% 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{mt62.tex}
\versiondate{15.12.16}
\copyrightdate{2017}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind

\def\chaptername{Martingales}

\newchapter{62}

The centre of the theory of stochastic integration,
since {\smc It\^o 44}, has been integrals
$\int\pmb{u}\,d\pmb{v}$ where $\pmb{v}$ is a martingale.
In \S621 I give a number of inequalities
involving {\it finite} martingales which will make it possible to go
straight to the general case in \S622.   In \S622 we have to check some algebra
concerning conditional expectations in order to make sense of the idea of
`fully adapted martingale', but the theorem that martingales are local
integrators (622G) is a straightforward consequence of 621Hf.

It is not in general the case that an indefinite integral with respect
to a martingale is again a martingale.   For a full-strength theorem in this
direction I think we need to turn to `virtually local' martingales and do some
hard work (623O).   To use It\^o's formula (619C) in its original form,
in which the integrator was Brownian motion, we need of course to know the
quadratic variation of Brownian motion, which I come to at last in 624F.

The next three sections are directed towards a structure theory for integrators
in \S627.   This volume is devoted to structures based on probability algebras
$(\frak A,\bar\mu)$.   The concepts of Chapter 61 are generally
law-independent in the sense that while the existence of the functional
$\bar\mu$ is essential, its replacement by another functional $\bar\nu$ such
that $(\frak A,\bar\nu)$ is still a probability algebra makes no difference.
However nearly everything involving martingales is shaken up by a change in law.
\S625 examines such changes, and we find, remarkably, that we do not change the
\sm{s} (625F).   In \S626 I introduce submartingales and previsible variations,
with the Doob-Meyer theorem on the expression of submartingales as \sm{s}.
In \S627 I apply this to supermartingales, and show that local integrators are
\sm{s}.

The essential inequality in 621Hf is proved by ordinary martingale methods in
\S621.   There is an alternative route, incidentally yielding a better constant,
which depends on a kind of interpolation;  I present this in \S628.

\discrpage




\frfilename{mt621.tex}
\versiondate{20.12.17/11.4.18}
\copyrightdate{2012}

\def\chaptername{Martingales}
\def\sectionname{Finite martingales}

\newsection{621}

I have to justify my repeated assertion that martingales are
integrators.   This, together with the developments it leads to, will need
some non-trivial facts about finite martingales which are most easily
described in advance of their applications.   However,
the complexity of some of
the lemmas below may be more bearable if you can see what they're for.
So you may wish to treat this section as an appendix, and disentangle the
ideas when you find them being called on in \S\S622, 624 and 626.

\leader{621A}{Notation}\cmmnt{ This section will be almost independent of the
work in Chapter 61, and will be based rather on the ideas of \S275,
interpreted as always in the language of Chapter 36.   Once again,}
$(\frak A,\bar\mu)$ will be a
probability algebra, and for $1\le p<\infty$,
$L^p_{\bar\mu}=L^p(\frak A,\bar\mu)\subseteq L^0(\frak A)$ will be the
associated
$L^p$-space\cmmnt{ $\{w:w\in L^0(\frak A)$, $\|w\|_p<\infty\}$ (\S366)}, while
$\Expn$ refers to the integral on $L^1_{\bar\mu}$\cmmnt{ (613Aa)}.


%621B
\leader{621B}{Uniform integrability} \cmmnt{We are going to need the
following results from Volumes 2 and 3.

\medskip

}{\bf (a)} Recall that a set
$A\subseteq L^1_{\bar\mu}$ is {\bf uniformly integrable} if for every
$\epsilon>0$ there is an $M\ge 0$ such that
$\Expn((|u|-M\chi 1)^+)\le\epsilon$ for every $u\in A$\cmmnt{ (246Ab, 354P,
365T\formerly{3{}65U})};  equivalently, if $A$ is $\|\,\|_1$-bounded and
for every $\epsilon>0$ there is a
$\delta>0$ such that $\Expn(|u|\times\chi a)\le\epsilon$ whenever
$u\in A$, $a\in\frak A$ and $\bar\mu a\le\delta$ (246Ca, 246G).

\spheader 621Bb A non-empty set $A\subseteq L^0=L^0(\frak A)$
is uniformly integrable iff

\Centerline{$\lim_{\alpha\to\infty}
  \sup_{u\in A}\Expn(|u|\times\chi\Bvalue{|u|>\alpha})=0$}

\noindent\cmmnt{ (246I)}.

\spheader 621Bc Suppose that $A$, $B\subseteq L^1_{\bar\mu}$ are
uniformly integrable.

\medskip

\quad{\bf (i)} Every subset of $A$ is uniformly integrable;  $\alpha A$ is
uniformly integrable for every $\alpha\in\Bbb R$;  $A+B$ is uniformly
integrable;  the solid hull of $A$ is uniformly integrable\prooflet{
(246C, 354Ra)}.

\medskip

\quad{\bf (ii)} The $\frak T$-closure $\overline{A}$ of $A$ is uniformly
integrable, where
$\frak T$ is the topology of convergence in measure on $L^0$\cmmnt{ (613B)}, and
$\frak T$ agrees with the norm topology of $L^1_{\bar\mu}$ on
$\overline{A}$\prooflet{ (246J)}.

\spheader 621Bd A subset of $L^1_{\bar\mu}$ is uniformly integrable iff it
is relatively compact for the weak topology
$\frak T_s(L^1_{\bar\mu},L^{\infty}(\frak A))$\prooflet{
(243Gb and 247C, or 365T(a-v))}.

\spheader 621Be\cmmnt{ The following useful fact was left in the
exercises for \S246.}   If $p>1$, then any
$\|\,\|_p$-bounded subset $A$ of $L^0$ is uniformly integrable.
\prooflet{\Prf\ Suppose that $\|u\|_p\le\gamma$ for $u\in A$.   Given
$\epsilon>0$, there is an $M>0$ such that
$\gamma^p\le\epsilon M^{p-1}$.   Now for any $u\in A$,

\Centerline{$(|u|-M\chi 1)^+\le\Bover1{M^{p-1}}|u|^p$,
\quad$\Expn((|u|-M\chi 1)^+)\le\Bover{\gamma^p}{M^{p-1}}\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $A$ is uniformly integrable.\ \Qed}

\leader{621C}{Conditional expectations} \cmmnt{Of course we cannot talk
about martingales without speaking of conditional expectations, and this
volume will call on the
full resources developed in Volumes 2 and 3, which I now recapitulate.

\medskip

}{\bf (a)} Following the definitions in \S365, we find that
if $\frak B$ is a closed subalgebra of $\frak A$ then
$L^0(\frak B)\cap L^1_{\bar\mu}
=L^1(\frak B,\bar\mu\restrp\frak B)$\cmmnt{ (365Qa\formerly{3{}65Ra})},
and we have a unique positive linear operator
$P_{\frak B}:L^1_{\bar\mu}\to L^0(\frak B)\cap L^1_{\bar\mu}$ such that
$\Expn(P_{\frak B}u\times\chi b)=\Expn(u\times\chi b)$ whenever
$u\in L^1_{\bar\mu}$ and $b\in\frak B$\cmmnt{ (365Q)}.   Counting
$\|u\|_p$ as $\infty$ if $u\in L^0(\frak A)\setminus L^p(\frak A,\bar\mu)$,
$\|P_{\frak B}u\|_p\le\|u\|_p$ for every
$u\in L^1_{\bar\mu}$ and $p\in[1,\infty]$\cmmnt{ (366J)}.

\spheader 621Cb If $\frak B$ and $\frak C$ are closed subalgebras of
$\frak A$ and $\frak B\subseteq\frak C$, then

\Centerline{$P_{\frak B}P_{\frak C}=P_{\frak C}P_{\frak B}=P_{\frak B}$.}

\noindent\prooflet{\Prf\ (Cf.\ 458M.)
Take any $u\in L^1_{\bar\mu}$.
(i) $P_{\frak B}P_{\frak C}u\in L^0(\frak B)$ and

\Centerline{$\Expn(P_{\frak B}P_{\frak C}u\times\chi b)
=\Expn(P_{\frak C}u\times\chi b)=\Expn(u\times\chi b)
=\Expn(P_{\frak B}u\times\chi b)$}

\noindent for every $b\in\frak B\subseteq\frak C$, so
$P_{\frak B}P_{\frak C}u=P_{\frak B}u$.   (ii)
$P_{\frak B}u\in L^0(\frak B)\subseteq L^0(\frak C)$ so
$P_{\frak C}P_{\frak B}u=P_{\frak B}u$.\ \Qed}

\spheader 621Cc If $\frak B$ is a closed subalgebra of $\frak A$,
$u\in L^1_{\bar\mu}$,
$v\in L^0(\frak B)$ and $u\times v\in L^1_{\bar\mu}$, then
$P_{\frak B}(u\times v)=P_{\frak B}u\times v$\cmmnt{ (233K,
365Qa)}.   So if $u$, $v$, $u\times P_{\frak B}v$ and $P_{\frak B}u\times v$
all belong to $L^1_{\bar\mu}$,

\Centerline{$P_{\frak B}(u\times P_{\frak B}v)
=P_{\frak B}u\times P_{\frak B}v=P_{\frak B}(P_{\frak B}u\times v)$}

\noindent and $\Expn(u\times P_{\frak B}v)=\Expn(P_{\frak B}u\times v)$.

\spheader 621Cd\cmmnt{ (`Jensen's inequality')}
Let $h:\Bbb R\to\Bbb R$ be a
convex function and $\bar h:L^0\to L^0$ the corresponding
map\cmmnt{ (612Ac)}.   If
$u$ and $\bar h(u)$ both belong to $L^1_{\bar\mu}$,
$\bar h(P_{\frak B}u)\le P(\bar h_{\frak B}(u))$ for every closed
subalgebra $\frak B$ of $\frak A$\cmmnt{ (365Qb)}.

\spheader 621Ce When $p=2$, we have a sharper result:  if
$u\in L^2_{\bar\mu}$ and $\frak B$ is a closed subalgebra of
$\frak A$, then $\|u\|_2^2=\|P_{\frak B}u\|_2^2+\|u-P_{\frak B}u\|^2$.
\prooflet{\Prf\ In the Hilbert space $L^2_{\bar\mu}$,

\Centerline{$\innerprod{P_{\frak B}u}{u}
=\Expn(u\times P_{\frak B}u)
=\Expn(P_{\frak B}(u\times P_{\frak B}u))
=\Expn(P_{\frak B}u\times P_{\frak B}u)$}

\noindent so

\Centerline{$\|u-P_{\frak B}u\|_2^2
=\|u\|_2^2-2\innerprod{u}{P_{\frak B}u}+\|P_{\frak B}u\|_2^2
=\|u\|_2^2-\|P_{\frak B}u\|_2^2$.   \Qed}}

\spheader 621Cf If $A\subseteq L^1_{\bar\mu}$ is uniformly integrable, then
$\{P_{\frak B}u:u\in A$, $\frak B$ is a closed subalgebra of $\frak A\}$ is
uniformly integrable\cmmnt{ (246D, 365Tb)}.

\spheader 621Cg{\bf (i)}
If $\Bbb B$ is a non-empty downwards-directed family of closed
subalgebras of $\frak A$ with intersection $\frak C$, and
$u\in L^1=L^1_{\bar\mu}$, then
$P_{\frak C}u$ is the $\|\,\|_1$-limit
$\1lim_{\frak B\downarrow\Bbb B}P_{\frak B}u$\cmmnt{ (367Qa)}.

\medskip

\quad{\bf (ii)} If $\Bbb B$ is a non-empty upwards-directed family of closed
subalgebras of $\frak A$, $\frak C$ is the closed subalgebra generated by
$\bigcup\Bbb B$ and $u\in L^1$, then
$P_{\frak C}u$ is the $\|\,\|_1$-limit
$\1lim_{\frak B\uparrow\Bbb B}P_{\frak B}u$\cmmnt{ (367Qb)}.

\leader{621D}{Definitions} For the rest of this section,
we shall be looking at a non-decreasing finite sequence
$\langle\frak A_i\rangle_{i\le n}$ of closed subalgebras of $\frak A$\cmmnt{;
that is, a filtration in the sense of 611A in which the totally ordered set $T$
is $\{0,\ldots,n\}$ for some integer $n$}.   In this context, I will write
$P_i:L^1_{\bar\mu}\to L^1_{\bar\mu}$ for the conditional expectation
operator associated with $\frak A_i$.
\cmmnt{Note that} $P_iP_j=P_{\min(i,j)}$
for all $i$, $j\le n$\cmmnt{, by 621Cb}.
Let $\pmb v=\langle v_i\rangle_{i\le n}$ be a finite
sequence in $L^1_{\bar\mu}$.

\spheader 621Da $\pmb v$ is a {\bf martingale}
adapted to $\langle\frak A_i\rangle_{i\le n}$
if $v_i=P_iv_j$ whenever $i\le j\le n$\cmmnt{;  equivalently, if
$v_n\in L^0(\frak A_n)$ and $v_i=P_iv_{i+1}$ for every $i<n$}.

\spheader 621Db $\pmb v$ is a {\bf submartingale}
adapted to $\langle\frak A_i\rangle_{i\le n}$
if $v_i\in L^0(\frak A_i)$ and $v_i\le P_iv_j$ whenever
$i\le j\le n$\cmmnt{;  equivalently, if $v_i\in L^0(\frak A_i)$ for every
$i\le n$ and $v_i\le P_iv_{i+1}$ for every $i<n$}.

\cmmnt{(Cf.\ 275A, 275Yg, 626B.)}

\leader{621E}{Doob's maximal inequality} If $\langle v_i\rangle_{i\le n}$
is a martingale adapted to $\langle\frak A_i\rangle_{i\le n}$,
and $\bar v=\sup_{i\le n}|v_i|$, then

\Centerline{$t\bar\mu\Bvalue{\bar v>t}
\le\Expn(|v_n|\times\chi\Bvalue{\bar v>t})
\le\|v_n\|_1$,}

\Centerline{$t\bar\mu\Bvalue{\bar v\ge t}
\le\Expn(|v_n|\times\chi\Bvalue{\bar v\ge t})
\le\|v_n\|_1$}

\noindent for every $t\ge 0$.

\proof{ (This is a small modification of 275D.)   Set
$a=\Bvalue{\bar v>t}$,
$b^+_i=\Bvalue{v_i>t}$,
$c^+_i=b^+_i\Bsetminus\sup_{j<i}b^+_j$ for $i\le n$ and
$a^+=\sup_{i\le n}b^+_i=\Bvalue{\sup_{i\le n}v_i>t}$ (364La).   Then

$$\eqalignno{t\bar\mu a^+
&=\sum_{i=0}^nt\bar\mu c^+_i
\le\sum_{i=0}^n\Expn(v_i\times\chi c^+_i)
=\sum_{i=0}^n\Expn(v_n\times\chi c^+_i)
\cr\displaycause{because $c^+_i\in\frak A_i$ and $v_i$ is the conditional
expectation of $v_n$ on $v_i$}&
=\Expn(v_n\times\chi a^+)
\le\Expn((v_n\vee 0)\times\chi a).
\cr}$$

\noindent Similarly, setting
$a^-=\sup_{i\le n}\Bvalue{-v_{\tau_i}>t}=\Bvalue{\inf_{i\le n}v_i<t}$,
we have $t\bar\mu a^-\le\Expn(((-v_n)\vee 0)\times\chi a)$.
Since $a^+\Bcup a^-$,

\Centerline{$t\bar\mu a\le\Expn(((v_n\vee 0)+((-v_n)\vee 0))\times\chi a)
=\Expn(|v_n|\times\chi a)\le\Expn(|v_n|)=\|v_n\|_1$,}

\noindent as claimed.

For the second version, the result is trivial if $t=0$, and for
$t>0$ we have $\Bvalue{\bar v\ge t}=\inf_{0\le s<t}\Bvalue{\bar v>s}$, so

$$\eqalign{t\bar\mu\Bvalue{\bar v\ge t}&
=\lim_{s\uparrow t}s\bar\mu\Bvalue{\bar v>s}\cr&
\le\lim_{s\uparrow t}\Expn(|v_n|\times\chi\Bvalue{\bar v>s})
=\Expn(|v_n|\times\chi\Bvalue{\bar v\ge t}).\cr}$$
}%end of proof of 621E

\leader{621F}{Lemma} Suppose that $\ofamily{i}{n}{u_i}$ and
$\langle v_i\rangle_{i\le n}$ are such that
$u_i\in L^{\infty}(\frak A_i)$ and $\|u_i\|_{\infty}\le 1$ for
every $i<n$ and $\langle v_i\rangle_{i\le n}$ is a martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$.
Set $z=\sum_{i=0}^{n-1}u_i\times(v_{i+1}-v_i)$.
Then $\|z\|_2\le\|v_n\|_2$.

\proof{ If $v_n$ is not square-integrable, that is, $\|v_n\|_2=\infty$,
this is trivial.   So let us suppose that $v_n\in L^2_{\bar\mu}$.   In this
case $v_i$ is square-integrable for every $i$ (621Cd), so
$u_i\times(v_{i+1}-v_i)$ is square-integrable for every $i<n$.   Now
if $i<j<n$,

$$\eqalign{&\Expn(u_i\times(v_{i+1}-v_i)\times u_j\times(v_{j+1}-v_j))\cr
&\mskip100mu
=\Expn(P_j(u_i\times(v_{i+1}-v_i)\times u_j\times(v_{j+1}-v_j)))\cr
&\mskip100mu
=\Expn(u_i\times(v_{i+1}-v_i)\times u_j\times P_j(v_{j+1}-v_j))
=0\cr}$$

\noindent by 621Cc.   At the same time, of course,
$\Expn((v_{i+1}-v_i)\times(v_{j+1}-v_j))=0$.   So

$$\eqalignno{\Expn(z^2)
&=\sum_{i=0}^{n-1}\Expn(u_i^2\times(v_{i+1}-v_i)^2)
\le\sum_{i=0}^{n-1}\Expn((v_{i+1}-v_i)^2)
=\sum_{i=0}^{n-1}\Expn(v_{i+1}^2-v_i^2)
\Displaycause{621Ce}
\le\Expn(v_n^2)\cr}$$

\noindent and $\|z\|_2\le\|v_n\|_2$.
}%end of proof of 621F

\leader{621G}{Proposition} Suppose that
$\pmb v=\langle v_i\rangle_{i\le n}$ is a submartingale adapted to
$\langle\frak A_i\rangle_{i\le n}$.   Then there are a non-decreasing
process $\pmb v^{\#}=\langle v_i^{\#}\rangle_{i\le n}$ and a martingale
$\hat{\pmb v}$ adapted to $\langle\frak A_i\rangle_{i\le n}$ such that
$\pmb v=\pmb v^{\#}+\hat{\pmb v}$ and $v^{\#}_0=0$.
If $-\chi 1\le v_i\le 0$ for every
$i\le n$, $\|\hat v_n\|_2^2\le\|v_n\|_1+2\|v_0\|_1$.

\proof{{\bf (a)} Set $w_j=P_jv_{j+1}-v_j$, so that $w_j\in L^0(\frak A_j)$
and $w_j\ge 0$ for $j<n$;  set
$v^{\#}_j=\sum_{i=0}^{j-1}w_i$ for $j\le n$, so that
$v^{\#}_j\in L^0(\frak A_j)$ for each $j$, $v^{\#}_0=0$ and
$\pmb v^{\#}=\langle v_i^{\#}\rangle_{i\le n}$ is non-decreasing.
Set $\hat{\pmb v}=\langle\hat v_i\rangle_{i\le n}$
where $\hat v_i=v_i-v^{\#}_i$ for each $i$, so that
$\pmb v=\pmb v^{\#}+\hat{\pmb v}$ and $\hat v_i\in L^0(\frak A_i)$ for
each $i$.   Also, of course, $v^{\#}_i$ and $\hat v_i$ belong to
$L^1_{\bar\mu}$ for every $i$.

\medskip

{\bf (b)} For $i<n$,

\Centerline{$\hat v_{i+1}-\hat v_i
=v_{i+1}-v_i-v^{\#}_{i+1}+v^{\#}_i
=v_{i+1}-v_i-P_iv_{i+1}+v_i
=v_{i+1}-P_iv_{i+1}$,}

\noindent so

\Centerline{$P_i\hat v_{i+1}-\hat v_i
=P_i(\hat v_{i+1}-\hat v_i)
=P_i(v_{i+1}-P_iv_{i+1})=0$;}

\noindent thus $\hat{\pmb v}$ is a martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$.

\medskip

{\bf (c)} Now suppose that $-\chi 1\le v_i\le 0$ for $i\le n$.
In this case all the $v_i$, $P_iv_j$, $v_i^{\#}$ and $\hat v_i$ belong to
$L^{\infty}(\frak A)\subseteq L^2_{\bar\mu}$.   If $i<n$ then

$$\eqalignno{\Expn(\hat v_{i+1}^2-\hat v_i^2)&
=\Expn(\hat v_{i+1}^2-(P_i\hat v_{i+1})^2)
=\Expn((\hat v_{i+1}-P_i\hat v_{i+1})^2)
\Displaycause{621Ce}
=\Expn((\hat v_{i+1}-\hat v_i)^2)
=\Expn((v_{i+1}-P_iv_{i+1})^2)
\Displaycause{by (b)}
=\Expn(v_{i+1}^2-(P_iv_{i+1})^2)
\Displaycause{621Ce again}
=\Expn(v_{i+1}^2-v_i^2)+\Expn(v_i^2-P_iv_{i+1}^2)\cr&
=\Expn(v_{i+1}^2-v_i^2)+\Expn((v_i-P_iv_{i+1})\times(v_i+P_iv_{i+1})))\cr&
\le\Expn(v_{i+1}^2-v_i^2)+2\Expn(|v_i-P_iv_{i+1}|)
\Displaycause{because $v_i$ and $P_iv_{i+1}$ both lie between
$-\chi 1$ and $0$}
=\Expn(v_{i+1}^2-v_i^2)+2\Expn(P_iv_{i+1}-v_i)
=\Expn(v_{i+1}^2-v_i^2)+2\Expn(v_{i+1}-v_i).\cr}$$

\noindent Summing over $i$,

$$\eqalignno{\Expn(\hat v_n^2)&
\le\Expn(\hat v_0^2)+\Expn(v_n^2)-\Expn(v_0^2)+2\Expn(v_n)-2\Expn(v_0)\cr&
\le\Expn(|v_n|)-2\Expn(v_0)
\Displaycause{because $v^{\#}=0$ so $\hat v_0=v_0$, while
$v_n^2\le|v_n|$ and $v_n\le 0$}
=\|v_n\|_1+2\|v_0\|_1.\cr}$$
}%end of proof of 621G

\leader{621H}{Lemma} Let $\pmb v=\langle v_i\rangle_{i\le n}$ be a finite
sequence in $L^1_{\bar\mu}$ such that $v_i\in L^0(\frak A_i)$ for $i\le n$.
Suppose that
$\langle\alpha_j\rangle_{j\le m}$, $\langle u_{ji}\rangle_{j\le m,i<n}$
are such that

\Centerline{$\alpha_j\ge 0$ for $j\le m$,
\quad$\sum_{j=0}^m\alpha_j=1$,}

\Centerline{$u_{ji}\in L^0(\frak A_i)$,
\quad$\|u_{ji}\|_{\infty}\le 1$ for $i<n$, $j\le m$.}

\noindent Set
$z=\sum_{j=0}^m\alpha_j|\sum_{i=0}^{n-1}u_{ji}\times(v_{i+1}-v_i)|$.

(a) If $\pmb v$ is non-negative and non-decreasing, then
$\bar\mu\Bvalue{z>1}\le\|v_n\|_1$.

(b) If $\pmb v$ is a martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$
then $\bar\mu\Bvalue{z>1}\le\|v_n\|_2^2$.

(c) If $\pmb v$ is a submartingale adapted to
$\langle\frak A_i\rangle_{i\le n}$ and $-\chi 1\le v_i\le 0$ for every
$i\le n$, then $\bar\mu\Bvalue{z>2}\le 3\|v_0\|_1$.

(d) If $\pmb v$ is a non-negative martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$, then
$\bar\mu\Bvalue{z>2}\le 4\Expn(v_n)$.

(e) If $\pmb v$ is a martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$, then
$\bar\mu\Bvalue{z>4}\le 4\|v_n\|_1$.

(f)(cf.\ {\smc Burkholder 66} %Theorem 6
and 628D below) If $\pmb v$ is a martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$, then
$\bar\mu\Bvalue{z>\gamma}\le\Bover{16}{\gamma}\|v_n\|_1$ for every
$\gamma>0$.

(g) If $\pmb v$ is a submartingale adapted to
$\langle\frak A_i\rangle_{i\le n}$, then
$\bar\mu\Bvalue{z>\gamma}
\le\Bover{66}{\gamma}\|v_n\|_1-\Bover{34}{\gamma}\Expn(v_0)$
for every $\gamma>0$.

\cmmnt{\medskip

\noindent{\bf Remark} For most applications (there is an important exception in
627M) it will be enough to consider the case $m=0$, so that we are looking at
$z=|\sum_{i=0}^{n-1}u_i\times(v_{i+1}-v_i)|$ where $u_i\in L^0(\frak A_i)$
and $\|u_i\|_{\infty}\le 1$ for every $i$;  this simplifies the formulae,
but seems to make no difference to the ideas required.}

\proof{{\bf (a)}

$$\eqalignno{\bar\mu\Bvalue{z>1}
&\le\Expn(z)
=\sum_{j=0}^m\alpha_j\sum_{i=0}^{n-1}\Expn(|u_{ji}|\times(v_{i+1}-v_i))\cr&
\le\sum_{j=0}^m\alpha_j\sum_{i=0}^{n-1}\Expn(v_{i+1}-v_i)
=\sum_{j=0}^m\alpha_j\Expn(v_n-v_0)\cr&
=\Expn(v_n-v_0)
\le\Expn(v_n)
=\|v_n\|_1.
\cr}$$

\medskip

{\bf (b)} For $j\le n$, set $z_j=|\sum_{i=0}^nu_{ji}\times(v_{i+1}-v_i)|$.
Then $\|z_j\|_2\le\|v_n\|_2$, by 621F.   Accordingly

\Centerline{$\|z\|_2
=\|\sum_{j=0}^m\alpha_jz_j\|_2
\le\sum_{j=0}^m\alpha_j\|z_j\|_2\le\|v_n\|_2$,}

\noindent and

\Centerline{$\bar\mu\Bvalue{z>1}
\le\Expn(z^2)=\|z\|_2^2\le\|v_n\|_2^2$.}

\medskip

{\bf (c)} By 621G, we can express $\pmb v$ as $\pmb v^{\#}+\hat{\pmb v}$
where $\pmb v^{\#}=\langle v^{\#}_i\rangle_{i\le n}$ is non-decreasing,
$\hat{\pmb v}=\langle\hat v_i\rangle_{i\le n}$ is a martingale,
$v^{\#}_0=0$ and $\|\hat v_n\|_2^2\le\|v_n\|_1+2\|v_0\|_1$.

Set

\Centerline{$z^{\#}=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_{ji}\times(v^{\#}_{i+1}-v^{\#}_i)|$,
\quad$\hat z=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_{ji}\times(\hat v_{i+1}-\hat v_i)|$.}

\noindent Then $z\le z^{\#}+\hat z$, so
$\Bvalue{z>2}\Bsubseteq\Bvalue{z^{\#}>1}\Bcup\Bvalue{\hat z>1}$ and

$$\eqalignno{\bar\mu\Bvalue{z>2}
&\le\bar\mu\Bvalue{z^{\#}>1}+\bar\mu\Bvalue{\hat z>1}
\le\|v_n^{\#}\|_1+\|\hat v_n\|_2^2
\Displaycause{by (a) and (b) above}
\le\Expn(v_n^{\#}-v^{\#}_0)+2\|v_0\|_1+\|v_n\|_1
=\Expn(v_n-v_0)+2\|v_0\|_1+\|v_n\|_1
\Displaycause{because $\pmb v-\pmb v^{\#}$ is a martingale}
=3\|v_0\|_1.
\cr}$$

\medskip

{\bf (d)} Set $\pmb{\tilde v}=\langle\tilde v_i\rangle_{i\le n}$ where
$\tilde v_i=-(v_i\wedge\chi 1)$ for each $i$.   Then $\pmb{\tilde v}$ is a
submartingale
adapted to $\langle\frak A_i\rangle_{i\le n}$.   \Prf\ For each $i$,
$\tilde v_i$ belongs to $L^0(\frak A_i)$ because $v_i$ does.   If $i<n$,
then

$$\eqalignno{\tilde v_i
&=-(v_i\wedge\chi 1)
=-(P_iv_{i+1}\wedge\chi 1)
\le P_i(-(v_{i+1}\wedge\chi 1))
\Displaycause{621Cd, with $h(\alpha)=-\min(\alpha,1)=\max(-\alpha,-1)$
for $\alpha\in\Bbb R$}
=P_i\tilde v_{i+1}.  \text{ \Qed}\cr}$$

\noindent Of course $-\chi 1\le\tilde v_i\le 0$ for each $i$ because
$v_i\ge 0$.  Set

\Centerline{$\tilde z=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_{ji}\times(\tilde v_{i+1}-\tilde v_i)|
=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_{ji}\times(-\tilde v_{i+1}+\tilde v_i)|$.}

\noindent Then

$$\eqalignno{\bar\mu\Bvalue{\tilde z>2}
&\le 3\|\tilde v_0\|_1&\text{((c) above)}\cr&
\le 3\|v_0\|_1.
\cr}$$

On the other hand,

\Centerline{$\Bvalue{z\ne\tilde z}
\Bsubseteq\sup_{i\le n}\Bvalue{v_i\ne-\tilde v_i}
=\sup_{i\le n}\Bvalue{v_i>1}
=\Bvalue{\sup_{i\le n}|v_i|>1}$,}

\noindent so

$$\eqalign{\bar\mu\Bvalue{z>2}
&\le\bar\mu\Bvalue{\tilde z>2}+\bar\mu\Bvalue{z\ne-\tilde z'}
\le 3\|v_0\|_1+\bar\mu\Bvalue{\sup_{i\le n}|v_i|>1}
\cr&\le\|v_n\|_1+3\|v_0\|_1
=\Expn(v_n+3v_0)
=4\Expn(v_n)
\cr}$$

\noindent by Doob's maximal inequality (621E).

\medskip

{\bf (e)} This time, set

\Centerline{$v'_i=\Bover12P_i(|v_n|+v_n)$,
\quad$v''_i=\Bover12P_i(|v_n|-v_n)$}

\noindent for $i\le n$, and

\Centerline{$z'=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_i\times(v'_{i+1}-v'_i)|$,
\quad$z''=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_i\times(v''_{i+1}-v''_i)|$.}

\noindent Then $\langle v'_i\rangle_{i\le n}$ is a martingale adapted to
$\langle\frak A_i\rangle_{i\le n}$, because

\Centerline{$P_iv'_{i+1}
=\Bover12P_iP_{i+1}(|v_n|+v_n)
=\Bover12P_i(|v_n|+v_n)
=v'_i$}

\noindent for every $i<n$, and of course $v'_i\ge 0$ for every $i\le n$.
Similarly, $\langle v''_i\rangle_{i\le n}$ is a non-negative martingale
adapted to $\langle\frak A_i\rangle_{i\le n}$.   Since
$v'_i-v''_i=P_iv_n=v_i$ for every $i$, $z\le z'+z''$.   So
$\Bvalue{z>4}\Bsubseteq\Bvalue{z'>2}\Bcup\Bvalue{z''>2}$ and

$$\eqalignno{\bar\mu\Bvalue{z>4}
&\le\bar\mu\Bvalue{z'>2}+\bar\mu\Bvalue{z''>2}
\le 4\Expn(v'_n+v''_n)&\text{((d) above)}\cr&
=4\Expn(v_n)
=4\|v_n\|_1,\cr}$$

\noindent as claimed.

\medskip

{\bf (f)} We now have

\Centerline{$\bar\mu\Bvalue{z>\gamma}
=\bar\mu\Bvalue{\Bover4{\gamma}z>4}
\le 4\|\Bover4{\gamma}v_n\|_1
=\Bover{16}{\gamma}\|v_n\|_1$}

\noindent for every $\gamma>0$.

\medskip

{\bf (g)} Again using 621G, express $\pmb v$ as $\pmb v^{\#}+\hat{\pmb v}$
where $\pmb v^{\#}=\langle v^{\#}_i\rangle_{i\le n}$ is non-decreasing,
$\hat{\pmb v}=\langle\hat v_i\rangle_{i\le n}$ is a martingale and
$v^{\#}_0=0$.   Set

\Centerline{$\hat z=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_i\times(\hat v_{i+1}-\hat v_i)|$,
\quad$z^{\#}=\sum_{j=0}^m\alpha_j
  |\sum_{i=0}^{n-1}u_i\times(v^{\#}_{i+1}-v^{\#}_i)|$.}

\noindent Applying (a) to $\Bover2{\gamma}\pmb v^{\#}$
and $\Bover2{\gamma}z^{\#}$, we see that
$\bar\mu\Bvalue{z^{\#}>\bover12\gamma}\le\Bover2{\gamma}\|v_n^{\#}\|_1$;
while (f) tells us that
$\bar\mu\Bvalue{\hat z>\bover12\gamma}\le\Bover{32}{\gamma}\|\hat v_n\|_1$.
Since $z\le z^{\#}+\hat z$,

\Centerline{$\bar\mu\Bvalue{z>\gamma}
\le\bar\mu\Bvalue{z^{\#}>\bover12\gamma}+\bar\mu\Bvalue{\hat z>\bover12\gamma}
\le\Bover2{\gamma}\|v_n^{\#}\|_1+\Bover{32}{\gamma}\|\hat v_n\|_1$.}

\noindent Now we know that

\Centerline{$\|v_n^{\#}\|_1=\Expn(v^{\#}_n)=\Expn(v_n)-\Expn(\hat v_n)
=\Expn(v_n)-\Expn(v_0)\le\|v_n\|_1-\Expn(v_0)$}

\noindent while

\Centerline{$\|\hat v_n\|_1\le\|v_n\|_1+\|v_n^{\#}\|_1
\le 2\|v_n\|_1-\Expn(v_0)$.}

\noindent So we get

\Centerline{$\bar\mu\Bvalue{z>\gamma}
\le\Bover{66}{\gamma}\|v_n\|_1-\Bover{34}{\gamma}\Expn(v_0)$.}
}%end of proof of 621H

\leader{621I}{Lemma} Suppose that $\langle v_i\rangle_{i\le n}$ is a
non-negative martingale adapted to $\langle\frak A_i\rangle_{i\le n}$, and that
$M\ge 0$ is such that $\Bvalue{v_i>M}\Bsubseteq\Bvalue{v_j=v_n}$ whenever
$i\le j\le n$.   Suppose that $u_i\in L^{\infty}(\frak A_i)$ and
$\|u_i\|_{\infty}\le 1$ for $i<n$, and set
$z=\sum_{i=0}^{n-1}u_i\times(v_{i+1}-v_i)$.   Take any $\delta>0$.

(a) $z$ is expressible as $z'+z''$ where $z'$, $z''\in L^0(\frak A_n)$ and

\Centerline{$\|z'\|_1
\le(2+\Bover{M}{\delta})\|(v_n-M\chi 1)^+\|_1$,}

\Centerline{$\|z''\|_2^2\le\delta^2+\|v_n\wedge M\chi 1\|_2^2$.}

(b) $\|z\|_1
\le\delta+(2+\Bover{M}{\delta})\|v_n\|_1+\sqrt{M\|v_n\|_1}$.

\proof{{\bf (a)} Induce on $n$.

\medskip

\quad{\bf (i)} The case $n=0$ is trivial.   For the
inductive step to $n\ge 1$, set $\beta=2+\Bover{M}{\delta}$ and
$z_0=\sum_{i=0}^{n-2}u_i\times(v_{i+1}-v_i)$;  since

\Centerline{$\Bvalue{v_i>M}
\Bsubseteq\Bvalue{v_j=v_n}\Bcap\Bvalue{v_{n-1}=v_n}
\Bsubseteq\Bvalue{v_j=v_{n-1}}$}

\noindent whenever $i\le j\le n-1$, the inductive
hypothesis tells us that we can express $z_0$ as $z_0'+z_0''$ where
$z'_0$, $z''_0\in L^0(\frak A_{n-1})$ and

\Centerline{$\|z_0'\|_1
\le\beta\|(v_{n-1}-M\chi 1)^+\|_1$,}

\Centerline{$\|z_0''\|_2^2\le\delta^2+\|v_{n-1}\wedge M\chi 1\|_2^2$.}

Write $P$ for $P_{n-1}$
and $\hat v$ for $(v_n-M\chi 1)^+-(v_{n-1}-M\chi 1)^+$.   Because

\Centerline{$\Bvalue{(v_{n-1}-M\chi 1)^+\ne 0}=\Bvalue{v_{n-1}>M}
\Bsubseteq\Bvalue{v_{n-1}=v_n}\Bsubseteq\Bvalue{\hat v=0}$,}

\noindent $\hat v=(v_n-M\chi 1)^+\times\chi\Bvalue{v_{n-1}\le M}\ge 0$,
and

$$\eqalign{v_{n-1}\wedge M\chi 1-P(v_n\wedge M\chi 1)
&=v_{n-1}-(v_{n-1}-M\chi 1)^+-Pv_n+P(v_n-M\chi 1)^+\cr
&=P(v_n-M\chi 1)^+-(v_{n-1}-M\chi 1)^+
=P\hat v
\ge 0.\cr}$$

\medskip

\quad{\bf (ii)} There is a $\tilde z\in L^0(\frak A_{n-1})$ such that
$\|\tilde z\|_2^2\le\delta^2+\|P(v_n\wedge M\chi 1)\|_2^2$
and $\|z_0''-\tilde z\|_1
\le\Bover{M}{\delta}\|\hat v\|_1$.   \Prf\
If $\|z_0''\|_2^2\le\delta^2+\|P(v_n\wedge M\chi 1)\|_2^2$, set
$\tilde z=z_0''$.  Otherwise, the function

\Centerline{$\gamma
\mapsto\|\med(-\gamma\chi 1,z_0'',\gamma\chi 1)\|_2^2$}

\noindent is continuous, so there is a
$\gamma\ge\delta$ such that
$\|\tilde z\|_2^2=\delta^2+\|P(v_n\wedge M\chi 1)\|_2^2$ where
$\tilde z=\med(-\gamma\chi 1,z_0'',\gamma\chi 1)$.  In this case,

\Centerline{$\|z_0''\|_2^2-\|\tilde z\|_2^2
=\Expn((|z_0''|+|\tilde z|)\times(|z_0''|-|\tilde z|))
\ge 2\delta\Expn(|z_0''|-|\tilde z|)
=2\delta\|z_0''-\tilde z\|_1$}

\noindent because $|z''_0|=|\tilde z|+|z''_0-\tilde z|$ and

\Centerline{$\Bvalue{z_0''\ne\tilde z}
=\Bvalue{|z_0''|>\gamma}
\Bsubseteq\Bvalue{|z_0''|\ge\delta}\Bcap\Bvalue{|\tilde z|\ge\delta}$,}

\noindent and similarly

$$\eqalign{&\|v_{n-1}\wedge M\chi 1\|_2^2
  -\|P(v_n\wedge M\chi 1)\|_2^2\cr
&\mskip50mu
=\Expn\bigl((v_{n-1}\wedge M\chi 1+P(v_n\wedge M\chi 1))
  \times(v_{n-1}\wedge M\chi 1-P(v_n\wedge M\chi 1))\bigr)\cr
&\mskip50mu
\le 2M\|v_{n-1}\wedge M\chi 1-P(v_n\wedge M\chi 1)\|_1
=2M\|P\hat v\|_1
=2M\|\hat v\|_1\cr}$$

\noindent because $\hat v\ge 0$.   So

$$\eqalign{\|z_0''-\tilde z\|_1
&\le\Bover1{2\delta}(\|z_0''\|_2^2-\|\tilde z\|_2^2)\cr
&\le\Bover1{2\delta}(\delta^2+\|v_{n-1}\wedge M\chi 1\|_2^2
  -\delta^2-\|P(v_n\wedge M\chi 1)\|_2^2)
\le\Bover{M}{\delta}\|\hat v\|_1.  \text{ \Qed}\cr}$$

\medskip

\quad{\bf (iii)}
Set $z''=\tilde z+u_{n-1}\times(v_n\wedge M\chi 1-P(v_n\wedge M\chi 1))$.
Since
$\tilde z\times u_{n-1}\in L^0(\frak A_{n-1})\cap L^2_{\bar\mu}$
and $v_n\wedge M\chi 1-P(v_n\wedge M\chi 1)\in L^{\infty}(\frak A)$,

$$\eqalign{0
&=\Expn(\tilde z\times u_{n-1}\times(v_n\wedge M\chi 1-P(v_n\wedge M\chi 1)))
  \cr
&=\Expn(P(v_n\wedge M\chi 1)
  \times(v_n\wedge M\chi 1-P(v_n\wedge M\chi 1)))\cr}$$

\noindent and

$$\eqalign{\|z''\|_2^2
&=\|\tilde z\|_2^2
   +\|u_{n-1}\times(v_n\wedge M\chi 1-P(v_n\wedge M\chi 1))\|_2^2\cr
&\le\delta^2+\|P(v_n\wedge M\chi 1)\|_2^2
   +\|v_n\wedge M\chi 1-P(v_n\wedge M\chi 1)\|_2^2
=\delta^2+\|v_n\wedge M\chi 1\|_2^2.\cr}$$

\medskip

\quad{\bf (iv)} Set

$$\eqalign{z'
&=z-z''\cr
&=z_0+u_{n-1}\times(v_n-v_{n-1})-\tilde z
  -u_{n-1}\times(v_n\wedge M\chi 1-P(v_n\wedge M\chi 1))\cr
&=z_0'+z_0''-\tilde z\cr
&\mskip30mu
+u_{n-1}\times((v_n-M\chi 1)^+-(v_{n-1}-M\chi 1)^+
  -v_{n-1}\wedge M\chi 1+P(v_n\wedge M\chi 1))\cr
&=z_0'+z_0''-\tilde z
  +u_{n-1}\times(\hat v-P\hat v),\cr}$$

\noindent so that		

$$\eqalign{\|z'\|_1
&\le\|z_0'\|_1+\|z_0''-\tilde z\|_1
  +\|\hat v\|_1+\|P\hat v\|_1\cr
&\le\beta\|(v_{n-1}-M\chi 1)^+\|_1+\Bover{M}{\delta}\|\hat v\|_1
  +2\|\hat v\|_1\cr
&=\beta(\|(v_{n-1}-M\chi 1)^+\|_1+\|\hat v\|_1)
=\beta\|(v_n-M\chi 1)^+\|_1,\cr}$$

\noindent and the induction proceeds.

\medskip

{\bf (b)} follows at once, because

\Centerline{$\|z''\|_1^2\le\|z''\|_2^2
\le\delta^2+\|v_n\wedge M\chi 1\|_2^2
\le\delta^2+M\|v_n\|_1$}

\noindent and $\|z''\|_1\le\delta+\sqrt{M\|v_n\|_1}$.
}%end of proof of 621I

\leader{621J}{Lemma} Suppose that
$\langle v_i\rangle_{i\le n}$ is a non-negative submartingale
adapted to $\langle\frak A_i\rangle_{i\le n}$ and
$z=\sum_{i=0}^{n-1}P_iv_{i+1}-v_i$.   Set
$f(\delta)=\inf_{\beta\ge 0}\beta\delta+\Expn((v_n-\beta\chi 1)^+)$
for $\delta\ge 0$.
Then $\Expn(z\times\chi\Bvalue{z>2\alpha})
\le 3f(\bover1{\alpha}\Expn(v_n))$ for every $\alpha>0$.

%can we manage a version for non-negative supermartingales?  query

\proof{ ({\smc Karatzas \& Shreve 91}){\bf (a)} For $i\le n$, set
$u_i=v_i-P_iv_n\le 0$;  then $u_n=0$ and

\Centerline{$u_i\in L^0(\frak A_i)$,
\quad$u_i=v_i-P_iv_n\le P_iv_{i+1}-P_iP_{i+1}v_n=P_iu_{i+1}$}

\noindent for $i<n$, so $\langle u_i\rangle_{i\le n}$ is a submartingale.
Observe that

\Centerline{$P_iv_n=v_i-u_i\ge-u_i$,}

\Centerline{$P_iu_{i+1}-u_i
=P_iv_{i+1}-P_iv_n-v_i+P_iv_n=P_iv_{i+1}-v_i\in L^0(\frak A_i)$}

\noindent for $i<n$.   Set

\Centerline{$z_i=\sum_{j=0}^{i-1}P_jv_{j+1}-v_j
=\sum_{j=0}^{i-1}P_ju_{j+1}-u_j
\in L^0(\frak A_{i-1})$}

\noindent for $i\le n$ (counting $\frak A_{-1}$ as $\{0,1\}$).   Then
$z_0=0$ and $z_n=z$.   For $i\le n$,

$$\eqalign{z_i-P_iz
&=P_i(z_i-z)
=-P_i(\sum_{j=i}^{n-1}(P_ju_{j+1}-u_j))
=-P_i(\sum_{j=i}^{n-1}P_j(u_{j+1}-u_j))\cr
&=-P_i(\sum_{j=i}^{n-1}u_{j+1}-u_j)
=-P_i(u_n-u_i)
=P_iu_i
=u_i.\cr}$$

For $\alpha>0$, set

\Centerline{$a_{\alpha k}=\Bvalue{z_{k+1}>\alpha}\Bsetminus\Bvalue{z_k>\alpha}$
for $k<n$,
\quad$a_{\alpha n}=1\Bsetminus\Bvalue{z>\alpha}$;}

\noindent then $\langle a_{\alpha k}\rangle_{k\le n}$ is a partition of
unity in $\frak A$, and $a_{\alpha k}\in\frak A_k$ for every $k$.
Set $b_{\alpha}=\Bvalue{z>\alpha}=\sup_{k<n}a_{\alpha k}$, and

\Centerline{$u'_{\alpha}=\sum_{k=0}^nu_k\times\chi a_{\alpha k}$.}

\medskip

{\bf (b)} For $\alpha>0$, consider

\Centerline{$\frak B_{\alpha}
=\{a:a\in\frak A$, $a\Bcap a_{\alpha k}\in\frak A_k$ for
every $k\le n\}$,}

\Centerline{$u''_{\alpha}=\sum_{k=0}^nP_kv_n\times\chi a_{\alpha k}
\ge\sum_{k=0}^n(-u_k)\times\chi a_{\alpha k}=-u'_{\alpha}$.}

\noindent Then $\frak B_{\alpha}$ is a closed subalgebra of $\frak A$,
$u''_{\alpha}\in L^0(\frak B_{\alpha})$ and if
$a\in\frak B_{\alpha}$

$$\eqalign{\Expn(\chi a\times u''_{\alpha})
&=\sum_{k=0}^n\Expn(\chi(a\Bcap a_{\alpha k})\times P_kv_n)\cr
&=\sum_{k=0}^n\Expn(\chi(a\Bcap a_{\alpha k})\times v_n)
=\Expn(\chi a\times v_n).\cr}$$

\noindent So if $Q_{\alpha}$
is the conditional expectation operator associated with $\frak B_{\alpha}$,
$u''_{\alpha}=Q_{\alpha}v_n$.   But this means that, for any
$a\in\frak A$ and $\beta\ge 0$,

$$\eqalign{\Expn(\chi a\times u''_{\alpha})
&=\Expn(\chi a\times Q_{\alpha}v_n)
=\Expn(\chi a\times Q_{\alpha}(v_n\wedge\beta\chi 1))
   +\Expn(\chi a\times Q_{\alpha}(v_n-\beta\chi 1)^+)\cr
&\le\Expn(\chi a\times Q_{\alpha}(\beta\chi 1))
   +\Expn(Q_{\alpha}(v_n-\beta\chi 1)^+)\cr
&=\Expn(\chi a\times \beta\chi 1)
   +\Expn((v_n-\beta\chi 1)^+)
=\beta\bar\mu a
   +\Expn((v_n-\beta\chi 1)^+).\cr}$$

\noindent As $\beta$ is arbitrary, and $-u'_{\alpha}\le u''_{\alpha}$,

\Centerline{$-\Expn(\chi a\times u'_{\alpha})
\le\Expn(\chi a\times u''_{\alpha})
\le f(\bar\mu a)$.}

\medskip

{\bf (c)}

\Centerline{$u'_{\alpha}\times\chi a_{\alpha k}
=u_k\times\chi a_{\alpha k}
=(z_k-P_kz)\times\chi a_{\alpha k}
\le(\alpha\chi 1-P_kz)\times\chi a_{\alpha k}$}

\noindent for $k<n$, so

$$\eqalign{\Expn(\chi b_{\alpha}\times u'_{\alpha})
&=\sum_{k=0}^{n-1}\Expn(\chi a_{\alpha k}\times u'_{\alpha})
\le\sum_{k=0}^{n-1}
  \alpha\bar\mu a_{\alpha k}-\Expn(\chi a_{\alpha k}\times P_kz)\cr
&=\sum_{k=0}^{n-1}\alpha\bar\mu a_{\alpha k}
   -\Expn(\chi a_{\alpha k}\times z)
=\alpha\bar\mu b_{\alpha}-\Expn(\chi b_{\alpha}\times z).\cr}$$

\noindent Consequently

$$\eqalignno{-\Expn(\chi b_{\alpha}\times u'_{\alpha})
&\ge\Expn(\chi b_{\alpha}\times(z-\alpha\chi 1))
\ge\Expn(\chi b_{2\alpha}\times(z-\alpha\chi 1))\cr
\displaycause{because $b_{2\alpha}\Bsubseteq b_{\alpha}
=\Bvalue{z>\alpha}$}
&\ge\alpha\bar\mu b_{2\alpha};\cr}$$

\noindent it follows that

$$\eqalignno{\Expn(\chi b_{2\alpha}\times z)
&\le 2\alpha\bar\mu b_{2\alpha}-\Expn(\chi b_{2\alpha}\times u'_{2\alpha})
  \cr
&\le -2\Expn(\chi b_{\alpha}\times u'_{\alpha})
   -\Expn(\chi b_{2\alpha}\times u'_{2\alpha})
\le 2f(\bar\mu b_{\alpha})+f(\bar\mu b_{2\alpha})\cr
\displaycause{by (b)}
&\le 3f(\bar\mu b_{\alpha})\cr}$$

\noindent because $f$ is non-decreasing.

\medskip

{\bf (d)} On the other hand,

\Centerline{$\Expn(z)
=\sum_{i=0}^{n-1}\Expn(P_iv_{i+1})-\Expn(v_i)
=\sum_{i=0}^{n-1}\Expn(v_{i+1})-\Expn(v_i)
=\Expn(v_n)-\Expn(v_0)
\le\Expn(v_n)$,}

\noindent so

\Centerline{$\bar\mu b_{\alpha}=\bar\mu\Bvalue{z>\alpha}
\le\Bover1{\alpha}\Expn(z)\le\Bover1{\alpha}\Expn(v_n)$}

\noindent and

\Centerline{$\Expn(z\times\chi\Bvalue{z>2\alpha})
=\Expn(z\times\chi b_{2\alpha})
\le 3f(\bar\mu b_{\alpha})
\le 3f(\Bover1{\alpha}\Expn(v_n))$,}

\noindent as required.
}%end of proof of 621J

\exercises{\leader{621X}{Basic exercises (a)}
%\spheader 621Xa
Suppose that we think of 621D-621J
  % 621D 621E 621F 621G 621H 621I 621J
as applying to a stochastic integration structure
\Asissix\ such that $T=\{0,\ldots,n\}$.
Show that if $\pmb v=\family{\sigma}{\Cal T}{v_{\sigma}}$ is any fully adapted
process, then

$$\eqalign{Q_{\Cal T}(d\pmb v)&
=\{\sum_{i=0}^{n-1}u_i\times(v_{(i+1)\check{\phantom i}}-v_{\check\imath}):
u_i\in L^0(\frak A_i),\,
\|u_i\|_{\infty}\le 1\text{ for every }i<n\}.\cr}$$
%621F

\spheader 621Xb In the context of Lemma
621H, show that, given $M\ge 0$, there are $n$, $\pmb v$ and
$\ofamily{i}{n}{u_i}$ such that $\pmb v$ is a martingale and
$\|z\|_1\ge M\|v_n\|_1$.   \Hint{$v_i=\gamma^i\chi a_i$,
$u_i=(-1)^i\chi 1$ where $1=a_0\Bsupseteq\ldots\Bsupseteq a_n$.}

%\leader{621Y}{Further exercises (a)}
}%end of exercises


\endnotes{
\Notesheader{621} There are some curious
formulae here, which is why I think
readers may wish to look ahead to see what they are supposed to do.
But 621H, 621I and 621J, as well as 621E, are of the same
kind;  they seek to bound quantities calculated from sequences
$\langle u_i\rangle_{i\le n}$ and
$\langle v_i\rangle_{i\le n}$ in terms of a quantity determined by
the final term $v_n$
alone.   The same is true of the fundamental inequalities 275D and 275F.

The constant $16$ in 621Hf is far from best possible;  in fact it is the
case there that $\bar\mu\Bvalue{|z|>\gamma}\le\Bover2{\gamma}\|v_n\|_1$ (628D).
(I do not know whether the inequality can be improved further.)   However
the proof in \S628 is substantially longer than that in 621G-621H and
demands techniques from measure theory which I do not think we shall need
elsewhere in this volume.   The argument above is a better preparation for
what will come later.   For instance,
the idea in 621G will reappear, in much more
general form, in the `previsible variations' of \S626.

The formulae in 621I-621J are bound to seem odd.
In 621I we are presented with $M$ but anticipate that
$\delta$ and $\|v_n\|_1$ can be forced to be small, so that
$\|z\|_1$ is small;  this will be used in one of the main results of the
chapter (623O).   In 621J we see that
$\lim_{\alpha\to\infty}f(\bover1{\alpha}\Expn(v_n))=0$, while
a bound on $\Expn(z\times\chi\Bvalue{z>\alpha})$ is something
we look for if we want to prove that a set is uniformly integrable.   See
part (a) of the proof of 626O.

The notation of this section is a little clumsier than it might be, because
from 621F onwards I
repeatedly speak of elements $z=\sum_{i=0}^{n-1}u_i\times(v_{i+1}-v_i)$ without
making the obvious association with $Q_{\Cal S}(d\pmb v)$ (621Xa).
I am in fact avoiding any appeal to the ideas introduced in
Chapter 61;  the material here can be regarded as a development of \S275 in
the language of Chapter 36, quite apart from its applications to stochastic
integration, even though it is manifestly directed to those applications.
}%end of notes



\discrpage

\frfilename{mt622.tex}
\versiondate{29.7.20}
\copyrightdate{2012}

\def\wlim{\mathop{\text{wlim}}}

\def\chaptername{Martingales}
\def\sectionname{Fully adapted martingales}

\newsection{622}

I come now to the promised central fact of the theory:  martingales are
local integrators.
The first step is to establish a concept of `martingale' for
fully adapted processes (622C), which involves us in the properties of
conditional expectations with respect to stopping-time algebras (622B).
Elementary facts about martingales are in 622D-622F. %622D 622E 622F
The theorem that every martingale is a local integrator is
now easy (622H);  of course it depends on non-trivial ideas from \S621.
In 622L I check that Brownian motion\cmmnt{, as defined in 612T,} is a local
martingale.
The rest of the section is a miscellany of results which will be needed later.

\leader{622A}{Notation} As in Chapter 61,
\Asissix\ will be a stochastic integration
structure\cmmnt{, with $\Cal T_f$ and $\Cal T_b$ the corresponding
lattices of finite and bounded stopping times}.
\cmmnt{For $t\in T$, $\check t$ is the constant stopping time at $t$.
As before, I write
$\theta(w)=\Expn(|w|\wedge\chi 1)$ for $w\in L^0(\frak A)$.
For a sublattice $\Cal S$ of $\Cal T$, $\Cal I(\Cal S)$ will be the set of
finite sublattices of $\Cal S$.

}For $\tau\in\Cal T$,
$P_{\tau}:L^1_{\bar\mu}\to L^0(\frak A_{\tau})\cap L^1_{\bar\mu}$, where
$L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$,
will be the conditional expectation operator
associated with the closed subalgebra $\frak A_{\tau}$ of
$\frak A$\cmmnt{ (621C)}.

\leader{622B}{}\cmmnt{ We need something to match the familiar rule on
composition of conditional expectation operators in 621Cb.

\medskip

\noindent}{\bf Proposition} Suppose that $\sigma$, $\tau\in\Cal T$.

(a) $P_{\sigma}P_{\tau}=P_{\sigma\wedge\tau}$.

(b) $\Bvalue{\sigma=\tau}\Bsubseteq\Bvalue{P_{\sigma}u=P_{\tau}u}$ for
every $u\in L^1_{\bar\mu}$.

\proof{{\bf (a)(i)} Set $a=\Bvalue{\sigma\le\tau}$, so that
$a\in\frak A_{\sigma\wedge\tau}=\frak A_{\sigma}\cap\frak A_{\tau}$
(611H(c-ii)), and $u\in L^0(\frak A_{\sigma\wedge\tau})$ whenever
$u\in L^0(\frak A_{\sigma})$ and $u=u\times\chi a$ (612C).
Now if $u\in L^1_{\bar\mu}$ and $u=u\times\chi a$, we see that

\Centerline{$(P_{\sigma}P_{\tau}u)\times\chi a
=P_{\sigma}(P_{\tau}u\times\chi a)
=P_{\sigma}P_{\tau}(u\times\chi a)
=P_{\sigma}P_{\tau}u\in L^0(\frak A_{\sigma})$}

\noindent so that $P_{\sigma}P_{\tau}u\in L^0(\frak A_{\sigma\wedge\tau})$.
And if $b\in\frak A_{\sigma\wedge\tau}$ we surely have

\Centerline{$\Expn(P_{\sigma}P_{\tau}u\times\chi b)
=\Expn(P_{\sigma}(P_{\tau}u\times\chi b))
=\Expn(P_{\tau}u\times\chi b)
=\Expn(u\times\chi b)$.}

\noindent So $P_{\sigma}P_{\tau}u=P_{\sigma\wedge\tau}u$.

\medskip

\quad{\bf (ii)} Next, setting $a'=\Bvalue{\tau\le\sigma}$,
we have  $u\in L^0(\frak A_{\sigma\wedge\tau})$ whenever
$u\in L^0(\frak A_{\tau})$ and $u=u\times\chi a'$.
So now, if $u\in L^1_{\bar\mu}$ and $u=u\times\chi a'$, we have
$P_{\tau}u=P_{\tau}u\times\chi a'\in L^0(\frak A_{\sigma\wedge\tau})$ and
$P_{\sigma}P_{\tau}u=P_{\tau}u\in L^0(\frak A_{\sigma\wedge\tau})$.
As in (i),

\Centerline{$\Expn(P_{\sigma}P_{\tau}u\times\chi b)
=\Expn(P_{\tau}u\times\chi b)
=\Expn(u\times\chi b)$}

\noindent whenever $b\in\frak A_{\sigma\wedge\tau}$.
So $P_{\sigma}P_{\tau}u=P_{\sigma\wedge\tau}u$ in this case also.

\medskip

\quad{\bf (iii)} Assembling these, and noting that
$1\Bsetminus a\Bsubseteq a'$,

$$\eqalign{P_{\sigma}P_{\tau}u
&=P_{\sigma}P_{\tau}(u\times\chi a)
   +P_{\sigma}P_{\tau}(u\times\chi(1\Bsetminus a))\cr
&=P_{\sigma\wedge\tau}(u\times\chi a)
   +P_{\sigma\wedge\tau}(u\times\chi(1\Bsetminus a))
=P_{\sigma\wedge\tau}u\cr}$$

\noindent for every $u\in L^1_{\bar\mu}$, and
$P_{\sigma}P_{\tau}=P_{\sigma\wedge\tau}$.

\medskip

{\bf (b)} Set $c=a\Bcap a'=\Bvalue{\sigma=\tau}$.   Then
$c\in\frak A_{\sigma\wedge\tau}$ and
$(u\times\chi c)\times\chi a'=u\times\chi c$, so

$$\eqalignno{P_{\tau}u\times\chi c
&=P_{\tau}(u\times\chi c)
=P_{\sigma}P_{\tau}(u\times\chi c)\cr
\displaycause{see (a-ii) above}
&=P_{\sigma\wedge\tau}(u\times\chi c).\cr}$$

\noindent Similarly,

\Centerline{$P_{\sigma}u\times\chi c=P_{\tau\wedge\sigma}(u\times\chi c)
=P_{\sigma\wedge\tau}(u\times\chi c)=P_{\tau}u\times\chi c$.}

\noindent So $\Bvalue{P_{\sigma}u=P_{\tau}u}\Bsupseteq c$.
}%end of proof of 622B

\leader{622C}{Fully adapted martingales}
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v =\family{\sigma}{\Cal S}{v_{\sigma}}$ a fully adapted process.

\spheader 622Ca\cmmnt{ $\pmb v $ is an {\bf $L^1$-process} if
$v_{\sigma}\in L^1_{\bar\mu}$ for every $\sigma\in\Cal S$.
Generally,} $\pmb v $ is an {\bf $L^p$-process}, for $1\le p<\infty$, if
$v_{\sigma}\in L^p(\frak A,\bar\mu)$\cmmnt{ (366A)} for every $\sigma\in\Cal S$,
and an {\bf $L^{\infty}$-process} if $v_{\sigma}\in L^{\infty}(\frak A)$
\cmmnt{(363A, 364J)} for every $\sigma\in\Cal S$.

For $1\le p\le\infty$, $\pmb v$ is {\bf $\|\,\|_p$-bounded} if
$\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_p$ is finite (counting the supremum
as $0$ if $\Cal S$ is empty).   \cmmnt{Note that} $\pmb v$ is
$\|\,\|_{\infty}$-bounded
iff it is order-bounded and $\sup|\pmb v|$\cmmnt{, as defined in 614Ea,} is in
$L^{\infty}(\frak A)$.

\spheader 622Cb $\pmb v $ is a {\bf martingale} if it is an $L^1$-process
and $v_{\sigma}=P_{\sigma}v_{\tau}$ whenever $\sigma\le\tau$ in $\Cal S$.

\spheader 622Cc $\pmb v $ is a {\bf local martingale} if there is a
covering ideal $\Cal S'$ of $\Cal S$\cmmnt{ (611N)} such that
$\pmb v\restr\Cal S'$ is a martingale.

\cmmnt{Taking $\Cal S'=\Cal S$ we see that every martingale is a local
martingale.   For classic examples of local martingales which are not
martingales, see 622Xe and 632N.

Note that I do {\it not} say `$\pmb v $ is a local martingale if
$\pmb v\restr\Cal S\wedge\tau$ is a martingale for every $\tau\in\Cal S$';
the `local' in `local martingale' is not the same as the `local' in
`local integrator' or `locally order-bounded' or `locally of bounded
variation' or `locally moderately oscillatory'.
}

\spheader 622Ce{\bf (i)} $\pmb v $ is {\bf uniformly integrable} if it is an
$L^1$-process and $\{v_{\sigma}:\sigma\in\Cal S\}$ is uniformly integrable.

\medskip

\quad{\bf (ii)} It will be convenient to use the phrase `$L^p$-martingale' to
mean a martingale which is also an $L^p$-process.

\leader{622D}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\Mfa=\Mfa(\Cal S)$ the Riesz space of fully adapted processes with domain
$\Cal S$.   Let $\Cal S'$ be a sublattice of $\Cal S$.

(a) For any $p\in[1,\infty]$, the set of $L^p$-processes with domain
$\Cal S$ is a solid linear subspace of $\Mfa$, and $\pmb v\restr\Cal S'$ is
an $L^p$-process whenever $\pmb v\in\Mfa$ is an $L^p$-process.

(b)(i) The set of martingales with domain $\Cal S$ is a linear subspace of
$\Mfa$.

\quad(ii) If $\pmb v\in\Mfa$ is a martingale then $\pmb v\restr\Cal S'$
is a martingale.

(c) The set of local
martingales with domain $\Cal S$ is a linear subspace of
$\Mfa$.   If $\Cal S'$ is an ideal of $\Cal S$, then
$\pmb v\restr\Cal S'$ is a local martingale for every local martingale
$\pmb v\in\Mfa$.

(d) The set of uniformly integrable processes with domain
$\Cal S$ is a solid linear subspace of $\Mfa$, and $\pmb v\restr\Cal S'$ is
uniformly integrable whenever $\pmb v\in\Mfa$ is uniformly integrable.

\proof{{\bf (a)} This is immediate from the definitions (622Ca).

\medskip

{\bf (b)(i)-(ii)} These too are immediate from the definitions.

\medskip

{\bf (c)} Applying (b-i) to appropriate ideals $\Cal S\wedge\tau$, we see that
the set of local martingales with domain $\Cal S$ is a linear subspace of
$\Mfa$.   Concerning ideals of $\Cal S$, we need to know that if
$\Cal S'$ is an ideal of $\Cal S$ and $\Cal S_1$ is a covering ideal of
$\Cal S$, then
$\Cal S_1\cap\Cal S'$ is a covering ideal of $\Cal S'$.   \Prf\ If
$\tau\in\Cal S'$, then

\Centerline{$\sup_{\sigma\in\Cal S_1\cap\Cal S'}\Bvalue{\tau=\sigma}
\Bsupseteq\sup_{\sigma\in\Cal S_1}\Bvalue{\tau=\tau\wedge\sigma}
\Bsupseteq\sup_{\sigma\in\Cal S_1}\Bvalue{\tau=\sigma}=1$.  \Qed}

\medskip

\quad{\bf (d)} All we need to know is that subsets,
sums and scalar multiples and solid
hulls of uniformly integrable sets are uniformly integrable, as declared in
621B(c-i).
}%end of proof of 622D

\leader{622E}{Elementary facts} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb u=\sfamilysS{u}$ a fully adapted process.

\spheader 622Ea If $\pmb u$ is constant with a value in $L^1_{\bar\mu}$, then
$\pmb u$ is a uniformly integrable martingale.

\spheader 622Eb{\bf (i)} If $\tau\in\Cal S$,
then $\pmb u$ is a martingale iff $\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are martingales.
\prooflet{\Prf\ If $\pmb u$ is a martingale then
$\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are martingales by 622Db.   If
$\pmb u\restr\Cal S\wedge\tau$ and
$\pmb u\restr\Cal S\vee\tau$ are martingales, then we have

\Centerline{$u_{\sigma}=u_{\sigma\vee\tau}+u_{\sigma\wedge\tau}-u_{\tau}
\in L^1_{\bar\mu}$}

\noindent for every $\sigma\in\Cal S$ (612Df).   So $\pmb u$ is an
$L^1$-process.   If $\sigma\le\sigma'$ in $\Cal S$ then

\Centerline{$P_{\sigma\wedge\tau}u_{\sigma'\vee\tau}
=P_{\sigma}P_{\tau}u_{\sigma'\vee\tau}
=P_{\sigma}u_{\tau}$}

\noindent (using 622Ba for the first equality), so

$$\eqalignno{\Bvalue{\sigma\le\tau}&
=\Bvalue{\sigma=\sigma\wedge\tau}
   \Bcap\Bvalue{\sigma=\sigma\wedge\sigma'\wedge\tau}
\cr&
\mskip100mu
   \Bcap\Bvalue{P_{\sigma}u_{\sigma'\vee\tau}
                    =P_{\sigma\wedge\tau}u_{\sigma'\vee\tau}}
   \Bcap\Bvalue{P_{\sigma}u_{\sigma'\wedge\tau}
                    =P_{\sigma\wedge\sigma'\wedge\tau}u_{\sigma'\wedge\tau}}
\Displaycause{611E(a-ii-$\beta$, 622Bb}
\Bsubseteq\Bvalue{\sigma=\sigma\wedge\sigma'\wedge\tau}
   \Bcap\Bvalue{P_{\sigma}u_{\sigma'\vee\tau}=P_{\sigma}u_{\tau}}\cr&
\mskip100mu
   \Bcap\Bvalue{P_{\sigma}u_{\sigma'\wedge\tau}
                    =u_{\sigma\wedge\sigma'\wedge\tau}}
   \Bcap\Bvalue{u_{\sigma}=u_{\sigma\wedge\sigma'}\wedge\tau}\cr&
\Bsubseteq\Bvalue{P_{\sigma}u_{\sigma'\vee\tau}=P_{\sigma}u_{\tau}}
   \Bcap\Bvalue{P_{\sigma}u_{\sigma'\wedge\tau}=u_{\sigma}}
\cr&
\Bsubseteq\Bvalue{P_{\sigma}(u_{\sigma'\wedge\tau}+u_{\sigma'\vee\tau}-u_{\tau})
                       =u_{\sigma}}
=\Bvalue{P_{\sigma}u_{\sigma'}=u_{\sigma}}.
\cr}$$

\noindent On the other hand,

$$\eqalign{\Bvalue{\tau\le\sigma}&
\Bsubseteq\Bvalue{\sigma=\sigma\vee\tau}\Bcap\Bvalue{\sigma'\wedge\tau=\tau}\cr&
\Bsubseteq\Bvalue{\sigma=\sigma\vee\tau}\Bcap\Bvalue{\sigma'\wedge\tau=\tau}
    \Bcap\Bvalue{P_{\sigma}u_{\tau}=P_{\sigma\vee\tau}u_{\tau}}\cr&
\mskip100mu
    \Bcap\Bvalue{P_{\sigma}u_{\sigma'\vee\tau}
                   =P_{\sigma\vee\tau}u_{\sigma'\vee\tau}}
    \Bcap\Bvalue{P_{\sigma}u_{\sigma'\wedge\tau}
                   =P_{\sigma\vee\tau}u_{\sigma'\wedge\tau}}\cr&
\Bsubseteq\Bvalue{\sigma=\sigma\vee\tau}\Bcap\Bvalue{\sigma'\wedge\tau=\tau}
    \Bcap\Bvalue{P_{\sigma}u_{\tau}=u_{\tau}}\cr&
\mskip100mu
    \Bcap\Bvalue{P_{\sigma}u_{\sigma'\vee\tau}=u_{\sigma\vee\tau}}
    \Bcap\Bvalue{P_{\sigma}u_{\sigma'\wedge\tau}=u_{\sigma'\wedge\tau}}\cr&
\Bsubseteq\Bvalue{P_{\sigma}u_{\tau}=u_{\tau}}
    \Bcap\Bvalue{P_{\sigma}u_{\sigma'\vee\tau}=u_{\sigma}}
    \Bcap\Bvalue{P_{\sigma}u_{\sigma'\wedge\tau}=u_{\tau}}
\Bsubseteq\Bvalue{P_{\sigma}u_{\sigma'}=u_{\sigma}}.
\cr}$$

\noindent So in fact $\Bvalue{P_{\sigma}u_{\sigma'}=u_{\sigma}}=1$ and
$P_{\sigma}u_{\sigma'}=u_{\sigma}$.   As $\sigma$ and $\sigma'$ are arbitrary,
$\pmb u$ is a martingale.\ \Qed}

\medskip

\quad{\bf (ii)} If $\tau\in\Cal S$, 
$\pmb u\restr\Cal S\wedge\tau$ is a martingale
and $\pmb u\restr\Cal S\vee\tau$ is constant, then $\pmb u$ is a martingale.
\prooflet{(For the constant value of $\pmb u\restr\Cal S\vee\tau$ must be
$u_{\tau}$, which is a value of $\pmb u\restr\Cal S\wedge\tau$ so
is in $L^1_{\bar\mu}$.)}

\spheader 622Ec If for every $\epsilon>0$ there is a martingale
$\pmb v=\sfamilysS{v}$ such that $\|u_{\sigma}-v_{\sigma}\|_1\le\epsilon$ for
every $\sigma\in\Cal S$, then $\pmb u$ is a martingale.   \prooflet{\Prf\
Note first that as there is an $L^1$-process $\pmb v$ such that
$\pmb u-\pmb v$ is an $L^1$-process, $\pmb u$ also is an $L^1$-process.   Now
suppose that $\tau\le\tau'$ in $\Cal S$ and $\epsilon>0$.   Let
$\pmb v=\sfamilysS{v}$ be a martingale such that
$\|u_{\sigma}-v_{\sigma}\|_1\le\bover12\epsilon$ for every $\sigma\in\Cal S$.
Then

\Centerline{$\|P_{\tau}u_{\tau'}-u_{\tau}\|_1
\le\|P_{\tau}u_{\tau'}-P_{\tau}v_{\tau'}\|_1+\|P_{\tau}v_{\tau'}-v_{\tau}\|_1
    +\|v_{\tau}-u_{\tau}\|_1
\le\epsilon$}

\noindent (using 621Ca to see that $\|P_{\tau}(u_{\tau'}-v_{\tau'})\|_1
\le\|u_{\tau'}-v_{\tau'}\|_1$).   As $\epsilon$ is arbitrary,
$P_{\tau}u_{\tau'}=u_{\tau}$;  as $\tau$ and $\tau'$ are arbitrary,
$\pmb u$ is a martingale.\ \Qed}

\spheader 622Ed If $\pmb u$ is a martingale and $A\subseteq\Cal S$ is non-empty
and downwards-directed, then the $\|\,\|_1$-limit
$z=\1lim_{\sigma\downarrow A}u_{\sigma}$ is defined and is the limit
$\lim_{\sigma\downarrow A}u_{\sigma}$ for the topology of convergence in
measure;  and if $\tau\in A$ then $z$ is the conditional
expectation of $u_{\tau}$ on $\bigcap_{\sigma\in A}\frak A_{\sigma}$.
\prooflet{\Prf\ By 367Qa,
$\1lim_{\sigma\downarrow A}u_{\sigma}
=\1lim_{\sigma\downarrow A}P_{\sigma}u_{\tau}$ is defined and is the conditional
expectation of $u_{\tau}$ on $\bigcap_{\sigma\in A}\frak A_{\sigma}$.
By 613B(d-i), this is also the limit $\lim_{\sigma\downarrow A}u_{\sigma}$.\
\Qed}

In particular, if $\Cal S$ is non-empty, then the starting value
$\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined and belongs to
$L^1_{\bar\mu}$.

\leader{622F}{Proposition} Take any $u\in L^1_{\bar\mu}$.

(a) $\pmb Pu=\family{\tau}{\Cal T}{P_{\tau}u}$ is a uniformly integrable
martingale.

(b) Suppose that $\sigma$, $\tau\in\Cal T$
and $\Bvalue{u\ne 0}\Bsubseteq\Bvalue{\sigma=\tau}$.
Then $P_{\sigma}u=P_{\tau}u$.

\proof{{\bf (a)} By 622Bb, $\pmb Pu$ is fully adapted;  by 622Ba, it is
a martingale;  by 621Cf, it is uniformly integrable.

\medskip

{\bf (b)} As in the proof of 622Ba, set $a=\Bvalue{\sigma\le\tau}$.   Then
$u=u\times\chi a$ and $a\in\frak A_{\sigma}$ so

\Centerline{$P_{\sigma}u=P_{\sigma}(u\times\chi a)
=P_{\sigma}u\times\chi a$;}

\noindent since $P_{\sigma}u\in L^0(\frak A_{\sigma})$,
$P_{\sigma}u\in L^0(\frak A_{\tau})$ (612C again) and

\Centerline{$P_{\sigma}u=P_{\tau}P_{\sigma}u=P_{\tau\wedge\sigma}u$.}

\noindent Similarly,

\Centerline{$P_{\tau}u=P_{\sigma\wedge\tau}u=P_{\tau\wedge\sigma}u
=P_{\sigma}u$,}

\noindent as claimed.
}%end of proof of 622F

\leader{622G}{}\cmmnt{ I have a great deal more to say about both
martingales and local martingales.   But I will move directly to the most
important result in this section.

\medskip

\noindent}{\bf Theorem} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v =\family{\sigma}{\Cal S}{v_{\sigma}}$ a $\|\,\|_1$-bounded
martingale.   Then $\pmb v $ is an integrator, therefore
moderately oscillatory and order-bounded.

\proof{ If $\Cal S$ is empty, this is trivial, so let us suppose that
$\Cal S\ne\emptyset$.   Then
$Q_{\Cal S}(d\pmb v )$ is topologically bounded.
\Prf\ Let $\epsilon>0$.   Let $\delta>0$ be such that
$\delta\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1\le\epsilon^2$.
Take $z\in Q_{\Cal S}(d\pmb v)$.   Then there are a
non-empty finite sublattice $I$ of $\Cal S$ and a fully adapted process
$\pmb u=\family{\sigma}{I}{u_{\sigma}}$ with
$\|\pmb u\|_{\infty}\le 1$ such that $z=S_I(\pmb u,d\pmb v)$.
Let $(\tau_0,\ldots,\tau_n)$
linearly generate the $I$-cells (611L).   Then
$z=\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})$.   Now
$\langle v_{\tau_i}\rangle_{i\le n}$ is a martingale in the classical sense
of 621Da.   So

$$\eqalignno{\theta(\delta z)
&\le\epsilon+\bar\mu\Bvalue{\delta|z|>\epsilon}
\le\epsilon+\bar\mu\Bvalue{|z|>\Bover{\epsilon}{\delta}}
\le\epsilon+\Bover{16\delta}{\epsilon}\|v_{\tau_n}\|_1
\Displaycause{621Hf}
\le 17\epsilon.
\cr}$$

\noindent As $\epsilon$ is arbitrary, $Q_{\Cal S}(d\pmb v )$ is
topologically bounded.\ \Qed

Thus $\pmb v $ is an integrator.   By 616Ib, it is moderately oscillatory and
order-bounded.
}%end of proof of 622G

\leader{622H}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$.
If $\pmb v =\family{\sigma}{\Cal S}{v_{\sigma}}$ is
a local martingale, then it is a local integrator, therefore locally moderately
oscillatory.

\proof{ Let $\Cal S'$ be a covering ideal of $\Cal S$ such that
$\pmb v\restr\Cal S'$ is a martingale.
Suppose that $\tau\in\Cal S$ and $\epsilon>0$.   Then
there is a $\tau_1\in\Cal S'$ such that
$c=\Bvalue{\tau\le\tau_1}$ has measure at least $1-\epsilon$ (611Mh).
Set $\pmb v'=\pmb Pv_{\tau_1}\restr\Cal S\wedge\tau$.
As $\pmb Pv_{\tau_1}$ is a
uniformly integrable martingale (622Fa), it is an integrator (622G), so
$\pmb v'$ is an integrator (616P(b-ii)).

Now if $\sigma\in\Cal S\wedge\tau$,

$$\eqalignno{\Bvalue{v_{\sigma}=v'_{\sigma}}
&=\Bvalue{v_{\sigma}=P_{\sigma}v_{\tau_1}}
=\Bvalue{v_{\sigma}=P_{\sigma}P_{\tau_1}v_{\tau_1}}
=\Bvalue{v_{\sigma}=P_{\sigma\wedge\tau_1}v_{\tau_1}}
=\Bvalue{v_{\sigma}=v_{\sigma\wedge\tau_1}}
\Displaycause{because $\sigma\wedge\tau_1$ belongs to $\Cal S'$
and $\pmb v\restr\Cal S'$ is a martingale}
\Bsupseteq\Bvalue{\sigma\wedge\tau_1=\sigma}
=\Bvalue{\sigma\le\tau_1}
\Bsupseteq\Bvalue{\tau\le\tau_1}
=c.\cr}$$

\noindent So

\Centerline{$\Bvalue{\pmb v'\ne\pmb v\restr\Cal S\wedge\tau}
=\sup_{\sigma\in\Cal S\wedge\tau}
  \Bvalue{v_{\sigma}\ne v'_{\sigma}}$}

\noindent is disjoint from $c$ and has measure at most $\epsilon$.
As $\epsilon$ is arbitrary,
$\pmb v\restr\Cal S\wedge\tau$ is an integrator (616P(b-iii)).
As $\tau$ is arbitrary, $\pmb v$ is a local integrator.\ \Qed
}%end of proof of 622H

\leader{622I}{}\cmmnt{ The principal martingale theorems (see \S275)
take slightly
different forms in the present context, so I take the space to spell one
of them out.

\medskip

\noindent}{\bf Doob's maximal inequality (second form)}
Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a martingale.   Then $\pmb v$ is
locally order-bounded, and

\Centerline{$\bar\mu(\sup_{\sigma\in\Cal S\wedge\tau}
  \Bvalue{|v_{\sigma}|>\gamma})
=\bar\mu(\Bvalue{\sup_{\sigma\in\Cal S\wedge\tau}|v_{\sigma}|>\gamma})
\le\Bover1{\gamma}\Expn(|v_{\tau}|)$}

\noindent for every $\tau\in\Cal S$ and $\gamma>0$.

\proof{ If $A\subseteq\Cal S\wedge\tau$ is a non-empty finite
set, then

\Centerline{$\bar\mu(\Bvalue{\sup_{\sigma\in A}|v_{\sigma}|>\gamma})
\le\Bover1{\gamma}\Expn(|v_{\tau}|)$.}

\noindent\Prf\ Let $I$ be the sublattice of $\Cal S$ generated by $A$, and take
$\tau_0\le\ldots\le\tau_n$ linearly generating the $I$-cells (611L again).
Then $(v_{\tau_0},\ldots,v_{\tau_n},v_{\tau})$ is a finite martingale
adapted to $(\frak A_{\tau_0},\ldots,\frak A_{\tau_n},\frak A_{\tau})$,
so

\Centerline{$\bar\mu(\Bvalue{\sup_{i\le n}|v_{\tau_i}|>\gamma})
\le\Bover1{\gamma}\Expn(|v_{\tau}|)$}

\noindent by 621E.   Write $v$ for $\sup_{i\le n}|v_{\tau_i}|$.
If $\sigma\in A$, then

\Centerline{$\Bvalue{|v_{\sigma}|\le v}
\Bsupseteq\sup_{i\le n}\Bvalue{v_{\sigma}=v_{\tau_i}}
\Bsupseteq\sup_{i\le n}\Bvalue{\sigma=\tau_i}
=1$}

\noindent by 611Ke.   So $v_{\sigma}\le v$.   Thus
$\Bvalue{\sup_{\sigma\in A}|v_{\sigma}|>\gamma}
\Bsubseteq\Bvalue{v>\gamma}$ has measure at most
$\Bover1{\gamma}\Expn(|v_{\tau}|)$.\ \Qed

Accordingly $c_{\gamma}
=\sup_{\sigma\in\Cal S\wedge\tau}\Bvalue{|v_{\sigma}|>\gamma}$ has
measure at most $\Bover1{\gamma}\Expn(|v_{\tau}|)$, by 321D.   Since this
tends to $0$ as $\gamma$ increases to $\infty$,
$\{|v_{\sigma}|:\sigma\in\Cal S\wedge\tau\}$ is bounded above in
$L^0(\frak A)$, and

\Centerline{$\bar\mu(\Bvalue{\sup_{\sigma\in\Cal S\wedge\tau}
   |v_{\sigma}|>\gamma})
=\bar\mu c_{\gamma}
\le\Bover1{\gamma}\Expn(|v_{\tau}|)$}

\noindent for every $\gamma>0$ (364L(a-ii)), as required.
}%end of proof of 622I

\leader{622J}{Proposition} Let $\Cal S$ be a
non-empty sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a uniformly integrable martingale.

(a) The $\|\,\|_1$-limit
$v=\1lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is defined in $L^1_{\bar\mu}$,
and $v$ is also the limit $\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ for the
topology of convergence in measure.

(b) $\pmb v=\pmb Pv\restr\Cal S$ is order-bounded, and
$\inf_{\tau\in\Cal S}\sup_{\sigma\in\Cal S\vee\tau}|v-v_{\sigma}|=0$.

\proof{{\bf (a)(i)}
$\{v_{\sigma}:\sigma\in\Cal S\}$ is relatively weakly compact
in $L^1_{\bar\mu}$ (621Bd).   Let $\Cal F$ be an ultrafilter on $\Cal S$
containing $\Cal S\vee\sigma$ for every $\sigma\in\Cal S$.   Then the
limit $u=\w1lim_{\sigma\to\Cal F}u_{\sigma}$ for the weak topology on
$L^1_{\bar\mu}$ is defined (2A3R).   If $\tau\in\Cal S$, then
$P_{\tau}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ is a norm-continuous linear operator,
so is weakly continuous (3A5Ec), and

\Centerline{$P_{\tau}u=\w1lim_{\sigma\to\Cal F}P_{\tau}v_{\sigma}=v_{\tau}$}

\noindent because $\Cal S\vee\tau\in\Cal F$ and
$P_{\tau}v_{\sigma}=v_{\tau}$ for every $\tau\in\Cal S\vee\sigma$.
So $\pmb v=\pmb Pu\restr\Cal S$.

\medskip

\quad{\bf (ii)} Let $\frak C$ be the closed subalgebra of $\frak A$
generated by $\bigcup_{\sigma\in\Cal S}\frak A_{\sigma}$ and set
$v=P_{\frak C}u$.   By 621C(g-ii),

\Centerline{$v=\1lim_{\sigma\uparrow\Cal S}P_{\sigma}u
=\1lim_{\sigma\uparrow\Cal S}v_{\sigma}$.}

\noindent By 613B(d-i) again, $v=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$.

\medskip

{\bf (b)} If $\tau\in\Cal S$, then

\Centerline{$P_{\tau}v=\1lim_{\sigma\uparrow\Cal S}P_{\tau}v_{\sigma}
=v_{\tau}$;}

\noindent accordingly $\pmb v=\pmb Pv\restr\Cal S$.   Being uniformly
integrable, $\pmb v$ is $\|\,\|_1$-bounded, so 622G tells us that it is
moderately oscillatory;  by 615Ga,
$\inf_{\tau\in\Cal S}\sup_{\sigma\in\Cal S\vee\tau}|v-v_{\sigma}|=0$.
}%end of proof of 622J

\leader{622K}{Lemma} Let $\Cal S$ be a finitely full sublattice of $\Cal T$, and
$\pmb u=\sfamilysS{u}$ an $L^1$-process such that
$\Expn(u_{\sigma})=\Expn(u_{\tau})$ for all $\sigma$, $\tau\in\Cal S$.
Then $\pmb u$ is a martingale.

\proof{ Take $\sigma\le\tau\in\Cal S$ and
$a\in\frak A_{\sigma}\subseteq\frak A_{\tau}$.   Then there is a
$\tau'\in\Cal T$ such that $a\Bsubseteq\Bvalue{\tau'=\tau}$ and
$1\Bsetminus a\Bsubseteq\Bvalue{\tau'=\sigma}$ (611I).   Now we have

$$\eqalign{\Expn(u_{\sigma}\times\chi a)
  +\Expn(u_{\sigma}\times\chi(1\Bsetminus a))&
=\Expn(u_{\sigma})
=\Expn(u_{\tau'})\cr&
=\Expn(u_{\tau'}\times\chi a)+\Expn(u_{\tau'}\times\chi(1\Bsetminus a))\cr&
=\Expn(u_{\tau}\times\chi a)+\Expn(u_{\sigma}\times\chi(1\Bsetminus a))\cr}$$

\noindent so

\Centerline{$\Expn(u_{\sigma}\times\chi a)=\Expn(u_{\tau}\times\chi a)$.}

\noindent As $a$ is arbitrary and
$u_{\sigma}\in L^1_{\bar\mu}\cap L^0(\frak C_{\sigma})$,
$u_{\sigma}$ is the conditional expectation of $u_{\tau}$ on $\frak C_{\sigma}$.
As $\sigma$ and $\tau$ are arbitrary, $\pmb u$ is a martingale.
}%end of proof of 622K

\leader{622L}{Brownian motion:  Theorem} Let
$\pmb w$ be Brownian motion, and
$\pmb{\iota}$ the corresponding identity process.   Then
$\pmb w$ and $\pmb w^2-\pmb{\iota}$ are local martingales, and
$\pmb w\restr\Cal T_b$ is a martingale.

\proof{ As in 612T and 612F, I regard
$\pmb w=\family{\sigma}{\Cal T_f}{w_{\sigma}}$ and
$\pmb{\iota}=\family{\sigma}{\Cal T_f}{\iota_{\sigma}}$
as processes on the real-time stochastic integration
structure $(\frak C,\bar\nu,\coint{0,\infty},\langle\frak C_t\rangle_{t\ge 0},
\Cal T,\family{\tau}{\Cal T}{\frak C_{\tau}})$, where $(\frak C,\bar\nu)$
is the measure algebra of $(\Omega,\Sigma,\nu)$,
$\Omega=C(\coint{0,\infty})_0$, $\nu$ is one-dimensional Wiener measure and
$\Sigma$ is its domain.

\medskip

{\bf (a)} For $n\in\Bbb N$, let $h_n:\Omega\to[0,\infty]$ be the Brownian
exit time from $\ooint{-n,n}$ (477I), so that $h_n$ is a stopping time;  write
$\tau_n$ for the corresponding stopping time in $\Cal T$ (612H).   Because
$h$ is finite $\nu$-almost everywhere (478Ma), $\tau_n\in\Cal T_f$.   Now
$\Expn(w_{\tau})=\Expn(w^2_{\tau}-\iota_{\tau})=0$ for every
$\tau\in\Cal T\wedge\tau_n$.   \Prf\ I use Dynkin's formula (478K).
Express $\tau$ as $h^{\ssbullet}$ where
$h\le h_n$ is a stopping time.   By 478Jd\Latereditions, there is a smooth
function $f:\Bbb R\to\Bbb R$
with compact support such that $f(x)=x$ for $x\in[-n-1,n+1]$.
In the language of 612H and 478K,

$$\eqalignno{\Expn(w_{\tau})&
=\Expn(X_h)
=\Expn(f(X_h))
\Displaycause{because $h(\omega)\le h_n(\omega)$, so
$|X_h(\omega)|=|\omega(h(\omega))|\le n$ for almost every $\omega\in\Omega$}
=f(0)+\Bover12\Expn(\int_0^h(\nabla^2f)(X_s)ds
=0\cr}$$

\noindent because $\nabla^2f(\xi)=0$ for $|\xi|\le n$, while
$|X_s(\omega)|=|\omega(s)|\le n$ whenever $s\in h(\omega)$.   Similarly,

$$\eqalignno{\Expn(w^2_{\tau}-\iota_{\tau})&
=\Expn(X^2_h-h)
=\Expn(f^2(X_h)-h)\cr&
=f^2(0)+\Bover12\Expn(\int_0^h(\nabla^2f^2)(X_s)-2\,ds
=0\cr}$$

\noindent because $(\nabla^2f^2)(\xi)=2$ for $|\xi|\le n$.\ \Qed

\medskip

{\bf (b)} By 622K, $\pmb w\restr\Cal T\wedge\tau_n$ and
$(\pmb w^2-\pmb{\iota})\restr\Cal T\wedge\tau_n$ are martingales for each
$n\in\Bbb N$.   Since $\sequencen{\tau_n}$ is non-decreasing,
$\pmb w\restr\Cal S$ and
$(\pmb w^2-\pmb{\iota})\restr\Cal S$ are martingales, where
$\Cal S=\bigcup_{n\in\Bbb N}\Cal T\wedge\tau_n$ is an ideal in $\Cal T_f$.
If $\tau$ is any member of $\Cal T_f$, it can be represented as $h^{\ssbullet}$
where $h$ is a finite-valued stopping time;  now $\omega$ is bounded on
$[0,h(\omega)]$ for every $\omega\in\Omega$, so
$\Omega=\bigcup_{n\in\Bbb N}\{\omega:h(\omega)\le h_n(\omega)\}$ and
$\sup_{n\in\Bbb N}\Bvalue{\tau\le\tau_n}=1$.   Accordingly $\Cal S$ is a
covering ideal of $\Cal T_f$ and
$\pmb w$ and $\pmb w^2-\pmb{\iota}$ are local martingales.

\medskip

{\bf (c)} To see that $\pmb w\restr\Cal T_b$ is a martingale, take any $t\ge 0$.
If $\sigma\in\Cal S\wedge\check t$,

\Centerline{$\Expn(w_{\sigma}^2)=\Expn(\iota_{\sigma})=\Expn(\sigma)\le t$,}

\noindent so $\pmb w\restr\Cal S\wedge\check t$ is $\|\,\|_2$-bounded, therefore
uniformly integrable, as well as being a martingale.   By 622J, it is of the
form $\pmb Pv\restr\Cal S\wedge\check t$ for some $v\in L^1_{\bar\mu}$.   But
$\Cal S$ covers $\Cal T_f$, so $\Cal S\wedge\check t$ covers
$\Cal T_f\wedge\check t$ (611M(g-ii)) and
$\pmb w\restr\Cal T_f\wedge\check t=\pmb Pv\restr\Cal T_f\wedge\check t$ is a
martingale.   As $t$ is arbitrary,
$\pmb w\restr\Cal T_b=\pmb w\restr\bigcup_{t\ge 0}\Cal T_f\wedge\check t$ is a
martingale.
}%end of proof of 622L

\cmmnt{\medskip

\noindent{\bf Remark} It is also the case that
$(\pmb w^2-\pmb\iota)\restr\Cal T_b$ is a martingale;  see 632Xe.}

\leader{622M}{}\cmmnt{ You will find various more or less elementary facts
about martingales in the exercises.   One which will be useful later
in this chapter is the following.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v =\family{\sigma}{\Cal S}{v_{\sigma}}$
a fully adapted process.
Then $\Cal S'=\{\tau:\tau\in\Cal S$, $\pmb v\restr\Cal S\wedge\tau$ is a
martingale$\}$ is an ideal in $\Cal S$, and $\pmb v\restr\Cal S'$ is a
martingale.

\proof{{\bf (a)} If $\tau\in\Cal S$, then $\tau\in\Cal S'$ iff
$v_{\sigma}=P_{\sigma}v_{\tau}$ whenever $\sigma\in\Cal S$ and
$\sigma\le\tau$.   \Prf\

$$\eqalign{\tau\in\Cal S'
&\iff v_{\sigma}=P_{\sigma}v_{\sigma'}\text{ whenever }
\sigma,\,\sigma'\in\Cal S\text{ and }\sigma\le\sigma'\le\tau\cr
&\,\,\Longrightarrow v_{\sigma}=P_{\sigma}v_{\tau}
\text{ whenever }
\sigma\in\Cal S\text{ and }\sigma\le\tau\cr
&\,\,\Longrightarrow v_{\sigma}=P_{\sigma}v_{\tau}=P_{\sigma}P_{\sigma'}v_{\tau}
=P_{\sigma}v_{\sigma'}
\text{ whenever }
\sigma,\,\sigma'\in\Cal S\text{ and }\sigma\le\sigma'\le\tau.
\text{ \Qed}\cr}$$

\medskip

{\bf (b)} If $\tau\in\Cal S$ and $\tau\le\tau'\in\Cal S'$ then
$\tau\in\Cal S'$.   \Prf\ If $\sigma\in\Cal S$ and $\sigma\le\tau$ then

\Centerline{$v_{\sigma}=P_{\sigma}v_{\tau'}=P_{\sigma}P_{\tau}v_{\tau'}
=P_{\sigma}v_{\tau}$.\  \Qed}

\medskip

{\bf (c)} If $\tau$, $\tau'\in\Cal S'$, $\sigma\in\Cal S$ and
$\tau\le\sigma\le\tau\vee\tau'$ then
$v_{\sigma}=P_{\sigma}v_{\tau\vee\tau'}$.   \Prf\ Set
$a=\Bvalue{\sigma\le\tau'}\in\frak A_{\sigma}$ (611H(c-i)).
We have

\Centerline{$a
=\Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma\le\tau'}
\Bsubseteq\Bvalue{\tau\le\tau'}
=\Bvalue{\tau'=\tau'\vee\tau}
\Bsubseteq\Bvalue{v_{\tau'}=v_{\tau'\vee\tau}}$}

\noindent (using 611E(c-iv-$\alpha$) and (a-ii-$\beta$)) and

\Centerline{$a=\Bvalue{\sigma=\sigma\wedge\tau'}
\Bsubseteq\Bvalue{P_{\sigma}v_{\tau'}=P_{\sigma\wedge\tau'}v_{\tau'}}
  \Bcap\Bvalue{v_{\sigma\wedge\tau'}=v_{\sigma}}$,}

\noindent (611E(a-ii-$\beta$) again and 622Bb) so

$$\eqalign{P_{\sigma}(v_{\tau\vee\tau'}\times\chi a)
&=P_{\sigma}(v_{\tau'}\times\chi a)
=P_{\sigma}v_{\tau'}\times\chi a
=P_{\sigma\wedge\tau'}v_{\tau'}\times\chi a\cr
&=v_{\sigma\wedge\tau'}\times\chi a
=v_{\sigma}\times\chi a.\cr}$$

Next, setting

$$\eqalignno{a'
&=\Bvalue{\tau'<\sigma}
\Bsubseteq\Bvalue{\tau'<\tau\vee\tau'}
=\Bvalue{\tau'<\tau}\Bcup\Bvalue{\tau'<\tau'}
\Displaycause{611Eb}
=\Bvalue{\tau'<\tau}
\Bsubseteq\Bvalue{\tau'\le\tau}
=\Bvalue{\tau\vee\tau'=\tau}
\Bsubseteq\Bvalue{\sigma\le\tau}
=\Bvalue{\sigma\vee\tau=\tau}
=\Bvalue{\sigma=\tau}\cr
&\Bsubseteq\Bvalue{v_{\tau\vee\tau'}=v_{\tau}}
  \Bcap\Bvalue{P_{\sigma}v_{\tau}=P_{\tau}v_{\tau}}
  \Bcap\Bvalue{v_{\tau}=v_{\sigma}},\cr}$$

\noindent we see that

\Centerline{$P_{\sigma}(v_{\tau\vee\tau'}\times\chi a')
=P_{\sigma}(v_{\tau}\times\chi a')
=P_{\sigma}(v_{\sigma}\times\chi a')
=v_{\sigma}\times\chi a'$.}

\noindent Adding, we have
$P_{\sigma}v_{\tau\vee\tau'}=v_{\sigma}$, as claimed.\ \Qed

\medskip

{\bf (d)} If $\tau$, $\tau'\in\Cal S'$, $\sigma\in\Cal S$ and
$\sigma\le\tau\vee\tau'$ then $v_{\sigma}=P_{\sigma}v_{\sigma\vee\tau}$.
\Prf\ Set

$$\eqalign{a
&=\Bvalue{\sigma\le\tau}=\Bvalue{\sigma\vee\tau=\tau}
=\Bvalue{\sigma\wedge\tau=\sigma}\cr
&\Bsubseteq\Bvalue{v_{\sigma\vee\tau}=v_{\sigma}}
  \Bcap\Bvalue{P_{\sigma\wedge\tau}v_{\tau}=P_{\sigma}v_{\tau}}
  \Bcap\Bvalue{v_{\sigma\wedge\tau}=v_{\sigma}}.\cr}$$

\noindent Then

$$\eqalign{P_{\sigma}(v_{\sigma\vee\tau}\times\chi a)
&=P_{\sigma}(v_{\tau}\times\chi a)
=P_{\sigma}v_{\tau}\times\chi a
=P_{\sigma}P_{\tau}v_{\tau}\times\chi a\cr
&=P_{\sigma\wedge\tau}v_{\tau}\times\chi a
=v_{\sigma\wedge\tau}\times\chi a
=v_{\sigma}\times\chi a.\cr}$$

\noindent And setting

\Centerline{$a'=\Bvalue{\tau<\sigma}
\Bsubseteq\Bvalue{\sigma\vee\tau=\sigma}
\Bsubseteq\Bvalue{v_{\sigma\vee\tau}=v_{\sigma}}$,}

\noindent we have

\Centerline{$P_{\sigma}(v_{\sigma\vee\tau}\times\chi a')
=P_{\sigma}(v_{\sigma}\times\chi a')
=v_{\sigma}\times\chi a'$,}

\noindent so $P_{\sigma}v_{\sigma\vee\tau}=v_{\sigma}$.\ \Qed

Now we have $\tau\le\sigma\vee\tau\le\tau\vee\tau'$, so (c) tells us that

\Centerline{$v_{\sigma}=P_{\sigma}v_{\sigma\vee\tau}
=P_{\sigma}P_{\sigma\vee\tau}v_{\tau\vee\tau'}
=P_{\sigma}v_{\tau\vee\tau'}$.}

\noindent As $\sigma$ is arbitrary, $\tau\vee\tau'\in\Cal S'$.
As $\tau$ and $\tau'$ are arbitrary, (a) tells us that $\Cal S'$ is an
ideal of $\Cal S$.

\medskip

{\bf (e)} Finally, we have $v_{\sigma}=P_{\sigma}v_{\tau}$ whenever
$\sigma$, $\tau\in\Cal S'$ and $\sigma\le\tau$, so $\pmb v\restr\Cal S'$
is a martingale.
}%end of proof of 622M

\leader{622N}{Extensions to covered envelopes:  Proposition}
Let $\Cal S$ be a sublattice of $\Cal T$ with covered envelope $\hat{\Cal S}$,
and $\pmb u=\sfamilysS{u}$ a fully adapted process with fully adapted extension
$\hatpmb u=\family{\sigma}{\hat{\Cal S}}{\hat u_{\sigma}}$.

(a) If $\hatpmb u$ is a martingale then $\pmb u$ is a martingale.

(b) If $\pmb u$ is a local martingale then $\hatpmb u$ is a local martingale.

(c) $\hatpmb u$ is a uniformly integrable martingale iff $\pmb u$ is a uniformly
integrable martingale.

\proof{{\bf (a)} This is a special case of 622Db.

\medskip

{\bf (b)} Take $\tau\in\hat{\Cal S}$ and
$\epsilon>0$.   We know that $\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}=1$, so
there is a non-empty finite subset $I$ of $\Cal S$ such that
$\bar\mu a\ge 1-\epsilon$, where
$a=\sup_{\sigma\in I}\Bvalue{\tau=\sigma}$.   Set $\tilde\tau=\sup I\in\Cal S$.
Then

\Centerline{$\Bvalue{\tau\le\tilde\tau}
\Bsupseteq\sup_{\sigma\in I}\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma\le\tilde\tau}
=a$}

\noindent so $\bar\mu\Bvalue{\tilde\tau<\tau}\le\epsilon$.   Because $\pmb u$ is a
local martingale, there is a $\tau'\in\Cal S$ such that
$\pmb u\restr\Cal S\wedge\tau'$ is a martingale and
$\bar\mu\Bvalue{\tau'<\tilde\tau}\le\epsilon$.   Consider the martingale
$\pmb Pu_{\tau'}$ (622F).   This is defined everywhere on $\Cal T$ and agrees
with $\pmb u$ on $\Cal S\wedge\tau'$.   Now $\hat\Cal S\wedge\tau'$ is the
covered envelope of $\Cal S\wedge\tau$ (611M(e-i)), while
$\pmb Pu_{\tau'}\restr\hat{\Cal S}\wedge\tau'$ is fully adapted and extends
$\pmb u\restr\Cal S\wedge\tau'$, so must be equal to
$\hatpmb u\restr\hat{\Cal S}\wedge\tau'$.   Thus
$\hatpmb u\restr\hat{\Cal S}\wedge\tau'$ is a martingale, while
$\Bvalue{\tau'<\tau}
\Bsubseteq\Bvalue{\tau'<\tilde\tau}\Bcup\Bvalue{\tilde\tau<\tau}$
has measure at most $2\epsilon$.   As $\tau$ and $\epsilon$ are arbitrary,
$\hatpmb u$ is a local martingale.

\medskip

{\bf (c)} If $\hatpmb u$ is a uniformly integrable martingale then
$\pmb u=\hatpmb u\restr\Cal S$ must be a uniformly integrable martingale.
If $\pmb u$ is a uniformly integrable martingale, it is of the form
$\pmb Pu\restr\Cal S$ for some $u\in L^1_{\bar\mu}$ (622J), and now
$\pmb Pu\restr\hat\Cal S$ is a uniformly integrable martingale.   But
$\pmb Pu\restr\hat\Cal S$ is a fully adapted process extending $\pmb u$, so must
be $\hatpmb u$, and $\hatpmb u$ is a uniformly integrable martingale.
}%\end of proof of 622N

\leader{622O}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a martingale.

(a) If $\Cal S_1$ is the ideal of $\Cal T$ generated by $\Cal S$, there
is a unique martingale $\pmb v=\family{\tau}{\Cal S_1}{v_{\tau}}$ extending
$\pmb u$.

(b) If $\Cal S_2$ is the full ideal of $\Cal T$ generated by $\Cal S$,
there is a
local martingale $\hatpmb v=\family{\tau}{\Cal S_2}{\hat v_{\tau}}$ extending
$\pmb u$.
%622Xh

\proof{{\bf (a)} Since $\Cal S$ is upwards-directed,
$\bigcup_{\sigma\in\Cal S}\Cal T\wedge\sigma$ is an ideal
of $\Cal T$ and must be $\Cal S_1$.   If $\tau\in\Cal S_1$ and $\sigma$,
$\sigma'\in\Cal S$ are such that $\tau\le\sigma$ and $\tau\le\sigma'$, then

\Centerline{$P_{\tau}u_{\sigma}=P_{\tau}P_{\sigma}u_{\sigma\vee\sigma'}
=P_{\tau}u_{\sigma\vee\sigma'}=P_{\tau}u_{\sigma'}$,}

\noindent so we can define $v_{\tau}\in L^0(\frak A)$ by saying that
$v_{\tau}=P_{\tau}v_{\sigma}$ whenever $\tau\le\sigma\in\Cal S$.   If
$\tau\le\tau'\in\Cal S_1$, there is a $\sigma\in\Cal S$ such that
$\tau'\le\sigma$ and

\Centerline{$P_{\tau}v_{\tau'}=P_{\tau}P_{\tau'}u_{\sigma}
=P_{\tau}v_{\sigma}=v_{\tau}$.}

\noindent In particular, $P_{\tau}v_{\tau}=v_{\tau}$ and
$v_{\tau}\in L^0(\frak A_{\tau})$.   Also

\Centerline{$\Bvalue{\tau=\tau'}
\Bsubseteq\Bvalue{P_{\tau}v_{\tau'}=P_{\tau'}v_{\tau'}}
=\Bvalue{v_{\tau}=v_{\tau'}}$}

\noindent by 622Bb.   So $\pmb v$ is fully adapted (612Db) and is a martingale.
If $\sigma\in\Cal S$ then $v_{\sigma}=P_{\sigma}u_{\sigma}=u_{\sigma}$, so
$\pmb v$ extends $\pmb u$.

As for uniqueness, if $\pmb v'=\family{\tau}{\Cal S_1}{v'_{\tau}}$ is another
martingale with domain $\Cal S_1$ extending $\pmb u$, then

\Centerline{$v'_{\tau}=P_{\tau}v'_{\sigma}=P_{\tau}u_{\sigma}=v_{\tau}$}

\noindent whenever $\tau\le\sigma\in\Cal S$, so $\pmb v'=\pmb v$.

\medskip

{\bf (b)} The covered envelope $\hat\Cal S_1$ is an ideal of $\Cal T$.   \Prf\
We know that it is a full sublattice (611M(b-i), 611M(c-ii)).
If $\tau\in\Cal T$ and $\tau\le\tau'\in\hat\Cal S_1$, then

$$\eqalignno{\sup_{\sigma\in\Cal S_1}\Bvalue{\tau=\sigma}&
\Bsupseteq\sup_{\sigma\in\Cal S_1}\Bvalue{\tau=\sigma\wedge\tau}
\Displaycause{because $\sigma\wedge\tau\in\Cal S_1$ for every
$\sigma\in\Cal S_1$}
=\sup_{\sigma\in\Cal S_1}\Bvalue{\tau\le\sigma}
\Bsupseteq\sup_{\sigma\in\Cal S_1}
   \Bvalue{\tau\le\sigma}\Bcap\Bvalue{\sigma=\tau'}\cr&
=\sup_{\sigma\in\Cal S_1}
   \Bvalue{\tau\le\tau'}\Bcap\Bvalue{\sigma=\tau'}
=\sup_{\sigma\in\Cal S_1}\Bvalue{\sigma=\tau'}
=1\cr}$$

\noindent and $\tau\in\hat\Cal S_1$.\ \Qed

So $\hat\Cal S_1$ is a full ideal of $\Cal T$, and of course it includes
$\Cal S$;  while any full ideal including $\Cal S$ must include $\Cal S_1$ and
$\hat\Cal S_1$.   Thus $\Cal S_2=\hat\Cal S_1$.   Now the fully adapted
extension $\hatpmb v$ of $\pmb v$ has domain $\Cal S_2$, extends $\pmb u$ and
is a local martingale (622Nb).
}%end of proof of 622O

\leader{622P}{Proposition} Let $\Cal S$ be a non-empty sublattice of
$\Cal T$, $\pmb u\cmmnt{\mskip5mu=\sfamilysS{u}}$
a moderately oscillatory process
and $\pmb v=\sfamilysS{v}$ a martingale.   Then
$\|\int_{\Cal S}\pmb u\,d\pmb v\|_2
\le\|\pmb u\|_{\infty}\sup_{\tau\in\Cal S}\|v_{\sigma}\|_2$.

\proof{ The formulae here depend on my conventions for the use of $\infty$, so
perhaps I should restate these.   $0\cdot\infty=0$, so if either
$\pmb u$ or $\pmb v$ is $\tbf 0$, the result is trivial.   Otherwise,
if $\pmb u$ is not $\|\,\|_{\infty}$-bounded,
then $\|\pmb u\|_{\infty}=\infty$;  if $\pmb v$ is not $\|\,\|_2$-bounded, then
$\sup_{\tau\in\Cal S}\|v_{\sigma}\|_2=\infty$;  both cases are trivial.
So we can suppose that $M=\|\pmb u\|_{\infty}$ and
$M'=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_2$ are both finite and non-zero.

As $\pmb v$ is $\|\,\|_2$-bounded it is $\|\,\|_1$-bounded, therefore an
integrator (622G), and $\int_{\Cal S}\pmb u\,d\pmb v$ is defined.
The point is that if $I\in\Cal I(\Cal S)$ then
$\|S_I(\pmb u,d\pmb v)\|_2\le MM'$.
\Prf\ If $I=\emptyset$ or $M=0$ this is trivial.   Otherwise,
take $(\tau_0,\ldots,\tau_n)$ linearly generating the $I$-cells.   Then

$$\eqalignno{\|S_I(\pmb u,d\pmb v)\|_2
&=M\|\sum_{i=0}^{n-1}
   \Bover1Mu_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})\|_2
\le M\|v_{\tau_n}\|_2
\Displaycause{by 621F, applied to the martingale
$(v_{\tau_0},\ldots,v_{\tau_n})$}
\le MM'.  \text{ \Qed}\cr}$$

Since $\|\,\|_2$-balls are closed in $L^0(\frak A)$ (613Bc),

\Centerline{$\|\int_{\Cal S}\pmb u\,d\pmb v\|_2
=\|\lim_{I\uparrow\Cal I(\Cal S)}S_I(\pmb u,d\pmb v)\|_2
\le MM'$,}

\noindent as claimed.
}%end of proof of 622P

\leader{622Q}{Proposition} Let $\Cal S$ be a sublattice of
$\Cal T$, and $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$,
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ fully adapted
processes such that $\pmb u$ is locally moderately oscillatory and
$\pmb v$ is a martingale.   Then
$\|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v\|_2
\le\|\pmb u\restr\Cal S\wedge\tau\|_{\infty}\|v_{\tau}\|_2$ for
every $\tau\in\Cal S$, and if the right-hand side is always finite,
the indefinite integral $\iivu$ is a martingale.

\proof{{\bf (a)} We know from 622H
that $\pmb v$ is a local integrator so the process $\iivu$ is defined
everywhere on $\Cal S$.   For $\tau\in\Cal S$ write $z_{\tau}$ for
$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v$.   If $\tau\in\Cal S$ and
$\sigma\in\Cal S\wedge\tau$, then $\|v_{\sigma}\|_2\le\|v_{\tau}\|_2$ by
366J, so 622P tells us that
$\|z_{\tau}\|_2\le\|\pmb u\restr\Cal S\wedge\tau\|_{\infty}\|v_{\tau}\|_2$.
At the same time, we see from the argument in 622P that
$\{S_I(\pmb u,d\pmb v):I\in\Cal I(\Cal S\wedge\tau)\}$
is $\|\,\|_2$-bounded, therefore uniformly integrable (621Be).

\medskip

{\bf (b)} Now suppose that
$\|\pmb u\restr\Cal S\wedge\tau\|_{\infty}\|v_{\tau}\|_2$ is finite for every
$\tau\in\Cal S$.   Take $\tau\le\tau'$ in $\Cal S$.   Because
$\{S_I(\pmb u,d\pmb v):I\in\Cal I(\Cal S\wedge\tau')\}
  \cup\{z_{\tau'}-z_{\tau}\}$
is uniformly integrable, and
$z_{\tau'}-z_{\tau}=\int_{\Cal S\cap[\tau,\tau']}\pmb u\,d\pmb v$
is the limit
$\lim_{I\uparrow\Cal I(\Cal S\cap[\tau,\tau'])}S_I(\pmb u,d\pmb v)$
for the topology
of convergence in measure, it is also the limit for $\|\,\|_1$
(621B(c-ii)).   As $P_{\tau}$ is $\|\,\|_1$-continuous, this means that

\Centerline{$P_{\tau}z_{\tau'}-z_{\tau}
=P_{\tau}(z_{\tau'}-z_{\tau})
=\lim_{I\uparrow\Cal I([\tau,\tau'])}P_{\tau}S_I(\pmb u,d\pmb v)
=0$}

\noindent because if $\tau\le\sigma\le\sigma'\le\tau'$ then

$$\eqalign{P_{\tau}(u_{\sigma}\times(v_{\sigma'}-v_{\sigma}))
&=P_{\tau}P_{\sigma}(u_{\sigma}\times(v_{\sigma'}-v_{\sigma}))
=P_{\tau}(u_{\sigma}\times P_{\sigma}(v_{\sigma'}-v_{\sigma}))\cr
&=P_{\tau}(u_{\sigma}\times(P_{\sigma}v_{\sigma'}-v_{\sigma}))
=0.\cr}$$

\noindent So $P_{\tau}z_{\tau'}=z_{\tau}$;  as $\tau$ and $\tau'$ are
arbitrary, $\iivu$ is a martingale.
}%end of proof of 622Q

\cmmnt{\leader{622R}{Law-independence} I remarked in 613I that the
Riemann-sum integral,
like the topology of convergence in measure on $L^0(\frak A)$, does not depend
on the measure $\bar\mu$ assigned to $\frak A$;
if $\bar\nu$ is any strictly positive totaly finite countably
additive functional on $\frak A$, the stochastic integration
structure $(\frak A,\bar\nu,T,\family{t}{T}{\frak A_t})$ will behave exactly
like the original structure \Asisfour.   This is the case for most of the rest
of Chapter 61.  In a formal sense there are a great
many definitions to check.   Already in 615B I defined the ucp topology on
a space $\Mob(\Cal S)$ in terms of an F-norm $\widehat{\theta}$ defined in terms
of the standard F-norm $\theta$ on $L^0(\frak A)$, and $\theta$ does depend on
the measure.   But if we take this into account, and speak of
$\theta_{\bar\mu}$ and $\theta_{\bar\nu}$ giving rise to
$\widehat\theta_{\bar\mu}$ and $\widehat\theta_{\bar\nu}$, then we know that

\Centerline{for every $\epsilon>0$ there is a $\delta>0$ such that
$\bar\nu(a)\le\epsilon$ whenever $\bar\mu(b)\le\delta$}

\noindent and consequently

\Centerline{for every $\epsilon>0$ there is a $\delta>0$ such that
$\theta_{\bar\nu}(u)\le\epsilon$ whenever $\theta_{\bar\mu}(u)\le\delta$,}

\Centerline{for every $\epsilon>0$ there is a $\delta>0$ such that
$\widehat\theta_{\bar\nu}(\pmb u)\le\epsilon$
whenever $\widehat\theta_{\bar\mu}(\pmb u)\le\delta$.}

\noindent Since this works equally well with $\bar\mu$ and $\bar\nu$ exchanged,
the $\bar\mu$-ucp topology and uniformity on $\Mob(\Cal S)$ are the same as the
$\bar\nu$-ucp topology and uniformity.

Continuing through `moderately oscillatory' processes (614O),
`integrating interval functions' and `integrators' (616F), these are defined
in terms of the topology of $L^0(\frak A)$, so are the same in both structures.
`Bounded variation' (614J), `cumulative variation' (614O) and
`(residual) oscillation' (618B) can,
with a little care, be defined in ways which do not call on the measure at all,
but only on the Riesz space structure of $L^0(\frak A)$,
so the same is true of `jump-free process'.
`Covariations' and `quadratic variations' (617H) are based on
the Riemann-sum integral which doesn't change.   So the way to It\^o's formula
(\S619) is clear throughout.

Of course the examples, Brownian motion (612T, 622L)
and Poisson processes (612U), are based on
explicitly defined measures, and make no sense without them.   But it is only
in the present section that we have come to a general class of processes for
which we really need to know which measure we are using.   We do not expect
anything to do with $L^p_{\bar\mu}$ or $\|\,\|_p$, for $p<\infty$, to be stable
in the way that $L^0(\frak A)$, $L^{\infty}(\frak A)$ and $\|\,\|_{\infty}$ are.
In particular, conditional expectations (621C) and martingales (622C) are
dependent on the exact measure we have in hand.   So in the present chapter our
expectations are reversed.   I will return to an occasion in which,
surprisingly, we {\it do} have law-independence, in 625F.
}%end of cmmnt

%\leader{622S}{}
%\leader{622T}{}
%\leader{622U}{}
%\leader{622V}{}
%\leader{622W}{}

\exercises{\leader{622X}{Basic exercises $\pmb>$(a)}
%\spheader 622Xa
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v =\family{\sigma}{\Cal S}{v_{\sigma}}$ a martingale.   Let
$\tau\in\Cal T$ and set $\Cal S'=\{\sigma:\sigma\in\Cal T$,
$\sigma\wedge\tau\in\Cal S\}$.   Show that
$\sigma\mapsto v_{\sigma\wedge\tau}:\Cal S'\to L^0(\frak A)$ is a martingale.
%622D

\spheader 622Xb Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$ a
process with domain $\Cal S$, and $z$ an element of
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Show that if $\pmb u$ is a martingale and
$z\pmb u$ is an $L^1$-process then $z\pmb u$ is a
martingale.
%622D

\spheader 622Xc Let $\pmb v $ be a local martingale defined on a sublattice
$\Cal S$ of $\Cal T$.   Show that $\pmb v\restr\Cal S\vee\tau$
is a local martingale for every $\tau\in\Cal S$.
%622D

\spheader 622Xd Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 616Xa and 615Xf.   Let $f:\coint{0,\infty}\to\Bbb R$ be a
function.   (i) Show that $f$ corresponds to a martingale with domain
$\Cal T_f$ iff it corresponds to a local martingale iff it is constant.
(ii) Show that $f$ corresponds to
a uniformly integrable process iff it is bounded.
%622D

\sqheader 622Xe
Let $(\frak A,\bar\mu)$ be the measure algebra of Lebesgue
measure on $[0,1]$.   (i) Set $T=[0,1]$ and for $t\in T$ set
$a_t=[t,1]^{\ssbullet}\in\frak A$,
$\frak A_t=\{a:a\Bcap a_t$ is either $0$ or $a_t\}$;
show that $\family{t}{T}{\frak A_t}$ is a
filtration in the sense of 611B.   (ii) Let $\Cal T$ be the associated
family of stopping times.   Show that for any $\tau\in\Cal T$ there is a
least $s_{\tau}\in[0,1]$ such that
$\Bvalue{\tau>s_{\tau}}\Bcap a_{s_{\tau}}=0$, and that
$\tau\mapsto s_{\tau}:\Cal T\to[0,1]$ is a lattice homomorphism.
(iii) Show that there is a $\tau^*\in\Cal T$ defined by saying that
$\Bvalue{\tau^*>t}=a_t$ for every $t\in[0,1]$, and that
$s_{\tau^*}=1$.   (iv) Set $\Cal S=\{\tau:\tau\le\tau^*\}$,
$\Cal S'=\{\tau:\tau\le\tau^*$, $s_{\tau}<1\}$;  show that
$\Cal S'=\Cal S\setminus\{\tau^*\}$ is a covering ideal of $\Cal S$.
(v) Set $u_{\tau}=\Bover1{1-s_{\tau}}\chi a_{s_{\tau}}$ for $\tau\in\Cal S'$,
$u_{\tau^*}=0$;  show that
$\pmb u=\family{\tau}{\Cal S}{u_{\tau}}$ is jump-free,
that $\pmb u\restr\Cal S'$ is a martingale, and that $\pmb u$ is
a local martingale which is not a martingale.   *(vi) Show that
$\family{t}{T}{\frak A_t}$ is right-continuous in the sense of 632B below.
%622D
%local martingale not martingale

\spheader 622Xf Let $\phi:[0,1]\to\coint{0,\infty}$ be such that
$\lim_{\delta\downarrow 0}\phi(\delta)=0$.   Let $\Cal S$ be a
sublattice of $\Cal T$ and $M_{\text{mart},\phi}$
the set of martingales $\pmb u=\sfamilysS{u}$ such that
$\Expn(|u_{\sigma}|\times\chi a)\le\phi(\bar\mu a)$ for every
$\sigma\in\Cal S$ and
$a\in\frak A$.   Show that $M_{\text{mart},\phi}$ is a subset of the space
$\Mob(\Cal S)$ of order-bounded processes and is closed in the ucp topology.
%622E

\sqheader 622Xg Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v =\family{\sigma}{\Cal S}{v_{\sigma}}$ a family in $L^1_{\bar\mu}$
such that
$P_{\sigma}v_{\tau}=v_{\sigma}$ whenever $\sigma$, $\tau\in\Cal S$ and
$\sigma\le\tau$.   Show that $\pmb v $ is fully adapted.
%622F

\spheader 622Xh Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v$ a local martingale with domain $\Cal S$.   Set
$\Cal S'=\{\tau:\tau\in\Cal T$,
$\inf_{\sigma\in\Cal S}\Bvalue{\sigma<\tau}=0$.   Show (i) that $\Cal S'$ is
the full ideal of $\Cal T$ generated by $\Cal S$ (ii) that there is a unique
local martingale on $\Cal S'$ extending $\pmb u$.
%622O  what about `virtually local'?

\spheader 622Xi
Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
and $\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$,
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ fully adapted
processes such that $\pmb v$ is a martingale and
$z=\int_{\Cal S}\pmb u\,d\pmb v$ is defined.   Show that
$\|z\|_2\le\|\pmb u\|_{\infty}\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_2$.
%622P

\sqheader 622Xj Let $(\Omega,\Sigma,\mu)$
be the interval $\ocint{0,1}$ with Lebesgue measure, and $T=\coint{1,\infty}$.
For $t\ge 1$, set $\Sigma_t=\{E:E\in\Sigma$, either $[0,\bover1t]\cap E$ or
$[0,\bover1t]\setminus E$ is negligible$\}$.    Set
$X_t(\omega)=\Bover1{\sqrt{\omega}}$ if $\omega t\ge 1$, $2\sqrt{t}$ if
$\omega t<1$, and $X_{\infty}(\omega)=\Bover1{\sqrt{\omega}}$ for every
$\omega\in\Omega$.
Set $Z_t(\omega)=-2$ if $\omega t\ge 1$, $2t-2$ if
$\omega t<1$, and $Z_{\infty}(\omega)=-2$ for every
$\omega\in\Omega$.    Let $(\frak A,\bar\mu)$ be the measure algebra of $\mu$,
and $\langle\frak A_t\rangle_{t\ge 1}$ the filtration corresponding to
$\langle\Sigma_t\rangle_{t\ge 1}$.   In $L^0(\frak A)$,
set $v=X_{\infty}^{\ssbullet}$ and let $\pmb v$ be the uniformly
integrable martingale
$\pmb Pv$;  let $\pmb z=\family{\tau}{\Cal T}{z_{\tau}}$ be the indefinite
integral $\iiv(\pmb v)$.
(i) Set $h(\omega)=\bover1{\omega}$ for $\omega\in\ocint{0,1}$;  show that $h$
is a stopping time adapted to $\langle\Sigma_t\rangle_{t\ge 1}$.
(ii) Let $\tau\in\Cal T$ be the stopping time associated with $h$.   Show that
$\Cal T\wedge\tau=\{\check t\wedge\tau:t\ge 1\}\cup\{\tau\}$.
(iii) Show that $P_{\check{t}}v=X_t^{\ssbullet}$ for every $t\ge 1$.
(iv) Show that $z_{\check{t}}=Z_t^{\ssbullet}$ for every $t\ge 1$.
(v) Show that $z_{\max\Cal T}=Z_{\infty}^{\ssbullet}$.
(vi) Show that $\pmb z$ is a jump-free local martingale and an $L^1$-process
but not a martingale.
*(vii) Show that $\langle\frak A_t\rangle_{t\ge 1}$ is right-continuous.
%622Xe 622F 622Q

\spheader 622Xk Give an example of a stochastic integration structure \Asissix,
a sublattice $\Cal S$ of $\Cal T$ and a uniformly integrable process with domain
$\Cal S$ such that its fully adapted extension to the covered envelope of 
$\Cal S$ is not an $L^1$-process.
%622Nc

\leader{622Y}{Further exercises (a)}
%\spheader 622Ya
Let $\pmb w=\family{\sigma}{\Cal T_f}{w_{\sigma}}$ be Brownian motion.
Show that
there is a sublattice $\Cal S$ of $\Cal T_f$, with covered envelope
$\hat\Cal S$, such that $\pmb w\restr\hat\Cal S$ is a local martingale, but
$\pmb w\restr\Cal S$ is not.
%622L mt62bits
}%end of exercises

\endnotes{
\Notesheader{622} Compared with the martingales of \S275 and \S621, the
point of `fully adapted' martingales is that they are defined on a lattice
which is not totally ordered in any case in which the theory here
is appropriate.   So we have some
new questions to ask.   In particular, it is not quite obvious that
$P_{\sigma}P_{\tau}$ will be $P_{\sigma\wedge\tau}$ (458M,
622Ba), or that every element of $L^1_{\bar\mu}$ will generate a
fully adapted martingale (622Fa).   622M is another result which is easy in
the totally ordered case, but demands finesse in the general context.

If you look at 612H and 632L below, you will see that some
measure-theoretic considerations enter the argument, in particular the
notion of progressive measurability, which have no direct parallel in the
theory of stochastic processes in $L^0$.   When eventually we come to
applications of the theory here, they will generally be based on such
processes as those examined in \S455, where the measure theory is
essential.   But for the moment we can leave this to look after itself.

Innumerable variations on the concept of `martingale' have been
investigated.   Here I have looked
only at `local' martingales;  `approximately local' 
and `virtually local' martingales will come in the
next section.   %\S623
Submartingales will reappear in \S626, and
supermartingales and quasimartingales in \S627.

By far the most important martingale in mathematics is Brownian motion,
which here appears in 622L.   In the proof I appeal to
Dynkin's formula, a fundamental result in the theory of harmonic functions
in \S478.
I did warn you, in the introduction to this volume, that applications might
depend on `further non-trivial ideas'.   But I ought to confess at once that
there is an alternative proof, not dependent on anything in \S478, using ideas
in \S632 below.

Once we have observed that martingales are local integrators (622H), it is
clear that we should try to understand indefinite integrals with respect to
martingales, which is indeed where stochastic integration began, with It\^o's
formula for integrals with respect to Brownian motion.   Most of the rest of
this chapter will be about integration with respect to martingales.   Here I
begin with a couple of baby steps, 622P-622Q, about $L^2$-martingales.   Since
jump-free processes
will often (subject to an appropriate interpretation of the word `locally')
be locally $\|\,\|_{\infty}$-bounded, 622Q will take us a long way
with jump-free martingales.
}%end of notes

\discrpage

\frfilename{mt623.tex}
\versiondate{10.12.21/21.8.23}
\copyrightdate{2018}

\def\chaptername{Martingales}
\def\sectionname{Approximately and virtually local martingales}
\def\alm{approximately local martingale}
\def\vlm{virtually local martingale}

\newsection{623}

I have presented a number of contexts in which an indefinite integral $\iivu$
can be expected to share properties with the integrator $\pmb v$
(614D, 614U, 616J, 618Q).   In contrast with this pattern,
we can have a martingale with a corresponding indefinite integral
which is not a martingale (622Xj), and this occurs in some of the central
examples of the theory (631Ya).   However the indefinite integral is often
`almost' a martingale in some
sense.   In this section I give what I think is the most important result
in this direction for the Riemann-sum indefinite integral (623O).  In the
generality here, we need to go a good deal deeper than in \S622, with what I
call `virtually local' martingales (623J).   These depend, in turn, on a
special class
of operators on spaces of locally moderately oscillatory processes (623B).

\cmmnt{
\leader{623A}{Notation} \Asissix\ will be a stochastic integration structure.
If $\Cal S$ is a sublattice of $\Cal T$,
$\Mfa(\Cal S)$, $\Mob(\Cal S)$, $\Mmo(\Cal S)$
and $\Mlmo(\Cal S)$ will be the spaces of
fully adapted, order-bounded, moderately oscillatory and
locally moderately oscillatory
processes with domain $\Cal S$, $\Mbv(\Cal S)\subseteq\Mfa(\Cal S)$
will be the space of processes with bounded variation, and
$\Cal I(\Cal S)$ will be the set of finite
sublattices of $\Cal S$.   If $h:\Bbb R\to\Bbb R$ is a Borel
function, $\bar h$ will denote either of the corresponding operators on
$L^0(\frak A)$ or $\Mlmo(\Cal S)$ (612Ac, 612Ia, 615Fb).   $L^1$ will be
$L^1(\frak A,\bar\mu)$.   For $\tau\in\Cal T$,
$P_{\tau}:L^1\to L^1\cap L^0(\frak A_{\tau})$
will be the conditional expectation associated with the closed subalgebra
$\frak A_{\tau}$, and if $z\in L^1\,\,\pmb Pz$ will
be the martingale $\family{\tau}{\Cal T}{P_{\tau}z}$.
}

\leader{623B}{The operators $R_A$:  Proposition}
Let $\Cal S$ be a sublattice of $\Cal T$ and
$A\subseteq\Cal S$ a non-empty downwards-directed set.

(a) We have an
$f$-algebra homomorphism $R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ defined by setting

\Centerline{$R_A(\sfamilysS{u})
=\family{\sigma}{\Cal S}{\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}}$}

\noindent whenever $\sfamilysS{u}\in\Mlmo(\Cal S)$, and
if $\pmb u\in\Mmo(\Cal S)$ then $R_A(\pmb u)\in\Mmo(\Cal S)$.

(b) $\bar hR_A=R_A\bar h:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ for every continuous
function $h:\Bbb R\to\Bbb R$.

(c) Take $\pmb u=\sfamilysS{u}\in\Mlmo(\Cal S)$ and express
$\pmb u'=R_A(\pmb u)$ as $\sfamilysS{u'}$.

\quad(i) The starting values $\lim_{\sigma\downarrow\Cal S}u'_{\sigma}$ and
$\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ are defined and equal.

\quad(ii) If $\pmb u$ is $\|\,\|_1$-bounded then $\pmb u'$ is
$\|\,\|_1$-bounded and
$\sup_{\sigma\in\Cal S}\|u'_{\sigma}\|_1
\le\sup_{\sigma\in\Cal S}\|u_{\sigma}\|_1$.

(d) Write $\hat\Cal S$ for the covered envelope of $\Cal S$.   If 
$\pmb u\in\Mlmo(\Cal S)$ has fully adapted extension $\hatpmb u$ to
$\hat\Cal S$, then $R_A(\hatpmb u)$ is the fully adapted extension of
$R_A(\pmb u)$ to $\hat\Cal S$.

\proof{{\bf (a)(i)} If $\pmb u=\sfamilysS{u}$ is locally moderately oscillatory,
$\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}$ is defined
(615Gb) and belongs to $L^0(\frak A_{\sigma})$ for every $\sigma\in\Cal S$
(613Bj).   If $\sigma$, $\tau\in\Cal S$ then

\Centerline{$\Bvalue{\sigma=\tau}
\Bsubseteq\inf_{\rho\in A}\Bvalue{u_{\sigma\wedge\rho}=u_{\tau\wedge\rho}}
\Bsubseteq\Bvalue{\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}
   =\lim_{\rho\downarrow A}u_{\tau\wedge\rho}}$,}

\noindent so $R_A(\pmb u)\in\Mfa(\Cal S)$.

\medskip

\quad{\bf (ii)} Because addition, multiplication and modulus are continuous
functions on $L^0(\frak A)$, $R_A:\Mlmo(\Cal S)\to\Mfa(\Cal S)$
is an $f$-algebra homomorphism.

\medskip

\quad{\bf (iii)} If $\pmb u=\sfamilysS{u}$ is moderately oscillatory, then
$|u_{\sigma\wedge\rho}|\le\sup|\pmb u|$ whenever $\sigma\in\Cal S$ and
$\rho\in A$, so $|\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}|\le\sup|\pmb u|$
whenever $\sigma\in\Cal S$, and $R_A(\pmb u)$ is order-bounded, with
$\sup|R_A(\pmb u)|\le\sup|\pmb u|$.   So we have an operator
$R_A:\Mmo(\Cal S)\to\Mob(\Cal S)$ which is continuous for the ucp topology.

\medskip

\quad{\bf (iv)} If $\pmb u=\sfamilysS{u}$ is non-decreasing, non-negative and
order-bounded, then whenever $\sigma\le\tau$ in $\Cal S$ we shall have

\Centerline{$0\le u_{\sigma\wedge\rho}\le u_{\tau\wedge\rho}\le\sup|\pmb u|$}

\noindent for every $\rho\in A$, so

\Centerline{$0\le\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}
\le\lim_{\rho\downarrow A}u_{\tau\wedge\rho}\le\sup|\pmb u|$.}

\noindent Thus $R_A(\pmb u)$ is also non-decreasing, non-negative and
order-bounded.

\medskip

\quad{\bf (v)} It follows that if $\pmb u\in\Mbv(\Cal S)$, that is, $\pmb u$ is
expressible as the difference of two non-decreasing non-negative
order-bounded processes. then $R_A(\pmb u)\in\Mbv(\Cal S)$.   Now
as $R_A:\Mmo(\Cal S)\to\Mob(\Cal S)$ is continuous,

\Centerline{$R_A[\Mmo(\Cal S)]=R_A[\overline{\Mbv(\Cal S)}]
\subseteq\overline{R_A[\Mbv(\Cal S)]}\subseteq\overline{\Mbv(\Cal S)}
=\Mmo(\Cal S)$,}

\noindent and $R_A(\pmb u)$ is moderately oscillatory whenever
$\pmb u\in\Mmo(\Cal S)$.

\medskip

\quad{\bf (vi)} Generally, if $\pmb u\in\Mlmo(\Cal S)$, take any
$\tau\in\Cal S$.   Then $A\wedge\tau=\{\rho\wedge\tau:\rho\in A\}$ is
a non-empty downwards-directed subset of $\Cal S\wedge\tau$, and
$\pmb u\restr\Cal S\wedge\tau$ is moderately oscillatory, so
$R_{A\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)$ is moderately oscillatory, by
(i)-(v) above.   And if $\sigma\in\Cal S\wedge\tau$ then

\Centerline{$\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}
=\lim_{\rho\downarrow A}u_{\sigma\wedge\tau\wedge\rho}
=\lim_{\upsilon\downarrow A\wedge\tau}u_{\sigma\wedge\upsilon}$,}

\noindent so $R_A(\pmb u)\restr\Cal S\wedge\tau
=R_{A\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)$ is moderately oscillatory.
As $\tau$ is arbitrary, $R_A(\pmb u)$ is locally moderately oscillatory, as
claimed.

\medskip

{\bf (b)} If $h:\Bbb R\to\Bbb R$ is continuous,
$\bar h:L^0(\frak A)\to L^0(\frak A)$ is
continuous (613Bb) and $\bar h(\lim_{\rho\downarrow A}u_{\tau\wedge\rho})
  =\lim_{\rho\downarrow A}\bar h(u_{\tau\wedge\rho})$
for every locally moderately
oscillatory $\pmb u=\sfamilysS{u}$ and $\tau\in\Cal S$.

\medskip

{\bf (c)(i)} By 615Gb,
$u_{\downarrow}=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined.   Now if
$\epsilon>0$, there is a $\tau\in\Cal S$ such that
$\theta(u_{\sigma}-u_{\downarrow})\le\epsilon$ for every
$\sigma\in\Cal S\wedge\tau$.   In this case, if $\sigma\in\Cal S\wedge\tau$,
$\theta(u_{\sigma\wedge\rho}-u_{\downarrow})\le\epsilon$ for every
$\rho\in A$, so $\theta(u'_{\sigma}-u_{\downarrow})\le\epsilon$.   As $\epsilon$
is arbitrary, $\lim_{\sigma\downarrow\Cal S}u'_{\sigma}=u_{\downarrow}$.

\medskip

\quad{\bf (ii)} Writing $\gamma$ for $\sup_{\sigma\in\Cal S}\|u_{\sigma}\|_1$,
$\{x:x\in L^0(\frak A)$, $\|x\|_1\le\gamma\}$ is closed (613Bc), so contains
$\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}$ for every $\sigma\in\Cal S$.

\medskip

{\bf (d)} We know that $\hatpmb u$ is locally moderately oscillatory
(615F(a-vi)), while of course $A$ is a non-empty downwards-directed subset of
$\hat\Cal S$, so we can speak of $R_A(\hatpmb u)$.   Looking at the formula in
(a), we see that if we express $\pmb u$ as $\sfamilysS{u}$ and
$\hatpmb u$ as $\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$, then we have
$R_A(\pmb u)=\sfamilysS{u'}$ and 
$R_A(\hatpmb u)=\family{\sigma}{\hat\Cal S}{\hat u'_{\sigma}}$ where

\Centerline{$u'_{\sigma}
=\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}$ for $\sigma\in\Cal S$,}

\Centerline{$\hat u'_{\sigma}
=\lim_{\rho\downarrow A}\hat u_{\sigma\wedge\rho}$ for $\sigma\in\hat\Cal S$,}

\noindent so that $\hat u'_{\sigma}=u'_{\sigma}$ for $\sigma\in\Cal S$ and
$R_A(\hatpmb u)$ extends $R_A(\pmb u)$.
Since $R_A(\hatpmb u)$ is fully adapted ((a-i) above), it must be the fully
adapted extension of $R_A(\pmb u)$

}%end of proof of 623B

\cmmnt{\medskip

\noindent{\bf Remark} The elementary case in which $A=\{\rho\}$ is a singleton,
so that $R_A(\pmb u)=\family{\sigma}{\Cal S}{u_{\sigma\wedge\rho}}$ (612Ib),
will be a useful guide.   But I introduce the idea here primarily for the sake
of applications based on the construction in 623I below.}

\leader{623C}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.
For a non-empty downwards-directed set $A\subseteq\Cal S$ let
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the operator described in 623B.
Let $A$, $B\subseteq\Cal S$ be non-empty downwards-directed sets.

(a) Setting $A\vee B=\{\rho\vee\rho':\rho\in A$, $\rho'\in B\}$ and
$A\wedge B=\{\rho\wedge\rho':\rho\in A$, $\rho'\in B\}$,
$R_{A\vee B}+R_{A\wedge B}=R_A+R_B$.

(b) $R_{A\wedge B}=R_AR_B=R_BR_A$.

(c) If $B\subseteq A$, then $R_AR_B=R_A$;  in particular,
$R_A^2=R_A$.

(d) If $B$ is a coinitial subset of $A$, then $R_B=R_A$.

\proof{{\bf (a)} Of course both $A\vee B$ and $A\wedge B$ are non-emptu
downwards-directed sets, so we can speak of $R_{A\vee B}$ and $R_{A\wedge B}$.
Take a moderately oscillatory process $\pmb u=\sfamilysS{u}$.   If
$\sigma\in\Cal S$, then

$$\eqalignno{
  \lim_{(\rho,\rho')\downarrow A\times B}u_{(\rho\vee\rho')\wedge\sigma}
    +\lim_{(\rho,\rho')\downarrow A\times B}u_{(\rho\wedge\rho')\wedge\sigma}&
=\lim_{(\rho,\rho')\downarrow A\times B}
        u_{(\rho\wedge\sigma)\vee(\rho'\wedge\sigma)}
        +u_{(\rho\wedge\sigma)\wedge(\rho'\wedge\sigma)}
\Displaycause{because $\Cal T$ is a distributive lattice, by 611Ca}
=\lim_{(\rho,\rho')\downarrow A\times B}
        u_{\rho\wedge\sigma}+u_{\rho'\wedge\sigma}
\Displaycause{612D(f-i)}
=\lim_{\rho\downarrow A}
        u_{\rho\wedge\sigma}+\lim_{\rho'\downarrow B}u_{\rho'\wedge\sigma}.
\cr}$$

\noindent As $\sigma$ is arbitrary,
$R_{A\vee B}(\pmb u)+R_{A\wedge B}(\pmb u)=R_A(\pmb u)+R_B(\pmb u)$;  as
$\pmb u$ is arbitrary, $R_{A\vee B}+R_{A\wedge B}=R_A+R_B$.

\quad

{\bf (b)} Take
$\pmb u=\sfamilysS{u}\in\Mlmo(\Cal S)$ and $\sigma\in\Cal S$, and set

\Centerline{$v_{\sigma}=\lim_{\rho\downarrow A\wedge B}u_{\sigma\wedge\rho}$,
\quad$w_{\sigma}=\lim_{\rho\downarrow A}\lim_{\rho'\downarrow B}
   u_{(\sigma\wedge\rho)\wedge\rho'}$.}

\noindent Let $\epsilon>0$.   Then there are $\tilde\rho\in A$,
$\tilde\rho'\in B$ such that
$\theta(v_{\sigma}-u_{\sigma\wedge\rho\wedge\rho'})\le\epsilon$ whenever
$\rho\in A$, $\rho'\in B$, $\rho\le\tilde\rho$ and $\rho'\le\tilde\rho'$.
Next, there are $\tau\in A$, $\tau'\in B$ such that

\Centerline{$\tau\le\tilde\rho$,
\quad
$\theta(w_{\sigma}-\lim_{\rho'\downarrow B}u_{\sigma\wedge\tau\wedge\rho'})
\le\epsilon$,}

\Centerline{$\tau'\le\tilde\rho'$,
\quad$\theta(u_{\sigma\wedge\tau\wedge\tau'}
       -\lim_{\rho'\downarrow B}u_{\sigma\wedge\tau\wedge\rho'})\le\epsilon$.}

\noindent Since we also have
$\theta(v_{\sigma}-u_{\sigma\wedge\tau\wedge\tau'})\le\epsilon$, we see that
$\theta(v_{\sigma}-w_{\sigma})\le 3\epsilon$.   As $\epsilon$ is arbitrary,
$v_{\sigma}=w_{\sigma}$.   As $\sigma$ is arbitrary,

\Centerline{$R_{A\wedge B}(\pmb u)=\sfamilysS{v}=\sfamilysS{w}
=R_AR_B(\pmb u)$.}

As $\pmb u$ is arbitrary, $R_{A\wedge B}=R_AR_B$.   Similarly,
$R_BR_A=R_{B\wedge A}=R_{A\wedge B}$.

\medskip

{\bf (c)} If $B\subseteq A$, consider
$A^*=\{\rho:\rho\in\Cal S$, $\rho'\le\rho$ for some $\rho'\in A\}$.
Then $A^*$ is closed under $\wedge$ and $A$ is coinitial with $A^*$;
as $B\subseteq A$, $A\wedge B\subseteq A^*$ is also coinitial with
$A^*$.   Now if $\pmb u=\sfamilysS{u}$ is locally moderately oscillatory,
$\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}$,
$\lim_{\rho\downarrow A^*}u_{\sigma\wedge\rho}$ and
$\lim_{\rho\downarrow A\wedge B}u_{\sigma\wedge\rho}$
are all defined and equal for every $\sigma\in\Cal S$.
As $\pmb u$ and $\sigma$ are arbitrary,
$R_A=R_{A^*}=R_{A\wedge B}$.

\medskip

{\bf (d)} If $B$ is a coinitial subset of $A$ then of course
$\lim_{\rho\downarrow B}u_{\sigma\wedge\rho}
=\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}$ whenever the latter is defined.
}%end of proof of 623C

\leader{623D}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$A\subseteq\Cal S$ a non-empty downwards-directed set.   Let
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the operator described in 623B.
If $\pmb v=\sfamilysS{v}$ is a (local) integrator, $R_A(\pmb v)$ is a
(local) integrator.

\proof{ Express $\pmb v'=R_A(\pmb v)$ as $\sfamilysS{v'}$.

\medskip

{\bf (a)} Consider first the case in which $\pmb v$ is an integrator.

\medskip

\quad{\bf (i)} If $\rho\in A$ then $Q_{\Cal S}(d(R_{\{\rho\}}(\pmb v)))
\subseteq Q_{\Cal S}(d\pmb v)$.   \Prf\ Take
$z\in Q_{\Cal S}(dR_{\{\rho\}}(\pmb v))$.   If $\Cal S=\emptyset$ then surely
$z=0\in Q_{\Cal S}(d\pmb v)$.   Otherwise, 616C(iii) tells us that
there are a fully adapted process
$\tildepmb u=\sfamilysS{\tilde u}$ such that
$\|\tildepmb u\|_{\infty}\le 1$ and
$\tau_0\le\ldots\le\tau_n$ such that

\Centerline{$z
=\sum_{i=0}^{n-1}\tilde u_{\tau_i}
  \times(v_{\tau_{i+1}\wedge\rho}-v_{\tau_i\wedge\rho})
=\sum_{i=0}^{n-1}\tilde u_{\tau_i\wedge\rho}
  \times(v_{\tau_{i+1}\wedge\rho}-v_{\tau_i\wedge\rho})$}

\noindent because if $i<n$ then

\Centerline{$\Bvalue{\tilde u_{\tau_i\wedge\rho}\ne\tilde u_{\tau_i}}
\Bsubseteq\Bvalue{\tau_i\wedge\rho<\tau_i}
\Bsubseteq\Bvalue{\tau_i\wedge\rho=\tau_{i+1}\wedge\rho}
\Bsubseteq\Bvalue{v_{\tau_{i+1}\wedge\rho}=v_{\tau_i\wedge\rho}}$.}

\noindent So $z\in Q_{\Cal S}(\pmb v)$, by 616C(ii).\ \Qed

\medskip

\quad{\bf (ii)} $Q_{\Cal S}(d\pmb v')
\Bsubseteq\overline{Q_{\Cal S}(\pmb u,d\pmb v)}$.   \Prf\ If
$z\in Q_{\Cal S}(d\pmb v')\setminus\{0\}$, express it as
$S_I(\pmb u,d\pmb v')$ where $I\in\Cal I(\Cal S)$,
$\pmb u=\family{\sigma}{I}{u_{\sigma}}\in\Mfa(I)$ and
$\|\pmb u\|_{\infty}\le 1$.   Let $(\tau_0,\ldots,\tau_n)$ linearly generate
the $I$-cells.   Then

$$\eqalign{z&
=\sum_{i=0}^{n-1}u_{\tau_i}\times(v'_{\tau_{i+1}}-v'_{\tau_i})
=\lim_{\rho\downarrow A}\sum_{i=0}^{n-1}u_{\tau_i}
   \times(v_{\tau_{i+1}\wedge\rho}-v_{\tau_i\wedge\rho})\cr&
=\lim_{\rho\downarrow A}S_I(\pmb u,dR_{\{\rho\}}(\pmb v))
\in\overline{Q_{\Cal S}(\pmb u,d\pmb v)}\cr}$$

\noindent by (i).\ \Qed

\medskip

\quad{\bf (iii)} Since $Q_{\Cal S}(d\pmb v)$ is topologically bounded, so is its
closure (613B(f-iii)), and $\pmb v'$ is an integrator.

\medskip

{\bf (b)} Now suppose that $\pmb v$ is a local integrator.   Take
$\tau\in\Cal S$.   Set $B=\{\tau\wedge\rho:\rho\in A\}$;  then $B$ is a
non-empty downwards-directed subset of $\Cal S\wedge\tau$, so we have a
corresponding operator $R_B$ on $\Mlmo(\Cal S\wedge\tau)$.   As $\pmb v$ is a
local integrator, $\pmb v\restr\Cal S\wedge\tau$ is an integrator and
$R_B(\pmb v\restr\Cal S\wedge\tau)$ is an integrator, by (a).   Express
$R_B(\pmb v\restr\Cal S\wedge\tau)$ as
$\family{\sigma}{\Cal S\wedge\tau}{w'_{\sigma}}$.   If
$\sigma\in\Cal S\wedge\tau$,

\Centerline{$v'_{\sigma}=\lim_{\rho\downarrow A}v_{\sigma\wedge\rho}
=\lim_{\rho\downarrow A}v_{\sigma\wedge\tau\wedge\rho}
=\lim_{\rho\downarrow B}v_{\sigma\wedge\rho}=w'_{\sigma}$.}

\noindent So
$\pmb v'\restr\Cal S\wedge\tau=R_B(\pmb v\restr\Cal S\wedge\tau)$ is an
integrator.   As $\tau$ is arbitrary, $\pmb v'$ is a local integrator.
}%end of proof of 623D

\leader{623E}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$A\subseteq\Cal S$ a non-empty downwards-directed set.   Let
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the operator described in 623B.
If $\pmb u=\sfamilysS{u}$ is a martingale, $R_A(\pmb u)$ is a martingale.

\proof{ $\pmb u$ is locally moderately oscillatory (622H) so
$R_A(\pmb u)$ is defined;  express it as $\sfamilysS{u'}$.   The point is that
if $\sigma\in\Cal S$ then $u'_{\sigma}$ is the limit
$\1lim_{\rho\downarrow A}P_{\rho}u_{\sigma}$ for the norm topology of $L^1$.
\Prf\ For $\rho\in A$,

\Centerline{$u_{\sigma\wedge\rho}=P_{\sigma\wedge\rho}u_{\sigma}
=P_{\rho}P_{\sigma}u_{\sigma}=P_{\rho}u_{\sigma}$}

\noindent (622Ba).   Next, $\1lim_{\rho\downarrow A}P_{\rho}z$ is defined for
every $z\in L^0(\frak A)$, by 621Cg, and this must also be the limit
$\lim_{\rho\downarrow A}P_{\rho}z$ for the topology of convergence in measure
(613B(d-i)).   So we have

\Centerline{$u'_{\sigma}=\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}
=\lim_{\rho\downarrow A}P_{\rho}u_{\sigma}
=\1lim_{\rho\downarrow A}P_{\rho}u_{\sigma}$,}

\noindent as claimed.\ \Qed

If now $\sigma\le\tau$ in $\Cal S$,

$$\eqalignno{P_{\sigma}u'_{\tau}&
=P_{\sigma}(\1lim_{\rho\downarrow A}P_{\rho}u_{\tau})
=\1lim_{\rho\downarrow A}P_{\sigma}P_{\rho}u_{\tau}
\Displaycause{because $P_{\sigma}:L^1\to L^1$
is $\|\,\|_1$-continuous}
=\1lim_{\rho\downarrow A}P_{\rho}P_{\sigma}u_{\tau}
=\1lim_{\rho\downarrow A}P_{\rho}u_{\sigma}
=u'_{\sigma},\cr}$$

\noindent and $R_A(\pmb u)=\sfamilysS{u'}$ is a martingale.
}%end of proof of 623E

\leader{623F}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$A\subseteq\Cal S$ a non-empty downwards-directed set and
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ the operator described in 623B.
Suppose that $\pmb u=\sfamilysS{u}$ is locally moderately oscillatory.

(a) If $\pmb u$ is order-bounded,
the residual oscillation $\Osclln(R_A(\pmb u))$ is at most $\Osclln(\pmb u)$.

(b) If $\pmb u$ is (locally) jump-free, then $R_A(\pmb u)$
is (locally) jump-free.

\proof{ Express $\pmb u'=R_A(\pmb u)$ as $\sfamilysS{u'}$.

\medskip

{\bf (a)(i)} As

\Centerline{$|u'_{\sigma}|\le\sup_{\rho\in A}|u_{\sigma\wedge\rho}|
\le\sup|\pmb u|$}

\noindent for every $\sigma\in\Cal S$, $\pmb u'$ is order-bounded and
$\Osclln(\pmb u')$ is defined.

\medskip

\quad{\bf (ii)} If $I\in\Cal I(\Cal S)$ then, in the language
of 618B,
$\Osclln_I(\pmb u')\le\Osclln^*_I(\pmb u)$.   \Prf\ If $I$ is empty, this is
trivial.   Otherwise, take $(\tau_0,\ldots,\tau_n)$ linearly generating the
$I$-cells.   Take $\rho\in A$ and $i<n$.   Then

$$\eqalign{\Bvalue{\rho\le\tau_i}&
\Bsubseteq\Bvalue{\tau_i\wedge\rho=\rho}\Bcap\Bvalue{\tau_{i+1}\wedge\rho=\rho}
\Bsubseteq\Bvalue{u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}=0}\cr&
\Bsubseteq\Bvalue{|u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}|
   \le\Osclln^*_I(\pmb u)},\cr}$$

$$\eqalign{\Bvalue{\tau_i\le\rho}\Bcap\Bvalue{\rho\le\tau_{i+1}}&
\Bsubseteq\Bvalue{\tau_i\wedge\rho=\tau_i}
   \Bcap\Bvalue{\tau_{i+1}\wedge\rho=\med(\tau_i,\rho,\tau_{i+1})}\cr&
\Bsubseteq\Bvalue{u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}
     =u_{\med(\tau_i,\rho,\tau_{i+1})}-u_{\tau_i}}\cr&
\Bsubseteq\Bvalue{|u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}|
   \le\Osclln^*_I(\pmb u)}\cr}$$

\noindent (618Ca),

$$\eqalign{\Bvalue{\tau_{i+1}\le\rho}&
\Bsubseteq\Bvalue{\tau_i\wedge\rho=\tau_i}
   \Bcap\Bvalue{\tau_{i+1}\wedge\rho=\tau_{i+1}}\cr&
\Bsubseteq\Bvalue{u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}
  =u_{\tau_{i+1}}-u_{\tau_i}}
\Bsubseteq\Bvalue{|u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}|
  \le\Osclln^*_I(\pmb u)}.\cr}$$

\noindent So in fact
$|u_{\tau_{i+1}\wedge\rho}-u_{\tau_i\wedge\rho}|\le\Osclln^*_I(\pmb u)$.
Taking the limit as $\rho\downarrow A$,
$|u'_{\tau_{i+1}}-u'_{\tau_i}|\le\Osclln^*_I(\pmb u)$.
As $i$ is arbitrary, $\Osclln_I(\pmb u')\le\Osclln^*_I(\pmb u)$ (618Ba).\ \Qed

\medskip

\quad{\bf (iii)} Now

$$\eqalign{\Osclln(\pmb u')&
=\inf_{I\in\Cal I(\Cal S)}\sup_{I\subseteq J\in\Cal I(\Cal S)}
   \Osclln_J(\pmb u')\cr&
\le\inf_{I\in\Cal I(\Cal S)}\sup_{I\subseteq J\in\Cal I(\Cal S)}
    \Osclln^*_J(\pmb u)\cr&
=\inf_{I\in\Cal I(\Cal S)}\sup_{I\subseteq J\subseteq K\in\Cal I(\Cal S)}
    \Osclln_K(\pmb u)\cr&
=\inf_{I\in\Cal I(\Cal S)}\sup_{I\subseteq K\in\Cal I(\Cal S)}
    \Osclln_K(\pmb u)
=\Osclln(\pmb u).\cr}$$

\medskip

{\bf (b)(i)} If $\pmb u$ is jump-free, then it is order-bounded (618B(b-ii))
and $\Osclln(\pmb u)=0$, so $\Osclln(\pmb u')=0$ and $\pmb u'$ is jump-free.

\medskip

\quad{\bf (ii)} If $\pmb u$ is locally jump-free, then we can apply (i) to
$\pmb u\restr\Cal S\wedge\tau$ and $A\wedge\tau=\{\rho\wedge\tau:\rho\in A\}$
to see that $\pmb u'\restr\Cal S\wedge\tau$ is jump-free for every
$\tau\in\Cal S$, so that $\pmb u'$ is locally jump-free.
}%end of proof of 623F

\leader{623G}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$A\subseteq\Cal S$ a non-empty downwards-directed set.   Let
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the operator described in 623B.
If $\pmb u=\sfamilysS{u}$ is locally moderately oscillatory and
$\pmb v=\sfamilysS{v}$ is a local integrator,

\Centerline{$R_A(\iivu)=ii_{R_A(\pmb v)}(\pmb u)
=ii_{R_A(\pmb v)}(R_A(\pmb u))$.}

\proof{{\bf (a)} Since $\pmb u'=R_A(\pmb u)$ is locally moderately oscillatory
(623Ba) and $\pmb v'=R_A(\pmb v)$ is a local integrator (623D),
all the indefinite integrals are defined everywhere on $\Cal S$;
while $\iivu$ also, being a local integrator (616Q(c-i)), is
locally moderately oscillatory
(616Ib), so we can speak of $\pmb w'=R_A(\iivu)$.

For $I\in\Cal I(\Cal S)$ and $\rho\in\Cal S$ write
$I\wedge\rho$ for $\{\sigma\wedge\rho:\sigma\in I\}$.
Express $\pmb u'$, $\pmb v'$,
$\pmb w'$ and $\pmb z'=ii_{\pmb v'}(\pmb u')=ii_{R_A(\pmb v)}(R_A(\pmb u))$ as
$\sfamilysS{u'}$, $\sfamilysS{v'}$,
$\sfamilysS{w'}$ and $\sfamilysS{z'}$ respectively.

\medskip

{\bf (b)} If $\sigma\le\tau$ in $\Cal S$ and $\rho\in\Cal S$,

\Centerline{$u_{\sigma\wedge\rho}
  \times(v_{\tau\wedge\rho}-v_{\sigma\wedge\rho})
=u_{\sigma}
  \times(v_{\tau\wedge\rho}-v_{\sigma\wedge\rho})$.}

\noindent\Prf\

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

\noindent so $(u_{\sigma\wedge\rho}-u_{\sigma})
  \times(v_{\tau\wedge\rho}-v_{\sigma\wedge\rho})=0$.\ \Qed

Letting $\rho\downarrow A$, we see that

\Centerline{$u'_{\sigma}\times(v'_{\tau}-v'_{\sigma})
=u_{\sigma}\times(v'_{\tau}-v'_{\sigma})$.}

\noindent It follows at once that

\Centerline{$S_I(\pmb u',d\pmb v')=S_I(\pmb u,d\pmb v')$}

\noindent for every $I\in\Cal I(\Cal S)$, and therefore that
$ii_{\pmb v'}(\pmb u)=ii_{\pmb v'}(\pmb u')=\pmb z'$.

\medskip

{\bf (c)} Take $\tau\in\Cal S$ and $\epsilon>0$.   There is a
$J_0\in\Cal I(\Cal S\wedge\tau)$ such that
$\theta(S_{I\wedge\tau\wedge\rho}(\pmb u,d\pmb v)
-\int_{\Cal S\wedge\tau\wedge\rho}\pmb u\,d\pmb v)
\le\epsilon$ whenever $\rho\in\Cal S$ and $I\in\Cal I(\Cal S\wedge\tau)$
includes
$J_0$ (613V(ii-$\beta$)), and a $J_1\in\Cal I(\Cal S\wedge\tau)$ such that
$\theta(S_I(\pmb u',d\pmb v')-\int_{\Cal S\wedge\tau}\pmb u'\,d\pmb v')
\le\epsilon$ whenever $I\in\Cal I(\Cal S\wedge\tau)$ includes
$J_1$.   Let $I$ be the sublattice generated by $J_0\cup J_1\cup\{\tau\}$.
Let $(\tau_0,\ldots,\tau_n)$ linearly generate the $I$-cells.   If
$\rho\in A$, then $(\tau_0\wedge\rho,\ldots,\tau_n\wedge\rho)$ linearly
generates the $(I\wedge\rho)$-cells (611Kg), while of course
$I\wedge\rho=I\wedge\tau\wedge\rho$ because $\tau=\max I$.   Accordingly

\Centerline{$S_{I\wedge\tau\wedge\rho}(\pmb u,d\pmb v)
=\sum_{i=0}^{n-1}u_{\tau_i\wedge\rho}
 \times(v_{\tau_{i+1}\wedge\rho}-v_{\tau_i\wedge\rho})$}

\noindent so

\Centerline{$\theta(\sum_{i=0}^{n-1}u_{\tau_i\wedge\rho}
 \times(v_{\tau_{i+1}\wedge\rho}-v_{\tau_i\wedge\rho})
   -\int_{\Cal S\wedge\tau\wedge\rho}\pmb u\,d\pmb v)\le\epsilon$.}

\noindent Taking the limit as $\rho\downarrow A$,

\Centerline{$\theta(\sum_{i=0}^{n-1}u'_{\tau_i}
    \times(v'_{\tau_{i+1}}-v'_{\tau_i})
  -w'_{\tau})\le\epsilon$,}

\noindent that is,

\Centerline{$\theta(S_I(\pmb u',d\pmb v')-w'_{\tau})\le\epsilon$.}

\noindent Since we also have $J_1\subseteq I\in\Cal I(\Cal S\wedge\tau)$,
$\theta(S_I(\pmb u',d\pmb v')-z'_{\tau})\le\epsilon$ and we conclude that
$\theta(w'_{\tau}-z'_{\tau})\le 2\epsilon$.   As $\tau$ and $\epsilon$ are
arbitrary,

\Centerline{$\pmb w'=\pmb z'=ii_{\pmb v'}(\pmb u')$,}

\noindent and we know from (b) that the last is equal to
$ii_{\pmb v'}(\pmb u)$.   So we have
$R_A(\iivu)=ii_{R_A(\pmb v)}(R_A(\pmb u))=ii_{R_A(\pmb v)}(\pmb u)$,
as claimed.
}%end of proof of 623G

\leader{623H}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$,
$A\subseteq\Cal S$ a non-empty downwards-directed set and
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ the operator described in 623B.
If $\pmb v=\sfamilysS{v}$ is a local integrator with quadratic variation
$\pmb v^*$, then $R_A(\pmb v^*)$ is the quadratic variation of $R_A(\pmb v)$.

\proof{ We know that $\pmb v$ is locally moderately oscillatory (616Ib again).
Writing $\pmb v'$ for $R_A(\pmb v)$, 623D and
623G tell us that $\pmb v'$ is a local
integrator and $R_A(\iiv(\pmb v))=ii_{\pmb v'}(\pmb v')$.   Now

$$\eqalignno{(\pmb v')^*&
=(\pmb v')^2-2ii_{\pmb v'}(\pmb v')-(v'_{\downarrow})^2\tbf 1
\Displaycause{where $v'_{\downarrow}$ is the starting value of $\pmb v'$}
=R_A(\pmb v^2)-2R_A(\iiv(\pmb v))-v_{\downarrow}^2\tbf 1
\Displaycause{623Ba, 623B(c-i)}
=R_A(\pmb v^2-2\iiv(\pmb v)-v_{\downarrow}^2\tbf 1)
\Displaycause{because $R_A(\tbf 1\restr\Cal S)=\tbf 1\restr\Cal S$}
=R_A(\pmb v)^*.\cr}$$

}%end of proof of 623H

\leader{623I}{Lemma} Let $\Cal S$ be a finitely full sublattice of
$\Cal T$\cmmnt{ (definition:  611O)}.
Suppose that $\pmb u=\sfamilysS{u}$ is a moderately oscillatory
process, $\tau^*\in\Cal S$ and $M\ge 0$.

(a) Set

\Centerline{$A=\{\rho:\rho\in\Cal S,\,\Bvalue{\rho<\tau^*}
\Bsubseteq\Bvalue{|u_{\rho}|\ge M}\}$.}

\noindent Then $\tau^*\in A$ and $\rho\wedge\rho'\in A$ whenever
$\rho$, $\rho'\in A$.

(b) Let $R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the operator described in 623B.
Suppose that $\pmb v=\sfamilysS{v}$ is a moderately oscillatory process such
that $R_A(\pmb v)=\pmb v$.

\quad(i) $\Bvalue{|u_{\sigma}|\ge M}\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau}}$
whenever $\sigma\le\tau$ in $\Cal S\wedge\tau^*$.

\quad(ii) Expressing $R_A(\pmb u)$ as $\sfamilysS{u'}$,
$\Bvalue{|u'_{\sigma}|>M}\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau^*}}$
for every $\sigma\in\Cal S$.   In particular,
$\Bvalue{|u'_{\sigma}|>M}\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\tau^*}}$.

\proof{{\bf (a)} $\tau^*\in A$ just because $\Bvalue{\tau^*<\tau^*}=0$.
To see that $A$ is downwards-directed, repeat the formula in
the proof of 615Ma:  if $\rho$, $\rho'\in A$, then

$$\eqalign{\Bvalue{\rho\wedge\rho'<\tau^*}&
=(\Bvalue{\rho\le\rho'}\Bcap\Bvalue{\rho<\tau^*})
    \Bcup(\Bvalue{\rho'\le\rho}\Bcap\Bvalue{\rho'<\tau^*})\cr&
\Bsubseteq(\Bvalue{\rho\le\rho'}\Bcap\Bvalue{|u_{\rho}|\ge M})
  \Bcup(\Bvalue{\rho'\le\rho}\Bcap\Bvalue{|u_{\rho'}|\ge M})\cr&
=(\Bvalue{\rho\le\rho'}
    \Bcap\Bvalue{|u_{\rho\wedge\rho'}|\ge M})
  \Bcup(\Bvalue{\rho'\le\rho}
    \Bcap\Bvalue{|u_{\rho\wedge\rho'}|\ge M})\cr&
=\Bvalue{|u_{\rho\wedge\rho'}|\ge M}\cr}$$

\noindent and $\rho\wedge\rho'\in A$.

\medskip

{\bf (b)(i)} If $\sigma\le\tau$ in $\Cal S\wedge\tau^*$, set
$a=\Bvalue{|u_{\sigma}|\ge M}$.   Then
$a\in\frak A_{\sigma}\subseteq\frak A_{\tau^*}$ so there is a
$\rho\in\Cal T$ such that $a\Bsubseteq\Bvalue{\rho=\sigma}$ and
$1\Bsetminus a\Bsubseteq\Bvalue{\rho=\tau^*}$.   As $\Cal S$ is finitely full,
$\rho\in\Cal S$.   Now

\Centerline{$\Bvalue{\rho<\tau^*}\Bsubseteq a
=\Bvalue{\rho=\sigma}\Bcap\Bvalue{|u_{\sigma}|\ge M}
\Bsubseteq\Bvalue{|u_{\rho}|\ge M}$}

\noindent and $\rho\in A$.

If $\rho'\in A$ and $\rho'\le\rho$, then

$$\eqalignno{a&
\Bsubseteq\Bvalue{\rho=\sigma\wedge\rho}
\Bsubseteq\Bvalue{\rho'=\sigma\wedge\rho'}
\Bsubseteq\Bvalue{\tau\wedge\rho'\le\sigma\wedge\rho'}
=\Bvalue{\tau\wedge\rho'=\sigma\wedge\rho'}
\Displaycause{because we are supposing that $\sigma\le\tau$}
\Bsubseteq\Bvalue{v_{\tau\wedge\rho'}=v_{\sigma\wedge\rho'}}.\cr}$$

\noindent As $\rho$ is arbitrary,

\Centerline{$a
\Bsubseteq\Bvalue{\lim_{\rho'\downarrow A}v_{\tau\wedge\rho'}
                        =\lim_{\rho'\downarrow A}v_{\sigma\wedge\rho'}}
=\Bvalue{v_{\tau}=v_{\sigma}}$}

\noindent because $R_A(\pmb v)=\pmb v$.

\medskip

\quad{\bf (ii)} For any $\rho\in A\wedge\tau^*$,
$\sigma\wedge\rho\le\tau^*$ so
$\Bvalue{|u_{\sigma\wedge\rho}|\ge M}
\Bsubseteq\Bvalue{v_{\sigma\wedge\rho}=v_{\tau^*}}$.   But we are
supposing that

$$\eqalignno{\pmb v&
=R_A(\pmb v)=R_{\{\rho\}}R_A(\pmb v)
\Displaycause{623Cc, with $B=\{\rho\}$}
=R_{\{\rho\}}(\pmb v),\cr}$$

\noindent so $v_{\sigma\wedge\rho}=v_{\sigma}$ and
$\Bvalue{|u_{\sigma\wedge\rho}|\ge M}\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau^*}}$.
Now

$$\eqalignno{\Bvalue{|u'_{\sigma}|>M}&
=\Bvalue{|\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}|>M}
\Bsubseteq\sup_{\rho\in A\wedge\tau^*}\Bvalue{|u_{\sigma\wedge\rho}|>M}
\Displaycause{because $\tau^*\in A$, so $A\wedge\tau^*$ is coinitial with $A$}
\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau^*}}.\cr}$$

\noindent As $R_A(R_A(\pmb u))=R_A(\pmb u)$, we can
apply this with $\pmb v=R_A(\pmb u)$ and get
$\Bvalue{|u'_{\sigma}|>M}\Bsubseteq\Bvalue{u'_{\sigma}=u'_{\tau^*}}$.
}%end of proof of 623I

\leader{623J}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u$ a locally moderately oscillatory process.
Let $\hat\Cal S$ be the covered envelope of $\Cal S$ and
$\hatpmb u=\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$ the fully adapted
extension of $\pmb u$ to $\hat\Cal S$.   Recall that
$\hatpmb u$ is locally moderately oscillatory\cmmnt{ (615F(b-v))}.   I will say
that $\pmb u$ is an {\bf approximately local martingale} if for every
$\sigma\in\Cal S$ and  $\epsilon>0$ there is a non-empty downwards-directed set
$A\subseteq\Cal S$
such that $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\sigma}\le\epsilon$ and
$R_A(\pmb u)$, as defined in 623B, is a martingale;
while $\pmb u$ is a {\bf virtually local martingale} if $\hatpmb u$ is an
approximately local martingale.

\cmmnt{\medskip

\noindent{\bf Remarks} Note that as the covered envelope of $\hat\Cal S$ is
itself, $\pmb u$ is a virtually local martingale iff $\hatpmb u$ is.
And if $\Cal S$ has a greatest element we can drop the `for every $\sigma$';
$\pmb u$ will be an approximately local martingale iff for every
$\epsilon>0$ there is a non-empty downwards-directed set
$A\subseteq\Cal S$
such that $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\max\Cal S}\le\epsilon$ and
$R_A(\pmb u)$ is a martingale.

In the context here\cmmnt{, since $A$ is downwards-directed,
$\{\Bvalue{\rho<\tau}:\rho\in A\}$ is upwards-directed, so
that} $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\sigma}$ will always be
$\bar\mu(\sup_{\rho\in A}\Bvalue{\rho<\sigma})$\cmmnt{ (321C)}.
}

\leader{623K}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.

(a)(i) The space $\Malm(\Cal S)$ of approximately local martingales on $\Cal S$ is
a linear subspace of $\Mlmo(\Cal S)$.

\quad(ii) The space $\Mvlm(\Cal S)$ of virtually local martingales on 
$\Cal S$ is a linear subspace of $\Mlmo(\Cal S)$.

(b)(i) A local martingale on $\Cal S$ is an approximately local martingale.

\quad(ii) An approximately local martingale on $\Cal S$ is a
virtually local martingale.

\quad(iii) If $\Cal S$ is finitely full, a virtually local martingale on
$\Cal S$ is an approximately local martingale.

(c) If $\pmb u\in\Mvlm(\Cal S)$ and $A\subseteq\Cal S$ is a non-empty
downwards-directed set, $R_A(\pmb u)$, as defined in 623B, is a virtually
local martingale.

(d) Every virtually local martingale on $\Cal S$ is a local integrator,
therefore locally moderately oscillatory.

(e)(i) If $\pmb u\in\Mfa(\Cal S)$, then $\pmb u$ is an
approximately local martingale
iff $\pmb u\restr\Cal S\wedge\tau$ is an approximately
local martingale for every $\tau\in\Cal S$.

\quad(ii) If $\pmb u\in\Mfa(\Cal S)$, 
then $\pmb u$ is a virtually local martingale
iff $\pmb u\restr\Cal S\wedge\tau$ is a virtually local martingale for every
$\tau\in\Cal S$.

(f) A uniformly integrable approximately local martingale on $\Cal S$
is a martingale.

% query:  what about uniformly integrable virtually local martingales?

(g) If $\Cal S\ne\emptyset$ and
$\pmb u=\sfamilysS{u}$ is a virtually local martingale, then
$\lim_{\sigma\downarrow\Cal S}u_{\sigma}$ is defined and belongs to
$L^1$.

(h) If $\pmb u=\sfamilysS{u}\in\Mvlm(\Cal S)$ and $\tau\in\Cal S$ then
$(\pmb u-u_{\tau}\tbf 1)\restr\Cal S\vee\tau$ is a virtually local martingale.

\proof{ Let $\hat\Cal S$ be the covered envelope of $\Cal S$;  for a
fully adapted process $\pmb u$ with domain $\Cal S$, write
$\hatpmb u=\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$ for its
fully adapted extension to $\hat\Cal S$.

\medskip

{\bf (a)(i)} It is built into the definition in 623J that a virtually local
martingale is locally moderately oscillatory.
If $\pmb u$, $\pmb v\in\Malm(\Cal S)$, $\tau\in\Cal S$ and
$\epsilon>0$, let $A$, $B\subseteq\Cal S$
be non-empty downwards-directed sets such
that $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\bover12\epsilon$,
$R_A(\pmb u)$ is a martingale,
$\sup_{\rho\in B}\bar\mu\Bvalue{\rho<\tau}\le\bover12\epsilon$ and
$R_B(\pmb v)$ is a martingale.   Then $A\wedge B$ is a non-empty
downwards-directed subset of $\Cal S$ and
$R_{A\wedge B}(\pmb u)=R_BR_A(\pmb u)$,
$R_{A\wedge B}(\pmb v)=R_AR_B(\pmb v)$ are martingales
(623Cb, 623E).   So $R_{A\wedge B}(\pmb u+\pmb v)$ is a martingale
(622Db), while

$$\eqalign{\sup_{\rho\in A\wedge B}\bar\mu\Bvalue{\rho<\tau}
=\sup_{\Atop{\rho\in A}{\rho'\in B}}\bar\mu\Bvalue{\rho\wedge\rho'<\tau}
=\sup_{\Atop{\rho\in A}{\rho'\in B}}
   \bar\mu(\Bvalue{\rho<\tau}\Bcup\Bvalue{\rho'<\tau})
\le\epsilon.\cr}$$

\noindent As $\tau$ and $\epsilon$ are arbitrary, $\pmb u+\pmb v$
is an approximately local martingale.

If $\pmb u\in\Malm(\Cal S)$ and $\alpha\in\Bbb R$, then for any
$\tau\in\hat\Cal S$ and $\epsilon>0$, we have a non-empty
downwards-directed set $A\subseteq\Cal S$ such
that $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$ and
$R_A(\pmb u)$ is a martingale.   Now $R_A(\alpha\pmb u)=\alpha R_A(\pmb u)$ 
is a martingale.   So $\alpha\pmb u\in\Malm(\Cal S)$.

\medskip

\quad{\bf (ii)} By definition, $\Mvlm(\Cal S)=\{\pmb u:\pmb u\in\Mlmo(\Cal S)$,
$\hatpmb u\in\Malm(\hat\Cal S)\}$, where $\hat\Cal S$ is the covered envelope of
$\Cal S$ and $\hatpmb u$ is the fully adapted extension of $\pmb u$ to 
$\hat\Cal S$.   Since $\Malm(\hat\Cal S)$ is a linear subspace of
$\Mlmo(\hat\Cal S)$, by (i), and 
$\pmb u\mapsto\hatpmb u:\Mlmo(\Cal S)\to\Mlmo(\hat\Cal S)$ is a linear operator
(615F(b-v), 612Qb), $\Mvlm(\Cal S)$ is a linear subspace of 
$\Mlmo(\Cal S)$.

\medskip

{\bf (b)(i)} Suppose that $\pmb u=\sfamilysS{u}$ is a local martingale.
Take any $\tau\in\hat\Cal S$ and
$\epsilon>0$.   Then there is a $\tau'\in\Cal S$ such that
$\pmb u\restr\hat\Cal S\wedge\tau'$ is a martingale and
$\bar\mu\Bvalue{\tau'<\tau}\le\epsilon$.   
Of course
$A=\{\tau'\}$ is a non-empty downwards-directed subset of $\Cal S$ and
$\sup_{\rho\in A}\Bvalue{\rho<\tau}=\Bvalue{\tau'<\tau}$ has measure at most
$\epsilon$.   Now

\Centerline{$R_{A}(\hatpmb u)
=\family{\sigma}{\Cal S}{\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}}
=\family{\sigma}{\Cal S}{u_{\sigma\wedge\tau'}}$}

\noindent agrees with $\pmb u$ on $\Cal S\wedge\tau$ and is constant on
$\Cal S\vee\tau$, so is a martingale (622E(b-ii)).
As $\epsilon$ is arbitrary, $\pmb u$ is an approximately local martingale.

\medskip

\quad{\bf (ii)} Now suppose that $\pmb u$ is an approximately local martingale.
Take $\tau\in\hat\Cal S$ and $\epsilon>0$.   As in part (b) of the proof of
622N, there is a $\tilde\tau\in\Cal S$ such that 
$\bar\mu\Bvalue{\tilde\tau<\tau}\le\epsilon$.   Now there is a non-empty
downwards-directed set $A\subseteq\Cal S$ such that 
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tilde\tau}\le\epsilon$ and $R_A(\pmb u)$
is a martingale.   Of course $A\wedge\tilde\tau$ is now a non-empty
downwards-directed subset of $\Cal S$ and

$$\eqalign{\sup_{\rho\in A\wedge\tilde\tau}\bar\mu\Bvalue{\rho<\tau}&
=\sup_{\rho\in A}\bar\mu\Bvalue{\rho\wedge\tilde\tau<\tau}
\le\sup_{\rho\in A}
   \bar\mu(\Bvalue{\rho<\tilde\tau}\Bcup\Bvalue{\tilde\tau<\tau})\cr&
\le\sup_{\rho\in A}
   (\bar\mu\Bvalue{\rho<\tilde\tau}+\bar\mu\Bvalue{\tilde\tau<\tau})
\le 2\epsilon,\cr}$$

\noindent while $R_{A\wedge\tilde\tau}(\pmb u)=R_{\{\tilde\tau\}}R_A(\pmb u)$ 
(623Cb) is a martingale (623E) and $R_{A\wedge\tilde\tau}(\hatpmb u)$ 
is the fully adapted extension of
$R_{A\wedge\tilde\tau}(\pmb u)$ (623Bd) and is therefore a local martingale 
(622Nb).   Accordingly 
$R_{A\wedge\tilde\tau}(\hatpmb u)\restr\hat\Cal S\wedge\tilde\tau$ is a 
martingale.  On the other side of $\tilde\tau$,

\Centerline{$R_{A\wedge\tilde\tau}(\hatpmb u)\restr\hat\Cal S\vee\tilde\tau
=R_AR_{\{\tilde\tau\}}(\hatpmb u)\restr\hat\Cal S\vee\tilde\tau$}

\noindent is constant, so $R_{A\wedge\tilde\tau}(\hatpmb u)$ is a martingale 
(622E(b-ii) again).

\medskip

\quad{\bf (iii)} This time, suppose that $\pmb u$ is a virtually local
martingale, so that $\hatpmb u$ is an approximately local martingale, and that
$\Cal S$ is finitely full.   Take $\tau\in\Cal S$ and $\epsilon>0$.   Then there
is a non-empty downwards-directed set $A\subseteq\hat\Cal S$ such that
$R_A(\hatpmb u)$ is a martingale and 
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$.   
As we can replace $A$ with $A\wedge\tau$ we can suppose that 
$A\subseteq\hat\Cal S\wedge\tau$.   Set
$B=\bigcup_{\rho\in A}\{\sigma:\sigma\in\Cal S$, $\rho\le\sigma\}$.
Then $B$ is a downwards-directed subset of $\Cal S$ and

\Centerline{$\sup_{\sigma\in B}\bar\mu\Bvalue{\sigma<\tau}
\le\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}
\le\epsilon$.}

\noindent The point is that $R_{B}(\pmb u)=R_A(\hatpmb u)\restr\Cal S$.
\Prf\ Express $\hatpmb u$, $R_A(\hatpmb u)$ and $R_{B}(\pmb u)$ as
$\family{\sigma}{\hat\Cal S}{\hat u_{\sigma}}$,
$\family{\sigma}{\hat\Cal S}{\hat u'_{\sigma}}$ and
$\family{\sigma}{\Cal S}{\tilde u_{\sigma}}$ respectively.

Take any $\sigma\in\Cal S$ and $\eta>0$;  then 
$\hat u'_{\sigma}=\lim_{\rho\downarrow A}\hat u_{\sigma\wedge\rho}$ and
$\tilde u_{\sigma}=\lim_{\rho\downarrow B}u_{\sigma\wedge\rho}$.
Let $\tilde\rho\in B$ be such that 
$\theta(\tilde u_{\sigma}-u_{\sigma\wedge\rho})\le\eta$ whenever
$\rho\in B$ and $\rho\le\tilde\rho$.   As there is a $\rho\in A$ such
that $\rho\le\tilde\rho$, we can find a $\hat\rho\in A$ such that
$\hat\rho\le\tilde\rho$ and 
$\theta(\hat u'_{\sigma}-\hat u_{\sigma\wedge\hat\rho})\le\eta$.
Because $\hat\rho\in\hat\Cal S$, 
$\sup_{\upsilon\in\Cal S}\Bvalue{\hat\rho=\upsilon}=1$ and there is a finite set
$I\subseteq\Cal S$ such that $\bar\mu a\ge 1-\eta$ where
$a=\sup_{\upsilon\in I}\Bvalue{\hat\rho=\upsilon}$.   Now 
$a\in\frak A_{\hat\rho}\subseteq\frak A_{\tau}$ so there is a $\rho^*\in\Cal T$
such that $a\Bsubseteq\Bvalue{\rho^*=\hat\rho}$ and 
$1\Bsetminus a\Bsubseteq\Bvalue{\rho^*=\tau}$  (611I).   We have

\Centerline{$\sup_{\upsilon\in I\cup\{\tau\}}\Bvalue{\rho^*=\upsilon}
\Bsupseteq\sup_{\upsilon\in I}
  (\Bvalue{\rho^*=\hat\rho}\Bcap\Bvalue{\hat\rho=\upsilon})
     \Bcup\Bvalue{\rho^*=\tau}
\Bsupseteq a\Bcup(1\Bsetminus a)=1$;}

\noindent as $\Cal S$ is finitely full, $\rho^*\in\Cal S$.   Consider
$\rho=\rho^*\wedge\tilde\rho$.   As
$\hat\rho=\hat\rho\wedge\tau\le\rho^*$ (611I), $\hat\rho\le\rho$ and $\rho\in B$.   Also

\Centerline{$a\Bsubseteq\Bvalue{\rho^*=\hat\rho}
=\Bvalue{\rho^*=\hat\rho\wedge\tilde\rho}
=\Bvalue{\rho^*=\rho}$,}

\noindent so 

\Centerline{$a\Bsubseteq\Bvalue{\rho=\hat\rho}
\Bsubseteq\Bvalue{\sigma\wedge\rho=\sigma\wedge\hat\rho}
\Bsubseteq\Bvalue{u_{\sigma\wedge\rho}=\hat u_{\sigma\wedge\hat\rho}}$}

\noindent and 

\Centerline{$\theta(u_{\sigma\wedge\rho}-\hat u_{\sigma\wedge\hat\rho})
\le\bar\mu(1\Bsetminus a)\le\eta$.}

\noindent But since also
$\theta(\tilde u_{\sigma}-u_{\sigma\wedge\rho})\le\eta$ because
$\rho\in B$ and $\rho\le\tilde\rho$, while 
$\theta(\hat u'_{\sigma}-\hat u_{\sigma\wedge\hat\rho})\le\eta$
by the choice of $\hat\rho$, we see that

\Centerline{$\theta(\tilde u_{\sigma}-\hat u'_{\sigma})
\le\theta(\tilde u_{\sigma}-u_{\sigma\wedge\rho})
   +\theta(u_{\sigma\wedge\rho}-\hat u_{\sigma\wedge\hat\rho})
   +\theta(\hat u'_{\sigma}-\hat u_{\sigma\wedge\hat\rho})
\le 3\eta$.}

\noindent As $\sigma$ and $\eta$ are arbitrary, 
$R_A(\hatpmb u)\restr\Cal S=R_{B}(\pmb u)$.\ \Qed

Consequently $R_B(\pmb u)$ is a martingale (622D(b-ii)).   As $\tau$ and
$\epsilon$ are arbitrary, $\pmb u$ is an approximately local martingale.

\medskip

{\bf (c)} As observed in (a-i) of the proof of
623B, we can regard $R_A$ either as an operator on
$\Mlmo(\Cal S)$ or as an operator on $\Mlmo(\hat\Cal S)$.
Again take $\tau\in\hat\Cal S$ and $\epsilon>0$, and let
$B\subseteq\hat\Cal S$ be a non-empty downwards-directed set such that
$\sup_{\rho\in B}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$ and
$R_B(\hatpmb u)$ is a martingale.   Then
$R_BR_A(\hatpmb u)=R_AR_B(\hatpmb u)$ (623Cb) is a martingale (623E).
As $\tau$ and $\epsilon$ are arbitrary, $R_A(\pmb u)$ is a martingale.

\medskip

{\bf (d)} If $\pmb u=\sfamilysS{u}$ is a virtually local martingale and
$\tau\in\hat\Cal S$, let $\epsilon>0$.   Then there is a
non-empty downwards-directed set $A\subseteq\hat\Cal S$ such that
$R_A(\hatpmb u)$ is a martingale and $\bar\mu a\le\epsilon$, where
$a=\sup_{\rho\in A}\Bvalue{\rho<\tau}$.   Express the martingale
$R_A(\hatpmb u)$
as $\pmb u'=\family{\sigma}{\hat\Cal S}{u'_{\sigma}}$.
This is a local integrator
(622H), so $\pmb u'\restr\hat\Cal S\wedge\tau$ is an integrator.   If
$\sigma\in\hat\Cal S\wedge\tau$, then

\Centerline{$1\Bsetminus a\Bsubseteq\Bvalue{\tau\le\rho}
\Bsubseteq\Bvalue{\sigma\le\rho}\Bsubseteq\Bvalue{\sigma\wedge\rho=\sigma}
\Bsubseteq\Bvalue{\hat u_{\sigma\wedge\rho}=\hat u_{\sigma}}$}

\noindent for every $\rho\in A$, so

\Centerline{$u'_{\sigma}\times\chi(1\Bsetminus a)
=\lim_{\rho\downarrow A}\hat u_{\sigma\wedge\rho}\times\chi(1\Bsetminus a)
=\lim_{\rho\downarrow A}\hat u_{\sigma}\times\chi(1\Bsetminus a)
=u_{\sigma}\times\chi(1\Bsetminus a)$.}

\noindent But this means that
$\Bvalue{u'_{\sigma}\ne\hat u_{\sigma}}\Bsubseteq a$;  as
$\sigma$ is arbitrary,
$\Bvalue{\pmb u'\restr\hat\Cal S\wedge\tau\ne\hatpmb u\restr\Cal S\wedge\tau}
\Bsubseteq a$ has measure at most $\epsilon$.   As $\epsilon$ is arbitrary,
$\hatpmb u\restr\Cal S\wedge\tau$ is an integrator (616P(b-iii));
as $\tau$ is arbitrary, $\hatpmb u$ is a local integrator.   Now
616Q(b-i) tells us that $\pmb u$ is a local integrator,
therefore locally moderately oscillatory (616Ib once more).

\medskip

{\bf (e)(i)}\grheada\ 
If $\pmb u$ is an approximately local martingale and $\tau\in\Cal S$,
let $\epsilon>0$.
Then there is a non-empty downwards-directed $A\subseteq\Cal S$ such that
$\bar\mu\Bvalue{\rho<\tau}\le\epsilon$ for every $\rho\in A$ and
$R_A(\hatpmb u)$ is a martingale.   We have

\Centerline{$\bar\mu\Bvalue{\rho\wedge\tau<\tau}
=\bar\mu\Bvalue{\rho<\tau}\le\epsilon$}

\noindent for every $\rho\in A$.   Write
$A\wedge\tau$ for $\{\rho\wedge\tau:\rho\in A\}$;  then
$A\wedge\tau$ is a non-empty downwards-directed subset of
$\Cal S\wedge\tau$
and $\bar\mu\Bvalue{\rho<\tau}\le\epsilon$ for every $\rho\in A\wedge\tau$.
Now

$$\eqalign{R_{A\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)&
=\family{\sigma}{\Cal S\wedge\tau}
    {\lim_{\rho\downarrow A\wedge\tau}u_{\sigma\wedge\rho}}
=\family{\sigma}{\Cal S\wedge\tau}
    {\lim_{\rho\downarrow A}u_{\sigma\wedge\rho\wedge\tau}}\cr&
=\family{\sigma}{\Cal S\wedge\tau}
    {\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}}
=R_A(\pmb u)\restr\Cal S\wedge\tau\cr}$$

\noindent is a martingale.   As $\epsilon$ is arbitrary,
$\pmb u\restr\Cal S\wedge\tau$ is an approximately
virtually local martingale (see the remarks in 623J).

\medskip

\qquad\grheadb\ If $\pmb u\restr\Cal S\wedge\tau$ is an approximately local
martingale for every $\tau\in\Cal S$, take $\tau\in\Cal S$ and $\epsilon>0$.   
Then there is a non-empty downwards-directed set
$A\subseteq\Cal S\wedge\tau$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$
and $R_A(\pmb u\restr\Cal S\wedge\tau)$ is a martingale.   As $\tau$ and
$\epsilon$ are arbitrary, $\pmb u$ is an approximately local martingale.

\medskip

{\bf (ii)}\grheada\ 
If $\pmb u$ is a virtually local martingale and $\tau\in\Cal S$, then
$\hatpmb u$ is an approximately local martingale.   By (i), 
$\hatpmb u\restr\hat\Cal S\wedge\tau$ is an approximately local martingale.
But $\hatpmb u\restr\hat\Cal S\wedge\tau$ is the fully adapted extension of
$\pmb u\restr\Cal S\wedge\tau$ to the covered envelope of $\Cal S\wedge\tau$,
so $\pmb u\restr\Cal S\wedge\tau$ is a virtually local martingale.

\medskip

\qquad\grheadb\ If $\pmb u\restr\Cal S\wedge\tau$ is a virtually local
martingale for every $\tau\in\Cal S$, that is,
$\hatpmb u\restr\hat\Cal S\wedge\tau$ is an approximately local martingale
for every $\tau\in\Cal S$, take any $\sigma\in\hat\Cal S$ and
$\epsilon>0$.   Let $\tau\in\Cal S$ be such that
$\bar\mu\Bvalue{\tau<\sigma}\le\bover12\epsilon$ (611Mh).
Then there is a non-empty downwards-directed set
$A\subseteq\hat\Cal S\wedge\tau$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\bover12\epsilon$
and $R_A(\hatpmb u\restr\hat\Cal S\wedge\tau)$ is a martingale.
By 623Cb, 
$R_{A\wedge\sigma\wedge\tau}(\hatpmb u\restr\hat\Cal S\wedge\tau)
=R_{\{\sigma\wedge\tau\}}R_A(\hatpmb u\restr\hat\Cal S\wedge\tau)$ 
is a martingale and
$R_{A\wedge\sigma\wedge\tau}(\hatpmb u\restr\hat\Cal S\wedge\sigma\wedge\tau)$ 
is a martingale;  as 
$R_{A\wedge\sigma\wedge\tau}(\hatpmb u\restr\hat\Cal S\vee(\sigma\wedge\tau))$ 
is constant, 
$R_{A\wedge\sigma\wedge\tau}(\hatpmb u)$ 
is a martingale (622E(b-ii) once more).   Now

\Centerline{$\bar\mu\Bvalue{\rho\wedge\sigma\wedge\tau<\sigma}
\le\bar\mu\Bvalue{\rho<\sigma\wedge\tau}+\bar\mu\Bvalue{\tau<\sigma}
\le\epsilon$}

\noindent for every $\rho\in A$, that is, 
$\bar\mu\Bvalue{\rho<\sigma}\le\epsilon$ for every 
$\rho\in A\wedge\sigma\wedge\tau$.   As $\sigma$ and $\epsilon$ are arbitrary,
$\hatpmb u$ is an approximately local martingale and $\pmb u$ is a virtually
local martingale.

\medskip

{\bf (f)} Let $\pmb u=\sfamilysS{u}$ be a uniformly integrable approximately 
local
martingale.   Set $C=\{u_{\sigma}:\sigma\in\Cal S\}$, so that $C$ is uniformly
integrable.   Then its closure $\overline{C}$ (for the topology of convergence
in measure) is also uniformly integrable (621B(c-ii)).
Take $\tau\le\tau'$ in $\Cal S$ and $\epsilon>0$.   Let $\delta>0$ be such that
$\|u\times\chi a\|_1\le\epsilon$ whenever $u\in\overline{C}$ and
$\bar\mu a\le\delta$ (621Ba).
Then there is a downwards-directed set $A\subseteq\Cal S$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau'}\le\delta$ and
$R_A(\pmb u)=\sfamilysS{u'}$ is a martingale.   For $\sigma\in\Cal S$, we have
$u'_{\sigma}=\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}\in\overline{C}$
and if $\sigma\le\tau'$ then

\Centerline{$a=\Bvalue{u'_{\sigma}\ne u_{\sigma}}
\Bsubseteq\sup_{\rho\in A}\Bvalue{\sigma\wedge\rho\ne\sigma}
\Bsubseteq\sup_{\rho\in A}\Bvalue{\rho<\tau'}$}

\noindent has measure at most $\delta$, so

\Centerline{$\|u'_{\sigma}-u_{\sigma}\|_1
=\|u'_{\sigma}\times\chi a-u_{\sigma}\times\chi a\|_1\le 2\epsilon$.}

\noindent Consequently

$$\eqalign{\|P_{\tau}u_{\tau'}-u_{\tau}\|_1&
\le\|P_{\tau}u_{\tau'}-P_{\tau}u'_{\tau'}\|_1
   +\|P_{\tau}u'_{\tau'}-u'_{\tau}\|_1+\|u'_{\tau}-u_{\tau}\|_1\cr&
\le\|u_{\tau'}-u'_{\tau'}\|_1 + 0 + 2\epsilon
\le 4\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $P_{\tau}u_{\tau'}=u_{\tau}$;
as $\tau$ and $\tau'$ are arbitrary, $\pmb u$ is a martingale.

\medskip

{\bf (g)} By 615H, $\lim_{\sigma\downarrow\Cal S}u_{\sigma}
=\lim_{\sigma\downarrow\hat\Cal S}\hat u_{\sigma}$.   
Because $\hat\Cal S$ is non-empty. 
there is a non-empty downwards-directed $A\subseteq\hat\Cal S$ such that 
$\family{\sigma}{\hat\Cal S}{u'_{\sigma}}=R_A(\hatpmb u)$ is a martingale.
Take any $\tau\in\hat\Cal S$.
By 623B(c-i), $\lim_{\sigma\downarrow\hat\Cal S}u'_{\sigma}$ 
is defined and equal to
$\lim_{\sigma\downarrow\hat\Cal S}\hat u_{\sigma}
=\lim_{\sigma\downarrow\Cal S}u_{\sigma}$;  
by 622Ed, this common value belongs to $L^1$.   

\medskip

{\bf (h)(i)} To begin with, suppose that $\Cal S$ is full.
If $\tau'\in\Cal S\vee\tau$ and $\epsilon>0$, then there is a non-empty
downwards-directed set $A\subseteq\Cal S$ such that
$\bar\mu\Bvalue{\rho<\tau'}\le\epsilon$ for every $\rho\in A$
and $R_A(\pmb u)$ is a martingale.
Now $R_{\{\tau\}\wedge A}(\pmb u)=R_{\{\tau\}}R_A(\pmb u)$ is a martingale,
by 623Cc and 623D, so $R_{\{\tau\}\vee A}(\pmb u)-R_{\{\tau\}}(\pmb u)
=R_A(\pmb u)-R_{\{\tau\}\wedge A}(\pmb u)$ (623Ca) is a difference of
martingales, therefor a martingale (622D(b-i)).

It follows at once that
$R_{\{\tau\}\vee A}(\pmb u)\restr\Cal S\vee\tau
  -R_{\{\tau\}}(\pmb u)\restr\Cal S\vee\tau$ is a martingale.   But as
$\{\tau\}\vee A$ and $\{\tau\}$ are non-empty downwards-directed subsets of
$\Cal S\vee\tau$, we can speak of
$R_{\{\tau\}\vee A}(\pmb u\restr\Cal S\vee\tau)$ and
$R_{\{\tau\}}(\pmb u\restr\Cal S\vee\tau)$.   Now
$R_{\{\tau\}\vee A}(\pmb u\restr\Cal S\vee\tau)
=R_{\{\tau\}\vee A}(\pmb u)\restr\Cal S\vee\tau$ and

\Centerline{$R_{\{\tau\}}(\pmb u\restr\Cal S\vee\tau)
=u_{\tau}\tbf 1\restr\Cal S\vee\tau
=R_{\{\tau\}\vee A}(u_{\tau}\tbf 1\restr\Cal S\vee\tau)$.}

\noindent So we see that
$R_{\{\tau\}\vee A}((\pmb u-u_{\tau}\tbf 1)\restr\Cal S\vee\tau)$
is a martingale.   And of course $\bar\mu\Bvalue{\rho<\tau'}\le\epsilon$ for
every $\rho\in\{\tau\}\vee A$.   As $\tau'$ and $\epsilon$ are arbitrary,
$(\pmb u-u_{\tau}\tbf 1)\restr\Cal S\vee\tau$ is a virtually local martingale.

\medskip

\quad{\bf (ii)} For the general case, given that $\pmb u\in\Mvlm(\Cal S)$ and
$\tau\in\Cal S$, we know that $\hatpmb u\in\Mvlm(\hat\Cal S)$, so (i) tells us
that $(\hatpmb u-\hat u_{\tau}\tbf 1)\restr\hat\Cal S\vee\tau$ is a
virtually local martingale;  but this is just the fully adapted extension of
$(\pmb u-u_{\tau}\tbf 1)\restr\Cal S\vee\tau$ to the covered envelope
$\hat\Cal S\vee\tau$ of $\Cal S\vee\tau$ (611M(e-i)), so
$(\pmb u-u_{\tau}\tbf 1)\restr\Cal S\vee\tau$ is a virtually local martingale.
}%end of proof of 623K

\leader{623L}{Theorem} Let $\Cal S$ be a non-empty sublattice of $\Cal T$, and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a $\|\,\|_1$-bounded
approximately local martingale.   Write $\gamma$ for
$\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$.

(a) $\pmb v$ is an integrator, therefore moderately oscillatory, and
$\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is defined.

(b) $\bar v=\sup_{\sigma\in\Cal S}|v_{\sigma}|$ is defined in $L^0(\frak A)$,
and $\theta(\bar v)\le 2\sqrt{\gamma}$.

% qauery what if $\pmb v$ is a virtually local martingale?

\proof{{\bf (a)} If $z\in Q_{\Cal S}(d\pmb v)$, then
$\theta(\delta z)\le\delta+17\sqrt{\delta\gamma}$
for every $\delta>0$.   \Prf\ Express $z$ as
$\sum_{i=0}^{n-1}u_i\times(v_{\tau_{i+1}}-v_{\tau_i})$ where
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and
$u_{\tau_i}\in L^0(\frak A_{\tau_i})$, $\|u_{\tau_i}\|_{\infty}\le 1$ for each
$i\le n$.   Then there is a
non-empty, downwards-directed $A\subseteq\Cal S$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau_n}\le\delta$ and
$\sfamilysS{v'}=R_A(\pmb v)$ is a martingale.   Now
$\|v'_{\tau}\|_1\le\gamma$ for every $\tau\in\Cal S$, just because
the ball $\{z:\|z\|_1\le\gamma\}$ is closed for the topology of convergence in
measure (613Bc once more)
and $v'_{\tau}=\lim_{\rho\downarrow A}v_{\tau\wedge\rho}$ belongs to
$\overline{\{v_{\sigma}:\sigma\in\Cal S\}}$.

Setting $z'=\sum_{i=0}^{n-1}u_i\times(v'_{\tau_{i+1}}-v'_{\tau_i})$,
$\Bvalue{z'\ne z}\Bsubseteq\sup_{\rho\in A}\Bvalue{\rho<\tau_n}$
has measure at most $\delta$, so $\theta(\delta z)\le\theta(\delta z')+\delta$.
But we see from 621Hf that

\Centerline{$\theta(\delta z')
\le\sqrt{\delta\gamma}+\bar\mu\Bvalue{\delta|z'|>\sqrt{\delta\gamma}}
\le\sqrt{\delta\gamma}+\Bover{16\delta}{\sqrt{\delta\gamma}}\|v'_{\tau_n}\|_1
\le 17\sqrt{\delta\gamma}$}

\noindent and $\theta(\delta z)\le\delta+17\sqrt{\delta\gamma}$.\ \Qed

As $\delta$ is arbitrary, $Q_{\Cal S}(d\pmb v)$ is bounded and $\pmb v$ is an
integrator.   Accordingly it is moderately oscillatory (616Ib, as always) and
$\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is defined (615Ga).

\medskip

{\bf (b)} If $B\subseteq\Cal S$ is a finite set, then

\Centerline{$\bar\mu(\Bvalue{\sup_{\sigma\in B}|v_{\sigma}|>\beta})
\le\Bover{\gamma}{\beta}$}

\noindent for every $\beta>0$.  \Prf\ If $B$ is empty, this is trivial.
Otherwise, let $\delta>0$.   Write $I$ for the sublattice of $\Cal S$
generated by $B$, and take
$\tau_0\le\ldots\le\tau_n$ linearly generating the $I$-cells (611L).
Let $A\subseteq\Cal S$ be a non-empty downwards-directed set such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau_n}\le\delta$ and
$\sfamilysS{v'}=R_A(\pmb v)$ is a martingale.   Then
$(v'_{\tau_0},\ldots,v'_{\tau_n})$ is a finite martingale
adapted to $(\frak A_{\tau_0},\ldots,\frak A_{\tau_n})$,
so

\Centerline{$\bar\mu(\Bvalue{\sup_{i\le n}|v'_{\tau_i}|>\beta})
\le\Bover1{\beta}\|v'_{\tau_n}\|_1$}

\noindent by 621E, while $\|v'_{\tau_n}\|\le\gamma$, as in (a) above.
Again,
$\Bvalue{v_{\tau_i}\ne v'_{\tau_i}}
\Bsubseteq\sup_{\rho\in A}\Bvalue{\rho<\tau_n}$ for each $i$, so
$\bar\mu(\Bvalue{\sup_{i\le n}|v_{\tau_i}|>\beta})
\le\Bover{\gamma}{\beta}+\delta$.   But if we write
$v$ for $\sup_{i\le n}|v_{\tau_i}|$, then

\Centerline{$\Bvalue{|v_{\sigma}|\le v}
\Bsupseteq\sup_{i\le n}\Bvalue{v_{\sigma}=v_{\tau_i}}
\Bsupseteq\sup_{i\le n}\Bvalue{\sigma=\tau_i}
=1$}

\noindent for any $\sigma\in B$, by 611Ke.   So
$\sup_{\sigma\in B}|v_{\sigma}|\le v$ and
$\Bvalue{\sup_{\sigma\in B}|v_{\sigma}|>\beta}
\Bsubseteq\Bvalue{v>\gamma}$ has measure at most
$\Bover{\gamma}{\beta}+\delta$.   As $\delta$ is arbitrary,
$\bar\mu\Bvalue{\sup_{\sigma\in B}|v_{\sigma}|>\beta}
\le\Bover{\gamma}{\beta}$.\ \Qed

Accordingly $c_{\gamma}
=\sup_{\sigma\in\Cal S}\Bvalue{|v_{\sigma}|>\gamma}$ has
measure at most $\Bover{\gamma}{\beta}$, by 321D.   Since this
tends to $0$ as $\beta$ increases to $\infty$,
$\bar v=\sup_{\sigma\in\Cal S}|v_{\sigma}|$ is defined in $L^0(\frak A)$, and

\Centerline{$\bar\mu\Bvalue{\bar v>\beta}
=\bar\mu c_{\gamma}\le\Bover{\gamma}{\beta}$}

\noindent for every $\beta>0$ (364L(a-ii)).   Consequently

\Centerline{$\theta(\bar v)
\le\sqrt{\gamma}+\bar\mu\Bvalue{\bar v>\sqrt{\gamma}}
\le 2\sqrt{\gamma}$.}
}%end of proof of 623L

\leader{623M}{Doob's quadratic maximal inequality:  Proposition}
If $\Cal S$ is a non-empty sublattice of $\Cal T$,
$\pmb{v}=\sfamilysS{v}$ is an approximately local martingale, and
$\gamma=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_2$ is finite,
then $\pmb v$ is order-bounded and $\|\sup|\pmb v|\|_2\le 2\gamma$.

\proof{{\bf (a)} It will simplify things if we note at once that as
$\|\,\|_2$-bounded sets are uniformly integrable (621Be), $\pmb v$ is a
uniformly integrable approximately
local martingale and is actually a martingale (623Kf).

\medskip

{\bf (b)(i)} Let us
suppose to begin with that $\Cal S$ is finite and totally
ordered;  let $\langle\tau_i\rangle_{i\le n}$ be its increasing
enumeration.   Write $\bar v$ for $\sup|\pmb v|$ and $v_n$ for
$v_{\tau_n}$;
set $a_t=\Bvalue{\bar v>t}$ for $t\ge 0$.   Then
$t\bar\mu a_t\le\Expn(|v_n|\times\chi a_t)$ for every $t$, by 621E.

\medskip

\quad{\bf (ii)} We need to know that if $u\ge 0$ in $L^0(\frak A)$ then
$\Expn(u\times\bar v)=\int_0^{\infty}\Expn(u\times\chi a_t)dt$, where
$\int\ldots dt$ is integration with respect to Lebesgue measure.   \Prf\
If $u=\chi c$, then

$$\eqalignno{\Expn(u\times\bar v)
&=\int_0^{\infty}\bar\mu\Bvalue{u\times\bar v>t}dt
\cr\displaycause{by the definition of integration in
$L^1(\frak A,\bar\mu)$, see 365A and 365Da}&
=\int_0^{\infty}\bar\mu(c\Bcap a_t)dt
=\int_0^{\infty}\Expn(u\times\chi a_t)dt.
\cr}$$

\noindent Because $\Expn$ is a linear functional,
$\Expn(u\times\bar v)=\int_0^{\infty}\Expn(u\times\chi a_t)dt$ for
every $u$ in $S(\frak A)$ as defined in \S361.   Generally, given
$u\in L^0(\frak A)^+$, there is a non-decreasing sequence
$\sequencen{u_n}$
in $S(\frak A)^+$ with supremum $u$, and now
$\{u_n\times\bar v:w\in S(\frak A)^+$, $n\in\Bbb N\}$ is a
non-decreasing sequence with supremum $u\times\bar v$
(353Pa\formerly{3{}53Oa}), so

\Centerline{$\Expn(u\times \bar v)
=\sup_{n\in\Bbb N}\Expn(u_n\times \bar v)
=\sup_{n\in\Bbb N}\int_0^{\infty}\Expn(u_n\times\chi a_t)dt
=\int_0^{\infty}\Expn(u\times\chi a_t)dt$}

\noindent because
$\Expn(u\times\chi a_t)=\sup_{n\in\Bbb N}\Expn(u_n\times\chi a_t)$ for
every $t$, so we can use B.Levi's theorem at the last step.\ \Qed

\medskip

\quad{\bf (iii)} Note also that $v_{\tau_i}^2\le P_i(v^2_n)$ for every
$i\le n$, where $P_i$ is the conditional expectation associated with
$\frak A_{\tau_i}$, by Jensen's inequality (621Cd).   So
$\Expn(v_{\tau_i}^2)\le\Expn(v_n^2)$ is finite for every $i$.

\medskip

\quad{\bf (iv)} We see now that

$$\eqalignno{\|\bar v\|_2^2
&=\Expn(\bar v^2)
=\int_0^{\infty}\bar\mu\Bvalue{\bar v^2>s}ds
=\int_0^{\infty}\bar\mu\Bvalue{\bar v>\sqrt s}ds
=\int_0^{\infty}2t\bar\mu\Bvalue{\bar v>t}dt\cr&
=2\int_0^{\infty}t\bar\mu a_tdt
\le 2\int_0^{\infty}\Expn(|v_n|\times\chi a_t)dt
=2\Expn(|v_n|\times\bar v)
\Displaycause{by (i) and (ii)}
\le 2\|v_n\|_2\|\bar v\|_2
\cr}$$

\noindent by Cauchy's inequality (244Eb).   Since we know that

\Centerline{$\|\bar v\|_2\le\sum_{i=0}^n\|v_{\tau_i}\|_2
\le(n+1)\gamma$}

\noindent is finite,
$\|\bar v\|_2\le 2\|v_n\|_2$.

\medskip

{\bf (c)} If $\Cal S$ is any non-empty finite sublattice of $\Cal T$, let
$(\sigma_0,\ldots,\sigma_n)$ linearly generate the $\Cal S$-cells;
then $\bar v=\sup_{i\le n}|v_{\sigma_i}|$ (612Dd)
and we can apply (b) to see that
$\|\bar v\|_2\le 2\|v_{\max\Cal S}\|_2$.
In general, setting
$\bar v_I=\sup_{\sigma\in I}|v_{\sigma}|$ when
$I\in\Cal I(\Cal S)$, starting with $\bar v_{\emptyset}=0$,
$\|\bar v_I\|_2\le 2\gamma$ for every $I$, while
$\langle\bar v_I^2\rangle_{I\in\Cal I(\Cal S)}$ is upwards-directed,
so

$$\eqalignno{\Expn(\bar v^2)
&=\Expn(\sup_{I\in\Cal I(\Cal S)}\bar v_I^2)
=\sup_{I\in\Cal I(\Cal S)}\Expn(\bar v_I^2)
\cr\displaycause{365Df}&
\le\sup_{I\in\Cal I(\Cal S)}4\gamma^2;
\cr}$$

\noindent taking square roots, we have the result in the general case.
}%end of proof of 623M

\cmmnt{\medskip

\noindent{\bf Remark} See 275Yd.}
	
\leader{623N}{Theorem} Let $\Cal S$ be a
non-empty sublattice of $\Cal T$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ an approximately local martingale.

(a) $\pmb v$ is a martingale iff $\pmb v\restr\Cal S\wedge\tau$ is
uniformly integrable for every $\tau\in\Cal S$.

(b) The following are equiveridical:

\quad(i) $\pmb v$ is uniformly integrable;

\quad(ii) there is a $z\in L^1$ such that
$\pmb v=\pmb Pz\restr\Cal S$;

\quad(iii) $\{v_{\sigma}:\sigma\in\Cal S\}$ is $\|\,\|_1$-bounded
and $\|v_{\uparrow}\|_1\ge\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$, where
$v_{\uparrow}=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$;

\quad(iv) $\pmb v$ is a martingale and the limit
$\1lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is defined in $L^1$.

\proof{{\bf (a)(i)} If $\pmb v$ is a martingale and $\tau\in\Cal S$, then
$\pmb v\restr\Cal S\wedge\tau=\pmb Pv_{\tau}\restr\Cal S\wedge\tau$
is uniformly integrable by 622Fa.

\medskip

\quad{\bf (ii)} Now suppose that $\pmb v\restr\Cal S\wedge\tau$
is uniformly integrable for every $\tau\in\Cal S$.
If $\sigma\le\tau$ in $\Cal S$, $\pmb v\restr\Cal S\wedge\tau$ is a
uniformly integrable approximately local martingale (623Ke) so is a martingale
(623Kf) and $v_{\sigma}=P_{\sigma}v_{\tau}$.   As $\sigma$ and $\tau$ are
arbitrary, $\pmb v$ is a martingale.

\medskip

{\bf (b)(i)$\Rightarrow$(iv)} If (i) is true, then (a) tells us that
$\pmb v$ is a martingale.   Of course
$A=\{v_{\sigma}:\sigma\in\Cal S\}$ is
$\|\,\|_1$-bounded, so 623La tells us that
$z=\lim_{\sigma\uparrow\Cal S}v_{\sigma}$ is defined
and belongs to $L^1$.
Now $A\cup\{z\}$ is uniformly integrable, so the
topology of convergence in measure and the topology defined by $\|\,\|_1$
agree on $A\cup\{z\}$, and
$z=\1lim_{\sigma\uparrow\Cal S}v_{\sigma}$.   Thus (iv) is true.

\medskip

\quad{\bf (iv)$\Rightarrow$(iii)} Suppose that (iv) is true.   Observe
first that

\Centerline{$\|v_{\sigma}\|_1=\|P_{\sigma}v_{\tau}\|_1
\le\|v_{\tau}\|_1$}

\noindent whenever $\sigma\le\tau$ in $\Cal S$.   Now we are supposing that
$z=\1lim_{\sigma\uparrow\Cal S}v_{\sigma}$.
Because the embedding $L^1\embedsinto L^0$ is continuous
(613B(d-i)), $z$ is also $\lim_{\sigma\uparrow\Cal S}v_{\sigma}$.
Moreover,

\Centerline{$\|z\|_1=\1lim_{\sigma\uparrow\Cal S}\|v_{\sigma}\|_1
=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$}

\noindent because $\sigma\mapsto\|v_{\sigma}\|_1:\Cal S\to\Bbb R$ is
non-decreasing.   So (iii) is true.

\medskip

\quad{\bf (iii)$\Rightarrow$(ii)} Suppose that (iii) is true.
Set $\gamma=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$.   By 623La, the
limit $v_{\uparrow}$ is defined;
we are supposing that $\|v_{\uparrow}\|_1\ge\gamma$.   In fact we must have
equality, because $\|\,\|_1$-balls are closed;  in particular,
$v_{\uparrow}\in L^1_{\bar\mu}$.

\medskip

\qquad\grheada\ $\pmb v$ is a martingale.
\Prf\ Take $\sigma_0\le\sigma_1$ in $\Cal S$, and $\epsilon>0$.   Since
$v_{\uparrow}\in\overline{\{v_{\tau}:\tau\in\Cal S\vee\sigma_1\}}$,
$\|v_{\uparrow}\|_1\le\sup_{\tau\in\Cal S\vee\sigma_1}\|v_{\tau}\|_1$ and there
is a $\tau\in\Cal S$ such that $\sigma_1\le\tau$ and
$\|v_{\tau}\|_1\ge\|v_{\uparrow}\|_1-\epsilon\ge\gamma-\epsilon$.

Let $\delta>0$ be such that $\Expn(\|v_{\sigma}\times\chi a\|_1)\le\epsilon$
whenever $\sigma\in\{\sigma_0,\sigma_1,\tau\}$ and $\bar\mu a\le\delta$.
Let $A\subseteq\Cal S$ be a non-empty downwards-directed set such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\delta$ and
$\sfamilysS{v'}=R_A(\pmb v)$ is a martingale.   Set
$a=\sup_{\rho\in A}\Bvalue{\rho<\tau}$, so that $\bar\mu a\le\delta$.
Since $v'_{\tau}=\lim_{\rho\downarrow A}v_{\tau\wedge\rho}$,

\Centerline{$\|v'_{\tau}\|_1
\le\sup_{\rho\in A}\|v_{\tau\wedge\rho}\|_1\le\gamma$.}

\noindent
On the other hand, $1\Bsetminus a\Bsubseteq\Bvalue{v'_{\tau}=v_{\tau}}$, so

\Centerline{$\Expn(|v'_{\tau}\times\chi(1\Bsetminus a)|)
=\Expn(|v_{\tau}\times\chi(1\Bsetminus a)|)
=\|v_{\tau}\|_1-\Expn(|v_{\tau}\times\chi a|)
\ge\gamma-2\epsilon$.}

\noindent So $\Expn(|v'_{\tau}\times\chi a|)\le 2\epsilon$.   Next, for
$i$ either $0$ or $1$,

\Centerline{$a_i=\Bvalue{v'_{\sigma_i}\ne v_{\sigma_i}}\Bsubseteq a$,
\quad$\bar\mu a_i\le\delta$}

\noindent and $a_i\in\Sigma_{\sigma_i}$.   So

$$\eqalign{\Expn(|v'_{\sigma_i}|\times\chi a_i)&
=\Expn(|P_{\sigma_i}v'_{\tau}|\times\chi a_i)
\le\Expn(P_{\sigma_i}|v'_{\tau}|\times\chi a_i)\cr&
=\Expn(P_{\sigma_i}(|v'_{\tau}|\times\chi a_i))
=\Expn(|v'_{\tau}|\times\chi a_i)
\le 2\epsilon,\cr}$$

\noindent while also $\Expn(|v_{\sigma_i}|\times\chi a_i)\le\epsilon$,
so $\|v_{\sigma_i}-v'_{\sigma_i}\|_1\le 3\epsilon$.

Now we see that

\Centerline{$\|v_{\sigma_0}-P_{\sigma_0}v_{\sigma_1}\|_1
\le 6\epsilon+\|v'_{\sigma_0}-P_{\sigma_0}v'_{\sigma_1}\|_1
=6\epsilon$.}

\noindent As $\sigma_0$, $\sigma_1$ and $\epsilon$ are arbitrary,
$\pmb v$ is a martingale.\ \Qed

\medskip

\qquad\grheadb\ Now note that $v_{\uparrow}=\1lim_{\tau\uparrow\Cal S}v_{\tau}$.
\Prf\ Let $\epsilon>0$.   By 613D(b-iv), there is a $\delta>0$ such that
$\|z-v_{\uparrow}\|_1\le\epsilon$ whenever $\|z\|_1\le\|v_{\uparrow}\|_1$
and $\theta(z-v_{\uparrow})\le\delta$.   But there is a $\tau\in\Cal S$ such
that $\theta(v_{\sigma}-v_{\uparrow})\le\delta$ whenever
$\sigma\in\Cal S\vee\tau$, so $\|v_{\sigma}-v_{\uparrow}\|_1\le\epsilon$ for
every $\sigma\in\Cal S\vee\tau$.\ \QeD\  But now we see that

\Centerline{$v_{\sigma}
=\1lim_{\tau\uparrow\Cal S}P_{\sigma}v_{\tau}
=P_{\sigma}(\1lim_{\tau\uparrow\Cal S}v_{\tau})
=P_{\sigma}v_{\uparrow}$}

\noindent for every $\sigma\in\Cal S$.   So (ii) is true.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} As in (a-i), this is immediate from 622Fa.
}%end of proof of 623N

\leader{623O}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb u=\sfamilysS{u}$ a locally moderately oscillatory process
and $\pmb v=\sfamilysS{v}$ a virtually local martingale.
Then $\iivu$ is a virtually local martingale.

\proof{ It is worth noting straight away that because $\pmb u$ is
locally moderately oscillatory
and $\pmb v$ is a local integrator (623Kd), $\iivu$ is certainly defined
everywhere on $\Cal S$.   If $A\subseteq\Cal S$ is non-empty and
downwards-directed, $R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ will be the corresponding
operator as described in 623B.

\medskip

\noindent{\bf part A} Suppose for the time being that $\Cal S$ is full and
has a greatest
element, $\pmb v$ is a martingale, and $\|\sup|\pmb u|\|_{\infty}\le 1$.
Then $\pmb v=\pmb{P}v_{\max\Cal S}\restr\Cal S$ is an integrator and is
order-bounded.

\medskip

{\bf (a)} Let $\epsilon>0$.   Consider the $\|\,\|_1$-bounded
martingale $\tildepmb v=\sfamilysS{\tilde v}$ where
$\tilde v_{\sigma}=P_{\sigma}|v_{\max\Cal S}|$ for $\sigma\in\Cal S$.
This is order-bounded (622G);  let $M\ge 0$ be such that $\bar\mu a\le\epsilon$
where $a=\Bvalue{\sup|\tildepmb v|\ge M}$.   Set

\Centerline{$A=\{\rho:\rho\in\Cal S$,
$\Bvalue{\rho<\max\Cal S}\Bsubseteq\Bvalue{|\tilde v_{\rho}|\ge M}\}$}

\noindent as in 623I.   Note that

\Centerline{$\Bvalue{\pmb w\ne R_A(\pmb w)}
\Bsubseteq\sup_{\rho\in A}\Bvalue{\rho<\max\Cal S}\Bsubseteq a$}

\noindent for every $\pmb w\in\Mlmo(\Cal S)$.
Express $\pmb v'=R_A(\pmb v)$ as $\sfamilysS{v'}$
and $\tildepmb v'=R_A(\tildepmb v)$ as $\sfamilysS{\tilde v'}$.

\medskip

{\bf (b)} Now $ii_{\pmb v'}(\pmb u)$ is a martingale.

\medskip

\Prf{\bf (i)} Let $\delta>0$.  Then there is an $\eta>0$ such that
$(2+\bover{M}{\delta})\eta+\sqrt{M\eta}\le\delta$.   Let $M'\ge M$ be such
that $\|(|v_{\max\Cal S}|-M'\chi 1)^+\|_1\le\eta$.   Set
$w=\med(-M'\chi 1,v_{\max\Cal S},M'\chi 1)$,
$w_1=(v_{\max\Cal S}-M'\chi 1)^+$ and $w_2=(-v'_{\max\Cal S}-M'\chi 1)^+$;
then
$v_{\max\Cal S}=w+w_1-w_2$, $w$ is square-integrable, $w_1\ge 0$, $w_2\ge 0$,
$\|w_1\|_1\le\eta$ and $\|w_2\|_1\le\eta$.   Consider the martingales
$\pmb w=\family{\sigma}{\Cal S}{P_{\sigma}w}$,
$\pmb w_1=\family{\sigma}{\Cal S}{P_{\sigma}w_1}$,
$\pmb w_2=\family{\sigma}{\Cal S}{P_{\sigma}w_2}$ and the associated
martingales $\pmb w'=R_A(\pmb w)=\sfamilysS{w'}$,
$\pmb w_1'=R_A(\pmb w_1)=\family{\sigma}{\Cal S}{w'_{1\sigma}}$
and $\pmb w'_2=R_A(\pmb w_2)=\family{\sigma}{\Cal S}{w'_{2\sigma}}$.

\medskip

\quad{\bf (ii)} We have $\|P_{\sigma}w\|_2\le\|w\|_2\le M'$ for every
$\sigma$ (244M, 366H(b-iii)), so
$\|w'_{\sigma}\|_2=\|\lim_{\rho\downarrow A}P_{\sigma\wedge\rho}w\|_2\le M'$
for every $\sigma\in\Cal S$ (613Bc).   Thus $\pmb w'$ is a $\|\,\|_2$-bounded
martingale, while we are supposing that $\pmb u$ is
moderately oscillatory, so $ii_{\pmb w'}(\pmb u)$ is a martingale (622Q).

\medskip

\quad{\bf (iii)} Observe that

$$\eqalign{0\le w'_{1\sigma}&
=\lim_{\rho\downarrow A}P_{\sigma\wedge\rho}w_1
\le\lim_{\rho\downarrow A}P_{\sigma\wedge\rho}|v_{\max\Cal S}|\cr&
=\lim_{\rho\downarrow A}\tilde v_{\sigma\wedge\rho}
=\tilde v'_{\sigma}\cr}$$

\noindent for every $\sigma\in\Cal S$.
Now $R_A(\pmb w'_1)=\pmb w'_1$ (623Cc), so if $\sigma\le\tau$ in $\Cal S$,

\Centerline{$\Bvalue{w'_{1\sigma}>M}
\Bsubseteq\Bvalue{\tilde v'_{\sigma}>M}
   \Bsubseteq\Bvalue{w'_{1\sigma}=w'_{1\tau}}$}

\noindent by 623I(b-i).   It follows that
if $\tau_0\le\ldots\le\tau_n$ in
$\Cal S$, then $(w'_{1\tau_0},\ldots,w'_{1\tau_n})$ is a non-negative martingale
and

\Centerline{$\Bvalue{w'_{1\tau_i}>M}
\Bsubseteq\Bvalue{w'_{1\tau_i}=w'_{1\tau_j}}
   \Bcap\Bvalue{w'_{1\tau_i}=w'_{1\tau_n}}
\Bsubseteq\Bvalue{w'_{1\tau_j}=w'_{1\tau_n}}$}

\noindent whenever $i\le j\le n$.
So $(\frak A_{\tau_0},\ldots,\frak A_{\tau_n})$,
$(w'_{1\tau_0},\ldots,w'_{1\tau_n})$
and $(u_{\tau_0},\ldots,u_{\tau_n})$
satisfy the conditions of 621I.   Moreover,

$$\eqalignno{\|w'_{1\tau_n}\|_1&
\le\sup_{\sigma\in\Cal S}\|P_{\sigma}w_1\|_1
\Displaycause{623B(c-ii)}
\le\|w_1\|_1
\le\eta.\cr}$$

\noindent So 621I tells us that

$$\eqalign{\|\sum_{i=0}^{n-1}
  u_{\tau_i}\times(w'_{1\tau_{i+1}}-w'_{1\tau_i})\|_1&
\le\delta+(2+\Bover{M}{\delta})\|w'_{1\tau_n}\|_1
  +\sqrt{M\|w'_{1\tau_n}\|_1}\cr&
\le\delta+(2+\Bover{M}{\delta})\eta+\sqrt{M\eta}
\le 2\delta\cr}$$

\noindent by the choice of $\eta$.

Re-expressing this in the standard form I am using for Riemann sums, we
have

\Centerline{$\|S_I(\pmb u,d\pmb w_1')\|_1\le 2\delta$}

\noindent for every finite sublattice $I$ of $\Cal S$.
Again because $\|\,\|_1$-balls are closed for the topology of convergence in
measure, it follows that
$\|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb w_1'\|_1\le 2\delta$ for every
$\tau\in\Cal S$.

\medskip

\quad{\bf (iv)} Similarly.
$\|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb w_2'\|_1\le 2\delta$ for every
$\tau\in\Cal S$.   But now recall that $v_{\max\Cal S}=w+w_1-w_2$
so $\pmb{P}v_{\max\Cal S}=\pmb{P}w+\pmb{P}w_1-\pmb{P}w_2$,
$\pmb v=\pmb w+\pmb w_1-\pmb w_2$,
$\pmb v'=\pmb w'+\pmb w'_1-\pmb w'_2$ and

\Centerline{$\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v'
=\int_{\Cal S\wedge\tau}\pmb u\,d\pmb w'
+\int_{\Cal S\wedge\tau}\pmb u\,d\pmb w'_1
-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb w'_2$}

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

\Centerline{$\|\int_{\Cal S\wedge\tau}\pmb u\,d\pmb v'
-\int_{\Cal S\wedge\tau}\pmb u\,d\pmb w'\|_1\le 4\delta$}

\noindent for every $\tau\in\Cal S$.   Since $ii_{\pmb w'}(\pmb u)$ is a
martingale, by (ii) above, and $\delta$ is arbitrary,
$ii_{\pmb v'}(\pmb u)$ is a martingale (622Ec).\ \Qed

\medskip

{\bf (c)} Recall now that

\Centerline{$ii_{\pmb v'}(\pmb u)=ii_{R_A(\pmb v)}(\pmb u)
=R_A(\iivu)$}

\noindent by 623G.   So $R_A(\iivu)$ is a martingale.
Since $\epsilon$ was arbitrary,
$\iivu$ is a virtually local martingale (see the remarks in 623J).

\medskip

\noindent{\bf part B} I set out to strip away the extra hypotheses demanded in
part A.

\medskip

{\bf (a)} Of course it will be enough to suppose that $\Cal S$ is full and
has a greatest element, $\pmb v$ is a martingale
and $\|\sup|\pmb u|\|_{\infty}$ is finite, since then we can apply (A) to
a non-zero multiple of $\pmb u$.

\medskip

{\bf (b)} Now suppose that $\Cal S$ is full and
has a greatest element and that $\pmb v$ is a martingale.   In this case,
$\pmb u$ is still moderately oscillatory and order-bounded.   Let $\epsilon>0$,
and take $M\ge 0$ such that
$\bar\mu\Bvalue{\sup|\pmb u|\ge M}\le\bover12\epsilon$.   Set

\Centerline{$A=\{\rho:\rho\in\Cal S$,
$\Bvalue{\rho<\max\Cal S}\Bsubseteq\Bvalue{|u_{\rho}|\ge M}\}$}

\noindent as in 623I.   Then $A$ is non-empty and downwards-directed and
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\max\Cal S}\le\bover12\epsilon$.
Set $\pmb u'=\sfamilysS{u'}=R_A(\pmb u)$,
$\pmb v'=\sfamilysS{v'}=R_A(\pmb v)$ and
$\tildepmb u=\sfamilysS{\tilde u}=\med(-M\tbf 1,\pmb u',M\tbf 1)$.
Then $\pmb u'$ and $\tildepmb u$ are moderately oscillatory,
$\pmb v'$ is an integrator (623D) and $\|\sup|\tildepmb u|\|_{\infty}\le M$ is
finite.   If $\sigma\le\tau$ in $\Cal S$, then

\Centerline{$\Bvalue{\tilde u_{\sigma}\ne u'_{\sigma}}
\Bsubseteq\Bvalue{|u'_{\sigma}|>M}
\Bsubseteq\Bvalue{v_{\sigma}=v_{\tau}}$}

\noindent (623I(b-i) again), so
$\tilde u_{\sigma}\times(v'_{\tau}-v'_{\sigma})
=u'_{\sigma}\times(v'_{\tau}-v'_{\sigma})$.   Accordingly
$S_I(\tildepmb u,d\pmb v')=S_I(\pmb u',d\pmb v')$ for every $I\in\Cal I(\Cal S)$
and $ii_{\pmb v'}(\pmb u')=ii_{\pmb v'}(\tildepmb u)$.   But $\pmb v'$ is a
martingale (623E again), so (a) just above tells us that
$ii_{\pmb v'}(\tildepmb u)$ is a virtually local martingale and
accordingly $ii_{\pmb v'}(\pmb u')$ is a virtually local martingale.

Let $B\subseteq\Cal S$ be a non-empty downwards-directed set such that
$\sup_{\rho\in B}\bar\mu\Bvalue{\rho<\max\Cal S}\le\bover12\epsilon$
and $R_B(ii_{\pmb v'}(\pmb u'))$ is a martingale.   Setting
$\pmb v''=R_B(\pmb v')$ and $\pmb u''=R_B(\pmb u')$,
623G tells us that $ii_{\pmb v''}(\pmb u'')=R_B(ii_{\pmb v'}(\pmb u'))$ is a
martingale.   But $\pmb v''=R_BR_A(\pmb v)=R_{A\wedge B}(\pmb v)$ (623Cb)
and similarly $\pmb u''=R_{A\wedge B}(\pmb u)$.   Applying 623G again, we see
that

\Centerline{$R_{A\wedge B}(\iivu)
=ii_{R_{A\wedge B}(\pmb v)}(R_{A\wedge B}(\pmb u))
=ii_{\pmb v''}(\pmb u'')$}

\noindent is a martingale.   And for any $\sigma\in\Cal S$ we have

$$\eqalign{\sup_{\rho\in A\wedge B}\bar\mu\Bvalue{\rho<\sigma}&
\le\sup_{\rho\in A\wedge B}\bar\mu\Bvalue{\rho<\max\Cal S}
=\sup_{\Atop{\rho\in A}{\rho'\in B}}\bar\mu\Bvalue{\rho\wedge\rho'<\max\Cal S}\cr&
=\sup_{\Atop{\rho\in A}{\rho'\in B}}
   \bar\mu(\Bvalue{\rho<\max\Cal S}\Bcup\Bvalue{\rho'<\max\Cal S})\cr&
\le\sup_{\Atop{\rho\in A}{\rho'\in B}}
   (\bar\mu\Bvalue{\rho<\max\Cal S}+\bar\mu\Bvalue{\rho'<\max\Cal S})
\le\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\iivu$ is a virtually local martingale.

\medskip

{\bf (c)} Thirdly, consider the case in which $\Cal S$ is full
and has a greatest element but $\pmb v$ is only a virtually local martingale.
Take $\epsilon>0$.   Then we have a non-empty downwards-directed set
$A\subseteq\Cal S$ such that
$\bar\mu\Bvalue{\rho<\max\Cal S}\le\bover12\epsilon$ for every $\rho\in A$
and $R_A(\pmb v)$ is a martingale.   By (b), $ii_{R_A(\pmb v)}(\pmb u)$ is
a virtually local martingale.   Let $B\subseteq\Cal S$ be a non-empty
downwards-directed set such that
$\sup_{\rho\in B}\bar\mu\Bvalue{\rho<\max\Cal S}\le\bover12\epsilon$
and $R_B(ii_{R_A(\pmb v)}(\pmb u))$ is a
martingale.   By 623G and 623Cb once more,

\Centerline{$R_{A\wedge B}(\iivu)=ii_{R_{A\wedge B}(\pmb v)}(\pmb u)
=ii_{R_BR_A(\pmb v)}(\pmb u)
=R_B(ii_{R_A(\pmb v)}(\pmb u))$}

\noindent is a martingale.   As in (b), we have
$\sup_{\rho\in A\wedge B}\bar\mu\Bvalue{\rho<\max\Cal S}\le\epsilon$.
As $\epsilon$ is arbitrary, $\iivu$ is a virtually local martingale.

\medskip

{\bf (d)} If we suppose only that $\Cal S$ is full, then for each
$\tau\in\Cal S$ we have a moderately oscillatory process
$\pmb u\restr\Cal S\wedge\tau$ and a virtually local martingale
$\pmb v\restr\Cal S\wedge\tau$ (623Ke again),
while $\Cal S\wedge\tau$ is full and has a greatest element.   By (c),
$ii_{\pmb v\restr\Cal S\wedge\tau}(\pmb u\restr\Cal S\wedge\tau)$ is a virtually
local martingale.   But this is just $\iivu\restr\Cal S\wedge\tau$.
So $\iivu$ is a virtually local martingale, by 623Ke in the other direction.

\medskip

{\bf (e)} Finally, in the general case, the covered envelope $\hat\Cal S$ of
$\Cal S$ is full, the fully adapted extension
$\hatpmb u$ of $\pmb u$ to $\hat\Cal S$ is locally
moderately oscillatory (615F(b-v))
and the fully adapted extension $\hatpmb v$ of $\pmb v$
is a virtually local martingale (623J).
So (d) tells us that $ii_{\hatpmb v}(\hatpmb u)$ is a virtually local
martingale.   But  $ii_{\hatpmb v}(\hatpmb u)$ is the fully adapted extension of
$\iivu$ (apply 613Uc to $\Cal S\wedge\tau$ for $\tau\in\Cal S$),
so $\iivu$ itself is a virtually local martingale (623J again).

This completes the proof.
}%end of proof of 623O

%\leader{623P}{}
%\leader{623Q}{}
%\leader{623R}{}
%\leader{623S}{}
%\leader{623T}{}
%\leader{623U}{}
%\leader{623V}{}
%\leader{623W}{}

\exercises{\leader{623X}{Basic exercises (a)}
%\spheader 623Xa
Let $\Cal S$ be a sublattice of $\Cal T$, $A\subseteq\Cal S$
a non-empty downwards-directed set and $\pmb u$ a locally moderately oscillatory
process with domain $\Cal S$.
(i) Show that if $\pmb u$ is order-bounded then $R_A(\pmb u)$ is order-bounded
and $\sup|R_A(\pmb u)|\le\sup|\pmb u|$,
(ii) Show that if $\pmb u$ is locally order-bounded then $R_A(\pmb u)$ is
locally order-bounded,
(iii) Show that if $\pmb u$ is of
bounded variation, then $R_A(\pmb u)$ is of bounded variation and
$\int_{\Cal S}|dR_A(\pmb u)|\le\int_{\Cal S}|d\pmb u|$.
(iv) Show that if $\pmb u$ is locally of
bounded variation, then $R_A(\pmb u)$ is locally of bounded variation.
(v) Show that if $1\le p\le\infty$ and $\pmb u$ is $\|\,\|_p$-bounded
(622Ca), then $R_A(\pmb u)$ is $\|\,\|_p$-bounded.
%623B

\spheader 623Xb Let $\Cal S$ be a sublattice of $\Cal T$ and
$A\subseteq\Cal S$ a non-empty downwards-directed set.   Let
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the operator described in 623B.
Show that if $\pmb u=\sfamilysS{u}$ is a local martingale then
$R_A(\pmb u)$ is a local martingale.
%623E

\spheader 623Xc Show that Brownian motion $\pmb w$ on $\Cal T_f$, as described
in 612T, is not an $L^1$-process, and that there is a $\tau\in\Cal T_f$ such
that $\pmb w\restr\Cal T_f\vee\tau$ is not a virtually local martingale.
%622L 632M

\spheader 623Xe Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a virtually local martingale.
Show that $\pmb u\restr\Cal S'$ is a virtually local martingale for
any ideal $\Cal S'$ of $\Cal S$.
%623K

\spheader 623Xf Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a virtually local martingale.   Show that
$u_{\sigma}=u_{\tau}$ whenever $\sigma$, $\tau\in\Cal S$ and
$\frak A_{\sigma}=\frak A_{\tau}$.
%623K out of order query

\spheader 623Xg Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ a fully adapted process.   Show that the
following are equiveridical:
($\alpha$) $\pmb u$ is a virtually local martingale and $u_{\tau}\in L^1$;
($\beta$) $\pmb u\restr\Cal S\wedge\tau$ and $\pmb u\restr\Cal S\vee\tau$
are virtually local martingales.
%623K

\spheader 623Xh Let $\Cal S$ be a sublattice of $\Cal T$.   Suppose that
$\pmb u\in\Mfa(\Cal S)$ and for every $\tau\in\Cal S$ and $\epsilon>0$ there is
a $\sigma\in\Cal S$ such that $\bar\mu\Bvalue{\sigma<\tau}\le\epsilon$ and
$\pmb u\restr\Cal S\wedge\sigma$ is a virtually local martingale.   Show that
$\pmb u$ is a virtually local martingale.
%623K 623Xe

\leader{623Y}{Further exercises (a)}
%\spheader 623Ya
Let $\Cal S$ be a sublattice
of $\Cal T$ and $\pmb u=\sfamilysS{u}$
a virtually local martingale.   Suppose that
$A\subseteq\Cal S$ is non-empty and downwards-directed,
$\inf A\in\Cal S$ and $\frak A_{\inf A}=\bigcap_{\sigma\in A}\frak A_{\sigma}$.
Show that $u_{\inf A}=\lim_{\sigma\downarrow A}u_{\sigma}$.
%623K 622G 622Xj mt62bits

}%end of exercises


\endnotes{
\Notesheader{623} At the price of a substantial effort, we have a theorem on
indefinite integrals with respect to virtually local martingales which
matches the form of the corresponding results on processes
of bounded variation (614U), integrators (616J)
and jump-free integrators (618Q).   But 623O
is especially important because martingales,
and in particular Brownian motion, are
central to any theory of stochastic integration.    And its difficulty lies
largely in the fact that an indefinite integral with respect to a martingale
is {\it not} necessarily a martingale (631Ya).   In the framework I have settled
on for this volume so far, `virtually local martingale' is the best I can do.
I ought to tell you that in the more conventional framework of right-continuous
filtrations (\S632), virtually local martingales on ideals of $\Cal T$
are actually local martingales (632Ib), and that
the corresponding special case of 623O is a good deal easier to prove, while
being a sufficient foundation for the standard theory.

Observe that 623La covers Doob's martingale convergence theorem (275G, 367Ja).
If we have a
$\|\,\|_1$-bounded virtually local martingale $\sfamilysS{v}$, it is moderately
oscillatory, so $\lim_{\sigma\uparrow\Cal S}v_{\sigma}=v_{\uparrow}$ is
defined, and
$\inf_{\tau\in\Cal S}\sup_{\sigma\in\Cal S\vee\tau}|v_{\sigma}-v_{\uparrow}|=0$
(615Ga), corresponding to order*-convergence in $L^0$ as defined in 367A, that
is, to
almost-everywhere pointwise convergence of sequences of measurable functions
(367F).

The main theorems of this section (623L-623O) %623L 623M 623N 623O
involve awkward shifts between `approximately local' and `virtually local',
which will have echoes later.   %624H
As long as we restrict ourselves to finitely full lattices, there is no
difference (623K(b-iii)).   I am reluctant to impose such a restriction
generally because the Riemann-sum integral does not insist on it (613T), and 
many of the ideas of this volume can be effectively expressed in terms of
lattices of constant stopping times.   Indeed applications often begin with
processes defined on such a lattice, as in 612H.
When we come to the structure theory
of integrators, the concept of `virtually local' martingale will provide
a particularly striking formulation of the main theorem (627Q).
}%end of notes

\discrpage



\frfilename{mt624.tex}
\versiondate{21.2.23/22.8.23}
\copyrightdate{2018}

\def\2lim{\mathop{2\mskip-1mu\text{lim}}}

\def\chaptername{Martingales}
\def\sectionname{Quadratic variation}

\newsection{624}

We are at last ready to determine the quadratic variation of Brownian motion
(624F).   I take the opportunity to tidy up some simple consequences of
results in \S\S617 and 623 (624B-624E), %624B 624C 624D 624E
and to give useful facts about $L^2$-martingales (624G-624I). %624G 624H 624I

\cmmnt{
\leader{624A}{Notation} \Asissix\ will be a stochastic integration structure.
I write $L^1$, $L^2$, $L^{\infty}$ for
$L^1(\frak A,\bar\mu)$, $L^2(\frak A,\bar\mu)$ and $L^{\infty}(\frak A)$
respectively.   $\Expn$ will be the integral on $L^1$
and $\theta$ will be the
F-norm defining the topology of convergence in measure on $L^0(\frak A)$, as in
613B.   If $\pmb v$ and $\pmb w$ are local integrators with the same
domain, $\covar{\pmb v}{\pmb w}$ will be their covariation (617H);
$\pmb v^*=\covar{\pmb v}{\pmb v}$ is the quadratic variation of $\pmb v$.
}

\leader{624B}{Theorem} Let $\Cal S$ be a non-empty sublattice of $\Cal T$
and $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$,
$\pmb w=\family{\sigma}{\Cal S}{w_{\sigma}}$ virtually local martingales
such that $v_{\downarrow}\times w_{\downarrow}\in L^1$ where
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ and
$w_{\downarrow}=\lim_{\sigma\downarrow\Cal S}w_{\sigma}$.
Then $\pmb v\times\pmb w-\covar{\pmb v}{\pmb w}$ is a
virtually local martingale.

\proof{ We know that $\pmb v$ and $\pmb w$ are local integrators (623Kd), so
the covariance $\covar{\pmb v}{\pmb w}$ is defined everywhere on $\Cal S$
(617Hb).   We have

\Centerline{$\pmb v\times\pmb w-\covar{\pmb v}{\pmb w}
=ii_{\pmb w}(\pmb v)+\iiv(\pmb w)+z\tbf{1}$}

\noindent where $z=v_{\downarrow}\times w_{\downarrow}
\in L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$;
as $z\in L^1$, $z\tbf{1}$ is a martingale.   But
$ii_{\pmb w}(\pmb v)$ and $\iiv(\pmb w)$ are virtually local
martingales, by 623O, so
$\pmb v\times\pmb w-\covar{\pmb v}{\pmb w}$ is a virtually local martingale.
}%end of proof of 624B

\leader{624C}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$, $\pmb w=\sfamilysS{w}$ local integrators.
If one of them is locally jump-free and one is locally
of bounded variation, then
$\covar{\pmb v}{\pmb w}=0$.   In particular, if $\pmb v$ is
locally jump-free and locally of bounded variation, then $\pmb v^*=0$.

\proof{{\bf (a)} Suppose that $\pmb v$ is locally jump-free and
$\pmb w$ is locally of
bounded variation.   Take any $\tau\in\Cal S$.
Set $\bar w=\int_{\Cal S\wedge\tau}|d\pmb w|$.
Take any $\epsilon>0$.   Let
$\delta>0$ be such that $\theta(z\times\bar w)\le\epsilon$ whenever
$\theta(z)\le\delta$.   Let $I\in\Cal I(\Cal S\wedge\tau)$ be such that
$\theta(\Osclln^*_I(\pmb v\restr\Cal S\wedge\tau))\le\delta$ (618B).
If $J\in\Cal I(\Cal S\wedge\tau)$,
$J\supseteq I$ and $e$ is a $J$-cell, express $e$ as $c(\sigma,\tau)$ where
$\sigma\le\tau$ in $J$;  then

\Centerline{$|\Delta_e(\tbf 1,d\pmb v\,d\pmb w)|
=|v_{\tau}-v_{\sigma}|\times|w_{\tau}-w_{\sigma}|
\le\Osclln_J(\pmb v)\times\Delta_e(\tbf 1,|d\pmb w|)$.}

\noindent Summing over the $J$-cells,

\Centerline{$|S_J(\tbf 1,d\pmb v\,d\pmb w)|
\le\Osclln_J(\pmb v)\times S_J(\tbf 1,|d\pmb w|)
\le\Osclln^*_I(\pmb v\restr\Cal S\wedge\tau)\times\bar w$}

\noindent and $\theta(S_J(\tbf 1,d\pmb v\,d\pmb w))\le\epsilon$.
As $\epsilon$ is arbitrary,

\Centerline{$\int_{\Cal S\wedge\tau}d\pmb v\,d\pmb w
=\lim_{J\uparrow\Cal I(\Cal S\wedge\tau)}S_J(\tbf 1,d\pmb v\,d\pmb w))=0$.}

\noindent This is true for every $\tau\in\Cal S$, so
$\covar{\pmb v}{\pmb w}=0$.

Of course the same arguments will apply if $\pmb v$ is locally of bounded
variation and $\pmb w$ is locally jump-free.

\medskip

{\bf (b)} If $\pmb v$ is both locally jump-free and locally of bounded
variation, then (a) tells us that $\pmb v^*=\covar{\pmb v}{\pmb v}=0$.
By 617M, it follows that $\covar{\pmb v}{\pmb w}=0$.   Similarly,
$\covar{\pmb v}{\pmb w}=0$ if $\pmb w$ is locally jump-free and locally of
bounded variation.
}%end of proof of 624C

\leader{624D}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$ a virtually local martingale.
Then the following are equiveridical:

(i) $\pmb v$ is constant;

(ii) $\pmb v$ is locally jump-free and locally of bounded variation;

(iii) the quadratic variation of $\pmb v$ is zero.

[\query\ For 624E/653G, want 
$\Bvalue{\pmb v\ne v_{\downarrow}\tbf 1}\Bsubseteq\Bvalue{\pmb v^*\ne\tbf 0}$]

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

\medskip

{\bf (ii)$\Rightarrow$(iii)} is immediate from 624C.

\medskip

{\bf (iii)$\Rightarrow$(i)} The result is trivial if $\Cal S$ is empty, so let
us suppose otherwise.   By 623Kd again, $\pmb v$ is locally integrable, so its
quadratic variation is defined everywhere on $\Cal S$.   By 623Kg,
$v_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v_{\sigma}$ is defined
and belongs to $L^1(\frak A,\bar\mu)$.

\medskip

\quad\grheada\ Suppose to begin with that $\Cal S$ is full and that
$v_{\downarrow}=0$.   Because the quadratic variation of $\pmb v$ is zero,
$\pmb v^2=2\iiv(\pmb v)$ is a virtually local martingale (623O again).
Take $\tau\in\Cal S$ and
$\epsilon>0$.   Let $A\subseteq\Cal S$ be a non-empty downwards-directed set
such that $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$ and
$\pmb z=R_A(\pmb v^2)$, as defined in 623B, is a martingale.   Express
$\pmb z$ as $\sfamilysS{z}$.   Then
$\lim_{\sigma\downarrow\Cal S}z_{\sigma}
=\lim_{\sigma\downarrow\Cal S}v^2_{\sigma}=0$ (623B(c-i)).   The set
$\{z_{\sigma}:\sigma\in\Cal S\wedge\tau\}
=\{P_{\sigma}z_{\tau}:\sigma\in\Cal S\wedge\tau\}$ is uniformly integrable
(621Cf), so the $\|\,\|_1$-limit $\1lim_{\sigma\downarrow\Cal S}z_{\sigma}$
is zero (621B(c-ii)) and

\Centerline{$\Expn(z_{\tau})
=\lim_{\sigma\downarrow\Cal S}\Expn(P_{\sigma}z_{\tau})
=\lim_{\sigma\downarrow\Cal S}\Expn(z_{\sigma})=0$.}

\noindent As $z_{\tau}=\lim_{\rho\downarrow A}v^2_{\tau\wedge\rho}\ge 0$,
$z_{\tau}=0$.

Now note that

\Centerline{$\Bvalue{v_{\tau}\ne 0}
=\Bvalue{v^2_{\tau}\ne z_{\tau}}
\Bsubseteq\sup_{\rho\in A}\Bvalue{v^2_{\tau}\ne v^2_{\tau\wedge\rho}}
\Bsubseteq\sup_{\rho\in A}\Bvalue{\rho<\tau}$}

\noindent has measure at most $\epsilon$ (because $A$ is downwards-directed).
As $\epsilon$ is arbitrary, $v_{\tau}=0$;  as $\tau$ is arbitrary,
$\pmb v=\tbf 0$ is constant.

\medskip

\quad\grheadb\ Next, suppose just that $\Cal S$ is full.
Set $w_{\sigma}=v_{\sigma}-v_{\downarrow}$ for $\sigma\in\Cal S$ and
$\pmb w=\sfamilysS{w}$.  Since
$v_{\downarrow}\in L^1(\frak A,\bar\mu)$,
$v_{\downarrow}\tbf 1$ is a martingale and $\pmb w=\pmb v-v_{\downarrow}\tbf 1$
is a virtually local martingale.   Next,
$(w_{\tau}-w_{\sigma})^2=(v_{\tau}-v_{\sigma})^2$ whenever $\sigma\le\tau$ in
$\Cal S$, so
$\int_{\Cal S\wedge\tau}(d\pmb w)^2=\int_{\Cal S\wedge\tau}(d\pmb v)^2$
for every $\tau\in\Cal S$, and the quadratic variation of $\pmb w$ is equal to
$\pmb v^*=\tbf 0$.   By ($\alpha$), $\pmb w=\tbf 0$ and $\pmb v$ is constant.

\medskip

\quad\grheadc\ Finally, for the general case, let $\hat{\Cal S}$ be the covered
envelope of $\Cal S$ and $\hatpmb v$ the fully adapted extension of $\pmb v$ to
$\hat{\Cal S}$.   As noted in 623J,
$\hatpmb v$ is a virtually local martingale.   Its quadratic variation is the
fully adapted extension of $\pmb v^*$ (617N) so is zero.   By ($\beta$),
$\hatpmb v$ is constant, so $\pmb v$ is.
}%end of proof of 624D

\leader{624E}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v=\sfamilysS{v}$ a virtually local martingale with
quadratic variation $\sfamilysS{v^*}$.
If $\tau$, $\tau'\in\Cal S$ are such that
$v^*_{\tau}=v^*_{\tau'}$, then $\pmb v$ is constant on
$\Cal S\cap[\tau\wedge\tau',\tau\vee\tau']$.

\query we need 
$\Bvalue{v_{\tau}\ne v_{\tau'}}\Bsubseteq\Bvalue{v_{\tau}^*\ne v_{\tau'}^*}$
for 653G

\proof{{\bf (a)} To begin with,
suppose that $\tau\le\tau'$.   Write $\Cal S_0$ for $\Cal S\cap[\tau,\tau']$.
As $\pmb v^*$ is
non-decreasing, $v^*_{\sigma}=v^*_{\tau}$, that is,
$\int_{\Cal S\wedge\sigma}(d\pmb v)^2=\int_{\Cal S\wedge\tau}(d\pmb v)^2$,
for every $\sigma\in\Cal S\cap[\tau,\tau']$.
By 623Kh and 623K(e-ii), $(\pmb v-v_{\tau}\tbf 1)\restr\Cal S\vee\tau$ and
$\pmb w=(\pmb v-v_{\tau}\tbf 1)\restr\Cal S_0$ are virtually
local martingales.   Now $\Delta\pmb w=\Delta\pmb v\restr\Cal S_0^{2\uparrow}$,
so

\Centerline{$\int_{\Cal S_0\wedge\sigma}(d\pmb w)^2
=\int_{\Cal S_0\wedge\sigma}(d\pmb v)^2
=\int_{\Cal S\wedge\sigma}(d\pmb v)^2-\int_{\Cal S'\wedge\tau}(d\pmb v)^2
=0$}

\noindent for every $\sigma\in\Cal S$, that is, the quadratic variation of
$\pmb w$ is zero.   By 624D, $\pmb w$ and $\pmb v=\pmb w+v_{\tau}\tbf 1$
are constant on $\Cal S_0$.

\medskip

{\bf (b)} This deals with the case $\tau\le\tau'$.   But for the general case,
given that $v^*_{\tau}=v^*_{\tau'}$, we have

$$\eqalign{\Bvalue{\tau\le\tau'}&
\Bsubseteq\Bvalue{\tau\wedge\tau'=\tau}\Bcap\Bvalue{\tau\vee\tau'=\tau'}\cr&
\Bsubseteq\Bvalue{v^*_{\tau\wedge\tau'}=v^*_{\tau}}
   \Bcap\Bvalue{v^*_{\tau\vee\tau'}=v^*_{\tau'}}
\Bsubseteq\Bvalue{v^*_{\tau\wedge\tau'}=v^*_{\tau\vee\tau'}}\cr}$$

\noindent and similarly
$\Bvalue{\tau'\le\tau}
\Bsubseteq\Bvalue{v^*_{\tau\wedge\tau'}=v^*_{\tau\vee\tau'}}$.
So $v^*_{\tau\wedge\tau'}=v^*_{\tau\vee\tau'}$ and (a) shows that
$\pmb v$ is constant on $\Cal S\cap[\tau\wedge\tau',\tau\vee\tau']$.
}%end of proof of 624E

\leader{624F}{Theorem} Let
$\pmb w=\family{\tau}{\Cal T_f}{w_{\tau}}$ be Brownian motion.   Then its
quadratic variation $\pmb w^*$ is the identity process
$\pmb{\iota}$.

\proof{ By 622L, $\pmb w$ and $\pmb w^2-\pmb\iota$ are local martingales.
Next, $\pmb w^2-\pmb w^*=2ii_{\pmb w}(\pmb w)$ is a virtually
local martingale, by 624B, so $\pmb w^*-\pmb{\iota}$ is a
virtually local martingale.
Because both $\pmb w^*$ and $\pmb{\iota}$ are non-decreasing,
$\pmb w^*-\pmb{\iota}$ is locally of bounded variation.
Now recall that $\pmb w$ is locally jump-free (618Jc), so
$\pmb w^*$ is locally jump-free (618T), while of course $\pmb{\iota}$
is locally jump-free (618Ja).   So $\pmb w^*-\pmb{\iota}$ is a virtually local
martingale, locally
jump-free and locally of bounded variation, and must be constant (624D).
Since it starts from $0$ at $\min\Cal T$, it is zero everywhere and
$\pmb w^*=\pmb{\iota}$, as claimed.
}%end of proof of 624F

\leader{624G}{}\cmmnt{ We saw in 622Q that indefinite integrals with respect to
$L^2$-martingales are particularly easy to analyze.   Here I give a
characterization and a striking property of such martingales.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ an $L^2$-martingale
and $\pmb v^*=\sfamilysS{v^*}$ its quadratic variation.   Then
$\Expn(v^*_{\tau})\le\Expn(v_{\tau}^2)$ for every $\tau\in\Cal S$.

\proof{ As before, $\pmb v^*$ is defined everywhere on $\Cal S$.   Now
$\Expn(S_I(\tbf 1,(d\pmb v)^2))\le\Expn(v_{\max I}^2)$ for
every non-empty $I\in\Cal I(\Cal S)$.   \Prf\ Take $(\tau_0,\ldots,\tau_n)$
linearly generating the $I$-cells.   If $i<n$, then

\Centerline{$P_{\tau_i}(v_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i}))
=v_{\tau_i}\times P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})=0$,}

\noindent so

\Centerline{$\Expn(v_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i}))
=\Expn(P_{\tau_i}(v_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})))
=0$.}

\noindent Consequently

$$\eqalign{\Expn(S_I(\tbf 1,(d\pmb v)^2))&
=\Expn(\sum_{i=0}^{n-1}(v_{\tau_{i+1}}-v_{\tau_i})^2)\cr&
=\sum_{i=0}^{n-1}\Expn(v_{\tau_{i+1}})-\Expn(v_{\tau_i})^2)
   -2\Expn(v_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i}))\cr&
=\Expn(v_{\tau_n}^2)-\Expn(v_{\tau_0}^2)
\le\Expn(v_{\max I}^2).  \text{ \Qed}\cr}$$

Now if $\tau\in\Cal S$,

\Centerline{$v^*_{\tau}
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\tbf 1,(d\pmb v)^2)
\in\overline{\{x:\|x\|_1\le\Expn(v_{\tau}^2)\}}$.}

\noindent Bu $\|\,\|_1$-balls are closed for the topology of convergence in
measure (613Bc), so $\Expn(v^*_{\tau})=\|v^*_{\tau}\|_1\le\Expn(v_{\tau}^2)$, as
clsimed.
}%end of proof of 624G

\leader{624H}{Proposition} Let $\Cal S$ be a non-empty sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ a virtually local martingale with starting value 
$0$, and $\pmb v^*=\sfamilysS{v^*}$ its quadratic variation.

(a) $\|v_{\tau}\|_2\le\sqrt{\|v^*_{\tau}\|_1}$ for every $\tau\in\Cal S$.

(b) If moreover $\pmb v$ is an approximately local martingale 
and $\pmb v^*$ is an $L^1$-process, then $\pmb v$ and $\iiv(\pmb v)$ are
martingales, and $\|v_{\tau}\|_2=\sqrt{\|v^*_{\tau}\|_1}$ for every
$\tau\in\Cal S$.

\cmmnt{\medskip

\noindent{\bf Remark} Recall that I count $\|v_{\tau}\|_2$ as $\infty$ if
$v_{\tau}\in L^0$ is not square-integrable, and $\|v^*_{\tau}\|_1$ as $\infty$
if $v^*_{\tau}\in(L^0)^+$ does not have finite expectation;  while of course
$\sqrt{\infty}$ is to be interpreted as $\infty$.}

\proof{{\bf (a)(i)} To begin with, suppose that $\Cal S$ is full.
As in 624B $\pmb v^*$ is well-defined and $\pmb v^*-\pmb v^2$ is a
virtually local martingale.   Take $\tau\in\Cal S$ and $\epsilon>0$.
Then there is a non-empty downwards-directed subset $A$ of $\Cal S$ such that
$a=\sup_{\rho\in A}\Bvalue{\rho<\tau}$ has measure at most $\epsilon$ and
$R_A(\pmb v^*-\pmb v^2)$ is a martingale, where
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ is defined as in 623B.   Now
$R_A(\pmb v)^*=R_A(\pmb v^*)$, by 623H, so
$\pmb z=R_A(\pmb v^*)-R_A(\pmb v)^2$ is a martingale.   The starting value of
$R_A(\pmb v)$ is $0$, by 623B(c-i), as is the starting value of
$R_A(\pmb v^*)$, so the starting value of $\pmb z$ is $0$;  expressing
$\pmb z$ as $\sfamilysS{z}$, $0$ is the conditional expectation of
$z_{\tau}$ on $\bigcap_{\sigma\in\Cal S}\frak A_{\sigma}$ (622Ed), so
$0=\Expn(z_{\tau})=\Expn(v^*_{A\tau}-v_{A\tau}^2)$, where
$v_{A\tau}=\lim_{\rho\downarrow A}v_{\rho\wedge\tau}$ and
$v^*_{A\tau}=\lim_{\rho\downarrow A}v^*_{\rho\wedge\tau}$.   Accordingly

\Centerline{$\|v_{A\tau}^2\|_1=\Expn(v_{A\tau}^2)=\Expn(v^*_{A\tau})
=\|v^*_{A\tau}\|_1\le\|v^*_{\tau}\|_1$}

\noindent because $0\le v^*_{A\tau}\le v^*_{\tau}$.

Now observe that

\Centerline{$\Bvalue{v_{A\tau}^2\ne v_{\tau}^2}
\Bsubseteq\Bvalue{v_{A\tau}\ne v_{\tau}}
\Bsubseteq\sup_{\rho\in A}\Bvalue{v_{\rho\wedge\tau}\ne v_{\tau}}
\Bsubseteq\sup_{\rho\in A}\Bvalue{\rho<\tau}=a$}

\noindent so $\theta(v^2_{\tau}-v^2_{A\tau})\le\epsilon$.   As $\epsilon$ is
arbitrary, $v_{\tau}^2$ belongs to the closure of
$\{x:\|x\|_1\le\|v^*_{\tau}\|_1\}$.   But this is a closed set for the topology
of convergence in measure (613Bc again),
so $\|v_{\tau}^2\|_1\le\|v^*_{\tau}\|_1$,
that is, $\|v_{\tau}\|_2\le\sqrt{\|v^*_{\tau}\|_1}$.

\medskip

\quad{\bf (ii)} For the general case, let $\hat\Cal S$ be the covered envelope
of $\Cal S$ and $\hatpmb v$ the fully adapted extension of $\pmb v$ to
$\hat\Cal S$,   Then $\hat\Cal S$ is full and $\hatpmb v$ is a virtually local
martingale (623J) with starting value $0$ (615H, as $\pmb v$ and $\hatpmb v$ are
locally moderately oscillatory).   Express $\hatpmb v$ as
$\family{\sigma}{\hat\Cal S}{\hat v_{\sigma}}$ and its quadratic variation
$\hatpmb v^*$ as $\family{\sigma}{\hat\Cal S}{\hat v^*_{\sigma}}$.
By 617N, $\hatpmb v^*$ extends $\pmb v^*$.   So (a) tells us that

\Centerline{$\|v_{\tau}\|_2=\|\hat v_{\tau}\|_2
\le\sqrt{\|\hat v^*_{\tau}\|_1}=\sqrt{\|v^*_{\tau}\|_1}$}

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

\medskip

{\bf (b)} Now suppose that $\pmb v$ is an approximately local martingale and
$\pmb v^*$ is an $L^1$-process, that is,
$\|v^*_{\tau}\|_1$ and therefore $\|v_{\tau}\|_2$ are finite
for every $\tau\in\Cal S$.

\medskip

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

\Centerline{$\|v_{\sigma}\|_2\le\sqrt{\|v^*_{\sigma}\|_1}
\le\sqrt{\|v^*_{\tau}\|_1}$}

\noindent for every $\sigma\in\Cal S\wedge\tau$, and
$\{v_{\sigma}:\sigma\in\Cal S\wedge\tau\}$ is $\|\,\|_2$-bounded, therefore
uniformly integrable (621Be), that is, $\pmb v\restr\Cal S\wedge\tau$ is
uniformly integrable.   By 623Na, $\pmb v$ is a martingale.

As $\pmb v$ is also an $L^2$-process, 624G tells us that
$\sqrt{\|v^*_{\tau}\|_1}\le\|v_{\tau}\|_2$ for every $\tau\in\Cal S$,
so we have equality.

\medskip

\quad{\bf (ii)} As for $\iiv(\pmb v)$, take $\tau\le\tau'$ in $\Cal S$ and
$c\in\frak A_{\tau}$.   For $\sigma\in\Cal S\vee\tau$ set

\Centerline{$w_{\sigma}=P_{\sigma}((v_{\tau'}-v_{\tau})\times\chi c)$}

\noindent so that $\pmb w=\family{\sigma}{\Cal S\vee\tau}{w_{\sigma}}$
is a martingale.
Write $\pmb w^*=\family{\sigma}{\Cal S\vee\tau}{w^*_{\sigma}}$
for its quadratic variation.   We see that
if $\sigma\in\Cal S\vee\tau$ then $c\in\frak A_{\sigma}$ so

\Centerline{$w_{\sigma}=P_{\sigma}(v_{\tau'}-v_{\tau})\times\chi c
=(v_{\sigma}-v_{\tau})\times\chi c$}

\noindent because $\pmb v$ is a martingale;
in particular, the starting value $w_{\tau}$ of $\pmb w$ is $0$.
As $(v_{\tau'}-v_{\tau})\times\chi c\in L^2$, $\pmb w$ is an $L^2$-martingale
and $\Expn(w^*_{\sigma})\le\Expn(w_{\sigma}^2)$ is finite for every
$\sigma\in\Cal S\vee\tau$, by 624G.   But from (a) we know that
$\Expn(w_{\sigma}^2)\le\Expn(w^*_{\sigma})$, so we have equality.

Now $w^*_{\tau'}-w^*_{\tau}=(v^*_{\tau'}-v^*_{\tau})\times\chi c$.   \Prf\
If $\tau_0\le\ldots\le\tau_n$ in $\Cal S\cap[\tau,\tau']$,

\Centerline{$\sum_{i=0}^n(w_{\tau_{i+1}}-w_{\tau_i})^2
=\sum_{i=0}^n(v_{\tau_{i+1}}-v_{\tau_i})^2\times\chi c$.}

\noindent So $S_I(\tbf 1,(d\pmb w)^2)=S_I(\tbf 1,(d\pmb v)^2)\times\chi c$
for every finite sublattice $I$ of $\Cal S\cap[\tau,\tau']$.   Taking the
limit as $I\uparrow\Cal I(\Cal S\cap[\tau,\tau'])$,

$$\eqalign{w^*_{\tau'}&
=\int_{\Cal S\cap[\tau,\tau']}(d\pmb w)^2
=\lim_{I\uparrow\Cal I(\Cal S\cap[\tau,\tau'])}S_I(\tbf 1,(d\pmb w)^2)\cr&
=\lim_{I\uparrow\Cal I(\Cal S\cap[\tau,\tau'])}
  S_I(\tbf 1,(d\pmb v)^2)\times\chi c\cr&
=\int_{\Cal S\cap[\tau,\tau']}(d\pmb v)^2\times\chi c
=(v^*_{\tau'}-v^*_{\tau})\times\chi c.  \text{ \Qed}\cr}$$

Turning to expectations, we have

$$\eqalign{\Expn((v^*_{\tau'}-v^*_{\tau})\times\chi c)&
=\Expn(w^*_{\tau'})
=\Expn(w_{\tau'}^2)
=\Expn((v_{\tau'}-v_{\tau})^2\times\chi c)\cr&
=\Expn((v_{\tau'}^2-v_{\tau}^2)\times\chi c)
   +2\Expn(v_{\tau}\times(v_{\tau'}-v_{\tau})\times\chi c).\cr}$$

\noindent But

$$\eqalign{\Expn(v_{\tau}\times(v_{\tau'}-v_{\tau})\times\chi c)&
=\Expn(P_{\tau}(v_{\tau}\times(v_{\tau'}-v_{\tau})\times\chi c))\cr&
=\Expn(v_{\tau}\times P_{\tau}(v_{\tau'}-v_{\tau})\times\chi c)
=0\cr}$$

\noindent so

\Centerline{$\Expn((v^*_{\tau'}-v^*_{\tau})\times\chi c)
=\Expn((v_{\tau'}^2-v_{\tau}^2)\times\chi c)$,}

\noindent that is to say,

\Centerline{$\Expn((v^*_{\tau'}-v_{\tau'}^2)\times\chi c)
=\Expn((v^*_{\tau}-v_{\tau}^2)\times\chi c)$.}

\noindent As $c$ is arbitrary,
$P_{\tau}(v^*_{\tau'}-v_{\tau'}^2)=v^*_{\tau}-v_{\tau}^2$.
As $\tau$ and $\tau'$ are arbitrary, $\pmb v^*-\pmb v^2$ is a
martingale.   But as $\pmb v$ has starting value $0$,

\Centerline{$\iiv(\pmb v)=\Bover12(\pmb v^2-\pmb v^*)$}

\noindent is a martingale, and the proof is complete.
}%end of proof of 624H

\leader{624I}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ a $\|\,\|_2$-bounded martingale with quadratic
variation $\pmb v^*$, and
$\pmb u=\sfamilysS{u}$ a $\|\,\|_{\infty}$-bounded moderately oscillatory
process.   Then
$\Expn((\int_{\Cal S}\pmb u\,d\pmb v)^2)$ and
$\Expn(\int_{\Cal S}\pmb u^2d\pmb v^*)$ are finite and equal.

\proof{ By 622Q, $\pmb z=\iivu$ is a $\|\,\|_2$-bounded martingale,
and of course its starting value is $0$ (613J(f-i)).   Writing
$\pmb z^*=\sfamilysS{z^*}$ for its quadratic variation, and
expressing $\pmb z$ as $\sfamilysS{z}$, we have
$\Expn(z^*_{\tau})=\Expn(z_{\tau}^2)$ for every $\tau\in\Cal S$, by 624G and
624Hb.   We know also that

\Centerline{$z^*_{\tau}
=\int_{\Cal S\wedge\tau}d\pmb z^*
=\int_{\Cal S\wedge\tau}\pmb u^2d\pmb v^*$}

\noindent for every $\tau\in\Cal S$, by 617Qb.

If $\Cal S$ has a greatest member, we just take $\tau=\max\Cal S$.   In general,
we need to check the limits as $\tau\uparrow\Cal S$.   By 622Q again,
$\Expn(z_{\tau}^2)\le\|\pmb u\|_{\infty}^2\Expn(v_{\tau}^2)$ for every
$\tau\in\Cal S$.   Since $\pmb v$ is $\|\,\|_2$-bounded,

\Centerline{$\beta
=\sup_{\tau\in\Cal S}\Expn(z_{\tau}^*)=\sup_{\tau\in\Cal S}\Expn(z^2_{\tau})$}

\noindent is finite.   Now $\pmb z^*$ is non-decreasing, so

$$\eqalign{\Expn(\int_{\Cal S}\pmb u^2d\pmb v^*)&
=\Expn(\lim_{\tau\uparrow\Cal S}\int_{\Cal S\wedge\tau}\pmb u^2d\pmb v^*)
=\Expn(\lim_{\tau\uparrow\Cal S}z^*_{\tau})\cr&
=\Expn(\sup_{\tau\in\Cal S}z^*_{\tau})
=\sup_{\tau\in\Cal S}\Expn(z_{\tau}^*)
=\beta\cr}$$

\noindent (613B(d-iii)).   On the other side, we know that if $\sigma\le\tau$ in
$\Cal S$ then

\Centerline{$\Expn((z_{\tau}-z_{\sigma})^2)
=\Expn((z_{\tau}^2-z_{\sigma}^2-2z_{\sigma}\times(z_{\tau}-z_{\sigma}))
=\Expn(z_{\tau}^2)-\Expn(z_{\sigma}^2)$}

\noindent as in (a-i) of the proof of 624H.   Generally, for $\sigma$,
$\tau\in\Cal S$,

$$\eqalignno{\|z_{\tau}-z_{\sigma}\|_2^2&
=\Expn((z_{\tau}-z_{\sigma})^2)
=\Expn((z_{\sigma\vee\tau}-z_{\sigma\wedge\tau})^2)
\Displaycause{612D(f-ii)}
=\Expn(z_{\sigma\vee\tau}^2)-\Expn(z_{\sigma\wedge\tau}^2)
\le\beta-\Expn(z_{\sigma\wedge\tau}^2)
\to 0\cr}$$

\noindent as $\sigma$, $\tau\uparrow\Cal S$.   So the $\|\,\|_2$-limit
$\2lim_{\tau\uparrow\Cal S}z_{\tau}$ is defined and must be equal to the limit
$\lim_{\tau\uparrow\Cal S}z_{\tau}=\int_{\Cal S}\pmb u\,d\pmb v$ for the
topology of convergence in measure (613B(d-i));  moreover,

$$\eqalign{\Expn((\int_{\Cal S}\pmb u\,d\pmb v)^2)&
=\|\int_{\Cal S}\pmb u\,d\pmb v\|_2^2
=\|\2lim_{\tau\uparrow\Cal S}z_{\tau}\|_2^2\cr&
=\lim_{\tau\uparrow\Cal S}\|z_{\tau}\|_2^2
=\beta
=\Expn(\int_{\Cal S}\pmb u^2d\pmb v^*),\cr}$$

\noindent as claimed.
}%end of proof of 624I

%\leader{624J}{}
%\leader{624K}{}
%\leader{624L}{}
%\leader{624M}{}
%\leader{624N}{}
%\leader{624O}{}
%\leader{624P}{}
%\leader{624Q}{}
%\leader{624R}{}
%\leader{624S}{}
%\leader{624T}{}
%\leader{624U}{}
%\leader{624V}{}
%\leader{624W}{}

\exercises{\leader{624X}{Basic exercises (a)}
%\spheader 624Xa
(i) Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a martingale.   Suppose that
$\tau$, $\tau'\in\Cal S$ are such that $v_{\tau}=v_{\tau'}$.   Show that
$\pmb v$ is constant on $\Cal S\cap[\tau\wedge\tau',\tau\vee\tau']$.
(ii) Give an example in which there
is a local martingale $\pmb v=\family{\sigma}{\Cal T}{v_{\sigma}}$, with
quadratic variation $\pmb v^*=\family{\sigma}{\Cal T}{v^*_{\sigma}}$,
such that $v_{\max\Cal T}=v_{\min\Cal T}$ but
$v^*_{\max\Cal T}\ne v^*_{\min\Cal T}$.
%624E mt62bits

\spheader 624Xb Let $\Cal S$ be a finitely full sublattice of $\Cal T$
and $\pmb v$, $\pmb w\,\,L^2$-martingales on $\Cal S$.
Show that $\pmb v\times\pmb w-\covar{\pmb v}{\pmb w}$ is a martingale.
\Hint{624B, 624G, 623K(b-iii), 623Kf.}

\spheader 624Xc Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v$ a $\|\,\|_2$-bounded martingale with domain $\Cal S$,
and $\pmb u$ a locally moderately
osciallatory process with domain $\Cal S$.   Write $\pmb v^*$ for the quadratic
variation of $\pmb v$.   Show that
$\Expn((\int_{\Cal S}\pmb u\,d\pmb v)^2)=\Expn(\int_{\Cal S}\pmb u^2d\pmb v^*)$
if either is finite.
%624I

\leaveitout{\leader{624Y}{Further exercises (a)}
}%end of leaveitout
}%end of exercises

\endnotes{\Notesheader{624}
Nothing in this section is surprising, but to get complete arguments
I think a little care is needed.   It is easy to believe that $\|v_{\tau}\|_2$
and $\|v^*_{\tau}\|_1$ in 624Ha are related, but not so simple to find exact
conditions on the process $\pmb v$ which will ensure this in the cases we might
encounter.   Of course there are many paths through the
forest.   In 624Xb, for instance, I sketch an alternative route to the result of
622L, not relying on Dynkin's formula.

}%end of notes

\discrpage
\frfilename{mt625.tex}
\versiondate{6.3.24}
\copyrightdate{2012}

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

\def\chaptername{Martingales}
\def\sectionname{Changing the measure}

\newsection{625}

I give essential formulae for calculating the effect of replacing a
given probability measure $\bar\mu$ with an equivalent probability measure
$\bar\nu$ (625B-625C).   
Semi-martingales (625D) remain \sm{s} under any such change (625F).

\leader{625A}{Notation} I continue in the framework developed in
Chapter 61.   \Asissix\ will be a stochastic integration structure, and
$\Bbb E_{\bar\mu}$ the integral corresponding to $\bar\mu$.
For $\tau\in\Cal T$, $P_{\tau}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ will be
the conditional expectation operator associated with $\frak A_{\tau}$;  if
$z\in L^1_{\bar\mu}$, $\pmb{P}z=\family{\tau}{\Cal T}{P_{\tau}z}$ will be
the martingale derived from $z$\cmmnt{ (622F)}.   If $\Cal S$ is a sublattice of
$\Cal T$, $\Mfa(\Cal S)$ will be the space of fully adapted processes with
domain $\Cal S$, and $\Mlmo(\Cal S)$ the space of locally moderately oscillatory
processes with domain $\Cal S$.

\leader{625B}{Change of law:  Theorem} Let $\bar\nu$ be a second functional
such that $(\frak A,\bar\nu)$ is a probability algebra;  write
$\Expn_{\bar\nu}$ and $L^1_{\bar\nu}$ for the corresponding integral and
$L^1$-space.

(a)(i) There is a unique
$z\in L^1_{\bar\mu}$ such that $\bar\nu a=\Expn_{\bar\mu}(z\times\chi a)$
for every $a\in\frak A$.

\quad(ii) $\Bvalue{z>0}=1$ and $z$ has a multiplicative inverse $\bover1z$
in $L^0$.

\quad(iii) For $w\in L^0$, $\Expn_{\bar\nu}(w)=\Expn_{\bar\mu}(w\times z)$
if either is defined in $[-\infty,\infty]$.

\quad(iv) $\bover1z\in L^1_{\bar\nu}$ and
$\bar\mu a=\Expn_{\bar\nu}(\bover1z\times\chi a)$ for every
$a\in\frak A$.

\quad(v) For $w\in L^0$,

\Centerline{$w\in L^1_{\bar\nu}\iff w\times z\in L^1_{\bar\mu}$,
\quad$w\in L^1_{\bar\mu}\iff w\times\bover1z\in L^1_{\bar\nu}$.}

\quad(vi) $\Bvalue{P_{\tau}z>0}=1$ for every $\tau\in\Cal T$.

\quad(vii) If $\tau\in\Cal T$ and $w\in L^0(\frak A_{\tau})$, then
$w\in L^1_{\bar\nu}$ iff $w\times P_{\tau}z\in L^1_{\bar\mu}$.

\quad(viii) We have a fully adapted process
$\pmb u=\family{\tau}{\Cal T}{u_{\tau}}$ defined by saying that
$u_{\tau}=\Bover1{P_{\tau}z}$
is the multiplicative inverse of $P_{\tau}z$ for every $\tau\in\Cal T$.

(b) For $\tau\in\Cal T$, let
$Q_{\tau}:L^1_{\bar\nu}\to L^1_{\bar\nu}\cap L^0(\frak A_{\tau})$
be the conditional expectation operator
with respect to the closed subalgebra $\frak A_{\tau}$
for the probability $\bar\nu$.

\quad (i) If $w\in L^1_{\bar\nu}$,
$P_{\tau}(w\times z)=Q_{\tau}w\times P_{\tau}z$.

\quad(ii) If $w\in L^1_{\bar\mu}$,
$Q_{\tau}(w\times\bover1z)=P_{\tau}w\times Q_{\tau}(\bover1z)$.

\quad(iii) $P_{\tau}(z)\times Q_{\tau}(\bover1z)=\chi 1$.

(c) Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb w=\family{\sigma}{\Cal S}{w_{\sigma}}$ a fully adapted process.
Write $\pmb{Q}(\bover1z)$ for the
$\bar\nu$-martingale $\family{\tau}{\Cal T}{Q_{\tau}(\bover1z)}$.

\quad(i) $\pmb w$ is
is a $\bar\nu$-martingale iff $\pmb w\times\pmb{P}z$ is a
$\bar\mu$-martingale.   In particular, $\pmb u$ in (a-viii)
is a $\bar\nu$-martingale.

\quad(ii) $\pmb w$ is
is a local $\bar\nu$-martingale iff $\pmb w\times\pmb{P}z$ is a
local $\bar\mu$-martingale.

\proof{{\bf (a)} Parts (i)-(iii) are in 365S\formerly{3{}65T}.   Now (iv) and
(v) follow immediately.

For (vi), we just have to note that

\Centerline{$\Expn_{\bar\nu}(P_{\tau}z\times\chi a)
=\Expn_{\bar\nu}(z\times\chi a)=\bar\nu a>0$}

\noindent for every non-zero $a\in\frak A_{\tau}$, because $\Bvalue{z>0}=1$.
It follows at once that $u_{\tau}=\Bover1{P_{\tau}z}$ is defined for every
$\tau$ (364N).   We see also that $P_{\tau}z$ is the
Radon-Nikod\'ym derivative of $\bar\nu\restrp\frak A_{\tau}$ with respect to
$\bar\mu\restrp\frak A_{\tau}$.   So, just as in (iii),
given $w\in L^0(\frak A_{\tau})$,
$\Expn_{\bar\nu}(w)=\Expn_{\bar\mu}(w\times P_{\tau}z)$ if either is
defined, and $w\in L^1_{\bar\nu}$ iff $w\times P_{\tau}z\in L^1_{\bar\mu}$.
This deals with (vii).

For any $\tau\in\Cal T$, $u_{\tau}\in L^0(\frak A_{\tau})$ because
$P_{\tau}z\in L^0(\frak A_{\tau})$.
If $\sigma$, $\tau\in\Cal T$ and $c=\Bvalue{\sigma=\tau}$, then
$P_{\sigma}z\times\chi c=P_{\tau}z\times\chi c$ (622Bb), so

\Centerline{$u_{\sigma}\times\chi c
=u_{\sigma}\times u_{\tau}\times P_{\tau}z\times\chi c
=u_{\sigma}\times u_{\tau}\times P_{\sigma}z\times\chi c
=u_{\tau}\times\chi c$}

\noindent and $c\Bsubseteq\Bvalue{u_{\sigma}=u_{\tau}}$.
Thus $\pmb u$ is fully adapted and (viii) is true.

\medskip

{\bf (b)(i)} We know from (a-v) that $w\times z\in L^1_{\bar\mu}$
so $P_{\tau}(w\times z)$ is defined.   Take any $a\in\frak A_{\tau}$.
We know that $Q_{\tau}w\in L^1_{\bar\nu}$ so
$Q_{\tau}w\times z\in L^1_{\bar\mu}$.   Since
$Q_{\tau}w\times\chi a\in L^0(\frak A_{\tau})$,
621Cc tells us that
$P_{\tau}(Q_{\tau}w\times\chi a\times z)
=Q_{\tau}w\times\chi a\times P_{\tau}z$.
But this means that

$$\eqalign{\Expn_{\bar\mu}(P_{\tau}(w\times z)\times\chi a)
&=\Expn_{\bar\mu}(w\times z\times\chi a)
=\Expn_{\bar\nu}(w\times\chi a)
=\Expn_{\bar\nu}(Q_{\tau}w\times\chi a)\cr
&=\Expn_{\bar\mu}(Q_{\tau}w\times z\times\chi a)
=\Expn_{\bar\mu}(P_{\tau}(Q_{\tau}w\times z\times\chi a))\cr
&=\Expn_{\bar\mu}(Q_{\tau}w\times P_{\tau}z\times\chi a).\cr}$$

\noindent As $a$ is arbitrary, and
$P_{\tau}(w\times z)\in L^1_{\bar\mu}\cap L^0(\frak A_{\tau})$, while
$Q_{\tau}w\times P_{\tau}z\in L^0(\frak A_{\tau})$,
$P_{\tau}(w\times z)=Q_{\tau}w\times P_{\tau}z$.

\medskip

\quad{\bf (ii)} Exchange $\bar\mu$ and $\bar\nu$ in (i).

\medskip

\quad{\bf (iii)} Set $w=z$ in (ii).

\medskip

{\bf (c)(i)} By (a-vii), $\pmb w$ is an $L^1_{\bar\nu}$-process iff
$\pmb w\times\pmb{P}z$ is an $L^1_{\bar\mu}$-process.   So we can suppose
that both of these are the case.   Now, for $\sigma\le\tau$ in $\Cal S$,

$$\eqalignno{w_{\sigma}=Q_{\sigma}w_{\tau}
&\iff w_{\sigma}\times P_{\sigma}z=Q_{\sigma}w_{\tau}\times P_{\sigma}z\cr
\displaycause{because $\Bvalue{P_{\sigma}z>0}=1$}
&\iff w_{\sigma}\times P_{\sigma}z=P_{\sigma}(w_{\tau}\times z)\cr
\displaycause{by (b-i)}
&\iff w_{\sigma}\times P_{\sigma}z=P_{\sigma}P_{\tau}(w_{\tau}\times z)\cr
\displaycause{because $P_{\sigma}P_{\tau}=P_{\sigma}$, by 622Ba}
&\iff w_{\sigma}\times P_{\sigma}z
=P_{\sigma}(w_{\tau}\times P_{\tau}z)\cr}$$

\noindent by 621Cc again.   So

$$\eqalign{\pmb w\text{ is a }\bar\nu\text{-martingale}
&\iff w_{\sigma}=Q_{\sigma}w_{\tau}
   \text{ whenever }\sigma\le\tau\text{ in }\Cal S\cr
&\iff w_{\sigma}\times P_{\sigma}z=P_{\sigma}(w_{\tau}\times P_{\tau}z)
   \text{ whenever }\sigma\le\tau\text{ in }\Cal S\cr
&\iff\pmb w\times\pmb{P}z\text{ is a }\bar\mu\text{-martingale}.\cr}$$

\medskip

\quad{\bf (ii)}

$$\eqalign{\pmb w\text{ is a local }\bar\nu\text{-martingale}
&\iff\text{ there is a covering ideal }\Cal S'\text{ of }\Cal S\cr
&\mskip100mu
   \text{ such that }\pmb w\restr\Cal S'
   \text{ is a }\bar\nu\text{-martingale}\cr
&\iff\text{ there is a covering ideal }\Cal S'\text{ of }\Cal S\cr
&\mskip100mu
   \text{ such that }(\pmb w\restr\Cal S')\times\pmb{P}z
   \text{ is a }\bar\mu\text{-martingale}\cr
&\iff\text{ there is a covering ideal }\Cal S'\text{ of }\Cal S\cr
&\mskip100mu
   \text{ such that }(\pmb w\times\pmb{P}z)\restr\Cal S'
   \text{ is a }\bar\mu\text{-martingale}\cr
&\iff\pmb w\times\pmb{P}z
   \text{ is a local }\bar\mu\text{-martingale}.\cr}$$
}%end of proof of 625B

\leader{625C}{}\cmmnt{ The next fact belongs with 625Bc, but the proof demands
new ideas, as well as being rather long, so I have separated it out.

\medskip

\noindent}{\bf Proposition} As in 625B, let $\bar\nu$ be
such that $(\frak A,\bar\nu)$ is a probability algebra;  write
$\Expn_{\bar\nu}$ and $L^1_{\bar\nu}$ for the corresponding integral and
$L^1$-space, and let
$z\in L^1_{\bar\mu}$ be such that $\bar\nu a=\Expn_{\bar\mu}(z\times\chi a)$
for every $a\in\frak A$.   Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb w=\sfamilysS{w}$ a fully adapted process.   Then $\pmb w$ is an
approximately local $\bar\nu$-martingale iff $\pmb w\times\pmb Pz$ is an
approximately local $\bar\mu$-martingale, and $\pmb w$ is a virtually
local $\bar\nu$-martingale iff $\pmb w\times\pmb Pz$ is a virtually local
$\bar\mu$-martingale.

\proof{{\bf (a)(i)} For a non-empty downwards-directed set
$A\subseteq\Cal S$, write $\frak A_A$ for
$\bigcap_{\sigma\in A}\frak A_{\sigma}$ and
$P_A:L^1_{\bar\mu}\to L^1_{\bar\mu}\cap L^0(\frak A_A)$
for the corresponding conditional expectation.   If $w\in L^1_{\bar\mu}$, then
$P_Aw=\lim_{\rho\downarrow A}P_{\rho}w=\1lim_{\rho\downarrow A}P_{\rho}w$ is the
limit for the norm topology of $L^1_{\bar\mu}$ (621C(g-i) again).   
Consequently

$$\eqalign{P_{\tau}P_Aw&
=\1lim_{\rho\downarrow A}P_{\tau}P_{\rho}w
=\1lim_{\rho\downarrow A}P_{\tau\wedge\rho}w
=\1lim_{\rho\downarrow A\wedge\tau}P_{\tau\wedge\rho}w
=P_{A\wedge\tau}w\cr&
=\1lim_{\rho\downarrow A}P_{\rho}P_{\tau}w
=P_AP_{\tau}w\cr}$$

\noindent for any $\tau\in\Cal S$.   Thus
$P_{\tau}P_A=P_AP_{\tau}=P_{A\wedge\tau}$.   Similarly, writing
$Q_A:L^1_{\bar\nu}\to L^1_{\bar\nu}\cap L^0(\frak A_A)$
for the conditional expectation corresponding to $\frak A_A$ with respect to the
probability measure $\bar\nu$, $Q_{\tau}Q_A=Q_AQ_{\tau}=Q_{A\wedge\tau}$ for
every $\tau\in\Cal S$.

\medskip

\quad{\bf (ii)} $Q_Aw\times P_Az=P_A(w\times z)$ whenever $w\in L^1_{\bar\nu}$,
that is, whenever $w\times z\in L^1_{\bar\mu}$.   \Prf\

$$\eqalignno{Q_A(w)\times P_Az&
=\lim_{\rho\downarrow A}Q_{\rho}w\times P_{\rho}z
=\lim_{\rho\downarrow A}P_{\rho}(w\times z)&\text{(625B(b-i))}\cr&
=P_A(w\times z).  \text{ \Qed}\cr}$$

\medskip

{\bf (b)} Let $\frak B$ be a closed subalgebra of $\frak A$ and
$P:L^1_{\bar\mu}\to L^1_{\bar\mu}\cap L^0(\frak B)=L^1_{\bar\mu\restr\frak B}$
the associated conditional expectation.   Then

\Centerline{$\bar\nu b=\Expn_{\bar\mu}(z\times\chi b)
=\Expn_{\bar\mu}(P(z\times\chi b))
=\Expn_{\bar\mu\restr\frak B}(Pz\times\chi b)$,}

\noindent for every $b\in\frak B$, that is, $Pz$ is the conditional expectation
of $\bar\nu\restr\frak B$ on $\bar\mu\restr\frak B$.   
So for $w\in L^0(\frak B)$

\Centerline{$w\in L^1_{\bar\nu}
\iff w\in L^1_{\bar\nu\restr\frak B}
\iff w\times Pz\in L^1_{\bar\mu\restr\frak B}
\iff w\times Pz\in L^1_{\bar\mu}$.}

\noindent Also $\Bvalue{Pz>0}=1$, just as in 625B(a-ii).

\medskip

{\bf (c)} Suppose that
$A\subseteq\Cal S$ is a non-empty downwards-directed set.   Let
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ be the corresponding
operator defined in 623B.   Then
$R_A(\pmb w)$ is a $\bar\nu$-martingale iff $R_A(\pmb w\times\pmb Pz)$ is
a $\bar\mu$-martingale.    \Prf\ Express $R_A(\pmb w)$ as $\sfamilysS{w'}$.
We have

\Centerline{$R_A(\pmb Pz)
=\family{\sigma}{\Cal S}{\lim_{\rho\downarrow A}P_{\sigma\wedge\rho}z}
=\family{\sigma}{\Cal S}{P_{A\wedge\sigma}(z)}$.}

\noindent So

$$\eqalignno{R_A(\pmb w\times\pmb Pz)&
=R_A(\pmb w)\times R_A(\pmb Pz)&\text{(623Ba)}\cr&
=\family{\sigma}{\Cal S}{w'_{\sigma}\times P_{A\wedge\sigma}z}.\cr}$$

\noindent If $\tau\in\Cal S$, then

\Centerline{$w'_{\tau}=\lim_{\rho\downarrow A}w_{\tau\wedge\rho}
=\lim_{\rho\downarrow A\wedge\tau}w_{\rho}\in L^0(\frak A_{A\wedge\tau})$}

\noindent by 613Bj.   So $w'_{\tau}\in L^1_{\bar\nu}$ iff
$w'_{\tau}\times P_{A\wedge\tau}z\in L^1_{\bar\mu}$, by (b) above.
Thus $R_A(\pmb w)$ is an $L^1_{\bar\nu}$-process iff
$R_A(\pmb w\times\pmb Pz)$ is an $L^1_{\bar\mu}$-process.

Suppose that this is the case.   If $\sigma\le\tau$ in $\Cal S$,

$$\eqalignno{Q_{\sigma}(w'_{\tau})\times P_{A\wedge\sigma}z&
=Q_{\sigma}Q_A(w'_{\tau})\times P_{A\wedge\sigma}z
\Displaycause{because
$w'_{\tau}\in L^0(\frak A_{A\wedge\tau})\subseteq L^0(\frak A_A)$}
=Q_{A\wedge\sigma}(w'_{\tau})\times P_{A\wedge\sigma}z
=P_{A\wedge\sigma}(w'_{\tau}\times z)
\Displaycause{by (a)}
=P_{A\wedge\sigma\wedge\tau}(w'_{\tau}\times z)
=P_{\sigma}P_{A\wedge\tau}(w'_{\tau}\times z)
=P_{\sigma}(w'_{\tau}\times P_{A\wedge\tau}z).\cr}$$

\noindent But now we see that

$$\eqalignno{R_A(\pmb w)\text{ is a }\bar\nu\text{-martingale}&
\iff Q_{\sigma}w'_{\tau}=w'_{\sigma}\text{ whenever }\sigma\le\tau\cr&
\iff Q_{\sigma}w'_{\tau}\times P_{A\wedge\sigma}z
   =w'_{\sigma}\times P_{A\wedge\sigma}z\text{ whenever }\sigma\le\tau
\Displaycause{because $\Bvalue{P_{A\wedge\sigma}z>0}=1$, as noted in (b)}
\iff P_{\sigma}(w'_{\tau}\times P_{A\wedge\tau}z)
   =w'_{\sigma}\times P_{A\wedge\sigma}z\text{ whenever }\sigma\le\tau\cr&
\iff R_A(\pmb w\times\pmb Pz)\text{ is a }\bar\mu\text{-martingale}\cr}$$

\noindent which is what I set out to prove.\ \Qed

\medskip

{\bf (d)(i)} If $\pmb w$ is an approximately local
$\bar\nu$-martingale then $\pmb w\times\pmb Pz$ is an approximately local
$\bar\mu$-martingale.   \Prf\ If $\tau\in\Cal S$ and $\epsilon>0$, there is a
$\delta>0$ such that $\bar\mu a\le\epsilon$ whenever $a\in\frak A$ and
$\bar\nu a\le\delta$.   Now there is a non-empty
downwards-directed set $A\subseteq\Cal S$ such that
$\sup_{\rho\in A}\bar\nu\Bvalue{\rho<\tau}\le\delta$ and $R_A(\pmb w)$ is a
$\bar\nu$-martingale (623J).   By (c), $R_A(\pmb w\times\pmb Pz)$ is a
$\bar\mu$-martingale, while 
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$;
as $\tau$ and $\epsilon$ are arbitrary,
$\pmb w\times\pmb Pz$ is an approximately local $\bar\mu$-martingale.\ \Qed

\medskip

\quad{\bf (ii)} If $\pmb w\times\pmb Pz$ is an approximately local
$\bar\mu$-martingale then $\pmb w\times\pmb Pz\times\pmb Q(\bover1z)$ 
is an approximately local
$\bar\nu$-martingale, where 
$\pmb Q(\bover1z)=\family{\tau}{\Cal T}{Q_{\tau}(\bover1z)}$ as in
625Bc.   \Prf\ Apply (i) with $(\bar\mu,\bar\nu,\bover1z,\pmb w\times\pmb Pz)$ 
in place of $(\bar\nu,\bar\mu,z,\pmb w)$.\ \QeD\   
But $\pmb Pz\times\pmb Q(\bover1z)=\tbf 1$, by
625B(b-iii), so $\pmb w$ is an approximately local $\bar\nu$-martingale.

\medskip

\quad{\bf (iii)} Thus $\pmb w$ is an approximately local
$\bar\nu$-martingale iff $\pmb w\times\pmb Pz$ is an approximately local
$\bar\mu$-martingale. 

\medskip

{\bf (e)} As for virtually local martingales, 
let $\hat{\Cal S}$ be the covered envelope of
$\Cal S$ and $\hatpmb w$ the fully adapted extension of $\pmb w$ to
$\hat{\Cal S}$.   Recalling that I interpret $\pmb w\times\pmb Pz$ to be defined
on $\dom\pmb w\cap\dom\pmb Pz=\Cal S\cap\Cal T=\Cal S$, so that it is
$\pmb w\times(\pmb Pz\restr\Cal S)$, then it is clear that
$\pmb Pz\restr\hat{\Cal S}$, being fully adapted, must be the fully adapted
extension of $\pmb Pz\restr\Cal S$ to $\hat{\Cal S}$, and that
$\hatpmb w\times\pmb Pz$ is the extension of $\pmb w\times\pmb Pz$ (612Qb).
Now we have

$$\eqalignno{&\pmb w\text{ is a virtually local }\bar\nu\text{-martingale}\cr&
\mskip50mu\iff\hatpmb w\text{ is an approximately local }
  \bar\nu\text{-martingale}  &\text{(623J)}\cr&
\mskip50mu\iff\hatpmb w\times\pmb Pz
    \text{ is an approximately local }\bar\mu\text{-martingale}
  &\text{((d) above)}\cr&
\mskip50mu\iff\pmb w\times\pmb Pz
    \text{ is a virtually local }\bar\mu\text{-martingale}.\cr}$$

\noindent This completes the proof.
}%end of proof of 625C

\leader{625D}{Definition} Let $\Cal S\subseteq\Cal T$ be a sublattice.
A process with domain $\Cal S$ is a {\bf \sm} if it is expressible as the sum of
a virtually local martingale and a process which is
locally of bounded variation\cmmnt{ (both, of course, with domain $\Cal S$)}.

\cmmnt{\medskip

\noindent{\bf Warning!} The standard definition of `semimartingale' (no hyphen)
is a process which is the sum of a local martingale, in the sense of 622Cc,
and a process which is locally of bounded variation.   Most presentations of the
theory take it for granted that the conditions of 632Ib below will
be satisfied, so that the distinction vanishes.   Nevertheless,
to limit the opportunities for confusion, I will try to be
consistent in hyphenating `\sm' when I have the
`virtually local martingale' form in mind.
}

\leader{625E}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.
The set of \sm{}s with domain $\Cal S$ is a linear subspace of the space
of local integrators with domain $\Cal S$.   In particular, every \sm\ is
locally moderately oscillatory.
%622H 622Xe 622Xi

\proof{ Since every virtually local martingale and every process of locally
bounded variation is a local integrator (623Kd, 616Ra) and the sum of two
local integrators is a local integrator (616Qa), every \sm\ is a local
integrator.   Since sums and scalar multiples of virtually local martingales are
virtually local martingales (623K(a-ii))
and sums and scalar multiples of processes which are locally of
bounded variation are again locally of bounded variation (614Q(b-iii)),
the set of \sm{}s is a linear subspace of $L^0(\frak A)^{\Cal S}$.
}%end of proof of 625E

\leader{625F}{Theorem} A \sm\ remains a \sm\ under any change of law.

\proof{ Let $\bar\nu$ be a
functional such that $(\frak A,\bar\nu)$ is a probability algebra.
Suppose that $\pmb v$ is a $\bar\nu$-\sm.   Express it as
$\pmb v_1+\pmb v_2$ where $\pmb v_1$ is a virtually local
$\bar\nu$-martingale and $\pmb v_2$ is locally of bounded variation.
Let $z\in L^1_{\bar\mu}$ be
the Radon-Nikod\'ym derivative of $\bar\nu$ with respect to
$\bar\mu$, as in 625B.   As in 625B, write
$Q_{\tau}:L^1_{\bar\nu}\to L^1_{\bar\nu}\cap L^0(\frak A_{\tau})$ for the
conditional expectation associated with $\frak A_{\tau}$ with respect to
$\bar\nu$, and $\pmb Q(\bover1z)$ for the $\bar\nu$-martingale
$\family{\tau}{\Cal T}{Q_{\tau}(\bover1z)}$.   As in (d-ii) of the proof of
625C, $\pmb Q(\bover1z)\times\pmb Pz=\tbf 1$.

Set $\pmb w=\pmb v_1\times\pmb Q(\bover1z)$.
Applying 625C with $(\bar\mu,\bar\nu,z,\pmb P)$ replaced by
$(\bar\nu,\bar\mu,\bover1z,\pmb Q)$, we see that
$\pmb w$ is a virtually local $\bar\mu$-martingale and

\Centerline{$\pmb v_1=\pmb v_1\times\pmb Q(\bover1z)\times\pmb Pz
=\pmb w\times\pmb Pz$.}

\noindent Writing $w_{\downarrow}$ and $P_{\downarrow}z$ for the starting values
of the local integrators $\pmb w$ (625E) and $\pmb Pz\restr\Cal S$, and
$\covar{\pmb w}{\pmb Pz}$ for the covariation of $\pmb w$ and $\pmb Pz$,

$$\eqalignno{\pmb v&
=\pmb w\times\pmb Pz+\pmb v_2\cr&
=\covar{\pmb w}{\pmb Pz}+ii_{\pmb Pz}(\pmb w)+ii_{\pmb w}(\pmb Pz)
   +(w_{\downarrow}\times P_{\downarrow}z)\tbf 1+\pmb v_2\cr}$$

\noindent by 617Ka.   But
$ii_{\pmb Pz}(\pmb w)$ and $ii_{\pmb w}(\pmb Pz)$ are virtually local
$\bar\mu$-martingales, by 623O, while
$\covar{\pmb w}{\pmb Pz}$ is locally of bounded variation (617L).
Thus $\pmb v$ is the sum of three processes which are locally
of bounded variation and two
virtually local $\bar\mu$-martingales, so is a $\bar\mu$-\sm.

Similarly, any $\bar\mu$-\sm\ is a $\bar\nu$-\sm.
}%end of proof of 625F


%\leader{625G}{}
%\leader{625H}{}
%\leader{625I}{}
%\leader{625J}{}
%\leader{625K}{}
%\leader{625L}{}
%\leader{625M}{}
%\leader{625N}{}
%\leader{625O}{}
%\leader{625P}{}
%\leader{625Q}{}
%\leader{625R}{}
%\leader{625S}{}
%\leader{625T}{}
%\leader{625U}{}
%\leader{625V}{}
%\leader{625W}{}

\exercises{\leader{625X}{Basic exercises (a)}
%\spheader 625Xa
Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa and 622Xd.   Let $f:\coint{0,\infty}\to\Bbb R$ be a
function.   Show that $f$ corresponds to a \sm\ iff $f\restr[0,t]$ is of
bounded variation for every $t\ge 0$.
%625E

\spheader 625Xb Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a \sm.   Show that $\pmb v\restr\Cal S'$ is a \sm\ for
any ideal $\Cal S'$ of $\Cal S$.
%623Xe 625E

\spheader 625Xc Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb{u}$ a
process with domain $\Cal S$, and $z$ an element of
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Show that if $\pmb{u}$ is a \sm\ then $z\pmb{u}$ is a \sm.
%622C %622Xb

\spheader 625Xd Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb u$ a locally
moderately oscillatory process and $\pmb v$ a \sm, both with domain $\Cal S$.
Show that $\iivu$ is a \sm.
%625E  616L 623O

\spheader 625Xe Let $\pmb v$, $\pmb w$ be \sm{}s with the same domain.   Show
that $\pmb v\times\pmb w$ is a \sm.
%625F 625Xd

%\leader{625Y}{Further exercises (a)}
}%end of exercises


\endnotes{
\Notesheader{625} I have said repeatedly that stochastic integration is
law-independent.   At the same time it is intimately entwined with the
theory of martingales, which are emphatically not law-independent.
625B is
a brisk run through the formulae we need if we are to move freely between
different probability measures on a fixed algebra.   In 625F we find that
the concept of `\sm' again turns out, remarkably, to be law-independent.
The ideas on stochastic processes required to state the result are not
trivial, but they do not mention any kind of integration;  while the
proof depends on an excursion through most of the theory of the Riemann-sum
integral so far developed.   I do not know whether there is anything one could
call an `elementary' proof of 625F.

There is a great deal more to be said about \sm{s}.   In 
625X I offer a handful of tasters.   
But these will be dramatically upstaged by 627Q below.
}%end of notes

\discrpage



\frfilename{mt626.tex}
\versiondate{26.10.17/23.8.23}
\copyrightdate{2012}

\def\2lim{\mathop{2\text{lim}}}

\def\chaptername{Martingales}
\def\sectionname{Submartingales and previsible variations}

\newsection{626}

Turning to submartingales, I give the elementary theory
(626B-626F), %626B 626C 626D 626E 626F
with a brief introduction to `finitely-covered envelopes' of sublattices
of $\Cal T$ (626G-626I).  %626G 626H 626I
Serious work begins with what I call `previsible variations' (626L-626M),
based on a new adapted interval function $P\Delta\pmb v$ (626J-626K).
Now the final formula of \S621 gives us the
celebrated Doob-Meyer decomposition theorem (626O, 626Q).   The computation of
previsible variations can be difficult, but I give some basic special cases
(626S, 626U and 626V).

\cmmnt{\leader{626A}{Notation} \Asissix\ is a stochastic integration structure.
For $t\in T$, $\check t$ is the constant
stopping time at $t$\cmmnt{ (611A(b-ii))}.   If $\Cal S$ is a sublattice of
$\Cal T$, $\Cal I(\Cal S)$ is the set of finite sublattices of $\Cal S$.
If $\sigma\le\tau$ in
$\Cal T$, $c(\sigma,\tau)$ is the corresponding stopping time
interval\cmmnt{ (611J)}.

On the $L$-space
$L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$ we have the integral $\Expn$ defined
by $\bar\mu$, giving rise to the functional
$\theta(w)=\Expn(|w|\wedge\chi 1)$ for $w\in L^0(\frak A)$, so that
$\theta$ defines the topology of convergence in measure on
$L^0(\frak A)$.   For $\tau\in\Cal T$,
$P_{\tau}$ is the conditional
expectation associated with $\frak A_{\tau}$.   If $w\in L^1_{\bar\mu}$,
$\pmb Pw$ is the martingale $\family{\tau}{\Cal T}{P_{\tau}w}$.

If $\Cal S$ is a sublattice of $\Cal T$, $\Mfa(\Cal S)$ will be the space of
fully adapted processes with domain $\Cal S$, and
$\Mlmo(\Cal S)\subseteq\Mfa(\Cal S)$ the space of locally moderately oscillatory
processes.

\cmmnt{We shall need to look at the norm and weak topologies on
$L^1_{\bar\mu}$ as
well as the topology of convergence in measure on $L^0(\frak A)$.   It will
therefore be helpful to have a notation which distinguishes between the
three corresponding notions of limit.}   I will use `$\lim$' for limits in
$L^0(\frak A)$ for the topology of convergence in measure, `$\1lim$' for limits in
$L^1_{\bar\mu}$ for the norm topology defined by $\|\,\|_1$, and
`$\w1lim$' for limits in $L^1_{\bar\mu}$ for the weak topology
$\frak T_s(L^1_{\bar\mu},L^{\infty}(\frak A))$\cmmnt{
(365Lc\formerly{3{}65Mc})}.
}

\leader{626B}{Definition}\cmmnt{ (Compare 621Db.)}
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a fully adapted process.
$\pmb v$ is a {\bf submartingale} if it is an
$L^1$-process\cmmnt{ (definition:  622Ca)}
and $v_{\sigma}\le P_{\sigma}v_{\tau}$ whenever
$\sigma\le\tau$ in $\Cal S$.

Clearly every martingale is a submartingale, and every non-decreasing
$L^1$-process is a submartingale\cmmnt{ (because if $\sigma\le\tau$ and
$v_{\sigma}\le v_{\tau}$, then
$P_{\sigma}v_{\tau}-v_{\sigma}=P_{\sigma}(v_{\tau}-v_{\sigma})\ge 0$)}.

\leader{626C}{Elementary facts} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a submartingale.

\spheader 626Ca
$\Expn(v_{\sigma})\le\cmmnt{\Expn(P_{\sigma}v_{\tau})=\mskip5mu}\Expn(v_{\tau})$
whenever $\sigma\le\tau$ in $\Cal S$.

\spheader 626Cb If $\Cal S'$ is a sublattice of $\Cal S$, then
$\pmb v\restr\Cal S'$ is a submartingale.   \prooflet{(Immediate from
626B.)}

\spheader 626Cc If $h:\Bbb R\to\Bbb R$ is a convex function,
$\bar h(\pmb v)=\family{\sigma}{\Cal S}{\bar h(v_{\sigma})}$\cmmnt{
(612B)} is an $L^1$-process and
{\it either} $h$ is non-decreasing {\it or} $\pmb v$ is a martingale, then
$\bar h(\pmb v)$ is a submartingale.   \prooflet{\Prf\ If $\sigma\le\tau$
in $\Cal S$ then
$\bar h(P_{\sigma}v_{\tau})\le P_{\sigma}(\bar h(v_{\tau}))$ by
Jensen's inequality (621Cd).   If $h$ is non-decreasing, then
$\bar h(v_{\sigma})\le\bar h(P_{\sigma}v_{\tau})$
because $v_{\sigma}\le P_{\sigma}v_{\tau}$;  if $\pmb v$ is a martingale,
then $\bar h(v_{\sigma})=\bar h(P_{\sigma}v_{\tau})$.   So in either case
we have $\bar h(v_{\sigma})\le P_{\sigma}(\bar h(v_{\tau}))$.\ \Qed}

\cmmnt{In particular, }$\alpha\pmb v$
is a submartingale for every $\alpha\ge 0$.

\spheader 626Cd If
$\pmb u=\family{\sigma}{\Cal S}{u_{\sigma}}$ is another submartingale,
$\pmb u+\pmb v$ is a submartingale.   \prooflet{(Immediate from
626B, because conditional expectations are linear operators.)}

\leader{626D}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a submartingale.

(a) If $\pmb v$ is $\|\,\|_1$-bounded, it is an integrator.

(b) If $\Cal S$ has a greatest element and
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below, $\pmb v$ is
$\|\,\|_1$-bounded.

(c) If $\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below, $\pmb v$ is a
local integrator.

\proof{{\bf (a)} (Cf.\ 622G) If $\Cal S$ is empty, this is trivial.
Otherwise, set $\beta=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$.
If $z\in Q_{\Cal S}(\pmb v)$,
it is expressible as $\sum_{i=0}^{n-1}u_i\times(v_{\tau_{i+1}}-v_{\tau_i})$
where $\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and
$u_i\in L^0(\frak A_{\tau_i})$ and $\|u_i\|_{\infty}\le 1$ for every $i<n$
(616C(ii)).   By 621Hg,

\Centerline{$\bar\mu\Bvalue{|z|>\gamma}
\le\Bover{66}{\gamma}\|v_{\tau_n}\|_1-\Bover{34}{\gamma}\Expn(v_{\tau_0})
\le\Bover{100}{\gamma}\beta$.}

So

\Centerline{$\inf_{\gamma>0}\sup_{z\in Q_{\Cal S}(\pmb v)}
\bar\mu\Bvalue{|z|>\gamma}=0$}

\noindent and $Q_{\Cal S}(\pmb v)$ is topologically bounded (613B(f-ii)),
that is, $\pmb v$ is an integrator.

\medskip

{\bf (b)}
If $\alpha=\inf_{\sigma\in\Cal S}\Expn(v_{\sigma})$, and $\sigma\in\Cal S$, then

\Centerline{$v_{\sigma}\le P_{\sigma}v_{\max\Cal S}$,
\quad$v_{\sigma}^+\le(P_{\sigma}v_{\max\Cal S})^+
\le P_{\sigma}v_{\max\Cal S}^+\le P_{\sigma}|v_{\max\Cal S}|$}

\noindent and

\Centerline{$\|v_{\sigma}\|_1
=\Expn(v_{\sigma}^+)+\Expn(v_{\sigma}^-)
=2\Expn(v_{\sigma}^+)-\Expn(v_{\sigma})
\le 2\Expn(P_{\sigma}|v_{\max\Cal S}|)-\alpha
=2\Expn(|v_{\max\Cal S}|)-\alpha$.}

\medskip

{\bf (c)} Take $\tau\in\Cal S$.
Since of course $\pmb v\restr\Cal S\wedge\tau$ is a submartingale
(626Cb), (b) tells us that $\pmb v\restr\Cal S\wedge\tau$ is $\|\,\|_1$-bounded
and (a) tells us that $\pmb v\restr\Cal S\wedge\tau$ is an
integrator.   As $\tau$ is arbitrary, $\pmb v$ is a local integrator.
}%end of proof of 626D

\leader{626E}{}\cmmnt{ From 626D we see that a submartingale $\pmb v$
will often be moderately oscillatory (616Ib),
so will have limits along directed sets
(615G).   But for downwards-directed sets we can look for more.

\medskip

\noindent}{\bf Proposition} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$
a submartingale and $A\subseteq\Cal S$ a non-empty downwards-directed set
such that $\{\Expn(v_{\sigma}):\sigma\in A\}$ is bounded
below.   Then the
$\|\,\|_1$-limit $\1lim_{\sigma\downarrow A}v_{\sigma}$ is defined and equal to
the limit $\lim_{\sigma\downarrow A}v_{\sigma}$ for the topology of convergence
in measure.

\proof{ Let $\epsilon>0$.   Then there is a $\tau\in A$ such that
$\|v_{\sigma}-v_{\tau}\|_1\le 3\epsilon$ whenever
$\sigma\in A$ and $\sigma\le\tau$.   \Prf\ Set
$\gamma=\inf_{\sigma\in A}\Expn(v_{\sigma})$.   Let $\tau_0\in A$ be such that
$\Expn(v_{\tau_0})\le\gamma+\epsilon$.   Then
$\1lim_{\sigma\downarrow A}P_{\sigma}v_{\tau_1}$ is defined (621C(g-i));
let $\tau\in A$ be such that $\tau\le\tau_1$ and
$\|P_{\sigma}v_{\tau_1}-P_{\tau}v_{\tau_1}\|_1\le\epsilon$ whenever
$\sigma\in A$ and $\sigma\le\tau$.   In this case, if
$\sigma\in A$ and $\sigma\le\tau$, we have

$$\eqalign{\|v_{\sigma}-v_{\tau}\|_1&
\le\|v_{\sigma}-P_{\sigma}v_{\tau_1}\|_1
      +\|P_{\sigma}v_{\tau_1}-P_{\tau}v_{\tau_1}\|_1
      +\|P_{\tau}v_{\tau_1}-v_{\tau}\|_1\cr&
\le\Expn(P_{\sigma}v_{\tau_1}-v_{\sigma})+\epsilon
      +\Expn(P_{\tau}v_{\tau_1}-v_{\tau})\cr&
=\Expn(v_{\tau_1}-v_{\sigma})+\epsilon+\Expn(v_{\tau_1}-v_{\tau})
\le 2\Expn(v_{\tau_1}-\gamma)+\epsilon
\le 3\epsilon.  \text{ \Qed}\cr}$$

\noindent As $L^1_{\bar\mu}$ is complete under $\|\,\|_1$,
the limit $\1lim_{\sigma\downarrow A}v_{\sigma}$ is defined.   As the embedding
$L^1_{\bar\mu}\embedsinto L^0(\frak A)$ is continuous, this is also the limit
$\lim_{\sigma\downarrow A}v_{\sigma}$.
}%end of proof of 626E

\leader{626F}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$ a submartingale such that
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below.   Let
$A\subseteq\Cal S$ be a non-empty downwards-directed set and
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ the corresponding operator as described in
623B.   Then $R_A(\pmb v)$ is defined and is a submartingale.

\proof{ Because $\pmb v$ is a local integrator (626Dc), it is locally moderately
oscillatory (616Ib) and $R_A(\pmb v)$ is defined;  express it as
$\sfamilysS{v'}$.   If $\sigma\le\tau$ in $\Cal S$ and $\rho\in A$,
$\sigma\wedge\rho\le\tau\wedge\rho$ and

\Centerline{$v_{\sigma\wedge\rho}
\le P_{\sigma\wedge\rho}v_{\tau\wedge\rho}
=P_{\sigma}P_{\rho}v_{\tau\wedge\rho}
=P_{\sigma}v_{\tau\wedge\rho}$.}

\noindent Now (because $\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below)
626E tells us that

\Centerline{$v'_{\sigma}=\lim_{\rho\downarrow A}v_{\sigma\wedge\rho}
=\1lim_{\rho\downarrow A}v_{\sigma\wedge\rho}$,
\quad$v'_{\tau}=\1lim_{\rho\downarrow A}v_{\tau\wedge\rho}$}

\noindent and therefore

\Centerline{$P_{\sigma}v'_{\tau}
=\1lim_{\rho\downarrow A}P_{\sigma}v_{\tau\wedge\rho}$.}

\noindent Accordingly

\Centerline{$P_{\sigma}v'_{\tau}-v'_{\sigma}
=\1lim_{\rho\downarrow A}P_{\sigma}v_{\tau\wedge\rho}-v_{\sigma\wedge\rho}
\ge 0$}

\noindent and $v'_{\sigma}\le P_{\sigma}v'_{\tau}$.   As $\sigma$ and $\tau$ are
arbitrary, $R_A(\pmb v)$ is a submartingale.
}%end of proof of 626F

\leader{626G}{Definition} Let $\Cal S$ be a sublattice of $\Cal T$.   The
{\bf finitely-covered envelope} of $\Cal S$ is the set of those $\tau\in\Cal T$
for which there is a finite subset $J\subseteq\Cal S$ such that
$\sup_{\sigma\in J}\Bvalue{\tau=\sigma}=1$.

\leader{626H}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and $\hat\Cal S_f$
its finitely-covered envelope.

(a) $\hat\Cal S_f$ is a sublattice of the covered envelope $\hat\Cal S$ of
$\Cal S$.

(b) If $\tau\in\hat\Cal S_f$, there are $\sigma_0\le\ldots\le\sigma_n$
in $\Cal S$ such that $\sup_{i\le n}\Bvalue{\tau=\sigma_i}=1$.

(c) $\Cal S$ is both coinitial and cofinal with $\hat\Cal S_f$.

(d) If $\pmb u=\family{\tau}{\hat\Cal S}{u_{\tau}}$ is a fully adapted process
and $\tau\in\hat\Cal S$, $u_{\tau}$ belongs to the closure of
$\{u_{\sigma}:\sigma\in\hat\Cal S_f\}$ for the topology of convergence in
measure.

(e) $\hat\Cal S_f$ is finitely full.

(f) If $\tau\in\hat\Cal S$ and $\epsilon>0$, there is a $\tau'\in\hat\Cal S_f$
such that $\bar\mu\Bvalue{\tau=\tau'}\ge 1-\epsilon$.

\proof{{\bf (a)} (Cf.\ 611M(b-i)) Of course $\hat\Cal S_f\subseteq\hat\Cal S$.
If $\tau$, $\tau'\in\hat\Cal S_f$, there is a finite set
$J\subseteq\Cal S$ such that
$\sup_{\sigma\in J}\Bvalue{\tau=\sigma}=1$;  now

\Centerline{$\Bvalue{\tau\le\tau'}
=\Bvalue{\tau\wedge\tau'=\tau}
=\sup_{\sigma\in J}\Bvalue{\tau\wedge\tau'=\tau}\Bcap\Bvalue{\tau=\sigma}
\Bsubseteq\sup_{\sigma\in J}\Bvalue{\tau\wedge\tau'=\sigma}$.}

\noindent Similarly, there is a finite set $J'\subseteq\Cal S$ such that
$\Bvalue{\tau'\le\tau}
\Bsubseteq\sup_{\sigma\in J'}\Bvalue{\tau\wedge\tau'=\sigma}$
But now $J\cup J'$ is finite and
$\sup_{\sigma\in J\cup J'}\Bvalue{\tau\wedge\tau'=\sigma}=1$.
Thus $\tau\wedge\tau'\in\hat\Cal S_f$.   The same method shows that
$\tau\vee\tau'\in\hat\Cal S_f$.   As $\tau$ and $\tau'$ are arbitrary,
$\hat\Cal S_f$ is a sublattice.

\medskip

{\bf (b)} There is a finite set $J\subseteq\Cal S$ such that
$\sup_{\sigma\in J}\Bvalue{\tau=\sigma}=1$.   Let $I$ be the sublattice of
$\Cal S$ generated by $J$;  then $I$ is finite and not empty, so there is a
non-decreasing sequence $(\sigma_0,\ldots,\sigma_n)$ linearly generating the
$I$-cells (611L).   Now

$$1=
\sup_{\sigma\in J}\Bvalue{\tau=\sigma}
\Bsubseteq\sup_{\sigma\in I}\Bvalue{\tau=\sigma}
=\sup_{\Atop{\sigma\in I}{i\le n}}
   \Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma=\sigma_i}
\Bsubseteq\sup_{i\le n}\Bvalue{\tau=\sigma_i}.$$

\medskip

{\bf (c)} If $\tau\in\hat\Cal S_f$ and $J\subseteq\Cal S$ is a finite set such
that $\sup_{\sigma\in J}\Bvalue{\tau=\sigma}=1$, we have $\sup J\in\Cal S$ and

\Centerline{$1
=\sup_{\sigma\in J}\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma\le\sup J}
\Bsubseteq\Bvalue{\tau\le\sup J}$}

\noindent and $\tau\le\sup J$.   As $\tau$ is arbitrary, $\Cal S$ is cofinal
with $\hat\Cal S_f$.   Similarly, $\Cal S$ is coinitial with $\hat\Cal S_f$.

\medskip

{\bf (d)} Let $\epsilon>0$.   There is a
non-empty finite $J\subseteq\Cal S$ such that
$a=\sup_{\sigma\in J}\Bvalue{\tau=\sigma}$ has measure at least $1-\epsilon$.
Define $\tau'\in\Cal T$ by saying that

\Centerline{$a\Bsubseteq\Bvalue{\tau'=\tau}$,
\quad$1\Bsetminus a\Bsubseteq\Bvalue{\tau'=\sup J}$.}

\noindent Then $\sup_{\sigma\in J\cup\{\sup J\}}\Bvalue{\tau'=\sigma}=1$, so
$\tau'\in\hat\Cal S_f$, while

\Centerline{$\theta(|u_{\tau}-u_{\tau'})\le\bar\mu\Bvalue{u_{\tau}\ne u_{\tau'}}
\le\bar\mu(1\Bsetminus a)\le\epsilon$.}

\medskip

{\bf (e)} If $\tau\in\Cal T$, $I\subseteq\hat\Cal S_f$ is finite, and
$\sup_{\sigma\in I}\Bvalue{\tau=\sigma}=1$, then for each $\sigma\in I$ take a
finite set $J_{\sigma}\subseteq\Cal S$ such that
$\sup_{\rho\in J_{\sigma}}\Bvalue{\sigma=\rho}=1$.   Then
$J=\bigcup_{\sigma\in I}J_{\sigma}$ is a finite subset of $\Cal S$ and

\Centerline{$\sup_{\rho\in J}\Bvalue{\tau=\rho}
\Bsupseteq\sup_{\sigma\in I,\rho\in J}
  \Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma=\rho}
=\sup_{\sigma\in I}\Bvalue{\tau=\sigma}=1$,}

\noindent so $\tau\in\hat\Cal S_f$.

\medskip

{\bf (f)} Since $\sup_{\sigma\in\Cal S}\Bvalue{\tau=\sigma}=1$, there is a
non-empty finite set $J\subseteq\Cal S$ such that
$\bar\mu(\sup_{\sigma\in J}\Bvalue{\tau=\sigma})\ge 1-\epsilon$.
Let $I$ be the sublattice of $\Cal S$ generated by $J$, and
$(\sigma_0,\ldots,\sigma_n)$ a sequence linearly generating the $I$-cells.
For $i\le n$ set $a_i=\Bvalue{\tau=\sigma_i}$ and
$b_i=a_i\Bsetminus\sup_{j<i}a_j$;  then $a_i$ and $b_i$ belong to
$\frak A_{\sigma_i}$ for each $i$, so

\Centerline{$b_{n+1}=1\Bsetminus\sup_{i\le n}b_i=1\Bsetminus\sup_{i\le n}a_i$}

\noindent belongs to $\frak A_{\sigma_n}$.   Now $(b_0,\ldots,b_{n+1})$ is a
partition of unity in $\frak A$, so by 611I there is a $\tau'\in\Cal T$ such
that $b_i\Bsubseteq\Bvalue{\tau'=\sigma_i}$ for each $i\le n$ and
$b_{n+1}\Bsubseteq\Bvalue{\tau'=\sigma_n}$.   We see that $\tau\in\hat\Cal S_f$.
and

$$\eqalign{\sup_{\sigma\in J}\Bvalue{\tau=\sigma}&
=\sup_{\sigma\in J,i\le n}\Bvalue{\tau=\sigma}\Bcap\Bvalue{\sigma=\sigma_i}
\Bsubseteq\sup_{i\le n}\Bvalue{\tau=\sigma_i}
=\sup_{i\le n}a_i\cr&
=\sup_{i\le n}b_i
=\sup_{i\le n}a_i\Bcap b_i
\Bsubseteq\sup_{i\le n}\Bvalue{\tau=\sigma_i}\Bcap\Bvalue{\tau'=\sigma_i}
\Bsubseteq\Bvalue{\tau=\tau'}.\cr}$$

\noindent So

\Centerline{$\bar\mu\Bvalue{\tau=\tau'}
\ge\bar\mu(\sup_{\sigma\in J}\Bvalue{\tau=\sigma})\ge 1-\epsilon$,}

\noindent as required.
}%end of proof of 626H

\leader{626I}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a submartingale.   Let
$\hatpmb v=\family{\tau}{\hat\Cal S}{\hat v_{\tau}}$ be the fully adapted
extension of $\pmb v$ to the covered envelope $\hat\Cal S$ of $\Cal S$, and
$\hat\Cal S_f$ the finitely-covered envelope of $\Cal S$.

(a) $\hatpmb v\restr\hat\Cal S_f$ is a submartingale.

(b) If $\pmb v$ is $\|\,\|_1$-bounded then
$\hatpmb v$ is $\|\,\|_1$-bounded.

(c) If $\Cal S$ has a greatest element
and $\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$
is bounded below, then $\hatpmb v$ is a submartingale.

\proof{{\bf (a)(i)} $\hatpmb v\restr\hat\Cal S_f$ is an $L^1$-process.
\Prf\ If $\tau\in\hat\Cal S_f$,
there is a finite set $J\subseteq\Cal S$ such that

\Centerline{$1=\sup_{\sigma\in J}\Bvalue{\tau=\sigma}
\Bsubseteq\sup_{\sigma\in J}\Bvalue{\hat v_{\tau}=v_{\sigma}}
\Bsubseteq\Bvalue{|\hat v_{\tau}|\le\sup_{\sigma\in J}|v_{\sigma}|}$}

\noindent so $\|\hat v_{\tau}\|_1\le\sum_{\sigma\in J}\|v_{\sigma}\|_1$
is finite.\ \Qed

\medskip

\quad{\bf (ii)} If $\tau\in\hat\Cal S_f$, $\sigma_0\le\ldots\le\sigma_n$ in
$\Cal S$ and $\sup_{i\le n}\Bvalue{\tau=\sigma_i}=1$, then
$\hat v_{\sigma\wedge\tau}\le P_{\sigma\wedge\tau}\hat v_{\tau}$ for every
$\sigma\in\Cal S$.   \Prf\ Induce on $n$.
The induction starts with $n=0$ and $\tau=\sigma_0$ and
$v_{\sigma\wedge\sigma_0}\le P_{\sigma\wedge\sigma_0}v_{\sigma_0}$.

\medskip

\qquad\grheada\  For the
inductive step to $n\ge 1$, set $\tau'=\tau\wedge\sigma_{n-1}$ and
$d=\Bvalue{\sigma_{n-1}<\tau}=\Bvalue{\tau'<\tau}$.   Then
$d\in\frak A_{\sigma_{n-1}}$ and
$d\Bsubseteq\Bvalue{\tau'=\sigma_{n-1}}\Bcap\Bvalue{\tau=\sigma_n}$.   Now

\Centerline{$\sup_{i<n}\Bvalue{\tau'=\sigma_i}
\Bsupseteq\sup_{i<n}\Bvalue{\tau=\sigma_i}\Bcup d=1$.}

\noindent So the inductive hypothesis assures us that
$\hat v_{\sigma\wedge\tau'}\le P_{\sigma\wedge\tau'}\hat v_{\tau'}$.
Next,

$$\eqalign{P_{\sigma_{n-1}}\hat v_{\tau}-\hat v_{\tau'}&
=P_{\sigma_{n-1}}(\hat v_{\tau}-\hat v_{\tau'})
=P_{\sigma_{n-1}}((\hat v_{\tau}-\hat v_{\tau'})\times\chi d)\cr&
=P_{\sigma_{n-1}}((v_{\sigma_n}-v_{\sigma_{n-1}})\times\chi d)
=P_{\sigma_{n-1}}(v_{\sigma_n}-v_{\sigma_{n-1}})\times\chi d\cr&
=(P_{\sigma_{n-1}}v_{\sigma_n}-v_{\sigma_{n-1}})\times\chi d
\ge 0\cr}$$

\noindent and

\Centerline{$P_{\sigma\wedge\tau'}\hat v_{\tau}
=P_{\sigma\wedge\tau'\wedge\sigma_{n-1}}\hat v_{\tau}
=P_{\sigma\wedge\tau'}P_{\sigma_{n-1}}\hat v_{\tau}
\ge P_{\sigma\wedge\tau'}\hat v_{\tau'}
\ge\hat v_{\sigma\wedge\tau'}$.}

\medskip

\qquad\grheadb\ Set $b=\Bvalue{\sigma\le\sigma_{n-1}}$.
Then

$$\eqalign{b&
\Bsubseteq\Bvalue{\sigma\wedge\tau=\sigma\wedge\sigma_{n-1}\wedge\tau}
=\Bvalue{\sigma\wedge\tau=\sigma\wedge\tau'}\cr&
\Bsubseteq\Bvalue{\hat v_{\sigma\wedge\tau}=\hat v_{\sigma\wedge\tau'}}
   \Bcap\Bvalue{P_{\sigma\wedge\tau}\hat v_{\tau}
       =P_{\sigma\wedge\tau'}\hat v_{\tau}}\cr&
\Bsubseteq\Bvalue{\hat v_{\sigma\wedge\tau}=\hat v_{\sigma\wedge\tau'}}
   \Bcap\Bvalue{P_{\sigma\wedge\tau}\hat v_{\tau}\ge\hat v_{\sigma\wedge\tau'}}
\Bsubseteq\Bvalue{P_{\sigma\wedge\tau}\hat v_{\tau}
   \ge\hat v_{\sigma\wedge\tau}}.\cr}$$

\medskip

\qquad\grheadc\ Set

\Centerline{$b'=\Bvalue{\sigma_{n-1}\le\sigma}\Bcap\Bvalue{\sigma<\tau}
\in\frak A_{\sigma}$.}

\noindent Then

$$\eqalignno{P_{\sigma\wedge\tau}\hat v_{\tau}\times\chi b'&
=P_{\sigma}P_{\tau}\hat v_{\tau}\times\chi b'
=P_{\sigma}\hat v_{\tau}\times\chi b'
=P_{\sigma}(\hat v_{\tau}\times\chi b')
\Displaycause{because $b'\in\frak A_{\sigma}$}
=P_{\sigma}(v_{\sigma_n}\times\chi b')
\Displaycause{because
$b'\Bsubseteq d\Bsubseteq\Bvalue{\tau=\sigma_n}$}
=P_{\sigma}v_{\sigma_n}\times\chi b'
=P_{\sigma}P_{\sigma_n}v_{\sigma_n}\times\chi b'\cr&
=P_{\sigma\wedge\sigma_n}v_{\sigma_n}\times\chi b'
\ge v_{\sigma\wedge\sigma_n}\times\chi b'
=\hat v_{\sigma\wedge\tau}\times\chi b'\cr}$$

\noindent because $b'\Bsubseteq\Bvalue{\sigma\wedge\sigma_n=\sigma\wedge\tau}$.

\medskip

\qquad\grheadd\ Set $b''=\Bvalue{\tau\le\sigma}$.  Then

\Centerline{$b''
=\Bvalue{\sigma\wedge\tau=\tau}
\Bsubseteq\Bvalue{\hat v_{\sigma\wedge\tau}=\hat v_{\tau}}
      \Bcap\Bvalue{P_{\sigma\wedge\tau}\hat v_{\tau}=P_{\tau}\hat v_{\tau}}
\Bsubseteq\Bvalue{\hat v_{\sigma\wedge\tau}
     =P_{\sigma\wedge\tau}\hat v_{\tau}}$.}

\medskip

\qquad\grheade\ Now observe that $b\Bcup b'\Bcup b''=1$.   So
$\Bvalue{\hat v_{\sigma\wedge\tau}\le P_{\sigma\wedge\tau}\hat v_{\tau}}=1$,
$\hat v_{\sigma\wedge\tau}\le P_{\sigma\wedge\tau}\hat v_{\tau}$
and the induction proceeds.\ \Qed

\medskip

\quad{\bf (iii)} With 626Hb, this tells us that
$\hat v_{\sigma\wedge\tau}\le P_{\sigma\wedge\tau}\hat v_{\tau}$ whenever
$\sigma\in\Cal S$ and $\tau\in\hat\Cal S_f$.
Now suppose that $\tau'\le\tau$ in $\hat\Cal S_f$.   Then

$$\eqalign{\Bvalue{\tau'=\sigma}&
\Bsubseteq\Bvalue{\sigma\wedge\tau=\tau'}\cr&
\Bsubseteq\Bvalue{\hat v_{\sigma\wedge\tau}=\hat v_{\tau'}}
    \Bcap\Bvalue{P_{\sigma\wedge\tau}\hat v_{\tau}
                   =P_{\tau'}\hat v_{\tau}}
\Bsubseteq\Bvalue{\hat v_{\tau'}
                   \le P_{\tau'}\hat v_{\tau}}\cr}$$

\noindent because
$\hat v_{\sigma\wedge\tau}\le P_{\sigma\wedge\tau}\hat v_{\tau}$, by
(ii).   Taking the supremum over $\sigma$,
$\Bvalue{v_{\tau'}\le P_{\tau'}v_{\tau}}=1$ and
$v_{\tau'}\le P_{\tau'}v_{\tau}$.   As $\tau'$ and $\tau$ are arbitrary,
$\hatpmb v\restr\hat\Cal S_f$ is a submartingale.

\medskip

{\bf (b)(i)} Set $\gamma=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$.
Then $\|\hat v_{\tau}\|_1\le 3\gamma$ for every $\tau\in\hat\Cal S_f$.
\Prf\ If $\tau\in\hat\Cal S_f$, there are $\sigma$, $\sigma'\in\Cal S$ such that
$\sigma\le\tau\le\sigma'$, by 626Hc.   Now
$\hat v_{\tau}\le P_{\tau}v_{\sigma'}$ so

\Centerline{$\Expn(\hat v_{\tau}^+)\le\|P_{\tau}v_{\sigma'}\|_1
\le\|v_{\sigma'}\|_1$.}

\noindent On the other side, $v_{\sigma}\le P_{\sigma}\hat v_{\tau}$
so

\Centerline{$\Expn(\hat v_{\tau})=\Expn(P_{\sigma}\hat v_{\tau})
\ge\Expn(v_{\sigma})$}

\noindent and

\Centerline{$\|\hat v_{\tau}\|_1=2\Expn(\hat v_{\tau}^+)-\Expn(v_{\tau})
\le 2\|v_{\sigma'}\|_1-\Expn(v_{\sigma})
\le 2\|v_{\sigma'}\|_1+\|v_{\sigma}\|_1\le 3\gamma$.  \Qed}

\medskip

\quad{\bf (ii)} Since $\{\hat v_{\tau}:\tau\in\hat\Cal S\}
\subseteq\overline{\{\hat v_{\tau}:\tau\in\hat\Cal S_f\}}$ (626Hd, applied to
$\hatpmb v$), and $\|\,\|_1$-balls are closed in $L^0(\frak A)$ (613Bc),
$\sup_{\tau\in\hat\Cal S}\|\hat v_{\tau}\|_1\le 3\gamma$ is finite
and $\hatpmb v$ is $\|\,\|_1$-bounded.

\medskip

{\bf (c)(i)} We know from 626Db and (b) here
that $\hatpmb v$ is an $L^1$-process.
Suppose that $\tau\in\hat\Cal S$ and $\sigma\in\Cal S$.   Then
$\hat v_{\tau\wedge\sigma}\le P_{\tau\wedge\sigma}\hat v_{\tau}$.   \Prf\
Let $\epsilon>0$.   Then there is a $\delta\in\ocint{0,\epsilon}$ such that
$\Expn(|v_{\max\Cal S}-\hat v_{\tau}|\times\chi a)\le\epsilon$ whenever
$\bar\mu a\le\delta$.   As $\sup_{\rho\in\Cal S}\Bvalue{\tau=\rho}=1$, there
is a finite set $J\subseteq\Cal S$ such that
$a=1\Bsetminus\sup_{\rho\in J}\Bvalue{\tau=\rho}$ has measure at most $\delta$.
Since $a\in\frak A_{\tau}\subseteq\frak A_{\max\Cal S}$, there is a $\tau'$ such
that

\Centerline{$1\Bsetminus a\Bsubseteq\Bvalue{\tau'=\tau}$,
\quad$a\Bsubseteq\Bvalue{\tau'=\max\Cal S}$.}

\noindent Now

\Centerline{$\|v_{\tau'}-v_{\tau}\|_1
=\Expn(\chi a\times|v_{\max\Cal S}-v_{\tau}|\le\epsilon$}

\noindent and $\tau'\in\hat\Cal S_f$, so
$\hat v_{\sigma\wedge\tau'}\le P_{\sigma\wedge\tau'}\hat v_{\tau'}$
and

\Centerline{$1\Bsetminus a\Bsubseteq\Bvalue{\hat v_{\sigma\wedge\tau}
           \le P_{\sigma\wedge\tau}\hat v_{\tau'}}$,
\quad$\theta((\hat v_{\sigma\wedge\tau}-P_{\sigma\wedge\tau}\hat v_{\tau'})^+)
\le\bar\mu a\le\epsilon$.}

\noindent On the other hand,

\Centerline{$\theta(P_{\sigma\wedge\tau}\hat v_{\tau'}
                -P_{\sigma\wedge\tau}\hat v_{\tau})
\le\|P_{\sigma\wedge\tau}\hat v_{\tau'}
                -P_{\sigma\wedge\tau}\hat v_{\tau}\|_1
\le\|\hat v_{\tau'}-\hat v_{\tau}\|_1\le\epsilon$,}

\noindent so

\Centerline{
$\theta((\hat v_{\sigma\wedge\tau}-P_{\sigma\wedge\tau}\hat v_{\tau})^+)
\le\theta((\hat v_{\sigma\wedge\tau}-P_{\sigma\wedge\tau}\hat v_{\tau'})^+
   +|P_{\sigma\wedge\tau}\hat v_{\tau'}
                -P_{\sigma\wedge\tau}\hat v_{\tau}|)
\le 2\epsilon$.}

As $\epsilon$ is arbitrary,
$\hat v_{\sigma\wedge\tau}\le P_{\sigma\wedge\tau}\hat v_{\tau}$.\ \Qed

\medskip

\quad{\bf (ii)} Repeating the argument of (a-iii), we now see that if
$\tau'\in\hat\Cal S$ and $\tau'\le\tau$ then
$\Bvalue{\tau'=\sigma}
\Bsubseteq\Bvalue{\hat v_{\tau'}\le P_{\tau'}\hat v_{\tau}}$
for every $\sigma\in\Cal S$ and $\hat v_{\tau'}\le P_{\tau'}\hat v_{\tau}$.
So $\hatpmb v$ is a submartingale.
}%end of proof of 626I

\leader{626J}{Proposition} Suppose that
$\Cal S$ is a sublattice of $\Cal T$ and
$\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ is an
$L^1$-process.   Then we have an adapted interval function\cmmnt{ (definition:
613C)} $P\Delta\pmb v$ defined by saying that
$(P\Delta\pmb v)(\sigma,\tau)=P_{\sigma}v_{\tau}-v_{\sigma}$
whenever $\sigma\le\tau$ in $\Cal S$.

\proof{ Of course
$P_{\sigma}v_{\tau}-v_{\sigma}\in L^0(\frak A_{\sigma}\subseteq\frak A_{\tau}$
whenever $\sigma\le\tau$ in $\Cal S$, and $P_{\sigma}v_{\sigma}-v_{\sigma}=0$
for every $\sigma\in\Cal S$.
Suppose that $\sigma$, $\sigma'$, $\tau$, $\tau'\in\Cal S$,
$\sigma\le\sigma'\le\tau'\le\tau$ and $b\in\frak A_{\sigma}$ is such that
$b\Bsubseteq\Bvalue{\sigma=\sigma'}\Bcap\Bvalue{\tau'=\tau}$.
Then
$b\Bsubseteq\Bvalue{P_{\sigma}v_{\tau'}=P_{\sigma'}v_{\tau'}}$ by 622Bb.
At the same time,

\Centerline{$\chi b\times P_{\sigma}v_{\tau}
=P_{\sigma}(\chi b\times v_{\tau})
=P_{\sigma}(\chi b\times v_{\tau'})
=\chi b\times P_{\sigma}v_{\tau'}$}

\noindent so $b\Bsubseteq\Bvalue{P_{\sigma}v_{\tau}=P_{\sigma}v_{\tau'}}$
and $b\Bsubseteq\Bvalue{P_{\sigma}v_{\tau}=P_{\sigma'}v_{\tau'}}$.
Since we also have
$b\Bsubseteq\Bvalue{v_{\sigma}=v_{\sigma'}}\Bcap\Bvalue{v_{\tau'}=v_{\tau}}$,
$b\Bsubseteq
\Bvalue{P_{\sigma}v_{\tau}-v_{\sigma}=P_{\sigma'}v_{\tau'}-v_{\sigma'}}$.
So the conditions of 613C(a-i) are satisfied by $P\Delta\pmb v$.
}%end of proof of 626J

%these are "previsible adapted interval functions"

\leader{626K}{Definitions} Corresponding to the interval functions
$P\Delta\pmb v$ of 626J,\cmmnt{ I will} write
$\Delta_e(\pmb u,Pd\pmb v)$,
$S_I(\pmb u,Pd\pmb v)$
%$\int_{\tau}^{\tau'}\pmb u\,Pd\pmb v$,
and $Q_{\Cal S}(Pd\pmb v)$
for $\Delta_e(\pmb u,d(P\Delta\pmb v))$
$S_I(\pmb u,d(P\Delta\pmb v))$
%$\int_{\tau}^{\tau'}\pmb u\,d(P\Delta\pmb v)$,
and $Q_{\Cal S}(d(P\Delta\pmb v))$
respectively\cmmnt{, as in 613F and 616B}.
\cmmnt{Similarly,}
$\Delta_e(\pmb u,|Pd\pmb v|)$,
$S_I(\pmb u,|Pd\pmb v|)$
%$\int_{\tau}^{\tau'}\pmb u\,|Pd\pmb v|$,
and $Q_{\Cal S}(|Pd\pmb v|)$
will mean
$\Delta_e(\pmb u,d|P\Delta\pmb v|)$,
$S_I(\pmb u,d|P\Delta\pmb v|)$
%$\int_{\tau}^{\tau'}\pmb u\,d|P\Delta\pmb v|$
and $Q_{\Cal S}(d|P\Delta\pmb v|)$.

Note that an $L^1$-process $\pmb v$ is a submartingale iff $P\Delta\pmb v\ge 0$.

\leader{626L}{Previsible variations} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v=\sfamilysS{v}$ an $L^1$-process.   If the weak limit

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

\noindent is defined in $L^1_{\bar\mu}$ for every $\tau\in\Cal S$,
I will say that
$\pmb v^{\#}=\sfamilysS{v^{\#}}$ is the {\bf previsible variation} of
$\pmb v$.

\leader{626M}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$.
Write $\Mdm=\Mdm(\Cal S)$ for the set of $L^1$-processes
with domain $\Cal S$ which have previsible variations.

(a) If $\pmb v\in\Mdm$ then $\pmb v^{\#}$ is an $L^1$-process\cmmnt{ (in
particular, it is fully adapted)} and
$\pmb v-\pmb v^{\#}$ is a martingale.

(b) $\Mdm$ is a linear subspace of $\Mfa(\Cal S)$, and the map
$\pmb v\mapsto\pmb v^{\#}:\Mdm\to\Mfa(\Cal S)$ is linear.

(c) If $\pmb v$ is a martingale with domain $\Cal S$, then
$\pmb v\in\Mdm$ and $\pmb v^{\#}=0$.

(d) Suppose that $\pmb v\in\Mdm$.

\quad(i) $\pmb v$ is locally moderately oscillatory iff $\pmb v^{\#}$ is
locally moderately oscillatory.

\quad(ii) $\pmb v$ is a local integrator iff $\pmb v^{\#}$ is a local
integrator.

\quad(iii) $\pmb v$ is a submartingale iff $\pmb v^{\#}$ is a submartingale.

\quad(iv) $\pmb v$ is a martingale iff $\pmb v^{\#}$ is a martingale.

(e) If $\pmb v\in\Mdm$ then $\pmb v^{\#}\in\Mdm$ and
$(\pmb v^{\#})^{\#}=\pmb v^{\#}$.

(f) Suppose that $\pmb v\in\Mdm$ and $\rho\in\Cal S$.   Express $\pmb v^{\#}$ as
$\sfamilysS{v^{\#}}$.

\quad(i) $\pmb v\restr\Cal S\wedge\rho$ has a previsible variation, which is
$\pmb v^{\#}\restr\Cal S\wedge\rho$.

\quad(ii) $\pmb v\restr\Cal S\vee\rho$ has a previsible variation, which is
$\family{\sigma}{\Cal S\vee\rho}{v^{\#}_{\sigma}-v^{\#}_{\rho}}$.

(g) If $\pmb v\in\Mfa(\Cal S)$ is such that $\pmb v\restr\Cal S\wedge\rho$ has a
previsible variation for every $\rho\in\Cal S$, then $\pmb v\in\Mdm$.

\proof{{\bf (a)} Express $\pmb v^{\#}$ as $\sfamilysS{v^{\#}}$.

\medskip

\quad{\bf (i)} If $\sigma\le\sigma'\le\tau$ in $\Cal S$ then
$P_{\sigma}v_{\sigma'}-v_{\sigma}\in L^0(\frak A_{\tau})$, so
$S_I(\tbf 1,Pd\pmb v)\in L^0(\frak A_{\tau})$ whenever
$I\in\Cal I(\Cal S\wedge\tau)$.   Since
$L^0(\frak A_{\tau})\cap L^1_{\bar\mu}
=L^1(\frak A_{\tau},\bar\mu\restr\frak A_{\tau})$ is a norm-closed subspace of
$L^1_{\bar\mu}$, therefore weakly closed, it contains
$\w1lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\tbf 1,Pd\pmb v)=v^{\#}_{\tau}$.

Suppose that $\tau\le\tau'$ in $\Cal S$ and $a=\Bvalue{\tau=\tau'}$.   Then
$a\in\frak A_{\tau}$ and $a\Bsubseteq\Bvalue{\tau=\sigma}$ whenever
$\sigma\in[\tau,\tau']$.   Accordingly

\Centerline{$\chi a\times P_{\tau}(P_{\sigma}v_{\sigma'}-v_{\sigma})
=\chi a\times P_{\tau}(v_{\sigma'}-v_{\sigma})
=P_{\tau}(\chi a\times(v_{\sigma'}-v_{\sigma})
=0$}

\noindent whenever $\tau\le\sigma\le\sigma'\le\tau'$.   But this means that
$\chi a\times S_I(\tbf 1,Pd\pmb v)=0$ whenever
$I\in\Cal I(\Cal S\cap[\tau,\tau'])$.   Now if $I\in\Cal I(\Cal S\wedge\tau')$
contains $\tau$, we shall have

\Centerline{$\chi a\times(S_I(\tbf 1,Pd\pmb v)-S_{I\wedge\tau}(\tbf 1,Pd\pmb v))
=\chi a\times S_{I\vee\tau}(\tbf 1,Pd\pmb v)
=0$}

\noindent (613G(a-i)).
As $u\mapsto\chi a\times u:L^1_{\bar\mu}\to L^1_{\bar\mu}$ is
linear and norm-continuous, therefore weakly continuous,

$$\eqalign{\chi a\times(v^{\#}_{\tau'}-v_{\tau})&
=\chi a\times\w1lim_{I\uparrow\Cal I(\Cal S\wedge\tau')}
    (S_I(\tbf 1,Pd\pmb v)-S_{I\wedge\tau}(\tbf 1,Pd\pmb v))\cr&
=\w1lim_{I\uparrow\Cal I(\Cal S\wedge\tau')}
    \chi a\times(S_I(\tbf 1,Pd\pmb v)-S_{I\wedge\tau}(\tbf 1,Pd\pmb v))
=0\cr}$$

\noindent and $a\Bsubseteq\Bvalue{v^{\#}_{\tau}=v^{\#}_{\tau'}}$.

Thus $\pmb v^{\#}$ is fully adapted.   Since $v^{\#}_{\tau}$ is defined as a
weak limit in $L^1_{\bar\mu}$ for every $\tau$, $\pmb v^{\#}$ is an
$L^1$-process.

\medskip

\quad{\bf (ii)} If $\tau\le\sigma\le\sigma'\le\tau'$ in $\Cal S$, then
$P_{\tau}(P_{\sigma}v_{\sigma'}-v_{\sigma})=P_{\tau}(v_{\sigma'}-v_{\tau})$;
consequently $P_{\tau}S_I(\tbf 1,Pd\pmb v)=P_{\tau}v_{\tau'}-v_{\tau}$ whenever
$I\in\Cal I(\Cal S\cap[\tau,\tau'])$.   Since
$P_{\tau}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ is norm-continuous, therefore weakly
continuous,
$P_{\tau}(v^{\#}_{\tau'}-v^{\#}_{\tau})=P_{\tau}(v_{\tau'}-v_{\tau}$, that is,
$P_{\tau}(v_{\tau'}-v^{\#}_{\tau'})=v_{\tau}-v^{\#}_{\tau}$.   As $\tau$ and
$\tau'$ are arbitrary, $\pmb v-\pmb v^{\#}$ is a martingale.

\medskip

{\bf (b)} If $\pmb u$, $\pmb v\in\Mdm$ and $\alpha\in\Bbb R$,
$\pmb u+\pmb v$ and $\alpha\pmb u$ are $L^1$-processes.   So
if $I\in\Cal I(\Cal S)$, $S_I(\tbf 1,Pd(\pmb u+\pmb v))$ and
$S_I(\tbf 1,Pd(\alpha\pmb u)$ are defined and equal to
$S_I(\tbf 1,Pd\pmb u)+S_I(\tbf 1,Pd\pmb v))$ and
$\alpha S_I(\tbf 1,Pd\pmb u$ respectively.   As the
weak topology on $L^1_{\bar\mu}$ is a linear space topology.
the weak limits

\Centerline{$\w1lim_{I\uparrow\Cal S\wedge\tau}S_I(\tbf 1,Pd(\pmb u+\pmb v))$,
\quad$\w1lim_{I\uparrow\Cal S\wedge\tau}S_I(\tbf 1,Pd(\alpha\pmb u))$}

\noindent are defined and equal to

\Centerline{$\w1lim_{I\uparrow\Cal S\wedge\tau}S_I(\tbf 1,Pd\pmb u)
  +\w1lim_{I\uparrow\Cal S\wedge\tau}S_I(\tbf 1,Pd\pmb v)$,
\quad$\alpha\w1lim_{I\uparrow\Cal S\wedge\tau}S_I(\tbf 1,Pd\pmb u)$}

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

\medskip

{\bf (c)} If $\pmb v=\sfamilysS{v}$ is a martingale, then
$P_{\sigma}v_{\tau}-v_{\sigma}$ whenever $\sigma\le\tau$ in $\Cal S$, so
$S_I(\tbf 1,Pd\pmb v)=0$ for every $I\in\Cal I(\Cal S)$ and $v^{\#}_{\tau}$ is
defined and zero for every $\tau\in\Cal S$.

\medskip

{\bf (d)} We know that $\pmb v^{\#}-\pmb v$ is a martingale, therefore a local
integrator and locally moderately oscillatory (622H), and
$\pmb v^{\#}-\pmb v$ is a submartingale.   Accordingly
$\pmb v^{\#}=\pmb v+(\pmb v^{\#}-\pmb v)$ will have any of these properties iff
$\pmb v$ does (615F(b-iii), 616Qa, 626Cd, 622Db).

\medskip

{\bf (e)} $(\pmb v^{\#})^{\#}=\pmb v^{\#}-(\pmb v-\pmb v^{\#})^{\#}$ is defined
and equal to $\pmb v^{\#}$.

\medskip

{\bf (f)(i)} If $\sigma\in\Cal S\wedge\rho$ then the calculation

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

\noindent yields the same result if we interpret the right-hand side as

\Centerline{$\w1lim_{I\uparrow\Cal I((\Cal S\wedge\rho)\wedge\sigma)}
   S_I(\tbf 1,Pd(\pmb v\restr\Cal S\wedge\rho))$.}

\medskip

\quad{\bf (ii)} If $\sigma\in\Cal S\vee\rho$ then for any
$I\in\Cal I(\Cal S\wedge\sigma)$ containing $\rho$ we have

$$\eqalignno{v^{\#}_{\sigma}&
=\w1lim_{I\uparrow\Cal I(\Cal S\wedge\sigma)}S_I(\tbf 1,Pd\pmb v)
=\w1lim_{I\uparrow\Cal I(\Cal S\wedge\sigma)}S_{I\wedge\rho}(\tbf 1,Pd\pmb v)
  +\w1lim_{I\uparrow\Cal I(\Cal S\wedge\sigma)}S_{I\vee\rho}(\tbf 1,Pd\pmb v)
\Displaycause{613G(a-i)}
=\w1lim_{I\uparrow\Cal I(\Cal S\wedge\rho)}S_I(\tbf 1,Pd\pmb v)
  +\w1lim_{I\uparrow\Cal I((\Cal S\wedge\sigma)\vee\rho)}S_I(\tbf 1,Pd\pmb v)
\Displaycause{613K}
=v^{\#}_{\rho}
  +\w1lim_{I\uparrow\Cal I((\Cal S\vee\rho)\wedge\sigma)}
      S_I(\tbf 1,Pd(\pmb v\restr\Cal S\vee\rho)).
  \cr}$$

\medskip

{\bf (g)} As with (f-i), this is immediate from the definition in 626L.
}%end of proof of 626M

\leader{626N}{}\cmmnt{ In this context, it is worth having an elementary fact
set out in quotable form.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a sublattice of $\Cal T$,
$\pmb v=\sfamilysS{v}$ an $L^1$-process and $z$ a member of
$L^{\infty}(\frak A)$.   Then

\Centerline{$\Expn(z\times S_I(\tbf 1,Pd\pmb v))=\Expn(S_I(\pmb Pz,d\pmb v))$}

\noindent for every $I\in\Cal I(\Cal S)$.

\proof{ If $\sigma\le\tau$ in $\Cal S$,

\Centerline{$\Expn(z\times(P_{\sigma}v_{\tau}-v_{\sigma}))
=\Expn(z\times P_{\sigma}(v_{\tau}-v_{\sigma}))
=\Expn(P_{\sigma}z\times(v_{\tau}-v_{\sigma}))$}

\noindent by 621Cb; that is, $\Expn(z\times\Delta_e(\tbf 1,Pd\pmb v))
=\Expn(\Delta_e(\pmb Pz,d\pmb v))$ for every stopping-time interval $e$ with
endpoints in $\Cal S$.   Summing over the $I$-cells,
$\Expn(z\times S_I(\tbf 1,Pd\pmb v))=\Expn(S_I(\pmb Pz,d\pmb v))$
}%end of proof of 626N


\leader{626O}{The Doob-Meyer theorem:  first form} Let $\Cal S$ be a sublattice
of $\Cal T$ and $\pmb v=\sfamilysS{v}$ a non-negative submartingale.    Then
$\pmb v\in\Mdm(\Cal S)$ and the previsible variation $\pmb v^{\#}$ is
non-negative and non-decreasing, with starting value $0$.

\proof{{\bf (a)} If $\sigma\le\sigma'$ in $\Cal S$, then (in the language of
613C and 613E)

\Centerline{$\Delta_{c(\sigma,\sigma')}(\tbf 1,Pd\pmb v)
=P_{\sigma}v_{\sigma'}-v_{\sigma}\ge 0$}

\noindent so $S_I(\tbf 1,Pd\pmb v)\ge 0$ for every $I\in\Cal I(\Cal S)$.
The key fact is that if $\tau\in\Cal S$ then
$\{S_I(\tbf 1,Pd\pmb v):I\in\Cal I(\Cal S\wedge\tau)\}$ is
uniformly integrable.   \Prf\ Let $\epsilon>0$.
Let $\beta>0$ be such that $\Expn((v_{\tau}-\beta\chi 1)^+)\le\epsilon$
and $\alpha>0$ such that
$\Bover{2\beta}{\epsilon}\Expn(v_{\tau})\le\alpha$.   If
$I\in\Cal I(\Cal S\wedge\tau)$ and $z=S_I(\tbf 1,Pd\pmb v)$,
there are $\sigma_0\le\ldots\le\sigma_n$ in
$\Cal S\wedge\tau$ such that

\Centerline{$z=\sum_{i=0}^{n-1}
v_{\sigma_i}\times(P_{\sigma_i}v_{\sigma_{i+1}}-v_{\sigma_i})$.}

\noindent Now $(v_{\sigma_0},\ldots,v_{\sigma_n})$ is a
non-negative submartingale adapted to
$(\frak A_{\sigma_0},\ldots,\frak A_{\sigma_n})$.   So
621J tells us that

\Centerline{$\Expn(z\times\chi\Bvalue{z>\alpha})
\le 3f(\bover2{\alpha}\Expn(v_{\sigma_n}))$}

\noindent where

\Centerline{$f(\delta)\le\beta\delta+\Expn((v_{\sigma_n}-\beta\chi 1)^+)$}

\noindent for $\delta\ge 0$.   But

\Centerline{$(v_{\sigma_n}-\beta\chi 1)^+
\le(P_{\sigma_n}v_{\tau}-\beta\chi 1)^+
=(P_{\sigma_n}(v_{\tau}-\beta\chi 1))\vee 0
\le P_{\sigma_n}((v_{\tau}-\beta\chi 1)\vee 0)$}

\noindent so

\Centerline{$\Expn((v_{\sigma_n}-\beta\chi 1)^+)
\le\Expn(P_{\sigma_n}((v_{\tau}-\beta\chi 1)^+))
=\Expn((v_{\tau}-\beta\chi 1)^+)
\le\epsilon$}

\noindent and

\Centerline{$\Bover2{\alpha}\Expn(v_{\sigma_n})
\le\Bover2{\alpha}\Expn(v_{\tau})
\le\Bover{\epsilon}{\beta}$.}

\noindent So $f(\bover2{\alpha}\Expn(v_{\sigma_n}))\le 2\epsilon$
and $\Expn(z\times\chi\Bvalue{z>\alpha})\le 6\epsilon$.   Of course
$z\ge 0$, so $\Expn(|z|\times\chi\Bvalue{|z|>\alpha})\le 6\epsilon$, and this is
true whenever $z\in\{S_I(\tbf 1,Pd\pmb v):I\in\Cal I(\Cal S\wedge\tau)\}$.
As $\epsilon$ is arbitrary,
$\{S_I(\tbf 1,Pd\pmb v):I\in\Cal I(\Cal S\wedge\tau)\}$ is uniformly integrable
(621Bb).\ \Qed

\medskip

{\bf (b)} Let $\Cal F$ be an ultrafilter on $\Cal I(\Cal S)$ such that
$\{I:\sigma\in I\in\Cal I(\Cal S)\}$ belongs to $\Cal F$ for every
$\sigma\in\Cal S$.

\medskip

\quad{\bf (i)} For $\tau\in\Cal S$ and $I\in\Cal I(\Cal S)$ write
$I\wedge\tau$ for $\{\sigma\wedge\tau:\sigma\in I\}\in\Cal I(\Cal S\wedge\tau)$.
Because $\{S_I(\tbf 1,Pd\pmb v):I\in\Cal I(\Cal S\wedge\tau)\}$ is uniformly
integrable, therefore relatively weakly compact in $L^1_{\bar\mu}$ (621Bd),
the weak limit
$v^{\#}_{\tau}=\w1lim_{I\to\Cal F}S_{I\wedge\tau}(\tbf 1,Pd\pmb v)$
is defined in $L^1_{\bar\mu}$.

\medskip

\quad{\bf (ii)} The arguments of the proof of 626Ma, replacing every
`$\lim_{I\uparrow\Cal I(\Cal S)}$' by `$\lim_{I\to\Cal F}$', now show that
$\pmb v^{\#}$ is an $L^1$-process and that $\pmb v-\pmb v^{\#}$ is a martingale.

\medskip

\quad{\bf (iii)} If $\sigma\le\tau$ in $\Cal S$ and $I\in\Cal I(\Cal S)$ contains
both $\sigma$ and $\tau$, then

\Centerline{$S_{I\wedge\tau}(\tbf 1,Pd\pmb v)
=S_{I\wedge\sigma}(\tbf 1,Pd\pmb v)+S_{I\cap[\sigma,\tau]}(\tbf 1,Pd\pmb v)
\ge S_{I\wedge\sigma}(\tbf 1,Pd\pmb v)$}

\noindent (613G(a-i) again).   Taking the limit,

\Centerline{$v^{\#}_{\tau}=\w1lim_{I\to\Cal F}S_{I\wedge\tau}(\tbf 1,Pd\pmb v)
\ge\w1lim_{I\to\Cal F}S_{I\wedge\tau}(\tbf 1,Pd\pmb v)=v^{\#}_{\sigma}$}

\noindent because the weak topology of $L^1_{\bar\mu}$ is a linear space
topology for which the positive cone $\{u:u\ge 0\}$ is closed.   Similarly,
$v^{\#}_{\tau}\ge 0$ for every $\tau\in\Cal S$.
Thus $\pmb v^{\#}=\sfamilysS{v^{\#}}$ is non-negative and non-decreasing.

\medskip

{\bf (c)} Now take any $z\in L^{\infty}(\frak A)$ and consider the martingale
$\pmb Pz=\family{\sigma}{\Cal S}{P_{\sigma}z}$.

\medskip

\quad{\bf (i)} If $I\in\Cal I(\Cal S)$ then

\Centerline{$\Expn(z\times S_I(\tbf 1,Pd\pmb v))
=\Expn(S_I(\pmb Pz,d\pmb v))
=\Expn(S_I(\pmb Pz,d\pmb v^{\#}))$}

\noindent\Prf\ If $\sigma\le\tau$ in $\Cal S$ then

\Centerline{$P_{\sigma}(v_{\tau}-v^{\#}_{\tau})
=v_{\sigma}-v^{\#}_{\sigma}$,
\quad$P_{\sigma}v_{\tau}-v_{\sigma}
=P_{\sigma}v^{\#}_{\tau}-v^{\#}_{\sigma}$}

\noindent because $\pmb v-\pmb v^{\#}$ is a martingale.   So
$P\Delta\pmb v=P\Delta\pmb v^{\#}$ and

\Centerline{$\Expn(S_I(\pmb Pz,d\pmb v))
=\Expn(z\times S_I(\tbf 1,Pd\pmb v))
=\Expn(z\times S_I(\tbf 1,Pd\pmb v^{\#}))
=\Expn(S_I(\pmb Pz,d\pmb v^{\#}))$}

\noindent by 626N.\ \Qed

\medskip

\quad{\bf (ii)} Suppose that $\tau\in\Cal S$.
Taking the limit as $I\rightarrow\Cal F$,

$$\eqalign{\Expn(z\times v_{\tau}^{\#})&
=\Expn(z\times\1lim_{I\to\Cal F}S_{I\wedge\tau}(\tbf 1,Pd\pmb v))
   \cr&
=\lim_{I\to\Cal F}\Expn(z\times S_{I\wedge\tau}(\tbf 1,Pd\pmb v))
=\lim_{I\to\Cal F}
   \Expn(z\times S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#})).\cr}$$

\noindent But if $\sigma_0\le\ldots\le\sigma_n\le\tau$ in $\Cal S$,

$$\eqalignno{|\sum_{i=0}^{n-1}P_{\sigma_i}z
     \times(v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i})|&
\le\sum_{i=0}^{n-1}\|P_{\sigma_i}z\|_{\infty}
    |v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i}|
\le\sum_{i=0}^{n-1}\|z\|_{\infty}(v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i})
\Displaycause{because $\pmb v^{\#}$ is non-decreasing}
=\|z\|_{\infty}(v^{\#}_{\sigma_n}-v^{\#}_{\sigma_0})
\le\|z\|_{\infty}v^{\#}_{\tau}.\cr}$$

\noindent This shows that
$\{S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#}):I\in\Cal S\wedge\tau\}$ is
order-bounded in $L^1_{\bar\mu}$, therefore uniformly integrable.   Now we know
also that $\pmb Pz$ is moderately oscillatory (622G), while
$\pmb v^{\#}$ is a local integrator (616Ra), so

\Centerline{$\int_{\Cal S\wedge\tau}\pmb Pz\,d\pmb v^{\#}
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\pmb Pz,d\pmb v^{\#})
=\lim_{I\uparrow\Cal I(\Cal S)}S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#})$}

\noindent (see 613K) is defined, and must be equal to
$\1lim_{I\uparrow\Cal I(\Cal S)}S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#})$, by
621B(c-ii).   But as $\Cal F$ includes the filter generated by
$\{\{I:\sigma\in I\in\Cal I(\Cal S)\}:\sigma\in\Cal S\}$,
this is also
$\1lim_{I\to\Cal F}S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#})$.
Accordingly

$$\eqalign{\Expn(z\times v_{\tau}^{\#})&
=\lim_{I\to\Cal F}
   \Expn(z\times S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#}))
=\Expn(z\times\1lim_{I\to\Cal F}S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#}))\cr&
=\Expn(z\times\1lim_{I\uparrow\Cal I(\Cal S)}
   S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#}))
=\lim_{I\uparrow\Cal I(\Cal S)}
   \Expn(z\times S_{I\wedge\tau}(\pmb Pz,d\pmb v^{\#}))\cr&
=\lim_{I\uparrow\Cal I(\Cal S)}
   \Expn(S_{I\wedge\tau}(\pmb Pz,d\pmb v))
=\lim_{I\uparrow\Cal I(\Cal S)}
   \Expn(z\times S_{I\wedge\tau}(\tbf 1,Pd\pmb v))\cr&
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau}
   \Expn(S_I(\pmb Pz,d\pmb v))
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}
   \Expn(z\times S_I(\tbf 1,Pd\pmb v)).\cr}$$

\noindent Since this is true for every
$z\in L^{\infty}(\frak A)\cong(L^1_{\bar\mu})'$,

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

\noindent Thus $\pmb v^{\#}$ is the previsible variation of $\pmb v$
as defined in 626L.

\medskip

{\bf (d)} To find the starting value
$v^{\#}_{\downarrow}=\lim_{\sigma\downarrow\Cal S}v^{\#}_{\sigma}$ (613Bk),
consider the case $z=\chi 1$ in the formula of (c-i) above.   We have
$\pmb Pz=\tbf 1$ so $\Expn(S_I(\tbf 1,Pd\pmb v)=\Expn(S_I(\tbf 1,d\pmb v))$
for every $I\in\Cal I(\Cal S)$.   Taking the limit as in 626L,

$$\eqalign{\Expn(v^{\#}_{\tau})&
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}\Expn(S_I(\tbf 1,d\pmb v))
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}\Expn(v_{\max I}-v_{\min I})\cr&
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}\Expn(v_{\tau}-v_{\min I})
=\Expn(v_{\tau})-\lim_{\sigma\downarrow\Cal S}\Expn(v_{\sigma})
=\Expn(v_{\tau})-\inf_{\sigma\in\Cal S}\Expn(v_{\sigma})\cr}$$

\noindent for every $\tau\in\Cal S$.   But this means that
$\lim_{\tau\downarrow\Cal S}\Expn(v^{\#}_{\tau})=0$, and as
$v^{\#}_{\tau}\ge 0$ for every $\tau\in\Cal S$ and $\pmb v^{\#}$ is
non-decreasing,
$\lim_{\tau\downarrow\Cal S}v^{\#}_{\tau}
=\inf_{\tau\in\Cal S}v^{\#}_{\tau}=0$.
}%end of proof of 626O

\leader{626P}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$,
$A\subseteq\Cal S$ a non-empty downwards-directed set, and
$R_A:\Mlmo(\Cal S)\to\Mlmo(\Cal S)$ the associated operator.   If
$\pmb v=\sfamilysS{v}$ is a non-negative submartingale with previsible variation
$\pmb v^{\#}=\sfamilysS{v^{\#}}$, then the previsible
variation $R_A(\pmb v)^{\#}$ of $R_A(\pmb v)$ is $R_A(\pmb v^{\#})$.

\proof{{\bf (a)} By 626F, $R_A(\pmb v)=\pmb v'=\sfamilysS{v'}$ is defined and
a submartingale and by 623Ba it is
non-negative, so the previsible variations $\pmb v^{\#}$ and
$\pmb v^{\prime\#}=\sfamilysS{v^{\prime\#}}$ are defined (626O).   Also
$\pmb v^{\#}$ is a non-negative submartingale
so $R_A(\pmb v^{\#})=\pmb v^{\#\prime}=\sfamilysS{v^{\#\prime}}$ is defined.

\medskip

{\bf (b)} Consider first the case in which $A=\{\rho\}$ is a singleton, so that
$v'_{\sigma}=v_{\sigma\wedge\rho}$ and
$v^{\#\prime}_{\sigma}=v^{\#}_{\sigma\wedge\rho}$ for every $\sigma\in\Cal S$.
In this case, $\pmb v\restr\Cal S\wedge\rho=\pmb v'\restr\Cal S\wedge\rho$ and

\Centerline{$\pmb v^{\#\prime}\restr\Cal S\wedge\rho
=\pmb v^{\#}\restr\Cal S\wedge\rho=\pmb v^{\prime\#}\restr\Cal S\wedge\rho$}

\noindent by 626M(f-i).   On the other side, $\pmb v'\restr\Cal S\vee\rho$ is
constant so $(\pmb v'\restr\Cal S\vee\rho)^{\#}=0$ and
$\pmb v^{\prime\#}\restr\Cal S\vee\rho$ is constant with value
$v^{\prime\#}_{\rho}$ (626M(f-ii)).   At the same time, $\pmb v^{\#\prime}$ is
constant with value $v^{\#\prime}_{\rho}=v^{\prime\#}_{\rho}$ on
$\Cal S\vee\rho$.   Thus $\pmb v^{\prime\#}$ and $\pmb v^{\#\prime}$ agree on
both $\Cal S\wedge\rho$ and $\Cal S\vee\rho$;  as they are fully adapted
processes, they agree on $\Cal S$ and are identical.

\medskip

{\bf (c)} Still supposing that $A=\{\rho\}$, take $\tau\in\Cal S$ and
$I\in\Cal I(\Cal S\wedge\tau)$ such that $\tau\wedge\rho\in I$.   Then
$\|S_I(\tbf 1,Pd\pmb v^{\#})-S_I(\tbf 1,Pd\pmb v^{\#\prime}\|_1
\le\|v^{\#}_{\tau}-v^{\#}_{\tau\wedge\rho}\|_1$.
\Prf\ Let $(\sigma_0,\ldots,\sigma_n)$ be a sequence linearly generating the
$I$-cells such that $\tau\wedge\rho=\sigma_k$ for some $k\le n$.   For
$i<k$, $\sigma_{i+1}\le\rho$ so

\Centerline{$P_{\sigma_i}v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i}
=P_{\sigma_i}v^{\#}_{\sigma_{i+1}\wedge\rho}-v^{\#}_{\sigma_i\wedge\rho}$;}

\noindent summing over $i<k$,
$S_{I\wedge\tau\wedge\rho}(\tbf 1,Pd\pmb v^{\#})
=S_{I\wedge\tau\wedge\rho}(\tbf 1,Pd\pmb v^{\#\prime})$.   On the other
side, if $k\le i\le n$, we have $\tau\wedge\rho\le\sigma_i\le\tau$ so
$\sigma_i\wedge\rho=\tau\wedge\rho$.   Now if $k\le i<n$,

\Centerline{$P_{\sigma_i}v^{\#}_{\sigma_{i+1}\wedge\rho}
=\tau\wedge\rho=v^{\#}_{\sigma_i\wedge\rho}$;}

\noindent it follows that
$S_{I\vee(\tau\wedge\rho)}(\tbf 1,Pd\pmb v^{\#\prime})=0$.   So

$$\eqalign{&\|S_I(\tbf 1,Pd\pmb v^{\#})
    -S_I(\tbf 1,Pd\pmb v^{\#\prime}\|_1\cr&
\mskip100mu
=\|S_{I\vee(\tau\wedge\rho)}(\tbf 1,Pd\pmb v^{\#})
   -S_{I\vee(\tau\wedge\rho)}(\tbf 1,Pd\pmb v^{\#\prime}\|_1\cr&
\mskip100mu
=\|S_{I\vee(\tau\wedge\rho)}(\tbf 1,Pd\pmb v^{\#})\|_1
=\|\sum_{i=k}^{n-1}P_{\sigma_i}v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i})\|_1\cr&
\mskip100mu
=\sum_{i=k}^{n-1}\Expn(P_{\sigma_i}v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i})
=\sum_{i=k}^{n-1}\Expn(v^{\#}_{\sigma_{i+1}}-v^{\#}_{\sigma_i})\cr&
\mskip100mu
=\Expn(v^{\#}_{\sigma_n}-v^{\#}_{\sigma_k})
\le\Expn(v^{\#}_{\tau}-v^{\#}_{\tau\wedge\rho})
=\|v^{\#}_{\tau}-v^{\#}_{\tau\wedge\rho}\|_1.  \text{ \Qed}\cr}$$

\medskip

{\bf (d)} Returning to the case of general $A$, we have
$\|v^{\#}_{\tau}-v^{\prime\#}_{\tau}\|_1
\le\|v^{\#}_{\tau}-v^{\#\prime}_{\tau}\|_1$ for every $\tau\in\Cal S$.
\Prf\ Whenever $\sigma\le\sigma'$ in $\Cal S\wedge\tau$,

$$\eqalignno{P_{\sigma}v'_{\sigma'}&
=P_{\sigma}(\lim_{\rho\downarrow A}v_{\sigma'\wedge\rho})
=P_{\sigma}(\1lim_{\rho\downarrow A}v_{\sigma'\wedge\rho})
\Displaycause{626E}
=\1lim_{\rho\downarrow A}P_{\sigma}v_{\sigma'\wedge\rho}\cr}$$

\noindent because $P_{\sigma}:L^1_{\bar\mu}\to L^1_{\bar\mu}$ is
$\|\,\|_1$-continuous.   Similarly,
$v'_{\sigma}=\1lim_{\rho\downarrow A}v_{\sigma\wedge\rho}$ so
$P_{\sigma}v'_{\sigma}-v_{\sigma}
=\1lim_{\rho\downarrow A}P_{\sigma}v_{\sigma'\wedge\rho}-v_{\sigma\wedge\rho}$.
Another way of expressing this is to say that if $e$ is a stopping time
interval with endpoints in $\Cal S\wedge\tau$,

\Centerline{$\Delta_e(\tbf 1,Pd\pmb v')
=\1lim_{\rho\downarrow A}\Delta_e(\tbf 1,Pd\pmb v')$.}

\noindent It follows that if $I\in\Cal I(\Cal S\wedge\tau)$ then

$$\eqalignno{S_I(\tbf 1,Pd\pmb v')&
=\1lim_{\rho\downarrow A}S_I(\tbf 1,PdR_{\rho}(\pmb v))\cr&
=\1lim_{\rho\downarrow A}S_I(\tbf 1,Pd(R_{\rho}(\pmb v)^{\#}))
=\1lim_{\rho\downarrow A}S_I(\tbf 1,PdR_{\rho}(\pmb v^{\#}))\cr}$$

\noindent by (b).   Consequently

$$\eqalign{\|S_I(\tbf 1,Pd\pmb v)-S_I(\tbf 1,Pd\pmb v')\|_1&
\le\sup_{\rho\in A}\|S_I(\tbf 1,Pd\pmb v)
     -S_I(\tbf 1,Pd\pmb v')\|_1\cr&
\le\sup_{\rho\in A}\|v^{\#}_{\tau}-v^{\#}_{\tau\wedge\rho}\|_1
=\|v^{\#}_{\tau}-v^{\#\prime}_{\tau}\|_1\cr}$$

\noindent because
$v^{\#\prime}_{\tau}=\1lim_{\rho\downarrow A}v^{\#}_{\tau\wedge\rho}$, by
626E again.\ \Qed

\medskip

{\bf (e)} Now take $\tau\in\Cal S$ and $\epsilon>0$.   Then there is a
$\rho\in A$ such that
$\|v^{\#}_{\tau\wedge\rho}-v^{\#\prime}_{\tau}\|_1\le\epsilon$.   Write
$\pmb w=R_{\{\rho\}}(\pmb v)$;  since $\pmb w$ is a non-negative submartingale,
everything above can be applied to $\pmb w$.
Since $R_AR_{\rho}=R_A$ (623Cc), $\pmb w'=R_A(\pmb w)$ is equal to
$\pmb v'$.   From (b) we see that the previsible variation $\pmb w^{\#}$
is $R_{\{\rho\}}(\pmb v^{\#})$ and therefore
$\pmb w^{\#\prime}=R_A(\pmb w^{\#})$ is equal to $\pmb v^{\#\prime}$.
Now (d) tells us that

\Centerline{$\|w^{\#}_{\tau}-w^{\prime\#}_{\tau}\|_1
\le\|w^{\#}_{\tau}-w^{\#\prime}_{\tau}\|_1$}

\noindent that is, that

\Centerline{$\|v^{\#}_{\tau\wedge\rho}-v^{\prime\#}_{\tau}\|_1
\le\|v^{\#}_{\tau\wedge\rho}-v^{\#\prime}_{\tau}\|_1\le\epsilon$}

\noindent and

\Centerline{$\|v^{\prime\#}_{\tau}-v^{\#\prime}_{\tau}\|_1\le 2\epsilon$.}

\noindent As $\tau$ and $\epsilon$ are arbitrary,
$\pmb v^{\prime\#}=\pmb v^{\#\prime}$, that is,
$R_A(\pmb v)^{\#}=R_A(\pmb v^{\#})$.
}%end of proof of 626P

\leader{626Q}{The Doob-Meyer theorem:  second form} Let $\Cal S$ be a sublattice
of $\Cal T$ and $\pmb v=\sfamilysS{v}$ a submartingale such that
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below.   Then $\pmb v$ is
expressible as the sum of a non-negative non-decreasing fully adapted process
and a virtually local martingale.

\proof{{\bf (a)} For the time being (down to the end of (d) below),
suppose that $\Cal S$ is finitely full.

\medskip

\quad{\bf (i)} For $\tau\in\Cal S$ and $M\ge 0$ set

\Centerline{$A_{M\tau}=\{\rho:\rho\in\Cal S,\,\Bvalue{\rho<\tau}
\Bsubseteq\Bvalue{|v_{\rho}|\ge M}\}$,}
As

\Centerline{$|u'_{\sigma}|\le\sup_{\rho\in A}|u_{\sigma\wedge\rho}|
\le\sup|\pmb u|$}

\noindent for every $\sigma\in\Cal S$, $\pmb u'$ is order-bounded
\noindent so that $\tau\in A_{\tau M}$ and $A_{\tau M}$ is closed under $\wedge$
(623I;  this is where we need to suppose that $\Cal S$ is finitely full).
Set $\Cal A=\{A_{M\tau}:\tau\in\Cal S$, $M\ge 0\}$.

\medskip

\quad{\bf (ii)} If $\tau\le\tau'$ in $\Cal S$ and $0\le M\le M'$, then

\Centerline{$A_{M'\tau'}
=\{\rho:\Bvalue{\tau'\le\rho}\Bcup\Bvalue{|v_{\rho}|\ge M'}\}
\Bsubseteq\{\rho:\Bvalue{\tau\le\rho}\Bcup\Bvalue{|v_{\rho}|\ge M}\}
=A_{M\tau}$}

\noindent so $\Cal A$ is downwards-directed.

\medskip

\quad{\bf (iii)} By 626F, $R_A(\pmb v)$ is defined and is a submartingale
for every $A\in\Cal A$.

\medskip

\quad{\bf (iv)} If $\tau\in\Cal S$ and $\epsilon>0$, there is an $A\in\Cal A$
such that $\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le\epsilon$.   \Prf\
Because $\pmb v$ is a local integrator (626Dc), $\pmb v\restr\Cal S\wedge\tau$
is order-bounded;  set $\bar v=\sup_{\sigma\in\Cal S\wedge\tau}|v_{\sigma}|$.
Let $M\ge 0$ be such that $\bar\mu\Bvalue{\bar v\ge M}\le\epsilon$, and set
$A=A_{M\tau}$.   If $\rho\in A$, then

\Centerline{$\Bvalue{\rho<\tau}
\Bsubseteq\Bvalue{|v_{\rho}|\ge M}\Bcap\Bvalue{v_{\rho\wedge\tau}=v_{\rho}}
\Bsubseteq\Bvalue{|v_{\rho\wedge\tau}|\ge M}
\Bsubseteq\Bvalue{\bar v\ge M}$}

\noindent has measure at most $\epsilon$.\ \Qed

\medskip

{\bf (b)} Take any $A\in\Cal A$.

\medskip

\quad{\bf (i)} There is a martingale $\pmb w$ such that
$R_A(\pmb v)+\pmb w\ge 0$.   \Prf\ Express $A$ as $A_{M\tau}$ where
$M\ge 0$ and $\tau\in\Cal S$, and $R_A(\pmb v)$ as $\sfamilysS{v'}$.
By 626F, this is a submartingale, therefore an $L^1$-process, and
$v'_{\tau}\in L^1_{\bar\mu}$.
Set $w_{\sigma}=P_{\sigma}|v'_{\tau}|+M\chi 1$ for $\sigma\in\Cal S$, so that
$\pmb w=\sfamilysS{w}$ is a martingale.

If $\sigma\in\Cal S$ and $a=\Bvalue{v'_{\sigma}<-M}$, then
$a\in\frak A_{\sigma}$ and $a\Bsubseteq\Bvalue{v'_{\sigma}=v'_{\tau}}$
(623I(b-ii)), so

\Centerline{$a
\Bsubseteq\Bvalue{P_{\sigma}v'_{\sigma}=P_{\sigma}v'_{\tau}}
\Bsubseteq\Bvalue{v'_{\sigma}+P_{\sigma}|v'_{\tau}|\ge 0}
\Bsubseteq\Bvalue{v'_{\sigma}+w_{\sigma}\ge 0}$,}

\noindent while of course $1\Bsetminus a$ also is included in
$\Bvalue{v'_{\sigma}+M\chi 1\ge 0}
\Bsubseteq\Bvalue{v'_{\sigma}+w_{\sigma}\ge 0}$.
Thus $R_A(\pmb v)+\pmb w$ is non-negative.\ \Qed

\medskip

\quad{\bf (ii)} $R_A(\pmb v)\in\Mdm(\Cal S)$ and its previsible variation
$R_A(\pmb v)^{\#}$ is non-negative and non-decreasing, with starting value $0$.
\Prf\ $R_A(\pmb v)+\pmb w$ is a submartingale (626Cd) and we have just seen that
it is non-negative.   By 626O, it has a previsible variation
$(R_A(\pmb v)+\pmb w)^{\#}$ with these properties.   But now
$R_A(\pmb v)^{\#}$ is defined and equal to $(R_A(\pmb v)+\pmb w)^{\#}$, by
626Mb-626Mc.\ \Qed

\medskip

\quad{\bf (iii)} Writing $\gamma$ for $\inf_{\sigma\in\Cal S}\Expn(v_{\sigma})$,
$\Expn(v^{\#}_{A\tau})\le\Expn(v_{\tau})-\gamma$ for every $\tau\in\Cal S$.
\Prf\ Repeating ideas from the proof of 626O,
expressing $R_A(\pmb v)$ as $\sfamilysS{v'}$.   For $\sigma\in\Cal S$,

\Centerline{$v'_{\sigma}
=\lim_{\rho\downarrow A}v_{\sigma\wedge\rho}
=\1lim_{\rho\downarrow A}v_{\sigma\wedge\rho}$}

\noindent so

\Centerline{$\Expn(v'_{\sigma})
=\lim_{\rho\downarrow A}\Expn(v_{\sigma\wedge\rho})
=\inf_{\rho\in A}\Expn(v_{\sigma\wedge\rho})$}

\noindent and $\gamma\le\Expn(v'_{\sigma})\le\Expn(v_{\sigma})$.   Next,

\Centerline{$\1lim_{\sigma\downarrow\Cal S}v^{\#}_{A\sigma}
=\lim_{\sigma\downarrow\Cal S}v^{\#}_{A\sigma}=0$}

\noindent so $\lim_{\sigma\downarrow\Cal S}\Expn(v^{\#}_{A\sigma})=0$.
Now $R_A(\pmb v)^{\#}-R_A(\pmb v)$ is a martingale, so

\Centerline{$\Expn(v^{\#}_{A\tau})
=\lim_{\sigma\downarrow\Cal S}\Expn(v^{\#}_{A\tau}-v^{\#}_{A\sigma})
=\lim_{\sigma\downarrow\Cal S}\Expn(v'_{\tau}-v'_{\sigma})
\le\Expn(v_{\tau})-\gamma$.  \Qed}

\medskip

\quad{\bf (iv)} If now $B\subseteq\Cal S$ is non-empty and downwards-directed,
$R_B(R_A(\pmb v))\in\Mdm(\Cal S)$ and
$R_B(R_A(\pmb v))^{\#}=R_B(R_A(\pmb v)^{\#})$.   \Prf\ By 626P,

$$\eqalign{R_B(R_A(\pmb v)^{\#})&
=R_B((R_A(\pmb v)+\pmb w)^{\#})
=R_B(R_A(\pmb v)+\pmb w)^{\#}\cr&
=(R_B(R_A(\pmb v))+R_B(\pmb w))^{\#}
=R_B(R_A(\pmb v))^{\#}\cr}$$

\noindent because $R_B(\pmb w)$ is a martingale (623E).   In particular,
if $B\supseteq A$ then $R_B(R_A(\pmb v)^{\#})=R_B(\pmb v)^{\#}$ (623Cc).\ \Qed

\medskip

{\bf (c)} Thus we have a family $\langle R_A(\pmb v)^{\#}\rangle_{A\in\Cal A}$
of previsible variations, all non-decreasing and non-negative, and
$R_B(\pmb v)^{\#}=R_B(R_A(\pmb v)^{\#})$ whenever $A$, $B\in\Cal A$ and
$A\subseteq B$.

\medskip

\quad{\bf (i)} Expressing $R_A(\pmb v)^{\#}$ as
$\family{\sigma}{\Cal S}{v^{\#}_{A\sigma}}$ for $A\in\Cal A$, we see that if
$A\subseteq B$ in $\Cal A$ then

\Centerline{$v^{\#}_{B\sigma}=\lim_{\rho\downarrow B}v^{\#}_{A,\sigma\wedge\rho}
\le v^{\#}_{A\sigma}$}

\noindent for every $\sigma$.   We know also that $v^{\#}_{A\sigma}\ge 0$ and
$\Expn(v^{\#}_{A\sigma})\le\Expn(v_{\sigma})-\gamma$ for every $A\in\Cal A$.
As $\Cal A$ is downwards-directed,
$\{v^{\#}_{A\sigma}:A\in\Cal A\}$ is upwards-directed and $\|\,\|_1$-bounded,
so $\1lim_{A\downarrow\Cal A}v^{\#}_{A\sigma}
=\sup_{A\in\Cal A}v^{\#}_{A\sigma}$ is defined for every $\sigma\in\Cal S$;   I
will call it $v^{\#}_{\sigma}$.

\medskip

\quad{\bf (ii)} $\pmb v^{\#}=\sfamilysS{v^{\#}}$ is a fully adapted process.
\Prf\ If $\sigma\in\Cal S$, $v^{\#}_{A\sigma}\in L^0(\frak A_{\sigma})$ for
every $A\in\Cal A$, so $v^{\#}_{\sigma}=\sup_{A\in\Cal A}v^{\#}_{A\sigma}$
belongs to $L^0(\frak A_{\sigma})$ (612B(e-i)).   If $\sigma$, $\tau\in\Cal S$
and $a=\Bvalue{\sigma=\tau}$, then
$\chi a\times v^{\#}_{A\sigma}=\chi a\times v^{\#}_{A\tau}$ for every
$A\in\Cal A$, so $\chi a\times v^{\#}_{\sigma}=\chi a\times v^{\#}_{\tau}$, that
is, $a\Bsubseteq\Bvalue{v^{\#}_{\sigma}=v^{\#}_{\tau}}$.\ \Qed

\medskip

\quad{\bf (iii)} I am {\it not} claiming that
$\pmb v^{\#}=\sfamilysS{v^{\#}}$ is actually a previsible variation in the sense
of 626L, or even an $L^1$-process.
But surely it is non-negative and non-decreasing, because
$0\le v^{\#}_{A\sigma}\le v^{\#}_{A\tau}$ whenever $A\in\Cal A$ and
$\sigma\le\tau$, so $0\le v^{\#}_{\sigma}\le v^{\#}_{\tau}$ whenever
$\sigma\le\tau$ in $\Cal S$.   Consequently it is locally moderately oscillatory
(616Ra again).

\medskip

\quad{\bf (iv)} If $\tau\in\Cal S$ and $\epsilon>0$, there are an
$a\in\frak A$ and a $B\in\Cal A$ such that $\bar\mu a\le\epsilon$ and
$\Bvalue{v^{\#}_{\sigma}\ne v^{\#}_{A\sigma}}\Bsubseteq a$ whenever
$\sigma\in\Cal S\wedge\tau$, $A\in\Cal A$ and $A\subseteq B$.   \Prf\
By (a-iv), there is a $B\in\Cal A$ such that
$a=\sup_{\rho\in B}\Bvalue{\rho<\tau}$ has measure at most $\epsilon$.
If $A\in\Cal A$,  $A\subseteq B$ and $\sigma\le\tau$ in $\Cal S$, then
$R_B(\pmb v)^{\#}=R_B(R_A(\pmb v^{\#}))$ so
$v^{\#}_{B\sigma}=\lim_{\rho\downarrow B}v^{\#}_{A,\sigma\wedge\rho}$.
But for each $\rho\in B$,

\Centerline{$\Bvalue{v^{\#}_{A,\sigma\wedge\rho}\ne v^{\#}_{A\sigma}}
\Bsubseteq\Bvalue{\rho<\sigma}\Bsubseteq\Bvalue{\rho<\tau}\Bsubseteq a$,}

\noindent so $\Bvalue{v^{\#}_{B\sigma}\ne v^{\#}_{A\sigma}}\subseteq a$.
Taking the limit as $A\downarrow\Cal A$,
$\Bvalue{v^{\#}_{B\sigma}\ne v^{\#}_{\sigma}}\subseteq a$.
But now returning to an arbitrary $A\in\Cal A$ included in $B$, we also have
$\Bvalue{v^{\#}_{A\sigma}\ne v^{\#}_{\sigma}}\subseteq a$.\ \Qed

\medskip

\quad{\bf (v)} If $B\in\Cal A$, then $R_B(\pmb v^{\#})=R_B(\pmb v)^{\#}$.
\Prf\ Take $\tau\in\Cal S$.   For any $\epsilon>0$ there are an
$a\in\frak A$ and an $A^*\in\Cal A$ such that $\bar\mu a\le\epsilon$ and
$\Bvalue{v^{\#}_{\sigma}\ne v^{\#}_{A\sigma}}\Bsubseteq a$ whenever
$\sigma\in\Cal S\wedge\tau$, $A\in\Cal A$ and $A\subseteq A^*$;  take
$A\in\Cal A$ such that $A\subseteq B\cap A^*$.   Since
$R_B(\pmb v)^{\#}=R_B(R_A(\pmb v^{\#})$,

$$\eqalign{\Bvalue{\lim_{\rho\downarrow B}v^{\#}_{\tau\wedge\rho}
              \ne v^{\#}_{B\tau}}&
=\lim_{\rho\downarrow B}v^{\#}_{\tau\wedge\rho}
              \ne\lim_{\rho\downarrow B}v^{\#}_{A,\tau\wedge\rho}\cr&
\Bsubseteq\sup_{\rho\in B}
     \Bvalue{v^{\#}_{\tau\wedge\rho}\ne v^{\#}_{A,\tau\wedge\rho}}
\Bsubseteq a\cr}$$

\noindent has measure at most $\epsilon$.   As $\epsilon$ is arbitrary,
$\lim_{\rho\downarrow B}v^{\#}_{\tau\wedge\rho}=v^{\#}_{B\tau}$;  as
$\tau$ is arbitrary, $R_B(\pmb v^{\#})=R_B(\pmb v)^{\#}$.\ \Qed

\medskip

{\bf (d)} Set $\pmb w=\pmb v-\pmb v^{\#}$.   Because $\pmb v$ and $\pmb v^{\#}$
are locally moderately oscillatory, so is $\pmb w$.   For every $A\in\Cal A$,

\Centerline{$R_A(\pmb w)=R_B(\pmb v)-R_A(\pmb v^{\#})
=R_A(\pmb v)-R_A(\pmb v)^{\#}$}

\noindent is a martingale because $R_A(\pmb v)^{\#}$ is the previsible variation
of $R_A(\pmb v)$.   Note that $R_A(\pmb w)$ is actually a uniformly integrable
martingale, because if $\rho\in A$ then $R_A(\pmb w)$ is constant on
$\Cal S\vee\rho$.   It follows that $\pmb w$ is a virtually local martingale.
\Prf\
The definition in 623J referred to the fully adapted extension $\hatpmb w$ of
$\pmb w$ to the covered envelope $\hat\Cal S$ of $\Cal S$, and our assumption
here is only that $\Cal S$ is finitely full, that is, that it is equal to its
finitely-covered envelope $\hat\Cal S_f$.   But each $A\in\Cal A$ is a non-empty
downwards-directed subset of $\hat\Cal S$, so we
can speak of $R_A$ as an operator
from $\Mlmo(\hat\Cal S)$ to itself, and $\hatpmb w$ is locally moderately
oscillatory (615F(b-v)), so $R_A(\hatpmb w)$ is defined as a fully adapted
process on $\hat\Cal S$.   But
$R_A(\hatpmb w)$ extends $R_A(\pmb w)$, so must be the fully adapted extension
of $R_A(\pmb w)$, and is again a uniformly integrable martingale (622Nc).

Suppose now that $\tau\in\hat\Cal S$ and $\epsilon>0$.   Then there is a
$\sigma\in\hat\Cal S_f$ such that $\bar\mu\Bvalue{\sigma=\tau}\ge 1-\epsilon$
(626Hf).   There is an $A\in\Cal A$ such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\sigma}\le\epsilon$.   But now
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau}\le 2\epsilon$.   As $\tau$ and
$\epsilon$ are arbitrary, $\{R_A:A\in\Cal A\}$ is a sufficient family of
operators to ensure that $\pmb w$ is a virtually local martingale.\ \Qed

Accordingly $\pmb v=\pmb v^{\#}+\pmb w$ is expressed as the sum of a
non-negative non-decreasing fully adapted process and a virtually local
martingale, as required.

\medskip

{\bf (e)} This deals with the case in which $\Cal S$ is finitely full.   For the
general case, writing $\hatpmb v$ for the fully adapted extension of $\pmb v$ to
the covered envelope $\hat\Cal S$ of $\Cal S$, and $\hat\Cal S_f$ for the
finitely-covered envelope of $\Cal S$,
$\hatpmb v\restr\hat\Cal S_f$ is a
submartingale (626Ia).   By (a)-(d) above, $\hatpmb v\restr\hat\Cal S_f$
is expressible in the form
$\pmb v^{\#}+\pmb w$ where $\pmb v^{\#}$ is non-negative and non-decreasing and
$\pmb w$ is a virtually local martingale.   Now
$\pmb v=(\pmb v^{\#}\restr\Cal S)+(\pmb w\restr\Cal S)$.
$\pmb v^{\#}\restr\Cal S$ is surely non-negative and non-decreasing.   But
$\pmb w$ and $\pmb w\restr\Cal S$ have the same fully adapted extension
$\hatpmb w$ to $\hat\Cal S$, and $\hatpmb w$ and $\pmb w\restr\Cal S$ will be
virtually local martingales, as noted in 623J.   Thus we have a decomposition of
$\pmb v$ of the type claimed.
}%end of proof of 626Q

\leader{626R}{Corollary} If $\Cal S$ is a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ is a submartingale such that
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below, then
$\pmb v$ is a semi-martingale.
%If $\pmb v$ is non-negative it is a semimartingale

\proof{ We just have to look at the definition in 625D and remember that
non-negative non-decreasing processes are locally of bounded variation
(614K(c-i)).
}%end of proof of 626R

\leader{626S}{}\cmmnt{ We know that the previsible variation of a martingale
is always zero.
Otherwise, it seems that the calculation of previsible
variations is not a trivial matter, even in the most basic cases.
I go through the argument for one of my leading examples.

\medskip

\noindent}{\bf Proposition} Suppose that $T=\coint{0,\infty}$ and that
$\pmb{\iota}=\family{\tau}{\Cal T_f}{\iota_{\tau}}$ is the identity process as
described in 612F.
Then the previsible variation of $\pmb{\iota}\restr\Cal T_b$ is itself.

\proof{{\bf (a)} Take $\tau\in\Cal T_b$ and $\epsilon>0$.
Let $m\in\Bbb N$ be such that $\tau$ is less than or equal to the constant
stopping time $(m\epsilon)\varBspcheck$ and
let $J$ be the finite sublattice of $\Cal T_b$ generated by
$\{\tau\wedge(k\epsilon)\varBspcheck:k\le m\}$.   Note that $\tau=\max J$
and $\check 0=\min\Cal T_b=\min J$.   Suppose that
$I\in\Cal I(\Cal T_b\wedge\tau)$ and $J\subseteq I$.
If $e\in\Sti_0(I\wedge\tau)$,
there is a $k<m$ such that $e$ is included in the stopping-time
interval
$c(\tau\wedge(k\epsilon)\varBspcheck,\tau\wedge((k+1)\epsilon)\varBspcheck)$
and $e=c(\sigma,\sigma')$ where
$\tau\wedge(k\epsilon)\varBspcheck
\le\sigma\le\sigma'\le\tau\wedge((k+1)\epsilon)\varBspcheck$, so that
$\iota_{\sigma'}-\iota_{\sigma}\le\epsilon\chi 1$.

\medskip

{\bf (b)} Let $\langle\sigma_i\rangle_{i\le n}$ linearly generate the
$(I\wedge\tau)$-cells.   Then
$0\le\iota_{\sigma_{i+1}}-\iota_{\sigma_i}\le\epsilon\chi 1$ for
every $i<n$, $\sigma_0=\check 0$ and $\sigma_n=\tau$.   Set

\Centerline{$v_i
=\sum_{j=0}^{i-1}P_{\sigma_j}\iota_{\sigma_{j+1}}-\iota_{\sigma_j}$,
\quad$w_i=\iota_{\sigma_i}-v_i$}

\noindent for $i\le n$.   Then $v_0=0$, $v_{i+1}\in L^0(\frak A_{\sigma_i})$
and $v_i\le v_{i+1}$ for $i<n$.   So, for $i<n$,

\Centerline{$P_{\sigma_i}w_{i+1}
=P_{\sigma_i}(\iota_{\sigma_{i+1}}-v_i
    -P_{\sigma_i}\iota_{\sigma_{i+1}}+P_{\sigma_i}\iota_{\sigma_i})
=\iota_{\sigma_i}-v_i=w_i$;}

\noindent thus $\langle w_i\rangle_{i\le n}$ is a
martingale adapted to $\langle\frak A_{\sigma_i}\rangle_{i\le n}$, starting
from $w_0=\iota_{\sigma_0}$.   We have

$$\eqalign{\Expn(\sum_{i=0}^{n-1}(v_{i+1}-v_i)^2)&
=\Expn(\sum_{i=0}^{n-1}(v_{i+1}-v_i)
   \times P_{\sigma_i}(\iota_{\sigma_{i+1}}-\iota_{\sigma_i}))\cr&
\le\Expn(\sum_{i=0}^{n-1}(v_{i+1}-v_i)\times\epsilon\chi 1)
=\epsilon\Expn(\sum_{i=0}^{n-1}v_{i+1}-v_i)\cr&
=\epsilon\Expn(\sum_{i=0}^{n-1}
                 P_{\sigma_i}\iota_{\sigma_{i+1}}-\iota_{\sigma_i})
=\epsilon\Expn(\sum_{i=0}^{n-1}
                 \iota_{\sigma_{i+1}}-\iota_{\sigma_i})\cr&
=\epsilon\Expn(\iota_{\tau}-\iota_{\check 0})
=\epsilon\Expn(\iota_{\tau}),\cr}$$

\noindent and similarly

\Centerline{$\Expn(\sum_{i=0}^{n-1}
  (\iota_{\sigma_{i+1}}-\iota_{\sigma_i})^2)
\le\Expn(\sum_{i=0}^{n-1}
  (\iota_{\sigma_{i+1}}-\iota_{\sigma_i})\times\epsilon\chi 1)
\le\epsilon\Expn(\iota_{\tau})$.}

\noindent So

$$\eqalignno{\Expn((\iota_{\tau}-S_I(\tbf 1,Pd\pmb{\iota}))^2)
&=\Expn((\iota_{\tau}-v_n)^2)
=\Expn(w_n^2)
=\sum_{i=0}^{n-1}\Expn((w_{i+1}-w_i)^2)\cr
\displaycause{because $\langle w_i\rangle_{i\le n}$ is a martingale, so
$\Expn((w_{i+1}-w_i)\times(w_{j+1}-w_j))=0$ when $i\ne j$}
&\le 2\sum_{i=0}^{n-1}\Expn((\iota_{\sigma_{i+1}}-\iota_{\sigma_i})^2
   +(v_{i+1}-v_i)^2)
\le 4\epsilon\Expn(\iota_{\tau}).\cr}$$

\medskip

{\bf (c)} This is true whenever $I\in\Cal I(\Cal T_b\wedge\tau)$ includes $J$.
As $\epsilon$ is arbitrary, $\iota_{\tau}$ is the limit
$\2lim_{I\uparrow\Cal I(\Cal T_b\wedge\tau)}S_{I\wedge\tau}(\tbf 1,Pd\pmb v)$
for the norm topology of $L^2_{\bar\mu}$.   It is therefore also the limit
\penalty-100
$\w1lim_{I\uparrow\Cal I(\Cal T_b\wedge\tau)}S_I(\tbf 1,Pd\pmb v)$ for
the weak topology of $L^1_{\bar\mu}$.

\medskip

{\bf (d)} As $\tau$ is arbitrary, the previsible variation
$(\pmb{\iota}\restr\Cal T_b)^{\#}$ agrees with $\pmb{\iota}$ on $\Cal T_b$.
}%end of proof of 626S

\leader{626T}{Lemma} Let $\Cal S$ be a
full sublattice of $\Cal T$ with greatest and least members,
and $\pmb u=\sfamilysS{u}$ a
non-decreasing non-negative jump-free $L^1$-process.   Then for every
$\epsilon>0$ there is an $I\in\Cal I(\Cal S)$, containing $\min\Cal S$ and
$\max\Cal S$, such that
$\|S_I(\tbf 1,Pd\pmb u)-u_{\max\Cal S}+u_{\min\Cal S}\|_1\le\epsilon$.

\proof{ Since $0\le u_{\sigma}\le u_{\max\Cal S}\in L^1_{\bar\mu}$ for every
$\sigma\in\Cal S$, $\pmb u$ is uniformly integrable.   Set
$\delta=\bover17\epsilon$.
Construct $\sequence{i}{D_i}$, $\sequence{i}{y_i}$ and $\sequence{i}{d_i}$
from $\pmb u$ and $\delta$ as in 615M.   Note that $\min D_0=\min\Cal S$ so
$y_0=u_{\min\Cal S}$ (615Ma).   If $i\in\Bbb N$, then
$D_i\subseteq\Cal S$ is closed under $\wedge$ and
$y_i=\lim_{\sigma\downarrow D_i}u_{\sigma}$, while
$d_{i+1}\Bsubseteq d_i\Bcap\Bvalue{|y_{i+1}-y_i|\ge\delta}$ (615Mc) and
$|y_{i+1}-y_i|\le\delta\chi 1$ (618N, because $\pmb u$ is
jump-free).   Because $0\le y_i\le u_{\max\Cal S}$, or otherwise,
$y_i\in L^1_{\bar\mu}$.
Set $\frak B_i=\bigcap_{\sigma\in D_i}\frak A_{\sigma}$ and write
$Q_i:L^1_{\bar\mu}\to L^1_{\bar\mu}\cap L^0(\frak B_i)$
for the associated conditional expectation.   As $y_i\in L^0(\frak B_i)$
(615Mb), $Q_iy_i=y_i$.   Because $\pmb u$ is uniformly integrable,
$y_i=\1lim_{\sigma\downarrow D_i}u_{\sigma}$ (621B(c-ii)).

If $i\in\Bbb N$ and $\sigma\in D_{i+1}$, there is a $\sigma'\in D_i$ such that
$\sigma'\le\sigma$ (615Ma);  now $\sigma''\le\sigma'$ and
$u_{\sigma''}\le u_{\sigma'}$ for every $\sigma''\in D_i\wedge\sigma'$, so
$y_i\le u_{\sigma}$;  as $\sigma$ is arbitrary, $y_i\le y_{i+1}$.
So in fact $d_{i+1}\Bsubseteq d_i\Bcap\Bvalue{y_{i+1}-y_i=\delta}$ and
$y_i\le y_{i+1}\le y_i+\delta\chi d_i$.   Now

\Centerline{$y_i=Q_iy_i\le Q_iy_{i+1}\le Q_i(y_i+\delta\chi d_i)
=y_i+\delta\chi d_i$}

\noindent (because $d_i\in\frak B_i$), and
$(Q_iy_{i+1}-y_{i+1})\times\chi(1\Bsetminus d_i)=0$.   At the same time,
$y_{i+1}\ge y_i+\delta\chi d_{i+1}$, so

\Centerline{$(Q_iy_{i+1}-y_{i+1})^+
\le((y_i+\delta\chi d_i)-(y_i+\delta\chi d_{i+1}))^+
=\delta\chi(d_i\Bsetminus d_{i+1})$}

\noindent and

\Centerline{$\|Q_iy_{i+1}-y_{i+1}\|_1
=2\Expn((Q_iy_{i+1}-y_{i+1})^+)-\Expn(Q_iy_{i+1}-y_{i+1})
\le 2\delta\bar\mu(d_i\Bsetminus d_{i+1})$.}

\noindent Summing over $i$,

\Centerline{$\sum_{i=0}^{\infty}\|Q_iy_{i+1}-y_{i+1}\|_1\le 2\delta$.}

By 615M(c-ii), there is an $m\ge 1$ such that
$\Expn(u_{\max\Cal S}\times\chi d_m)\le\delta$.
Set $\eta=\Bover1m\delta$.
Again because every member of $D_{i+1}$
dominates some member of $D_i$, we can choose
$\sigma_m,\sigma_{m-1},\ldots,\sigma_0\in\Cal S$ such that $\sigma_m\in D_m$,
$\|u_{\sigma_m}-y_m\|_1\le\eta$ and

\Centerline{$\sigma_i\in D_i$,
\quad$\sigma_i\le\sigma_{i+1}$,
\quad$\|u_{\sigma_i}-y_i\|_1\le\eta$,
\quad$\|P_{\sigma_i}y_{i+1}-Q_iy_{i+1}\|_1\le\eta$}

\noindent for $i<m$ (using 621C(g-i) for the last clause),
while $\sigma_0=\min\Cal S$.   Since
$\Bvalue{\sigma_m<\max\Cal S}\Bsubseteq d_m$ and
$0\le u_{\sigma_m}\le u_{\max\Cal S}$,

\Centerline{$\|u_{\max\Cal S}-u_{\sigma_m}\|_1
\le\Expn(u_{\max\Cal S}\times\chi d_m)\le\delta$.}

Set $I=\{\sigma_i:i\le m\}\cup\{\max\Cal S\}$.   Then

$$\eqalign{\|S_I(\tbf 1,&Pd\pmb u)-u_{\max\Cal S}+u_{\min\Cal S}\|_1\cr&
=\|P_{\sigma_m}u_{\max\Cal S}-u_{\sigma_m}-u_{\max\Cal S}+u_{\sigma_m}
  +\sum_{i=0}^{m-1}
    P_{\sigma_i}u_{\sigma_{i+1}}-u_{\sigma_i}-u_{\sigma_{i+1}}+u_{\sigma_i}\|_1
    \cr&
\le\|P_{\sigma_m}(u_{\max\Cal S}-u_{\sigma_m})\|_1
  +\|u_{\max\Cal S}-u_{\sigma_m}\|_1
  +\sum_{i=0}^{m-1}
       \|P_{\sigma_i}u_{\sigma_{i+1}}-u_{\sigma_{i+1}}\|_1\cr&
\le 2\|u_{\max\Cal S}-u_{\sigma_m}\|_1
  +\sum_{i=0}^{m-1}\|Q_iy_{i+1}-y_{i+1}\|_1\cr&
\mskip50mu
  +\sum_{i=0}^{m-1}\|P_{\sigma_i}u_{\sigma_{i+1}}-u_{\sigma_{i+1}}
                       -Q_iy_{i+1}+y_{i+1}\|_1\cr&
\le 2\delta+2\delta
   +\sum_{i=0}^{m-1}\|P_{\sigma_i}u_{\sigma_{i+1}}-P_{\sigma_i}y_{i+1}\|_1
   +\sum_{i=0}^{m-1}\|P_{\sigma_i}y_{i+1}-Q_iy_{i+1}\|_1\cr&
\mskip50mu
   +\sum_{i=0}^{m-1}\|y_{i+1}-u_{\sigma_{i+1}}\|_1\cr&
\le 4\delta+m\eta+m\eta+m\eta
=\epsilon.\cr}$$
}%end of proof of 626T

\leader{626U}{Proposition} Let $\Cal S$ be a non-empty
full sublattice of $\Cal T$, and $\pmb v=\sfamilysS{v}$ a
non-decreasing locally jump-free $L^1$-process starting from $0$.   Then
it is equal to its previsible variation.

\proof{{\bf (a)} Because $\pmb v$ is moderately oscillatory (618Gb), its
starting value $v_{\downarrow}$ is well-defined (615G).   We are supposing
that $v_{\downarrow}$ is zero;  as $\pmb v$ is a non-decreasing $L^1$-process it
is a non-negative submartingale and has a previsible variation
$\pmb v^{\#}=\sfamilysS{v^{\#}}$.
I need to show that
$\w1lim_{I\uparrow(\Cal S\wedge\tau)}S_I(\tbf 1,Pd\pmb v)=v_{\tau}$ for every
$\tau\in\Cal S$.   Clearly it is enough to consider the case in which
$\Cal S=\Cal S\wedge\tau$.   Take $w\in L^{\infty}(\frak A)$ and $\epsilon>0$.

\medskip

{\bf (b)} If $J\in\Cal I(\Cal S)$ is non-empty, there is an $I\in\Cal I(\Cal S)$
such that $J\subseteq I$ and
$\|S_I(\tbf 1,Pd\pmb v)-v_{\max J}+v_{\min J}\|_1\le\epsilon$.   \Prf\ If
$J$ is a singleton, set $I=J$.   Otherwise, take
$(\tau_0,\ldots,\tau_k)$ linearly generating the $J$-cells, so that
$\tau_0=\min J$ and $\tau_k=\max J$.   If $j<k$,
$\pmb v\restr\Cal S\cap[\tau_j,\tau_{j+1}]$ is a non-decreasing non-negative
$L^1$-process, and is jump-free by 618Gc, while $\Cal S\cap[\tau_j,\tau_{j+1}]$
is full (see 611Md and 611Me).   So 626T tells us that there is a finite
sublattice $I_j$ of $\Cal S\cap[\tau_j,\tau_{j+1}]$, containing $\tau_j$ and
$\tau_{j+1}$, such that
$\|S_{I_j}(\tbf 1,Pd\pmb v)-v_{\tau_{j+1}}+v_{\tau_j}\|_1\le\Bover1k\epsilon$.
Let $I$ be the sublattice of $\Cal S$ generated by $J\cup\bigcup_{j<k}I_j$.
Then $\min I=\tau_0=\min J$, $\max I=\tau_k=\max J$ and
$I\cap[\tau_j,\tau_{j+1}]=I_j$ for each $j$.   So

\Centerline{$S_I(\tbf 1,Pd\pmb v)=\sum_{j=0}^{k-1}S_{I_j}(\tbf 1,Pd\pmb v)$,
\quad$v_{\max I}-v_{\min I}=\sum_{j=0}^kv_{\tau_{j+1}}-v_{\tau_j}$}

\noindent and

\Centerline{$\|S_I(\tbf 1,Pd\pmb v)-v_{\max J}+v_{\min J}\|_1
\le\sum_{j=0}^{k-1}\|S_{I_j}(\tbf 1,Pd\pmb v)-v_{\tau_{j+1}}+v_{\tau_j}\|_1
\le\epsilon$.  \Qed}

\medskip

{\bf (c)} If $J\in\Cal I(\Cal S)$ is non-empty, there is an $I\in\Cal I(\Cal S)$
such that $J\subseteq I$ and
$\|S_I(\tbf 1,Pd\pmb v)-v_{\max\Cal S}\|_1\le 2\epsilon$.
\Prf\ As in the proof of 626T, $\pmb v$ here is uniformly integrable, so
$0=\1lim_{\sigma\downarrow\Cal S}v_{\sigma}$.
Let $\sigma\in\Cal S$ be such that $\sigma\le\min J$ and
$\|v_{\sigma}\|_1\le\epsilon$.   By (b), there is an
$I\in\Cal I(\Cal S)$ such that $I\supseteq J\cup\{\sigma,\tau\}$ and
$\|S_I(\tbf 1,Pd\pmb v)-v_{\tau}+v_{\sigma}\|_1\le\epsilon$, so that
$\|S_I(\tbf 1,Pd\pmb v)-v_{\tau}\|_1\le 2\epsilon$.\ \Qed

\medskip

{\bf (d)} We know that there is a non-empty $J\in\Cal I(\Cal S)$ such that
$|\Expn(w\times(S_I(\tbf 1,Pd\pmb v)-v^{\#}_{\tau}))|\le\epsilon$ whenever
$I\in\Cal I(\Cal S)$ includes $J$.   By (c), there is such an $I$ with
$\|S_I(\tbf 1,Pd\pmb v)-v_{\tau}\|_1\le 2\epsilon$, so that
$|\Expn(w\times(S_I(\tbf 1,Pd\pmb v)-v_{\tau}))|\le 2\epsilon\|w\|_{\infty}$
and $|\Expn(w\times(v^{\#}_{\tau}-v_{\tau}))|\le\epsilon(1+2\|w\|_{\infty})$.
As $\epsilon$ and $w$ are arbitrary, $v^{\#}_{\tau}=v_{\tau}$, which is what we
needed to prove.
}

\leader{626V}{Proposition} Let $\Cal S$ be a full sublattice of
$\Cal T$, and $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$ a locally jump-free
$L^2$-martingale.   Then the quadratic variation
$\pmb v^*=\family{\sigma}{\Cal S}{v^*_{\sigma}}$ of $\pmb v$ is the
previsible variation $(\pmb v^2)^{\#}$ of the submartingale $\pmb v^2$.

\proof{ If $\Cal S$ is empty, this is trivial;  suppose otherwise.   By 622H,
$\pmb v$ is locally moderately oscillatory, so its starting value
$v_{\downarrow}$ is defined (615G again).

\medskip

{\bf (a)} Suppose to begin with that $v_{\downarrow}=0$.   By Jensen's
inequality (621Cd), $\pmb v^2$ is a submartingale.   Also $\pmb v^*$ is an
$L^1$-process (624G) so
$\pmb v^2-\pmb v^*=2\iiv(\pmb v)$ (617Ja) is a martingale (624Hb) and
$(\pmb v^2)^{\#}=(v^*)^{\#}$ (626Mc).   Now
$\pmb v^*$ is an $L^1$-process, non-decreasing and starting from $0$
(617Jb), while $\pmb v^2$ and $\iiv(\pmb v)$ are locally jump-free
(618Ga, 618R).   By 626U, $\pmb v^*=(\pmb v^*)^{\#}=(\pmb v^2)^{\#}$.

\medskip

{\bf (b)} For the general case, set
$\pmb w=\pmb v-v_{\downarrow}\tbf 1$.   Then
$\pmb w$ is a locally jump-free martingale and an $L^2$-process starting at $0$,
so $(\pmb w^2)^{\#}=\pmb w^*$.   Now, expressing $\pmb w^*$ as
$\sfamilysS{w^*}$,

$$\eqalignno{w^*_{\tau}&
=\int_{\Cal S\wedge\tau}d\pmb w^*
=\int_{\Cal S\wedge\tau}(d\pmb w)^2
\Displaycause{617I}
=\int_{\Cal S\wedge\tau}(d\pmb v)^2
\Displaycause{because the interval functions $\Delta\pmb w$, $\Delta\pmb v$ are
equal}
=v^*_{\tau}\cr}$$

\noindent for every $\tau\in\Cal S$.   At the same time,

\Centerline{$\pmb v^2-\pmb w^2=2\pmb v-v_{\downarrow}^2\tbf 1$}

\noindent is a martingale, so

\Centerline{$(\pmb v^2)^{\#}=(\pmb w^2)^{\#}=\pmb w^*=\pmb v^*$.}
}%end of proof of 626V

%\leader{626W}{}

\exercises{\leader{626X}{Basic exercises (a)}
%\spheader 626Xa
Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa, 617Xb, 618Xa and 622Xd.
Let $f:\coint{0,\infty}\to\Bbb R$
be a function and $\pmb u$ the corresponding process on $\Cal T_f$.
(i) Show that $\pmb u$ is a submartingale iff $f$ is non-decreasing.
(ii) Show that this case the previsible variation of
$\pmb u$ corresponds to the function $t\mapsto f(t)-f(0)$.
%626B

\spheader 626Xb Let $\Cal S$ be a sublattice of $\Cal T$,
$\Cal S'$ a covering ideal of $\Cal S$ and $\pmb u=\sfamilysS{u}$ a fully
adapted process.  Show that $\{u_{\sigma}:\sigma\in\Cal S\}
\subseteq\overline{\{u_{\sigma}:\sigma\in\Cal S'\}}$.
%626H

\spheader 626Xc Let $\Cal S$ be a sublattice, $\pmb v$ a fully adapted
process with domain $\Cal S$ and $\tau\in\Cal S$.
Show that $\pmb v$ is a submartingale iff
$\pmb v\restr\Cal S\wedge\tau$ and $\pmb v\restr\Cal S\vee\tau$ are both
submartingales.
%626I

\spheader 626Xd Let $\Cal S$ be a sublattice of $\Cal T$, $\pmb v$ a
non-negative submartingale with domain $\Cal S$,
and $z$ a non-negative member of
$L^0(\bigcap_{\sigma\in\Cal S}\frak A_{\sigma})$.
Show that if $z\pmb v$ (612D(e-ii)) is an
$L^1$-process then it is a submartingale, and that $z\pmb v^{\#}$ is the
previsible variation of $z\pmb v$.
%626O

\spheader 626Xe Suppose that $T=\Bbb N$ and that $\frak A$ is atomless.
Let $u\in L^0(\frak A)$ be such that $\bar\mu\Bvalue{u>1-\alpha}=\alpha$ for
$\alpha\in[0,1]$, and suppose that

\Centerline{$\frak A_n=\{a:a\in\frak A$,
   $a\Bcap\Bvalue{u>1-2^{-n}}\in\{0,\Bvalue{u>1-2^{-n}}\}$
   for every $n\in\Bbb N\}$.}

\noindent Show that the martingale $\pmb Pu$ is of bounded variation.
\Hint{611Xh.}
%626Q

\spheader 626Xf Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a submartingale such that
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded below.   Show that $\pmb v$
is locally moderately oscillatory.   \Hint{626Ic or 626R.}
%626I

%\leader{626Y}{Further exercises (a)}
}%end of exercises

\endnotes{
\Notesheader{626} The previsible variations of 626L-626M can be thought of as
indefinite integrals of interval functions $Pd\pmb v$ based on the weak topology
of $L^1_{\bar\mu}$ instead of the topology of convergence in measure on
$L^0(\frak A)$.   I do not really wish to go farther along this route.   But
there are obvious questions to ask about the applicability of the ideas of
\S613 in this context.

In both 626O and 626Q, we start with a submartingale $\pmb v$ and seek to
express it as $\pmb v^{\#}+\pmb w$ where $\pmb v^{\#}$ is non-decreasing and
$\pmb w$ is more or less a martingale.   In 626O we have an explicit formula,
with $\pmb v^{\#}$ the `previsible variation' as defined in 626L, so that we
have picked out a particular solution.
In 626Q the statement of the theorem makes no claim that the solution is unique,
and of course it is not, because a constant process can always be added to one
term and subtracted from the other.   Indeed it is not difficult to show that
there are non-trivial martingales of bounded variation (626Xe) and therefore
that there are many non-negative non-decreasing processes which are not their
own previsible variations.   The construction in the proof of 626Q gives rise to
a particular pair $\pmb v^{\#}$, $\pmb w$ but I do not know of a simple
characterization of the processes $\pmb v^{\#}$ which can arise in this way.

The calculation in 626S is rather elaborate, but I do not know of an essentially
more direct method.   Concerning the other two examples in \S612, Brownian
motion has previsible variation $0$ just because it is a martingale, and the
Poisson process turns out to have previsible variation $\pmb\iota$;  but for a
proof of this we shall have to wait for some more of the general theory (632Mb).
}%end of notes

\discrpage



\frfilename{mt627.tex}
\versiondate{27.3.21}
\copyrightdate{2012}


\def\chaptername{Martingales}
\def\sectionname{Integrators and \sm{s}}

\newsection{627}

This section is devoted to a kind of structure theory for integrators
(627I-627J, 627L, 627Q);  I take a route which
passes some further important classes of stochastic process
(627B) and ideas from the theory of linear topological spaces
(627F-627G).

\cmmnt{\leader{627A}{Notation} As always, \Asissix\ is a stochastic integration
structure.
For a sublattice $\Cal S$ of $\Cal T$, $\Cal I(\Cal S)$ is the set of
finite sublattices of $\Cal S$;  $\Mfa(\Cal S)$ and $\Mlmo(\Cal S)$ are the
spaces of fully adapted and locally moderately oscillatory processes with domain
$\Cal S$.   If $\pmb u=\sfamilysS{u}$ is order-bounded,
$\sup|\pmb u|=\sup(\{0\}\cup\{|u_{\sigma}|:\sigma\in\Cal S\})$ (614E).

$\tbf{1}$ is the constant process with
value $\chi 1$.   For $I\subseteq\Cal T$ and
$\tau\in\Cal T$, $I\wedge\tau$ and $I\vee\tau$ are
$\{\sigma\wedge\tau:\sigma\in I\}$, $\{\sigma\vee\tau:\sigma\in I\}$
respectively.
$\Expn$ is the standard integral on $L^1_{\bar\mu}=L^1(\frak A,\bar\mu)$,
and $\theta(w)=\Expn(|w|\wedge\chi 1)$ for $w\in L^0=L^0(\frak A)$.
For $\tau\in\Cal T$,
$P_{\tau}:L^1_{\bar\mu}\to L^1_{\bar\mu}\cap L^0(\frak A_{\tau})$ is
the associated conditional expectation.   If $z\in L^1_{\bar\mu}$,
$\pmb Pz=\family{\tau}{\Cal T}{P_{\tau}z}$.
}

\leader{627B}{Definitions} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a fully adapted process with domain $\Cal S$.

\spheader 627Ba
$\pmb v$ is a {\bf supermartingale} if $-\pmb v$ is a
submartingale\cmmnt{ (626B), that is, $\pmb v$ is an $L^1$-process and
$P_{\sigma}v_{\tau}\le v_{\sigma}$ whenever $\sigma\le\tau$ in $\Cal S$}.
\cmmnt{(Mnemonic:  $Pd\pmb v\le 0$.)}

\spheader 627Bb
$\pmb v$ is a {\bf quasimartingale} if $\pmb v$ is an
$L^1$-process
and $\{\Expn(w):w\in Q_{\Cal S}(d\pmb v)\}$\cmmnt{ (definition:  616B)}
is bounded in $\Bbb R$,
%Protter III.4

\spheader 627Bc
I will say that $\pmb v$ is a {\bf strong integrator} if whenever $\epsilon>0$
there are a uniformly integrable martingale $\pmb w$
and a fully adapted process $\pmb w'$ of bounded variation, both
with domain $\Cal S$, such that
$\bar\mu\Bvalue{\pmb v\ne\pmb w+\pmb w'}\le\epsilon$.

\leader{627C}{Elementary facts (a)(i)} If $\pmb v$ is a supermartingale,
so is $\pmb v\restr\Cal S'$ for any sublattice $\Cal S'$ of $\dom\pmb v$.
If $\pmb v$ and $\pmb w$ are supermartingales, so is $\pmb v+\pmb w$.
\prooflet{(Immediate from the definition.)}

\medskip

\quad{\bf (ii)} If $h:\Bbb R\to\Bbb R$ is concave and non-decreasing,
$\pmb v\cmmnt{\mskip5mu=\family{\sigma}{\Cal S}{v_{\sigma}}}$ is a
supermartingale and $\bar h\pmb v$ is an $L^1$-process,
then $\bar h\pmb v$ is a supermartingale.
\prooflet{\Prf\ Set $g(x)=-h(-x)$ for $x\in\Bbb R$.   Then $g$ is convex
and non-decreasing,
so $\bar g\frsmallcirc(-\pmb v)$ is a submartingale (626Cc) and
$\bar h\pmb v=-\bar g\frsmallcirc(-\pmb v)$ is a supermartingale.\ \Qed}

\medskip

\quad{\bf (iii)} A supermartingale $\family{\sigma}{\Cal S}{v_{\sigma}}$ is
a martingale iff $\Expn(v_{\sigma})=\Expn(v_{\tau})$ whenever
$\sigma\le\tau$ in $\Cal S$.   \prooflet{(For in this case we have
$P_{\sigma}v_{\tau}\le v_{\sigma}$ while $\Expn(P_{\sigma}v_{\tau})
=\Expn(v_{\tau})=\Expn(v_{\sigma})$, so
$P_{\sigma}v_{\tau}=v_{\sigma}$.)}

\spheader 627Cb Every
martingale\cmmnt{ $\family{\sigma}{\Cal S}{v_{\sigma}}$} is a
quasimartingale.  \prooflet{\Prf\ If $\sigma\in\tau$ in $\Cal S$ and
$u\in L^{\infty}(\frak A_{\sigma})$, then

\Centerline{$\Expn(u\times(v_{\tau}-v_{\sigma}))
=\Expn(P_{\sigma}(u\times(v_{\tau}-v_{\sigma})))
=\Expn(u\times(P_{\sigma}v_{\tau}-v_{\sigma}))
=0$,}

\noindent so $\Expn (w)=0$ for every $w\in Q_{\Cal S}(d\pmb v)$.\ \Qed}

\spheader 627Cc{\bf (i)} A strong integrator is an integrator.
\prooflet{\Prf\ By 622G, 616Ra and 616Pa, the sum of a uniformly integrable
martingale
and a fully adapted process of bounded variation is an integrator.
Now 616P(b-iii) shows that a strong integrator is an integrator.\ \Qed}

\medskip

\quad{\bf (ii)}
A linear combination of strong integrators is a strong
integrator.    \prooflet{(We just have to recall that sums and scalar multiples
of martingales
are martingales, sums and scalar multiples of uniformly integrable sets
are uniformly integrable
and sums and scalar multiples of processes of bounded variation have bounded
variation.)}

\medskip

\quad{\bf (iii)} If $\pmb v$ is a strong integrator with domain $\Cal S$
and $\Cal S'$ is a sublattice of $\Cal S$, then $\pmb v\restr\Cal S'$ is a
strong integrator.   \prooflet{\Prf\ Given $\epsilon>0$, there are a
uniformly integrable
martingale $\pmb w$ and a process $\pmb w'$ of bounded variation, both
with domain $\Cal S$, such that $\Bvalue{\pmb v\ne\pmb w+\pmb w'}$ has
measure at most $\epsilon$.   Now $\pmb w\restr\Cal S'$ is a
uniformly integrable martingale
(622Dd), $\pmb w\restr\Cal S'$ is of bounded variation (614Lb) and
$\Bvalue{\pmb v\restr\Cal S'
\ne\pmb w\restr\Cal S'+\pmb w'\restr\Cal S'}$ is included in
$\Bvalue{\pmb v\ne\pmb w+\pmb w'}$, so has measure at most $\epsilon$.\
\Qed}

\medskip

\quad{\bf (iv)} If $\pmb v$ is a fully adapted process with domain
$\Cal S$ and for every $\epsilon>0$ there is a strong
integrator $\pmb v'$ with domain $\Cal S$, such that
$\Bvalue{\pmb v\ne\pmb v'}$ has
measure at most $\epsilon$, then $\pmb v$ is a strong integrator.
\prooflet{(Immediate from the definition of `strong integrator' and from
the fact that
$\Bvalue{\pmb v\ne\pmb v''}
\Bsubseteq\Bvalue{\pmb v\ne\pmb v'}\Bcup\Bvalue{\pmb v'\ne\pmb v''}$
for any fully adapted processes $\pmb v$, $\pmb v'$ and $\pmb v''$ with
domain $\Cal S$.)}

\leader{627D}{Proposition} Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v$ a non-negative fully adapted process with domain $\Cal S$.

(a) If $\pmb v$ is a virtually local martingale, it
is a $\|\,\|_1$-bounded supermartingale.
%for 651D

(b)  If $\pmb v$ is a $\|\,\|_1$-bounded supermartingale it is order-bounded.

\proof{ If $\Cal S$ is empty, both parts are trivial, so let us suppose that
$\Cal S\ne\emptyset$.    Express $\pmb v$ as $\sfamilysS{v}$.

\medskip

{\bf (a)(i)} The fully adapted extension
$\hatpmb v=\family{\sigma}{\hat\Cal S}{\hat v_{\sigma}}$ of $\pmb v$ to the
covered envelope $\hat\Cal S$ of $\Cal S$ is an approximately local martingale
(623J).   Let $\Cal A$ be the family of non-empty downwards-directed subsets
$A$ of $\hat\Cal S$ such that $R_A(\hatpmb v)$, as defined in 623B, is a
martingale, and for $A\in\Cal A$ express $R_A(\hatpmb v)$ as
$\family{\sigma}{\hat\Cal S}{\hat v_{A\sigma}}$.

\medskip

\quad{\bf (ii)} If $A\in\Cal A$, the starting values of
$R_A(\hatpmb v)$ and $\hatpmb v$ are the
same (623B(c-i));  so we have a common starting value
$v_{\downarrow}=\lim_{\sigma\downarrow\hat\Cal S}v_{A\sigma}$ for
all the $R_A(\hatpmb v)$.   Now $v_{\downarrow}$ is always
the $\|\,\|_1$-limit $\1lim_{\sigma\downarrow\hat\Cal S}\hat v_{A\sigma}$
(626E, since of course $R_A(\hatpmb v)$ is a submartingale).
Since all the expectations $\Expn(\hat v_{A\sigma})$, for
$\sigma\in\Cal S$, must be the same, this is also $\Expn(v_{\downarrow})$.
This is so for every $A\in\Cal A$, so we have
$\Expn(\hat v_{A\sigma})=\Expn(v_{\downarrow})$ for every $A\in\Cal A$ and
$\sigma\in\hat\Cal S$.

\medskip

\quad{\bf (iii)} Since $v_{\sigma}\ge 0$ for every $\sigma\in\Cal S$,
$\hat v_{\sigma}\ge 0$ for every $\sigma\in\hat\Cal S$,
$\hat v_{A\sigma}\ge 0$ for every $A\in\Cal A$ and $\sigma\in\hat\Cal S$, and
$v_{\downarrow}\ge 0$.   So in fact we have
$\|\hat v_{A\sigma}\|_1=\|v_{\downarrow}\|_1$ for every $A\in\Cal A$ and
$\sigma\in\hat\Cal S$.

Consequently $\|\hat v_{\sigma}\|_1\le\|v_{\downarrow}\|_1$ for every
$\sigma\in\hat\Cal S$.   \Prf\ Because $\hatpmb v$ is a virtually local
martingale, there is for every $\epsilon>0$ an $A\in\Cal A$ such that
$\bar\mu\Bvalue{\hat v_{A\sigma}\ne\hat v_{\sigma}}\le\epsilon$.   So
$v_{\sigma}$ belongs to the closure
$\overline{\{\hat v_{A\sigma}:A\in\Cal A\}}$ for the topology of convergence in
measure.   Because $\|\,\|_1$-balls are closed for this topology (613Bc),

\Centerline{$\|\hat v_{\sigma}\|_1\le\sup_{A\in\Cal A}\|\hat v_{A\sigma}\|_1
=\|v_{\downarrow}\|_1$.   \Qed}

\noindent In particular, $\hatpmb v$ is an $L^1$-process.

\medskip

\quad{\bf (iv)} \Quer\ Suppose, if possible, that $\hatpmb v$ is not a
supermartingale.   Then there are $\sigma\le\tau$ in $\hat\Cal S$ such that
$P_{\sigma}\hat v_{\tau}\not\le\hat v_{\sigma}$, that is,
$c=\Bvalue{\hat v_{\sigma}<P_{\sigma}\hat v_{\tau}}$ is non-zero.
For each $n\in\Bbb N$ we can find an $A_n\in\Cal A$ such that
$a_n=\sup_{\rho\in A}\Bvalue{\rho<\tau}$ has measure at most
$2^{-n-2}\bar\mu c$.   Setting $b_n=\sup_{\rho\in A}\Bvalue{\rho<\sigma}$ for
each $n$, $b_n\in\frak A_{\sigma}$ and

\Centerline{$\sum_{n=0}^{\infty}\bar\mu b_n
\le\sum_{n=0}^{\infty}\bar\mu a_n<\bar\mu c$,}

\noindent so $c'=c\Bsetminus\sup_{n\in\Bbb N}b_n$ is a non-zero member of
$\frak A_{\sigma}$.   Consider $\Expn(v_{\tau}\times\chi c')$.   For each
$n\in\Bbb N$,

\Centerline{$\theta(v_{\tau}-v_{A_n\tau})
\le\bar\mu\Bvalue{v_{\tau}\ne v_{A_n\tau}}\le\bar\mu a_n\to 0$}

\noindent as $n\to\infty$, so $v_{\tau}=\lim_{n\to\infty}v_{A_n\tau}$
and $v_{\tau}\times\chi c'=\lim_{n\to\infty}v_{A_n\tau}\times\chi v'$.
Now

$$\eqalignno{\Expn(v_{\sigma}\times\chi c')&
<\Expn(P_{\sigma}v_{\tau}\times\chi c')
\Displaycause{because $0\ne c'\Bsubseteq\Bvalue{v_{\sigma}<P_{\sigma}v_{\tau}}$}
=\Expn(P_{\sigma}(v_{\tau}\times\chi c'))
\Displaycause{because $c'\in\frak A_{\sigma}$}
=\Expn(v_{\tau}\times\chi c')
=\|v_{\tau}\times\chi c'\|_1
\le\sup_{n\in\Bbb N}\|v_{A_n\tau}\times\chi c'\|_1
\Displaycause{because $\|\,\|_1$-balls are closed for the
topology of convergence in measure}
=\sup_{n\in\Bbb N}\Expn(v_{A_n\tau}\times\chi c').\cr}$$

\noindent There is therefore an $n\in\Bbb N$ such that
$\Expn(v_{A_n\tau}\times\chi c')>\Expn(v_{\sigma}\times\chi c')$.
But $c'$ is disjoint from $b_n$, so
$v_{\sigma}\times\chi c'=v_{A_n\sigma}\times\chi c'$ and

$$\eqalign{\Expn(v_{A_n\sigma}\times\chi c')&
<\Expn(v_{A_n\tau}\times\chi c')
=\Expn(P_{\sigma}(v_{A_n\tau}\times\chi c'))\cr&
=\Expn(P_{\sigma}v_{A_n\tau}\times\chi c')
=\Expn(v_{A_n\sigma}\times\chi c')\cr}$$

\noindent because $R_{A_n}(\pmb v)$ is a martingale.   But this is absurd.\
\Bang

\medskip

\quad{\bf (v)} Thus $\hatpmb v$ is a supermartingale.   But it follows at once
that $\pmb v=\hatpmb v\restr\Cal S$ is a supermartingale.   And we saw in
(iii) that $\hatpmb v$ is $\|\,\|_1$-bounded, so $\pmb v$ also is
$\|\,\|_1$-bounded.

\medskip

{\bf (b)(i)} Set $\gamma=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$.
If $\tau_0\le\tau_1\le\ldots\le\tau_n$ in
$\Cal S$ there is a $v\ge 0$ such that
$P_{\tau_i}v\ge v_{\tau_i}$ for every $i\le n$ and
$\Expn(v)\le\gamma$. \Prf\ Induce on $n$.   If
$n=0$ we can set  $v=v_{\tau_0}$.   For the inductive step to $n+1>0$,
take $v'\ge 0$ such that $\Expn(v')\le\gamma$ and
$P_{\tau_i}v'\ge v_{\tau_i}$ for $i\le n$.   Then
$P_{\tau_n}v_{\tau_{n+1}}\le v_{\tau_n}\le P_{\tau_n}v'$;  set
$v=v_{\tau_{n+1}}+P_{\tau_n}v'-P_{\tau_n}v_{\tau_{n+1}}$.   In this case,

\Centerline{$\Expn(v)=\Expn(v')\le\gamma$,}

\noindent while

\Centerline{$P_{\tau_{n+1}}v=v\ge v_{\tau_{n+1}}\ge 0$}

\noindent and for $i\le n$

\Centerline{$P_{\tau_i}v
=P_{\tau_i}v_{\tau_{n+1}}+P_{\tau_i}v'-P_{\tau_i}v_{\tau_{n+1}}
\ge v_{\tau_i}$.  \Qed}

So if $w=\sup_{i\le n}v_{\tau_i}$,

\Centerline{$\bar\mu\Bvalue{w\ge M}
\le\bar\mu\Bvalue{\sup_{i\le n}P_{\tau_i}v\ge M}
\le\Bover1M\|P_{\tau_n}v\|_1
\le\Bover{\gamma}M$}

\noindent for every $M>0$, by Doob's maximal inequality (621E).

\medskip

\quad{\bf (ii)} If $A\subseteq\Cal S$ is finite and not empty and
$z_A=\sup_{\sigma\in A}v_{\sigma}$, then
$\bar\mu\Bvalue{z_A\ge M}\le\Bover{\gamma}M$ for every $M\ge 0$.
\Prf\ Let $I$ be the
sublattice of $\Cal S$ generated by $A$, and $\tau_0\le\ldots\le\tau_n$ a
sequence in $I$ linearly generating the $I$-cells as in 611L.   Set
$z'=\sup_{i\le n}v_{\tau_i}$.   Because
$\sup_{i\le n}\Bvalue{\tau=\tau_i}=1$, $v_{\tau}\le z'$ for every
$\tau\in I$, $z_A\le z'$ and

\Centerline{$\bar\mu\Bvalue{z_A\ge M}
\le\bar\mu\Bvalue{z'\ge M}\le\Bover{\gamma}M$}

\noindent by (i).\ \Qed

\medskip

\quad{\bf (iii)} For $n\in\Bbb N$, set
$c_n=\sup\{\Bvalue{z_A\ge n}:\emptyset\ne A\in[\Cal S]^{<\omega}\}$.
Then $\bar\mu c_n\le\bover{\gamma}n$ for every $n>1$.   By 364L(a-i),
$\{z_A:\min\Cal S\in A\in[\Cal S]^{<\omega}\}$ is bounded above
in $L^0$, that is, $\{v_{\sigma}:\sigma\in\Cal S\}$ is bounded above.
Since it is also bounded below, $\pmb v$ is order-bounded.
}%end of proof of 627D

\leader{627E}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v=\family{\sigma}{\Cal S}{v_{\sigma}}$
a quasimartingale.

(a) There is a non-negative $\|\,\|_1$-bounded supermartingale $\pmb w$ such
that $\pmb v+\pmb w$ is a supermartingale.

(b) If $\Cal S$ has a greatest element then $\pmb v$ is expressible as the
difference of two non-negative supermartingales.

(c) If $\pmb v$ is $\|\,\|_1$-bounded then it is a \sm
\cmmnt{ (definition:  625D)}.

\proof{{\bf (a)(i)} Set $\gamma=\sup\{\Expn(w):w\in Q_{\Cal S}(d\pmb v)\}$.
Then $\Expn(w)\le\gamma$ for every $w\in Q_{\Cal S}(Pd\pmb v)$.   \Prf\
If $w=0$ this is trivial, as $0\in Q_{\Cal S}(d\pmb v)$, so
$\gamma\ge 0$.   Otherwise, there are a
non-empty $I\in\Cal I(\Cal S)$ and
a fully adapted process $\pmb u=\family{\sigma}{I}{u_{\sigma}}$ such that
$\|\pmb u\|_{\infty}\le 1$ and $w=S_I(\pmb u,Pd\pmb v)$.   Let
$\langle\tau_i\rangle_{i\le n}$ be a sequence linearly generating the
$I$-cells.   Then

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

\noindent because $S_I(\pmb u,d\pmb v)\in Q_{\Cal S}(d\pmb v)$.\ \Qed

\medskip

\quad{\bf (ii)} If $I\in\Cal I(\Cal S)$, then
$S_I(\tbf{1},|Pd\pmb v|)\in Q_{\Cal S}(Pd\pmb v)$.
\Prf\ If $I$ is empty this is trivial.   Otherwise, let
$\tau_0\le\ldots\le\tau_n$ linearly generate the $I$-cells.
For each $i<n$, set

\Centerline{$a_i=\Bvalue{P_{\tau_i}v_{\tau_{i+1}}-v_{\tau_i}\ge 0}$,
\quad$u_i=\chi a_i-\chi(1\Bsetminus a_i)$.}

\noindent Then $a_i\in\frak A_{\tau_i}$,
$u_i\in L^{\infty}(\frak A_{\tau_i})$ and
$\|u_i\|_{\infty}\le 1$.   So

\Centerline{$S_I(\tbf{1},|Pd\pmb v|)
=\sum_{i=0}^{n-1}|P_{\tau_i}v_{\tau_{i+1}}-v_{\tau_i}|
=\sum_{i=0}^{n-1}u_i\times(P_{\tau_i}v_{\tau_{i+1}}-v_{\tau_i})
\in Q_{\Cal S}(Pd\pmb v)$}

\noindent by 616C(ii).\ \Qed

\medskip

\quad{\bf (iii)} For $\tau\in\Cal S$ set

\Centerline{$A_{\tau}
=\{P_{\tau}S_I(\tbf{1},|Pd\pmb v|):
I\in\Cal I(\Cal S\vee\tau)\}$.}

\medskip

\qquad\grheada\ If $\tau\le\sigma_0\le\sigma_1\le\sigma_2$ in $\Cal S$,

$$\eqalign{|P_{\sigma_0}(v_{\sigma_2}-v_{\sigma_0})|
&=|P_{\sigma_0}(v_{\sigma_1}-v_{\sigma_0})
  +P_{\sigma_0}P_{\sigma_1}(v_{\sigma_2}-v_{\sigma_1})|\cr
&\le|P_{\sigma_0}(v_{\sigma_1}-v_{\sigma_0}|
  +P_{\sigma_0}|P_{\sigma_1}(v_{\sigma_2}-v_{\sigma_1})|,\cr
P_{\tau}|P_{\sigma_0}(v_{\sigma_2}-v_{\sigma_0})|
&\le P_{\tau}|P_{\sigma_0}(v_{\sigma_1}-v_{\sigma_0})|
  +P_{\tau}|P_{\sigma_1}(v_{\sigma_2}-v_{\sigma_1})|.\cr}$$

\noindent So if $\tau\le\tau_0\le\ldots\le\tau_n$ in $\Cal S$,

\Centerline{$P_{\tau}|P_{\tau_0}(v_{\tau_n}-v_{\tau_0})|
\le P_{\tau}\sum_{i=0}^{n-1}|P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|$.}

\noindent\Prf\ Induce on $n$.   The case $n=1$ is trivial.   For the
inductive step to $n+1>1$, we have

$$\eqalign{P_{\tau}|P_{\tau_0}(v_{\tau_{n+1}}-v_{\tau_0})|
&\le P_{\tau}|P_{\tau_0}(v_{\tau_n}-v_{\tau_0})|
  +P_{\tau}|P_{\tau_n}(v_{\tau_{n+1}}-v_{\tau_n})|\cr
&\le P_{\tau}\sum_{i=0}^{n-1}|P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|
  +P_{\tau}|P_{\tau_n}(v_{\tau_{n+1}}-v_{\tau_n})|\cr
&=P_{\tau}\sum_{i=0}^n|P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|.
\text{ \Qed}\cr}$$

\medskip

\qquad\grheadb\ If $\tau\in\Cal S$ and
$I\in\Cal I(\Cal S\vee\tau)$ is non-empty, then
$P_{\tau}|P_{\min I}v_{\max I}-v_{\min I}|
\le P_{\tau}S_I(\tbf{1},|Pd\pmb v|)$.   \Prf\ Apply ($\alpha$) with
$\tau_0,\ldots,\tau_n$ a sequence
linearly generating the $I$-cells.\ \Qed

\medskip

\qquad\grheadc\ If $\tau\in\Cal S$,
$I$, $J\in\Cal I(\Cal S\vee\tau)$ and $J\subseteq I$,
then $P_{\tau}S_J(\tbf{1},|Pd\pmb v|)
\le P_{\tau}S_I(\tbf{1},|Pd\pmb v|)$.   \Prf\ If $J$ is empty this is
trivial.   Otherwise, let
$\tau_1\le\ldots\le\tau_n$ be a sequence linearly generating the
$J$-cells;  set $\tau_0=\min I$ and $\tau_{n+1}=\max I$.   For $j\le n$ set
$I_j=I\cap[\tau_j,\tau_{j+1}]$.   Then

$$\eqalign{P_{\tau}S_J(\tbf{1},|Pd\pmb v|)
&=\sum_{j=1}^{n-1}P_{\tau}|P_{\tau_j}v_{\tau_{j+1}}-v_{\tau_j}|
\le\sum_{j=0}^nP_{\tau}|P_{\tau_j}v_{\tau_{j+1}}-v_{\tau_j}|\cr
&\le\sum_{j=0}^nP_{\tau}S_{I_j}(\tbf{1},|Pd\pmb v|)
=P_{\tau}\sum_{j=0}^nS_{I_j}(\tbf{1},|Pd\pmb v|)
=P_{\tau}S_I(\tbf{1},|Pd\pmb v|)\cr}$$

\noindent by 613G(a-ii).\ \Qed

\medskip

\qquad\grheadd\ Thus $A_{\tau}$ is upwards-directed.   If
$w\in A_{\tau}$, then $w=P_{\tau}w'$ for some $w'\in Q_{\Cal S}(Pd\pmb v)$,
by (ii);  in which case

\Centerline{$\|w\|_1=\Expn(w)=\Expn(w')\le\gamma$.}

\noindent It follows that
$w_{\tau}=\sup A_{\tau}$ is defined in $L^1_{\bar\mu}$, and belongs to the
$\|\,\|_1$-closure of $A_{\tau}$ (613B(d-iii)).   As
$A_{\tau}\subseteq L^0(\frak A_{\tau})$,
$w_{\tau}\in L^0(\frak A_{\tau})$.   Also
$\|w_{\tau}\|_1\le\sup_{w\in A_{\tau}}\|w\|_1\le\gamma$.

\medskip

\qquad\grheade\ Suppose that $\tau$, $\tau'\in\Cal S$ and
$\Bvalue{\tau=\tau'}=a$.   If $w\in A_{\tau}$, then
$w\times\chi a\le w_{\tau'}\times\chi a$.   \Prf\ Let
$\tau_0\le\ldots\le\tau_n$ in $\Cal S\vee\tau$ be such that
$w=\sum_{i=0}^{n-1}P_{\tau}|P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|$;
set $\tau'_i=\tau_i\vee\tau'$ for $i\le n$, and
$w'=\sum_{i=0}^{n-1}P_{\tau'}|P_{\tau'_i}(v_{\tau'_{i+1}}-v'_{\tau_i})|$,
so that $w'\in A_{\tau'}$.   Now, for $i\le n$,

\Centerline{$\Bvalue{\tau'_i=\tau_i}
=\Bvalue{\tau_i\vee\tau'=\tau_i\vee\tau}
\Bsupseteq a$}

\noindent by 611E(c-v-$\beta$),
so $\Bvalue{v_{\tau'_i}=v_{\tau_i}}\Bsupseteq a$ and
$\Bvalue{P_{\tau'_i}v=P_{\tau_i}v}\Bsupseteq a$ for every
$v\in L^1_{\bar\mu}$ (622Bb);  also, of course
$\Bvalue{P_{\tau}v=P_{\tau'}v}\Bsupseteq a$ for every $v\in L^1_{\bar\mu}$.
Moreover, $a\in\frak A_{\tau\wedge\tau'}$ (611H(c-i)).   Accordingly

$$\eqalignno{w\times\chi a
&=\sum_{i=0}^{n-1}\chi a\times P_{\tau}
   |P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|
=\sum_{i=0}^{n-1}\chi a\times P_{\tau'}
   |P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|\cr
&=\sum_{i=0}^{n-1}
   P_{\tau'}(\chi a\times|P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|)\cr
\displaycause{because $\chi a\in L^{\infty}(\frak A_{\tau'})$}
&=\sum_{i=0}^{n-1}
   P_{\tau'}|\chi a\times P_{\tau_i}(v_{\tau_{i+1}}-v_{\tau_i})|
=\sum_{i=0}^{n-1}
   P_{\tau'}|\chi a\times P_{\tau'_i}(v_{\tau_{i+1}}-v_{\tau_i})|\cr
&=\sum_{i=0}^{n-1}
   P_{\tau'}|P_{\tau'_i}(\chi a\times (v_{\tau_{i+1}}-v_{\tau_i}))|\cr
\displaycause{because $\chi a\in L^{\infty}(\frak A_{\tau'_i})$ for every
$i$}
&=\sum_{i=0}^{n-1}
   P_{\tau'}|P_{\tau'_i}(\chi a\times (v_{\tau'_{i+1}}-v_{\tau'_i}))|
=w'\times\chi a\cr
\displaycause{following a parallel path back}
&\le w_{\tau'}\times\chi a.  \text{ \Qed}\cr}$$

As $w$ is arbitrary,
$w_{\tau}\times\chi a\le w_{\tau'}\times\chi a$.
Similarly,
$w'_{\tau}\times\chi a\le w_{\tau}\times\chi a$ and
$a\Bsubseteq\Bvalue{w_{\tau}=w_{\tau'}}$.   As $\tau$, $\tau'$ are
arbitrary, $\pmb w=\family{\tau}{\Cal S}{w_{\tau}}$ is a fully
adapted process.   From ($\delta$), we know that it is $\|\,\|_1$-bounded.

\medskip

\quad{\bf (iv)}\grheada\ If $\tau\le\tau'$ in $\Cal S$ and
$I\in\Cal I(\Cal S\vee\tau')$, then

\Centerline{$P_{\tau}P_{\tau'}S_I(\tbf{1},|Pd\pmb v|)
=P_{\tau}S_I(\tbf{1},|Pd\pmb v|)
\le w_{\tau}$;}

\noindent that is, $P_{\tau}w\le w_{\tau}$ for every $w\in A_{\tau'}$.
As $P_{\tau}$ is $\|\,\|_1$-continuous, and $w_{\tau'}$ belongs to the
$\|\,\|_1$-closure of $A_{\tau'}$, $P_{\tau}w_{\tau'}$ also is less than or
equal to $w_{\tau}$.   Thus $\pmb w$ is a
supermartingale and of course it is non-negative.

\medskip

\qquad\grheadb\ If $\tau\le\tau'$ in $\Cal S$ and
$I\in\Cal I(\Cal S\vee\tau')$, then

$$\eqalign{P_{\tau}(v_{\tau'}+P_{\tau'}S_I(\tbf{1},|Pd\pmb v|))
&\le v_{\tau}+|P_{\tau}(v_{\tau'}-v_{\tau})|
   +P_{\tau}S_I(\tbf{1},|Pd\pmb v|))\cr
&=v_{\tau}+P_{\tau}S_{I\cup\{\tau\}}(\tbf{1},|Pd\pmb v|))
\le v_{\tau}+w_{\tau}.\cr}$$

\noindent So $P_{\tau}(v_{\tau'}+w_{\tau'})\le v_{\tau}+w_{\tau}$.
This shows that $\pmb v+\pmb w$ is a supermartingale.

\medskip

{\bf (b)} If $\Cal S$ has a greatest element, set

\Centerline{$\pmb u=\pmb v+\pmb w+\pmb P|v_{\max\Cal S}|$,
\quad$\pmb u'=\pmb w+\pmb P|v_{\max\Cal S}|$.}

\noindent Then $\pmb u$
and $\pmb u'$ are sums of supermartingales, therefore themselves
supermartingales.   As
their final values $v_{\max\Cal S}+w_{\max\Cal S}+|v_{\max\Cal S}|$,
$w_{\max\Cal S}+|v_{\max\Cal S}|$ are both greater than or
equal to $0$, $\pmb u$ and $\pmb u'$ are both non-negative, and
their difference $\pmb v$ is expressed in the required form.

\medskip

{\bf (c)} Since $\pmb w$ is a $\|\,\|_1$-bounded supermartingale, $-\pmb w$ is a
$\|\,\|_1$-bounded submartingale, so is a \sm, by 626R.   Thus $\pmb w$ is a
\sm.   Also we are now supposing that $\pmb v$ is $\|\,\|_1$-bounded, so
$\pmb v+\pmb w$ is a $\|\,\|_1$-bounded supermartingale, and it too is a \sm.
Accordingly $\pmb v=(\pmb v+\pmb w)-\pmb w$ is a difference of \sm{}s and is a
\sm.
}%end of proof of 627E

\leader{627F}{}\cmmnt{ For the next step, we need a couple of facts from
the theory of linear topological spaces.

\medskip

\noindent}{\bf Lemma} Let $U$ be a Banach space, $C$ a
convex subset of $U$ and $K$ a non-empty
weak*-compact convex subset of the dual $U^*$ of $U$.   Suppose that
$\gamma\ge 0$ is such that for every $u\in C$ there is an $f\in K$
such that $f(u)\le\gamma$.   Then there is a $g\in K$ such that
$g(u)\le\gamma$ for every $u\in C$.

\proof{{\bf (a)} For each finite $I\subseteq C$ set
$K_I=\{f:f\in K$, $f(u)\le\gamma$ for every $u\in I\}$.   Then
$K_I\ne\emptyset$.   \Prf\Quer\ Otherwise, $I$ is certainly
non-empty.   Set
$Tf=\family{u}{I}{f(u)}$ for $f\in U^*$.
Then $T:U^*\to\BbbR^I$ is a linear operator which is continuous for
the weak* topology on $U^*$, so $T[K]$ is a convex compact subset of
$\BbbR^I$;  and as $K_I$ is empty, $T[K]$ does not meet the closed
convex set
$F=\{v:v\in\BbbR^I$, $v(u)\le\gamma$ for every $u\in I\}$.

The set $D=T[K]-F$ is convex (2A5Ea) and closed (4A5Ef)
and does not contain $0$.
We therefore have a linear functional $h:\BbbR^I\to\Bbb R$
such that $\inf_{v\in D}h(v)>0$ (3A5Cb), that is,
$\sup_{v\in F}h(v)<\inf_{v'\in T[K]}h(v')$.
For $u\in I$ write $e_u$ for the corresponding unit vector in $\BbbR^I$,
and set $\alpha_u=h(e_u)$, so that
$h(v)=\sum_{u\in I}\alpha_uv(u)$ for every $v\in\BbbR^I$.
Because $F$ contains $\beta e_u$ for every $\beta\le 0$,
and $\sup_{v\in F}h(v)$ is finite, $\alpha_u$ is at least $0$,
for every $u\in I$.   Also $h$ cannot be zero, so
not every $\alpha_u$ is zero.   Set $\alpha=\sum_{u\in I}\alpha_u$ and
consider $\tilde u=\bover1{\alpha}\sum_{u\in I}\alpha_uu$, so that
$\tilde u\in C$ and there is a $\tilde f\in K$ such that
$\tilde f(\tilde u)\le\gamma$.   In this case, $T\tilde f\in T[K]$ and

\Centerline{$h(T\tilde f)
=\sum_{u\in I}\alpha_u\tilde f(u)
=\tilde f(\sum_{u\in I}\alpha_uu)
=\alpha\tilde f(\tilde u)
\le\alpha\gamma
=h(\sum_{u\in I}\gamma e_u)$.}

\noindent But $\sum_{u\in I}\gamma e_u\in F$, so this
is impossible, by the choice of $h$.\ \Bang\Qed

\medskip

{\bf (b)} Now $\{K_I:I\in[C]^{<\omega}\}$ is a downwards-directed
family of non-empty closed subsets of the compact set $K$, so has
non-empty intersection, and any member of the intersection will serve
for $g$.
}%end of proof of 627F

\leader{627G}{Lemma} Suppose that $C\subseteq L^1_{\bar\mu}$ is a
non-empty topologically bounded convex set.   Then there is a
$w\in\cmmnt{L^{\infty}=\mskip5mu}L^{\infty}(\frak A)$
such that
$\Bvalue{w>0}=1$ and $\sup_{u\in C}\Expn(u\times w)$ is finite.

\proof{{\bf (a)} For any $\epsilon\in\ocint{0,1}$ there is a
$w\in L^{\infty}$  such that
$0\le w\le\chi 1$, $\bar\mu\Bvalue{w=0}\le\epsilon$ and
$\sup_{u\in C}\Expn(u\times w)$ is finite.   \Prf\ Let $K$ be the set
$\{w:0\le w\le\chi 1$, $\Expn(w)\ge 1-\epsilon\}$.   Then $K$ is convex
and is closed for the topology $\frak T_s(L^{\infty},L^1_{\bar\mu})$,
so can be regarded as a convex subset of $(L^1_{\bar\mu})^*$ which is
compact for the weak* topology (365Lc\formerly{3{}65Mc}).
Let $\gamma\ge 1$ be such that
$\theta(\bover1{\gamma}u)\le\epsilon$ for every $u\in C$;  then
$\bar\mu\Bvalue{u\ge\gamma}\le\epsilon$ for every $u\in C$.
So for every $u\in C$ there is a $w\in K$ such that
$\Expn(u\times w)\le\gamma$ (we can take $w=\chi\Bvalue{u\le\gamma}$).
By 627F, there is a $w\in K$ such that
$\Expn(u\times w)\le\gamma$ for every $u\in C$.
As $\Expn(w)\ge 1-\epsilon$, $\bar\mu\Bvalue{w=0}\le\epsilon$.\ \Qed

\medskip

{\bf (b)} We can therefore find sequences $\sequencen{w_n}$ in
$L^{\infty}$ and $\sequencen{\gamma_n}$ in $\coint{0,\infty}$ such
that $0\le w_n\le\chi 1$, $\bar\mu\Bvalue{w_n=0}\le 2^{-n}$ and
$\Expn(u\times w_n)\le\gamma_n$ for every $n\in\Bbb N$ and
$u\in C$.   Now $w=\sum_{n=0}^{\infty}\Bover{2^{-n}}{\gamma_n+1}w_n$
is defined in $L^{\infty}$, $\Bvalue{w>0}=1$ and
$\Expn(u\times w)\le 2$ for every $u\in C$.
}%end of proof of 627G

\leader{627H}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v$ a fully adapted process with domain $\Cal S$.

(a) If $I\in\Cal I(\Cal S)$, $\pmb u\in\Mfa(I)$ and
$\|\pmb u\|_{\infty}\le 1$, then there is a $\pmb w\in\Mfa(\Cal S)$
such that $\|\pmb w\|_{\infty}\le 1$ and
$S_J(\pmb w,d\pmb v)=S_I(\pmb u,d\pmb v)$ whenever
$I\subseteq J\in\Cal I(\Cal S)$, so that
$\int_{\Cal S}\pmb w\,d\pmb v$ is defined and equal to
$S_I(\pmb u,d\pmb v)$.

(b) $Q_{\Cal S}(d\pmb v)$ is convex.

\proof{{\bf (a)} (The key.)   Express $\pmb u$ as
$\family{\sigma}{I}{u_{\sigma}}$.
If $I$ is empty then of course we can take $\pmb w$ to be the zero process with
domain $\Cal S$.   Otherwise, let $\tau_0\le\ldots\le\tau_n$ linearly
generate the $I$-cells.
By 612Ka, applied to $(u_{\tau_0},\ldots,u_{\tau_{n-1}},0)$ and $u_*=0$,
there is a simple fully adapted process
$\pmb w=\family{\sigma}{\Cal S}{w_{\sigma}}$ such that

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

\noindent for $i<n$, while

\Centerline{$\Bvalue{w_{\sigma}=0}
\Bsupseteq\Bvalue{\sigma<\tau_0}\Bcup\Bvalue{\tau_n\le\sigma}$,}

\noindent for every $\sigma\in\Cal S$.   Evidently
$\|\pmb w\|_{\infty}\le\sup_{i\le n}\|u_{\tau_i}\|_{\infty}\le 1$.
Now suppose that $I\subseteq J\in\Cal I(\Cal S)$.   Then,
expressing $\pmb v$ as $\sfamilysS{v}$,

$$\eqalignno{S_J(\pmb w,d\pmb v)
&=\sum_{e\in\Sti_0(J)}\Delta_e(\pmb w,d\pmb v)\cr
&=\sum_{e\in\Sti_0(J\wedge\tau_0)}\Delta_e(\pmb w,d\pmb v)
  +\sum_{e\in\Sti_0(J\vee\tau_n)}\Delta_e(\pmb w,d\pmb v)\cr&
\mskip150mu
  +\sum_{i=0}^{n-1}\sum_{e\in\Sti_0(J\cap[\tau_i,\tau_{i+1}])}
      \Delta_e(\pmb w,d\pmb v)
\Displaycause{611J(e-iii)}
=0+0+\sum_{i=0}^{n-1}\sum_{e\in\Sti_0(J\cap[\tau_i,\tau_{i+1}])}
     u_{\tau_i}\times\Delta_e(\tbf{1},d\pmb v)
\Displaycause{because for $\sigma\le\tau$ in $J$, if $\tau\le\tau_0$ then
$\Bvalue{v_{\sigma}\ne v_{\tau}}
\Bsubseteq\Bvalue{\sigma<\tau_0}\Bsubseteq\Bvalue{w_{\sigma}=0}$;
if $\tau_n\le\sigma$ then $w_{\sigma}=0$;
and if $\tau_i\le\sigma\le\tau\le\tau_{i+1}$, then
$\Bvalue{v_{\sigma}\ne v_{\tau}}
\Bsubseteq\Bvalue{\tau_i\le\sigma}\Bcap\Bvalue{\sigma<\tau_{i+1}}
\Bsubseteq\Bvalue{w_{\sigma}=u_{\tau_i}}$}
=\sum_{i=0}^{n-1}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})\cr
\displaycause{613L(b-i)}
&=S_I(\pmb u,d\pmb v)\cr}$$

\noindent (613Ec).

Taking the limit as $J$ increases through $\Cal I(\Cal S)$,
$\int_{\Cal S}\pmb w\,d\pmb v=S_I(\pmb u,d\pmb v)$.

\medskip

{\bf (b)} Suppose that $z$, $z'\in Q_{\Cal S}(d\pmb v)$
and $\alpha\in[0,1]$.
Then there are $I$, $I'\in\Cal I(\Cal S)$ and $\pmb u\in\Mfa(I)$,
$\pmb u'\in\Mfa(I')$ such that $\|\pmb u\|_{\infty}\le 1$,
$\|\pmb u'\|_{\infty}\le 1$, $z=S_I(\pmb u,d\pmb v)$ and
$z'=S_{I'}(\pmb u',d\pmb v)$.   Let
$\pmb w$, $\pmb w'\in\Mfa(\Cal S)$ be
as in (a), starting from $\pmb u$, $\pmb u'$ respectively.
Set $\tilde{\pmb w}=\alpha\pmb w+(1-\alpha)\pmb w'$;  then
$\|\tilde{\pmb w}\|_{\infty}\le 1$.   If $J=I\sqcup I'$
is the sublattice generated by $I\cup I'$,

$$\eqalign{\alpha z+(1-\alpha)z'
&=\alpha S_I(\pmb u,d\pmb v)+(1-\alpha)S_{I'}(\pmb u,d\pmb v)\cr
&=\alpha S_J(\pmb w,d\pmb v)+(1-\alpha)S_J(\pmb w,d\pmb v)
=S_J(\tilde{\pmb w},d\pmb v)\in Q_{\Cal S}(d\pmb v).\cr}$$

\noindent As $z$, $z'$ and $\alpha$ are arbitrary,
$Q_{\Cal S}(d\pmb v)$ is convex.
}%end of proof of 627H

\leader{627I}{Theorem}\cmmnt{ ({\smc Bichteler 79},
{\smc Dellacherie \& Meyer 82}, \S VIII.4)}
Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v$ an integrator with domain $\Cal S$.
Then there is a $\bar\nu$ such that $(\frak A,\bar\nu)$ is
a probability algebra and
$\pmb v$ is a uniformly integrable quasimartingale
with respect to $\bar\nu$.

\proof{{\bf (a)} Express $\pmb v$ as $\sfamilysS{v}$.
$\pmb v$ is order-bounded, by 616Ib;  set $\bar w=\sup|\pmb v|$.

\medskip

{\bf (b)} Suppose to begin with that $\bar w\in L^1_{\bar\mu}$.
Set $C=Q_{\Cal S}(d\pmb v)$, so that
$C$ is topologically bounded in $L^0$.   By 627H, $C$ is convex.
Of course $0\in C$, and $C\subseteq L^1_{\bar\mu}$ because
$u\times(v_{\tau}-v_{\sigma})\in L^1_{\bar\mu}$ whenever
$\sigma\le\tau$ in $\Cal S$ and $u\in L^{\infty}(\frak A_{\sigma})$.
Let $C'$ be the linear sum
$C+[-\bar w,\bar w]$;  then $C'$ is convex, non-empty,
topologically bounded and included in $L^1_{\bar\mu}$.
By 627G, there is a $w\in L^{\infty}(\frak A)$ such that
$\Bvalue{w>0}=1$ and $\sup_{z\in C'}\Expn_{\bar\mu}(z\times w)$ is
finite.   Adjusting $w$ by a scalar factor if necessary, we can
arrange that $\Expn_{\bar\mu}(w)=1$.   In this case, we have a
strictly positive probability $\bar\nu$ on $\frak A$ defined by
saying that $\bar\nu a=\Expn_{\bar\mu}(w\times\chi a)$
for every $a\in\frak A$,
and now $\Expn_{\bar\nu}(z)=\Expn_{\bar\mu}(w\times z)$ for every
$z\in L^0$ for which either expectation is defined (625B(a-iii)).

It follows that $\sup_{z\in C}\Expn_{\bar\nu}(z)$ is finite.   Since
$-z\in Q_{\Cal S}(d\pmb v)$ whenever $z\in Q_{\Cal S}(d\pmb v)$,
$-z\in C$ whenever $z\in C$.  So $\sup_{z\in C}|\Expn_{\bar\nu}(z)|$
is finite, and $\pmb v$ is a quasimartingale with respect to $\bar\nu$.
At the same time, $\Expn_{\bar\nu}(\bar w)=\Expn_{\bar\mu}(\bar w\times w)$
is finite, so $\pmb v$ is actually order-bounded in $L^1_{\bar\nu}$ and is
surely uniformly integrable.

\medskip

{\bf (c)} In general,
let $w'$ be a scalar multiple of $\Bover1{\bar w+\chi 1}$ such that
$\Expn_{\bar\mu}(w')=1$, and set
$\bar\lambda a=\Expn_{\bar\mu}(w'\times\chi a)$ for $a\in\frak A$.
Then $(\frak A,\bar\lambda)$ is a probability algebra, and
$\bar w\in L^1_{\bar\lambda}$.
By (b), we now have a $\bar\nu$ of the kind required.
}%end of proof of 627I

\leader{627J}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v$ an integrator with domain $\Cal S$.   Then $\pmb v$ is a \sm.

\proof{ By 627I, there is a probability measure $\bar\nu$ such that $\pmb v$ is
a $\bar\nu$-uniformly integrable $\bar\nu$-quasimartingale.   By 627Ec,
$\pmb v$ is a $\bar\nu$-\sm.   By 625F, $\pmb v$ is a $\bar\mu$-\sm.
}%end of proof of 627J

\leader{627K}{Lemma} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb{v}$ an integrator with domain $\Cal S$.
Set $\Cal S'=\Cal S\cup\{\min\Cal T,\max\Cal T\}$.   Then there is an
integrator $\pmb{v}'$ with domain $\Cal S'$ extending $\pmb v$.

\proof{{\bf (a)} Express $\pmb v$ as $\sfamilysS{v}$, and set
$\bar v=\sup|\pmb v|$.   Let $\pmb w=\family{\tau}{\Cal T}{w_{\tau}}$
be an extension of $\pmb{v}$ to a fully adapted process defined on the
whole of $\Cal T$ as described in 612P.   Then

$$\eqalign{\Bvalue{|w_{\tau}|\le\bar v}
&\Bsupseteq\Bvalue{w_{\tau}=0}
  \Bcup\sup_{\sigma\in\Cal S}(\Bvalue{w_{\tau}=w_{\sigma}}
  \Bcap\Bvalue{w_{\sigma}=v_{\sigma}}\Bcap\Bvalue{|v_{\sigma}|\le\bar v})\cr
&\Bsupseteq\Bvalue{w_{\tau}=0}
  \Bcup\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau}
=1,\cr}$$

\noindent so $|w_{\tau}|\le\bar v$, for every $\tau\in\Cal T$.

Of course $\Cal S'$ is a sublattice of $\Cal T$.   Set
$\pmb{v}'=\pmb w\restr\Cal S'$;
then $\pmb{v}'$ is fully adapted and extends $\pmb{v}$.

\medskip

{\bf (b)} Now
$Q_{\Cal S'}(d\pmb{v}')\Bsubseteq Q_{\Cal S}(d\pmb{v})+[-4\bar v,4\bar v]$.
\Prf\ Take $z\in Q_{\Cal S'}(d\pmb{v}')$.   Express $z$ as
$S_I(\pmb u,d\pmb v')$ where $I\in\Cal I(\Cal S')$,
$\pmb u=\family{\sigma}{I}{u_{\sigma}}$ is fully adapted and
$\|\pmb u\|_{\infty}\le 1$.   If $I=\emptyset$ then $z=0$ certainly belongs to
$Q_{\Cal S}(d\pmb{v})+[-4\bar v,4\bar v]$.   Otherwise, let
$\langle\tau_i\rangle_{i\le n}$ be the increasing enumeration of a maximal
totally ordered subset of $I$.   If $n=0$ then again
$z=0$ belongs to $Q_{\Cal S}(d\pmb{v})+[-4\bar v,4\bar v]$.   If $n=1$ then

\Centerline{$|z|=|u_{\tau_0}\times(w_{\tau_1}-w_{\tau_0})|\le 2\bar v$}

\noindent and $z\in Q_{\Cal S}(d\pmb{v})+[-4\bar v,4\bar v]$.   If $n=2$ then

\Centerline{$|z|=|u_{\tau_0}\times(w_{\tau_1}-w_{\tau_0})
   +u_{\tau_1}\times(w_{\tau_2}-w_{\tau_1})|\le 4\bar v$}

\noindent and $z\in Q_{\Cal S}(d\pmb{v})+[-4\bar v,4\bar v]$.   If $n\ge 3$
then, because $\tau_0\le\tau_1\le\ldots\le\tau_{n-1}\le\tau_n$ are all
different, $\tau_1,\ldots,\tau_{n-1}$ must all belong to $\Cal S$, so
$\sum_{i=1}^{n-2}u_{\tau_i}\times(w_{\tau_{i+1}}-w_{\tau_i})
=\sum_{i=1}^{n-2}u_{\tau_i}\times(v_{\tau_{i+1}}-v_{\tau_i})$ belongs to
$Q_{\Cal S}(d\pmb v)$, while

\Centerline{$u_{\tau_0}\times(w_{\tau_1}-w_{\tau_0})
   +u_{\tau_{n-1}}\times(w_{\tau_n}-w_{\tau_{n-1}})
\in[-4\bar v,4\bar v]$,}

\noindent so

$$\eqalign{z&
=u_{\tau_0}\times(w_{\tau_1}-w_{\tau_0})
   +\sum_{i=1}^{n-2}u_{\tau_i}\times(w_{\tau_{i+1}}-w_{\tau_i})
   +u_{n-1}\times(w_{\tau_n}-w_{\tau_{n-1}})\cr&
\in Q_{\Cal S}(d\pmb{v})+[-4\bar v,4\bar v].  \text{ \Qed}\cr}$$

\noindent Since $Q_{\Cal S}(d\pmb{v})$ and $[-4\bar v,4\bar v]$ are
topologically bounded and the sum of topologically bounded sets is
topologically bounded, $Q_{\Cal S'}(d\pmb{v}')$ is topologically bounded
and $\pmb{v}'$ is an integrator.
}%end of proof of 627K

\leader{627L}{Theorem} Let $\Cal S$ be a sublattice of $\Cal T$
and $\pmb v$ a fully adapted process with domain $\Cal S$.

(a) If $\Cal S$ has a greatest element and
$\pmb v$ is a non-negative submartingale, $\pmb v$ is a strong
integrator.

(b) If $\Cal S$ has greatest and least elements and
$\pmb v$ is a non-negative supermartingale,
$\pmb v$ is a strong integrator.

(c) If $\Cal S$ has greatest and least elements and
$\pmb v$ is a quasimartingale, $\pmb v$ is a strong integrator.

(d) The following are equiveridical:

\quad(i) $\pmb v$ is an integrator;

\quad(ii) there is a functional
$\bar\nu$ such that $(\frak A,\bar\nu)$ is a probability
algebra and $\pmb v$ is a strong
integrator with respect to $\bar\nu$.

\proof{{\bf (a)} By the Doob-Meyer theorem (626O), $\pmb v$ has a previsible
variation $\pmb v^{\#}$ which is non-negative and non-decreasing, therefore of
bounded variation (because $\Cal S$ has a greatest element).   Now
$\pmb w=\pmb v-\pmb v^{\#}$ is a martingale (626Ma);  again because $\Cal S$ has
a greatest element, $\pmb w$ is uniformly integrable.   So $\pmb v$ is a strong
integrator.

\medskip

{\bf (b)}  Express $\pmb v$ as $\sfamilysS{v}$.
For any $\sigma\in\Cal S$,

\Centerline{$\|v_{\sigma}\|_1=\Expn(v_{\sigma})
=\Expn(P_{\min\Cal S}v_{\sigma})\le\Expn(v_{\min\Cal S})$,}

\noindent so $\pmb v$ is $\|\,\|_1$-bounded, therefore order-bounded (627Db).
Write $\bar v$ for $\sup|\pmb v|$.

Take any $\epsilon>0$.   Let $M\ge 0$ be such that
$\bar\mu\Bvalue{\bar v\ge M}\le\epsilon$.   Then
$M\tbf{1}\wedge\pmb v$ is a supermartingale (627C(a-ii)) and
$\pmb u=M\tbf{1}-M\tbf{1}\wedge\pmb v$ is a
non-negative submartingale, therefore a strong integrator, by (a).
It follows
that $\pmb v'=M\tbf{1}-\pmb u$ is a strong integrator.   But

\Centerline{$\Bvalue{\pmb v'\ne\pmb v}
\Bsubseteq\Bvalue{\bar v\ge M}$}

\noindent has measure at most $\epsilon$.   As $\epsilon$
is arbitrary, $\pmb v$ is a strong integrator, by 627C(c-iv).

\medskip

{\bf (c)} follows at once from 627Eb, (b) here and 627C(c-ii).

\medskip

{\bf (d)(ii)$\Rightarrow$(i)} is immediate from
627C(c-i) and the definition 616Fc, which shows that the property of being an
integrator depends only on the linear space topology of $L^0$, not on the
measure.

\medskip

\quad{\bf (ii)$\Rightarrow$(i)} Suppose that $\pmb v$ is an integrator.
By 627K, there is an integrator $\pmb v'$ with domain
$\Cal S\cup\{\min\Cal T,\max\Cal T\}$ extending $\pmb v$.
627I tells us that there is a probability measure $\bar\nu$ such that
$\pmb v'$ is a $\bar\nu$-quasimartingale, therefore a
$\bar\nu$-strong integrator, by (c) just above.   Consequently $\pmb v$ is a
$\bar\nu$-strong integrator, by 627C(c-iii).
}%end of proof of 627L

\leader{627M}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$, and
$\pmb v$ an integrator with domain $\Cal S$.   Then the solid convex
hull of $Q_{\Cal S}(d\pmb v)$ is topologically bounded.

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

\medskip

{\bf (a)} I should begin by noting that the set of those
$z\in L^0$ for which there are
$\alpha_0,\ldots,\alpha_m\ge 0$ and $z_0,\ldots,z_m\in Q_{\Cal S}(d\pmb v)$
such that $\sum_{j=0}^m\alpha_j=1$ and $|z|\le\sum_{j=0}^m\alpha_j|z_j|$
is a solid convex set including $Q_{\Cal S}(d\pmb v)$;  moreover, as noted
in 613B(f-iv), the solid hull of a topologically bounded set in $L^0$ is
topologically bounded.   It will therefore be enough to show that

\Centerline{$C(d\pmb v)
=\{\sum_{j=0}^m\alpha_j|z_j|:\alpha_j\ge 0$, $z_j\in Q_{\Cal S}(d\pmb v)$
for every $j\le m$, $\sum_{j=0}^m\alpha_j=1\}$}

\noindent is topologically bounded.   Moreover, for this it will be enough
to show that
$\inf_{\gamma>0}\sup_{z\in C(d\pmb v)}\bar\mu\Bvalue{z>\gamma}=0$
(613B(f-ii)).

Next, if $z\in C(d\pmb v)$, there are $\alpha_0,\ldots,\alpha_m\ge 0$,
$\sigma_0\le\ldots\le\sigma_n$ in $\Cal S$ and a family
$\langle u_{ji}\rangle_{j\le m,i<n}$ in $L^0$ such that
$u_{ji}\in L^0(\frak A_{\sigma_i})$ and $\|u_{ji}\|_{\infty}\le 1$
for all $j$ and $i$,
$\sum_{j=1}^m\alpha_j=1$ and $z=\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(v_{\sigma_{i+1}}-v_{\sigma_i})|$.   \Prf\
Let $\alpha_0,\ldots,\alpha_m,z_0,\ldots,z_m$ be such that
$\alpha_j\ge 0$ and $z_j\in Q_{\Cal S}(d\pmb v)$ for every $j$,
while $\sum_{j=0}^m\alpha_j=1$ and $z=\sum_{j=0}^m\alpha_j|z_j|$.
For each $j\le m$, let $I_j\in\Cal I(\Cal S)$ be such that
$z_j\in Q_{I_j}(d\pmb v)$ (616Da).   Let $I$ be a non-empty
finite sublattice of $\Cal S$ including $\bigcup_{j\le m}I_j$.
Then $z_j\in Q_I(d\pmb v)$ for every $j$ (616Dd).   Let
$\langle\sigma_i\rangle_{i\le n}$ linearly generate the $I$-cells.   Then
for each $j$ we can find $u_{j0},\ldots,u_{j,n-1}$ such that
$u_{ji}\in L^0(\frak A_{\sigma_i})$ and $\|u_{ji}\|_{\infty}\le 1$
for each $i<n$ and
$z_j=\sum_{i=0}^{n-1}u_{ji}\times(v_{\sigma_{i+1}}-v_{\sigma_i})$
(616C(ii) again).   Now $z=\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(v_{\sigma_{i+1}}-v_{\sigma_i})|$, as
required.\ \Qed

\medskip

{\bf (b)} Suppose that $\pmb v$ is a uniformly integrable
martingale.   Then $C(d\pmb v)$ is topologically bounded.
\Prf\ By 621Ba, $\beta=\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$ is
finite.   Take $z\in C(d\pmb v)$ and $\gamma>0$.   Take
$\alpha_0,\ldots,\alpha_m\ge 0$,
$\sigma_0\le\ldots\le\sigma_n$ in $\Cal S$ and a family
$\langle u_{ji}\rangle_{j\le m,i<n}$ in $L^0$ such that
$u_{ji}\in L^0(\frak A_{\sigma_i})$ and $\|u_{ji}\|_{\infty}\le 1$
for all $j$ and $i$,
$\sum_{j=1}^m\alpha_j=1$ and $z=\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(v_{\sigma_{i+1}}-v_{\sigma_i})|$.
By 621Hf, in its full strength,

\Centerline{$\bar\mu\Bvalue{|z|>\gamma}=\bar\mu\Bvalue{z>\gamma}
\le\Bover{16}{\gamma}\|v_{\sigma_n}\|_1
\le\Bover{16\beta}{\gamma}$.}

\noindent As $z$ is arbitrary,

\Centerline{$\inf_{\gamma>0}\sup_{z\in C(d\pmb v)}
  \bar\mu\Bvalue{|z|>\gamma}
\le\inf_{\gamma>0}\Bover{16\beta}{\gamma}=0$}

\noindent and $C(d\pmb v)$ is topologically bounded.\ \Qed

\medskip

{\bf (c)} Suppose that $\pmb v$ is order-bounded and non-decreasing.
Then $C(d\pmb v)$ is topologically bounded.   \Prf\ Set
$v_{\uparrow}=\sup_{\sigma\in\Cal S}v_{\sigma}$ and
$v_{\downarrow}=\inf_{\sigma\in\Cal S}v_{\sigma}$.   If
$z\in C(d\pmb v)$, express it as
$\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(v_{\sigma_{i+1}}-v_{\sigma_i})|$ where
$\alpha_0,\ldots,\alpha_m\ge 0$,
$\sigma_0\le\ldots\le\sigma_n$ in $\Cal S$ and
$\langle u_{ji}\rangle_{j\le m,i<n}$ are such that
$\sum_{j=0}^m\alpha_j=1$ and
$u_{ji}\in L^0(\frak A_{\sigma_i})$ and $\|u_{ji}\|_{\infty}\le 1$ for all $j$
and $i$.   Then

$$\eqalign{z
&\le\sum_{j=0}^m\alpha_j
  \sum_{i=0}^{n-1}|u_{ji}|\times(v_{\sigma_{i+1}}-v_{\sigma_i})
\cr&\le\sum_{j=0}^m\alpha_j
  \sum_{i=0}^{n-1}v_{\sigma_{i+1}}-v_{\sigma_i}
=v_{\sigma_n}-v_{\sigma_0}
\le v_{\uparrow}-v_{\downarrow}.
\cr}$$

\noindent Thus $C(d\pmb v)\subseteq[0,v_{\uparrow}-v_{\downarrow}]$ is
order-bounded, therefore topologically bounded.\ \Qed

\medskip

{\bf (d)} Suppose that $\pmb v$ is a strong integrator.   Then $C(d\pmb v)$
is topologically bounded.   \Prf\ Let $\epsilon>0$.   Then there are a
uniformly integrable martingale $\pmb w$ and non-decreasing, order-bounded
processes $\pmb w'$, $\pmb w''$, all with domain $\Cal S$, such that
$\bar\mu\Bvalue{\pmb v\ne\pmb w+\pmb w'-\pmb w''}\le\epsilon$.
By (b)-(c), $C(d\pmb w)$, $C(d\pmb w')$ and $C(d\pmb w'')$ are all
topologically bounded, so their algebraic sum is topologically bounded and
there is a $\gamma>0$ such that
$\bar\mu\Bvalue{z+z'+z''>\gamma}\le\epsilon$ whenever
$z\in C(d\pmb w)$, $z'\in C(d\pmb w')$ and $z''\in C(d\pmb w'')$.
Now suppose that $z^*\in C(d\pmb v)$.   Express $z^*$ as
$\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(v_{\sigma_{i+1}}-v_{\sigma_i})|$ where
$\alpha_0,\ldots,\alpha_m\ge 0$,
$\sigma_0\le\ldots\le\sigma_n$ in $\Cal S$ and
$\langle u_{ji}\rangle_{j\le m,i<n}$ are such that
$\sum_{j=0}^m\alpha_j=1$ and
$u_{ji}\in L^0(\frak A_{\sigma_i})$ and $\|u_{ji}\|_{\infty}\le 1$ for all $j$
and $i$.   Set $\tilde{\pmb v}=\pmb w+\pmb w'-\pmb w''$,

\Centerline{$z=\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(w_{\sigma_{i+1}}-w_{\sigma_i})|
\in C(d\pmb w)$,}

\Centerline{$z'=\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(w'_{\sigma_{i+1}}-w'_{\sigma_i})|
\in C(d\pmb w')$,}

\Centerline{$z''=\sum_{j=0}^m\alpha_j
|\sum_{i=0}^{n-1}u_{ji}\times(w''_{\sigma_{i+1}}-w''_{\sigma_i})|
\in C(d\pmb w'')$}

\noindent and

\Centerline{$\tilde z=\sum_{j=0}^m\alpha_j|\sum_{i=0}^{n-1}u_{ji}
\times(\tilde v_{\sigma_{i+1}}-\tilde v_{\sigma_i})|$}

\noindent where $\pmb v=\sfamilysS{v}$, etc.   Then

\Centerline{$\tilde z\le z+z'+z''$,
\quad$\Bvalue{z^*\ne\tilde z}\Bsubseteq\Bvalue{\pmb v\ne\tilde{\pmb v}}$,}

\noindent so

$$\eqalign{\bar\mu\Bvalue{|z^*|>\gamma}
&=\bar\mu\Bvalue{z^*>\gamma}
\le\bar\mu\Bvalue{\tilde z>\gamma}+\bar\mu\Bvalue{\pmb v\ne\tilde{\pmb v}}
\cr&\le\bar\mu\Bvalue{z+z'+z''>\gamma}+\epsilon
\le 2\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $C(d\pmb v)$ is topologically
bounded.   By (a), the solid convex hull of $Q_{\Cal S}(d\pmb v)$ is
topologically bounded.\ \Qed

\medskip

{\bf (e)} For the general case, we know from 627Ld that $\pmb v$ is a
strong integrator with respect to an alternative law, which gives the same
topologically bounded sets in $L^0$, so the
solid convex hull of $Q_{\Cal S}(d\pmb v)$ is
topologically bounded in this case also.
}%end of proof of 627M

%what about local integrators?

\leader{*627N}{}\cmmnt{ I star the next couple of lemmas because I am sure
that most readers will be interested primarily in the case
$T=\coint{0,\infty}$, for which they are essentially irrelevant, in view of
627R.

\medskip

\noindent}{\bf Lemma} Let $\Cal S$ be a non-empty finitely
full sublattice of $\Cal T$
and $\pmb u=\sfamilysS{u}$ a fully adapted process such that
$\lim_{\sigma\uparrow A}u_{\sigma}$ is defined in $L^0(\frak A)$ for every
non-empty upwards-directed set $A\subseteq\Cal S$ with an upper bound in
$\Cal S$.   Then there are a non-decreasing sequence $\sequencen{\tau_n}$
in $\Cal S$ and a non-decreasing sequence $\sequencen{d_n}$ in $\frak A$
such that

\Centerline{$d_n\in\frak A_{\tau_n}$,
\quad$d_n\Bsubseteq\Bvalue{\tau_{n+1}=\tau_n}$}

\noindent for every $n\in\Bbb N$, and

\Centerline{$\sup_{n\in\Bbb N}(d_n\Bcup\Bvalue{\tau\le\tau_n})=1$,
\quad$u_{\tau}=\lim_{n\to\infty}u_{\tau\wedge\tau_n}$}

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

\proof{{\bf (a)} For $\sigma\in\Cal S$ and $k\in\Bbb N$, set

\Centerline{$a_{\sigma k}
=\sup_{\tau\in\Cal S\vee\sigma}
   \Bvalue{|u_{\tau}-u_{\sigma}|\ge 2^{-k}}$;}

\noindent for $\sigma\in\Cal S$ set

\Centerline{$b_{\sigma}=\sup_{k\in\Bbb N}a_{\sigma k}
=\sup_{\tau\in\Cal S\vee\sigma}\Bvalue{u_{\tau}\ne u_{\sigma}}$,
\quad$c_{\sigma}=1\Bsetminus\upr(b_{\sigma},\frak A_{\sigma})$.}

\noindent (Here $\upr(b_{\sigma},\frak A_{\sigma})
=\inf\{a:a\in\frak A_{\sigma}$, $b_{\sigma}\Bsubseteq a\}$, the upper envelope
of $b_{\sigma}$ in $\frak A_{\sigma}$, as in 313S.)
Set $d=\sup_{\sigma\in\Cal S}c_{\sigma}$.

\medskip

{\bf (b)} If $\sigma\in\Cal S$ and $k\in\Bbb N$, then
$\inf_{\tau\in\Cal S\vee\sigma}
\bar\mu(a_{\sigma k}\Bsetminus\Bvalue{|u_{\tau}-u_{\sigma}|\ge 2^{-k}})
=0$.   \Prf\ For $\tau\in\Cal S\vee\sigma$
set $e_{\tau}=\Bvalue{|u_{\tau}-u_{\sigma}|\ge 2^{-k}}$, so that
$a_{\sigma k}=\sup_{\tau\in\Cal S\vee\sigma}e_{\tau}$.

\medskip

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

$$\eqalign{e_{\tau}\Bcap\Bvalue{\tau=\tau'}&
=\Bvalue{|u_{\tau}-u_{\sigma}|\ge 2^{-k}}
   \Bcap\Bvalue{\tau=\tau'}\cr&
=\Bvalue{|u_{\tau'}-u_{\sigma}|\ge 2^{-k}}\Bcap\Bvalue{\tau=\tau'}
=e_{\tau'}\Bcap\Bvalue{\tau=\tau'},\cr}$$

\Centerline{$e_{\tau}\Bcap\Bvalue{\tau\le\tau'}
=e_{\tau}\Bcap\Bvalue{\tau=\tau\wedge\tau'}
\Bsubseteq e_{\tau\wedge\tau'}$,}

\Centerline{$e_{\tau}\Bcap\Bvalue{\tau'\le\tau}
=e_{\tau}\Bcap\Bvalue{\tau=\tau\vee\tau'}
\Bsubseteq e_{\tau\vee\tau'}$,}

\noindent and
$e_{\tau}\Bsubseteq e_{\tau\wedge\tau'}\Bcup e_{\tau\vee\tau'}$.
Similarly,
$e_{\tau'}\Bsubseteq e_{\tau\wedge\tau'}\Bcup e_{\tau\vee\tau'}$.

Now $e_{\tau\wedge\tau'}
\in\frak A_{\tau\wedge\tau'}\Bsubseteq\frak A_{\tau\vee\tau'}$,
so there is a $\tau''\in\Cal T$ such that

\Centerline{$e_{\tau\wedge\tau'}
\Bsubseteq\Bvalue{\tau''=\tau\wedge\tau'}$,
\quad$1\Bsetminus e_{\tau\wedge\tau'}
\Bsubseteq\Bvalue{\tau''=\tau\vee\tau'}$}

\noindent (611I).
Because $\Cal S$ is finitely full, $\tau''\in\Cal S$, and of course
$\sigma\le\tau\wedge\tau'\le\tau''$.    Now

\Centerline{$e_{\tau''}
\Bsupseteq\Bvalue{\tau''=\tau\wedge\tau'}\Bcap e_{\tau\wedge\tau'}
=e_{\tau\wedge\tau'}$}

\noindent and also

\Centerline{$e_{\tau''}
\Bsupseteq\Bvalue{\tau''=\tau\vee\tau'}\Bcap e_{\tau\vee\tau'}
\Bsupseteq e_{\tau\vee\tau'}\Bsetminus e_{\tau\wedge\tau'}$,}

\noindent so

\Centerline{
$e_{\tau''}\Bsupseteq e_{\tau\wedge\tau'}\Bcup e_{\tau\vee\tau'}
\Bsupseteq e_{\tau}\Bcup e_{\tau'}$.}

\medskip

\quad{\bf (ii)} Thus $\{e_{\tau}:\tau\in\Cal S\wedge\sigma\}$ is
upwards-directed and

\Centerline{$\sup_{\tau\in\Cal S\wedge\sigma}\bar\mu e_{\tau}
=\bar\mu(\sup_{\tau\in\Cal S\wedge\sigma}e_{\tau})
=\bar\mu a_{\sigma k}$.}

\noindent Taking complements in $a_{\sigma k}$,

\Centerline{$\inf_{\tau\in\Cal S\wedge\sigma}
\bar\mu(a_{\sigma_k}\Bsetminus e_{\tau})
=0$,}

\noindent as required.\ \Qed

\medskip

{\bf (c)(i)} For any $\sigma\in\Cal S$, $c_{\sigma}\in\frak A_{\sigma}$
and $c_{\sigma}\Bcap a_{\sigma k}=0$ for every $k\in\Bbb N$, so
$c_{\sigma}\Bsubseteq\Bvalue{u_{\tau}=u_{\sigma}}$ for every
$\tau\in\Cal S\vee\sigma$.   Generally, for $\tau\in\Cal S$,

$$\eqalign{\Bvalue{u_{\tau}=u_{\tau\wedge\sigma}}
&\Bsupseteq\Bvalue{\tau=\tau\wedge\sigma}
   \Bcup(\Bvalue{\tau=\tau\vee\sigma}
      \Bcap\Bvalue{u_{\sigma}=u_{\tau\vee\sigma}})\cr
&\Bsupseteq\Bvalue{\tau\le\sigma}
   \Bcup(\Bvalue{\sigma\le\tau}\Bcap c_{\sigma})
\Bsupseteq c_{\sigma}.\cr}$$

\medskip

\quad{\bf (ii)} If $\sigma\le\sigma'\le\tau$ in $\Cal S$,

\Centerline{$\Bvalue{u_{\tau}\ne u_{\sigma'}}
\Bsubseteq\Bvalue{u_{\tau}\ne u_{\sigma}}
  \Bcup\Bvalue{u_{\sigma'}\ne u_{\sigma}}
\Bsubseteq b_{\sigma}$;}

\noindent taking the supremum over $\tau$,

\Centerline{$b_{\sigma'}\Bsubseteq b_{\sigma}
\Bsubseteq\upr(b_{\sigma},\frak A_{\sigma})\in\frak A_{\sigma}
\Bsubseteq\frak A_{\sigma'}$}

\noindent and

\Centerline{$\upr(b_{\sigma'},\frak A_{\sigma'})
\Bsubseteq\upr(b_{\sigma},\frak A_{\sigma})$,}

\noindent that is, $c_{\sigma}\Bsubseteq c_{\sigma'}$.

\medskip

{\bf (d)(i)} There is a non-empty countable subset $A_0\subseteq\Cal S$
such that $d=\sup_{\sigma\in A_0}c_{\sigma}$.   Now there is a
countable sublattice $\Cal S_0$ of $\Cal S$, including $A_0$,
such that whenever $\sigma\in\Cal S_0$ and $k\in\Bbb N$ then
$\inf_{\tau\in\Cal S_0\vee\sigma}
\bar\mu(a_{\sigma k}\Bsetminus\Bvalue{|u_{\tau}-u_{\sigma}|\ge 2^{-k}})
=0$.
\Prf\ For $k\in\Bbb N$ and $\sigma\in\Cal S$ there is a sequence
$\sequence{i}{\tau_{\sigma ki}}$ in $\Cal S\vee\sigma$
such that $\inf_{i\in\Bbb N}
\bar\mu(a_{\sigma k}
   \Bsetminus\Bvalue{|u_{\tau_{\sigma ki}}-u_{\sigma}|\ge 2^{-k}})
=0$, by (b).   Set $\Cal S_0=\bigcup_{n\in\Bbb N}A_n$ where

\Centerline{$A_{n+1}
=\{\sigma\wedge\sigma':\sigma,\,\sigma'\in A_n\}
   \cup\{\sigma\vee\sigma':\sigma,\,\sigma'\in A_n\}
   \cup\{\tau_{\sigma ki}:\sigma\in A_n,\,k,\,i\in\Bbb N\}$}

\noindent for each $n\in\Bbb N$.\ \Qed

\medskip

\quad{\bf (ii)}
Let $\sequencen{\sigma'_n}$ be a sequence running over $\Cal S_0$ and set
$\sigma_n=\sup_{i\le n}\sigma'_i$ for $n\in\Bbb N$;  then
$\sequencen{\sigma_n}$ is a non-decreasing sequence in $\Cal S_0$ and
$\{\sigma_n:n\in\Bbb N\}$ is cofinal with $\Cal S_0$.   Set
$d_n=c_{\sigma_n}$ for $n\in\Bbb N$;  by (c),

\Centerline{$d=\sup_{\sigma\in A_0}c_{\sigma}
\Bsubseteq\sup_{n\in\Bbb N}c_{\sigma'_n}
\Bsubseteq\sup_{n\in\Bbb N}c_{\sigma_n}
=\sup_{n\in\Bbb N}d_n\Bsubseteq d$}

\noindent and $d=\sup_{n\in\Bbb N}d_n$.   Note that
$d_n\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\sigma_n}}$ for every
$n\in\Bbb N$ and $\tau\in\Cal S$, by (c-i) above, while $d_n\Bsubseteq d_{n+1}$
for every $n$ by (c-ii).

\medskip

{\bf (e)} If $\tau\in\Cal S$ then
$\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}\Bsubseteq c_{\tau}$.
\Prf\Quer\ Otherwise,

\Centerline{$\upr(b_{\tau},\frak A_{\tau})
\Bcap\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}\ne 0$.}

\noindent Because
$\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}\in\frak A_{\tau}$,

\Centerline{$b_{\tau}
\Bcap\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}\ne 0$}

\noindent and there is a $k\ge 1$ such that

\Centerline{$a_{\tau,k-1}
\Bcap\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}\ne 0$;}

\noindent finally, there is a $\tau'\in\Cal S\vee\tau$ such that

\Centerline{$a=\Bvalue{|u_{\tau'}-u_{\tau}|\ge 2^{-k+1}}
\Bcap\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}$}

\noindent is non-zero.

If $\sigma\in\Cal S_0$ then

$$\eqalign{a
&\Bsubseteq(\Bvalue{\sigma<\tau}
   \Bcap\Bvalue{|u_{\tau'}-u_{\tau}|\ge 2^{-k+1}})\cr
&\Bsubseteq(\Bvalue{\sigma\le\tau'}
    \Bcap\Bvalue{|u_{\tau'}-u_{\sigma}|\ge 2^{-k}})
  \Bcup(\Bvalue{\sigma\le\tau}
    \Bcap\Bvalue{|u_{\tau}-u_{\sigma}|\ge 2^{-k}})\cr
&\Bsubseteq\Bvalue{|u_{\sigma\vee\tau'}-u_{\sigma}|\ge 2^{-k}})
  \Bcup\Bvalue{|u_{\sigma\vee\tau}-u_{\sigma}|\ge 2^{-k}})
\Bsubseteq a_{\sigma k}.\cr}$$

\noindent Consequently

\Centerline{$\inf_{\rho\in\Cal S_0\vee\sigma}
   \bar\mu(a\Bsetminus\Bvalue{|u_{\rho}-u_{\sigma}|\ge 2^{-k}})
\le\inf_{\rho\in\Cal S_0\vee\sigma}
   \bar\mu(a_{\sigma k}\Bsetminus\Bvalue{|u_{\rho}-u_{\sigma}|\ge 2^{-k}})
=0$.}

\noindent We can therefore choose inductively a non-decreasing
sequence $\sequencen{\rho_n}$ in $\Cal S_0$ such that

\Centerline{$\bar\mu(a\Bsetminus
   \Bvalue{|u_{\rho_{n+1}}-u_{\rho_n}|\ge 2^{-k}})
\le\Bover12\bar\mu a$}

\noindent for every $n\in\Bbb N$.   But this means that

$$\eqalign{\theta(u_{\tau\wedge\rho_{n+1}}-u_{\tau\wedge\rho_n})
&\ge 2^{-k}
 \bar\mu\Bvalue{|u_{\tau\wedge\rho_{n+1}}-u_{\tau\wedge\rho_n}|\ge 2^{-k}})
 \cr
&\ge 2^{-k}
 \bar\mu(\Bvalue{\rho_{n+1}<\tau}
    \Bcap\Bvalue{|u_{\rho_{n+1}}-u_{\rho_n}|\ge 2^{-k}})\cr
&\ge 2^{-k}
 \bar\mu(a\Bcap\Bvalue{|u_{\rho_{n+1}}-u_{\rho_n}|\ge 2^{-k}})
\ge 2^{-k-1}\bar\mu a\cr}$$

\noindent for every $n$;  which contradicts our hypothesis that
$\lim_{\sigma\uparrow A}u_{\sigma}$ is defined in $L^0(\frak A)$ for every
non-empty upwards-directed set $A\subseteq\Cal S$ with an upper bound in
$\Cal S$.\ \Bang\Qed

\medskip

{\bf (f)} So for any $\tau\in\Cal S$

$$\eqalignno{\sup_{n\in\Bbb N}(d_n\Bcup\Bvalue{\tau\le\sigma_n})
&=d\Bcup(1\Bsetminus\inf_{n\in\Bbb N}\Bvalue{\sigma_n<\tau})
=d\Bcup(1\Bsetminus\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau})\cr
\displaycause{because if $\sigma\in\Cal S_0$ there is an
$n\in\Bbb N$ such that $\sigma\le\sigma_n$ and
$\Bvalue{\sigma_n<\tau}\Bsubseteq\Bvalue{\sigma<\tau}$}
&=1\cr}$$

\noindent because $\inf_{\sigma\in\Cal S_0}\Bvalue{\sigma<\tau}
\Bsubseteq c_{\tau}\Bsubseteq d$.

\medskip

{\bf (g)} We have most of what we want.   But as there is no reason why
$d_n$ should be included in $\Bvalue{\sigma_{n+1}=\sigma_n}$, we have to make a
further adjustment.

\medskip

\quad{\bf (i)} We can define a non-decreasing sequence
$\sequencen{\tau_n}$ in $\Cal S$ inductively by
saying that $\tau_0=\sigma_0$ and that

\Centerline{$d_n\Bsubseteq\Bvalue{\tau_{n+1}=\tau_n}$,
\quad$1\Bsetminus d_n\Bsubseteq\Bvalue{\tau_{n+1}=\sigma_{n+1}}$}

\noindent for $n\in\Bbb N$.   \Prf\ The point is that $\tau_n\le\sigma_n$
and $d_n\in\frak A_{\tau_n}$ for every $n$.   To see this, we know that
we have $\tau_0\le\sigma_0$ and

\Centerline{$d_0=c_{\sigma_0}\in\frak A_{\sigma_0}=\frak A_{\tau_0}$.}

\noindent At the inductive step, given that
$\tau_n\le\sigma_n$ and $d_n\in\frak A_{\tau_n}$, 611I tells us that
$\tau_{n+1}$ is well-defined in $\Cal T$ and that
$\tau_n\le\tau_{n+1}\le\sigma_{n+1}$;   $\tau_{n+1}\in\Cal S$ because
$\Cal S$ is finitely full.   To see that $d_{n+1}\in\frak A_{\tau_{n+1}}$, note
first that, because $\sigma_n\le\sigma_{n+1}$,

\Centerline{$d_n=c_{\sigma_n}\Bsubseteq c_{\sigma_{n+1}}
=d_{n+1}$}

\noindent by (c-ii) again.   We know that
$d_n\in\frak A_{\tau_n}\subseteq\frak A_{\tau_{n+1}}$, while
$d_{n+1}\Bsetminus d_n$ belongs to $\frak A_{\sigma_{n+1}}$ and is included
in $\Bvalue{\tau_{n+1}=\sigma_{n+1}}$, so belongs to
$\frak A_{\tau_{n+1}}$, by 611H(c-iii).
Accordingly $d_{n+1}=d_n\Bcup(d_{n+1}\Bsetminus d_n)$ belongs to
$\frak A_{\tau_{n+1}}$.   So the induction proceeds.\ \Qed

\medskip

\quad{\bf (ii)} We now certainly have $d_n\in\frak A_{\tau_n}$ and
$d_n\Bsubseteq\Bvalue{\tau_{n+1}=\tau_n}$ for every $n$.
If $\tau\in\Cal S$,

$$\eqalignno{1&
=\sup_{n\in\Bbb N}(d_n\Bcup\Bvalue{\tau\le\sigma_n})
\Displaycause{by (f)}
\Bsubseteq\sup_{n\in\Bbb N}(d_n\Bcup\Bvalue{\tau\le\sigma_{n+1}})
=\sup_{n\in\Bbb N}(d_n
   \Bcup(\Bvalue{\tau\le\sigma_{n+1}}\Bsetminus d_n))\cr&
\Bsubseteq\sup_{n\in\Bbb N}(d_n
   \Bcup(\Bvalue{\tau\le\sigma_{n+1}}\Bcap\Bvalue{\sigma_{n+1}=\tau_{n+1}}))\cr&
\Bsubseteq\sup_{n\in\Bbb N}(d_{n+1}\Bcup\Bvalue{\tau\le\tau_{n+1}})
\Bsubseteq\sup_{n\in\Bbb N}(d_n\Bcup\Bvalue{\tau\le\tau_n}).\cr}$$

\medskip

\quad{\bf (iii)}
If $n\in\Bbb N$ and $\tau\in\Cal S$, then
$d_n\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\tau_n}}$.   \Prf\ Induce on
$n$.   If $n=0$ we just have to recall from (d-ii) that
$d_0\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\sigma_0}}$.   For the
inductive step to $n+1\ge 1$, we have

$$\eqalignno{d_n
&\Bsubseteq\Bvalue{\tau_n=\tau_{n+1}}
   \Bcap\Bvalue{u_{\tau}=u_{\tau\wedge\tau_n}}
\Bsubseteq\Bvalue{\tau\wedge\tau_n=\tau\wedge\tau_{n+1}}
   \Bcap\Bvalue{u_{\tau}=u_{\tau\wedge\tau_n}}\cr
\displaycause{611E(c-v-$\alpha$)}
&\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\tau_{n+1}}},\cr}$$

\noindent while $d_{n+1}\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\sigma_{n+1}}}$
by (d-ii), so

\Centerline{$d_{n+1}\Bsetminus d_n
\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\sigma_{n+1}}}
   \Bcap\Bvalue{\tau_{n+1}=\sigma_{n+1}}
\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\tau_{n+1}}}$.}

\noindent Putting these together,
$d_{n+1}\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\tau_{n+1}}}$, as required
to continue the induction.\ \Qed

\medskip

\quad{\bf (iv)}
It follows that $u_{\tau}=\lim_{n\to\infty}u_{\tau\wedge\tau_n}$ for every
$\tau\in\Cal S$.   \Prf\ By (iii), we have

\Centerline{$d_n\Bcup\Bvalue{\tau\le\tau_n}
=d_n\Bcup\Bvalue{\tau=\tau\wedge\tau_n}
\Bsubseteq\Bvalue{u_{\tau}=u_{\tau\wedge\tau_n}}$,}

\noindent while $\sequencen{d_n\Bcup\Bvalue{\tau\le\tau_n}}$ is a
non-decreasing sequence in $\frak A$ with supremum $1$, so

\Centerline{$\theta(u_{\tau}-u_{\tau\wedge\tau_n})
\le\bar\mu(1\Bsetminus(d_n\Bcup\Bvalue{\tau\le\tau_n}))\to 0$}

\noindent as $n\to\infty$.\ \QeD\
So $\sequencen{d_n}$ and $\sequencen{\tau_n}$ have all the listed
properties.
}%end of proof of 627N

\leader{*627O}{Lemma} Suppose that we are given a sublattice
$\Cal S$ of $\Cal T$, a non-decreasing sequence $\sequencen{\tau_n}$ in
$\Cal S$ and a non-decreasing sequence $\sequencen{d_n}$ in $\frak A$ such
that

\Centerline{$d_n\in\frak A_{\tau_n}$,
\quad$d_n\Bsubseteq\Bvalue{\tau_{n+1}=\tau_n}$}

\noindent for every $n\in\Bbb N$, and

\Centerline{$1=\sup_{n\in\Bbb N}d_n\Bcup\Bvalue{\tau\le\tau_n}$}

\noindent for every $\tau\in\Cal S$.   Set
$\Cal S_0=\bigcup_{n\in\Bbb N}\Cal S\wedge\tau_n$ and suppose that
$\pmb u=\family{\sigma}{\Cal S_0}{u_{\sigma}}$ is a fully adapted process.

(a) There is a fully adapted process
$\tildepmb u=\family{\tau}{\Cal S}{\tilde u_{\tau}}$ such that

\quad(i) $\tilde u_{\tau}=\lim_{n\to\infty}u_{\tau\wedge\tau_n}$
for every $\tau\in\Cal S$,

\quad(ii) $d_n\Bcup\Bvalue{\tau\le\tau_n}
\Bsubseteq\Bvalue{\tilde u_{\tau}=u_{\tau\wedge\tau_n}}$ for every
$\tau\in\Cal S$ and $n\in\Bbb N$,

\quad(iii) $\tildepmb u$ extends $\pmb u$.

(b) Write $\hat\Cal S$ for the covered envelope of $\Cal S$,
$\hat\Cal S_0$ for $\bigcup_{n\in\Bbb N}\hat\Cal S\wedge\tau_n$ and
$\hatpmb u=\family{\sigma}{\hat\Cal S_0}{\hat u_{\sigma}}$ for the fully adapted
extension of $\pmb u$ to $\hat\Cal S_0$.   Set
$\tildehat u_{\tau}=\lim_{n\to\infty}\hat u_{\tau\wedge\tau_n}$ for every
$\tau\in\hat\Cal S$.   Then
$\tildehatpmb u=\family{\tau}{\hat\Cal S}{\tildehat u_{\tau}}$
is the fully adapted extension of $\tildepmb u$ to $\hat\Cal S$.

(c) If $\pmb u$ is locally moderately oscillatory,
$\tildepmb u$ is locally moderately oscillatory.

(d) If $\pmb u$ is a virtually local martingale, $\tildepmb u$ is a
virtually local martingale.

(e) If $\pmb u$ is locally of bounded variation,
$\tildepmb u$ is locally of bounded variation.

(f) If $\pmb u$ is locally order-bounded and
$\bar w=\sup_{n\in\Bbb N}\Osclln(\pmb u\restr\Cal S_0\wedge\tau_n)$
is defined, then
$\bar w\ge\sup_{\tau\in\Cal S}
\Osclln(\tildepmb u\restr\hbox{$\Cal S\wedge\tau$})$.

(g) If $\pmb u$ is a \sm, $\tildepmb u$ is a \sm.

\proof{{\bf (a)(i)}\grheada\ If $\tau\in\Cal S$ and $k\le n\in\Bbb N$, then

\Centerline{$d_k\Bsubseteq\Bvalue{\tau_k=\tau_n}
\Bsubseteq\Bvalue{u_{\tau\wedge\tau_k}=u_{\tau\wedge\tau_n}}$}

\noindent while similarly

\Centerline{$\Bvalue{\tau\le\tau_k}
\Bsubseteq\Bvalue{\tau\wedge\tau_k=\tau\wedge\tau_n}
\Bsubseteq\Bvalue{u_{\tau\wedge\tau_k}=u_{\tau\wedge\tau_n}}$.}

\noindent So if $k\le m\le n$ in $\Bbb N$,
$d_k\Bcup\Bvalue{tau\le\tau_k}
    \Bsubseteq\Bvalue{u_{\tau\wedge\tau_m}=u_{\tau\wedge\tau_n}}$ and

\Centerline{$\theta(u_{\tau\wedge\tau_m}-u_{\tau\wedge\tau_n})
\le\bar\mu(1\Bsetminus(d_k\Bcup\Bvalue{\tau\le\tau_k}))
\to 0$}

\noindent as $k\to\infty$.   Thus
$\sequencen{u_{\tau\wedge\tau_n}}$ is a Cauchy sequence and
$\tilde u_{\tau}=\lim_{n\to\infty}u_{\tau\wedge\tau_n}$ is defined in
$L^0(\frak A)$.   Moreover, since
$u_{\tau\wedge\tau_n}\in L^0(\frak A_{\tau\wedge\tau_n})
\subseteq L^0(\frak A_{\tau})$ for every $n$,
$\tilde u_{\tau}\in L^0(\frak A_{\tau})$.

\medskip

\qquad\grheadb\ If $\tau$, $\tau'\in\Cal S$ and $c=\Bvalue{\tau=\tau'}$,
then $c\Bsubseteq\Bvalue{\tau\wedge\tau_n=\tau'\wedge\tau_n}$ for every
$n$, so

\Centerline{$\tilde u_{\tau}\times\chi c
=\lim_{n\to\infty}u_{\tau\wedge\tau_n}\times\chi c
=\lim_{n\to\infty}u_{\tau'\wedge\tau_n}\times\chi c
=\tilde u_{\tau'}\times\chi c$}

\noindent and $c\Bsubseteq\Bvalue{\tilde u_{\tau}=\tilde u_{\tau'}}$.
As $\tau$ and $\tau'$ are arbitrary, $\tildepmb u$ is fully adapted.

\medskip

\quad{\bf (ii)} Set $d_n'=d_n\Bcup\Bvalue{\tau\le\tau_n}$.
For any $k\ge n$,

\Centerline{$d_n'
\Bsubseteq(\Bvalue{\tau_k=\tau_n}
   \Bcup(\Bvalue{\tau\le\tau_n}\Bcap\Bvalue{\tau_n\le\tau_k})
\Bsubseteq\Bvalue{\tau\wedge\tau_k=\tau\wedge\tau_n}
\Bsubseteq\Bvalue{u_{\tau\wedge\tau_k}=u_{\tau\wedge\tau_n}}$.}

\noindent So

\Centerline{$\tilde u_{\tau}\times\chi d_n'
=\lim_{k\to\infty}u_{\tau\wedge\tau_k}\times\chi d_n'
=u_{\tau\wedge\tau_n}\times\chi d_n'$}

\noindent and $d_n'\Bsubseteq\Bvalue{\tilde u_{\tau}=u_{\tau\wedge\tau_n}}$.

\medskip

\quad{\bf (iii)} If $\tau\in\Cal S_0$ then $\tau\wedge\tau_n=\tau$ for all
$n$ large enough, so
$\tilde u_{\tau}=\lim_{n\to\infty}u_{\tau\wedge\tau_n}=u_{\tau}$.

\medskip

{\bf (b)} We just have to note that $\tildehatpmb u$ is fully adapted,
by (a-i) applied to $\hat\Cal S$ and $\hatpmb u$, and extends $\tildepmb u$.

\medskip

{\bf (c)(i)} Consider first the case in which $\Cal S$ is full.   Suppose that
$\tau\in\Cal S$ and $\sequence{i}{\sigma_i}$ is a monotonic sequence in
$\Cal S\wedge\tau$.   For $n\in\Bbb N$ set
$d'_n=d_n\Bcup\Bvalue{\tau\le\tau_n}$.   Then

\Centerline{$d'_n\Bsubseteq d_n\Bcup\Bvalue{\sigma_i\le\tau_n}
\Bsubseteq\Bvalue{u_{\sigma_i}=u_{\sigma_i\wedge\tau_n}}$}

\noindent for every $i\in\Bbb N$, and

\Centerline{$\sequence{i}{\tilde u_{\sigma_i}\times\chi d'_n}
=\sequence{i}{u_{\sigma_i\wedge\tau_n}\times\chi d'_n}$}

\noindent is convergent.   Since
$\theta(\tilde u_{\sigma_i}-\tilde u_{\sigma_i}\times\chi d'_n)\le\bar\mu(1\Bsetminus d'_n)$
for all $i$ and $n$, and $\lim_{n\to\infty}\bar\mu(1\Bsetminus d'_n)=0$,
$\sequence{i}{\tilde u_{\sigma_i}}$ is convergent.   As $\sequence{i}{\sigma_i}$ is
arbitrary, $\tildepmb u\restr\Cal S\wedge\tau$ is moderately oscillatory;
as $\tau$ is arbitrary, $\tildepmb u$ is locally moderately oscillatory.

\medskip

\quad{\bf (ii)} In general, take $\hat\Cal S$, $\hat\Cal S_0$,
$\hatpmb u$ and $\tildehatpmb u$ as in (b).
For each $n\in\Bbb N$,
$\hat\Cal S_0\wedge\tau_n=\hat\Cal S\wedge\tau_n$ is the covered envelope of
$\Cal S\wedge\tau_n$ (611M(e-i)), and
$\pmb u\restr\Cal S\wedge\tau_n$ is moderately oscillatory,
so $\hatpmb u\restr\hat\Cal S\wedge\tau_n$ is moderately oscillatory
(615F(a-i));   as
$\{\tau_n:n\in\Bbb N\}$ is cofinal with $\hat\Cal S_0$, $\hatpmb u$ is locally
moderately oscillatory.   By (i), $\tildehatpmb u$
is locally moderately oscillatory,
so $\tildepmb u$ is locally moderately oscillatory (615F(b-v)).

\medskip

{\bf (d)} We know from (c) that $\tildepmb u$ is locally moderately
oscillatory.

\medskip

\quad{\bf (i)} Consider first the case in which $\Cal S$ is full.

\medskip

\qquad\grheada\ Take
$\tau\in\Cal S$ and $\epsilon>0$.   Then there is an $n\in\Bbb N$ such that
$\bar\mu(d_n\Bcup\Bvalue{\tau\le\tau_n})\ge 1-\epsilon$.   By 623Ke,
$\pmb u\restr\Cal S\wedge\tau_n=\pmb u\restr\Cal S_0\wedge\tau_n$
is a virtually local martingale.
Let $A\subseteq\Cal S\wedge\tau_n$ be a non-empty
downwards-directed set such that
$\sup_{\rho\in A}\bar\mu\Bvalue{\rho<\tau_n}<\epsilon$ and
$R_A(\pmb u\restr\Cal S\wedge\tau_n)$, as defined in 623B, is a martingale.
If $\rho\in A$, then
$a_{\rho}=d_n\Bcap\Bvalue{\tau_n\le\rho\wedge\tau}$ belongs to
$\frak A_{\rho\wedge\tau}$, by 611H(c-iii) again.
So we have a $\rho'\in\Cal T$ such that

\Centerline{$1\Bsetminus a_{\rho}\Bsubseteq\Bvalue{\rho'=\rho\wedge\tau}$,
\quad$a_{\rho}\Bsubseteq\Bvalue{\rho'=\tau}$,}

\noindent and $\rho'\in\Cal S$.   If $\rho_0\le\rho_1$ in $A$ then
$\rho_0'\le\rho_1'$.   \Prf\ $a_{\rho_0}\Bsubseteq a_{\rho_1}$.   Now

$$\eqalign{a_{\rho_0}&
\Bsubseteq\Bvalue{\rho_0'=\tau}\Bcap\Bvalue{\rho_1'=\tau}
\Bsubseteq\Bvalue{\rho_0'\le\rho_1'},\cr
a_{\rho_1}\Bsetminus a_{\rho_0}
&\Bsubseteq\Bvalue{\rho_0'=\rho_0\wedge\tau}\Bcap\Bvalue{\rho_1'=\tau}
\Bsubseteq\Bvalue{\rho_0'\le\rho_1'},\cr
1\Bsetminus a_{\rho_1}
&\Bsubseteq\Bvalue{\rho_0'=\rho_0\wedge\tau}
\Bcap\Bvalue{\rho_1'=\rho_1\wedge\tau}
\Bsubseteq\Bvalue{\rho_0'\le\rho_1'}.\cr}$$

\noindent So $\rho_0'\le\rho_1'$.\ \Qed

\medskip

\qquad\grheadb\ If $\rho\in A$ and $\sigma\in\Cal S$,
$\tilde u_{\sigma\wedge\rho'}=u_{\sigma\wedge\tau\wedge\rho}$.
\Prf\ By (a-ii),
$d_n\Bsubseteq\Bvalue{\tilde u_{\sigma}=u_{\sigma\wedge\tau_n}}$ for every
$\sigma\in\Cal S$.   So

$$\eqalign{a_{\rho}&
\Bsubseteq d_n\Bcap\Bvalue{\tau_n\le\tau\wedge\rho}\Bcap\Bvalue{\rho'=\tau}
\Bsubseteq\Bvalue{\tilde u_{\sigma\wedge\rho'}=\tilde u_{\sigma\wedge\rho'\wedge\tau_n}}
   \Bcap\Bvalue{\rho=\tau_n}\Bcap\Bvalue{\rho'=\tau}\cr&
\Bsubseteq\Bvalue{\tilde u_{\sigma\wedge\rho'}=\tilde u_{\sigma\wedge\tau\wedge\rho}}
=\Bvalue{\tilde u_{\sigma\wedge\rho'}=u_{\sigma\wedge\tau\wedge\rho}}\cr}$$

\noindent because $\sigma\wedge\tau\wedge\rho\in\Cal S\wedge\tau_n
\subseteq\Cal S_0$.   On the other hand,

\Centerline{$1\Bsetminus a_{\rho}
\Bsubseteq\Bvalue{\rho'=\tau\wedge\rho}
\Bsubseteq\Bvalue{\tilde u_{\sigma\wedge\rho'}
   =\tilde u_{\sigma\wedge\tau\wedge\rho}}
=\Bvalue{\tilde u_{\sigma\wedge\rho'}=u_{\sigma\wedge\tau\wedge\rho}}$.}

\noindent Putting these together, we have the result.\ \Qed

\medskip

\qquad\grheadc\ Set $A'=\{\rho':\rho\in A\}$;  then $A'\subseteq\Cal S$ is
downwards-directed and not empty.   Now
$\sup_{\rho\in A'}\bar\mu\Bvalue{\rho<\tau}
=\sup_{\rho\in A}\bar\mu\Bvalue{\rho'<\tau}$ is at most $2\epsilon$.
\Prf\ If $\rho\in A$, then

$$\eqalign{\Bvalue{\rho'<\tau}&
=\Bvalue{\rho\wedge\tau<\tau}\Bsetminus a_{\rho}
=\Bvalue{\rho<\tau}\Bsetminus(d_n\Bcap\Bvalue{\tau_n\le\rho\wedge\tau})\cr&
\Bsubseteq\Bvalue{\rho<\tau_n}
  \Bcup(\Bvalue{\rho=\tau_n}\Bcap\Bvalue{\tau_n<\tau}
          \Bsetminus(d_n\Bcap\Bvalue{\tau_n\le\rho\wedge\tau}))\cr&
=\Bvalue{\rho<\tau_n}\Bcup(\Bvalue{\rho=\tau_n}\Bcap\Bvalue{\tau_n<\tau}
                                   \Bsetminus d_n)\cr&
\Bsubseteq\Bvalue{\rho<\tau_n}\Bcup
  (1\Bsetminus(d_n\Bcup\Bvalue{\tau\le\tau_n}))\cr}$$

\noindent has measure at most $2\epsilon$.\ \Qed

\medskip

\qquad\grheadd\ Because $\tildepmb u$ is locally moderately oscillatory,
$R_{A'}(\tildepmb u)$ is defined.   Furthermore, it is a martingale.
\Prf\ Express the martingale $R_A(\pmb u\restr\Cal S\wedge\tau_n)$ as
$\langle u'_{\sigma}\rangle_{\sigma\in\Cal S\wedge\tau_n}$, so that
$u'_{\sigma}=\lim_{\rho\downarrow A}u_{\sigma\wedge\rho}$ belongs to
$L^1_{\bar\mu}$ for $\sigma\in\Cal S\wedge\tau_n$, and
$P_{\sigma}u'_{\sigma'}=u'_{\sigma}$ whenever
$\sigma\le\sigma'$ in $\Cal S\wedge\tau_n$.

For $\sigma\in\Cal S$, write $v'_{\sigma}$ for

\Centerline{$\lim_{\rho\downarrow A'}\tilde u_{\sigma\wedge\rho}
=\lim_{\rho\downarrow A}\tilde u_{\sigma\wedge\rho'}
=\lim_{\rho\downarrow A}u_{\sigma\wedge\tau\wedge\rho}
=\lim_{\rho\downarrow A}u_{\sigma\wedge\tau_n\wedge\tau\wedge\rho}
=u'_{\sigma\wedge\tau_n\wedge\tau}$;}

\noindent this belongs to $L^1_{\bar\mu}$.
And if $\sigma\le\sigma'$ in $\Cal S$,

$$\eqalign{P_{\sigma}(v'_{\sigma'})
=P_{\sigma}u'_{\sigma'\wedge\tau_n\wedge\tau}
=P_{\sigma\wedge\tau_n\wedge\tau}u'_{\sigma'\wedge\tau_n\wedge\tau}
=u'_{\sigma\wedge\tau_n\wedge\tau}
=v'_{\sigma}.\cr}$$

\noindent So $\sfamilysS{v'}=R_{A'}(\tildepmb u)$ is a martingale.\ \Qed

\medskip

\qquad\grheade\ Since $\tau$ and $\epsilon$ are arbitrary, $\tildepmb u$ is a
virtually local martingale.

\medskip

\quad{\bf (ii)} Generally, take
$\hat\Cal S$, $\hat\Cal S_0$, $\tildehatpmb u$ and $\hatpmb u$ as in (b) and
(c-ii).    Because
$\hatpmb u\restr\hat\Cal S_0\wedge\tau_n$ is the fully adapted extension of
$\pmb u\restr\Cal S\wedge\tau_n$, it is a virtually local martingale for every
$n\in\Bbb N$;  because $\{\tau_n:n\in\Bbb N\}$ is cofinal with $\hat\Cal S_0$,
$\hatpmb u$ is a virtually local martingale;  by (i) here, $\tildehatpmb u$
is a
virtually local martingale;  it follows that $\tildepmb u$ is a virtually local
martingale.

\bigskip

{\bf (e)} Take $\tau\in\Cal S$.   For $n\in\Bbb N$,
$\bar z_n=\int_{\Cal S\wedge\tau_n}|d\pmb u|$ is defined in
$L^0(\frak A)$.   Set
$d'_{-1}=0$, $d'_n=d_n\Bcup\Bvalue{\tau\le\tau_n}$ for $n\ge 0$ and

\Centerline{$\bar z
=\sum_{n=0}^{\infty}(\bar z_n\times\chi(d'_n\Bsetminus d'_{n-1}))
\in L^0$.}

\noindent
Now suppose that $\sigma_0\le\ldots\le\sigma_k\le\tau$ in $\Cal S$, and set
$z=\sum_{i=0}^{k-1}|\tilde u_{\sigma_{i+1}}-\tilde u_{\sigma_i}|$.   Then

$$\eqalign{d'_n&
\Bsubseteq\inf_{i\le k}\Bvalue{\tilde u_{\sigma_i}=u_{\sigma_i\wedge\tau_n}}\cr&
\Bsubseteq\Bvalue{z
   =\sum_{i=0}^{k-1}|u_{\sigma_{i+1}\wedge\tau_n}-u_{\sigma_i\wedge\tau_n}|}
\Bsubseteq\Bvalue{z\le\bar z_n}\cr}$$

\noindent so

\Centerline{$d'_n\Bsetminus d'_{n-1}
\Bsubseteq\Bvalue{z\le\bar z_n}\Bcap\Bvalue{\bar z_n\le\bar z}
\Bsubseteq\Bvalue{z\le\bar z}$}

\noindent for every $n\in\Bbb N$.   Since $\sequencen{d'_n}$ is non-decreasing
and has supremum $1$, $z\le\bar z$.   As $\sigma_0,\ldots,\sigma_k$ are
arbitrary, $\tildepmb u\restr\Cal S\wedge\tau$ is of bounded variation.

\medskip

{\bf (f)} Take $\tau^*\in\Cal S$.   I need to show that
$\Osclln(\tildepmb u\restr\Cal S\wedge\tau^*)\le\bar w$.
We know from (c) that
$\tildepmb u\restr\Cal S\wedge\tau^*$ is order-bounded;  set
$\bar w'=\Osclln(\tildepmb u\restr\Cal S\wedge\tau^*)$.   Take any
$n\in\Bbb N$ and $\epsilon>0$.   Set $\tilde\tau=\tau^*\wedge\tau_n$.
Note that by 618Da,

\Centerline{$\Osclln(\pmb u\restr\Cal S\wedge\tilde\tau)
\le\Osclln(\pmb u\restr\Cal S\wedge\tau_n)
\le\bar w$.}

Suppose that $I\subseteq J\in\Cal I(\Cal S\wedge\tau^*)$.
If $\tilde\tau\le\tau\le\tau'\le\tau^*$ in $\Cal S$, then
$\tau\wedge\tau_n=\tilde\tau=\tau'\wedge\tau_n$, so

\Centerline{$d_n\Bcup\Bvalue{\tau^*\le\tau_n}
\Bsubseteq\Bvalue{\tilde u_{\tau}=u_{\tau\wedge\tau_n}}
  \Bcap\Bvalue{\tilde u_{\tau'}=u_{\tau'\wedge\tau_n}}
\Bsubseteq\Bvalue{\tilde u_{\tau}=\tilde u_{\tau'}}$.}

\noindent As $\tau$ and $\tau'$ are arbitrary,

$$\eqalignno{d_n\Bcup\Bvalue{\tau^*\le\tau_n}
&\Bsubseteq\Bvalue{\Osclln_{J\vee\tilde\tau}(\tildepmb u)=0}
\Bsubseteq\Bvalue{\Osclln_J(\tildepmb u)=\Osclln_{J\wedge\tilde\tau}(\tildepmb u)}
  \cr
&=\Bvalue{\Osclln_J(\tildepmb u)=\Osclln_{J\wedge\tilde\tau}(\pmb u)}\cr
\displaycause{because $\tildepmb u\restr\Cal S\wedge\tilde\tau
=\pmb u\restr\Cal S\wedge\tilde\tau$}
&\Bsubseteq\Bvalue{\Osclln_J(\tildepmb u)
   \le\Osclln^*_I(\pmb u\restr\Cal S\wedge\tilde\tau)}.\cr}$$

\noindent As $J$ is arbitrary,

$$\eqalign{d_n\Bcup\Bvalue{\tau^*\le\tau_n}&
\Bsubseteq\Bvalue{\Osclln^*_I(\tildepmb u\restr\Cal S\wedge\tau^*)
   \le\Osclln^*_I(\pmb u\restr\Cal S\wedge\tilde\tau)}\cr&
\Bsubseteq\Bvalue{\bar w'
   \le\Osclln^*_I(\pmb u\restr\Cal S\wedge\tilde\tau)}.\cr}$$

\noindent As $I$ is arbitrary,

\Centerline{$d_n\Bcup\Bvalue{\tau^*\le\tau_n}
\Bsubseteq\Bvalue{\bar w'\le\Osclln(\pmb u\restr\Cal S\wedge\tilde\tau)}
\Bsubseteq\Bvalue{\bar w'\le\bar w}$.}

\noindent As $n$ is arbitrary,
$\Osclln(\tildepmb u\restr\Cal S\wedge\tau^*)=\bar w'\le\bar w$, as required.

\medskip

{\bf (g)} Express $\pmb u$ as $\pmb v+\pmb w$ where
$\pmb v=\family{\sigma}{\Cal S_0}{v_{\sigma}}$ is a
virtually local martingale and $\pmb w=\family{\sigma}{\Cal S_0}{w_{\sigma}}$
is locally of bounded variation.
Let $\tildepmb v=\family{\tau}{\Cal S}{\tilde v_{\tau}}$ and
$\tildepmb w=\family{\tau}{\Cal S}{\tilde w_{\tau}}$ be the corresponding
extensions, so that $\tildepmb v$ is a virtually local martingale and
$\tildepmb w$ is locally of bounded variation ((d) and (e) above).   Then

\Centerline{$\tilde u_{\tau}
=\lim_{n\to\infty}u_{\tau\wedge\tau_n}
=\lim_{n\to\infty}v_{\tau\wedge\tau_n}
   +\lim_{n\to\infty}w_{\tau\wedge\tau_n}
=\tilde v_{\tau}+\tilde w_{\tau}$}

\noindent for every $\tau\in\Cal S$, so that
$\tildepmb u=\tildepmb v+\tildepmb w$ is a \sm.
}%end of proof of 627O

\leader{627P}{Corollary} Let $\Cal S$ be a sublattice of $\Cal T$,
$\tau$ a member of $\Cal S$, and $\pmb u=\sfamilysS{u}$,
$\pmb w=\family{\sigma}{\Cal S\wedge\tau}{w_{\sigma}}$ two fully adapted
processes.   Set

\Centerline{$v_{\sigma}=u_{\sigma}-u_{\sigma\wedge\tau}+w_{\sigma\wedge\tau}$}

\noindent for $\sigma\in\Cal S$.   Then $\pmb v=\sfamilysS{v}$ is fully adapted,
and is locally moderately oscillatory, or a virtually local martingale, or
locally of bounded variation, or a \sm\ if $\pmb u$ and $\pmb w$ both are, while

\Centerline{$v_{\sigma}=w_{\sigma}$ if $\sigma\in\Cal S\wedge\tau$,
\quad$v_{\sigma}=u_{\sigma}-u_{\tau}+w_{\tau}$ if $\sigma\in\Cal S\vee\tau$.}

\proof{ Apply 627O with $d_n=1$, $\tau_n=\tau$ for every $n\in\Bbb N$, so that
$\Cal S_0=\Cal S\wedge\tau$ and $\sigma\mapsto u_{\sigma\wedge\tau}$,
$\sigma\mapsto w_{\sigma\wedge\tau}$ are the extensions $\tildepmb u$,
$\tildepmb w$ of
$\pmb u\restr\Cal S\wedge\tau$ and $\pmb w$ as described in 627O.   Now for each
of the four
properties listed, $\pmb u\restr\Cal S\wedge\tau$ will have the property
if $\pmb u$ does;  see 623Ke for virtually local martingales, and the others are
almost immediate.   So 627O tells us that $\tildepmb u$ and $\tildepmb w$
also have
the property considered, so that $\pmb v=\pmb u-\tildepmb u+\tildepmb w$
also does, using 614Q(b-iii), 615F(b-iii) or 623Ka.
}%end of proof of 627P

\leader{627Q}{Theorem} A fully adapted process is a \sm\ iff it is a local
integrator.

\proof{{\bf (a)} We saw in 625E that a \sm\ is a local integrator.   So it will
be enough to show that if $\Cal S$ is a sublattice of $\Cal T$ and
$\pmb u=\sfamilysS{u}$ is a local integrator, then it is a \sm.

\medskip

{\bf (b)} To begin with, suppose that $\Cal S$ is full.   Since
$\pmb u$ is locally moderately oscillatory (616Ib), the hypotheses of 627N are
satisfied;  let $\sequencen{d_n}$ and $\sequencen{\tau_n}$ be as described
there.   For each $n\in\Bbb N$, $\pmb u\restr\Cal S\wedge\tau_n$ is an
integrator, therefore a \sm\ (627J);  let
$\pmb v_n=\family{\sigma}{\Cal S\wedge\tau_n}{v_{n\sigma}}$ be a virtually local
martingale and
$\pmb w_n=\family{\sigma}{\Cal S\wedge\tau_n}{w_{n\sigma}}$ a process of
bounded variation such that
$\pmb v_n+\pmb w_n=\pmb u_n\restr\Cal S\wedge\tau_n$.   Define
$\pmb v'_n=\family{\sigma}{\Cal S\wedge\tau_n}{v'_{n\sigma}}$ and
$\pmb w'_n=\family{\sigma}{\Cal S\wedge\tau_n}{w'_{n\sigma}}$ inductively by
saying that

\Centerline{$\pmb v'_0=\pmb v_0$,\quad$\pmb w'_0=\pmb w_0$,}

\Centerline{$v'_{n+1,\sigma}=v_{n+1,\sigma}-v_{n+1,\sigma\wedge\tau_n}
                                +v'_{n,\sigma\wedge\tau_n}$,}

\Centerline{$w'_{n+1,\sigma}=w_{n+1,\sigma}-w_{n+1,\sigma\wedge\tau_n}
                                +w'_{n,\sigma\wedge\tau_n}$}

\noindent for $n\in\Bbb N$ and $\sigma\in\Cal S\wedge\tau_{n+1}$.   We see by
induction, using 627P, that $\pmb v'_n$ is a virtually local martingale and
$\pmb w'_n$ is of bounded variation, while $\pmb v'_{n+1}$ extends $\pmb v'_n$,
$\pmb w'_{n+1}$ extends $\pmb w'_n$ and $\pmb v'_n+\pmb w'_n=\pmb u_n$, for
every $n\in\Bbb N$.   We therefore have processes
$\pmb v=\bigcup_{n\in\Bbb N}\pmb v'_n$ and
$\pmb w=\bigcup_{n\in\Bbb N}\pmb w'_n$, both with domain
$\Cal S_0=\bigcup_{n\in\Bbb N}\Cal S\wedge\tau_n$, such that $\pmb v$ is a
virtually local martingale (623Ke again) and $\pmb w$ is locally of
bounded variation, while $\pmb v+\pmb w=\pmb u\restr\Cal S_0$.
Thus $\pmb u\restr\Cal S_0$ is a \sm.   But as the construction in
627N arranges that $u_{\tau}=\lim_{n\to\infty}u_{\tau\wedge\tau_n}$ for every
$\tau\in\Cal S$, $\pmb u$ itself is a \sm, by 627Og.

\medskip

{\bf (c)} For general $\Cal S$, let $\hat\Cal S$ be the covered
envelope of $\Cal S$, and write $\hatpmb u$ for the fully adapted
extension of $\pmb u$ to $\hat\Cal S$.   By 616Ia, $\hatpmb u$ is a local
integrator,   By (b) here, $\hatpmb u$ is a \sm, and is expressible as the sum
of a virtually local martingale $\hatpmb v$ and a process $\hatpmb w$ which is
locally of bounded variation.   Setting $\pmb v=\hatpmb v\restr\Cal S$ and
$\pmb w=\hatpmb w\restr\Cal S$, $\pmb u=\pmb v+\pmb w$, while $\pmb v$ is a
virtually local martingale (623J) and $\pmb w$ is locally of bounded variation
(614L(b-ii)).   So $\pmb u$ is a \sm, which is what we need to know.
}%end of proof of 627Q

\leader{627R}{}\cmmnt{ In the cases we really care about we can escape
most of the work in Lemmas 627N and 627O by using the following.

\medskip

\noindent}{\bf Proposition} Suppose that $T$ is separable in its order
topology.   If $\Cal S$ is any sublattice of $\Cal T$, there is a non-decreasing
sequence $\sequencen{\tau_n}$ in $\Cal S$ such that
$\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}=1$ for every $\sigma\in\Cal S$.

\proof{{\bf (a)} Set $\tau^*=\sup\Cal S$ in $\Cal T$.
Then there is a countable set $T_0\subseteq T$ such that
$\Bvalue{\tau^*>t}=\sup_{s\in T_0,s\ge t}\Bvalue{\tau^*>s}$ for every $t\in T$.
\Prf\ Take $T_1$ to be a countable dense subset of $T$ and set

\Centerline{$T_2=\{t:\Bvalue{\tau^*>t}\ne\sup_{s>t}\Bvalue{\tau^*>s}\}
=\{t:\bar\mu\Bvalue{\tau^*>t}>\sup_{s>t}\bar\mu\Bvalue{\tau^*>s}\}$,}

\Centerline{$T_3=\{t:\inf\{s:s>t\}\in T_2\}$,}

\noindent so that $T_2$ and $T_3$ are countable;
then $T_0=T_1\cup T_2\cup T_3$ is countable.
If $t\in T_0$ then of course
$\Bvalue{\tau^*>t}=\sup_{s\in T_0,s\ge t}\Bvalue{\tau^*>s}$.
If $t\in T\setminus T_0$ and $\epsilon>0$, then $t\notin T_2$ so
there is an $s_0>t$ such that
$\bar\mu\Bvalue{\tau^*>t}\le\bar\mu\Bvalue{\tau^*>s_0}+\epsilon$;
as $t\notin T_3$, we can suppose that $s_0\notin T_2$, so that there is an
$s_1>s_0$ such that
$\bar\mu\Bvalue{\tau^*>s_0}\le\bar\mu\Bvalue{\tau^*>s_1}+\epsilon$.
Now the open
interval $\ooint{\tau^*,s_1}$ is non-empty, so meets $T_1$, and there is an
$s\in T_1$ such that
$\bar\mu\Bvalue{\tau^*>t}\le\bar\mu\Bvalue{\tau^*>s}+2\epsilon$.   As
$\epsilon$ is arbitrary,
$\Bvalue{\tau^*>t}=\sup_{s\in T_0,s\ge t}\Bvalue{\tau^*>s}$ in this case
also.\ \Qed

\medskip

{\bf (b)} Set $a=\sup_{\sigma\in\Cal S}\Bvalue{\sigma=\tau^*}$.
Then we have a countable $C\subseteq\Cal S$ such that
$a=\sup_{\tau\in C}\Bvalue{\tau=\tau^*}$.   Next, for each
$t\in T_0$, let $D_t$ be a countable subset of $\Cal S$ such that

\Centerline{$\sup_{\tau\in D_t}\Bvalue{\tau>t}
=\sup_{\sigma\in\Cal S}\Bvalue{\sigma>t}
=\Bvalue{\tau^*>t}$}

\noindent(611Cb).   Set
$D=C\cup\bigcup_{t\in T_0}D_t\in[\Cal S]^{\le\omega}$.

\medskip

{\bf (c)} $\sup D=\tau^*$.   \Prf\ For any $t\in T$,

$$\Bvalue{\tau^*>t}
=\sup_{\Atop{s\in T_0}{s\ge t}}\Bvalue{\tau^*>s}
=\sup_{\Atop{s\in T_0}{\Atop{s\ge t}{\tau\in D_s}}}\Bvalue{\tau>s}
\Bsubseteq\sup_{\Atop{s\in T_0}{\tau\in D_s}}\Bvalue{\tau>t}
\Bsubseteq\Bvalue{\sup D>t}$$

\noindent so $\tau^*\le\sup D$;  and of course $\sup D\le\tau^*$.\ \Qed

\medskip

{\bf (d)} Take any $\sigma\in\Cal S$.   We have

\Centerline{$\Bvalue{\sigma<\tau^*}=\Bvalue{\sigma<\sup D}
=\sup_{\tau\in D}\Bvalue{\sigma<\tau}$,}

\noindent while $\Bvalue{\sigma=\tau^*}=\sup_{\tau\in D}\Bvalue{\sigma=\tau}$
because $C\subseteq D$.   Consequently

\Centerline{$\sup_{\tau\in D}\Bvalue{\sigma\le\tau}
=\Bvalue{\sigma\le\tau^*}=1$.}

\medskip

{\bf (e)} Let $\sequencen{\tau'_n}$ be a sequence running over $D$ and set
$\tau_n=\sup_{i\le n}\tau'_i$ for $n\in\Bbb N$.   Then
$\sup_{n\in\Bbb N}\Bvalue{\sigma\le\tau_n}
=\sup_{\tau\in D}\Bvalue{\sigma\le\tau}=1$, as required.
}%end of proof of 627R

\cmmnt{\medskip

\noindent{\bf Remark} Accordingly we have the result of 627N
in a much stronger form, with every $d_n$ equal to $0$, and with the sequence
$\sequencen{\tau_n}$ independent of the process $\pmb u$.
}

%\leader{627S}{}
%\leader{627T}{}
%\leader{627U}{}
%\leader{627V}{}
%\leader{627W}{}

\exercises{\leader{627X}{Basic exercises (a)}
%\spheader 627Xa
Suppose that $T=\coint{0,\infty}$ and $\frak A=\{0,1\}$, as
in 613W, 615Xf, 616Xa, 617Xb, 618Xa, 622Xd and 626Xa,
Let $f:\coint{0,\infty}\to\Bbb R$
be a function and $\pmb u$ the corresponding process on $\Cal T_f$.
(i) Show that $\pmb u$ is a supermartingale iff $f$ is non-increasing.
(ii) Show that $\pmb u$ is a quasimartingale iff it is a strong
integrator iff $f$ is of bounded variation.
%627B 

\spheader 627Xb
Let $\Cal S$ be a sublattice of $\Cal T$ and
$\pmb v=\sfamilysS{v}$ a supermartingale.   Write $\hatpmb v$ for the
fully adapted extension of $\pmb v$ to the covered envelope $\hat\Cal S$ of
$\Cal S$, and $\hat\Cal S_f$ for the finitely-covered envelope of $\Cal S$.
Show that

\quad (i) if $\pmb v$ is $\|\,\|_1$-bounded, it is an integrator;  %626Da

\quad(ii) if $\Cal S$ has a greatest element and
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded above, $\pmb v$ is
$\|\,\|_1$-bounded:  %626Db

\quad(iii) if $A\subseteq\Cal S$ is non-empty and downwards-directed and
$\{\Expn(v_{\sigma}):\sigma\in A\}$ is bounded above, then ($\alpha$)
$\lim_{\sigma\downarrow A}v_{\sigma}$ and $\1lim_{\sigma\downarrow A}v_{\sigma}$
are defined and equal  %626E
($\beta$) $R_A(\pmb v)$ is defined and is a supermartingale;  %626F

\quad(iv)($\alpha$) $\hatpmb v\restr\hat\Cal S_f$ is a supermartingale %626Ia
($\beta$) if $\pmb v$ is $\|\,\|_1$-bounded then $\hatpmb v$ is
$\|\,\|_1$-bounded %626Ib
($\gamma$) if $\Cal S$ has a greatest element and
$\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded above, then
$\hatpmb v$ is a supermartingale;  %626Ic

\quad(v) if $\pmb v$ has a previsible variation $\pmb v^{\#}$
(definition:  626L), then $\pmb v^{\#}$ is non-increasing, therefore
a supermartingale;  %626O

\quad(vi) if $\{\Expn(v_{\sigma}):\sigma\in\Cal S\}$ is bounded above, then
$\pmb v$ is expressible as the sum of a non-increasing fully adapted process and
a virtually local martingale.  %626Q
%627C

\spheader 627Xc
Show that sums and scalar multiples of quasimartingales
are quasimartingales.
%627C

\spheader 627Xd
Let $\pmb v$ be an $L^1$-process defined on
a sublattice $\Cal S$ of $\Cal T$.   (i) Show that
$Q_{\Cal S}(|Pd\pmb v|)\Bsubseteq Q_{\Cal S}(Pd\pmb v)$.
(ii) Show that the following are
equiveridical:  ($\alpha$) $\pmb v$ is a quasimartingale;  ($\beta$)
$Q_{\Cal S}(Pd\pmb v)$ is $\|\,\|_1$-bounded;   ($\gamma$)
$Q_{\Cal S}(|Pd\pmb v|)$ is $\|\,\|_1$-bounded;   ($\delta$)
$\{\Expn(S_I(\tbf{1},|Pd\pmb v|)):I\in\Cal I(\Cal S)\}$
is bounded.   \Hint{for ($\alpha)\Rightarrow(\beta$), recall that weakly
bounded sets in Banach spaces are bounded.}
%627E

\spheader 627Xe Let $\Cal S$ be a sublattice of $\Cal T$ with greatest and
least elements.   Show that a submartingale with domain $\Cal S$ is a
quasimartingale.
%627Xd 627E

%\leader{627Y}{Further exercises (a)}

}%end of exercises

\endnotes{
\Notesheader{627} The target of this section is Theorem 627Q.
The traditional approach to stochastic integration has been integration with
respect to \sm{s}, typically in contexts which ensure that they can be
described in terms of local martingales rather than virtually local martingales.
so that {\smc Protter 05}, for instance, uses the word `semimartingale' for
what I call `integrators', and the phrase `classical semimartingale'
for a sum of a process of
bounded variation and a local martingale.   Protter's Theorem 47
corresponds to my
627J.   The first steps are to prove that a \sm\ is a local
integrator (625E) and that an integrator is a \sm.   To show that
every local integrator is a \sm\ we need some more technique, using
627N-627O or 627R.

Clearly supermartingales are going to be like submartingales in many ways.
But it does not at all
follow that non-negative supermartingales, as in 627D-627E,
are going to behave like non-negative submartingales, as in 626O.   The idea of
quasimartingales is to get back to something symmetric, like integrators and
\sm{s}.

The principal results here (627I, 627J, 627L)
depend on the theory of convex sets in
locally convex linear topological spaces (627F-627G), and our difficulty is that
the topology of convergence in measure is not locally convex in the interesting
cases.   So we have to negotiate carefully to ensure that we have topologically
bounded convex sets.
If $\pmb v$ is an integrator, $Q_{\Cal S}(d\pmb v)$ is convex,
by 627Hb, and its
solid hull is topologically bounded by 613B(f-iv).   But it does not
seem to follow directly that the convex hull of its solid hull will be
bounded.   So for 627M we need to refer back to the methods used in
621H, and these work best on what I am calling `strong integrators'.   In fact
627L-627M will be used rarely in this volume.   But I think we need them to get
a proper idea of what an integrator is.

I remarked in the notes to \S625 that the law-independence of the class
of \sm s (625F) was surprising.   Using that independence, we have come to a
law-independent characterization of \sm s as local integrators
(627Q).   There are no coincidences in mathematics.   But there are many
deep structures with unexpected outcrops.
}%end of notes

\discrpage

\frfilename{mt628.tex}
\versiondate{31.12.17}
\copyrightdate{2017}

\def\doubleheadrightarrow{\rightarrow\mskip-14mu\rightarrow}


\def\chaptername{Martingales}
\def\sectionname{Refining a martingale inequality}

\newsection{*628}

I remarked in \S621 that the constant $16$ in the inequality 621Hf
can be reduced to $2$ if we are willing to use some rather more advanced measure
theory.   This treatise is not about finding best constants.   But 2 is a much
prettier number than 16 and the method I have devised passes through a
construction (628C) which may have other uses, as in 628F-628G,
so I present it here.

\leader{628A}{Lemma} Let $(\frak A,\bar\mu)$ be a probability algebra, and
$\frak A_0$ a closed subalgebra of $\frak A$;  write
$P_0:L^1_{\bar\mu}\to L^1_{\bar\mu}$ for the corresponding conditional
expectation operator.   Suppose that $v\in L^1_{\bar\mu}$.   Set $v_0=P_0v$.
Then there are a probability algebra $(\frak B,\bar\nu)$,
closed subalgebras $\frak B_0\subseteq\frak B_1\subseteq\frak B$, and
a measure-preserving Boolean homomorphism $\pi:\frak A\to\frak B$ such that
$\pi[\frak A_0]=\frak B_0$ and
if $T=T_{\pi}:L^1_{\bar\mu}\to L^1_{\bar\nu}$ is the associated
embedding\cmmnt{ (365N\formerly{3{}65O})} and $w_i$ is the conditional
expectation of $Tv$ on $\frak B_i$ for both $i$, then

\Centerline{$w_0=Tv_0$,}

$$\eqalign{\Bvalue{|w_1|=1}\Bcap\Bvalue{|w_0|<1}
&=\Bvalue{|w_1|\ge 1}\Bcap\Bvalue{|w_0|<1}\cr
&\Bsupseteq\Bvalue{|Tv|\ge 1}\Bcap\Bvalue{|w_0|<1}.\cr}$$

\proof{{\bf (a)} Let $(\frak A',\bar\mu')$ be a homogeneous probability
algebra of infinite Maharam type at least equal to the relative Maharam
type of $\frak A$ over $\frak A_0$ (333Aa).   Let
$(\frak C,\bar\lambda)$ be the probability algebra free product of
$(\frak A_0,\bar\mu\restrp\frak A_0)$ and $(\frak A',\bar\mu')$
(325K).   Then there is a measure-preserving
homomorphism $\pi_0:\frak A\to\frak C$ extending the canonical homomorphism
$a\mapsto a\otimes 1:\frak A_0\to\frak A_0\tensorhat\frak A'=\frak C$
(333Fa).   Set $\frak C_0=\pi_0[\frak A_0]=\{a\otimes 1:a\in\frak A_0\}$;
let $T_0:L^1_{\bar\mu}\to L^1_{\bar\lambda}$ be the embedding
associated with
$\pi_0$, and $Q_0:L^1_{\bar\lambda}\to L^1_{\bar\lambda}$ the
conditional expectation operator corresponding to $\frak C_0$.
Then $Q_0T_0=T_0P_0$, by 365Qd\formerly{3{}65Rd}.

\medskip

{\bf (b)} Let $(\Omega,\Sigma,\mu)$ and $(\Omega',\Sigma',\mu')$ be
probability spaces with measure algebras isomorphic to
$(\frak A_0,\bar\mu\restrp\frak A_0)$ and $(\frak A',\bar\mu')$
respectively (321J).   Let $\lambda$ be the product measure
on $\Omega\times\Omega'$.   Then the measure algebra of $\lambda$ is
isomorphic to $(\frak C,\bar\lambda)$ (325I), and if $E\in\Sigma$ then
$(E\times\Omega')^{\ssbullet}$ corresponds to
$E^{\ssbullet}\otimes 1$, so
that $\pi_0\restrp\frak A_0:\frak A_0\to\frak C$ corresponds to the map
$E\mapsto E\times\Omega'$, and $\frak C_0$ to the $\sigma$-algebra
$\{E\times\Omega':E\in\Sigma\}$.   Accordingly the conditional expectation
operator $Q_0$
can be defined by saying that if $g\in\eusm L^1(\lambda)$ then
$Q_0(g^{\ssbullet})=f^{\ssbullet}$, where
$f(\omega,\tilde\omega')=\int g(\omega,\omega')d\omega'$ whenever
this is defined (253H), the integral here being with respect to $\mu'$.

\medskip

{\bf (c)} Let $g\in\eusm L^1(\lambda)$ be
such that $g^{\ssbullet}=T_0v$;  we may suppose that $g$ is defined
everywhere on $\Omega\times\Omega'$ and is
$\Sigma\tensorhat\Sigma'$-measurable
(because $\lambda$ is the completion of
its restriction to $\Sigma\tensorhat\Sigma'$, by 251K).
Moreover, adjusting $g$ on a negligible set of the form $E\times\Omega'$ if
necessary, we can arrange that $\int g(\omega,\omega')d\omega'$ is
defined for every $\omega\in\Omega$ (252B).   Set
$E=\{\omega:-1<\int g(\omega,\omega')d\omega'<1\}$;  by
252P, $E\in\Sigma$.

\medskip

{\bf (d)} For $\omega\in E$ set

\Centerline{$F_{\omega}=\{\omega':g(\omega,\omega')\le -1\}$,
\quad$F'_{\omega}=\{\omega':g(\omega,\omega')\ge 1\}$.}

\noindent Let me explain where I think I am going.
I seek to define measurable sets $G_{\omega}$,
$G'_{\omega}\subseteq\Omega'\times[0,1]$ with the properties that

\Centerline{$G_{\omega}\cap G'_{\omega}=\emptyset$,
\quad$G_{\omega}\cup G'_{\omega}
\supseteq(F_{\omega}\cup F'_{\omega})\times[0,1]$,}

\Centerline{$\iint_{G_{\omega}}g(\omega,\omega')d\omega'd\omega''
=-(\mu'\times\mu'')(G_{\omega})$,}

\Centerline{$\iint_{G'_{\omega}}g(\omega,\omega')d\omega'd\omega''
=(\mu'\times\mu'')(G'_{\omega})$,}

\noindent where $\mu''$ is Lebesgue measure on $[0,1]$ and the integrals
$\int\ldots d\omega''$ are taken with respect to $\mu''$;  moreover, I wish to do this in such a way that

\Centerline{$G=\{(\omega,\omega',\omega''):\omega\in E$,
$(\omega',\omega'')\in G_{\omega}\}$,}

\Centerline{$G'=\{(\omega,\omega',\omega''):\omega\in E$,
$(\omega',\omega'')\in G'_{\omega}\}$}

\noindent belong to $\Sigma\tensorhat\Sigma'\tensorhat\Sigma''$,
where $\Sigma''$ is the Borel $\sigma$-algebra of $[0,1]$.

\medskip

{\bf (e)} Set

\Centerline{$E_0
=\{\omega:\omega\in E$, $\mu'F_{\omega}=\mu'F'_{\omega}=0\}$,}

\Centerline{$E_1
=\{\omega:\omega\in E\setminus E_0$,
$\int_{F_{\omega}\cup F'_{\omega}}g(\omega,\omega')d\omega'
\le-\mu'(F_{\omega}\cup F'_{\omega})\}$,}

\Centerline{$E_2
=\{\omega:\omega\in E\setminus E_0$,
$\int_{F_{\omega}\cup F'_{\omega}}g(\omega,\omega')d\omega'
\ge\mu'(F_{\omega}\cup F'_{\omega})\}$,}

\Centerline{$E_3
=\{\omega:\omega\in E\setminus E_0$,
$-\mu'(F_{\omega}\cup F'_{\omega})<
  \int_{F_{\omega}\cup F'_{\omega}}g(\omega,\omega')d\omega'
  <\mu'(F_{\omega}\cup F'_{\omega})\}$.}

\noindent Because $\omega\mapsto\mu'F_{\omega}$,
$\omega\mapsto\mu'F'_{\omega}$,
$\omega\mapsto\int_{F_{\omega}}g(\omega,\omega')d\omega'$ and
$\omega\mapsto\int_{F'_{\omega}}g(\omega,\omega')d\omega'$
are $\Sigma$-measurable (252P again), $(E_0,E_1,E_2,E_3)$
is a partition of $E$ into members of $\Sigma$.

\medskip

{\bf (f)(i)} If $\omega\in E_0$,
set $G_{\omega}=F_{\omega}\times[0,1]$ and
$G'_{\omega}=F'_{\omega}\times[0,1]$.

\medskip

\quad{\bf (ii)} If $\omega\in E_1$, then
(because $\int_{\Omega'}g(\omega,\omega')d\omega'>-1$)
there is exactly one $\alpha_{\omega}\in\coint{0,1}$ such that

\Centerline{$(1-\alpha_{\omega})
  \int_{F_{\omega}\cup F'_{\omega}}g(\omega,\omega')d\omega'
+\alpha_{\omega}\int_{\Omega'}g(\omega,\omega')d\omega'
=-(1-\alpha_{\omega})
   \mu'(F_{\omega}\cup F'_{\omega})-\alpha_{\omega}$;}

\noindent set

\Centerline{$G_{\omega}
=((F_{\omega}\cup F'_{\omega})\times[0,1])
   \cup(\Omega'\times[0,\alpha_{\omega}])$,
\quad$G'_{\omega}=\emptyset$.}

\noindent Then

$$\eqalign{\iint_{G_{\omega}}g(\omega,\omega')d\omega'd\omega''
&=(1-\alpha_{\omega})
  \int_{F_{\omega}\cup F'_{\omega}}g(\omega,\omega')d\omega'
+\alpha_{\omega}\int_{\Omega'}g(\omega,\omega')d\omega'\cr
&=-(1-\alpha_{\omega})
  \mu'(F_{\omega}\cup F'_{\omega})-\alpha_{\omega}
=-(\mu'\times\mu'')(G_{\omega}),\cr}$$

\Centerline{$\iint_{G'_{\omega}}g(\omega,\omega')d\omega'd\omega''
 =0=(\mu'\times\mu'')(G'_{\omega})$.}

\noindent Because $\omega\mapsto\alpha_{\omega}$ is expressible as
a rational combination of measurable functions, it is $\Sigma$-measurable
and

\Centerline{$\{(\omega,\omega',\omega''):\omega\in E_1$,
$(\omega',\omega'')\in G_{\omega}\}$}

\noindent belongs to $\Sigma\tensorhat\Sigma'\tensorhat\Sigma''$, while of
course

\Centerline{$\{(\omega,\omega',\omega''):\omega\in E_1$,
$(\omega',\omega'')\in G'_{\omega}\}=\emptyset$}

\noindent also does.

\medskip

\quad{\bf (iii)} Similarly, if $\omega\in E_2$, then
there is exactly one $\alpha_{\omega}\in\ocint{0,1}$ such that

\Centerline{$(1-\alpha_{\omega})
  \int_{F_{\omega}\cup F'_{\omega}}g(\omega,\omega')d\omega'
+\alpha_{\omega}\int_{\Omega'}g(\omega,\omega')d\omega'
=(1-\alpha_{\omega})
  \mu'(F_{\omega}\cup F'_{\omega})+\alpha_{\omega}$,}

\noindent and we can set

\Centerline{$G_{\omega}=\emptyset$,
\quad$G'_{\omega}
=((F_{\omega}\cup F'_{\omega})\times[0,1])
   \cup(\Omega'\times[0,\alpha_{\omega}])$.}

\medskip

\quad{\bf (iv)} Now consider the case in which $\omega\in E_3$.
In this case, because

\Centerline{$\int_{F_{\omega}}g(\omega,\omega')d\omega'
\le -\mu'F_{\omega}$,
\quad$\int_{F_{\omega}}g(\omega,\omega')d\omega'
+\int_{F'_{\omega}}g(\omega,\omega')d\omega'
>-\mu'F_{\omega}-\mu'F'_{\omega}$,}

\noindent there is a unique $\alpha_{\omega}\in\coint{0,1}$ such that

\Centerline{$\int_{F_{\omega}}g(\omega,\omega')d\omega'
+\alpha_{\omega}\int_{F'_{\omega}}g(\omega,\omega')d\omega'
=-\mu'F_{\omega}-\alpha_{\omega}\mu'F'_{\omega}$.}

\noindent Next, because

\Centerline{$(1-\alpha_{\omega})\int_{F'_{\omega}}g(\omega,\omega')d\omega'
\ge(1-\alpha_{\omega})\mu'F'_{\omega}$,}

\Centerline{$\int_{F_{\omega}}g(\omega,\omega')d\omega'
+\int_{F'_{\omega}}g(\omega,\omega')d\omega'
<\mu'F_{\omega}+\mu'F'_{\omega}$,}

\noindent there is a unique $\beta_{\omega}\in\ocint{0,1}$ such that

$$\eqalign{(1-\beta_{\omega})\int_{F_{\omega}}g(\omega,\omega')d\omega'
&+(1-\alpha_{\omega}\beta_{\omega})
  \int_{F'_{\omega}}g(\omega,\omega')d\omega'\cr
&=(1-\beta_{\omega})\mu'F_{\omega}
  +(1-\alpha_{\omega}\beta_{\omega})\mu'F'_{\omega}.\cr}$$

\noindent So if we set

\Centerline{$G_{\omega}=(F_{\omega}\times[0,\beta_{\omega}])
 \cup
 (F'_{\omega}\times[0,\alpha_{\omega}\beta_{\omega}])$,}

\Centerline{$G'_{\omega}=(F_{\omega}\times\ocint{\beta_{\omega},1})
 \cup (F'_{\omega}
   \times\ocint{\alpha_{\omega}\beta_{\omega},1})$,}

\noindent we shall have $G_{\omega}\cap G'_{\omega}=\emptyset$,
$G_{\omega}\cup G'_{\omega}=(F_{\omega}\cup F'_{\omega})\times[0,1]$,

$$\eqalign{\iint_{G_{\omega}}g(\omega,\omega')d\omega'd\omega''
&=\beta_{\omega}\int_{F_{\omega}}g(\omega,\omega')d\omega'
  +\alpha_{\omega}\beta_{\omega}
     \int_{F'_{\omega}}g(\omega,\omega')d\omega'\cr
&=\beta_{\omega}(\int_{F_{\omega}}g(\omega,\omega')d\omega'
  +\alpha_{\omega}\int_{F'_{\omega}}g(\omega,\omega')d\omega')\cr
&=-\beta_{\omega}(\mu'F_{\omega}+\alpha_{\omega}\mu'F'_{\omega})
=-(\mu'\times\mu'')(G_{\omega}),\cr}$$

\noindent and

$$\eqalign{\iint_{G'_{\omega}}g(\omega,\omega')d\omega'd\omega''
&=(1-\beta_{\omega})\int_{F_{\omega}}g(\omega,\omega')d\omega'
  +(1-\alpha_{\omega}\beta_{\omega})
     \int_{F'_{\omega}}g(\omega,\omega')d\omega'\cr
&=(1-\beta_{\omega})\mu'F_{\omega}
   +(1-\alpha_{\omega}\beta_{\omega})\mu'F'_{\omega}
=(\mu'\times\mu'')(G'_{\omega}).\cr}$$

\noindent Once again, $\omega\mapsto\alpha_{\omega}$ and
$\omega\mapsto\beta_{\omega}$ are $\Sigma$-measurable, so
$\{(\omega,\omega',\omega''):\omega\in E_3$,
$(\omega',\omega'')\in G_{\omega}\}$
and $\{(\omega,\omega',\omega''):\omega\in E_3$,
$(\omega',\omega'')\in G'_{\omega}\}$ belong to
$\Sigma\tensorhat\Sigma'\tensorhat\Sigma''$.

\medskip

{\bf (g)} Thus the project set out in (d) has been accomplished.
We need one more refinement:   define $\tilde G_{\omega}$ and
$\gamma_{\omega}$, for $\omega\in E$, by setting

\Centerline{$\tilde G_{\omega}
=(\Omega'\times\Omega'')\setminus(G_{\omega}\cup G'_{\omega})$,}

$$\eqalign{\gamma_{\omega}
&=0\text{ if }(\mu'\times\mu'')(\tilde G_{\omega})=0,\cr
&=\Bover1{(\mu'\times\mu'')(\tilde G_{\omega})}
  \iint_{\tilde G_{\omega}}g(\omega,\omega')d\omega'd\omega''
  \text{ otherwise}.\cr}$$

\noindent Note that as $-1<g(\omega,\omega')<1$ whenever
$(\omega',\omega'')\in\tilde G_{\omega}$, $\gamma_{\omega}\in\ooint{-1,1}$
for every $\omega$;  and by the same arguments as in (f),
$\tilde G=\{(\omega,\omega',\omega''):\omega\in E$,
$(\omega',\omega'')\in\tilde G_{\omega}\}$ belongs to
$\Sigma\tensorhat\Sigma'\tensorhat\Sigma''$ and
$\omega\mapsto\gamma_{\omega}$ is $\Sigma$-measurable.  (Of course
$\tilde G=(E\times\Omega'\times\Omega'')\setminus(G\cup G')$.)

\medskip

{\bf (h)}
Now, in $\Omega\times\Omega'\times\Omega''$, consider the product measure
$\nu=\mu\times\mu'\times\mu''$, the $\sigma$-algebras $\Tau=\dom\nu$ and
$\Tau_0=\{E\times\Omega'\times[0,1]:E\in\Sigma\}$, and the $\sigma$-algebra
$\Tau_1$ generated by $\Tau_0\cup\{G,G'\}$.    For
$\omega\in\Omega$, $\omega'\in\Omega'$ and $\omega''\in[0,1]$
set

\Centerline{$h(\omega,\omega',\omega'')=g(\omega,\omega')$,}

\Centerline{$h_0(\omega,\omega',\omega'')
=\int g(\omega,\omega')d\omega'
=\iint h(\omega,\omega',\omega'')d\omega'd\omega''$,}

$$\eqalign{h_1(\omega,\omega',\omega'')
&=-1\text{ if }(\omega,\omega',\omega'')\in G,\cr
&=1\text{ if }(\omega,\omega',\omega'')\in G',\cr
&=\gamma_{\omega}\text{ if }(\omega,\omega,\omega'')\in\tilde G,\cr
&=\int g(\omega,\omega')d\omega'
  \text{ if }\omega\in\Omega\setminus E.\cr}$$

\noindent Then $h_0$ is a conditional expectation of $h$ on $\Tau_0$, by
253H again.   The point is that $h_1$ is a conditional
expectation of $h$ on $\Tau_1$.   \Prf\ Of course $h_1$ is
$\Tau_1$-measurable because $G$, $G'$ and $\tilde G$ belong to $\Tau_1$ and
$\Tau_0\subseteq\Tau_1$.   Now any element of $\Tau_1$ is of the
form $W^*=W\cup W'\cup\tilde W\cup W_0$ where

\Centerline{$W=(H\times\Omega'\times\Omega'')\cap G$,
\quad$W'=(H'\times\Omega'\times\Omega'')\cap G'$,}

\Centerline{$\tilde W=(\tilde H\times\Omega'\times\Omega'')\cap\tilde G$,
\quad$W_0=H_0\times\Omega'\times\Omega''$,}

\noindent $H$, $H'$, $\tilde H$ and $H_0$ belong to $\Sigma$, and
$H_0\cap E=\emptyset$.   Now

$$\eqalignno{\int_Wh\,d\nu
&=\int_{H\cap E}\iint_{G_{\omega}}g(\omega,\omega')d\omega'd\omega''d\omega
  \cr
&=-\int_{H\cap E}(\mu'\times\mu'')(G_{\omega})d\omega
=-\nu W
=\int_Wh_1d\nu,\cr
\int_{W'}h\,d\nu
&=\int_{H\cap E}\iint_{G'_{\omega}}
  g(\omega,\omega')d\omega'd\omega''d\omega\cr
&=\int_{H\cap E}(\mu'\times\mu'')(G'_{\omega})d\omega
=\nu W'
=\int_{W'}h_1d\nu,\cr
\int_{\tilde W}h\,d\nu
&=\int_{H\cap E}\iint_{\tilde G_{\omega}}
  g(\omega,\omega')d\omega'd\omega''d\omega\cr
&=\int_{H\cap E}\gamma_{\omega}(\mu'\times\mu'')(\tilde G_{\omega})d\omega
=\int_Wh_1d\nu,\cr
\int_{W_0}h\,d\nu&
=\int_{H_0}\int g(\omega,\omega')d\omega'd\omega
=\int_{W_0}h_1d\nu.\cr}$$

\noindent Adding, $\int_{W^*}h\,d\nu=\int_{W^*}h_1d\nu$;  as $W^*$ is
arbitrary, $h_1$ is a conditional expectation of $h$ on $\Tau_1$.\ \Qed

Just because $\Tau_0\subseteq\Tau_1$, it follows at once that $h_0$ is a
conditional expectation of $h_1$ on $\Tau_0$.

\medskip

{\bf (i)} We are almost home.   Let $(\frak B,\bar\nu)$ be the measure
algebra of $\nu$, $\frak B_0=\{W^{\ssbullet}:W\in\Tau_0\}$ and
$\frak B_1=\{W^{\ssbullet}:W\in\Tau_1\}$, so that
$\frak B_0\subseteq\frak B_1$ are closed subalgebras of $\frak B$.
Set $w=h^{\ssbullet}$, $w_0=h_0^{\ssbullet}$ and $w_1=h_1^{\ssbullet}$;
then $w_0$, $w_1$ are the conditional expectations of $w$ on $\frak B_0$,
$\frak B_1$ respectively.   Next,

\Centerline{$\Bvalue{|w_0|<1}
=(E\times\Omega'\times\Omega'')^{\ssbullet}
=(G\cup G'\cup\tilde G)^{\ssbullet}$,}

\Centerline{$\Bvalue{|w_1|\ge 1}
=\{(\omega,\omega',\omega''):
  |h_1(\omega,\omega',\omega'')|\ge 1\}^{\ssbullet}$,}

\noindent so

\Centerline{$\Bvalue{|w_1|\ge 1}\Bcap\Bvalue{|w_0|<1}
=\Bvalue{|w_1|=1}\Bcap\Bvalue{|w_0|<1}
=(G\cup G')^{\ssbullet}$.}

\noindent Moreover, if we set

\Centerline{$F=\{(\omega,\omega',\omega''):\omega\in E$,
$h(\omega,\omega',\omega'')\le -1\}$,}

\Centerline{$F'=\{(\omega,\omega',\omega''):\omega\in E$,
$h(\omega,\omega',\omega'')\ge 1\}$,}

\noindent then $F\cup F'\subseteq G\cup G'$ because
$(F_{\omega}\cup F'_{\omega})\times[0,1]
\subseteq G_{\omega}\cup G'_{\omega}$ for every $\omega\in E$.   So

\Centerline{$\Bvalue{|w|\ge 1}\Bcap\Bvalue{|w_0|<1}
=(F\cup F')^{\ssbullet}
\Bsubseteq\Bvalue{|w_1|=1}\Bcap\Bvalue{|w_0|<1}$.}

\medskip

{\bf (j)} To complete the pattern demanded in the statement of the
lemma, I must describe the homomorphism $\pi$.   Let
$\varepsilon:\frak C\to\frak B$ be the canonical map
corresponding to the \imp\ function
$(\omega,\omega',\omega'')\mapsto(\omega,\omega')$, and
$\pi=\varepsilon\pi_0:\frak A\to\frak B$, so that $\pi$ is a
measure-preserving Boolean homomorphism;  let
$T:L^0(\frak A)\to L^0(\frak B)$ be the associated embedding, corresponding
to the \imp\ function $(\omega,\omega',\omega'')\to\omega$.
Because $g^{\ssbullet}=T_0v$, $w=h^{\ssbullet}=Tv$;  and
$\pi[\frak A_0]=\varepsilon[\frak C_0]=\frak B_0$, while
$w_0=Tv_0$.

This ends the proof.
}%end of proof of 628A

\leader{628B}{Lemma} Suppose that $(\frak A,\bar\mu)$ is a probability algebra
and $\langle u_i\rangle_{i\le n}$ is a martingale adapted to
a non-decreasing finite sequence $\langle\frak A_i\rangle_{i\le n}$
of closed subalgebras of $\frak A$.
Then there are a probability algebra $(\frak B,\bar\nu)$,
closed subalgebras
$\frak B_0\subseteq\ldots\subseteq\frak B_{2n}$ of $\frak B$, a martingale
$\langle w_j\rangle_{j\le 2n}$ adapted to
$\langle\frak B_j\rangle_{j\le 2n}$ and
a measure-preserving Boolean homomorphism $\pi:\frak A\to\frak B$
such that
if $T=T_{\pi}:L^1_{\bar\mu}\to L^1_{\bar\nu}$
is the associated embedding then

\Centerline{$\pi[\frak A_i]\subseteq\frak B_{2i}$,
$w_{2i}=Tu_i$ for $i\le n$,}

$$\eqalign{\Bvalue{|w_j|=1}\Bcap\Bvalue{|w_{j-1}|<1}
&=\Bvalue{|w_j|\ge 1}\Bcap\Bvalue{|w_{j-1}|<1}\cr
&\Bsupseteq\Bvalue{|w_{j+1}|\ge 1}\Bcap\Bvalue{|w_{j-1}|<1}\cr}$$

\noindent for odd $j<2n$.

\proof{ Induce on $n$.   If $n=0$ the result is trivial (we can take
$\frak B_0=\frak A_0$, $\frak B=\frak A$).

\medskip

{\bf (a)} For the inductive step to $n+1$,
we suppose that we have a probability algebra $(\frak A,\bar\mu)$,
closed subalgebras
$\frak A_0\subseteq\ldots\subseteq\frak A_{n+1}$,
and a martingale $\langle u_i\rangle_{i\le n+1}$ adapted to
$\langle\frak A_i\rangle_{i\le n+1}$.
The inductive hypothesis tells us that there are
a probability algebra $(\frak C,\bar\lambda)$,
closed subalgebras $\frak C_0\subseteq\ldots\subseteq\frak C_{2n}$ of
$\frak C$, a martingale $\langle v_j\rangle_{j\le 2n}$ adapted to
$\langle\frak C_j\rangle_{j\le 2n}$,
and a measure-preserving Boolean homomorphism
$\phi:\frak A\to\frak C$
such that,
writing $T_{\phi}$ for the associated embedding of
$L^0(\frak A)$ into $L^0(\frak C)$,

\Centerline{$\phi[\frak A_i]\subseteq\frak C_{2i}$,
$v_{2i}=T_{\phi}u_i$ for $i\le n$,}

$$\eqalign{\Bvalue{|v_j|=1}\Bcap\Bvalue{|v_{j-1}|<1}
&=\Bvalue{|v_j|\ge 1}\Bcap\Bvalue{|v_{j-1}|<1}\cr
&\Bsupseteq\Bvalue{|v_{j+1}|\ge 1}\Bcap\Bvalue{|v_{j-1}|<1}\cr}$$

\noindent for every odd $j<2n$.

\medskip

{\bf (b)}
Let $(\frak D,\bar\zeta,\varepsilon_1,\varepsilon_2)$ be the relative
free product of
$(\frak A,\bar\mu,\iota)$ and $(\frak C,\bar\lambda,\phi)$ over
$(\frak A_n,\bar\mu\restrp\frak A_n)$ in the sense of
458N-458O, where $\iota:\frak A_n\embedsinto\frak A$ is the
identity map;  so that $(\frak D,\bar\zeta)$ is a probability algebra,
$\varepsilon_1:\frak A\to\frak D$ and $\varepsilon_2:\frak C\to\frak D$
are measure-preserving homomorphisms,
$\varepsilon_1\restrp\frak A_n=\varepsilon_2\phi\restrp\frak A_n$, and
$\varepsilon_1[\frak A]$, $\varepsilon_2[\frak C]$ are relatively
independent over

\Centerline{$\frak D'
=\varepsilon_1[\frak A_n]=\varepsilon_2[\phi[\frak A_n]]
\subseteq\varepsilon_2[\frak C_{2n}]$.}

\noindent It follows that $T_{\varepsilon_1}u_n$
is the conditional expectation of
$T_{\varepsilon_1}u_{n+1}$ on $\varepsilon_2[\frak C_{2n}]$.
\Prf\ $T_{\varepsilon_1}u_n$ is certainly
the conditional expectation of $T_{\varepsilon_1}u_{n+1}$ on $\frak D'$,
by 365Qd again.   By 458Fb or 458M,
$T_{\varepsilon_1}u_n$ is the conditional expectation
of $T_{\varepsilon_1}u_{n+1}$ on $\varepsilon_2[\frak C_{2n}]$,
just because $\varepsilon_1[\frak A]$ and
$\varepsilon_2[\frak C_{2n}]\subseteq\varepsilon_2[\frak C]$
are relatively independent over
$\frak D'\subseteq\varepsilon_2[\frak C_{2n}]$.\ \Qed

Note that as $\varepsilon_1$ and $\varepsilon_2\phi$ agree on $\frak A_n$,
$T_{\varepsilon_1}u=T_{\varepsilon_2\phi}u=T_{\varepsilon_2}T_{\phi}u$
(364Pe) for every $u\in L^0(\frak A_n)$;  in particular,
$T_{\varepsilon_1}u_i=T_{\varepsilon_2}v_{2i}$ for every $i\le n$.

\medskip

{\bf (c)} Apply 628A to $(\frak D,\bar\zeta)$, the closed
subalgebra $\varepsilon_2[\frak C_{2n}]$ of $\frak D$
and the element $T_{\varepsilon_1}u_{n+1}$ of
$L^1_{\bar\zeta}$ to find a probability algebra $(\frak B,\bar\nu)$,
a measure-preserving Boolean homomorphism
$\psi:\frak D\to\frak B$, and
a closed subalgebra $\frak B_{2n+1}$ of $\frak B$ such that
$\frak B_{2n}=\psi[\varepsilon_2[\frak C_{2n}]]\subseteq\frak B_{2n+1}$
and if $w_{2n+2}=T_{\psi}T_{\varepsilon_1}u_{n+1}$,
$w_{2n}$ is the conditional expectation of $w_{2n+2}$ on
$\frak B_{2n}$ and $w_{2n+1}$ is the conditional
expectation of $w_{2n+2}$ on $\frak B_{2n+1}$, then

\Centerline{$w_{2n}=T_{\psi}T_{\varepsilon_1}u_n$,}

$$\eqalign{\Bvalue{|w_{2n+1}|=1}\Bcap\Bvalue{|w_{2n}|<1}
&=\Bvalue{|w_{2n+1}|\ge 1}\Bcap\Bvalue{|w_{2n}|<1}\cr
&\Bsupseteq\Bvalue{|w_{2n+2}|\ge 1}\Bcap\Bvalue{|w_{2n}|<1}.\cr}$$

For $j<2n$, set $\frak B_j=\psi\varepsilon_2[\frak C_j]$ and
$w_j=T_{\psi\varepsilon_2}v_j$.    By 365Qd once more,
$w_j$ is the conditional expectation of

\Centerline{$T_{\psi\varepsilon_2}v_{2n}
=T_{\psi}T_{\varepsilon_2}v_{2n}
=T_{\psi}T_{\epsilon_1}u_n=w_{2n}$}

\noindent on $\frak B_j$.   So if we set $\frak B_{2n+2}=\frak B$,
$\langle w_j\rangle_{j\le 2n+2}$ is a
martingale adapted to $\langle\frak B_j\rangle_{j\le 2n+2}$.

\medskip

{\bf (d)} Set $\pi=\psi\varepsilon_1:\frak A\to\frak B$.   Then

\Centerline{$\pi[\frak A_i]=\psi[\varepsilon_1[\frak A_i]]
=\psi[\varepsilon_2\phi[\frak A_i]]
\subseteq\psi\varepsilon_2[\frak C_{2i}]
=\frak B_{2i}$,}

\Centerline{$T_{\pi}u_i=T_{\psi}T_{\varepsilon_1}u_i
=T_{\psi}T_{\varepsilon_2}v_{2i}=w_{2i}$}

\noindent for $i\le n$.   Next, for odd $j<2n$,

$$\eqalign{\Bvalue{|w_j|=1}\Bcap\Bvalue{|w_{j-1}|<1}
&=\psi\varepsilon_2(\Bvalue{|v_j|=1}\Bcap\Bvalue{|v_{j-1}|<1})\cr
&=\psi\varepsilon_2(\Bvalue{|v_j|\ge 1}\Bcap\Bvalue{|v_{j-1}|<1})\cr
&=\Bvalue{|w_j|\ge 1}\Bcap\Bvalue{|w_{j-1}|<1},\cr}$$

$$\eqalign{\Bvalue{|w_j|=1}\Bcap\Bvalue{|w_{j-1}|<1}
&\Bsupseteq
  \psi\varepsilon_2(\Bvalue{|v_{j+1}|\ge 1}\Bcap\Bvalue{|v_{j-1}|<1})\cr
&=\Bvalue{|w_{j+1}|\ge 1}\Bcap\Bvalue{|w_{j-1}|<1}.\cr}$$

\noindent At the last step,

\Centerline{$\pi[\frak A_{n+1}]\subseteq\frak B=\frak B_{2n+2}$,}

\Centerline{$w_{2n+2}=T_{\psi}T_{\varepsilon_1}u_{n+1}
=T_{\pi}u_{n+1}$,}

$$\eqalign{\Bvalue{|w_{2n+1}|=1}\Bcap\Bvalue{|w_{2n}|<1}
&=\Bvalue{|w_{2n+1}|\ge 1}\Bcap\Bvalue{|w_{2n}|<1}\cr
&\Bsupseteq\Bvalue{|w_{2n+2}|\ge 1}\Bcap\Bvalue{|w_{2n}|<1}\cr}$$

\noindent by the choice of $\frak B$, $w_{2n}$, $w_{2n+1}$ and $w_{2n+2}$.
So $\frak B$, $\bar\nu$, $\frak B_0,\ldots,\frak B_{2n+2}$ and $\pi$
witness that the induction proceeds.
}%end of proof of 628B

\leader{628C}{Corollary} Suppose that
$(\frak A,\bar\mu)$ is a probability algebra
and $\langle u_i\rangle_{i\le n}$ is a martingale adapted to
a non-decreasing finite sequence $\langle\frak A_i\rangle_{i\le n}$
of closed subalgebras of $\frak A$.
Then there are a probability algebra $(\frak B,\bar\nu)$,
closed subalgebras
$\frak C_0\subseteq\ldots\subseteq\frak C_n$ of $\frak B$, a martingale
$\langle v_i\rangle_{i\le n}$ adapted to
$\langle\frak C_i\rangle_{i\le n}$ and
a measure-preserving Boolean homomorphism $\pi:\frak A\to\frak B$
such that

\Centerline{$\pi[\frak A_i]\subseteq\frak C_i$,
\quad$\|v_i\|_{\infty}\le 1$,
\quad$\|v_i\|_1\le\|u_i\|_1$}

\noindent for every $i\le n$, and

\Centerline{$\bar\nu(\sup_{i\le n}\Bvalue{v_i\ne T_{\pi}u_i})
\le\|u_n\|_1$.}

\proof{{\bf (a)} Let $\frak B$, $\bar\nu$, $\frak B_0,\ldots,\frak B_{2n}$,
$w_0,\ldots,w_{2n}$ and $\pi$ be as in 628B.
For $j\le 2n$ write $Q_j$ for the conditional expectation operator associated
with $\frak B_j$, and
set $c_j=\inf_{k\le j}\Bvalue{|w_k|<1}\in\frak B_j$;  for $1\le j\le 2n$ set

$$\eqalign{b_j
&=c_{j-1}\Bsetminus c_j
=\Bvalue{|w_j|\ge 1}\Bcap\inf_{k<j}\Bvalue{|w_k|<1}\cr
&=0\text{ if }0<j\le 2n\text{ and }j\text{ is even},\cr
&\Bsubseteq\Bvalue{|w_j|=1}\text{ if }j<2n\text{ is odd}.\cr}$$

\noindent Now set

\Centerline{$\hat w_j
=w_j\times\chi c_j+\sum_{k=1}^jw_k\times\chi b_k
\in L^0(\frak B_j)$.}

\noindent Then $\langle\hat w_j\rangle_{j\le 2n}$ is a martingale adapted
to $\langle\frak B_j\rangle_{j\le 2n}$.   \Prf\ If $j<2n$,

\Centerline{$Q_j(\hat w_{j+1}-\hat w_j)
=Q_j(w_{j+1}\times\chi c_j-w_j\times\chi c_j)
=Q_j(w_{j+1}-w_j)\times\chi c_j=0$.\ \Qed}

Observe that because
$b_j\Bsubseteq\Bvalue{|w_j|=1}$ for $0<j\le 2n$, while
$c_j\Bsubseteq\Bvalue{|w_j|<1}$ for every $j$, we have
$\|\hat w_j\|_{\infty}\le 1$ for every $j$.   We also see that
$\hat w_j\times\chi c_{2n}=w_j\times\chi c_{2n}$ for every $j$.

\medskip

{\bf (b)} This means that if we set $\frak C_i=\frak B_{2i}$ and
$v_i=\hat w_{2i}$ for $i\le n$, we shall have a martingale
$\langle v_i\rangle_{i\le n}$ adapted to
$\langle\frak C_i\rangle_{i\le n}$ with

\Centerline{$\pi[\frak A_i]\subseteq\frak C_i$,
\quad$\|v_i\|_{\infty}\le 1$}

\noindent for each $i\le n$, while

\Centerline{$\sup_{i\le n}\Bvalue{T_{\pi}u_i\ne v_i}
=\sup_{i\le n}\Bvalue{w_{2i}\ne\hat w_{2i}}
\Bsubseteq 1\Bsetminus c_{2n}
=\sup_{j\le 2n}\Bvalue{|w_j|\ge 1}$}

\noindent has measure at most $\|w_{2n}\|_1=\|u_n\|_1$, by
621E.

\medskip

{\bf (c)} Note also that $\|\hat w_j\|_1\le\|w_j\|_1$ for $j\le 2n$.
\Prf\ Since $c_j\in\frak B_j$ and $b_k\in\frak B_k$ for $1\le k\le j$,

$$\eqalign{\Expn(|\hat w_j|)&
=\Expn(|w_j|\times\chi c_j)+\sum_{k=1}^j\Expn(|w_k|\times\chi b_k)
=\Expn(|w_j|\times\chi c_j)+\sum_{k=1}^j\Expn(|Q_kw_j|\times\chi b_k)\cr&
\le\Expn(|w_j|\times\chi c_j)+\sum_{k=1}^j\Expn(Q_k|w_j|\times\chi b_k)\cr&
=\Expn(|w_j|\times\chi c_j)+\sum_{k=1}^j\Expn(|w_j|\times\chi b_k)
\le\Expn(|w_j|).  \text{ \Qed}\cr}$$

\noindent So

\Centerline{$\|v_i\|_1=\|\hat w_{2i}\|_1\le\|w_{2i}\|_1=\|u_i\|_1$}

\noindent for every $i\le n$.
}%end of proof of 628C

\leader{628D}{Proposition} Suppose that
$(\frak A,\bar\mu)$ is a probability algebra
and $\langle v_i\rangle_{i\le n}$ is a martingale adapted to
a non-decreasing finite sequence $\langle\frak A_i\rangle_{i\le n}$
of closed subalgebras of $\frak A$.   Let
$\langle\alpha_j\rangle_{j\le m}$, $\langle u_{ji}\rangle_{j\le m,i<n}$
be such that

\Centerline{$\alpha_j\ge 0$ for $j\le m$,
\quad$\sum_{j=0}^m\alpha_j=1$,}

\Centerline{$u_{ji}\in L^0(\frak A_i)$,
\quad$\|u_{ji}\|_{\infty}\le 1$ for $i<n$, $j\le m$.}

\noindent Set
$z=\sum_{j=0}^m\alpha_j|\sum_{i=0}^{n-1}u_{ji}\times(v_{i+1}-v_i)|$.
Then $\bar\mu\Bvalue{z>\gamma}\le\Bover2{\gamma}\|v_n\|_1$ for every
$\gamma>0$.

\proof{{\bf (a)} By 628C, applied to the martingale
$\langle v_i\rangle_{i\le n}$,
there are a probability algebra $(\frak B,\bar\nu)$,
closed subalgebras
$\frak B_0\subseteq\ldots\subseteq\frak B_n$ of $\frak B$, a martingale
$\langle w_i\rangle_{i\le n}$ adapted to
$\langle\frak B_i\rangle_{i\le n}$ and
a measure-preserving Boolean homomorphism $\pi:\frak A\to\frak B$
such that

\Centerline{$\pi[\frak A_i]\subseteq\frak B_i$,
\quad$\|w_i\|_{\infty}\le 1$}

\noindent for every $i\le n$, and

\Centerline{$\bar\nu(\sup_{i\le n}\Bvalue{w_i\ne T_{\pi}v_i})
\le\|v_n\|_1$}

\noindent where $T_{\pi}:L^0(\frak A)\to L^0(\frak B)$ is the $f$-algebra
homomorphism corresponding to $\pi$.   Now

\Centerline{$T_{\pi}z
=\sum_{j=0}^m\alpha_j|\sum_{i=0}^{n-1}T_{\pi}u_{ji}
   \times(v_{i+1}-v_i)|$}

\noindent and if we set

\Centerline{$z'
=\sum_{j=0}^m\alpha_j|\sum_{i=0}^{n-1}T_{\pi}u_{ji}\times(w_{i+1}-w_i)|$.}

\noindent then $\Bvalue{z'\ne T_{\pi}z}
\Bsubseteq\sup_{i\le n}\Bvalue{w_i\ne T_{\pi}v_i}$ and
$\bar\nu\Bvalue{z'\ne T_{\pi}z}\le\|v_n\|_1$.

For $j\le m$, set $z'_j=\sum_{i=0}^{n-1}T_{\pi}u_{ji}\times(w_{i+1}-w_i)$.
Because
$\|T_{\pi}u_{ji}\|_{\infty}=\|u_{ji}\|_{\infty}\le 1$ and
$\pi[\frak A_i]\subseteq\frak B_i$ so $T_{\pi}u_{ji}\in L^0(\frak B_i)$ for
every $i$, 621F tells us that $\|z'_j\|_2\le\|w_n\|_2$.
As $z'=\sum_{j=0}^n\alpha_j|z'_j|$,

\Centerline{$\|z'\|_2\le\sum_{j=0}^m\alpha_j\|w_n\|_2=\|w_n\|_2
\le\|w_n\|_{\infty}\|w_n\|_1\le\|w_n\|_1=\|v_n\|_1$.}

Now we see that

\Centerline{$\bar\mu\Bvalue{z>1}
=\bar\nu\Bvalue{T_{\pi}z>1}
\le\bar\nu\Bvalue{z'>1}+\bar\nu\Bvalue{z'\ne T_{\pi}z}
\le\|v_n\|_1+\|v_n\|_1=2\|v_n\|_1$.}

\medskip

{\bf (b)} For the general case, observe that

\Centerline{$\Bover1{\gamma}z
=\sum_{j=0}^m\alpha_j|\sum_{i=0}^{n-1}u_{ji}
   \times(\Bover1{\gamma}v_{i+1}-\Bover1{\gamma}v_i)|$,}

\noindent while $\langle\bover1{\gamma}v_i\rangle_{i\le n}$ is a martingale.
So (a) tells us that

\Centerline{$\bar\nu\Bvalue{z>\gamma}
=\bar\nu\Bvalue{\bover1{\gamma}z>1}
\le 2\|\bover1{\gamma}v_n\|_1
=\Bover2{\gamma}\|v_n\|_1$.}
}%end of proof of 628D

\leader{628E}{Corollary} Let \Asissix\ be a stochastic integration structure,
$\Cal S$ a non-empty sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$ a martingale.   Then

\Centerline{$\bar\mu\Bvalue{|z|>\gamma}
\le\Bover1{\gamma}\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$}

\noindent whenever $z\in Q_{\Cal S}(\pmb v)$ and $\gamma>0$.

\proof{ $z$ is expressible as
$\sum_{i=0}^{n-1}u_i\times(v_{\tau_{i+1}}-v_{\tau_i})$ where
$\tau_0\le\ldots\le\tau_n$ in $\Cal S$ and $u_i\in\frak A_{\tau_i}$ for
every $i<n$.
Now apply 628D to the martingale $\langle v_{\tau_i}\rangle_{i\le n}$
adapted to $\langle\frak A_{\tau_i}\rangle_{i\le n}$.
}%end of proof of 628E

\leader{628F}{} \cmmnt{An argument along the same lines as
that in 628D gives a similar result for quadratic variations.

\medskip

\noindent}{\bf Proposition} Suppose that $(\frak A,\bar\mu)$ is a probability
algebra and $\langle u_i\rangle_{i\le n}$ is a martingale adapted to a
non-decreasing finite sequence $\langle\frak A_i\rangle_{i\le n}$ of closed
subalgebras of $\frak A$.   Set $u^*=\sum_{i=0}^{n-1}(u_{i+1}-u_i)^2$.
Then $\bar\mu\Bvalue{u^*>\gamma^2}\le\Bover2{\gamma}\|u_n\|_1$ for every
$\gamma>0$.

\proof{{\bf (a)} If $\|u_n\|_{\infty}\le 1$, then $\|u^*\|_1\le\|u_n\|_2^2$.
\Prf\ This is a greatly simplified version of 624G.  For a direct argument,
note that every $u_i$ is square-integrable and
$(u_{i+1}-u_i)^2=u_{i+1}^2-u_i^2-2(u_{i+1}-u_i)\times u_i$;  take the
expectation of both sides and sum over $i$.\ \Qed

\medskip

{\bf (b)} Take $\frak B$, $\bar\nu$, $\frak C_0,\ldots,\frak C_n$,
$\langle v_i\rangle_{i\le n}$, $\pi:\frak A\to\frak B$ and
$T_{\pi}:L^0(\frak A)\to L^0(\frak B)$ as in 628C.
Setting $v^*=\sum_{i=0}^{n-1}(v_{i+1}-v_i)^2$, we see that

\Centerline{$\bar\nu\Bvalue{v^*\ne T_{\pi}u^*}
\le\bar\nu(\sup_{i\le n}\Bvalue{v_i\ne T_{\pi}})\le\|u_n\|_1$}

\noindent because $T_{\pi}$ is an $f$-algebra homomorphism.   Now

\Centerline{$\bar\nu\Bvalue{v^*>1}\le\|v^*\|_1\le\|v_n\|^2\le\|v_n\|_1
\le\|u_n\|_1$.}

\noindent Accordingly

\Centerline{$\bar\mu\Bvalue{u^*>1}
=\bar\nu\Bvalue{T_{\pi}u^*>1}
\le\bar\nu\Bvalue{v^*>1}+\bar\nu\Bvalue{v^*\ne T_{\pi}u^*}
\le 2\|u_n\|_1$.}

\medskip

{\bf (c)} This deals with the case $\gamma=1$.   For the general case,
look at the martingale $\langle\hat u_i\rangle_{i\le n}$ where
$\hat u_i=\Bover1{\gamma}u_i$ for $i\le n$.   Setting

\Centerline{$\hat u^*=\sum_{i=0}^{n-1}(\hat u_{i+1}-\hat u_i)^2
=\Bover1{\gamma^2}u^*$,}

\noindent we have

\Centerline{$\bar\mu\Bvalue{u^*>\gamma^2}
=\bar\mu\Bvalue{\hat u^*>1}
\le 2\|\hat u_n\|_1
=\Bover{2}{\gamma}\|u_n\|_1$,}

\noindent as claimed.
}%end of proof of 628F

\leader{628G}{Proposition} Let \Asissix\ be a stochastic integration structure,
$\Cal S$ a sublattice of $\Cal T$, and $\pmb v=\sfamilysS{v}$ a
martingale.   Let $\pmb v^*=\sfamilysS{v}$ be its quadratic variation.   Then

\Centerline{$\bar\mu\Bvalue{v^*_{\tau}>\gamma^2}
\le\Bover2{\gamma}\|v_{\tau}\|_1$}

\noindent whenever $\gamma>0$ and $\tau\in\Cal S$.

\proof{ If $I$ belongs to $\Cal I(\Cal S\wedge\tau)$,
the set of finite sublattices of $\Cal S$ bounded above by $\tau$, and
$(\sigma_0,\ldots,\sigma_n)$ linearly generates the $I$-cells, then

$$\eqalignno{\bar\mu\Bvalue{S_I(\tbf 1,(d\pmb v)^2))>\gamma^2}&
=\bar\mu\Bvalue{\sum_{i=0}^{n-1}(v_{\sigma_{i+1}}-v_{\sigma_i})^2>\gamma^2}
\le\Bover2{\gamma}\|v_{\sigma_n}\|
\Displaycause{by 612F applied to the martingale
$\langle v_{\sigma_i}\rangle_{i\le n}$ adapted to
$\langle\frak A_{\sigma_i}\rangle_{i\le n}$}
\le\Bover2{\gamma}\|v_{\tau}\|_1.\cr}$$

\noindent Now $\{u:\bar\mu\Bvalue{u>\gamma^2}\le\Bover2{\gamma}\|v_{\tau}\|_1\}$
is closed for the topology of convergence in measure, by 613Bo, and

\Centerline{$v^*_{\tau}=\int_{\Cal S\wedge\tau}(d\pmb v)^2
=\lim_{I\uparrow\Cal I(\Cal S\wedge\tau)}S_I(\tbf 1,(d\pmb v)^2)$,}

\noindent so we have $\bar\mu\Bvalue{v^*_{\tau}>\gamma^2}
\le\Bover2{\gamma}\|v_{\tau}\|_1$.
}%end of proof of 628G



\exercises{\leader{628X}{Basic exercises (a)}
%\spheader 628Xa
Suppose that $\langle u_i\rangle_{i\le n}$ is a martingale, and
$\alpha<\beta$ in $\Bbb R$.
Show that there is a martingale $\langle w_i\rangle_{i\le 2n}$ (possibly
on a different probability algebra) such that $(w_0,w_2,w_4,\ldots,w_{2n})$
has the same joint distribution as $(u_0,\ldots,u_n)$ (definition:
364Yo, 653B) and

$$\eqalign{\Bvalue{\alpha<w_{j-1}<\beta}\Bsetminus\Bvalue{\alpha<w_j<\beta}
&=\Bvalue{\alpha<w_{j-1}<\beta}
  \Bcap(\Bvalue{w_j=\alpha}\Bcup\Bvalue{w_j=\beta})\cr
&\Bsupseteq\Bvalue{\alpha<w_{j-1}<\beta}
  \Bsetminus\Bvalue{\alpha<w_{j+1}<\beta}\cr}$$

\noindent for odd $j<2n$.
%628B

\spheader 628Xb Let \Asissix\ be a stochastic integration structure,
$\Cal S$ a non-empty sublattice of $\Cal T$, and
$\pmb v=\sfamilysS{v}$ a virtually
local martingale.   Show that
$\bar\mu\Bvalue{|z|>\gamma}
\le\Bover1{\gamma}\sup_{\sigma\in\Cal S}\|v_{\sigma}\|_1$ whenever $\gamma>0$
and $z\in Q_{\Cal S}(\pmb v)$.
%628E 616Ia

%\leader{628Y}{Further exercises (a)}

}%end of exercises


\endnotes{
\Notesheader{628}
Both 628C and 621I are addressed to the same question.  We have a
martingale $\langle u_i\rangle_{i\le n}$
and would very much rather it was $\|\,\|_{\infty}$-bounded, with
$-\chi 1\le u_i\le\chi 1$ for every $i$.   The natural approach is to stop it as
soon as it leaves the interval $\ooint{-1,1}$.   In 621I this has been
done, and we are looking at the stopped martingale.   Here a crude, if
complex, inequality is enough for the application in 623O.   In 622G,
however, we need something more like
621Hf or 628C.   For the latter, I offer a method of
approximating the given martingale by a martingale which really never leaves
the interval $[-1,1]$.
I have no application in mind for 628F-628G, but I include them as a further
motive for mastering the technique of 628A-628C.  %628A 628B 628C

The proof I give of 628C depends on 628B, which is a kind of interpolation
theorem.   Given a martingale $\langle u_i\rangle_{i\le n}$ and the interval
$\ooint{-1,1}$, we can
interpolate terms to convert $\langle u_i\rangle_{i\le n}$
into the even terms of a martingale
$\langle w_j\rangle_{j\le 2n}$ in which the first
exit from this interval takes one of the end-point values.   We shall now
have a stopping time $\tau$ such that the martingale
$\langle w_{\tau\wedge j}\rangle_{j\le 2n}$ either starts outside
$\ooint{-1,1}$ or runs to full time inside
$\ooint{-1,1}$ or stops at $\pm 1$ precisely,
which is something we expect of a continuous
martingale but not of a discrete martingale.   To achieve
this, of course, we
have to enlarge our probability algebra.   You will see that the proof I
give of 628B depends on the one-step case 628A, and that here I
abandon the abstract formulation in terms of
probability algebras and move to
ordinary probability spaces, for which the key calculations in part (f) of
the proof of 628A are elementary and reasonably natural.


}%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{}
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\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



