% The following TeXfile comprises part of

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

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

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

%        %   \amstexsupportedfalse

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

%        \input amstex
%        \documentstyle{amsppt}

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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


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

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

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

   \Centerline{#1}

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

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

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

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

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

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

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

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

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

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

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

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

%End of lines for missing AmSTeX

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

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

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

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

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


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

%hyphenations
\hyphenation{Archi-me-dean}

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

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

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

%Headline assignments

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

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

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

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

\newdimen\pagetotalplus
\newdimen\pagetotalminus

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\Sous{\Cal S\text{ous}}  %used in vol 6
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\tinyplus{{\vrule height 1.6pt depth -1.3pt width 2.5pt}
    \hskip-1.3pt{\vrule height 2.6pt depth -0.1pt width 0.3pt}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

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

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

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

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

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

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

\frfilename{mt63.tex}
\versiondate{8.3.24}
\copyrightdate{2012}

\def\chaptername{Structural alterations}

\newchapter{63}

One of the daunting things about stochastic calculus is the amount of
preliminary material required before one can approach the main theorems,
let alone the interesting applications.   Before proceeding to
Chapter 64 we must do some more work more or less
at the level of Chapters 61-62,
and this is what I will try to deal with in the present chapter.

I start with a quick run through the properties of near-simple processes
(\S631) which got pushed out of Chapter 61 due to shortage of space.
The real work of the chapter begins in \S632, with `right-continuous'
filtrations $\family{t}{T}{\frak A_t}$.   For such stochastic integration
structures, which include the most important examples (632D), there are useful
simplifications in the theory (632C, 632F, 632I, 632J), so it is not surprising
that most presentations of this material take right-continuity of the
filtration as a standard hypothesis.

The integral $\int_{\Cal S}\pmb u\,d\pmb v$, as I have defined it,
depends on an elaborate structure:  a probability algebra
$(\frak A,\bar\mu)$, a filtration $\family{t}{T}{\frak A_t}$ and the
sublattice $\Cal S$.   We anticipate that changing any of these will change
the value of the integral.   But there are many cases in which this doesn't
happen.   The simplest of these is `change of law'.   As long as we have a
strictly positive countably additive measure on the given algebra
$\frak A$, we shall have the same integrals, as I pointed out right at the
beginning in 613I.
Next, there are important classes of pairs $\Cal S'$, $\Cal S$ of lattices
for which we can expect equality of the integrals $\int_{\Cal S'}$ and
$\int_{\Cal S}$.   For these we have to work fairly hard, since
it is certainly not enough just to have $\min\Cal S'=\min\Cal S$ and
$\max\Cal S'=\max\Cal S$.   In \S633 I explore sufficient conditions to
make a sublattice $\Cal S'$ of $\Cal S$ behave as if it had full outer
(Riemann) measure, so that an integral over $\Cal S$ will be the same
when taken over $\Cal S'$ (633K).

We are now in a position to look at the effect of replacing
$(\frak A,\bar\mu,\family{t}{T}{\frak A_t})$ with
$(\frak B,\bar\mu\restrp\frak B,\family{t}{T}{\frak B\cap\frak A_t})$ where
$\frak B$ is a subalgebra of $\frak A$.   As long as we are just looking at
integrals, we need only quote from \S633;  but if we want to understand
martingales (634I), we need the theory of relative independence from Chapter 48.
And there is a yet more radical change
which we can consider, where the filtration $\family{t}{T}{\frak A_t}$ is
replaced by $\family{r}{R}{\frak A_{\pi_r}}$ for some family
$\family{r}{R}{\pi_r}$ of stopping times.   This is what I do in \S635.

\discrpage
\frfilename{mt631.tex}
\versiondate{4.5.21/21.1.22}
\copyrightdate{2018}

\def\chaptername{Structural variations}
\def\sectionname{Near-simple processes}

\def\halfquad{\hskip.5em\relax}

\newsection{631}

My presentation so far has focused on `moderately oscillatory' integrands,
with regular mentions of `simple' processes, and an excursion into `jump-free'
processes in \S\S618-619.   Later on, however, there will be many important
results applying to an intermediate class, the `near-simple' processes.

\cmmnt{\leader{631A}{Notation}
We shall need a good many of the formulae introduced in
Chapter 61, and as we are starting a new chapter I will give a particularly
detailed list.   Throughout, \Asissix\ will be a stochastic integration
structure (\S613 {\it notes}).
For $u\in L^0(\frak A)$, $\theta(u)$ is $\Expn(|u|\wedge\chi 1)$ (613B).
If $t\in T$, $\check t$ is the constant stopping time at $t$ (611A).
If $h:\Bbb R\to\Bbb R$ is a Borel function, $\bar h$ is the associated
function from $L^0(\frak A)$ to itself (612A).   If $\Cal S$ is a sublattice of
$\Cal T$ and $\tau\in\Cal T$, then
$\Cal S\wedge\tau=\{\sigma\wedge\tau:\sigma\in\Cal S\}$,
$\Cal S\vee\tau=\{\sigma\vee\tau:\sigma\in\Cal S\}$ and $\Cal I(\Cal S)$ is the
set of finite sublattices of $\Cal S$.   $\Bvalue{\,\,}$ will appear in
formulae of the type $\Bvalue{u>\alpha}$, $\Bvalue{u\le v}$ where
$u$, $v\in L^0(\frak A)$ and $\alpha\in\Bbb R$, as in Chapter 36;  in
formulae of the type $\Bvalue{\sigma>t}$ and $\Bvalue{\sigma<\tau}$, where
$\sigma$, $\tau\in\Cal T$ and $t\in T$, as in 611A and 611D;  and in formulae of
the type
$\Bvalue{\pmb u\ne\tbf 0}=\sup_{\sigma\in\Cal S}\Bvalue{u_{\sigma}\ne 0}$ where
$\pmb u=\sfamilysS{u}\in L^0(\frak A)^{\Cal S}$, as in 612S.

For a sublattice $\Cal S$ of $\Cal T$, $\Mfa(\Cal S)$, $\Msimp(\Cal S)$,
$\Mob(\Cal S)$ and $\Mlob(\Cal S)$ and are the $f$-algebras of, respectively,
the fully adapted processes (612I), the simple processes
(612J), the order-bounded processes (614E) and the locally order-bounded
processes (614E) with domain $\Cal S$.
I will use the symbol $\tbf 1$ for
constant processes with value $\chi 1\in L^0(\frak A)$ (361D, 612De).
If $\pmb u=\sfamilysS{u}\in\Mfa(\Cal S)$ and
$z\in L^0(\frak A)\cap\bigcap_{\sigma\in\Cal S}L^0(\frak A_{\sigma})$, then
$z\pmb u$ is the process $\sfamilysS{z\times u}$ (612De).

For an order-bounded process $\pmb u=\sfamilysS{u}$,
$\sup|\pmb u|=\sup_{\sigma\in\Cal S}|u_{\sigma}|\in L^0(\frak A)$.
For a finite sublattice $I$ of $\Cal T$ and fully adapted processes
$\pmb u$, $\pmb v$ with domains including $I$,
$S_I(\pmb u,d\pmb v)$ is the Riemann sum defined in 613E-613F.
$\int_{\Cal S}\pmb u\,d\pmb v$ and $\int_{\Cal S}|d\pmb v|$ will
denote Riemann-sum integrals as in 613L, while $\iivu$ will be an indefinite
integral as defined in 613O.
}

\leader{631B}{Definitions} Let $\Cal S$ be a
sublattice of $\Cal T$.

\spheader 631Ba A fully adapted process $\pmb u$ with domain $\Cal S$
is {\bf near-simple} if it is in the closure of $\Msimp(\Cal S)$ for the ucp
topology on $\Mob(\Cal S)$;  that is, it is order-bounded and
for every $\epsilon>0$ there
is a simple process $\pmb v$ with domain $\Cal S$ such
that $\theta(\sup|\pmb u-\pmb v|)\le\epsilon$.

\spheader 631Bb A fully adapted process
$\pmb u$ with domain $\Cal S$ is {\bf locally near-simple}
if $\pmb u\restr\Cal S\wedge\tau$ is near-simple for every
$\tau\in\Cal S$.

\cmmnt{\medskip

\noindent{\bf Remarks} Note that in this definition it is essential
that the approximating simple process $\pmb v$ should have the same
domain as the process $\pmb u$.   We have no general assurance that
if $\pmb v$ is a simple process, and $\Cal S$ is a sublattice of
$\dom\pmb v$, then $\pmb v\restr\Cal S$ is near-simple (see
631Xb).
}%end of comment

\leader{631C}{Proposition} (a)
(Locally) near-simple processes are (locally) moderately oscillatory.

(b) (Locally) jump-free processes are (locally) near-simple.

\proof{{\bf (a)} We know that simple processes are moderately oscillatory
(615E) and that the space of moderately oscillatory processes is closed in the
space of order-bounded processes for the ucp topology (615F(a-iv)), so
near-simple processes are moderately oscillatory.   It follows at once that
locally near-simple processes are locally moderately oscillatory.

\medskip

{\bf (b)} As declared in 612Ja, the empty process counts as
simple, so we need only to look at non-trivial sublattices.
Re-reading part (i) of the proof of 618Gb,
we see that if $\Cal S$ is a non-empty sublattice of $\Cal T$, $\pmb u$ is a
jump-free process with domain $\Cal S$, and $\epsilon>0$, there is a simple
process $\pmb u'$ with domain $\Cal S$ such that
$\theta(\sup|\pmb u-\pmb u'|)\le\epsilon$.   As $\epsilon$ is arbitrary,
$\pmb u$ is near-simple.   As in (a) just above,
it follows immediately that locally jump-free processes are locally simple.
}%end of proof of 631C

\leader{631D}{Where near-simple processes come from:  Theorem}
Let $(\Omega,\Sigma,\mu)$ be a complete probability space and
$\langle\Sigma_t\rangle_{t\ge 0}$ a filtration
of $\sigma$-subalgebras of $\Sigma$, 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 $t$ and
$t\mapsto X_t(\omega):\coint{0,\infty}\to\Bbb R$ is
\cadlag\cmmnt{ (in the sense of 4A2A)} for
every $\omega\in\Omega$.

In this case,
$\langle X_t\rangle_{t\ge 0}$ is progressively measurable, and if
\Asissixt\ and
$\family{\sigma}{\Cal T_f}{x_{\sigma}}$ are defined as in 612H, then
$\pmb x=\family{\sigma}{\Cal T_f}{x_{\sigma}}$ is locally near-simple.

\proof{ Just as 613Ub corresponds to 618Gb, the proof here follows that of 618H.
Corresponding to the weaker hypothesis and less ambitious objective,
some modifications are necessary.   To begin with, the
formulae can be copied exactly, but rather than keep you turning back and forth,
I repeat the details.

\medskip

{\bf (a)} As before, 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

$$\eqalignno{&\{(s,\omega):s\le t,\,X_s(\omega)>\alpha\}\cr&
\mskip80mu
=\{(s,\omega):s\le t,\,
   \limsup_{q\downarrow Q\cap[s,t]}X_q(\omega)>\alpha\}
\Displaycause{because $s\mapsto X_s(\omega)$ is \cadlag\ for every $\omega$}
\mskip80mu
=\bigcup_{k\in\Bbb N}\bigcap_{q\in Q}\bigcup_{\Atop{q'\in Q}{q'\le q}}
    \{(s,\omega):s\le t,\,s>q\text{ or }s\le q'\text{ and }
        X_{q'}(\omega)\ge\alpha+2^{-k}\}\cr&
\mskip80mu
\in\Cal B([0,t])\tensorhat\Sigma_t\cr}$$

\noindent where $\Cal B([0,t])$ is the Borel $\sigma$-algebra of
$[0,t]$.\ \Qed

We can therefore again 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
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)} 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 (Once again I can write $\min$ rather than $\inf$ because
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 \cadlag.   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 \cadlag.   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 $g_n(\omega)=0$,

----- $g_n(\omega)\le h(\omega)$,

----- $|X_t(\omega)-X_{g_n}(\omega)|<\epsilon$ whenever
$g_n(\omega)\le t<g_{n+1}(\omega)$,

----- if $g_{n+1}(\omega)<h(\omega)$ then
$|X_{g_{n+1}}(\omega)-X_{g_n}(\omega)|\ge\epsilon$.}

\noindent It is at this point that the argument begins to diverge from that of
618H.   But we still see that 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 \Asissix, 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{----- if $n=0$ then $\tau_n=\check 0=\min\Cal T_f$,

----- $\tau_n\le\tau$,

----- $\Bvalue{\tau_n\le\sigma}\Bcap\Bvalue{\sigma<\tau_{n+1}}
\Bsubseteq\Bvalue{|x_{\sigma}-x_{\tau_n}|<\epsilon}$ for every
$\sigma\in\Cal T_f$,}

%----- $\Bvalue{\tau_{n+1}<\tau}
%\Bsubseteq\Bvalue{|x_{\tau_{n+1}}-x_{\tau_n}|\ge\epsilon}$,}

\noindent and

\Centerline{$\sup_{n\in\Bbb N}\Bvalue{\tau_n=\tau}=1$.}

\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
take $\pmb v=\family{\sigma}{\Cal T\wedge\tau}{v_{\sigma}}$
to be the simple function defined from $\tau_0,\ldots,\tau_n$ and
$x_{\tau_0},\ldots,x_{\tau_n}$, then
$\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{|x_{\sigma}-v_{\sigma}|<\epsilon}$ for every
$\sigma\in\Cal T\wedge\tau$ and $i<n$.   So
$\Bvalue{|x_{\sigma}-v_{\sigma}|\ge\epsilon}\Bsubseteq c$ for every
$\sigma\in\Cal T\wedge\tau$, and
$\theta(\sup|(\pmb x\restr\Cal T\wedge\tau)-\pmb v|)\le 2\epsilon$.
As $\epsilon$ is arbitrary, $\pmb x\restr\Cal T\wedge\tau$ is near-simple;
as $\tau$ is arbitrary, $\pmb x$ is locally near-simple.
}%end of proof of 631D

\leader{631E}{Proposition} (a) If $T=\coint{0,\infty}$, then the identity
process on $\Cal T_f$ is locally near-simple.

(b) Brownian motion is locally near-simple.

(c) The Poisson process is locally near-simple.

\proof{ By 618Ja and 618Jc, the identity process and Brownian motion are locally
jump-free, so by 631Cb they are locally near-simple.   As for the Poisson
process $\pmb v$, I defined it in 612Ub in terms of a probability
measure on $\Omega=\Cdlg$ and a
process $\langle X_t\rangle_{t\ge 0}$ with
$X_t(\omega)=\omega(t)$ for every $\omega$, so that $t\mapsto X_t(\omega)$ is
surely \cadlag\ for every $\omega$, and 631D tells us that $\pmb v$ is locally
near-simple.
}%end of proof of 631E

\leader{631F}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.

(a) Write $\Mns\cmmnt{\mskip5mu=\Mns(\Cal S)}$
for the set of near-simple processes with domain $\Cal S$.

\quad(i) If $h:\Bbb R\to\Bbb R$ is continuous, then
$\bar h\pmb u\in\Mns$ for every $\pmb u\in\Mns$.

\quad(ii) $\Mns$ is an $f$-subalgebra of $\Mob=\Mob(\Cal S)$.

\quad(iii) $\Mns$ is complete for the ucp uniformity.

\quad(iv) If $\tau\in\Cal S$ and $\pmb u\in\Mfa(\Cal S)$, then
$\pmb u$ is near-simple iff $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are both near-simple.

\quad(v) If $\pmb u\in\Mns$ and
$z\in L^0(\frak A\cap\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$, then
$z\pmb u$ belongs to $\Mns$.

\quad(vi) If $\pmb u$ is near-simple it is locally near-simple.

(b) Write $\Mlns\cmmnt{\mskip5mu=\Mlns(\Cal S)}$ for the set of
locally near-simple processes with domain $\Cal S$.

\quad(i) If $h:\Bbb R\to\Bbb R$ is continuous, then
$\bar h\pmb u\in\Mlns$ for every $\pmb u\in\Mlns$.

\quad(ii) $\Mlns$ is an $f$-subalgebra of the space $\Mlob=\Mlob(\Cal S)$ of
locally order-bounded processes with domain $\Cal S$.

\quad(iii) If $\pmb u\in\Mfa(\Cal S)$ and $\tau\in\Cal S$, then $\pmb u$
is locally near-simple iff $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are both locally near-simple.

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

\quad(v) If $\pmb u\in\Mfa(\Cal S)$ and
$\{\sigma:\sigma\in\Cal S$, $\pmb u\restr\Cal S\wedge\sigma$ is near-simple$\}$
covers $\Cal S$, then $\pmb u\in\Mlns$.

(c) Suppose that $\pmb u$ is a moderately oscillatory process with domain
$\Cal S$.

\quad(i) If $\pmb u\restr\Cal S\cap[\tau,\tau']$ is near-simple whenever
$\tau\le\tau'$ in $\Cal S$, then $\pmb u$ is near-simple.

\quad(ii) If $\pmb u$ is locally near-simple it is near-simple.

\proof{{\bf (a)(i)} By 615Ca,
$\pmb u\mapsto\bar h\pmb u:\Mob(\Cal S)\to\Mob(\Cal S)$
is continuous, while $\bar h\pmb u\in\Msimp(\Cal S)$ for every
$\pmb u\in\Msimp(\Cal S)$, by
612La.   Since $\Mns(\Cal S)$ is the closure of $\Msimp(\Cal S)$,
$\bar h\pmb v\in\Mns(\Cal S)$ for every $\pmb v\in\Mns(\Cal S)$.

\medskip

\quad{\bf (ii)} Similarly, as $\Msimp(\Cal S)$ is closed under the operations
$+$, $\times$ and $\pmb u\mapsto|\pmb u|$, and these are
continuous on $\Mob(\Cal S)$, $\Mns(\Cal S)$ also is closed under these
operations, and is an $f$-subalgebra of $\Mob(\Cal S)$.

\medskip

\quad{\bf (iii)} By definition, $\Mns$ is a closed subset of $\Mob$.   As $\Mob$
is complete under its ucp uniformity (615Cc), so is $\Mns$
(3A4Fd\Latereditions).

\medskip

\quad{\bf (iv)}\grheada\ The operation
$\pmb u\mapsto\pmb u\restr\Cal S\wedge\tau:
\Mob(\Cal S)\to\Mob(\Cal S\wedge\tau)$ is linear, and also continuous, as

\Centerline{$\theta(\sup|\pmb u\restr\Cal S\wedge\tau|)
\le\theta(\sup|\pmb u|)$}

\noindent for every $\pmb u\in\Mob(\Cal S)$.   Since
$\pmb u\restr\Cal S\wedge\tau\in\Msimp(\Cal S\wedge\tau)$ for every
$\pmb u\in\Msimp(\Cal S)$ (612K(d-ii)),
$\pmb u\restr\Cal S\wedge\tau\in\overline{\Msimp(\Cal S\wedge\tau)}
=\Mns(\Cal S)$ for every
$\pmb u\in\overline{\Msimp(\Cal S)}=\Mns(\Cal S)$.
Similarly, using 612K(d-iii),
$\pmb u\restr\Cal S\vee\tau\in\Mns(\Cal S\vee\tau)$ for every
$\pmb u\in\Mns(\Cal S)$.

\medskip

\qquad\grheadb\ Now suppose that $\pmb u\in\Mfa(\Cal S)$ is such that
$\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are near-simple.   Let $\epsilon>0$.
Then there are simple processes
$\pmb u'=\family{\sigma}{\Cal S\wedge\tau}{u'_{\sigma}}$ and
$\pmb u''=\family{\sigma}{\Cal S\vee\tau}{u''_{\sigma}}$ such that
$\theta(\bar u')$, $\theta(\bar u'')$ are both less than or equal to
$\epsilon$, where $\bar u'=\sup|\pmb u'-\pmb u\restr\Cal S\wedge\tau|$ and
$\bar u''=\sup|\pmb u''-\pmb u\restr\Cal S\vee\tau|$.

Let $(\tau'_0,\ldots,\tau'_m)$ and $(\tau''_0,\ldots,\tau''_n)$ be
breakpoint strings for $\pmb u'$, $\pmb u''$ respectively;  we can suppose
that $\tau'_m=\tau''_0=\tau$.   Let $\pmb v$ be the simple process defined
by the formula in 612Ka from the breakpoint string
$(\tau'_0,\ldots,\tau,\ldots,\tau''_n)$ and the values
$u'_{\downarrow},u'_{\tau'_0},\ldots,u'_{\tau'_{m-1}},
u''_{\tau},\ldots,u''_{\tau''_n}$, where $u'_{\downarrow}$
is the starting value of $\pmb u'$ (614Ba).   If $\sigma\in\Cal S$ then

$$\eqalign{\Bvalue{\sigma<\tau}&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau}}
   \Bcap\Bvalue{u'_{\sigma\wedge\tau}=v_{\sigma\wedge\tau}}
   \Bcap\Bvalue{v_{\sigma\wedge\tau}=v_{\sigma}}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le\bar u'},\cr
\Bvalue{\tau\le\sigma}
&\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\vee\tau}}
   \Bcap\Bvalue{u'_{\sigma\vee\tau}=v_{\sigma\vee\tau}}
   \Bcap\Bvalue{v_{\sigma\vee\tau}=v_{\sigma}}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le\bar u''},\cr}$$

\noindent so

\Centerline{$\theta(\sup|\pmb u-\pmb v|)
\le\theta(\bar u'\vee\bar u'')\le 2\epsilon$;}

\noindent as $\epsilon$ is arbitrary, $\pmb u$ is near-simple.

\medskip

\quad{\bf (v)} This follows from (ii) because $z\tbf 1\restr\Cal S$ is simple
and $z\pmb u=(z\tbf 1\restr\Cal S)\times\pmb u$.

\medskip

\quad{\bf (vi)} This is immediate from (iv).

\medskip

{\bf (b)(i)} follows from (a-i) because
$(\bar h\pmb u)\restr\Cal S\wedge\tau=\bar h(\pmb u\restr\Cal S\wedge\tau)$
for every $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)} Similarly, restriction respects
the algebraic and lattice operations on $\Mob(\Cal S)$.

\medskip

\quad{\bf (iii)}\grheada\ If $\pmb u$ is locally near-simple, then
$\pmb u\restr\Cal S\wedge\tau$ is simple, therefore locally near-simple.
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 near-simple, so $\pmb u\restr(\Cal S\vee\tau)$ is locally
near-simple.

\medskip

\qquad\grheadb\ Suppose that $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are locally near-simple.   Take any
$\tau'\in\Cal S\vee\tau$.   Then
$(\pmb u\restr\Cal S\wedge\tau')\restr(\Cal S\wedge\tau')\wedge\tau
=\pmb u\restr\Cal S\wedge\tau$ 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 near-simple, so $\pmb u\restr\Cal S\wedge\tau'$ is near-simple.

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 near-simple, so $\pmb u$ is locally near-simple.

\medskip

\quad{\bf (iv)} follows from (ii) here or from (a-v).

\medskip

\quad{\bf (v)} Write $A$ for $\{\sigma:\sigma\in\Cal S$,
$\pmb u\restr\Cal S\wedge\sigma\}$ is near-simple$\}$.

\medskip

\qquad\grheada\ If $\sigma\in A$ then
$\pmb v_{\sigma}=\langle u_{\rho\wedge\sigma}\rangle_{\rho\in\Cal S}$
is near-simple.
\Prf\ $\pmb v_{\sigma}\restr\Cal S\wedge\sigma=\pmb u\restr\Cal S\wedge\sigma$
is near-simple, by (a-iv), while $\pmb v_{\sigma}\restr\Cal S\vee\sigma$ is
constant with value
$u_{\sigma}$, therefore simple.   By (a-iv) in the other direction,
$\pmb v_{\sigma}$ is near-simple.\
\Qed

\medskip

\qquad\grheadb\ $A$ is closed under $\vee$.   \Prf\ If $\sigma$, $\tau\in A$,
then $u_{\rho\wedge(\sigma\vee\tau)}
=u_{\rho\wedge\sigma}+u_{\rho\wedge\tau}-u_{\rho\wedge\sigma\wedge\tau}$
for every $\rho\in\Cal S\wedge(\sigma\vee\tau)$, by 612D(f-i), that is,
$\pmb v_{\sigma\vee\tau}
=\pmb v_{\sigma}+\pmb v_{\tau}-\pmb v_{\sigma\wedge\tau}$.   By ($\alpha$)
here, $\pmb v_{\sigma}$, $\pmb v_{\tau}$
and $\pmb v_{\sigma\wedge\tau}$ are all near-simple, so the
linear combination $\pmb v_{\sigma\vee\tau}$ and
the restriction $\pmb u\restr\Cal S\wedge(\sigma\vee\tau)=
\pmb v_{\sigma\vee\tau}\restr\Cal S\wedge(\sigma\vee\tau)$ is near-simple.
So $\sigma\vee\tau\in A$.\ \Qed

\medskip

\qquad\grheadc\ Now suppose that $\tau\in\Cal S$.   Then $\{\tau\}$ is
covered by $A$ so for any
$\epsilon>0$ there are $\sigma_0,\ldots,\sigma_n\in A$ such that
$\sup_{i\le n}\Bvalue{\sigma_i=\tau}$ has measure at least $1-\epsilon$.   Now
$\sigma=\sup_{i\le n}\sigma_i$ belongs to $A$, by ($\beta$), and
$a=\Bvalue{\sigma<\tau}$ has measure at most $\epsilon$.   By ($\alpha$),
$\pmb v_{\sigma}$ is near-simple, while for $\rho\in\Cal S\wedge\tau$

\Centerline{$\Bvalue{u_{\rho}\ne v_{\sigma\rho}}
\Bsubseteq\Bvalue{\rho\ne\rho\wedge\sigma}=\Bvalue{\sigma<\rho}
\Bsubseteq\Bvalue{\sigma<\tau}=a$.}

\noindent Thus
$\Bvalue{\pmb u\restr\Cal S\wedge\tau\ne\pmb v_{\sigma}\restr\Cal S\wedge\tau}
\Bsubseteq a$ has measure at most $\epsilon$, while
$\pmb v_{\sigma}\restr\Cal S\wedge\tau$ is near-simple.   In particular,
$\pmb v_{\sigma}\restr\Cal S\wedge\tau$ is order-bounded;  as $\epsilon$ is
arbitrary, $\pmb u\restr\Cal S\wedge\tau$ is order-bounded, by 613B(p-ii).
But now we see from the same formula that for every $\epsilon>0$ there is a
$\pmb v\in\Mns(\Cal S\wedge\tau)$ such that
$\theta(\sup|\pmb u\restr\Cal S\wedge\tau-\pmb v|)\le\epsilon$;  as
$\Mns(\Cal S\wedge\tau)$ is closed in $\Mob(\Cal S\wedge\tau)$
in the ucp topology, $\pmb u\restr\Cal S\wedge\tau$ is near-simple.   As $\tau$
is arbitrary, $\pmb u$ is locally near-simple, as claimed.

\medskip

{\bf (c)(i)} If $\Cal S$ is empty, the result is trivial.   Otherwise,
since $\pmb u$ is moderately oscillatory, it is surely
order-bounded.   Let $\epsilon>0$.   By 615Ga,
$u_{\uparrow}=\lim_{\sigma\uparrow\Cal S}u_{\sigma}$ is
defined and there is a $\tau'\in\Cal S$ such that $\theta(w')\le\epsilon$ where
$w'=\sup_{\sigma\in\Cal S\vee\tau'}|u_{\sigma}-u_{\uparrow}|$.
Similarly, applying 615Gb to $A=\Cal S\wedge\tau'$,
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined
and there is a
$\tau\in\Cal S\wedge\tau'$ such that $\theta(w)\le\epsilon$ where
$w=\sup_{\sigma\in\Cal S\wedge\tau}|u_{\sigma}-u_{\downarrow}|$.

Now $\pmb u\restr\Cal S\cap[\tau,\tau']$ is near-simple,
so there is a simple process
$\pmb v'=\family{\sigma}{\Cal S\wedge\tau}{v'_{\sigma}}$ such that
$\theta(w'')\le\epsilon$ where
$w''=\sup_{\sigma\in\Cal S\wedge\tau}|u_{\sigma}-v'_{\sigma}|$.   Take a
breakpoint string $(\tau_0,\ldots,\tau_n)$ for $\pmb v'$ starting with
$\tau_0=\tau$ and ending with $\tau_n=\tau'$ (see 612Kb), and let
$\pmb v=\sfamilysS{v}$ be the simple process with domain $\Cal S$ based on
the starting value $u_{\downarrow}$, the breakpoint string
$(\tau_0,\ldots,\tau_n)$
and the values $(v'_{\tau_0},\ldots,v'_{\tau_n})$.   If $\sigma\in\Cal S$ then
$|u_{\sigma}-v_{\sigma}|\le w+2w'+2w''$.   \Prf\

\Centerline{$\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{v_{\sigma}=u_{\downarrow}}
   \Bcap\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau}}
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le w}$,}

\Centerline{$\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma\le\tau'}
\Bsubseteq\Bvalue{u_{\sigma}=u_{\med(\tau,\sigma,\tau')}}
    \Bcap\Bvalue{v_{\sigma}=v'_{\med(\tau,\sigma,\tau')}}
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le w''}$,}

$$\eqalign{\Bvalue{\tau'\le\sigma}&
\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\vee\tau'}}
   \Bcap\Bvalue{v_{\sigma}=v'_{\tau'}}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|
     \le|u_{\sigma\vee\tau'}-u_{\uparrow}|+|u_{\tau'}-u_{\uparrow}|
                 +|u_{\tau'}-v'_{\tau'}|}\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le 2w'+w''}.  \text{ \Qed}\cr}$$

Accordingly $\theta(\sup|\pmb u-\pmb v|)\le\theta(w+2w'+2w'')\le 5\epsilon$.
As $\epsilon$ is arbitrary. $\pmb u$ is near-simple.

\medskip

\quad{\bf (ii)} If $\pmb u$ is locally near-simple and $\tau\le\tau'$ in
$\Cal S$, then $\pmb u\restr\Cal S\wedge\tau'$ is near-simple so
$\pmb u\restr\Cal S\cap[\tau,\tau']=\pmb u\restr(\Cal S\wedge\tau')\vee\tau$ is
near-simple ((a-iv) above).   As $\tau$ and $\tau'$ are arbitrary,
(i) here tells us that $\pmb u$ is near-simple.
}%end of proof of 631F

\leader{631G}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$\hat\Cal S$ its covered envelope, $\pmb u=\sfamilysS{u}$ a fully adapted
process, and $\hatpmb u$ its fully adapted extension to $\hat\Cal S$.

(a) $\pmb u$ is near-simple iff $\hatpmb u$ is near-simple.

(b) $\pmb u$ is locally near-simple iff $\hatpmb u$ is locally near-simple.
%formerly 6{}36Gc  6{}36Xe

\proof{ If $\Cal S$ is empty both parts are trivial, so suppose otherwise.

\medskip

{\bf (a)(i)}  Suppose that $\pmb u$ is near-simple.
Then it is order-bounded, so
$\hatpmb u$ is order-bounded (612Q(d-i)),   Let $\epsilon>0$.
Then there is a simple process $\pmb v=\sfamilysS{v}$ such that
$\theta(\sup|\pmb u-\pmb v|)\le\epsilon$.   By 612Qf, the fully adapted
extension $\hatpmb v$ of $\pmb v$ to $\hat\Cal S$ is simple, while
$\sup|\hatpmb u-\hatpmb v|=\sup|\pmb u-\pmb v|$ (612Qb, 612Q(d-i)).
So $\theta(\sup|\hatpmb u-\hatpmb v|)\le\epsilon$;  as $\epsilon$ is arbitrary,
$\hatpmb u$ is near-simple.

\medskip

\quad{\bf (ii)} Suppose that $\hatpmb u$ is near-simple.

\medskip

\qquad\grheada\ $\hatpmb u$ is order-bounded,
so $\pmb u=\hatpmb u\restr\Cal S$ is order-bounded.   Let $\epsilon>0$.
Then there is a simple
process $\pmb w=\family{\tau}{\hat\Cal S}{w_{\tau}}$ such that
$\theta(\sup|\hatpmb u-\pmb w|)\le\epsilon$.   Let $w_{\downarrow}$
be the starting value of $\pmb w$ and
$(\tau_0,\ldots,\tau_n)$ a breakpoint sequence for $\pmb w$.   For each
$i\le n$, $\sup_{\sigma\in\Cal S}\Bvalue{\tau_i=\sigma}=1$, so there is a
finite set $A_i\subseteq\Cal S$ such that
$\bar\mu(\sup_{\sigma\in A_i}\Bvalue{\tau_i=\sigma})
\ge 1-\Bover{\epsilon}{n+1}$.   Let $I$ be a finite sublattice of $\Cal S$
including $\bigcup_{i\le n}A_i$, and take $(\rho_0,\ldots,\rho_m)$ linearly
generating the $I$-cells.   If $i\le n$,

$$\eqalignno{\sup_{\sigma\in A_i}\Bvalue{\tau_i=\sigma}&
\Bsubseteq\sup_{\sigma\in I}\Bvalue{\tau_i=\sigma}
=\sup_{\sigma\in I,j\le m}\Bvalue{\tau_i=\sigma}\Bcap\Bvalue{\sigma=\rho_j}
\Displaycause{611L}
\Bsubseteq\sup_{j\le m}\Bvalue{\tau_i=\rho_j},\cr}$$

\noindent so

\Centerline{$a=\inf_{i\le n}\sup_{j\le m}\Bvalue{\tau_i=\rho_j}$}

\noindent has measure at least $1-\epsilon$.   Note that

\Centerline{$a
\Bsubseteq\sup_{j\le m}\Bvalue{\rho_j\le\tau_0}
     \Bcap\sup_{j\le m}\Bvalue{\tau_n\le\rho_j}
\Bsubseteq\Bvalue{\rho_0\le\tau_0}\Bcap\Bvalue{\tau_n\le\rho_m}$.}

\noindent In between,

\Centerline{$a\Bcap\Bvalue{\tau_i<\rho_{j+1}}
  \Bsubseteq\Bvalue{\tau_i\le\rho_j}$}

\noindent whenever $i\le n$ and $j<m$.   \Prf\

$$\eqalign{a\Bcap\Bvalue{\tau_i<\rho_{j+1}}&
\Bsubseteq\sup_{k\le m}\Bvalue{\tau_i<\rho_{j+1}}\Bcap\Bvalue{\tau_i=\rho_k}
=\sup_{k\le m}\Bvalue{\rho_k<\rho_{j+1}}\Bcap\Bvalue{\tau_i=\rho_k}\cr&
\Bsubseteq\sup_{k\le j}\Bvalue{\tau_i=\rho_k}
\Bsubseteq\Bvalue{\tau_i\le\rho_j}.  \text{ \Qed}\cr}$$

\medskip

\qquad\grheadb\
Now let $\pmb v=\sfamilysS{v}$ be the simple process with starting value
$w_{\downarrow}$, breakpoint string $(\rho_0,\ldots,\rho_m)$ and values
$v_{\rho_j}=w_{\rho_j}$ for $j\le m$.   Take $\sigma\in\Cal S$.   Let $\frak B$
be the finite subalgebra of $\frak A$ generated by

\Centerline{$\{\Bvalue{\tau_i=\rho_j}:i\le n$, $j\le m\}
  \cup\{\Bvalue{\tau_i\le\sigma}:i\le n\}
  \cup\{\Bvalue{\rho_j\le\sigma}$, $j\le m\}$;}

\noindent note that $a\in\frak B$.

\medskip

\qquad\grheadc\ Let $b$ be a atom of $\frak B$ included in
$a$.   Then $b\Bsubseteq\Bvalue{v_{\sigma}=w_{\sigma}}$.
\Prf\ Since

\Centerline{$(\Bvalue{\sigma<\rho_0},
\Bvalue{\rho_0\le\sigma}\Bcap\Bvalue{\sigma<\rho_1},
\ldots,\Bvalue{\rho_{m-1}\le\sigma}\Bcap\Bvalue{\sigma<\rho_m},
\Bvalue{\rho_m\le\sigma})$}

\noindent is a partition of unity in $\frak A$, one of these includes $b$;  and
similarly either $b\Bsubseteq\Bvalue{\sigma<\tau_0}$ or
$b\Bsubseteq\Bvalue{\tau_n\le\sigma}$ or there is an $i<n$ such that
$b\Bsubseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma\le\tau_{i+1}}$.
Now $b\Bsubseteq a\Bsubseteq\Bvalue{\rho_0\le\tau_0}$ so if
$b\Bsubseteq\Bvalue{\sigma<\rho_0}$ then $b\Bsubseteq\Bvalue{\sigma<\tau_0}$ and

\Centerline{$b
\Bsubseteq\Bvalue{v_{\sigma}=w_{\downarrow}}
   \Bcap\Bvalue{w_{\sigma}=w_{\downarrow}}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\sigma}}$.}

\noindent Similarly, $b\Bsubseteq a\Bsubseteq\Bvalue{\tau_n\le\rho_m}$ so if
$b\Bsubseteq\Bvalue{\rho_m\le\sigma}$ then $b\Bsubseteq\Bvalue{\tau_n\le\sigma}$
and

\Centerline{$b
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_m}}
   \Bcap\Bvalue{w_{\rho_m}=w_{\tau_n}}\Bcap\Bvalue{w_{\sigma}=w_{\tau_n}}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\sigma}}$.}

\noindent Otherwise, let $j<m$ be such that

\Centerline{$b\Bsubseteq\Bvalue{\rho_j\le\sigma}\Bcap\Bvalue{\sigma<\rho_{j+1}}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_j}}$.}

\noindent If $b\Bsubseteq\Bvalue{\sigma<\tau_0}$ then

$$\eqalign{b&
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_j}}
     \Bcap\Bvalue{\rho_j<\tau_0}\Bcap\Bvalue{w_{\sigma}=w_{\downarrow}}\cr&
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_j}}\Bcap\Bvalue{w_{\rho_j}=w_{\sigma}}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\sigma}}.\cr}$$

\noindent If $b\Bsubseteq\Bvalue{\tau_n\le\sigma}$ then

$$\eqalignno{b&
\Bsubseteq a\Bcap\Bvalue{v_{\sigma}=w_{\rho_j}}
     \Bcap\Bvalue{\tau_n<\rho_{j+1}}\Bcap\Bvalue{w_{\sigma}=w_{\tau_n}}\cr&
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_j}}
     \Bcap\Bvalue{\tau_n\le\rho_j}\Bcap\Bvalue{w_{\sigma}=w_{\tau_n}}
\Displaycause{because
$a\Bcap\Bvalue{\tau_n<\rho_{j+1}}\Bsubseteq\Bvalue{\tau_n\le\rho_j}$}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_j}}
     \Bcap\Bvalue{w_{\rho_j}=w_{\tau_n}}\Bcap\Bvalue{w_{\sigma}=w_{\tau_n}}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\sigma}}.\cr}$$

\noindent And in between, if $i<n$ and
$b\Bsubseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}$, then

\Centerline{$b
\Bsubseteq a\Bcap\Bvalue{\tau_i<\rho_{j+1}}\Bcap\Bvalue{\rho_j<\tau_{i+1}}
\Bsubseteq\Bvalue{\tau_i\le\rho_j}\Bcap\Bvalue{\rho_j<\tau_{i+1}}
\Bsubseteq\Bvalue{w_{\rho_j}=w_{\tau_i}}$}

\noindent so

\Centerline{$b
\Bsubseteq\Bvalue{v_{\sigma}=w_{\rho_j}}\Bcap\Bvalue{w_{\sigma}=w_{\tau_i}}
    \Bcap\Bvalue{w_{\rho_j}=w_{\tau_i}}
\Bsubseteq\Bvalue{v_{\sigma}=w_{\sigma}}$}

\noindent in this case also, and every possibility is covered.\ \Qed

\medskip

\qquad\grheadd\ As $b$ is arbitrary,
$\Bvalue{v_{\sigma}=w_{\sigma}}\Bsupseteq a$, and this is true for every
$\sigma\in\Cal S$.   But now we see that

\Centerline{$|u_{\sigma}-v_{\sigma}|\times\chi a
=|u_{\sigma}-w_{\sigma}|\times\chi a
\le\sup|\hatpmb u-\pmb w|$}

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

\Centerline{$\sup|\pmb u-\pmb v|\times\chi a\le\sup|\hatpmb u-\pmb w|$,
\quad$\theta(\sup|\pmb u-\pmb v|)
\le\theta(\sup|\hatpmb u-\pmb w|)+\bar\mu(1\Bsetminus a)\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\pmb u$ is near-simple, as claimed.

\medskip

{\bf (b)(i)} Now suppose that $\pmb u$ is locally near-simple.
If $\sigma\in\Cal S$, then $\hat\Cal S\wedge\sigma$ is the covered envelope of
$\Cal S\wedge\sigma$ (611M(e-i)) so
$\hatpmb\restr\hat\Cal S\wedge\sigma$ is the fully
adapted extension of $\pmb u\restr\Cal S\wedge\sigma$ to its covered envelope
and is near-simple, by (a).   As $\Cal S$ covers $\hat\Cal S$, 631F(b-v) tells
us that $\hatpmb u$ is locally near-simple.

\medskip

\quad{\bf (ii)} Finally, suppose that $\hatpmb u$ is locally
near-simple and that $\sigma\in\Cal S$.   Again because
the fully adapted extension of $\pmb u\restr\Cal S\wedge\sigma$ to the covered
envelope of $\Cal S\wedge\sigma$ is
$\hatpmb u\restr\hat\Cal S\wedge\sigma$, which is near-simple;  so
(a) in the other direction
tells us that $\pmb u\restr\Cal S\wedge\sigma$ is near-simple.   As $\sigma$ is
arbitrary, $\pmb u$ is locally near-simple.
}%end of proof of 631G

\leader{631H}{}\cmmnt{ Many of the arguments of \S\S614-617 are substantially
simplified if we restrict attention to near-simple processes, and some curious
new patterns emerge.   A striking one, which will be useful in \S641, %641W
is the following.

\medskip

\noindent}{\bf Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$ a process of bounded variation with domain $\Cal S$.

(a)(i) $\int_{\Cal S}\pmb u\,d\pmb v$ is defined for every
$\pmb v\in\Mns(\Cal S)$.

\quad(ii) $\pmb v\mapsto\int_{\Cal S}\pmb u\,d\pmb v:\Mns(\Cal S)\to L^0$
is continuous for the ucp topology on $\Mns(\Cal S)$
and the topology of convergence in measure on $L^0$.

(b)(i) The indefinite integral
$\iiv(\pmb u)$ is near-simple for every $\pmb v\in\Mns(\Cal S)$.

\quad(ii) $\pmb v\mapsto\iiv(\pmb u):\Mns(\Cal S)\to\Mns(\Cal S)$
is continuous for the ucp topology.

\proof{{\bf (a)(i)}\grheada\ Of course $\pmb u$ is order-bounded (614La).
Write $\bar u$ for $2\sup|\pmb u|+\int_{\Cal S}|d\pmb u|$.
The key is the following fact:  for any $\pmb v\in\Mob(\Cal S)$,

\Centerline{$|S_I(\pmb u,d\pmb v)|
\le\bar u\times\sup|\pmb v|$}

\noindent for every $I\in\CalIS$.   \Prf\ If $I$ is empty, this is trivial.
Otherwise, take a sequence $(\tau_0,\ldots,\tau_n)$ linearly generating the
$I$-cells.   Expressing $\pmb u$ as $\sfamilysS{u}$ and $\pmb v$ as
$\sfamilysS{v}$,

$$\eqalign{S_I(\pmb u,d\pmb v)&
=\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})\cr&
=\sum_{i=0}^{n-1}(u_{\tau_i}-u_{\tau_{i+1}})\times v_{\tau_{i+1}}
    -u_{\tau_0}\times v_{\tau_0}+u_{\tau_n}\times v_{\tau_n}\cr}$$

\noindent so

$$\eqalign{|S_I(\pmb u,d\pmb v)|&
\le\sum_{i=0}^{n-1}|u_{\tau_i}-u_{\tau_{i+1}}|\times|v_{\tau_{i+1}}|
    +|u_{\tau_0}|\times|v_{\tau_0}|+|u_{\tau_n}|\times|v_{\tau_n}|\cr&
\le\bar u\times\sup|\pmb v|\cr}$$

\noindent (see 614I).\ \Qed

\medskip

\qquad\grheadb\ If $\pmb v\in\Msimp(\Cal S)$ then $\pmb v$ is of bounded
variation (614Q(a-iii)), therefore an integrator (616Ra), while of course
$\pmb u$ is also an integrator and therefore moderately oscillatory.   So in
this case $\int_{\Cal S}\pmb u\,d\pmb v$ is defined.   Now suppose that
$\pmb v\in\Mns(\Cal S)$.   Let $\epsilon>0$.   Then there
is a $\delta>0$ such that $\theta(\bar u\times x)\le\epsilon$ whenever
$x\in L^0(\frak A)$ and $\theta(x)\le\delta$.   Take
$\pmb v'\in\Msimp(\Cal S)$ such that $\theta(\sup|\pmb v'-\pmb v|)\le\delta$ and
$J\in\CalIS$ such that
$\theta(S_I(\pmb u,d\pmb v')-\int_{\Cal S}\pmb u\,d\pmb v')\le\epsilon$ whenever
$J\subseteq I\in\CalIS$.   In this case, if $J\subseteq I\in\CalIS$,

$$\eqalign{\theta(S_I(\pmb u,d\pmb v)-\int_{\Cal S}\pmb u\,d\pmb v')&
\le\theta(S_I(\pmb u,d\pmb v)-S_I(\pmb u,d\pmb v'))
   +\theta(S_I(\pmb u,d\pmb v')-\int_{\Cal S}\pmb u\,d\pmb v')\cr&
\le\theta(S_I(\pmb u,d(\pmb v-\pmb v')))+\epsilon
\le\theta(\bar u\times\sup|\pmb v-\pmb v'|)+\epsilon
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\int_{\Cal S}\pmb u\,d\pmb v=\lim_{I\uparrow\CalIS}S_I(\pmb u,d\pmb v)$
is defined.

\medskip

\quad{\bf (ii)} If $\pmb v\in\Mns(\Cal S)$ then

\Centerline{$\theta(\int_{\Cal S}\pmb u\,d\pmb v)
\le\sup_{I\in\CalIS}\theta(S_I(\pmb u,d\pmb v))
\le\theta(\bar u\times\sup|\pmb v|)$,}

\noindent so the linear operator
$\pmb v\mapsto\int_{\Cal S}\pmb u\,d\pmb v:\Mns(\Cal S)\to L^0$
is continuous at $0$, therefore continuous.

\medskip

{\bf (b)(i)} If $\pmb v\in\Msimp(\Cal S)$ then $\iivu$ is simple, by
614D.   Now suppose that $\pmb v\in\Mns(\Cal S)$ and $\epsilon>0$.
Again take $\delta>0$ such that $\theta(\bar u\times x)\le\epsilon$ whenever
$\theta(x)\le\delta$, and $\pmb v'\in\Msimp(\Cal S)$ such that
$\theta(\sup|\pmb v'-\pmb v|)\le\delta$.   For any $\tau\in\Cal S$,

\Centerline{$|\int_{\Cal S\wedge\tau}\pmb u\,d(\pmb v-\pmb v')|
\le\sup_{I\in\Cal I(\Cal S\wedge\tau)}|S_I(\pmb u,d(\pmb v-\pmb v'))|
\le\bar u\times\sup|\pmb v-\pmb v'|$,}

\noindent so

\Centerline{$\theta(\sup|\iivu-ii_{\pmb v'}(\pmb u)|)
\le\theta(\bar u\times\sup|\pmb v-\pmb v'|)
\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$\iivu$ is approximated in the ucp topology by simple processes
and is near-simple.

\medskip

\quad{\bf (ii)} Finally, the same arguments tell us that for $\pmb v$,
$\pmb v'\in\Mns(\Cal S)$,

\Centerline{$\theta(\sup|\iivu-ii_{\pmb v'}(\pmb u)|)
\le\theta(\bar u\times\sup|\pmb v-\pmb v'|)$}

\noindent so that $\pmb v\mapsto\iivu$ is continuous.
}%end of proof of 631H

\leader{631I}{}\cmmnt{ We have another result showing that an indefinite
integral will share a property with the corresponding integrator, as in
614U, 615Rb, 616J, 616Q(c-i) and 618Q-618R.

\medskip

\noindent}{\bf Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u$ a moderately oscillatory process and
$\pmb v$ a near-simple integrator, both with domain $\Cal S$.
Then $\iivu$ is near-simple.

\proof{{\bf (a)} Suppose to begin with that $\Cal S$ is full and has a greatest
element.   Take $\epsilon>0$.   Let $\delta>0$ be such that
$\theta(\sup|\iiv(\pmb w)|)\le\epsilon$ whenever $\pmb w$ is a moderately
oscillatory process with domain $\Cal S$ such that
$\theta(\sup|\pmb w|)\le\delta$ (616J).   Let $\pmb u'$ be a process of bounded
variation such that $\theta(\sup|\pmb u-\pmb u'|)\le\delta$.
Then $\iiv(\pmb u')$ is near-simple, by 631H(b-i), while

\Centerline{$\theta(\sup|\iivu-\iiv(\tildepmb u)|)
=\theta(\sup|\iiv(\pmb u-\tildepmb u)|)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\iivu$ belongs to the closure of
$\Mns(\Cal S)$ in $\Mob(\Cal S)$ and is near-simple, by the definition
in 631B.

\medskip

{\bf (b)} Now suppose just that $\Cal S$ has a greatest member.   Let
$\hatpmb u$ and $\hatpmb v$ be the fully adapted extensions of $\pmb u$ and
$\pmb v$ to the covered envelope $\hat\Cal S$ of $\Cal S$.   Then
$\hatpmb u$ is moderately oscillatory (615F(a-vi)) and $\hatpmb v$ is a
near-simple
integrator (631Ga, 616Ia) so $ii_{\hatpmb v}(\hatpmb u)$ is near-simple, by
(a).   But this is just the fully adapted extension of $\iivu$ to
$\hat\Cal S$ (616Q(c-ii)).   So $\iivu$ is near-simple, by 631Ga in the other
direction.

\medskip

{\bf (c)} For the general case, we can apply (b) to
$\pmb u\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\wedge\tau$ to see that
$\iivu\restr\Cal S\wedge\tau$ is near-simple for every $\tau\in\Cal S$, that
is. $\iivu$ is locally near-simple.
We know also that $\iivu$ is an integrator (616J), therefore
moderately oscillatory (616Ib).
By 631F(c-ii), $\iivu$ is near-simple.
}%end of proof of 631I

\leader{631J}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.

(a) If $\pmb v$, $\pmb w$ are near-simple integrators with domain $\Cal S$,
then $\covar{\pmb v}{\pmb w}$ and $\pmb v^*$ are near-simple.

(b) If $\pmb v$, $\pmb w$ are locally near-simple local
integrators with domain $\Cal S$,
then $\covar{\pmb v}{\pmb w}$ and $\pmb v^*$ are locally near-simple.

\proof{{\bf (a)} By 617I, applied in $\Cal S\wedge\tau$ with
$\pmb u=\tbf 1$ for $\tau\in\Cal S$,

\Centerline{$\covar{\pmb v}{\pmb w}
=\pmb v\times\pmb w-(v_{\downarrow}\times w_{\downarrow})\tbf 1
   -\iiv(\pmb w)-ii_{\pmb w}(\pmb v)$}

\noindent where $v_{\downarrow}$ and $w_{\downarrow}$ are the starting values of
$\pmb v$ and $\pmb w$.   Now
$\iiv(\pmb w)$ and $ii_{\pmb w}(\pmb v)$ are near-simple by 631I, and
$(v_{\downarrow}\times w_{\downarrow})\tbf 1$ is simple;
so 631F(a-ii) tells us that $\covar{\pmb v}{\pmb w}$ is near-simple.
Of course it follows at once that $\pmb v^*=\covar{\pmb v}{\pmb v}$ is
near-simple.

\medskip

{\bf (b)} Apply (a) to $\Cal S\wedge\tau$ for $\tau\in\Cal S$.
}%end of proof of 631J

\leader{631K}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v$ a near-simple process of bounded variation.
Then its cumulative variation
$\pmb v^{\uparrow}$\cmmnt{ (614O)}
is near-simple.

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

\medskip

{\bf (a)(i)} Let
$\epsilon>0$.   Let $\pmb u'=\sfamilysS{u'}$ be a simple process such that
$\theta(\bar w_0)\le\epsilon$, where $\bar w_0=\sup|\pmb v-\pmb u'|$.
Let $I_0$ be a non-empty
finite sublattice of $\Cal S$ including a breakpoint string for $\pmb u'$.
Take $I\in\Cal I(\Cal S)$ such that $I_0\subseteq I$ and
$\theta(\bar w_1)\le\epsilon$, where
$\bar w_1=\int_{\Cal S}|d\pmb v|-S_I(\tbf 1,|d\pmb v|)$.   Take
$(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells;  note that
$(\tau_0,\ldots,\tau_n)$ is a breakpoint string for $\pmb u'$ (612Kb).
Let $u'_{\downarrow}$ be the starting value of $\pmb u'$, and
$v_{\downarrow}$ the starting value of $\pmb v$.   Of course

\Centerline{$|v_{\downarrow}-u'_{\downarrow}|
=\lim_{\sigma\downarrow\Cal S}|v_{\sigma}-u'_{\sigma}|
\le\bar w_0$.}

\medskip

\quad{\bf (ii)} Express $\pmb v$ as $\sfamilysS{v}$.
Note that if $\sigma\in\Cal S$ then

\Centerline{$\Bvalue{\sigma<\tau_0}
\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\downarrow}}
\Bsubseteq\Bvalue{|v_{\sigma}-u'_{\downarrow}|\le\bar w_0}
\Bsubseteq\Bvalue{|v_{\sigma}-v_{\downarrow}|\le 2\bar w_0}$,}

$$\eqalign{\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
&\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\tau_i}}
\Bsubseteq\Bvalue{|v_{\sigma}-u'_{\tau_i}|\le\bar w_0}
  \Bcap\Bvalue{|v_{\tau_i}-u'_{\tau_i}|\le\bar w_0}
\cr&\Bsubseteq\Bvalue{|v_{\sigma}-v_{\tau_i}|\le 2\bar w_0}\cr}$$

\noindent for $i<n$,

\Centerline{$\Bvalue{\tau_n\le\sigma}
\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\tau_n}}
\Bsubseteq\Bvalue{|v_{\sigma}-v_{\tau_n}|\le 2\bar w_0}$.}

\medskip

{\bf (b)} Set $\bar w=\bar w_1+2\bar w_0$.   Take any $\tau\in\Cal S$.

\medskip

\quad{\bf (i)} $\Bvalue{\tau<\tau_0}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}|\le\bar w}$.
\Prf\ Set $\sigma=\tau\wedge\tau_0$.   Then
$0\le v^{\uparrow}_{\sigma}\le|v_{\sigma}-v_{\downarrow}|+\bar w_1$
(614P(c-ii-$\alpha$)) and

$$\eqalign{\Bvalue{\tau<\tau_0}
&\Bsubseteq\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma<\tau_0}
\Bsubseteq\Bvalue{v^{\uparrow}_{\tau}=v^{\uparrow}_{\sigma}}\Bcap
  \Bvalue{|v_{\sigma}-v_{\downarrow}|\le 2\bar w_0}
\cr&\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}|\le\bar w_1+2\bar w_0}
=\Bvalue{|v^{\uparrow}_{\tau}|\le\bar w}.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (ii)} If $i<n$ then
$\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_i}|\le\bar w}$.
\Prf\ Set $\sigma=\med(\tau_i,\tau,\tau_{i+1})$.
Then $0\le v^{\uparrow}_{\sigma}-v^{\uparrow}_{\tau_i}
\le|v_{\sigma}-v_{\tau_i}|+\bar w_1$
(614P(c-ii-$\beta$)).   Now

$$\eqalign{\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}
&\Bsubseteq\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}\cr&
\Bsubseteq
   \Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_i}|
        \le|v_{\sigma}-v_{\tau_i}|+\bar w_1}
   \Bcap\Bvalue{|v_{\sigma}-v_{\tau_i}|\le 2\bar w_0}
\cr&\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_i}|
        \le\bar w_1+2\bar w_0}.
\text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)}
$\Bvalue{\tau_n\le\tau}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_n}|\le\bar w}$.
\Prf\ Set $\sigma=\tau\vee\tau_n$.   Then
$0\le v^{\uparrow}_{\sigma}-v^{\uparrow}_{\tau_n}
\le|v_{\sigma}-v_{\tau_n}|+\bar w_1$
(614P(c-ii-$\gamma$)).   So

\Centerline{$\Bvalue{\tau_n\le\tau}
\Bsubseteq\Bvalue{\tau=\sigma}
  \Bcap\Bvalue{|v_{\sigma}-v_{\tau_n}|\le 2\bar w_0}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_n}|
\le 2\bar w_0+\bar w_1}$.
\Qed}

\medskip

{\bf (c)} So if we take
$\pmb v'=\family{\tau}{\Cal S}{v'_{\tau}}$
to be the simple process
with breakpoint string $(\tau_0,\ldots,\tau_n)$,
$v'_{\tau_i}=v^{\uparrow}_{\tau_i}$
for $i\le n$ and $\Bvalue{\tau<\tau_0}\Bsubseteq\Bvalue{v'_{\tau}=0}$ for
every $\tau\in\Cal S$, we shall have, for any $\tau\in\Cal S$,

\Centerline{$\Bvalue{\tau<\tau_0}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}|\le\bar w}
  \Bcap\Bvalue{v'_{\tau}=0}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v'_{\tau}|\le\bar w}$,}

\Centerline{$\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_i}|\le\bar w}
   \Bcap\Bvalue{v'_{\tau}=v^{\uparrow}_{\tau_i}}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v'_{\tau}|\le\bar w}$}

\noindent for $i<n$,

\Centerline{$\Bvalue{\tau_n\le\tau}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v^{\uparrow}_{\tau_n}|\le\bar w}
   \Bcap\Bvalue{v'_{\tau}=v^{\uparrow}_{\tau_n}}
\Bsubseteq\Bvalue{|v^{\uparrow}_{\tau}-v'_{\tau}|\le\bar w}$;}

\noindent assembling these, we see that
$|v^{\uparrow}_{\tau}-v'_{\tau}|\le\bar w$.

Thus $\sup|\pmb v^{\uparrow}-\pmb v'|\le\bar w$,
while $\theta(\bar w)\le 2\theta(w_0)+\theta(w_1)\le 3\epsilon$ and
$\pmb v'$ is simple.   As $\epsilon$ was arbitrary, $\pmb v^{\uparrow}$ is
near-simple.
}%end of proof of 631K

\leader{631L}{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 near-simple and of bounded variation
iff it is expressible as the difference of two
non-negative non-decreasing near-simple processes.

\proof{ Given that $\pmb v=\sfamilysS{v}$ is of bounded variation, let
$\pmb v^{\uparrow}$ be its cumulative variation
and set $v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$;
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 near-simple
if $\pmb v$ is (631K).   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 non-decreasing near-simple processes,
these are order-bounded (by the definition 631Ba) so
$\pmb v$ is of bounded variation by 614I and near-simple
by 631F(a-ii).
}%proof of 631L

\leader{631M}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\Cal S'$ a sublattice of $\Cal S$ which is coinitial with $\Cal S$.

(a)(i) There is a
unique function $\Phi:\Msimp(\Cal S')\to\Msimp(\Cal S)$ such that,
for every $\pmb u\in\Msimp(\Cal S')$, $\Phi(\pmb u)$ extends $\pmb u$
and has a breakpoint string in $\Cal S'$.

\quad (ii) $\Phi(\bar h\pmb u)=\bar h\Phi(\pmb u)$ for every
Borel measurable $h:\Bbb R\to\Bbb R$ and every $\pmb u\in\Msimp(\Cal S')$.

\quad (iii) $\Phi$ is a multiplicative Riesz homomorphism.

\quad (iv) If $\Cal S'\ne\emptyset$ then
$\pmb u$ and $\Phi(\pmb u)$ have the same starting value for every
$\pmb u\in\Msimp(\Cal S')$.

\quad (v) $\sup|\Phi(\pmb u)|=\sup|\pmb u|$ for every
$\pmb u\in\Msimp(\Cal S')$.

(b)(i) There is a
unique function $\Psi:\Mns(\Cal S')\to\Mns(\Cal S)$ such that $\Psi$
extends $\Phi$ and is continuous with respect to the ucp topologies on
$\Mns(\Cal S')$ and $\Mns(\Cal S)$.

\quad (ii) $\Psi(\pmb u)\restr\Cal S'=\pmb u$ and
$\sup|\Psi(\pmb u)|=\sup|\pmb u|$
for every $\pmb u\in\Mns(\Cal S')$.

\quad (iii) $\Psi$ is a multiplicative
Riesz homomorphism and $\Psi(\bar h\pmb u)=\bar h\Psi(\pmb u)$ for
every continuous $h:\Bbb R\to\Bbb R$ and every $\pmb u\in\Mns(\Cal S')$.
%648D 648G 652Z 
% include:    615C(b-ii)

\quad(iv) For $\pmb u\in\Mns(\Cal S')$,
$\Psi(\pmb u)$ is non-decreasing iff $\pmb u$ is non-decreasing.
%648D 649F

\quad(v) For $\pmb u\in\Mns(\Cal S')$,
$\sup|\Psi(\pmb u)|=\sup|\pmb u|$, so
$\Bvalue{\Psi(\pmb u)\ne\tbf 0}=\Bvalue{\pmb u\ne\tbf 0}$.
%648G

(c) Now suppose that $\sup_{\tau\in\Cal S'}\Bvalue{\sigma\le\tau}=1$ for every
$\sigma\in\Cal S$.

\quad (i) If $\pmb v$ is an integrator with domain $\Cal S$,
then $\int_{\Cal S'}\pmb u\,d\pmb v=\int_{\Cal S}\Psi(\pmb u)\,d\pmb v$
for every $\pmb u\in\Mns(\Cal S')$.

\quad (ii) There
is a unique function $\Psi^*:\Mlns(\Cal S')\to\Mlns(\Cal S)$
extending the map $\Phi:\Msimp(\Cal S')\to\Msimp(\Cal S)$ and such that
$\sup|\Psi^*(\pmb u)\restr\Cal S\wedge\tau|=\sup|\pmb u\restr\Cal S'\wedge\tau|$
whenever $\pmb u\in\Mlns(\Cal S)$ and $\tau\in\Cal S'$.

\quad(iii) $\Psi^*(\pmb u)\restr\Cal S'=\pmb u$ for every
$\pmb u\in\Mlns(\Cal S')$, $\Psi^*$ is a multiplicative Riesz homomorphism,
$\Psi^*(\bar h\pmb u)=\bar h\Psi^*(\pmb u)$ for
every continuous $h:\Bbb R\to\Bbb R$ and every $\pmb u\in\Mlns(\Cal S')$, and
$\Psi^*(\pmb u)$ is non-decreasing whenever $\pmb u\in\Mlns(\Cal S')$ is
non-decreasing.

\proof{ If $\Cal S'$ is empty so is $\Cal S$ and everything is trivial, so
suppose otherwise.

\medskip

{\bf (a)(i)}\grheada\ If $\pmb u\in\Msimp(\Cal S')$, there is a
$\pmb v\in\Msimp(\Cal S)$ such that $\pmb v$ extends $\pmb u$ and
$\pmb v$ has a breakpoint string consisting of members of $\Cal S'$.
\Prf\ Let $(\tau_0,\ldots,\tau_n)$ be a breakpoint string for
$\pmb u$, and $u_{\downarrow}$ its starting value.
Then $u_{\downarrow}\in L^0(\bigcap_{\sigma\in\Cal S'}\frak A_{\sigma})$;
because $\Cal S'$ is coinitial with $\Cal S$,
$\bigcap_{\sigma\in\Cal S'}\frak A_{\sigma}
=\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$.   We therefore have a
simple process $\pmb v=\sfamilysS{v}$ defined from
$\tau_0,\ldots,\tau_n,u_{\downarrow},u_{\tau_0},\ldots,u_{\tau_n}$ as in 612Ka.
Take $\sigma\in\Cal S'$.   Then $\Bvalue{u_{\sigma}=v_{\sigma}}$ includes

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

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

\noindent and

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

\noindent so $u_{\sigma}=v_{\sigma}$.   As $\sigma$ is arbitrary,
$\pmb v$ extends $\pmb u$, and of course it has a breakpoint string
$(\tau_0,\ldots,\tau_n)$ in $\Cal S'$.\ \Qed

\medskip

\qquad\grheadb\ If $\pmb u\in\Msimp(\Cal S')$, there is at most one
$\pmb v\in\Msimp(\Cal S)$ such that $\pmb v$ extends $\pmb u$ and
$\pmb v$ has a breakpoint string consisting of members of $\Cal S'$.
\Prf\ Suppose that $\pmb v=\sfamilysS{v}$ and
$\pmb v'=\sfamilysS{v'}$ are two such processes.
Then there is a finite sublattice of $\Cal S'$ which includes
breakpoint strings for both $\pmb v$ and $\pmb v'$, and now a string
$(\tau_0,\ldots,\tau_n)$ which linearly generates the $I$-cells will be a
breakpoint string for both $\pmb v$ and $\pmb v'$ (612Kb).   Observe
also that $\pmb v$, $\pmb v'$ can be defined from
$v_{\downarrow},v_{\tau_0},\ldots,v_{\tau_n}$ and
$v'_{\downarrow},v'_{\tau_0},\ldots,v'_{\tau_n}$ where

\Centerline{$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$,
\quad$v'_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v'_{\sigma}$}

\noindent by 614Ba.   But as $\Cal S'$ is coinitial with $\Cal S$,

\Centerline{$v_{\downarrow}
=\lim_{\sigma\downarrow\Cal S}v_{\sigma}
=\lim_{\sigma\downarrow\Cal S'}v_{\sigma}
=\lim_{\sigma\downarrow\Cal S'}u_{\sigma}
=u_{\downarrow}$,}

\noindent and similarly $v'_{\downarrow}=u_{\downarrow}$.   And of course
$v'_{\tau_i}=u_{\tau_i}=v_{\tau_i}$ for every $i\le n$.   So
$\pmb v=\pmb v'$.\ \Qed

We can therefore define
$\Phi:\Msimp(\Cal S')\to\Msimp(\Cal S)$ in the way claimed.

\medskip

\quad{\bf (ii)} If $\pmb u\in\Msimp(\Cal S')$ and $h:\Bbb R\to\Bbb R$ is Borel
measurable, then $\bar h\pmb u$ is simple so
$\bar h\Phi(\pmb u)$ is simple and extends $\bar h\pmb u$;
moreover, any breakpoint string for $\pmb u$ is a breakpoint string for
$\Phi(\pmb u)$ and $\bar h\Phi(\pmb u)$ (612La).
So $\bar h\Phi(\pmb u)=\Phi(\bar h\pmb u)$.

\medskip

\quad{\bf (iii)} If $\pmb u$,
$\pmb u'\in\Msimp(\Cal S')$, then
$\Phi(\pmb u+\pmb u')=\Phi(\pmb u)+\Phi(\pmb u')$.   \Prf\
$\Phi(\pmb u)+\Phi(\pmb u')$ belongs to $\Msimp(\Cal S)$, extends
$\pmb u+\pmb u'$ and has a breakpoint string in $\Cal S'$ (see the
argument in (i-$\beta$) above), so must be equal to $\Phi(\pmb u+\pmb u')$.\
\QeD\

It follows from (ii) just above that

\Centerline{$\Phi(\alpha\pmb u)=\alpha\Phi(\pmb u)$,
\quad$\Phi(\pmb u^2)=(\Phi(\pmb u))^2$,
\quad$\Phi|\pmb u|=|\Phi(\pmb u)|$}

\noindent whenever $\pmb u\in\Msimp(\Cal S')$ and $\alpha\in\Bbb R$, so
that $\Phi$ is a multiplicative Riesz homomorphism (cf.\ 612Bc).

\medskip

\quad{\bf (iv)} I noted this in (i-$\beta$) above.

\medskip

\quad{\bf (v)} Since $\pmb u$ and $\Phi(\pmb u)$ have the same starting value
$u_{\downarrow}$
and agree on a common breakpoint string $(\tau_0,\ldots,\tau_n)$, 614Ec tells us
that

\Centerline{$\sup|\pmb u|=|u_{\downarrow}\vee\sup_{i\le n}|u_{\tau_i}|
=\sup|\Phi(\pmb u)|$.}

{\bf (b)(i)} In the notation of 615B,

\Centerline{$\widehat{\theta}\Phi(\pmb u)
=\theta(\sup|\Phi(\pmb u)|)=\widehat{\theta}(\pmb u)$}

\noindent for every
$\pmb u\in\Msimp(\Cal S')$, and $\Phi$, regarded as a linear operator from
$\Msimp(\Cal S')$ to $\Mob(\Cal S)$ is continuous for the ucp topologies.
Because
$\Mns(\Cal S')$ is the closure of $\Msimp(\Cal S')$ for the ucp topology on
$\Mob(\Cal S')$ (631Ba), and $\Mob(\Cal S)$ is complete as a linear
topological space (615Cc), $\Phi$ has a unique extension to a continuous
function $\Psi:\Mns(\Cal S')\to\Mob(\Cal S)$
(use 3A4Cf\Latereditions\ and 3A4G).

\medskip

\quad{\bf (ii)} If $\tau\in\Cal S'$ and we set

\Centerline{$\pi'_{\tau}(\family{\sigma}{\Cal S'}{u_{\sigma}})
=u_{\tau}$,
\quad$\pi_{\tau}(\family{\sigma}{\Cal S}{v_{\sigma}})
=v_{\tau}$,}

\noindent we have continuous maps $\pi_{\tau}':\Mns(\Cal S')\to L^0$ and
$\pi_{\tau}:\Mns(\Cal S)\to L^0$;  as $\pi_{\tau}\Psi$ and $\pi_{\tau}'$
agree on $\Msimp(\Cal S')$ they agree on $\Mns(\Cal S')$;  as $\tau$ is
arbitrary, this means that $\Psi(\pmb u)\restr\Cal S'=\pmb u$ for every
$\pmb u\in\Mns(\Cal S')$, that is, that $\Psi(\pmb u)$ always extends
$\pmb u$.   Similarly, the functions
$\pmb u\mapsto\sup|\pmb u|:\Mob(\Cal S')\to L^0(\frak A)$
and $\pmb u\mapsto\sup|\Psi(\pmb u)|:\Mns(\Cal S')\to L^0(\frak A)$
are continuous (615C(b-ii)) and agree on $\Msimp(\Cal S')$, so agree on
$\Mns(\Cal S')$.

\medskip

\quad{\bf (iii)} If $h:\Bbb R\to\Bbb R$ is continuous, then
$\pmb u\mapsto\bar h\pmb u:\Mns(\Cal S')\to\Mns(\Cal S')$ and
$\pmb v\mapsto\bar h\pmb v:\Mns(\Cal S)\to\Mns(\Cal S)$ are continuous (615Ca).
Since $\bar h\Psi(\pmb u)=\Psi(\bar h\pmb u)$ for $\pmb u\in\Msimp(\Cal S)$,
$\bar h\Psi(\pmb u)=\Psi(\bar h\pmb u)$ for $\pmb u\in\Mns(\Cal S)$.
Now we see that

\Centerline{$\Psi(\pmb u^2)=(\Phi(\pmb u))^2$,
\quad$\Psi|\pmb u|=|\Phi(\pmb u)|$}

\noindent whenever $\pmb u\in\Mns(\Cal S')$, so
that $\Psi$ is a multiplicative Riesz homomorphism, as in (a-iii).

\medskip

\quad{\bf (iv)} If $\pmb u\in\Mns(\Cal S')$ and
$\Psi(\pmb u)$ is non-decreasing then of course
$\pmb u=\Psi(\pmb u)\restr\Cal S'$ is non-decreasing.   Conversely,
looking at the method in (a-i) above, we see that if
$\pmb u\in\Msimp(\Cal S')$ is non-decreasing then $\Phi(\pmb u)$ is
non-decreasing.   Next, if $\pmb u=\family{\sigma}{\Cal S'}{u_{\sigma}}$ is
a non-decreasing near-simple process,
then for any $\epsilon>0$ there is a non-decreasing simple process
$\pmb v$ such that $\theta(\sup|\pmb u-\pmb v|)\le\epsilon$.
\Prf\ There is certainly a simple process $\pmb v'$ such that
$\theta(\sup|\pmb u-\pmb v'|)\le\bover12\epsilon$.
Let $(\tau_0,\ldots,\tau_n)$
be a breakpoint sequence for $\pmb v'$.   Consider the simple process
$\pmb v=\family{\sigma}{\Cal S'}{v_{\sigma}}$ with starting value equal
to the starting value of $\pmb u$, with breakpoint string
$(\tau_0,\ldots,\tau_n)$, and with $v_{\tau_i}=u_{\tau_i}$ for
$i\le n$.   Then $\pmb v$ is non-decreasing and
$\theta(\sup|\pmb v-\pmb v'|)\le\bover12\epsilon$, so
$\theta(\sup|\pmb v-\pmb u|)\le\epsilon$.\ \Qed

Now $\Psi(\pmb v)=\Phi(\pmb v)$ is non-decreasing and

\Centerline{$\theta(\sup|\Psi(\pmb u)-\Psi(\pmb v)|)
=\theta(\sup|\Psi(\pmb u-\pmb v)|)
=\theta(\sup|\pmb u-\pmb v|)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\Psi u$ must be non-decreasing.

\medskip

\quad{\bf (v)} As $\pmb u\mapsto\sup|\pmb u|:\Mob(\Cal S')\to L^0(\frak A)$
and $\pmb u\mapsto\sup|\Psi(\pmb u)|:\Mns(\Cal S')\to L^0(\frak A)$ are
continuous (615C(b-ii)), it follows from (a-v) that
$\sup|\Psi(\pmb u)|=\sup|\pmb u|$ for every $\pmb u\in\Mns(\Cal S')$.   Now

\Centerline{$\Bvalue{\Psi(\pmb u)\ne 0}=\Bvalue{\sup|\Psi(\pmb u)|>0}
=\Bvalue{\sup|\pmb u|>0}=\Bvalue{\pmb u\ne 0}$.}

\medskip

{\bf (c)} Write $\Cal S'_1$ for
$\bigcup_{\tau\in\Cal S'}\Cal S\wedge\tau$,
the ideal of $\Cal S$ generated by $\Cal S'$.

\medskip

\quad{\bf (i)}\grheada\ We know that $\pmb v$ is moderately oscillatory
(616Ib again), so $v_{\uparrow}=\lim_{\tau\uparrow\Cal S}v_{\tau}$
and $v_{\downarrow}=\lim_{\tau\downarrow\Cal S}v_{\tau}$
are defined (615G).   Now if $\epsilon>0$ there are a
$\tau\in\Cal S$ such that $\theta(v_{\tau'}-v_{\uparrow})\le\bover12\epsilon$
whenever $\tau'\in\Cal S\vee\tau$, and a $\sigma\in\Cal S'$ such that
$\bar\mu\Bvalue{\sigma<\tau}\le\bover12\epsilon$.   Now if
$\sigma'\in\Cal S'\vee\sigma$,

$$\eqalign{\theta(v_{\sigma'\vee\tau}-v_{\uparrow})&
\le\bar\mu\Bvalue{v_{\sigma'}\ne v_{\sigma'\vee\tau}}
\le\bar\mu\Bvalue{\sigma'\ne\sigma'\vee\tau}\cr&
=\bar\mu\Bvalue{\sigma'<\tau}
\le\bar\mu\Bvalue{\sigma<\tau}
\le\Bover12\epsilon,\cr}$$

\noindent so

\Centerline{$\theta(v_{\sigma'}-v_{\uparrow})
\le\theta(v_{\sigma'}-v_{\sigma'\vee\tau})
   +\theta(v_{\sigma'\vee\tau}-v_{\uparrow})
\le\bover12\epsilon+\bover12\epsilon
=\epsilon$.}

\noindent As $\epsilon$ is arbitrary,
$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S'}v_{\sigma}$.

\medskip

\qquad\grheadb\ Because $\Cal S'$ is coinitial with $\Cal S$, we
see that $v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ is also
$\lim_{\sigma\downarrow\Cal S'}v_{\sigma}$.   We also know that
$\pmb v'=\pmb v\restr\Cal S'$ is an integrator (616P(b-ii)).   Now
$\int_{\Cal S'}\pmb u\,d\pmb v'=\int_{\Cal S}\Phi(\pmb u)\,d\pmb v$
for every $\pmb u\in\Msimp(\Cal S')$.   \Prf\
Let $(\tau_0,\ldots,\tau_n)$ be a breakpoint string for
$\pmb u$ and $u_{\downarrow}$ its starting value.   Then these are also a
breakpoint string and starting value for $\Phi(\pmb u)$ as described in the
proof of (a).   So

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

\noindent by 614C.\ \Qed

\medskip

\qquad\grheadc\ Now the operators

\Centerline{$\pmb u\mapsto\int_{\Cal S'}\pmb u\,d\pmb v'$,
\quad$\pmb u\mapsto\int_{\Cal S}\Psi(\pmb u)\,d\pmb v$}

\noindent from $\Mns(\Cal S')$ to $L^0$ are continuous for the ucp topology on
$\Mns(\Cal S')$ and the topology of convergence in measure on $L^0$,
by 616J and (b) above.   By ($\beta$), they agree on $\Msimp(\Cal S')$,
which is dense in $\Mns(\Cal S')$, so agree everywhere on
$\Mns(\Cal S')$, as claimed.

\medskip

\quad{\bf (ii)}\grheada\ For $\tau\in\Cal S'$,
(a)-(b) here tell us that we have unique functions
$\Phi_{\tau}:\Msimp(\Cal S'\wedge\tau)\to\Msimp(\Cal S\wedge\tau)$ and
$\Psi_{\tau}:\Mns(\Cal S'\wedge\tau)\to\Mns(\Cal S\wedge\tau)$
such that $\Phi_{\tau}(\pmb u)$ is a simple
process extending $\pmb u$ and with a breakpoint string in $\Cal S'\wedge\tau$
for every $\pmb u\in\Msimp(\Cal S'\wedge\tau)$, while $\Psi_{\tau}$ extends
$\Phi_{\tau}$ and is continuous with
respect to the ucp topologies.   If $\tau\le\tau'$ in
$\Cal S'$ and $\pmb u\in\Mns(\Cal S'\wedge\tau')$, then

\Centerline{$\pmb u\restr\Cal S'\wedge\tau
   =\pmb u\restr(\Cal S'\wedge\tau')\wedge\tau
   \in\Mns((\Cal S'\wedge\tau')\wedge\tau)
   =\Mns(\Cal S'\wedge\tau)$}

\noindent (using 631F(a-iv)), while similarly
$\Psi_{\tau'}(\pmb u)\restr\Cal S\wedge\tau\in\Mns(\Cal S\wedge\tau)$.
If $\pmb u\in\Msimp(\Cal S'\wedge\tau')$, then
$\Psi_{\tau'}(\pmb u)\in\Msimp(\Cal S\wedge\tau')$ has a breakpoint string
in $\Cal S'\wedge\tau'$, so
$\Psi_{\tau'}(\pmb u)\restr\Cal S\wedge\tau\in\Msimp(\Cal S\wedge\tau)$
has a breakpoint string in $\Cal S'\wedge\tau$ (612K(d-ii)) and extends
$\pmb u\restr\Cal S'\wedge\tau$, so must be equal to
$\Psi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)$, by (a-i) here.

If now $\pmb u\in\Mlns(\Cal S')$, $\Psi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)
=\Psi_{\tau'}(\pmb u\restr\Cal S'\wedge\tau')\restr\Cal S\wedge\tau$ whenever
$\tau\le\tau'$ in $\Cal S'$.   We therefore have a unique process
$\pmb v=\Psi_0^*(\pmb u)$ with domain $\Cal S'_1$ such that
$\pmb v\restr\Cal S\wedge\tau=\Psi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)$
for every $\tau\in\Cal S'$, and $\pmb v\in\Mlns(\Cal S'_1)$.
Now $\pmb v$ has a unique fully adapted extension $\hatpmb v$
to the covered envelope $\hat\Cal S'_1$ of $\Cal S'_1$.

At this point, observe that our hypothesis
`$\sup_{\tau\in\Cal S'}\Bvalue{\sigma\le\tau}=1$ for every
$\sigma\in\Cal S$' ensures that $\Cal S'_1$ covers $\Cal S$ in the sense of
611M, that is, that $\Cal S\subseteq\hat\Cal S'_1$.   As
$\Cal S'_1\subseteq\Cal S\subseteq\hat\Cal S'_1$, $\hat\Cal S'_1$ is also the
covered envelope of $\Cal S$.   Using 631Gb in both directions, we see that
$\hatpmb v$ and $\hatpmb v\restr\Cal S$ are both locally near-simple.
I will say that $\Psi^*(\pmb u)=\hatpmb v\restr\Cal S$, the unique fully adapted
extension of $\Psi_0(\pmb u)$ to $\Cal S$.

\medskip

\qquad\grheadb\ Just as $\Phi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)
=\Phi_{\tau'}(\pmb u\restr\Cal S'\wedge\tau')\restr\Cal S\wedge\tau$ whenever
$\pmb u\in\Msimp(\Cal S')$ and $\tau\le\tau'$ in $\Cal S'$, we have
$\Phi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)
=\Phi(\pmb u)\restr\Cal S\wedge\tau$ whenever $\pmb u\in\Msimp(\Cal S')$ and
$\tau\in\Cal S'$.   Consequently $\Psi_0^*(\pmb u)=\Phi(\pmb u)\restr\Cal S'_1$
and $\Psi^*(\pmb u)=\Phi(\pmb u)$ whenever $\pmb u\in\Msimp(\Cal S')$.

\medskip

\qquad\grheadc\ I noted in (a-v) that
$\sup|\Phi_{\tau}(\pmb u)|=\sup|\pmb u|$ whenever $\tau\in\Cal S'$ and
$\pmb u\in\Msimp(\Cal S'\wedge\tau)$;  by 615C(b-ii),
$\sup|\Psi_{\tau}(\pmb u)|=\sup|\pmb u|$ whenever $\tau\in\Cal S'$ and
$\pmb u\in\Mns(\Cal S'\wedge\tau)$, so

$$\eqalign{\sup|\Psi^*(\pmb u)\restr\Cal S\wedge\tau|&
=\sup|\Psi_0^*(\pmb u)\restr\Cal S\wedge\tau|\cr&
=\sup|\Psi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)|
=\sup|\pmb u\restr\Cal S'\wedge\tau|\cr}$$

\noindent for every $\tau\in\Cal S'$ and $\pmb u\in\Mlns(\Cal S')$.

\medskip

\qquad\grheadd\
As for the uniqueness of the function $\Psi^*$, suppose that
$\Psi_1^*:\Mlns(\Cal S')\to\Mlns(\Cal S)$ extends $\Phi$ and is such that
$\sup|\Psi_1^*(\pmb u)\restr\Cal S\wedge\tau|
=\sup|\pmb u\restr\Cal S'\wedge\tau|$
whenever $\pmb u\in\Mlns(\Cal S)$ and $\tau\in\Cal S'$.
Then $\Psi_1^*(\pmb u)\restr\Cal S\wedge\tau
=\Phi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)$
whenever $\pmb u\in\Msimp(\Cal S')$ and $\tau\in\Cal S'$, so
$\Psi_1^*(\pmb u)\restr\Cal S\wedge\tau
=\Psi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)$
whenever $\pmb u\in\Mlns(\Cal S')$ and $\tau\in\Cal S'$,
$\Psi_1^*(\pmb u)\restr\Cal S'_1=\Psi_0^*(\pmb u)$
whenever $\pmb u\in\Mlns(\Cal S')$ and
$\Psi_1^*(\pmb u)=\Psi^*(\pmb u)$
whenever $\pmb u\in\Mlns(\Cal S')$.

\medskip

\quad{\bf (iii)} If $\tau\in\Cal S'$ and $\pmb u\in\Mlns(\Cal S')$,

\Centerline{$\Psi^*(\pmb u)\restr\Cal S'\wedge\tau
=\Psi_0^*(\pmb u)\restr\Cal S'\wedge\tau
=\Psi_{\tau}(\pmb u\restr\Cal S\wedge\tau)\restr\Cal S'\wedge\tau
=\pmb u\restr\Cal S'\wedge\tau$}

\noindent by (b-ii);  as $\tau$ is arbitrary, $\Psi^*(\pmb u)$ extends
$\pmb u$.   Next, it follows at once from (b-iii) that $\Psi_0^*$
is a multiplicative Riesz homomorphism and
$\Psi_0^*(\bar h\pmb u)=\bar h\Psi_0^*(\pmb u)$ for
every continuous $h:\Bbb R\to\Bbb R$ and every $\pmb u\in\Mlns(\Cal S')$.
By 612Qb, we have the same result for $\Psi^*$.   Finally, if
$\pmb u\in\Mlns(\Cal S)$ is non-decreasing, then
$\Psi_{\tau}(\pmb u\restr\Cal S'\wedge\tau)$ is non-decreasing for every
$\tau\in\Cal S'$, by (b-iv), so $\Psi_0^*(\pmb u)$ is non-decreasing and
$\Psi^*(\pmb u)$ is non-decreasing by 612Qg.
}%end of proof of 631M

\leader{631N}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a locally near-simple process,
and $A\subseteq\Cal S$ a non-empty set such that $\inf A\in\Cal S$.   If
$u_{\rho}=0$ for every $\rho\in A$, then $u_{\inf A}=0$.

\proof{ Write $\tau$ for $\inf A$.

\medskip

{\bf (a)} To begin with, suppose that $\Cal S=[\tau,\max\Cal S]$ is a closed
interval in $\Cal T$ with least element $\tau$,
$u_{\max\Cal S}=0$ and $\pmb u\ge\tbf 0$.
In this case, $\pmb u$ is near-simple.

\medskip

\quad{\bf (i)} If $\sigma\in\Cal S$, $a=\Bvalue{u_{\sigma}>0}$ and $\epsilon>0$,
there are a $\sigma_1\in\Cal S$ and a $c\in\frak A$ such that
$\sigma_1\ge\sigma$, $a\Bsubseteq\Bvalue{\sigma<\sigma_1}$,
$\bar\mu c\le\epsilon$ and

\Centerline{$a
\Bsubseteq c\Bcup\Bvalue{\rho<\sigma}
   \Bcup\Bvalue{\sigma_1\le\rho}\Bcup\Bvalue{u_{\rho}>0}$}

\noindent whenever $\rho\in\Cal S$.   \Prf\ Let
$\eta>0$ be such that $c_0=a\Bcap\Bvalue{u_{\sigma}\le 2\eta})$ is at most
$\bover12\epsilon$.   Then there
is a simple process $\pmb v$ with domain $\Cal S$ such that, writing
$\bar v$ for $\sup|\pmb v-\pmb u|$,
$\theta(\bar v)\le\bover12\eta\epsilon$, so that
$c_1=\Bvalue{\bar v>\eta}$ has measure at most $\bover12\epsilon$.
Set $c=c_0\Bcup c_1$.
As $\pmb v\restr\Cal S\vee\sigma$ is simple (612K(d-iii)), it has a breakpoint
string $(\sigma_0,\ldots,\sigma_n)$ such that $\sigma_0=\sigma$,
$\sigma_n=\max\Cal S$ and
$\Bvalue{\sigma<\sigma_1}=\Bvalue{\sigma<\max\Cal S}$ (612M).   Since
$u_{\max\Cal S}=0$, $a\Bsubseteq\Bvalue{\sigma<\sigma_1}$.   Next, if
$\rho\in\Cal S$, then

\Centerline{$\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\sigma_1}
\Bsubseteq\Bvalue{v_{\rho}=v_{\sigma}}
\Bsubseteq\Bvalue{|u_{\rho}-u_{\sigma}|\le 2\bar v}$}

\noindent and

$$\eqalign{a\Bcap\Bvalue{u_{\rho}=0}\Bcap&\Bvalue{u_{\sigma}>2\eta}
    \Bcap\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\sigma_1}\cr&
\Bsubseteq a\Bcap\Bvalue{|u_{\rho}-u_{\sigma}|>2\eta}
    \Bcap\Bvalue{|u_{\rho}-u_{\sigma}|\le 2\bar v}
\Bsubseteq\Bvalue{\bar v>\eta}
=c_1.\cr}$$

\noindent So

\Centerline{$a\Bsubseteq c_0\Bcup(a\Bcap\Bvalue{u_{\sigma}>2\eta})
\Bsubseteq c_0\Bcup c_1\Bcup\Bvalue{\rho<\sigma}
   \Bcup\Bvalue{\sigma_1\le\rho}\Bcup\Bvalue{u_{\rho}>0}$,}

\noindent as required.\ \Qed

\medskip

\quad{\bf (ii)} For $t\in T$, set $b_t
=\inf\{\Bvalue{u_{\sigma}>0}\Bcup\Bvalue{\sigma>t}:\sigma\in\Cal S\}$.

\medskip

\qquad\grheada\ $b_t\in\frak A_t$ for every $t\in T$, because
$\Bvalue{u_{\sigma}>0}\in\frak A_{\sigma}$ so

\Centerline{$\Bvalue{u_{\sigma}>0}\Bcup\Bvalue{\sigma>t}
=(\Bvalue{u_{\sigma}>0}\Bcap\Bvalue{\sigma\le\check t})\Bcup\Bvalue{\sigma>t}
\in\frak A_{\check t}=\frak A_t$}

\noindent for every $\sigma\in\Cal T$, using 611H(c-iii).

\medskip

\qquad\grheadb\ If $s\le t$ then $b_t\Bsubseteq b_s$ because
$\Bvalue{\sigma>t}\Bsubseteq\Bvalue{\sigma>s}$ for every $\sigma$.

\medskip

\qquad\grheadc\ If $t\in T$ is not isolated on the right,
$b_t=\sup_{s>t}b_s$.   \Prf\ Take $\epsilon>0$.   Write $\upsilon$ for
$\med(\tau,\check t,\max\Cal S)\in\Cal S$.
By (i) above, there are a $\sigma_1\in\Cal S$ and a $c\in\frak A$
such that

\Centerline{$\upsilon\le\sigma_1$,
\quad
$\Bvalue{u_{\upsilon}>0}\Bsubseteq\Bvalue{\upsilon<\sigma_1}$,
\quad$\bar\mu c\le\epsilon$,}

\noindent and

\Centerline{$\Bvalue{u_{\upsilon}>0}
\Bsubseteq c\Bcup\Bvalue{\sigma<\upsilon}
   \Bcup\Bvalue{\sigma_1\le\sigma}\Bcup\Bvalue{u_{\sigma}>0}$}

\noindent for every $\sigma\in\Cal S$.
Let $s>t$ be such that $\bar\mu d\le\epsilon$ where
$d=(\Bvalue{\tau>t}\Bsetminus\Bvalue{\tau>s})
  \Bcup(\Bvalue{\sigma_1>t}\Bsetminus\Bvalue{\sigma_1>s})$.

Take any $\sigma\in\Cal S$.   We have

\Centerline{$b_t
  \Bsubseteq\Bvalue{u_{\upsilon}>0}\Bcup\Bvalue{\upsilon>t}
      =\Bvalue{u_{\upsilon}>0}\Bcup\Bvalue{\tau>t}$,}

\vskip-20pt

$$\eqalignno{b_t\Bcap\Bvalue{\sigma<\upsilon}&
  \Bsubseteq(\Bvalue{u_{\sigma}>0}\Bcup\Bvalue{\sigma>t})
      \Bcap\Bvalue{\sigma<\check t}
\Displaycause{because $\Bvalue{\sigma<\tau}=0$}
  \Bsubseteq\Bvalue{u_{\sigma}>0},&(*)\cr}$$

\Centerline{$\Bvalue{u_{\upsilon}>0}\Bcap\Bvalue{\upsilon\le\sigma}
      \Bcap\Bvalue{\sigma<\sigma_1}
  \Bsubseteq c\Bcup\Bvalue{u_{\sigma}>0}$,}

\Centerline{$\Bvalue{\tau>t}
  \Bsubseteq d\Bcup\Bvalue{\tau>s}\Bsubseteq d\Bcup\Bvalue{\sigma>s}$,}

\vskip-20pt

$$\eqalignno{b_t\Bcap\Bvalue{\upsilon\le\sigma}
      \Bcap\Bvalue{\sigma<\sigma_1}&
  \Bsubseteq c\Bcup d\Bcup\Bvalue{u_{\sigma}>0}\Bcup\Bvalue{\sigma>s},&(*)\cr}$$

\vskip-24pt

$$\eqalignno{\Bvalue{\sigma_1\le\sigma}&
  \Bsubseteq\Bvalue{\sigma>s}\Bcup\Bvalue{\sigma_1\le\check s}\cr&
  \Bsubseteq d\Bcup\Bvalue{\sigma>s}\Bcup\Bvalue{\sigma_1\le\check t}\cr&
  \Bsubseteq d\Bcup\Bvalue{\sigma>s}
     \Bcup(\Bvalue{\tau\le\check t}\Bcap\Bvalue{u_{\upsilon}=0})\cr}$$

\vskip-4pt

\noindent (because $\Bvalue{\sigma_1\le\upsilon}$ is disjoint from
$\Bvalue{u_{\upsilon}>0}$),

\vskip-16pt

$$\eqalignno{b_t\Bcap\Bvalue{\sigma_1\le\sigma}&
\Bsubseteq d\Bcup\Bvalue{\sigma>s}.&(*)\cr}$$

\noindent Collecting the three lines marked ($*$), we see that
$b_t\Bsubseteq c\Bcup d\Bcup\Bvalue{u_{\sigma}>0}\Bcup\Bvalue{\sigma>s}$.
As $\sigma$ is arbitrary, $b_t\Bsubseteq c\Bcup d\Bcup b_s$ and
$\bar\mu(b_t\Bsetminus b_s)\le\bar\mu c+\bar\mu d\le 2\epsilon$.
As $\epsilon$ is arbitrary, $b_t=\sup_{s>t}b_s$.\ \Qed

\medskip

\quad{\bf (iii)} Accordingly there is a $\tau'\in\Cal T$ such that
$\Bvalue{\tau'>t}=b_t$ for every $t\in T$.   If $t\in T$,
$\Bvalue{\tau>t}\Bsubseteq\Bvalue{\sigma>t}$ for every $\sigma\in\Cal S$,
so $\Bvalue{\tau>t}\Bsubseteq b_t$;  thus $\tau\le\tau'$.   On the other hand,
if $\sigma\in A$, then $b_t\Bsubseteq\Bvalue{\sigma>t}$ for every $t$, so
$\tau'\le\sigma$;  as $\tau=\inf A$, $\tau'=\tau$.   Thus
$b_t=\Bvalue{\tau>t}$ for every $t\in T$.

\medskip

\quad{\bf (iv)} \Quer\ If $u_{\tau}\ne 0$, set $a=\Bvalue{u_{\tau}>0}$.
By (i), we have a $\sigma_1\in\Cal S$
and a $c\in\frak A$ such that $a\Bsubseteq\Bvalue{\tau<\sigma_1}$,
$\bar\mu c\le\bover12\bar\mu a$ and

\Centerline{$a
\Bsubseteq c\Bcup\Bvalue{\sigma_1\le\rho}\Bcup\Bvalue{u_{\rho}>0}$}

\noindent for every $\rho\in\Cal S$.   Since

\Centerline{$0\ne a\Bsetminus c\Bsubseteq\Bvalue{\tau<\sigma_1}
=\sup_{t\in T}(\Bvalue{\sigma_1>t}\Bsetminus\Bvalue{\tau>t})$,}

\noindent there is a $t\in T$ such that

\Centerline{$0
\ne (a\Bsetminus c)\Bcap\Bvalue{\sigma_1>t}\Bsetminus\Bvalue{\tau>t}
=(a\Bsetminus c)\Bcap\Bvalue{\sigma_1>t}\Bcap\Bvalue{\tau\le\check t}$.}

\noindent Next, we know that

\Centerline{$\Bvalue{\tau>t}=b_t
=\inf_{\rho\in\Cal S}(\Bvalue{u_{\rho}>0}\Bcup\Bvalue{\rho>t})$;}

\noindent taking complements,

\Centerline{$\Bvalue{\tau\le\check t}
=\sup_{\rho\in\Cal S}(\Bvalue{u_{\rho}=0}\Bcap\Bvalue{\rho\le\check t})$.}

\noindent There is therefore a $\rho\in\Cal S$ such that

$$\eqalign{0&
\ne(a\Bsetminus c)\Bcap\Bvalue{\sigma_1>t}\Bcap\Bvalue{u_{\rho}=0}
       \Bcap\Bvalue{\rho\le\check t}\cr&
\Bsubseteq(a\Bsetminus c)\Bcap\Bvalue{u_{\rho}=0}
       \Bcap\Bvalue{\rho<\sigma_1}
=(a\Bsetminus c)
       \Bsetminus(\Bvalue{\sigma_1\le\rho}\Bcup\Bvalue{u_{\rho}\ne 0})\cr}$$

\noindent and

\Centerline{$a
\notBsubseteq c\Bcup\Bvalue{\sigma_1\le\rho}\Bcup\Bvalue{u_{\rho}>0}$}

\noindent contrary to the choice of $\sigma_1$ and $c$.\ \Bang

\medskip

\quad{\bf (v)} So $u_{\tau}=0$, and we have the result in the case when
$\Cal S=[\tau,\max\Cal S]$, $u_{\max\Cal S}=0$ and $\pmb u\ge\tbf 0$.

\medskip

{\bf (b)} For the general case, take any $\rho_0\in A$.   We know that
$|\pmb u|$ is locally near-simple (631F(b-ii)) so
$|\pmb u|\restr\Cal S\wedge\rho_0$ and
$|\pmb u|\restr\Cal S\cap[\tau,\rho_0]$ are near-simple (631F(a-iv)).
By 631Mb, there is an extension of $|\pmb u|\restr\Cal S\cap[\tau,\rho_0]$ to a
non-negative near-simple process
$\pmb u'=\family{\sigma}{[\tau,\rho_0]}{u'_{\sigma}}$.   Now
$A\wedge\rho_0=\{\rho\wedge\rho_0:\rho\in A\}$ is a subset of
$[\tau,\rho_0]$, and if $\rho\in A$ then

\Centerline{$\Bvalue{\rho\le\rho_0}
\Bsubseteq\Bvalue{u_{\rho\wedge\rho_0}=u_{\rho}}
\Bsubseteq\Bvalue{u_{\rho\wedge\rho_0}=0}$,
\quad$\Bvalue{\rho_0\le\rho}
\Bsubseteq\Bvalue{u_{\rho\wedge\rho_0}=u_{\rho_0}}
\Bsubseteq\Bvalue{u_{\rho\wedge\rho_0}=0}$}

\noindent so $u_{\rho\wedge\rho_0}=0$.   Thus $u'_{\rho}=|u_{\rho}|=0$ for
every $\rho\in A\wedge\rho_0$.   Also, of course, $\tau=\inf(A\wedge\rho_0)$.
So (a) tells us that $|u_{\tau}|=u'_{\tau}=0$ and $u_{\tau}=0$.   Thus the
result is true in the general case too.
}%end of proof of 631N

\leader{631O}{Witnessing sequences}\cmmnt{ Both `near-simple' and `jump-free'
processes
can be thought of as processes approximated in the right way by simple
processes with finite breakpoint strings.   In appropriate circumstances they
can be characterized by the existence of suitable (infinite) non-decreasing
sequences of stopping times, as follows.}
Let $\Cal S$ be a sublattice of $\Cal T$ with greatest and least elements, and
$\pmb u=\sfamilysS{u}$ a fully adapted process.

\spheader 631Oa $\SL_1(\pmb u)$ is the statement

\inset{for every $\delta>0$ there is a
non-decreasing sequence $\sequence{i}{\tau_i}$ in $\Cal S$ such that

\quad($\alpha$) $\tau_0=\min\Cal S$,

\quad($\beta$) $\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|<\delta}
\Bsubseteq\Bvalue{\tau_{i+1}=\max\Cal S}$ for every $i\in\Bbb N$,

\quad($\gamma$) $\inf_{i\in\Bbb N}\Bvalue{\tau_i<\max\Cal S}=0$,

\quad($\delta$) $\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_i}|<\delta}$ whenever
$i\in\Bbb N$ and $\sigma\in\Cal S\cap[\tau_i,\tau_{i+1}]$.}

\spheader 631Ob $\SL_2(\pmb u)$ is the statement

\inset{for every $\delta>0$ there is a
non-decreasing sequence $\sequence{i}{\tau_i}$ in $\Cal S$ such that

\quad($\alpha$) $\tau_0=\min\Cal S$,

\quad($\beta$) $\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|<\delta}
\Bsubseteq\Bvalue{\tau_{i+1}=\max\Cal S}$ for every $i\in\Bbb N$,

\quad($\gamma$) $\inf_{i\in\Bbb N}\Bvalue{\tau_i<\max\Cal S}=0$,

\quad($\delta$) $\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_i}|<\delta}$ whenever
$i\in\Bbb N$ and $\sigma\in\Cal S\cap[\tau_i,\tau_{i+1}]$,

\quad($\epsilon$)
$|u_{\tau_{i+1}}-u_{\tau_i}|\le\delta$ for every $i\in\Bbb N$.}

\leader{631P}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ with
greatest and least elements and $\pmb u$ a fully adapted process with domain
$\Cal S$.

(a) If $\SL_1(\pmb u)$ is true, then $\pmb u$ is near-simple.

(b) If $\SL_2(\pmb u)$ is true, then $\pmb u$ is jump-free.

\proof{ Express $\pmb u$ as $\sfamilysS{u}$.

\medskip

{\bf (a)(i)} For every $\delta>0$
there is a simple process $\pmb v=\sfamilysS{v}$ such that

\Centerline{$\bar\mu(\sup_{\sigma\in\Cal S}
\Bvalue{|u_{\sigma}-v_{\sigma}|>\delta})\le\delta$.}

\noindent\Prf\ Let $\sequence{i}{\tau_i}$ be as in $\SL_1(\pmb u)$.   As
$\sequence{i}{\Bvalue{\tau_i<\max\Cal S}}$ is a non-increasing sequence with
infimum $0$, there is an $n\in\Bbb N$ such that $\bar\mu d\le\delta$,
where $d=\Bvalue{\tau_n<\max\Cal S}$.
Let $\pmb v=\sfamilysS{v}$ be the simple process such that

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

\noindent for every $\sigma\in\Cal S$.   Then, for any $\sigma\in\Cal S$,

$$\eqalignno{1\Bsetminus d&
\Bsubseteq\Bvalue{\tau_n=\sigma}
  \Bcup\sup_{i<n}\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Displaycause{because $\tau_0=\min\Cal S$ and
$\Bvalue{\tau_n\le\sigma}\Bsetminus d
=\Bvalue{\tau_n=\max\Cal S}\Bcap\Bvalue{\sigma=\max\Cal S}$}
\Bsubseteq(\Bvalue{v_{\sigma}=u_{\tau_n}}\Bcap\Bvalue{u_{\sigma}=u_{\tau_n}})
  \Bcup\sup_{i<n}(\Bvalue{v_{\sigma}=u_{\tau_i}}
     \Bcap\Bvalue{|u_{\sigma}-u_{\tau_i}|<\delta})\cr&
\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|<\delta}.\cr}$$

\noindent So
$\sup_{\sigma\in\Cal S}\Bvalue{|u_{\sigma}-v_{\sigma}|>\delta}
\Bsubseteq d$ has measure at most $\delta$.\ \Qed

\medskip

\quad{\bf (ii)} It follows that $\pmb u$ is order-bounded.
\Prf\ For any $\delta>0$ there is a simple process $\pmb v=\sfamilysS{v}$
such that $\bar\mu a\ge 1-\delta$ and

\Centerline{$a\Bsubseteq\Bvalue{|u_{\sigma}-v_{\sigma}|\le\delta}
\Bsubseteq\Bvalue{|u_{\sigma}|\le\sup|\pmb v|+\delta\chi 1}$}

\noindent for every $\sigma\in\Cal S$.   But this means that
$\{|u_{\sigma}\times\chi a:\sigma\in\Cal S\}$ is bounded above in $L^0$.
As $\delta$ is arbitrary, 613C(p-i) tells us that
$\{|u_{\sigma}|:\sigma\in\Cal S\}$ is bounded above in $L^0$, that is,
$\pmb u$ is order-bounded.\ \Qed

\medskip

\quad{\bf (iii)} Now we can re-phrase (i) as saying that

\Centerline{`For every $\delta>0$ there is a simple process
$\pmb v$ such that $\bar\mu(\Bvalue{\sup|\pmb v-\pmb u|)>\delta})\le\delta$'}

\noindent so that for every $\delta>0$ there is a simple process $\pmb v$ such
that $\theta(\sup|\pmb v-\pmb u|)\le 2\delta$,
and $\pmb u$ is near-simple.

\medskip

{\bf (b)} Of course $\SL_2(\pmb u)$ implies $\SL_1(\pmb u)$, so (a) tells us that
$\pmb u$ is near-simple, therefore order-bounded;  set
$\bar u=\sup|\pmb u|$.

Let $\delta>0$, and take $\sequence{i}{\tau_i}$ as in the statement of
$\SL_2(\pmb u)$.
Let $n\ge 1$ be such that $\bar\mu\Bvalue{\tau_n<\max\Cal S}\le\delta$;  set
$a=\Bvalue{\tau_n<\max\Cal S}$.   Consider
$I=\{\tau_0,\ldots,\tau_n\}$.   If $i<n$ and
$\sigma$, $\sigma'\in[\tau_i,\tau_{i+1}]$ then

\Centerline{$\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_i}|<\delta}$,
\quad$\Bvalue{\sigma=\tau_{i+1}}
\Bsubseteq\Bvalue{|u_{\sigma}-u_{\tau_i}|\le\delta}$,}

\noindent so $|u_{\sigma}-u_{\tau_i}|\le\delta\chi 1$.   Similarly,
$|u_{\sigma'}-u_{\tau_i}|\le\delta\chi 1$ and
$|u_{\sigma}-u_{\sigma'}|\le 2\delta\chi 1$.

If $\sigma$, $\sigma'\in[\tau_n,\max\Cal S]$ then
$\Bvalue{u_{\sigma}\ne u_{\sigma'}}
\Bsubseteq\Bvalue{\tau_n<\max\Cal S}=a$, while
$|u_{\sigma}-u_{\sigma'}|\le 2\bar u$, so
$|u_{\sigma}-u_{\sigma'}|\le 2\bar u\times\chi a$.
By 618Ca,

\Centerline{$\Osclln(\pmb u)
\le\Osclln^*_I(\pmb u)
\le 2\delta\chi 1+2\bar u\times\chi a$}

\noindent and

\Centerline{$\theta(\Osclln(\pmb u))
\le\theta(2\delta\chi 1+2\bar u\times\chi a)
\le 2\delta+\bar\mu a
\le 3\delta$.}

\noindent As $\delta$ is arbitrary, $\Osclln(\pmb u)=0$ and
$\pmb u$ is jump-free.
}%end of proof of 631P

\leader{631Q}{Lemma} Let $\Cal S$ be a finitely full sublattice of $\Cal T$
with a greatest member such that
$\inf A\in\Cal S$ for every non-empty $A\subseteq\Cal S$, and
$\pmb u$ a near-simple process with domain $\Cal S$.   Take $\delta>0$ and
construct $\sequence{i}{D_i}$ and $\sequence{i}{y_i}$ from $\pmb u$ and
$\delta$ as in 615M\cmmnt{;  this is possible because $\pmb u$ is moderately
oscillatory (631Ca)}.   Then $\inf D_i\in D_i$ and $u_{\inf D_i}=y_i$ for every
$i\in\Bbb N$.

\proof{ Induce on $i$.   At the bottom, $\min\Cal S=\min D_0\in D_0$
and $y_0=u_{\min D_0}$.   For the inductive step to $i+1$, write $\tau$ for
$\inf D_i$ and set $\Cal S'=\Cal S\vee\tau$.
Define a process $\pmb v=\family{\sigma}{\Cal S'}{v_{\sigma}}$
by setting

\Centerline{$v_{\sigma}
=((|u_{\sigma}-y_i|\wedge\delta\chi 1)\vee\delta w_{\sigma})
   -\delta\chi 1$}

\noindent for $\sigma\in\Cal S'$, where
$w_{\sigma}=\chi\Bvalue{\sigma=\max\Cal S}$
for $\sigma\in\Cal S'$.   Because $\Mns(\Cal S')$ is closed under linear
and lattice operations and contains $\pmb u\restr\Cal S'$ (631Fa)
and the simple processes $y_i\tbf 1\restr\Cal S'$ and
$\pmb w=\family{\sigma}{\Cal S'}{w_{\sigma}}$ (612Jb),
$\pmb v$ is near-simple, with $v_{\max\Cal S}=0$.

Because $\tau\in D_i$, $\tau=\min D_i$ and

\Centerline{$D_{i+1}=\{\sigma:\sigma\in\Cal S'$,
$\Bvalue{\sigma<\max\Cal S}
  \Bsubseteq\Bvalue{|u_{\sigma}-y_i|\ge\delta}\}
=\{\sigma:\sigma\in\Cal S'$, $v_{\sigma}=0\}$.}

\noindent Write $\tau'$ for $\inf D_{i+1}$.   Then 631N tells us that
$v_{\tau'}=0$, that is, that $\tau'\in D_{i+1}$.
It follows at once that
$y_{i+1}=\lim_{\sigma\downarrow D_{i+1}}u_{\sigma}=u_{\tau'}$, and the
induction continues.
}%end of proof of 631Q

\leader{631R}{Stopping Lemmas:  Theorem}
Let $\Cal S$ be a finitely full sublattice of $\Cal T$
with greatest and least members such that
$\inf A\in\Cal S$ for every non-empty $A\subseteq\Cal S$, and
$\pmb u$ a moderately oscillatory process with domain $\Cal S$.

(a) $\pmb u$ is near-simple iff $\SL_1(\pmb u)$ is true.

(b) $\pmb u$ is jump-free iff $\SL_2(\pmb u)$ is true.

\proof{ Express $\pmb u$ as $\sfamilysS{u}$.

\medskip

{\bf (a)} We know from 631Pa that if $\SL_1(\pmb u)$ is true then
$\pmb u$ is near-simple, so it will be enough to show the converse.   Suppose
that $\pmb u$ is near-simple, and take any $\delta>0$.

\medskip

\quad{\bf (i)} Construct $\sequence{i}{D_i}$,
$\sequence{i}{y_i}$, $\sequence{i}{d_i}$ and
$\langle c_{i\sigma}\rangle_{i\in\Bbb N,\sigma\in\Cal S}$
from $\pmb u$ and $\delta$ as in 615M.   Set $\tau_i=\inf D_i$ for each $i$.
By 631Q, $\tau_i=\min D_i$ and $u_{\tau_i}=y_i$ for every $i$.   Because every
member of $D_{i+1}$ is greater than or equal to some member of $D_i$,
$\tau_i\le\tau_{i+1}$ for every $i$.   Of course $\tau_0=\min D_0=\min\Cal S$.

\medskip

\quad{\bf (ii)} Now $d_i=\Bvalue{\tau_i<\max\Cal S}$ for each $i$,
so $\inf_{i\in\Bbb N}\Bvalue{\tau_i<\max\Cal S}=0$.   Since
$\tau_{i+1}\in D_{i+1}$,

\Centerline{$\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|<\delta}
=\Bvalue{|u_{\tau_{i+1}}-y_i|<\delta}
\Bsubseteq\Bvalue{\tau_{i+1}=\max\Cal S}$}

\noindent for every $i\in\Bbb N$.

If $i\in\Bbb N$ and $\sigma\in\Cal S\cap[\tau_i,\tau_{i+1}]$ then

\Centerline{$c_{i\sigma}=\Bvalue{\tau_i\le\sigma}=1$,
\quad$c_{i+1,\sigma}=\Bvalue{\tau_{i+1}\le\sigma}
=1\Bsetminus\Bvalue{\sigma<\tau_{i+1}}$.}

\noindent So

$$\eqalignno{\Bvalue{\sigma<\tau_{i+1}}&
=c_{i\sigma}\Bsetminus c_{i+1,\sigma}
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|<\delta}
\Displaycause{615M(d-v)}
=\Bvalue{|u_{\sigma}-u_{\tau_i}|<\delta}.\cr}$$

\noindent Thus $\sequence{i}{\tau_i}$ satisfies ($\alpha$)-($\delta$)
of $\SL_1(\pmb u)$.   As $\delta$ was arbitrary, $\SL_1(\pmb u)$ is true.

\medskip

{\bf (b)} Here we know from 631Pb that if $\SL_2(\pmb u)$ is true then
$\pmb u$ is jump-free, so again it will be enough to show the converse.
Suppose that $\pmb u$ is jump-free.
By 631Cb, $\pmb u$ is near-simple.   By (a) just above, $\SL_1(\pmb u)$
is true.   Take $\delta>0$, and let $\sequence{i}{\tau_i}$ be a sequence
as in the statement of $\SL_1(\pmb u)$.

Take any $i\in\Bbb N$.   \Quer\ Suppose that
$\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}>\delta}$ is non-zero.
Then there is an $\eta>0$ such that
$a=\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}>\delta+\eta}\ne 0$.
Let $I\in\Cal I(\Cal S)$ be such that
$\theta(\Osclln^*_I(\pmb u))<\eta\bar\mu a$;  we can suppose that
$\tau_i$, $\tau_{i+1}\in I$.   Set
$b=\Bvalue{\Osclln^*_I(\pmb u)\le\eta}$, so that
$a\Bcap b\ne 0$.   Let $\sigma_0\le\ldots\le\sigma_m$
linearly generate the $I\cap[\tau_i,\tau_{i+1}]$-cells.   Then

$$\eqalign{a\Bcap b\Bcap\Bvalue{\sigma_j<\tau_{i+1}}
&\Bsubseteq a\Bcap b\cap\Bvalue{u_{\sigma_j}\le u_{\tau_i}+\delta}\cr
&\Bsubseteq\Bvalue{u_{\sigma_{j+1}}\le u_{\sigma_j}+\eta}
  \Bcap\Bvalue{u_{\sigma_j}<u_{\tau_{i+1}}-\eta}
\Bsubseteq\Bvalue{\sigma_{j+1}<\tau_{i+1}}\cr}$$

\noindent for every $j<m$.   Inducing on $j$, we see that
$a\Bcap b\Bsubseteq\Bvalue{\sigma_j<\tau_{i+1}}$ for every $j\le m$.
But $\sigma_m=\tau_{i+1}$, so this is impossible.\ \Bang

Similarly, $\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}<-\delta}$
cannot be non-zero, and
$|u_{\tau_{i+1}}-u_{\tau_i}|\le\delta\chi 1$, for every $i$.
Thus $\sequence{i}{\tau_i}$ has the property ($\epsilon$)
in the statement of $\SL_2(\pmb u)$ as well as the properties
($\alpha$)-($\delta$) there.   As $\delta$ is arbitrary, $\SL_2(\pmb u)$ is true.
}%end of proof of 631R

\leader{631S}{Proposition}
Let $\Cal S$ be a finitely full sublattice of $\Cal T$
with greatest and least members
and $\pmb u=\sfamilysS{u}$ a moderately oscillatory process.
If $\inf A\in\Cal S$ and $u_{\inf A}=\lim_{\sigma\downarrow A}u_{\sigma}$
for every non-empty downwards-directed $A\subseteq\Cal S$, then
$\pmb u$ is near-simple.

%\query use 631F(c-ii) to remove hypothesis of \max\Cal S

\proof{ Take any $\delta>0$.

\medskip

{\bf (a)} Construct $\sequence{i}{D_i}$,
$\sequence{i}{y_i}$, $\sequence{i}{d_i}$ and
$\langle c_{i\sigma}\rangle_{i\in\Bbb N,\sigma\in\Cal S}$
from $\pmb u$ and $\delta$ as in 615M.   Set $\tau_i=\inf D_i$ for each $i$;
then $\tau_i\in\Cal S$.   Since every member of $D_{i+1}$ is greater than or
equal to some member of $D_i$, $\tau_i\le\tau_{i+1}$ for every $i$.

\medskip

{\bf (b)} $\tau_i\in D_i$ for every $i\in\Bbb N$.
\Prf\ Induce on $i$.   At the bottom, $\tau_0=\min\Cal S=\min D_0\in D_0$.
For the inductive step to $i+1$, we have

\Centerline{$y_i=\lim_{\sigma\downarrow D_i}u_{\sigma}=u_{\tau_i}
\in L^0(\frak A_{\tau_i})$.}

\noindent Set

\Centerline{$a=\Bvalue{|u_{\tau_{i+1}}-y_i|\ge\delta}\in\frak A_{\tau_{i+1}}
\subseteq\frak A_{\max\Cal S}$.}

\noindent Then there is a $\tau\in\Cal T$ such that
$a\Bsubseteq\Bvalue{\tau=\tau_{i+1}}$ and
$1\Bsetminus a\Bsubseteq\Bvalue{\tau=\max\Cal S}$.   Because $\Cal S$ is
finitely full, $\tau\in\Cal S\vee\tau_{i+1}$ and

\Centerline{$\Bvalue{\tau<\max\Cal S}\Bsubseteq a
=\Bvalue{\tau=\tau_{i+1}}\Bcap\Bvalue{|u_{\tau_{i+1}}-y_i|\ge\delta}
\Bsubseteq\Bvalue{|u_{\tau}-y_i|\ge\delta}$;}

\noindent as also $\tau_i\in D_i$ and $\tau_i\le\tau$, $\tau\in D_{i+1}$.

If $\sigma$, $\sigma'\in D_{i+1}$ and $\sigma\le\sigma'$, then

\Centerline{$\Bvalue{\sigma'<\max\Cal S}\Bsubseteq\Bvalue{\sigma<\max\Cal S}
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|\ge\delta}$}

\noindent and $|u_{\sigma}-y_i|\ge\delta\chi\Bvalue{\sigma'<\max\Cal S}$.
Taking the limit as $\sigma\downarrow D_{i+1}$,

\Centerline{$|u_{\tau_{i+1}}-y_i|\ge\delta\chi\Bvalue{\sigma'<\max\Cal S}$,
\quad$\Bvalue{\sigma'<\max\Cal S}\Bsubseteq a$.}

\noindent But now we have

\Centerline{$a\Bsubseteq\Bvalue{\tau=\tau_{i+1}}
\Bsubseteq\Bvalue{\tau\le\sigma'}$,
\quad$1\Bsetminus a\Bsubseteq\Bvalue{\sigma'=\max\Cal S}
\Bsubseteq\Bvalue{\tau\le\sigma'}$,}

\noindent so $\tau\le\sigma'$.   As $\sigma'$ is arbitrary,
$\tau\le\inf D_{i+1}=\tau_{i+1}$ and $\tau_{i+1}=\tau\in D_{i+1}$.\ \Qed

\medskip

{\bf (c)} So $\tau_i=\min D_i$ and $y_i=u_{\tau_i}\in L^0(\frak A_{\tau_i})$
for each $i$.   Since $\Bvalue{\tau_i<\max\Cal S}=d_i$ for each $i$,
$\inf_{i\in\Bbb N}\Bvalue{\tau_i<\max\Cal S}=0$.   Since
$\tau_{i+1}\in D_{i+1}$,

\Centerline{$\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|<\delta}
=\Bvalue{|u_{\tau_{i+1}}-y_i|<\delta}
\Bsubseteq\Bvalue{\tau_{i+1}=\max\Cal S}$}

\noindent for every $i\in\Bbb N$.

If $i\in\Bbb N$ and $\sigma\in\Cal S\cap[\tau_i,\tau_{i+1}]$ then

\Centerline{$c_{i\sigma}=\Bvalue{\tau_i\le\sigma}=1$,
\quad$c_{i+1,\sigma}=\Bvalue{\tau_{i+1}\le\sigma}
=1\Bsetminus\Bvalue{\sigma<\tau_{i+1}}$.}

\noindent So

$$\eqalignno{\Bvalue{\sigma<\tau_{i+1}}&
=c_{i\sigma}\Bsetminus c_{i+1,\sigma}
\Bsubseteq\Bvalue{|u_{\sigma}-y_i|<\delta}
\Displaycause{615M(d-v) again}
=\Bvalue{|u_{\sigma}-u_{\tau_i}|<\delta}.\cr}$$

\noindent Thus $\sequence{i}{\tau_i}$ satisfies ($\alpha$)-($\delta$)
of $\SL_1(\pmb u)$.   As $\delta$ was arbitrary, $\SL_1(\pmb u)$ is true;
by 631Pa, $\pmb u$ is near-simple.
}%end of proof of 631S

\leader{631T}{}\cmmnt{ So far we have been looking at questions arising
naturally from the ideas of Chapter 61.   I come now to something which
will be important in \S644.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$C$ the set of non-negative non-decreasing order-bounded near-simple
processes with
domain $\Cal S$.   If $\pmb u\cmmnt{\mskip5mu=\sfamilysS{u}}$ belongs to $C$ and
$\Osclln(\pmb u)\ne 0$, there is a non-zero simple process $\pmb v\in C$
such that $\pmb u-\pmb v\in C$.
%for 631U

\proof{{\bf (a)} Since we are supposing that $\Osclln(\pmb u)\ne 0$, we can
be sure that $\Cal S$ is non-empty.   Set

\Centerline{$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}
=\inf_{\sigma\in\Cal S}u_{\sigma}$,
\quad$u_{\uparrow}=\lim_{\sigma\uparrow\Cal S}u_{\sigma}
=\sup_{\sigma\in\Cal S}u_{\sigma}$}

\noindent (631C, 613Ba).

\medskip

{\bf (b)} Set $\epsilon=\Bover1{9}\theta(\Osclln(\pmb u))$.
Let $\pmb w=\sfamilysS{w}$ be a simple process such that
$\theta(\bar w)\le\epsilon^2$, where $\bar w=\sup|\pmb u-\pmb w|$,
and $\tau_*$, $\tau^*\in\Cal S$ such
that $\theta(u_{\tau_*}-u_{\downarrow})\le\epsilon$ and
$\theta(u_{\uparrow}-u_{\tau^*})\le\epsilon$;  let $I$ be a finite
sublattice of $\Cal S$ containing $\tau_*$ and $\tau^*$ and including a
breakpoint string for $\pmb w$.   Take $(\tau_0,\ldots,\tau_n)$ linearly
generating the $I$-cells;  then $(\tau_0,\ldots,\tau_n)$ is a breakpoint
string for $\pmb w$ (612Kb again),
while $\tau_0\le\tau_*$ and $\tau^*\le\tau_n$.   Now

\Centerline{$\Osclln(\pmb u)\le\Osclln^*_I(\pmb u)
=(u_{\tau_0}-u_{\downarrow})
\vee\sup_{i<n}(u_{\tau_{i+1}}-u_{\tau_i})\vee(u_{\uparrow}-u_{\tau_n})$}

\noindent (618Cb, because $\pmb u$ is non-decreasing), so

$$\eqalign{\theta(\sup_{i<n}(u_{\tau_{i+1}}-u_{\tau_i}))
&\ge\theta(\Osclln(\pmb u))-\theta(u_{\tau_0}-u_{\downarrow})
   -\theta(u_{\uparrow}-u_{\tau_n})
\cr&\ge\theta(\Osclln(\pmb u))-\theta(u_{\tau_0}-u_{\tau_*})
   -\theta(u_{\uparrow}-u_{\tau^*})
\ge 7\epsilon.\cr}$$

\noindent Consequently

\Centerline{$a=\Bvalue{\sup_{i<n}u_{\tau_{i+1}}-u_{\tau_i}>4\epsilon}
=\sup_{i<n}\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}>4\epsilon}$}

\noindent has measure
at least $2\epsilon$.   On the other hand,
$\Bvalue{\bar w\ge\epsilon}$ has measure at most $\epsilon$,
so $a\Bcap\Bvalue{\bar w\le\epsilon}$ is non-zero.
There is therefore an $i<n$ such that

\Centerline{$b=\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}>4\epsilon}
   \Bcap\Bvalue{\bar w\le\epsilon}$}

\noindent is non-zero.

\medskip

{\bf (c)} Consider

\Centerline{$c=\Bvalue{u_{\tau_{i+1}}\le\epsilon}
   \Bcup\sup_{\sigma\in\Cal S}(\Bvalue{\sigma<\tau_{i+1}}
      \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon})$.}

\noindent Then $c\in\frak A_{\tau_{i+1}}$.   \Prf\ For any
$\sigma\in\Cal T$, $\Bvalue{\sigma<\tau_{i+1}}\in\frak A_{\tau_{i+1}}$
(611H(c-i)) while $\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
\in\frak A_{\tau_{i+1}\vee\sigma}$, so

$$\eqalignno{\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
   \Bcap\Bvalue{\sigma\le\tau_{i+1}}
&=\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
   \Bcap\Bvalue{\sigma\vee\tau_{i+1}=\tau_{i+1}}\cr
\displaycause{611E(a-ii-$\beta$) again}
&=\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
   \Bcap\Bvalue{\sigma\vee\tau_{i+1}\le\tau_{i+1}}
\in\frak A_{\tau_{i+1}}\cr}$$

\noindent (611H(c-iii)),

\Centerline{$\Bvalue{\sigma<\tau_{i+1}}
      \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
=\Bvalue{\sigma<\tau_{i+1}}
      \Bcap\Bvalue{\sigma\le\tau_{i+1}}
      \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
\in\frak A_{\tau_{i+1}}$.}

\noindent Taking the supremum over $\sigma$,

\Centerline{$\sup_{\sigma\in\Cal S}(\Bvalue{\sigma<\tau_{i+1}}
      \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon})
      \in\frak A_{\tau_{i+1}}$;}

\noindent as surely
$\Bvalue{u_{\tau_{i+1}}\le\epsilon}\in\frak A_{\tau_{i+1}}$,
$c\in\frak A_{\tau_{i+1}}$.\ \Qed

Next, $c\Bcap b=0$.   \Prf\ Of course

\Centerline{$\Bvalue{u_{\tau_{i+1}}\le\epsilon}\Bcap b
\Bsubseteq\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}\le 4\epsilon}\Bcap b
=0$.}

\noindent If $\sigma\in\Cal S$ then

$$\eqalign{\Bvalue{\sigma<\tau_{i+1}}
&=\Bvalue{\sigma<\tau_0}
  \Bcup\sup_{j\le i}\Bvalue{\tau_j\le\sigma}\Bcap\Bvalue{\sigma<\tau_{j+1}}
\cr&\Bsubseteq\Bvalue{u_{\sigma}=u_{\sigma\wedge\tau_0}}
  \Bcup\sup_{j\le i}\Bvalue{w_{\sigma}=w_{\tau_j}}
\cr&\Bsubseteq\Bvalue{u_{\sigma}\le u_{\tau_0}}
  \Bcup\sup_{j\le i}\Bvalue{u_{\sigma}\le u_{\tau_j}+2\bar w}
\Bsubseteq\Bvalue{u_{\sigma}\le u_{\tau_i}+2\bar w}.\cr}$$

\noindent So

$$\eqalign{&b\Bcap\Bvalue{\sigma<\tau_{i+1}}
      \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
\cr&\mskip100mu
\Bsubseteq\Bvalue{u_{\sigma}\le u_{\tau_i}+2\bar w}
   \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
   \Bcap\Bvalue{\bar w\le\epsilon}
   \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}\ge 4\epsilon}
\cr&\mskip100mu
\Bsubseteq\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}\le 3\epsilon}
   \Bcap\Bvalue{u_{\tau_{i+1}}-u_{\tau_i}\ge 4\epsilon}
=0.\cr}$$

\noindent Taking the supremum over $\sigma$, $b\Bcap c=0$.\ \Qed

\medskip

{\bf (d)} Since $c\in\frak A_{\tau_{i+1}}$, we have a simple process
$\pmb v=\sfamilysS{u}$, with singleton breakpoint string $(\tau_{i+1})$,
defined by saying that

\Centerline{$\Bvalue{\sigma<\tau_{i+1}}\Bsubseteq\Bvalue{v_{\sigma}=0}$,
\quad$\Bvalue{\tau_{i+1}\le\sigma}
\Bsubseteq\Bvalue{v_{\sigma}=\epsilon\chi(1\Bsetminus c)}$}

\noindent for every $\sigma\in\Cal S$.   Evidently $\pmb v$ is near-simple
and non-negative, while if $\sigma\le\sigma'$ in $\Cal S$,

$$\eqalign{\Bvalue{\sigma'<\tau_{i+1}}\Bcup\Bvalue{\tau_{i+1}\le\sigma}
   \Bcup c
&\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma'}}
\cr&\Bsubseteq\Bvalue{v_{\sigma}\le v_{\sigma'}}
   \Bcap\Bvalue{u_{\sigma}-v_{\sigma}\le u_{\sigma'}-v_{\sigma'}},\cr}$$

\Centerline{$\Bvalue{\sigma<\tau_{i+1}}\Bcup c
\Bsubseteq\Bvalue{v_{\sigma}=0}
\Bsubseteq\Bvalue{u_{\sigma}-v_{\sigma}\ge 0}$,}

$$\eqalign{&\Bvalue{\sigma<\tau_{i+1}}
   \Bcap\Bvalue{\tau_{i+1}\le\sigma'}\Bsetminus c
\cr&\mskip100mu
\Bsubseteq\Bvalue{v_{\sigma}=0}\Bcap\Bvalue{v_{\sigma'}=\epsilon}
   \Bcap\Bvalue{u_{\tau_{i+1}}\le u_{\sigma'}}
   \Bsetminus\Bvalue{u_{\tau_{i+1}}-u_{\sigma}\le\epsilon}
\cr&\mskip100mu
\Bsubseteq\Bvalue{v_{\sigma}\le v_{\sigma'}}
   \Bcap\Bvalue{v_{\sigma'}-v_{\sigma}\le\epsilon}
   \Bcap\Bvalue{u_{\sigma'}-u_{\sigma}\ge\epsilon}
\cr&\mskip100mu
\Bsubseteq\Bvalue{v_{\sigma}\le v_{\sigma'}}
   \Bcap\Bvalue{u_{\sigma}-v_{\sigma}\le u_{\sigma'}-v_{\sigma'}},\cr}$$

\Centerline{$\Bvalue{\tau_{i+1}\le\sigma}\Bsetminus c
\Bsubseteq\Bvalue{u_{\tau_{i+1}}\le u_{\sigma}}
   \Bcap\Bvalue{u_{\tau_{i+1}}\ge\epsilon}
   \Bcap\Bvalue{v_{\sigma}\le\epsilon}
\Bsubseteq\Bvalue{u_{\sigma}-v_{\sigma}\ge 0}$.}

\noindent So $\pmb v$ and $\pmb u-\pmb v$ are both non-decreasing, while
$\pmb u-\pmb v$ is non-negative.   Finally,

\Centerline{$v_{\tau_{i+1}}=\epsilon\chi(1\Bsetminus c)
\ge\epsilon\chi b\ne 0$}

\noindent and $\pmb v$ is non-zero.
Thus we have an appropriate $\pmb v$.
}%end of proof of 631T

\leader{631U}{Theorem} Let $\Cal S$ be a non-empty sublattice of $\Cal T$
and $\pmb u$ a non-negative, order-bounded, non-decreasing near-simple process
with domain
$\Cal S$.   Then for any $\epsilon>0$ there are non-negative,
non-decreasing processes $\pmb v$, $\pmb w$ with domain $\Cal S$
such that
$\pmb v$ is simple, $\pmb w$ is jump-free,
$\pmb u-\pmb v-\pmb w$ is non-negative and non-decreasing, and
$\theta(\sup|\pmb u-\pmb v-\pmb w|)\le\epsilon$.

\proof{{\bf (a)} As in 631T, let $C$ be the set of non-negative
non-decreasing near-simple order-bounded processes with domain $\Cal S$.
Note that if $\pmb v=\sfamilysS{v}$ is a non-negative non-decreasing
order-bounded process,

\Centerline{$\sup|\pmb v|=\sup_{\sigma\in\Cal S}v_{\sigma}
=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$.}

\noindent (I think the formulae will look cleaner if I write
$\sup\pmb v$ rather than $\sup|\pmb v|$ in this case.)   It follows that
if $\pmb v$, $\pmb v'\in C$ then
$\sup(\pmb v+\pmb v')=\sup\pmb v+\sup\pmb v'$.

\medskip

{\bf (b)} Write $C'$ for the set of simple processes
$\pmb v\in C$ such that $\pmb u-\pmb v\in C$.   For $\pmb v\in C'$,
set

\Centerline{$q(\pmb v)
=\sup\{\theta(\sup\pmb v'):\pmb v'\in C$,
  $\pmb v+\pmb v'\in C'\}$.}

\noindent Choose $\sequencen{\pmb v_n}$ inductively such that

\Centerline{$\pmb v_n\in C'$,
\quad$\pmb v_{n+1}-\pmb v_n\in C$,}

\Centerline{$\theta(\sup(\pmb v_{n+1}-\pmb v_n))
  \ge q(\pmb v_n)-2^{-n}$}

\noindent for every $n\in\Bbb N$.   Then
$\sequencen{\pmb v_n}$ is a non-decreasing sequence of processes
bounded above by $\pmb u$.   Set
$\pmb v'=\sup_{n\in\Bbb N}\pmb v_n$;   by 612Ia,
$\pmb v'$ is a fully adapted process.   Evidently $\pmb v'$ is
non-decreasing and non-negative;  moreover,
$\pmb v_m-\pmb v_n\in C$ whenever $m\ge n$, so
$\pmb v'-\pmb v_n$ is non-decreasing and non-negative for every $n$.

\medskip

{\bf (c)} $\lim_{n\to\infty}\theta(\sup(\pmb v_{n+1}-\pmb v_n))=0$.   \Prf\
$\sequencen{\sup\pmb v_n}$ is non-decreasing and bounded
above by $\sup\pmb u$, so is convergent for the topology of measure
(613Ba);  accordingly, using (a),

$$\eqalign{\lim_{n\to\infty}
  \theta(\sup(\pmb v_{n+1}-\pmb v_n))
&=\lim_{n\to\infty}
  \theta(\sup\pmb v_{n+1}-\sup\pmb v_n)
=0.  \text{ \Qed}\cr}$$

It follows that $\lim_{n\to\infty}q(\pmb v_n)=0$.   Note also
that $\sup\pmb v'=\sup_{n\in\Bbb N}\sup\pmb v_n$, so
that

\Centerline{$\lim_{n\to\infty}\theta(\sup(\pmb v'-\pmb v_n))
=\lim_{n\to\infty}\theta(\sup\pmb v'-\sup\pmb v_n)
=0$;}

\noindent as every $\pmb v_n$ is simple, $\pmb v'$ is near-simple and
belongs to $C$.

\medskip

{\bf (d)} Set $\pmb w=\pmb u-\pmb v'$.   By (c), $\pmb w$ is near-simple;
moreover, as $\pmb u-\pmb v_n$ is non-decreasing and non-negative for
every $n$, so is $\pmb w$.   Now $\pmb w$ is jump-free.
\Prf\Quer\ Otherwise, 631T
tells us that there is a non-zero simple process $\pmb v\in C$ such that
$\pmb w-\pmb v\in C$.   Consequently

\Centerline{$\pmb v_n+\pmb v$,
\quad$\pmb u-\pmb v_n-\pmb v
=(\pmb u-\pmb v'-\pmb v)+(\pmb v'-\pmb v_n)$}

\noindent belong to $C$ for every $n$, and
$q(\pmb v_n)\ge\theta(\sup\pmb v)>0$ for every $n$;  which is
impossible.\ \Bang\Qed

\medskip

{\bf (e)} Now observe that

\Centerline{$\theta(\sup(\pmb u-\pmb v_n-\pmb w))
=\theta(\sup\pmb v'-\sup\pmb v_n)\to 0$}

\noindent as $n\to\infty$.   So $\pmb v_n$ will serve for
$\pmb v$ for any sufficiently large $n$.
}%end of proof of 631U

\leader{631V}{Corollary} Let $\Cal S$ be a non-empty sublattice of $\Cal T$ and
$\pmb u$ a near-simple process of bounded variation with domain
$\Cal S$.   Then for any $\epsilon>0$ there are processes
$\pmb v$, $\pmb w$ with domain $\Cal S$ such that
$\pmb v$ is simple, $\pmb w$ is jump-free and of bounded variation
and $\theta(\int_{\Cal S}|d(\pmb u-\pmb v-\pmb w)|)\le\epsilon$.

\proof{ By 631L, $\pmb u$ is expressible as $\pmb u_0-\pmb u_1$ where
$\pmb u_0$, $\pmb u_1$ are non-negative non-decreasing near-simple order-bounded
processes.   By 631U, we have simple processes $\pmb v_0$,
$\pmb v_1$ and jump-free processes of bounded variation
$\pmb w_0$, $\pmb w_1$ such that
$\pmb z_0=\pmb u_0-\pmb v_0-\pmb w_0$ and
$\pmb z_1=\pmb u_1-\pmb v_1-\pmb w_1$ are non-decreasing and
non-negative and $\theta(\sup|\pmb z_0|)$, $\theta(\sup|\pmb z_1|)$ are both
at most $\bover12\epsilon$.

Set $\pmb v=\pmb v_0-\pmb v_1$,
$\pmb w=\pmb w_0-\pmb w_1$ and $\pmb z=\pmb z_0-\pmb z_1$,
so that $\pmb v$ is simple,
$\pmb w$ is jump-free and of bounded variation, and
$\pmb u-\pmb v-\pmb w=\pmb z$.   Now
$\int_{\Cal S}|d\pmb z|\le\sup|\pmb z_0|+\sup|\pmb z_1|$.
\Prf\ Expressing $\pmb z$, $\pmb z_0$ and $\pmb z_1$ as
$\family{\sigma}{\Cal S}{z_{\sigma}}$,
$\family{\sigma}{\Cal S}{z_{0\sigma}}$ and
$\family{\sigma}{\Cal S}{z_{1\sigma}}$, we see that

\Centerline{$|z_{\tau}-z_{\sigma}|
\le|z_{0\tau}-z_{0\sigma}|+|z_{1\tau}-z_{1\sigma}|
=(z_{0\tau}-z_{0\sigma})+(z_{1\tau}-z_{1\sigma})$}

\noindent whenever $\sigma\le\tau$ in $\Cal S$.   Consequently

\Centerline{$S_I(\tbf 1,|d\pmb z|)
\le S_I(\tbf 1,d\pmb z_0)+S_I(\tbf 1,d\pmb z_1)$}

\noindent for every $I\in\Cal I(\Cal S)$, and

$$\eqalignno{\int_{\Cal S}|d\pmb z|
&\le\int_{\Cal S}d\pmb z_0+\int_{\Cal S}d\pmb z_1
=\lim_{\sigma\uparrow\Cal S}(z_{0\sigma}+z_{1\sigma})
   -\lim_{\sigma\downarrow\Cal S}(z_{0\sigma}+z_{1\sigma})\cr
\displaycause{613N}
&\le\lim_{\sigma\uparrow\Cal S}z_{0\sigma}
  +\lim_{\sigma\uparrow\Cal S}z_{1\sigma}
=\sup|\pmb z_0|+\sup|\pmb z_1|.  \text{ \Qed}\cr}$$

So we have

\Centerline{$\theta(\int_{\Cal S}|d\pmb z|)
\le\theta(\sup|\pmb z_0|)+\theta(\sup|\pmb z_1|)
\le\epsilon$}

\noindent as required.
}%end of proof of 631V

%\leader{631W}{}

\exercises{\leader{631X}{Basic exercises (a)}
\spheader 631Xj Let $\Cal S$ be a sublattice of $\Cal T$, and give
$(L^0)^{\Cal S}$ its product topology when each factor is given the
topology of convergence in measure.   Let $\Msimp\subseteq(L^0)^{\Cal S}$
be the set of simple processes with domain $\Cal S$.   Show that the
topological closure of $\Msimp$ in $(L^0)^{\Cal S}$ is the set of all
fully adapted processes with domain $\Cal S$.
%631B

%\spheader 631Xa
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$ a near-simple process with domain $\Cal S$.   Show that
$\pmb u\restr\Cal S\cap[\tau,\tau']$ is near-simple whenever $\tau\le\tau'$
in $\Cal S$.
%631Fa

\spheader 631Xb Give an example of a simple process $\pmb u$ and a
sublattice $\Cal S'$ of $\dom\pmb u$ such that $\pmb u\restr\Cal S'$ is
not near-simple.   \Hint{take $\Cal S'$ the lattice of constant stopping times.}
%631F

\spheader 631Xc Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb u$ a near-simple process with domain $\Cal S$.   Show that
for any $\epsilon>0$ there is a simple process $\pmb u'$ with domain
$\Cal S$ such that
$\theta(\sup|\pmb u-\pmb u'|)\le\epsilon$ and
$\sup|\pmb u'|\le\sup|\pmb u|$.
%631F mt61bits

\spheader 631Xd Let $\Cal S$ be a sublattice of $\Cal T$ and
$\sfamilysS{u}$ a near-simple process.   Set
$v_{\tau}=\sup_{\sigma\in\Cal S\wedge\tau}u_{\sigma}$ for $\tau\in\Cal S$
(cf.\ 614Fb).
Show that $\family{\tau}{\Cal S}{v_{\tau}}$ is a near-simple process.
%631F 614F 614Xh

\spheader 631Xe In 631Mb, show that $\pmb v\in\Mns(\Cal S')$ is of bounded
variation iff $\Psi(\pmb v)\in\Mns(\Cal S)$ is of bounded variation, and in this
case $\Psi(\pmb v^{\uparrow})=\Psi(\pmb v^{\uparrow})$.
%631M

\spheader 631Xf In 631Mb, show that if $\pmb v$ is a local integrator with
domain $\Cal S$, then
$ii_{\pmb v\restr\Cal S'}(\pmb u)=\iiv(\Psi(\pmb u))\restr\Cal S'$ for every
$\pmb u\in\Mns(\Cal S')$.
%631M

\spheader 631Xh Let $\Cal S$ be a finitely full sublattice of $\Cal S$
such that $\inf A\in\Cal S$ for every non-empty
$A\subseteq\Cal S$.   Show that $\Cal S\wedge\tau$ is full for every
$\tau\in\Cal S$.
%631R
%if \tau' covered by \Cal S\wedge\tau then for J\in\Cal I(\Cal S\wedge\tau)
%set  a_J=\sup_{\sigma\in J}\Bvalue{\tau'=\sigma}  and define take  \sigma_J
%by  a_J\Bsubseteq\Bvalue{\sigma=\tau'}  and
%1\Bsetminus a_J\Bsubseteq\Bvalue{\sigma=\tau} , then
%\tau=\inf_{J\in\Cal I(\Cal S\wedge\tau}\sigma_j

\spheader 631Xi Let $\pmb v$ be
a non-negative, non-decreasing near-simple process.
(i) Show that there are a non-negative, non-decreasing jump-free process
$\pmb w$ and a sequence $\sequencen{\pmb u_n}$ of non-negative
non-decreasing simple processes such that
$\pmb v=\pmb w+\sum_{n=0}^{\infty}\pmb u_n$.   (ii) Show that for
every $\epsilon>0$ there is a non-decreasing simple process $\pmb u$ such
that $\pmb v-\pmb u$ is non-decreasing and
$\theta(\Osclln(\pmb v-\pmb u))\le\epsilon$.
%631U

\spheader 631Xk In 631Mb, show that $\Osclln(\Psi(\pmb u))=\Osclln(\pmb u)$ for
every $\pmb u\in\Mns(\Cal S')$, so that $\Psi(\pmb u)$ is jump-free iff
$\pmb u$ is jump-free.
%631M out of order query 63bits

\spheader 631Xl Let $\Cal S$ be a non-empty finitely full sublattice of 
$\Cal T$, and $\pmb u=\sfamilysS{u}$ a fully adapted process.   Show that
$\pmb u$ is near-simple iff $\pmb u$ is locally near-simple and moderately
oscillatory iff $\pmb u$ is locally near-simple and order-bounded
and $\lim_{\sigma\uparrow\Cal S}u_{\sigma}$ is defined in $L^0(\frak A)$.
%631F(a-vi) 615Da out of order query

\leader{631Y}{Further exercises (a)}
%\spheader 631Ya
Let $\pmb w=\family{\sigma}{\Cal T_f}{w_{\sigma}}$ be Brownian motion
based on the real-time stochastic integration structure
$(\frak C,\bar\nu,\coint{0,\infty},\langle\frak C_t\rangle_{t\ge 0})$, as
described in 612T.   Set $\Cal S=\Cal T\wedge\check 1$ where $\check 1$ is the
constant stopping time with value $1$.   Show that there is a near-simple
process $\pmb u=\sfamilysS{u}$ of bounded variation such that
$ii_{\pmb w}(\pmb u)$ is not a martingale.
%631B 623O mt63bits %622L $w$ a local martingale


\spheader 631Yb Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a near-simple process.   Show that if $\Cal S'$ is a
sublattice of $\Cal S$ which is order-convex in $\Cal S$ (that is,
$\Cal S\cap[\sigma,\tau]\subseteq\Cal S'$ whenever $\sigma\le\tau$ in
$\Cal S'$), then $\pmb u\restr\Cal S'$ is near-simple.
%631F 615G

\spheader 631Yc Let $\pmb u_1,\ldots,\pmb u_k$ be (locally) near-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 $\bar h\pmb U$ (619G) is (locally) near-simple.
%631F(a-ii) 619G 619Yb

\spheader 631Ye Let $\Cal S$ be a sublattice of $\Cal T$.   Show that
$\Mlns(\Cal S)$ is closed in $\Mlob(\Cal S)$ for the local ucp topology of
615Xb.
%631F \query out of order

\spheader 631Yd
Suppose that $\Cal S=[\min\Cal S,\max\Cal S]$ is a closed interval
in $\Cal T$ and that $\pmb u=\sfamilysS{u}$ is a jump-free process.
Show that for every $z\in L^0(\frak A_{\min\Cal S})$
there is a $\tau\in\Cal S$ such that
$\Bvalue{u_{\min\Cal S}\le z}\Bcap\Bvalue{z\le u_{\max\Cal S}}
\Bsubseteq\Bvalue{u_{\tau}=z}$.
%631R mt61bits

\spheader 631Yf Let $(\frak A,\bar\mu)$ be the Lebesgue measure algebra and
$\sequencen{e_n}$ a stochastically independent sequence of members of $\frak A$
of measure $\bover12$.   For $n\in\Bbb N$ let $\frak A_{2n}$ be the subalgebra
of $\frak A$ generated by $\{e_i:i<n\}$ and set $\frak A_{2n+1}=\frak A_{2n+2}$
for each $n$.   In the stochastic integration structure
$(\frak A,\bar\mu,\Bbb N,\sequencen{\frak A_n},\Cal T,
\family{\tau}{\Cal T}{\frak A_{\tau}})$, take
$\Cal S=\{\check n:n\in\Bbb N\}$ and
$\Cal S'=\{\sigma_n:n\in\Bbb N\}$ where $\sigma_n=\check{2n}$ for each $n$.
Set $v_{\sigma_n}=\sum_{0\le i<n}\Bover1{i+1}(2\chi e_i-\chi 1)$
for $n\in\Bbb N$.   Show that $\pmb v=\sequencen{v_{\sigma_n}}$ is a
$\|\,\|_2$-bounded martingale and a near-simple integrator with domain
$\Cal S'$, but that if we  define $\Psi(\pmb v)\in\Mns(\Cal S)$ as in 631Mb,
$\Psi(\pmb v)$ is not an integrator.
%631M

\leaveitout{\spheader 631Y? \query
Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb{u}=\sfamilysS{u}$ a locally near-simple process
and $\pmb v=\sfamilysS{v}$ a locally near-simple $L^2$-process
which is a local martingale.   Suppose that
$\int_{\Cal S}\pmb{u}^2\,d\pmb v^*$ is defined and belongs to $L^1_{\bar\mu}$.
Show that $\int_{\Cal S}\pmb u\,d\pmb v$ is defined and that
$\|\int_{\Cal S}\pmb{u}\,d\pmb v\|_2^2
\le\|\int_{\Cal S}\pmb{u}^2\,d\pmb v^*\|_1$.
}

}%end of exercises


\endnotes{
\Notesheader{631}

From the beginning, the theory of continuous-time stochastic processes has given
special prominence to those which have continuous or \cadlag\ sample paths;  see
\S455.   Just as the jump-free processes of \S618 correspond to continuous
sample paths (618H), near-simple processes correspond to \cadlag\ sample paths
(631D).   In comparison, the definition I offer of `near-simple' process (631B)
is agreeably more straightforward than the definition of `jump-free' process in
618B.   The only thing to remember is that,
as with simple processes and jump-free processes,
the restriction of a near-simple process to a sublattice need not
be near-simple.   So we have to take care when specifying
the domain of any near-simple process we want to think about.   But the journey
through the elementary properties of the space $\Mns$ of near-simple processes
(631C, 631D-631J) %631D 631E 631F 631G 631H 631I 631J
gives no difficulties.   631K calls for a little ingenuity, but all the
ingredients are in \S614.   Perhaps this is a good place to remark that
near-simple processes are defined in terms of the ucp topology, so remain
near-simple under any change of law.

In 631M I give a wide-ranging result on the extension of near-simple processes
to larger sublattices.   Like the previous extension theorem 612P,
it is (in essence) too facile to offer much insight.
But in the proofs of 612Qa and 631N these general propositions give us a helpful
lift over potentially awkward obstacles.

The point of the stopping lemma $\SL_1$ in 631O-631S %631Oa 631Pa 631Ra 631S
is that we often have a description of (locally) near-simple processes
which is of a quite different kind from that in 631B.   Furthermore,
this alternative description, when applicable, leads us to a strikingly stronger
approximation by simple processes than that promised by 631B.   If we take
a sequence $\sequencen{\tau_n}$ as constructed in the proof of 631Ra, and
truncate it appropriately, we find ourselves with a breakpoint string
$(\tau_0,\ldots,\tau_n)$ from which we can define a simple process
$\pmb u'\,\,\epsilon$-approximating the near-simple process $\pmb u$
except in the small region $\Bvalue{\tau_n<\max\Cal S}$.   Now we knew that we
had an approximation which would be $\epsilon$-close to $\pmb u$ except
perhaps in a region of measure at most $\epsilon$.   But the Stopping Lemma
gives us a progressive description of the approximation, each successive
breakpoint being chosen in terms of a criterion observable at the time of
the jump.   Once again, a simple modification of $\SL_1$ gives us a matching
description $\SL_2$ of jump-free processes.

I have tried to set these out in a way which exhibits one of the essential
difficulties in the theory of stochastic processes.   If a process is
significant at all, stopping times defined by that process are likely to be
important.   (See \S\S477-479, for instance, or the jump-times of the Poisson
process in 612Uf.)   For jump-free processes like Brownian motion, especially if
we have a representation with continuous sample paths, we expect
hitting times to be easy to investigate, though there can be real
surprises (see 652M-652N below).   In fact we can have a kind of Darboux
continuity, as in 631Yd.   But with \cadlag\ paths we never know quite
what will happen next;  as we watch the process evolve, it might at any moment
flip into a new state unforeseeable from anything we have seen before.
I will go into this further in \S643, with the notion
of `accessibility' of a stopping time.

As usual, my presentation is ahistorical.   Near-simple processes, as
I have defined them, are close enough to the classical concept of \cadlag\
process for the classical theory (see 633R below)
to have results corresponding to 631R.
The `moderately oscillatory' processes of \S614, and the fundamental facts in
615M, are my own invention.   But you will have no difficulty in seeing how
615M could be inspired by a proof of 631Ra.

There are other ways in which a process may be expressible as the limit of a
more or less special sequence of simple processes.
I give 631U as a `theorem' because this
is the form I shall want to quote in \S644.   But 631V is an
alternative form of the same idea.   In the version of 631Xi(i),
we see that $\pmb v$ is expressed as the sum of
a jump-free part and a `saltus' part, recalling the Lebesgue decomposition
of a function of bounded variation into a continuous part and a saltus
part, as in 226Ca.   In 631V I suggest a different measure of
approximation,
using $\theta(\int_{\Cal S}|d\pmb u|)$ in place of $\theta(\sup|\pmb u|)$.
And in 643M there will be a (much deeper) result on martingales which also
seeks to express a given process as the sum of a simpler process and a
process with small residual oscillation.
}%end of notes

\discrpage

\frfilename{mt632.tex}
\versiondate{14.8.20/6.12.21}
\copyrightdate{2018}

\def\chaptername{Structural variations}
\def\sectionname{Right-continuous filtrations}

\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind

\newsection{632}

Up to this point, we have been able (with some effort)
to work in the full generality
of stochastic integration structures \Asisfour\ as described in \S\S611-613.
We are now approaching territory in which we shall
need to have filtrations which are
`right-continuous' in the sense of 632B.   These include the standard
examples (632D).   The results I present here are a quick run through new
features of the structures developed in \S\S611-612 (632C) and an important
characterization of near-simple processes on infimum-closed
full sublattices (632F).   With this in hand,
we see that in the most familiar contexts
local martingales will be locally near-simple (632I) and we have a useful
test for being a martingale (632J).
In 632N I describe a classic example of a local martingale.

\cmmnt{\leader{632A}{Notation}\cmmnt{ As usual,} \Asissix\
will be a stochastic integration structure.
If $A\subseteq\Cal T$ and $\tau\in\Cal T$, I write $A\wedge\tau$ for
$\{\sigma\wedge\tau:\sigma\in A\}$.   The $f$-algebra
$L^0=L^0(\frak A)$ will be
given its linear space topology of convergence in
measure.   $\Cal T_b$, $\Cal T_f\subseteq\Cal T$ will be the ideals of bounded
and finite-valued stopping times\cmmnt{, as in 611A(b-iii)}.
For $\sigma\in\Cal T$,
$P_{\sigma}$ will be the conditional expectation operator from
$L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$ to itself
associated with the closed subalgebra
$\frak A_{\sigma}$;  for $z\in L^1_{\bar\mu}$, $\pmb Pz$ will be the
martingale $\family{\sigma}{\Cal T}{P_{\sigma}z}$.   I will write
$\1lim$ for $\|\,\|_1$-limits in $L^1_{\bar\mu}$.
}

\leader{632B}{Definition} I will say that $\family{t}{T}{\frak A_t}$ or
\Asisfour\ or \Asissix\ is {\bf right-continuous}
if $\frak A_t=\bigcap_{s>t}\frak A_s$ whenever $t\in T$ is not isolated
on the right.

\cmmnt{ The notion of `right-continuity' is of course
a phenomenon which can appear anywhere in the theory of
ordered sets, and will recur intermittently in this volume in a variety of
contexts.   I have therefore chosen this phrase for general use, as
follows.}   If $P$ and $Q$ are partially ordered sets, an order-preserving
function $f:P\to Q$ is {\bf right-continuous} if $\inf f[C]$ is defined and
equal to $f(\inf C)$ whenever $C\subseteq P$ is non-empty and
downwards-directed and has an infimum in $P$.
\cmmnt{When $P$ is totally ordered,
this will be the case iff $\inf_{q>p}f(q)=f(p)$ whenever $p\in P$ is not
isolated on the right.
So a filtration $\family{t}{T}{\frak A_t}$
of closed subalgebras of $\frak A$ is right-continuous if it is
right-continuous when considered as an order-preserving function from
$T$ to the set of closed subalgebras of $\frak A$ (or, of course, when
considered as a function from $T$ to $\Cal P\frak A$).   Similarly, a
filtration $\family{t}{\coint{0,\infty}}{\Sigma_t}$ of $\sigma$-algebras is
right-continuous if $\Sigma_t=\bigcap_{s>t}\Sigma_s$ for every
$t\ge 0$.   See
632C(a-iii) for another example, and 632C(a-ii) for a kind of inverted example.}

\leader{632C}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.

(a) Suppose that $C\subseteq\Cal T$ is non-empty.

\quad(i)

$$\eqalign{\Bvalue{\inf C>t}
&=\inf_{\tau\in C}\Bvalue{\tau>t}\text{ if }
   t\text{ is isolated on the right},\cr
&=\sup_{s>t}\inf_{\tau\in C}\Bvalue{\tau>s}\text{ otherwise}.\cr}$$

\quad(ii)
$\Bvalue{\inf C<\tau}=\sup_{\upsilon\in C}\Bvalue{\upsilon<\tau}$ for
every $\tau\in\Cal T$.

\quad(iii)
$\frak A_{\inf C}=\bigcap_{\tau\in C}\frak A_{\tau}$.

(b)\cmmnt{ (Compare 611Cd.)} If $C$, $D\subseteq\Cal T$ are non-empty, then
$\inf C\vee\inf D=\inf\{\sigma\vee\tau:\sigma\in C$, $\tau\in D\}$.

(c) If $\Cal S$ 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 $u=\lim_{\sigma\downarrow A}u_{\sigma}$ is defined in $L^0$, then
$u\in L^0(\frak A_{\inf A})$.

\proof{{\bf (a)} Write $\sigma$ for $\inf C$.

\medskip

\quad{\bf (i)} If $s'>t$ in $T$, then

\Centerline{$\sup_{s>t}\inf_{\tau\in C}\Bvalue{\tau>s}
=\sup_{s'\ge s>t}\inf_{\tau\in C}\Bvalue{\tau>s}\in\frak A_{s'}$,}

\noindent so

\Centerline{$\sup_{s>t}\inf_{\tau\in C}\Bvalue{\tau>s}
\in\bigcap_{s>t}\frak A_s=\frak A_t$}

\noindent whenever $t\in T$ is not isolated on the right, and the result
follows immediately from 611F.

\medskip

\quad{\bf (ii)} By 611E(c-iv-$\beta$),
$\Bvalue{\sigma<\tau}\Bsupseteq\sup_{\upsilon\in C}\Bvalue{\upsilon<\tau}$.
Let $t\in T$.   If $t$ is isolated on the right, then

$$\eqalignno{\Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
&=\Bvalue{\tau>t}\Bsetminus\inf_{\upsilon\in C}\Bvalue{\upsilon>t}\cr
\displaycause{(i) above}
&=\sup_{\upsilon\in C}\Bvalue{\tau>t}\Bsetminus\Bvalue{\upsilon>t}
\Bsubseteq\sup_{\upsilon\in C}\Bvalue{\upsilon<\tau};\cr}$$

\noindent otherwise, again using (i),

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

\noindent So

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

\noindent and we have equality.

\medskip

\quad{\bf (iii)} Now suppose that $\sigma=\inf C$ in $\Cal T$.   By 611H(c-ii),
$\frak A_{\sigma}\subseteq\bigcap_{\tau\in C}\frak A_{\tau}$.
Conversely, suppose that $a\in\bigcap_{\tau\in C}\frak A_{\tau}$ and that
$t\in T$.
For $s\in T$ set $a_s=\inf_{\tau\in C}\Bvalue{\tau>s}$;  then
$a\Bsetminus a_s=\sup_{\tau\in C}a\Bsetminus\Bvalue{\tau>s}\in\frak A_s$ (611G).
By (a), $\Bvalue{\sigma>t}=a_t$ if $t$ is isolated on the
right, and is $\sup_{s>t}a_s$ otherwise.   So

$$\eqalign{a\Bsetminus\Bvalue{\sigma>t}
&=a\Bsetminus a_t\in\frak A_t
   \text{ if }t\text{ is isolated on the right,}\cr
&=a\Bsetminus\sup_{s>t}a_s=\inf_{s>t}a\Bsetminus a_s
\in\bigcap_{t'>t}\frak A_{t'}=\frak A_t\text{ otherwise}\cr}$$

\noindent because

\Centerline{$\inf_{s>t}a\Bsetminus a_s
=\inf_{t'\ge s>t}a\Bsetminus a_s\in\frak A_{t'}$}

\noindent whenever $t'>t$.   As $t$ is arbitrary, $a\in\frak A_{\sigma}$.   As
$a$ is arbitrary,
$\frak A_{\sigma}\supseteq\bigcap_{\tau\in C}\frak A_{\tau}$ and we
have equality.

\medskip

{\bf (b)} Set $E=\{\sigma\vee\tau:\sigma\in C$, $\tau\in D\}$,
$\phi_C(t)=\inf_{\sigma\in C}\Bvalue{\sigma>t}$,
$\phi_D(t)=\inf_{\sigma\in D}\Bvalue{\sigma>t}$ and
$\phi_E(t)=\inf_{\sigma\in E}\Bvalue{\sigma>t}$ for $t\in T$.
Then

$$\eqalignno{\phi_E(t)&
=\inf_{\sigma\in C,\tau\in D}\Bvalue{\sigma\vee\tau>t}
=\inf_{\sigma\in C,\tau\in D}\Bvalue{\sigma>t}\Bcup\Bvalue{\tau>t}
=\inf_{\sigma\in C}\Bvalue{\sigma>t}\Bcup\inf_{\tau\in D}\Bvalue{\tau>t}
\Displaycause{313Bd}
=\phi_C(t)\Bcup\phi_D(t)\cr}$$

\noindent for every $t\in T$.   If $t$ is isolated on the right, we have

\Centerline{$\Bvalue{\inf E>t}=\phi_E(t)=\phi_C(t)\Bcup\phi_D(t)
=\Bvalue{\inf C>t}\Bcup\Bvalue{\inf D>t}$.}

\noindent If $t$ is not isolated on the right, then

$$\eqalign{\Bvalue{\inf E>t}&
=\sup_{s>t}\phi_E(s)
=\sup_{s>t}\phi_C(s)\Bcup\phi_D(s)
=\sup_{s,s'>t}\phi_C(s)\Bcup\phi_D(s')\cr&
=\sup_{s>t}\phi_C(s)\Bcup\sup_{s'>t}\phi_D(s')
=\Bvalue{\inf C>t}\Bcup\Bvalue{\inf D>t}.\cr}$$

\noindent Thus

\Centerline{$\Bvalue{\inf E>t}=\Bvalue{\inf C>t}\Bcup\Bvalue{\inf D>t}
=\Bvalue{\inf C\vee\inf D>t}$}

\noindent for every $t$, and $\inf E=\inf C\vee\inf D$, as claimed.

\medskip

{\bf (c)} Put 613Bj and (a-iii) here together.

\medskip

}%end of proof of 632C

\leader{632D}{Examples (a)} In the construction of Brownian motion
in 612T, $\langle\frak C_t\rangle_{t\ge 0}$ is right-continuous.
\prooflet{

\Prf\ Suppose that $t\ge 0$ and that $c\in\bigcap_{s>t}\frak C_s$.
For each $k\in\Bbb N$ let $E_k\in\Sigma_{t+2^{-k}}$ be such that
$E_k^{\ssbullet}=c$.   Since $E_j\symmdiff E_k$ is negligible for all
$j$ and $k$, $c=E^{\ssbullet}$ where
$E=\bigcap_{k\in\Bbb N}\bigcup_{j\ge k}E_j$.   But now
$E\in\bigcap_{s>t}\Sigma_s$;  as 477Hc tells us that
$\bigcap_{s>t}\hat\Sigma_s=\hat\Sigma_t$, $E\in\hat\Sigma_t$
and $c\in\frak C_t$.\ \Qed}

\spheader 632Db In the construction of the standard Poisson process
in 612U, $\langle\frak A_t\rangle_{t\ge 0}$ is right-continuous.
\prooflet{\Prf\ From the formulae

\Centerline{$\Sigma_t=\{F:F\subseteq\Omega$, $F\in\hat{\ddot\Sigma}_t\}$}

\noindent and

\Centerline{$\hat{\ddot\Sigma}_t=\bigcap_{s>t}\hat{\ddot\Sigma}_s$}

\noindent in 612Uc, it is clear that $\Sigma_t=\bigcap_{s>t}\Sigma_s$
and therefore $\frak A_t=\bigcap_{s>t}\frak A_t$ for every $t\ge 0$.\ \Qed}

\leader{632E}{Lemma}\cmmnt{ (Compare 631N.)}
Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.    Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a locally near-simple process,
and $A\subseteq\Cal S$ a non-empty downwards-directed set
such that $\tau=\inf A$ belongs to $\Cal S$.
Then $u_{\tau}=\lim_{\sigma\downarrow A}u_{\sigma}$, and in fact
for every $\epsilon>0$ there is a $\sigma\in A$ such that
$\theta(\sup_{\rho\in\Cal S\cap[\tau,\sigma]}|u_{\rho}-u_{\tau}|)\le\epsilon$.

\proof{{\bf (a)}
 Take any $\tau'\in A$.   Then $\pmb u\restr\Cal S\cap[\tau,\tau']$ is
near-simple (631F(iv)), so there is a simple process
$\pmb v=\family{\rho}{\Cal S\cap[\tau,\tau']}{v_{\rho}}$ such that
$\theta(\sup|\pmb v-\pmb u\restr\Cal S\cap[\tau,\tau'])\le\bover13\epsilon$.
$\pmb v$ has
starting value $v_{\tau}$ and a breakpoint string $(\sigma_0,\ldots,\sigma_n)$
say.   For each $i\le n$,
$\Bvalue{\tau<\sigma_i}=\sup_{\sigma\in A}\Bvalue{\sigma<\sigma_i}$
(632C(a-ii));  as $A$ is downwards-directed, there is a $\sigma\in A$ such that
$\sigma\le\tau'$ and
$\bar\mu(\Bvalue{\tau<\sigma_i}\Bsetminus\Bvalue{\sigma<\sigma_i})
\le\Bover{\epsilon}{3(n+1)}$ for every $i$.   Set
$b=\sup_{i\le n}\Bvalue{\tau<\sigma_i}\Bsetminus\Bvalue{\sigma<\sigma_i}$, so
that $\bar\mu b\le\bover13\epsilon$ and

\Centerline{$\Bvalue{v_{\rho}\ne v_{\tau}}
\Bsubseteq\sup_{i\le n}\Bvalue{\sigma_i\le\rho}\Bcap\Bvalue{\tau<\sigma_i}
\Bsubseteq b$,}

\Centerline{$1\Bsetminus b\Bsubseteq\Bvalue{v_{\rho}=v_{\tau}}
\Bsubseteq\Bvalue{|u_{\rho}-u_{\tau}|\le\bover23\epsilon}$}

\noindent for every $\rho\in\Cal S\cap[\tau,\sigma]$.
Now $\pmb u\restr\Cal S\cap[\tau,\tau']$ is order-bounded so
$\bar u=\sup|\pmb u\restr\Cal S\cap[\tau,\tau']|$ is defined in $L^0$, and now
$|u_{\rho}-u_{\tau}|\le\Bover23\epsilon\chi 1+\bar u\times\chi b$ for every
$\rho\in\Cal S\cap[\tau,\sigma]$, so

\Centerline{$\theta(\sup_{\rho\in\Cal S\cap[\tau,\sigma]}|u_{\rho}-u_{\tau}|)
\le\theta(\Bover23\epsilon\chi 1)+\theta(\bar u\times\chi b)
\le\Bover23\epsilon+\bar\mu b
\le\epsilon$.}

\medskip

{\bf (b)} Of course it follows at once that
$u_{\tau}=\lim_{\sigma\downarrow A}u_{\sigma}$.
}%end of proof of 632E

\leader{632F}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and has a
lower bound in $\Cal S$.
If $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ is a fully adapted process,
then $\pmb u$ is locally near-simple iff it is locally moderately oscillatory
and

$$\eqalign{(\dagger)\,\,u_{\inf A}=\lim_{\sigma\downarrow A}&
u_{\sigma}\text{ for every
non-empty downwards-directed }A\subseteq\Cal S\cr&
\mskip250mu\text{ with a lower bound in }\Cal S.\cr}$$

\proof{{\bf (a)} Suppose that $\pmb u$ satisfies the conditions.

\medskip

\quad{\bf (i)} If $\Cal S$ has a least element, we can apply
631S to see that $\pmb u\restr\Cal S\wedge\tau$ is
near-simple for every $\tau\in\Cal S$, that is, that $\pmb u$ is locally
near-simple.

\medskip

\quad{\bf (ii)} Generally, we can apply (i) to see that
$\pmb u\restr\Cal S\cap[\tau,\tau']$ is near-simple whenever $\tau\le\tau'$ in
$\Cal S$.   Now if $\tau'\in\Cal S$, $\pmb u\restr\Cal S\wedge\tau'$ is
moderately oscillatory and 631F(c-i) assures us that
$\pmb u\restr\Cal S\wedge\tau'$ is near-simple.   Accordingly $\pmb u$ is
locally near-simple.

\medskip

{\bf (b)} Now suppose that $\pmb u$ is locally near-simple.
By 631Ca, $\pmb u$ is surely locally moderately oscillatory, so
it will be enough to show that it satisfies the condition ($\gamma$).   Take a
non-empty downwards-directed set $A\subseteq\Cal S$ with a lower bound in
$\Cal S$;  by hypothesis, $\inf A\in\Cal S$;  by 632E,
$u_{\inf A}=\lim_{\sigma\downarrow A}u_{\sigma}$;  so
$\pmb u$ satisfies ($\dagger$), and we're done.
}%end of proof of 632F

\leader{632G}{Corollary} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely
full sublattice of $\Cal T$ such that $\inf A\in\Cal S$ for
every non-empty $A\subseteq\Cal S$ with a lower bound in $\Cal S$,
and $a\in\frak A$.   Set
$u_{\sigma}=\chi(\upr(a,\frak A_{\sigma}))$ for $\sigma\in\Cal S$,
and $\pmb u=\sfamilysS{u}$.   Then $\pmb u$ is near-simple.

\proof{ For $\sigma\in\Cal S$, write $a_{\sigma}$ for
$\upr(a,\frak A_{\sigma})$.   If $\sigma\le\tau$ in $A$,
$\frak A_{\sigma}\subseteq\frak A_{\tau}$ so
$a_{\tau}\Bsubseteq a_{\sigma}$ and
$\chi 1\ge u_{\sigma}\ge u_{\tau}\ge 0$.   Accordingly
$\pmb u$ is of bounded variation and moderately
oscillatory.   If $A\subseteq\Cal S$ is non-empty and downwards-directed and
$\tau=\inf A$, then
$a_{\tau}=\sup_{\sigma\in A}a_{\sigma}$.   \Prf\
Set $c=\sup_{\sigma\in A}a_{\sigma}$.   If $\sigma\in A$,
$c=\sup_{\sigma'\in A,\sigma'\le\sigma}a_{\sigma'}$ belongs to
$\frak A_{\sigma}$.   So
$c\in\bigcap_{\sigma\in A}\frak A_{\sigma}=\frak A_{\tau}$, by 632C(a-iii).
As $a\Bsubseteq c$, $a_{\tau}\Bsubseteq c$.   But of course
$c\Bsubseteq a_{\tau}$, so we have equality.\ \QeD\
This means that
$a_{\tau}=\lim_{\sigma\downarrow A}a_{\sigma}$ for the measure-algebra topology
on $\frak A$ (323D(a-ii)) and

\Centerline{$u_{\tau}=\chi(a_{\tau})=\lim_{\sigma\downarrow A}\chi(a_{\sigma})
=\lim_{\sigma\downarrow A}u_{\sigma}$}

\noindent (367Ra).   As $A$ is arbitrary, 632F tells us that $\pmb u$ is
locally near-simple, and by 631F(c-ii) it is actually near-simple.
}%end of proof of 632G

\leader{632H}{Corollary} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.  Let $\Cal S$ be a finitely
full sublattice of $\Cal T$ such that $\inf A\in\Cal S$ for
every non-empty $A\subseteq\Cal S$ with a lower bound in $\Cal S$,
and $\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$
a locally jump-free non-decreasing process.
Then $\pmb v:\Cal S\to L^0$ is an order-continuous lattice homomorphism.

\proof{ By 614Ka, $\pmb v$ is a lattice homomorphism.
If $A\subseteq\Cal S$ is upwards-directed and $\sup A\in\Cal S$, then

\Centerline{$v_{\sup A}=\lim_{\tau\uparrow A}v_{\tau}
=\sup_{\tau\in A}v_{\tau}$}

\noindent by 618I and 613Ba.   On the other side, $\pmb v$ is
locally near-simple
(631Cb), so if $A\subseteq\Cal S$ is downwards-directed and
$\inf A\in\Cal S$, then

\Centerline{$v_{\inf A}=\lim_{\tau\downarrow A}v_{\tau}
=\inf_{\tau\in A}v_{\tau}$}

\noindent by 632F.   Accordingly $\pmb v$ is order-continuous in the sense
of 313H.
}%end of proof of 632H

\leader{632I}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely
full sublattice of $\Cal T$ such that $\inf A\in\Cal S$ for
every non-empty $A\subseteq\Cal S$.   Then a virtually local
martingale with domain $\Cal S$ is a locally near-simple local martingale.

\proof{{\bf (a)} Let $\pmb u=\sfamilysS{u}$
be a virtually local martingale with domain $\Cal S$.   By 623K(b-iii),
it is an approximately local martingale.   Write $\Cal S'$ for
$\{\sigma:\sigma\in\Cal S$, $\pmb u\restr\Cal S\wedge\sigma$ is a
martingale$\}$, so that $\Cal S'$ is an ideal of $\Cal S$ (622M).
Take $\tau\in\Cal S$ and $\epsilon>0$.   Then there is a
non-empty downwards-directed $A\subseteq\Cal S\wedge\tau$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$ and
$R_A(\pmb u)$, as defined in 623B, is a martingale (623J).   Now
$\inf A\in\Cal S$, while
$\Bvalue{\inf A<\tau}=\sup_{\rho\in A}\Bvalue{\rho<\tau}$, by 632C(a-ii).
So $\bar\mu\Bvalue{\inf A<\tau}=\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}$
is at most $\epsilon$.   Moreover, for $\sigma\in\Cal S\wedge\inf A$,
$\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}=u_{\sigma}$, so
$\pmb u\restr\Cal S\wedge\inf A
=R_A(\pmb u)\restr\Cal S\wedge\inf A$ is a
martingale, and $\inf A\in\Cal S'$, while
$\Bvalue{\tau=\inf A}\ge 1-\epsilon$.   As $\tau$ and $\epsilon$ are
arbitrary, $\Cal S'$ covers $\Cal S$ and $\pmb u$ is a local martingale.

\medskip

{\bf (b)} We know that $\pmb u$ is locally moderately oscillatory
(622H).   If $A\subseteq\Cal S$ is non-empty and downwards-directed,
take any $\upsilon\in A$ and $\epsilon>0$.  Then there is
a $\rho\in\Cal S$ such that $\bar\mu\Bvalue{\rho<\upsilon}\le\epsilon$ and
$\pmb u\restr\Cal S\wedge\rho$ is a martingale.   Now
$A\wedge\rho$ is downwards-directed and has infimum $\rho\wedge\inf A$ (611Ch),
so $\frak A_{\rho\wedge\inf A}=\bigcap_{\sigma\in A}\frak A_{\rho\wedge\sigma}$
(632C(a-iii)) and

$$\eqalignno{u_{\rho\wedge\inf A}&
=P_{\rho\wedge\inf A}u_{\rho\wedge\upsilon}
=\1lim_{\sigma\downarrow A}P_{\rho\wedge\sigma}u_{\rho\wedge\upsilon}
\Displaycause{621C(g-i)}
=\lim_{\sigma\downarrow A}P_{\rho\wedge\sigma}u_{\rho\wedge\upsilon}
\Displaycause{613B(d-i)}
=\lim_{\sigma\downarrow A}u_{\rho\wedge\sigma}.\cr}$$

\noindent There is therefore a $\sigma_0\in A\wedge\upsilon$ such that
$\theta(u_{\rho\wedge\sigma}-u_{\rho\wedge\inf A})\le\epsilon$ whenever
$\sigma\in A$ and $\sigma\le\sigma_0$.   But now we have, whenever
$\sigma\in A$ and $\sigma\le\sigma_0$,

\Centerline{$\theta(u_{\rho\wedge\sigma}-u_{\sigma})
\le\bar\mu\Bvalue{\rho<\sigma}
\le\bar\mu\Bvalue{\rho<\upsilon}\le\epsilon$}

\noindent and similarly $\theta(u_{\rho\wedge\inf A}-u_{\inf A})\le\epsilon$,
so $\theta(u_{\sigma}-u_{\inf A})\le 3\epsilon$.   As $\epsilon$ is arbitrary,
$u_{\inf A}=\lim_{\sigma\downarrow A}u_{\sigma}$.
Thus condition ($\dagger$) of 632F is satisfied and $\pmb u$ is
locally near-simple.
}%end of proof of 632I

\leader{632J}{Where martingales come \dvrocolon{from}}\cmmnt{ Coming back
to the basic theory of martingales, we have the following.

\medskip

\noindent}{\bf Proposition} Suppose that
$\family{t}{T}{\frak A_t}$ is right-continuous.   Let
$\family{t}{T}{u_t}$ be a martingale in the sense that
$u_t\in L^1_{\bar\mu}$ and $u_s$ is the conditional expectation of $u_t$ on
$\frak A_s$ whenever $s\le t$ in $T$.   If
$\pmb w=\family{\tau}{\Cal T_b}{w_{\tau}}$ is a locally near-simple
process such that $w_{\check{t}}=u_t$ for every $t\in T$, then
$\pmb w$ is a martingale.

\proof{{\bf (a)} Fix $t\in T$ for the moment.   Set
$C=\{\sigma:\sigma\in\Cal T_b$, $w_{\sigma}=P_{\sigma}u_t\}$.

\medskip

\quad{\bf (i)} If $s\le t$, then $\check s\in C$, just because
$\frak A_s=\frak A_{\check s}$, so $w_{\check s}=u_s=P_{\check s}u_t$.

\medskip

\quad{\bf (ii)} If $\sigma_0$, $\sigma_1\in C$ and $\tau\in\Cal T_b$ is such
that $\Bvalue{\tau=\sigma_0}\Bcup\Bvalue{\tau=\sigma_1}=1$, then
$\tau\in C$.   \Prf\

$$\eqalignno{
\Bvalue{P_{\tau}u_t=w_{\tau}}
&\Bsupseteq\Bvalue{\sigma_0=\tau}
   \Bcap\Bvalue{P_{\sigma_0}u_t=w_{\sigma_0}}
   \Bcap\Bvalue{P_{\tau}u_t=P_{\sigma_0}u_t}
   \Bcap\Bvalue{w_{\tau}=w_{\sigma_0}}
\cr&=\Bvalue{\sigma_0=\tau}
   \Bcap\Bvalue{P_{\tau}u_t=P_{\sigma_0}u_t}
\Displaycause{because $\sigma_0\in C$ and $\pmb w$ is fully adapted}
=\Bvalue{\sigma_0=\tau}
\cr}$$

\noindent by 622Bb.   Similarly,
$\Bvalue{P_{\tau}u_t=w_{\tau}}\Bsupseteq\Bvalue{\tau=\sigma_1}$ and

\Centerline{$\Bvalue{P_{\tau}u_t=w_{\tau}}
\Bsupseteq\Bvalue{\tau=\sigma_0}\Bcup\Bvalue{\tau=\sigma_1}=1$,}

\noindent so $P_{\tau}u_t=w_{\tau}$, that is,
$\tau\in C$.\ \Qed

\medskip

\quad{\bf (iii)} In particular, putting (ii) here and 611E(a-ii-$\gamma$)
together, $\sigma\wedge\tau\in C$ whenever $\sigma$, $\tau\in C$.

\medskip

\quad{\bf (iv)} If $A\subseteq C$ is non-empty and downwards-directed,
with infimum $\tau$, then $\tau\in C$.   \Prf\
$\pmb{P}u_t=\family{\sigma}{\Cal T}{P_{\sigma}u_t}$ is a martingale
(622Fa), so

$$\eqalignno{P_{\tau}u_t&
=\lim_{\sigma\downarrow A}P_{\sigma}u_t
\Displaycause{put 632I and 632F together, or look at (a) of the proof of
632I}
=\lim_{\sigma\downarrow A}w_{\sigma}
=w_{\tau}\cr}$$

\noindent by 632F again.\ \Qed

\medskip

\quad{\bf (v)} If $\tau\le\check t$ then $\tau\in C$.   \Prf\
Set $A=\{\sigma:\sigma\in C$, $\tau\le\sigma\}$.   Then $A$ is
downwards-directed, by (iii), and non-empty, because it contains
$\check t$.
Because $\Cal T$ is Dedekind complete, $\tau^*=\inf A$ is defined, and
$\tau^*\in C$ by (iv).   Of course $\tau\le\tau^*$.
\Quer\ If $\tau\ne\tau^*$, then $\Bvalue{\tau<\tau^*}\ne 0$ and there is an
$s\in T$ such that $\Bvalue{\tau^*>s}\Bsetminus\Bvalue{\tau>s}\ne 0$.
As $\Bvalue{\tau>s}\ne 0$, $s$ must be less than or equal to $t$, and
$\check s\in C$
by (i).   Now consider $c=\Bvalue{\tau\le\check s}$.   We have

\Centerline{$c\in\frak A_{\check s}\Bcap\frak A_{\tau}
\Bsubseteq\frak A_{\check s}\Bcap\frak A_{\tau^*}$}

\noindent by 611H(c-ii).   So there is a $\tau'\in\Cal T$ such that
$\Bvalue{\tau'=\check s}\Bsupseteq c$ and
$\Bvalue{\tau'=\tau^*}\Bsupseteq 1\Bsetminus c$ (611I), and
$\tau'\in C$ by (ii).

Consider

$$\eqalign{\Bvalue{\tau\le\tau'}
&\Bsupseteq(\Bvalue{\tau\le\check s}\Bcap\Bvalue{\tau'=\check s})
  \Bcup(\Bvalue{\tau\le\tau^*}\Bcap\Bvalue{\tau'=\tau^*})
\cr&\Bsupseteq(\Bvalue{\tau\le\check s}\Bcap c)
  \Bcup(\Bvalue{\tau\le\tau^*}\Bsetminus c)
=c\Bcup(1\Bsetminus c)
=1.
\cr}$$

\noindent Thus $\tau\le\tau'$, so $\tau'\in A$ and $\tau^*\le\tau'$.
On the other hand,

$$\eqalignno{0
&\ne\Bvalue{\tau^*>s}\Bsetminus\Bvalue{\tau>s}
=\Bvalue{\check s<\tau^*}\Bcap\Bvalue{\tau\le\check s}
\Displaycause{611E(a-i-$\delta$)}
\Bsubseteq\Bvalue{\check s<\tau^*}\Bcap\Bvalue{\tau'=\check s}
\Bsubseteq\Bvalue{\tau'<\tau^*}
\cr}$$

\noindent (611E(c-iii-$\gamma$), so this is impossible.\ \Bang

Accordingly $\tau=\tau^*$ belongs to $C$, as claimed.\ \Qed

\medskip

{\bf (b)} Thus $w_{\tau}=P_{\tau}u_t$ whenever $t\in T$ and
$\tau\le\check t$.   Now, of course, if $\sigma\le\tau$ in $\Cal T_b$,
there is a $t\in T$ such that $\tau\le\check t$, and

\Centerline{$P_{\sigma}w_{\tau}=P_{\sigma}P_{\tau}u_t=P_{\sigma}u_t
=w_{\sigma}$.}

\noindent As $\sigma$ and $\tau$ are arbitrary, $\pmb w$ is a martingale.
}%end of proof of 632J

\leader{632K}{Lemma} Let $(\Omega,\Sigma,\mu)$ be a probability space with
measure algebra $(\frak A,\bar\mu)$, and $\langle\Sigma_t\rangle_{t\ge 0}$ a
right-continuous filtration of $\sigma$-subalgebras of $\Sigma$.   If we set
$\frak A_t=\{E^{\ssbullet}:E\in\Sigma_t\}$ for $t\ge 0$, then
$\langle\frak A_t\rangle_{t\ge 0}$ is a right-continuous filtration.

\proof{ Take $t\ge 0$ and $a\in\bigcap_{s>t}\frak A_s$.   Then for each
$n\in\Bbb N$ there is an $E_n\in\Sigma_{t+2^{-n}}$ such that
$E_n^{\ssbullet}=a$.   Set $E=\bigcup_{n\in\Bbb N}\bigcap_{i\ge n}E_i$;  then
$E\in\bigcap_{s>t}\Sigma_s=\Sigma_t$ and $a=E^{\ssbullet}\in\frak A_t$.
}%end of proof of 632K

\leader{632L}{Proposition}
Let $(\Omega,\Sigma,\mu)$ be a complete probability space and
$\family{t}{\coint{0,\infty}}{\Sigma_t}$ a right-continuous filtration
of $\sigma$-subalgebras of $\Sigma$, all containing every negligible
subset of $\Omega$.
Suppose that we are given a family $\langle X_t\rangle_{t\ge 0}$
of measurable real-valued functions on $\Omega$ such that
$t\mapsto X_t(\omega)$ is \cadlag\ for every $\omega$ and
$X_s$ is a conditional expectation of $X_t$ on $\Sigma_s$ whenever
$0\le s\le t$.
Define $(\frak A,\bar\mu,\langle\frak A_t\rangle_{t\ge 0})$ and
$\pmb u=\family{\tau}{\Cal T_f}{u_{\tau}}$ as in 612H and 631D.   Then
$\pmb u\restr\Cal T_b$ is a martingale and $\pmb u$ is a local
martingale.

\proof{ By 631D, $\pmb u$ is well-defined and is
a locally near-simple fully adapted process.
We have $u_{\check{t}}=X_t^{\ssbullet}$ for $t\in T$, so
$u_{\check{s}}$ is the conditional expectation of $u_{\check{t}}$ on
$\frak A_s=\{E^{\ssbullet}:E\in\Sigma_s\}$ whenever $s\le t$.
Moreover,
$\family{t}{\coint{0,\infty}}{\frak A_t}$ is right-continuous, by 632K.
But now 632J tells us that $\pmb u\restr\Cal T_b$ is a martingale, so that
$\pmb u$ is a local martingale.
}%end of proof of 632L

\leader{632M}{Proposition} Let $\pmb v=\family{\tau}{\Cal T_f}{v_{\tau}}$
be the Poisson process\cmmnt{ as defined in 612U}.

(a) If $\pmb\iota$ is the identity process,
$(\pmb v-\pmb\iota)\restr\Cal T_b$ is a martingale, so that
$\pmb v-\pmb\iota$ is a local martingale.

(b) The previsible variation of $\pmb v\restr\Cal T_b$ is
$\pmb\iota\restr\Cal T_b$.

\proof{{\bf (a)(i)} In order to use results from \S455,
I will in fact take a step back from the structure
$(\Omega,\Sigma,\mu,\family{t}{\coint{0,\infty}}{\Sigma_t})$ introduced in
612Uc, and for most of the argument work with the structure
$(\Cdlg,\ddot\Sigma,\ddot\mu,
\family{t}{\coint{0,\infty}}{\ddot\Sigma_t})$ from which it was
defined, with the associated $\sigma$-algebras
$\ddot\Sigma_t^+=\bigcap_{s>t}\ddot\Sigma_s$;  here $\Cdlg$ is the space of
all \cadlag\ functions from $\coint{0,\infty}$ to $\Bbb R$.
For $t\ge 0$ and $\omega\in\Cdlg$, set
$Y_t(\omega)=\omega(t)-t$.   If $0\le s\le t$, then
$Y_s$ is a conditional expectation of $Y_t$ on $\ddot\Sigma^+_s$.   \Prf\
This is trivial if $s=t$, just because $Y_s$ is
$\ddot\Sigma_s$-measurable.
If $s<t$, then we can use 455O and 455S, as follows.   Let
$h:\Cdlg\to\coint{0,\infty}$ be the constant stopping time with
value $s$.   Then the $\sigma$-algebra $\ddot\Sigma_h^+$ of 455Ob is

\Centerline{
$\{F:F\in\ddot\Sigma,\,F\cap\{\omega:h(\omega)\le s'\}\in\ddot\Sigma_{s'}^+$
for every $s'\ge 0\}
=\ddot\Sigma_s^+$.}

\noindent So if we define $\ddot\mu_{\omega s}$, for $\omega\in\Cdlg$, as
in 455O, and set

\Centerline{$g(\omega)=\int_{\Cdlg}Y_t(\omega')\,
\ddot\mu_{\omega s}(d\omega')$}

\noindent whenever this is defined, then $g$ will be a conditional
expectation of $Y_t$ on $\ddot\Sigma_s^+$, by 455Ob.
Next, $\phi_{\omega s}:\Cdlg\to\Cdlg$ is \imp\ for $\ddot\mu$ and
$\ddot\mu_{\omega s}$, where

$$\eqalign{\phi_{\omega s}(\omega')(s')
&=\omega(s')\text{ if }s'<s,\cr
&=\omega(s)+\omega'(s'-s)\text{ if }s\le s'\cr}$$

\noindent (455Sc).   So

$$\eqalignno{g(\omega)
&=\int_{\Cdlg}Y_t(\phi_{\omega s}(\omega'))\ddot\mu(d\omega')
=\int_{\Cdlg}\phi_{\omega s}(\omega')(t)-t\,\ddot\mu(d\omega')\cr
&=\int_{\Cdlg}\omega(s)+\omega'(t-s)-t\,\ddot\mu(d\omega')\cr
&=\omega(s)-t+\int_{\Cdlg}\omega'(t-s)\ddot\mu(d\omega')
=\omega(s)-t+\Expn(\lambda_{t-s})\cr
\displaycause{where $\lambda_{t-s}$ is the distribution of the random
variable $\omega'\mapsto\omega'(t-s)$}
&=\omega(s)-s
=Y_s(\omega)\cr}$$

\noindent because $\lambda_{t-s}$ is the Poisson distribution with
expectation $t-s$, as chosen in 612U.

This shows that $Y_s$ is a conditional expectation of $Y_t$ on
$\ddot\Sigma_s^+$.\ \Qed

\medskip

\quad{\bf (ii)} We know also from 455T that $\ddot\Sigma_s^+$ is included in

\Centerline{$\hat{\ddot\Sigma}_s
=\{F\symmdiff A:F\in\ddot\Sigma_s$, $A$ is $\ddot\mu$-negligible$\}$.}

\noindent It follows at once that
$Y_s$ is a conditional expectation of $Y_t$ on the $\sigma$-algebra
$\hat{\ddot\Sigma}_s$, and therefore that $Y_s\restr\Omega$
is a conditional expectation of
$Y_t\restr\Omega$ on $\Sigma_s$, because $\Omega$ is a conegligible subset
of $\Cdlg$.   And this is true whenever $s\le t$ in $\coint{0,\infty}$.

\medskip

\quad{\bf (iii)} As noted in 632Db,
$\family{t}{\coint{0,\infty}}{\Sigma_t}$ is a
right-continuous filtration, and it follows at once that
$\family{t}{\coint{0,\infty}}{\frak A_t}$ is right-continuous.
And we know that $\pmb v$ and $\pmb\iota$ are locally near-simple
processes (631E), so (ii) and
632Jc tell us that $(\pmb v-\pmb\iota)\restr\Cal T_b$ is a martingale.
Of course this means that $\pmb v-\pmb\iota$ will be a local martingale on
$\Cal T_f$.

\medskip

{\bf (b)} Since $(\pmb v-\pmb\iota)\restr\Cal T_b$ is a martingale,
its previsible variation is zero (626Mc), so the previsible variation of
$\pmb v\restr\Cal T_b$ is the same as the previsible variation of
$\pmb\iota\restr\Cal T_b$ (using 626Mb),
which is $\pmb\iota\restr\Cal T_b$ (626S).
}%end of proof of 632M

\leader{*632N}{}\cmmnt{ I described local martingales which are not
martingales in 622Xe and 622Xj, and plenty of other examples are provided by
632L;
for instance, `Brownian motion', as described in 612T, is a martingale if
you take it to be defined on $\Cal T_b$ (622L), but only a local
martingale if you take it to be defined everywhere on $\Cal T_f$ (632Ye).
There is a classic example which is more interesting,
because it arises naturally in the context of \S\S477-479.

\medskip

\noindent}{\bf Example (a)} Let $\mu=\mu_W$ be three-dimensional
Wiener measure\cmmnt{ (477D)} on
$\Omega=C(\coint{0,\infty};\BbbR^3)_0$, $\Sigma$ its domain,
and $(\frak A,\bar\mu)$ its measure
algebra.   For $t\ge 0$ set

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

\Centerline{$\Sigma_t=\{F\symmdiff H:F\in\Sigma'_t$,
$\mu H=0\}$,}

\Centerline{$\frak A_t=\{F^{\ssbullet}:F\in\Sigma_t\}$.}

\noindent\cmmnt{By 477Hc in three dimensions,
$\Sigma_t=\bigcap_{s>t}\Sigma_s$, so} $\langle\frak A_t\rangle_{t\ge 0}$
is right-continuous.

\spheader 632Nb Let $e$ be a unit vector in $\BbbR^3$.
Set $\Omega'=\{\omega:\omega\in\Omega$, $e$ is not a value of
$\omega\}$\cmmnt{;
by 478Mc, $\Omega'$ is conegligible, so belongs to $\Sigma_t$ for every
$t$}.   For $t\ge 0$ and $\omega\in\Omega$ set

$$\eqalign{Y_t(\omega)
&=\Bover1{\|\omega(t)-e\|}\text{ if }\omega\in\Omega',\cr
&=0\text{ otherwise }.\cr}$$

\noindent Then $Y_t$ is $\Sigma_t$-measurable for every $t$ and
$t\mapsto Y_t(\omega)$ is continuous for every $\omega$, so
we have a corresponding locally jump-free process
$\pmb v\cmmnt{\mskip5mu=\family{\tau}{\Cal T_f}{v_{\tau}}}$\cmmnt{ (618H)}.

\spheader 632Nc $\lim_{t\to\infty}\Expn(Y_t)=0$.
\prooflet{\Prf\ For $t$, $R>0$,
the probability density function
of the random variable $\omega\mapsto\omega(t)$ is
$x\mapsto\Bover1{(\sqrt{2\pi t})^3}e^{-\|x\|^2/2t}$ (put 274Ad and
272I together), so by 271Ic

$$\eqalignno{\Expn(Y_t)
&=\int_{\BbbR^3}
  \Bover1{(\sqrt{2\pi t})^3}e^{-\|x\|^2/2t}\Bover1{\|x-e\|}dx\cr
\displaycause{where the integral is taken with respect to Lebesgue measure
on $\BbbR^3$}
&\le\Bover1R+
   \int_{B(e,R)}
     \Bover1{(\sqrt{2\pi t})^3}e^{-\|x\|^2/2t}\Bover1{\|x-e\|}dx\cr
&\le\Bover1R+\Bover1{(\sqrt{2\pi t})^3}
   \int_{B(e,R)}\Bover1{\|x-e\|}dx
=\Bover1R+\Bover1{(\sqrt{2\pi t})^3}
   \int_{B(0,R)}\Bover1{\|x\|}dx\cr
&=\Bover1R+\Bover1{(\sqrt{2\pi t})^3}
   \int_0^R4\pi s^2\cdot\Bover1s\,ds
=\Bover1R+\Bover{2\pi R^2}{(\sqrt{2\pi t})^3}.\cr}$$

\noindent So $\limsup_{t\to\infty}\Expn(Y_t)\le\Bover1R$ for every $R>0$
and must be $0$.\ \Qed}
\cmmnt{

Since
$\Expn(Y_0)=1$, $\langle Y_t\rangle_{t\ge 0}$ is not a martingale and}
$\pmb v$ is not a martingale.

%\query is it the sum of a martingale and a locally BV process?

\spheader 632Nd If $n\in\Bbb N$ and
$h_n$ is the Brownian hitting time to the ball $B(e,2^{-n})$\cmmnt{ (477I)},
then $h_n$\cmmnt{ is adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$ (477I(c-iii)), so}
represents a stopping time $\tau_n$
adapted to $\langle\frak A_t\rangle_{t\ge 0}$\cmmnt{ (612Ha)}.   Set
$\Cal S=\bigcup_{n\in\Bbb N}\{\tau:\tau\in\Cal T_f$, $\tau\le\tau_n\}$.
\cmmnt{Because $\sequencen{\tau_n}$ is non-decreasing, $\Cal S$ is an ideal of
$\Cal T_f$.   Now} $\Cal S$ is a covering ideal\cmmnt{ in the sense of 611N}.
\prooflet{\Prf\
(This is where we need to be in three or more dimensions.)
For each $n\in\Bbb N$ and $t\ge 0$,

$$\eqalignno{\bar\mu\Bvalue{\tau_n>t}
&=\mu_W\{\omega:h_n(\omega)>t\}\cr
&\ge\mu_W\{\omega:B(e,2^{-n})\cap\omega[\,\coint{0,\infty}\,]=\emptyset\}
=1-\hp(B(e,2^{-n}))
\Displaycause{where $\hp$ is `Brownian hitting probability', see 477Ia}
=1-\tilde W_{B(e,2^{-n})}(0)
\Displaycause{where $\tilde W$ is `equilibrium potential', 479Cb and
479Pb}
=1-2^{-n}\cr}$$

\noindent by 479Da.   But this means that if $\tau\in\Cal T_f$ and
$n\in\Bbb N$ there is a $t\ge 0$ such that
$\bar\mu\Bvalue{\tau>t}\le 2^{-n}$, and now

\Centerline{$\Bvalue{\tau>\tau_n}
\Bsubseteq\Bvalue{\tau>t}\Bcup\Bvalue{\tau_n\le t}$}

\noindent has measure at most $2^{-n+1}$.   So

\Centerline{$\sup_{\sigma\in\Cal S}\bar\mu\Bvalue{\tau=\sigma}
\ge\bar\mu\Bvalue{\tau=\tau\wedge\tau_n}
=1-\bar\mu\Bvalue{\tau>\tau_n}\ge 1-2^{-n+1}$.}

\noindent As $n$ is arbitrary,
$\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}=1$;  as $\tau$ is arbitrary,
$\Cal S$ is a covering ideal of $\Cal T_f$.\ \Qed}

\spheader 632Ne
At the same time, for each $n\in\Bbb N$,
$\langle v_{\sigma}\rangle_{\sigma\le\tau_n}$ is a martingale.
\prooflet{\Prf\
Apply 478V with $G=\{x:\|x-e\|>2^{-n}\}$ and $f(x)=\Bover1{\|x-e\|}$,
so that $h_n$ is the Brownian exit time from $G$.   Again because we are
in at least three dimensions, $G$ has few wandering paths (478N), so 478Vb
tells us that if $g\le h_n$ is a stopping time representing $\sigma\le\tau_n$
then $Y_g$ is the conditional expectation of $Y_{h_n}$ on $\Sigma_g$, that is,
$v_{\sigma}=P_{\sigma}v_{\tau_n}$.\ \Qed

Thus} $\pmb v$ is a local martingale.

%\leader{632O}
%\leader{632P}
%\leader{632Q}
%\leader{632R}
%\leader{632S}
%\leader{632T}
%\leader{632U}
%\leader{632V}
%\leader{632W}

\exercises{\leader{632X}{Basic exercises (a)}
%\spheader 632Xa
Suppose that $\family{t}{T}{\frak A_t}$ is not right-continuous.
Show that in this case there
are a simple process $\pmb u=\family{\sigma}{\Cal T}{u_{\sigma}}$
and a non-empty downwards-directed set $A\subseteq\Cal T$ such
that $\lim_{\sigma\downarrow A}u_{\sigma}\ne u_{\inf A}$.
\Hint{reduce to the case in which $a\in\bigcap_{s>t}\frak A_s\setminus\frak A_t$
does not include any non-zero member of $\frak A_t$.   Try
$A=\{\sigma:\sigma\ge\check t$, $\Bvalue{\sigma=\max\Cal T}\Bsupseteq a\}$,
$u_{\check t}=0$, $u_{\max\Cal T}=\chi 1$.}
%632C mt62bits

\spheader 632Xb
Suppose that $T=\coint{0,\infty}$.   Show that if
$\langle\frak A_t\rangle_{t\ge 0}$ is
right-continuous, and we identify $\Cal T_f$ with a subset of $L^0$ as in
611Xa, then $\Cal T_f$ is order-closed in $L^0$ (definition:  313Da).
%632C

\spheader 632Xc  Suppose that $\family{t}{T}{\frak A_t}$ is right-continuous.
Show that the function $\sigma\mapsto\sigma\vee\tau:\Cal T\to\Cal T$ is
order-continuous for every $\tau\in\Cal T$.
%632C mt63bits

\spheader 632Xd Suppose that \Asissix\ is the structure described in 612U, and
$\pmb v$ the standard Poisson process.   Show that
$(\pmb v-\pmb\iota)^2-\pmb\iota$ is a local martingale.
%632M

\spheader 632Xe Let $\pmb w$ be Brownian motion (612T, 622L) and $\pmb\iota$ the
corresponding identity process.   Use the method of 632M to show that
$\pmb w\restr\Cal T_b$ and $\pmb w^2-\pmb\iota\restr\Cal T_b$
are martingales, without appealing to Dynkin's formula.
%612T 622L 632M

\leader{632Y}{Further exercises (a)}
%\spheader 632Ya
Suppose that $\frak A$ has countable Maharam type.
Show that $\{t:t\in T$, $\frak A_t\ne\frak A\cap\bigcap_{s>t}\frak A_s\}$ is
countable.
%632B

\spheader 632Yb Define a family $\family{t}{T}{\frak A_t^+}$ of closed
subalgebras of $\frak A$ by setting

$$\eqalign{\frak A_t^+
&=\frak A_t\text{ if }t\in T\text{ is isolated on the right,}\cr
&=\bigcap_{s>t}\frak A_s\text{ otherwise}.\cr}$$

\noindent(i) Show that $\family{t}{T}{\frak A_t^+}$ is a
right-continuous filtration.   (ii) Let $\Cal T^+$ be the set of stopping
times with respect to $\family{t}{T}{\frak A_t^+}$.   Show that if we think
of $\Cal T$ and $\Cal T^+$ as subsets of $\frak A^T$, as suggested in
611Ac, then
$\Cal T$ is a sublattice of $\Cal T^+$.   (iii) Show that every member of
$\Cal T^+$ is the infimum in $\Cal T^+$ of a subset of $\Cal T$.   (iv)
Show that if $A$ is a non-empty subset of $\Cal T$, its supremum in
$\Cal T^+$ belongs to $\Cal T$.   (v) Show that for every $\tau\in\Cal T^+$
there is a greatest $\sigma\in\Cal T$ such that $\sigma\le\tau$ in
$\Cal T^+$.
%632C

\spheader 632Yc Dropping the measure $\bar\mu$, and supposing only that
$\frak A$ is a Dedekind complete Boolean algebra and $\family{t}{T}{\frak A_t}$
is a filtration of order-closed subalgebras of $\frak A$, as in 611A, show that
parts (a)-(b) of 632C are still valid.

\spheader 632Yd Suppose that $\frak A$ is a complete \wsid\ Boolean algebra
(e.g., a probability algebra;  see 316G and 322F) and
$\family{t}{T}{\frak A_t}$ is a right-continuous filtration of order-closed
subalgebras of $\frak A$.   (i)
Let $\sequencen{C_n}$ be a sequence of ideals of $\Cal T$.
Show that $\sup(\bigcap_{n\in\Bbb N}C_n)=\inf_{n\in\Bbb N}\sup C_n$.
(ii) Let
$\Cal S$ be a sublattice of $\Cal T$ such that $\inf D\in\Cal S$ for
every non-empty countable set $D\subseteq\Cal S$.
Show that the intersection of a sequence
of covering ideals of $\Cal S$ is a covering ideal.
%632Yc 632C

\spheader 632Ye Let $(\frak C,\bar\nu,\coint{0,\infty},
   \penalty-100\langle\frak C_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak C_{\tau}})$
be the real-time stochastic integration structure of 612T, and
$\pmb w=\family{\sigma}{\Cal T_f}{w_{\sigma}}$
Brownian motion.   Show that for every $z\in L^0(\frak C)$ and
$\sigma\in\Cal T_f$ there is a $\tau\in\Cal T_f$ such that
$\sigma\le\tau$ and $w_{\tau}=z$;  so that $\pmb w$ is not a martingale.
%632D

\spheader 632Yf
Let $\mu_L$ be Lebesgue measure on $\coint{0,\infty}$, and
$\nu$ the corresponding Poisson point process with intensity $1$
as described in \S495, so that $\nu$ is a
complete probability measure on $\Cal P\coint{0,\infty}$.   Let
$\Omega$ be the set $\{\omega:\omega\subseteq\ooint{0,\infty}$,
$\omega\cap[0,a]$ is finite for every $a\ge 0\}$.   Show that
$\nu\Omega=1$;  let $\mu$ be the subspace measure on $\Omega$ and $\Sigma$
its domain.   For $t\ge 0$ let $\Sigma_t$ be the $\sigma$-algebra generated
by sets of the form
$\{\omega:\omega\in\Omega$, $\#(\omega)\cap[0,s]=n\}\symmdiff E$
where $n\in\Bbb N$, $s\le t$ and $\mu E=0$;  show that
$\langle\Sigma_t\rangle_{t\ge 0}$ is a right-continuous filtration of
$\sigma$-subalgebras of $\Sigma$.   For $t\ge 0$, $\omega\in\Omega$ set
$X_t(\omega)=\#(\omega\cap[0,t])$;  show that $\langle X_t\rangle_{t\ge 0}$
is adapted to $\langle\Sigma_t\rangle_{t\ge 0}$, and that
$t\mapsto X_t(\omega)$ is \cadlag\ for every $\omega$.
Show that if we take the stochastic process defined from
$(\Omega,\nu,\langle\Sigma_t\rangle_{t\ge 0},\langle X_t\rangle_{t\ge 0})$
by the method of 612H/631D, we obtain a structure
$(\frak A,\bar\mu,\langle\frak A_t\rangle_{t\ge 0},
\family{\tau}{\Cal T_f}{v_{\tau}})$ isomorphic to that of 612U.
%632D

\spheader 632Yg Give an example of near-simple
processes $\pmb u$, $\pmb v$ defined on $\Cal T$,
in a real-time structure with a right-continuous filtration,
such that $\int_{\check T}\pmb u\,d\pmb v=0$, where
$\check T$ is the set of constant stopping times, but
$\int_{\Cal T_s}\pmb u\,d\pmb v$ is undefined, where $\Cal T_s$ is
the finitely-covered envelope of $\check T$.
%613S

\leaveitout{\spheader 632Y? Suppose that $\family{t}{T}{\frak A_t}$
is right-continuous
and that $\Cal S$ is a sublattice of $\Cal T$ such that the infimum of any
non-empty downwards-directed family $\Cal S$ belongs to $\Cal S$.
Show that a process with domain $\Cal S$ is a local martingale iff
it is almost a martingale.
%632C(a-ii) out of order query

\spheader 632Y? Say $\pmb $ is {\bf nearly a martingale} if for every
$\epsilon>0$ there is a martingale $\pmb v$ such that
$\bar\mu{\pmb w\n\pmb v}\le\epsilon$.   Give an example of a right-continuous
structure with a near-martingale on $\Cal T_f$ which is not a local
martingale.   Hint:  $T=\{0,1\}$, $\frak A_0=\{0,1\}$, $\tau_{\check 0}=0$,
$\tau_{\check 1}=\pm 2^n$ on elements of measure $2^{-n-2}$.
}}%end of exercises

\endnotes{
\Notesheader{632} The aim of this section is to show the kind of simplification
which is achieved by assuming right-continuity.   After the basic list in 632C,
the most important general facts are 632F and 632I.

As with 612U, we don't really need \S455 in the proof of 632Ma.   Instead of
expressing the Poisson process in terms of a L\'evy process as described in
\S455, we can start from a probability space better adapted to
the problem (632Yf).   But we still have some work to do, because the
Markov property has got to come in somewhere, and the Poisson point
processes of \S495 aren't trivial.   The method above
also provides an alternative route
to 622L, missing out the harmonic analysis, and giving a slightly stronger
result (632Xe).

}%end of notes

\discrpage

\frfilename{mt633.tex}
\versiondate{18.12.20/30.10.23}
\copyrightdate{2013}

\def\Asissixprime{$(\frak A,\bar\mu,T\strprime,
    \family{r}{T\strprime}{\frak A_r},\Cal T\strprime,
    \family{\rho}{\Cal T\strprime}{\frak A_{\rho}})$}
\def\Tni{T_{\text{ni}}}

\def\chaptername{Structural variations}
\def\sectionname{Separating sublattices}

\newsection{633}

At various points, I have looked at relations between a process
$\pmb u=\sfamilysS{u}$ and its fully adapted extension to the covered envelope
of $\Cal S$;   turning these round,
we find connections between the properties of $\pmb u$ and its restriction to a
covering sublattice of $\Cal S$.   When the filtration is
right-continuous and we have a near-simple process
we can go much farther, and an effective concept is that of
`separating' sublattice (633B).   Once again we have a useful result on
equality of integrals (633K) and many correspondences between properties of
$\pmb u$ and $\pmb u\restr\Cal S'$ (633O, 633P).

\cmmnt{\leader{633A}{Notation} Once again, \Asissix\ will be a stochastic
integration structure.   If $\Cal S$ is a sublattice of $\Cal T$,
$\CalIS$ is the set of finite sublattices of $\Cal S$, and $\Mlns(\Cal S)$ the
space of locally near-simple processes with domain $\Cal S$.   For $t\in T$,
$\check t$ will be the constant stopping time at $t$.
$L^0=L^0(\frak A)$ will be given the linear space topology of convergence
in measure.   For $w\in L^0$, $\theta(w)$ will be
$\Expn(|w|\wedge\chi 1)$, where
$\Expn$ is integration with respect to $\bar\mu$;  for an order-bounded
fully adapted process $\pmb u=\sfamilysS{u}$,
$\sup|\pmb u|$ will be $\sup(\{0\}\cup\{|u_{\sigma}|:\sigma\in\Cal S\})$.
If $I$, $J\subseteq\Cal T$ are sublattices, I
write $I\sqcup J$ for the sublattice of $\Cal T$ generated by $I\cup J$.
}

\leader{633B}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$A$, $B$ subsets of $\Cal S$.

\spheader 633Ba   I will say that $A$ {\bf separates}
$B$ if $\Bvalue{\sigma<\tau}=\sup_{\rho\in A}
(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\sigma<\tau})$ for all $\sigma$,
$\tau\in B$.

\spheader 633Bb If $\pmb v=\sfamilysS{v}$ is a fully adapted process,
I will say that $A$ {\bf $\pmb v$-separates} $B$
if whenever $\sigma$, $\tau\in B$ then
$\Bvalue{\sigma<\tau}\Bcap\Bvalue{v_{\sigma}\ne v_{\tau}}
\Bsubseteq\sup_{\rho\in A}(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau})$.

\leader{633C}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and $A$, $B$, $C$, $D$  subsets of $\Cal S$.

(a) If $A\subseteq B$, $C\subseteq D$ and $A$ separates $D$ then $B$ separates $C$.

(b) $A$ separates its covered envelope.

(c) If $A$ separates $B$ and $B$ separates $C$ then $A$ separates $C$.

(d) If $A$ separates $B$ and $\tau^*$ is an upper bound of $B$ in $\Cal T$ then
$A\wedge\tau^*=\{\sigma\wedge\tau^*:\sigma\in A\}$
separates $B$.

(e) $A$ separates $B$ iff $A\,\,\pmb v$-separates $B$ for every
fully adapted process $\pmb v$ with domain $\Cal S$.

%635J

\proof{{\bf (a)} Immediate from 633Ba.

\medskip

{\bf (b)} If $\sigma$, $\tau$ belong to the covered envelope of $A$,

$$\eqalign{\Bvalue{\sigma<\tau}&
=\sup_{\sigma',\tau'\in A}
  \Bvalue{\sigma<\tau}\Bcap\Bvalue{\sigma'=\sigma}\Bcap\Bvalue{\tau'=\tau}\cr&
=\sup_{\sigma',\tau'\in A}
  \Bvalue{\sigma'=\sigma}\Bcap\Bvalue{\tau'=\tau}
      \Bcap\Bvalue{\sigma'<\tau'}\cr&
=\sup_{\Atop{\sigma',\tau'\in\Cal S}{\rho\in A}}
  \Bvalue{\sigma'=\sigma}\Bcap\Bvalue{\tau'=\tau}
      \Bcap\Bvalue{\sigma'\le\rho}\Bcap\Bvalue{\rho<\tau'}\cr&
\Bsubseteq\sup_{\rho\in A}
  \Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}
\Bsubseteq\Bvalue{\sigma<\tau},\cr}$$

\noindent so $\Bvalue{\sigma<\tau}=\sup_{\rho\in A}
  \Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}$.

\medskip

{\bf (c)} If $\sigma$, $\tau\in C$, then

$$\eqalign{\Bvalue{\sigma<\tau}&
\Bsubseteq\sup_{\sigma'\in B}
 (\Bvalue{\sigma\le\sigma'}\Bcap\Bvalue{\sigma'<\tau})\cr&
=\sup_{\sigma'\in B,\rho\in A}
 (\Bvalue{\sigma\le\sigma'}\Bcap\Bvalue{\sigma'<\tau}\Bcap\Bvalue{\sigma'\le\rho}
     \Bcap\Bvalue{\rho<\tau}\cr&
\Bsubseteq\sup_{\rho\in A}
    (\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}
\Bsubseteq\Bvalue{\sigma<\tau}.\cr}$$

\medskip

{\bf (d)} If $\sigma$, $\tau\in A$ then

$$\eqalign{\Bvalue{\sigma\wedge\tau^*<\tau\wedge\tau^*}&
=\Bvalue{\sigma<\tau}\Bcap\Bvalue{\sigma<\tau^*}
=\sup_{\rho\in A}
  (\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}\Bcap\Bvalue{\sigma<\tau^*})\cr&
\Bsubseteq\sup_{\rho\in A}
  (\Bvalue{\sigma\wedge\tau^*\le\rho\wedge\tau^*}
     \Bcap\Bvalue{\rho\wedge\tau^*<\tau\wedge\tau^*})\cr
}$$

\medskip

{\bf (e)} Immediately from the definitions in 633B we see that if $A$
separates $B$ then $A\,\,\pmb v$-separates $B$ for every
$\pmb v\in\Mfa(\Cal S)$.   Conversely, if $A\,\,\pmb v$-separates $B$ for
every $\pmb v\in\Mfa(\Cal S)$ and $\sigma$, $\tau\in B$, set
$v_{\rho}=\chi\Bvalue{\rho<\tau}$ for $\rho\in\Cal S$.   Then
$\pmb v=\family{\rho}{\Cal S}{v_{\rho}}$ is fully adapted, being a simple
process with breakpoint string $(\tau)$ (612J).   And

\Centerline{$\Bvalue{\sigma<\tau}
=\Bvalue{\sigma<\tau}\Bcap\Bvalue{v_{\sigma}\ne v_{\tau}}
\Bsubseteq\sup_{\rho\in A}(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau})
\Bsubseteq\Bvalue{\sigma<\tau}$}

\noindent so $\Bvalue{\sigma<\tau}
=\sup_{\rho\in A}(\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau})$.
}%end of proof of 633C

\leader{633D}{Proposition}  Let $\Cal S$ be a sublattice of $\Cal T$.

(a) If $T_0\subseteq T$ is dense for the order topology of
$T$\cmmnt{ (4A2R)} and contains every point of $T$
which is isolated on the right in $T$, then
$\check T_0=\{\check t:t\in T_0\}$
separates $\Cal S$.
%634F 643I 652E 633M 653F

(b)(i) If $A\subseteq\Cal T$ separates $\Cal S$, $B\subseteq\Cal S$ is
coinitial with $\Cal S$ and $C\subseteq\Cal S$ is cofinal with $\Cal S$,
then
$A'=\{\med(\tau,\sigma,\tau'):\tau\in B$, $\sigma\in A$, $\tau'\in C\}$
separates $\Cal S$.
%634F

\quad(ii) If $\pmb v$ is a fully adapted process with domain
$\Cal S$, $A\subseteq\Cal T\,\,\pmb v$-separates $\Cal S$,
$B\subseteq\Cal S$ is
coinitial with $\Cal S$ and $C\subseteq\Cal S$ is cofinal with $\Cal S$,
then $\Cal S$ is $\pmb v$-separated by
$A'=\{\med(\tau,\sigma,\tau'):\tau\in B$, $\sigma\in A$, $\tau'\in C\}$.
%635I

\proof{{\bf (a)} Suppose that $\tau$, $\tau'$ in $\Cal S$ and
$t\in T$.   Then
$\Bvalue{\tau'>t}\Bsetminus\Bvalue{\tau>t}
\Bsubseteq\sup_{s\in T_0}(\Bvalue{\tau'>s}\Bsetminus\Bvalue{\tau>s})$.
\Prf\ If $t$ is isolated on the right in $T$, then $t\in T_0$ and the result is
trivial.   Otherwise,

\Centerline{$\Bvalue{\tau'>t}\Bsetminus\Bvalue{\tau>t}
=\sup_{s'>t}(\Bvalue{\tau'>s'}\Bsetminus\Bvalue{\tau>t})
\Bsubseteq\sup_{s\in T_0}(\Bvalue{\tau'>s}\Bsetminus\Bvalue{\tau>s})$}

\noindent because if $s'>t$ then the open interval $\ooint{t,s'}$ is
non-empty and meets $T_0$, and if $s\in T_0\cap\ooint{t,s'}$ then
$\Bvalue{\tau'>s'}\Bsubseteq\Bvalue{\tau'>s}$ and
$\Bvalue{\tau>s}\Bsubseteq\Bvalue{\tau>t}$.\ \Qed

Consequently

$$\eqalign{\Bvalue{\tau<\tau'}&
=\sup_{t\in T}(\Bvalue{\tau'>t}\Bsetminus\Bvalue{\tau>t})\cr&
=\sup_{s\in T_0}(\Bvalue{\tau'>s}\Bsetminus\Bvalue{\tau>s})
=\sup_{s\in T_0}(\Bvalue{\tau\le\check s}\Bcap\Bvalue{\check s<\tau})\cr}$$

\noindent by 611E(a-i-$\delta$).   As $\tau$ and $\tau'$ are arbitrary,
$\check T_0$ separates $\Cal S$.

\medskip

{\bf (b)(i)} Suppose that $\tau$, $\tau'$ in $\Cal S$ and
$\Bvalue{\tau<\tau'}\ne 0$.   Let $\tau_1\in B$, $\tau_1'\in C$ be such
that $\tau_1\le\tau\le\tau'\le\tau_1'$.   We know that
$\Bvalue{\tau<\tau'}
=\sup_{\sigma\in A}\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'}$.
For $\sigma\in A$ set
$\hat\sigma=\med(\tau_1,\sigma,\tau_1')\in A'$.
We have

$$\eqalignno{\Bvalue{\tau\le\hat\sigma}\Bcap\Bvalue{\hat\sigma<\tau'}
&\Bsupseteq\Bvalue{\tau\le\tau_1'\wedge\sigma}
   \Bcap\Bvalue{\tau_1\vee\sigma<\tau'}
\Displaycause{because
$\tau_1'\wedge\sigma\le\hat\sigma\le\tau_1\vee\sigma$}
=\Bvalue{\tau\le\tau_1'}\Bcap\Bvalue{\tau\le\sigma}
   \Bcap\Bvalue{\tau_1<\tau'}\Bcap\Bvalue{\sigma<\tau'}
\Displaycause{611E(c-i-$\alpha$) and (c-ii-$\alpha$)}
\Bsupseteq\Bvalue{\tau\le\sigma}
   \Bcap\Bvalue{\tau<\tau'}\Bcap\Bvalue{\sigma<\tau'}
=\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'}\cr}$$

\noindent (611E(c-iii-$\gamma$)).   So

$$\eqalignno{\sup_{\sigma\in A'}
   (\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'})&
=\sup_{\sigma\in A}
   (\Bvalue{\tau\le\hat\sigma}\Bcap\Bvalue{\hat\sigma<\tau'})\cr&
\Bsupseteq\sup_{\sigma\in A}(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'})
=\Bvalue{\tau<\tau'}\cr}$$

\noindent because $A$ separates $\Cal S$.
\noindent As $\tau$ and $\tau'$ are arbitrary, $A'$ separates
$\Cal S$.

\medskip

\quad{\bf (ii)} Use the same argument, but looking only at the case
$\tau\le\tau'$ and reading
'$\Bvalue{v_{\tau}\ne v_{\tau'}}$' for
`$\Bvalue{\tau<\tau'}$'.
}%end of proof of 633D

\leader{633E}{Lemma} Let $\Cal S$ be a finitely full sublattice of
$\Cal T$ such that $\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty
and has a lower bound in $\Cal S$,
$\Cal S'$ a sublattice of $\Cal S$,
$\hat\Cal S'_f$ the finitely-covered envelope\cmmnt{ (626G)}
of $\Cal S'$, and $\tau$ an element of
$\bigcup_{\sigma\in\Cal S'}\Cal S\wedge\sigma$.

(a)
$A=\{\sigma:\tau\le\sigma\in\hat\Cal S'_f\}$ is non-empty and downwards-directed,
and $\inf A\in\Cal S$.

(b) If $\Cal S'$ separates $\Cal S$ then $\inf A=\tau$.
%634G 643I

(c) If $\pmb v=\sfamilysS{v}$ is fully adapted  and
$\Cal S'\,\,\pmb v$-separates $\Cal S$, then $v_{\inf A}=v_{\tau}$.

\proof{{\bf (a)}  By the hypothesis on $\tau$,
$A$ is non-empty, and of course it is
downwards-directed.   Because $\Cal S$ is finitely full, $A\subseteq\Cal S$, and
$\tau$ is a lower bound for $A$.   So $\tau^*=\inf A$ belongs to $\Cal S$, and
$\tau\le\tau^*$.

Note that if $\sigma\in\Cal S'$ and
$a=\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau^*}$, then $a=0$.   \Prf\
Take any $\sigma_0\in\Cal S'$ such that $\tau\le\sigma_0$.   Then we have
a $\sigma^*\in\Cal T$ such that $\Bvalue{\sigma^*=\sigma}\Bsupseteq a$,
while $\Bvalue{\sigma^*=\sigma_0}\Bsupseteq 1\Bsetminus a$ (611I).
In this case,
$\sigma^*\in A$ so $\Bvalue{\sigma^*<\tau^*}=0$ and $a=0$.\ \Qed

\medskip

{\bf (b)} It follows at once that if $\Cal S'$ separates $\Cal S$ then
$\tau=\tau^*$, that is, $\inf A=\tau$.

\medskip

{\bf (c)} Similarly, if $\Cal S'\,\,\pmb v$-separates $\Cal S$, then
$v_{\tau}=v_{\tau^*}$.
}%end of proof of 633E

\leader{633F}{Proposition} Let $\Cal S$ be a
sublattice of $\Cal T$ and $\pmb u$,
$\pmb v$ locally near-simple processes with domain $\Cal S$.   Suppose that
$C\subseteq\Cal S$ separates $\Cal S$ and that
$\inf_{\sigma\in C}\Bvalue{\sigma<\tau}=0$ for every $\tau\in\Cal S$.   If
$\pmb u\restr C=\pmb v\restr C$ then $\pmb u=\pmb v$.

\proof{{\bf (a)}
Set $\pmb w=\pmb u-\pmb v$, and express $\pmb w$ as $\sfamilysS{w}$.
Then $\pmb w$ is near-simple (631F(a-ii)), $\pmb w$ is zero on $C$
and I need to show that $\pmb w$ is zero on $\Cal S$.

\medskip

{\bf (a)} Consider first the case in which $\Cal S$
has a greatest element which belongs to $C$.
If $\pmb z=\sfamilysS{z}$ is a simple process then
$|\sup w|\le 2\sup|\pmb z-\pmb w|$.   \Prf\
Let $(\sigma_0,\ldots,\sigma_n)$ be a breakpoint string for $\pmb z$ with
$\sigma_n=\max\Cal S$, and write $z_{\downarrow}$ for the starting value of
$\pmb z$ and $\bar z$ for $\sup|\pmb z-\pmb w|$.   Take any $\tau\in\Cal S$.
Then

$$\eqalignno{\Bvalue{\tau<\tau_0}&
=\sup_{\sigma\in C}(\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau_0})
\Bsubseteq\sup_{\sigma\in C}(\Bvalue{\tau>\tau_0}\Bcap\Bvalue{\sigma<\tau_0})
  \cr&
\Bsubseteq\sup_{\sigma\in C}(\Bvalue{z_{\tau}=z_{\downarrow}}
   \Bcap\Bvalue{z_{\sigma}=z_{\downarrow}})
\Bsubseteq\sup_{\sigma\in C}\Bvalue{z_{\tau}=z_{\sigma}}
\Bsubseteq\sup_{\sigma\in C}\Bvalue{|z_{\tau}|=|z_{\sigma}-w_{\sigma}|}
\Displaycause{because $w_{\sigma}=0$ for every $\sigma\in C$}
\Bsubseteq\Bvalue{|z_{\tau}|\le\bar z},\cr}$$

\noindent while if $i\le n$ then

$$\eqalignno{\Bvalue{\sigma_i\le\tau}\Bcap\Bvalue{\tau<\sigma_{i+1}}&
=\sup_{\sigma\in C}(\Bvalue{\sigma_i\le\tau}\Bcap\Bvalue{\tau<\sigma_{i+1}}
   \Bcap\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\sigma_{i+1}})\cr&
\Bsubseteq\sup_{\sigma\in C}(\Bvalue{\sigma_i\le\tau}
   \Bcap\Bvalue{\tau<\sigma_{i+1}}
   \Bcap\Bvalue{\sigma_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}})\cr&
\Bsubseteq\sup_{\sigma\in C}\Bvalue{z_{\tau}=z_{\sigma}}
\Bsubseteq\Bvalue{|z_{\tau}|\le\bar z}\cr}$$

\noindent and

\Centerline{$\Bvalue{\sigma_n\le\tau}
\Bsubseteq\Bvalue{z_{\tau}=z_{\sigma_n}-w_{\sigma_n}}
\Bsubseteq\Bvalue{|z_{\tau}|\le\bar z}$.}

\noindent Thus $|z_{\tau}|\le\bar z$;  as $\tau$ is arbitrary,
$\sup|\pmb z|\le\bar z$.\ \Qed

Now we see that

\Centerline{$\sup|\pmb w|\le\sup|\pmb z|+\sup|\pmb z-\pmb w|
\le 2\sup|\pmb z-\pmb w|$;}

\noindent since $\pmb w$ is near-simple, there is
for any $\epsilon>0$ a simple process $\pmb z$
such that $\theta(\sup|\pmb z-\pmb w|)\le\epsilon$ and
$\theta(\sup|\pmb w|)\le 2\epsilon$.   We conclude that $\pmb w=\tbf 0$,
as required.

\medskip

{\bf (b)} In the general case, take any $\tau\in\Cal S$ and
$\tau^*\in C$.   Then $C$ and $C\wedge\tau^*$
separate $\Cal S\wedge\tau^*$ (633Ca, 633Cd), while
$\{\sigma,\tau^*\}$ covere $\{\sigma\wedge\tau^*\}$ (611M(b-i)) so
$|w_{\sigma\wedge\tau^*}|\le|w_{\sigma}|\vee|w_{\tau^*}|=0$ (614Ga) for every
$\sigma\in C$.   By (a), applied to
$\pmb w\restr\Cal S\wedge\tau^*$, $w_{\tau\wedge\tau^*}=0$ and
$\Bvalue{\tau\le\tau^*}\Bsubseteq\Bvalue{w_{\tau}=0}$.   As
$\inf_{\tau^*\in C}\Bvalue{\tau^*<\tau}=0$, $w_{\tau}=0$ and
$\pmb w=\tbf 0$ in this case also.
}% end of proof of 633F

\leader{633G}{Lemma} Let $\Cal S\subseteq\Cal T$ be a sublattice
and $D\subseteq\Cal S$ a cofinal finitely full set which separates
$\Cal S$ and is such that $\inf A\in D$ for every non-empty
downwards-directed $A\subseteq D$ with a lower bound in $\Cal S$.
Then $D=\Cal S$.
%653G

\proof{ Take any $\tau\in\Cal S$, and set
$A=\{\sigma:\sigma\in D$, $\tau\le\sigma\}$.   Because $D$ is cofinal with
$\Cal S$, $A$ is non-empty;  because $D$ is finitely full, it is closed
under $\wedge$ (626Ha), so $A$ also is closed under $\wedge$, and is
downwards-directed, while $\tau$ is a lower bound of $A$ belonging to $\Cal S$.
Set $\tau^*=\inf A$;  then $\tau^*\in D$ and $\tau\le\tau^*$.

\Quer\ If $\tau\ne\tau^*$, there is a $\sigma\in D$ such that
$\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau^*}\ne 0$, because $D$
separates $\Cal S$.   Set
$c=\Bvalue{\tau\le\sigma}$;  then $c\in\frak A_{\sigma}$ and
$1\Bsetminus c\in\frak A_{\tau}\subseteq\frak A_{\tau^*}$.
So there is a $\tau'\in\Cal T$ such that
$c\Bsubseteq\Bvalue{\tau'=\sigma}$ and
$1\Bsetminus c\Bsubseteq\Bvalue{\tau'=\tau^*}$ (611I again).   Since
$\Bvalue{\tau'=\sigma}\Bcup\Bvalue{\tau'=\tau^*}=1$, and $D$ is finitely
full, $\tau'\in D$;  also

\Centerline{$c=\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\tau'=\sigma}
\Bsubseteq\Bvalue{\tau\le\tau'}$,
\quad$1\Bsetminus c\Bsubseteq\Bvalue{\tau'=\tau^*}
\Bsubseteq\Bvalue{\tau\le\tau'}$}

\noindent so $\tau\le\tau'$.   Accordingly $\tau'\in A$ and
$\tau^*\le\tau'$.   But

\Centerline{$\Bvalue{\tau'<\tau^*}
\Bsupseteq\Bvalue{\tau'=\sigma}\Bcap\Bvalue{\sigma<\tau^*}
\Bsupseteq\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau^*}
\ne 0$}

\noindent so this is impossible.\ \Bang

Thus $\tau=\tau^*$ belongs to $D$;  as $\tau$ is arbitrary, $D=\Cal S$.
}%end of proof of 633G

\leader{633H}{Lemma} Suppose that $I$, $J\in\Cal I(\Cal T)$, with
$J\subseteq I$.   Then there are totally ordered sets $J_0\subseteq J$ and
$I_0\subseteq I$ such that $J_0$ covers $J$, $I_0$ covers $I$ and
$J_0\subseteq I_0$.

\proof{ We know from 611Ke that there is a totally ordered set
$J_0\subseteq J$ covering $J$.   If $J_0=\emptyset$, all we have to do is
to use 611Ke again to find a totally ordered set $I_0\subseteq I$ covering
$I$.   Otherwise, enumerate $J_0$ in ascending order as
$\langle\sigma_i\rangle_{i\le n}$.   Set

\Centerline{$K_0=I\cap[\min\Cal T,\sigma_0]$,
\quad$K_i=I\cap[\sigma_{i-1},\sigma_i]$ for $1\le i\le n$,
\quad$K_{n+1}=I\cap[\sigma_n,\max\Cal T]$.}

\noindent For each $i\le n+1$, let $K'_i\subseteq K_i$
be a totally ordered set covering
the finite sublattice $K_i$;  now set $I_0=J_0\cup\bigcup_{i\le n+1}K'_i$,
so that $I_0$ is a totally ordered subset of $I$, including $J_0$, which
covers $K=\bigcup_{i\le n+1}K_i$.

At the same time, $K$ covers $I$.   \Prf\ Let $\tau\in I$.   Set

\Centerline{$\tau_0=\tau\wedge\sigma_0$,
\quad$\tau_i=\med(\sigma_{i-1},\tau,\sigma_i)$ for $1\le i\le n$,
\quad$\tau_{n+1}=\tau\vee\sigma_n$.}

\noindent Then $\tau_i\in K_i\subseteq K$ for every $i\le n+1$.   Also
$\Bvalue{\tau=\tau_0}=\Bvalue{\tau\le\sigma_0}$ and
$\Bvalue{\tau=\tau_{n+1}}=\Bvalue{\sigma_n\le\tau}$,
by 611E(a-ii-$\beta$).   As for the middle terms, if $1\le i\le n$ then %OK

$$\eqalignno{\Bvalue{\tau=\tau_i}
&=\Bvalue{\tau=(\tau\vee\sigma_{i-1})\wedge\sigma_i}
\Bsupseteq\Bvalue{\tau=\tau\vee\sigma_{i-1}}
  \Bcap\Bvalue{\tau\vee\sigma_{i-1}=(\tau\vee\sigma_{i-1})\wedge\sigma_i}
  \cr
&=\Bvalue{\sigma_{i-1}\le\tau}\Bcap\Bvalue{\tau\vee\sigma_{i-1}\le\sigma_i}
=\Bvalue{\sigma_{i-1}\le\tau}\Bcap\Bvalue{\tau\le\sigma_i}
   \Bcap\Bvalue{\sigma_{i-1}\le\sigma_i}\cr
\displaycause{611Eb}
&=\Bvalue{\sigma_{i-1}\le\tau}\Bcap\Bvalue{\tau\le\sigma_i}.\cr}$$

\noindent So

$$\eqalign{\sup_{\sigma\in K}\Bvalue{\tau=\sigma}
&\Bsupseteq\sup_{i\le n+1}\Bvalue{\tau=\tau_i}\cr
&\Bsupseteq\Bvalue{\tau\le\sigma_0}
  \Bcup\sup_{1\le i\le n}
    (\Bvalue{\sigma_{i-1}\le\tau}\Bcap\Bvalue{\tau\le\sigma_i})
  \Bcup\Bvalue{\sigma_n\le\tau}
=1\cr}$$

\noindent (611Ed),
and $K$ covers $\{\tau\}$.   As $\tau$ is arbitrary, $K$ covers $I$.\ \Qed

Consequently $I_0$ covers $I$, by 611M(g-i) as usual.
}%end of proof of 633H

\leader{633I}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, $\Cal S'$ a
finitely full sublattice of $\Cal S$
and $\psi$ a strictly adapted interval function defined on
$\Cal S^{2\uparrow}$.   Suppose that $J\in\CalIS$ and
$\family{\tau}{J}{\sigma_{\tau}}$ are such that
$\tau\le\sigma_{\tau}\in\Cal S'$ and

\Centerline{$\bar u_{\tau}
=\sup_{\sigma\in\Cal S\wedge\tau,\tau'\in\Cal S'\cap[\tau,\sigma_{\tau}]}
   |\psi(\sigma,\tau')-\psi(\sigma,\tau)|$}

\noindent is defined in $L^0$ for each $\tau\in J$.
Let $I\in\Cal I(\Cal S')$ be such that
$\sigma_{\tau}\in I$ for every $\tau\in J$.   Then

\Centerline{$|S_{I\sqcup J}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|
\le 2\sum_{\tau\in J}\bar u_{\tau}$.}

\proof{{\bf (a)} To begin with, suppose that $J=\{\tau\}$ is a singleton.

\medskip

\quad{\bf (i)} Write $\hat I$ for the covered envelope of $I$.
There is a $\tau'\in\hat I$ such that $\tau\le\tau'$ and
$\Bvalue{\rho<\tau'}=\Bvalue{\rho<\tau}$ for every $\rho\in\hat I$.
\Prf\ For each $\sigma\in I$,
$\Bvalue{\tau\le\sigma}\in\frak A_{\sigma}\cap\frak A_{\sigma_{\tau}}$ so there is a
$\sigma'\in\Cal T$ such that
$\Bvalue{\tau_j\le\sigma}\Bsubseteq\Bvalue{\sigma'=\sigma}$ and
$\Bvalue{\sigma<\tau_j}\Bsubseteq\Bvalue{\sigma'=\sigma_{\tau}}$.   Now
$\tau\le\sigma'\in\hat I$ for each $\sigma\in I$, so

\Centerline{$\tau\le\tau'=\inf_{\sigma\in I}\sigma'\in\hat I
\subseteq\Cal S'$.}

\noindent Take $\rho\in\hat I$.   Then certainly
$\Bvalue{\rho<\tau}\Bsubseteq\Bvalue{\rho<\tau'}$.
\Quer\ If $a=\Bvalue{\rho<\tau'}\Bsetminus\Bvalue{\rho<\tau}$ is
non-zero then there is a $\sigma\in I$ such that
$a'=a\Bcap\Bvalue{\rho=\sigma}$ is non-zero.   In this case,

$$\eqalign{0\ne a'&
\Bsubseteq\Bvalue{\sigma<\tau'}\Bsetminus\Bvalue{\sigma<\tau}
\Bsubseteq\Bvalue{\sigma<\tau'}\Bcap\Bvalue{\tau\le\sigma}\cr&
\Bsubseteq\Bvalue{\sigma<\tau'}\Bcap\Bvalue{\sigma'=\sigma}
\Bsubseteq\Bvalue{\sigma<\tau'}\Bcap\Bvalue{\tau'\le\sigma}\cr}$$

\noindent because $\tau'\le\sigma'$;  but this is impossible.\ \BanG\ So
$\Bvalue{\rho<\tau}=\Bvalue{\rho<\tau'}$, as required.\ \Qed

Of course $0=\Bvalue{\sigma_{\tau}<\tau}=\Bvalue{\sigma_{\tau}<\tau'}$ so
$\tau'\le\sigma_{\tau}$.

\medskip

\quad{\bf (ii)} Write $I'=I\sqcup\{\tau'\}$ and $K=I'\sqcup\{\tau\}$.
Then $\{\rho:\rho\in\Cal T$, $\rho\vee\tau'\in I'\}$ is a sublattice including
$I'\cup\{\tau\}$ so it includes $K$ and $K\vee\tau'\subseteq I'$;  consequently
$K\vee\tau'=I'\vee\tau'$ and
$S_{K\vee\tau'}(\tbf 1,d\psi)=S_{I'\vee\tau'}(\tbf 1,d\psi)$.

\medskip

\quad{\bf (iii)}
$|S_{K\wedge\tau'}(\tbf 1,d\psi)-S_{I'\wedge\tau'}(\tbf 1,d\psi)|
\le 2\bar u_{\tau}$.
\Prf\ Take $(\sigma_0,\ldots,\sigma_m)$ linearly generating the
$(I'\wedge\tau')$-cells;  note that every $\sigma_i$ belongs to
$\hat I$.   Then $(\sigma_0\wedge\tau,\ldots,\sigma_m\wedge\tau)$
linearly generates the $(K\wedge\tau')$-cells and
$(\sigma_0\vee\tau,\ldots,\sigma_m\vee\tau)$
linearly generates the $(K\cap[\tau,\tau'])$-cells.   So

$$\eqalign{&S_{K\wedge\tau'}(\tbf 1,d\psi)-S_{I'\wedge\tau'}(\tbf 1,d\psi)\cr&
\mskip50mu
=\sum_{j=0}^{m-1}\psi(\sigma_i\wedge\tau,\sigma_{j+1}\wedge\tau)
   +\psi(\sigma_j\vee\tau,\sigma_{j+1}\vee\tau)
   -\psi(\sigma_j,\sigma_{j+1}).\cr}$$

\noindent Set

$$\eqalign{b_i&=\Bvalue{\tau<\sigma_0}\text{ if }i=-1,\cr
&=\Bvalue{\sigma_i\le\tau}\Bcap\Bvalue{\tau<\sigma_{i+1}}
   \text{ if }0\le i<m,\cr
&=\Bvalue{\tau=\sigma_m}\text{ if }i=m;\cr}$$

\noindent then $b_{-1},\ldots,b_m$ is a partition of unity in $\frak A$.   Now
observe that if $-1\le i<j<m$ then

$$\eqalign{b_i&
\Bsubseteq\Bvalue{\tau<\sigma_{i+1}}
\Bsubseteq\Bvalue{\tau<\sigma_j}\cr&
\Bsubseteq\Bvalue{\sigma_j\wedge\tau=\sigma_{j+1}\wedge\tau}
   \Bcap\Bvalue{\sigma_j\vee\tau=\sigma_j}
   \Bcap\Bvalue{\sigma_{j+1}\vee\tau=\sigma_{j+1}}\cr&
\Bsubseteq\Bvalue{\psi(\sigma_j\wedge\tau,\sigma_{j+1}\wedge\tau)=0}
    \Bcap\Bvalue{\psi(\sigma_j\vee\tau,\sigma_{j+1}\vee\tau)
          =\psi(\sigma_j,\sigma_{j+1})}\cr&
\Bsubseteq\Bvalue{\psi(\sigma_i\wedge\tau,\sigma_{j+1}\wedge\tau)
   +\psi(\sigma_j\vee\tau,\sigma_{j+1}\vee\tau)
   -\psi(\sigma_j,\sigma_{j+1})=0},\cr}$$

\noindent while if $0\le j<i\le m$ then

$$\eqalign{b_i&
\Bsubseteq\Bvalue{\sigma_i\le\tau}
\Bsubseteq\Bvalue{\sigma_{j+1}\le\tau}
\Bsubseteq\Bvalue{\sigma_j\vee\tau=\sigma_{j+1}\wedge\tau}
   \Bcap\Bvalue{\sigma_j\wedge\tau=\sigma_j}
   \Bcap\Bvalue{\sigma_{j+1}\wedge\tau=\sigma_{j+1}}\cr&
\Bsubseteq\Bvalue{\psi(\sigma_i\vee\tau,\sigma_{j+1}\vee\tau)=0}
    \Bcap\Bvalue{\psi(\sigma_i\wedge\tau,\sigma_{j+1}\wedge\tau)
          =\psi(\sigma_i,\sigma_{j+1})}\cr&
\Bsubseteq\Bvalue{\psi(\sigma_j\wedge\tau,\sigma_{j+1}\wedge\tau)
   +\psi(\sigma_j\vee\tau,\sigma_{j+1}\vee\tau)
   -\psi(\sigma_j,\sigma_{j+1})=0}.\cr}$$

\noindent So

$$\eqalignno{&|S_{K\wedge\tau'}(\tbf 1,d\psi)
    -S_{I'\wedge\tau'}(\tbf 1,d\psi)|\cr&
\mskip30mu
=|\sum_{j=0}^{m-1}\sum_{i=-1}^m
   \bigl(\psi(\sigma_j\wedge\tau,\sigma_{j+1}\wedge\tau)
     +\psi(\sigma_j\vee\tau,\sigma_{j+1}\vee\tau)
     -\psi(\sigma_j,\sigma_{j+1})\bigr)\times\chi b_i|\cr&
\mskip30mu
=|\sum_{j=0}^{m-1}
   \bigl(\psi(\sigma_j\wedge\tau,\sigma_{j+1}\wedge\tau)
     +\psi(\sigma_j\vee\tau,\sigma_{j+1}\vee\tau)
     -\psi(\sigma_j,\sigma_{j+1})\bigr)\times\chi b_j|\cr&
\mskip30mu
\le\sum_{j=0}^{m-1}
   |\psi(\sigma_j\wedge\tau,\tau)
     +\psi(\tau,\sigma_{j+1}\vee\tau)
     -\psi(\sigma_j\wedge\tau,\sigma_{j+1}\vee\tau)|\times\chi b_j\cr&
\mskip30mu
\le\sum_{j=0}^{m-1}
   \bigl(|\psi(\sigma_j\wedge\tau,\sigma_{j+1}\vee\tau)
        -\psi(\sigma_j\wedge\tau,\tau)|
     +|\psi(\tau,\sigma_{j+1}\vee\tau)-\psi(\tau,\tau)|\bigr)\times\chi b_j\cr&
\mskip30mu
=\sum_{j=0}^{m-1}
   \bigl(|\psi(\sigma_j\wedge\tau,\tau')
        -\psi(\sigma_j\wedge\tau,\tau)|
     +|\psi(\tau,\tau')-\psi(\tau,\tau)|\bigr)\times\chi b_j
\Displaycause{because $b_j\Bsubseteq\Bvalue{\tau\le\sigma_{j+1}}
=\Bvalue{\tau'\le\sigma_{j+1}}=\Bvalue{\tau'=\sigma_{j+1}}
=\Bvalue{\tau'=\sigma_{j+1}\vee\tau}$ for every $j<m$}
\mskip30mu
\le\sum_{j=0}^{m-1}2\bar u_{\tau}\times\chi b_j
\Displaycause{because $\tau'\in\Cal S'$ and $\tau\le\tau'\le\sigma_{\tau}$}
\mskip30mu
\le 2\bar u_{\tau}\cr}$$

\noindent as claimed.\ \Qed

\medskip

\quad{\bf (iv)} Putting (ii) and (iii) together, we see that
$|S_K(\tbf 1,d\psi)-S_{I'}(\tbf 1,d\psi)|\le 2\bar u_{\tau}$.   But $I\sqcup J$ is a
sublattice of $K$ covering $I'$ and $J$, so it covers $K$, and
$S_{I\sqcup J}(\tbf 1,d\psi)=S_K(\tbf 1,d\psi)$ by 613S.   Similarly,
$S_{I'}(\tbf 1,d\psi)=S_I(\tbf 1,d\psi)$, and
$|S_{I\sqcup J}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|\le 2\bar u_{\tau}$, as required.

\medskip

{\bf (b)} For the general case, induce on $\#(J)$.   If $J$ is empty, the result
is trivial, and if $\#(J)=1$ it is covered by (a) above.   For the inductive
step to $\#(J)=n\ge 2$, if $J$ is not totally ordered then it has a
totally ordered subset $J'$ which covers $J$ (611Ke), and

$$\eqalignno{|S_{I\sqcup J}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|&
=|S_{I\sqcup J'}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|
\le 2\sum_{\tau\in J'}\bar u_{\tau}
\Displaycause{by the inductive hypothesis}
\le 2\sum_{\tau\in J}\bar u_{\tau}.\cr}$$

\noindent If $J$ is totally ordered, set $J'=J\setminus\{\max J\}$;  then

$$\eqalignno{|S_{I\sqcup J}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|&
\le|S_{I\sqcup J}(\tbf 1,d\psi)-S_{I\sqcup J'}(\tbf 1,d\psi)|
   +|S_{I\sqcup J'}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|\cr&
\le|S_{(I\sqcup J')\sqcup\{\max J\}})(\tbf 1,d\psi)
          -S_{I\sqcup J'}(\tbf 1,d\psi)|
   +2\sum_{\tau\in J'}u_{\tau}
\Displaycause{by the inductive hypothesis}
\le 2\sum_{\tau\in J}\bar u_{\tau}\cr}$$

\noindent by (a).   Thus the induction continues in either case.
}%end of proof of 633I

\leader{633J}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, $\Cal S'$ a
finitely full cofinal sublattice of $\Cal S$ and $\psi$ an order-bounded
strictly adapted interval function defined on $\Cal S^{2\uparrow}$.   For
$\tau\in\Cal S$ set $A_{\tau}=\{\sigma:\tau\le\sigma\in\Cal S'\}$, and for
$\tau\in\Cal S$, $\tau'\in A_{\tau}$ set

\Centerline{$u_{\tau\tau'}
=\sup_{\sigma\in\Cal S\wedge\tau,\rho\in A_{\tau}\wedge\tau'}
     |\psi(\sigma,\rho)-\psi(\sigma,\tau)|$.}

\noindent Suppose
that $z=\int_{\Cal S}d\psi$ is defined and that
$\inf_{\tau'\in A_{\tau}}u_{\tau\tau'}=0$
for every $\tau\in\Cal S$.   Then $\int_{\Cal S'}d\psi$ is defined and
equal to $z$.

\proof{ Let $\epsilon>0$.   Then there is a $J\in\CalIS$ such that
$\theta(z-S_K(\tbf 1,d\psi))\le\epsilon$ whenever $J\subseteq K\in\Cal IS$.
For each $\tau\in J$, $\tau'\mapsto u_{\tau\tau'}:A_{\tau}\to(L^0)^+$ is
order-preserving so $\lim_{\tau'\downarrow\tau}u_{\tau\tau'}=0$ and there is a
$\sigma_{\tau}\in A_{\tau}$ such that
$\theta(u_{\tau\sigma_{\tau}})\le\Bover{\epsilon}{1+\#(J)}$.
Set $I_0=\{\sigma_{\tau}:\tau\in J\}$.

Suppose that $I_0\subseteq I\in\Cal I(\Cal S')$.   Then
$|S_{I\sqcup J}(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|
\le\sum_{\tau\in J}u_{\tau\sigma_{\tau}}$ by 633I.   But now we see that

\Centerline{$\theta(z-S_I(\tbf 1,d\psi))
\le\theta(z-S_{I\sqcup J}(\tbf 1,d\psi))
   +\sum_{\tau\in J}\theta(u_{\tau\sigma_{\tau}})
\le\epsilon+2\epsilon$}

\noindent by the choice of $J$ and $\family{\tau}{J}{\sigma_J}$.
As $\epsilon$ is arbitrary, $\int_{\Cal S'}d\psi$ is defined and equal to $z$.
}%end of proof of 633J

\leader{633K}{Theorem} (a) Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full sublattice of $\Cal T$ such
that $\inf A\in\Cal S$ for every non-empty subset $A$ of $\Cal S$ with
a lower bound in $\Cal S$, and $\pmb u$, $\pmb v$ fully adapted processes
with domain $\Cal S$ such that $\pmb u$ is order-bounded and $\pmb v$ is
locally near-simple.
Let $\Cal S'$ be a sublattice of $\Cal S$, cofinal with $\Cal S$, which
$\pmb v$-separates $\Cal S$.   If
$z=\int_{\Cal S}\pmb u\,d\pmb v$ is defined then
$\int_{\Cal S'}\pmb u\,d\pmb v$ is defined and equal to $z$.
%634F 635F 635I 653G

(b) Suppose that $\Cal S$, $\Cal S'$ are sublattices of
$\Cal T$ such that $\Cal S'$ is finitely full and is included in $\Cal S$.
Let $\pmb u$, $\pmb v$ be
fully adapted processes defined on $\Cal S$.   For $\tau\in\Cal S$ set
$A_{\tau}=\{\sigma:\tau\le\sigma\in\Cal S'\}$.   Suppose that $A_{\tau}$ is
non-empty and

\Centerline{$u_{\tau}=\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}$,
\quad$v_{\tau}=\lim_{\sigma\downarrow A_{\tau}}v_{\sigma}$}

\noindent for every $\tau\in\Cal S$.   If $z=\int_{\Cal S'}\pmb u\,d\pmb v$
is defined then $\int_{\Cal S}\pmb u\,d\pmb v$ is defined and equal to $z$.

\proof{{\bf (a)(i)} Suppose to begin with that $\Cal S'$ is finitely full.
For $\tau\in\Cal S$,
set $A_{\tau}=\{\sigma:\tau\le\sigma\in\hat\Cal S'_f\}$.
Because $\Cal S'$ is cofinal with $\Cal S$, $A_{\tau}\ne\emptyset$ and
$\inf A_{\tau}\in\Cal S$, while $v_{\tau}=v_{\inf A_{\tau}}$ by 633Ec.
Because $\family{t}{T}{\frak A_t}$ is right-continuous and
$\pmb v$ is locally near-simple,

\Centerline{$\lim_{\sigma\downarrow A_{\tau}}
       \sup_{\rho\in A_{\tau},\rho\le\sigma}|v_{\rho}-v_{\tau}|
=\lim_{\sigma\downarrow A_{\tau}}\sup_{\rho\in A_{\tau},\rho\le\sigma}
    |v_{\rho}-v_{\inf A}|
=0$}

\noindent (632E).   Now set $\psi=\pmb u\Delta\pmb v$ (613Cc, 613Dd), so that
$\psi(\sigma,\tau)=u_{\sigma}\times(v_{\tau}-v_{\sigma})$ when
$\sigma\le\tau$ in $\Cal S$, and $\psi$ is strictly adapted and order-bounded.
Write $\bar u$ for $\sup|\pmb u|$.   Then whenever $\tau\in\Cal S$ we have
$v_{\tau}=v_{\inf A_{\tau}}$ (633Ec), so

$$\lim_{\tau'\downarrow A_{\tau}}
   \sup_{\sigma\in\Cal S\wedge\tau,\rho\in A_{\tau}\wedge\tau'}
      |\psi(\sigma,\rho)-\psi(\sigma,\tau)|
\le\lim_{\tau'\downarrow A_{\tau}}
   \sup_{\rho\in A_{\tau}\wedge\tau'}
      \bar u\times|v_{\rho}-v_{\tau}|
=0.$$

\noindent Thus $\psi$ satisfies the conditions of 633J and

$$\eqalignno{\int_{\Cal S}\pmb u\,d\pmb v&
=\int_{\Cal S}d\psi
\Displaycause{613Hc}
=\int_{\Cal S'}d\psi
\Displaycause{633J}
=\int_{\Cal S'}\pmb u\,d\pmb v\cr}$$

\noindent as required.

\medskip

{\bf (ii)} In general, the finitely-covered envelope $\hat\Cal S'_f$ of
$\Cal S'$ is a cofinal finitely full sublattice of $\Cal S$ which
$\pmb v$-separates $\Cal S$, so

\Centerline{$\int_{\Cal S}\pmb u\,d\pmb v
=\int_{\hat\Cal S'_f}\pmb u\,d\pmb v
=\int_{\Cal S'}\pmb u\,d\pmb v$}

\noindent by 613T.

\medskip

{\bf (b)} Let $\epsilon>0$.   Let $J\in\Cal I(\Cal S')$ be such that
$\theta(z-S_I(\pmb u,d\pmb v))\le\epsilon$ whenever
$J\subseteq I\in\Cal I(\Cal S')$.   Now suppose that $I$ is any finite
sublattice of $\Cal S$ including $J$.
Then there are a totally ordered subset $J_0$ of $J$ covering $J$
and a totally ordered subset $I_0$ of $I$,
including $J_0$, which covers $I$ (633H).
Let $\langle\tau_i\rangle_{i\le n}$ be the
increasing enumeration of $I_0$.   Then we can find $\tau'_i\in A_{\tau_i}$, for
$i\le n$, such that

\inset{----- whenever $i$, $j\le n$, $\sigma\in A_{\tau_i}$,
$\sigma'\in A_{\tau_j}$, $\sigma\le\tau'_i$ and $\sigma'\le\tau'_j$ then
$\theta(u_{\sigma}\times v_{\sigma'}-u_{\tau_i}\times v_{\tau_j})
\le\Bover{\epsilon}{n+1}$,

----- if $i\le n$ and $\tau_i\in\Cal S'$ then $\tau'_i=\tau_i$.}

Set $\sigma_i=\inf_{n\ge j\ge i}\tau'_j$ for $i\le n$.  Then
$\sigma_0\le\ldots\le\sigma_n$;  for each $i\le n$, $\sigma_i\in\Cal S'$
and $\tau_i\le\sigma_i\le\tau'_i$;  and $\sigma_i=\tau_i$ if $\tau_i\in J_0$.
So if we set $I'_0=\{\sigma_i:i\le n\}$, we have

$$\eqalign{\theta(S_{I'_0}(\pmb u,d\pmb v)-S_I(\pmb u,d\pmb v))
&=\theta\bigl(\sum_{i=0}^{n-1}u_{\sigma_i}\times v_{\sigma_{i+1}}
   -u_{\tau_i}\times v_{\tau_{i+1}}
   -u_{\sigma_i}\times v_{\sigma_i}
   +u_{\tau_i}\times v_{\tau_i}\bigr)\cr
&\le 2\epsilon.\cr}$$

\noindent
Because $J_0$ covers $J$ and $J_0\subseteq I'_0$, the covered envelope of
$I'_0$ includes $I'=I'_0\sqcup J$ (611M(b-i)) and
$S_{I'_0}(\pmb u,d\pmb v)=S_{I'}(\pmb u,d\pmb v)$ (613T again).   But
$I'\supseteq J$ so
$\theta(z-S_{I'}(\pmb u,d\pmb v))\le\epsilon$.   Consequently
$\theta(z-S_I(\pmb u,d\pmb v))\le 3\epsilon$;  and this is true whenever
$J\subseteq I\in\CalIS$.

As $\epsilon$ is arbitrary,
$\int_{\Cal S}\pmb u\,d\pmb v$ is defined and equal to $z$.
}%end of proof of 633K

\leader{633L}{Corollary} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous, and $\tau\le\tau'$ in $\Cal T$.   Let
$\pmb u$ be a near-simple process and
$\pmb v$ a near-simple integrator, both defined on $[\tau,\tau']$.
Suppose that $T_0\subseteq T$ is a dense set for the
order topology containing every point of $T$ which is isolated on the right.
Set $\Cal S'=\{\tau'\}\cup\{\med(\tau,\check{t},\tau'):t\in T_0\}$.
Then $\int_{\Cal S'}\pmb u\,d\pmb v$ is defined and equal to
$\int_{[\tau,\tau']}\pmb u\,d\pmb v$.

\proof{ By 616K $\int_{[\tau,\tau']}\pmb u\,d\pmb v$ is
defined.   By 633D(b-i), $\Cal S'$ separates $[\tau,\tau']$, and it is
cofinal with $[\tau,\tau']$ just because it contains $\tau'$.
We can therefore apply 633Ka to see that
$\int_{\Cal S'}\pmb u\,d\pmb v=\int_{[\tau,\tau']}\pmb u\,d\pmb v$.
}%end of proof of 633L

\leader{*633M}{}\cmmnt{ I star the next couple of paragraphs because they are
here for a very special application in \S652.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a
moderately oscillatory process with domain $\Cal S$.

(a) For every $\epsilon>0$ and $\beta>0$, there are a $b\in\frak A$ and a
$\gamma\ge 0$ such that $\bar\mu b\ge 1-\epsilon$ and

\Centerline{$\sum_{i=0}^{n-1}
   \bar\mu(b\Bcap\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\beta})
  \le\gamma$}

\noindent whenever $\tau_0\le\ldots\le\tau_n$ in $\Cal S$,

(b)(i) For $\sigma\le\tau$ in $\Cal S$, set

\Centerline{$\psi(\sigma,\tau)
 =\med(-\chi 1,u_{\tau}-u_{\sigma},\chi 1)$,
\quad$\psi'(\sigma,\tau)
 =u_{\tau}-u_{\sigma}-\psi(\sigma,\tau)$.}

\noindent Then $\psi$ and $\psi'$ are strictly adapted interval functions
on $\Cal S$.

\quad(ii) If $\sigma\le\tau$ and $\sigma'\le\tau'$ in $\Cal S$, then
$|\psi(\sigma,\tau)-\psi(\sigma',\tau')|$ and
$|\psi'(\sigma,\tau)-\psi'(\sigma',\tau')|$ are both at most
$|u_{\sigma}-u_{\sigma'}|+|u_{\tau}-u_{\tau'}|$.

(c) $\int_{\Cal S}d\psi'$ and $\int_{\Cal S}d\psi$ are defined.

(d) $\pmb w'=ii_{\psi'}(\tbf 1)$ is of bounded variation and
$\pmb w=ii_{\psi}(\tbf 1)$ is moderately oscillatory.

(e) $\Osclln(\pmb w)\le\chi 1$.

(f) Express $\pmb w$ as $\family{\tau}{\Cal S}{w_{\tau}}$.
If $\tau\le\tau'$ in $\Cal S$, then
$w_{\tau'}-w_{\tau}\in L^0(\frak D_{\tau})$, where $\frak D_{\tau}$ is the
closed subalgebra of $\frak A$ generated by
$\{u_{\sigma'}-u_{\sigma}:\sigma$, $\sigma'\in\Cal S\cap[\tau,\tau']$,
$\sigma\le\sigma'\}$.

(g) If $\pmb u$ is near-simple, $\pmb w$ is near-simple.

\proof{ If $\Cal S=\emptyset$ all of this is true for trivial reasons, so in the
following arguments I will take it that $\Cal S$ is non-empty.

\medskip

{\bf (a)} (Compare (i)$\Rightarrow$(iv) in the proof of 615N.)
It is enough to deal with the case in which $\beta\le 1$.
We know that there is a process $\pmb v=\sfamilysS{v}$ of bounded variation such
that $\theta(\sup|\pmb u-\pmb v|)<\Bover16\beta\epsilon$.   Set
$b_0=\Bvalue{\sup|\pmb u-\pmb v|<\bover13\beta}$;  then
$\bar\mu b_0\ge 1-\bover12\epsilon$.   Writing $\bar v=\int_{\Cal S}|d\pmb v|$,
we have a $\gamma\ge 0$ such that $\bar\mu b_1\ge 1-\bover12\epsilon$ where
$b_1=\Bvalue{\bar v<\bover13\gamma\beta}$.   Set $b=b_0\Bcap b_1$;  then
$\bar\mu b\ge 1-\epsilon$.

If $\sigma\le\tau$ in $\Cal S$ then

\Centerline{$\Bvalue{|u_{\tau}-u_{\sigma}|\ge\beta}
\Bsubseteq\Bvalue{|u_{\tau}-v_{\tau}|\ge\bover13\beta}
  \Bcup\Bvalue{|v_{\tau}-v_{\sigma}|\ge\bover13\beta}
  \Bcup\Bvalue{|v_{\tau}-u_{\tau}|\ge\Bover13\beta}$,}

\noindent so $b\Bcap\Bvalue{|u_{\tau}-u_{\sigma}|\ge\beta}
\Bsubseteq b\Bcap\Bvalue{|v_{\tau}-v_{\sigma}|\ge\bover13\beta}$.
Now take any $\tau_0\le\ldots\le\tau_n$ in $\Cal S$.   Then

$$\eqalign{\sum_{i=0}^{n-1}
    \bar\mu(b\Bcap\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\beta})
&\le\sum_{i=0}^{n-1}
    \bar\mu(b_1\Bcap\Bvalue{|v_{\tau_{i+1}}-v_{\tau_i}|\ge\bover13\beta})\cr&
\le\sum_{i=0}^{n-1}
    \Bover3{\beta}\Expn(\chi b_1\times|v_{\tau_{i+1}}-v_{\tau_i}|)\cr&
\le\Bover3{\beta}\Expn(\chi b_1\times\bar v)
\le\Bover3{\beta}\cdot\Bover13\gamma\beta
\le\gamma.\cr}$$

\medskip

{\bf (b)(i)} Since $\alpha\mapsto\med(-1,\alpha,1):\Bbb R\to\Bbb R$ is Borel
measurable and $(\sigma,\tau)\mapsto u_{\tau}-u_{\sigma}$
is strictly adapted
(613Cc), $\psi$ and $\psi'$ are strictly adapted (613Db-613Dc).

\medskip

\quad{\bf (ii)} For $x$, $y$, $z$, $x'$, $y'$, $z'\in L^0$,

\Centerline{$x'=x+(x'-x)\le(x\vee y)+(x'-x)\vee(y'-y)$,}

\Centerline{$x'\vee y'\le(x\vee y)+(x'-x)\vee(y'-y)$,
\quad$|x'\vee y'-x\vee y|\le|x'-x|\vee|y'-y|$,}

\Centerline{$|x'\vee y'\vee z'-x\vee y\vee z|
\le|x'\vee y'-x\vee y|\vee|z'-z|
\le|x'-x|\vee|y'-y|\vee|z'-z|$;}

\noindent  similarly,
$|x'\wedge y'-x\wedge y|\le|x'-x|\vee|y'-y|$.   Consequently

$$\eqalign{&|\med(x',y',z')-\med(x,y,z)|\cr&
\mskip100mu
=|(x'\wedge y')\vee(x'\wedge z')\vee(y'\wedge z')
   -(x\wedge y)\vee(x\wedge z)\vee(y\wedge z)|\cr&
\mskip100mu
\le|x'\wedge y'-x\wedge y|\vee|x'\wedge z'-x\wedge z|
   \vee|y'\wedge z'-y\wedge z|\cr&
\mskip100mu
\le|x'-x|\vee|y'-y|\vee|z'-z|.\cr}$$

\noindent In particular,
$|\med(-\chi 1,x',\chi 1)-\med(-\chi 1,x,\chi 1)|\le|x'-x|$, so

\Centerline{$|\psi(\sigma,\tau)-\psi(\sigma',\tau')|
\le|u_{\tau}-u_{\sigma}-u_{\tau'}+u_{\sigma'}|
\le|u_{\sigma}-u_{\sigma'}|+|u_{\tau}-u_{\tau'}|$.}

Next, for any $x\in L^0$,
$\med(-\chi 1,x,\chi 1)-x=\med(-\chi 1-x,0,\chi 1-x)$, so

$$\eqalign{&|(\med(-\chi 1,y,\chi 1)-y)-(\med(-\chi 1,x,\chi 1)-x)|\cr&
\mskip100mu
\le|-\chi 1-x+\chi 1+y|\vee|\chi 1-y-\chi 1+x|=|y-x|\cr}$$

\noindent for all $x$, $y\in L^0$.   And

\Centerline{$|\psi'(\sigma,\tau)-\psi'(\sigma',\tau')|
\le|u_{\tau}-u_{\sigma}-u_{\tau'}+u_{\sigma'}|
\le|u_{\sigma}-u_{\sigma'}|+|u_{\tau}-u_{\tau'}|$.}

\medskip

{\bf (c)} To begin
with (down to the end of (v) below), suppose that $\Cal S$ is finitely
full.   Let $\epsilon>0$.

\medskip

\quad{\bf (i)} By 615Ga,
$u_{\uparrow}=\lim_{\sigma\uparrow\Cal S}u_{\sigma}$ is defined and there is a
$\tau^*\in\Cal S$ such that that $\theta(z^*)\le\epsilon$ where
$z^*=\sup_{\tau\in\Cal S\vee\tau^*}|u_{\tau}-u_{\uparrow}|$.   Then
$\bar\mu\Bvalue{z^*\ge\bover14}\le 4\epsilon$.

\medskip

\quad{\bf (ii)} By (a) above, there are $b_0\in\frak A$ and
$\gamma\ge 0$ such that $\bar\mu b_0\ge 1-\epsilon$ and

\Centerline{$\sum_{i=0}^{n-1}
   \bar\mu(b_0\Bcap\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14})
  \le\gamma$}

\noindent whenever $\tau_0\le\ldots\le\tau_n$ in $\Cal S$.   Set
$\delta=\min(\Bover18,\Bover{\epsilon}{\gamma})$.

\medskip

\quad{\bf (iii)} By 615F(a-i) $\pmb u\restr\Cal S\wedge\tau^*$ is moderately
oscillatory, therefore
\updowncgt (615G/615J), while $\Cal S\wedge\tau^*$ is full (611Me).
Construct $\sequence{i}{D_i}$, $\sequence{i}{y_i}$
and $\sequence{i}{d_i}$
from $\pmb u\restr\Cal S\wedge\tau^*$ and $\delta$ as in 615M.   For
$i\in\Bbb N$ set $D_i^*=\bigcup_{\sigma\in D_i}\Cal S\vee\sigma$.
There is an $m\in\Bbb N$ such that $\bar\mu d_m\le\epsilon$ (615M(c-ii)).
Choose $\tau_m,\ldots,\tau_0$ in such a way that for each $i\le m$

\inset{$\tau_i\in D_i$,

$\theta(z_i)\le\Bover{\epsilon\delta}{m+1}$ where
$z_i=\sup\{|u_{\sigma}-y_i|:\sigma\in D_i^*$, $\sigma\le\tau_i\}$,

if $i<m$ then $\tau_i\le\tau_{i+1}$.}

\noindent (By 615Gb, when we come to choose $\tau_k$ anything far enough down
$D_k$ will serve.)   Set $z=\sup_{k\le m}z_k$;  then
$\theta(z)\le\epsilon\delta$ so $\bar\mu\Bvalue{z\ge\delta}\le\epsilon$.
Set $b=b_0\Bcap\Bvalue{z<\delta}\Bcap\Bvalue{z^*<\bover14}\Bsetminus d_m$;
then $\bar\mu b\ge 1-7\epsilon$.

\medskip

\quad{\bf (iv)} Take $i<m$ and a finite sublattice $K$ of
$\Cal S\cap[\tau_i,\tau_{i+1}]$ containing $\tau_i$ and $\tau_{i+1}$,

\medskip

\qquad\grheada\ Let $\langle\sigma_j\rangle_{j\le n}$ be the increasing
enumeration of a maximal totally ordered subset of $K$, so that
$\tau_i=\sigma_0$ and $\tau_{i+1}=\sigma_n$.   Define
$\langle a_j\rangle_{j\le n}$ inductively by saying that

\Centerline{$a_j
=\Bvalue{|u_{\sigma_j}-y_i|\ge\delta}
   \Bsetminus\sup_{j'<j}a_{j'}$}

\noindent for each $j$ and $a^*=1\Bsetminus\sup_{j\le n}a_j$.
Then the $a_j$ are disjoint and
$a_j\in\frak A_{\sigma_j}\subseteq\frak A_{\tau_{i+1}}$ for each $j$, so we have
a $\tau'_i\in\Cal T$ such that
$a_j\Bsubseteq\Bvalue{\tau'_i=\sigma_j}$ for
$j\le n$, while $a^*\Bsubseteq\Bvalue{\tau'_i=\tau_i}$.   As $\Cal S$ is
supposed to be full, $\tau'_i\in\Cal S$.   Now
$\sup_{j\le n}\Bvalue{\tau'_i=\sigma_j}=1$,
$\tau_i\le\tau'_i\le\tau_{i+1}$ and
$\Bvalue{|u_{\tau'_i}-y_i|<\delta}$ is disjoint from every $a_j$ and from
$\Bvalue{|u_{\tau_{i+1}}-y_i|\ge\delta}$ so is included in
$\Bvalue{\tau'_i=\tau_{i+1}}\Bcap\Bvalue{\tau_{i+1}=\tau^*}
\Bsubseteq\Bvalue{\tau'_i=\tau^*}$.   So $\tau'_i\in D_{i+1}$.

\medskip

\qquad\grheadb\ For $l\le n$,

\Centerline{$|S_K(\tbf 1,d\psi')-\psi'(\tau_i,\tau_{i+1})|\times\chi(b\Bcap a_l)
\le 4\delta\chi\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14}$.}

\noindent\Prf\ Concerning the case $l=0$, observe that
$a_0=\Bvalue{|u_{\tau_i}-y_i|\ge\delta}\Bsubseteq\Bvalue{z_i\ge\delta}$ is
disjoint from $b$ and the result is trivial.    Otherwise, we know that
$S_K(\tbf 1,d\psi')=\sum_{j=0}^{n-1}\psi'(\sigma_j,\sigma_{j+1})$
(613Ec).   Now if $0\le j\le l-2$ then

\Centerline{$a_l\Bsubseteq\Bvalue{|u_{\sigma_j}-y_i|<\delta}
  \Bcap\Bvalue{|u_{\sigma_{j+1}}-y_i|<\delta}
\Bsubseteq\Bvalue{|u_{\sigma_{j+1}}-u_{\sigma_j}|<2\delta}
\Bsubseteq\Bvalue{\psi'(\sigma_j,\sigma_{j+1})=0}$}

\noindent and if $l\le j<n$ then
$\tau'_i\le\tau'_i\vee\sigma_j\le\tau'_i\vee\sigma_{j+1}\le\tau_{i+1}$
so $\tau'_i\vee\sigma_j$ and $\tau'_i\vee\sigma_{j+1}$ belong to
$D^*_{i+1}$ and

$$\eqalign{a_l&
\Bsubseteq\Bvalue{\sigma_j=\tau'_i\vee\sigma_j}
   \Bcap\Bvalue{\sigma_{j+1}=\tau'_i\vee\sigma_{j+1}}\cr&
\mskip100mu
   \Bcap\Bvalue{|u_{\tau'_i\vee\sigma_j}-u_{\tau_{i+1}}|\le z_{i+1}}
   \Bcap\Bvalue{|u_{\tau'_i\vee\sigma_{j+1}}-u_{\tau_{i+1}}|\le z_{i+1}}\cr&
\Bsubseteq\Bvalue{|u_{\sigma_{j+1}}-u_{\sigma_j}|\le 2z}\cr}$$

\noindent and

\Centerline{$a_l\cap b
\Bsubseteq\Bvalue{|u_{\sigma_{j+1}}-u_{\sigma_j}|\le\bover12}
\Bsubseteq\Bvalue{\psi'(\sigma_j,\sigma_{j+1})=0}$.}

So

\Centerline{$|S_K(\tbf 1,d\psi')-\psi'(\tau_i,\tau_{i+1})|\times\chi(b\Bcap a_l)
=|\psi'(\sigma_{l-1},\sigma_l)-\psi'(\tau_i,\tau_{i+1})|\times\chi(b\Bcap a_l)$.}

\noindent Now

\Centerline{$a_l\Bsubseteq\Bvalue{|u_{\sigma_{l-1}}-y_i|\le\delta}
       \Bcap\Bvalue{|u_{\sigma_0}-y_i|\le\delta}
\Bsubseteq\Bvalue{|u_{\sigma_{l-1}}-u_{\tau_i}|\le 2\delta}$}

\noindent and also

\Centerline{$a_l\Bsubseteq\Bvalue{\sigma_l=\sigma_l\vee\tau'_i}
\Bsubseteq\Bvalue{|u_{\sigma_l}-u_{\tau_{i+1}}|\le 2z_{i+1}}$}

\noindent so

$$\eqalign{b\Bcap a_l&
  \Bsubseteq\Bvalue{|u_{\sigma_{l-1}}-u_{\tau_i}|+|u_{\sigma_l}-u_{\tau_{i+1}}|
                              \le 4\delta}\cr&
  \Bsubseteq\Bvalue{|\psi'(\sigma_{l-1},\sigma_l)-\psi'(\tau_i,\tau_{i+1})|
                              \le 4\delta}\cr}$$

\noindent ((b-ii) above).   So

\Centerline{$|\psi'(\sigma_{l-1},\sigma_l)-\psi'(\tau_i,\tau_{i+1})|
     \times\chi(b\Bcap a_l)
   \le 4\delta\chi 1$.}

\noindent At the same time,

$$\eqalign{b\Bcap a_l\Bcap\Bvalue{|u_{\tau_i}-u_{\tau_{i+1}}|\le\bover14}&
   \Bsubseteq\Bvalue{\psi'(\tau_i,\tau_{i+1})=0}
      \Bcap\Bvalue{|u_{\sigma_{l-1}}-u_{\sigma_l}|\le\bover14+4\delta}\cr&
   \Bsubseteq\Bvalue{\psi'(\tau_i,\tau_{i+1})=0}
      \Bcap\Bvalue{\psi'(\sigma_{l-1},\sigma_l)=0},\cr}$$

\noindent and in fact

$$\eqalign{|S_K(\tbf 1,d\psi')-\psi'(\tau_i,\tau_{i+1})|\times\chi(b\Bcap a_l)&
=|\psi'(\sigma_{l-1},\sigma_l)-\psi'(\tau_i,\tau_{i+1})|\times\chi(b\Bcap a_l)\cr&
\le 4\delta\chi\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14}.
\text{ \Qed}\cr}$$

\medskip

\qquad\grheadc\ Concerning $a^*$, we have

$$\eqalign{a^*&
\Bsubseteq\inf_{j\le n}\Bvalue{|u_{\sigma_j}-y_i|<\delta}\cr&
\Bsubseteq\inf_{j\le k\le n}\Bvalue{|u_{\sigma_k}-u_{\sigma_j}|\le 1}\cr&
\Bsubseteq\Bvalue{\psi'(\tau_i,\tau_{i+1})=0}
   \Bcap\inf_{j<n}\Bvalue{\psi'(\sigma_j,\sigma_{j+1})=0}\cr&
\Bsubseteq\Bvalue{S_K(\tbf 1,d\psi')=\psi'(\tau_i,\tau_{i+1})}.\cr}$$

\medskip

\qquad\grheadd\ Since $a^*\Bcup\sup_{j\le n}a_j=1$, we see that

\Centerline{$|S_K(\tbf 1,d\psi')-\psi'(\tau_i,\tau_{i+1})|\times\chi b
\le 4\delta\chi\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14}$.}

\noindent And this is true whenever $i<m$ and $K$ is a finite sublattice of
$\Cal S\cap[\tau_i,\tau_{i-1}]$ containing $\tau_i$ and $\tau_{i+1}$.

\medskip

\quad{\bf (v)} Now suppose that $J\in\Cal I(\Cal S)$ includes
$I=\{\tau_0,\ldots,\tau_m\}$.

\medskip

\qquad\grheada\ $S_{J\wedge\tau_0}(\tbf 1,d\psi')\times\chi b=0$.   \Prf\ If
$\sigma\le\tau\le\tau_0$ in $\Cal S$, then $\sigma$ and $\tau$ both belong to
$\Cal S\wedge\tau^*=D_0$, so $|u_{\tau}-u_{\sigma}|\le 2z_i\le 2z$ and

\Centerline{$b\Bsubseteq\Bvalue{|u_{\tau}-u_{\sigma}|\le 2\delta}
\Bsubseteq\Bvalue{\psi'(\sigma,\tau)=0}$,}

\noindent that is, $\psi'(\sigma,\tau)\times\chi b=0$.   As $\sigma$ and $\tau$
are arbitrary, $S_{J\wedge\tau_0}(\tbf 1,d\psi')\times\chi b=0$.\ \Qed

Similarly, $|u_{\tau}-u_{\sigma}|\le 2z^*$ and

\Centerline{$b\Bsubseteq\Bvalue{|u_{\tau}-u_{\sigma}|\le\bover12}
\Bsubseteq\Bvalue{\psi'(\sigma,\tau)=0}$}

\noindent whenever $\tau^*\le\sigma\le\tau$ in $\Cal S$, so
$S_{J\vee\tau^*}(\tbf 1,d\psi')\times\chi b=0$.   We see also that if
$\tau_m\le\sigma\le\tau\le\tau^*$ then

\Centerline{$\Bvalue{\psi'(\sigma,\tau)\ne 0}
\Bsubseteq\Bvalue{\sigma<\tau}\Bsubseteq\Bvalue{\tau_m<\tau^*}\Bsubseteq d_m$}

\noindent (615M(c-i)), so $\psi'(\sigma,\tau)\times\chi b=0$, and we
conclude that $S_{J\cap[\tau_m,\tau^*]}(\tbf 1,d\psi')\times\chi b=0$.
So in fact we have

$$\eqalign{S_J(\tbf 1,d\psi')\times\chi b&
=\bigl(S_{J\wedge\tau_0}(\tbf 1,d\psi')
   +\sum_{i=0}^{m-1}S_{J\cap[\tau_i,\tau_{i+1}]}\tbf 1,d\psi')\cr&
\mskip150mu
   +S_{J\cap[\tau_m,\tau^*]}(\tbf 1,d\psi')
   +S_{J\vee\tau^*}(\tbf 1,d\psi')\bigr)\times\chi b\cr&
=\sum_{i=0}^{m-1}S_{J\cap[\tau_i,\tau_{i+1}]}\tbf 1,d\psi')\times\chi b,\cr}$$

$$\eqalign{|S_J(\tbf 1,d\psi')-S_I(\tbf 1,d\psi')|\times\chi b&
\le\sum_{i=0}^{m-1}|S_{J\cap[\tau_i,\tau_{i+1}]}(\tbf 1,d\psi')
   \psi'(\tau_i,\tau_{i+1}|\times\chi b\cr&
\le\sum_{i=0}^{m-1}
     4\delta\chi\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14}\cr}$$

\noindent by (iv-$\delta$) above, and

$$\eqalign{\theta(|S_J(\tbf 1,d\psi')-S_I(\tbf 1,d\psi')|\times\chi b)&
\le\Expn(\sum_{i=0}^{m-1}
   4\delta\chi\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14})\cr&
=4\delta\sum_{i=0}^{m-1}\bar\mu\Bvalue{|u_{\tau_{i+1}}-u_{\tau_i}|\ge\bover14}
\le 4\delta\gamma
\le 4\epsilon}$$

\noindent by (ii) above.   Since $\bar\mu b\ge 1-7\epsilon$,
$\theta(S_J(\tbf 1,d\psi')-S_I(\tbf 1,d\psi'))\le 11\epsilon$,
and this is true for every finite sublattice of $\Cal S$ including $I$.
As $\epsilon$ is arbitrary, $\int_{\Cal S}d\psi'$ is defined.

\medskip

\quad{\bf (vi)} All this has been on the assumption that $\Cal S$ was full.
But for the general case, we can take the covered envelope
$\hat\Cal S$ of $\Cal S$ and the
fully adapted extension $\hatpmb u=\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$
of $\pmb u$ to $\hat\Cal S$ (612Q).   Setting
$\hat\psi'(\sigma,\tau) =\hat u_{\tau}-\hat u_{\sigma}
-\med(-\chi 1,\hat u_{\tau}-\hat u_{\sigma},\chi 1)$ when $\sigma\le\tau$ in
$\hat\Cal S$, (i)-(v) tell us that $\int_{\hat\Cal S}d\hat\psi'$ is defined.
And now we see from 613T that $\int_{\Cal S}d\psi'$ is defined.

\medskip

\quad{\bf (vii)} In the notation of 613Cc, $\psi=\Delta\pmb u-\psi'$.   Because
$\pmb u$ is moderately oscillatory,
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined (615Gb) so
$\int_{\Cal S}d\pmb u$ is defined (613N) and $\int_{\Cal S}d\psi$ is defined.

\medskip

{\bf (d)(i)} As observed in 613O(b-i),
$\pmb w=\sfamilysS{w}=\family{\sigma}{\Cal S}{\int_{\Cal S\wedge\sigma}d\psi}$
and
$\pmb w'=\sfamilysS{w'}=\family{\sigma}{\Cal S}{\int_{\Cal S\wedge\sigma}d\psi'}$
are well-defined, while

\Centerline{$\pmb u
=u_{\downarrow}\tbf 1+\family{\sigma}{\Cal S}{\int_{\Cal S\wedge\sigma}d\pmb u}
=u_{\downarrow}\tbf 1+\pmb w+\pmb w'$.}

\noindent Set $A=\{S_I(\tbf 1,d|\psi'|):I\in\Cal I(\Cal S)\}$.

\medskip

\quad{\bf (ii)} Let $\epsilon>0$.   Then there is a process
$\pmb v=\sfamilysS{v}$ of bounded variation such that
$\theta(\sup|\pmb u-\pmb v|)\le\Bover13\epsilon$.   Set
$b=\Bvalue{\sup|\pmb v-\pmb u|\le\bover13}$;  then $\bar\mu b\ge 1-\epsilon$.
If $\sigma\le\tau$ in $\Cal S$, then

$$\eqalign{|\psi'(\sigma,\tau)|\times\chi b&
\le|u_{\tau}-u_{\sigma}|\times\chi(b\Bcap\Bvalue{|u_{\tau}-u_{\sigma}|>1})\cr&
\le(|v_{\tau}-v_{\sigma}|+\Bover23\chi 1)
      \times\chi\Bvalue{|v_{\tau}-v_{\sigma}|>\bover13}
\le 3|v_{\tau}-v_{\sigma}|.\cr}$$

\noindent So

\Centerline{$S_I(\tbf 1,d|\psi'|)\times\chi b\le 3S_I(\tbf 1,|d\pmb v|)
\le 3\int_{\Cal S}|d\pmb v|$}

\noindent for every $I\in\Cal I(\Cal S)$, and $\{x\times\chi b:x\in A\}$ is
bounded above in $L^0$.   As $\epsilon$ is arbitrary,
$A$ is bounded above in $L^0$ (613B(p-i)).   Set $\bar z=\sup A$.

\medskip

\quad{\bf (iii)} If $\tau_0\le\ldots\le\tau_n$ in $\Cal S$,
$\sum_{i=0}^{n-1}|w'_{\tau_{i+1}}-w'_{\tau_i}|\le\bar z$.   \Prf\
Take $\epsilon>0$.   Then for each $i<n$ there is a
$J_i\in\Cal I(\Cal S\cap[\tau_i,\tau_{i+1}]$, containing $\tau_i$ and
$\tau_{i+1}$, such that $\theta(z_i)\le\bover{\epsilon}n$ where

$$\eqalign{z_i&
=|\int_{\Cal S\cap[\tau_i,\tau_{i+1}]}d\psi'-S_{J_i}(\tbf 1,d\psi')|\cr&
=|w'_{\tau_{i+1}}-w'_{\tau_i}-S_{J_i}(\tbf 1,d\psi')|
\ge|w'_{\tau_{i+1}}-w'_{\tau_i}|-|S_{J_i}(\tbf 1,d\psi')|\cr}$$

\noindent Now

\Centerline{$|w'_{\tau_{i+1}}-w'_{\tau_i}|\le z_i+|S_{J_i}(\tbf 1,d\psi')|
\le z_i+S_{J_i}(\tbf 1,|d\psi'|)$}

\noindent for each $i$.   Setting $I=\bigcup_{i<n}J_i$,

\Centerline{$\sum_{i=0}^{n-1}|w'_{\tau_{i+1}}-w'_{\tau_i}|
\le S_I(\tbf 1,|d\psi'|)+\sum_{i=0}^{n-1}z_i
\le\bar z+\sum_{i=0}^{n-1}z_i$}

\noindent while $\theta(\sum_{i=0}^{n-1}z_i)\le\epsilon$.   As $\epsilon$ is
arbitrary, $\sum_{i=0}^{n-1}|w'_{\tau_{i+1}}-w'_{\tau_i}|\le\bar z$.\ \Qed

As $\tau_0,\ldots,\tau_n$ are arbitrary, $\pmb w'$ is of bounded variation.

\medskip

\quad{\bf (iv)} In particular, $\pmb w'$ is moderately oscillatory, so
$\pmb w=\pmb u-u_{\downarrow}\tbf 1-\pmb w'$ is moderately oscillatory.

\medskip

{\bf (e)} Now the
arguments of (a)-(d) apply equally to the fully adapted extension
$\hatpmb u$ of $\pmb u$
to $\hat\Cal S$ and the associated strictly adapted interval function
$\hat\psi$, which must be the adapted extension of $\psi$ to
$\hat\Cal S^{2\uparrow}$.   The corresponding indefinite integral
$\hatpmb w=ii_{\hat\psi}(\tbf 1)
=\family{\tau}{\hat\Cal S}{\int_{\hat\Cal S\wedge\tau}d\hat\psi}$ extends
$\pmb w$ (613T) so must be the fully adapted extension of $\pmb u$ to
$\hat\Cal S$.   Now
$\Osclln(\pmb w)=\Osclln(\hatpmb w)$ (618La) and
$\Osclln(\hatpmb w)$ is at most $\Osclln(\hat\psi)$ (618Pc), which is at most
$\chi 1$, as can be seen directly from the formulae in 618O.

\medskip

{\bf (f)} We know that $w_{\tau'}-w_{\tau}=\int_{\Cal S\cap[\tau,\tau']}d\psi$
(613J(c-i)).   Now $\psi(\sigma,\sigma')\in L^0(\frak D_{\tau})$ for every
$(\sigma,\sigma')\in(\Cal S\vee\tau)^{2\uparrow}$, so
$S_I(\tbf 1,d\psi)\in L^0(\frak D_{\tau})$ for every
$I\in\Cal I(\Cal S\vee\tau)$.   As $L^0(\frak D_{\tau})$ is closed for the
topology of convergence in measure (613B(i-i)),
$w_{\tau'}-w_{\tau}
=\lim_{I\uparrow\Cal I(\Cal S\cap[\tau,\tau'])}S_I(\tbf 1,d\psi)$ belongs to
$L^0(\frak D_{\tau})$.

\medskip

{\bf (g)} Let $\epsilon>0$.   Then there is a simple process
$\pmb v=\sfamilysS{v}$ such that
$\theta(\sup|\pmb u-\pmb v|)\le\bover12\epsilon$.
Set $d=\Bvalue{\sup|\pmb u-\pmb v|>\bover12}$;  then
$\bar\mu d\le\epsilon$.   Let $(\tau_0,\ldots,\tau_n)$ be a
breakpoint string for $\pmb v$, and consider the simple process
$\pmb z=\sfamilysS{z}$ define by saying that if $\sigma\in\Cal S$ then

\Centerline{$\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{z_{\sigma}=0}$,
\quad$\Bvalue{\tau_n\le\sigma}\Bsubseteq\Bvalue{z_{\sigma}=w_{\tau_n}}$,}

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

\noindent Now $\Bvalue{\pmb w\ne\pmb z}\Bsubseteq d$.   \Prf\ Take
$\tau\in\Cal S$ and write $a$ for $\Bvalue{w_{\tau}\ne z_{\tau}}$.
Then

$$\eqalign{a\Bcap\Bvalue{\tau<\tau_0}&
=\Bvalue{w_{\tau}\ne 0}\Bcap\Bvalue{\tau<\tau_0}
\Bsubseteq\sup\{\Bvalue{\psi(\sigma,\sigma')\ne 0}
    \Bcap\Bvalue{\tau<\tau_0}:\sigma\le\sigma'\le\tau\}
    \cr&
\Bsubseteq\sup\{\Bvalue{\psi(\sigma,\sigma')\ne 0}
    \Bcap\Bvalue{\sigma<\tau_0}\Bcap\Bvalue{\sigma'<\tau_0}:\sigma\le\sigma'\}
    \cr&
\Bsubseteq\sup\{\Bvalue{|u_{\sigma'}-u_{\sigma}|>1}
    \Bcap\Bvalue{v_{\sigma}=v_{\sigma'}}:\sigma,\,\sigma'\in\Cal S\}\cr&
\Bsubseteq\sup\{\Bvalue{|u_{\sigma}-v_{\sigma}|>\hbox{$\bover12$}}:
    \sigma\in\Cal S\}
=d;\cr}$$

$$\eqalign{a\Bcap\Bvalue{\tau_n\le\tau}&
=\Bvalue{w_{\tau\vee\tau_n}\ne w_{\tau_n}}
\Bsubseteq\sup\{\Bvalue{\psi(\sigma,\sigma')\ne 0}:
    \tau_n\le\sigma\le\sigma'\le\tau\vee\tau_n\}\cr&
\Bsubseteq\sup\{\Bvalue{|u_{\sigma'}-u_{\sigma}|>1}
    \Bcap\Bvalue{\tau_n\le\sigma}\Bcap\Bvalue{\tau_n\le\sigma'}:
    \sigma\le\sigma'\}\cr&
\Bsubseteq\sup\{\Bvalue{|u_{\sigma}-v_{\sigma}|>\hbox{$\bover12$}}
    :\sigma\in\Cal S\}
=d;\cr}$$

\noindent and for $i<n$

$$\eqalign{a\Bcap\Bvalue{\tau_i\le\tau}\Bcap\Bvalue{\tau<\tau_{i+1}}&
=\Bvalue{w_{\tau\vee\tau_i}\ne w_{\tau_i}}\Bcap\Bvalue{\tau<\tau_{i+1}}\cr&
\Bsubseteq\sup\{\Bvalue{\psi(\sigma,\sigma')\ne 0}
    \Bcap\Bvalue{\sigma'<\tau_{i+1}}:\tau_i\le\sigma\le\sigma'\}\cr&
\Bsubseteq\sup\{\Bvalue{|u_{\sigma'}-u_{\sigma}|>1}
    \Bcap\Bvalue{\sigma'<\tau_{i+1}}:\tau_i\le\sigma\le\sigma'\}\cr&
\Bsubseteq\sup\{\Bvalue{|u_{\sigma}-v_{\sigma}|>\hbox{$\bover12$}}:
    \sigma\in\Cal S\}
=d.\cr}$$

\noindent Thus in fact $a\Bsubseteq d$;  as $\tau$ is arbitrary,
$\Bvalue{\pmb w\ne\pmb z}\Bsubseteq d$.\ \QeD\

Now $\theta(\pmb w-\pmb z)\le\bar\mu d\le\epsilon$.   As $\epsilon$ is
arbitrary, $\pmb w$ is near-simple.
}%end of proof of 633M

\leader{*633N}{Lemma} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and has a
lower bound in $\Cal S$, and
$\pmb u=\sfamilysS{u}$ a near-simple process.   As in 633M, set

\Centerline{$\psi(\sigma,\tau)
=\med(-\chi 1,u_{\tau}-u_{\sigma},\chi 1)$}

\noindent when $\sigma\le\tau$ in $\Cal S$.
Let $\Cal S'$ be a  cofinal sublattice of $\Cal S$ which separates
$\Cal S$. %633Ba
Then $\int_{\Cal S'}d\psi=\int_{\Cal S}d\psi$.

\proof{ We can use the arguments of 633I-633K.

\medskip

{\bf (a)} To begin with, suppose that $\Cal S'$ is finitely full.
The point is that if $\sigma\le\tau\le\tau'$
then $|\psi(\sigma,\tau')-\psi(\sigma,\tau)|\le|u_{\tau'}-u_{\tau}|$
(633M(b-ii)).   Let $\Cal S'$ be the finitely covered envelope of $\Cal S_0$.
If $\tau\in\Cal S$, $A_{\tau}=\Cal S'\cap(\Cal T\vee\tau)$
is downwards-directed with infimum $\tau$ (633Eb) and
$\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}=u_{\tau}$ (632E);  as $A_{\tau}$ is
finitely full (611O(b-i)),

$$\lim_{\tau'\downarrow A_{\tau}}
    \sup_{\Atop{\sigma\in\Cal S\wedge\tau}{\rho\in\Cal S'\cap[\tau,\tau']}}
        |\psi(\sigma,\rho)-\psi(\sigma,\tau)|
=\lim_{\tau'\downarrow A_{\tau}}
    \sup_{\rho\in A_{\tau}\wedge\tau'}|u_{\rho}-u_{\tau}|
=0$$

\noindent by 615Db.

Now we know from 633Mc that $z=\int_{\Cal S}d\psi$ is defined.
Let $\epsilon>0$.
Let $J\in\CalIS$ be such that
$\theta(z-S_K(\tbf 1,d\psi))\le\epsilon$ whenever
$J\subseteq K\in\CalIS$.   For each $\tau\in J$, let
$\sigma_{\tau}\in A_{\tau}$ be such that
$\theta(\bar u_{\tau})\le\Bover{\epsilon}{1+\#(J)}$ where

\Centerline{$\bar u_{\tau}
=\sup_{\sigma\in\Cal S\wedge\tau,\rho\in\Cal S'\cap[\tau,\tau']}
        |\psi(\sigma,\rho)-\psi(\sigma,\tau)|$.}

\noindent If $I\in\Cal I(\Cal S')$ contains $\sigma_{\tau}$ for every
$\tau\in J$,
consider $K=I\sqcup J$.   By the choice of $J$,
$\theta(S_K(\tbf 1,d\psi)-z)\le\epsilon$.   By 633I,
$|S_K(\tbf 1,d\psi)-S_I(\tbf 1,d\psi)|\le 2\sum_{\tau\in J}\bar u_{\tau}$
and $\theta(S_K(\tbf 1,d\psi)-S_I(\tbf 1,d\psi))\le 2\epsilon$.   But now

$$\eqalign{\theta(z-S_I(\tbf 1,d\psi))&
\le\theta(z-S_K(\tbf 1,d\psi))
   +\theta(S_K(\tbf 1,d\psi)-S_I(\tbf 1,d\psi))\cr&
\le\epsilon+2\epsilon
=3\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\int_{\Cal S'}d\psi$ is defined and equal to $z$.

\medskip

{\bf (b)} In general, writing $\hat\Cal S'_f$ for the finitely-covered envelope
of $\Cal S$, $\hat\Cal S'_f$ is a cofinal finitely full sublattice of $\Cal S$
which separates $\Cal S$, so $\int_{\hat\Cal S'_f}d\psi=z$.   But now
613T tells us that $\int_{\Cal S'}d\psi=z$.
}%end of proof of 633N

\leader{633O}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\inf A\in\Cal S$ for every non-empty $A\subseteq\Cal S$ with a
lower bound in $\Cal S$,
$\pmb u=\sfamilysS{u}$ an order-bounded locally near-simple process,
and $\Cal S'$ a sublattice of $\Cal S$ which is cofinal and coinitial
with $\Cal S$ and $\pmb u$-separates $\Cal S$.
Write $\pmb u'$ for $\pmb u\restr\Cal S'$.

(a) $\Osclln(\pmb u)=\Osclln(\pmb u')$.

(b) $\pmb u$ is jump-free iff $\pmb u'$ is jump-free.
%634F

\proof{ This is trivial if $\Cal S$ is empty;  suppose otherwise.   Of course
$\Cal S'$ must also be non-empty.

\medskip

{\bf (a)(i)} If $\hat\Cal S'_f$ is the finitely-covered
envelope of $\Cal S'$, then
$\hat\Cal S'_f\subseteq\Cal S$, and of
course $\hat\Cal S'_f$ is cofinal with $\Cal S$ and
$\pmb u$-separates $\Cal S$.
Since $\Osclln(\pmb u')=\Osclln(\pmb u\restr\hat\Cal S'_f)$
(618L), it is enough to consider the case in which $\Cal S'$ itself is
finitely full.

For $\tau\in\Cal S$, set $A_{\tau}=\{\sigma:\tau\le\sigma\in\Cal S'\}$
and $\tau^{\#}=\inf A_{\tau}$, so
that $\tau^{\#}\in\Cal S$, $u_{\tau^{\#}}=u_{\tau}$ (633Ec) and
(because $\family{t}{T}{\frak A_t}$
is right-continuous and $\pmb u$ is locally
near-simple) $u_{\tau}=\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}$.
In fact, $\pmb u$ is constant on $\Cal S\cap[\tau,\tau^{\#}]$.   \Prf\
If $\tau'\in\Cal S\cap[\tau,\tau^{\#}]$ then $A_{\tau'}=A_{\tau}$ so
$u_{\tau'}=\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}=u_{\tau}$.\ \Qed

\medskip

\quad{\bf (ii)} If $I\in\Cal I(\Cal S')$ is non-empty then
$\Osclln^*_I(\pmb u)\le\Osclln^*_I(\pmb u')$.   \Prf\ Let
$\sigma_0\le\ldots\le\sigma_n$ linearly
generate the $I$-cells, and set $\sigma_{-1}=\min\Cal T$,
$\sigma_{n+1}=\max\Cal T$.   Write $w$ for $\Osclln^*_I(\pmb u')$.
\Quer\ Suppose that $-1\le i\le n$, $\tau$, $\tau'\in\Cal S$ and
$\sigma_i\le\tau\le\tau'\le\sigma_{i+1}$,
$|u_{\tau}-u_{\tau'}|\not\le w$.   Writing
$A_{\tau'}\wedge\sigma_{i+1}$ for
$\{\sigma\wedge\sigma_{i+1}:\sigma\in A_{\tau'}\}$,
$A_{\tau'}\wedge\sigma_{i+1}$ is a coinitial subset of $A_{\tau'}$, so
$u_{\tau'}=\lim_{\sigma\downarrow A_{\tau'}\wedge\sigma_{i+1}}u_{\sigma}$
and there is a $\sigma'\in\Cal S'\cap[\tau',\sigma_{i+1}]$ such that
$|u_{\sigma'}-u_{\tau}|\not\le w$.   Similarly, there is a
$\sigma\in\Cal S'\cap[\tau,\sigma']$ such that
$|u_{\sigma'}-u_{\sigma}|\not\le w$.   But now
$\sigma_i\le\sigma\le\sigma'\le\sigma_{i+1}$, so 618B(b-iv) and 618Ca
tell us that

\Centerline{$|u_{\sigma'}-u_{\sigma}|\le\Osclln^*_I(\pmb u')=w$. \Bang}

\noindent Thus $|u_{\tau}-u_{\tau'}|\le w$ whenever $-1\le i\le n$, $\tau$,
$\tau'\in\Cal S$ and $\sigma_i\le\tau\le\tau'\le\sigma_{i+1}$.
By 618Ca in the other direction, $\Osclln^*_I(\pmb u)\le w$.\ \Qed

Accordingly

$$\eqalign{\Osclln(\pmb u)&
=\inf_{I\in\CalIS}\Osclln^*_I(\pmb u)
\le\inf_{I\in\Cal I(\Cal S')\setminus\{\emptyset\}}\Osclln^*_I(\pmb u)\cr&
\le\inf_{I\in\Cal I(\Cal S')\setminus\{\emptyset\}}\Osclln^*_I(\pmb u')
=\lim_{I\uparrow\Cal I(\Cal S')}\Osclln^*_I(\pmb u')
=\Osclln(\pmb u').\cr}$$

\medskip

\quad{\bf (iii)}\grheada\ Take $\epsilon>0$, a non-empty $I\in\CalIS$
and $(\tau_0,\ldots,\tau_n)$
linearly generating the $I$-cells.   Set $\tau_{-1}=\min\Cal T$ and
$\tau_{n+1}=\max\Cal T$.

\medskip

\qquad\grheadb\ We know that $\pmb u$ is locally moderately oscillatory
(631Ca).   So we can choose $\sigma_n\ge\ldots\ge\sigma_0$, in that order,
such that $\sigma_i\in A_{\tau_i}$ and
$\theta(\sup\{|u_{\sigma'}-u_{\tau_i}|:
\sigma'\in A_{\tau_i},\sigma'\le\sigma_i\})\le\Bover{\epsilon}{n+1}$ for each
$i$ (apply 615Gb to
$\pmb u\restr\Cal S\wedge(\sigma^*\wedge\inf_{i<j\le n}\sigma_j)$
where $\sigma^*\in A_{\tau_n}$).   Set
$w_i=\sup_{\sigma'\in A_{\tau_i},\sigma'\le\sigma_i}|u_{\sigma'}-u_{\tau_i}|$
for $0\le i\le n$ and $w=\sup_{0\le i\le n}w_i$, so that
$\theta(w)\le\epsilon$.   Set
$K=\{\sigma_0,\ldots,\sigma_n\}\in\Cal I(\Cal S')$.

Set $\sigma_{-1}=\min\Cal T$ and
$\sigma_{n+1}=\max\Cal T$.
If $0\le i\le n+1$, $\sigma\in\Cal S'$ and $\sigma\le\sigma_i$ then
$|u_{\sigma}-u_{\sigma\wedge\tau_i}|\le w$.   \Prf\ If $i=n+1$ this is
trivial.   If $i\le n$, then
$\Bvalue{\tau_i\le\sigma}\in\frak A_{\sigma}\Bsubseteq\frak A_{\sigma_i}$ so
we can define $\sigma'\in\Cal T$ by saying that

\Centerline{$\Bvalue{\tau_i\le\sigma}\Bsubseteq\Bvalue{\sigma'=\sigma}$,
\quad$\Bvalue{\sigma<\tau_i}\Bsubseteq\Bvalue{\sigma'=\sigma_i}$}

\noindent (611I).   Because $\Cal S'$ is finitely full, $\sigma'\in\Cal S'$,
while of course $\tau_i\le\sigma'$, so $\sigma'\in A_{\tau_i}$.   Now

$$\eqalignno{|u_{\sigma}-u_{\sigma\wedge\tau_i}|&
=|u_{\sigma\vee\tau_i}-u_{\tau_i}|
\Displaycause{612D(f-i)}
=|u_{\sigma}-u_{\tau_i}|\times\chi\Bvalue{\tau_i\le\sigma}
\le|u_{\sigma'}-u_{\tau_i}|
\le w_i
\le w.  \text{ \Qed}\cr}$$

Now suppose that $-1\le i\le n$ and $\sigma$, $\sigma'\in\Cal S'$ are such that
$\sigma_i\le\sigma\le\sigma'\le\sigma_{i+1}$.   Then
$\tau_i\le\sigma\wedge\tau_{i+1}\le\sigma'\wedge\tau_{i+1}\le\tau_{i+1}$.
So

$$\eqalign{|u_{\sigma'}-u_{\sigma}|&
\le|u_{\sigma'}-u_{\sigma'\wedge\tau_{i+1}}|
     +|u_{\sigma'\wedge\tau_{i+1}}-u_{\sigma\wedge\tau_{i+1}}|
     +|u_{\sigma\wedge\tau_{i+1}}-u_{\sigma}|\cr&
\le\Osclln_I^*(\pmb u)+2w\cr}$$

\noindent (using 618Ca again).
Taking the supremum over $i$, $\sigma$ and $\sigma'$
and applying 618Ca in the other direction,

\Centerline{$\Osclln(\pmb u')
\le\Osclln_K^*(\pmb u')
\le\Osclln_I^*(\pmb u)+2w$}

\noindent and

\Centerline{$\theta((\Osclln(\pmb u')-\Osclln_I^*(\pmb u))^+)
\le\theta(2w)\le 2\epsilon$.}

\medskip

\qquad\grheadc\ As $\epsilon$ is arbitrary,
$\theta((\Osclln(\pmb u')-\Osclln_I^*(\pmb u))^+)=0$ and
$\Osclln(\pmb u')\le\Osclln_I^*(\pmb u)$.   As $I$ is arbitrary,
$\Osclln(\pmb u')\le\Osclln(\pmb u)$.   With (ii) above, this shows that
$\Osclln(\pmb u)=\Osclln(\pmb u')$.

\medskip

{\bf (b)} In particular, $\Osclln(\pmb u)=0$ iff
$\Osclln(\pmb u')=0$, that is, $\pmb u$ is jump-free iff $\pmb u'$ is jump-free.
}%end of proof of 633O

\leader{633P}{}\cmmnt{ 633K is an attempt to reproduce the success of
613T for sublattices which are separating rather than covering.
We can try to follow other results about covering envelopes in the same way.
%634F

\medskip

\noindent}{\bf Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be a finitely full sublattice of $\Cal T$
such that $\inf A\in\Cal S$ for every non-empty $A\subseteq\Cal S$ with a lower
bound in $\Cal S$,
$\Cal S'$ a sublattice of $\Cal S$ which is cofinal and coinitial with $\Cal S$
and separates $\Cal S$,
and $\pmb u=\sfamilysS{u}$ a locally near-simple fully adapted process.
Write $\pmb u'$ for $\pmb u\restr\Cal S'$.

%\query how far can we take $\pmb u$-separating instead of separating?

(a) If $\pmb u'$ is simple then $\pmb u$ is simple.
%634F

(b) If $\pmb u'$ is near-simple then $\pmb u$ is near-simple.
%634F

(c) $\pmb u$ is order-bounded iff $\pmb u'$ is order-bounded.
%634F

(d) $\pmb u$ is (locally) of bounded variation iff $\pmb u'$ is
(locally) of bounded variation.
%634F 652K

(e) $\pmb u$ is an integrator iff $\pmb u'$ is an integrator.
%634F

(f) $\pmb u$ is a martingale iff $\pmb u'$ is a martingale.

(g) If $\pmb u'$ is a local martingale then $\pmb u$ is a local martingale.

(h) Suppose that $\pmb u'$ is a local integrator.
Then $\pmb u$ is a
local integrator and the quadratic variation of $\pmb u'$ is
$\pmb u^*\restr\Cal S'$, where $\pmb u^*$ is the quadratic variation of
$\pmb u$.
%634F 634N

\proof{ Throughout the proof, write $\hat\Cal S'_f$ for the finitely-covered
envelope of $\Cal S'$, so that $\hat\Cal S'_f\subseteq\Cal S$.
For $\tau\in\Cal S$, set $A_{\tau}=\{\sigma:\tau\le\sigma\in\hat\Cal S'_f\}$, so
that $\inf A_{\tau}=\tau$ (633Eb) and (because $\family{t}{T}{\frak A_t}$
is right-continuous and $\pmb u$ is locally
near-simple) $u_{\tau}=\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}$ (632F once
more).
Everything is trivial if $\Cal S'$ is empty, so I shall suppose otherwise.

\medskip

{\bf (a)} By 612Qf, $\pmb u\restr\hat\Cal S'_f$ is simple.
We have a starting value $u_*$ and breakpoint string $(\tau_0,\ldots,\tau_n)$
for $\pmb u\restr\hat\Cal S'_f$.   Because $\Cal S'$ is coinitial with $\Cal S$,
$\bigcap_{\sigma\in\hat\Cal S'_f}\frak A_{\sigma}
=\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$ and
$u^*\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
By 612Ka, there is a simple fully adapted process
$\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ defined by saying that

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

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

\noindent for every $\tau\in\Cal S$.   Evidently $\pmb v$ extends
$\pmb u\restr\hat\Cal S'_f$.   Because $\family{t}{T}{\frak A_t}$ is
right-continuous,

\Centerline{$v_{\tau}=\lim_{\sigma\downarrow A_{\tau}}v_{\sigma}
=\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}=u_{\tau}$}

\noindent for every $\tau\in\Cal S$ (632F yet again),
and $\pmb u=\pmb v$ is simple.

\medskip

{\bf (b)} By 631Ga, $\pmb u\restr\hat\Cal S'_f$ is near-simple, because
the covered envelopes of $\Cal S'$ and $\hat\Cal S'_f$ are the same.
Let $\epsilon>0$.   Then there is a simple process
$\pmb u'=\family{\sigma}{\hat\Cal S'_f}{u'_{\sigma}}$ such that
$\theta(\bar u)\le\epsilon$, where
$\bar u=\sup_{\sigma\in\hat\Cal S'_f}|u_{\sigma}-u'_{\sigma}|$.
As in (a) just above, we have a simple process
$\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ extending $\pmb u'$.
Now if $\tau\in\Cal S$,

\Centerline{$|u_{\tau}-v_{\tau}|
=\lim_{\sigma\downarrow A_{\tau}}|u_{\sigma}-v_{\sigma}|
=\lim_{\sigma\downarrow A_{\tau}}|u_{\sigma}-u'_{\sigma}|
\le\bar u$}

\noindent because $A_{\tau}\subseteq\hat\Cal S'_f$.   Thus

\Centerline{$\theta(\sup_{\tau\in\Cal S}|u_{\tau}-v_{\tau}|)
=\theta(\bar u)\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\pmb u$ is near-simple.

\medskip

{\bf (c)} If $\pmb u$ is order-bounded then $\pmb u'$ is
certainly order-bounded, by 614Fa.
If $\pmb u'$ is order-bounded, so is its fully adapted extension
$\hatpmb u=\family{\sigma}{\hat\Cal S'}{\hat u'_{\sigma}}$ to
the covered envelope of $\Cal S'$ (612Q(d-i)).   Writing
$\bar u$ for $\sup|\hatpmb u'|$,

\Centerline{$|u_{\tau}|
=\lim_{\sigma\downarrow A_{\tau}}|u_{\sigma}|\le\bar u$}

\noindent for every $\tau\in\Cal S$, so $\pmb u$ is order-bounded.

\medskip

{\bf (d)(i)} If $\pmb u$ is (locally) of bounded variation then
$\pmb u'$ is (locally) of bounded variation by 614Lb and 614Q(b-i).

\medskip

\quad{\bf (ii)}\grheada\ If $\pmb u'$ is of bounded variation, so is
$\pmb u\restr\hat\Cal S'_f$, by 614Lb;  set
$z=\int_{\hat\Cal S'_f}|d\pmb u|$.   Suppose that $\tau_0\le\ldots\le\tau_n$ in
$\Cal S$ and $\epsilon>0$.   Then there are $\sigma_i\in A_{\tau_i}$, for
$i\le n$, such that $\theta(u_{\sigma}-u_{\tau_i})\le\epsilon$ whenever
$\sigma\in A_{\tau_i}$ and $\sigma\le\sigma_i$.   Set
$\sigma'_i=\inf_{i\le j\le n}\sigma_j$ for $i\le n$;  then
$\sigma'_i\in A_{\tau_i}$ and $\sigma'_i\le\sigma_i$ for each $i$, and
$\sigma'_0\le\ldots\le\sigma'_n$.   Now

$$\eqalign{\theta((\sum_{i=0}^{n-1}|u_{\tau_{i+1}}-u_{\tau_i}|-z)^+)
&\le\theta((\sum_{i=0}^{n-1}|u_{\sigma'_{i+1}}-u_{\sigma'_i}|-z)^+)
   +2\sum_{i=0}^n\theta(u_{\sigma'_i}-u_{\tau_i})\cr
&\le 2(n+1)\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary,
$\sum_{i=0}^{n-1}|u_{\tau_{i+1}}-u_{\tau_i}|\le z$.   As
$\tau_0,\ldots,\tau_n$ are arbitrary, $\pmb u$ is of bounded variation.

\medskip

\qquad\grheadb\ If $\pmb u'$ is locally of bounded variation, and
$\tau\in\Cal S$, take $\sigma\in\Cal S'$ such that $\tau\le\sigma$, and consider
$\Cal S\wedge\sigma$ and $\Cal S'\wedge\sigma$.
Evidently
$\Cal S\wedge\sigma$ is a finitely full sublattice of $\Cal T$ such that
$\inf A\in\Cal S\wedge\sigma$ whenever $A\subseteq\Cal S\wedge\sigma$ is
non-empty and has a lower bound in $\Cal S\wedge\sigma$, and
$\Cal S'\wedge\sigma$ separates
$\Cal S\wedge\sigma$ by 633C(b-iii), while
$(\pmb u\restr\Cal S\wedge\sigma)\restr(\Cal S'\wedge\sigma)
=\pmb u'\restr\Cal S'\wedge\sigma$ is of bounded variation.   So ($\alpha$)
tells us that $\pmb u\restr\Cal S\wedge\sigma$ is of bounded variation.   It
follows at once that $\pmb u\restr\Cal S\wedge\tau$ is of bounded variation;
as $\tau$ is arbitrary, $\pmb u$ is locally of bounded variation.

\medskip

{\bf (e)(i)} By 616P(b-ii), $\pmb u'$ is surely an integrator
if $\pmb u$ is.   So let us suppose that $\pmb u'$ is an
integrator and seek to show that $\pmb u$ is.

\medskip

\quad{\bf (ii)} By 616Ia, $\pmb u\restr\hat\Cal S'_f$ is an integrator.   Now
$Q_{\Cal S}(d\pmb u)\subseteq\overline{Q_{\hat\Cal S'_f}(d\pmb u)}$.   \Prf\
Take $w\in Q_{\Cal S}(d\pmb u)$ and $\epsilon>0$.
There are $\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and
$v_i\in L^{\infty}(\frak A_{\tau_i})$, for $i<n$, such that
$\|v_i\|_{\infty}\le 1$ for every $i$ and
$w=\sum_{i=0}^{n-1}v_i\times(u_{\tau_{i+1}}-u_{\tau_i})$.
For $i\le n$ there is a $\sigma_i\in A_{\tau_i}$ such that
$\theta(u_{\sigma}-u_{\tau_i})\le\Bover{\epsilon}{n+1}$ whenever
$\sigma\in A_{\tau_i}$ and $\sigma\le\sigma_i$.   Set
$\sigma'_i=\inf_{i\le j\le n}\sigma_j$ for $i\le n$, so that
$\sigma'_0\le\ldots\le\sigma'_n$, while $\tau_i\le\sigma'_i\in\hat\Cal S'_f$
and $\theta(u_{\sigma'_i}-u_{\tau_i})\le\Bover{\epsilon}{n+1}$ for each
$i$.

Because $\tau_i\le\sigma'_i$,
$\frak A_{\tau_i}\subseteq\frak A_{\sigma'_i}$
and $v_i\in L^{\infty}(\frak A_{\sigma'_i})$ for each
$i$.   By 616C(ii),
$w'=\sum_{i=0}^nv_i\times(u_{\sigma'_{i+1}}-u_{\sigma'_i})$
belongs to $Q_{\hat\Cal S'_f}(d\pmb u)$.   But

$$\eqalign{\theta(w-w')
&=\theta(|w-w'|)
\le\theta(\sum_{i=0}^{n-1}|v_i|
  \times(|u_{\tau_{i+1}}-u_{\sigma'_{i+1}}|
     +|u_{\tau_i}-u_{\sigma'_i}|))\cr
&\le\theta(2\sum_{i=0}^n|u_{\tau_i}-u_{\sigma'_i}|)
\le 2\sum_{i=0}^n\theta(|u_{\tau_i}-u_{\sigma'_i}|)
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ and $w$ are arbitrary,
$Q_{\Cal S}(d\pmb u)
\subseteq\overline{Q_{\hat\Cal S'_f}(d\pmb u)}$.\ \Qed

As the closure of a topologically bounded set is topologically bounded,
$Q_{\Cal S}(d\pmb u)$ is topologically bounded.

\medskip

{\bf (f)(i)} If $\pmb u$ is a martingale then of course
$\pmb u'$ is a martingale (622Db).

\medskip

\quad{\bf (ii)} If $\pmb u'$ is a martingale then
there is a martingale
$\pmb v=\family{\tau}{\Cal S_1}{v_{\tau}}$ extending $\pmb u'$, where
$\Cal S_1$ is the ideal of $\Cal T$ generated by $\Cal S'$ (622Oa).
Since $\Cal S'$ is cofinal with $\Cal S$, $\Cal S\subseteq\Cal S_1$ and
$\pmb v\restr\Cal S$ is a martingale, while $\pmb v$ agrees with $\pmb u$ on
$\Cal S'$ and therefore on $\hat\Cal S'_f$.

If $\tau\in\Cal S$, then $\Cal S\vee\tau$ is a finitely full sublattice of
$\Cal T$ containing $\inf A$ whenever $A\subseteq\Cal S\vee\tau$ is
non-empty. so $\pmb v\restr\Cal S\vee\tau$ is locally near-simple (632Ia).
By 632F, as always,

\Centerline{$v_{\tau}=\lim_{\sigma\downarrow A_{\tau}}v_{\sigma}
=\lim_{\sigma\downarrow A_{\tau}}u_{\sigma}=u_{\tau}$.}

\noindent So $\pmb u=\pmb v\restr\Cal S$ is a martingale.

\medskip

{\bf (g)} If $\pmb u'$ is a local martingale, there is a covering ideal
$\Cal S'_1$ of $\Cal S'$ such that $\pmb u'\restr\Cal S'_1$ is a martingale.
Let $\Cal S_1$ be the ideal of $\Cal S$ generated by $\Cal S'_1$.   Then
$\Cal S_1$ covers $\Cal S'$ which is cofinal with $\Cal S$, so $\Cal S_1$ covers
$\Cal S$.   By 633C(b-ii), $\Cal S'_1$ separates $\Cal S_1$.
Also $\Cal S_1$ is a finitely full sublattice of $\Cal T$ such that
$\inf A\in\Cal S_1$ for every non-empty $A\subseteq\Cal S_1$ with a lower bound
in $\Cal S_1$, while $\pmb u\restr\Cal S_1$ is locally near-simple (631Gb, since
$\Cal S_1$ and $\Cal S$ have the same covered envelope).   Moreover,
$\Cal S'_1$ is cofinal and coinitial with $\Cal S_1$.   Since
$\pmb u\restr\Cal S'_1=\pmb u'\restr\Cal S'_1$ is a martingale, (f) tells us
that $\pmb u\restr\Cal S_1$ is a martingale, so that $\pmb u$ is a local
martingale.

\medskip

{\bf (h)} We have to check that (e) still works in  a `local'
form.   The point is that if $\tau\in\Cal S'$
then $\Cal S'\wedge\tau$ separates
$\Cal S\wedge\tau$ (633C(b-iii)).   So if
$\pmb u'$ is a local integrator and $\tau\in\Cal S$,
there is a $\tau'\in\Cal S'$ such
that $\tau\le\tau'$, and now $\pmb u\restr\Cal S\wedge\tau'$
is an integrator because $\pmb u'\restr\Cal S'\wedge\tau'$ is an
integrator;  by 616P(b-ii), $\pmb u\restr\Cal S\wedge\tau
=(\pmb u\restr\Cal S\wedge\tau')\restr\Cal S\wedge\tau$
is an integrator.   As $\tau$ is arbitrary, $\pmb u$ is a
local integrator.

If $\tau\in\Cal S'$, then $\Cal S'\wedge\tau$ separates
$\Cal S\wedge\tau$ and therefore $\pmb u$-separates
$\Cal S\wedge\tau$;  and of course
$\Cal S'\wedge\tau$ is cofinal and coinitial with $\Cal S\wedge\tau$.
Setting $u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$, this is
also $\lim_{\sigma\downarrow\Cal S'}u_{\sigma}$ because $\Cal S'$ is coinitial
with $\Cal S$.   So

\Centerline{$u^*_{\tau}
=u_{\tau}^2-u_{\downarrow}^2-2\int_{\Cal S\wedge\tau}\pmb u\,d\pmb u
=u_{\tau}^2-u_{\downarrow}^2-2\int_{\Cal S'\wedge\tau}\pmb u\,d\pmb u$}

\noindent (633Ka).   As $\tau$ is arbitrary,
$\pmb u^*\restr\Cal S'$ is the quadratic variation of $\pmb u'$.
}%end of proof of 633P

\leader{633S}{Proposition}
Suppose that $\family{t}{T}{\frak A_t}$ is right-continuous.
Let $\Cal S\subseteq\Cal T$ be a finitely full sublattice such that
$\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and has a lower bound
in $\Cal S$, and $A$ a subset of $\Cal S$.

(a) If $\pmb v=\sfamilysS{v}$ is fully adapted and whenever
$\sigma\le\tau$ in $\Cal S$ and $v_{\sigma}\ne v_{\tau}$ there is a
$\rho\in A$ such that
$\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}\ne 0$, then
$A$ $\pmb v$-separates $\Cal S$.

(b) If whenever
$\sigma\le\tau$ in $\Cal S$ and $\Bvalue{\sigma<\tau}\ne 0$ there is a
$\rho\in A$ such that
$\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}\ne 0$, then
$A$ separates $\Cal S$.

\proof{{\bf (a)(i)} If $A$ is empty then $\pmb v$ is constant
and the result is trivial, so suppose otherwise.   For $\rho\in A$ set $b_{\rho}=\Bvalue{\sigma\le\rho}\Bcap\Bvalue{\rho<\tau}$.   Then
$b_{\rho}\Bsubseteq\Bvalue{\sigma\wedge\tau\le\rho}\Bcap\Bvalue{\rho\le\tau}$ and
$b_{\rho}\in\frak A_{\rho}$ so $b_{\rho}\in\frak A_{\tau}$.   There is therefore
a $\rho'\in\Cal T$ such that $b_{\rho}\Bsubseteq\Bvalue{\rho'=\rho}$ and
$1\Bsetminus b_{\rho}\Bsubseteq\Bvalue{\rho'=\tau}$.   Because $\Cal S$ is finitely full, $\rho'\in\Cal S$.   We see that
$\sigma\wedge\tau\le\rho'\le\tau$, and that $b_{\rho}=\Bvalue{\rho'<\tau}$.

\medskip

\quad{\bf (ii)} Set $\tau'=\inf A$.   By hypothesis, $\tau'\in\Cal S$, and
$\sigma\wedge\tau\le\tau'\le\tau$.   By 632C(a-ii),
$\Bvalue{\tau'<\tau}=\sup_{\rho\in A}\Bvalue{\rho'<\tau}
=\sup_{\rho\in A}b_{\rho}$.

\Quer\ If $v_{\tau'}\ne v_{\sigma\wedge\tau}$, there is a $\rho\in A$ such that
$c=\Bvalue{\sigma\wedge\tau\le\rho}\Bcap\Bvalue{\rho<\tau'}$ is non-zero.
As $\tau'\le\tau$,
$c\Bsubseteq b_{\rho}\Bsubseteq\Bvalue{\rho=\rho'}
\Bsubseteq\Bvalue{\tau'\le\rho}$;  but this is impossible.\ \Bang

\medskip

\quad{\bf (iii)} Thus $v_{\tau'}=v_{\sigma\wedge\tau}$ and

$$\eqalignno{\Bvalue{v_{\sigma}\ne v_{\tau}}\Bcap\Bvalue{\sigma<\tau}&
\Bsubseteq\Bvalue{v_{\sigma\wedge\tau}\ne v_{\tau}}
\Bsubseteq\Bvalue{v_{\tau'}\ne v_{\tau}}\cr&
\Bsubseteq{\tau'<\tau}
=\sup_{\rho\in A}b_{\rho}.\cr}$$

As $\sigma$  and $\tau$ are arbitrary, $A$ separates $\Cal S$.

\medskip

{\bf (b)} This now follows from 633Ce.
}%end of proof of 633S

%633E
\leader{633Q}{Continuous \dvrocolon{time}}\cmmnt{ Using
some of the ideas above, we can
build stochastic integration structures in which no time is isolated on the
right.

\medskip

\noindent}{\bf Proposition}
Let \Asissix\ be a right-continuous stochastic integration structure.   Then
there are a right-continuous stochastic integration structure
$(\frak A,\bar\mu,T\strprime,
\discretionary{}{}{}\family{r}{T\strprime}{\frak A_r},\Cal T\strprime,
\family{\rho}{\Cal T\strprime}{\frak A_{\rho}})$, based on the same probability
algebra $(\frak A,\bar\mu)$, and a lattice homomorphism
$\sigma\mapsto\sigma\strprime:\Cal T\to\Cal T\strprime$
such that

\inset{$T\strprime$ has no points isolated on the right,

$\frak A_{\sigma\strprime}=\frak A_{\sigma}$ and
$\Bvalue{\sigma\strprime<\tau\strprime}=\Bvalue{\sigma<\tau}$ for all
$\sigma$, $\tau\in\Cal T$,

%$\bigcap_{r\in T\strprime}\frak A_r=\bigcap_{t\in T}\frak A_t$

for every $\rho\in\Cal T\strprime$ there is a $\sigma\in\Cal T$ such that
$\sigma\strprime\le\rho$ and $\frak A_{\sigma}=\frak A_{\rho}$.}

\prooflet{
\medskip

\noindent{\bf Construction (a)} It will be enough to deal with the case in which
$T$ is disjoint from $T\strprime=T\times\coint{0,1}$.   Give $T\strprime$ its
lexicographic ordering.   Then $T\strprime$ is
a non-empty totally ordered set with no points isolated on the right.
For $t\in T$ and $\alpha\in\coint{0,1}$ set
$\frak A_{(t,\alpha)}=\frak A_t$.   Then
$\family{r}{T\strprime}{\frak A_r}$ is a right-continuous filtration of closed
subalgebras.   Let
$(\frak A,\bar\mu,T\strprime,
\discretionary{}{}{}\family{r}{T\strprime}{\frak A_r},\Cal T\strprime,
\family{\rho}{\Cal T\strprime}{\frak A_{\rho}})$ be the corresponding stochastic
integration structure.   Note that $\Cal T\subseteq\frak A^T$ and
and $\Cal T\strprime\subseteq\frak A^{T\strprime}$ are disjoint.

\medskip

{\bf (b)} For $\sigma\in\Cal T$,
define $\sigma\strprime\in\Cal T\strprime$ by setting

\Centerline{$\Bvalue{\sigma\strprime>(t,\alpha)}=\Bvalue{\sigma>t}$}

\noindent whenever $t\in T$ and $\alpha\in\coint{0,1}$.   Then
$\sigma\mapsto\sigma\strprime:\Cal T\to\Cal T\strprime$
is a lattice homomorphism,
and $(\min\Cal T)\strprime=\min\Cal T\strprime$,
$(\max\Cal T)\strprime=\max\Cal T\strprime$ (use the formulae
in 611C).   If $\sigma$, $\tau\in\Cal T$,

$$\eqalign{\Bvalue{\sigma\strprime<\tau\strprime}&
=\sup_{t\in T,\alpha\in\coint{0,1}}
   \Bvalue{\tau\strprime>(t,\alpha)}\Bsetminus\Bvalue{\sigma\strprime>(t,\alpha)}\cr&
=\sup_{t\in T}
   \Bvalue{\tau>t}\Bsetminus\Bvalue{\sigma>t}
=\Bvalue{\sigma<\tau}.\cr}$$

\noindent Consequently $\Bvalue{\sigma\strprime\le\tau\strprime}=\Bvalue{\sigma\le\tau}$ and
$\Bvalue{\sigma\strprime=\tau\strprime}=\Bvalue{\sigma=\tau}$ for all $\sigma$,
$\tau\in\Cal T$.   Next, for $\sigma\in\Cal T$,

$$\eqalign{\frak A_{\sigma\strprime}&
=\{a:a\Bsetminus\Bvalue{\sigma\strprime>(t,\alpha)}\in\frak A_{(t,\alpha)}
   \text{ whenever }t\in T\text{ and }\alpha\in\coint{0,1}\}\cr&
=\{a:a\Bsetminus\Bvalue{\sigma>t}\in\frak A_t
   \text{ whenever }t\in T\}
=\frak A_{\sigma}.\cr}$$

\noindent If $t\in T$ and $\check t\in\Cal T$ is the
constant stopping time at $t$, and $(s,\beta)\in T\strprime$, then

$$\eqalign{\Bvalue{\check t\strprime>(s,\beta)}&
=\Bvalue{\check t>s}\cr&
=1\text{ if }s<t\text{, that is, if }(s,\beta)<(t,0),\cr&
=0\text{ if }s\ge t\text{, that is, if }(t,0)\le(s,\beta).\cr}$$

\noindent So $\check t\strprime\in\Cal T\strprime$ is the constant
stopping time at $(t,0)$.

\medskip

{\bf (c)} Consider the set

\Centerline{$\Cal S
=\{\rho:\rho\in\Cal T\strprime$ and there is a $\sigma\in\Cal T$ such that
$\sigma\strprime\le\rho$ and $\frak A_{\sigma}=\frak A_{\rho}\}$.}

\medskip

\quad{\bf (i)} The constant stopping time $\check r$ belongs to $\Cal S$ for
every $r\in T\strprime$.   \Prf\ If $r=(t,\alpha)$, then
$r\ge(t,0)$ so $\check r\ge(t,0)\varCspcheck=\check t\strprime$, while
$\frak A_{\check t}=\frak A_t=\frak A_{(t,\alpha)}=\frak A_r$.\ \QeD\   And of
course $\max\Cal T\strprime=(\max\Cal T)\strprime$ belongs to $\Cal S$ (see
611Xb).

\medskip

\quad{\bf (ii)} $\Cal S$ is finitely full.   \Prf\ If $\rho\in\Cal T\strprime$
and there is a finite set $J\subseteq\Cal S$ such that
$\sup_{\rho'\in J}\Bvalue{\rho'=\rho}=1$, we may take it that $J$ is a
sublattice of $\Cal S$.   Taking $(\rho_0,\ldots,\rho_n)$ linearly generating
the $J$-cells, we have $\rho_0\le\ldots\le\rho_n$ and
$\sup_{i\le n}a_i=1$ where $a_i=\Bvalue{\rho_i=\rho}$ for $i\le n$.
Set
$b_i=a_i\Bsetminus\sup_{j<i}a_j$ for $i\le n$;  then
$b_i\in\frak A_{\rho_i}$ and $b_i\Bsubseteq\Bvalue{\rho_i=\rho}$ for each $i$,
while $\langle b_i\rangle_{i\le n}$ is a partition of unity in $\frak A$.

For each $i\le n$, let $\sigma_i\in\Cal T$ be such that
$\sigma_i\strprime\le\rho_i$ and $\frak A_{\sigma_i}=\frak A_{\rho_i}$.
As $b_i\in\frak A_{\sigma_i}$ for each $i$, there is a $\sigma\in\Cal T$ such
that $b_i\Bsubseteq\Bvalue{\sigma=\sigma_i}$ for each $i$ (611I yet again).
Now

\Centerline{$b_i
\Bsubseteq\Bvalue{\sigma=\sigma_i}\Bcap\Bvalue{\rho_i=\rho}
=\Bvalue{\sigma\strprime=\sigma_i\strprime}
   \Bcap\Bvalue{\sigma_i\strprime\le\rho_i}
   \Bcap\Bvalue{\rho_i=\rho}
\Bsubseteq\Bvalue{\sigma\strprime\le\rho}$}

\noindent for each $i$, so $\sigma\strprime\le\rho$.   Thus $\rho\in\Cal S$;  as
$\rho$ is arbitrary, $\Cal S$ is finitely full.\ \Qed

\medskip

\quad{\bf (iii)} $\inf A\in\Cal S$ for every non-empty $A\subseteq\Cal S$.
\Prf\ Set

\Centerline{$B=\{\sigma:\sigma\in\Cal T$ and there is a $\rho\in A$ such that
$\frak A_{\sigma}=\frak A_{\rho}\}$,
\quad$\tau=\inf B$.}

\noindent Then for every $\rho\in A$ there is a $\sigma\in B$ such that
$\sigma\strprime\le\rho$ and $\frak A_{\sigma}=\frak A_{\rho}$, in which case
$\tau\strprime\le\sigma\strprime\le\rho$.   Accordingly
$\tau\strprime\le\inf A$.   On the other hand, because
$\family{t}{T}{\frak A_t}$ and $\family{r}{T\strprime}{\frak A_r}$
are both right-continuous, 632C(a-iii) tells us that

\Centerline{$\frak A_{\tau}=\bigcap_{\sigma\in B}\frak A_{\sigma}
\supseteq\bigcap_{\rho\in A}\frak A_{\rho}=\frak A_{\inf A}
\supseteq\frak A_{\tau\strprime}=\frak A_{\tau}$}

\noindent as observed in (b).
So $\tau$ witnesses that $\inf A\in\Cal S$.\ \Qed

\medskip

\quad{\bf (iv)} Consequently $\Cal S=\Cal T\strprime$.   \Prf\ By (i),
$\Cal S$ separates $\Cal T\strprime$ and is cofinal with $\Cal T\strprime$;
by (ii), $\Cal S$ is equal to its finitely-covered envelope.   So 633E tells us
that if $\rho\in\Cal T\strprime$ there is a non-empty $A\subseteq\Cal S$ such
that $\inf A=\rho$ and $\rho\in\Cal S$.\ \Qed

Thus we have a suitable stochastic integration structure
$(\frak A,\bar\mu,T\strprime,
\discretionary{}{}{}\family{r}{T\strprime}{\frak A_r},\Cal T\strprime,
\family{\rho}{\Cal T\strprime}{\frak A_{\rho}})$.
}%end of proof of 633Q

\leader{633R}{Theorem} If \Asissixt\ is a real-time stochastic
integration structure and $\pmb u\in\Mlns(\Cal T_f)$, then $\pmb u$ can be
represented by a process with \cadlag\ sample paths as in Theorem 631D.

%should be 631Q?  no, relies on 633D-633F

\proof{ Express $\pmb u$ as $\family{\sigma}{\Cal T_f}{u_{\sigma}}$.

\medskip

{\bf (a)} By 321J, we can suppose that the probability algebra
$(\frak A,\bar\mu)$ is the measure algebra of a complete probability space
$(\tilde\Omega,\tilde\Sigma,\tilde\mu)$.   Writing
$\tilde\Sigma_t=\{E:E\in\tilde\Sigma$, $E^{\ssbullet}\in\frak A_t\}$ for
$t\ge 0$, we have a filtration $\langle\tilde\Sigma_t\rangle_{t\ge 0}$ of
$\sigma$-subalgebras of $\tilde\Sigma$, all containing every
$\tilde\mu$-negligible set.   Write $Q$ for $\Bbb Q\cap\coint{0,\infty}$, and
for $q\in Q$ choose a $\tilde\Sigma$-measurable function
$\tilde X_q:\tilde\Omega\to\Bbb R$ such that its equivalence class
$\tilde X_q^{\ssbullet}$ in $L^0(\tilde\mu)\cong L^0(\frak A)$ is equal to
$u_{\check q}$.

For each $n\in\Bbb N$, let
$\pmb v_n=\family{\sigma}{\Cal T\wedge\check n}{v_{n\sigma}}$ be a simple
process with domain
$\Cal T\wedge\check n$ such that $\theta(z_n)\le 4^{-n}$, where
$z_n=\sup_{\sigma\in\Cal T\wedge\check n}|u_{\sigma}-v_{n\sigma}|$,
and let $(\tau_{n0},\ldots,\tau_{nm_n})$ be a breakpoint sequence for
$\pmb v_n$ starting from $\tau_{n0}=\check 0$ and ending with
$\tau_{nm_n}=\check n$.   Choose stopping times
$\tilde g_{n0},\ldots,\tilde g_{nm_n}$, adapted to
$\langle\tilde\Sigma_t\rangle_{t\ge 0}$, representing
$\tau_{n0},\ldots,\tau_{nm_n}$ respectively in the way described in 612Ha, so
that
$\{\omega:\tilde g_{ni}(\omega)>t\}^{\ssbullet}=\Bvalue{\tau_{ni}>t}$ for every
$t\ge 0$ and $i\le m_n$.   Adjusting the $\tilde g_{ni}$ on a negligible set if
necessary, we can arrange that $0=\tilde g_{n0}(\omega)\le\tilde g_{n1}(\omega)
\le\ldots\le\tilde g_{nm_n}(\omega)=n$ for every $\omega\in\tilde\Omega$.
Finally, choose a $\tilde\Sigma$-measurable function
$\tilde h_n:\tilde\Omega\to\coint{0,\infty}$ such that
$\tilde h_n^{\ssbullet}=z_n$ in $L^0(\frak A)$.

\medskip

{\bf (b)} As $\theta(z_n)\le 4^{-n}$,
$\tilde\mu\{\omega:\omega\in\tilde\Omega$, $\tilde h_n(\omega)\ge 2^{-n}\}
\le 2^{-n}$ for every $n\in\Bbb N$;  consequently

\Centerline{$F=\{\omega:\omega\in\tilde\Omega$,
$\limsup_{n\to\infty}\tilde h_n(\omega)>0\}
\subseteq\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}
    \{\omega:\tilde h_m(\omega)\ge 2^{-m}\}$}

\noindent is negligible.   Next, for $n\in\Bbb N$, $i<m_n$ and $q$, $q'\in Q$
such that $q\le q'\le n$, set

\Centerline{$F_{niqq'}
=\{\omega:\omega\in\tilde\Omega$,
$\tilde g_{ni}{\omega)}\le q\le q'<\tilde g_{n,i+1}(\omega)$,
$|\tilde X_{q'}(\omega)-\tilde X_q(\omega)|>2\tilde h_n(\omega)\}$;}

\noindent then

$$\eqalignno{F_{niqq'}^{\ssbullet}&
=\Bvalue{\tau_{ni}\le\check q}\Bcap\Bvalue{\check q'<\tau_{n,i+1}}
   \Bcap\Bvalue{|u_{\check q'}-u_{\check q}|>2z_n}\cr&
\Bsubseteq\Bvalue{|u_{\check q'}-v_{n\check q'}|>z_n}
      \Bcup\Bvalue{|u_{\check q}-v_{n\check q}|>z_n}
\Displaycause{because
$\Bvalue{\tau_{ni}\le\check q}\Bcap\Bvalue{\check q'<\tau_{n,i+1}}
\Bsubseteq\Bvalue{v_{n\check q'}=v_{n\tau_{ni}}}
   \Bcap\Bvalue{v_{n\check q}=v_{n\tau_{ni}}}$}
=0,\cr}$$

\noindent so $F_{niqq'}$ is negligible.   Set

\Centerline{$\Omega
=\Omega'\setminus(F\cup\bigcup_{n\in\Bbb N,i<m_n,q,q'\in Q,q\le q'\le n}
    F_{niqq'}$;}

\noindent then $\Omega$ is a conegligible subset of $\tilde\Omega$.

\medskip

{\bf (c)} Fix $\omega\in\Omega$ for the moment.

\medskip

\quad{\bf (i)} For every $t\ge 0$,
$\lim_{q\downarrow Q\cap\coint{t,\infty}}\tilde X_q(\omega)$ is defined in
$\Bbb R$.
\Prf\ If $\epsilon>0$, then (because $\omega\notin F$) there is an
$n>t$ such that $\tilde h_n(\omega)\le\bover12\epsilon$.   Since
$0=\tilde g_{n0}(\omega)\le\ldots\le\tilde g_{nm_n}(\omega)=n$, there is an
$i<m_n$ such that $\tilde g_{ni}(\omega)\le t<\tilde g_{n,i+1}(\omega)$.
If now $q$, $q'\in Q$ and $t\le q\le q'<\tilde g_{n,i+1}(\omega)$,
we know that $\omega\notin F_{niqq'}$ so
$|\tilde X_{q'}(\omega)-\tilde X_q(\omega)|\le 2\tilde h_n(\omega)\le\epsilon$.
As $\epsilon$ is arbitrary (and $Q$ is dense in $\coint{0,\infty}$)
$\lim_{q\downarrow Q\cap\coint{t,\infty}}\tilde X_q(\omega)$ is defined.\ \Qed

\medskip

\quad{\bf (ii)} Set
$X_t(\omega)=\lim_{q\downarrow Q\cap\coint{t,\infty}}\tilde X_q(\omega)$ for
$t\ge 0$.   Note that $X_q(\omega)=\tilde X_q(\omega)$ for
$q\in Q$.

\medskip

\quad{\bf (iii)} $\langle X_t(\omega)\rangle_{t\ge 0}$ is \cadlag.
\Prf\ We just have to rerun the argument for (i) above.   Given
$t\ge 0$ and $\epsilon>0$, there is a $t'=\tilde g_{n,i+1}(\omega)$ such that
$t<t'$ and $|\tilde X_{q'}(\omega)-\tilde X_q(\omega)|\le\epsilon$ whenever
$q$, $q'\in Q$ and $t\le q\le q'<t'$.   It follows at once that
$|X_s(\omega)-X_t(\omega)|\le\epsilon$ whenever $t\le s<t'$.   As $t$ and
$\epsilon$ are arbitrary, $\langle X_t(\omega)\rangle_{t\ge 0}$ is \cadlag.\
\Qed

\medskip

{\bf (d)} Because $\Omega$ is conegligible in $\tilde\Omega$ and $\mu$ is
complete, the subspace $\sigma$-algebra $\Sigma=\{E\cap\Omega:E\in\Sigma\}$
is just $\Sigma\cap\Cal P\Omega$, and the subspace measure
$\mu=\tilde\mu\restr\Sigma$ is a probability measure, with measure algebra
isomorphic to $\frak A$ (322Jb).   We can identify $L^0(\mu)$ with
$L^0(\tilde\mu)$ (indeed, in the formulations of Chapter 24
$\eusm L^0(\mu)$ is actually a subset of $\eusm L^0(\tilde\mu)$), and as every
$\tilde\Sigma_t$ contains $\tilde\Omega\setminus\Omega$, we have a filtration
$\langle\Sigma_t\rangle_{t\ge 0}$ of $\sigma$-subalgebras of $\Sigma$ given by
the formula

\Centerline{$\Sigma_t=\{E\cap\Omega:E\in\Sigma_t\}=\Sigma_t\cap\Cal P\Omega$}

\noindent for $t\ge 0$.   If $\tilde g:\tilde\Omega\to\coint{0,\infty}$ is a
stopping time adapted to $\langle\tilde\Sigma_t\rangle_{t\ge 0}$, then
$\tilde g\restr\Omega$ is a
stopping time adapted to $\langle\Sigma_t\rangle_{t\ge 0}$;  and conversely, if
$g:\Omega\to\coint{0,\infty}$ is a stopping time adapted to
$\langle\Sigma_t\rangle_{t\ge 0}$, then any extension of $g$ to a real-valued
function defined on $\tilde\Omega$ is a stopping time adapted to
$\langle\tilde\Sigma_t\rangle_{t\ge 0}$.   What this means is that
\Asisfourt\ is represented by
$(\Omega,\Sigma,\mu,\langle\Sigma_t\rangle_{t\ge 0})$ just as well as by
$(\tilde\Omega,\tilde\Sigma,\tilde\mu,\langle\tilde\Sigma_t\rangle_{t\ge 0})$.

\medskip

{\bf (e)} Now fix $t\ge 0$ for a bit.

\medskip

\quad{\bf (i)} The first thing to note is that if $t<n\in\Bbb N$ then
$v_{n\check t}=\lim_{q\downarrow Q\cap\coint{t,\infty}}v_{n\check q}$ in
$L^0$.   \Prf\ For any $s<n$,

$$\eqalign{v_{n\check s}&
=\sum_{i=0}^{m_n-1}v_{n\tau_{ni}}
  \times\chi(\Bvalue{\tau_{ni}\le\check s}\Bcap\Bvalue{\check s<\tau_{n,i+1}})
  \cr&
=\sum_{i=0}^{m_n-1}v_{n\tau_{ni}}
  \times(\chi\Bvalue{\tau_{n,i+1}>s}-\chi\Bvalue{\tau_{ni}>s}).\cr}$$

\noindent But

\Centerline{$\lim_{q\downarrow Q\cap\coint{t,\infty}}\chi\Bvalue{\tau_{ni}>q}
=\chi\Bvalue{\tau_{ni}>t}$}

\noindent for all $n\in\Bbb N$ and $i\le m_n$, just because
$\Bvalue{\tau{ni}>t}=\sup_{s>t}\Bvalue{\tau_{ni}>s}$ (611A(b-i)).
So $v_{n\check t}=\lim_{q\downarrow Q\cap\coint{t,\infty}}v_{n\check q}$. \Qed

\medskip

\quad{\bf (ii)} It follows that
$u_{\check t}=\lim_{q\downarrow Q\cap\coint{t,\infty}}u_{\check q}$.
\Prf\ If $n>t$ then we have

\Centerline{
$\limsup_{q\downarrow Q\cap\coint{t,\infty}}|u_{\check q}-u_{\check t}|
\le 2z_n+\limsup_{q\downarrow Q\cap\coint{t,\infty}}
    |v_{n\check q}-v_{n\check t}|
=2z_n$,}

\noindent so

\Centerline{
$\limsup_{q\downarrow Q\cap\coint{t,\infty}}|u_{\check q}-u_{\check t}|
\le 2\inf_{n>t}z_n=0$}

\noindent and
$u_{\check t}=\lim_{q\downarrow Q\cap\coint{t,\infty}}u_{\check q}$.\ \Qed

\medskip

\quad{\bf (iii)} Since $Q\cap\coint{t,\infty}$ is countable,
$u_{\check q}=\tilde X_q^{\ssbullet}$ for every $q\in Q$ and
$X_t\eae\lim_{q\downarrow Q\cap\coint{t,\infty}}\tilde X_q$,

\Centerline{$X_t^{\ssbullet}
=\lim_{q\downarrow Q\cap\coint{t,\infty}}\tilde X_q^{\ssbullet}
=\lim_{q\downarrow Q\cap\coint{t,\infty}}u_{\check q}
=u_{\check t}$.}

\noindent As $u_{\check t}\in L^0(\frak A_t)$ and
$\Sigma_t=\{E:E\subseteq\Omega$, $E^{\ssbullet}\in\frak A_t\}$,
$X_t$ is $\Sigma_t$-measurable.

\medskip

{\bf (e)} Thus $\Omega$, $\Sigma$, $\mu$, $\langle\Sigma_t\rangle_{t\ge 0}$ and
$\langle X_t\rangle_{t\ge 0})$ satisfy the conditions of 631D, and provide a
locally near-simple process
$\pmb x=\langle x_{\sigma}\rangle_{\sigma\in\Cal T_f}$ in the corresponding
stochastic integration structure, which we are identifying with
\Asissixt.   Now $\pmb u$ and $\pmb x$ are locally near-simple processes with
domain $\Cal T_f$ and $u_{\check t}=X_t^{\ssbullet}=x_{\check t}$ for every
$t\ge 0$.   As $\{\check t:t\ge 0\}$ separates $\Cal T_f$ (633Da) and
$\inf_{t\ge 0}\Bvalue{\check t<\tau}=\inf_{t\ge 0}\Bvalue{\tau>t}=0$ for every
$\tau\in\Cal T_f$, 633F tells us that $\pmb u=\pmb x$.   So we have a
representation of $\pmb u$ of the kind we seek.
}%end of proof of 633R

%\leader{633T}{}
%\leader{633U}{}
%\leader{633V}{}
%\leader{633W}{}

\exercises{\leader{633X}{Basic exercises (a)}
%\spheader 633Xa
Suppose that $\family{t}{T}{\frak A_t}$
is right-continuous.   Let $\Cal S$ be a sublattice of $\Cal T$ which
separates its order-convex hull $\Cal S^{\sim}$,
$\pmb u$ a near-simple process with
domain $\Cal S$, and $\pmb v$ a martingale with domain $\Cal S$.
Show that $\iivu$ is a local martingale.
\Hint{reduce to the case in which $\Cal S$ has a least member.
Let $\pmb v^{\sim}$ be the
martingale with domain $\Cal S^{\sim}$ extending $\pmb v$, and
$\pmb u^{\sim}$ the near-simple process with domain $\Cal S^{\sim}$
extending $\pmb u$ (use 631Mb).   Show that
$\iivu=ii_{\pmb v^{\sim}}(\pmb u^{\sim})\restr\Cal S$, and apply
623O, 632Ib and 622Dc.}
%633E

\spheader 633Xb
Suppose that $T=\coint{0,\infty}$, $(\frak A,\bar\mu)$ is
the measure algebra of Lebesgue measure $\mu$ on $[0,1]$, and
$\frak A_t=\frak A$ for every $t\in T$.   Define $X_t(\omega)$,
for $t\in T$ and $\omega\in[0,1]$,
by setting $X_t(\omega)=\bover1{\omega-t}$ if
$\omega\ne t$, $0$ if $\omega=t$.   Let $\check{T}\subseteq\Cal T$
be the set of constant stopping times,
so that $\Cal S=\check T\cup\{\max\Cal T\}$ is a cofinal and coinitial
sublattice which separates $\Cal T$.   Define
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ by setting
$u_{\max\Cal T}=0$
and $u_{\check{t}}=X_t^{\ssbullet}$ for $t\in T$.
(i) Show that $u_{\inf A}=\lim_{\sigma\downarrow A}u_{\sigma}$ and
$\lim_{\sigma\uparrow A}u_{\sigma}=u_{\sup A}$ for
every non-empty $A\subseteq\Cal S$.
(ii) Show that if we take $\Sigma_t=\dom\mu$ for every $t\in T$, then
$\family{t}{T}{X_t}$ is progressively measurable, so that we
have a natural extension of $\pmb u$ to a fully adapted process defined on
$\Cal T$, %612H
while \Asisfour\ is right-continuous.
(iii) Show that $\pmb u$ is not locally order-bounded.
%633P 631G 632F

\spheader 633Xc Show that every subset of $\Cal T$ separates its covered
envelope.
%633C out of order query

\leader{633Y}{Further exercises (a)}
%\spheader 633Ya
Let \Asissix\ be a stochastic integration structure.
Write $\Tni$ for
$\{t:t\in T$ is not isolated on the right in $T\}$.   Take an
injective function $t\mapsto t^+$ from $T$ to a set disjoint from $T$.
Set $T\strplus=T\cup\{t^+:t\in\Tni\}$.
On $T\strplus$ take the relation

$$\eqalign{&\{(s,t):s,\,t\in T,\,s\le t\}
  \cup\{(s,t^+):s\in T,\,t\in\Tni,\,s\le t\}\cr&
\mskip50mu
  \cup\{(s^+,t):s\in\Tni,\,t\in T,\,s<t\}
  \cup\{(s^+,t^+):s,\,t\in\Tni,\,s\le t\}.\cr}$$

\noindent (i) Show that $T\strplus$ is totally ordered.
(ii) Set
$\frak A_{t^+}=\bigcap_{s\in T,s>t}\frak A_s$ for $t\in T_{ni}$;  show that
$\family{r}{T\strplus}{\frak A_r}$ is a
right-continuous filtration of closed subalgebras
of $\frak A$.   Let $(\frak A,\bar\mu,T\strplus,
\family{r}{T\strplus}{\frak A_{r}},\discretionary{}{}{}\Cal T\strplus,
\family{\rho}{\Cal T\strplus}{\frak A_{\rho}})$ be the corresponding
stochastic integration structure.
(iii) Show that we have a lattice homomorphism
$\sigma\mapsto\sigma\strplus:\Cal T\to\Cal T\strplus$ defined by saying that
$\Bvalue{\sigma\strplus>t}=\Bvalue{\sigma>t}$ for $t\in T$ and
$\Bvalue{\sigma\strplus>t^+}=\Bvalue{\sigma>t}$ for $t\in T_{ni}$.
(iv) Show that $\frak A_{\sigma\strplus}=\frak A_{\sigma}$ and
$\Bvalue{\sigma\strplus<\tau\strplus}=\Bvalue{\sigma<\tau}$ for all
$\sigma$, $\tau\in\Cal T$.
(v) Show that $\{\sigma\strplus:\sigma\in\Cal T\}$ separates $\Cal T\strplus$.
(vi) Suppose that $\pmb w=\family{\rho}{\Cal T\strplus}{w_{\rho}}$
is a fully adapted
process.   Show that $\pmb u=\family{\sigma}{\Cal T}{w_{\sigma\strplus}}$ is
fully adapted, that $\pmb u$
is a martingale if $\pmb w$ is a martingale, and that
$\Osclln(\pmb u)\le 2\Osclln(\pmb w)$ if $\pmb w$ is near-simple.
%633Q mt63bits

\spheader 633Yb
Let $(\frak A,\bar\mu,T\strplus,\family{r}{T\strplus}{\frak A_r},
\Cal T\strplus,
\family{\rho}{\Cal T\strplus}{\frak A_{\rho}})$ be the
right-continuous version of
\Asissix, as in 633Ya.   Let $\Cal S$ be an order-convex sublattice of $\Cal T$ and
$\Cal R\subseteq\Cal T\strplus$ the order-convex hull of
$\Cal R'=\{\sigma\strplus:\sigma\in\Cal S\}$.   (i) Show that
$\Cal R'$ separates $\Cal R$.  %633Qg 633D(b-i)
(ii) Suppose that $\pmb u=\sfamilysS{u}$ is a near-simple process.   Show that
there is a near-simple process
$\pmb v=\family{\rho}{\Cal R}{v_{\rho}}$ such
that $v_{\sigma\strplus}=u_{\sigma}$ for every $\sigma\in\Cal S$.
%633Ya 633Q 631M

\spheader 633Yc Suppose that \Asissixt\ is a real-time stochastic
integration structure and $\pmb u$ is a locally jump-free process with domain
$\Cal T_f$.   Show that $\pmb u$ can be
represented by a process with continuous sample paths as in 618H.
%633R follow same proof?

\spheader 633Yd Suppose that $\family{t}{T}{\frak A_t}$ is right-continuous.
Let $\Cal S\subseteq\Cal T$ be a finitely full sublattice such that
$\inf A\in\Cal S$ whenever $A\subseteq\Cal S$ is non-empty and has a lower bound
in $\Cal S$, and $\Cal S'$ a sublattice of $\Cal S$ such that whenever
$\tau$, $\tau'\in\Cal S$ and $\Bvalue{\tau<\tau'}\ne 0$ there is a
$\sigma\in\Cal S'$ such that
$\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma<\tau'}\ne 0$.   Show that
$\Cal S'$ separates $\Cal S$.
%633E
}%end of exercises

\endnotes{
\Notesheader{633} This volume began with two sections devoted to
the algebra of lattices of stopping times and fully adapted processes.   It
would all have been much easier if we could have worked throughout with
totally ordered sets of stopping times.   Even for integrals of the form
$\int_{[\check{s},\check{t}]}$, however, this is inadequate (632Yg).
The problem is not so much in the calculation of the integral
as in its definition.   In 633K-633L we see that if an integral
$\int_{\Cal S}\pmb u\,d\pmb v$ is
defined on the rules of \S613, and subject to
mild conditions on the processes $\pmb u$ and $\pmb v$, then we can hope
to find a totally ordered sublattice $\Cal S'$ of $\Cal S$ such that
$\int_{\Cal S'}\pmb u\,d\pmb v=\int_{\Cal S}\pmb u\,d\pmb v$.
But the nearest thing to a converse result, in 633Kb, looks at finitely full
sublattices $\Cal S'$, which will not be totally ordered except in trivial
cases.

It is true that we can reduce a Riemann
sum $S_I$ on an arbitrary finite sublattice $I$ to the corresponding sum
on a totally ordered sublattice $I_0$ of $I$ (613Ec), and in
the calculations so far
this is what I have done more often than not.   But whenever
we want to look at sums $S_I$, $S_J$ on two different sublattices, this
method becomes problematic;  there is no reason to suppose that there will
be compatible totally ordered sublattices $I_0\subseteq I$ and
$J_0\subseteq J$ which will be suitable.   Even when $J\subseteq I$ we have
to do some work (633H).   What we want is a totally ordered sublattice of
$\Cal T$ which will deal with all integrals of interest simultaneously;
and even in the most favourable case (633L), we should have to restrict
ourselves to integrals between constant stopping times, which are
inadequate for many of the most important applications of the theory.

I have given a number of results showing that classical stochastic processes,
based on probability spaces and filtrations of $\sigma$-algebras, can be
associated with processes dealt with in this volume (612H, 615P, 614R, 618H,
632L).   I have not spent much time on converses. looking at cases in which
a process defined by the properties considered here can be represented by one of
the classical expressions.   But in 633R we have a comfortingly
straightforward result which seems entitled to a couple of pages.
}%end of notes

\leaveitout{
\medskip

\quad{\bf (iii)} Observe also that $T$ is dense in $T\strplus$
for the order-topology of $T\strplus$.
\prooflet{\Prf\ If $r<r'<r''$ in $T\strplus$, then either $r'\in T$ and
surely $T\cap\ooint{r,r''}\ne\emptyset$, or $r'=t^+$ for some $t\in\Tni$.
In this case, $r''$ is either $s_0$ or $s_0^+$ for some $s_0\in T$ such
that $t<s_0$;  but as $t\in\Tni$, there is an $s\in T$ such that
$t<s<s_0$, in which case $r<s<r''$ and again $T$ meets $\ooint{r,r''}$.\ \Qed}

}%end of leaveitout

\frfilename{mt634.tex}
\versiondate{25.1.22/28.9.23}
\copyrightdate{2012}

\def\chaptername{Structural alterations}
\def\sectionname{Changing the algebra}

\newsection{634}

If $(\frak A,\bar\mu)$ is a probability algebra with a filtration
$\family{t}{T}{\frak A_t}$ and $\frak B$ is a closed subalgebra of
$\frak A$, then we have a probability algebra
$(\frak B,\bar\mu\restrp\frak B)$ with a filtration
$\family{t}{T}{\frak B\cap\frak A_t}$.   In this section I examine
elementary connexions between stochastic calculus in the two structures,
with notes on lattices of stopping times (634C) and stochastic processes
(634E).   The case in which
$\frak B$ and $\frak A_t$ are relatively independent over their
intersection for every $t$ is particularly important
(634F-634I). %634F 634G 634H 634I
I end the section with a product construction adapted to the
representation of families of independent stochastic processes
(634K-634M) %634K 634L 634M
and a worked example on independent Poisson processes (634N).

\leader{634A}{Notation} \cmmnt{
$\Bbb A=(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t},\Cal T_{\Bbb A},
\family{\tau}{\Cal T_{\Bbb A}}{\frak A_{\tau}})$ will be a
stochastic integration structure,
and $\theta=\theta_{\bar\mu}$ the usual functional defining the topology of
convergence in measure on $L^0(\frak A)$\cmmnt{, as in 613Ba}.
If $\Cal S$ is a sublattice of $\Cal T_{\Bbb A}$,
then $\Cal I(\Cal S)$ is the set of finite sublattices of $\Cal S$.

If $\frak B$ is a closed subalgebra of $\frak A$,
$P_{\frak B}:L^1_{\bar\mu}\to L^1_{\bar\mu}\cap L^0(\frak B)$ will be the
conditional expectation associated with $\frak B$.
For $\tau\in\Cal T_{\Bbb A}$, I will write $P_{\tau}$ for $P_{\frak A_{\tau}}$.

Recall that if $\frak B$ is an order-closed subalgebra of $\frak A$, we can
identify $L^0(\frak B)$ with an order-closed $f$-subalgebra of
$L^0(\frak A)$\cmmnt{ (612Ae)}.   In particular, a subset of
$L^0(\frak B)$ is order-bounded in $L^0(\frak B)$ iff it is order-bounded
in $L^0(\frak A)$.   We have
$L^1(\frak B,\bar\mu\restrp\frak B)
=L^0(\frak B)\cap L^1(\frak A,\bar\mu)=L^0(\frak B)\cap L^1_{\bar\mu}$,
and the expectation $\Expn_{\bar\mu\restrp\frak B}$ on
$L^1(\frak B,\bar\mu\restrp\frak B)$ is just the restriction
of $\Expn=\Expn_{\bar\mu}$ to $L^1(\frak B,\bar\mu\restrp\frak B)$\cmmnt{
(see 365Oa\formerly{3{}65Pa}, 365Qa\formerly{3{}65Ra})}.   Consequently
$\theta_{\bar\mu\restrp\frak B}=\theta_{\bar\mu}\restr L^0(\frak B)$, and
the topology of convergence in measure on $L^0(\frak B)$ is the
subspace topology induced by the topology of convergence in measure on
$L^0(\frak A)$.   It will also be useful to note that a subset of
$L^0(\frak B)\cap L^1(\frak A,\bar\mu)$ is uniformly integrable in
$L^1(\frak B,\bar\mu\restrp\frak B)$ iff it is uniformly integrable in
$L^1_{\bar\mu}$.

}% end of comment
For a family $\family{k}{K}{\frak C_k}$ of closed subalgebras of $\frak A$,
I write $\bigvee_{k\in K}\frak C_k$ for the closed subalgebra
generated by $\bigcup_{k\in K}\frak C_k$.
Similarly, for closed subalgebras $\frak C$, $\frak C'$ I will write
$\frak C\vee\frak C'$ for the closed subalgebra generated by
$\frak C\cup\frak C'$.   \cmmnt{(Cf.\ 458Ad.)}   Note that\cmmnt{ as
$\frak A$ is
ccc (322G),} a subalgebra of $\frak A$ is closed iff it is a
$\sigma$-subalgebra\cmmnt{ (331G)}.

\leader{634B}{}\cmmnt{ I begin with a general result on morphisms
of the structures here.

\medskip

\noindent}{\bf Proposition} Let $\frak B$ be a Dedekind complete
algebra and $\family{t}{T}{\frak B_t}$ a filtration of order-closed subalgebras
of $\frak B$;  write $\Cal T_{\Bbb B}$ and
$\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}}$
for the associated lattice of stopping times and family of
closed subalgebras of $\frak B$\cmmnt{, as in \S611}.
Suppose that $\phi:\frak B\to\frak A$ is an order-continuous Boolean
homomorphism such that $\phi[\frak B_t]\subseteq\frak A_t$ for every
$t\in T$.

(a) We have a lattice homomorphism
$\hat\phi:\Cal T_{\Bbb B}\to\Cal T_{\Bbb A}$ defined by saying that
$\Bvalue{\hat\phi(\sigma)>t}=\phi\Bvalue{\sigma>t}$ for every
$\sigma\in\Cal T_{\Bbb B}$.

(b)(i) $\hat\phi(\min\Cal T_{\Bbb B})=\min\Cal T_{\Bbb A}$,
$\hat\phi(\max\Cal T_{\Bbb B})=\max\Cal T_{\Bbb A}$.
If $t\in T$ and $\check t$ is the constant stopping time at $t$
in $\Cal T_{\Bbb B}$, then
$\hat\phi(\check t)$ is the constant stopping time at $t$ in
$\Cal T_{\Bbb A}$.

\quad(ii) If $C\subseteq\Cal T_{\Bbb B}$ then
$\hat\phi(\sup C)=\sup\hat\phi[C]$.

\quad(iii) $\hat\phi[\Cal T_{\Bbb B b}]\subseteq\Cal T_{\Bbb A b}$
and $\hat\phi[\Cal T_{\Bbb Bf}]\subseteq\Cal T_{\Bbb Af}$.

(c)(i) If $\sigma$, $\sigma'\in\Cal T_{\Bbb B}$ then

\Centerline{$\Bvalue{\hat\phi(\sigma)<\hat\phi(\sigma')}
=\phi\Bvalue{\sigma<\sigma'}$,
\quad$\Bvalue{\hat\phi(\sigma)\le\hat\phi(\sigma')}
=\phi\Bvalue{\sigma\le\sigma'}$,}

\Centerline{$\Bvalue{\hat\phi(\sigma)=\hat\phi(\sigma')}
=\phi\Bvalue{\sigma=\sigma'}$.}

\quad(ii) If $\phi$ is injective, so is $\hat\phi$.

(d) $\phi[\frak B_{\sigma}]\subseteq\frak A_{\hat\phi(\sigma)}$ for every
$\sigma\in\Cal T_{\Bbb B}$.

(e) If $\family{t}{T}{\frak B_t}$ is right-continuous, then
$\hat\phi$ is order-continuous,

(f) If $T=\coint{0,\infty}$ and we define the identity processes
$\family{\sigma}{\Cal T_{\Bbb Bf}}{\iota_{\sigma}}$,
$\family{\tau}{\Cal T_{\Bbb Af}}{\iota_{\tau}}$ as in
612F, then $\iota_{\hat\phi(\sigma)}=T_{\phi}\iota_{\sigma}$ for
every $\sigma\in\Cal T_{\Bbb Bf}$, where
$T_{\phi}:L^0(\frak B)\to L^0(\frak A)$ is the $f$-algebra homomorphism
associated with $\phi$\cmmnt{ (612Af)}.

\proof{{\bf (a)(i)} If $\sigma\in\Cal T_{\Bbb B}$, then

\inset{$\phi\Bvalue{\sigma>t}\in\phi[\frak B_t]\subseteq\frak A_t$
for every $t\in T$,

if $s\le t$ then $\Bvalue{\sigma>t}\Bsubseteq\Bvalue{\sigma>s}$ so
$\phi\Bvalue{\sigma>t}\Bsubseteq\phi\Bvalue{\sigma>s}$,

if $t\in T$ is not isolated on the right then
$\Bvalue{\sigma>t}=\sup_{s>t}\Bvalue{\sigma>s}$, and because $\phi$ is
order-continuous, $\phi\Bvalue{\sigma>t}=\sup_{s>t}\phi\Bvalue{\sigma>s}$.}

\noindent Thus the function $t\mapsto\phi\Bvalue{\sigma>t}$ satisfies the
conditions of 611A(b-i), and defines a member of $\Cal T_{\Bbb A}$
which we may call $\hat\phi(\sigma)$.

\medskip

\quad{\bf (ii)} If $C\subseteq\Cal T_{\Bbb B}$ is non-empty, then

$$\eqalignno{\Bvalue{\hat\phi(\sup C)>t}
&=\phi\Bvalue{\sup C>t}
=\phi(\sup_{\sigma\in C}\Bvalue{\sigma>t})
\Displaycause{611Cb}
=\sup_{\sigma\in C}\phi\Bvalue{\sigma>t}
\Displaycause{again because $\phi$ is order-continuous}
=\sup_{\sigma\in C}\Bvalue{\hat\phi(\sigma)>t}
=\Bvalue{\sup_{\sigma\in C}\hat\phi(\sigma)>t}
\cr}$$

\noindent for every $t\in T$, so $\hat\phi(\sup C)=\sup\hat\phi[C]$.
If $\sigma$, $\sigma'\in\Cal T_{\Bbb B}$, then

$$\eqalignno{\Bvalue{\hat\phi(\sigma\wedge\sigma')>t}
&=\phi\Bvalue{\sigma\wedge\sigma'>t}
=\phi(\Bvalue{\sigma>t}\Bcap\Bvalue{\sigma'>t})
\Displaycause{611Cc}
=\phi\Bvalue{\sigma>t}\Bcap\phi\Bvalue{\sigma'>t}
\cr&=\Bvalue{\hat\phi(\sigma)>t}\Bcap\Bvalue{\hat\phi(\sigma')>t}
=\Bvalue{\hat\phi(\sigma)\wedge\hat\phi(\sigma')>t}
\cr}$$

\noindent for every $t\in T$, and
$\hat\phi(\sigma\wedge\sigma')=\hat\phi(\sigma)\wedge\hat\phi(\sigma')$.
Putting these together, we see that $\hat\phi$ is a lattice homomorphism.

\medskip

{\bf (b)(i)} Look at the descriptions of $\max\Cal T$ and $\min\Cal T$ in
611Cf, and remember that $\phi 0_{\frak B}=0_{\frak A}$ and
$\phi 1_{\frak B}=1_{\frak A}$.   Concerning constant stopping times, if
$t\in T$ then

$$\eqalign{\Bvalue{\hat\phi(\check t)>s}=\phi\Bvalue{\check t>s}
&=\phi 1_{\frak B}=1_{\frak A}\text{ if }s<t,\cr
&=\phi 0_{\frak B}=0_{\frak A}\text{ if }s\ge t,\cr}$$

\noindent so $\hat\phi(\check t)$ is the constant stopping time at $t$ in the
structure $(\frak A,\family{t}{T}{\frak A_t})$.

\medskip

\quad{\bf (ii)} This is covered by (a-ii) above.

\medskip

\quad{\bf (iii)} Because $\hat\phi$ is order-preserving and takes constant
stopping times to constant stopping times,
$\hat\phi[\Cal T_{\Bbb B b}]\subseteq\Cal T_{\Bbb A b}$.
Now if $\sigma\in\Cal T_{\Bbb Bf}$,
$\sigma=\sup_{t\in T}\sigma\wedge\check t$, so (ii) tells us that

\Centerline{$\hat\phi(\sigma)
=\sup_{t\in T}\hat\phi(\sigma)\wedge\check t$}

\noindent (now interpreting $\check t$ as a constant stopping
time in $\Cal T_{\Bbb A}$) and belongs to $\Cal T_{\Bbb Af}$.

\medskip

{\bf (c)(i)} By the definition in 611D

$$\eqalign{\phi\Bvalue{\sigma<\sigma'}
&=\phi(\sup_{t\in T}\Bvalue{\sigma'>t}\Bsetminus\Bvalue{\sigma>t})
=\sup_{t\in T}(\phi\Bvalue{\sigma'>t}\Bsetminus\phi\Bvalue{\sigma>t})
\cr&=\sup_{t\in T}(\Bvalue{\hat\phi(\sigma')>t}
   \Bsetminus\Bvalue{\hat\phi(\sigma)>t})
=\Bvalue{\hat\phi(\sigma)<\hat\phi(\sigma')}.
\cr}$$

\noindent The other equalities follow at once.

\medskip

\quad{\bf (ii)} If $\phi$ is injective and
$\sigma$, $\sigma'\in\Cal T_{\Bbb B}$ are different,
then $\Bvalue{\sigma=\sigma'}\ne 1_{\frak B}$ so

\Centerline{$\Bvalue{\hat\phi(\sigma)=\hat\phi(\sigma')}
=\phi\Bvalue{\sigma=\sigma'}\ne\phi(1_{\frak B})=1_{\frak A}$}

\noindent and $\hat\phi(\sigma)\ne\hat\phi(\sigma')$.

\medskip

{\bf (d)} Suppose that $b\in\frak B_{\sigma}$.   Then for any $t\in T$,
$b\Bsetminus\Bvalue{\sigma>t}\in\frak B_t$, so

\Centerline{$\phi b\Bsetminus\Bvalue{\hat\phi(\sigma)>t}
=\phi b\Bsetminus\phi\Bvalue{\sigma>t}
=\phi(b\Bsetminus\Bvalue{\sigma>t})
\in\phi[\frak B_t]\subseteq\frak A_t$.}

\noindent Thus $\phi b\in\frak A_{\hat\phi(\sigma)}$.

\medskip

{\bf (e)} If $C\subseteq\Cal T_{\Bbb B}$ is a non-empty set with
has infimum $\sigma_*$, set $\tau_*=\inf\hat\phi[C]$.   As
$\hat\phi$ is order-preserving, $\hat\phi(\sigma_*)$ is a lower bound of
$\hat\phi[C]$ and $\hat\phi(\sigma_*)\le\tau_*$.

If $t\in T$ is isolated on the right, then

\Centerline{$\Bvalue{\tau_*>t}
\Bsubseteq\inf_{\sigma\in C}\Bvalue{\hat\phi(\sigma)>t}
=\phi(\inf_{\sigma\in C}\Bvalue{\sigma>t})
=\phi\Bvalue{\sigma_*>t}$}

\noindent by 632C(a-i).   If $t\in T$ is not isolated on the right, then

$$\eqalign{\Bvalue{\tau_*>t}&
=\sup_{s>t}\Bvalue{\tau_*>s}
\Bsubseteq\sup_{s>t}\inf_{\sigma\in C}\Bvalue{\hat\phi(\sigma)>s}\cr&
=\phi(\sup_{s>t}\inf_{\sigma\in C}\Bvalue{\sigma>s})
=\phi\Bvalue{\sigma_*>t}\cr}$$

\noindent by the other formula in 632C(a-i).   So

\Centerline{$\Bvalue{\tau_*>t}\Bsubseteq\phi\Bvalue{\sigma_*>t}
=\Bvalue{\hat\phi(\sigma_*)>t}$}

\noindent for every $t$, and $\tau_*\le\hat\phi(\sigma_*)$.
Thus $\inf\hat\phi[C]=\hat\phi(\inf C)$.   We know already from (b-ii)
that $\sup\hat\phi[C]=\sup\phi(\inf C)$.   As $C$ is arbitrary,
$\hat\phi$ is order-continuous in the sense of 313Ha.

\medskip

{\bf (f)} By (b-iii), $\hat\phi(\sigma)\in\Cal T_{\Bbb Af}$.
Now, for $t\in\Bbb R$,

$$\eqalign{\Bvalue{\iota_{\hat\phi(\sigma)}>t}
&=\Bvalue{\hat\phi(\sigma)>t}=\phi\Bvalue{\sigma>t}
   =\phi\Bvalue{\iota_{\sigma}>t}=\Bvalue{T_{\phi}\iota_{\sigma}>t}
   \text{ if }t\ge 0,\cr
&=1_{\frak A}=\phi 1_{\frak B}=\phi\Bvalue{\iota_{\sigma}>t}
   =\Bvalue{T_{\phi}\iota_{\sigma}>t}
   \text{ otherwise},\cr}$$

\noindent so $\iota_{\hat\phi(\sigma)}=T_{\phi}\iota_{\sigma}$.

}%end of proof of 634B

\leader{634C}{Proposition} Suppose that $\frak B$ is an
(order-)closed subalgebra of $\frak A$.

(a) Set $\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.   Then
$\family{t}{T}{\frak B_t}$ is a filtration of closed subalgebras of
$\frak B$.

(b) Let $\Cal T_{\Bbb B}$ be the set of stopping times defined from
$(\frak B,\family{t}{T}{\frak B_t})$ by the formula of 611A(b-i).

\quad (i) $\Cal T_{\Bbb B}$ is a sublattice of $\Cal T_{\Bbb A}$
containing $\min\Cal T_{\Bbb A}$, $\max\Cal T_{\Bbb A}$
and all constant stopping times.

\quad (ii) If $C\subseteq\Cal T_{\Bbb B}$ is non-empty then its
supremum is the same
whether calculated in $\Cal T_{\Bbb B}$ or in $\Cal T_{\Bbb A}$.

\quad(iii) We can identify the order-ideals $\Cal T_{\Bbb Bb}$ and
$\Cal T_{\Bbb Bf}$ of bounded and finite stopping times in
$\Cal T_{\Bbb B}$ with
$\Cal T_{\Bbb B}\cap\Cal T_{\Bbb Ab}$ and
$\Cal T_{\Bbb B}\cap\Cal T_{\Bbb Af}$ respectively.

(c) If $\sigma$, $\tau\in\Cal T_{\Bbb B}$ then the regions
$\Bvalue{\sigma<\tau}$,
$\Bvalue{\sigma\le\tau}$ and $\Bvalue{\sigma=\tau}$, when defined by the
formulae of 611D interpreted in either $(\frak A,\family{t}{T}{\frak A_t})$ or
$(\frak B,\family{t}{T}{\frak B_t})$, are the same, and
belong to $\frak B$.

(d) If $\tau\in\Cal T_{\Bbb B}$, and we define corresponding algebras
$\frak A_{\tau}$ and $\frak B_{\tau}$ by the formula of 611G interpreted in
$(\frak A,\family{t}{T}{\frak A_t})$, $(\frak B,\family{t}{T}{\frak B_t})$
respectively, then $\frak B_{\tau}=\frak B\cap\frak A_{\tau}$.

(e) Suppose that $\Cal S$ is a sublattice of $\Cal T_{\Bbb B}$.

\quad(i) If $\hat\Cal S_{\Bbb A}$ is the covered envelope of $\Cal S$ in
$\Cal T_{\Bbb A}$, then $\hat\Cal S_{\Bbb A}\cap\Cal T_{\Bbb B}$ is the
covered envelope $\hat\Cal S_{\Bbb B}$ of $\Cal S$ in $\Cal T_{\Bbb B}$.

\quad(ii) A family
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ in $L^0(\frak B)$ is fully
adapted to $\family{t}{T}{\frak B_t}$ iff it is fully adapted to
$\family{t}{T}{\frak A_t}$.

\quad(iii) If $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ is fully adapted to
$\family{t}{T}{\frak B_t}$ and $\hatpmb u$ is the extension of $\pmb u$ to
$\hat\Cal S_{\Bbb A}$ which is fully adapted to
$\family{t}{T}{\frak A_t}$, then $\hatpmb u\restr\hat\Cal S_{\Bbb B}$ is the
extension of $\pmb u$ to $\hat\Cal S_{\Bbb B}$ which is fully adapted to
$\family{t}{T}{\frak B_t}$.

(f) Suppose that $\family{t}{T}{\frak A_t}$ is right-continuous.

\quad(i) $\family{t}{T}{\frak B_t}$ is right-continuous.

\quad(ii) $\Cal T_{\Bbb B}$ is order-closed in $\Cal T_{\Bbb A}$.

(g) If $T=\coint{0,\infty}$ and we write
$\pmb\iota=\family{\sigma}{\Cal T_{\Bbb Af}}{\iota_{\sigma}}$ for the
identity process in the structure
$(\frak A,\family{t}{T}{\frak A_t},\Cal T_{\Bbb A})$, then
$\pmb\iota\restr\Cal T_{\Bbb Bf}$
is the identity process in the structure
$(\frak B,\family{t}{T}{\frak B_t},\Cal T_{\Bbb B})$.

\proof{{\bf (a)} Each $\frak B_t$, being the intersection of order-closed
subalgebras of $\frak A$, is an order-closed subalgebra of $\frak A$ and
therefore an order-closed subalgebra of $\frak B$.   If $s\le t$ then
$\frak A_s\subseteq\frak A_t$ so $\frak B_s\subseteq\frak B_t$.

We are now in the special case of 634B in which $\phi b=b$ for every
$b\in\frak B$.   Consequently $\hat\phi(\sigma)=\sigma$ for every
$\sigma\in\Cal T_{\Bbb B}$, and $\Cal T_{\Bbb B}$ is actually a subset of
$\Cal T_{\Bbb A}$.   Because the identity map $\hat\phi$ is a lattice
homomorphism (634B(b-i)), $\Cal T_{\Bbb B}$ is a sublattice of
$\Cal T_{\Bbb A}$.

\medskip

{\bf (b)} We just have to adapt 634Bb to this special case, in which
$\Cal T_{\Bbb B}\subseteq\Cal T_{\Bbb A}$ and $\hat\phi$ is the identity
embedding.

\medskip

{\bf (c)} Similarly, this is just the form now taken by 634B(c-i).

\medskip

{\bf (d)} Going back to the formula in 611G,

$$\eqalignno{\frak B_{\tau}&
=\bigcap_{t\in T}\{b:b\in\frak B,\,b\Bsetminus\Bvalue{\tau>t}\in\frak B_t\}\cr&
=\bigcap_{t\in T}\{b:b\in\frak B,\,b\Bsetminus\Bvalue{\tau>t}
    \in\frak B\cap\frak A_t\}
=\bigcap_{t\in T}\{b:b\in\frak B,\,b\Bsetminus\Bvalue{\tau>t}
    \in\frak A_t\}
\Displaycause{because $\Bvalue{\tau>t}\in\frak B$ for every $t\in T$}
=\frak B\cap\bigcap_{t\in T}\{a:a\in\frak A,\,a\Bsetminus\Bvalue{\tau>t}
    \in\frak A_t\}
=\frak B\cap\frak A_{\tau}.\cr}$$

\medskip

{\bf (e)(i)} If $\tau\in\Cal T_{\Bbb B}$, then

\Centerline{$\tau\in\hat\Cal S_{\Bbb A}
\iff\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}=1
\iff\tau\in\hat\Cal S_{\Bbb B}$.}

\noindent So

\Centerline{$\hat\Cal S_{\Bbb A}\cap\Cal T_{\Bbb B}
=\hat\Cal S_{\Bbb B}\cap\Cal T_{\Bbb B}
=\hat\Cal S_{\Bbb B}$.}

\medskip

\quad{\bf (ii)} Immediate from 612Da, (c) and (d), if we recall that

\Centerline{$L^0(\frak B\cap\frak A_{\tau})
=L^0(\frak B)\cap L^0(\frak A_{\tau})$}

\noindent for every $\tau$ (612A(e-i)).

\medskip

\quad{\bf (iii)} The point is that if
$\hatpmb u=\family{\tau}{\hat\Cal S_{\Bbb A}}{\hat u_{\tau}}$, then
$\hat u_{\tau}\in L^0(\frak B)$ for every $\tau\in\hat\Cal S_{\Bbb B}$.
\Prf\ If $\tau\in\hat\Cal S_{\Bbb B}$ and $\alpha\in\Bbb R$ then

$$\eqalign{\Bvalue{\hat u_{\tau}>\alpha}&
=\Bvalue{\hat u_{\tau}>\alpha}
   \Bcap\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}
=\sup_{\sigma\in\Cal S}
   (\Bvalue{\hat u_{\tau}>\alpha}\Bcap\Bvalue{\tau=\sigma})\cr&
=\sup_{\sigma\in\Cal S}
   (\Bvalue{u_{\sigma}>\alpha}\Bcap\Bvalue{\tau=\sigma})
\in\frak B.\cr}$$

\noindent (Of course I am relying on (c) for assurance that
$\Bvalue{\tau=\sigma}$ has a consistent interpretation.)\ \QeD\
Now (ii) just above tells us that $\hatpmb u\restr\hat\Cal S_{\frak B}$ is fully
adapted to $\family{t}{T}{\frak B_t}$, while it surely extends $\pmb u$.

\medskip

{\bf (f)(i)} Look at the definition in 632B:  as $\family{t}{T}{\frak A_t}$ is
right-continuous,

\Centerline{$\frak B_t=\frak B\cap\frak A_t
=\frak B\cap\bigcap_{s>t}\frak A_s
=\bigcap_{s>t}\frak B\cap\frak A_s
=\bigcap_{s>t}\frak B_s$}

\noindent whenever $t\in T$ is not isolated on the right.

\medskip

\quad{\bf (ii)} If $C\subseteq\Cal T_{\Bbb B}$ is non-empty,
then $\sup C$, calculated in $\Cal T_{\Bbb A}$, belongs to $\Cal T_{\Bbb B}$
by (b-ii) above.   As for its infimum, this is given by the formula

$$\eqalign{\Bvalue{\inf C>t}
&=\inf_{\tau\in C}\Bvalue{\tau>t}\text{ if }
   t\text{ is isolated on the right},\cr
&=\sup_{s>t}\inf_{\tau\in C}\Bvalue{\tau>s}\text{ otherwise}\cr}$$

\noindent of 632C(a-i), which (because $\frak B$ is order-closed in $\frak A$)
is the same whether calculated in $\frak B$ or $\frak A$.   So $\sup C$ and
$\inf C$, taken in $\Cal T_{\Bbb A}$, both belong to $\Cal T_{\Bbb B}$;  as
$C$ is arbitrary, $\Cal T_{\Bbb B}$ is an order-closed sublattice of
$\Cal T_{\Bbb A}$ as defined in 313Da.

\medskip

{\bf (g)} This is just the form taken by 634Bf in the present context.
}%end of proof of 634C

\cmmnt{\medskip

\noindent{\bf Remark} If we think of $\Cal T_{\Bbb A}$ as a sublattice of
$\prod_{t\in T}\frak A_t$, as suggested in 611Ac, then
$\Cal T_{\Bbb B}=\Cal T_{\Bbb A}\cap\prod_{t\in T}\frak B_t$, while
$\prod_{t\in T}\frak B_t$ is an order-closed subalgebra of
$\prod_{t\in T}\frak A_t$ (315Xc).
}%end of comment

\leader{634D}{Notation} In the context of 634Ce, we shall be able to
regard a
process $\pmb u\in L^0(\frak B)^{\Cal S}$ as being fully adapted either in
the structure
$\Bbb A=(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t},\Cal T_{\Bbb A},
\family{\tau}{\Cal T_{\Bbb A}}{\frak A_{\tau}})$
$(\frak A,\family{t}{T}{\frak A_t})$
or in the structure
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.
Consequently we shall be able to test $\pmb u$ against the definitions
in this volume in two different ways.   We have to check which properties are
`absolute', in the sense that $\pmb u$ will have them in one structure iff
it has them in the other, and which are not.   \cmmnt{Most
of the checks are very easy, just as 634C is.}   While working through the list,
it will save a great many words if I use
abbreviated expressions of the type `$\pmb u$ is $\Bbb A$-simple',
`$\pmb u$ is a $\Bbb B$-integrator' to mean
`interpreted in the structure $\Bbb A$, $\pmb u$ is a simple process',
`interpreted in the structure $\Bbb B$,
$\pmb u$ is an integrator', and so forth.   \cmmnt{But we have already seen in
634C(e-ii) that if $\pmb u$ is a process such that its domain is a sublattice
of $\Cal T_{\Bbb B}$ and its values all belong to $L^0(\frak B)$ -- these
properties being intrinsic to $\pmb u$ -- then $\pmb u$ is
$\Bbb A$-fully-adapted iff it is $\Bbb B$-fully-adapted, so in this context I
can say simply that `$\pmb u$ is fully adapted' with little danger of
confusion.   A useful number of other concepts are equally easily handled, as in
the next proposition.}

\leader{634E}{Proposition} Let $\frak B$ be a closed subalgebra of
$\frak A$, and
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\penalty-100\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$
the corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.
Suppose that $\Cal S$ is a
sublattice of $\Cal T_{\Bbb B}$ and that $\pmb u\in L^0(\frak B)^{\Cal S}$ is
fully adapted.

(a) $\pmb u$ is $\Bbb B$-simple iff it is $\Bbb A$-simple.

(b) $\pmb u$ is $\Bbb B$-(locally)-near-simple iff it is
$\Bbb A$-(locally)-near-simple.

(c) $\pmb u$ is %(locally)
$\Bbb B$-order-bounded iff it is %(locally)
$\Bbb A$-order-bounded.

(d) $\pmb u$ is %(locally)
of $\Bbb B$-bounded variation iff it is %(locally)
of $\Bbb A$-bounded variation.

(e) $\pmb u$ is $\Bbb B$-(locally)-moderately-oscillatory iff it is
$\Bbb A$-(locally)-moderately-oscillatory.

(f) $\pmb u$ is %(locally)
$\Bbb A$-jump-free iff it is %(locally)
$\Bbb B$-jump-free.

(g) If $\pmb v\in L^0(\frak B)^{\Cal S}$
is another fully adapted process,
then the integral $\int_{\Cal S}\pmb u\,d\pmb v$ is defined for
$(\frak B,\bar\mu\restrp\frak B,\family{t}{T}{\frak B_t})$
iff it is defined for
$(\frak A,\bar\mu,\family{t}{T}{\frak A_t})$, with the same value;  that
is, $\Bbb B\int_{\Cal S}\pmb u\,d\pmb v
=\Bbb A\int_{\Cal S}\pmb u\,d\pmb v$ if either is defined.

\proof{ Really all we have to do is to look at the definitions.

\medskip

{\bf (a)} Whether a process $\sfamilysS{u}$ is simple depends only on the
ordering of
$\Cal S$ and the values of $\Bvalue{\sigma<\tau}$, $\Bvalue{\sigma\le\tau}$ for
$\sigma$, $\tau\in\Cal S$;  given that $\Cal S\subseteq\Cal T_{\Bbb B}$.
634C(b-i) and 634Cc assure us that we can calculate these either in $\Bbb A$ or
$\Bbb B$ and get the same results.

\medskip

{\bf (b)} The definition of `near-simple'
process calls on expressions of the form
$\widehat{\theta}(\pmb u-\pmb u')
=\theta(\sup_{\sigma\in\Cal S}|u_{\sigma}-u'_{\sigma}|)$ (631Ba, 615B);
but the supremum here will have the same value whether interpreted in
$L^0(\frak A)$ or $L^0(\frak B)$, and I noted in 634A that there is no
dispute about the value of $\theta(u)$ for $u\in L^0(\frak B)$.   So
$\pmb u$ is $\Bbb B$-near-simple iff it is $\Bbb A$-near-simple;  applying this
to $\pmb u\restr\Cal S\wedge\tau$ for $\tau\in\Cal S$, we see that
$\pmb u$ is $\Bbb B$-locally-near-simple iff it is $\Bbb A$-locally-near-simple.

\medskip

{\bf (c)} Here we just need to remember that a subset of $L^0(\frak B)$ is
order-bounded in $L^0(\frak B)$ iff it is order-bounded in
$L^0(\frak A)$, as noted in 634A.

\medskip

{\bf (d)} Satisfaction of any of the conditions (i)-(iii) in 614I is independent
of the structure we are working in.

\medskip

{\bf (e)} Here it seems that we need a more sophisticated approach, using both
characterizations of `moderately oscillatory'.

\medskip

\quad{\bf (i)}\grheada If $\pmb u$ is $\Bbb B$-moderately-oscillatory, then for
every $\epsilon>0$ there is a fully adapted process
$\pmb u'\in(L^0(\frak B))^{\Cal S}$ of bounded variation (determined in either
system, see (d)) such that
$\theta(\sup|\pmb u-\pmb u'|)\le\epsilon$, so $\pmb u$ is
$\Bbb A$-moderately-oscillatory.

\medskip

\qquad\grheadb\ If $\pmb u$ is $\Bbb A$-moderately-oscillatory,
write $\hat\Cal S_{\Bbb A}\subseteq\Cal T_{\Bbb A}$
for the $\Bbb A$-covered envelope of $\Cal S$, and
$\hatpmb u=\family{\sigma}{\hat\Cal S_{\Bbb A}}{\hat u_{\sigma}}$ for the
corresponding extension of $\pmb u$.   By 615N,
$\family{n}{\Bbb N}{u_{\sigma_n}}$ is
convergent in $L^0(\frak A)$ for every monotonic sequence $\sequencen{\sigma}$
in $\Cal S$;  but $L^0(\frak B)$ is a
topologically closed set in $L^0(\frak A)$ (367Rc), so
$\family{n}{\Bbb N}{u_{\sigma_n}}$ is
convergent in $L^0(\frak B)$ for every monotonic sequence $\sequencen{\sigma_n}$
in $\Cal S$, and $\pmb u$ is
$\Bbb B$-moderately-oscillatory, by 615N in the other direction.

\medskip

\quad{\bf (ii)} Now

$$\eqalignno{&\pmb u\text{ is }\Bbb B\text{-locally-moderately-oscillatory}\cr&
\mskip50mu
\iff\pmb u\restr\Cal S\wedge\tau\text{ is }\Bbb B
   \text{-moderately-oscillatory for every }\tau\in\Cal S\cr&
\mskip50mu
\iff\pmb u\restr\Cal S\wedge\tau\text{ is }\Bbb A
   \text{-moderately-oscillatory for every }\tau\in\Cal S\cr&
\mskip50mu
\iff\pmb u\text{ is }\Bbb A\text{-locally-moderately-oscillatory}.\cr}$$

\medskip

{\bf (f)} See 618B.
For $I\in\Cal I(\Cal S)$, $\Osclln_I(\pmb u)$ and $\Osclln^*_I(\pmb u)$
are calculated in terms of suprema in $L^0$;  again because
$L^0(\frak B)$ is order-closed in $L^0(\frak A)$, and
$\Delta_e(\tbf{1},|d\pmb u|)\in L^0(\frak B)$ whenever $e$ is a
stopping-time interval with endpoints in $\Cal S$, we will get the same
values for $\Osclln_I(\pmb u)$,
$\Osclln^*_I(\pmb u)
=\sup_{I\subseteq J\in\Cal I(\Cal S)}\Osclln_J(\pmb u)$ and
$\Osclln(\pmb u)=\inf_{I\in\Cal I(\Cal S)}\Osclln^*_I(\pmb u)$ on either
interpretation.

\medskip

{\bf (g)} Again because $L^0(\frak B)$ is a
topologically closed set in $L^0(\frak A)$,
convergence in $L^0(\frak A)$ of a filter on $L^0(\frak B)$ implies
convergence in $L^0(\frak B)$.   And for $I\in\Cal I(\Cal S)$, the evaluation of
$S_I(\pmb u,d\pmb v)\in L^0(\frak B)$ is the same in either structure.   So

\Centerline{$\Bbb B\int_{\Cal S}\pmb u\,d\pmb v
=\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb v)
=\Bbb A\int_{\Cal S}\pmb u\,d\pmb v$}

if either form of the integral is defined.
}%end of proof of 634E

\cmmnt{\medskip

\noindent{\bf Remarks} Thus I shall be able from now on, in the context of
processes in $L^0(\frak B)^{\Cal S}$ where $\Cal S\subseteq\Cal T_{\Bbb B}$, to
use the phrases `simple', `near-simple', `order-bounded',
'bounded variation', `moderately oscillatory', 'jump-free', and the formula
$\int_{\Cal S}\pmb u\,d\pmb v$, without specifying whether I am thinking in the
structure $\Bbb A$ or the structure $\Bbb B$.
}

\leader{634F}{Relative \dvrocolon{independence}}\cmmnt{
You may have noticed an omission in 634E above.   There
is no mention of `martingales'.   In general, the shift
from an algebra $\frak A$ to a subalgebra $\frak B$ need not respect
conditional expectations in the right way.   In many cases, however (for
instance, if $\frak A$ is based on a product probability measure, and
$\frak B$ on one of the factors), we do have strong connexions between
martingales for $(\frak A,\bar\mu,\family{t}{T}{\frak A_t})$ and
martingales for $(\frak B,\bar\mu\restr\frak B,\family{t}{T}{\frak B_t})$.
The key concept is that of `relative independence', already used in \S628.
In that section I was relying heavily on the work of \S458, so did not
separate the more elementary ideas;  perhaps it will help if I set them
out now.

\medskip

\noindent}{\bf Definitions (a)}\cmmnt{ (see 458La)}
If $\frak B$, $\frak C$ and $\frak D$ are closed
subalgebras of $\frak A$, \cmmnt{I say that} $\frak B$ and $\frak C$ are
{\bf relatively (stochastically) independent} over $\frak D$ if
$P_{\frak D}\chi(b\Bcap c)=P_{\frak D}\chi b\times P_{\frak D}\chi c$ for
all $b\in\frak B$ and $c\in\frak C$.

\spheader 634Fb I\cmmnt{ will} say that a closed subalgebra $\frak B$ of
$\frak A$ is {\bf coordinated} with the filtration
$\family{t}{T}{\frak A_t}$ if
$\frak B$ and $\frak A_t$ are relatively independent over
$\cmmnt{\frak B_t=\mskip5mu}\frak B\cap\frak A_t$ for every $t\in T$.

\leader{634G}{Proposition}
If $\frak B$, $\frak C$ are closed
subalgebras of $\frak A$, the following are equiveridical:

(i) $\frak B$ and $\frak C$ are relatively independent over
$\frak B\cap\frak C$;

(ii) $P_{\frak B\cap\frak C}(u\times v)
=P_{\frak B\cap\frak C}u\times P_{\frak B\cap\frak C}v$
for all $u\in L^{\infty}(\frak B)$, $v\in L^{\infty}(\frak C)$;

(iii) $P_{\frak B}P_{\frak C}=P_{\frak B\cap\frak C}$;

(iv) $P_{\frak B}P_{\frak C}=P_{\frak C}P_{\frak B}$;

(v) $P_{\frak B}u\in L^0(\frak C)$ whenever
$u\in L^1_{\bar\mu}\cap L^0(\frak C)$.

\proof{458M.}

\leader{634H}{Lemma} Let $\frak B$ be a closed subalgebra of
$\frak A$ which is coordinated with $\family{t}{T}{\frak A_t}$,
and $\Bbb B=(\frak B,\bar\mu\restr\frak B,T,
\penalty-100\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.   Then
$\frak B$ and $\frak A_{\sigma}$ are relatively independent over
$\frak B_{\sigma}$ for every $\sigma\in\Cal T_{\Bbb B}$.

\proof{{\bf (a)} Writing $\check{t}$ for the constant stopping time at $t$
(611A(b-ii)), we have $\frak A_{\check{t}}=\frak A_t$ (611Hb);  by
634G(v), the hypothesis `$\frak B$  is coordinated with
$\family{t}{T}{\frak A_t}$' amounts to saying that

\Centerline{$P_{\check{t}}u\in L^0(\frak B)$ for every
$t\in T$ and $u\in L^1_{\bar\mu}\cap L^0(\frak B)$.}

\medskip

{\bf (b)} For $t\in T$ and $c\in\frak B_t$, define $\check{t}c$
by saying that

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

\noindent It is easy to check that (because
$c\in\frak B_t$) the conditions of
611B(b-i) are satisfied, so that $\check{t}c\in\Cal T_{\Bbb B}$.
We see that

$$\eqalign{\frak A_{\check{t}c}
&=\{a:a\in\frak A,\,a\Bsetminus\Bvalue{\check{t}c>s}\in\frak A_s
   \text{ for every }s\in T\}\cr
&=\{a:a\Bsetminus c\in\frak A_s
   \text{ for every }s\ge t\}
=\{a:a\Bsetminus c\in\frak A_t\}.\cr}$$

Now $\check{t}\le\check{t}c$ and

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

\noindent So $\Bvalue{\check{t}c=\check{t}}=1\Bsetminus c$.
If $u\in L^1_{\bar\mu}$,

$$\eqalignno{P_{\check{t}c}u
&=P_{\check{t}c}(u\times\chi c)
   +P_{\check{t}c}(u\times\chi(1\Bsetminus c))
=u\times\chi c+(P_{\check{t}c}u)\times\chi(1\Bsetminus c)\cr
\displaycause{because $u\times\chi c\in L^0(\frak A_{\check{t}c})$ and
$1\Bsetminus c\in\frak A_{\check{t}c}$}
&=u\times\chi c+(P_{\check{t}}u)\times\chi(1\Bsetminus c)\cr}$$

\noindent by 622Bb.   It follows that if
$u\in L^1_{\bar\mu}\cap L^0(\frak B)$, then
$P_{\check{t}c}u\in L^0(\frak B)$.

\medskip

{\bf (c)} For $\sigma\in\Cal T_{\Bbb B}$, set

\Centerline{$C_{\sigma}=\{\check{t}c:t\in T$, $c=\Bvalue{\sigma>t}\}$,}

\Centerline{$D_{\sigma}=\{\tau_0\wedge\ldots\wedge\tau_n:
\tau_i\in C_{\sigma}$ for $i\le n\}$.}

\noindent Then

\Centerline{$\frak A_{\sigma}
=\bigcap_{t\in T}\{a:a\Bsetminus\Bvalue{\sigma>t}\in\frak A_t\}
=\bigcap_{\tau\in C_{\sigma}}\frak A_{\tau}
=\bigcap_{\tau\in D_{\sigma}}\frak A_{\tau}$}

\noindent by 611Eb.   We know from (b) that
$P_{\tau}u\in L^0(\frak B)\cap L^1_{\bar\mu}$
for every $\tau\in C_{\sigma}$;  now if
$\tau_0,\ldots,\tau_n\in C_{\sigma}$ and
$\tau=\tau\wedge\ldots\wedge\tau_n$,

\Centerline{$P_{\tau}u=P_{\tau_0}\ldots P_{\tau_n}u
\in L^0(\frak B)\cap L^1_{\bar\mu}$}

\noindent for every $u\in L^0(\frak B)\cap L^1_{\bar\mu}$, by
622Ba.   Thus $P_{\tau}u\in L^0(\frak B)$ whenever
$\tau\in D_{\sigma}$ and $u\in L^0(\frak B)\cap L^1_{\bar\mu}$.

Now observe that $\{\frak A_{\tau}:\tau\in D_{\sigma}\}$ is
downwards-directed and has intersection $\frak A_{\sigma}$.   So
$P_{\sigma}u=\lim_{\tau\downarrow D_{\sigma}}P_{\tau}u$ for every
$u\in L^1_{\bar\mu}$, by 621C(g-i).   In 621Cg I called this a
a $\|\,\|_1$-limit, but for our purposes here it is enough to think of it
as a limit for the topology of convergence in measure, because
$L^0(\frak B)$ is closed for the
topology of convergence in measure, and all we need
to know is that if $u\in L^0(\frak B)\cap L^1_{\bar\mu}$ then
$P_{\sigma}u\in L^0(\frak B)$.   Thus $\frak A_{\sigma}$ and $\frak B$
satisfy 634G(v), and are relatively independent over
$\frak A_{\sigma}\cap\frak B=\frak B_{\sigma}$ (634Cd).
}%end of proof of 634H

\leader{634I}{Theorem} Let $\frak B$ be a closed subalgebra of
$\frak A$ which is coordinated with $\family{t}{T}{\frak A_t}$,
and $\Bbb B=(\frak B,\bar\mu\restr\frak B,T,\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.
Let $\Cal S$ be a sublattice of $\Cal T_{\Bbb B}$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process such that
$u_{\sigma}\in L^0(\frak B)$ for every $\sigma\in\Cal S$.

(a) $\pmb u$ is a (local) $\Bbb B$-martingale
iff it is a (local) $\Bbb A$-martingale.

(b) $\pmb u$ is a (local) $\Bbb B$-integrator iff it is a
(local) $\Bbb A$-integrator, and in this case its $\Bbb B$-quadratic
variation is the same as its $\Bbb A$-quadratic variation.

\proof{{\bf (a)} For $\sigma\in\Cal T_{\Bbb B}$, write
$Q_{\sigma}:L^1(\frak B,\bar\mu\restrp\frak B)
\to L^1(\frak B,\bar\mu\restrp\frak B)$ for the conditional expectation
associated with the closed subalgebra $\frak B_{\sigma}$.   Suppose that
$u\in L^0(\frak B)\cap L^1_{\bar\mu}=L^1(\frak B,\bar\mu\restrp\frak B)$.
Then $\frak B$ and $\frak A_{\sigma}$ are
relatively independent over $\frak B_{\sigma}=\frak A\cap\frak B_{\sigma}$
(634H), so
$P_{\sigma}u\in L^0(\frak B)\cap L^0(\frak A_{\sigma})
=L^0(\frak B_{\sigma})$, and we have
$\Expn(P_{\sigma}u\times\chi b)=\Expn(u\times\chi b)$ for every
$b\in\frak A_{\sigma}$ and therefore for every $b\in\frak B_{\sigma}$.
But this means that $P_{\sigma}u$ satisfies the property defining
$Q_{\sigma}u$, and $P_{\sigma}u=Q_{\sigma}u$.

Now we see that

$$\eqalign{&\pmb u\text{ is a }\frak B\text{-martingale}\cr
&\mskip50mu
\iff Q_{\sigma}u_{\tau}=u_{\sigma}
  \text{ whenever }\sigma\le\tau\text{ in }\Cal S\cr
&\mskip50mu
\iff P_{\sigma}u_{\tau}=u_{\sigma}
  \text{ whenever }\sigma\le\tau\text{ in }\Cal S\cr
&\mskip50mu
\iff\pmb u\text{ is a }\frak A\text{-martingale}.\cr}$$

Observe next that a sublattice $\Cal S'$ of $\Cal S$ is a
$\Bbb B$-covering ideal of
$\Cal S$ iff it is an $\Bbb A$-covering ideal of $\Cal S$, by 634Cc.
So we have

$$\eqalign{&\pmb u\text{ is a }\frak B\text{-local martingale}\cr
&\mskip50mu
\iff\text{there is a covering ideal }\Cal S'\text{ of }\Cal S\cr
&\mskip200mu\text{ such
that }\pmb u\restr\Cal S'\text{ is a }\frak B\text{-martingale}\cr
&\mskip50mu
\iff\text{there is a covering ideal }\Cal S'\text{ of }\Cal S\cr
&\mskip200mu\text{ such
that }\pmb u\restr\Cal S'\text{ is a }\frak A\text{-martingale}\cr
&\mskip50mu
\iff\pmb u\text{ is an }\frak A\text{-local martingale}.\cr}$$


\medskip

{\bf (b)(i)}\grheada\ Suppose that $\pmb u$ is a $\Bbb B$-integrator.
Then there is a $\bar\nu_0$ such that $(\frak B,\bar\nu_0)$
is a probability
algebra and $\pmb u$ is a $(\frak B,\bar\nu_0)$-strong integrator
(627Ld).   Let
$z\in L^1_{\bar\mu}\cap L^0(\frak B)$ be such that
$\bar\nu_0b=\Expn_{\bar\mu}(z\times\chi b)$ for every $b\in\frak B$
(625B;  note that in this sentence
we do not need to distinguish between
$\Expn_{\bar\mu}$ and $\Expn_{\bar\mu\restrp\frak B}$).
Set $\bar\nu a=\Expn_{\bar\mu}(z\times\chi a)$ for every $a\in\frak A$;
because $\Bvalue{z>0}=1$, $(\frak A,\bar\nu)$ is a probability algebra.

If $\tau\in\Cal T_{\Bbb B}$,
$\frak B$ and $\frak A_{\tau}$ are relatively independent over
$\frak B\cap\frak A_{\tau}$ for $\bar\nu$ as well as for $\bar\mu$.
\Prf\ Let $Q_{\tau}:L^1_{\bar\nu}\to L^1_{\bar\nu}\cap L^0(\frak B_{\tau})$
be the conditional expectation associated with $\frak B_{\tau}$ for
$\bar\nu$.   If $w\in L^1_{\bar\nu}\cap L^0(\frak B)$, then

\Centerline{$Q_{\tau}w=P_{\tau}(w\times z)\times\Bover1{P_{\tau}z}$}

\noindent (625B(b-i);  of course $\Bvalue{P_{\tau}z>0}=1$ by 625B(a-vi), so
there is no problem with the reciprocal).
But $w\times z\in L^1_{\bar\mu}\cap L^0(\frak B)$, because
$z\in L^0(\frak B)$, so $P_{\tau}(w\times z)\in L^0(\frak B)$, using
634H.   Similarly, $P_{\tau}z\in L^0(\frak B)$, so
$Q_{\tau}w\in L^0(\frak B)$.   By 634G(v),
$\frak B$ and $\frak A_{\tau}$ are relatively independent over
$\frak B\cap\frak A_{\tau}$ for $\bar\nu$.\ \Qed

Now suppose that $\epsilon>0$.   Then we have processes
$\pmb{w}$, $\pmb{w}'$ with domain $\Cal S$, $\Bbb B$-fully adapted,
such that
$\bar\nu\Bvalue{\pmb u\ne\pmb{w}+\pmb{w}'}\ne 0$,
$\pmb{w}$ is a $(\frak B,\bar\nu_0)$-uniformly integrable martingale,
and $\pmb{w}'$ is of $\Bbb B$-bounded variation.   By (a) above,
$\pmb{w}$ is an $(\frak A,\bar\nu)$-martingale, and $\pmb{w}'$ is of
$\Bbb A$-bounded variation, as noted in 634Ed.
Also, of course, $\pmb{w}$ is still uniformly integrable in
$L^1_{\bar\nu}$.
As $\epsilon$ is arbitrary, $\pmb u$ is an
$(\frak A,\bar\nu)$-strong integrator.
It follows at once that $\pmb u$ is an $\Bbb A$-integrator.

\medskip

\qquad\grheadb\ If $\pmb u$ is a $\Bbb B$-local-integrator, then we can
apply (i) to see that $\pmb u\restr\Cal S\wedge\tau$ is an
$\Bbb A$-integrator whenever
$\tau\in\Cal S$, so that $\pmb u$ is an $\Bbb A$-local-integrator.

\medskip

\quad{\bf (ii)} In the other direction, if $\pmb u$ is an
$\Bbb A$-integrator, then it is a
$\Bbb B$-integrator.  \Prf\ Going back to the definition of `integrator'
in 616Fc, we see that $z\in\Bbb BQ_{\Cal S}(\pmb u)$ there are
$I\in\Cal I(\Cal S)$ and $\pmb w\in L^0(\frak B)^I$ such that $\pmb w$ is
$\Bbb B$-fully-adapted, $\|\pmb w\|_{\infty}\le 1$ and
$z=S_I(\pmb w,d\pmb u)$.   Now, of course,
$\pmb w\in L^0(\frak A)^I$ is $\Bbb A$-fully-adapted, so
$z\in\Bbb AQ_{\Cal S}(\pmb u)$.   Thus
$\Bbb BQ_{\Cal S}(\pmb u)\subseteq\Bbb AQ_{\Cal S}(\pmb u)\cap L^0(\frak B)$
is topologically bounded in $L^0(\frak A)$ and therefore also in
$L^0(\frak B)$, since the linear space topology of $L^0(\frak B)$ is that
induced by the linear space topology of $L^0(\frak A)$.   So
$\pmb u$ is a $\Bbb B$-integrator.\ \Qed

Just as in (i-$\beta$) above, it follows at once that if $\pmb u$ is an
$\Bbb A$-local-integrator it is a $\Bbb B$-local-integrator.

\medskip

\quad{\bf (iii)} If $\pmb u$ is a local integrator (either in $\Bbb A$ or in
$\Bbb B$), set
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$.
For any $\tau\in\Cal S$,

\Centerline{$u_{\tau}^2-u_{\downarrow}^2-2\int_{\Cal S}\pmb u\,d\pmb v$}

\noindent is the same whether calculated in $\Bbb A$ or $\Bbb B$, as
noted in 634Eg.   By 617Ka, the quadratic variation
$\family{\tau}{\Cal S}{u^*_{\tau}}$ of $\pmb u$ is the same in either structure.
}%end of proof of 634I

\leader{634J}{}\cmmnt{ One of the most important operations in probability
theory is the use of product measures to simultaneously represent
independent random variables.   In order to do the same for the
stochastic processes we have been studying here, we have to pay attention
to filtrations.   The next lemma provides one of the tools we need.

\medskip

\noindent}{\bf Lemma}
Let $(\Omega,\Sigma,\mu)$ and $(\Omega',\Sigma',\mu')$
be probability spaces, and $(\Xi,\le)$
a non-empty downwards-directed partially ordered
set.   Let $\family{\xi}{\Xi}{\Sigma_{\xi}}$ and
$\family{\xi}{\Xi}{\Sigma'_{\xi}}$ be families of $\sigma$-subalgebras of
$\Sigma$, $\Sigma'$ respectively such that
$\Sigma_{\xi}\subseteq\Sigma_{\eta}$ and
$\Sigma'_{\xi}\subseteq\Sigma'_{\eta}$ whenever $\xi\le\eta$, while
every $\Sigma_{\xi}$ contains every $\mu$-negligible set and every
$\Sigma'_{\xi}$ contains every $\mu'$-negligible set.
Write $\Tau=\bigcap_{\xi\in\Xi}\Sigma_{\xi}$ and
$\Tau'=\bigcap_{\xi\in\Xi}\Sigma'_{\xi}$.   Let $\lambda$ be the product
probability measure on $\Omega\times\Omega'$, and $\Lambda$ its domain.

(a) Suppose that $W\in\Lambda$ is such that for every
$\xi\in\Xi$ there is a $W_{\xi}\in\Sigma_{\xi}\tensorhat\Sigma'$ such that
$W\symmdiff W_{\xi}$ is $\lambda$-negligible.   Then there is a
$W'\in\Tau\tensorhat\Sigma'$ such that $W\symmdiff W'$ is
$\lambda$-negligible.

(b) Suppose that $W\in\Lambda$ is such that for every
$\xi\in\Xi$ there is a $W_{\xi}\in\Sigma_{\xi}\tensorhat\Sigma'_{\xi}$
such that
$W\symmdiff W_{\xi}$ is $\lambda$-negligible.   Then there is a
$W'\in\Tau\tensorhat\Tau'$ such that $W\symmdiff W'$ is
$\lambda$-negligible.

(c) Suppose that $W\in\bigcap_{\xi\in\Xi}\Sigma_{\xi}\tensorhat\Sigma'$,
$E\in\Sigma$ and $\epsilon\ge 0$ are such that
$\lambda(W\symmdiff(E\times\Omega'))\le\epsilon$.   Then there is an
$E_1\in\Tau$ such that
$\lambda(W\symmdiff(E_1\times\Omega'))\le 3\epsilon$.

\proof{{\bf (a)} Since there is certainly a member of
$\Sigma\tensorhat\Sigma'$ differing from $W$ by a $\lambda$-negligible
set (251Ib),
we can suppose that $W$ itself belongs to $\Sigma\tensorhat\Sigma'$.
Let $\frak A'$ be the measure algebra of $(\Omega',\Sigma',\mu')$.
Then 418Ta tells us that we have a function $f:\Omega\to\frak A'$ defined
by setting $f(\omega)=W[\{\omega\}]^{\ssbullet}$ for $\omega\in\Omega$, and
that $f[\Omega]$ is a separable subset of $\frak A'$
for the measure-algebra topology of $\frak A'$ and $f$ is
$\Sigma$-measurable for this topology.   Similarly, for any $\xi\in\Xi$, we
have a $\Sigma_{\xi}$-measurable $f_{\xi}:\Omega\to\frak A'$ such that
$f_{\xi}(\omega)=W_{\xi}[\{\omega\}]^{\ssbullet}$ for every $\omega$.
But since $W\symmdiff W_{\xi}$ is $\lambda$-negligible,
$W[\{\omega\}]\symmdiff W_{\xi}[\{\omega\}]$ is $\mu'$-negligible for
$\mu$-almost every $\omega$, by Fubini's theorem (252D).   So
$f\eae f_{\xi}$.   As we are supposing that every $\mu$-negligible set
belongs to $\Sigma_{\xi}$, $f$ is $\Sigma_{\xi}$-measurable.

This is true for every $\xi\in\Xi$, so $f$ is actually $\Tau$-measurable.
But now we can use 418Ta in the other direction to see that there is a
$W'\in\Tau\symmdiff\Sigma'$ such that
$W'[\{\omega\}]^{\ssbullet}=f(\omega)$
for every $\omega\in\Omega$.   By Fubini's theorem again,
$\lambda(W\symmdiff W')=0$, as required.

\medskip

{\bf (b)} For each $\eta\in\Xi$, we can apply (a) to
$(\Omega,\Sigma,\mu)$,
$(\Omega',\Sigma'_{\eta},\mu'_{\eta}\restr\Sigma'_{\eta})$
and $\langle W_{\xi}\rangle_{\xi\in\Xi,\xi\le\eta}$ to see that there is a
$W'_{\eta}\in\Tau\tensorhat\Sigma'_{\eta}$ such that
$\lambda_{\eta}(W\symmdiff W'_{\eta})=0$, where
$\lambda_{\eta}$ is the product of $\mu$ and $\mu'_{\eta}$.
Of course it follows that $\lambda(W\symmdiff W'_{\eta})=0$ and also
that $\tilde\lambda(W'_{\eta}\symmdiff W'_{\zeta})=0$ whenever $\eta$,
$\zeta\in\Xi$, where $\tilde\lambda$ is the product of
$\mu\restr\Tau$ and $\mu'$.   Fixing $\zeta\in\Xi$, we can apply (a) again to
$(\Omega',\Sigma',\mu')$, $(\Omega,\Tau,\mu\restr\Tau)$ and
$\family{\eta}{\Xi}{W'_{\eta}}$ to see that there is a
$W'\in\Tau\tensorhat\Tau'$ such that
$\tilde\lambda(W'\symmdiff W'_{\zeta})=0$, in which case
$\lambda(W'\symmdiff W)$ will be zero.

\medskip

{\bf (c)} Try
$E_1=\{\omega:\omega\in\Omega$, $\mu'W[\{\omega\}]\ge\bover12\}$.
By 252P, $E_1\in\Sigma_{\xi}$ for every $\xi\in\Xi$, so $E_1\in\Tau$.
We have

\Centerline{$\Bover12\mu(E_1\setminus E)
\le\int\mu'((W\setminus(E\times\Omega'))[\{\omega\}]\mu(d\omega)
=\lambda(W\setminus(E\times\Omega'))$,}

\Centerline{$\Bover12\mu(E\setminus E_1)
\le\int\mu'((E\times\Omega')\setminus W)[\{\omega\}]\mu(d\omega)
=\lambda((E\times\Omega')\setminus W)$}

\noindent so

\Centerline{$\Bover12\mu(E_1\symmdiff E)
\le\lambda(W\symmdiff(E\times\Omega'))\le\epsilon$}

\noindent and

\Centerline{$\lambda(W\symmdiff(E_1\times\Omega'))
\le\mu(E_1\symmdiff E)+\lambda(W\symmdiff(E\times\Omega'))
\le 3\epsilon$.}
}%end of proof of 634J

\leader{634K}{Theorem} Let $\familyiI{\frak A_i}$ be a stochastically
independent family of closed subalgebras of $\frak A$\cmmnt{ (see 325L)}.
Suppose that $(\Xi,\le)$
is a non-empty downwards-directed partially ordered
set and that for each $i\in I$ we have a non-decreasing
family $\family{\xi}{\Xi}{\frak B_{i\xi}}$ of closed subalgebras of
$\frak A_i$ with intersection $\frak B_i$.   Set
$\frak D=\bigvee_{i\in I}\frak B_i$, and for $\xi\in\Xi$
set $\frak D_{\xi}=\bigvee_{i\in I}\frak B_{i\xi}$.   Then
$\frak D=\bigcap_{\xi\in\Xi}\frak D_{\xi}$.

\proof{{\bf (a)} If $I$ is empty or a singleton,
the result is trivial.   Note also that for any $I$ we surely have
$\frak D\subseteq\bigcap_{\xi\in\Xi}\frak D_{\xi}$, so it will be enough to
show that $\bigcap_{\xi\in\Xi}\frak D_{\xi}\subseteq\frak D$.


\medskip

{\bf (b)} Suppose that $I=\{i,j\}$ has two members.   Let
$(\Omega,\Sigma,\mu)$ and $(\Omega',\Sigma',\mu')$ be probability spaces
with measure algebras which can be identified with
$(\frak A_i,\bar\mu\restrp\frak A_i)$,
$(\frak A_{j},\bar\mu\restrp\frak A_{j})$ respectively (321J);  set
$\Sigma_{\xi}=\{E:E\in\Sigma$, $E^{\ssbullet}\in\frak B_{i\xi}\}$,
$\Sigma'_{\xi}=\{F:E\in\Sigma'$, $F^{\ssbullet}\in\frak B_{j\xi}\}$ for
$\xi\in\Xi$.   Note that every $\Sigma_{\xi}$ and $\Sigma'_{\xi}$ is a
$\sigma$-algebra containing all negligible sets, and that
$\family{\xi}{\Xi}{\Sigma_{\xi}}$ and
$\family{\xi}{\Xi}{\Sigma'_{\xi}}$ are non-decreasing.

We can identify the measure algebra $(\frak C,\bar\lambda)$
of the product $(\Omega\times\Omega',\Lambda,\lambda)$
with the probability algebra free product of
$(\frak A_i,\bar\mu\restrp\frak A_i)$ and
$(\frak A_{j},\bar\mu\restrp\frak A_{j})$ (325D or 325I),
which in turn embeds naturally in $\frak A$ (325J, 325L);
write $\phi:\frak C\to\frak A$ for the embedding.
If $E\in\Sigma$ and $F\in\Sigma'$, then
$\phi((E\times F)^{\ssbullet})=E^{\ssbullet}\Bcap F^{\ssbullet}$.
Now, for $\xi\in\Xi$,
$\phi[\{W^{\ssbullet}:W\in\Sigma_{\xi}\tensorhat\Sigma'_{\xi}\}]$
is the $\sigma$-subalgebra of $\frak A$ generated by
$\frak B_{i\xi}\cup\frak B_{j\xi}$, that is,
$\frak D_{\xi}$.   So if $d\in\bigcap_{\xi\in\Xi}\frak D_{\xi}$,
we have for each
$\xi\in\Xi$ a $W_{\xi}\in\Sigma_{\xi}\tensorhat\Sigma'_{\xi}$ such that
$\phi(W_{\xi}^{\ssbullet})=d$.   It follows that
$\lambda(W_{\xi}\symmdiff W_{\eta})=0$ for all $\xi$, $\eta\in\Xi$.
By 634Jb, there is a $W\in\Tau\tensorhat\Tau'$ such that
$\lambda(W_{\xi}\symmdiff W)=0$ for every $\xi\in\Xi$, that is,
$\phi W^{\ssbullet}=d$, where $\Tau=\bigcap_{\xi\in\Xi}\Sigma_{\xi}$
and $\Tau'=\bigcap_{\xi\in\Xi}\Sigma'_{\xi}$.   But now
$\phi W^{\ssbullet}$ belongs to the
closed subalgebra of $\frak A$ generated by
$\{E^{\ssbullet}:E\in\Tau\}\cup\{F^{\ssbullet}:F\in\Tau'\}
=\frak B_i\cup\frak B_{j}$, which is $\frak D$.   As $d$ is arbitrary,
$\bigcap_{\xi\in\Xi}\frak D_{\xi}\subseteq\frak D$, as required.

\medskip

{\bf (c)} Suppose that $I$ is finite.   Then
$\frak D=\bigcap_{\xi\in\Xi}\frak D_{\xi}$.   \Prf\ Induce on $\#(I)$.
I have already dealt with
the cases $\#(I)\le 2$.   For the inductive step to
$\#(I)\ge 3$, take any $j\in I$ and set $I'=I\setminus\{j\}$.
Set $\frak A'=\bigvee_{i\in I'}\frak A_i$;
then $\frak A'$ and $\frak A_j$ are independent, as in 272K or 458Le.

For $\xi\in\Xi$, set $\frak D'_{\xi}=\bigvee_{i\in I'}\frak B_{i\xi}$.
By the
inductive hypothesis, $\frak D'=\bigvee_{i\in I'}\frak B_i$ is
$\bigcap_{\xi\in\Xi}\frak D'_{\xi}$.
Now, of course, $\family{\xi}{\Xi}{\frak D'_{\xi}}$ is non-decreasing, and
$\frak D_{\xi}=\frak B_{j\xi}\vee\frak D'_{\xi}$ for each $\xi$, while
$\frak D=\frak B_j\vee\frak D'$.
So (b) tells us that $\frak D=\bigcap_{\xi\in\Xi}\frak D_{\xi}$.
Thus the induction proceeds.\ \Qed

\medskip

{\bf (d)} Now consider the case of arbitrary $I$.   Suppose that
$d\in\bigcap_{\xi\in\Xi}\frak D_{\xi}$ and $0<\epsilon\le\bover12$.
Because the topological closure of the subalgebra
$\bigcup_{J\subseteq I\text{ is finite}}\bigvee_{i\in J}\frak A_i$
is $\bigvee_{i\in I}\frak A_i$ (323J), which contains $d$, there are a
finite $J\subseteq I$ and an element $e$ of $\bigvee_{i\in J}\frak A_i$
such that $\bar\lambda(d\Bsymmdiff e)\le\epsilon$.
This time, observe that $\tilde{\frak A}=\bigvee_{i\in J}\frak A_i$ and
$\tilde{\frak A}'=\bigvee_{i\in I\setminus J}\frak A_i$ are
independent.   Setting
$\tilde{\frak B}_{\xi}=\bigvee_{i\in J}\frak B_{i\xi}$ for $\xi\in\Xi$, we
see from (c) that
$\bigcap_{\xi\in\Xi}\tilde{\frak B}_{\xi}=\bigvee_{i\in J}\frak D_i$.

Now, following the pattern of (b), take
probability spaces $(\Omega,\Sigma,\mu)$ and $(\Omega',\Sigma',\mu')$
with measure algebras isomorphic to $\tilde{\frak A}$ and
$\tilde{\frak A}'$
respectively, and set
$\Sigma_{\xi}=\{E:E^{\ssbullet}\in\tilde{\frak B}_{\xi}\}$ for $\xi\in\Xi$.
Translating 634Jc into terms of the measure algebras, we
see that there is an $e_1\in\bigvee_{i\in J}\frak D_i$ such that
$\bar\mu(d\Bsymmdiff e_1)\le 3\epsilon$.   Of course
$e_1\in\frak D$.

As $\epsilon$ is arbitrary, $d$ belongs to the closure of $\frak D$ for the
measure-algebra topology;  but $\frak D$ is topologically closed, so
$d\in\frak D$.   Thus we have the required result in this case also, and
the proof is complete.
}%end of proof of 634K

\leader{634L}{Theorem} Let $\familyiI{\frak B_i}$ be a
stochastically independent family of closed
subalgebras of $\frak A$.   Suppose that for each
$i\in I$ we have a filtration $\family{t}{T}{\frak B_{it}}$ of closed
subalgebras of $\frak B_i$.   For each $t\in T$ set
$\frak C_t=\bigvee_{i\in I}\frak B_{it}$.

(a) $\family{t}{T}{\frak C_t}$ is a filtration.

(b) For $i\in I$ and $t\in T$,
$\frak B_i\cap\frak C_t=\frak B_{it}$ and
$\frak B_i$ and $\frak C_t$ are relatively independent over
$\frak B_{it}$.

(c) If $\family{t}{T}{\frak B_{it}}$ is right-continuous for every
$i\in I$, then $\family{t}{T}{\frak C_t}$ is right-continuous.

\proof{{\bf (a)} Immediate from the definitions of `filtration' and
`$\bigvee$'.

\medskip

{\bf (b)} Of course
$\frak B_i\cap\frak C_t\supseteq\frak B_{it}$.   Next,
$\frak B_i$ and $\bigvee_{j\in I\setminus\{i\}}\frak B_j$ are
independent, that is, they are relatively independent over $\{0,1\}$;
by 458Ld, they are relatively independent over
$\frak B_{it}\subseteq\frak B_i$.   {\it A fortiori},
$\frak B_i$ and $\bigvee_{j\ne i}\frak B_{jt}$ are relatively independent
over $\frak B_{it}$, and therefore
$\frak B_i$ and $\frak B_{it}\vee\bigvee_{j\ne i}\frak B_{jt}=\frak C_t$
are relatively independent over $\frak B_{it}$ (458Ld again).
But this means that
$\frak B_i\cap\frak C_t$ is relatively independent of itself over
$\frak B_{it}$, and must be equal to $\frak B_{it}$.

\medskip

{\bf (c)} If $t\in T$ is not isolated on the right, apply Theorem 634K with
$\Xi=\{s:s>t\}$ to see that $\frak C_t=\bigcap_{s>t}\frak C_s$.
}%end of proof of 634L

\leader{634M}{Corollary} Suppose that $\familyiI{(\frak A_i,\bar\mu_i)}$
is a family of probability algebras, and that $\family{t}{T}{\frak A_{it}}$
is a filtration in $\frak A_i$ for each $i$.   Then there are a probability
algebra $(\frak C,\bar\lambda)$ with a filtration
$\family{t}{T}{\frak C_t}$ and a stochastically independent
family $\familyiI{\frak B_i}$ of closed
subalgebras such that $\frak B_i$ is coordinated with $\family{t}{T}{\frak C_t}$
and $(\frak B_i,\bar\lambda\restrp\frak B_i,
\family{t}{T}{\frak B_i\cap\frak C_t})$ is isomorphic to
$(\frak A_i,\bar\mu_i,\family{t}{T}{\frak A_{it}})$ for every $i\in I$.
If every $\family{t}{T}{\frak A_{it}}$ is
right-continuous, we can arrange that
$\family{t}{T}{\frak C_t}$ should be right-continuous.

\proof{ Take $(\frak C,\bar\lambda,\familyiI{\varepsilon_i})$ to be the
probability algebra free product of $\familyiI{(\frak A_i,\bar\mu_i)}$
(325K), and set $\frak B_i=\varepsilon_i[\frak A_i]$ for $i\in I$,
$\frak C_t=\bigvee_{i\in I}\varepsilon_i[\frak A_{it}]$ for $t\in T$.
As noted in 325L, $\familyiI{\frak B_i}$ is an independent family of
closed subalgebras of $\frak C$, while
$(\frak B_i,\bar\lambda\restrp\frak B_i)\cong(\frak A_i,\bar\mu_i)$ for
every $i$, so 634L gives the result.
}%end of proof of 634M

\leader{634N}{Example:
independent Poisson \dvrocolon{processes}}\cmmnt{ To show how these
ideas may be used, I work through an important special case.

\medskip

}{\bf (a)} Let $(\frak B,\bar\nu,\langle\frak B_t\rangle_{t\ge 0},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb Bf}}{u_{\sigma}})$ be the
standard Poisson process\cmmnt{ of 612U} in its measure-algebra form.   Let
$(\frak A,\bar\mu)$ be the probability algebra free product of
$(\frak B,\bar\nu)$ with itself, with associated embeddings
$\varepsilon_1:\frak B\to\frak A$, $\varepsilon_2:\frak B\to\frak A$;
write $\frak B^{(i)}$ for $\varepsilon_i[\frak B]$ for each $i$.   Set
$\frak A_t=\varepsilon_1[\frak B_t]\vee\varepsilon_2[\frak B_t]$ for
$t\ge 0$, so that $\langle\frak A_t\rangle_{t\ge 0}$ is a
right-continuous filtration\cmmnt{ (612Uc, 632Db, 634Lc)}, while
$\varepsilon_i[\frak B_t]=\frak B^{(i)}\cap\frak A_t$ for both $i$
and every $t$, and each $\frak B^{(i)}$ is coordinated with
$\langle\frak A_t\rangle_{t\ge 0}$\cmmnt{ (634Lb)}.

\spheader 634Nb For each $i$, $\varepsilon_i$ is an isomorphism between
$(\frak B,\bar\nu,\langle\frak B_t\rangle_{t\ge 0})$ and
$(\frak B^{(i)},\bar\mu\restrp\frak B^{(i)},
\family{t}{\coint{0,\infty}}{\frak B^{(i)}\cap\frak A_t})$, so matches
$\Cal T_{\Bbb Bf}$ with
$\Cal T_{\Bbb B^{(i)}f}
=\Cal T_{\Bbb B^{(i)}}\cap\Cal T_{\Bbb Af}$\cmmnt{ (634Cb)}.
Now $\pmb u
=\family{\sigma}{\Cal T_{\Bbb Bf}}{u_{\sigma}}$ is matched with
$\pmb u_i
=\family{\sigma}{\Cal T_{\Bbb B^{(i)}f}}{u_{i\sigma}}$, and
$\pmb u_i$ is locally near-simple\cmmnt{ because $\pmb u$ is (634Eb)}.
Because $\Cal T_{\Bbb B^{(i)}f}$ contains all the constant processes,
$\sup\{\Bvalue{\tau\le\sigma}:\sigma\in\Cal T_{\Bbb B^{(i)}f}\}=1$ for every
$\tau\in\Cal T_{\Bbb Af}$, $\pmb u_i$
has an extension to a locally near-simple process
defined on $\Cal T_{\Bbb Af}$\cmmnt{ (631M(c-ii))}.   Because
$\Cal T_{\Bbb B^{(i)}f}$ contains the constant processes,\cmmnt{ it separates
$\Cal T_{\Bbb Af}$ (633D), so} the extension is unique\cmmnt{ (633F)};
I will call it $\pmb v_i$.

\spheader 634Nc For each $i$, $\pmb v_i-\pmb\iota_{\frak A}$ is a local
martingale.   \prooflet{\Prf\ By 632Ma, $\pmb u-\pmb\iota_{\frak B}$
(where $\pmb\iota_{\frak B}$ as
the identity process on $\Cal T_{\Bbb B f}$) is a local martingale, so
its copy $(\pmb v_i-\pmb\iota_{\frak B_i^{(f)}})\restr\Cal T_{\Bbb B^{(i)}f}$
is a local martingale (634Ia) and $\pmb v_i-\pmb\iota_{\frak A}$
is a local martingale
(633Pg;  of course $\Cal T_{\Bbb B^{(i)}f}$ separates
$\Cal T_{\Bbb Af}$ because it contains all constant processes).\ \QeD\
It follows that} $\pmb w=\pmb v_1-\pmb v_2$ is a local martingale.

\spheader 634Nd{\bf (i)}
$\pmb v_i^*=\pmb v_i$\cmmnt{, because $\pmb u^*=\pmb u$
(617Ob) so $\pmb u_i^*=\pmb u_i$, and we can apply 633Ph to see that
$\pmb v_i^*$ and $\pmb v_i$ agree on $\Cal T_{\Bbb Bf}$, and therefore on
$\Cal T_{\Bbb Af}$ (633F)}.

\medskip

\quad{\bf (ii)} The covariation $\covar{\pmb v_1}{\pmb v_2}$ is zero.
\prooflet{\Prf\ Express $\pmb v_i$ as
$\family{\sigma}{\Cal T_{\Bbb Af}}{v_{i\sigma}}$ for each $i$.
If $0\le s\le t$, then, writing $\check s$ and $\check t$ for the constant
stopping times at $s$ and $t$ respectively,
$v_{i\check t}-v_{i\check s}$ has a Poisson
distribution with mean $t-s$, so

\Centerline{$\bar\mu\Bvalue{v_{i,\check t}\ne v_{i,\check s}}=1-e^{-(t-s)}
\le t-s$}

\noindent for each $i$.   Now suppose that $t\ge 0$, $n\ge 1$ and that
$I$ is a sublattice of
$[\check 0,\check t]$ containing $\check s_j$ for every $j\le n$, where
$s_j=jt/n$.   In this case

$$\eqalignno{\theta(S_I(\tbf 1,d\pmb v_1d\pmb v_2))
&\le\bar\mu\Bvalue{S_I(\tbf 1,d\pmb v_1d\pmb v_2)\ne 0}
\cr&\le\bar\mu(\sup_{e\in\Sti_0(I)}
  \Bvalue{\Delta_e(\tbf 1,d\pmb v_1d\pmb v_2)\ne 0})
\Displaycause{where $\Sti_0(I)$ is the set of $I$-cells}
\le\bar\mu(\sup_{j<n}
  \Bvalue{v_{1,\check s_{j+1}}\ne v_{1,\check s_j}}
  \Bcap\Bvalue{v_{2,\check s_{j+1}}\ne v_{2,\check s_j}})
\Displaycause{because both $\pmb v_1$ and $\pmb v_2$ are non-decreasing}
\le\sum_{j=0}^{n-1}
  \bar\mu(\Bvalue{v_{1,\check s_{j+1}}\ne v_{1,\check s_j}}
    \Bcap\Bvalue{v_{2,\check s_{j+1}}\ne v_{2,\check s_j}})
\cr&
=\sum_{j=0}^{n-1}\bar\mu(\Bvalue{v_{1,\check s_{j+1}}\ne v_{1,\check s_j}})
  \cdot\bar\mu(\Bvalue{v_{2,\check s_{j+1}}\ne v_{2,\check s_j}})
\Displaycause{because $\Bvalue{v_{i,\check s_{j+1}}\ne v_{i,\check s_j}}
\in\frak B^{(i)}$ for all $i$ and $j$, and $\frak B^{(1)}$ and
$\frak B^{(2)}$ are stochastically independent}
\le\sum_{j=0}^{n-1}(s_{j+1}-s_j)^2
=\Bover{t^2}n.
}$$

\noindent Taking the limit as $I\uparrow\Cal S([\check 0,\check t])$,
$\int_{[\check 0,\check t]}d\pmb v_1d\pmb v_2=0$.
As $t$ is arbitrary, the covariation $\covar{\pmb v_1}{\pmb v_2}$ is zero
at every constant stopping time.
Because the constant stopping times separate $\Cal T_{\Bbb A}$,
and $\covar{\pmb v_1}{\pmb v_2}$ is locally near-simple
(631Jb), $\covar{\pmb v_1}{\pmb v_2}=0$, by 633F.\ \Qed}

\medskip

\quad{\bf (iii)} $\pmb{w}^*=\pmb v_1+\pmb v_2$.
\prooflet{

\Centerline{\Prf\ $\pmb w^*
=\covar{\pmb v_1-\pmb v_2}{\pmb v_1-\pmb v_2}
=\covar{\pmb v_1}{\pmb v_1}+\covar{\pmb v_2}{\pmb v_2}
=\pmb v_1^*+\pmb v_2^*
=\pmb v_1+\pmb v_2$.  \Qed}
}

\medskip

\quad{\bf (iv)}
The previsible variation of $\pmb w^2\restr\Cal T_{\Bbb Ab}$ is
$2\pmb\iota\restr\Cal T_{\Bbb Ab}$.
\prooflet{\Prf\ We know that $\pmb w$ is a local martingale, so
$\pmb w^2-\pmb w^*$ is a virtually local martingale (624B), therefore a
local martingale (632I);  now
(iii) just above, with (c), tells us that $\pmb w^2-2\pmb\iota_{\frak A}$ is a
local martingale, so $\pmb w^2\restr\Cal T_{\Bbb Ab}$ and
$2\pmb\iota\restr\Cal T_{\Bbb Ab}$
have the same previsible variation (as in the proof of 632Mb), which is
$2\pmb\iota\restr\Cal T_{\Bbb Ab}$, by 626S.\ \Qed}

\medskip

\spheader 634Ne \cmmnt{Note that}
$\pmb w$ corresponds to a L\'evy process\cmmnt{ (455Q, \S652)} derived from
the family $\langle\lambda'_t\rangle_{t\ge 0}$ where

\Centerline{$\lambda'_t(\{n\})
=e^{-2t}t^n\sum_{k=\max(-n,0)}^{\infty}\Bover{t^{2k}}{k!(k+n)!}$}

\noindent for $n\in\Bbb Z$\cmmnt{,
because $\pmb v_1$ and $\pmb v_2$ can be determined from
$\pmb w$ and $\pmb w^*$}.

%\leader{634Q}{}
%\leader{634R}{}
%\leader{634S}{}
%\leader{634T}{}
%\leader{634U}{}
%\leader{634V}{}
%\leader{634W}{}

\exercises{\leader{634X}{Basic exercises (a)}
%\spheader 634Xa
Let $\frak B$ be a closed subalgebra of $\frak A$, and
$\Bbb B=(\frak B,\bar\mu\restr\frak B,T,
\penalty-100\langle\frak B_t\rangle_{t\in T},
\Cal T_{\Bbb B},\family{\sigma}{\Cal T_{\Bbb B}}{\frak B_{\sigma}})$ the
corresponding stochastic integration structure, where
$\frak B_t=\frak B\cap\frak A_t$ for $t\in T$.
Suppose that $\Cal S$ is a
sublattice of $\Cal T_{\Bbb B}$ and that $\pmb u\in L^0(\frak B)^{\Cal S}$ is
fully adapted.
Show that (i) if $\pmb u$ is an
$\Bbb A$-integrator it is a $\Bbb B$-integrator
(ii) if $\pmb u$ is a local
$\Bbb A$-integrator it is a local $\Bbb B$-integrator and its
$\Bbb B$-quadratic variation is the same as its
$\Bbb A$-quadratic variation.
%634I

\leader{634Y}{Further exercises (a)}
%\spheader 634Ya
Let $\frak A$ be an eight-element Boolean algebra with
atoms $a$, $b$, $c$, and $\frak B$ the subalgebra generated by $\{b\}$.
Let $\frak A_0$ be the subalgebra generated by $\{a\}$ and for $t>0$ let
$\frak A_t$ be the subalgebra generated by $\{a,b\}$.   For $t>0$ set
$\tau_t=\check t(a\Bcup c)$ in the language of part (b) of
the proof of 634H.
Show that the infimum $\tau=\inf_{s>0}\tau_s$, taken in
$\Cal T_{\Bbb A}$, is such that $\Bvalue{\tau>0}=a$, so is not the
infimum of $\{\tau_s:s>0\}$ in $\Cal T_{\Bbb B}$.
%634C

\spheader 634Yb Suppose that $\Cal S$ is a sublattice of
$\Cal T_{\Bbb A}$ and that
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$
is a fully adapted process which is not of bounded variation.
(i) Show that there are $\frak C$, $\bar\lambda$,
$\family{t}{T}{\frak C_t}$, $\frak B$, $\family{t}{T}{\frak B_t}$
and $\pmb v'$ such that $(\frak C,\bar\lambda)$ is a probability
algebra, $\family{t}{T}{\frak C_t}$ is a filtration
of closed subalgebras of $\frak C$, $\frak B$ is a closed subalgebra of
$\frak C$, $\frak B_t=\frak B\cap\frak C_t$ for every $t\in T$,
$(\frak A,\bar\mu,\family{t}{T}{\frak A_t},\pmb v)$
is isomorphic to
$(\frak B,\bar\lambda\restrp\frak B,\family{t}{T}{\frak B_t},\pmb v')$,
but $\pmb v'$ is not an integrator
for the structure $(\frak C,\family{t}{T}{\frak C_t})$.
(ii) Show that if
$\family{t}{T}{\frak A_t}$ is right-continuous we can arrange
that $\family{t}{T}{\frak C_t}$ is right-continuous.
%634I mt63bits

\spheader 634Yc Give an example of a probability algebra
$(\frak A,\bar\mu)$ with a closed subalgebra $\frak C$ and a non-increasing
sequence $\sequencen{\frak B_n}$ of closed subalgebras such that
$\bigcap_{n\in\Bbb N}\frak B_n=\{0,1\}$ but
$\bigcap_{n\in\Bbb N}(\frak C\vee\frak B_n)\ne\frak C$.
%mt63bits 634K

\spheader 634Yd (`Emery's example', see {\smc Protter 05}, IV.2)
% p 178
In the structure defined in 634N, let
$\pmb x=\family{\tau}{\Cal T_{\Bbb Af}}{x_{\tau}}$ be the fully adapted
process defined by saying that $x_{\check 0}=0$ and
$x_{\tau}=\Bover1{\iota_{\tau}}$ if $\tau\in\Cal T_{\Bbb Af}$ and
$\Bvalue{\tau>0}=1$, where $\pmb\iota$ is the identity process.
(i) Show that $\int_{[\check 0,\tau]}\pmb x\,d\pmb w$ is defined for
every $\tau\in\Cal T_{\Bbb Af}$.   (ii) Show that
the indefinite integral $\pmb z=ii_{\pmb w}(\pmb x)$ is not a local
martingale, but that for every $\epsilon>0$ there is a local martingale
$\pmb z'$ with domain $\Cal T_{\Bbb Af}$ such that
$\bar\mu\Bvalue{\pmb z\ne\pmb z'}\le\epsilon$.
%634N
}%end of exercises

\endnotes{
\Notesheader{634} Even by the standards of this volume, I have taken
634B-634E %634B 634A 634C 634E
extremely slowly.   The point is that nearly everything up to this point
was written on the understanding that we are starting from a settled
structure $(\frak A,\bar\mu,\family{t}{T}{\frak A_t})$.   I have remarked
many times that the topology of convergence in measure is more important
than the measure itself, but this is fairly easy to incorporate into one's
intuitive picture, because so many of the definitions and arguments refer
directly to the topology.   In this section I am exploring a much more
radical change.   Substituting $\frak B$ for $\frak A$, in the
measure-algebra version of probability theory I am working in here,
corresponds to a
deliberate closing of the eyes to some of the randomness in the `outer'
model $(\frak A,\bar\mu)$.   Probability theory would of course be
impossible without this.   We never suppose that we have grasped all the
possible stochastic elements of a real-world situation, and must always be
ready to extend our model by elaborating on our probability space or
probability algebra.   I have discussed this aspect of probability
theory in the introduction to \S275.   But this means that we must at some
point check every concept and theorem for the transformations which will be
appropriate when the framework shifts.

Many of the checks are so
elementary that they hardly need mentioning, and you may feel that I have
laboured unnecessarily.   If so, do feel at
liberty to write `obvious' in the margin of your copy.
In particular, it is not clear that we need to spell 634B out explicitly, rather
than proceed directly to the more intuitively appealing 634C.   But I fear that
the tempting intuitions here are not perfectly safe, and that pulling the
algebras $\frak B$, $\frak A$ apart as in 634B forces a necessary extra clarity,
if you can face the extra elaboration in the notation.

We come to some new phenomena when we investigate relative independence
(634F-634G).   The point is that the concepts of `martingale' and
`integrator' are less absolute than `fully adapted', `bounded variation'
and so on.   Even Brownian motion, unsuitably embedded, can become a
process which is not an integrator (634Yb).   Without checking,
we cannot be sure that the conditional expectation associated
with $\frak B_t=\frak B\cap\frak A_t$ is appropriately related to the
conditional expectation associated with $\frak A_t$.
When it is, we have a much closer relationship between the structure
$(\frak B,\bar\mu\restrp\frak B,\family{t}{T}{\frak B_t},
\Cal T_{\Bbb B})$ and its attendant space $M_{\text{fa}\Bbb B}$ of
fully adapted processes, regarded in isolation,
and its alternative realization as a subspace of $M_{\text{fa}\Bbb A}$
(634H, 634I)..

Product measures provide a tool for representing arbitrary independent
families of random variables (272J).   The corresponding construction for
probability algebras is the probability algebra free product of \S325.
To use this, we need to be able to define a filtration on the product from
given filtrations on the factors.   To begin with, this is easy
(634La-634Lb), at least if you know the right things about relative
independence.   But elsewhere in this section we repeatedly depend on
right-continuity, and to be sure that the elementary construction I offer
produces a right-continuous filtration
on the product we have to look hard at product measures (634J).
}%end of notes

\discrpage

\frfilename{mt635.tex}
\versiondate{6.3.23/29.9.23}
\copyrightdate{2012}

\def\chaptername{Structural alterations}
\def\sectionname{Changing the filtration}
\def\Tni{T_{\text{ni}}}

\newsection{635}

In this section I introduce the elementary theory of
`local times'.   In the principal applications, we have a process
which is easier to handle if we replace the standard clock
$T$ with a variable-speed clock
$\family{r}{R}{\pi_r}$ where the clock-times are now a totally ordered
family of stopping times.   I will come to such applications in
Chapter 65.   Here I want to set up a language to discuss the
transformation in which
a process $\family{\tau}{\Cal T}{u_{\tau}}$ is mapped to
$\family{\rho}{\Cal R}{u_{\pi(\rho)}}$, where
$\Cal R$ is the lattice of $R$-based stopping times and
$\pi(\rho)\in\Cal T$ corresponds to $\rho\in\Cal R$.
Starting from the construction in 635B, we have basic algebraic properties
(corresponding to ideas in \S611) in 635C and can then follow a programme along
the same lines as elsewhere, looking at the usual kinds of process and
Riemann-sum integrals.


\cmmnt{\leader{635A}{Notation} \Asissix\ will be a stochastic integration
structure, and $\theta$ the usual F-norm on $L^0(\frak A)$ (613Ba).
For $\tau\in\Cal T$, $P_{\tau}$ will be the conditional expectation operator
associated with $\frak A_{\tau}$.  If $\Cal S$ is a
sublattice of $\Cal T$,
$\Cal I(\Cal S)$ will be the family of finite sublattices of $\Cal S$.

Now for the new idea.}

\leader{635B}{Construction} For the whole of this section,
$(R,\le)$ will be a new (non-empty) totally ordered set,
and $\family{r}{R}{\pi_r}$ a non-decreasing family in $\Cal T$.
For $r\in R$, I will write $\frak B_r$ for $\frak A_{\pi_r}$, so that
$\family{r}{R}{\frak B_r}$ is a filtration of closed subalgebras of
$\frak A$, and we shall have a corresponding stochastic integration structure
$(\frak A,\bar\mu,R,R,\family{r}{R}{\frak B_r},\Cal R,
\family{\rho}{\Cal R}{\frak B_{\rho}})$.

For $\rho\in\Cal R$, $\pi(\rho)\in L^0(\frak A)^T$ will be defined by
saying that

$$\eqalign{\Bvalue{\pi(\rho)>t}
&=\inf_{r\in R}(\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t})
\text{ if }t\in T\text{ is isolated on the right in }T,\cr
&=\sup_{s>t}\inf_{r\in R}(\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>s})
\text{ for other }t\in T.\cr}$$

\leader{635C}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.

(a) For every $\rho\in\Cal R$,
$\pi(\rho)$, as defined in 635B, belongs
to $\Cal T$, and $\frak B_{\rho}=\frak A_{\pi(\rho)}$.

(b)(i) The map $\pi:\Cal R\to\Cal T$ is a lattice homomorphism.

\quad (ii) $\pi(\min\Cal R)=\inf_{r\in R}\pi_r$ in $\Cal T$,
$\pi(\max\Cal R)=\max\Cal T$.

\quad(iii) If $r\in R$ and $\check{r}\in\Cal R$ is the corresponding
constant stopping time, then $\pi(\check{r})=\pi_r$.

\quad(iv) If $\rho\in\Cal R_f$ then
$\pi(\rho)\le\sup_{r\in R}\pi_r$.

(c)
$\Bvalue{\pi(\rho)<\pi(\rho')}\Bsubseteq\Bvalue{\rho<\rho'}$,
$\Bvalue{\pi(\rho)\le\pi(\rho')}
\Bsupseteq\Bvalue{\rho\le\rho'}$ and
$\Bvalue{\rho=\rho'}\Bsubseteq\Bvalue{\pi(\rho)=\pi(\rho')}$
for all $\rho$, $\rho'\in\Cal R$.

(d) Suppose that $\family{r}{R}{\pi_r}$ is {\bf right-continuous} in the sense
that $\pi_r=\inf_{q\in R,q>r}\pi_q$ in $\Cal T$ whenever $r\in R$ is not
isolated on the right in $R$.   Then

\quad(i) $\family{r}{R}{\frak B_r}$ is right-continuous;

\quad(ii) $\pi$ is right-continuous\cmmnt{ (see 632B)}.

\proof{{\bf (a)(i)} For $\rho\in\Cal R$ and $r\in R$ there is a
$\pi(\rho,r)\in\Cal T$ defined by saying that
$\Bvalue{\pi(\rho,r)>t}
=\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}$ for $t\in T$.
\Prf\ $\Bvalue{\rho>r}\in\frak B_r=\frak A_{\pi_r}$;  looking at the
definition of $\frak A_{\pi_r}$ (611G), we see that
$\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}\in\frak A_t$ for every
$t\in T$ .   If $s\le t$ in $T$ then
$\Bvalue{\pi_r>t}\Bsubseteq\Bvalue{\pi_r>s}$ so
$\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}
\Bsubseteq\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>s}$.   If $t\in T$ is not
isolated on the right,
$\Bvalue{\pi_r>t}=\sup_{s>t}\Bvalue{\pi_r>s}$ so
$\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}
=\sup_{s>t}\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>s}$.
So the function $t\mapsto\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}$
satisfies the conditions of 611A(b-i).\ \Qed

\medskip

\quad{\bf (ii)} Comparing the definition of $\pi(\rho)$ in
635B with the formula in 632C(a-i) (which is applicable because
$\family{t}{T}{\frak A_t}$ is right-continuous), we see that
$\pi(\rho)$ is just $\inf_{r\in R}\pi(\rho,r)$ in $\Cal T$,
and certainly belongs to $\Cal T$.

\medskip

\quad{\bf (iii)} Take any $\rho\in\Cal R$.
For $a\in\frak A$,

$$\eqalign{a\in\frak B_{\rho}
&\iff a\Bsetminus\Bvalue{\rho>r}\in\frak B_r=\frak A_{\pi_r}
  \text{ for every }r\in R\cr
&\iff a\Bsetminus(\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t})\in\frak A_t
  \text{ for every }r\in R\text{ and }t\in T\cr
&\iff a\Bsetminus\Bvalue{\pi(\rho,r)>t}\in\frak A_t
  \text{ for every }r\in R\text{ and }t\in T\cr
&\iff a\in\frak A_{\pi(\rho,r)}
  \text{ for every }r\in R
\iff a\in\bigcap_{r\in R}\frak A_{\pi(\rho,r)}
=\frak A_{\pi(\rho)}
  \cr}$$

\noindent by 632C(a-iii), again using the general hypothesis that
$\family{t}{T}{\frak A_t}$ is right-continuous.   So
$\frak B_{\rho}=\frak A_{\pi(\rho)}$.

\medskip

{\bf (b)(i)}\grheada\ For $r\in R$,
$\rho\mapsto\pi(\rho,r):\Cal R\to\Cal T$ is a
lattice homomorphism.   \Prf\ Use 611Cb and 611Cc;
we just have to note that $\rho\mapsto\Bvalue{\rho>r}$ is a
lattice homomorphism, so that $\rho\mapsto\Bvalue{\pi(\rho,r)>t}$ is a
lattice homomorphism for every $t\in T$.\ \Qed

\medskip

\qquad\grheadb\
$\pi(\rho\wedge\rho')=\pi(\rho)\wedge\pi(\rho')$
for $\rho$, $\rho'\in\Cal R$.   \Prf\ By ($\alpha$), both are equal to
$\inf_{r\in R}(\pi(\rho,r)\wedge\pi(\rho',r))$.\ \Qed

\medskip

\qquad\grheadc\
$\pi(\rho)\vee\pi(\rho')\le\pi(\rho\vee\rho')$ for
all $\rho$, $\rho\in\Cal R$.   (By ($\beta$), or otherwise,
$\pi$ is order-preserving.)

\medskip

\qquad\grheadd\
$\pi(\rho\vee\rho')\le\pi(\rho)\vee\pi(\rho')$ for
all $\rho$, $\rho'\in\Cal R$.   \Prf\ For any $t\in T$,

$$\eqalign{&\inf_{r\in R}\Bvalue{\pi(\rho,r)>t}
   \Bcup\inf_{r\in R}\Bvalue{\pi(\rho',r)>t}
=\inf_{r,r'\in R}
   (\Bvalue{\pi(\rho,r)>t}\Bcup\Bvalue{\pi(\rho',r')>t})\cr
&\mskip100mu
=\inf_{r,r'\in R}(\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}
   \Bcup\Bvalue{\rho'>r'}\Bcup\Bvalue{\pi_{r'}>t})\cr
&\mskip100mu
\Bsupseteq\inf_{r,r'\in R}
   (\Bvalue{\rho>\max(r,r')}\Bcup\Bvalue{\rho'>\max(r,r')}
      \Bcup\Bvalue{\pi_r\vee\pi_{r'}>t})\cr
&\mskip100mu
=\inf_{r,r'\in R}
  (\Bvalue{\rho\vee\rho'>\max(r,r')}\Bcup\Bvalue{\pi_{\max(r,r')}>t}
=\inf_{r\in R}\Bvalue{\pi(\rho\vee\rho',r)>t}.\cr}$$

\noindent So if $t$ is isolated on the right,

$$\eqalignno{\Bvalue{\pi(\rho\vee\rho')>t}
&=\inf_{r\in R}\Bvalue{\pi(\rho\vee\rho',r)>t}\cr
\displaycause{632C(a-i) again}
&\Bsubseteq\inf_{r\in R}\Bvalue{\pi(\rho,r)>t}
   \Bcup\inf_{r\in R}\Bvalue{\pi(\rho',r)>t}
\cr&=\Bvalue{\pi(\rho)>t}\Bcup\Bvalue{\pi(\rho')>t}
=\Bvalue{\pi(\rho)\vee\pi(\rho')>t},\cr}$$

\noindent while if $t$ is not isolated on the right,

$$\eqalignno{\Bvalue{\pi(\rho\vee\rho')>t}
&=\sup_{s>t}\inf_{r\in R}\Bvalue{\pi(\rho\vee\rho',r)>s}
\Bsubseteq\sup_{s>t}(\inf_{r\in R}\Bvalue{\pi(\rho,r)>s}
   \Bcup\inf_{r\in R}\Bvalue{\pi(\rho',r)>s})\cr
&=\sup_{s>t}\inf_{r\in R}\Bvalue{\pi(\rho',r)>s}
   \Bcup\sup_{s>t}\inf_{r\in R}\Bvalue{\pi(\rho',r)>s}\cr
&=\Bvalue{\pi(\rho)>t}\Bcup\Bvalue{\pi(\rho')>t}
=\Bvalue{\pi(\rho)\vee\pi(\rho')>t}.\cr}$$

\noindent Accordingly
$\pi(\rho\vee\rho')\le\pi(\rho)\vee\pi(\rho')$.\ \Qed

\medskip

\qquad\grheade\
Thus $\pi:\Cal R\to\Cal T$ is a lattice homomorphism.

\medskip

\quad{\bf (ii)} If $r\in R$, then

\Centerline{$\Bvalue{\pi(\min\Cal R,r)>t}=\Bvalue{\pi_r>t}$,
\quad$\Bvalue{\pi(\max\Cal R,r)>t}=1$}

\noindent for every $t\in T$, so $\pi(\min\Cal R,r)=\pi_r$ and
$\pi(\max\Cal R,r)=\max\Cal T$.   Accordingly

\Centerline{$\pi(\min\Cal R)
=\inf_{r\in R}\pi(\min\Cal R,r)=\inf_{r\in R}\pi_r$,}

\Centerline{$\pi(\max\Cal R)=\inf_{r\in R}\pi(\max\Cal R,r)
=\max\Cal T$.}

\medskip

\quad{\bf (iii)} For $q\in R$ and $t\in T$,

$$\eqalign{\Bvalue{\pi(\check{r},q)>t}
&=\Bvalue{\pi_q>t}\text{ if }r\le q,\cr
&=1\text{ otherwise},\cr}$$

\noindent so

$$\eqalign{\pi(\check{r},q)
&=\pi_q\text{ if }r\le q,\cr
&=\max\Cal T\text{ otherwise}.\cr}$$

\noindent Taking the infimum over $r$,
$\pi(\check{r})=\inf_{q\ge r}\pi_q=\pi_r$.

\medskip

\quad{\bf (iv)} Write $\tau$ for the supremum $\sup_{r\in R}\pi_r$ taken in
$\Cal T$.   If $\rho\in\Cal R_f$ then

$$\eqalign{\inf_{r\in R}(\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t})
&\Bsubseteq\inf_{r\in R}(\Bvalue{\rho>r}\Bcup\Bvalue{\tau>t})\cr
&=\Bvalue{\tau>t}\Bcup\inf_{r\in R}\Bvalue{\rho>r}
=\Bvalue{\tau>t}.\cr}$$

\noindent So

$$\eqalign{\Bvalue{\pi(\rho)>t}
&=\inf_{r\in R}(\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t})
\Bsubseteq\Bvalue{\tau>t}
\text{ if }t\in T\text{ is isolated on the right in }T,\cr
&=\sup_{s>t}\inf_{r\in R}(\Bvalue{\rho>s}\Bcup\Bvalue{\pi_s>t})
\Bsubseteq\sup_{s>t}\Bvalue{\tau>s}=\Bvalue{\tau>t}
\text{ for other }t\in T,\cr}$$

\noindent and $\pi(\rho)\le\tau$, as claimed.

\medskip

{\bf (c)(i)}
Set $a=\Bvalue{\rho<\rho'}$.   If $t\in T$ and $r\in R$, then

\Centerline{$\Bvalue{\pi(\rho',r)>t}
=\Bvalue{\rho'>r}\Bcup\Bvalue{\pi_r>t}
\Bsubseteq a\Bcup\Bvalue{\rho>r}\Bcup\Bvalue{\pi_r>t}
=a\Bcup\Bvalue{\pi(\rho,r)>t}$.}

\noindent If $t$ is isolated on the right in $T$, then

$$\eqalign{\Bvalue{\pi(\rho')>t}
&=\inf_{r\in R}\Bvalue{\pi(\rho',r)>t}
\Bsubseteq\inf_{r\in R}(a\Bcup\Bvalue{\pi(\rho,r)>t})\cr
&=a\Bcup\inf_{r\in R}\Bvalue{\pi(\rho,r)>t}
=a\Bcup\Bvalue{\pi(\rho)>t},\cr}$$

\noindent while if $t$ is not isolated on the right,

$$\eqalign{\Bvalue{\pi(\rho')>t}
&=\sup_{s>t}\inf_{r\in R}\Bvalue{\pi(\rho',r)>s}
\Bsubseteq\sup_{s>t}\inf_{r\in R}a\Bcup\Bvalue{\pi(\rho,r)>s}\cr
&=a\Bcup\sup_{s>t}\inf_{r\in R}\Bvalue{\pi(\rho,r)>s}
=a\Bcup\Bvalue{\pi(\rho)>t}.\cr}$$

\noindent So

\Centerline{$\Bvalue{\pi(\rho)<\pi(\rho')}
=\sup_{t\in T}(\Bvalue{\pi(\rho')>t}\Bsetminus\Bvalue{\pi(\rho)>t})
\Bsubseteq a$.}

\medskip

\quad{\bf (ii)} Similarly,
$\Bvalue{\pi(\rho')<\pi(\rho)}\Bsubseteq\Bvalue{\rho'<\rho}$;
taking complements, $\Bvalue{\pi(\rho)\le\pi(\rho')}
\Bsupseteq\Bvalue{\rho\le\rho'}$.

\medskip

\quad{\bf (iii)} Now

$$\eqalign{\Bvalue{\pi(\rho)=\pi(\rho')}
&=\Bvalue{\pi(\rho)\le\pi(\rho')}
   \Bcap\Bvalue{\pi(\rho')\le\pi(\rho)}\cr
&\Bsupseteq\Bvalue{\rho\le\rho'}\Bcap\Bvalue{\rho'\le\rho}
=\Bvalue{\rho=\rho'}.\cr}$$

\medskip

{\bf (d)(i)} If $r\in R$ is not isolated on the right, then
$\pi_r=\inf_{q>r}\pi_q$, so

\Centerline{$\frak B_r=\frak A_{\pi_r}
=\bigcap_{q>r}\frak A_{\pi_q}=\bigcap_{q>r}\frak B_q$}

\noindent (632C(a-iii) again).  %OK
Thus $\family{r}{R}{\frak B_r}$ is right-continuous.

\medskip

\quad{\bf (ii)} Set

\Centerline{$\tau=\inf_{\rho\in D}\pi(\rho)
=\inf_{\rho\in D,r\in R}\pi(\rho,r)$,}

\noindent taken in $\Cal T$.

\medskip

\qquad\grheada\ If $r\in R$ is isolated on the right, then
$\tau\le\pi(\inf D,r)$.   \Prf\ For any $t\in T$,

$$\eqalignno{\Bvalue{\pi(\inf D,r)>t}&
=\Bvalue{\inf D>r}\Bcup\Bvalue{\pi_r>t}
=(\inf_{\rho\in D}\Bvalue{\rho>r})\Bcup\Bvalue{\pi_r>t}
\Displaycause{applying 632C(a-i) to $\family{r}{\Cal R}{\frak B_r}$}
=\inf_{\rho\in D}\Bvalue{\pi(\rho,r)>t}
\Bsupseteq\Bvalue{\inf_{\rho\in D}\pi(\rho,r)>t}
\Bsupseteq\Bvalue{\tau>t}.   \text{ \Qed}\cr}$$

\medskip

\qquad\grheadb\ If $r\in R$ is not isolated on the right,
then $\tau\le\pi(\inf D,r)$.   \Prf\ We have
$\pi_r=\inf_{q>r}\pi_q$.   If $t\in T$ is isolated on the right,

$$\eqalign{\Bvalue{\pi(\inf D,r)>t}&
=\Bvalue{\inf D>r}\Bcup\Bvalue{\pi_r>t}
=(\sup_{q>r}\inf_{\rho\in D}\Bvalue{\rho>q})
   \Bcup\Bvalue{\inf_{q>r}\pi_q>t}\cr&
=(\sup_{q>r}\inf_{\rho\in D}\Bvalue{\rho>q})
   \Bcup(\inf_{q>r}\Bvalue{\pi_q>t})
\Bsupseteq\inf_{q>r}(\Bvalue{\pi_q>t}
   \Bcup\inf_{\rho\in D}\Bvalue{\rho>q})\cr&
=\inf_{q>r}\inf_{\rho\in D}(\Bvalue{\rho>q}
   \Bcup\Bvalue{\pi_q>t})
=\inf_{q>r}\inf_{\rho\in D}\Bvalue{\pi(\rho,q)>t}
\Bsupseteq\Bvalue{\tau>t}.\cr}$$

\noindent If $t\in T$ is not isolated on the right, then

$$\eqalign{\Bvalue{\pi(\inf D,r)>t}&
=\Bvalue{\inf D>r}\Bcup\Bvalue{\pi_r>t}
=(\sup_{q>r}\inf_{\rho\in D}\Bvalue{\rho>q})
     \Bcup(\sup_{s>t}\inf_{q>r}\Bvalue{\pi_q>s})\cr&
=\sup_{s>t}((\sup_{q>r}\inf_{\rho\in D}\Bvalue{\rho>q})
     \Bcup(\inf_{q>r}\Bvalue{\pi_q>s}))\cr&
\Bsupseteq\sup_{s>t}\inf_{q>r}(\inf_{\rho\in D}\Bvalue{\rho>q}
     \Bcup\Bvalue{\pi_q>s})\cr&
%&=\sup_{s>t}\inf_{q>r}\inf_{\rho\in D}(\Bvalue{\rho>q}
%     \Bcup\Bvalue{\pi_q>s})\cr
=\sup_{s>t}\inf_{q>r}\inf_{\rho\in D}\Bvalue{\pi(\rho,q)>s}
\Bsupseteq\sup_{s>t}\Bvalue{\tau>s}
=\Bvalue{\tau>t}.\cr}$$

\noindent Thus $\Bvalue{\tau>t}\Bsubseteq\Bvalue{\pi(\inf D,r)>t}$
for every $t$ and $\tau\le\pi(\inf D,r)$.\ \Qed

\medskip

\qquad\grheadc\ Putting these together,
$\tau\le\pi(\inf D,r)$ for
every $r$ and $\tau\le\pi(\inf D)$.   On the other hand,

\Centerline{$\pi(\inf D)=\inf_{r\in R}\pi(\inf D,r)
\le\inf_{\rho\in D,r\in R}\pi(\rho,r)=\tau$.}

\noindent So in fact
$\pi(\inf D)=\tau=\inf_{\rho\in D}\pi(\rho)$,
as required.
}%end of proof of 635C

%\query do we have a uniqueness result in 635B-635C?

\leader{635D}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal Q$ be a sublattice of $\Cal R$.

(a)(i) $\pi[\Cal Q]$ is a sublattice of $\Cal T$.

\quad (ii) If $\pmb u=\family{\tau}{\pi[\Cal Q]}{u_{\tau}}$ is a process fully
adapted to $\family{t}{T}{\frak A_t}$, then
$\pmb u\pi=\family{\rho}{\Cal Q}{u_{\pi(\rho)}}$ is fully adapted to
$\family{r}{R}{\frak B_r}$.

\quad (iii) Let
$\psi:\pi[\Cal Q]^{2\uparrow}\to L^0(\frak A)$ be an adapted interval
function\cmmnt{ (613C)}, and set
$\psi_{\pi}(\rho,\rho')=\psi(\pi(\rho),\pi(\rho'))$ when
$\rho\le\rho'$ in $\Cal Q$.   Then $\psi_{\pi}$ is an adapted interval
function.

\quad (iv) In (iii), if $\psi$ is strictly adapted
then $\psi_{\pi}$ is strictly adapted.

(b) Now suppose that $\pmb u=\family{\tau}{\pi[\Cal Q]}{u_{\tau}}$ is
fully adapted and that $\psi$ is an adapted interval function
on $\pi[\Cal Q]$.
Then $\int_{\Cal Q}\pmb u\pi\,d\psi_{\pi}=\int_{\pi[\Cal Q]}\pmb u\,d\psi$
if either is defined.

\proof{{\bf (a)(i)} This is just because $\pi$ is a
lattice homomorphism (635C(b-i)).

\medskip

\quad{\bf (ii)} If $\rho\in\Cal Q$ then
$u_{\pi(\rho)}\in L^0(\frak A_{\pi(\rho)})
=L^0(\frak B_{\rho})$
by 635Ca.   If $\rho$, $\rho\in\Cal Q$ then

\Centerline{$\Bvalue{u_{\pi(\rho)}=u_{\pi(\rho)}}
\Bsupseteq\Bvalue{\pi(\rho)=\pi(\rho)}
\Bsupseteq\Bvalue{\rho=\rho}$}

\noindent by 635Cc.   So
$\family{\rho}{\Cal Q}{u_{\pi(\rho)}}$ satisfies both the conditions of
612Da.

\medskip

\quad{\bf (iii)}\grheada\ If $\rho\le\rho'$ in $\Cal Q$ then
$\pi(\rho)\le\pi(\rho')$ in $\pi[\Cal Q]$ (by 635C(b-i) again),
so $\psi(\pi(\rho),\pi(\rho'))$ is defined and belongs to
$L^0(\frak A_{\pi(\rho')})=L^0(\frak B_{\rho'})$ (635Ca).   Thus
$\psi_{\pi}$ is a well-defined function on $\Cal Q^{2\uparrow}$.
And of course $\psi_{\pi}(\rho,\rho)=\psi(\pi(\rho),\pi(\rho))=0$
for every $\rho\in\Cal Q$.

\medskip

\qquad\grheadb\ If $\rho$, $\rho'$, $\sigma$, $\sigma'\in\Cal Q$,
$\rho\le\rho'\le\sigma'\le\sigma$, $b\in\frak B_{\rho}$ and
$b\Bsubseteq\Bvalue{\rho=\rho'}\Bcap\Bvalue{\sigma'=\sigma}$, then
$\pi(\rho)\le\pi(\rho')\le\pi(\sigma')\le\pi(\sigma)$ in $\pi[\Cal Q]$,
$b\in\frak A_{\pi(\rho)}$ and
$b\Bsubseteq\Bvalue{\pi(\rho)=\pi(\rho')}
    \Bcap\Bvalue{\pi(\sigma')=\pi(\sigma)}$ (635Cc), so
$b\Bsubseteq\Bvalue{\psi(\pi(\rho),\pi(\sigma))=\psi(\pi(\rho'),\pi(\sigma'))}$.
With ($\alpha$), this shows that $\psi_{\pi}$ is an
adapted interval function.

\medskip

\quad{\bf (iv)} Continuing from (iii), suppose that $\psi$ is strictly adapted.
Then

$$\eqalign{\Bvalue{\rho=\rho'}\Bcap\Bvalue{\sigma'=\sigma}&
\Bsubseteq\Bvalue{\pi(\rho)=\pi(\rho')}\Bcap\Bvalue{\pi(\sigma')=\pi(\sigma)}
  \cr&
\Bsubseteq\Bvalue{\psi(\pi(\rho),\pi(\sigma))=\psi(\pi(\rho'),\pi(\sigma'))}
  \cr}$$

\noindent so $\psi_{\pi}$ is  strictly adapted.

\medskip

{\bf (b)(i)} We know from (a) that $\pmb u\pi$ is a fully adapted
process and
$\psi_{\pi}$ is an adapted interval function.   If $J\in\Cal I(\Cal Q)$,
then $\pi[J]\in\Cal I(\pi[\Cal Q])$ because $\pi$ is a lattice homomorphism.   Now
$S_J(\pmb u\pi,d\psi_{\pi})=S_{\pi[J]}(\pmb u,d\psi)$.   \Prf\ If $J$ is empty,
this is trivial.   Otherwise, let $(\rho_0,\ldots,\rho_n)$
linearly generate the $J$-cells (611L).   If $\rho\in J$, then

\Centerline{$\sup_{i\le n}\Bvalue{\pi(\rho)=\pi(\rho_i)}
\Bsupseteq\sup_{i\le n}\Bvalue{\rho=\rho_i}
=1$.}

\noindent Consider the string $(\tau_0,\ldots,\tau_n)$ where
$\tau_i=\pi(\rho_i)$ for $i\le n$.
This is a non-decreasing sequence in $\pi[J]$.   If $\tau\in\pi[J]$, there must
be a $\rho\in J$ such that $\pi(\rho)=\tau$;  now
$\rho_0\le\rho\le\rho_n$, so
$\tau_0\le\tau\le\tau_n$.   Thus $\tau_0=\min\pi[J]$ and
$\tau_n=\max\pi[J]$.   If $i<n$ and
$\tau\in\pi[J]\cap[\tau_i,\tau_{i+1}]$, let $\rho\in J$ be such that
$\tau=\pi(\rho)$;  setting
$\rho'=\med(\rho_i,\rho,\rho_{i+1})$,
$\pi(\rho')=\med(\tau_i,\tau,\tau_{i+1})=\tau$ while
$\rho'\in\{\rho_i,\rho_{i+1}\}$, so that
$\tau\in\{\tau_i,\tau_{i+1}\}$.   But this means that
$\{\tau_0,\ldots,\tau_n\}$ is a maximal totally ordered subset of
$\pi[J]$, so that
$(\tau_0,\ldots,\tau_n)$ linearly generates the $\pi[J]$-cells.

Consequently

$$\eqalign{S_{\pi[J]}(\pmb u,d\psi)
=\sum_{i=0}^{n-1}u_{\tau_i}
  \times\psi(\tau_i,\tau_{i+1})
=\sum_{i=0}^{n-1}u_{\pi(\rho_i)}\times\psi_{\pi}(\rho_i,\rho_{i+1})
=S_J(\pmb u\pi,d\psi_{\pi}).   \text{ \Qed}\cr}$$

\medskip

\quad{\bf (ii)} Suppose that $z=\int_{\pi[\Cal Q]}\pmb u\,d\psi$ is defined.
Let $\epsilon>0$.   Then there is an $I_0\in\Cal I(\pi[\Cal Q])$ such that
$\theta(z-S_I(\pmb u,d\psi))\le\epsilon$ whenever
$I_0\subseteq I\in\Cal I(\pi[\Cal Q])$.   Let $A$ be a finite subset of
$\Cal Q$ such that $I_0=\pi[A]$, and
$J_0\in\Cal I(\Cal Q)$ the sublattice generated by $A$.   If
$J_0\subseteq J\in\Cal I(\Cal Q)$, then
$\pi[J]\supseteq\pi[J_0]\supseteq I_0$, so

\Centerline{$\theta(z-S_J(\pmb u\pi,d\psi_{\pi}))
=\theta(z-S_{\pi[J]}(\pmb u,d\psi))\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\int_{\Cal Q}\pmb u\pi\,d\psi_{\pi}$
is defined and equal to $z$.

\medskip

\quad{\bf (iii)} Suppose that $z=\int_{\Cal Q}\pmb u\pi\,d\psi_{\pi}$ is
defined.   Let $\epsilon>0$.   Then there is a $J_0\in\Cal I(\Cal Q)$ such that
$\theta(z-S_J(\pmb u\pi,d\psi_{\pi}))\le\epsilon$ whenever
$J_0\subseteq J\in\Cal I(\Cal Q)$.   If
$\pi[J_0]\subseteq I\in\Cal I(\pi[\Cal Q])$, let $A$ be a finite subset of
$\Cal Q$ such that $I=\pi[A]$, and let
$J$ be the sublattice of $\Cal Q$ generated by $J_0\cup A$;  then
$J_0\subseteq J\in\Cal I(\Cal Q)$ and $\pi[J]=I$, so

\Centerline{$\theta(z-S_I(\pmb u,d\psi))
=\theta(z-S_J(\pmb u\pi,d\psi_{\pi}))\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\int_{\pi[\Cal Q]}\pmb u\,d\psi$ is
defined and equal to $z$.   This completes the proof.
}%end of proof of 635D

\leader{635E}{Proposition} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S\subseteq\Cal T$ be a sublattice and
$\pmb u=\family{\tau}{\Cal S}{u_{\tau}}$ a fully adapted process.   Set
$\Cal Q=\pi^{-1}[\Cal S]=\dom(\pi\pmb u)$.

(a) $\pmb u\pi$ is order-bounded iff $\pmb u\restr\pi[\Cal Q]$ is
order-bounded.

(b) $\pmb u\pi$ is of bounded variation iff $\pmb u\restr\pi[\Cal Q]$
is of bounded variation.

(c) $\pmb u\pi$ is an integrator for the structure
$(\frak A,\bar\mu,R,\family{r}{R}{\frak B_r})$
iff $\pmb u\restr\pi[\Cal Q]$ is an integrator for the structure
$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$.

(d) $\pmb u\pi$ is a martingale for the structure
$(\frak A,\bar\mu,R,\family{r}{R}{\frak B_r})$ iff
$\pmb u\restr\pi[\Cal Q]$ is a martingale for the structure
$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$.

(e) $\pmb u\pi$ is jump-free iff $\pmb u\restr\pi[\Cal Q]$ is jump-free.

(f) Let $\hat\Cal S$ be the covered envelope of $\Cal S$ in $\Cal T$,
$\hat\Cal Q$ the covered envelope of $\Cal Q$ in $\Cal R$
and $\hatpmb u$ the fully
adapted extension of $\pmb u$ to $\hat\Cal S$.   Then
$\pi[\hat\Cal Q]\subseteq\hat\Cal S$ and
$\hatpmb u\pi\restr\hat\Cal Q$ is the fully adapted extension of
$\pmb u\pi$ to $\hat\Cal Q$.

(g) If $\pmb u$ is moderately oscillatory then $\pmb u\pi$ is moderately
oscillatory.

(h)(i) If $\Cal S$ is order-convex in $\Cal T$ then $\Cal Q$ is order-convex
in $\Cal R$.

\quad (ii) Suppose that $\family{r}{R}{\pi_r}$ is
right-continuous\cmmnt{ (635Cd)}.
If $\Cal S$ is order-convex and $\pmb u$ is near-simple,
then $\pmb u\pi$ is near-simple.

\proof{ If $\Cal Q$ is empty, then $\pi[\Cal Q]$ will be empty
and everything is trivial.    So suppose that $\Cal Q$ is non-empty.
Because $\pi:\Cal R\to\Cal T$ is a lattice homomorphism (635C(b-i)),
$\pi[\Cal Q]$ is a sublattice of $\Cal T$.

\medskip

{\bf (a)} We have only to note that
$\{u_{\pi(\rho)}:\rho\in\Cal Q\}=\{u_{\rho}:\rho\in\pi[\Cal Q]\}$.

\medskip

{\bf (b)(i)} If $\pmb u\restr\pi[\Cal Q]$ is of bounded variation
and $\rho_0\le\ldots\le\rho_n$ in $\Cal Q$ then

\Centerline{$\sum_{i=0}^{n-1}|u_{\pi(\rho_{i+1})}-u_{\pi(\rho_i)}|
\le\int_{\pi[\Cal Q]}|d\pmb u|$,}

\noindent so $\pmb u\pi$ is of bounded variation.

\medskip

\quad{\bf (ii)} If $\pmb u\pi$ is of bounded variation and
$\tau_0\le\ldots\le\tau_n$ in $\pi[\Cal Q]$,
take $\rho'_i\in\Cal Q$ such that
$\pi(\rho'_i)=\tau_i$ for each $i$.
Set $\rho_i=\sup_{j\le i}\rho'_j$
for $i\le n$, so that $\rho_0\le\ldots\le\rho_n$,
$\pi(\rho_i)=\tau_i$ for $i\le n$, and

\Centerline{$\sum_{i=0}^{n-1}|u_{\tau_{i+1}}-u_{\tau_i}|
=\sum_{i=0}^{n-1}|u_{\pi(\rho_{i+1})}-u_{\pi(\rho_i)}|
\le\int_{\Cal Q}|d(\pmb u\pi)|$.}

\noindent As $\tau_0,\ldots,\tau_n$ are arbitrary,
 $\pmb u\restr\pi[\Cal Q]$ is of bounded variation.

\medskip

{\bf (c)(i)} $Q_{\pi[\Cal Q]}(d\pmb u)\subseteq Q_{\Cal Q}(d(\pmb u\pi))$.
\Prf\ Suppose that $I\in\Cal I(\pi[\Cal Q])$ is non-empty and that
$\pmb{w}=\langle w_{\tau}\rangle_{\tau\in I}$ is a fully adapted
process with $\|\pmb{w}\|_{\infty}\le 1$.   Let $(\tau_0,\ldots,\tau_n)$
linearly generate the $I$-cells.   As in (b-ii) above, we have
$\rho_0\le\ldots\le\rho_n$ in $\Cal Q$ such that $\pi(\rho_i)=\tau_i$
for each $i$.   Now
$w_{\tau_i}\in L^0(\frak A_{\pi(\rho_i)})=L^0(\frak B_{\rho_i})$ and
$\|w_{\tau_i}\|_{\infty}\le 1$.   So

$$\eqalign{S_I(\pmb{w},d\pmb u)&
=\sum_{i=0}^{n-1}w_{\tau_i}\times(u_{\tau_{i+1}}-u_{\tau_i})\cr&
=\sum_{i=0}^{n-1}w_{\tau_i}\times(u_{\pi(\rho_{i+1})}-u_{\pi(\rho_i)})
\in Q_{\Cal Q}(d(\pmb u\pi))\cr}$$

\noindent (616C(ii)).   As $I$ and $\pmb{w}$ are arbitrary,
$Q_{\pi[\Cal Q]}(d\pmb u)\subseteq Q_{\Cal Q}(d(\pmb u\pi))$.\ \Qed

So if $\pmb u\pi$ is an integrator, $\pmb u\restr\pi[\Cal Q]$ is an integrator
(616Fc, 613B(f-iii)).

\medskip

\quad{\bf (ii)}
$Q_{\Cal Q}(d(\pmb u\pi))\subseteq Q_{\pi[\Cal Q]}(d\pmb u)$.
\Prf\ Suppose that $I\in\Cal I(\Cal Q)$ is non-empty and that
$\pmb{w}=\langle w_{\tau}\rangle_{\tau\in I}$ is a fully adapted
process with $\|\pmb{w}\|_{\infty}\le 1$.
Let $(\rho_0,\ldots,\rho_n)$ linearly generate the $I$-cells.
Now $\pi(\rho_0)\le\ldots\le\pi(\rho_n)$ and
$w_{\rho_i}\in L^0(\frak A_{\pi(\rho_i)})$ and
$\|w_{\rho_i}\|_{\infty}\le 1$ for each $i$.   So

\Centerline{$S_I(\pmb{w},d(\pmb u\pi))
=\sum_{i=0}^{n-1}w_{\rho_i}\times(u_{\pi(\rho_{i+1})}-u_{\pi(\rho_i)})
\in Q_{\pi[\Cal Q]}(d\pmb u)$.}

\noindent As $I$ and $\pmb{w}$ are arbitrary,
$Q_{\Cal Q}(d(\pmb u\pi))\subseteq Q_{\pi[\Cal Q]}(d\pmb u)$.\ \Qed

So if $\pmb u\restr\pi[\Cal Q]$ is an integrator,
$\pmb u\pi$ is an integrator.

\medskip

{\bf (d)(i)} Suppose that $\pmb u\pi$ is a martingale.
Take $\tau\le\tau'$
in $\pi[\Cal Q]$.   Again as in (b-ii) of this proof, there are
$\rho\le\rho'$ in $\Cal Q$ such that $\tau=\pi(\rho)$ and
$\tau'=\pi(\rho')$.   Now $\frak A_{\tau}=\frak B_{\rho}$, so

\Centerline{$P_{\tau}u_{\tau'}
=P_{\tau}u_{\pi(\rho')}
=u_{\pi(\rho)}
=u_{\tau}$.}

\noindent As $\tau$ and $\tau'$ are arbitrary, $\pmb u\restr\pi[\Cal Q]$ is a
martingale.

\medskip

\quad{\bf (ii)} Suppose that $\pmb u\restr\pi[\Cal Q]$ is a martingale.
Take $\rho\le\rho'$ in $\Cal Q$.  Then
$\pi(\rho)\le\pi(\rho')$ in $\pi[\Cal Q]$, and
$\frak B_{\rho}=\frak A_{\pi(\rho)}$.   So the conditional
expectation of $u_{\pi(\rho')}$ on $\frak B_{\rho}$ is
$P_{\pi(\rho)}u_{\pi(\rho')}=u_{\pi(\rho)}$.
As $\rho$ and $\rho'$ are arbitrary, $\pmb u\pi$ is a
martingale.

\medskip

{\bf (e)} By (a), we know that if either $\pmb u\pi$ or
$\pmb u\restr\pi[\Cal Q]$ is jump-free, then both are order-bounded.   So we
can suppose throughout that this is so.

\medskip

\quad{\bf (i)} Suppose that $\rho_0\le\ldots\le\rho_n$ in $\Cal Q$.
Set $\rho_{-1}=\min\Cal R$, $\rho_{n+1}=\max\Cal R$,
$\tau_{-1}=\min\Cal T$, $\tau_{n+1}=\max\Cal T$ and
$\tau_i=\pi(\rho_i)$ for $0\le i\le n$.   Then

$$\eqalign{\tilde w&=\sup\{|u_{\pi(\rho')}-u_{\pi(\rho)}|:
   \rho,\,\rho'\in\Cal Q\text{ and there is an }i\cr
&\mskip150mu\text{such that }-1\le i\le n\text{ and }
   \rho_i\le\rho\le\rho'\le\rho_{i+1}\},\cr
w&=\sup\{|u_{\tau'}-u_{\tau}|:
   \tau,\,\tau'\in\pi[\Cal Q]\text{ and there is an }i\cr
&\mskip150mu\text{such that }-1\le i\le n\text{ and }
   \tau_i\le\tau\le\tau'\le\tau_{i+1}\}\cr}$$

\noindent are equal.

\medskip

\quad\Prf\grheada\ If $\tau$, $\tau'\in\pi[\Cal Q]$ are such
that $\tau_i\le\tau\le\tau'\le\tau_{i+1}$ where $-1\le i\le n+1$,
then we can express $\tau$, $\tau'$ as $\pi(\rho)$, $\pi(\rho')$ for
$\rho$, $\rho'\in\Cal Q$;  replacing $\rho$ by
$\med(\rho_i,\rho,\rho_{i+1})$ if necessary, we can suppose that
$\rho_i\le\rho\le\rho_{i+1}$;  replacing $\rho'$ by
$\med(\rho,\rho',\rho_{i+1})$ if necessary, we can suppose that
$\rho\le\rho'\le\rho_{i+1}$.   Now

\Centerline{$|u_{\tau'}-u_{\tau}|=|u_{\pi(\rho')}-u_{\pi(\rho)}|\le\tilde w$.}

\noindent As $\tau$ and $\tau'$ are arbitrary, $w\le\tilde w$.

\medskip

\qquad\grheadb\ If $w$ is defined and $\rho$, $\rho'\in\Cal Q$ are such
that $\rho_i\le\rho\le\rho'\le\rho_{i+1}$ where $-1\le i\le n+1$,
then $\tau_i\le\pi(\rho)\le\pi(\rho')\le\tau_{i+1}$,
so $|u_{\pi(\rho')}-u_{\pi(\rho)}|\le w$.
As $\rho$ and $\rho'$ are arbitrary, $\tilde w\le w$.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $\rho_0\le\ldots\le\rho_n$ in $\Cal Q$
and that $I\in\Cal I(Q)$, $I'\in\Cal I(\pi[\Cal Q])$ are such that
$(\rho_0,\ldots,\rho_n)$ linearly generates the $I$-cells, while
$(\pi(\rho_0),\ldots,\pi(\rho_n))$ linearly generates the $I'$-cells.
Then $\Osclln^*_I(\pmb u\pi)=\Osclln^*_{I'}(\pmb u\restr\pi[\Cal Q])$.
\Prf\ Put (i) here together with 618Ca.\ \Qed

\medskip

\quad{\bf (iii)} Suppose that $\pmb u\pi$ is jump-free.
Let $\epsilon>0$.
Then there is a non-empty $I\in\Cal I(\Cal Q)$ such that
$\Osclln^*_I(\pmb u\pi)\le\epsilon$.   Let $(\rho_0,\ldots,\rho_n)$
linearly generate the $I$-cells.   Set
$I'=\{\pi(\rho_0),\ldots,\pi(\rho_n)\}\in\Cal I(\pi[\Cal Q])$.   Then
$(\pi(\rho_0),\ldots,\pi(\rho_n))$ linearly generates the
$I'$-cells.   By (ii),

\Centerline{$\theta(\Osclln^*_{I'}(\pmb u\restr\pi[\Cal Q]))
=\theta(\Osclln^*_I(\pmb u\pi))\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\pmb u\restr\pi[\Cal Q]$ is
jump-free.

\medskip

\quad{\bf (iv)} Suppose that $\pmb u\restr\pi[\Cal Q]$ is jump-free.
Let $\epsilon>0$.
Then there is a non-empty $I'\in\Cal I(\pi[\Cal Q])$ such that
$\Osclln^*_{I'}(\pmb u\restr\pi[\Cal Q])\le\epsilon$.
Let $(\tau_0,\ldots,\tau_n)$
linearly generate the $I'$-cells.   Take $\rho_0\le\ldots\le\rho_n$ in
$\Cal Q$ such that $\pi(\rho_i)=\tau_i$ for each $i$.
Set
$I=\{\rho_0,\ldots,\rho_n\}\in\Cal I(\Cal Q)$.   Then
$(\rho_0,\ldots,\rho_n)$ linearly generates the
$I$-cells, so

\Centerline{$\theta(\Osclln^*_I(\pmb u\pi))
=\theta(\Osclln^*_{I'}(\pmb u\restr\pi[\Cal Q]))\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\pmb u\pi$ is jump-free.

\medskip

{\bf (f)} If $\rho\in\hat\Cal Q$ then

\Centerline{$\sup_{\tau\in\Cal S}\Bvalue{\pi(\rho)=\tau}
\Bsupseteq\sup_{\rho'\in\Cal Q}\Bvalue{\pi(\rho)=\pi(\rho')}
\Bsupseteq\sup_{\rho'\in\Cal Q}\Bvalue{\rho=\rho'}  %635Cc
=1$,}

\noindent so $\pi(\rho)\in\hat\Cal S$.   Now $\hatpmb u\pi\restr\hat\Cal Q$
is fully adapted and extends $\pmb u\pi$, so is the fully adapted extension of
$\pmb u\pi$.

\medskip

{\bf (g)} For any $\epsilon>0$, there is a process
$\pmb v=\family{\tau}{\Cal S}{v_{\tau}}$ of bounded variation such that
$\theta(\sup|\pmb u-\pmb v|)\le\epsilon$.   Now $\pmb v\pi$ is of bounded
variation ((b) above) and

\Centerline{$\sup|\pmb u\pi-\pmb v\pi|
=\sup_{\rho\in\Cal Q}|u_{\pi(\rho)}-v_{\pi(\rho)}|
\le\sup_{\tau\in\Cal S}|u_{\tau}-v_{\tau}|
=\sup|\pmb u-\pmb v|$,}

\noindent so $\theta(\sup|\pmb u\pi-\pmb v\pi|)\le\epsilon$.   As $\epsilon$ is
arbitrary, $\pmb u\pi$ is moderately oscillatory.

\medskip

{\bf (h)(i)} If $\rho_0$, $\rho_1\in\Cal Q$, $\rho\in\Cal R$ and
$\rho_0\le\rho\le\rho_1$, then
$\pi(\rho_0)\le\pi(\rho)\le\pi(\rho_1)$, while
$\pi(\rho_0)$ and $\pi(\rho_1)$ belong to $\Cal S$;  so
$\pi(\rho)\in\Cal S$ and $\rho\in\Cal Q$.   Thus $\Cal Q$ is
order-convex.

\medskip

\quad{\bf (ii)}
Suppose that $A\subseteq\Cal Q$ is non-empty and downwards-directed and has a
lower bound $\rho_*\in\Cal Q$.   Then $\inf A$, the infimum of $A$ in
$\Cal R$, lies between $\rho_*$ and any member of $A$, so belongs to $\Cal Q$.
Next, $B=\pi[A]$ is non-empty and
downwards-directed and has a lower bound in $\Cal S$, so $\inf B\in\Cal S$,
while $\inf B=\pi(\inf A)$ by 635C(d-ii).   Given $\epsilon>0$,
there is a $\tau_0\in B$ such that $\theta(u_{\tau}-u_{\inf B})\le\epsilon$
whenever $\tau\in B$ and $\tau\le\tau_0$ (632E).
Take $\rho_0\in A$ such that $\tau_0=\pi(\rho_0)$;  then
$\theta(u_{\pi(\rho)}-u_{\pi(\inf A)})\le\epsilon$ whenever $\rho\in A$ and
$\rho\le\rho_0$.   As $A$ and $\epsilon$ are arbitrary,
$\pmb u\pi$ satisfies $(\dagger)$ of 632F.

As $\pmb u$ is near-simple, it is moderately oscillatory, so $\pmb u\pi$ is
moderately oscillatory.   By 632F, $\pmb u\pi$ is locally near-simple, and in
fact near-simple, by 631F(c-ii).
}%end of proof of 635E

\leader{635F}{Theorem} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T$, and $\pmb v=\sfamilysS{v}$ a near-simple integrator;   let
$\Cal Q$ be a sublattice of $\Cal R$ such that $\pi[\Cal Q]$ is a cofinal
sublattice of $\Cal S$ which $\pmb v$-separates $\Cal S$\cmmnt{ (633Bb)}.
If $\pmb u=\family{\tau}{\Cal S}{u_{\tau}}$ is a moderately oscillatory process,
then $\int_{\Cal Q}\pmb u\pi\,d(\pmb v\pi)$ is defined and equal to
$\int_{\Cal S}\pmb u\,d\pmb v$.

\proof{ By 633Ka, $\int_{\pi[\Cal Q]}\pmb u\,d\pmb v$ is defined and equal to
$\int_{\Cal S}\pmb u\,d\pmb v$.   By 635Db,
this is also $\int_{\Cal Q}\pmb u\pi\,d(\pmb v\pi)$,
since of course the adapted interval
function $\Delta(\pmb v\pi)$ (613Cc) is precisely
$(\Delta\pmb v)_{\pi}$ as defined in 635Db.
}%end of proof of 635F

\leader{635G}{Corollary} Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T$, and
$\pmb v=\sfamilysS{v}$ a near-simple integrator, with quadratic variation
$\pmb v^*$.   If $\Cal Q$ is a sublattice of $\pi^{-1}[\Cal S]$ such that
$\pi[\Cal Q]\,\,\pmb v$-separates $\Cal S$,
then the quadratic variation of $\pmb v\pi\restr\Cal Q$ is
$\pmb v^*\pi\restr\Cal Q$.

\proof{ If $\Cal Q$ is empty, this is trivial;  so suppose otherwise.
By 635Ec, $\pmb v\pi$ is an integrator, so
$\pmb v\pi\restr\Cal Q$ is an integrator and has a quadratic variation.
Take any $\rho\in\Cal Q$.   Then $\pi[\Cal Q\wedge\rho]\,\,\pmb v$-separates
$\Cal S\wedge\pi(\rho)$.   \Prf\ If $\tau$, $\tau'\in\Cal S\wedge\pi(\rho)$ and
$\sigma\in\Cal Q$, then

\Centerline{$\Bvalue{\tau\le\pi(\sigma\wedge\rho)}
=\Bvalue{\tau\le\pi(\sigma)\wedge\pi(\rho)}
=\Bvalue{\tau\le\pi(\sigma)}\Bcap\Bvalue{\tau\le\pi(\rho)}
=\Bvalue{\tau\le\pi(\sigma)}$,}

\Centerline{$\Bvalue{\pi(\sigma\wedge\rho)<\tau'}
=\Bvalue{\pi(\sigma)\wedge\pi(\rho)<\tau'}
=\Bvalue{\pi(\sigma)<\tau'}\Bcup\Bvalue{\pi(\rho)<\tau'}
=\Bvalue{\pi(\sigma)<\tau'}$,}

\Centerline{$\Bvalue{\tau\le\pi(\sigma\wedge\rho)}
  \Bcap\Bvalue{\pi(\sigma\wedge\rho)<\tau'}
=\Bvalue{\tau\le\pi(\sigma)}\Bcap\Bvalue{\pi(\sigma)<\tau'}$.}

\noindent So if $\tau\le\tau'$, then

$$\eqalign{\Bvalue{v_{\tau}\ne v_{\tau'}}&
\Bsubseteq\sup_{\sigma\in\Cal Q}
  (\Bvalue{\tau\le\pi(\sigma)}\Bcap\Bvalue{\pi(\sigma)<\tau'})\cr&
=\sup_{\sigma\in\Cal Q}(\Bvalue{\tau\le\pi(\sigma\wedge\rho)}
  \Bcap\Bvalue{\pi(\sigma\wedge\rho)<\tau'}).\cr}$$

\noindent As $\tau$ and $\tau'$ are
arbitrary, $\pi[\Cal Q\wedge\rho]\,\,\pmb v$-separates $\Cal S\wedge\pi(\rho)$.\
\Qed

Of course $\Cal S\wedge\pi(\rho)$ is still order-convex, and
$\pi[\Cal Q\wedge\rho]$ is cofinal with $\Cal S\wedge\pi(\rho)$ because it
contains $\pi(\rho)$.   So
635F tells us that $\int_{\Cal Q\wedge\rho}\pmb v\pi\,d(\pmb v\pi)
=\int_{\Cal S\wedge\pi(\rho)}\pmb v\,d\pmb v$.   Next, $\pmb v^2$ is also a
near-simple integrator (631F(a-i), 616N), and clearly
$\pmb v^2\pi=(\pmb v\pi)^2$, so

\Centerline{$\int_{\Cal Q\wedge\rho}d((\pmb v\pi)^2)
=\int_{\Cal Q\wedge\rho}d(\pmb v^2\pi)
=\int_{\Cal S\wedge\pi(\rho)}d(\pmb v^2)$}

\noindent by 635F again.   Accordingly

\Centerline{$\int_{\Cal Q\wedge\rho}d((\pmb v\pi)^2)
   -2\int_{\Cal Q\wedge\rho}\pmb v\pi\,d(\pmb v\pi)
=\int_{\Cal S\wedge\pi(\rho)}d(\pmb v^2)
   -2\int_{\Cal S\wedge\pi(\rho)}\pmb v\,d\pmb v
=v^*_{\pi(\rho)}$}

\noindent (617Kb).
As $\rho$ is arbitrary, the quadratic variation of $\pmb v\pi\restr\Cal Q$
is the function $\rho\mapsto v^*_{\pi(\rho)}:\Cal Q\to L^0(\frak A)$, that is,
$\pmb v^*\pi\restr\Cal Q$.
}%end of proof of 635G

%\leader{635H}{}
%\leader{635I}{}
%\leader{635J}{}
%\leader{635K}{}
%\leader{635L}{}
%\leader{635M}{}
%\leader{635N}{}
%\leader{635O}{}
%\leader{635P}{}
%\leader{635Q}{}
%\leader{635R}{}
%\leader{635S}{}
%\leader{635T}{}
%\leader{635U}{}
%\leader{635V}{}
%\leader{635W}{}

\exercises{\leader{635X}{Basic exercises (a)}
Suppose that $\family{t}{T}{\frak A_t}$ is
right-continuous.   Let $\Cal S$ be an order-convex sublattice of
$\Cal T$, and $\pmb v$, $\pmb w$
local integrators with domain $\Cal S$, with covariation
$\covar{\pmb v}{\pmb w}$.   Show that
if $\Cal Q$ is a sublattice of $\pi^{-1}[\Cal S]$
such that $\pi[\Cal Q]$ separates $\Cal S$,
then $\covar{\pmb v\pi\restr\Cal Q}{\pmb w\pi\restr\Cal Q}$ is defined and
equal to $\covar{\pmb v}{\pmb w}\pi\restr\Cal Q$.
%635G

\leader{635Y}{Further exercises (a)}
Set $T^{\|}=\{t^-:t\in T\}\cup\{t^+:t\in T\}$
with the total ordering defined by saying that, for $s$, $t\in T$,

\Centerline{$s^+\le t^+\iff s^-\le t^-\iff s^-\le t^+\iff s\le t$,
\quad$s^+\le t^-\iff s<t$.}

\noindent(i) Show that $T^{\|}$ is totally ordered.   (ii) For $t\in T$,
set $\frak B_{t^-}=\frak A_t$, $\frak B_{t^+}=\frak A$ if $t=\max T$,
and $\frak B_{t^+}=\bigcap_{s>t}\frak A_s$ otherwise.   Show that
$\family{r}{T^{\|}}{\frak B_r}$ is a right-continuous filtration.
Write $\Cal T^{\|}$ for the corresponding lattice of stopping times.
(iii) For $t\in T$ set $\pi_t=\check{t}^-\in\Cal T^{\|}$.
For $\rho\in\Cal T$ let $\pi(\rho)\in\Cal T^{\|}$ be given by the
formula in 635B.   Show that

\Centerline{$\Bvalue{\pi(\rho)>t^+}=\Bvalue{\pi(\rho)>t^-}
=\Bvalue{\rho>t}$}

\noindent for every $t\in T$.   (iv) Show that
$\Bvalue{\pi(\rho)<\pi(\tau)}
=\Bvalue{\rho<\tau}$ for all $\rho$, $\tau\in\Cal T$, and that
$\pi(\rho)=\pi(\tau)$ iff $\rho=\tau$.
(v) Show that if $\tau\in\Cal T^{\|}$
then $\frak B_{\tau}=\bigcap\{\frak A_{\rho}:\rho\in\Cal T$,
$\pi(\rho)\ge\tau\}$.
%mt63bits
}%end of exercises


\endnotes{
\Notesheader{635} In this section,
right-continuity of the filtration $\family{t}{T}{\frak A_t}$ is a
generally ruling
hypothesis;  this is because it seems to be necessary for the basic algebra
of the lattice homomorphism $\pi:\Cal R\to\Cal T$ (635Ca).
With the elementary properties of this homomorphism in hand, the
programme of 635D-635G %635D 635E 635F 635G
is obvious, and we have just to make sure that it is watertight.
In 653G-653J %653G 653I 653J
I will explain how the ideas behind 635F can sometimes be
used to replace an integral with respect to an unfamiliar jump-free
martingale by an integral with respect to Brownian motion.
}%end of notes

\discrpage



\references{
\frfilename{mt6r.tex}
\versiondate{10.2.23}
\copyrightdate{2022}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]


\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}


\def\bsdot{.\penalty-100}

\centerline{\bf References for Volume 6}

\medskip%A\medskip%B

Bain A.\ [09] {\it Stochastic Calculus.}
{\tt http://www.chiark.greenend.org.uk/$^{\sim}$alanb/}, 25.5.09.
%bain.pdf

Bichteler K.\ [79] `Stochastic integrators',
Bull.\ Amer.\ Math.\ Soc.\ (2) 1 (1979) 761-765.
%http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&page=toc&handle=euclid.bams/1183544712
\cmmnt{[627I.]}

Bichteler K.\ [81] `Stochastic integration and $L^p$ theory of
martingales', Ann.\ Prob.\ 9 (1981) 49-89.
%https://projecteuclid.org/euclid.aop/1176994509
\cmmnt{[649C.]}

Black F.\ \& Scholes M.\ [73] `The pricing of options and corporate
liabilities', J.\ Political Economy 81 (1973) 637-659.
\cmmnt{[\S655 {\it intro.}]}

Burkholder D.L.\ [66] `Martingale Transforms',
Ann.\ of Math.\ Stat.\ 37 (1966) 1494-1504.
\cmmnt{[621H.]}
%burkholder.66.pdf

%\medskip%C
\medskip%D

\leaveitout{
Delbaen F.\ \& Schachermeyer W.\ [06] {\it The Mathematics of Arbitrage.}
Springer, 2006.
ISBN-13: 978-3540219927
HG 6024.A3D4  0620 6436 3
}

\leaveitout{
Dellacherie C.\ \& Meyer P.-A.\ [79]
Probabilities and Potential, Volume 1
North-Holland, 1979  (Mathematics Studies).
% ISBN: 072040701X chapters I-VI
}

Dellacherie C.\ \& Meyer P.-A.\ [82]
{\it Probabilities and Potential B. Theory of Martingales.}
Elsevier, 1982 (North-Holland Mathematics Studies 72).
%ISBN: 0444865268 chapters V-VIII
%translator J.P.Wilson
\cmmnt{[627I.]}

\leaveitout{
Dellacherie C.\ \& Meyer P.-A.\ [88]
{\it Probabilities and Potential C: Potential Theory for Discrete and
Continuous Semigroups.}
Elsevier, 1988
%ISBN: 0444703861 (pt.C)
}

\leaveitout{
Dellacherie C., Meyer P.-A.\ \& Weil M.\ [77] (eds.)
{\it S\'eminaire de Probabilit\'es XI.} Springer, 1977
(Lecture Notes in Mathematics 581).
% ISBN 978-3-540-08145-6
}

\leaveitout{
Dol\'eans-Dade C.\ [76] `On the existence and unicity of solutions of
stochastic differential equations', Z.\ Wahrscheinlichkeitstheorie verw.\
Gebiete 36 (1976) 93-101.
%? fundamental theorem of martingales
% ILL 15.9.12
}

%\medskipE\medskip%F

\leaveitout{
Fremlin D.H.\ [n11] `Vector-valued gauge integrals', note of 12.7.11
({\tt http:{\bsp}/www1.essex.ac.uk{\bsp}maths{\bsp}people{\bsp}fremlin{\bsp}preprints.htm}).
%n11527
}

%\medskipG\medskip%H
\medskip%I

It\^o K.\ [44] `Stochastic Integral', Proc.\ Imperial Acad.\ Tokyo 20
(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



