inf_{i\in\Bbb N}u_{i<\tau}\ge\epsilon}$.   But also we have

\Centerline{$\bar\mu a=\inf_{n\in\Bbb N}\bar\mu a_n\ge\epsilon>0$}

\noindent so $\inf_{i\in\Bbb N}u_{i<\tau}\ne 0$, contrary to hypothesis.\ \Bang

\medskip

\quad{\bf (iv)} Thus we see that if $\Cal S$ has greatest and least elements and
$\pmb v$ is jump-free,
$\inf_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v=0$.

\medskip

{\bf (b)} Now suppose only that $\pmb v\in\Mnsu(\Cal S)$, while still assuming
that $\Cal S$ has greatest and least elements.

\medskip

\quad{\bf (i)} Consider first the case in which $\pmb v$ is actually simple.
Let $(\tau_0,\ldots,\tau_m)$ be a breakpoint string for
$\pmb v$ starting from $\tau_0=\min\Cal S$ and finishing with
$\tau_m=\max\Cal S$.   Then

\Centerline{$\int_{\Cal S}\pmb u_n\,d\pmb v
=\sum_{j=0}^{m-1}u_{n<\tau_{j+1}}
   \times(v_{\tau_{j+1}}-v_{\tau_j})$}

\noindent for every $n\in\Bbb N$ (641J).   So

\Centerline{$\lim_{n\to\infty}\int_{\Cal S}\pmb u_n\,d\pmb v
=\sum_{j=0}^{m-1}\lim_{n\to\infty}u_{n<\tau_{j+1}}
   \times(v_{\tau_{j+1}}-v_{\tau_j})=0$.}

\medskip

\quad{\bf (ii)} If $\pmb v$ is just near-simple, take $\epsilon>0$.   Set
$\bar u=\sup|\pmb u_0|$.   Let
$\delta>0$ be such that $\theta(v\times\bar u)\le\epsilon$ whenever
$\theta(v)\le\delta$.   By 631U, there are non-negative non-decreasing
processes $\pmb v'$, $\pmb w$, $\pmb v''$, all with domain $\Cal S$,
such that $\pmb v'$ is simple,
$\pmb w$ is jump-free, $\pmb v=\pmb v'+\pmb w+\pmb v''$ and
$\theta(\sup\pmb v'')\le\delta$.   In this case, we see that

$$\eqalignno{\int_{\Cal S}\pmb u_n\,d\pmb v''&
\le\sup_{I\in\Cal I(\Cal S)}S_I(\pmb u_n,d\pmb v'')
\le\bar u\times\sup_{I\in\Cal I(\Cal S)}S_I(\tbf 1,d\pmb v'')
\Displaycause{because, expressing $\pmb v''$ as $\sfamilysS{v''}$,
$u_{n\sigma}\times(v''_{\tau}-v''_{\sigma})
\le\bar u\times(v''_{\tau}-v''_{\sigma})$ whenever $\sigma\le\tau$ in $\Cal S$}
\le\bar u\times\sup\pmb v''\cr}$$

\noindent and $\theta(\sup\pmb v'')\le\delta$, so that
$\theta(\int_{\Cal S}\pmb u_n\,d\pmb v'')\le\epsilon$, for every
$n\in\Bbb N$.   From (a) above
we know that $\lim_{n\to\infty}\int_{\Cal S}\pmb u_n\,d\pmb w=0$,
and from (i) here we see that
$\lim_{n\to\infty}\int_{\Cal S}\pmb u_n\,d\pmb v'=0$.
So

\Centerline{$\limsup_{n\to\infty}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)
=\limsup_{n\to\infty}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v'')
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lim_{n\to\infty}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)=0$.

\medskip

{\bf (c)} Finally, for the general case,
if $\Cal S$ is empty the result is of course
trivial.   Otherwise, given $\epsilon>0$, there are $\tau$, $\tau'\in\Cal S$
such that $\tau\le\tau'$,
$\theta(\int_{\Cal S\wedge\tau}\pmb u_0d\pmb v)\le\epsilon$ and
$\theta(\int_{\Cal S\vee\tau'}\pmb u_0d\pmb v)\le\epsilon$.   Consequently
$\theta(\int_{\Cal S\wedge\tau}\pmb u_nd\pmb v)\le\epsilon$ and
$\theta(\int_{\Cal S\vee\tau'}\pmb u_nd\pmb v)\le\epsilon$ for every $n$, and

\Centerline{$\inf_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_nd\pmb v)
\le 2\epsilon+\inf_{n\in\Bbb N}\theta(\int_{\Cal S'}\pmb u_nd\pmb v)
=2\epsilon+\theta(\inf_{n\in\Bbb N}\int_{\Cal S'}\pmb u_nd\pmb v)$}

\noindent where $\Cal S'=\Cal S\cap[\tau,\tau']$.   If $n\in\Bbb N$,
$\pmb u_n\restr\Cal S'$ is non-negative and moderately oscillatory
(615F(a-i)), and
$0\le(\pmb u_n\restr\Cal S')_<\le\pmb u_{n<}\restr\Cal S'$ (641Gc).   So
$\inf_{n\in\Bbb N}(\pmb u_n\restr\Cal S')_<=0$, while of course
$\sequencen{\pmb u_n\restr\Cal S'}$ is non-increasing.   As
$\pmb v\restr\Cal S'=\pmb v\restr(\Cal S\vee\tau)\wedge\tau'$
is non-decreasing and near-simple (631F(a-iv)),
(b) above tells us that
$\inf_{n\in\Bbb N}\int_{\Cal S'}\pmb u_nd\pmb v=0$.   Consequently
$\theta(\inf_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v)\le 2\epsilon$;  as
$\epsilon$ is arbitrary, $\inf_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v=0$,
and the proof is complete.
}%end of proof of 644C

\leader{644D}{Lemma} Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\sup D\in\Cal S$ whenever $D\subseteq\Cal S$ is countable, non-empty
and bounded above in $\Cal S$.
Let $\sequencen{\pmb u_n}
=\sequencen{\langle u_{n\sigma}\rangle_{\sigma\in\Cal S}}$ be a
uniformly order-bounded sequence of moderately oscillatory processes such that
$\sequencen{\pmb u_{n<}}$ is order*-convergent in
$(L^0)^{\Cal S}$\cmmnt{ (definition:  642B)}.
If $\pmb v\in\Mnsu(\Cal S)$, $\lim_{n\to\infty}\int_{\Cal S}\pmb u_nd\pmb v$ is
defined in $L^0$ for the topology of convergence in measure.

\proof{ Let $\epsilon>0$.

\medskip

{\bf (a)} Writing $\bar u$ for
$\sup_{n\in\Bbb N,\sigma\in\Cal S}|u_{n\sigma}|$, we have

\Centerline{$|\int_{\Cal S}\pmb u\,d\pmb v|
\le\sup_{I\in\Cal I(\Cal S)}|S_I(\pmb u,d\pmb v)|
\le\bar u\times\sup_{I\in\Cal I(\Cal S)}S_I(\tbf 1,d\pmb v)
=\bar u\times\int_{\Cal S}d\pmb v$}

\noindent whenever $\pmb u$ is a moderately oscillatory
process with domain $\Cal S$ and
$\sup|\pmb u|\le\bar u$.   In particular, if $n\in\Bbb N$, the
monotonic sequences
$\langle\int_{\Cal S}
\inf_{n\le i\le m}\pmb u_id\pmb v\rangle_{m\ge n}$,
$\langle\int_{\Cal S}
\sup_{n\le i\le m}\pmb u_id\pmb v\rangle_{m\ge n}$ are
order-bounded in
$L^0$ and have limits $\bar w_n$, $\bar z_n$ respectively in $L^0$ (613Ba).
If we take $k_n\ge n$ such that

\Centerline{$\theta(\int_{\Cal S}\inf_{n\le i\le k_n}\pmb u_id\pmb v-\bar w_n)
\le 2^{-n}\epsilon,
\quad\theta(\bar z_n-\int_{\Cal S}\sup_{n\le i\le k_n}\pmb u_id\pmb v)
\le 2^{-n}\epsilon$,}

\noindent and set $\pmb w_n=\inf_{n\le i\le k_n}\pmb u_i$,
$\pmb z_n=\sup_{n\le i\le k_n}\pmb u_i$, then for any $m\ge n$

$$\eqalign{\theta(\int_{\Cal S}(\pmb w_n-\pmb u_m)^+d\pmb v)&
\le\theta(\int_{\Cal S}(\pmb w_n-\inf_{i\le\max(m,k_n)}\pmb u_i)d\pmb v)\cr&
\le\theta(\int_{\Cal S}\pmb w_nd\pmb v-\bar w_n)
\le 2^{-n}\epsilon\cr}$$

\noindent and similarly

\Centerline{$\theta(\int_{\Cal S}(\pmb u_m-\pmb z_n)^+d\pmb v)
\le 2^{-n}\epsilon$.}

\medskip

{\bf (b)}
Define $\sequencen{\pmb x_n}$, $\sequencen{\pmb y_n}$ inductively by setting
$\pmb x_0=\pmb w_0$, \quad$\pmb y_0=\pmb z_0$ and

\Centerline{$\pmb x_{n+1}=\med(\pmb x_n,\pmb w_{n+1},\pmb y_n)$,
\quad$\pmb y_{n+1}=\med(\pmb x_n,\pmb z_{n+1},\pmb y_n)$}

\noindent for every $n\in\Bbb N$.   Because $\pmb w_n\le\pmb u_n\le\pmb z_n$ for
each $n$, $\pmb x_n\le\pmb x_{n+1}\le\pmb y_{n+1}\le\pmb y_n$ for each $n$.
We also have $\pmb y_n-\pmb x_n\le\pmb z_n-\pmb w_n$ for every $n\in\Bbb N$.
\Prf\ If $n=0$ this is immediate.   For $n>0$,

$$\eqalign{\pmb y_n-\pmb x_n&
=(\pmb x_{n-1}\vee \pmb z_n)\wedge\pmb y_{n-1}
   -(\pmb x_{n-1}\vee \pmb w_n)\wedge\pmb y_{n-1}\cr&
\le(\pmb x_{n-1}\vee \pmb w_n+\pmb x_{n-1}\vee \pmb z_n-\pmb x_{n-1}\vee \pmb w_n)
    \wedge(\pmb y_{n-1}+\pmb x_{n-1}\vee \pmb z_n-\pmb x_{n-1}\vee \pmb w_n)\cr&
\mskip100mu
    -(\pmb x_{n-1}\vee \pmb w_n)\wedge \pmb y_{n-1}\cr&
=\pmb x_{n-1}\vee \pmb z_n-\pmb x_{n-1}\vee \pmb w_n\cr&
\le(\pmb x_{n-1}+\pmb z_n-\pmb w_n)\vee(\pmb w_n+\pmb z_n-\pmb w_n)-\pmb x_{n-1}\vee \pmb w_n
=\pmb z_n-\pmb w_n.  \text{ \Qed}\cr}$$

\medskip

{\bf (c)}
Next, $\theta(\int_{\Cal S}(\pmb x_n-\pmb u_m)^+d\pmb v)\le(2-2^{-n})\epsilon$
whenever $n\le m$.   \Prf\ Induce on $n$.   For $n=0$ we have

\Centerline{$\theta(\int_{\Cal S}(\pmb x_0-\pmb u_m)^+d\pmb v)
=\theta(\int_{\Cal S}(\pmb w_0-\pmb u_m)^+d\pmb v)\le\epsilon$}

\noindent by (a).   For the inductive step to $n+1\le m$, we have
$\pmb x_{n+1}\le\pmb x_n\vee\pmb w_{n+1}$ so

$$\eqalign{\theta(\int_{\Cal S}(\pmb x_{n+1}-\pmb u_m)^+d\pmb v)&
\le\theta(\int_{\Cal S}((\pmb x_n-\pmb u_m)^+
   +(\pmb w_{n+1}-\pmb u_m)^+)d\pmb v)\cr&
\le\theta(\int_{\Cal S}(\pmb x_n-\pmb u_m)^+d\pmb v)
   +\theta(\int_{\Cal S}(\pmb w_{n+1}-\pmb u_m)^+d\pmb v)\cr&
\le(2-2^{-n})\epsilon+2^{-n-1}\epsilon
=(2-2^{-n-1})\epsilon.  \text{ \Qed}\cr}$$

\noindent Similarly,
$\theta(\int_{\Cal S}(\pmb u_m-\pmb y_n)^+d\pmb v)\le(2-2^{-n})\epsilon$
whenever $n\le m$.

\medskip

{\bf (d)} Set $\pmb u'_n=\med(\pmb x_n,\pmb u_n,\pmb y_n)$ for $n\in\Bbb N$.
Then

\Centerline{$(\pmb u_n-\pmb u'_n)^+\le\pmb u_n-\pmb u_n\wedge\pmb y_n
=(\pmb u_n-\pmb y_n)^+$}

\noindent and similarly $(\pmb u'_n-\pmb u_n)^+\le(\pmb x_n-\pmb u_n)^+$, so

$$\eqalign{\theta(\int_{\Cal S}\pmb u_nd\pmb v-\int_{\Cal S}\pmb u'_nd\pmb v)&
\le\theta(\int_{\Cal S}|\pmb u_n-\pmb u'_n|d\pmb v)
=\theta(\int_{\Cal S}((\pmb u_n-\pmb u'_n)^++(\pmb u'_n-\pmb u_n)^+)d\pmb v)\cr&
\le\theta(\int_{\Cal S}(\pmb u_n-\pmb y_n)^+d\pmb v)
  +\theta(\int_{\Cal S}(\pmb x_n-\pmb u_n)^+d\pmb v)
\le 4\epsilon\cr}$$

\noindent for every $n$.

\medskip

{\bf (e)} Observe next that all the $\pmb w_n$, $\pmb z_n$, $\pmb x_n$ and
$\pmb y_n$ are moderately oscillatory processes, and that
$\sequencen{\pmb y_n-\pmb x_n}$ is a non-increasing sequence of non-negative
moderately oscillatory processes.   Now for every $n$ we have

\Centerline{$\inf_{m\ge n}\pmb u_{m<}
\le\inf_{n\le i\le k_n}\pmb u_{m<}
=(\inf_{n\le i\le k_n}\pmb u_m)_<
=\pmb w_{n<}$}

\noindent because $\pmb u\mapsto\pmb u_<$ is a Riesz homomorphism (641G(e-i)),
and similarly $\pmb z_{n<}\le\sup_{m\ge n}u_{m<}$.   Consequently

$$\eqalignno{\inf_{n\in\Bbb N}(\pmb y_{n<}-\pmb x_{n<})&
\le\inf_{n\in\Bbb N}(\pmb z_{n<}-\pmb w_{n<})
\Displaycause{by (b)}
\le\inf_{n\in\Bbb N}(\sup_{m\ge n}\pmb u_{m<}-\inf_{m\ge n}\pmb u_{m<})
=0\cr}$$

\noindent because $\sequencen{\pmb u_{n<}}$ is order*-convergent,
so $\inf_{n\in\Bbb N}\sup_{m\ge n}\pmb u_{m<}
=\sup_{n\in\Bbb N}\inf_{m\ge n}\pmb u_{m<}$ (642B).

\medskip

{\bf (f)} By 644C,
$\lim_{n\to\infty}\int_{\Cal S}(\pmb y_n-\pmb x_n)d\pmb v=0$.   As $\pmb v$ is
non-decreasing, $\pmb u\mapsto\int_{\Cal S}\pmb u\,d\pmb v$ is a positive linear
operator, so $\lim_{n\to\infty}\int_{\Cal S}\pmb y_nd\pmb v$
and $\lim_{n\to\infty}\int_{\Cal S}\pmb x_nd\pmb v$ exist and are equal.
Since $\pmb x_n\le\pmb u'_n\le\pmb y_n$ for every $n$, the common limit of these
integrals is also $\lim_{n\to\infty}\int_{\Cal S}\pmb u'_nd\pmb v$.
Accordingly

$$\eqalign{&\limsup_{m,n\to\infty}
    \theta(\int_{\Cal S}\pmb u_md\pmb v-\int_{\Cal S}\pmb u_nd\pmb v)\cr&
\mskip50mu
\le\sup_{m\in\Bbb N}
    \theta(\int_{\Cal S}\pmb u_md\pmb v-\int_{\Cal S}\pmb u'_md\pmb v)
  +\limsup_{m,n\to\infty}
     \theta(\int_{\Cal S}\pmb u'_md\pmb v-\int_{\Cal S}\pmb u'_nd\pmb v)\cr&
\mskip150mu
  +\sup_{n\in\Bbb N}
     \theta(\int_{\Cal S}\pmb u'_nd\pmb v-\int_{\Cal S}\pmb u_nd\pmb v)\cr&
\mskip50mu
\le 4\epsilon+0+4\epsilon
=8\epsilon\cr}$$

\noindent by (d).
As $\epsilon$ is arbitrary, $\sequencen{\int_{\Cal S}\pmb u_nd\pmb v}$
is Cauchy, therefore convergent.
}%end of proof of 644D

\leader{644E}{Corollary} Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\sup D\in\Cal S$ whenever $D\subseteq\Cal S$ is countable, non-empty
and bounded above in $\Cal S$,
and $\sequencen{\pmb u_n}$ a uniformly order-bounded
sequence of moderately oscillatory processes with domain $\Cal S$
such that $\sequencen{\pmb u_{n<}}$ is order*-convergent to $\tbf 0$ in
$(L^0)^{\Cal S}$.
Then $\lim_{n\to\infty}\int_{\Cal S}\pmb u_nd\pmb v=0$
for $\pmb v\in\Mnsu(\Cal S)$.

\proof{ Apply 644D to the sequence $\sequencen{\pmb w_n}$ where
$\pmb w_{2n}=\pmb u_n$ and $\pmb w_{2n+1}=0$ for every $n$.
}%end of proof of 644E

\leader{644F}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, $A$ a uniformly
order-bounded subset of $\Mmo(\Cal S)$, and $\pmb v$ an integrator with domain
$\Cal S$.

(a)(i) $\{\int_{\Cal S}\pmb u\,d\pmb v:\pmb u\in A\}$ is
topologically bounded in $L^0$.

\quad(ii) if $\Cal S$ is non-empty,

\Centerline{$\lim_{\tau\uparrow\Cal S}\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\vee\tau}\pmb u\,d\pmb v)
=\lim_{\tau\downarrow\Cal S}\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)
=0$.}

(b) If $\pmb v$ is non-decreasing, then
$\{\int_{\Cal S}\pmb u\,d\pmb v:\pmb u\in A\}$
is order-bounded in $L^0$.

%could be in \S616 query
%note:  we know that $\iivu$ is an integrator, therefore moderately oscillatory,
%  for each moderately oscillatory $\pmb u$.   This result says that
%  $\{\iivu:\pmb u\in A\}$ is in a sense `uniformly moderately oscillatory'.

\proof{{\bf (a)} Let $\bar u\in L^0$ be such that $\sup|\pmb u|\le\bar u$ for
every $\pmb u\in A$.

\medskip

\quad{\bf (i)} Take any $\epsilon>0$.   Then there are an $M>0$ such that
$\Bvalue{\bar u\ge M}$ has measure at most $\epsilon$, and a $\delta>0$
such that $\theta(\delta z)\le\epsilon$ for every
$z\in Q_{\Cal S}(d\pmb v)$.   Suppose that $\pmb u\in A$ and
set $\pmb u'=\med(-M\tbf 1^{(\Cal S)},\pmb u,M\tbf 1^{(\Cal S)})$.
If $I\in\Cal I(\Cal S)$
then $S_I(\bover1M\pmb u',d\pmb v)\in Q_{\Cal S}(d\pmb v)$
so $\theta(\bover{\delta}MS_I(\pmb u',d\pmb v))\le\epsilon$.
Also

$$\eqalignno{\Bvalue{S_I(\pmb u,d\pmb v)\ne S_I(\pmb u,d\pmb v)}
&\Bsubseteq\Bvalue{\pmb u\ne\pmb u'}\cr
\displaycause{613Gd}
&=\Bvalue{\sup|\pmb u|>M}
\Bsubseteq\Bvalue{\bar u\ge M}\cr}$$

\noindent has measure at most $\epsilon$, so

\Centerline{$\theta(\Bover{\delta}MS_I(\pmb u,d\pmb v))
\le\epsilon+\theta(\Bover{\delta}MS_I(\pmb u',d\pmb v))
\le 2\epsilon$.}

\noindent Taking the limit as $I\uparrow\Cal S$,
$\theta(\bover{\delta}M\int_{\Cal S}\pmb u\,d\pmb v)\le 2\epsilon$;
and this is true for every $\pmb u\in A$.
As $\epsilon$ is arbitrary,
$\{\int_{\Cal S}\pmb u\,d\pmb v:\pmb u\in A\}$ is
topologically bounded.

\medskip

\quad{\bf (ii)}\grheada\ \Quer\ Suppose, if possible, that
such that $\epsilon=\bover13\limsup_{\tau\uparrow\Cal S}\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\vee\tau}\pmb u\,d\pmb v)$ is greater than $0$.
As just above, take $M\ge 1$ such that
$\Bvalue{\bar u\ge M}$ has measure at most $\epsilon$, and
set $\eta=\Bover{\epsilon}{4M}$.
Take $\gamma>0$, $m\ge 1$, $r\ge m$ and $k\ge 1$ such that

\Centerline{$\theta(\bover1{\gamma}z)<\eta$ for every
$z\in Q_{\Cal S}(d\pmb v)$,}

\Centerline{$m\eta\ge 2\gamma$,
\quad$1-\Bover{r!}{r^m(r-m)!}\le\Bover12\eta^m$,
\quad$2k\eta^m\ge\eta$,}

\noindent and set $n=rk$.

Choose $\langle\tau_i\rangle_{i\le n}$ inductively, as follows.   Start from any
$\tau_0\in\Cal S$ such that $\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\vee\tau_0}\pmb u\,d\pmb v)>2\epsilon$.
Given that $i<n$, $\tau_i\in\Cal S$ and
$\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\vee\tau_i}\pmb u\,d\pmb v)>2\epsilon$, take
$\pmb u_i\in A$ such that
$\theta(\int_{\Cal S\vee\tau_i}\pmb u_i\,d\pmb v)>2\epsilon$, and set
$\pmb u'_i=\med(-M\tbf 1^{(\Cal S)},\pmb u_i,M\tbf 1^{(\Cal S)})$.  As
before,

\Centerline{$\theta(\int_{\Cal S\vee\tau_i}\pmb u'_i\,d\pmb v)
\ge\theta(\int_{\Cal S\vee\tau_i}\pmb u'_i\,d\pmb v)
    -\bar\mu\Bvalue{\pmb u'_i\ne\pmb u_i}
\ge\theta(\int_{\Cal S\vee\tau_i}\pmb u'_i\,d\pmb v)
    -\bar\mu\Bvalue{\bar u\ge M}
>\epsilon$.}

\noindent Let $I_i\in\Cal I(\Cal S\vee\tau_i)$ be such that
$\theta(S_{I_i}(\pmb u'_i,d\pmb v))>\epsilon$, and take
$\tau_{i+1}\in\Cal S\vee\max I_i$ such that
$\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\vee\tau_{i+1}}\pmb u\,d\pmb v)>2\epsilon$.   Continue.

Then $\tau_0\le\ldots\le\tau_n$ in $\Cal S$,
and if $i<n$ then
$I_i\in\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])$ for each $i<n$ and

\Centerline{$\sup\{\theta(z):z\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\pmb v)\}
\ge\theta(S_{I_i}(\bover1M\pmb u'_i,d\pmb v))
\ge\Bover1M\theta(S_{I_i}(\pmb u'_i,d\pmb v))
>\Bover{\epsilon}M
=4\eta$.}

\noindent So by 616Hc there is a $z\in Q_{\Cal S}(d\pmb v)$ such that

\Centerline{$\eta\le\bar\mu\Bvalue{|z|\ge\gamma}
\le\theta(\bover1{\gamma}w)$,}

\noindent and we chose $\gamma$ so that this would not be possible.\ \Bang

Thus $\lim_{\tau\uparrow\Cal S}\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\vee\tau}\pmb u\,d\pmb v)=0$.

\medskip

\qquad\grheadb\ The same argument works downwards.   \Quer\ If
$\epsilon=\bover13\limsup_{\tau\downarrow\Cal S}\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)$ is greater than $0$,
take $M\ge 1$, $\eta>0$, $\gamma>0$, $r\ge m\ge 1$, $k\ge 1$ and $n=mk$
as before, and choose $\tau_n\ge\ldots\ge\tau_0$,
$\pmb u_{n-1},\ldots,\pmb u_0$ and $I_{n-1},\ldots,I_0$
such that for $n\ge i>0$

\Centerline{$\pmb u_{i-1}\in A$,
\quad$\theta(\int_{\Cal S\wedge\tau_i}\pmb u_{i-1}\,d\pmb v)>2\epsilon$,
\quad$I_{i-1}\in\Cal I(\Cal S\wedge\tau_i)$,}

\Centerline{$\theta(S_{I_{i-1}}
   (\med(-M\tbf 1^{(\Cal S)},\pmb u_{i-1},M\tbf 1^{(\Cal S)}),d\pmb v))
>\epsilon$,
\quad$\tau_{i-1}\le\min I_{i-1}$.}

Once again we finish with $\tau_0\le\ldots\le\tau_n$ in $\Cal S$,
and $I_i\in\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])$ for each $i<n$, while

\Centerline{$\sup\{\theta(w):w\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\pmb v)\}
\ge 4\eta$}

\noindent for each $i<n$, which is impossible.\ \BanG\  So
$\lim_{\tau\downarrow\Cal S}\sup_{\pmb u\in A}
   \theta(\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)=0$.

\medskip

{\bf (b)} $A$ is upwards-directed;  because $\pmb v$ is non-decreasing,
$B=\{\int_{\Cal S}\pmb u\,d\pmb v:\pmb u\in A\}$ is upwards-directed;
being topologically bounded, by (a), it is bounded above in $L^0$
(613B(f-v)).   Now $-B=B$ is bounded above, $B$ is bounded below
and $B$ is order-bounded.
}%end of proof of 644F

\leader{644G}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, and $\Cal S$ is a non-empty
order-convex subset of $\Cal T$.
On the space $\Mmo=\Mmo(\Cal S)$ of moderately oscillatory processes,
we have a linear space topology $\frak S$
defined by functionals of the form
$\pmb u\mapsto\theta(\int_{\Cal S}|\pmb u|d\pmb v)$
where $\pmb v\in\Mnsu(\Cal S)$.   Let
$\pmb v=\sfamilysS{v}$ be a near-simple integrator.
Then $\pmb u\mapsto\int_{\Cal S}\pmb u\,d\pmb v:\Mmo\to L^0$ is
uniformly continuous, for the uniformity induced by $\frak S$, on any
uniformly order-bounded set in $\Mmo$.

\proof{ For an integrator $\pmb v$ with
domain $\Cal S$ and $u\in\Mmo$, write
$\psi_{\pmb v}(u)=\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

{\bf (a)} I should begin by noting straight away that if
$\pmb v\in\Mnsu(\Cal S)$ and we set
$\phi_{\pmb v}(\pmb u)=\theta(\psi_{\pmb v}|\pmb u|)$
for $\pmb u\in\Mmo$, then
$\phi_{\pmb v}$ is an F-seminorm.   \Prf\ (i) If
$\pmb u$, $\pmb u'\in\Mmo$, then
$|\pmb u+\pmb u'|\le|\pmb u|+|\pmb u'|$.   As $\pmb v$ is
non-decreasing,

\Centerline{$\phi_{\pmb v}(\pmb u+\pmb u')
=\psi_{\pmb v}(|\pmb u+\pmb u'|)
\le\psi_{\pmb v}(|\pmb u|+|\pmb u'|)
=\psi_{\pmb v}(|\pmb u|)+\psi_{\pmb v}(|\pmb u'|)
=\phi_{\pmb v}(\pmb u)+\phi_{\pmb v}(\pmb u')$,}

\noindent (ii) If
$\pmb u\in\Mmo$ and $\alpha\in\Bbb R$, then
$\phi_{\pmb v}(\alpha\pmb u)=|\alpha|\phi_{\pmb v}(\pmb u)$, so
$\lim_{\alpha\to 0}\phi_{\pmb v}(\alpha\pmb u)=0$ and
$\phi_{\pmb v}(\alpha\pmb u)\le\phi_{\pmb v}(\pmb u)$ if
$|\alpha|\le 1$.\ \Qed

Accordingly we have a linear space topology $\frak S$ on $\Mmo$ defined by
$\{\phi_{\pmb v}:\pmb v\in\Mnsu(\Cal S)\}$.

It is worth noting that if $\pmb v$, $\pmb v'\in\Mnsu(\Cal S)$ then
$\pmb v+\pmb v'\in\Mnsu(\Cal S)$, while
$\phi_{\pmb v+\pmb v'}=\phi_{\pmb v}+\phi_{\pmb v'}$.
So if $G$ is any $\frak S$-neighbourhood of $0$, there are
$\pmb v\in\Mnsu(\Cal S)$ and
$\delta>0$ such that
$\{\pmb u:\pmb u\in\Mmo$, $\phi_{\pmb v}(\pmb u)\le\delta\}\subseteq G$.

\medskip

{\bf (b)} From now on, I take it that $A\subseteq\Mmo$ is a uniformly
order-bounded set and $\pmb v\in\Mns(\Cal S)$ is an integrator.   Let
$\bar u\in L^0$ be such that
$\sup|\pmb u|\le\bar u$ for every $\pmb u\in A$.   The argument proceeds by
looking at a succession of special cases.

If $\pmb v$ is actually non-negative and non-increasing, then
of course $\psi_{\pmb v}:A\to L^0$ is uniformly
continuous, just because

\Centerline{$|\psi_{\pmb v}(\pmb u)-\psi_{\pmb v}(\pmb u')|
=|\psi_{\pmb v}(\pmb u-\pmb u')|
\le\psi_{\pmb v}(|\pmb u-\pmb u'|)
=\phi_{\pmb v}(\pmb u-\pmb u')$.}

\noindent So if $\pmb v$ is of bounded variation, therefore expressible as the
difference of members of $\Mnsu(\Cal S)$ (631L),
then $\psi_{\pmb v}$ is a difference of uniformly
continuous processes, therefore uniformly continuous, on $A$.

\medskip

{\bf (c)} (The key.)   Suppose that $\Cal S$ has greatest and least elements,
$M=\|\bar u\|_{\infty}$ is
finite and $\pmb v$ is an $L^{\infty}$-martingale.   Then
$\|v_{\sigma}\|_2\le\|v_{\max\Cal S}\|_2\le\|v_{\max\Cal S}\|_{\infty}$
for every
$\sigma\in\Cal S$ (621Ca) and $\pmb v$ is actually $\|\,\|_2$-bounded.
Let $\pmb v^*=\sfamilysS{v^*}$
be the quadratic variation of $\pmb v$.   Applying 624I
with $\pmb u=\tbf 1$,

\Centerline{$\Expn(v^*_{\max\Cal S})=\Expn((v_{\max\Cal S}-v_{\min\Cal S})^2)
\le\Expn(v_{\max\Cal S}^2)<\infty$.}

\noindent Take any $\epsilon>0$.   Let $\delta\in\ocint{0,1}$ be such that
$(2M+1)(\delta+2M\Expn(v^*_{\max\Cal S}\times\chi c))\le\epsilon^2$
whenever $\bar\mu c\le\delta$.
Take $\pmb u$, $\pmb w\in A$ such that
$\theta(\int_{\Cal S}|\pmb u-\pmb w|d\pmb v^*)\le\delta^2$.
In this case, setting $c=\Bvalue{\int_{\Cal S}|\pmb u-\pmb w|d\pmb v^*>\delta}$,
$\bar\mu c\le\delta$.   Consequently

$$\eqalignno{\|\int_{\Cal S}|\pmb u-\pmb w|d\pmb v^*\|_1
&\le\delta+\Expn(\chi c\times\int_{\Cal S}|\pmb u-\pmb w|d\pmb v^*)
\le\delta+\Expn(\chi c\times 2M\int_{\Cal S}d\pmb v^*)\cr
\displaycause{because $|\pmb u-\pmb w|\le 2M\tbf 1$}
&=\delta+\Expn(\chi c\times 2Mv^*_{\max\Cal S})
\le\Bover{\epsilon^2}{2M+1}.\cr}$$

\noindent We see that

\Centerline{$(\pmb u-\pmb w)^2
=|\pmb u-\pmb w||\pmb u+\pmb w|
\le 2M|\pmb u-\pmb w|$,}

\noindent so
$\|\int_{\Cal S}(\pmb u-\pmb w)^2d\pmb v^*\|_1\le\epsilon^2$.
Now

$$\eqalignno{\theta(\int_{\Cal S}\pmb u\,d\pmb v
   -\int_{\Cal S}\pmb w\,d\pmb v)
&=\theta(\int_{\Cal S}(\pmb u-\pmb w)d\pmb v)
\le\|\int_{\Cal S}(\pmb u-\pmb w)d\pmb v\|_1\cr
&\le\|\int_{\Cal S}(\pmb u-\pmb w)d\pmb v\|_2
=\sqrt{\|\int_{\Cal S}(\pmb u-\pmb w)^2d\pmb v^*\|_1}\cr
\displaycause{624I again}
&\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary (and $\pmb v^*$ is non-negative and
non-decreasing and near-simple, by 631Ja),
$\psi_{\pmb v}$ is uniformly continuous on $A$.

\medskip

{\bf (d)} Next, suppose that $\Cal S$ has greatest and least elements and
$\|\bar u\|_{\infty}$ is finite.   Take $\epsilon>0$.   Since
$\Cal S=[\min\Cal S,\max\Cal S]$ is finitely full, order-closed in $\Cal T$ and
has a greatest member,
there are an $L^{\infty}$-martingale $\tildepmb v$ and a near-simple process
$\pmb v'$ of bounded variation
such that $\bar\mu\Bvalue{\pmb v\ne\tildepmb v+\pmb v'}\le\epsilon$ (643O).
Now

$$\eqalign{\theta(\psi_{\pmb v}(\pmb u)
  -\psi_{\tildepmb v}(\pmb u)-\psi_{\pmb v'}(\pmb u))&
\le\bar\mu\Bvalue{\psi_{\pmb v}(\pmb u)
  \ne\psi_{\tildepmb v}(\pmb u)+\psi_{\pmb v'}(\pmb u)}\cr&
\le\bar\mu\Bvalue{\pmb v\ne\tildepmb v+\pmb v'}
\le\epsilon\cr}$$

\noindent for every $\pmb u\in A$, while $\psi_{\tildepmb v}+\psi_{\pmb v'}$
is uniformly continuous on $A$, by (b) and (c).   As $\epsilon$ is arbitrary,
$\psi_{\pmb v}$ is uniformly continuous on $A$.

\medskip

{\bf (e)} Suppose just that $\Cal S$ has greatest and least elements,
while $\bar u$ is defined in $L^0$ and
$\pmb v$ is a near-simple integrator.   Let $\epsilon>0$.   Then there is
an $M\ge 0$ such that $\bar\mu\Bvalue{\bar u>M}\le\epsilon$.
Set $A_1=\{\pmb u:\pmb u\in\Mmo$, $\|\pmb u\|_{\infty}\le M\}$ and
$\bar h(\pmb u)=\med(-M\tbf 1,\pmb u,M\tbf 1)$ for
$\pmb u\in A$.
Then $|\bar h(\pmb u)-\bar h(\pmb w)|\le|\pmb u-\pmb w|$ for all
$\pmb u$, $\pmb w\in A$, so $\bar h:A\to A_1$ is uniformly continuous;
by (d), $\psi_{\pmb v}\restr A_1$ is uniformly continuous, so
the composition $\psi_{\pmb v}\bar h$ is uniformly continuous.   Next,

\Centerline{$\theta(\psi_{\pmb v}(\pmb u)-\psi_{\pmb v}(\bar h(\pmb u)))
\le\bar\mu\Bvalue{\psi_{\pmb v}(\pmb u)\ne\psi_{\pmb v}(\bar h(\pmb u))}
\le\bar\mu\Bvalue{\pmb u\ne\bar h(\pmb u)}
\le\bar\mu\Bvalue{\bar u>M}
\le\epsilon$}

\noindent for every $\pmb u\in A$.   As $\epsilon$ is arbitrary,
$\psi_{\pmb v}$ is uniformly approximated on $A$ by uniformly continuous
functions and is itself uniformly continuous.

\medskip

{\bf (f)} Finally, for the general case, suppose only that $\Cal S$ is
order-convex, $A\subseteq\Mmo$ is uniformly order-bounded and that $\pmb v$ is a
near-simple integrator with domain $\Cal S$.   If $\Cal S$ is empty then
$\#(A)\le 1$ and trivially we have a
uniformly continuous function.    Otherwise, let $\epsilon>0$.   By 644F(a-ii),
there is a $\tau\in\Cal S$ such that
$\sup_{\pmb u\in A}\theta(\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)\le\epsilon$;
applying the other half of 644F(a-ii) to $\pmb v\restr\Cal S\vee\tau$ and
$\{\pmb u\restr\Cal S\vee\tau:\pmb u\in A\}$, we see that there is a
$\tau'\in\Cal S\vee\tau$ such that
$\sup_{\pmb u\in A}\theta(\int_{\Cal S\vee\tau'}\pmb u\,d\pmb v)\le\epsilon$.
(Of course this step depends on knowing that $\pmb v\restr\Cal S\vee\tau$ is a
near-simple integrator, by 616P(b-ii) and 631F(a-iv), while
$\pmb u\restr\Cal S\vee\tau$ is moderately oscillatory for every
$\pmb u\in\Mmo$, by 615F(a-i) again;   clearly
$\{\pmb u\restr\Cal S\vee\tau:\pmb u\in A\}$ will be uniformly
order-bounded if $A$ is.)

Now consider $\Cal S'=\Cal S\cap[\tau,\tau']=[\tau,\tau']$,
$\pmb v\restr\Cal S'$ and $\{\pmb u\restr\Cal S':\pmb u\in A\}$.   From (e) and
the last remark in (a) above we
know that there are a non-negative, non-decreasing process
$\tildepmb v'=\langle\tilde v'_{\sigma}\rangle_{\sigma\in\Cal S'}$
and $\delta>0$ such that
$\theta(\int_{\Cal S'}\pmb u-\pmb u'\,d\pmb v)\le\epsilon$ whenever
$\pmb u$, $\pmb u'\in A$ and
$\int_{\Cal S'}|\pmb u-\pmb u'|d\tildepmb v'\le\delta$.   Define
$\tildepmb v=\sfamilysS{\tilde v}$ by setting

\Centerline{$\tilde v_{\sigma}
  =\tilde v'_{\med(\tau,\sigma,\tau')}
        \times\chi(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'})
     +\tilde v'_{\tau'}\times\chi\Bvalue{\tau'\le\sigma}$}

\noindent for $\sigma\in\Cal S$.   It is straightforward to check that
$\tildepmb v$ is fully adapted (use 612C), non-negative, non-decreasing and
order-bounded (with greatest value $v'_{\tau'}$), while
$\tildepmb v\restr\Cal S'=\tildepmb v'$;  and by 631F(a-iv) again $\tildepmb v$
is near-simple.

Suppose that $\pmb u\in A$.   Then

\Centerline{$\theta(\int_{\Cal S}\pmb u\,d\pmb v-\int_{\Cal S'}\pmb u\,d\pmb v)
=\theta(\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
         +\int_{\Cal S\vee\tau'}\pmb u\,d\pmb v)
\le 2\epsilon$}

\noindent by the choice of $\tau$ and $\tau'$.
Now if $\pmb u$, $\pmb u'\in A$ and
$\theta(\int_{\Cal S}|\pmb u-\pmb u'|d\tildepmb v)\le\delta$,
we shall have

\Centerline{$\theta(\int_{\Cal S'}|\pmb u-\pmb u'|d\tildepmb v')
=\theta(\int_{\Cal S'}|\pmb u-\pmb u'|d\tildepmb v)\le\delta$,}

\noindent so that
$\theta(\int_{\Cal S'}(\pmb u-\pmb u')d\pmb v)\le\epsilon$ and
$\theta(\int_{\Cal S}(\pmb u-\pmb u')d\pmb v)\le 5\epsilon$.
As $\epsilon$ is arbitrary,
$\pmb u\mapsto\int_{\Cal S}\pmb u\,d\pmb v:A\to L^0$ is uniformly continuous,
and the proof is complete.
}%end of proof of 644G

\leader{644H}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T$ and $\sequencen{\pmb u_n}$
a uniformly order-bounded sequence of moderately oscillatory processes with
domain $\Cal S$
such that $\sequencen{\pmb u_{n<}}$ is order*-convergent in $(L^0)^{\Cal S}$.
Then $\lim_{n\to\infty}\int_{\Cal S}\pmb u_nd\pmb v$ is defined
for every near-simple integrator $\pmb v$ with domain $\Cal S$.
If $\sequencen{\pmb u_{n<}}$ is order*-convergent to $\pmb u_<$, where
$\pmb u$ is moderately oscillatory, then
$\lim_{n\to\infty}\int_{\Cal S}\pmb u_nd\pmb v
=\int_{\Cal S}\pmb u\,d\pmb v$.

\proof{ If $\sequence{k}{n_k}$ is strictly increasing and we set
$\pmb w_k=|\pmb u_{n_{k+1}}-\pmb u_{n_k}|$ for each $k$,
$\sequence{k}{\pmb w_k}$ is a uniformly order-bounded sequence of
moderately oscillatory processes such that $\sequence{k}{\pmb w_{<k}}$ is
order*-convergent to $0$ (apply the last sentence of 642Ba with
$g(\alpha,\beta)=|\alpha-\beta|$).   So 644E tells us that
$\lim_{k\to\infty}\int_{\Cal S}\pmb w_k\,d\pmb v=0$ for every
$\pmb v\in\Mnsu(\Cal S)$.   As $\sequence{k}{n_k}$ is arbitrary,
$\lim_{m,n\to\infty}\theta(\int_{\Cal S}|\pmb u_n-\pmb u_m|d\pmb v)=0$ for
every $\pmb v\in\Mnsu(\Cal S)$.   By 644G,
$\lim_{m,n\to\infty}\int_{\Cal S}\pmb u_n-\pmb u_md\pmb v=0$
for every near-simple integrator $\pmb v$ with domain $\Cal S$, that is,
$\lim_{n\to\infty}\int_{\Cal S}\pmb u_nd\pmb v$ is defined
for every near-simple integrator $\pmb v$ with domain $\Cal S$.

If we know that there is a moderately oscillatory
process $\pmb u$ such that $\sequencen{\pmb u_{n<}}$ order*-converges to
$\pmb u_<$, then we can apply the trick of 644E to a sequence alternating
between $\pmb u_n-\pmb u$ and zero to see that
$\lim_{n\to\infty}\int_{\Cal S}\pmb u_n\,d\pmb v
=\int_{\Cal S}\pmb u\,d\pmb v$.
}%\end of proof of 644H

%\leader{644P}{}
%\leader{644Q}{}
%\leader{644R}{}
%\leader{644S}{}
%\leader{644T}{}
%\leader{644U}{}
%\leader{644V}{}
%\leader{644W}{}

\exercises{\leader{644X}{Basic exercises (a)}
%\spheader 644Xa
Let $\Cal S$ be a sublattice of $\Cal T$, and
$\sequencen{\pmb u_n}=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$
a uniformly order-bounded sequence in $(L^0)^{\Cal S}$ which is
order*-convergent to $\pmb u\in(L^0)^{\Cal S}$.   Show that
$\pmb u$ is order-bounded.
%644B

\sqheader 644Xb Give an example of a simple process $\pmb v$ and a
non-increasing sequence $\sequencen{\pmb u_n}$ of
simple processes, all these processes having the same domain
$\Cal S$, such that $\inf_{n\in\Bbb N}\pmb u_n=\tbf 0$ but
$\sequencen{\int_{\Cal S}\pmb u_nd\pmb v}$ is not convergent to $0$.
\Hint{take $T=\coint{0,\infty}$, $\frak A=\{0,1\}$.}
%644D

\spheader 644Xc In 644G, show that the topology $\frak S$ described there
is coarser than the ucp topology.
%644G

\spheader 644Xd(i) Suppose that $U$ and $V$ are linear topological spaces
and $T:U\to V$ is a linear operator.   Let $A\subseteq U$ be a non-empty
set such that $T\restr(A-A)$ is continuous at $0$.   Show that
$T\restr A$ is uniformly continuous.   (ii) Use this to simplify the
formulae in the proof of 644G.
%644G

\spheader 644Xe In 644G,
suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa, 617Xb, 618Xa, 622Xd, 626Xa, 627Xa and 642Xd.
Show that the topology $\frak S$ on $\Mob(\Cal T_f)$ corresponds to the
topology on the space of bounded functions in $\tildeClll$ (615Xf)
generated by
functionals of the form $f\mapsto\int|f_<|d\nu$ where $\nu$ is a totally
finite Radon measure on $\coint{0,\infty}$, taking $f_<(0)=0$ and
$f_<(t)=\lim_{s\uparrow t}f(s)$ for $t>0$.
%644G

%\leader{644Y}{Further exercises (a)}

}%end of exercises


\endnotes{
\Notesheader{644} The essence of the Lebesgue integral lies in the convergence
theorems of \S123, all concerning limits of integrals of sequences of functions.
Here for the first time in this volume we have a corresponding result for
limits of Riemann-sum integrals.   The real content lies in the case
in which the sequence $\sequencen{\pmb u_n}$ of integrands
is a non-increasing sequence of non-negative processes such that
$\inf_{n\in\Bbb N}\pmb u_{n<}=0$ and the integrator is non-decreasing (644C).
Moving to the case of a uniformly order-bounded sequence
$\sequencen{\pmb u_n}$ such
that $\sequencen{\pmb u_{n<}}$ is order*-convergent
in $(L^0)^{\Cal S}$ (644D) is not quite trivial,
but is really a result about positive linear operators on Riesz spaces.
The point here is that
$\Mmo(\Cal S)$ is a Riesz subspace of $(L^0)^{\Cal S}$, but is not as a rule
sequentially order-closed.

To handle general integrators, the
first step is to deal with $L^2$-martingales (part (c) of the proof of 644G).
At this point it seems that we have to move up a gear,
assuming right-continuity of
the filtration and order-convexity of the sublattice $\Cal S$ (644G) so as to
apply the Fundamental Theorem of Martingales.
But under these conditions we get a result more or less corresponding to the
Lebesgue's Dominated Convergence Theorem (644H).   When adapting this
to define a second kind of integral (\S645), it will be helpful to refer to
a new uniformity on the space of moderately oscillatory processes (644G).
The point is that this is much coarser than the uniformity
corresponding to the ucp topology, but
still makes integration uniformly continuous on uniformly order-bounded sets.

}%end of notes

\discrpage

\frfilename{mt645.tex}
\versiondate{29.5.21}
\copyrightdate{2013}

\def\chaptername{The S-integral}
\def\sectionname{Construction of the S-integral}

\newsection{645}

We are now in a position to define a sequentially smooth integral which
corresponds, in a sense, to Lebesgue-Stieltjes integrals on the real line.
The objective is to integrate bounded previsible processes with respect to
near-simple integrators, and I set this up as a kind of Daniell
integral\cmmnt{ (see 436Ya)} based on ideas in \S644.   Since simple and
near-simple and mderately oscillatory processes,
as I have defined them in this volume, are often not
previsible, we need to deal throughout with their previsible versions;
and as our integrals take values in $L^0$ rather than in $\Bbb R$ or
$\Bbb C$, we have to calculate with the functional $\theta$ rather than
with a modulus or norm.

The key to the programme is really Lemma 644G.   We saw there that
(subject to certain conditions) integration with respect to an arbitrary
near-simple integrator is controlled by integration with respect to
appropriate non-decreasing processes.   We can therefore do nearly all
the work of the present section with non-decreasing integrators, which are
very much easier to handle, even though our real aim is to understand
integration with respect to martingales.   With a non-decreasing
integrator, as with an ordinary non-negative measure, integration is a
positive linear operator.   This makes it possible to consider upper
integrals, which are what, in effect, we have in Definition 645Bb.
Based on the functionals $\widehat{\theta}^{\#}_{\pmb v}$ there, we
have a linear space topology $\TSi$ on a large space $\Mpob$ of
order-bounded processes (645F).   As with ordinary integration,
unbounded sequences of integrands can be uncontrollable, so we have to
find types of domination -- preferably weaker than simply assuming
uniform $\|\,\|_{\infty}$-boundedness -- which will be adequate to
ensure convergence of sequences of integrals.
(See 645G-645L.) %645G 645L
These bring us to a definition of what I call the `S-integral' in
645P, in a form which makes it easy to check that it is bilinear in
integrand and integrator (645Rb), and with the tools to show that it is
sequentially smooth in the integrand (645T).

\cmmnt{\leader{645A}{Notation} This section brings together many ideas,
and there is a correspondingly long list of notations to
recall.   As always, \Asissix\ will be a stochastic integration structure.
For a sublattice $A$ of $\Cal T$ and $\tau\in\Cal T$,
$A\wedge\tau=\{\sigma\wedge\tau:\sigma\in A\}$ and
$A\vee\tau=\{\sigma\vee\tau:\sigma\in A\}$.
$L^0=L^0(\frak A)$ (612A) and
$\theta(w)=\Expn(|w|\wedge\chi 1)$ for $w\in L^0$ (613Ba).
If $k\ge 1$ and $h:\Bbb R^k\to\Bbb R$ is Borel measurable, I will write
$\bar h$ for any of the associated functions from $(L^0)^k$ to $L^0$
(612A, 619E)
and from $((L^0)^{\Cal S})^k$ to $(L^0)^{\Cal S}$ (612B, 619F).

If $\Cal S$ is a sublattice of $\Cal T$, $\pmb u$, $\pmb v$ are processes
with domain $\Cal S$, and $I$ is a finite
sublattice of $\Cal S$, then
$S_I(\pmb u,d\pmb v)$ will be the corresponding Riemann sum (613Fb),
$Q_{\Cal S}(d\pmb v)$ the capped-stake variation set (616B) and
$\int_{\Cal S}\pmb u\,d\pmb v$ (if it exists) the
Riemann-sum integral defined in
613H/613Na.   $\tbf 1^{(\Cal S)}$ will be the constant process with domain
$\Cal S$ and value $\chi 1$.

If $\Cal S$ is a sublattice of $\Cal T$,
$\Mob(\Cal S)$ is the $f$-algebra of order-bounded processes with domain
$\Cal S$ (614Fc);  for $\pmb u=\sfamilysS{u}\in\Mob(\Cal S)$,
$\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.
$\Mmo(\Cal S)$ is the $f$-algebra of moderately oscillatory processes with
domain $\Cal S$ (615Fa), and $\Mmo(\Cal S)^+$ its positive cone.
If $\pmb u=\sfamilysS{u}\in\Mmo(\Cal S)$,
$\pmb u_<=\family{\sigma}{\Cal S}{u_{<\sigma}}$ will always be its previsible
version (641L).   $\Mfa(\Cal S)$, $\Msimp(\Cal S)$ and $\Mns(\Cal S)$
are the spaces of fully adapted,
simple and near-simple processes with domain $\Cal S$ (612I, 612L, 631Fa).
$\Mnsu(\Cal S)$ is
the cone of non-negative non-decreasing near-simple processes with domain
$\Cal S$ (644Bb), and $\Mbv(\Cal S)$ is the space of processes of bounded
variation with domain $\Cal S$ (614J).
}

\leader{645B}{Definitions} Let $\Cal S$ be a sublattice of $\Cal T$.

\spheader 645Ba{\bf (i)}
I will say that a fully adapted process $\pmb x$ with
domain $\Cal S$ is {\bf previsibly order-bounded} if there is a
non-negative $\pmb u\in\Mmo(\Cal S)$ such that
$|\pmb x|\le\pmb u_<$.
$\Mpob(\Cal S)$ will be the set of previsibly order-bounded
fully adapted processes $\pmb x$ with domain $\Cal S$.

\medskip

\quad{\bf (ii)} I will say that a set $A\subseteq\Mpob(\Cal S)$ is
{\bf uniformly previsibly order-bounded} if there is a
non-negative $\pmb u\in\Mmo(\Cal S)$ such that
$|\pmb x|\le\pmb u_<$ for every $\pmb x\in A$.   A uniformly
previsibly order-bounded set is uniformly order-bounded\cmmnt{ in the
sense of 644Bb, by 641G(a-vii-$\gamma$)}.

\spheader 645Bb For $\pmb x\in\Mpob(\Cal S)$
and $\pmb v\in\Mnsu(\Cal S)$,
write $\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$ for

$$\eqalign{
&\inf\{\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v):
\sequencen{\pmb u_n}
\text{ is a uniformly order-bounded non-decreasing sequence}\cr
&\mskip200mu\text{of non-negative processes in }\Mmo(\Cal S)\text{ and }
|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}\}.\cr}$$

\cmmnt{\noindent For the avoidance of doubt,
perhaps I should say that the supremum
here is to be taken in $(L^0)^{\Cal S}$, so
that (in the notation of 641F) we have $|x_{\sigma}|
\le\sup_{n\in\Bbb N}u_{n<\sigma}$
for every $\sigma\in\Cal S$.}

\spheader 645Bc If $\pmb v=\sfamilysS{v}$, $\pmb v'=\sfamilysS{v'}$
are two fully adapted processes with domain $\Cal S$, I will write
$\pmb v\preccurlyeq\pmb v'$ if $\pmb v'-\pmb v$ is non-decreasing\cmmnt{, that
is, $v_{\tau}-v_{\sigma}\le v'_{\tau}-v'_{\sigma}$ whenever $\sigma\le\tau$
in $\Cal S$, that is, $\Delta\pmb v\le\Delta\pmb v'$ where
$\Delta\pmb v$ and $\Delta\pmb v'$ are the associated adapted
interval functions\cmmnt{ (613Cc)}}.

In this case, if $\pmb u$ is any non-negative fully adapted process with
domain $\Cal S$,
$\int_{\Cal S}\pmb u\,d\pmb v\le\int_{\Cal S}\pmb u\,d\pmb v'$ if both
integrals are defined\cmmnt{ (616R(b-i), applied to $\pmb v'-\pmb v$)}.

\leader{645C}{}\cmmnt{ A standard fragment of real analysis didn't quite
get into \S4A3.

\medskip

\noindent}{\bf Lemma} (a) Let $X$ be a metrizable space.   Then the set
of Borel measurable real-valued functions on $X$ is the smallest
subset $U$ of $\BbbR^X$ which contains every bounded continuous
real-valued function and is such that $\lim_{n\to\infty}h_n\in U$ whenever
$\sequencen{h_n}$ is a sequence in $U$ which has a limit in $\Bbb R$
at every point and which is either non-decreasing or non-increasing.

(b) If $k\ge 1$ and $h:\BbbR^k\to\Bbb R$ is a locally bounded function,
then there is a continuous non-decreasing function
$g:\Bbb R\to\coint{0,\infty}$ such that $|h(x)|\le g(\|x\|)$ for every
$x\in\BbbR^k$.

(c) If $k\ge 1$ and $U$ is a set of real-valued functions on $\BbbR^k$
such that ($\alpha$) every continuous function belongs to $U$ ($\beta$)
$\lim_{n\to\infty}f_n\in U$ whenever $\sequencen{f_n}$ is a pointwise
convergent sequence in $U$ and $\sup_{n\in\Bbb N}|f_n|$ is locally bounded,
then every locally bounded Borel measurable function on $\BbbR^k$ belongs
to $U$.

\proof{{\bf (a)(i)}
We know that the set of Borel measurable functions is
closed under pointwise limits of sequences
(121E-121F) and that every continuous function is Borel measurable
(4A3Cd), so every member of $U$ is Borel measurable.

\medskip

\quad{\bf (ii)} In the other direction, we can argue as follows.
Write $C_b$ for the space of bounded continuous functions on $X$, and
let $\Bbb V$ be the family of subsets $V$ of $\BbbR^X$
such that $C_b(X)\subseteq V$ and $V$ is closed under monotone sequential
limits.   Then $U=\bigcap\Bbb V$.

\medskip

\qquad\grheada\ If $g\in C_b$, then it is easy to check that
$\{h:h\in\BbbR^X$, $g+h\in U\}$ belongs to $\Bbb V$, so includes $U$.
Thus $g+h\in U$ for every $g\in C_b$ and $h\in U$.   Next,
$\{g:g\in\BbbR^X$, $g+h\in U$ for every $h\in U\}$ belongs to $\Bbb V$, so
$U$ is closed under addition.   And if $\alpha\in\Bbb R$,
$\{h:h\in\BbbR^X$, $\alpha h\in U\}\in\Bbb V$, so $\alpha U\subseteq U$.
Thus $U$ is a linear subspace of $\BbbR^X$.

\medskip

\qquad\grheadb\ If $G\subseteq X$ is open then
$\alpha\chi G\in U$.   \Prf\
If $G=X$ then $\chi G\in C_b$ and we can stop.   Otherwise,
let $\rho$ be a metric on $X$ defining its topology, and set
$h_n(x)=\min(1,2^n\rho(x,X\setminus G))$ for $n\in\Bbb N$ and $x\in X$;
then $\sequencen{h_n}$ is a non-decreasing sequence in $C_b$ with limit
$\chi G$, so $\chi G\in U$.\ \Qed

\medskip

\qquad\grheadc\ Set $\Cal E=\{E:E\subseteq X$, $\chi E\in U\}$.
Then $\Cal E$ satisfies (i) of 136A, so is a Dynkin class;  as it
contains every open set, it contains every Borel set, by the Monotone Class
Theorem (136B).

\medskip

\qquad\grheadd\ Because $U$ is a linear subspace of $\BbbR^X$, it contains
any linear combination of indicator functions of Borel sets.   But any
non-negative Borel measurable
function is expressible as the limit of a non-decreasing
sequence of such simple Borel measurable functions, so belongs to $U$;
and now every Borel measurable function is the difference of non-negative
Borel measurable functions, so belongs to $U$.

\medskip

{\bf (b)} For $n\in\Bbb N$, set
$\gamma_n=\sup\{|h(x)|:x\in\BbbR^k$, $\|x\|\le n\}$;
define $g:\Bbb R\to\coint{0,\infty}$ by setting

$$\eqalign{g(\alpha)
&=\gamma_1\text{ if }\alpha\le 0,\cr
&=(\alpha-n)\gamma_{n+2}+(n+1-\alpha)\gamma_{n+1}
\text{ if }n\in\Bbb N\text{ and }n\le\alpha\le n+1.\cr}$$

\noindent Then $g$ is continuous and non-decreasing and
$|h(x)|\le g(\|x\|)$ for every $x$.

\medskip

{\bf (c)(i)} Take any continuous $g:\BbbR^k\to\coint{0,\infty}$ and set
$V_g=\{f:f\in\BbbR^{\BbbR^k}$, $\med(-g,f,g)\in U\}$.
Then $V_g$ contains all
continuous functions and is closed under sequential pointwise convergence,
so contains every Borel measurable function on $\BbbR^k$, by (a).

\medskip

\quad{\bf (ii)} If now $h:\BbbR^k\to\Bbb R$
is any locally bounded Borel measurable
function, there is a continuous $g$ such that $|h|\le g$, by (b), while
$h\in V_g$, so that
$h=\med(-g,h,g)$ belongs to $U$.
}%end of proof of 645C

\leader{645D}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$.

(a)(i) If $\pmb x_0,\ldots,\pmb x_{k-1}\in\Mpob=\Mpob(\Cal S)$ and
$h:\BbbR^k\to\Bbb R$ is a locally bounded Borel
measurable function, then
$\bar h(\pmb x_0,\ldots,\pmb x_{k-1})\times\tbfoneminus{\Cal S}
\in\Mpob$;
and if $h(0,\ldots,0)=0$, then
$\bar h(\pmb x_0,\ldots,\pmb x_{k-1})\in\Mpob$.

\quad(ii) $\Mpob$ is an $f$-subalgebra of $\Mob(\Cal S)$.

\quad(iii) $\pmb u_<\in\Mpob$ for every
$\pmb u\in\Mob(\Cal S)$.

\quad(iv) If
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ then
$z\pmb x$\cmmnt{ (definition:  612De)} belongs to $\Mpob$ for every
$\pmb x\in\Mpob$.

(b) Suppose that $\pmb v\in\Mnsu(\Cal S)$.
Then $\widehat{\theta}^{\#}_{\pmb v}$ is an
F-seminorm and if $\pmb x$, $\pmb x'\in\Mpob$, $|\pmb x|\le|\pmb x'|$
and $\alpha\in\Bbb R$ then
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x')$ and
$\widehat{\theta}^{\#}_{\pmb v}(\alpha\pmb x)
\le\max(1,|\alpha|)\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$.

(c) If now we have another $\pmb v'\in\Mnsu(\Cal S)$
and $\pmb v\preccurlyeq\pmb v'$ in the sense of 645Bc,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v'}(\pmb x)$ for every
$\pmb x\in\Mpob(\Cal S)$.

\proof{{\bf (a)(i)}
Let $\pmb u\in\Mmo^+$ be such that $\sum_{i=0}^{k-1}|\pmb x_i|\le\pmb u_<$.
By 645Cb, there is a continuous non-decreasing
$g:\Bbb R\to\coint{0,\infty}$ such that $|h(x)|\le g(\|x\|)$ for every
$x\in\Bbb R^k$;  it follows that

\Centerline{$|\bar h(\pmb x_0,\ldots,\pmb x_{k-1})
  \times\tbfoneminus{\Cal S}|
\le\bar g\frsmallcirc\pmb u_<\times\tbfoneminus{\Cal S}
=(\bar g\pmb u)_<$}

\noindent (641Gd) and
$\bar h(\pmb x_0,\ldots,\pmb x_{k-1})\times\tbfoneminus{\Cal S}\in\Mpob$.
If in addition $h(0,\ldots,0)=0$, then

$$\eqalignno{\bar h(\pmb x_0,\ldots,\pmb x_{k-1})&
=\bar h(\pmb x_0\times\tbfoneminus{\Cal S},\ldots,
     \pmb x_{k-1}\times\tbfoneminus{\Cal S})\cr&
=\bar h(\pmb x_0,\ldots,\pmb x_{k-1})\times\tbfoneminus{\Cal S}
\Displaycause{by 619G(e-ii), because
$\tbfoneminus{\Cal S}\times\tbfoneminus{\Cal S}
=(\tbf 1^{\Cal S}\times\tbf 1^{\Cal S})_<
=\tbfoneminus{\Cal S}$}
\in\Mpob.\cr}$$

\medskip

\quad{\bf (ii)}\grheada\ If $\pmb x\in\Mpob$, there is a
non-negative
$\pmb u\in\Mmo=\Mmo(\Cal S)$ such that $|\pmb x|\le\pmb u_<$.
We know that $\pmb u$ is order-bounded, while
$\sup|\pmb u_<|\le\sup|\pmb u|$, as in
641G(a-vii), so $\pmb u_<$ and $\pmb x$ are order-bounded.

\medskip

\qquad\grheadb\ As noted in 612Bc, (i) just above is enough to ensure that
$\Mpob$ is an $f$-aslgebra.

\medskip

\qquad\grheadd\ As in 612Bc, ($\beta$) and ($\gamma$) are enough to ensure
that $\Mpob$ is an $f$-algebra.

\medskip

\quad{\bf (ii)} This is elementary;  all we need to know is that
$|\pmb u|$ is moderately oscillatory and $|\pmb u_<|=|\pmb u|_<$ (641G(e-i)).

\medskip

\quad{\bf (iii)} There is a $\pmb u\in\Mmo^+$ such that
$|\pmb x|\le\pmb u_<$;  now $|z|\pmb u=|z|\tbf 1\times\pmb u$
is moderately oscillatory (615F(a-iii)) and non-negative, and
$|z\pmb x|\le|z|\pmb u_<=(|z|\pmb u)_<$ (641G(a-iv)), so
$z\pmb x$ is previsibly order-bounded.

\medskip

\quad{\bf (iv)} This was covered in (i-$\gamma$) above.

\medskip

{\bf (b)(i)} $\widehat{\theta}^{\#}_{\pmb v}(\pmb x+\tilde{\pmb x})
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
+\widehat{\theta}^{\#}_{\pmb v}(\tilde{\pmb x})$ for all
$\pmb x$, $\tilde{\pmb x}\in \Mpob$.
\Prf\ For any $\epsilon>0$, there are non-decreasing uniformly order-bounded
sequences
$\sequencen{\pmb u_{n}}$, $\sequencen{\tildepmb u_{n}}$ of
non-negative moderately oscillatory processes, all with domain $\Cal S$,
such that

\Centerline{$|\pmb x|
\le\sup_{n\in\Bbb N}\pmb u_{n<}$,
\quad$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)+\epsilon$,}

\Centerline{$|\tilde{\pmb x}|
\le\sup_{n\in\Bbb N}\tildepmb u_{n<}$,
\quad$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\tildepmb u_{n}\,d\pmb v)
\le\widehat{\theta}^{\#}_{\pmb v}(\tilde{\pmb x})+\epsilon$.}

\noindent Now
$\sequencen{\pmb u_{n}+\tildepmb u_{n}}$ is a non-decreasing
uniformly order-bounded sequence of non-negative moderately oscillatory
processes, $|\pmb x+\tilde{\pmb x}|
\le\sup_{n\in\Bbb N}(\pmb u_{n}+\tildepmb u_{n})_<$ and

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v}(\pmb x+\tilde{\pmb x})
&\le\sup_{n\in\Bbb N}
  \theta(\int_{\Cal S}(\pmb u_{n}+\tildepmb u_{n})\,d\pmb v)\cr
&\le\sup_{n\in\Bbb N}
  \bigl(\theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v)
    +\theta(\int_{\Cal S}\tildepmb u_{n})\,d\pmb v\bigr)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
    +\widehat{\theta}^{\#}_{\pmb v}(\tilde{\pmb x})
    + 2\epsilon.\cr}$$

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

\medskip

\quad{\bf (ii)} Suppose that $\pmb x\in\Mpob$.
Let $\pmb u\in\Mmo^+$ be such that $|\pmb x|\le\pmb u_<$.
Then

\Centerline{$|\alpha\pmb x|\le|\alpha|\pmb u_<=(|\alpha|\pmb u)_<$}

\noindent for every $\alpha$, so

$$\eqalign{\limsup_{\alpha\to 0}
  \widehat{\theta}^{\#}_{\pmb v}(\alpha\pmb x)
&\le\limsup_{\alpha\to 0}
  \theta(\int_{\Cal S}|\alpha|\pmb u\,d\pmb v)
\cr&=\limsup_{\alpha\to 0}
  \theta(\alpha\int_{\Cal S}\pmb u\,d\pmb v)
=0.\cr}$$

\noindent So $\widehat{\theta}^{\#}_{\pmb v}$ is an F-seminorm.

\medskip

\quad{\bf (iii)} Immediately from
the definition in 645Bb, we see that
if $\pmb x$, $\pmb x'\in \Mpob$ and
$|\pmb x|\le|\pmb x'|$ then
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x')$.

\medskip

\quad{\bf (iv)} If $\sequencen{\pmb u_{n}}$ is a non-decreasing
sequence in $\Mmo^+$ such that
$|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$, then
$\sequencen{|\alpha|\pmb u_{n}}$ is a non-decreasing
sequence in $\Mmo^+$ and
$|\alpha\pmb x|\le\sup_{n\in\Bbb N}|\alpha|\pmb u_{n<}$, so

$$\eqalignno{\widehat{\theta}^{\#}_{\pmb v}(\alpha\pmb x)
&\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}|\alpha|\pmb u_{n}\,d\pmb v)
  \cr
&=\sup_{n\in\Bbb N}\theta(|\alpha|\int_{\Cal S}\pmb u_{n}\,d\pmb v)
\le\sup_{n\in\Bbb N}\max(1,|\alpha|)
   \theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v)
\Displaycause{613Ba}
=\max(1,|\alpha|)\sup_{n\in\Bbb N}
   \theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v).\cr}$$

\noindent As $\sequencen{\pmb u_{n}}$ is arbitrary,
$\widehat{\theta}^{\#}_{\pmb v}(\alpha\pmb x)
\le\max(1,|\alpha|)\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$;  in particular,
$\widehat{\theta}^{\#}_{\pmb v}(\alpha\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$ if $|\alpha|\le 1$.

\medskip

\quad{\bf (v)} If $\pmb x\in\Mpob$,
take a non-negative $\pmb u\in\Mmo$ such that
$|\pmb x|\le\pmb u_<$.   For any $\alpha\in\Bbb R$,
$|\alpha\pmb x|\le|\alpha|\pmb u_<$ so

\Centerline{$\limsup_{\alpha\to 0}\widehat{\theta}^{\#}_{\pmb v}(\alpha\pmb x)
\le\limsup_{\alpha\to 0}\theta(\int_{\Cal S}|\alpha|\pmb u\,d\pmb v)
=\limsup_{\alpha\to 0}\theta(|\alpha|\int_{\Cal S}\pmb u\,d\pmb v)
=0$.}

\noindent So all the conditions of 2A5B are satisfied, and
$\widehat{\theta}^{\#}_{\pmb v}$ is an F-seminorm.

\medskip

{\bf (c)} If $\sequencen{\pmb u_{n}}$ is a uniformly order-bounded
non-decreasing sequence of non-negative moderately oscillatory
processes with domain
$\Cal S$ and $|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$, then
$0\le\int_{\Cal S}\pmb u_{n}\,d\pmb v
\le\int_{\Cal S}\pmb u_{n}\,d\pmb v'$, so
$\theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v)
\le\theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v')$ for every $n$;
consequently

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_{n}\,d\pmb v')$.}

\noindent As $\sequencen{\pmb u_{n}}$ is arbitrary,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v'}(\pmb x)$.
}%end of proof of 645D

\leader{645E}{The topology $\TSi$:  Proposition}
Let $\Cal S$ be a sublattice of $\Cal T$.

(a)(i) We have a linear space topology
$\TSi$ on $\Mpob=\Mpob(\Cal S)$ defined by
the functionals $\widehat{\theta}^{\#}_{\pmb v}$ as $\pmb v$ runs over
$\Mnsu(\Cal S)$.

\quad(ii) If $\pmb x\in G\in\TSi$, there are $\pmb v\in\Mnsu(\Cal S)$
and a $\delta>0$ such that
$\{\pmb x':\pmb x'\in\Mpob$,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x'-\pmb x)\le\delta\}\subseteq G$.

\quad(iii) For any $\tau\in\Cal S$, the coordinate projection
$\sfamilysS{x}\mapsto x_{\tau}:\Mpob\to L^0$ is continuous for $\TSi$ and
the topology of convergence in measure on $L^0$.

\quad(iv) $\TSi$ is Hausdorff.

\quad(v)($\alpha$) For any $\pmb x\in\Mpob$, the map
$\pmb x'\mapsto\pmb x'\times\pmb x:\Mpob\to\Mpob$ is continuous.

\qquad($\beta$) If $A\subseteq\Mpob$ is uniformly previsibly
order-bounded, then
$(\pmb x,\pmb x')\mapsto\pmb x\times\pmb x':A\times A\to\Mpob$
is uniformly continuous.

\quad(vi) If
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ then
$\pmb x'\mapsto z\pmb x':\Mpob\to\Mpob$ is continuous.

(b) Let $\frak S$ be the linear space topology on
$\Mmo=\Mmo(\Cal S)$ defined
by the F-seminorms $\pmb u\mapsto\theta(\int_{\Cal S}|\pmb u|\,d\pmb v)$
as $\pmb v$ runs over $\Mnsu(\Cal S)$.   Then
$\pmb u\mapsto\pmb u_<:\Mmo\to\Mpob$ is continuous for $\frak S$ and
$\TSi$.   Consequently
$\pmb u\mapsto\pmb u_<:\Mmo\to\Mpob$ is continuous for the ucp topology
on $\Mmo$ and $\TSi$.

\proof{{\bf (a)(i)} By 645Db and 2A5B,
the functionals $\widehat{\theta}^{\#}_{\pmb v}$, for
$\pmb v\in\Mnsu=\Mnsu(\Cal S)$,
define a linear space topology on $\Mpob$.

\medskip

\quad{\bf (ii)} By the definition of $\TSi$ (2A5B, 2A3Fc) there are
$\pmb v_0,\ldots,\pmb v_n\in\Mnsu$ and a $\delta>0$ such that

\Centerline{$\{\pmb x':\pmb x'\in\Mpob$,
$\max_{i\le n}\widehat{\theta}^{\#}_{\pmb v_i}(\pmb x'-\pmb x)
  \le\delta\}\subseteq G$.}

\noindent Set $\pmb v=\sum_{i=0}^n\pmb v_i$;  then $\pmb v\in\Mnsu$ and
$\pmb v_j\preccurlyeq\pmb v$ so
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x')
\ge\widehat{\theta}^{\#}_{\pmb v_j}(\pmb x')$ for every
$\pmb x'\in\Mpob$ and $j\le n$ (645Dc).   Accordingly
$\{\pmb x':\pmb x'\in\Mpob$,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x'-\pmb x)\le\delta\}\subseteq G$.

\medskip

\quad{\bf (iii)} Let $\pmb v$ be the simple process
defined on $\Cal S$ by the formula in 612Ka, taking the
breakpoint string to be $(\tau)$, the base value $v_*$ to be $0$, and
$v_0=\chi 1$.   Observe that the starting value
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ is
$\chi(1\Bsetminus e_{\tau})$ where
$e_{\tau}=\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau}$ (614Ba).   Of course
$\pmb v$ is non-negative and non-decreasing.

If $\pmb u=\sfamilysS{u}$ is a moderately oscillatory process, then

$$\eqalignno{\int_{\Cal S}\pmb u\,d\pmb v
&=u_{<\tau}\times(v_{\tau}-v_{\downarrow})\cr
\displaycause{641J}
&=u_{<\tau}\times\chi e_{\tau}
=u_{<\tau}.\cr}$$

\noindent Let $\sequencen{\pmb u_n}
=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$
be any non-decreasing sequence
in $\Mmo^+$ such that $|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$.
Then $|x_{\tau}|\le\sup_{n\in\Bbb N}u_{n<\tau}$ so

\Centerline{$\theta(x_{\tau})
\le\sup_{n\in\Bbb N}\theta(u_{n<\tau})
=\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)$.}

\noindent As $\sequencen{\pmb u_n}$ is arbitrary,
$\theta(x_{\tau})\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$.
This shows that $\sfamilysS{x}\mapsto x_{\tau}$ is continuous at $\tbf 0$.
But as it is a linear operator, it is continuous everywhere in $\Mpob$.

\medskip

\quad{\bf (iv)} It follows at once that $\TSi$ is Hausdorff, being finer
than the topology induced on $\Mpob$ by the product topology on
$(L^0)^{\Cal S}$.

\medskip

\quad{\bf (v)}\grheada\ I noted in 645D(a-ii) that $\Mpob$ is closed under
multiplication.   Take $\pmb v\in\Mnsu$
and $\epsilon>0$.   There is a $\pmb u\in\Mmo^+$ such that
$|\pmb x|\le\pmb u_<$.   Let $M\ge 1$ be such that
$\bar\mu\Bvalue{\sup|\pmb u|\ge M}\le\epsilon$;  then

$$\eqalignno{\Bvalue{\int_{\Cal S}\pmb u'\times\pmb u\,d\pmb v
  >M\int_{\Cal S}\pmb u'\,d\pmb v}&
=\Bvalue{\int_{\Cal S}(\pmb u'\times\pmb u-M\pmb u')d\pmb v>0}
\Bsubseteq\Bvalue{\int_{\Cal S}(\pmb u'\times\pmb u-M\pmb u')^+d\pmb v>0}\cr&
\Bsubseteq\Bvalue{(\pmb u'\times\pmb u-M\pmb u')^+\ne\tbf 0}\cr&
\Displaycause{613Jd}
\Bsubseteq\Bvalue{\sup|\pmb u|\ge M}\cr}$$

\noindent for every $\pmb u'\in\Mmo^+$, so

\Centerline{$\theta(\int_{\Cal S}\pmb u'\times\pmb u\,d\pmb v)
\le\epsilon+M\theta(\int_{\Cal S}\pmb u'\,d\pmb v)$}

\noindent for every $\pmb u'\in\Mmo^+$.   Now suppose that
$\pmb x'\in\Mmo^+$ and
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x')<\Bover{\epsilon}M$.
Then there is a non-decreasing uniformly order-bounded sequence
$\sequencen{\pmb u_n}$ in $\Mmo^+$ such that
$|\pmb x'|\le\sup_{n\in\Bbb N}\pmb u_{n<}$ and
$\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)\le\Bover{\epsilon}M$ for every
$n$.   In this case,
$\sequencen{\pmb u_n\times\pmb u}$
is a non-decreasing uniformly order-bounded sequence
in $\Mns^+$ such that

\Centerline{$|\pmb x'\times\pmb x|
\le|\pmb x'|\times\pmb u_<
\le\sup_{n\in\Bbb N}\pmb u_{n<}\times\pmb u_<
=\sup_{n\in\Bbb N}(\pmb u_n\times\pmb u)_<$,}

\noindent so

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb x'\times\pmb x)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\times\pmb u\,d\pmb v)
\le 2\epsilon$.}

\noindent As $\pmb v$ and $\epsilon$ are arbitrary, the linear operator
$\pmb x'\mapsto\pmb x'\times\pmb x$ is continuous at $\tbf 0$, therefore
continuous.

\medskip

\qquad\grheadb\ Let
$\pmb u\in\Mmo^+$ be such that $|\pmb x|\le\pmb u_<$ for every
$\pmb x\in A$.   Let $G$ be a $\TSi$-neighbourhood of $\tbf 0$.
By (ii) above, there are a $\pmb v\in\Mnsu$
and a $\delta>0$ such that $\pmb x\in G$ whenever $\pmb x\in\Mpob$ and
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)\le\delta$.
By ($\alpha$), $\pmb x\mapsto\pmb x\times\pmb u_<$ is
continuous, so there is a $\TSi$-neighbourhood $G'$ of $0$ such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x\times\pmb u_<)\le\bover12\delta$
for every $\pmb x\in G'$.   Now suppose that
$\pmb x_1$, $\pmb x_2$, $\pmb x'_1$, $\pmb x'_2\in A$ and both
$\pmb x_1-\pmb x_2$ and $\pmb x_1'-\pmb x'_2$ belong to $G'$.
Then

$$\eqalign{|(\pmb x_1\times\pmb x'_1)-(\pmb x_2-\pmb x'_2)|
&\le|\pmb x_1|\times|\pmb x'_1-\pmb x'_2|
  +|\pmb x_1-\pmb x_2|\times|\pmb x'_2|
\cr&\le|\pmb x'_1-\pmb x'_2|\times\pmb u_<
  +|\pmb x_1-\pmb x_2|\times\pmb u_<
\cr}$$

\noindent and

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}
  ((\pmb x_1\times\pmb x'_1)-(\pmb x_2\times\pmb x'_2))
\le\widehat{\theta}^{\#}_{\pmb v}((\pmb x'_1-\pmb x'_2)\times\pmb u_<)
  +\widehat{\theta}^{\#}_{\pmb v}((\pmb x_1-\pmb x_2)\times\pmb u_<)
\le\delta$,}

\noindent so $(\pmb x_1\times\pmb x'_1)-(\pmb x_2\times\pmb x'_2)\in G$.
As $G$ is arbitrary, $(\pmb x,\pmb x')\mapsto\pmb x\times\pmb x'$ is
uniformly continuous on $A\times A$.

\medskip

\quad{\bf (vi)} This is the special case of (v-$\alpha$)
in which $\pmb x=z\tbfoneminus{\Cal S}$.

\medskip

{\bf (b)} We have pnly to look at the definition in 645Bb;
$\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<)
\le\theta(\int_{\Cal S}|\pmb u|d\pmb v)$ for all relevant $\pmb u$ and
$\pmb v$, so we shall be able to apply 2A3H to see that
$\pmb u\mapsto\pmb u_<$ is continuous for $\frak S$ and
$\TSi$.   As for the ucp topology, this is finer than $\frak S$, because
if $\pmb v\in\Mnsu$ then
$\theta(\int_{\Cal S}\pmb u\,d\pmb v)
\le\theta(\sup|\pmb u|\times\sup|\pmb v|)$.   So
$\pmb u\mapsto\pmb u_<$ is continuous for the ucp topology and $\TSi$.
}%end of proof of 645E

\leader{645F}{Definitions} Let $\Cal S$ be a sublattice of $\Cal T$.

\spheader 645Fa I will call the topology $\TSi$ defined in 645E the
{\bf S-integration topology} on $\Mpob(\Cal S)$.   As it is a linear space
topology\cmmnt{ (645E(a-i))}, there is an associated uniformity\cmmnt{ (3A4Ad)}
which I will call the {\bf S-integration uniformity}.

\spheader 645Fb $\MSi^0(\Cal S)$ will be the $\TSi$-closure of
$\{\pmb u_<:\pmb u\in\Mmo(\Cal S)\}$ in $\Mpob(\Cal S)$.

\spheader 645Fc $\MSi(\Cal S)$ will be the set of fully adapted processes
$\pmb x$ with domain $\Cal S$ such that
$\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0(\Cal S)$.

\leader{645G}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\TSi$ the S-integration topology on $\cmmnt{\Mpob=\mskip5mu}\Mpob(\Cal S)$.
If $\sequencen{\pmb x_n}$ is a uniformly previsibly order-bounded
$\TSi$-Cauchy sequence in $\Mpob$, then it is $\TSi$-convergent.

\proof{{\bf (a)} By the definition of `uniformly previsibly order-bounded',
there must be a non-negative
$\pmb u\in\Mmo=\Mmo(\Cal S)$ such that $|\pmb x_n|\le\pmb u_<$ for
every $n\in\Bbb N$.   Express each $\pmb x_n$ as
$\family{\sigma}{\Cal S}{x_{n\sigma}}$.   Because the coordinate projections from
$\Mpob$ to $L^0$ are continuous linear operators, therefore uniformly
continuous (4A5Hd),
$\sequencen{x_{n\sigma}}$ is a Cauchy sequence in $L^0$ for each
$\sigma\in\Cal S$ (4A2Ji),
with a limit $x_{\sigma}$ because $L^0$ is a complete linear topological
space (613Bh).   Now $\pmb x=\sfamilysS{x}$ is fully adapted
(613Bl) and $|\pmb x|\le\pmb u_<$ (613Bm), so $\pmb x\in\Mpob$.

\medskip

{\bf (b)} Take $\pmb v\in\Mnsu(\Cal S)$ and $\epsilon>0$.
Then we have a non-decreasing sequence $\sequence{k}{n_k}$ such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x_m-\pmb x_{n_k})
\le 2^{-k}\epsilon$ whenever $k$, $m\in\Bbb N$ and $n_k\le m$.   For each
$k\in\Bbb N$ choose a non-decreasing uniformly order-bounded sequence
$\sequence{i}{\pmb u_{ki}}$ in $\Mmo^+$ such that
$|\pmb x_{n_{k+1}}-\pmb x_{n_k}|
\le\sup_{i\in\Bbb N}\pmb u_{ki<}$ and
$\sup_{i\in\Bbb N}\theta(\int_{\Cal S}\pmb u_{ki}\,d\pmb v)
\le 2^{-k+1}\epsilon$.   Set
$\pmb u'_m=2\pmb u\wedge\sum_{k=0}^m\pmb u_{km}$ for
$m\in\Bbb N$.   Then $\sequence{m}{\pmb u'_m}$ is a
non-decreasing uniformly order-bounded sequence in $\Mmo^+$.
The point is that
$|\pmb x-\pmb x_{n_0}|
\le\sup_{m\in\Bbb N}\pmb u'_{m<}$.
\Prf\ Note first that for each $\sigma\in\Cal S$,
$x_{\sigma}-x_{n_0\sigma}
=\lim_{k\to\infty}x_{n_k\sigma}-x_{n_0\sigma}$.
This is a limit for the topology of convergence in measure, but as we know
that $|\pmb x_{n_k}|\le\pmb u_<$ for every $k$, we must certainly have

\Centerline{$|x_{\sigma}-x_{n_0\sigma}|
\le\sup_{k\in\Bbb N}2u_{<\sigma}
  \wedge|x_{n_k\sigma}-x_{n_0\sigma}|$}

\noindent for every $\sigma$.   (Here I am thinking of $\pmb u$ as
$\sfamilysS{u}$ and its previsible version
$\pmb u_<$ as $\family{\sigma}{\Cal S}{u_{<\sigma}}$,
in the manner of 641E-641F.)   Thus we have

$$\eqalignno{|\pmb x-\pmb x_{n_0}|
&\le\sup_{k\ge 1}2\pmb u_<\wedge|\pmb x_{n_k}-\pmb x_{n_0}|
\le\sup_{k\ge 1}2\pmb u_<
   \wedge\sum_{j=0}^{k-1}|\pmb x_{n_{j+1}}-\pmb x_{n_j}|\cr
&\le\sup_{k\ge 1}2\pmb u_<
   \wedge\sum_{j=0}^{k-1}\sup_{i\ge k}\pmb u_{ji<}
=\sup_{k\ge 1}\sup_{i\ge k}2\pmb u_<
   \wedge\sum_{j=0}^{k-1}\pmb u_{ji<}
\Displaycause{because $\sequence{i}{\pmb u_{ji<}}$
is non-decreasing for each $j$}
\le\sup_{k\ge 1}\sup_{i\ge k}\pmb u'_{i<}
=\sup_{m\in\Bbb N}\pmb u'_{m<}.  \text{ \Qed}\cr}$$

\medskip

{\bf (c)} Accordingly, if $n\ge n_0$,

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb x_n)
&\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x_n-\pmb x_{n_0})
  +\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb x_{n_0})\cr
&\le\epsilon
  +\sup_{m\in\Bbb N}\theta(\int_{\Cal S}\pmb u'_m\,d\pmb v)
\le\epsilon+\sup_{m\in\Bbb N}
    \sum_{k=0}^m\theta(\int_{\Cal S}\pmb u_{km}\,d\pmb v)
\le 5\epsilon.\cr}$$

\noindent As $\epsilon$ and $\pmb v$ are arbitrary, $\pmb x$ is the
$\TSi$-limit of $\sequencen{\pmb x_n}$.
}%end of proof of 645G

\leader{645H}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$.
Suppose that $\sequencen{\pmb x_n}$ is a
uniformly previsibly order-bounded sequence in
$\cmmnt{\MSi^0=\mskip5mu}\MSi^0(\Cal S)$
which is order*-convergent to $\pmb x\in(L^0)^{\Cal S}$.   Then
$\pmb x\in\MSi^0$ and $\sequencen{\pmb x_n}$ converges to $\pmb x$
for the S-integration topology\cmmnt{ $\TSi$}.

\proof{{\bf (a)} The first thing to note is that if a sequence
$\sequencen{\pmb x_n}=\sequencen{\langle x_{n\sigma}\rangle_{\sigma\in\Cal S}}$
in $\Mfa=\Mfa(\Cal S)$ is order*-convergent in $(L^0)^{\Cal S}$ to
$\pmb x=\sfamilysS{x}$, and $\tau\in\Cal S$, then
$\sequencen{x_{n\tau}}$ is order*-convergent
to $x_{\tau}$ in $L^0$, therefore convergent in the topology of convergence in
measure (367Ma).   So $\sequencen{\pmb x_n}\to\pmb x$ for the product
topology on $(L^0)^{\Cal S}$, and $\pmb x\in\Mfa$ (613Bl).
If moreover
$\sequencen{\pmb x_n}$ is uniformly previsibly order-bounded, that is, there
is a $\pmb u\in\Mmo^+=\Mmo(\Cal S)^+$
such that $|\pmb x_n|\le\pmb u_<$ for every
$n$, then $|\pmb x|\le\pmb u_<$ so $\pmb x\in\Mpob=\Mpob(\Cal S)$.

Secondly, if a sequence in $\Mpob$ is simultaneously order*-convergent and
$\TSi$-convergent, the limits must be the same, because the coordinate
projections are $\TSi$-continuous (645E(a-iii)), that is, $\TSi$ is finer than
the topology induced by the product topology on $(L^0)^{\Cal S}$.   And if
a sequence in $\MSi^0$ is $\TSi$-convergent in $\Mpob$, its
limit belongs to $\MSi^0$, because $\MSi^0$ is defined to be a
$\TSi$-closure.

\medskip

{\bf (b)} So all we need to prove is that, under the conditions of the
theorem, $\sequencen{\pmb x_n}$ is $\TSi$-convergent in $\Mpob$;
and by 645G it will in fact be enough to show that it is $\TSi$-Cauchy.
I seek to do this successively more complex
sequences $\sequencen{\pmb x_n}$.   First, suppose it is of the form
$\sequencen{\pmb u_{n<}}$ where $\sequencen{\pmb u_n}$
is a non-decreasing sequence in $\Mmo^+$ with an upper bound
$\pmb u\in\Mmo$.   Then $\sequencen{\pmb x_n}$ is a non-decreasing
sequence with an upper bound $\pmb u_<$, so is order*-convergent to
its supremum $\pmb x$ in $(L^0)^{\Cal S}$,
which is a fully adapted process (612Ia).

As $0\le\pmb x\le\pmb u_<$, $\pmb x\in\Mpob$.
If $\pmb v\in\Mnsu(\Cal S)$,
then $\sequencen{\int_{\Cal S}\pmb u_n\,d\pmb v}$
is a non-decreasing sequence in $L^0$ with an upper bound
$\int_{\Cal S}\pmb u\,d\pmb v$, so has a limit in $L^0$ (613Ba) and
is Cauchy.   For any $m\in\Bbb N$,

\Centerline{$0\le\pmb x-\pmb x_m
\le\sup_{n\ge m}\pmb u_{n<}-\pmb u_{m<}$,}

\noindent so

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb x_m)
\le\sup_{n\ge m}
  \theta(\int_{\Cal S}\pmb u_n-\pmb u_m\,d\pmb v)$.}

\noindent Thus

\Centerline{$\lim_{m\to\infty}
   \widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb x_m)
\le\lim_{m\to\infty}\sup_{n\ge m}
  \theta(\int_{\Cal S}\pmb u_n-\pmb u_m\,d\pmb v)
=0$;}

\noindent as $\pmb v$ is arbitary,
$\sequencen{\pmb x_n}$ is $\TSi$-convergent to $\pmb x$.

\medskip

{\bf (c)} Suppose next that $\sequencen{\pmb x_n}$ is a non-decreasing
sequence in $(\MSi^0)^+$, and that
it is bounded above by $\pmb u_<$ where $\pmb u\in\Mmo^+$.   Then
again it is
order*-convergent to its supremum $\pmb x$, and $\pmb x\in\Mpob$.
Take $\pmb v\in\Mnsu(\Cal S)$. and
$\epsilon>0$.   For each $n\in\Bbb N$, let $\pmb u_n\in\Mmo$ be such
that $\widehat{\theta}^{\#}_{\pmb v}(\pmb x_n-\pmb u_{n<})
\le 2^{-n}\epsilon$.   For each $n$,
set $\pmb u'_n=\med(\tbf 0,\sup_{i\le n}\pmb u_i,\pmb u)$;  then

$$\eqalignno{|\pmb x_n-\pmb u'_{n<}|
&=|\med(\tbf 0,\sup_{i\le n}\pmb x_i,\pmb u_<)
   -\med(\tbf 0,\sup_{i\le n}\pmb u_{i<},\pmb u_<)|
\Displaycause{641Ge}
\le|\sup_{i\le n}\pmb x_i-\sup_{i\le n}\pmb u_{i<}|
\le\sum_{i=0}^n|\pmb x_i-\pmb u_{i<}|,\cr}$$

\noindent so

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}
   (\pmb x_n-\pmb u'_{n<})
\le\sum_{i=0}^n\widehat{\theta}^{\#}_{\pmb v}
   (\pmb x_i-\pmb u_{i<})
\le 2\epsilon$.}

\noindent Now we know from (b) that $\sequencen{\pmb u'_{n<}}$ is
$\TSi$-convergent, so there is an $m\in\Bbb N$ such that
$\widehat{\theta}^{\#}_{\pmb v}
   (\pmb u'_{n<}-\pmb u'_{m<})\le\epsilon$ for every
$n\ge m$;  in which case $\widehat{\theta}^{\#}_{\pmb v}
   (\pmb x_{n<}-\pmb x_{m<})\le 5\epsilon$ for every
$n\ge m$.   As $\pmb v$ and $\epsilon$ are arbitrary,
$\sequencen{\pmb x_n}$ is Cauchy, therefore convergent, and the limit
must be $\pmb x$.

\medskip

{\bf (d)} Applying (c) to $\sequencen{\pmb x_n-\pmb x_{0}}$, we see
that if $\sequencen{\pmb x_n}$ is a non-decreasing previsibly order-bounded
sequence in $\MSi^0$, it will be $\TSi$-convergent;
it follows at once that if
$\sequencen{\pmb x_n}$ is a non-increasing previsibly order-bounded
sequence in $\MSi^0$, it is $\TSi$-convergent.

\medskip

{\bf (e)} Now let $\sequencen{\pmb x_n}$ be any previsibly order-bounded
sequence in $\MSi^0$ which is order*-convergent to
$\pmb x\in (L^0)^{\Cal S}$.   For each $n\in\Bbb N$, set
$\overline{\pmb x}_n=\sup_{i\ge n}\pmb x_i$, the supremum being
taken in $(L^0)^{\Cal S}$.   Then $\overline{\pmb x}_n$ is the
order*-convergent limit of the non-decreasing sequence
$\sequence{m}{\sup_{i\le m}\pmb x_{n+i}}$ in $\MSi^0$, which of course is
bounded above in $\Mpob$ because $\sequence{i}{\pmb x_i}$ is,
so $\overline{\pmb x}_n\in\MSi^0$, by (d).    We now see that
$\sequencen{\overline{\pmb x}_n}$ is a non-increasing sequence in
$\MSi^0$, bounded below in $\Mpob$, which is order*-convergent to
$\pmb x$.   So $\pmb x\in\MSi^0$ and is the $\TSi$-limit of
$\sequencen{\overline{\pmb x}_n}$.

Similarly,
$\underline{\pmb x}_n=\inf_{i\ge n}\pmb x_i$ belongs to $\MSi^0$
for every $n$, and $\pmb x$ is the $\TSi$-limit of
$\sequencen{\underline{\pmb x}_n}$.

To see that $\pmb x$ is in fact the $\TSi$-limit of the original sequence
$\sequencen{\pmb x_n}$, take any
$\pmb v\in\Mnsu(\Cal S)$.   If $n\in\Bbb N$, then
$|\pmb x_n-\pmb x|
\le\overline{\pmb x}_n-\underline{\pmb x}_n$, so

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v}(\pmb x_n-\pmb x)
&\le\widehat{\theta}^{\#}_{\pmb v}
    (\overline{\pmb x}_n-\underline{\pmb x}_n)\cr
&\le\widehat{\theta}^{\#}_{\pmb v}(\overline{\pmb x}_n-\pmb x)
  +\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\underline{\pmb x}_n)
\to 0\cr}$$

\noindent as $n\to\infty$.   As $\pmb v$ is arbitrary,
$\sequencen{\pmb x_n}$ is topologically convergent to $\pmb x$.
This completes the proof.
}%end of proof of 645H

\leader{645I}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$.
If $\pmb x\in\Mpob(\Cal S)$ is a previsible process\cmmnt{ (642C)},
then $\pmb x\in\MSi^0\cmmnt{\mskip5mu=\MSi^0(\Cal S)}$.

\proof{ Let $\pmb u\in\Mmo(\Cal S)^+$ be such that $|\pmb x|\le\pmb u_<$.
The set

\Centerline{$\{\pmb x':
\pmb x'\in(L^0)^{\Cal S}$, $\med(-\pmb u_<,\pmb x',\pmb u_<)
\in\MSi^0\}$}

\noindent contains $\pmb u'_<$ for every $\pmb u'\in\Mns(\Cal S)$
and is closed under order*-convergence, by 645H, so contains
$\pmb x=\med(-\pmb u_<,\pmb x,\pmb u_<)$.
}%end of proof of 645I

\leader{645J}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$k\ge 1$ an integer and $h:\BbbR^k\to\Bbb R$ a locally bounded Borel
measurable function.   Write $\MSi^0$, $\MSi$ for $\MSi^0(\Cal S)$,
$\MSi(\Cal S)$.

(a) If $\pmb X\in(\MSi^0)^k$, then
$\bar h\pmb X\times\tbfoneminus{\Cal S}\in\MSi^0$;
if $h(0,\ldots,0)=0$, then $\bar h\pmb X\in\MSi^0$.

(b) $\bar h\pmb X\in\MSi$ for every $\pmb X\in\MSi^k$.

\proof{ Express $\pmb X$ as $\ofamily{i}{k}{\pmb x_i}$.   For any
$\pmb x\in\Mfa(\Cal S)$, write $R(\pmb x)$ for
$\pmb x\times\tbfoneminus{\Cal S}$.
By 645D(a-i), $R(\bar h\pmb X)\in\Mpob=\Mpob(\Cal S)$.

\medskip

{\bf (a)(i)} Let $g:\BbbR^k\to\Bbb R$ be a continuous function.

\medskip

\qquad\grheada\ If $\pmb v\in\Mnsu(\Cal S)$ and
$\epsilon>0$, there is an $M\ge 0$ such that
$\widehat{\theta}^{\#}_{\pmb v}(R(\bar g\pmb X)-R(\bar g\pmb X'))
\le\epsilon$, where $\pmb X'=\ofamily{i}{k}
  {\med(-M\tbfoneminus{\Cal S},\pmb x_i,M\tbfoneminus{\Cal S})}$.
\Prf\ There is a $\pmb u\in\Mmo^+=\Mmo(\Cal S)^+$ such that
$|\pmb x_i|\le\pmb u_<$ for every $i<k$.   Set
$\bar u=\sup|\pmb u|$, and let $M\ge 0$ be such
that $\bar\mu\Bvalue{\bar u\ge M}\le\epsilon$.   In this case, setting
$\pmb x_i'=\med(-M\tbfoneminus{\Cal S},\pmb x_i,M\tbfoneminus{\Cal S})$,
we have

\Centerline{$|\pmb x_i-\pmb x_i'|=(|\pmb x_i|-M\tbfoneminus{\Cal S})^+
\le(\pmb u_<-M\tbfoneminus{\Cal S})^+
=((\pmb u-M\tbf 1^{(\Cal S)})^+)_<$.}

\noindent So, writing $\pmb X'$ for $\ofamily{i}{k}{\pmb x'_i}$,

$$\eqalignno{\Bvalue{R(\bar g\pmb X)\ne R(\bar g\pmb X')}
&\Bsubseteq\Bvalue{\bar g\pmb X\ne\bar g\pmb X'}
\Bsubseteq\sup_{i<k}\Bvalue{\pmb x_i\ne\pmb x_i'}\cr
\displaycause{612Sc, 619Ec}
&\Bsubseteq\Bvalue{((\pmb u-M\tbf 1^{(\Cal S)})^+)_<\ne 0}.\cr}$$

\noindent Next, $|\pmb x'_i|\le|\pmb x_i|$ so $\pmb x'_i\in\Mpob$ for
every $i$,
$R(\bar g\pmb X')\in\Mpob$, $R(\bar g\pmb X)-R(\bar g\pmb X')\in\Mpob$
and there is a
$\tilde{\pmb u}$ in $\Mmo^+$ such that
$|R(\bar g\pmb X)-R(\bar g\pmb X')|
\le\tilde{\pmb u}_<$.   Setting
$\tildepmb u_n
=\tilde{\pmb u}\wedge n(\pmb u-M\tbf 1^{(\Cal S)})^+$ for
$n\in\Bbb N$, $\sequencen{\tildepmb u_n}$ is a non-decreasing
uniformly order-bounded sequence in $\Mmo^+$ and

\Centerline{$|R(\bar g\pmb X)-R(\bar g\pmb X')|
\le\sup_{n\in\Bbb N}\tilde{\pmb u}_<
   \wedge n(\pmb u_<-M\tbfoneminus{\Cal S})^+
=\sup_{n\in\Bbb N}\tildepmb u_{n<}$.}

\noindent Accordingly

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v}
  (R(\bar g\pmb X)-R(\bar g\pmb X'))
&\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\tildepmb u_n\,d\pmb v)
\le\sup_{n\in\Bbb N}
  \bar\mu\Bvalue{\int_{\Cal S}\tildepmb u_n\,d\pmb v\ne 0}
\cr&\le\sup_{n\in\Bbb N}
  \bar\mu\Bvalue{\tildepmb u_n\ne\tbf 0}
\le\bar\mu\Bvalue{(\pmb u-M\tbf 1^{(\Cal S)})^+\ne 0}
=\bar\mu\Bvalue{\bar u>M}
\le\epsilon,\cr}$$

\noindent as required.\ \Qed

\medskip

\qquad\grheadb\ $R(\bar g\pmb X)\in\MSi^0$.   \Prf\
Take $\pmb v\in\Mnsu(\Cal S)$ and
$\epsilon>0$.   By ($\alpha$), there is an $M\ge 0$ such that, setting
$\pmb X'=\ofamily{i}{k}{\pmb x_i'}$ where
$\pmb x_i'=\med(-M\tbfoneminus{\Cal S},\pmb x_i,M\tbfoneminus{\Cal S})$ for
$i<k$, we have
$\widehat{\theta}^{\#}_{\pmb v}(R(\bar g\pmb X)-R(\bar g\pmb X'))
\le\epsilon$.   Next, let $\delta>0$ be such that
$\widehat{\theta}^{\#}_{\pmb v}(\delta\tbfoneminus{\Cal S})\le\epsilon$
(645Db), and $\eta>0$ such that $|g(\beta)-g(\beta')|\le\delta$ whenever
$\beta$, $\beta'\in[-M,M]^k$ and $\|\beta-\beta'\|_{\infty}\le\eta$;
setting $K=1+\Bover{2}{\eta}\sup_{\|\beta\|_{\infty}\le M}|g(\beta)|$,
we shall have
$|g(\beta)-g(\beta')|\le\delta+K\|\beta-\beta'\|_{\infty}$ whenever
$\beta$, $\beta'\in[-M,M]^k$, while $K\ge 1$.
For each $i<k$, because $\pmb x_i\in\MSi^0$, there is a
$\pmb u_i\in\Mmo=\Mmo(\Cal S)$ such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x_i-\pmb u_{i<})
\le\Bover{\epsilon}K$;  set $\pmb U'=\ofamily{i}{k}{\pmb u_i'}$
and $\pmb U'_<=\ofamily{i}{k}{\pmb u'_{i<}}$ where
$\pmb u_i'=\med(-M\tbf 1^{(\Cal S)},\pmb u_i,M\tbf 1^{(\Cal S)})$ for
$i<k$.   Note that
$\pmb u_{i<}'
=\med(-M\tbfoneminus{\Cal S},\pmb u_{i<},M\tbfoneminus{\Cal S})$
for $i<k$.

Since $|\pmb x_i'-\pmb u'_{i<}|\le|\pmb x_i-\pmb u_{i<}|$,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x_i'-\pmb u'_{i<})
\le\Bover{\epsilon}K$ for each $i$;  since $\|\pmb x_i'\|_{\infty}$
and $\|\pmb u'_{i<}\|_{\infty}$ are both less than or equal to
$M$ for each $i$,

$$\eqalignno{|R(\bar g\pmb X')-\bar g\pmb U'_<|
&=|R(\bar g\pmb X')-R(\bar g\pmb U'_<)|
\Displaycause{641Gd}
=R(|\bar g\pmb X'-\bar g\pmb U'_<|)
\le R(\delta\pmb 1^{(\Cal S)}+K\sum_{i=0}^{k-1}|\pmb x_i'-\pmb u_{i<}|)
\cr&\le\delta\pmb 1^{(\Cal S)}_<+K\sum_{i=0}^{k-1}|\pmb x_i'-\pmb u_{i<}|
\cr}$$

\noindent and

$$\eqalignno{\widehat{\theta}^{\#}_{\pmb v}
  (R(\bar g\pmb X)-\bar g\pmb U')_<)
&\le\widehat{\theta}^{\#}_{\pmb v}(R(\bar g\pmb X)-R(\bar g\pmb X'))
  +\widehat{\theta}^{\#}_{\pmb v}(R(\bar g\pmb X)-\bar g\pmb U'_<)
\cr&\le\epsilon
  +\widehat{\theta}^{\#}_{\pmb v}(\delta\tbfoneminus{\Cal S})
  +K\sum_{i=0}^{k-1}\widehat{\theta}^{\#}_{\pmb v}(\pmb x_i'-\pmb u_{i<})
\cr&\le\epsilon+\epsilon+k\epsilon
=(k+2)\epsilon.\cr}$$

\noindent As $\pmb v$ and $\epsilon$ are
arbitrary, $R(\bar g\pmb X)\in\MSi^0$.\ \Qed

\medskip

\quad{\bf (ii)} Now consider the space $U$ of Borel measurable functions
$f:\BbbR^k\to\Bbb R$ such that $R(\bar f\pmb X)\in\MSi^0$.   Suppose that
$\sequencen{f_n}$ is a pointwise convergent sequence in $U$ with limit $f$, and
that $g=\sup_{n\in\Bbb N}|f_n|$ is locally bounded.
In this case, $|\bar f_n\pmb X|\le\bar g\pmb X$ for
each $n$, while $R(\bar g\pmb X)\in\Mpob$ (645D(a-i)), so
$\sequencen{R(\bar f_n\pmb X)}$ is a uniformly previsibly order-bounded
sequence in $\MSi^0$.   Also $\sequencen{\bar f_n\pmb X}$ is
order*-convergent in $(L^0)^{\Cal S}$ to $\bar f\pmb X$, by 642Bd applied at
each coordinate in $\Cal S$.   So
$\sequencen{R(\bar f_n\pmb X)}$ is order*-convergent to $R(\bar f\pmb X)$,
and 645H tells us that $R(\bar f\pmb X)\in\MSi^0$, that is, $f\in U$.

Since we know from (i) that every continuous function from $\BbbR^k$ to
$\Bbb R$ belongs to $U$, we see that our locally bounded Borel measurable
function $h$ belongs to $U$, by 645Cc, and
$\bar h\pmb X\times\tbfoneminus{\Cal S}\in\MSi^0$.

\medskip

\quad{\bf (iii)} If $h(0,\ldots,0)=0$ then, as in part (a-i) of the proof of
645D, we can use 619Ge to see that

\Centerline{$\bar h\pmb X=\bar hR(\pmb X)=R(\bar h\pmb X)
=R(\bar hR(\pmb X))=R(\bar h\pmb X)\in\MSi^0$.}

\medskip

{\bf (b)} If $\pmb X\in\MSi^k$ then $R(\pmb X)
=(\pmb x_1\times\tbfoneminus{\Cal S},\ldots,\pmb x_k\times\tbfoneminus{\Cal S})$
belongs to $(\MSi^0)^k$ and
$R(\bar hR(\pmb X))\in\MSi^0$, by (a);  but now 619G(e-i) tells us that
$R(\bar h\pmb X)=R(\bar hR(\pmb X))$, so $\bar h\pmb X\in\MSi$.
}%end of proof of 645J

\leader{645K}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$k\ge 1$ an integer, $h:\BbbR^k\to\Bbb R$ a locally bounded Borel
measurable function
and $z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.

(a) $\MSi^0=\MSi^0(\Cal S)$ is an $f$-subalgebra of
$\Mpob(\Cal S)$ and
$z\pmb x\in\MSi^0$ for every $\pmb x\in\MSi^0$.

(b) $\MSi=\MSi(\Cal S)$ is an $f$-subalgebra of
$\Mob(\Cal S)$ and $z\pmb x\in\MSi$ for
every $\pmb x\in\MSi$.

\proof{{\bf (a)} Because the S-integration topology on $\Mpob$
is a linear space topology (645E(a-i)), and
$\{\pmb u_<:\pmb u\in\Mmo(\Cal S)\}$ is a linear subspace of
$\Mpob=\Mpob(\Cal S)$, its closure $\MSi^0$ is a linear subspace (2A5Ec).
Now 645Ja and 612Bc, as usual, show
that $\MSi^0$ is an $f$-subalgebra of $\Mpob$.
Since $z\tbf 1^{\Cal S}\in\Mmo(\Cal S)$,
$z\tbfoneminus{\Cal S}=(z\tbf 1^{\Cal S})_<$ (641G(a-iv)) belongs to
$\MSi^0$, and if $\pmb x\in\MSi^0$ then

\Centerline{$z\pmb x=(z\tbf 1^{\Cal S})\times\pmb x
=z\tbf 1^{\Cal S}\times(\tbfoneminus{\Cal S}\times\pmb x)
=z\tbfoneminus{\Cal S}\times\pmb x
\in\MSi^0$}

\noindent because $\MSi^0$ is closed under $\times$.

\medskip

{\bf (b)(i)} If $\pmb x=\sfamilysS{x}$ belongs to $\MSi$, then
$\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0\subseteq\Mpob$ is order-bounded.
Setting $e_{\sigma}=\sup_{\sigma'\in\Cal S}\Bvalue{\sigma'<\sigma}$ for
$\sigma\in\Cal S$, as in 641Gb, we have

\Centerline{
$\sup_{\sigma\in\Cal S}|x_{\sigma}\times\chi(1\Bsetminus e_{\sigma})|
=\sup|\pmb x\times\tbfoneminus{\Cal S}|=\bar w$}

\noindent say defined in $L^0$.   Set $e=\inf_{\sigma\in\Cal S}e_{\sigma}$.

Take any $\epsilon>0$.   If $\sigma$, $\tau\in\Cal S$
and $\sigma\le\tau$, then $e_{\sigma}\Bsubseteq e_{\tau}$, so there is
a $\tau\in\Cal S$ such that $\bar\mu(e_{\tau}\Bsetminus e)\le\epsilon$.
Now if $\sigma\in\Cal S$,

\Centerline{$\Bvalue{\tau\le\sigma}\Bcup e
\Bsubseteq\Bvalue{\sigma=\tau}\Bcup e_{\sigma}
\Bsubseteq\Bvalue{|x_{\sigma}|=|x_{\tau}|}\Bcup\Bvalue{|x_{\sigma}|\le w}
\Bsubseteq\Bvalue{|x_{\sigma}|\le|x_{\tau}|\vee w}$.}

\Centerline{$\Bvalue{\sigma<\tau}\Bsetminus e
\Bsubseteq e_{\tau}\Bsetminus e$.}

\noindent But this means that
$\{x_{\sigma}\times\chi(1\Bsetminus(e_{\tau}\Bsetminus e)):\sigma\in\Cal S\}$
is order-bounded in $L^0$, while
$\bar\mu(1\Bsetminus(e_{\tau}\Bsetminus e))\ge 1-\epsilon$.   As $\epsilon$ is
arbitrary, $\{x_{\sigma}:\sigma\in\Cal S\}$ is order-bounded, by 613B(p-i), and
$\pmb x\in\Mob(\Cal S)$.

\medskip

\quad{\bf (ii)} $\MSi=\{\pmb x:\pmb x\in\Mfa(\Cal S)$,
$\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0\}$
is an $f$-subalgebra just because
$\pmb y\mapsto\pmb y\times\tbfoneminus{\Cal S}:(L^0)^{\Cal S}\to(L^0)^{\Cal S}$
is a multiplicative
Riesz homomorphism and $\MSi^0$ is an $f$-subalgebra of $\Mpob$.
Similarly, if $\pmb x=\sfamilysS{x}\in\MSi$,
$(z\pmb x)\times\tbfoneminus{\Cal S}
=z(\pmb x\times\tbfoneminus{\Cal S})$ belongs to $\MSi^0$ so
$z\pmb x\in\MSi$.
}%end of proof of 645K

\leader{645L}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$.
Give $\Mpob(\Cal S)$ its S-integration topology $\TSi$.
Suppose that $\pmb x\in\MSi^0(\Cal S)$.

(a) If $\pmb u^*\in\Mmo=\Mmo(\Cal S)$ is such that
$|\pmb x|\le\pmb u^*_<$, then
$A=\{\pmb u:\pmb u\in\Mmo$, $|\pmb u|\le\pmb u^*\}$
is uniformly order-bounded and

\Centerline{$\pmb x\in\overline{\{\pmb u_<:\pmb u\in A\}}
\subseteq\overline{\{\pmb u_<:\pmb u\in\Mmo,\,
\sup|\pmb u|\le\sup|\pmb u^*|\}}$.}

(b) There is a $\pmb w^*\in\Mmo$ such that $\pmb x
\in\overline{\{\pmb u_<:\pmb u\in\Msimp(\Cal S),\,|\pmb u|\le\pmb w^*\}}$.

\proof{{\bf (a)} Setting
$\bar u=\sup|\pmb u^*|$, of course
$\bar u=\sup_{\pmb u\in A}\sup|\pmb u|$, so $A$ is uniformly
order-bounded.

If $G$ is a neighbourhood of $\pmb x$ in $\Mpob=\Mpob(\Cal S)$, there are a
$\pmb v\in\Mnsu=\Mnsu(\Cal S)$ and a $\delta>0$ such that

\Centerline{$\{\pmb x':\pmb x'\in\Mpob$,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x'-\pmb x)\le\delta\}$}

\noindent is included in $G$ (645E(a-ii)).
Since $\pmb x\in\MSi^0=\MSi^0(\Cal S)$, there is a $\pmb u\in\Mmo$ such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<-\pmb x)\le\delta$.   Consider
$\pmb u'=\med(-\pmb u^*,\pmb u,\pmb u^*)$.   Then
$\pmb u'\in\Mmo$ and $\pmb u'\in A$.
Next, because the operation of taking previsible version is a lattice
homomorphism (641Ge again),
$\pmb u'_<=\med(-\pmb u^*_<,\pmb u_<,\pmb u^*_<)$.
Since $|\pmb x|\le\pmb u^*_<$,
$|\pmb x-\pmb u'_<|\le|\pmb x-\pmb u_<|$ and

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u'_<)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u_<)
\le\delta$,}

\noindent so $\pmb u'_<\in G$.

Thus $G$ meets $\{\pmb u_<:\pmb u\in A\}$.   As $G$ is arbitrary,
$\pmb x\in\overline{\{\pmb u_<:\pmb u\in A\}}$.

\medskip

{\bf (b)} As $\pmb x\in\Mpob$, there is certainly a $\pmb u^*\in\Mmo$ such that
$|\pmb x|\le\pmb u^*$;  again write $\bar u$ for $\sup|\pmb u^*|$, and now let
$\pmb w^*$ be the non-negative non-decreasing process defined from $\bar u$ as
in 614Ke, so that $|\pmb u|\le\pmb w^*$ whenever $\pmb u\in\Mfa(\Cal S)$ and
$\sup|\pmb u|\le\bar u$.   Take any $\pmb v\in\Mnsu$ and $\epsilon>0$.   By (a),
there is a $\pmb u\in\Mmo$ such that $|\pmb u|\le\pmb u^*$ and
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u_<)\le\epsilon$.

Let $\delta>0$ be such that $\theta(\int_{\Cal S}|\pmb z|d\pmb v)\le\epsilon$
whenever $\pmb z\in\Mmo$ and $\theta(\sup|\pmb z|)\le\delta$ (616J, applied to
$\Delta\pmb v$).
By 615O, we have a process $\pmb w\in\Mbv(\Cal S)$ such that
$\theta(\sup|\pmb w-\pmb u|)\le\delta$ and $\sup|\pmb w|\le\sup|\pmb u|$.
We can express $\pmb w$ as $\pmb w'-\pmb w''$ where $\pmb w$, $\pmb w''$ are
order-bounded non-negative non-decreasing processes (614I).

Turn now to the construction in 617B.   For any $I\in\Cal I(\Cal S)$ we have
a simple process $\pmb w_I=\family{\sigma}{\Cal S}{w_{I\sigma}}$
defined by saying that $\pmb w_I$ has a breakpoint string in $I$, $\pmb w_I$ and
$\pmb w$ agree on $I$, and
$1\Bsetminus\sup_{\tau\in I}\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{w_{I\sigma}=0}$ for every
$\sigma\in\Cal S$;  and we have corresponding processes $\pmb w'_I$,
$\pmb w''_I$.   Evidently $\pmb w_I=\pmb w'_I-\pmb w''_I$.   Now because
$\pmb w'$ is non-negative and non-decreasing, we must have
$0\le\pmb w'_I\le\pmb w'$ for every
$I\in\Cal I(\Cal S)$.   Next, 617B(b-ii) tells us that

\Centerline{$\int_{\Cal S}\pmb w'd\pmb v
=\lim_{I\uparrow\Cal I}\int_{\Cal S}\pmb w'_Id\pmb v$,}

\noindent that is,

\Centerline{$0=\lim_{I\uparrow\Cal I(\Cal S)}
   \int_{\Cal S}\pmb w'-\pmb w'_I\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}|\pmb w'-\pmb w'_I|d\pmb v$.}

\noindent Consequently

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}|\pmb w-\pmb w_I|d\pmb v
\le\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}
   (|\pmb w'-\pmb w'_I|+|\pmb w''-\pmb w''_I|)d\pmb v
=0$,}

\noindent and there is an $I\in\Cal I(\Cal S)$ such that
$\int_{\Cal S}|\pmb w-\pmb w_I|d\pmb v\le\epsilon$,
while $\pmb w_I$ is a simple process.

At this point note that $\sup|\pmb w_I|\le\sup|\pmb w\restr I|\le\sup|\pmb u|$,
so $|\pmb w_I|\le\pmb w^*$, while

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb w_{I<})&
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u_<)
     +\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<-\pmb w_{I<})
\le\epsilon+\int_{\Cal S}|\pmb u-\pmb w_I|d\pmb v\cr&
\le\epsilon+\int_{\Cal S}|\pmb u-\pmb w|d\pmb v
     +\int_{\Cal S}|\pmb w-\pmb w_I|d\pmb v
\le 2\epsilon+\int_{\Cal S}|\pmb u-\pmb w|d\pmb v
\le 3\epsilon\cr}$$

\noindent by the choice of $\delta$ and $\pmb w$.   As $\epsilon$ and $\pmb v$
are arbitrary,
$\pmb x
\in\overline{\{\pmb u_<:\pmb u\in\Msimp(\Cal S),\,|\pmb u|\le\pmb w^*\}}$.
}%end of proof of 645L

\leader{645N}{}\cmmnt{ I have spent all this time on $\MSi^0$ and $\TSi$
because these can be described and investigated in a very general context.
But for the new integral, we need to restrict ourselves in a way which is
already familiar.

\medskip

\noindent}{\bf Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T$.   If $\pmb u\in\Mmo(\Cal S)$ and
$\pmb v\in\Mnsu(\Cal S)$, then
$\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<)
=\theta(\int_{\Cal S}|\pmb u|d\pmb v)$.

\proof{{\bf (a)} Setting $\pmb u_n=|\pmb u|$ for every $n\in\Bbb N$,
$\sequencen{\pmb u_n}$ is a uniformly order-bounded non-decreasing sequence in
$\Mmo^+$ and $|\pmb u_<|\le\sup_{n\in\Bbb N}\pmb u_{n<}$, so

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_nd\pmb v)
=\theta(\int_{\Cal S}|\pmb u|d\pmb v)$.}

\medskip

{\bf (b)} Now suppose that $\sequencen{\pmb u_n}$ is a non-decreasing
uniformly order-bounded sequence in $\Mmo^+$ such that
$|\pmb u_<|\le\sup_{n\in\Bbb N}\pmb u_{n<}$.   Then
$\sequencen{\pmb u_{n<}}$ is non-decreasing, so
$\sequencen{|\pmb u_<|\wedge\pmb u_{n<}}$ order*-converges to
$|\pmb u_<|$, that is,
$\sequencen{(|\pmb u|\wedge\pmb u_n)_<}$
order*-converges to
$|\pmb u|_<$.   Now $\sequencen{|\pmb u|\wedge\pmb u_n}$ is
uniformly order-bounded, so
Theorem 644H tells us that $\int_{\Cal S}|\pmb u|\,d\pmb v
=\lim_{n\to\infty}\int_{\Cal S}|\pmb u|\wedge\pmb u_n\,d\pmb v$ and

\Centerline{$\theta(\int_{\Cal S}|\pmb u|\,d\pmb v)
=\lim_{n\to\infty}
   \theta(\int_{\Cal S}|\pmb u|\wedge\pmb u_n\,d\pmb v)
\le\lim_{n\to\infty}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)$.}

\noindent As $\sequencen{\pmb u_n}$ is arbitrary,
$\theta(\int_{\Cal S}|\pmb u|d\pmb v)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<)$.
}%end of proof of 645N

\leader{645O}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T$, and give $\Mpob(\Cal S)$
its S-integration topology $\TSi$.   If
$\pmb x\in\MSi(\Cal S)$
and $\pmb v\in\Mns(\Cal S)$ is an integrator,
then there is a unique $z\in L^0$ such that whenever
$A\subseteq\Mmo(\Cal S)$
is uniformly order-bounded and $\epsilon>0$
there is a $\TSi$-neighbourhood $G$ of $\pmb x\times\tbfoneminus{\Cal S}$
such that $\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$ whenever
$\pmb u\in A$ and $\pmb u_<\in G$.

\proof{ Write $R(\pmb x)\in\MSi^0(\Cal S)$ for
$\pmb x\times\tbfoneminus{\Cal S}$.

\medskip

{\bf (a)} Let $\Cal A$ be the family of uniformly order-bounded
subsets of $\Mmo=\Mmo(\Cal S)$ such that
$R(\pmb x)\in\overline{\{\pmb u_<:\pmb u\in A\}}$.   By 645L,
$\Cal A$ is not empty.

Define $T:\Mmo\to L^0$
by setting $T\pmb u=\int_{\Cal S}\pmb u\,d\pmb v$ for $\pmb u\in\Mmo$.
For $A\in\Cal A$, let $\Cal F_A$ be the filter on $\Mmo$ generated
by sets of the form $\{\pmb u:\pmb u\in A$, $\pmb u_<\in G\}$ where
$G$ is a $\TSi$-neighbourhood of $R(\pmb x)$.   Then
the image filter $T[[\Cal F_A]]$ on $L^0$ is Cauchy.   \Prf\ Let
$\epsilon>0$.   By 644G, $T\restr A$ is uniformly
continuous for the uniformity induced by the topology $\frak S$ on
$\Mmo$ described there,
so there are a $\pmb v'\in\Mnsu(\Cal S)$ and a $\delta>0$ such that
$\theta(T\pmb u-T\pmb u')\le\epsilon$ whenever $\pmb u$, $\pmb u'\in A$
and $\int_{\Cal S}|\pmb u-\pmb u'|d\pmb v'\le\delta$.   Next,
$G=\{\pmb y:\pmb y\in\Mpob=\Mpob(\Cal S)$,
$\widehat{\theta}^{\#}_{\pmb v'}(\pmb y-R(\pmb x))\le\bover12\delta\}$
is a neighbourhood of $R(\pmb x)$, so
$F=\{\pmb u:\pmb u\in A$, $\pmb u_<\in G\}$ belongs to $\Cal F_A$.
If $\pmb u$, $\pmb u'\in F$, then

$$\eqalignno{\int_{\Cal S}|\pmb u-\pmb u'|d\pmb v'
&=\widehat{\theta}^{\#}_{\pmb v'}((\pmb u-\pmb u')_<)\cr
\displaycause{645N}
&=\widehat{\theta}^{\#}_{\pmb v'}(\pmb u_<-\pmb u'_<)
\le\delta,\cr}$$

\noindent and $\theta(T\pmb u-T\pmb u')\le\epsilon$.   This shows that
$T[[\Cal F_A]]$ contains a set of diameter at most $\epsilon$ for the
metric
defined by $\theta$.   As $\epsilon$ is arbitrary, $T[[\Cal F_A]]$ is
Cauchy.\ \Qed

\medskip

{\bf (b)} It follows that

\Centerline{$z_A=\lim T[[\Cal F_A]]
=\lim_{\pmb u\to\Cal F_A}\int_{\Cal S}\pmb u\,d\pmb v$}

\noindent is defined.   If $A$, $A'\in\Cal A$ and $A\subseteq A'$,
then $\Cal F_{A'}\subseteq\Cal F_A$ and
$\lim_{\pmb u\to\Cal F_{A'}}\int_{\Cal S}\pmb u\,d\pmb v$ is defined, so
this is also $\lim_{\pmb u\to\Cal F_A}\int_{\Cal S}\pmb u\,d\pmb v$, that is,
$z_{A'}=z_A$.   If $A$, $A'$ are any two members of
$\Cal A$, then $A\cup A'\in\Cal A$ so $z_A=z_{A\cup A'}=z_{A'}$.   We can
therefore define $z\in L^0$ by saying that $z=z_A$ for every $A\in\Cal A$.

\medskip

{\bf (c)} Suppose that $\epsilon>0$ and
$A\subseteq\Mmo$ is uniformly order-bounded.   If $A\notin\Cal A$,
then $G=\Mpob\setminus\{\pmb u_<:\pmb u\in A\}$ is a neighbourhood
of $R(\pmb x)$, and certainly
$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$ whenever
$\pmb u\in A$ and $\pmb u_<\in G$.   If $A\in\Cal A$, then there is
an $F\in\Cal F_A$ such that
$\theta(z_A-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$ whenever
$\pmb u\in F$.   Now $F$ must include some set of the form
$\{\pmb u:\pmb u\in A$, $\pmb u_<\in G\}$ where $G$ is a neighbourhood
of $R(\pmb x)$, so we see that

\Centerline{$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)
=\theta(z_A-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$}

\noindent whenever $\pmb u\in A$ and $\pmb u_<\in G$.

\medskip

{\bf (d)} Thus we have found a $z$ with the given properties.   To see that
it is unique, recall from (a) that there is an $A\in\Cal A$, and observe
that $z=\lim_{\pmb u\to\Cal F_A}\int_{\Cal S}\pmb u\,d\pmb v$.
}%end of proof of 645O

\leader{645P}{Definition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is
an order-convex sublattice of $\Cal T$.

\spheader 645Pa If $\pmb x\in\MSi(\Cal S)$ and
$\pmb v\in\Mns(\Cal S)$ is an integrator, I will say that
the element $z$ of $L^0$ defined as in Theorem 645O is
$\Sint_{\Cal S}\pmb x\,d\pmb v$, the
{\bf S-integral} of $\pmb x$ with respect to $\pmb v$.

\spheader 645Pb In these circumstances I will say that members of
$\MSi(\Cal S)$ are {\bf S-integrable}.

\cmmnt{\spheader 645Pc Note that if $\pmb x$ is a fully adapted process
with domain $\Cal S$, $\pmb x$ is S-integrable iff
$\pmb x\times\tbfoneminus{\Cal S}$ is S-integrable, and in this case
$\Sint_{\Cal S}\pmb x\,d\pmb v
=\Sint_{\Cal S}\pmb x\times\tbfoneminus{\Cal S}\,d\pmb v$ for every
near-simple integrator $\pmb v$ with domain $\Cal S$.
}

\spheader 645Pd If $\pmb x$ is a fully adapted process with domain
$\Cal S$, I will say that it is {\bf locally S-integrable} if
$\pmb x\restr\Cal S\wedge\tau\in\MSi(\Cal S\wedge\tau)$ for every
$\tau\in\Cal S$.

\spheader 645Pe\cmmnt{ Following my practice with the Riemann-sum integral
(613H),} I shall allow myself to write $\Sint_{\Cal S}\pmb x\,d\pmb v$
for $\Sint_{\Cal S}(\pmb x\restr\Cal S)\,d(\pmb v\restr\Cal S)$
whenever $\pmb x$, $\pmb v$ are fully adapted processes such that
$\Cal S\subseteq\dom\pmb x\cap\dom\pmb v$,
$\pmb x\restr\Cal S$ is S-integrable and $\pmb v\restr\Cal S$ is a
near-simple integrator.

\cmmnt{\leader{645Q}{Law-independence} It is a while since I mentioned
law-independence, but the question of which features of a structure \Asisfour\
really depend on the measure $\bar\mu$, rather than just the measurable algebra
$\frak A$ and the filtration $\family{t}{T}{\frak A_t}$, never goes away.   I
have done my best to express concepts and theorems in terms which make it easy
to see that they are law-independent.   I have found however that some results
seem most naturally expressed in terms of the functionals $\theta$ of 613Ba,
and these need to be checked.   In many cases there is an obvious re-statement
of a theorem in terms of the topology of convergence in measure, which is
safely law-independent.
%631U 631V
In the second half of \S634, I introduced `coordinated' subalgebras,
which are surely not law-independent, but they have been practically invisible
since.   I remind you that anything involving martingales is not expected to be
law-independent (though the property of being a \sm\ is), but the regions of
accessibility and approachability in 643C are law-independent.

In 645E, however, we have a new formula involving $\theta$.
Just as ucp topologies are defined in terms of functionals
$\pmb u\mapsto\widehat{\theta}(\pmb u)=\theta(\sup|\pmb u|)$ (615B),
645E uses functionals

\Centerline{$\pmb u\mapsto\widehat{\theta}^{\#}_{\pmb v}(\pmb u)
=\inf_{\sequencen{\pmb u_n}}
   \sup_n\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)$.}

\noindent But changing the measure on $\frak A$ just produces a new F-norm
$\vartheta$ on $L^0$ which is equivalent to
$\theta$, so the F-norms $\widehat{\theta}^{\#}_{\pmb v}$ and
$\widehat{\vartheta}^{\#}_{\pmb v}$ will be equivalent and induce the same
topology $\TSi$ on $\Mpob$.   Thus $\TSi$,
the spaces $\MSi^0$ and $\MSi$ and the S-integral are law-independent.
}

\leader{645R}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$.

(a) Suppose that $\pmb u\in\Mmo(\Cal S)$ and
$\pmb v\in\Mns=\Mns(\Cal S)$
is an integrator.

\quad(i) $\pmb u_<\in\MSi^0(\Cal S)$
and $\Sint_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.

\quad(ii) If {\it either} $\pmb v$ is jump-free {\it or} $T$ has no points
isolated on the right,
$\Sint_{\Cal S}\pmb u_<d\pmb v=\int_{\Cal S}\pmb u_<\,d\pmb v$.

(b) If $\pmb x$, $\pmb x'\in\MSi=\MSi(\Cal S)$,
$\pmb v$, $\pmb v'\in\Mns$ are
integrators, and $\alpha\in\Bbb R$, then

\Centerline{$\Sint_{\Cal S}\pmb x+\pmb x'\,d\pmb v
=\Sint_{\Cal S}\pmb x\,d\pmb v+\Sint_{\Cal S}\pmb x'\,d\pmb v$,
\quad$\Sint_{\Cal S}\pmb x\,d(\pmb v+\pmb v')
=\Sint_{\Cal S}\pmb x\,d\pmb v+\Sint_{\Cal S}\pmb x\,d\pmb v'$,}

\Centerline{$\Sint_{\Cal S}\alpha\pmb x\,d\pmb v
=\Sint_{\Cal S}\pmb x\,d(\alpha\pmb v)
=\alpha\Sint_{\Cal S}\pmb x\,d\pmb v$.}

(c)(i) If $\pmb x\in\MSi$, $\pmb v\in\Mnsu(\Cal S)$ and $\pmb x\ge 0$, then
$\Sint_{\Cal S}\pmb x\,d\pmb v\ge 0$;

\quad(ii) if $\pmb x\in\MSi$ and $\pmb v$ is a constant process with domain
$\Cal S$, then $\Sint_{\Cal S}\pmb x\,d\pmb v=0$.

\proof{{\bf (a)(i)} Directly from the definition in 645Fb we see that
$\pmb u_<\in\MSi^0=\MSi^0(\Cal S)$.
Write $\TSi$ for the S-integration topology
on $\Mpob=\Mpob(\Cal S)$.
Set $z=\Sint_{\Cal S}\pmb u_<\,d\pmb v$.
If $\epsilon>0$, then $A=\{\pmb u\}$ is surely a uniformly order-bounded
subset of $\Mmo=\Mmo(\Cal S)$,
so there is a $\TSi$-neighbourhood $G$ of $\pmb u_<$ in $\Mpob$
such that $\theta(z-\int_{\Cal S}\pmb u'\,d\pmb v)\le\epsilon$ whenever
$\pmb u'\in A$ and $\pmb u'_<\in G$;  that is,
$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$.
As $\epsilon$ is arbitrary, $z=\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

\quad{\bf (ii)} Now 641T and 641W tell us that this will be equal to
$\int_{\Cal S}\pmb u_<d\pmb v$ if either $\pmb v$ is jump-free or
$T$ has no points isolated on the right.

\medskip

{\bf (b)} For $\pmb x\in\MSi$ write $R(\pmb x)$ for
$\pmb x\times\tbfoneminus{\Cal S}$.

\medskip

\quad{\bf (i)} We know from 645Kb that $\alpha\pmb x+\pmb x'\in\MSi$.
Set

\Centerline{$z=\Sint_{\Cal S}\pmb x\,d\pmb v$,
\quad$z'=\Sint_{\Cal S}\pmb x'\,d\pmb v$,
\quad$\tilde z=\Sint_{\Cal S}\alpha\pmb x+\pmb x'\,d\pmb v.$}

\noindent By 645L, there are uniformly order-bounded sets
$A$, $A'\subseteq\Mns$ such that
$R(\pmb x)\in\overline{\{\pmb u_<:\pmb u\in A\}}$ and
$R(\pmb x')\in\overline{\{\pmb u_<:\pmb u\in A'\}}$.   In this case,
$\alpha A+A'$ is uniformly order-bounded and
$R(\alpha\pmb x+\pmb x')
\in\overline{\{\pmb u_<:\pmb u\in\alpha A+A'\}}$,
because $R$ is a linear operator and $\TSi$ is a linear space topology.

Take any $\epsilon>0$.   Then there are neighbourhoods
$G$, $G'$ and $\tilde G$ of $R(\pmb x)$, $R(\pmb x')$ and
$R(\alpha\pmb x+\pmb x')$ respectively such that

\Centerline{$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$
for every $\pmb u\in A$ such that $\pmb u_<\in G$,}

\Centerline{$\theta(z'-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$
for every $\pmb u\in A'$ such that $\pmb u_<\in G'$,}

\Centerline{$\theta(\tilde z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$
for every $\pmb u\in\alpha A+A'$ such that $\pmb u_<\in\tilde G$.}

\noindent Again because $\TSi$ is a linear space topology, we can suppose,
shrinking $G$ and $G'$ if necessary, that $\alpha G+G'\subseteq\tilde G$.
Now we know that there are $\pmb u\in A$, $\pmb u'\in A'$ such that
$\pmb u_<\in G$ and $\pmb u'_<\in G'$.   Set
$\tilde{\pmb u}=\alpha\pmb u+\pmb u'$;
then $\tilde{\pmb u}\in\alpha A+A'$ and
$\tilde{\pmb u}_<\in\alpha G+G'\subseteq\tilde G$.   So

\Centerline{$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$,
\quad$\theta(z'-\int_{\Cal S}\pmb u'\,d\pmb v)\le\epsilon$,
\quad$\theta(\tilde z-\int_{\Cal S}\tilde{\pmb u}\,d\pmb v)\le\epsilon$.}

\noindent But $\int_{\Cal S}\tilde{\pmb u}\,d\pmb v
=\alpha\int_{\Cal S}\pmb u\,d\pmb v+\int_{\Cal S}\pmb u'\,d\pmb v$,
so $\theta(\tilde z-\alpha z-z')\le 3\epsilon$.
As $\epsilon$ is arbitrary,
$\tilde z=\alpha z+z'$, that is,
$\Sint_{\Cal S}\alpha\pmb x+\pmb x'\,d\pmb v
=\alpha\Sint_{\Cal S}\pmb x\,d\pmb v+\Sint_{\Cal S}\pmb x'\,d\pmb v$.

\medskip

\quad{\bf (ii)} To see that $\Sint_{\Cal S}$
is linear in the integrator as well as
in the integrand, repeat the method, with slight variations, as follows.
Start with

\Centerline{$z=\Sint_{\Cal S}\pmb x\,d\pmb v$,
\quad$z'=\Sint_{\Cal S}\pmb x\,d\pmb v'$,
\quad$\tilde z=\Sint_{\Cal S}\pmb x\,d(\alpha\pmb v+\pmb v')$.}

\noindent Let $A\subseteq\Mns$ be a uniformly order-bounded set
such that $R(\pmb x)\in\overline{\{\pmb u_<:\pmb u\in A\}}$.

Take any $\epsilon>0$.   Then there is a neighbourhood
$G$ of $R(\pmb x)$ such that

\Centerline{$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$ ,
\quad$\theta(z'-\int_{\Cal S}\pmb u\,d\pmb v')\le\epsilon$,
\quad$\theta(\tilde z-\int_{\Cal S}\pmb u\,d(\alpha\pmb v+\pmb v'))
\le\epsilon$}

\noindent for every $\pmb u\in A$ such that $\pmb u_<\in G$.
Since there is a $\pmb u\in A$ such that $\pmb u_<\in G$, and

\Centerline{$\alpha\int_{\Cal S}\pmb u\,d\pmb v
  +\int_{\Cal S}\pmb u\,d\pmb v'
=\int_{\Cal S}\pmb u\,d(\alpha\pmb v+\pmb v')$,}

\noindent $\theta(\tilde z-\alpha z-z')\le 3\epsilon$.
As $\epsilon$ is arbitrary,
$\tilde z=\alpha z+z'$, that is,
$\Sint_{\Cal S}\pmb x\,d(\alpha\pmb v+\pmb v')
=\alpha\Sint_{\Cal S}\pmb x\,d\pmb v+\Sint_{\Cal S}\pmb x\,d\pmb v'$.
So we have both halves of the result claimed.

\medskip

{\bf (c)(i)} Since $\pmb x\times\tbf 1^{(\Cal S)}_<\ge 0$, and
$\Sint_{\Cal S}\pmb x\,d\pmb v
=\Sint_{\Cal S}\pmb x\times\tbf 1^{(\Cal S)}_<d\pmb v$, it will be enough to
deal with the case in which $\pmb x\in\MSi^0$.   Let
$\pmb u^*\in\Mmo^+$ be such that $|\pmb x|\le\pmb u^*_<$, and set
$A=\{\pmb u:\pmb u\in\Mmo$, $|\pmb u|\le\pmb u^*\}$.   By 645L,
$\pmb x\in\overline{\{\pmb u_<:\pmb u\in A\}}$.   Let $\epsilon>0$.   Then there
are a $\pmb w\in\Mnsu(\Cal S)$ and a $\delta>0$ such that
$\theta(\Sint_{\Cal S}\pmb x\,d\pmb v-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$
whenever $\pmb u\in A$ and
$\widehat\theta^{\#}_{\pmb w}(\pmb x-\pmb u_<)\le\delta$.
Now $|\pmb u|\in A$ and

\Centerline{$|\pmb x-|\pmb u|_<|
=||\pmb x|-|\pmb u_<||\le|\pmb x-\pmb u_<|$,
\quad$\widehat\theta^{\#}_{\pmb w}(\pmb x-|\pmb u|_<)
\le\widehat\theta^{\#}_{\pmb w}(\pmb x-\pmb u_<)\le\delta$}

\noindent (645Db), so
$\theta(\Sint_{\Cal S}\pmb x\,d\pmb v-\int_{\Cal S}|\pmb u|d\pmb v)
\le\epsilon$.   But $\int_{\Cal S}|\pmb u|d\pmb v\in(L^0)^+$ (616R(b-i) again).
As $\epsilon$ is arbitrary and $(L^0)^+$ is closed (613Ba),
$\Sint_{\Cal S}\pmb x\,d\pmb v\in(L^0)^+$, as claimed.

\medskip

\quad{\bf (ii)} If $\pmb v$ is constant, it is an
integrator (616P(b-i)), and $\Sint_{\Cal S}\pmb x\,d\pmb v$ is defined.   Also
$\int_{\Cal S}\pmb u\,d\pmb v=0$ for every $\pmb u\in\Mmo$ (613Lc), so
$\Sint_{\Cal S}\pmb x\,d\pmb v=0$.
}%end of proof of 645R

\leader{645S}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex sublattice of
$\Cal T$.   Give $\MSi^0=\MSi^0(\Cal S)$ and $L^0$ the
S-integration topology $\TSi$ and the topology of convergence in measure
respectively, with their associated uniformities.
If $\pmb v\in\Mns(\Cal S)$ is an integrator,
then $\pmb x\mapsto\Sint_{\Cal S}\pmb x\,d\pmb v:\MSi^0\to L^0$ is
uniformly continuous on any uniformly previsibly order-bounded subset of
$\MSi^0$.

\proof{ Let $A\subseteq\MSi^0$ be uniformly previsibly order-bounded.
Then there is a $\pmb u^*\in\Mns(\Cal S)^+$ such that
$A\subseteq\{\pmb x:|\pmb x|\le\pmb u^*_<\}$.   Set
$B=\{\pmb u:\pmb u\in\Mns(\Cal S)$, $|\pmb u|\le\pmb u^*\}$;  note that
$B$ and $B-B$ are uniformly order-bounded.   Take
$\epsilon>0$.   Then there is a $\TSi$-neighbourhood $G$ of $\tbf 0$
in $\MSi^0$ such that
$\theta(\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$ whenever
$\pmb u\in B-B$ and $\pmb u_<\in G$
(645O).   Let $H$ be a neighbourhood of $0$ in $\MSi^0$ such that
$H+H-H\subseteq G$.   If $\pmb x$, $\pmb x'\in A$ and
$\pmb x'-\pmb x\in H$, then there are $\pmb u$,
$\pmb u'\in B$ such that

\Centerline{$\pmb x-\pmb u_<\in H$,
\quad$\theta(\biggerSint_{\Cal S}\pmb x\,d\pmb v
   -\int_{\Cal S}\pmb u\,d\pmb v)
\le\epsilon$,}

\Centerline{$\pmb x'-\pmb u'\in H$,
\quad$\theta(\biggerSint_{\Cal S}\pmb x'\,d\pmb v
   -\int_{\Cal S}\pmb u'\,d\pmb v)
\le\epsilon$}

\noindent (645L, 645O).   In this case, $\pmb u-\pmb u'\in B-B$ and
$\pmb u_<-\pmb u'_<\in G$, so

\Centerline{$\theta(\biggerSint_{\Cal S}\pmb x'\,d\pmb v
          -\biggerSint_{\Cal S}\pmb x\,d\pmb v)
\le 2\epsilon
   +\theta(\int_{\Cal S}\pmb u'\,d\pmb v-\int_{\Cal S}\pmb u\,d\pmb v)
=2\epsilon
   +\theta(\int_{\Cal S}\pmb u'-\pmb u\,d\pmb v)
\le 3\epsilon$.}

\noindent As $\epsilon$ is arbitrary, S-integration with respect to
$\pmb v$ is uniformly continuous on $A$.
}%end of proof of 645S

\leader{645T}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$.
Suppose that $\sequencen{\pmb x_n}$ is a sequence in $\MSi(\Cal S)$ such that
$\sequencen{\pmb x_n\times\tbfoneminus{\Cal S}}$ is
uniformly previsibly order-bounded and $\sequencen{\pmb x_n}$ is
order*-convergent to $\pmb x$ in $(L^0)^{\Cal S}$.   Then
$\pmb x$ is S-integrable and $\Sint_{\Cal S}\pmb x\,d\pmb v
=\lim_{n\to\infty}\Sint_{\Cal S}\pmb x_n\,d\pmb v$ for every
integrator $\pmb v\in\Mns(\Cal S)$.

\proof{ Write $R(\pmb x)$ for $\pmb x\times\tbfoneminus{\Cal S}$, etc., as
usual.   By 645H, $\sequencen{R(\pmb x_n)}$ is
$\TSi$-convergent to $R(\pmb x)$.   Since
$\{R(\pmb x)\}\cup\{R(\pmb x_n):n\in\Bbb N\}$ is uniformly previsibly
order-bounded, 645S tells us that

\Centerline{$\biggerSint_{\Cal S}\pmb x\,d\pmb v
=\biggerSint_{\Cal S}R(\pmb x)\,d\pmb v
=\lim_{n\to\infty}\biggerSint_{\Cal S}R(\pmb x_n)\,d\pmb v
=\lim_{n\to\infty}\biggerSint_{\Cal S}\pmb x_n\,d\pmb v$.}
}%end of proof of 645T

%\leader{645V}{}
%\leader{645W}{}

\exercises{\leader{645X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 645Xa
Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, and that $\Cal S$ is an order-convex sublattice of
$\Cal T$.
Show that if $A\subseteq\Mns(\Cal S)$ is uniformly
order-bounded, then it is bounded above and below in $\Mns(\Cal S)$.
\Hint{show that if $\bar u\ge 0$ in $L^0$, and we set
$u_{\sigma}=\sup\{u:u\in L^0(\frak A_{\sigma})$, $u\le\bar u\}$ for
$\sigma\in\Cal S$, then $\sfamilysS{u}$ satisfies the conditions of
632F.}
%645B

\spheader 645Xb Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa, 617Xb, 618Xa, 622Xd, 626Xa, 627Xa, 642Xd and
644Xe.   (i) Show that $\Mpob(\Cal T_f)$ corresponds to the space $V$ of
bounded functions $f:\coint{0,\infty}\to\Bbb R$ such that $f(0)=0$.
(ii) Show that the S-integration topology
on $\Mpob(\Cal T_f)$ corresponds to
the topology on $V$ generated by functionals
$f\mapsto\overline{\int}f\,d\nu$ where $\nu$ is a totally finite Radon
measure on $\coint{0,\infty}$.
(iii) Show that $\MSi^0(\Cal T_f)$ corresponds to the space
$W=\{f:f\in V$, $f$ is universally measurable$\}$.   (iv) Show that
if $f\in W$ corresponds to $\pmb x\in\MSi^0$,
$g:\coint{0,\infty}\to\Bbb R$ is a non-decreasing
function of bounded variation which is
continuous on the right and $\pmb v$ is the process corresponding to
$g$, then $\Sint_{\Cal T_f}\pmb x\,d\pmb v=\int f\,d\nu_g$ where
$\nu_g$ is the Lebesgue-Stieltjes measure on $\coint{0,\infty}$
defined from $g$.
%645F 645P

\spheader 645Xc Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice,
and $\sequencen{\pmb u_n}$ a uniformly
previsibly order-bounded sequence of
jump-free processes with domain $\Cal S$ which is order*-convergent to
a jump-free process $\pmb u$.   Show that
$\lim_{n\to\infty}\int_{\Cal S}\pmb u_nd\pmb v
=\int_{\Cal S}\pmb u\,d\pmb v$ for every near-simple integrator $\pmb v$
with domain $\Cal S$.
%641O 645R(a-i) 645T

\leader{645Y}{Further exercises (a)}
%\spheader 645Ya
Let $\Cal S$ be a sublattice of $\Cal T$.   (i) Show that $\preccurlyeq$ (645Bc)
is transitive and reflexive, that is, is a pre-order on $\Mfa=\Mfa(\Cal S)$
in the sense of 511A.   (ii) Show that if $\equiv$ is the associated equivalence
relation, then the set $\Mfa/\hbox{$\equiv$}$
of equivalence classes can be thought of
as a partially ordered linear space in the sense of \S351.
(iii) Taking $\Mbv\subseteq\Mfa$ to be the space of processes of bounded
variation, show that if $\pmb v\in\Mbv$ and $\pmb v'\equiv\pmb v$ then
$\pmb v'\in\Mbv$.   (iv) Show that the image of $\Mbv$ in
$M/\hbox{$\equiv$}$
is a Dedekind complete Riesz space in the sense of \S353.
%645B

\spheader 645Yb Find an example of a sublattice $\Cal S$ with a
least element and an order-bounded
fully adapted process $\pmb x=\sfamilysS{x}$ such that
$x_{\min\Cal S}=0$ but $\pmb x\notin\Mpob(\Cal S)$.
%mt64bits 645B
}%end of exercises

\endnotes{
\Notesheader{645} In 645P we have at last arrived at something like the
standard
stochastic integral as described in {\smc Rogers \& Williams 00} and
{\smc Protter 05}.   In particular, we have a dominated convergence theorem
(645T).   The S-integral is {\it not} an extension of the Riemann-sum stochastic
integral of \S613.   They are linked by the formula
$\Sint\pmb u_<\,d\pmb v=\int\pmb u\,d\pmb v$ of 645Ra.
But integrands for the Riemann-sum
integral need not be previsible, and integrators need not be near-simple.
It is true that we shall often have
$\Sint\pmb u_<\,d\pmb v=\int\pmb u_<\,d\pmb v$ (645R(a-ii)), but this is a
touch accidental, and the theorems I have offered concerning the
Riemann-sum integral are mostly unconcerned with this phenomenon.

I said in the introduction to this section
that I was trying to define an integral which
would look like a Lebesgue-Stieltjes integral.   For a non-decreasing
integrator, the S-integral is indeed of this type, at least if
$T=\coint{0,\infty}$.   But the purpose of this volume is to
look at integration with respect to martingales, and here the analogy is
too weak to be useful.   Only if we suppress martingales altogether, as in
645Xb or 649L, do we get a direct correspondence.   The wonder is that we can
devise any kind of sequentially smooth integral with a martingale
integrator.

The definition of $\MSi^0$ in 645F corresponds to a kind of universal
measurability, as in 645Xb, but with an extra boundedness condition.
Typically, theorems involving the S-integral, when they speak of
uniformly order-bounded families, require families included in sets
$\{\pmb x:|\pmb x|\times\tbf 1_<\le\pmb u_<\}$
where $\pmb u$ is moderately oscillatory, rather
than just $\{\pmb x:\sup|\pmb x|\le\bar w\}$.

This section is based on a discussion of the properties of previsible versions
$\pmb u_<$ of moderately oscillatory processes $\pmb u$.   It is therefore
worth noting that in the context of the principal results here, it would have
been enough to start from near-simple processes $\pmb u$ (642N).
}%end of notes

\discrpage

\frfilename{mt646.tex}
\versiondate{21.2.22}
\copyrightdate{2013}

\def\chaptername{The S-integral}
\def\sectionname{Basic properties of the S-integral}

\newsection{646}

Having defined the S-integral as an adaptation of the Riemann-sum integral
for previsible processes\cmmnt{ (645O, 645Pa, 645I)}, it is natural to
look for parallels to the properties of the Riemann-sum integral set out in
Chapters 61-63.   After a few easy remarks (646B-646D), %646B 646C 646D
I embark
on the question of splitting a domain $\Cal S$ into $\Cal S\wedge\tau$ and
$\Cal S\vee\tau$ (646J).   This leads naturally to an examination of
indefinite S-integrals (646K), which I approach through a result on
capped-stake variation sets for martingale integrators (646P).
We have a change-of-variable theorem (646R), a formula for jumps in an
indefinite S-integral (646S) and a version of It\^o's formula (646T).

\cmmnt{
\leader{646A}{Notation} As in \S645, I shall be calling on a substantial
part of the special notation developed in this volume.
\Asissix\ will be a stochastic integration structure, with its
associated Riesz space $L^0=L^0(\frak A)$, endowed with the linear space
topology of
convergence in measure and the defining F-seminorm $\theta$ where
$\theta(w)=\Expn(|w|\wedge\chi 1)$ for $w\in L^0$ (613Ba).   I write
$L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$ for the $L$-space of members of $L^0$ with
finite expectation.   For $\sigma\in\Cal T$,
$P_{\sigma}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ will be the conditional expectation
associated with  $\frak A_{\sigma}$.   We have the algebras
$\frak A_{\Cal S<\tau}$ for sublattices $\Cal S$ of $\Cal T$ and
$\tau\in\Cal T$ (641B).
If $\Cal S$ is a sublattice of $\Cal T$ and $\tau\in\Cal S$,
I write $\Cal S\wedge\tau$ and $\Cal S\vee\tau$ for
$\{\sigma\wedge\tau:\sigma\in\Cal S\}$ and
$\{\sigma\vee\tau:\sigma\in\Cal S\}$ respectively, and
$\Cal I(\Cal S)$ for the upwards-directed set of finite sublattices of
$\Cal S$.   $\tbf 1^{(\Cal S)}$ will be the constant process with domain
$\Cal S$ and value $\chi 1$.

For a process $\pmb u=\sfamilysS{u}\in(L^0)^{\Cal S}$,
I write $\Bvalue{\pmb u\ne\tbf 0}$ for
$\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}$ (612Sb);  if
$\pmb u$ is order-bounded, I write
$\sup|\pmb u|$ for $\sup_{\sigma\in\Cal S}|u_{\sigma}|$;
if $\sup|\pmb u|\in L^{\infty}(\frak A)$, I write $\|\pmb u\|_{\infty}$
for
$\|\sup|\pmb u|\|_{\infty}=\sup_{\sigma\in\Cal S}\|u_{\sigma}\|_{\infty}$.
If $\pmb u$ and $\pmb v$ are fully adapted processes and $I$ is a finite
sublattice of $\dom\pmb u\cap\dom\pmb v$, then
$S_I(\pmb u,d\pmb v)$ is the Riemann sum described in
613Fb.   If $\pmb v$ is a fully adapted process and $\Cal S$ is a sublattice
of $\dom\pmb v$, $Q_{\Cal S}(d\pmb v)$ is the capped-stake variation set
described in 616B.   For processes $\pmb v$, $\pmb v'$
with the same domain
I say that $\pmb v\preccurlyeq\pmb v'$ if $\pmb v'-\pmb v$
is non-decreasing (645Bc),
and in this context $\pmb v\equiv\pmb v'$ if $\pmb v'-\pmb v$ is constant.
An $L^2$-process is a process $\pmb u=\sfamilysS{u}$ such that
$u_{\sigma}^2\in L^1_{\bar\mu}$ for every $\sigma\in\Cal S$ (622Ca).
If $\pmb u=\sfamilysS{u}$ is a process and
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$,
then $z\pmb u=\sfamilysS{z\times u}$ (612De).
For a moderately oscillatory process $\pmb u$,
$\pmb u_<$ is its previsible version (641L).

For a sublattice $\Cal S$ of $\Cal T$, $\Mfa(\Cal S)$ is the space of
fully adapted processes with domain $\Cal S$,
$\Msimp(\Cal S)\subseteq\Mfa(\Cal S)$ the space of simple processes (612L),
$\Mob(\Cal S)$ the space of order-bounded processes (614Fc),
$\Mmo(\Cal S)$ the space of moderately oscillatory processes (615Fa),
$\Mns(\Cal S)$ the space of near-simple processes (631Fa),
$\Mnsu(\Cal S)$ the cone of non-negative non-decreasing near-simple processes
(644Bb),
$\Mjf(\Cal S)$ the space of jump-free processes (618G) and
$\Mpob(\Cal S)$ the space of previsibly order-bounded processes
(645Ba).   On $\Mob(\Cal S)$ we have the ucp topology (615B) and on
$\Mpob(\Cal S)$ we have the S-integration topology $\TSi$;
$\MSi^0(\Cal S)$ is the $\TSi$-closure of $\{\pmb u_<:\pmb u\in\Mmo(\Cal S)\}$
(645F), and
$\MSi(\Cal S)$ is $\{\pmb x:\pmb x\in\Mob(\Cal S)$,
$\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0(\Cal S)\}$ (645Fc).
If $\pmb v\in\Mnsu(\Cal S)$, $\widehat{\theta}^{\#}_{\pmb v}$ is the
corresponding F-seminorm on $\Mpob(\Cal S)$ (645Bb).

Finally, we shall be looking at both Riemann-sum integrals
$\int_{\Cal S}\pmb u\,d\pmb v$ (613L) and
S-integrals $\Sint_{\Cal S}\pmb x\,d\pmb v$ (645P).
For the former case, our default assumption is that $\pmb u$ is moderately
oscillatory and $\pmb v$ is an integrator (616K);  in the latter, that
$\family{t}{T}{\frak A_t}$ is right-continuous (632B), $\Cal S$ is
order-convex, $\pmb x\in\MSi(\Cal S)$ and
$\pmb v$ is a near-simple integrator.
}

\leader{646B}{Lemma} Let $\Cal S$ be a sublattice of
$\Cal T$.   If $\pmb x=\sfamilysS{x}\in\MSi^0(\Cal S)$ then
$x_{\tau}\in L^0(\frak A_{\Cal S<\tau})$ for every $\tau\in\Cal S$.

\proof{ Define $\pmb v=\sfamilysS{v}$ by saying that
$v_{\sigma}=\chi\Bvalue{\tau\le\sigma}$ for $\sigma\in\Cal S$.   Then
$\pmb v$ is a non-negative order-bounded
non-decreasing simple process with breakpoint string $(\tau)$,
so belongs to $\Mnsu(\Cal S)$.   If $\epsilon>0$, there is a
$\pmb u=\sfamilysS{u}\in\Mmo(\Cal S)$ such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u_<)<\epsilon$.
Now there is a uniformly order-bounded non-decreasing sequence
$\sequencen{\pmb u_n}=\sequencen{\family{\sigma}{\Cal S}{u_{n\sigma}}}$ of
non-negative moderately oscillatory processes such that
$|\pmb x-\pmb u_<|\le\sup_{n\in\Bbb N}\pmb u_{n<}$ and
$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)\le\epsilon$.
In this case,
$|x_{\tau}-u_{<\tau}|\le\sup_{n\in\Bbb N}u_{n<\tau}$, while
$\sequencen{u_{n<\tau}}$ is a non-decreasing sequence
and $\sup_{n\in\Bbb N}\theta(u_{n<\tau})\le\epsilon$ (using 641J).
So $\theta(x_{\tau}-u_{<\tau})\le\epsilon$.   We know also
that $u_{<\tau}\in L^0(\frak A_{\Cal S<\tau})$ (641G(a-i)).   As $\epsilon$ is
arbitrary and $L^0(\frak A_{\Cal S<\tau})$ is closed,
$x_{\tau}\in L^0(\frak A_{\Cal S<\tau})$.
}%end of proof of 646B

\leader{646C}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$,
$\pmb x$ an S-integrable process and
$\pmb v$ a near-simple integrator, both with domain $\Cal S$.
Then $\Bvalue{\Sint_{\Cal S}\pmb x\,d\pmb v\ne 0}
\Bsubseteq\Bvalue{\pmb v\ne\tbf 0}$.

\proof{ Set $a=\inf_{\sigma\in\Cal S}\Bvalue{v_{\sigma}=0}$
and $z=\Sint_{\Cal S}\pmb x\,d\pmb v
=\Sint_{\Cal S}\pmb x\times\tbfoneminus{\Cal S}\,d\pmb v$.
By the definition of the
S-integral in 645O-645P, there is for any $\epsilon>0$
a $\pmb u\in\Mmo(\Cal S)$ such that
$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$, so
that $\theta(z\times\chi a-\chi a\times\int_{\Cal S}\pmb u\,d\pmb v)
\le\epsilon$.
But $\chi a\times\int_{\Cal S}\pmb u\,d\pmb v=0$ (613Ld), so
$\theta(z\times\chi a)\le\epsilon$.   As $\epsilon$ is arbitrary,
$\theta(z\times\chi a)=0$ and
$\Bvalue{z\ne 0}\Bsubseteq 1\Bsetminus a=\Bvalue{\pmb v\ne\tbf 0}$.
}%end of proof of 646C

\cmmnt{\medskip

\noindent{\bf Remark} It seems to be harder to match the other half of
613Ld;  see 647J.   But we have the following easy remark.
}

\leader{646D}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$,
$\pmb x$ a member of $\MSi(\Cal S)$, and
$\pmb v$ a near-simple integrator with domain $\Cal S$.   If
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$, then

\Centerline{$\biggerSint_{\Cal S}z\pmb x\,d\pmb v
=\biggerSint_{\Cal S}\pmb x\,d(z\pmb v)
=z\times\biggerSint_{\Cal S}\pmb x\,d\pmb v$.}

\proof{{\bf (a)} To begin with, suppose that
$\pmb x\in\MSi^0=\MSi^0(\Cal S)$.   In this case,
$z\pmb x\in\MSi^0$ (645Ka), and if $\pmb u^*\in\Mmo^+=\Mmo(\Cal S)^+$ is such
that $|\pmb x|\le\pmb u^*_<$, then $|z\pmb x|\le(|z|\pmb u^*)_<$ (see
the proof of 645D(a-iv)).   Also $z\pmb v$ is a near-simple integrator
(631F(a-v), 616P(b-iv)).   Let $\epsilon>0$.
Then there is a $\delta>0$ such that
$\theta(z\times u)\le\epsilon$ whenever $\theta(u)\le\delta$.   Setting
$A=\{\pmb u:\pmb u\in\Mmo$, $|\pmb u|\le\pmb u^*\}$ and
$A'=\{\pmb u:\pmb u\in\Mmo$, $|\pmb u|\le|z|\pmb u^*\}$,
we have $\TSi$-neighbourhoods $G$, $G'$ of $\pmb x$, $z\pmb x$ respectively
such that

\Centerline{$\theta(\biggerSint_{\Cal S}\pmb x\,d(z\pmb v)
   -\int_{\Cal S}\pmb u\,d(z\pmb v))
\le\epsilon$ whenever $\pmb u\in A$ and $\pmb u_<\in G$,}

\Centerline{$\theta(\biggerSint_{\Cal S}\pmb x\,d\pmb v
   -\int_{\Cal S}\pmb u\,d\pmb v)
\le\delta$ whenever $\pmb u\in A$ and $\pmb u_<\in G$,}

\Centerline{$\theta(\biggerSint_{\Cal S}z\pmb x\,d\pmb v
   -\int_{\Cal S}\pmb u\,d\pmb v)
\le\epsilon$ whenever $\pmb u\in A'$ and $\pmb u_<\in G'$;}

\noindent and because $\pmb x'\mapsto z\pmb x'$ is continuous (645E(a-vi)),
we can suppose that $z\pmb x'\in G'$ whenever $\pmb x'\in G$.   Now there
is a $\pmb u\in A$ such that $\pmb u_<\in G$, in which case we shall have

\Centerline{$\theta(\biggerSint_{\Cal S}\pmb x\,d(z\pmb v)
   -\int_{\Cal S}\pmb u\,d(z\pmb v))
\le\epsilon$,}

\Centerline{$\theta(z\times\biggerSint_{\Cal S}\pmb x\,d\pmb v
   -z\times\int_{\Cal S}\pmb u\,d\pmb v)
\le\epsilon$,}

\Centerline{$\theta(\biggerSint_{\Cal S}z\pmb x\,d\pmb v
   -\int_{\Cal S}z\pmb u\,d\pmb v)
\le\epsilon$,}

\noindent and moreover

\Centerline{$\int_{\Cal S}\pmb u\,d(z\pmb v)
=z\times\int_{\Cal S}\pmb u\,d\pmb v
=\int_{\Cal S}z\pmb u\,d\pmb v$}

\noindent (613L(b-ii)).   So

\Centerline{$\theta(\biggerSint_{\Cal S}\pmb x\,d(z\pmb v)
   -z\times\biggerSint_{\Cal S}\pmb x\,d\pmb v)\le 2\epsilon$,
\quad
$\theta(\biggerSint_{\Cal S}z\pmb x\,d\pmb v
   -z\times\biggerSint_{\Cal S}\pmb x\,d\pmb v)\le 2\epsilon$;}

\noindent as $\epsilon$ is arbitrary, the three expressions are equal.

\medskip

{\bf (b)} For the general case of $\pmb x\in\MSi(\Cal S)$, we have

\Centerline{$\Sint_{\Cal S}z\pmb x\,d\pmb v
=\Sint_{\Cal S}(z\pmb x)\times\tbfoneminus{\Cal S}\,d\pmb v
=\Sint_{\Cal S}z(\pmb x\times\tbfoneminus{\Cal S})d\pmb v$,}

\Centerline{$\Sint_{\Cal S}\pmb x\,d(z\pmb v)
=\Sint_{\Cal S}\pmb x\times\tbfoneminus{\Cal S}\,d(z\pmb v)$,
\quad
$z\times\Sint_{\Cal S}\pmb x\,d\pmb v
=z\times\Sint_{\Cal S}\pmb x\times\tbfoneminus{\Cal S}\,d\pmb v$,}

\noindent while $\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0$, so
(a) tells us that these three terms are equal.
}%end of proof of 646D

\leader{646E}{}\cmmnt{ It will be convenient to have a
partial expression of the
continuity of S-integration in a more precise form than those offered in \S645.

\medskip

\noindent}{\bf Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, $\Cal S$ is an order-convex sublattice of $\Cal T$, and
$\pmb v\in\Mnsu(\Cal S)$.   If $\pmb x\in\MSi^0(\Cal S)$, then
$\theta(\Sint_{\Cal S}\pmb x\,d\pmb v)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$.
%for 649

\proof{ Let $\epsilon>0$.   Then there is an order-bounded non-decreasing
sequence $\sequencen{\pmb u_n}$ in $\Mmo(\Cal S)^+$ such that
$|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$ and
$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_nd\pmb v)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)+\epsilon$.   We are supposing that
$\pmb x$ is previsibly order-bounded, so there is a $\pmb u\in\Mmo(\Cal S)^+$
such that $|\pmb x|\le\pmb u_<$.   Set
$\pmb u'_n=\pmb u\wedge\pmb u_n$ for each $n\in\Bbb N$.   Then

\Centerline{$|\pmb x|\le\pmb u_<\wedge\sup_{n\in\Bbb N}\pmb u_{n<}
=\sup_{n\in\Bbb N}(\pmb u_<\wedge\pmb u_{n<})
=\sup_{n\in\Bbb N}\pmb u'_{n<}$}

\noindent (641G(e-i)), so

$$\eqalignno{0&
\le|\Sint_{\Cal S}\pmb x\,d\pmb v|
\le\Sint_{\Cal S}|\pmb x|d\pmb v
\le\Sint_{\Cal S}(\sup_{n\in\Bbb N}\pmb u'_{n<})d\pmb v
\Displaycause{because $\Sint_{\Cal S}..d\pmb v$ is a positive linear
operator, by 645Rc}
=\sup_{n\in\Bbb N}\Sint_{\Cal S}\pmb u'_{n<}d\pmb v
\Displaycause{645T, because $\sequencen{\pmb u'_{n<}}$ is uniformly
previsibly order-bounded and order*-convergent to its supremum}
=\sup_{n\in\Bbb N}\int_{\Cal S}\pmb u'_nd\pmb v
\le\sup_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v\cr}$$

\noindent and

\Centerline{$\theta(\Sint_{\Cal S}\pmb x\,d\pmb v)
\le\theta(\sup_{n\in\Bbb N}\int_{\Cal S}\pmb u_nd\pmb v)
=\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_nd\pmb v)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)+\epsilon$.}

\noindent As $\epsilon$ is arbitrary, we have the result.
}%end of proof of 646E

\leader{646F}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\tau$ a member of $\Cal S$.   Suppose
that $\pmb u'=\family{\sigma}{\Cal S\wedge\tau}{u'_{\sigma}}$
and $\pmb u''=\family{\sigma}{\Cal S\vee\tau}{u''_{\sigma}}$
are families in $L^0$.   Define
$R(\pmb u',\pmb u'')\in(L^0)^{\Cal S}$ by saying that
$R(\pmb u',\pmb u'')=\sfamilysS{u}$ where

\Centerline{$u_{\sigma}
=u'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}
   +u''_{\sigma\vee\tau}\times\chi\Bvalue{\tau\le\sigma}$}

\noindent for $\sigma\in\Cal S$.

(a)(i) $R(\pmb u',\pmb u'')\restr\Cal S\wedge\tau=\pmb u'+\pmb z$, where
$\pmb z=\langle(u''_{\tau}-u'_{\tau})\times\chi\Bvalue{\sigma=\tau}
\rangle_{\sigma\in\Cal S\wedge\tau}$.

\quad(ii) $R(\pmb u',\pmb u'')\restr\Cal S\vee\tau=\pmb u''$.

(b) Regarded as an operator from
$(L^0)^{\Cal S\wedge\tau}\times(L^0)^{\Cal S\vee\tau}$ to
$(L^0)^{\Cal S}$, $R$ is linear, positive and order-continuous.

(c) If $\pmb u'$ and $\pmb u''$ are fully adapted, then
$R(\pmb u',\pmb u'')$ is fully adapted.

(d) If $\pmb u'$ and $\pmb u''$ are order-bounded, then
$R(\pmb u',\pmb u'')$ is order-bounded and
$\sup|R(\pmb u',\pmb u'')|\le\sup|\pmb u'|\vee\sup|\pmb u''|$.

(e) Suppose that $\pmb u'$ and $\pmb u''$ are moderately oscillatory.

\quad(i) $R(\pmb u',\pmb u'')$ is moderately oscillatory.

\quad(ii) $\pmb u'_<=R(\pmb u',\pmb u'')_<\restr\Cal S\wedge\tau$.

(f) If $\pmb u'$ and $\pmb u''$ are near-simple, $R(\pmb u',\pmb u'')$ is
near-simple.

(g) Suppose that $\pmb u'$ and $\pmb u''$ are moderately oscillatory, and
that $\pmb v$ is an integrator with domain $\Cal S$.   Then

\Centerline{$\int_{\Cal S}R(\pmb u',\pmb u'')d\pmb v
=\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v
  +\int_{\Cal S\vee\tau}\pmb u''\,d\pmb v$.}

\proof{ Throughout the proof I will write $\pmb u$ for
$R(\pmb u',\pmb u'')$.

\medskip

{\bf (a)(i)} For $\sigma\in\Cal S\wedge\tau$,

$$\eqalign{u'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}&
   +u''_{\sigma\vee\tau}\times\chi\Bvalue{\tau\le\sigma}\cr&
=u'_{\sigma}
   +(u''_{\tau}-u'_{\tau})\times\chi\Bvalue{\sigma=\tau}.\cr}$$

\medskip

\quad{\bf (ii)} For $\sigma\in\Cal S\vee\tau$,

\Centerline{$u'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}
   +u''_{\sigma\vee\tau}\times\chi\Bvalue{\tau\le\sigma}
=u''_{\sigma}$.}

\medskip

{\bf (b)} We just have to observe that, for any $\sigma\in\Cal S$,

\Centerline{$(\pmb u',\pmb u'')
\mapsto(u'_{\sigma\wedge\tau},u''_{\sigma\vee\tau}):
(L^0)^{\Cal S\wedge\tau}\times(L^0)^{\Cal S\vee\tau}
\to L^0\times L^0$}

\noindent is linear, positive and order-continuous, and so is

\Centerline{$(u',u'')
\mapsto u'\times\chi\Bvalue{\sigma<\tau}
   +u''\times\chi\Bvalue{\tau\le\sigma}:L^0\times L^0\to L^0$.}

\medskip

{\bf (c)(i)} If $\sigma\in\Cal S$, then
$u'_{\sigma\wedge\tau}\in L^0(\frak A_{\sigma\wedge\tau})
\subseteq L^0(\frak A_{\sigma})$;
$\Bvalue{\sigma<\tau}\in\frak A_{\sigma}$ so
$\chi\Bvalue{\sigma<\tau}\in L^0(\frak A_{\sigma})$;  and

$$\eqalignno{u''_{\sigma\vee\tau}\times\chi\Bvalue{\tau\le\sigma}
&=u''_{\sigma\vee\tau}\times\chi\Bvalue{\sigma\vee\tau=\sigma}\cr
\displaycause{611E(a-ii-$\beta$)}
&\in L^0(\frak A_{\sigma})\cr}$$

\noindent by 612C.   So $u_{\sigma}\in L^0(\frak A_{\sigma})$.

\medskip

\quad{\bf (ii)} If $\sigma$, $\sigma'\in\Cal S$ then

$$\eqalignno{\Bvalue{\sigma=\sigma'}
&\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma'\wedge\tau}
   \Bcap\Bvalue{\sigma\vee\tau=\sigma'\vee\tau}\cr
&\mskip100mu
   \Bsetminus((\Bvalue{\sigma<\tau}\Bsymmdiff\Bvalue{\sigma'<\tau})
      \Bcup(\Bvalue{\tau\le\sigma}\Bsymmdiff\Bvalue{\tau\le\sigma'}))\cr
\displaycause{611Ec, {\it passim}}
% passim:  "scattered through the text"
&\Bsubseteq\Bvalue{u'_{\sigma\wedge\tau}=u'_{\sigma'\wedge\tau}}
   \Bcap\Bvalue{u''_{\sigma\vee\tau}=u''_{\sigma'\vee\tau}}\cr
&\mskip100mu
   \Bcap\Bvalue{\chi\Bvalue{\sigma<\tau}=\chi\Bvalue{\sigma'<\tau}}
   \Bcap\Bvalue{\chi\Bvalue{\tau\le\sigma}=\chi\Bvalue{\tau\le\sigma'}}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma'}}.\cr}$$

\noindent So $R(\pmb u',\pmb u'')=\sfamilysS{u}$
is fully adapted.

\medskip

{\bf (d)} Setting $\bar u=\sup|\pmb u'|\vee\sup|\pmb u''|$, we have

$$\eqalign{|u_{\sigma}|&
=|u'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}
   +u''_{\sigma\vee\tau}\times\chi\Bvalue{\tau\le\sigma}|\cr&
=(|u'_{\sigma\wedge\tau}|\times\chi\Bvalue{\sigma<\tau})
   \vee(|u''_{\sigma\vee\tau}|\times\chi\Bvalue{\tau\le\sigma})\cr&
\le(\bar u\times\chi\Bvalue{\sigma<\tau})
   \vee(\bar u\times\chi\Bvalue{\tau\le\sigma})
=\bar u\cr}$$

\noindent for every $\sigma\in\Cal S$.

\medskip

{\bf (e)(i)} Use (a).  The process $\pmb z=
\langle(u''_{\tau}-u'_{\tau})\times\chi\Bvalue{\sigma=\tau}
\rangle_{\sigma\in\Cal S\wedge\tau}$ is simple (612Ka), so
$\pmb u\restr\Cal S\wedge\tau=\pmb u'+\pmb z$ is moderately
oscillatory, while $\pmb u\restr\Cal S\vee\tau=\pmb u''$ is also moderately
oscillatory.   By 615F(a-v), $\pmb u$ is moderately oscillatory.

\medskip

\quad{\bf (ii)} By the formula in 641Ia with $\tau_0=\tau$, $\pmb z_<=\tbf 0$,
so

$$\eqalignno{\pmb u_<\restr\Cal S\wedge\tau&
=(\pmb u\restr\Cal S\wedge\tau)_<
\Displaycause{641G(c-ii)}
=\pmb u'_<+\pmb z_<
\Displaycause{641G(e-i)}
=\pmb u'.\cr}$$

\medskip

{\bf (f)} Again, if $\pmb u'$ and $\pmb u''$ are near-simple, so are
$\pmb u\restr\Cal S\vee\tau=\pmb u''$ and
$\pmb u\restr\Cal S\wedge\tau=\pmb u'+\pmb z$, because
$\pmb z$ is simple;  so $R(\pmb u',\pmb u'')$ is near-simple, by
631F(a-iv).

\medskip

{\bf (g)} Since $\pmb u$ is moderately oscillatory,
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined and equal to
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
  +\int_{\Cal S\vee\tau}\pmb u\,d\pmb v$ (613J(c-i)).
Now $\int_{\Cal S\vee\tau}\pmb u\,d\pmb v
=\int_{\Cal S\vee\tau}\pmb u''\,d\pmb v$ because
$\pmb u''=\pmb u\restr\Cal S\vee\tau$.   As for
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v$, observe that if
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ then

$$\eqalignno{\int_{\Cal S\wedge\tau}\pmb z\,d\pmb v&
=\lim_{\sigma\downarrow\Cal S}u''_{\tau}\times\chi\Bvalue{\sigma=\tau}
    \times(v_{\tau}-v_{\downarrow})
  +(u''_{\tau}-u'_{\tau})\times(v_{\tau}-v_{\tau})
\Displaycause{614C}
=u''_{\tau}\times\chi(\inf_{\sigma\in\Cal S}\Bvalue{\sigma=\tau})
    \times(v_{\tau}-v_{\downarrow})\cr&
=0\cr}$$

\noindent because
$\inf_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}
\Bsubseteq\Bvalue{v_{\downarrow}=v_{\tau}}$.   So

\Centerline{$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
=\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v+\int_{\Cal S\wedge\tau}\pmb z\,d\pmb v
=\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v$}

\noindent and

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v
=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
  +\int_{\Cal S\vee\tau}\pmb u\,d\pmb v
=\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v
  +\int_{\Cal S\vee\tau}\pmb u''\,d\pmb v$,}

\noindent as claimed.
}%end of proof of 646F

\leader{646G}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\tau$ a member of $\Cal S$.   Suppose that
$\pmb v'=\family{\sigma}{\Cal S\wedge\tau}{v'_{\sigma}}
\in\Mnsu(\Cal S\wedge\tau)$
and $\pmb v''=\family{\sigma}{\Cal S\vee\tau}{v''_{\sigma}}
\in\Mnsu(\Cal S\vee\tau)$.

(a) There is a $\pmb v=\sfamilysS{v}\in\Mnsu(\Cal S)$
such that $\pmb v''=\pmb v\restr\Cal S\vee\tau$.

(b) There is a $\pmb v=\sfamilysS{v}\in\Mnsu(\Cal S)$ such that

\Centerline{$\pmb v'=\pmb v\restr\Cal S\wedge\tau$,
\quad$\pmb v''\equiv\pmb v\restr\Cal S\vee\tau$.}

(c) If $\pmb w\in\Mnsu(\Cal S)$,
there is a $\pmb v^*\in\Mnsu(\Cal S)$ such that

\Centerline{$\pmb w\preccurlyeq\pmb v^*$,
\quad$\pmb v'\preccurlyeq\pmb v^*\restr\Cal S\wedge\tau$,
\quad$\pmb v''\preccurlyeq\pmb v^*\restr\Cal S\vee\tau$.}

\proof{{\bf (a)} In the language of 646F, set
$\pmb v=R(\tbf{0},\pmb v'')$, where $\tbf{0}$ is the zero process
with domain $\Cal S\wedge\tau$.   By 646Ff,
$\pmb v$ is a near-simple process;  the defining formula

\Centerline{$v_{\sigma}
=v''_{\sigma\vee\tau}\times\Bvalue{\tau\le\sigma}$}

\noindent shows that $\pmb v\restr\Cal S\vee\tau=\pmb v''$ and that
$\pmb v$ is non-negative and non-decreasing.   Now we know that
$\pmb v''$ is order-bounded (616Ib), so
$\sup|\pmb v|=\sup|\pmb v''|$ is defined.   Thus
$\pmb v$ is order-bounded;  being monotonic, it is an integrator (616Ra).

\medskip

{\bf (b)} Set

\Centerline{$\tildepmb v'
=\pmb v''+(v'_{\tau}-v''_{\tau})\tbf{1}\restr\Cal S\vee\tau
=\family{\sigma}{\Cal S\vee\tau}{v'_{\tau}+v''_{\sigma}-v''_{\tau}}$,}

\noindent so
that $\tildepmb v'\in\Mnsu(\Cal S\vee\tau)$.

Now set $\pmb v=R(\pmb v',\tilde{\pmb v}')$, where
$R:(L^0)^{\Cal S\wedge\tau}\times(L^0)^{\Cal S\vee\tau}\to(L^0)^{\Cal S}$
is defined as in 646F.    By 646Ff, $\pmb v$ is near-simple.
$\pmb v$ is
non-negative because $\pmb v'$ and $\tilde{\pmb v}'$ are
non-negative.   $\pmb v$ is
non-decreasing because if $\sigma\le\sigma'$ in $\Cal S$,
$a=\Bvalue{\sigma'<\tau}$, $b=\Bvalue{\sigma<\tau}\Bsetminus a$ and
$c=\Bvalue{\tau\le\sigma}$ then

$$\eqalign{v_{\sigma}
&=v'_{\sigma\wedge\tau}\times\chi a
   +v'_{\sigma\wedge\tau}\times\chi b
   +(v'_{\tau}-v''_{\tau}+v''_{\sigma\vee\tau})
      \times\chi c\cr
&\le v'_{\sigma'\wedge\tau}\times\chi a
   +v'_{\tau}\times\chi b
   +(v'_{\tau}-v''_{\tau}+v''_{\sigma'\vee\tau})
      \times\chi c\cr
&\le v'_{\sigma'\wedge\tau}\times\chi a
   +(v'_{\tau}-v''_{\tau}+v''_{\sigma'\vee\tau})\times\chi b
   +(v'_{\tau}-v''_{\tau}+v''_{\sigma'\vee\tau})
      \times\chi c
=v_{\sigma'}.\cr}$$

\noindent If $\sigma\in\Cal S\wedge\tau$, then

$$\eqalign{v_{\sigma}
&=v'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}
   +(v'_{\tau}-v''_{\tau}+v''_{\sigma\vee\tau})
     \times\chi\Bvalue{\tau\le\sigma}\cr
&=v'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}
   +(v'_{\tau}-v''_{\tau}+v''_{\sigma\vee\tau})
     \times\chi\Bvalue{\tau=\sigma}\cr
&=v'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma<\tau}
   +v'_{\tau}\times\chi\Bvalue{\tau=\sigma}
=v'_{\sigma},\cr}$$

\noindent so $\pmb v'=\pmb v\restr\Cal S\wedge\tau$.   If
$\sigma\le\sigma'$ in $\Cal S\vee\tau$ then of course

\Centerline{$v_{\sigma'}-v_{\sigma}
=(v'_{\tau}-v''_{\tau}+v''_{\sigma'})
   -(v'_{\tau}-v''_{\tau}+v''_{\sigma})
=v''_{\sigma'}-v''_{\sigma}$,}

\noindent so $\pmb v\restr\Cal S\vee\tau\equiv\pmb v''$.
Finally, as in (a), $\pmb v$ is order-bounded, therefore an
integrator.

\medskip

{\bf (c)} Take $\pmb v$ as in (b) and set $\pmb v^*=\pmb v+\pmb w$.
}%end of proof of 646G

\leader{646H}{Lemma}
Let $\Cal S$ be a sublattice of $\Cal T$, and $\tau$ a member
of $\Cal S$.   For $\pmb u=\sfamilysS{u}\in(L^0)^{\Cal S}$
define $R^*(\pmb u)\in(L^0)^{\Cal S\vee\tau}$ by saying that

\Centerline{$R^*(\pmb u)=\family{\sigma}{\Cal S\vee\tau}{u_{\sigma}
   \times\chi\Bvalue{\tau<\sigma}}
=(\pmb u\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}$.}

(a)(i) $R^*:(L^0)^{\Cal S}\to(L^0)^{\Cal S\vee\tau}$ is an
order-continuous $f$-algebra homomorphism.

\quad(ii) If $\pmb u\in(L^0)^{\Cal S}$ is fully adapted, then
$R^*(\pmb u)$ is fully adapted.

\quad(iii) If $\pmb u\in\Mob(\Cal S)$, then
$R^*(\pmb u)\in\Mob(\Cal S\vee\tau)$ and
$\sup|R^*(\pmb u)|\le\sup|\pmb u|$.

\quad(iv) If $\pmb u\in\Mmo(\Cal S)$ then
$R^*(\pmb u)\in\Mmo(\Cal S\vee\tau)$.

\quad(v) If $\pmb x$, $\pmb u\in(L^0)^{\Cal S}$,
$\pmb x\restr\Cal S\wedge\tau\le\pmb u\restr\Cal S\wedge\tau$ and
$R^*(\pmb x)\le\pmb u\restr\Cal S\vee\tau$, then $\pmb x\le\pmb u$.

(b) If $\pmb u\in\Mmo(\Cal S)$, then
$(\pmb u\restr\Cal S\vee\tau)_<=R^*(\pmb u_<)$.

(c) Suppose that $\pmb x\in\Mpob(\Cal S)$.   Write $\pmb x'$ for
$\pmb x\restr\Cal S\wedge\tau$.

\quad(i) $\pmb x'\in\Mpob(\Cal S\wedge\tau)$
and $R^*(\pmb x)\in\Mpob(\Cal S\vee\tau)$.

\quad(ii) If $\pmb v\in\Mnsu(\Cal S)$ , and we set
$\pmb v'=\pmb v\restr\Cal S\wedge\tau$ and
$\pmb v''=\pmb v\restr\Cal S\vee\tau$, then

\Centerline{$\max(\widehat{\theta}^{\#}_{\pmb v'}(\pmb x'),
\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb x))
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v'}(\pmb x')
  +\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb x)$.}

(d) If $\pmb x\in\MSi(\Cal S)$ then
$\pmb x\restr\Cal S\wedge\tau\in\MSi(\Cal S\wedge\tau)$ and
$\pmb x\restr\Cal S\vee\tau\in\MSi(\Cal S\vee\tau)$.

\proof{{\bf (a)(i)} Immediate from the facts that
$\Bvalue{\tau<\sigma}\in\frak A_{\sigma}$ for all $\tau$ and $\sigma$,
and if $a\in\frak A$
then $u\mapsto u\times\chi a$ is an order-continuous
$f$-algebra homomorphism from $L^0$ to itself.

\medskip

\quad{\bf (ii)} If $\pmb u$ is fully adapted, then
$\pmb u\restr\Cal S\vee\tau$ is fully
adapted (612Dc).   Next,
$\pmb x=\family{\sigma}{\Cal S\vee\tau}{\chi\Bvalue{\tau<\sigma}}$ is
fully adapted, because if $\sigma$, $\sigma'\in\Cal S\vee\tau$ then

\Centerline{$\Bvalue{\chi\Bvalue{\tau<\sigma}\ne\chi\Bvalue{\tau<\sigma'}}
=\Bvalue{\tau<\sigma}\Bsymmdiff\Bvalue{\tau<\sigma'}$}

\noindent does not meet $\Bvalue{\sigma=\sigma'}$, by 611E(c-iv-$\beta$).   So
$R^*(\pmb u)=(\pmb u\restr\Cal S\vee\tau)\times\pmb x$ is fully adapted.

\medskip

\quad{\bf (iii)} This is just because
$|u_{\sigma}\times\chi\Bvalue{\tau<\sigma}|\le|u_{\sigma}|\le\sup|\pmb u|$
for every $\sigma\in\Cal S\wedge\tau$.

\medskip

\quad{\bf (iv)} $\tbf 1^{(\Cal S\vee\tau)}$ is constant therefore moderately
oscillatory, so $\tbfoneminus{\Cal S\vee\tau}$ is moderately oscillatory (641L),
while $\pmb u\restr\Cal S\vee\tau$ is moderately oscillatory
(615F(a-i));  so the product
$R^*(\pmb u)=(\pmb u\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}$ is
moderately oscillatory (615F(a-iii)).

\medskip

\quad{\bf (v)} Express $\pmb x$, $\pmb u$ as $\sfamilysS{x}$ and
$\sfamilysS{u}$.   For $\sigma\in\Cal S$,

\Centerline{$x_{\sigma}\times\chi\Bvalue{\sigma\le\tau}
=x_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma\le\tau}
\le u_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma\le\tau}
=u_{\sigma}\times\chi\Bvalue{\sigma\le\tau}$}

\noindent and

$$\eqalignno{x_{\sigma}\times\chi\Bvalue{\tau<\sigma}
&=x_{\sigma\vee\tau}\times\chi\Bvalue{\tau<\sigma}
\le u_{\sigma\vee\tau}\times\chi\Bvalue{\tau<\sigma}\cr
\displaycause{because $R^*(\pmb x)\le\pmb u\restr\Cal S\vee\tau$}
&=u_{\sigma}\times\chi\Bvalue{\tau<\sigma},\cr}$$

\noindent so $x_{\sigma}\le u_{\sigma}$.

\medskip

{\bf (b)} This is covered by 641G(c-ii).

\medskip

{\bf (c)(i)} Let $\pmb u\in\Mmo(\Cal S)^+$
be such that $|\pmb x|\le\pmb u_<$.   Then
$\pmb u\restr\Cal S\wedge\tau\in\Mmo(\Cal S\wedge\tau)$
and $\pmb u\restr\Cal S\vee\tau\in\Mmo(\Cal S\vee\tau)$ (615F(a-i) again),
while
$(\pmb u\restr\Cal S\wedge\tau)_<=\pmb u_<\restr\Cal S\wedge\tau$ and
$(\pmb u\restr\Cal S\vee\tau)_<=R^*(\pmb u_<)$, by 641Gc.   So

\Centerline{
$|\pmb x\restr\Cal S\wedge\tau|\le(\pmb u\restr\Cal S\wedge\tau)_<$,
\quad$\pmb x\restr\Cal S\wedge\tau\in\Mpob(\Cal S\wedge\tau)$,}

\Centerline{
$|R^*(\pmb x)|=R^*(|\pmb x|)\le R^*(\pmb u_<)
  =(\pmb u\restr\Cal S\vee\tau)_<$,
\quad$R^*(\pmb x)\in\Mpob(\Cal S\vee\tau)$.}

\medskip

\quad{\bf (ii)} Observe first that $\pmb v'$ and $\pmb v''$ are
non-negative non-decreasing near-simple integrators (see 631F(a-iv) and
616P(b-ii)).

\medskip

\qquad\grheada\ Let $\sequencen{\pmb u_n}$ be a uniformly
order-bounded
non-decreasing sequence of non-negative moderately oscillatory processes
with domain
$\Cal S$ such that $|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$.
Setting $\pmb x'=\pmb x\restr\Cal S\wedge\tau$,
$\pmb v'=\pmb v\restr\Cal S\wedge\tau$ and
$\pmb u'_n=\pmb u_n\restr\Cal S\wedge\tau$ for each $n$,
we have $\pmb u'_{n<}=\pmb u_{n<}\restr\Cal S\wedge\tau$ for
each $n$ (641Gc again), so

\Centerline{$|\pmb x'|
\le\sup_{n\in\Bbb N}\pmb u_{n<}\restr\Cal S\wedge\tau
=\sup_{n\in\Bbb N}\pmb u'_{n<}$,}

\noindent while $\sequencen{\pmb u'_n}$ is a non-decreasing
sequence of non-negative moderately oscillatory processes
(615F(a-i) once more);  also
$\sup|\pmb u'_n|\le\sup|\pmb u_n|$ for each $n$, so
$\sequencen{\pmb u'_n}$ is uniformly order-bounded.
Accordingly

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v'}(\pmb x')
&\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S\wedge\tau}
   \pmb u'_n\,d\pmb v')\cr
&=\sup_{n\in\Bbb N}\theta(\int_{\Cal S\wedge\tau}\pmb u_n\,d\pmb v)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)\cr}$$

\noindent because $\pmb v$ is non-decreasing and $\pmb u_n$ is
non-negative, so

\Centerline{$0
\le\int_{\Cal S\wedge\tau}\pmb u_n\,d\pmb v
=\int_{\Cal S}\pmb u_n\,d\pmb v
  -\int_{\Cal S\vee\tau}\pmb u_n\,d\pmb v
\le\int_{\Cal S}\pmb u_n\,d\pmb v$}

\noindent for every $n$.    As $\sequencen{\pmb u_n}$ is arbitrary,
$\widehat{\theta}^{\#}_{\pmb v'}(\pmb x')
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$.

\medskip

\qquad\grheadb\ The same line of argument works in $\Cal S\vee\tau$.
Again, let $\sequencen{\pmb u_n}$ be a uniformly
order-bounded
non-decreasing sequence of non-negative moderately oscillatory
processes with domain
$\Cal S$ such that $|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$.
This time, setting
$\pmb v''=\pmb v\restr\Cal S\vee\tau$ and
$\pmb u''_n=\pmb u_n\restr\Cal S\vee\tau$ for each $n$,
we have $\pmb u''_{n<}=R^*(\pmb u_{n<})$ for
each $n$ ((b) above), so

$$\eqalignno{|R^*(\pmb x)|
&\le\sup_{n\in\Bbb N}R^*(\pmb u_{n<})\cr
\displaycause{recall that $R^*$ is an order-continuous
lattice homomorphism with domain $(L^0)^{\Cal S}$}
&=\sup_{n\in\Bbb N}\pmb u''_{n<}.\cr}$$

\noindent As in ($\alpha$),
$\sequencen{\pmb u''_n}$ is a uniformly order-bounded
non-decreasing
sequence of non-negative near-simple processes.   Accordingly

$$\eqalign{\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb x)
&\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S\vee\tau}
   \pmb u''_n\,d\pmb v'')\cr
&=\sup_{n\in\Bbb N}\theta(\int_{\Cal S\vee\tau}\pmb u_n\,d\pmb v)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v).\cr}$$

\noindent As $\sequencen{\pmb u_n}$ is arbitrary,
$\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb x)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$.

\medskip

\qquad\grheadc\ Thus

\Centerline{$\max(\widehat{\theta}^{\#}_{\pmb v'}(\pmb x'),
\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb x))
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x)$.}

\noindent As for the other inequality, let $\sequencen{\pmb u'_n}$,
$\sequencen{\pmb u''_n}$ be
uniformly order-bounded non-decreasing sequences in
$\Mmo(\Cal S\wedge\tau)^+$, $\Mmo(\Cal S\vee\tau)^+$ respectively such that
$|\pmb x'|\le\sup_{n\in\Bbb N}\pmb u'_{n<}$ and
$|R^*(\pmb x)|\le\sup_{n\in\Bbb N}\pmb u''_{n<}$.   Set
$\pmb u_n=R(\pmb u'_n,\pmb u''_n)$ for each $n$, where $R$ is the operator
defined in 646F;
then $\sequencen{\pmb u_n}$ is a uniformly order-bounded non-decreasing
sequence in $\Mmo(\Cal S)$ (646Fb, 646Fd, 646F(e-i)).
Now $|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$.   \Prf\ Express
$\pmb x$ as $\sfamilysS{x}$, $\pmb u_n$ as
$\family{\sigma}{\Cal S}{u_{n\sigma}}$,
$\pmb u_{n<}$ as $\family{\sigma}{\Cal S}{u_{n<\sigma}}$ and
$\pmb u''_{n<}$ as $\family{\sigma}{\Cal S\wedge\tau}{u''_{n<\sigma}}$
for each $n$.   If $\sigma\in\Cal S$, then

$$\eqalign{|x_{\sigma}|\times\chi\Bvalue{\sigma\le\tau}
&=|x_{\sigma\wedge\tau}|\times\chi\Bvalue{\sigma\le\tau}
\le\sup_{n\in\Bbb N}u_{n<(\sigma\wedge\tau)}
  \times\chi\Bvalue{\sigma\le\tau}
\cr&=\sup_{n\in\Bbb N}u_{n<\sigma}
  \times\chi\Bvalue{\sigma\le\tau}\cr}$$

\noindent because

\Centerline{$|\pmb x'|\le\sup_{n\in\Bbb N}\pmb u'_{n<}
=\sup_{n\in\Bbb N}\pmb u_{n<}\restr\Cal S\wedge\tau$}

\noindent(641Gc once more) and $\pmb u_{n<}$ is fully adapted.   We also have

$$\eqalignno{|x_{\sigma}|\times\chi\Bvalue{\tau<\sigma}
&=|x_{\sigma\vee\tau}|\times\chi\Bvalue{\tau<\sigma}
\le\sup_{n\in\Bbb N}u''_{n<(\sigma\vee\tau)}
  \times\chi\Bvalue{\tau<\sigma}
\Displaycause{because $|R^*(\pmb x)|\le\sup_{n\in\Bbb N}\pmb u''_{n<}$}
=\sup_{n\in\Bbb N}u''_{n<(\sigma\vee\tau)}
  \times\chi\Bvalue{\tau<\sigma}
=\sup_{n\in\Bbb N}u''_{n<\sigma}
  \times\chi\Bvalue{\tau<\sigma}
\cr}$$

\noindent because $\pmb u''_{n<}=R^*(\pmb u_{n<})$ for every $n$, by
(b) above.   So
$|x_{\sigma}|\le\sup_{n\in\Bbb N}u_{n<\sigma}$;  as $\sigma$ is
arbitrary, $|\pmb x|\le\sup_{n\in\Bbb N}\pmb u_{n<}$.\ \Qed

Accordingly

$$\eqalignno{\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
&\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_n\,d\pmb v)
\le\sup_{n\in\Bbb N}\bigl(\theta(\int_{\Cal S\wedge\tau}\pmb u'_n\,d\pmb v)
  +\theta(\int_{\Cal S\wedge\tau}\pmb u''_n\,d\pmb v)\bigr)\cr
\displaycause{646Fg}
&=\sup_{n\in\Bbb N}
   \bigl(\theta(\int_{\Cal S\wedge\tau}\pmb u'_n\,d\pmb v')
  +\theta(\int_{\Cal S\wedge\tau}\pmb u''_n\,d\pmb v'')\bigr)
\cr&=\sup_{n\in\Bbb N}
     \theta(\int_{\Cal S\wedge\tau}\pmb u'_n\,d\pmb v')
  +\sup_{n\in\Bbb N}
     \theta(\int_{\Cal S\wedge\tau}\pmb u''_n\,d\pmb v'')\cr}$$

\noindent because these are all non-decreasing sequences.
As $\sequencen{\pmb u'_n}$ and $\sequencen{\pmb u''_n}$ are
arbitrary,

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb x)
  \le\widehat{\theta}^{\#}_{\pmb v'}(\pmb x')
    +\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb x)$.}

\medskip

{\bf (d)(i)} Here write $\pmb y$ for $\pmb x\times\tbf 1^{(\Cal S)}_<$ and
$\pmb y'$ for

\Centerline{$\pmb y\restr\Cal S\wedge\tau
=(\pmb x\restr\Cal S\wedge\tau)
   \times(\tbf 1^{(\Cal S)}_<\restr\Cal S\wedge\tau)
=(\pmb x\restr\Cal S\wedge\tau)\times\tbf 1^{(\Cal S\wedge\tau)}_<$}

\noindent (using 641G(c-ii) again).
As $\pmb y\in\MSi^0(\Cal S)$,
we know from (c-i) that $\pmb y'\in\Mpob(\Cal S\wedge\tau)$.   Take
$\pmb v'\in\Mnsu(\Cal S\wedge\tau)$ and $\epsilon>0$.   Then we have
a $\pmb v\in\Mnsu(\Cal S)$ such that $\pmb v'=\pmb v\restr\Cal S\wedge\tau$
(646Gb).   Let $\pmb u\in\Mmo(\Cal S)$ be such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb y-\pmb u_<)
\le\epsilon$, and set $\pmb u'=\pmb u\restr\Cal S\wedge\tau$.
Then (c-ii) tells us that

\Centerline{$\widehat{\theta}^{\#}_{\pmb v'}(\pmb y'-\pmb u'_<)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb y-\pmb u_<)
\le\epsilon$.}

\noindent As $\pmb v'$ and $\epsilon$ are arbitrary,
$\pmb y'\in\MSi^0(\Cal S\wedge\tau)$ and
$\pmb x\restr\Cal S\wedge\tau\in\MSi(\Cal S\wedge\tau)$.

\medskip

\quad{\bf (ii)} As for
$(\pmb x\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}
=R^*(\pmb x)$,
the point is that this is equal to $R^*(\pmb x\times\tbfoneminus{\Cal S})$.
\Prf\

$$\eqalignno{R^*(\pmb x\times\tbfoneminus{\Cal S})
&=R^*(\pmb x)\times R^*(\tbfoneminus{\Cal S})
=R^*(\pmb x)\times\tbfoneminus{\Cal S\vee\tau}
\Displaycause{(b) above}
=R^*(\pmb x).  \text{ \Qed}\cr}$$

\noindent Since $\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0(\Cal S)$,
(c-i) tells us that $R^*(\pmb x)\in\Mpob(\Cal S\vee\tau)$.
Take $\pmb v''\in\Mnsu(\Cal S\vee\tau)$ and $\epsilon>0$.   Then we have
a $\pmb v\in\Mnsu(\Cal S)$ such that
$\pmb v''=\pmb v\restr\Cal S\vee\tau$ (646Ga).
Let $\pmb u\in\Mmo(\Cal S)$ be such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u_<)
\le\epsilon$.   Then $\pmb u''=\pmb u\restr\Cal S\vee\tau$ is
moderately oscillatory and

\Centerline{$\widehat{\theta}^{\#}_{\pmb v''}(R^*(\pmb x)-\pmb u''_<)
=\widehat{\theta}^{\#}_{\pmb v''}(R^*(\pmb x)-R^*(\pmb u_<))
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb x-\pmb u_<)
\le\epsilon$}

\noindent by (b) and (c-ii).   As $\pmb v''$ and $\epsilon$ are arbitrary,
$R^*(\pmb x)\in\MSi^0(\Cal S\vee\tau)$, so
$\pmb x\restr\Cal S\vee\tau\in\MSi(\Cal S\vee\tau)$.
}%end of proof of 646H

\leader{646I}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$
and $\tau$ a member of $\Cal S$.   Define
$R^*:(L^0)^{\Cal S}\to(L^0)^{\Cal S\vee\tau}$ as in 646H.
Take $\pmb x\in(L^0)^{\Cal S}$.

(a) If
$\pmb x\restr\Cal S\wedge\tau\in\Mpob(\Cal S\wedge\tau)$ and
$R^*(\pmb x)\in\Mpob(\Cal S\vee\tau)$, then $\pmb x\in\Mpob(\Cal S)$.

(b) If
$\pmb x\restr\Cal S\wedge\tau\in\MSi^0(\Cal S\wedge\tau)$ and
$R^*(\pmb x)\in\MSi^0(\Cal S\vee\tau)$, then $\pmb x\in\MSi^0(\Cal S)$.

(c) If
$\pmb x\restr\Cal S\wedge\tau\in\MSi(\Cal S\wedge\tau)$ and
$\pmb x\restr\Cal S\vee\tau\in\MSi(\Cal S\vee\tau)$,
then $\pmb x\in\MSi(\Cal S)$.

\proof{ Write $\pmb x'$ for $\pmb x\restr\Cal S\wedge\tau$.

\medskip

{\bf (a)} There are $\pmb u'\in\Mmo(\Cal S\wedge\tau)^+$ and
$\pmb u''\in\Mmo(\Cal S\vee\tau)^+$ such that
$|\pmb x'|\le\pmb u'_<$
and $|R^*(\pmb x)|\le\pmb u''_<$.   Set $\pmb u=R(\pmb u',\pmb u'')$, as
defined in 646F.   Then $\pmb u\in\Mmo(\Cal S)^+$ (646F(e-i)),

\Centerline{$|\pmb x'|\le\pmb u'_<=\pmb u_<\restr\Cal S\wedge\tau$}

\noindent (646F(e-ii)) and

\Centerline{$|R^*(\pmb x)|\le\pmb u''_<
=(\pmb u\restr\Cal S\vee\tau)_<
=R^*(\pmb u_<)
\le\pmb u_<\restr\Cal S\vee\tau$}

\noindent (646Fa, 646Hb).   Putting these together,
$|\pmb x|\le\pmb u_<$ (646H(a-v)) and $\pmb x\in\Mpob(\Cal S)$.

\medskip

{\bf (b)} Take $\pmb v\in\Mnsu(\Cal S)$ and $\epsilon>0$.   Then
$\pmb v'=\pmb v\restr\Cal S\wedge\tau$ and
$\pmb v''=\pmb v\restr\Cal S\vee\tau$ are non-negative non-decreasing
near-simple integrators, so there are
$\pmb u'\in\Mmo(\Cal S\wedge\tau)$ and
$\pmb u''\in\Mmo(\Cal S\vee\tau)$ such that

\Centerline{$\widehat{\theta}^{\#}_{\pmb v'}(\pmb u'_<-\pmb x')
\le\epsilon$,
\quad$\widehat{\theta}^{\#}_{\pmb v''}(\pmb u''_<-R^*(\pmb x))
\le\epsilon$.}

\noindent Set $\pmb u=R(\pmb u',\pmb u'')$.   Then
$\pmb u'_<=\pmb u_<\restr\Cal S\wedge\tau$ (646F(e-ii)) and
$R^*(\pmb u_<)=\pmb u''_<$ (646Hb), so

$$\eqalignno{\widehat{\theta}^{\#}_{\pmb v}(\pmb u_<-\pmb x)
&\le\widehat{\theta}^{\#}_{\pmb v'}
     ((\pmb u_<-\pmb x)\restr\Cal S\wedge\tau)
  +\widehat{\theta}^{\#}_{\pmb v''}R^*(\pmb u_<-\pmb x)\cr
\displaycause{646H(c-ii)}
&=\widehat{\theta}^{\#}_{\pmb v'}(\pmb u'_<-\pmb x')
  +\widehat{\theta}^{\#}_{\pmb v''}(\pmb u''_<-R^*(\pmb x))
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\pmb x\in\MSi^0(\Cal S)$.

\medskip

{\bf (c)} $(\pmb x\times\tbf 1^{(\Cal S)}_<)\restr\Cal S\wedge\tau
=\pmb x'\times\tbf 1^{(\Cal S\wedge\tau)}_<$ belongs to
$\MSi^0(\Cal S\wedge\tau)$ and
$R^*(\pmb x\times\tbf 1^{(\Cal S)}_<)\in\MSi^0(\Cal S\vee\tau)$.
\Prf\ Setting
$e_{\sigma}=\sup_{\sigma'\in\Cal S}\Bvalue{\sigma'<\sigma}$ for
$\sigma\in\Cal S$, $\tbf 1^{(\Cal S)}_<
=\family{\sigma}{\Cal S}{\chi e_{\sigma}}$ (641Gb), and

\Centerline{$R^*(\tbf 1^{(\Cal S)}_<)
=\family{\sigma}{\Cal S\vee\tau}{\chi e_{\sigma}\times\chi\Bvalue{\tau<\sigma}}
=\family{\sigma}{\Cal S\vee\tau}{\chi\Bvalue{\tau<\sigma}}
=\tbf 1^{(\Cal S\vee\tau)}_<$}

\noindent because $\Bvalue{\tau<\sigma}\Bsubseteq e_{\sigma}$ for every
$\sigma$.   Now

$$\eqalignno{R^*(\pmb x\times\tbf 1^{(\Cal S)}_<)
&=R^*(\pmb x)\times R^*(\tbf 1^{(\Cal S)}_<)
\Displaycause{646H(a-i)}
=(\pmb x\restr\Cal S\vee\tau)\times\tbf 1^{(\Cal S\vee\tau)}_<
   \times\tbf 1^{(\Cal S\vee\tau)}_<
=(\pmb x\restr\Cal S\vee\tau)\times\tbf 1^{(\Cal S\vee\tau)}_<\cr}$$

\noindent belongs to $\MSi^0(\Cal S\vee\tau)$.\ \Qed

Now (b) tells us that $\pmb x\times\tbf 1^{(\Cal S)}_<\in\MSi^0(\Cal S)$, that
is, $\pmb x\in\MSi(\Cal S)$.
}%end of proof of 646I

\leader{646J}{Theorem}
Suppose that $\family{t}{T}{\frak A_t}$ is right-continuous.   Let
$\Cal S$ be an order-convex sublattice of $\Cal T$,
and $\tau$ a member of $\Cal S$.   If
$\pmb x\in(L^0)^{\Cal S}$, then $\pmb x$ is an S-integrable process
iff $\pmb x\restr\Cal S\wedge\tau$ and $\pmb x\restr\Cal S\vee\tau$
are both S-integrable, and in this case

\Centerline{$\biggerSint_{\Cal S}\pmb x\,d\pmb v
=\biggerSint_{\Cal S\wedge\tau}\pmb x\,d\pmb v
    +\biggerSint_{\Cal S\vee\tau}\pmb x\,d\pmb v$}
	
\noindent for every near-simple integrator $\pmb v$ with domain $\Cal S$.
	
\proof{{\bf (a)} By 646Hd and 646Ic, $\pmb x$ is S-integrable iff
$\pmb x\restr\Cal S\wedge\tau$ and $\pmb x\restr\Cal S\vee\tau$ are
S-integrable.

\medskip

{\bf (b)(i)} Now suppose that $\pmb x$, $\pmb x'=\pmb x\restr\Cal S\wedge\tau$
and $\pmb x''=\pmb x\restr\Cal S\vee\tau$ are all
S-integrable, and that $\pmb v$ is a near-simple integrator with domain
$\Cal S$.   Then
$\pmb v'=\pmb v\restr\Cal S\wedge\tau$ and
$\pmb v''=\pmb v\restr\Cal S\vee\tau$ are near-simple integrators
(631F(a-iv) and 616P(b-ii) again), so we can form the integrals

\Centerline{$z=\Sint_{\Cal S}\pmb x\,d\pmb v$,
\quad$z'=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$,
\quad$z''=\Sint_{\Cal S\vee\tau}\pmb x\,d\pmb v$,}

\noindent and setting

\Centerline{$\pmb y=x\times\tbfoneminus{\Cal S}$.
\quad$\pmb y'=\pmb y\restr\Cal S\wedge\tau
=(\pmb x\restr\Cal S\wedge\tau)\times\tbfoneminus{\Cal S\wedge\tau}$,}

\Centerline{$R^*(\pmb y)
=(\pmb y\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}
=(\pmb x\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}$}

\noindent as in 646H-646I, we have

\Centerline{$z=\Sint_{\Cal S}\pmb y\,d\pmb v$,
\quad$z'=\Sint_{\Cal S\wedge\tau}\pmb y'\,d\pmb v'$,
\quad$z''=\Sint_{\Cal S\vee\tau}R^*(\pmb y)d\pmb v''$.}

By 645La, there is a non-negative $\bar u\in L^0$ such that
$\pmb y$ belongs to the closure of
$A=\{\pmb u_<:\pmb u\in\Mmo(\Cal S)$,
$\sup|\pmb u|\le\bar u\}$ for the S-integration topology
on $\MSi^0(\Cal S)$.   Set
$A'=\{\pmb u_<:\pmb u\in\Mmo(\Cal S\wedge\tau)$,
$\sup|\pmb u|\le\bar u\}$ and
$A''=\{\pmb u_<:\pmb u\in\Mmo(\Cal S\vee\tau)$,
$\sup|\pmb u|\le\bar u\}$.

\medskip

\quad{\bf (ii)} Take $\epsilon>0$.
Then we have $\pmb v$, $\pmb v'$,
$\pmb v''$ and $\delta>0$ such that
$\pmb v=\sfamilysS{v}$,
$\pmb v'=\family{\sigma}{\Cal S\wedge\tau}{v'_{\sigma}}$
and $\pmb v''=\family{\sigma}{\Cal S\vee\tau}{v''_{\sigma}}$
are all non-negative non-decreasing near-simple integrators,

\Centerline{$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$ whenever
$\pmb u\in A$ and $\widehat{\theta}^{\#}_{\pmb v}(\pmb y-\pmb u_<)\le\delta$,}

\Centerline{$
\theta(z'-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v')\le\epsilon$ whenever
$\pmb u\in A'$ and
$\widehat{\theta}^{\#}_{\pmb v'}(\pmb y'-\pmb u_<)\le\delta$,}

\Centerline{
$\theta(z''-\int_{\Cal S\vee\tau}\pmb u\,d\pmb v'')\le\epsilon$
whenever $\pmb u\in A''$ and
$\widehat{\theta}^{\#}_{\pmb v''}(R^*(\pmb y)-\pmb u''_<)
\le\delta$.}

\noindent Now there is a $\pmb w=\sfamilysS{w^*}\in\Mnsu(\Cal S)$ such that

\Centerline{$\pmb v\preccurlyeq\pmb w$,
\quad $\pmb v'\preccurlyeq\pmb w\restr\Cal S\wedge\tau$,
\quad $\pmb v''\preccurlyeq\pmb w\restr\Cal S\vee\tau$}

\noindent(646Gc).   We chose
$\bar u$ so that there would be a $\pmb u\in A$ such that
$\widehat{\theta}^{\#}_{\pmb w}(\pmb y-\pmb u_<)\le\delta$.
In this case, setting

\Centerline{$\pmb w'=\pmb w\restr\Cal S\wedge\tau$,
\quad$\pmb w''=\pmb w\restr\Cal S\vee\tau$,}

\Centerline{$\pmb u'=\pmb u\restr\Cal S\wedge\tau\in A'$,
\quad$\pmb u''=\pmb u\restr\Cal S\vee\tau\in A''$,}

\noindent we have

\Centerline{$\widehat{\theta}_{\pmb v}(\pmb y-\pmb u_<)
\le\widehat{\theta}^{\#}_{\pmb w}(\pmb y-\pmb u_<)\le\delta$}

\noindent (645Dc),

\Centerline{$\widehat{\theta}_{\pmb v'}(\pmb y'-\pmb u'_<)
\le\widehat{\theta}^{\#}_{\pmb w'}(\pmb y'-\pmb u'_<)
\le\widehat{\theta}^{\#}_{\pmb w}(\pmb y-\pmb u_<)
\le\delta$}

\noindent (646H(c-ii)) and

\Centerline{$\widehat{\theta}_{\pmb v''}
  (R^*(\pmb y)-\pmb u''_<)
\le\widehat{\theta}^{\#}_{\pmb w''}(R^*(\pmb y)-\pmb u''_<)
\le\widehat{\theta}^{\#}_{\pmb w}(\pmb y-\pmb u_<)
\le\delta$}

\noindent by the other half of 646H(c-ii).   So

\Centerline{$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$,
\quad$\theta(z'
  -\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v')\le\epsilon$,
\quad$\theta(z''
  -\int_{\Cal S\vee\tau}\pmb u''\,d\pmb v'')\le\epsilon$.}

\noindent But

$$\eqalignno{\int_{\Cal S}\pmb u\,d\pmb v
&=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
  +\int_{\Cal S\vee\tau}\pmb u\,d\pmb v\cr
\displaycause{613J(c-i) again}
&=\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v'
  +\int_{\Cal S\vee\tau}\pmb u''\,d\pmb v'',\cr}$$

\noindent so $\theta(z-z'-z'')\le 3\epsilon$.   As $\epsilon$ is
arbitrary, $z=z'+z''$, that is,

\Centerline{$\biggerSint_{\Cal S}\pmb x\,d\pmb v
=\biggerSint_{\Cal S\wedge\tau}\pmb x\,d\pmb v
    +\biggerSint_{\Cal S\vee\tau}\pmb x\,d\pmb v$,}

\noindent as claimed.	
}%end of proof of 646J

\leader{646K}{Indefinite S-integrals} Suppose that
$\family{t}{T}{\frak A_t}$ is right-continuous and that
$\Cal S$ is an order-convex sublattice of $\Cal T$.

\spheader 646Ka Suppose that $\pmb x$ is a locally S-integrable process
and that $\pmb v$ is a
locally near-simple local integrator, both with domain $\Cal S$.
Then we can define
$z_{\tau}=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$ for
$\tau\in\Cal S$\cmmnt{ (646Hd, 646J)}.
Now the {\bf indefinite S-integral of $\pmb x$ with respect to $\pmb v$}
is $\Siiv(\pmb x)=\family{\tau}{\Cal S}{z_{\tau}}$.

\spheader 646Kb It will more than once be useful to note that, in the
context of (a) just above,
$\Siiv(\pmb x)=\Siiv(\pmb x\times\tbfoneminus{\Cal S})$.   \prooflet{\Prf\
If $\tau\in\Cal S$ then

$$\eqalignno{\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v
&=\Sint_{\Cal S\wedge\tau}\pmb x\times\tbfoneminus{\Cal S\wedge\tau}
  \,d\pmb v
=\Sint_{\Cal S\wedge\tau}
  \pmb x\times\tbfoneminus{\Cal S}\times\tbfoneminus{\Cal S\wedge\tau}
  \,d\pmb v
\Displaycause{because $\tbfoneminus{\Cal S}\restr\Cal S\wedge\tau=
\tbfoneminus{\Cal S\wedge\tau}$}
=\Sint_{\Cal S\wedge\tau}
  \pmb x\times\tbfoneminus{\Cal S}\,d\pmb v.  \text{ \Qed}
\cr}$$
}

\spheader 646Kc If $\pmb u$ is a locally moderately oscillatory
process with domain $\Cal S$ then
$\Siiv(\pmb u_<)=\iivu$.   \prooflet{\Prf\ If $\tau\in\Cal S$ then

$$\eqalignno{\Sint_{\Cal S\wedge\tau}\pmb u_<d\pmb v
&=\Sint_{\Cal S\wedge\tau}(\pmb u_<\restr\Cal S\wedge\tau)d\pmb v
=\Sint_{\Cal S\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)_<\,d\pmb v
\Displaycause{641G(c-ii) once more}
=\int_{\Cal S\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)d\pmb v\cr
\displaycause{645R(a-i)}
&=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v.  \text{ \Qed}\cr}$$
}

\leader{646L}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex sublattice of
$\Cal T$.   Let $\pmb v$ be a near-simple integrator
and $\pmb x$ an S-integrable process, both with domain $\Cal S$.
For $\tau\in\Cal S$, set
$z_{\tau}=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$.
Suppose that $\pmb u^*\in\Mmo(\Cal S)^+$ is such that
$|\pmb x|\le\pmb u^*$.   Then for any $\epsilon>0$ there is a
$\TSi$-neighbourhood $G$ of $\pmb x$ such that
$\theta(z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)\le\epsilon$
whenever $\pmb u\in\Mmo(\Cal S)$, $|\pmb u|\le\pmb u^*$,
$\pmb u_<\in G$ and $\tau\in\Cal S$.

\proof{{\bf (a)} I should begin by noting that $z_{\tau}$ is defined for
every $\tau\in\Cal S$, by 646J.   Define
$R_{\tau}:(L^0)^{\Cal S}\to(L^0)^{\Cal S}$, for $\tau\in\Cal S$, by setting

\Centerline{$R_{\tau}(\sfamilysS{u})
=\family{\sigma}{\Cal S}{u_{\sigma}\times\chi\Bvalue{\sigma\le\tau}}$}

\noindent for $\sfamilysS{u}\in(L^0)^{\Cal S}$.   As in 646F and
646H, $R_{\tau}$ is an order-continuous $f$-algebra homomorphism and
$R_{\tau}(\pmb u)$ is fully adapted whenever $\pmb u\in(L^0)^{\Cal S}$ is
fully adapted.   If $\pmb y$ is any S-integrable process with domain
$\Cal S$ and $\tau\in\Cal S$ then

\Centerline{$R_{\tau}(\pmb y)\restr\Cal S\wedge\tau
=\pmb y\restr\Cal S\wedge\tau$,
\quad$(R_{\tau}(\pmb y)\restr\Cal S\vee\tau)\times\tbfoneminus{\Cal S\vee\tau}
=\tbf 0$,}

\noindent so

$$\eqalignno{\Sint_{\Cal S}R_{\tau}(\pmb y)d\pmb v
&=\Sint_{\Cal S\wedge\tau}R_{\tau}(\pmb y)d\pmb v
  +\Sint_{\Cal S\vee\tau}R_{\tau}(\pmb y)d\pmb v
\Displaycause{646J}
=\Sint_{\Cal S\wedge\tau}R_{\tau}(\pmb y)d\pmb v
  +\Sint_{\Cal S\vee\tau}R_{\tau}(\pmb y)\times\tbfoneminus{\Cal S\vee\tau}
     \,d\pmb v\cr
&=\Sint_{\Cal S\wedge\tau}\pmb y\,d\pmb v.\cr}$$

\noindent In particular, $z_{\tau}=\Sint_{\Cal S}R_{\tau}(\pmb x)d\pmb v$.

If $\pmb u\in\Mmo(\Cal S)$ then

$$\eqalignno{\Sint_{\Cal S}R_{\tau}(\pmb u_<)d\pmb v
&=\Sint_{\Cal S\wedge\tau}(\pmb u_<\restr\Cal S\wedge\tau)d\pmb v
=\Sint_{\Cal S\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)_<\,d\pmb v\cr&
=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
\cr}$$

\noindent by 645R(a-i) again.

\medskip

\quad{\bf (b)} Let $\epsilon>0$.   By 645S, there are a
$\pmb v^*\in\Mnsu(\Cal S)$ and a $\delta>0$ such that
$\theta(\Sint_{\Cal S}\pmb y\,d\pmb v-\Sint_{\Cal S}\pmb y'\,d\pmb v)
\le\epsilon$ whenever $\pmb y$, $\pmb y'\in\MSi^0(\Cal S)$,
$|\pmb y|\le\pmb u^*_<$, $|\pmb y'|\le\pmb u^*_<$ and
$\widehat{\theta}^{\#}_{\pmb v^*}(\pmb y-\pmb y')\le\delta$.
Set $G=\{\pmb y:\pmb y\in\Mpob(\Cal S)$,
$\widehat{\theta}^{\#}_{\pmb v^*}(\pmb y-\pmb x)\le\delta\}$,
so that $G$ is a $\TSi$-neighbourhood of $\pmb x$.
If $\pmb u\in\Mns(\Cal S)$,
$|\pmb u|\le\pmb u^*$ and $\pmb u_<\in G$, then,
for any $\tau\in\Cal S$,

\Centerline{$|R_{\tau}(\pmb x)-R_{\tau}(\pmb u_<)|
=|R_{\tau}(\pmb x-\pmb u_<)|\le|\pmb x-\pmb u_<|$,}

\noindent so

\Centerline{$\widehat{\theta}^{\#}_{\pmb v^*}
  (R_{\tau}(\pmb x)-R_{\tau}(\pmb u_<))
\le\widehat{\theta}^{\#}_{\pmb v^*}(\pmb x-\pmb u_<)\le\delta$}

\noindent and

\Centerline{$\theta(z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)
=\theta(\biggerSint_{\Cal S}R_{\tau}(\pmb x)d\pmb v
   -\biggerSint_{\Cal S}R_{\tau}(\pmb u_<)d\pmb v)
\le\epsilon$,}

\noindent as required.
}%end of proof of 646L

\leader{646M}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is a non-empty order-convex sublattice of
$\Cal T$.   Let $\pmb v$ be a near-simple integrator
and $\pmb x$ an S-integrable process, both with domain $\Cal S$;  set
$z=\Sint_{\Cal S}\pmb x\,d\pmb v$ and
$z_{\tau}=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$ for $\tau\in\Cal S$.
Then $\lim_{\tau\uparrow\Cal S}z_{\tau}=z$ and
$\lim_{\tau\downarrow\Cal S}z_{\tau}=0$.

\proof{ By 646Kb,
$\Siiv(\pmb x)=\Siiv(\pmb x\times\tbfoneminus{\Cal S})$,
while $\Sint_{\Cal S}\pmb x\,d\pmb v
=\Sint_{\Cal S}\pmb x\times\tbfoneminus{\Cal S}\,d\pmb v$;  so we may
assume that $\pmb x$ itself belongs to $\MSi^0(\Cal S)$.
Let $\pmb u^*\in\Mmo^+=\Mmo(\Cal S)^+$ be such that $|\pmb x|\le\pmb u^*_<$, and
take $\epsilon>0$.   By 646L, there is a $\TSi$-neighbourhood $G$ of $\pmb x$
such that $\theta(z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)\le\epsilon$
whenever $\tau\in\Cal S$, $\pmb u\in\Mmo$, $|\pmb u|\le\pmb u^*$ and
$\pmb u_<\in G$.   At the same time, there is a $\TSi$-neigbourhood
$G'$ of $\pmb x$
such that $\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$
whenever $\pmb u\in\Mmo$, $|\pmb u|\le\pmb u^*$ and
$\pmb u_<\in G'$.   By 645La, there is a $\pmb u\in\Mmo$ such that
$|\pmb u|\le\pmb u^*$ and $\pmb u_<\in G\cap G'$.   So now we have

\Centerline{$\theta(z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)\le\epsilon$
whenever $\tau\in\Cal S$,
\quad$\theta(z-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$.}

\noindent Accordingly

$$\eqalignno{\limsup_{\tau\uparrow\Cal S}\theta(z-z_{\tau})&
\le\theta(z-\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$})
 +\limsup_{\tau\uparrow\Cal S}
  \theta(\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}
    -\hbox{$\biggerint_{\Cal S\wedge\tau}\pmb{u}\,d\pmb{v}$})
\cr&
\mskip200mu
 +\limsup_{\tau\uparrow\Cal S}
  \theta(\hbox{$\biggerint_{\Cal S\wedge\tau}\pmb{u}\,d\pmb{v}$}-z_{\tau})\cr&
\le\epsilon+0+\epsilon
\Displaycause{613J(f-ii)}
=2\epsilon,\cr
\limsup_{\tau\downarrow\Cal S}\theta(z_{\tau})&
\le\limsup_{\tau\downarrow\Cal S}
  \theta(\hbox{$\biggerint_{\Cal S\wedge\tau}\pmb{u}\,d\pmb{v}$})
   +\limsup_{\tau\downarrow\Cal S}
  \theta(\hbox{$\biggerint_{\Cal S\wedge\tau}\pmb{u}\,d\pmb{v}$}-z_{\tau})\cr&
\le 0+\epsilon
\Displaycause{613J(f-i)}
=\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, we have the result.
}%end of proof of 646M

\leader{646N}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex sublattice of
$\Cal T$.   Let $\pmb v$ be a near-simple integrator
and $\pmb x$ an S-integrable process, both with domain $\Cal S$.
Then $\Siiv(\pmb x)$ is a near-simple integrator.

\proof{{\bf (a)} As in 646M, we may
assume that $\pmb x$ belongs to $\MSi^0(\Cal S)$.   Again set
$z_{\tau}=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$ for $\tau\in\Cal S$, and take
$\epsilon>0$ and $\pmb u^*\in\Mmo^+=\Mmo(\Cal S)^+$ such that
$|\pmb x|\le\pmb u^*_<$.   Also as in 646M, using 646L and
645L, we have a $\pmb u\in\Mmo$ such that $|\pmb u|\le\pmb u^*$ and
$\theta(z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)\le\epsilon$
for every $\tau\in\Cal S$.
Since $\Cal S$ is finitely full (611Ob),
$\theta(\sup|\Siiv(\pmb x)-\iivu|)\le 2\sqrt{\epsilon}$ (615Db);
while $\iivu$ is near-simple, by 631I.   As $\epsilon$ is arbitrary,
$\Siiv(\pmb x)$ is fully adapted (by 613Bl) and near-simple.

\medskip

{\bf (b)} To see that $\Siiv(\pmb x)$ is an integrator, we can use
617E, as follows.   Consider $C=\{\int\pmb u\,d\pmb v:\pmb u\in\Mmo$,
$|\pmb u|\le\pmb u^*\}$.   Then $C$ is topologically bounded in $L^0$.
\Prf\ Set $\bar u=\sup|\pmb u^*|$.   Let $\epsilon>0$.   Then there is
an $M>0$ such that $\bar\mu\Bvalue{\bar u\ge M}\le\epsilon$.   Let
$\delta>0$ be such that $\theta(M\delta z)\le\epsilon$ for every
$z\in Q_{\Cal S}(d\pmb v)$.   If $\pmb u\in\Mmo$,
$I\in\Cal I(\Cal S)$ and
$\|\pmb u\|_{\infty}\le M$, then
$\Bover1MS_I(\pmb u,\,d\pmb v)\in Q_{\Cal S}(d\pmb v)$
so $\theta(\delta S_I(\pmb u,d\pmb v))\le\epsilon$;  taking the limit
as $I\uparrow\Cal I(\Cal S)$,
$\theta(\delta\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$.
Generally, if $\pmb u\in\Mmo$ and
$|\pmb u|\le\pmb u^*$, set
$\pmb u'=\med(-M\tbf 1^{(\Cal S)},\pmb u,M\tbf 1^{(\Cal S)})$;
then $\theta(\delta\int_{\Cal S}\pmb u'\,d\pmb v)\le\epsilon$
and
$\Bvalue{\int_{\Cal S}\pmb u\,d\pmb v\ne\int_{\Cal S}\pmb u'\,d\pmb v}
\Bsubseteq\Bvalue{\pmb u\ne\pmb u'}$ has measure at most $\epsilon$,
so $\theta(\delta\int_{\Cal S}\pmb u\,d\pmb v)\le 2\epsilon$.
Thus $\theta(\delta z)\le 2\epsilon$ for every $z\in C$.   As $\epsilon$ is
arbitrary, $C$ is topologically bounded.\ \Qed

\medskip

{\bf (c)} Now $Q_{\Cal S}(d\Siiv(\pmb x))$ is included in the topological
closure $\overline{C}$ of $C$.
\Prf\ If $z\in Q_{\Cal S}(d\Siiv(\pmb x))$,
there are an $I\in\Cal I(\Cal S)$ and
a $\pmb u\in\Mfa(I)$
such that $\|\pmb u\|_{\infty}\le 1$ and $z=S_I(\pmb u,d\Siiv(\pmb x))$.
Take $\epsilon>0$.   Then there is a $\pmb u'\in\Mmo$ such that
$|\pmb u'|\le\pmb u^*$ and
$\theta(z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v)
\le\Bover{\epsilon}{1+\#(I)}$ for every $\tau\in\Cal S$, as in (a) above,
that is, such that
$\theta(z_{\tau}-z'_{\tau})\le\Bover1{1+\#(I)}\epsilon$ for every
$\tau$, where $\family{\tau}{\Cal S}{z'_{\tau}}=\pmb z'$
is the indefinite
integral $\iiv(\pmb u')$.   In this case,
$\theta(S_I(\pmb u,d\Siiv(\pmb x))-S_I(\pmb u,d\pmb z'))\le\epsilon$.
By 627Ha, there is a $\pmb w\in\Msimp(\Cal S)$ such that
$\|\pmb w\|_{\infty}\le 1$ and
$S_I(\pmb u,d\pmb z')=\int_{\Cal S}\pmb w\,d\pmb z'$.
Now we know from 617E that $\int_{\Cal S}\pmb w\,d\pmb z'
=\int_{\Cal S}\pmb w\times\pmb u'\,d\pmb v$, which belongs to
$C$ because $|\pmb w\times\pmb u'|\le|\pmb u'|\le\pmb u^*$.
And $\theta(z-\int_{\Cal S}\pmb w\,d\pmb z')\le\epsilon$.   As
$z$ and $\epsilon$ are arbitrary,
$Q_{\Cal S}(d\Siiv(\pmb x))\subseteq\overline{C}$.\ \Qed

As the closure of a topologically bounded set in $L^0$ is topologically
bounded (613B(f-iii)),
$Q_{\Cal S}(d\Siiv(\pmb x))$ is topologically bounded and
$\Siiv(\pmb x)$ is an integrator.
}%end of proof of 646N

\leader{646O}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex
sublattice of $\Cal T$.   Let $\pmb v$ be a
near-simple integrator with domain $\Cal S$, and
$A\subseteq\cmmnt{\MSi^0=\mskip5mu}\MSi^0(\Cal S)$ a
uniformly previsibly order-bounded set.   Then $\Siiv\restr A$ is
uniformly continuous with respect to the $S$-integration uniformity on $A$ and
and the ucp uniformity on $\Mob(\Cal S)$.

\proof{{\bf (a)} It is enough to consider the case in which
$A=\{\pmb x:\pmb x\in\MSi^0$, $|\pmb x|\le\pmb u^*_<\}$ where
$\pmb u^*\in\Mmo(\Cal S)^+$.
By 646N, we know that $\Siiv(\pmb x)\in\Mns(\Cal S)$ for every
$\pmb x\in A$.

\medskip

{\bf (b)} Take $\epsilon>0$.   Then there is a $\TSi$-neighbourhood
$G$ of $0$ in $\MSi^0$ such that
$\theta(\Sint_{\Cal S}\pmb x\,d\pmb v
-\Sint_{\Cal S}\pmb y\,d\pmb v)\le\epsilon^2$ whenever
$\pmb x$, $\pmb y\in A$ and $\pmb x-\pmb y\in G$ (645S);
and we may suppose that $G$ is of the form $\{\pmb x:\pmb x\in\MSi^0$,
$\widehat{\theta}^{\#}_{\pmb w}(\pmb x)\le\delta\}$ where
$\delta>0$ and $\pmb w\in\Mnsu(\Cal S)$.
Fix $\tau\in\Cal S$ for the moment.   Defining
$R_{\tau}:(L^0)^{\Cal S}\to (L^0)^{\Cal S}$ as in the proof of 646L,
$R_{\tau}(\pmb x)\in\MSi^0$
and $|R_{\tau}(\pmb x)|\le|\pmb x|$ for $\pmb x\in\MSi^0$, so
$R_{\tau}(\pmb x)\in A$ for every $\pmb x\in A$;  also
$R_{\tau}(\pmb x)\in G$ for every $\pmb x\in G$.
Consequently
$\theta(\Sint_{\Cal S}R_{\tau}(\pmb x)d\pmb v
-\Sint_{\Cal S}R_{\tau}(\pmb y)d\pmb v)\le\epsilon^2$ whenever
$\pmb x$, $\pmb y\in A$, $\pmb x-\pmb y\in G$ and $\tau\in\Cal S$.
But we also have $\Sint_{\Cal S}R_{\tau}(\pmb x)d\pmb v
=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$ for
$\pmb x\in\MSi^0$ and
$\tau\in\Cal S$, as observed in part (a) of the proof of 646L.
So we get

\Centerline{$\theta(\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v
-\Sint_{\Cal S\wedge\tau}\pmb y\,d\pmb v)\le\epsilon^2$}

\noindent whenever $\pmb x$, $\pmb y\in A$, $\pmb x-\pmb y\in G$ and
$\tau\in\Cal S$.

\medskip

{\bf (c)} Now we know that $\Siiv(\pmb x)$ is near-simple, therefore
order-bounded, for every $\pmb x\in\MSi^0$;  while $\Cal S$ is
order-convex, therefore finitely full.   So we can apply
615Db again to see that
$\theta(\sup|\Siiv(\pmb x)-\Siiv(\pmb y)|)\le 2\epsilon$
whenever $\pmb x$, $\pmb y\in A$ and $\pmb x-\pmb y\in G$.
As $\epsilon$ is arbitrary, $\Siiv\restr A$ is uniformly continuous.
}%end of proof of 646O

\leader{646P}{Lemma} Suppose that
$\family{t}{T}{\frak A_t}$ is right-continuous and that
$\Cal S$ is an order-convex sublattice of $\Cal T$
with a least element.   Let $\pmb v=\sfamilysS{v}$ be a martingale.
Then $\Cal S_{\pmb v}=\{\tau:\tau\in\Cal S$,
$Q_{\Cal S\wedge\tau}(d\pmb v)$ is uniformly integrable$\}$
is a covering ideal of $\Cal S$.

\proof{{\bf (a)} Note first that if $\tau\in\Cal S_{\pmb v}$ and
$\sigma\in\Cal S\wedge\tau$, then
$Q_{\Cal S\wedge\sigma}(d\pmb v)\subseteq Q_{\Cal S\wedge\tau}(d\pmb v)$
is uniformly integrable, so $\sigma\in\Cal S_{\pmb v}$.
If $\tau$, $\tau'\in\Cal S_{\pmb v}$, then
$Q_{\Cal S\wedge(\tau\vee\tau')}(d\pmb v)
\subseteq Q_{\Cal S\wedge\tau}(d\pmb v)
+Q_{\Cal S\wedge\tau'}(d\pmb v)$.   \Prf\
If $z\in Q_{\Cal S\wedge(\tau\vee\tau')}(d\pmb v)$
there is a simple process
$\pmb u=\family{\sigma}{\Cal S\wedge(\tau\vee\tau')}{u_{\sigma}}$ such that
$\|\pmb u\|_{\infty}\le 1$ and
$z=\int_{\Cal S\wedge(\tau\vee\tau')}\pmb u\,d\pmb v$ (627Ha again).
By 612K(d-i), there is a breakpoint string for $\pmb u$ of the form
$(\tau_0,\ldots,\tau_m,\ldots,\tau_n)$ where
$\tau_0=\min\Cal S$, $\tau_m=\tau$ and $\tau_n=\tau\vee\tau'$.
Now

$$\eqalignno{z
&=\int_{\Cal S\wedge(\tau\vee\tau')}\pmb u\,d\pmb v
=\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})\cr
\displaycause{614C, because $\tau_0=\min\Cal S$ and
$\tau_n=\tau\vee\tau'$}
&=\sum_{i=0}^{m-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})
  +\sum_{i=m}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})
\in Q_{\Cal S\wedge\tau}(d\pmb v)+Q_{\Cal S\wedge\tau'}(d\pmb v).
\text{ \Qed}\cr}$$

\noindent So $Q_{\Cal S\wedge(\tau\vee\tau')}(d\pmb v)$ is included in the
sum of two uniformly integrable sets and is uniformly integrable
(621B(c-i)).   As $\tau$ and $\tau'$ are arbitrary,
$\Cal S_{\pmb v}$ is closed under $\vee$ and is an ideal in $\Cal S$.

\medskip

{\bf (b)} Suppose that $\Cal S$ has a greatest
element and $\pmb v=\sfamilysS{v}$ is an $L^2$-martingale.   Then
$\|v_{\sigma}\|_2\le\|v_{\max\Cal S}\|_2$ for every $\sigma\in\Cal S$ and
$\pmb v$ is $L^2$-bounded.
Let $\pmb v^*=\sfamilysS{v^*}$ be its quadratic variation.
If $\pmb u$ is a simple process with domain $\Cal S$ and
$\|\pmb u\|_{\infty}\le 1$, then

\Centerline{$\|\int_{\Cal S}\pmb u\,d\pmb v\|_2^2
\le\|\int_{\Cal S}\pmb u^2\,d\pmb v^*\|_1
\le\|\int_{\Cal S}d\pmb v^*\|_1
=\|v^*_{\max\Cal S}\|_1<\infty$}

\noindent by 624I and 624G.   So

\Centerline{$Q_{\Cal S}(d\pmb v)
\subseteq\{\int_{\Cal S}\pmb u\,d\pmb v:\pmb u\in\Msimp(\Cal S),
  \,\|\pmb u\|_{\infty}\le 1\}$}

\noindent (627Ha once more) is a $\|\,\|_2$-bounded subset of
$L^2$ and is uniformly integrable (621Be).

\medskip

{\bf (c)} Next, suppose that $\Cal S$ has a greatest element
and that $\pmb v$ is such that $v_{\min\Cal S}=0$ and
$\Osclln(\pmb v)\le\chi 1$.   As $\pmb v$ is locally near-simple (632Ia),
therefore near-simple,
there is a non-decreasing sequence $\sequencen{\tau_n}$ in
$\Cal S$ such that $\tau_0=\min\Cal S$,
$\inf_{n\in\Bbb N}\Bvalue{\tau_n<\max\Cal S}=0$ and
$\Bvalue{|v_{\sigma}-v_{\tau_n}|\ge 1}\subseteq\Bvalue{\sigma=\tau_{n+1}}$
whenever $n\in\Bbb N$ and $\sigma\in[\tau_n,\tau_{n+1}]$ (631Ra).   Now
$|v_{\tau}-v_{\tau_n}|\le 2\chi 1$ whenever
$n\in\Bbb N$ and $\tau_n\le\tau\le\tau_{n+1}$.
\Prf\ Applying the method of 641E to
$\pmb v'=\pmb v\restr[\tau_n,\tau_{n+1}]$, we
see that if $\tau\in[\tau_n,\tau_{n+1}]$ then
$|v_{I<\tau}-v_{\tau_n}|\times\chi\Bvalue{\tau_n<\tau}\le\chi 1$ for every
finite sublattice $I$ of $[\tau_n,\tau]$ containing $\tau_n$,
so that $|v_{<\tau}-v_{\tau_n}|\times\chi\Bvalue{\tau_n<\tau}\le\chi 1$;
while
$|v_{\tau}-v_{<\tau}|\times\chi\Bvalue{\tau_n<\tau}\le\Osclln(\pmb v')$,
by 641Na. So

\Centerline{$|v_{\tau}-v_{\tau_n}|\times\chi\Bvalue{\tau_n<\tau}
\le\chi 1+\Osclln(\pmb v')\le\chi 1+\Osclln(\pmb v)
\le 2\chi 1$}

\noindent (using 618D(b-ii)), whenever
$\tau_n\le\tau\le\tau_{n+1}$.   It follows at once that
$|v_{\tau}-v_{\tau_n}|\le 2\chi 1$ for every $\tau\in[\tau_n,\tau_{n+1}]$.\
\Qed

Consequently $|v_{\tau}|\le 2n\chi 1$ whenever $n\in\Bbb N$ and
$\min\Cal S\le\tau\le\tau_n$, and
$\pmb v\restr[\min\Cal S,\tau_n]$ is an
$L^2$-martingale for every $n$.   By (b), $\tau_n\in\Cal S_{\pmb v}$ for
every $n$;  since

\Centerline{$\sup_{n\in\Bbb N}\Bvalue{\tau\le\tau_n}
\Bsupseteq\sup_{n\in\Bbb N}\Bvalue{\max\Cal S=\tau_n}=1$}

\noindent for every $\tau\in\Cal S$, $\Cal S_{\pmb v}$ is a
covering ideal of $\Cal S$.

\medskip

{\bf (d)} If $\Cal S$ has a greatest element
and $\pmb v$ is such that $\Osclln(\pmb v)\le\chi 1$, but we are told
nothing about $v_{\min\Cal S}$, then we can apply (c) to
$\pmb v'=\family{\sigma}{\Cal S}{v_{\sigma}-v_{\min\Cal S}}$
to see that
$\Cal S_{\pmb v}=\Cal S_{\pmb v'}$ is a covering ideal of $\Cal S$
because $Q_{\Cal S\wedge\tau}(d\pmb v)=Q_{\Cal S\wedge\tau}(d\pmb v')$ for
every $\tau$.

\medskip

{\bf (e)} Now suppose that $\pmb v$ is of bounded variation and
$\bar v=\int_{\Cal S}|d\pmb v|$ has finite expectation.
If $\pmb u\in\Msimp(\Cal S)$ and $\|\pmb u\|_{\infty}\le 1$, then
$|S_I(\pmb u,d\pmb v)|\le S_I(\tbf 1^{(\Cal S)},|d\pmb v|)\le\bar v$.   So
$|z|\le\bar v$ for every $z\in Q_{\Cal S}(d\pmb v)$ and
$Q_{\Cal S}(d\pmb v)$ is uniformly integrable.

\medskip

{\bf (f)} Suppose that $\Cal S$ has a greatest element and
$\pmb v$ is of bounded variation, but we do not know
whether $\bar v=\int_{\Cal S}|d\pmb v|$ has finite expectation.
Set $z_{\tau}=\int_{\Cal S\wedge\tau}|d\pmb v|$ for $\tau\in\Cal S$.
Because $\pmb v$ is a martingale, it is near-simple (632Ia again),
and $\pmb z=\family{\tau}{\Cal S}{z_{\tau}}$ is near-simple
(631K).   By 631Ra again, there is a non-decreasing sequence
$\sequencen{\tau_n}$ in $\Cal S$ such that $\tau_0=\min\Cal S$,
$\inf_{n\in\Bbb N}\Bvalue{\tau_n<\max\Cal S}=0$ and
$\Bvalue{|z_{\sigma}-z_{\tau_n}|\ge 1}\subseteq\Bvalue{\sigma=\tau_{n+1}}$
whenever $n\in\Bbb N$ and $\sigma\in[\tau_n,\tau_{n+1}]$.
In this case,

\Centerline{$z_{\tau_{n+1}}-z_{\tau_n}
\le 2\chi 1+|v_{\tau_{n+1}}-v_{\tau_n}|$}

\noindent for every
$n$.   \Prf\ Suppose that
$\tau_n=\sigma_0\le\ldots\le\sigma_k=\tau_{n+1}$.   For $j\le k$,
set

\Centerline{$a_j=\Bvalue{\sum_{i=0}^{j-1}|v_{\sigma_{i+1}}-v_{\sigma_i}|
  >2\chi 1+|v_{\tau_{n+1}}-v_{\tau_n}|}$,
\quad$b_j=a_{j+1}\Bsetminus a_j$ if $j<k$.}

\noindent Then, for any $j<k$,

$$\eqalign{b_j
&\Bsubseteq\Bvalue{v_{\sigma_{j+1}}\ne v_{\sigma_j}}
\Bsubseteq\Bvalue{\sigma_j<\tau_{n+1}}
\cr&\Bsubseteq\Bvalue{z_{\sigma_j}-z_{\tau_n}<\chi 1}
\Bsubseteq\Bvalue{|v_{\sigma_j}-v_{\tau_n}|<\chi 1}
   \Bcap\Bvalue{\sum_{i=0}^{j-1}|v_{\sigma_{i+1}}-v_{\sigma_i}|
      <\chi 1};\cr}$$

\noindent since also $b_j\Bsubseteq a_{j+1}$,

\Centerline{$b_j\Bsubseteq\Bvalue{|v_{\sigma_{j+1}}-v_{\sigma_j}|
  >\chi 1+|v_{\tau_{n+1}}-v_{\tau_n}|}$.}

\noindent Now observe that

\Centerline{$\Bvalue{\sigma_{j+1}<\tau_{n+1}}
\Bsubseteq\Bvalue{z_{\sigma_{j+1}}-z_{\tau_n}<\chi 1}
\Bsubseteq\Bvalue{\sum_{i=0}^j|v_{\sigma_{i+1}}-v_{\sigma_i}|<\chi 1}$}

\noindent is disjoint from $a_{j+1}$ and therefore from $b_j$, so

\Centerline{$b_j\Bsubseteq\Bvalue{\sigma_{j+1}=\tau_{n+1}}
\Bsubseteq\Bvalue{|v_{\sigma_{j+1}}-v_{\sigma_j}|
   =|v_{\tau_{n+1}}-v_{\sigma_j}|}$.}

\noindent Putting these together,

$$\eqalign{b_j
&\Bsubseteq\Bvalue{|v_{\sigma_j}-v_{\tau_n}|<\chi 1}
   \Bcap\Bvalue{|v_{\sigma_{j+1}}-v_{\sigma_j}|
                  >\chi 1+|v_{\tau_{n+1}}-v_{\tau}|}
\cr&\mskip100mu
   \Bcap\Bvalue{|v_{\sigma_{j+1}}-v_{\sigma_j}|
                  =|v_{\tau_{n+1}}-v_{\sigma_j}|}
\cr&\Bsubseteq\Bvalue{|v_{\sigma_j}-v_{\tau_n}|<\chi 1}
   \Bcap\Bvalue{|v_{\tau_{n+1}}-v_{\sigma_j}|
                  >\chi 1+|v_{\tau_{n+1}}-v_{\tau_n}|}
=0.\cr}$$

This is true for every $j<k$, so $a_k=0$ and 		
$\sum_{i=0}^{j-1}|v_{\sigma_{i+1}}-v_{\sigma_i}|
  \le 2\chi 1+|v_{\tau_{n+1}}-v_{\tau_n}|$.   As $\sigma_0,\ldots,\sigma_k$
are arbitrary,
$z_{\tau_{n+1}}-z_{\tau_n}\le 2\chi 1+|v_{\tau_{n+1}}-v_{\tau_n}|$.\ \Qed

Since $\pmb v$, being a martingale, is an $L^1$-process, we see that
$z_{\tau_n}\in L^1_{\bar\mu}$ for every $n$.   Consequently (e) tells us that
$Q_{\Cal S\wedge\tau_n}(d\pmb v)$ is uniformly integrable
and $\tau_n\in\Cal S_{\pmb v}$ for every $n$.   As
$\sup_{n\in\Bbb N}\Bvalue{\tau_n=\max\Cal S}=1$, $\Cal S_{\pmb v}$ is a
covering ideal of $\Cal S$.

\medskip

{\bf (g)} Next, suppose only that $\Cal S$ has a greatest member.
By the fundamental theorem of martingales (643L), there is a local martingale
$\hatpmb v$ with domain $\Cal S$ such that
$\Osclln(\hatpmb v)\le\chi 1$ and $\tildepmb v=\pmb v-\hatpmb v$ has
bounded variation.   Now, for any $\tau\in\Cal S$,

\Centerline{$Q_{\Cal S\wedge\tau}(d\pmb v)
\subseteq Q_{\Cal S\wedge\tau}(d\hatpmb v)
   +Q_{\Cal S\wedge\tau}(d\tildepmb v)$}

\noindent (616Dc), so
$\Cal S_{\pmb v}\supseteq\Cal S_{\hatpmb v}\cap\Cal S_{\tildepmb v}$
includes the intersection of two covering ideals and (being itself an
ideal) is a covering ideal of $\Cal S$ (611Nc).

\medskip

{\bf (h)} Finally, if we suppose only that $\Cal S$ is order-convex and has
a least member, we know from (g) that $\Cal S_{\pmb v}\cap(\Cal S\wedge\tau)$
covers $\Cal S\wedge\tau$ for every $\tau\in\Cal S$, so
$\Cal S_{\pmb v}$ covers $\Cal S$, and we're done.
}%end of proof of 646P

\leader{646Q}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex
sublattice of $\Cal T$ with a least element.   Let $\pmb v$ be a
near-simple integrator and $\pmb x$ an S-integrable process, both
with domain $\Cal S$.

(a) If $\pmb v$ is a martingale, then $\Siiv(\pmb x)$ is a local
martingale.

(b) If $\pmb v$ is jump-free, then $\Siiv(\pmb x)$ is jump-free.

(c) If $\pmb v$ is of bounded variation, then $\Siiv(\pmb x)$ is of bounded
variation.

\proof{ Since $\Siiv(\pmb x)=\Siiv(\pmb x\times\tbfoneminus{\Cal S})$
(646Kb), and
$\pmb x\times\tbfoneminus{\Cal S}$ belongs to $\MSi^0(\Cal S)$, we can
suppose throughout that $\pmb x\in\MSi^0(\Cal S)$.    Set
$z_{\tau}=\Sint_{\Cal S\wedge\tau}\pmb x\,d\pmb v$ for $\tau\in\Cal S$.

\medskip

{\bf (a)(i)} Write

\Centerline{$\Cal S_{\pmb v}
=\{\tau:\tau\in\Cal S$, $Q_{\Cal S\wedge\tau}(d\pmb v)$ is uniformly
integrable$\}$,}

\Centerline{$\Cal S'
=\{\tau:\tau\in\Cal S$,
$\|\pmb x\restr\Cal S\wedge\tau\|_{\infty}<\infty\}$.}

\noindent Then $\Cal S'$ is an ideal of $\Cal S$, because
$(\Cal S\wedge\tau)\cup(\Cal S\wedge\tau')$ covers
$\Cal S\wedge(\tau\vee\tau')$ so
$\sup|\pmb x\restr\Cal S\wedge(\tau\vee\tau')|
=\sup|\pmb x\restr\Cal S\wedge\tau|
\vee\sup|\pmb x\restr\Cal S\wedge\tau'|$ for all $\tau$,
$\tau'\in\Cal S$.   As in part (a) of the proof of 646P,
$\Cal S_{\pmb v}$ is an ideal in $\Cal S$,
so $\Cal S_{\pmb v}\cap\Cal S'$ also is.

\medskip

\quad{\bf (ii)} $\Siiv(\pmb x)\restr\Cal S_{\pmb v}\cap\Cal S'$ is a martingale.
\Prf\ Take $\sigma\le\tau$ in $\Cal S_{\pmb v}\cap\Cal S'$ and work in
$\Cal S\wedge\tau$.   There is an
$M\ge 0$ such that $|\pmb x\restr\Cal S\wedge\tau|\le M\tbf 1$;  now
$|\pmb x\restr\Cal S\wedge\tau|\le M\tbfoneminus{\Cal S\wedge\tau}$.
By 645La once more, $\pmb x\restr\Cal S\wedge\tau$ belongs to the
$\frak T_{\text{S-i}}$-closure of $\{\pmb u_<:\pmb u\in A\}$ where
$A=\{\pmb u:\pmb u\in\Mmo(\Cal S\wedge\tau)$, $|\pmb u|\le M\tbf 1\}$;
by 646L there is for every $\delta>0$ a $\pmb u\in A$
such that
$\theta(z_{\sigma'}-\int_{\Cal S\wedge\sigma'}\pmb u\,d\pmb v)\le\delta$
for every $\sigma'\in\Cal S\wedge\tau$.
Set $C=MQ_{\Cal S\wedge\tau}(d\pmb v)$, so that $C$ is uniformly
integrable.   If
$\pmb u\in\Mmo(\Cal S\wedge\tau)$ and $\|\pmb u\|_{\infty}\le M$, then
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\pmb u,d\pmb v)$
belongs to the closure
$\overline{C}$ of $C$ for the topology of convergence in measure.
But now we see that $z_{\tau}$ also belongs to $\overline{C}$;  and the
same applies to $z_{\sigma}$, since of course
$C\supseteq MQ_{\Cal S\wedge\sigma}(d\pmb v)$.

Because $C$ is uniformly integrable, $\overline{C}\subseteq L^1_{\bar\mu}$
and the norm topology of $L^1_{\bar\mu}$
agrees with the topology of convergence in
measure on $C$ (621B(c-ii)).   Given $\epsilon>0$, there is a $\delta>0$
such that

\Centerline{$\|z_{\tau}-u\|_1\le\epsilon$ whenever $u\in\overline{C}$ and
$\theta(z_{\tau}-u)\le\delta$,}

\Centerline{$\|z_{\sigma}-u\|_1\le\epsilon$ whenever $u\in\overline{C}$ and
$\theta(z_{\sigma}-u)\le\delta$.}

\noindent We therefore have a $\pmb u\in\Mmo(\Cal S\wedge\tau)$ such that

\Centerline{$\|z_{\tau}-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v\|_1
   \le\epsilon$,
\quad$\|z_{\sigma}-\int_{\Cal S\wedge\sigma}\pmb u\,d\pmb v\|_1
   \le\epsilon$.}

At this point, we repeat the trick in a different way.   As
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\pmb u,d\pmb v)$
for the topology of convergence in measure, and
$S_I(\pmb u,d\pmb v)\in C$ for every $I\in\Cal I(\Cal S\wedge\tau)$,
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
=\1lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\pmb u,d\pmb v)$, the limit
for the norm topology on $L^1_{\bar\mu}$.   There is therefore an
$I_0\in\Cal I(\Cal S\wedge\tau)$ such that

\Centerline{$\|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
  -S_I(\pmb u,d\pmb v)\|_1
\le\epsilon$}

\noindent whenever $I_0\subseteq I\in\Cal I(\Cal S\wedge\tau)$.
Similarly, there is a $J_0\in\Cal I(\Cal S\wedge\sigma)$ such that

\Centerline{$\|\int_{\Cal S\wedge\sigma}\pmb u\,d\pmb v
  -S_J(\pmb u,d\pmb v)\|_1
\le\epsilon$}

\noindent whenever $J_0\subseteq J\in\Cal I(\Cal S\wedge\sigma)$.
Let $K$ be the sublattice of $\Cal S\wedge\tau$ generated by
$I_0\cup J_0\cup\{\tau\}$.   Then $K\wedge\sigma\supseteq J_0$ so

\Centerline{$\|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
  -S_K(\pmb u,d\pmb v)\|_1
\le\epsilon$,
\quad$\|\int_{\Cal S\wedge\sigma}\pmb u\,d\pmb v
  -S_{K\wedge\sigma}(\pmb u,d\pmb v)\|_1
\le\epsilon$}

\noindent and

\Centerline{$\|z_{\tau}-S_K(\pmb u,d\pmb v)\|_1\le 2\epsilon$,
\quad$\|z_{\sigma}-S_{K\wedge\sigma}(\pmb u,d\pmb v)\|_1\le 2\epsilon$.}

\noindent Now

$$\eqalign{P_{\sigma}(u_{\sigma'}\times(v_{\tau'}-v_{\sigma'}))&
=P_{\sigma}P_{\sigma'}(u_{\sigma'}\times(v_{\tau'}-v_{\sigma'}))
=P_{\sigma}(u_{\sigma'}\times(P_{\sigma'}v_{\tau'}-P_{\sigma'}v_{\sigma'}))\cr&
=P_{\sigma}(u_{\sigma'}\times(v_{\sigma'}-v_{\sigma'}))
=0\cr}$$

\noindent whenever $\sigma\le\sigma'\le\tau'\le\tau$, so
$P_{\sigma}S_{K\vee\sigma}(\pmb u,d\pmb v)=0$ and

\Centerline{$P_{\sigma}S_K(\pmb u,d\pmb v)
=P_{\sigma}S_{K\wedge\sigma}(\pmb u,d\pmb v)
=S_{K\wedge\sigma}(\pmb u,d\pmb v)$}

\noindent (using 613G(a-ii)).   Since $P_{\sigma}$ is $\|\,\|_1$-reducing,

\Centerline{$\|P_{\sigma}z_{\tau}-z_{\sigma}\|_1\le 4\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $P_{\sigma}z_{\tau}=z_{\sigma}$;
as $\sigma$ and $\tau$ are arbitrary, $\Siiv(\pmb x)\restr\Cal S'$ is a
martingale.\ \Qed

\medskip

\quad{\bf (iii)} $\Cal S'$ is a covering ideal of $\Cal S$.
\Prf\ There is a non-negative moderately oscillatory process
$\pmb u^*=\sfamilysS{u^*}$ such that $|\pmb x|\le\pmb u^*_<$, and by
642N we can suppose that $\pmb u^*$ is near-simple.
Take $\tau\in\Cal S$ and $\epsilon>0$.   There is an $M>0$ such that
$\bar\mu\Bvalue{\sup|\pmb u^*|\ge M}\le\epsilon$.
Because $\pmb u^*$ is
near-simple, there is a $\tau'\in\Cal S\wedge\tau$ such that
$\Bvalue{u^*_{\sigma}\ge M}\Bsubseteq\Bvalue{\sigma=\tau'}$ for
every $\sigma\in\Cal S\wedge\tau'$ and
$\Bvalue{\tau'<\tau}\Bsubseteq\Bvalue{u^*_{\tau}\ge M}$ (631Ra once more,
applied in $\Cal S\wedge\tau$ with $\delta=M$).
So $\bar\mu\Bvalue{\tau'<\tau}\le\epsilon$.   Next, for
$\sigma\in\Cal S\wedge\tau'$, $u^*_{I<\sigma}\le M\chi 1$ for every
$I\in\Cal I(\Cal S\wedge\sigma)$, because
$\Bvalue{\sigma'<\sigma}\Bsubseteq\Bvalue{\sigma'<\tau'}
\Bsubseteq\Bvalue{u^*_{\sigma'}\le M}$ for every
$\sigma'\in\Cal S\wedge\sigma$.   So $u^*_{<\sigma}\le M\chi 1$ for
$\sigma\le\tau'$, and

\Centerline{$\sup|\pmb x\restr\Cal S\wedge\tau'|
\le\sup|\pmb u^*_<\restr\Cal S\wedge\tau'|
\le M\chi 1$.}

\noindent Thus $\tau'\in\Cal S'$;  as $\epsilon$ is arbitrary,
$\Cal S'$ covers $\{\tau\}$;  as $\tau$ is arbitrary,
$\Cal S'$ covers $\Cal S$.\ \Qed

\medskip

\quad{\bf (iv)} By 646P, $\Cal S_{\pmb v}$ is a covering ideal of
$\Cal S$.   So $\Cal S_{\pmb v}\cap\Cal S'$ covers $\Cal S$ (611Nc again).
Since $\Siiv(\pmb x)\restr\Cal S_{\pmb v}\cap\Cal S'$ is a martingale,
$\Siiv(\pmb x)$ is a local martingale.

\medskip

{\bf (b)} Let $\pmb u^*\in\Mmo=\Mmo(\Cal S)^+$ be such that
$|\pmb x|\le\pmb u^*_<$ and set $A=\{\pmb y:\pmb y\in\MSi^0(\Cal S)$,
$|\pmb y|\le\pmb u^*_<\}$, $B=\{\pmb u:\pmb u\in\Mmo$,
$|\pmb u|\le\pmb u^*\}$.   Then
$\pmb x\in\overline{\{\pmb u_<:\pmb u\in B\}}$,
taking the closure for the S-integration topology
(645La yet again).   By 646O, $\Siiv(\pmb x)$ belongs to the closure of
$\{\Siiv(\pmb u_<):\pmb u\in B\}$ for the ucp topology on $\Mob(\Cal S)$.
But observe that if $\pmb u\in\Mmo$ then

\Centerline{$\Sint_{\Cal S\wedge\tau}(\pmb u_<\restr\Cal S\wedge\tau)\,
  d\pmb v
=\Sint_{\Cal S\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)_<\,
  d\pmb v
=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v$}

\noindent so $\Siiv(\pmb u_<)=\iivu$.   And we know that $\iivu$ is
jump-free for every $\pmb u\in\Mmo$, by 618Q.   So
$\Siiv(\pmb x)$ belongs to the closure of $\Mjf(\Cal S)$ in $\Mob(\Cal S)$
and is jump-free by 618Ga.

\medskip

{\bf (c)} Take $\pmb u^*$, $A$ and $B$ as in (b) just above, and set
$\bar u=\sup|\pmb u^*|\times\int_{\Cal S}|d\pmb v|$.
Then $\iivu$ is of bounded variation, with

\Centerline{$\int_{\Cal S}|d(\iivu)|
\le\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|
\le\bar u$}

\noindent for every $\pmb u\in B$ (614U).   As before,

\Centerline{$\Siiv(\pmb x)\in\overline{\{\Siiv(\pmb u_<):\pmb u\in B\}}
=\overline{\{\iivu:\pmb u\in B\}}$}

\noindent where the closures are taken in the ucp topology.   But this
means that $\Siiv(\pmb x)$ belongs to the closure of
$\{\pmb u:\pmb u\in\Mbv(\Cal S)$, $\int_{\Cal S}|d\pmb u|\le\bar u\}$ for
the product topology on $(L^0)^{\Cal S}$, and $\Siiv(\pmb x)$ is of bounded
variation, by 614N.
}%end of proof of 646Q

\leader{646R}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex
sublattice of $\Cal T$.   Let $\pmb v$ be a
near-simple integrator and $\pmb x$, $\pmb x'$ two S-integrable processes,
all with domain $\Cal S$;  write $\pmb z$ for the indefinite S-integral
$\Siiv(\pmb x)$.   Then
$\Sint_{\Cal S}\pmb x'\,d\pmb z
=\Sint_{\Cal S}\pmb x'\times\pmb x\,d\pmb v$.

\proof{{\bf (a)} Of course it will be enough to deal with the case of non-empty
$\Cal S$.   We know from 646N that $\pmb z$ is a near-simple integrator, and
from 645Kb that $\pmb x\times\pmb x'$ is S-integrable, so
the integrals here are well-defined.   Since

\Centerline{$\pmb z=\Siiv(\pmb x)=\Siiv(\pmb x\times\tbfoneminus{\Cal S})$,
\quad$\Sint_{\Cal S}\pmb x'\,d\pmb z
=\Sint_{\Cal S}\pmb x'\times\tbfoneminus{\Cal S}\,d\pmb z$}

\noindent and

\Centerline{$\Sint_{\Cal S}\pmb x'\times\pmb x\,d\pmb v
=\Sint_{\Cal S}\pmb x'\times\pmb x\times\tbfoneminus{\Cal S}\,d\pmb v
=\Sint_{\Cal S}(\pmb x'\times\tbfoneminus{\Cal S})
\times(\pmb x\times\tbfoneminus{\Cal S})d\pmb v$}

\noindent (646Kb again),
we can suppose throughout that $\pmb x$ and $\pmb x'$ belong to
$\MSi^0=\MSi^0(\Cal S)$.   There is a $\pmb u^*\in\Mmo^+=\Mmo(\Cal S)^+$
such that
$|\pmb x|\vee|\pmb x'|\le\pmb u^*_<$;  by 642N and 645Lb, we can suppose that
$\pmb u^*\in\Mns(\Cal S)^+$ and that $\pmb x'$ belongs to the $\TSi$-closure
of $\{\pmb u_<:\pmb u\in\Msimp(\Cal S)$, $|\pmb u|\le\pmb u^*\}$.

\medskip

{\bf (b)} Take any simple process $\pmb u=\sfamilysS{u}$ and $\epsilon>0$.
Write $\bar u$ for $\sup|\pmb u|$.

\medskip

\quad{\bf (i)} There is a $\TSi$-neighbourhood
$G_0$ of $\pmb x$ such that whenever $\checkpmb u\in\Mmo$ is such that
$|\checkpmb u|\le\pmb u^*$ and $\checkpmb u_<\in G_0$ then
$\theta(\int_{\Cal S}\pmb u\,d\pmb z
  -\int_{\Cal S}\pmb u\,d\checkpmb z)\le\epsilon$,
where $\checkpmb z=\iiv(\checkpmb u)$.   \Prf\
Let $(\tau_0,\ldots,\tau_n)$ be a breakpoint string for $\pmb u$.
Take $\delta>0$ such that
$\theta(\bar u\times z)\le\Bover{\epsilon}{2n+2}$ whenever
$\theta(z)\le\delta$.
By 646L, there is a $\TSi$-neighbourhood $G_0$ of $\pmb x$ such that
whenever
$\checkpmb u\in\Mmo$ is such that $|\checkpmb u|\le\pmb u^*$ and
$\checkpmb u_<\in G_0$ then
$\theta(\check z_{\tau}-z_{\tau})\le\delta$ for every $\tau\in\Cal S$, where
$\family{\tau}{\Cal S}{\check z_{\tau}}=\checkpmb z=\iiv(\checkpmb u)$
and $\family{\tau}{\Cal S}{z_{\tau}}=\pmb z$.   In this case,

$$\eqalign{\int_{\Cal S}\pmb u\,d\pmb z-\int_{\Cal S}\pmb u\,d\checkpmb z
&=u_{\downarrow}\times(z_{\tau_0}-\check z_{\tau_0})
   +\sum_{i=0}^{n-1}u_{\tau_i}
      \times((z_{\tau_{i+1}}-\check z_{\tau_{i+1}})
           -(z_{\tau_i}-\check z_{\tau_i}))
\cr&\mskip100mu +u_{\tau_n}\times((z_{\uparrow}-\check z_{\uparrow})
   -(z_{\tau_n}-\check z_{\tau_n}))\cr}$$

\noindent where $u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$,
$z_{\uparrow}=\lim_{\sigma\uparrow\Cal S}z_{\sigma}$ and
$\check z_{\uparrow}=\lim_{\sigma\uparrow\Cal S}\check z_{\sigma}$
(614C again, since $\lim_{\sigma\downarrow\Cal S}z_{\sigma}
=\lim_{\sigma\downarrow\Cal S}\check z_{\sigma}=0$,
by 646M and 613J(f-i) again).   Consequently

$$\eqalign{\theta(\int_{\Cal S}\pmb u\,d\pmb z
   -\int_{\Cal S}\pmb u\,d\checkpmb z)
&\le(2n+2)\sup_{\sigma,\tau\in\Cal S}
   \theta(u_{\sigma}\times(z_{\tau}-\check z_{\tau}))
\cr& \le(2n+2)\sup_{\tau\in\Cal S}
   \theta(\bar u\times(z_{\tau}-\check z_{\tau}))
\le\epsilon.\cr}$$

\noindent So this $G_0$ will serve.\ \Qed

\medskip

\quad{\bf (ii)} There is a $\TSi$-neighbourhood $G_1$ of
$\pmb x$ such that whenever
$\checkpmb u\in\Mmo$ is such that $|\checkpmb u|\le\pmb u^*$ and
$\checkpmb u_<\in G_1$ then
$\theta(\Sint_{\Cal S}\pmb u_<\times\checkpmb u_<\,d\pmb v
  -\Sint_{\Cal S}\pmb u_<\times\pmb x\,d\pmb v)\le\epsilon$.
\Prf\ Set $C=\{\pmb y:\pmb y\in\MSi^0$, $|\pmb y|\le\pmb u^*_<\}$ and
$D=\{\pmb y:\pmb y\in\MSi^0$, $|\pmb y|\le(|\pmb u|\times\pmb u^*)_<\}$.
The operators $\pmb y\mapsto\pmb u_<\times\pmb y:C\to D$ and
$\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d\pmb v:D\to L^0$ are continuous
when $C$ and $D$ are given the S-integration topology and $L^0$
is given the topology of convergence in measure (645E(a-v-$\alpha$), 645S), so
$\pmb y\mapsto\Sint_{\Cal S}\pmb u_<\times\pmb y\,d\pmb v:C\to L^0$ is
continuous and there is a neighbourhood $G_1$ of $\pmb x$ such that
$\theta(\Sint_{\Cal S}\pmb u_<\times\pmb y\,d\pmb v
  -\Sint_{\Cal S}\pmb u_<\times\pmb x\,d\pmb v)\le\epsilon$
whenever $\pmb y\in G_1\cap C$;  this $G_1$ serves.\ \Qed

\medskip

\quad{\bf (iii)} Now there is a $\checkpmb u\in\Mmo$ such that
$|\checkpmb u|\le\pmb u^*$ and $\checkpmb u_<\in G_0\cap G_1$
(645La, as always).
In this case, writing $\checkpmb z=\iiv(\checkpmb u)$ as before,
$\theta(\int_{\Cal S}\pmb u\,d\checkpmb z-\int_{\Cal S}\pmb u\,d\pmb z)
\le\epsilon$ and
$\theta(\Sint_{\Cal S}\pmb u_<\times\checkpmb u_<\,d\pmb v
  -\Sint_{\Cal S}\pmb u_<\times\pmb x\,d\pmb v)\le\epsilon$.
But again we know from 617E that
$\int_{\Cal S}\pmb u\,d\checkpmb z
=\int_{\Cal S}\pmb u\times\checkpmb u\,d\pmb v$, and from
645R(a-i) that
$\Sint_{\Cal S}\pmb u_<\times\checkpmb u_<\,d\pmb v
=\int_{\Cal S}\pmb u\times\checkpmb u\,d\pmb v$.   So we have

$$\eqalign{&\theta(\int_{\Cal S}\pmb u\,d\pmb z
  -\Sint_{\Cal S}\pmb u_<\times\pmb x\,d\pmb v)
\cr&\mskip70mu
\le\theta(\int_{\Cal S}\pmb u\,d\pmb z-\int_{\Cal S}\pmb u\,d\checkpmb z)
  +\theta(\Sint_{\Cal S}\pmb u_<\times\checkpmb u_<\,d\pmb v
     -\Sint_{\Cal S}\pmb u_<\times\pmb x\,d\pmb v)
\le 2\epsilon.\cr}$$

\medskip

\quad{\bf (iv)} As $\epsilon$ is arbitrary,
$\int_{\Cal S}\pmb u\,d\pmb z=\Sint_{\Cal S}\pmb u_<\times\pmb x\,d\pmb v$.

\medskip

{\bf (c)} We know that, setting
$C=\{\pmb y:\pmb y\in\MSi^0$, $|\pmb y|\le\pmb u^*_<\}$
and $D=\{\pmb y:\pmb y\in\MSi^0$,
$|\pmb y|\le(\pmb u^*\times\pmb u^*)_<\}$,

\Centerline{$\pmb y\mapsto\pmb y\times\pmb x:C\to D$}

\noindent is $(\TSi,\TSi)$-continuous and

\Centerline{$\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d\pmb z:C\to L^0$,
\quad$\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d\pmb v:D\to L^0$}

\noindent are $\TSi$-continuous (645E(a-v-$\alpha$), 645S again), so

\Centerline{$B=\{\pmb y:\pmb y\in C$,
  $\Sint_{\Cal S}\pmb y\times\pmb x\,d\pmb v
      =\Sint_{\Cal S}\pmb y\,d\pmb z\}$}

\noindent is (relatively) closed in $C$ for $\TSi$;  and it includes
$C'=\{\pmb u_<:\pmb u\in\Msimp(\Cal S)$, $|\pmb u|\le\pmb u^*\}$, by
(b).   But we chose $\pmb u^*$ so that $\pmb x'$ belongs to $C$ and
is in the $\TSi$-closure of $C'$, so $\pmb x'\in B$, as required.
}%end of proof of 646R

\medskip

\noindent{\bf Mnemonic} $d(\Siiv(\pmb x))=\pmb x\,d\pmb v$.

\leader{646S}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex
sublattice of $\Cal T$.   Let $\pmb v$ be a
near-simple integrator and $\pmb x$ an S-integrable process, both with
domain $\Cal S$.   Then
$\Siiv(\pmb x)-\Siiv(\pmb x)_<
=\pmb x\times(\pmb v-\pmb v_<)\times\tbf 1^{(\Cal S)}_<$.

\proof{{\bf (a)} To begin with, suppose that
$\pmb x\in\MSi^0=\MSi^0(\Cal S)$.   By 646N, $\Siiv(\pmb x)$ is near-simple,
so we can speak of its previsible version $\Siiv(\pmb x)_<$.
Let $\pmb u^*\in\Mmo^+=\Mmo(\Cal S)^+$ be such that $|\pmb x|\le\pmb u^*_<$.
Set $C=\{\pmb y:\pmb y\in\MSi^0$, $|\pmb y|\le\pmb u^*_<\}$,
$C'=\{\pmb u_<:\pmb u\in\Mmo$, $|\pmb u|\le\pmb u^*\}$
and $B=\{\pmb y:\pmb y\in C$,
$\Siiv(\pmb y)-\Siiv(\pmb y)_<=\pmb y\times(\pmb v-\pmb v_<)\}$.
If $\pmb u\in\Mmo$ and $|\pmb u|\le\pmb u^*$, then

$$\eqalignno{\Siiv(\pmb u_<)-\Siiv(\pmb u_<)_<
&=\iiv(\pmb u)-\iiv(\pmb u)_<\cr
\displaycause{646Kc}
&=\pmb u_<\times(\pmb v-\pmb v_<)\cr}$$

\noindent by 641Q, so $\pmb u_<\in B$;  thus $C'\subseteq B$.
We know that
$\Siiv:C\to\Mns(\Cal S)$ is continuous for $\TSi$ and the ucp topology
(646O), while $\pmb u\mapsto\pmb u_<:\Mmo\to\Mmo$ is
continuous for the ucp topology (641G(e-ii)), so
$\pmb y\mapsto\Siiv(\pmb y)-\Siiv(\pmb y)_<$ is continous for
$\TSi$ and the ucp topology and therefore for $\TSi$ and the product
topology on $(L^0)^{\Cal S}$.   At the same time,
the embedding $\MSi^0\embedsinto(L^0)^{\Cal S}$ is
continuous (645E(a-iii)), so
$\pmb y\mapsto\pmb y\times(\pmb v-\pmb v_<):
\MSi^0\to(L^0)^{\Cal S}$ is continuous and
$B$ is closed in $C$.   Since $B$ includes the dense subset $C'$ of $C$ (see
645La), $B=C$, so

\Centerline{$\Siiv(\pmb x)-\Siiv(\pmb x)_<
=\pmb x\times(\pmb v-\pmb v_<)
=\pmb x\times\tbf 1^{(\Cal S)}_<\times(\pmb v-\pmb v_<)$.}

\medskip

{\bf (b)} For the general case of $S$-integrable $\pmb x$, set
$\pmb y=\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0(\Cal S)$;  then
$\Siiv(\pmb x)=\Siiv(\pmb y)$, so

\Centerline{$\Siiv(\pmb x)-\Siiv(\pmb x)_<
=\Siiv(\pmb y)-\Siiv(\pmb y)_<
=\pmb y\times(\pmb v-\pmb v_<)
=\pmb x\times(\pmb v-\pmb v_<)\times\tbfoneminus{\Cal S}$.}
}%end of proof of 646S

\leader{646T}{It\^o's Formula, fourth form}
Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, and that $\Cal S$ is an order-convex
sublattice of $\Cal T$ with a least element.   Let
$\pmb v$ be a jump-free integrator with domain $\Cal S$, and
$\pmb v^*$ its quadratic variation;  let $h:\Bbb R\to\Bbb R$ be a
differentiable function such that its derivative $h'$ is locally
Lipschitz\cmmnt{, that is, Lipschitz on every bounded set}.
If $h'':\Bbb R\to\Bbb R$ is a locally bounded Borel measurable function
Lebesgue-almost-everywhere equal to the derivative of $h'$, then

\Centerline{$\biggerSint_{\Cal S}\pmb x\,d(\bar h\pmb v)
=\biggerSint_{\Cal S}\pmb x\times\bar h'\pmb v\,d\pmb v
+\Bover12\biggerSint_{\Cal S}\pmb x\times\bar h''\pmb v\,d\pmb v^*$}

\noindent for every $\pmb x\in\cmmnt{\MSi=\mskip5mu}\MSi(\Cal S)$.

\proof{{\bf (a)} To begin with (down to the end of (b)), suppose that
$h'$ is continuously differentiable and that its derivative is everywhere
equal to $h''$.    As before, we need only to prove the result for
$\pmb x\in\MSi^0(\Cal S)$.

We need to know that if $\pmb y\in\Mpob(\Cal S)$ then
$\pmb y\times\bar h'\pmb v=\pmb y\times(\bar h'\pmb v)_<$ and
$\pmb y\times\bar h''\pmb v=\pmb y\times(\bar h''\pmb v)_<$.
\Prf\ By 618Ga, $\bar h'\pmb v$ is jump-free.   Set
$e_{\sigma}=\sup_{\tau\in\Cal S}\Bvalue{\tau<\sigma}$ for $\sigma\in\Cal S$.
Expressing $\bar h'\pmb v$ as
$\sfamilysS{v'}$ and $\pmb y$ as $\sfamilysS{y}$, we see from 641Na that
$(v'_{\sigma}-v'_{<\sigma})\times\chi e_{\sigma}=0$, while
$y_{\sigma}\times\chi(1\Bsetminus e_{\sigma})=0$,
for every $\sigma\in\Cal S$;  so
$y_{\sigma}\times v'_{\sigma}=y_{\sigma}\times v'_{\sigma_<}$ for every
$\sigma$, and
$\pmb y\times\bar h'\pmb v=\pmb y\times(\bar h'\pmb v)_<$.
Similarly,
$\pmb y\times\bar h''\pmb v=\pmb y\times(\bar h''\pmb v)_<$.\ \Qed

It follows that if $\pmb u\in\Mmo=\Mmo(\Cal S)$ then

\Centerline{$(\pmb u\times\bar h'\pmb v)_<
=\pmb u_<\times(\bar h'\pmb v)_<
=\pmb u_<\times\bar h'\pmb v$,
\quad$(\pmb u\times\bar h''\pmb v)_<
=\pmb u_<\times\bar h''\pmb v$.}

\medskip

{\bf (b)} Now for the main argument.   Let $\pmb u\in\Mmo^+$ be such that
$|\pmb x|\le\pmb u_<$, and set

\Centerline{$\pmb u^*
=\pmb u\times(|\bar h'\pmb v|\vee|\bar h''\pmb v|)$,}

\Centerline{$A=\{\pmb y:\pmb y\in\MSi^0$, $|\pmb y|\le\pmb u_<\}$,
\quad
$A^*=\{\pmb y:\pmb y\in\MSi^0$, $|\pmb y|\le\pmb u^*_<\}$.}

\noindent By 645Ja,
$\pmb y\times\bar h'\pmb v=\pmb y\times(\bar h'\pmb v)_<$
belongs to $\MSi^0$ for every $\pmb y\in A$, and
$\pmb y\mapsto\pmb y\times\bar h'\pmb v:\MSi^0\to\MSi^0$
is $\TSi$-continuous (645E(a-v)), while
$\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d\pmb v$ is continuous on $A^*$
(645S).   So
$\pmb y\mapsto\Sint_{\Cal S}\pmb y\times\bar h'\pmb vd\pmb v:
A\to L^0$ is continuous.   Similarly,
$\pmb y\mapsto\Sint_{\Cal S}\pmb y\times\bar h''\pmb vd\pmb v:
A\to L^0$  and
$\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d(\bar h\pmb v):A\to L^0$ are
continuous (recalling from 616O that $\bar h\pmb v$ is an integrator, and
from 631F(a-i) that it is near-simple).

Now we know from 619D that if $\pmb u'\in\Mmo$ then

\Centerline{$\int_{\Cal S}\pmb u'\,d(\bar h\pmb v)
=\int_{\Cal S}\pmb u'\times\bar h'\pmb v\,d\pmb v
+\Bover12\int_{\Cal S}\pmb u'\times\bar h''\pmb v\,d\pmb v^*$,}

\noindent so that

\Centerline{$\biggerSint_{\Cal S}\pmb u'_<\,d(\bar h\pmb v)
=\biggerSint_{\Cal S}\pmb u'_<\times\bar h'\pmb vd\pmb v
+\Bover12\biggerSint_{\Cal S}
   \pmb u'_<\times\bar h''\pmb vd\pmb v^*$.}

\noindent And we know also that $\pmb x$ belongs to the closure of
$A'=\{\pmb u'_<:\pmb u'\in\Mmo$, $|\pmb u'|\le\pmb u\}$ in $\MSi^0$.
Since we are looking
at continuous operators into a Hausdorff space, we can conclude that

\Centerline{$\biggerSint_{\Cal S}\pmb x\,d(\bar h\pmb v)
=\biggerSint_{\Cal S}\pmb x\times\bar h'\pmb v\,d\pmb v
+\Bover12\biggerSint_{\Cal S}\pmb x\times\bar h''\pmb v\,d\pmb v^*$.}

\medskip

{\bf (c)} Now turn everything round, and start from locally bounded
Borel measurable functions $g:\Bbb R\to\Bbb R$.   For such a function,
define $I(g):\Bbb R\to\Bbb R$ by setting

$$\eqalign{I(g)(\alpha)
&=\int_0^{\alpha}g(\beta)d\beta\text{ if }\alpha\ge 0,\cr
&=-\int_{\alpha}^0g(\beta)d\beta\text{ if }\alpha\le 0,\cr}$$

\noindent the integrals here being with respect to Lebesgue measure, of
course.   Then $I(g)$ will be continuous, indeed
locally Lipschitz, and its derivative will be
almost everywhere equal to $g$;  $I(g)$ will be the
derivative of $I^2(g)$ everywhere.   Fix an $\pmb x\in\MSi$, and let
$\Phi$ be the set of locally bounded Borel measurable $g:\Bbb R\to\Bbb R$
such that

\Centerline{$\biggerSint_{\Cal S}\pmb x\,d(\overline{I^2(g)}\pmb v)
=\biggerSint_{\Cal S}\pmb x\times\overline{I(g)}\pmb v\,d\pmb v
+\Bover12\biggerSint_{\Cal S}\pmb x\times\bar g\pmb v\,d\pmb v^*$.}

\noindent Then (a)-(b) tell us that $\Phi$ contains all continuous
functions from $\Bbb R$ to itself.   Now the point is that if
$\sequencen{g_n}$ is a sequence in $\Phi$ converging pointwise to
a function $g:\Bbb R\to\Bbb R$, and if $\sequencen{g_n}$ is locally
uniformly bounded in the sense that
$\sup_{n\in\Bbb N,|\alpha|\le M}|g_n(\alpha)|$ is finite for every
$M\ge 0$, then $g\in\Phi$.   \Prf\ Because
$g(\alpha)=\lim_{n\to\infty}g_n(\alpha)$ for every $\alpha$,
$\sequencen{\pmb x\times\bar g_n\pmb v}$ is order*-convergent to
$\pmb x\times\bar g\pmb v$.   Next, $\sup_{n\in\Bbb N}|g_n|$ is
locally bounded, so is bounded above by a continuous function (645Cb), and
$\{|\bar g_n\pmb v|:n\in\Bbb N\}$ is bounded above by a
jump-free process (618Ga again).   Accordingly
$\{\bar g_n\pmb v\times\tbfoneminus{\Cal S}:n\in\Bbb N\}$ is uniformly
previsibly order-bounded (use 641O),
so $\{\pmb x\times\bar g_n\pmb v\times\tbfoneminus{\Cal S}:n\in\Bbb N\}$
is uniformly previsibly order-bounded (because $\Mpob(\Cal S)$ is closed
under multiplication, by 645D(a-ii)).   By 645T,

$$\eqalign{\Sint_{\Cal S}\pmb x\times\bar g\pmb v\,d\pmb v^*
&=\Sint_{\Cal S}
  \pmb x\times\bar g\pmb v\times\tbfoneminus{\Cal S}\,d\pmb v^*
\cr&=\lim_{n\to\infty}\Sint_{\Cal S}
  \pmb x\times\bar g_n\pmb v\times\tbfoneminus{\Cal S}\,d\pmb v^*
=\lim_{n\to\infty}\Sint_{\Cal S}
  \pmb x\times\bar g_n\pmb v\,d\pmb v^*.
\cr}$$

Next, $\sequencen{I(g_n)}$ and $\sequencen{I^2(g_n)}$ are also locally
uniformly bounded sequences converging pointwise to $I(g)$ and $I^2(g)$
respectively.   So

\Centerline{$\biggerSint_{\Cal S}\pmb x\times\overline{I(g)}\pmb v\,d\pmb v
=\lim_{n\to\infty}
  \biggerSint_{\Cal S}\pmb x\times\overline{I(g_n)}\pmb v\,d\pmb v$,}

$$\eqalign{\Sint_{\Cal S}\pmb x\times\overline{I^2(g)}\pmb v\,d\pmb v
&=\lim_{n\to\infty}
  \Sint_{\Cal S}\pmb x\times\overline{I^2(g_n)}\pmb v\,d\pmb v
\cr&=\lim_{n\to\infty}
  \Sint_{\Cal S}\pmb x\times\overline{I(g_n)}\pmb v\,d\pmb v
  +\Bover12\Sint_{\Cal S}\pmb x\times\bar g_n\pmb v\,d\pmb v^*
\cr&=\Sint_{\Cal S}\pmb x\times\overline{I(g)}\pmb v\,d\pmb v
  +\Bover12\Sint_{\Cal S}\pmb x\times\bar g\pmb v\,d\pmb v^*
\cr}$$

\noindent and $g\in\Phi$.\ \Qed

Consequently $\Phi$ contains all locally bounded Borel measurable functions
(645Cc).

\medskip

{\bf (d)} Returning to the stated hypothesis, in which $\pmb x\in\MSi$,
$h:\Bbb R\to\Bbb R$ is differentiable, its derivative $h'$ is locally
Lipschitz, and $h''$ is a locally bounded Borel measurable function
equal almost everywhere to the derivative of $h'$, set $\beta=h(0)$ and
$\gamma=h'(0)$;  then

\Centerline{$h'(\alpha)=\gamma+I(h'')(\alpha)$,
\quad$h(\alpha)=\beta+\gamma\alpha+I^2(h'')(\alpha)$}

\noindent for every $\alpha$.   Setting $g=h''$, we have

\Centerline{$\bar h'\pmb v=\gamma\tbf 1+\overline{I(g)}\pmb v$,
\quad$\bar h\pmb v=\beta\tbf 1+\gamma\pmb v+\overline{I^2(g)}\pmb v$.}

\noindent Now

$$\eqalignno{\Sint_{\Cal S}\pmb x\,d(\bar h\pmb v)&
=\beta\Sint_{\Cal S}\pmb x\,d\tbf 1
  +\gamma\Sint_{\Cal S}\pmb x\,d\pmb v
  +\Sint_{\Cal S}\pmb x\,d(\overline{I^2(g)}\pmb v)\cr&
=0+\gamma\Sint_{\Cal S}\pmb x\,d\pmb v
  +\Sint_{\Cal S}\pmb x\times\overline{I(g)}\pmb v\,d\pmb v
  +\Bover12\Sint_{\Cal S}\pmb x\times\bar g\pmb v\,d\pmb v^*
\Displaycause{using 645R(c-ii) and (c) above}
=\Sint_{\Cal S}\pmb x\times\bar h'\pmb v\,d\pmb v
  +\Bover12\Sint_{\Cal S}\pmb x\times\bar h''\pmb v\,d\pmb v^*
\cr}$$

\noindent as required.
}%end of proof of 646T

%\leader{646U}{}
%\leader{646V}{}
%\leader{646W}{}

\exercises{\leader{646X}{Basic exercises (a)}
%\spheader 646Xa
In 646F, show that
$R(\bar h\pmb u',\bar h\pmb u'')=\bar hR(\pmb u',\pmb u'')$ whenever
$\pmb u'\in(L^0)^{\Cal S\wedge\tau}$,
$\pmb u''\in(L^0)^{\Cal S\vee\tau}$
and $h:\Bbb R\to\Bbb R$ is a Borel measurable function.
%646F

\spheader 646Xb
In 646H, show that
$R^*(\bar h\pmb u)=\bar hR^*(\pmb u)$ whenever
$\pmb u\in(L^0)^{\Cal S}$ and $h:\Bbb R\to\Bbb R$ is a Borel measurable
function with $h(0)=0$.
%646H

\spheader 646Xc Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, $\Cal S$ is an order-convex sublattice of $\Cal T$
and $\bar u\in(L^0)^+$.   Show that
$\{\pmb u:\pmb u\in\Mns(\Cal S)$, $\sup|\pmb u|\le\bar u\}$ has a greatest
member defined by setting
$u_{\sigma}=\sup\{u:u\in L^0(\frak A_{\sigma})$, $u\le\bar u\}$ for
$\sigma\in\Cal S$.   \Hint{use 611H and 632F.}
%611H 632F

\spheader 646Xd In 646R, show that
$\Sii_{\pmb z}(\pmb x)=\Sii_{\pmb v}(\pmb x'\times\pmb x)$.
%646R

\spheader 646Xe In 646T, show that
$\Sii_{\bar h\pmb v}(\pmb x)=\Sii_{\pmb v}(\pmb x\times\bar h'\pmb v)
  +\bover12\Sii_{\pmb v^*}(\pmb x\times\bar h''\pmb v)$.
%\leader{646Y}{Further exercises (a)}

}%end of exercises

\endnotes{
\Notesheader{646} Working through the basic properties of the Riemann-sum
integral as set out in Chapter 61, we find that (as we naturally hope)
most of them seem to be shared by the S-integral.   I checked bilinearity
in 645Rb.   Some of the others are straightforward (646B, 646C, 646M, 646Qc.
646S, 646T), some take more thought (645K, 646N, 646Qb, 646R) and
some are apparently much harder (646J, 646Qa, 647J).

Even in terms of the concepts used in this presentation, I have not
taken the shortest road to 646J, and surely it is going to need a proper
look at the definition of the S-integral, but all the same it seems to take
more pages than it should.   The difficulty is that in the S-integral we
are committed to a special class of
integrands, so that when we split the sublattice $\Cal S$ into
$\Cal S\wedge\tau$ and $\Cal S\vee\tau$ the breakpoint $\tau$ has to be
analysed from both sides;  this is what the operators $R$ of 646F and
its approximate inverse
$\pmb x\mapsto(\pmb x\restr\Cal S\wedge\tau,R^*(\pmb x))$ are doing, and the
problem is that these are not true inverses of each other.

Coming to the properties of indefinite S-integrals (646N, 646Q),
we find ourselves facing interesting questions.   When showing
that an indefinite
integral with respect to a martingale is a virtually local martingale, there
were already significant obstacles for the Riemann-sum integral,
but the methods of \S621 turned out to be adequate.   For the S-integral, I
think we need to go much deeper, with another appeal to the fundamental
theorem of martingales.  %646P prf g 643L 644G prf d
In fact this allows us to bypass
Theorem 623O altogether, as well as giving a striking new fact about
martingale integrators in 646P.   Concerning the value of
$\Bvalue{\Sint_{\Cal S}\pmb x\,d\pmb v\ne 0}$, we have an easy argument to
show that it is included in $\Bvalue{\pmb v\ne 0}$ (646C), but to show that
it is included in $\Bvalue{\pmb x\ne 0}$ (corresponding to the result for
the Riemann-sum integral in 613Ld), we apparently need a special
construction, which I will come to in the next section.

Naturally, the proof I give of the change-of-variable result 646R
is based on the corresponding
result 617E for the Riemann-sum integral;  the application of 617E is hidden in
(b-iii) of the proof of 646R.   But here again we have a not-quite-trivial check
to make on the limiting process we need to use.   It goes a bit more smoothly if
we start from 645Lb rather than 645La.   As for It\^o's formula, the obvious
extension (part (b) of the proof of 646T) is straightforward in view of what has
been done before.   But with a little bit of analysis we can use the
sequential convergence properties of the S-integral to extend the result to a
larger class of functions $h$.   I am rather supposing that you recognise the
relevance of Rademacher's theorem (262Q);  if
$h':\Bbb R\to\Bbb R$ is locally Lipschitz, it will be differentiable almost
everywhere, and its derivative will have an extension to a
locally bounded Borel measurable function (225J).
%262Bc 262Yd

The definition of $\MSi^0$ in 645F leaves the nature of S-integrable
processes unclear.   Of course you could say the same for
universal measurability as defined in 434D;  it is markedly less
easy to form a mental picture of a `typical' universally measurable
function than of a typical Lebesgue integrable function, because
universal measurability is not easily described in terms of universally
negligible sets (439Xe).   But we can say a little bit, as in 646B.
}%end of notes

\discrpage

\frfilename{mt647.tex}
\versiondate{29.3.23}
\copyrightdate{2014}

\def\chaptername{The S-integral}
\def\sectionname{Changing the filtration II}

\def\AMfa{\Bbb A\Mfa}
\def\AMbv{\Bbb A\Mbv}
\def\AMns{\Bbb A\Mns}
\def\AMmo{\Bbb A\Mmo}
\def\AMob{\Bbb A\Mob}
\def\AMSizero{\Bbb A\MSi^0}
\def\AMSi{\Bbb A\MSi}
\def\AMsimp{\Bbb A\Msimp}
\def\ATSi{\Bbb A\TSi}
\def\ASint{\Bbb A\mskip-5mu\Sint}

\def\BMfa{\Bbb B\Mfa}
\def\BMbv{\Bbb B\mskip-1mu\Mbv}
\def\BMns{\Bbb B\mskip-1mu\Mns}
\def\BMmo{\Bbb B\mskip-1mu\Mmo}
\def\BMob{\Bbb B\mskip-1mu\Mob}
\def\BMSizero{\Bbb B\mskip-1mu\MSi^0}
\def\BMSi{\Bbb B\mskip-1mu\MSi}
\def\BMsimp{\Bbb B\mskip-1mu\Msimp}
\def\BTSi{\Bbb B\TSi}
\def\BSint{\Bbb B\mskip-5mu\Sint}

\newsection{647}

The answer (647J) to a natural question left over from \S646
leads us to a new construction to
add to those in the second half of Chapter 63.

\cmmnt{\leader{647A}{Notation}
As usual, $(\frak A,\bar\mu)$ will be a probability algebra, and
$\family{t}{T}{\frak A_t}$ a filtration of closed subalgebras of $\frak A$,
with the associated lattice $\Cal T$ of stopping times and family
$\family{\tau}{\Cal T}{\frak A_{\tau}}$ of subalgebras.
If $\frak C$ is a closed subalgebra of $\frak A$ and $a\in\frak A$,
$\upr(a,\frak C)$ is the upper envelope of $a$ in $\frak C$.
$L^0=L^0(\frak A)$ will be given its topology of convergence in measure.

If $\Cal S$ is a sublattice of $\Cal T$ and $\tau\in\Cal S$,
$\Cal S\wedge\tau=\{\sigma\wedge\tau:\sigma\in\Cal S\}$ and
$\Cal I(\Cal S)$ is the set of finite sublattices of $\Cal S$.
We shall be looking at some of the usual spaces of fully adapted processes;
if $\Cal S$ is a sublattice of $\Cal T$, then $\Mfa(\Cal S)$ is the space of
fully adapted processes with domain $\Cal S$,
$\Msimp$ the space of simple processes, $\Mob$ the space of order-bounded
processes, $\Mmo$ the space of moderately oscillatory processes,
$\Mns$ the space of near-simple processes,
$\Mnsu$ the space of non-negative non-decreasing near-simple processes,
$\Mbv$ the space of processes of finite variation,
$\Mpob$ the space of previsibly order-bounded processes (645B),
$\MSi^0$ the closure of $\{\pmb u_<:\pmb u\in\Mmo\}$ for the
S-integration topology $\TSi$ on $\Mpob$, and
$\MSi$ the space $\{\pmb x:\pmb x\times\tbfoneminus{\Cal S}\in\MSi^0\}$ (645F).
If $\pmb u=\sfamilysS{u}$ is order-bounded,
$\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.
If $\pmb u$ and $\pmb v$ are fully
adapted processes, I write $\pmb u\preccurlyeq\pmb v$ if
$\pmb v-\pmb u$ is non-decreasing.
For $\pmb u\in\Mmo$,
$\pmb u_<\in\Mpob$ is its previsible version.
$\int$ will denote the Riemann-sum integral of \S613
and $\Sint$ the S-integral of \S645.
}

\leader{647B}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\frak T$ the linear space topology on $\Mpob=\Mpob(\Cal S)$
defined by the F-seminorms
$\widehat{\theta}^{\#}_{\pmb v}$\cmmnt{ (645Db)}
where $\pmb v\in\Mnsu(\Cal S)$ is $\|\,\|_{\infty}$-bounded.
Then $\frak T$ is the S-integration topology on $\Mpob$.

\proof{ Because the S-integration topology $\TSi$ is defined by a larger
family of F-seminorms, $\frak T$ is coarser than $\TSi$.   In the other
direction, let
$G$ be a $\TSi$-neighbourhood of $0$ in $\Mpob$.
Then there are a $\pmb v\in\Mnsu(\Cal S)$ and an $\epsilon>0$ such that
$G$ includes $\{\pmb w:\pmb w\in\Mpob$,
$\widehat{\theta}^{\#}_{\pmb v}(\pmb w)\le 3\epsilon\}$ (645E(a-ii)).
Set $\bar v=\sup|\pmb v|$, let $M\ge 0$ be
such that $\bar\mu\Bvalue{\bar v\ge M}\le\epsilon$ and set
$\pmb v'=\pmb v\wedge M\tbf 1$.   Then
$G'=\{\pmb w:\pmb w\in\Mpob$,
$\widehat{\theta}^{\#}_{\pmb v'}(\pmb w)\le\epsilon\}$
is a $\frak T$-neighbourhood of $0$.   Now $G'\subseteq G$.
\Prf\ The point is that, for any
$\pmb u\in\Mmo=\Mmo(\Cal S)$,

$$\eqalignno{
\Bvalue{\int_{\Cal S}\pmb u\,d\pmb v'\ne\int_{\Cal S}\pmb u\,d\pmb v}
&\Bsubseteq\Bvalue{\pmb v'\ne\pmb v}\cr
\displaycause{613Jd}
&=\Bvalue{\bar v>M}\cr}$$

\noindent has measure at most $\epsilon$, and

\Centerline{$\theta(\int_{\Cal S}\pmb u\,d\pmb v)
\le\epsilon+\theta(\int_{\Cal S}\pmb u\,d\pmb v')$.}

\noindent If $\pmb w\in G'$, there is a uniformly order-bounded
non-decreasing sequence $\sequencen{\pmb u^{(n)}}$ in
$\Mmo^+$ such that $|\pmb w|\le\sup_{n\in\Bbb N}\pmb u^{(n)}_<$ and
$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u^{(n)}d\pmb v')
\le 2\epsilon$.   In this case,

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb w)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u^{(n)}d\pmb v)
\le 3\epsilon$}

\noindent and $\pmb w\in G$.   As $\pmb w$ is arbitrary,
$G'\subseteq G$.\ \Qed

Thus $G$ is a $\frak T$-neighbourhood of $0$, and $0$ has the same
neighbourhoods for $\frak T$ and $\TSi$.   As these are both linear space
topologies, they coincide.
}%end of proof of 647B

\leader{647C}{}\cmmnt{ I give a couple of perfectly elementary facts
which were not spelt out in Volume 3.

\medskip

\noindent}{\bf Lemma} Suppose that $\frak D$ is a closed subalgebra of
$\frak A$, and $b\in\frak A$;  let $\frak B$ be the closed subalgebra of
$\frak A$ generated by $\{b\}\cup\frak D$.

(a) If $c\in\frak B$, then
$b\Bcap c=b\Bcap\upr(b\Bcap c,\frak D)$.

(b) If $u\in L^0(\frak B)$, there are $u'$, $u''\in L^0(\frak D)$ such that
$u=u'\times\chi b+u''\times\chi(1\Bsetminus b)$.

\proof{{\bf (a)}
Of course $b\Bcap c\Bsubseteq b\Bcap\upr(b\Bcap c,\frak D)$.
In the other direction, we know that
$\frak B=\{(a\Bcap b)\Bcup(a'\Bsetminus b):a$, $a'\in\frak D\}$
(314Ja), so that there is an $a\in\frak D$ such that
$b\Bcap c=b\Bcap a$.   In this case,

$$\eqalignno{b\Bcap\upr(b\Bcap c,\frak D)
&=b\Bcap\upr(b\Bcap a,\frak D)
=b\Bcap a\Bcap\upr(b,\frak D)\cr
\displaycause{313Sc}
&=b\Bcap a
=b\Bcap c,\cr}$$

\noindent as claimed.

\medskip

{\bf (b)} Let $\frak C$ be the principal ideal of $\frak B$ generated by
$b$.   Then the maps $d\mapsto d\Bcap b:\frak D\to\frak C$ and
$d\mapsto d\Bcap b:\frak B\to\frak C$ are both order-continuous
surjective Boolean
homomorphisms.   So the corresponding Riesz homomorphisms
$T':L^0(\frak D)\to L^0(\frak C)$ and $T:L^0(\frak B)\to L^0(\frak C)$
(364P) are surjective, and there is a $u'\in L^0(\frak D)$ such that
$T'u'=Tu$.   Now if $\alpha\in\Bbb R$,

$$\eqalignno{b\Bcap\Bvalue{u'>\alpha}
&=\Bvalue{T'u'>\alpha}
\Displaycause{by the defining formula for $T'$ in 364Pa}
=\Bvalue{Tu>\alpha}
=b\Bcap\Bvalue{u>\alpha}.\cr}$$

\noindent So

$$\eqalign{\Bvalue{u'\times\chi b>\alpha}
&=b\Bcap\Bvalue{u'>\alpha}=\Bvalue{u\times\chi b>\alpha}
\text{ if }\alpha\ge 0,\cr
&=b\Bcap\Bvalue{u'>\alpha}\Bcup(1\Bsetminus b)
=\Bvalue{u\times\chi b>\alpha}
\text{ if }\alpha<0,\cr}$$

\noindent and $u'\times\chi b=u\times\chi b$.

Repeating the argument with $1\Bsetminus b$ in the place of $b$,
we obtain an appropriate $u''$.
}%end of proof of 647C

\leader{647D}{Construction} For most of the rest of this section,
$b$ will be a fixed member of $\frak A$.
For $t\in T$, let $\frak B_t$ be the
subalgebra of $\frak A$ generated by $\{b\}\cup\frak A_t$;  then
$\frak B_t\cmmnt{\mskip5mu=\{(a\Bcap b)\Bcup(a'\Bsetminus b):a,\,a'\in\frak A\}}$ is a
closed subalgebra\cmmnt{ (312N, 314J)}.
\cmmnt{If $s\le t$ then
$\{b\}\cup\frak A_s\subseteq\frak B_t$ so $\frak B_s\subseteq\frak B_t$;
accordingly }$\family{t}{T}{\frak B_t}$ is a filtration.

\leader{647E}{Notation} From now on\cmmnt{, therefore}, we shall have
the two filtrations $\family{t}{T}{\frak A_t}$ and
$\family{t}{T}{\frak B_t}$, giving stochastic integration structures
$\Bbb A=(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T_{\Bbb A},
   \family{\sigma}{\Cal T_{\Bbb A}}{\frak A_{\sigma}})$
and
$\Bbb B=(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak B_t},
   \penalty-100\Cal T_{\Bbb B},
   \family{\tau}{\Cal T_{\Bbb B}}{\frak B_{\tau}})$.
For the various spaces of processes, I\cmmnt{ will} write
$\AMSizero(\Cal S)$, $\BMmo(\Cal S)$, $\Bbb A\Mpob(\Cal S)$ etc.
When we come to S-integration, I\cmmnt{ will} talk of F-seminorms
$\Bbb A\widehat{\theta}^{\#}_{\pmb v}$, the
S-integration topology $\BTSi$ and S-integrals $\ASint$.
\cmmnt{(As we shall see in
647Fe, there is no need for such distinctions in regard to
Riemann-sum integrals.)}

\leader{647F}{Proposition} (a)(i) $\Cal T_{\Bbb A}$ is a sublattice of
$\Cal T_{\Bbb B}$.

\quad(ii) $\min\Cal T_{\Bbb A}=\min\Cal T_{\Bbb B}$ and
$\max\Cal T_{\Bbb A}=\max\Cal T_{\Bbb B}$.

(b) For any $\sigma\in\Cal T_{\Bbb A}$,
$\frak B_{\sigma}$ is the subalgebra of $\frak A$
generated by $\{b\}\cup\frak A_{\sigma}$.

(c) If $\sigma$, $\tau\in\Cal T_{\Bbb A}$, then
$\Bvalue{\sigma<\tau}$ and $\Bvalue{\sigma=\tau}$ are the same in either
structure.

(d) Let $\Cal S$ be a sublattice of $\Cal T_{\Bbb A}$.

\quad(i) If $\pmb u$ is an $\Bbb A$-fully adapted process with domain
$\Cal S$, it is $\Bbb B$-fully adapted.

\quad(ii) $\AMsimp(\Cal S)\subseteq\BMsimp(\Cal S)$.

\quad(iii) $\AMob(\Cal S)\subseteq\BMob(\Cal S)$, and the
ucp topology on $\AMob(\Cal S)$ is the subspace topology induced by
the ucp topology on $\BMob(\Cal S)$.

\quad(iv) $\AMns(\Cal S)\subseteq\BMns(\Cal S)$.

\quad(v) $\AMbv(\Cal S)\subseteq\BMbv(\Cal S)$.

\quad(vi) $\AMmo(\Cal S)\subseteq\BMmo(\Cal S)$.

(e) If $\pmb u$, $\pmb v$ are $\Bbb A$-fully adapted processes with domain
$\Cal S$, and
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined in either of the structures
$\Bbb A$, $\Bbb B$, then it is defined in the other, with the same value.

(f) If $\Cal S$ is a sublattice of $\Cal T_{\Bbb A}$
and $\pmb v$ an $\Bbb A$-integrator with domain $\Cal S$, then
$\pmb v$ is a $\Bbb B$-integrator.

(g) If $\Cal S$ is a sublattice of $\Cal T_{\Bbb A}$ and
$\pmb u$ belongs to $\AMmo(\Cal S)\subseteq\BMmo(\Cal S)$,
then its previsible version
$\pmb u_<$ is the same when calculated in either of the structures
$\Bbb A$, $\Bbb B$.

\proof{{\bf (a)(i)}
The definition in 611A(b-i) makes it plain
that $\Cal T_{\Bbb A}\subseteq\Cal T_{\Bbb B}$ just because
$\frak A_t\subseteq\frak B_t$ for every $t$.   The formulae of 611Cb and
611Cc show that $\Cal T_{\Bbb A}$ and $\Cal T_{\Bbb B}$ can both be
regarded as sublattices of $\frak A^T$, so that $\Cal T_{\Bbb A}$ is a
sublattice of $\Cal T_{\Bbb B}$.

\medskip

\quad{\bf (ii)} Immediate from the formulae in 611Cf.

\medskip

{\bf (b)} Write $\frak B'_{\sigma}$ for the subalgebra
of $\frak A$ generated by $\{b\}\cup\frak A_{\sigma}$.
If $a\in\frak A_{\sigma}$, then
$a\Bsetminus\Bvalue{\sigma>t}\in\frak A_t\subseteq\frak B_t$ for every
$t\in T$, so $a\in\frak B_{\sigma}$.   Next,
$b$ and $\Bvalue{\sigma>t}$, and therefore $b\Bsetminus\Bvalue{\sigma>t}$,
belong to $\frak B_t$ for every $t$, so
$b\in\frak B_{\sigma}$;  accordingly
$\frak B'_{\sigma}\subseteq\frak B_{\sigma}$.

In the other direction, take $c\in\frak B_{\sigma}$.
For each $t\in T$, set
$a_t=\upr(b\Bcap c\Bsetminus\Bvalue{\sigma>t},\frak A_t)$.   Because
$c\Bsetminus\Bvalue{\sigma>t}\in\frak B_t$,
$b\Bcap a_t=b\Bcap c\Bsetminus\Bvalue{\sigma>t}$ (647Ca).
So $b\Bcap a_s\Bsubseteq b\Bcap a_t$ if
$s\le t$.   Also, because $\Bvalue{\sigma>t}\in\frak A_t$,
$a_t=\upr(b\Bcap c,\frak A_t)\Bsetminus\Bvalue{\sigma>t}$ is disjoint from
$\Bvalue{\sigma>t}$ for each $t$.   Set $a=\sup_{t\in T}a_t$.   If
$s\in T$, then

$$\eqalignno{a\Bsetminus\Bvalue{\sigma>s}
&=\sup_{t\in T}(\upr(b\Bcap c,\frak A_t)\Bsetminus\Bvalue{\sigma>t})
    \Bsetminus\Bvalue{\sigma>s}
\cr&=\sup_{t<s}(\upr(b\Bcap c,\frak A_t)\Bsetminus\Bvalue{\sigma>t})
    \Bcup\sup_{t\ge s}(\upr(b\Bcap c,\frak A_t)\Bsetminus\Bvalue{\sigma>s})
\cr&=\sup_{t\le s}(\upr(b\Bcap c,\frak A_t)\Bsetminus\Bvalue{\sigma>t})\cr
\displaycause{because $\frak A_s\subseteq\frak A_t$ so
$\upr(b\Bcap c,\frak A_t)\Bsubseteq\upr(b\Bcap c,\frak A_s)$ if
$t\ge s$}
&\in\frak A_s.\cr}$$

\noindent Thus $a\in\frak A_{\sigma}$ and $b\Bcap a\in\frak B'_{\sigma}$, while

\Centerline{$b\Bcap a
=\sup_{t\in T}b\Bcap c\Bsetminus\Bvalue{\sigma>t}
=b\Bcap c\Bsetminus\inf_{t\in T}\Bvalue{\sigma>t}$.}

\noindent On the other hand,

\Centerline{$(b\Bcap c\Bcap\inf_{t\in T}\Bvalue{\sigma>t})
   \Bsetminus\Bvalue{\sigma>s}=0\in\frak A_s$}

\noindent for every $s\in T$, so
$b\Bcap c\Bcap\inf_{t\in T}\Bvalue{\sigma>t}$
belongs to $\frak A_{\sigma}\subseteq\frak B'_{\sigma}$.
Accordingly $b\Bcap c\in\frak B'_{\sigma}$.

Similarly, $c\Bsetminus b\in\frak B'_{\sigma}$.   So $c\in\frak B'_{\sigma}$.
As $c$ is arbitrary, $\frak B'_{\sigma}=\frak B_{\sigma}$, as claimed.

\medskip

{\bf (c)} The defining formulae

\Centerline{$\Bvalue{\sigma<\tau}
=\sup_{t\in T}\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}$,
\quad$\Bvalue{\sigma=\tau}
=1\Bsetminus(\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau<\sigma})$}

\noindent do not refer to the filtrations.

\medskip

{\bf (d)(i)} $\Cal S$ is a sublattice of $\Cal T_{\Bbb B}$ because
$\Cal T_{\Bbb A}$ is.   Expressing $\pmb u$ as $\sfamilysS{u}$,
$u_{\sigma}\in L^0(\frak A_{\sigma})\subseteq L^0(\frak B_{\sigma})$
for every $\sigma\in\Cal S$.   If $\sigma$, $\tau\in\Cal S$, then
$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$ in either
structure, so $\pmb u$ is $\Bbb B$-fully adapted.

\medskip

\quad{\bf (ii)} Of course
$\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}
\subseteq\bigcap_{\sigma\in\Cal S}\frak B_{\sigma}$.   So if $\pmb u$
satisfies the conditions of 612J with respect to
$\sfamilysS{\frak A}$, it satisfies them with respect to
$\sfamilysS{\frak B}$.

\medskip

\quad{\bf (iii)} `Order-boundedness' of a process depends only on the
structure of $L^0$, not on the filtration (614Ea), so
$\AMob(\Cal S)\subseteq\BMob(\Cal S)$.   Similarly, if we think of
$\widehat{\theta}$ (615B) as an F-seminorm defined on
the space of order-bounded families in $(L^0)^{\Cal S}$, the ucp topologies
on $\AMob(\Cal S)$, $\BMob(\Cal S)$ are defined by
the restrictions of $\widehat{\theta}$ to these spaces, so must
agree on the smaller space $\AMob(\Cal S)$.

\medskip

\quad{\bf (iv)} Now the closure $\AMns(\Cal S)$ of $\AMsimp(\Cal S)$
in $\AMob(\Cal S)$ (631Ba) must be included in the closure
$\BMns(\Cal S)$ of $\BMsimp(\Cal S)$ in $\BMob(\Cal S)$ just because
$\AMsimp(\Cal S)\subseteq\BMsimp(\Cal S)$.

\medskip

\quad{\bf (v)} As in (iii), the definition of `bounded variation' in
614I-614J refers to the lattice structure of $\Cal S$, but not to
the filtration.

\medskip

\quad{\bf (vi)} And now, as in (iv), the closure $\AMmo(\Cal S)$ of
$\AMbv(\Cal S)$ (615E) must be included in the closure $\BMmo(\Cal S)$ of
$\BMbv(\Cal S)$.

\medskip

{\bf (e)} Again, working through the definitions (\S613), we see that the
filtration is nowhere referred to, and that $\int_{\Cal S}\pmb u\,d\pmb v$
is determined by the lattice structure of $\Cal S$, the processes
$\pmb u$, $\pmb v\in(L^0)^{\Cal S}$ and the topology of convergence in
measure on $L^0$.

\medskip

{\bf (f)} We know that the capped-stake variation set
$B=Q^{(\Bbb A)}_{\Cal S}(d\pmb v)$, calculated with reference to the
family $\sfamilysS{\frak A}$, is topologically
bounded.   Now

\Centerline{$Q^{(\Bbb B)}_{\Cal S}(d\pmb v)
\subseteq\{z'\times\chi b+z''\times\chi(1\Bsetminus b):
z'$, $z''\in B\}$.}

\noindent\Prf\ If $\Cal S$ is empty, this is trivial.   Otherwise,
if $z\in Q^{(\Bbb B)}_{\Cal S}(d\pmb v)$, there are
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and $u_0,\ldots,u_{n-1}$ such that
$u_i\in L^0(\frak B_{\tau_i})$ and $\|u_i\|_{\infty}\le 1$ for each $i$,
and $z=\sum_{i=0}^{n-1}u_i\times(v_{\tau_{i+1}}-v_{\tau_i})$.
For each $i$, $\frak B_{\tau_i}$ is the algebra generated by
$\frak A_{\tau_i}\cup\{b\}$, so
there are $u'_i$, $u''_i\in L^0(\frak A_{\tau_i})$ such that
$u_i=u'_i\times\chi b+u''_i\times\chi(1\Bsetminus b)$ (647Cb),
and replacing these by $\med(-\chi 1,u'_i,\chi 1)$ and
$\med(-\chi 1,u''_i,\chi 1)$ if necessary, we can suppose that
$|u'_i|$, $|u''_i|\le\chi 1$.   In this case,

\Centerline{$z'=\sum_{i=0}^{n-1}u'_i\times(v_{\tau_{i+1}}-v_{\tau_i})$,
\quad$z''=\sum_{i=0}^{n-1}u''_i\times(v_{\tau_{i+1}}-v_{\tau_i})$}

\noindent belong to $B$ (616C(ii)).   And
$z=z'\times\chi b+z''\times\chi(1\Bsetminus b)$.\ \Qed

Since $z'\mapsto z'\times\chi b:L^0\to L^0$ is linear and continuous,
$\{z'\times\chi b:z'\in B\}$ is topologically bounded (3A5N(b-v)).
Similarly,
$\{z'\times\chi(1\Bsetminus b):z'\in B\}$ is topologically bounded;
so $Q^{(\Bbb B)}_{\Cal S}(d\pmb v)$ is included in the sum of two
topologically bounded sets and is topologically bounded, that is,
$\pmb v$ is a $\Bbb B$-integrator.

\medskip

{\bf (g)} Once again, the definition in 641E-641F depends only on the lattice
structure of $\Cal S$, the regions $\Bvalue{\sigma<\tau}$ for $\sigma$,
$\tau\in\Cal S$ and the topology of convergence in measure on $L^0$,
which are unchanged by the change in filtration.
}%end of proof of 647F

\leader{647G}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.

(a) $\family{t}{T}{\frak B_t}$ is right-continuous.

(b) $\Cal T_{\Bbb A}$ is an order-closed sublattice of $\Cal T_{\Bbb B}$.

(c) If $\tau\in\Cal T_{\Bbb B}$, there are $\sigma$,
$\sigma'\in\Cal T_{\Bbb A}$ such that $b\Bsubseteq\Bvalue{\tau=\sigma}$
and $1\Bsetminus b\Bsubseteq\Bvalue{\tau=\sigma'}$.
\cmmnt{In particular, }$\Cal T_{\Bbb A}$ covers $\Cal T_{\Bbb B}$.
%in fact finitely covers

\proof{{\bf (a)} If $t\in T$ is not
isolated on the right, take $c\in\bigcap_{s>t}\frak B_s$.   Then for every
$s>t$ there are $a_s$, $a'_s\in\frak A_s$ such that
$c\Bcap b=a_s\Bcap b$ and $c\Bsetminus b=a'_s\Bsetminus b$.
Set

\Centerline{$a=\sup_{s'>t}\inf_{t<s\le s'}a_s$,
\quad$a'=\sup_{s'>t}\inf_{t<s\le s'}a'_s$;}

\noindent then $a\in\bigcap_{s'>t}\frak A_s=\frak A_t$ and

$$\eqalignno{c\Bcap b
&=\sup_{s'>t}\inf_{t<s\le s'}(a_s\Bcap b)
=(\sup_{s'>t}\inf_{t<s\le s'}a_s)\Bcap b\cr
\displaycause{by the distributive laws 313B, as usual}
&=a\Bcap b,\cr}$$

\noindent and similarly $a'\in\frak A_t$ and
$c\Bsetminus b=a'\Bsetminus b$, so
$c=(a\Bcap b)\Bcup(a'\Bsetminus b)$ belongs to $\frak B_t$.   As $c$ and
$t$ are arbitrary, $\family{t}{T}{\frak B_t}$ is right-continuous.

\medskip

{\bf (b)} Use 611Cb in its full strength and 632C(a-i) to see that if
$A\subseteq\Cal T_{\Bbb A}$ is non-empty, we get formulae
defining $\sup A$ and $\inf A$ in both $\Cal T_{\Bbb A}$ and
$\Cal T_{\Bbb B}$.

\medskip

{\bf (c)} For $t\in T$, set
$a_t=\upr(\Bvalue{\tau>t}\Bcap b,\frak A_t)$;  then
$a_t\Bcap b=\Bvalue{\tau>t}\Bcap b$, by 647Ca, because
$\Bvalue{\tau>t}\in\frak B_t$.   If $s\le t$ in $T$, then

\Centerline{$\Bvalue{\tau>t}\Bcap b\Bsubseteq\Bvalue{\tau>s}\Bcap b
\Bsubseteq a_s\in\frak A_s\subseteq\frak A_t$,}

\noindent so $a_t\Bsubseteq a_s$.   If $t\in T$ is not isolated on the
right in $T$, set $a=\sup_{s>t}a_s$;  then
$a=\sup_{t<s\le s'}a_s\in\frak A_{s'}$ for every $s'>t$, so
(because the filtration is right-continuous) $a\in\frak A_t$,
while

\Centerline{$\Bvalue{\tau>t}\Bcap b=\sup_{s>t}\Bvalue{\tau>s}\Bcap b
\Bsubseteq a$}

\noindent so $a_t\Bsubseteq a$ and $a_t=\sup_{s>t}a_s$.
But this means that the conditions of 611A(b-i) are satisfied
by $\family{t}{T}{a_t}$ and
we have a $\sigma\in\Cal T_{\Bbb A}$ such that
$\Bvalue{\sigma>t}=a_t$ for every $t$.   Next, for each $t$,

\Centerline{$\Bvalue{\sigma>t}\Bcap b
=a_t\Bcap b=\Bvalue{\tau>t}\Bcap b$.}

\noindent So if we calculate $\Bvalue{\sigma<\tau}\Bcap b$
from the formulae in 611D, we see that it is

\Centerline{$
\sup_{t\in T}(b\Bcap\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})
=\sup_{t\in T}((\Bvalue{\tau>t}\Bcap b)\Bsetminus(\Bvalue{\sigma>t}\Bcap b))
=0$,}

\noindent and similarly $\Bvalue{\tau<\sigma}\Bcap b=0$, so
$b\Bsubseteq\Bvalue{\tau=\sigma}$.

Similarly, we have a $\sigma'\in\Cal T_{\Bbb A}$ defined by setting
$\Bvalue{\sigma'>t}=\upr(\Bvalue{\tau>t}\Bsetminus b,\frak A_t)$ for every
$t\in T$, and $1\Bsetminus b\Bsubseteq\Bvalue{\tau=\sigma'}$.
}%end of proof of 647G

\leader{647H}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, and that $\Cal S$ is an order-convex sublattice of
$\Cal T_{\Bbb A}$.   Let $\pmb v\in\Bbb B\Mnsu(\Cal S)$ be
$\|\,\|_{\infty}$-bounded.   Then there is a $\|\,\|_{\infty}$-bounded
$\pmb w\in\Bbb A\Mnsu(\Cal S)$
such that $\pmb v\preccurlyeq\pmb w$.

\cmmnt{\medskip

\noindent{\bf Remark} Perhaps I should make it clear that when I write
`$\Cal S$ is an order-convex sublattice of $\Cal T_{\Bbb A}$'
I mean that whenever $\tau$, $\tau'\in\Cal S$,
$\sigma\in\Cal T_{\Bbb A}$ and $\tau\le\sigma\le\tau'$, then
$\sigma\in\Cal S$.   We do not expect $\Cal S$ to be order-convex when
regarded as a sublattice of $\Cal T_{\Bbb B}$.
}

\proof{{\bf (a)} Express $\pmb v$ as $\family{\tau}{\Cal S}{v_{\tau}}$.
For the time being (down to the end of (f) below)
suppose that $\Bvalue{\pmb v\ne\tbf 0}\Bsubseteq b$.
Set $M=\|\pmb v\|_{\infty}$.   For $\tau\in\Cal S$ set
$\tilde v_{\tau}=\min\{v:v\in L^0(\frak A_{\tau})$, $v\ge v_{\tau}\}$;
this is
defined because $\{v:v\in L^0(\frak A_{\tau})$, $v\ge v_{\tau}\}$ is
non-empty (it contains $M\chi 1$, for instance) and downwards-directed, and
$L^0(\frak A_{\tau})$ is order-closed in $L^0$.

\medskip

{\bf (b)(i)} Of course
$0\le\tilde v_{\tau}\le M\chi(\upr(b,\frak A_{\tau}))$, so
$\Bvalue{\tilde v_{\tau}>0}\Bsubseteq\upr(b,\frak A_{\tau})$,
for every $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)}
Next, $\Bvalue{v_{\tau}=\tilde v_{\tau}}\Bsupseteq b$ for each $\tau$.
\Prf\ (Cf.\ 121I, 364Xp.)  If $\alpha\ge 0$, then
$\Bvalue{v_{\tau}>\alpha}$ belongs to $\frak B_{\tau}$ and is included in
$b$, so by 647Fb is of the form $b\Bcap a$ for some $a\in\frak A_{\tau}$.
In this case, $M\chi a+\alpha\chi(1\Bsetminus a)$ belongs to
$L^0(\frak A_{\tau})$ and is greater than or equal to $v_{\tau}$, so is
greater than or equal to $\tilde v_{\tau}$, and
$\Bvalue{\tilde v_{\tau}>\alpha}\Bsubseteq a$.   Consequently

\Centerline{$\Bvalue{\tilde v_{\tau}\times\chi b>\alpha}
\Bsubseteq b\Bcap a=\Bvalue{v_{\tau}>\alpha}
=\Bvalue{v_{\tau}\times\chi b>\alpha}$.}

\noindent Of course

\Centerline{$\Bvalue{\tilde v_{\tau}\times\chi b>\alpha}
=1=\Bvalue{v_{\tau}\times\chi b>\alpha}$}

\noindent for every $\alpha<0$, because $v_{\tau}\ge 0$.
As $\alpha$ is arbitrary,
$\tilde v_{\tau}\times\chi b\le v_{\tau}\times\chi b$.   Since
$v_{\tau}\le\tilde v_{\tau}$,
$\tilde v_{\tau}\times\chi b\le v_{\tau}\times\chi b$ and
$\tilde v_{\tau}\times\chi b=v_{\tau}\times\chi b$, that is,
$b\Bsubseteq\Bvalue{\tilde v_{\tau}=v_{\tau}}$.\ \Qed

\medskip

{\bf (c)} $\tildepmb v=\family{\tau}{\Cal S}{\tilde v_{\tau}}$ is
$\Bbb A$-fully adapted.   \Prf\ We arranged in (a) that
$\tilde v_{\tau}\in L^0(\frak A_{\tau})$ for every $\tau\in\Cal S$.
Suppose that $\sigma$, $\tau$ in $\Cal S$ and
$a=\Bvalue{\sigma=\tau}$.   Then $a\in\frak A_{\sigma}\cap\frak A_{\tau}$
(611H(c-i)) and $a\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau}}$.   Now

\Centerline{$\tilde v_{\sigma}\times\chi a
\ge v_{\sigma}\times\chi a
=v_{\tau}\times\chi a$}

\noindent so
$\tilde v_{\sigma}\times\chi a+M\chi(1\Bsetminus a)\ge v_{\tau}$;  at the
same time, $\chi(1\Bsetminus a)$ and
$\tilde v_{\sigma}\times\chi a
=\tilde v_{\sigma}\times\chi\Bvalue{\sigma\le\tau}\times\chi a$ belong
to $L^0(\frak A_{\tau})$ (612C), so
$\tilde v_{\sigma}\times\chi a+M\chi(1\Bsetminus a)
\in L^0(\frak A_{\tau})$ and
$\tilde v_{\sigma}\times\chi a+M\chi(1\Bsetminus a)\ge\tilde v_{\tau}$.
Thus $\tilde v_{\sigma}\times\chi a\ge\tilde v_{\tau}\times\chi a$.
Similarly, $\tilde v_{\tau}\times\chi a\ge\tilde v_{\sigma}\times\chi a$
and the two are equal, that is,
$a\Bsubseteq\Bvalue{\tilde v_{\sigma}=\tilde v_{\tau}}$.   As $\sigma$ and
$\tau$ are arbitrary, $\tildepmb v$ is fully adapted.\ \Qed

\medskip

{\bf (d)} $\tildepmb v$ is of bounded variation.   \Prf\
Suppose that $\tau_0\le\ldots\le\tau_n$ in $\Cal S$.
For $1\le k\le n$, set
$a_k=\Bvalue{\tilde v_{\tau_k}<\tilde v_{\tau_{k-1}}}$.   If $1\le k\le i\le n$,
then $a_k\in\frak A_{\tau_i}$ and

\Centerline{$a_k\Bcap b
\Bsubseteq\Bvalue{\tilde v_{\tau_k}=v_{\tau_k}}
  \Bcap\Bvalue{\tilde v_{\tau_k}<\tilde v_{\tau_{k-1}}}
  \Bcap\Bvalue{\tilde v_{\tau_{k-1}}=v_{\tau_{k-1}}}
\Bsubseteq\Bvalue{v_{\tau_k}<v_{\tau_{k-1}}}
=0$,}

\noindent so $a_k\Bcap\upr(b,\frak A_{\tau_i})=0$ and
$a_k\Bsubseteq\Bvalue{\tilde v_{\tau_i}=0}$;  consequently
$a_k\Bcap a_i=0$ if $k<i$.   Turning this round, we see that
$a_k\Bcap a_i=0$ if $1\le i<k$, so that

$$\eqalign{a_k
&\Bsubseteq\inf_{1\le i<k}\Bvalue{\tilde v_{\tau_{i-1}}\le\tilde v_{\tau_i}}
\Bsubseteq\Bvalue{\sum_{i=1}^{k-1}|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
   =\tilde v_{\tau_{k-1}}-\tilde v_{\tau_0}}
\cr&
\Bsubseteq\Bvalue{\sum_{i=1}^{k-1}|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
   \le M\chi 1},\cr}$$

\noindent \noindent while on the other side

\Centerline{$a_k
\Bsubseteq\inf_{k\le i\le n}\Bvalue{\tilde v_{\tau_i}=0}
\Bsubseteq\Bvalue{\sum_{i=k+1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
  =0}$.}

\noindent Putting these together,

\Centerline{$a_k
\Bsubseteq\Bvalue{\sum_{i=1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
  \le M\chi 1+|\tilde v_{\tau_k}-\tilde v_{\tau_{k-1}}|}
\Bsubseteq\Bvalue{\sum_{i=1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
  \le 2M\chi 1}$.}

On the other hand,

$$\eqalign{1\Bsetminus\sup_{1\le i\le n}a_i
&\Bsubseteq\inf_{1\le i\le n}
  \Bvalue{\tilde v_{\tau_{i-1}}\le\tilde v_{\tau_i}}
\Bsubseteq\Bvalue{\sum_{i=1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
  =\tilde v_{\tau_n}-\tilde v_{\tau_0}}
\cr&
\Bsubseteq\Bvalue{\sum_{i=1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
  \le M\chi 1},\cr}$$

\noindent so in fact we have

\Centerline{$\Bvalue{\sum_{i=1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|
  \le 2M\chi 1}=1$}

\noindent
and $\sum_{i=1}^n|\tilde v_{\tau_i}-\tilde v_{\tau_{i-1}}|\le 2M\chi 1$.
As $\tau_0,\ldots,\tau_n$ are arbitrary,
$\tildepmb v$ is of bounded variation, with
$\int_{\Cal S}|d\tildepmb v|\le 2M\chi 1$.\ \Qed

\medskip

{\bf (e)} $\tildepmb v$ is $\Bbb A$-near-simple.   \Prf\
Suppose that $\tau_0\le\tau_1$ in $\Cal S$.
If $A\subseteq\Cal S\cap[\tau_0,\tau_1]$ is non-empty and downwards-directed and
has infimum $\tau$ in $\Cal T_{\Bbb A}$, then $\tau\in A$ because $A$ is
order-convex in $\Cal T_{\Bbb A}$.   We know that
$v=\lim_{\sigma\downarrow A}\tilde v_{\sigma}$ is defined, because
$\tildepmb v$ is of $\Bbb A$-bounded variation, therefore
$\Bbb A$-moderately-oscillatory (616Ra).

Of course
$v\ge 0$.   Now $v\in\bigcap_{\sigma\in A}L^0(\frak A_{\sigma})
=L^0(\frak A_{\tau})$ (632C(a-iii)), while

\Centerline{$v\ge v\times\chi b
=\lim_{\sigma\downarrow A}\tilde v_{\sigma}\times\chi b
=\lim_{\sigma\downarrow A}v_{\sigma}
=v_{\tau}$}

\noindent by 632F, because $\pmb v$ is $\Bbb B$-near-simple and
$\Cal T_{\Bbb A}$ is order-closed in $\Cal T_{\Bbb B}$, so
$\tau$ is still the infimum of $A$ in $\Cal T_{\Bbb B}$.
It follows that $\tilde v_{\tau}\le v$, while

\Centerline{$b\Bsubseteq\Bvalue{v=v_{\tau}}\Bcap\Bvalue{v_{\tau}=\tilde v_{\tau}}
\Bsubseteq\Bvalue{v=\tilde v_{\tau}}$.}

Set $a=\Bvalue{\tilde v_{\tau}<v}\Bsubseteq\Bvalue{v>0}$.
Then $a\in\frak A_{\tau}$ and
$a\Bcap b=0$.   But this means that, for every $\sigma\in A$,
$a$ is disjoint from $\upr(b,\frak A_{\sigma})$ and
$\tilde v_{\sigma}\times\chi a=0$.   Consequently
$v\times\chi a=\lim_{\sigma\downarrow A}\tilde v_{\sigma}\times\chi a
=0$, and $a$ must be $0$.   Thus $v\le\tilde v_{\tau}$ and
$v=\tilde v_{\tau}$.   As $\tildepmb v\restr\Cal S\cap[\tau_0,\tau_1]$
is $\Bbb A$-moderately-oscillatory,
it is $\Bbb A$-locally-near-simple, by 632F again, and in fact
$\Bbb A$-near-simple,
since $\max(\Cal S\cap[\tau_0,\tau_1])=\tau_1$
belongs to $\Cal S\cap[\tau_0,\tau_1]$.

Since the whole process $\tildepmb v$ is
$\Bbb A$-moderately-oscillatory, it is $\Bbb A$-near-simple, by 631Fc.\ \Qed

\medskip

{\bf (f)} Setting
$w_{\tau}=\int_{\Cal S\wedge\tau}|d\tildepmb v|$ for $\tau\in\Cal S$,
$\pmb w=\family{\tau}{\Cal S}{w_{\tau}}$ is
$\Bbb A$-near-simple (631K), and of course it is non-negative and
non-decreasing;
moreover, $\|\pmb w\|_{\infty}=\|\int_{\Cal S}|d\tildepmb v|\|_{\infty}$
is finite.   If $\sigma\le\tau$ in $\Cal S$, then

\Centerline{$b
\Bsubseteq\Bvalue{v_{\tau}-v_{\sigma}=\tilde v_{\tau}-\tilde v_{\sigma}}$,
\quad$1\Bsetminus b
\Bsubseteq\Bvalue{v_{\tau}-v_{\sigma}=0}$}

\noindent so

\Centerline{$v_{\tau}-v_{\sigma}\le|\tilde v_{\tau}-\tilde v_{\sigma}|
\le w_{\tau}-w_{\sigma}$;}

\noindent as $\sigma$ and $\tau$ are arbitrary,
$\pmb v\preccurlyeq\pmb w$.

\medskip

{\bf (g)} This deals with the case when $\Bvalue{\pmb v\ne\tbf 0}\Bsubseteq b$.
Of course just the same arguments apply when
$\Bvalue{\pmb v\ne\tbf 0}\Bsubseteq 1\Bsetminus b$.   For the general case,
consider
$\pmb v'=(\chi b)\pmb v=\family{\tau}{\Cal S}{\chi b\times v_{\tau}}$
and $\pmb v''=\pmb v-\pmb v'$.   These are both
$\Bbb B$-near-simple non-decreasing
non-negative processes bounded above by a multiple of $\chi 1$,
while $\Bvalue{\pmb v'\ne\tbf 0}\Bsubseteq b$ and
$\Bvalue{\pmb v''\ne\tbf 0}\Bsubseteq 1\Bsetminus b$.
So we have $\|\,\|_{\infty}$-bounded processes
$\pmb w'$, $\pmb w''\in\Bbb A\Mnsu(\Cal S)$
such that $\pmb v'\preccurlyeq\pmb w'$ and
$\pmb v''\preccurlyeq\pmb w''$.   Setting
$\pmb w=\pmb w'+\pmb w''$, $\pmb w\in\Bbb A\Mnsu(\Cal S)$
is $\|\,\|_{\infty}$-bounded and $\pmb v\preccurlyeq\pmb w$, as required.
}%end of proof of 647H

\leader{647I}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex
sublattice of $\Cal T_{\Bbb A}$, and
$\hat\Cal S^{(\Bbb B)}$ the covered envelope of $\Cal S$ in
$\Cal T_{\Bbb B}$.

(a)(i) $\hat\Cal S^{(\Bbb B)}$ is order-convex in $\Cal T_{\Bbb B}$.

\quad(ii) For any $\Bbb A$-fully adapted process $\pmb x$ with domain
$\Cal S$, there is a unique $\Bbb B$-fully adapted process $\hatpmb x$
with domain $\hat\Cal S^{(\Bbb B)}$ extending $\pmb x$.

(b)(i) If $\pmb x\in\Bbb A\Mob(\Cal S)$ then
$\hatpmb x\in\Bbb B\Mob(\hat\Cal S^{(\Bbb B)})$ and
$\sup|\hatpmb x|=\sup|\pmb x|$.  %647F(d-iii)

\quad(ii) If $\pmb v\in\Bbb A\Mns(\Cal S)$ then
$\hatpmb v\in\Bbb B\Mns(\hat\Cal S^{(\Bbb B)})$.   %647F(d-iv)

\quad(iii) If $\pmb v\in\Bbb A\Mbv(\Cal S)$ then
$\hatpmb v\in\Bbb B\Mbv(\hat\Cal S^{(\Bbb B)})$.  %647F(d-v)

\quad(iv) If $\pmb v$ is an $\Bbb A$-integrator with domain $\Cal S$, then
$\hatpmb v$ is a $\Bbb B$-integrator.   %647Ff

\quad(v) If $\pmb u\in\Bbb A\Mmo(\Cal S)$ then
$\hatpmb u\in\Bbb B\Mmo(\hat\Cal S^{(\Bbb B)})$  %647F(d-vi)
and the previsible version of $\hatpmb u$ is the image of the previsible
version of $\pmb u$.  %647Fg

(c)(i) If $\pmb w\in\Bbb A\Mpob(\Cal S)$, then
$\hatpmb w\in\Bbb B\Mpob(\hat\Cal S^{(\Bbb B)})$.

\quad(ii) The map $\pmb w\mapsto\hatpmb w:
\Bbb A\Mpob(\Cal S)\to\Bbb B\Mpob(\hat\Cal S^{(\Bbb B)})$ is continuous for the
S-integration topologies $\Bbb A\TSi$ and $\Bbb B\TSi$.

\quad(iii) If $\pmb x\in\AMSi(\Cal S)$, then
$\hatpmb x\in\BMSi(\hat\Cal S^{(\Bbb B)})$.

(d) If $\pmb x\in\AMSi(\Cal S)$, then
$\BSint_{\hat\Cal S^{(\Bbb B)}}\hatpmb x\,d\hatpmb v
=\ASint_{\Cal S}\pmb x\,d\pmb v$ for
every $\Bbb A$-integrator $\pmb v\in\AMns(\Cal S)$.

\proof{{\bf (a)(i)} Suppose that $\tau$, $\tau'\in\hat\Cal S^{(\Bbb B)}$ and
$\tilde\tau\in\Cal T_{\Bbb B}$ are such that $\tau\le\tilde\tau\le\tau'$.
Take $a\in\frak A\setminus\{0\}$.   Because $\Cal T_{\Bbb A}$ covers
$\Cal T_{\Bbb B}$ (647Gc), there are
$\sigma$, $\sigma'\in\Cal S$ and $\tilde\sigma\in\Cal T_{\Bbb A}$ such that
$a'=a\Bcap\Bvalue{\sigma=\tau}\Bcap\Bvalue{\sigma'=\tau'}
\Bcap\Bvalue{\tilde\sigma=\tilde\tau}$ is non-zero.   Now
$\med(\sigma,\tilde\sigma,\sigma')\in\Cal S$, because $\Cal S$ is
order-convex in $\Cal T_{\Bbb A}$ and
$\sigma\wedge\sigma'\le\med(\sigma,\tilde\sigma,\sigma')
\le\sigma\vee\sigma'$, and

\Centerline{$0\ne a'\Bsubseteq a\Bcap\Bvalue{\med(\tau,\tilde\tau,\tau')
=\med(\sigma,\tilde\sigma,\sigma')}
\Bsubseteq a\Bcap\Bvalue{\tilde\tau=\med(\sigma,\tilde\sigma,\sigma')}$.}

\noindent As $a$ is arbitrary, $\tilde\tau\in\hat\Cal S^{(\Bbb B)}$;  thus
$\hat\Cal S^{(\Bbb B)}$ is order-convex in $\Cal T_{\Bbb B}$.

\medskip

\quad{\bf (ii)} We just have to remember that $\pmb u$ is $\Bbb B$-fully
adapted, as noted in 647F(d-i), and apply 612Qa.

\medskip

{\bf (b)(i)} By 647F(d-iii), $\pmb x$ is $\Bbb B$-order-bounded, so
$\hatpmb x$ is $\Bbb B$-order-bounded, by 614Ga.

\medskip

\quad{\bf (ii)} Argue as in (i), but using 647F(d-iv) for the first step and
631Ga for the second.

\medskip

\quad{\bf (iii)} Use 647F(d-v) and 614Q(a-iv-$\beta$).

\medskip

\quad{\bf (iv)} 647Ff and 616I.

\medskip

\quad{\bf (v)} By 647F(d-vi) and 615F(a-vi), $\hatpmb u$ is
$\Bbb B$-moderately-oscillatory.   Concerning its previsible version
$\hatpmb u_<$, we know from 641G(a-v), applied in $\Bbb B$,
that $\hatpmb u_<\restr\Cal S$ is the
previsible version of $\pmb u\restr\Cal S$, which is the same whether
calculated in $\Bbb A$ or $\Bbb B$, as noted in 647Fg;  so
$\hatpmb u_<$ must be the $\Bbb B$-fully-adapted extension of the
$\Bbb A$-previsible version $\pmb u_<$ of $\pmb u$, that is.
$\hatpmb u_<$ is the image of $\pmb u_<$ in
$\Bbb B\Mfa(\hat\Cal S^{(\Bbb B)})$.

\medskip

{\bf (c)(i)} $\pmb w$ is $\Bbb A$-fully adapted, so $\hatpmb w$ is
$\Bbb B$-fully adapted, by (a-ii) just above.   Let
$\pmb u\in\AMmo(\Cal S)^+$ be such that $|\pmb w|\le\pmb u_<$, and
consider $\hatpmb u$.   We saw in (b-v) that, writing $\hatpmb u_<$ for the
previsible version of $\hatpmb u$, we have
$\hatpmb u_<\restr\Cal S=\pmb u_<$, so
$0=(|\pmb w|-\pmb u_<)^+=(|\hatpmb w|-\hatpmb u_<)^+\restr\Cal S$ (612Qb),
and $|\hatpmb w|\le\hatpmb u_<$.
Thus $\hatpmb w\in\Bbb B\Mpob(\Cal S)$, as claimed.

\medskip

\quad{\bf (ii)} Let $G$ be a $\BTSi$-neighbourhood of $\tbf 0$ in
$\Bbb B\Mpob(\hat\Cal S^{(\Bbb B)})$.   Then there are a
$\|\,\|_{\infty}$-bounded
$\pmb v\in\Bbb B\Mnsu(\hat\Cal S^{(\Bbb B)})$ and an $\epsilon>0$
such that $\pmb w\in G$ whenever
$\pmb w\in\Bbb B\Mpob(\hat\Cal S^{(\Bbb B)})$ and
$\Bbb B\widehat{\theta}^{\#}_{\pmb v}(\pmb w)\le\epsilon$ (put 645E(a-ii)
and 647B together).   Now $\hatpmb v$ is $\Bbb B$-near-simple ((b-ii) above)
and is non-negative and non-decreasing (614Q(a-iv-$\alpha$)) and
bounded above by a constant process, so by 647H there is a
$\tildepmb v\in\Bbb A\Mnsu(\Cal S)$
such that $\pmb v\restr\Cal S\preccurlyeq\tildepmb v$.
In this case, $G'=\{\pmb w:\pmb w\in\Bbb A\Mpob(\Cal S)$,
$\Bbb A\widehat{\theta}^{\#}_{\tildepmb v}(\pmb w)<\epsilon\}$ is
an $\ATSi$-neighbourhood of $\tbf 0$ in $\Bbb A\Mpob(\Cal S)$.
If $\pmb w\in G'$, $\hatpmb w\in G$.   \Prf\ There is a non-decreasing
uniformly order-bounded
sequence $\sequencen{\pmb u^{(n)}}$ in $\AMmo(\Cal S)^+$ such that
$|\pmb w|\le\sup_{n\in\Bbb N}\pmb u^{(n)}_<$ and
$\theta(\int_{\Cal S}\pmb u^{(n)}d\tildepmb v)\le\epsilon$
for every $n\in\Bbb N$.
Because $\pmb v\restr\Cal S\preccurlyeq\tildepmb v$,

$$\eqalignno{\theta(\int_{\hat\Cal S^{(\Bbb B)}}\hatpmb u^{(n)}d\pmb v)&
=\theta(\int_{\Cal S}\pmb u^{(n)}d\pmb v)
\Displaycause{613T, work8ing in $\Bbb B$}
\le\theta(\int_{\Cal S}\pmb u^{(n)}d\tildepmb v)
\Displaycause{645Bc, working in $\Bbb A$}
\le\epsilon\cr}$$

\noindent for every $n$.   Next, I said in 645Bb that the supremum
$\sup_{n\in\Bbb N}\pmb u^{(n)}_<$ is to be taken in $(L^0)^{\Cal S}$;
but of course the space $\Bbb B\Mfa(\Cal S)$
of $\Bbb B$-fully adapted processes with domain $\Cal S$ is order-closed
in $(L^0)^{\Cal S}$ (612Ia), so
$\sup_{n\in\Bbb N}\pmb u^{(n)}_<\in\Bbb B\Mfa(\Cal S)$ and is the
supremum of $\{\pmb u^{(n)}_<:n\in\Bbb N\}$ in $\Bbb B\Mfa(\Cal S)$.
Now the map $\pmb u\mapsto\hatpmb u$ is an order-isomorphism between
$\Bbb B\Mfa(\Cal S)$ and the space
$\Bbb B\Mfa(\hat\Cal S^{(\Bbb B)})$ of
$\Bbb B$-fully adapted processes with domain $\hat\Cal S^{(\Bbb B)}$
(612Qb again), and
$\sup_{n\in\Bbb N}(\pmb u^{(n)}_<)\sphat\
=(\sup_{n\in\Bbb N}\pmb u^{(n)}_<)\sphat\ $.   Consequently

\Centerline{$|\hatpmb w|
\le(\sup_{n\in\Bbb N}\pmb u^{(n)}_<)\sphat\
=\sup_{n\in\Bbb N}(\pmb u^{(n)}_<)\sphat\
=\sup_{n\in\Bbb N}\hatpmb u^{(n)}_<,$}

\noindent while $\sequencen{\hatpmb u^{(n)}}$ is a non-decreasing
uniformly order-bounded sequence
in $\BMmo(\hat\Cal S^{(\Bbb B)})^+$.   Accordingly
$\Bbb B\widehat{\theta}^{\#}_{\pmb v}(\hatpmb w)\le\epsilon$
and $\hatpmb w\in G$.\ \Qed

As $G$ is arbitrary, $\pmb w\mapsto\hatpmb w$ is continuous at $0$;  being
a linear operator, it is continuous.

\medskip

\quad{\bf (iii)}\grheada\
Suppose to begin with that $\pmb x\in\AMSizero(\Cal S)$.
Because $\hatpmb u\in\BMmo(\hat\Cal S^{(\Bbb B)})$ and
$(\pmb u_<)\sphat\ =(\hatpmb u)_<$
for every $\pmb u\in\AMmo(\Cal S)$, and
$\pmb y\mapsto\hatpmb y:
\Bbb A\Mpob(\Cal S)\to\Bbb B\Mpob(\hat\Cal S^{(\Bbb B)})$ is continuous,
$\hatpmb x$ will belong to

\Centerline{$\overline{\{\hatpmb u_<:\pmb u\in\AMmo(\Cal S)\}}
\subseteq\overline{\{\pmb u'_<:\pmb u'\in\BMmo(\hat\Cal S^{(\Bbb B)})\}}
=\BMSizero(\hat\Cal S^{(\Bbb B)})$.}

\qquad\grheadb\ Now take any $\pmb x\in\AMSi(\Cal S)$.   Then
$\pmb x\times\tbfoneminus{\Cal S}\in\AMSizero(\Cal S)$, and
$(\pmb x\times\tbfoneminus{\Cal S})\mskip-1.5mu\hat{\phantom{x}}
\in\BMSizero(\hat\Cal S^{(\Bbb B)})$.  The point is that
$\tbfoneminus{\Cal S}=\tbfoneminus{\hat\Cal S^{(\Bbb B)}}\restr\Cal S$, just
because $\tbf 1^{(\hat\Cal S^{(\Bbb B)})}
=(\tbf 1^{(\Cal S)})\mskip-1.5mu\hat{\phantom{x}}$.
It follows that $\hatpmb x\times\tbfoneminus{\hat\Cal S^{(\Bbb B)}}$ is a
$\Bbb B$-fully adapted process with domain $\hat\Cal S^{(\Bbb B)}$ extending
$\pmb x\times\tbfoneminus{\Cal S}$ and must be equal to
$(\pmb x\times\tbfoneminus{\Cal S})\mskip-1.5mu\hat{\phantom{x}}
\in\BMSizero(\hat\Cal S^{(\Bbb B)})$.   Consequently
$\hatpmb x\in\BMSi(\hat\Cal S^{(\Bbb B)})$ in
this case too.

\medskip

{\bf (d)(i)} Because
$\hat\Cal S^{(\Bbb B)}$ is order-convex in $\Cal T_{\Bbb B}$, by (a-i),
and $\family{t}{T}{\frak B_t}$ is right-continuous (647Ga),
we can speak of both S-integrals (645P).
Write $z$ for $\BSint_{\hat\Cal S^{(\Bbb B)}}\hatpmb x\,d\hatpmb v$ and
$z'$ for $\ASint_{\Cal S}\pmb x\,d\pmb v$.

\medskip

\quad{\bf (ii)} Suppose to begin with that $\pmb x\in\AMSizero(\Cal S)$.
Let $\epsilon>0$.
We know that there is a $\pmb u^*\in\AMmo(\Cal S)^+$ such that
$|\pmb x|\le\pmb u^*_<$,
in which case $\hatpmb u^*\in\BMmo(\hat\Cal S^{(\Bbb B)})^+$
and $|\hatpmb x|\le\hatpmb u^*_<$.   There is a neighbourhood
$G$ of $\hatpmb x$ in $\BMSizero(\hat\Cal S^{(\Bbb B)})$ such that
$\theta(z-\int_{\hat\Cal S^{(\Bbb B)}}\tildepmb u\,d\hatpmb v)\le\epsilon$
whenever $\tildepmb u\in\BMmo(\hat\Cal S^{(\Bbb B)})$,
$|\tildepmb u|\le\hatpmb u^*$ and $\tildepmb u_<\in G$;
similarly, there is a neighbourhood $G'$ of
$\pmb x$ in in $\AMSizero(\Cal S)$ such that
$\theta(z'-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$
whenever $\pmb u\in\AMmo(\hat\Cal S^{(\Bbb B)})$,
$|\pmb u|\le\pmb u^*$ and $\pmb u_<\in G'$.  And since the
map $\pmb x'\mapsto\hatpmb x'$ is continuous, we can suppose
that $\hatpmb x'\in G$ whenever $\pmb x'\in G'$.

By 645La, there is a $\pmb u\in\AMmo(\Cal S)$ such that
$|\pmb u|\le\pmb u^*$ and $\pmb u_<\in G'$.   In this case,
$\theta(z-\int_{\hat\Cal S^{(\Bbb B)}}\hatpmb u\,d\hatpmb v)\le\epsilon$ and
$\theta(z'-\int_{\Cal S}\pmb u\,d\pmb v)\le\epsilon$, while
$\int_{\Cal S}\pmb u\,d\pmb v
=\int_{\hat\Cal S^{(\Bbb B)}}\hatpmb u\,d\hatpmb v$ by 613T.
So $\theta(z-z')\le 2\epsilon$.   As $\epsilon$ is arbitrary,
$z=z'$, as required.

\medskip

\quad{\bf (iii)} And of course we now have

\Centerline{$\BSint_{\hat\Cal S^{(\Bbb B)}}\hatpmb x\,d\hatpmb v
=\BSint_{\hat\Cal S^{(\Bbb B)}}
   \hatpmb x\times\tbfoneminus{\hat\Cal S^{(\Bbb B)}}\,d\hatpmb v
=\ASint_{\Cal S}\pmb x\times\tbfoneminus{\Cal S}\,d\hatpmb v
=\ASint_{\Cal S}\pmb x\,d\hatpmb v$}

\noindent for every $\pmb x\in\AMSi(\Cal S)$.
}%end of proof of 647I

\leader{647J}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and $\Cal S$ is an order-convex sublattice of
$\Cal T$.   If $\pmb w$ is an S-integrable process
with domain $\Cal S$, and $\pmb v$ is a near-simple integrator with domain
$\Cal S$, then
$\Bvalue{\Sint_{\Cal S}\pmb w\,d\pmb v\ne 0}
\Bsubseteq\Bvalue{\pmb w\ne\tbf 0}$.

\proof{ I stopped speaking of $\Cal T_{\Bbb A}$ and $\AMSizero$ and
$\ASint$ because I wish to regard this result as a fact about
S-integration expressible in the language of \S\S645-646.   But of course
the idea is to apply the ideas of 647D-647I,
%647D 647E 647F 647G 647H 647I
so I immediately set $b=\Bvalue{\pmb w\ne\tbf 0}$ and move to the structures
considered in 647D-647E.   By 647Id,

\Centerline{$\Sint_{\Cal S}\pmb w\,d\pmb v
=\ASint_{\Cal S}\pmb w\,d\pmb v
=\BSint_{\hat\Cal S^{(\Bbb B)}}\hatpmb w\,d\hatpmb v$,}

\noindent where $\hat\Cal S^{(\Bbb B)}$ is the covered envelope of $\Cal S$ in
$\Cal T_{\Bbb B}$ and $\pmb u\mapsto\hatpmb u$
is the canonical isomorphism
between $\BMfa(\Cal S)$ and $\BMfa(\hat\Cal S^{(\Bbb B)})$
described in 612Qb.   Now $\Bvalue{\hatpmb w\ne\tbf 0}=\Bvalue{\pmb w\ne\tbf 0}$
(612S(c-ii)).   At the same time,
$\chi b\in L^0(\frak B_{\min\hat\Cal S^{(\Bbb B)}})$.   So

\Centerline{$\BSint_{\hat\Cal S^{(\Bbb B)}}\hatpmb w\,d\hatpmb v
=\BSint_{\hat\Cal S^{(\Bbb B)}}(\chi b)\hatpmb w\,d\hatpmb v
=\chi b\times\BSint_{\hat\Cal S^{(\Bbb B)}}\hatpmb w\,d\hatpmb v$}

\noindent(646D), and

\Centerline{$\Bvalue{\ASint_{\Cal S}\pmb w\,d\pmb v\ne 0}
=\Bvalue{\chi b\times\BSint_{\Cal S^{(\Bbb B)}}\hatpmb w\,d\hatpmb v\ne 0}
\Bsubseteq b$,}

\noindent as claimed.
}%end of proof of 647J

%\leader{647L}{}
%\leader{647M}{}
%\leader{647N}{}
%\leader{647O}{}
%\leader{647P}{}
%\leader{647Q}{}
%\leader{647R}{}
%\leader{647S}{}
%\leader{647T}{}
%\leader{647U}{}
%\leader{647V}{}
%\leader{647W}{}

\exercises{\leader{647X}{Basic exercises (a)}
%\spheader 647Xa
Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous and that $\Cal S$ is an order-convex
sublattice of $\Cal T$.   Let $\pmb v$ be a
near-simple process of bounded variation and $\pmb w$ an
S-integrable process, both with domain $\Cal S$.
Show that
$\int_{\Cal S}|d(\Siiv(\pmb w))|
\le\sup|\pmb w|\times\int_{\Cal S}|d\pmb v|$.
\Hint{apply 647J to $\pmb w-\med(-M\tbf 1_<,\pmb w,M\tbf 1_<)$ for
$M\ge 0$.}
%647J

%\leader{647Y}{Further exercises (a)}

}%end of exercises


\endnotes{
\Notesheader{647} In \S646 we found that most of the standard properties of the
Riemann-sum integral transferred without much trouble to the
S-integral;  only the result
$\int_{\Cal S}=\int_{\Cal S\wedge\tau}+\int_{\Cal S\vee\tau}$ gave
difficulty, and this was attributable to a technical problem at the
least member $\tau$ of $\Cal S\vee\tau$.   We did of course need to know a
little more about martingales (646P), and this fact seems to depend on the
fundamental theorem of martingales.   However,
there was a curious gap when we came to look at
$\Bvalue{\Sint\pmb w\,d\pmb v\ne 0}$.   A straightforward calculation
(613Jd) showed that, for the Riemann-sum integral,
$\Bvalue{\int\pmb u\,d\pmb v\ne 0}
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\pmb v\ne\tbf 0}$.   There was no
difficulty in showing that
$\Bvalue{\Sint\pmb w\,d\pmb v\ne 0}\Bsubseteq\Bvalue{\pmb v\ne\tbf 0}$ (646C).
But it seems to be much harder to confirm that
\penalty-150$\Bvalue{\Sint\pmb w\,d\pmb v\ne 0}
\penalty-100\Bsubseteq\Bvalue{\pmb w\ne\tbf 0}$ (647J).

}%end of notes

\discrpage



\frfilename{mt648.tex}
\versiondate{4.5.20/29.9.23}
\copyrightdate{2020}

\def\chaptername{The S-integral}
\def\sectionname{Changing the algebra II}
%\def\arctanh{\mathop{\text{arctanh}}\nolimits}

\newsection{648}

In \S634, I looked at questions involving pairs $(\frak A,\frak B)$ where
$\frak B$ is a closed subalgebra of $\frak A$, and the corresponding stochastic
integration structures $\Bbb A$ and $\Bbb B$.
In particular, we can relate Riemann-sum integrals
calculated in the two structures (634Eg).   Unsurprisingly, there is a
corresponding result for S-integration (648G), though it seems to need a good
deal more work.

\leader{648A}{Notation} As usual,
$\Bbb A=(\frak A,\bar\mu,T,\langle\frak A_t\rangle_{t\in T},
\Cal T_{\Bbb A},\family{\sigma}{\Cal T_{\Bbb A}}{\frak A_{\sigma}})$ will be a
stochastic
integration structure.   For nearly the whole section, we shall have a closed
subalgebra $\frak B$ of $\frak A$ with the associated structure
$\Bbb B=(\frak B,\bar\mu\restrp\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ as in
634C.   As in \S634, I will use formulations like `$\pmb x$ is
$\Bbb A$-previsibly-order-bounded', `$\pmb v'\in\Bbb B\Mns(\Cal S')$',
`$\ASint_{\Cal S}\pmb x\,d\pmb v$' to
indicate which structure is being considered at any particular moment.

If $E\subseteq\Bbb R$ is a Borel set and $h:E\to\Bbb R$ is a Borel
measurable function, $\bar h$ is the corresponding function from
$\{u:u\in L^0(\frak A)$, $\Bvalue{u\in E}=1\}$ to $L^0(\frak A)$ (612Ac).
If $\pmb u$ is a moderately oscillatory process, $\pmb u_<$ will denote its
previsible version (641F).   I use the symbol $\int$ for Riemann-sum integrals
(613H, 613L) and $\Sint$ for S-integrals (645P);  $\Expn$ will be the ordinary
integral on $L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$, and $\theta$ the associated
F-norm on $L^0(\frak A)$ (613Ba).   The F-seminorms
$\widehat{\theta}^{\#}_{\pmb v}$ will be those of 645B.   $S_I(\pmb u,d\pmb v)$
will be the Riemann sum (613Fb).   $\tbf 1^{(\Cal S)}$ will be the constant
process on $\Cal S$ with value $\chi 1$.

We shall have the usual spaces of processes:
$\Mfa$ (fully adapted, 612I),
$\Mmo$  (moderately oscillatory, 614O),
$\Mns$ (near-simple, 631Ba),
$\Mpob$ (previsibly order-bounded, 645Ba),
$\Mnsu$ (non-decreasing non-negative near-simple, 644Bb),
$\MSi^0$ and $\MSi$ (S-integrable, 645F).
If $\Cal S$ is a sublattice of $\Cal T_{\Bbb A}$, $\Cal I(\Cal S)$ will be the
directed set of finite sublattices of $\Cal S$.

\leader{648B}{Lemma} Let $E$ be a Borel subset of $\Bbb R$;  write $Q_E$ for
$\{u:u\in L^0(\frak A)$, $\Bvalue{u\in E}=1\}$.   Let $h:E\to\Bbb R$ be a
continuous function.   Suppose that $\Cal S$ is a finitely full sublattice of
$\Cal T_{\Bbb A}$ and $\pmb u=\sfamilysS{u}$ a moderately oscillatory process
such that
$\overline{\{u_{\sigma}:\sigma\in\Cal S\}}\subseteq Q_E$, the closure being for
the topology of convergence in measure on $L^0(\frak A)$.   Then
$\bar h\pmb u=\family{\sigma}{\Cal S}{\bar h(u_{\sigma})}$ is a moderately
oscillatory process.

\proof{ (Compare 615F(a-ii).)

\medskip

{\bf (a)(i)} If $\sigma\in\Cal S$ and $\alpha\in\Bbb R$,
$\Bvalue{\bar h(u_{\sigma})>\alpha}
=\Bvalue{u_{\sigma}\in h^{-1}[\,\ooint{\alpha,\infty}\,]}\in\frak A_{\sigma}$, so
$\bar h(u_{\sigma})\in L^0(\frak A_{\sigma})$.

\medskip

\quad{\bf (ii)} If $\sigma$, $\tau\in\Cal S$,

\Centerline{$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}
\Bsubseteq\Bvalue{\bar h(u_{\sigma})=\bar h(u_{\tau})}$}

\noindent by 612A(d-iii).   Thus $\bar h\pmb u$ is fully adapted.

\medskip

{\bf (b)} If $\sequencen{\sigma_n}$ is a monotonic sequence in $\Cal S$,
$u=\lim_{n\to\infty}u_{\sigma_n}$ is defined in $L^0(\frak A)$ and belongs to
$\overline{\{u_{\sigma}:\sigma\in\Cal S\}}\subseteq Q_E$.   Since
$\bar h:Q_E\to L^0(\frak A)$ is continuous (367S/613Bb),
$\lim_{n\to\infty}\bar h(u_{\sigma_n})=\bar h(u)$ is defined.   As $\Cal S$ is
finitely full, this is enough to ensure that $\bar h\pmb u$ is moderately
oscillatory (615N(iii)).
}%end of proof of 648B

\leader{648C}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T_{\Bbb A}$,
$\pmb w$ a $\|\,\|_2$-bounded martingale with domain $\Cal S$, and
$\pmb x$ a $\|\,\|_{\infty}$-bounded S-integrable process with domain $\Cal S$.
If $\pmb w^*$ is the quadratic variation of $\pmb w$,
$\|(\Sint_{\Cal S}\pmb x\,d\pmb w)^2\|_1
\le\|\Sint_{\Cal S}\pmb x^2d\pmb w^*\|_1
<\infty$.

%\query maybe equal?

\proof{ This is trivial if $\Cal S=\emptyset$,
so suppose otherwise.   Express $\pmb w$ as $\sfamilysS{w}$.

\medskip

{\bf (a)} I had better run through the check that the S-integrals are defined.
Since $\pmb w$ is $\|\,\|_2$-bounded, it is $\|\,\|_1$-bounded and an
integrator and moderately oscillatory (622G).   Because the filtration is
right-continuous, $\pmb w$ is locally near-simple (632Ia),
therefore near-simple (631F(c-ii)), and $\pmb w^*$ is also near-simple
(631J) as well as being an integrator (617I).   And $\pmb x^2\in\MSi^0(\Cal S)$
by 645Ka.    So we can form $\Sint_{\Cal S}\pmb x\,d\pmb w$ and
$\Sint_{\Cal S}\pmb x^2d\pmb w^*$.

\medskip

{\bf (b)} For the time being (down to the end of (h) below), suppose that
$\pmb x\in\MSi^0(\Cal S)$.   Set $M=\|\pmb x\|_{\infty}$,
$K=\sup_{\sigma\in\Cal S}\|w_{\sigma}\|_2$,
$A=\{\pmb y:\pmb y\in\MSi^0(\Cal S)$, $\|\pmb y\|_{\infty}\le M\}$,
$A_2=\{\pmb y:\pmb y\in\MSi^0(\Cal S)$, $\|\pmb y\|_{\infty}\le M^2\}$.
Note that if $\pmb y\in A$ then $|\pmb y|\le M\tbf 1^{(\Cal S)}_<$.   \Prf\
We know that $|\pmb y|\le M\tbf 1^{(\Cal S)}$ and also that there
is a non-negative $\pmb u\in\Mmo(\Cal S)$ such that $|\pmb y|\le\pmb u_<$.
Now

\Centerline{$\pmb u_<=(\pmb u\times\tbf 1^{\Cal S)})_<
=\pmb u_<\times\tbf 1^{(\Cal S)}_<$,}

\noindent so

\Centerline{$|\pmb y|\times(\tbf 1^{(\Cal S)}-\tbf 1^{\Cal S}_<)
\le\pmb u_<\times(\tbf 1^{(\Cal S)}-\tbf 1^{\Cal S}_<)=0$}

\noindent and

\Centerline{$|\pmb y|=|\pmb y|\times\tbf 1^{(\Cal S)}
=|\pmb y|\times\tbf 1^{(\Cal S)}_<
\le M\tbf 1^{\Cal S)}\times\tbf 1^{(\Cal S)}_<
=M\tbf 1^{(\Cal S)}_<$.\ \Qed}

\noindent So $A$ is uniformly previsibly order-bounded.   Similarly, $A_2$ is
uniformly previsibly order-bounded.

\medskip

{\bf (c)} By 645S,
$\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d\pmb w:A\to L^0(\frak A)$
and $\pmb y\mapsto\Sint_{\Cal S}\pmb y\,d\pmb w^*:A_2\to L^0(\frak A)$ are
continuous when $\MSi^0(\Cal S)$ is given its S-integration topology and
$L^0(\frak A)$ is given the topology of convergence in measure.
Since multiplication on $L^0(\frak A)$ is continuous,
$\pmb y\mapsto(\Sint_{\Cal S}\pmb y\,d\pmb w)^2:A\to L^0(\frak A)$
is continuous.
Since $\pmb y\mapsto\pmb y^2:A\mapsto A_2$ is continuous (645E(a-v-$\beta$)),
$\pmb y\mapsto\Sint_{\Cal S}\pmb y^2d\pmb w^*:A\to L^0(\frak A)$
is continuous.

\medskip

{\bf (d)} If $\pmb u\in\Mmo(\Cal S)$ is $\|\,\|_{\infty}$-bounded,

$$\eqalignno{\|(\int_{\Cal S}\pmb u\,d\pmb w)^2\|_1&
=\Expn((\int_{\Cal S}\pmb u\,d\pmb w)^2)
=\Expn(\int_{\Cal S}\pmb u^2d\pmb w^*)
\Displaycause{624I}
=\|\int_{\Cal S}\pmb u^2d\pmb w^*\|_1\cr}$$

\noindent by 614E(b-i), because $\pmb u^2$ is non-negative and $\pmb w^*$
is non-decreasing.   And by 622P we have
$\|\int_{\Cal S}\pmb u\,d\pmb w\|_2\le K\|\pmb u\|_{\infty}$,
so $\|(\int_{\Cal S}\pmb u\,d\pmb w)^2\|_1\le K^2\|\pmb u\|_{\infty}^2$.

\medskip

{\bf (e)} Set $B=\{\pmb u:\pmb u\in\Mmo(\Cal S)$, $\|\pmb u\|_{\infty}\le M\}$.
Then $A$ is the
closure of $\{\pmb u_<:\pmb u\in B\}$ for the S-integration topology of
$\MSi^0(\Cal S)$.   \Prf\ If $\pmb u\in B$,
$-M\tbf 1^{(\Cal S)}\le\pmb u\le M\tbf 1^{\Cal S}$ so
$-M\tbf 1^{(\Cal S)}_<\le\pmb u_<\le M\tbf 1^{\Cal S}_<$ and
$\pmb u_<\in A$.   On the other hand, if $\pmb y\in A$,
$\pmb v\in\Mnsu(\Cal S)$ and $\epsilon>0$,
there is certainly a $\pmb u\in\Mmo(\Cal S)$ such that
$\widehat{\theta}^{\#}_{\pmb v}(\pmb y-\pmb u_<)\le\epsilon$, by the definition
of $\MSi^0(\Cal S)$.   Now, setting
$\tildepmb u=\med(-M\tbf 1^{(\Cal S)},\pmb u,M\tbf 1^{(\Cal S)})$,

\Centerline{$|\pmb y-\tildepmb u_<|
=|\med(-M\tbf 1^{(\Cal S)}_<,\pmb y,M\tbf 1^{(\Cal S)}_<)
   -\med(-M\tbf 1^{(\Cal S)}_<,\pmb u_<,M\tbf 1^{(\Cal S)}_<)|
\le|\pmb y-\pmb u_<|$}

\noindent so

\Centerline{$\widehat{\theta}^{\#}_{\pmb v}(\pmb y-\tildepmb u_<)
\le\widehat{\theta}^{\#}_{\pmb v}(\pmb y-\pmb u_<)
\le\epsilon$}

\noindent (645Db), while $\tildepmb u\in B$.   As $\pmb y$, $\pmb v$ and
$\epsilon$ are
arbitrary, $A\subseteq\overline{\{\pmb u_<:\pmb u\in B\}}$ (645E(a-ii)).\ \Qed

\medskip

{\bf (f)} It follows that if $\pmb y\in A$, then
$(\Sint_{\Cal S}\pmb y\,d\pmb w)^2$ belongs to the closure, for the topology of
convergence in measure, of

\Centerline{$\{(\Sint_{\Cal S}\pmb u_<d\pmb w)^2:\pmb u\in B\}
=\{(\int_{\Cal S}\pmb u\,d\pmb w)^2:\pmb u\in B\}
\subseteq C$}

\noindent where $C=\{z:z\in L^0(\frak A)$, $\|z\|_1\le K^2M^2\}$.
So $(\Sint_{\Cal S}\pmb y\,d\pmb w)^2\in\overline{C}=C$ (613Bc).
Because $\|\,\|_1:L^0(\frak A)\to[0,\infty]$ is lower semi-continuous
(613Bc), $\pmb y\mapsto\|(\Sint_{\Cal S}\pmb y\,d\pmb w)^2\|_1:A\to\Bbb R$
is lower semi-continuous (4A2B(d-i)).

\medskip

{\bf (g)} Concerning
$\Sint_{\Cal S}\pmb y^2d\pmb w^*$ we have a little more.   Note first that,
setting $\pmb u=\tbf 1^{(\Cal S)}$ in (d), we see that

\Centerline{$\|\int_{\Cal S}d\pmb w^*\|_1=\|(\int_{\Cal S}d\pmb w)^2\|_1
\le K^2$}

\noindent is finite.   Now if $\pmb u\in B$,
$0\le\pmb u^2\le M^2\tbf 1^{(\Cal S)}$, so

\Centerline{$0\le\int_{\Cal S}\pmb u^2d\pmb w^*\le M^2\bar w$}

\noindent where $\bar w=\int_{\Cal S}d\pmb w^*$.   Setting
$D=\{z:z\in L^0(\frak A)$, $0\le z\le M^2\bar w\}$,
$D$ is closed for the topology
of convergence in measure and includes

\Centerline{$\{\int_{\Cal S}\pmb u^2d\pmb w^*:u\in B\}
=\{\Sint_{\Cal S}(\pmb u^2)_<d\pmb w^*:u\in B\}
=\{\Sint_{\Cal S}\pmb u_<^2d\pmb w^*:u\in B\}$}

\noindent so contains $\Sint_{\Cal S}\pmb y^2d\pmb w^*$ for every $y\in A$.
In particular, $\Sint_{\Cal S}\pmb x^2d\pmb w^*\in D$, so

\Centerline{$\|\Sint_{\Cal S}\pmb x^2d\pmb w^*\|_1\le M^2\|\bar w\|_1
\le M^2K^2<\infty$.}

\noindent
Observe next that $D$ is uniformly integrable, so the $\|\,\|_1$-topology on
$D$ agrees with the topology of convergence in measure (621B(c-ii)), and
$z\mapsto\|z\|_1:D\to\Bbb R$ is continuous.   But this means that
$\pmb y\mapsto\|\Sint_{\Cal S}\pmb y^2d\pmb w^*\|_1:A\to\Bbb R$ is continuous.

\medskip

{\bf (h)} Putting (f) and (g) together, we see that

\Centerline{$\pmb y\mapsto\|(\Sint_{\Cal S}\pmb y\,d\pmb w)^2\|_1
          -\|\Sint_{\Cal S}\pmb y^2d\pmb w^*\|_1:A\to\Bbb R$}

\noindent is lower semi-continuous (4A2B(d-iii)).   And as

\Centerline{$\|(\Sint_{\Cal S}\pmb u_<\,d\pmb w)^2\|_1
          -\|\Sint_{\Cal S}\pmb u_<^2d\pmb w^*\|_1
=\|(\int_{\Cal S}\pmb u\,d\pmb w)^2\|_1
          -\|\int_{\Cal S}\pmb u^2d\pmb w^*\|_1
=0$}

\noindent for every $\pmb u\in B$,
$\|(\Sint_{\Cal S}\pmb y\,d\pmb w)^2\|_1
          -\|\Sint_{\Cal S}\pmb y^2d\pmb w^*\|_1\le 0$ for every
$\pmb y\in A$.   In particular,
$\|(\Sint_{\Cal S}\pmb x\,d\pmb w)^2\|_1
          \le\|\Sint_{\Cal S}\pmb x^2d\pmb w^*\|_1$.

\medskip

{\bf (i)} This deals with the result when $\pmb x\in\MSi^0(\Cal S)$.
For the general case in which $\pmb x\in\MSi(\Cal S)$, apply
(b)-(h) to $\pmb x\times\tbf 1^{(\Cal S)}_<$.
}%end of proof of 648C

\leader{648D}{}\cmmnt{ We come now to the main work of the section in which we
are working in both the original stochastic integration structure $\Bbb A$ and
an embedded structure $\Bbb B$.

\medskip

\noindent}{\bf Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let
$\frak B$ be a closed subalgebra of $\frak A$ which is coordinated with
$\family{t}{T}{\frak A_t}$, and
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.   Let $\Cal S'$ be a relatively
order-convex sublattice of $\Cal T_{\Bbb B}$ and $\Cal S$ its order-convex
hull in $\Cal T_{\Bbb A}$.
If $\pmb x\in\Bbb A\MSi^0(\Cal S)$ and
$\pmb x\restr\Cal S'\in L^0(\frak B)^{\Cal S'}$, then
$\pmb x\restr\Cal S'\in\Bbb B\MSi^0(\Cal S)$.

\proof{{\bf (a)} We need the following basic facts:
$\Cal S'$ is $\Bbb B$-finitely full (611Ob),
$\Cal S'$ separates $\Cal S$ (633Da and 633D(b-i),
because constant processes belong to $\Cal T_{\Bbb B}$),
$\Cal S'$ is cofinal with $\Cal S$ and
$\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and
has a lower bound in $\Cal S$.   Recall also that we have a multiplicative
Riesz homomorphism $\Psi:\Bbb A\Mns(\Cal S')\to\Bbb A\Mns(\Cal S)$ such that
$\Psi(\pmb u')$ extends $\pmb u'$ for every $\pmb u'\in\Bbb A\Mns(\Cal S)$
(631Mb), while
$\Bbb B\Mns(\Cal S')=\Bbb A\Mns(\Cal S')\cap L^0(\frak B)^{\Cal S'}$ (634Eb).
It follows that
$\Bbb B\Mnsu(\Cal S')=\Bbb A\Mnsu(\Cal S')\cap L^0(\frak B)^{\Cal S'}$.

\medskip

{\bf (b)(i)} Let $P:L^1_{\bar\mu}\to L^1_{\bar\mu}$ be the conditional
expectation operator associated with $\frak B$.   If
$\pmb y=\sfamilysS{y}$ belongs to $(L^1_{\bar\mu})^{\Cal S}$ set
$P'(\pmb y)=\family{\sigma}{\Cal S'}{P(y_{\sigma)}}$.

\medskip

\quad{\bf (ii)} If $\pmb u\in\Bbb A\Mfa(\Cal S)$ is $\|\,\|_{\infty}$-bounded
then $P'(\pmb u)\in\Bbb B\Mfa(\Cal S')$.   \Prf\ Express $\pmb u$ as
$\sfamilysS{u}$.   If $\sigma\in\Cal S'$ then
$u_{\sigma}\in L^0(\frak A_{\sigma})$.   Writing $P_{\sigma}$ for the
conditional expectation associated with the closed subalgebra
$\frak A_{\sigma}$, we know that $\frak B$ and $\frak A_{\sigma}$ are relatively
independent over their intersection $\frak B_{\sigma}$ (634H) so that
$PP_{\sigma}$ is the conditional expectation operator corresponding to
$\frak B_{\sigma}$.   Now
$P(u_{\sigma})=PP_{\sigma}(u_{\sigma})\in L^0(\frak B_{\sigma})$.   Thus
$P'(\pmb u)\in\prod_{\sigma\in\Cal S'}L^0(\frak B_{\sigma})$.

If $\sigma$, $\tau\in\Cal S'$ and $b=\Bvalue{\sigma=\tau}$, then
$b\in\frak B$, so

\Centerline{$P(u_{\sigma})\times\chi b=P(u_{\sigma}\times\chi b)
=P(u_{\tau}\times\chi b)=P(u_{\tau})\times\chi b$}

\noindent and $b\Bsubseteq\Bvalue{P(u_{\sigma})=P(u_{\tau})}$.   So
$P'(\pmb u)$ is $\Bbb B$-fully adapted.\ \Qed

\medskip

\quad{\bf (iii)} If $\pmb u\in\Bbb A\Mmo(\Cal S)$ is $\|\,\|_{\infty}$-bounded,
then $P'(\pmb u)\in\Bbb B\Mmo(\Cal S')$ and $P'(\pmb u)_<=P'(\pmb u_<)$.   \Prf\
We have just seen that $P'(\pmb u)$ is $\Bbb B$-fully adapted.
Write $M$ for $\|\pmb u\|_{\infty}$.
On the uniformly integrable set
$A=\{z:z\in L^0(\frak A)$, $\|z\|_{\infty}\le M\}$
the topology of convergence in measure coincides with the norm topology of
$L^1_{\bar\mu}$;  as $P:A\to A$ is $\|\,\|_1$-continuous, it is continuous for
the topology of convergence in measure.   If $\sequencen{\sigma_n}$ is a
monotonic sequence in $\Cal S'$, $\sequencen{u_{\sigma_n}}$ is a sequence in $A$
which converges for the topology of convergence in measure to a member of $A$,
so $\sequencen{(P(u_{\sigma_n})}$ converges for the topology of convergence in
measure.   As $\sequencen{\sigma_n}$ is arbitrary and $\Cal S'$ is
$\Bbb B$-finitely-full, $P'(\pmb u)$ is $\Bbb B$-moderately-oscillatory
(615N(iii) again).

Express $\pmb u$ as $\sfamilysS{u}$ and $P'(\pmb u)$ as
$\family{\sigma}{\Cal S'}{w_{\sigma}}$.   For $\tau\in\Cal S'$ and
non-empty
$I\in\Cal I(\Cal S')$, define $u_{I<\tau}$ from $\sfamilysS{u}$ and
$w_{I<\tau}$ from $\family{\sigma}{\Cal S'}{w_{\sigma}}$ as in 641E.
Now for $\sigma\in I$ set

\Centerline{$b_{\sigma}=\Bvalue{\sigma<\tau}
\Bsetminus\sup_{\sigma'\in I}\Bvalue{\sigma<\sigma'}\Bcap\Bvalue{\sigma'<\tau}$,
}

\noindent and set $b=\Bvalue{\tau\le\min I}$.   Because
$\Cal S'\subseteq\Cal T_{\Bbb B}$, $b_{\sigma}$ and $b$ belong to $\frak B$,
while

\Centerline{$b_{\sigma}
\Bsubseteq\Bvalue{u_{I<\tau}=u_{\sigma}}\Bcap\Bvalue{w_{I<\tau}=w_{\sigma}}$,}

\Centerline{$b
\Bsubseteq\Bvalue{u_{I<\tau}=0}\Bcap\Bvalue{w_{I<\tau}=0}$.}

\noindent Accordingly

\Centerline{$b_{\sigma}
\Bsubseteq\Bvalue{P(u_{I<\tau})=w_{\sigma}}\Bcap\Bvalue{w_{I<\tau}=w_{\sigma}}
\Bsubseteq\Bvalue{P(u_{I<\tau})=w_{I<\tau}}$,}

\Centerline{$b\Bsubseteq\Bvalue{P(u_{I<\tau})=w_{I<\tau}}$.}

\noindent As $b\Bcup\sup_{\sigma\in I}b_{\sigma}=1$,
$w_{I<\tau}=P(u_{I<\tau})$.   As $I$ is arbitrary and $P$ is continuous on
$\{z:z\in L^0(\frak A)$, $|z|\le M\chi 1\}$,

\Centerline{$w_{<\tau}=\lim_{I\uparrow\Cal I(\Cal S')}w_{I<\tau}
=P(\lim_{I\uparrow\Cal I(\Cal S')}u_{I<\tau})$;}

\noindent as $\tau$ is arbitrary, $(P'(\pmb u))_<=P(\pmb u\restr\Cal S')_<$.
But $\Cal S'$ separates $\Cal S$, so
$(\pmb u\restr\Cal S')_<=\pmb u_<\restr\Cal S'$ (641Hb) and

\Centerline{$(P'(\pmb u))_<=P(\pmb u_<\restr\Cal S')=P'(\pmb u)_<$. \Qed}

\medskip

{\bf (c)} $\pmb x\restr\Cal S'\in\Bbb B\Mpob(\Cal S')$.

\Prf\ Let $\pmb u\in\Bbb A\Mmo(\Cal S)^+$ be such that $|\pmb x|\le\pmb u_<$.
Consider the functions $g:\Bbb R\to\ooint{-1,1}$, $h:\ooint{-1,1}$ given by the
formulae

\Centerline{$g(\alpha)=\Bover{\alpha}{1+|\alpha|}$ for $\alpha\in\Bbb R$,
\quad$h(\beta)=\Bover{\beta}{1-|\beta|}$ for $\beta\in\ooint{-1,1}$.}

\noindent Then $g=h^{-1}$, $h=g^{-1}$ are the two halves of an order-preserving
homeomorphism between $\Bbb R$ and $\ooint{-1,1}$.   Set

\Centerline{$Q_{\frak A}=\{u:u\in L^0(\frak A)$, $\Bvalue{|u|<1}=1\}$,
\quad $Q_{\frak B}=\{u:u\in L^0(\frak B)$, $\Bvalue{|u|<1}=1\}$.}

\noindent I will write

\Centerline{$\bar g:L^0(\frak A)^{\Cal S}\to Q_{\frak A}^{\Cal S}$,
\quad$\bar g':L^0(\frak B)^{\Cal S'}\to Q_{\frak B}^{\Cal S'}$
\quad$\bar h:Q_{\frak B}^{\Cal S'}\to L^0(\frak B)^{\Cal S'}$}

\noindent for the induced functions as in 612A-612B, so that
$\bar h$ is the inverse of $\bar g'$.

By 615F(a-ii), $\bar g\pmb u$ is $\frak A$-moderately-oscillatory, and we have

$$\eqalignno{|\bar g\pmb x|&
=\bar g|\pmb x|
\Displaycause{because $|g(\alpha)|=g(|\alpha|)$ for $\alpha\in\Bbb R$}
\le\bar g\pmb u_<
\Displaycause{by 612A(d-iii), because $g$ is order-preserving}
=(\bar g\pmb u)_<\cr}$$

\noindent by 641Gd, while setting $\bar u=\sup|\pmb u|$,
$\sup\bar g\pmb u\le\bar g(\bar u)$.   Since
$\bar g\pmb u\in Q_{\frak A}^{\Cal S}$ is
$\|\,\|_{\infty}$-bounded, we can form $P'(\bar g\pmb u)$;  by (b),
$P'(\bar g\pmb u)$ is $\Bbb B$-moderately-oscillatory and
$(P'(\bar g\pmb u))_<=P'((\bar g\pmb u)_<)$.
Note that $P(u)\in Q_{\frak B}$ for every $u\in Q_{\frak A}$, so
$P'(\bar g\pmb u)\in Q_{\frak B}^{\Cal S'}$ and
$\bar g\pmb u\in[-P\bar g(\bar u),P\bar g(\bar u)]^{\Cal S'}$, while
$[-P\bar g(\bar u),P\bar g(\bar u)]$ is a topologically closed subset of
$Q_{\frak B}$.   By 648B, applied in $\Bbb B$,
$\bar hP'(\bar g\pmb u)\in L^0(\frak B)^{\Cal S'}$ is a
$\Bbb B$-moderately-oscillatory process.

Still working in $\Bbb B$, we know that the previsible version
$(\bar hP'(\bar g\pmb u))_<$ is defined.   To compute it, note that
we have
$\bar g'(\bar hP'(\bar g\pmb u))_<=(\bar g'\bar hP'(\bar g\pmb u))_<
=(P'(\bar g\pmb u))_<$.   So

\Centerline{$(\bar hP'(\bar g\pmb u))_<
=\bar h\bar g'((\bar hP'(\bar g\pmb u))_<)
=\bar hP'(\bar g\pmb u)_<$.}

\noindent Now we see that

$$\eqalignno{|\pmb x\restr\Cal S'|&
=|\bar h\bar g'\pmb x\restr\Cal S'|
=|\bar hP'(\bar g\pmb x)|
\Displaycause{because $\pmb x\restr\Cal S'\in L^0(\frak B)^{\Cal S'}$
and $(\bar g\pmb x)\restr\Cal S'\in L^{\infty}(\frak B)^{\Cal S'}$}
\le\bar hP'(|\bar g\pmb x|)
=\bar hP'(\bar g|\pmb x|)
\le\bar hP'(\bar g\pmb u_<)
\Displaycause{because $P'$ is a positive linear operator and $\bar h$ is
order-preserving}
=\bar hP'(\bar g\pmb u_<)
=(\bar hP'(\bar g\pmb u))_<\cr}$$

\noindent while $\bar hP'(\bar g\pmb u)$ is non-negative and
$\Bbb B$-moderately-oscillatory.   So $\pmb x\restr\Cal S'$ is
$\Bbb B$-previsibly-order-bounded.\ \Qed

\medskip

{\bf (d)} Suppose that $\pmb v=\sfamilysS{v}$ is a non-negative non-decreasing
$\|\,\|_{\infty}$-bounded $\Bbb A$-fully-adapted process such that
$v_{\sigma}\in L^0(\frak B)$ for every $\sigma\in\Cal S'$, and that
$\pmb u=\sfamilysS{u}$ is a $\|\,\|_{\infty}$-bounded
$\Bbb A$-moderately-oscillatory process.   Then
$\|\int_{\Cal S}\pmb u\,d\pmb v\|_{\infty}
\le\|\pmb u\|_{\infty}\|\pmb v\|_{\infty}$ and
$P(\int_{\Cal S}\pmb u\,d\pmb v)=\int_{\Cal S'}P'(\pmb u)d\pmb v$.

\Prf\ Because $\pmb u$ is $\Bbb A$-moderately-oscillatory,
$P'(\pmb u)$ is $\Bbb B$-moderately oscillatory, while $\pmb v$ is
$\Bbb A$-fully adapted, $\pmb v'=\pmb v\restr\Cal S'$ is
$\Bbb B$-fully adapted and both are of bounded variation, the integrals are
defined.   As $\Cal S'$ is cofinal with $\Cal S$ and separates $\Cal S$,
while $\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and bounded
below in $\Cal S$, 633K tells us that
$\int_{\Cal S'}\pmb u\,d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.

Write $M$, $M'$ for $\|\pmb u\|_{\infty}$ and $\|\pmb v\|_{\infty}$.
If $\sigma_0\le\ldots\le\sigma_n$ in $\Cal S'$,

$$\eqalignno{\|\sum_{i=0}^{n-1}
   u_{\sigma_i}\times(v_{\sigma_{i+1}}-v_{\sigma_i})\|_{\infty}&
\le\|\sum_{i=0}^{n-1}
   |u_{\sigma_i}|\times|v_{\sigma_{i+1}}-v_{\sigma_i}|\|_{\infty}\cr&
\le\|\sum_{i=0}^{n-1}
   M|v_{\sigma_{i+1}}-v_{\sigma_i}|\|_{\infty}\cr&
=M\|v_{\sigma_n}-v_{\sigma_0}\|_{\infty}
\le M\|v_{\sigma_n}\|_{\infty}
\le MM'.\cr}$$

\noindent At the same time

\Centerline{$P(\sum_{i=0}^{n-1}
   u_{\sigma_i}\times(v_{\sigma_{i+1}}-v_{\sigma_i}))
=\sum_{i=0}^{n-1}P(u_{\sigma_i})\times(v_{\sigma_{i+1}}-v_{\sigma_i})$}

\noindent because $v_{\sigma_{i+1}}-v_{\sigma_i}\in L^{\infty}(\frak B)$ for
every $i$.

Accordingly
$\|S_I(\pmb u,d\pmb v)\|_{\infty}\le MM'$ and
$P(S_I(\pmb u,d\pmb v))=S_I(P\pmb u,d\pmb v)$
for every $I\in\Cal I(\Cal S')$.
Taking the limit as $I\uparrow\Cal I(\Cal S')$,
$\|\int_{\Cal S'}\pmb u\,d\pmb v\|_{\infty}\le MM'$.   At the same time, since
$P$ is continuous on $\{y:\|y\|_{\infty}\le MM'\}$,

$$\eqalign{P(\int_{\Cal S}\pmb u\,d\pmb v)&
=P(\int_{\Cal S'}\pmb u\,d\pmb v)
=\lim_{I\uparrow\Cal I(\Cal S')}P(S_I(\pmb u,d\pmb v))\cr&
=\lim_{I\uparrow\Cal I(\Cal S')}S_I(P\pmb u,d\pmb v)
=\int_{\Cal S'}P\pmb u\,d\pmb v
=\int_{\Cal S'}P'(\pmb u)\,d\pmb v.  \text{ \Qed}\cr}$$

\medskip

{\bf (e)} Suppose that $\pmb y\in\Bbb A\Mpob(\Cal S)$ is
$\|\,\|_{\infty}$-bounded, that $\pmb v\in\Bbb A\Mnsu(\Cal S)$ is
$\|\,\|_{\infty}$-bounded, and that
$\pmb v'=\pmb v\restr\Cal S'$
belongs to $L^0(\frak B)^{\Cal S'}$.
Then $P'(\pmb y)\in\Bbb B\Mpob(\Cal S')$, $\pmb v'\in\Bbb B\Mns(\Cal S')$
and $\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(P'(\pmb y))
\le\max(1,\|\pmb y\|_{\infty}\|\pmb v\|_{\infty})
   \Bbb A\widehat{\theta}_{\pmb v}^{\#}(\pmb y)$.

\Prf\ Set $M=\|\pmb y\|_{\infty}$, $M'=\|\pmb v\|_{\infty}$ and
$\gamma=\Bbb A\widehat{\theta}^{\#}_{\pmb v}(\pmb y)$.   Then
$|\pmb y|\le M\tbf 1^{(\Cal S)}_<$ so

\Centerline{$|P'(\pmb y)|\le P'(|\pmb y|)\le MP'(\tbf 1^{(\Cal S)}_<)
=MP'(\tbf 1^{(\Cal S)})_<=M\tbf 1^{(\Cal S')}_<$}

\noindent by (b-iii).   So $\pmb y'$ is $\Bbb B$-previsibly-order-bounded.

Let $\epsilon>0$.   Then there is a uniformly order-bounded non-decreasing
sequence $\sequencen{\pmb u_n}$ in $\Bbb A\Mmo(\Cal S)^+$ such that
$|\pmb y|\le\sup_{n\in\Bbb N}\pmb u_{n<}$ and
$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_nd\pmb v)\le\gamma+\epsilon$.
Replacing $\pmb u_n$ by $\pmb u_n\wedge M\tbf 1^{(\Cal S)}$ if necessary,
we can arrange that $\|\pmb u_n\|_{\infty}\le M$ for every $n$.
Expressing $\pmb y$ as $\sfamilysS{y}$ and $\pmb u_{n<}$ as
$\family{\sigma}{\Cal S}{u_{n<\sigma}}$ for each $n$, we see that for
$\sigma\in\Cal S'$

$$\eqalignno{|P(y_{\sigma})|&
\le P(|y_{\sigma}|)
\le P(\sup_{n\in\Bbb N}u_{n<\sigma})
=P(\1lim_{n\to\infty}u_{n<\sigma})
\Displaycause{taking the $\|\,\|_1$-limit}
=\1lim_{n\to\infty}P(u_{n<\sigma})
=\sup_{n\in\Bbb N}P(u_{n<\sigma}).\cr}$$

\noindent So

\Centerline{$|P'(\pmb y)|\le\sup_{n\in\Bbb N}P'(\pmb u_{n<})
=\sup_{n\in\Bbb N}P'(\pmb u_n)_<$}

\noindent by (b-iii) above, while $\sequencen{P'(\pmb u_n)}$ is a uniformly
order-bounded non-decreasing sequence of non-negative
$\Bbb B$-moderately-oscillatory processes.   So

$$\eqalignno{\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(P'(\pmb y))&
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S'}P'(\pmb u_n)d\pmb v')
=\sup_{n\in\Bbb N}\theta(P(\int_{\Cal S}\pmb u_nd\pmb v))
\Displaycause{by (d) above}
\le\sup_{n\in\Bbb N}\Expn(P(\int_{\Cal S}\pmb u_nd\pmb v))
=\sup_{n\in\Bbb N}\Expn(\int_{\Cal S}\pmb u_nd\pmb v)\cr&
\le\max(1,MM')\sup_{n\in\Bbb N}
  \Expn(\Bover1{\max(1,MM')}\int_{\Cal S}\pmb u_nd\pmb v)\cr&
=\max(1,MM')\sup_{n\in\Bbb N}
  \theta(\Bover1{\max(1,MM')}\int_{\Cal S}\pmb u_nd\pmb v)
\Displaycause{because $\|\int_{\Cal S}\pmb u_nd\pmb v\|_{\infty}\le MM'$,
also noted in (d)}
\le\max(1,MM')\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb u_nd\pmb v)
\le\max(1,MM')(\gamma+\epsilon).\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(P'(\pmb y))\le\max(1,MM')\gamma$, as
claimed.\ \Qed

\medskip

{\bf (f)} If $\pmb x$ is
$\|\,\|_{\infty}$-bounded, then $\pmb x'\in\Bbb B\MSi^0(\Cal S)$.
\Prf\ By (c), or otherwise, $\pmb x'$ is $\Bbb B$-previsibly-order-bounded.
Take $\pmb v'\in\Bbb B\Mnsu(\Cal S')$ and
$\epsilon>0$.   Then there is an $M'\ge 0$ such that
$\bar\mu\Bvalue{\sup\pmb v'>M'}\le\bover12\epsilon$.   Set
$\pmb w'=\pmb v'\wedge M'\tbf 1^{(\Cal S')}$, so that
$\pmb w'\in\Bbb B\Mnsu(\Cal S')$ and $\|\pmb w'\|_{\infty}\le M'$, while
$\bar\mu\Bvalue{\pmb w'\ne\pmb v'}\le\bover12\epsilon$.
As noted in (a), $\pmb w'$ belongs to $\Bbb A\Mns(\Cal S')$ and has an extension
$\pmb w=\Psi(\pmb w')$ belonging to $\Bbb A\Mns(\Cal S)$.   Because
$\Psi$ is a Riesz homomorphism,
$0\le\pmb w\le M'\tbf 1^{(\Cal S)}$, and by 631M(b-iv) $\pmb w$ is
non-decreasing.   So $\pmb w\in\Bbb A\Mnsu(\Cal S)$.

We can therefore speak of the F-seminorm
$\Bbb A\widehat{\theta}^{\#}_{\pmb w}$.   Set $M=\|\pmb x\|_{\infty}$, and
let $\delta>0$ be such that $\max(1,2MM')\delta<\bover12\epsilon$.
As $\pmb x\in\Bbb A\MSi^0(\Cal S)$,
there is a $\pmb u\in\Bbb A\Mmo(\Cal S)$ such that
$\Bbb A\widehat{\theta}^{\#}_{\pmb w}(\pmb x-\pmb u_<)\le\delta$.
Replacing $\pmb u$ by $\med(-M\tbf 1^{(\Cal S)}.\pmb u,M\tbf 1^{(\Cal S)})$ if
necessary, we can arrange that $\|\pmb u\|_{\infty}\le M$, so that
$\|\pmb x-\pmb u_<\|_{\infty}\le 2M$.   Now $P'(\pmb u)\in\Bbb B\Mmo(\Cal S')$
and

$$\eqalignno{\Bbb B\widehat{\theta}^{\#}_{\pmb w'}(\pmb x'-(P'(\pmb u))_<)&
=\Bbb B\widehat{\theta}^{\#}_{\pmb w'}(\pmb x'-P'(\pmb u_<))
=\Bbb B\widehat{\theta}^{\#}_{\pmb w'}(P'(\pmb x-\pmb u_<))\cr&
\le(1+2MM')\Bbb A\widehat{\theta}^{\#}_{\pmb w}(\pmb x-\pmb u_<)
\Displaycause{by (e) above}
<(1+2MM')\delta
\le\Bover12\epsilon.\cr}$$

\noindent Working in $\Bbb B$, we have a uniformly previsibly
order-bounded non-decreasing sequence
$\sequencen{\pmb u'_n}$ in $\Bbb B\Mmo(\Cal S')^+$ such that
$|\pmb x'-(P'(\pmb u))_<|\le\sup_{n\in\Bbb N}\pmb u'_{n<}$ and
$\sup_{n\in\Bbb N}\theta(\int_{\Cal S'}\pmb u'_nd\pmb w')\le\bover12\epsilon$.
For each $n$,

$$\eqalignno{\theta(\int_{\Cal S'}\pmb u'_nd\pmb v'
        -\int_{\Cal S'}\pmb u'_nd\pmb w')&
\le\bar\mu\Bvalue{\int_{\Cal S'}\pmb u'_nd\pmb v'
    \ne\int_{\Cal S'}\pmb u'_nd\pmb w'}
\le\bar\mu\Bvalue{\pmb v'\ne\pmb w'}
\Displaycause{613Ld}
\le\Bover12\epsilon,\cr}$$

\noindent so

\Centerline{$\theta(\int_{\Cal S'}\pmb u'_nd\pmb v')
\le\Bover12\epsilon+\theta(\int_{\Cal S'}\pmb u'_nd\pmb w')
\le\epsilon$.}

\noindent Accordingly

\Centerline{$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb x'-P'(\pmb u)_<)
\le\sup_{n\in\Bbb N}\theta(\int_{\Cal S'}\pmb u'_nd\pmb v')
\le\epsilon$.}

\noindent As $\pmb v'$ and $\epsilon$ are arbitrary,
$\pmb x'\in\Bbb B\MSi^0(\Cal S')$.\ \Qed

\medskip

{\bf (g)} Finally, in the general case in which $\pmb x\in\Bbb A\MSi^0(\Cal S)$
and $\pmb x'\in L^0(\frak B)^{\Cal S'}$, set
$\pmb x_n=\med(-M\tbf 1^{(\Cal S)}_<,\pmb x,M\tbf 1^{(\Cal S)}_<)$
and $\pmb x'_n=\pmb x_n\restr\Cal S'$ for each
$n\in\Bbb N$.   Then $\pmb x_n\in\Bbb A\MSi^0(\Cal S)$ is
$\|\,\|_{\infty}$-bounded and
$\pmb x'_n=\med(-M\tbf 1^{(\Cal S')}_<,\pmb x',M\tbf 1^{(\Cal S')}_<)$
belongs to
$L^0(\frak B)^{\Cal S'}$;  by (f), $\pmb x'_n\in\Bbb B\MSi^0(\Cal S')$,
for every $n\in\Bbb N$.   Writing $\pmb x'$ for $\pmb x\restr\Cal S'$,
$\pmb x'$ is $\Bbb B$-previsibly-order-bounded
(by (c)) and $|\pmb x'_n|\le|\pmb x'|$ for every $n$, so
$\sequencen{\pmb x'_n}$ is
uniformly $\Bbb B$-previsibly-order-bounded, while
it is order*-convergent to $\pmb x'$ in
$L^0(\frak B)^{\Cal S'}$;  by 645H,
$\pmb x'\in\Bbb B\MSi^0(\Cal S')$.
}%end of proof of 648D

\leader{648E}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let
$\frak B$ be a closed subalgebra of $\frak A$ which is coordinated with
$\family{t}{T}{\frak A_t}$, and
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.   Let $\Cal S'$ be a relatively
order-convex sublattice of $\Cal T_{\Bbb B}$ and $\Cal S$ its order-convex
hull in $\Cal T_{\Bbb A}$.   If $\pmb u\in\Bbb A\Mns(\Cal S)$ and
$\pmb u\restr\Cal S'\in L^0(\frak B)^{\Cal S'}$ then
$\pmb u'=\pmb u\restr\Cal S'$ is near-simple in either $\Bbb A$ or $\Bbb B$.

\proof{ $\pmb u$ is $\Bbb A$-moderately-oscillatory (631Ca),
and $\Cal S$ is $\Bbb A$-finitely-full (611Ob), so $\pmb u$ is \updowncgt\
(615N) and $\pmb u\restr\Cal S'$ is \updowncgt, working in either $\Bbb A$ or
$\Bbb B$.   If $A\subseteq\Cal S'$ is non-empty and has a lower bound in
$\Cal S'$, then $A$ has a greatest lower bound in $\Cal T_{\Bbb B}$ which is
also its infimum in $\Cal T_{\Bbb A}$, because $\Cal T_{\Bbb B}$ is an
order-closed sublattice of
$\Cal T_{\Bbb A}$ (634C(f-ii));  and because $\Cal S'$ is order-convex in
$\Cal T_{\Bbb B}$, this common infimum belongs to $\Cal S$.

We know that $\pmb u'$ is $\Bbb A$-moderately-oscillatory (615F(a-i)), therefore
$\Bbb B$-moderately-oscillatory (634Ee).   If $A\subseteq\Cal S'$ is
downwards-directed and has a lower bound in $\Cal S'$, then (as we have just
seen) there is a common value $\inf A\in\Cal S'$ of its infimum taken in either
$\Cal T_{\Bbb B}$ or $\Cal T_{\Bbb A}$.   By 632H,
$u_{\inf A}=\lim_{\sigma\downarrow A}u_{\sigma}$, because we are supposing that
$\pmb u$ is $\Bbb A$-near-simple.   Finally,
the filtration $\family{t}{T}{\frak B_t}$ is right-continuous (634C(f-i)).
As $A$ is arbitary, we can apply 632H in the other direction, working in the
structure $\Bbb B$, to see that $\pmb u'$ is $\Bbb B$-near-simple, therefore
$\Bbb A$-near-simple (634Eb).

}%end of proof of 648E

\leader{648F}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let
$\frak B$ be a closed subalgebra of $\frak A$ which is coordinated with
$\family{t}{T}{\frak A_t}$, and
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.   Let $\Cal S'$ be a relatively
order-convex sublattice of $\Cal T_{\Bbb B}$ and $\Cal S$ its order-convex
hull in $\Cal T_{\Bbb A}$.   Let $\pmb x\in\Bbb A\MSi^0(\Cal S)$,
$\pmb v\in\Bbb A\Mnsu(\Cal S)$ be such that
$\pmb x'=\pmb x\restr\Cal S'$ and $\pmb v'=\pmb v\restr\Cal S'$ belong to
$L^0(\frak B)^{\Cal S'}$.   If
$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb x')=0$ then
$\ASint_{\Cal S}\pmb x\,d\pmb v=0$.

\proof{ We know that $\Cal S'\Bsupseteq\{\med(\sigma,\check t,\tau):\sigma$,
$\tau\in\Cal S'$, $t\in T\}\,\,\Bbb A$-separates $\Cal S$ (633D).
By 648D, $\pmb x'\in\Bbb B\MSi^0(\Cal S')$,
while $\pmb v'$ is non-negative, non-decreasing and order-bounded, and by
648E is near-simple in either structure.
Thus $\pmb v'\in\Bbb B\Mnsu(\Cal S')$,
and $\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb x')$ is surely defined.

\Quer\ Suppose, if possible, that
$\epsilon=\bover14\theta(\Bbb A\Sint_{\Cal S}\pmb x\,d\pmb v)$ is
greater than $0$.
Working in $\Bbb B$, we have an order-bounded
non-decreasing sequence $\sequencen{\pmb u'_n}$ in
$\Bbb B\Mmo(\Cal S')^+$ such that $\pmb x'\le\sup_{n\in\Bbb N}\pmb u'_{n<}$
and $\sup_{n\in\Bbb N}\theta(\int_{\Cal S'}\pmb u'_nd\pmb v')\le\epsilon$.
Next, working in $\Bbb A$,
there are a $\pmb w\in\Bbb A\Mmo(\Cal S)^+$ such that
$\int_{\Cal S}\pmb w\,d\pmb v\ge 3\epsilon$ and
$\Bbb A\widehat{\theta}^{\#}_{\pmb v}(|\pmb x-\pmb w_<|)<\epsilon$,
so there is an order-bounded
non-decreasing sequence $\sequencen{\pmb w_n}$ in $\Bbb A\Mmo(\Cal S)^+$
such that
$|\pmb x-\pmb w_<|\le\sup_{n\in\Bbb N}\pmb w_{n<}$ and
$\sup_{n\in\Bbb N}\theta(\int_{\Cal S}\pmb w_nd\pmb v)\le\epsilon$.
Observe that

\Centerline{$\pmb x\ge\pmb w_<-|\pmb x-\pmb w_<|
\ge\inf_{n\in\Bbb N}(\pmb w_<-\pmb w_{n<})
=\inf_{n\in\Bbb N}(\pmb w-\pmb w_n)_<$.}

\noindent
Setting $\pmb w'=\pmb w\restr\Cal S'$, $\pmb w'_n=\pmb w_n\restr\Cal S'$ we have
$(\pmb w'-\pmb w'_n)_<=(\pmb w-\pmb w_n)_<\restr\Cal S'$ for each $n$
(641Hb again).   So

\Centerline{$\inf_{n\in\Bbb N}(\pmb w'-\pmb w'_n)_<\le\pmb x'
\le\sup_{n\in\Bbb N}\pmb u'_n$,
\quad$\pmb w_<'\le\sup_{n\in\Bbb N}\pmb w'_{n<}+\pmb u'_{n<}$,}

\noindent
$\sequencen{(\pmb w'\wedge(\pmb w'_n+\pmb u'_n))_<}$ is order*-convergent to
$\pmb w'_<$ in $L^0(\frak B)^{\Cal S'}$ and $\int_{\Cal S'}\pmb w'd\pmb v'
=\lim_{n\to\infty}\int_{\Cal S'}(\pmb w'\wedge(\pmb w'_n+\pmb u'_n))d\pmb v'$ by
644H.   On the other hand, 633Ka tells us that

\Centerline{$\int_{\Cal S'}\pmb w'd\pmb v'=\int_{\Cal S}\pmb w\,d\pmb v$,
\quad$\int_{\Cal S'}\pmb w'_nd\pmb v'=\int_{\Cal S}\pmb w_nd\pmb v$}

\noindent for every $n\in\Bbb N$, so
$\theta(\int_{\Cal S'}\pmb w'd\pmb v')\ge 3\epsilon$ while

$$\eqalign{\theta(\int_{\Cal S'}(\pmb w'\wedge(\pmb w'_n+\pmb u_n))d\pmb v')&
\le\theta(\int_{\Cal S'}\pmb w'_nd\pmb v')
   +\theta(\int_{\Cal S'}\pmb u_nd\pmb v')\cr&
\le\theta(\int_{\Cal S}\pmb w_nd\pmb v)+\epsilon
\le 2\epsilon\cr}$$

\noindent for every $n$.\ \BanG\   Thus
$\ASint_{\Cal S}\pmb x\,d\pmb v=0$.
}%end of proof of 648F

\leader{648G}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let
$\frak B$ be a closed subalgebra of $\frak A$ which is coordinated with
$\family{t}{T}{\frak A_t}$, and
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.   Let $\Cal S'$ be a relatively
order-convex sublattice of $\Cal T_{\Bbb B}$ and $\Cal S$ its order-convex
hull in $\Cal T_{\Bbb A}$.   Let $\pmb x\in\Bbb A\MSi(\Cal S)$ and an
$\Bbb A$-near-simple $\Bbb A$-integrator $\pmb w$ with domain $\Cal S$
be such that
$\pmb x'=\pmb x\restr\Cal S'$, $\pmb w'=\pmb w\restr\Cal S'$ belong to
$L^0(\frak B)^{\Cal S'}$.
Then $\BSint_{\Cal S'}\pmb x'd\pmb w'=\ASint_{\Cal S}\pmb x\,d\pmb w$.

\proof{{\bf (a)} For the time being (down to the end of (h) below),
suppose that $\Cal S'$ has a greatest member and
$\pmb x\in\Bbb A\MSi^0(\Cal S)$ is
$\|\,\|_{\infty}$-bounded.

\medskip

\quad{\bf (i)} We know from 648D above that
$\pmb x'\in\Bbb B\MSi^0(\Cal S')$, while $\pmb w'$ is a
$\Bbb B$-near-simple $\Bbb B$-integrator (648E, 634Ib).
So the integral $\BSint_{\Cal S'}\pmb x'd\pmb w'$ is defined.

\medskip

\quad{\bf (ii)} Conversely, by 634Eb again,
$\Bbb B\Mns(\Cal S')=\Bbb A\Mns(\Cal S')\cap L^0(\frak B)^{\Cal S'}$ is an
$f$-subalgebra of $\Bbb A\Mns(\Cal S')$.   By
631M we have a multiplicative Riesz homomorphism
$\Psi:\Bbb A\Mns(\Cal S')\to\Bbb A\Mns(\Cal S)$ such that
$\Psi(\pmb u)$ extends $\pmb u$ for every $\pmb u\in\Bbb A\Mns(\Cal S')$.
By 633F, $\Psi(\pmb u)$ is the only $\Bbb A$-near-simple process with domain
$\Cal S$ extending $\pmb u$.

\medskip

{\bf (b)} Let $M\ge 0$ be such that
$-M\tbf 1^{(\Cal S)}_<\le\pmb x\le M\tbf 1^{(\Cal S)}_<$.
Write $Y$ for
the set of those $\pmb y\in\Bbb A\MSi^0(\Cal S)$ such that
$|\pmb y|\le 3M\tbf 1^{(\Cal S)}_<$,
$\pmb y\restr\Cal S'\in L^0(\frak B)^{\Cal S'}$ and
$\BSint_{\Cal S'}(\pmb y\restr\Cal S')d(\pmb z\restr\Cal S')
=\ASint_{\Cal S}\pmb y\,d\pmb z$
whenever $\pmb z\in\Bbb A\Mns(\Cal S)$ is an $\Bbb A$-integrator and
$\pmb z\restr\Cal S'\in L^0(\frak B)^{\Cal S'}$.

\medskip

{\bf (c)} $\pmb u_<\in Y$ whenever $\pmb u\in\Bbb A\Mmo(\Cal S)$,
$|\pmb u|\le 3M\tbf 1^{(\Cal S)}$ and
$\pmb u'=\pmb u\restr\Cal S'$ belongs to $L^0(\frak B)^{\Cal S'}$.
\Prf\ Certainly
$\pmb u_<\in\Bbb A\MSi^0(\Cal S)$ and $|\pmb u_<|\le 3M\tbf 1^{(\Cal S)}_<$.
By 641Hb once more,
$\pmb u_<\restr\Cal S'=\pmb u'_<$ belongs to
$L^0(\frak B)^{\Cal S'}$.   If $\pmb z$ is an $\Bbb A$-near-simple
$\Bbb A$-integrator with domain $\Cal S$ and
$\pmb z'=\pmb z\restr\Cal S'$ belongs to $L^0(\frak B)^{\Cal S'}$,
then $\pmb z'$ is a
$\Bbb B$-near-simple $\Bbb B$-integrator (648E, 634Ib as before).   Now

$$\eqalignno{\BSint_{\Cal S'}(\pmb u_<\restr\Cal S')d\pmb z'&
=\BSint_{\Cal S'}\pmb u'_<d\pmb z'
=\Bbb B\int_{\Cal S'}\pmb u'd\pmb z'
\Displaycause{645R(a-i)}
=\Bbb A\int_{\Cal S'}\pmb u\,d\pmb z
\Displaycause{634Eg}
=\Bbb A\int_{\Cal S}\pmb u\,d\pmb z
\Displaycause{633Ka, because $\Cal S'$ separates $\Cal S$, by 633D(b-i)}
=\ASint_{\Cal S}\pmb u_<d\pmb z.\cr}$$

\noindent As $\pmb z$ is arbitrary, $\pmb u_<\in Y$.\ \Qed

\medskip

{\bf (d)}
If $\sequencen{\pmb y_n}$ is a sequence in $Y$ which is
order*-convergent to $\pmb y$ in $L^0(\frak A)^{\Cal S}$, then
$\pmb y\in Y$.   \Prf\ Because
$|\pmb y_n|\le 3M\tbf 1^{(\Cal S)}_<$ for every $n$,
$\sequencen{\pmb y_n}$ is uniformly $\Bbb A$-previsibly-order-bounded
and $|\pmb y|\le 3M\tbf 1^{(\Cal S)}_<$.
Writing $\pmb y'=\pmb y\restr\Cal S'$,
$\pmb y'_n=\pmb y_n\restr\Cal S'$ for each $n\in\Bbb N$, we see that
$\sequencen{\pmb y'_n}$ is order*-convergent to $\pmb y'$ in
$L^0(\frak A)^{\Cal S'}$;  as $\pmb y'_n\in L^0(\frak B)^{\Cal S'}$ for every
$n$ and $L^0(\frak B)$ is topologically closed in $L^0(\frak A)$,
$\pmb y'\in L^0(\frak B)^{\Cal S'}$.   Also
$|\pmb y'_n|\le 3M\tbf 1^{(\Cal S')}_<$ for every $n$, so
$\sequencen{\pmb y'_n}$ is uniformly $\Bbb B$-previsibly-order-bounded.
By 645H, working first in $\Bbb A$ and then in $\Bbb B$,
$\pmb y\in\Bbb A\MSi^0(\Cal S)$ and
$\pmb y'\in\Bbb B\MSi^0(\Cal S')$.   Moreover,
if $\pmb z$ is an
$\Bbb A$-near-simple $\Bbb A$-integrator with domain $\Cal S$ and
$\pmb z\restr\Cal S'\in L^0(\frak B)^{\Cal S'}$, then

$$\eqalignno{\BSint_{\Cal S'}\pmb y'd(\pmb z\restr\Cal S')&
=\lim_{n\to\infty}\BSint_{\Cal S'}\pmb y'_nd(\pmb z\restr\Cal S')
\Displaycause{645S}
=\lim_{n\to\infty}\ASint_{\Cal S}\pmb y_nd\pmb z
=\ASint_{\Cal S}\pmb y\,d\pmb z.\cr}$$

\noindent So $\pmb y\in Y$.\ \Qed

\medskip

{\bf (e)} If
$\pmb v'\in\Bbb B\Mns(\Cal S')^+$ is non-decreasing and order-bounded, there is a
$\pmb y\in Y$ such that
$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb x'-\pmb y\restr\Cal S')=0$.

\medskip

\Prf\ {\bf (i)} For each $n\in\Bbb N$ take
$\acute{\pmb u}'_n\in\Bbb B\Mmo(\Cal S')$
such that
$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(|\pmb x'-\acute{\pmb u}'_{n<}|)
\le 2^{-n}$, and a non-decreasing sequence
$\sequence{m}{\acute{\pmb u}'_{nm}}$ in $\Bbb B\Mmo(\Cal S')^+$ such that
$\theta(\int_{\Cal S'}\acute{\pmb u}'_{nm}d\pmb v')\le 2^{-n+1}$ for every $m$
and $|\pmb x'-\acute{\pmb u}'_{n<}|\le\sup_{m\in\Bbb N}\acute{\pmb u}'_{nm<}$.
By 642N, working in $\Bbb B$, we have $\grave{\pmb u}'_n$,
$\grave{\pmb u}'_{nm}$ in $\Bbb B\Mns(\Cal S')$ such that
$\grave{\pmb u}'_{n<}=\acute{\pmb u}'_{n<}$,
$\grave{\pmb u}'_{nm<}=\acute{\pmb u}'_{nm}$ for
all $n$, $m\in\Bbb N$.   Setting

\Centerline{$\pmb u'_n
=\med(-M\tbf 1^{(\Cal S')},\grave{\pmb u}'_n,M\tbf 1^{(\Cal S')})$,
\quad$\pmb u'_{nm}
=(\sup_{i\le m}\grave{\pmb u}'_{ni})^+\wedge 2M\tbf 1^{(\Cal S')}$}

\noindent for $m$, $n\in\Bbb N$, we see that
$\pmb u'_n$, $\pmb u'_{nm}\in\Bbb B\Mns(\Cal S')$,
$0\le\pmb u'_n\le 2M\tbf 1^{(\Cal S')}$ and

$$\eqalign{|\pmb x'-\pmb u'_{n<}|&
\le|\pmb x'-\grave{\pmb u}'_{n<}|\wedge 2M\tbf 1^{(\Cal S')}_<
=|\pmb x'-\acute{\pmb u}'_{n<}|\wedge 2M\tbf 1^{(\Cal S')}_<\cr&
\le\sup_{m\in\Bbb N}\acute{\pmb u}'_{nm<}\wedge 2M\tbf 1^{(\Cal S')}_<
=\sup_{m\in\Bbb N}\grave{\pmb u}'_{nm<}\wedge 2M\tbf 1^{(\Cal S')}_<
=\sup_{m\in\Bbb N}\pmb u'_{nm<}\cr}$$

\noindent for all $m$, $n\in\Bbb N$.
Moreover, because $\pmb v'$ is non-decreasing,

$$\eqalignno{\int_{\Cal S'}\pmb u'_{nm}d\pmb v'&
\le\int_{\Cal S'}(\sup_{i\le n}\grave{\pmb u}'_{ni})^+d\pmb v'
=\int_{\Cal S'}(\sup_{i\le n}\acute{\pmb u}'_{ni})^+d\pmb v'
\Displaycause{using 645R(a-i),
because $(\sup_{i\le n}\grave{\pmb u}'_{ni})^+_<
=(\sup_{i\le n}\acute{\pmb u}'_{ni})^+_<$}
=\int_{\Cal S'}\acute{\pmb u}'_{nm}d\pmb v'
\le 2^{-n+1}\cr}$$

\noindent for all $m$, $n\in\Bbb N$.

\medskip

\quad{\bf (ii)} Looking at the function $\Psi$ of 631M and (a-ii) above,
$\Psi(\tbf 1^{(\Cal S')})$ must be $\tbf 1^{(\Cal S)}$.
We can speak of $\pmb u_n=\Psi(\pmb u'_n)$ and
$\pmb u_{nm}=\Psi(\pmb u'_{nm})$ for $n$, $m\in\Bbb N$, and we shall have
$|\pmb u_n|\le M\tbf 1^{(\Cal S)}$, $0\le\pmb u_{nm}\le 2M\tbf 1^{(\Cal S)}$ for
all $n$ and $m$, while $\sequence{m}{\pmb u_{nm}}$ is non-decreasing for each
$n$.   Now consider

\Centerline{$\pmb z_{nm}
=\inf_{i\le n}(\pmb u_i+\pmb u_{im})\in\Bbb A\Mns(\Cal S)$}

\noindent for $n$, $m\in\Bbb N$.   Then $\pmb z_{nm}$ always lies between
$-M\tbf 1^{(\Cal S)}$ and $3M\tbf 1^{(\Cal S)}$, while
$\sequence{m}{\pmb z_{nm}}$ is non-decreasing for each $n$ and
$\sequencen{\pmb z_{nm}}$ is non-increasing for each $m$.   So if we set

\Centerline{$\pmb y_n=\sup_{m\in\Bbb N}\pmb z_{nm<}\in\Bbb A\MSi^0(\Cal S)$}

\noindent for each $n$, $\pmb y_n$ will be the order*-limit of
$\sequence{m}{\pmb z_{nm<}}$.   By (c)-(d) above, $\pmb y_n\in Y$ for each
$n$, while $\sequencen{\pmb y_n}$ is non-increasing, so
$\pmb y=\inf_{n\in\Bbb N}\pmb y_n$ also belongs to $Y$.

\medskip

\quad{\bf (iii)} Write $\pmb z'_{nm}=\pmb z_{nm}\restr\Cal S'$,
$\pmb y'_n=\pmb y_n\restr\Cal S'$ and $\pmb y'=\pmb y\restr\Cal S'$.
We always have $\pmb x'\le\pmb u'_{n<}+\sup_{m\in\Bbb N}\pmb u'_{nm<}$.
So if $n\in\Bbb N$,

$$\eqalignno{\pmb x'&
\le\inf_{i\le n}(\pmb u'_{i<}+\sup_{m\in\Bbb N}\pmb u'_{im<})
=\sup_{m\in\Bbb N}\inf_{i\le n}(\pmb u'_{i<}+\pmb u'_{im<})
\Displaycause{using the distributive law 352Ea repeatedly}
=\sup_{m\in\Bbb N}\pmb z'_{nm<}
=\pmb y'_n.\cr}$$

\noindent (The point here is that, by 641Hb yet again,
$(\pmb z_{nm}\restr\Cal S')_<=(\pmb z_{nm})_<\restr\Cal S'$, so we can use the
formula $\pmb z'_{nm<}$ without inhibitions.)
Accordingly $\pmb x'\le\pmb y'$.   Now, for any $n\in\Bbb N$,

$$\eqalign{0&
\le\pmb y'-\pmb x'
\le\pmb y'_n-\pmb u'_{n<}+|\pmb u'_{n<}-\pmb x'|\cr&
=\sup_{m\in\Bbb N}(\pmb w'_{nm<}-\pmb u'_{n<})+|\pmb u'_{n<}-\pmb x'|
\le\sup_{m\in\Bbb N}\pmb u'_{nm<}+|\pmb u'_{n<}-\pmb x'|\cr}$$

\noindent so

$$\eqalign{\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb y'-\pmb x')&
\le\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\sup_{m\in\Bbb N}\pmb u'_{nm<})
  +\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb u_{n<}'-\pmb x')\cr&
\le\sup_{m\in\Bbb N}\int_{\Cal S'}\pmb u'_{nm}d\pmb v'+2^{-n}
\le 2^{-n+1}+2^{-n}.\cr}$$

\noindent As $n$ is arbitrary,
$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb y'-\pmb x')=0$, as required.\ \Qed

\medskip

{\bf (f)} If $\pmb v\in\Mnsu(\Cal S)$ and $\pmb v'=\pmb v\restr\Cal S'$
belongs to $L^0(\frak B)^{\Cal S'}$, then
$\BSint_{\Cal S'}\pmb x'd\pmb v'=\ASint_{\Cal S'}\pmb x\,d\pmb v$.   \Prf\
By 648E again,
$\pmb v'$ is $\Bbb B$-near-simple, and of course it is non-negative and
non-decreasing.   By (e), there is a $\pmb y\in Y$ such that
$\Bbb B\widehat{\theta}^{\#}_{\pmb v'}(\pmb x'-\pmb y')=0$, writing
$\pmb y'$ for $\pmb y\restr\Cal S'$.   Now

\inset{$\BSint_{\Cal S'}\pmb x'-\pmb y'\,d\pmb v'=0$ by 646E,

$\BSint_{\Cal S'}\pmb y'd\pmb v'=\ASint_{\Cal S}\pmb y\,d\pmb v$ because
$\pmb y\in Y$,

$\ASint_{\Cal S}\pmb x-\pmb y\,d\pmb v=0$ by 648F,}

\noindent so $\BSint_{\Cal S'}\pmb x'd\pmb v'=\ASint_{\Cal S}\pmb x\,d\pmb v$.\
\Qed

\medskip

{\bf (g)} If $\pmb z$ is an $\Bbb A$-$L^2$-martingale with domain $\Cal S$ and
$\pmb z'=\pmb z\restr\Cal S'$ belongs to $L^0(\frak B)^{\Cal S'}$, then
$\BSint_{\Cal S'}\pmb x'd\pmb z'=\ASint_{\Cal S'}\pmb x\,d\pmb z$.   \Prf\
We are supposing that $\Cal S'$ has a greatest member, which will also be the
greatest member of $\Cal S$.   By 632Ia again, $\pmb z$ is
$\Bbb A$-locally-near-simple,
therefore $\Bbb A$-near-simple;  by 366J, it is $\|\,\|_2$-bounded, therefore
$\|\,\|_1$-bounded and an $\Bbb A$-integrator (622G).
Write $\pmb z^*$ for the $\Bbb A$-quadratic variation of $\pmb z$,
which is non-negative,
non-decreasing, locally near-simple (631Jb) and an integrator (617I again),
therefore
belongs to $\Bbb A\Mnsu(\Cal S)$.   The $\Bbb A$-quadratic variation of
$\pmb z'$ is $\pmb z^*\restr\Cal S'$ (633Ph), and this is equal to its
$\Bbb B$-quadratic variation (634Ib).

By (e) again, there is a $\pmb y\in Y$ such that,
setting $\pmb y'=\pmb y\restr\Cal S'$,
$\Bbb B\widehat{\theta}^{\#}_{\pmb z^*\restr\Cal S'}(\pmb x'-\pmb y')=0$.
Since $|\pmb x-\pmb y|\le 4M\tbf 1^{(\Cal S)}$,
$(\pmb x'-\pmb y')^2\le 4M|\pmb x'-\pmb y'|$ and
$\Bbb B\widehat{\theta}^{\#}_{\pmb z^*\restr\Cal S'}(\pmb x'-\pmb y')^2=0$.
By 646E and 648F,

\Centerline{$\BSint_{\Cal S'}(\pmb x'-\pmb y')^2d(\pmb z^*\restr\Cal S')
=\ASint_{\Cal S}(\pmb x-\pmb y)^2d\pmb z^*=0$;}

\noindent by 648C,

\Centerline{$\|(\BSint_{\Cal S'}(\pmb x'-\pmb y')d\pmb z')^2\|_1
=\|(\ASint_{\Cal S}(\pmb x-\pmb y)d\pmb z)^2\|_1=0$,}

\Centerline{$\BSint_{\Cal S'}(\pmb x'-\pmb y')d\pmb z'
=\ASint_{\Cal S}(\pmb x-\pmb y)d\pmb z=0$}

\noindent and

\Centerline{$\BSint_{\Cal S'}\pmb x'd\pmb z'
=\BSint_{\Cal S'}\pmb y'd\pmb z'
=\ASint_{\Cal S}\pmb y\,d\pmb z
=\ASint_{\Cal S}\pmb x\,d\pmb z$.  \Qed}

\medskip

{\bf (h)} Now turn to the given $\Bbb A$-near-simple $\Bbb A$-integrator
$\pmb w$.   Let $\epsilon>0$.   Working in
$\Bbb B$, 643O tells us that
we have an $L^{\infty}$-bounded $\Bbb B$-martingale $\tildepmb w'$ and a
$\Bbb B$-near-simple
process $\pmb v'$ of bounded variation, both with domain $\Cal S'$, such that
$a=\Bvalue{\pmb w'\ne\pmb v'+\tildepmb w'}$ has measure at most $\epsilon$.
Now $\pmb v'$ can be expressed as $\pmb v'_1-\pmb v'_2$
where $\pmb v'_1$, $\pmb v'_2\in\Bbb B\Mnsu(\Cal S')$ (631L).   Let
$\pmb v_1=\Psi(\pmb v'_1)$, $\pmb v_2=\Psi(\pmb v'_2)$
be the $\Bbb A$-near-simple processes extending
$\pmb v'_1$ and $\pmb v'_2$ to $\Cal S$.
Then $\pmb v_1$ and $\pmb v_2$ belong to $\Bbb A\Mnsu(\Cal S)$.
Since $\tildepmb w'$ is also an
$\Bbb A$-martingale (634Ia), and $\Cal S'$ is cofinal with $\Cal S$, there is a
$\Bbb A$-martingale $\tildepmb w$ with domain $\Cal S$ which extends
$\tildepmb w'$.   As in (g), $\tildepmb w$ is $\Bbb A$-near-simple,
so must be
$\Psi(\tildepmb w')$, and in particular is $\|\,\|_{\infty}$-bounded.

Writing $\pmb z$ for $\pmb v_1-\pmb v_2+\tildepmb w$ and $\pmb z'$ for
$\pmb z\restr\Cal S'=\pmb v'+\tildepmb w'$, we have

$$\eqalignno{\BSint_{\Cal S'}\pmb x'd\pmb z'&
=\BSint_{\Cal S'}\pmb x'd\pmb v'_1-\BSint_{\Cal S'}\pmb x'd\pmb v'_2
     +\BSint_{\Cal S'}\pmb x'd\tildepmb w'\cr&
=\ASint_{\Cal S}\pmb x\,d\pmb v_1-\ASint_{\Cal S}\pmb x\,d\pmb v_2
     +\ASint_{\Cal S}\pmb x\,d\tildepmb w
\Displaycause{by (f) and (g)}
=\ASint_{\Cal S}\pmb x\,d\pmb z.\cr}$$

\noindent So

$$\eqalignno{\Bvalue{\BSint_{\Cal S'}\pmb x'd\pmb w'
    \ne\ASint_{\Cal S'}\pmb x\,d\pmb w}&
\Bsubseteq\Bvalue{\BSint_{\Cal S'}\pmb x'd\pmb w'
      \ne\BSint_{\Cal S'}\pmb x'd\pmb z'}
   \Bcup\Bvalue{\ASint_{\Cal S}\pmb x\,d\pmb w
      \ne\ASint_{\Cal S}\pmb x\,d\pmb z}\cr&
\Bsubseteq\Bvalue{\pmb w'\ne\pmb z'}
   \Bcup\Bvalue{\pmb w\ne\pmb z}
\Displaycause{646C}
=\Bvalue{\pmb w'-\pmb z'\ne 0}
   \Bcup\Bvalue{\Psi(\pmb w'-\pmb z')\ne 0}
=\Bvalue{\pmb w'-\pmb z'\ne 0}
\Displaycause{631M(b-v)}
=a\cr}$$

\noindent has measure at most $\epsilon$.   As $\epsilon$ is arbitrary,
$\BSint_{\Cal S'}\pmb x'd\pmb w'=\ASint_{\Cal S'}\pmb x\,d\pmb w$.

\medskip

{\bf (i)} This deals with the case in which $\Cal S'$ has a greatest member and
$\pmb x\in\Bbb A\MSi^0(\Cal S)$ is $\|\,\|_{\infty}$-bounded.   If we know just
that $\Cal S'$ has a greatest member, set

\Centerline{$\pmb x_n=\med(-n\tbf 1^{(\Cal S)}_<,\pmb x,n\tbf 1^{(\Cal S)}_<)$}

\noindent for each $n$.   Then each $\pmb x_n\in\Bbb A\MSi^0(\Cal S)$
is $\|\,\|_{\infty}$-bounded.   Because
$\pmb x\times\tbf 1^{(\Cal S)}_<\in\Bbb A\MSi^0(\Cal S)$ is
$\Bbb A$-previsibly-order-bounded,
$\sequencen{\pmb x_n}$ is $\Bbb A$-uniformly-previsibly-order-bounded, while
$\sequencen{\pmb x_n}$ is order*-convergent to
$\pmb x\times\tbf 1^{(\Cal S)}_<$.   By 648D,
$(\pmb x\times\tbf 1^{(\Cal S)}_<)\restr\Cal S'\in\Bbb B\MSi^0(\Cal S')$ is
$\Bbb B$-previsibly-order-bounded so $\sequencen{\pmb x_n\restr\Cal S'}$ is
$\Bbb B$-uniformly-previsibly-order-bounded, while it is
order*-convergent to $(\pmb x\times\tbf 1^{(\Cal S)}_<)\restr\Cal S'$.
Accordingly

$$\eqalignno{\ASint_{\Cal S}\pmb x\,d\pmb w&
=\ASint_{\Cal S}\pmb x\times\tbf 1^{(\Cal S)}_<\,d\pmb w
=\lim_{n\to\infty}\ASint_{\Cal S}\pmb x_nd\pmb w
\Displaycause{645T}
=\lim_{n\to\infty}\BSint_{\Cal S}(\pmb x_n\restr\Cal S')d\pmb w'
\Displaycause{by (a)-(h)}
=\BSint_{\Cal S'}(\pmb x\restr\Cal S')\times\tbf 1^{(\Cal S')}_<d\pmb w'
=\BSint_{\Cal S'}\pmb x'd\pmb w'.\cr}$$

\medskip

{\bf (j)} Finally, for the general case of a relatively order-convex sublattice
$\Cal S'$ of $\Cal T_{\Bbb B}$, we know from 646M that

\Centerline{$\BSint_{\Cal S'}\pmb x'd\pmb w'
=\lim_{\tau\uparrow\Cal S'}\BSint_{\Cal S'\wedge\tau}\pmb x'd\pmb w'$,}

\Centerline{$\ASint_{\Cal S}\pmb x\,d\pmb w'
=\lim_{\tau\uparrow\Cal S}\ASint_{\Cal S\wedge\tau}\pmb x\,d\pmb w
=\lim_{\tau\uparrow\Cal S'}\ASint_{\Cal S\wedge\tau}\pmb x\,d\pmb w$}

\noindent because $\Cal S'$ is cofinal with $\Cal S$.   But for every
$\tau\in\Cal S'$, $\Cal S\wedge\tau$ is the order-convex hull of
$\Cal S'\wedge\tau$, so we can apply (b)-(i) to see that
$\BSint_{\Cal S'\wedge\tau}\pmb x'd\pmb w'
=\ASint_{\Cal S\wedge\tau}\pmb x\,d\pmb w$.   In the limit, we have
$\BSint_{\Cal S'}\pmb x'd\pmb w'=\ASint_{\Cal S}\pmb x\,d\pmb w$.
}%end of proof of 648G

%\leader{648L}{}
%\leader{648M}{}
%\leader{648N}{}
%\leader{648O}{}
%\leader{648P}{}
%\leader{648Q}{}
%\leader{648R}{}
%\leader{648S}{}
%\leader{648T}{}
%\leader{648U}{}
%\leader{648V}{}
%\leader{648W}{}

\exercises{\leader{648X}{Basic exercises (a)} Let $(\frak A,\bar\mu)$ be a
probability algebra and $\frak B$ a closed subalgebra of $\frak A$.   Show that
there is a continuous order-preserving projection from $L^0(\frak A)$ onto
$L^0(\frak B)$.
%648D

%\leader{648Y}{Further exercises (a)}

}%end of exercises

\leader{648Z}{Problem} In 648G, can we drop the hypothesis that `$\frak B$ is
coordinated with $\family{t}{T}{\frak A_t}$'?

\endnotes{
\Notesheader{648} We have just gone through an uncommonly dense argument aiming
at an expected result.   Some of the complications (e.g., the shift from
the moderately oscillatory processes $\acute{\pmb u}'_n$ to
the near-simple processes $\grave{\pmb u}'_n$ in part (e) of the proof of 648G)
arise from the idiosyncratic formulations I have chosen.   Others come from my
extension of the S-integral from previsible processes, as described in \S642, to
the S-integrable processes of \S645;
in effect, this is part (f) of the proof of 648G, relying on 648F.   I note also
that part (d) of the proof of 648D duplicates the idea of 647B.

Periodically, in this volume, I have looked at the question of law-independence.
I noted in 645Q that the S-integral is law-independent.   In 648G the
hypotheses and conclusion are all law-independent {\it except} for the
requirement that $\frak B$ should be coordinated with
$\family{t}{T}{\frak A_t}$.   It would be sufficient to suppose that

\inset{there is a $\bar\nu$ such that $(\frak A,\bar\nu)$ is a probability
algebra and $\frak B$ is coordinated with
$(\frak A,\bar\nu,\family{t}{T}{\frak A_t})$,}

\noindent obtaining a version which demands only that $\frak A$ should be a
measurable algebra.   But an affirmative answer to 648Z would show that no such
manoeuvre is called for.

}%end of notes

\discrpage



\frfilename{mt649.tex}
\versiondate{30.9.14/28.1.20}
\copyrightdate{2014}

\def\chaptername{The S-integral}
\def\sectionname{Pathwise integration}

\newsection{649}

The integrals of \S613 and \S645 are defined in terms of convergence in
$L^0$.   The most important applications are associated with processes of the
form $\langle X_t(\omega)\rangle_{t\ge 0,\omega\in\Omega}$ with paths
$\langle X_t(\omega)\rangle_{t\ge 0}$.
It turns out that in the case of the Riemann-sum integral, we can often,
with some effort, define integrals `pathwise'.
I do not think that this approach gives a good picture of the theory as a
whole, but it is surely worth knowing what can be done.

The S-integral is rather different;  I do not see any way of giving a
pathwise description of the S-integral with respect to Brownian motion, for
instance.   But for non-decreasing integrators we have an effective
approach through Stieltjes integrals, which I have hinted at in earlier
sections.   I now give a detailed account of the method (649H, 649L).

\cmmnt{\leader{649A}{Notation} We need only fragments
of the standard framework.   \Asissix, if not explicitly introduced,
will be a stochastic integration
structure.   $L^0$ will be the
Riesz space $L^0(\frak A)$ with the topology of convergence in measure.
If $\Cal S$ is a sublattice of $\Cal T$,
$\Cal I(\Cal S)$ will be the set of finite sublattices of $\Cal S$,
$\Msimp(\Cal S)$ will be the space of simple processes with domain
$\Cal S$ and $\Mns(\Cal S)$ will be the space of near-simple processes with
domain $\Cal S$;  if $\tau\in\Cal S$ then $\Cal S\wedge\tau$ will be
$\{\sigma\wedge\tau:\sigma\in\Cal S$;  if
$\pmb v$ is a fully adapted
process with domain including $\Cal S$,
$Q_{\Cal S}(d\pmb v)$ will be the capped-stake variation set of
$\pmb v\restr\Cal S$.   When $\pmb v$ is near-simple, $\pmb v_<$ will be
its previsible version.   I use $\int$ to denote Riemann-sum integrals and
$\Sint$ to denote S-integrals.
}

\leader{649B}{Theorem} Suppose that $\Cal S$ is a sublattice of $\Cal T$
with a least element.   Let
$\pmb{v}=\sfamilysS{v}$ be an integrator and $\pmb{u}=\sfamilysS{u}$
a near-simple process.
Suppose that we have, for each $n\in\Bbb N$,
a non-decreasing sequence $\sequence{i}{\tau_{ni}}$ in $\Cal S$
such that $\tau_{n0}=\min\Cal S$,
$\inf_{i\in\Bbb N}\Bvalue{\tau_{ni}<\sup\Cal S}=0$ and, for each
$i\in\Bbb N$,

\Centerline{$\Bvalue{\sigma<\tau_{n,i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_{n,i}}|\le 2^{-n}}$ for every
$\sigma\in[\tau_{n,i},\tau_{n,i+1}]$,}

\Centerline{$\Bvalue{\tau_{n,i+1}<\sup\Cal S}
\Bsubseteq\Bvalue{|u_{\tau_{n,i+1}}-u_{\tau_{n,i}}|\ge 2^{-n}}$.}

\noindent Then

\Centerline{$z_n=\lim_{k\to\infty}\sum_{i=0}^{k-1}
u_{\tau_{n,i}}\times(v_{\tau_{n,i+1}}-v_{\tau_{n,i}})$}

\noindent is defined for each $n$, and $\sequencen{z_n}$
is order*-convergent to $\int_{\Cal S}\pmb u\,d\pmb v$.

\proof{{\bf (a)} Let $n\in\Bbb N$.

\medskip

\quad{\bf (i)} Set $z_{nk}=\sum_{i=0}^{k-1}
u_{\tau_{ni}}\times(v_{\tau_{n,i+1}}-v_{\tau_{ni}})$ for $k\in\Bbb N$.
If $k\le l$, then

$$\eqalign{\Bvalue{z_{nk}\ne z_{nl}}
&\Bsubseteq\sup_{k\le i<l}\Bvalue{v_{\tau_{n,i+1}}\ne v_{\tau_{ni}}}
\Bsubseteq\sup_{k\le i<l}\Bvalue{\tau_{n,i+1}\ne\tau_{ni}}
\cr&\Bsubseteq\sup_{k\le i\le l}\Bvalue{\tau_{ni}\ne\sup\Cal S}
\Bsubseteq\Bvalue{\tau_{nk}<\sup\Cal S}.
\cr}$$

\noindent But as $\sequence{k}{\Bvalue{\tau_{nk}<\sup\Cal S}}$ is a
non-increasing sequence with infimum $0$, this means that
$\sequence{k}{z_{nk}}$ is actually order*-convergent to some
$z_n\in L^0$, with
$\Bvalue{z_{nk}\ne z_n}\Bsubseteq\Bvalue{\tau_{nk}<\sup\Cal S}$ for every
$k$.   Thus the topological limit $z_n$ is always defined.

\medskip

\quad{\bf (ii)} Define
$\pmb u_n=\family{\sigma}{\Cal S}{u_{n\sigma}}$ by saying that,
for $\sigma\in\Cal S$,

\Centerline{$\Bvalue{\tau_{ni}=\sigma}
   \Bcup(\Bvalue{\tau_{ni}\le\sigma}\Bcap\Bvalue{\sigma<\tau_{n,i+1}})
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_{ni}}}$}

\noindent for every $i\in\Bbb N$;  because

\Centerline{$\sequence{i}{\Bvalue{\tau_{ni}=\sigma}
   \Bcup(\Bvalue{\tau_{ni}\le\sigma}\Bcap\Bvalue{\sigma<\tau_{n,i+1}})}$}

\noindent is always disjoint, with supremum

\Centerline{$\sup_{i\in\Bbb N}\Bvalue{\sigma\le\tau_{ni}}
\Bsupseteq\sup_{i\in\Bbb N}\Bvalue{\tau_{ni}=\sup\Cal S}=1$,}

\noindent $\pmb u_n$ is a fully adapted process.   Next,
$\pmb u_n\restr\Cal S\wedge\tau_{nk}$ is simple, with breakpoint string
$(\tau_{n0},\ldots,\tau_{nk})$, and
$z_{nk}=\int_{\Cal S\wedge\tau_{nk}}\pmb u_nd\pmb v$
for each $k$ (614C).   We know also that if $k\in\Bbb N$ and
$\sigma\in\Cal S\wedge\tau_{nk}$,

\Centerline{$1
=\sup_{i<k}(\Bvalue{\tau_{ni}\le\sigma}\Bcap\Bvalue{\sigma<\tau_{n,i+1}})
    \Bcup\Bvalue{\sigma=\tau_{nk}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{n\sigma}|\le 2^{-n}}$.}

\noindent But this means that if $I\in\Cal I(\Cal S\wedge\tau_{nk})$,
$S_I(\pmb u-\pmb u_n,d\pmb v)\in 2^{-n}Q_{\Cal S}(d\pmb v)$.

\medskip

\quad{\bf (iii)} The integral $\int_{\Cal S}\pmb u\,d\pmb v$ is
$\lim_{k\to\infty}\int_{\Cal S\wedge\tau_{nk}}\pmb u\,d\pmb v$.
\Prf\ We know that $\int_{\Cal S}\pmb u\,d\pmb v
=\lim_{\sigma\uparrow\Cal S}\int_{\Cal S\wedge\sigma}\pmb u\,d\pmb v$
(613J(f-ii)).   Similarly, if $k\in\Bbb N$,
$\int_{\Cal S\wedge\tau_{nk}}\pmb u\,d\pmb v=\lim_{\sigma\uparrow\Cal S}
\int_{\Cal S\wedge\tau_{nk}\wedge\sigma}\pmb u\,d\pmb v$.
But this means that

$$\eqalignno{\Bvalue{\int_{\Cal S}\pmb u\,d\pmb v
  \ne\int_{\Cal S\wedge\tau_{nk}}\pmb u\,d\pmb v}
&\Bsubseteq\sup_{\sigma\in\Cal S}
  \Bvalue{\int_{\Cal S\wedge\sigma}\pmb u\,d\pmb v
    \ne\int_{\Cal S\wedge\tau_{nk}\wedge\sigma}\pmb u\,d\pmb v}
\cr&\Bsubseteq\sup_{\sigma\in\Cal S}
  \Bvalue{\sigma\ne\tau_{nk}\wedge\sigma}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\tau_{nk}<\sigma}
=\Bvalue{\tau_{nk}<\sup\Cal S}
\cr}$$

\noindent (611Eb).   As
$\inf_{k\in\Bbb N}\sup_{l\ge k}\Bvalue{\tau_{nk}<\sup\Cal S}=0$,
$\int_{\Cal S}\pmb u\,d\pmb v
=\lim_{k\to\infty}\int_{\Cal S\wedge\tau_{nk}}\pmb u\,d\pmb v$.\ \Qed

\medskip

{\bf (b)} Let $C$ be the solid convex hull of
$Q_{\Cal S}(d\pmb v)$, and $\overline{C}$ its topological
closure.   Then $C$ is topologically bounded (627M) so $\overline{C}$ is
topologically bounded (3A5N(b-ii));  also $\overline{C}$ is convex
(2A5Eb) and solid (613B(f-vi)).   Now we see from (a-ii) that if
$n$, $k\in\Bbb N$ then
$S_I(\pmb u-\pmb u_n,d\pmb v)\in 2^{-n}C$ for every
$I\in\Cal I([\min\Cal S,\tau_{nk}])$, so

\Centerline{$\int_{\Cal S\wedge\tau_{nk}}\pmb u\,d\pmb v-z_{nk}
=\int_{\Cal S\wedge\tau_{nk}}\pmb u-\pmb u_nd\pmb v
\in 2^{-n}\overline{C}$;}

\noindent letting $k\to\infty$, and using (a-iii),

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v-z_n
\in 2^{-n}\overline{C}$.}

\noindent Because $\overline{C}$ is solid and convex,

\Centerline{$\sup_{l\le n\le m}|z_n-\int_{\Cal S}\pmb u\,d\pmb v|
\in\sum_{n=l}^m2^{-n}\overline{C}\subseteq 2^{-l+1}\overline{C}$}

\noindent whenever $l\le m$.   By 613B(f-v),
$\{\sup_{l\le n\le m}|z_n-\int_{\Cal S}\pmb u\,d\pmb v|:m\ge l\}$
is bounded above in $L^0$, and its supremum $w_l$ belongs to
$2^{-l+1}\overline{C}$ (613Ba).

Repeating the argument, $w'_p=\sup_{l\ge p}w_l$ is defined, and belongs to
$2^{-p+2}\overline{C}$, for every $p\in\Bbb N$.   As $\overline{C}$ is
topologically bounded, $\lim_{p\to\infty}\theta(w'_p)=0$ and
$\inf_{p\in\Bbb N}w'_p=0$.   But
$|z_n-\int_{\Cal S}\pmb u\,d\pmb v|\le w_p\le w'_p$ whenever
$n\ge p$, while $\sequence{p}{w'_p}$ is non-increasing and has infimum $0$,
so $\sequencen{z_n}$ is order*-convergent to
$\int_{\Cal S}\pmb u\,d\pmb v$, as claimed.
}%end of proof of 649B

\leader{649C}{}\cmmnt{ From 649B, we see that, under the hypotheses there,
we have a chance of expressing a Riemann-sum integral as an order*-limit
of order*-limits of explicitly defined Riemann sums.   The hypotheses are
elaborate, but they correspond to a version of $\SL_1$ in 631Oa,
so they will be satisfied in a useful number of cases.   The really
important feature of the result, however, is that (under appropriate
conditions) the $\tau_{ni}$, and hence the $z_n$, can be determined
pathwise, as in the following form of the theorem.

\medskip

\noindent}{\bf Corollary}\cmmnt{ ({\smc Bichteler 81}, 7.14, or
{\smc Karandikar 95})}
Let $(\Omega,\Sigma,\mu)$ be a complete probability space and
$\family{t}{\coint{0,\infty}}{\Sigma_t}$ a
right-continuous filtration of $\sigma$-subalgebras of $\Sigma$,
all containing every negligible subset of $\Omega$;
suppose that $(\frak A,\bar\mu)$ and $\langle\frak A_t\rangle_{t\ge 0}$
are the
corresponding measure algebra and filtration of closed subalgebras.
Let $\langle U_t\rangle_{t\ge 0}$,
$\langle V_t\rangle_{t\ge 0}$ be stochastic processes
such that $t\mapsto U_t(\omega):\coint{0,\infty}\to\Bbb R$ is \cadlag\
for every $\omega\in\Omega$, and
$(t,\omega)\mapsto V_t(\omega):\coint{0,\infty}\times\Omega\to\Bbb R$
is progressively measurable;  let $\pmb{u}$, $\pmb{v}$ be the
corresponding fully adapted processes with domain $\Cal T_f$\cmmnt{, as in
612H and 631D}.   Suppose that $\pmb{v}$ is a local integrator.

Let $h:\Omega\to\coint{0,\infty}$ be a stopping time, and
$\tau^*=h^{\ssbullet}$ the corresponding stopping time in $\Cal T_f$.
For $n\in\Bbb N$ and $\omega\in\Omega$, define
$h_{ni}(\omega)$,
for $i\in\Bbb N$, by setting $h_{n0}(\omega)=0$ and then

\Centerline{$h_{n,i+1}(\omega)=\inf(\{h(\omega)\}
  \cup\{t:t\ge h_{ni}(\omega),\,
    |U_t(\omega)-U_{h_{ni}(\omega)}(\omega)|>2^{-n}\})$}

\noindent for $i\in\Bbb N$.

In this case,

(a) every $h_{ni}$ is a stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$,

(b)

\Centerline{$f_n(\omega)=\sum_{i=0}^{\infty}
   U_{h_{ni}}(\omega)
     \bigl(V_{h_{n,i+1}}(\omega)-V_{h_{ni}}(\omega)\bigr)$}

\noindent is defined for all $n\in\Bbb N$ and $\omega\in\Omega$,

(c)
$f=\lim_{n\to\infty}f_n$ is defined in $\Bbb R$ almost everywhere in
$\Omega$, and $f^{\ssbullet}=\int_{[\check 0,\tau^*]}\pmb{u}\,d\pmb{v}$.

\proof{{\bf (a)} Induce on $i$.   If $i=0$ this is trivial.
For the inductive step to $i+1$, given $t\ge 0$,

$$\eqalign{&\{\omega:h_{n,i+1}(\omega)<t\}
\cr&\mskip50mu
=\{\omega:h(\omega)<t\}
  \cup\bigcup_{s<t}\{\omega:h_{ni}(\omega)\le s,\,
    |U_s(\omega)-U_{h_{ni}(\omega)}(\omega)|>2^{-n}\}
\cr&\mskip50mu
=\{\omega:h(\omega)<t\}
  \cup\bigcup_{q\in\Bbb Q,q<t}\{\omega:h_{ni}(\omega)\le q,\,
    |U_q(\omega)-U_{h_{ni}(\omega)}(\omega)|>2^{-n}\}
\cr}$$

\noindent because $s\mapsto U_s(\omega)$ is \cadlag\ for every $\omega$.
Now the inductive hypothesis tells us that $h_{ni}$ is a stopping time,
and we saw in 631D that $(t,\omega)\mapsto U_t(\omega)$ is progressively
measurable, so $U_{h_{ni}}$ is $\Sigma_{h_{ni}}$-measurable
(455Le), where I write $U_{h_{ni}}$ for
$\omega\mapsto U_{h_{ni}(\omega)}(\omega)$, as in 455L and 612H.
For $\alpha\in\Bbb R$, set

\Centerline{$E_{\alpha}=\{\omega:U_{h_{ni}(\omega)}(\omega)>\alpha\}
=\{\omega:U_{h_{ni}}(\omega)>\alpha\}\in\Sigma_{h_{ni}}$;}

\noindent if $q<t$, then $E_{\alpha}\cap\{\omega:h_{ni}(\omega)\le q\}$
belongs to $\Sigma_q\subseteq\Sigma_t$.   But this means that
$U_{h_{ni}}\times\chi\{\omega:h_{ni}(\omega)\le q\}$, and therefore
$(U_q-U_{h_{ni}})\times\chi\{\omega:h_{ni}(\omega)\le q\}$, are
$\Sigma_t$-measurable for every $q<t$, and
$\bigcup_{q\in\Bbb Q,q<t}\{\omega:h_{ni}(\omega)\le q,\,
    |U_q(\omega)-U_{h_{ni}(\omega)}(\omega)|>2^{-n}\}\in\Sigma_t$.

It follows at once that $\{\omega:h_{n,i+1}(\omega)<t\}$ belongs to
$\Sigma_t$ for every $t$.   But we are supposing that
$\langle\Sigma_t\rangle_{t\ge 0}$ is right-continuous, so
$h_{ni}$ is a stopping time adapted to $\langle\Sigma_t\rangle_{t\ge 0}$,
by 455Lb.

\medskip

{\bf (b)} Accordingly I will allow myself to use the notations
$U_{h_{ni}}$, $V_{h_{ni}}$ for all $n$, $i\in\Bbb N$.

\medskip

\quad{\bf (i)}
$|U_{h_{n,i+1}}(\omega)-U_{h_{ni}}(\omega)|\ge 2^{-n}$ whenever
$n$, $i\in\Bbb N$, $\omega\in\Omega$ and $h_{n,i+1}(\omega)<h(\omega)$.
\Prf\Quer\ Otherwise, because $t\mapsto U_t(\omega)$ is
everywhere continuous on the right, there is a $t>h_{n,i+1}(\omega)$ such
that $|U_s(\omega)-U_{h_{ni}}(\omega)|\le 2^{-n}$ for every
$s\in[h_{n,i+1}(\omega),t]$;  but now $h_{n,i+1}(\omega)$ must be greater
than or equal to $\min(t,h(\omega))$.\ \Bang\Qed

\medskip

\quad{\bf (ii)} Note also that, for each $\omega\in\Omega$ and
$n\in\Bbb N$,
$h_{ni}(\omega)=h(\omega)$ for all but finitely many $i$.   \Prf\Quer\
Otherwise, $\sequence{i}{h_{ni}(\omega)}$ is a strictly increasing sequence
in $[0,h(\omega)]$, while $\sequence{i}{U_{h_{ni}(\omega)}(\omega)}$
is not convergent.\ \Bang\QeD\  Accordingly

\Centerline{$f_n(\omega)=\sum_{i=0}^{\infty}
   U_{h_{ni}(\omega)}
     \bigl(V_{h_{n,i+1}}(\omega)-V_{h_{ni}}(\omega)\bigr)$}

\noindent is defined, as required.

\medskip

{\bf (c)(i)} Set $\Cal S=[\check 0,\tau^*]\subseteq\Cal T$.   We know from
631D that $\pmb u$ is locally near-simple, so
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined.   For
$n$, $i\in\Bbb N$, set $\tau_{ni}=h_{ni}^{\ssbullet}\in\Cal T_f$, as in
612Ha.   Then $\langle\tau_{ni}\rangle_{n,i\in\Bbb N}$ satisfies the
hypotheses of 649B.   \Prf\ Take $n\in\Bbb N$.
Because $\sequence{i}{h_{ni}}$ is non-decreasing, so is
$\sequence{i}{\tau_{ni}}$.   Of course $\tau_{n0}=\check 0=\min\Cal S$.
We have $\sup S=\tau^*=h^{\ssbullet}$, so

\Centerline{$\inf_{i\in\Bbb N}\Bvalue{\tau_{ni}<\sup\Cal S}
=(\bigcap_{i\in\Bbb N}\{\omega:h_{ni}(\omega)<h(\omega)\})^{\ssbullet}
=0$}

\noindent by (b-ii).   If $i\in\Bbb N$ and
$\sigma\in[\tau_{ni},\tau_{n,i+1}]$
then we can express $\sigma$ as $g^{\ssbullet}$ where $g$ is a stopping
time and $h_{ni}\le g\le h_{n,i+1}$;  in this case
$u_{\sigma}=U_g^{\ssbullet}$ (612H(b-i)) and
$|U_g(\omega)-U_{h_{ni}}(\omega)|\le 2^{-n}$ whenever
$g(\omega)<h_{n,i+1}(\omega)$, so
$\Bvalue{\sigma<\tau_{n,i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_{ni}}|\le 2^{-n}}$.   And (b) above
tells us that $\Bvalue{\tau_{n,i+1}<\tau^*}
\Bsubseteq\Bvalue{|u_{\tau_{n,i+1}}-u_{\tau_{ni}}|\ge 2^{-n}}$.\ \Qed

\medskip

\quad{\bf (ii)} If $n\in\Bbb N$,

\Centerline{$f_n(\omega)=\lim_{k\to\infty}\sum_{i=0}^{k-1}
   U_{h_{ni}}(\omega)
     \bigl(V_{h_{n,i+1}}(\omega)-V_{h_{ni}}(\omega)\bigr)$}

\noindent for every $\omega$.   So

\Centerline{$f_n^{\ssbullet}=\lim_{k\to\infty}\sum_{i=0}^{k-1}
   u_{\tau_{ni}}\times(v_{\tau_{n,i+1}}-v_{\tau_{ni}})$}

\noindent can be identified with $z_n$ as described in 649B.   As
$\sequencen{z_n}\to^*\int_{\Cal S}\pmb u\,d\pmb v$, 642Ba tells us that
$\sequencen{f_n}$ is convergent a.e., and if $f=\lim_{n\to\infty}f_n$
then $f^{\ssbullet}=\int_{\Cal S}\pmb u\,d\pmb v$, as required.
}%end of proof of 649C

\leader{649D}{}\cmmnt{ In the presence of a special kind of filter on
$\Bbb N$, we have a quite different way of calculating stochastic integrals
by looking at one path at a time.

\medskip

}{\bf (a) Definition} A filter $\Cal F$ on $\Bbb N$
is {\bf measure-converging}\cmmnt{ (538Ag)} if
whenever $(\Omega,\Sigma,\mu)$ is a probability space,
$\sequencen{E_n}$ is a sequence in
$\Sigma$, and $\lim_{n\to\infty}\mu E_n=1$, then
$\bigcup_{A\in\Cal F}\bigcap_{n\in A}E_n$ is conegligible.

\spheader 649Db Suppose that $\Cal F$ is a measure-converging
filter on $\Bbb N$, $(\Omega,\Sigma,\mu)$ is a probability space,
and $\sequencen{f_n}$ is a sequence in $\eusm L^0=\eusm L^0(\mu)$
which converges in measure to
$f\in\eusm L^0$.   Then $\lim_{n\to\Cal F}f_n\eae f$\prooflet{
(538N(a-iii))}.

\cmmnt{\medskip

\noindent{\bf Remark} It seems still to be unknown whether ZFC is enough to
prove the existence of a measure-converging filter (see 538Z).   However,
the continuum hypothesis, for instance, is more than sufficient to ensure
that measure-converging filters exist (538Ng).
}

\leader{649E}{Proposition}\cmmnt{ (cf.\ {\smc Nutz p11})}
Suppose that $\Cal F$ is a measure-converging filter on $\Bbb N$.
Let $(\Omega,\Sigma,\mu)$ be a complete probability space,
$\langle\Sigma_t\rangle_{t\ge 0}$ a right-continuous filtration of
$\sigma$-subalgebras of $\Sigma$ all containing every negligible set;
suppose that $(\frak A,\bar\mu)$,
$\langle\frak A_t\rangle_{t\ge 0}$ are
the corresponding probability algebra
and filtration of closed subalgebras.   Let
$\langle U_t\rangle_{t\ge 0}$,
$\langle V_t\rangle_{t\ge 0}$ be stochastic
processes on $\Omega$, adapted to $\langle\Sigma_t\rangle_{t\ge 0}$,
such that the paths $t\mapsto U_t(\omega)$,
$t\mapsto V_t(\omega)$ are \cadlag\ for every $\omega$;  let
$\pmb u$, $\pmb{v}$ be the corresponding locally near-simple
processes defined on $\Cal T_f$\cmmnt{ (631D)},
and suppose that $\pmb{v}$ is a local integrator.
Let $h$, $h':\Omega\to\coint{0,\infty}$ be stopping times corresponding to
$\tau$, $\tau'\in\Cal T_f$, with $h(\omega)\le h'(\omega)$ for every
$\omega$.
Enumerate $\Bbb Q\cap\coint{0,\infty}$ as $\sequencen{q_n}$,
starting with $q_0=0$, and for $n\in\Bbb N$ let
$\langle q_{ni}\rangle_{i\le n}$
be the increasing enumeration of $\{q_i:i\le n\}$.   Set

\Centerline{$f_n(\omega)
=\sum_{i=0}^{n-1}U_{\med(h(\omega),q_{ni},h'(\omega))}
(\omega)\bigl(V_{\med(h(\omega),q_{n,i+1},h'(\omega))}(\omega)
  -V_{\med(h(\omega),q_{ni},h'(\omega))}(\omega)\bigr)$}

\noindent for $\omega\in\Omega$.   Then
$f(\omega)=\lim_{n\to\Cal F}f_n(\omega)$ is defined for almost every
$\omega$, $f$ is $\Sigma$-measurable and
$f^{\ssbullet}=\int_{[\tau,\tau']}\pmb u\,d\pmb{v}$.

\proof{ Write $\Cal S'$ for
$\{\med(\tau,\check q,\tau'):q\in\Bbb Q\cap\coint{0,\infty}\}
\cup\{\tau'\}$, as in 633L.    For $n\in\Bbb N$, set
$I_n=\{\med(\tau,\check{q_i},\tau'):i\le n\}$,
$I'_n=I_n\cup\{\tau'\}$.   Note that
$\tau=\med(\tau,\check q_0,\tau')$ belongs to $I_n$,
and that if we set

\Centerline{$a_n=\sup_{i\le n}\Bvalue{\tau'=\med(\tau,\check{q_i},\tau')}
\Bsupseteq\sup_{i\le n}\Bvalue{\tau'\le\check{q_i}}$;}

then $a_n\Bsubseteq\sup_{\sigma\in I_n}\Bvalue{\rho=\sigma}$ for every
$\rho\in I'_n$.   Because $\tau'\in\Cal T_f$, $\lim_{n\to\infty}\bar\mu a_n=1$.

Now

$$\eqalignno{\int_{[\tau,\tau]}\pmb u\,d\pmb{v}
&=\int_{\Cal S'}\pmb u\,d\pmb{v}\cr
\displaycause{633L}
&=\lim_{n\to\infty}S_{I'_n}(\pmb u,d\pmb{v})\cr
\displaycause{because if $J\subseteq\Cal S'$ is finite there is an
$m\in\Bbb N$ such that $J\subseteq I'_n$ for every $n\ge m$}
&=\lim_{n\to\infty}S_{I_n}(\pmb u,d\pmb{v})\cr
\displaycause{because
$\Bvalue{S_{I'_n}(\pmb u,d\pmb{v})=S_{I_n}(\pmb u,d\pmb{v})}
\Bsupseteq a_n$, by 613S}
&=\lim_{n\to\infty}f_n^{\ssbullet}
=(\lim_{n\to\Cal F}f_n)^{\ssbullet}\cr}$$

\noindent because $\Cal F$ is a measure-converging filter.
}%end of proof of 649E

\leader{649F}{}\cmmnt{ I have given a number of results (612H, 614S, 631D,
649C, 649E) on the ways in which classical stochastic processes,
based on probability spaces, give rise to the processes considered in this
volume, based on probability algebras.   To get full value from these, we
need to know which processes can be represented in this way.   I have held
off so far because there are technical complications which I feel are
irrelevant to the ideas I really want to express, but I think it is time I
gave a result which handles a reasonable proportion of cases.   I begin
with a minor extension of ideas from \S\S631 and 633.

\medskip

\noindent}{\bf Lemma}
Suppose that $\family{t}{T}{\frak A_t}$ is right-continuous.
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a locally near-simple process.
Let $\tilde{\Cal S}$ be
the full ideal of $\Cal T$ generated by $\Cal S$.   Then there is a locally
near-simple process $\tildepmb u$
with domain $\tilde{\Cal S}$ extending $\pmb u$.   If $\pmb u$ is
non-negative and non-decreasing, we can arrange that $\tildepmb u$
should be non-negative and non-decreasing.

\proof{ If $\Cal S=\emptyset$ then $\tilde{\Cal S}=\emptyset$ and there
is nothing to prove.   So I suppose from now on that $\Cal S$ is non-empty.

\medskip

{\bf (a)} Set $\sigma_1=\inf\Cal S$ and
$\Cal S_1=\Cal S\cup\{\sigma_1\}$.   Then
$\Cal S_1$ is a sublattice of $\Cal S$ and the starting value
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined and belongs to
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ (631Ca, applied
to $\Cal S\wedge\tau$ for any $\tau\in\Cal S$).   Because
$\family{t}{T}{\frak A_t}$ is right-continuous,
$u_{\downarrow}\in L^0(\frak A_{\sigma_1})$ (632C(a-iii));  moreover, if
$\tau\in\Cal S$, then $\Bvalue{\sigma_1<\tau}
=\sup_{\sigma\in\Cal S\wedge\tau}\Bvalue{\sigma<\tau}$
(632C(a-ii)), so

\Centerline{$\Bvalue{\sigma_1=\tau}
=\inf_{\sigma\in\Cal S\wedge\tau}\Bvalue{\sigma=\tau}
\Bsubseteq\inf_{\sigma\in\Cal S\wedge\tau}\Bvalue{u_{\sigma}=u_{\tau}}
\Bsubseteq\Bvalue{u_{\downarrow}=u_{\tau}}$.}

\noindent So we have a fully adapted process
$\pmb u_1=\family{\sigma}{\Cal S_1}{u_{1\sigma}}$ defined by saying that
$u_{1,\sigma_1}=u_{\downarrow}$ and $u_{1\sigma}=u_{\sigma}$ for every
$\sigma\in\Cal S$.

Next, $\pmb u_1$ is locally near-simple.   \Prf\ Take any $\tau\in\Cal S$
and $\epsilon>0$.   Then there is a simple process
$\pmb u'=\family{\sigma}{\Cal S\wedge\tau}{u'_{\sigma}}$ such that
$\theta(\pmb u)\le\epsilon$ where
$\pmb u=\sup_{\sigma\in\Cal S\wedge\tau}|u_{\sigma}-u'_{\sigma}|)$.
Let $(\tau_0,\ldots,\tau_n)$ be a breakpoint string for
$\pmb u'$ ending with $\tau_n=\tau$.   Consider the simple process
$\pmb u'_1=\family{\sigma}{\Cal S_1\wedge\tau}{u'_{1\sigma}}$ with
breakpoint string $(\sigma_1,\tau_0,\ldots,\tau_n)$ and

\Centerline{$\Bvalue{u'_{1\sigma}=u'_{\tau_i}}
\Bsupseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}$ for
$i<n$,}

\Centerline{$\Bvalue{u'_{1\sigma}=u_{\downarrow}}\Bsupseteq\Bvalue{\sigma<\tau_0}$}

\noindent for every $\sigma\in\Cal S_1$, while
$u'_{1\tau_n}=u'_{\tau_n}$.
Let $u'_{\downarrow}$ be the starting value of $\pmb u'$ (614Ba).   Since
$u'_{\downarrow}=\lim_{\sigma\downarrow\Cal S\wedge\tau}u'_{\sigma}$ and
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S\wedge\tau}u_{\sigma}$,
$|u'_{\downarrow}-u_{\downarrow}|\le\bar u$.
Now we see that, for $\sigma\in\Cal S$,

$$\eqalign{\Bvalue{\sigma<\tau_0}
&\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\downarrow}}
   \Bcap\Bvalue{u'_{1\sigma}=u_{\downarrow}}
   \Bcap\Bvalue{|u_{\sigma}-u'_{\sigma}|\le\bar u}
\cr&\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\downarrow}}
   \Bcap\Bvalue{u'_{1\sigma}=u_{\downarrow}}
   \Bcap\Bvalue{|u_{\sigma}-u'_{\sigma}|\le\bar u}
   \Bcap\Bvalue{u_{1\sigma}=u_{\sigma}}
\Bsubseteq\Bvalue{|u_{1\sigma}-u'_{1\sigma}|\le 2\bar u},
\cr}$$

\noindent while

$$\eqalign{\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
&\Bsubseteq\Bvalue{u'_{1\sigma}=u'_{\tau_i}}
   \Bcap\Bvalue{u'_{\sigma}=u'_{\tau_i}}
   \Bcap\Bvalue{u_{1\sigma}=u_{\sigma}}
   \Bcap\Bvalue{|u_{\sigma}-u'_{\sigma}|\le\bar u}
\cr&\Bsubseteq\Bvalue{|u_{1\sigma}-u'_{1\sigma}|\le\bar u}
\cr}$$

\noindent for $i<n$ and, of course,

\Centerline{$|u_{1\tau}-u'_{1\tau}|=|u_{\tau}-u'_{\tau}|\le\bar u$.}

\noindent Thus we have $|u_{1\sigma}-u'_{1\sigma}|\le 2\bar u$ for every
$\sigma\in\Cal S$.   At the bottom end,

\Centerline{$\Bvalue{\sigma_1<\tau_0}
\Bsubseteq\Bvalue{u'_{1\sigma_1}=u_{\downarrow}}
=\Bvalue{u'_{1\sigma_1}=u_{1\sigma_1}}$,}

\Centerline{$\Bvalue{\sigma_1=\tau_0}
\Bsubseteq\Bvalue{u'_{1\sigma_1}=u'_{\tau_0}}
  \Bcap\Bvalue{u_{1\sigma_1}=u_{\tau_0}}
\Bsubseteq\Bvalue{|u_{1\sigma_1}-u'_{1\sigma_1}|\le\bar u}$.}

\noindent Assembling these,

\Centerline{$\theta(\sup_{\sigma\in\Cal S_1}|u_{1\sigma}-u'_{1\sigma}|)
\le\theta(2\bar u)\le 2\epsilon$.}

\noindent As $\tau$ and
$\epsilon$ is arbitrary, $\pmb u_1$ is locally near-simple.\ \Qed

\medskip

{\bf (b)} Write $\sigma_2$ for $\min\Cal T$ and $\Cal S_2$ for the
lattice $\Cal S_1\cup\{\sigma_2\}=\Cal S\cup\{\sigma_2,\sigma_1\}$.
We have a fully adapted process
$\pmb u_2=\family{\sigma}{\Cal S_2}{u_{2\sigma}}$ defined by saying that
$u_{2\sigma}=u_{1\sigma}$ for $\sigma\in\Cal S_1$ and

\Centerline{$\Bvalue{u_{2\sigma_2}=0}\Bsupseteq\Bvalue{\sigma_2<\sigma_1}$,
\quad$\Bvalue{u_{2\sigma_2}=u_{\downarrow}}
\Bsupseteq\Bvalue{\sigma_2=\sigma_1}$.}

\noindent Since $\pmb u_2\restr\Cal S_2\wedge\sigma_1$ is simple while
$\pmb u_2\restr\Cal S_2\vee\sigma_1=\pmb u_1$ is locally near-simple,
$\pmb u_2$ is locally near-simple (631F(a-iv)).   And of course
$\pmb u_2$ extends $\pmb u$.   Moreover, if $\pmb u$ is non-negative and
non-decreasing,

\Centerline{$0\le u_{2\sigma_2}\le u_{\downarrow}
=\inf_{\sigma\in\Cal S}u_{\sigma}$}

\noindent so $\pmb u_2$ is non-negative and non-decreasing.

\medskip

{\bf (c)} We are now in a position to turn to 631M.
$\Cal S_2$ is coinitial with
$\tilde\Cal S$, just because $\min\tilde\Cal S=\min\Cal T$ belongs to
$\Cal S_2$.   So we have
a function $\Psi^*:\Mlns(\Cal S_2)\to\Mlns(\tilde\Cal S)$ as described in
631Mc.   Set $\tildepmb u=\Psi^*(\pmb u_2)$.   Then $\tildepmb u$ is locally
near-simple and extends $\pmb u_2$ (631M(c-iii)), so extends $\pmb u$.
If $\pmb u$, and therefore $\pmb u_2$, are non-negative and non-decreasing,
so is $\tildepmb u$, by 631M(c-iii) again.
So we have an extension of the kind we need.
}%end of proof of 649F

\leader{649G}{Lemma} Suppose that \Asissixt\ is a real-time integration
structure and
$\Cal S$ is a non-empty sublattice of $\Cal T$.
There is a non-decreasing sequence $\sequencen{\tau_n}$
in $\Cal S$ such that $\sup_{n\in\Bbb N}\Bvalue{\tau\le\tau_n}=1$
for every $\tau\in\Cal S$.

\proof{ For $q\in\Bbb Q\cap\coint{0,\infty}$,
$\Bvalue{\sup\Cal S>q}=\sup_{\sigma\in\Cal S}\Bvalue{\sigma>q}$ (611Cb), so
there is a countable set $A_q\subseteq\Cal S$ such that
$\Bvalue{\sup\Cal S>q}=\sup_{\sigma\in A_q}\Bvalue{\sigma>q}$, because
$\frak A$ is ccc (322G, 316E).   Accordingly there is a
(non-empty) countable set $A\subseteq\Cal S$ such that
$\Bvalue{\sup\Cal S>q}=\sup_{\sigma\in A}\Bvalue{\sigma>q}$ for every
rational $q\ge 0$.   At the same time, there is a countable
$B\subseteq\Cal S$ such that
$\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\sup\Cal S}
=\sup_{\sigma\in B}\Bvalue{\sigma=\sup\Cal S}$.   Taking a sequence
$\sequencen{\sigma_n}$ running over $A\cup B$,
and setting $\tau_n=\sup_{i\le n}\sigma_i$ for $n\in\Bbb N$,
$\sequencen{\tau_n}$ is a non-decreasing sequence in $\Cal S$ and

\Centerline{$\Bvalue{\sup\Cal S>q}=\sup_{n\in\Bbb N}\Bvalue{\tau_n>q}$
for every $q\in\Bbb Q\cap\coint{0,\infty}$,}

\Centerline{$\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\sup\Cal S}
=\sup_{n\in\Bbb N}\Bvalue{\tau_n=\sup\Cal S}$.}

Now take any $\tau\in\Cal S$ and non-zero $a\in\frak A$.   If
$a'=a\Bcap\Bvalue{\tau=\sup\Cal S}$ is non-zero, there is an $n\in\Bbb N$
such that $a'\Bcap\Bvalue{\tau_n=\sup\Cal S}\ne 0$, and now
$a\Bcap\Bvalue{\tau_n=\tau}\ne 0$.   Otherwise, there is a $t\ge 0$ such
that
$a''=a\Bcap\Bvalue{\sup\Cal S>t}\Bsetminus\Bvalue{\tau>t}$ is non-zero.
Let $q>t$ be rational and such that $a''\Bcap\Bvalue{\sup\Cal S>q}\ne 0$;
then there is an $n\in\Bbb N$ such that

\Centerline{$0\ne a''\Bcap\Bvalue{\tau_n>q}
\Bsubseteq\Bvalue{\tau<\tau_n}$}

\noindent and $a\Bcap\Bvalue{\tau\le\tau_n}\ne 0$.

As $a$ is arbitrary, $\sup_{n\in\Bbb N}\Bvalue{\tau\le\tau_n}=1$ and we
have a suitable sequence $\sequencen{\tau_n}$.
}%end of proof of 649G

\leader{649H}{Theorem} Suppose that \Asissixt\ is a
right-continuous real-time stochastic integration structure.

(a)(i) There is a complete probability space $(\Omega,\Sigma,\mu)$ such
that $(\frak A,\bar\mu)$ can be identified with the measure algebra of
$(\Omega,\Sigma,\mu)$.

\quad(ii) For $E\in\Sigma$, write $E^{\ssbullet}$ for the corresponding
member of $\frak A$;  for $t\ge 0$ set
$\Sigma_t=\{E:E\in\Sigma$, $E^{\ssbullet}\in\frak A_t\}$.   Then
$\langle\Sigma_t\rangle_{t\ge 0}$ is a right-continuous filtration of
$\sigma$-algebras all containing every negligible subset of $\Omega$.

\quad(iii) Members of $\Cal T$ can be represented by stopping times
$h:\Omega\to[0,\infty]$ as in 612H, with the corresponding identification
of the algebras $\frak A_{\tau}$ as in 612H(a-iii).

(b) Now suppose that $\pmb u=\sfamilysS{u}$
is a locally near-simple process with
non-empty domain $\Cal S\subseteq\Cal T_f$.
Then there are a progressively measurable stochastic process
$\langle U_t\rangle_{t\ge 0}$ and a non-decreasing sequence
$\sequencen{h_n}$ of finite-valued stopping times, all adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$, such that

\inset{($\alpha$) $h_n$ represents a stopping time $\tau_n\in\Cal S$
for every
$n\in\Bbb N$, and $\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}=1$ for every
$\sigma\in\Cal S$,

($\beta$) $U_g^{\ssbullet}=u_{\sigma}$
whenever $g:\Omega\to\coint{0,\infty}$ is a stopping time representing
$\sigma\in\Cal S$,

($\gamma$) $t\mapsto U_t(\omega):[0,h_n(\omega)]\to\Bbb R$ is
\cadlag\ for every $\omega\in\Omega$ and $n\in\Bbb N$.}

\proof{{\bf (a)} Really this is just the Loomis-Sikorski theorem
(321J);  I remarked in 321K that the construction there always gives
a complete measure space, and in this context, of course, it gives a
probability space.   To see that $\langle\Sigma_t\rangle_{t\ge 0}$ is
a right-continuous filtration we need only look at the definitions in
611Aa and 632B, and surely every negligible set belongs to every
$\Sigma_t$.   Now 612H tells us all we need to know.

I remark that because every $\Sigma_t$ contains every negligible set, the
same is true of $\Sigma_h$, as defined in 612H(a-iii),
for every stopping time $h$.

\medskip

{\bf (b)(i)} For the time being (down to the end of (v)) I will
suppose that $\Cal S$ is an ideal of $\Cal T_f$.
By 649G, we have a non-decreasing sequence
$\sequencen{\tau_n}$ in $\Cal S$, starting with
$\tau_0=\min\Cal T=\check 0$, such that
$\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}=1$ for every
$\sigma\in\Cal S$.   Now each $\tau_n$ can be represented by a
stopping
time $h_n$;  since $\sequencen{h_n(\omega)}$ must be a non-decreasing
sequence in $\coint{0,\infty}$ for almost every $\omega$, and negligible
sets all belong to $\Sigma_0$, we can adjust the $h_n$, if necessary, to
arrange that $\sequencen{h_n}$ is a non-decreasing sequence of finite
stopping times.

\medskip

\quad{\bf (ii)} Set $S^*=\bigcup_{k\ge 1}\Bbb N^k$.   Choose
$\family{r}{S^*}{\tau_r}$, $\family{r}{S^*}{h_r}$ and
$\family{r}{S^*}{Y_r}$ inductively, as
follows.   Start with $\tau_{\fraction{n}}=\tau_n$ and
$h_{\fraction{n}}=h_n$, for $n\in\Bbb N$, where I write
$\fraction{n}$ for the member of $\Bbb N^1$ with value $n$.
Now choose a $\Sigma_{h_n}$-measurable real-valued $Y_{\fraction{n}}$
such that $Y_{\fraction{n}}^{\ssbullet}=u_{\tau_n}$;  do this in such a way
that $Y_{\fraction{n+1}}(\omega)=Y_{\fraction{n}}(\omega)$ whenever
$h_{n+1}(\omega)=h_n(\omega)$.

Given $r\in\Bbb N^k$ let $r'$ be its successor in the lexicographic
ordering of $\Bbb N^k$, so that $r'\restr k-2=r\restr k-2$ and
$r'(k-1)=r(k-1)+1$;  let $\sequencen{\tau_{r^{\smallfrown}\fraction{n}}}$
be a non-decreasing sequence in $[\tau_r,\tau_{r'}]$ such that

\Centerline{$\tau_{r^{\smallfrown}\fraction{0}}=\tau_r$,
\quad$\sup_{n\in\Bbb N}
  \Bvalue{\tau_{r^{\smallfrown}\fraction{n}}=\tau_{r'}}=1$,}

\Centerline{$\Bvalue{\sigma<\tau_{r^{\smallfrown}\fraction{n+1}}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_{r^{\smallfrown}\fraction{n+1}}}|
  \le 2^{-n}}$ for every
$\sigma\in[\tau_{r^{\smallfrown}\fraction{n}},
  \tau_{r^{\smallfrown}\fraction{n+1}}]$}

\noindent for each $n\in\Bbb N$ (631Ra).   Now choose a sequence
$\sequencen{h_{r^{\smallfrown}\fraction{n}}}$ of stopping times such
that $h_{r^{\smallfrown}\fraction{n}}$ represents
$\tau_{r^{\smallfrown}\fraction{n}}$ for each $n$;  as we shall necessarily
have

\Centerline{$h_{r^{\smallfrown}\fraction{0}}\eae h_r$,
\quad$\sup_{n\in\Bbb N}h_{r^{\smallfrown}\fraction{n}}\eae h_{r'}$,}

\Centerline{$h_{r^{\smallfrown}\fraction{n}}
\leae h_{r^{\smallfrown}\fraction{n+1}}$ for each $n$,}

\noindent we can adjust the functions on a negligible set so that

\Centerline{$h_{r^{\smallfrown}\fraction{}}=h_r$,
\quad$\sup_{n\in\Bbb N}h_{r^{\smallfrown}\fraction{n}}=h_{r'}$,}

\Centerline{$h_{r^{\smallfrown}\fraction{n}}
\le h_{r^{\smallfrown}\fraction{n+1}}$ for each $n$.}

\noindent Finally, choose $\sequencen{Y_{r^{\smallfrown}\fraction{n}}}$
such that
$Y_{r^{\smallfrown}\fraction{0}}=Y_r$ and, for each $n$,

\inset{$Y_{r^{\smallfrown}\fraction{n}}:\Omega\to\Bbb R$ is a
$\Sigma_{h_{r^{\smallfrown}\fraction{n}}}$-measurable function
and $Y_{r^{\smallfrown}\fraction{n}}^{\ssbullet}
=u_{\tau_{r^{\smallfrown}\fraction{n}}}$,

$Y_{r^{\smallfrown}\fraction{n+1}}(\omega)
=Y_{r^{\smallfrown}\fraction{n}}(\omega)$ whenever
$h_{r^{\smallfrown}\fraction{n+1}}(\omega)
=h_{r^{\smallfrown}\fraction{n}}(\omega)$,

$Y_{r^{\smallfrown}\fraction{n}}(\omega)=Y_{r'}(\omega)$ whenever
$h_{r^{\smallfrown}\fraction{n}}(\omega)=h_{r'}(\omega)$,

$|Y_{r^{\smallfrown}\fraction{n}}(\omega)-Y_r(\omega)|
\le 2^{-k}$ whenever
$h_{r^{\smallfrown}\fraction{n}}(\omega)<h_{r'}(\omega)$.}

\noindent (The point here is that
$\frak A_{\tau_{r^{\smallfrown}\fraction{n}}}$ is always
$\{E^{\ssbullet}:E\in\Sigma_{r^{\smallfrown}\fraction{n}}$, by
612H(a-iii), so that we can manage the first clause in this list,
and the rest can be achieved by adjustments on negligible sets.)

Note that the construction ensures that if $r_0$, $r_1\in S^*$ and
$h_{r_0}(\omega)=h_{r_1}(\omega)$, we shall have
$Y_{r_0}(\omega)=Y_{r_1}(\omega)$.

\medskip

\quad{\bf (iii)} At the end of the induction, define $U_t(\omega)$,
for $t\ge 0$ and $\omega\in\Omega$, as follows.
If $h_r(\omega)<t$ for every $r\in S^*$, set
$U_t(\omega)=0$.   Otherwise, we have a sequence $\gamma\in\NN$ such
that $h_{\gamma\restr k}(\omega)\le t<h_{(\gamma\restr k)'}(\omega)$
for every $k\ge 1$.   In this case,
$|Y_{\gamma\restr k+1}(\omega)-Y_{\gamma\restr k}(\omega)|\le 2^{-k}$
for every $k$, so $\lim_{k\to\infty}Y_{\gamma\restr k}(\omega)$ is defined;
take this for $U_t(\omega)$.   Note that this will ensure that
$|U_t(\omega)-Y_r(\omega)|\le 2^{-k-1}$ whenever $k\ge 1$,
$r\in\Bbb N^k$ and $h_r(\omega)\le t<h_{r'}(\omega)$.

Now $(t,\omega)\mapsto U_t(\omega)$ is progressively measurable.
\Prf\ Take $t_0\in\coint{0,\infty}$ and $\alpha\in\Bbb R$, and set
$H=\{(t,\omega):t\le t_0$, $\omega\in\Omega$, $U_t(\omega)>\alpha\}$.
Then

\Centerline{$H=H_0
   \cup\bigcup_{k\ge 1}\bigcup_{r\in\Bbb N^k}\{(t,\omega):t\le t_0$,
     $h_r(\omega)\le t<h_{r'}(\omega)$, $Y_r(\omega)>t+2^{-k-1}\}$}

\noindent where
$H_0=\{(t,\omega):h_r(\omega)<t\le t_0\}$ if
$\alpha<0$, else $\emptyset$.   Now if $r\in S^*$ and we set
$E_r=\{\omega:h_r(\omega)\le t_0\}$, $E_r\in\Sigma_{t_0}$
and $h_r\restr E_r$ is $\Sigma_{t_0}$-measurable, so
$\{(t,\omega):h_r(\omega)\le t\le t_0\}$ and
$\{(t,\omega):h_r(\omega)<t\le t_0\}$ belong to
$\Cal B\tensorhat\Sigma_{t_0}$,
where $\Cal B$ is the Borel $\sigma$-algebra of $\Bbb R$.
It follows that $H_0\in\Cal B\tensorhat\Sigma_{t_0}$.
Next, given $r\in S^*$, $Y_r$ is $\Sigma_{h_r}$-measurable, so
$Y_r\restr E_r$ is $\Sigma_{t_0}$-measurable, since
$E\cap E_r\in\Sigma_{t_0}$ for every $E\in\Sigma_{h_r}$.
Next, if $k\in\Bbb N$ and $r\in\Bbb N^k$

\Centerline{$\{(t,\omega):h_r(\omega)\le t\le t_0$,
$Y_r(\omega)>t+2^{-k-1}\}$}

\noindent belongs to $\Cal B\tensorhat\Sigma_{h_r}$ and is included in
$[0,t_0]\times E_r$, it belongs to $\Cal B\tensorhat\Sigma_{t_0}$;  while

\Centerline{$\{(t,\omega):h_{r'}(\omega)<t\le t_0$}

\noindent also belongs to $\Cal B\tensorhat\Sigma_{t_0}$, so the difference

\Centerline{$\{(t,\omega):t\le t_0$,
     $h_r(\omega)\le t<h_{r'}(\omega)$, $Y_r(\omega)>t+2^{-k-1}\}$}

\noindent belongs to $\Sigma_{t_0}$.   Taking the union of these,
$H\in\Sigma_{t_0}$.   As $t_0$ and $\alpha$ are arbitrary,
$(t,\omega)\mapsto U_t(\omega)$ is progressively measurable.\ \Qed

\medskip

\quad{\bf (iv)} Accordingly we have a fully adapted process
$\pmb v=\family{\sigma}{\Cal T_f}{v_{\sigma}}$ defined by saying that
$v_{\sigma}=U_g^{\ssbullet}$ whenever $g:\Omega\to\coint{0,\infty}$ is a
stopping time representing $\sigma\in\Cal T_f$.   Since
$U_{h_r(\omega)}=Y_r(\omega)$ whenever $r\in S^*$ and $\omega\in\Omega$,
$U_{h_r}=Y_r$ and $v_{\tau_r}=u_{\tau_r}$ for every $r\in S^*$.

We know also that if $k\ge 1$, $r\in\Bbb N^k$ and
$g$ is a finite-valued stopping time representing $\sigma\in\Cal T$, then

\Centerline{$|U_{g(\omega)}(\omega)-Y_r(\omega)|\le 2^{-k-1}$
whenever $h_r(\omega)\le g(\omega)<h_{r'}(\omega)$.}

\noindent Translating this into terms of $\Cal T$ and $\pmb v$,

\Centerline{$\Bvalue{\tau_r\le\sigma}\Bcap\Bvalue{\sigma<\tau_{r'}}
\Bsubseteq\Bvalue{|v_{\sigma}-u_{\tau_r}|\le 2^{-k-1}}$}

\noindent Since, by the choice of the $\tau_r$, we also have

\Centerline{$\Bvalue{\tau_r\le\sigma}\Bcap\Bvalue{\sigma<\tau_{r'}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_r}|\le 2^{-k}}$,}

\noindent we get

\Centerline{$\Bvalue{\tau_r\le\sigma}\Bcap\Bvalue{\sigma<\tau_{r'}}
\Bsubseteq\Bvalue{|v_{\sigma}-u_{\sigma}|\le 2^{-k-2}}$.}

\noindent If $\sigma\in\Cal S$, so that
$\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}=1$, then

\Centerline{$1
=\sup_{r\in\Bbb N^k}\Bvalue{\sigma=\tau_r}
  \Bcup\sup_{r\in\Bbb N^k}
    (\Bvalue{\tau_r<\sigma}\Bcap\Bvalue{\sigma<\tau_{r'}})
\Bsubseteq\Bvalue{|v_{\sigma}=u_{\sigma}|\le 2^{-k-2}}$}

\noindent for every $k\in\Bbb N$, so that $v_{\sigma}=u_{\sigma}$.

Thus $\pmb v$ extends $\pmb u$.

\medskip

\quad{\bf (v)} If $n\in\Bbb N$ and $\omega\in\Omega$ then
$t\mapsto U_t(\omega):[0,h_n(\omega)]\to\Bbb R$ is \cadlag.

\medskip

\Prf\grheada\ Suppose that $\sequence{i}{t_i}$ is a strictly decreasing
sequence in $[0,h_n(\omega)]$.   Set $t=\lim_{i\to\infty}t_i$.   For each
$k\ge 1$, there is an $r\in\Bbb N^k$ such that
$h_r(\omega)\le t<h_{r'}(\omega)$.   But in this case there is a
$j\in\Bbb N$ such that $h_r(\omega)\le t_i<h_{r'}(\omega)$ for every
$i\ge j$, so that $|U_{t_i}(\omega)-Y_r(\omega)|\le 2^{-k-1}$ for
every $i\ge j$ and $|U_t(\omega)-Y_r(\omega)|\le 2^{-k-1}$.   As $k$ is
arbitrary, $U_t(\omega)=\lim_{i\to\infty}U_{t_i}(\omega)$.

\medskip

\qquad\grheadb\ If now $\sequence{i}{t_i}$ is a strictly increasing
sequence in $[0,h_n(\omega)]$, then by the choice of $\family{r}{S^*}{h_r}$
we find that

\inset{for every $k\ge 1$ there are an $r\in\Bbb N^k$ and a $j\in\Bbb N$
such that $h_r(\omega)\le t_i<h_{r'}(\omega)$ for every $i\ge j$.}

\noindent But this means that for every $k\ge 1$ there are an
$r\in\Bbb N^k$ and a $j\in\Bbb N$ such that
such that $|U_{t_i}(\omega)-Y_r(\omega)|\le 2^{-k-1}$ for every $i\ge j$;
it follows at once that $\lim_{i\to\infty}U_{t_i}(\omega)$ is defined.
Putting these together,
$t\mapsto U_t(\omega):[0,h_n(\omega)]\to\Bbb R$ is \cadlag.\ \Qed

\medskip

\quad{\bf (vi)} This proves the result when $\Cal S$ is an ideal.   In
general, we still have a non-decreasing sequence $\sequencen{\tau_n}$ in
$\Cal S$ such that $\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}=1$ for every
$\sigma\in\Cal S$.   Now
$\tilde{\Cal S}=\{\sigma:\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}=1\}$
is a full ideal of $\Cal T$ included in $\Cal T_f$,
so is the full ideal generated by $\Cal S$, and we have a locally
near-simple process $\tildepmb u$ with domain $\tilde{\Cal S}$ extending
$\pmb u$ (649F).   Applying (i)-(v) to $\tildepmb u$, we get a suitable
process $\langle U_t\rangle_{t\ge 0}$.
}%end of proof of 649H

\leader{649I}{Scholium} If, in 649Hb, $\pmb u$ is a
non-negative non-decreasing process,
then we can arrange that $\langle U_t\rangle_{t\ge 0}$ is non-decreasing.
\prooflet{\Prf\ Working through the proof, we see that in part (b-ii), when
we come to choose $\family{r}{S^*}{Y_r}$, we have
$0\le u_{\tau_n}\le u_{\tau_{n+1}}$, so can require that
$0\le Y_{\fraction{n}}\le Y_{\fraction{n+1}}$, for each $n$.   Similarly,
when we come to choose $\sequencen{Y_{r^{\smallfrown}\fraction{n}}}$, we
shall have $Y_r\le Y_{r'}$ and

\Centerline{$u_{\tau_r}\le u_{\tau_{r^{\smallfrown}\fraction{n}}}
\le u_{\tau_{r^{\smallfrown}\fraction{n+1}}}\le u_{\tau_{r'}}$}

\noindent for each $n$, so we can arrange that

\Centerline{$Y_r\le Y_{r^{\smallfrown}\fraction{n}}
\le Y_{r^{\smallfrown}\fraction{n+1}}\le Y_{r'}$}

\noindent for each $n$.   Looking at the definition of $U_t$ in (b-iii),
we see that we shall now necessarily get a non-negative non-decreasing
process.

Of course I ought to check also that we can still use the idea in (b-vi).
But I noted in 649F that if we start from a non-negative
non-decreasing process $\pmb u$, then the extension $\tildepmb u$
can be made to be non-negative and non-decreasing.   So this part of the
argument also works.\ \Qed}

\leader{649J}{Lemma} Let $(\Omega,\Sigma,\mu)$ be a complete probability
space, $\langle\Sigma_t\rangle_{t\ge 0}$ a right-continuous
filtration of $\sigma$-subalgebras of $\Sigma$ all containing every
negligible set,
and $\langle U_t\rangle_{t\ge 0}$ a progressively measurable stochastic
process.   Let $h:\Omega\to\coint{0,\infty}$ be a stopping time such that
$t\mapsto U_t(\omega):[0,h(\omega)]\to\Bbb R$ is \cadlag\ for every
$\omega\in\Omega$.   For $\omega\in\Omega$ and $t\ge 0$ set

$$\eqalign{U_{<t}(\omega)
&=\lim_{s\uparrow t}U_s(\omega)\text{ if }0<t\le h(\omega),\cr
&=0\text{ otherwise}.\cr}$$

(a) $\langle U_{<t}\rangle_{t\ge 0}$ is a previsibly measurable
stochastic process, therefore progressively measurable.

(b) Let \Asissixt\ be the real-time stochastic integration structure
defined from $(\Omega,\Sigma,\mu,\langle\Sigma_t\rangle_{t\ge 0})$, and
$\tau\in\Cal T_f$ the stopping time represented by $h$\cmmnt{ (612H(a-i))}.
If $\pmb u$, $\pmb z$ are the fully adapted processes defined from $U$ and
$U_<$\cmmnt{ as in 612Hb}, then $\pmb u\restr\Cal T\wedge\tau$ is near-simple
and its previsible version is $\pmb z\restr\Cal T\wedge\tau$.

(c) Now suppose that $\langle V_t\rangle_{t\ge 0}$ is another
progressively measurable stochastic process, this time
non-negative and non-decreasing, such
that $t\mapsto V_t(\omega):[0,h(\omega)]\to\Bbb R$
is \cadlag\ for every $\omega\in\Omega$.   Let $\pmb v$ be the process
defined by $\langle V_t\rangle_{t\ge 0}$.
For $\omega\in\Omega$ let $\nu_{\omega}$ be the Radon measure on
$[0,h(\omega)]$ such that $\nu_{\omega}[0,t]=V_t(\omega)$ for every
$t\ge 0$, and set

\Centerline{$e(\omega)=\int_{[0,h(\omega)]}U_{<t}(\omega)\nu_{\omega}(dt)$.}

\noindent Then $e:\Omega\to\Bbb R$ is $\Sigma_h$-measurable and
$e^{\ssbullet}=\int_{\Cal T\wedge\tau}\pmb u\,d\pmb v$.

\proof{{\bf (a)} If $\alpha\in\Bbb R$, then

$$\eqalignno{&\{(t,\omega):U_{<t}(\omega)>\alpha\}
\cr&\mskip100mu=H_0\cup\bigcup_{\Atop{q\in\Bbb Q}{q>\alpha}}
\,\bigcup_{q'\in\Bbb Q}
\,\bigcap_{\Atop{q''\in\Bbb Q}{q''\ge q'}}
\cr&\mskip150mu\{(t,\omega):q'<t\le h(\omega),\,
  \text{ either }t\le q''\text{ or }U_{q''}(\omega)\ge q\}
&(*)
\cr}$$

\noindent where

$$\eqalignno{H_0
&=(\{0\}\times\Omega)\cup\{(t,\omega):h(\omega)<t\}
\text{ if }\alpha<0,
&(**)\cr
&=\emptyset\text{ if }\alpha\ge 0.\cr}$$

\noindent Now as $U_{q''}$ is always $\Sigma_{q''}$-measurable, the sets
$\{(t,\omega):q'<t$, $U_{q''}(\omega)\ge q\}$ always belong to the
previsible $\sigma$-algebra $\Lambdapv$ (642Ha).   Also

\Centerline{$\{(t,\omega):h(\omega)<t\}
=\bigcup_{q\in\Bbb Q}
   \coint{q,\infty}\times\{\omega:h(\omega)\le q\}$}

\noindent belongs to $\Lambdapv$, so $\{(t,\omega):t\le h(\omega)\}$ also
does, and of course $\coint{0,\infty}\times\Omega$ and
$\{0\}\times\Omega$ also do.
So all the elements of the formulae (*) and (**) correspond to sets in
$\Lambdapv$ and $\{(t,\omega):U_{<t}(\omega)>\alpha\}\in\Lambdapv$.
As $\alpha$ is arbitrary,
$\langle U_{<t}\rangle_{t\ge 0}$ is previsibly measurable, therefore
progressively measurable (642I).

\medskip

{\bf (b)(i)} $\pmb u\restr\Cal T\wedge\tau$ is near-simple.
\Prf\ This can be proved
by a simple adaptation of the argument in 631D;  or alternatively, we can
set

$$\eqalign{U'_t(\omega)
&=U_t(\omega)\text{ if }t<h(\omega),\cr
&=U_{h(\omega)}(\omega)\text{ if }t\ge h(\omega).\cr}$$

\noindent Then, for any $t\ge 0$ and $\alpha\in\Bbb R$,

\Centerline{$\{\omega:U'_t(\omega)>\alpha\}
=\{\omega:h(\omega)>t,\,U_t(\omega)>\alpha\}
   \cup\{\omega:h(\omega)\le t,\,U_{h(\omega)}(t)>\alpha\}$}

\noindent belongs to $\Sigma_t$.   Evidently $t\mapsto U'_t(\omega)$ is
\cadlag\ for every $\omega$.   We can therefore apply 631D as written
to show that the
process $\pmb u'$ defined from $\langle U'_t\rangle_{t\ge 0}$ is
locally near-simple, so that
$\pmb u\restr\Cal T\wedge\tau=\pmb u'\restr\Cal T\wedge\tau$ is
near-simple.\ \Qed

\medskip

\quad{\bf (ii)} Let $\epsilon>0$.   Then there is a simple process
$\pmb w=\family{\sigma}{\Cal T\wedge\tau}{w_{\sigma}}$ such that
$\theta(\bar u)\le\epsilon$ where
$\bar u=\sup|\pmb w-\pmb u\restr\Cal T\wedge\tau|)$.   Take
a breakpoint sequence $(\sigma_0,\ldots,\sigma_n)$ for $\pmb w$ such that
$\check 0=\sigma_0\le\ldots\le\sigma_n=\tau$, and choose stopping times
$g_0,\ldots,g_n$ such that $0=g_0\le\ldots\le g_n=h$ and $g_i$ represents
$\sigma_i$ for each $i\le n$;  for each $i$,
let $f_i$ be a $\Sigma_{g_i}$-measurable
function such that $f_i^{\ssbullet}=w_{\sigma_i}$.   Set

$$\eqalign{W_t(\omega)
&=f_i(\omega)\text{ if }i<n\text{ and }g_i(\omega)\le t<g_{i+1}(\omega),\cr
&=f_n(\omega)\text{ if }g_n(\omega)\le t.\cr}$$

\noindent Then $\langle W_t\rangle_{t\ge 0}$ is a progressively
measurable stochastic process representing $\pmb w$, and we have

$$\eqalign{W_{<t}(\omega)
&=0\text{ if }t = 0,\cr
&=f_i(\omega)\text{ if }i<n\text{ and }
  g_i(\omega)<t\le g_{i+1}(\omega),\cr
&=g_n(\omega)\text{ if }g_n(\omega)<t.\cr}$$

\noindent Like $\langle U_{<t}\rangle_{t\ge 0}$, this is a previsibly
measurable process;  let $\pmb w'$ be the process it represents.   Then
$\pmb w'\restr\Cal T\wedge\tau=\pmb w_<$, by 641Ia.

Now consider the previsibly measurable process
$\langle U_{<t}- W_{<t}\rangle_{t\ge 0}$.
If $g:\Omega\to\coint{0,\infty}$ is a
stopping time and $g\le h$, and we write $U_{<g}(\omega)$,
$ W_{<g}(\omega)$ for
$U_{<g(\omega)}(\omega)$ and $ W_{<g(\omega)}(\omega)$, then

$$\eqalign{U_{<g}(\omega)- W_{<g}(\omega)
&=\lim_{s\uparrow g(\omega)}U_s(\omega)- W_s(\omega)
   \text{ if }g(\omega)>0,\cr
&=0\text{ otherwise.}\cr}$$

\noindent If $\bar f:\Omega\to\coint{0,\infty}$ is such
that $\bar f^{\ssbullet}=\bar u$, then

\Centerline{$\{\omega:
   |U_{<g}(\omega)- W_{<g}(\omega)|>\bar f(\omega)\}
\subseteq\bigcup_{q\in\Bbb Q,q\ge 0}
  \{\omega:|U_{q\wedge h}(\omega)- W_{q\wedge h}(\omega)|
  >\bar f(\omega)\}$}

\noindent where I write $q\wedge h$ for the stopping time
$\omega\mapsto\min(q,h(\omega))$, so is negligible, and
$|U_{<g}^{\ssbullet}- W_{<g}^{\ssbullet}|
\le\bar u$.   Thus

\Centerline{$\sup|\pmb z-\pmb w_<|
=\sup|\pmb z-\pmb w'\restr\Cal T\wedge\tau|\le\bar u$.}

\noindent On the other hand,

\Centerline{$\sup|\pmb w_<-(\pmb u\restr\Cal T\wedge\tau)_<|
\le\sup|\pmb w-\pmb u\restr\Cal T\wedge\tau|
\le\bar u$}

\noindent by 641G(a-vii).    So

\Centerline{$\theta(\sup|\pmb z-(\pmb u\restr\Cal T\wedge\tau)_<|)
\le\theta(2\bar u)\le 2\epsilon$;}

\noindent as $\epsilon$ is arbitrary,
$\pmb z=(\pmb u\restr\Cal T\wedge\tau)_<$.

\medskip

{\bf (c)(i)} Of course $\nu_{\omega}$
is the Stieltjes measure associated with the
non-decreasing function $t\mapsto V_t(\omega)$.
The description in 114Xa is fully adequate for our needs here, and
you should have no difficulty in filling in the arguments sketched there.
If you are willing to use the full
resources of the numbered theorems in previous volumes, however, and the
fact that the function is \cadlag, the quickest
route to the present case may be to apply 416K to the finitely
additive functional $\nu'_{\omega}$ defined on the ring $\Tau$ of subsets
of $[0,h(\omega)]$ generated by the half-open intervals of the form
$\coint{0,\alpha}$ by saying that

\Centerline{$\nu'_{\omega}\coint{0,\alpha}
=\lim_{t\uparrow\alpha}V_t(\omega)$}

\noindent if $0<\alpha\le h(\omega)$

Because
$t\mapsto U_{<t}(\omega)$ is bounded and Borel measurable on
$[0,h(\omega)]$, the integral $e(\omega)$ is always defined.
Let $f:\Omega\to\Bbb R$ be a $\Sigma$-measurable function such that
$f^{\ssbullet}=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v$.

\medskip

\quad{\bf (ii)} Much as in (b-ii), take $\epsilon>0$ and a simple process
$\pmb w=\family{\sigma}{\Cal T\wedge\tau}{w_{\sigma}}$
such that $\theta(\bar u)\le\epsilon^2$ where
$\bar u=\sup|\pmb w-\pmb u\restr\Cal T\wedge\tau|$.
Let $M>0$ be such that $F=\{\omega:V_h(\omega)>M\}$ has measure at
most $\epsilon$, and take
a finite sublattice $I$ of $\Cal T\wedge\tau$ such that
$\theta(S_J(\pmb u,d\pmb v)-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)
\le\Bover{\epsilon^2}M$ whenever $I\subseteq J\in\Cal I(\Cal T\wedge\tau)$.
Let $J\supseteq I\cup\{\check 0,\tau\}$
be a finite sublattice of $\Cal T\wedge\tau$ which
includes a breakpoint string for $\pmb w$;  let
$(\sigma_0,\ldots,\sigma_n)$ linearly generate the $J$-cells, so that
$\check 0=\sigma_0\le\ldots\le\sigma_n=\tau$,
$(\sigma_0,\ldots,\sigma_n)$ is a breakpoint sequence for $\pmb w$
(612Kb), and $S_J(\pmb u,d\pmb v)
=\sum_{i=1}^nu_{\sigma_{i-1}}\times(v_{\sigma_i}-v_{\sigma_{i-1}})$.
Now choose $g_0,\ldots,g_n,f_0,\ldots,f_n$ and define
$\langle W_{<t}\rangle_{t\ge 0}$ as in (b-ii).

This time, we calculate

$$\eqalignno{\hat f(\omega)
&=\int_{[0,h(\omega)]} W_{<t}(\omega)\nu_{\omega}(dt)
=\sum_{i=1}^n
  f_i(\omega)\nu_{\omega}(\ocint{g_{i-1}(\omega),g_i(\omega)})
\Displaycause{because $f_0(\omega)=0$}
=\sum_{i=1}^nf_i(\omega)(V_{g_i}(\omega)-V_{g_{i-1}}(\omega))\cr}$$

\noindent for each $\omega$.
So $\hat f^{\ssbullet}=S_J(\pmb u\,d\pmb v)$ and
$\int_{\Omega}\min(1,|f(\omega)-\hat f(\omega)|)\mu(d\omega)\le\epsilon^2$;
accordingly $F'=\{\omega:|f(\omega)-\hat f(\omega)|\ge\epsilon\}$ has
measure at most $\epsilon$.

At the same time, again taking $\bar f:\Omega\to\coint{0,\infty}$ such
that $\bar f^{\ssbullet}=\bar u$, the set

\Centerline{$E
=\{\omega:|U_{<t}(\omega)- W_{<t}(\omega)|\le\bar f(\omega)$
for every $t\in[0,h(\omega)]$}

\noindent is conegligible.   \Prf\ Set

$$\eqalign{E'
&=\{\omega:|U_q(\omega)-f_i(\omega)|\le\bar f(\omega)\text{ whenever }
 i<n\text{ and }q\in\Bbb Q\cap\coint{g_i(\omega),g_{i+1}(\omega)}
\cr&\supseteq\{\omega:|U_{q\wedge h}(\omega)-U'_{q\wedge h}(\omega)|
  \le\bar f(\omega)
  \text{ for every }q\in\Bbb Q\cap\coint{0,\infty}\}.\cr}$$

\noindent
Because $|u_{\check q\wedge\tau}-w_{\check q\wedge\tau}|\le\bar u$ for
every $q\ge 0$, $E'$ is conegligible.   If $\omega\in E'$ and
$t\in[0,h(\omega)]$ then

$$\eqalign{|U_{<t}(\omega)- W_{<t}(\omega)|
&=0\text{ if }t=0,\cr
&=\lim_{q\in\Bbb Q,q\uparrow t}|U_q(\omega)-f_i(\omega)|
    \text{ if }i<n\text{ and }g_i(\omega)<t\le g_{i+1}(\omega),\cr}$$

\noindent and in either case is at most $\bar f(\omega)$.   So
$E\supseteq E'$ is conegligible.\ \Qed

We also know that
$\int_{\Omega}\min(1,\bar f(\omega))\mu(d\omega)\le\Bover{\epsilon^2}M$,
so that $F''=\{\omega:\bar f(\omega)>\Bover{\epsilon}M\}$ has measure at
most $\epsilon$.
Now take any $\omega\in E\setminus(F\cup F'\cup F'')$.   Then
$|U_{<t}(\omega)- W_{<t}(\omega)|\le\bar f(\omega)$ for every
$t\in[0,h(\omega)]$, so

$$\eqalign{|e(\omega)-f(\omega)|
&\le|e(\omega)-\hat f(\omega)|+|\hat f(\omega)-f(\omega)|
\le\bar f(\omega)\nu_{\omega}([0,h(\omega)])+\epsilon
\cr&\le M\bar f(\omega)+\epsilon
\le 2\epsilon,\cr}$$

\noindent while $F\cup F'\cup F''$ has measure at most $3\epsilon$.
As $\epsilon$ is arbitrary, $e\eae f$ and
$e^{\ssbullet}=f^{\ssbullet}=\int_{\Cal T\wedge\tau}\pmb u\,d\pmb v$.
}%end of proof of 649J

\leader{649K}{Lemma} Let $(\Omega,\Sigma,\mu)$ be a probability space and
$\langle\Sigma_t\rangle_{t\ge 0}$ a filtration of $\sigma$-subalgebras of
$\Sigma$.   Let $\Lambdapv$ be the corresponding previsible
$\sigma$-algebra and write $\eusm L$ for the smallest subset of
$\Bbb R^{\coint{0,\infty}\times\Omega}$ such that

\inset{constant functions belong to $\eusm L$,

scalar multiples of functions in $\eusm L$ belong to $\eusm L$,

if $\phi\in\eusm L$ and $\psi\in\Bbb R^{\coint{0,\infty}\times\Omega}$
and $|\phi|\wedge|\psi|=0$, then $\phi+\psi\in\eusm L$ iff
$\psi\in\eusm L$,

$\chi(\ooint{s,\infty}\times E)\in\eusm L$
whenever $s\ge 0$ and $E\in\Sigma_s$,

$\lim_{n\to\infty}\phi_n\in\eusm L$ whenever $\sequencen{\phi_n}$ is a
pointwise convergent sequence in $\eusm L$.
}

\noindent Then $\eusm L$ is the set of
all $\Lambdapv$-measurable real-valued functions on
$\coint{0,\infty}\times\Omega$.

\proof{{\bf (a)} Writing $\eusm L^0(\Lambdapv)$ for the set of
$\Lambdapv$-measurable functions, this is a linear space containing
$\ooint{s,\infty}\times E$ whenever $s\ge 0$ and $E\in\Sigma_s$ and is
closed under sequential pointwise convergence, so includes $\eusm L$.

\medskip

{\bf (b)} In the other direction, write $\Lambda$ for
$\{W:W\subseteq\coint{0,\infty}\times\Omega$, $\chi W\in\eusm L\}$.
Then $\Lambda$ is a Dynkin class (136A).   Also
$\Cal I=\{\ooint{s,\infty}\times E:s\ge 0$, $E\in\Sigma_s\}$ is a subset of
$\Lambda$ closed under finite intersections (if $E\in\Sigma_s$ and
$F\in\Sigma_t$ then $E\cap F\in\Sigma_{\max(s,t)}$ and
$(\ooint{s,\infty}\times E)\cap(\ooint{t,\infty}\times F)
=\ooint{\max(s,t),\infty}\times(E\cap F)$), so $\Lambda$ includes the
$\sigma$-algebra generated by $\Cal I$ (136B), which is $\Lambdapv$.
Thus $\chi W\in\eusm L$ for every $W\in\Lambdapv$.

It follows at once that $\sum_{i=0}^n\alpha_i\chi W_i\in\eusm L$ whenever
$\alpha_0,\ldots,\alpha_n\in\Bbb R$ and $W_0,\ldots,W_n\in\Lambdapv$,
therefore that $f\in\eusm L$ whenever $f:\Omega\to\coint{0,\infty}$ is
$\Lambdapv$-measurable, and finally that
$\eusm L^0(\Lambdapv)\subseteq\eusm L$.
So we have equality.
}%end of proof of 649K

\leader{649L}{Theorem} Let $(\Omega,\Sigma,\mu)$ be a complete probability
space, $\langle\Sigma_t\rangle_{t\ge 0}$ a right-continuous
filtration of $\sigma$-subalgebras of $\Sigma$ all containing every
negligible set, \Asissixt\ the corresponding real-time stochastic integration
structure,
$\langle X_t\rangle_{t\ge 0}$ a previsibly measurable stochastic
process and $\langle V_t\rangle_{t\ge 0}$ a non-negative
non-decreasing stochastic process.
Let $h:\Omega\to\coint{0,\infty}$ be a stopping time such that
$t\mapsto X_t(\omega)$ is bounded on $[0,h(\omega)]$ and
$t\mapsto V_t(\omega):[0,h(\omega)]\to\Bbb R$
is \cadlag\ for every $\omega\in\Omega$, and write $\tau$ for the corresponding
stopping time in $\Cal T$.   Let $\pmb x$, $\pmb v$
be the processes
defined by $\langle X_t\rangle_{t\ge 0}$ and $\langle V_t\rangle_{t\ge 0}$.
For $\omega\in\Omega$ let $\nu_{\omega}$ be the Radon measure on
$[0,h(\omega)]$ such that $\nu_{\omega}[0,t]=V_t(\omega)$ for every
$t\ge 0$, and set

\Centerline{$e(\omega)
=\int_{\ocint{0,h(\omega)}}X_t(\omega)\nu_{\omega}(dt)$.}

\noindent Then $e:\Omega\to\Bbb R$ is $\Sigma$-measurable and
$e^{\ssbullet}=\Sint_{\Cal T\wedge\tau}\pmb x\,d\pmb v$.

\proof{{\bf (a)}
Let $\Lambdapv$ be the previsible $\sigma$-algebra derived from
$\langle\Sigma_t\rangle_{t\ge 0}$, and $\eusm L^0=\eusm L^0(\Lambdapv)$ the
space of $\Lambdapv$-measurable real-valued functions on
$\coint{0,\infty}\times\Omega$.   By the definition in 642Hb, we have a
one-to-one correspondence between $\eusm L^0$ and the space of
previsibly measurable processes, matching $\phi\in\eusm L^0$ with
$\langle X_t\rangle_{t\ge 0}$ where $X_t(\omega)=\phi(t,\omega)$ for all
$t$ and $\omega$;  and in this case $\langle X_t\rangle_{t\ge 0}$
corresponds to the process $\pmb x_{\phi}$ as defined in 642L.
Recall from 642L that $\phi\mapsto\pmb x_{\phi}$ is an $f$-algebra
homomorphism taking pointwise convergent sequences to order*-convergent
sequences.

\medskip

{\bf (b)} For $M\ge 0$, $\phi\in\eusm L^0$ and $\omega\in\Omega$, set

\Centerline{$e_{\phi M}(\omega)
=\int_{\ocint{0,h(\omega)}}\med(-M,\phi(t,\omega),M)\nu_{\omega}(dt)$.}

\noindent(This is always defined because $t\mapsto\phi(t,\omega)$ is always
Borel measurable.)   Let $\Cal L$ be the set of those $\phi\in\eusm L^0$
such that $e_{\phi M}$ is measurable and

\Centerline{$e_{\phi M}^{\ssbullet}
=\biggerSint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v$}

\noindent for every $M\ge 0$.   Then $\Cal L$ satisfies the conditions of
649K.

\Prf{\bf (i)} If $\phi$ is constant with value $\alpha$, then

\Centerline{$e_{\phi M}(\omega)=\med(-M,\alpha,M)
  (V_{h(\omega)}(\omega)-V_0(\omega))$}

\noindent for every $\omega$, so

\Centerline{$e_{\phi M}^{\ssbullet}
=\med(-M,\alpha,M)(V_h^{\ssbullet}-V_0^{\ssbullet})
=\med(-M,\alpha,M)v_{\tau}-v_{\check 0}$,}

\noindent while

$$\eqalign{\Sint_{\Cal T\wedge\tau}
  \med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v
&=\Sint_{\Cal T\wedge\tau}\med(-M,\alpha,M)\tbf 1_<\,d\pmb v
\cr&=\int_{\Cal T\wedge\tau}\med(-M,\alpha,M)\tbf 1\,d\pmb v
\cr&=\med(-M,\alpha,M)(v_{\tau}-v_{\check 0}).
\cr}$$

\noindent So these are equal, for every $M$, and $\phi\in\eusm L$.

\medskip

\quad{\bf (ii)} Of course a scalar multiple of a member of $\eusm L$ belongs to
$\eusm L$ because $\phi\mapsto e_{\phi}$ and
$\Siiv$ are linear.   Similarly, if $|\phi|\wedge|\psi|=0$ then
$|\pmb x_{\phi}|\wedge|\pmb x_{\psi}|=0$, so

\Centerline{$e_{\phi+\psi,M}=e_{\phi M}+e_{\psi M}$,}

\Centerline{$\med(-M\tbf 1_<,\pmb x_{\phi+\psi},M\tbf 1_<)
=\med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)
   +\med(-M\tbf 1_<,\pmb x_{\psi},M\tbf 1_<)$}

\noindent and if $\phi\in\eusm L$ then $\phi+\psi\in\eusm L$ iff
$\psi\in\eusm L$.

\medskip

\quad{\bf (iii)} As in (i), if $\phi=\chi(\ooint{s,\infty}\times E)$, we
can compute

$$\eqalign{e_{\phi M}(\omega)
&=\min(1,M)(V_{h(\omega)}(\omega)-V_s(\omega))
   \text{ if }\omega\in E\text{ and }h(\omega)\le s,\cr
&=0\text{ otherwise},\cr}$$

\noindent so

\Centerline{$e_{\phi M}^{\ssbullet}
=\min(1,M)\chi a\times(v_{\tau}-v_{\tau\wedge\check s})$}

\noindent where $a=E^{\ssbullet}$.
At the same time, $\pmb x_{\phi}=\pmb u_<$ where $\pmb u$ is the
simple process with breakpoint $\check s$ and value $0$ below $\check s$,
value $\chi a$ from $\check s$ onwards.   So

$$\eqalignno{
\Sint_{\Cal T\wedge\tau}\med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v
&=\Sint_{\Cal T\wedge\tau}\min(1,M)\pmb u_<d\pmb v
=\int_{\Cal T\wedge\tau}\min(1,M)\pmb u\,d\pmb v
\Displaycause{645R(a-i)}
=\int_{(\Cal T\wedge\tau)\vee\check s}\min(1,M)\tbf 1\,d\pmb v
\cr&=\min(1,M)\chi a\times(v_{\tau}-v_{\tau\wedge\check s})
\cr}$$

\noindent and again we have equality, so $\phi\in\eusm L$.

\medskip

\quad{\bf (iv)} And finally, for the key point, if $\sequencen{\phi_n}$ is
a sequence in $\eusm L$ converging pointwise to $\phi$, then
$\sequencen{\pmb x_{\phi_n}}$ is order*-convergent to $\pmb x_{\phi}$
(642L(c-i)), so $\sequencen{\med(-M\tbf 1_<,\pmb x_{\phi_n},M\tbf 1_<)}$ is
always a previsibly order-bounded sequence which is order*-convergent
to $\med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)$, and

\Centerline{$\biggerSint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v
=\lim_{n\to\infty}\biggerSint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi_n},M\tbf 1_<)d\pmb v$}

\noindent by 645T.   At the same time, of course,
$\sequencen{\med(-M,\phi_n(t,\omega),M)}\to\med(-M,\phi(t,\omega),M)$ for
every $t$ and $\omega$, so by the ordinary dominated convergence theorem
$e_{\phi M}(\omega)=\lim_{n\to\infty}e_{\phi_nM}(\omega)$ for every
$\omega$ and

$$\eqalign{e_{\phi M}^{\ssbullet}
&=\lim_{n\to\infty}e_{\phi_nM}^{\ssbullet}
=\lim_{n\to\infty}\Sint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi_n},M\tbf 1_<)d\pmb v
\cr&=\Sint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v,\cr}$$

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

\medskip

{\bf (c)} By 649K, $\eusm L=\eusm L^0$.   Now suppose that
$\langle X_t\rangle_{t\ge 0}$ is a previsibly measurable stochastic
such that $t\mapsto X_t(\omega)$ is bounded on $[0,h(\omega)]$ for each
$\omega$, and that
$e(\omega)=\int_{\ocint{0,h(\omega)}}X_t(\omega)\nu_{\omega}(dt)$ for every
$\omega$.   Set $\phi(t,\omega)=X_t(\omega)$ for $t\ge 0$ and
$\omega\in\Omega$;  then $\phi\in\eusm L^0$ so $\phi\in\eusm L$.
If $M\ge 0$, then
$\{(t,\omega):t\le h(\omega),\,|\phi(t,\omega)|>M\}$ belongs to
$\Lambdapv$, so its projection
$F_M=\{\omega:|\phi(t,\omega)|>M$ for some $t\le h(\omega)\}$
belongs to $\Sigma$ (642Jb).   Because
$t\mapsto X_t(\omega)$ is bounded on $[0,h(\omega)]$ for each
$\omega$, $\bigcap_{M\in\Bbb N}F_M=\emptyset$.

If $M\in\Bbb N$ and $\omega\in\Omega\setminus F_M$,

\Centerline{$e_{\phi M}(\omega)
=\int_{\ocint{0,h(\omega)}}\phi(t,\omega)\nu_{\omega}(dt)
=e(\omega)$.}

\noindent So $e=\lim_{M\to\infty}e_{\phi M}$ is measurable.
Similarly,

\Centerline{$\Bvalue{\med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)
   \ne\pmb x_{\phi}\times\tbf 1_<}
\Bsubseteq F_M^{\ssbullet}$}

\noindent for every $M$, so

\Centerline{$\Bvalue{\Sint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v
  \ne\Sint_{\Cal T\wedge\tau}\pmb x_{\phi}\times\tbf 1_<\,d\pmb v}$}

\noindent is included in $F_M^{\ssbullet}$, for every $M$ (647J).
But $\inf_{M\in\Bbb N}F_M^{\ssbullet}=0$, so

$$\eqalignno{\Sint_{\Cal T\wedge\tau}\pmb x_{\phi}d\pmb v
&=\Sint_{\Cal T\wedge\tau}\pmb x_{\phi}\times\tbf 1_<\,d\pmb v
\Displaycause{645Pc}
=\lim_{M\to\infty}\Sint_{\Cal T\wedge\tau}
   \med(-M\tbf 1_<,\pmb x_{\phi},M\tbf 1_<)d\pmb v
=\lim_{M\to\infty}e_{\phi M}^{\ssbullet}
=e^{\ssbullet},
\cr}$$

\noindent as required.
}%end of proof of 649L

%\leader{649N}{}
%\leader{649O}{}
%\leader{649P}{}
%\leader{649Q}{}
%\leader{649R}{}
%\leader{649S}{}
%\leader{649T}{}
%\leader{649U}{}
%\leader{649V}{}
%\leader{649W}{}

\exercises{\leader{649X}{Basic exercises (a)} In 649H, show that if
$\pmb u$ is locally of bounded variation then we can arrange that
$t\mapsto U_t(\omega):[0,h_n(t)]\to\Bbb R$ is of bounded variation for
every $\omega\in\Omega$ and every $n$.

\spheader 649Xb State and prove a form of 649L which will cover
S-integrals with respect to processes which are locally
of bounded variation.
\Hint{\S437.}

\leader{649Y}{Further exercises (a)} In 649C, show that the formula

\Centerline{$h_{n,i+1}(\omega)=\inf(\{h(\omega)\}
  \cup\{t:t\ge h_{ni}(\omega),\,
    |U_t(\omega)-U_{h_{ni}(\omega)}(\omega)|\ge 2^{-n}\})$}

\noindent also works.

}%end of exercises


\endnotes{
\Notesheader{649} Both 649C and 649E refer to the Riemann-sum integral;
there seem to be insuperable difficulties in devising any kind of
path-by-path definition for the general S-integral.
Bichteler's construction (649C)
depends on some deep arguments, not here but in the proof of 627M.
The method of 649E demands rather less understanding of stochastic
processes (in a formal sense, I don't think we have to know even that
martingales are local integrators), but in the context of this volume
measure-converging filters are black magic;  even the full axiom of choice
is a bit strong for what we want to do here, and while I expect most
readers would prefer to do measure theory with dependent choice at least,
I think the techniques of Chapter 56 ought to be enough for a sufficiently
determined purist to set out nearly all the ideas here in a form based on
ZF alone.   The advantage of 649E, if there is one, is that (at least for
integrals between constant stopping times) we have a formula in terms of
$U_t$, $V_t$ alone, without even seeking to define $U_h$ and $V_h$ for
non-constant stopping times $h$.
I suppose that in principle this offers an
opportunity to use the formula as a definition of $\int_t^{t'}U\,dV$
without troubling about right-continuity of the filtration
$\langle\Sigma_t\rangle_{t\ge 0}$ or any hypothesis on the paths of
the processes $U$ and $V$.   But I do not see much use for
a stochastic integral
which does not allow for integration over intervals determined by arbitrary
stopping times, and for these I think we need something like the hypotheses
of 649E.

Most of this section, by page-count, has been devoted to an elaborate
analysis of the relationship between S-integration with respect to
a non-decreasing process and pathwise Lebesgue-Stieltjes integration.
Of course this should be regarded as one of
the starting points for the theory of
the S-integral, not its culmination.   The whole point of the S-integral is
that it provides a common extension of the natural integral
with respect to integrators
of bounded variation and It\^o's integral with respect to Browniam motion.
}%end of notes

\discrpage

\references{
\frfilename{mt6r.tex}
\versiondate{10.2.23}
\copyrightdate{2022}

% 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}


\def\bsdot{.\penalty-100}

\centerline{\bf References for Volume 6}

\medskip%A\medskip%B

Bain A.\ [09] {\it Stochastic Calculus.}
{\tt http://www.chiark.greenend.org.uk/$^{\sim}$alanb/}, 25.5.09.
%bain.pdf

Bichteler K.\ [79] `Stochastic integrators',
Bull.\ Amer.\ Math.\ Soc.\ (2) 1 (1979) 761-765.
%http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&page=toc&handle=euclid.bams/1183544712
\cmmnt{[627I.]}

Bichteler K.\ [81] `Stochastic integration and $L^p$ theory of
martingales', Ann.\ Prob.\ 9 (1981) 49-89.
%https://projecteuclid.org/euclid.aop/1176994509
\cmmnt{[649C.]}

Black F.\ \& Scholes M.\ [73] `The pricing of options and corporate
liabilities', J.\ Political Economy 81 (1973) 637-659.
\cmmnt{[\S655 {\it intro.}]}

Burkholder D.L.\ [66] `Martingale Transforms',
Ann.\ of Math.\ Stat.\ 37 (1966) 1494-1504.
\cmmnt{[621H.]}
%burkholder.66.pdf

%\medskip%C
\medskip%D

\leaveitout{
Delbaen F.\ \& Schachermeyer W.\ [06] {\it The Mathematics of Arbitrage.}
Springer, 2006.
ISBN-13: 978-3540219927
HG 6024.A3D4  0620 6436 3
}

\leaveitout{
Dellacherie C.\ \& Meyer P.-A.\ [79]
Probabilities and Potential, Volume 1
North-Holland, 1979  (Mathematics Studies).
% ISBN: 072040701X chapters I-VI
}

Dellacherie C.\ \& Meyer P.-A.\ [82]
{\it Probabilities and Potential B. Theory of Martingales.}
Elsevier, 1982 (North-Holland Mathematics Studies 72).
%ISBN: 0444865268 chapters V-VIII
%translator J.P.Wilson
\cmmnt{[627I.]}

\leaveitout{
Dellacherie C.\ \& Meyer P.-A.\ [88]
{\it Probabilities and Potential C: Potential Theory for Discrete and
Continuous Semigroups.}
Elsevier, 1988
%ISBN: 0444703861 (pt.C)
}

\leaveitout{
Dellacherie C., Meyer P.-A.\ \& Weil M.\ [77] (eds.)
{\it S\'eminaire de Probabilit\'es XI.} Springer, 1977
(Lecture Notes in Mathematics 581).
% ISBN 978-3-540-08145-6
}

\leaveitout{
Dol\'eans-Dade C.\ [76] `On the existence and unicity of solutions of
stochastic differential equations', Z.\ Wahrscheinlichkeitstheorie verw.\
Gebiete 36 (1976) 93-101.
%? fundamental theorem of martingales
% ILL 15.9.12
}

%\medskipE\medskip%F

\leaveitout{
Fremlin D.H.\ [n11] `Vector-valued gauge integrals', note of 12.7.11
({\tt http:{\bsp}/www1.essex.ac.uk{\bsp}maths{\bsp}people{\bsp}fremlin{\bsp}preprints.htm}).
%n11527
}

%\medskipG\medskip%H
\medskip%I

It\^o K.\ [44] `Stochastic Integral', Proc.\ Imperial Acad.\ Tokyo 20
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\cmmnt{[Vol.\ 6 {\it intro.}, Chap.\ 62 {\it intro.}]}

\leaveitout{
\medskip%J

Jacod J.\ [79] {\it Calcul Stochastique et Probl\`emes de Martingales.}
Springer, 1979 (Lecture Notes in Mathematics 714).
}

\medskip%K

Karandikar R.L.\ [95] `On pathwise stochastic integration', Stochastic
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\cmmnt{[649C.]}

Karatzas I.\ \& Shreve S.\ [91] {\it Brownian Motion and Stochastic
Calculus.}  Springer, 1991.
\cmmnt{[621J.]}

\leaveitout{
Kazamaki N.\ [77] `On a problem of Girsanov',
T\^ohoku Math.\ J.\ (2) 29 (1977) 597-600.
}

\leaveitout{
Kunita H.\ \& Watanabe S.\ [67] `On square integrable martingales', Nagoya
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}

\medskip%L

Lowenstein R.\ [01] {\it When Genius Failed: the rise and fall of
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%HG 4928.5,  0120 9001 8,  0520 6473 5
\cmmnt{[\S655 {\it notes\/}.]}

\leaveitout{
\medskip%M

Meyer P.A.\ [76] `Un cours sur les int\'egrales stochastiques',
pp.\ 245-400 in {\smc Dellacherie Meyer \& Weil 77}.
}

\medskip%N

\leaveitout{
Novikov A.A.\ [72] `On an identity for stochastic integrals',
Theory Prob.\ Appl.\ 17 (1972) 717-720.
}

Nutz M.\ [p11]  `Pathwise construction of stochastic integrals',
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\cmmnt{[649E.]}

\medskip%O\medskip%P

Protter P.E.\ [05] {\it Stochastic Integration and Differential Equations.}
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\cmmnt{[\S627 {\it notes\/}, 634Yd, \S64 {\it intro.\/}, \S643 {\it notes\/}, 
\S645 {\it notes\/}, \S654 {\it notes\/}.]}
 % ISBN 978-3-540-00313-7 419 pp

\medskip%Q\medskip%R

Rogers L.C.J.\ \& Williams D.\ [00] {\it Diffusions, Markov Processes
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\cmmnt{[\S641 {\it notes\/}, \S645 {\it notes\/}.]}

\medskip%S

Sato K.-I.\ [13] {\it L\'evy processes and infinitely divisible distributions.} 
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\cmmnt{[\S652 {\it notes\/}.]} 
% Cambridge studies in advanced mathematics 68. 

%Seifried F.\ {\tt seifried\@mathematik.uni-kl.de}
%\medskip%T\medskip%U\medskip%V

\leaveitout{
\medskip%W

Weizs\"acker H.von \& Winkler G,\ [90] {\it Stochastic Integrals.}
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}

\medskip%X\medskip%Y\medskip%Z

\discrpage

}%end of references
%https://almostsure.wordpress.com/2009/11/30/predictable-stopping-times/
%https://almostsure.wordpress.com/  George Lowther
%   george.lowther@blueyonder.co.uk   (2009)

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

\end



