% 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{mt61.tex}
\versiondate{12.1.24}
\copyrightdate{2012}

\def\chaptername{The Riemann-sum integral}

\newchapter{61}

I begin with an attempt to give a coherent and complete description of the
principal
form of stochastic integration which will be investigated in this volume.

As elsewhere in probability theory, it is customary to set this
material out in terms of ordinary random variables, that is, measurable
functions defined on probability spaces.   We find immediately, however,
that while integrands and integrators may well present themselves most
naturally in this form, the integrals we construct are defined, in the
cases for which this theory has been developed, in terms
of convergence in $\|\,\|_1$ or $\|\,\|_2$ or in measure, and
therefore correspond not to explicit functions, but to equivalence classes
of functions.   Moreover, integrands and integrators can be changed on
negligible sets without affecting the values of the corresponding
integrals.   I believe that the theory becomes clearer and cleaner if we
move directly to operations on evolving families in $L^0$.   While this
demands an initial investment by the reader in a more abstract framework
for the ideas of elementary probability theory, the translation is not
difficult, and a full exposition can be found in Chapter 36.

Again, stochastic processes are usually expressed as families
$\family{t}{T}{X_t}$ of random variables, indexed by a set $T$ of `times'.
There are very good reasons for this.   However, to
describe the stochastic integral in reasonable generality we need, as a
first step, to discuss the random variable $X_{\tau}$ for a stopping time
$\tau$.   The measure theory to make this possible (the notion of
`progressively measurable' process) is well understood and has been
described in \S455.   %455L
When we come, following my principle above, to look
at $\langle X_t^{\ssbullet}\rangle_{t\in T}$, we find that we can have
$X_t^{\ssbullet}=Y_t^{\ssbullet}$, that is, $X_t\eae Y_t$, for every $t$,
while $X_{\tau}^{\ssbullet}\ne Y_{\tau}^{\ssbullet}$.   This is just a
nuisance.
For our purposes here, it makes better sense to start from a
family $\family{\tau}{\Cal S}{u_{\tau}}$ where $\Cal S$ is a set of
stopping times and $u_{\tau}\in L^0$ for every $\tau\in\Cal S$.   The
construction of such families from processes
$\family{t}{T}{X_t}$ is important and interesting, but has nothing to do
with the very substantial difficulties of the basic theory of stochastic
integration.

Of course I now have to look at filtrations and stopping times,
and these too are not
best described in terms of $\sigma$-algebras of sets and real-valued
functions.   In the formulation I wish to use
here, we don't even have a probability space for the functions to be
defined on.   Instead of thinking of a filtration as a family
$\family{t}{T}{\Sigma_t}$ of $\sigma$-subalgebras of the domain $\Sigma$ of
a probability measure $\mu$, I look at the corresponding family
of subalgebras of the measure algebra $\frak A$ of $\mu$.
This is easy (at least,
if you have read Chapter 32;  and this is my last apology for insisting
that you know something of Volume 3).   A stopping time $\tau$ now becomes
defined in terms of elements $\Bvalue{\tau>t}\in\frak A$, `the region where
$\tau>t$'.   We need to develop a theory of regions $\Bvalue{\sigma<\tau}$,
$\Bvalue{\sigma=\tau}$ in $\frak A$, and subalgebras $\frak A_{\tau}$ of
$\frak A$, for stopping times $\sigma$, $\tau$;  and now the
processes $\family{\tau}{\Cal S}{u_{\tau}}$ we work with must be such that
`$u_{\sigma}=u_{\tau}$ whenever $\sigma=\tau$', that is,
$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$.
Setting up these structures takes the greater part of \S\S611-612,
which come to about a quarter of the chapter.   It happens that
nearly everything in these two sections
can be done without mentioning `measure' at all.

I say again that none of this is difficult, but it does take quite a long
time;  there are some new kinds of algebra to get a solid basis in,
particularly the theory of stopping-time intervals (611E, 611J-611K)
and fully adapted processes (612D).
With this established, however, we are within reach of a
direct definition of a stochastic integral as a limit of Riemann sums
(\S613).   As long as we do not enquire about when the integral is actually
defined, this is very straightforward and can be done in great generality.
The next three sections are devoted to finding the basic cases
of processes $\pmb u$, $\pmb v$ for which
we shall have a well-defined integral $\int\pmb u\,d\pmb v$.
Concerning $\pmb u$, we have
`simple' and `moderately oscillatory' processes (612J, 615E).
Concerning $\pmb v$, we have the concept of `integrator' (616Fc), which is
well adapted to the basic theorem 616K, but is otherwise obscure.
It is easy enough to find a definition of `bounded variation' for
stochastic processes (614I) and to show that processes of bounded variation
are integrators (616R), but this is not what the stochastic integral is
for;  in this case we have much more direct methods available.

Now we are ready, at least in a formal sense,
for some proper stochastic calculus in \S\S617 and 619.
Here I set out useful general manipulations.   Some of them reproduce
patterns familiar from the ordinary Riemann integral (616J),
but others are radically different (617I, 619C).   On the way to the latter
(`It\^o's formula') we need to understand `jump-free' processes, corresponding
to processes with continuous sample paths (\S618).

The theory here involves a large number of constructions.   Many of these
have no short descriptions in terms of the concepts developed in Volumes
1-4, and correspondingly require new terminology and notation.   I have
tried to arrange the material in such a way that, within any individual
section, substantial parts of the basic framework can be taken to be
constant.   From \S614 on, these are indicated in introductory paragraphs
headed `Notation'.   These paragraphs
are highly repetitive.   But until you are very
familiar with my language, it is likely that opening at a random page,
and scanning for the next `Theorem', will lead you to something totally
mysterious.   Sometimes a check in the index for terminology will help.
But sometimes there will be a baffling symbol, and then it will be worth
while turning to the beginning of the section to see if the symbol appears
there.   It seems to me that while this expands the volume by several pages
in total, it is kinder than referring you each time to a complete list of the
terminological quirks of this presentation.

\discrpage

\frfilename{mt611.tex}
\versiondate{11.12.17/13.4.21}

\def\Mlob{M_{\text{lob}}}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Stopping times}

\newsection{611}

The first step is to describe the structures within which the work of this
volume will proceed.   While everything really important will have to be
based on probability algebras, I start with ideas
which can be applied to arbitrary Dedekind complete Boolean algebras.
This section introduces filtrations of subalgebras, the lattice of
stopping times, the algebras associated with stopping times,
stopping-time intervals and covered envelopes.

\leader{611A}{Filtrations} Throughout this volume, $\frak A$ will denote a
Dedekind complete Boolean algebra\cmmnt{ (definitions:  311A, 314A)}, with
Boolean operations $\Bsymmdiff$, $\Bcap$, $\Bcup$ and $\Bsetminus$, zero $0$ and
multiplicative identity $1$.

\spheader 611Aa Let $T$ be a non-empty totally ordered
set.   A {\bf filtration of order-closed subalgebras of $\frak A$}
will be a non-decreasing family $\family{t}{T}{\frak A_t}$ of order-closed
subalgebras of $\frak A$\cmmnt{ (definition:  313D)}.

\spheader 611Ab{\bf (i)} A {\bf stopping time $\tau$ adapted to
$\family{t}{T}{\frak A_t}$} is a family
$\family{t}{T}{\Bvalue{\tau>t}}$ such that

\inset{$\Bvalue{\tau>t}\in\frak A_t$ for every $t\in T$,

if $s\le t$ in $T$ then $\Bvalue{\tau>t}\Bsubseteq\Bvalue{\tau>s}$,

if $t\in T$ is not isolated on the right\cmmnt{, that is, $t$ is neither the
greatest element of $T$ nor the lower endpoint of a gap in $T$, that is,
$\{s:s>t\}$ is non-empty and has infimum $t$,} then
$\Bvalue{\tau>t}=\sup_{s>t}\Bvalue{\tau>s}$.}

\cmmnt{\noindent (Compare 364A.)}

\medskip

\quad{\bf (ii)}
It will be worth checking each concept against the constant stopping
times, where for $t\in T$
the {\bf constant stopping time at $t$}, $\check{t}$, is given by
setting

$$\eqalign{\Bvalue{\check{t}>s}
&=1\text{ if }s<t,\cr
&=0\text{ if }s\ge t.\cr}$$

\medskip

\quad{\bf (iii)} I will say that a stopping time $\tau$ is

\inset{----- {\bf finite-valued} if
$\inf_{t\in T}\Bvalue{\tau>t}=0$,

% and either $T$ has a least element or
%$\sup_{t\in T}\Bvalue{\tau>t}=1$,

----- {\bf bounded} if there is a $t\in T$ such that
$\Bvalue{\tau>t}=0$.}

%, and either $T$ has a least element or there is a
%$t'\in T$ such that $\Bvalue{\tau>t'}=1$.}

\noindent Constant stopping times are
bounded, and bounded stopping times are finite-valued.

\medskip

\quad{\bf (iv)}
I will write $\Cal T$ for the set of stopping times adapted to
$\family{t}{T}{\frak A_t}$, $\Cal T_f\subseteq\Cal T$
for the set of finite-valued stopping times, and
$\Cal T_b\subseteq\Cal T_f$
for the set of bounded stopping times.

\spheader 611Ac It is convenient to think of a stopping time $\tau\in\Cal T$ as
the element $\family{t}{T}{\Bvalue{\tau>t}}$ of
the simple product algebra
$\prod_{t\in T}\frak A_t$\cmmnt{ (definition:  315A)}.\cmmnt{   But a warning!
while this represents $\Cal T$ as a sublattice of
$\prod_{t\in T}\frak A_t$
(611Ca-611Cc) below), $\Cal T$ is not as a rule order-closed (see 611F and
632C(a-i)).}

\cmmnt{\spheader 611Ad We are going to have to think of $\Cal T$ as disjoint
from $T$ (see, for instance, 611G below).   Subject to the Axiom of Foundation
({\smc Kunen 80}, \S III.4), and on any ordinary formalization of the concepts of
`ordered pair' and `function', $T$ and $\prod_{t\in T}\frak A_t$
will automatically
be disjoint.   I will therefore suppose without further comment
that this is always the case.}

\leader{611B}{The partial ordering of stopping times}
If $\sigma$, $\tau\in\Cal T$, say that $\sigma\le\tau$ if
$\Bvalue{\sigma>t}\Bsubseteq\Bvalue{\tau>t}$ for every $t\in T$, that
is, $\sigma\Bsubseteq\tau$ in $\prod_{t\in T}\frak A_t$.
This defines a partial order on $\Cal T$.

\leader{611C}{Proposition} (a) $\Cal T$ is a Dedekind complete
distributive lattice.
Consequently any finite subset of $\Cal T$ is included in a finite
sublattice of $\Cal T$.

(b) If $C\subseteq\Cal T$ is non-empty, then
$\sup C$ is defined by saying that

\Centerline{$\Bvalue{\sup C>t}=\sup_{\tau\in C}\Bvalue{\tau>t}$}

\noindent for every $t\in T$, that is, the supremum of $C$ in $\Cal
T$ is the same as the supremum of $C$ in $\prod_{t\in T}\frak A_t$.

(c) If $\sigma$, $\tau\in\Cal T$,
then $\sigma\wedge\tau$ is defined by saying that

\Centerline{$\Bvalue{\sigma\wedge\tau>t}
=\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t}$}

\noindent for every $t\in T$, that is, $\sigma\wedge\tau$ in $\Cal T$
corresponds to $\sigma\Bcap\tau$ in $\prod_{t\in T}\frak A_t$.

(d) If $C$, $C'\subseteq\Cal T$ are non-empty, then
$\sup C\wedge\sup C'=\sup\{\sigma\wedge\sigma':\sigma\in C$,
$\sigma'\in C'\}$.

(e) Writing $\check{t}$ for the constant stopping time at $t$, the
map $t\mapsto\check{t}:T\to\Cal T$ is an
order-continuous lattice homomorphism, which is injective if $\frak A\ne\{0\}$.

(f) $\Cal T$ has greatest and least elements defined by saying that

\Centerline{$\Bvalue{\max\Cal T>t}=1$,
\quad$\Bvalue{\min\Cal T>t}=0$}

\noindent for every $t\in T$, that is, they correspond to the
greatest and
least elements $1$ and $0$ of $\prod_{t\in T}\frak A_t$.
If $T$ has a least element $\min T$, then $\min\Cal T$ is
the constant stopping time at $\min T$.

(g) $\Cal T_f$ and $\Cal T_b$ are ideals\footnote{If $P$ is a lattice, an
{\bf ideal} of $P$ is a set $Q\subseteq P$ such that $p\vee q\in Q$ for all
$p$, $q\in Q$ and $p\in Q$ whenever $q\in Q$ and $p\le q$ in $P$.
In this context I do not insist that $Q$ should be non-empty.} in
$\Cal T$.

(h) The function $\sigma\mapsto\sigma\wedge\tau:\Cal T\to\Cal T$ is
order-continuous\cmmnt{ (definition:  313Ha)} for every $\tau\in\Cal T$.

\proof{{\bf (a)(i)} I start with a direct verification of (b).
Setting
$a_t=\sup_{\tau\in C}\Bvalue{\tau>t}$ for $t\in T$, we see that

\inset{$a_t\in\frak A_t$ for every $t\in T$, because $\frak A_t$ is
order-closed in $\frak A$,

if $s\le t$ in $T$ then $a_t\Bsubseteq a_s$,

if $t\in T$ is not isolated on the right then

\Centerline{$a_t
=\sup_{\tau\in C}\Bvalue{\tau>t}
=\sup_{\tau\in C,s>t}\Bvalue{\tau>s}
=\sup_{s>t}a_s$.}}

\noindent So we have a stopping time $\sigma$ defined by writing
$\Bvalue{\sigma>t}=a_t$ for every $t$.   Now it is easy to see that
$\tau\le\sigma$ for every $\tau\in C$ and that $\sigma$ is the least
such stopping time.   Thus our formula defines $\sup C$ in the partially
ordered set $\Cal T$.

\medskip

\quad{\bf (ii)}
This is enough to show that $\Cal T$ is a Dedekind complete lattice
(314Aa, 314Bb).

\medskip

\quad{\bf (iii)} Similarly, we can check the formula in (c).
Set $b_t=\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t}$ for $t\in T$.   Then

\inset{$b_t\in\frak A_t$ for every $t\in T$,

if $s\le t$ then $b_t\Bsubseteq b_s$,

if $t\in T$ is not isolated on the right, then}

$$\eqalignno{\sup_{s>t}b_s
&=\sup_{s>t}\Bvalue{\sigma>s}\Bcap\Bvalue{\tau>s}
=\sup_{s,s'>t}\Bvalue{\sigma>s}\Bcap\Bvalue{\tau>s'}\cr
&=\sup_{s>t}\Bvalue{\sigma>s}\Bcap\sup_{s'>t}\Bvalue{\tau>s'}\cr
\displaycause{313Bc}
&=\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t}
=b_t.\cr}$$

\noindent So we have a stopping time $\upsilon$ such that
$\Bvalue{\upsilon>t}=\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t}$ for
every $t$.   But now it is easy to see that
$\upsilon=\sigma\wedge\tau$.

\medskip

\quad{\bf (iv)} Putting these formulae together, we see that
$\sigma\mapsto\Bvalue{\sigma>t}:\Cal T\to\frak A$ is
a lattice homomorphism for every $t\in T$, so
$\sigma\mapsto\family{t}{T}{\Bvalue{\sigma>t}}$ is an
injective lattice homomorphism from $\Cal T$ into the distributive
lattice
$\prod_{t\in T}\frak A_t$, and identifies $\Cal T$ with a sublattice
of $\prod_{t\in T}\frak A_t$, which must be distributive.
And it is true in any distributive lattice that finitely-generated
sublattices are finite (3A1I(c-iii)).

\medskip

{\bf (b)-(c)} have been dealt with in (a-i) and (a-iii) above.

\medskip

{\bf (d)} Translated into $\prod_{t\in T}\frak A_t$, as in (b)-(c) just
above, this becomes

\Centerline{$\sup C\Bcap\sup C'=\sup\{\sigma\Bcap\sigma':\sigma\in C$,
$\sigma'\in C'\}$,}

\noindent which is 313Bc.

\medskip

{\bf (e)(i)}
Concerning constant stopping times, we see at once from the
formulae in 611B-611B that
$\check{s}\le\check{t}$ when $s\le t$, so that $t\mapsto\check{t}$
is order-preserving;  because $T$ is totally ordered, it is a lattice
homomorphism.

\medskip

\quad{\bf (ii)} If $A\subseteq T$ is a non-empty set with supremum
$t$ in $T$, then

\Centerline{$\Bvalue{\check{t}>t'}=0
=\sup_{s\in A}\Bvalue{\check{s}>t'}$}

\noindent whenever $t'\ge t$, while if $t'<t$ there is an $s'\in A$
such that $t'<s'$, so that

\Centerline{$\Bvalue{\check{t}>t'}=1=\Bvalue{\check{s}'>t'}
=\sup_{s\in A}\Bvalue{\check{s}>t'}$.}

\noindent By (b), $\check{t}$ is the supremum $\sup_{s\in A}\check{s}$
in $\Cal T$.

\medskip

\quad{\bf (iii)} If $\emptyset\ne A\subseteq T$ and $t=\inf A$,
then $\check{t}\le\check{s}$ for every $s\in A$, as noted in (i).
Now suppose that $\tau\in\Cal T$ and
$\check{t}\le\tau\le\check{s}$ for every $s\in A$.   If $t$ is isolated on the
right in $T$, then $t\in A$ so surely $\tau=\check t$.   Otherwise,

\Centerline{$\Bvalue{\tau>t}
=\sup_{s>t}\Bvalue{\tau>s}
=\sup_{s\in A}\Bvalue{\tau>s}
\Bsubseteq\sup_{s\in A}\Bvalue{\check{s}>s}
=0$.}

\noindent Of course we now have

\Centerline{$\Bvalue{\tau>s}
\Bsubseteq\Bvalue{\tau>t}=0=\Bvalue{\check t>s}$}

\noindent for $s\ge t$, while

\Centerline{$\Bvalue{\tau>s}\Bsupseteq\Bvalue{\check{t}>s}=1$}

\noindent for $s<t$, so $\tau=\check{t}$.   As $\tau$ is arbitrary,
$\check{t}=\inf_{s\in A}\check{s}$.

Thus $t\mapsto\check{t}$ is order-continuous.

\medskip

\quad{\bf (iv)} If $\frak A\ne\{0\}$, that is, $0\ne 1$ in $\frak A$, and $s<t$
in $T$, then

\Centerline{$\Bvalue{\check s>s}=0\ne 1=\Bvalue{\check t>s}$}

\noindent so $\check s\ne\check t$;  thus $t\mapsto\check t$ is injective.

\medskip

{\bf (f)} The formulae offered for
$\max\Cal T$ and $\min\Cal T$ describe stopping times
corresponding to $1$ and $0$ in $\prod_{t\in T}\frak A_t$, so give us
the greatest and least elements of $\Cal T$.   If $\min T$ is
defined, the
formula for $\min\Cal T$ agrees with that for the constant
stopping time $(\min T)\check{\phantom a}$.

\medskip

{\bf (g)(i)} If $\sigma$, $\tau\in\Cal T^f$,

$$\eqalignno{\inf_{t\in T}\Bvalue{\sigma\vee\tau>t}
&=\inf_{t\in T}\Bvalue{\sigma>t}\Bcup\Bvalue{\tau>t}
=\inf_{s,t\in T}\Bvalue{\sigma>s}\Bcup\Bvalue{\tau>t}\cr&
=\inf_{s\in T}\Bvalue{\sigma>s}\Bcup\inf_{t\in T}\Bvalue{\tau>t}\cr
\displaycause{313Bd}
&=0.\cr}$$

\noindent So $\sigma\vee\tau\in\Cal T_f$.   If $\sigma\le\tau$ in $\Cal T$
and $\tau\in\Cal T_f$, then

\Centerline{$\inf_{t\in T}\Bvalue{\sigma>t}
\Bsubseteq\inf_{t\in T}\Bvalue{\tau>t}=0$,}

\noindent so $\sigma\in\Cal T_f$.   Thus $\Cal T_f$ is an ideal in
$\Cal T$.

\medskip

\quad{\bf (ii)} As for $\Cal T_b$, observe that $\tau\in\Cal T$
is bounded iff there is a constant stopping time $\check{t}$
such that $\tau\le\check{t}$.
Now we have just seen that the set of constant
stopping times is totally ordered, so $\Cal T_b$ is an ideal in $\Cal T$.

\medskip

{\bf (h)(i)} If $A\subseteq\Cal T$ is a non-empty
downwards-directed set, then

\Centerline{$\inf_{\sigma\in A}\tau\wedge\sigma=\inf(\{\tau\}\cup A)
=\tau\wedge\inf A$.}

\medskip

\quad{\bf (ii)} If $A\subseteq\Cal T$ is a non-empty upwards-directed with
supremum $\sigma^*$, and $\sigma^*_1=\sup_{\sigma\in A}\sigma\wedge\tau$,
then, for any $t\in T$,

$$\eqalignno{\Bvalue{\sigma^*_1>t}
&=\sup_{\sigma\in A}\Bvalue{\tau\wedge\sigma>t}\cr
\displaycause{by (b)}
&=\sup_{\sigma\in A}\Bvalue{\tau>t}\Bcap\Bvalue{\sigma>t}\cr
\displaycause{by (c)}
&=\Bvalue{\tau>t}\Bcap\sup_{\sigma\in A}\Bvalue{\sigma>t}\cr
\displaycause{313Bc again}
&=\Bvalue{\tau>t}\Bcap\Bvalue{\sigma^*>t}
=\Bvalue{\tau\wedge\sigma^*>t},\cr}$$

\noindent so $\sigma^*_1=\tau\wedge\sigma^*$.
}%end of proof of 611C

\medskip

\noindent{\bf Remark}\cmmnt{ (The following applies in any lattice.)}
If $A\subseteq\Cal T$ and $\tau\in\Cal T$, I will write
$A\vee\tau$ for $\{\sigma\vee\tau:\sigma\in A\}$ and $A\wedge\tau$ for
$\{\sigma\wedge\tau:\sigma\in A\}$.   Note that if $\Cal S$ is a sublattice of
$\Cal T$ and $\tau\in\Cal S$, then

\Centerline{$\Cal S\vee\tau=\{\sigma:\sigma\in\Cal S$, $\tau\le\sigma\}$,
\quad$\Cal S\wedge\tau=\{\sigma:\sigma\in\Cal S$, $\sigma\le\tau\}$.}

\noindent So if $\Cal S$ is a sublattice of $\Cal T$, $\tau$, $\tau'\in\Cal S$
and $\tau\le\tau'$,

\Centerline{$\Cal S\cap[\tau,\tau']
=\{\sigma:\sigma\in\Cal S,\,\tau\le\sigma\le\tau'\}
=\{\sigma:\sigma\in\Cal S\vee\tau,\,\sigma\le\tau'\}
=(\Cal S\vee\tau)\wedge\tau'$}

\noindent because $\Cal S\vee\tau=\{\sigma:\sigma\in\Cal S$, $\tau\le\sigma\}$
is a sublattice of $\Cal T$.

\leader{611D}{The region where $\sigma<\tau$} If $\sigma$, $\tau\in\Cal T$ set

\Centerline{$\Bvalue{\sigma<\tau}
=\sup_{t\in T}(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})$,}

\Centerline{$\Bvalue{\sigma\le\tau}
=1\Bsetminus\Bvalue{\tau<\sigma}
=\inf_{t\in T}(\Bvalue{\tau>t}\Bcup(1\Bsetminus\Bvalue{\sigma>t}))$,}

\Centerline{$\Bvalue{\sigma=\tau}
=\Bvalue{\sigma\le\tau}\Bcap\Bvalue{\tau\le\sigma}
=1\Bsetminus\sup_{t\in T}
  (\Bvalue{\sigma>t}\Bsymmdiff\Bvalue{\tau>t})$.}

\leader{611E}{}\cmmnt{ Analysts commonly think of algebra as trivial, and
so it is.   But the algebra of stopping times and regions
$\Bvalue{\sigma<\tau}$ is a very rich structure, with a
large number of elementary identities;  and as the definition of
$\Bvalue{\sigma<\tau}$ includes the supremum of an infinite set, there is
room for surprises.   Consequently the fluency necessary for
effective use of these ideas requires a good deal of
practice.   In the next theorem I have collected a more or less
comprehensive list of facts which will be useful in one way or another.
It is a very long list, and correspondingly few readers will be inclined to
work through it systematically.   I recommend rather that you treat it as a
running buffet, to be sampled from time to time.   I hope that there are
enough cross-references later for you to know
when you have to return for another fragment.

\medskip

\noindent}{\bf Theorem} (a) Let $\sigma$, $\tau\in\Cal T$.

\quad(i)($\alpha$)
$(\Bvalue{\sigma<\tau},\Bvalue{\sigma=\tau},\Bvalue{\tau<\sigma})$ is
a partition of unity in $\frak A$. %@a

\qquad($\beta$) $\Bvalue{\sigma>t}\Bcap\Bvalue{\sigma=\tau}
=\Bvalue{\tau>t}\Bcap\Bvalue{\sigma=\tau}$ for every $t\in T$. %@a

\qquad($\gamma$) $\Bvalue{\sigma<\tau}=0$ iff $\Bvalue{\tau\le\sigma}=1$ iff
$\tau\le\sigma$;  $\Bvalue{\sigma=\tau}=1$ iff $\sigma=\tau$. %@a

\qquad($\delta$) Writing $\check{t}$ for the constant stopping time
at $t$,
$\Bvalue{\check{t}<\tau}=\Bvalue{\tau>t}$ for every $t\in T$. %@a

\qquad($\epsilon$) $\Bvalue{\min\Cal T<\max\Cal T}=1$.

\qquad($\zeta$) If $s<t$ in $T$, then $\Bvalue{\check s<\check t}=1$;
$\Bvalue{\check s<\max\Cal T}=1$ for every $s\in T$.

\quad(ii)%formerly 6{}11G(a-ii)
($\alpha$) $\Bvalue{\sigma<\tau}=\Bvalue{\sigma\wedge\tau<\tau}
=\Bvalue{\sigma<\sigma\vee\tau}$.   %@bii

\qquad($\beta$) $\Bvalue{\sigma\le\tau}=\Bvalue{\sigma=\sigma\wedge\tau}
=\Bvalue{\tau=\sigma\vee\tau}$.

\qquad($\gamma$)
$\Bvalue{\sigma\wedge\tau=\sigma}\Bcup\Bvalue{\sigma\wedge\tau=\tau}
=\Bvalue{\sigma\vee\tau=\sigma}\Bcup\Bvalue{\sigma\vee\tau=\tau}=1$.
%@biii

(b) If $\sigma\in\Cal T$ and $C\subseteq\Cal T$ is non-empty
then $\Bvalue{\sigma<\sup C}=\sup_{\tau\in C}\Bvalue{\sigma<\tau}$ and
$\Bvalue{\sup C\le\sigma}=\inf_{\tau\in C}\Bvalue{\tau\le\sigma}$.

%@cii

(c) Let $\sigma$, $\tau$, $\upsilon\in\Cal T$.

\quad(i)($\alpha$) $\Bvalue{\sigma\wedge\tau<\upsilon}
=\Bvalue{\sigma<\upsilon}\Bcup\Bvalue{\tau<\upsilon}$,  %@ci
$\Bvalue{\upsilon\le\sigma\wedge\tau}
=\Bvalue{\upsilon\le\sigma}\Bcap\Bvalue{\upsilon\le\tau}$.

\qquad($\beta$) $\Bvalue{\upsilon<\sigma\wedge\tau}
=\Bvalue{\upsilon<\sigma}\Bcap\Bvalue{\upsilon<\tau}$,    %@ciii
$\Bvalue{\sigma\wedge\tau\le\upsilon}
=\Bvalue{\sigma\le\upsilon}\Bcup\Bvalue{\tau\le\upsilon}$.

\quad(ii)($\alpha$) $\Bvalue{\sigma\vee\tau<\upsilon}
=\Bvalue{\sigma<\upsilon}\Bcap\Bvalue{\tau<\upsilon}$,   %@civ
$\Bvalue{\upsilon\le\sigma\vee\tau}
=\Bvalue{\upsilon\le\sigma}\Bcup\Bvalue{\upsilon\le\tau}$.

\qquad($\beta$) $\Bvalue{\upsilon<\sigma\vee\tau}
=\Bvalue{\upsilon<\sigma}\Bcup\Bvalue{\upsilon<\tau}$,
$\Bvalue{\sigma\vee\tau\le\upsilon}
=\Bvalue{\sigma\le\upsilon}\Bcap\Bvalue{\tau\le\upsilon}$.

\quad(iii)($\alpha$) $\Bvalue{\sigma<\upsilon}
\Bsubseteq\Bvalue{\sigma<\tau}\Bcup\Bvalue{\sigma\vee\tau<\upsilon}
\Bsubseteq\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau<\upsilon}$.  %cv

\qquad($\beta$) $\Bvalue{\sigma\le\upsilon}
\Bsubseteq\Bvalue{\sigma\le\tau}\Bcup\Bvalue{\tau<\upsilon}$.

\qquad($\gamma$) $\Bvalue{\sigma<\upsilon}\Bcap\Bvalue{\upsilon\le\tau}
\Bsubseteq\Bvalue{\sigma<\tau}$,
$\Bvalue{\sigma\le\upsilon}\Bcap\Bvalue{\upsilon<\tau}
\Bsubseteq\Bvalue{\sigma<\tau}$.

\quad(iv)($\alpha$) $\Bvalue{\sigma\le\tau}\Bcap\Bvalue{\tau\le\upsilon}
\Bsubseteq\Bvalue{\sigma\le\upsilon}$.

\qquad($\beta$) $\Bvalue{\sigma\le\tau}\Bcap\Bvalue{\tau<\upsilon}
\Bsubseteq\Bvalue{\sigma<\upsilon}$.

\qquad($\gamma$) $\Bvalue{\sigma=\tau}\Bcap\Bvalue{\tau=\upsilon}
=\Bvalue{\sigma=\tau}\Bcap\Bvalue{\sigma=\upsilon}
\Bsubseteq\Bvalue{\sigma=\upsilon}$.   %@cvii

\quad(v)($\alpha$) $\Bvalue{\sigma\wedge\upsilon=\tau\wedge\upsilon}
\Bsupseteq\Bvalue{\sigma=\tau}$.  %@cviii

\qquad($\beta$) $\Bvalue{\sigma\vee\upsilon=\tau\vee\upsilon}
\Bsupseteq\Bvalue{\sigma=\tau}$.

\quad(vi) If $\sigma\le\tau\le\upsilon$, then
$\Bvalue{\sigma<\upsilon}
=\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau<\upsilon}$.  %@cvi

(d) If $\tau_0\le\ldots\le\tau_n$ in $\Cal T$ and $\sigma\in\Cal T$, then

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

\noindent is a partition of unity in $\frak A$.

%zz1Q
\proof{{\bf (a)(i)}\grheada\ We have only to check that
$\Bvalue{\sigma<\tau}\Bcap\Bvalue{\tau<\sigma}=0$.   But

$$\eqalignno{(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})
  \Bcap(\Bvalue{\sigma>s}\Bsetminus\Bvalue{\tau>s})
&\Bsubseteq\Bvalue{\tau>t}\Bsetminus\Bvalue{\tau>s}=0
   \text{ if }s\le t,\cr
&\Bsubseteq\Bvalue{\sigma>s}\Bsetminus\Bvalue{\sigma>t}=0
   \text{ if }t\le s,\cr}$$

\noindent so

$$\eqalign{\Bvalue{\sigma<\tau}\Bcap\Bvalue{\tau<\sigma}
&=\sup_{t\in T}(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})
  \Bcap\sup_{s\in T}(\Bvalue{\sigma>s}\Bsetminus\Bvalue{\tau>s})\cr
&=\sup_{s,t\in T}(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})
  \Bcap(\Bvalue{\sigma>s}\Bsetminus\Bvalue{\tau>s})
=0.\cr}$$

\medskip

\qquad{\bf\grheadb-\grheadc} are immediate from the definitions in
611B and 611D.

\medskip

\qquad\grheadd\ $\Bvalue{\check{t}<\tau}
=\sup_{s\in T}\Bvalue{\tau>s}\Bsetminus\Bvalue{\check{t}>s}
=\sup_{s\ge t}\Bvalue{\tau>s}
=\Bvalue{\tau>t}$.

\medskip

\qquad\grheade\ Recall that we are assuming that $T$ is not empty.
And $\Bvalue{\max\Cal T>t}\Bsetminus\Bvalue{\min\Cal T>t}=1$ for any
$t\in T$.

\medskip

\qquad\grheadz\ $\Bvalue{\check s<\check t}
\Bsupseteq\Bvalue{\check t>s}\Bsetminus\Bvalue{\check s>s}=1$;
$\Bvalue{\check s<\max\Cal T}
\Bsupseteq\Bvalue{\max\Cal T>s}\Bsetminus\Bvalue{\check s>s}=1$.

\medskip

\quad{\bf (ii)}\grheada\

$$\eqalignno{\Bvalue{\sigma\wedge\tau<\tau}
&=\sup_{t\in T}\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma\wedge\tau>t}\cr
&=\sup_{t\in T}
  \Bvalue{\tau>t}\Bsetminus(\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t})\cr
\displaycause{611Cc}
&=\sup_{t\in T}
  \Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
=\Bvalue{\sigma<\tau},\cr
\Bvalue{\sigma<\sigma\vee\tau}
&=\sup_{t\in T}\Bvalue{\sigma\vee\tau>t}\Bsetminus\Bvalue{\sigma>t}\cr
&=\sup_{t\in T}
   (\Bvalue{\sigma>t}\Bcup\Bvalue{\tau>t})\Bsetminus\Bvalue{\sigma>t}\cr
\displaycause{611Cb}
&=\sup_{t\in T}
  \Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
=\Bvalue{\sigma<\tau}.\cr}$$

\medskip

\qquad\grheadb\

$$\eqalignno{\Bvalue{\sigma=\sigma\wedge\tau}&
=\Bvalue{\sigma\le\sigma\wedge\tau}\Bcap\Bvalue{\sigma\wedge\tau\le\sigma}
=(1\Bsetminus\Bvalue{\sigma\wedge\tau<\sigma})\Bcap 1
\Displaycause{by (i-$\gamma$)}
=1\Bsetminus\Bvalue{\tau<\sigma}
\Displaycause{by ($\alpha$) just above}
=\Bvalue{\sigma\le\tau},\cr}$$

\noindent and similarly

\Centerline{$\Bvalue{\tau=\sigma\vee\tau}
=\Bvalue{\tau\le\sigma\vee\tau}\Bcap\Bvalue{\sigma\vee\tau\le\tau}
=1\Bsetminus\Bvalue{\tau<\sigma\vee\tau}
=1\Bsetminus\Bvalue{\tau<\sigma}
=\Bvalue{\sigma\le\tau}$.}

\medskip

\qquad\grheadc\ Using both parts of ($\beta$), we have

$$\eqalignno{\Bvalue{\sigma\wedge\tau=\sigma}
   \Bcup\Bvalue{\sigma\wedge\tau=\tau}&
=\Bvalue{\tau\le\sigma}\Bcup\Bvalue{\sigma\le\tau}=1\cr&
=\Bvalue{\sigma\vee\tau=\sigma}\Bcup\Bvalue{\sigma\vee\tau=\tau}.\cr}$$

\medskip

{\bf (b)}

$$\eqalignno{\sup_{t\in T}\Bvalue{\sup C>t}\Bsetminus\Bvalue{\sigma>t}
&=\sup_{t\in T}(\sup_{\tau\in C}\Bvalue{\tau>t})
  \Bsetminus\Bvalue{\sigma>t}\cr
\displaycause{611Cb again}
&=\sup_{t\in T}
  \sup_{\tau\in C}(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})
=\sup_{\tau\in C}\Bvalue{\sigma<\tau}.\cr}$$

\noindent Taking complements,

\Centerline{$\Bvalue{\sup C\le\sigma}
=\inf_{\tau\in C}\Bvalue{\tau\le\sigma}$.}

\medskip

{\bf (c)(i)}\grheada\
$\Bvalue{\sigma\wedge\tau>t}=\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t}$
for every $t$, so

$$\eqalign{\Bvalue{\sigma\wedge\tau<\upsilon}
&=\sup_{t\in T}
\Bvalue{\upsilon>t}\Bsetminus(\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t})\cr
&=\sup_{t\in T}
(\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\sigma>t})
  \Bcup(\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\tau>t})
=\Bvalue{\sigma<\upsilon}\Bcup\Bvalue{\tau<\upsilon}.\cr}$$

\noindent Taking complements,

\Centerline{$\Bvalue{\upsilon\le\sigma\wedge\tau}
=\Bvalue{\upsilon\le\sigma}\Bcap\Bvalue{\upsilon\le\tau}$.}

\medskip

\qquad\grheadb

$$\eqalignno{\Bvalue{\upsilon<\sigma}\Bcap\Bvalue{\upsilon<\tau}
&=(\sup_{s\in T}\Bvalue{\sigma>s}\Bsetminus\Bvalue{\upsilon>s})
   \Bcap(\sup_{t\in
T}\Bvalue{\tau>t}\Bsetminus\Bvalue{\upsilon>t})\cr
&=\sup_{s,t\in T}(\Bvalue{\sigma>s}\Bsetminus\Bvalue{\upsilon>s})
   \Bcap(\Bvalue{\tau>t}\Bsetminus\Bvalue{\upsilon>t})\cr
&=\sup_{s,t\in T}\Bvalue{\sigma>s}\Bcap\Bvalue{\tau>t}
   \Bsetminus(\Bvalue{\upsilon>s}\Bcup\Bvalue{\upsilon>t})\cr
&=\sup_{s,t\in T}\Bvalue{\sigma>s}\Bcap\Bvalue{\tau>t}
   \Bsetminus\Bvalue{\upsilon>\min(s,t)}\cr
&=\sup_{t'\in T}\sup_{s\ge t',t\ge t'}
  \Bvalue{\sigma>s}\Bcap\Bvalue{\tau>t}\Bsetminus\Bvalue{\upsilon>t'}\cr
&=\sup_{t'\in T}
  \Bvalue{\sigma>t'}\Bcap\Bvalue{\tau>t'}\Bsetminus\Bvalue{\upsilon>t'}\cr
&=\sup_{t'\in T}
   \Bvalue{\sigma\wedge\tau>t'}\Bsetminus\Bvalue{\upsilon>t'}
=\Bvalue{\upsilon<\sigma\wedge\tau}.\cr}$$

\noindent Taking complements,

\Centerline{$\Bvalue{\sigma\wedge\tau\le\upsilon}
=\Bvalue{\sigma\le\upsilon}\Bcup\Bvalue{\tau\le\upsilon}$.}

\medskip

\quad{\bf (ii)}

$$\eqalignno{\Bvalue{\sigma<\upsilon}\Bcap\Bvalue{\tau<\upsilon}
&=(\sup_{s\in T}\Bvalue{\upsilon>s}\Bsetminus\Bvalue{\sigma>s})
   \Bcap(\sup_{t\in T}\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\tau>t})\cr
&=\sup_{s,t\in T}(\Bvalue{\upsilon>s}\Bsetminus\Bvalue{\sigma>s})
   \Bcap(\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\tau>t})\cr
&=\sup_{s,t\in T}\Bvalue{\upsilon>s}\Bcap\Bvalue{\upsilon>t}
   \Bsetminus(\Bvalue{\sigma>s}\Bcup\Bvalue{\tau>t})\cr
&=\sup_{s,t\in T}\Bvalue{\upsilon>\max(s,t)}
   \Bsetminus(\Bvalue{\sigma>s}\Bcup\Bvalue{\tau>t})\cr
&=\sup_{t'\in T}\sup_{s\le t',t\le t'}
  \Bvalue{\upsilon>t'}\Bsetminus(\Bvalue{\sigma>s}\Bcup\Bvalue{\tau>t})\cr
&=\sup_{t'\in T}\Bvalue{\upsilon>t'}
   \Bsetminus(\Bvalue{\sigma>t'}\Bcup\Bvalue{\tau>t'})\cr
&=\sup_{t'\in T}
  \Bvalue{\upsilon>t'}\Bsetminus\Bvalue{\sigma\vee\tau>t'}
=\Bvalue{\sigma\vee\tau<\upsilon}.\cr}$$

\noindent Taking complements,

\Centerline{$\Bvalue{\upsilon\le\sigma\vee\tau}
=\Bvalue{\upsilon\le\sigma}\Bcup\Bvalue{\upsilon\le\tau}$.}

\medskip

\quad{\bf (iii)}\grheada\ For every $t\in T$,

\Centerline{$\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\sigma>t}
\Bsubseteq(\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\tau>t})
\Bcup(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})$.}

\noindent So $\Bvalue{\sigma<\upsilon}
\Bsubseteq\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau<\upsilon}$.
But equally we must also have

\Centerline{$\Bvalue{\sigma<\upsilon}
\Bsubseteq\Bvalue{\sigma<\sigma\vee\tau}
   \Bcup\Bvalue{\sigma\vee\tau<\upsilon}
=\Bvalue{\sigma<\tau}
   \Bcup\Bvalue{\sigma\vee\tau<\upsilon}$}

\noindent by (a-ii-$\alpha$).   And
$\Bvalue{\sigma<\tau}\Bcup\Bvalue{\sigma\vee\tau<\upsilon}
\Bsubseteq\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau<\upsilon}$ by (ii-$\alpha$)
just above.

\medskip

\qquad\grheadb\ Now

\Centerline{$1
=\Bvalue{\sigma\le\tau}\Bcup\Bvalue{\tau<\sigma}
\Bsubseteq\Bvalue{\sigma\le\tau}
  \Bcup\Bvalue{\tau<\upsilon}\Bcup\Bvalue{\upsilon<\sigma}$}

\noindent by ($\alpha$), so

\Centerline{$\Bvalue{\sigma\le\upsilon}
=1\Bsetminus\Bvalue{\upsilon<\sigma}
\Bsubseteq\Bvalue{\sigma\le\tau}\Bcup\Bvalue{\tau<\upsilon}$.}

\medskip

\qquad\grheadc\ Using ($\alpha$) twice,

\Centerline{$\Bvalue{\sigma<\upsilon}\Bcap\Bvalue{\upsilon\le\tau}
\Bsubseteq(\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau<\upsilon})
   \Bcap\Bvalue{\upsilon\le\tau}
\Bsubseteq\Bvalue{\sigma<\tau}$,}

\Centerline{$\Bvalue{\sigma\le\upsilon}\Bcap\Bvalue{\upsilon<\tau}
\Bsubseteq\Bvalue{\sigma\le\upsilon}
   \Bcap(\Bvalue{\upsilon<\sigma}\Bcup\Bvalue{\sigma<\tau})
\Bsubseteq\Bvalue{\sigma<\tau}$.}

\medskip

\quad{\bf (iv)}\grheada\ Take complements in (iii-$\alpha$) and
exchange the names $\sigma$, $\upsilon$.

\medskip

\qquad\grheadb\ Take complements in (iii-$\beta$) and
exchange the names $\sigma$, $\upsilon$.

\medskip

\qquad\grheadc\ Use ($\alpha$) twice to prove the first equality;  the second
is trivial.

\medskip

\quad{\bf (v)}\grheada\

\Centerline{$\Bvalue{\sigma\wedge\upsilon<\tau\wedge\upsilon}
=\Bvalue{\sigma<\tau\wedge\upsilon}
  \Bcup\Bvalue{\upsilon<\tau\wedge\upsilon}
\Bsubseteq\Bvalue{\sigma<\tau}$}

\noindent by (i) above.   Similarly,
$\Bvalue{\tau\wedge\upsilon<\sigma\wedge\upsilon}
\Bsubseteq\Bvalue{\tau<\sigma}$, and both are disjoint from
$\Bvalue{\sigma=\tau}$.

\medskip

\qquad\grheadb\ As ($\alpha$), but starting from

\Centerline{$\Bvalue{\sigma\vee\upsilon<\tau\vee\upsilon}
=\Bvalue{\sigma\vee\upsilon<\tau}
  \Bcap\Bvalue{\sigma\vee\upsilon<\upsilon}
\Bsubseteq\Bvalue{\sigma<\tau}$.}

\medskip

\quad{\bf (vi)} In this case, for every $t\in T$,

\Centerline{$\Bvalue{\sigma>t}\Bsubseteq\Bvalue{\tau>t}
\Bsubseteq\Bvalue{\upsilon>t}$,}

\noindent so
$\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\sigma>t}
=(\Bvalue{\upsilon>t}\Bsetminus\Bvalue{\tau>t})
\Bcup(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})$;  taking the supremum
over $t$, we have the result.

\medskip

{\bf (d)} Induce on $n$.   If $n=0$ the list reduces to

\Centerline{($\Bvalue{\sigma<\tau_0}$, $\Bvalue{\tau_0\le\sigma}$)}

\noindent which is a partition of unity by the definition in 611D.

For the inductive step to $n+1>1$, we know that

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

\noindent and

\Centerline{($\Bvalue{\sigma<\tau_{n+1}}$, $\Bvalue{\tau_{n+1}\le\sigma}$)}

\noindent are partitions of unity.   So

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

\noindent is a partition of unity.   And

\Centerline{$\Bvalue{\tau_n\le\sigma}\Bcap\Bvalue{\tau_{n+1}\le\sigma}
=\Bvalue{\tau_n\vee\tau_{n+1}\le\sigma}
=\Bvalue{\tau_{n+1}\le\sigma}$}

\noindent by (c-ii-$\gamma$).   So our partition of unity reduces to

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

\noindent as required for the inductive step.
}%end of proof of 611E

\cmmnt{\medskip

\noindent{\bf Remark} I have taken the trouble to spell out
direct proofs based on the definitions in 611A, 611B and 611D.
But you will observe that every formula here corresponds to the case in
which $\sigma$, $\tau$ and $\upsilon$ are real-valued functions defined on
$[0,1]$, with $\frak A=\Cal P[0,1]$,

\Centerline{$(\sigma\wedge\tau)(x)=\min(\sigma(x),\tau(x))$,
\quad$(\sigma\vee\tau)(x)=\max(\sigma(x),\tau(x))$}

\noindent for $x\in[0,1]$, and

\Centerline{$\Bvalue{\sigma<\tau}=\{x:\sigma(x)<\tau(x)\}$,}

\Centerline{$\Bvalue{\sigma=\tau}=\{x:\sigma(x)=\tau(x)\}$,
\quad$\Bvalue{\sigma\le\tau}=\{x:\sigma(x)\le\tau(x)\}$.}

\noindent In (b) we have a bit of luck -- the formulae for infima are more
complicated (see 632C(a-ii)).   Elsewhere, with only finitely many
stopping times involved, we are perfectly safe, though I do not attempt to
state and prove an appropriate metatheorem.
}

\leader{611F}{Infima in $\Cal T$:  Proposition} Let $A\subseteq\Cal T$ be a
non-empty set such that

\Centerline{$\sup_{s>t}\inf_{\sigma\in A}\Bvalue{\sigma>s}$}

\noindent belongs to $\frak A_t$ whenever $t\in T$ is not isolated on the
right.

(a)

$$\eqalign{\Bvalue{\inf A>t}
&=\inf_{\sigma\in A}\Bvalue{\sigma>t}\text{ if }t\in T\text{ is isolated on the
right}\cr
&=\sup_{s>t}\inf_{\sigma\in A}\Bvalue{\sigma>s}
  \text{ for other }t\in T.\cr}$$

(b) $\Bvalue{\inf A<\tau}=\sup_{\sigma\in A}\Bvalue{\sigma<\tau}$ for every
$\tau\in\Cal T$.
%out of order \query

\proof{{\bf (a)} For $t\in T$, set

\Centerline{$a_t=\inf_{\sigma\in A}\Bvalue{\sigma>t}\in\frak A_t$,}

$$\eqalign{b_t
&=a_t\text{ if }t\in T\text{ is isolated on the right},\cr
&=\inf_{s>t}a_s\text{ for other }t\in T.\cr}$$

\noindent By hypothesis, $b_t\in\frak A_t$ for every $t\in T$.
Now $a_s\Bsubseteq a_t$ whenever $t\le s$, so
$b_s\Bsubseteq b_t$ whenever $t\le s$.   If $t\in T$ is not isolated on the
right,

$$\eqalign{\inf\{b_s:s>t\}&
\inf\{a_{s'}:\text{ there is an }s\text{ such that }s'>s>t\}\cr&
=\inf\{a_{s'}:s'>t\}=b_t.\cr}$$

\noindent Accordingly there is a $\tau_0\in\Cal T$ such that
$\Bvalue{\tau_0>t}=b_t$ for every $t\in T$.

If $\sigma\in A$,

\Centerline{$\Bvalue{\tau_0>t}
=b_t\Bsubseteq a_t\Bsubseteq\Bvalue{\sigma>t}$}

\noindent for every $t\in T$, so $\tau_0\le\sigma$;  as $\sigma$ is arbitrary,
$\tau_0\le\inf A$.   On the other hand, for any $t\in T$,
$\Bvalue{\inf A>t}\Bsubseteq\Bvalue{\sigma>t}$ for every $\sigma\in A$, so
$\Bvalue{\inf A>t}\Bsubseteq a_t$;  and if $t$ is not isolated on the right,

\Centerline{$\Bvalue{\inf A>t}
=\sup_{s>t}\Bvalue{\inf A>s}\Bsubseteq\sup_{s>t}a_s=b_t$.}

\noindent Thus $\Bvalue{\inf A>t}\Bsubseteq b_t=\Bvalue{\tau_0>t}$
for every $t$, and $\inf A\le\tau_0$.

Accordingly $\inf A=\tau_0$ satisfies the formula claimed.

\medskip

{\bf (b)} If $\sigma\in A$,
$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\inf A<\tau}$ by 611E(c-iv-$\beta$);
so $\sup_{\sigma\in A}\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\inf A<\tau}$.
In the other direction, take any $t\in T$.
If $t$ is isolated on the right, then

$$\eqalignno{\Bvalue{\tau>t}\Bsetminus\Bvalue{\inf A>t}
&=\Bvalue{\tau>t}\Bsetminus\inf_{\sigma\in A}\Bvalue{\sigma>t}\cr
&=\sup_{\sigma\in A}\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
\Bsubseteq\sup_{\sigma\in A}\Bvalue{\sigma<\tau};\cr}$$

\noindent otherwise,

$$\eqalign{\Bvalue{\tau>t}\Bsetminus\Bvalue{\inf A>t}
&=\sup_{s>t}(\Bvalue{\tau>s}
  \Bsetminus\sup_{s'>t}\inf_{\sigma\in A}\Bvalue{\sigma>s'})\cr
&\Bsubseteq\sup_{s>t}
  (\Bvalue{\tau>s}\Bsetminus\inf_{\sigma\in A}\Bvalue{\sigma>s})\cr
&=\sup_{s>t}\sup_{\sigma\in A}
  (\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s})
\Bsubseteq\sup_{\sigma\in A}\Bvalue{\sigma<\tau}.\cr}$$

\noindent So

\Centerline{$\Bvalue{\inf A<\tau}
=\sup_{t\in T}\Bvalue{\tau>t}\Bsetminus\Bvalue{\inf A>t}
\Bsubseteq\sup_{\sigma\in C}\Bvalue{\sigma<\tau}$}

\noindent and we have equality.
}%end of proof of 611F

\leader{611G}{The algebra defined by a stopping time:  Definition}
%\formerly 6{}11D
If $\tau\in\Cal T$, write $\frak A_{\tau}$ for

\Centerline{$\{a:a\in\frak A$,
$a\Bsetminus\Bvalue{\tau>t}\in\frak A_t$ for every $t\in T\}$.}

\noindent Then $\frak A_{\tau}$ is\cmmnt{ an intersection of order-closed
subalgebras, so is itself} an order-closed subalgebra of $\frak A$.

\leader{611H}{Proposition} (a) %formerly 6{}11Ea
Suppose that $\tau\in\Cal T$ and $t\in T$.

\quad(i) If
$b\in\bigcap_{s>t}\frak A_s$ and $b\Bsubseteq\Bvalue{\tau>t}$, then
$b\in\frak A_{\tau}$.
\cmmnt{In particular, }$\Bvalue{\tau>t}$ and\cmmnt{ therefore}
$1\Bsetminus\Bvalue{\tau>t}$
belong to $\frak A_{\tau}$.

\quad(ii) If $b\in\frak A_t$ and
$b\Bsubseteq\Bvalue{\tau>s}$ for every $s<t$, then
$b\in\frak A_{\tau}$.

\quad(iii) If $b\in\frak A_{\tau}$ and
$b\Bcap\Bvalue{\tau>t}=0$, then $b\in\frak A_t$.

(b) %formerly 6{}11Ec
If $\check{t}$ is the constant stopping time at $t$, then
$\frak A_{\check{t}}=\frak A_t$.

(c) Suppose that $\sigma$, $\tau\in\Cal T$.

\quad (i) %formerly 6{}11G(a-i-\alpha)
$\Bvalue{\sigma<\tau}$, $\Bvalue{\sigma=\tau}$ and
$\Bvalue{\tau<\sigma}$
belong to $\frak A_{\sigma}\cap\frak A_{\tau}$.

\quad (ii) %formerly 6{}11Eb
$\frak A_{\sigma\wedge\tau}=\frak A_{\sigma}\cap\frak A_{\tau}$;
in particular,
$\frak A_{\sigma}\subseteq\frak A_{\tau}$ if $\sigma\le\tau$.

\quad (iii) %formerly 6{}11G(a-ii-beta)
If $a\in\frak A_{\tau}$ then
$a\Bcap\Bvalue{\tau\le\sigma}
=a\Bsetminus\Bvalue{\sigma<\tau}\in\frak A_{\sigma\wedge\tau}$.
%@bi\beta

\quad (iv) %formerly 6{}11H
$\frak A_{\sigma\vee\tau}$ is the subalgebra of $\frak A$
generated by $\frak A_{\sigma}\cup\frak A_{\tau}$.

\proof{{\bf (a)(i)} If $s\le t$,

\Centerline{$b\Bsetminus\Bvalue{\tau>s}
\Bsubseteq\Bvalue{\tau>t}\Bsetminus\Bvalue{\tau>s}=0$}

\noindent and $b\Bsetminus\Bvalue{\tau>s}\in\frak A_s$.
If $s>t$ then $b$ and $\Bvalue{\tau>s}$ both belong to
$\frak A_s$, so $b\Bsetminus\Bvalue{\tau>s}\in\frak A_s$.

\medskip

\quad{\bf (ii)} If $s<t$,
$b\Bsetminus\Bvalue{\tau>s}=0\in\frak A_s$.
If $s\ge t$ then $b$ and $\Bvalue{\tau>s}$ both belong to
$\frak A_s$, so $b\Bsetminus\Bvalue{\tau>s}\in\frak A_s$.

\medskip

\quad{\bf (iii)} $b=b\Bsetminus\Bvalue{\tau>t}$.

\medskip

{\bf (b)} Use (a-ii) and (a-iii).

\medskip

{\bf (c)(i)} For every $t\in T$,
$\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}$ belongs to
$\frak A_t$ and therefore to $\frak A_{\tau}$, by (a-i) or (a-ii);
now the supremum of these, $\Bvalue{\sigma<\tau}$,
will also belong to $\frak A_{\tau}$.   Moreover, for every $t$,

$$\eqalign{\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\sigma>t}
&=\sup_{s\in T}(\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s})
  \Bsetminus\Bvalue{\sigma>t}\cr
&=\sup_{s\ge t}(\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>t})
  \Bcup\sup_{s\le t}(\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s})\cr
&=(\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t})
  \Bcup\sup_{s\le t}(\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s})\cr
&=\sup_{s\le t}\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s}
\in\frak A_t.\cr}$$

\noindent So $\Bvalue{\sigma<\tau}$ also belongs to $\frak A_{\sigma}$.

In the same way,
$\Bvalue{\tau<\sigma}\in\frak A_{\sigma}\cap\frak A_{\tau}$;  by
611E(a-i-$\alpha$),
$\Bvalue{\sigma=\tau}\in\frak A_{\sigma}\cap\frak A_{\tau}$.

\medskip

\quad{\bf (ii)}\grheada\
If $\sigma\le\tau$, $a\in\frak A_{\sigma}$ and $t\in T$,
then

\Centerline{$a\Bsetminus\Bvalue{\tau>t}
=(a\Bsetminus\Bvalue{\sigma>t})\Bsetminus\Bvalue{\tau>t}
\in\frak A_t$}

\noindent because $a\Bsetminus\Bvalue{\sigma>t}$ and
$\Bvalue{\tau>t}$ both belong to $\frak A_t$.   As $a$ and $t$ are
arbitrary, $\frak A_{\sigma}\subseteq\frak A_{\tau}$.

\medskip

\qquad\grheadb\ So for any $\sigma$, $\tau\in\Cal T$ we shall have
$\frak A_{\sigma\wedge\tau}\subseteq\frak A_{\sigma}\cap\frak A_{\tau}$.
Conversely, if $a\in\frak A_{\sigma}\cap\frak A_{\tau}$ and $t\in T$,
then

$$\eqalignno{a\Bsetminus\Bvalue{\sigma\wedge\tau>t}
&=a\Bsetminus(\Bvalue{\sigma>t}\Bcap\Bvalue{\tau>t})\cr
\displaycause{611Cc}
&=(a\Bsetminus\Bvalue{\sigma>t})\Bcup(a\Bsetminus\Bvalue{\tau>t})
\in\frak A_t\cr}$$

\noindent because both $a\Bsetminus\Bvalue{\sigma>t}$ and
$a\Bsetminus\Bvalue{\tau>t}$ belong to $\frak A_t$.   As $a$ and $t$
are arbitrary,
$\frak A_{\sigma}\cap\frak A_{\tau}\supseteq\frak A_{\sigma\wedge\tau}$
and we have equality.

\medskip

\quad{\bf (iii)} If $t\in T$, then
$\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}\Bsubseteq\Bvalue{\sigma<\tau}$,
so
$\Bvalue{\tau>t}\Bsubseteq\Bvalue{\sigma>t}\Bcup\Bvalue{\sigma<\tau}$.
Since $a\Bsetminus\Bvalue{\tau>t}$, $\Bvalue{\sigma>t}$,
$\Bvalue{\tau>t}$ and
$\Bvalue{\sigma<\tau}\Bsetminus\Bvalue{\tau>t}$ all belong to
$\frak A_t$ (put (a-i), (a-iii) and (i) just above together for the last),
so do $\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau>t}$,
$\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau>t}\Bcup\Bvalue{\sigma>t}$ and

$$\eqalign{&(a\Bsetminus\Bvalue{\tau>t})
  \Bsetminus(\Bvalue{\sigma<\tau}
     \Bcup\Bvalue{\sigma>t}\Bcup\Bvalue{\tau>t})\cr
&\mskip100mu
=a\Bsetminus
  (\Bvalue{\sigma<\tau}\Bcup\Bvalue{\tau>t}\Bcup\Bvalue{\sigma>t})\cr
&\mskip100mu
=a\Bsetminus(\Bvalue{\sigma<\tau}\Bcup\Bvalue{\sigma>t})
=(a\Bsetminus\Bvalue{\sigma<\tau})\Bsetminus\Bvalue{\sigma>t}.\cr}$$

\noindent As $t$ is arbitrary,
$a\Bsetminus\Bvalue{\sigma<\tau}\in\frak A_{\sigma}$.   But also
$a$ and $\Bvalue{\sigma<\tau}$ belong to $\frak A_{\tau}$, so

\Centerline{$a\Bsetminus\Bvalue{\sigma<\tau}
\in\frak A_{\sigma}\cap\frak A_{\tau}
=\frak A_{\sigma\wedge\tau}$}

\noindent by (i) again.

\medskip

\quad{\bf (iv)} Let $\frak B$ be the subalgebra generated by
$\frak A_{\sigma}\cup\frak A_{\tau}$.   By (i),
$\frak B\subseteq\frak A_{\sigma\vee\tau}$.   On the other hand, suppose
that $a\in\frak A_{\sigma\vee\tau}$.   Then

\Centerline{$a\Bsetminus\Bvalue{\sigma<\tau}
=a\Bsetminus\Bvalue{\sigma<\sigma\vee\tau}
\in\frak A_{\sigma}\subseteq\frak B$}

\noindent by 611E(a-ii-$\alpha$) and (iii).   Similarly
$a\Bsetminus\Bvalue{\tau<\sigma}\in\frak B$.   As $\Bvalue{\sigma<\tau}$
and $\Bvalue{\tau<\sigma}$ are disjoint, $a\in\frak B$.   As $a$ is
arbitrary, $\frak A_{\sigma\vee\tau}\subseteq\frak B$ and we have equality.
}%end of proof of 611H

\leader{611I}{Lemma} Suppose that $\familyiI{\tau_i}$ is a family in
$\Cal T$ and $\familyiI{a_i}$ is a partition of unity in $\frak A$ such that
$a_i\in\frak A_{\tau_i}$ for every $i\in I$.   Then there is a
unique $\sigma\in\Cal T$ such that
$\Bvalue{\sigma=\tau_i}\Bsupseteq a_i$ for every $i\in I$, and
$\inf_{i\in I}\tau_i\le\sigma\le\sup_{i\in I}\tau_i$.

\proof{ For $t\in T$, set
$b_t=\sup_{i\in I}a_i\Bcap\Bvalue{\tau_i>t}$.   Because
$\familyiI{a_i}$ is a partition of unity,

\Centerline{$1\Bsetminus b_t
=\sup_{i\in I}a_i\Bsetminus b_t
=\sup_{i\in I}a_i\Bsetminus\Bvalue{\tau_i>t}$}

\noindent belongs to $\frak A_t$, so $b_t\in\frak A_t$.   If $t$ is not
isolated on the right, then

\Centerline{$\sup_{s>t}b_s
=\sup_{i\in I,s>t}a_i\Bcap\Bvalue{\tau_i>s}
=\sup_{i\in I}a_i\Bcap\Bvalue{\tau_i>t}
=b_t$.}

\noindent So $\family{t}{T}{b_t}$ satisfies the conditions of
611A(b-i) and we have a stopping time $\sigma\in\Cal T$ such that
$\Bvalue{\sigma>t}=b_t$ for every $t$.   Now, for $i\in I$ and $t\in T$,
$\Bvalue{\sigma>t}\Bcap a_i=\Bvalue{\tau_i>t}\Bcap a_i$, so
$\Bvalue{\sigma<\tau_i}$ and $\Bvalue{\tau_i<\sigma}$
are both disjoint from $a_i$, and $a_i\Bsubseteq\Bvalue{\sigma=\tau_i}$.

To see that $\sigma$ is unique, suppose that $\sigma'\in\Cal T$ has the
same property;  then

\Centerline{$\Bvalue{\sigma>t}\Bcap a_i
=\Bvalue{\tau_i>t}\Bcap a_i=\Bvalue{\sigma'>t}\Bcap a_i$}

\noindent for every $i\in I$, so $\Bvalue{\sigma>t}=\Bvalue{\sigma'>t}$
for every $t\in T$, and $\sigma'=\sigma$.

To see that $\inf_{i\in I}\tau_i\le\sigma$, observe that

\Centerline{$\Bvalue{\inf_{i\in I}\tau_i\le\sigma}
\Bsupseteq\Bvalue{\inf_{i\in I}\tau_i\le\tau_j}
   \Bcap\Bvalue{\tau_j\le\sigma}
\Bsupseteq a_j$}

\noindent for every $j\in I$, so
$\Bvalue{\inf_{i\in I}\tau_i\le\sigma}=1$ and
$\inf_{i\in I}\tau_i\le\sigma$.   Similarly,
$\sigma\le\sup_{i\in I}\tau_i$.
}%end of proof of 611I

\leader{611J}{Dissections by stopping times (a)}
Recall\cmmnt{ from 611B-611C} that
if we regard a stopping time $\tau=\family{t}{T}{\Bvalue{\tau>t}}$ as
a member of the algebra $\prod_{t\in T}\frak A_t$, then
the partial order $\le$ and the lattice operations $\vee$, $\wedge$ on
$\Cal T$ correspond to the
Boolean relation and operations $\Bsubseteq$, $\Bcup$, $\Bcap$ on
$\prod_{t\in T}\frak A_t$, and
moreover that arbitrary suprema in $\Cal T$ correspond to suprema
in $\prod_{t\in T}\frak A_t$\cmmnt{ (611Cb),
though there can be complications for general infima (632C(a-i))}.

In view of this representation it is natural to consider\cmmnt{ other
Boolean
operations on members of $\Cal T$, in particular,} set difference.
I will in fact prefer the notation

\Centerline{$c(\sigma,\tau)
=\family{t}{T}{\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}}$,}

\noindent rather than writing $\tau\setminus\sigma$\cmmnt{,
as perhaps leaving less scope for confusion, and carrying the notion
of an `interval' from $\sigma$ to $\tau$}.
I will say that $c(\sigma,\tau)$ is a {\bf stopping time
interval} with {\bf endpoints} $\sigma$, $\tau$.\cmmnt{
(Warning!  the endpoints
are not uniquely defined;  but see (d) here, and also 613Cc below.)}

If $\Cal S$ is a sublattice of $\Cal T$, $\Sti(\Cal S)$ will be the set of
stopping-time intervals expressible as $c(\sigma,\tau)$ where
$\sigma\le\tau$ in $\Cal S$.

\spheader 611Jb\cmmnt{ However, the Boolean interpretation of
$c(\sigma,\tau)$,
combined with the formulae in 611C and
the distributive laws of Boolean algebra, leads us
directly to such elementary facts as}

\Centerline{$c(\sigma,\tau)\Bcap c(\sigma',\tau')
=c(\sigma\vee\sigma',\tau\wedge\tau')$}

\noindent for all $\sigma$, $\sigma'$, $\tau$, $\tau'\in\Cal T$\cmmnt{,
corresponding to the formula

\Centerline{$(b\Bsetminus a)\Bcap(b'\Bsetminus a')
=(b\Bcap b')\Bsetminus(a\Bcup a')$}

\noindent of Boolean algebra}, and

\Centerline{$c(\sigma\wedge\tau,\sigma\wedge\tau')
\Bsubseteq c(\tau,\tau')$\dvro{.}{,}}

\noindent\cmmnt{corresponding to
$(a\Bcap b')\Bsetminus(a\Bcap b)\Bsubseteq b'\Bsetminus b$. }Similarly,

\Centerline{$c(\sigma,\sup C)
=\sup_{\tau\in C}c(\sigma,\tau)$,
\quad$c(\sigma\wedge\sigma',\tau)
=c(\sigma,\tau)\Bcup c(\sigma',\tau)$}

\noindent for $\sigma$, $\sigma'$, $\tau\in\Cal T$ and
$C\subseteq\Cal T$,
and if $\sigma\le\upsilon\le\tau$, then

\Centerline{$c(\sigma,\upsilon)\Bcup c(\upsilon,\tau)=c(\sigma,\tau)$,
\quad$c(\sigma,\upsilon)\Bcap c(\upsilon,\tau)=0$.}

\noindent\cmmnt{Of course }$c(\sigma,\tau)=0$ iff $\tau\le\sigma$.

\spheader 611Jc\cmmnt{ We can now interpret $\Bvalue{\sigma<\tau}$,
as defined in 611D, as a kind of projection of
$c(\sigma,\tau)$, so that, for instance,}
$\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\sigma'<\tau'}$ whenever
$c(\sigma,\tau)\Bsubseteq c(\sigma',\tau')$.   More precisely,
if $\sigma$, $\tau$, $\sigma'$, $\tau'\in\Cal T$ then
$c(\sigma,\tau)\Bsubseteq c(\sigma',\tau')$ iff

\Centerline{$\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\sigma'\le\sigma}\Bcap\Bvalue{\tau\le\tau'}$.}

\noindent\prooflet{\Prf\ We have

$$\eqalignno{\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\sigma'\le\sigma}
&\iff 0=\Bvalue{\sigma<\tau}\Bcap\Bvalue{\sigma<\sigma'}
=\Bvalue{\sigma<\tau\wedge\sigma'}\cr
\displaycause{611E(c-i-$\beta$)}
&\iff\tau\wedge\sigma'\le\sigma,\cr}$$

$$\eqalignno{\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\tau\le\tau'}
&\iff 0=\Bvalue{\sigma<\tau}\Bcap\Bvalue{\tau'<\tau}
=\Bvalue{\sigma\vee\tau'<\tau}\cr
\displaycause{611E(c-ii)}
&\iff\tau\le\sigma\vee\tau'.\cr}$$

\noindent Next, for elements $a$, $b$, $a'$, $b'$ of any Boolean
algebra,

$$\eqalign{b\Bsetminus a\Bsubseteq b'\Bsetminus a'
&\iff b\Bsetminus a\Bsubseteq b'
  \text{ and }(b\Bsetminus a)\Bcap b'\Bcap a'=0\cr
&\iff b\Bsetminus a\Bsubseteq b'
  \text{ and }(b\Bsetminus a)\Bcap a'=0\cr
&\iff b\Bsubseteq a\Bcup b'
  \text{ and }b\Bcap a'\Bsubseteq a.\cr}$$

\noindent Translating this into terms of
$c(\sigma,\tau)=\tau\Bsetminus\sigma$ and
$c(\sigma',\tau')=\tau'\Bsetminus\sigma'$ in $\prod_{t\in T}\frak
A_t$,

$$\eqalignno{c(\sigma,\tau)\Bsubseteq c(\sigma',\tau')
&\iff\tau\Bsubseteq\sigma\Bcup\tau'
  \text{ and }\tau\Bcap\sigma'\Bsubseteq\sigma\cr
&\iff\tau\le\sigma\vee\tau'
  \text{ and }\tau\wedge\sigma'\le\sigma\cr
\displaycause{translating into terms of $\le$, $\wedge$ and $\vee$ in
the lattice $\Cal T$}
&\iff\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\tau\le\tau'}
  \text{ and
}\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\sigma'\le\sigma}\cr
&\iff\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\sigma'\le\sigma}\Bcap\Bvalue{\tau\le\tau'}.
\text{ \Qed}\cr}$$
}

\spheader 611Jd Similarly,
if $\sigma$, $\tau$, $\sigma'$, $\tau'\in\Cal T$ then
$c(\sigma,\tau)=c(\sigma',\tau')$ iff

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

\prooflet{\noindent\Prf\ If the condition is satisfied, then
(c) shows at once that $c(\sigma,\tau)=c(\sigma',\tau')$.   Conversely, if
$c(\sigma,\tau)=c(\sigma',\tau')$ then the first remark in (c) tells us that
$\Bvalue{\sigma<\tau}=\Bvalue{\sigma'<\tau'}$, and now both are included in

\Centerline{$\Bvalue{\sigma'\le\sigma}\Bcap\Bvalue{\tau'\le\tau}
\Bcap\Bvalue{\sigma\le\sigma'}\Bcap\Bvalue{\tau\le\tau'}
=\Bvalue{\sigma'=\sigma}\Bcap\Bvalue{\tau=\tau'}$.\
\Qed}}

\spheader 611Je{\bf (i)} For a finite sublattice $I$ of $\Cal T$,
an {\bf $I$-cell} will be a
minimal non-zero stopping time interval of the
form $c(\sigma,\tau)$ where $\sigma$, $\tau\in I$.\cmmnt{   (If $I$ is
non-empty and we think of it as
a sublattice of $\prod_{t\in T}\frak A_t$, then an $I$-cell is an atom of
the subalgebra of $\prod_{t\in T}\frak A_t$ generated by $I$
 which is included in
$c(\min I,\max I)$.)}

\medskip

\quad{\bf (ii)} Let $I$ be a finite sublattice of $\Cal T$,
$\Sti_0(I)$ the set of $I$-cells, and $\tau\in I$.   If we write

\Centerline{$I\wedge\tau=\{\sigma\wedge\tau:\sigma\in I\}$,
\quad $I\vee\tau=\{\sigma\vee\tau:\sigma\in I\}$,}

\noindent then
$\Sti_0(I\wedge\tau)$, $\Sti_0(I\vee\tau)$ are disjoint sets with union
$\Sti_0(I)$.
\prooflet{\Prf\ We can think of $\tau$ and $1\Bsetminus\tau$
as complementary elements of
$\prod_{t\in T}\frak A_t$, and if $\sigma$, $\sigma'\in\Cal T$, then

\Centerline{$c(\sigma,\sigma')\Bcap\tau
=c(\sigma\wedge\tau,\sigma'\wedge\tau)$,
\quad$c(\sigma,\sigma')\Bsetminus\tau
=c(\sigma\vee\tau,\sigma'\vee\tau)$}

\noindent are stopping time intervals determined by endpoints in $I$.
So if $c(\sigma,\sigma')\in\Sti_0(I)$, it must be equal either to
$c(\sigma\wedge\tau,\sigma'\wedge\tau)$
or $c(\sigma\vee\tau,\sigma\vee\tau')$, and
belong to $\Sti_0(I\wedge\tau)$ or $\Sti_0(I\vee\tau)$ accordingly.\ \Qed}

\medskip

\quad{\bf (iii)} More generally, if $I$ is a non-empty finite sublattice of
$\Cal T$ and $\tau_0\le\ldots\le\tau_n$ in $I$, then
setting

\Centerline{$I_{-1}=I\wedge\tau_0$,
\quad$I_j=I\cap[\tau_j,\tau_{j+1}]$ for $j<n$,
\quad$I_n=I\vee\tau_n$,}

\noindent $\langle\Sti_0(I_j)\rangle_{-1\le j\le n}$ is
a partition of $\Sti_0(I)$.   \prooflet{(Induce on $n$, noting that
$(I\vee\tau_{n-1})\wedge\tau_n=I\cap[\tau_{n-1},\tau_n]$,
$(I\vee\tau_{n-1})\vee\tau_n=I\vee\tau_n$ if $n>0$.)}

\leader{611K}{}\cmmnt{ The following facts will be extremely useful.

\medskip

\noindent}{\bf Lemma} Let $I\subseteq\Cal T$ be a non-empty finite
sublattice, and $\Sti_0(I)$ the set of $I$-cells.   Let $I_0$ be a maximal
totally ordered subset of $I$, and
$\langle\tau_i\rangle_{i\le n}$ the increasing enumeration of $I_0$.

(a) $\tau_0=\min I$, $\tau_1=\max I$.

(b) If $i<n$ then $I\cap[\tau_i,\tau_{i+1}]=\{\tau_i,\tau_{i+1}\}$.

(c) $\Sti_0(I)=\{c(\tau_i,\tau_{i+1}):i<n\}$.

(d) $\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}=0$ whenever
$i<n$ and $\tau\in I$.

(e) $\sup_{i\le n}\Bvalue{\tau=\tau_i}=1$ for every $\tau\in I$.

*(f) If $\sigma\in\Cal T$ then

$$\eqalign{J_0&
=\{\sigma\wedge\tau_0,\tau_0,\med(\tau_0,\sigma,\tau_1),
\tau_1,\med(\tau_1,\sigma,\tau_2),\cr&
\mskip150mu
\ldots,\tau_{n-1},\med(\tau_{n-1},\sigma,\tau_n),\tau_n,\sigma\vee\tau_n\}\cr}$$

\noindent is a maximal totally ordered subset of the sublattice
$I\sqcup\{\sigma\}$ of $\Cal T$ generated by $I\cup\{\sigma\}$.\footnote{In a
distributive lattice, $\med(p,q,r)=(p\wedge q)\vee(p\wedge r)\vee(q\wedge r)$;
see 3A1Ic.}

*(g) If $\sigma\in\Cal T$, then $I\wedge\sigma=\{\tau\wedge\sigma:\tau\in I\}$
is a sublattice of $\Cal T$, and
$\{\tau_0\wedge\sigma,\ldots,\tau_n\wedge\sigma\}$ is a maximal totally ordered
subset of $I\wedge\sigma$.

*(h) If $\tau_0\le\sigma_0\le\ldots\le\sigma_m\le\tau_n$ in $\Cal T$, and
$K$ is the sublattice of
$\Cal T$ generated by $I\cup\{\sigma_0,\ldots,\sigma_m\}$, then
$J_j=\{\med(\sigma_j,\tau_i,\sigma_{j+1}):i\le n\}$ is a maximal totally ordered
subset of $K\cap[\sigma_j,\sigma_{j+1}]$, for every $j<m$.

\proof{{\bf (a)}
$I_0\cup\{\min I,\max I\}$ is a totally ordered subset of $I$ so must be
equal to $I_0$.

\medskip

{\bf (b)} If $i<n$, $\tau\in I$ and
$\tau_i\le\tau\le\tau_{i+1}$, then $I_0\cup\{\tau\}$ is totally ordered and
$\tau\in I_0$;  thus $I\cap[\tau_i,\tau_{i+1}]=\{\tau_i,\tau_{i+1}\}$ for
every $i<n$.

\medskip

{\bf (c)} Writing

\Centerline{$I\wedge\tau_i=\{\sigma\wedge\tau_i:\sigma\in I\}
=\{\sigma:\sigma\in I$, $\sigma\le\tau_i\}$}

\noindent as in 611Je above, we see that

\Centerline{$\Sti_0(I\wedge\tau_{i+1})
=\Sti_0(I\wedge\tau_i)\cup\Sti_0((I\wedge\tau_{i+1})\vee\tau_i)
=\Sti_0(I\wedge\tau_i)\cup\Sti_0(I\cap[\tau_i,\tau_{i+1}])$}

\noindent for each $i<n$.   Since $I\wedge\tau_0$ is the singleton
$\{\tau_0\}$, $\Sti_0(I\wedge\tau_0)=\emptyset$;  since
$I\cap[\tau_i,\tau_{i+1}]=\{\tau_i,\tau_{i+1}\}$,
$\Sti_0(I\cap[\tau_i,\tau_{i+1}])=\{c(\tau_i,\tau_{i+1})\}$ for each $i<n$.
Inducing on $m$, $\Sti_0(I\wedge\tau_m)=\{c(\tau_i,\tau_{i+1}):i<m\}$
for each $m\le n$, and

\Centerline{$\Sti_0(I)=\Sti_0(I\wedge\tau_n)
=\{c(\tau_i,\tau_{i+1}):i<n\}$.}

\medskip

{\bf (d)} Set

\Centerline{$\tau'=\med(\tau_i,\tau,\tau_{i+1})
=\tau_i\vee(\tau\wedge\tau_{i+1})=(\tau_i\vee\tau)\wedge\tau_{i+1}$.}

\noindent Then $\tau_i\le\tau'\le\tau_{i+1}$ and
$I_0\cup\{\tau'\}$ is a totally ordered subset of $I$,
so either $\tau_i=\tau'$ or $\tau'=\tau_{i+1}$.   Accordingly

$$\eqalignno{0&
=\Bvalue{\tau_i<\tau'}\Bcap\Bvalue{\tau'<\tau_{i+1}}
=\Bvalue{\tau_i<\tau_i\vee(\tau\wedge\tau_{i+1})}
   \Bcap\Bvalue{(\tau_i\vee\tau)\wedge\tau_{i+1}<\tau_{i+1}}\cr&
=(\Bvalue{\tau_i<\tau_i}
   \Bcup(\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_i<\tau_{i+1}}))
  \Bcap((\Bvalue{\tau_i<\tau_{i+1}}\Bcap\Bvalue{\tau<\tau_{i+1}})
    \Bcup\Bvalue{\tau_{i+1}<\tau_{i+1}})
\Displaycause{611E(c-i) and (c-ii)}
=\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau_i<\tau_{i+1}}
   \Bcap\Bvalue{\tau<\tau_{i+1}}
=\Bvalue{\tau_i<\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}\cr}$$

\noindent (611C(c-iii-$\gamma$)), as required.

\medskip

{\bf (e)} If $\tau\in I$ then
$\Bvalue{\tau\le\tau_m}=\sup_{i\le m}\Bvalue{\tau=\tau_i}$ for every
$m\le n$.   \Prf\ Induce on $m$.   If $m=0$ we have $\tau_0\le\tau$ so
$\Bvalue{\tau\le\tau_0}=\Bvalue{\tau\wedge\tau_0=\tau_0}=\Bvalue{\tau=\tau_0}$.
For the inductive step to $m+1\le n$,

$$\eqalignno{\Bvalue{\tau\le\tau_{m+1}}&
=(\Bvalue{\tau\le\tau_m}\Bcap\Bvalue{\tau\le\tau_{m+1}})
   \Bcup(\Bvalue{\tau_m<\tau}\Bcap\Bvalue{\tau\le\tau_{m+1}})
\Displaycause{611E(a-i-$\alpha$)}
=\Bvalue{\tau\le\tau_m\wedge\tau_{m+1}}
   \Bcup(\Bvalue{\tau_m<\tau}\Bcap\Bvalue{\tau<\tau_{m+1}})
   \Bcup(\Bvalue{\tau_m<\tau}\Bcap\Bvalue{\tau=\tau_{m+1}})
\Displaycause{611E(c-i-$\alpha$)}
\Bsubseteq\Bvalue{\tau\le\tau_m}\Bcup\Bvalue{\tau=\tau_{m+1}}
\Displaycause{(d) above}
\Bsubseteq\sup_{i\le m}\Bvalue{\tau=\tau_i}\Bcup\Bvalue{\tau=\tau_{m+1}}
\Displaycause{by the inductive hypothesis}
=\sup_{i\le m+1}\Bvalue{\tau=\tau_i}.  \text{ \Qed}\cr}$$

\noindent At the end of the induction,

\Centerline{$1=\Bvalue{\tau\le\tau_n}=\sup_{i\le n}\Bvalue{\tau=\tau_i}$}

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

\medskip

{\bf (f)} The set

$$\eqalign{\{\tau:\tau\in\Cal T,\,
  &\tau\wedge\tau_0\in\{\sigma\wedge\tau_0,\tau_0\},\,
  \tau\vee\tau_n\in\{\tau_n,\sigma\vee\tau_n\},
  \cr&\med(\tau_i,\tau,\tau_{i+1})
    \in\{\tau_i,\med(\tau_i,\sigma,\tau_i),\tau_{i+1}\}
    \text{ for every }i<n\}\cr}$$

\noindent is a sublattice of $\Cal T$ (because all the operations
$\tau\mapsto\tau\wedge\tau_0$, $\tau\mapsto\med(\tau_i,\tau,\tau_{i+1})$,
$\tau\mapsto\tau\vee\tau_n$ are lattice homomorphisms), containing $\sigma$
(obviously) and including $I$ (because $I\wedge\tau_0=\{\tau_0\}$,
$I\vee\tau_n=\{\tau_n\}$ and
$I\cap[\tau_i,\tau_{i+1}]=\{\tau_i,\tau_{i+1}\}$ for $i<n$);  so it
includes $I\sqcup\{\sigma\}$.   But this means that there is no member of
$I\sqcup\{\sigma\}$ lying strictly between any two terms of the string
defining $J_0$, while $\min(I\sqcup\{\sigma\})=\sigma\wedge\tau_0$ is the
first member of $J_0$ and $\max(I\sqcup\{\sigma\})=\sigma\vee\tau_n$
is the last.   Thus $J_0$ is a maximal totally ordered subset of
$I\sqcup\{\sigma\}$, as claimed.

\medskip

{\bf (g)} $I\wedge\sigma$ is a sublattice of $\Cal T$ because $\Cal T$ is a
distributive lattice.
Since $\tau_0=\min I$ and $\tau_n=\max I$, $\tau_0\wedge\sigma$ and
$\tau_n\wedge\sigma$ are the least and greatest members of $I\wedge\sigma$.
Suppose that $\tau\in I$ and $i<n$ are such that
$\tau_i\wedge\sigma\le\tau\wedge\sigma\le\tau_{i+1}\wedge\sigma$.
Then $\tau'=\med(\tau_i,\tau,\tau_{i+1})\in I$ and

\Centerline{$\tau\wedge\sigma
=((\tau\vee\tau_i)\wedge\tau_{i+1})\wedge\sigma
=((\tau\wedge\sigma)\vee(\tau_i\wedge\sigma))\wedge\tau_{i+1})\wedge\sigma
=\tau'\wedge\sigma$.}

\noindent But as $\tau'\in I\cap[\tau_i,\tau_{i+1}]$ and
$I_0$ is a maximal totally ordered subset of $I$,
$\tau'\in\{\tau_i,\tau_{i+1}\}$ and
$\tau\wedge\sigma\in\{\tau_i\wedge\sigma,\tau_{i+1}\wedge\sigma\}$.   So
$\{\tau_i\wedge\sigma:i\le n\}$ is a maximal totally ordered subset of
$I\wedge\sigma$.

\medskip

{\bf (h)} As $J_j\subseteq K$ and

\Centerline{$\sigma_j=\med(\sigma_j,\tau_0,\sigma_{j+1})
\le\med(\sigma_j,\tau_1,\sigma_{j+1})\le\ldots
\le\med(\sigma_j,\tau_n,\sigma_{j+1})=\sigma_{j+1}$,}

\noindent $J_j$ is a totally ordered subset of $K\cap[\sigma_j,\sigma_{j+1}]$
containing $\sigma_j$ and $\sigma_{j+1}$.   As in (f), the set

\Centerline{$\{\rho:\rho\in\Cal T$,
$\med(\sigma_j,\rho,\sigma_{j+1})
  \in\{\med(\sigma_j,\tau,\sigma_{j+1}):\tau\in I\}\}$}

\noindent is a sublattice of $\Cal T$ including
$\{\sigma_0,\ldots,\sigma_m\}\cup I$ and therefore includes $K$.   Now suppose
that $\rho\in K\cap[\sigma_j,\sigma_{j+1}]$ and $J_j\cup\{\rho\}$ is totally
ordered.    Then there is an $i<n$ such that

\Centerline{$\med(\sigma_j,\tau_i,\sigma_{j+1})
\le\rho\le\med(\sigma_j,\tau_{i+1},\sigma_{j+1})$.}

\noindent Let $\tau\in I$ be such that
$\med(\sigma_j,\rho,\sigma_{j+1})=\med(\sigma_j,\tau,\sigma_{j+1})$ and
consider $\tau'=\med(\tau_i,\tau,\tau_{i+1})$.   Then

$$\eqalignno{\med(\sigma_j,\tau',\sigma_{j+1})&
=\med(\med(\sigma_j,\tau_i,\sigma_{j+1}),\med(\sigma_j,\tau,\sigma_{j+1}),
        \med(\sigma_j,\tau_{i+1},\sigma_{j+1}))
\Displaycause{because $\sigma\mapsto\med(\sigma_j,\sigma,\sigma_{j+1})$
is a lattice homomorphism}
=\med(\med(\sigma_j,\tau_i,\sigma_{j+1}),\med(\sigma_j,\rho,\sigma_{j+1}),
        \med(\sigma_j,\tau_{i+1},\sigma_{j+1}))\cr&
=\med(\med(\sigma_j,\tau_i,\sigma_{j+1}),\rho,
        \med(\sigma_j,\tau_{i+1},\sigma_{j+1}))
\Displaycause{because $\sigma_j\le\med(\sigma_j,\tau_i,\sigma_{j+1})
\le\rho\le\med(\sigma_j,\tau_{i+1},\sigma_{j+1})\le\sigma_{j+1}$}
=\rho.\cr}$$

\noindent But as $\tau'\in I$ and $\{\tau_k:k\le n\}$ is a maximal totally
ordered subset of $I$, either $\tau'=\tau_i$ and
$\rho=\med(\sigma_j,\tau_i,\sigma_{j+1})$ or
$\tau'=\tau_{i+1}$ and
$\rho=\med(\sigma_j,\tau_{i+1},\sigma_{j+1})$;  in either case
$\rho\in J_j$.   Thus
$J_j$ is a maximal totally ordered subset of $K\cap[\sigma_j,\sigma_{j+1}]$, as
claimed.
}%end of proof of 611K

\leader{611L}{Definition} If $I$ is a finite sublattice of $\Cal T$, I will
say that a sequence $\langle\tau_i\rangle_{i\le n}$ in $I$
{\bf linearly generates the $I$-cells} if it is non-decreasing and
$\{\tau_i:i\le n\}$ is a maximal totally ordered subset of $I$.
\cmmnt{(It will be convenient not to insist that the sequence be strictly
increasing.   But we shall always have
$\tau_0=\min I$, $\tau_n=\max I$,
$I\cap[\tau_i,\tau_{i+1}]=\{\tau_i,\tau_{i+1}\}$ for every $i<n$,
$\Sti_0(I)=\{c(\tau_i,\tau_{i+1}):i<n$, $\tau_i\ne\tau_{i+1}\}$,
$\sup_{i\le n}\Bvalue{\sigma=\tau_i}=1$ for every $\sigma\in I$,
$\Bvalue{\tau_i<\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}=0$ whenever
$\sigma\in I$ and $i<n$, and

\Centerline{$(\sigma\wedge\tau_0,\tau_0,\med(\tau_0,\sigma,\tau_1),
\tau_1,\ldots,
\tau_{n-1},\med(\tau_{n-1},\sigma,\tau_n),\tau_n,\sigma\vee\tau_n)$}

\noindent will linearly generate the $(I\sqcup\{\sigma\})$-cells for
every $\sigma\in\Cal T$.)}

\leader{611M}{Covering and full sublattices (a)(i)}
If $A$, $B\subseteq\Cal T$,
\cmmnt{ I will say that} {\bf $A$ covers $B$} if
$\sup_{\sigma\in A}\Bvalue{\tau=\sigma}=1$ for every $\tau\in B$.
\cmmnt{(The formula is to be interpreted as
including a promise that, except in the trivial cases $\frak A=\{0\}$ and
$B=\emptyset$, $A$ is non-empty, following the rule that
$\sup\emptyset=0$ in any Boolean algebra.)}

\medskip

\quad{\bf (ii)} If $A\subseteq\Cal T$, the {\bf covered envelope} of
$A$ will be the set
$\{\tau:\tau\in\Cal T$, $\sup_{\sigma\in A}\Bvalue{\tau=\sigma}=1\}$\cmmnt{,
that is, the largest subset of $\Cal T$ covered by $A$}.
Of course $A$ covers
itself, that is, the covered envelope of $A$ includes $A$.

\spheader 611Mb{\bf (i)}
If $A\subseteq\Cal T$ and $a\in\frak A$, the set

\Centerline{$\Cal S=\{\tau:\tau\in\Cal T$,
$a\Bsubseteq\sup_{\sigma\in A}\Bvalue{\sigma=\tau}\}$}

\noindent is a sublattice of $\Cal T$.
\prooflet{\Prf\ If $\tau$, $\tau'\in\Cal S$, then

$$\eqalignno{\sup_{\sigma\in A}\Bvalue{\tau\vee\tau'=\sigma}&
\Bsupseteq\sup_{\sigma\in A}
  (\Bvalue{\tau\vee\tau'=\tau}\Bcap\Bvalue{\tau=\sigma})
     \Bcup\sup_{\sigma\in A}
  (\Bvalue{\tau\vee\tau'=\tau}\Bcap\Bvalue{\tau'=\sigma})
\Displaycause{611E(c-iv-$\gamma$)}
\Bsupseteq(a\Bcap\Bvalue{\tau'\le\tau})\Bcup(a\Bcap\Bvalue{\tau\le\tau'})
\Displaycause{611E(a-ii-$\beta$)}
=a\cr}$$

\noindent (611D).   Similarly

$$\eqalignno{\sup_{\sigma\in A}\Bvalue{\tau\wedge\tau'=\sigma}&
\Bsupseteq\sup_{\sigma\in A}
  (\Bvalue{\tau\wedge\tau'=\tau}\Bcap\Bvalue{\tau=\sigma})
     \Bcup\sup_{\sigma\in A}
  (\Bvalue{\tau\wedge\tau'=\tau}\Bcap\Bvalue{\tau'=\sigma})\cr&
\Bsupseteq(a\Bcap\Bvalue{\tau\le\tau'})\Bcup(a\Bcap\Bvalue{\tau'\le\tau})
=a.\cr}$$

\noindent So $\tau\vee\tau'$ and $\tau\wedge\tau'$ belong to $\Cal S$.
As $\tau$ and $\tau'$ are arbitrary, $\Cal S$ is a sublattice of $\Cal T$.\
\Qed}

In particular, the covered envelope $\hat A$ of $A$ is a sublattice of $\Cal T$.

\medskip

\quad{\bf (ii)} If $\rho$ is an upper bound for $A$ in $\Cal T$, then
$\rho$ is an upper bound for $\hat A$.   \prooflet{\Prf\ If $\tau\in\hat A$,
then

\Centerline{$\Bvalue{\tau\le\rho}
\Bsupseteq\sup_{\sigma\in A}\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma\le\rho}
\Bsupseteq\sup_{\sigma\in A}\Bvalue{\tau=\sigma}=1$}

\noindent and $\tau\le\rho$.\ \QeD}   Similarly, if $\rho$ is a lower bound
for $A$, it is a lower bound for $\hat A$.

\medskip

\quad{\bf (iii)} Since $A\subseteq\hat A$, it follows that if $A$ has a greatest
member then this is also the greatest member of $\hat A$, and that
if $A$ has a least member then this is also the least member of $\hat A$.

\medskip

\quad{\bf (iv)} Note that if $\sigma$, $\tau\in\Cal T$ then $\{\sigma,\tau\}$
covers $\{\sigma\wedge\tau,\sigma\vee\tau\}$\cmmnt{ (by (i) above)} and also
$\{\sigma\wedge\tau,\sigma\vee\tau\}$ covers $\{\sigma,\tau\}$
\ifresultsonly{.}\else{ (because

\Centerline{$\Bvalue{\sigma\le\tau}
\Bsubseteq\Bvalue{\sigma=\sigma\wedge\tau}
   \Bcap\Bvalue{\tau=\sigma\vee\tau}$,
\quad$\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{\sigma=\sigma\vee\tau}
   \Bcap\Bvalue{\tau=\sigma\wedge\tau}$).}}\fi

\medskip

\spheader 611Mc I will say that a sublattice of $\Cal T$ is {\bf full} if
it is equal to its covered envelope.

\medskip

\quad{\bf (i)} The intersection of any non-empty
family of full sublattices of $\Cal T$ is full.   \prooflet{\Prf\ If
$\Bbb S$ is a non-empty family of full sublattices of $\Cal T$, and
$\tau$ belongs to the covered envelope of $\Cal S^*=\bigcap\Bbb S$, then for any
$\Cal S\in\Bbb S$ we have

\Centerline{$\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
\Bsupseteq\sup_{\sigma\in\Cal S^*}\Bvalue{\tau=\sigma}
=1$}

\noindent so $\tau\in\Cal S$.   As $\Cal S$ is arbitrary,
$\tau\in\Cal S^*$;  as $\tau$ is arbitrary, $\Cal S^*$ is full.\ \Qed}

\medskip

\quad{\bf (ii)} If $A\subseteq\Cal T$, its covered envelope $\hat A$ is full.
\prooflet{\Prf\ If $\rho$ belongs to the covered envelope of $\hat A$ then

\Centerline{$\sup_{\sigma\in A}\Bvalue{\rho=\sigma}
\Bsupseteq\sup_{\sigma\in A,\tau\in\Cal S}
   \Bvalue{\rho=\tau}\Bcap\Bvalue{\tau=\sigma}
=\sup_{\tau\in\Cal S}\Bvalue{\rho=\tau}
=1$}

\noindent and $\rho\in\hat A$.\ \Qed}

\spheader 611Md For any $\rho\in\Cal T$, $\Cal T\wedge\rho$ is full.
\prooflet{\Prf\ If $\tau$ belongs to the covered envelope of
$\Cal T\wedge\rho$, then

$$\eqalignno{\Bvalue{\tau\le\rho}&
\Bsupseteq\sup_{\sigma\in\Cal T\wedge\rho}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma\le\rho}
\Displaycause{611E(c-iv-$\alpha$)}
=\sup_{\sigma\in\Cal T\wedge\rho}\Bvalue{\tau=\sigma}
=1,\cr}$$

\noindent so $\tau\in\Cal T\wedge\rho$ (611E(a-i-$\gamma$)).\ \QeD}   Similarly,
$\Cal T\vee\rho$ is full.   Putting these together,
$[\rho,\rho']=(\Cal T\wedge\rho')\cap(\Cal T\vee\rho)$ is full whenever
$\rho\le\rho'$ in $\Cal T$.

\spheader 611Me{\bf (i)}
If $\Cal S$ is a sublattice of $\Cal T$ with covered envelope
$\hat{\Cal S}$, and $\rho\in\Cal S$, then $\hat{\Cal S}\wedge\rho$ is the
covered envelope of $\Cal S\wedge\rho$ and $\hat{\Cal S}\vee\rho$ is the
covered envelope of $\Cal S\vee\rho$.   \prooflet{\Prf\ Since $\hat{\Cal S}$ is
a sublattice of $\Cal T$ ((b-i) above) and $\rho\in\hat{\Cal S}$,

\Centerline{$\hat{\Cal S}\wedge\rho=\{\tau:\tau\in\hat{\Cal S}$, $\tau\le\rho\}
=\hat{\Cal S}\cap(\Cal T\wedge\rho)$}

\noindent(see the remark following 611C) and is full
(putting (c) and (d) together).   As it includes $\Cal S\wedge\rho$,
it includes the covered envelope of $\Cal S\wedge\rho$.   In the other
direction, if $\tau\in\hat{\Cal S}\wedge\rho$, then

$$\eqalignno{\sup_{\sigma\in\Cal S\wedge\rho}\Bvalue{\tau=\sigma}&
=\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma\wedge\rho}
\Displaycause{because $\Cal S\wedge\rho=\{\sigma\wedge\rho:\sigma\in\Cal S\}$}
=\sup_{\sigma\in\Cal S}\Bvalue{\tau\wedge\rho=\sigma\wedge\rho}
\Displaycause{because $\tau=\tau\wedge\rho$}
\Bsupseteq\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
\Displaycause{611E(c-v-$\alpha$)}
=1\cr}$$

\noindent because $\rho\in\hat{\Cal S}$ and $\hat{\Cal S}$
is a sublattice of $\Cal T$, so $\tau\in\hat{\Cal S}$.   Thus $\tau$ belongs to
the covered envelope of $\Cal S\wedge\rho$, and this is the whole of
$\hat{\Cal S}\wedge\rho$.

Replacing every $\wedge$ above by $\vee$, we see that $\hat{\Cal S}\vee\rho$
is the covered envelope of $\Cal S\vee\rho$.\ \Qed}

\medskip

\quad{\bf (ii)} If $\Cal S$ is a sublattice of $\Cal T$, $\rho$,
$\rho'\in\Cal S$ and $\rho\le\rho'$, then
the covered envelope of

\Centerline{$\Cal S\cap[\rho,\rho']=(\Cal S\vee\rho)\wedge\rho'
=\{\med(\rho,\sigma,\rho'):\sigma\in\Cal S\}$}

\noindent is $(\hat{\Cal S}\vee\rho)\wedge\rho'=\hat{\Cal S}\cap[\rho,\rho']$.
\prooflet{ (Because $\Cal T$ is a distributive lattice, $\Cal S\vee\rho$ is a
sublattice of $\Cal T$, so we can apply (i) twice.)}

\spheader 611Mf If $\Cal S$ is a sublattice of $\Cal T$ with covered envelope
$\hat{\Cal S}$, then $\bigcap_{\tau\in\hat{\Cal S}}\frak A_{\tau}
=\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$.   \prooflet{\Prf\ Write
$\frak B$ for $\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$.   Then
$\bigcap_{\tau\in\hat{\Cal S}}\frak A_{\tau}\subseteq\frak B$ just because
$\Cal S\subseteq\hat{\Cal S}$.   In the other direction, if $b\in\frak B$ and
$\tau\in\hat{\Cal S}$, then for any $\sigma\in\Cal S$ we have
$b\in\frak a_{\sigma}$ so $b\Bcap\Bvalue{\tau=\sigma}\in\frak A_{\tau}$
(611H(c-iii));  accordingly
$b=\sup_{\sigma\in\Cal S}b\Bcap\Bvalue{\tau=\sigma}$ belongs to
$\frak A_{\tau}$.   As $b$ and $\tau$ are arbitrary,
$\frak B\subseteq\bigcap_{\tau\in\hat{\Cal S}}\frak A_{\tau}$ and we have
equality.\ \Qed}

\spheader 611Mg Suppose that $A$, $B\subseteq\Cal T$ and $A$ covers $B$.

\medskip

\quad{\bf (i)}
$A$ covers the covered envelope of $B$\prooflet{, because $A$ covers its own
covered envelope which is a full sublattice including $B$}.

\medskip

\quad{\bf (ii)} If $\tau\in\Cal T$, then
$A\wedge\tau=\{\sigma\wedge\tau:\sigma\in A\}$ covers
$B\wedge\tau=\{\sigma\wedge\tau:\sigma\in B\}$.   \prooflet{\Prf\
If $\sigma\in B$, then

\Centerline{$1=\sup_{\sigma'\in A}\Bvalue{\sigma=\sigma'}
\Bsubseteq\sup_{\sigma'\in A}\Bvalue{\sigma\wedge\tau=\sigma'\wedge\tau}$}

\noindent by 611E(c-v-$\alpha$).\ \Qed}

\spheader 611Mh Suppose now that
$\bar\mu$ is such that $(\frak A,\bar\mu)$ is a
probability algebra.   If $\Cal S$ is a sublattice of $\Cal T$,
$\tau$ belongs to the covered envelope of $\Cal S$ and $\epsilon>0$, then
there is a $\tau'\in\Cal S$ such that
$\bar\mu\Bvalue{\tau<\tau'}\le\epsilon$.   \prooflet{\Prf\ We know that
$\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}=1$ in $\frak A$, so there
is a non-empty finite $I\subseteq\Cal S$ such that
$\bar\mu(\sup_{\sigma\in I}\Bvalue{\sigma=\tau})\ge 1-\epsilon$.
Setting $\tau'=\sup I\in\Cal S$,

$$\eqalignno{\Bvalue{\tau\le\tau'}&
\Bsupseteq\sup_{\sigma\in I}\Bvalue{\sigma\le\tau'}\Bcap\Bvalue{\sigma=\tau}
\Displaycause{611E(c-iv-$\alpha$)}
\Bsupseteq\sup_{\sigma\in I}\Bvalue{\sigma=\tau}\cr}$$

\noindent has measure at least $1-\epsilon$, so its complement
$\Bvalue{\tau'<\tau}$ has measure at most $\epsilon$.\ \Qed}

\leader{611N}{Covering ideals}
Let $\Cal S$ be a sublattice of $\Cal T$.

\spheader 611Na {\bf Definition} I will
say that a {\bf covering ideal} of $\Cal S$ is an ideal $\Cal S'$ of
$\Cal S$ which covers $\Cal S$\cmmnt{ in the sense of 611M}.

\spheader 611Nb{\bf (i)} If $\tau\in\Cal S$ and $\Cal S'$ is an ideal
of $\Cal S$, then
$\{\Bvalue{\sigma=\tau}:\sigma\in\Cal S'\}$ is upwards-directed.
\prooflet{\Prf\ If $\sigma$, $\sigma'\in\Cal S'$ then
$\upsilon=(\sigma\vee\sigma')\wedge\tau$ belongs to $\Cal S'$, and
$\Bvalue{\upsilon=\tau}
\Bsupseteq\Bvalue{\sigma=\tau}\Bcup\Bvalue{\sigma'=\tau}$.\ \Qed}

\medskip

\quad{\bf (ii)} If $\tau\in\Cal S$ and $\Cal S'$ is an ideal of $\Cal S$,
then $\sup_{\sigma\in\Cal S'}\Bvalue{\sigma=\tau}
=\sup_{\sigma\in\Cal S'}\Bvalue{\tau\le\sigma}$.   \prooflet{\Prf\
For any $\sigma\in\Cal S'$, $\sigma\wedge\tau\in\Cal S'$ and

\Centerline{$\Bvalue{\sigma=\tau}
\Bsubseteq\Bvalue{\tau\le\sigma}
=\Bvalue{\tau=\sigma\wedge\tau}$.  \Qed}}

\spheader 611Nc If $\Cal S$ is a sublattice of $\Cal T$ and
$\Cal S_1$, $\Cal S_2$ are two covering ideals of $\Cal S$, then
$\Cal S_0=\Cal S_1\cap\Cal S_2$ is a covering ideal of $\Cal S$.

\prooflet{\Prf\ Certainly $\Cal S_0$ is an ideal of $\Cal S$.
Take $\tau\in\Cal S$ and
$a\in\frak A\setminus\{0\}$.   Then there is a $\sigma_1\in\Cal S_1$
such that
$a_1=a\Bcap\Bvalue{\tau=\sigma_1}\ne 0$.
Next, there is a $\sigma_2\in\Cal S_2$ such that
$a_2=a_1\Bcap\Bvalue{\tau=\sigma_2}\ne 0$.
In this case, $\sigma=\sigma_1\wedge\sigma_2$ belongs to $\Cal S_0$,
a

\Centerline{$\Bvalue{\tau=\sigma}
\Bsupseteq\Bvalue{\tau=\sigma_1}\Bcap\Bvalue{\tau=\sigma_2}
\Bsupseteq a_2$}

\noindent meets $a$.   As $a$ and $\tau$ are arbitrary, $\Cal S_0$ is
a covering ideal of $\Cal S$.\ \Qed}

\spheader 611Nd
If $\Cal S'$ is a covering ideal of $\Cal S$ and $\Cal S''$
is a covering ideal of $\Cal S'$, then $\Cal S''$ is a covering ideal of
$\Cal S$.
\prooflet{\Prf\ It is elementary to check that $\Cal S''$ is an ideal
of $\Cal S$.   If $\tau\in\Cal S$ and $a\in\frak A\setminus\{0\}$, there
is a $\sigma\in\Cal S'$ such that $b=a\Bcap\Bvalue{\sigma=\tau}$ is
not $0$.   Now there is an $\upsilon\in\Cal S''$ such that
$c=b\Bcap\Bvalue{\upsilon=\sigma}$ is non-zero.   But now
$c\Bsubseteq\Bvalue{\upsilon=\tau}$, so
$a\Bcap\Bvalue{\upsilon=\tau}\ne 0$.   As $a$ is arbitrary,
$\sup_{\upsilon\in\Cal S''}\Bvalue{\upsilon=\tau}=0$;  as $\tau$ is
arbitrary, $\Cal S''$ is covering in $\Cal S$.\ \Qed}

\spheader 611Ne{\bf (i)} $\Cal T_f$ is full.   \prooflet{\Prf\ If
$\tau\in\Cal T$ and $\sup_{\sigma\in\Cal T_f}\Bvalue{\tau=\sigma}=1$,
take any non-zero $a\in\frak A$.   Then there are a $\sigma\in\Cal T_f$ such
that $a\Bcap\Bvalue{\tau=\sigma}\ne 0$ and a $t\in T$ such that

\Centerline{$a\Bcap\Bvalue{\tau=\sigma}
\notBsubseteq\Bvalue{\sigma>t}\Bcap\Bvalue{\tau=\sigma}
=\Bvalue{\tau>t}\Bcap\Bvalue{\tau=\sigma}$}

\noindent (611E(i-$\beta$)),
so $a\notBsubseteq\Bvalue{\tau>t}$.   As $a$ is arbitrary,
$\inf_{t\in T}\Bvalue{\tau>t}=0$ and $\tau\in\Cal T_f$.   Thus $\Cal T_f$ is its
own covered envelope and is full.\ \Qed}

\medskip

\quad{\bf (ii)} $\Cal T_b$ is a covering ideal of $\Cal T_f$.
\prooflet{\Prf\ I observed in 611Cg that $\Cal T_b$ is an ideal in $\Cal T$
and therefore in $\Cal T_f$.
If $\tau\in\Cal T_f$ and $t\in T$, then $\check{t}$ and
$\tau\wedge\check{t}$ belong to $\Cal T_b$, while

$$\eqalignno{\Bvalue{\tau=\tau\wedge\check{t}}
&=1\Bsetminus\Bvalue{\check{t}<\tau}\cr
\displaycause{611E(a-ii-$\alpha$)}
&=1\Bsetminus\sup_{s\in T}\Bvalue{\tau>s}\Bsetminus\Bvalue{\check{t}>s}
=1\Bsetminus\sup_{s\ge t}\Bvalue{\tau>s}
=1\Bsetminus\Bvalue{\tau>t}.\cr}$$

\noindent So

\Centerline{$\sup_{\sigma\in\Cal T_f}\Bvalue{\tau=\sigma}
\Bsupseteq\sup_{t\in T}\Bvalue{\tau=\tau\wedge\check{t}}
=1\Bsetminus\inf_{t\in T}\Bvalue{\tau>t}
=1$}

\noindent because $\tau\in\Cal T_f$.   As $\tau$ is arbitrary, $\Cal T_b$
is covering in $\Cal T_f$.\ \Qed}

\leader{611O}{(a) Definition}\cmmnt{ A variation on the concept of `full'
sublattice will be relevant in \S614, and
important in \S626.}   I will say that a sublattice $\Cal S$ of $\Cal T$
is {\bf finitely full} if $\tau\in\Cal S$ whenever $\tau\in\Cal T$ and there is
a finite set $I\subseteq\Cal S$ such that
$\sup_{\sigma\in I}\Bvalue{\tau=\sigma}=1$.

\medskip

{\bf (b)} Two useful facts.

\medskip

\quad{\bf (i)} The intersection of any non-empty family $\pmb S$
of finitely full
sublattices of $\Cal T$ is finitely full.
\prooflet{\Prf\ (Cf.\ 611M(c-i).)   If $I$ is a finite
subset of $\bigcap\Bbb S$ then its covered envelope is included in every member
of $\Bbb S$ so is included in $\bigcap\Bbb S$.\ \Qed}

\medskip

\quad{\bf (ii)} If $\frak A\ne\{0\}$ and
$\Cal S$ is a sublattice of $\Cal T$ which is order-convex\cmmnt{
(that is, $\tau\in\Cal S$ whenever $\sigma\le\tau\le\sigma'$ in $\Cal T$ and
$\sigma$, $\sigma'\in\Cal S$)}, then $\Cal S$ is finitely full.
\prooflet{\Prf\ If $\tau\in\Cal T$ and
there is a finite set $I\subseteq\Cal S$ such that
$\sup_{\sigma\in I}\Bvalue{\tau=\sigma}=1$, then $I$ is non-empty, so we can
speak of $\inf I$ and $\sup I$, which both belong to $\Cal S$.   Now

\Centerline{$\Bvalue{\inf I\le\tau}
\Bsubseteq\sup_{\sigma\in I}\Bvalue{\inf I\le\sigma}\Bcap\Bvalue{\tau=\sigma}
=\sup_{\sigma\in I}\Bvalue{\tau=\sigma}
=1$}

\noindent so $\inf I\le\tau$;  similarly, $\tau\le\sup I$.   As $\Cal S$ is
order-convex, $\tau\in\Cal S$.\ \Qed}

%\leader{611P}{}
%\leader{611Q}{}
%\leader{611R}{}
%\leader{611S}{}
%\leader{611T}{}
%\leader{611U}{}
%\leader{611V}{}
%\leader{611W}{}

\exercises{\leader{611X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 611Xa
Suppose that $T=\coint{0,\infty}$.   (i) Show that $\Cal T_f$
can be identified with a sublattice of $L^0(\frak A)^+$
which is closed under
addition and multiplication by scalars greater than or equal to $1$.
(See \S364 or 612A for the space $L^0=L^0(\frak A)$.)
(ii) Show that $\Cal T_b$ becomes identified with
$\Cal T_f\cap L^{\infty}(\frak A)$.
(iii) Show that a set $C\subseteq\Cal T_f$ is bounded above
in $\Cal T_f$ iff it is bounded above in $L^0$, and that in this case
its supremum taken in $\Cal T_f$ is the same as its supremum taken in
$L^0$.
%611C

\sqheader 611Xb Show that $\frak A_{\max\Cal T}=\frak A$ and that
$\frak A_{\min\Cal T}=\bigcap_{t\in T}\frak A_t$.
%611G

\sqheader 611Xc
In 611J, let $Z$ be the Stone space of $\frak A$, so that
$\frak A$ can be identified with the algebra $\Cal E$ of open-and-closed
subsets of $Z$, and $\frak A^T$ with the family

\Centerline{$\{W:W\subseteq Z\times T$, $W^{-1}[\{t\}]\in\Cal E$ for every
$t\in T\}$.}

\noindent(i) Show that a stopping time $\tau\in T$ corresponds to an
ordinate set $W_{\tau}\subseteq Z\times T$ such that
$(z,t)\in W_{\tau}$ whenever $(z,s)\in W_{\tau}$ and $t\le s$.   (ii)
Show that
if $\sigma$, $\tau\in T$, then $c(\sigma,\tau)$ corresponds to a subset
$W$ of $Z\times T$ in which all vertical sections are intervals, and
$\Bvalue{\sigma<\tau}$ is now
the interior of the closure of the projection $W^{-1}[T]$ of
$W$ onto $Z$.
%611J

\spheader 611Xd
Let $\Cal S$ be a sublattice of $\Cal T$.   Show that
$\Cal S$ is order-convex (4A2A)
iff $\med(\tau,\sigma,\tau')$ belongs to $\Cal S$
whenever $\tau$, $\tau'\in\Cal S$ and $\sigma\in\Cal T$.
%611K footnote

\spheader 611Xe Suppose that $I$ is a non-empty sublattice of $\Cal T$, and
that $(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells.
Show that if $\sigma\in\Cal T$ then

\Centerline{$(\tau_0\wedge\sigma,\ldots,\tau_n\wedge\sigma,\sigma,
\tau_0\vee\sigma,\ldots,\tau_n\vee\sigma)$}

\noindent linearly generates the $(I\sqcup\{\sigma\})$-cells, where
$I\sqcup\{\sigma\}$ is the sublattice of $\Cal T$ generated by
$I\cup\{\sigma\}$.
%611K

\spheader 611Xh Suppose that $T=\Bbb N$.   (i) Show that if $\sigma\in\Cal T$
and $n\in\Bbb N$ then $\Bvalue{\sigma>n}$ cannot meet
$\Bvalue{(n+1)\varBspcheck>\sigma}$, so that
$\Bvalue{\sigma>\check n}=\Bvalue{\sigma\ge(n+1)\varBspcheck}$.
%611E
(ii) Show that $\Cal T$ is the covered envelope of
$\{\check n:n\in\Bbb N\}\cup\{\max\Cal T\}$.
%611M

\spheader 611Xg In 611M(b-i), show that
$I=\{\tau:a\Bsubseteq\sup_{\sigma\in A}\Bvalue{\tau=\sigma}\}$ is a full
sublattice of $\Cal T$.
%611M out of order query

\spheader 611Xf Let $\Cal S$ be a finitely full sublattice of $\Cal T$ and
$\Cal S_0$ an ideal of $\Cal S$.   Show that $\Cal S_0$ is finitely full.
%611O

\leader{611Y}{Further exercises (a)}
%\spheader 611Ya
Suppose that $\frak A$ is ccc (definition:  316A).   Show that if
$A\subseteq\Cal T$ is non-empty, there is a countable $B\subseteq A$ such
that $\sup B=\sup A$ in $\Cal T$.   \Hint{remember to cover the case
$T=\omega_1$.}
%611C 611Xa

\spheader 611Yb
Let $\frak A=\{0,a,1\Bsetminus a,1\}$ be a four-element
Boolean algebra, and take $T=\coint{0,\infty}$.
Set $\frak A_0=\{0,1\}$ and $\frak A_t=\frak A$ for
$t>0$.   Show that for $s>0$ there is a stopping time $\tau_s$, adapted
to $\langle\frak A_t\rangle_{t\ge 0}$, defined by saying that
$\Bvalue{\tau_s>t}=a$ if $t\ge s$, $1$ if $t<s$.   (i) Show that
$\inf_{s>0}\tau_s$ is the constant stopping time
$\check{0}=\min\Cal T$, and is not defined by the formula in
611F.   (ii) Show that
$\bigcap_{s>0}\frak A_{\tau_s}\ne\frak A_{\check 0}$.   (iii) Show that
$\sup_{s>0}\Bvalue{\tau_s<\max\Cal T}\ne\Bvalue{\check 0<\max\Cal T}$.
%611E 611H 611C

\spheader 611Yc
Show that if $A\subseteq\Cal T$ is finite then the
covered envelope of $A$ is order-closed in $\Cal T$.
%611M

\spheader 611Yd Let $\Cal S$ be a sublattice of $\Cal T$, and write $\Cal I$ for
the set of totally ordered finite subsets of $\Cal S$.
For $I$, $J\in\Cal I$ say that
$I\sqsubseteq J$ if $J$ covers $I$.   Show that $\sqsubseteq$ is a pre-order on
$\Cal I$ (511A) under which $\Cal I$ is upwards-directed.
%611K 611M

}%end of exercises

\endnotes{
\Notesheader{611} Stochastic calculus is ordinarily presented in terms of
probability spaces and random variables.  I followed this line myself in
the brief introduction to stochastic processes in \S455.   To go farther,
however, I believe that (as with the ergodic theory of Chapter 38)
the essential ideas can be expressed more clearly in terms of
probability algebras $(\frak A,\bar\mu)$ and processes in the associated
spaces $L^0(\frak A)$, $L^1(\frak A,\bar\mu)$ and $L^2(\frak A,\bar\mu)$.
The machinery for this has already been developed in Chapter 36, so we can
go directly to the new ideas in 611B, 611G and 612D.   I give a
translation from the standard framework in 612H.

In this context, let me recall what I wrote in the introduction to Chapter 27.
The primary concept of probability theory is not a measure space of measure 1.
Rather, it is `random variable' with its associated distribution.   `Probability
spaces', as I use the phrase in this treatise, can be regarded as models for a
theory of random variables.   But in any statistical question, it is the
variable itself which we try to measure and make predictions for.   If we knew
where it came from, we'd study that.   Now `probability algebras', in my
terminology, provide a perfectly adequate model for distributions, including
joint distributions (see 653B and 653Xc below),
while evading some of the technical problems
associated with the arbitrary nature of any choice of probability space.   I
ought of course to admit that they simultaneously obscure some important
sources of intuition.

Again, it is normal to think of filtrations and stochastic processes as
based on real-valued times, so that the totally ordered
set $T$ of 611A is $\coint{0,\infty}$ or
something very like it.   (See 611Xa for a note on how to look at this
case.)   I have nothing interesting to say about other
cases, but there is very little extra work involved in the shift to an arbitrary
totally ordered set, and this enables me to avoid an occasional shuffle
when $\ooint{0,\infty}$ or $[0,1]$ seems a more appropriate parameter
space.   In fact the extra steps needed in the general case, dealing with
gaps in the parameter space (see the last clause in the definition
611B(b-i)), are already needed in the relatively elementary case
$T=\Bbb N$.   But it will become plain in Chapter 63 that some of the most
important ideas of the theory apply only in contexts essentially excluding
or erasing gaps in time.

I have deliberately cast the principal definitions in forms which make them
applicable in such cases as $T=\Bbb Z$ or $T=\Bbb R$.   But I note that
the phrases `finite-valued' and `bounded' in 611A(b-iii) could be
misleading when $T$ has no least member.   Of course we always have the
option of adding a least member $-\infty$ to $T$ and setting
$\frak A_{-\infty}=\bigcap_{t\in T}\frak A_t$.   And then we might be
tempted to add yet another element $-\infty'<-\infty$ to $T$ and set
$\frak A_{-\infty'}=\{0,1\}$.   Such manoeuvres can make no difference to
the mathematical content of the work here, but they will sometimes smooth
the task of adapting general results to specific applications.

The calculations in 611C, 611E and 611H
are elaborate but fundamentally elementary.
I ought to offer a word on the intuition behind the concept of
`stopping time'.   The requirement `$\Bvalue{\tau>t}\in\frak A_t$' (or, if
you prefer, `$\Bvalue{\tau\le t}\in\frak A_t$') is a declaration that the
decision whether to continue beyond
time $t$ must be based on what can be observed
at that time, the potential observations being those which can be
represented by members of $\frak A_t$.   In 611I we have a direct
expression of this idea:  the stopping time $\sigma$ corresponds to
waiting until we reach a stopping time $\tau\in A$, checking whether
we are in the region $a_{\tau}$, and if so halting.
This can be done only if the decision is based on something which will be
observable when we reach $\tau$, that is, if $a_{\tau}\in\frak A_{\tau}$.

The point at which I think an imaginative effort is required is in
611J-611K.   Here I expect that most people will find the expression in
611Xc useful in visualizing the sets $\Sti_0(I)$ of $I$-cells for finite
sublattices $I$;  they look like patchwork quilts.   The same
formulation can be used to help with 611C-611E.
}%end of notes


\discrpage

\frfilename{mt612.tex}
\versiondate{24.12.16/21.10.22}

%\def\Mlob{M_{\text{lob}}}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Fully adapted processes}

\newsection{612}

The next step is to introduce the processes which this volume is devoted to
studying.   These are an abstract version of the real-valued stochastic
processes $\langle X_t\rangle_{t\ge 0}$ of \S\S455 and 477.   Instead of
starting from $\Sigma_t$-measurable functions $X_t:\Omega\to\Bbb R$ and
then showing that it is possible to define $\Sigma_h$-measurable functions
$X_h$ for stopping times $h:\Omega\to\coint{0,\infty}$, I move directly to
families $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ of equivalence
classes of measurable functions where $\Cal S$ is a sublattice of the
lattice $\Cal T$ of stopping times discussed in \S611.   A `fully adapted
process' is one satisfying the essential measurability and consistency
requirements of 612D.   Among these, the `simple' processes (612J), those
which are constant between finitely many break points,
and `order-bounded' processes (614E) are particularly important.
I end with descriptions of Brownian motion (612T) and the standard Poisson
process (612U) in this language.

\leader{612A}{$\frak A$ and $L^0(\frak A)$ (a)} Given a Dedekind complete
Boolean algebra, we have a Dedekind complete
$f$-algebra $L^0=L^0(\frak A)$ as described in \S364.
\cmmnt{(For exact statements of the
algebraic relationships between linear structure, lattice structure and
multiplication which go to make an `$f$-algebra', see
351A, 352A, 352D and 352W.)   \S364 was dedicated to setting
up a coherent description of $L^0(\frak A)$ from $\frak A$
in logically primitive terms, so that, in particular, it would be visibly
free of any dependence on the axiom of choice.   But if you are willing to
relax this discipline, then I think that the easiest way to approach
these formulae is to think of $\frak A$ as a quotient $\Sigma/\Cal I$,
where $\Sigma$ is a $\sigma$-algebra of subsets of a set $\Omega$ and
$\Cal I$ is a $\sigma$-ideal of $\Sigma$ (314M), and to
recall that $L^0$ can now be regarded as a space of
equivalence classes of $\Sigma$-measurable functions (364C), with the
natural definitions of addition, multiplication and lattice operations.
}

\spheader 612Ab
In \S364 I introduced the formulae $\Bvalue{u>\alpha}$, $\Bvalue{u\in E}$,
where $u\in L^0$, $\alpha\in\Bbb R$ and $E\subseteq\Bbb R$ is a Borel
set\cmmnt{,
to represent `the region where $u$ is greater than $\alpha$' or `the region
where $u$ lies in $E$' (364A, 364G)}.   I mentioned formulae
$\Bvalue{u\ge\alpha}$, $\Bvalue{u<0}$ and $\Bvalue{u\ne 0}$,
and\cmmnt{ in the exercise 364Yb, I suggested a way of interpreting}
$\Bvalue{(u_1,\ldots,u_n)\in E}$ when $E$ is a Borel subset of
$\BbbR^n$.   Here it will be
convenient to extend the notation to such formulae as $\Bvalue{u\ne v}$,
meaning\cmmnt{, if you like,} $\Bvalue{|u-v|>0}$.
In terms of the representation
of $L^0$ as a space of equivalence classes of functions, we have

\Centerline{$\Bvalue{(f_1^{\ssbullet},\ldots,f_n^{\ssbullet})\in E}
=\{\omega:(f_1(\omega),\ldots,f_n(\omega))\in E\}^{\ssbullet}$}

\noindent for all $\Sigma$-measurable functions
$f_1,\ldots,f_n:\Omega\to\Bbb R$.   \cmmnt{Similarly,}
$\Bvalue{f_1^{\ssbullet}\ne f_2^{\ssbullet}}$ can be interpreted as
$\{x:f_1(\omega)\ne f_2(\omega)\}^{\ssbullet}$\cmmnt{,
and while this interpretation skates over some technical issues, it gives
clear signposts to such basic identities as

\Centerline{$\Bvalue{u=v}\Bcap\Bvalue{v=w}\Bsubseteq\Bvalue{u=w}$}

\noindent without any real danger of your being led astray}.

\spheader 612Ac Let $E\subseteq\Bbb R$ be a Borel set and
$h:E\to\Bbb R$ a Borel measurable function.   Set

\Centerline{$Q_E=\{u:u\in L^0$, $\Bvalue{u\in E}=1\}
=\{f^{\ssbullet}:f:\Omega\to E$ is measurable$\}$.}

\noindent If $u\in Q_E$,we have an $\bar h(u)\in L^0$ defined by saying that
$\Bvalue{\bar h(u)\in F}=\Bvalue{u\in h^{-1}[F]}$ for every Borel set
$F\subseteq\Bbb R$\cmmnt{ (364H)}.   If $u$, $u'\in Q_E$ then
$\Bvalue{u=u'}\Bsubseteq\Bvalue{\bar h(u)=\bar h(u')}$.   Observe that if
$h_1$, $h_2:\Bbb R\to\Bbb R$ are both Borel measurable, we now have
$\overline{h_1h_2}(u)=\bar h_1(\bar h_2(u))$ for all
$h_1$, $h_2$ and $u\in L^0$\prooflet{, because

\Centerline{$\Bvalue{u\in(h_1h_2)^{-1}[F]}
=\Bvalue{u\in h_2^{-1}[h_1^{-1}[F]]}
=\Bvalue{\bar h_2(u)\in h_1^{-1}[F]}
=\Bvalue{\bar h_1(\bar h_2(u))\in F}$}

\noindent for every Borel set $F$}.   \cmmnt{Also, of course, }$\bar h(u)=u$ if
$E=\Bbb R$ and
$h(\alpha)=\alpha$ for every $\alpha\in\Bbb R$\cmmnt{, that is, $h$ is the
identity of the semigroup $H$ of Borel measurable functions from
$\Bbb R$ to itself under the operation of composition of functions}.
So we have a semigroup action $\action$ of $H$ on
$L^0$ defined by saying that $h\action u=\bar h(u)$ for $h\in H$ and
$u\in L^0$.

\spheader 612Ad{\bf (i)}\cmmnt{ The following elementary facts are easy to
check.}

\inset{($\alpha$) If $\gamma\in\Bbb R$ and $h(\alpha)=\gamma\alpha$ for
$\alpha\in\Bbb R$, then $\bar h(u)=\gamma u$ for every $u\in L^0$.

($\beta$) If $h(\alpha)=|\alpha|$ for
$\alpha\in\Bbb R$, then $\bar h(u)=|u|$ for every $u\in L^0$.

($\gamma$) If $h(\alpha)=\alpha^2$ for
$\alpha\in\Bbb R$, then $\bar h(u)=u\times u=u^2$ for every $u\in L^0$.

($\delta$) If $h(\alpha)=1$ for
$\alpha\in\Bbb R$, then $\bar h(u)=\chi 1$ is the multiplicative identity
of $L^0$ for every $u\in L^0$.}

($\epsilon$) If $h:\Bbb R\to\Bbb R$ is non-decreasing, then 
$\bar h(u)\le\bar h(v)$ whenever $u\le v$ in $L^0$.

\medskip

\quad{\bf (ii)} It follows that if $V\subseteq L^0$ is such that
$u+v\in V$
for all $u$, $v\in V$ and $\bar h(u)\in V$ for every convex function
$h:\Bbb R\to\Bbb R$ such that $h(0)=0$,
then $V$ is an {\bf $f$-subalgebra} of $L^0$,
that is, a Riesz subspace closed under multiplication.
\prooflet{\Prf\  By (i-$\alpha$), it is closed
under scalar multiplication so is a linear subspace.   By (i-$\beta$),
$|u|\in V$ for every $u\in V$, so $V$ is a Riesz subspace (352Ic).
If $u$, $v\in V$ then
$u\times v=\bover12((u+v)^2-u^2-v^2)$ belongs to $V$ by (i-$\gamma$),
so $V$ is closed under multiplication and is an $f$-subalgebra.\ \Qed}
{\it A fortiori}, if $V$ is such that $u+v\in V$
for all $u$, $v\in V$ and $\bar h(u)\in V$ for every continuous function
$h:\Bbb R\to\Bbb R$ such that $h(0)=0$,
then $V$ is an $f$-subalgebra of $L^0$.

\medskip

\quad{\bf (iii)} Continuing from (c) above,
it will be important also to note that, for any
$u$, $v\in Q_E$, $\Bvalue{u\le v}\Bsubseteq\Bvalue{\bar h(u)\le\bar h(v)}$.
\prooflet{\Prf\ Identifying $Q_E$ with a set of equivalence classes of
real-valued measurable functions from $\Omega$ to $E$,
we find that $\bar h(f^{\ssbullet})=(hf)^{\ssbullet}$
for all such functions $f$ (364Ib), so that

$$\eqalign{\Bvalue{\bar h(f^{\ssbullet})\le\bar h(g^{\ssbullet})}
&=\Bvalue{(hf)^{\ssbullet}\le(hg)^{\ssbullet}}
=\{\omega:hf(\omega)\le hg(\omega)\}^{\ssbullet}\cr
&\Bsupseteq\{\omega:f(\omega)\le g(\omega)\}^{\ssbullet}
=\Bvalue{f^{\ssbullet}\le g^{\ssbullet}}\cr}$$

\noindent for all measurable $f$, $g:\Omega\to\Bbb R$.\ \Qed}
\cmmnt{It follows at once that }
$\Bvalue{u=v}\Bsubseteq\Bvalue{\bar h(u)=\bar h(v)}$.

\medskip

\quad{\bf (iv)} Take any $u\in L^0$.
Again writing $H$ for the space of Borel measurable functions
from $\Bbb R$ to itself, $H$ is an $f$-subalgebra of the $f$-algebra
$\BbbR^{\Bbb R}$ as well as a sub-semigroup under composition.
Treating $H$ as an $f$-algebra, the map $h\mapsto\bar h(u):H\to L^0$ is a
multiplicative Riesz homomorphism.   \prooflet{\Prf\
It is not especially
hard to prove this directly from the formula in (c), but you may prefer to
use the alternative description of $\bar h$ in 364Ib:  expressing
$\frak A$ with $\Sigma/\Cal I$, where $\Sigma$ is a $\sigma$-algebra of
subsets of a set $\Omega$
and $\Cal I$ is a $\sigma$-ideal of $\Sigma$, as in
(a), so that $u$ can be thought of as the equivalence class of a
$\Sigma$-measurable function $f:\Omega\to\Bbb R$ and $\bar h(u)$ becomes
the equivalence class of the composition $hf$, then
$h\mapsto hf$ is a multiplicative Riesz
homomorphism, so $h\mapsto(hf)^{\ssbullet}=\bar h(u)$ also is.\ \Qed}

\medskip

\quad{\bf (v)} It will happen more than once that we have two Dedekind
complete Boolean algebras $\frak A$ and $\frak B$, $f$-subalgebras
$V$, $W$ of $L^0(\frak A)$ and $L^0(\frak B)$ respectively, and
a linear operator $Q:V\to W$ such that $Q|v|=|Qv|$ and $Q(v^2)=(Qv)^2$
for all $v\in V$.   In this case, $Q$ will be an {\bf $f$-algebra
homomorphism}, that is, a multiplicative Riesz homomorphism.
\prooflet{(Use the ideas of (ii).)}

\spheader 612Ae{\bf (i)}
Now suppose that $\frak B$ is an order-closed subalgebra of
$\frak A$.   In this case we can think of $L^0(\frak B)$ as being the
subspace

\Centerline{$\{u:u\in L^0(\frak A)$, $\Bvalue{u>\alpha}\in\frak B$ for
every $\alpha\in\Bbb R\}$.}

\noindent The arguments of 364F show that this is equal to

\Centerline{$\{u:u\in L^0(\frak A)$, $\Bvalue{u\in E}\in\frak B$ for
every Borel set $E\subseteq\Bbb R\}$}

\noindent and\cmmnt{ therefore} that $\bar h(u)\in L^0(\frak B)$ whenever
$h\in H$ and $u\in L^0(\frak B)$.   Looking at this a little
more deeply, we see that
if $h\in H$ we have two different functions
$\bar h_{\frak A}:L^0(\frak A)\to L^0(\frak A)$ and
$\bar h_{\frak B}:L^0(\frak B)\to L^0(\frak B)$, but that
$\bar h_{\frak B}=\bar h_{\frak A}\restr L^0(\frak B)$, so that we
can\cmmnt{ fairly safely} use the same symbol $\bar h$ for either.

\cmmnt{Note also that if $A\subseteq L^0(\frak B)$ is non-empty
and has a supremum
$v$ in $L^0(\frak A)$, then $v\in L^0(\frak B)$.   \prooflet{\Prf\ For any
$\alpha\in\Bbb R$, $\Bvalue{v>\alpha}=\sup_{u\in A}\Bvalue{u>\alpha}$, by
364L(a-ii), and this belongs to $\frak B$ because $\frak B$ is order-closed
in $\frak A$.\ \Qed}   It follows that if
$A\subseteq L^0(\frak B)$ is non-empty and has an infimum
$v$ in $L^0(\frak A)$, then $v\in L^0(\frak B)$\prooflet{ (because
$-v=\sup(-A)$)}.   So }$L^0(\frak B)$ is an order-closed sublattice of
$L^0(\frak A)$.

\medskip

\quad{\bf (ii)} If $\familyiI{\frak B_i}$ is a non-empty
family of order-closed
subalgebras of $\frak A$ with intersection $\frak B$, then $\frak B$ is an
order-closed subalgebra of $\frak A$
and \ifresultsonly$L^0(\frak B)=\bigcap_{i\in I}L^0(\frak B_i)$.\else

$$\eqalign{L^0(\frak B)
&=\{u\in L^0(\frak A),\,\Bvalue{u>\alpha}\in\frak B\text{ for
every }\alpha\in\Bbb R\}
\cr&=\{u\in L^0(\frak A),\,\Bvalue{u>\alpha}\in\frak B_i\text{ for
every }\alpha\in\Bbb R\text{ and }i\in I\}
=\bigcap_{i\in I}L^0(\frak B_i).\cr}$$
\fi

\medskip

\quad{\bf (iii)} For any $u\in L^0(\frak A)$, the set
$\{\Bvalue{u\in E}:E\subseteq\Bbb R$ is Borel$\}$ is a $\sigma$-subalgebra
of $\frak A$, the smallest $\sigma$-subalgebra $\frak B$ of $\frak A$
such that $u\in L^0(\frak B)$;  it is the $\sigma$-subalgebra generated by
$\{\Bvalue{u>\alpha}:\alpha\in\Bbb R\}$.   \cmmnt{Following 272C,}
I will say that it is the
$\sigma$-subalgebra of $\frak A$ defined by $\pmb u$.   Similarly,
if $A\subseteq L^0(\frak A)$, I will say that the $\sigma$-subalgebra of
$\frak A$ generated by $\{\Bvalue{u>\alpha}:u\in A$, $\alpha\in\Bbb R\}$
is the $\sigma$-subalgebra defined by $A$.

\spheader 612Af Let $\frak C$ be another Dedekind complete Boolean algebra,
and $\phi:\frak A\to\frak C$ an order-continuous Boolean homomorphism.
Then we have a unique order-continuous $f$-algebra homomorphism
$T_{\phi}:L^0(\frak A)\to L^0(\frak C)$ such that

\inset{$\Bvalue{T_{\phi}u>\alpha}=\phi\Bvalue{u>\alpha}$ for every
$u\in L^0(\frak A)$ and $\alpha\in\Bbb R$,

$T_{\phi}(\chi a)=\chi(\phi a)$ for every $a\in\frak A$,

$\Bvalue{T_{\phi}u\in E}=\phi\Bvalue{u\in E}$ for every Borel set
$E\subseteq\Bbb R$,

$T_{\phi}\bar h_{\frak A}=\bar h_{\frak C}T_{\phi}$ for every
Borel measurable $h:\Bbb R\to\Bbb R$,

$T_{\phi}$ is injective or surjective iff $\phi$ is\ifresultsonly .\else\fi
}

\cmmnt{\noindent (364P).}

\leader{612B}{Products and processes} For the rest of this section\cmmnt{, and
indeed for nearly all the rest of the volume, }$\frak A$ will be a Dedekind
complete Boolean algebra, $T$ a totally ordered set,
$\family{t}{T}{\frak A_t}$ a filtration of closed subalgebras of $\frak A$,
$\Cal T$ the associated lattice of stopping times, and
$\family{\tau}{\Cal T}{\frak A_{\tau}}$ the corresponding family of
order-closed subalgebras\cmmnt{ (611B-611C, 611G)}.   For
$\sigma$, $\tau\in\Cal T$, $\Bvalue{\sigma<\tau}$, $\Bvalue{\sigma\le\tau}$
$\Bvalue{\sigma=\tau}$ will be the regions defined in 611D.

\spheader 612Ba If $\Cal S$ is a sublattice of $\Cal T$, we can form the
family $\family{\sigma}{\Cal S}{L^0(\frak A_{\sigma})}$.   If we take the
natural product linear space, lattice and multiplicative structures,
we get an $f$-algebra
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$\cmmnt{ (364R)}.
Moreover, writing $H$ for the semigroup of Borel measurable functions from
$\Bbb R$ to itself\cmmnt{ as in 612Ac}, we have a natural action of $H$ on
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$ defined by setting

\Centerline{$h\action\family{\sigma}{\Cal S}{u_{\sigma}}
=\family{\sigma}{\Cal S}{h\action u_{\sigma}}$}

\noindent whenever $h\in H$ and $u_{\sigma}\in L^0(\frak A_{\sigma})$ for
every $\sigma\in\Cal S$.

\spheader 612Bb Another way of looking at
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$ is to identify it
with $L^0(\frak C)$, where $\frak C$ is the simple Boolean algebra product
$\prod_{\sigma\in\Cal S}\frak A_{\sigma}$\cmmnt{ (315A, 364R)}.
Once again, it is easy to see that if $h\in H$ then
$\bar h_{\frak C}:L^0(\frak C)\to L^0(\frak C)$ matches the
$\bar h:\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})
\to\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$ described in (a).

\spheader 612Bc From (b) and 612A(d-ii)\cmmnt{, or otherwise,}
we now see that if $V$ is a subset of
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$ such that
$\pmb u+\pmb v\in V$ and $\bar h\pmb u\in V$ whenever $\pmb u$,
$\pmb v\in V$, $h:\Bbb R\to\Bbb R$ is convex and $h(0)=0$, then
$V$ is an $f$-subalgebra of $\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$.

\leader{612C}{}\cmmnt{ Before going farther, I give the
following fragment complementing the results of \S611.

\medskip

\noindent}{\bf Lemma} If $\sigma$, $\tau\in\Cal T$ and
$u\in L^0(\frak A_{\tau})$ then
$u\times\chi\Bvalue{\tau\le\sigma}$ and
$u\times\chi\Bvalue{\tau=\sigma}$ and $u\times\chi\Bvalue{\tau<\sigma}$
belong to $L^0(\frak A_{\sigma\wedge\tau})$.

\proof{ It is enough to consider the case $u\ge 0$.   In this case,
for $\alpha\in\Bbb R$,

$$\eqalignno{\Bvalue{u\times\chi\Bvalue{\tau\le\sigma}>\alpha}
&=\Bvalue{u>\alpha}\Bcap\Bvalue{\tau\le\sigma}
=\Bvalue{u>\alpha}\Bsetminus\Bvalue{\sigma<\tau}
  \in\frak A_{\sigma\wedge\tau}\text{ if }\alpha\ge 0\cr
\displaycause{611H(c-iii))}
&=1\in\frak A_{\sigma\wedge\tau}\text{ if }\alpha<0,\cr}$$

\noindent so
$u\times\chi\Bvalue{\tau\le\sigma}\in L^0(\frak A_{\sigma\wedge\tau})$.
As for the other parts, $\Bvalue{\tau=\sigma}$ belongs to
$\frak A_{\sigma\wedge\tau}$ so

\Centerline{$u\times\chi\Bvalue{\tau=\sigma}
=u\times\chi\Bvalue{\tau\le\sigma}\times\chi\Bvalue{\tau=\sigma}$}

\noindent belongs to $L^0(\frak A_{\sigma\wedge\tau})$;  while
$u\times\chi\Bvalue{\tau<\sigma}$ is the difference of the other two.
}%end of proof of 612C

\leader{612D}{Fully adapted processes} To continue the real work of this
section, let $\Cal S$ be a sublattice of
$\Cal T$ and $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ a family in
$L^0(\frak A)$.

\spheader 612Da {\bf Definition} I will say that
$\pmb u$ is {\bf fully adapted} to $\family{t}{T}{\frak A_t}$
if $u_{\sigma}\in L^0(\frak A_{\sigma})$ and
$\Bvalue{\sigma=\tau}\discretionary{}{}{}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$
%$\Bvalue{\sigma=\tau}\VVdash\,u_{\sigma}=u_{\tau}$
whenever $\sigma$, $\tau\in\Cal S$.

\spheader 612Db Note that if $u_{\tau}\in L^0(\frak A_{\tau})$ and
$\Bvalue{\sigma=\tau}\discretionary{}{}{}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$
%$\Bvalue{\sigma=\tau}\VVdash\,u_{\sigma}=u_{\tau}$
whenever $\sigma\le\tau\in\Cal S$, then $\pmb u$ is fully adapted.
\prooflet{\Prf\ For general $\sigma$, $\tau\in\Cal S$ we now have

$$\eqalignno{\Bvalue{\sigma=\tau}
&\Bsubseteq\Bvalue{\sigma=\sigma\wedge\tau}
  \Bcap\Bvalue{\tau=\sigma\wedge\tau}\cr
\displaycause{611E(a-ii-$\beta$)}
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau}}
  \Bcap\Bvalue{u_{\tau}=u_{\sigma\wedge\tau}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}.  \text{ \Qed}\cr}$$
}

\spheader 612Dc If $\pmb u$ is fully adapted
and $\Cal S'$ is a sublattice of $\Cal S$, then\cmmnt{ of course}
$\pmb u\restr\Cal S'$ is still a fully adapted process.

\spheader 612Dd If $\pmb u$ is fully adapted, $I$ is a finite sublattice
of $\Cal S$, and $(\tau_0\,\ldots,\tau_n)$ linearly
generates the $I$-cells\cmmnt{ as in
611K-611L}, then for any $\sigma\in I$ we have

\Centerline{$\sup_{i\le n}\Bvalue{u_{\sigma}=u_{\tau_i}}
\Bsupseteq\sup_{i\le n}\Bvalue{\sigma=\tau_i}=1$}

\noindent\cmmnt{by 611Ke.  }So\cmmnt{, for instance,} if
$\bar u=\sup_{i\le n}u_{\tau_i}$,\cmmnt{

\Centerline{$\Bvalue{u_{\sigma}\le\bar u}
\Bsupseteq\sup_{i\le n}\Bvalue{u_{\sigma}=u_{\tau_i}}=1$}

\noindent and $u_{\sigma}\le\bar u$;  thus}
$\bar u=\sup_{\sigma\in I}u_{\sigma}$.   \cmmnt{Similarly,}
$\sup_{\sigma\in I}|u_{\sigma}|=\sup_{i\le n}|u_{\tau_i}|$.

%In this context I will normally write $u_t$ for $u_{\tau}$ when
%$\tau$ is the constant stopping time at $t\in T$, that is,
%when

%$$\eqalign{\Bvalue{\tau\le s}
%&=0\text{ if }s<t,\cr
%&=1\text{ if }s\ge t.\cr}$$

\spheader 612De{\bf (i)}
Note that if $\pmb u$ is constant, say $u_{\sigma}=z$ for
every $\sigma\in\Cal S$, then $\pmb u$ is fully adapted iff
$z\in\bigcap_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$;
if $\Cal S$ has a least
element, this will be so iff $z\in L^0(\frak A_{\min\Cal S})$.
For any $z\in L^0(\frak A)$, I will write $z\tbf{1}$ for the fully adapted
process $\family{\sigma}{\Cal S}{z}$ where $\Cal S$ is the sublattice
$\{\sigma:\sigma\in\Cal T$, $z\in L^0(\frak A_{\sigma})\}$.
When $z=\chi 1$ and $\Cal S=\Cal T$ I will write just $\tbf 1$;   similarly,
$\tbf 0$ will be the constant process with value $0\in L^0(\frak A)$.

\medskip

\quad{\bf (ii)} Generally, if $z\in L^0(\frak A)$, I will write $z\pmb u$
for the process
$z\tbf 1\times\pmb u=\family{\sigma}{\Cal S'}{z\times u_{\sigma}}$, where
$\Cal S'=\{\sigma:\sigma\in\Cal S$, $z\in L^0(\frak A_{\sigma})\}$.
Then $\Cal S'$ is a sublattice of $\Cal S$ and $z\pmb u$ is fully adapted.
\prooflet{\Prf\ If $\sigma$, $\tau\in\Cal S'$ then
$\frak A_{\sigma\wedge\tau}=\frak A_{\sigma}\cap\frak A_{\tau}$
and $\frak A_{\sigma\vee\tau}\supseteq\frak A_{\sigma}$ (611H(c-ii)),
so $L^0(\frak A_{\sigma\wedge\tau})
=L^0(\frak A_{\sigma})\cap L^0(\frak A_{\tau})$
and $L^0(\frak A_{\sigma\vee\tau})\supseteq L^0(\frak A_{\sigma})$
(612A(e-ii)) both contain $z$, and $\sigma\wedge\tau$ and $\sigma\vee\tau$
belong to $\Cal S'$;  thus $\Cal S'$ is a sublattice of $\Cal S$.
If $\sigma\in\Cal S'$, then $z$ and $u_{\sigma}$ both belong to
$L^0(\frak A_{\sigma})$, so $z\times u_{\sigma}\in L^0(\frak A_{\sigma})$.
If $\sigma$, $\tau\in\Cal S'$, then

\Centerline{$\Bvalue{z\times u_{\sigma}=z\times u_{\tau}}
=\Bvalue{z=0}\Bcup\Bvalue{u_{\sigma}=u_{\tau}}
\Bsupseteq\Bvalue{\sigma=\tau}$,}

\noindent so $z\pmb u$ is fully adapted.\ \Qed}

\medskip

\spheader 612Df Suppose that $\pmb u$ is fully adapted.

\medskip

\quad{\bf (i)}
$u_{\sigma\wedge\tau}+u_{\sigma\vee\tau}=u_{\sigma}+u_{\tau}$
and $u_{\sigma\wedge\tau}\vee u_{\sigma\vee\tau}=u_{\sigma}\vee u_{\tau}$
for all $\sigma$, $\tau\in\Cal S$.  \prooflet{\Prf\

$$\eqalignno{\Bvalue{\sigma\le\tau}&
=\Bvalue{\sigma\wedge\tau=\sigma}
   \Bcap\Bvalue{\sigma\vee\tau=\tau}\cr
\Displaycause{611E(a-ii-$\beta$)}
\Bsubseteq\Bvalue{u_{\sigma\wedge\tau}=u_{\sigma}}
   \Bcap\Bvalue{u_{\sigma\vee\tau}=u_{\tau}}\cr&
\Bsubseteq
   \Bvalue{u_{\sigma\wedge\tau}+u_{\sigma\vee\tau}=u_{\sigma}+u_{\tau}}
   \Bcap\Bvalue{u_{\sigma\wedge\tau}\vee u_{\sigma\vee\tau}
      =u_{\sigma}\vee u_{\tau}},\cr}$$

\noindent and similarly

\Centerline{$\Bvalue{\tau\le\sigma}
\Bsubseteq
  \Bvalue{u_{\sigma\wedge\tau}+u_{\sigma\vee\tau}=u_{\sigma}+u_{\tau}}
  \Bcap\Bvalue{u_{\sigma\wedge\tau}\vee u_{\sigma\vee\tau}
      =u_{\sigma}\vee u_{\tau}}$;}

\noindent accordingly

\Centerline{$
\Bvalue{u_{\sigma\wedge\tau}+u_{\sigma\vee\tau}=u_{\sigma}+u_{\tau}}
  \Bcap\Bvalue{u_{\sigma\wedge\tau}\vee u_{\sigma\vee\tau}
      =u_{\sigma}\vee u_{\tau}}
\Bsupseteq\Bvalue{\sigma\le\tau}\Bcup\Bvalue{\tau\le\sigma}
=1$}

\noindent and
$u_{\sigma\wedge\tau}+u_{\sigma\vee\tau}=u_{\sigma}+u_{\tau}$,
$u_{\sigma\wedge\tau}\vee u_{\sigma\vee\tau}=u_{\sigma}\vee u_{\tau}$.\ \Qed}

\medskip

\quad{\bf (ii)}
$|u_{\tau}-u_{\sigma}|=|u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}|$ for all
$\sigma$, $\tau\in\Cal S$.   \prooflet{\Prf\

$$\eqalign{\Bvalue{\sigma\le\tau}&
\Bsubseteq\Bvalue{u_{\sigma\wedge\tau}=u_{\sigma}}
   \Bcap\Bvalue{u_{\sigma\vee\tau}=u_{\tau}}\cr&
\Bsubseteq\Bvalue{u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}
   =u_{\tau}-u_{\sigma}}
\Bsubseteq\Bvalue{|u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}|
   =|u_{\tau}-u_{\sigma}|}\cr}$$

\noindent and similarly

\Centerline{$\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{|u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}|
   =|u_{\sigma}-u_{\tau}|}
=\Bvalue{|u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}|
   =|u_{\tau}-u_{\sigma}|}$}

\noindent so $\Bvalue{|u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}|
   =|u_{\tau}-u_{\sigma}|}=1$ and
$|u_{\tau}-u_{\sigma}|=|u_{\sigma\vee\tau}-u_{\sigma\wedge\tau}|$.\ \Qed}

\medskip

\quad{\bf (iii)} $|u_{\sigma}-u_{\rho}|
\le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
  +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|$
for all $\rho$, $\sigma$, $\tau\in\Cal S$.   \prooflet{\Prf\

$$\eqalign{\Bvalue{\rho\le\tau}\Bcap\Bvalue{\sigma\le\tau}&
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\rho}|
    =|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\rho}|
  \le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|},\cr}$$

$$\eqalign{\Bvalue{\tau\le\rho}\Bcap\Bvalue{\tau\le\sigma}&
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\rho}|
    =|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\rho}|
  \le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|},\cr}$$

$$\eqalign{\Bvalue{\rho\le\tau}\Bcap\Bvalue{\tau\le\sigma}&
\Bsubseteq\Bvalue{|u_{\tau}-u_{\rho}|+|u_{\sigma}-u_{\tau}|
  =|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\rho}|
  \le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|},\cr}$$

\noindent and similarly

\Centerline{$\Bvalue{\sigma\le\tau}\Bcap\Bvalue{\rho\le\sigma}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\rho}|
  \le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|}$.}

\noindent Assembling these,

\Centerline{$\Bvalue{|u_{\sigma}-u_{\rho}|
  \le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|}\Bsupseteq 1$,}

\noindent that is, $|u_{\sigma}-u_{\rho}|
  \le|u_{\sigma\wedge\tau}-u_{\rho\wedge\tau}|
    +|u_{\sigma\vee\tau}-u_{\rho\vee\tau}|$.\ \Qed}

\leader{612E}{Where fully adapted processes come from} In
applications, one commonly starts from a family $\family{t}{T}{X_t}$ of
random variables, corresponding to a family
$\family{t}{T}{u_t}\in\prod_{t\in T}L^0(\frak A_t)$.
\cmmnt{In Chapter 63 I will look
at general rules for converting such families into fully adapted
processes.   For the moment, I describe a couple of special cases.
Another of the same kind is in 612R.}

\spheader 612Ea If $T$ is finite and not empty, with least value $\min T$,
then for $\tau\in\Cal T$ and $t\in T$ set

\Centerline{$a_{\tau t}
=(\inf_{s<t}\Bvalue{\tau>s})\Bsetminus\Bvalue{\tau>t}$}

\noindent(counting $\inf\emptyset$ as $1$, as usual, so that
$a_{\tau,\min T}=1\Bsetminus\Bvalue{\tau>\min T}$).
Then $\family{t}{T}{a_{\tau t}}$ is a partition of unity in $\frak A$,
and $a_{\tau t}\in\frak A_t$ for every $t$.  Now set

\Centerline{$u'_{\tau}
=\sum_{t\in T}u_t\times\chi a_{\tau t}$.}
	
\prooflet{\noindent If $t\in T$ and $\alpha\in\Bbb R$ then

$$\eqalignno{\Bvalue{u'_{\tau}>\alpha}\Bsetminus\Bvalue{\tau>t}
&=\sup_{s\in T}\Bvalue{u_s>\alpha}
  \Bcap a_{\tau s}\Bsetminus\Bvalue{\tau>t}
=\sup_{s\le t}\Bvalue{u_s>\alpha}\Bcap a_{\tau s}\Bsetminus\Bvalue{\tau>t}
  \cr
\displaycause{because $a_{\tau s}\Bsubseteq\Bvalue{\tau>t}$ if $s>t$}
&\in\frak A_t\cr}$$

\noindent because $\Bvalue{\tau>t}\in\frak A_t$ and
$\Bvalue{u_s>\alpha}\Bcap a_{\tau s}\in\frak A_s\subseteq\frak A_t$
for $s\le t$.   As $t$ is arbitrary,
$\Bvalue{u'_{\tau}>\alpha}\in\frak A_{\tau}$;  as $\alpha$ is
arbitrary, $u'_{\tau}\in L^0(\frak A_{\tau})$.

If $\sigma$, $\tau\in\Cal T$ and $\Bvalue{\sigma=\tau}=a$, then
$a\Bcap\Bvalue{\sigma>t}=a\Bcap\Bvalue{\tau>t}$ for every $t$
(611E(a-i-$\beta$), $a\Bcap a_{\sigma t}=a\Bcap a_{\tau t}$ for every $t$,
$u'_{\sigma}\times\chi a=u'_{\tau}\times\chi a$ and
$a\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\tau}}$.   So we have a fully
adapted
family.

If $\tau=\check{t}$ then $a_{\check{t},t}=1$ so
$u'_{\check{t}}=u_t$,
and we have
the required correspondence between $\family{\tau}{\Cal T}{u'_{\tau}}$ and
$\family{t}{T}{u_t}$.}

\spheader 612Eb If $T$ is well-ordered and not empty,
we can use essentially the same formulae for
$\tau\in\Cal T_f$.   \prooflet{We
need to check that $\family{t}{T}{a_{\tau t}}$ is a partition of
unity.
\Prf\ Certainly it is disjoint.   If $a\in\frak A\setminus\{0\}$, then
there is a $t\in T$ such that $a'=a\Bsetminus\Bvalue{\tau>t}$ is
non-zero, because $\tau$
is finite-valued;  because $T$ is well-ordered, we may suppose that
$t$ is minimal;  now $0\ne a'\Bsubseteq a\Bcap a_{\tau t}$.\ \Qed

In defining $u'_{\tau}$, if you do not like an infinite sum, simply
declare that $\Bvalue{u'_{\tau}>\alpha}
=\sup_{t\in T}\Bvalue{u_t>\alpha}\Bcap a_{\tau t}$ for every
$\alpha\in\Bbb R$;  then the rest of the argument proceeds as before.}

\leader{612F}{The identity process} \cmmnt{In the leading special cases of
this theory, in which $T=\coint{0,\infty}$, we have special processes
based on the similarity between the formula defining `stopping time' in
611A and that defining $L^0$ in 364A.

\medskip

}{\bf (a)} Suppose that $T=\coint{0,\infty}$.   For $\tau\in\Cal T_f$,
we can define $\iota_{\tau}\in L^0(\frak A)$ by saying that, for
$t\in\Bbb R$,

$$\eqalign{\Bvalue{\iota_{\tau}>t}
&=\Bvalue{\tau>t}\text{ if }t\ge 0,\cr
&=1\text{ if }t<0.\cr}$$

\prooflet{\noindent\Prf\
Since no member of $T$ is isolated on the right, the conditions
of 611A(b-i) imply that

\inset{$\Bvalue{\iota_{\tau}>t}\in\frak A$ for every $t\in\Bbb R$,

$\Bvalue{\iota_{\tau}>t}=\sup_{s>t}\Bvalue{\iota_{\tau}>s}$
for every $t\in\Bbb R$,

$\sup_{t\in\Bbb R}\Bvalue{\iota_{\tau}>t}=1$.}

\noindent Since $\tau\in\Cal T_f$, we also have

\inset{$\inf_{t\in\Bbb R}\Bvalue{\iota_{\tau}>t}=0$.}

\noindent So all the conditions ($\alpha$)-($\gamma$) of 364Aa are
satisfied, and $\iota_{\tau}\in L^0(\frak A)$.\ \Qed}

\spheader 612Fb $\pmb{\iota}=\family{\tau}{\Cal T_f}{\iota_{\tau}}$
is a fully adapted process.   \prooflet{\Prf\ If $\tau\in\Cal T_f$,

$$\eqalign{\Bvalue{\iota_{\tau}>s}
&=1\in\frak A_{\tau}\text{ if }s<0,\cr
&=\Bvalue{\tau>s}\in\frak A_{\tau}\text{ if }s\ge 0,\cr}$$

\noindent so $\iota_{\tau}\in L^0(\frak A_{\tau})$.   If $\sigma$,
$\tau\in\Cal T_f$ and $a=\Bvalue{\sigma=\tau}$, then

$$\eqalign{\Bvalue{\chi a\times\iota_{\sigma}>t}
&=a\Bcap\Bvalue{\iota_{\tau}>t}
=a\Bcap\Bvalue{\tau>t}
=a\Bcap\Bvalue{\sigma>t}
=\Bvalue{\chi a\times\iota_{\sigma}>t}\text{ if }t\ge 0,\cr
&=1=\Bvalue{\chi a\times\iota_{\sigma}>t}
  \text{ if }t<0,\cr}$$

\noindent so
$\chi a\times\iota_{\sigma}=\chi a\times\iota_{\tau}$ and
$a\Bsubseteq\Bvalue{\iota_{\sigma}=\iota_{\tau}}$.\ \Qed}

\spheader 612Fc $\iota_{\check t}=t\chi 1$ for every $t\ge 0$.
\prooflet{\Prf\ If $s\ge 0$,

$$\eqalign{\Bvalue{\iota_{\check t}>s}=\Bvalue{\check t>s}
&=\chi 1=\Bvalue{t\chi 1>s}\text{ if }t>s,\cr
&=0=\Bvalue{t\chi 1>s}\text{ otherwise}.  \text{ \Qed}\cr}$$
}

I will call $\pmb{\iota}$ the {\bf identity process} for the structure
$(\frak A,\langle\frak A_t\rangle_{t\ge 0})$.


\cmmnt{\leader{612G}{} However, the fully adapted families of greatest
importance to us will be those which can be constructed in the following
way.   I repeat some definitions from 455L.

\medskip

\noindent{\bf Definitions}
Let $(\Omega,\Sigma,\mu)$ be a probability space.

\medskip

{\bf (a)} A family $\langle\Sigma_t\rangle_{t\ge 0}$ of
$\sigma$-subalgebras of $\Sigma$ is a {\bf filtration} if
$\Sigma_s\subseteq\Sigma_t$ whenever $0\le s\le t$.

\medskip

{\bf (b)} A function $h:\Omega\to[0,\infty]$ is a {\bf stopping time}
(adapted to the filtration $\langle\Sigma_t\rangle_{t\ge 0}$)
if $\{\omega:h(\omega)\le t\}\in\Sigma_t$ for every $t\ge 0$.
In this case, I will write $\Sigma_h$ for the $\sigma$-algebra

\Centerline{$\{E:E\in\Sigma$,
$E\cap\{\omega:h(\omega)\le t\}\in\Sigma_t$ for every
$t\ge 0\}$}
\noindent (see 455L(c-iii)).

\medskip

{\bf (c)} A family
$\langle X_t\rangle_{t\ge 0}$ of real-valued functions on $\Omega$ is a
{\bf progressively measurable process}
(with respect to the filtration $\langle\Sigma_t\rangle_{t\ge 0}$) if
$(s,\omega)\mapsto X_s(\omega):[0,t]\times\Omega\to Y$ is
$\Cal B([0,t])\tensorhat\Sigma_t$-measurable for every $t\ge 0$, where
$\Cal B([0,t])$ is the Borel $\sigma$-algebra of $[0,t]$ for each $t$.
}%end of comment

\leader{612H}{Theorem} Let $(\Omega,\Sigma,\mu)$ be a complete
probability space, and $\langle\Sigma_t\rangle_{t\ge 0}$ a filtration of
$\sigma$-subalgebras of $\Sigma$ such that
every $\mu$-negligible set belongs to every $\Sigma_t$.
Let $(\frak A,\bar\mu)$ be the measure algebra of $\mu$ and set
$\frak A_t=\{E^{\ssbullet}:E\in\Sigma_t\}$ for each $t\ge 0$;  then
$\langle\frak A_t\rangle_{t\ge 0}$ is a filtration\cmmnt{ in the sense of
611Aa}.   Let $\Cal T$ be the associated family of stopping times.

(a)(i) If $h:\Omega\to[0,\infty]$ is a stopping time,
we have a stopping time $\tau\in\Cal T$ defined by saying that
$\Bvalue{\tau>t}=\{\omega:h(\omega)>t\}^{\ssbullet}$ for every
$t\ge 0$;  \cmmnt{in this case,} I will say that $h$ {\bf represents} $\tau$.

\quad(ii) Conversely, if $\tau\in\Cal T$, there is a stopping time
$h:\Omega\to [0,\infty]$ representing $\tau$.

\quad(iii) If $h$ represents $\tau$, then
$\Sigma_h=\{E:E\in\Sigma$, $E^{\ssbullet}\in\frak A_{\tau}\}$ and
$\frak A_{\tau}=\{E^{\ssbullet}:E\in\Sigma_h\}$.

\quad(iv) If $g$, $h:\Omega\to [0,\infty]$ are stopping
times representing $\sigma$, $\tau\in\Cal T$, then

\Centerline{$\Bvalue{\sigma<\tau}
=\{\omega:g(\omega)<h(\omega)\}^{\ssbullet}$,}

\Centerline{$\Bvalue{\sigma\le\tau}
=\{\omega:g(\omega)\le h(\omega)\}^{\ssbullet}$,
\quad$\Bvalue{\sigma=\tau}
=\{\omega:g(\omega)=h(\omega)\}^{\ssbullet}$.}

\noindent\cmmnt{So }$g$ and $h$ represent the same member of $\Cal T$ iff
they are equal almost everywhere.

\quad (v) If $h$ represents $\tau$, then $\tau\in\Cal T_f$ iff
$h(\omega)<\infty$ for almost every $\omega$;  \cmmnt{so} $\tau\in\Cal T_f$
iff it can be represented by a stopping time $h:\Omega\to\coint{0,\infty}$.

\quad(vi) If $t\ge 0$, then the constant function with value $t$
represents the constant stopping time at $t$.

(b) Now suppose that $\langle X_t\rangle_{t\ge 0}$ is a progressively
measurable process on $\Omega$.

\quad(i) For every $\tau\in\Cal T_f$ we have an
$x_{\tau}\in L^0(\frak A)\cong L^0(\mu)$ defined by saying that
$x_{\tau}$ is the equivalence class of the function
$X_h$, where $X_h(\omega)=X_{h(\omega)}(\omega)$ for
$\omega\in h^{-1}[\,\coint{0,\infty}\,]$,
whenever $h$ represents $\tau$\cmmnt{ in the sense of (a-ii)}.

\quad(ii) The family $\family{\tau}{\Cal T_f}{x_{\tau}}$ is fully
adapted to $\langle\frak A_t\rangle_{t\ge 0}$.

\proof{{\bf (a)(i)} If $h:\Omega\mapsto [0,\infty]$ is a
stopping time, and $t\ge 0$, then $\{\omega:h(\omega)>t\}^{\ssbullet}$
belongs to $\frak A_t$ and is equal to

$$\eqalignno{(\bigcup_{s\in\Bbb Q,s>t}\{\omega:h(\omega)>s\})^{\ssbullet}&
=\sup_{s\in\Bbb Q,s>t}\{\omega:h(\omega)>s\}^{\ssbullet}\cr&
=\sup_{t'>t}\{\omega:h(\omega)>t'\}^{\ssbullet}.\cr}$$

\noindent So $t\mapsto\{\omega:h(\omega)>t\}^{\ssbullet}$ satisfies
the conditions of 611A(b-i), and defines a member of $\Cal T$.

\medskip

\quad{\bf (ii)} Write $\Bbb Q^+$ for $\Bbb Q\cap\coint{0,\infty}$.
Take any $\tau\in\Cal T$.
For each $s\in\Bbb Q^+$, $\Bvalue{\tau>s}\in\frak A_s$, so we can choose
$E_s\in\Sigma_s$ such that
$E_s^{\ssbullet}=\Bvalue{\tau>s}$.   Next, for $s\in\Bbb Q^+$ set
$F_s=\bigcap_{s'\in\Bbb Q\cap[0,s]}E_{s'}$;  because $\Bbb Q$ is countable,
$F_s^{\ssbullet}=\inf_{s'\in\Bbb Q\cap[0,s]}\Bvalue{\tau>s'}
=\Bvalue{\tau>s}$, while $\family{s}{\Bbb Q^+}{F_s}$ is non-increasing.
Now, for $\omega\in\Omega$, set
$h(\omega)=\sup\{s:s\in\Bbb Q^+$, $\omega\in F_s\}$, counting
$\sup\emptyset$ as $0$.   Then, for any $t\ge 0$,

$$\eqalign{\{\omega:h(\omega)>t\}^{\ssbullet}
&=(\bigcup_{s\in\Bbb Q^+,s>t}F_s)^{\ssbullet}
=\sup_{s\in\Bbb Q^+,s>t}F_s^{\ssbullet}\cr
&=\sup_{s\in\Bbb Q^+,s>t}\Bvalue{\tau>s}
=\sup_{s>t}\Bvalue{\tau>s}
=\Bvalue{\tau>t}
\in\frak A_t.\cr}$$

\noindent So there is an $E\in\Sigma_t$ such that
$\{\omega:h(\omega)>t\}\symmdiff E$ is negligible;  as $\Sigma_t$ contains
every negligible set, $\{\omega:h(\omega)>t\}$ and
$\{\omega:h(\omega)\le t\}$ belong to $\Sigma_t$.   This is true for every
$t\ge 0$, so $h$ is a stopping time in the conventional sense;  and as
$\{\omega:h(\omega)>t\}^{\ssbullet}=\Bvalue{\tau>t}$ for every $t$,
$h$ represents $\tau$ in the sense here.

\medskip

\quad{\bf (iii)} If $E\in\Sigma_h$, then

\Centerline{$E\setminus\{\omega:h(\omega)>t\}
=E\cap\{\omega:h(\omega)\le t\}\in\Sigma_t$,}

\Centerline{$E^{\ssbullet}\Bsetminus\Bvalue{\tau>t}
=(E\setminus\{\omega:h(\omega)>t\})^{\ssbullet}
\in\frak A_t$}

\noindent for every $t\ge 0$, and $E^{\ssbullet}\in\frak A_{\tau}$.
Conversely, if $a\in\frak A_{\tau}$, take $E\in\Sigma$ such that
$E^{\ssbullet}=a$.   Then, for any $t\ge 0$,

\Centerline{$(E\cap\{\omega:h(\omega)\le t\})^{\ssbullet}
=E^{\ssbullet}\Bsetminus\Bvalue{\tau>t}
\in\frak A_t$}

\noindent so there is an $F\in\Sigma_t$ such that
$\mu((E\cap\{\omega:h(\omega)>t\})\symmdiff F)=0$;
again because $\Sigma_t$ contains all negligible sets,
$E\cap\{\omega:h(\omega)>t\}$ belongs to $\Sigma_t$.   As
$t$ is arbitrary, $E\in\Sigma_h$ and we have a suitable
representative of $a$.

\medskip

\quad{\bf (iv)} The point is that $\Bvalue{\sigma<\tau}
\Bsubseteq\sup_{s\in\Bbb Q^+}\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s}$.
\Prf\ For any $t\ge 0$, $t=\inf\{s:s\in\Bbb Q$, $s\ge t\}$, so

\Centerline{$\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
\Bsubseteq\sup_{s\in \Bbb Q,s\ge t}\Bvalue{\tau>s}
   \Bsetminus\Bvalue{\sigma>t}
\Bsubseteq\sup_{s\in\Bbb Q^+}
  \Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s}$.  \Qed}

\noindent So

$$\eqalign{\Bvalue{\sigma<\tau}
&=\sup_{t\ge 0}\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
=\sup_{s\in\Bbb Q^+}\Bvalue{\tau>s}\Bsetminus\Bvalue{\sigma>s}\cr
&=\sup_{s\in\Bbb Q^+}(\{\omega:h(\omega)>s\}
  \setminus\{\omega:g(\omega)>s\})^{\ssbullet}\cr
&=(\bigcup_{s\in\Bbb Q^+}\{\omega:h(\omega)>s\}
  \setminus\{\omega:g(\omega)>s\})^{\ssbullet}
=\{\omega:g(\omega)<h(\omega)\}^{\ssbullet}.\cr}$$

\noindent It follows at once that

\Centerline{$\Bvalue{\sigma\le\tau}=1\Bsetminus\Bvalue{\tau<\sigma}
=(\{\omega:g(\omega)\le h(\omega)\})^{\ssbullet}$}

\noindent and therefore that

\Centerline{$\Bvalue{\sigma=\tau}
=(\{\omega:g(\omega)=h(\omega)\})^{\ssbullet}$,}

\noindent so that

\Centerline{$\sigma=\tau
\iff\Bvalue{\sigma=\tau}=1
\iff g\eae h$.}

\medskip

\quad{\bf (v)}

$$\eqalign{\tau\in\Cal T_f
&\iff\inf_{t\ge 0}\Bvalue{\tau>t}=0\cr
&\iff\inf_{s\in\Bbb Q^+}\Bvalue{\tau>s}=0\cr
&\iff\bigcap_{s\in\Bbb Q^+}\{\omega:h(\omega)>s\}\text{ is negligible},\cr
&\iff\{\omega:h(\omega)=\infty\}\text{ is negligible}.\cr}$$

\noindent In this case, because we are supposing that negligible sets
belong to
$\Sigma_t$ for every $t$, we can adjust $h$ on the negligible set
$\Omega\setminus h^{-1}[\,\coint{0,\infty}\,]$,
if necessary, to get a stopping time with
finite values which represents $\tau$.

\medskip

\quad{\bf (vi)} Immediate from the definition of `constant stopping
time' in 611A(b-ii).

\medskip

{\bf (b)} For the identification of $L^0(\frak A)$ with $L^0(\mu)$,
see 364Ic.   Note that as we are assuming that every negligible set belongs to
every $\Sigma_t$, $(\Omega,\Sigma_t,\mu\restr\Sigma_t)$ is a complete
probability space and $\Sigma_t$ is closed under Souslin's operation (431A),
for every $t\ge 0$.

\medskip

\quad{\bf (i)} By (a-v), every $\tau\in\Cal T_f$ can be represented by a
stopping time which takes finite values.
If $h:\Omega\to\coint{0,\infty}$ represents $\tau$, then $X_h$ is defined
everywhere;  because the process is progressively measurable,
$X_h$ is $\Sigma_h$-measurable (455Le), and $X_h^{\ssbullet}$ is defined in
$L^0(\mu\restr\Sigma_h)\cong L^0(\frak A_{\tau})$ ((a-iii) above).

We note also from (a-iv) that if $g$, $h$ are stopping times both
representing $\tau$, then $g\eae h$ so $X_g\eae X_h$ and
$X_g^{\ssbullet}=X_h^{\ssbullet}$.   So we have a well-defined member
$x_{\tau}$ of $L^0(\frak A)$.

\medskip

\quad{\bf (ii)} We have already seen that
$x_{\tau}\in L^0(\frak A_{\tau})$ for every $\tau\in\Cal T_f$.
If $\sigma$, $\tau\in\Cal T_f$ are represented
by stopping times $g$ and $h$, then by (a-iv)

\Centerline{$
\Bvalue{\sigma=\tau}=\{\omega:g(\omega)=h(\omega)\}^{\ssbullet}
\Bsubseteq\{\omega:X_g(\omega)=X_h(\omega)\}^{\ssbullet}
=\Bvalue{X_g^{\ssbullet}=X_h^{\ssbullet}}
=\Bvalue{x_{\sigma}=x_{\tau}}$.}

\noindent So both conditions of 612Da are satisfied and
$\family{\tau}{\Cal T_f}{x_{\tau}}$ is fully adapted to
$\langle\frak A_t\rangle_{t\ge 0}$.
}%end of proof of 612H

\cmmnt{\medskip

\noindent{\bf Remarks} There will be a variation on this result in 649H.

Note that the representation of stopping times in (a)
above corresponds to an identification of $\Cal T_f$ with the set

\Centerline{$\{u:u\ge 0$, $\Bvalue{u>t}\in\frak A_t$ for every
$t\ge 0\}\subseteq L^0(\frak A)$,}

\noindent and that the lattice operations $\vee$, $\wedge$ on $\Cal T_f$
agree with those on $L^0(\frak A)$, as do arbitrary suprema (611C(b)-(c),
364L(a-ii) and 364L(b-i)).
}%

\leader{612I}{}\cmmnt{ We do not need to spend much time on general fully
adapted processes, but the following elementary facts are fundamental.

\medskip

\noindent}{\bf Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\Mfa(\Cal S)\subseteq\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$
the set of fully adapted processes with domain $\Cal S$.

(a) $\Mfa(\Cal S)$ is an order-closed $f$-subalgebra of the $f$-algebra
$L^0(\frak A)^{\Cal S}$, and if
$h:\Bbb R\to\Bbb R$ is a Borel measurable function then
$\bar h\pmb u\in \Mfa(\Cal S)$ for every $\pmb u\in \Mfa(\Cal S)$.
\cmmnt{Regarded as a Riesz space in its own right, }$\Mfa(\Cal S)$ is Dedekind
complete.

(b) Suppose that $\family{\sigma}{\Cal S}{u_{\sigma}}\in \Mfa(\Cal S)$ and
$\tau\in\Cal T$.   Set $\Cal S'=\{\sigma:\sigma\in\Cal T$,
$\sigma\wedge\tau\in\Cal S\}$.   Then
$\family{\sigma}{\Cal S'}{u_{\sigma\wedge\tau}}\in \Mfa(\Cal S')$.

\proof{{\bf (a)(i)} If $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ belong to $\Mfa(\Cal S)$, then

\inset{----- $u_{\sigma}+v_{\sigma}\in L^0(\frak A_{\sigma})$ for every
$\sigma\in\Cal S$,

----- if $\sigma$, $\tau\in\Cal S$, then

\Centerline{$\Bvalue{u_{\sigma}+v_{\sigma}=u_{\tau}+v_{\tau}}
\Bsupseteq\Bvalue{u_{\sigma}=v_{\sigma}}\Bcap\Bvalue{u_{\tau}=v_{\tau}}
\Bsupseteq\Bvalue{\sigma=\tau}$.}}

\noindent So
$\pmb u+\pmb v=\family{\sigma}{\Cal S}{u_{\sigma}+v_{\sigma}}
\in \Mfa(\Cal S)$.

Similarly, if $h:\Bbb R\to\Bbb R$ Borel measurable,

\Centerline{$\Bvalue{\bar h(u_{\sigma})=\bar h(u_{\tau})}
\Bsupseteq\Bvalue{\sigma=\tau}$}

\noindent for all $\sigma$, $\tau\in\Cal S$.   \Prf\ Set
$c=\Bvalue{u_{\sigma}=u_{\tau}}$,
and write $\frak A_c$ for the principal ideal of
$\frak A$ generated by $c$.   Setting $\phi a=a\Bcap c$ for $a\in\frak A$,
we have an order-continuous Boolean homomorphism from $\frak A$ to
$\frak A_c$, and the corresponding Riesz homomorphism
$T_{\phi}$ from $L^0(\frak A)$ to $L^0(\frak A_c)$ is defined by saying that
$\Bvalue{T_{\phi}w>\alpha}=\Bvalue{w>\alpha}\Bcap c$ for every
$w\in L^0(\frak A)$ (364P).   But this means that
$T_{\phi}u_{\sigma}=T_{\phi}u_{\tau}$, so

\Centerline{$T_{\phi}\bar h(u_{\sigma})=\bar h(T_{\phi}u_{\sigma})
=\bar h(T_{\phi}u_{\tau})=T_{\phi}\bar h(u_{\tau})$}

\noindent (612Af), and

\Centerline{$\Bvalue{\bar h(u_{\sigma})=\bar h(u_{\tau})}\Bsupseteq c
\Bsupseteq\Bvalue{\sigma=\tau}$.\ \Qed}

\noindent Thus $\bar h\pmb u\in \Mfa(\Cal S)$ for every
$\pmb u\in \Mfa(\Cal S)$.

Putting these together, $\Mfa(\Cal S)$ is an $f$-subalgebra of
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$, by 612Bc.

\medskip

\quad{\bf (ii)} To see that
$\Mfa(\Cal S)$ is order-closed in $L^0(\frak A)^{\Cal S}$,
take a non-empty set $A\subseteq \Mfa(\Cal S)$ with a
supremum $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ in
$L^0(\frak A)^{\Cal S}$.   For each $\sigma\in\Cal S$,
$v_{\sigma}=\sup_{\pmb u\in A}u_{\sigma}$ belongs to
$L^0(\frak A_{\sigma})$ (612Ae).   If $\sigma$,
$\tau\in\Cal S$, set $a=\Bvalue{\sigma=\tau}$;  then
$u_{\sigma}\times\chi a=u_{\tau}\times\chi a$ for every
$\pmb u\in A$ (taking $\pmb u$ to be
$\family{\sigma}{\Cal S}{u_{\sigma}}$ for every $\pmb u$).
Because multiplication by $\chi a$ is order-continuous in $L^0(\frak A)$
(353Pa),

\Centerline{$v_{\sigma}\times\chi a
=\sup_{\pmb u\in A}u_{\sigma}\times\chi a
=\sup_{\pmb u\in A}u_{\tau}\times\chi a
=v_{\tau}\times\chi a$,}

\noindent and $\Bvalue{v_{\sigma}=v_{\tau}}\Bsupseteq a$.   As $\sigma$
and $\tau$ are arbitrary, $\pmb v\in \Mfa(\Cal S)$.

Similarly, or applying the argument above to $\{-\pmb u:\pmb u\in A\}$,
$\inf A\in \Mfa(\Cal S)$ whenever $A$ is a non-empty subset of
$\Mfa(\Cal S)$ with an infimum in
$L^0(\frak A)^{\Cal S}$.   So $\Mfa(\Cal S)$ is order-closed in
$L^0(\frak A)^{\Cal S}$ in the sense of 313Da.

\medskip

\quad{\bf (iii)} We know that $L^0(\frak A)$ is Dedekind complete.
For every $\sigma\in\Cal S$,
$L^0(\frak A_{\sigma})$ is order-closed in $L^0(\frak A)$ (612Ae), so
in itself is Dedekind complete (353K(b-ii)).   So the product
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$ is Dedekind complete
(315D(e-i)) and its order-closed subspace $\Mfa(\Cal S)$ is Dedekind complete.

\medskip

{\bf (b)} Because $\Cal T$ is a distributive lattice, $\Cal S'$ is a
sublattice of $\Cal T$.   For any $\sigma\in\Cal S'$,
$u_{\sigma\wedge\tau}\in L^0(\frak A_{\sigma\wedge\tau})
\subseteq L^0(\frak A_{\sigma})$ (611H(c-ii)).
If $\sigma$, $\sigma'\in\Cal S'$, then by 611E(c-v-$\alpha$)

\Centerline{$\Bvalue{\sigma=\sigma'}
\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma'\wedge\tau}
\Bsubseteq\Bvalue{u_{\sigma\wedge\tau}=u_{\sigma'\wedge\tau}}$.}

\noindent So $\family{\sigma}{\Cal S'}{u_{\sigma\wedge\tau}}$ is fully
adapted.
}%end of proof of 612I

\leader{612J}{Simple processes (a) Definition} Let $\Cal S$ be a sublattice
of $\Cal T$.    A fully adapted process
$\family{\sigma}{\Cal S}{u_{\sigma}}$ is
{\bf simple} if either $\Cal S$ is empty or there are
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and
$u_*\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$
such that for every $\sigma\in\Cal S$

\Centerline{$\Bvalue{\sigma<\tau_0}
\Bsubseteq\Bvalue{u_{\sigma}=u_*}$,
\quad$\Bvalue{\tau_n\le\sigma}\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n}}$,}

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}$ for every $i<n$.}

\noindent In this case I will say that
$(\tau_0,\ldots,\tau_n)$ is a {\bf breakpoint string} for $\pmb u$.

\spheader 612Jb As a particularly elementary example, if $\Cal S$ is a
sublattice of $\Cal T$ and $\tau\in\Cal S$, then
$\pmb u=\family{\sigma}{\Cal S}{\chi\Bvalue{\tau\le\sigma}}$
is a simple process.   \prooflet{\Prf\
Setting $u_{\sigma}=\chi\Bvalue{\tau\le\sigma}$ for $\sigma\in\Cal S$
and $u_*=0$, we have $u_{\tau}=\chi 1$ and

\Centerline{$\Bvalue{\sigma<\tau}=\Bvalue{u_{\sigma}=0}
=\Bvalue{u_{\sigma}=u_*}$,
\quad$\Bvalue{\tau\le\sigma}=\Bvalue{u_{\sigma}=\chi 1}
=\Bvalue{u_{\sigma}=u_{\tau}}$}

\noindent for every $\sigma\in\Cal S$, so $(\tau)$ is a breakpoint string for
$\pmb u$.\ \Qed}

\medskip

\cmmnt{\noindent{\bf Warning!} most authors use the phrase
`simple process' for what I call a `previsibly simple process';  see 612Ye.
As we shall see shortly,
a simple process will normally have many breakpoint strings.   The
following ideas will be useful.}


\leader{612K}{Lemma}
Let $\Cal S$ be a non-empty sublattice of $\Cal T$.
Write $\frak B$ for $\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$.

(a) Suppose that $\tau_0\le\ldots\le\tau_n$ in
$\Cal S$, $u_i\in L^0(\frak A_{\tau_i})$ for $i\le n$ and
$u_*\in L^0(\frak B)$.   Then there is a
unique simple fully adapted process
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ such that whenever
$\sigma\in\Cal S$ then

\Centerline{$\Bvalue{v_{\sigma}=u_i}
\Bsupseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}$,}

\noindent for $i<n$, while

\Centerline{$\Bvalue{v_{\sigma}=u_*}
\Bsupseteq\Bvalue{\sigma<\tau_0}$,
\quad$\Bvalue{v_{\sigma}=u_n}\Bsupseteq\Bvalue{\tau_n\le\sigma}$;}

\noindent and $(\tau_0,\ldots,\tau_n)$ is a breakpoint string for
$\pmb v$.

(b) Suppose that $I$ is a non-empty finite sublattice of $\Cal S$ and
$(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells.   If
a simple process $\pmb u$ with domain $\Cal S$ has a breakpoint string in
$I$, then $(\tau_0,\ldots,\tau_n)$ is a breakpoint string for $\pmb u$.

(c) Suppose that $K$ is a finite set
and $\pmb u_k$ is a simple process with domain $\Cal S$
for each $k\in K$.   Then
there is a single string $(\tau_0,\ldots,\tau_n)$ in $\Cal S$ which is a
breakpoint string for every $\pmb u_k$.
%@612L

(d) Suppose that $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$
is a simple process with breakpoint string
$(\tau_0,\ldots,\tau_n)$ in $\Cal S$, and $\tau\in\Cal S$.

\quad(i) $(\tau_0\wedge\tau,\ldots,\tau_0\wedge\tau_n,\tau,
\tau_0\vee\tau,\ldots,\tau_n\vee\tau)$ is a breakpoint string for
$\pmb u$.

\quad(ii) Writing $\Cal S\wedge\tau$ for
$\{\sigma\wedge\tau:\sigma\in\Cal S\}=\Cal S\cap[\min\Cal T,\tau]$,
$\pmb u\restr\Cal S\wedge\tau$ is simple, with breakpoint string
$(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau,\tau)$.

\quad(iii) Writing $\Cal S\vee\tau$ for $\{\sigma\vee\tau:\sigma\in\Cal S\}
=\Cal S\cap[\tau,\max\Cal T]$,
$\pmb u\restr\Cal S\vee\tau$ is simple, with breakpoint string
$(\tau_0\vee\tau,\ldots,\tau_n\vee\tau)$.

(e) Suppose that $\pmb u$ is a fully adapted process with domain $\Cal S$, and
that $\tau\in\Cal S$.   If $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are simple processes with breakpoint strings
$(\tau_0,\ldots,\tau_m)$ and $(\tau'_0,\ldots,\tau'_n)$ respectively, then
$\pmb u$ is simple, with breakpoint string
$(\tau_0,\ldots,\tau_m,\tau,\tau'_0,\ldots,\tau'_n)$.

\proof{{\bf (a)} Set $\tau_{-1}=\min\Cal T$, $u_{-1}=u_*$.
For $\sigma\in\Cal S$, set
$b_{\sigma i}=\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}$ for
$-1\le i<n$, and $b_{\sigma n}=\Bvalue{\tau_n\le\sigma}$;
by 611H(c-i), $b_{\sigma i}\in\frak A_{\sigma}$ for each $i$.   Set

\Centerline{$v_{\sigma}
=\sum_{i=-1}^nu_i\times\chi b_{\sigma i}$.}

\noindent Then $v_{\sigma}\in L^0(\frak A_{\sigma})$.   \Prf\
For each $i$,
$u_i\times\chi\Bvalue{\tau_i\le\sigma}\in L^0(\frak A_{\sigma})$
(612C), so
$u_i\times\chi(\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}})$
belongs to $L^0(\frak A_{\sigma})$ if $i<n$, while also
$u_n\times\chi\Bvalue{\tau_n\le\sigma}$ belongs to
$L^0(\frak A_{\sigma})$.\ \Qed

If $\sigma$, $\sigma'\in\Cal S$ and $a=\Bvalue{\sigma=\sigma'}$, then
$a\Bcap b_{\sigma i}=a\Bcap b_{\sigma'i}$ for each $i$
(611E(c-iv-$\alpha$) and (c-iv-$\beta$) twice), so

\Centerline{$v_{\sigma}\times\chi a
=\sum_{i=-1}^nu_i\times\chi(a\Bcap b_{\sigma i})
=v_{\sigma'}\times\chi a$.}

\noindent Thus
$a\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma'}}$.
As $\sigma$ and $\sigma'$ are arbitrary, $\pmb v$ is fully
adapted.

The definition of $\pmb v$ makes it plain that, for any $\sigma\in\Cal S$,

\Centerline{$\Bvalue{\tau_n\le\sigma}
=b_{\sigma_n}\Bsubseteq\Bvalue{v_{\sigma}=u_n}$,}

\noindent while for $-1\le i<n$,

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
=b_{\sigma i}
\Bsubseteq\Bvalue{v_{\sigma}=u_i}$.}

\noindent Now we see that
$\Bvalue{\tau_i<\tau_{i+1}}=b_{\tau_i,i}\Bsubseteq\Bvalue{v_{\tau_i}=u_i}$
for $-1\le i<n$, while $b_{\tau_n,n}=1$ and $v_{\tau_n}=u_n$.   So for any
$\sigma\in\Cal S$,

$$\eqalignno{\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
&\Bsubseteq\Bvalue{v_{\sigma}=u_i}\Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr
\displaycause{611E(c-iii-$\gamma$)}
&\Bsubseteq\Bvalue{v_{\sigma}=u_i}\Bcap\Bvalue{v_{\tau_i}=u_i}
\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau_i}}\cr}$$

\noindent whenever $-1\le i<n$, while

\Centerline{$\Bvalue{\tau_n\le\sigma}\Bsubseteq\Bvalue{v_{\sigma}=u_n}
=\Bvalue{v_{\sigma}=v_{\tau_n}}$.}

\noindent So $\pmb v$ is a simple process and $(\tau_0,\ldots,\tau_n)$ is
a breakpoint string for $\pmb v$.

Concerning the uniqueness of $\pmb v$,
we just have to note that for every $\sigma\in\Cal S$,
$\sup_{-1\le i\le n}b_{\sigma i}=1$,
so that the given formula defines $v_{\sigma}$ uniquely.

\medskip

{\bf (b)} Express $\pmb u$ as $\sfamilysS{u}$.
Let $(\sigma_0,\ldots,\sigma_m)$ be a breakpoint string for
$\pmb u$ in $I$, and take $u_*\in L^0(\frak B)$ as in 612J.

Set $\tau_{-1}=\sigma_{-1}=\min\Cal T$,
$\tau_{n+1}=\sigma_{m+1}=\max\Cal T$.
Take any $\sigma\in\Cal S$.   Since $\tau_0\le\sigma_0$
and $\sigma_m\le\tau_n$,

\Centerline{$\Bvalue{u_{\sigma}=u_*}
\Bsupseteq\Bvalue{\sigma<\sigma_0}\Bsupseteq\Bvalue{\sigma<\tau_0}$,}

$$\eqalign{\Bvalue{u_{\sigma}=u_{\tau_n}}
&\Bsupseteq\Bvalue{u_{\sigma}=u_{\sigma_m}}
  \Bcap\Bvalue{u_{\tau_n}=u_{\sigma_m}}
\Bsupseteq\Bvalue{\sigma_m\le\sigma}
  \Bcap\Bvalue{\sigma_m\le\tau_n}\cr
&=\Bvalue{\sigma_m\le\sigma}
\Bsupseteq\Bvalue{\tau_n\le\sigma}.\cr}$$

\noindent Next, suppose that $i<n$.   Then the stopping-time interval
$c(\tau_i,\tau_{i+1})$ is either $0$ or an $I$-cell;  taking $J$ to be
$\{\sigma_i:-1\le i\le m+1\}$, $c(\tau_i,\tau_{i+1})$ must be included in a
$J$-cell, which must be of the form $c(\sigma_j,\sigma_{j+1})$ where
$-1\le j\le m$.   In this case
$\Bvalue{\tau_i<\tau_{i+1}}
\Bsubseteq\Bvalue{\sigma_j\le\tau_i}
   \Bcap\Bvalue{\tau_{i+1}\le\sigma_{j+1}}$ (611Jc).   So

$$\eqalignno{\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
&=\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
   \Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr
\displaycause{611E(c-iii-$\gamma$)}
&=\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
   \Bcap\Bvalue{\tau_i<\tau_{i+1}}
   \Bcap\Bvalue{\sigma_j\le\tau_i}
   \Bcap\Bvalue{\tau_{i+1}\le\sigma_{j+1}}\cr
&\Bsubseteq\Bvalue{\sigma_j\le\sigma}\Bcap\Bvalue{\sigma<\sigma_{j+1}}
   \Bcap\Bvalue{\sigma_j\le\tau_i}
   \Bcap\Bvalue{\tau_i<\sigma_{j+1}}\cr
\displaycause{611E(c-iv-$\alpha$), 611E(c-iii-$\gamma$)}
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma_j}}
  \Bcap\Bvalue{u_{\tau_i}=u_{\sigma_j}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}.}$$

\noindent So $u_*,\tau_0,\ldots,\tau_n$ satisfy the definition in 612J and
$\tau_0,\ldots,\tau_n$ is a breakpoint string for $\pmb u$.

\medskip

{\bf (c)} Because breakpoint strings are finite, there is a finite set
$A\subseteq\Cal S$ such that every $\pmb u_k$ has a breakpoint
string in $A$.   Let $I$ be the sublattice of $\Cal S$ generated
by $A$, so that $I$ is finite (611Ca) and there is a sequence
$(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   By (b),
$(\tau_0,\ldots,\tau_n)$ is a breakpoint string for $\pmb u_k$, for
any $k\in K$.

\medskip

{\bf (d)(i)} Of course
$\tau_0\wedge\tau\le\ldots\le\tau_n\wedge\tau\le\tau
\le\tau_0\vee\tau\le\ldots\le\tau_n\vee\tau$.   Let $u_*\in L^0(\frak B)$ be
such that $\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_*}$ for
every $\sigma\in\Cal S$.

Take any $\sigma\in\Cal S$.   Then

\Centerline{$\Bvalue{\sigma<\tau_0\wedge\tau}
\Bsubseteq\Bvalue{\sigma<\tau_0}
\Bsubseteq\Bvalue{u_{\sigma}=u_*}$.}

\noindent If $i<n$,

$$\eqalignno{
\Bvalue{\tau_i\wedge\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}\wedge\tau}
&=(\Bvalue{\tau_i\le\sigma}\Bcup\Bvalue{\tau\le\sigma})
    \Bcap\Bvalue{\sigma<\tau_{i+1}}\Bcap\Bvalue{\sigma<\tau}\cr
\displaycause{611E(c-i-$\beta$)}
&=\Bvalue{\tau_i\le\sigma}
    \Bcap\Bvalue{\sigma<\tau_{i+1}}\Bcap\Bvalue{\sigma<\tau}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}\Bcap\Bvalue{\tau_i<\tau}
   \Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr
\displaycause{611E(c-iii-$\gamma$)}
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}
   \Bcap\Bvalue{\tau_i=\tau_i\wedge\tau}
   \Bcap\Bvalue{\tau_i<\tau_{i+1}\wedge\tau}\cr
\displaycause{611E(a-ii-$\alpha$)}
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}
   \Bcap\Bvalue{\tau_i=\tau_i\wedge\tau}
   \Bcap\Bvalue{\tau_i\wedge\tau<\tau_{i+1}}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}
   \Bcap\Bvalue{u_{tau_i\wedge\tau}=u_{\tau_i}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i\wedge\tau}}.\cr}$$

\noindent Next,

$$\eqalign{\Bvalue{\tau_n\wedge\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau}
&=(\Bvalue{\tau_n\le\sigma}\Bcup\Bvalue{\tau\le\sigma})
    \Bcap\Bvalue{\sigma<\tau}\cr
&=\Bvalue{\tau_n\le\sigma}
    \Bcap\Bvalue{\sigma<\tau}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n}}\Bcap\Bvalue{\tau_n\le\tau}\cr
&=\Bvalue{u_{\sigma}=u_{\tau_n}}
  \Bcap\Bvalue{\tau_n\le\tau_n\wedge\tau}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n}}
   \Bcap\Bvalue{u_{\tau_n\wedge\tau}=u_{\tau_n}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n\wedge\tau}}\cr}$$

\noindent and

$$\eqalignno{\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau_0\vee\tau}
&=\Bvalue{\tau\le\sigma}
   \Bcap(\Bvalue{\sigma<\tau_0}\Bcup\Bvalue{\sigma<\tau})\cr
\displaycause{611Eb}
&=\Bvalue{\tau\le\sigma}
   \Bcap\Bvalue{\sigma<\tau_0}\cr
&\Bsubseteq\Bvalue{\tau<\tau_0}
   \Bcap\Bvalue{\sigma<\tau_0}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_*}
   \Bcap\Bvalue{u_{\tau}=u_*}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}.\cr}$$

\noindent Continuing, if $i<n$ then

$$\eqalignno{
\Bvalue{\tau_i\vee\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}\vee\tau}
&=\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\tau\le\sigma}
   \Bcap(\Bvalue{\sigma<\tau_{i+1}}\Bcup\Bvalue{\sigma<\tau})\cr
&=\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\tau\le\sigma}
   \Bcap\Bvalue{\sigma<\tau_{i+1}}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}\Bcap\Bvalue{\tau<\tau_{i+1}}
   \Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr
&=\Bvalue{u_{\sigma}=u_{\tau_i}}
   \Bcap\Bvalue{\tau_i\vee\tau<\tau_{i+1}}\cr
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}
   \Bcap\Bvalue{u_{\tau_i\vee\tau}=u_{\tau_i}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i\vee\tau}}.\cr}$$

\noindent And finally

\Centerline{$\Bvalue{\tau_n\vee\tau\le\sigma}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n}}
   \Bcap\Bvalue{u_{\tau_n\vee\tau}=u_{\tau_n}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n\vee\tau}}$.}

\noindent So the string $(\tau_0\wedge\tau,\ldots,\tau_0\wedge\tau_n,\tau,
\tau_0\vee\tau,\ldots,\tau_n\vee\tau)$ passes the test.

\medskip

\quad{\bf (ii)} The formulae in (i) tell us that, for any
$\sigma\in\Cal S\wedge\tau$,

\Centerline{$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_*}$,
\quad$\Bvalue{\tau_n\wedge\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n\wedge\tau}}$,}

\Centerline{$\Bvalue{\tau_i\wedge\tau\le\sigma}
  \Bcap\Bvalue{\sigma<\tau_{i+1}\wedge\tau}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i\wedge\tau}}$ for every $i<n$;}

\noindent and of course

\Centerline{$\Bvalue{\tau\le\sigma}
=\Bvalue{\tau=\sigma}\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$.}

\noindent Also
$u_*\in L^0(\bigcap_{\sigma\in\Cal S\wedge\tau}\frak A_{\sigma}$).
So $u_*,\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau,\tau)$ witness that
$\pmb u\restr\Cal S\wedge\tau$ is simple, with breakpoint string
$(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau,\tau)$.

\medskip

\quad{\bf (iii)} Similarly, for any
$\sigma\in\Cal S\vee\tau$,

\Centerline{$\Bvalue{\sigma<\tau_0\vee\tau}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$,
\quad$\Bvalue{\tau_n\vee\tau\le\sigma}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n\vee\tau}}$,}

\Centerline{$\Bvalue{\tau_i\vee\tau\le\sigma}
  \Bcap\Bvalue{\sigma<\tau_{i+1}\vee\tau}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i\vee\tau}}$ for every $i<n$.}

\noindent So
$\pmb u\restr\Cal S\vee\tau$ is simple, with breakpoint string
$(\tau_0\vee\tau,\ldots,\tau_n\vee\tau)$.

\medskip

{\bf (e)} Of course

\Centerline{$\tau_0\le\ldots\le\tau_m\le\tau\le\tau'_0\le\ldots\le\tau'_m$.}

\noindent
Let $u_*\in L^0(\bigcap_{\sigma\in\Cal S\wedge\tau}\frak A_{\sigma})$ be
such that $\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_*}$ for every
$\sigma\in\Cal S\wedge\tau$.   Then
$u_*\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma\wedge\tau})
\subseteq L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Now, for $\sigma\in\Cal S$, $i<m$ and $j<n$,

$$\eqalign{\Bvalue{\sigma<\tau_0}&
\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma}
   \Bcap\Bvalue{\sigma\wedge\tau<\tau_0}\cr&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau}}
   \Bcap\Bvalue{u_{\sigma\wedge\tau}=u_*}
\Bsubseteq\Bvalue{u_{\sigma}=u_*},\cr
\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}&
\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma}
  \Bcap\Bvalue{\tau_i\le\sigma\wedge\tau}
  \Bcap\Bvalue{\sigma\wedge\tau<\tau_{i+1}}\cr&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau}}
   \Bcap\Bvalue{u_{\sigma\wedge\tau}=u_{\tau_i}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}},\cr
\Bvalue{\tau_m\le\sigma}\Bcap\Bvalue{\sigma<\tau}&
\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma}
   \Bcap\Bvalue{\tau_m\le\sigma\wedge\tau}\cr&
\Bsubseteq\Bvalue{u_{\sigma\wedge\tau}=u_{\sigma}}
   \Bcap\Bvalue{u_{\sigma\wedge\tau}=u_{\tau_m}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_m}},\cr
\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'_0}&
\Bsubseteq\Bvalue{\tau\vee\sigma=\sigma}
   \Bcap\Bvalue{\tau\vee\sigma<\tau'_0}
   \Bcap\Bvalue{\tau<\tau'_0}\cr&
\Bsubseteq\Bvalue{u_{\tau\vee\sigma}=u_{\sigma}}
   \Bcap\Bvalue{u_{\tau\vee\sigma}=u_{\tau}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}},\cr
\Bvalue{\tau'_j\le\sigma}\Bcap\Bvalue{\sigma<\tau'_{j+1}}&
\Bsubseteq\Bvalue{\sigma\vee\tau=\sigma}
  \Bcap\Bvalue{\tau'_j\le\sigma\vee\tau}
  \Bcap\Bvalue{\sigma\vee\tau<\tau'_{j+1}}\cr&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\vee\tau}}
   \Bcap\Bvalue{u_{\sigma\vee\tau}=u_{\tau'_j}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau'_j}},\cr
\Bvalue{\tau'_n\le\sigma}&
\Bsubseteq\Bvalue{\sigma\vee\tau=\sigma}
   \Bcap\Bvalue{\tau'_n\le\sigma\vee\tau}\cr&
\Bsubseteq\Bvalue{u_{\sigma\vee\tau}=u_{\sigma}}
   \Bcap\Bvalue{u_{\sigma\vee\tau}=u_{\tau'_n}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau'_n}}.\cr}$$

\noindent So $(\tau_0,\ldots,\tau_m,\tau,\tau'_0,\ldots,\tau'_m)$ is indeed a
breakpoint string for $\pmb u$ and witnesses that $\pmb u$ is simple.
}%end of proof of 612K

\leader{612L}{Proposition} Let $\Cal S$ be a sublattice of
$\Cal T$.   Write $\Msimp=\Msimp(\Cal S)$ for the set of simple processes
with domain $\Cal S$.

(a) If $h:\Bbb R\to\Bbb R$ is a Borel measurable function and
$\pmb u\in\Msimp$, then $\bar h\pmb u\in\Msimp$ and any breakpoint string for
$\pmb u$ is a breakpoint string for $\bar h\pmb u$.

(b) $\Msimp$ is
an $f$-subalgebra of $\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$.

(c) If $z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ and
$\pmb u\in\Msimp$, then $z\pmb u\in\Msimp$.

\proof{ If $\Cal S$ is empty, this is trivial;  suppose otherwise.
Set $\frak B=\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$.

\medskip

{\bf (a)} Express $\pmb u$ as $\family{\sigma}{\Cal S}{u_{\sigma}}$;  let
$(\tau_0,\ldots,\tau_n)$ be a breakpoint string for $\pmb u$, and
take $u_*\in L^0(\frak B)$ such
that $\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_*}$ for every
$\sigma\in\Cal S$.   Then $\bar h(u_*)\in L^0(\frak B)$.
If $\sigma\in\Cal S$ then

\Centerline{$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_*}
\Bsubseteq\Bvalue{\bar h(u_{\sigma})=\bar h(u_*)}$,
\quad
$\Bvalue{\tau_n\le\sigma}\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n}}
\Bsubseteq\Bvalue{\bar h(u_{\sigma})=\bar h(u_{\tau_n})}$,}

\noindent and for $i<n$

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}
\Bsubseteq\Bvalue{\bar h(u_{\sigma})=\bar h(u_{\tau_i})}$.}

\noindent So $\bar h(u_*),\tau_0,\ldots,\tau_n$ witness that
$\bar h\pmb u$ is simple.

\medskip

{\bf (b)} If $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ are simple processes,
then there is a string $(\tau_0,\ldots,\tau_n)$
in $\Cal S$ which is a breakpoint string for both, by 612Kc.   As above,
take $u_*$, $v_*\in L^0(\frak B)$ such that

\Centerline{$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_*}$,
\quad$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{v_{\sigma}=v_*}$}

\noindent for every $\sigma\in\Cal S$.   Now $u_*+v_*\in L^0(\frak B)$ and

\Centerline{$\Bvalue{u_{\sigma}+v_{\sigma}=u_*+v_*}
\Bsupseteq\Bvalue{u_{\sigma}=u_*}
  \Bcap\Bvalue{v_{\sigma}=v_*}
\Bsupseteq\Bvalue{\sigma<\tau_0}$,}

\Centerline{$\Bvalue{u_{\sigma}+v_{\sigma}=u_{\tau_i}+v_{\tau_i}}
\Bsupseteq\Bvalue{u_{\sigma}=u_{\tau_i}}
  \Bcap\Bvalue{v_{\sigma}=v_{\tau_i}}
\Bsupseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}$}

\noindent for $i<n$, and

\Centerline{$\Bvalue{u_{\sigma}+v_{\sigma}=u_{\tau_n}+v_{\tau_n}}
\Bsupseteq\Bvalue{u_{\sigma}=u_{\tau_n}}
  \Bcap\Bvalue{v_{\sigma}=v_{\tau_n}}
\Bsupseteq\Bvalue{\tau_n\le\sigma}$.}

\noindent So $\pmb u+\pmb v$ is a simple process with a breakpoint string
$(\tau_0,\ldots,\tau_n)$.

By 612Bc this, together with (a), implies that
the set of simple processes with domain $\Cal S$ is
an $f$-subalgebra of $\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$.

\medskip

{\bf (c)} Follow the argument of (a) above, but with
$z\times u_*$, $z\times u_{\sigma}$ in place of
$\bar h(u_*)$, $\bar h(u_{\sigma})$, etc.
}%end of proof of 612L

\leader{612M}{Lemma} Let $\Cal S=[\min\Cal S,\max\Cal S]$
be a closed interval in $\Cal T$, and
$\pmb u$ a simple process with domain $\Cal S$.   Then
there is a breakpoint string $(\tau_0,\ldots,\tau_n)$ for $\pmb u$ such that
$\tau_0=\min\Cal S$, $\tau_n=\max\Cal S$ and
$\Bvalue{\tau_i<\tau_{i+1}}=\Bvalue{\tau_i<\max\Cal S}$ for every $i<n$.

\proof{ Applying 612K(d-i) twice, we see that $\pmb u$ has a breakpoint string
$(\sigma_0,\ldots,\sigma_m)$ starting with $\sigma_0=\min\Cal S$ and ending
with $\sigma_m=\max\Cal S$.   I seek to show by induction on $m$ that the result
is true in this case.   If $m=0$ then we can (and must) take $n=0$ and
$\tau_0=\min\Cal S=\max\Cal S$.

For the inductive step to $m\ge 1$, set

\Centerline{$a_i=\Bvalue{\min\Cal S<\sigma_i}$ for $i\le m$,}

\noindent so that

\Centerline{$0=a_0\Bsubseteq\ldots\Bsubseteq a_m
=\Bvalue{\min\Cal S<\max\Cal S}$}

\noindent and $a_i\in\frak A_{\min\Cal S}$ for every $i$.
Now set $b_m=1\Bsetminus a_m$ and

\Centerline{$b_i=a_{i+1}\Bsetminus a_i
  =\Bvalue{\min\Cal S<\sigma_{i+1}}\Bcap\Bvalue{\min\Cal S=\sigma_i}$
for $i<m$,}

\noindent  so that $\langle b_i\rangle_{i\le m}$
is a partition of unity in $\frak A$
and $b_i\in\frak A_{\min\Cal S}$ for
$i\le m$.   By 611I, there is a $\tau\in\Cal T$ such that
$b_i\Bsubseteq\Bvalue{\tau=\sigma_{i+1}}$ for $i<m$ and
$b_m\Bsubseteq\Bvalue{\tau=\sigma_m}$, while
$\sigma_1\le\tau\le\max\Cal S$, so $\tau\in\Cal S$.   Now

$$\eqalign{\Bvalue{\min\Cal S<\tau}&
\Bsupseteq\sup_{i<m}
  \Bvalue{\tau=\sigma_{i+1}}\Bcap\Bvalue{\min\Cal S<\sigma_{i+1}}\cr&
\Bsupseteq\sup_{i<m}b_i\Bcap a_{i+1}
=\sup_{i<m}b_i
=a_m
=\Bvalue{\min\Cal S<\max\Cal S}.\cr}$$

\noindent Also, for any $\sigma\in\Cal S$ and $i<m$,

$$\eqalign{b_i\Bcap\Bvalue{\sigma<\tau}&
\Bsubseteq\Bvalue{\min\Cal S=\sigma_i}
   \Bcap\Bvalue{\tau=\sigma_{i+1}}\Bcap\Bvalue{\sigma<\tau}\cr&
\Bsubseteq\Bvalue{\min\Cal S=\sigma_i}\Bcap\Bvalue{\sigma_i\le\sigma}
   \Bcap\Bvalue{\sigma<\sigma_{i+1}}\cr&
\Bsubseteq\Bvalue{\min\Cal S=\sigma_i}\Bcap\Bvalue{u_{\sigma}=u_{\sigma_i}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\min\Cal S}},\cr}$$

\noindent while

\Centerline{$b_m\Bcap\Bvalue{\sigma<\tau}
=\Bvalue{\min\Cal S=\max\Cal S}\Bcap\Bvalue{\sigma<\tau}=0$.}

\noindent So

\Centerline{$\Bvalue{\sigma<\tau}=\sup_{i\le m}b_i\Bcap\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\min\Cal S}}$.}

Next, as $\sigma_1\le\tau$,
$(\sigma_1\vee\tau,\ldots,\sigma_m\vee\tau)$ is a breakpoint string for
$\pmb u\restr\Cal S\vee\tau$ (612K(d-iii)) of length $m$.   By the inductive
hypothesis, $\pmb u\restr\Cal S\vee\tau=\pmb u\restr[\tau,\max\Cal S]$ has
a breakpoint string $(\tau_0,\ldots,\tau_n)$ starting with $\tau_0=\tau$,
ending with $\tau_n=\max\Cal S$, and such that
$\Bvalue{\tau_i<\tau_{i+1}}=\Bvalue{\tau_i<\max\Cal S}$ for every $i<n$.

We now find that $(\min\Cal S,\tau_0,\ldots,\tau_n)$ is a breakpoint sequence
for $\pmb u$.   \Prf\ Surely we have $\min\Cal S\le\tau_0\le\ldots\le\tau_n$.
If $\sigma\in\Cal S$, then

\Centerline{$\Bvalue{\min\Cal S\le\sigma}\Bcap\Bvalue{\sigma<\tau_0}
=\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{u_{\sigma}=u_{\min\Cal S}}$,}

\noindent while

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_i}}$}

\noindent for $i<n$, and

\Centerline{$\Bvalue{\tau_n\le\sigma}=\Bvalue{\tau_n=\sigma}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_n}}$.  \Qed}

And as $\Bvalue{\min\Cal S<\tau}=\Bvalue{\min\Cal S<\max\Cal S}$,
$(\min\Cal S,\tau_0,\ldots,\tau_n)$ is a breakpoint string of the right kind for
$\pmb u$, and the induction proceeds.
}%end of proof of 612M

\leader{612P}{}\cmmnt{ The next lemma is a bit of a sledgehammer, and
in the form given here will be used only at the end of
Chapter 22;  %627K
but most of the ideas are required for the important result 612Qa.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ a fully adapted process.
Then there is a %unique
fully adapted process
$\pmb v=\family{\tau}{\Cal T}{v_{\tau}}$, extending $\pmb u$, such that

\Centerline{$\Bvalue{v_{\tau}\ne 0}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}$}

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

\proof{{\bf (a)}
For $\tau\in\Cal T$, $\alpha\in\Bbb R$ set

\Centerline{$a_{\tau}
=1\Bsetminus\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}$,
\quad$b_{\tau\alpha}=\sup_{\sigma\in\Cal S}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}$.}

\noindent Then $b_{\tau\alpha}=\sup_{\beta>\alpha}b_{\tau\beta}$ for every
$\alpha\in\Bbb R$.   \Prf\

$$\eqalign{\sup_{\beta>\alpha}\sup_{\sigma\in\Cal S}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\beta}
&=\sup_{\sigma\in\Cal S}(\Bvalue{\tau=\sigma}
   \Bcap\sup_{\beta>\alpha}\Bvalue{u_{\sigma}>\beta})\cr
&=\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
   \Bcap\Bvalue{u_{\sigma}>\alpha}.  \text{ \Qed}\cr}$$

\noindent Next, $\inf_{\alpha\ge 0}b_{\tau\alpha}=0$.   \Prf\ If
$a\in\frak A\setminus\{0\}$, either $a\Bcap a_{\tau}\ne 0$ and
$a\notBsubseteq b_{\tau 0}$, or there are a $\sigma_0\in\Cal S$
such that $c=a\Bcap\Bvalue{\tau=\sigma_0}$ is non-zero and an
$\alpha\in\Bbb R$ such that $d=c\Bsetminus\Bvalue{u_{\sigma_0}>\alpha}$
is non-zero.   Now

$$\eqalign{d\Bcap\sup_{\sigma\in\Cal S}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}
&\Bsubseteq d\Bcap\sup_{\sigma\in\Cal S}
   \Bvalue{\sigma_0=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}\cr
&\Bsubseteq d\Bcap\Bvalue{u_{\sigma_0}>\alpha}
=0.\cr}$$

\noindent Thus in either case $a\Bsetminus\inf_{\alpha\ge 0}b_{\tau\alpha}$
is non-zero.   As $a$ is arbitrary,
$\inf_{\alpha\ge 0}b_{\tau\alpha}=0$.\ \Qed

On the other side,

\Centerline{$\sup_{\alpha<0}b_{\tau\alpha}
=\sup_{\sigma\in\Cal S}(\Bvalue{\tau=\sigma}
   \Bcap\sup_{\alpha<0}\Bvalue{u_{\sigma}>\alpha})
=\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
=1\Bsetminus a_{\tau}$.}

\medskip

{\bf (b)} This means that if $\tau\in\Cal T$ and we set

$$\eqalign{b'_{\tau\alpha}
&=b_{\tau\alpha}\text{ if }\alpha\ge 0,\cr
&=a_{\tau}\Bcup b_{\tau\alpha}\text{ if }\alpha<0,\cr}$$

\noindent we shall have

$$\eqalign{\sup_{\beta>\alpha}b'_{\tau\beta}
&=\sup_{\beta>\alpha}b_{\tau\beta}=b_{\tau\alpha}=b'_{\tau\alpha}
   \text{ if }\alpha\ge 0,\cr
&=\sup_{\beta>\alpha}a_{\tau}\Bcup b_{\tau\beta}=
a_{\tau}\Bcup b_{\tau\alpha}=b'_{\tau\alpha}
   \text{ if }\alpha<0,\cr}$$

\noindent while

\Centerline{$\inf_{\alpha\in\Bbb R}b'_{\tau\alpha}
\Bsubseteq\inf_{\alpha\ge 0}b_{\tau\alpha}
=0$,}

\Centerline{$\sup_{\alpha\in\Bbb R}b'_{\tau\alpha}
\Bsupseteq a_{\tau}\Bcup\sup_{\alpha<0}b_{\tau\alpha}
=1$.}

\noindent So $\alpha\mapsto b'_{\tau\alpha}$ satisfies all the conditions
of 364Aa, and we have an element $v_{\tau}$ of $L^0$ defined by saying that
$\Bvalue{v_{\tau}>\alpha}=b'_{\tau\alpha}$ for every $\alpha\in\Bbb R$.

\medskip

{\bf (c)} $v_{\tau}\in L^0(\frak A_{\tau})$ for every
$\tau\in\Cal T$.   \Prf\ For any $\alpha\in\Bbb R$ and
$\sigma\in\Cal S$, $\Bvalue{u_{\sigma}>\alpha}\in\frak A_{\sigma}$ so
$\Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}\in\frak A_{\tau}$, by
612C.   Accordingly $b_{\tau\alpha}=\sup_{\sigma\in\Cal S}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}$ belongs to
$\frak A_{\tau}$.   On the other hand, because
$\Bvalue{\sigma=\tau}\in\frak A_{\tau}$ for every $\sigma$ (611H(c-i) again),
$a_{\tau}\in\frak A_{\tau}$.   So
$\Bvalue{v_{\tau}>\alpha}=b'_{\tau\alpha}$ belongs to $\frak A_{\tau}$ for
every $\alpha$, and $v_{\tau}\in L^0(\frak A_{\tau})$.\ \Qed

\medskip

{\bf (d)} If $\tau\in\Cal T$ and $\sigma\in\Cal S$ then
$\Bvalue{v_{\tau}=u_{\sigma}}\Bsupseteq\Bvalue{\tau=\sigma}$.
\Prf\  If $\alpha\in\Bbb R$ then

$$\eqalignno{\Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}
&\Bsubseteq\Bvalue{\tau=\sigma}\Bcap b_{\tau\alpha}
=\Bvalue{\tau=\sigma}\Bcap\Bvalue{v_{\tau}>\alpha}\cr
\displaycause{because $a_{\tau}\Bcap\Bvalue{\tau=\sigma}=0$}
&=\sup_{\sigma'\in\Cal S}\Bvalue{\tau=\sigma}\Bcap\Bvalue{\tau=\sigma'}
  \Bcap\Bvalue{u_{\sigma'}>\alpha}\cr
&\Bsubseteq\sup_{\sigma'\in\Cal S}\Bvalue{\sigma=\sigma'}
  \Bcap\Bvalue{u_{\sigma'}>\alpha}\cr
&\Bsubseteq\sup_{\sigma'\in\Cal S}\Bvalue{u_{\sigma}=u_{\sigma'}}
  \Bcap\Bvalue{u_{\sigma'}>\alpha}
\Bsubseteq\Bvalue{u_{\sigma}>\alpha},\cr}$$

\noindent so in fact

\Centerline{$\Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}
=\Bvalue{\tau=\sigma}\Bcap\Bvalue{v_{\tau}>\alpha}$.}

\noindent As $\alpha$ is arbitrary,
$\Bvalue{\tau=\sigma}\Bsubseteq\Bvalue{v_{\tau}=u_{\sigma}}$.\ \Qed

In particular, if $\sigma\in\Cal S$ then $v_{\sigma}=u_{\sigma}$.

\medskip

{\bf (e)} If $\tau$, $\tau'\in\Cal T$ then
$\Bvalue{\tau=\tau'}\Bsubseteq\Bvalue{v_{\tau}=v_{\tau'}}$.
\Prf\ For any $\alpha\in\Bbb R$,

$$\eqalign{\Bvalue{\tau=\tau'}\Bcap b_{\tau\alpha}
&=\sup_{\sigma\in\Cal S}\Bvalue{\tau=\tau'}
   \Bcap\Bvalue{\tau=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}\cr
&=\sup_{\sigma\in\Cal S}\Bvalue{\tau=\tau'}
   \Bcap\Bvalue{\tau'=\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha}
=\Bvalue{\tau=\tau'}\Bcap b_{\tau'\alpha}.\cr}$$

\noindent At the same time,

\Centerline{$\Bvalue{\tau=\tau'}
   \Bcap\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
=\sup_{\sigma\in\Cal S}\Bvalue{\tau=\tau'}\Bcap\Bvalue{\tau=\sigma}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\tau'=\sigma}$,}

\noindent so in fact

\Centerline{$\Bvalue{\tau=\tau'}
   \Bcap\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
=\Bvalue{\tau=\tau'}\Bcap\sup_{\sigma\in\Cal S}\Bvalue{\tau'=\sigma}$}

\noindent and
$\Bvalue{\tau=\tau'}\Bcap a_{\tau}=\Bvalue{\tau=\tau'}\Bcap a_{\tau'}$.
Accordingly

$$\eqalign{\Bvalue{\tau=\tau'}\Bcap\Bvalue{v_{\tau}>\alpha}
&=\Bvalue{\tau=\tau'}\Bcap b_{\tau\alpha}
=\Bvalue{\tau=\tau'}\Bcap b_{\tau'\alpha}\cr
&\mskip100mu
=\Bvalue{\tau=\tau'}\Bcap\Bvalue{v_{\tau'}>\alpha}
\text{ if }\alpha\ge 0,\cr
&=\Bvalue{\tau=\tau'}\Bcap(a_{\tau}\Bcup b_{\tau\alpha})
=\Bvalue{\tau=\tau'}\Bcap(a_{\tau'}\Bcup b_{\tau'\alpha})\cr
&\mskip100mu
=\Bvalue{\tau=\tau'}\Bcap\Bvalue{v_{\tau'}>\alpha}
\text{ if }\alpha<0.\cr}$$

\noindent As $\alpha$ is arbitrary,
$\Bvalue{\tau=\tau'}\Bsubseteq\Bvalue{v_{\tau}=v_{\tau'}}$.\ \QeD\
As $\tau$ and $\tau'$ are arbitrary,
$\pmb v=\family{\tau}{\Cal T}{v_{\tau}}$ is fully adapted, and we
have already seen that it extends $\pmb u$.

\medskip

{\bf (f)} Observe that, for any $\tau\in\Cal T$,

\Centerline{$\Bvalue{v_{\tau}>0}=b_{\tau 0}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}
=1\Bsetminus a_{\tau}$,}

\noindent while

\Centerline{$\Bvalue{v_{\tau}>\alpha}=b'_{\tau\alpha}\Bsupseteq a_{\tau}$}

\noindent for every $\alpha<0$.   But this means that
$a_{\tau}\Bsubseteq\Bvalue{v_{\tau}=0}$, that is, that
$\Bvalue{v_{\tau}\ne 0}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}$, as required.
\leaveitout{
\medskip

{\bf (g)} Finally, to see that $\pmb v$ is uniquely defined, suppose that
$\pmb v'=\family{\tau}{\Cal T}{w'_{\tau}}$ is another fully adapted
process extending $\pmb u$ and such that
$\Bvalue{w'_{\tau}\ne 0}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}$ for every $\tau$.
Then, for any $\tau$,

$$\eqalign{\Bvalue{w'_{\tau}=v_{\tau}}
&\Bsupseteq(\Bvalue{w'_{\tau}=0}\Bcap\Bvalue{v_{\tau}=0})
  \Bcup\sup_{\sigma\in\Cal S}
    (\Bvalue{w'_{\tau}=u_{\sigma}}\Bcap\Bvalue{v_{\tau}=u_{\sigma}})\cr
&\Bsupseteq a_{\tau}
  \Bcup\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
=1.\cr}$$

\noindent So $w'_{\tau}=v_{\tau}$ for every $\tau\in\Cal T$ and
$\pmb v'=\pmb v$.}
}%end of proof of 612P

\leader{612Q}{Proposition}
Suppose that $\Cal S$ is a sublattice of $\Cal T$,
$\hat\Cal S$ its covered envelope\cmmnt{ (611M)} and
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ a fully adapted process.

(a) $\pmb u$ has a unique extension to a fully adapted process
$\hatpmb u=\sfamilysS{\hat u}$ with domain $\hat\Cal S$.

(b) The map $\pmb u\mapsto\hatpmb u$ is an isomorphism
from the $f$-algebra $\Mfa(\Cal S)$
of fully adapted processes with domain $\Cal S$ to
the $f$-algebra $\Mfa(\hat\Cal S)$, and
$\bar h\hatpmb u=(\bar h\pmb u)\hbox{\^{\null}}$
whenever $\pmb u\in \Mfa(\Cal S)$
and $h:\Bbb R\to\Bbb R$ is Borel measurable.

%\quad(iii) $\Bvalue{\hatpmb u\ne 0}=\Bvalue{\pmb u\ne 0}$.

(c) If $\tau\in\Cal S$, then $\hatpmb u\restr\hat\Cal S\wedge\tau$ is the fully
adapted extension of $\pmb u\restr\Cal S\wedge\tau$ to the covered envelope of
$\Cal S\wedge\tau$.

(e) If $z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ then
$z\hatpmb u$ is the fully adapted extension of $z\pmb u$.

(f) If $\pmb u$ is simple, with a witnessing string
$(u_*,\tau_0,\ldots,\tau_n)$ as in 612Ja, and $\Cal S'$ is a sublattice of
$\hat\Cal S$ including $\Cal S$, then $\hatpmb u\restr\Cal S'$ is simple,
with the same witnessing string.

(g) If $\pmb u$ is non-decreasing, so is $\hatpmb u$.

\proof{{\bf (a)} By 612P, there is a fully adapted process
$\pmb v=\family{\tau}{\Cal T}{v_{\tau}}$ extending $\pmb u$;  set
$\hat u_{\tau}=v_{\tau}$ for $\tau\in\hat\Cal S$, so that
$\hatpmb u=\family{\tau}{\hat\Cal S}{\hat u_{\tau}}$ is a fully adapted
process with domain $\hat\Cal S$ extending $\pmb u$.   If
$\family{\tau}{\hat\Cal S}{w_{\tau}}$ is any
fully adapted process extending $\family{\sigma}{\Cal S}{u_{\sigma}}$,
then

$$\eqalign{\Bvalue{\hat u_{\tau}=w_{\tau}}
&\Bsupseteq\Bvalue{\hat u_{\tau}=u_{\sigma}}
   \Bcap\Bvalue{w_{\tau}=u_{\sigma}}\cr
&=\Bvalue{\hat u_{\tau}=\hat u_{\sigma}}\Bcap\Bvalue{w_{\tau}=w_{\sigma}}
\Bsupseteq\Bvalue{\tau=\sigma}\cr}$$

\noindent for every $\sigma\in\Cal S$, so $\hat u_{\tau}=w_{\tau}$, for every
$\tau\in\hat\Cal S$.   Thus the extension is unique.

\medskip

{\bf (b)} The point is just that $\pmb u\mapsto\hatpmb u$ is the
inverse of $\pmb v\mapsto\pmb v\restr\Cal S:\Mfa(\hat\Cal S)\to\Mfa(\Cal S)$,
which has the declared properties.

\medskip

{\bf (c)} We saw in 611M(e-i) that $\hat\Cal S\wedge\tau$ is the covered
envelope of $\Cal S\wedge\tau$.   Now $\hatpmb u\restr\hat\Cal S\wedge\tau$ is
fully adapted and extends $\pmb u\restr\Cal S\wedge\tau$, so must be the fully
adapted extension of $\pmb u\restr\Cal S\wedge\tau$, by the uniqueness noted in
(a) above.

\medskip

{\bf (e)} Because
$\bigcap_{\tau\in\hat\Cal S}\frak A_{\tau}
=\bigcap_{\sigma\in\hat\Cal S}\frak A_{\sigma}$ (611Mf),
$z\hatpmb u$ is defined in $\Mfa(\hat\Cal S)$, and of course it
extends $z\pmb u$.

\medskip

{\bf (f)} Again because $\bigcap_{\tau\in\hat\Cal S}\frak A_{\tau}
=\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$,
$u_*\in L^0(\bigcap_{\tau\in\hat\Cal S}\frak A_{\tau})$.   Now if
$\tau\in\hat\Cal S$,

$$\eqalign{\Bvalue{\tau<\tau_0}&
=\sup_{\sigma\in\Cal S}\Bvalue{\tau<\tau_0}\Bcap{\tau=\sigma}
=\sup_{\sigma\in\Cal S}\Bvalue{\sigma<\tau_0}\Bcap{\tau=\sigma}\cr&
\Bsubseteq\sup_{\sigma\in\Cal S}
   \Bvalue{u_{\sigma}=u_*}\Bcap\Bvalue{\hat u_{\tau}=u_{\sigma}}
\Bsubseteq\Bvalue{u_{\tau}=u_*},\cr}$$

$$\eqalign{\Bvalue{\tau_n\le\tau}&
=\sup_{\sigma\in\Cal S}\Bvalue{\tau_n\le\tau}\Bcap{\tau=\sigma}
=\sup_{\sigma\in\Cal S}\Bvalue{\tau_n\le\sigma}\Bcap{\tau=\sigma}\cr&
\Bsubseteq\sup_{\sigma\in\Cal S}
   \Bvalue{u_{\sigma}=u_{\tau_n}}\Bcap\Bvalue{\hat u_{\tau}=u_{\sigma}}
\Bsubseteq\Bvalue{\hat u_{\tau}=u_{\tau_n}}
=\Bvalue{\hat u_{\tau}=\hat u_{\tau_n}},\cr}$$

\noindent and for $i<n$

$$\eqalign{\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}&
=\sup_{\sigma\in\Cal S}
  \Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}\Bcap\Bvalue{\tau=\sigma}
  \cr&
=\sup_{\sigma\in\Cal S}
  \Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}\Bcap{\tau=\sigma}\cr&
\Bsubseteq\sup_{\sigma\in\Cal S}
   \Bvalue{u_{\sigma}=u_{\tau_i}}\Bcap\Bvalue{\hat u_{\tau}=u_{\sigma}}
\Bsubseteq\Bvalue{\hat u_{\tau}=\hat u_{\tau_i}}.\cr}$$

\noindent So $\hatpmb u$ is simple, with the declared witnessing string.

As for $\hatpmb u\restr\Cal S'$, we have
$\bigcap_{\tau\in\hat\Cal S}\frak A_{\tau}
\Bsubseteq\bigcap_{\sigma\in\Cal S'}\frak A_{\sigma}$, so
$u_*\in L^0(\bigcap_{\tau\in\Cal S'}\frak A_{\tau})$, while
$\tau_0,\ldots,\tau_n$ belong to $\Cal S'$.   Now the formulae
of 612Ja show immediately that $(u_*,\tau_0,\ldots,\tau_n)$ witnesses that
$\hatpmb u\restr\Cal S'$ is simple.

\medskip

{\bf (g)} If $\tau\le\tau'$ in $\hat\Cal S$ and $a\in\frak A\setminus\{0\}$,
there are $\sigma$, $\sigma'\in\Cal S$ such that
$b=a\Bcap\Bvalue{\tau=\sigma}$ and $c=b\Bcap\Bvalue{\tau'=\sigma'}$ are
non-zero.   Now

$$\eqalign{c&
\Bsubseteq\Bvalue{\tau=\sigma}\Bcap\Bvalue{\tau'=\sigma'}
     \Bcap\Bvalue{\tau\le\tau'}
\Bsubseteq\Bvalue{\hat u_{\tau}=u_{\sigma}}
     \Bcap\Bvalue{\hat u_{\tau'}=u_{\sigma'}}
     \Bcap\Bvalue{\sigma\le\sigma'}\cr&
\Bsubseteq\Bvalue{\hat u_{\tau}=u_{\sigma}}
     \Bcap\Bvalue{\hat u_{\tau'}=u_{\sigma'}}
     \Bcap\Bvalue{\sigma=\sigma\wedge\sigma'}
     \Bcap\Bvalue{\sigma'=\sigma\vee\sigma'}\cr&
\Bsubseteq\Bvalue{\hat u_{\tau}=u_{\sigma\wedge\sigma'}}
     \Bcap\Bvalue{\hat u_{\tau'}=u_{\sigma\vee\sigma'}}
\Bsubseteq\Bvalue{\hat u_{\tau}\le\hat u_{\tau'}}\cr}$$

\noindent because $u_{\sigma\wedge\sigma'}\le u_{\sigma\vee\sigma'}$,
while $0\ne c\Bsubseteq a$.   Thus
$a\Bcap\Bvalue{\hat u_{\tau}\le\hat u_{\tau'}}\ne 0$;
as $a$ is arbitrary, $\Bvalue{\hat u_{\tau}\le\hat u_{\tau'}}=1$ and
$\hat u_{\tau}\le\hat u_{\tau'}$;  as $\tau$ and $\tau'$ are arbitrary,
$\hatpmb u$ is non-decreasing.
}%end of proof of 612Q

\leader{612R}{Corollary} Suppose that $\Cal S$ is a sublattice of
$\Cal T$ and $\Cal S'$ is a sublattice of $\Cal S$ covering $\Cal S$.
Then any fully adapted process
$\pmb u=\family{\sigma}{\Cal S'}{u_{\sigma}}$
has a unique extension to a
fully adapted process $\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$.

\proof{ Let $\hatpmb u$ be the fully adapted extension of $\pmb u$ to the
covered envelope $\hat\Cal S'$ of $\Cal S'$;  then
$\pmb v=\hatpmb u\restr\Cal S$ is fully adapted and extends $\pmb u$.
To see that $\pmb v$ is unique, repeat the argument of part (a-iii) of the proof
of 612Q;  if $\family{\tau}{\Cal S}{v'_{\tau}}$ is any
fully adapted process extending $\family{\sigma}{\Cal S'}{u_{\sigma}}$,
then

$$\eqalign{\Bvalue{v_{\tau}=v'_{\tau}}
&\Bsupseteq\Bvalue{v_{\tau}=u_{\sigma}}
   \Bcap\Bvalue{v'_{\tau}=u_{\sigma}}\cr
&=\Bvalue{v_{\tau}=v_{\sigma}}\Bcap\Bvalue{v'_{\tau}=v'_{\sigma}}
\Bsupseteq\Bvalue{\tau=\sigma}\cr}$$

\noindent for every $\sigma\in\Cal S'$, so $v_{\tau}=v'_{\tau}$, for every
$\tau\in\Cal S$.   Thus the extension is unique.
}%end of proof of 612R

\leader{612S}{Two more definitions}\cmmnt{ We shall have uses for
the following ideas.}   Let $\Cal S$ be a sublattice of $\Cal T$.

\spheader 612Sa For a fully adapted process
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$, write
$\|\pmb u\|_{\infty}
=\sup_{\sigma\in\Cal S}\|u_{\sigma}\|_{\infty}$\cmmnt{;
counting the supremum as $0$ if $\Cal S$ is empty, and
$\|u_{\sigma}\|_{\infty}$ as $\infty$ if
$u_{\sigma}$ does not belong to $L^{\infty}(\frak A)$ when this is
identified as a subspace of $L^0(\frak A)$ as in 364J}.

\spheader 612Sb For fully adapted processes
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$,
write $\Bvalue{\pmb u\ne\pmb v}$ for
$\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne v_{\sigma}}$, and
$\Bvalue{\pmb u\ne\tbf 0}=\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}$.

\spheader 612Sc Suppose that $\pmb u=\sfamilysS{u}$ and $\pmb v=\sfamilysS{v}$
are fully adapted processes.

\medskip

\quad{\bf (i)} If $h:\Bbb R\to\Bbb R$ is Borel
measurable, then
$\Bvalue{\bar h\pmb u\ne\bar h\pmb v}
\Bsubseteq\Bvalue{\pmb u\ne\pmb v}$.   \prooflet{(Use 612A(d-iii).)}

\medskip

\quad{\bf (ii)} If $\hat{\Cal S}$ is the covered envelope of $\Cal S$ and
$\hatpmb u=\family{\tau}{\hat{\Cal S}}{\hat u_{\tau}}$,
$\hatpmb v=\family{\tau}{\hat{\Cal S}}{\hat v_{\tau}}$ are the fully adapted
extensions of $\pmb u$, $\pmb v$ to $\hat{\Cal S}$\cmmnt{ (612Q)}, then
$\Bvalue{\hatpmb u\ne\hatpmb v}=\Bvalue{\pmb u\ne\pmb v}$.
\prooflet{\Prf\ If $\tau\in\hat{\Cal S}$ then

$$\eqalign{\Bvalue{\hat u_{\tau}\ne\hat v_{\tau}}&
=\sup_{\sigma\in\Cal S}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{\hat u_{\tau}\ne\hat v_{\tau}}\cr&
=\sup_{\sigma\in\Cal S}
   \Bvalue{\hat u_{\tau}=\hat u_{\sigma}}
   \Bcap\Bvalue{\hat v_{\tau}=\hat v_{\sigma}}
   \Bcap\Bvalue{\hat u_{\tau}\ne\hat v_{\tau}}\cr&
=\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{\hat u_{\sigma}\ne\hat v_{\sigma}}
=\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne v_{\sigma}}
=\Bvalue{\pmb u\ne\pmb v};\cr}$$

\noindent as $\tau$ is arbitrary,
$\Bvalue{\hatpmb u\ne\hatpmb v}\Bsubseteq\Bvalue{\pmb u\ne\pmb v}$;  and of
course

\Centerline{$\Bvalue{\pmb u\ne\pmb v}
=\sup_{\sigma\in\Cal S}\Bvalue{\hat u_{\sigma}\ne\hat v_{\sigma}}
\Bsubseteq\Bvalue{\hatpmb u\ne\hatpmb v}$,}

\noindent so we have equality.\ \Qed}

\leader{612T}{}\cmmnt{ The construction in 612H gives us direct routes to
some of the leading examples of stochastic process, and most sections of
this volume will introduce concepts which should be tested against these
examples.

\medskip

\noindent}{\bf Example:  Brownian motion (a)} Let
$\Omega=C(\coint{0,\infty})_0$ be the set of continuous functions
$\omega:\coint{0,\infty}\to\Bbb R$ such that $\omega(0)=0$, and
$\nu$ one-dimensional Wiener measure on
$\Omega$\cmmnt{ (477D)}, with $\Sigma$ its domain.
Recall that $\nu$ is a Radon
measure with respect to the topology $\frak T_c$ of uniform convergence on
compact sets\cmmnt{ (477B)}.
Let $(\frak C,\bar\nu)$ be the measure algebra of
$\nu$.   For $t\ge 0$, write $\Sigma_t$ for

\Centerline{$\{F:F\in\Sigma$, $\omega'\in F$ whenever $\omega\in F$,
$\omega'\in\Omega$ and $\omega'\restr[0,t]=\omega\restr[0,t]\}$,}

\noindent and let $\hat\Sigma_t$ be
$\{F\symmdiff A:F\in\Sigma_t$, $\nu A=0\}$\cmmnt{ (cf.\ 477H)};  set
$\frak C_t=\{F^{\ssbullet}:F\in\hat\Sigma_t\}
=\{F^{\ssbullet}:F\in\Sigma_t\}$
and $X_t(\omega)=\omega(t)$ for $t\ge 0$ and $\omega\in\Omega$.
Then $(s,\omega)\mapsto X_s(\omega):[0,t]\times\Omega\to\Bbb R$ is
continuous\cmmnt{, therefore $\Cal B([0,t])\tensorhat\Sigma_t$-measurable
and $\Cal B([0,t])\tensorhat\hat\Sigma_t$-measurable
(4A3Q(c-i)),} for every $t\ge 0$,
and $\langle X_t\rangle_{t\ge 0}$ is progressively measurable with respect
to $\langle\hat\Sigma_t\rangle_{t\ge 0}$.
We\cmmnt{ can therefore apply 612H to see that we} have a process
$\pmb w=\family{\tau}{\Cal T_f}{w_{\tau}}$ fully adapted to
$\langle\frak C_t\rangle_{t\ge 0}$.   In this volume I will use the phrase
{\bf Brownian motion} to mean the process $\pmb w$\cmmnt{, rather than the
process $\langle X_t\rangle_{t\ge 0}$ as in Chapter 47;  I hope that this
will not lead to any confusion}.

\spheader 612Td\cmmnt{ It will be important to know that} $\pmb w$ determines
$\frak C$ and $\langle\frak C_t\rangle_{t\ge 0}$, in that

\inset{(i) $\frak C$ is the closed subalgebra $\frak D$
of itself generated by
$\{\Bvalue{w_{\check t}>\alpha}:t\ge 0$, $\alpha\in\Bbb R\}$,

(ii) $\frak C_t$ is the closed subalgebra generated by
$\{\Bvalue{w_{\check s}>\alpha}:s\in[0,t]$, $\alpha\in\Bbb R\}$ for every
$t\ge 0$.}

\prooflet{\medskip

\Prf{\bf (i)} If $t\ge 0$ and $\alpha\in\Bbb R$,
$\Bvalue{w_{\check t}>\alpha}=\{\omega:\omega(t)>\alpha\}^{\ssbullet}$, so
$\frak D=\{E^{\ssbullet}:E\in\Tau_{\coint{0,\infty}}\}$ where
$\Tau_{\coint{0,\infty}}$ is the
$\sigma$-algebra of subsets of $\Omega=C(\coint{0,\infty})_0$ generated by
$\{\{\omega:\omega(t)>\alpha\}:t\ge 0$, $\alpha\in\Bbb R\}$.

Consider the family $\Cal V=\frak T_c\cap\Tau_{\coint{0,\infty}}$.
This is closed
under union and intersection and the topology on $\Omega$
it generates is Hausdorff.   It follows that if $K$, $L$ are
disjoint $\frak T_c$-compact sets, then there are disjoint
$U$, $V\in\Cal V$ such that $K\subseteq U$ and $L\subseteq V$.   But
$\nu$ is inner regular with respect to the $\frak T_c$-compact sets, so if
$E\in\dom\nu$ and $\epsilon>0$ there is a $V\in\Cal V$ such that
$\nu(E\symmdiff V)\le\epsilon$.   Consequently
$\frak D$ is dense in $\frak C$ for the
measure-algebra topology;  as it is also closed, it is the whole of
$\frak C$, as claimed.

\medskip

\quad{\bf (ii)} Now take any $t\ge 0$;  write
$\Tau_{[0,t]}$ for the $\sigma$-algebra of subsets of $\Omega$ generated by
$\{\{\omega:\omega(s)>\alpha\}:s\in[0,t]$, $\alpha\in\Bbb R\}$, and
$\frak D_t$ for the closed subalgebra generated by
$\{\Bvalue{w_{\check s}>\alpha}:s\in[0,t]$, $\alpha\in\Bbb R\}$, so that
$\frak D_t=\{E^{\ssbullet}:E\in\Tau_{[0,t]}\}$.
Consider $\Sigma_t$ as defined in (a) above.   Of
course $\Tau_{[0,t]}\subseteq\Sigma_t$, so $\frak D_t\subseteq\frak C_t$.
On the other hand, given $F\in\Sigma_t$ and $\epsilon>0$, there are
$\frak T_c$-compact sets $K\subseteq F$ and $L\subseteq\Omega\setminus F$
such that $\nu(K\cup L)\ge\epsilon$.   If we set
$\Cal V_t=\frak T_c\cap\Tau_t$, then for any $\omega\in K$, $\omega'\in L$
there is an $s\le t$ such that $\omega(s)\ne\omega'(s)$, so that
there are disjoint $U$, $V\in\Cal V_t$ such that $\omega\in U$ and
$\omega'\in V$;  because $\Cal V_t$ is closed under finite unions and
intersections, there are disjoint $U$, $V\in\Cal V_t$ such that
$K\subseteq U$ and $L\subseteq V$, so that
$\nu(F\symmdiff U)\le\epsilon$, while $U\in\Tau_t$.   As $\epsilon$ is
arbitrary, $F^{\ssbullet}\in\frak D_t$;  as $F$ is arbitrary,
$\frak C_t\subseteq\frak D_t$.\ \Qed}

\spheader 612Te\ifresultsonly{ Every}\else{ In order to apply 612H directly,
I have cast the discussion
above in terms of stopping times adapted to the filtration
$\langle\hat\Sigma_t\rangle_{t\ge 0}$.   However it will make it easier to
call on further results from \S477 if I remark that every}\fi\ member of
$\Cal T_f$ can be represented by a stopping time adapted to
$\langle\Sigma^+_t\rangle_{t\ge 0}$, where $\Sigma^+_t=\bigcap_{s>t}\Sigma_s$
for $t\ge 0$.   \prooflet{\Prf\ Recall that $\langle\Sigma^+_t\rangle_{t\ge 0}$
is a filtration of $\sigma$-algebras (455L) and that

\Centerline{$\Sigma^+_t\subseteq\hat\Sigma_t
=\hat\Sigma^+_t=\bigcap_{s>t}\hat\Sigma_s$}

\noindent for every $t$ (477Hc).   If $\tau\in\Cal T_f$
there is an $h:\Omega\to\coint{0,\infty}$ representing $\tau$
which is adapted to $\langle\hat\Sigma_t\rangle_{t\ge 0}
=\langle\hat\Sigma^+_t\rangle_{t\ge 0}$;  by 455L(e-iii), there is
a stopping time $g$, adapted to $\langle\Sigma^+_t\rangle_{t\ge 0}$,
which is equal to $h$ almost everywhere.   Of course
$g$ is now adapted to $\langle\hat\Sigma_t\rangle_{t\ge 0}$ so represents a
stopping time in $\Cal T_f$ in the sense here, and by 612H(a-iv) $g$ represents
$\tau$.\ \Qed}

\leader{612U}{Example:  the Poisson process (a)} For $t>0$ let $\lambda_t$
be the Poisson distribution with
expectation $t$, that is, the Radon probability
measure on $\Bbb R$ such that $\lambda_t\{n\}=e^{-t}t^n/n!$ for every
$n\in\Bbb N$\cmmnt{ (495Aa)}.   Then the convolution
$\lambda_s*\lambda_t$ is equal to
$\lambda_{s+t}$ whenever $s$, $t>0$\cmmnt{ (495Ab)}, and
$\lim_{t\downarrow 0}\lambda_tG=1$ for every open set $G\subseteq\Bbb R$
including $0$.   So\cmmnt{ 455Pc
tells us that} we have an associated probability measure $\ddot\mu$
on the space $\Cdlg$ of \cadlag\ real-valued functions defined on
$\coint{0,\infty}$.   This measure is\cmmnt{ described in 455P as} the subspace
measure on $\Cdlg$ induced by a
complete measure on $\BbbR^{\coint{0,\infty}}$
defined in terms of transitional probabilities\cmmnt{, following 455E}.
The formula of 455E tells us that if $0=t_0<\ldots<t_n$ in $\Bbb R$ and
$0=k_0\le\ldots\le k_n$ in $\Bbb N$, then
the measure of $\{\omega:\omega(t_i)=k_i$ for $i\le n\}$ is

\Centerline{$\prod_{i=1}^n\lambda_{t_i-t_{i-1}}\{k_i-k_{i-1}\}
=e^{-t_n}\prod_{i=1}^n
\Bover{(t_i-t_{i-1})^{k_i-k_{i-1}}}{(k_i-k_{i-1})!}$.}

\spheader 612Ub\cmmnt{ As in 455K,} $\ddot\mu$ is a completion regular
quasi-Radon
measure on $\Cdlg$ if we give $\Cdlg$ the topology of pointwise
convergent inherited from $\BbbR^{\coint{0,\infty}}$.   \cmmnt{Now the set}

\Centerline{$\Omega=\{\omega:\omega\in\Cdlg$ is non-decreasing,
$\omega(t)\in\Bbb N$ for every $t$ and $\omega(0)=0\}$}

\noindent is the support of $\ddot\mu$.   \prooflet{\Prf\
$\Omega$ is a closed subset of $\Cdlg$.   The formula in (a) tells us that
$\ddot\mu\{\omega:\omega(0)=0\}=1$ and

\Centerline{$\ddot\mu\{\omega:\omega(t)\in\Bbb N\}
=\sum_{k=0}^{\infty}e^{-t}\Bover{t^k}{k!}=1$}

\noindent for every $t>0$, while

\Centerline{$\ddot\mu\{\omega:\omega(s)\le\omega(t)\}
=e^{-t}\sum_{k_0=0}^{\infty}\sum_{k_1=0}^{\infty}
   \Bover{s^{k_0}}{k_0!}\,\Bover{(t-s)^{k_1}}{k_1!}=1$}

\noindent whenever $0<s<t$.   Thus $\Omega$ is expressible as the
intersection of a family of conegligible closed sets and is itself a
conegligible closed set.   If $G\subseteq\Cdlg$ is an open set meeting
$\Omega$, there are a $\tilde\omega\in G\cap\Omega$ and $t_0<\ldots<t_n$
such that $t_0=0$ and
$\{\omega:\omega(t_i)=\tilde\omega(t_i)$ for every $i\le n\}\subseteq G$;
in this case $\ddot\mu(\Omega\cap G)>0$.   So $\Omega$ is the support of
$\ddot\mu$.\ \Qed}

\spheader 612Uc Let $\mu$ be the subspace measure on $\Omega$ induced by
$\ddot\mu$ and $\Sigma$ its domain,
so that $\mu$ is a quasi-Radon probability measure on
$\Omega$\cmmnt{ (415B)}.   For $t\ge 0$, set

\Centerline{$\ddot\Sigma_t=\{F:F\in\dom\ddot\mu$, $\omega'\in F$ whenever
$\omega'\in\Cdlg$, $\omega\in F$ and
$\omega'\restr[0,t]=\omega\restr[0,t]\}$}

\noindent (see 455O) and

\Centerline{$\hat{\ddot\Sigma}_t
=\{F\symmdiff A:F\in\ddot\Sigma_t$, $\ddot\mu A=0\}$.}

\noindent Then $\hat{\ddot\Sigma}_t=\bigcap_{s>t}\hat{\ddot\Sigma}_s$ for
every $t$\cmmnt{ (455T)}.   So if\cmmnt{ we set}

\Centerline{$\Sigma_t=\{F:F\subseteq\Omega$, $F\in\hat{\ddot\Sigma}_t\}$}

\noindent for $t\ge 0$, $\langle\Sigma_t\rangle_{t\ge 0}$ will be a
filtration of $\sigma$-algebras.   Consequently, if we
take $(\frak A,\bar\mu)$ to be the measure algebra of $\mu$, and set
$\frak A_t=\{F^{\ssbullet}:F\in\Sigma_t\}$ for each $t$,
$\langle\frak A_t\rangle_{t\ge 0}$ will be a filtration of closed subalgebras of
$\frak A$.

\spheader 612Ud For $\omega\in\Omega$ and $t\ge 0$ set
$X_t(\omega)=\omega(t)$.
Then $X_t$ has a Poisson distribution with expectation $t$.
Now $\langle X_t\rangle_{t\ge 0}$ is progressively measurable\cmmnt{
(4A3Q(c-i) again)}.
We\cmmnt{ therefore} have a corresponding fully adapted process
$\pmb v=\family{\tau}{\Cal T_f}{v_{\tau}}$\cmmnt{ defined as in 612Hb};
in this volume I will call this the {\bf standard Poisson process}.

\spheader 612Ue{\bf (i)} For each $n\in\Bbb N$ and $\omega\in\Omega$, set

\Centerline{$g_n(\omega)
=\inf\{t:t\in\coint{0,\infty}$, $\omega(t)\ge n\}$,}

\noindent counting $\inf\emptyset$ as $\infty$.   Then $g_0(\omega)=0$ for
every $\omega$.
If $g_n(\omega)$ is finite, then $\omega(g_n(\omega))\ge n$\cmmnt{,
because $\omega$ is \cadlag}.   \cmmnt{Note that

\Centerline{$\mu\{\omega:g_n(\omega)>t\}
\le\Pr(X_t<n)
=e^{-t}\sum_{i=0}^{n-1}\Bover{t^i}{i!}
\to 0$}

\noindent as $t\to\infty$, so} $g_n$ is finite a.e.
\cmmnt{Of course}
$g_n\le g_{n+1}$ for every $n$.   In fact, for almost every $\omega$,
$\sequencen{g_n(\omega)}$ is strictly increasing.   \prooflet{\Prf\
If $0\le s<t$, then

\Centerline{$\mu\{\omega:\omega(t)-\omega(s)\ge 2\}
\le 1-e^{-(t-s)}(1+t-s)\le(t-s)^2$.}

Suppose that $\gamma>0$, $n\in\Bbb N$, $m\ge 1$ and $\omega\in\Omega$ are
such that $g_n(\omega)=g_{n+1}(\omega)\le\gamma$.
Set $t_i=\Bover{i\gamma}m$ for $i\le m$.   Then there is a first $i\le m$ such
that $g_n(\omega)\le t_i$, that is, $\omega(t_i)\ge n$.   Since
$g_{n+1}(\omega)\le t_i$, $\omega(t_i)\ge n+1$.   By the definition of $\Omega$,
$\omega(t_0)=0$, so $i>0$ and we can speak of $\omega(t_{i-1})$, which must be
less than $n$.   But $\omega(t_{i-1})$ is an integer, so it is at most $n-1$,
and $\omega(t_i)\ge\omega(t_{i-1})+2$.

This shows that

$$\eqalign{&\mu\{\omega:g_n(\omega)=g_{n+1}(\omega)\le\gamma\}\cr
&\mskip100mu
\le\mu\{\omega:\text{there is an }i\text{ such that }1\le i\le m\text{ and }
   \omega(t_i)-\omega(t_{i-1})\ge 2\}\cr
&\mskip100mu
\le m(\Bover{\gamma}m)^2
=\Bover{\gamma^2}m.\cr}$$

\noindent Letting $m\to\infty$, we see that
$\{\omega:g_n(\omega)=g_{n+1}(\omega)\le\gamma\}$ is negligible;  letting
$\gamma\to\infty$, $\{\omega:g_n(\omega)=g_{n+1}(\omega)\}$ is negligible
and its complement is conegligible.\ \Qed}

Observe that, for any $n\in\Bbb N$,
$\omega(g_n(\omega))=n$ for almost every $\omega$.
\prooflet{\Prf\ If $g_n(\omega)<g_{n+1}(\omega)$, then
$n\le\omega(g_n(\omega))<n+1$ so $\omega(g_n(\omega))=n$;  and this is the
case for almost every $\omega$.\ \Qed}

\medskip

\quad{\bf (ii)} For $n\in\Bbb N$\cmmnt{ and $t>0$},\cmmnt{

\Centerline{$\{\omega:g_n(\omega)<t\}
=\{\omega:\text{ there is a rational }q<t\text{ such that }\omega(q)\ge n\}
\in\Sigma_t$,}

\noindent so} $g_n$ is a stopping time adapted to
\cmmnt{$\langle\Sigma_t^+\rangle_{t\ge 0}$ (455Lb), which here is just}
$\langle\Sigma_t\rangle_{t\ge 0}$.
Let $\tau_n=g_n^{\ssbullet}$ be the corresponding stopping time in
$\Cal T_f$.   \cmmnt{Since
$\lim_{n\to\infty}g_n(\omega)=\infty$ for every $\omega$,
$\sup_{n\in\Bbb N}\Bvalue{\tau_n>t}=1$ for every $t$, and}
$\sup_{n\in\Bbb N}\tau_n=\max\Cal T$.
\cmmnt{Since $X_{g_n(\omega)}(\omega)=n$
for almost every $\omega$,} $v_{\tau_n}=n\chi 1$, for every $n\in\Bbb N$.

\spheader 612Uf If $\tau\in\Cal T_f$, then

\Centerline{$\Bvalue{v_{\tau}\in\Bbb N}=1$,
\quad$\Bvalue{v_{\tau}=v_{\tau_n}}
=\Bvalue{v_{\tau}=n}
=\Bvalue{\tau_n\le\tau}\Bcap\Bvalue{\tau<\tau_{n+1}}$ for every
$n\in\Bbb N$.}

\prooflet{\noindent\Prf\ As in 612H, we have a stopping time
$h:\Omega\to\coint{0,\infty}$
representing $\tau$, and $v_{\tau}=X_h^{\ssbullet}$.   Now
$X_{h(\omega)}(\omega)=\omega(h(\omega))\in\Bbb N$ for
every $\omega$, so
$\Bvalue{v_{\tau}\in\Bbb N}=1$.   For any particular $n\in\Bbb N$,

$$\eqalign{\Bvalue{v_{\tau}=n}
&=\{\omega:\omega(h(\omega))=n\}^{\ssbullet}\cr
&=\{\omega:g_n(\omega)\le h(\omega)<g_{n+1}(\omega)\}^{\ssbullet}
=\Bvalue{\tau_n\le\tau}\Bcap\Bvalue{\tau<\tau_{n+1}}.  \text{ \Qed}\cr}$$
}

\spheader 612Ug\cmmnt{ Because every $\omega$ is non-negative and
non-decreasing,
$0\le X_g\le X_h$ whenever $g$, $h$ are stopping times and $g\le h$,
and $0\le v_{\sigma}\le v_{\tau}$ whenever $\sigma\le\tau$ in $\Cal T_f$.
It follows immediately that} $\pmb v$ is locally order-bounded.

%\leader{612N}{}
%\leader{612O}{}
%\leader{612V}{}
%\leader{612W}{}

\exercises{\leader{612X}{Basic exercises (a)}
%\spheader 612Xa
In 612A(d-iv), show that if $u\ge 0$ then
$h\mapsto\bar h(u):H\to L^0$ is order-preserving and sequentially
order-continuous.
%612A

\spheader 612Xb
Suppose that $\Cal S$ is a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process.   Show that

\Centerline{$u_{\sigma}+u_{\tau}+u_{\upsilon}
=u_{\sigma\wedge\tau\wedge\upsilon}
+u_{\med(\sigma,\tau,\upsilon)}
+u_{\sigma\vee\tau\vee\upsilon}$}

\noindent for all $\sigma$, $\tau$, $\upsilon\in\Cal S$.
%612D

\spheader 612Xc
In 612Hb, show that we can have a progressively measurable
family $\langle X_t\rangle_{t\ge 0}$ such that $X_t\eae 0$ for every
$\tau$, but there is a $\tau\in\Cal T_b$ such that $u_{\tau}\ne 0$.
%612H

\spheader 612Xd Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$ a simple process with domain $\Cal S$.   Show that
$\pmb u\restr\Cal S\vee\tau=\pmb u\restr\Cal S\cap[\tau,\max\Cal T]$
and $\pmb u\restr\Cal S\cap[\tau,\tau']$ are simple processes whenever
$\tau\le\tau'$ in $\Cal S$.
%612K

\sqheader 612Xe Let $I$ be a finite sublattice of $\Cal T$ and
$\pmb u$ a fully adapted process with domain $I$.   Show that there is a
simple process with domain $\Cal T$ extending $\pmb u$ and with a
breakpoint string in $I$.   How far is this extension unique?
%612K

\sqheader 612Xg Suppose that $\Cal S$ is a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ is a fully adapted process, and
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Show that if $\pmb u$ is
order-bounded then $z\pmb u$ is order-bounded.
%612De 612J 614E

\spheader 612Xi Let $A\subseteq\Cal T$ be any set, and $\Cal S$ its covered
envelope (611Mb).   (i) Show that if $A$ is an ideal
of $\Cal T$ then $\Cal S$ is an ideal of $\Cal T$.
(ii) Show that if
$\pmb u=\family{\sigma}{A}{u_{\sigma}}
\in\prod_{\sigma\in A}L^0(\frak A_{\sigma})$ is such that
$\Bvalue{u_{\sigma}=u_{\tau}}\Bsupseteq\Bvalue{\sigma=\tau}$ for all
$\sigma$, $\tau\in A$, then there is a unique fully adapted process
$\hatpmb u$ with domain $\Cal S$ extending $\pmb u$.
(iii) In (ii), show that
$\|\hatpmb u\|_{\infty}=\sup_{\sigma\in A}\|u_{\sigma}\|_{\infty}$,
$\Bvalue{\hatpmb u\ne 0}=\sup_{\sigma\in A}\Bvalue{u_{\sigma}\ne 0}$ and that
$\hatpmb u$ is order-bounded iff $\{u_{\sigma}:\sigma\in A\}$ is an
order-bounded subset of $L^0$.
%612Q

\spheader 612Xj Suppose that $\Cal S$ is a sublattice of $\Cal T$,
$\hat\Cal S$ is its covered envelope, and that $\pmb u$ is a fully
adapted process with domain $\Cal S$ with corresponding extension
$\hatpmb u$ to a fully adapted process with domain $\hat\Cal S$.
Show that if $\pmb u$ is simple then $\hatpmb u$ is simple.
%612Q 612Xd used in 631G

\spheader 612Xk Show that if $\tau\in\Cal T$ then
$\family{\sigma}{\Cal T}{\chi\Bvalue{\sigma=\tau}}$ is a fully adapted process.
%612D out of order query

\spheader 612Xl Show that if $\Cal S$ is a sublattice of $\Cal T$ with covered
envelope $\hat\Cal S$, and $\pmb u$ is a non-decreasing fully adapted process
with domain $\Cal S$, then its fully adapted extension to $\hat\Cal S$ is
non-decreasing.
%612Q

\sqheader 612Xm Let $\Cal S$ be a sublattice of $\Cal T$, and $\tau$ a member of
$\Cal S$.   Suppose that
$\pmb u=\langle u_{\sigma}\rangle_{\sigma\in\Cal S\wedge\tau}$ and
$\pmb v=\langle v_{\sigma}\rangle_{\sigma\in\Cal S\vee\tau}$ are fully adapted
processes such that $u_{\tau}=v_{\tau}$.   Show that there is a unique
fully adapted process with domain $\Cal S$ extending both $\pmb u$ and
$\pmb v$.
%612K 612R used in 615O, probably elsewhere


\leader{612Y}{Further exercises (a)}
%\spheader 612Ya
Show that the ideas of 612H can be applied to any totally ordered set $T$
which is Polish in its order topology, in place of $\coint{0,\infty}$.
(First tackle the case $T=\Bbb N$ to establish rules when $T$ has gaps.
You may have to use $T$ or
$T\cup\{-\infty\}$ or $T\cup\{-\infty,\infty\}$ in place of $[0,\infty]$.)
%612H mt61x

\spheader 612Yc Let $\family{\sigma}{\Cal S}{u_{\sigma}}$ be a simple fully
additive process, and $\tau\in\Cal S$.   Show that
$\family{\sigma}{\Cal S}{u_{\sigma\wedge\tau}}$ is a simple fully
additive process.
%612K

\spheader 612Ye I will say that a fully adapted process
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ on a
non-empty sublattice $\Cal S$ of
$\Cal T$ is {\bf previsibly simple}
if there are $\tau_0\le\ldots\le\tau_n$ in $\Cal S$,
$u_i\in L^0(\frak A_{\tau_i})$ for $i\le n$ and a
$u_*\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$
such that, for every $\tau\in\Cal S$,

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

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

\noindent Formulate and prove results for previsibly simple processes
corresponding to the facts listed in 612K.
%612K

\spheader 612Yf Give an example in which $\Cal S$ is a sublattice of $\Cal T$
and $\pmb u$ is a fully adapted process with domain $\Cal S$ such that the
fully adapted extension of $\pmb u$ to the covered envelope of $\Cal S$ is
simple, but $\pmb u$ is not simple.
%612Xj
}%end of exercises

\endnotes{
\Notesheader{612} In 612H, I look at filtrations of $\sigma$-algebras of
measurable sets which are supposed all to contain every negligible set.
In the most natural representations of the most important stochastic
processes, the filtrations don't have this property;  see 477H.
An incidental advantage of working with measure algebras is that such
questions disappear until we turn to specific examples.

Many special spaces of fully adapted processes will be important in the
work below;  here I mention only the simple processes and the locally
order-bounded processes (612J, 612Xd, 614E),
as those which have descriptions accessible from our present position.
More interesting is the use of
`covering ideals' (611N, 612R).   We shall have many cases in which a
restriction to an
appropriate covering ideal will render a process more amenable -- e.g., by
making it a martingale.

You may have noticed that there is no mention of `measure' or `probability'
in this section except in the construction described in 612H, and there is
no `$\epsilon>0$' anywhere.
We are still working through the foothills, with
essentially algebraic arguments.   There are some suprema of infinite sets,
but in so far as there is anything non-trivial here, it is a reflection of
the work in \S364 and \S611.   `Simple' processes, however, demand a
bit of attention.   They are supposed to be a stochastic representation of
step-functions on $\Bbb R$ (226Xb, 242O) and will play a similar
role when we come to the theory of integration (see \S614), but the extra
complication of working on a lattice rather than on a totally ordered set
makes some essential points (e.g., 612Kc) trickier.

If you have spent any time with Volumes 1-5 of this treatise, you will know
that I consider a function to be inadequately defined if there is any doubt
about its domain.   This is a demanding discipline which is more important
in some places than others.   A point at which we can be relatively relaxed
is in the definition of integration in 613Hb,
where I shall insist only that $\dom\pmb u$ and
$\dom\psi$ should be large enough for $S_I(\pmb u,d\psi)$ to be defined
for every finite sublattice $I$ of $\Cal S$.   A point at which we have to
be more careful is in the definition of `simple process' in 612J, where
we have to know the exact domain of a process $\pmb u$ before we can
confirm that the proposed breakpoints belong to that domain.   Manoeuvres like
the proof of 612Qf will often be required.

Referring you to \S455 in the course of 612U is unkind.   It ought to be
much easier than this, and indeed it is.   You should have no real
difficulty in finding your own way to a proof of the really important bit,
which is that the formula in 612Ua defines a probability measure on the
space $\Omega$ of non-decreasing \cadlag\ functions from $\coint{0,\infty}$
to $\Bbb N$ starting at $0$.   But we are going to need the fact that
the filtration $\langle\Sigma_t\rangle_{t\ge 0}$ is right-continuous
(632Da below), and this seems to demand thought.   More generally, the processes
considered in \S455 furnish many other important examples for the theory here.

The construction $z\pmb u$ of 612De will be one of the leitmotivs of this
volume.   For the theory here, we can expect $z\pmb u$ to behave like a
scalar multiple of $\pmb u$;  in effect, if $\Cal S$ is a
non-empty sublattice of
$\Cal T$, the $f$-algebra $\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$
can be thought of as an
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$-module.
The idea is that a fully adapted process with domain $\Cal S$ is supposed
to represent the evolution of a system over time, and that
$\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$ is the algebra of
events observable from the beginning of the process;  so that if we think of
$z$ as a function rather than a member of $L^0$, its values are determinate
scalars, and any feature of the process preserved by scalar multiplication
ought to be preserved by multiplication by $z$.   I shall give a fair bit
of space, in total, to such calculations as 612Lc,
but they will nearly always be
elementary adaptations of ideas already indicated.
}%end of notes

\discrpage

\frfilename{mt613.tex}
\versiondate{15.7.20/12.10.22}

\def\zz{}
\def\Gint{\mathchoice{\hbox to 0pt{\hskip 3.1pt G\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm G}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm G}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm G}\hss}}
     \intop\nolimits}
\def\Mbv{\mathop{M_{\text{bv}}}}
\def\Mlbv{\mathop{M_{\text{lbv}}}}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Definition of the integral}

\newsection{613}

I come now to the definition of a stochastic integral which will be used
for the next three chapters.
We are looking for an effective way to interpret the formula
$\int_{\tau}^{\tau'}\pmb u\,d\pmb v$ where $\tau\le\tau'$ are stopping
times and $\pmb u$, $\pmb v$ are fully adapted processes defined on
an interval $[\tau,\tau']$ in $\Cal T$.
I will define this as a kind of Riemann-Stieltjes
integral, a limit of `Riemann sums' of the
form $\sum_{i=0}^nu_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})$ where
$\tau=\tau_0\le\ldots\le\tau_n=\tau'$.   For this we need a notion of
convergence, for which `convergence in measure'\cmmnt{ (\S\S245, 367)}
turns out to be suitable, and a particular limiting process, to be
described in 613Hb.   Because our processes are defined on a lattice
$\Cal T$ of stopping times, rather than a totally ordered set, there are
some technical obstacles to clear out of the way;  I aim to do this in
613C-613G.  %613C 613E 613G
The rest of the section is devoted to elementary properties of
this new integral.

\leader{613A}{Probability algebras (a)} For the rest of this volume,
$(\frak A,\bar\mu)$ will denote a probability algebra.
$L^1(\frak A,\bar\mu)$ or $L^1_{\bar\mu}$ will be its $L^1$ space\cmmnt{ as
described in \S365, a linear subspace of $L^0(\frak A)$}.
For $w$ in $L^0=L^0(\frak A)$,
I will write $\Expn(w)=\Expn_{\bar\mu}(w)=\Expn(w^+)-\Expn(w^-)$
for its integral with respect to $\bar\mu$\cmmnt{ as defined in 365D},
provided that at most one of $\Expn(w^+)$,
$\Expn(w^-)$ is infinite.   \cmmnt{(I am reserving the symbol $\int$ for
the stochastic integral to be defined in 613H.)}

\spheader 613Ab\cmmnt{ As in \S\S611-612,} $T$ will be a totally ordered set and
$\family{t}{T}{\frak A_t}$ a filtration of order-closed subalgebras of
$\frak A$.   \cmmnt{Recall from 316Fb and 323H that in
this context a subalgebra of $\frak A$ is order-closed iff it is
a $\sigma$-subalgebra of $\frak A$ iff it is
topologically closed in the measure-algebra
topology of $\frak A$, which is that of the measure
metric $(a,b)\mapsto\bar\mu(a\Bsymmdiff b)$;  so we can safely call such
subalgebras `closed' without specifying which aspect we have primarily
in mind.}   $\Cal T$ will be the set of stopping
times adapted to $\family{t}{T}{\frak A_t}$.  For $\tau\in\Cal T$,
$\frak A_{\tau}$ will be the closed subalgebra corresponding to
$\tau$\cmmnt{ (611G)}.   When I say that a process is `fully adapted'
I shall always mean that it is `fully adapted to
$\family{t}{T}{\frak A_t}$'.

\leader{613B}{Convergence in measure }\cmmnt{I begin with notes on two
fundamental concepts from probability theory.
The first is essential for the notion of `integral' to be defined in this
section.

\medskip

}{\bf (a)}
\cmmnt{Recall that }$L^0\cmmnt{\mskip5mu=L^0(\frak A)}$
now has a topology $\frak T$ of {\bf convergence in measure}
which can be defined by the F-norm $\theta$ where

\Centerline{$\theta(w)=\Expn(|w|\wedge\chi 1)$ for every
$w\in L^0$\dvro{.}{}}

\noindent\cmmnt{(245Da, 2A5B, 367L).   }This is a
complete Hausdorff linear space topology for which
multiplication and the lattice operations $\vee$, $\wedge$ and
$|\,\,|$ are
continuous\cmmnt{ (367M)}.
\cmmnt{Because $(\frak A,\bar\mu)$ is always
isomorphic to the measure algebra of some probability space (321J),
we can apply all the results proved in \S\S245-246 for the topology of
convergence
in measure on spaces $L^0(\mu)$, as well as those spelt out in \S367.}
In particular, the positive cone $(L^0)^+$ is closed.

\cmmnt{Concerning the functional $\theta$, it is subadditive\cmmnt{
(that is,

$$\eqalign{\theta(w_1+w_2)
&=\Expn(|w_1+w_2|\wedge\chi 1)
\le\Expn((|w_1|+|w_2|)\wedge\chi 1)\cr
&\le\Expn(|w_1|\wedge\chi 1+|w_2|\wedge\chi 1)
=\theta(w_1)+\theta(w_2)\cr}$$

\noindent for all $w_1$, $w_2\in L^0$)}.   It is
not a seminorm except in
trivial cases, but it does have the property that}
$\theta(\alpha w)\le\alpha\theta(w)$ if $w\in L^0$ and $\alpha\ge 1$.
\prooflet{\Prf\
$\Expn(\alpha|w|\wedge\chi 1)\le\Expn(\alpha|w|\wedge\alpha\chi 1)
=\alpha\Expn(|w|\wedge\chi 1)$.\ \Qed}   \cmmnt{Also, of course,}
$\theta(v)\le\theta(w)$ whenever $|v|\le|w|$\cmmnt{ (cf.\ 354A)}.
\cmmnt{Finally, because
$\Expn$ is order-continuous (365Da;  cf.\ 354Dc),}
$\lim_{w\downarrow A}\theta(w)=0$ whenever
$A\subseteq L^0$ is a
non-empty downwards-directed family with infimum $0$\cmmnt{ (367Na)}, so that
$\sup A\in\overline{A}$ and
$\lim_{w\uparrow A}\theta(w)=\theta(\sup A)$ whenever $A\subseteq L^0$
is a non-empty upwards-directed set with an upper bound in $L^0$;
similarly, if $A\subseteq L^0$
is a non-empty downwards-directed set with a lower bound in $L^0$,
$\lim_{w\downarrow A}\theta(w)=\theta(\inf A)$.

\spheader 613Bb If $E\subseteq\Bbb R$ is a Borel set and
$Q_E=\{u:u\in L^0$, $\Bvalue{u\in E}=1\}$, then for any continuous
$h:E\to\Bbb R$ the corresponding function $\bar h:Q_E\to L^0$ is
continuous\cmmnt{ (367S)}.

\spheader 613Bc If $1\le p\le\infty$, all the $\|\,\|_p$-balls
$\{u:u\in L^0$, $\|u\|_p\le\alpha\}$ are
$\frak T$-closed\cmmnt{ (245J(b-i), 245Xk)}.
Consequently the $\frak T$-closure
of a $\|\,\|_p$-bounded set is again $\|\,\|_p$-bounded, and
$\|\,\|_p:L^0\to[0,\infty]$ is lower semi-continuous\cmmnt { (4A2A)}.

\spheader 613Bd{\bf (i)} For any $p\in[1,\infty]$,
the embedding $L^p_{\bar\mu}\embedsinto L^0$ is
continuous for the norm topology of $L^p_{\bar\mu}$ and
$\frak T$\cmmnt{ (245G)}.

\medskip

\quad{\bf (ii)} If $A\subseteq L^1_{\bar\mu}$ is non-empty
and downwards-directed and $\inf A=0$ in $L^1_{\bar\mu}$, then
$\inf_{u\in A}\|u\|_1=\lim_{u\downarrow A}\|u\|_1=0$\cmmnt{ (365C)}.

\medskip

\quad{\bf (iii)} If
$A\subseteq(L^1_{\bar\mu})^+$ is non-empty and upwards-directed and
$\gamma=\sup_{u\in A}\|u\|_1$ is finite, then $A$ is bounded above in
$L^1_{\bar\mu}$, $\sup A$ belongs to the $\|\,\|_1$-closure of $A$
and $\|\sup A\|_1=\gamma$\cmmnt{ (365C again)}.

\medskip

\quad{\bf (iv)} If $u\in L^1_{\bar\mu}$ and $\epsilon>0$, there is a
$\delta>0$ such that $\|u-v\|_1\le\epsilon$ whenever $v\in L^1_{\bar\mu}$,
$\|v\|_1\le\|u\|_1+\delta$ and $\theta(u-v)\le\delta$\cmmnt{ (245H(b-i))}.

\spheader 613Be If $A\subseteq L^0$ and $v\in\overline{A}$ then
$\Bvalue{v>\alpha}\subseteq\sup_{u\in A}\Bvalue{u>\alpha}$
for every $\alpha\in\Bbb R$.   \prooflet{\Prf\ Setting
$a=1\Bsetminus\sup_{u\in A}\Bvalue{u>\alpha}$ we see that
$u\times\chi a\le\alpha\chi a$ for every $u\in A$, so
$v\times\chi a\le\alpha\chi a$
and $\Bvalue{v>\alpha}$ does not meet $a$.\ \Qed}

\spheader 613Bf{\bf (i)}\cmmnt{ Because
$\frak T$ is a linear space topology, there is a
corresponding notion of bounded set
(3A5N).}   I will say that a set $A\subseteq L^0$ is
{\bf topologically bounded} if for every neighbourhood $G$ of $0$ in $L^0$
there is an $n\in\Bbb N$ such that $A\subseteq nG$;  equivalently, if for
every $\epsilon>0$ there is a $\delta>0$ such that
$\theta(\delta u)\le\epsilon$ for every $u\in A$.

\medskip

\quad{\bf (ii)} If $A\subseteq L^0$ is non-empty, then $A$ is topologically
bounded iff $\inf_{\gamma>0}\sup_{u\in A}\bar\mu\Bvalue{|u|>\gamma}=0$.
\prooflet{\Prf\grheada\ If $A$ is topologically bounded and $0<\epsilon\le 1$,
let $\delta>0$ be such that $\theta(\delta u)\le\epsilon^2$ for every $u\in A$.
If $u\in A$ then

\Centerline{$\epsilon\bar\mu\Bvalue{|u|>\bover1{\delta}}
=\epsilon\bar\mu\Bvalue{|\delta u|>1}
\le\theta(\delta u)\le\epsilon^2$}

\noindent so $\bar\mu\Bvalue{|u|>\bover1{\delta}}\le\epsilon$.
As $\epsilon$ is
arbitrary, $\inf_{\gamma>0}\sup_{u\in A}\bar\mu\Bvalue{|u|>\gamma}=0$.
\grheadb\ If $\inf_{\gamma>0}\sup_{u\in A}\bar\mu\Bvalue{|u|>\gamma}=0$,
take $\epsilon>0$.   Let $\gamma>0$ be such that
$\bar\mu\Bvalue{|u|>\gamma}\le\epsilon$ for every $u\in A$, and set
$\delta=\Bover{\epsilon}{\gamma}$.   If $u\in A$, then

\Centerline{$\theta(\delta u)
\le\epsilon+\bar\mu\Bvalue{|\delta u|>\epsilon}
\le\epsilon+\bar\mu\Bvalue{|u|>\gamma}
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $A$ is topologically bounded.\ \Qed}

\medskip

\quad{\bf (iii)} If $A$, $B\subseteq L^0$ are topologically bounded, so are
$A+B$ and $\alpha A$ for any
$\alpha\in\Bbb R$, the closure $\overline{A}$
of $A$ for the topology of convergence in measure\cmmnt{ (3A5Nb)}, and any
subset of $A$.

\medskip

\quad{\bf (iv)} If $A\subseteq L^0$ is topologically bounded, so is its
{\bf solid hull} $\{u:u\in L^0,\,\Exists v\in A$, $|u|\le|v|\}$.
\prooflet{(For if $|u|\le|v|$, then $\theta(\delta u)\le\theta(\delta v)$.)}
In particular, an order-bounded subset of $L^0$ is topologically bounded.

\medskip

\quad{\bf (v)} An
upwards-directed topologically bounded set is bounded above.
\prooflet{\Prf\ If $A\subseteq L^0$ is an upwards-directed set which is not
bounded above, then $c=\inf_{n\in\Bbb N}\sup_{u\in A}\Bvalue{u>n}$ is
non-zero (364La).    If $n\in\Bbb N$, then
$\{c\Bcap\Bvalue{u>n}:u\in A\}$ is upwards-directed and has supremum $c$,
so there is a $u_n\in A$ such that
$\bar\mu(c\Bsetminus\Bvalue{u_n>n})\le 2^{-n-2}\bar\mu c$.
Set $d=c\Bsetminus\sup_{n\in\Bbb N}\Bvalue{u_n>n}$;  then
$\bar\mu d>0$.   But now observe that $\theta(\delta u_{n+1})\ge\bar\mu d$
whenever $n\in\Bbb N$ and $\delta\ge\bover1{n+1}$, so $A$ is not
topologically bounded.\ \Qed}

\medskip

\quad{\bf (vi)} If $A\subseteq L^0$ is solid, so is
$\overline{A}$.   \prooflet{\Prf\ If $v\in\overline{A}$, $|u|\le |v|$ and
$\epsilon>0$, there is a $v'\in A$ such that $\theta(v'-v)\le\epsilon$;
now $u'=\med(-|v'|,u,|v'|)$ belongs to $A$, while
$u=\med(-|v|,u,|v|)$, so $|u-u'|\le|v-v'|$ and $\theta(u-u')\le\epsilon$.
As $\epsilon$ is arbitrary, $u\in\overline{A}$;  as $u$ and $v$ are
arbitrary, $\overline{A}$ is solid.\ \Qed}

\spheader 613Bg If $\bar\nu:\frak A\to[0,1]$ is any functional such that
$(\frak A,\bar\nu)$ is a probability algebra, then $\bar\mu$ and $\bar\nu$
are mutually absolutely continuous, that is,

\inset{----- for every $\epsilon>0$ there is a $\delta>0$ such that
$\max(\bar\mu a,\bar\nu a)\le\epsilon$ whenever $a\in\frak A$ and
$\min(\bar\mu a,\bar\nu a)\le\delta$\cmmnt { (393F)}.}

\noindent\cmmnt{(Compare 232Ba.)   Consequently }$\frak T$ is still the
topology of convergence in measure on $L^0$ if we apply the formulae of (a)
with the integral $\Expn_{\bar\nu}$ defined from $\bar\nu$ in place of
$\Expn=\Expn_{\bar\mu}$\cmmnt{ (see 367T)}, and if we set
$\theta_{\bar\nu}(w)=\Expn_{\bar\nu}(|w|\wedge\chi 1)$ for $w\in L^0$, then

\inset{----- for every $\epsilon>0$ there is a $\delta>0$ such that
$\max(\theta_{\bar\mu}(w),\theta_{\bar\nu}(w))\le\epsilon$
whenever $\min(\theta_{\bar\mu}(w),\theta_{\bar\nu}(w))\le\delta$.}

\cmmnt{This will be a recurring theme in the rest of this volume, so}
I introduce a code
phrase\cmmnt{ here}:
the topology of convergence in measure is {\bf law-independent},
since replacing the `law' $\bar\mu$ by the law $\bar\nu$ leaves it unchanged.


\spheader 613Bh $(L^0,\theta)$ is a complete metric space;  that is,
$L^0$ is complete when regarded as a linear topological
space\prooflet{ (367Mc)}.

\spheader 613Bi Now suppose that $\frak B$ is a
closed subalgebra of $\frak A$.

\medskip

\quad{\bf (i)} $L^0(\frak B)$,
regarded as a subset of $L^0(\frak A)$\cmmnt{ (612Ae)},
is closed for the topology of
convergence in measure\cmmnt{ (367Rc)}.

\medskip

\quad{\bf (ii)}\cmmnt{ As the embedding $L^1_{\bar\mu}\embedsinto L^0(\frak A)$
is continuous for the norm topology of $L^1_{\bar\mu}$ ((d-i) above),}
$L^1_{\bar\mu}\cap L^0(\frak B)$ is $\|\,\|_1$-closed in $L^1_{\bar\mu}$;
\cmmnt{being a linear subspace,} it is also closed for the weak topology of
$L^1_{\bar\mu}$\cmmnt{ (4A4Ed)}.

\spheader 613Bj\cmmnt{ The following is a useful consequence of (i).}
Suppose that $\Cal S$ is a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ is a
fully adapted process, and $A\subseteq\Cal S$ is a non-empty
downwards-directed
set such that the limit $u_*=\lim_{\sigma\downarrow A}u_{\sigma}$ is
defined.   Then
$u_*\in\bigcap_{\sigma\in A}L^0(\frak A_{\sigma})
=L^0(\bigcap_{\sigma\in A}\frak A_{\sigma})$.   \prooflet{\Prf\ If
$\sigma\in A$, then $u_*$ belongs to the closure of
$\{u_{\sigma'}:\sigma'\in A$, $\sigma'\le\sigma\}
\subseteq L^0(\frak A_{\sigma})$;  as $L^0(\frak A_{\sigma})$ is closed,
$u_*\in L^0(\frak A_{\sigma})$.   As $\sigma$ is arbitrary,
$u_*\in\bigcap_{\sigma\in A}L^0(\frak A_{\sigma})$.   For the other expression,
write $\frak B$ for $\bigcap_{\sigma\in A}\frak A_{\sigma}$;  being the
intersection of closed subalgebras of $\frak A$, $\frak B$ is a closed
subalgebra of $\frak A$.   Now, for $v\in L^0$,

$$\eqalign{v\in L^0(\frak B)
&\iff\Bvalue{v>\alpha}\in\frak B\text{ for every }\alpha\in\Bbb R\cr
&\iff\Bvalue{v>\alpha}\in\frak A_{\sigma}\text{ for every }\alpha\in\Bbb R
  \text{ and }\sigma\in A\cr
&\iff v\in L^0(\frak A_{\sigma})\text{ for every }\sigma\in A,\cr}$$

\noindent so $L^0(\frak B)=\bigcap_{\sigma\in A}L^0(\frak A_{\sigma})$, and in
particular contains $u_*$.\ \Qed}

\spheader 613Bk\cmmnt{ A particular manifestation of the idea in (j) above
appears often enough to be given a name.}
If $\Cal S$ is non-empty and $\pmb u=\sfamilysS{u}$
is a fully adapted process such that the topological limit
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined
in $L^0(\frak A)$, I will call
$u_{\downarrow}
\cmmnt{\mskip5mu\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})}$
the {\bf starting value} of $\pmb u$.

\spheader 613Bl If $\Cal S$ is a sublattice of $\Cal T$, then we can give
$(L^0)^{\Cal S}$ its product topology, under which it is a linear
topological space\cmmnt{ (4A4Bb)}.   Now the space\cmmnt{ $\Mfa(\Cal S)$}
of fully adapted processes with domain $\Cal S$ is a closed subspace of
$(L^0)^{\Cal S}$.   \prooflet{\Prf\

$$\eqalign{\Mfa(\Cal S)
&=(L^0)^{\Cal S}\cap\bigcap_{\tau,\tau'\in\Cal S}
   \{\sfamilysS{u}:\sfamilysS{u}\in(L^0)^{\Cal S},\,\cr
&\mskip200mu
u_{\tau}\in L^0(\frak A_{\tau}),\,
      \Bvalue{\tau=\tau'}\Bsubseteq\Bvalue{u_{\tau}=u_{\tau'}}\}.\cr}$$

\noindent Accordingly $L^0(\frak A_{\tau})$ is closed in $L^0$ for every
$\tau$, by
(i-i) above.   Moreover, if $\tau$, $\tau'\in\Cal S$ and $a\in\frak A$, then

\Centerline{$\{(u,v):u,\,v\in L^0,\,a\Bsubseteq\Bvalue{u=v}\}
=\{(u,v):u,\,v\in L^0,\,u\times\chi a=v\times\chi a\}$}

\noindent is closed in $(L^0)^2$ because multiplication in $L^0$ is
continuous and the topology of $L^0$ is Hausdorff.   So $\Mfa(\Cal S)$ is an
intersection of closed sets and is closed.\ \Qed}

\spheader 613Bm Because the lattice operations on $L^0$ are continuous,
and the topology is Hausdorff, sets of the form
$\{|u|:u\le\bar u\}=\{u:|u|\vee\bar u=\bar u\}$ are closed for any
$\bar u\in L^0$.   Consequently, in a product space $(L^0)^{\Cal S}$, sets
of the form $\{\pmb u:\pmb u\in(L^0)^{\Cal S}$,
$|\pmb u|\le\bar{\pmb u}\}$, where $\bar{\pmb u}\in(L^0)^{\Cal S}$,
are closed for the product topology.


\spheader 613Bn Now suppose that $(\frak B,\bar\nu)$ is another probability
algebra, and $\phi:\frak A\to\frak B$ is a measure-preserving Boolean
homomorphism.   Then\cmmnt{ $\phi$ is order-continuous (324Kb), so}
we have a corresponding injective $f$-algebra homomorphism
$T_{\phi}:L^0(\frak A)\to L^0(\frak B)$\cmmnt{ (612Af)}.   Writing
$\Expn_{\bar\mu}$, $\Expn_{\bar\nu}$ for expectations in
$L^1_{\bar\mu}$, $L^1_{\bar\nu}$ respectively, and $\theta_{\bar\mu}$,
$\theta_{\bar\nu}$ for the corresponding functionals on $L^0(\frak A)$ and
$L^0(\frak B)$, $\Expn_{\bar\nu}(T_{\phi}u)=\Expn_{\bar\mu}(u)$ for every
$u\in L^1_{\bar\mu}$\cmmnt{ (365Nb\formerly{3{}65Ob})};
\cmmnt{as $T_{\phi}$ is a Riesz
homomorphism and $T_{\phi}(\chi 1_{\frak A})=\chi 1_{\frak B}$,}
$\theta_{\bar\nu}(T_{\phi}u)=\theta_{\bar\mu}(u)$ for every
$u\in L^0(\frak A)$, and $T_{\phi}$ is continuous for the topologies of
convergence in measure.

\spheader 613Bo For any $\alpha\in\Bbb R$, the function
$u\mapsto\bar\mu\Bvalue{u>\alpha}:L^0\to[0,1]$ is lower semi-continuous.
\prooflet{\Prf\ Suppose that $u\in L^0=L^0(\frak A)$
and $\bar\mu\Bvalue{u>\alpha}>t$.
Then there is a $\delta\in\ocint{0,1}$ such that
$\bar\mu\Bvalue{u>\alpha+\delta}>t+\delta$.   Suppose that $v\in L^0$ is such
that $\theta(v-u)\le\delta^2$.   Then $\bar\mu\Bvalue{|v-u|\ge\delta}\le\delta$.
Now

\Centerline{$\Bvalue{v>\alpha}
\Bsupseteq\Bvalue{u>\alpha+\delta}\Bcap\Bvalue{v-u\ge-\delta}$,}

\Centerline{$\bar\mu\Bvalue{v>\alpha}
\ge\bar\mu\Bvalue{u>\alpha+\delta}-\bar\mu\Bvalue{|v-u|>\delta}>t$.}

\noindent This shows that $\{u:\bar\mu\Bvalue{u>\alpha}>t\}$ is open;
as $t$ is arbitrary,
$u\mapsto\bar\mu\Bvalue{u>\alpha}:L^0\to[0,1]$ is lower semi-continuous.\
\Qed}

\spheader 613Bp{\bf (i)}
Suppose that $A\subseteq L^0$ and that for every $\epsilon>0$
there is an $a\in\frak A$ such that $\{u\times\chi a:u\in A\}$ is order-bounded
in $L^0$ and $\bar\mu a\ge 1-\epsilon$.   Then $A$ is order-bounded in $L^0$.
\prooflet{\Prf\ For each $n\in\Bbb N$ set
$c_n=\sup_{u\in A}\Bvalue{|u|>n}$.   Then $\sequencen{c_n}$ is
a non-increasing sequence in $\frak A$.   Given $\epsilon>0$,
there is an $a\in\frak A$ such that
$\bar\mu a\ge 1-\bover12\epsilon$ and $\bar u=\sup_{u\in A}|u\times\chi a|$ is
defined.   Let $n\in\Bbb N$ be such that $b=\Bvalue{\bar u>n}$ has measure at
most $\bover12\epsilon$.   If $u\in A$,

\Centerline{$\Bvalue{|u|>n}
\Bsubseteq(1\Bsetminus a)\Bcup\Bvalue{|u\times\chi a|>n}
\Bsubseteq(1\Bsetminus a)\Bcup b$.}

\noindent So if $m\ge n$,

\Centerline{$c_m\Bsubseteq c_n\Bsubseteq(1\Bsetminus a)\Bcup b$,
\quad$\bar\mu c_m\le(1-\bar\mu a)+\bar\mu b\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lim_{n\to\infty}\bar\mu c_n=0$.

Consequently $\inf_{n\in\Bbb N}c_n=0$ and $\{|u|:u\in A\}$ is
bounded above in $L^0$ (364L(a-i)), that is, $A$ is order-bounded in
$L^0$.\ \Qed}

\medskip

\quad{\bf (ii)} If $\pmb u=\sfamilysS{u}$ is a fully adapted process
and for every $\epsilon>0$ there is an order-bounded process
$\pmb v=\sfamilysS{v}$ such that $\bar\mu\Bvalue{\pmb u\ne\pmb v}\le\epsilon$,
then $\pmb u$ is order-bounded.   \prooflet{\Prf\ Set
$A=\{u_{\sigma}:\sigma\in\Cal S\}$.   If $\epsilon>0$, there is an order-bounded
process $\pmb v=\sfamilysS{v}$ such that
$\bar\mu\Bvalue{\pmb u\ne\pmb v}\le\epsilon$.   Set
$a=1\Bsetminus\Bvalue{\pmb u\ne\pmb v}$.   Then $\bar\mu a\ge 1-\epsilon$ and

\Centerline{$|u_{\sigma}\times\chi a|=|v_{\sigma}\times\chi a|\le\sup|\pmb v|$}

\noindent for every $\sigma\in\Cal S$;  so $\{u\times\chi a:u\in A\}$ is
order-bounded in $L^0$.   By (i) just above, $A$ is order-bounded in $L^0$,
that is, $\pmb u$ is order-bounded.\ \Qed}

\leader{613C}{Interval functions}\cmmnt{ Now for a new idea.

\medskip

} {\bf (a)} Let $\Cal S$ be a sublattice of $\Cal T$.
I will write $\Cal S^{2\uparrow}$ for
$\{(\sigma,\tau):\sigma$, $\tau\in\Cal S$, $\sigma\le\tau\}$.

\medskip

\quad{\bf (i)} I
say that a function $\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ is an
{\bf adapted interval function} on $\Cal S$ if

\Centerline{$\psi(\sigma,\tau)\in L^0(\frak A_{\tau})$,
\quad$\psi(\sigma,\sigma)=0$,
\quad$b\Bsubseteq\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}$}

\noindent whenever $\sigma\le\sigma'\le\tau'\le\tau$ in $\Cal S$,
$b\in\frak A_{\sigma}$
and $b\Bsubseteq\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}$.

\medskip

\quad{\bf (ii)} In this case, I say that $\psi$ is a
{\bf strictly adapted interval function} if

\Centerline{$\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}$}

\noindent whenever $\sigma\le\sigma'\le\tau'\le\tau$ in $\Cal S$.

\spheader 613Cb Let $\Cal S$ be a sublattice of $\Cal T$ and
$\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ an adapted interval function.

\medskip

\quad{\bf (i)} $\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{\psi(\sigma,\tau)=0}$
whenever
$\sigma\le\tau$ in $\Cal S$.   \prooflet{\Prf\ Setting $\sigma'=\tau'=\sigma$,
we have

\Centerline{$\Bvalue{\sigma=\tau}
=\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}\in\frak A_{\sigma}$}

\noindent so

\Centerline{$\Bvalue{\sigma=\tau}
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=0}$.  \Qed}}

\noindent\cmmnt{Taking complements,}
$\Bvalue{\psi(\sigma,\tau)\ne 0}\Bsubseteq\Bvalue{\sigma<\tau}$
whenever $\sigma\le\tau$ in $\Cal S$.

\medskip

\quad{\bf (ii)} $\psi(\sigma,\sigma\vee\tau)=\psi(\sigma\wedge\tau,\tau)$ for
all $\sigma$, $\tau\in\Cal S$.   \prooflet{\Prf\
$\Bvalue{\sigma\le\tau}\in\frak A_{\sigma\wedge\tau}$ (611Hc) and

$$\eqalign{\Bvalue{\sigma\le\tau}&
=\Bvalue{\sigma\wedge\tau=\sigma}\Bcap\Bvalue{\tau=\sigma\vee\tau}\cr&
\Bsubseteq
\Bvalue{\psi(\sigma\wedge\tau,\sigma\vee\tau)=\psi(\sigma,\sigma\vee\tau)}
 \Bcap\Bvalue{\psi(\sigma\wedge\tau,\sigma\vee\tau)=\psi(\sigma\wedge\tau,\tau)}
 \cr&
\Bsubseteq
\Bvalue{\psi(\sigma,\sigma\vee\tau)=\psi(\sigma\wedge\tau,\tau)},\cr}$$

$$\eqalign{\Bvalue{\tau\le\sigma}&
=\Bvalue{\sigma\wedge\tau=\tau}\Bcap\Bvalue{\sigma=\sigma\vee\tau}\cr&
\Bsubseteq
\Bvalue{\psi(\sigma\wedge\tau,\tau)=0}
 \Bcap\Bvalue{\psi(\sigma,\sigma\vee\tau)=0}
\Bsubseteq
\Bvalue{\psi(\sigma,\sigma\vee\tau)=\psi(\sigma\wedge\tau,\tau)};\cr}$$

\noindent putting these together,
$\Bvalue{\psi(\sigma,\sigma\vee\tau)=\psi(\sigma\wedge\tau,\tau)}=1$
and $\psi(\sigma,\sigma\vee\tau)=\psi(\sigma\wedge\tau,\tau)$.\ \Qed}

\medskip

\quad{\bf (iii)} If $\Cal S_0$ is any sublattice of $\Cal S$, then
$\psi\restr\Cal S_0^{2\uparrow}$ is an adapted interval function on
$\Cal S_0$, and is strictly adapted if $\psi$ is.   \prooflet{(Immediate
from the definitions in (a).)}

\medskip

\quad{\bf (iv)} If $\psi$ is strictly adapted then
$\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}$ whenever
$\sigma\le\tau$ and $\sigma'\le\tau'$ in $\Cal S$.   \prooflet{\Prf\

$$\eqalign{\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}&
=\Bvalue{\sigma\wedge\sigma'=\sigma}\Bcap\Bvalue{\tau=\tau\vee\tau'}
  \Bcap\Bvalue{\sigma\wedge\sigma'=\sigma'}\Bcap\Bvalue{\tau'=\tau\vee\tau'}\cr&
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=\psi(\sigma\wedge\sigma',\tau\vee\tau')}
  \Bcap\Bvalue{\psi(\sigma',\tau')=\psi(\sigma\wedge\sigma',\tau\vee\tau')}\cr&
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}.  \text{ \Qed}\cr}$$
}

\spheader 613Cc\cmmnt{ Much the most important
examples of such functions are the following.}  If $\Cal S\subseteq\Cal T$ is a
sublattice and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ is a fully adapted process,
we have a function
$\Delta\pmb v:\Cal S^{2\uparrow}\to L^0(\frak A)$
defined by saying that

\Centerline{$(\Delta\pmb v)(\sigma,\tau)=v_{\tau}-v_{\sigma}$}

\noindent whenever $\sigma\le\tau$ in $\Cal S$, and $\Delta\pmb v$ is a
strictly adapted interval function on $\Cal S$.   \prooflet{\Prf\
If $\sigma\le\sigma'\le\tau'\le\tau$ in $\Cal S$, then
$v_{\tau}-v_{\sigma}\in L^0(\frak A_{\tau})$ because $\sigma\le\tau$,
$\Bvalue{v_{\tau}-v_{\sigma}\ne 0}\Bsubseteq\Bvalue{\sigma\ne\tau}
=\Bvalue{\sigma<\tau}$, and

\Centerline{$\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}
\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma'}}\Bcap\Bvalue{v_{\tau'}=v_{\tau}}
\Bsubseteq\Bvalue{v_{\tau}-v_{\sigma}=v_{\tau'}-v_{\sigma'}}$.  \Qed}}

\leader{613D}{Constructions for interval functions} Let $\Cal S$
be a sublattice of $\Cal T$
and $\psi$, $\psi'$ (strictly) adapted interval functions on $\Cal S$.

(a) $\psi+\psi'$ and $\psi\times\psi'$ are (strictly) adapted interval
functions.

(b) If $h:\Bbb R\to\Bbb R$ is a Borel measurable function, then the
composition $\bar h\psi$ is a (strictly) adapted interval function.

(c)\cmmnt{ In particular,} $\psi^2$ and $|\psi|$ and $\alpha\psi$,
for any $\alpha\in\Bbb R$, are (strictly) adapted interval functions;
the space of (strictly)
adapted interval functions on $\Cal S$ is an $f$-subalgebra of
$L^0(\frak A)^{\Cal S^{2\uparrow}}$.

(d) If $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$
is a fully adapted process,
then we have a (strictly) adapted interval function $\pmb u\psi$ on $\Cal S$
defined by setting
$(\pmb u\psi)(\sigma,\tau)=u_{\sigma}\times\psi(\sigma,\tau)$
whenever $\sigma\le\tau$ in $\Cal S$.

\proof{{\bf (a)-(c)} are immediate consequences of the definitions in
613Ca.

\medskip

{\bf (d)} Write $\psi'(\sigma,\tau)=u_{\sigma}\times\psi(\sigma,\tau)$.
If $\sigma\le\tau$ in $\Cal S$, then
$\psi'(\sigma,\tau)\in L^0(\frak A_{\tau})$ because
$u_{\sigma}$ and $\psi(\sigma,\tau)$ both belong to $L^0(\frak A_{\tau})$; and

\Centerline{$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{\psi(\sigma,\tau)=0}
\Bsubseteq\Bvalue{\psi'(\sigma,\tau)=0}$.}

\noindent Next, as in 613Cc, take $\sigma$, $\tau$, $\sigma'$, $\tau'\in\Cal S$
such that $\sigma\le\sigma'\le\tau'\le\tau$ and $b\in\frak A_{\sigma}$ such that
$b\Bsubseteq\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}$.  Then

\Centerline{$b
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma'}}
    \Bcap\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}
\Bsubseteq\Bvalue{\psi'(\sigma,\tau)=\psi'(\sigma',\tau')}$.}

\noindent If $\psi$ is strictly adapted, then the same formula will be valid
with $b=\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}$
}%end of proof of 613D

\cmmnt{\medskip

\noindent{\bf Remark} Another class of adapted interval functions
will be based on conditional expectations;  see 626J.}

\leader{613E}{Riemann sums} Let $\Cal S\subseteq\Cal T$
be a sublattice, $\psi$ an adapted interval function on $\Cal S$, and
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ a fully adapted process.
\cmmnt{For $\sigma\le\tau$ in $\Cal S$ set

\Centerline{$c(\sigma,\tau)
=\family{t}{T}{\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}}$}

\noindent as in 611J.}

\spheader 613Ea For a stopping-time interval $e\in\Sti(\Cal S)$, we can define
$\Delta_e(\pmb u,d\psi)$ by saying that
$\Delta_e(\pmb u,d\psi)=u_{\sigma}\times\psi(\sigma,\tau)$ whenever
$e=c(\sigma,\tau)$ with $\sigma\le\tau$ in $\Cal S$.   \prooflet{\Prf\
I need to confirm that if $e=c(\sigma,\tau)=c(\sigma',\tau')$ then
$u_{\sigma}\times\psi(\sigma,\tau)=u_{\sigma'}\times\psi(\sigma',\tau')$.

\medskip

\quad{\bf (i)}
Consider first the case in which $\sigma\le\sigma'\le\tau'\le\tau$.
Set $b=\Bvalue{\sigma<\tau}\in\frak A_{\sigma}$.   Then

\Centerline{$b=\Bvalue{\sigma'<\tau'}
\Bsubseteq\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}$}

\noindent (611Jd), so

$$\eqalignno{b&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma'}}
   \Bcap\Bvalue{\psi(\sigma,\tau)=\psi(\sigma',\tau')}
\Bsubseteq\Bvalue{u_{\sigma}\times\psi(\sigma,\tau)
   =u_{\tau'}\times\psi(\sigma',\tau')}.\cr}$$

\noindent On the other hand,

$$\eqalign{1\Bsetminus b&
\Bsubseteq\Bvalue{\psi(\sigma,\tau)=0}\Bcap\Bvalue{\psi(\sigma',\tau')=0}
\Bsubseteq\Bvalue{u_{\sigma}\times\psi(\sigma,\tau)=0}
   \Bcap\Bvalue{u_{\tau'}\times\psi(\sigma',\tau')=0}\cr&
\Bsubseteq\Bvalue{u_{\sigma}\times\psi(\sigma,\tau)
   =u_{\tau'}\times\psi(\sigma',\tau')}.\cr}$$

\noindent So
$u_{\sigma}\times\psi(\sigma,\tau)=u_{\tau'}\times\psi(\sigma',\tau')$.

\medskip

\quad{\bf (ii)} Generally, if $c(\sigma,\tau)=c(\sigma',\tau')$ then

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

\noindent so

\Centerline{$\Bvalue{\sigma\wedge\sigma'<\tau\vee\tau'}=\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\sigma\wedge\sigma'=\sigma}\Bcap\Bvalue{\tau=\tau\vee\tau'}$}

\noindent and $c(\sigma\wedge\sigma',\tau\vee\tau')=c(\sigma,\tau)$.
By (i),
$u_{\sigma'\wedge\tau'}\times\psi(\sigma\wedge\sigma',\tau\vee\tau')=
u_{\sigma}\times\psi(\sigma,\tau)$.   But the same argument shows that
$u_{\sigma'\wedge\tau'}\times\psi(\sigma\wedge\sigma',\tau\vee\tau')=
u_{\sigma'}\times\psi(\sigma',\tau')$, so
$u_{\sigma}\times\psi(\sigma,\tau)=u_{\sigma'}\times\psi(\sigma',\tau')$, as
required.\ \Qed}

\spheader 613Eb If $I\subseteq\Cal S$ is a finite sublattice
and $\Sti_0(I)\subseteq\Sti(\Cal S)$ is the set of
$I$-cells\cmmnt{ (611Je)}, write

\Centerline{$S_I(\pmb u,d\psi)
=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,d\psi)$.}

\spheader 613Ec\cmmnt{ In this context, I will repeatedly use 611L, in the
following way.}   If $I\subseteq\Cal S$ is a non-empty finite sublattice,
then there is a string $(\tau_0,\ldots,\tau_n)$ in $I$
linearly generating the $I$-cells.   \cmmnt{In this case}
$S_I(\pmb u,d\psi)$ will be
$\sum_{i=0}^{n-1}u_{\tau_i}\times\psi(\tau_i,\tau_{i+1})$.
\cmmnt{It will not
be necessary to check that the sequence $\langle\tau_i\rangle_{i\le n}$ is
strictly increasing, because if $\tau_i=\tau_{i+1}$ then
$\psi(\tau_i,\tau_{i+1})$ will be $0$.}

\spheader 613Ed Now suppose that
$\psi=\Delta\pmb v$ for some fully adapted
process $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$.   If
$I\subseteq\Cal S$ is any non-empty sublattice, then
$S_I(\tbf{1},d(\Delta\pmb v))=v_{\max I}-v_{\min I}$.   \prooflet{\Prf\
(Recall from 612D that $\tbf 1$ is the constant process with value $\chi 1$.)
Take $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   Then

$$\eqalign{S_I(\tbf{1},d(\Delta\pmb v))
&=\sum_{i=0}^{n-1}\chi 1\times(\Delta\pmb v)(c(\tau_i,\tau_{i+1}))
=\sum_{i=0}^{n-1}v_{\tau_{i+1}}-v_{\tau_i}\cr
&=v_{\tau_n}-v_{\tau_0}
=v_{\max I}-v_{\min I}.  \text{ \Qed}\cr}$$}

\spheader 613Ee If $\psi=\pmb u\psi'$,
where $\psi'$ is another adapted interval
function with domain $\Cal S^{2\uparrow}$ and $\pmb u=\sfamilysS{u}$ is a fully
adapted process\cmmnt{ (613Dd)},
then $S_I(\tbf 1,d\psi)=S_I(\pmb u,d\psi')$ for every
finite sublattice $I$ of $\Cal S$.   \prooflet{\Prf\

\Centerline{$\Delta_e(\tbf 1,d\psi)
=\psi(\sigma,\tau)=u_{\sigma}\times\psi'(\sigma,\tau)
=\Delta_e(\pmb u,d\psi')$}

\noindent whenever $e=c(\sigma,\tau)$ is a stopping-time interval with
endpoints $\sigma$, $\tau$ in $\Cal S$.   So

\Centerline{$S_I(\tbf 1,d\psi)
=\sum_{e\in\Sti_0(I)}\Delta_e(\tbf 1,d\psi)
=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,d\psi')=S_I(\pmb u,d\psi')$. \Qed}}

\leader{613F}{Notation}\cmmnt{ I will use abbreviations for some of
the most important interval functions.}   Let
$\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$,
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ and
$\pmb w=\family{\sigma}{\Cal S}{w_{\sigma}}$
fully adapted processes.

\spheader 613Fa If $\sigma\le\tau$ in $\Cal S$ and
$e=c(\sigma,\tau)$, then I write

\Centerline{$\Delta_e(\pmb u,d\pmb v)
=\Delta_e(\pmb u,d(\Delta\pmb v))
=u_{\sigma}\times(v_{\tau}-v_{\sigma})$,}

\Centerline{$\Delta_e(\pmb u,d\pmb vd\pmb w)
=\Delta_e(\pmb u,d(\Delta\pmb v\times\Delta\pmb w))
=u_{\sigma}\times(v_{\tau}-v_{\sigma})\times(w_{\tau}-w_{\sigma})$,}

\Centerline{$\Delta_e(\pmb u,|d\pmb v|)
=\Delta_e(\pmb u,d|\Delta\pmb v|)
=u_{\sigma}\times|v_{\tau}-v_{\sigma}|$.}

\spheader 613Fb Now if $I\subseteq\Cal S$ is a finite sublattice
and $\Sti_0(I)$ is the set of $I$-cells, write

\Centerline{$S_I(\pmb u,d\pmb v)
=S_I(\pmb u,d(\Delta\pmb v))
=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,d\pmb v)$,}

\Centerline{$S_I(\pmb u,d\pmb vd\pmb w)
=S_I(\pmb u,d(\Delta\pmb v\times\Delta\pmb w))
=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,d\pmb vd\pmb w)$,}

\Centerline{$S_I(\pmb u,|d\pmb v|)
=S_I(\pmb u,d|\Delta\pmb v|)
=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,|d\pmb v|)$.}

\cmmnt{\noindent I hope
that these expressions will make the formulae below more appealing,
and perhaps offer some hints of the manipulations which will be possible.}

\leader{613G}{Proposition} Suppose that $I$ is a finite
sublattice of $\Cal T$,
$\psi:I^{2\uparrow}\to L^0(\frak A)$ is an adapted interval function and
$\pmb u=\family{\tau}{I}{u_{\tau}}$ is a fully adapted process.

(a)(i) If $\tau\in I$\cmmnt{ and we set
$I\vee\tau=\{\sigma\vee\tau:\sigma\in I\}
=\{\sigma:\tau\le\sigma\in I\}$ and
$I\wedge\tau=\{\sigma\wedge\tau:\sigma\in I\}
=\{\sigma:\sigma\in I$, $\sigma\le\tau\}$,} then
$S_I(\pmb u,d\psi)
=S_{I\wedge\tau}(\pmb u,d\psi)+S_{I\vee\tau}(\pmb u,d\psi)$.

\quad (ii) If $\tau_0,\ldots,\tau_n\in I$ and
$\min I=\tau_0\le\tau_1\le\ldots\le\tau_n=\max I$, then
$S_I(\pmb u,d\psi)
=\sum_{i=0}^{n-1}S_{I\cap[\tau_i,\tau_{i+1}]}(\pmb u,d\psi)$.

(b) For $\tau\in I$ set $z_{\tau}=S_{I\wedge\tau}(\pmb u,d\psi)$.   Then
$\family{\tau}{I}{z_{\tau}}$ is a fully adapted process.

(c) If $\tau$, $\tau'\in I$ then
$S_{I\wedge\tau}(\pmb u,d\psi)+S_{I\wedge\tau'}(\pmb u,d\psi)
=S_{I\wedge(\tau\vee\tau')}(\pmb u,d\psi)
+S_{I\wedge(\tau\wedge\tau')}(\pmb u,d\psi)$.

(d) $\Bvalue{S_I(\pmb u,d\psi)\ne 0}
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\min I<\max I}$.

(e) If $\pmb v=\family{\tau}{I}{v_{\tau}}$ is another fully adapted
process, then $S_I(\pmb u,d(\pmb v\psi))=S_I(\pmb u\times\pmb v,d\psi)$.

\proof{{\bf (a)(i)} We have only to recall that $\Sti_0(I)$ is the disjoint union of
$\Sti_0(I\wedge\tau)$ and $\Sti_0(I\vee\tau)$ (611J(e-ii)).

\medskip

\quad{\bf (ii)} Use 611J(e-iii).

\medskip

{\bf (b)(i)} If $\tau\in I$ and
$e\in\Sti_0(I\wedge\tau)$ then $e$ is expressible as
$c(\sigma,\sigma')$ where $\sigma\le\sigma'\le\tau$, in which case

\Centerline{$u_{\sigma}\times\psi(\sigma,\sigma')\in L^0(\frak A_{\sigma'})
\subseteq L^0(\frak A_{\tau})$.}

\noindent So $z_{\tau}\in L^0(\frak A_{\tau})$.

\medskip

\quad{\bf (ii)} Suppose that $\tau\le\tau'\in I$.   Then

$$\eqalign{z_{\tau'}&
=S_{I\wedge\tau'}(\pmb u,d\psi)
=S_{(I\wedge\tau')\wedge\tau}(\pmb u,d\psi)
   +S_{(I\wedge\tau')\vee\tau}(\pmb u,d\psi)\cr&
=S_{I\wedge\tau}(\pmb u,d\psi)
   +S_{I\cap[\tau,\tau']}(\pmb u,d\psi)
=z_{\tau}+S_{I\cap[\tau,\tau']}(\pmb u,d\psi).\cr}$$

\noindent Now if $e\in\Sti_0(I\cap[\tau,\tau'])$ there are $\sigma$, $\sigma'$
such that $\tau\le\sigma\le\sigma'\le\tau'$ and $e=c(\sigma,\sigma')$.   In this
case

\Centerline{$\Bvalue{\tau=\tau'}\Bsubseteq\Bvalue{\sigma=\sigma'}
\Bsubseteq\Bvalue{\psi(\sigma,\sigma')=0}
\Bsubseteq\Bvalue{\Delta_e(\pmb u,d\psi)=0}$}

\noindent (using 613C(b-i)).   So

\Centerline{$\Bvalue{\tau=\tau'}
\Bsubseteq\Bvalue{\sum_{e\in\Sti_0(I\cap[\tau,\tau'])}\Delta_e(\pmb u,d\psi)=0}
=\Bvalue{S_{I\cap[\tau,\tau']}(\pmb u,d\psi)=0}
=\Bvalue{z_{\tau}=z_{\tau'}}$.}

\noindent By 612Db, $\family{\tau}{I}{z_{\tau}}$ is fully adapted.

\medskip

{\bf (c)} Put (b) and 612D(f-i) together.

\medskip

{\bf (d)(i)} Setting
$a=\Bvalue{\pmb u\ne 0}=\sup_{\sigma\in I}\Bvalue{u_{\sigma}\ne 0}$, we
have

\Centerline{$\Delta_e(\pmb u,d\psi)
=u_{\sigma}\times\psi(\sigma,\tau)
=u_{\sigma}\times\chi a\times\psi(\sigma,\tau)
=\Delta_e(\pmb u,d\psi)\times\chi a$}

\noindent whenever $e=c(\sigma,\tau)$ is an $I$-cell, so
$S_I(\pmb u,d\psi)=S_I(\pmb u,d\psi)\times\chi a$
and $\Bvalue{S_I(\pmb u,d\psi)\ne 0}\Bsubseteq a$.

\medskip

\quad{\bf (ii)} On the other hand, if $e=c(\sigma,\tau)$ is an $I$-cell,

\Centerline{$\Bvalue{\Delta_e(\pmb u,d\psi)\ne 0}
\Bsubseteq\Bvalue{\psi(\sigma,\tau)\ne 0}
\Bsubseteq\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\min I<\max I}$,}

\noindent by 613Cb, and

\Centerline{$\Bvalue{S_I(\pmb u,d\psi)\ne 0}
\Bsubseteq\sup_{e\in\Sti_0}\Bvalue{\Delta_e(\pmb u,d\psi)\ne 0}
\Bsubseteq\Bvalue{\min I<\max I}$.}

\medskip

{\bf (e)} We have only to note that if $\sigma\le\tau$ in $I$ and
$e=c(\sigma,\tau)$, then

\Centerline{$\Delta_e(\pmb u,d(\pmb v\psi))
=u_{\sigma}\times v_{\sigma}\times\psi(\sigma,\tau)
=\Delta_e(\pmb u\times\pmb v,d\psi)$;}

\noindent summing over $e\in\Sti_0(I)$, we get the result.
}%end of proof of 613G

\leader{613H}{Definitions (a)} For a lattice $\Cal S$,
write $\Cal I(\Cal S)$ for the family of finite sublattices of $\Cal S$.

\spheader 613Hb Let $\Cal S$ be a sublattice of
$\Cal T$, $\pmb u$ a fully adapted process with domain including $\Cal S$
and $\psi$ an adapted interval function
defined\cmmnt{ (at least)} on $\Cal S^{2\uparrow}$.
Then I define the {\bf integral} of $\pmb u$ over $\Cal S$ with
respect to $\psi$ to be

\Centerline{$\int_{\Cal S}\pmb u\,d\psi
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\psi)$}

\noindent if the limit is defined for the topology of convergence in
measure\cmmnt{, that is, $z=\int_{\Cal S}\pmb u\,d\psi$ if for
every $\epsilon>0$ there is a $J\in\Cal I(\Cal S)$
such that $\theta(S_I(\pmb u,d\psi)-z)\le\epsilon$
whenever $I\in\Cal I(\Cal S)$ includes $J$}.

%When $\Cal S=[\tau,\tau']$ is an interval in the lattice $\Cal T$, I may
%write $\int_{\tau}^{\tau'}\pmb u\,d\psi$ in place of
%$\int_{[\tau,\tau']}\pmb u\,d\psi$.


\spheader 613Hc Note that if, in (b),
we set $\psi'=\pmb u\psi$\cmmnt{, as in 613Dd},
then\cmmnt{ 613Ee tells us that}

\Centerline{$\int_{\Cal S}d\psi'
=\int_{\Cal S}\tbf 1\,d\psi'
\cmmnt{\mskip5mu=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\tbf 1,d\psi')
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\psi)}
=\int_{\Cal S}\pmb u\,d\psi$}

\noindent if either integral is defined.

\cmmnt{\spheader 613Hd {\bf Remarks (i)} I see that I am writing
`$\lim_{I\uparrow\Cal I(\Cal S)}$'
where in Volume 3 I might have written
`$\lim_{I\to\Cal F(\Cal I(\Cal S)\closeuparrow)}$'.

\medskip

\quad{\bf (ii)} The notation $\int_{\Cal S}\pmb u\,d\psi$ is supposed to convey
the fact that the existence and value of the integral depend only on
$\pmb u\restr\Cal S$ and $\psi\restr\Cal S^{2\uparrow}$.  But a
warning!  regarded as a function of $\Cal S$, it does not behave like an
indefinite integral except in special circumstances. as in 613O below.

\medskip

\quad{\bf (iii)} In this volume, the integral defined in (b) above will be the
basic concept of `stochastic integral', and $\int_{\Cal S}$ should
always be interpreted in the way described here unless there is some clear
indication to the contrary.   But in \S\S645-646
I will introduce a different integral,
the `S-integral', and it will then sometimes be helpful to be able to
declare explicitly that I mean the integral I have
just defined;  so I will keep the
phrase {\bf Riemann-sum integral} in reserve to signify the integral of
this chapter.

\medskip

\quad{\bf (iv)} Of course a Riemann sum $S_I(\pmb u,d\psi)$ can also be thought of
as $\int_I\pmb u\,d\psi$.
}

\leader{613I}{Invariance under change of law}\cmmnt{ The following point is
perfectly elementary, but it is so important that I give it a numbered
paragraph to itself.}   The integral $\int_{\Cal S}\pmb u\,d\psi$
depends\cmmnt{, of course,} on the process $\pmb u$,
the interval function $\psi$ and the lattice $\Cal S$;
behind these declared variables lie
the undeclared structure $(\frak A,T,\family{t}{T}{\frak A_t})$ and the
derived objects $L^0=L^0(\frak A)$ and $\Cal T$.   But
we do not really need the measure $\bar\mu$.   What we use is the
topology of convergence in measure on $L^0$.
Now this topology\cmmnt{ (though
not the functionals $\Bbb E$ and $\theta$ of 613Aa)}
can be defined in terms of the\cmmnt{ Riesz space structure of
$L^0$, which in turn depends only on the} Boolean algebra structure of
$\frak A$\cmmnt{ (613Bg)}.   \cmmnt{In
place of introducing $(\frak A,\bar\mu)$ as a probability
algebra, I could just as well have said that in this section $\frak A$
would be a measurable algebra in the sense of 391Ba.   (Indeed, the ideas
so far would work perfectly if I asked only that $\frak A$ should be a
Maharam algebra in the sense of 393Ea.   But this extension would fail
when we came to the real meat of the theory in \S622.)}

So the Riemann-sum integral is law-independent,
and we shall always be at liberty to replace the measure\cmmnt{ (or `law')}
$\bar\mu$ by another strictly positive countably additive functional on
$\frak A$\cmmnt{, if that seems to make calculations easier}.
\cmmnt{As I have
arranged my exposition, this will remain the case throughout this chapter.
I will return to the issue in 622R.}

\leader{613J}{Theorem} Let $\Cal S$ be a sublattice of
$\Cal T$, $\pmb u=\family{\sigma}{\dom\pmb u}{u_{\sigma}}$
a fully adapted process with $\Cal S\subseteq\dom\pmb u$,
and $\psi$ an adapted interval function
defined\cmmnt{ (at least)} on $\Cal S^{2\uparrow}$.

(a) Suppose that
for every $\epsilon>0$ there are a
$z\in L^0(\frak A)$ and a $J\in\Cal I(\Cal S)$ such that
$\theta(S_I(\pmb u,d\psi)-z)\le\epsilon$ whenever
$J\subseteq I\in\Cal I(\Cal S)$.   Then
$\int_{\Cal S}\pmb u\,d\psi$ is defined.

(b) If $\pmb u'$ is another fully adapted process defined on $\Cal S$,
$\psi'$ is another adapted interval function defined on
$\Cal S^{\uparrow 2}$, and $\int_{\Cal S}\pmb u\,d\psi$,
$\int_{\Cal S}\pmb u'\,d\psi$ and
$\int_{\Cal S}\pmb u\,d\psi'$ are all
defined, then $\int_{\Cal S}\pmb u+\pmb u'\,d\psi$ and
$\int_{\Cal S}\pmb u\,d(\psi+\psi')$ are defined and

\Centerline{$\int_{\Cal S}\pmb u+\pmb u'\,d\psi
=\int_{\Cal S}\pmb u\,d\psi+\int_{\Cal S}\pmb u'\,d\psi$,
\quad$\int_{\Cal S}\pmb u\,d(\psi+\psi')
=\int_{\Cal S}\pmb u\,d\psi
+\int_{\Cal S}\pmb u\,d\psi'$.}

\noindent Similarly, for any $\alpha\in\Bbb R$,
$\int_{\Cal S}\alpha\pmb u\,d\psi$ and
$\int_{\Cal S}\pmb u\,d(\alpha\psi)$
are defined and equal to $\alpha\int_{\Cal S}\pmb u\,d\psi$.

(c)(i) Suppose that $\tau\in\Cal S$.   Then

\Centerline{$\int_{\Cal S}\pmb u\,d\psi
=\int_{\Cal S\wedge\tau}\pmb u\,d\psi+\int_{\Cal S\vee\tau}\pmb u\,d\psi$}

\noindent if either side is defined.

\quad(ii) Suppose that $\tau_0\le\ldots\le\tau_n$ in $\Cal S$.   Then

\Centerline{$\int_{\Cal S}\pmb u\,d\psi
=\int_{\Cal S\wedge\tau_0}\pmb u\,d\psi
 +\sum_{i=0}^{n-1}\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb u\,d\psi
 +\int_{\Cal S\vee\tau_n}\pmb u\,d\psi$}

\noindent if either side is defined.

(d) If $z=\int_{\Cal S}\pmb u\,d\psi$ is defined, then

$$\eqalign{\Bvalue{z\ne 0}&
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}
  \Bcap\sup_{(\sigma,\tau)\in\Cal S^{2\uparrow}}\Bvalue{\psi(\sigma,\tau)\ne 0}
  \cr&
\Bsubseteq\sup_{\sigma,\tau\in\Cal S}
   (\Bvalue{u_{\sigma}\ne 0}\Bcap\Bvalue{\sigma<\tau})
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}.\cr}$$

(e) Set $\Cal S'=\{\tau:\tau\in\Cal S$,
$\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ is defined$\}$.

\quad(i) $\Cal S'$ is an ideal of $\Cal S$.

\quad(ii) Setting
$z_{\tau}=\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ for
$\tau\in\Cal S'$, $\family{\tau}{\Cal S'}{z_{\tau}}$ is fully adapted.

\quad(iii) If $\tau\in\Cal S$ and
$\sup_{\tau'\in\Cal S'}\Bvalue{\tau'=\tau}=1$, then $\tau\in\Cal S'$.

(f) Suppose that $\Cal S\ne\emptyset$
and $z=\int_{\Cal S}\pmb u\,d\psi$ is defined.   Set
$z_{\tau}=\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ for $\tau\in\Cal S$.

\quad(i) The starting value $\lim_{\tau\downarrow\Cal S}z_{\tau}$ is $0$.

\quad(ii) $\lim_{\tau\uparrow\Cal S}\int_{\Cal S\vee\tau}\pmb u\,d\psi=0$,
$\lim_{\tau\uparrow\Cal S}z_{\tau}=z$.

(g) Let $\pmb v$ be another fully adapted process with domain $\Cal S$.
Then $\int_{\Cal S}\pmb u\,d(\pmb v\psi)
=\int_{\Cal S}\pmb u\times\pmb v\,d\psi$ in the sense that if one is
defined so is the other, and they are then equal.

\proof{{\bf (a)} For $J\in\Cal I(\Cal S)$ set
$C_J=\{S_I(\pmb u,d\psi):J\subseteq I\in\Cal I(\Cal S)\}$.
The hypothesis
guarantees that the filter $\Cal F$ on $L^0(\frak A)$ generated by
$\{C_J:J\in\Cal I(\Cal S)\}$ is Cauchy for the uniformity of
convergence in measure;  since $L^0(\frak A)$ is complete
(367Mc again), $\Cal F$ is convergent, and
its limit is $\int_{\Cal S}\pmb u\,d\psi$.

\medskip

{\bf (b)} All we need to know is that the operators
$S_I$ are bilinear and that the linear space operations on $L^0(\frak A)$ are
continuous (367Ma).

\medskip

{\bf (c)(i)}\grheada\
Suppose that $z_0=\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ and
$z_1=\int_{\Cal S\vee\tau}\pmb u\,d\psi$ are defined.
For any $\epsilon>0$, there are $J_0\in\Cal I(\Cal S\wedge\tau)$
and $J_1\in\Cal I(\Cal S\vee\tau)$ such that

\Centerline{$\theta(S_I(\pmb u,d\psi)-z_0)\le\epsilon$ whenever
$J_0\subseteq I\in\Cal I(\Cal S\wedge\tau)$,}

\Centerline{$\theta(S_I(\pmb u,d\psi)-z_1)\le\epsilon$ whenever
$J_1\subseteq I\in\Cal I(\Cal S\vee\tau)$.}

\noindent Set $J=J_0\cup\{\tau\}\cup J_1$, so that
$J\in\Cal I(\Cal S)$.   Suppose that
$J\subseteq I\in\Cal I(\Cal S)$.
Set

\Centerline{$I\wedge\tau=\{\sigma\wedge\tau:\sigma\in I\}
=I\cap[\min\Cal T,\tau]$,}

\Centerline{$I\vee\tau=\{\sigma\vee\tau:\sigma\in I\}
=I\cap[\tau,\max\Cal T]$,}

\noindent so that $J_0\subseteq I\wedge\tau\in\Cal I(\Cal S\wedge\tau)$ and
$J_1\subseteq I\vee\tau\in\Cal I(\Cal S\vee\tau)$.   By 613Ga,
$S_I(\pmb u,d\psi)
=S_{I\wedge\tau}(\pmb u,d\psi)+S_{I\vee\tau}(\pmb u,d\psi)$,
while $\theta(S_{I\wedge\tau}(\pmb u,d\psi)-z_0)\le\epsilon$ and
$\theta(S_{I\vee\tau}(\pmb u,d\psi)-z_1)\le\epsilon$.   Accordingly
$\theta(S_I(\pmb u,d\psi)-(z_0+z_1))\le 2\epsilon$;  and this is true
whenever $J\subseteq I\in\Cal I(\Cal S)$.   As $\epsilon$ is
arbitrary, $\int_{\Cal S}\pmb u\,d\psi$ is defined and equal to
$z_0+z_1$.

\medskip

\qquad\grheadb\ Now suppose that $z=\int_{\Cal S}\pmb u\,d\psi$ is
defined.   Let $\epsilon>0$.   Then there is a finite sublattice
$J$ of $\Cal S$ such that
$\theta(S_I(\pmb u,d\psi)-z)\le\epsilon$ whenever
$J\subseteq I\in\Cal I(\Cal S)$.   Of course we can suppose that
$\tau\in J$.   Take any $K\in\Cal I(\Cal S\wedge\tau)$ including
$J\wedge\tau$.
Let $I$ be the (finite) sublattice of $\Cal S$ generated by
$J\cup K$.   Since

\Centerline{$\{\sigma:\sigma\in\Cal S$, $\sigma\wedge\tau\in K$,
$\sigma\vee\tau\in J\}$}

\noindent is a
sublattice of $\Cal S$ including $J\cup K$, it includes $I$,
so that $I\in\Cal I(\Cal S)$, $I\wedge\tau=K$ and $I\vee\tau=J\vee\tau$.
Now

$$\eqalign{\theta(S_K(\pmb u,d\psi)&-S_{J\wedge\tau}(\pmb u,d\psi))
=\theta(S_{I\wedge\tau}(\pmb u,d\psi)-S_{J\wedge\tau}(\pmb u,d\psi))\cr&
=\theta(S_I(\pmb u,d\psi)-S_{I\vee\tau}(\pmb u,d\psi)
   -S_J(\pmb u,d\psi)+S_{J\vee\tau}(\pmb u,d\psi))\cr&
=\theta(S_I(\pmb u,d\psi)-S_J(\pmb u,d\psi))\cr&
\le\theta(S_I(\pmb u,d\psi)-z)+\theta(z-S_J(\pmb u,d\psi))
\le 2\epsilon.\cr}$$

\noindent By (a) above,
$\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ is defined.

\medskip

\qquad\grheadc\ For $\int_{\Cal S\vee\tau}\pmb u\,d\psi$ we can
repeat the argument, inverted.   Again
suppose that $z=\int_{\Cal S}\pmb u\,d\psi$ is
defined, and take $\epsilon>0$ and $J\in\Cal I(\Cal S)$ such that $\tau\in J$
and $\theta(S_I(\pmb u,d\psi)-z)\le\epsilon$ whenever
$J\subseteq I\in\Cal I(\Cal S)$.   This time, take any
$K\in\Cal I(\Cal S\vee\tau)$ including
$J\vee\tau$.
Let $I$ be the (finite) sublattice of $\Cal S$ generated by
$J\cup K$.   Since

\Centerline{$\{\sigma:\sigma\in\Cal S$, $\sigma\vee\tau\in K$,
$\sigma\wedge\tau\in J\}$}

\noindent is a sublattice of $\Cal S$ including $J\cup K$, it includes $I$,
so that $I\in\Cal I(\Cal S)$, $I\vee\tau=K$ and $I\wedge\tau=J\wedge\tau$.
Now

$$\eqalign{\theta(S_K(\pmb u,d\psi)&-S_{J\vee\tau}(\pmb u,d\psi))
=\theta(S_{I\vee\tau}(\pmb u,d\psi)-S_{J\vee\tau}(\pmb u,d\psi))\cr&
=\theta(S_I(\pmb u,d\psi)-S_{I\wedge\tau}(\pmb u,d\psi)
   -S_J(\pmb u,d\psi)+S_{J\wedge\tau}(\pmb u,d\psi))\cr&
=\theta(S_I(\pmb u,d\psi)-S_J(\pmb u,d\psi))
\le 2\epsilon.\cr}$$

\noindent So $\int_{\Cal S\vee\tau}\pmb u\,d\psi$ is defined.

\medskip

\quad{\bf (ii)} Induce on $n$, using (i) for the inductive step.

\medskip

\Centerline{$\Bvalue{z\ne 0}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}
  \Bcap\sup_{(\sigma,\tau)\in\Cal S^{2\uparrow}}\Bvalue{\psi(\sigma,\tau)\ne 0}
\Bsubseteq\sup_{\sigma,\tau\in\Cal S}
   (\Bvalue{u_{\sigma}\ne 0}\Bcap\Bvalue{\sigma<\tau})$.}

{\bf (d)} Setting

\Centerline{$a=1\Bsetminus(\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}
   \Bcap\sup_{(\sigma,\tau)\in\Cal S^{2\uparrow}}
      \Bvalue{\psi(\sigma,\tau)\ne 0})$,}

\noindent $a\Bsubseteq\Bvalue{\Delta_e(\pmb u,d\psi)=0}$ for every stopping-time interval
$e$ with endpoints in $\Cal S$, so
$S_I(\pmb u,d\psi)\times\chi a=0$ for every $I\in\Cal I(\Cal S)$;
because multiplication in $L^0(\frak A)$ is continuous, $z\times\chi a=0$, that is,

\Centerline{$\Bvalue{z\ne 0}\Bsubseteq 1\Bsetminus a
=\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}
   \Bcap\sup_{(\sigma,\tau)\in\Cal S^{2\uparrow}}
      \Bvalue{\psi(\sigma,\tau)\ne 0}$.}

\noindent For the second inequality, observe just that
$\Bvalue{\psi(\sigma,\tau)\ne 0}\Bsubseteq\Bvalue{\sigma<\tau}$ whenever
$\sigma\le\tau$ in $\Cal S$, by 613C(b-i);  and the third inequality is
immediate from the definition in 612Sb.

\medskip

{\bf (e)(i)}\grheada\ If $\tau\in\Cal S'$, $\sigma\in\Cal S$ and
$\sigma\le\tau$ then

\Centerline{$\int_{\Cal S\wedge\sigma}\pmb u\,d\psi
=\int_{(\Cal S\wedge\tau)\wedge\sigma}
   \pmb u\,d\psi$}

\noindent is defined, by (c).

\medskip

\qquad\grheadb\ Suppose that $\tau$, $\tau'\in\Cal S'$.   Then
$z_{\tau}=\int_{\Cal S\wedge\tau}\pmb u\,d\psi$
and $z_{\tau'}=\int_{\Cal S\wedge\tau'}\pmb u\,d\psi$ and
$z_{\tau\wedge\tau'}=\int_{\Cal S\wedge(\tau\wedge\tau')}\pmb u\,d\psi$
are defined.   We know also that if $I$ is a sublattice of
$\Cal S\wedge(\tau\vee\tau')$ containing both $\tau$ and $\tau'$,
then

\Centerline{$S_I(\pmb u,d\psi)=S_{I\wedge\tau}(\pmb u,d\psi)
    +S_{I\wedge\tau'}(\pmb u,d\psi)
    -S_{I\wedge\tau\wedge\tau'}(\pmb u,d\psi)$}

\noindent by 613Gc.   Looking at the limits as
$I\uparrow\Cal I(\Cal S\wedge(\tau\vee\tau'))$, we see that

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S\wedge(\tau\vee\tau'))}
   S_{I\wedge\tau}(\pmb u,d\psi)
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}
   S_I(\pmb u,d\psi)=z_{\tau}$,}

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S\wedge(\tau\vee\tau'))}
   S_{I\wedge\tau'}(\pmb u,d\psi)
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau')}
   S_I(\pmb u,d\psi)=z_{\tau'}$,}

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S\wedge(\tau\vee\tau')}
   S_{I\wedge(\tau\wedge\tau')}(\pmb u,d\psi)
=\lim_{I\uparrow\Cal I(\Cal S\wedge(\tau\wedge\tau')}
   S_I(\pmb u,d\psi)=z_{\tau\wedge\tau'}$,}

\noindent and therefore that

\Centerline{$\int_{\Cal S\wedge(\tau\vee\tau')}\pmb u\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S\wedge(\tau\vee\tau'))}
   S_I(\pmb u,d\psi)
=z_{\tau}+z_{\tau'}-z_{\tau\wedge\tau'}$}

\noindent is defined.   In particular, $\tau\vee\tau'\in\Cal S'$.
With ($\alpha$), this shows that $\Cal S'$ is an ideal of $\Cal S$.

\medskip

\quad{\bf (ii)} If $\sigma\le\sigma'$ in $\Cal S$, then
$u_{\sigma}\in L^0(\frak A_{\sigma})$ and
$\psi(\sigma,\sigma')\in L^0(\frak A_{\sigma'})$, so
$\Delta_{c(\sigma,\sigma')}(\pmb u,d\psi)\in L^0(\frak A_{\sigma'})$;
consequently $S_I(\pmb u,d\psi)\in L^0(\frak A_{\tau})$ whenever
$\tau\in\Cal S$ and $I\in\Cal I(\Cal S\wedge\tau)$.
Since $L^0(\frak A_{\tau})$ is closed in
$L^0(\frak A)$ for the topology of convergence in measure
(367Rc again), the limiting value
$z_{\tau}$ belongs to $L^0(\frak A_{\tau})$, for every
$\tau\in\Cal S'$.

If $\sigma\le\tau$ in $\Cal S'$, then

\Centerline{$z_{\tau}-z_{\sigma}
=\int_{(\Cal S\wedge\tau)\vee\sigma}\pmb u\,d\psi
\int_{\Cal S\cap[\sigma,\tau]}\pmb u\,d\psi$,}

\noindent by (c).
This time, for any finite sublattice $I$ of $\Cal S\cap[\sigma,\tau]$,

$$\eqalign{\Bvalue{S_I(\pmb u,d\psi)\ne 0}
&\Bsubseteq\sup_{\rho\le\rho'\text{ in }
  \Cal S\cap[\sigma,\tau]}
  \Bvalue{\psi(\rho,\rho')\ne 0}
  \cr
&\Bsubseteq\sup_{\rho\le\rho'\text{ in }
  \Cal S\cap[\sigma,\tau]}
  \Bvalue{\rho<\rho'}
\Bsubseteq\Bvalue{\sigma<\tau}.\cr}$$

\noindent Taking the limit as $I$ increases,
$\Bvalue{z_{\sigma}\ne z_{\tau}}\Bsubseteq\Bvalue{\sigma<\tau}$,
that is,
$\Bvalue{z_{\sigma}=z_{\tau}}\Bsupseteq\Bvalue{\sigma=\tau}$.
By 612Db again, this shows that
$\family{\sigma}{\Cal S'}{z_{\sigma}}$ is fully adapted.

\medskip

\quad{\bf (iii)} Let $\epsilon>0$.   Then there is a $\tau'\in\Cal S'$ such that
$\bar\mu\Bvalue{\tau=\tau'}\ge 1-\bover12\epsilon$;  as
$\tau'\wedge\tau\in\Cal S$
and $\Bvalue{\tau=\tau'\wedge\tau}\Bsupseteq\Bvalue{\tau=\tau'}$, we may suppose
that $\tau'\le\tau$.   We have $\bar\mu\Bvalue{\tau'<\tau}\le\bover12\epsilon$.
There is a $J'\in\Cal I(\Cal S\wedge\tau')$ such that
$\theta(z_{\tau'}-S_{I'}(\pmb u,d\psi))\le\bover12\epsilon$ whenever
$J'\subseteq I'\in\Cal I(\Cal S\wedge\tau')$;  we may suppose that
$\tau'\in J'$.   Now take any $I\in\Cal I(\Cal S\wedge\tau)$ such that
$J'\subseteq I$.   In this case
$J'\subseteq I\wedge\tau'\in\Cal I(\Cal S\wedge\tau')$ and

$$\eqalignno{\theta(z_{\tau'}-S_I(\pmb u,d\psi))&
=\theta(z_{\tau'}-S_{I\wedge\tau'}(\pmb u,d\psi)-S_{I\vee\tau'}(\pmb u,d\psi))
\Displaycause{613Ga}
\le\theta(z_{\tau'}-S_{I\wedge\tau'}(\pmb u,d\psi))
   +\theta(S_{I\vee\tau'}(\pmb u,d\psi))\cr&
\le\Bover12\epsilon+\bar\mu\Bvalue{S_{I\vee\tau'}(\pmb u,d\psi)\ne 0}
\le\Bover12\epsilon+\bar\mu\Bvalue{\tau'<\sup I}
\Displaycause{613Gd}
\le\Bover12\epsilon+\bar\mu\Bvalue{\tau'<\tau}
\le\epsilon.\cr}$$

\noindent By (a) above, $\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ is defined
and $\tau\in\Cal S'$, as claimed.

\medskip

{\bf (f)} Of course (c-i) assures us that $z_{\tau}$ is defined for every
$\tau\in\Cal S$.

\medskip

\quad{\bf (i)} Let $\epsilon>0$.   Then there is a
non-empty $J\in\Cal I(\Cal S)$
such that $\theta(S_I(\pmb u,d\psi)-z)\le\epsilon$ whenever
$J\subseteq I\in\Cal S$.   Take any $\tau\in\Cal S$ such that
$\tau\le\min J$.   Then $\theta(S_K(\pmb u,d\psi))\le 2\epsilon$ whenever
$\tau\in K\in\Cal I(\Cal S\wedge\tau)$.   \Prf\ $I=K\cup J$ and
$I'=\{\tau\}\cup J$ are both finite sublattices of $\Cal S$ including
$J$.   So

$$\eqalignno{\theta(S_K(\pmb u,d\psi))
&=\theta(S_{I\wedge\tau}(\pmb u,d\psi))
=\theta(S_I(\pmb u,d\psi)-S_{I\vee\tau}(\pmb u,d\psi))\cr
\displaycause{613Ga}
&=\theta(S_I(\pmb u,d\psi)-S_{I'}(\pmb u,d\psi))
\le 2\epsilon.  \text{ \Qed}\cr}$$

\noindent As $K$ is arbitrary,

\Centerline{$\theta(z_{\tau})
=\lim_{K\uparrow\Cal I(\Cal S\wedge\tau)}\theta(S_K(\pmb u,d\psi))
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lim_{\tau\downarrow\Cal S}z_{\tau}=0$.

\medskip

\quad{\bf (ii)} The argument is essentially the same.
Let $\epsilon>0$.   Again, there is a non-empty $J\in\Cal I(\Cal S)$
such that $\theta(S_I(\pmb u,d\psi)-z)\le\epsilon$ whenever
$J\subseteq I\in\Cal S$.   This time, take $\tau\in\Cal S$ such that
$\max J\le\tau$.   Then $\theta(S_K(\pmb u,d\psi))\le 2\epsilon$
whenever
$\tau\in K\in\Cal I(\Cal S\vee\tau)$.   \Prf\ $I=K\cup J$ and
$I'=\{\tau\}\cup J$ are both finite sublattices of $\Cal S$ including
$J$.   So

$$\eqalignno{\theta(S_K(\pmb u,d\psi))
&=\theta(S_{I\vee\tau}(\pmb u,d\psi))
=\theta(S_I(\pmb u,d\psi)-S_{I\wedge\tau}(\pmb u,d\psi))\cr
\displaycause{613Ga}
&=\theta(S_I(\pmb u,d\psi)-S_{I'}(\pmb u,d\psi))
\le 2\epsilon.  \text{ \Qed}\cr}$$

\noindent As $K$ is arbitrary,

\Centerline{$\int_{\Cal S\vee\tau}\pmb u\,d\psi
=\lim_{K\uparrow\Cal I(\Cal S\vee\tau)}\theta(S_K(\pmb u,d\psi))
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lim_{\tau\uparrow\Cal S}\int_{\Cal S\vee\tau}\pmb u\,d\psi=0$.   By (c-i),

\Centerline{$\lim_{\tau\uparrow\Cal S}z_{\tau}
=\lim_{\tau\uparrow\Cal S}(z-\int_{\Cal S\vee\tau}\pmb u\,d\psi)
=z$.}

\medskip

{\bf (g)} Immediate from 613Ge.
}%end of proof of 613J

\cmmnt{\leader{613K}{Remark} The key step in the proof of 613Jc is that if
$\tau^*\in J\in\Cal I(\Cal S)$ and
$J\wedge\tau^*\subseteq I\in\Cal I(\Cal S\wedge\tau^*)$, then there is a
$K\in\Cal I(\Cal S)$ such that $K\supseteq J$ and $K\wedge\tau^*=I\wedge\tau^*$.
It follows that if $\phi$ is any function from $\Cal I(\Cal S\wedge\tau^*)$ to
a Hausdorff topological space $Y$,
$\lim_{I\uparrow\Cal I(\Cal S\wedge\tau^*)}\phi(I)
=\lim_{I\uparrow\Cal I(\Cal S)}\phi(I\wedge\tau^*)$ if either limit is defined.
The same argument applies with $\vee$ in the place of $\wedge$, so that
if $\phi$ is any function from $\Cal I(\Cal S\vee\tau^*)$ to
a Hausdorff topological space $Y$, then
$\lim_{I\uparrow\Cal I(\Cal S\vee\tau^*)}\phi(I)
=\lim_{I\uparrow\Cal I(\Cal S)}\phi(I\vee\tau^*)$ if either limit is defined.
Mostly we shall be dealing with the case in which $Y$ is $L^0$ with the
topology of convergence in measure, but I shall also have an application in
which $Y=L^1_{\bar\mu}$ with the weak topology (626L).
}%end of comment

\leader{613L}{More easy bits (a)}
If $\Cal S$ is a sublattice
of $\Cal T$ and $\pmb u$, $\pmb v$ and $\pmb w$
are fully adapted processes defined on
$\Cal S$, I will write

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v
=\int_{\Cal S}\pmb u\,d(\Delta\pmb v)
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb v)$,}

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb vd\pmb w
=\int_{\Cal S}\pmb u\,d(\Delta\pmb v\times\Delta\pmb w)
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb vd\pmb w)$,}

\Centerline{$\int_{\Cal S}\pmb u\,|d\pmb v|
=\int_{\Cal S}\pmb u\,d|\Delta\pmb v|
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,|d\pmb v|)$}

\noindent when the limits exist in $L^0(\frak A)$.

\spheader 613Lb Three trivial calculations:
if $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ and $\pmb u$
are fully adapted processes with domain a
sublattice $\Cal S$ of $\Cal T$, then

\quad(i) $S_I(\tbf{1},d\pmb v)=v_{\max I}-v_{\min I}$
for every non-empty finite sublattice $I$ of $\Cal S$\cmmnt{ (613Ed)}, so
$\int_{\Cal S\cap[\tau,\tau']}\tbf{1}\,d\pmb v=v_{\tau'}-v_{\tau}$
whenever $\tau\le\tau'$ in $\Cal S$;

\quad(ii) if $\pmb v$ is constant then $S_I(\pmb u,d\pmb v)=0$ for every
$I\in\Cal I(\Cal S)$, so $\int_{\Cal S}\pmb u\,d\pmb v$ is defined and equal to
$0$;

\quad(iii) if $z\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$, then
\cmmnt{(in the language of 612De)}
$S_I(z\pmb u,d\pmb v)=S_I(\pmb u,d(z\pmb v))=z\times S_I(\pmb u,d\pmb v)$
for every $I\in\Cal I(\Cal S)$, so
$\int_{\Cal S}z\pmb u\,d\pmb v$ and $\int_{\Cal S}\pmb u\,d(z\pmb v)$
are defined and equal to $z\times\int_{\Cal S}\pmb u\,d\pmb v$ if the last
integral is defined.

\spheader 613Lc\cmmnt{ The
following straightforward fact, refining 613Jd, will frequently be useful.}
Suppose that $I$ is a finite sublattice of $\Cal T$ and
$\pmb u=\family{\sigma}{I}{u_{\sigma}}$,
$\pmb u'=\family{\sigma}{I}{u'_{\sigma}}$,
$\pmb v=\family{\sigma}{I}{v_{\sigma}}$ and
$\pmb v'=\family{\sigma}{I}{v'_{\sigma}}$ are fully adapted processes.
Set $d=\sup_{\sigma\in I}\Bvalue{u_{\sigma}\ne u'_{\sigma}}
  \Bcup\Bvalue{v_{\sigma}\ne v'_{\sigma}}$.
\cmmnt{For any stopping-time interval
$e=c(\sigma,\tau)$ where $\sigma\le\tau$ in $I$,

\Centerline{$\Bvalue{\Delta_e(\pmb u,d\pmb v)
  \ne\Delta_e(\pmb u',d\pmb v')}
\Bsubseteq\Bvalue{u_{\sigma}\ne u'_{\sigma}}
   \Bcup\Bvalue{v_{\sigma}\ne v'_{\sigma}}
   \Bcup\Bvalue{v_{\tau}\ne v'_{\tau}}
\Bsubseteq d$,}

\noindent so }$\Bvalue{S_I(\pmb u,d\pmb v)\ne S_I(\pmb u',d\pmb v')}
\Bsubseteq d$.

Similarly,\cmmnt{ of course,}

\Centerline{
$\Bvalue{S_I(\pmb u,(d\pmb v)^2)\ne S_I(\pmb u',(d\pmb v')^2)}
\Bsubseteq d$,
\quad  $\Bvalue{S_I(\pmb u,|d\pmb v|)\ne S_I(\pmb u',|d\pmb v'|)}
\Bsubseteq d$.}

\noindent Indeed, if $\psi$, $\psi'$ are any adapted interval
functions defined on $\Cal I^{2\uparrow}$, and we set

\Centerline{$d=\sup_{\sigma\in I}\Bvalue{u_{\sigma}\ne u'_{\sigma}}
   \Bcup\sup_{\sigma\le\tau\text{ in }I}
     \Bvalue{\psi(\sigma,\tau)\ne\psi'(\sigma,\tau)}$,}

\noindent then
$\Bvalue{S_I(\pmb u,d\psi)\ne S_I(\pmb u',d\psi')}\Bsubseteq d$.

\spheader 613Ld It follows that if $\Cal S$ is any sublattice of
$\Cal T$ and $\pmb u=\sfamilysS{u}$, $\pmb v=\sfamilysS{v}$ are fully adapted
processes such that $z=\int_{\Cal S}\pmb u\,d\pmb v$ is defined, then
\ifresultsonly$\Bvalue{z\ne 0}
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\pmb v\ne\tbf 0}$.
\else

$$\eqalign{\Bvalue{z\ne 0}&
\Bsubseteq\sup_{I\in\Cal I(\Cal S)}\Bvalue{S_I(\pmb u,d\pmb v)\ne 0}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}
   \Bcap\sup_{\sigma,\tau\in\Cal S}\Bvalue{v_{\sigma}\ne v_{\tau}}\cr&
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\pmb v\ne\tbf 0}.\cr}$$
\fi

\leader{613M}{}\cmmnt{ The next calculation is perfectly elementary,
but will be helpful more than once
below, and is a useful exercise in the definitions here.

\medskip

\noindent}{\bf Lemma}
Let $\Cal S$ be a sublattice of $\Cal T$ and $\pmb u$, $\pmb v$,
$\pmb w$ fully adapted processes defined on $\Cal S$.   Then

$$\eqalign{S_I(\pmb u,d\pmb vd\pmb w)
&=S_I(\pmb u,d(\pmb v\times\pmb w))
 -S_I(\pmb u\times\pmb v,d\pmb w)-S_I(\pmb u\times\pmb w,d\pmb v)\cr
&=\Bover12\bigl(S_I(\pmb u,(d(\pmb v+\pmb w))^2)
 -S_I(\pmb u,(d\pmb v)^2)-S_I(\pmb u,(d\pmb w)^2)\bigr)\cr}$$

\noindent for every finite sublattice $I$ of $\Cal S$.   Consequently

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb vd\pmb w
=\int_{\Cal S}\pmb u\,d(\pmb v\times\pmb w)
 -\int_{\Cal S}\pmb u\times\pmb v\,d\pmb w
 -\int_{\Cal S}\pmb u\times\pmb w\,d\pmb v$}

\noindent if any three of the four integrals are defined, and

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb vd\pmb w
=\Bover12\bigl(\int_{\Cal S}\pmb u\,(d(\pmb v+\pmb w))^2
 -\int_{\Cal S}\pmb u\,(d\pmb v)^2
 -\int_{\Cal S}\pmb u\,(d\pmb w)^2\bigr)$}

\noindent if any three of the integrals are defined.

\proof{{\bf (a)} Express $\pmb u$, $\pmb v$ and $\pmb w$ as
$\family{\sigma}{\dom\pmb u}{u_{\sigma}}$, etc.

If $\sigma\le\tau$ in $\Cal S$ then

$$\eqalign{u_{\sigma}\times(v_{\tau}-v_{\sigma})\times(w_{\tau}-w_{\sigma})
&=u_{\sigma}
  \times(v_{\tau}\times w_{\tau}-v_{\sigma}\times w_{\sigma})\cr
&\mskip50mu
-u_{\sigma}\times v_{\sigma}\times(w_{\tau}-w_{\sigma})
-u_{\sigma}\times w_{\sigma}\times(v_{\tau}-v_{\sigma})\cr
&=\Bover12\bigl(u_{\sigma}\times(v_{\tau}+w_{\tau}-v_{\sigma}-w_{\sigma})^2
  \cr
&\mskip50mu
  -u_{\sigma}\times(v_{\tau}-v_{\sigma})^2
  -u_{\sigma}\times(w_{\tau}-w_{\sigma})^2\bigr),\cr}$$

\noindent that is,

$$\eqalign{\Delta_e(\pmb u,d\pmb vd\pmb w)
&=\Delta_e(\pmb u,d(\pmb v\times\pmb w))
-\Delta_e(\pmb u\times\pmb v,d\pmb w)
-\Delta_e(\pmb u\times\pmb w,d\pmb v)\cr
&=\Bover12\bigl(\Delta_e(\pmb u,(d(\pmb v+\pmb w))^2)
 -\Delta_e(\pmb u,(d\pmb v)^2)-\Delta_e(\pmb u,(d\pmb w)^2)\bigr)\cr}$$

\noindent for every $e\in\Sti(\Cal S)$.
So, writing $\Sti_0(I)$ for the set of $I$-cells, as usual,

$$\eqalign{S_I(\pmb u,d\pmb vd\pmb w)
&=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,d\pmb vd\pmb w)\cr&
=\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,d(\pmb v\times\pmb w))
 -\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u\times\pmb v,d\pmb w)\cr&
\mskip100mu-\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u\times\pmb w,d\pmb v)\cr&
=\Bover12\bigl(\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,(d(\pmb v+\pmb w))^2)
 -\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,(d\pmb v)^2)\cr&
\mskip100mu -\sum_{e\in\Sti_0(I)}\Delta_e(\pmb u,(d\pmb w)^2)\bigr)\cr&
=\Bover12\bigl(S_I(\pmb u,(d(\pmb v+\pmb w))^2)
 -S_I(\pmb u,(d\pmb v)^2)-S_I(\pmb u,(d\pmb w)^2)\bigr).\cr}$$

\medskip

{\bf (b)} Taking limits as $I\uparrow\Cal I(\Cal S)$, we get the
corresponding identities for the integrals, as long as all but one is defined.
}%end of proof of 613M

\leader{613N}{Proposition}
Let $\Cal S$ be a non-empty sublattice of $\Cal T$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a fully adapted process.
Interpreting $\int_{\Cal S}d\pmb v$ as $\int_{\Cal S}\tbf{1}\,d\pmb v$
where $\tbf{1}$ is the constant process with value $\chi 1$,
$\int_{\Cal S}d\pmb v$ is defined iff
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ and
$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ are defined,
and in this case
$\int_{\Cal S}d\pmb v=v_{\uparrow}-v_{\downarrow}$.

\proof{{\bf (a)} Suppose that $z=\int_{\Cal S}d\pmb v$ is defined.   For
$\tau\in\Cal S$ set $z_{\tau}=\int_{\Cal S\wedge\tau}d\pmb v$;  by
613Jc, this is defined.   If $\tau\le\tau'$ in $\Cal S$,

$$\eqalignno{z_{\tau'}-z_{\tau}
&=\int_{\Cal S\wedge\tau'}d\pmb v
   -\int_{\Cal S\wedge\tau'\wedge\tau}d\pmb v
=\int_{(\Cal S\wedge\tau')\vee\tau}d\pmb v\cr
\displaycause{613J(c-i) again}
&=\int_{\Cal S\cap[\tau,\tau']}d\pmb v
=v_{\tau'}-v_{\tau}\cr}$$

\noindent by 613Lb.   Now we know that
$\lim_{\tau\downarrow\Cal S}z_{\tau}=0$ and
$\lim_{\tau\uparrow\Cal S}z_{\tau}=z$ (613Jf).
So, starting from any $\tau^*\in\Cal S$,

$$\eqalign{v_{\downarrow}
&=\lim_{\tau\downarrow\Cal S}v_{\tau}
=v_{\tau^*}-\lim_{\tau\downarrow\Cal S}(v_{\tau^*}-v_{\tau})\cr
&=v_{\tau^*}-\lim_{\tau\downarrow\Cal S}(z_{\tau^*}-z_{\tau})
=v_{\tau^*}-z_{\tau^*}\cr}$$

\noindent and

$$\eqalign{v_{\uparrow}
&=\lim_{\tau'\uparrow\Cal S}v_{\tau'}
=v_{\tau^*}+\lim_{\tau'\uparrow\Cal S}v_{\tau'}-v_{\tau^*}\cr
&=v_{\tau^*}+\lim_{\tau'\uparrow\Cal S}z_{\tau'}-z_{\tau^*}
=v_{\tau^*}+z-z_{\tau^*}\cr}$$

\noindent are defined, and $v_{\uparrow}-v_{\downarrow}=z$.

\medskip

{\bf (b)} Now suppose that the limits $v_{\uparrow}$ and $v_{\downarrow}$
are defined.   Let $\epsilon>0$.   Then there are
$\tau_0$, $\tau_1\in\Cal S$ such that
$\theta(v_{\tau'}-v_{\uparrow})+\theta(v_{\tau'}-v_{\downarrow})
\le\epsilon$
whenever $\tau$, $\tau'\in\Cal S$, $\tau\le\tau_0$ and $\tau_1\le\tau'$.
Now suppose that $I\in\Cal I(\Cal S)$ includes
$\{\tau_0\wedge\tau_1,\tau_0\vee\tau_1\}$.   Then

$$\eqalignno{\theta(v_{\uparrow}-v_{\downarrow}-S_I(\tbf{1},d\pmb v))
&=\theta(v_{\uparrow}-v_{\downarrow}-v_{\max I}+v_{\min I})\cr
\displaycause{613Lb again}
&\le\theta(v_{\uparrow}-v_{\max I})
  +\theta(v_{\downarrow}-v_{\min I})
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\int_{\Cal S}d\pmb v=\int_{\Cal S}\tbf{1}\,d\pmb v$ is defined and equal
to $v_{\uparrow}-v_{\downarrow}$.
}%end of proof of 613N

\leader{613O}{Indefinite integrals (a) Definition}
Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$ a fully adapted
process with domain $\Cal S$, and $\psi$ an adapted interval function with
domain $\Cal S^{2\uparrow}$.   Set
$\Cal S'=\{\tau:\tau\in\Cal S$,
$\int_{\Cal S\wedge\tau}\pmb u\,d\psi$
is defined$\}$;\cmmnt{  by 613J(e-i),} $\Cal S'$ is an ideal of
$\Cal S$.   The
{\bf indefinite integral} of $\pmb u$ with respect to $\psi$ is the process
$ii_{\psi}(\pmb u)=\family{\tau}{\Cal S'}
{\int_{\Cal S\wedge\tau}\pmb u\,d\psi}$;\cmmnt{ by
613J(e-ii),} this is a fully adapted process.

When $\psi$ is of the form $\Delta\pmb v$ for a fully adapted process $\pmb v$,
I will write $\iivu=\family{\tau}{\Cal S'}
{\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v}$ for the
indefinite integral of $\pmb u$ with respect to $\pmb v$.

\spheader 613Ob{\bf (i)} Note that if $\int_{\Cal S}\pmb u\,d\psi$ is defined,
the domain $\Cal S'$ of $ii_{\psi}(\pmb u)$ is the whole of
$\Cal S$\cmmnt{, by 613J(c-i)}.

\medskip

\quad{\bf (ii)} It is obvious from the definition, but perhaps it is worth
stating formally that if $\tau\in\Cal S$ and
$\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ is defined then

$$\eqalign{ii_{\psi}(\pmb u)\restr\Cal S\wedge\tau&
=\family{\sigma}{\Cal S\wedge\tau}{\int_{\Cal S\wedge\sigma}\pmb u\,d\psi}\
=\family{\sigma}{\Cal S\wedge\tau}{\int_{\Cal S\wedge\sigma}
  (\pmb u\restr\Cal S\wedge\tau)d(\psi\restr(\Cal S\wedge\tau)^{2\uparrow})}\cr&
=ii_{\psi\restr(\Cal S\wedge\tau)^{2\uparrow}}(\pmb u\restr\Cal S\wedge\tau).
\cr}$$

\medskip

\quad{\bf (iii)} On the other side,
if $\Cal S'=\Cal S$ and $\tau\in\Cal S$, then
$ii_{\psi\restr(\Cal S\vee\tau)^{2\uparrow}}(\pmb u\restr\Cal S\vee\tau)$
is defined on the whole of $\Cal S\vee\tau$ and is equal to
$(ii_{\psi}(\pmb u)\restr\Cal S\vee\tau)
-(\int_{\Cal S\wedge\tau}\pmb u\,d\psi)\tbf 1$.
\prooflet{\Prf\
Expanding the definitions, all this is saying is that if
$\tau'\in\Cal S\vee\tau$ then
$\int_{(\Cal S\vee\tau)\wedge\tau'}\pmb u\,d\psi$ is defined and equal to
$\int_{\Cal S\wedge\tau'}\pmb u\,d\psi
-\int_{\Cal S\wedge\tau}\pmb u\,d\psi$;  since
$(\Cal S\vee\tau)\wedge\tau'=(\Cal S\wedge\tau')\vee\tau$, this is immediate
from 613J(c-i).\ \Qed}

\cmmnt{\spheader 613Oc I have put the
definition in (a) in a form which can accommodate cases in which
$\Cal S'=\dom ii_{\psi}(\pmb u)$ is not
the whole of $\Cal S$, leaving open the possibility
that $\Cal S'$ is actually empty.   But we know that $\Cal S'$
is an ideal of $\Cal S$, and from 613Jf that if $\Cal S'$ is non-empty
then

\Centerline{$\lim_{\tau\downarrow\Cal S}
  \int_{\Cal S\wedge\tau}\pmb u\,d\psi
=\lim_{\tau\downarrow\Cal S'}\int_{\Cal S\wedge\tau}\pmb u\,d\psi
=0$.}

\noindent In particular, if $\Cal S$ has a least element $\min\Cal S$,
then $\min\Cal S\in\Cal S'$ and the value
of $ii_{\psi}(\pmb u)$ there is equal to $0$.}

\leader{613R}{Proposition} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\pmb u$, $\pmb v$ fully adapted
processes with domain $\Cal S$.   Then\cmmnt{ (in the notation of 612Sb)}
$\Bvalue{\iivu\ne\tbf 0}
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\pmb v\ne\tbf 0}$.

\proof{ Setting
$c=1\Bsetminus(\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\pmb v\ne\tbf 0})$, we see
from 613Lc that
$\chi c\times S_I(\pmb u,d\pmb v)=0$ for every finite sublattice $I$ of
$\Cal S$;  because multiplication in $L^0(\frak A)$ is continuous,
$\chi c\times\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v=0$
whenever $\tau\in\Cal S$ and the integral is defined, that is,

\Centerline{$\Bvalue{\iivu\ne 0}\Bsubseteq 1\Bsetminus c
\Bsubseteq\Bvalue{\pmb u\ne\tbf 0}\Bcap\Bvalue{\pmb v\ne\tbf 0}$.}
}%end of proof of 613R

\leader{613S}{}\cmmnt{ It is a striking and very convenient fact that we can
often approach an integral $\int_{\Cal S}\tbf 1\,d\psi$ over a general
sublattice $\Cal S$ through an integral over the covered envelope of $\Cal S$,
which is a full sublattice of $\Cal T$.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$
and $\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ a
strictly %query do we need this?
adapted interval function.   Suppose that $I$, $J\in\Cal I(\Cal S)$,
$J\subseteq I$ and $a\Bsubseteq\sup_{\sigma\in J}\Bvalue{\tau=\sigma}$
for every $\tau\in I$.   Then
$a\Bsubseteq\Bvalue{S_I(\tbf 1,d\psi)=S_J(\tbf 1,d\psi)}$.
In particular, if $J$ covers $I$\cmmnt{ (definition:  611Ma)}
then $S_I(\tbf 1,d\psi)=S_J(\tbf 1,d\psi)$.

\proof{{\bf (a)} The case $a=0$ is trivial;  so is the case
$a\ne 0$ and $J=\emptyset$, as then $I$ also must be empty and

\Centerline{$S_I(\tbf 1,d\psi)=S_J(\tbf 1,d\psi)=0$.}

\noindent So suppose otherwise.
Let $(\sigma_0,\ldots,\sigma_n)$ linearly generate the
$J$-cells.   If $\tau\in I$ then

$$\eqalignno{a\Bsubseteq\sup_{\sigma\in J}\Bvalue{\tau=\sigma}&
=\sup_{\sigma\in J}\sup_{j\le n}
   (\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma=\sigma_j})
\Displaycause{611Ke}
\Bsubseteq\sup_{j\le n}\Bvalue{\tau=\sigma_j}
\Bsubseteq\Bvalue{\sigma_0\le\tau}\Bcap\Bvalue{\tau\le\sigma_n}.\cr}$$

\noindent Set $I_{-1}=I\wedge\sigma_0$,
$I_j=I\cap[\sigma_j,\sigma_{j+1}]$ for $0\le j<n$,
$I_n=I\vee\sigma_n$.

\medskip

{\bf (b)} If $\tau\le\tau'$ in $I_{-1}$, then

$$\eqalignno{a\Bcap\Bvalue{\psi(\tau,\tau')\ne 0}&
\Bsubseteq a\Bcap\Bvalue{\tau<\tau'}
\Displaycause{613C(b-i)}
\Bsubseteq a\Bcap\Bvalue{\tau<\sigma_0}
=0,\cr}$$

\noindent so

\Centerline{$a\Bcap\Bvalue{\Delta_e(\tbf 1,d\psi)\ne 0}=0$}

\noindent for every $I_{-1}$-cell $e$.   Summing over $e$,
$S_{I_{-1}}(\tbf 1,d\psi)\times\chi a=0$.

\medskip

{\bf (c)} In the same way, if $\tau\le\tau'$ in $I_n$, then

\Centerline{$a\Bcap\Bvalue{\psi(\tau,\tau')\ne 0}
\Bsubseteq a\Bcap\Bvalue{\tau<\tau'}
\Bsubseteq a\Bcap\Bvalue{\sigma_n<\tau'}=0$,}

\noindent so

\Centerline{$S_{I_n}(\tbf 1,d\psi)\times\chi a=0$.}

\medskip

{\bf (d)} If $0\le j<n$,
$a\Bsubseteq\Bvalue{S_{I_j}(\tbf 1,d\psi)
=\psi(\sigma_j,\sigma_{j+1})}$.
\Prf\ For every $\tau\in I_j$,

$$\eqalign{a\Bcap\Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau<\sigma_{j+1}}&
\Bsubseteq\sup_{k\le n}(\Bvalue{\tau=\sigma_k}
   \Bcap\Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau<\sigma_{j+1}})\cr&
\Bsubseteq\sup_{k\le n}
   (\Bvalue{\sigma_j<\sigma_k}\Bcap\Bvalue{\sigma_k<\sigma_{j+1}})\cr&
\Bsubseteq(\sup_{k\le j}\Bvalue{\sigma_j<\sigma_k})
   \Bcup(\sup_{j+1\le k}\Bvalue{\sigma_k<\sigma_{j+1}})
=0.\cr}$$

\noindent Take $(\tau_0,\ldots,\tau_m)$ linearly generating the
$I_j$-cells;  then $\sigma_j=\tau_0\le\ldots\le\tau_m=\sigma_{j+1}$.
For $i<m$, set $b_i=\Bvalue{\tau_i<\tau_{i+1}}$, so that
$\sup_{i<m}b_i=\Bvalue{\sigma_j<\sigma_{j+1}}$.   For $i<m$,

$$\eqalignno{a\Bcap b_i&
=a\Bcap\Bvalue{\tau_i<\tau_{i+1}}
=a\Bcap\Bvalue{\sigma_j<\tau_{i+1}}
    \Bcap\Bvalue{\tau_i<\sigma_{j+1}}\Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr&
\Bsubseteq a\Bcap\Bvalue{\tau_{i+1}=\sigma_{j+1}}
    \Bcap\Bvalue{\tau_i=\sigma_j}\Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr&
\Bsubseteq\Bvalue{\psi(\tau_i,\tau_{i+1})=\psi(\sigma_j,\sigma_{j+1})}
    \Bcap\inf_{k<i}\Bvalue{\tau_k=\tau_{k+1}}
    \Bcap\inf_{i<k<m}\Bvalue{\tau_k=\tau_{k+1}}
    \Bcap\Bvalue{u_{\tau_i}=u_{\sigma_j}}
\Displaycause{because $\psi$ is strictly adapted}
\Bsubseteq\Bvalue{\psi(\tau_i,\tau_{i+1})=\psi(\sigma_j,\sigma_{j+1})}
    \Bcap\inf_{\Atop{k<m}{k\ne i}}\Bvalue{\psi(\tau_k,\tau_{k+1})=0}
    \Bcap\Bvalue{u_{\tau_i}=u_{\sigma_j}}
\Displaycause{using 613C(b-i)}
\Bsubseteq\Bvalue{\psi(\tau_i,\tau_{i+1})=
        \psi(\sigma_j,\sigma_{j+1})}
    \Bcap\Bvalue{S_{I_j}(\tbf 1,d\psi)=\psi(\tau_i,\tau_{i+1})}
        \cr&
\Bsubseteq\Bvalue{S_{I_j}(\tbf 1,d\psi)
         =\psi(\sigma_j,\sigma_{j+1}}.\cr}$$

\noindent Taking the supremum over $i$,

\Centerline{$a\Bcap\Bvalue{\sigma_j<\sigma_{j+1}}
   \Bsubseteq\Bvalue{S_{I_j}(\tbf 1,d\psi)
         =\psi(\sigma_j,\sigma_{j+1})}$.}

\noindent But

$$\eqalign{\Bvalue{\sigma_j=\sigma_{j+1}}&
=\inf_{k<m}\Bvalue{\tau_k=\tau_{k+1}}\Bcap\Bvalue{\sigma_j=\sigma_{j+1}}\cr&
\Bsubseteq\Bvalue{S_{I_j}(\tbf 1,d\psi)=0}
     \Bcap\Bvalue{\psi(\sigma_j,\sigma_{j+1})=0}\cr&
\Bsubseteq\Bvalue{S_{I_j}(\tbf 1,d\psi)
     =\psi(\sigma_j,\sigma_{j+1})},\cr}$$

\noindent so
$a\Bsubseteq\Bvalue{S_{I_j}(\tbf 1,d\psi)
=\psi(\sigma_j,\sigma_{j+1})}$.\ \Qed

\medskip

{\bf (e)} Assembling these,

$$\eqalign{S_I(\tbf 1,d\psi)\times\chi a
&=\sum_{j=-1}^nS_{I_j}(\tbf 1,d\psi)\times\chi a\cr
&=\sum_{j=0}^{n-1}u_{\sigma_j}\times\psi(\sigma_j,\sigma_{j+1})
   \times\chi a
=S_J(\tbf 1,d\psi)\times\chi a\cr}$$

\noindent and
$a\Bsubseteq\Bvalue{S_I(\tbf 1,d\psi)=S_J(\tbf 1,d\psi)}$.

\medskip

{\bf (f)} If $J$ actually covers $I$ then we can take $a=1$ and conclude
that $S_I(\tbf 1,d\psi)=S_J(\tbf 1,d\psi)$.
}%end of proof of 613S

\leader{613T}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, $\Cal S'$ a
sublattice of $\Cal S$ which covers $\Cal S$,
$\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ a strictly adapted interval function
and $\pmb u:\Cal S\to L^0(\frak A)$ a fully adapted process.
If $\int_{\Cal S}\pmb u\,d\psi$ is defined, so is $\int_{\Cal S'}\pmb u\,d\psi$,
and the integrals are equal.

\proof{{\bf (a)}
Consider first the case in which $\pmb u=\tbf 1$, so that
$z=\int_{\Cal S}d\hat\psi$ is defined.
Let $\epsilon>0$.   Let $J\in\Cal I(\Cal S)$ be such that
$\theta(z-S_I(\tbf 1,d\hat\psi))\le\epsilon$ whenever
$J\subseteq I\in\Cal I(\Cal S)$.   Let $A\subseteq\Cal S'$ be a finite set
such that $a=\inf_{\tau\in J}\sup_{\sigma\in A}\Bvalue{\sigma=\tau}$ has measure
at least $1-\epsilon$.
Let $J_0$ be the sublattice of $\Cal S'$ generated by
$A$, so that $a\Bsubseteq\sup_{\sigma\in J_0}\Bvalue{\tau=\sigma}$ for every
$\tau\in J_0$.
If $J_0\subseteq K\in\Cal I(\Cal S')$, consider the sublattice
$J\sqcup K$ of $\Cal S$ generated by $J\cup K$.   Since
$a\Bsubseteq\sup_{\sigma\in K}\Bvalue{\tau=\sigma}$
for every $\tau\in J$,
$a\Bsubseteq\sup_{\sigma\in K}\Bvalue{\tau=\sigma}$ for every $\tau\in J\cup K$
and therefore for every $\tau\in J\sqcup K$, by 611M(b-i).   By 613S,
$\Bvalue{S_K(\tbf 1,d\hat\psi)=S_{J\sqcup K}(\tbf 1,d\hat\psi)}
\Bsupseteq a$.

Consequently

\Centerline{$\theta(z-S_K(\tbf 1,d\psi))
=\theta(z-S_K(\tbf 1,d\hat\psi))
\le\theta(z-S_{J\sqcup K}(\tbf 1,d\hat\psi))
   +\bar\mu(1\Bsetminus a)\le 2\epsilon$.}

\noindent And this is true whenever $J_0\subseteq K\in\Cal I(\Cal S')$.

As $\epsilon$ is arbitrary, $\int_{\Cal S'}d\psi$ is defined and
equal to $z$.

\medskip

{\bf (b)} For the general case, apply (i) to $\pmb u\psi$, using
613Jg to see that

\Centerline{$\int_{\Cal S'}\pmb u\,d\psi=\int_{\Cal S'}d(\pmb u\psi)
=\int_{\Cal S}d((\pmb u\psi)\varCsphat)
=\int_{\Cal S}\hatpmb u\,d\hat\psi$.}
}%end of proof of 613T

\leader{613U}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
and $\hat\Cal S$ its covered envelope\cmmnt{ (611M)}.

(a) For every strictly adapted interval function
$\psi:\Cal S^{\uparrow 2}\to L^0\cmmnt{\mskip5mu=L^0(\frak A)}$
there is a unique extension of $\psi$ to a strictly adapted
interval function $\hat\psi:\hat\Cal S^{2\uparrow}\to L^0$.

(b)(i) The function $\psi\mapsto\hat\psi$ is an $f$-algebra homomorphism from
the space of strictly adapted interval functions on $\Cal S$ to the space of
strictly adapted interval functions on $\hat\Cal S$.

\quad (ii) If $\psi$ is a strictly adapted interval function on $\Cal S$ and
$h:\Bbb R\to\Bbb R$ is Borel measurable, then
$(\bar h\psi)\varCsphat=\bar h\hat\psi$.

\quad (iii) If $\psi$ is a strictly adapted interval function on $\Cal S$ and
$\pmb u$ is a fully adapted process with domain $\Cal S$, then
$(\pmb u\psi)\varCsphat=\hatpmb u\hat\psi$, where $\hatpmb u$ is the fully
adapted extension of $\pmb u$ to $\hat\Cal S$.

\proof{ (Compare 612Q.)

\medskip

{\bf (a)(i)} The first thing to note is that if
$(\sigma,\tau)\in\hat{\Cal S}^{2\uparrow}$ and
$c=\sup_{(\sigma',\tau')\in\Cal S^{2\uparrow}}
(\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'})$ then $c=1$.   \Prf\ If
$a\in\frak A$ is non-zero, there is a $\sigma''\in\Cal S$ such
that $a'=a\Bcap\Bvalue{\sigma=\sigma''}\ne 0$, and now there is a
$\tau'\in\Cal S$ such
that $a''=a'\Bcap\Bvalue{\tau=\tau'}\ne 0$.   Set
$\sigma'=\sigma''\wedge\tau'$;  then $(\sigma',\tau')\in\Cal S^{2\uparrow}$ and

$$\eqalign{a''&
\Bsubseteq\Bvalue{\sigma=\sigma''}\Bcap\Bvalue{\tau=\tau'}
     \Bcap\Bvalue{\sigma\le\tau}
\Bsubseteq\Bvalue{\sigma=\sigma''}\Bcap\Bvalue{\tau=\tau'}
     \Bcap\Bvalue{\sigma''\le\tau'}\cr&
\Bsubseteq\Bvalue{\sigma=\sigma''}\Bcap\Bvalue{\tau=\tau'}
     \Bcap\Bvalue{\sigma''=\sigma'}
\Bsubseteq\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}
\Bsubseteq c.\cr}$$

\noindent So $a\Bcap c\Bsupseteq a''$ is non-zero;  as $a$ is arbitrary,
$c=1$.\ \Qed

\medskip

\quad{\bf (ii)} Take $(\sigma,\tau)\in\hat{\Cal S}^{2\uparrow}$.
For $\alpha\in\Bbb R$, set

\Centerline{$A_{\sigma\tau\alpha}
=\{\Bvalue{\psi(\sigma',\tau')>\alpha}\Bcap\Bvalue{\sigma=\sigma'}
  \Bcap\Bvalue{\tau=\tau'}:\sigma'\le\tau'$ in $\Cal S\}$,
\quad$a_{\sigma\tau\alpha}=\sup A_{\sigma\tau\alpha}$.}

\noindent Then

$$\eqalign{a_{\sigma\tau\alpha}&
=\sup_{(\sigma',\tau')\in\Cal S^{2\uparrow}}(\Bvalue{\psi(\sigma',\tau')>\alpha}
     \Bcap\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'})\cr&
=\sup_{\Atop{(\sigma',\tau')\in\Cal S^{2\uparrow}}{\beta>\alpha}}
  (\Bvalue{\psi(\sigma',\tau')>\beta}
     \Bcap\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'})
=\sup_{\beta>\alpha}a_{\sigma\tau\beta}.\cr}$$

\noindent Moreover, if $a\in\frak A\setminus\{0\}$, there are a pair
$(\sigma',\tau')\in\Cal S^{2\uparrow}$ such that
$a'=a\Bcap\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}$ is non-zero,
by (i), and $\alpha$, $\beta\in\Bbb R$ such that
$a''=a'\Bcap\Bvalue{\psi(\sigma',\tau')>\alpha}
  \Bcap\Bvalue{\psi(\sigma',\tau')\le\beta}$ is
non-zero;  in which case, whenever $(\sigma'',\tau'')\in\Cal S^{2\uparrow}$,

$$\eqalign{&a''\Bcap\Bvalue{\psi(\sigma'',\tau'')>\beta}
   \Bcap\Bvalue{\sigma''=\sigma}\Bcap\Bvalue{\tau''=\tau}\cr&
\mskip 100mu
\Bsubseteq\Bvalue{\psi(\sigma'',\tau'')\ne\psi(\sigma',\tau'}
   \Bcap\Bvalue{\sigma''=\sigma'}\Bcap\Bvalue{\tau''=\tau'}
=0.\cr}$$

\noindent But this means that $a''\Bcap a_{\sigma\tau\beta}=0$, while on the
other hand $a''\Bsubseteq a_{\sigma\tau\alpha}$.   As $a$ is arbitrary,
$\inf_{\beta\in\Bbb R}a_{\sigma\tau\beta}=0$ and
$\sup_{\alpha\in\Bbb R}a_{\sigma\tau\alpha}=1$.

Accordingly we have a $\hat\psi(\sigma,\tau)\in L^0(\frak A)$ such that
$\Bvalue{\hat\psi(\sigma,\tau)>\alpha}=a_{\sigma\tau\alpha}$ for every
$\alpha\in\Bbb R$.
Observe next that $A_{\sigma\tau\alpha}\Bsubseteq\frak A_{\tau}$ for every
$\alpha\in\Bbb R$ (use 611H(c-iii)),
so $a_{\sigma\tau\alpha}\in\frak A_{\tau}$ for every
$\alpha$, and $\hat\psi(\sigma,\tau)\in L^0(\frak A_{\tau})$.

\medskip

\quad{\bf (iii)} $\hat\psi$ extends $\psi$.   \Prf\ If
$(\sigma,\tau)\in\Cal S^{2\uparrow}$ and $\alpha\in\Bbb R$, then
$\Bvalue{\psi(\sigma,\tau)>\alpha}\in A_{\sigma\tau\alpha}$
so $\Bvalue{\psi(\sigma,\tau)>\alpha}\Bsubseteq a_{\sigma\tau\alpha}$.
If $(\sigma',\tau')\in\Cal S^{2\uparrow}$ then

$$\eqalign{\Bvalue{\psi(\sigma',\tau')>\alpha}&
\Bcap\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}\cr&
=\Bvalue{\psi(\sigma,\tau)>\alpha}
  \Bcap\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}
  \Bsubseteq\Bvalue{\psi(\sigma,\tau)>\alpha},\cr}$$

\noindent so
$a_{\sigma\tau\alpha}\Bsubseteq\Bvalue{\psi(\sigma,\tau)>\alpha}$.
Thus $\Bvalue{\hat\psi(\sigma,\tau)>\alpha}=\Bvalue{\psi(\sigma,\tau)>\alpha}$.
As $\alpha$ is arbitrary, $\hat\psi(\sigma,\tau)=\psi(\sigma,\tau)$.\ \Qed

\medskip

\quad{\bf (iv)}
If $(\sigma,\tau)$, $(\tilde\sigma,\tilde\tau)\in\hat{\Cal S}^{2\uparrow}$,
$c=\Bvalue{\sigma=\tilde\sigma}\Bcap\Bvalue{\tau=\tilde\tau}$ and
$\alpha\in\Bbb R$,
then $c\Bcap a_{\sigma\tau\alpha}\Bsubseteq a_{\tilde\sigma\tilde\tau\alpha}$.
\Prf\ If $a\in A_{\sigma\tau\alpha}$, express it as
$\Bvalue{\psi(\sigma',\tau')>\alpha}
  \Bcap\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}$ where
$(\sigma',\tau')\in\Cal S^{2\uparrow}$.   Then
$c\Bcap a\Bsubseteq\Bvalue{\tilde\sigma=\sigma'}\Bcap\Bvalue{\tilde\tau=\tau'}$
(611E(c-iv-$\gamma$)), so
$c\Bcap a\Bsubseteq a_{\tilde\sigma\tilde\tau\alpha}$.   It follows that

\Centerline{$c\Bcap a_{\sigma\tau\alpha}
=c\Bcap\sup A_{\sigma\tau\alpha}
=\sup_{a\in A_{\sigma\tau\alpha}}c\Bcap a
\Bsubseteq a_{\tilde\sigma\tilde\tau\alpha}$.  \Qed}

\noindent Similarly,
$c\Bcap a_{\tilde\sigma,\tilde\tau,\alpha}\Bsubseteq a_{\sigma\tau\alpha}$,
so $c\Bcap\Bvalue{\hat\psi(\sigma,\tau)>\alpha}
=c\Bcap\Bvalue{\hat\psi(\tilde\sigma,\tilde\tau)>\alpha}$.
As this is true for every $\alpha$,
$c\Bsubseteq\Bvalue{\hat\psi(\sigma,\tau)=\hat\psi(\tilde\sigma,\tilde\tau)}$.

\medskip

\quad{\bf (v)} If $\sigma\in\hat{\Cal S}$ then
$\hat\psi(\sigma,\sigma)=0$.   \Prf\ If $\sigma'\in\Cal S$
then

\Centerline{$\Bvalue{\sigma=\sigma'}
\Bsubseteq\Bvalue{\hat\psi(\sigma,\sigma)=\psi(\sigma',\sigma')}
\Bsubseteq\Bvalue{\hat\psi(\sigma,\sigma)=0}$;}

\noindent since $\sigma\in\hat{\Cal S}$,
$\sup_{\sigma'\in\Cal S}\Bvalue{\sigma=\sigma'}=1$ and
$\hat\psi(\sigma,\sigma)=0$.\ \QeD

It follows that $\hat\psi$ is a strictly adapted interval function.   \Prf\
I have already checked that $\hat\psi(\sigma,\tau)\in L^0(\frak A_{\tau})$
whenever $(\sigma,\tau)\in\hat{\Cal S}^{2\uparrow}$.
Now (iv) tells us that
$\hat\psi$ is a strictly adapted interval function.\ \Qed

\medskip

\quad{\bf (vi)} Finally, if $\psi':\hat{\Cal S}^{2\uparrow}\to L^0(\frak A)$
is any strictly adapted interval function
process extending $\psi$, and $(\sigma,\tau)\in\hat{\Cal S}^{2\uparrow}$, we
shall have

$$\eqalign{\Bvalue{\hat\psi(\sigma,\tau)=\psi'(\sigma,\tau)}&
\Bsupseteq\sup_{(\sigma',\tau')\in\Cal S^{2\uparrow}}
  \Bvalue{\hat\psi(\sigma,\tau)=\hat\psi(\sigma',\tau')}
  \Bcap\Bvalue{\psi'(\sigma,\tau)=\psi'(\sigma',\tau')}\cr&
\Bsupseteq\sup_{(\sigma',\tau')\in\Cal S^{2\uparrow}}
  \Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau=\tau'}
=1\cr}$$

\noindent by (i) again, so $\hat\psi(\sigma,\tau)=\psi'(\sigma,\tau)$.   Thus
$\hat\psi=\psi'$ and $\hat\psi$ is the unique strictly adapted interval
function extending $\psi$.

\medskip

{\bf (b)} Really this is immediate from the definitions and the fact that
$\hat\psi$ is always the unique strictly adapted extension of $\psi$, just as
(in (iii)) $\hatpmb u$ is the unique fully adapted extension of $\pmb u$.   Of
course we do need to know that $\hat\psi+\hat\psi'$, $\hat\psi\times\hat\psi'$,
$\bar h\hat\psi$ and $\hatpmb u\hat\psi$ are always strictly adapted, as noted
in 613D.
}%end of proof of 613U

\cmmnt{\medskip

\noindent{\bf Remark} Note that if $\pmb v$ is a fully adapted
process with domain $\Cal S$ and fully adapted extension $\hatpmb v$
to $\hat\Cal S$, then
\hbox{$(\Delta\pmb v)\mskip-4mu\hat{\phantom{\i}}=\Delta\hatpmb v$}.
}

\leader{613V}{}\cmmnt{ The definition of the integral means that we shall
always be able to approximate an integral $\int_{\Cal S}\pmb u\,d\psi$ by
Riemann sums $S_I(\pmb u,d\psi)$.   There will be occasions when it is
very useful to have simultaneous approximations for integrals over parts of
$\Cal S$.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$
a fully adapted process with domain $\Cal S$, and
$\psi$ an
adapted interval function with domain $\Cal S^{2\uparrow}$ such that
$\int_{\Cal S}\pmb u\,d\psi$ is defined.   Let
$I\in\Cal I(\Cal S)$ and $\epsilon>0$ be such that
$\theta(S_J(\pmb u,d\psi)-S_K(\pmb u,d\psi))\le\epsilon$ whenever
$J$, $K\in\Cal I(\Cal S)$ include $I$.

(i) If $\tau_0\le\tau'_0\le\tau_1\le\tau'_1\le\ldots\le\tau_n\le\tau'_n$ in
$I$, then

\Centerline{$\theta(\sum_{i=0}^n(S_{I\cap[\tau_i,\tau'_i]}(\pmb u,d\psi)
    -\int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb u\,d\psi))\le\epsilon$.}

(ii)($\alpha$) If $\tau\in I$ then
$\theta(S_{I\wedge\tau}(\pmb u,d\psi)-\int_{\Cal S\wedge\tau}\pmb u\,d\psi)
\le\epsilon$.

\quad($\beta$) For any $\tau\in\Cal S$,
$\theta(S_{I\wedge\tau}(\pmb u,d\psi)-\int_{\Cal S\wedge\tau}\pmb u\,d\psi)
\le 2\epsilon$.

\proof{{\bf (a)} Note straight away that by 613Jc all the integrals
$\int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb u\,d\psi$,
$\int_{\Cal S\wedge\tau}\pmb u\,d\psi$ will be defined.   Observe also that

\Centerline{$\theta(S_J(\pmb u,d\psi)-\int_{\Cal S}\pmb u\,d\psi)
=\lim_{K\uparrow\Cal I(\Cal S)}
   \theta(S_J(\pmb u,d\psi)-S_K(\pmb u,d\psi))\le\epsilon$}

\noindent whenevever $I\subseteq J\in\Cal I(\Cal S)$.

\medskip

{\bf (b)} For the time being, suppose that $\pmb u$ is the constant process
$\tbf 1$.

\medskip

\quad{\bf (i)} Take any $\eta>0$.   Then we have $J_0,\ldots,J_{n+1}$ such
that

$$\eqalignno{&J_0\in\Cal I(\Cal S\wedge\tau_0),\cr&
\mskip50mu
  \theta(S_K(\tbf 1,d\psi)-\int_{\Cal S\wedge\tau_0}d\psi)
\le\eta\text{ whenever }J_0\subseteq K\in\Cal I(\Cal S\wedge\tau_0),\cr&
J_i\in\Cal I(\Cal S\cap[\tau'_{i-1},\tau_i]),\cr&
\mskip50mu
  \theta(S_K(\tbf 1,d\psi)
   -\int_{\Cal S\cap[\tau'_{i-1},\tau_i]}d\psi)
\le\eta\text{ whenever }
J_i\subseteq K\in\Cal I(\Cal S\cap[\tau'_{i-1},\tau_i])\cr
\noalign{for $1\le i\le n$,}
&J_{n+1}\in\Cal I(\Cal S\vee\tau'_n),\cr&
\mskip50mu
  \theta(S_K(\tbf 1,d\psi)-\int_{\Cal S\vee\tau'_n}d\psi)
\le\eta\text{ whenever }J_{n+1}\subseteq K\in\Cal I(\Cal S\vee\tau_n).\cr}$$

\noindent Now take $K$ to be the sublattice generated by
$I\cup\bigcup_{i\le n+1}J_i$.   Observe that if $i\le n$ then
$\{\sigma:\sigma\in\Cal S$, $\med(\tau_i,\sigma,\tau'_i)\in I\}$ is a sublattice
of $\Cal S$ including $I\cup\bigcup_{i\le n+1}J_i$ so includes $K$, and
$K\cap[\tau_i,\tau'_i]=I\cap[\tau_i,\tau'_i]$.   We see also that

\Centerline{$\theta(S_K(\tbf 1,d\psi)-\int_{\Cal S}d\psi)
=\lim_{L\uparrow\Cal I(\Cal S)}
     \theta(S_K(\tbf 1,d\psi)-S_L(\tbf 1,d\psi))
\le\epsilon$}

\noindent because $K\supseteq I$.   Now

$$\eqalignno{S_K(\tbf 1,d\psi)&
=S_{K\wedge\tau_0}(\tbf 1,d\psi)
   +\sum_{i=0}^nS_{K\cap[\tau_i,\tau'_i]}(\tbf 1,d\psi)\cr&
\mskip50mu   +\sum_{i=1}^nS_{K\cap[\tau'_{i-1},\tau_i]}(\tbf 1,d\psi)
   +S_{K\vee\tau'_n}(\tbf 1,d\psi)\cr}$$

\noindent (613G(a-ii)),

\Centerline{$\int_{\Cal S}d\psi
=\int_{\Cal S\wedge\tau_0}d\psi
  +\sum_{i=0}^n\int_{\Cal S\cap[\tau_i,\tau'_i]}d\psi
  +\sum_{i=1}^n\int_{\Cal S\cap[\tau'_{i-1},\tau_i]}d\psi
  +\int_{\Cal S\vee\tau'_n}d\psi$}

\noindent (613J(c-ii)).   So

$$\eqalign{\theta(\sum_{i=0}^nS_{I\cap[\tau_i,\tau'_i]}&(\tbf 1,d\psi)
                     -\int_{\Cal S\cap[\tau_i,\tau'_i]}d\psi)\cr&
\le\theta(S_K(\tbf 1,d\psi)-\int_{\Cal S}d\psi)
   +\theta(S_{K\wedge\tau_0}(\tbf 1,d\psi)
                     -\int_{\Cal S\wedge\tau_0}d\psi)\cr&
\mskip50mu
   +\sum_{i=1}^n\theta(S_{K\cap[\tau'_{i-1},\tau_i]}(\tbf 1,d\psi)
                     -\int_{\Cal S\cap[\tau'_{i-1},\tau_i]}d\psi)\cr&
\mskip50mu
   +\theta(S_{K\vee\tau'_n}(\tbf 1,d\psi)
                     -\int_{\Cal S\vee\tau'_n}d\psi)\cr&
\le\epsilon+(n+2)\eta.\cr}$$

\noindent As $\eta$ is arbitrary,

\Centerline{$\theta(\sum_{i=0}^nS_{I\cap[\tau_i,\tau'_i]}(\tbf 1,d\psi)
                     -\int_{\Cal S\cap[\tau_i,\tau'_i]}d\psi)
\le\epsilon$,}

\noindent as claimed.

\medskip

\quad{\bf (ii)}\grheada\ The argument is the same as that of (i), but simpler.
For any
$\eta>0$, there is a $J\in\Cal I(\Cal S\vee\tau)$ such that
$\theta(S_K(\tbf 1,d\psi)-\int_{\Cal S\vee\tau}d\psi)\le\eta$ whenever
$K\in\Cal I(\Cal S\vee\tau)$ includes $J$.   Let $K$ be the sublattice generated
by $I\cup J$.   Since $I\cup J$ is included in the sublattice
$\{\sigma:\sigma\wedge\tau\in I\}$, $K\wedge\tau=I\wedge\tau$, while
$K\vee\tau\supseteq J$.   So

$$\eqalign{\theta(S_{I\wedge\tau}(\tbf 1,d\psi)
    -\int_{\Cal S\wedge\tau}d\pmb v)&
=\theta(S_{K\wedge\tau}(\tbf 1,d\pmb v)
    -\int_{\Cal S\wedge\tau}d\pmb v)\cr&
\le\theta(S_K(\tbf 1,d\pmb v)
    -\int_{\Cal S}d\pmb v))
  +\theta(S_{K\vee\tau}(\tbf 1,d\pmb v)
    -\int_{\Cal S\vee\tau}d\pmb v)\cr&
\le\epsilon+\eta;\cr}$$

\noindent as $\eta$ is arbitrary,
$\theta(S_{I\wedge\tau}(\tbf 1,d\pmb v)-\int_{\Cal S\wedge\tau}d\pmb v)
\le\epsilon$.

\medskip

\qquad\grheadb\ If $I$ is empty, then
$\theta(S_J(\tbf 1,d\psi))\le\epsilon$ for every $J\in\Cal I(\Cal S)$, so

\Centerline{$\theta(S_{I\wedge\tau}(\tbf 1,d\psi)
   -\int_{\Cal S\wedge\tau}d\psi)
=\theta(\int_{\Cal S\wedge\tau}d\psi)\le\epsilon$.}

\noindent Otherwise, let $J$ be the sublattice of $\Cal S$ generated by
$I\cup\{\tau\}$, and write $\tau^*$ for $\max I$.   Then

\Centerline{$I\wedge\tau=I\wedge\tau\wedge\tau^*
\subseteq J\wedge\tau\wedge\tau^*$,}

\noindent while $\{\sigma:\sigma\wedge\tau\wedge\tau^*\in I\wedge\tau\}$ is a
sublattice of $\Cal S$ including $I\cup\{\tau\}$, so includes $J$, and
$I\wedge\tau=J\wedge\tau\wedge\tau^*$.   Next,
$\theta(S_L(\tbf 1,d\psi)-S_{L'}(\tbf 1,d\psi))\le\epsilon$ whenever
$L$, $L'\in\Cal I(\Cal S)$ include $J$, so

\Centerline{$\theta(S_{I\wedge\tau}(\tbf 1,d\psi)
   -\int_{\Cal S\wedge\tau\wedge\tau^*}d\psi)
=\theta(S_{J\wedge\tau\wedge\tau^*}(\tbf 1,d\psi)
   -\int_{\Cal S\wedge\tau\wedge\tau^*}d\psi)
\le\epsilon$}

\noindent by ($\alpha$).   Because $\sigma\mapsto\int_{\Cal S\wedge\sigma}d\psi$
is fully adapted (613J(e-ii)),

$$\eqalignno{\int_{\Cal S\cap[\tau\wedge\tau^*,\tau]}d\psi&
=\int_{\Cal S\wedge\tau}d\psi-\int_{\Cal S\wedge\tau\wedge\tau^*}d\psi
=\int_{\Cal S\wedge(\tau\vee\tau^*)}d\psi-\int_{\Cal S\wedge\tau^*}d\psi
\Displaycause{612D(f-i)}
=\int_{\Cal S\cap[\tau^*,\tau\vee\tau^*]}d\psi
=\lim_{K\uparrow\Cal I(\Cal S\cap[\tau^*,\tau\vee\tau^*])}S_K(\tbf 1,d\psi).
   \cr}$$

\noindent But if $K\in\Cal I(\Cal S\cap[\tau^*,\tau\vee\tau^*])$ contains
$\tau^*$, then $I\cup K\in\Cal I(\Cal S)$,

$$\eqalign{S_K(\tbf 1,d\psi)=
&S_{(I\cup K)\vee\tau^*}(\tbf 1,d\psi)\cr&
=S_{I\cup K}(\tbf 1,d\psi)-S_{(I\cup K)\wedge\tau^*}(\tbf 1,d\psi)
=S_{I\cup K}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)\cr}$$

\noindent and

\Centerline{$\theta(S_K(\tbf 1,d\psi))
=\theta(S_{I\cup K}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi))\le\epsilon$;}

\noindent as $K$ is arbitrary,
$\theta(\int_{\Cal S\cap[\tau\wedge\tau^*,\tau]}d\psi)\le\epsilon$.
Accordingly

$$\eqalign{\theta(S_{I\wedge\tau}(\tbf 1,d\psi)-\int_{\Cal S\wedge\tau}d\psi)&
=\theta(S_{I\wedge\tau}(\tbf 1,d\psi)
   -\int_{\Cal S\wedge\tau\wedge\tau^*}d\psi
   -\int_{\Cal S\cap[\tau\wedge\tau^*,\tau]}d\psi)\cr&
\le\theta(S_{I\wedge\tau}(\tbf 1,d\psi)
   -\int_{\Cal S\wedge\tau\wedge\tau^*}d\psi)
   +\theta(\int_{\Cal S\cap[\tau\wedge\tau^*,\tau]}d\psi)\cr&
\le 2\epsilon,\cr}$$

\noindent as required.

\medskip

{\bf (c)} For the general case, recall that $\pmb u\psi$, as defined in
613Dd, is an adapted interval function, and that every sum or integral
$S_J(\pmb u,d\psi)$ or $\int_{\Cal S'}\pmb u\,d\psi$ can be interpreted as
$S_J(\tbf 1,d(\pmb u\psi))$ or $\int_{\Cal S'}d(\pmb u\psi)$ (613Ee,
613Hc).   So we get the general result at once.
}%end of proof of 613V

\leader{613W}{The one-dimensional case (a)}
\cmmnt{There are real difficulties in the
theory of stochastic integration, and it will be a long time before you can
expect it to feel natural and familiar.   For the rest of this chapter we
shall be teasing out more or less special cases in which the integral of
613H is defined.   But there is a particular special case which we can
approach immediately.}   Suppose that $(\frak A,\bar\mu)$ is the trivial
probability algebra in which $\frak A=\{0,1\}$.   Then
$L^0(\frak A)=\{\alpha\chi 1:\alpha\in\Bbb R\}$ can be identified, as
$f$-algebra, with $\Bbb R$;  of course we have
$\theta(\alpha\chi 1)=\min(1,|\alpha|)$ for every $\alpha$, so the topology
of convergence in measure on $L^0(\frak A)$ corresponds to the usual topology of
$\Bbb R$.   Necessarily, $\frak A_t=\frak A$ for every $t\in T$, so the
filtration is trivial.   If it is also the case that $T$ has no points isolated
on the right, then every stopping time except $\max\Cal T$ and possibly
$\min\Cal T$ will be a constant stopping time as described in 611A(b-ii),
every subset of $\Cal T$
is a sublattice, and every real-valued function $f$ defined on a subset $S$ of
$T$ corresponds to a fully adapted process $\{(\check s,f(s)\chi 1):s\in S\}$.

\spheader 613Wb If also $T$ has a least element, we can identify $\Cal T_f$ with $T$
and $\Mfa(\Cal T_f)=(L^0)^{\Cal T_f}$ with $\BbbR^T$.   Under this
identification, if $f:T\to\Bbb R$ and $g:T\to\Bbb R$ represent processes
$\pmb u$, $\pmb v$ with domain $\Cal T_f$, and $I\subseteq\Cal T_f$
is a non-empty finite set, there are $t_0\le\ldots\le t_n$ in $T$ such that
$I=\{\check t_i:i\le n\}$, and

\Centerline{$S_I(\pmb u,d\pmb v)
=\bigl(\sum_{i=0}^{n-1}f(t_i)(g(t_{i+1})-g(t_i))\bigr)\chi 1$.}

\spheader 613Wd What this amounts to is that we have a kind of
Riemann-Stieltjes integral on $T$,
I spell this out in detail here partly because there are well-known
Stieltjes integrals on the real line,
of which the most important, from the point of view of my treatise as a
whole, is integration with respect to Lebesgue-Stieltjes measures described
in exercises from 114Xa onwards.   Here we suppose that
$g:\Bbb R\to\Bbb R$
is non-decreasing, so that there is a Radon measure $\nu_g$ on $\Bbb R$ with
$\nu_g\coint{a,b}=\lim_{x\uparrow b}g(x)-\lim_{x\uparrow a}g(x)$ whenever
$a<b$ in $\Bbb R$.   Now the point I need to make here is that if
$\Cal S=\{\check s:s\in\Bbb R\}$ then the integral
$\int_{\Cal S}\pmb u\,d\pmb v$ is {\it not} the same as the
Lebesgue-Stieltjes integral $\int f\,d\nu_g$\cmmnt{, even in some of the most
elementary situations}.   Consider\cmmnt{, for instance,} the case in which
$f=g=\chi\coint{0,\infty}$.   \cmmnt{In this case,}
$\nu_g$ is the Dirac measure
concentrated at $0$, so that $\int f\,d\nu_g=f(0)=1$.   But when we look at
sums $S_I(\pmb u,d\pmb v)$ where $I=\{\check t_0,\ldots,\check t_n\}$
is a finite subset of $\Cal S$, and supposing that $t_0\le\ldots\le t_n$, we get

$$\eqalign{f(t_i)(g(t_{i+1})-g(t_i))
&=0(g(t_{i+1})-g(t_i))=0\text{ if }t_i<0,\cr
&=f(t_i)(1-1)=0\text{ if }t_i\ge 0,\cr}$$

\noindent so $S_I(f,dg)=0$;  as this is true for every $I$,
$\int_{\Cal S}f\,dg=0$.   In\cmmnt{ effect, the jump in $g$ at the time $0$
is necessarily applied to a value of $f$ calculated before the time $0$;
in} the language of Lebesgue-Stieltjes integration, we are calculating
$\int f_-\,d\nu_g$ where $f_-(x)=\lim_{y\uparrow x}f(y)$ for each $x$.

In my view,\cmmnt{ there are excellent reasons (especially in view of its
applications to financial mathematics) why} a theory of stochastic
integration should insist on calculating integrals $\int\pmb u\,d\pmb v$
in terms of products $u_{\sigma}\times(v_{\tau}-v_{\sigma})$ where
$\sigma\le\tau$ (rather than $u_{\tau}\times(v_{\tau}-v_{\sigma})$, for
instance).   We are going to have to return to this point from
time to time, because it is one on which my presentation of the theory
differs from that of most authors.

\leader{*613S}{Example} If $T=\coint{0,\infty}$, $(\frak A,\bar\mu)$ is the
measure algebra of Lebesgue measure on $[0,1]$ and $\frak A_t=\frak A$ for every
$t\ge 0$, then there are a sublattice
$\Cal S$ of $\Cal T$ and fully adapted processes $\pmb u$, $\pmb v$ with
domain $\Cal S$ such that $\int_{\Cal S}\pmb u\,d\pmb v$ is defined but
$\int_{\hat\Cal S}\hatpmb u\,d\hatpmb v$ is not, where $\hat\Cal S$ is the
covered envelope of $\Cal S$ and $\hatpmb u$, $\hatpmb v$ are the fully adapted
extensions of $\pmb u$ and $\pmb v$ to $\hat\Cal S$.

\ifwithproofs{\medskip

\noindent{\bf Construction (a)} For integers $k\in\Bbb N$ and $i$, $j<2^k$ write
$Q_{kij}$ for the half-open square
$\coint{2^{-k}i,2^{-k}(i+1)}\times\coint{2^{-k}j,2^{-k}(j+1)}$, so that
$\{Q_{kij}:i$, $j<2^k\}$ is a partition of $Q_{000}=\coint{0,1}^2$.   For
$k\ge 1$ set

$$\eqalign{\Cal Q_k&
=\{Q_{k,2i,2i+1}:i<2^{k-1}\}\cup\{Q_{k,2i+1,2i}:i<2{k-1}\}\cr&
\subseteq\bigcup_{i<2^{k-1}}Q_{k-1,i,i}\setminus\bigcup_{i<2^k}Q_{kii}.\cr}$$

\noindent Then

\inset{if $Q$, $Q'\in\Cal Q_k$ are distinct, their horizontal projections are
disjoint and their vertical projections are disjoint,

the horizontal and vertical projections of
$\bigcup\Cal Q_k$ are both $\coint{0,1}$,

$\bigcup\Cal Q_j\cap\bigcup\Cal Q_k=\emptyset$ whenever $j\ne k$.}

\noindent So $\Cal Q=\bigcup_{k\ge 1}\Cal Q_k$ is a disjoint family.

\medskip

{\bf (b)} Next, for $Q\in\Cal Q$, take the $k\ge 1$ and $i$, $j<2^k$
such that $Q=Q_{kij}$ and set

\Centerline{$R(Q)
=\coint{2^{-k}i,2^{-k}(i+1)}\times\coint{2^{-k}(j+\bover34,2^{-k}(j+1)}$,}

$$\eqalign{R_l(Q)
&=\coint{2^{-k}(i+2^{-k}l),2^{-k}(i+2^{-k}(l+1))}\cr
&\mskip150mu
   \times\coint{2^{-k}(j+2^{-k-2}l),2^{-k}(j+2^{-k-2}(l+1))}\cr}$$

\noindent for $l<2^k$, so that $R(Q)$ is the top quarter of $Q$ and the $R_l(Q)$ are
rectangles stepping up along the diagonal of the bottom quarter of $Q$.   Now
for $\omega\in\coint{0,1}$ and $t\ge 0$ define

$$\eqalign{X_t(\omega)
&=\Bover1{\sqrt{k}}\text{ if }k\ge 1,\,Q\in\Cal Q_k,\,l<2^{-k}
   \text{ are such that }(\omega,t)\in R_l(Q),\cr
&=0\text{ otherwise},\cr
Y_t(\omega)
&=\Bover1{\sqrt{k}}\text{ if }k\ge 1,\,Q\in\Cal Q_k
   \text{ are such that }(\omega,t)\in R(Q),\cr
&=0\text{ otherwise}.\cr}$$

\medskip

{\bf (c)} Taking $(\Omega,\Sigma,\mu)$ to be Lebesgue measure on $\coint{0,1}$
and $\Sigma_t$ to be $\Sigma$ for every $t\ge 0$,
$(\Omega,\mu,\langle\Sigma_t\rangle_{t\ge 0})$ represents
$(\frak A,\bar\mu,\langle\frak A_t\rangle_{t\ge 0}$ is the manner of Theorem
612H, while $\langle X_t\rangle_{t\ge 0}$ and $\langle Y_t\rangle_{t\ge 0}$
are progressively measurable,
so we have corresponding fully adapted processes
$\pmb u=\family{\sigma}{\Cal T_f}{u_{\sigma}}$ and
$\pmb v=\family{\sigma}{\Cal T_f}{v_{\sigma}}$ (612Hb).

\medskip

{\bf (d)(i)} Since $\Cal Q$ is disjoint,
$Y_t(\omega)=0$ whenever $X_t(\omega)>0$, and
$S_I(\pmb{u},d\pmb{v})\ge\Delta_e(\pmb{u},d\pmb{v})\ge 0$ whenever $I$ is a
finite sublattice of $\Cal T_f$ and $e$ is an $I$-cell.

\medskip

\quad{\bf (ii)} Suppose that $m\ge 1$, $Q\in\Cal Q_m$, $(\omega,s)\in Q$,
$0\le s\le t$ and
$X_s(\omega)(Y_t(\omega)-Y_s(\omega))\ne 0$.   Then we must have
$X_s(\omega)=\Bover1{\sqrt{m}]}$ and $Y_t(\omega)>0$.   In this case, there are
$i$, $j$, $l<2^m$ such that $Q=Q_{mij}$ and
$(\omega,s)\in R_l(Q_{mij})$, that is,

\Centerline{$2^{-m}(i+2^{-m}l)\le\omega<2^{-m}(i+2^{-m}(l+1)$,}

\Centerline{$2^{-m}(j+2^{-m-2}l)\le s<2^{-m}(j+2^{-m-2}(l+1))$.}

\noindent Now $Y_{t'}(\omega)=0$ whenever $(\omega,t')$ belongs to the bottom
three-quarters of $Q$, that is, whenever
$2^{-m}j\le t'<2^{-m}(j+\bover34)$, and $t$ must be at least
$2^{-m}(j+\bover34)$, while $s<2^{-m}(j+\bover14)$.   Accordingly
$t-s>2^{-m-1}$.   Also, of course, $t<1$.

\medskip

{\bf (e)(i)} Now suppose that $0\le s_0<\ldots<s_n$ and
$k\ge 1$ is such that $s_{p+1}-s_p\le 2^{-k}$ for every $p<n$.
For $m\ge 1$ and $Q\in\Cal Q_m$, set

\Centerline{$E_{mQ}=\bigcup_{p<n}
     \{\omega:(\omega,s_p)\in Q,\,X_{s_p}(\omega)Y_{s_{p+1}}(\omega)>0\}$.}

\medskip

\quad{\bf (ii)} If $m<k$ then $E_{mQ}=\emptyset$ for every $Q\in\Cal Q_m$.
\Prf\ If $p<n$ then $s_{p+1}-s_p\le 2^{-m+1}$, so by (d-ii) there can be no
$(\omega,s_p)$ such that $X_{s_p}(\omega)Y_{s_{p+1}}(\omega)>0$.\ \Qed

\medskip

\quad{\bf (iii)} If $m\ge k$ and $Q\in\Cal Q_m$ then $\mu E_{mQ}\le 2^{-2m}$.
\Prf\ We have $Q=Q_{mij}$ where $i$, $j<2^{-m}$.
If $p<n$ and there is any $\omega$ such that $(\omega,s_p)\in Q$
and $X_{s_p}(\omega)Y_{s_{p+1}}(\omega)>0$, then
$s_p<2^{-m}(j+\bover14)$ and $2^{-m}(j+\bover34)\le s_{p+1}$, by (d-ii).
Now there can be at most one such $p$.   So if $E_{mQ}\ne\emptyset$ there is a
$p<n$ such that

\Centerline{$E_{mQ}
=\{\omega:(\omega,s_p)\in Q,\,X_{s_p}(\omega)Y_{s_{p+1}}(\omega)>0\}
\subseteq\{\omega:(\omega,s_p)\in R_l(Q)\}$}

\noindent where $l=\lfloor 2^{2m+1}(s_p-2^{-m}j)\rfloor<2^m$.   Accordingly, in
this case, $\mu E_{mQ}\le 2^{-2m}$, as required.\ \Qed

\medskip

\quad{\bf (iv)} Putting (ii) and (iii) together,

$$\eqalign{&
\{\omega:\sum_{p=0}^{n-1}X_{s_p(\omega)}(Y_{s_{p+1}}(\omega)-Y_{s_p}(\omega))
       \ne 0\}\cr&
\mskip100mu=\bigcup_{p<n}\{\omega:X_{s_p}(\omega)Y_{s_{p+1}}(\omega)>0\}\cr&
\mskip100mu=\bigcup_{m\ge 1,Q\in\Cal Q_m}E_{mQ}
=\bigcup_{m\ge k,Q\in\Cal Q_m}E_{mQ}\cr}$$

\noindent has measure at most

\Centerline{$\sum_{m=k}^{\infty}2^{-2m}\#(\Cal Q_m)
=\sum_{m=k}^{\infty}2^{-m}=2^{-k+1}$.}

\medskip

{\bf (f)} Writing $\Cal T_c$ for the set $\{\check t:t\ge 0\}$ of constant
stopping times, $\int_{\Cal T_c}\pmb u\,d\pmb v$ is defined and equal to $0$.
\Prf\ Given $\epsilon>0$, take $k\ge 1$ such that $2^{-k+1}\le\epsilon$,
and let $I\in\Cal I(\Cal T_c)$ be such that
$(2^{-k}i)\check{\phantom{g}}\in I$ for every $i<2^k$.   Enumerate $I$ in
increasing order as $(\check s_0,\ldots,\check s_n)$.   If $p<n$ and
$s_p\ge 1$ then certainly
$X_{s_p(\omega)}(Y_{s_{p+1}}(\omega)-Y_{s_p}(\omega))=0$ for every $\omega$.
So if $n'$ is such that $s_{n'}=1$,

$$\eqalign{&\{\omega:\sum_{p=0}^{n-1}
   X_{s_p(\omega)}(Y_{s_{p+1}}(\omega)-Y_{s_p}(\omega))\ne 0\}\cr&
\mskip100mu
=\{\omega:\sum_{p=0}^{n'-1}X_{s_p(\omega)}(Y_{s_{p+1}}(\omega)-Y_{s_p}(\omega))
       \ne 0\}\cr}$$

\noindent has measure at most  $2^{-k+1}\le\epsilon$, and

$$\eqalign{\theta(S_I(\pmb u,\pmb v))&
\le\bar\mu\Bvalue{S_I(\pmb u,d\pmb v)\ne 0}\cr&
=\mu\{\omega:\sum_{p=0}^{n-1}
   X_{s_p(\omega)}(Y_{s_{p+1}}(\omega)-Y_{s_p}(\omega))>00\}
\le\epsilon.\cr}$$

\noindent And this is true whenever $I\in\Cal I(\Cal T_c)$ includes
$\{(2^{-k}i)\check{\phantom{g}}:i<2^k\}$.   As $\epsilon$ is arbitrary,
$\int_{\Cal T_c}\pmb u\,d\pmb v=0$.

\medskip

{\bf (g)} I turn now to the covered envelope $\hat{\Cal T}_c$ of $\Cal T_c$.

\medskip

\quad{\bf (i)} For each $k\ge 1$ define stopping times $h_k$,
$h'_k:\coint{0,1}\to\coint{0,\infty}$ by setting

\Centerline{$h_k(\omega)=2^{-k}(j+2^{-k-2}l)$,
\quad $h'_k(\omega)=2^{-k}(j+\bover34)$}

\noindent if $i$, $j$, $l<2^k$ are such that
$Q_{kij}\in\Cal Q_k$ and $2^{-k}(i+2^{-k}l)\le\omega<2^{-k}(i+2^{-k}(l+1))$.
Then we have $h_k(\omega)<h'_k(\omega)$ and

\Centerline{$X_{h_k}(\omega)=X_{h_k(\omega)}(\omega)=Y_{h'_k}(\omega)
=\Bover1{\sqrt{k}}$,
\quad$X_{h_k}(\omega)(Y_{h'_k}(\omega)-Y_{h_k}(\omega))=\Bover1k$}

\noindent for every $\omega$.   If $k$, $m\ge 1$ are different, then for every
$\omega$ there is a $Q\in\Cal Q_k$ such that

\Centerline{$[h_k(\omega),h'_k(\omega)]\subseteq\{s:(\omega,s)\in Q\}
\subseteq\{s:(\omega,s)\in\bigcup\Cal Q_k\}$;}

\noindent as $\bigcup\Cal Q_k$ is disjoint from $\bigcup\Cal Q_m$,
$[h_k(\omega),h'_k(\omega)]\cap[h_m(\omega),h'_m(\omega)]=\emptyset$.

\medskip

\quad{\bf (ii)} For $k>1$, write $\tau_k$, $\tau'_k$ for the stopping times in
$\Cal T_f$ associated with $h_k$ and $h'_k$.   Since
all the values of $h_k$ and $h'_k$ belong to the countable set
$I=\{2^{-k}(j+2^{-k-2}l):j$, $l<2^k\}\cup\{2^{-k}(j+\bover34):j<2^k\}$,

\Centerline{$\sup_{t\in I}\Bvalue{\tau_k=\check t}
=\sup_{t\in I}\{\omega:h(\omega)=t\}^{\ssbullet}
=\{\omega:h_k(\omega)\in I\}^{\ssbullet}=1$}

\noindent and $\tau_k\in\hat{\Cal T}_c$;  similarly,
$\tau'_k\in\hat{\Cal T}_c$;

Next, for each $k$, 612Hb tells us that

\Centerline{$\Bvalue{\tau_k<\tau'_k}=1$,
\quad$u_{\tau_k}=X_{h_k}^{\ssbullet}=\Bover1{\sqrt{k}}\chi 1$,
\quad$v_{\tau'_k}=Y_{h'_k}^{\ssbullet}=\Bover1{\sqrt{k}}\chi 1$,
\quad$v_{\tau_k}=Y_{h_k}^{\ssbullet}=0$,}

\noindent so
$u_{\tau_k}\times(v_{\tau'_k}-v_{\tau_k})=\Bover1k\chi 1$;  moreover, if
$k\ne m$, then for every $\omega\in\coint{0,1}$ either
$h'_k(\omega)<h_m(\omega)$ or $h'_m(\omega)<h_k(\omega)$, so
$\Bvalue{\tau'_k<\tau_m}\Bcup\Bvalue{\tau'_m<\tau_k}=1$.

It follows that if $\sigma\in\Cal T_f$ then

\Centerline{$(\Bvalue{\tau_k<\sigma}\Bcap\Bvalue{\sigma<\tau'_k})
   \Bcap(\Bvalue{\tau_m<\sigma}\Bcap\Bvalue{\sigma<\tau'_m})=0$}

\noindent whenever $k$, $m$ are distinct.

\medskip

{\bf (h)(i)} Now suppose that $I\in\Cal I(\hat{\Cal T}_c)$.   Let $n>\#(I)+1$ be
so large that $\sum_{k=\#(I)+2}^n\Bover1k\ge 1$.   Let $J$ be the sublattice of
$\hat{\Cal T}_c$ generated by
$I\cup\{\tau_k:1\le k\le n\}\cup\{\tau'_k:1\le k\le n\}$, and $B$ the
set of atoms of the subalgebra $\frak B$ of $\frak A$ generated by
$\{\Bvalue{\sigma<\tau}:\sigma$, $\tau\in J\}$.   Take any $b\in B$.
If $\sigma$, $\tau\in J$ then $b$ is either included in or disjoint from
$\Bvalue{\sigma<\tau}$.   Now if $\sigma\in I$, there can be at most
one $k\le n$ such that
$b\Bsubseteq\Bvalue{\tau_k<\sigma}\Bcap\Bvalue{\sigma<\tau'_k}$, so the
set

$$\eqalign{K&
=\{k:\,\Exists\sigma\in I,\,
   b\Bsubseteq\Bvalue{\tau_k<\sigma}\Bcap\Bvalue{\sigma<\tau'_k}\}\cr&
=\{k:\,\Exists\sigma\in I,\,
   b\Bcap\Bvalue{\tau_k<\sigma}\Bcap\Bvalue{\sigma<\tau'_k}\ne 0\}\cr}$$

\noindent has at most $\#(I)$ members.   Set
$K'=\{k:1\le k\le n$, $k\notin K\}$, so that $\sum_{k\in K'}\Bover1k\ge 1$.
Now

\Centerline{$\{\sigma:
   b\Bcap\Bvalue{\tau_k<\sigma}\cap\Bvalue{\sigma<\tau_k}=0$
for every $k\in K'\}$}

\noindent is a sublattice of $\Cal T$ including
$I\cup\{\tau_k:1\le k\le n\}\cup\{\tau'_k:1\le k\le n\}$ and therefore including
$J$.

\medskip

\quad{\bf (ii)}  Fix $k\in K'$ for the moment.
If we think of intervals $c(\sigma,\tau)$, for $\sigma\le\tau$ in $\Cal T$, as
members of $\prod_{t\ge 0}\frak A_t=\frak A^{\coint{0,\infty}}$, as in 611J,
then we see that

\Centerline{$c(\tau_k,\tau'_k)=\sup\{e:e\in\Sti_0(J)$,
$e\Bsubseteq c(\tau_k,\tau'_k)\}$}

\noindent where $\Sti_0(J)$ is the set of $J$-cells (611J(e-i)).   Consequently

$$1=\Bvalue{\tau_k,\tau'_k}=\sup_{t\ge 0}c(\tau_k,\tau'_k)(t)
=\sup_{t\ge 0}\sup_{\Atop{e\in\Sti_0(J)}{e\Bsubseteq c(\tau_k,\tau'_k)}}e(t)$$

\noindent and there must be an $e_k\in\Sti_0(J)$ such that
$e_k\Bsubseteq c(\tau_k,\tau'_k)$ and $b\Bcap\sup_{t\ge 0}e_k(t)\ne 0$.   Expressing
$e_k$ as $c(\sigma,\tau)$ where $\sigma\le\tau$ in $J$, we have
$\sup_{t\ge 0}e_k(t)=\Bvalue{\sigma<\tau}$.   As this
belongs to $\frak B$ and meets $b$, it includes $b$, and
$b\Bsubseteq\Bvalue{\sigma<\tau}$.   At the same time,
$\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\tau_k\le\sigma}\Bcap\Bvalue{\tau\le\tau'_k}$ (611Jc) so
$b\Bsubseteq\Bvalue{\sigma\vee\tau_k<\tau'_k}$;  as $k\in K'$,
$b\Bcap\Bvalue{\tau_k<\sigma\vee\tau_k}=0$ and
$b\Bsubseteq\Bvalue{\sigma=\tau_k}$.   Similarly,
$b\Bsubseteq\Bvalue{\tau=\tau_{k+1}}$.   But this means that

\Centerline{$b\Bsubseteq\Bvalue{u_{\sigma}\times(v_{\tau}-v_{\sigma})
    =u_{\tau_k}\times(v_{\tau'_k}-v_{\tau_k})}
\Bsubseteq\Bvalue{\Delta_{e_k}(\pmb u,d\pmb v)=\Bover1k}$.}

\medskip

\quad{\bf (iii)}
This is true for every $k\in K'$.   If $k$, $m\in K'$ are distinct, then
either $\tau'_m\le\tau_k$ or $\tau'_k\le\tau_m$, and in either case
$c(\tau_k,\tau'_k)\Bcap c(\tau_m,\tau'_m)=0$ in $\frak A^{\coint{0,\infty}}$ and
$e_k\ne e_m$.   As we also know that $\Delta_e(\pmb u,d\pmb v)\ge 0$ for every
stopping-time interval with endpoints in $\Cal T_f$ ((d-i) above),
we have

\Centerline{$S_J(\pmb u,d\pmb v)=\sum_{e\in\Sti_0(J)}\Delta_e(\pmb u,d\pmb v)
\ge\sum_{k\in K'}\Delta_{e_k}(\pmb u,d\pmb v)$}

\noindent and

\Centerline{$b\Bsubseteq\Bvalue{S_J(\pmb u,d\pmb v)\ge\sum_{k\in K'}\Bover1k}
\Bsubseteq\Bvalue{S_J(\pmb u,d\pmb v)\ge 1}$.}

\medskip

\quad{\bf (iv)} As $b$ is arbitrary, $1\Bsubseteq\Bvalue{S_J(\pmb u,d\pmb v)\ge 1}$
and $S_J(\pmb u,d\pmb v)\ge\chi 1$, while
$I\subseteq J\in\Cal I(\hat{\Cal T}_c)$.   As $I$ is arbitrary,
$\lim_{J\uparrow\Cal I(\hat{\Cal T}_c)}S_J(\pmb u,d\pmb v)$ either does not
exist or exists and is not equal to $0$.
But the latter is impossible, by 613T, because
we saw in (f) that $\int_{\Cal T_c}\pmb u\,d\pmb v=0$.   So we find that
$\lim_{J\uparrow\Cal I(\hat{\Cal T}_c)}S_J(\pmb u,d\pmb v)$ is undefined, that
is, that $\int_{\hat{\Cal T}_c}\pmb u\,d\pmb v$ is undefined.

\medskip

\noindent{\bf Remark} The processes $\pmb u$ and $\pmb v$ have a number of
special properties which will be discussed in this volume;  see in particular
615Yc and 618Yd.
}\else\fi  %end of construction 613S

\exercises{\leader{613X}{Basic exercises (a)}
%\spheader 613Xa
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a fully adapted process defined on $\Cal S$.
Show that $\int_{\Cal S}\pmb v\,d\pmb v$ is defined whenever
$\int_{\Cal S}d\pmb v$ and
$\int_{\Cal S}(d\pmb v)^2$ are defined, and that in this case

\Centerline{$2\int_{\Cal S}\pmb v\,d\pmb v
    +\int_{\Cal S}(d\pmb v)^2
=\lim_{\tau\uparrow\Cal S}v_{\tau}^2
-\lim_{\tau\downarrow\Cal S}v_{\tau}^2$.}
%613J 613N

\spheader 613Xc Let $\Cal S$ be a full sublattice of
$\Cal T$, $\pmb u$ a fully adapted
process with domain $\Cal S$, and $\psi$ an adapted interval function with
domain $\Cal S^{2\uparrow}$.   Write $\Cal S'$ for the domain of
the indefinite integral $ii_{\psi}(\pmb u)$.   Show that
$\Cal S'$ is full.
%613O

\spheader 613Xd In 613V, show that if $\tau$, $\tau'\in I$ and
$I\subseteq J\in\Cal I(\Cal S)$ then
$\theta(S_{J\wedge\tau}(\pmb u,d\psi)
   -\int_{\Cal S\wedge\tau}\pmb u\,d\psi)$,
$\theta(S_{J\vee\tau'}(\pmb u,d\psi)
   -\int_{\Cal S\vee\tau'}\pmb u\,d\psi)$
and $\theta(S_{J\cap[\tau,\tau']}(\pmb u,d\psi)
   -\int_{\Cal S\cap[\tau,\tau']}\pmb u\,d\psi)$ are all at most $\epsilon$.
%613V

\spheader 613Xe Let $\Cal S$ be a sublattice of $\Cal T$ and $\Cal I_t$ the
set of finite totally ordered subsets of $\Cal S$.   Suppose that
$\pmb u$ is a fully
adapted process with domain $\Cal S$, $\psi$ is
an adapted interval function with
domain $\Cal S^{2\uparrow}$ and $z$ is a member of $L^0(\frak A)$.   Show that
$\int_{\Cal S}\pmb u\,d\psi=z$ iff for every $\epsilon>0$ and $I\in\Cal I_t$
there is a $J\in\Cal I_t$, including $I$, such that
$\theta(S_K(\pmb u,d\psi)-z)\le\epsilon$ whenever $K\in\Cal I_t$ includes
$J$.
%613Ec 613G(a-ii)

\leader{613Y}{Further exercises (a)}
%\spheader 613Ya
Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$ a fully adapted process defined on $\Cal S$ and $\psi$ an
adapted interval function on $\Cal S^{2\uparrow}$.   Show that if
$\tau$, $\tau'\in\Cal S$, then
$\int_{\Cal S\vee(\tau\wedge\tau')}\pmb u\,d\psi$ is
defined iff both $\int_{\Cal S\vee\tau}\pmb u\,d\psi$ and
$\int_{\Cal S\vee\tau'}\pmb u\,d\psi$ are defined, and in this case

\Centerline{$\int_{\Cal S\vee(\tau\wedge\tau')}\pmb u\,d\psi
+\int_{\Cal S\vee\tau\vee\tau'}\pmb u\,d\psi
=\int_{\Cal S\vee\tau}\pmb u\,d\psi
+\int_{\Cal S\vee\tau'}\pmb u\,d\psi$.}
%613J

\spheader 613Yb Suppose that $\Cal S$ is a sublattice of $\Cal T$ and
$\Cal S'$ a covering ideal of $\Cal S$.   Let $\pmb u$ be a fully
adapted process defined on $\Cal S$ and $\psi$ an adapted interval
function defined on $\Cal S^{2\uparrow}$.   Show that
$\int_{\Cal S}\pmb u\,d\psi$ is defined iff
$\int_{\Cal S'}\pmb u\,d\psi$ is defined, and the integrals are then
equal.
%613Je  613U prf (a-i)

\spheader 613Yc
Let $\Cal C$ be the set of all bounded intervals in $\Bbb R$, and
$T^*$ the straightforward set of tagged partitions
generated by $Q=\{(a,C):C\in\Cal C\setminus\{\emptyset\}$, $a=\inf C\}$
(see 481B).
For a finite set $I\subseteq\Bbb R$, let $\delta_I$ be the set
$\{\pmb{t}:\pmb{t}\in T^*$,
$I\cap\interior C=\emptyset$ whenever $(a,C)\in\pmb{t}\}$;  for
$a\le b$ in $\Bbb R$, let $\Cal R_{ab}$ be
$\{\emptyset\}\cup\{\Bbb R\setminus[c,d]:c\le a$, $b\le d\}$.
Set $\Delta=\{\delta_I:I\in[\Bbb R]^{<\omega}\}$ and
$\frak R=\{\Cal R_{ab}:a\le b\}$ (see 481K).   (i) Show that
$(\Bbb R,T^*,\Delta,\frak R)$
is a tagged-partition structure allowing subdivisions, witnessed by
$\Cal C$ (see 481G).
(ii) For a function $g:\coint{0,\infty}\to\Bbb R$ define
$\nu_g:\Cal C\to\Bbb R$ by saying that $\nu_g\emptyset=0$,
$\nu_gC=g(\sup C)-g(\inf C)$ for non-empty $C\in\Cal C$.
Let $\Cal F(T^*,\Delta,\frak R)$ be the filter on $T^*$ defined from
$\Delta$ and $\frak R$ as in 481F, and for $f:\Bbb R\to\Bbb R$ set
$I_{\nu_g}(f)
=\lim_{\pmb{t}\to\Cal F(T^*,\Delta,\frak R)}S_{\pmb{t}}(f,\nu_g)$ when this
is defined, as in 481C.   Next, for functions $f$, $g:\Bbb R\to\Bbb R$,
define $\int_{\Cal S}f\,dg$ as in 613W, taking $\Cal S$ to be the set of
constant stopping times when $\frak A=\{0,1\}$ and $T=\Bbb R$, and $f$ and
$g$ are interpreted as functions from $\Cal S$ to $L^0(\frak A)$.
Show that $I_{\nu_g}(f)=\int_{\Cal S}f\,dg$ when either is defined.
%613W

\spheader 613Yd In 613S, show that $\int_{\Cal S}\pmb u\,d\pmb v$ is undefined
for any sublattice $\Cal S$ of $\Cal T_f$ including $\hat{\Cal T}_c$.
%613S

\spheader 613Ye Give an example of a strictly adapted interval function $\psi$
on a sublattice $\Cal S$ such that $\int_{\Cal S}d\psi$ is defined, but
$\int_{\hat\Cal S}d\hat\psi$ is not, where $\hat\psi$ is the strictly adapted
extension of $\psi$ on the covered envelope $\hat\Cal S$ of $\Cal S$.
%613S  %613D %613T \psi = \pmb u\Delta\pmb v

\spheader 613Yf Show that for strictly adapted interval functions $\psi$ we can
re-work this section in terms of a definition of
$S_I(\pmb u,d\psi)$ restricted to finite totally ordered sets $I$, as in
613Ec, taking $\int_{\Cal S}\pmb u\,d\psi$ to be the limit
$\lim_{I\uparrow\Cal I_t(\Cal S)}S_I(\pmb u,d\psi)$ where
$\Cal I_t(\Cal S)$ is the set of finite totally ordered subsets of $\Cal S$ with
the pre-order $\sqsubseteq$ of 611Yd.   (Begin by showing that
$S_I(\pmb u,d\psi)=S_J(\pmb u,d\psi)$ whenever
$I\sqsubseteq J$ and $J\sqsubseteq I$.)

\leaveitout{\spheader 613Yg\query
Re-work this section using the criterion of 613Xe to define
the Riemann-sum integral.   (Begin by showing that
$S_J(\pmb u,d\psi)=S_K(\pmb u,d\psi)$ whenever
$I$ is a finite sublattice of $\Cal S$ and $J$, $K$ are both maximal totally
ordered subsets of $\Cal I$.)
%hence dodging I-cells?
}
}%end of exercises


\endnotes{
\Notesheader{613} My objective in this section has been to reach a
formally adequate definition of a stochastic integral as quickly as
possible, with just enough of its properties to serve as a foundation
for the rest of the volume.

The definition in 613H is a `gauge integral' of the type
examined in \S\S481-482.   I cannot quote the results there because it is a
vector-valued, rather than scalar-valued integral;  and even on the basis
of a vector-valued version of the material, which is easy to develop
(there is a hint in 613Yc),
the ideas of \S482 do not directly illuminate the theory here.
(Though there is an echo of the Saks-Henstock Lemma, 482B, in 613V.)   The
disadvantage of this headlong approach is that we have no real idea of
which pairs $\pmb u$, $\pmb v$ will match to give integrals
$\int_{\Cal S}\pmb u\,d\pmb v$.   The case of simple processes (614B-614C)
is plainly elementary  -- indeed, 613L(b-i) doesn't belong to the
topological theory at all, and could (along with
613C-613G) %613C 613D 613E 613F 613G
have been expressed in the context of \S\S611-612,
with an arbitrary Dedekind
complete algebra $\frak A$ rather than a probability algebra.
The integral $\int_{\Cal S}|d\pmb v|$ in 614I below can also
be expressed in terms of a notion of convergence
which does not involve the topology of convergence in measure,
but is off the line of the main argument, as well as being elementary.
The centre of the theory is really occupied by
martingale integrators $\pmb v$, and we have a fair
amount of work to do before these become accessible in \S622.

All the main work of this chapter will be done with strictly adapted interval
functions, starting with the basic examples $\Delta\pmb v$ as in 613Cc.   I
include the more general formulation of `adapted interval function' in 613C(a-i)
only because the language of 613E will be useful in \S626 when talking about a
quite different kind of integral.

The point of 614Ba is that in the formula of 612J defining `simple process'
there is a canonical choice of the starting value $u_*$ definable from the
process $\pmb u$.   I delayed the result to this point because
it speaks of `limits', and it seems convenient to treat this as a limit for
the topology of convergence in measure;  but if you look at the proof
you will see that the limit required is really an order-limit, and
could have been described in the context of an arbitrary Dedekind complete
Boolean algebra $\frak A$, as in \S612.

While I have, I hope, given an exact definition of the integration process which
will dominate the next three chapters, this section suffers from
a singular lack of calculation of particular examples.   In fact it is by no
means trivial to show that even in the cases of our three leading examples (the
identity process, Brownian motion and the Poisson process) we have a full set of
indefinite integrals.   There is an effective description of indefinite
integrals of simple processes in 614Xb, but for other integrands, even when
integrating with respect to simple processes as in 614D, we aren't yet in a
position to get a formula.   One will appear in 641J.

I ought to remark that the integral I
have defined does not coincide with everyone's.   I will return to this point in
Chapter 64, in the course of defining what I call the `S-integral'.
Commonly the primary definition of the integral is based on what I call
`previsibly simple' processes, and the formula of 641Yd(ii)
is used to define $\int_{[\tau,\tau']}\pmb u\,d\pmb v$.   As will be noted in
641Yd, this in itself won't affect the principal cases.

I will present a large number of results showing that processes share properties
with their fully adapted extensions to the covered envelopes of their domains.
In those cases in which we have a good match between properties of
$\pmb u$ and $\hatpmb u$, starting with the isomorphism between $\Mfa(\Cal S)$
and $\Mfa(\hat\Cal S)$ (612Qb), we shall have a similar match between properties
of $\pmb u$ and of its alternative realisation $\hatpmb u\restr\Cal S'$ on any
sublattice $\Cal S'$ with the same covered envelope, that is, such that
$\Cal S$ covers $\Cal S'$ and $\Cal S'$ covers $\Cal S$.
These offer powerful methods for reducing problems to questions in which
processes are defined on full sublattices of $\Cal T$.
It is correspondingly
important to recognise cases in which the correspondence may not be exact.   One
is in 613T:  if $\int_{\hat\Cal S}\hatpmb u\,d\hat\psi$ is defined, then
$\int_{\Cal S}\pmb u\,d\psi$ is defined, with the same value;  but the
converse is not universally true (613S).   (In the opposite direction we have
612Xj and 612Yf.)   I have starred 613S because it is
hard work and the methods I use do not seem to contribute much to our toolkit.
Ordinarily I would leave it as a `further exercise'.   But like a rocky outcrop
in the bank of a river it affects the flow of the arguments in this volume, so
I have taken three pages %\query
to include a detailed solution.

Nearly everything in the rest of this volume will be expressed in
terms of structures \Asisfour\ where
$(\frak A,\bar\mu)$ is a probability algebra, $T$ is a non-empty totally
ordered set, and $\family{t}{T}{\frak A_t}$ is a filtration of closed
subalgebras of $\frak A$;  for definiteness, it will often be useful to
simultaneously declare names for the lattice of stopping times adapted to
$\family{t}{T}{\frak A_t}$ and the associated family of closed subalgebras
of $\frak A$ to get a sextuple \Asissix.
I will call such quadruples or sextuples
{\bf stochastic integration structures}.   When $T=\coint{0,\infty}$, as in
612F and 612H, so that we have a structure
\Asisfourt\ or \Asissixt, I will
call it a {\bf real-time} stochastic integration structure.

Perhaps I should note explicitly that a stochastic integration structure
\Asissix\ carries a lot of undeclared baggage.   The symbol $\frak A$ involves
not just the set $\frak A$ but its Boolean operations $\Bsymmdiff$, $\Bcap$,
$\Bcup$ and $\Bsetminus$, with the induced relation $\Bsubseteq$, and its
greatest and least elements $1$ and $0$.
The symbol $T$ is accompanied by its total ordering
$\le$ and the associated relation $<$.   With $\Cal T$ we have the
functions $(\sigma,\tau)\mapsto\Bvalue{\sigma<\tau}$,
$\Bvalue{\sigma\le\tau}$ and $\Bvalue{\sigma=\tau}$.   We are all well used to
such things being omitted from hypotheses, but in the rest of this volume
we shall have to be ready for a richer hidden substructure than is usual.

In the introduction to this volume
I said I would try to explain why I am working
with structures $(\frak A,\bar\mu)$ and $L^0(\frak A)$ rather than with
probability spaces and measurable functions.   The move from
$(\Omega,\Sigma,\mu,\langle\Sigma_t\rangle_{t\ge 0})$ to
$(\frak A,\bar\mu,\family{t}{T}{\frak A_t})$ is really no more than a
change in notation.   Every probability algebra, in the sense here,
can be represented by a classical probability space (321J),
and while the move from $\coint{0,\infty}$ to
an arbitrary non-empty totally ordered space is a generalisation, I shall
have nothing significant to say about the new cases.   But the move from
real-valued measurable functions to their equivalence classes in $L^0$ is forced
on us by the definition of the integral in 613H.   The integral is defined as a
limit in $L^0$ for the topology of convergence in measure;  and as this is a
Hausdorff topology, the limit, when it exists, is unique.   There are variations
of the theory using different topologies -- in \S626, for instance, there will
be limits for the weak topology on $L^1$ -- but none of them give functions
as limits;  they all provide members of $L^0$ or $L^1$ or $L^2$, corresponding
to equivalence classes of functions.   Starting in Chapter 62, conditional
expectations will be enormously important, and if we want to speak of
well-defined conditional expectation operators, we again need to work in $L^1$,
not in an associated space of functions.

Of course, we could set up a theory in which we accepted that an integral
$\int X_t\,dY_t$ would belong to $L^0$, while all the $X_t$, $Y_t$ were
real-valued functions.   But when we get to stochastic calculus in \S617 we
shall want to look at integration with respect to indefinite integrals.   So our
theory needs to be able to deal with integrals $\int d\pmb v$ where
$\pmb v$ is a process taking values in $L^0$.

Concerning stopping times, the argument takes a different form.   Since the
values of a process $\sfamilysS{u}$ are going to be in $L^0$, and
changing $\sigma$ on a set of measure zero must not affect $u_{\sigma}$, it is
to my mind more natural to define a stopping time $\sigma\in\Cal T$
in such a way that
it corresponds to an equivalence class of stopping times $h:X\to[0,\infty]$,
even if this complicates the definition of the region $\Bvalue{\sigma<\tau}$
in 611D and renders the algebra of 611E less transparent.   And there is another
issue.   If you look at the leading examples in 612T-612U, or at the general
formulations in 612H and 615P, you will see that
I appeal to results in Volume 4 at several points.   The intuitions behind
Brownian motion and the Poisson process naturally lead to processes
$\langle X_t\rangle_{t\ge 0}$, and in order to go farther we need to look at the
corresponding processes $\langle X_h\rangle_{h\text{ is a stopping time}}$.
This step demands deeper ideas than anything else in the present chapter.   By
going directly to processes $\sfamilysS{u}$, where $\Cal S$ may be any
sublattice of $\Cal T$, I can skate over these difficulties while still giving
you something to cut your teeth on.   Moreover, there are technical advantages;
because $\Cal T$ is Dedekind complete, I can speak uninhibitedly of infima, in
such results as 611Ch and 618C,
without needing any formula for calculating them.

}%end of notes


\discrpage

\frfilename{mt614.tex}
\versiondate{9.11.20/1.11.23}
\copyrightdate{2019}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Simple and order-bounded processes and bounded variation}

\newsection{614}

In \S613 I gave a definition of an integral with no very useful indication
of where it might be applicable.    This section and the next two will be
devoted to teasing out the basic case in
which Riemann-sum integrals $\int_{\Cal S}\pmb u\,d\pmb v$ are defined:
$\pmb u$ should be `moderately oscillatory'
(615E) and $\pmb v$ should be an `integrator'
(616K).   Before we come to either of these notions, however, it will
be helpful to look at three simpler concepts:  `simple' processes
(614B), `order-bounded' processes
(614E) and processes of `bounded variation' (614I-614J).

\cmmnt{\leader{614A}{Notation} \Asissix\ will be a stochastic integration
structure, as described in the notes to \S613, with
regions $\Bvalue{\sigma<\tau}$, $\Bvalue{\sigma\le\tau}$ and
$\Bvalue{\sigma=\tau}$ (611D) for stopping times $\sigma$, $\tau\in\Cal T$.
If $\sigma\le\tau$ in $\Cal T$, $c(\sigma,\tau)$ will be the corresponding
stopping-time interval (611J).   If $\Cal S$ is a sublattice of $\Cal T$,
then $\Sti(\Cal S)$ will be the set of stopping-time intervals with
endpoints in $\Cal S$ and when $I$ is a finite sublattice of
$\Cal T\,\,\Sti_0(I)$ will be the set of $I$-cells (611Je).
If $\Cal S$ is a sublattice of $\Cal T$ and $\tau\in\Cal S$, then
$\Cal S\wedge\tau=\{\sigma\wedge\tau:\sigma\in\Cal S\}$.
$\Cal S\vee\tau=\{\sigma\vee\tau:\sigma\in\Cal S\}$, $\Cal I(\Cal S)$ is
the upwards-directed set of finite sublattices of $\Cal S$ and
$\Cal S^{2\uparrow}=\{(\sigma,\tau):\sigma$, $\tau\in\Cal S$,
$\sigma\le\tau\}$.   $\Mfa(\Cal S)$ will be the space of fully adapted
processes with domain $\Cal S$ (612I).
I write $\tbf 1$ for the constant process with value
$\chi 1$, and if $z\in L^0$ then $z\tbf 1$ will be the constant
process with value $z$, defined on $\{\sigma:z\in L^0(\frak A_{\sigma})$
(612De).

$L^0=L^0(\frak A)$ (612A), and if
$\pmb u=\sfamilysS{u}\in(L^0)^{\Cal S}$,
I write $\sup|\pmb u|$ for $\sup_{\sigma\in\Cal S}|u_{\sigma}|$ if this is
defined in $L^0$.   $\theta$ will be the standard
F-seminorm defining the topology of convergence in measure on $L^0$ (613Ba),
and limits will be taken with respect to
this topology.   If $\Cal S$ is a sublattice of $\Cal T$,
$e\in\Sti(\Cal S)$, $I\in\Cal I(\Cal S)$, $\pmb u$ and $\pmb v$ are
processes defined on $\Cal S$ and $\psi$ is an interval
function defined on $\Cal S^{2\uparrow}$, we shall have
$\Delta_e(\pmb u,d\psi)$, $\Delta_e(\pmb u,d\pmb v)$,
$\Delta_e(\pmb u,|d\pmb v|)$ and the Riemann sums
$S_I(\pmb u,d\psi)$, $S_I(\pmb u,d\pmb v)$ and $S_I(\pmb u,|d\pmb v|)$
as defined in 613E-613F, with the Riemann-sum integrals
$\int_{\Cal S}\pmb u\,d\psi$,
$\int_{\Cal S}\pmb u\,d\pmb v$ and
$\int_{\Cal S}\pmb u\,|d\pmb v|$ (when they are defined) as in
613H.
}%end of comment

\leader{614B}{} \cmmnt{If either $\pmb u$ or $\pmb v$ is a simple process, then
$\int\pmb u\,d\pmb v$ is particularly straightforward.

\medskip

\noindent}{\bf Proposition}
Suppose that $\Cal S$ is a non-empty sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a simple fully adapted process with a breakpoint string
$(\tau_0,\ldots,\tau_n)$\cmmnt{ (612J)}.

(a) The starting value
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined, and
$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{u_{\sigma}=u_{\downarrow}}$ for
every $\sigma\in\Cal S$.

(b) Suppose that $\psi:\Cal S^{2\uparrow}\to L^0\cmmnt{\mskip5mu=L^0(\frak A)}$
is an adapted interval function such that
$\int_{\Cal S}d\psi\cmmnt{\mskip5mu=\int_{\Cal S}\tbf 1\,d\psi}$ is defined.
Then $\int_{\Cal S}\pmb u\,d\psi$ is defined and equal to

\Centerline{$u_{\downarrow}\times v_{\tau_0}
+\sum_{i=0}^{n-1}u_{\tau_i}
  \times(v_{\tau_{i+1}}-v_{\tau_i})
  +u_{\tau_n}\times(v_{\uparrow}-v_{\tau_n})$}

\noindent where $v_{\tau}=\int_{\Cal S\wedge\tau}d\psi$ for $\tau\in\Cal S$, and
$v_{\uparrow}=\int_{\Cal S}d\psi$.

\proof{{\bf (a)} For $\sigma\in\Cal S$, set
$a_{\sigma}=\Bvalue{\sigma<\tau_0}$;  set
$a=\sup_{\sigma\in\Cal S}a_{\sigma}$.   Note that
$a_{\sigma}\Bsupseteq a_{\tau}$ if $\sigma\le\tau$ in $\Cal S$, so
$a$ is the limit $\lim_{\sigma\downarrow\Cal S}a_{\sigma}$ for the
measure-algebra topology of $\frak A$ (323D(b-ii)), and
$\chi a=\lim_{\sigma\downarrow\Cal S}\chi a_{\sigma}$ for the
topology of convergence in measure (367Ra).   Next,
we know that there is a $u_*\in L^0(\frak B)$ such that
$a_{\sigma}\Bsubseteq\Bvalue{u_{\sigma}=u_*}$ for every $\sigma\in\Cal S$.
If $\sigma\in\Cal S$ and $\sigma\le\tau_0$, then
$1\Bsetminus a_{\sigma}=\Bvalue{\sigma=\tau_0}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau_0}}$, so

\Centerline{$u_{\sigma}
=u_*\times\chi a_{\sigma}+u_{\tau_0}\times(\chi 1-\chi a_{\sigma})$.}

\noindent But this means that

$$\eqalign{u_{\downarrow}
&=\lim_{\sigma\downarrow\Cal S}u_{\sigma}
=\lim_{\sigma\downarrow\Cal S}
  u_*\times\chi a_{\sigma}+u_{\tau_0}\times(\chi 1-\chi a_{\sigma})\cr&
=u_*\times\chi a+u_{\tau_0}\times(\chi 1-\chi a).\cr}$$

Finally, for any $\sigma\in\Cal S$,

$$\eqalignno{\Bvalue{\sigma<\tau_0}&
\Bsubseteq a\Bcap\Bvalue{u_{\sigma}=u_*}
\Bsubseteq\Bvalue{u_*=u_{\downarrow}}\Bcap\Bvalue{u_{\sigma}=u_*}
\Displaycause{because $u_*\times\chi a=u_{\downarrow}\times\chi a$}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\downarrow}}.\cr}$$

\medskip

{\bf (b)(i)} By 613J(c-i), $v_{\tau}$ is defined for every $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)}
$\int_{\Cal S\wedge\tau_0}\pmb u\,d\psi=u_{\downarrow}\times v_{\tau_0}$.
\Prf\ If $\sigma$, $\tau\in\Cal S$ and $\sigma\le\tau\le\tau_0$, then

$$\eqalignno{u_{\sigma}\times\psi(\sigma,\tau)&
=u_{\sigma}\times\chi\Bvalue{\sigma<\tau}\times\psi(\sigma,\tau)
\Displaycause{because
$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{\psi(\sigma,\tau)=0}$, by
613C(b-i)}
=u_{\sigma}\times\chi\Bvalue{\sigma<\tau_0}
   \times\chi\Bvalue{\sigma<\tau}\times\psi(\sigma,\tau)\cr&
=u_{\downarrow}\times\chi\Bvalue{\sigma<\tau_0}
   \times\chi\Bvalue{\sigma<\tau}\times\psi(\sigma,\tau)
=u_{\downarrow}\times\psi(\sigma,\tau).\cr}$$

\noindent So $\Delta_e(\pmb u,d\psi)=u_{\downarrow}\times\Delta_e(\tbf 1,d\psi)$
for every stopping-time interval with endpoints in $\Cal S\wedge\tau_0$, and
$S_I(\pmb u,d\psi)=u_{\downarrow}\times S_I(\tbf 1,d\psi)$
for every $I\in\Cal I(\Cal S\wedge\tau_0)$.   Accordingly

$$\eqalign{\int_{\Cal S\wedge\tau_0}\pmb u\,d\psi&
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau_0)}S_I(\pmb u,d\psi)
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau_0)}
    u_{\downarrow}\times S_I(\tbf 1,d\psi)\cr&
=u_{\downarrow}
      \times\lim_{I\uparrow\Cal I(\Cal S\wedge\tau_0)}S_I(\tbf 1,d\psi)
=u_{\downarrow}\times v_{\tau_0}.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)} If $i<n$, then
$\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb u\,d\psi
=u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})$.   \Prf\
If $\sigma$, $\tau\in\Cal S$ and $\tau_i\le\sigma\le\tau\le\tau_{i+1}$, then

$$\eqalignno{u_{\sigma}\times\psi(\sigma,\tau)&
=u_{\sigma}\times\chi\Bvalue{\sigma<\tau}\times\psi(\sigma,\tau)
=u_{\sigma}\times\chi\Bvalue{\sigma<\tau_{i+1}}
   \times\chi\Bvalue{\sigma<\tau}\times\psi(\sigma,\tau)\cr&
=u_{\tau_i}\times\chi\Bvalue{\sigma<\tau_{i+1}}
   \times\chi\Bvalue{\sigma<\tau}\times\psi(\sigma,\tau)
=u_{\tau_i}\times\psi(\sigma,\tau).\cr}$$

\noindent So $\Delta_e(\pmb u,d\psi)=u_{\tau_i}\times\Delta_e(\tbf 1,d\psi)$
for every stopping-time interval with endpoints in
$\Cal S\cap[\tau_i,\tau_{i+1}]$, and
$S_I(\pmb u,d\psi)=u_{\tau_i}\times S_I(\tbf 1,d\psi)$
for every $I\in\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])$.   Accordingly

$$\eqalignno{\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb u\,d\psi&
=\lim_{I\uparrow\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])}S_I(\pmb u,d\psi)
=\lim_{I\uparrow\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])}
   u_{\tau_i}\times S_I(\tbf 1,d\psi)\cr&
=u_{\tau_i}
   \times\lim_{I\uparrow\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])}S_I(\tbf 1,d\psi)
=u_{\tau_i}\times\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}d\psi\cr&
=u_{\tau_i}\times(\int_{\Cal S\wedge\tau_{i+1}}d\psi
   -\int_{\Cal S\wedge\tau_i}d\psi)
\Displaycause{using 613J(c-i) again}
=u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i}).  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iv)} $\int_{\Cal S\vee\tau_n}\pmb u\,d\psi
=u_{\tau_n}\times(v_{\uparrow}-v_{\tau_n})$.
\Prf\ If $\sigma$, $\tau\in\Cal S$ and $\tau_n\le\sigma\le\tau$, then
$u_{\sigma}\times\psi(\sigma,\tau)=u_{\tau_n}\times\psi(\sigma,\tau)$.
So $\Delta_e(\pmb u,d\psi)=u_{\tau_n}\times\Delta_e(\tbf 1,d\psi)$
for every stopping-time interval with endpoints in
$\Cal S\vee\tau_n$, and
$S_I(\pmb u,d\psi)=u_{\tau_n}\times S_I(\tbf 1,d\psi)$
for every $I\in\Cal I(\Cal S\vee\tau_n)$ and

$$\eqalign{\int_{\Cal S\vee\tau_n}\pmb u\,d\psi&
=\lim_{I\uparrow\Cal I(\Cal S\vee\tau_n)}S_I(\pmb u,d\psi)
=\lim_{I\uparrow\Cal I(\Cal S\vee\tau_n)}
   u_{\tau_n}\times S_I(\tbf 1,d\psi)\cr&
=u_{\tau_n}
   \times\lim_{I\uparrow\Cal I(\Cal S\vee\tau_n)}S_I(\tbf 1,d\psi)
=u_{\tau_i}\times\int_{\Cal S\vee\tau_n}d\psi\cr&
=u_{\tau_i}\times(\int_{\Cal S}d\psi
   -\int_{\Cal S\wedge\tau_n}d\psi)
=u_{\tau_i}\times(v_{\uparrow}-v_{\tau_n}).  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (v)} Assembling these, as in 613J(c-ii), $\int_{\Cal S}\pmb u\,d\psi$
is defined and equal to

$$\eqalign{&\int_{\Cal S\wedge\tau_0}\pmb u\,d\psi
 +\sum_{i=0}^{n-1}\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb u\,d\psi
 +\int_{\Cal S\vee\tau_n}\pmb u\,d\psi\cr&
\mskip100mu=u_{\downarrow}\times v_{\tau_0}
+\sum_{i=0}^{n-1}u_{\tau_i}
  \times(v_{\tau_{i+1}}-v_{\tau_i})
  +u_{\tau_n}\times(v_{\uparrow}-v_{\tau_n}),\cr}$$

\noindent as claimed.
}%end of proof of 614B

\cmmnt{\medskip

\noindent{\bf Remark} I didn't have to say so in the course of the proof above,
but of course $\int_{\Cal S}d\psi=\lim_{\tau\uparrow\Cal S}v_{\tau}$
(613J(f-ii)), so the formula $v_{\uparrow}$ here matches the usage in 613N.}

\leader{614C}{Corollary}
Suppose that $\Cal S$ is a non-empty sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ is a simple fully
adapted process with a breakpoint string
$(\tau_0,\ldots,\tau_n)$, and
$\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ is a fully adapted process such that
$v_{\uparrow}=\lim_{\tau\uparrow\Cal S}v_{\tau}$
and $v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ are
defined.   Then
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined and equal to

\Centerline{$u_{\downarrow}\times(v_{\tau_0}-v_{\downarrow})
+\sum_{i=0}^{n-1}u_{\tau_i}
  \times(v_{\tau_{i+1}}-v_{\tau_i})
  +u_{\tau_n}\times(v_{\uparrow}-v_{\tau_n})$.}

\proof{ Apply 614B with $\psi=\Delta\pmb v$;  the point being just that
$\int_{\Cal S\wedge\tau}d\pmb v=v_{\tau}-v_{\downarrow}$ for every
$\tau\in\Cal S$, by 613N applied to $\pmb v\restr\Cal S\wedge\tau$,
while $\int_{\Cal S}d\pmb v=v_{\uparrow}-v_{\downarrow}$.
}%end of proof of 614C

\leader{614D}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ a simple process, and $\pmb u=\sfamilysS{u}$ a fully
adapted process such that $\int_{\Cal S}\pmb u\,d\pmb v$ is defined.   Then
$\iivu$ is simple.

\proof{{\bf (a)}
If $\pmb v$ is constant, this is trivial, as $\Delta\pmb v=0$ and
$\iivu$ is constant with value $0$.   Otherwise, let $(\tau_0,\ldots,\tau_n)$ be
a breakpoint string for $\pmb v$, and $v_{\downarrow}$
the starting value of $\pmb v$.   For $\tau\in\Cal S$ write $w_{\tau}$ for
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v$, so that
$\family{\tau}{\Cal S}{w_{\tau}}=\iivu$.

\medskip

{\bf (b)} If $\tau\in\Cal S\wedge\tau_0$ then
$\Bvalue{\tau<\tau_0}\Bsubseteq\Bvalue{w_{\tau}=0}$.   \Prf\
By 613Ld,

$$\eqalign{\Bvalue{w_{\tau}\ne 0}&
\Bsubseteq\sup_{\sigma,\sigma'\in\Cal S\wedge\tau}
    \Bvalue{v_{\sigma}\ne v_{\sigma'}}
\Bsubseteq\sup_{\sigma\in\Cal S\wedge\tau}\Bvalue{v_{\sigma}\ne v_{\downarrow}}
   \cr&
\Bsubseteq\sup_{\sigma\in\Cal S\wedge\tau}\Bvalue{\sigma=\tau_0}
\Bsubseteq\Bvalue{\tau=\tau_0}\cr}$$

\noindent so $\Bvalue{\tau<\tau_0}\Bsubseteq\Bvalue{w_{\tau}=0}$.\ \QeD\
Generally, if $\tau\in\Cal S$,

$$\eqalign{\Bvalue{\tau<\tau_0}&
\Bsubseteq\Bvalue{\tau=\tau\wedge\tau_0}\Bcap\Bvalue{\tau\wedge\tau_0<\tau_0}
   \cr&
\Bsubseteq\Bvalue{w_{\tau}=w_{\tau\wedge\tau_0}}
   \Bcap\Bvalue{w_{\tau\wedge\tau_0}=0}
\Bsubseteq\Bvalue{w_{\tau}=0}\cr}$$

\noindent because $\iivu$ is fully adapted (613Oa).

\medskip

{\bf (c)} If $i\le n$ and $\tau\in\Cal S\cap[\tau_i,\tau_{i+1}]$ then
$\Bvalue{\tau<\tau_{i+1}}\Bsubseteq\Bvalue{w_{\tau}=w_{\tau_i}}$.
\Prf\

$$\eqalignno{\Bvalue{w_{\tau}\ne w_{\tau_i}}&
=\Bvalue{\int_{\Cal S\cap[\tau_i,\tau]}\pmb u\,d\pmb v\ne 0}
\Displaycause{613Jc}
\Bsubseteq\sup_{\sigma,\sigma'\in\Cal S\cap[\tau_i,\tau]}
   \Bvalue{v_{\sigma}\ne v_{\sigma'}}
\Bsubseteq\sup_{\sigma\in\Cal S\cap[\tau_i,\tau]}
   \Bvalue{v_{\sigma}\ne v_{\tau_i}}\cr&
\Bsubseteq\sup_{\sigma\in\Cal S\cap[\tau_i,\tau]}
   \Bvalue{\sigma=\tau_{i+1}}
\Bsubseteq\Bvalue{\tau=\tau_{i+1}},\cr}$$

\noindent so $\Bvalue{\tau<\tau_{i+1}}\Bsubseteq\Bvalue{w_{\tau}=w_{\tau_i}}$.\
\QeD\   Generally, if $\tau\in\Cal S$,

$$\eqalign{\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}&
\Bsubseteq\Bvalue{\tau=\med(\tau_i,\tau,\tau_{i+1})}
   \Bcap\Bvalue{\med(\tau_i,\tau,\tau_{i+1})<\tau_{i+1})}\cr&
\Bsubseteq\Bvalue{w_{\tau}=w_{\med(\tau_i,\tau,\tau_{i+1})}}
   \Bcap\Bvalue{w_{\med(\tau_i,\tau,\tau_{i+1})}=w_{\tau_i}}\cr&
\Bsubseteq\Bvalue{w_{\tau}=w_{\tau_i}}.\cr}$$

\medskip

{\bf (d)} If $\tau\in\Cal S\vee\tau_n$ then $w_{\tau}=w_{\tau_n}$.   \Prf\

$$\eqalign{\Bvalue{w_{\tau}\ne w_{\tau_n}}&
=\Bvalue{\int_{\Cal S\cap[\tau_n,\tau]}\pmb u\,d\pmb v\ne 0}\cr&
\Bsubseteq\sup_{\sigma,\sigma'\in\Cal S\cap[\tau_n,\tau]}
   \Bvalue{v_{\sigma}\ne v_{\sigma'}}
\Bsubseteq\sup_{\sigma\in\Cal S\vee\tau_n}
   \Bvalue{v_{\sigma}\ne v_{\tau_n}}
=0.  \text{ \Qed}\cr}$$

\noindent Generally, if $\tau\in\Cal S$,

\Centerline{$\Bvalue{\tau_n\le\tau}
\Bsubseteq\Bvalue{\tau=\tau\vee\tau_n}
\Bsubseteq\Bvalue{w_{\tau}=w_{\tau\vee\tau_n}}
\Bsubseteq\Bvalue{w_{\tau}=w_{\tau_n}}$.}

\noindent Thus $\iivu$ is simple, with breakpoint string
$(\tau_0,\ldots,\tau_n)$.
}%end of proof of 614D

\leader{614E}{Order-bounded \dvrocolon{processes}}\cmmnt{ Now for a much
larger class of processes.

\medskip

\noindent}{\bf Definitions} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ a fully adapted process.

\spheader 614Ea $\pmb u$ is {\bf order-bounded} if
$\{u_{\sigma}:\sigma\in\Cal S\}$ is bounded above and below in
$L^0$.   In this case,\cmmnt{ if
$\Cal S\ne\emptyset$, $\sup_{\sigma\in\Cal S}|u_{\sigma}|$ is defined in
$L^0$, because $L^0$ is Dedekind complete.   It will be convenient
to write} $\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|$,
taking the supremum in $(L^0)^+$, so that
$\sup|\pmb u|=0$ if $\Cal S=\dom\pmb u$ is empty.

\spheader 614Eb $\pmb u$ is {\bf locally order-bounded} if
$\pmb u\restr\Cal S\wedge\tau
=\langle u_{\sigma}\rangle_{\sigma\in\Cal S,\sigma\le\tau}$ is
order-bounded for every $\tau\in\Cal S$.

\spheader 614Ec Suppose that $\pmb u$ is simple, with breakpoint string
$(\tau_0,\ldots,\tau_n)$ and starting value $u_{\downarrow}$.
Then $\pmb u$ is order-bounded and
$\sup|\pmb u|=|u_{\downarrow}|\vee\sup_{i\le n}|u_{\tau_i}|$.
\prooflet{\Prf\
Set $\bar u=|u_{\downarrow}|\vee\sup_{i\le n}|u_{\tau_i}|$.
If $\sigma\in\Cal S$, then

$$\eqalign{\Bvalue{|u_{\sigma}|\le\bar u}&
\Bsupseteq\Bvalue{u_{\sigma}=u_{\downarrow}}
   \Bcup\sup_{i\le n}\Bvalue{u_{\sigma}=u_{\tau_i}}\cr&
\Bsupseteq\Bvalue{\sigma<\tau_0}
   \Bcup\sup_{i<n}(\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}})
   \Bcup\Bvalue{\tau_n\le\sigma}
=1\cr}$$

\noindent and $|u_{\sigma}|\le\bar u$.   As $\sigma$ is arbitrary,
$\pmb u$ is order-bounded and $\sup|\pmb u|\le\bar u$.   On the other hand,

\Centerline{$|u_{\downarrow}|=\lim_{\sigma\downarrow\Cal S}|u_{\sigma}|
\le\sup|\pmb u|$}

\noindent so $\bar u\le\sup|\pmb u|$ and we have equality.\ \Qed}

\leader{614F}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.

(a)(i) If $\pmb u$ is an order-bounded process with domain $\Cal S$, then
$\pmb u\restr\Cal S'$ is order-bounded for any sublattice $\Cal S'$ of
$\Cal S$;  in particular, $\pmb u$ is locally order-bounded.

\quad (ii) If $\pmb u$ is a locally order-bounded process
with domain $\Cal S$, then
$\pmb u\restr\Cal S'$ is locally order-bounded for any sublattice
$\Cal S'$ of $\Cal S$.

(b) Suppose that $\pmb u=\sfamilysS{u}$ is a locally order-bounded
process.   Set $v_{\tau}=\sup_{\sigma\in\Cal S\wedge\tau}|u_{\sigma}|$ for
$\tau\in\Cal S$.   Then $\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ is a
non-decreasing fully adapted process.

(c) Write $\Mob=\Mob(\Cal S)$
for the set of order-bounded fully adapted processes with domain $\Cal S$.

\quad(i) If $h:\Bbb R\to\Bbb R$ is a Borel measurable function
which is bounded on every bounded interval in $\Bbb R$, then
$\bar h\pmb u\in\Mob$ for every $\pmb u\in\Mob$.

\quad(ii) $\Mob$ is an $f$-subalgebra of
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$.

\quad(iii) If
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ then
$z\pmb u\in\Mob$, with $\sup|z\pmb u|=|z|\times\sup|\pmb u|$,
for every $\pmb u\in\Mob$.

(d) Write $\Mlob=\Mlob(\Cal S)$ for the set of locally
order-bounded fully adapted processes with
domain $\Cal S$.

\quad(i) If $h:\Bbb R\to\Bbb R$ is a Borel measurable function
which is bounded on every bounded interval in $\Bbb R$, then
$\bar h\pmb u\in\Mlob$ for every $\pmb u\in\Mlob$.

\quad(ii) $\Mlob$ is an $f$-subalgebra of
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$.

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

\medskip

{\bf (b)} For each $\tau\in\Cal S$,
$\{|u_{\sigma}|:\sigma\in\Cal S\wedge\tau\}$ is bounded above
in the Dedekind
complete lattice $L^0(\frak A)$, so $v_{\tau}$ is defined.
Moreover,
$|u_{\sigma}|\in L^0(\frak A_{\sigma})\subseteq L^0(\frak A_{\tau})$ for
every $\sigma\in\Cal S\wedge\tau$;  as $L^0(\frak A_{\tau})$ is an
order-closed sublattice of
$L^0(\frak A)$ (612A(e-i)), $v_{\tau}\in L^0(\frak A_{\tau})$.
Of course $v_{\tau}\le v_{\tau'}$ whenever $\tau\le\tau'$ in $\Cal S$,
just because $\Cal S\wedge\tau\subseteq\Cal S\wedge\tau'$.

Now suppose that $\tau$, $\tau'\in\Cal S$ and $c=\Bvalue{\tau=\tau'}$.
If $\sigma\in\Cal S\wedge\tau$, then
$\sigma\wedge\tau'\in\Cal S\wedge\tau'$
and

\Centerline{$c\Bsubseteq\Bvalue{\sigma\le\tau'}
=\Bvalue{\sigma=\sigma\wedge\tau'}
\Bsubseteq\Bvalue{|u_{\sigma}|=|u_{\sigma\wedge\tau'}|}
\Bsubseteq\Bvalue{|u_{\sigma}|\le v_{\tau'}}$.}

\noindent So $|u_{\sigma}|\times\chi c\le v_{\tau'}$.   Since
$u\mapsto u\times\chi c:L^0(\frak A)\to L^0(\frak A)$ is
an order-continuous lattice homomorphism,

\Centerline{$v_{\tau}\times\chi c
=\sup_{\sigma\in\Cal S\wedge\tau}|u_{\sigma}|\times\chi c
\le v_{\tau'}\times\chi c$,}

\noindent and
$c\Bsubseteq\Bvalue{v_{\tau}\le v_{\tau'}}$.
Similarly, $c\Bsubseteq\Bvalue{v_{\tau'}\le v_{\tau}}$ and
$\Bvalue{\tau=\tau'}=c\Bsubseteq\Bvalue{v_{\tau}=v_{\tau'}}$.
As $\tau$ and $\tau'$ are arbitrary, $\pmb v$ is fully adapted.

\medskip

{\bf (c)(i)} For $x\in\Bbb R$, set $g(x)=\sup_{|y|\le|x|}|h(y)|$.
Then $g$ is monotonic on each of $\ocint{-\infty,0}$ and
$\coint{0,\infty}$, so is Borel measurable, and $|\bar h(v)|\le\bar g(|u|)$
whenever $u$, $v\in L^0$ and $|v|\le u$.
So if $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$
is an order-bounded fully adapted process, and $u$ is an upper bound of
$\{|u_{\sigma}|:\sigma\in\Cal S\}$, $\bar g(u)$ will be an upper bound of
$\{|\bar h(u_{\sigma})|:\sigma\in\Cal S\}$, and $\bar h\pmb u$ is
order-bounded.

\medskip

\quad{\bf (ii)} If $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ are order-bounded, let
$u$, $v$ be upper bounds of $\{|u_{\sigma}|:\sigma\in\Cal S\}$,
$\{|v_{\sigma}|:\sigma\in\Cal S\}$ respectively;  then
$u+v$ is an upper bound of $\{|u_{\sigma}+v_{\sigma}|:\sigma\in\Cal S\}$
and $\pmb u+\pmb v$ is order-bounded.   Thus $\Mob$ is closed under
addition.   By 612Bc as usual, $\Mob$ is an $f$-subalgebra of
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$.

\medskip

\quad{\bf (iii)} The map $u\mapsto|z|\times u:L^0\to L^0$ is an
order-continuous Riesz homomorphism, so if $\pmb u=\sfamilysS{u}$ belongs
to $\Mob$ we shall have

\Centerline{$|z|\times\sup|\pmb u|
=|z|\times\sup_{\sigma\in\Cal S}|u_{\sigma}|
=\sup_{\sigma\in\Cal S}|z|\times|u_{\sigma}|
=\sup_{\sigma\in\Cal S}|z\times u_{\sigma}|$}

\noindent and $z\pmb u$ is order-bounded, with
$\sup|z\pmb u|=|z|\times\sup|\pmb u|$.

\medskip

{\bf (d)} Apply (c) to $\pmb u\restr\Cal S\wedge\tau$ for each
$\tau\in\Cal S$.
}%end of proof of 614F

\leader{614G}{Proposition} Suppose that $\Cal S$ is a sublattice of $\Cal T$
and
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ a fully adapted process.

(a) If $A$, $B\subseteq\Cal S$, $A$ covers $B$ and $\{u_{\sigma}:\sigma\in A\}$
is order-bounded, then $\{u_{\sigma}:\sigma\in B\}$ is order-bounded and
$\sup_{\sigma\in B}|u_{\sigma}|\le\sup_{\sigma\in A}|u_{\sigma}|$.

(b) If $\Cal S'$ is a sublattice of $\Cal S$ which covers $\Cal S$

\quad (i) $\pmb u$ is order-bounded iff $\pmb u\restr\Cal S'$ is order-bounded, and
in this case $\sup|\pmb u|=\sup|\pmb u\restr\Cal S'|$,

\quad (ii) $\pmb u$ is locally order-bounded iff $\pmb u\restr\Cal S'$ is
locally order-bounded.
%formerly 612Qd

\proof{{\bf (a)} Write $\bar u$ for $\sup_{\sigma\in A}|u_{\sigma}|$, counting the supremum of the empty set as $0$.   If $\tau\in B$ then

\Centerline{$\Bvalue{|u_{\tau}|\le\bar u}
\Bsupseteq\sup_{\sigma\in A}\Bvalue{u_{\tau}=u_{\sigma}}
   \Bcap\Bvalue{|u_{\sigma}|\le\bar u}
\Bsupseteq\sup_{\sigma\in A}\Bvalue{\tau=\sigma}
=1$,}

\noindent so $|u_{\tau}|\le\bar u$.   As $\tau$ is arbitrary,
$\pmb u\restr B$ is order-bounded and 
$\sup|\pmb u\restr B|\le\bar u=\sup|\pmb u\restr A|$.

\medskip

{\bf (b)(i)} Immediate from (a), as each of $\Cal S$, $\Cal S'$ covers the other.

\medskip

\quad{\bf (ii)} If $\pmb u$ is locally order-bounded and
$\sigma^*\in\Cal S'$, $\{u_{\sigma}:\sigma\in\Cal S'\wedge\sigma^*\}
\subseteq\{u_{\tau}:\tau\in\Cal S\wedge\sigma^*\}$ is order-bounded in
$L^0(\frak A)$, so $\pmb u\restr\Cal S'$ is locally order-bounded.

Now suppose that $\pmb u$ is locally order-bounded,
$\tau^*\in\Cal S$ and $a\in\frak A$ is non-zero.   Since
$\sup_{\sigma\in\Cal S'}\Bvalue{\tau^*=\sigma}=1$, there is a
$\sigma\in\Cal S'$ such that $a'=a\Bcap\Bvalue{\tau^*=\sigma}$ is non-zero.
Now $\Cal S\wedge\sigma$ is covered by $\Cal S'\wedge\sigma$ (611M(e-i))
and $\pmb u\restr\Cal S'\wedge\sigma$ is order-bounded, so
$\pmb u\restr\Cal S\wedge\sigma$ is order-bounded, by (i) above.
There is therefore a $\alpha\ge 0$ such that
$a'=a\Bsetminus\sup_{\tau\in\Cal S\wedge\sigma}
\Bvalue{|u_{\tau}|>\alpha}$ is non-zero.   But in this case

$$\eqalign{a'\Bcap\Bvalue{|u_{\tau\wedge\tau^*}|>\alpha}&
=a'\Bcap\Bvalue{|u_{\tau\wedge\tau^*}|>\alpha}
   \Bcap\Bvalue{\tau^*=\sigma}\cr&
\Bsubseteq a'\Bcap\Bvalue{|u_{\tau\wedge\sigma}|>\alpha}
=0\cr}$$

\noindent for every $\tau\in\Cal S$, that is,
$a'\Bcap\Bvalue{|u_{\tau}|>\alpha}=0$ for every
$\tau\in\Cal S\wedge\tau^*$.   So
$\sup_{\tau\in\Cal S\wedge\tau^*}\Bvalue{|u_{\tau}|>\alpha}$ does not
include $a$;  as $a$ is arbitrary,
$\inf_{\gamma>0}\sup_{\tau\in\Cal S\wedge\tau^*}
  \Bvalue{|u_{\tau}|>\gamma}=0$ and
$\{u_{\tau}:\tau\in\Cal S\wedge\tau^*\}$ is bounded above in
$L^0(\frak A)$ (364L(a-ii)).   As $\tau^*$ is arbitrary, $\pmb u$ is locally
order-bounded.
}%end of proof of 614G

\leader{614H}{Proposition} Brownian motion, as described in 612T, is
locally order-bounded.

\proof{ I follow the notation of 612T.
If $\tau\in\Cal T_f$, let $h:\Omega\to\coint{0,\infty}$ be a stopping time
representing $\tau$.   Then $f(\omega)=\sup_{t\in[0,h(\omega)]}|\omega(t)|$
is finite for every $\omega\in\Omega$, because $\omega$ is continuous.
Moreover, again because every $\omega$ is continuous,
$f(\omega)=\sup_{q\in\Bbb Q}f_q(\omega)$ for every $\omega$, where
$f_q(\omega)=|\omega(q)|$ if $q\le h(\omega)$, $0$ otherwise;  as every
$f_q$ is measurable, so is $f$.   Now we see that if $\sigma\le\tau$ in
$\Cal T_f$ there is a stopping time $g:\Omega\to\coint{0,\infty}$ such that
$g$ represents $\sigma$ and $g\le h$, so that $|X_g|\le f$ and
$|w_{\sigma}|=|X_g^{\ssbullet}|\le f^{\ssbullet}$ in $L^0(\frak C)$.
Thus $\{w_{\sigma}:\sigma\le\tau\}$ is order-bounded;  as
$\tau$ is arbitrary, $\pmb w$ is locally order-bounded.
}%end of proof of 614H

\leader{614I}{Bounded \dvrocolon{variation}}\cmmnt{ The third class of processes
I wish to discuss is intermediate between the other two.

\medskip

\noindent}{\bf Theorem} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\pmb v=\sfamilysS{v}$ a fully adapted process.   Then the
following are equiveridical:

\inset{(i) $\pmb v$ is expressible as the difference of two order-bounded
non-negative non-decreasing fully adapted processes,

(ii) $\{\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|:\tau_0\le\ldots\le\tau_n$ in
$\Cal S\}$ is bounded above in $L^0$,

(iii) $\int_{\Cal S}|d\pmb v|$ is defined;}

\noindent and in this case

\Centerline{$\int_{\Cal S}|d\pmb v|
=\sup\{\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|:\tau_0\le\ldots\le\tau_n$ in
$\Cal S\}$}

\noindent if we count $\sup\emptyset$ as $0$.

\proof{ If $\Cal S$ is empty, then (interpreting $\sup\emptyset$ as $0$ in the
last clause) the result is true for trivial reasons,
so let us suppose that $\Cal S\ne\emptyset$.

\medskip

{\bf (i)$\Rightarrow$(ii)} If $\pmb v=\pmb v'-\pmb v''$ where
$\pmb v'=\sfamilysS{v'}$ and $\pmb v''=\sfamilysS{v''}$ are
non-decreasing, non-negative and order-bounded, set
$\bar v=\sup_{\sigma\in\Cal S}v'_{\sigma}
  +\sup_{\sigma\in\Cal S}v''_{\sigma}$.   Then

$$\eqalign{\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
&\le\sum_{i=0}^{n-1}|v'_{\tau_{i+1}}-v'_{\tau_i}|
  +\sum_{i=0}^{n-1}|v''_{\tau_{i+1}}-v''_{\tau_i}|\cr
&=\sum_{i=0}^{n-1}v'_{\tau_{i+1}}-v'_{\tau_i}
  +\sum_{i=0}^{n-1}v''_{\tau_{i+1}}-v''_{\tau_i}\cr
&=v'_{\tau_n}-v'_{\tau_0}+v''_{\tau_n}-v'_{\tau_0}
\le\bar v\cr}$$

\noindent whenever $\tau_0\le\ldots\le\tau_n$ in $\Cal S$.
So (ii) is true.

\medskip

{\bf (ii)$\Rightarrow$(iii)}\grheada\ The key fact is that
$S_J(\tbf{1},|d\pmb v|)\le S_I(\tbf{1},|d\pmb v|)$ whenever
$J\subseteq I$ in $\Cal I(\Cal S)$.
\Prf\ If $J$ is empty this is trivial.
Otherwise, let $(\tau_0,\ldots,\tau_n)$ linearly generate the
$J$-cells, so that

\Centerline{$S_J(\tbf{1},|d\pmb v|)
=\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|$}

\noindent (613Ec),

\Centerline{$S_I(\tbf{1},|d\pmb v|)
=S_{I\wedge\tau_0}(\tbf{1},|d\pmb v|)
  +\sum_{i=0}^{n-1}S_{I\cap[\tau_i,\tau_{i+1}]}(\tbf{1},|d\pmb v|)
+S_{I\vee\tau_n}(\tbf{1},|d\pmb v|)$.}

\noindent Now for any stopping-time interval
$e=c(\sigma,\tau)$ with endpoints in $\Cal S$,

\Centerline{$\Delta_e(\tbf{1},|d\pmb v|)=|v_{\tau}-v_{\sigma}|
=|\Delta_e(\tbf{1},d\pmb v)|$,}

\noindent so for any $K\in\Cal I(\Cal S)$

$$\eqalign{|S_K(\tbf{1},d\pmb v)|
&=|\sum_{e\in\Sti_0(K)}\Delta_e(\tbf{1},d\pmb v)|
\le\sum_{e\in\Sti_0(K)}|\Delta_e(\tbf{1},d\pmb v)|\cr
&=\sum_{e\in\Sti_0(K)}\Delta_e(\tbf{1},|d\pmb v|)
=S_K(\tbf{1},|d\pmb v)|.\cr}$$

\noindent In particular, for $i<n$,

$$\eqalignno{|v_{\tau_{i+1}}-v_{\tau_i}|
&=|S_{I\cap[\tau_i,\tau_{i+1}]}(\tbf{1},d\pmb v)|\cr
\displaycause{613Ed}
&\le S_{I\cap[\tau_i,\tau_{i+1}]}(\tbf{1},|d\pmb v|).\cr}$$

\noindent Summing over $i$,

\Centerline{$S_J(\tbf{1},|d\pmb v|)
=\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
\le\sum_{i=0}^{n-1}S_{I\cap[\tau_i,\tau_{i+1}]}(\tbf{1},|d\pmb v|)
\le S_I(\tbf{1},|d\pmb v|)$.  \Qed}

\medskip

\quad\grheadb\ Setting
$A=\{\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|:\tau_0\le\ldots\le\tau_n$ in
$\Cal S\}$, $A'=\{S_I(\tbf{1},|d\pmb v|):I\in\Cal I(\Cal S)\}$, we have
$A'\subseteq A$, and we know from ($\alpha$) that $A'$ is upwards-directed;
moreover,

\Centerline{$\int_{\Cal S}|d\pmb v|
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\tbf 1,|d\pmb v|)
=\lim_{z\uparrow A'}z=\sup A'$}

\noindent if any of these is defined in
$L^0$.   If we assume that (ii) is true, so that $A$ and $A'$
are bounded above, then $\sup A'$ is defined and (iii) is true.

In fact, of course, $A\subseteq A'$, because if
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ then
$\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|=S_I(\tbf{1},|d\pmb v|)$ where
$I=\{\tau_0,\ldots,\tau_n\}$.   So we must have
$\sup A=\int_{\Cal S}|d\pmb v|$, as claimed in the final clause of the
statement of this proposition.

\medskip

{\bf (iii)$\Rightarrow$(i)} Suppose that $\int_{\Cal S}|d\pmb v|$ is defined.

\medskip

\quad\grheada\ $v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ is
defined.   \Prf\ Let $\epsilon>0$.   Then there is a
non-empty $J\in\Cal I(\Cal S)$ such that
$\theta(S_I(\tbf{1},|d\pmb v|)-S_J(\tbf{1},|d\pmb v|))\le\epsilon$ whenever
$I\in\Cal I(\Cal S)$ includes $J$.   If $\sigma\in\Cal S\wedge\min J$, set
$I=J\cup\{\sigma\}$; then

$$\eqalign{\epsilon
&\ge\theta(S_I(\tbf{1},|d\pmb v|)-S_J(\tbf{1},|d\pmb v|))
=\theta(\Delta_{c(\sigma,\min J)}(\tbf{1},|d\pmb v|))\cr
&=\theta(|v_{\min J}-v_{\sigma}|)
=\theta(v_{\min J}-v_{\sigma}).\cr}$$

\noindent Because $L^0$ is complete, this is enough to ensure that
$v_{\uparrow}$ is defined.\ \Qed

\medskip

\quad\grheadb\ $v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is
defined.   \Prf\ Argue as in ($\alpha$).   Let $\epsilon>0$.   Then there is a
non-empty $J\in\Cal I(\Cal S)$ such that
$\theta(S_I(\tbf{1},|d\pmb v|)-S_J(\tbf{1},|d\pmb v|))\le\epsilon$ whenever
$I\in\Cal I(\Cal S)$ includes $J$.   Now

\Centerline{$\theta(v_{\sigma}-v_{\max J})
=\theta(S_{J\cup\{\sigma\}}(\tbf{1},|d\pmb v|)-S_J(\tbf{1},|d\pmb v|)
\le\epsilon$}

\noindent whenever $\max J\le\sigma\in\Cal S$.
Because $L^0$ is complete, this is enough to ensure that
$v_{\downarrow}$ is defined.\ \Qed

\medskip

\quad\grheadc\ Set
$v'_{\tau}=v_{\downarrow}^++\int_{\Cal S\wedge\tau}|d\pmb v|$ for
$\tau\in\Cal S$.   Because $v_{\downarrow}\in L^0(\frak A_{\tau})$ for every
$\tau\in\Cal S$ (613Bj), $\pmb v'=\family{\tau}{\Cal S}{v'_{\tau}}$ is fully
adapted.   If $\sigma\le\tau$ in $\Cal S$,
$v'_{\tau}-v'_{\sigma}=\int_{\Cal S\cap[\sigma,\tau]}|d\pmb v|\ge 0$, so
$\pmb v'$ is
non-decreasing.   Because $v_{\downarrow}^+\ge 0$, $\pmb v'$ is non-negative.
As
$\lim_{\tau\uparrow\Cal S}v'_{\tau}=v^+_{\downarrow}+\int_{\Cal S}|d\pmb v|$ is
defined (613J(f-ii)), this is $\sup_{\tau\in\Cal S}v'_{\tau}$ and
$\pmb v'$ is order-bounded.

\medskip

\quad\grheadd\ Set
$v''_{\tau}=v'_{\tau}-v_{\tau}$ for $\tau\in\Cal S$, so that
$\pmb v''=\family{\tau}{\Cal S}{v''_{\tau}}$ is fully adapted.   If
$\sigma\le\tau$ in $\Cal S$,

$$\eqalignno{v''_{\tau}-v''_{\sigma}
&=\int_{\Cal S\wedge\tau}|d\pmb v|-\int_{\Cal S\wedge\sigma}|d\pmb v|
   -v_{\tau}+v_{\sigma}\cr
&=\int_{\Cal S\cap[\sigma,\tau]}|d\pmb v|-v_{\tau}+v_{\sigma}
\ge|v_{\tau}-v_{\sigma}|-v_{\tau}+v_{\sigma}\cr
\displaycause{as in ($\alpha$) of (ii)$\Rightarrow$(iii) above}
&\ge 0.\cr}$$

\noindent So $\pmb v''$ is non-decreasing.   We have

$$\eqalignno{\lim_{\tau\downarrow\Cal S}v''_{\tau}
&=v_{\downarrow}^+
  +\lim_{\tau\downarrow\Cal S}\int_{\Cal S\wedge\tau}|d\pmb v|
  -v_{\downarrow}
=v_{\downarrow}^+-v_{\downarrow}
\Displaycause{613J(f-i)}
\ge 0,\cr}$$

\noindent so $\pmb v''$ is non-negative.   At the other end,

\Centerline{$\lim_{\tau\uparrow\Cal S}v''_{\tau}
=v_{\downarrow}^+
  +\lim_{\tau\uparrow\Cal S}\int_{\Cal S\wedge\tau}|d\pmb v|
  -\lim_{\tau\uparrow\Cal S}v_{\tau}
=v_{\downarrow}^++\int_{\Cal S}|d\pmb v|-v_{\uparrow}$}

\noindent is defined, so this must be an upper bound of
$\{v''_{\tau}:\tau\in\Cal S\}$, and $\pmb v''$ is order-bounded.

\medskip

\quad\grheade\  Accordingly $\pmb v'-\pmb v''=\pmb v$ witnesses that
(i) is true.
}%end of proof of 614I

\leader{614J}{Definitions} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\pmb v$ a fully adapted process with domain $\Cal S$.

\spheader 614Ja $\pmb v$ is {\bf of bounded variation} if
it satisfies the conditions of Theorem 614I.

\spheader 614Jb $\pmb v$ is {\bf locally of bounded variation} if
$\pmb v\restr\Cal S\wedge\tau$ is of bounded variation for every
$\tau\in\Cal S$.

\leader{614K}{Non-decreasing processes}\cmmnt{ I pause for some nearly
trivial remarks.}   Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb{v}=\sfamilysS{v}$ a non-decreasing fully adapted process.

\spheader 614Ka $\pmb v$ is a lattice homomorphism.   \prooflet{\Prf\
If $\sigma$, $\tau\in\Cal S$, then $v_{\sigma\wedge\tau}\le v_{\sigma\vee\tau}$
and

$$\eqalignno{\Bvalue{\sigma\le\tau}&
=\Bvalue{\sigma=\sigma\wedge\tau}\Bcap\Bvalue{\tau=\sigma\vee\tau}
\Displaycause{611E(a-ii-$\beta$)}
\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma\wedge\tau}}
   \Bcap\Bvalue{v_{\tau}=v_{\sigma\vee\tau}}
\Bsubseteq\Bvalue{v_{\sigma}\wedge v_{\tau}=v_{\sigma\wedge\tau}}
   \Bcap\Bvalue{v_{\sigma}\vee v_{\tau}=v_{\sigma\vee\tau}}
\cr}$$

\noindent and similarly

\Centerline{$\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{v_{\sigma}\wedge v_{\tau}=v_{\sigma\wedge\tau}}
   \Bcap\Bvalue{v_{\sigma}\vee v_{\tau}=v_{\sigma\vee\tau}}$,}

\noindent so
$v_{\sigma}\wedge v_{\tau}=v_{\sigma\wedge\tau}$ and
$v_{\sigma}\vee v_{\tau}=v_{\sigma\vee\tau}$.   As
$\sigma$ and $\tau$ are arbitrary, $\pmb v$ is a lattice homomorphism.\
\Qed}

\spheader 614Kb\cmmnt{ As in the proof of (a) just above,}

\Centerline{$\Bvalue{\sigma\le\tau}
\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma\wedge\tau}}
  \Bcap\Bvalue{v_{\tau}=v_{\sigma\vee\tau}}
\Bsubseteq\Bvalue{v_{\sigma}\le v_{\tau}}$}

\noindent for all $\sigma$, $\tau\in\Cal S$.

\spheader 614Kc{\bf (i)} If $\pmb v$ is non-negative it is
locally order-bounded.
\prooflet{\Prf\ If $\tau\in\Cal S$ then $0\le v_{\sigma}\le v_{\tau}$ for every
$\sigma\in\Cal S\wedge\tau$.\ \Qed}

\medskip

\qquad{\bf (ii)} If $\pmb v$ is order-bounded it is of bounded variation.
\prooflet{\Prf\ If $\Cal S=\emptyset$ this is trivial.   Otherwise, set
$\bar v=\sup|\pmb v|$.   If $\tau_0\le\ldots\le\tau_n$ in $\Cal S$, then

\Centerline{$\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
=\sum_{i=0}^{n-1}v_{\tau_{i+1}}-v_{\tau_i}
=v_{\tau_n}-v_{\tau_0}
\le\bar v-(-\bar v)=2\bar v$.}

\noindent So $\pmb v$ satisfies condition (ii) of 614I.\ \Qed}

\spheader 614Kd If $\Cal S\ne\emptyset$ and $\pmb v$ is order-bounded, then

\Centerline{$\int_{\Cal S}|d\pmb{v}|=\int_{\Cal S}d\pmb{v}
=v_{\uparrow}-v_{\downarrow}$}

\noindent where $v_{\uparrow}=\sup_{\sigma\in\Cal S}v_{\sigma}
=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ and
$v_{\downarrow}=\inf_{\sigma\in\Cal S}v_{\sigma}
=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$.
\prooflet{\Prf\ To identify $\inf_{\sigma\in\Cal S}v_{\sigma}$ with
$\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ and
$\sup_{\sigma\in\Cal S}v_{\sigma}$ with
$\lim_{\sigma\uparrow\Cal S}v_{\sigma}$, use the last sentence of 613Ba.
Now 613N tells us that $\int_{\Cal S}d\pmb v=v_{\uparrow}-v_{\downarrow}$,
and since the interval functions $\Delta\pmb v$ and $|\Delta\pmb v|$
(613Cc) are equal, this is also $\int_{\Cal S}|d\pmb v|$.\ \Qed}

\spheader 614Ke\cmmnt{ There is a special kind of non-decreasing process which
it will sometimes be useful to remember.}   Suppose that $w\in (L^0)^+$.
For each $\sigma\in\Cal S$, set
$w_{\sigma}=\sup\{x:x\in L^0(\frak A_{\sigma}$, $x\le w\}$.   Now
$\pmb w=\sfamilysS{w}$ is a non-negative non-decreasing fully adapted process.
\prooflet{\Prf\ If $\sigma\in\Cal S$, then
$L^0(\frak A_{\sigma})$ is order-closed in
$L^0=L^0(\frak A)$ (612A(e-i)), so $w_{\sigma}\in L^0(\frak A_{\sigma})$, and
of course $0\le w_{\sigma}$.   If
$\sigma\le\tau$ in $\Cal S$ then $\frak A_{\sigma}\subseteq\frak A_{\tau}$ so
$L^0(\frak A_{\sigma})\subseteq L^0(\frak A_{\tau})$ and
$w_{\sigma}\le w_{\tau}$.   Finally, if $\sigma$, $\tau\in\Cal S$ then
$w_{\tau}\times\chi\Bvalue{\sigma=\tau}\in L^0(\frak A_{\sigma})$ (612C)
and $w_{\tau}\times\chi\Bvalue{\sigma=\tau}\le w$ so
$w_{\tau}\times\chi\Bvalue{\sigma=\tau}\le w_{\sigma}$;  similarly,
$w_{\sigma}\times\chi\Bvalue{\sigma=\tau}\le w_{\tau}$, so
$w_{\tau}\times\chi\Bvalue{\sigma=\tau}
=w_{\sigma}\times\chi\Bvalue{\sigma=\tau}$, that is,
$\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{w_{\sigma}=w_{\tau}}$.   Thus
$\sfamilysS{w}$ is fully adapted.\ \Qed}

\cmmnt{Observe also that }$|\pmb u|\le\pmb w$
whenever $\pmb u=\sfamilysS{u}$ is fully
adapted and $|\sup\pmb u|\le w$\cmmnt{, just because
$|u_{\sigma}|\in L^0(\frak A_{\sigma})$ and $|u_{\sigma}|\le w$ for every
$\sigma\in\Cal S$}.

\spheader 614Kf If $h:\Bbb R\to\Bbb R$ is non-decreasing, then $\bar h\pmb v$
is non-decreasing.   \prooflet{(Use 612A(d-i-$\epsilon$).)}

\leader{614L}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a fully adapted process with domain $\Cal S$.

(a) If $\pmb v$ is (locally) of bounded variation it is (locally) order-bounded.

(b)(i) If $\pmb v$ is of bounded variation and $\Cal S'$ is a sublattice
of $\Cal S$, then $\pmb v\restr\Cal S'$ is of bounded variation and
$\int_{\Cal S'}|d\pmb v|\le\int_{\Cal S}|d\pmb v|$.

\quad(ii) If $\pmb v$ is locally of bounded variation and $\Cal S'$ is a
sublattice
of $\Cal S$, then $\pmb v\restr\Cal S'$ is locally of bounded variation.

(c) If $\tau\in\Cal S$, then $\pmb v$ is (locally) of
bounded variation iff $\pmb v\restr\Cal S\wedge\tau$ and
$\pmb v\restr\Cal S\vee\tau$ are both (locally) of bounded variation.

\proof{{\bf (a)} Since $\Mob(\Cal S)$ is closed under subtraction (614F(c-ii)),
it follows at once from 614I(i) and 614Kc that if $\pmb v$ is of bounded
variation then it is order-bounded.   Applying this to $\Cal S\wedge\tau$ for
$\tau\in\Cal S$, we see that if $\pmb v$ is locally of bounded variation it is
locally order-bounded.

\medskip

{\bf (b)(i)} $\pmb v$ satisfies condition (i) of 614I, so $\pmb v\restr\Cal S'$
also does, and $\int_{\Cal S'}|d\pmb v|\le\int_{\Cal S'}|d\pmb v|$ by the
last formula in 614I.

\medskip

\quad{\bf (ii)}
If $\pmb v$ is locally of bounded variation and $\tau\in\Cal S'$,
then $\pmb v\restr\Cal S\wedge\tau$ is of bounded variation and
$\Cal S'\wedge\tau$ is a
sublattice of $\Cal S\wedge\tau$, so $\pmb v\restr\Cal S'\wedge\tau$ is of
bounded variation;  as $\tau$ is arbitrary, $\pmb v\restr\Cal S'$ is locally of
bounded variation.

\medskip

{\bf (c)} If $\pmb v$ is (locally) of bounded variation, then
$\pmb v\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\vee\tau$ are (locally) of
bounded variation because $\Cal S\wedge\tau$ and $\Cal S\vee\tau$ are
sublattices of $\Cal S$.

If $\pmb v\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\vee\tau$ are
of bounded variation then $\int_{\Cal S\wedge\tau}|d\pmb v|$ and
$\int_{\Cal S\vee\tau}|d\pmb v|$ are defined, so
$\int_{\Cal S}|d\pmb v|$ is defined (613J(c-i)) and $\pmb v$ is of bounded
variation.

If $\pmb v\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\vee\tau$ are
locally of bounded variation and $\sigma\in\Cal S$, then
$\pmb v\restr(\Cal S\wedge\sigma)\wedge\tau
=\pmb v\restr(\Cal S\wedge\tau)\wedge\sigma$ and
$\pmb v\restr(\Cal S\wedge\sigma)\vee\tau
=\pmb v\restr(\Cal S\vee\tau)\wedge(\sigma\vee\tau)$ are of bounded variation,
so $\pmb v\restr\Cal S\wedge\sigma$ is of bounded variation;  accordingly
$\pmb v$ is locally of bounded variation.
}%end of proof of 614L

\leader{614M}{Proposition} The Poisson process, as described in 612U,
is locally of bounded variation.

\proof{ It is non-negative and non-decreasing, so 614Kc applies.
}%end of proof of 614M

\leader{614N}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\bar u\in(L^0)^+$.   Then
$\{\pmb v:\pmb v\in\Mbv(\Cal S)$, $\int_{\Cal S}|d\pmb v|\le\bar u\}$ is
closed in $(L^0)^{\Cal S}$ for its product topology.

\proof{ The point is just that if $I\in\Cal I(\Cal S)$, then
$\pmb v\mapsto S_I(\tbf 1,|d\pmb v|):(L^0)^{\Cal S}\to L^0$ is continuous,
so

\Centerline{$A=\{\pmb v:\pmb v\in(L^0)^{\Cal S}$,
$S_I(\tbf 1,|d\pmb v|)\le\bar u$ for every $I\in\Cal I(\Cal S)\}$}

\noindent is closed in $(L^0)^{\Cal S}$.   Now the set $\Mfa(\Cal S)$ of fully
adapted processes with domain $\Cal S$ is closed in $(L^0)^{\Cal S}$
(613Bl), so
$\{\pmb v:\pmb v\in\Mbv(\Cal S)$, $\int_{\Cal S}|d\pmb v|\le\bar u\}
=A\cap \Mfa(\Cal S)$ is closed.
}%end of proof of 614N


\leader{614O}{Cumulative variation} Let $\Cal S$ be a sublattice of $\Cal T$,
and $\pmb{v}$ a process with domain $\Cal S$
which is locally of bounded variation.   Then
$v^{\uparrow}_{\tau}=\int_{\Cal S\wedge\tau}|d\pmb v|$ is
defined for every $\tau\in\Cal S$\cmmnt{ (614I-614J)}, and
$\pmb v^{\uparrow}
=\family{\tau}{\Cal S}{\int_{\Cal S\wedge\tau}v^{\uparrow}_{\tau}}$ is fully
adapted\cmmnt{ (613J(e-ii))}.   I will call
$\pmb v^{\uparrow}$ the {\bf cumulative variation} of $\pmb v$.

\leader{614P}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ a process which is locally of bounded variation,
and $\pmb v^{\uparrow}=\family{\tau}{\Cal S}{v^{\uparrow}_{\tau}}$ its
cumulative variation.

(a)(i) If $\sigma\le\tau$ in $\Cal S$, then

\Centerline{$v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}
=\int_{\Cal S\cap[\sigma,\tau]}|d\pmb v|
\ge|v_{\tau}-v_{\sigma}|\ge 0$.}

\quad(ii)
$\pmb v^{\uparrow}$ is non-negative and non-decreasing and has starting
value $0$.

\quad(iii) $\pmb v^{\uparrow}+\pmb v$ and $\pmb v^{\uparrow}-\pmb v$ are
non-decreasing.

\quad(iv) $\pmb v$ is of bounded variation iff
$\lim_{\tau\uparrow\Cal S}v^{\uparrow}_{\tau}=\sup|\pmb v^{\uparrow}|$ is
defined, and in this case the limit is $\int_{\Cal S}|d\pmb v|$.

(b) If $\tau\in\Cal S$ then, writing
$(\pmb v\restr\Cal S\wedge\tau)^{\uparrow}$ and
$(\pmb v\restr\Cal S\vee\tau)^{\uparrow}$  for the cumulative variations of
$\pmb v\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\vee\tau$,

\Centerline{$(\pmb v\restr\Cal S\wedge\tau)^{\uparrow}
=\pmb v^{\uparrow}\restr\Cal S\wedge\tau$,
\quad$(\pmb v\restr\Cal S\vee\tau)^{\uparrow}
=\pmb v^{\uparrow}\restr\Cal S\vee\tau-v^{\uparrow}_{\tau}\tbf 1$.}

(c) Suppose that $I\in\Cal I(\Cal S)$ is not empty and
$(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells\cmmnt{ (611L)}.

\quad(i)

\Centerline{$v^{\uparrow}_{\tau_{i+1}}-v^{\uparrow}_{\tau_i}
\le|v_{\tau_{i+1}}-v_{\tau_i}|
   +v^{\uparrow}_{\max I}-v^{\uparrow}_{\min I}-S_I(\tbf{1},|d\pmb v|)$}

\noindent for every $i<n$.
%for 618U

\quad(ii) If $\pmb v$ is of bounded variation, write
$w$ for $\int_{\Cal S}|d\pmb v|-S_I(\tbf 1,|d\pmb v|)$, and let
$v_{\downarrow}$ be the starting value of $\pmb v$.

\qquad($\alpha$) If $\tau\in\Cal S\wedge\tau_0$, then
$v^{\uparrow}_{\tau}\le|v_{\tau}-v_{\downarrow}|+w$.

\qquad($\beta$) If $i<n$ and $\tau\in\Cal S\cap[\tau_i,\tau_{i+1}]$ then
$v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_i}\le|v_{\tau}-v_{\tau_i}|+w$.

\qquad($\gamma$) If $\tau\in\Cal S\vee\tau_n$ then
$v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_n}\le|v_{\tau}-v_{\tau_n}|+w$.

\proof{{\bf (a)(i)} Immediate from 613J(c-i) and 614I.

\medskip

\quad{\bf (ii)} By 614I, $\pmb v^{\uparrow}$ is non-negative;  by
(i) here, it is non-decreasing;  by
613J(f-i) its starting value is $0$.

\medskip

\quad{\bf (iii)} If $\sigma\le\tau$ in $\Cal S$, then

\Centerline{$(v^{\uparrow}_{\tau}\pm v_{\tau})
     -(v^{\uparrow}_{\sigma}\pm v_{\sigma})
=v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}\mp(v_{\tau}-v_{\sigma})
\ge v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}-|v_{\tau}-v_{\sigma}|
\ge 0$}

\noindent by (i) above.

\medskip

\quad{\bf (iv)} Because $\pmb v^{\uparrow}$ is non-negative and non-decreasing,
$\lim_{\tau\uparrow\Cal S}v^{\uparrow}_{\tau}=\sup|\pmb v^{\uparrow}|$ if either
is defined in $L^0(\frak A)$.   If $\pmb v$ is of bounded variation,

\Centerline{$\int_{\Cal S}|d\pmb v|
=v^{\uparrow}_{\tau}+\int_{\Cal S\vee\tau}|d\pmb v|\ge v^{\uparrow}_{\tau}$}

\noindent for every $\tau\in\Cal S$, so $\sup|\pmb v^{\uparrow}|$ is defined
and less than or equal to $\int_{\Cal S}|d\pmb v|$.  If
$\sup|\pmb v^{\uparrow}|$ is defined, then whenever $\tau_0\le\ldots\le\tau_n$
in $\Cal S$,

\Centerline{$\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
\le\int_{\Cal S\wedge\tau_n}|d\pmb v|\le\sup|\pmb v^{\uparrow}|$,}

\noindent so $\int_{\Cal S}|d\pmb v|$ is defined and is at most
$\sup|\pmb v^{\uparrow}|$.   Thus in either case we have equality.

\medskip

{\bf (b)} Of course

\Centerline{$(\pmb v\restr\Cal S\wedge\tau)^{\uparrow}
=\family{\sigma}{\Cal S\wedge\tau}{\int_{\Cal S\wedge\sigma}|d\pmb v|}
=\pmb v^{\uparrow}\restr\Cal S\wedge\tau$.}

\noindent On the other hand, for $\sigma\in\Cal S\vee\tau$,

$$\eqalignno{v^{\uparrow}_{\sigma}&
=\int_{\Cal S\wedge\sigma}|d\pmb v|
=\int_{(\Cal S\wedge\sigma)\wedge\tau}|d\pmb v|
   +\int_{(\Cal S\wedge\sigma)\vee\tau}|d\pmb v|
\Displaycause{613G(a-i)}
=v^{\uparrow}_{\tau}+\int_{(\Cal S\vee\tau)\wedge\sigma}|d\pmb v|,\cr}$$

\noindent so

\Centerline{$\pmb v^{\uparrow}\restr\Cal S\vee\tau
=v^{\uparrow}_{\tau}\tbf 1+(\pmb v\restr\Cal S\vee\tau)^{\uparrow}$.}

\medskip

{\bf (c)(i)}

$$\eqalign{S_I(\tbf 1,|d\pmb v|)&
=|v_{\tau_{i+1}}-v_{\tau_i}|
   +\sum_{\Atop{j<n}{j\ne i}}|v_{\tau_{j+1}}-v_{\tau_j}|
\le|v_{\tau_{i+1}}-v_{\tau_i}|
   +\sum_{\Atop{j<n}{j\ne i}}v^{\uparrow}_{\tau_{j+1}}-v^{\uparrow}_{\tau_j}\cr&
=|v_{\tau_{i+1}}-v_{\tau_i}|+S_I(\tbf 1,d\pmb v^{\uparrow})
   -v^{\uparrow}_{\tau_{i+1}}+v^{\uparrow}_{\tau_i}\cr&
=|v_{\tau_{i+1}}-v_{\tau_i}|+v^{\uparrow}_{\max I}-v^{\uparrow}_{\min I}
   -v^{\uparrow}_{\tau_{i+1}}+v^{\uparrow}_{\tau_i};\cr}$$

\noindent rearranging, we have the result.

\medskip

\quad{\bf (ii)}\grheada\ If $\sigma\in\Cal S\wedge\tau$ then
$v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}\le|v_{\tau}-v_{\sigma}|+w$.
\Prf\
Set $J=\{\sigma,\tau,\tau_0,\ldots,\tau_n\}$.   Then
$J\in\Cal I(\Cal S)$ and $(\sigma,\tau,\tau_0,\ldots,\tau_n)$
linearly generates the $J$-cells, so

$$\eqalign{v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}&
\le|v_{\tau}-v_{\sigma}|+v^{\uparrow}_{\tau_n}-v^{\uparrow}_{\sigma}
    -S_J(\tbf 1,|d\pmb v|)
\le|v_{\tau}-v_{\sigma}|+\int_{\Cal S}|d\pmb v|
  -\sum_{i=0}^n|v_{\tau_{i+1}}-v_{\tau_i}|\cr&
=|v_{\tau}-v_{\sigma}|+\int_{\Cal S}|d\pmb v|
  -S_I(\tbf 1,|d\pmb v|)
=|v_{\tau}-v_{\sigma}|+w.  \text{ \Qed}\cr}$$

\noindent Now

\Centerline{$v^{\uparrow}_{\tau}
=\lim_{\sigma\downarrow\Cal S}v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}
\le\lim_{\sigma\downarrow\Cal S}|v_{\tau}-v_{\sigma}|+w
=|v_{\tau}-v_{\downarrow}|+w$.}

\medskip

\qquad\grheadb\ Set
$J=\{\tau_0,\ldots,\tau_i,\tau,\tau_{i+1},\ldots,\tau_n\}$.   Then

\Centerline{$v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_i}
\le|v_{\tau}-v_{\tau_i}|+v^{\uparrow}_{\tau_n}-v^{\uparrow}_{\tau_0}-S_J(\tbf 1,|d\pmb v|)
\le|v_{\tau}-v_{\tau_i}|+w$.}

\medskip

\qquad\grheadc\ This time, set $J=\{\tau_0,\ldots,\tau_n,\tau\}$, so that

\Centerline{$v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_n}
\le|v_{\tau}-v_{\tau_n}|+v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_0}-S_J(\tbf 1,|d\pmb v|)
\le|v_{\tau}-v_{\tau_n}|+w$.}
}%end of proof of 614P

\leader{614Q}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\hat\Cal S$ its covered envelope.

(a) Write
$\Mbv=\Mbv(\Cal S)$ for the set of fully adapted processes of bounded variation
with domain $\Cal S$.

\quad(i) $\bar h\pmb v\in\Mbv$ whenever $\pmb v\in\Mbv$ and
$h:\Bbb R\to\Bbb R$ is Lipschitz on every bounded interval.

\quad(ii) $\Mbv$ is an $f$-subalgebra of $\Mob(\Cal S)$.

\quad(iii) The space $\Msimp$ of simple processes with domain $\Cal S$ is an
$f$-subalgebra of $\Mbv$ closed under $\bar h$ for every Borel measurable
$h:\Bbb R\to\Bbb R$.

\quad(iv) If $\pmb v\in\Mfa(\Cal S)$ and $\hatpmb v$ is its fully adapted
extension to $\hat\Cal S$, then

\qquad($\alpha$) $\pmb v$ is non-decreasing iff $\hatpmb v$ is non-decreasing,

\qquad($\beta$) $\pmb v$ is of bounded variation
iff $\hatpmb v$ is of bounded variation, and in this case
$\int_{\hat\Cal S}|d\hatpmb v|=\int_{\Cal S}|d\pmb v|$ and
the cumulative variation $\hatpmb v^{\uparrow}$
of $\hatpmb v$ is the fully adapted extension of the
cumulative variation $\pmb v^{\uparrow}$ of $\pmb v$.

(b) Write
$\Mlbv=\Mlbv(\Cal S)$ for the set of fully adapted processes with domain
$\Cal S$ which are locally of bounded variation.

\quad(i) If $\pmb v\in\Mlbv(\Cal S)$ then
$\pmb v\restr\Cal S'$ is locally of bounded variation
for every sublattice $\Cal S'$ of $\Cal S$.

\quad(ii) $\bar h\pmb v\in\Mlbv$ whenever $\pmb v\in\Mlbv$ and
$h:\Bbb R\to\Bbb R$ is Lipschitz on every bounded interval.

\quad(iii) $\Mlbv$ is an $f$-subalgebra of $\Mlob(\Cal S)$.

\quad(iv) If $\pmb v\in\Mfa(\Cal S)$ then $\pmb v$ is locally of bounded
variation iff it is expressible as the difference of two non-negative
non-decreasing fully adapted processes.

\quad(v) If $\pmb u\in\Mfa(\Cal S)$, then $\pmb u$ is locally of
bounded variation iff its fully adapted extension to
$\hat\Cal S$ is locally of bounded variation, and in this case
the cumulative variation of $\hatpmb v$ is the fully adapted extension of the
cumulative variation of $\pmb v$.

\proof{{\bf (a)(i)} Define $g:\Bbb R\to\Bbb R$ by setting
$g(0)=0$ and

\Centerline{$g(x)=\sup_{-|x|\le y<y'\le|x|}\Bover{|h(y')-h(y)|}{y'-y}$}

\noindent for $x\ne 0$.   Then $g$ is monotonic on both $\ocint{-\infty,0}$ and
$\coint{0,\infty}$, so is Borel measurable.   As

\Centerline{$|h(y')-h(y)|\le g(x)|y'-y|$
whenever $|y|$, $|y'|\le x$ in $\Bbb R$,}

\noindent we shall have

\Centerline{$|\bar h(v')-\bar h(v)|\le\bar g(u)\times|v'-v|$
whenever $|v|$, $|v'|\le u$ in $L^0$.}

Now suppose that $\pmb v=\sfamilysS{v}$ is of bounded variation.
By (ii), it is order-bounded;  let $u$ be an upper bound of
$\{v_{\sigma}:\sigma\in\Cal S\}$.   If
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$, then

$$\eqalign{\sum_{i=0}^{n-1}|\bar h(v_{\tau_{i+1}})-\bar h(v_{\tau_i})|
&\le\sum_{i=0}^{n-1}\bar g(u)\times|v_{\tau_{i+1}}-v_{\tau_i}|\cr
&=\bar g(u)\times\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
\le\bar g(u)\times\int_{\Cal S}|d\pmb v|\cr}$$

\noindent (614I).
Thus $\bar h\pmb v$ satisfies (ii) of 614I and is of bounded
variation.

\medskip

\quad{\bf (ii)} If $\pmb v$, $\pmb v'$ are non-negative non-decreasing
processes defined on $\Cal S$, then of course $\pmb v+\pmb v'$ is non-negative
and non-decreasing.   Condition (i) of 614I now makes it plain that the sum of
processes of bounded variation is of bounded variation.   From (i) we see
that $\bar h\pmb v\in\Mbv$ whenever $\pmb v\in\Mbv$ and $h:\Bbb R\to\Bbb R$
is convex, so 612Bc tells us that $\Mbv$ is an $f$-subalgebra of
$\prod_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$ and therefore of
$\Mob$.

\medskip

\quad{\bf (iii)} Let $\pmb v=\sfamilysS{v}$ be a simple process.
If $\pmb v$ is constant, then surely it is of bounded variation.
Otherwise, let $(\tau_0,\ldots,\tau_n)$ be a breakpoint sequence for $\pmb v$
and $v_*$ its starting value.   Consider the simple process
$\pmb w=\sfamilysS{w}$ with the
same breakpoint sequence defined by saying that

\Centerline{$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{w_{\sigma}=v_*}$,
\quad$\Bvalue{\tau_n\le\sigma}\Bsubseteq\Bvalue{w_{\sigma}
=v_*+|v_{\tau_0}-v_*|+\sumop_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|}$,}

\Centerline{$\Bvalue{\tau_j\le\sigma}\Bcap\Bvalue{\sigma<\tau_{j+1}}
\Bsubseteq\Bvalue{w_{\sigma}
=v_*+|v_{\tau_0}-v_*|+\sumop_{i=0}^{j-1}|v_{\tau_{i+1}}-v_{\tau_i}|}$}

\noindent for $0\le j<n$.   Since $(v_*,w_{\tau_0},\ldots,w_{\tau_n})$ and
$(0,w_{\tau_0}-v_{\tau_0},\ldots,w_{\tau_n}-v_{\tau_n})$ are both
non-decreasing, $\pmb w$ and $\pmb w-\pmb v$ are both non-decreasing,
so their difference $\pmb v$ is of bounded variation.

Thus $\Msimp\subseteq\Mbv$.   Now refer to 612L for its other properties.

\medskip

\quad{\bf (iv)} Express $\pmb v$ and $\hatpmb v$ as
$\sfamilysS{v}$ and $\family{\sigma}{\hat\Cal S}{\hat v_{\sigma}}$.

\medskip

\qquad\grheada\ If $\hatpmb v$ is non-decreasing then of course
$\pmb v=\hatpmb v\restr\Cal S$ is non-decreasing.   If $\pmb v$ is
non-decreasing and $\tau_0\le\tau_1$ in $\hat\Cal S$,
take any non-zero $a\in\frak A$.   Then
there are a $\sigma_0\in\Cal S$ such that $a_0=a\Bcap\Bvalue{\tau_0=\sigma_0}$
is non-zero, and a $\sigma_1\in\Cal S$ such that
$a_1=a_0\Bcap\Bvalue{\tau_1=\sigma_1}$ is non-zero.   Now

$$\eqalign{a_1&
\Bsubseteq\Bvalue{\hat v'_{\tau_0}=v'_{\sigma_0}}
   \Bcap\Bvalue{\hat v'_{\tau_1}=v'_{\sigma_1}}
   \Bcap\Bvalue{\sigma_0\le\sigma_1}\cr&
\Bsubseteq\Bvalue{\hat v'_{\tau_0}=v'_{\sigma_0}}
   \Bcap\Bvalue{\hat v'_{\tau_1}=v'_{\sigma_0\vee\sigma_1}}
\Bsubseteq\Bvalue{\hat v'_{\tau_0}\le\hat v'_{\tau_1}}.\cr}$$

\noindent As $a$ is arbitrary,
$\hat v'_{\tau_0}\le\hat v'_{\tau_1}$.   So $\hatpmb v'$ non-decreasing.

\medskip

\qquad\grheadb\ If $\hatpmb v$ is of bounded variation then
$\pmb v=\hatpmb v\restr\Cal S$ is of bounded variation, by (i) above.
If $\pmb v$ is of bounded variation, express it as $\pmb v'-\pmb v''$ where
$\pmb v'=\sfamilysS{v'}$, $\pmb v''$
are order-bounded non-negative non-decreasing fully
adapted processes.   Let $\hatpmb v'$,
$\hatpmb v''$ be their fully adapted
extensions to $\hat\Cal S$, so that $\hatpmb v=\hatpmb v'-\hatpmb v''$.
By ($\alpha$), $\hatpmb v'$ and $\hatpmb v''$ are non-decreasing;
by 612Q(d-i) they are order-bounded;  and by 612Qb they are non-negative.
So they witness that $\hatpmb v$ is of bounded variation.

Now 613T, with $\pmb u=\tbf 1$ and
$\psi(\sigma,\tau)=|\hat v_{\tau}-\hat v_{\sigma}|$ for
$\sigma\le\tau$ in $\hat\Cal S$, tells us that 
$\int_{\Cal S}|d\pmb v|=\int_{\hat\Cal S}|d\hatpmb v|$.   And if 
$\tau\in\Cal S$ then $\hat\Cal S\wedge\tau$ is the covered
envelope of $\Cal S\wedge\tau$ (611M(e-i)) and
$\int_{\Cal S\wedge\tau}|d\pmb v|=\int_{\hat\Cal S\wedge\tau}|d\hatpmb v|$,
so $\pmb v^{\uparrow}=\hatpmb v^{\uparrow}\restr\Cal S$
and $\hatpmb v^{\uparrow}$ is the fully adapted extension of
$\pmb v^{\uparrow}$.

\medskip

{\bf (b)(i)-(iii)} These follow immediately from 614L and (a-i).

\medskip

\quad{\bf (iv)}\grheada\ If $\pmb v=\sfamilysS{v}$ is non-negative and
non-decreasing and $\tau\in\Cal S$, then
$\pmb v\restr\Cal S\wedge\tau$ takes all its values in
$[0,v_{\tau}]$ so is order-bounded and of bounded variation;  as $\tau$ is
arbitrary, $\pmb v$ is locally of bounded variation.   By (iii), if
$\pmb v$ is the difference of two non-negative non-decreasing processes
then it is locally of bounded variation.

\medskip

\qquad\grheadb\ Conversely, if $\pmb v$ is locally of bounded variation, then
it has a cumulative variation $\pmb v^{\uparrow}=\sfamilysS{v^{\uparrow}}$,
which is non-negative and
non-decreasing and has starting value $0$ (614P(a-ii)).
Because $\pmb v$ is locally moderately oscillatory (or otherwise),
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ is defined.
We have

\Centerline{$v^{\uparrow}_{\tau}
=\lim_{\sigma\downarrow\Cal S}v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}
\ge\lim_{\sigma\downarrow\Cal S}|v_{\tau}-v_{\sigma}|
=|v_{\tau}-v_{\downarrow}|
\ge v_{\tau}-v_{\downarrow}$}

\noindent so $v^{\uparrow}_{\tau}+v_{\downarrow}^+\ge v_{\tau}$,
for every $\tau\in\Cal S$.
Now if we set $\pmb w=\pmb v^{\uparrow}+v_{\downarrow}^+\tbf 1$,
both $\pmb w$
and $\pmb w-\pmb v$ will be non-negative and non-decreasing, and
$\pmb v=\pmb w-(\pmb w-\pmb v)$ is expressed as a difference of non-negative
non-decreasing processes.

\medskip

\quad{\bf (v)} We can follow the proof of (a-iv), using (iv) just above in
place of 614I(i).
}%end of proof of 614Q

\leader{614R}{}\cmmnt{ Following on from 612H, I give a
result on the construction of order-bounded processes and
processes of bounded variation.

\medskip

\noindent}{\bf Proposition} Let $(\Omega,\Sigma,\mu)$ be a complete
probability space, and $\langle\Sigma_t\rangle_{t\ge 0}$ a filtration of
$\sigma$-subalgebras of $\Sigma$ such that
every $\mu$-negligible set belongs to every $\Sigma_t$.
Let \Asissixt\ be the associated stochastic integration structure as in
612H;  for a stopping time $h:\Omega\to[0,\infty]$ let $h^{\ssbullet}$ be the
corresponding member of $\Cal T$.
Suppose that $\langle X_t\rangle_{t\ge 0}$ is a progressively
measurable process on $\Omega$, with associated fully adapted process
$\pmb x=\family{\tau}{\Cal T_f}{x_{\tau}}$.

(a) If $\{X_s(\omega):s\ge 0\}$ is bounded for
almost every $\omega\in\Omega$, then $\pmb x$ is order-bounded.

(b) If
$s\mapsto X_s(\omega):\coint{0,\infty}\to\Bbb R$ is of bounded variation for
almost every $\omega\in\Omega$, then $\pmb x$ is of bounded variation.

\proof{{\bf (a)(i)} We need the following fact which was left as an exercise in
\S364.  %364Xn
Suppose that $V:\Omega\to\coint{0,\infty}$ is any function.
Let $F$ be the set of $\Sigma$-measurable functions $f:\Omega\to\Bbb R$ such
that $f\le V$.   Then $A=\{f^{\ssbullet}:f\in F\}$ is bounded above in
$L^0$.   \Prf\ For $n\in\Bbb N$ set $c_n=\sup_{u\in A}\Bvalue{u>n}$.
$\frak A$ is ccc (322G) so there is a countable set $I_n\subseteq A$ such that
$c_n=\sup_{u\in I_n}\Bvalue{u>n}$ (316E).
Let $\sequencen{f_n}$ be a sequence in $F$ such that
$\bigcup_{n\in\Bbb N}I_n=\{f_n^{\ssbullet}:n\in\Bbb N\}$.
Set $f(x)=\sup_{n\in\Bbb N}f_n(x)$ for $x\in X$;  then $f\in F$ and
$f_n^{\ssbullet}\le f^{\ssbullet}$ for every $n\in\Bbb N$.
Setting $v=f^{\ssbullet}\in L^0$, $u\le v$ for every
$u\in\bigcup_{n\in\Bbb N}I_n$, so $c_n\Bsubseteq\Bvalue{v>n}$ for every $n$, and
$\inf_{n\in\Bbb N}c_n=0$.   By 364L(a-ii), $A$ is bounded above in $L^0$.\ \Qed

\medskip

\quad{\bf (ii)} Set $\Omega'=\{\omega:\{X_s(\omega):s\ge 0\}$ is bounded$\}$,
and define

$$\eqalign{V(\omega)
&=\sup_{s\ge 0}|X_s(\omega)|\text{ if }\omega\in\Omega',\cr
&=0\text{ if }\omega\in\Omega\setminus\Omega'.\cr}$$

Let $A$ be the corresponding order-bounded subset of $L^0$ as defined in (i).
If $h:\Omega\to\coint{0,\infty}$ is a stopping time, then $X_h$ is measurable
and $|X_h(\omega)|=|X_{h(\omega)}(\omega)|\le V(\omega)$ for every
$\omega\in\Omega'$,
so $|X_h^{\ssbullet}|\in A$.   But this means that $|x_{\tau}|\in A$ for every
$\tau\in\Cal T_f$ and $\pmb x$ is order-bounded.

\medskip

{\bf (b)} Set

\Centerline{$\Omega'
=\{\omega:s\mapsto X_s(\omega):\coint{0,\infty}\to\Bbb R$ is of
bounded variation$\}$,}

\noindent so that $\Omega'$ is conegligible.   For $\omega\in\Omega$
let $V(\omega)\in\coint{0,\infty}$ be the total
variation of $s\mapsto X_s(\omega):\coint{0,\infty}\to\Bbb R$ (224A) if
$\omega\in\Omega'$, and zero otherwise.
As in (a-i), let $F$ be the set of $\Sigma$-measurable functions
$f:\Omega\to\Bbb R$ such that $f\le V$, and
$A=\{f^{\ssbullet}:f\in F\}$;  then $A$ is bounded above in $L^0$.

Suppose that $\tau_0\le\ldots\le\tau_n$ in $\Cal T_f$.
Then there are stopping times $h_0,\ldots,h_n:\Omega\to\coint{0,\infty}$ such
that $h_i^{\ssbullet}=\tau_i$ for $i\le n$ (612H(a-ii)), and
$h_i\leae h_{i+1}$ for $i<n$;  replacing each $h_i$ by
$\sup_{j\le i}h_j\times\chi\Omega'$ if
necessary, we can arrange that $h_0\le\ldots\le h_n$ and
$h_i(\omega)=0$ whenever $i\le n$ and $\omega\in\Omega\setminus\Omega'$.
In this case,
$\sum_{i=0}^{n-1}|X_{h_{i+1}(\omega)}(\omega)-X_{h_i(\omega)}(\omega)|
\le V(\omega)$ for each $\omega\in\Omega$.   But this means that
$\sum_{i=0}^{n-1}|x_{\tau_{i+1}}-x_{\tau_i}|\in A$.   Thus

\Centerline{$\{\sum_{i=0}^{n-1}|x_{\tau_{i+1}}-x_{\tau_i}|:
\tau_0\le\ldots\le\tau_n$ in $\Cal T_f\}\subseteq A$}

\noindent is bounded above in $L^0(\frak A)$ and $\pmb x$ is of bounded
variation.
}%end of proof of 614R

\leader{614S}{Lemma} If $I\in\Cal I(\Cal T)$ is non-empty and
$\pmb u=\family{\sigma}{I}{u_{\sigma}}$,
$\pmb v=\family{\sigma}{I}{v_{\sigma}}$ are fully adapted processes,
then

\Centerline{$|S_I(\pmb u,d\pmb v)|
\le\min(\sup|\pmb u|\times\int_I|d\pmb v|,
          \sup|\pmb v|\times(\int_I|d\pmb u|+2\sup|\pmb u|)$.}

\proof{ If $\#(I)\le 1$ this is trivial.   Otherwise,
take $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   Then

$$\eqalign{|S_I(\pmb u,d\pmb v)|&
=|\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})|
\le\sum_{i=0}^{n-1}|u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})|\cr&
\le\sup|\pmb u|\times\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
=\sup|\pmb u|\times\int_I|d\pmb v|\cr}$$

\noindent and also

$$\eqalign{|S_I(\pmb u,d\pmb v)|&
=|\sum_{i=0}^{n-1}(u_{\tau_i}-u_{\tau_{i+1}})\times v_{\tau_{i+1}})
    +u_{\tau_n}\times v_{\tau_n}-u_{\tau_0}\times v_{\tau_0}|\cr&
\le\sup|\pmb v|\times(\int_I|d\pmb u|+|u_{\tau_n}|+|u_{\tau_0}|)
\le\sup|\pmb v|\times(\int_I|d\pmb u|+2\sup|\pmb u|).\cr}$$
}%end of proof of 614S

\leader{614T}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u=\sfamilysS{u}$, $\pmb v=\sfamilysS{v}$
two processes of bounded variation with domain $\Cal S$.
Then $\int_{\Cal S}\pmb u\,d\pmb v$ is defined and

\Centerline{$|\int_{\Cal S}\pmb u\,d\pmb v|
\le\min(\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|,
\sup|\pmb v|\times(\int_{\Cal S}|d\pmb u|+2\sup|\pmb u|))$.}

%query:  second half of inequality to 614X ?

\proof{ Note first that $\pmb u$ and $\pmb v$ are order-bounded
(614L), so $\sup|\pmb u|$ and $\sup|\pmb v|$ are defined.

\medskip

{\bf (a)} To begin with, consider the case in which both $\pmb u$ and
$\pmb v$ are non-decreasing and non-negative.

\medskip

\quad{\bf (i)} If $I\in\Cal I(\Cal S)$ is non-empty and
$(\tau_0,\ldots,\tau_n)$ linearly generate the $I$-cells, then

$$\eqalign{0&
\le u_{\tau_0}\times(v_{\tau_n}-v_{\tau_0})
=u_{\tau_0}\times\sum_{i=0}^{n-1}(v_{\tau_{i+1}}-v_{\tau_i})
\le\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})\cr&
=S_I(\pmb u,d\pmb v)
\le\sum_{i=0}^{n-1}u_{\tau_n}\times(v_{\tau_{i+1}}-v_{\tau_i})
=u_{\tau_n}\times(v_{\tau_n}-v_{\tau_0})
\le\sup|\pmb u|\times\sup|\pmb v|.\cr}$$

\medskip

\quad{\bf (ii)} If $I\subseteq J$ in $\Cal I(\Cal S)$, then
$S_I(\pmb u,d\pmb v)\le S_J(\pmb u,d\pmb v)$.   \Prf\ If $I$ is empty
this is trivial.   Otherwise, take $(\tau_0,\ldots,\tau_n)$ linearly generating
the $I$-cells.   Then

$$\eqalignno{S_J(\pmb u,d\pmb v)&
=S_{J\cap[\min J,\tau_0]}(\pmb u,d\pmb v)
   +\sum_{i=0}^{n-1}S_{J\cap[\tau_i,\tau_{i+1}]}(\pmb u,d\pmb v)
   +S_{J\cap[\tau_n,\max J]}(\pmb u,d\pmb v)
\Displaycause{613G(a-ii)}
\ge 0+\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})+0
\Displaycause{by (i) above}
=S_I(\pmb u,d\pmb v).  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)} Thus $\{S_I(\pmb u,d\pmb v):I\in\Cal I(\Cal S)\}$
is upwards-directed.   As it is also bounded above in $L^0$, by (i),
it has a supremum, which is
$\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb v)
=\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

{\bf (c)} So the integral is defined when both $\pmb u$ and $\pmb v$ are
non-negative and non-decreasing;  as integration is bilinear (613Jb),
the integral is defined for all processes $\pmb u$, $\pmb v$ of bounded
variation.   Concerning the bound for $|\int_{\Cal S}\pmb u\,d\pmb v|$, we see
from 614S that if $I\in\Cal I(\Cal S)$ then

$$\eqalign{|S_I(\pmb u,d\pmb v)|&
\le\min(\sup|\pmb u\restr I|\times\int_I|d\pmb v|,
          \sup|\pmb v\restr I|\times(\int_I|d\pmb u|+2\sup|\pmb u\restr I|))\cr&
\le\min(\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|,
          \sup|\pmb v|\times(\int_{\Cal S}|d\pmb u|+2\sup|\pmb u|))\cr}$$

\noindent (using 614L(b-i)).   Accordingly

$$\eqalign{|\int_{\Cal S}\pmb u\,d\pmb v|&
=|\lim_{I\uparrow\Cal S}S_I(\pmb u,d\pmb v)|
=\lim_{I\uparrow\Cal S}|S_I(\pmb u,d\pmb v)|\cr&
\le\min(\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|,
          \sup|\pmb v|\times(\int_\Cal S|d\pmb u|+2\sup|\pmb u|)).\cr}$$

}%end of proof of 614T

\leader{614U}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u$, $\pmb v$ fully adapted processes with domain $\Cal S$ such that
$\pmb u$ is order-bounded,
$\pmb v$ is of bounded variation and $\int_{\Cal S}\pmb u\,d\pmb v$ is defined.
Then the indefinite integral $\iivu$\cmmnt{ (613O)}
is of bounded variation, and
$\int_{\Cal S}|d(\iivu)|\le\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|$.

\proof{ As noted in 613Oa and 613O(b-i), $\iivu$ is fully adapted and has domain
$\Cal S$.
Now suppose that $\tau_0\le\ldots\le\tau_n$ in $\Cal S$.   For $i<n$ write
$\Cal S_i$ for
$\Cal S\cap[\tau_i,\tau_{i+1}]=(\Cal S\wedge\tau_{i+1})\vee\tau_i$.   Then

$$\eqalignno{\sum_{i=0}^{n-1}|z_{\tau_{i+1}}-z_{\tau_i}|&
=\sum_{i=0}^{n-1}|\int_{\Cal S_i}\pmb u\,d\pmb v|
\le\sum_{i=0}^{n-1}\sup|u\restr\Cal S_i|\times\int_{\Cal S_i}|d\pmb v|
\Displaycause{614T}
\le\sup|\pmb u|
   \times(\int_{\Cal S\wedge\tau_0}|d\pmb v|
     +\sum_{i=0}^{n-1}\int_{\Cal S_i}|d\pmb v|+\int_{\Cal S\vee\tau_n}|d\pmb v|)
  \cr&
=\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|\cr}$$

\noindent (613J(c-ii)).
Thus $\iivu$ satisfies the condition 614I(ii) and is of bounded
variation, with
$\int_{\Cal S}|d(\iivu)|\le\sup|\pmb u|\times\int_{\Cal S}|d\pmb v|$.
}%end of proof of 614U

%\leader{614V}{}
%\leader{614W}{}

\exercises{\leader{614X}{Basic exercises (a)}
%\spheader 614Xa
Suppose that $\frak A=\{0,1\}$.   Let $S$ be a non-empty subset of
$T$, $f:S\to\Bbb R$ a function and $\pmb u=\{(\check t,f(t)\chi 1)\}$ the
corresponding process with domain $\Cal S=\{\check t:t\in S\}$, as
in 613W.

\quad(i) Show that $\pmb u$ is a
simple process with breakpoint string $(\check t_0,\ldots,\check t_n)$ iff
$f$ is constant on each of the intervals

\Centerline{$S\cap\ooint{-\infty,t_0}$,
\quad$\Cal S\cap\coint{t_i,t_{i+1}}$ for $i<n$,
\quad$\Cal S\cap{t_n,\infty}$.}

\quad(ii) Now suppose that $g:S\to\Bbb R$ is another function
such that $g_{\downarrow}=\lim_{s\downarrow S}g(s)$ and
$g_{\uparrow}=\lim_{s\uparrow S}g(s)$ are defined.   Let $\pmb v$ be the
process corresponding to $g$.   Show that

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v
=\bigl(f_{\downarrow}(g(t_0)-g_{\downarrow})
   +\sum_{i=0}^{n-1}f(t_i)(g(t_{i+1})-g(t_i))
   +f(t_n)(g_{\uparrow}-g(t_n))\bigr)\chi 1$\dvro{.}{}}

\cmmnt{\noindent (614C).}
%614C

\spheader 614Xb Suppose that $\Cal S$ is a non-empty sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ is a simple fully
adapted process with a breakpoint string
$(\tau_0,\ldots,\tau_n)$, and
$\pmb v=\sfamilysS{v}$ is a fully adapted process such that
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ is
defined.   Show that
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v$ is defined and equal to

\Centerline{$u_{\downarrow}\times(v_{\tau_0\wedge\tau}-v_{\downarrow})
+\sum_{i=0}^{n-1}u_{\tau_i}
  \times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
  +u_{\tau_n}\times(v_{\tau}-v_{\tau_n\wedge\tau})$}

\noindent for every $\tau\in\Cal S$.
%614C

\spheader 614Xc
In 614Fb, show that if $\pmb u$ is simple then $\pmb v$ is simple.
%614F

\spheader 614Xd Suppose that $\frak A=\{0,1\}$, as
in 613W, and that $T$ is not empty.  Let $f:T\to\Bbb R$ be a function.
(i) Show that $f$ represents a simple process
order-bounded process iff it is bounded, and a process of bounded variation iff
it is of bounded variation.
%613W 614E 614J

\spheader 614Xe Suppose that $T=\coint{0,\infty}$.
Show that the identity process (612F) and the standard Poisson process
(612U) are locally of bounded variation, and that Brownian motion (612T) is
locally order-bounded, but that none of these is order-bounded.
%614K

\spheader 614Xf Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb v$ a fully
adapted process with domain $\Cal S$, and $\tau\in\Cal S$.   Show that
$\pmb v$ is (locally) of bounded variation iff $\pmb v\restr\Cal S\wedge\tau$
and $\pmb v\restr\Cal S\vee\tau$ are.
%614P

\spheader 614Xg
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$ a simple process with domain $\Cal S$.   (i) Show that
$\pmb u$ is of bounded variation.   (ii) Show that the cumulative variation of
$\pmb u$ is simple.
%614P

\spheader 614Xh In 614Rb, show that if we set $f_{\omega}(s)=X_s(\omega)$ for
$s\ge 0$ and $\omega\in\Omega$, then $(t,\omega)\mapsto\Var_{[0,t]}f_{\omega}$
represents the cumulative variation of $\pmb x$.
%614R

\leader{614Y}{Further exercises (a)}
%\spheader 614Ya
Show that previsibly simple processes %(612Ye)
are of bounded variation.
%614Xg 614P

\spheader 614Yb Show that Brownian motion %(612T)
is not locally of bounded
variation.   \Hint{for $n\ge 1$, $i<n$ set $t_{in}=i/n$;  for $i<n$ set
$z_{ni}=|w_{\check t_{i+1,n}}-w_{\check t_{in}}|$,
$z_n=\sum_{i=0}^{n-1}z_{ni}$.   Show that
$z_{ni}$ has expectation $\Bover{2}{\sqrt{2n\pi}}$ and
variance $\Bover1n(1-\Bover{2}{\pi})$, and hence, using Lindeberg's
theorem (274F), that
$\lim_{n\to\infty}\bar\mu\Bvalue{z_n\ge\bover{2n}{\sqrt{2n\pi}}}=\Bover12$.}
%614Q 614Xe mt61bits

\spheader 614Yc In the construction described in 613S, show that $\pmb u$ and
$\pmb v$ are order-bounded but not of bounded variation.
%613C 614E 614I
}%end of exercises




\endnotes{
\Notesheader{614} This volume is supposed to be about stochastic integration.
However a very large part of the work we need to do will concern the structure
of various types of stochastic process.   Necessarily I started with fully
adapted processes (\S612).   These were sufficient for the definition of the
Riemann-sum integral in \S613.   Now we have three further classes.
Integration of, and with respect to, simple processes really is simple
(614B-614D).  %614B 614C 614D
From 614E to 614R, `integration' hardly appears except in the formula
$\int_{\Cal S}|d\pmb v|$, which is really just an abbreviation for

\Centerline{$\sup\{\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|:
   \tau_0\le\ldots\le\tau_n$ in $\Cal S\}$}

\noindent (614I).   Again, in \S615, to follow, the bulk of the section will be
devoted to properties of the space of `moderately oscillatory' processes.   Only
in \S616 will integration again move to front centre stage.   But as we move
through the foundations of the theory, I will drip-feed results about
integrals involving the special classes of stochastic process being examined, as
in 614T-614U,

I hope that you will recognise 614I(i)$\Leftrightarrow$(ii)
as an elaborate form of the classical result 224D concerning real functions of
bounded variation.   The difference now is that I have set up a
definition of an integral $\int\,|d\pmb v|$.
We shall find that there are many reasons why Brownian motion is not
locally of bounded variation.
But the fact is of such importance that in 614Yb I suggest a method of proof
based on ideas already covered in Volume 2.

It will transpire that often an indefinite integral
inherits properties of the integrator.   So far we have two cases:
an indefinite integral with respect to a simple process is simple (614D)
and an indefinite integral of an order-bounded process
with respect to a process of
bounded variation is again of bounded variation (614U).
In both cases the hypothesis has to include a clause `$\int\pmb u\,d\pmb v$ is
defined' because we are far away from any clear idea of which processes
$\pmb u$ can be expected to appear in applications.

In 614S you will see that a key step is an equality

\Centerline{$\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})
=-\sum_{i=0}^{n-1}v_{\tau_{i+1}}\times(u_{\tau_{i+1}}-u_{\tau_i})
    -v_{\tau_0}\times u_{\tau_0}+v_{\tau_n}\times u_{\tau_n}$,}

\noindent which is a manipulation which I expect you have seen before.
But this is {\it not} a relationship between $S_I(\pmb u,d\pmb v)$ and
$S_I(\pmb v,d\pmb u)$, because the latter would have to look at
$\sum_{i=0}^{n-1}v_{\tau_i}\times(u_{\tau_{i+1}}-u_{\tau_i})$;  the Riemann-sum
integral of \S613 insists on taking tags at the left-hand ends of intervals
rather than the right-hand ends.   The difference is of the form
$\sum_{i=0}^{n-1}(v_{\tau_{i+1}}-v_{\tau_i})\times(u_{\tau_{i+1}}-u_{\tau_i})$,
which will appear in \S617 associated with the `covariation' of
$\pmb u$ and $\pmb v$, and need not vanish in the limit.

This volume is devoted to the thesis that stochastic processes, at least in
respect to stochastic integration,
are best regarded as abstract fully adapted processes in
the sense of \S612.   But most of the ideas of the theory were of course
developed in the context of classical processes $\langle X_t\rangle_{t\ge 0}$ of
real-valued functions on probability spaces.   I have therefore given a couple
of results (612H, 614R and 614Xa(i)) showing that the types of process I
have introduced here correspond, in some sense, to natural conditions on
sample paths $t\mapsto X_t(\omega)$, as in 614Xd.
You will see that all the non-trivial results are
one-way;  given a property of typical sample paths, we can deduce a property of
the corresponding abstract process.   There are theorems in the other
direction (e.g., 633R below), but they need
different techniques, starting of course with the Stone representation
of measure algebras (321J).
}%end of notes

\discrpage

\frfilename{mt615.tex}
\versiondate{28.10.21/9.8.22}
\copyrightdate{2018}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Moderately oscillatory processes}
\newsection{615}

I come now to the class of integrands in the basic theorem 616K, the
`moderately oscillatory' processes.   I have chosen a path which starts with a
natural linear space topology
on the space of order-bounded processes, the ucp topology (615B).
This gives a straightforward definition of the space of moderately oscillatory
processes (615E) with their elementary properties
(615F-615H).  %615F  615G  615H
When the domain is finitely full, we have an alternative definition in terms of
convergence along monotonic sequences of stopping times
(615I-615N). % 615I 615J 615K 615L 615M 615N
Classical stochastic processes with \cadlag\ sample paths give rise to
locally moderately oscillatory processes (615P).

\cmmnt{\leader{615A}{Notation} \Asissix\ will be a stochastic integration
structure, as defined in the notes to \S613.   For
$A\subseteq\Cal T$ and $\tau\in\Cal T$,
$A\wedge\tau=\{\sigma\wedge\tau:\sigma\in A\}$
and $A\vee\tau=\{\sigma\vee\tau:\sigma\in A\}$.   For $w\in L^0=L^0(\frak A)$,
$\theta(w)=\Expn(|w|\wedge\chi 1)$ as in 613Ba;  limits in $L^0$ will be taken
with respect to the topology of convergence in measure.   If $\Cal S$ is a
sublattice of $\Cal T$, $\Mfa(\Cal S)$ is the space of fully adapted processes
with domain $\Cal S$ (612I), $\Mob(\Cal S)\subseteq\Mfa(\Cal S)$
is the space of order-bounded processes (614F) and
$\Mbv(\Cal S)\subseteq\Mob(\Cal S)$ is the space of processes of bounded
variation (614Q).   For an order-bounded process $\pmb =\sfamilysS{u}$,
$\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.
$\tbf 1$ will be the process with constant value $\chi 1$.
}

\leader{615B}{The ucp topology} Let $\Cal S$ be a sublattice of $\Cal T$.

\spheader 615Ba For $\pmb u\in\Mob(\Cal S)$, set

\Centerline{$\widehat{\theta}(\pmb u)=\theta(\sup|\pmb u|)$.}

\spheader 615Bb $\widehat{\theta}$ is an F-norm on $\Mob(\Cal S)$.
\prooflet{\Prf\ The point is that

\Centerline{$\sup|\pmb u+\pmb v|\le\sup|\pmb u|+\sup|\pmb v|$,
\quad$\sup|\alpha\pmb u|=|\alpha|\sup|\pmb u|$}

\noindent for all $\pmb u$, $\pmb v\in\Mob=\Mob(\Cal S)$.   Since $\theta$ is an
F-norm on $L^0$, and $\theta(u)\le\theta(v)$ whenever $|u|\le|v|$, it follows at
once that

\Centerline{$\widehat{\theta}(\pmb u+\pmb v)
\le\widehat{\theta}(\pmb u)+\widehat{\theta}(\pmb v)$,
\quad$\widehat{\theta}(\alpha\pmb u)\le\widehat{\theta}(\pmb u)$}

\noindent whenever $\pmb u$, $\pmb v\in\Mob$ and $|\alpha|\le 1$, and that

\Centerline{$\lim_{\alpha\to 0}\widehat{\theta}(\alpha\pmb u)=0$}

\noindent for every $\pmb u\in\Mob$.   Finally, if
$\widehat{\theta}(\pmb u)=0$, then $\theta(u_{\sigma})=0$ and $u_{\sigma}=0$ for
every $\sigma\in\Cal S$, so $\pmb u=0$.\ \Qed}

\spheader 615Bc\cmmnt{ Accordingly} $\widehat{\theta}$ defines a
metrizable linear space topology\cmmnt{ (2A5B)}.   I will call this
the {\bf ucp topology} on $\Mob(\Cal S)$ and the
associated uniformity the {\bf ucp uniformity} on $\Mob(\Cal S)$.

\cmmnt{{\bf Warning!} The phrase `ucp topology' is commonly used to mean
something closer to the local ucp topology described in 615Xb.}

\leader{615C}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
give $\Mob=\Mob(\Cal S)$ its ucp topology.

(a) If $h:\Bbb R\to\Bbb R$ is continuous, then
$\bar h\pmb u\in\Mob$ for every $\pmb u\in\Mob$, and
$\pmb u\mapsto\bar h\pmb u:\Mob\to\Mob$ is continuous.

(b)(i) $(\pmb u,\pmb v)\mapsto\pmb u\times\pmb v:\Mob\times\Mob\to\Mob$ is
continuous.

\quad(ii) $\pmb u\mapsto\sup|\pmb u|:\Mob\to L^0$
is uniformly continuous.

(c) $\Mob$ is complete as linear topological space.
%ucp topology is locally solid, but this not in mti.tex

\proof{{\bf (a)} If $\Cal S=\emptyset$ then $\Mob=\{\emptyset\}$ and the
result is trivial.   So let us suppose that $\Cal S$ is non-empty.

\medskip

\quad{\bf (i)} For $x\in\Bbb R$, set $g(x)=\sup\{|h(y)|:|y|\le|x|\}$;  then
$g$ is continuous, and $|\bar h(v)|\le g(w)$ whenever $v$, $w\in L^0$ and
$|v|\le w$.

Suppose that $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$
belongs to $\Mob$.   Set $\bar u=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.
Then $|\bar h(u_{\sigma})|\le g(\bar u)$ for every $\sigma\in\Cal S$, so
$\{\bar h(u_{\sigma}):\sigma\in\Cal S\}$ is order-bounded in $L^0$, and
$\bar h\pmb u\in\Mob$.

\medskip

\quad{\bf (ii)} Now take
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}\in\Mob$ and $\epsilon>0$.
Set $\bar v=\sup_{\sigma\in\Cal S}|v_{\sigma}|$,
and let $M\ge 0$ be such that $\bar\mu\Bvalue{\bar v>M}\le\epsilon$.
Let $\delta\in\ocint{0,1}$ be such that
$|h(x)-h(y)|\le\epsilon$ whenever $y\in[-M-1,M+1]$ and
$|x-y|\le\delta$.   Then for any $w$, $w'\in L^0$,

\Centerline{$\Bvalue{|\bar h(w)-\bar h(w')|>\epsilon}
\Bsubseteq\Bvalue{|w|>M}\Bcup\Bvalue{|w-w'|>\delta}$.}

Take any $\pmb u\in\Mob$ such that
$\widehat{\theta}(\pmb v-\pmb u)\le\delta\epsilon$.
Set $\bar u=\sup_{\sigma\in\Cal S}|v_{\sigma}-u_{\sigma}|$ and
$\bar w=\sup_{\sigma\in\Cal S}|\bar h(v_{\sigma})-\bar h(u_{\sigma})|$.
Then $\bar\mu\Bvalue{\bar u>\delta}\le\epsilon$, so

$$\eqalign{\Bvalue{\bar w>\epsilon}
&=\sup_{\sigma\in\Cal S}
   \Bvalue{|\bar h(v_{\sigma})-\bar h(u_{\sigma})|>\epsilon}\cr
&\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{|v_{\sigma}|>M}
   \Bcup\sup_{\sigma\in\Cal S}\Bvalue{|v_{\sigma}-u_{\sigma}|>\delta}
=\Bvalue{\bar v>M}\Bcup\Bvalue{\bar u>\delta}\cr}$$

\noindent has measure at most $2\epsilon$, and

\Centerline{$\widehat{\theta}(\bar h\pmb u-\bar h\pmb v)
=\theta(\bar w)
\le 3\epsilon$.}

\noindent As $\pmb v$ and $\epsilon$ are arbitrary, $\bar h:\Mob\to\Mob$
is continuous.

\medskip

{\bf (b)(i)} As the ucp topology is a linear space topology, addition and
subtraction are certainly continuous.   By (a), the operation
$\pmb u\mapsto\pmb u^2$ is continuous.   But this means that

\Centerline{$(\pmb u,\pmb v)
\mapsto\Bover12((\pmb u+\pmb v)^2-\pmb u^2-\pmb v^2)
=\pmb u\times\pmb v$}

\noindent is continuous.

\medskip

\quad{\bf (ii)} If $\pmb u=\sfamilysS{u}$, $\pmb v=\sfamilysS{v}$
belong to $\Mob$, then

\Centerline{$|u_{\sigma}|\le|v_{\sigma}|+|u_{\sigma}-v_{\sigma}|
\le\sup|\pmb v|+\sup|\pmb u-\pmb v|=\sup|\pmb v|+\sup|\pmb v-\pmb u|$}

\noindent for every $\sigma\in\Cal S$, so
$\sup|\pmb u|\le\sup|\pmb v|+\sup|\pmb u-\pmb v|$;  similarly,
$\sup|\pmb v|\le\sup|\pmb u|+\sup|\pmb u-\pmb v|$ and
$|\sup|\pmb v|-\sup|\pmb u||\le\sup|\pmb u-\pmb v|$.   So

\Centerline{$\theta(\sup|\pmb u|-\sup|\pmb v|)\le\theta(\sup|\pmb u-\pmb v|)
=\widehat\theta(\pmb u-\pmb v)$.}

\noindent As $\theta$, $\widehat\theta$ are F-norms
defining the linear space
topologies of $L^0$ and $\Mob$, $\pmb u\mapsto\sup|\pmb u|$ is uniformly
continuous.

\medskip

{\bf (c)} Again, if $\Cal S=\emptyset$ this is trivial, so suppose
otherwise.   Let $\Cal F$ be a filter on $\Mob$ which is Cauchy for the ucp
topology.

\medskip

\quad{\bf (i)} For each $\sigma\in\Cal S$, define
$T_{\sigma}:\Mob\to L^0$ by saying that $T_{\sigma}\pmb u=u_{\sigma}$
whenever
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}\in\Mob$;  then $T_{\sigma}$ is
linear, and it is also continuous, because
$\theta(T_{\sigma}\pmb u)
\le\widehat\theta(\pmb u)$ for every
$\pmb u\in\Mob$.   $T_{\sigma}$ is therefore uniformly continuous for
the uniformities associated with the linear space topologies here
(3A4Cf\Latereditions).   Accordingly the image filter
$T_{\sigma}[[\Cal F]]$ (2A1Ib)
is Cauchy (4A2Ji), and has a limit $w_{\sigma}$ say in $L^0$,
because $L^0$ is complete as linear topological space (613Bh).

\medskip

\quad{\bf (ii)} $\pmb{w}=\family{\sigma}{\Cal S}{w_{\sigma}}$ is fully
adapted.   \Prf\ Suppose that $\sigma$, $\tau\in\Cal S$ and
$a=\Bvalue{\sigma=\tau}$.   Then
$T_{\sigma}\pmb u\times\chi a=T_{\tau}\pmb u\times\chi a$ for every
$\pmb u\in\Mob$.   As $u\mapsto u\times\chi a:L^0\to L^0$ is continuous,

$$\eqalign{w_{\sigma}\times\chi a
&=(\lim_{\pmb u\to\Cal F}T_{\sigma}\pmb u)\times\chi a
=\lim_{\pmb u\to\Cal F}(T_{\sigma}\pmb u\times\chi a)\cr
&=\lim_{\pmb u\to\Cal F}T_{\tau}\pmb u\times\chi a
=w_{\tau}\times\chi a,\cr}$$

\noindent and $a\Bsubseteq\Bvalue{w_{\sigma}=w_{\tau}}$.   As $\sigma$ and
$\tau$ are arbitrary, $\pmb{w}$ is fully adapted.\ \Qed

\medskip

\quad{\bf (iii)} $\pmb{w}$ is order-bounded.   \Prf\ For $n\in\Bbb N$ set
$c_n=\sup_{\sigma\in\Cal S}\Bvalue{|w_{\sigma}|>n}$.   Given
$\epsilon>0$, there is a set
$A\subseteq\Cal F$ such that
$\widehat{\theta}(\pmb u-\pmb v)\le\epsilon$ for all $\pmb u$,
$\pmb v\in A$.   Fix $\pmb v\in A$, and set
$\bar v=\sup_{\sigma\in\Cal S}|T_{\sigma}\pmb v|$;  let $n\in\Bbb N$ be
such that $\bar\mu\Bvalue{\bar v\ge n}\le\epsilon$.
For $\pmb u\in\Mob$, set
$\bar v_{\pmb u}=\sup_{\sigma\in\Cal S}|T_{\sigma}\pmb u-T_{\sigma}\pmb v|$,
so that

\Centerline{$\theta(\bar v_{\pmb u})
=\widehat{\theta}(\pmb u-\pmb v)\le\epsilon$}

\noindent for $\pmb u\in A$.

Let $I$ be any finite subset of $\Cal S$.   Then

\Centerline{$\sup_{\sigma\in I}|w_{\sigma}|
=\lim_{\pmb u\to\Cal F}\sup_{\sigma\in I}|T_{\sigma}(\pmb u)|$}

\noindent because the lattice operations on $L^0$ are continuous (613Ba), so
$\family{\sigma}{I}{y_{\sigma}}\mapsto\sup_{\sigma\in I}|y_{\sigma}|$
is a continuous function from $(L^0)^I$ to $L^0$.
There is therefore a $\pmb u\in A$ such that
$\theta(\sup_{\sigma\in I}|w_{\sigma}|
  -\sup_{\sigma\in I}|T_{\sigma}\pmb u|)\le\epsilon$ and

$$\eqalign{\bar\mu\Bvalue{\sup_{\sigma\in I}|w_{\sigma}|\ge n+2}
&\le\bar\mu\Bvalue{\sup_{\sigma\in I}|w_{\sigma}|
        -\sup_{\sigma\in I}|T_{\sigma}\pmb u|\ge 1}\cr
&\mskip50mu
    +\bar\mu\Bvalue{\sup_{\sigma\in I}
        |T_{\sigma}\pmb u-T_{\sigma}\pmb v|\ge 1}
    +\bar\mu\Bvalue{\sup_{\sigma\in I}
        |T_{\sigma}\pmb v|\ge n}\cr
&\le
\theta(\sup_{\sigma\in I}|w_{\sigma}|
      -\sup_{\sigma\in I}|T_{\sigma}\pmb u|)
    +\bar\mu\Bvalue{\bar v_{\pmb u}\ge 1}
    +\bar\mu\Bvalue{\bar v\ge n}\cr
&\le\epsilon+\theta(\bar v_{\pmb u})+\epsilon
=2\epsilon+\widehat{\theta}(\pmb u-\pmb v)
\le 3\epsilon.\cr}$$

\noindent As $I$ is arbitrary, $\bar\mu c_{n+2}\le 3\epsilon$.
As $\epsilon$ is arbitrary, $\inf_{n\in\Bbb N}c_n=0$ and
$\{w_{\sigma}:\sigma\in\Cal S\}$ is order-bounded.\ \Qed

\medskip

\quad{\bf (iv)} Thus $\pmb{w}\in\Mob$.   Now, given $\epsilon>0$, again
take $A\in\Cal F$ such that
$\widehat{\theta}(\pmb u-\pmb v)\le\epsilon$ for all $\pmb u$,
$\pmb v\in A$.   Then
$\widehat{\theta}(\pmb{w}-\pmb v)\le\epsilon$ for all
$\pmb v\in A$.   \Prf\ For any finite $I\subseteq\Cal S$,

\Centerline{$\sup_{\sigma\in I}|w_{\sigma}-T_{\sigma}\pmb v|
=\lim_{\pmb u\to\Cal F}
   \sup_{\sigma\in I}|T_{\sigma}\pmb u-T_{\sigma}\pmb v|$}

\noindent and

$$\eqalign{\theta(\sup_{\sigma\in I}|w_{\sigma}-T_{\sigma}\pmb v|)
&=\lim_{\pmb u\to\Cal F}
   \theta(\sup_{\sigma\in I}|T_{\sigma}\pmb u-T_{\sigma}\pmb v|)\cr
&\le\sup_{\pmb u\in A}\widehat{\theta}(\pmb u-\pmb v)
\le\epsilon.\cr}$$

\noindent So

$$\eqalignno{\widehat{\theta}(\pmb{w}-\pmb v)
&=\theta(\sup_{\sigma\in\Cal S}|w_{\sigma}-T_{\sigma}\pmb v|)
=\sup_{I\in[\Cal S]^{<\omega}}
   \theta(\sup_{\sigma\in I}|w_{\sigma}-T_{\sigma}\pmb v|)
\Displaycause{613Ba}
\le\epsilon.  \text{ \Qed}\cr}$$

As $\epsilon$ is arbitrary, $\pmb{w}=\lim\Cal F$ for the ucp topology.
As $\Cal F$ is arbitrary, $\Mob$ is complete.
}%end of proof of 615C

\leader{615D}{}\cmmnt{ When we have a finitely
full lattice, there is an alternative approach to the ucp topology.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a non-empty
finitely full sublattice of $\Cal T$,
and $\pmb u=\sfamilysS{u}$ a fully adapted process.

(a)  If $\gamma>0$ then

\Centerline{$\bar\mu(\sup_{\tau\in\Cal S}\Bvalue{|u_{\tau}|>\gamma})
=\sup_{\tau\in\Cal S}\bar\mu\Bvalue{|u_{\tau}|>\gamma}$.}

(b) If $\pmb u$ is order-bounded,
$\theta(\sup|\pmb u|)\le 2\sqrt{\sup_{\sigma\in\Cal S}\theta(u_{\sigma})}$.

\proof{{\bf (a)} If $\gamma>0$ then

\Centerline{$\bar\mu\Bvalue{\sup|\pmb u|>\gamma}
=\sup_{\tau\in\Cal S}\bar\mu\Bvalue{|u_{\tau}|>\gamma}$.}

\noindent\Prf\ We have

$$\eqalignno{\bar\mu\Bvalue{\sup|\pmb u|>\gamma}&
=\bar\mu(\sup_{\sigma\in\Cal S}\Bvalue{|u_{\sigma}|>\gamma})
\Displaycause{364L(a-ii)}
=\sup_{I\in[\Cal S]^{<\omega}}
    \bar\mu(\sup_{\sigma\in I}\Bvalue{|u_{\sigma}|>\gamma})
=\sup_{I\in\Cal I(\Cal S)}
    \bar\mu(\sup_{\sigma\in I}\Bvalue{|u_{\sigma}|>\gamma}).\cr}$$

\noindent Take any $\alpha<\bar\mu\Bvalue{\sup|\pmb u|>\gamma}$.
Let $I\in\Cal I(\Cal S)$ be such that
$\bar\mu(\sup_{\sigma\in I}\Bvalue{|u_{\sigma}|>\gamma})>\alpha$;  as $\Cal S$
is non-empty, we can arrange that $I$ should be non-empty.
Let $(\tau_0,\ldots,\tau_n)$ be a maximal totally ordered subset
of $I$.   Then

\Centerline{$\Bvalue{|u_{\sigma}|>\gamma}
\Bsubseteq\sup_{i\le n}\Bvalue{\sigma=\tau_i}\Bcap\Bvalue{|u_{\tau_i}|>\gamma}
\Bsubseteq\sup_{i\le n}\Bvalue{|u_{\tau_i}|>\gamma}$}

\noindent for every $\sigma\in I$ (611Ke), so
$\alpha<\bar\mu(\sup_{i\le n}\Bvalue{|u_{\tau_i}|>\gamma})$.

For $i\le n$, set $a_i=\Bvalue{|u_{\tau_i}|>\gamma}$ and
$b_i=a_i\Bsetminus\sup_{j<i}a_j$;  set $b'=1\Bsetminus\sup_{i\le n}a_i$.
Then $a_i$ and $b_i$ belong to $\frak A_{\tau_i}$ for every $i\le n$, and
$b'\in\frak A_{\tau_n}$.   So there is a $\tau\in\Cal T$ such that

\Centerline{$b_i\Bsubseteq\Bvalue{\tau=\tau_i}$ for $i\le n$,
\quad$b'\Bsubseteq\Bvalue{\tau=\tau_n}$}

\noindent (611I), and $\tau\in\Cal S$ because $\Cal S$ is finitely full.
Now

$$\eqalign{\Bvalue{|u_{\tau}|>\gamma}&
\Bsupseteq\sup_{i\le n}b_i\Bcap\Bvalue{|u_{\tau_i}|>\gamma}
=\sup_{i\le n}b_i
=\sup_{i\le n}a_i\cr&
=\sup_{i\le n}\Bvalue{|u_{\tau_i}|>\gamma}
=\sup_{\sigma\in I}\Bvalue{|u_{\sigma}|>\gamma}\cr}$$

\noindent and $\alpha<\bar\mu\Bvalue{|u_{\tau}|>\gamma}$.
As $\alpha$ is arbitrary, this gives the result.\ \Qed

\medskip

{\bf (b)} Write $\alpha$ for $\sup_{\tau\in\Cal S}\theta(|u_{\tau}|)$.
If $\gamma<\bover12\theta(\sup|\pmb u|)$ we must have
$\bar\mu\Bvalue{\sup|\pmb u|>\gamma}>\gamma$, so there is a $\tau\in\Cal S$
such that $\bar\mu\Bvalue{|u_{\tau}|>\gamma}>\gamma$,
$\gamma^2<\theta(|u_{\tau}|)\le\alpha$ and $\gamma\le\sqrt{\alpha}$.
As $\gamma$ is arbitrary, $\theta(\sup|\pmb u|)\le 2\sqrt{\alpha}$, as
claimed.
}%end of proof of 615D

\cmmnt{\medskip

\noindent{\bf Remark} What this means is that the F-norm

\Centerline{$\pmb u\mapsto\sup_{\sigma\in\Cal S}\theta(u_{\sigma}):
\Mob(\Cal S)\to[0,1]$}

\noindent also defines the ucp topology on $\Mob(\Cal S)$, since of course
$\sup_{\sigma\in\Cal S}\theta(u_{\sigma})\le\widehat{\theta}(\pmb u)$.
For a general sublattice $\Cal S$, we could use the isomorphism
$\pmb u\mapsto\hatpmb u:\Mob(\Cal S)\to\Mob(\hat\Cal S)$ and the
F-norm

\Centerline{$\pmb u\mapsto\sup_{\sigma\in\hat\Cal S}\theta(\hat u_{\sigma}):
\Mob(\Cal S)\to[0,1]$,}

\noindent since

\Centerline{$\widehat{\theta}(\hatpmb u)=\theta(\sup|\hatpmb u|)
=\theta(\sup|\pmb u|)=\widehat{\theta}(\pmb u)$}

\noindent for every $\pmb u\in\Mob(\Cal S)$.
}

\leader{615E}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$.

\medskip

{\bf (a)} I will call a process with domain $\Cal S$
{\bf moderately oscillatory} if it
is in the closure of $\Mbv(\Cal S)$ in $\Mob(\Cal S)$ for the
ucp topology.

\medskip

{\bf (b)} A process $\pmb u$ with domain $\Cal S$ is {\bf locally moderately
oscillatory} if $\pmb u\restr\Cal S\wedge\tau$ is moderately oscillatory for
every $\tau\in\Cal S$.

\medskip

\noindent{\bf Remark} The definitions imply directly that (locally) moderately
oscillatory processes are (locally) order-bounded.
Of course processes of bounded variation (e.g., simple
processes, 614Q(a-iii), and in particular constant processes)
are moderately oscillatory, and processes which are locally of
bounded variation are locally moderately oscillatory.

\leader{615F}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\hat\Cal S$ its covered envelope.

(a) Write $\Mmo\cmmnt{\mskip5mu=\Mmo(\Cal S)}$
for the set of moderately oscillatory processes with domain $\Cal S$.

\quad(i) If $\Cal S'$ is a sublattice of $\Cal S$ then
$\pmb u\restr\Cal S'$ is moderately oscillatory for every
$\pmb u\in\Mmo$.

\quad(ii) If $h:\Bbb R\to\Bbb R$ is continuous, then
$\bar h\pmb u\in\Mmo$ for every $\pmb u\in\Mmo$.

\quad(iii) $\Mmo$ is an $f$-subalgebra of $\Mob=\Mob(\Cal S)$.

\quad(iv) $\Mmo$ is closed in $\Mob(\Cal S)$ for the ucp topology, so is
complete for the ucp uniformity.

\quad(v) If $\tau\in\Cal S$, then a fully adapted process $\pmb u$ with domain
$\Cal S$ is moderately oscillatory iff $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are both moderately oscillatory.

\quad(vi) If $\pmb u\in\Mmo$, then its fully adapted extension to $\hat\Cal S$
is moderately oscillatory.

(b) Write $\Mlmo\cmmnt{\mskip5mu=\Mlmo(\Cal S)}$ for the set of
locally moderately oscillatory processes with domain $\Cal S$.

\quad(i) $\Mmo\subseteq\Mlmo$.

\quad(ii) If $h:\Bbb R\to\Bbb R$ is continuous, then
$\bar h\pmb u\in\Mlmo$ for every $\pmb u\in\Mlmo$.

\quad(iii) $\Mlmo$ is an $f$-subalgebra of the space $\Mlob=\Mlob(\Cal S)$ of
locally order-bounded processes with domain $\Cal S$.

\quad(iv) If $\tau\in\Cal S$, then a fully adapted process $\pmb u$ with domain
$\Cal S$ is locally moderately oscillatory iff $\pmb u\restr\Cal S\wedge\tau$
and $\pmb u\restr\Cal S\vee\tau$ are both locally moderately oscillatory.

\quad(v) If $\pmb u\in\Mlmo$, then its fully adapted extension $\hatpmb u$ to
$\hat\Cal S$ is locally moderately oscillatory.

\proof{{\bf (a)(i)} For any $\epsilon>0$ there is a process
$\pmb v\in\Mbv(\Cal S)$ such that $\theta(\sup|\pmb u-\pmb v|)\le\epsilon$.
Now $\pmb u\restr\Cal S'$ is order-bounded (614F(a-i)),
$\pmb v\restr\Cal S\wedge\tau$ is of bounded variation (614Lb) and
$\sup|\pmb u\restr\Cal S'-\pmb v\restr\Cal S'|
\le\sup|(\pmb u-\pmb v)\restr\Cal S'|$, so

\Centerline{$\theta(\sup|\pmb u\restr\Cal S'-\pmb v\restr\Cal S'|)
\le\theta(\sup|(\pmb u-\pmb v)\restr\Cal S'|)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\pmb u\restr\Cal S'\in\Mmo(\Cal S')$.

\medskip

\quad{\bf (ii)}\grheada\ The point is that if $M\ge 0$ and $\epsilon>0$, there
is a Lipschitz function $g:\Bbb R\to\Bbb R$ such that
$|h(x)-g(x)|\le\epsilon$ whenever $|x|\le M$.   \Prf\ $h\restr[-M,M]$ is
continuous, therefore uniformly continuous, and there is an $m\ge 1$ such that
$|h(x)-h(y)|\le\bover12\epsilon$ whenever $-M\le x\le y\le M$ and
$y\le x+\bover{M}{m}$.   Set
$x_k=Mk/m$ for $-m\le k\le m$, and define $g$ by saying that

$$\eqalign{g(x)&=h(-M)\text{ if }x\le -M,\cr
&=h(M)\text{ if }x\ge M,\cr
&=\Bover{m}{M}((x-x_k)h(x_{k+1})+(x_{k+1}-x)h(x_k))\cr&
\mskip200mu\text{ if }-m\le k\le m-1\text{ and }x_k\le x\le x_{k+1}.\cr}$$

\noindent Then $g$ is Lipschitz and

$$\eqalign{|h(x)-g(x)|&
\le |h(x)-h(x_k)|+|g(x)-h(x_k)|\cr&
\le |h(x)-h(x_k)|+|h(x_{k+1})-h(x_k)|
\le\epsilon\cr}$$

\noindent whenever $-m\le k\le m-1$ and $x_k\le x\le x_{k+1}$.   So
$|h(x)-g(x)|\le\epsilon$ whenever $|x|\le M$.\ \Qed

\medskip

\qquad\grheadb\ Now $h$ is Borel measurable and bounded on bounded subsets of
$\Bbb R$, so $\bar h\pmb u$ is order-bounded (614F(c-i)).   Take $\epsilon>0$.
Set $\eta=\min(1,\bover{\epsilon}4)$.
Then there is an $M\ge 0$ such that
$\bar\mu\Bvalue{\sup|\pmb u|>M}\le\eta$.   By ($\alpha$), there is a
Lipschitz function $g:\Bbb R\to\Bbb R$ such that
$|h(x)-g(x)|\le\eta$ whenever $|x|\le M$.   Expressing
$\pmb u$ as $\sfamilysS{u}$ and setting $a=\Bvalue{\sup|\pmb u|\le M}$,
$a\Bsubseteq\Bvalue{|\bar h(u_{\sigma})-\bar g(u_{\sigma})|\le\eta\chi 1}$
for every $\sigma\in\Cal S$, so
$a\Bsubseteq\Bvalue{\sup|\bar h\pmb u-\bar g\pmb u|\le\eta\chi 1}$.
But this means that

\Centerline{$\theta(\sup|\bar h\pmb u-\bar g\pmb u|)
\le\eta+\bar\mu(1\Bsetminus a)\le 2\eta$,}

Let $\gamma\ge 1$ be such that $g$ is $\gamma$-Lipschitz.   Because $\pmb u$ is
moderately oscillatory, there is a process $\pmb v=\sfamilysS{v}$ such that
$\theta(\sup|\pmb u-\pmb v|)\le\eta^2/\gamma$, so that
$\bar\mu\Bvalue{\sup|\pmb u-\pmb v|\ge\bover{\eta}{\gamma}}\le\eta$.   Setting
$b=\Bvalue{\sup|\pmb u-\pmb v|\le\bover{\eta}{\gamma}}$, we have
$|\bar g(u_{\sigma})-g(v_{\sigma})|\le\gamma|u_{\sigma}-v_{\sigma}|$ for every
$\sigma\in\Cal S$, so
$\sup|\bar g\pmb u-\bar g\pmb v|\le\gamma\sup|\pmb u-\pmb v|$,
$b\Bsubseteq\Bvalue{\sup|\bar g\pmb u-\bar g\pmb v|\le\eta}$ and

\Centerline{$\theta(\sup|\bar g\pmb u-\bar g\pmb v|)
\le\eta+\bar\mu(1\Bsetminus b)\le 2\eta$,
\quad$\theta(\sup|\bar h\pmb u-\bar g\pmb v|)\le 4\eta\le\epsilon$.}

\noindent And we know from 614Q(a-i) that $\bar g\pmb v$ is of bounded
variation.   As $\epsilon$ is arbitrary, $\bar h\pmb u$ is moderately
oscillatory,

\medskip

\quad{\bf (iii)} As $\Mbv=\Mbv(\Cal S)$ is an $f$-subalgebra of $\Mob$
(614Q(a-ii)) and the operations $(\pmb u,\pmb v)\to\pmb u+\pmb v$,
$(\alpha,\pmb u)\mapsto\alpha\pmb u$, $(\pmb u,\pmb v)\mapsto\pmb u\times\pmb v$
and $\pmb u\mapsto|\pmb u|$ are continuous for the ucp topology on $\Mob$
(615Ca, 615Cb), $\Mmo=\overline{\Mbv}$ must be closed under these operations,
that is, is an $f$-subalgebra of $\Mob$.

\medskip

\quad{\bf (iv)} $\Mmo$ is defined as the closure of $\Mbv(\Cal S)$, so is surely
closed.   Being a closed subspace of the complete linear topological space
$\Mob$ (615Cc), $\Mmo$ is complete (3A4Fd\Latereditions).

\medskip

\quad{\bf (v)} Write $\Cal S'$ for $\Cal S\wedge\tau$ and $\Cal S''$ for
$\Cal S\vee\tau$.   If $\pmb u$ is moderately oscillatory, so are
$\pmb u'=\pmb u\restr\Cal S'$ and $\pmb u''=\pmb u\restr\Cal S''$, by
(i) above.   If $\pmb u'$ and $\pmb u''$
are moderately oscillatory, take any $\epsilon>0$.   Then there are processes
$\pmb v'=\family{\sigma}{\Cal S'}{v'_{\sigma}}$ and
$\pmb v''=\family{\sigma}{\Cal S''}{v''_{\sigma}}$, both of bounded
variation, such that $\theta(\sup|\pmb v'-\pmb u'|)$
and $\theta(\sup|\pmb v''-\pmb u''|)$ are both at most
$\bover{\epsilon}3$.   Define $\pmb v=\sfamilysS{v}$ by saying that

\Centerline{$v_{\sigma}=v'_{\sigma\wedge\tau}\times\chi\Bvalue{\sigma\le\tau}
   +(v''_{\sigma\vee\tau}-v''_{\tau}+v'_{\tau})
       \times\chi\Bvalue{\tau<\sigma}$}

\noindent for $\sigma\in\Cal S$.   Using 611E, 611Hc and 612C, it is easy to
check that $\pmb v$ is fully adapted, while $\pmb v\restr\Cal S'=\pmb v'$ and
$\pmb v\restr\Cal S''=\pmb v''+(v_{\tau}-v''_{\tau})\tbf 1$ are both of
bounded variation, so $\pmb v\in\Mbv$ (614Lc).   Now

$$\eqalign{|u_{\sigma}-v_{\sigma}|
&=|u_{\sigma}-v'_{\sigma}|\le\sup|\pmb v'-\pmb u'|
   \text{ if }\sigma\in\Cal S\wedge\tau,\cr
&\le|u_{\sigma}-v''_{\sigma}|+|v''_{\tau}-u_{\tau}|+|u_{\tau}-v'_{\tau}|
   \le\sup|\pmb v'-\pmb u'|
       +2\sup|\pmb v''-\pmb u''|
   \text{ if }\sigma\in\Cal S'',\cr}$$

\noindent and

\Centerline{$\sup_{\sigma\in\Cal S'\cup\Cal S''}|u_{\sigma}-v_{\sigma}|
\le\sup|\pmb v'-\pmb u'|+2\sup|\pmb v''-\pmb u''|$.}

\noindent As $\Cal S'\cup\Cal S''$ covers $\Cal S$ (611M(b-iv)),

$$\eqalignno{\theta(\sup|\pmb u-\pmb v|)&
=\theta(\sup_{\sigma\in\Cal S'\cup\Cal S''}|u_{\sigma}-v_{\sigma}|)
\Displaycause{614Ga}
\le\theta(\sup|\pmb v'-\pmb u'|)+2\theta(\sup|\pmb v''-\pmb u''|)
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\pmb u$ is moderately oscillatory.

\medskip

\quad{\bf (vi)} For $\pmb u\in\Mfa(\Cal S)$, let $\hatpmb u$ be its fully
adapted extension to $\hat\Cal S$.   Then
$\pmb u\mapsto\hatpmb u:\Mfa(\Cal S)\to\Mfa(\hat\Cal S)$ is linear, while
$\hatpmb u\in\Mob(\hat\Cal S)$ and $\sup|\hatpmb u|=\sup|\pmb u|$ for every
$\pmb u\in\Mob$ (614Ga);  so $\pmb u\mapsto\hatpmb u:\Mob\to\Mob(\hat\Cal S)$ is
continuous for the ucp topologies.   Also $\hatpmb u\in\Mbv(\hat\Cal S)$ for every
$\pmb u\in\Mbv$ (614Q(a-iv)), so if $\pmb u\in\Mmo=\overline{\Mbv}$ then
$\hatpmb u\in\overline{\Mbv(\hat\Cal S)}=\Mmo(\hat\Cal S)$.

\medskip

{\bf (b)(i)} Immediate from (a-i).

\medskip

\quad{\bf (ii)} follows from (a-ii) because
$(\bar h\pmb u)\restr\Cal S\wedge\tau=\bar h(\pmb u\restr\Cal S\wedge\tau)$.

\medskip

\quad{\bf (iii)} Similarly, restriction respects
the algebraic and lattice operations on $\Mmo(\Cal S)$, so we czn use (a-iii).

\medskip

\quad{\bf (iv)}\grheada\ If $\pmb u$ is locally moderately oscillatory, then
$\pmb u\restr\Cal S\wedge\tau$ is moderately oscillatory,
therefore locally moderately oscillatory.
Also, if $\tau'\in\Cal S\vee\tau$,

\Centerline{$(\pmb u\restr\Cal S\vee\tau)\restr(\Cal S\vee\tau)\wedge\tau'
=(\pmb u\restr\Cal S\wedge\tau')\restr(\Cal S\wedge\tau')\vee\tau$}

\noindent is moderately oscillatory,
so $\pmb u\restr(\Cal S\vee\tau)$ is locally moderately oscillatory.

\medskip

\qquad\grheadb\ Suppose that $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are locally moderately oscillatory.   Take any
$\tau'\in\Cal S\vee\tau$.   Then
$(\pmb u\restr\Cal S\wedge\tau')\restr(\Cal S\wedge\tau')\wedge\tau
=(\pmb u\restr\Cal S\wedge\tau)\restr(\Cal S\wedge\tau)\wedge\tau'$ and
$(\pmb u\restr\Cal S\vee\tau')\restr(\Cal S\wedge\tau')\vee\tau
=(\pmb u\restr\Cal S\vee\tau)\restr(\Cal S\vee\tau)\wedge\tau'$
are moderately oscillatory, so
$\pmb u\restr\Cal S\wedge\tau'$ is moderately oscillatory.

In general, if $\tau'$ is an arbitrary member of $\Cal S$,

\Centerline{$\pmb u\restr\Cal S\wedge\tau'
=(\pmb u\restr\Cal S\wedge(\tau'\vee\tau))
  \restr(\Cal S\wedge(\tau'\vee\tau))\wedge\tau'$}

\noindent is moderately oscillatory,
so $\pmb u$ is locally moderately oscillatory.

\medskip

\quad{\bf (v)} Take any $\tau^*\in\hat\Cal S$ and $\epsilon\in\ocint{0,1}$.
Then $\hatpmb u\restr\hat\Cal S\wedge\tau^*$ is order-bounded (614Gb)
and there is a finite $I\subseteq\Cal S$ such that
$\bar\mu(\sup_{\sigma\in I}\Bvalue{\tau^*=\sigma})\ge 1-\bover12\epsilon$;
setting $\sigma^*=\sup I$,
$\bar\mu\Bvalue{\sigma^*<\tau^*}\le\bover12\epsilon$.   As
$\hat\Cal S\wedge\sigma^*$ is the covered envelope of $\Cal S\wedge\sigma^*$
(611M(e-i)), (a-vi) tells us that $\hatpmb u\restr\hat\Cal S\wedge\sigma^*$ is
moderately oscillatory, and there is a process
$\pmb v=\family{\tau}{\hat\Cal S\wedge\sigma^*}{v_{\tau}}$ of bounded
variation such that $\theta(\bar u)\le\bover12\epsilon$, where
$\bar u=\sup|\pmb v-\hatpmb u\restr\hat\Cal S\wedge\sigma^*|$.
Setting $\hatpmb v=\family{\tau}{\hat\Cal S}{v_{\tau\wedge\sigma^*}}$,
$\hatpmb v$ is fully adapted (612Ib),
$\hatpmb v\restr\hat\Cal S\wedge\sigma^*=\pmb v$ is of bounded variation, and
$\hatpmb v\restr\hat\Cal S\vee\sigma^*$ is constant;  so $\hatpmb v$ is of
bounded variation (614Lc).   Now if $\tau\in\hat\Cal S\wedge\tau^*$

$$\eqalign{\Bvalue{\tau^*\le\sigma^*}&
\Bsubseteq\Bvalue{\tau\le\sigma^*}\cr&
\Bsubseteq\Bvalue{\hat v_{\tau}=v_{\tau}}
   \Bcap\Bvalue{|\hat u_{\tau}-\hat v_{\tau}|\le\bar u}\cr}$$

\noindent so

\Centerline{$\Bvalue{\tau^*\le\sigma^*}
\Bsubseteq\Bvalue{\sup|\hatpmb v-\hatpmb u\restr\hat\Cal S\wedge\tau^*|
                       \le\bar u}$}

\noindent and

\Centerline{$\theta(\sup|\hatpmb v-\hatpmb u\restr\hat\Cal S\wedge\tau^*|)
   \le\theta(\bar u)+\bar\mu(1\Bsetminus\Bvalue{\tau*\le\sigma^*}
   \le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\hatpmb u\restr\hat\Cal S\wedge\tau^*$
is moderately oscillatory;  as $\tau^*$ is arbitrary, $\hatpmb u$ is locally
moderately oscillatory.
}%end of proof of 615F

\leader{615G}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb u=\sfamilysS{u}$ a fully adapted process.

(a) Suppose that $\pmb u$ is moderately oscillatory and
$A\subseteq\Cal S$ is non-empty and upwards-directed.
Then $w=\lim_{\sigma\uparrow A}u_{\sigma}$ is defined.
Setting $A_*=\{\rho:\rho\in\Cal S$,
$\sup_{\sigma\in A}\Bvalue{\rho\le\sigma}=1\}$,

\Centerline{$\lim_{\sigma\uparrow A}\sup_{\rho\in A_*\vee\sigma}
  |u_{\rho}-w|
=0$.}

(b) Suppose that $\pmb u$ is locally moderately oscillatory and
$A\subseteq\Cal S$ is non-empty and downwards-directed.
Then $w=\lim_{\sigma\downarrow A}u_{\sigma}$ is defined.
Setting $A^*=\{\rho:\rho\in\Cal S$,
$\sup_{\sigma\in A}\Bvalue{\sigma\le\rho}=1\}$,

\Centerline{$\lim_{\sigma\downarrow A}\sup_{\rho\in A^*\wedge\sigma}
  |u_{\rho}-w|
=0$.}

\proof{{\bf (a)(i)} To begin with, consider the case in which $\pmb u$ is
non-negative, non-decreasing and order-bounded.   In this case,
$B=\{u_{\sigma}:\sigma\in A\}\subseteq(L^0)^+$ is upwards-directed and bounded
above, so has a supremum $w$ in $L^0$, and

\Centerline{$w=\lim_{x\uparrow B}x=\lim_{\sigma\uparrow A}u_{\sigma}$}

\noindent (613Ba).   If $\sigma\in A$ and $\rho\in A_*\vee\sigma$,
$u_{\sigma}\le u_{\rho}$ so $w-u_{\rho}\le w-u_{\sigma}$ and

$$\eqalignno{1&
=\sup_{\sigma'\in A}\Bvalue{\rho\le\sigma'}
=\sup_{\sigma\le\sigma'\in A}\Bvalue{\rho\le\sigma'}
\Displaycause{because $A$ is upwards-directed}
=\sup_{\sigma\le\sigma'\in A}\Bvalue{\rho=\sigma'\wedge\rho}
\Bsubseteq\sup_{\sigma\le\sigma'\in A}
   \Bvalue{u_{\rho}=u_{\sigma'\wedge\rho}}\cr&
\Bsubseteq\sup_{\sigma\le\sigma'\in A}\Bvalue{u_{\rho}\le u_{\sigma'}}
\Bsubseteq\Bvalue{u_{\rho}\le w}
\Bsubseteq\Bvalue{|w-u_{\rho}|\le w-u_{\sigma}}\cr}$$

\noindent and $\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-w|\le w-u_{\sigma}$;  as
$\sigma\in A_*\vee\sigma$, we have equality.
So $\lim_{\sigma\uparrow A}\sup_{\rho\in A_*\vee\sigma}
=\lim_{\sigma\in A}w-u_{\sigma}=0$.

\medskip

\quad{\bf (ii)} Now suppose that $\pmb u$ is of bounded variation.   Then it can
be expressed as $\pmb u'-\pmb u''$ where $\pmb u'=\sfamilysS{u'}$ and
$\pmb u''=\sfamilysS{u''}$ are non-negative, non-decreasing and order-bounded.
Using (i), we see that $w'=\lim_{\sigma\uparrow A}u'_{\sigma}$ and
$w''=\lim_{\sigma\uparrow A}u''_{\sigma}$ are defined, so that
$w=w'-w''$ is $\lim_{\sigma\uparrow A}u_{\sigma}$ and

\Centerline{$\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-w|
\le\sup_{\rho\in A_*\vee\sigma}(|u'_{\rho}-w'|+|u''_{\rho}-w''|)$,}

\Centerline{$\theta(\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-w|)
\le\theta(\sup_{\rho\in A_*\vee\sigma}|u'_{\rho}-w'|)
  +\theta(\sup_{\rho\in A_*\vee\sigma}|u''_{\rho}-w''|)
\to 0$}

\noindent as $\sigma\uparrow A$.

\medskip

\quad{\bf (iii)} Now suppose just that $\pmb u$ is moderately oscillatory.

\medskip

\qquad\grheada\ $w=\lim_{\sigma\uparrow A}u_{\sigma}$ is defined,
\Prf\ Given $\epsilon>0$, there is a process $\pmb v=\sfamilysS{v}$ of
bounded variation such that $\theta(\bar u)\le\bover13\epsilon$ where
$\bar u=\sup|\pmb u-\pmb v|$.   There is a $\sigma\in A$ such that
$\theta(v_{\tau}-v_{\sigma})\le\bover13\epsilon$ whenever $\sigma\le\tau\in A$;
now $\theta(u_{\tau}-u_{\sigma})\le\epsilon$ whenever $\sigma\le\tau\in A$,
As $L^0$ is complete, $\lim_{\sigma\uparrow A}u_{\sigma}$ is defined.\ \Qed

\medskip

\qquad\grheadb\ Now take any $\epsilon>0$ and
a process $\pmb v=\sfamilysS{v}$ of
bounded variation such that $\theta(\bar u)\le\bover12\epsilon$ where
$\bar u=\sup|\pmb u-\pmb v|$.   We know from (ii) that
$w'=\lim_{\sigma\uparrow A}v_{\sigma}$ is defined and that
$\lim_{\sigma\in A}\sup_{\rho\in A_*\vee\sigma}|v_{\rho}-w'|=0$.
Now

\Centerline{$|w-w'|=\lim_{\sigma\uparrow A}|u_{\sigma}-v_{\sigma}|
\le\sup|\pmb u-\pmb v|$,}

\noindent and if $\sigma\in A$ then

$$\eqalign{\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-w|&
\le\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-v_{\rho}|+|v_{\rho}-w'|+|w'-w|\cr&
\le 2\sup|\pmb u-\pmb v|+\sup_{\rho\in A_*\vee\sigma}|v_{\rho}-w'|\cr}$$

\noindent so

$$\eqalign{\limsup_{\sigma\uparrow A}
   \theta(\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-w|)&
\le 2\theta(\sup|\pmb u-\pmb v|)
   +\limsup_{\sigma\uparrow A}\theta(\sup_{\rho\in A_*\vee\sigma}|v_{\rho}-w'|)
   \cr&
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\lim_{\sigma\uparrow A}
   \theta(\sup_{\rho\in A_*\vee\sigma}|u_{\rho}-w|)=0$.

\medskip

{\bf (b)(i)} If $\pmb u$ is actually moderately oscillatory,
we can follow the same argument, inverted at the right points;  for
instance, the key sentence in (a-i) becomes

\inset{If $\sigma\in A$ and $\rho\in A^*\wedge\sigma$,
$u_{\rho}\le u_{\sigma}$ so $u_{\rho}-w\le u_{\sigma}-w$ and

$$\eqalignno{1&
=\sup_{\sigma'\in A}\Bvalue{\sigma'\le\rho}
=\sup_{\sigma'\in A,\sigma'\le\sigma}\Bvalue{\sigma'\le\rho}\cr&
=\sup_{\sigma'\in A,\sigma'\le\sigma}\Bvalue{\rho=\sigma'\vee\rho}
\Bsubseteq\sup_{\sigma'\in A,\sigma'\le\sigma}
   \Bvalue{u_{\rho}=u_{\sigma'\vee\rho}}\cr&
\Bsubseteq\sup_{\sigma'\in A,\sigma'\le\sigma}\Bvalue{u_{\sigma'}\le u_{\rho}}
\Bsubseteq\Bvalue{w\le u_{\rho}}
\Bsubseteq\Bvalue{|u_{\rho}-w|\le u_{\sigma}-w}\cr}$$

\noindent and $\sup_{\rho\in A^*\wedge\sigma}|u_{\rho}-w|\le u_{\sigma}-w$;  as
$\sigma\in A^*\wedge\sigma$, we have equality.}

\medskip

\quad{\bf (ii)}
For the general case, in which $\pmb u$ is just locally moderately oscillatory,
we can take any $\tau\in A$ and apply (i) here to $\pmb u\restr\Cal A\wedge\tau$
and $\{\sigma:\sigma\in A$, $\sigma\le\tau\}$.
}%end of proof of 615G

\leader{615H}{Corollary} Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
$\hat\Cal S$ its covered envelope, $\pmb u=\sfamilysS{u}$ a locally moderately
oscillatory process, and
$\hatpmb u=\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$
its fully adapted extension to $\hat\Cal S$.   Then $\pmb u$ and $\hatpmb u$
have starting values, which are the same.

\proof{{\bf (a)} By 615F(b-v), $\hatpmb u$ is locally moderately oscillatory.
So 615Gb tells us that

\Centerline{$w=\lim_{\sigma\downarrow\hat\Cal S\wedge\tau}\hat u_{\sigma}
=\lim_{\sigma\downarrow\hat\Cal S}\hat u_{\sigma}$}

\noindent is defined and is the starting value of $\hatpmb u$.

\medskip

{\bf (b)} To see that $w$ is also the starting value of $\pmb u$,
take any $\epsilon>0$.   Then there is a $\tau\in\hat\Cal S$ such that
$\theta(\hat u_{\sigma}-w)\le\epsilon$ whenever
$\sigma\in\hat\Cal S\wedge\tau$.   Next,
$\bar\mu(\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma})=1$, so there is a
non-empty finite set $I\subseteq\Cal S$ such that
$\bar\mu(\sup_{\sigma\in I}\Bvalue{\tau=\sigma})\ge 1-\epsilon$ and
$\bar\mu\Bvalue{\tau<\min I}\le\epsilon$.   Now if $\sigma\in\Cal S$ and
$\sigma\le\min I$, $\theta(\hat u_{\sigma\wedge\tau}-w)\le\epsilon$;
but

\Centerline{$\Bvalue{\hat u_{\sigma\wedge\tau}\ne\hat u_{\sigma}}
\Bsubseteq\Bvalue{\tau<\sigma}\Bsubseteq\Bvalue{\tau<\min I}$}

\noindent has measure at most $\epsilon$, so
$\theta(\hat u_{\sigma}-\hat u_{\sigma\wedge\tau})\le\epsilon$ and

\Centerline{$\theta(u_{\sigma}-w)=\theta(\hat u_{\sigma}-w)\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $w$ is also the starting value of
$\pmb u$.
}%end of proof of 615H

\leader{615I}{Definition}
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process.
I will say that $\pmb u$ is {\bf \updowncgt} if

\doubleinset{($\updown$)
$\lim_{n\to\infty}u_{\sigma_n}$ is defined whenever
$\sequencen{\sigma_n}$ is a sequence in $\Cal S$
which is either non-increasing or non-decreasing.}

\leader{615J}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a moderately oscillatory process.   Then $\pmb u$ is
\updowncgt.

\proof{ 615G.
}%end of proof of 615J

\leader{615K}{Lemma} Let $\Cal S$ be a finitely full
sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ an \updowncgt\ process.   Then
$\pmb u$ is order-bounded.

\proof{ I will work towards the contrapositive, that is,
assuming that $\pmb u$ is not
order-bounded I will show that it is not \updowncgt.

\medskip

{\bf (a)} To begin with (down to the end of (d) below) I will take it that
$u_{\sigma}\ge 0$ for every
$\sigma\in\Cal S$.   A couple of preliminary remarks will be useful.

\medskip

\quad{\bf (i)} For any $\alpha\in\Bbb R$,
$\{\Bvalue{u_{\sigma}>\alpha}:\sigma\in\Cal S\}$ is
upwards-directed.   \Prf\ (This is where we need to know that $\Cal S$ is
finitely full.)   Given $\sigma$, $\tau\in\Cal S$, set
$b=\Bvalue{u_{\sigma\wedge\tau}>\alpha}$, take $\rho\in\Cal S$ such that
$b\Bsubseteq\Bvalue{\rho=\sigma\wedge\tau}$ and
$1\Bsetminus b\Bsubseteq\Bvalue{\rho=\sigma\vee\tau}$ (611I again);
then

$$\eqalign{\Bvalue{u_{\rho}>\alpha}&
=b\Bcup(\Bvalue{u_{\sigma\vee\tau}>\alpha}\Bsetminus b)
=\Bvalue{u_{\sigma\wedge\tau}>\alpha}\Bcup\Bvalue{u_{\sigma\vee\tau}>\alpha}\cr&
\Bsupseteq(\Bvalue{\sigma\le\tau}\Bcap\Bvalue{u_{\sigma}>\alpha})
    \Bcup(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{u_{\sigma}>\alpha})
=\Bvalue{u_{\sigma}>\alpha},\cr}$$

\noindent and similarly
$\Bvalue{u_{\rho}>\alpha}\Bsupseteq\Bvalue{u_{\tau}>\alpha}$.\ \Qed

\medskip

\quad{\bf (ii)} For any $\alpha\in\Bbb R$ and $b\in\frak A$,
$\{\sigma:\sigma\in\Cal S$, $b\Bsubseteq\Bvalue{u_{\sigma}\le\alpha}\}$
is a sublattice of $\Cal S$.   \Prf\ If $\sigma$, $\tau\in\Cal S$,
$b\Bsubseteq\Bvalue{u_{\sigma}\le\alpha}$ and
$b\Bsubseteq\Bvalue{u_{\tau}\le\alpha}$, then

\Centerline{$\Bvalue{u_{\sigma\vee\tau}\le\alpha}
=(\Bvalue{\sigma\le\tau}\Bcap\Bvalue{u_{\tau}\le\alpha})
  \Bcup(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{u_{\sigma}\le\alpha})
\Bsupseteq b$}

\noindent and

\Centerline{$\Bvalue{u_{\sigma\wedge\tau}\le\alpha}
=(\Bvalue{\sigma\le\tau}\Bcap\Bvalue{u_{\sigma}\le\alpha})
  \Bcup(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{u_{\tau}\le\alpha})
\Bsupseteq b$.  \Qed}

\medskip

\quad{\bf (iii)} Similarly,
$\{\sigma:\sigma\in\Cal S$, $b\Bsubseteq\Bvalue{u_{\sigma}\ge\alpha}\}$
is a sublattice of
$\Cal S$ for any $\alpha\in\Bbb R$ and $b\in\frak A$.

\medskip

{\bf (b)} We are supposing that $\pmb u$ is not order-bounded, so $\Cal S$ is
surely not empty, and $\{u_{\sigma}:\sigma\in\Cal S\}$ has no upper bound in
$L^0(\frak A)$.   By 364L(a-i),
$a=\inf_{n\in\Bbb N}\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}>n}$ is non-zero.
If $n\in\Bbb N$ and $\epsilon>0$, there must be a finite set
$K\subseteq\Cal S$ such that
$\bar\mu(\sup_{\sigma\in K}\Bvalue{u_{\sigma}>n})\ge\bar\mu a-\epsilon$;
by (a-i) above, there is a single $\sigma\in\Cal S$ such that
$\bar\mu\Bvalue{u_{\sigma}>n}\ge\bar\mu a-\epsilon$.   We can therefore
choose $\sequencen{m_n}$, $\sequencen{\sigma_n}$ inductively such that

\Centerline{$m_0=0$,}

\Centerline{$\sigma_n\in\Cal S$,
\quad$\bar\mu\Bvalue{u_{\sigma_n}>m_n+1}\ge(1-2^{-n-2})\bar\mu a$,}

\Centerline{$m_{n+1}>m_n$,
\quad$\sum_{i=0}^n\bar\mu(\Bvalue{u_{\sigma_i}>m_{n+1}})
\le 2^{-n-2}\bar\mu a$.}

\noindent Set

\Centerline{$b=\inf_{n\in\Bbb N}\Bvalue{u_{\sigma_n}>m_n+1}
    \Bsetminus\sup_{i<n\in\Bbb N}\Bvalue{u_{\sigma_i}>m_n}$}

\noindent so that $\bar\mu b\ge\bover12\bar\mu a>0$ and
$b\Bsubseteq\Bvalue{u_{\sigma_i}\le m_n}\Bcap\Bvalue{u_{\sigma_j}\ge m_n+1}$
whenever $i<n\le j$.   By (a-ii) and (a-iii),

\Centerline{$b\Bsubseteq\Bvalue{u_{\sigma}\le m_n}
  \Bcap\Bvalue{u_{\tau}\ge m_n+1}$}

\noindent whenever $\sigma$ is in the lattice generated by $\{\sigma_i:i\le n\}$
and $\tau$ is in the lattice generated by $\{\sigma_j:j>n\}$.

\medskip

{\bf (c) case 1} Suppose that there are $k\in\Bbb N$ and $\epsilon>0$
such that for every $l>k$

\Centerline{$\bar\mu(b\Bcap\sup_{j>l}\Bvalue{\sigma_j>\sup_{k<i\le l}\sigma_i})
>\epsilon$.}

\noindent Then we can choose inductively a strictly increasing sequence
$\sequence{r}{k_r}$ in $\Bbb N$ such that $k_0=k+1$ and

\Centerline{$\bar\mu(b\Bcap\sup_{k_r<j\le k_{r+1}}
  \Bvalue{\sigma_j>\sup_{k<i\le k_r}\sigma_i})
>\epsilon$}

\noindent for every $r$.   For each $r$, set
$\tau'_r=\sup_{k_r<i\le k_{r+1}}\sigma_i$ and
$\tau_r=\sup_{k<i\le k_r}\sigma_j$ for
each $i$;  then we always have $\tau_{r+1}=\tau_r\vee\tau'r$, so

\Centerline{$\Bvalue{\tau_{r+1}=\tau'_r}
\Bsupseteq\Bvalue{\tau'_r>\tau_r}
\Bsupseteq\sup_{k_r<j\le k_{r+1}}\Bvalue{\sigma_j>\tau_r}$}

\noindent and $\bar\mu(b\Bcap\Bvalue{\tau_{r+1}=\tau'_r})\ge\epsilon$.

On the other hand,

\Centerline{$b\Bsubseteq\Bvalue{u_{\tau_r}\le m_{k_r}}
   \Bcap\Bvalue{u_{\tau'_r}\ge m_{k_r}+1}
\Bsubseteq\Bvalue{u_{\tau'_r}-u_{\tau_r}\ge 1}$.}

\noindent So

$$\eqalign{\theta(u_{\tau_{r+1}}-u_{\tau_r})&
\ge\bar\mu\Bvalue{u_{\tau_{r+1}}-u_{\tau_r}\ge 1}
\ge\bar\mu(\Bvalue{u_{\tau'_r}-u_{\tau_r}\ge 1}
     \Bcap\Bvalue{\tau_{r+1}=\tau'_r}\cr&
\ge\bar\mu(b\Bcap\Bvalue{\tau_{r+1}=\tau'_r})
\ge\epsilon\cr}$$

\noindent for every $r$, and $\sequence{r}{u_{\tau_r}}$ is not convergent, while
$\sequence{r}{\tau_r}$ is a non-decreasing sequence in $\Cal S$.   So in this
case $\pmb u$ is not moderately oscillatory.

\medskip

{\bf (d) case 2} Suppose that for every $k\in\Bbb N$ and $\epsilon>0$
there is an $l>k$ such that

\Centerline{$\bar\mu(b\Bcap\sup_{j>l}
   \Bvalue{\sigma_j>\sup_{k<i\le l}\sigma_i})
\le\epsilon$.}

\noindent Then we can choose a strictly increasing sequence $\sequence{r}{k_r}$
such that

\Centerline{$\bar\mu(b\Bcap\sup_{j>k_{r+1}}
   \Bvalue{\sigma_j>\sup_{k_r<i\le k_{r+1}}\sigma_i})\le 2^{-r-1}\bar\mu b$}

\noindent for each $r$.   Set $\tau'_r=\sup_{k_r<i\le k_{r+1}}\sigma_i$ for each
$r$;  then
$\bar\mu(b\Bcap\Bvalue{\tau'_{r+1}>\tau'_r})\le 2^{-r-1}\bar\mu b$ for each
$r$, and $c=b\Bsetminus\sup_{r\in\Bbb N}\Bvalue{\tau'_{r+1}>\tau'_r}$ is
non-zero.   Now $c\Bsubseteq\Bvalue{\tau'_{r+1}\le\tau'_r}$ for every
$r$, so if we set $\tau_r=\inf_{i\le r}\tau'_i$, we shall have
$c\Bsubseteq\Bvalue{\tau_r=\tau'_r}$ for every $r$.   \Prf\ Induce on $r$;
for the inductive step to $r+1$ we have

$$\eqalignno{c\Bcap\Bvalue{\tau_{r+1}=\tau'_{r+1}}&
=c\Bcap\Bvalue{\tau_r\wedge\tau'_{r+1}=\tau'_{r+1}}
=c\Bcap\Bvalue{\tau'_{r+1}\le\tau_r}
=c\Bcap\Bvalue{\tau'_{r+1}\le\tau'_r}
\Displaycause{by the inductive hypothesis}
=c.  \text{ \Qed}\cr}$$

On the other hand,

\Centerline{$b\Bsubseteq\Bvalue{u_{\tau'_r}\ge m_{k_r+1}}
   \Bcap\Bvalue{u_{\tau'_r}\le m_{k_{r+1}}}$}

\noindent for each $r$, so

\Centerline{$c\Bsubseteq\Bvalue{u_{\tau'_r}\le m_{k_{r+1}}}
   \Bcap\Bvalue{u_{\tau'_{r+1}}\ge m_{k_{r+1}+1}}
\Bsubseteq\Bvalue{u_{\tau'_{r+1}}-u_{\tau'_r}\ge 1}$;}

\noindent consequently $c\Bsubseteq\Bvalue{u_{\tau_{r+1}-u_{\tau_r}}\ge 1}$
and $\theta(u_{\tau_{r+1}}-u_{\tau_r})\ge\bar\mu c$ for every $r$, while
$\sequence{r}{\tau_r}$ is non-increasing.   So in this case also
$\pmb u$ is not moderately oscillatory.

\medskip

{\bf (e)} Since case 1 and case 2 are exhaustive, this deals with the case in
which $u_{\sigma}\ge 0$ for every $\sigma$.
If $\pmb u$ is not necessarily
positive. then $|\pmb u|$ is not order-bounded, so cannot be \updowncgt,
and there is a monotonic sequence in $\Cal S$ such that
$\lim_{n\to\infty}|u_{\sigma_n}|$ is undefined.   But
$|\,\,|:L^0(\frak A)\to L^0(\frak A)$ is continuous (613Ba), so
$\lim_{n\to\infty}u_{\sigma_n}$ must be undefined, and $\pmb u$ also is not
\updowncgt.
}%end of proof of 615K

\leader{615L}{Lemma} Let $\Cal S$ be a non-empty
finitely full sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ an \updowncgt\ process.
Suppose that $A\subseteq\Cal S$ is non-empty and downwards-directed.
Then $w=\lim_{\sigma\downarrow A}u_{\sigma}$ is defined.
Setting $A^*=\{\rho:\rho\in\Cal S$,
$\sup_{\sigma\in A}\Bvalue{\sigma\le\rho}=1\}$,

\Centerline{$\lim_{\sigma\downarrow A}\sup_{\rho\in A^*\wedge\sigma}
  |u_{\rho}-w|
=0$.}

\proof{{\bf (a)} Note first that $A^*$ is finitely full.
\Prf\ Suppose that $\tau\in\Cal T$ and that $\{\tau\}$ is covered by a finite
subset of $A^*$.   Take any
$a\in\frak A\setminus\{0\}$.   As $\Cal S$ is finitely full, $\tau\in\Cal S$.
There are $\rho\in A^*$ and $\sigma\in A$ such that
$b=a\Bcap\Bvalue{\tau=\rho}$ and $c=b\Bcap\Bvalue{\rho\le\sigma}$ are non-zero.
Now $0\ne c\Bsubseteq a\Bcap\Bvalue{\tau\le\sigma}$.   As $a$
is arbitrary, $\tau\in A^*$;  as $\tau$ is arbitrary, $A^*$ is
finitely full.\ \Qed

\medskip

{\bf (b)} In particular, $A^*$ is closed under $\wedge$.
Consequently $w=\lim_{\sigma\downarrow A^*}u_{\sigma}$ is
defined.   \Prf\Quer\ Otherwise, there is an $\epsilon>0$ such that for
every $\sigma\in A^*$ there is a $\tau\in A^*\wedge\sigma$ such that
$\theta(u_{\tau}-u_{\sigma})\ge\epsilon$.   Now we can choose a sequence
$\sequencen{\sigma_n}$ in $A^*$ such that

\Centerline{$\sigma_{n+1}\le\sigma_n$,
\quad$\theta(u_{\sigma_{n+1}}-u_{\sigma_n})\ge\epsilon$}

\noindent for every $n\in\Bbb N$, and $\sequencen{\sigma_n}$ is a monotonic
sequence in $\Cal S\subseteq\hat\Cal S$ such that
$\sequencen{\hat u_{\sigma_n}}$ is not convergent.\ \Bang\Qed

\medskip

{\bf (c)}
$\lim_{\sigma\downarrow A}\sup_{\rho\in A^*\wedge\sigma}
  \theta(u_{\rho}-w)=0$.   \Prf\ Let $\epsilon>0$.
Then there is a $\tau'\in A^*$ such that $\theta(u_{\rho}-w)\le\epsilon$
whenever
$\rho\in A^*\vee\tau'$.   Next, $\sup_{\sigma\in A}\Bvalue{\sigma\le\tau'}=1$
and $A$ is downwards-directed, so there is a $\sigma_0\in A$ such that
$\bar\mu\Bvalue{\sigma_0\le\tau}\ge 1-\epsilon$.   If we now take $\sigma\in A$
and $\rho\in A^*$ such that $\rho\le\sigma\le\sigma_0$, we see that
$\theta(u_{\tau'\wedge\rho}-w)\le\epsilon$ and

\Centerline{$\theta(u_{\rho}-u_{\tau'\wedge\rho})
\le\bar\mu\Bvalue{\tau'<\rho}
\le\bar\mu\Bvalue{\tau'<\sigma_0}\le\epsilon$,}

\noindent so $\theta(u_{\rho}-w)\le 2\epsilon$.\ \Qed

Of course it follows that
$\lim_{\sigma\downarrow A}\sup_{\rho\in A,\rho\le\sigma}
  \theta(u_{\rho}-w)=0$, and $w=\lim_{\sigma\downarrow A}u_{\sigma}$.

\medskip

{\bf (d)} In fact
$\lim_{\sigma\downarrow A}\theta(\sup_{\rho\in A^*\wedge\sigma}
  |u_{\rho}-w|)=0$.   \Prf\ Again take $\epsilon>0$.   By (iii), there
is a $\sigma\in A$ such that
$\theta(u_{\rho}-w)\le\min(\Bover12\epsilon,\Bover1{16}\epsilon^2)$
for every $\rho\in A^*\wedge\sigma$.   Set $v_{\rho}=u_{\rho}-u_{\sigma}$
for $\rho\in A^*\wedge\sigma$, so that

\Centerline{$\pmb v
=\family{\rho}{A^*\wedge\sigma}{v_{\rho}}
=(\pmb u-u_{\sigma}\tbf 1)\restr A^*\wedge\sigma$}

\noindent is a fully adapted process on the
finitely full sublattice $A^*\vee\sigma$ and
$\theta(v_{\rho})\le\Bover1{16}\epsilon^2$ for every
$\rho\in A^*\vee\sigma$.   Also $\pmb u$ is order-bounded (615K), so
$\pmb v$ also is.   By 615Db,
$\theta(\sup|\pmb v|)\le\bover12\epsilon$.   Now

\Centerline{$\sup_{\rho\in A^*\vee\sigma}|u_{\rho}-w|
\le\sup_{\rho\in A^*\vee\sigma}|u_{\rho}-u_{\sigma}|
   +|u_{\sigma}-w|
=\sup|\pmb v|+|u_{\sigma}-w|$,}

\noindent so

\Centerline{$\theta(\sup_{\rho\in A^*\vee\sigma}|u_{\rho}-w|)
\le\theta(\sup|\pmb v|)+\theta(u_{\sigma}-w)
\le\Bover12\epsilon+\Bover12\epsilon=\epsilon$,}

\noindent and this is true whenever $\sigma\in A$ and $\sigma\ge\sigma_0$.\ \Qed
}%end of proof of 615L

\leader{615M}{Construction}
Let $\Cal S$ be a finitely full sublattice of $\Cal T$ with
a greatest member, $\pmb u=\sfamilysS{u}$ an \updowncgt\ process, and
$\delta>0$.   Then there are sequences $\sequence{i}{D_i}$, $\sequence{i}{y_i}$,
$\sequence{i}{d_i}$, a family
$\langle c_{i\sigma}\rangle_{i\in\Bbb N,\sigma\in\Cal S}$ and a process
$\tildepmb u=\sfamilysS{\tilde u}$ with the following properties.

(a) $D_0=\Cal S$;   for every $i\in\Bbb N$, $\max\Cal S\in D_i\subseteq\Cal S$,
$D_i$ is closed under $\wedge$, $y_i=\lim_{\sigma\downarrow D_i}u_{\sigma}$ and

$$\eqalign{D_{i+1}&
=\{\sigma:\sigma\in\Cal S,\,\Bvalue{\sigma<\max\Cal S}
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|\ge\delta}\cr&
\mskip150mu\text{and there is a $\sigma'\in D_i$ such that }\sigma'\le\sigma\}.
\cr}$$

(b) %!$y_0=u_{\min\Cal S}$ and
$y_i\in\bigcap_{\sigma\in D_i}L^0(\frak A_{\sigma})$ for every $i\in\Bbb N$.

(c)(i) For every $i\in\Bbb N$,

\inset{$d_i=\sup_{\sigma\in D_i}\Bvalue{\sigma<\max\Cal S}$,

$d_i\in\bigcap_{\sigma\in D_i}\frak A_{\sigma}$,

$d_{i+1}\Bsubseteq d_i$,

$d_{i+1}\Bsubseteq\Bvalue{|y_{i+1}-y_i|\ge\delta}$,

$1\Bsetminus d_i
\Bsubseteq\Bvalue{y_i=u_{\max\Cal S}}\Bcap\Bvalue{y_i=y_{i+1}}$.}

\quad(ii) $\inf_{i\in\Bbb N}d_i=0$.

(d)(i) If $\sigma\in\Cal S$ and $i\in\Bbb N$,

\Centerline{$c_{i\sigma}=\sup_{\tau\in D_i}\Bvalue{\tau\le\sigma}$,
\quad$c_{i+1,\sigma}\Bsubseteq c_{i\sigma}$,
\quad$\Bvalue{\sigma=\max\Cal S}\Bsubseteq c_{i\sigma}
\Bsubseteq\Bvalue{\sigma=\max\Cal S}\Bcup d_i$.}

\quad(ii) If $i\in\Bbb N$ and
$\sigma\in D_i$ then $c_{i\sigma}=1$;  \cmmnt{in particular,}
$c_{0\sigma}=1$ for every $\sigma\in\Cal S$.

\quad(iii) If $\sigma$, $\sigma'$ in $\Cal S$ then
$\Bvalue{\sigma\le\sigma'}\Bcap c_{i\sigma}\Bsubseteq c_{i\sigma'}$
for every $i\in\Bbb N$.

\quad(iv) $\inf_{i\in\Bbb N}c_{i\sigma}=\Bvalue{\sigma=\max\Cal S}$ for
every $\sigma\in\Cal S$.

\quad(v) If $\sigma\in\Cal S$ and $i\in\Bbb N$ then
$c_{i\sigma}\Bsetminus c_{i+1,\sigma}
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|<\delta}$.

(e) If $\sigma\in\Cal S$ then

\Centerline{$c_{i\sigma}\Bsetminus c_{i+1,\sigma}
\Bsubseteq\Bvalue{\tilde u_{\sigma}=y_i}$}

\noindent for every $i\in\Bbb N$, and
$\Bvalue{\sigma=\max\Cal S}\Bsubseteq\Bvalue{\tilde u_{\sigma}=u_{\max\Cal S}}$.

(f) $\tildepmb u$ is fully adapted, $\sup|\tildepmb u|\le\sup|\pmb u|$ and
$\sup|\pmb u-\tildepmb u|\le\delta\chi 1$.

\leaveitout{(g)%\query never used
If $I\in\Cal I(\Cal S)$ contains $\max\Cal S$
there is a sequence
$\sequence{i}{\tau_i}$ such that $\tau_0=\min I$ and, for each $i\in\Bbb N$,

\quad(i) $\tau_i$ belongs to the covered envelope $\hat I$ of $I$,

\quad(ii) $\tau_i\le\tau_{i+1}$,

\quad(iii) $\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{\tilde u_{\sigma}=y_i}$
whenever $\sigma\in\hat I$,

\quad(iv) $\Bvalue{\tau_i<\max\Cal S}\Bsubseteq d_i$,

\quad(v) $\tau_i\le\tau$ whenever $i\in\Bbb N$ and $\tau\in\hat I\cap D_i$,

\quad(vi) $c_{i\tau_i}=1$.
}

(g) $\tildepmb u$ is of bounded variation.

\proof{{\bf (a)} We start the induction with $D_0=\Cal S$.
The inductive hypothesis we need is just that $D_i$ is
non-empty and closed under $\wedge$.   Given this,
$y_i=\lim_{\sigma\downarrow D_i}u_{\sigma}$ is defined (615L).
Looking at the formula for $D_{i+1}$ we see at once that
$\max\Cal S\in D_{i+1}$.   If $\sigma$, $\tau\in D_{i+1}$, then

$$\eqalign{\Bvalue{\sigma\wedge\tau<\max\Cal S}&
=(\Bvalue{\sigma\le\tau}\Bcap\Bvalue{\sigma<\max\Cal S})
    \Bcup(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\tau<\max\Cal S})\cr&
\Bsubseteq(\Bvalue{\sigma\le\tau}\Bcap\Bvalue{|u_{\sigma}-y_i|\ge\delta})
  \Bcup(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{|u_{\tau}-y_i|\ge\delta})\cr&
\Bsubseteq(\Bvalue{\sigma\le\tau}
    \Bcap\Bvalue{|u_{\sigma\wedge\tau}-y_i|\ge\delta})
  \Bcup(\Bvalue{\tau\le\sigma}
    \Bcap\Bvalue{|u_{\sigma\wedge\tau}-y_i|\ge\delta})\cr&
\Bsubseteq\Bvalue{|u_{\sigma\wedge\tau}-y_i|\ge\delta};\cr}$$

\noindent at the same time, there are $\sigma'$, $\tau'\in D_i$ such that
$\sigma'\le\sigma$ and $\tau'\le\tau$, so that $\sigma'\wedge\tau'\in D_i$ and
$\sigma'\wedge\tau'\le\sigma\wedge\tau$.   Thus $\sigma\wedge\tau\in D_{i+1}$;
as $\sigma$ and $\tau$ are arbitrary, $D_{i+1}$ is closed under $\wedge$ and the
induction proceeds.

\medskip

{\bf (b)}
%!Because $D_0=\Cal S$ has a least element $\min\Cal S$,
%!$y_0=u_{\min\Cal S}$.
For every $i\in\Bbb N$, 613Bj tells us that
$y_i\in L^0(\bigcap_{\sigma\in D_i}\frak A_{\sigma})$.

\medskip

{\bf (c)(i)}\grheada\ Define $d_i$ as the supremum
$\sup_{\sigma\in D_i}\Bvalue{\sigma<\max\Cal S}$ for each $i$.
Since $D_i$ is closed under $\wedge$,

\Centerline{$d_i=\sup_{\tau\in D_i\wedge\sigma}\Bvalue{\tau<\max\Cal S}
\in\frak A_{\sigma}$}

\noindent for every $\sigma\in D_i$, and
$d_i\in\bigcap_{\sigma\in D_i}\frak A_{\sigma}$.

\medskip

\qquad\grheadb\ For any $i\in\Bbb N$ and $\sigma\in D_{i+1}$,
there is a $\sigma'\in D_i$ such that $\sigma'\le\sigma$, in which case
$\Bvalue{\sigma<\max\Cal S}\Bsubseteq\Bvalue{\sigma'<\max\Cal S}\Bsubseteq d_i$;
taking the supremum over such $\sigma$, $d_{i+1}\Bsubseteq d_i$.

\medskip

\qquad\grheadc\ Whenever $\tau\le\sigma$ in $D_{i+1}$,

\Centerline{$\Bvalue{\sigma<\max\Cal S}
\Bsubseteq\Bvalue{\tau<\max\Cal S}
\Bsubseteq\Bvalue{|u_{\tau}-y_i|\ge\delta}$.}

\noindent So

\Centerline{$\Bvalue{\sigma<\max\Cal S}
\Bsubseteq\inf_{\tau\in D_{i+1},\tau\le\sigma}\Bvalue{|u_{\tau}-y_i|\ge\delta}
\Bsubseteq\Bvalue{|y_{i+1}-y_i|\ge\delta}$}

\noindent for every $\sigma\in D_{i+1}$;  taking the supremum,
$d_{i+1}\Bsubseteq\Bvalue{|y_{i+1}-y_i|\ge\delta}$.

\medskip

\qquad\grheadd\ For any $\sigma\in D_i$,

\Centerline{$1\Bsetminus d_i\Bsubseteq\Bvalue{\sigma=\max\Cal S}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\max\Cal S}}$,}

\noindent so $1\Bsetminus d_i\Bsubseteq\Bvalue{y_i=u_{\max\Cal S}}$.
Now we also have

\Centerline{$1\Bsetminus d_i\Bsubseteq 1\Bsetminus d_{i+1}
\Bsubseteq\Bvalue{y_{i+1}=u_{\max\Cal S}}$,}

\noindent so $1\Bsetminus d_i\Bsubseteq\Bvalue{y_i=y_{i+1}}$.

\medskip

\quad{\bf (ii)} For $i\in\Bbb N$, write
$D_i^*=\bigcup_{\sigma\in D_i}\Cal S\vee\sigma$;  now for
$\sigma\in D_i$, set

\Centerline{$a_{i\sigma}
=\inf_{\tau\in D_i^*,\tau\le\sigma}\Bvalue{|u_{\tau}-y_i|\le\bover13\delta}$.}

\noindent Of course $a_{i\sigma'}\Bsubseteq a_{i\sigma}$ whenever
$\sigma\le\sigma'$ in $D_i$, because $a_{i\sigma'}$ is the infimum of a larger
set.   Also $\sup_{\sigma'\in D_i}a_{i\sigma'}=1$.   \Prf\ Given $\eta>0$, there
is a $\sigma\in D_i$ such that
$\theta(\sup_{\tau\in D_i^*\wedge\sigma}|u_{\tau}-y_i|)
\le 3\eta\min(\delta,1)$ (615L again).   So
$b=\Bvalue{\sup_{\tau\in D_i^*\wedge\sigma}|u_{\tau}-y_i|>\bover13\delta}$
has measure at most $\eta$.   But
$\Bvalue{|u_{\tau}-y_i|\le\bover13\delta}\Bcup b=1$ whenever
$\tau\in D_i^*$ and $\tau\le\sigma$, so $a_{i\sigma}\Bsupseteq b$.
Thus $\sup_{\sigma'\in D_i}a_{i\sigma'}$ has measure at least $1-\eta$;  as
$\eta$ is arbitrary, $\sup_{\sigma'\in D_i}a_{i\sigma'}=1$.\ \Qed

\Quer\ If $d=\inf_{i\in\Bbb N}d_i$ is non-zero, choose $\sigma_i\in D_i$, for
each $i\in\Bbb N$, such that

\Centerline{$\bar\mu(d_i\Bsetminus(a_{i\sigma_i}
    \Bcap\Bvalue{\sigma_i<\max\Cal S}))
\le 2^{-i-2}\bar\mu d$.}

\noindent (Any element far enough down $D_i$ will serve.)   Then

\Centerline{$a
=\inf_{i\in\Bbb N}\Bvalue{\sigma_i<\max\Cal S}\Bcap a_{i\sigma_i}$}

\noindent has measure at least $\bover12\bar\mu d$ and is non-zero.
For each $i\in\Bbb N$,
$\sigma_i\wedge\sigma_{i+1}\in D_i^*$ (because there is some member of
$D_i$ less than or equal to $\sigma_{i+1}$) and

$$\eqalignno{a\Bcap\Bvalue{\sigma_{i+1}\le\sigma_i}&
\Bsubseteq\Bvalue{\sigma_{i+1}<\max\Cal S}
  \Bcap\Bvalue{u_{\sigma_{i+1}}=u_{\sigma_i\wedge\sigma_{i+1}}}\cr&
\mskip100mu
  \Bcap\Bvalue{|u_{\sigma_i\wedge\sigma_{i+1}}-y_i|\le\bsover13\delta}
  \Bcap\Bvalue{|y_{i+1}-y_i|\ge\delta}
\Displaycause{because $a_{i\sigma_i}
\Bsubseteq\Bvalue{|u_{\sigma_i\wedge\sigma_{i+1}}-y_i|\le\bsover13\delta}$
and $\Bvalue{\sigma_{i+1}<\max\Cal S}\Bsubseteq\Bvalue{|y_{i+1}-y_i|\ge\delta}$}
\Bsubseteq\Bvalue{|u_{\sigma_{i+1}}-y_i|\le\bsover13\delta}
  \Bcap\Bvalue{|u_{\sigma_{i+1}}-y_{i+1}|\le\bsover13\delta}
  \Bcap\Bvalue{|y_{i+1}-y_i|\ge\delta}
=0.\cr}$$

\noindent Set $\rho_i=\sup_{j\le i}\sigma_j$ for each
$i\in\Bbb N$.   Then

\Centerline{$a\Bsubseteq\inf_{j<i}\Bvalue{\sigma_j<\sigma_{j+1}}
\Bsubseteq\Bvalue{\sigma_i=\rho_i}\Bsubseteq\Bvalue{u_{\rho_i}=u_{\sigma_i}}$}

\noindent for every $i\in\Bbb N$.   But we also have

\Centerline{$a\Bsubseteq\Bvalue{|u_{\sigma_i}-y_i|\le\bover13\delta}
    \Bcap\Bvalue{|y_{i+1}-y_i|\ge\delta}$}

\noindent for each $i$,
so $a\Bsubseteq\Bvalue{|u_{\rho_{i+1}}-u_{\rho_i}|\ge\bover13\delta}$
for each $i$ and $\lim_{i\to\infty}u_{\rho_i}$ is not defined.\ \Bang

Thus $\inf_{i\in\Bbb N}d_i=0$.

\medskip

{\bf (d)(i)} As every member of
$D_{i+1}$ dominates a member of $D_i$, $c_{i+1,\sigma}\Bsubseteq c_{i\sigma}$.
As $\max\Cal S\in D_i$,
$\Bvalue{\sigma=\max\Cal S}\Bsubseteq c_{i\sigma}$.   If $\tau\in D_i$,

\Centerline{$\Bvalue{\tau\le\sigma}
\Bsubseteq(\Bvalue{\tau=\max\Cal S}\Bcap\Bvalue{\tau\le\sigma})
   \Bcup\Bvalue{\tau<\max\Cal S}
\Bsubseteq\Bvalue{\sigma=\max\Cal S}\Bcup d_i$;}

\noindent taking the supremum over $\tau$,
$c_{i\sigma}\Bsubseteq\Bvalue{\sigma=\max\Cal S}\Bcup d_i$.

\medskip

\quad{\bf (ii)} If $\sigma\in D_i$ then
$c_{i\sigma}\Bsupseteq\Bvalue{\sigma\le\sigma}=1$.   As $D_0=\Cal S$,
$c_{0\sigma}=1$ for every $\sigma\in\Cal S$.

\medskip

\quad{\bf (iii)}

\Centerline{$\Bvalue{\sigma\le\sigma'}\Bcap c_{i\sigma}
=\sup_{\tau\in D_i}\Bvalue{\sigma\le\sigma'}\Bcap\Bvalue{\tau\le\sigma}
\Bsubseteq\sup_{\tau\in D_i}\Bvalue{\tau\le\sigma'}
=c_{i\sigma'}$.}

\medskip

\quad{\bf (iv)} By (i),

\Centerline{$\Bvalue{\sigma=\max\Cal S}\Bsubseteq\inf_{i\in\Bbb N}c_{i\sigma}
\Bsubseteq\Bvalue{\sigma=\max\Cal S}\Bcup\inf_{i\in\Bbb N}d_i
=\Bvalue{\sigma=\max\Cal S}$}

\noindent by (c-iii).

\medskip

\quad{\bf (v)} Suppose that $\tau\in D_i$.   Then

\Centerline{$b
=\Bvalue{|u_{\sigma}-y_i|\ge\delta}\Bcap\Bvalue{\tau\le\sigma}
   \Bcap\Bvalue{\sigma<\max\Cal S}$}

\noindent belongs to $\frak A_{\sigma}$, because
$y_i\times\chi\Bvalue{\tau\le\sigma}\in\frak A_{\sigma}$, as observed in
(i) above.   We therefore have a $\tau'\in\Cal T$ such that

\Centerline{$b\Bsubseteq\Bvalue{\tau'=\sigma}$,
\quad$1\Bsetminus b\Bsubseteq\Bvalue{\tau'=\max\Cal S}$}

\noindent and $\tau'\in\Cal S$ because $\Cal S$ is finitely full.   Now

\Centerline{$b\Bsubseteq\Bvalue{\tau'=\sigma}\Bcap\Bvalue{\tau\le\sigma}
\Bsubseteq\Bvalue{\tau\le\tau'}$,
\quad$b\Bsubseteq\Bvalue{\tau'=\max\Cal S}\Bcap\Bvalue{\tau\le\max\Cal S}
\Bsubseteq\Bvalue{\tau\le\tau'}$}

\noindent so $\tau\le\tau'$;  and

\Centerline{$\Bvalue{\tau'<\max\Cal S}\Bsubseteq b
\Bsubseteq\Bvalue{\tau'=\sigma}\Bcap\Bvalue{|u_{\sigma}-y_i|\ge\delta}
\Bsubseteq\Bvalue{|u_{\tau'}-y_i|\ge\delta}$}

\noindent so $\tau'\in D_{i+1}$.   But this means that

\Centerline{$c_{i+1,\sigma}\Bsupseteq\Bvalue{\tau'\le\sigma}
\Bsupseteq b
=\Bvalue{|u_{\sigma}-y_i|\ge\delta}\Bcap\Bvalue{\tau\le\sigma}
   \Bcap\Bvalue{\sigma<\max\Cal S}$.}

\noindent As $\tau$ is arbitrary,

\Centerline{$c_{i+1,\sigma}
\Bsupseteq\Bvalue{|u_{\sigma}-y_i|\ge\delta}\Bcap c_{i\sigma}
   \Bcap\Bvalue{\sigma<\max\Cal S}$}

\noindent and

$$\eqalign{c_{i\sigma}\Bsetminus c_{i+1,\sigma}&
\Bsubseteq c_{i\sigma}\Bcap\Bvalue{\sigma<\max\Cal S}\Bsetminus c_{i+1,\sigma}
  \cr&
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|<\delta}\Bcap\Bvalue{\tilde u_{\sigma}=y_i}
\Bsubseteq\Bvalue{|u_{\sigma}-\tilde u_{\sigma}|<\delta}.\cr}$$

\medskip

{\bf (e)} Since
$\sequence{i}{c_{i\sigma}\Bsetminus c_{i+1,\sigma}}$ is disjoint with supremum
$1\Bsetminus\Bvalue{\sigma=\max\Cal S}$ ((d-iv) above),
the formula here defines a member $\tilde u_{\sigma}$ of $L^0(\frak A)$.

\medskip

{\bf (f)(i)} Take $\sigma\in\Cal S$.   Then
$\Bvalue{\tau\le\sigma}\in\frak A_{\sigma}$ for every $\tau\in\Cal S$,
so $c_{i\sigma}\in\frak A_{\sigma}$ for every $i\in\Bbb N$;  also, of course,
$\Bvalue{\sigma=\max\Cal S}\in\frak A_{\sigma}$.   Next,
$y_i\in L^0(\frak A_{\tau})$ for every $\tau\in D_i$, so
$y_i\times\chi\Bvalue{\tau\le\sigma}\in L^0(\frak A_{\sigma})$ for every
$\tau\in D_i$ (612C) and
$y_i\times\chi c_{i\sigma}
=\lim_{\tau\downarrow D_i}y_i\times\chi\Bvalue{\tau\le\sigma}$ belongs to
$L^0(\frak A_{\sigma})$ for every $i\in\Bbb N$.   Finally,
$u_{\max\Cal S}\times\chi\Bvalue{\sigma=\max\Cal S}\in L^0(\frak A_{\sigma})$.
So $\tilde u_{\sigma}\in L^0(\frak A_{\sigma})$.

\medskip

\quad{\bf (ii)} Suppose that $\sigma$, $\sigma'\in\Cal S$ and
$b=\Bvalue{\sigma=\sigma'}$.   Then
$b\Bcap\Bvalue{\tau\le\sigma}=b\Bcap\Bvalue{\tau\le\sigma'}$ for every
$\tau\in\Cal S$ (611E(c-iv-$\alpha$)) so
$b\Bcap c_{i\sigma}=b\Bcap c_{i\sigma'}$ for every $i\in\Bbb N$ and
$b\Bcap\Bvalue{\sigma=\max\Cal S}=a\Bcap\Bvalue{\sigma'=\max\Cal S}$.
Accordingly

\Centerline{$b\Bcap(c_{i\sigma}\Bsetminus c_{i+1,\sigma})
\Bsubseteq\Bvalue{\tilde u_{\sigma}=y_i}\Bcap\Bvalue{\tilde u_{\sigma'}=y_i}
\Bsubseteq\Bvalue{\tilde u_{\sigma}=\tilde u_{\sigma'}}$}

\noindent for every $i\in\Bbb N$, and

\Centerline{$b\Bcap\Bvalue{\sigma=\max\Cal S}
\Bsubseteq\Bvalue{\tilde u_{\sigma}=u_{\max\Cal S}}
    \Bcap\Bvalue{\tilde u_{\sigma'}=u_{\max\Cal S}}
\Bsubseteq\Bvalue{\tilde u_{\sigma}=\tilde u_{\sigma'}}$.}

\noindent So $b\Bsubseteq\Bvalue{\tilde u_{\sigma}=\tilde u_{\sigma'}}$.
Thus $\tilde u$ is fully adapted.

\medskip

\quad{\bf (iii)} From (d-v) we see that

\Centerline{$c_{i\sigma}\Bsetminus c_{i+1,\sigma}
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|<\delta}\Bcap\Bvalue{\tilde u_{\sigma}=y_i}
\Bsubseteq\Bvalue{|u_{\sigma}-\tilde u_{\sigma}|\le\delta}$}

\noindent for each $i$.  At the top end, we surely have

\Centerline{$\Bvalue{\sigma=\max\Cal S}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\max\Cal S}}
   \Bcap\Bvalue{\tilde u_{\sigma}=u_{\max\Cal S}}
\Bsubseteq\Bvalue{|u_{\sigma}-\tilde u_{\sigma}|\le\delta}$.}

\noindent Once again, these parts assemble into

\Centerline{$\Bvalue{|u_{\sigma}-\tilde u_{\sigma}|\le\delta}=1$}

\noindent and $|u_{\sigma}-\tilde u_{\sigma}|\le\delta\chi 1$.   As $\sigma$ is
arbitrary, $\sup|\pmb u-\tildepmb u|\le\delta\chi 1$.

\medskip

\quad{\bf (iv)} Writing $\bar u$ for $\sup|\pmb u|$, we have

\Centerline{$|y_i|=|\lim_{\sigma\downarrow D_i}u_{\sigma}|
=\lim_{\sigma\downarrow D_i}|u_{\sigma}|\le\bar u$}

\noindent for every $i\in\Bbb N$.   So if $\sigma\in\Cal S$,

$$\eqalignno{\Bvalue{|\tilde u_{\sigma}|\le\bar u}&
\Bsupseteq\Bvalue{\tilde u_{\sigma}=u_{\max\Cal S}}
   \Bcup\sup_{i\in\Bbb N}\Bvalue{\tilde u_{\sigma}=y_i}
\Bsupseteq\Bvalue{\sigma=\max\Cal S}
   \Bcup\sup_{i\in\Bbb N}(c_{i\sigma}\Bsetminus c_{i+1,\sigma})
\Displaycause{by (e)}
\Bsupseteq(\inf_{i\in\Bbb N}c_{i\sigma})
   \Bcup(c_{0\sigma}\Bsetminus\inf_{i\in\Bbb N}c_{i\sigma})
\Displaycause{by (d-iv)}
=1\cr}$$

\noindent by (d-ii).   As $\sigma$ is arbitrary, $\sup|\tildepmb u|\le\bar u$.

\medskip

\leaveitout{{\bf (g)(i)} Let $\langle\sigma_j\rangle_{j\le m}$ be a maximal totally ordered
subset of $I$.   Then for each $i\in\Bbb N$ we have
$c_{i\sigma_0}\Bsubseteq\ldots\Bsubseteq c_{i\sigma_m}=1$ and
$c_{i\sigma_k}\in\frak A_{\sigma_j}$ whenever $k\le j\le m$, so
we can define $\tau_i\in\hat I$ by saying that

\Centerline{$\Bvalue{\tau_i=\sigma_j}
\Bsupseteq c_{i\sigma_j}\Bsetminus\sup_{k<j}c_{i\sigma_k}$}

\noindent for every $j\le m$.   Observe that $\sigma_0=\min I$ and
$\Bvalue{\tau_0=\sigma_0}\Bsupseteq c_{0\sigma_0}=1$ by (d-ii), so
$\tau_0=\min I$.

Inducing on $j$ we see that
$c_{i\sigma_j}\Bsubseteq\Bvalue{\tau_i\le\sigma_j}$ for every $j\le m$.
In fact we have equality.   \Prf\Quer\ If
$a=\Bvalue{\tau_i\le\sigma_j}\Bsetminus c_{i\sigma_j}$ is non-zero, let
$k\le m$ be such that
$a'=a\Bcap c_{i\sigma_k}\Bsetminus\sup_{k'<k}c_{i\sigma{k'}}$ is non-zero.
Then $a'\Bsubseteq\Bvalue{\tau_i=\sigma_k}\Bcap\Bvalue{\tau_i\le\sigma_j}
\Bsubseteq\Bvalue{\sigma_k\le\sigma_j}$;  as $a'\Bsubseteq c_{i\sigma_k}$,
$a'\Bsubseteq c_{i\sigma_j}$, by (d-ii), $a'\Bsubseteq c_{i\sigma_j}$;
but $a$ is disjoint from $c_{i\sigma_j}$.\ \Bang\Qed

\medskip

\quad{\bf (ii)} It follows that $\tau_i\le\tau_{i+1}$ for $i\in\Bbb N$.   \Prf\

$$\eqalign{\Bvalue{\tau_{i+1}<\tau_i}&
=\sup_{j,k\le m}\Bvalue{\tau_{i+1}=\sigma_j}\Bcap\Bvalue{\tau_i=\sigma_k}
   \Bcap\Bvalue{\sigma_j<\sigma_k}\cr&
\Bsubseteq\sup_{j\le m}\Bvalue{\tau_{i+1}\le\sigma_j}
   \Bsetminus\Bvalue{\tau_i\le\sigma_j}
=\sup_{j\le m}c_{i+1,\sigma_j}\Bsetminus c_{i,\sigma_j}
=0.  \text{ \Qed}\cr}$$

\noindent Thus $\sequence{i}{\tau_i}$ is non-decreasing.

\medskip

\quad{\bf (iii)} If $\sigma\in\hat I$

$$\eqalignno{\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}&
=\sup_{j\le m}\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
   \Bcap\Bvalue{\sigma=\sigma_j}
\Displaycause{because $\{\sigma_j:j\le m\}$ covers $I$ and therefore covers
$\hat I$}
=\sup_{j\le m}
  (\Bvalue{\tau_i\le\sigma_j}\Bsetminus\Bvalue{\tau_{i+1}\le\sigma_j})
   \Bcap\Bvalue{\sigma=\sigma_j}\cr&
=\sup_{j\le m}
  (c_{i\sigma_j}\Bsetminus c_{i+1,\sigma_j})\Bcap\Bvalue{\sigma=\sigma_j}\cr&
\Bsubseteq\sup_{j\le m}
  \Bvalue{\tilde u_{\sigma_j}=y_i}\Bcap\Bvalue{\sigma=\sigma_j}
\Bsubseteq\Bvalue{\tilde u_{\sigma}=y_i}.\cr&
\cr}$$

\medskip

\quad{\bf (iv)} For any $i\in\Bbb N$

$$\eqalignno{\Bvalue{\tau_i<\max\Cal S}&
=\sup_{j\le m}\Bvalue{\tau_i=\sigma_j}\Bcap\Bvalue{\sigma_j<\max\Cal S}
\Bsubseteq\sup_{j\le m}c_{i\sigma_j}\Bcap\Bvalue{\sigma_j<\max\Cal S}
\Displaycause{(i) above}
\Bsubseteq d_i\cr}$$

\noindent by (d-i).

\medskip

\quad{\bf (v)} Now suppose that $i\in\Bbb N$ and $\tau\in\hat I\cap D_i$.
In this case, for any $j\le m$,

$$\eqalignno{\Bvalue{\tau=\sigma_j}&
=c_{i\sigma_j}\Bcap\Bvalue{\tau=\sigma_j}
\Displaycause{by the formula defining $c_{i\sigma_j}$}
=\Bvalue{\tau_i\le\sigma_j}\Bcap\Bvalue{\tau=\sigma_j}
\Displaycause{by (i) above}
\Bsubseteq\Bvalue{\tau_i\le\tau}.\cr}$$

\noindent As $\sup_{j\le m}\Bvalue{\tau=\sigma_j}=1$,
$\tau_i\le\tau$.

\medskip

\quad{\bf (vi)} If $a\in\frak A\setminus\{0\}$ there is a $j\le m$ such that
$a'=a\Bcap c_{i\sigma_j}\Bsetminus\sup_{k<j}c_{i\sigma_k}$ is non-zero.
Now $a'\Bsubseteq c_{i\sigma_j}$, so there is a
$\sigma\in D_i$ such that $a''=a'\Bcap\Bvalue{\sigma\le\sigma_j}$ is non-zero,
while

\Centerline{$a''
\Bsubseteq a\Bcap\Bvalue{\sigma\le\sigma_j}\Bcap\Bvalue{\sigma_j=\tau_i}
\Bsubseteq a\Bcap\Bvalue{\sigma\le\tau_i}
\Bsubseteq a\Bcap c_{i\tau_i}$.}

\noindent As $a$ is arbitrary, $c_{i\tau_i}=1$.

\medskip
}%end of former 615Mg

{\bf (g)(i)} For $n\in\Bbb N$ set
$x_n=\sum_{i=0}^{n-1}|y_{i+1}-y_i|+\sup_{i<n}|u_{\max\Cal S}-y_i|$, and write
$x$ for $x_0\times\chi(1\Bsetminus d_0)
   \vee\sup_{n\in\Bbb N}x_{n+1}\times\chi(d_n\Bsetminus d_{n+1})$;  this is
defined because $\sequencen{d_n}$ is non-increasing (c-i), so
$\sequencen{d_n\Bsetminus d_{n+1}}$ is disjoint.
Take $\sigma_0\le\ldots\le\sigma_m$ in $\Cal S$ such that $\sigma_m=\max\Cal S$.

\medskip

\quad{\bf (ii)} Let $n\in\Bbb N$.   Then

\Centerline{$\Bvalue{1\Bsetminus d_n}
\Bsubseteq\Bvalue{\sum_{j=0}^{m-1}|\tilde u_{\sigma_{j+1}}-\tilde u_{\sigma_j}|
              \le x_n}$.}

\noindent\Prf\ Let $\frak B$ be the (finite) subalgebra of $\frak A$ generated
by

\Centerline{$\{c_{i\sigma_j}:i\le n$, $j\le m\}
   \cup\{\Bvalue{\sigma_j<\sigma_{j+1}}:j<m\}\cup\{d_n\}$,}

\noindent and $b$ an atom of $\frak B$ disjoint from $d_n$.   For
$j<m$ either

\Centerline{$b\Bsubseteq\Bvalue{\sigma_j=\max\Cal S}
\Bsubseteq\Bvalue{\tilde u_{\sigma_j}=\tilde u_{\sigma_{j+1}}}$}

\noindent or there is just one $i<n$ such that
$b\Bsubseteq c_{i\sigma_j}\Bsetminus c_{i+1,\sigma_j}$.
So if we set

$$\eqalign{K&
=\{j:j<m,\,b\Bcap\Bvalue{\tilde u_{\sigma_j}\ne\tilde u_{\sigma_{j+1}}}\ne 0\}
  \cr&
=\{j:j<m,\,b\Bcap\Bvalue{\sigma_j<\sigma_{j+1}}\ne 0\}
=\{j:j<m,\,b\Bsubseteq\Bvalue{\sigma_j<\sigma_{j+1}}\}\cr}$$

\noindent there is for each $j\in K$ an $i_j<n$ such that

\Centerline{$b\Bsubseteq c_{i_j\sigma_j}\Bsetminus c_{i_j+1,\sigma_j}
\Bsubseteq\Bvalue{\tilde u_{\sigma_j}=y_{i_j}}$}

\noindent and we must have $i_j<i_k$ whenever $j$, $k\in K$ and $j<k$.
For $j\in K$ let $j'$ be the next member of $K\cup\{m\}$ above
$j$;  then $i_{j'}>i_j$ for every $j\in K\setminus\{\max K\}$, so

$$\eqalign{\sum_{j=0}^{m-1}
   |\tilde u_{\sigma_{j+1}}-\tilde u_{\sigma_j}|\times\chi b&
=\sum_{j\in K}|\tilde u_{\sigma_{j+1}}-\tilde u_{\sigma_j}|\times\chi b
=\sum_{j\in K}|\tilde u_{\sigma_{j'}}-\tilde u_{\sigma_j}|\times\chi b\cr&
\le\sum_{j\in K\setminus\{\max K\}}|y_{i_{j'}}-y_{i_j}|
     +|u_{\max\Cal S}-y_{i_{\max K}}|
\le x_n.\cr}$$

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

\medskip

\quad{\bf (iii)} Writing $w$ for $\sum_{j=0}^{m-1}|\tilde u_{j+1}-\tilde u_j|$,
we have

$$\eqalignno{w&
=w\times\chi(1\Bsetminus d_0)
   \vee\sup_{n\in\Bbb N}w\times\chi(d_n\Bsetminus d_{n+1})
\Displaycause{because $\sequencen{d_n}$ is non-increasing and
$\inf_{n\in\Bbb N}d_n=0$, by 615Mc}
\le x_0\times\chi(1\Bsetminus d_0)
   \vee\sup_{n\in\Bbb N}x_{n+1}\times\chi(d_n\Bsetminus d_{n+1})
=x.\cr}$$

\noindent As $\sigma_0,\ldots,\sigma_m$ are arbitrary, $\tilde u$ is of bounded
variation.
}%end of proof of 615M

\leader{615N}{Theorem} Let $\Cal S$ be a finitely full sublattice of $\Cal T$,
and $\pmb u=\sfamilysS{u}$ a fully adapted process.   Then the following
are equiveridical:

(i) $\pmb u$ is moderately oscillatory;

(ii) $\pmb u$ is \updowncgt;

\vskip-2.5pt

(iii) $\sequencen{u_{\sigma_n}}$ is Cauchy for every monotonic sequence
$\sequencen{\sigma_n}$ in $\Cal S$;

(iv) for every $\epsilon>0$ there is an $m\ge 1$ such that whenever
$\sigma_0\le\ldots\le\sigma_m$ in $\Cal S$ there is a $j<m$ such that
$\theta(u_{\sigma_j}-u_{\sigma_{j+1}})\le\epsilon$.

\proof{{\bf (i)$\Rightarrow$(iv)} Suppose that $\pmb u$ is moderately
oscillatory.   Take $\epsilon>0$ and set $\eta=\bover{\epsilon}4$.
Let $\pmb v=\sfamilysS{v}$ be a process of
bounded variation such that $\theta(\sup|\pmb u-\pmb v|)\le\eta$.
Write $\bar v$ for $\int_{\Cal S}|d\pmb v|$, and let $\gamma\ge 0$ be
such that $\bar\mu(\bar v>\gamma)\le\eta$.   Set
$m=\lceil\gamma/\eta\rceil$.   If $\sigma_0\le\ldots\le\sigma_m$ in
$\Cal S$, then $\sum_{i=0}^{m-1}|v_{\sigma_{i+1}}-v_{\sigma_i}|\le\bar v$
(614I).   If $b=\Bvalue{\bar v\le\gamma}$,
$\sum_{i=0}^{m-1}\Expn(|v_{\sigma_{i+1}}-v_{\sigma_i}|\times\chi b)\le\gamma$
so there is a $j<m$ such that

\Centerline{$\Bover{\gamma}m\ge\Expn(|v_{\sigma_{j+1}}-v_{\sigma_j}|\times\chi b)
\ge\theta(|v_{\sigma_{j+1}}-v_{\sigma_j}|\times\chi b)$.}

\noindent Now

$$\eqalign{\theta(|u_{\sigma_{j+1}}-u_{\sigma_j}|)&
\le\theta(|v_{\sigma_{j+1}}-v_{\sigma_j}|)+2\theta(\sup|\pmb u-\pmb v|)\cr&
\le\theta(|v_{\sigma_{j+1}}-v_{\sigma_j}|\times\chi b)+\bar\mu(1\Bsetminus b)
  +2\eta
\le\Bover{\gamma}{m}+3\eta
\le 4\eta
=\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, (ii) is true.

\medskip

{\bf (iv)$\Rightarrow$(iii)} Suppose that (iv) is true.
If $\sequencen{\tau_n}$ is a
sequence in $\Cal S$ such that $\sequencen{u_{\tau_n}}$ is not
Cauchy, there are an $\epsilon>0$ and a strictly
increasing sequence $\sequence{k}{n_k}$ in $\Bbb N$ such that
$\theta(\tau_{n_{k+1}}-\tau_{n_k})>\epsilon$ for every $k\in\Bbb N$.
By (iv), we have an $m\ge 1$ such that whenever
$\sigma_0\le\ldots\le\sigma_m$ in $\Cal S$ there is a $j<m$ such that
$\theta(u_{\sigma_j}-u_{\sigma_{j+1}})\le\epsilon$.   So we cannot have either
$\tau_{n_0}\le\ldots\le\tau_{n_m}$ or $\tau_{n_m}\le\ldots\le\tau_{n_0}$, and
$\sequencen{\tau_n}$ is not monotonic.   Turning this round,
$\sequencen{u_{\sigma_n}}$ is convergent whenever $\sequencen{\sigma_n}$ is a
monotonic sequence in $\Cal S$, and $\pmb u$ is \updowncgt.

\medskip

{\bf (iii)$\Rightarrow$(ii)} because $L^0$ is a complete linear topological
space.

\medskip

{\bf (ii)$\Rightarrow$(i)}\grheada\ To begin with, suppose that $\Cal S$ has a
greatest member.   Then 615M tells us that for every $\delta>0$ there is a
process $\tildepmb u$ such that $\sup|\pmb u-\tildepmb u|\le\delta\chi 1$, while
$\tildepmb u$ is of bounded variation (615Mg).   So $\pmb u$ is moderately
oscillatory.

\medskip

\qquad\grheadb\ In general, the result is trivial if $\Cal S$ is empty, so
suppose otherwise.   Given $\epsilon>0$, there is a $\tau\in\Cal S$ such
that $\theta(u_{\sigma}-u_{\tau})\le\epsilon$ for every
$\sigma\in\Cal S\vee\tau$.   \Prf\Quer\ Otherwise, starting from any
$\tau_0\in\Cal S$, we can choose a non-decreasing sequence $\sequencen{\tau_n}$
in $\Cal S$ such that $\theta(u_{\tau_{n+1}}-u_{\tau_n})>\epsilon$ for every
$n\in\Bbb N$, in which case $\sequencen{u_{\tau_n}}$ cannot be convergent.\
\Bang\Qed\   Now consider
$\pmb u'=\family{\sigma}{\Cal S}{u_{\sigma\wedge\tau}}$.   This is fully adapted
(612Ib).   Since $\pmb u$ is order-bounded (615K), so are $\pmb u'$ and
$\pmb u-\pmb u'$, while
$\pmb u'\restr\Cal S\wedge\tau=\pmb u\restr\Cal S\wedge\tau$ is \updowncgt\
therefore moderately oscillatory, by ($\alpha$) just above, and
$\pmb u'\restr\Cal S\vee\tau$ is constant, so also moderately oscillatory.
By 615F(a-v), $\pmb u'$ is moderately oscillatory.   Now the choice of $\tau$
ensures that

$$\eqalignno{\theta(u_{\sigma}-u_{\sigma\wedge\tau})&
=\theta(u_{\sigma\vee\tau}-u_{\tau})
\Displaycause{612D(f-i)}
\le\epsilon\cr}$$

\noindent for every $\sigma\in\Cal S$,   As $\Cal S$ is finitely full,
$\theta(\sup|\pmb u-\pmb u'|)\le 2\sqrt\epsilon$ (615Db).   But
$\pmb u'\in\Mmo(\Cal S)$ and $\epsilon$ was arbtirary, so $\pmb u$ belongs to
the closure of $\Mmo(\Cal S)$ for the ucp topology on $\Mob(\Cal S)$, and is
itself moderately oscillatory (615F(a-iv)).
}%end of proof of 615N

\leader{615O}{Proposition} Suppose that $\Cal S$ is a sublattice of $\Cal T$,
$\pmb u\in\Mmo(\Cal S)$ and $\epsilon>0$.   Then there is a
$\pmb u'\in\Mbv(\Cal S)$ such that $\theta(\sup|\pmb u'-\pmb u|)\le\epsilon$ and
$\sup|\pmb u'|\le\sup|\pmb u|$.

\proof{{\bf (a)} If $\Cal S$ is full and has a greatest member, this follows
immediately from 615M.   We know that $\pmb u$ is \updowncgt;  taking
$\delta=\epsilon$ in 615M, we get a process $\tildepmb u$ of bounded variation
which will serve for $\pmb u'$, by 615Mf and 615Mg.

\medskip

{\bf (b)} If we just know that $\Cal S$ has a greatest member, consider the
fully adapted extension $\hatpmb u$ of $\pmb u$ to the covered envelope
$\hat\Cal S$ of $\Cal S$.   By 615F(a-vi), $\hatpmb u$ is moderately
oscillatory;  by (a) here, there is a process $\hatpmb u'\in\Mbv(\hat\Cal S)$
such that $\theta(\sup|\hatpmb u'-\hatpmb u|)\le\epsilon$ and
$\sup|\hatpmb u'|\le\sup|\hatpmb u|$;  now $\pmb u'=\hatpmb u'\restr\Cal S$ is
of bounded variation (614L(b-i)),
$\sup|\pmb u'-\pmb u|\le\sup|\hatpmb u'-\hatpmb u|$ so
$\theta(\sup|\pmb u'-\pmb u|)
\le\theta(\sup|\hatpmb u'-\hatpmb u|)\le\epsilon$;  and

\Centerline{$\sup|\pmb u'|\le\sup|\hatpmb u'|\le\sup|\hatpmb u|=\sup|\pmb u|$}

\noindent by 614Ga.

\medskip

{\bf (c)} For the general case, if $\Cal S$ is empty the result is trivial.
Otherwise, express $\pmb u$ as $\sfamilysS{u}$.   By 615Ga,
$u_{\uparrow}=\lim_{\sigma\uparrow\Cal S}u_{\sigma}$ is defined, and
there is a $\tau\in\Cal S$ such that
$\theta(\sup_{\sigma\in\Cal S\vee\tau}|u_{\sigma}-u_{\uparrow}|)
\le\bover14\epsilon$;  now
$\theta(\sup_{\sigma\in\Cal S\vee\tau}|u_{\sigma}-u_{\tau}|)
\le\bover12\epsilon$.   Of course $\pmb u\restr\Cal S\wedge\tau$ is moderately
oscillatory (615F(a-i));  by (b), we have a process
$\pmb v=\langle v_{\sigma}\rangle_{\sigma\in\Cal S\wedge\tau}$ of bounded
variation such that
$\theta(\sup_{\sigma\in\Cal S\wedge\tau}|v_{\sigma}-u_{\sigma}|)
\le\bover14\epsilon$ and $\sup|\pmb v|\le\sup|\pmb u\restr\Cal S\wedge\tau$.
It follows that
$\theta(\sup_{\sigma\in\Cal S\vee\tau}|u_{\sigma}-v_{\tau}|\le\bover34\epsilon$.

Write $\Cal S'$ for $(\Cal S\wedge\tau)\cup(\Cal S\vee\tau)$.   Since
$\sigma\le\sigma'$ whenever $\sigma\in\Cal S\wedge\tau$ and
$\sigma'\in\Cal S\vee\tau$, $\Cal S'$ is a sublattice of $\Cal S$.   If
$\sigma\in\Cal S$, then

\Centerline{$\Bvalue{\sigma=\sigma\wedge\tau}\Bcup\Bvalue{\sigma=\sigma\vee\tau}
\Bsupseteq\Bvalue{\sigma\le\tau}\Bcup\Bvalue{\tau\le\sigma}
=1$,}

\noindent so $\Cal S'$ covers $\Cal S$.   Because
$(\Cal S\wedge\tau)\cap(\Cal S\vee\tau)=\{\tau\}$, we have a family
$\langle u'_{\sigma}\rangle_{\sigma\in\Cal S'}$ defined by setting
$u'_{\sigma}=v_{\sigma}$ for $\sigma\in\Cal S\wedge\tau$ and
$u'_{\sigma}=v_{\tau}$ for $\sigma\in\Cal S\vee\tau$.
If $\sigma\in\Cal S\wedge\tau$ and $\sigma'\in\Cal S\vee\tau$, then
$\sigma\le\tau\le\sigma'$ so

\Centerline{$\Bvalue{\sigma=\sigma'}
=\Bvalue{\sigma=\tau}\Bcap\Bvalue{\tau=\sigma'}
\Bsubseteq\Bvalue{u'_{\sigma}=u'_{tau}}\Bcap\Bvalue{u'_{\tau}=u'_{\sigma'}}$;}

\noindent now it is easy to check that
$\langle u'_{\sigma}\rangle_{\sigma\in\Cal S'}$ is fully adapted, and therefore
has a unique fully adapted extension $\pmb u'=\sfamilysS{u'}$ (612R).

Now $\pmb u'\restr\Cal S\wedge\tau=\pmb v$ is of bounded variation,
$\pmb u'\restr\Cal S\vee\tau$ is constant, and

$$\eqalignno{\sup|\pmb u'-\pmb u|&
=\sup_{\sigma\in\Cal S'}|u'_{\sigma}-u_{\sigma}|
\Displaycause{614Ga}
=\sup_{\sigma\in\Cal S\wedge\tau}|v_{\sigma}-u_{\sigma}|
   \vee\sup_{\sigma\in\Cal S\vee\tau}|u_{\sigma}-v_{\tau}|.\cr}$$

\noindent Accordingly $\pmb u'$ is of bounded variation (614Lc),
$\theta(\sup|\pmb u'-\pmb u|)\le\bover34\epsilon+\bover14\epsilon
=\epsilon$, and

\Centerline{$\sup|\pmb u'|=\sup|\pmb v|\le\sup|\pmb u|$,}

\noindent as required.
}%end of proof of 615O

\leader{615P}{Where moderately oscillatory processes come from} There is an
easy condition on the structure in 612H which will ensure that the process
generated there is moderately oscillatory.

\medskip

\noindent{\bf Proposition} Let $(\Omega,\Sigma,\mu)$ be a complete
probability space, and $\langle\Sigma_t\rangle_{t\ge 0}$ a filtration of
$\sigma$-subalgebras of $\Sigma$ such that
every $\mu$-negligible set belongs to every $\Sigma_t$.
Let $(\frak A,\bar\mu)$ be the measure algebra of $\mu$ and set
$\frak A_t=\{E^{\ssbullet}:E\in\Sigma_t\}$ for each $t\ge 0$;  then we have
a real-time stochastic integration structure
$(\frak A,\bar\mu,\coint{0,\infty},\langle\frak A_r\rangle_{t\ge 0},
\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$.
Let $\langle X_t\rangle_{t\ge 0}$ be a progressively
measurable process on $\Omega$, and $\pmb x=\family{\tau}{\Cal T_f}{x_{\tau}}$
the corresponding fully adapted process as described in 612H.   Suppose that
$\lim_{n\to\infty}X_{t_n}(\omega)$ is defined in $\Bbb R$ for every
bounded monotonic sequence $\sequencen{t_n}$ in $\coint{0,\infty}$ and every
$\omega\in\Omega$.
Then $\pmb x$ is locally moderately oscillatory.

\proof{ Take any $\sigma\in\Cal T_f$.

\medskip

{\bf (a)} We have a stopping time $h:\Omega\to\coint{0,\infty}$ adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$ which represents $\sigma$ in the sense that
$\Bvalue{\sigma>t}=\{\omega:h(\omega)>t\}^{\ssbullet}$ for every $t\ge 0$
(612H(a-v)).
Now $\{X_t(\omega):t\le h(\omega)\}$ is bounded for every $\omega\in\Omega$.
\Prf\Quer\ Otherwise, there is a sequence $\sequencen{t_n}$ in
$[0,h(\omega)]$ such that $\lim_{n\to\infty}|X_{t_n}(\omega)|=\infty$.
Setting $s_0=\sup\{s:s\ge 0$, $\{n:t_n\le s\}$ is finite$\}$, there is a a
monotonic subsequence $\sequencen{t'_n}$ of $\sequencen{t_n}$ converging to
$s$, and $\sequencen{X_{t'_n}(\omega)}$ has no limit in $\Bbb R$.\ \Bang\Qed

\medskip

{\bf (b)} Next, setting $f(\omega)=\sup_{t\le h(\omega)}|X_t(\omega)|$ for
$\omega\in\Omega$, $f:\Omega\to\Bbb R$ is measurable.   \Prf\
For any $\alpha\in\Bbb R$ and $n\in\Bbb N$,

\Centerline{$\{(t,\omega):t\le n$, $|X_t(\omega)|>\alpha\}
\in\Cal B([0,n])\tensorhat\Sigma_n
\subseteq\Cal B(\coint{0,\infty})\tensorhat\Sigma$}

\noindent where $\Cal B([0,n])$ is the Borel $\sigma$-algebra of $[0,n]$
and $\Cal B(\coint{0,\infty})$ is the Borel $\sigma$-algebra of
$\coint{0,\infty}$.   Because $h$ is $\Sigma$-measurable,

\Centerline{$\{(t,\omega):t\le h(\omega)\}
=\bigcup_{q\in\Bbb Q}[0,q]\times\{\omega:q\le h(\omega)\}
\in\Cal B(\coint{0,\infty})\tensorhat\Sigma$}

\noindent and

$$\eqalign{W=&
\{(t,\omega):t\le h(\omega),\,|X_t(\omega)|>\alpha\}\cr&
=\bigcup_{n\in\Bbb N}\{(t,\omega):t\le h(\omega)\}
   \cap\{(t,\omega):t\le n,\,|X_t(\omega)|>\alpha\}\cr}$$

\noindent belongs to $\Cal B(\coint{0,\infty})\tensorhat\Sigma$.
Since $\mu$ is complete, $\Sigma$ is closed under
Souslin's operation (431A) and contains the projection of $W$ onto
$\Omega$ (423O\Latereditions).   But this is just $\{\omega:f(\omega)>\alpha\}$.
As $\alpha$ is arbitrary, $f$ is $\Sigma$-measurable.\ \Qed

It follows that $\pmb x\restr\Cal T\wedge\sigma$ is order-bounded.   \Prf\
If $\tau\in\Cal T$ and $\tau\le\sigma$, there is a stopping time
$g:\Omega\to\coint{0,\infty}$ representing $\tau$ (612H(a-v) again) and
$g\leae h$ (612H(a-iv)), so $\min(g,h)$ still represents $\tau$.   Since
$|X_{\min(g(\omega),h(\omega))}(\omega)|\le f(\omega)$ for every
$\omega\in\Omega$, $x_{\tau}\le f^{\ssbullet}$ in $L^0(\frak A)$.
Thus $\sup|\pmb x\restr\Cal T\wedge\sigma|$ is defined and at most
$f^{\ssbullet}$.\ \Qed

\medskip

{\bf (c)} Suppose that $\sequencen{\tau_n}$ is a non-decreasing sequence in
$\Cal T\wedge\sigma$.   Choose $g_n:\Omega\to\coint{0,\infty}$ representing
$\tau_n$ for each $n$, and set $g_n'=\min(h,\sup_{i\le n}g_i)$ for $n\in\Bbb N$;
then $g'_n$ represents $\tau_n$ for each $n$, and $\sequencen{g'_n(\omega)}$ is
a non-decreasing sequence in $[0,h(\omega)]$ for each $\omega\in\Omega$, so
$g(\omega)=\lim_{n\to\infty}X_{g'_n(\omega)}(\omega)$ is defined in $\Bbb R$ for
every $\omega\in\Omega$.   But now $g^{\ssbullet}=\lim_{n\to\infty}x_{\tau_n}$
in $L^0(\frak A)$ (245Ca).   So $\lim_{n\to\infty}x_{\tau_n}$ is defined.

Similarly, if $\sequencen{\tau_n}$ is a non-increasing sequence in
$\Cal T\wedge\sigma$, $g_n:\Omega\to\coint{0,\infty}$ represents
$\tau_n$ for each $n$, and $g_n'=\min(h,\inf_{i\le n}g_i)$ for $n\in\Bbb N$,
then $g'_n$ represents $\tau_n$ for each $n$, $\sequencen{g'_n(\omega)}$ is
a non-increasing sequence in $[0,h(\omega)]$ for each $\omega\in\Omega$, and
$g(\omega)=\lim_{n\to\infty}X_{g'_n(\omega)}(\omega)$ is defined in $\Bbb R$ for
every $\omega\in\Omega$.   Once again,
$\lim_{n\to\infty}x_{\tau_n}$ is defined and equal to $g^{\ssbullet}$.

Putting these together with (b), $\pmb x\restr\Cal T\wedge\sigma$ is
\updowncgt, therefore moderately oscillatory, since $\Cal T\wedge\sigma$ is
(finitely) full;  as $\sigma$ is arbitrary,
$\pmb x$ is locally moderately oscillatory on $\Cal T_f$.
}%end of proof of 615P


\leader{615Q}{Proposition} The identity process\cmmnt{ (as described in 612F)},
Brownian motion\cmmnt{ (612T)} and the Poisson process\cmmnt{ (612U)}
are all locally moderately oscillatory.

\proof{ In the case of the identity process
$\pmb\iota=\family{\tau}{\Cal T_f}{\iota_{\tau}}$ and the Poisson process
$\pmb v=\family{\tau}{\Cal T_f}{v_{\tau}}$ it is enough to know that they
start at
$\tbf 0$ and are non-decreasing, so that if $\tau\in\Cal T_f$ and
$\sequencen{\sigma_n}$ is a monotonic sequence in $\Cal T_f\wedge\tau$,
$\sequencen{\iota_{\sigma_n}}$ and $\sequencen{v_{\sigma_n}}$ are monotonic
sequences in $[0,\iota_{\tau}]$ and $[0,v_{\tau}]$ respectively, and must be
convergent in $L^0$ (613Ba).   (Of course $\Cal T_f\wedge\tau$ is full, so we
can work directly from 614O.)   As for Brownian motion, I based this on
$\Omega=C(\coint{0,\infty})_0$ and
$X_t(\omega)=\omega(t)$ for $t\ge 0$ and $\omega\in\Omega$.   So the
sample paths $t\mapsto X_t(\omega)$ are continuous, and
615P gives the result at once.
}%end of proof of 615Q

\leader{615R}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$ a process of bounded variation with domain $\Cal S$.

(a) If $\pmb v\in\Mmo=\Mmo(\Cal S)$ then $\int_{\Cal S}\pmb u\,d\pmb v$ is
defined and $|\int_{\Cal S}\pmb u\,d\pmb v|
\le(\int_{\Cal S}|d\pmb u|+2\sup|\pmb u|)\times\sup|\pmb v|$.

(b) $\iivu\in\Mmo$ for every $\pmb v\in\Mmo$, and
$\pmb v\mapsto\iivu:\Mmo\to\Mmo$ is continuous for the ucp topology on $\Mmo$.

\proof{ Write $w$ for $\int_{\Cal S}|d\pmb u|+2\sup|\pmb u|$.

\medskip

{\bf (a)} Let $\epsilon>0$.    Then there is a $\delta>0$ such that
$\theta(x\times w)\le\epsilon$ whenever $\theta(x)\le\delta$.   Next, there is a
process $\pmb v'$ of bounded variation such that $\theta(\sup|v-v'|)\le\delta$.
By 614T, $\int_{\Cal S}\pmb u\,d\pmb v'$ is defined.   Let
$I\in\Cal I(\Cal S)$ be such that
$\theta(S_I(\pmb u,d\pmb v')-S_J(\pmb u,d\pmb v'))\le\epsilon$ whenever
$I\subseteq J\in\Cal I(\Cal S)$.   Now for any $J\in\Cal I(\Cal S)$

$$\eqalignno{|S_J(\pmb u,d\pmb v)-S_J(\pmb u,d\pmb v'))|&
=|S_J(\pmb u,d(\pmb v-\pmb v'))|\cr&
\le\sup(|\pmb v-\pmb v'|\restr J)\times(\int_J|d\pmb u|+2\sup(|\pmb u\restr J|))
\Displaycause{614S}
\le\sup(|\pmb v-\pmb v'|)\times(\int_{\Cal S}|d\pmb u|+2\sup(|\pmb u|))
\Displaycause{using 614Lb}
=\sup(|\pmb v-\pmb v'|\times w),\cr}$$

\noindent so

\Centerline{$\theta(S_J(\pmb u,d\pmb v)-S_J(\pmb u,d\pmb v'))\le\epsilon$.}

\noindent By the choice of $I$,
$\theta(S_I(\pmb u,d\pmb v)-S_J(\pmb u,d\pmb v))\le 3\epsilon$ whenever
$I\subseteq J\in\Cal I(\Cal S)$.   As $\epsilon$ is arbitrary,
$\int_{\Cal S}\pmb u\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb v)$ is defined.

At the same time, we see that
$|S_J(\pmb u,d\pmb v)|\le\sup|\pmb v|\times w$ for every $J\in\Cal I(\Cal S)$,
so in the limit we have
$|\int_{\Cal S}\pmb u\,d\pmb v|\le\sup|\pmb v|\times w$.

\medskip

{\bf (b)} If $\pmb v\in\Mmo$ then $\int_{\Cal S}\pmb u\,d\pmb v$ is defined,
by (a), so $\iivu$ is defined everywhere on $\Cal S$.
Applying (a) to $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb v\restr\Cal S\wedge\tau$ for $\tau\in\Cal S$, we see that

$$\eqalign{\sup|\iivu|&
=\sup_{\tau\in\Cal S}|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v|\cr&
\le\sup_{\tau\in\Cal S}
\bigl((\int_{\Cal S\wedge\tau}|d\pmb u|+2\sup|\pmb u\restr\Cal S\wedge\tau|)
  \times\sup|\pmb v\restr\Cal S\wedge\tau|\bigr)\cr&
\le(\int_{\Cal S}|d\pmb u|+2\sup|\pmb u|)\times\sup|\pmb v|,\cr}$$

\noindent so $\iivu$ is order-bounded.   The same formula shows that
$\pmb v\mapsto\iivu:\Mmo\to\Mob=\Mob(\Cal S)$ is continuous for the ucp
topologies on $\Mmo$ and $\Mob$.   We know that $\iivu\in\Mbv$ whenever
$\pmb v\in\Mbv$  (614U), so $\iivu\in\overline{\Mbv}=\Mmo$ whenever
$\pmb v\in\Mmo$, and $\pmb v\mapsto\iivu:\Mmo\to\Mmo$ is continuous.
}%end of proof of 615R

%\leader{}{615S}
%\leader{}{615T}
%\leader{}{615U}
%\leader{}{615V}
%\leader{}{615W}

\exercises{\leader{615X}{Basic exercises (a)}
%\spheader 615Xa
Let $\bar\nu:\frak A\to[0,1]$ be any functional such that
$(\frak A,\bar\nu)$ is a probability algebra.   Show that the ucp topology and
uniformity on $\Mob(\Cal S)$
defined from the associated F-norm $\theta_{\bar\nu}$ (613Bg) are the same as
those defined from $\bar\mu$ and $\theta=\theta_{\bar\mu}$.

\spheader 615Xb
Let $\Cal S$ be a sublattice of $\Cal T$.   Write
$\Mlob=\Mlob(\Cal S)$ for the space of locally order-bounded fully adapted
processes with domain $\Cal S$ (615Fb).   For $\tau\in\Cal S$ and
$\pmb u\in\Mlob$ set $\widehat{\theta}_{\tau}(\pmb u)
=\theta(\sup_{\sigma\in\Cal S\wedge\tau}|u_{\sigma}|)$.   (i) Show that
$\{\widehat{\theta}_{\tau}:\tau\in\Cal S\}$
defines a complete Hausdorff linear space topology
on $\Mlob$ for which multiplication and
the operations $\vee$, $\wedge$ are continuous;  I will call this the {\bf
local ucp topology} on $\Mlob$.   (ii) Show that $\Mlmo(\Cal S)$ is closed for
the local ucp topology on $\Mlob$.
%615B

\spheader 615Xc Let $\Cal S$ be a sublattice of $\Cal T$ and $z$ a member of
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.   Show that, in the
language of 612D(e-ii), $\pmb u\mapsto z\pmb u:
\Mob(\Cal S)\to\Mob(\Cal S)$ is continuous for the ucp topology.
%615C

\spheader 615Xd Let $\Cal S$ be a finitely full sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process such that
$\{u_{\sigma}:\sigma\in\Cal S\}$ is topologically bounded in the sense of
613Bf.   Show that $\pmb u$ is order-bounded.
%615D

\spheader 615Xe Give examples of (i) an \updowncgt\
fully adapted process which is not order-bounded
(ii) an \updowncgt\ order-bounded
process which is not moderately oscillatory.
\Hint{$\Cal S=\{\check n:n\in\Bbb N\}$.}
%615K

\spheader 615Xf Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W.   Write $\tildeClll$ for the set of functions
$f:\coint{0,\infty}\to\Bbb R$ such that $\lim_{s\downarrow t}f(s)$ is
defined for every $t\ge 0$ and $\lim_{s\uparrow t}f(s)$ is defined for every
$t>0$ (cf.\ 438P-438Q).   Let $f:\coint{0,\infty}\to\Bbb R$ be a function and
$\pmb x\in\Mfa(\Cal T_f)$ the corresponding process.
(i) Show that the ucp topology on $\Mob(\Cal T_f)$ corresponds to the norm
topology on $\ell^{\infty}(\coint{0,\infty})$.
(ii) Show that
$\pmb x$ is locally moderately oscillatory iff $f\in\tildeClll$, and moderately
oscillatory iff $f\in\tildeClll$ and $\lim_{t\to\infty}f(t)$ is defined in
$\Bbb R$.
%613W 615N

\spheader 615Xg Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$ a process of bounded variation, and $\pmb v$ a moderately oscillatory
process, both with domain $\Cal S$.   Show that $\iivu$ is moderately
oscillatory.
%615R

\spheader 615Xh Let $\Cal S$ be a finitely full sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process such that
$\{u_{\sigma}:\sigma\in\Cal S\}$ is topologically bounded (613Bf).
Show that $\pmb u$ is order-bounded.
%615D out of order query

\leader{615Y}{Further exercises (a)}
%\spheader 615Ya
Let $\Cal S$ be a sublattice of $\Cal T$.   Show that the
ucp topology on the Riesz space
$\Mob(\Cal S)$ is {\bf locally solid} in the sense that for every
neighbourhood $G$ of $0$ there is a solid neighbourhood $H$ of $0$ included
in $G$.
%615B

\spheader 615Yb Let $\Cal S$ be a sublattice of $\Cal T$.
For $\pmb u$, $\pmb v\in\Mfa(\Cal S)$ set
$\rho(\pmb u,\pmb v)=\min(1,\|\pmb u-\pmb v\|_{\infty})$ (612Sa).
Show that $\rho$ is a metric on $\Mfa(\Cal S)$ under which $\Mfa(\Cal S)$
is complete and addition is continuous.   Show that if $\Cal S$ is finitely full
then $\Mmo(\Cal S)$ is the closure of $\Mbv(\Cal S)$ for the topology defined by
$\rho$.
%615Xd 615D 615N

\spheader 615Yc In the construction described in 613S, show that the processes
$\pmb u$ and $\pmb v$ are moderately oscillatory.
%613S 615P
}%end of exercises

\endnotes{
\Notesheader {615}
Note that the  concepts of `ucp topology', `ucp uniformity' and
`moderately oscillatory process', like that
of `Riemann-sum integral' (613I), depend on the structure
$(\frak A,L^0,\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$
and on the topology of convergence in measure on $L^0$, but
otherwise are independent of the measure $\bar\mu$ (615Xa).

So many processes turn out to be locally moderately oscillatory
that we can quite happily regard them as the norm, at
least to begin with.   In Theorem 455G I showed that the most important
real-valued Markov processes on $\coint{0,\infty}$ have
representations in which all paths are \cadlag\ functions, and
615P shows that these correspond to
locally moderately oscillatory processes.   There are enough examples
(see 612T-612U) and preservation results (e.g., 615F)
to ensure that we can hope to remain in this territory for a long time.

The words `local' and `locally' are going to appear repeatedly in this
volume.   The point is that basic theorems are often most simply
expressed in terms of `global' concepts,
as in 614F, 614G, 614Q, 615C, 615Fa, 615G and 616R.
But the most important applications tend
to present themselves in `local' terms;  thus Brownian motion is locally
order-bounded (614H) and the Poisson process is locally of bounded variation
(614M).

From Theorem 615N we see that, at least if the sublattice $\Cal S$ is finitely
full, we have two possible definitions of `moderately oscillatory process';  one
in terms of approximation by processes of bounded variation, as in 615E, and one
in terms of a kind of sequential convergence, as in 615I.   The former is easier
to handle (compare 615G and 615L), but the latter gives a path to a fundamental
fact (616Ib).
In particular, 615M can be regarded as a method of constructing a process
$\tildepmb u$ from a given moderately oscillatory process $\pmb u$ which is not
only of bounded variation but approximates $\pmb u$ in a much finer topology
than the ucp topology (615Yb).   615O looks like an insignificant variation on
the definition in 615E, but I do not see a quicker way to it, and it will be
useful later. %645L

As in \S614, I conclude the section with a note on a special type of integral,
providing another case in which an indefinite integral inherits a property from
an integrator.   It looks esoteric, but happens to be a useful
step towards one of the principal theorems of the next section (616K).
}%end of notes

\discrpage

\frfilename{mt616.tex}
\versiondate{31.1.23/3.2.23}
\copyrightdate{2019}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Integrating interval functions}

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\newsection{616} In this section I present a fundamental theorem on the
existence of Riemann-sum integrals (616M),
dealing with the case of moderately oscillatory
integrands and integrating interval functions (616F).   The most important
integrating interval functions are those defined by integrators (616Fc, 616I).
The integrators on a lattice $\Cal S$ form an $f$-subalgebra of the
space of moderately oscillatory processes with domain $\Cal S$ (616P).

\cmmnt{\leader{616A}{Notation} \Asissix\ will be a stochastic integration
structure.   For a sublattice
$\Cal S$ of $\Cal T$, $\Cal I(\Cal S)$ is the set of finite sublattices of
$\Cal S$, and if $\tau\in\Cal T$,
$\Cal S\wedge\tau=\{\sigma\wedge\tau:\sigma\in\Cal S\}$
and $\Cal S\vee\tau=\{\sigma\vee\tau:\sigma\in\Cal S\}$.
%If $I\in\Cal I(\Cal T)$, $\Sti_0(I)$ is the set of
%$I$-cells (611Je).
If $\pmb u$ and $\pmb v$ are fully adapted processes
defined (at least) on a finite sublattice $I$ of $\Cal T$ and
$\psi$ is an adapted interval function defined (at least)
on $I^{2\uparrow}
=\{(\sigma,\tau):\sigma$, $\tau\in I$, $\sigma\le\tau\}$,
then $S_I(\pmb u,d\psi)$ and $S_I(\pmb u,d\pmb v)$ will be the
Riemann sums defined in 613Eb and 613Fb.   For $w\in L^0=L^0(\frak A)$,
$\theta(w)=\Expn(|w|\wedge\chi 1)$ as in 613Ba.   If $\Cal S$ is a
sublattice of $\Cal T$, $\Mfa(\Cal S)$,
$\Mob(\Cal S)$, $\Mlob(\Cal S)$, $\Mbv(\Cal S)$ and $\Mmo(\Cal S)$ are
the spaces of fully adapted processes, order-bounded processes,
locally order-bounded processes, processes of bounded variation
and moderately oscillatory processes with domain $\Cal S$.
For $\pmb u\in\Mob(\Cal S)$,
$\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.
% and $\widehat{\theta}(\pmb u)=\theta(\sup|\pmb u|)$.
$\tbf{1}$ will be the process with constant value $\chi 1$.
If $\pmb u=\sfamilysS{u}$ and $\pmb v=\sfamilysS{v}$ are fully adapted
processes with the same domain $\Cal S$,
$\Bvalue{\pmb u\ne\pmb v}
=\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne v_{\sigma}}$.
}

\leader{616B}{Definition} Let $\Cal S$ be a sublattice of
$\Cal T$.   If $\psi$ is an adapted interval function\cmmnt{ (613C)}
defined\cmmnt{ (at least)} on $\Cal S^{2\uparrow}$,
the {\bf capped-stake variation set of $\psi$
over $\Cal S$} is the set $Q_{\Cal S}(d\psi)$ of Riemann sums
$S_I(\pmb u,d\psi)$ where $I\in\Cal I(\Cal S)$,
$\pmb u$ is a fully adapted process with domain $I$
and $\sup|\pmb u|\le\chi 1$.

If $\pmb v$, $\pmb w$ are
fully adapted processes defined\cmmnt{ (at least)}  on $\Cal S$ then,
corresponding to the basic interval functions of 613F,
I will write $Q_{\Cal S}(d\pmb v)$, $Q_{\Cal S}(d\pmb vd\pmb w)$,
$Q_{\Cal S}(|d\pmb v|)$ for $Q_{\Cal S}(d(\Delta\pmb v))$,
$Q_{\Cal S}(d(\Delta\pmb v\times\Delta\pmb w))$ and
$Q_{\Cal S}(d|\Delta\pmb v|)$.

\leader{616C}{}\cmmnt{ The following elementary facts will be
useful.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
$\psi$ an adapted interval function defined on
$\Cal S^{2\uparrow}$, and $z$ an element of $L^0(\frak A)$.
Then the following are equiveridical:

\quad(i) $z\in Q_{\Cal S}(d\psi)$;

\quad(ii) there are
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and $u_0,\ldots,u_{n-1}$ such that
$u_i\in L^{\infty}(\frak A_{\tau_i})$ and $|u_i|\le\chi 1$ for every
$i<n$ and $z=\sum_{i=0}^{n-1}u_i\times\psi(\tau_i,\tau_{i+1})$;

\quad(iii) there are
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and an order-bounded process
$\pmb u=\sfamilysS{u}$
such that $\sup|\pmb u|\le\chi 1$ and
$z=\sum_{i=0}^{n-1}u_{\tau_i}\times\psi(\tau_i,\tau_{i+1})$.

\proof{{\bf (i)$\Rightarrow$(ii)} If $z=0$ we can take $n=0$ and any
$\tau_0\in\Cal S$.   Otherwise, let $I\in\Cal I(\Cal S)$ and
$\pmb u=\family{\sigma}{I}{u_{\sigma}}\in\Mfa(I)$ be such that
$\sup|\pmb u|\le\chi 1$ and
$z=S_I(\pmb u,d\psi)$.   Take a sequence $(\tau_0,\ldots,\tau_n)$ linearly
generating the $I$-cells.   Then

\Centerline{$z=S_I(\pmb u,d\psi)
=\sum_{i=0}^{n-1}u_{\tau_i}\times\psi(\tau_i,\tau_{i+1})$}

\noindent (613Ec), while $\tau_0\le\ldots\le\tau_n$ and
$u_{\tau_i}\in L^0(\frak A_{\tau_i})$ and $|u_{\tau_i}|\le\chi 1$
for every $i<n$.

\medskip

\quad{\bf (ii)$\Rightarrow$(iii)}
By 612Ka, there is a fully adapted process
$\pmb u'=\family{\sigma}{\Cal S}{u'_{\sigma}}$ such that, for
$\sigma\in\Cal S$,

\Centerline{$\Bvalue{u'_{\sigma}=u_i}
\Bsupseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}$,}

\noindent for every $i<n$, while

\Centerline{$\Bvalue{u'_{\sigma}=0}
\Bsupseteq\Bvalue{\sigma<\tau_0}$,
\quad$\Bvalue{u'_{\sigma}=u_{\tau_n}}\Bsupseteq\Bvalue{\tau_n\le\sigma}$.}

\noindent Observe that $|u'_{\sigma}|\le\chi 1$ for every $\sigma$
(because

\Centerline{$\Bvalue{\sigma<\tau_0}\Bcup\Bvalue{\tau_n\le\sigma}
\Bcup\sup_{i<n}\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
=1$),}

\noindent so $\pmb u$ is order-bounded and $\sup|\pmb u|\le\chi 1$.
Now if we set $I=\{\tau_0,\ldots,\tau_n\}$,

$$\eqalign{z&
=\sum_{i=0}^{n-1}u_i\times\psi(\tau_i,\tau_{i+1})
=\sum_{i=0}^{n-1}u_i\times\chi(\Bvalue{\tau_i<\tau_{i+1}})
  \times\psi(\tau_i,\tau_{i+1})\cr&
=\sum_{i=0}^{n-1}u'_{\tau_i}\times\chi(\Bvalue{\tau_i<\tau_{i+1}})
  \times\psi(\tau_i,\tau_{i+1})
=\sum_{i=0}^{n-1}u'_{\tau_i}\times\psi(\tau_i,\tau_{i+1}).\cr}$$

\medskip

\quad{\bf (iii)$\Rightarrow$(i)} Setting $I=\{\tau_0,\ldots,\tau_n\}$,
$z=S_I(\pmb u\restr I,d\psi)$.

}%end of proof of 616C

\leader{616D}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\psi$, $\psi'$ adapted interval functions defined on
$\Cal S^{2\uparrow}$.

(a) $Q_{\Cal S}(d\psi)=\bigcup_{I\in\Cal I(\Cal S)}Q_I(d\psi)$.

(b) $Q_{\Cal S}(d(\alpha\psi))=\alpha Q_{\Cal S}(d\psi)$
for every $\alpha\in\Bbb R$.

(c) $Q_{\Cal S}(d(\psi+\psi'))
\subseteq Q_{\Cal S}(d\psi)+Q_{\Cal S}(d\psi')$.

(d) If $\Cal S'$ is a sublattice of $\Cal S$ then
$Q_{\Cal S'}(d\psi)\subseteq Q_{\Cal S}(d\psi)$.

(e) If $w\in Q_{\Cal S}(d\psi)$,
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$ and
$|z|\le\chi 1$, then $z\times w\in Q_{\Cal S}(\psi)$.

(f) If $\tau\in\Cal S$ then $
Q_{\Cal S\wedge\tau}(d\psi)+Q_{\Cal S\vee\tau}(d\psi)
\subseteq Q_{\Cal S}(d\psi)$.

\proof{{\bf (a)} Immediate from the definition.

\medskip

{\bf (b)} $S_I(\pmb u,\alpha\psi)=\alpha S_I(\pmb u,d\psi)$
for all $I$ and $\pmb u$.

\medskip

{\bf (c)}
$S_I(\pmb u,d(\psi+\psi'))=S_I(\pmb u,d\psi)+S_I(\pmb u,d\psi')$ for all $I$ and
$\pmb u$.

\medskip

{\bf (d)} $\Cal I(\Cal S')\subseteq\Cal I(\Cal S)$.

\medskip

{\bf (e)} We can express $w$ as $S_I(\pmb u,d\psi)$ where
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and $\sup|\pmb u|\le\chi 1$.
Now $z\pmb u\in\Mfa(I)$ (612D(e-ii)) and
$\sup|z\pmb u|\le|z|\times\sup|\pmb u|\le\chi 1$, so
$Q_{\Cal S}(d\psi)$ contains $S_I(z\pmb u,d\psi)=z\times w$ (613L(b-iii)).

\medskip

{\bf (f)} If
$w\in Q_{\Cal S\wedge\tau}(d\psi)+Q_{\Cal S\vee\tau}(d\psi)$,
then by 616C(ii)
there are $\tau'_0\le\ldots\le\tau'_{n'}$ in $\Cal S\wedge\tau$,
$\tau''_0\le\ldots\le\tau''_{n''}$ in $\Cal S\vee\tau$,
$u'_i\in L^0(\frak A_{\tau'_i})$ such that $|u'_i|\le\chi 1$ for
$i<n'$,
$u''_j\in L^0(\frak A_{\tau'_j})$ such that $|u''_j|\le\chi 1$ for
$j<n''$. such that

\Centerline{$w=\sum_{i=0}^{n'-1}u'_i\times\psi(\tau'_i,\tau'_{i+1})
  +\sum_{j=0}^{n''-1}u''_i\times\psi(\tau''_i,\tau''_{i+1})$.}

\noindent Set $n=n'+n''+1$,

$$\eqalign{\tau_k
&=\tau'_k\text{ for }k\le n',\cr
&=\tau''_{k-n'-1}\text{ for }n'<k\le n'+n''+1,\cr
u_k
&=u'_k\text{ for }k<n',\cr
&=0\text{ for }k=n',\cr
&=u''_{k-n'-1}\text{ for }n'<k\le n'+n''.\cr}$$

\noindent Then $\tau_0\le\ldots\le\tau_n$ in $\Cal S$ while
$u_{\tau_k}\in L^0(\frak A_{\tau_k})$ and $|u_{\tau_k}|\le\chi 1$ for
$k<n$ and $w=\sum_{k=0}^{n-1}u_k\times\psi(\tau_k,\tau_{k+1})$, so
$w\in Q_{\Cal S}(d\psi)$.
}%end of proof of 616D

\leader{616E}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\psi$ an adapted interval function on $\Cal S$.
Then the following are equiveridical:

(i) $Q_{\Cal S}(d\psi)$ is topologically bounded;

(ii) for every $\epsilon>0$ there is a $\delta>0$ such that
$\theta(S_I(\pmb u,d\psi))\le\epsilon$ whenever $I\in\Cal I(\Cal S)$,
$\pmb u\in\Mfa(I)$ and $\theta(\sup|\pmb u|)\le\delta$;

(iii) for every $\epsilon>0$ there is a $\delta>0$ such that
$\theta(S_I(\pmb u,d\psi))\le\epsilon$ whenever $I\in\Cal I(\Cal S)$,
$\pmb u\in\Mob(\Cal S)$ and $\theta(\sup|\pmb u|)\le\delta$.

\proof{{\bf (i)$\Rightarrow$(ii)} Suppose that
$Q_{\Cal S}(d\psi)$ is topologically bounded
and $\epsilon>0$.   Let $\eta>0$ be such that
$\theta(\eta S_I(\pmb u,d\psi))\le\bover12\epsilon$ whenever
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and
$\sup|\pmb u|\le\chi 1$.   Set $\delta=\bover12\epsilon\eta$.
If $I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and
$\theta(\sup|\pmb u|)\le\delta$, set
$\pmb u'=\med(-\tbf{1},\bover1{\eta}\pmb u,\tbf{1})$.
Then

$$\eqalignno{\Bvalue{S_I(\pmb u,d\psi)\ne\eta S_I(\pmb u',d\psi)}
=\Bvalue{S_I(\pmb u-\eta\pmb u',d\psi)\ne 0}&
\Bsubseteq\Bvalue{\pmb u\ne\eta\pmb u'}
\Displaycause{613Gd}
=\Bvalue{\sup|\pmb u|>\eta}\cr}$$

\noindent has measure at most $\bover{\delta}{\eta}=\bover12\epsilon$.   So

\Centerline{$\theta(S_I(\pmb u,d\psi))
\le\Bover12\epsilon+\theta(\eta S_I(\pmb u',d\psi))
\le\epsilon$.}

\noindent Thus $\delta$ witnesses that (ii) is true.

\medskip

{\bf (ii)$\Rightarrow$(iii)} If (ii) is true and $\epsilon>0$, take $\delta$ as
in (ii).   If $I\in\Cal I(\Cal S)$, $\pmb u\in\Mob(\Cal S)$ and
$\theta(\sup|\pmb u|)\le\delta$, then $\pmb u\restr I\in\Mfa(I)$ and
$\sup|\pmb u\restr I|\le\sup|\pmb u|$, so
$\theta(\sup|\pmb u\restr I|)\le\delta$ and
$\theta(S_I(\pmb u,d\psi))=\theta(S_I(\pmb u\restr I,d\psi))\le\epsilon$.

\medskip

{\bf (iii)$\Rightarrow$(i)} Suppose that (iii) is true.
If $\Cal S$ is empty then
$Q_{\Cal S}(d\psi)=\{0\}$ is certainly topologically bounded.
Otherwise, given $\epsilon>0$, take $\delta$ as in
(iii).   Then for any $z\in Q_{\Cal S}(d\psi)$ there are
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and
$\pmb u=\sfamilysS{u}\in\Mob(\Cal S)$ such that
$\sup|\pmb u|\le\chi 1$ and
$z=\sum_{i=0}^{n-1}u_{\tau_i}\times\psi(\tau_i,\tau_{i+1})$ (616C(ii)).
But now $z=S_I(\pmb u,d\psi)$ where $I=\{\tau_0,\ldots,\tau_n\}$, while
$\theta(\sup|\delta\pmb u|)\le\theta(\delta\chi 1)\le\delta$ so
$\theta(\delta z)=\theta(S_I(\delta\pmb u,d\psi))\le\epsilon$.
As $\epsilon$ and $z$ are arbitrary, $Q_{\Cal S}(d\psi)$ is topologically
bounded.
}%end of proof of 616E

\leader{616F}{Definition} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\psi:\Cal S^{2\uparrow}\to L^0$ a function.

\spheader 616Fa\cmmnt{ I will say that} $\psi$ is an {\bf integrating
interval function} on $\Cal S$ if

\inset{($\alpha$) $\psi$ is a strictly adapted interval function;

($\beta$) writing $\hat\Cal S$ for the covered envelope of $\Cal S$ and
$\hat\psi:\hat\Cal S^{2\uparrow}\to L^0$ for the strictly adapted extension of
$\psi$\cmmnt{ (613U)},
$\int_{\hat\Cal S}d\hat\psi=\int_{\hat\Cal S}\tbf{1}\,d\hat\psi$
is defined\cmmnt{ in the sense of 613H};

%see 632Yg

($\gamma$)
$Q_{\hat\Cal S}(d\hat\psi)$ is topologically bounded in $L^0$.}


\spheader 616Fb $\psi$ is a {\bf locally integrating interval function}
if $\psi\restr(\Cal S\wedge\tau)^{2\uparrow}$
is an integrating interval function for every $\tau\in\Cal S$.

\spheader 616Fc A fully adapted process $\pmb v$ defined on $\Cal S$ is an
{\bf integrator} if $Q_{\Cal S}(d\pmb v)$ is topologically bounded in
$L^0$, and a {\bf local integrator} if $\pmb v\restr\Cal S\wedge\tau$ is an
integrator for every $\tau\in\Cal S$.

I will write $\Mir(\Cal S)$ for the set of integrators with domain $\Cal S$, and
$\Mlir(\Cal S)$ for the set of local integrators with domain $\Cal S$.

\medskip

\noindent{\bf Remarks} Evidently a strictly adapted interval function
$\psi$ on a sublattice $\Cal S$ is an integrating interval function iff its
adapted extension on the covered envelope of $\Cal S$ is an integrating
interval function.

\cmmnt{I have given a definition of `integrator' which does not
obviously correspond directly to the definition of `integrating interval
function'.   We shall see in 616I that in fact it matches exactly,
but it is convenient to work with the simpler formulation here for the moment.
Of course we can see already that if $\pmb v$ is fully adapted and
$\Delta\pmb v$ is a (locally) integrating interval function, then $\pmb v$ is a
(local) integrator.
}

\leader{616G}{Proposition} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\psi$, $\psi'$ integrating interval functions on $\Cal S$.

(a) $\psi+\psi'$ and $\alpha\psi$ are integrating interval functions on
$\Cal S$ for every $\alpha\in\Bbb R$.

(b) $\psi$ is a locally integrating interval function.

\proof{{\bf (a)} $\psi+\psi'$ and $\alpha\psi$ are strictly adapted interval
functions (613D) and

\Centerline{$\int_{\Cal S}d(\psi+\psi')
=\int_{\Cal S}d\psi+\int_{\Cal S}d\psi'$,
\quad$\int_{\Cal S}d(\alpha\psi)
=\alpha\int_{\Cal S}d\psi$}

\noindent are defined (613Jb).   Next, writing $\hat\psi$ and $\hat\psi'$ for
the strictly adapted extensions of $\psi$ and $\psi'$, as in 616Fa,
$\hat\psi+\hat\psi'$ is a strictly adapted interval
function extending $\psi+\psi'$, so is equal to
$(\psi+\psi')\varCsphat$, and

\Centerline{$\int_{\hat\Cal S}d((\psi+\psi')\varCsphat)
=\int_{\hat\Cal S}d(\hat\psi+\hat\psi')
=\int_{\hat\Cal S}d\hat\psi+\int_{\hat\Cal S}d\hat\psi'$}

\noindent is defined, while

\Centerline{$Q_{\hat\Cal S}(d(\psi+\psi')\varCsphat)
=Q_{\hat\Cal S}(d(\hat\psi+\hat\psi'))
\subseteq Q_{\hat\Cal S}(d\hat\psi)+Q_{\hat\Cal S}(d\hat\psi')$}

\noindent (616Dc) is topologically bounded (613B(f-iii)).   So
$\psi+\psi'$ is an integrating interval function.

Similarly,

\Centerline{$\int_{\hat\Cal S}d((\alpha\psi)\varCsphat)
=\alpha\int_{\hat\Cal S}d\hat\psi$,
\quad$Q_{\hat\Cal S}(d(\alpha\psi)\varCsphat)
=Q_{\hat\Cal S}(d(\alpha\hat\psi))
=\alpha Q_{\hat\Cal S}(d\hat\psi)$}

\noindent is topologically bounded and $\alpha\psi$ is an integrating interval
function.

\medskip

{\bf (b)} If $\tau\in\Cal S$, then the covered envelope of $\Cal S\wedge\tau$
is $\hat\Cal S\wedge\tau$ (611M(e-i)).
$\psi'=\psi\restr(\Cal S\wedge\tau)^{2\uparrow}$
is a strictly adapted interval function (613C(b-iii)) and
its strictly adapted extension $\hat\psi'$ to
$(\hat\Cal S\wedge\tau)^{2\uparrow}$ must be
$\hat\psi\restr(\hat\Cal S\wedge\tau)^{2\uparrow}$.   Now
$\int_{\hat\Cal S\wedge\tau}d\hat\psi'
=\int_{\hat\Cal S\wedge\tau}d\hat\psi$
is defined, by 613J(c-i), and

\Centerline{$Q_{(\Cal S\wedge\tau)\varCsphat}(d\hat\psi')
=Q_{\hat\Cal S\wedge\tau}(d\hat\psi')
=Q_{\hat\Cal S\wedge\tau}(d\hat\psi)
\subseteq Q_{\hat\Cal S}(d\hat\psi)$}

\noindent (616Dd) is topologically bounded (613B(f-iii) again).
}%end of proof of 616G

\leader{616H}{}\cmmnt{ The next theorem depends on some machinery.
%616H

\medskip

\noindent}{\bf Lemma} Suppose that

\Centerline{$\epsilon>0$,
\quad$\gamma\ge 0$,
\quad$m\ge 1$,
\quad$m\epsilon\ge 2\gamma$,}

\Centerline{$r\ge m$,
\quad$1-\Bover{r!}{r^m(r-m)!}\le\bover12\epsilon^m$,
\quad$k\ge 1$,
\quad$2k\epsilon^m\ge\epsilon$,
\quad$n=rk$.}

\noindent Let $\Cal S$ be a sublattice of $\Cal T$ and
$\psi$ an adapted interval function with domain $\Cal S^{2\uparrow}$.

(a) Let $\ofamily{i}{r}{a_i}$ be a family in $\frak A$
such that $\bar\mu a_i\ge\epsilon$ for every $i<r$.
Then there is a $J\in[r]^m$ such that
$\bar\mu(\inf_{i\in J}a_i)\ge\bover12\epsilon^m$.

(b) Let $\tau_0\le\ldots\le\tau_r$ in $\Cal S$ be such that
$\sup\{\theta(w):w\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\psi)\}>4\epsilon$
for every $i<r$, while
$z\in L^0(\frak A_{\tau_0})$ is such that
$\bar\mu\Bvalue{|z|\ge\gamma}\le\epsilon$.   Then there is a
$w\in Q_{\Cal S}(d\psi)$ such that
$\bar\mu\Bvalue{|z+w|\ge\gamma}
\ge\bar\mu\Bvalue{|z|\ge\gamma}+\bover12\epsilon^m$.

(c) Let $\tau_0\le\ldots\le\tau_n$ in $\Cal S$ be such that
$\sup\{\theta(w):w\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\psi)\}>4\epsilon$
for every $i<n$.   Then there is a $w\in Q_{\Cal S}(d\psi)$ such that
$\bar\mu\Bvalue{|w|\ge\gamma}\ge\epsilon$.

(d) Let $\tau_0\le\ldots\le\tau_n$ in $\Cal S$ be such that
$\theta(\psi(\tau_i,\tau_{i+1}))>4\epsilon$ for every $i<n$.
Then there is a $w\in Q_{\Cal S}(d\psi)$ such that
$\bar\mu\Bvalue{|w|\ge\gamma}\ge\epsilon$.

\proof{{\bf (a)} The case $m=1$ is trivial.   Otherwise, set
$u=\Bover1r\sum_{i=0}^{r-1}\chi a_i$ and $q=\Bover{m}{m-1}$.   Then

\Centerline{$(\int u^m)^{1/m}=\|u\|_m\|\chi 1\|_q\ge\int u\ge\epsilon$}

\noindent by H\"older's inequality (244Eb),
so if $F$ is the set of injective functions from $m$ to $r$,

$$\eqalign{r^m\epsilon^m&
\le r^m\int u^m
=\sum_{f\in r^m}\int\prod_{i<m}\chi a_{f(i)}
=\sum_{f\in r^m}\bar\mu(\inf_{i<m}a_{f(i)})\cr&
\le r^m-\#(F)+\sum_{f\in F}\bar\mu(\inf_{i<m}a_{f(i)})
\le r^m-\#(F)+\#(F)\sup_{f\in F}\bar\mu(\inf_{i<m}a_{f(i)})\cr}$$

\noindent and there is a $J\in[r]^m$ such that

$$\eqalign{\bar\mu(\inf_{i\in J}a_i)&
\ge\Bover1{\#(F)}(r^m\epsilon^m-r^m+\#(F))\cr&
=\Bover{r^m}{\#(F)}(\epsilon^m-(1-\Bover{\#(F)}{r^m}))
\ge\epsilon^m-(1-\Bover{r!}{r^m(r-m)!})
\ge\Bover12\epsilon^m,\cr}$$

\noindent as required.

\medskip

{\bf (b)} Set $c=\Bvalue{|z|\ge\gamma}\in\frak A_{\tau_0}$;  we are
supposing that $\bar\mu c\le\epsilon$.
For each $i<r$, let $w_i\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\psi)$ be such
that $\theta(w_i)\ge 4\epsilon$;  then

$$\eqalign{\bar\mu(\Bvalue{w_i\ge\epsilon}\Bsetminus c)
   +\bar\mu(\Bvalue{w_i\le-\epsilon}\Bsetminus c)&
\ge\bar\mu\Bvalue{|w_i|\ge\epsilon}-\bar\mu c\cr&
\ge\theta(w_i)-\epsilon-\epsilon
\ge 2\epsilon;\cr}$$

\noindent set

\Centerline{$K=\{i:i<r$,
  $\bar\mu(\Bvalue{w_i\ge\epsilon}\Bsetminus c)\ge\epsilon\}$,
\quad$K'=r\setminus K$,}

$$\eqalign{a_i
&=\Bvalue{w_i\ge\epsilon}\Bsetminus c\text{ for }i\in K,\cr
&=\Bvalue{-w_i\ge\epsilon}\Bsetminus c\text{ for }i\in K',\cr}$$

\noindent so that $\bar\mu a_i\ge\epsilon$ for every $i<r$.

By (a), there is a set $J\in[r]^m$ such that
$\bar\mu(\inf_{i\in J}a_i)\ge\Bover12\epsilon^m$.
Set $d=\inf_{i\in J}a_i$,

$$\eqalign{w'_i
&=w_i\times\chi(1\Bsetminus c)\text{ for }i\in J\cap K,\cr
&=-w_i\times\chi(1\Bsetminus c)\text{ for }i\in J\cap K',\cr
&=0\text{ for }i\in r\setminus J.\cr}$$

\noindent Then $w'_i\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\psi)$
for every $i$ (616De) so
$w=\sum_{i=0}^{r-1}w'_i$ belongs to $Q_{\Cal S}(d\psi)$
(use 616Df repeatedly).   Next,

$$\eqalign{w'_i\times\chi d
&\ge\epsilon\chi d\text{ for }i\in J,\cr
&=0\text{ for }i\in r\setminus J,\cr}$$

\noindent so $w\times\chi d\ge m\epsilon\chi d\ge 2\gamma\chi d$
while $w\times\chi c=0$.   But this means that

\Centerline{$\Bvalue{|w+z|\ge\gamma}
\Bsupseteq(\Bvalue{w=0}\Bcap\Bvalue{|z|\ge\gamma})
  \Bcup(\Bvalue{|w|\ge 2\gamma}\Bsetminus\Bvalue{|z|\ge\gamma})
\Bsupseteq c\Bcup d$,}

\noindent and

\Centerline{$\bar\mu\Bvalue{|w+z|\ge\gamma}
\ge\bar\mu c+\bar\mu d\ge\bar\mu\Bvalue{|z|\ge\gamma}+\Bover12\epsilon^m$,}

\noindent as required.

\medskip

{\bf (c)} For $j<k$ set $\Cal S_j=\Cal S\cap[\tau_{jr},\tau_{(j+1)r}]$.
Choose $\langle z_j\rangle_{j\le k}$, $\ofamily{j}{k}{w_j}$ as follows.
$z_0=0$.   Given that $j<k$,
$z_j\in L^0(\frak A_{\tau_{jr}})$ and that
$\bar\mu\Bvalue{|z_j|\ge\gamma}\ge\min(\epsilon,\bover12j\epsilon^m)$,
(b) tells us that if $\bar\mu\Bvalue{|z_j|\ge\gamma}\le\epsilon$ there
is a $w_j\in Q_{\Cal S_j}(d\psi)$ such that
$\bar\mu\Bvalue{|z_j+w_j|\ge\gamma}\ge\bover12(j+1)\epsilon^m$.
If $\bar\mu\Bvalue{|z_j|\ge\gamma}>\epsilon$ take $w_j=0$.
Of course $w_j\in L^0(\frak A_{\tau_{j+1}})$ and
$\Bvalue{|z_j+w_j|\ge\gamma}\ge\min(\epsilon,\bover12(j+1)\epsilon^m)$
in either case, so we
can set $z_{j+1}=z_j+w_j$, and continue.
At the end of the induction, set $w=z_k$.

Inducing on $j$, using 616Df again for the inductive step, we see that
$z_j\in Q_{\Cal S\wedge\tau_{jr}}(d\psi)$ for every $j\le k$.
So $w\in Q_{\Cal S}(d\psi)$, while
$\bar\mu\Bvalue{|w|\ge\gamma}\ge\min(\epsilon,\bover12k\epsilon^m)=\epsilon$, as
required.

\medskip

{\bf (d)} Since
$\psi(\tau_i,\tau_{i+1})\in Q_{\Cal S\cap[\tau_i,\tau_{i+1}]}(d\psi)$ for every
$i<n$, this is a special case of (c).
}%end of proof of 616H

\leader{616I}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$ a (local) integrator.

(a) The fully adapted extension of $\pmb v$ to the covered envelope
of $\Cal S$ is a (local) integrator.

(b) $\pmb v$ is (locally) moderately oscillatory, therefore (locally)
order-bounded.

(c) $\Delta\pmb v$ is a (locally) integrating interval function.

\proof{{\bf (a)(i)} The point is that $Q_{\hat\Cal S}(d\hatpmb v)$ is included
in the topological closure of $Q_{\Cal S}(d\pmb v)$.   \Prf\ Take
$z\in Q_{\hat\Cal S}(d\hatpmb v)$.   Express $z$ as $S_I(\pmb y,d\hatpmb v)$
where $I\in\Cal I(\hat\Cal S)$, $\pmb y=\family{\tau}{I}{y_{\tau}}\in\Mfa(I)$
and $\sup|\pmb y|\le\chi 1$.   If $I=\emptyset$ then
$z=0\in Q_{\Cal S}(d\pmb v)$.   Otherwise, let $(\tau_0,\ldots,\tau_n)$
enumerate a maximal totally ordered subset of $I$.   Let
$\pmb w=\family{\tau}{\hat\Cal S}{w_{\tau}}$ be the
simple process defined by saying that if $\tau\in\hat\Cal S$ then

\Centerline{$\Bvalue{\tau<\tau_0}\Bcup\Bvalue{\tau_n\le\tau}
\Bsubseteq\Bvalue{w_{\tau}=0}$,}

\Centerline{$\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}
\Bsubseteq\Bvalue{w_{\tau}=y_{\tau_i}}$ for $i<n$}

\noindent (612Ka once more).
Then $\int_{\hat\Cal S\wedge\tau_0}\pmb w\,d\hatpmb v=0$ because

\Centerline{$w_{\tau}\times(\hat v_{\tau'}-\hat v_{\tau})
=w_{\tau}\times(\hat v_{\tau'}-\hat v_{\tau})\times\chi\Bvalue{\tau<\tau'}
=0$}

\noindent whenever $\tau\le\tau'\le\tau_0$, and
$\int_{\hat\Cal S\vee\tau_n}\pmb w\,d\hatpmb v=0$
because $w_{\tau}=0$ whenever $\tau_n\le\tau$.   Accordingly

$$\eqalignno{\int_{\hat\Cal S}\pmb w\,d\hatpmb v&
=\int_{\hat\Cal S\wedge\tau_0}\pmb w\,d\hatpmb v
  +\int_{\hat\Cal S\cap[\tau_0,\tau_n]}\pmb w\,d\hatpmb v
  +\int_{\hat\Cal S\vee\tau_n}\pmb w\,d\hatpmb v
\Displaycause{613J(c-ii)}
=\int_{\hat\Cal S\cap[\tau_0,\tau_n]}\pmb w\,d\hatpmb v
=\sum_{i=0}^{n-1}y_{\tau_i}\times(\hat v_{\tau_{i+1}}-\hat v_{\tau_i})
\Displaycause{614Bb}
=S_I(\pmb y,d\hatpmb v)
\Displaycause{613Ec}
=z.\cr}$$

Consequently $\int_{\Cal S}(\pmb w\restr\Cal S)d\pmb v$ is defined and
equal to $z$ (613T).   So

\Centerline{$z=\lim_{J\uparrow\Cal I(\Cal S)}S_J(\pmb w,d\pmb v)$.}

\noindent But $\sup|\pmb w|\le\sup_{i\le n}|y_{\tau_i}|\le\chi 1$,
so $S_J(\pmb w,d\pmb v)\in Q_{\Cal S}(d\pmb v)$ for every $J\in\Cal I(\Cal S)$
and $z\in\overline{Q_{\Cal S}(d\pmb v)}$.   As $z$ is arbitrary, we have the
result.\ \Qed

\medskip

\quad{\bf (ii)} If $\pmb v$ is an integrator, then
$Q_{\Cal S}(d\pmb v)$ is topologically bounded,
so its closure is also topologically bounded (613B(f-iii))
and $\hatpmb v$ is an integrator.

\medskip

\quad{\bf (iii)} If $\pmb v$ is a local integrator and $\tau\in\hat\Cal S$,
take any $\epsilon>0$.   Then there is a $\sigma\in\Cal S$ such that
$\bar\mu\Bvalue{\sigma<\tau}\le\bover12\epsilon$.   We are supposing that
$\pmb v\restr\Cal S\wedge\sigma$ is an integrator, so
$\hatpmb v\restr\hat\Cal S\wedge\sigma$ is an integrator, and there is a
$\delta>0$ such that $\theta(\delta z)\le\bover12\epsilon$
whenever $z\in Q_{\hat\Cal S\wedge\sigma}(d\hatpmb v)$.   Now take
$z\in Q_{\hat\Cal S\wedge\tau}(d\hatpmb v)$.   By 616C(iii), there is a
fully adapted
process $\pmb u=\family{\sigma}{\hat\Cal S\wedge\tau}{u_{\sigma}}$ with
breakpoint string $(\tau_0,\ldots,\tau_n)$ such that $\sup|\pmb u|\le\chi 1$
and $z=\sum_{i=0}^{n-1}u_{\tau_i}\times(\hat v_{\tau_{i+1}}-\hat v_{\tau_i})$.
Consider

\Centerline{$z'
=\sum_{i=0}^{n-1}u_{\tau_i\wedge\sigma}
    \times(\hat v_{\tau_{i+1}\wedge\sigma}-\hat v_{\tau_i\wedge\sigma})$.}

\noindent By 616C(ii), $z'\in Q_{\hat\Cal S\wedge\sigma}(d\hatpmb v)$, while

$$\eqalign{\Bvalue{z\ne z'}&
\Bsubseteq\sup_{i<n}\Bvalue{u_{\tau_i}\ne u_{\tau_i\wedge\sigma}}
   \Bcup\sup_{i\le n}\Bvalue{\hat v_{\tau_i}\ne\hat v_{\tau_i\wedge\sigma}}\cr&
\Bsubseteq\sup_{i\le n}\Bvalue{\tau_i\wedge\sigma<\tau_i}
\Bsubseteq\Bvalue{\sigma<\tau}\cr}$$

\noindent has measure at most $\bover12\epsilon$.   So

\Centerline{$\theta(\delta z)\le\theta(\delta z')+\theta(\delta(z-z'))
\le\Bover12\epsilon+\Bover12\epsilon=\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $Q_{\hat\Cal S\wedge\tau}(d\hatpmb v)$
is topologically bounded;  as $\tau$ is arbitrary, $\hatpmb v$ is a local
integrator.

\medskip

{\bf (b)(i)} Suppose to begin with that $\pmb v$ is an integrator.   By (a),
$\hatpmb v$ is an integrator.
The idea is to use 616Hd with $\psi=\Delta\hatpmb v$, knowing that
$Q_{\hat\Cal S}(d\psi)=Q_{\hat\Cal S}(d\hatpmb v)$ is topologically
bounded.   Express $\hatpmb v$ as
$\family{\sigma}{\hat\Cal S}{\hat v_{\sigma}}$.   Let $\epsilon>0$.
Then there is a $\gamma>0$ such that $\bar\mu\Bvalue{|w|\ge\gamma}<\epsilon$ for
every $w\in Q_{\Cal S}(d\hatpmb v)$ (613B(f-ii)).   Take $m$, $r$, $k\ge 1$
such that $m\epsilon\ge 2\gamma$, $r\ge m$,
$1-\Bover{r!}{r^m(r-m)!}\le\Bover12\epsilon^m$ and $2k\epsilon^m\ge\epsilon$,
and set $n=rk$.   Then whenever $\tau_0\le\ldots\le\tau_n$ in $\hat\Cal S$,
there is an $i<n$ such that
$\theta(\hat v_{\tau_{i+1}}-\hat v_{\tau_i})\le 4\epsilon$,
by 616Hd.   Thus $\hatpmb v$ satisfies condition (iv) of 615N and is moderately
oscillatory.  It follows at once that $\pmb v$ is moderately
oscillatory (615F(a-i)).

\medskip

\quad{\bf (ii)} If $\pmb v$ is a local integrator, then we can apply
(i) to $\pmb v\restr\Cal S\wedge\tau$, for $\tau\in\Cal S$, to see that
$\pmb v$ is locally moderately oscillatory.

\medskip

{\bf (c)(i)} Again, suppose to begin with that $\pmb v$ is an integrator.
If $\Cal S$ is full, $\Delta\pmb v$ is certainly an integrating interval
function, because we know that
$Q_{\Cal S}(d\pmb v)=Q_{\Cal S}(d(\Delta\pmb v))$ is topologically bounded, and from (b) we know that
$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ and
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ are defined
(615G), so

\Centerline{$\int_{\Cal S}d(\Delta\pmb v)=\int_{\Cal S}d\pmb v
=v_{\uparrow}-v_{\downarrow}$}

\noindent (613N) is defined.   In general, (a) tells us that $\hatpmb v$
is an integrator, so $\Delta\hatpmb v$ is an integrating interval function;
but $\Delta\hatpmb v$ extends $\Delta\pmb v$,
so is the strictly adapted extension of $\Delta\pmb v$ to
$\hat\Cal S^{2\uparrow}$,
and $\Delta\pmb v$ is an integrating interval function, as remarked in 616F.

\medskip

\quad{\bf (ii)} As in (b), it follows at once that if $\pmb v$ is a local
integrator then $\Delta\pmb v$ is a locally integrating interval function.
}%end of proof of 616I

\leader{616J}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\psi$ an integrating interval function with domain $\Cal S^{2\uparrow}$.
Set

\Centerline{$M_{\psi}=\{\pmb u:\pmb u\in\Mob(\Cal S)$,
$\int_{\Cal S}\pmb u\,d\psi$ is defined$\}$.}

\noindent Then $M_{\psi}$ is a closed linear subspace of $\Mob(\Cal S)$
and we have an indefinite integral operator
$ii_{\psi}:M_{\psi}\to\Mir(\Cal S)$ which is linear and continuous for the
ucp topology on $\Mob(\Cal S)$.

\proof{{\bf (a)} Take any $\pmb u\in M_{\psi}$.

\medskip

\quad{\bf (i)} Setting $y_{\sigma}=\int_{\Cal S\wedge\sigma}\pmb u\,d\psi$
for $\sigma\in\Cal S$, and $\pmb y=\sfamilysS{y}$, $\pmb y$
is defined everywhere on $\Cal S$ and is fully adapted (613O(b-i)).
Let $\epsilon>0$.   Then there is a $\delta>0$ such that
$\theta(S_I(\pmb w,d\psi))\le\epsilon$ whenever $I\in\Cal I(\Cal S)$,
$\pmb w\in\Mfa(I)$ and $\theta(\sup|\pmb w|)\le\delta$ (616E).   We
are supposing that
$\pmb u$ is order-bounded;  write $\bar u$ for $\sup|\pmb u|$.
Let $\eta>0$ be such that $\theta(\bar u\times w)\le\delta$ whenever
$w\in L^0(\frak A)$ and $\theta(w)\le\eta$ (613Ba).

\medskip

\quad{\bf (ii)} Suppose that $I\in\Cal I(\Cal S)$, $\pmb w\in\Mfa(I)$ and
$\theta(\bar w)\le\eta$, where $\bar w=\sup|\pmb w|$.   Then
$\theta(S_I(\pmb w,d\pmb y))\le\epsilon$.   \Prf\ If $I=\emptyset$ this is
trivial.

\medskip

\qquad\grheada\ If $I$ is non-empty,
let $(\tau_0,\ldots,\tau_n)$ be the increasing enumeration
of a maximal totally ordered subset of $I$, so that
$S_I(\pmb w,d\pmb y)
=\sum_{i=0}^{n-1}w_{\tau_i}\times(y_{\tau_{i+1}}-y_{\tau_i})$.
We know that
$y_{\tau'}=y_{\tau}+\int_{\Cal S\cap[\tau,\tau']}\pmb u\,d\psi$ whenever
$\tau\le\tau'$ in $\Cal S$ (613Jc), so

\Centerline{$S_I(\pmb w,d\pmb y)
=\sum_{i=0}^{n-1}w_{\tau_i}
  \times\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb u\,d\psi$.}

\noindent Let $\tildepmb w=\sfamilysS{\tilde w}$
be the simple process with domain $\Cal S$ defined by saying that

\Centerline{$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{\tilde w_{\sigma}=0}$,
\quad$\Bvalue{\tau_n\le\sigma}\Bsubseteq\Bvalue{\tilde w_{\sigma}=w_{\tau_n}}$,}

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
   \Bsubseteq\Bvalue{\tilde w_{\sigma}=w_{\tau_i}}$}

\noindent for $\sigma\in\Cal S$ and $i<n$ (612Ka again);
then $\tildepmb w$ extends $\pmb w$
and $\sup|\tildepmb w|=\sup_{i\le n}|w_{\tau_i}|=\bar w$.   Consequently
$\sup|\pmb u\times\tildepmb w|\le\bar u\times\bar w$,
$\theta(\sup|\pmb u\times\tildepmb w|)\le\delta$ and
$\theta(S_J(\pmb u\times\tildepmb w),d\psi)\le\epsilon$ for every
$J\in\Cal I(\Cal S)$.

\medskip

\qquad\grheadb\ For each $i<n$ let $J_i$ be a finite sublattice of
$\Cal S\cap[\tau_i,\tau_{i+1}]$ containing $\tau_i$ and $\tau_{i+1}$.
Let $(\sigma_{i0},\ldots,\sigma_{im_i})$ be the increasing enumeration
of a maximal totally ordered subset of $J_i$.   Then

\Centerline{$u_{\sigma_{ij}}\times\tilde w_{\sigma_{ij}}
   \times\psi(\sigma_{ij},\sigma_{i,j+1})
=u_{\sigma_{ij}}\times w_{\tau_i}
   \times\psi(\sigma_{ij},\sigma_{i,j+1})$}

\noindent for every $j<m_i$, because

\Centerline{$\Bvalue{\psi(\sigma_{ij},\sigma_{i,j+1})\ne 0}
\Bsubseteq\Bvalue{\sigma_{ij}<\sigma_{i,j+1}}
%613C(b-i)
\Bsubseteq\Bvalue{\sigma_{ij}<\tau_{i+1}}
\Bsubseteq\Bvalue{\tilde w_{\sigma_{ij}}=w_{\tau_i}}$.}

\noindent Now

$$\eqalignno{S_{J_i}(\pmb u\times\tildepmb w,d\psi)&
=\sum_{j=0}^{m_i-1}u_{\sigma_{ij}}\times\tilde w_{\sigma_{ij}}
   \times\psi(\sigma_{ij},\sigma_{i,j+1})\cr&
=\sum_{j=0}^{m_i-1}u_{\sigma_{ij}}\times w_{\tau_i}
   \times\psi(\sigma_{ij},\sigma_{i,j+1})
=w_{\tau_i}\times S_{J_i}(\pmb u,d\psi).\cr}$$

\noindent Set $J=\bigcup_{i<n}J_i$, so that $J\in\Cal I(\Cal S)$ and
$J\cap[\tau_i,\tau_{i+1}]=J_i$ for each $i$.   Then

\Centerline{$\sum_{i=0}^{n-1}w_{\tau_i}\times S_{J_i}(\pmb u,d\psi)
=\sum_{i=1}^{n-1}S_{J_i}(\pmb u\times\tildepmb w,d\psi)
=S_J(\pmb u\times\tildepmb w,d\psi)$}

\noindent (613Ga) and
$\theta(\sum_{i=0}^{n-1}w_{\tau_i}\times S_{J_i}(\pmb u,d\psi))
\le\epsilon$.   Taking the limit as
$J_i\uparrow\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}])$ for each $i$,

\Centerline{$\theta(S_I(\pmb w,d\pmb y))
=\theta(\sum_{i=0}^{n-1}
  w_{\tau_i}\times\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb u\,d\psi)
\le\epsilon$.  \Qed}

\medskip

\quad{\bf (iii)} As $\epsilon$ is arbitrary, $Q_{\Cal S}(d\pmb y)$ is
topologically bounded, by 616E(iii)$\Rightarrow$(i), and $\pmb y$
is an integrator.

Thus $ii_{\psi}$ is a function from $M_{\psi}$ to $\Mir(\Cal S)$.
By 613Jb, $M_{\psi}$ is a linear subspace of $\Mob(\Cal S)$ and $ii_{\psi}$ is a
linear operator.   Recall from 616Ib that $\Mir(\Cal S)\subseteq\Mob(\Cal S)$.

\medskip

{\bf (b)} Suppose that $\pmb u\in\Mob(\Cal S)$ belongs to the closure of
$M_{\psi}$ for the ucp topology.   Let $\epsilon>0$.   By 616E, there is a
$\delta>0$ such that $\theta(S_I(\pmb u',d\psi))\le\epsilon$ whenever
$I\in\Cal I(\Cal S)$,
$\pmb u'\in\Mob(\Cal S)$ and $\theta(\sup|\pmb u'|)\le\delta$.
Now there are a $\pmb u'\in M_{\psi}$ such that
$\theta(\sup|\pmb u-\pmb u'|)\le\delta$, and a $J\in\Cal I(\Cal S)$ such that
$\theta(S_I(\pmb u',d\psi)-\int_{\Cal S}\pmb u'd\psi)\le\epsilon$
whenever $J\subseteq I\in\Cal I(\Cal S)$.   Suppose now that
$I\in\Cal I(\Cal S)$ includes $J$;  then

$$\eqalign{\theta(S_I(\pmb u,d\psi)
    -\hbox{$\biggerint_{\Cal S}\,$}\pmb u'd\psi)&
\le\theta(S_I(\pmb u,d\psi)-S_I(\pmb u',d\psi))
  +\theta(S_I(\pmb u',d\psi)-\hbox{$\biggerint_{\Cal S}\,$}\pmb u'd\psi)\cr&
\le\epsilon+\epsilon
=2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\int_{\Cal S}\pmb u\,d\psi$ is defined (613Ja), and
$\pmb u\in M_{\psi}$.   As $\pmb u$ is arbitrary, $M_{\psi}$ is closed for
the ucp topology.

\medskip

{\bf (c)(i)}
Let $\epsilon>0$.   Because $\psi$ is an integrating interval function
there is a $\delta>0$
such that $\theta(S_I(\pmb u,d\psi))\le\epsilon^2$ whenever
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and $\theta(\sup|\pmb u|)\le\delta$
(616E).   It follows that
$\bar\mu\Bvalue{|S_I(\pmb u,d\psi)|>\epsilon}\le\epsilon$ whenever
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and $\theta(\sup|\pmb u|)\le\delta$.

\medskip

\quad{\bf (ii)} (The key.) Suppose that $I\in\Cal I(\Cal S)$ is non-empty.   Let
$\pmb u=\family{\sigma}{I}{u_{\sigma}}\in\Mfa(\Cal I)$ be such that
$\theta(\sup|\pmb u|)\le\delta$.   Then
$\bar\mu\Bvalue{\sup_{\sigma\in I}|S_{I\wedge\sigma}(\pmb u,d\psi)|>\epsilon}
\le\epsilon$.

\medskip

\Prf\grheada\ Take $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.
For $i\le n$ write $z_i$ for $S_{I\wedge\tau_i}(\pmb u,d\psi)$.
Since $\tau_0,\ldots,\tau_i)$ linearly generates the $I\wedge\tau_i$-cella
(611Kg), $z_i=\sum_{j=0}^{i-1}u_{\tau_j}\times\psi(\tau_j,\tau_{j+1})$ (613Ec).
Next, set $a_i=\Bvalue{|z_i|>\epsilon}$,
$y_i=u_{\tau_i}\times\chi(1\Bsetminus\sup_{j\le i}a_j)$ and
$b_i=a_i\Bsetminus\sup_{j<i}a_j$ for $i\le n$.
Then $z_i\in L^0(\frak A_{\tau_i})$ (613Gb), $a_i\in\frak A_{\tau_i}$,
$y_i\in L^0(\frak A_{\tau_i})$
and $b_i\in\frak A_{\tau_i}$ for all $i\le n$, while
$\sup_{i\le n}|y_i|\le\sup_{\sigma\in I}|u_{\sigma}|$ so
$\theta(\sup_{i\le n}|y_i|)\le\delta$.   By 612Ka again,
there is a fully adapted
process $\pmb w=\family{\sigma}{I}{w_{\sigma}}$ such that

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{w_{\sigma}=y_i}$ for $i<n$,
\quad$\Bvalue{\tau_n=\sigma}\Bsubseteq\Bvalue{w_{\sigma}=\tau_n}$}

\noindent for every $\sigma\in I$.
(The point here is that $\tau_0\le\sigma\le\tau_n$ for every
$\sigma\in I$.)   Set

$$\eqalignno{z=S_I(\pmb w,d\psi)&
=\sum_{j=0}^{n-1}w_{\tau_j}\times\psi(\tau_j,\tau_{j+1})
\Displaycause{613Ec again}
=\sum_{j=0}^{n-1}y_j\times\psi(\tau_j,\tau_{j+1})\cr}$$

\noindent because
$\Bvalue{\psi(\tau_j,\tau_{j+1})\ne 0}\Bsubseteq\Bvalue{\tau_j<\tau_{j+1}}$ for
each $j$, by 613C(b-i).

Now if $\sigma\in I$ then $|w_{\sigma}|\le\sup_{i\le n}|y_i|$
so $\theta(\sup|\pmb w|)\le\delta$ and
$\bar\mu\Bvalue{|z|>\epsilon}\le\epsilon$.

\medskip

\qquad\grheadb\ If $i\le n$,

$$\eqalign{b_i
&\Bsubseteq\inf_{i\le j<n}\Bvalue{y_j=0}\Bcap\inf_{j<i}\Bvalue{y_j=u_{\tau_j}}
  \cr&
\Bsubseteq\Bvalue{\sum_{j=0}^{n-1}y_j\times\psi(\tau_j,\tau_{j+1})
   =\sum_{j=0}^{i-1}u_{\tau_j}\times\psi(\tau_j,\tau_{j+1})}
=\Bvalue{z=z_i};
\cr}$$

\noindent as $b_i\Bsubseteq\Bvalue{|z_i|>\epsilon}$,
$b_i\Bsubseteq\Bvalue{|z|>\epsilon}$.   Next,
$\family{\sigma}{I}{S_{I\wedge\sigma}(\pmb u,d\psi)}$ is fully adapted
(613Gb again).   If $\sigma\in I$ then
$\sup_{i\le n}\Bvalue{\sigma=\tau_i}=1$ (611Ke), so

$$\eqalign{\Bvalue{|S_{I\wedge\sigma}(\pmb u,d\psi)|>\epsilon}&
=\sup_{i\le n}\Bvalue{\sigma=\tau_i}
   \Bcap\Bvalue{|S_{I\wedge\sigma}(\pmb u,d\psi)|>\epsilon}\cr&
\Bsubseteq\sup_{i\le n}\Bvalue{|S_{I\wedge\tau_i}(\pmb u,d\psi)|>\epsilon}
=\sup_{i\le n}a_i
=\sup_{i\le n}b_i
\Bsubseteq\Bvalue{|z|>\epsilon}.\cr}$$

\noindent Accordingly

$$\eqalignno{
\Bvalue{\sup_{\sigma\in I}|S_{I\wedge\sigma}(\pmb u,d\psi)|>\epsilon}&
=\sup_{\sigma\in I}\Bvalue{|S_{I\wedge\sigma}(\pmb u,d\psi)|>\epsilon}
\Displaycause{364L(a-ii)}
\Bsubseteq\Bvalue{|z|>\epsilon}\cr}$$

\noindent has measure at most $\epsilon$.\ \Qed

\medskip

\quad{\bf (iii)} Take $\pmb u\in M_{\psi}$ such that $\theta(\sup|\pmb u|)\le\delta$.
Then
$\theta(\sup_{\sigma\in I}|S_{I\wedge\sigma}(\pmb u,d\psi)|)\le 2\epsilon$
for every $I\in\Cal I(\Cal S)$,
if we interpret the supremum as zero if $I$ is empty.   Write
$w_{\sigma}$ for $\int_{\Cal S\wedge\sigma}\pmb u\,d\psi$.

\medskip

\qquad\grheada\ If $I\in\Cal I(\Cal S)$ then
$\theta(\sup_{\sigma\in I}|w_{\sigma}|)\le 2\epsilon$.
\Prf\ Let $\eta>0$.   For each $\sigma\in I$ let
$J_{\sigma}\in\Cal I(\Cal S\wedge\sigma)$ be such that
$\theta(S_K(\pmb u,d\psi)-w_{\sigma})\le\eta$
whenever $J_{\sigma}\subseteq K\in\Cal I(\Cal S\wedge\sigma)$.   Let $K$ be the
sublattice of $\Cal S$ generated by $I\cup\bigcup_{\sigma\in I}J_{\sigma}$.
Then $J_{\sigma}\subseteq K\wedge\sigma\in\Cal I(\Cal S\wedge\sigma)$
for every $\sigma\in I$.   So

$$\eqalign{\theta(\sup_{\sigma\in I}|w_{\sigma}|)&
\le\theta(\sup_{\sigma\in I}|S_{K\wedge\sigma}(\pmb u,d\psi)|
  +\sup_{\sigma\in I}|w_{\sigma}-S_{K\wedge\sigma}(\pmb u,d\psi)|)\cr&
\le\theta(\sup_{\sigma\in K}|S_{K\wedge\sigma}(\pmb u,d\psi)|)
  +\sum_{\sigma\in I}\theta(w_{\sigma}-S_{K\wedge\sigma}(\pmb u,d\psi))
\le 2\epsilon+\eta\#(I).\cr}$$

\noindent As $\eta$ is arbitrary,
$\theta(\sup_{\sigma\in I}|w_{\sigma}|)\le 2\epsilon$.\ \Qed

\medskip

\qquad\grheadb\ We know that $ii_{\psi}(\pmb u)$ is order-bounded, so
$\bar w=\sup_{\sigma\in\Cal S}|w_{\sigma}|$ is defined in $L^0(\frak A)$.
Now

$$\eqalign{\theta(\sup|ii_{\psi}(\pmb u)|)&
=\theta(\bar w)
=\sup_{I\subseteq\Cal S\text{ is finite}}\theta(\sup_{\sigma\in I}|w_{\sigma}|)
  \cr&
=\sup_{I\in\Cal I(\Cal S)}\theta(\sup_{\sigma\in I}|w_{\sigma}|)
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ was arbitrary, $ii_{\psi}:M_{\psi}\to\Mob(\Cal S)$ is
continuous at $0$;  being a linear operator, it is continuous everywhere on
$M_{\psi}$, as claimed.
}%end of proof of 616J

\leader{616K}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$ a
moderately oscillatory process with domain $\Cal S$, and $\pmb v$ an integrator
with domain $\Cal S$.   Then $\int_{\Cal S}\pmb u\,d\pmb v$ is defined, and
$\iivu$ is an integrator.

\proof{ By 616I, $\pmb v$ is moderately oscillatory and $\Delta\pmb v$ is an
integrating interval function.  By 615Ra,
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined if $\pmb u$ is of bounded variation.
In the language of 616J, $\Mbv(\Cal S)\subseteq M_{\Delta\pmb v}$, while
$M_{\Delta\pmb v}$ is closed in $\Mob(\Cal S)$ for the ucp topology.   But
$\Mmo(\Cal S)$ is just the closure of $\Mbv(\Cal S)$ in $\Mob(\Cal S)$. so is
included in $M_{\Delta\pmb v}$, that is, $\int_{\Cal S}\pmb u\,d\pmb v$ is
defined.   As for the indefinite integral $\iivu$,
this is an integrator by 616J.
}%end of proof of 616K

\leader{616L}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$.
If $\pmb u$ is a locally moderately
oscillatory process and $\pmb v$ a fully adapted
process which is locally of bounded variation, both with domain $\Cal S$,
then $\iivu$ is locally of bounded variation.

\proof{ Apply 614U to $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb v\restr\Cal S\wedge\tau$ for each $\tau\in\Cal S$.
}%end or proof of 616L

\leader{616M}{Theorem} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\psi$ an integrating interval function on $\Cal S$.   Write
$\pmb v$ for
$ii_{\psi}(\tbf 1)=\family{\tau}{\Cal S}{\int_{\Cal S\wedge\tau}d\psi}$.
Then $\int_{\Cal S}\pmb u\,d\psi$ is defined and equal to
$\int_{\Cal S}\pmb u\,d\pmb v$ whenever
$\pmb u$ is a moderately oscillatory process with domain $\Cal S$.

%\query what if we had this just for full $\Cal S$?
% won't really do

\proof{{\bf (a)} I start by noting that $\pmb v$ is well-defined and is an
integrator, by 616J.   If $\Cal S$ is empty the result is trivial, so suppose
otherwise.   For the time being, I will suppose that $\pmb v=\tbf 0$ and
$\Cal S$ is full.
Note that the definition 616Fa requires $\int_{\Cal S}d\psi$ to be defined, and
we now have
$\int_{\Cal S}d\psi=\lim_{\tau\uparrow\Cal S}\int_{\Cal S\wedge\tau}d\psi
=0$ by 613J(f-ii).   Write $M'_{\psi}$ for

\Centerline{$\{\pmb u:\pmb u\in\Mob(\Cal S)$, $\int_{\Cal S}\pmb u\,d\psi$
is defined and equal to $0\}$.}

\noindent Then $M'_{\psi}$ is a linear subspace of $\Mob=\Mob(\Cal S)$, and
by 616J it is closed for the ucp topology.
Note that the constant process $\tbf 1$ with domain $\Cal S$ belongs to
$M'_{\psi}$.

\medskip

{\bf (b)} To begin with, suppose that $\Cal S$ is full, that
$\pmb v=\tbf 0$, that $\pmb u=\sfamilysS{u}$ is non-increasing
and that $\pmb u$ is $\{0,1\}$-valued,
that is, $\Bvalue{u_{\sigma}\in\{0,1\}}=1$ for every $\sigma\in\Cal S$.

\medskip

\quad{\bf (i)} If $I\in\Cal I(\Cal S)$ is non-empty, there is a
$\tau\in\Cal S$ such that
$S_I(\pmb u,d\psi)=S_{I\wedge\tau}(\tbf 1,d\psi)$.   \Prf\

\medskip

\qquad\grheada\ Take $(\tau_0,\ldots,\tau_n)$ linearly
generating the $I$-cells;  note that $\tau_n=\max I$.
For $i<n$, set $a_i=\Bvalue{u_{\tau_i}=0}$ and
$b_i=a_i\Bsetminus\sup_{j<i}a_j$;  set $b_n=1\Bsetminus a_{n-1}$.
Because $\pmb u$ is non-increasing,
$a_0\Bsubseteq a_1\Bsubseteq\ldots\Bsubseteq a_{n-1}$ and
$\langle b_i\rangle_{i\le n}$ is a partition of unity in $\frak A$.   Also
$a_i$ and $b_i$ belong to $\frak A_{\tau_i}$ for $i<n$ and
$b_n\in\frak A_{\tau_n}$.   By 611I there is a $\tau\in\Cal T$ such that
$b_i\Bsubseteq\Bvalue{\tau=\tau_i}$ for $i\le n$;  because $\Cal S$ is full,
$\tau\in\Cal S$.

\medskip

\qquad\grheadb\ If $j<i<n$ then

$$\eqalign{b_i\Bsubseteq\Bvalue{\tau=\tau_i}&
\Bsubseteq\Bvalue{\tau\wedge\tau_j=\tau_j}
   \Bcap\Bvalue{\tau\wedge\tau_{j+1}=\tau_{j+1}}\cr&
\Bsubseteq\Bvalue{\psi(\tau\wedge\tau_j,\tau\wedge\tau_{j+1})
    =\psi(\tau_j,\tau_{j+1})}\cr}$$

\noindent because $\psi$ is strictly adapted, and if $i\le j<n$ then

$$\eqalign{b_i\Bsubseteq\Bvalue{\tau=\tau_i}&
\Bsubseteq\Bvalue{\tau\wedge\tau_j=\tau}
   \Bcap\Bvalue{\tau\wedge\tau_{j+1}=\tau}\cr&
\Bsubseteq\Bvalue{\psi(\tau\wedge\tau_j,\tau\wedge\tau_{j+1})
    =\psi(\tau,\tau)}
\Bsubseteq\Bvalue{\psi(\tau\wedge\tau_j,\tau\wedge\tau_{j+1})=0}.\cr}$$

\medskip

\qquad\grheadc\ If $i<n$ then

$$\eqalignno{\chi b_i\times S_I(\pmb u,d\psi)&
=\sum_{j=0}^{n-1}\chi b_i\times u_{\tau_j}\times\psi(\tau_j,\tau_{j+1})
=\sum_{j=0}^{i-1}\chi b_i\times u_{\tau_j}\times\psi(\tau_j,\tau_{j+1})
\Displaycause{because if $i\le j$ then $b_i\Bsubseteq a_i\Bsubseteq a_j$ and
$b_i\Bsubseteq\Bvalue{u_{\tau_j}=0}$}
=\sum_{j=0}^{i-1}\chi b_i\times\psi(\tau\wedge\tau_j,\tau\wedge\tau_{j+1})
\Displaycause{as in ($\beta$) just above}
=\sum_{j=0}^{n-1}\chi b_i\times\psi(\tau\wedge\tau_j,\tau\wedge\tau_{j+1})
\Displaycause{because if $i\le j<n$ then
$b_i\Bsubseteq\Bvalue{\psi(\tau\wedge\tau_j,\tau\wedge\tau_{j+1})=0}$, by the
other half of ($\beta$)}
=\chi b_i\times S_{I\wedge\tau}(\tbf 1,d\psi)\cr}$$

\noindent because $(\tau\wedge\tau_0,\ldots,\tau\wedge\tau_n)$ linearly
generates the $(I\wedge\tau)$-cells (611Kg).   Also

$$\eqalign{b_n&
\Bsubseteq\Bvalue{\tau=\tau_n}\Bcap\inf_{i<n}\Bvalue{u_{\tau_i}=1}\cr&
\Bsubseteq\Bvalue{S_{I\wedge\tau}(\tbf 1,d\psi)=S_I(\tbf 1,d\psi)}
  \Bcap\Bvalue{S_I(\pmb u,d\psi)=S_I(\tbf 1,d\psi)}\cr&
\Bsubseteq\Bvalue{S_I(\pmb u,d\psi)=S_{I\wedge\tau}(\tbf 1,d\psi)},\cr}$$

\noindent so in fact $S_I(\pmb u,d\psi)=S_{I\wedge\tau}(\tbf 1,d\psi)$.\ \Qed

\medskip

\quad{\bf (ii)} Consequently $\pmb u\in M'_{\psi}$.   \Prf\
Let $\epsilon>0$.   Then there is an $I_0\in\Cal I(\Cal S)$,
containing $\max\Cal S$, such that
$\theta(S_J(\tbf 1,d\psi)-\theta(S_K(\tbf 1,d\psi))\le\epsilon$ whenever
$J$, $K\in\Cal I(\Cal S)$ include $I_0$.   If $I\in\Cal I(\Cal S)$ includes
$I_0$, there is a $\tau\in\Cal S$ such that
$S_I(\pmb u,d\psi)=S_{I\wedge\tau}(\tbf 1,d\psi)$, by (i).   At the same time,
of course,  $\theta(S_J(\tbf 1,d\psi)-\theta(S_K(\tbf 1,d\psi))\le\epsilon$
whenever $J$, $K\in\Cal I(\Cal S)$ include $I$.   By 613V(ii-$\beta$),

$$\eqalignno{2\epsilon&
\ge\theta(S_{I\wedge\tau}(\tbf 1,d\psi)-\int_{\Cal S\wedge\tau}d\psi)
=\theta(S_{I\wedge\tau}(\tbf 1,d\psi))
\Displaycause{because $\pmb v=\tbf 0$}
=\theta(S_I(\pmb u,d\psi)),\cr}$$

\noindent and this is true whenever $I$ includes $I_0$.   As $\epsilon$ is
arbitrary, $\int_{\Cal S}\pmb u\,d\psi=0$ and $\pmb u\in M'_{\psi}$.\ \Qed

\medskip

{\bf (c)} Still supposing that $\Cal S$ is full and $\pmb v=\tbf 0$,
I explore $M'_{\psi}$.

\medskip

\quad{\bf (i)} If $\pmb u\in\Mfa=\Mfa(\Cal S)$ is $\{0,1\}$-valued and
non-decreasing, then $\pmb u\in M'_{\psi}$.   \Prf\ $\tbf 1-\pmb u$ is
$\{0,1\}$-valued and non-increasing, so (b) tells us that it belongs to
$M'_{\psi}$;  as noted in (a), $\tbf 1\in M'_{\psi}$;  as
$M'_{\psi}$ is a linear subspace, $\pmb u\in M'_{\psi}$.\ \Qed

\medskip

\quad{\bf (ii)} If $n\ge 1$ and $\pmb u\in\Mfa$ is
non-decreasing and takes values in $\{0,1,\ldots,n\}$, then
$\pmb u\in M'_{\psi}$.   \Prf\ Induce on $n$.   If $n=1$ this is just (i) above.
For the inductive step, if $\pmb u$ takes values in $\{0,\ldots,n+1\}$ then
$\pmb u\wedge n\tbf 1$ is non-decreasing and takes values in $\{0,\ldots,n\}$,
so belongs to $M'_{\psi}$ by the inductive hypothesis, while
$\pmb u-\pmb u\wedge n\tbf 1=(\pmb u-n\tbf 1)^+$ is non-decreasing and takes
values in $\{0,1\}$;  so their sum $\pmb u$ belongs to $M'_{\psi}$ and the
induction continues.\ \Qed

It follows at once that if $n\ge 1$, $\delta>0$ and
$\pmb u\in\Mfa$ is
non-decreasing and takes values in $\{0,\delta,\ldots,n\delta\}$ then
$\pmb u\in M'_{\psi}$.

\medskip

\quad{\bf (iii)} If $\pmb u\in\Mob$ is non-negative and non-decreasing
it belongs to $M'_{\psi}$.   \Prf\ Write $\bar u$ for $\sup|\pmb u|$,
Let $\epsilon>0$.   Then there is an $n\ge 1$ such that
$\bar\mu\Bvalue{\bar u>n\epsilon}\le\epsilon$.   Define $h:\Bbb R\to\Bbb R$ by
saying that

$$\eqalign{h(\alpha)
&=n\epsilon\text{ if }\alpha\ge n\epsilon,\cr
&=i\epsilon
   \text{ if }i\le n\text{ and }i\epsilon\le\alpha<(i+1)\epsilon,\cr
&=\alpha\text{ if }\alpha\le 0.\cr}$$

\noindent Then $h$ is non-decreasing and Borel measurable,
$h(\alpha)\le\alpha$ for every $\alpha$ and $\alpha\le h(\alpha)+\epsilon$ if
$\alpha\le n\epsilon$.   So $\pmb u'=\bar h\pmb u$ is fully adapted and
non-decreasing;  $\pmb u'\le\pmb u$;
because $\pmb u$ is non-negative, $\pmb u'$ takes values in
$\{0,\epsilon,\ldots,n\epsilon\}$;  and
$\Bvalue{\pmb u\le n\epsilon}
\Bsubseteq\Bvalue{\pmb u\le\pmb u'+\epsilon\tbf 1}$.
By (ii) just above, $\pmb u'\in M'_{\psi}$.   And

\Centerline{$\tbf 0\le\pmb u-\pmb u'
\le\epsilon\tbf 1+\chi\Bvalue{\bar u>n\epsilon}\times\bar u$,
\quad
$\theta(\sup|\pmb u-\pmb u'|)\le\epsilon+\bar\mu\Bvalue{\bar u>n\epsilon}
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary and $M'_{\psi}$ is closed,
$\pmb u\in M'_{\psi}$.\ \Qed

\medskip

\quad{\bf (iv)} Because $M'_{\psi}$ is a closed
linear subspace of $\Mob$, it
includes $\Mbv=\Mbv(\Cal S)$ and its closure $\Mmo=\Mmo(\Cal S)$.   Thus we have the
required result if $\Cal S$ is full and $\pmb v=\tbf 0$.

\medskip

{\bf (d)} Still supposing that $\Cal S$ is full, take $\pmb u\in\Mmo$
and any integrating interval
function $\psi$ on $\Cal S$, and set $v_{\tau}=\int_{\Cal S\wedge\tau}d\psi$ for
$\tau\in\Cal S$,   Then $\pmb v=\sfamilysS{v}$ is an
integrator (616J) and $\Delta\pmb v$ is an integrating interval function
(616Ic), so $\psi'=\psi-\Delta\pmb v$ also is (616Ga).   For $\tau\in\Cal S$,

$$\eqalignno{\int_{\Cal S\wedge\tau}d\psi'&
=\int_{\Cal S\wedge\tau}d\psi-\int_{\Cal S\wedge\tau}d(\Delta\pmb v)
=v_{\tau}-\int_{\Cal S\wedge\tau}d\pmb v
=v_{\tau}-v_{\tau}+\lim_{\sigma\downarrow\Cal S}v_{\sigma}
\Displaycause{613N}
=0\cr}$$

\noindent (613J(f-i)).   So $\Cal S$ and $\psi'$ satisfy the conditions of
(a)-(c), and $\int_{\Cal S}\pmb u\,d\psi'$ is defined and equal to $0$.
But we know already from 616K that
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined, so $\int_{\Cal S}\pmb u\,d\psi$ is
defined and equal to
$\int_{\Cal S}\pmb u\,d\psi'+\int_{\Cal S}\pmb u\,d\pmb v
=\int_{\Cal S}\pmb u\,d\pmb v$.

\medskip

{\bf (e)} Finally, if $\Cal S$ is not full, write $\hat\Cal S$ for its covered
envelope, $\hatpmb u$ for the fully adapted extension of $\pmb u\in\Mmo$ to
$\hat\Cal S$, and $\hat\psi$ for the strictly adapted extension of $\psi$ to
$\hat\Cal S^{2\uparrow}$.   Then $\hat\psi$ is an integrating interval function
(616F) and $\hatpmb u$ is moderately oscillatory (615F(a-vi)).   Write
$\hatpmb v$ for the indefinite integral $ii_{\hat\psi}(\tbf 1)$.   If
$\tau\in\Cal S$, $\hat\Cal S\wedge\tau$ is the covered envelope of
$\Cal S\wedge\tau$ and $\int_{\hat\Cal S\wedge\tau}\tbf 1\,d\hat\psi$ is
defined, so is equal to $\int_{\Cal S\wedge\tau}\tbf 1\,d\psi$ (613T again);
thus $\pmb v=\hatpmb v\restr\Cal S$ is the indefinite integral
$ii_{\psi}(\tbf 1)$.   Now
$\int_{\hat\Cal S}\hatpmb u\,d\hat\psi$ is defined and equal to
$\int_{\hat\Cal S}\hatpmb u\,d\hatpmb v$, by (d);
by 613T once more, $\int_{\Cal S}\pmb u\,d\psi$ and
$\int_{\Cal S}\pmb u\,d\pmb v$ are defined and equal.   So we have the general
result.
}%end of proof of 616M

\leader{616N}{}\cmmnt{ So far, I have been working almost entirely with general
integrating interval functions.   But 616M makes it plain that we can expect
that usually it will be enough to look at integration with respect to
integrators, which is what most of the rest of this volume will be devoted to.
As we have already seen in 616I, integrators have some special properties, which
I will now set out to describe.

\medskip

\noindent}{\bf Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a (local) integrator.   If
$f:\Bbb R\to\Bbb R$ is convex, then $\bar f\pmb v$ is a (local) integrator.

%or maybe:  if $f$ Lipschitz on every bdd interval \query

\proof{{\bf (a)} To begin with (down to the end of (c) below) suppose
that $\pmb v$ is an integrator.   Note straight away that $f$ is continuous
(6A1Aa), so that $\bar f:L^0\to L^0$ is continuous (613Bb).
As $\pmb v$ is moderately oscillatory, therefore order-bounded (616I),
we see from 615G that
$\lim_{\sigma\downarrow\Cal S}\bar f(v_{\sigma})
=\bar f(\lim_{\sigma\downarrow\Cal S}v_{\sigma})$,
$\lim_{\sigma\uparrow\Cal S}\bar f(v_{\sigma})
=\bar f(\lim_{\sigma\uparrow\Cal S}v_{\sigma})$ are defined, while
$\bar f\pmb v$ is order-bounded (614F(c-i)).

Write $Q$ for $Q_{\Cal S}(d\pmb v)$ and
$Q^*$ for $Q_{\Cal S}(d(\bar f\pmb v))$.   We know that
$Q$ is topologically bounded, and we need to show that $Q^*$ is
topologically bounded.

Let $g$ be the right derivative of $f$, so that
$g$ is non-decreasing and
$(y-x)g(x)\le f(y)-f(x)$ for all $x$, $y\in\Bbb R$ (6A1Ab).
Consequently

\Centerline{$\bar g(v_{\sigma})\times(v_{\sigma'}-v_{\sigma})
\le\bar f(v_{\sigma'})-\bar f(v_{\sigma})$}

\noindent for all $\sigma$, $\sigma'\in\Cal S$.

By 616Ib and 614F(c-i), $\bar f\pmb v$ is order-bounded;
set $w=\sup|\bar f\pmb v|\vee\sup|\pmb v|$.

\medskip

{\bf (b)} Suppose for the moment that
$|g(x)|\le M$ for every $x\in\Bbb R$, where $M\ge 0$.   Because the
solid hull of a topologically bounded set is topologically bounded,
a scalar multiple of a topologically bounded set is topologically
bounded and the algebraic sum of two topologically bounded sets is
topologically bounded (613Bf), $A_0=[-2w,2w]+MQ$, the solid hull
$A_1$ of $A_0$ and $A=A_1+MQ$ are topologically bounded.   Now
$Q^*\subseteq A$.   \Prf\ Take $z\in Q^*$.   Then there are
a finite sublattice $I$ of
$\Cal S$ and a fully adapted family
$\pmb u=\family{\sigma}{I}{u_{\sigma}}$ such that
$|u_{\sigma}|\le\chi 1$ for $\sigma\in I$ and
$z=S_I(\pmb u,d(\bar f\pmb v))$.   If $I=\emptyset$ then
$z=0$ surely belongs to $A$.   Otherwise, let
$\tau_0\le\ldots\le\tau_n$ linearly generate the $I$-cells.
For $i\le n$ set

\Centerline{$w_i=\bar f(v_{\tau_i})-\bar f(v_{\tau_0})
   -\sum_{j=0}^{i-1}\bar g(v_{\tau_j})
        \times(v_{\tau_{j+1}}-v_{\tau_j})
\in [-2w,2w]+MQ=A_0$}

\noindent
because $|\bar f(v_{\tau_i})|\le w$, $|\bar f(v_{\tau_0})|\le w$ and
$|\bar g(v_{\tau_j}|\le M\chi 1$ for every $j$.	

For $i<n$,

\Centerline{$w_{i+1}-w_i
=\bar f(v_{\tau_{i+1}})-\bar f(v_{\tau_i})
   -\bar g(v_{\tau_i})\times(v_{\tau_{i+1}}-v_{\tau_i})
\ge 0$.}

\noindent Now we have

$$\eqalignno{z
&=\sum_{i=0}^{n-1}u_{\tau_i}
     \times(\bar f(v_{\tau_{i+1}})-\bar f(v_{\tau_i}))\cr
\displaycause{613Ec}
&=\sum_{i=0}^{n-1}u_{\tau_i}\times(w_{i+1}-w_i)
    +\sum_{i=0}^{n-1}u_{\tau_i}\times\bar g(v_{\tau_i})
        \times(v_{\tau_{i+1}}-v_{\tau_i}).\cr}$$
	
\noindent But

$$\eqalign{|\sum_{i=0}^{n-1}u_{\tau_i}\times(w_{i+1}-w_i)|
&\le\sum_{i=0}^{n-1}|u_{\tau_i}|\times|w_{i+1}-w_i|
\le\sum_{i=0}^{n-1}|w_{i+1}-w_i|\cr
&=\sum_{i=0}^{n-1}w_{i+1}-w_i
=w_n-w_0
=w_n
\in A_0,\cr}$$

\noindent so $\sum_{i=0}^{n-1}u_{\tau_i}\times(w_{i+1}-w_i)\in A_1$;
while

\Centerline{$\sum_{i=0}^{n-1}u_{\tau_i}\times\bar g(v_{\tau_i})
        \times(v_{\tau_{i+1}}-v_{\tau_i})\in MQ$}
	
\noindent because $|u_{\tau_i}\times\bar g(v_{\tau_i})|\le M\chi 1$
for every $i$.   So $z\in A_1+MQ=A$.   	As $z$ is arbitrary,
$Q^*\subseteq A$.\ \Qed

Thus in this case $Q^*$ is topologically bounded.

\medskip

{\bf (c)} For general $g$, take $\epsilon>0$.   Let $M\ge 0$
be such that
$\bar\mu\Bvalue{w>M}\le\epsilon$.   Define $f_1:\Bbb R\to\Bbb R$ by setting

$$\eqalign{f_1(x)
&=f(x)\text{ if }|x|\le M,\cr
&=f(-M)+(-M-x)g(-M)\text{ if }x\le -M,\cr
&=f(M)+(M-x)g(M)\text{ if }x\ge M.\cr}$$

\noindent Then $f_1$ is convex and its right derivative takes values in
$[g(-M),g(M)]$ so is bounded.

By (b), there is a $\delta>0$ such that
$\theta(\delta z_1)\le\epsilon$ for every
$z_1\in Q_{\Cal S}(\bar f_1\pmb v)$.   Now take
$z\in Q^*$.   Express $z$ as $S_I(\pmb u,d(\bar f\pmb v))$ where
$I$ is a finite sublattice of
$\Cal S$ and $\pmb u=\family{\sigma}{I}{u_{\sigma}}$
is a fully adapted family such that
$|u_{\sigma}|\le\chi 1$ for $\sigma\in I$.
Set $z_1=S_I(\pmb u,d(\bar f_1\pmb v))$.   Then
$z_1\in Q_{\Cal S}(\bar f_1\pmb v)$ and
$\theta(\delta z_1)\le\epsilon$.   But

$$\eqalignno{\Bvalue{z_1\ne z}
&\Bsubseteq\sup_{\sigma\in\Cal S}
   \Bvalue{\bar f_1(v_{\sigma})\ne\bar f(v_{\sigma})}
\Bsubseteq\sup_{\sigma\in\Cal S}
   \Bvalue{|v_{\sigma}|>M}\cr
\displaycause{because $f_1(x)=f(x)$ if $|x|\le M$}
&\Bsubseteq\Bvalue{w>M}\cr}$$

\noindent has measure at most $\epsilon$, and

\Centerline{$\theta(\delta z)
\le\theta(\delta z_1)+\bar\mu\Bvalue{z_1\ne z}
\le 2\epsilon$.}

\noindent This is true for every $z\in Q^*$.
As $\epsilon$ is arbitrary, $Q^*$ is topologically bounded, and
$\bar f\pmb v$ is an integrator.

\medskip

{\bf (d)} If $\pmb v$ is a local integrator, (a)-(c) show that
$(\bar f\pmb v)\restr\Cal S\wedge\tau
=\bar f(\pmb v\restr\Cal S\wedge\tau)$ is an integrator for every
$\tau\in\Cal S$, so
$\bar f\pmb v$ is a local integrator.
}%end of proof of 616N

\leader{616O}{Corollary}
If $\pmb v$ is a (local) integrator and $f:\Bbb R\to\Bbb R$
is a function, absolutely continuous on every bounded interval in $\Bbb R$,
such that its derivative $f'$ has bounded variation on every bounded interval,
then $\bar f\pmb v$ is a (local) integrator.

\proof{ $f$ is expressible as a difference of two convex functions
(6A1B), so $\bar f\pmb v$ is the difference of two (local)
integrators and is a (local) integrator.
}%end of proof of 616O

\leader{616P}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$.

(a) $\Mir(\Cal S)$ is an $f$-subalgebra of the space $\Mmo(\Cal S)$ of
moderately oscillatory processes with domain $\Cal S$.

(b)(i) Constant processes are integrators.

\quad(ii) If $\pmb v\in\Mir(\Cal S)$ then
$\pmb v\restr\Cal S'\in\Mir(\Cal S')$ for any sublattice $\Cal S'$ of
$\Cal S$.   In particular, $\pmb v$ is a local integrator.

\quad(iii) Suppose that $\pmb v\in\Mfa(\Cal S)$ and for every
$\epsilon>0$ there is a $\pmb v'\in\Mir(\Cal S)$ such
that $\bar\mu\Bvalue{\pmb v\ne\pmb v'}\le\epsilon$.
Then $\pmb v\in\Mir(\Cal S)$.

\quad(iv) If $\pmb v=\sfamilysS{v}\in\Mir(\Cal S)$ and
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$, then
$z\pmb v\cmmnt{\mskip5mu=\sfamilysS{z\times v}}\in\Mir(\Cal S)$.

\quad(v) If $\pmb v\in\Mfa(\Cal S)$ then $\pmb v\in\Mir(\Cal S)$ iff
$\pmb v\restr\Cal S\wedge\tau\in\Mir(\Cal S\wedge\tau)$ and
$\pmb v\restr\Cal S\vee\tau\in\Mir(\Cal S\vee\tau)$.

\proof{{\bf (a)} We saw in 616Ib that $\Mir(\Cal S)\subseteq\Mmo(\Cal S)$.
Since a sum of topologically bounded sets
is topologically bounded, 616Dc shows that $\Mir(\Cal S)$ is closed under
addition.   By 616N, $\bar h\pmb v\in\Mir(\Cal S)$ whenever
$h:\BbbR\to\BbbR$ is convex and $\pmb v\in\Mir(\Cal S)$,
so $\Mir(\Cal S)$ is an $f$-subalgebra of $\Mmo(\Cal S)$ (612Bc).

\medskip

{\bf (b)(i)}  If $\pmb v$ is constant then $Q_{\Cal S}(d\pmb v)=\{0\}$.

\medskip

\quad{\bf (ii)} As in 616Dd,
$Q_{\Cal S}(d(\pmb v\restr\Cal S'))\Bsubseteq Q_{\Cal S}(d\pmb v)$.

\medskip

\quad{\bf (iii)} Let $\epsilon>0$.   Then there are a
$\pmb v'\in\Mir(\Cal S)$ such
that $\bar\mu\Bvalue{\pmb v\ne\pmb v'}\le\bover12\epsilon$, and a $\delta>0$
such that $\theta(\delta z)\le\bover12\epsilon$ for every
$z\in Q_{\Cal S}(d\pmb v')$.   Now suppose that $z\in Q_{\Cal S}(d\pmb v)$.
Then there is an $I\in\Cal I(\Cal S)$ such that $z=S_I(\tbf 1,d\pmb v)$.
In this case, expressing $\pmb v$ and $\pmb v'$ as $\sfamilysS{v}$ and
$\sfamilysS{v'}$,

\Centerline{$\Bvalue{S_I(\tbf 1,d\pmb v')\ne z}
\Bsubseteq\sup_{\sigma\in I}\Bvalue{v'_{\sigma}\ne v_{\sigma}}
\Bsubseteq\Bvalue{\pmb v'\ne\pmb v}$}

\noindent and

$$\eqalign{\theta(\delta z)&
\le\theta(\delta z-\delta S_I(\tbf 1,d\pmb v'))
   +\theta(\delta S_I(\tbf 1,d\pmb v'))\cr&
\le\bar\mu\Bvalue{S_I(\tbf 1,d\pmb v')\ne z}+\Bover12\epsilon
\Bsubseteq\bar\mu\Bvalue{\pmb v'\ne\pmb v}+\Bover12\epsilon
\le\epsilon.\cr}$$

\noindent Accordingly $Q_{\Cal S}(d\pmb v)$ is topologically bounded and
$\pmb v\in\Mir(\Cal S)$.

\medskip

\quad{\bf (iv)}

$$\eqalignno{Q_{\Cal S}(d(z\pmb v))
&=\{S_I(\pmb u,d(z\pmb v)):I\in\Cal I(\Cal S),\,
  \pmb u\in\Mfa(I),\,\sup|\pmb u|\le\chi 1\}
\cr&=\{z\times S_I(\pmb u,d\pmb v):I\in\Cal I(\Cal S),\,
  \pmb u\in\Mfa(I),\,\sup|\pmb u|\le\chi 1\}\cr
\displaycause{613L(b-iii)}
&=\{z\times w:w\in Q_{\Cal S}(d\pmb v)\}.\cr}$$

\noindent As $Q_{\Cal S}(d\pmb v)$ is topologically bounded, and
$w\mapsto z\times w:L^0\to L^0$ is a continuous linear operator,
$Q_{\Cal S}(d(z\pmb v))=\{z\times w:w\in Q_{\Cal S}(d\pmb v)\}$ is
topologically bounded (3A5N(b-v)) and $z\pmb v$ is an integrator.

\medskip

\quad{\bf (v)}
If $\pmb v$ is an integrator, then $\pmb v\restr\Cal S\wedge\tau$ and
$\pmb v\restr\Cal S\vee\tau$ are integrators, by (ii) above.
On the other hand, $Q_{\Cal S}(d\pmb v)
\subseteq Q_{\Cal S\wedge\tau}(d\pmb v)+Q_{\Cal S\vee\tau}(d\pmb v)$.   \Prf\
Take $z\in Q_{\Cal S}(d\pmb v)$.   Then $z$ is expressible as
$\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})$ where
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and
$\pmb u=\sfamilysS{u}$ is a ful;ly adapted process such that
$\sup|\pmb u|\le\chi 1$ (616C(iii)).   Now for
each $i\le n$, $v_{\tau_i}=v_{\tau_i\wedge\tau}+v_{\tau_i\vee\tau}-v_{\tau}$
(612D(f-i)).   So if $i<n$ then

\Centerline{$v_{\tau_{i+1}}-v_{\tau_i}
=v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau}
     +v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau}$.}

\noindent Define $u'_i$, $u''_i$ by saying that

\Centerline{$u'_i=u_{\tau_i}\times\chi\Bvalue{\tau_i\le\tau}$,
\quad$u''_i
=u_{\tau_i}\times\chi\Bvalue{\tau<\tau_{i+1}}$.}

\noindent Evidently $|u'_i|\le\chi 1$ and $|u''_i|\le\chi 1$.
By 612C, $u'_i\in L^0(\frak A_{\tau_i\wedge\tau})$;  and since
$\Bvalue{\tau<\tau_{i+1}}\in\frak A_{\tau}\Bsubseteq\frak A_{\tau_i\vee\tau}$,
$u''_i\in L^0(\frak A_{\tau_i\vee\tau})$.   Next,

$$\eqalign{\Bvalue{\tau<\tau_i}&
\Bsubseteq\Bvalue{u'_i=0}
 \Bcap\Bvalue{\tau_i\vee\tau=\tau_i}\Bcap\Bvalue{\tau_{i+1}\vee\tau=\tau_{i+1}}
 \Bcap\Bvalue{u''_i=u_{\tau_i}}\cr&
\Bsubseteq\Bvalue{u'_i=0}
 \Bcap\Bvalue{u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})
        =u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})}\cr&
\Bsubseteq\Bvalue{u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
   +u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})
        =u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})},\cr
\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}&
\Bsubseteq[\![u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
   +u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})\cr&
\mskip100mu =u_{\tau_i}\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau}
                       +v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})]\!]\cr&
=\Bvalue{u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
   +u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})
        =u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})},\cr
\Bvalue{\tau_{i+1}\le\tau}&
\Bsubseteq\Bvalue{u'_i=u_{\tau_i}}
  \Bcap\Bvalue{\tau_i\wedge\tau=\tau_i}
  \Bcap\Bvalue{\tau_{i+1}\wedge\tau=\tau_{i+1}}
  \Bcap\Bvalue{u''_i=0}\cr&
\Bsubseteq\Bvalue{u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
        =u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})}
  \Bcap\Bvalue{u''_i=0}\cr&
\Bsubseteq\Bvalue{u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
   +u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})
        =u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})},\cr}$$

\noindent so

\Centerline{$u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
   +u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})
        =u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})$.}

\noindent Now if we write

\Centerline{$z'
=\sum_{i=0}^{n-1}u'_i\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})$,}

\Centerline{$z''
=\sum_{i=0}^{n-1}u''_i\times(v_{\tau_{i+1}\vee\tau}-v_{\tau_i\vee\tau})$,}

\noindent we have $z'\in Q_{\Cal S\wedge\tau}(d\pmb v)$ (616C(ii)),
$z''\in Q_{\Cal S\vee\tau}(d\pmb v)$ and $z=z'+z''$.
As $z$ is arbitrary, $Q_{\Cal S}(d\pmb v)
\subseteq Q_{\Cal S\wedge\tau}(d\pmb v)+Q_{\Cal S\vee\tau}(d\pmb v)$.\ \QeD

If $\pmb v\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\vee\tau$ are
integrators, $Q_{\Cal S\wedge\tau}(d\pmb v)$ and
$Q_{\Cal S\wedge\tau}(d\pmb v)$ are topologically bounded, so
$Q_{\Cal S}(d\pmb v)$ is topologically bounded (613B(f-iii) once more) and
$\pmb v$ is an integrator.
}%end of proof of 616P

\leader{616Q}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$.

(a) $\Mlir(\Cal S)$ is an $f$-subalgebra of the space $\Mlmo(\Cal S)$ of
locally moderately oscillatory processes with domain $\Cal S$.

(b)(i) If $\pmb v\in\Mlir(\Cal S)$ then
$\pmb v\restr\Cal S'\in\Mlir(\Cal S')$ for any sublattice $\Cal S'$ of
$\Cal S$.

\quad(ii) If $\pmb v\in\Mlir(\Cal S)$ and
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$, then
$z\pmb v\in\Mlir(\Cal S)$.

(c) Suppose that $\pmb u\in\Mlmo(\Cal S)$ and $\pmb v\in\Mlir(\Cal S)$.

\quad(i) The indefinite integral $\iivu$ belongs to $\Mlir(\Cal S)$.

\quad(ii) Let $\hat\Cal S$ be the covered envelope of $\Cal S$, and
$\hatpmb u$, $\hatpmb v$ the fully adapted extensions of
$\pmb u$, $\pmb v$ to $\hat\Cal S$.   Then
$\iivu=ii_{\hatpmb v}(\hatpmb u)\restr\Cal S$.

(d) Suppose that $\pmb v\in\Mfa(\Cal S)$ and $\Cal S'$ is a covering ideal of
$\Cal S$ such that $\pmb v\restr\Cal S'\in\Mlir(\Cal S')$.   Then
$\pmb v\in\Mlir(\Cal S)$.

\proof{{\bf (a)-(b)}
We just have to apply 616P to $\pmb v\restr\Cal S\wedge\tau$ for
$\tau\in\Cal S$.

\medskip

{\bf (c)(i)} Apply 616J to $\pmb u\restr\Cal S\wedge\tau$ and
$\psi=\Delta(\pmb v\restr\Cal S\wedge\tau)$ for $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)} By 615F(b-v) and 616Ia, $\hatpmb u\in\Mlmo(\hat\Cal S)$ and
$\hatpmb v\in\Mlir(\hat\Cal S)$, so $ii_{\hatpmb v}(\hatpmb u)$ is well-defined
on $\hat\Cal S$.   Now if $\tau\in\Cal S$,

\Centerline{$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
=\int_{\hat\Cal S\wedge\tau}\hatpmb u\,d\hatpmb v$}

\noindent as in 613T.

\medskip

{\bf (d)} If $\tau\in\Cal S$ and $\epsilon>0$, there is a $\sigma\in\Cal S'$
such that $\sigma\le\tau$ and $\bar\mu\Bvalue{\sigma<\tau}\le\epsilon$.
Consider the process
$\pmb v'=\family{\rho}{\Cal S\wedge\tau}{v_{\rho\wedge\sigma}}$.   By
612Ib, this is fully adapted, while

\Centerline{$\pmb v'\restr\Cal S\wedge\sigma=\pmb v\restr\Cal S\wedge\sigma
=\pmb v\restr\Cal S'\wedge\sigma$}

\noindent is an integrator, and $\pmb v'\restr(\Cal S\wedge\tau)\vee\sigma$ is
constant, therefore an integrator.   So $\pmb v'\in\Mir(\Cal S\wedge\tau)$
(616P(b-v)), while

\Centerline{$\Bvalue{\pmb v'\ne\pmb v\restr\Cal S\wedge\tau}
=\sup_{\rho\in\Cal S\wedge\tau}\Bvalue{v_{\rho\wedge\sigma}\ne v_{\rho}}
\Bsubseteq\sup_{\rho\in\Cal S\wedge\tau}\Bvalue{\rho\wedge\sigma<\rho}
=\Bvalue{\sigma<\tau}$}

\noindent has measure at most $\epsilon$.   As $\epsilon$ is arbitrary,
616P(b-iii) tells us that
$\pmb v\restr\Cal S\wedge\tau\in\Mir(\Cal S\wedge\tau)$.   As $\tau$ is
arbitrary, $\pmb v\in\Mlir(\Cal S)$.
}%end of proof of 616Q

\leader{616R}{Proposition} Suppose that $\Cal S$ is a
sublattice of $\Cal T$, and that a fully adapted process
$\pmb v$ with domain $\Cal S$ is (locally) of bounded variation.

(a) $\pmb v$ is a (local) integrator.

(b) Now suppose that $\pmb v$ is non-decreasing and that
$\pmb u$ is a non-negative moderately
oscillatory process with domain $\Cal S$.

\quad(i) If $\pmb v$ is of bounded variation then
$\int_{\Cal S}\pmb u\,d\pmb v\ge 0$.

\quad(ii) If $\pmb v$ is locally of bounded variation then $\iivu$ is
non-decreasing.

\proof{{\bf (a)(i)} Suppose that $\pmb v=\sfamilysS{v}$ is of bounded variation.
Then $\{S_I(\tbf{1},|d\pmb v|):I\in\Cal I(\Cal S)\}$ has an
upper bound $\bar z$ say.  If
$I$ is a finite sublattice of $\Cal S$ and
$\pmb u=\family{\sigma}{I}{u_{\sigma}}$
is a fully adapted process with $\sup|\pmb u|\le\chi 1$, then

\Centerline{$|u_{\sigma}\times(v_{\sigma'}-v_{\sigma})|
\le|v_{\sigma'}-v_{\sigma}|$}

\noindent whenever $\sigma\le\sigma'$ in $I$, so
$|S_I(\pmb u,d\pmb v)|\le S_I(\tbf{1},|d\pmb v|)\le\bar z$.
Thus $Q_{\Cal S}(d\pmb v)$ is order-bounded in $L^0$ and must be
bounded for the topology of convergence in measure.
So $\pmb v$ is an integrator.
By 616Ib, or otherwise, $\pmb v$ is moderately oscillatory and order-bounded.

\medskip

\quad{\bf (ii)} If $\pmb v$ is locally of bounded variation, apply
(a) to $\pmb v\restr\Cal S\wedge\tau$, for $\tau\in\Cal S$, to see that
$\pmb v$ is a local integrator and therefore locally moderately oscillatory and
locally order-bounded.

\medskip

{\bf (b)} Recall that by 616K the integral $\int_{\Cal S}\pmb u\,d\pmb v$ is
defined.

\medskip

\quad{\bf (i)} All the sums $S_I(\pmb u,d\pmb v)$, for $I\in\Cal I(\Cal S)$,
are non-negative, so the limit $\int_{\Cal S}\pmb u\,d\pmb v$ also is.

\medskip

\quad{\bf (ii)} If $\tau$, $\tau'\in\Cal S$ and $\tau\le\tau'$, then

\Centerline{$\int_{\Cal S\wedge\tau'}\pmb u\,d\pmb v
   -\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v
=\int_{\Cal S\cap[\tau,\tau']}\pmb u\,d\pmb v\ge 0$.}
}%end of proof of 616R

\leader{616S}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$ and $\pmb v$
a process of bounded variation with domain $\Cal S$.   Then $|\Delta\pmb v|$ is
an integrating interval function.

\proof{{\bf (a)} By 613Cc and 613Db, $|\Delta\pmb v|$ is a
strictly adapted interval function on $\Cal S$.   We know that
$\bar v=\int_{\Cal S}|\Delta\pmb v|=\int_{\Cal S}|d\pmb v|$
is defined and an upper bound of
$\{\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|:\tau_0\le\ldots\le\tau_n$
in $\Cal S\}$ (614I).   But now we see that
$Q_{\Cal S}(|d\pmb v|)\subseteq[-\bar v,\bar v]$.
\Prf\ Take $z\in Q_{\Cal S}(|d\pmb v|)$.   If $z=0$ then surely
$z\in[-\bar v,\bar v]$.   Otherwise, there are a non-empty
$I\in\Cal I(\Cal S)$ and a $\pmb u\in\Mfa(I)$ such that
$\sup|\pmb u|\le\chi 1$ and $z=S_I(\pmb u,|d\pmb v|)$.
Take $\tau_0\le\ldots\le\tau_n$ linearly generating the $I$-cells.
Then

$$\eqalign{|z|&
=|\sum_{i=0}^{n-1}u_{\tau_i}\times|v_{\tau_{i+1}}-v_{\tau_i}||\cr&
\le\sum_{i=0}^{n-1}|u_{\tau_i}|\times|v_{\tau_{i+1}}-v_{\tau_i}|
\le\sum_{i=0}^{n-1}|v_{\tau_{i+1}}-v_{\tau_i}|
\le\bar v\cr}$$

\noindent and again $z\in[-\bar v,\bar v]$.\ \QeD\
So $Q_{\Cal S}(|d\pmb v|)$ is order-bounded, therefore topologically bounded.

\medskip

{\bf (b)} If $\Cal S$ is full, this is already enough to check that
$|\Delta\pmb v|$ is an integrating interval function.
If $\Cal S$ is not full, then
by 614Q(a-iv) the adapted extension $\hatpmb v$ of $\pmb v$ to the
covered envelope $\hat\Cal S$ of $\Cal S$ is of bounded variation, and
$|\Delta\hatpmb v|$ must be the adapted extension of $|\Delta\pmb v|$ to
$\hat\Cal S^{2\uparrow}$.   Since $|\Delta\hatpmb v|$ is an integrating interval
function, so is $|\Delta\pmb v|$ (616F).
}%end of proof of 616S

\leader{616T}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u$, $\pmb v$ fully adapted processes with domain $\Cal S$.

(a) If $\pmb u$ is moderately oscillatory and $\pmb v$ is of bounded variation,
then $\int_{\Cal S}\pmb u\,|d\pmb v|$ is defined and equal to
$\int_{\Cal S}\pmb u\,d\pmb v^{\uparrow}$, where $\pmb v^{\uparrow}$ is the
cumulative variation of $\pmb v$.

(b) If $\pmb u$ is locally moderately oscillatory and $\pmb v$ is locally
of bounded variation, then the indefinite integrals
$ii_{|\Delta\pmb v|}(\pmb u)$ and $ii_{\pmb v^{\uparrow}}(\pmb u)$ are equal.

\proof{{\bf (a)} By the definition 614O,
$\pmb v^{\uparrow}=ii_{|\Delta\pmb v|}(\tbf 1)$,
so we can apply 616M with $\psi=|\Delta\pmb v|$.

\medskip

{\bf (b)} Apply (a) to $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb v\restr\Cal S\wedge\tau$ for $\tau\in\Cal S$.
}

\medskip

\noindent{\bf Mnemonic} $|d\pmb v|\sim d\pmb v^{\uparrow}$.

%\leader{616V}{}
%\leader{616W}{}

\exercises{\leader{616X}{Basic exercises (a)}
%\spheader 616Xa
Suppose that $T=\coint{0,\infty}$ and that
$\frak A=\{0,1\}$, as in 613W, 614Xd and 615Xf.
(i) Show that if $f:\coint{0,\infty}\to\Bbb R$ is interpreted as a fully
adapted process $\pmb v$ with domain $\Cal T_f$, then
$Q_{\Cal T_f}(d\pmb v)$ can be identified with either
$\ooint{-\Var(f),\Var(f)}$ or $[-\Var(f),\Var(f)]$ where $\Var(f)$ is the
total variation of $f$ (224A).
(ii) Show that $f:\coint{0,\infty}\to\Bbb R$ corresponds to an integrator
iff it corresponds to a process of bounded variation iff it is
itself of bounded variation in the sense of 224A.
%613W 616B 616Fc

\spheader 616Xb Let $\Cal S$ be a sublattice of $\Cal T$.   Show that any simple
process with domain $\Cal S$ is an integrator.
%616R 614Q(a-iii)

\spheader 616Xc Let $\Cal S$ be a sublattice of $\Cal T$ and $\pmb v$ a
fully adapted process with domain $\Cal S$.   Show that $\pmb v$ is of bounded
variation iff $Q_{\Cal S}(d\pmb v)$ is bounded above
iff $Q_{\Cal S}(|d\pmb v|)$ is bounded above, and that in this case
$\int_{\Cal S}|d\pmb v|=\sup Q_{\Cal S}(d\pmb v)=\sup Q_{\Cal S}(|d\pmb v|)$.
\Hint{351Dc.}
%616S 351Dc

\leader{616Y}{Further exercises $>$(a)}
%\spheader 616Ya
Let $\pmb w=\family{\sigma}{\Cal T_f}{w_{\sigma}}$ be Brownian
motion as described in 612T, based on the real-time stochastic integration
structure $(\frak C,\bar\nu,\langle\frak C_t\rangle_{t\ge 0},\Cal T,
\family{\tau}{\Cal T}{\frak C_{\tau}})$.   (i) Show that
$w_{\sigma}\in L^2(\bar\nu)$ for every $\sigma\in\Cal T_b$
({\it hint\/}:  477A).
(ii) Show that whenever $\sigma\le\tau$ in $\Cal T_b$ and
$u\in L^2(\bar\nu)\cap L^0(\frak C_{\sigma})$ then
$\Expn(u\times(w_{\tau}-w_{\sigma}))=0$ ({\it hint\/}:  477G).
(iii) Show that if $\sigma_0\le\ldots\le\sigma_n$ in
$\Cal T_b$ and $u_i\in L^{\infty}(\frak C_{\sigma_i})$ and
$|u_i|\le\chi 1$ for $i\le n$, then
($\alpha$) $\Expn(u_i\times u_j\times(w_{\sigma_{i+1}}-w_{\sigma_i})
  \times(w_{\tau_{j+1}}-w_{\tau_j})=0$ whenever $i<j<n$
($\beta$)

$$\eqalign{\Expn((\sum_{i=0}^{n-1}u_i\times(w_{\sigma_{i+1}}-w_{\sigma_i}))^2)&
=\Expn(\sum_{i=0}^{n-1}u_i^2\times(w_{\sigma_{i+1}}-w_{\sigma_i})^2)
\le\sum_{i=0}^{n-1}\Expn((w_{\sigma_{i+1}}-w_{\sigma_i})^2)\cr&
=\sum_{i=0}^{n-1}\Expn(w_{\sigma_{i+1}}^2-w_{\sigma_i}^2)
\le\|w_{\sigma_n}\|_2^2.\cr}$$

\noindent (iv) Show that if $\tau\in\Cal T_b$ then
$Q_{\Cal T\wedge\tau}(d\pmb w)$ is $\|\,\|_2$-bounded.   (v) Show that
$\pmb w\restr\Cal T_b$ is a local integrator ({\it hint\/}:  613B(f-ii)).
(vi) Show that $\pmb w$ is a local integrator ({\it hint\/}:  616Qd).

\spheader 616Yb Find an example in which $\Cal S$ is a sublattice of $\Cal T$,
$\psi:\Cal S^{2\uparrow}\to L^0$ is a strictly adapted interval function and
$Q_{\Cal S}(d\psi)$ is  topologically bounded but $Q_{\hat\Cal S}(d\hat\psi)$ is
not, where $\hat\Cal S$ is the covered envelope of $\Cal S$ and
$\hat\psi:\hat\Cal S^{2\uparrow}\to L^0$ is the strictly adapted extension of
$\psi$.

\leaveitout{$\psi(\check q_{ni},\check q'_{ni})=n\chi a_{ni}$ for $i<2^n$,
where $\bigcap_{n,i}\ooint{q_{ni},q'_{ni}}\ne\emptyset$ and
$\ofamily{i}{2^n}{a_{ni}}$ is a partition of unity,
and $\psi(\check s,\check t)=0$ for other $s\le t$.}
%616F

\spheader 616Yc Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$ a previsibly
simple process with domain $\Cal S$, % (612Ye)
and $\pmb v$ an integrator with
domain $\Cal S$.   Show that $\int_{\Cal S}\pmb u\,d\pmb v$ is defined.
%mt61bits 616Ib

\spheader 616Yd In the construction described in 613S, show that the process
$\pmb v$ is moderately oscillatory but not an integrator.
%613S 615Yc 616K
}%end of exercises

\endnotes{
\Notesheader{616}
You will have noticed that I have given practically no examples in this section.
It is in fact the case that all three of the central
examples in \S612 (the identity process, Brownian motion and Poisson processes)
are local integrators.   The first and third are easy (614Xe).   Brownian
motion, as usual, is more interesting and much more important.   Its natural
place in the structure of this volume is in Chapter 62 below (see 622L).
But as It\^o's formula (\S619 below) would be of no importance without
Brownian motion, or at least something very like it, I have sketched an
argument in 616Ya, which I hope you will tackle straight away.   Even if parts
are too difficult at the moment, it should be instructive.

We shall be spending a great deal of time teasing out
the nature of integrators and integrating interval functions.   An elementary
class of integrators was treated in \S614.   This will take us
to the statement and proof of It\^o's formula,
if not to its most important applications.   There will be some much deeper
results about integrators in \S622 and \S627.
But from the abstract definition, we can see immediately that
whether an adapted interval function is `integrating', or a process is an
`integrator', depends on the
topology of $L^0$, but not on the measure inducing this topology;  as with
integration (613I) and `moderately oscillatory' processes (615E), the property
is law-independent.

In 616K-616M, we have the central case in which
we can be sure that a Riemann-sum integral
$\int_{\Cal S}\pmb u\,d\psi$ or $\int_{\Cal S}\pmb u\,d\pmb v$ is defined;
it will be sufficient
to suppose that $\pmb u$ is moderately oscillatory and $\psi$ or $\Delta\pmb v$
is an integrating interval function.
Compared with 614C and 614T, we have a condition on
$\pmb u$ which allows very much more variety,
balanced by a strong new condition on $\pmb v$.

In 616M, we have a first example of a phenomenon which will be important on
many occasions.   Here we start with an integrating interval function $\psi$,
construct an integrator $\pmb v=ii_{\psi}(\tbf 1)$, and observe that
$\int_{\Cal S}\pmb u\,d\psi=\int_{\Cal S}\pmb u\,d\pmb v$ for every moderately
oscillatory process $\pmb u$ with domain $\Cal S$.   So in this sense $d\psi$
and $d\pmb v$ are interchangeable.   This is not at all because the interval
functions $\psi$ and $\Delta\pmb v$ are the same.   For
$(\sigma,\tau)\in\Cal S^{2\uparrow}$, we have
$\Delta\pmb v(\sigma,\tau)=\int_{\Cal S\cap[\sigma,\tau]}d\psi$, which only in
special cases will be equal to $\psi(\sigma,\tau)$.   But in the formulae of
this theory of integration, $d\psi$ and $d\pmb v$ are equivalent, and I will
write $d\psi\sim d\pmb v$ as an aide-memoire.

Let me try to explain the phrase `capped-stake variation set'.
One of the ways of interpreting a stochastic integral
$\int_{\Cal S}\pmb u\,d\psi$ is as the net gain of a gambler who at any
stopping time $\sigma$ chooses to wager a stake $u_{\sigma}$, and
whose winnings over an interval $[\sigma,\tau]$ are
$u_{\sigma}\times\psi(\sigma,\tau)$.   (So if $\psi=\Delta\pmb v$,
he gets $u_{\sigma}\times(v_{\tau}-v_{\sigma})$, representing the gain
on betting $u_{\sigma}$ units in a stock whose value changes from
$v_{\sigma}$ to $v_{\tau}$.)   On this formulation, a simple strategy for
$\pmb u$ is one which involves only finitely many stopping times declared in
advance, like a stop-loss order.   A Riemann sum
$S_I(\pmb u,d\psi)$ represents his winnings if he adjusts his stake to
follow his strategy $\pmb u$
whenever a stopping time in $I$ is reached;  and we count the integral
as defined if all sufficiently fine readjustment schedules give about
the same result with high probability.   The
terms $S_I(\pmb u,d\psi)$ calculated
in the formula for $Q_{\Cal S}(d\psi)$ are those corresponding to
stakes (positive or negative) capped by $\pm 1$.\footnote{I
see that the metaphor is creaking at this point, because few investors
cap their investments by the
number of shares they hold, rather than the value of those shares,
which would correspond to a bound on $u_{\sigma}\times v_{\sigma}$.
The Dow-Jones index is exceptional in this respect.}

}%end of notes

\discrpage

\frfilename{mt617.tex}
\versiondate{10.11.21}

\def\ucplim{\mathop{\text{ucplim}}}

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Integral identities and quadratic variations}

\newsection{617}

We come now to proper calculus, with change-of-variable theorems.
617D-617E is a stochastic-calculus
version of the result that if $\nu=f\mu$ is an indefinite-integral
measure, then $\int g\,d\nu=\int g\times f\,d\mu$ (235K).   Similar formulae
describe the cumulative variation of an indefinite integral with respect to
a process of bounded variation (617G).
The next theme is `quadratic variation' (617H).
Given two integrators $\pmb v$ and $\pmb w$, the
interval function corresponding to $d\pmb vd\pmb w$
gives the same integrals as a process
$\covar{\pmb v}{\pmb w}$ (617I) which is locally of bounded variation.
In particular, $(d\pmb v)^2$ mimics $d\pmb v^*$ where
the quadratic variation $\pmb v^*$ is a non-decreasing
process.   Based on this, we have a second change-of-variable theorem
(617P-617Q),
using approximations of moderately oscillatory processes by simple
processes (617B).

\cmmnt{\leader{617A}{Notation}\cmmnt{ As before,} \Asissix\ will be a
stochastic integration structure.
$L^0=L^0(\frak A)$ will be given its topology of convergence in measure, with
corresponding F-norm $\theta$ (613Ba).
If $A\subseteq\Cal T$ and $\tau\in\Cal T$,
$A\wedge\tau=\{\sigma\wedge\tau:\sigma\in A\}$ and
$A\vee\tau=\{\sigma\vee\tau:\sigma\in A\}$.
If $\Cal S$ is a sublattice of $\Cal T$, $\Cal I(\Cal S)$
is the set of finite
sublattices of $\Cal S$, $\Mfa(\Cal S)$ is the space of fully adapted
processes with domain $\Cal S$, $\Mob(\Cal S)\subseteq\Mfa(\Cal S)$
is the space of order-bounded processes and $\Mmo(\Cal S)\subseteq\Mob(\Cal S)$
is the space of moderately oscillatory processes.   If
$\pmb u=\sfamilysS{u}\in\Mob(\Cal S)$,
$\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.
$\tbf 1$ will denote the constant process with value $\chi 1$.
}

\leader{617B}{Lemma} Let $\Cal S$ be a finitely full sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a moderately oscillatory process.

(a) For each non-empty
$I\in\Cal I(\Cal S)$ there is a unique simple process
$\pmb u_I=\family{\sigma}{\Cal S}{u_{I\sigma}}$ such that
$\pmb u_I$ has a breakpoint string in $I$, $\pmb u_I$ and $\pmb u$ agree on
$I$, and $\Bvalue{\sigma<\min I}\Bsubseteq\Bvalue{u_{I\sigma}=0}$ for every
$\sigma\in\Cal S$.

(b) Complete the definition in (a) by setting $u_{\emptyset\sigma}=0$
for every $\sigma\in\Cal S$.
For every integrator $\pmb v$ with domain $\Cal S$,

\quad (i) the indefinite integral $\iivu$ is the limit
$\lim_{I\uparrow\Cal I(\Cal S)}\iiv(\pmb u_I)$ for the ucp topology,

\quad (ii) $\int_{\Cal S}\pmb u\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}\pmb u_Id\pmb v$ in $L^0$.

\proof{{\bf (a)} Take $(\tau_0,\ldots,\tau_n)$ linearly generating the
$I$-cells, and let $\pmb u_I=\family{\sigma}{\Cal S}{u_{I\sigma}}$
be the simple process defined by the formulae of
612Ka applied to $\tau_0,\ldots,\tau_n$, $u_{\tau_0},\ldots,u_{\tau_n}$ and
$u_*=0$.   Then $u_{I\tau_n}=u_{\tau_n}$.  For $i<n$,

$$\eqalign{\Bvalue{u_{I\tau_i}=u_{\tau_i}}&
\Bsupseteq\Bvalue{\tau_i<\tau_{i+1}}\Bcup
   (\Bvalue{\tau_i=\tau_{i+1}}\Bcap\Bvalue{u_{I\tau_{i+1}}=u_{\tau_i}})\cr&
=\Bvalue{\tau_i<\tau_{i+1}}\Bcup
   (\Bvalue{\tau_i=\tau_{i+1}}\Bcap\Bvalue{u_{I\tau_{i+1}}=u_{\tau_{i+1}}})
\cr}$$

\noindent so a downwards induction shows that
$\Bvalue{u_{I\tau_i}=u_{\tau_i}}=1$ for every
$i\le n$.   Since $\{\tau_i:i\le n\}$ covers $I$ (611Ke), $\pmb u_I$ agrees with
$\pmb u$ on $I$.   Also

\Centerline{$\Bvalue{\sigma<\min I}=\Bvalue{\sigma<\tau_0}
\Bsubseteq\Bvalue{u_{I\sigma}=0}$}

\noindent by the choice of $u_*$.

To see that $\pmb u_I$ is unique, use 612Kb to see that $(\tau_0,\ldots,\tau_n)$
must be a breakpoint string for any process satisfying the conditions.

\medskip

{\bf (b)(i)} If $\Cal S$ is empty, the result is trivial;  suppose otherwise.
Let $\epsilon>0$.   Let $\delta\in\ocint{0,1}$
be such that $4\sqrt\delta\le\epsilon$.
By 615F(a-iii), $\pmb u$ is order-bounded and $\sup|\pmb u|$ is
defined in $L^0$;  let $\eta>0$ be such that
$\theta(x\times\sup|\pmb u|)\le\delta$ whenever $x\in L^0$ and
$\theta(x)\le\eta$.   As $\pmb v$ is moderately oscillatory (616Ib),
$\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is defined;  let $\tau^*\in\Cal S$ be
such that $\theta(v_{\sigma}-v_{\sigma'})\le\eta$ whenever
$\sigma$, $\sigma'\in\Cal S\vee\tau^*$.   Since
$\int_{\Cal S}\pmb u\,d\pmb v=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb v)$
is defined (616K), there is a $K\in\Cal I(\Cal S)$ such that $\tau^*\in K$ and
$\theta(S_I(\pmb u,d\pmb v)-S_J(\pmb u,d\pmb v))\le\delta$ whenever
$I$, $J\in\Cal I(\Cal S)$ include $K$.

Take $I\in\Cal I(\Cal S)$ such that $K\subseteq I$, and $\tau\in\Cal S$.
Then
$\theta(S_{I\wedge\tau}(\pmb u,d\pmb v)-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)
\le 2\delta$ (613V(ii-$\beta$)).   Also
$\theta(S_{I\wedge\tau}(\pmb u,d\pmb v)
  -\int_{\Cal S\wedge\tau}\pmb u_I\,d\pmb v)\le\delta$.   \Prf\ Write
$u_{I\downarrow}$, $v_{\downarrow}$
for $\lim_{\sigma\downarrow\Cal S}u_{I\sigma}$,
$\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ respectively,   If
$(\tau_0,\ldots,\tau_n)$ linearly generates $I$, then
$(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau)$ linearly generates the
$(I\wedge\tau)$-cells
(611Kg) and $(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau,\tau)$ is a breakpoint
string for $\pmb u_I\restr\Cal S\wedge\tau$ (612K(d-ii)), so we have

\Centerline{$S_{I\wedge\tau}(\pmb u,d\pmb v)
=\sum_{i=0}^{n-1}u_{\tau_i\wedge\tau}
\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})$,}

$$\eqalign{\int_{\Cal S\wedge\tau}\pmb u_Id\pmb v&
=u_{I\downarrow}\times(v_{\tau_0\wedge\tau}-v_{\downarrow})
   +\sum_{i=0}^{n-1}u_{I,\tau_i\wedge\tau}
      \times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})\cr&
\mskip250mu  +u_{I,\tau_n\wedge\tau}\times(v_{\tau}-v_{\tau_n\wedge\tau})\cr}$$

\noindent (613Ec, 614C).   Now

$$\eqalign{u_{I\downarrow}\times(v_{\tau_0\wedge\tau}-v_{\downarrow})&
=\lim_{\sigma\downarrow\Cal S}
   u_{I\sigma}\times(v_{\tau_0\wedge\tau}-v_{\sigma})
=\lim_{\sigma\downarrow\Cal S\wedge\tau_0\wedge\tau}
   u_{I\sigma}\times(v_{\tau_0\wedge\tau}-v_{\sigma})\cr&
=\lim_{\sigma\downarrow\Cal S\wedge\tau_0\wedge\tau}
   u_{I\sigma}\times(v_{\tau_0\wedge\tau}-v_{\sigma})
             \times\chi\Bvalue{\sigma<\tau_0}
=0\cr}$$

\noindent because we chose $u_*$ to be $0$ in the definition of $\pmb u_I$.
Next, for each $i<n$,

$$\eqalign{\Bvalue{\tau_i<\tau}&
\Bsubseteq\Bvalue{u_{I,\tau_i\wedge\tau}=u_{I\tau_i}}
             \Bcap\Bvalue{u_{\tau_i\wedge\tau}=u_{\tau_i}}
\Bsubseteq\Bvalue{u_{I,\tau_i\wedge\tau}=u_{\tau_i\wedge\tau}},\cr
\Bvalue{\tau\le\tau_i}&
\Bsubseteq\Bvalue{\tau_{i+1}\wedge\tau=\tau_i\wedge\tau}
\Bsubseteq\Bvalue{v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau}=0},\cr}$$

\noindent so $u_{\tau_i\wedge\tau}
\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})
=u_{I,\tau_i\wedge\tau}
\times(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})$.   Accordingly

\Centerline{$\int_{\Cal S\wedge\tau}\pmb u_Id\pmb v
   -S_{I\wedge\tau}(\pmb u,d\pmb v)
=u_{I,\tau_n\wedge\tau}\times(v_{\tau}-v_{\tau_n\wedge\tau})
=u_{I,\tau_n\wedge\tau}\times(v_{\tau\vee\tau_n}-v_{\tau_n})$}

\noindent (612D(f-i)), and

\Centerline{$\theta(\int_{\Cal S\wedge\tau}\pmb u_Id\pmb v
   -S_{I\wedge\tau}(\pmb u,d\pmb v))
\le\theta(\sup|\pmb u|\times|v_{\tau\vee\tau_n}-v_{\tau_n}|)
\le\delta$}

\noindent because $\tau^*\le\tau_n\le\tau\vee\tau_n$ and
$\theta(v_{\tau\vee\tau_n}-v_{\tau})\le\eta$.\ \Qed

Consequently

$$\eqalign{\theta(\int_{\Cal S\wedge\tau}(\pmb u-\pmb u_I)d\pmb v)&
\le\theta(S_{I\wedge\tau}(\pmb u,d\pmb v)
               -\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v)
   +\theta(S_{I\wedge\tau}(\pmb u,d\pmb v)
           -\int_{\Cal S\wedge\tau}\pmb u_I\,d\pmb v)\cr&
\le 4\delta.\cr}$$

We know that $\iiv(\pmb u-\pmb u_I)$ is fully adapted and order-bounded
(613J(e-ii), 616J, 616Ib).
We are assuming that $\Cal S$ is finitely full, so if we set

\Centerline{$\bar z_I
   =\sup_{\tau\in\Cal S}|\int_{\Cal S\wedge\tau}(\pmb u-\pmb u_I)d\pmb v|$}

\noindent then $\theta(\bar z_I)\le 2\sqrt{4\delta}\le\epsilon$ (615Db).

This is true whenever $K\subseteq I\in\Cal I(\Cal S)$.   So

\Centerline{$\limsup_{I\uparrow\Cal S}\theta(\sup|\iivu-\iiv(\pmb u_I)|)
=\limsup_{I\uparrow\Cal S}\theta(\bar z_I)\le\epsilon$.}

\noindent Bu $\epsilon$ was arbitrary, so
$\iivu=\lim_{I\uparrow\Cal S}\iiv(\pmb u_I)$.

\medskip

\quad{\bf (ii)} Here we have only to intercept the argument of (i) to see that

$$\eqalignno{\theta(\int_{\Cal S}(\pmb u-\pmb u_I)d\pmb v)&
=\lim_{\tau\uparrow\Cal S}
   \theta(\int_{\Cal S\wedge\tau}(\pmb u-\pmb u_I)d\pmb v)
\Displaycause{613J(f-ii)}
\le 4\sqrt{\delta}
\le\epsilon\cr}$$

\noindent whenever $K\subseteq I\in\Cal I(\Cal S)$.    So
$\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}\pmb u_Id\pmb v
=\int_{\Cal S}\pmb u\,d\pmb v$.
}%end of proof of 617B

\leader{617D}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, $\psi$ an
integrating interval function on $\Cal S$, and $\pmb u$,
$\pmb z$ moderately oscillatory
processes with domain $\Cal S$.

(a) $\pmb z\psi$, as defined in 613D, is an
integrating interval function on $\Cal S$.

(b) Set $\pmb w=ii_{\psi}(\pmb z)$.   Then $\pmb w$ is an integrator and

\Centerline{$\int_{\Cal S}\pmb u\,d(\pmb z\psi)
=\int_{\Cal S}\pmb u\times\pmb z\,d\psi
=\int_{\Cal S}\pmb u\,d\pmb w$.}


\proof{{\bf (a)(i)} Suppose to begin with that $\Cal S$ is full.
Then $\pmb z\psi$ is strictly adapted (613D).   Now
$S_I(\pmb y,d(\pmb z\psi))=S_I(\pmb y\times\pmb z,d\psi)$ whenever
$I\in\Cal I(\Cal S)$ and $\pmb y=\family{\sigma}{I}{y_{\sigma}}$ is fully
adapted.   \Prf\
If $e=c(\sigma,\tau)$ is a stopping-time interval
with endpoints $\sigma\le\tau$ in $I$, then

\Centerline{$\Delta_e(\pmb y,d(\pmb z\psi))
=y_{\sigma}\times(\pmb z\psi)(\sigma,\tau)
=y_{\sigma}\times z_{\sigma}\times\psi(\sigma,\tau)
=\Delta_e(\pmb y\times\pmb z,d\psi)$;}

\noindent summing over the $I$-cells, we have the result.\ \Qed

In particular, $S_I(\tbf 1,d(\pmb z\psi))=S_I(\pmb z,d\psi)$ for every
$I\in\Cal I(\Cal S)$, and

\Centerline{$\int_{\Cal S}d(\pmb z\psi)
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\tbf 1,d(\pmb z\psi))
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb z,d\psi)
=\int_{\Cal S}\pmb z\,d\psi$}

\noindent is defined.   Write $\bar z$ for $\sup|\pmb z|$.   Given
$\epsilon>0$, there is a $\delta>0$ such that
$\theta(S_I(\pmb y,d\psi))\le\delta$ whenever $I\in\Cal I(\Cal S)$,
$\pmb y\in\Mfa(I)$ and $\theta(\sup|\pmb y|)\le\delta$.   Now there is an
$\eta>0$ such that
$\theta(\bar z\times x)\le\epsilon$ whenever $x\in L^0(\frak A)$ and
$\theta(x)\le\eta$.   If we now take $I\in\Cal I(\Cal S)$ and
$\pmb y\in\Mfa(I)$ such that $\theta(\sup|\pmb y|)\le\eta$, we have
$\sup|\pmb y\times\pmb z|\le\bar u\times\sup|\pmb y|$, so

\Centerline{$\theta(\sup|\pmb y\times\pmb z|)
\le\theta(\bar z\times\sup|\pmb y|)\le\delta$}

\noindent and

\Centerline{$\theta(S_I(\pmb y,d(\pmb z\psi)))
=\theta(S_I(\pmb y\times\pmb z,d\psi))\le\epsilon$.}

\noindent Thus $Q_{\Cal S}(d(\pmb z\psi))$ is topologically bounded, and
$\pmb z\psi$ is an integrating interval function.

\medskip

\quad{\bf (ii)} Generally, we have the adapted extension $\hat\psi$ on the
covered envelope $\hat\Cal S$ and the fully adapted extension $\hatpmb z$ to
$\hat\Cal S$;  as $\hat\psi$ is an integrating interval function and
$\hatpmb z$ is moderately oscillatory, (i) tells us that
$\hatpmb z\hat\psi$ is an integrating interval function.   But
$\hatpmb z\hat\psi$ extends $\pmb z\psi$, so is the adapted extension of
$\pmb z\psi$ to $\hat\Cal S^{2\uparrow}$, and $\pmb z\psi$ also is an
integrating interval function.

\medskip

{\bf (b)(i)} As noted in (a-i), we have
$S_I(\pmb u,d(\pmb z\psi))=S_I(\pmb u\times\pmb z,d\psi)$ for every
finite sublattice $I$ of $\Cal S$;  taking the limit as
$I\uparrow\Cal I(\Cal S)$,
$\int_{\Cal S}\pmb u\,d(\pmb z\psi)=\int_{\Cal S}\pmb u\times\pmb z\,d\psi$.
Applying this to $\Cal S\wedge\tau$ with $\pmb u=\tbf 1$,
$\int_{\Cal S\wedge\tau}\pmb z\,d\psi=\int_{\Cal S\wedge\tau}d(\pmb z\psi)$
for every $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)} By 616J again, $\pmb w$ is an integrator, so 616K
assures us that $\int_{\Cal S}\pmb u\,d\pmb w$ is defined.
Expressing $\pmb w$ as
$\sfamilysS{w}$, we see that
$w_{\downarrow}=\lim_{\sigma\downarrow\Cal S}w_{\sigma}$ is $0$
(613J(f-ii)) so

\Centerline{$\int_{\Cal S\wedge\tau}d\pmb w=w_{\tau}-w_{\downarrow}
=w_{\tau}=\int_{\Cal S\wedge\tau}\pmb z\,d\psi
=\int_{\Cal S\wedge\tau}d(\pmb z\psi)$}

\noindent for every $\tau\in\Cal S$.   By 616M,

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb w=\int_{\Cal S}\pmb u\,d(\pmb z\psi)$.}
}%end of proof of 617D

\cmmnt{\medskip

\noindent{\bf Mnemonic} $d(ii_{\psi}(\pmb z))\sim d(\pmb z\psi)$.
}

\leader{617E}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$, $\pmb z$ moderately oscillatory processes with domain $\Cal S$, and
$\pmb v$ an integrator with domain $\Cal S$.   Set
$\pmb w=\iiv(\pmb z)$.   Then
$\int_{\Cal S}\pmb u\,d\pmb w=\int_{\Cal S}\pmb u\times\pmb z\,d\pmb v$.

\cmmnt{\medskip

\noindent{\bf Mnemonic} $d(\iiv(\pmb z))\sim\pmb z\,d\pmb v$.
}

\leader{617F}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb z$ a moderately oscillatory process and
$\pmb v$ a process of bounded variation, both with domain $\Cal S$.
Write $\pmb w$ for $\iiv(\pmb z)$.   Then $\pmb w$ is of bounded variation and
$\int_{\Cal S}|d\pmb w|=\int_{\Cal S}|\pmb z||d\pmb v|$.

\proof{{\bf (a)} If $\Cal S$ is empty, this is trivial,
so let us suppose otherwise.   Express $\pmb v$, $\pmb z$ as $\sfamilysS{v}$ and
$\sfamilysS{z}$.   We know that $\pmb v$ is an integrator
(616Ra), so $\pmb w$ is well-defined;  for $\tau\in\Cal S$,
write $w_{\tau}=\int_{\Cal S\wedge\tau}\pmb z\,d\pmb v$.
By 616T, $\int_{\Cal S}|\pmb z||d\pmb v|$ is defined.   To get started, note
that if $\pmb z$ is non-negative and $\pmb v$ is non-decreasing, $\pmb w$ will
be non-decreasing (616R(b-ii)), and we shall have

\Centerline{$\int_{\Cal S}|d\pmb w|
=\int_{\Cal S}d\pmb w
=\int_{\Cal S}\pmb z\,d\pmb v
=\int_{\Cal S}|\pmb z||d\pmb v|$}

\noindent by 617E.

\medskip

{\bf (b)} We need a couple of elementary formulae.

\medskip

\quad{\bf (i)} If $U$ is a Riesz space and
$u$, $v\in U$, then, using identities in 352D,

\Centerline{$u+v=u\wedge v+(u-v)^++u\wedge v+(v-u)^+=2(u\wedge v)+|u-v|$,}

\Centerline{$|u-v|\le|u|+|v|=2(|u|\wedge|v|)+||u|-|v||
\le 2(|u|\wedge|v|)+|u-v|$.}

\medskip

\quad{\bf (ii)} If $\pmb x$, $\pmb y$ are processes of bounded variation with
domain $\Cal S$, then applying (i) in the Riesz space
$(L^0)^{\Cal S^{2\uparrow}}$,

\Centerline{$\Delta\pmb x\wedge\Delta\pmb y
=\Bover12(\Delta\pmb x+\Delta\pmb y-|\Delta\pmb x-\Delta\pmb y|)$.}

\noindent Now $\pmb x-\pmb y$ is of bounded variation (614Q(a-ii)) so
$\Delta\pmb x$, $\Delta\pmb y$ and $\Delta\pmb x-\Delta\pmb y$ are integrating
interval functions (616Ra, 616Ic).
By 616S, $|\Delta\pmb x-\Delta\pmb y|$ is an integrating interval function so
$\Delta\pmb x\wedge\Delta\pmb y
=\bover12(\Delta\pmb x+\Delta\pmb y-|\Delta\pmb x-\Delta\pmb y|)$ also is
(352D, 616Ga).

Writing
$d\pmb x\wedge d\pmb y$ for $d(\Delta\pmb x\wedge\Delta\pmb y)$, in the
spirit of 613F and 613I,

\Centerline{$S_I(\tbf 1,d(\pmb x+\pmb y))
   =S_I(\tbf 1,|d(\pmb x-\pmb y)|)+2S_I(\tbf 1,d\pmb x\wedge d\pmb y)$}

\noindent for every finite sublattice $I$ of $\Cal S$, and
$\int_{\Cal S}d\pmb x\wedge d\pmb y$ is defined and equal to
$\Bover12
(\int_{\Cal S}d\pmb x+\int_{\Cal S}d\pmb y-\int_{\Cal S}|d(\pmb x-\pmb y)|)$.

Similarly,

\Centerline{$|\Delta(\pmb x-\pmb y)|
\le|\Delta\pmb x|+|\Delta\pmb y|
\le|\Delta(\pmb x-\pmb y)|+2(|\Delta\pmb x|\wedge|\Delta\pmb y|)$}

\noindent and

\Centerline{$S_I(\tbf 1,|d(\pmb x-\pmb y)|)
\le S_I(\tbf 1,|d\pmb x|)+S_I(\tbf 1,|d\pmb y|)
\le S_I(\tbf 1,|d(\pmb x-\pmb y)|)+2S_I(\tbf 1,|d\pmb x|\wedge|d\pmb y|)$}

\noindent for every $I\in\Cal I(\Cal S)$, so if
$\int_{\Cal S}|d\pmb x|\wedge|d\pmb y|=0$ then
$\int_{\Cal S}|d(\pmb x-\pmb y)|=\int_{\Cal S}|d\pmb x|+\int_{\Cal S}|d\pmb y|$.

\medskip

{\bf (c)} Next, suppose that $\Cal S$ is full and
$\pmb v$ is non-negative and non-decreasing.

\medskip

\quad{\bf (i)} Writing $\pmb z^+=\sfamilysS{z^+}$
for $\pmb z\vee\tbf 0$ and
$\pmb z^-=\sfamilysS{z^-}$ for $(-\pmb z)\vee\tbf 0$,
both $\int_{\Cal S}\pmb z^+d\pmb v$ and
$\int_{\Cal S}\pmb z^-d\pmb v$ are defined.   Set $\pmb x=\iiv(\pmb z^+)$ and
$\pmb y=\iiv(\pmb z^-)$;  then $\pmb w=\pmb x-\pmb y$ while $\pmb x$ and
$\pmb y$ are of bounded variation.   Now $\int_{\Cal S}d\pmb x\wedge d\pmb y=0$.
\Prf\ Let $\epsilon>0$.   Then there is an $I\in\Cal I(\Cal S)$ such that

\Centerline{$\theta(S_J(\pmb z^+,d\pmb v)-S_K(\pmb z^+,d\pmb v))\le\epsilon$,
\quad$\theta(S_J(\pmb z^-,d\pmb v)-S_K(\pmb z^-,d\pmb v))\le\epsilon$}

\noindent whenever $I\subseteq J$, $K\in\Cal I(\Cal S)$.   Take any non-empty
$J\in\Cal I(\Cal S)$ including $I$, and a sequence $(\tau_0,\ldots,\tau_n)$
linearly generating the $J$-cells.   For each $i<n$,
$a_i=\Bvalue{z_{\tau_i}\ge 0}$ belongs to
$\frak A_{\tau_i}\subseteq\frak A_{\tau_{i+1}}$, so we have a
$\tau'_i\in\Cal T$ such that

\Centerline{$a_i\Bsubseteq\Bvalue{\tau'_i=\tau_i}$,
\quad$1\Bsetminus a_i\Bsubseteq\Bvalue{\tau'_i=\tau_{i+1}}$}

\noindent (611I);  as $\Cal S$ is full, $\tau'_i\in\Cal S$, while
$\tau_i\le\tau'_i\le\tau_{i+1}$.   Let $K$ be the (finite)
sublattice of $\Cal S$ generated by $J\cup\{\tau'_i:i<n\}$;  then $K$ is
included in the sublattice

\Centerline{$\{\sigma:\med(\tau_i,\sigma,\tau_{i+1})
\in\{\tau_i,\tau'_i,\tau_{i+1}\}$ for every $i<n\}$,}

\noindent so $\{\tau_0,\tau'_0,\tau_1,\ldots,\tau'_{n-1},\tau_n\}$ is a maximal
totally ordered subset of $K$ and
$(\tau_0,\tau'_0,\tau_1,\ldots,\tau'_{n-1},\tau_n)$ linearly generates the
$K$-cells.

For each $i<n$,

\Centerline{$\Bvalue{v_{\tau'_i}\ne v_{\tau_i}}
\Bsubseteq\Bvalue{\tau_i<\tau'_i}\Bsubseteq a_i
\Bsubseteq\Bvalue{z^-_{\tau_i}=0}$,}

\Centerline{$\Bvalue{v_{\tau_{i+1}}\ne v_{\tau'_i}}
\Bsubseteq\Bvalue{\tau'_i<\tau_{i+1}}
\Bsubseteq(1\Bsetminus a_i)\Bcap\Bvalue{\tau'_i=\tau_i}
\Bsubseteq\Bvalue{z^+_{\tau'_i}=0}$.}

\noindent So we have

$$\eqalignno{0&
\le S_J(\tbf 1,d\pmb x\wedge d\pmb y)
=S_K(\tbf 1,d\pmb x\wedge d\pmb y)
\Displaycause{because $J\subseteq K$ and $J$ covers $K$, see 613S}
=\sum_{i=0}^{n-1}
     \Delta_{c(\tau_i,\tau'_i)}(\tbf 1,d\pmb x\wedge d\pmb y)
  +\Delta_{c(\tau'_i,\tau_{i+1})}(\tbf 1,d\pmb x\wedge d\pmb y)\cr&
\le\sum_{i=0}^{n-1}\Delta_{c(\tau_i,\tau'_i)}(\tbf 1,d\pmb y)
  +\Delta_{c(\tau'_i,\tau_{i+1})}(\tbf 1,d\pmb x)\cr&
=\sum_{i=0}^{n-1}
    \int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb z^-d\pmb v
  +\int_{\Cal S\cap[\tau'_i,\tau_{i+1}]}\pmb z^+d\pmb v\cr&
=\sum_{i=0}^{n-1}
    \int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb z^-d\pmb v
        -z^-_{\tau_i}\times(v_{\tau'_i}-v_{\tau_i})\cr&
\mskip100mu  +\int_{\Cal S\cap[\tau'_i,\tau_{i+1}]}\pmb z^+d\pmb v
        -z^+_{\tau'_i}\times(v_{\tau_{i+1}}-v_{\tau'_i})\cr&
=\sum_{i=0}^{n-1}
    \int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb z^-d\pmb v
        -S_{K\cap[\tau_i,\tau'_i]}(\pmb z^-,d\pmb v)\cr&
\mskip100mu  +\int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb z^+d\pmb v
        -S_{K\cap[\tau_i,\tau'_i]}(\pmb z^+,d\pmb v)\cr}$$

\noindent and

$$\eqalign{\theta(S_J(\tbf 1,d\pmb x\wedge d\pmb y))&
\le\theta(\sum_{i=0}^{n-1}
    \int_{\Cal S\cap[\tau_i,\tau'_i]}\pmb z^-d\pmb v
        -S_{K\cap[\tau_i,\tau'_i]}(\pmb z^-,d\pmb v))\cr&
\mskip50mu  +\theta(\sum_{i=0}^{n-1}
    \int_{\Cal S\cap[\tau'_i,\tau_{i+1}]}\pmb z^+d\pmb v
        -S_{K\cap[\tau_i,\tau'_i]}(\pmb z^+,d\pmb v))\cr&
\le 2\epsilon\cr}$$

\noindent (613V(i)).   This is true whenever $I\subseteq J$ in
$\Cal I(\Cal S)$;  as $\epsilon$ is arbitrary,
$\int_{\Cal S}d\pmb x\wedge d\pmb y=0$.\ \Qed

\medskip

\quad{\bf (ii)} By (b),

$$\eqalign{\int_{\Cal S}|d\pmb w|&
=\int_{\Cal S}d\pmb x+\int_{\Cal S}d\pmb y-2\int_{\Cal S}d\pmb x\wedge d\pmb y
  \cr&
=\int_{\Cal S}\pmb z^+d\pmb v+\int_{\Cal S}\pmb z^-d\pmb v
=\int_{\Cal S}|\pmb z|d\pmb v
=\int_{\Cal S}|\pmb z||d\pmb v|,\cr}$$

\noindent and we have the required result when $\Cal S$ is full and
$\pmb v$ is non-decreasing.

\medskip

{\bf (d)} Next suppose only that $\Cal S$ is full.

\medskip

\quad{\bf (i)} Let $\pmb v^{\uparrow}$ be
the cumulative variation of $\pmb v$ (614O) and set

\Centerline{$\pmb v_1=\bover12(\pmb v^{\uparrow}+\pmb v)$,
\quad$\pmb x=ii_{\pmb v_1}(\pmb z)$,
\quad$\pmb v_2=\bover12(\pmb v^{\uparrow}-\pmb v)$,
\quad$\pmb y=ii_{\pmb v_2}(\pmb z)$.}

\noindent Then $\pmb v_1$ and $\pmb v_2$ are non-decreasing (614P(a-iii)), and

\Centerline{$\pmb v_1-\pmb v_2=\pmb v$,
\quad$\pmb w=\pmb x-\pmb y$,
\quad$\pmb v_1+\pmb v_2=\pmb v^{\uparrow}$.}

\noindent Express $\pmb v_1$ and $\pmb x$ as
$\family{\sigma}{\Cal S}{v_{1\sigma}}$, $\sfamilysS{x}$ respectively.

\medskip

\quad{\bf (ii)} By (b) above and 616Ta,

$$\eqalign{2\int_{\Cal S}d\pmb v_1\wedge d\pmb v_2&
=\int_{\Cal S}d\pmb v_1+\int_{\Cal S}d\pmb v_2-\int_{\Cal S}|d\pmb v|\cr&
=\int_{\Cal S}d\pmb v^{\uparrow}-\int_{\Cal S}|d\pmb v|
=0.\cr}$$

\noindent Set $\bar z=\sup|\pmb z|$.   If $\sigma\le\tau$ in $\Cal S$ and
$e=c(\sigma,\tau)$ is the corresponding stopping-time interval, then

\Centerline{$|\Delta_e(\pmb z,d\pmb v_1)|
=|z_{\sigma}|\times(v_{1\tau}-v_{1\sigma})
\le\bar z\times\Delta_e(\tbf 1,\pmb v_1)$.}

\noindent So $|S_I(\pmb z,d\pmb v_1)|\le\bar z\times S_I(\tbf 1,d\pmb v_1)$
for every $I\in\Cal I(\Cal S)$.   Again, if $e=c(\sigma,\tau)$ is a
stopping-time interval with endpoints in $\Cal S$,

$$\eqalign{|\Delta_e(\tbf 1,d\pmb x)|
&=|x_{\tau}-x_{\sigma}|
=|\int_{\Cal S\cap[\sigma,\tau]}\pmb z\,d\pmb v_1|\cr&
\le\bar z\times\int_{\Cal S\cap[\sigma,\tau]}d\pmb v_1
=\bar z\times(v_{1\tau}-v_{1\sigma})
=\bar z\times\Delta_e(\tbf 1,d\pmb v_1).\cr}$$

\noindent Thus $|\Delta\pmb x|\le\bar z\times\Delta\pmb v_1$.
Similarly, $|\Delta\pmb y|\le\bar z\times\Delta\pmb v_2$.   But this means that
$|\Delta\pmb x|\wedge|\Delta\pmb y|
\le\bar z\times(\Delta\pmb v_1\wedge\Delta\pmb v_2)$.   Consequently

\Centerline{$0\le S_I(\tbf 1,|d\pmb x|\wedge|d\pmb y|)
\le\bar z\times S_I(\tbf 1,d\pmb v_1\wedge d\pmb v_2)$}

\noindent for every $I\in\Cal I(\Cal S)$, and in the limit

\Centerline{$0\le\int_{\Cal S}|d\pmb x|\wedge|d\pmb y|
\le\bar z\times\int_{\Cal S}d\pmb v_1\wedge d\pmb v_2=0$.}

\medskip

\quad{\bf (iii)} Returning to the formulae of (b), we see now that

$$\eqalignno{\int_{\Cal S}|d\pmb w|&
=\int_{\Cal S}|d\pmb x|+\int_{\Cal S}|d\pmb y|
=\int_{\Cal S}|\pmb z|d\pmb v_1+\int_{\Cal S}|\pmb z|d\pmb v_2
\Displaycause{by (c)}
=\int_{\Cal S}|\pmb z|d\pmb v^{\uparrow}
=\int_{\Cal S}|\pmb z||d\pmb v|\cr}$$

\noindent by 616Ta again in its full strength.

\medskip

{\bf (e)} Finally, if $\Cal S$ is not full, let $\hat\Cal S$ be its covered
envelope and $\hatpmb z$, $\hatpmb v$ the fully adapted extensions of $\pmb z$
and $\pmb v$ to $\hat\Cal S$.   Then $\hatpmb z$ is moderately oscillatory
(615F(a-vi)) and $\hatpmb v$ is of bounded variation (614Q(a-iv-$\beta$)), 
while the
fully adapted extension $\hatpmb w$ of $\pmb w$ is $ii_{\hatpmb v}(\hatpmb z)$
(616Q(c-ii)).   Since $\Delta\hatpmb v$ is the strictly
adapted extension of the interval function $\Delta\pmb v$,
$|\Delta\hatpmb v|$ is the strictly adapted extension of $|\Delta\pmb v|$
(613U(b-ii));  similarly, $|\Delta\hatpmb w|$ is the extension of
$|\Delta\pmb w|$.   Now

$$\eqalignno{\int_{\Cal S}|d\pmb w|&
=\int_{\Cal S}d|\Delta\pmb w|
=\int_{\hat\Cal S}d|\Delta\hatpmb w|
\Displaycause{613T}
=\int_{\hat\Cal S}|d\hatpmb w|
=\int_{\hat\Cal S}|\hatpmb z||d\hatpmb v|
\Displaycause{(d) above}
=\int_{\Cal S}|\pmb z||d\pmb v|,\cr}$$

\noindent and we have the result in the general case.
}%end of proof of 617F

\leader{617G}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$ and $\pmb z$ moderately oscillatory processes and $\pmb v$ a process
of bounded variation, all with domain $\Cal S$.
Write $\pmb w$ for $\iiv(\pmb z)$, and
$\pmb v^{\uparrow}$, $\pmb w^{\uparrow}$ for the cumulative variations
of $\pmb v$ and $\pmb w$.   Then

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb w^{\uparrow}
=\int_{\Cal S}\pmb u\,|d\pmb w|
=\int_{\Cal S}\pmb u\times|\pmb z|\,|d\pmb v|
=\int_{\Cal S}\pmb u\times|\pmb z|\,d\pmb v^{\uparrow}$.}

\proof{ By 617F, $\pmb w$ is of bounded variation.
Let $\pmb v^{\uparrow}$, $\pmb w^{\uparrow}$ be the cumulative
variations of $\pmb v$ and $\pmb w$.
If $\tau\in\Cal S$, then

$$\eqalignno{\int_{\Cal S\wedge\tau}d\pmb w^{\uparrow}&
=\int_{\Cal S\wedge\tau}|d\pmb w|
\Displaycause{616Ta once more}
=\int_{\Cal S\wedge\tau}|\pmb z||d\pmb v|
\Displaycause{617F}
=\int_{\Cal S\wedge\tau}|\pmb z|d\pmb v^{\uparrow}.\cr}$$

\noindent So $\pmb w^{\uparrow}=ii_{\pmb v^{\uparrow}}(|\pmb z|)$.
By 617E, $\int_{\Cal S}\pmb u\,d\pmb w^{\uparrow}
=\int_{\Cal S}\pmb u\times|\pmb z|\,d\pmb v^{\uparrow}$, while
$\int_{\Cal S}\pmb u\,|d\pmb w|=\int_{\Cal S}\pmb u\,d\pmb w^{\uparrow}$ and
$\int_{\Cal S}\pmb u\times|\pmb z|\,|d\pmb v|
=\int_{\Cal S}\pmb u\times|\pmb z|\,d\pmb v^{\uparrow}$ by 616Ta yet again.
}%end of proof of 617G

\medskip

\noindent{\bf Mnemonic}
$d(\iiv(\pmb z)^{\uparrow})\sim|\pmb z|d\pmb v^{\uparrow}$.

\leader{617H}{Quadratic variation} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v$, $\pmb w$ local integrators with domain $\Cal S$.

\spheader 617Ha{\bf (i)} If $\pmb v$ and $\pmb w$ are integrators, then
the strictly adapted interval function $\Delta\pmb v\times\Delta\pmb w$
on $\Cal S$\cmmnt{ (613Da)} is an integrating interval function.
\prooflet{\Prf\
As observed in part (a) of the proof of 613M,

\Centerline{$\Delta\pmb v\times\Delta\pmb w
=\Delta(\pmb v\times\pmb w)-\pmb v\Delta\pmb w-\pmb w\Delta\pmb v$.}

\noindent Now $\pmb v\times\pmb w$ is an integrator (616Pa), so
$\Delta(\pmb v\times\pmb w)$ is an integrating interval function
(616Ic), while $\pmb v\Delta\pmb w$ and $\pmb w\Delta\pmb v$ are
integrating interval functions by 617Da.
So $\Delta\pmb v\times\Delta\pmb w$ also is, by 616Ga.\ \Qed}

\medskip

\quad{\bf (ii)} In any case, $\Delta\pmb v\times\Delta\pmb w$
is a locally integrating interval function\prooflet{ (apply (i) to
$\pmb v\restr\Cal S\wedge\tau$, $\pmb w\restr\Cal S\wedge\tau$ for
$\tau\in\Cal S$)}.

\spheader 617Hb The {\bf covariation} of $\pmb v$ and $\pmb w$ is the
indefinite integral

\Centerline{$\covar{\pmb v}{\pmb w}
=ii_{\Delta\pmb v\times\Delta\pmb w}(\tbf 1)$.}

\noindent When $\pmb w=\pmb v$,\cmmnt{ we say that}
$\pmb v^*=\covar{\pmb v}{\pmb v}=ii_{(\Delta\pmb v)^2}(\tbf 1)$
is the {\bf quadratic variation} of $\pmb v$.

\spheader 617Hc Note that as

\Centerline{$(\pmb v,\pmb w)\mapsto\Delta\pmb v\times\Delta\pmb w$}

\noindent is bilinear, so is
$(\pmb v,\pmb w)\mapsto\covar{\pmb v}{\pmb w}$.

\leader{617I}{Theorem} Let $\Cal S$ be a sublattice of
$\Cal T$, $\pmb v$, $\pmb w$ two integrators and
$\pmb u$ a moderately oscillatory process, all with domain $\Cal S$.
Then $\covar{\pmb v}{\pmb w}$ is an integrator and

\Centerline{$\int_{\Cal S}\pmb u\,d\covar{\pmb v}{\pmb w}$,
\quad$\int_{\Cal S}\pmb u\,d(\pmb v\times\pmb w)
       -\int_{\Cal S}\pmb u\times\pmb v\,d\pmb w
       -\int_{\Cal S}\pmb u\times\pmb w\,d\pmb v$,
\quad$\int_{\Cal S}\pmb u\,d\pmb vd\pmb w$}

\noindent are defined and equal.

\proof{ Because $\Delta\pmb v\times\Delta\pmb w$ is an integrating interval
function (617H(a-i)), its indefinite integral $\covar{\pmb v}{\pmb w}$ is an
integrator (616J once more).   Now

\Centerline{$\int_{\Cal S}\pmb u\,d(\pmb v\Delta\pmb w)
=\int_{\Cal S}\pmb u\times\pmb v\,d(\Delta\pmb w)
=\int_{\Cal S}\pmb u\times\pmb v\,d\pmb w$}

\noindent by 617Db.   Similarly, $\int_{\Cal S}\pmb u\,d(\pmb w\Delta\pmb v)
=\int_{\Cal S}\pmb u\times\pmb w\,d\pmb v$, while
$\int_{\Cal S}\pmb u\,d\covar{\pmb v}{\pmb w}
=\int_{\Cal S}\pmb u\,d(\Delta\pmb v\times\Delta\pmb w)$ by the other part of
617Db.   Of course $\int_{\Cal S}\pmb u\,d(\pmb v\times\pmb w)
=\int_{\Cal S}\pmb u\,d(\Delta(\pmb v\times\pmb w))$
and $\int_{\Cal S}\pmb u\,d\pmb vd\pmb w
=\int_{\Cal S}\pmb u\,d(\Delta\pmb v\times\Delta \pmb w)$ by the definitions
in 613L.   Since

\Centerline{$\Delta\pmb v\times\Delta\pmb w
=\Delta(\pmb v\times\pmb w)-\pmb v\Delta\pmb w-\pmb w\Delta\pmb v$,}

\Centerline{$\int_{\Cal S}\pmb u\,d\covar{\pmb v}{\pmb w}
=\int_{\Cal S}\pmb u\,d(\pmb v\times\pmb w)
       -\int_{\Cal S}\pmb u\times\pmb v\,d\pmb w
       -\int_{\Cal S}\pmb u\times\pmb w\,d\pmb v
=\int_{\Cal S}\pmb u\,d\pmb vd\pmb w$.}
}%end of proof of 617I

\cmmnt{\medskip

\noindent{\bf Mnemonic} $d\covar{\pmb v}{\pmb w}
\sim d(\pmb v\times\pmb w)-\pmb v\,d\pmb w-\pmb w\,d\pmb v
=d\pmb vd\pmb w$.
}

\leader{617J}{Corollary} Let $\Cal S$ be a non-empty sublattice of
$\Cal T$ and $\pmb v$ an integrator with domain $\Cal S$.   Let
$\pmb v^*$ be the quadratic variation of $\pmb v$.

(a) $\pmb v^*$ is an integrator,
and if $\pmb u$ is a moderately oscillatory process with domain $\Cal S$ then

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v^*$,
\quad$\int_{\Cal S}\pmb u\,d(\pmb v^2)
       -2\int_{\Cal S}\pmb u\times\pmb v\,d\pmb v$,
\quad$\int_{\Cal S}\pmb u\,(d\pmb v)^2$}

\noindent are defined and equal.

(b)(i) Expressing $\pmb v^*$ as $\family{\tau}{\Cal S}{v^*_{\tau}}$,
$\lim_{\tau\downarrow\Cal S}v^*_{\tau}=0$.

\quad(ii) $\pmb v^*$ is non-negative, non-decreasing and order-bounded.

(c) If $\pmb w$ is another integrator with domain $\Cal S$, then
$\covar{\pmb v}{\pmb w}$ is of bounded variation.

\proof{{\bf (a)} This is immediate from 617I and the definition of
$\pmb v^*$ as $\covar{\pmb v}{\pmb v}$.

\medskip

{\bf (b)(i)} Immediate from 613J(f-i).

\medskip

\quad{\bf (ii)} Express $\pmb v$ as $\sfamilysS{v}$.
If $\tau\le\tau'$ in $\Cal S$, then

\Centerline{$v^*_{\tau'}-v^*_{\tau}
=\int_{\Cal S\cap[\tau,\tau']}d\pmb v^*
=\int_{\Cal S\cap[\tau,\tau']}(d\pmb v)^2$.}

\noindent But looking back at the definitions in \S613, we see that

\Centerline{$(\Delta\pmb v)^2(c(\sigma,\sigma'))
=(v_{\sigma'}-v_{\sigma})^2\ge 0$}

\noindent for every stopping time interval $c(\sigma,\sigma')$ with endpoints in
$\Cal S$, so $S_I(\tbf 1,(d\pmb v)^2)\ge 0$ for every $I\in\Cal I(\Cal S)$
and $\int_{\Cal S\cap[\tau,\tau']}(d\pmb v)^2\ge 0$.   So
$v^*_{\tau}\le v^*_{\tau'}$.   As $\tau$ and $\tau'$ are arbitrary,
$\pmb v^*$ is non-decreasing.

As $\pmb v^*$ is an integrator,
$v^*_{\uparrow}=\lim_{\tau\uparrow\Cal S}v^*_{\tau}$ is defined;  as
$\pmb v^*$ is non-decreasing,
$v^*_{\downarrow}\le v_{\tau}\le v^*_{\uparrow}$ for every $\tau\in\Cal S$
and $\pmb v^*$ is non-negative and order-bounded.

\medskip

{\bf (c)} By 614I(i), $\pmb v^*$ is of bounded variation.
Similarly, $\pmb w^*$ and $(\pmb v+\pmb w)^*$ are of bounded variation.
Since covariation is bilinear (617Hc),

\Centerline{$\covar{\pmb v}{\pmb w}
=\Bover12(\covar{\pmb v+\pmb w}{\pmb v+\pmb w}
  -\covar{\pmb v}{\pmb v}-\covar{\pmb w}{\pmb w})
=\Bover12((\pmb v+\pmb w)^*-\pmb v^*-\pmb w^*)$}

\noindent is a linear combination of processes of bounded variation and is
of bounded variation (614Q(a-ii)).
}%end of proof of 617J

\cmmnt{\medskip

\noindent{\bf Mnemonic} $(d\pmb v)^2
=d(\pmb v^2)-2\pmb v\,d\pmb v\sim d\pmb v^*$.}

\leader{617K}{Remarks}
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$, $\pmb w$ local integrators with domain $\Cal S$.

\spheader 617Ka\cmmnt{ Applying 617I to $\Cal S\wedge\tau$, for $\tau\in\Cal S$,
we see that}

\Centerline{$ii_{\covar{\pmb v}{\pmb w}}(\pmb u)
=ii_{\pmb v\times\pmb w}(\pmb u)
       -ii_{\pmb w}(\pmb u\times\pmb v)
       -\iiv(\pmb u\times\pmb w)$}

\noindent for every locally moderately oscillatory $\pmb u$ with domain
$\Cal S$.   \cmmnt{Taking $\pmb u=\tbf 1$,}

\dvro{\Centerline{$\covar{\pmb v}{\pmb w}
=ii_{\pmb v\times\pmb w}(\tbf 1)-ii_{\pmb w}(\pmb v)-ii_{\pmb v}(\pmb w)$;}}
{$$\eqalignno{\covar{\pmb v}{\pmb w}&
=ii_{\covar{\pmb v}{\pmb w}}(\tbf 1)
\Displaycause{using 613N and 617J(b-i)}
=ii_{\pmb v\times\pmb w}(\tbf 1)-ii_{\pmb w}(\pmb v)-ii_{\pmb v}(\pmb w);\cr}$$}

\cmmnt{\noindent taking $\pmb v=\pmb w$,}

\Centerline{$\pmb v^*=ii_{\pmb v^2}(\tbf 1)-2ii_{\pmb v}(\pmb v)$.}

\noindent If $\Cal S$ is not empty,

\Centerline{$\covar{\pmb v}{\pmb w}
=\pmb v\times\pmb w-(v_{\downarrow}\times w_{\downarrow})\tbf 1
    -ii_{\pmb w}(\pmb v)-ii_{\pmb v}(\pmb w)$,
\quad$\pmb v^*=\pmb v^2-v_{\downarrow}^2\tbf 1-2ii_{\pmb v}(\pmb v)$,}

\noindent where $v_{\downarrow}$ and $w_{\downarrow}$ are the starting values
of $\pmb v$ and $\pmb w$\cmmnt{, by 613N again}.

\query

\spheader 617Kb\cmmnt{ Immediately from the definition in 617H,}
$\covar{\pmb v\restr\Cal S\wedge\tau}{\pmb w\restr\Cal S\wedge\tau}
=\covar{\pmb v}{\pmb w}\restr\Cal S\wedge\tau$ for every $\tau\in\Cal S$.
\cmmnt{We also have a formula for
$\covar{\pmb v\restr\Cal S\vee\tau}{\pmb w\restr\Cal S\vee\tau}$;
if $\tau'\in\Cal S\vee\tau$, then

\Centerline{$\int_{(\Cal S\vee\tau)\wedge\tau'}d\pmb vd\pmb w
=\int_{\Cal S\cap[\tau,\tau']}d\pmb vd\pmb w
=\int_{\Cal S\wedge\tau'}d\pmb vd\pmb w
  -\int_{\Cal S\wedge\tau}d\pmb vd\pmb w$,}

\noindent so}

\Centerline{$\covar{\pmb v\restr\Cal S\vee\tau}{\pmb w\restr\Cal S\vee\tau}
=(\covar{\pmb v}{\pmb w}\restr\Cal S\vee\tau)-z\tbf 1$}

\noindent where
$z=\int_{\Cal S\wedge\tau}d\pmb vd\pmb w\in L^0(\frak A_{\tau})$\cmmnt{ is the
final value of $\covar{\pmb v}{\pmb w}\restr\Cal S\wedge\tau$}.
\cmmnt{When $\pmb w=\pmb v$, this becomes}

\Centerline{$\pmb v^*\restr\Cal S\vee\tau
=v_{\tau}^*\tbf 1+(\pmb v\restr\Cal S\vee\tau)^*$}

\noindent where $v_{\tau}^*=\int_{\Cal S\wedge\tau}(d\pmb v)^2$.

\spheader 617Kc A perfectly elementary fact\cmmnt{ which it is worth having
out in the open} is that if $\pmb v-\pmb u$ is constant then\cmmnt{ the
interval functions $\Delta\pmb u$ and $\Delta\pmb v$ are equal, so
$(\Delta\pmb u)^2=(\Delta\pmb v)^2$ and} $\pmb u^*=\pmb v^*$.

\leader{617L}{Corollary} Let $\Cal S$ be a sublattice of
$\Cal T$ and $\pmb v$ a local integrator with domain $\Cal S$.   Let
$\pmb v^*$ be the quadratic variation of $\pmb v$.
Then $\pmb v^*$ is non-negative, non-decreasing and locally of bounded
variation.   If $\pmb w$ is another local integrator with domain $\Cal S$, then
$\covar{\pmb v}{\pmb w}$ is locally of bounded variation.
%covers part of old 617H

\proof{ Apply 617J to $\pmb v\restr\Cal S\wedge\tau$ for $\tau\in\Cal S$.
}%end of proof of 617L

\leader{617M}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$, $\pmb w$ local integrators with domain
$\Cal S$.   Then $\covar{\pmb v}{\pmb w}^2\le\pmb v^*\times\pmb w^*$.

\proof{ Of course this is just a form of Cauchy's inequality.
I think it relies, however, on something not spelt out in Volume 3.

\medskip

{\bf (a)(i)} Suppose that $\alpha$, $\beta$, $\gamma\in\Bbb R$ are such
that $\xi^2\alpha+2\xi\eta\beta+\eta^2\gamma\ge 0$ for all real $\xi$,
$\eta$.   Then $\alpha\ge 0$, $\gamma\ge 0$ and $\beta^2\le\alpha\gamma$.
\Prf\ Taking $\xi=1$, $\eta=0$ we see that $\alpha\ge 0$.   If $\alpha=0$
then $\eta^2\gamma\pm 2\eta\beta\ge 0$ for every $\eta>0$, so
$|\beta|\le\bover12\eta\gamma$ for every $\eta>0$ and
$\beta^2=\alpha\gamma=0$.   Similarly, $\gamma\ge 0$ and if $\gamma=0$ then
$\beta^2=\alpha\gamma$.

If $\alpha$, $\gamma>0$ then, taking $\xi=\sqrt{\gamma}$ and
$\eta=\pm\sqrt{\alpha}$, we have
$2\alpha\gamma\pm 2\sqrt{\alpha}\sqrt{\gamma}\beta\ge 0$, so
$|\sqrt{\alpha\gamma}\beta|\le\alpha\gamma$ and
$\beta^2\le\alpha\gamma$, as required.\ \Qed

\medskip

\quad{\bf (ii)} If $(\Omega,\Sigma,\mu)$ is a probability space and
$f$, $g$, $h:\Omega\to\Bbb R$ are measurable functions such that, for
any $\xi$, $\eta\in\Bbb R$,
$\xi^2f+2\xi\eta g+\eta^2h\ge 0$ almost everywhere, then
$f\ge 0$ a.e., $h\ge 0$ a.e.\ and $g^2\le f\times h$ a.e.   \Prf\ The set
$\Omega_0$ of $\omega\in\Omega$ such that
$\xi^2f(\omega)+2\xi\eta g(\omega)+\eta^2h(\omega)\ge 0$ for all
rational $\xi$ and $\eta$ is conegligible.   Now, for $\omega\in\Omega_0$,
$\xi^2f(\omega)+2\xi\eta g(\omega)+\eta^2h(\omega)\ge 0$ for all
real $\xi$ and $\eta$, so (i) tells us that
$f(\omega)\ge 0$, $h(\omega)\ge 0$ and $g(\omega)^2\le f(\omega)h(\omega)$.
As $\Omega_0$ is conegligible, $f\ge 0$, $h\ge 0$ and $g^2\le f\times h$
almost everywhere.\ \Qed

\medskip

\quad{\bf (iii)} If $u$, $v$, $w\in L^0$ are such that
$\xi^2u+2\xi\eta v+\eta^2w\ge 0$ for all real $\xi$, $\eta$, then
$u\ge 0$, $w\ge 0$ and $v^2\le u\times w$.   \Prf\ Express $\frak A$ as the
measure algebra of a probability space $(\Omega,\Sigma,\mu)$, identify
$L^0(\frak A)$ with $L^0(\mu)$, take measurable functions $f$, $g$, $h$
such that $f^{\ssbullet}=u$, $g^{\ssbullet}=v$ and $h^{\ssbullet}=w$, and
apply (ii).\ \Qed

\medskip

{\bf (b)} Now the proposition follows at once from (a-iii)
if we observe that, for any
$\tau\in\Cal S$ and $\xi$, $\eta\in\Bbb R$,

\Centerline{$\xi^2\int_{\Cal S\wedge\tau}(d\pmb v)^2
+2\xi\eta\int_{\Cal S\wedge\tau}d\pmb v\,d\pmb w
+\eta^2\int_{\Cal S\wedge\tau}(d\pmb w)^2
=\int_{\Cal S\wedge\tau}(d\pmb z)^2\ge 0$ in $L^0$}

\noindent where $\pmb z=\xi\pmb v+\eta\pmb w$ is a local integrator.
}%end of proof of 617M

\leader{617N}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v$, $\pmb w$ local integrators with domain $\Cal S$.   Let
$\hatpmb v$, $\hatpmb w$ be their fully adapted extensions to the covered
envelope $\hat{\Cal S}$ of $\Cal S$.   Then $\covar{\hatpmb v}{\hatpmb w}$ is
the fully adapted extension of $\covar{\pmb v}{\pmb w}$ to $\hat{\Cal S}$.
In particular, the quadratic variation of $\hatpmb v$ is the fully adapted
extension to $\hat{\Cal S}$ of the quadratic variation of $\pmb v$.

\proof{ By 616Ia, $\hatpmb v$ and $\hatpmb w$ are local integrators.
Of course $\Delta\hatpmb v\times\Delta\hatpmb w$ extends
$\Delta\pmb v\times\Delta\pmb w$ just because $\hatpmb v$ extends $\pmb v$ and
$\hatpmb w$ extends $\pmb w$.   If $\tau\in\Cal S$, then
$\int_{\hat\Cal S\wedge\tau}d(\Delta\hatpmb v\times\Delta\hatpmb w)$ is
defined, so is equal to
$\int_{\Cal S\wedge\tau}d(\Delta\pmb v\times\Delta\pmb w)$ (613T, as
$\hat\Cal S\wedge\tau$ is the covered envelope of $\Cal S\wedge\tau$).
Thus $\covar{\pmb v}{\pmb w}=\covar{\hatpmb v}{\hatpmb w}\restr\Cal S$ and
$\covar{\hatpmb v}{\hatpmb w}$ is
the fully adapted extension of $\covar{\pmb v}{\pmb w}$ to $\hat\Cal S$.
}%end of proof of 617N

\leader{617O}{Examples} Suppose that $T=\coint{0,\infty}$.

\spheader 617Oa Let
$\pmb{\iota}\cmmnt{\mskip5mu=\family{\tau}{\Cal T_f}{\iota_{\tau}}}$ be the
identity process\cmmnt{ (612F)}.   Then its quadratic variation
$\pmb{\iota}^*$ is zero.   \prooflet{\Prf\ $\pmb{\iota}$ is non-negative
and non-decreasing, therefore locally of bounded
variation (614K(c-i)) and a local integrator (616Ra).
So $\pmb{\iota}^*$ is
well-defined;  express it as $\family{\tau}{\Cal T_f}{\iota^*_{\tau}}$.
Take $t>0$ and $m\ge 1$, and set
%$\tau_i=\check{it/m}$,
$\tau_i=\check{\bover{it}{m}}$,
the constant stopping time at $\bover{it}m$, for $i\le m$;
let $I$ be any sublattice of
$\Cal T\wedge\check t$ containing $\tau_i$ for every $i\le m$.
If $e$ is an $I$-cell, then
$e$ is expressible as $c(\sigma,\tau)$ where
$\tau_i\le\sigma\le\tau\le\tau_{i+1}$ for some $i$.   Now

\Centerline{$\Delta_e(\tbf 1,(d\pmb{\iota})^2)
=(\iota_{\tau}-\iota_{\sigma})^2
\le\Bover{t}{m}(\iota_{\tau}-\iota_{\sigma})
=\Bover{t}{m}\Delta_e(\tbf 1,d\pmb{\iota})$.}

\noindent Summing over $e$,

\Centerline{$S_I(\tbf 1,(d\pmb{\iota})^2)
\le\Bover{t}{m}S_I(\tbf 1,d\pmb{\iota})
=\Bover{t}{m}\iota_{\check t}$.}

Taking the limit as $I\uparrow\Cal I(\Cal T\wedge\check t)$,

\Centerline{$\iota^*_{\check t}
=\int_{\Cal T\wedge\check t}(d\pmb{\iota})^2
\le\Bover{t}{m}\iota_{\check t}$.}

\noindent As $t$ and $m$ are arbitrary,
$\iota^*_{\tau}=0$ for every constant stopping time
$\tau$.   Since $\pmb{\iota}^*$ is non-decreasing,
$\iota^*_{\tau}=0$ for every $\tau\in\Cal T_b$.   As $\Cal T_b$ is a
covering ideal of $\Cal T_f$ (611Ne), $\pmb{\iota}^*$ must be the zero
process (612R).\ \Qed}

\spheader 617Ob  Let $\pmb v\cmmnt{\mskip5mu=\family{\tau}{\Cal T_f}{v_{\tau}}}$
be the standard Poisson process\cmmnt{, based on the structure
$(\Omega,\Sigma,\mu,\frak A,\bar\mu,\langle\frak A_t\rangle_{t\ge 0})$
described in 612U}.   Then $\pmb v$ is equal to its quadratic variation
$\pmb v^*$.   \prooflet{\Prf\
$\pmb v$ is non-negative and non-decreasing, so
$\pmb v^*$ is well-defined.   Express $\pmb v^*$ as
$\family{\tau}{\Cal T_f}{v^*_{\tau}}$.
Let $\sequencen{\tau_n}$ be the sequence of
jump times of $\pmb v$, as in 612Ue-612Uf, so that $\tau_0=\min\Cal T$,
$\sup_{n\in\Bbb N}\tau_n=\max\Cal T$ and

\Centerline{$\Bvalue{v_{\sigma}=n}
=\Bvalue{\tau_n\le\sigma}\Bcap\Bvalue{\sigma<\tau_{n+1}}$}

\noindent for every $n\in\Bbb N$ and $\sigma\in\Cal T_f$.

Given $\tau\in\Cal T_f$ and $\epsilon>0$, there is an $n\in\Bbb N$ such
that $a=\Bvalue{\tau>\tau_n}$ has measure at most $\epsilon$.   Set
$J=\{\tau\wedge\tau_i:i\le n\}$ and take any sublattice $I$
of $T\wedge\tau$ including $J$.   If $e$ is an $I$-cell, $e$ is
expressible as $c(\sigma,\sigma')$ where either
$\tau_i\wedge\tau\le\sigma\le\sigma'\le\tau_{i+1}\wedge\tau$ for some
$i<n$, or $\tau_n\wedge\tau\le\sigma\le\sigma'\le\tau$.   In the first
case,

$$\eqalign{\Bvalue{v_{\sigma}\ne v_{\sigma'}}
&\Bsubseteq\Bvalue{\sigma<\sigma'}
\Bsubseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
  \Bcap\Bvalue{\sigma'\le\tau_{i+1}}\cr
&\Bsubseteq\Bvalue{v_{\sigma}=i}\Bcap\Bvalue{v_{\sigma'}\le i+1}
\Bsubseteq\Bvalue{v_{\sigma'}-v_{\sigma}\in\{0,1\}}
=\Bvalue{(v_{\sigma'}-v_{\sigma})^2=v_{\sigma'}-v_{\sigma}}.\cr}$$

\noindent But this means that we actually have
$(v_{\sigma'}-v_{\sigma})^2=v_{\sigma'}-v_{\sigma}$ and
$\Delta_e(\tbf 1,(d\pmb v)^2)=\Delta_e(\tbf 1,d\pmb v)$.
In the second case,

\Centerline{$\Bvalue{v_{\sigma}\ne v_{\sigma'}}
\Bsubseteq\Bvalue{\tau_n<\tau}=a$}

\noindent and

\Centerline{$\Bvalue{\Delta_e(\tbf 1,(d\pmb v)^2)
   \ne\Delta_e(\tbf 1,d\pmb v)}\Bsubseteq a$.}

Thus we have

\Centerline{$\Bvalue{\Delta_e(\tbf 1,(d\pmb v)^2)
   \ne\Delta_e(\tbf 1,d\pmb v)}\Bsubseteq a$}

\noindent for every $I$-cell $e$, and consequently

\Centerline{$\Bvalue{S_I(\tbf 1,(d\pmb v)^2)
   \ne S_I(\tbf 1,d\pmb v)}\Bsubseteq a$.}

\noindent Taking the limit as $I\uparrow\Cal I(\Cal T\wedge\tau)$,

\Centerline{$\Bvalue{v^*_{\tau}\ne v_{\tau}}
=\Bvalue{\int_{\Cal T\wedge\tau}(d\pmb v)^2
    \ne\int_{\Cal T\wedge\tau}d\pmb v}
\Bsubseteq a$}

\noindent has measure at most $\epsilon$.   As $\epsilon$ is arbitrary,
$v^*_{\tau}=v_{\tau}$;  as $\tau$ is arbitrary, $\pmb v^*=\pmb v$.\ \Qed}

\leader{617P}{}\cmmnt{ A more elaborate result of the same kind as 617E
can be proved by the methods here.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a full sublattice of $\Cal T$
with a greatest element, $\pmb z$ a moderately oscillatory process and
$\pmb v$, $\pmb v'$ integrators, all with domain $\Cal S$.
Set $\pmb w=\iiv(\pmb z)$.   Then
$\int_{\Cal S}d\pmb w\,d\pmb v'=\int_{\Cal S}\pmb z\,d\pmb v\,d\pmb v'$.

\proof{ Express $\pmb z$, $\pmb v$, $\pmb v'$, $\pmb w$ as $\sfamilysS{z}$,
$\sfamilysS{v}$, $\sfamilysS{v'}$, $\sfamilysS{w}$.

\medskip

{\bf (a)} Consider first the case in which $\pmb z$ is simple, with starting
value $z_{\downarrow}$ and breakpoint string $(\tau_0,\ldots,\tau_n)$ where
$\tau_n=\max\Cal S$.   Then

$$\eqalign{w_{\sigma}
&=z_{\downarrow}\times(v_{\sigma}-v_{\downarrow})\text{ for }\sigma\le\tau_0,\cr
&=w_{\tau_i}+z_{\tau_i}\times(v_{\sigma}-v_{\tau_i})\text{ for }i<n
  \text{ and }\tau_i\le\sigma\le\tau_{i+1},\cr}$$

\noindent so if $\sigma\le\tau$ in $\Cal S$,

$$\eqalign{(\Delta\pmb w\times\Delta\pmb v')(\sigma,\tau)
&=z_{\downarrow}\times(\Delta\pmb v\times\Delta\pmb v')(\sigma,\tau)
=z_{\sigma}\times(\Delta\pmb v\times\Delta\pmb v')(\sigma,\tau)
  \text{ if }\tau\le\tau_0,\cr
&=z_{\tau_i}\times(\Delta\pmb v\times\Delta\pmb v')(\sigma,\tau)
=z_{\sigma}\times(\Delta\pmb v\times\Delta\pmb v')(\sigma,\tau)\cr&
\mskip150mu
\text{ if }i<n\text{ and }\tau_i\le\sigma\le\tau\le\tau_{i+1}.\cr}$$

\noindent Consequently

\Centerline{$\int_{\Cal S\wedge\tau_0}d\pmb w\,d\pmb v'
=\int_{\Cal S\wedge\tau_0}\pmb z\,d\pmb v\,d\pmb v'$,}

\Centerline{$\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}d\pmb w\,d\pmb v'
=\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}\pmb z\,d\pmb v\,d\pmb v'$}

\noindent for $i<n$, and

\Centerline{$\int_{\Cal S}d\pmb w\,d\pmb v'
=\int_{\Cal S}\pmb z\,d\pmb v\,d\pmb v'$.}

\medskip

{\bf (b)} Generally, for $I\in\Cal I(\Cal S)$, let
$\pmb z_I=\family{\sigma}{\Cal S}{z_{I\sigma}}$ be the simple process with
breakpoints in $I$ agreeing
with $\pmb z$ on $I$ as in 617B, and set $\pmb w_I=\iiv(\pmb z_I)$.
Then we know from 617I and (a) here that

\Centerline{$\int_{\Cal S}d(\pmb w_I\times\pmb v')
  -\int_{\Cal S}\pmb w_Id\pmb v'-\int_{\Cal S}\pmb v'd\pmb w_I
=\int_{\Cal S}d\pmb w_Id\pmb v'
=\int_{\Cal S}\pmb z_Id\pmb v\,d\pmb v'$.}

\medskip

{\bf (c)(i)} Since $0$ is the starting value of $\pmb w_I$ and therefore of
$\pmb w_I\times\pmb v'$,
$\int_{\Cal S}d(\pmb w_I\times\pmb v')
=w_{I\max\Cal S}\times v'_{\max\Cal S}$ for each
$I\in\Cal I(\Cal S)$, and

$$\eqalign{\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}d(\pmb w_I\times\pmb v')&
=v'_{\max\Cal S}\times\lim_{I\uparrow\Cal I(\Cal S)}w_{I\max\Cal S}
=v'_{\max\Cal S}\times\lim_{I\uparrow\Cal I(\Cal S)}
    \int_{\Cal S}\pmb z_Id\pmb v\cr&
=v'_{\max\Cal S}\times\int_{\Cal S}\pmb z\,d\pmb v
=v'_{\max\Cal S}\times w_{\max\Cal S}
=\int_{\Cal S}d(\pmb w\times\pmb v').\cr}$$

\noindent by 617B(b-ii).

\medskip

\quad{\bf (ii)} Next, we know from 617B(b-i)
that $\pmb w=\iiv(\pmb z)$ is the limit
$\ucplim_{I\uparrow\Cal I(\Cal S)}\iiv(\pmb z_I)$ for the ucp topology on
$\Mob(\Cal S)$.   Consequently
$\int_{\Cal S}\pmb w\,d\pmb v'
=\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}\pmb w_Id\pmb v'$ (616J).

\medskip

\quad{\bf (iii)} By 617E,

\Centerline{$\int_{\Cal S}\pmb v'd\pmb w_I
=\int_{\Cal S}\pmb v'd(\iiv(\pmb z_I))
=\int_{\Cal S}\pmb v'\times\pmb z_I\,d\pmb v
=\int_{\Cal S}\pmb z_Id(\iiv(\pmb v'))$,}

\noindent and similarly
$\int_{\Cal S}\pmb v'd\pmb w=\int_{\Cal S}\pmb z\,d(\iiv(\pmb v'))$.
But this means that

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}\pmb v'd\pmb w_I
=\int_{\Cal S}\pmb v'd\pmb w$}

\noindent by 617B(b-ii) again.

\medskip

\quad{\bf (iv)} On the other side, using the other part of 617I,

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S)}
   \int_{\Cal S}\pmb z_Id\pmb v\,d\pmb v'
=\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}\pmb z_Id\covar{\pmb v}{\pmb v'}
=\int_{\Cal S}\pmb zd\covar{\pmb v}{\pmb v'}
=\int_{\Cal S}\pmb z\,d\pmb v\,d\pmb v'$.}

\medskip

{\bf (d)} Assembling these, we find that

$$\eqalign{\int_{\Cal S}d\pmb w\,d\pmb v'&
=\int_{\Cal S}d(\pmb w\times\pmb v')
  -\int_{\Cal S}\pmb w\,d\pmb v'-\int_{\Cal S}\pmb v'd\pmb w\cr&
=\lim_{I\uparrow\Cal I(\Cal S)}\bigl(\int_{\Cal S}d(\pmb w_I\times\pmb v')
  -\int_{\Cal S}\pmb w_Id\pmb v'-\int_{\Cal S}\pmb v'd\pmb w_I\bigr)\cr&
=\lim_{I\uparrow\Cal I(\Cal S)}\int_{\Cal S}\pmb z_Id\pmb v\,d\pmb v'
=\int_{\Cal S}\pmb z\,d\pmb v\,d\pmb v',\cr}$$

\noindent as claimed.
}%end of proof of 617P

\leader{617Q}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$, $\pmb z$ and $\pmb z'$ locally moderately oscillatory processes
with domain $\Cal S$, and
$\pmb v$, $\pmb v'$
local integrators with domain $\Cal S$.   Set $\pmb w=\iiv(\pmb z)$,
$\pmb w'=ii_{\pmb v'}(\pmb z')$.

(a)(i) $\covar{\pmb w}{\pmb v'}=ii_{\covar{\pmb v}{\pmb v'}}(\pmb z)$,
\quad$ii_{\covar{\pmb w}{\pmb v'}}(\pmb u)
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb u\times\pmb z)$.

\quad(ii) $\covar{\pmb w}{\pmb w'}
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb z\times\pmb z')$,
\quad$ii_{\covar{\pmb w}{\pmb w'}}(\pmb u)
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb u\times\pmb z\times\pmb z')$.

\quad(iii) $\pmb w^*
=ii_{\pmb v^*}(\pmb z^2)$,
\quad$ii_{\pmb w^*}(\pmb u)
=ii_{\pmb v^*}(\pmb u\times\pmb z^2)$.

(b) If $\pmb u$, $\pmb z$ and $\pmb z'$ are moderately oscillatory
and $\pmb v$, $\pmb v'$ are integrators,

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb w\,d\pmb w'
=\int_{\Cal S}\pmb u\times\pmb z\times\pmb z'd\pmb v\,d\pmb v'$,
\quad$\int_{\Cal S}\pmb u\,d\pmb w^*
=\int_{\Cal S}\pmb u\times\pmb z^2d\pmb v^*$.}

\proof{{\bf (a)(i)}\grheada\ Suppose to begin with that $\Cal S$ is full.
Take any $\tau\in\Cal S$.  Then 617P tells us that

\Centerline{$\int_{\Cal S\wedge\tau}d\pmb w\,d\pmb v'
=\int_{\Cal S\wedge\tau}\pmb z\,d\pmb v\,d\pmb v'
=\int_{\Cal S\wedge\tau}\pmb z\,d\covar{\pmb v}{\pmb v'}$.}

\noindent But this means that the indefinite integrals
$\covar{\pmb w}{\pmb v'}$, $ii_{\covar{\pmb v}{\pmb v'}}(\pmb z)$ are
equal.

\medskip

\qquad\grheadb\ In general, take $\hat\Cal S$ to be the covered envelope of
$\Cal S$ and $\hatpmb z$, $\hatpmb v$, $\hatpmb v'$ and $\hatpmb w$
the fully adapted
extensions of $\pmb z$, $\pmb v$, $\pmb v'$ and $\pmb w$ to $\hat\Cal S$.   Then
$\hatpmb w=ii_{\hatpmb v}(\hatpmb z)$ (616Q(c-ii)), so

\Centerline{$\covar{\hatpmb w}{\hatpmb v'}
=ii_{\covar{\hatpmb v}{\hatpmb v'}}(\hatpmb z)$}

\noindent and

$$\eqalignno{\covar{\pmb w}{\pmb v'}
&=\covar{\hatpmb w}{\hatpmb v'}\restr\Cal S
\Displaycause{617N}
=ii_{\covar{\hatpmb v}{\hatpmb v'}}(\hatpmb z)\restr\Cal S
=ii_{\covar{\hatpmb v}{\hatpmb v'}\restr\Cal S}(\pmb z)
\Displaycause{613T again}
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb z).\cr}$$

\medskip

\qquad\grheadc\ By 617Db,

\Centerline{$\int_{\Cal S\wedge\tau}\pmb u\,d\covar{\pmb w}{\pmb v'}
=\int_{\Cal S\wedge\tau}\pmb u\,d(ii_{\covar{\pmb v}{\pmb v'}}(\pmb z))
=\int_{\Cal S\wedge\tau}\pmb u\times\pmb z\,d\covar{\pmb v}{\pmb v'})$}

\noindent for every $\tau\in\Cal S$, that is,
$ii_{\covar{\pmb w}{\pmb v'}}(\pmb u)
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb u\times\pmb z)$.

\medskip

\quad{\bf (ii)} By (i),

$$\eqalign{\covar{\pmb w}{\pmb w'}&
=\covar{\pmb w'}{\pmb w}
=ii_{\covar{\pmb v'}{\pmb w}}(\pmb z')
=ii_{\covar{\pmb w}{\pmb v'}}(\pmb z')\cr&
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb z'\times\pmb z)
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb z\times\pmb z').\cr}$$

\noindent It follows, just as in (i-$\gamma$) above, that

\Centerline{$ii_{\covar{\pmb w}{\pmb w'}}(\pmb u)
=ii_{\covar{\pmb v}{\pmb v'}}(\pmb u\times\pmb z\times\pmb z')$.}

\medskip

\quad{\bf (iii)} This is now just the special case $\pmb v'=\pmb v$,
$\pmb z'=\pmb z$ of (ii).

\medskip

{\bf (b)} Since $\covar{\pmb v}{\pmb v'}$,
$\pmb w$, $\pmb w'$, $\covar{\pmb w}{\pmb w'}$, $\pmb v^*$ and $\pmb w^*$ are
all integrators (616J yet again, 617I), all the integrals are defined and

$$\eqalign{\int_{\Cal S}\pmb u\,d\pmb w\,d\pmb w'&
=\int_{\Cal S}\pmb u\,d\covar{\pmb w}{\pmb w'}
=\lim_{\tau\uparrow\Cal S}
  \int_{\Cal S\wedge\tau}\pmb u\,d\covar{\pmb w}{\pmb w'}
=\lim_{\tau\uparrow\Cal S}
  \int_{\Cal S\wedge\tau}\pmb u\times\pmb z\times\pmb z'\,
      d\covar{\pmb v}{\pmb v'}\cr&
=\int_{\Cal S}\pmb u\times\pmb z\times\pmb z'\,d\covar{\pmb v}{\pmb v'}
=\int_{\Cal S}\pmb u\times\pmb z\times\pmb z'\,d\pmb v\,d\pmb v'.\cr}$$

\noindent Taking $\pmb v'=\pmb v$, $\pmb z'=\pmb z$ and $\pmb w'=\pmb w$ the
formulae simplify to
$\int_{\Cal S}\pmb u\,d\pmb w^*=\int_{\Cal S}\pmb u\times\pmb z^2d\pmb v^*$.
}%end of proof of 617Q

\cmmnt{\medskip

\noindent{\bf Mnemonic} $d\covar{\pmb w}{\pmb w'}
\sim d\pmb w\,d\pmb w'
\sim(\pmb z\,d\pmb v)(\pmb z'd\pmb v')
\sim\pmb z\times\pmb z'\,d\pmb v\,d\pmb v'
\sim\pmb z\times\pmb z'\,d\covar{\pmb v}{\pmb v'}$.
}

\leader{617R}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a process with domain $\Cal S$ which is locally of bounded variation.
Then $\pmb v$ and its cumulative variation have the same quadratic variation.

\proof{{\bf (a)} I begin with an elementary fact about
$f$-algebras which was left as an exercise in \S352\Latereditions. %352Yb (2016)
Suppose that $u$, $v\in L^0$.   Then
$u\times v=(u\wedge v)\times(u\vee v)$.   \Prf\

$$\eqalignno{u\times v
&=((u\wedge v)+(u-v)^+)\times((u\wedge v)+(v-u)^+)\cr
\displaycause{352D}
&=(u\wedge v)\times(u\wedge v+(u-v)^++(v-u)^+)\cr
\displaycause{because $(u-v)^+\wedge(v-u)^+=0$, by 352D again, so
$(u-v)^+\times(v-u)^+=0$, by 352W(b-i)}
&=(u\wedge v)\times(u\vee v)\cr}$$

\noindent (352D once more).\ \Qed

\medskip

{\bf (b)} Suppose for the moment that $\pmb v$ is actually of bounded variation.
Express $\pmb v$ as $\sfamilysS{v}$ and its cumulative variation as
$\pmb v^{\uparrow}=\sfamilysS{v^{\uparrow}}$.   Set

\Centerline{$\pmb v_1=\bover12(\pmb v^{\uparrow}+\pmb v)$,
\quad$\pmb v_2=\bover12(\pmb v^{\uparrow}-\pmb v)$,}

\noindent so that $\pmb v_1$ and $\pmb v_2$ are both
non-decreasing processes (614P(a-iii)), $\pmb v=\pmb v_1-\pmb v_2$ and
$\pmb v^{\uparrow}=\pmb v_1+\pmb v_2$.   Let $\psi$ be the
interval function $\Delta\pmb v^{\uparrow}-|\Delta\pmb v|$ with domain
$\Cal S^{2\uparrow}$;  then

\Centerline{$\int_{\Cal S\wedge\tau}d\psi
=\int_{\Cal S\wedge\tau}d\pmb v^{\uparrow}-\int_{\Cal S\wedge\tau}|d\pmb v|
=v^{\uparrow}_{\tau}-\int_{\Cal S\wedge\tau}|d\pmb v|
=0$}

\noindent for every $\tau\in\Cal S$ (614O).
Note that if $\sigma\le\tau$ in $\Cal S$ then

\Centerline{$\psi(\sigma,\tau)
=v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}-|v_{\tau}-v_{\sigma}|
\ge 0$}

\noindent (614P(a-i)) and, expressing $\pmb v_1$ as
$\family{\sigma}{\Cal S}{v_{1\sigma}}$ and $\pmb v_2$ as
$\family{\sigma}{\Cal S}{v_{2\sigma}}$,

\Centerline{$\psi(\sigma,\tau)
=(v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}+v_{\tau}-v_{\sigma})
   \wedge(v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}-v_{\tau}+v_{\sigma})
=(2v_{1\tau}-2v_{1\sigma})\wedge(2v_{2\tau}-2v_{2\sigma})$;}

\noindent consequently

$$\eqalignno{(v_{1\tau}-v_{1\sigma})\times(v_{2\tau}-v_{2\sigma})
&=((v_{1\tau}-v_{1\sigma})\wedge(v_{2\tau}-v_{2\sigma}))
   \times((v_{1\tau}-v_{1\sigma})\vee(v_{2\tau}-v_{2\sigma}))\cr
\displaycause{by (a)}
&\le((v_{1\tau}-v_{1\sigma})\wedge(v_{2\tau}-v_{2\sigma}))\times 2\bar v
=\bar v\times\psi(\sigma,\tau)\cr}$$

\noindent where $\bar v=\bover12(\sup|\pmb v_1|+\sup|\pmb v_2|)$.
But this means that

\Centerline{$0\le S_I(\tbf 1,d\pmb v_1d\pmb v_2)
\le\bar v\times S_I(\tbf 1,d\psi)$ for every $I\in\Cal I(\Cal S)$,}

\Centerline{$0\le\int_{\Cal S\wedge\tau}d\pmb v_1d\pmb v_2
\le\bar v\times\int_{\Cal S\wedge\tau}d\psi=0$ for every $\tau\in\Cal S$}

\noindent and $\covar{\pmb v_1}{\pmb v_2}=0$.   Accordingly the quadratic
variations

$$\eqalign{(\pmb v^{\uparrow})^*
&=\covar{\pmb v^{\uparrow}}{\pmb v^{\uparrow}}
=\covar{\pmb v_1+\pmb v_2}{\pmb v_1+\pmb v_2}
=\covar{\pmb v_1}{\pmb v_1}
   +2\covar{\pmb v_1}{\pmb v_2}
   +\covar{\pmb v_2}{\pmb v_2}\cr&
=\covar{\pmb v_1}{\pmb v_1}
   -2\covar{\pmb v_1}{\pmb v_2}
   +\covar{\pmb v_2}{\pmb v_2}
=\pmb v^*\cr}$$

\noindent are equal.

\medskip

{\bf (c)} Now the general result follows at once because

$$\eqalignno{(\pmb v^{\uparrow})^*\restr\Cal S\wedge\tau&
=(\pmb v^{\uparrow}\restr\Cal S\wedge\tau)^*
\Displaycause{617Kb}
=((\pmb v\restr\Cal S\wedge\tau)^{\uparrow})^*
\Displaycause{614Pb}
=(\pmb v\restr\Cal S\wedge\tau)^*
\Displaycause{by (b)}
=\pmb v^*\restr\Cal S\wedge\tau\cr}$$

\noindent for every $\tau\in\Cal S$.
}%end of proof of 617R

\medskip

\noindent{\bf Mnemonic} $|d\pmb v|^2=d\pmb v^2$.

\exercises{\leader{617X}{Basic exercises (a)}
\spheader 617Xa Let $\pmb z$ be a locally moderately oscillatory process and
$\pmb v$ a process locally of bounded variation with the same domain.
Show that $\iiv(\pmb z)^{\uparrow}=ii_{\pmb v^{\uparrow}}(|\pmb z|)$.
%617G

\spheader 617Xb Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf and 616Xa.   Let $f:\coint{0,\infty}\to\Bbb R$ be a
\cadlag\ function such that $f\restr[0,t]$
is of bounded variation for every $t\ge 0$.   Let $\pmb v$ be the
corresponding process on $\Cal T_f$ and $\pmb v^*$ its quadratic
variation.   Show that $\pmb v^*$ corresponds to the function
$g:\Bbb R\to\Bbb R$ defined by saying that
$g(t)=\sum_{0<s\le t}(f(s)-\lim_{s'\uparrow s}f(s'))^2$.
%617H

\spheader 617Xc Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
$\pmb v$ an integrator with domain $\Cal S$, and
$z$ a member of $L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Show that if $\pmb v^*$ is the quadratic variation of $\pmb v$, then
(using the language of 612D(e-ii))
$z^2\pmb v^*$ is the quadratic variation of $z\pmb v$.
%617H

\spheader 617Xd Supposing that $T=\coint{0,\infty}$, take
$\pmb{\iota}=\family{\tau}{\Cal T_f}{\iota_{\tau}}$ to be the
identity process.   Show that

\Centerline{$\int_{\Cal T\wedge\tau}\pmb{\iota}\,d\pmb{\iota}
=\Bover12\iota_{\tau}^2$}

\noindent for every $\tau\in\Cal T_f$.
%617O

\sqheader 617Xe Let $\pmb v=\family{\tau}{\Cal T_f}{v_{\tau}}$
be the standard Poisson process.   Show that

\Centerline{$\int_{\Cal T\wedge\tau}\pmb v\,d\pmb v
=\Bover12(v_{\tau}^2-v_{\tau})$}

\noindent for every $\tau\in\Cal T_f$.
%617O

%\leader{617S}{}
%\leader{617T}{}
%\leader{617U}{}
%\leader{617V}{}
%\leader{617W}{}

%\leader{617Y}{Further exercises (a)}
}%end of exercises

\endnotes{
\Notesheader{617} There is a lot of meat in this section.   We have already
seen the relatively straightforward, but obviously fundamental,
fact that indefinite
integrals are commonly local integrators (616J, 616Q(c-i)).   It is a general
principle, of which we shall see many examples, that an indefinite integral
$\iivu$ is likely to share any special properties of the (local) integrator
$\pmb v$;  as a first step, $\iivu$ will be locally of bounded variation if
$\pmb v$ is (616L).   And we know that an indefinite integral operator
$\iiv$ with respect to an integrator $\pmb v$ is continuous for the ucp
topology (616J).

In fact there is nothing here as difficult as
the basic theorem on existence of Riemann-sum integrals (616M).   Once we know
that the integrals (with respect to integrating interval functions, as well
as with respect to integrators) are defined, the analysis in \S613 takes us a
long way;  if you like, Riemann-sum integrability is a very restrictive
condition which we shall not escape until we come to the S-integral in Chapter
64.

In the `mnemonics' offered in this section,
we have to distinguish between $\sim$ and $=$.    When I wrote

\Centerline{$(d\pmb v)^2
=d(\pmb v^2)-2\pmb v\,d\pmb v\sim d\pmb v^*$}

\noindent (617J),
the equality corresponds to a pair of interval functions being
equal;  writing $\pmb v\Delta\pmb v$ for the interval function
$c(\sigma,\tau)\mapsto v_{\sigma}\times(v_{\tau}-v_{\sigma})$,
I am saying that

\Centerline{$(v_{\tau}-v_{\sigma})^2
=v_{\tau}^2-v_{\sigma}^2-2v_{\sigma}\times(v_{\tau}-v_{\sigma})$}

\noindent and therefore that

\Centerline{$(\Delta\pmb v)^2
=\Delta(\pmb v^2)-2\pmb v\Delta{\pmb v}$,}

\Centerline{$\Delta_e(\pmb u,(d\pmb v)^2)
=\Delta_e(\pmb u,d(\pmb v^2))-2\Delta_e(\pmb u\times\pmb v,d\pmb v)$,}

\Centerline{$S_I(\pmb u,(d\pmb v)^2)
=S_I(\pmb u,d(\pmb v^2))-2S_I(\pmb u\times\pmb v,d\pmb v)$,}

\Centerline{$\int_{\Cal S}\pmb u\,(d\pmb v)^2
=\int_{\Cal S}\pmb u\,d(\pmb v^2)
  -2\int_{\Cal S}\pmb u\times\pmb v\,d\pmb v$}

\noindent for all $e$, $I$, $\pmb u$ and $\Cal S$
for which these are defined.
We have to pause for a moment to decide when the integrals exist, but
everything else is elementary algebra.   But
`$d\pmb v^*\sim(d\pmb v)^2$' does not assert that
$v^*_{\tau}-v^*_{\sigma}$ is related to $(v_{\tau}-v_{\sigma})^2$ for any
particular $\sigma$ and $\tau$.   It is in the first place a claim that
$\int_{\Cal S}d\pmb v^*=\int_{\Cal S}(d\pmb v)^2$, and then a claim that
$\int_{\Cal S}\pmb u\,d\pmb v^*
=\int_{\Cal S}\pmb u\,(d\pmb v)^2$.   The point is that the
interval function $\Delta\pmb v^*$ is additive, but $(\Delta\pmb v)^2$
is not.   The exposition in
Chapter 48 of Volume 4 tacitly assumes that when forming gauge integrals
we rather
expect our measures to be at least finitely additive.   In the present
volume we are dealing with integrals with respect to `measures'
which in some cases are most definitely not additive.

I introduced `adapted interval functions' in \S613 partly in order to have a
direct definition for such expressions as
$\int_{\Cal S}\pmb u|d\pmb v|$ and
$\int_{\Cal S}\pmb u\,d\pmb vd\pmb v'$.
In the contexts here, they give what turn out
to be useful suggestions:  if $\pmb w=\iiv(\pmb z)$, then we have mnemonics
$d\pmb w\sim\pmb z\,d\pmb v$ for 617E, $|d\pmb w|\sim|\pmb z||d\pmb v|$ for
617G and
$d\pmb wd\pmb v'\sim\pmb z\,d\pmb vd\pmb v'$ for 617Q.    But these
simple-minded formulae give no hint of the pages
of detailed calculations in 617F and 617P between the two expressions.
The problem lies
in the fact that $\Delta\pmb w$ is not quite equal to $\pmb z\Delta\pmb v$,
so we cannot immediately relate $|\Delta\pmb w|$ to $|\pmb z||\Delta\pmb v|$ or
$\Delta\pmb w\times\Delta\pmb v'$ to
$\pmb z\Delta\pmb v\times\Delta\pmb v'$.

Of course all the work of the second half of the section is dependent on being
able to move between $d\pmb vd\pmb v'$ and $d\covar{\pmb v}{\pmb v'}$, where
$\covar{\pmb v}{\pmb v'}$ is not merely a local integrator but actually locally
of bounded variation;  you will see that from 617P on I am repeatedly employing
whichever expression is most immediately convenient.

In both parts of 617O, we find that the calculation of the quadratic
variation from the formula

\Centerline{$v^*_{\tau}=\int_{\Cal S\wedge\tau}(d\pmb v)^2$}

\noindent (617Hb) is elementary in the sense that we just have to work
carefully through the definitions, looking at very natural sublattices
$\{\check{it/m}:i\le m\}$ in 617Oa and $\{\tau\wedge\tau_i:i\le n\}$
in 617Ob.   Of course we could use the same ideas to give us direct
calculations of the indefinite integrals $ii_{\pmb{\iota}}(\pmb{\iota})$
and $\iiv(\pmb v)$ (617Xd-617Xe), but these would necessarily be a touch
more complex.   In 617Xe we have a first outright declaration of a
difference between the formulae of elementary calculus and the
corresponding formulae of stochastic calculus, a foretaste of It\^o's
formula (\S619).

You will note that 617O omits any description of the
quadratic variation of Brownian motion.   As has happened before in this
chapter, Brownian motion seems to be essentially more difficult and more
interesting, as well as more important, than the Poisson process.   Like the
identity process, Brownian motion has no jumps (618J), but even so its
quadratic variation is non-zero (624F);  the argument I will give in 622L
depends on Dynkin's formula in harmonic analysis (478K).

617R is based on an elementary manipulation, but I think it requires
validation using ideas from \S352.
}%end of notes

\discrpage

\frfilename{mt618.tex}
\versiondate{8.9.12/26.8.22}

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

\def\chaptername{The Riemann-sum integral}
\def\sectionname{Oscillations and jump-free processes}

\newsection{618}

For the work so far, moderately oscillatory processes have been
sufficiently regular for our needs.
But for the next development (It\^o's formula, 619C), we are going to
need a new concept.   In 618B I formulate a notion of `jump-free' process
corresponding to the idea of `process with continuous sample paths' (618H).

\cmmnt{\leader{618A}{Notation} As previously,
\Asissix\ will be a stochastic integration structure.
For a sublattice $\Cal S$ of $\Cal T$, $\Mob(\Cal S)$ will be the
space of order-bounded processes with domain $\Cal S$ (614Fc) and
$\Cal I(\Cal S)$ the set of finite sublattices of $\Cal S$.
For an order-bounded process $\pmb u=\sfamilysS{u}$, $\sup|\pmb u|$ will be
$\sup_{\sigma\in\Cal S}|u_{\sigma}|$ (614Ea).
$L^{\infty}$ will be $L^{\infty}(\frak A)$ with its norm
$\|\,\|_{\infty}$\cmmnt{ as defined in \S363}.
$L^0(\frak A)$\cmmnt{ (\S364)} will be endowed with its
topology and uniformity of convergence in measure,
with the defining F-norm $\theta$\cmmnt{ (613Ba)}.
$\tbf 1$ will be the constant process with value $\chi 1$.
If $h:\Bbb R\to\Bbb R$ is a continuous function, I will write
$\bar h$ for any of the derived functions from $L^0(\frak A)$ to itself
or from a space of fully adapted processes to itself
(612A, 612Ia).
}

\leader{618B}{Definitions (a)} Let $I$ be a finite sublattice of $\Cal T$,
and $\pmb u$ a fully adapted process defined (at least) on $I$.   The
{\bf $I$-oscillation of $\pmb u$} is

\Centerline{$\Osclln_I(\pmb u)
=\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\pmb u|)$\dvro{.}{,}}

\cmmnt{\noindent where $\Sti_0(I)$ is the set of $I$-cells (611Je)
and for a stopping-time interval $e=c(\sigma,\tau)$ I write
$\Delta_e(\tbf 1,|d\pmb u|)=|u_{\tau}-u_{\sigma}|$ (613Fa).
Take the supremum in $(L^0(\frak A))^+$, so that if $\#(I)\le 1$ and $\Sti_0(I)$
is empty, then $\Osclln_I(\pmb u)=0$.}

Note that if $(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells, then
$\Osclln_I(\pmb u)=\sup_{i<n}|u_{\tau_{i+1}}-u_{\tau_i}|$\cmmnt{, just because
$\Sti_0(I)$ is the set of stopping-time intervals
$\{c(\tau_i,\tau_{i+1}):i<n$, $\tau_i\ne\tau_{i+1}\}$ (611L)}.

\spheader 618Bb Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u =\sfamilysS{u}$ an
order-bounded process\cmmnt{ (definition:  614E)}.
Set $\bar u=\sup|\pmb u|$.

\medskip

\quad{\bf (i)}\cmmnt{ $\Delta_e(\tbf 1,|d\pmb u |)\le 2\bar u$
for every $e$ in the set
$\Sti(\Cal S)$ of stopping-time intervals with endpoints in $\Cal S$,
so} $\Osclln_J(\pmb u)\le 2\bar u$ for every $J\in\Cal I(\Cal S)$.
We\cmmnt{ can therefore} set

\Centerline{$\Osclln^*_I(\pmb u)
=\sup_{J\in\Cal I(\Cal S),J\supseteq I}\Osclln_J(\pmb u)
\le 2\bar u$}

\noindent for every $I\in\Cal I(\Cal S)$.

\medskip

\quad{\bf (ii)} The {\bf residual oscillation}
$\Osclln(\pmb u)$ is
$\inf_{I\in\Cal I(\Cal S)}\Osclln^*_I(\pmb u)\le 2\bar u$.
\cmmnt{I will say that }$\pmb u$ is {\bf jump-free} if
$\Osclln(\pmb u)=0$.   \cmmnt{For definiteness, I add that if,
in the rest of this volume, I say that a process is jump-free, I mean to imply
that it is order-bounded.}

\medskip

\quad{\bf (iii)}\cmmnt{ I will say that} $\pmb u$ is {\bf locally jump-free}
if $\pmb u \restr\Cal S\wedge\tau$ is jump-free for every $\tau\in\Cal S$.

\medskip

\quad{\bf (iv)}\dvro{ $\Osclln^*_{\emptyset}(\pmb u)
=\sup\{|u_{\sigma'}-u_{\sigma}|:\sigma,\,\sigma'\in\Cal S\}$.}{

$$\eqalign{\Osclln^*_{\emptyset}(\pmb u)&
=\sup_{J\in\Cal I(\Cal S)}\Osclln_J(\pmb u)
=\sup\{|u_{\sigma'}-u_{\sigma}|:
  \sigma,\,\sigma'\in\Cal S\text{ and }\sigma\le\sigma'\}\cr&
=\sup\{|u_{\sigma'\vee\sigma}-u_{\sigma'\wedge\sigma}|:
  \sigma,\,\sigma'\in\Cal S\}
=\sup\{|u_{\sigma'}-u_{\sigma}|:\sigma,\,\sigma'\in\Cal S\}\cr}$$

\noindent by 612D(f-ii).}

\medskip

\quad{\bf (v)}\cmmnt{ {\bf Remarks} \grheada\ Note that
$I\mapsto\Osclln^*_I(\pmb u):
\Cal I(\Cal S)\to L^0(\frak A)$ is non-increasing, so}
$\Osclln(\pmb u)$ is the limit
$\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(\pmb u)$\cmmnt{ for the topology of
convergence in measure,} and $\pmb u$ is jump-free iff

\Centerline{$\inf_{I\in\Cal I(\Cal S)}\theta(\Osclln^*_I(\pmb u))
=\lim_{I\uparrow\Cal I(\Cal S)}\theta(\Osclln^*_I(\pmb u))=0$.}

\cmmnt{\medskip

\qquad\grheadb\ In the formula

\Centerline{$\Osclln_I(\pmb u)
=\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\pmb u|)$,}

\noindent the arguments
$I$ and $\pmb u$ are explicit in the term
$\Osclln_I(\pmb u)$.   In the formula

\Centerline{$\Osclln^*_I(\pmb u)
=\sup_{J\in\Cal I(\Cal S),J\supseteq I}\Osclln_J(\pmb u)$,}

\noindent the lattice $\Cal S$ is not declared overtly in the expression
$\Osclln^*_I(\pmb v)$;  a purist would prefer

\Centerline{$\Osclln^*_I(\pmb u)
=\sup_{J\in\Cal I(\dom\pmb u),J\supseteq I}\Osclln_J(\pmb u)$,}

\noindent just as

\Centerline{$\Osclln(\pmb u)
=\inf_{I\in\Cal I(\dom\pmb u)}\Osclln^*_I(\pmb u)$.}

\medskip

\qquad\grheadc\ Observe that we have
$\Osclln_I(\pmb u)=\Osclln(\pmb u\restr I)$ for every $I\in\Cal I(\Cal S)$.
}%end of comment

\spheader 618Bc Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u =\sfamilysS{u}$, $\pmb v=\sfamilysS{v}$
order-bounded processes.

\medskip

\quad{\bf (i)} For any $\alpha\in\Bbb R$,

\Centerline{$\Osclln_I(\alpha\pmb u)=|\alpha|\Osclln_I(\pmb u)$ for every
$I\in\Cal I(\Cal S)$,}

\Centerline{$\Osclln^*_I(\alpha\pmb u)=|\alpha|\Osclln^*_I(\pmb u)$
for every $I\in\Cal I(\Cal S)$,}

\Centerline{$\Osclln(\alpha\pmb u)=|\alpha|\Osclln(\pmb u)$.}

\medskip

\quad{\bf (ii)}

\Centerline{$\Osclln_I(\pmb u +\pmb v)
\le\Osclln_I(\pmb u)+\Osclln_I(\pmb v)$ for every
$I\in\Cal I(\Cal S)$,}

\Centerline{$\Osclln^*_I(\pmb u +\pmb v)
\le\Osclln^*_I(\pmb u)+\Osclln^*_I(\pmb v)$ for every
$I\in\Cal I(\Cal S)$,}

\Centerline{$\Osclln(\pmb u +\pmb v)
\le\Osclln(\pmb u)+\Osclln(\pmb v)$.}

\medskip

\quad{\bf (iii)} Writing $\bar u$, $\bar v$ for $\sup|\pmb u|$ and
$\sup|\pmb v|$,

\Centerline{$\Osclln_I(\pmb u \times \pmb v)
\le\bar v\times\Osclln_I(\pmb u)+\bar u\times\Osclln_I(\pmb v)$ for every
$I\in\Cal I(\Cal S)$,}

\Centerline{$\Osclln^*_I(\pmb u \times\pmb v)
\le\bar v\times\Osclln^*_I(\pmb u)+\bar u\times\Osclln^*_I(\pmb v)$
for every $I\in\Cal I(\Cal S)$,}

\Centerline{$\Osclln(\pmb u \times\pmb v)
\le\bar v\times\Osclln(\pmb u)+\bar u\times\Osclln(\pmb v)$.}

\leader{618C}{}\cmmnt{ We shall not need them
immediately, but the following
descriptions of $\Osclln^*$ will be useful later on.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u =\sfamilysS{u}$ an order-bounded fully adapted process.
Let $I$ be a non-empty finite sublattice of $\Cal S$;  suppose that
$(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells.

(a) Set $\tau_{-1}=\inf\Cal S$ and $\tau_{n+1}=\sup\Cal S$ and

$$\eqalign{w
&=\sup\{|u_{\sigma'}-u_{\sigma}|:\sigma,\,\sigma'\in\Cal S
\text{ and there is an }i\cr
&\mskip100mu\text{such that }-1\le i\le n
\text{ and }\tau_i\le\sigma\le\sigma'\le\tau_{i+1}\},\cr
w'
&=\sup\{|u_{\sigma'}-u_{\sigma}|:\sigma,\,\sigma'\in\Cal S
\text{ and there is an }i\cr
&\mskip100mu\text{such that }-1\le i\le n
\text{ and }\sigma,\,\sigma'\in[\tau_i,\tau_{i+1}]\}.\cr}$$

\noindent Then $w=w'=\Osclln^*_I(\pmb u)$.

(b) Now suppose that $\pmb u$ is non-decreasing.   Set
$u_{\downarrow}=\inf_{\sigma\in\Cal S}u_{\sigma}$ and
$u_{\uparrow}=\sup_{\sigma\in\Cal S}u_{\sigma}$.   Then

\Centerline{$\Osclln^*_I(\pmb u)
=(u_{\tau_0}-u_{\downarrow})\vee\sup_{i<n}(u_{\tau_{i+1}}-u_{\tau_i})
\vee(u_{\uparrow}-u_{\tau_n})$.}

\proof{{\bf (a)(i)} Suppose that $J\in\Cal I(\Cal S)$, $I\subseteq J$ and
$e\in\Sti_0(J)$.   Then there is an $i$ such that $-1\le i\le n$ and
$e\in\Sti_0(J\cap[\tau_i,\tau_{i+1}])$, by 611J(e-iii).   In this case,
the stopping-time interval
$e$ is expressible as $c(\sigma,\sigma')$ where
$\tau_i\le\sigma\le\sigma'\le\tau_{i+1}$ and $\sigma$, $\sigma'\in\Cal S$.
Accordingly

\Centerline{$\Delta_e(\tbf 1,|d\pmb u |)=|u_{\sigma'}-u_{\sigma}|\le w'$.}

\noindent As $e$ is arbitrary, $\Osclln_J(\pmb u)\le w'$;  as
$J$ is arbitrary, $\Osclln^*_I(\pmb u)\le w$.

\medskip

\quad{\bf (ii)} Suppose that $-1\le i\le n$ and
$\tau_i\le\sigma\le\sigma'\le\tau_{i+1}$.   Then
$|u_{\sigma'}-u_{\sigma}|\le\Osclln^*_I(\pmb u)$.
\Prf\ If $\sigma=\sigma'$ this is trivial.   Otherwise,
set $J=I\cup\{\sigma,\sigma'\}$.   If $\tau\in I$ then
$\med(\tau_i,\tau,\tau_{i+1})$ belongs to $I$ (because
$\tau\vee\tau_{-1}=\tau\wedge\tau_{n+1}=\tau\in I$), and
must be either $\tau_i$ or $\tau_{i+1}$;  so either $\tau\le\sigma$ or
$\sigma'\le\tau$.   It follows that
$J$ is a sublattice of $\Cal S$ including $I$
and $c(\sigma,\sigma')$ is a $J$-cell.  So

\Centerline{$|u_{\sigma'}-u_{\sigma}|\le\Osclln_J(\pmb u)
\le\Osclln^*_I(\pmb u)$.  \Qed}

\noindent Thus $w\le\Osclln^*_I(\pmb u)$ and the two are equal.

\medskip

\quad{\bf (iii)} Of course $w\le w'$.   On the other hand, if
$\sigma$, $\sigma'$ both belong to $\Cal S\cap[\tau_i,\tau_{i+1}]$ for some $i$,
then so do $\sigma\wedge\sigma'$ and $\sigma\vee\sigma'$, and

\Centerline{$w\ge|u_{\sigma\vee\sigma'}-u_{\sigma\wedge\sigma'}|
=|u_{\sigma}-u_{\sigma'}|$}

\noindent (612D(f-ii)).   So $w\ge w'$ and we have equality.

\medskip

{\bf (b)} This is now elementary, because

$$\eqalign{\sup\{|u_{\sigma'}-u_{\sigma}|:\sigma,\,\sigma'\in\Cal S,\,
   \tau_i\le\sigma\le\sigma'\le\tau_{i+1}\}
&=u_{\tau_0}-u_{\downarrow}\text{ if }i=-1,\cr
&=u_{\tau_{i+1}}-u_{\tau_i}\text{ if }0\le i<n,\cr
&=u_{\uparrow}-u_{\tau_n}\text{ if }i=n.\cr}$$
}%end of proof of 618C

\leader{618D}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u =\sfamilysS{u}$ a locally order-bounded process.

(a) Set $v_{\tau}=\Osclln(\pmb u \restr\Cal S\wedge\tau)$ for
$\tau\in\Cal S$.   Then $\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$
is a non-decreasing fully adapted process.

(b) If $\pmb u$ is order-bounded, then

\quad(i) $\Osclln(\pmb u)=\Osclln(\pmb u \restr\Cal S\wedge\tau)
  \vee\Osclln(\pmb u \restr\Cal S\vee\tau)$
for every $\tau\in\Cal S$,

\quad(ii) $\Osclln(\pmb u\restr\Cal S\cap[\tau,\tau'])\le\Osclln(\pmb u)$
whenever $\tau\le\tau'$ in $\Cal S$.

\proof{{\bf (a)(i)}
Looking at the definitions in 618B, we see that, for any $\tau\in\Cal S$,

\Centerline{$\Osclln_I(\pmb u \restr\Cal S\wedge\tau)
\in L^0(\frak A_{\tau})$
for every $I\in\Cal I(\Cal S\wedge\tau)$,}

\noindent so that

\Centerline{$\Osclln^*_I(\pmb u \restr\Cal S\wedge\tau)
\in L^0(\frak A_{\tau})$ for every $I\in\Cal I(\Cal S\wedge\tau)$}

\noindent (because $L^0(\frak A_{\tau})$ is an order-closed sublattice of
$L^0(\frak A)$, by 612Ae) and

\Centerline{$v_{\tau}
=\inf_{I\in\Cal I(\Cal S\wedge\tau)}
   \Osclln^*_I(\pmb u \restr\Cal S\wedge\tau)
\in L^0(\frak A_{\tau})$.}

\medskip

\quad{\bf (ii)} If $\tau\le\tau'$ in $\Cal S$, then

\Centerline{$\Osclln_I(\pmb u)
=\Osclln_{I\wedge\tau}(\pmb u)\vee\Osclln_{I\vee\tau}(\pmb u)$
whenever $\tau\in I\in\Cal I(\Cal S)$}

\noindent\prooflet{(use 611J(e-ii))}.   Suppose that
$\tau\in I\in\Cal I(\Cal S\wedge\tau')$ and that
$I\wedge\tau\subseteq J\in\Cal I(\Cal S\wedge\tau)$.   Let $K$ be the
sublattice generated by $I\cup J$;  then $K\wedge\tau=J$, so

\Centerline{$\Osclln_J(\pmb u)\le\Osclln_K(\pmb u)
\le\Osclln^*_I(\pmb u \restr\Cal S\wedge\tau')$.}

\noindent As $J$ is arbitrary,

\Centerline{$v_{\tau}=\Osclln(\pmb u \restr\Cal S\wedge\tau)
\le\Osclln^*_{I\wedge\tau}(\pmb u \restr\Cal S\wedge\tau)
\le\Osclln^*_I(\pmb u \restr\Cal S\wedge\tau')$.}

\noindent As $I$ is arbitrary,

\Centerline{$v_{\tau}\le\Osclln(\pmb u \restr\Cal S\wedge\tau')
=v_{\tau'}$.}

\noindent Thus $\pmb v$ is non-decreasing.

\medskip

\quad{\bf (iii)} If $\tau\le\tau'$ in $\Cal S$ and $c=\Bvalue{\tau=\tau'}$,
then $\chi c\times|v_{\sigma'}-v_{\sigma}|=0$ whenever
$\tau\le\sigma\le\sigma'\le\tau'$, so
$\chi c\times\Osclln_{I\vee\tau}(\pmb u)=0$ and
$\chi c\times\Osclln_{I\wedge\tau}(\pmb u)
=\chi c\times\Osclln_I(\pmb u)$ whenever
$\tau\in I\in\Cal I(\Cal S\wedge\tau')$.   Consequently

$$\eqalign{\chi c\times v_{\tau}
&=\chi c\times\inf_{\tau\in I\in\Cal I(\Cal S)}
  \sup_{\Atop{J\in\Cal I(\Cal S)}{J\supseteq I}}
     \Osclln_{J\wedge\tau}(\pmb u)\cr
&=\inf_{\tau\in I\in\Cal I(\Cal S)}
  \sup_{\Atop{J\in\Cal I(\Cal S)}{J\supseteq I}}
     \chi c\times\Osclln_{J\wedge\tau}(\pmb u)\cr
&=\inf_{\tau\in I\in\Cal I(\Cal S)}
  \sup_{\Atop{J\in\Cal I(\Cal S)}{J\supseteq I}}
     \chi c\times\Osclln_J(\pmb u)
=\chi c\times v_{\tau'}\cr}$$

\noindent and $\Bvalue{\tau=\tau'}\Bsubseteq\Bvalue{v_{\tau}=v_{\tau'}}$.
As noted in 612Db, this is enough to show that $\pmb v$ is fully adapted.

\medskip

{\bf (b)(i)} If $\tau\in J\in\Cal I(\Cal S)$ then (as noted in (a-ii)
above)

\Centerline{$\Osclln_J(\pmb u)
  =\Osclln_{J\wedge\tau}(\pmb u)\vee\Osclln_{J\vee\tau}(\pmb u)$.}

\noindent So if $\tau\in I\in\Cal I(\Cal S)$ then

$$\eqalign{\Osclln^*_I(\pmb u)
&=\sup_{I\subseteq J\in\Cal I(\Cal S)}\Osclln_J(\pmb u)
=\sup_{I\subseteq J\in\Cal I(\Cal S)}
    \Osclln_{J\wedge\tau}(\pmb u)\vee\Osclln_{J\vee\tau}(\pmb u)\cr
&=\sup_{\Atop{I\wedge\tau\subseteq K\in\Cal I(\Cal S\wedge\tau)}
   {I\vee\tau\subseteq L\in\Cal I(\Cal S\vee\tau)}}
      \Osclln_K(\pmb u)\vee\Osclln_L(\pmb u)\cr
&=\Osclln^*_{I\wedge\tau}(\pmb u \restr\Cal S\wedge\tau)
   \vee\Osclln^*_{I\vee\tau}(\pmb u \restr\Cal S\vee\tau).\cr}$$

\noindent Taking the infimum over $I$ (using the distributive law
352Eb),

\Centerline{$\Osclln(\pmb u)=\Osclln(\pmb u \restr\Cal S\wedge\tau)
   \vee\Osclln(\pmb u \restr\Cal S\vee\tau)$}

\noindent as claimed.

\medskip

\quad{\bf (ii)} Now

\Centerline{$\Osclln(\pmb u\restr\Cal S\cap[\tau,\tau'])
  =\Osclln(\pmb u\restr(\Cal S\wedge\tau')\vee\tau)
  \le\Osclln(\pmb u\restr\Cal S\wedge\tau')
  \le\Osclln(\pmb u)$.}
}%end of proof of 618D

\leader{*618E}{}\cmmnt{ An elementary fact will turn out to be useful in Chapter
64. %644C

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a finitely full
sublattice of $\Cal T$ with a greatest element,
$\pmb u=\sfamilysS{u}$ a jump-free process, $\tau\in\Cal S$ and
$\epsilon>0$.   Then there is a $\tau'\in\Cal S\vee\tau$ such that
$\Bvalue{\tau<\tau'}=\Bvalue{\tau<\max\Cal S}$ and
$\theta(\sup_{\sigma\in\Cal S\cap[\tau,\tau']}|u_{\sigma}-u_{\tau}|)
\le\epsilon$.

\proof{ By 618D(b-i), $\pmb u'=\pmb u\restr\Cal S\vee\tau$ is jump-free.
Let $I$ be a finite sublattice of $\Cal S\vee\tau$ containing $\tau$ and
$\max\Cal S$ and such that $\theta(\Osclln^*_I(\pmb u'))\le\epsilon$.
Take $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   For
$i<n$ set $b_i=\Bvalue{\tau<\tau_{i+1}}\Bsetminus\Bvalue{\tau<\tau_i}$;
set $b_n=\Bvalue{\tau=\max\Cal S}$.   Then
$\langle b_i\rangle_{i\le n}$ is disjoint and has supremum $1$,
$b_i\in\frak A_{\tau_{i+1}}$ for $i<n$, and $b_n\in\frak A_{\max\Cal S}$.
By 611I there is a $\tau'\in\Cal T$ such that
$b_i\Bsubseteq\Bvalue{\tau'=\tau_{i+1}}$ for $i<n$ and
$b_n\Bsubseteq\Bvalue{\tau'=\max\Cal S}$.   Because $\Cal S$ is finitely full,
$\tau'\in\Cal S$, while $\tau\le\tau'$ and
$\Bvalue{\tau<\tau'}=\sup_{i<n}b_i=\Bvalue{\tau<\max\Cal S}$.   If
$\sigma\in\Cal S$ and $\tau\le\sigma\le\tau'$, then for $i<n$

$$\eqalign{b_i&
\Bsubseteq\Bvalue{\tau=\tau_i}\Bcap\Bvalue{\tau_i\le\sigma}
   \Bcap\Bvalue{\sigma\le\tau_{i+1}}
\Bsubseteq\Bvalue{\tau=\tau_i}
   \Bcap\Bvalue{\sigma=\med(\tau_i,\sigma,\tau_{i+1})}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau}|
    =|u_{\med(\tau_i,\sigma,\tau_{i+1})}-u_{\tau_i}|}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau}|\le\Osclln^*_I(\pmb u')}\cr}$$

\noindent by 618Ca.   At the same time,

\Centerline{$b_n\Bsubseteq\Bvalue{\tau=\sigma}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau}|=0}$.}

\noindent So $|u_{\sigma}-u_{\tau}|\le\Osclln^*_I(\pmb u')$.   As $\sigma$ is
arbitrary,

\Centerline{$\theta(\sup_{\sigma\in\Cal S\cap[\tau,\tau']}|u_{\sigma}-u_{\tau}|)
\le\theta(\Osclln^*_I(\pmb u'))\le\epsilon$.}
}%end of proof of 618E

\leader{618F}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.

(a) If $\pmb u$, $\pmb v$ are order-bounded processes with domain $\Cal S$, then
$|\Osclln(\pmb u)-\Osclln(\pmb v)|\le 2\sup|\pmb u-\pmb v|$.

(b) $\Osclln:\Mob(\Cal S)\to L^0(\frak A)$
is uniformly continuous if $\Mob(\Cal S)$ is given its
ucp uniformity.

\proof{{\bf (a)} By 618B(c-ii) and 618B(b-ii),

\Centerline{$\Osclln(\pmb u)\le\Osclln(\pmb v)+\Osclln(\pmb u-\pmb v)
\le\Osclln(\pmb v)+2\sup|\pmb u-\pmb v|$}

\noindent and similarly
$\Osclln(\pmb v)\le\Osclln(\pmb u)+2\sup|\pmb u-\pmb v|$.

\medskip

{\bf (b)} Consequently

\Centerline{$\theta(\Osclln(\pmb v)-\Osclln(\pmb u))
\le 2\theta(\sup|\pmb u-\pmb v|)
=2\widehat{\theta}(\pmb u-\pmb v)$}

\noindent where $\widehat{\theta}$ is the F-norm defining the ucp linear space
topology on $\Mob(\Cal S)$ (615B).   So $\Osclln:\Mob(\Cal S)\to L^0(\frak A)$
is uniformly continuous.
}%end of proof of 618F

\leader{618G}{Proposition}
Let $\Cal S$ be a sublattice of $\Cal T$.
Write $\Mjf(\Cal S)$ for the set of jump-free fully adapted processes with
domain $\Cal S$.

(a) The set $\Mjf(\Cal S)$ of jump-free fully adapted processes with domain
$\Cal S$ is a topologically closed $f$-subalgebra of $\Mob(\Cal S)$, and
$\bar h\pmb v\in\Mjf(\Cal S)$ whenever $\pmb v\in\Mjf(\Cal S)$ and
$h:\Bbb R\to\Bbb R$ is continuous.

(b) A (locally) jump-free fully adapted process on $\Cal S$ is
(locally) moderately oscillatory.

(c) If $\pmb v\in\Mjf(\Cal S)$, then $\pmb v\restr\Cal S\vee\tau$,
$\pmb v\restr\Cal S\wedge\tau'$ and $\pmb v\restr\Cal S\cap[\tau,\tau']$
are jump-free whenever $\tau\le\tau'$ in $\Cal S$.   In particular,
$\pmb v$ is locally jump-free.

\proof{{\bf (a)(i)} If $\pmb u$, $\pmb v\in\Mjf=\Mjf(\Cal S)$,
then $\Osclln(\pmb u +\pmb v)\le\Osclln(\pmb u)+\Osclln(\pmb v)=0$ (618B(c-ii))
so $\pmb u+\pmb v\in\Mjf$.

\medskip

\quad{\bf (ii)} Since $\{0\}$ is closed in $L^0(\frak A)$
and $\Osclln:\Mob(\Cal S)\to L^0(\frak A)$ is continuous (618Fb),
$\Mjf(\Cal S)=\{\pmb u:\pmb u\in\Mob(\Cal S)$, $\Osclln(\pmb u)=0\}$ is
closed in $\Mob(\Cal S)$.

\medskip

\quad{\bf (iii)} If $\pmb u\in\Mjf$ and $h:\Bbb R\to\Bbb R$ is
uniformly continuous, then $\bar h\pmb u\in\Mjf$.   \Prf\ By 614F(c-i),
$\bar h\pmb u$ is order-bounded.   For $x\in\Bbb R$, set
$g(x)=\sup\{|h(y)-h(y')|:|y-y'|\le|x|\}$;  then $g$ is Borel measurable,
$\lim_{x\to 0}g(x)=0$ and $|h(x)-h(y)|\le g(|x-y|)$ for all $x$,
$y\in\Bbb R$, so $|\bar h(u)-\bar h(v)|\le\bar g(|u-v|)$ for all $u$,
$v\in L^0(\frak A)$.   Accordingly
$\Delta_e(\tbf 1,|d\bar h\pmb u|)
\le\bar g(\Delta_e(\tbf 1,|d\pmb u |))$ for every $e\in\Sti(\Cal S)$,
$\Osclln_J(\bar h\pmb u)
\le\bar g(\Osclln_J(\pmb u))$ for every $J\in\Cal I(\Cal S)$
and $\Osclln^*_I(\bar h\pmb u)
\le\bar g(\Osclln^*_I(\pmb u))$ for every $I\in\Cal I(\Cal S)$.
Since $\inf_{I\in\Cal I(\Cal S)}\Osclln^*_I(\pmb u)=0$ and
$\lim_{x\downarrow 0}g(x)=0$,
$\inf_{I\in\Cal I(\Cal S)}\bar g(\Osclln^*_I(\pmb u))=0$ and
$\inf_{I\in\Cal I(\Cal S)}\Osclln^*_I(\bar h\pmb u)=0$ and
$\bar h\pmb u$ is jump-free.\ \Qed

\medskip

\quad{\bf (iv)} If $\pmb u =\family{\sigma}{\Cal S}{u_{\sigma}}$ is
jump-free and $\bar h:\Bbb R\to\Bbb R$ is
continuous, then $\bar h\pmb u$ is jump-free.   \Prf\ Again,
$\bar h\pmb u$ is order-bounded.   Set
$\bar u=\sup_{\sigma\in\Cal S}|u_{\sigma}|$.   Given $\epsilon>0$, let
$M\ge 0$ be such that $a=\Bvalue{\bar u\ge M}$ has measure at most
$\epsilon$.   Set

$$\eqalign{h_1(x)
&=h(-M)\text{ if }x\le -M,\cr
&=h(x)\text{ if }|x|\le M,\cr
&=h(M)\text{ if }x\ge M.\cr}$$

\noindent Then $h_1$ is uniformly continuous so
$\bar h_1(\pmb u)\in\Mjf$, by (ii).   But

\Centerline{$\Bvalue{\bar h\pmb u\ne\bar h_1\pmb u}
=\sup_{\sigma\in\Cal S}
   \Bvalue{\bar h(u_{\sigma})\ne\bar h_1(u_{\sigma})}
\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{|u_{\sigma}|>M}
=\Bvalue{\bar u>M}
\Bsubseteq a$,}

\noindent so

\Centerline{$\widehat{\theta}(\bar h\pmb u-\bar h_1\pmb u)
\le\bar\mu a\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary and $\Mjf$ is closed,
$\bar h\pmb u\in\Mjf$.\ \Qed

\medskip

\quad{\bf (v)} Putting (i) and (iv) together, $\Mjf$ is an
$f$-subalgebra of $\Mob(\Cal S)$, by 612Bc as usual.

\medskip

{\bf (b)(i)} Suppose that $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ is
jump-free.   If $\Cal S$ is empty, $\pmb u$ is certainly moderately oscillatory;
suppose otherwise.

\medskip

\qquad\grheada\ Set $\frak B=\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$.
Then the starting value
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is
defined and belongs to $L^0(\frak B)$.   \Prf\ Let $\epsilon>0$.   Let
$I\in\Cal I(\Cal S)$ be non-empty and such that
$\theta(\Osclln^*_I(\pmb u))\le\epsilon$.   If $\sigma\in\Cal S$ and
$\sigma\le\min I$, then $J=I\cup\{\sigma\}\in\Cal I(\Cal S)$ and
$|u_{\min I}-u_{\sigma}|\le\Osclln_J(\pmb u)\le\Osclln^*_I(\pmb u)$, so
$\theta(u_{\sigma}-u_{\min I})\le\epsilon$.   As $L^0(\frak A)$ is complete in
the linear space topology of convergence in measure, this is enough to show
that $u_{\downarrow}$ is defined.   By 613Bj, $u_{\downarrow}\in L^0(\frak B)$.\
\Qed

\medskip

\qquad\grheadb\ If $I\in\Cal I(\Cal S)$ is non-empty, there is a simple
process $\pmb u'$ such that
$\sup|\pmb u-\pmb u'|\le\Osclln^*_I(\pmb u)$.   \Prf\  Write $w$ for
$\Osclln^*_I(\pmb u)$.
Let $\tau_0\le\ldots\le\tau_n$ linearly generate the $I$-cells
(611L).   Let $\pmb u'=\sfamilysS{u'}$
be the simple process defined on $\Cal S$ by the
formulae of 612Ka from $\tau_0,\ldots,\tau_n$  and
$u_{\downarrow},u_{\tau_0},\ldots,u_{\tau_n}$.

If $\sigma'\le\sigma\in\Cal S\wedge\tau_0$, then

\Centerline{$|u_{\sigma}-u_{\sigma'}|
\le\Osclln_{I\cup\{\sigma',\sigma\}}(\pmb u)
\le w$;}

\noindent taking the limit as $\sigma\downarrow\Cal S$,
$|u_{\sigma}-u_{\downarrow}|\le w$.
If $i<n$ and $\sigma\in\Cal S\cap[\tau_i,\tau_{i+1}]$, then

\Centerline{$|u_{\sigma}-u_{\tau_i}|
\le\Osclln_{I\cup\{\sigma\}}(\pmb u)
\le w$;}

\noindent similarly, if $\sigma\in\Cal S\vee\tau_n$ then

\Centerline{$|u_{\sigma}-u_{\tau_n}|
\le\Osclln_{I\cup\{\sigma\}}(\pmb u)
\le w$.}

For any $\sigma\in\Cal S$ we see now that

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

$$\eqalign{\Bvalue{\tau_n\le\sigma}&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\vee\tau_n}}
   \Bcap\Bvalue{u'_{\sigma}=u_{\tau_n}}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-u'_{\sigma}|
   =|u_{\sigma\vee\tau_n}-u_{\tau_n}|}
\Bsubseteq\Bvalue{|u_{\sigma}-u'_{\sigma}|\le w},\cr}$$

\noindent and for any $i<n$

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

\noindent Since

\Centerline{$\Bvalue{\sigma<\tau_0}\Bcup\Bvalue{\tau_n\le\sigma}
 \Bcup\sup_{i<n}(\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}})
=1$,}

\noindent $|u_{\sigma}-u'_{\sigma}|\le w$.   As $\sigma$ is arbitrary,
$\sup|\pmb u-\pmb u'|\le w$, as claimed.\ \Qed

\medskip

\qquad\grheadc\ Now take any $\epsilon>0$.   As $\pmb u$ is jump-free,
there is a non-empty $I\in\Cal I(\Cal S)$ such that
$\theta(\Osclln^*_I(\pmb u))\le\epsilon$.   By ($\beta$), there is a
simple process $\pmb u'$ such that $\theta(\sup|\pmb u-\pmb u'|)\le\epsilon$.
But we know that simple processes are moderately oscillatory (615E).   As
$\epsilon$ is arbitrary, $\pmb u$ belongs to the closure of the set
$\Mmo(\Cal S)$ of moderately oscillatory processes for the ucp topology
on $\Mob(\Cal S)$.   But $\Mmo(Cal S)$ is closed
(615F(a-iv)), so $\pmb u$ is moderately oscillatory.

\medskip

\quad{\bf (ii)} If $\pmb u$ is locally jump-free, then applying (i)
we see that $\pmb u\restr\Cal S\wedge\tau$ is moderately oscillatory for every
$\tau\in\Cal S$, so $\pmb u$ is locally moderately oscillatory.

\medskip

{\bf (c)(i)} Given $I\in\Cal I(\Cal S)$, let $I'$ be the sublattice
generated by $I\cup\{\tau,\tau'\}$.   Then
$\Osclln^*_{I'\vee\tau}(\pmb v\restr\Cal S\vee\tau)
\le\Osclln^*_I(\pmb v)$.   \Prf\ If
$I'\vee\tau\subseteq J\in\Cal I(\Cal S\vee\tau)$, and $K$ is the
sublattice generated by $I'\cup J$, we shall have
$I\subseteq K\in\Cal I(\Cal S)$, $K\vee\tau=J$ and
$\Sti_0(J)\subseteq\Sti_0(K)$, so

\Centerline{$\Osclln_J(\pmb v)\le\Osclln_K(\pmb v)
\le\Osclln^*_I(\pmb v)$.}

\noindent As $J$ is arbitrary,

\Centerline{$\Osclln^*_{I'\vee\tau}(\pmb v\restr\Cal S\vee\tau)
\le\Osclln^*_I(\pmb v)$.  \Qed}

\noindent Consequently

\Centerline{$\inf_{I\in\Cal I(\Cal S\vee\tau)}
   \Osclln^*_I(\pmb v\restr\Cal S\vee\tau)
\le\inf_{I\in\Cal I(\Cal S)}\Osclln^*_I(\pmb v)=0$}

\noindent and $\pmb v\restr\Cal S\vee\tau$ is jump-free.

\medskip

\quad{\bf (ii)}
Replacing `$\vee\tau$' by `$\wedge\tau'$' throughout the argument above, we
see that $\pmb v\restr\Cal S\wedge\tau'$ is jump-free.   Putting these
together, $\pmb v\restr\Cal S\cap[\tau,\tau']
=(\pmb v\restr\Cal S\vee\tau)\restr\Cal S\wedge\tau'$ is jump-free.
And as $\tau'$ is arbitrary, $\pmb v$ is locally jump-free.
}%end of proof of 618G

\leader{618H}{Where jump-free processes come from:  Proposition}
Let $(\Omega,\Sigma,\mu)$ be a complete probability space and
$\langle\Sigma_t\rangle_{t\ge 0}$ a
family of $\sigma$-subalgebras of $\Sigma$, all containing every negligible
subset of $\Omega$.   Suppose that we are given a
family $\langle X_t\rangle_{t\ge 0}$ of real-valued functions on
$\Omega$ such that $X_t$ is $\Sigma_t$-measurable for every $t\ge 0$ and
$t\mapsto X_t(\omega):\coint{0,\infty}\to\Bbb R$ is continuous for
every $\omega\in\Omega$.   Then $\langle X_t\rangle_{t\ge 0}$ is progressively
measurable, and if
$(\frak A,\bar\mu,\langle\frak A_t\rangle_{t\ge 0},\Cal T,
\family{\tau}{\Cal T}{\frak A_{\tau}})$ and
$\family{\sigma}{\Cal T_f}{x_{\sigma}}$ are defined
as in 612H, $\pmb x=\family{\sigma}{\Cal T_f}{x_{\sigma}}$ is locally jump-free.

\proof{ The idea is to combine the approaches of 612H and 615P.

\medskip

{\bf (a)} I start by showing that we have a
progressively measurable process.   \Prf\
Take any $t\ge 0$ and $\alpha\in\Bbb R$.   Set
$Q=\{qt:q\in\Bbb Q\cap[0,1]\}$.   Then

$$\eqalign{&\{(s,\omega):s\le t,\,X_s(\omega)>\alpha\}
=\{(s,\omega):s\le t,\,
   \limsup_{q\downarrow Q\cap[s,t]}X_q(\omega)>\alpha\}\cr&
\mskip100mu
=\bigcup_{k\in\Bbb N}\bigcap_{q\in Q}\bigcup_{\Atop{q'\in Q}{q'\le q}}
    \{(s,\omega):s\le t,\,s>q\text{ or }s\le q'\text{ and }
        X_{q'}(\omega)\ge\alpha+2^{-k}\}\cr&
\mskip100mu
\in\Cal B([0,t])\tensorhat\Sigma_t\cr}$$

\noindent where $\Cal B([0,t])$ is the Borel $\sigma$-algebra of
$[0,t]$.\ \Qed

We can therefore apply the method of 612H to define, for each stopping time
$h:\Omega\to\coint{0,\infty}$, the $\sigma$-algebra $\Sigma_h$ and
the $\Sigma_h$-measurable function $X_h$, and we find ourselves with a
real-time stochastic integration structure \Asissixt\ and a fully adapted process
$\pmb x=\family{\sigma}{\Cal T_f}{x_{\sigma}}$ such that
$x_{h^{\ssbullet}}=X_h^{\ssbullet}$ for every $h$.

\medskip

{\bf (b)} The main part of the argument depends on the idea of `hitting time'
in a form similar to that of 445M.   Let $h:\Omega\to\coint{0,\infty}$
be a stopping time, and $\epsilon>0$.   For $\omega\in\Omega$ set

\Centerline{$f(\omega)=\min(\{h(\omega)\}
   \cup\{t:t\ge 0$, $|X_t(\omega)|\ge\epsilon\})$.}

\noindent (I can write $\min$ rather than $\inf$ because $t\mapsto X_t(\omega)$
is continuous, so if $\{t:|X_t(\omega)|\ge\epsilon\}$ is non-empty
it contains its
infimum.)   Then $f$ is a stopping time.   \Prf\ For any $t\ge 0$,
$\Sigma_t$ contains every $\mu$-negligible set, so
$(\Omega,\Sigma_t,\mu\restr\Sigma_t)$ is a complete probability space and
$\Sigma_t$ is closed under Souslin's operation (431A).   Next,

\Centerline{$\{(s,\omega):s\le t$, $|X_s(\omega)|\ge\epsilon\}
=\bigcap_{k\in\Bbb N}\{(s,\omega):s\le t$, $|X_s(\omega)|>\epsilon-2^{-k}\}$}

\noindent belongs to $\Cal B([0,t])\tensorhat\Sigma_t$, by (a) applied to the
process $(s,\omega)\mapsto|X_s(\omega)|$.
So its projection $E=\{\omega:\,\Exists s\in[0,t]$, $|X_s(\omega)|\ge\epsilon\}$
belongs to $\Sigma_t$ (423O\Latereditions).   Now

\Centerline{$\{\omega:f(\omega)\le t\}=\{\omega:h(\omega)\le t\}\cup E
\in\Sigma_t$.}

\noindent As $t$ is arbitrary, $f$ is a stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$.\ \Qed

\medskip

{\bf (c)} Again suppose that $h:\Omega\to\coint{0,\infty}$
is a stopping time and $\epsilon>0$.   Define $g_n$ and
$X^{(n)}_t$, for $n\in\Bbb N$ and $t\ge 0$, by setting

\Centerline{$g_0(\omega)=0$ for every $\omega\in\Omega$}

\noindent and

$$\eqalign{X^{(n)}_t(\omega)
&=0\text{ if }g_n(\omega)>t,\cr
&=X_t(\omega)-X_{g_n}(\omega)\text{ if }g_n(\omega)\le t,\cr}$$

\Centerline{$g_{n+1}(\omega)
  =\inf(\{h(\omega)\}\cup\{t:t\ge 0$, $|X^{(n)}_t(\omega)|\ge\epsilon\}$}

\noindent for $n\in\Bbb N$, $t\ge 0$ and $\omega\in\Omega$,   We see immediately
that $t\mapsto X^{(n)}_t(\omega)$ is always continuous.   Also
we can see by induction on $n$ that every $g_n$ is a stopping time and every
$X^{(n)}_t$ is $\Sigma_t$-measurable.
\Prf\ For $n=0$ this is trivial, since of course
$X_t-X_0$ is $\Sigma_t$-measurable and $t\mapsto X_t(\omega)-X_0(\omega)$ is
always continuous.   For the inductive step to $n\ge 1$,
$g_n$ is a stopping time, by (b) applied to
$\langle X^{(n-1)}_t\rangle_{t\ge 0}$.   Next, setting
$F=\{\omega:g_n(\omega)\le t\}$, we have

$$\eqalign{F\cap\{\omega:g_n(\omega)\le s\}
&=\{\omega:g_n(\omega)\le s\}\in\Sigma_s\subseteq\Sigma_t
   \text{ if }s\le t,\cr
&=F\in\Sigma_t\text{ if }t\le s\cr}$$

\noindent so $F\in\Sigma_{g_n}$.   If $E\in\Sigma_{g_n}$ then
$E\cap F\in\Sigma_t$, so $X_{g_n}\times\chi F$ is $\Sigma_t$-measurable;  while
$F\in\Sigma_t$ so $X_t\times\chi F$ is $\Sigma_t$-measurable.   Consequently
$X^{(n)}_t=(X_t-X_{g_n})\times\chi F$ is $\Sigma_t$-measurable, and the
induction continues.\ \Qed

We therefore have a non-decreasing sequence $\sequencen{g_n}$ of stopping times
such that, for any $n\in\Bbb N$ and $\omega\in\Omega$,

\inset{----- if $n=0$ then $g_n(\omega)=0$,

----- $g_n(\omega)\le h(\omega)$,

----- $|X_t(\omega)-X_{g_n}(\omega)|\le\epsilon$ whenever
$g_n(\omega)\le t\le g_{n+1}(\omega)$}

\noindent because $t\mapsto X_t(\omega)-X_{g_n}(\omega)$ is continuous,
so if $|X_t(\omega)-X_{g_n}(\omega)|>\epsilon$ there is an
$s\in\coint{g_n(\omega),t}$ such that $|X_s(\omega)-X_{g_n}(\omega)|\ge\epsilon$
and $g_{n+1}(\omega)\le s<t$.

Moreover, for any $\omega$, $\sequencen{g_n(\omega)}$ is a
non-decreasing sequence bounded above by $h(\omega)$, so

\Centerline{$\lim_{n\to\infty}X_{g_n}(\omega)
=\lim_{n\to\infty}X_{g_n(\omega)}(\omega)$}

\noindent is defined and finite, in which case there must be some $n$ such that
$g_{n+1}(\omega)=h(\omega)$.

\medskip

{\bf (d)} Translating (c) into the language of the stochastic integration
structure \Asissixt, we see that given any $\tau\in\Cal T_f$ and $\epsilon>0$
we have a non-decreasing sequence $\sequencen{\tau_n}$ in $\Cal T_f$ such that
for every $n\in\Bbb N$,

\inset{----- $\tau_n\le\tau$,

----- $\Bvalue{\tau_n\le\sigma}\Bcap\Bvalue{\sigma\le\tau_{n+1}}
\Bsubseteq\Bvalue{|x_{\sigma}-x_{\tau_n}|\le\epsilon}$ for every
$\sigma\in\Cal T_f$,}

\noindent and

\Centerline{$\tau_0=\check 0=\min\Cal T_f$,
\quad$\sup_{n\in\Bbb N}\Bvalue{\tau_n=\tau}=1$.}

\noindent But this means that if we take $n\in\Bbb N$ such that
$c=1\Bsetminus\Bvalue{\tau_n=\tau}$ has measure at most $\epsilon$, and
set $I=\{\tau_0,\ldots,\tau_n,\tau\}\in\Cal I(\Cal T_f\wedge\tau)$, then

$$\eqalignno{\Osclln^*_I(\pmb x\restr\Cal T_f\wedge\tau)&
\le\sup_{i<n}
   \sup\{|x_{\sigma'}-x_{\sigma}|:\tau_i\le\sigma\le\sigma'\le\tau_{i+1}\}\cr&
\mskip200mu
   \vee\sup\{|x_{\sigma'}-x_{\sigma}|:\tau_n\le\sigma\le\sigma'\le\tau\}
\Displaycause{618Ca}
\le 2\epsilon\chi 1\vee(2\sup|\pmb x\restr\Cal T_f\wedge\tau|\times\chi c)\cr}$$

\noindent and

\Centerline{$\theta(\Osclln^*_I(\pmb x\restr\Cal T_f\wedge\tau))
\le 2\epsilon+\bar\mu c\le 3\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\pmb x\restr\Cal T_f\wedge\tau$ is
jump-free;  as $\tau$ is arbitrary, $\pmb x$ is locally jump-free.
}%end of proof of 618H

\leader{618I}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u=\sfamilysS{u}$ a locally jump-free fully adapted process.
If $A\subseteq\Cal S$ is non-empty and upwards-directed and
$\sup A\in\Cal S$, then $u_{\sup A}=\lim_{\sigma\uparrow A}u_{\sigma}$.

\proof{ It is enough to consider the case in which $\sup A=\max\Cal S$, so
that $\pmb u$ is actually jump-free.

\medskip

{\bf (a)} (The key.)   Suppose that
$I\in\Cal I(\Cal S)$, that $\tau_0\le\ldots\le\tau_n$ linearly generate the
$I$-cells, that $\sigma\le\sigma'\in\Cal S$, and that

\Centerline{$a\Bcap\Bvalue{\sigma\le\tau_i}
=a\Bcap\Bvalue{\sigma'\le\tau_i}$}

\noindent for every $i\le n$.   Then
$|u_{\sigma}-u_{\sigma'}|\times\chi a\le\Osclln^*_I(\pmb u)$.   \Prf\
Set $\bar u=\Osclln^*_I(\pmb u)$,

\Centerline{$b_0=\Bvalue{\sigma\le\tau_0}$,
\quad$b_i=\Bvalue{\tau_{i-1}<\sigma}\Bcap\Bvalue{\sigma\le\tau_i}$
for $1\le i\le n$,
\quad$b_{n+1}=\Bvalue{\tau_n<\sigma}$.}

\noindent Then $\langle b_i\rangle_{i\le n+1}$ is a partition of unity in
$\frak A$.   Next, set

\Centerline{$\sigma_0=\sigma\wedge\tau_0$,
\quad$\sigma'_0=\sigma'\wedge\tau_0$,
\quad$\sigma_{n+1}=\sigma\vee\tau_n$,
\quad$\sigma'_{n+1}=\sigma'\vee\tau_n$,}

\Centerline{$\sigma_i=\med(\tau_{i-1},\sigma,\tau_i)$,
\quad$\sigma'_i=\med(\tau_{i-1},\sigma',\tau_i)$ for $1\le i\le n$.}

\noindent If $i\le n+1$ and $\sigma_i\ne\sigma'_i$,
the stopping-time
interval $c(\sigma_i,\sigma'_i)$ is a $J_i$-cell, where $J_i$ is the
sublattice generated by $I\cup\{\sigma,\sigma'\}$, so

\Centerline{$|u_{\sigma_i}-u_{\sigma'_i}|\le\Osclln_J(\pmb u)\le\bar u$.}

\noindent Next, $b_i\Bsubseteq\Bvalue{\sigma_i=\sigma}$ and
$a\Bcap b_i\Bsubseteq\Bvalue{\sigma'_i=\sigma'}$ for each $i$, so

\Centerline{$a
=\sup_{i\le n+1}a\Bcap b_i
\Bsubseteq\sup_{i\le n+1}
  \Bvalue{u_{\sigma_i}=u_{\sigma}}\Bcap\Bvalue{u_{\sigma'_i}=u_{\sigma'}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\sigma'}|\le\bar u}$}

\noindent and $|u_{\sigma}-u_{\sigma'}|\times\chi a\le\bar u$.\ \Qed

\medskip

{\bf (b)} Take any $\epsilon>0$, and let
$I\in\Cal I(\Cal S)$ be such that $\theta(\bar u)\le\epsilon$ where
$\bar u=\Osclln^*_I(\pmb u)$.   Let $(\tau_0,\ldots,\tau_n)$ linearly
generate the $I$-cells.   For each $i\le n$,
$\Bvalue{\sup A\le\tau_i}=\inf_{\sigma\in A}\Bvalue{\sigma\le\tau_i}$ for
each $i$ (611Eb), so there is a $\sigma_0\in A$ such that
$c=\sup_{i\le n}\Bvalue{\sigma_0\le\tau_i}
\Bsetminus\Bvalue{\sup A\le\tau_i}$ has measure at most $\epsilon$.
If now $\sigma\in A$ and $\sigma_0\le\sigma$, (a) tells us that

\Centerline{$|u_{\sup A}-u_{\sigma}|\times\chi(1\Bsetminus c)
\le\bar u$,
\quad$\theta(u_{\sup A}-u_{\sigma})\le\theta(\bar u)+\bar\mu c
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\lim_{\sigma\uparrow A}u_{\sigma}=u_{\sup A}$.
}%end of proof of 618I

\leader{618J}{Examples}\cmmnt{ Let
us look again at our three standard examples.}   Take $T=\coint{0,\infty}$.

\spheader 618Ja The identity process\cmmnt{
$\pmb{\iota}=\family{\tau}{\Cal T_f}{\iota_{\tau}}$ (612F)} is
locally jump-free.
\prooflet{\Prf\ We know that it is locally order-bounded (615Q, or otherwise).
Take $\tau\in\Cal T_f$ and $\epsilon>0$.   Let $m\in\Bbb N$ be such that
$a=\Bvalue{\tau>m\epsilon}$ has measure at most $\epsilon$.   Set
$I=\{\tau\wedge\check{i\epsilon}:i\le m\}$, where
$\check{i\epsilon}$ is the constant stopping time at
$i\epsilon$.   If $J$ is a sublattice of $\Cal T\wedge\tau$ including $I$
and $e$ is a $J$-cell, then

\inset{{\it either} there is an $i<m$ such that $e$ can be expressed as
$c(\sigma,\sigma')$ where
$\tau\wedge\check{i\epsilon}\le\sigma
\le\sigma'\le\tau\wedge\check{(i+1)\epsilon}$, in which case

\Centerline{$\Bvalue{\sigma<\sigma'}\Bsubseteq
\Bvalue{\check{i\epsilon}\le\sigma}
\Bcap\Bvalue{\sigma'\le\check{(i+1)\epsilon}}$}

\noindent and
$\Delta_e(\tbf 1,|d\pmb{\iota}|)=\iota_{\sigma}'-\iota_{\sigma}\le\epsilon\chi 1$,

{\it or} $e$ can be expressed as $c(\sigma,\sigma')$ where
$\tau\wedge\check{m\epsilon}\le\sigma\le\sigma'\le\tau$, in which case
$\Bvalue{\sigma<\sigma'}\Bsubseteq a$ and
$\Delta_e(\tbf 1,|d\pmb{\iota}|)=\iota_{\sigma}'-\iota_{\sigma}
\le\iota_{\tau}\times\chi a$.}

\noindent Thus $\Osclln^*_I(\pmb{\iota})
\le\epsilon\chi 1+\iota_{\tau}\times\chi a$ and
$\theta(\Osclln^*_I(\pmb{\iota}))\le\epsilon+\bar\mu a\le 2\epsilon$.
As $\epsilon$ is arbitrary, $\pmb{\iota}\restr\Cal T\wedge\tau$ is
jump-free;  as $\tau$ is arbitrary, $\pmb{\iota}$ is jump-free.\ \Qed}

\spheader 618Jb The standard Poisson process\cmmnt{
$\pmb v=\family{\tau}{\Cal T_f}{v_{\tau}}$ (612U)} is not locally
jump-free.   \prooflet{\Prf\ Let $\tau_1$ be the first jump time of
$\pmb v$, as in 612Ue-612Uf, so that

\Centerline{$\Bvalue{\sigma<\tau_1}=\Bvalue{v_{\sigma}=0}$,
\quad$\Bvalue{\tau_1\le\sigma}=\Bvalue{v_{\sigma}\ge 1}$}

\noindent for every $\sigma\in\Cal T_f$.   If $I$ is any sublattice of
$\Cal T\wedge\tau_1$, set $J=I\cup\{\min\Cal T,\tau_1\}$ and
let $\sigma_0\le\ldots\le\sigma_n$ linearly generate
the $J$-cells;  then

$$\eqalign{\Osclln^*_I(\pmb v)
&\ge\Osclln_J(\pmb v)
\ge\sup_{i<n}|v_{\sigma_{i+1}}-v_{\sigma_i}|\cr
&\ge\sup_{i<n}
  \chi(\Bvalue{\sigma_{i+1}=\tau_1}\Bcap\Bvalue{\sigma_i<\tau_1})\cr
&=\chi(\sup_{i<n}\Bvalue{\sigma_{i+1}=\tau_1}
   \Bsetminus\Bvalue{\sigma_i=\tau_1})
=\chi\Bvalue{\tau_1>0}
=\chi 1\cr}$$

\noindent and $\theta(\Osclln^*_I(\pmb v))=1$.   As $I$ is arbitrary,
$\pmb v\restr\Cal T\wedge\tau_1$ is not jump-free and $\pmb v$ is not
locally jump-free.\ \Qed}

\spheader 618Jc Brownian motion\cmmnt{
$\pmb{w}=\family{\tau}{\Cal T_f}{w_{\tau}}$ (612T)} is locally
jump-free.   \prooflet{\Prf\ This is immediate from 618H and the continuity
of Brownian sample paths, built into the description in 612Ta.\ \Qed}

\leader{618K}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$.   If
$I$, $J\in\Cal I(\Cal S)$ and $a\in\frak A$ are such that $J\subseteq I$
and $a\Bsubseteq\sup_{\sigma\in J}\Bvalue{\tau=\sigma}$ for every $\tau\in I$,
then $a\Bsubseteq\Bvalue{\Osclln_I(\pmb u)=\Osclln_J(\pmb u)}$
for every fully adapted process $\pmb u=\sfamilysS{u}$.

\proof{ The argument is essentially identical to that for 613S.

\medskip

{\bf (a)} The case $a=0$ is trivial;  so is the case
$a\ne 0$ and $J=\emptyset$, as then $I$ must also be empty and

\Centerline{$\Osclln_I(\pmb u)=\Osclln_J(\pmb u)=0$.}

\noindent So suppose otherwise.
Let $(\sigma_0,\ldots,\sigma_n)$ linearly generate the
$J$-cells.   If $\tau\in I$ then

$$\eqalign{a\Bsubseteq\sup_{\sigma\in J}\Bvalue{\tau=\sigma}&
=\sup_{\sigma\in J}\sup_{j\le n}\Bvalue{\tau=\sigma}
     \Bcap\Bvalue{\sigma=\sigma_j}\cr&
\Bsubseteq\sup_{j\le n}\Bvalue{\tau=\sigma_j}
\Bsubseteq\Bvalue{\sigma_0\le\tau}\Bcap\Bvalue{\tau\le\sigma_n}.\cr}$$

\noindent Set $I_{-1}=I\wedge\sigma_0$,
$I_j=I\cap[\sigma_j,\sigma_{j+1}]$ for $0\le j<n$,
$I_n=I\vee\sigma_n$.

\medskip

{\bf (b)} If $\tau\le\tau'$ in $I_{-1}$, then

\Centerline{$a\Bcap\Bvalue{u_{\tau'}\ne u_{\tau}}
\Bsubseteq a\Bcap\Bvalue{\tau<\tau'}
\Bsubseteq a\Bcap\Bvalue{\tau<\sigma_0}=0$,}

\noindent so

\Centerline{$a\Bcap\Bvalue{|\Delta_e(\tbf 1,d\pmb u)|\ne 0}=0$}

\noindent for every $I_{-1}$-cell $e$.   Taking the supremum over $e$,
$\Osclln_{I_{-1}}(\pmb u)\times\chi a=0$.

\medskip

{\bf (c)} In the same way, if $\tau\le\tau'$ in $I_n$, then

\Centerline{$a\Bcap\Bvalue{u_{\tau'}\ne u_{\tau}}
\Bsubseteq a\Bcap\Bvalue{\tau<\tau'}
\Bsubseteq a\Bcap\Bvalue{\sigma_n<\tau'}=0$,}

\noindent so

\Centerline{$\Osclln_{I_n}(\pmb u)\times\chi a=0$.}

\medskip

{\bf (d)} If $0\le j<n$,
$a\Bsubseteq\Bvalue{\Osclln_{I_j}(\pmb u)
=|u_{\sigma_{j+1}}-u_{\sigma_j}|}$.
\Prf\ For every $\tau\in I_j$,

$$\eqalign{a\Bcap\Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau<\sigma_{j+1}}&
\Bsubseteq\sup_{k\le n}\Bvalue{\tau=\sigma_k}
   \Bcap\Bvalue{\sigma_j<\tau}\Bcap\Bvalue{\tau<\sigma_{j+1}}\cr&
\Bsubseteq\sup_{k\le n}
   \Bcap\Bvalue{\sigma_j<\sigma_k}\Bcap\Bvalue{\sigma_k<\sigma_{j+1}}\cr&
\Bsubseteq(\sup_{k\le j}\Bcap\Bvalue{\sigma_j<\sigma_k})
   \Bcup(\sup_{j+1\le k}\Bvalue{\sigma_k<\sigma_{j+1}})
=0.\cr}$$

\noindent Take $(\tau_0,\ldots,\tau_m)$ linearly generating the
$I_j$-cells;  then $\sigma_j=\tau_0\le\ldots\le\tau_m=\sigma_{j+1}$.
For $i<m$, set $b_i=\Bvalue{\tau_i<\tau_{i+1}}$, so that
$\sup_{i<m}b_i=\Bvalue{\sigma_j<\sigma_{j+1}}$.   Now

$$\eqalignno{a\Bcap b_i&
=a\Bcap\Bvalue{\tau_i<\tau_{i+1}}
=a\Bcap\Bvalue{\sigma_j<\tau_{i+1}}
    \Bcap\Bvalue{\tau_i<\sigma_{j+1}}\Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr&
\Bsubseteq a\Bcap\Bvalue{\tau_{i+1}=\sigma_{j+1}}
    \Bcap\Bvalue{\tau_i=\sigma_j}\Bcap\Bvalue{\tau_i<\tau_{i+1}}\cr&
\Bsubseteq\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|
            =|u_{\sigma_{j+1}}-u_{\sigma_j}|}
    \Bcap\inf_{k<i}\Bvalue{\tau_k=\tau_{k+1}}
    \Bcap\inf_{i<k<m}\Bvalue{\tau_k=\tau_{k+1}}\cr&
\Bsubseteq\Bvalue{|u_{\tau_{j+1}}-u_{\tau_j}|=|u_{\sigma_{j+1}}-u_{\sigma_j}|}
    \Bcap\inf_{\Atop{k<m}{k\ne i}}\Bvalue{u_{\tau_k}=u_{\tau_{k+1}}}\cr&
\Bsubseteq\Bvalue{|u_{\tau_{j+1}}-u_{\tau_j}|=|u_{\sigma_{j+1}}-u_{\sigma_j}|}
    \Bcap\Bvalue{\Osclln_{I_j}(\pmb u)=|u_{\tau_{j+1}}-u_{\tau_j}|}\cr&
\Bsubseteq\Bvalue{\Osclln_{I_j}(\pmb u)
         =|u_{\sigma_{j+1}}-u_{\sigma_j}|}.\cr}$$

\noindent Taking the supremum over $i$,

\Centerline{$a\Bcap\Bvalue{\sigma_j<\sigma_{j+1}}
   \Bsubseteq\Bvalue{\Osclln_{I_j}(\pmb u)
         =|u_{\sigma_{j+1}}-u_{\sigma_j}|}$.}

\noindent Since $\Bvalue{\sigma_j=\sigma_{j+1}}
   \Bsubseteq\Bvalue{\Osclln_{I_j}(\pmb u)=0}$,
$a\Bsubseteq\Bvalue{\Osclln_{I_j}(\pmb u)
=|u_{\sigma_{j+1}}-u_{\sigma_j}|}$.\ \Qed

\medskip

{\bf (e)} Assembling these,

$$\eqalignno{\Osclln_I(\pmb u)\times\chi a
&=\sup_{-1\le j\le n}\Osclln_{I_j}(\pmb u)\times\chi a
\Displaycause{618D(b-i)}
=\sup_{0\le j<n}|u_{\sigma_{j+1}}-u_{\sigma_j}|
   \times\chi a
=\Osclln_J(\pmb u)\times\chi a\cr}$$

\noindent and
$a\Bsubseteq\Bvalue{\Osclln_I(\pmb u)=\Osclln_J(\pmb u)}$.
}%end of proof of 618K

\leader{618L}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a fully adapted process, $\hat{\Cal S}$ the covered
envelope of $\Cal S$, and
$\hatpmb u=\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$
the fully adapted extension of $\pmb u$ to $\hat{\Cal S}$.

(a) If either
$\pmb u$ or $\hatpmb u$ is order-bounded, so is the other, and in this case
$\Osclln(\hatpmb u)=\Osclln(\pmb u)$.

(b) In particular,
$\pmb u$ is jump-free iff $\hatpmb u$ is jump-free.

(c) If either $\sup_{\sigma\in\Cal S}\Osclln(\pmb u\restr\Cal S\wedge\sigma)$
or  $\sup_{\tau\in\hat\Cal S}\Osclln(\hatpmb u\restr\hat\Cal S\wedge\tau)$
is defined in $L^0(\frak A)$, so is the other, and they are equal.

\proof{ If $\Cal S$ is empty, this is trivial;  suppose otherwise.

\medskip

{\bf (a)(i)} Suppose that $\pmb u$ is
order-bounded;  it follows that $\hatpmb u$ is order-bounded (612Q(d-i)).
Let $\epsilon>0$.   Then there is a non-empty finite sublattice $I$ of
$\Cal S$ such that $\theta(\Osclln^*_I(\pmb u)-\Osclln(\pmb u))\le\epsilon$.
As in 618C,
take a sequence $(\tau_0,\ldots,\tau_n)$ linearly generating the
$I$-cells, and set $\tau_{-1}=\inf\Cal S=\inf\hat{\Cal S}$ and
$\tau_{n+1}=\sup\Cal S=\sup\hat{\Cal S}$.   \Quer\ Suppose that there are an
$i$ such that $-1\le i\le n$ and $\rho$, $\rho'\in\hat\Cal S$ such that
$\tau_i\le\rho\le\rho'\le\tau_{i+1}$ and
$|\hat u_{\rho'}-\hat u_{\rho}|\not\le\Osclln^*_I(\pmb u)$.
Set $c=\Bvalue{|\hat u_{\rho'}-\hat u_{\rho}|>\Osclln^*_I(\pmb u)}$,
so that $c\ne 0$.   Take $\sigma\in\Cal S$ such that
$c\Bcap\Bvalue{\rho=\sigma}\ne 0$, and $\sigma'\in\Cal S$ such that
$c'=c\Bcap\Bvalue{\rho=\sigma}\Bcap\Bvalue{\rho'=\sigma'}$ is non-zero.
Then

\Centerline{$c'
\Bsubseteq\Bvalue{\med(\tau_i,\sigma,\tau_{i+1})=\med(\tau_i,\rho,\tau_{i+1})}
\Bsubseteq\Bvalue{\med(\tau_i,\sigma,\tau_{i+1})=\rho}$}

\noindent and similarly
$c'\Bsubseteq\Bvalue{\med(\tau_i,\sigma',\tau_{i+1})=\rho'}$, so

\Centerline{$c'\Bsubseteq\Bvalue{|u_{\med(\tau_i,\sigma',\tau_{i+1})}
      -u_{\med(\tau_i,\sigma,\tau_{i+1})}|>\Osclln^*_I(\pmb u)}$;}

\noindent but
$|u_{\med(\tau_i,\sigma',\tau_{i+1})}
      -u_{\med(\tau_i,\sigma,\tau_{i+1})}|\le\Osclln^*_I(\pmb u)$, by
the second formula in 618Ca.   So this is impossible.\ \Bang

As $\rho$ and $\rho'$ are arbitrary, 618Ca tells us that
$\Osclln^*_I(\hatpmb u)\le\Osclln^*_I(\pmb u)$ and

\Centerline{$\theta((\Osclln(\hatpmb u)-\Osclln(\pmb u))^+)
\le\theta((\Osclln^*_I(\hatpmb u)-\Osclln(\pmb u))^+)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\Osclln(\hatpmb u)\le\Osclln(\pmb u)$.

It follows at once that if $\Osclln(\pmb u)=0$ then $\Osclln(\hatpmb u)=0$, that
is, if $\pmb u$ is jump-free then $\hatpmb u$ is jump-free.

\medskip

\quad{\bf (ii)} Suppose that $\hatpmb u$ is
order-bounded;  it follows that $\pmb u$ is order-bounded (612Q(d-i),
in the other direction).   Let $\epsilon>0$.
Then there is a non-empty finite sublattice
$I$ of $\hat{\Cal S}$ such that
$\theta(\Osclln^*_I(\hatpmb u)-\Osclln(\hatpmb u))\le\epsilon$.
Let $J\in\Cal I(\Cal S)$ be such that
$a=\inf_{\tau\in I}\sup_{\sigma\in J}\Bvalue{\tau=\sigma}$ has measure at least
$1-\epsilon$.   Suppose $K\in\Cal I(\Cal S)$ includes $J$, and write
$K\sqcup I$ for the sublattice generated by $K\cup I$.   Then
$a\Bsubseteq\sup_{\sigma\in K}\Bvalue{\tau=\sigma}$ for every $\tau\in K\cup I$
and therefore for every $\tau\in K\sqcup I$ (611M(b-i)).   By 618K,

$$\eqalign{a&
\Bsubseteq\Bvalue{\Osclln_K(\hatpmb u)=\Osclln_{K\sqcup I}(\hatpmb u)}\cr&
\Bsubseteq\Bvalue{\Osclln_K(\hatpmb u)\le\Osclln_I^*(\hatpmb u)}
=\Bvalue{\Osclln_K(\pmb u)\le\Osclln_I^*(\hatpmb u)},\cr}$$

\noindent and

\Centerline{$\Osclln_K(\pmb u)
\le\Osclln_I^*(\hatpmb u)+2\sup|\pmb u|\times\chi(1\Bsetminus a)$.}

\noindent As $K\supseteq J$ is arbitrary,

\Centerline{$\Osclln(\pmb u)\le\Osclln_J^*(\pmb u)
\le\Osclln_I^*(\hatpmb u)+2\sup|\pmb u|\times\chi(1\Bsetminus a)$}

\noindent and

$$\eqalign{\theta((\Osclln(\pmb u)-\Osclln(\hatpmb u))^+)&
\le\theta(\Osclln_I^*(\hatpmb u)-\Osclln(\hatpmb u)
   +2\sup|\pmb u|\times\chi(1\Bsetminus a))\cr&
\le\theta(\Osclln_I^*(\hatpmb u)-\Osclln(\hatpmb u))
   +\theta(2\sup|\pmb u|\times\chi(1\Bsetminus a))\cr&
\le\epsilon+\bar\mu(1\Bsetminus a)
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\theta((\Osclln(\pmb u)-\Osclln(\hatpmb u))^+)=0$ and
$\Osclln(\pmb u)\le\Osclln(\hatpmb u)$.

\medskip

{\bf (b)}
It follows immediately that if $\hatpmb u$ is jump-free, so is $\pmb u$.

\medskip

{\bf (c)(i)} If
$w=\sup_{\sigma\in\Cal S}\Osclln(\pmb u\restr\Cal S\wedge\sigma)$ is defined,
then, in particular, $\pmb u$ is locally order-bounded, so
$\hatpmb u$ is locally order-bounded (612Q(d-ii)).   If $\sigma\in\Cal S$ then
$\hatpmb u\restr\hat\Cal S\wedge\sigma$ is the fully adapted extension of
$\pmb u\restr\Cal S$ (612Qc), so
$\Osclln(\hatpmb u\restr\hat\Cal S\wedge\sigma)
=\Osclln(\pmb u\restr\Cal S\wedge\sigma)\le w$, by (a).   But
$\family{\tau}{\hat\Cal S}{\Osclln(\hatpmb u\restr\hat\Cal S\wedge\tau)}$ is
fully adapted (618Da), so

\Centerline{$
  \sup_{\tau\in\hat\Cal S}\Osclln(\hatpmb u\restr\hat\Cal S\wedge\tau)
=\sup_{\sigma\in\Cal S}\Osclln(\hatpmb u\restr\hat\Cal S\wedge\sigma)
\le w$}

\noindent by 612Q(d-i).

\medskip

\quad{\bf (ii)} If
$w=\sup_{\sigma\in\Cal S}\Osclln(\hatpmb u\restr\Cal S\wedge\sigma)$ is defined,
then $\hatpmb u$ is locally order-bounded, so
$\pmb u$ is locally order-bounded, by 612Q(d-ii) again.   And now

\Centerline{$\sup_{\sigma\in\Cal S}\Osclln(\pmb u\restr\Cal S\wedge\sigma)
=\sup_{\sigma\in\Cal S}\Osclln(\hatpmb u\restr\hat\Cal S\wedge\sigma)
\le w$.}
}%end of proof of 618L

\leader{618M}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u\cmmnt{\mskip5mu=\sfamilysS{u}}$ a moderately oscillatory process.   Then
$\Osclln(\pmb u)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb u)$.

\proof{ We know that $\pmb u$ is order-bounded (615F(a-iii)),
so $\Osclln(\pmb u)$ is
defined.   If $\Cal S=\emptyset$ the result is trivial;  suppose otherwise.

\medskip

{\bf (a)} To begin with, suppose that $\pmb u$
is of bounded variation.

\medskip

\quad{\bf (i)} The first thing to note is that if
$\Cal S'$ is a sublattice of $\Cal S$, then
$\Osclln_J(\pmb u\restr\Cal S')\le\int_{\Cal S'}|d\pmb u|$ for every
$J\in\Cal I(\Cal S)$.   \Prf\ Of course

\Centerline{$\Osclln_{\emptyset}(\pmb u\restr\Cal S')
=0\le\int_{\Cal S'}|d\pmb u|$.}

\noindent If $J\in\Cal I(\Cal S')$ is non-empty and $(\tau_0,\ldots,\tau_n)$
linearly generates the $J$-cells, then

\Centerline{$\Osclln_J(\pmb u)=\sup_{i<n}|u_{\tau_{i+1}}-u_{\tau_i}|
\le\sum_{i=0}^{n-1}|u_{\tau_{i+1}}-u_{\tau_i}|
\le\int_{\Cal S'}|d\pmb u|$}

\noindent as in 614I.\ \Qed

\medskip

\quad{\bf (ii)} Now, for any non-empty $I\in\Cal I(\Cal S)$, set
$z_I=\int_{\Cal S}|d\pmb u|-S_I(\tbf 1,|d\pmb u|)$.   Then $z_I\ge 0$ and
$\Osclln^*_I(\pmb u)\le\Osclln_I(\pmb u)+z_I$.   \Prf\ Take
$(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells, so that
$\Osclln_I(\pmb u)=\sup_{i<n}|u_{\tau_{i+1}}-u_{\tau_i}|$ and

\Centerline{$S_I(\tbf 1,|d\pmb u|)=\sum_{i=0}^n|u_{\tau_{i+1}}-u_{\tau_i}|
\le\int_{\Cal S}|d\pmb u|$.}

\noindent If
$J\in\Cal I(\Cal S)$ includes $I$, then

$$\eqalignno{\Osclln_J(\pmb u)&
=\Osclln(\pmb u\restr J)
\Displaycause{618B(b-v-$\gamma$)}
=\Osclln(\pmb u\restr J\wedge\tau_0)
   \vee\sup_{i<n}\Osclln(\pmb u\restr J\cap[\tau_i,\tau_{i+1}])
   \vee\Osclln(\pmb u\restr J\vee\tau_n)
\Displaycause{618D(b-i)}
=\Osclln_{J\wedge\tau_0}(\pmb u)
   \vee\sup_{i<n}\Osclln_{J\cap[\tau_i,\tau_{i+1}]}(\pmb u)
   \vee\Osclln_{J\vee\tau_n}(\pmb u)\cr&
\le\int_{\Cal S\wedge\tau_0}|d\pmb u|
   \vee\sup_{i<n}\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}|d\pmb u|
   \vee\int_{\Cal S\vee\tau_n}|d\pmb u|
\Displaycause{using (i) above}
\le\int_{\Cal S\wedge\tau_0}|d\pmb u|
      +\sum_{i=0}^{n-1}(\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}|d\pmb u|
               -|u_{\tau_{i+1}}-u_{\tau_i}|)\cr&
\mskip200mu
      +\sup_{i<n}|u_{\tau_{i+1}}-u_{\tau_i}|
      +\int_{\Cal S\vee\tau_n}|d\pmb u|\cr&
=z_I+\Osclln_I(\pmb u).
\cr}$$

\noindent So

\Centerline{$\Osclln^*_I(\pmb u)
=\sup_{I\subseteq J\in\Cal I(\Cal S)}\Osclln_J(\pmb u)
\le z_I+\Osclln_I(\pmb u)$.  \text{\Qed}}

Since we always have $\Osclln_I(\pmb u)\le\Osclln^*_I(\pmb u)$, we see that
$|\Osclln_I(\pmb u)-\Osclln^*_I(\pmb u)|\le z_I$.

\medskip

\quad{\bf (iii)} Now we know that

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(\pmb u)=\Osclln(\pmb u)$,
\quad$\lim_{I\uparrow\Cal I(\Cal S)}S_I(\tbf 1,|d\pmb u|)
=\int_{\Cal S}|d\pmb u|$,}

\noindent so $\lim_{I\uparrow\Cal I(\Cal S)}z_I=0$ and
$\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb u)=\Osclln(\pmb u)$,

\medskip

{\bf (b)} In general, if $\pmb u$ is just moderately oscillatory, let
$\epsilon>0$.   Then there is a process $\pmb v$ of bounded variation, with
domain $\Cal S$, such that $\theta(z)\le\epsilon$, where
$z=\sup|\pmb u-\pmb v|$.   Now

\Centerline{$|\Osclln_I(\pmb  u)-\Osclln_I(\pmb v)|
\le\Osclln_I(\pmb u-\pmb v)\le z$
for every $I\in\Cal I(\Cal S)$,}

\Centerline{$|\Osclln(\pmb  u)-\Osclln(\pmb v)|\le\Osclln(\pmb u-\pmb v)\le z$}

\noindent (618B(c-ii)).   So

\Centerline{$|\Osclln_I(\pmb u)-\Osclln(\pmb u)|
\le|\Osclln_I(\pmb v)-\Osclln(\pmb v)|+2z$}

\noindent for every $I$, and

$$\eqalign{&\limsup_{I\uparrow\Cal I(\Cal S)}
   \theta(\Osclln_I(\pmb u)-\Osclln(\pmb u))\cr&
\mskip100mu
\le\limsup_{I\uparrow\Cal I(\Cal S)}\theta(\Osclln_I(\pmb v)-\Osclln(\pmb v))
   +2\theta(z)
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\lim_{I\uparrow\Cal S}\Osclln_I(\pmb u)=\Osclln(\pmb u)$.
}%end of proof of 618M


\leader{618N}{}\cmmnt{ The difference between a `jump-free' process and a
`moderately oscillatory' process can be described in terms of the construction
in 615M.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a full sublattice of $\Cal T$ with
a greatest element, $\pmb u=\sfamilysS{u}$ a moderately oscillatory process, and
$\delta>0$.   Let $\sequence{i}{y_i}$ be the sequence
constructed from $\pmb u$ and $\delta$ as in 615M.   Then
$|y_{i+1}-y_i|\le\Osclln(\pmb u)+\delta\chi 1$ for every $i\in\Bbb N$.

\proof{ By 618Gb, $\pmb u$ is moderately oscillatory, therefore \updowncgt\
(615N), so we can indeed apply
the method of 615M.    Write $w$ for $\Osclln(\pmb u)$.
Take the sequence $\sequence{i}{D_i}$
of downwards-directed subsets of $\Cal S$ constructed
from $\pmb u$ and $\delta$ there.
\Quer\ Suppose that $a=\Bvalue{|y_{i+1}-y_i|-w>\delta}$ is non-zero.
Then there is an $\eta>0$ such that

\Centerline{$a_1=\Bvalue{|y_{i+1}-y_i|-w>\delta+2\eta}$}

\noindent has measure at least
$4\eta$.   Let $\tau\in D_{i+1}$ be such that

\Centerline{$\theta(\sup_{\tau'\in D_{i+1},\tau'\le\tau}
  |u_{\tau'}-y_{i+1}|)\le\eta^2$}

\noindent (615Gb), so that $\bar\mu a_2\ge 3\eta$ where

\Centerline{$a_2=a_1\Bsetminus
\Bvalue{\sup_{\tau'\in D_{i+1},\tau'\le\tau}|u_{\tau'}-y_{i+1}|>\eta}$.}

\noindent Let $\sigma\in D_i$ be such that $\sigma\le\tau$ and
$\theta(|u_{\sigma}-u_{y_i}|)\le\delta\eta$, so that $\bar\mu a_3\ge 2\eta$
where

\Centerline{$a_3=a_2\Bsetminus\Bvalue{|u_{\sigma}-y_i|\ge\delta}$.}

We have a finite sublattice $I$ of
$\Cal S$, containing $\sigma$ and $\tau$, such that
$\theta(w-\Osclln^*_I(\pmb u))\le\eta^2$;  write $w'$ for
$\Osclln^*_I(\pmb u)$.   Then $\bar\mu a_4\ge\eta$, where

\Centerline{$a_4=a_3\Bsetminus\Bvalue{|w-w'|\ge\eta}$.}

\noindent In particular, $a_4\ne 0$.
Let $(\rho_0,\ldots,\rho_n)$ linearly generate the
$(I\cap[\sigma,\tau])$-cells and define $b_0,\ldots,b_n$ by saying that

\Centerline{$b_j
=\Bvalue{|u_{\rho_j}-y_i|\ge\delta}\Bsetminus\sup_{j'<j}b_{j'}$}

\noindent for $j\le n$, and $b^*=1\Bsetminus\sup_{j\le n}b_j$.
Then $b_j\in\frak A_{\rho_j}$ for each $j\le n$ and $b^*\in\frak A_{\rho_n}$, so
there is a $\tau'\in\Cal T$ such that
$b_j\Bsubseteq\Bvalue{\tau'=\rho_j}\Bsubseteq b_j$ for
$j\le n$ and $b^*\Bsubseteq\Bvalue{\tau'=\rho_n}$.   Now $\tau'\in\Cal S$,
$\sigma\le\tau'\le\tau$ and

\Centerline{$\Bvalue{|u_{\tau'}-y_i|<\delta}
\Bsubseteq b^*\Bcap\Bvalue{|u_{\tau}-y_i|<\delta}
\Bsubseteq\Bvalue{\tau'=\max\Cal S}$}

\noindent so $\tau'\in D_{i+1}$.

Since $a_4\Bsubseteq a_3\subseteq 1\Bsetminus b_0$, there is a $j\ge 1$ such
that $b=a_4\cap\Bvalue{\tau'=\rho_j}$ is non-zero.   Now
$|u_{\rho_j}-u_{\rho_{j-1}}|\le w'$, but at the same time
$b\Bsubseteq\Bvalue{|u_{\rho_{j-1}}-y_i|<\delta}$ and
$b\Bsubseteq a_2\Bsubseteq\Bvalue{|u_{\rho_j}-y_{i+1}|\le\eta}$
and $b\Bsubseteq\Bvalue{|w-w'|\le\eta}$,
so $b\Bsubseteq\Bvalue{|y_{i+1}-y_i|-w\le\delta+2\eta}$, which is impossible, as
$b\Bsubseteq a_1$.\ \Bang

So $a=0$ and $|y_{i+1}-y_i|\le w+\delta\chi 1$, as claimed.
}%end of proof of 618N

\leader{618O}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ an adapted interval function which is
order-bounded\cmmnt{ in the sense that
$\{\psi(\sigma,\tau):(\sigma,\tau)\in\Cal S^{2\uparrow}\}$ is bounded above and
below in $L^0(\frak A)$}.   Following 618B, set

\Centerline{$\Osclln_I(\psi)
=\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\psi|)$}

\noindent for $I\in\Cal I(\Cal S)$ (counting $\sup\emptyset$ as $0$),

\Centerline{$\Osclln^*_I(\psi)
=\sup_{J\in\Cal I(\Cal S),J\supseteq I}\Osclln_J(\psi)$}

\noindent for $I\in\Cal I(\Cal S)$, and

\Centerline{$\Osclln(\psi)=\inf_{I\in\Cal I(\Cal S)}\Osclln^*_I(\psi)$.}

\noindent\cmmnt{Note that as
$\Osclln^*_{I'}(\psi)\le\Osclln^*_I(\psi)$ whenever $I\subseteq I'$ in
$\Cal I(\Cal S)$, }%
$\Osclln(\psi)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(\psi)$.
Moreover, if $\pmb u$ is an order-bounded fully adapted process and
$\psi=\Delta\pmb u$ the corresponding interval function,
$\Osclln_I(\psi)=\Osclln_I(\pmb u)$ for every $I\in\Cal I(\Cal S)$ and
$\Osclln(\psi)=\Osclln(\pmb u)$.

\leader{618P}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ a
strictly adapted interval function.
For $I\in\Cal I(\Cal S)$, set
$\pmb w_I=\family{\tau}{\Cal S}{w_{I\tau}}$ where
$w_{I\tau}=S_{I\wedge\tau}(\tbf 1,d\psi)$ for $\tau\in\Cal S$.

(a) For any $I\in\Cal I(\Cal S)$, $\pmb w_I$ is fully adapted.

(b) Suppose that $\Cal S$ is finitely full, $\psi$ is order-bounded and
$\int_{\Cal S}d\psi$ is defined.
Then $ii_{\psi}(\tbf 1)$ is order-bounded and
$\Osclln(ii_{\psi}(\tbf 1))\le 2\Osclln(\psi)$.

(c) Suppose tht $\psi$ is order-bounded and $\int_{\hat\Cal S}d\hat\psi$ is
defined, where $\hat\Cal S$ is the covered envelope of $\Cal S$ and
$\hat\psi$ is the adapted extension of $\psi$ to
$\hat\Cal S^{\uparrow}$\cmmnt{ (613U)}.   If
$ii_{\psi}(\tbf 1)$ is moderately oscillatory, then
$\Osclln(ii_{\psi}(\tbf 1))\le\Osclln(\psi)$.

(d) If $\psi$ is an order-bounded integrating interval function, then
$\Osclln(ii_{\psi}(\tbf 1))\le\Osclln(\psi)$.

\proof{ All four parts are trivial if $\Cal S$ is empty,
so suppose otherwise.

\medskip

{\bf (a)} If $I$ is empty this is trivial, since
$I\wedge\tau=\{\sigma\wedge\tau:\sigma\in I\}$ is empty for every $\tau$.
Otherwise, take a
string $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   If
$\tau\in\Cal S$, $(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau)$ linearly generates
the $(I\wedge\tau)$-cells (611Kg), and $w_{I\tau}
=\sum_{i=0}^{n-1}\psi(\tau_i\wedge\tau,\tau_{i+1}\wedge\tau)$ (613Ec).
Since $\psi(\tau_i\wedge\tau,\tau_{i+1}\wedge\tau)
\in L^0(\frak A_{\tau_{i+1}\wedge\tau})\subseteq L^0(\frak A_{\tau})$ for each
$i$, $w_{I\tau}\in L^0(\frak A_{\tau})$.
Now if $\sigma$, $\tau\in\Cal S$,

$$\eqalignno{\Bvalue{\sigma=\tau}&
\Bsubseteq\inf_{i\le n}\Bvalue{\tau_i\wedge\sigma=\tau_i\wedge\tau}
\Bsubseteq\inf_{i<n}\Bvalue{\psi(\tau_i\wedge\sigma,\tau_{i+1}\wedge\sigma)
    =\psi(\tau_i\wedge\tau,\tau_{i+1}\wedge\tau)}
\Displaycause{because $\psi$ is strictly adapted}
\Bsubseteq\Bvalue{w_{I\sigma}=w_{I\tau}}.\cr}$$

\noindent Thus
$\tau\mapsto w_{I\tau}:\Cal S\to L^0(\frak A)$
is fully adapted.

\medskip

{\bf (b)(i)} Write $\bar u$ for
$\sup_{(\sigma,\tau)\in\Cal S^{2\uparrow}}|\psi(\sigma,\tau)|$.
By 613O(b-i), $\pmb v=ii_{\psi}(\tbf 1)$ is defined everywhere on $\Cal S$ and
is fully adapted;  set $v_{\tau}=\int_{\Cal S\wedge\tau}d\psi$ and
$w_{I\tau}=w_{I\tau}$ for $\tau\in\Cal S$, and
$\pmb w_I=\family{\tau}{\Cal S}{w_{I\tau}}$,
If $I\in\Cal I(\Cal S)$ and $\tau\in\Cal S$,
$|w_{I\tau}|\le\#(\Sti_0(I))\bar u$;  so
$\family{\tau}{\Cal S}{w_{I\tau}}$ is order-bounded.

Let $\epsilon>0$.   Then there is an $I\in\Cal I(\Cal S)$ such
that $\theta(S_J(\tbf 1,d\psi)-S_K(\tbf 1,d\psi))\le\epsilon$ whenever
$J$, $K\in\Cal I(\Cal S)$ include $I$.   Now if $J\in\Cal I(\Cal S)$ includes
$I_0$, 613V(ii-$\beta$) tells us that
$\theta(w_{J\tau}-v_{\tau})\le 2\epsilon$ for every
$\tau\in\Cal S$;  so if $J$, $K\in\Cal I(\Cal S)$ include $I$,
$\theta(w_{J\tau}-w_{K\tau})
\le 4\epsilon$.   Now (a) tells us that
$\tau\mapsto w_{J\tau}-w_{K\tau}$
is fully adapted, and we have just seen that it is order-bounded, so 615Db tells
us that

\Centerline{$\theta(\sup|\pmb w_J-\pmb w_K|)
=\theta(\sup_{\tau\in\Cal S}|w_{J\tau}-w_{K\tau}|)
\le 2\sqrt{4\epsilon}=4\sqrt{\epsilon}$.}

As $\epsilon$ is arbitrary, this shows that the filter on the space
$\Mob=\Mob(\Cal S)$ generated by

\Centerline{$\{\{\pmb w_J:
I\subseteq J\in\Cal I(\Cal S)\}:I\in\Cal I(\Cal S)\}$}

\noindent is a Cauchy filter;  by 615Cc, it has a limit in $\Mob$.   But as
$\lim_{I\uparrow\Cal I(\Cal S)}w_{I\tau}=v_{\tau}$ for every
$\tau\in\Cal S$, this limit must be $\pmb v$, and $\pmb v$ is order-bounded.

\medskip

\quad{\bf (ii)} Now 618Fb tells us that
$\Osclln(\pmb v)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln(\pmb w_I)$.   If
$I\in\Cal I(\Cal S)$ is non-empty, take
$(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   If
$J\in\Cal I(\Cal S)$ includes $I$ and $e$ is a $J$-cell, then we can express
$e$ as $c(\sigma,\sigma')$ where either $\sigma\le\sigma'\le\tau_0$ or
$\tau_n\le\sigma\le\sigma'$ or there is an $i<n$ such that
$\tau_i\le\sigma\le\sigma'\le\tau_{i+1}$.   Now

$$\eqalign{w_{I\sigma'}-w_{I\sigma}&
=\sum_{j=0}^{n-1}\psi(\tau_j\wedge\sigma',\tau_{j+1}\wedge\sigma')
  -\psi(\tau_j\wedge\sigma,\tau_{j+1}\wedge\sigma)\cr&
=0\text{ if }\sigma'\le\tau_0\text{ or }\tau_n\le\sigma,\cr&
=\psi(\tau_i\wedge\sigma',\tau_{i+1}\wedge\sigma')
  -\psi(\tau_i\wedge\sigma,\tau_{i+1}\wedge\sigma)
\text{ if }\tau_i\le\sigma\le\sigma'\le\tau_{i+1}\cr}$$

\noindent and in any case $|w_{I\sigma'}-w_{I\sigma}|\le 2\Osclln^*_I(\psi)$.
So $\Osclln_J(\pmb w_I)\le 2\Osclln(\psi)$.   As $J$ is arbitrary,
$\Osclln_I^*(\pmb w_I)\le 2\Osclln^*_I(\psi)$ and

$$\eqalign{\theta((\Osclln(\pmb w_I)-2\Osclln(\psi))^+)&
\le\theta((\Osclln_I^*(\pmb w_I)-2\Osclln(\psi))^+)\cr&
\le\theta(2\Osclln_I^*(\psi)-2\Osclln(\psi)).\cr}$$

\noindent Taking the limit as $I$ increases through $\Cal I(\Cal S)$,

\Centerline{$\theta((\Osclln(\pmb w)-2\Osclln(\psi))^+)
\le 2\lim_{I\uparrow\Cal I(\Cal S)}\theta(\Osclln_I^*(\psi)-\Osclln(\psi))
=0$,}

\noindent and $\Osclln(\pmb w)\le 2\Osclln(\psi)$, as claimed.

\medskip

{\bf (c)(i)} Suppose to begin with that $\Cal S$ is full, so that we
are in the situation of (b), but knowing that $\pmb v$ is
moderately oscillatory, so that
$\Osclln(\pmb v)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb v)$ (618K).
As observed in (b-i) above,
$\lim_{I\uparrow\Cal I(\Cal S)}
\sup_{\tau\in\Cal S}\theta(w_{I\tau}-v_{\tau})=0$, and because $\Cal S$ is
finitely full, it follows that
$\lim_{I\uparrow\Cal I(\Cal S)}z_I=0$, where
$z_I=\sup|\pmb w_I-\pmb v|$
Next, if $I\in\Cal I(\Cal S$ and $(\tau_0,\ldots,\tau_n)$ linearly
generate the $I$-cells,

$$\eqalign{w_{I\tau_{i+1}}-w_{I\tau_i}&
=\sum_{j=0}^{n-1}\psi(\tau_j\wedge\tau_{i+1},\tau_{j+1}\wedge\tau_{i+1})
   -\sum_{j=0}^{n-1}\psi(\tau_j\wedge\tau_i,\tau_{j+1}\wedge\tau_i)\cr&
=\sum_{j=0}^i\psi(\tau_j\wedge\tau_{i+1},\tau_{j+1}\wedge\tau_{i+1})
   -\sum_{j=0}^{i-1}\psi(\tau_j\wedge\tau_i,\tau_{j+1}\wedge\tau_i)\cr&
=\psi(\tau_i,\tau_{i+1}),\cr}$$

\noindent so

\Centerline{$|v_{\tau_{i+1}}-v_{\tau_i}|
\le|w_{I\tau_{i+1}}-w_{I\tau_i}|+2z_I
\le|\psi(\tau_i,\tau_{i+1}|+2z_I$}

\noindent for every $i<n$.   Acccordingly

$$\eqalign{\Osclln_I(\pmb v)&
=\sup_{i<n}|v_{\tau_{i+1}}-v_{\tau_i}|
\le\sup_{i<n}|w_{I\tau_{i+1}}-w_{I\tau_i}|+2z_I\cr&
\le\sup_{i<n}|\psi(\tau_i,\tau_{i+1})|+2z_I
=\Osclln_I(\psi)+2z_I
\le\Osclln^*_I(\psi)+2z_I\cr}$$

\noindent and

\Centerline{$\Osclln(\pmb v)
\le\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(\psi)
   +\lim_{I\uparrow\Cal I(\Cal S)}z_I
=\Osclln(\psi)$.}

\medskip

\quad{\bf (ii)}
For the general case, we have the adapted extension $\hat\psi$ of
$\psi$ to the covered envelope $\hat\Cal S$ of $\Cal S$.

\medskip

\qquad\grheada\ $\hat\psi$ is order-bounded.   \Prf\ Write $\bar w$ for
$\sup_{(\sigma,\sigma')\in\Cal S^{2\uparrow}}|\psi(\sigma,\sigma')|$.
If $\tau\le\tau'$ in $\hat\Cal S$,

$$\eqalign{\Bvalue{|\hat\psi(\tau,\tau')|\le\bar w}&
\Bsupseteq\sup_{\sigma,\sigma'\in\Cal S}
   \Bvalue{\sigma=\tau'}\Bcap\Bvalue{\sigma\vee\sigma'=\tau'}
      \Bcap\Bvalue{|\psi(\sigma,\sigma\vee\sigma')|\le\bar w}\cr&
\Bsupseteq\sup_{\sigma,\sigma'\in\Cal S}
   \Bvalue{\sigma=\tau'}\Bcap\Bvalue{\sigma'=\tau'}
=1\cr}$$

\noindent and $|\hat\psi(\tau,\tau')|\le\bar w$.\ \Qed

\medskip

\qquad\grheadb\ $\Osclln(\hat\psi)\le\Osclln(\psi)$.   \Prf\ Follow parts (i)
and (ii) of the proof of 618Ca and part (a) of
the proof of 618L, replacing every $\hat u_{\rho'}-\hat u_{\rho}$ with
$\hat\psi(\rho,\rho')$ and every $u_{\sigma'}-u_{\sigma}$ with
$\psi(\sigma,\sigma')$, to see that

\inset{----- if $I\in\Cal I(\Cal S)$,
$(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells.
$\tau_{-1}=\inf\Cal S$ and $\tau_{n+1}=\sup\Cal S$ then

$$\eqalign{\Osclln^*_I(\psi)
&=\sup\{|\psi(\sigma,\sigma')|:\sigma,\,\sigma'\in\Cal S
\text{ and there is an }i\cr
&\mskip100mu\text{such that }-1\le i\le n
\text{ and }\tau_i\le\sigma\le\sigma'\le\tau_{i+1}\},\cr}$$

----- $\Osclln^*_I(\hat\psi)\le\Osclln^*_I(\psi)$
for every $I\in\Cal I(\Cal S)$}

\noindent and
therefore $\Osclln(\hat\psi)\le\Osclln(\psi)$.\ \Qed

\medskip

\qquad\grheadc\ Now recall that the fully adapted extension $\hatpmb v$ of
$\pmb v$ is $ii_{\hat\psi}(\tbf 1)$ (616Q(c-ii)) and is moderately oscillatory
(615F(a-i)).   So, using 618L itself and (b) above, we see that

\Centerline{$\Osclln(\pmb v)=\Osclln(\hatpmb v)\le\Osclln(\hat\psi)
\le\Osclln(\psi)$,}

\noindent as required.

\medskip

{\bf (d)} $\pmb v$ is an integrator (616J) therefore moderately oscillatory
(616Ib), and $\int_{\hat\Cal S}d\hat\psi$ exists by the definition in
616Fa, so this is a consequence of (c).
}%end of proof of 618P

\leader{618Q}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a moderately oscillatory process, and
$\pmb v=\sfamilysS{v}$ an integrator.   Then
$\Osclln(\iivu)\le\sup|\pmb u|\times\Osclln(\pmb v)$.   \cmmnt{In particular,}
$\iivu$ is jump-free if $\pmb v$ is.

\proof{ Consider the integrating interval functions $\Delta\pmb v$ and
$\psi=\pmb u\Delta\pmb v$ (617Da).   Write $\bar u$ for $\sup|\pmb u|$.
Comparing 618O with 618B, we see that
$\Delta\pmb v$ is order-bounded because $\pmb v$ is (616Ib), and that
$\Osclln(\Delta\pmb v)=\Osclln(\pmb v)$.   Next, if $\sigma\le\sigma'$ in
$\Cal S$,

\Centerline{$|\psi(\sigma,\sigma')|=|u_{\sigma}\times(v_{\sigma'}-v_{\sigma})|
\le\bar u\times|v_{\sigma'}-v_{\sigma}|$,}

\noindent so if $I\in\Cal I(\Cal S)$

$$\eqalign{\Osclln_I(\psi)&
=\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\psi|)\cr&
\le\bar u\times\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\pmb v|)
=\bar u\times\Osclln_I(\pmb v).\cr}$$

\noindent Accordingly $\Osclln^*_I(\psi)\le\bar u\times\Osclln^*_I(\pmb v)$ for
every $I\in\Cal I\Cal S)$ and $\Osclln(\psi)\le\bar u\times\Osclln(\pmb v)$.
But now 617Db tells us that $\iivu=ii_{\psi}(\tbf 1)$ and 618Pc tells us that

\Centerline{$\Osclln(\iivu)\le\Osclln(\psi)
\le\bar u\times\Osclln(\pmb v)$.}

\noindent In particular, $\Osclln(\iivu)$ will be zero if $\Osclln(\pmb v)$ is.
}%end of proof of 618Q

\leader{618R}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a locally moderately oscillatory process, and
$\pmb v=\sfamilysS{v}$ a locally jump-free local integrator.
Then\cmmnt{ the indefinite integral} $\iivu$ is locally jump-free.

\proof{ Apply 618Q to $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb v\restr\Cal S\wedge\tau$ for $\tau\in\Cal S$, using 615F(a-i),
616P(b-ii), 618Gc and 613O(b-ii).
}%end of proof of 618R

\leader{618S}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$, $\pmb w=\sfamilysS{w}$ two integrators.

(a)
$\Osclln(\covar{\pmb v}{\pmb w})\le\Osclln(\pmb v)\times\Osclln(\pmb w)$.

(b) $\Osclln(\pmb v^*)=(\Osclln(\pmb v))^2$.

\proof{{\bf (a)} Being integrators, $\pmb v$ and $\pmb w$ are order-bounded,
while
$\covar{\pmb v}{\pmb w}$ is of bounded variation (617Jc) therefore also
order-bounded.   Thus all three residual oscillations are well-defined.

Consider the integrating interval function
$\psi=\Delta\pmb v\times\Delta\pmb w$ (617Ha).   We have
$|\psi|=|\Delta\pmb v|\times|\Delta\pmb w|$ so

\Centerline{$\Delta_e(\tbf 1,|d\psi|)
   =\Delta_e(\tbf 1,|d\pmb v|)\times\Delta_e(\tbf 1,|d\pmb w|)$}

\noindent for every stopping-time interval $e$ with endpoints in $\Cal S$,

$$\eqalign{\Osclln_I(\psi)&
=\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\psi|)
\le\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\pmb v|)
  \times\sup_{e\in\Sti_0(I)}\Delta_e(\tbf 1,|d\pmb w|)\cr&
=\Osclln_I(\pmb v)\times\Osclln_I(\pmb w)\cr}$$

\noindent for every $I\in\Cal I(\Cal S)$,

$$\eqalign{\Osclln^*_I(\psi)&
=\sup_{I\subseteq J\in\Cal I(\Cal S)}\Osclln_J(\psi)
\le\sup_{I\subseteq J\in\Cal I(\Cal S)}\Osclln_J(d\pmb v)
   \times\sup_{I\subseteq J\in\Cal I(\Cal S)}\Osclln_J(d\pmb w)\cr&
=\Osclln^*_I(\pmb v)\times\Osclln^*_I(\pmb w)\cr}$$

\noindent for every $I\in\Cal I(\Cal S)$,

$$\eqalign{\Osclln(\psi)&
=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(\psi)
\le\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(d\pmb v)\times\Osclln^*_I(d\pmb w)
  \cr&
=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(d\pmb v)
  \times\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(d\pmb w)
=\Osclln(\pmb v)\times\Osclln(\pmb w)\cr}$$

\noindent and

\Centerline{$\Osclln(\covar{\pmb v}{\pmb w})
=\Osclln(ii_{\psi}(\tbf 1))
\le\Osclln(\psi)
\le\Osclln(\pmb v)\times\Osclln(\pmb w)$.}

\medskip

{\bf (b)} From (a) we see at once that
$\Osclln(\pmb v^*)\le\Osclln(\pmb v)^2$.
But for the reverse inequality it seems that we need a further idea from \S613.

\medskip

\quad{\bf (i)} To begin with (down to the end of (v) below), suppose that
$\Cal S$ is full and not empty.
Express $\pmb v$, $\pmb v^*$ as $\sfamilysS{v}$ and
$\sfamilysS{v^*}$.   We know that

\Centerline{$\Osclln(\pmb v)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb v)$,
\quad$\Osclln(\pmb v^*)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb v^*)$}

\noindent (618M),

\Centerline{$\int_{\Cal S}(d\pmb v)^2
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\tbf 1,(d\pmb v)^2)$ is defined,
%(617H(a-i))
\quad$v^*_{\tau}=\int_{\Cal S\wedge\tau}(d\pmb v)^2$ for $\tau\in\Cal S$}
%(617Hb)

\noindent (617H).   Now
$\Osclln(\pmb v)^2=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb v)^2$,
and given $\epsilon>0$ there is a non-empty $I\in\Cal I(\Cal S)$ such that

\Centerline{$\theta(\Osclln(\pmb v)^2-\Osclln_I(\pmb v)^2)\le\epsilon$,
\quad$\theta(\Osclln(\pmb v^*)-\Osclln_I(\pmb v^*))\le\epsilon$,}

\Centerline{$\theta(S_J(\tbf 1,(d\pmb v)^2)-S_K(\tbf 1,(d\pmb v)^2))\le\epsilon$
whenever $J$, $K\in\Cal I(\Cal S)$ include $I$.}

\noindent From 613V(ii-$\beta$) we see that
$\theta(v^*_{\tau}-S_{I\wedge\tau}(\tbf 1,(d\pmb v)^2))\le 2\epsilon$
for every $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)} Let $(\tau_0,\ldots,\tau_n)$ be a sequence linearly generating
the $I$-cells.   Translating the formulae above with the aid of the last
remark in 618Ba, we have

\Centerline{$
\theta(\Osclln(\pmb v)^2-\sup_{i<n}(v_{\tau_{i+1}}-v_{\tau_i})^2)
=\theta(\Osclln(\pmb v)^2-(\sup_{i<n}|v_{\tau_{i+1}}-v_{\tau_i}|)^2)
\le\epsilon$}

\noindent and

\Centerline{
$\theta(\Osclln(\pmb v^*)-\sup_{i<n}|v^*_{\tau_{i+1}}-v^*_{\tau_i}|)
\le\epsilon$,}

\noindent while

\Centerline{
$\theta(v^*_{\tau}-\sum_{i<n}(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})^2)
\le 2\epsilon$ for every $\tau\in\Cal S$}

\noindent because $(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau)$ linearly
generates the $(I\wedge\tau)$-cells (611Kg).

\medskip

\quad{\bf (iii)} The next thing to note is that

\Centerline{$\tau
\mapsto v^*_{\tau}-\sum_{i<n}(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})^2:
\Cal S\to L^0(\frak A)$}

\noindent is fully adapted and order-bounded,
just because $\tau\mapsto v^*_{\tau}$ and
$\tau\mapsto v_{\tau_i\wedge\tau}$ are fully adapted and order-bounded
for every $i\le n$ (612Ib).
Since $\Cal S$ is supposed to be full, $\theta(w)\le 2\sqrt{2\epsilon}$ where

\Centerline{$w=\sup_{\tau\in\Cal S}|v^*_{\tau}
  -\sum_{i<n}(v_{\tau_{i+1}\wedge\tau}-v_{\tau_i\wedge\tau})^2|$}

\noindent (615Db).   Now if $j\le n$,

\Centerline{$w
\ge|v^*_{\tau_j}
      -\sum_{i<n}(v_{\tau_{i+1}\wedge\tau_j}-v_{\tau_i\wedge\tau_j})^2|
=|v^*_{\tau_j}-\sum_{i<j}(v_{\tau_{i+1}}-v_{\tau_i})^2|$,}

\noindent so if $j<n$

$$\eqalign{2w&
\ge|v^*_{\tau_{j+1}}-\sum_{i\le j}(v_{\tau_{i+1}}-v_{\tau_i})^2|
      +|v^*_{\tau_j}-\sum_{i<j}(v_{\tau_{i+1}}-v_{\tau_i})^2|\cr&
\ge|v^*_{\tau_{j+1}}-v^*_{\tau_j}-(v_{\tau_{j+1}}-v_{\tau_j})^2|.\cr}$$

\medskip

\quad{\bf (iv)} Assembling these, we have

$$\eqalign{\theta(\Osclln(\pmb v^*)-\Osclln(\pmb v)^2)&
\le\theta(\Osclln(\pmb v^*)-\sup_{i<n}(v^*_{\tau_{i+1}}-v^*_{\tau_i}))\cr&
\mskip50mu +\theta(\sup_{i<n}(v^*_{\tau_{i+1}}-v^*_{\tau_i})
                     -\sup_{i<n}(v_{\tau_{i+1}}-v_{\tau_i})^2)\cr&
\mskip50mu +\theta(\Osclln(\pmb v)^2-\sup_{i<n}(v_{\tau_{i+1}}-v_{\tau_i})^2)
    \cr&
\le\epsilon
    +\theta(\sup_{i<n}|v^*_{\tau_{i+1}}-v^*_{\tau_i}
                  -(v_{\tau_{i+1}}-v_{\tau_i})^2|)
    +\epsilon\cr&
\le 2\epsilon+\theta(2w)
\le 2\epsilon+4\sqrt{2\epsilon}.
\cr}$$

\medskip

\quad{\bf (v)} As $\epsilon$ was arbitrary,
$\Osclln(\pmb v^*)=\Osclln(\pmb v)^2$, at least if $\Cal S$ is full and not
empty.

\medskip

\quad{\bf (vi)} In general, the case $\Cal S=\emptyset$ is trivial.
Otherwise, taking $\hat\Cal S$ to be the covered envelope of
$\Cal S$ and
$\hatpmb v$, $\hatpmb v^*$ to be the fully adapted extensions of $\pmb v$,
$\pmb v^*$ to $\hat\Cal S$, we see that $\hatpmb v^*$ is the quadratic variation
of $\pmb v$ (617N), while $\Osclln(\pmb v)=\Osclln(\hatpmb v)$ and
$\Osclln(\pmb v^*)=\Osclln(\hatpmb v^*)$ (618La).   So

\Centerline{$\Osclln(\pmb v^*)=\Osclln(\hatpmb v^*)
=\Osclln(\hatpmb v)^2=\Osclln(\pmb v)^2$}

\noindent in all cases.
}%end of proof of 618S

\leader{618T}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$, $\pmb w=\sfamilysS{w}$
two (local) integrators with domain
$\Cal S$ of which $\pmb v$ is (locally) jump-free.
Then the covariation $\covar{\pmb v}{\pmb w}$ and the quadratic
variation $\pmb v^*$ are (locally) jump-free.

\proof{ If $\pmb v$ and $\pmb v$ are integrators and $\pmb v$ is jump-free
then $\Osclln(\pmb v)=0$ so $\Osclln(\covar{\pmb v}{\pmb w})=0$ and the
covariation is jump-free.   If they are local integrators and $\pmb v$ is
locally jump-free then we apply this to their restrictions to
$\Cal S\wedge\tau$, for $\tau\in\Cal S$, to see that $\covar{\pmb v}{\pmb w}$ is
locally jump-free.   Taking $\pmb w=\pmb v$ we have the results for $\pmb v^*$.
}%end of proof of 618T

\leader{618U}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$ a process of bounded variation.   Let
$\pmb v^{\uparrow}\cmmnt{\mskip5mu=\family{\tau}{\Cal S}{v^{\uparrow}_{\tau}}}$
be its cumulative variation.   Then $\Osclln(\pmb v^{\uparrow})$ is equal to
$\Osclln(\pmb v)$;  \cmmnt{in particular,} $\pmb v$ is jump-free iff
$\pmb v^{\uparrow}$ is jump-free.

\proof{ Consider the interval functions $\psi=\Delta\pmb v$ and
$\psi'=|\Delta\pmb v|$.   Then $\pmb v=ii_{\psi}(\tbf 1)+v_{\downarrow}\tbf 1$,
where $v_{\downarrow}$ is the starting value of $\pmb v$, and
$\pmb v^{\uparrow}=ii_{\psi'}(\tbf 1)$.
As $\Delta\pmb v=\Delta ii_{\psi}(\tbf 1)$,

\Centerline{$\Osclln(\pmb v)=\Osclln(ii_{\psi}(\tbf 1)=\Osclln(\psi)$}

\noindent as remarked in 618O;  as $|\psi|=|\psi'|$,
$\Osclln(\psi')=\Osclln(\psi)$;  and by 618M,
$\Osclln(\pmb v^{\uparrow})\le\Osclln(\psi')$, so
$\Osclln(\pmb v^{\uparrow})\le\Osclln(\pmb v)$.

In the other direction, because
$|v_{\tau}-v_{\sigma}|\le v^{\uparrow}_{\tau}-v^{\uparrow}_{\sigma}$
whenever $\sigma\le\tau$ in $\Cal S$ (614P(a-i)), we see that

\Centerline{$\Osclln_I(\pmb v)\le\Osclln_I(\pmb v^{\uparrow})$,
\quad$\Osclln^*_I(\pmb v)\le\Osclln^*_I(\pmb v^{\uparrow})$}

\noindent for every $I\in\Cal I(\Cal S)$, so that
$\Osclln(\pmb v)\le\Osclln(\pmb v^{\uparrow})$.   Thus the residual oscillations
are equal.   Now

\Centerline{$\pmb v^{\uparrow}\text{ is jump-free}\iff\Osclln(\pmb v^{\uparrow})=0
\iff\Osclln(\pmb v)=0\iff\pmb v\text{ is jump-free}$.}
}%end of proof of 618U

\leader{618V}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a fully adapted process with domain $\Cal S$.
Then $\pmb v$ is jump-free and of bounded variation
iff it is expressible as the difference of two
non-negative non-decreasing order-bounded jump-free processes.

\proof{ Given that $\pmb v=\sfamilysS{v}$ is of bounded variation and jump-free,
let $\pmb v^{\uparrow}$ be its cumulative variation
and set $v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$
(614Kd);
then $\pmb v^{\uparrow}+|v_{\downarrow}|\tbf 1$ and
$\pmb v^{\uparrow}-\pmb v+|v_{\downarrow}|\tbf 1$ are order-bounded
non-negative
non-decreasing processes with difference $\pmb v$, and
are jump-free if $\pmb v$ is.   In the other direction, given that
$\pmb v=\pmb v_1-\pmb v_2$ where $\pmb v_1$ and $\pmb v_2$ are
non-negative order-bounded jump-free non-decreasing processes, then
$\pmb v$ is of bounded variation by 614I
and jump-free if $\pmb v_1$ and $\pmb v_2$ are
by 618Ga.   Of course I am asking you to remember that
jump-free processes are order-bounded (618B).
}%end of proof of 618V	

%\leader{618W}{}

\exercises{\leader{618X}{Basic exercises (a)}
%\spheader 618Xa
Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa and 617Xb.   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 bounded, then $\Osclln(\pmb u)$ can be identified
with $\max(\sup_{t\ge 0}\limsup_{s\downarrow t}|f(s)-f(t)|,
\sup_{t>0}\limsup_{s\uparrow t}|f(s)-f(t)|,
\inf_{t\ge 0}\limsup_{s\to\infty}|f(s)-f(t)|)$.   (ii) Show that
$\pmb u$ is jump-free iff $f$ is continuous and
$\lim_{t\to\infty}f(t)$ is defined.
%618B

\spheader 618Xb Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$ an
order-bounded fully adapted
process with domain $\Cal S$, and $z$ a member of
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.   Show that, in the
language of 612D(e-ii),
$\Osclln(z\pmb u)=|z|\times\Osclln(\pmb u)$, so that
$z\pmb u$ is jump-free if $\pmb u$ is.
%618B

\spheader 618Xd Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$, $\pmb v$ order-bounded processes with domain $\Cal S$.   Show that
$\Bvalue{\Osclln(\pmb u)\ne\Osclln(\pmb v)}
\Bsubseteq\Bvalue{\pmb u\ne\pmb v}$.
%618B

\spheader 618Xe
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u =\sfamilysS{u}$ an order-bounded process.   Set
$\pmb v=\family{\tau}{\Cal S}{\Osclln(\pmb u \restr\Cal S\wedge\tau)}$, as
in 618Da.   Show that $\Osclln(\pmb v)=\Osclln(\pmb u)$.
%618D

\spheader 618Xf
Give an example of a jump-free fully adapted process
$\pmb v$ with domain $\Cal S$ and a sublattice $\Cal S'$ of $\Cal S$ such
that $\pmb v\restr\Cal S'$ is not jump-free.   \Hint{try $\#(\Cal S')=2$.}
%618G

\spheader 618Xg Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a simple process with starting value
$u_{\downarrow}$, and $I$ a sublattice of $\Cal S$
including a breakpoint string for $\pmb u$.   Show that
$\Osclln(\pmb u)=\Osclln_I(\pmb u)\vee|u_{\min I}-u_{\downarrow}|$.
%618M

\spheader 618Xh Let $\Cal S$ be a sublattice of $\Cal T$, and $\pmb u$ a
moderately oscillatory process with domain $\Cal S$.   Show that
$\Osclln(\pmb u)=\sup_{\tau\in\Cal S}\Osclln(\pmb u\restr\Cal S\wedge\tau)$, so
that if $\pmb u$ is locally jump-free then it is jump-free.
%618N  615G

\spheader 618Xi Let $\Cal S$ be a full sublattice of $\Cal T$ with a greatest
element and $\pmb u=\sfamilysS{u}$ a moderately oscillatory process.   For
$\delta>0$ let $\sequence{i}{y_{\delta i}}$ be the sequence constructed from
$\pmb u$ and $\delta$ as in 615M.   Show that
$\Osclln(\pmb u)=\lim_{\delta\downarrow 0}
  \sup_{i\in\Bbb N}|y_{\delta,i+1}-y_{\delta}|$.
%618N

\leader{618Y}{Further exercises (a)}
%\spheader 618Ya
Let $\Cal S$ be a sublattice of $\Cal T$ and $\Cal S'$ a
sublattice of $\Cal S$ which is order-convex in $\Cal S$.   Let
$\pmb u=\sfamilysS{u}$ be an order-bounded process.   Show that
$\Osclln(\pmb u\restr\Cal S')\le\Osclln(\pmb u)$.
%618D  mt61bits

\spheader 618Yb Suppose that $T=\coint{0,\infty}$ and there
is a $t\ge 0$ such that $\frak A_t\ne\bigcap_{s>t}\frak A_s$.
Let $\pmb{\iota}=\family{\sigma}{\Cal T_f}{\iota_{\sigma}}$ be the identity
process.   Show there is a non-empty
downwards-directed set $A\subseteq\Cal T_f$ such
that $\lim_{\sigma\downarrow A}\iota_{\sigma}\ne\iota_{\inf A}$.
%618I

\spheader 618Yc Let $\Cal S$ be a finitely full sublattice of $\Cal T$, and
$\pmb{u}=\sfamilysS{u}$ an order-bounded fully adapted process.   Show that
$\pmb{u}$ is jump-free iff
$\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb{u})=0$ in $L^0(\frak A)$.
%618M mt61bits

\spheader 618Yd Show that with a modification of the construction in 613S, we
can find jump-free processes $\pmb u$ and $\pmb v$ with domain $\Cal T_c$ such
that $\int_{\Cal T_c}\pmb u\,d\pmb v$ is defined but the integral
$\int_{\hat{\Cal T}_c}\hatpmb u\,d\hatpmb v$ is not.
%613S 618H
}

\endnotes{
\Notesheader{618}
In fact rather little of the work of this section is directly
necessary for It\^o's lemma
as it will be stated in \S619;  only the definition is really essential, though
of course it makes no sense without 618H.   And since we shall need to know
when an integrator is jump-free, results like 618G, 618L,
618R, 618T and 618U are worth establishing.   I have taken a bit of extra
trouble to calculate residual oscillations;   I think that %in \query\
these give us a clearer notion of what is really going on.   And the heavy
labour of 618M leads to a fact which is not only remarkable in itself but
goes to the heart of the concept of `moderately oscillatory' process.   If
$\pmb u=\sfamilysS{u}$ is a simple process, then a straightforward calculation
(618Xg) tells us that
$\Osclln(\pmb u)=\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb u)$.   Similarly,
it is easy to see that processes of bounded variation are integrators
(616R).
The point of 616K-616M and 618M is that moderately oscillatory processes are
`nearly' of bounded variation.

Note that `jump-free', like `simple' but for different reasons, is
another property which is not automatically inherited by restrictions
to sublattices (618Xf).   But observe that once again we have a property of
stochastic processes which is independent of the measure chosen on the
underlying Boolean algebra.
}%end of notes

\discrpage

\leader{618zO}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\psi:\Cal S^{2\uparrow}\to L^0(\frak A)$ an order-bounded
integrating interval function with
indefinite integral $\pmb v=ii_{\psi}(\tbf 1)$.   Then

\Centerline{$\Osclln(\pmb v)\le\Osclln(\psi)$.}

\proof{ The result is trivial if $\Cal S$ is empty, so suppose otherwise.

\medskip

{\bf (a)} To begin with, consider the case in which $\Cal S$ is full.
Let $\epsilon>0$.

\medskip

\quad{\bf (i)} Since

\Centerline{$\lim_{I\uparrow\Cal I(\Cal S)}S_I(\tbf 1,d\psi)=\int_{\Cal S}d\psi$,
\quad$\lim_{I\uparrow\Cal I(\Cal S)}\Osclln^*_I(\psi)=\Osclln(\psi)$,}

\noindent there is a non-empty $I_0\in\Cal S$ such that whenever
$I\in\Cal I(\Cal S)$ includes $I_0$ then

\inset{($\alpha$) $\theta(S_J(\tbf 1,d\psi)-S_K(\tbf 1,d\psi))\le\epsilon$
whenever $J$, $K\in\Cal I(\Cal S)$ include $I$

($\beta$) $\theta(\Osclln^*_I(\psi)-\Osclln(\psi))\le\epsilon$.}

\noindent Take any such $I$.

\medskip

\quad{\bf (ii)} By 613V(ii-$\beta$),
$\theta(w_{I\tau}-v_{\tau})\le 2\epsilon$ for every
$\tau\in\Cal S$.   Now
$\tau\mapsto w_{I\tau}:\Cal S\to L^0$ is fully adapted
and order-bounded.   \Prf\ If $I$ is empty, this is trivial.   Otherwise, take a
string $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   If
$\tau\in\Cal S$, $(\tau_0\wedge\tau,\ldots,\tau_n\wedge\tau)$ linearly generates
the $(I\wedge\tau)$-cells (611Kg), and
$w_{I\tau}
=\sum_{i=0}^{n-1}\psi(\tau_i\wedge\tau,\tau_{i+1}\wedge\tau)$ (613Ec).
Since $\psi(\tau_i\wedge\tau,\tau_{i+1}\wedge\tau)
\in L^0(\frak A_{\tau_{i+1}\wedge\tau})\subseteq L^0(\frak A_{\tau})$ for each
$i$, $w_{I\tau}\in L^0(\frak A_{\tau})$.
Now if $\sigma$, $\tau\in\Cal S$,

$$\eqalignno{\Bvalue{\sigma=\tau}&
\Bsubseteq\inf_{i\le n}\Bvalue{\tau_i\wedge\sigma=\tau_i\wedge\tau}
\Bsubseteq\inf_{i<n}\Bvalue{\psi(\tau_i\wedge\sigma,\tau_{i+1}\wedge\sigma)
    =\psi(\tau_i\wedge\tau,\tau_{i+1}\wedge\tau)}
\Displaycause{because $\psi$ is strictly adapted}
\Bsubseteq\Bvalue{S_{I\wedge\sigma}(\tbf 1,d\psi)
   =w_{I\tau}}.\cr}$$

\noindent Thus $\tau\mapsto w_{I\tau}:\Cal S\to L^0$
is fully adapted.   As for order-boundedness, it is the sum of the $n$
order-bounded functions
$(\sigma,\tau)\mapsto\psi(\sigma\wedge\tau_i,\tau\wedge\tau_i)$ for $i<n$, so is
order-bounded.\ \Qed

\medskip

\quad{\bf (iii)} It follows that
$\tau\mapsto w_{I\tau}-v_{\tau}$ is fully adapted and
order-bounded, so that,
writing $w$ for $\sup_{\tau\in\Cal S}
  |w_{I\tau}-v_{\tau}|$,
$\theta(w)\le 2\sqrt{2\epsilon}$.   Now take $(\tau_0,\ldots,\tau_n)$
linearly generating the $I$-cells.   Then

$$\eqalign{\Osclln_I(\pmb v)&
=\sup_{i<n}|v_{\tau_{i+1}}-v_{\tau_i}|
\le w+\sup_{i<n}
   |S_{I\wedge\tau_{i+1}}(\tbf 1,d\psi)-S_{I\wedge\tau_i}(\tbf 1,d\psi)|\cr&
\le w+\sup_{i<n}
   |\sum_{j=0}^i\psi(\tau_j,\tau_{j+1})
       -\sum_{j=0}^{i-1}\psi(\tau_j,\tau_{j+1})|\cr&
=w+\sup_{i<n}|\psi(\tau_i,\tau_{i+1})|
=w+\Osclln_I(\psi)
\le w+\Osclln^*_I(\psi)\cr}$$

\noindent and

\Centerline{$\theta((\Osclln_I(\pmb v)-\Osclln(\psi))^+)
\le\theta(w)+\theta(\Osclln^*_I(\psi)-\Osclln(\psi))
\le 2\sqrt{2\epsilon}+\epsilon$.}

\medskip

\quad{\bf (iv)} This is true whenever $I_0\subseteq I\in\Cal I(\Cal S)$.
As $\epsilon$ is arbitrary,
$\lim_{I\uparrow\Cal I(\Cal S)}(\Osclln_I(\pmb v)-\Osclln(\psi))^+=0$.   But
because $\pmb v$ is an integrator (616J), therefore
moderately oscillatory (616Ib),
$\lim_{I\uparrow\Cal I(\Cal S)}\Osclln_I(\pmb v)=\Osclln(\pmb v)$ (618K).
So $(\Osclln(\pmb v)-\Osclln(\psi))^+=0$ and $\Osclln(\pmb v)\le\Osclln(\psi)$.

\medskip

{\bf (b)} For the general case, we have the adapted extension $\hat\psi$ of
$\psi$ to the covered envelope $\hat\Cal S$ of $\Cal S$, which is again an
integrating interval function (616F).

\medskip

\quad{\bf (i)} $\hat\psi$ is order-bounded.   \Prf\ Write $\bar w$ for
$\sup_{(\sigma,\sigma')\in\Cal S^{2\uparrow}}|\psi(\sigma,\sigma')|$.
If $\tau\le\tau'$ in $\hat\Cal S$,

$$\eqalign{\Bvalue{|\hat\psi(\tau,\tau')|\le\bar w}&
\Bsupseteq\sup_{\sigma,\sigma'\in\Cal S}
   \Bvalue{\sigma=\tau'}\Bcap\Bvalue{\sigma\vee\sigma'=\tau'}
      \Bcap\Bvalue{|\psi(\sigma,\sigma\vee\sigma')|\le\bar w}\cr&
\Bsupseteq\sup_{\sigma,\sigma'\in\Cal S}
   \Bvalue{\sigma=\tau'}\Bcap\Bvalue{\sigma'=\tau'}
=1\cr}$$

\noindent and $|\hat\psi(\tau,\tau')|\le\bar w$.\ \Qed

\medskip

\quad{\bf (ii)} $\Osclln(\hat\psi)\le\Osclln(\psi)$.   \Prf\ Follow parts (i)
and (ii) of the proof of 618Ca and part (a) of
the proof of 618L, replacing every $\hat u_{\rho'}-\hat u_{\rho}$ with
$\hat\psi(\rho,\rho')$ and every $u_{\sigma'}-u_{\sigma}$ with
$\psi(\sigma,\sigma')$, to see that

\inset{\grheada\ if $I\in\Cal I(\Cal S)$,
$(\tau_0,\ldots,\tau_n)$ linearly generates the $I$-cells.
$\tau_{-1}=\inf\Cal S$ and $\tau_{n+1}=\sup\Cal S$ then

$$\eqalign{\Osclln^*_I(\psi)
&=\sup\{|\psi(\sigma,\sigma')|:\sigma,\,\sigma'\in\Cal S
\text{ and there is an }i\cr
&\mskip100mu\text{such that }-1\le i\le n
\text{ and }\tau_i\le\sigma\le\sigma'\le\tau_{i+1}\},\cr}$$

\grheadb\ $\Osclln^*_I(\hat\psi)\le\Osclln^*_I(\psi)$
for every $I\in\Cal I(\Cal S)$}

\noindent and
therefore $\Osclln(\hat\psi)\le\Osclln(\psi)$.\ \Qed

\medskip

\quad{\bf (iii)} Now recall that the fully adapted extension $\hatpmb v$ of
$\pmb v$ is $ii_{\hat\psi}(\tbf 1)$ (616Q(c-ii)).
So, using 618L itself and (a) above, we see that

\Centerline{$\Osclln(\pmb v)=\Osclln(\hatpmb v)\le\Osclln(\hat\psi)
\le\Osclln(\psi)$,}

\noindent as required.
}%end of proof of 618zO

\frfilename{mt619.tex}
\versiondate{13.3.17/29.7.19}

\def\chaptername{Continuous processes and It\^o's formula}
\def\sectionname{It\^o's formula}

\newsection{619}

I give three versions of It\^o's formula (619C, 619D and 619J).   The last
depends on elementary facts about the action of functions of more than one
real variable on strings of processes (619E-619G).   %619E 619G

\cmmnt{\leader{619A}{Notation} As usual, \Asissix\ will be a stochastic integration
structure.   $L^{\infty}$ will be $L^{\infty}(\frak A)$ with its norm
$\|\,\|_{\infty}$.   $L^0$ will be $L^0(\frak A)$ endowed with its
topology of convergence in measure, and $\Expn$ will be
the integral defined from $\bar\mu$;  for $w\in L^0$,
$\theta(w)$ will be $\Expn(|w|\wedge\chi 1)$.
If $y\in L^0$, then $y\tbf 1$ will denote the constant
process on $\{\sigma:\sigma\in\Cal T$, $y\in L^0(\frak A_{\sigma})\}$
with value $y$.   For local integrators $\pmb v$ and $\pmb w$,
$\covar{\pmb v}{\pmb w}$ will be their covariation\cmmnt{ (617Hb)};
$\pmb v^*$ will be the quadratic variation of $\pmb v$.
For a sublattice $\Cal S$ of $\Cal T$, $\Mfa(\Cal S)$ will be the space of
fully adapted processes with domain $\Cal S$.
}

\leader{619B}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ an integrator.   Then for every $\epsilon>0$ there is a
$\delta>0$ such that $\theta(S_I(\pmb u,(d\pmb v)^2)\le\epsilon$ whenever
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(\Cal I)$ and
$\theta(\sup|\pmb u|)\le\delta$.

\proof{ As $\pmb v$ is an integrator, it is order-bounded (616Ib) and
there is a $\delta_0>0$ such that

\Centerline{$\theta(S_I(\pmb u,d\pmb v))\le\bover13\epsilon$ whenever
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(\Cal I)$ and
$\theta(\sup|\pmb u|)\le\delta_0$}

\noindent (616E).   Also $\pmb v^2$ is an integrator (616Pa),
so there is a $\delta>0$ such that

\Centerline{$\theta(S_I(\pmb u,d(\pmb v^2)))\le\bover13\epsilon$ whenever
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(\Cal I)$ and
$\theta(\sup|\pmb u|)\le\delta$}

\noindent and $\theta(x\times\sup|\pmb v|)\le\delta_0$ whenever
$\theta(x)\le\delta$.   Now suppose that
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(\Cal I)$ and
$\theta(\sup|\pmb u|)\le\delta$.   If $\sigma\le\tau$ in $I$,

\Centerline{$u_{\sigma}\times(v_{\tau}-v_{\sigma})^2
=u_{\sigma}\times(v_{\tau}^2-v_{\sigma}^2)
   -2u_{\sigma}\times v_{\sigma}\times(v_{\tau}-v_{\sigma})$.}

\noindent Working through the definitions in \S613, we see that

\Centerline{$S_I(\pmb u,(d\pmb v)^2)
=S_I(\pmb u,d(\pmb v^2))-2S_I(\pmb u\times\pmb v,d\pmb v)$}

\noindent while

\Centerline{$\theta(\sup|\pmb u\times\pmb v|)
\le\theta(\sup|\pmb u|\times\sup|\pmb v|)\le\delta_0$,}

\noindent so

\Centerline{$\theta(S_I(\pmb u,(d\pmb v)^2))
\le\theta(S_I(\pmb u,d(\pmb v^2)))+2\theta(S_I(\pmb u\times\pmb v,d\pmb v))
\le\Bover13\epsilon+\Bover23\epsilon
=\epsilon$,}

\noindent as required.
}%end of proof of 619B

\leader{619C}{It\^o's Formula, first form} Let $\Cal S$ be a
sublattice of $\Cal T$,
$\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ a jump-free integrator, and
$\pmb v^*$ its quadratic variation.
If $h:\Bbb R\to\Bbb R$ is a twice-differentiable
function with continuous second derivative, then

\Centerline{$\int_{\Cal S}\bar h'\pmb v\,d\pmb v
+\Bover12\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*$}

\noindent is defined and equal to
$\bar h(v_{\uparrow})-\bar h(v_{\downarrow})$, where

\Centerline{$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$,
\quad$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$.}

\medskip

\noindent{\bf Remark} In the formula above,
$\bar h':L^0\to L^0$ and $\bar h'':L^0\to L^0$ should
be read as $\overline{h'}$ and $\overline{h''}$.

\proof{{\bf (a)} We can note straight away that $\pmb v$ is moderately
oscillatory (616Ib), so $\bar h'\pmb v$ and
$\bar h''\pmb v$ are moderately oscillatory (615F(a-ii)),
while $\bar h\pmb v$ and $\pmb v^*$ are integrators (616O, 617I),
so that the integrals
$\int_{\Cal S}d(\bar h\pmb v)$,
$\int_{\Cal S}\bar h'\pmb v\,d\pmb v$ and
$\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*$ are defined
(616K);  moreover,

\Centerline{$\int_{\Cal S}d(\bar h\pmb v)
=\lim_{\sigma\uparrow\Cal S}\bar h(v_{\sigma})
   -\lim_{\sigma\downarrow\Cal S}\bar h(v_{\sigma})
=\bar h(v_{\uparrow})-\bar h(v_{\downarrow})$}

\noindent (613N, 613Bb), and

\Centerline{$\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*
=\int_{\Cal S}\bar h''\pmb v\,(d\pmb v)^2$}

\noindent (617I again).

\medskip

{\bf (b)} For the time being (down to the end of (c)), suppose that
%$\Cal S$ has greatest and least elements and
$h''$ is uniformly continuous.   Let $\epsilon>0$.
Let $\eta>0$ be such that
$\theta(S_I(\pmb u,(d\pmb v)^2))\le\epsilon$ whenever $I\in\Cal I(\Cal S)$,
$\pmb u\in\Mfa(I)$ and $\theta(\sup|\pmb u|)\le\eta$.
Then there is a
$\delta>0$ such that $|h''(\alpha)-h''(\beta)|\le\eta$ whenever
$|\alpha-\beta|\le 2\delta$.   Now take any such $\alpha$ and $\beta$.
By Taylor's theorem with remainder,
there is a $\gamma$ lying between $\alpha$ and $\beta$ such that

\Centerline{$h(\beta)=h(\alpha)+(\beta-\alpha)h'(\alpha)
+\Bover12(\beta-\alpha)^2h''(\gamma)$,}

\noindent so that

\Centerline{$|h(\beta)-h(\alpha)-(\beta-\alpha)h'(\alpha)
-\Bover12(\beta-\alpha)^2h''(\alpha)|\le\eta(\beta-\alpha)^2$.}

\noindent It follows that if $w$, $w'\in L^0$
then $\Bvalue{|w'-w|\le\delta}$ is included in

\Centerline{$\Bvalue{|\bar h(w')-\bar h(w)-\bar h'(w)\times(w'-w)
-\Bover12\bar h''(w)\times(w'-w)^2|\le\eta(w'-w)^2}$.}

\medskip

{\bf (c)} Let $J\in\Cal I(\Cal S)$ be such that
$\theta(z)\le\delta\epsilon$, where $z=\Osclln^*_J(\pmb v)$ (618Bb).
Take any $I\in\Cal I(\Cal S)$ such that $I\supseteq J$.   Then
$\Osclln_I(\pmb v)\le z$, and $a=\Bvalue{z\le\delta}$ has measure
at least $1-\epsilon$.   Now if $e=c(\sigma,\sigma')$ is an $I$-cell,
and we set

$$\eqalign{y_e
&=\Delta_e(\pmb{1},d(\bar h\pmb v))
   -\Delta_e(\bar h'\pmb v,d\pmb v)
   -\Bover12\Delta_e(\bar h''\pmb v,(d\pmb v)^2)\cr
&=\bar h(v_{\tau})-\bar h(v_{\sigma})
   -\bar h'(v_{\sigma})\times(v_{\tau}-v_{\sigma})
   -\Bover12\bar h''(v_{\sigma})\times(v_{\tau}-v_{\sigma})^2,\cr}$$

\noindent we have

\Centerline{$|v_{\tau}-v_{\sigma}|=\Delta_e(\tbf 1,|d\pmb v|)
\le\Osclln_I(\pmb v)\le z$}

\noindent and

\Centerline{$a\Bsubseteq\Bvalue{|v_{\tau}-v_{\sigma}|\le\delta}
\Bsubseteq\Bvalue{|y_e|\le\eta(v_{\tau}-v_{\sigma})^2}
=\Bvalue{|y_e|\le\Delta_e(\eta\tbf 1,(d\pmb v)^2}$.}

\noindent Summing over $e$,

$$\eqalign{a
&\Bsubseteq\Bvalue{\sum_{e\in\Sti_0(I)}|y_e|
\le\sum_{e\in\Sti_0(I)}\Delta_e(\eta\tbf 1,(d\pmb v)^2)}\cr
&\Bsubseteq\Bvalue{|\sum_{e\in\Sti_0(I)}y_e|
\le\sum_{e\in\Sti_0(I)}\Delta_e(\eta\tbf 1,(d\pmb v)^2)}\cr
&=\Bvalue{|S_I(\tbf 1,d(\bar h\pmb v))
  -S_I(\bar h'\pmb v,d\pmb v)-S_I(\bar h''\pmb v,(d\pmb v)^2|
 \le S_I(\eta\tbf 1,(d\pmb v)^2)},\cr}$$

\noindent and

\Centerline{$\theta\bigl(S_I(\tbf 1,d(\bar h\pmb v))
  -S_I(\bar h'\pmb v,d\pmb v)-S_I(\bar h''\pmb v,(d\pmb v)^2)\bigr)
\le\bar\mu(1\Bsetminus a)+\bar\theta(\eta S_I(\tbf 1,(d\pmb v)^2))
\le 2\epsilon$}

\noindent by the choice of $J$ and $\eta$.   And this is true
whenever $I\supseteq J$.   Accordingly, taking the limit as
$I\uparrow\Cal I(\Cal S)$, we have

$$\eqalign{\bar h(v_{\uparrow})-\bar h(v_{\downarrow})
   &-\int_{\Cal S}\bar h'\pmb v\,d\pmb v
   -\Bover12\int_{\Cal S}\bar h''\pmb v),(d\pmb v)^2\cr
&=\int_{\Cal S}d(\bar h\pmb v)
   -\int_{\Cal S}\bar h'\pmb v\,d\pmb v
   -\Bover12\int_{\Cal S}\bar h''\pmb v\,(d\pmb v)^2\cr
&=\lim_{I\uparrow\Cal I(\Cal S)}
  S_I(\tbf 1,d(\bar h\pmb v))
    -S_I(\bar h'\pmb v,d\pmb v)-S_I(\bar h''\pmb v,(d\pmb v)^2)\cr
&=0,\cr}$$

\noindent and

\Centerline{$\int_{\Cal S}\bar h'\pmb v\,d\pmb v
   +\Bover12\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*
=\int_{\Cal S}\bar h'\pmb v\,d\pmb v
   +\Bover12\int_{\Cal S}\bar h''\pmb v\,(d\pmb v)^2
=\bar h(v_{\uparrow})-\bar h(v_{\downarrow})$,}

\noindent as required.

\medskip

{\bf (d)} This deals with the case in which $h''$ is uniformly continuous.
For the general case, recall that there is a $\bar v\in L^0(\frak A)$ such
that $|v_{\sigma}|\le\bar v$ for every $\sigma\in\Cal S$ (616Ib).
Take any $M\ge 0$.   There is a smooth function
$f:\Bbb R\to\Bbb R$ with compact support such that $f(x)=1$ for every
$x\in[-M,M]$.   \Prf\ By 473Eb there is a smooth function
$\tilde h_0:\Bbb R\to\Bbb R$ such that $\tilde h_0(x)=0$ if $|x|\ge 1$ and
the Lebesgue integral $\int_{-1}^1\tilde h_0(t)dt$ is equal to $1$.
Take $f$ to be the convolution $\tilde h_0*\chi[-M-1,M+1]$;  by
473De, $f$ is smooth, while $f(x)=\int_{-1}^1\tilde h_0(t)dt=1$ if
$|x|\le M$ and $f(x)=0$ if $|x|\ge M+2$.\ \Qed

Consider $g=f\times h$.   Then $g$ is twice differentiable with
continuous second derivative, while
$g'(x)=h'(x)$ and $g''(x)=h''(x)$ whenever $|x|<M$.
Since $g''$, like $f$ and $g$, has compact
support, it is uniformly continuous (4A2Jf).   Accordingly

\Centerline{$\bar g(v_{\uparrow})-\bar g(v_{\downarrow})
=\int_{\Cal S}\bar g'\pmb v\,d\pmb v
  +\Bover12\int_{\Cal S}\bar g''\pmb v\,d\pmb v^*$.}

\noindent So

$$\eqalignno{&\Bvalue{\bar h(v_{\uparrow})-\bar h(v_{\downarrow})
  \ne\int_{\Cal S}\bar h'\pmb v\,d\pmb v
      +\Bover12\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*}\cr
&\mskip100mu\Bsubseteq\Bvalue{\bar h(v_{\uparrow})\ne\bar g(v_{\uparrow})}
    \Bcup\Bvalue{\bar h(v_{\downarrow})\ne\bar g(v_{\downarrow})}
   \Bcup\Bvalue{\int_{\Cal S}\bar h'\pmb v\,d\pmb v
      \ne\int_{\Cal S}\bar g'\pmb v\,d\pmb v}\cr
&\mskip250mu
   \Bcup\Bvalue{\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*
      \ne\int_{\Cal S}\bar g''\pmb v\,d\pmb v^*}\cr
&\mskip100mu\Bsubseteq\Bvalue{\bar h\pmb v\ne\bar g\pmb v}
   \Bcup\Bvalue{\bar h'\pmb v\ne\bar g'\pmb v}
   \Bcup\Bvalue{\bar h''\pmb v\ne\bar g''\pmb v}\cr
\displaycause{613Jd twice}
&\mskip100mu\Bsubseteq\sup_{\sigma\in\Cal S}\Bvalue{|v_{\sigma}|\ge M}
\Bsubseteq\Bvalue{\bar v\ge M}.\cr}$$

\noindent Letting $M\to\infty$,

\Centerline{$\Bvalue{\bar h(v_{\uparrow})-\bar h(v_{\downarrow})
  \ne\int_{\Cal S}\bar h'\pmb v\,d\pmb v
      +\Bover12\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*}
\Bsubseteq\inf_{M\ge 0}\Bvalue{\bar v\ge M}
=0$}

\noindent and

\Centerline{$\bar h(v_{\uparrow})-\bar h(v_{\downarrow})
  =\int_{\Cal S}\bar h'\pmb v\,d\pmb v
      +\Bover12\int_{\Cal S}\bar h''\pmb v\,d\pmb v^*$}

\noindent in this case also.
}%end of proof of 619C

\leader{619D}{It\^o's Formula, second form} Let $\Cal S$ be a
sublattice of $\Cal T$, and
$\pmb v$ a jump-free integrator with domain $\Cal S$
and quadratic variation $\pmb v^*$.
If $\pmb u$ is a moderately oscillatory process with domain $\Cal S$, and
$h:\Bbb R\to\Bbb R$ is a twice-differentiable
function with continuous second derivative, then

\Centerline{$\int_{\Cal S}\pmb u\,d(\bar h\pmb v)
=\int_{\Cal S}\pmb u\times\bar h'\pmb v\,d\pmb v
+\Bover12\int_{\Cal S}\pmb u\times\bar h''\pmb v\,d\pmb v^*$.}

\proof{ Applying 619C to $\Cal S\wedge\tau$, we have

\Centerline{$\bar h(v_{\tau})-\bar h(v_{\downarrow})
  =\int_{\Cal S\wedge\tau}\bar h'\pmb v\,d\pmb v
      +\Bover12\int_{\Cal S\wedge\tau}\bar h''\pmb v\,d\pmb v^*$}

\noindent for every $\tau\in\Cal S$;  that is, setting

\Centerline{$\pmb z=\bar h\pmb v
   -\iiv(\bar h'\pmb v)-\Bover12 ii_{\pmb v^*}(\bar h''\pmb v)$,}

\noindent $\pmb z=\bar h(v_{\downarrow})\tbf 1\restr\Cal S$.
But recalling that

\Centerline{$\bar h\pmb v$,
\quad$\iiv(\bar h'\pmb v)$,
\quad$ii_{\pmb v^*}(\bar h''\pmb v)$}

\noindent are all integrators (616O, 616J), it follows that

\Centerline{$\int_{\Cal S}\pmb u\,d(\bar h\pmb v)
   -\int_{\Cal S}\pmb u\,d(\iiv(\bar h'\pmb v))
   -\Bover12\int_{\Cal S}\pmb u\,d(ii_{\pmb v^*}(\bar h''\pmb v))
=\int_{\Cal S}\pmb u\,d\pmb z=0$.}

\noindent Now applying 617E this becomes

$$\eqalign{\int_{\Cal S}\pmb u\,d(\bar h\pmb v)&
=\int_{\Cal S}\pmb u\,d(\iiv(\bar h'\pmb v))
   +\Bover12\int_{\Cal S}\pmb u\,d(ii_{\pmb v^*}(\bar h''\pmb v))\cr&
=\int_{\Cal S}\pmb u\times\bar h'\pmb v\,d\pmb v
   +\Bover12\int_{\Cal S}\pmb u\times\bar h''\pmb v\,d\pmb v^*,\cr}$$

\noindent as claimed.
}%end of proof of 619D

\leader{619E}{}\cmmnt{ Of course It\^o's formula also has a
multidimensional version.   There are no really new ideas needed,
and ordinarily I leave such adaptations to the exercises, but there are
points below where we can use two-dimensional formulae to tell us
interesting things about one-dimensional processes, so I spell things out
here.   The first step will have to be to establish an interpretation of
$\bar h(v_1,\ldots,v_k)$ where $h:\BbbR^k\to\Bbb R$
is a Borel measurable function and
$v_1,\ldots,v_k\in L^0$.   Since I did leave this to the exercises in
\S364, I give the details now.

\medskip

\noindent}{\bf Proposition} Let $k\ge 1$ be an integer.

(a) Suppose that
$u_1,\ldots,u_k\in L^0$.   Let $\Cal B_k$ be the Borel $\sigma$-algebra of
$\BbbR^k$.   Then there is a unique sequentially order-continuous Boolean
homomorphism $\phi:\Cal B_k\to\frak A$ such that
$\phi\{(\xi_1,\ldots,\xi_k):\xi_i>\alpha\}=\Bvalue{u_i>\alpha}$ whenever
$1\le i\le k$ and $\alpha\in\Bbb R$.

In this context, write $\Bvalue{(u_1,\ldots,u_k)\in E}$ for
$\phi E$, for every Borel set $E\subseteq\BbbR^k$.

(b) Suppose that $h:\BbbR^k\to\Bbb R$ is a Borel measurable function.
Then there is a unique operator $\bar h:(L^0)^k\to L^0$ such that
$\Bvalue{\bar h(u_1,\ldots,u_k)\in F}
=\Bvalue{(u_1,\ldots,u_k)\in h^{-1}[F]}$
whenever $F\subseteq\Bbb R$ is a Borel set and $u_1,\ldots,u_k\in L^0$.

(c) If $u_1,\ldots,u_k,v_1,\ldots,v_k\in L^0$, then

\Centerline{$\inf_{1\le i\le k}\Bvalue{u_i=v_i}
\Bsubseteq\Bvalue{\bar h(u_1,\ldots,u_k)=\bar h(v_1,\ldots,v_k)}$.}

(d) If $h:\BbbR^k\to\Bbb R$ is continuous, then $\bar h:(L^0)^k\to L^0$ is
continuous for the topology of convergence in measure.

(e) Suppose that
$\Omega$ is a set, $\Sigma$ is a $\sigma$-algebra of subsets of
$\Omega$, $\Cal N$ is a $\sigma$-ideal of $\Sigma$, and $\frak A$ is
isomorphic to the
quotient Boolean algebra $\Sigma/\Cal N$.   Write $\eusm L^0$ for the
$f$-algebra of real-valued $\Sigma$-measurable functions on $\Omega$, and
$\eusm W$ for the ideal

\Centerline{$\{f:f\in\eusm L^0$, $\{\omega:f(\omega)\ne 0\}\in\Cal N\}$,}

\noindent so that $L^0$ can be identified with the $f$-algebra
quotient $\eusm L^0/\eusm W$\cmmnt{ (364Ib)}.
Write $E\mapsto E^{\ssbullet}:\Sigma\to\frak A$ and
$f\mapsto f^{\ssbullet}:\eusm L^0\to L^0$ for the homomorphisms
corresponding to the identifications $\frak A\cong\Sigma/\Cal N$ and
$L^0\cong\eusm L^0/\eusm W$.   Then if
$h:\BbbR^k\to\Bbb R$ is a Borel measurable function,

\Centerline{$\bar h(f_1^{\ssbullet},\ldots,f_k^{\ssbullet})
=(h(f_1,\ldots,f_k))^{\ssbullet}$}

\noindent for all $f_1,\ldots,f_k\in\eusm L^0$, defining the composition
$h(f_1,\ldots,f_k)$ by setting
$(h(f_1,\ldots,f_k))(\omega)=h(f_1(\omega),\ldots,f_k(\omega))$ for every
$\omega\in\Omega$.

(f) Suppose that $\sequencen{h_n}$ is a non-decreasing sequence of Borel
measurable functions from $\BbbR^k$ to $\Bbb R$, and that
$h(x)=\sup_{n\in\Bbb N}h_n(x)$ is finite for every $x\in\BbbR^k$.
Then $\sequencen{\bar h_n(u_1,\ldots,u_k)}$ is a non-decreasing sequence in
$L^0$ with supremum $\bar h(u_1,\ldots,u_k)$, for all
$u_1,\ldots,u_k\in L^0$.

(g) Now suppose that $(\frak C,\nu)$ is another probability algebra and
$\phi:\frak A\to\frak C$ is an order-continuous Boolean homomorphism.
Let $T_{\phi}:L^0(\frak A)\to L^0(\frak C)$ be the corresponding
$f$-algebra homomorphism\cmmnt{ (612Bf)}.
Take $u_1,\ldots,u_k\in L^0(\frak A)$.

\quad(i) If $E\in\Cal B_k$ is a Borel set, then
$\Bvalue{(T_{\phi}u_1,\ldots,T_{\phi}u_k)\in E}
=\phi\Bvalue{(u_1,\ldots,u_k)\in E}$.

\quad(ii) If $h:\BbbR^k\to\Bbb R$ is Borel measurable, then
$\bar h(T_{\phi}u_1,\ldots,T_{\phi}u_k)=T_{\phi}\bar h(u_1,\ldots,u_k)$.

\proof{ Since there is a measurable space with negligibles
$(\Omega,\Sigma,\Cal N)$, as in (e), such that $\frak A\cong\Sigma/\Cal N$
(314M), we can suppose from the beginning that $\frak A$ is actually such a
quotient;  at some point, of course, we shall have to check that it
won't matter which such representation is chosen.

\medskip

{\bf (a)(i)}
If $u_1,\ldots,u_k\in L^0$, let $f_1,\ldots,f_k\in\eusm L^0$ be
such that $f_i^{\ssbullet}=u_i$ for $1\le i\le k$.
If $E\in\Cal B_k$, then
$\{\omega:(f_1(\omega),\ldots,f_k(\omega))\in E\}\in\Sigma$
(121K).   So we can set
$\phi E=\{\omega:(f_1(\omega),\ldots,f_k(\omega))\in E\}^{\ssbullet}
\in\frak A$.

Clearly $E\mapsto\{\omega:(f_1(\omega),\ldots,f_k(\omega))\in E\}:
\Cal B_k\to\Sigma$ is a sequentially order-continuous Boolean homomorphism,
so $\phi$ also is.   And if $1\le i\le k$ and $\alpha\in\Bbb R$,

\Centerline{$\phi\{(\xi_1,\ldots,\xi_k):\xi_i>\alpha\}
=\{\omega:f_i(\omega)>\alpha\}^{\ssbullet}
=\Bvalue{u_i>\alpha}$}

\noindent (364Ib again).

\medskip

\quad{\bf (ii)} Now suppose that $\phi':\Cal B_k\to\frak A$ is another
homomorphism of the same kind.   Let $\Cal J$ be the family of sets of the
form $F=\prod_{i=1}^k\ooint{\alpha_i,\infty}$ where $\alpha_i\in\Bbb R$
for every $i$.   For such a set,

$$\eqalignno{\phi F
&=\{\omega:f_i(\omega)>\alpha_i\text{ for }1\le i\le k\}^{\ssbullet}
=\inf_{1\le i\le k}\{\omega:f_i(\omega)>\alpha_i\}^{\ssbullet}
=\inf_{1\le i\le k}\Bvalue{u_i>\alpha_i}\cr
&=\inf_{1\le i\le k}
  \phi'\{(\xi_1,\ldots,\xi_k):\xi_i>\alpha_i\}
=\phi'(\bigcap_{1\le i\le k}\{(\xi_1,\ldots,\xi_k):\xi_i>\alpha_i\})
=\phi' F.\cr}$$

\noindent So $\phi$ and $\phi'$ agree on $\Cal J$ and therefore on
$\Cal J'=\Cal J\cup\{\BbbR^k\}$.   Set
$\Cal A=\{E:E\in\Cal B_k$, $\phi E=\phi'E\}$.   Then
$\Cal J'\subseteq\Cal A$.   If $E$, $F\in\Cal A$ then

\Centerline{$\phi(E\setminus F)=\phi E\Bsetminus\phi F
=\phi'E\Bsetminus\phi'F=\phi'(E\setminus F)$}

\noindent and $E\setminus F\in\Cal A$;  if $\sequencen{E_n}$ is a
non-decreasing sequence in $\Cal A$ with union $E$,

\Centerline{$\phi E=\phi(\bigcup_{n\in\Bbb N}E_n)
=\sup_{n\in\Bbb N}\phi E_n=\sup_{n\in\Bbb N}\phi'E_n=\phi'E$}

\noindent and $E\in\Cal A$.
Since $\Cal J'$ is closed under intersections, the Monotone Class Theorem in the
form 313Gb tells us that $\Cal A$
must include the $\sigma$-subalgebra of $\Cal B_k$ generated by $\Cal J$.
But this is the whole of $\Cal B_k$, by 121J.   So $\phi'=\phi$.

This shows that $\phi$ is unique.

\medskip

{\bf (b)(i)} Write $\Cal B$ for the Borel $\sigma$-algebra of $\Bbb R$.
Suppose $u_1,\ldots,u_k\in L^0$.
The function $F\to h^{-1}[F]:\Cal B\to\Cal B_k$ is a sequentially
order-continuous Boolean homomorphism,
so $F\mapsto\Bvalue{(u_1,\ldots,u_k)\in h^{-1}[F]}$ also is.   By
364F, there is a unique member of $L^0$, which we may call
$\bar h(u_1,\ldots,u_k)$, such that
$\Bvalue{h(u_1,\ldots,u_k)\in F}=\Bvalue{(u_1,\ldots,u_k)\in h^{-1}[F]}$
for every $F\in\Cal B$.

\medskip

\quad{\bf (ii)} In the present context,
we have an alternative expression for $\bar h$, as follows.   If
$f_1,\ldots,f_k\in\eusm L^0$, then $g=h(f_1,\ldots,f_k)$ belongs to
$\eusm L^0$ (121Kb).   If $F\in\Cal B$ and $u_i=f_i^{\ssbullet}$ for each
$i$, then

$$\eqalignno{\Bvalue{\bar h(u_1,\ldots,u_k)\in F}
&=\Bvalue{(u_1,\ldots,u_k)\in h^{-1}[F]}\cr
&=\{\omega:(f_1(\omega),\ldots,f_k(\omega))\in h^{-1}[F]\}^{\ssbullet}\cr
\displaycause{by the construction in (a)}
&=\{\omega:g(\omega)\in F\}^{\ssbullet}
=\Bvalue{g^{\ssbullet}\in F}.\cr}$$

\noindent So
$\bar h(f_1^{\ssbullet},\ldots,f_k^{\ssbullet})=g^{\ssbullet}
=(h(f_1,\ldots,f_k))^{\ssbullet}$.

\medskip

{\bf (c)} Take $f_i$, $g_i\in\eusm L^0$ such that $f_i^{\ssbullet}=u_i$ and
$g_i^{\ssbullet}=v_i$ for $1\le i\le k$.   Set
$f=h(f_1,\ldots,f_k)$ and $g=h(g_1,\ldots,g_k)$.   Then, using the
expression for $\bar h$ in (b-ii),

$$\eqalign{\Bvalue{\bar h(u_1,\ldots,u_k)=\bar h(v_1,\ldots,v_k)}
&=\Bvalue{f^{\ssbullet}=g^{\ssbullet}}
=\{\omega:f(\omega)=g(\omega)\}^{\ssbullet}\cr
&\Bsupseteq(\bigcap_{1\le i\le k}
   \{\omega:f_i(\omega)=g_i(\omega)\})^{\ssbullet}\cr
&=\inf_{1\le i\le k}
   \{\omega:f_i(\omega)=g_i(\omega)\}^{\ssbullet}
=\inf_{1\le i\le k}\Bvalue{u_i=v_i}.\cr}$$

\medskip

{\bf (d)} Take $U=(u_1,\ldots,u_k)\in(L^0)^k$ and $\epsilon>0$.
Let $M\ge 0$ be such that $\Bvalue{|u_i|\ge M}$ has
measure at most $\epsilon$ for each $i$.
Because $h\restr[-M-1,M+1]^k$ must be
uniformly continuous, there is a $\delta\in\ocint{0,1}$ such that
$|h(y)-h(x)|\le\epsilon$ whenever $x$, $y\in[-M-1,M+1]^k$ and
$y-x\in[-\delta,\delta]^k$.

Suppose that
$V=(v_1,\ldots,v_k)\in(L^0)^k$ is such that
$\theta(v_i-u_i)\le\delta\epsilon$ for every $i\le k$.
Let $f_1,\ldots,f_k,g_1,\ldots,g_k\in\eusm L^0$ be such that
$f_i^{\ssbullet}=u_i$ and $g_i^{\ssbullet}=v_i$ for each $i$.
Then

$$\eqalign{\delta\mu\{\omega:|g_i(\omega)-f_i(\omega)|\ge\delta\}
&\le\Expn(\chi\Omega\wedge|g_i-f_i|)
=\Expn(\chi 1\wedge|v_i-u_i|)\cr
&=\theta(v_i-u_i)
\le\delta\epsilon,\cr}$$

\noindent so $\{\omega:|g_i(\omega)-f_i(\omega)|\ge\delta\}$ has measure
at most $\epsilon$, for each $i$.   Set

\Centerline{$H=\bigcap_{1\le i\le k}\{\omega:|f_i(\omega)|\le M$,
    $|g_i(\omega)-f_i(\omega)|\le\delta\}$.}

\noindent Then $\mu H\ge 1-2k\epsilon$.   For $\omega\in H$, both
$x=(f_1(\omega),\ldots,f_k(\omega))$ and
$y=(g_1(\omega),\ldots,g_k(\omega))$ belong to $[-M-1,M+1]^k$ (because
$\delta\le 1$), and $y-x\in[-\delta,\delta]^k$, so
$|h(x)-h(y)|\le\epsilon$.   But this means that

$$\eqalign{\theta(\bar h(V)-\bar h(U))
&=\Expn(\chi\Omega\wedge|h(f_1,\ldots,f_k)-h(g_1,\ldots,g_k)|)\cr
&\le\epsilon+\mu(\Omega\setminus H)
\le(2k+1)\epsilon.\cr}$$

As $U$ and $\epsilon$ are arbitrary, $\bar h$ is continuous.

\medskip

{\bf (e)} To see that any other isomorphism $\frak A\cong\Sigma'/\Cal N'$,
where $(\Omega',\Sigma',\Cal N')$ is another measurable space with
negligibles, would lead to the same operators $\bar h:(L^0)^k\to L^0$,
note that $\phi E=\Bvalue{(u_1,\ldots,u_k)\in E}$, for $E\in\Cal B_k$ and
$u_1,\ldots,u_k\in L^0$, is determined from the values
$\Bvalue{u_i>\alpha}$ for $1\le i\le k$ and $\alpha\in\Bbb R$, as
shown in (a).   Since (following 364A) each $u_i$ is neither more nor less
than the family $\family{\alpha}{\Bbb R}{\Bvalue{u_i>\alpha}}$, these are
necessarily independent of representations of $\frak A$.
Similarly, (b) above also provides a representation-invariant
description of $\bar h$.

\medskip

{\bf (f)} Taking $f_1,\ldots,f_k\in\eusm L^0$ such that
$f_i^{\ssbullet}=u_i$ for each $i$, we have
$\sequencen{h_n(f_1,\ldots,f_k)}\uparrow h(f_1,\ldots,f_k)$ in $\eusm L^0$,
so $\sequencen{\bar h_n(f_1,\ldots,f_k)}
\uparrow\bar h(f_1,\ldots,f_k)$ in $L^0$.

\medskip

{\bf (g)(i)} Set $\Cal E=\{E:E\in\Cal B_k$,
$\Bvalue{(T_{\phi}u_1,\ldots,T_{\phi}u_k)\in E}
=\phi\Bvalue{(u_1,\ldots,u_k)\in E}\}$.   If $E_i\subseteq\Bbb R$ is
a Borel set for $i\le k$, and $E=\prod_{1\le i\le k}E_i$, then

$$\eqalignno{\Bvalue{(T_{\phi}u_1,\ldots,T_{\phi}u_k)\in E}
&=\inf_{1\le i\le k}\Bvalue{T_{\phi}u_i\in E_i}
=\inf_{1\le i\le k}\phi\Bvalue{u_i\in E_i}
\Displaycause{612Af}
=\phi(\inf_{1\le i\le k}\Bvalue{u_i\in E_i})
=\phi\Bvalue{(u_1,\ldots,u_k)\in E}
\cr}$$

\noindent and $E\in\Cal E$.   Because
$E\mapsto\Bvalue{(u_1,\ldots,u_k)\in E}:\Cal B_k\to\frak A$,
$E\mapsto\Bvalue{(T_{\phi}u_1,\ldots,T_{\phi}u_k)\in E}:
\Cal B_k\to\frak C$ and
$E\mapsto\phi\Bvalue{(u_1,\ldots,u_k)\in E}\}:\Cal E\to\frak C$
are all sequentially order-continuous Boolean homomorphisms,
$\Cal E$ is a Dynkin class.   So
the Monotone Class Theorem (136B) tells us that $\Cal E$ includes
the $\sigma$-subalgebra of $\Cal B_k$ generated by products of Borel
subsets of $\Bbb R$, and is the whole of $\Cal B_k$, as claimed.

\medskip

\quad{\bf (ii)} For any Borel set $E\subseteq\Bbb R$,

$$\eqalignno{\Bvalue{\bar h(T_{\phi}u_1,\ldots,T_{\phi}u_k)\in E}
&=\Bvalue{(T_{\phi}u_1,\ldots,T_{\phi}u_k)\in h^{-1}[E]}
\Displaycause{612Ac}
=\phi\Bvalue{(u_1,\ldots,u_k)\in h^{-1}[E]}
\Displaycause{(i) just above}
=\phi\Bvalue{\bar h(u_1,\ldots,u_k)\in E}
=\Bvalue{T_{\phi}\bar h(u_1,\ldots,u_k)\in E};
\cr}$$

\noindent as $E$ is arbitrary,
$\bar h(T_{\phi}u_1,\ldots,T_{\phi}u_k)=T_{\phi}\bar h(u_1,\ldots,u_k)$.
}%end of proof of 619E

\leader{619F}{Definition} Let $\pmb u_1,\ldots,\pmb u_k$ be fully
adapted processes defined on sublattices $\Cal S_1,\ldots,\Cal S_k$
of $\Cal T$
and $h:\BbbR^k\to\Bbb R$ a Borel measurable function.   Regarding
$\pmb U=(\pmb u_1,\ldots,\pmb u_k)$ as the function
$\sigma\mapsto(u_{1\sigma},\ldots,u_{k\sigma}):\Cal S\to(L^0)^k$,
where $\pmb u_i=\family{\sigma}{\Cal S_i}{u_{i\sigma}}$ for each $i$ and
$\Cal S=\bigcap_{1\le i\le k}\Cal S_i$, we have a composition

\Centerline{$\bar h\pmb U
=\family{\sigma}{\Cal S}{\bar h(u_{1\sigma},\ldots,u_{k\sigma})}
$.}

\leader{619G}{Proposition} Let $\pmb u_1,\ldots,\pmb u_k$ be fully
adapted processes all with the same domain $\Cal S$,
and $h:\BbbR^k\to\Bbb R$ a Borel measurable function.
Write $\pmb{U}$ for $(\pmb u_1,\ldots,\pmb u_k)$.

(a) $\bar h\pmb{U}$ is fully adapted.

(b) If every $\pmb u_i$ is order-bounded and $h$ is {\bf locally bounded},
that is, bounded on bounded
subsets of $\BbbR^k$, then $\bar h\pmb{U}$ is order-bounded.

(c) If every $\pmb u_i$ is (locally) moderately oscillatory
and $h$ is continuous,
then $\bar h\pmb{U}$ is (locally) moderately oscillatory.

(d) If every $\pmb u_i$ is (locally) jump-free and $h$ is continuous, then
$\bar h\pmb{U}$ is (locally) jump-free.

*(e) If $\pmb z$ is a fully adapted process with domain $\Cal S$ and
$\pmb z^2=\pmb z$, then

\quad(i) $\pmb z\times\bar h(\pmb z\times\pmb u_1,\ldots,\pmb z\times\pmb u_k)
=\pmb z\times\bar h\pmb U$,

\quad(ii) and if $h(0,\ldots,0)=0$, then
$\bar h(\pmb z\times\pmb u_1,\ldots,\pmb z\times\pmb u_k)
=\pmb z\times\bar h\pmb U$.

\proof{ If $\Cal S$ is empty, all these are trivially true;  suppose that
$\Cal S\ne\emptyset$.   Express each $\pmb u_i$ as
$\family{\sigma}{\Cal S}{u_{i\sigma}}$ and set
$U_{\sigma}=(u_{1\sigma},\ldots,u_{k\sigma})$ for
$\sigma\in\Cal S$.

\medskip

{\bf (a)} If $\sigma$, $\tau\in\Cal S$,

$$\eqalignno{\Bvalue{\sigma=\tau}
&\Bsubseteq\inf_{1\le i\le k}\Bvalue{u_{i\sigma}=u_{i\tau}}
\Bsubseteq\Bvalue{\bar h(u_{1\sigma},\ldots,u_{k\sigma})
  =\bar h(u_{1\tau},\ldots,u_{k\tau})}\cr
\displaycause{619Ec}
&=\Bvalue{\bar h(U_{\sigma})=\bar h(U_{\tau})}.\cr}$$

\noindent So $\bar h\pmb{U}=\family{\sigma}{\Cal S}{\bar h(U_{\sigma})}$
is fully adapted.

\medskip

{\bf (b)} (Cf.\ 614F(c-i).) For $x\in\BbbR^k$, set
$g(x)=\sup_{\|y\|\le\|x\|}|h(y)|$.   Then $g$ is Borel
measurable.   If $\bar u_i=\sup_{\sigma\in\Cal S}|u_{i\sigma}|$ for
$1\le i\le k$, then $\bar g(\bar u_1,\ldots,\bar u_k)$ is an upper bound
for $\{\bar h(U_{\sigma}):\sigma\in\Cal S\}$.

\medskip

{\bf (c)} (Cf.\ 615F(a-ii))
Write $\hat\Cal S$ for the covered envelope of
$\Cal S$, and for $1\le i\le k$ let
$\hatpmb u_i=\family{\tau}{\hat\Cal S}{\hat u_{i\tau}}$
be the fully adapted extension
of $\pmb u_i$ to $\hat\Cal S$.   If every $\pmb u_i$ is moderately oscillatory
and $\sequencen{\tau_n}$ is a monotonic sequence in $\hat\Cal S$, then
$\lim_{n\to\infty}\hat u_{i\tau_n}$ exists for every $i$;  since
$\bar h:(L^0)^k\to L^0$ is continuous (619Ec),
$\lim_{n\to\infty}\bar h(\hat U_{\tau_n})$ exists, where
$\hat U_{\tau}=(\hat u_{1\tau},\ldots,\hat u_{k\tau})$ for $\tau\in\hat\Cal S$.
By 615N, $\bar h\hatpmb U=\family{\tau}{\hat\Cal S}{\bar h(\hat U_{i\tau})}$
is moderately oscillatory.
But $\bar h\hatpmb U$ extends $\bar h\pmb U$, so $\bar h\pmb U$ is moderately
oscillatory, by 615F(a-i).

The result for locally moderately oscillatory processes follows at once.

\medskip

{\bf (d)(i)} (Cf.\ 618Ga.) Suppose first
that every $\pmb u_i$ is jump-free.   We know from (b) that
$\bar h\pmb{U}$ is order-bounded.   Set

\Centerline{$\bar w
  =\sup_{\sigma\in\Cal S}\sup_{1\le i\le k}|u_{i\sigma}|$}

\Centerline{$\bar w'
  =\sup_{\sigma\in\Cal S}|\bar h(U_{\sigma})|$.}

Let $\epsilon>0$.   Let $M\ge 0$ be such
that $\bar\mu\Bvalue{\bar w\ge M}\le\epsilon$, and $\delta>0$ such that
$|h(y)-h(x)|\le\epsilon$ whenever $x$, $y\in[-M,M]^k$
and $x-y\in[-\delta,\delta]^k$.   For $1\le i\le k$, let
$I_i\in\Cal I(\Cal S)$ be such that
$\theta(\Osclln^*_{I_i}(\pmb u_i))\le\epsilon\delta$;  set

\Centerline{$a=\Bvalue{\bar w\ge M}\Bcup\sup_{1\le i\le k}
   \Bvalue{\Osclln^*_{I_i}(\pmb u_i))\ge\delta}$,}

\noindent so that $\bar\mu a\le(k+1)\epsilon$.   Let $I$ be the sublattice
generated by $\bigcup_{1\le i\le k}I_i$.   Then
$\Osclln^*_I(\bar h\pmb{U})\le\epsilon\chi 1+2\bar w'\times\chi a$.
\Prf\ Suppose that $J\in\Cal I(\Cal S)$ and $J\supseteq I$.   Take
$e=c(\sigma,\tau)\in\Sti_0(J)$.   Then

$$\eqalign{\Bvalue{|\bar h(U_{\tau})-\bar h(U_{\sigma})|>\epsilon}
&\Bsubseteq\Bvalue{\bar w\ge M}
  \Bcup\sup_{1\le i\le k}\Bvalue{|u_{i\tau}-u_{i\sigma}|\ge\delta}\cr
&\Bsubseteq\Bvalue{\bar w\ge M}
  \Bcup\sup_{1\le i\le k}\Bvalue{\Osclln_J(\pmb u_i)\ge\delta}\cr
&\Bsubseteq\Bvalue{\bar w\ge M}
  \Bcup\sup_{1\le i\le k}\Bvalue{\Osclln^*_I(\pmb u_i)\ge\delta}
=a,\cr}$$

\noindent and

\Centerline{$|\bar h(U_{\tau})-\bar h(U_{\sigma})|
  \le\epsilon\chi 1+2\bar w'\times\chi a$.}

\noindent As $e$ is arbitrary,
$\Osclln^*_J(\bar h\pmb{U})\le\epsilon\chi 1+2\bar w'\times\chi a$;
as $J$ is arbitrary,
$\Osclln^*_I(\bar h\pmb{U})\le\epsilon\chi 1+2\bar w'\times\chi a$.\
\Qed

It follows that

\Centerline{$\theta(\Osclln^*_I(\bar h\pmb{U}))\le\epsilon+\bar\mu a
\le(k+2)\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\bar h\pmb{U}$ is jump-free.

\medskip

\quad{\bf (ii)} As in (c), it follows at once that
if every $\pmb u_i$ is locally jump-free, then $\bar h\pmb U$ is locally
jump-free.

\medskip

{\bf (e)(i)} Express $\pmb z$ as $\sfamilysS{z}$.   Setting
$d_{\sigma}=\Bvalue{z_{\sigma}=1}$, we see that
that $1\Bsetminus d_{\sigma}=\Bvalue{z_{\sigma}=0}$ for each
$\sigma\in\Cal S$.   If $\sigma\in\Cal S$,

$$\eqalign{1\Bsetminus d_{\sigma}
&\Bsubseteq\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
u_{k\sigma}\times z_{\sigma})\times z_{\sigma}=0}
  \Bcap\Bvalue{\bar h(u_{1\sigma},\ldots,u_{k\sigma})
     \times z_{\sigma}=0}
\cr&\Bsubseteq\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
u_{k\sigma}\times z_{\sigma})\times z_{\sigma}
  =\bar h(u_{1\sigma},\ldots,u_{k\sigma})
     \times z_{\sigma}},
\cr d_{\sigma}
&\Bsubseteq\inf_{1\le i\le k}
  \Bvalue{u_{i\sigma}\times z_{\sigma}=u_{i\sigma}}
\cr&\Bsubseteq\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
  u_{k\sigma}\times z_{\sigma})=\bar h(u_{1\sigma},\ldots,u_{k\sigma})}.
\cr}$$

\noindent So
$\bar h(u_{1\sigma}\times z_{\sigma},\ldots,u_{k\sigma}\times z_{\sigma})
  \times z_{\sigma}
=\bar h(u_{1\sigma},\ldots,u_{k\sigma})\times z_{\sigma}$;  as
$\sigma$ is arbitrary,
$\pmb z\times\bar h(\pmb z\times\pmb u_1,\ldots,\pmb z\times\pmb u_k)
=\pmb z\times\bar h(\pmb u_1,\ldots,\pmb u_k)$.

\medskip

\quad{\bf (ii)} Now suppose in addition that $h(0,\ldots,0)=0$.   Then for each
$\sigma\in\Cal S$ we have

$$\eqalign{1\Bsetminus d_{\sigma}&
=\Bvalue{z_{\sigma}=0}
   \Bcap\inf_{1\le i\le k}\Bvalue{u_{i\sigma}\times z_{\sigma}=0}\cr&
=\Bvalue{z_{\sigma}=0}\Bcap\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
    u_{k\sigma}\times z_{\sigma})=0}\cr&
\Bsubseteq\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
      u_{k\sigma}\times z_{\sigma})
  =\bar h(u_{1\sigma},\ldots,u_{k\sigma})\times z_{\sigma}}\cr}$$

\noindent and

$$\eqalign{d_{\sigma}&
\Bsubseteq\Bvalue{z_{\sigma}=\chi 1}
   \Bcap\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
  u_{k\sigma}\times z_{\sigma})=\bar h(u_{1\sigma},\ldots,u_{k\sigma})}\cr&
\Bsubseteq\Bvalue{\bar h(u_{1\sigma}\times z_{\sigma},\ldots,
  u_{k\sigma}\times z_{\sigma})
   =\bar h(u_{1\sigma},\ldots,u_{k\sigma})\times z_{\sigma}},\cr}$$

\noindent so that
$\bar h(\pmb z\times\pmb u_1,\ldots,\pmb z\times\pmb u_k)
=\pmb z\times\bar h(\pmb u_1,\ldots,\pmb u_k)$.
}%end of proof of 619G

\leader{619H}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, $k\ge 1$
an integer and $h:\BbbR^k\to\Bbb R$ a continuous function.   Then
$\bar h:\Mob(\Cal S)^k\to\Mob(\Cal S)$ is continuous when $\Mob(\Cal S)$ is
given the ucp topology and $\Mob(\Cal S)^k$ the corresponding product topology.
%\query out of order

\proof{ (Cf.\ 615Ca.) Take $\pmb V=(\pmb v_1,\ldots,\pmb v_k)\in\Mob^k$, where
$\Mob=\Mob(\Cal S)$, and $\epsilon>0$.
Set $\bar v=\sup_{1\le i\le k}\sup|\pmb v_i|$, and let $M\ge 0$ be such that
$\bar\mu\Bvalue{\bar v>M}\le\epsilon$.   Let $\delta\in\ocint{0,1}$ be such that
$|h(x)-h(y)|\le\delta$ whenever $x$, $y\in\BbbR^k$, $\|x\|_{\infty}\le M+1$,
$\|y\|_{\infty}\le M+1$ and $\|x-y\|_{\infty}\le\delta$.   Take any
$\pmb U=(\pmb u_1,\ldots,\pmb u_k)\in\Mob^k$ such that
$\theta(\sup|\pmb u_i-\pmb v_i|)\le\epsilon\delta$ for every $i$.
Set $\bar u=\sup_{1\le i\le k}\sup|\pmb u_i-\pmb v_i|$ and
$\bar w=\sup|\bar h\pmb U-\bar h\pmb V|$.   Then

\Centerline{$\bar\mu\Bvalue{\bar u>\delta}
\le\sum_{i=1}^k\bar\mu\Bvalue{\sup|\pmb u_i-\pmb v_i|>\delta}
\le k\epsilon$,}

\noindent so, expressing $\pmb u_i$ as $\family{\sigma}{\Cal S}{u_{i\sigma}}$
and $\pmb v_i$ as $\family{\sigma}{\Cal S}{v_{i\sigma}}$ for each $i$.

$$\eqalign{\Bvalue{\bar w>\epsilon}&
=\sup_{\sigma\in\Cal S}
    \Bvalue{|\bar h(u_{1\sigma},\ldots,u_{k\sigma})
       -\bar h(v_{1\sigma},\ldots,v_{k\sigma})|>\epsilon}\cr&
\Bsubseteq\sup_{\Atop{1\le i\le k}{\sigma\in\Cal S}}
   \Bvalue{|v_{i\sigma}|>M}\Bcup\Bvalue{|u_{i\sigma}-v_{i\sigma}|>\delta}
=\Bvalue{\bar v>M}\Bcup\Bvalue{\bar u>\delta}\cr}$$

\noindent has measure at most $(k+1)\epsilon$. and

\Centerline{$\theta(\sup|\bar h\pmb U-\bar h\pmb V|)=\theta(\bar w)
\le(k+2)\epsilon$.}

\noindent As $\pmb V$ and $\epsilon$ are arbitrary, $\bar h:\Mob^k\to\Mob$ is
continuous.
}%end of proof of 619H

\leader{619I}{Theorem} Let $h:\BbbR^k\to\Bbb R$ be a differentiable
function;  write $h_1,\ldots,h_k$ for its partial derivatives.
Suppose that every $h_i$ is Lipschitz on every
bounded set in $\BbbR^k$.   Let $\pmb v_1,\ldots,\pmb v_k$
be integrators, all with the same domain $\Cal S$.
Then $\bar h(\pmb v_1,\ldots,\pmb v_k)$ is an integrator.

\proof{{\bf (a)} Suppose that $M$, $K\ge 0$ are such that
$|h_i(x)-h_i(y)|\le K\|x-y\|$ whenever $1\le i\le k$
and $x$, $y\in[-M,M]^k$.   Then

\Centerline{$|h(y)-h(x)-\sum_{i=1}^k(\eta_i-\xi_i)h_i(x)|
\le K\sqrt{k}\|y-x\|^2$}

\noindent whenever $x=(\xi_1,\ldots,\xi_k)$,
$y=(\eta_1,\ldots,\eta_k)\in[-M,M]^k$.
\Prf\ Set $g(t)=h(ty+(1-t)x)$ for $t\in\Bbb R$.   Then $g$ is
differentiable with

\Centerline{$g'(t)=\sum_{i=1}^k(\eta_i-\xi_i)h_i(ty+(1-t)x)$}

\noindent for every $t$.   By the Mean Value Theorem there
is a $t\in[0,1]$ such that $g(1)-g(0)=g'(t)$, that is,
$h(y)-h(x)=\sum_{i=1}^k(\eta_i-\xi_i)h_i(z)$ where $z=ty+(1-t)x$.   So

$$\eqalign{|h(y)-h(x)-\sum_{i=1}^k(\eta_i-\xi_i)h_i(x)|
&=|\sum_{i=1}^k(\eta_i-\xi_i)(h_i(z)-h_i(x))|\cr
&\le\sqrt{\sum_{i=1}^k(\eta_i-\xi_i)^2}
   \sqrt{\sum_{i=1}^k(h_i(z)-h_i(x))^2}\cr
&\le\sqrt{k}\|y-x\|\max_{1\le i\le k}|h_i(z)-h_i(x)|\cr
&\le K\sqrt{k}\|y-x\|\|z-x\|
\le K\sqrt{k}\|y-x\|^2.  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} Consequently, using 619Ee or otherwise,

$$\eqalign{&|\bar h(w'_1,\ldots,w'_k)-\bar h(w_1,\ldots,w_k)
   -\sum_{i=1}^k(w'_i-w_i)\times\bar h_i(w_1,\ldots,w_k)|\cr
&\mskip100mu
\le K\sqrt{k}\sum_{i=1}^k|w'_i-w_i|^2\cr}$$

\noindent whenever $w_1,\ldots,w_k,w'_1,\ldots,w'_k$ belong to
$L^{\infty}$ and $\|w_i\|_{\infty}\le M$,
$\|w'_i\|_{\infty}\le M$ for every $i$.

If now $I$ is a finite
sublattice of $\Cal T$ and $\pmb u=\family{\sigma}{I}{u_{\sigma}}$,
$\pmb w_i=\family{\sigma}{I}{w_{i\sigma}}$ are fully
adapted processes with $\|\pmb w_i\|_{\infty}\le M$ for $1\le i\le k$,
and we set $\pmb W=(\pmb w_1,\ldots,\pmb w_k)$,
$W_{\sigma}=(w_{1\sigma},\ldots,w_{k\sigma})$ for $\sigma\in I$, then

$$\eqalign{&|\Delta_e(\pmb u,d(\bar h\pmb W))
  -\sum_{i=1}^k\Delta_e(\pmb u\times\bar h_i\pmb W,d\pmb w_i)|\cr
&\mskip100mu
=|u_{\sigma}\times\bigl((\bar h(W_{\tau})-\bar h(W_{\sigma}))
  -\sum_{i=1}^k\bar h_i(W_{\sigma})\times(w_{i\tau}-w_{i\sigma})\bigr)|\cr
&\mskip100mu
\le K\sqrt{k}|u_{\sigma}|\times\sum_{i=1}^k(w_{i\tau}-w_{i\sigma})^2
=K\sqrt{k}\sum_{i=1}^k\Delta_e(|\pmb u|,(d\pmb w_i)^2)\cr}$$

\noindent
whenever $\sigma\le\tau$ in $I$ and $e$ is the stopping-time interval
$c(\sigma,\tau)$.   Summing over the $I$-cells,

\Centerline{$|S_I(\pmb u,d(\bar h\pmb W))
   -\sum_{i=1}^kS_I(\pmb u\times\bar h_i\pmb W,d\pmb w_i)|
\le K\sqrt{k}\sum_{i=1}^kS_I(|\pmb u|,(d\pmb w_i)^2)$.}

\medskip

{\bf (c)} Now suppose that $\epsilon>0$.   Set
$\bar v=\sup_{\sigma\in\Cal S,1\le i\le k}|v_{i\sigma}|$
and let $M\ge 0$ be such that $\bar\mu c\le\epsilon$ where
$c=\Bvalue{\bar v\ge M}$.   Set
$\pmb w_i=\med(-M\tbf 1,\pmb v_i,M\tbf 1)$ for each $i$, so that
$\|\pmb w_i\|_{\infty}\le M$ and
$\Bvalue{\pmb w_i\ne\pmb v_i}\Bsubseteq c$ for $1\le i\le k$;  write
$\pmb V=(\pmb v_1,\ldots,\pmb v_k)$
and $\pmb W=(\pmb w_1,\ldots,\pmb w_k)$.

Let $K\ge 0$ be such that $h_i\restr[-M,M]^k$ is $K$-Lipschitz for each $i$,
and $M'>0$ such that $|h_i(x)|\le M'$ whenever $x\in[-M.M]^k$ and
$1\le i\le k$;  then $\sup|\Bover1{M'}\bar h_i\pmb W|\le\chi 1$ for
each $i$.   Take $\delta>0$ such that

\Centerline{$M'\theta(S_I(\pmb u,d\pmb v_i))\le\epsilon$,
\quad$K\sqrt k\,\theta(S_I(\pmb u,(d\pmb w_i))^2)\le\epsilon$}

\noindent whenever $1\le i\le k$,
$I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and
$\theta(\sup|\pmb u|)\le\delta$ (using 619B again).

Take a finite sublattice $I$ of $\Cal S$ and $\pmb u\in\Mfa(I)$ such that
$\theta(\pmb u)\le\delta$.   Using 613Lc repeatedly, we have

\Centerline{$\Bvalue{S_I(\pmb u,d(\bar h\pmb V))
   \ne S_I(\pmb u,d(\bar h\pmb W)}
\Bsubseteq c$,}

\Centerline{$\Bvalue{S_I(\pmb u\times\bar h_i(\pmb W),d\pmb v_i)
  \ne S_I(\pmb u\times\bar h_i\pmb W,d\pmb w_i)}\Bsubseteq c$,}

\Centerline{$\Bvalue{S_I(|\pmb u|,(d\pmb v_i)^2)
\ne S_I(|\pmb u|,(d\pmb w_i)^2)}\Bsubseteq c$}

\noindent for every $i$, so

$$\eqalign{&\Bvalue{|S_I(\pmb u,d(\bar h\pmb V))
     -\sum_{i=1}^kS_I(\pmb u\times\bar h_i\pmb W,d\pmb v_i)|
  >K\sqrt{k}\sum_{i=1}^kS_I(|\pmb u|,(d\pmb v_i)^2)}\cr
&\mskip40mu
\Bsubseteq c\Bcup\Bvalue{|S_I(\pmb u,d(\bar h\pmb W))
     -\sum_{i=1}^kS_I(\pmb u\times\bar h_i\pmb W,d\pmb w_i)|
   >K\sqrt{k}\sum_{i=1}^kS_I(|\pmb u|,(d\pmb w_i)^2)}\cr
&\mskip40mu
=c,\cr}$$

\noindent and

$$\eqalign{\theta(S_I(\pmb u,d(\bar h\pmb V)))&
\le\sum_{i=1}^k
   M'\theta(S_I(\pmb u\times\Bover1{M'}\bar h_i\pmb W,d\pmb v_i))\cr&
\mskip100mu
  +\sum_{i=1}^kK\sqrt k\theta(S_I(|\pmb u|,(d\pmb v_i)^2))
  +\bar\mu c\cr
&\le (1+2k)\epsilon\cr}$$

\noindent by the choice of $\delta$, because
$\theta(\sup|\pmb u\times\Bover1{M'}\bar h_i\pmb W|)
\le\theta(\sup|\pmb u|)\le\delta$ for each $i$.
And this is true for every finite sublattice $I$ of
$\Cal S$.

As $\epsilon$ is arbitrary, $\bar h\pmb V$ is an integrator.
}%end of proof of 619I

\leader{619J}{It\^o's Formula, third form} Let $k\ge 1$ be an integer, and
$h:\BbbR^k\to\Bbb R$ a twice-differentiable function with continuous second
derivative.   Denote its first partial
derivatives by $h_1,\ldots,h_k$ and its second partial derivatives by
$h_{11},\ldots,h_{kk}$.   Let $\Cal S$ be a sublattice of
$\Cal T$, and $\pmb v_1,\ldots,\pmb v_k$ jump-free
integrators with domain $\Cal S$;  let $\pmb u$ be a moderately oscillatory
process with domain $\Cal S$.   Write
$\pmb V=(\pmb v_1,\ldots,\pmb v_k)$.   Then

\Centerline{$\int_{\Cal S}\pmb u\,d(\bar h\pmb V)
=\sum_{i=1}^k\int_{\Cal S}
   \pmb u\times\bar h_i\pmb V\,d\pmb v_i
      +\Bover12\sum_{i=1}^k\sum_{j=1}^k\int_{\Cal S}
        \pmb u\times\bar h_{ij}\pmb V\,
        d\covar{\pmb v_i}{\pmb v_j}$.}

\proof{ We can follow exactly the same lines as in 619C-619D.

\medskip

{\bf (a)} We know that all the $\pmb v_i$ and $\bar h_i\pmb V$ and
$\bar h_{ij}\pmb V$ are fully adapted and moderately oscillatory
(618Gb, 619Gc), while $\bar h\pmb V$ and
every $\covar{\pmb v_i}{\pmb v_j}$ is an integrator (619I, 617I);  so
all the integrals are well-defined.   Moreover,

\Centerline{$\int_{\Cal S}\pmb u\times\bar h_{ij}\pmb V\,
        d\covar{\pmb v_i}{\pmb v_j}
=\int_{\Cal S}\pmb u\times\bar h_{ij}\pmb V\,
        d\pmb v_id\pmb v_j$}
	
\noindent for $1\le i,j\le k$ (617I, as before).
Express each $\pmb v_i$ as
$\family{\sigma}{\Cal S}{v_{i\sigma}}$, and set
$V_{\sigma}=(v_{1\sigma},\ldots,v_{k\sigma})$ for
$\sigma\in\Cal S$.
	
\medskip

{\bf (b)} For the time being (down to the end of (d)) suppose that
all the $h_{ij}$ are uniformly continuous.   Let $\epsilon>0$.
Let $\delta_0>0$ be such that $\theta(S_I(\pmb z,(d\pmb v_i)^2))\le\epsilon$
whenever $1\le i\le k$, $I\in\Cal I(\Cal S)$, $\pmb z\in\Mfa(I)$ and
$\theta(\sup|\pmb z|)\le\delta_0$.
Then there is a $\delta>0$ such that

$$\eqalign{&|h(y)-h(x)-\sum_{i=1}^k(\eta_i-\xi_i)h_i(x)
  -\Bover12\sum_{i=1}^k\sum_{j=1}^k(\eta_i-\xi_i)(\eta_j-\xi_j)
     h_{ij}(x)|\cr
&\mskip250mu
\le\delta_0\sum_{i=1}^k(\eta_i-\xi_i)^2\cr}$$

\noindent whenever $x=(\xi_1,\ldots,\xi_k)$, $y=(\eta_1,\ldots,\eta_k)$
belong to $\BbbR^k$ and $y-x\in[-\delta,\delta]^k$.

\Prf\ Take $\delta$ such that $|h_{ij}(y)-h_{ij}(x)|\le\Bover{2\delta_0}k$
whenever $1\le i,j\le k$ and $y-x\in[-\delta,\delta]^k$.   Now suppose that
$x=(\xi_1,\ldots,\xi_k)$, $y=(\eta_1,\ldots,\eta_k)$
belong to $\BbbR^k$ and $y-x\in[-\delta,\delta]^k$.   Set
$g(t)=h(ty+(1-t)x)$ for $t\in\Bbb R$.   Then $g$ is twice differentiable
with

$$\eqalign{g'(t)&=\sum_{i=1}^k(\eta_i-\xi_i)h_i(ty+(1-t)x),\cr
g''(t)&=\sum_{i=1}^k\sum_{j=1}^k(\eta_i-\xi_i)(\eta_j-\xi_j)
  h_{ij}(ty+(1-t)x)\cr}$$

\noindent for every $t$.
By Taylor's theorem, there is a $t\in\ooint{0,1}$ such that

\Centerline{$g(1)=g(0)+g'(0)+\Bover12g''(t)$.}

\noindent Setting $z=ty+(1-t)x$, we get

$$\eqalignno{&|h(y)-h(x)-\sum_{i=1}^k(\eta_i-\xi_i)h_i(x)
  -\Bover12\sum_{i=1}^k\sum_{j=1}^k(\eta_i-\xi_i)(\eta_j-\xi_j)
     h_{ij}(x)|\cr
&\mskip50mu
=\Bover12|\sum_{i=1}^k\sum_{j=1}^k(\eta_i-\xi_i)(\eta_j-\xi_j)
     (h_{ij}(z)-h_{ij}(x))|
%     \cr
%&\mskip50mu
\le\Bover{\delta_0}{k}\sum_{i=1}^k\sum_{j=1}^k
  |\eta_i-\xi_i||\eta_j-\xi_j|\cr
\displaycause{because $\|z-x\|_{\infty}=t\|y-x\|_{\infty}\le\delta$}
&\mskip50mu
\le\Bover{\delta_0}{2k}\sum_{i=1}^k\sum_{j=1}^k
  ((\eta_i-\xi_i)^2+(\eta_j-\xi_j)^2)
=\delta_0\sum_{i=1}^k(\eta_i-\xi_i)^2,\cr}$$

\noindent as required.\ \Qed

It follows that if $w_1,\ldots,w_k,w'_0,\ldots,w'_k\in L^0$ then, setting
$W=(w_1,\ldots,w_k)$, $W'=(w'_1,\ldots,w'_k)$,

$$\eqalign{&\bar z=\bar h(W')-\bar h(W)
     -\sum_{i=1}^k(w'_i-w_i)\times\bar h_i(W)\cr&
\mskip170mu-\Bover12\sum_{i=1}^k\sum_{j=1}^k(w'_i-w_i)\times(w'_j-w_j)
     \times\bar h_{ij}(W),\cr}$$

\noindent we shall have

\Centerline{$\Bvalue{|\bar z|>\delta_0\sum_{i=1}^k(w'_i-w_i)^2}
\Bsubseteq\sup_{1\le i\le k}\Bvalue{|w'_i-w_i|>\delta}$.}

\medskip

{\bf (c)} For $1\le i\le k$ let $J_i\in\Cal I(\Cal S)$ be such that
$\theta(\Osclln^*_{J_i}(\pmb v_i))\le\delta\epsilon$,
and take $J$ to be the
sublattice generated by $\bigcup_{1\le i\le k}J_i$.   Set
$a=\sup_{1\le i\le k}\Bvalue{\Osclln^*_{J_i}(\pmb v_i)\ge\delta}$, so
that $\bar\mu a\le k\epsilon$.   If $I\in\Cal I(\Cal S)$, set

\Centerline{$\bar z_I=S_I(\tbf 1,d(\bar h\pmb V))
  -\sum_{i=1}^kS_I(\bar h_i\pmb V,d\pmb v_i)
  -\Bover12\sum_{i=1}^k\sum_{j=1}^k
     S_I(\bar h_{ij}\pmb V,d\pmb v_id\pmb v_j)$.}

\noindent Then

\Centerline{$\Bvalue{|\bar z_I|>\delta_0\sum_{i=1}^kS_I(\pmb{1},(d\pmb v_i)^2)}
\Bsubseteq a$}

\noindent whenever $J\subseteq I\in\Cal I(\Cal S)$.

\Prf\ If $e=c(\sigma,\tau)$ is an $I$-cell and we set

\Centerline{$\bar z_e=\Delta_e(\tbf 1,d(\bar h\pmb V))
  -\sum_{i=1}^k\Delta_e(\bar h_i\pmb V,d\pmb v_i)
  -\Bover12\sum_{i=1}^k\sum_{j=1}^k
     \Delta_e(\bar h_{ij}\pmb V,d\pmb v_id\pmb v_j)$,}

\noindent then

$$\eqalign{\bar z_e
&=\bar h(V_{\tau})-\bar h(V_{\sigma})
  -\sum_{i=1}^k\bar h_i(V_{\sigma})
       \times(v_{i\sigma'}-v_{i\sigma})\cr
&\mskip150mu
  -\Bover12\sum_{i=1}^k\sum_{j=1}^k\bar h_{ij}(V_{\sigma})
       \times(v_{i\tau}-v_{i\sigma})
       \times(v_{j\tau}-v_{j\sigma})\cr}$$

\noindent and

$$\eqalign{\Bvalue{|\bar z_e|
  >\delta_0\sum_{i=1}^k\Delta_e(\tbf 1,(d\pmb v_i)^2)}
&=\Bvalue{|\bar z_e|>\delta_0\sum_{i=1}^k(v_{i\tau}-v_{i\sigma})^2}\cr
&\Bsubseteq\sup_{1\le i\le k}\Bvalue{|v_{i\tau}-v_{i\sigma}|>\delta}
\Bsubseteq\sup_{1\le i\le k}\Bvalue{\Osclln_I(\pmb v_i)>\delta}\cr
&\Bsubseteq\sup_{1\le i\le k}\Bvalue{\Osclln^*_J(\pmb v_i)>\delta}
\Bsubseteq a.\cr}$$

\noindent Summing,

$$\eqalign{\Bvalue{|\bar z_I|>\delta_0\sum_{i=1}^kS_I(\tbf 1,(d\pmb v_i)^2)}
&\Bsubseteq\sup_{e\in\Sti_0(I)}\Bvalue{|\bar z_e|
  >\delta_0\sum_{i=1}^k\Delta_e(\tbf 1,(d\pmb v_i)^2)}\cr
&\Bsubseteq a.  \text{ \Qed}\cr}$$

\noindent Consequently

\Centerline{$\theta(\bar z_I)
\le\theta(\sum_{i=1}^kS_I(\delta_0\tbf 1,(d\pmb v_i)^2))+\bar\mu a
\le 2k\epsilon$}

\noindent because $\theta(\sup|\delta_0\tbf 1|)
=\theta(\delta_0\chi 1)\le\delta_0$.
And this is true whenever $I\in\Cal I(\Cal S)$ includes $J$.

\medskip

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

$$\eqalign{0&
=\lim_{I\uparrow\Cal I(\Cal S)}\bar z_I
=\int_{\Cal S}d(\bar h\pmb V)
   -\sum_{i=1}^k\int_{\Cal S}\bar h_i\pmb V\,d\pmb v_i
   -\Bover12\sum_{i=1}^k\sum_{j=1}^k\int_{\Cal S}
        \bar h_{ij}\pmb V\,d\pmb v_id\pmb v_j\cr&
=\int_{\Cal S}d(\bar h\pmb V)
   -\sum_{i=1}^k\int_{\Cal S}\bar h_i\pmb V\,d\pmb v_i
   -\Bover12\sum_{i=1}^k\sum_{j=1}^k\int_{\Cal S}
        \bar h_{ij}\pmb V\,d\covar{\pmb v_i}{\pmb v_j},\cr}$$

\noindent and the formula is valid when $\pmb u=\tbf 1$ and the
$h_{ij}$ are all uniformly continuous.

\medskip

{\bf (e)} As for the case in which not all the $h_{ij}$ are uniformly
continuous, let $\bar v$ be an upper bound for
$\{|v_{i\sigma}|:1\le i\le k$, $\sigma\in\Cal S\}$.   Take any
$\epsilon>0$.   Let $M\ge 0$ be such that $b=\Bvalue{\bar v\ge M}$ has
measure at most $\epsilon$.   There is a smooth function
$f:\Bbb R^k\to\Bbb R$ with compact support such that $f(x)=1$ whenever
$\|x\|_{\infty}\le M$.  \Prf\ Take a smooth function $f_0:\Bbb R\to\Bbb R$
with compact support such that $f_0(\xi)=1$ for every $\xi\in[-M,M]$, and
set $f(\xi_1,\ldots,\xi_k)=\prod_{i=1}^kf_0(\xi_i)$ for
$\xi_1,\ldots,\xi_k\in\Bbb R$.\ \QeD\   Now set $g=f\times h$.   Writing
$g_i$, $g_{ij}$ for the partial derivatives of $g$, these are all
uniformly continuous and

\Centerline{$\int_{\Cal S}d(\bar g\pmb V)
=\sum_{i=1}^k\int_{\Cal S}\bar g_i\pmb V\,d\pmb v_i
  +\Bover12\sum_{i=1}^k\sum_{j=1}^k
     \int_{\Cal S}
        \bar g_{ij}\pmb V\,d\pmb v_id\pmb v_j$.}

\noindent At the same time, $g(x)=h(x)$,
$g_i(x)=h_i(x)$ and $g_{ij}(x)=h_{ij}(x)$
whenever $1\le i,j\le k$ and $x\in\ooint{-M,M}^k$, so

\Centerline{$\Bvalue{\bar g(V_{\sigma})\ne\bar h(V_{\sigma})}
\Bsubseteq\sup_{1\le i\le k}\Bvalue{|v_{i\sigma}|\ge M}
\Bsubseteq b$}

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

\Centerline{$\Bvalue{\bar g_i(V_{\sigma})\ne\bar h_i(V_{\sigma})}
\Bsubseteq b$,
\quad$\Bvalue{\bar g_{ij}(V_{\sigma})\ne\bar h_{ij}(V_{\sigma})}
\Bsubseteq b$}

\noindent whenever $1\le i,j\le k$ and $\sigma\in\Cal S$.
Consequently

\Centerline{$\Bvalue{\int_{\Cal S}d(\bar h\pmb V)
\ne\sum_{i=1}^k\int_{\Cal S}\bar h_i\pmb V\,d\pmb v_i
  +\Bover12\sum_{i=1}^k\sum_{j=1}^k
     \int_{\Cal S}
        \bar h_{ij}\pmb V\,d\pmb v_id\pmb v_j}
\Bsubseteq b$}

\noindent has measure at most $\epsilon$.   As $\epsilon$ is arbitrary,

\Centerline{$\int_{\Cal S}d(\bar h\pmb V)
=\sum_{i=1}^k\int_{\Cal S}\bar h_i\pmb V\,d\pmb v_i
  +\Bover12\sum_{i=1}^k\sum_{j=1}^k
     \int_{\Cal S}
        \bar h_{ij}\pmb V\,d\pmb v_id\pmb v_j$}

\noindent in this case also.   Translating into terms of covariations, we
have

\Centerline{$\int_{\Cal S}d(\bar h\pmb V)
=\sum_{i=1}^k\int_{\Cal S}\bar h_i\pmb V\,d\pmb v_i
  +\Bover12\sum_{i=1}^k\sum_{j=1}^k
     \int_{\Cal S}
        \bar h_{ij}\pmb V\,d\covar{\pmb v_i}{\pmb v_j}$.}

\medskip

{\bf (f)} This deals with the case $\pmb u=\tbf 1$.   For other
$\pmb u$, we can use the same method as in 619D,
because the components of the formula can be translated into terms
of the integrators

\Centerline{$\bar h\pmb V$,
\quad$ii_{\pmb v_i}(\bar h_i\pmb V)$,
\quad$ii_{\covar{\pmb v_i}{\pmb v_j}}(\bar h_{ij}\pmb V)$.}

\noindent So we get the general result as before.
}%end of proof of 619J

\leader{619K}{}\cmmnt{ Applying this to $\Cal S\wedge\tau$ for
$\tau\in\Cal S$, we get the following version.

\medskip

\noindent}{\bf Corollary} Let $k\ge 1$ be an integer, and
$h:\BbbR^k\to\Bbb R$ a twice-differentiable function with continuous second
derivative.   Let $\Cal S$ be a sublattice of
$\Cal T$, and $\pmb v_1,\ldots,\pmb v_k$ locally jump-free
local integrators with domain $\Cal S$;  let $\pmb u$ be a locally moderately
oscillatory process with domain $\Cal S$.   Write
$\pmb V=(\pmb v_1,\ldots,\pmb v_k)$.   Then

\Centerline{$ii_{\bar h\pmb V}(\pmb u)
=\sum_{i=1}^kii_{\pmb v_i}(\pmb u\times\bar h_i\pmb V)
  +\Bover12\sum_{i=1}^k\sum_{j=1}^k
   ii_{\covar{\pmb v_i}{\pmb v_j}}(\pmb u\times\bar h_{ij}\pmb V)$.}

\exercises{\leader{619X}{Basic exercises (a)}
%\spheader 619Xa
Suppose that $h_1$, $h_2$ are Borel measurable functions from
$\BbbR^k\to\Bbb R$.   Show that if $u_1,\ldots,u_k\in L^0$ then
$\Bvalue{\bar h_1(u_1,\ldots,u_k)\le\bar h_2(u_1,\ldots,u_k)}
=\Bvalue{(u_1,\ldots,u_k)\in E}$, where
$E=\{x:x\in\BbbR^k$, $h_1(x)\le h_2(x)\}$.
%619E

\spheader 619Xb
Suppose that $l\ge 1$, that $h_1,\ldots,h_l$ are Borel measurable
functions from $\BbbR^k$ to $\Bbb R$, that $h:\BbbR^l\to\Bbb R$ is Borel
measurable, and that $g(x)=h(h_1(x),\ldots,h_l(x))$ for $x\in\BbbR^k$.
Show that
$\bar g(u)=\bar h(\bar h_1(u),\ldots,\bar h_l(u))$ for $u\in L^0$.
%619E

\leader{619Y}{Further exercises (a)}
%\spheader 619Ya
Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ a jump-free integrator, and
$\pmb v^*$ its quadratic variation.   Suppose that $G\subseteq\Bbb R$ is
an open set such that $\Bvalue{v_{\tau}\in G}=1$ for every
$\tau\in\Cal S$, and $h:G\to\Bbb R$ is a twice-differentiable
function with continuous second derivative.   Show that

\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 whenever $\pmb u$ is a moderately oscillatory
process with domain $\Cal S$.
%619D

\spheader 619Yb Let $\pmb u_1,\ldots,\pmb u_k$ be simple
processes all with the same domain $\Cal S$,
and $h:\BbbR^k\to\Bbb R$ a continuous function.
Write $\pmb{U}$ for $(\pmb u_1,\ldots,\pmb u_k)$.
Show that $\bar h\pmb U$ is simple.
%619G

\spheader 619Yc(i) Suppose that $I$ is a finite sublattice of $\Cal T$ and
$\pmb v\in\Mfa(I)$.   Show that
$S_I(\tbf 1,|(d\pmb v)^3|)\le\Osclln_I(\pmb v)\times S_I(\tbf 1,(d\pmb v)^2)$.
(ii) Suppose that $\Cal S$ is a sublattice of $\Cal T$ and $\pmb v$ is a
jump-free integrator with domain $\Cal S$.   Show that
$\int_{\Cal S}\pmb u\,(d\pmb v)^3=0$ for every order-bounded process
$\pmb u$ with domain $\Cal S$.
}%end of exercises


\endnotes{
\Notesheader{619} In the ordinary theory of the Riemann-Stieltjes
integral, we are
well accustomed to the formula $h(v(b))-h(v(a))=\int_a^bh'(v)dv$ for
continuous functions $v$ of bounded variation.   In 619C we see that we
have a correction term, which would correspond to a term
$\bover12\int_a^bh''(v)(dv)^2$.   In the real-variable case we expect the
term $(dv)^2$ to count as zero, as it must do if $\int_a^b|dv|$ is finite
and $v$ is continuous.   The point of stochastic integration is that we
integrating with respect to integrators which are {\it not} of bounded
variation (e.g., Brownian motion), but for which we can use
$\int_{\Cal S}(d\pmb v)^2=\int_{\Cal S}d\pmb v^*$
to give a general formula which will be the basis of almost every application
of the theory to jump-free processes.   In the
proof of 619C we see that this comes from the second-order Taylor expansion

\Centerline{$h(\beta)\bumpeq h(\alpha)
+(\beta-\alpha)h'(\alpha)+\Bover12(\beta-\alpha)^2h''(\alpha)$}

\noindent just as the familiar identity comes from the first-order
expansion

\Centerline{$h(\beta)\bumpeq h(\alpha)+(\beta-\alpha)h'(\alpha)$.}

\noindent The analysis is very much harder, and the welcome simplicity of
the formula depends on concepts (`moderately oscillatory process',
`jump-free integrator' and `quadratic variation') which are far from
elementary.   But in Chapter 65 we shall find that, taking
proper care, we can go a long way with formal manipulations.

If you look back at 617H, you will see that it amounts to a
special case of 619J:  we use the formula

\Centerline{$\int_{\Cal S}d(\pmb v\times\pmb w)
=\int_{\Cal S}\pmb v\,d\pmb w
+\int_{\Cal S}\pmb w\,d\pmb v
+\Bover12(\int_{\Cal S}d\covar{\pmb v}{\pmb w}
   +\int_{\Cal S}d\covar{\pmb w}{\pmb v})$}

\noindent to tell us what $\covar{\pmb v}{\pmb w}$ should be, at least if
covariation is to be symmetric.   What is remarkable is that this approach
extends so dramatically from multiplication
to general twice-continuously-differentiable functions.
In 619Yc I offer a sort of explanation of the fact that we
don't have to examine third derivatives in this context.

You can see why most textbooks leave the proof of
619J to their readers.   I have to admit that my formulation introduces
extra obstacles in the proofs of 619I and 619J, where we need to turn
inequalities concerning a function $h:\BbbR^k\to\Bbb R$ into
corresponding inequalities for $\bar h:(L^0)^k\to L^0$.
If we allow ourselves to think of $\bar h$ as being defined by the formula
$\bar h(f_1^{\ssbullet},\ldots,f_k^{\ssbullet})
=h(f_1,\ldots,f_k)^{\ssbullet}$, as in 619Ee, this step becomes elementary.
In \S364 I went to a good deal of trouble to describe the $f$-algebra
structure of $L^0$ in terms which did not depend on representations of this
kind.   In the context of the present volume this self-denial seems
unnecessary, since in the great majority of applications our probability
algebra is explicitly derived from some probability space.
I hope however that even if you are
impatient with measure algebras, and have been more or less frankly
translating everything into expressions concerning conventional
stochastic processes (as in 612H and elsewhere), you will agree that it is
important to remember that trivial changes to the probability space can't
affect the results we are looking at, and that it might be helpful to
remember that `trivial' can, for many of
our purposes here, be interpreted as `not
affecting the measure algebra'.   But to safely continue with this
assumption, we have to check from time to time that our manipulations
really are invariants of the measure algebras.   And this is what I am
trying to do in 619E-619G.

For those interested in the logical status of the theorems here, which in
my view ought to include those concerned with their application to
questions arising in the real world,
there is another reason to have qualms about the
proof of 619E given here:  it depends on the Loomis-Sikorski theorem
314M, which requires a strong form of the axiom of choice.   But
there are no new problems beyond those already
considered in \S364 and Chapter 56.

Of course there is a more important problem facing us.   I have offered no
effective means of calculating $\pmb v^*$ for any of the important
processes $\pmb v$ of the theory.   In 617O I looked at the identity
process, which is almost trivial, and the Poisson process, which is
irrelevant because it's not locally jump-free.   We don't seem to be ready
for Brownian motion (624F),
and this will be one of the aims of the next chapter.
}%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
(1944) 519-524.
\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
Processes and their Applications 57 (1995) 11-18.
\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
Math.\ J.\ 30 (1967) 209-245.
}

\medskip%L

Lowenstein R.\ [01] {\it When Genius Failed: the rise and fall of
Long-Term Capital Management.}  Fourth Estate, 2001.
%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',
12.6.12;
{\tt www.math.columbia.edu{\bsp}$^{\sim}$mnutz{\bsp}docs/pathwInt.pdf}.
\cmmnt{[649E.]}

\medskip%O\medskip%P

Protter P.E.\ [05] {\it Stochastic Integration and Differential Equations.}
Springer, 2005.
\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
and Martingales,} vol.\ II, {\it It\^o Calculus}.   Cambridge U.P, 2000.
\cmmnt{[\S641 {\it notes\/}, \S645 {\it notes\/}.]}

\medskip%S

Sato K.-I.\ [13] {\it L\'evy processes and infinitely divisible distributions.} 
Cambridge University Press, 2013.  
\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.}
Vieweg, 1990 (Vieweg Advanced Lectures).
}

\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



